question_id
string | site
string | title
string | body
string | link
string | tags
list | votes
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timestamp[s] | comments
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1
|
bioacoustics
|
How do animals adapt to a partially or fully deaf individual of their group?
|
In eusocial animals for which hearing is important to communicate between individuals, is there any species apart humans for which there are some evidence that the group adapts their acoustic communication to a fully or partially deaf individual of the group?
For instance, by systematically vocalizing louder/closer to a partially deaf individual, or switching the communication modality (e.g. acoustics to touching) for a fully deaf individual, or by showing more support to them when an alarm call is raised, etc.
I have not found any information after a quick search on the web and Google Scholar.
|
https://bioacoustics.stackexchange.com/questions/1235/how-do-animals-adapt-to-a-partially-or-fully-deaf-individual-of-their-group
|
[
"hearing",
"adaptation"
] | 6 | 2022-11-14T07:01:24 |
[
"Thanks @DanStowell, done!",
"Did you search for possible literature? If so it would be good to give some indication of that. Also, I think \"eurosocial\" should be \"eusocial\", right?"
] | 2 |
Science
| 0 |
2
|
codegolf
|
Minimum cost of solving the Eni-Puzzle
|
You're tasked with writing an algorithm to efficiently estimate cost of solving an [Eni-Puzzle](http://www.enipuzzles.com/buy/eni-braille-puzzle-with-bold-colors) from a scrambled state as follows:
You're given m lists of containing n elements each(representing the rows of the puzzle). The elements are numbers between 0 and n-1 inclusive (representing the colors of tiles). There are exactly m occurrences of each integers across all m lists (one for each list).
For example:
m=3, n=4 :
[[3, 0, 3, 1], [[1, 3, 0, 1],
[1, 0, 2, 2], or [0, 2, 3, 1],
[3, 0, 1, 2]] [0, 3, 2, 2]]
You can manipulate these lists in two ways:
1: Swapping two elements between circularly adjacent indices in (non circularly) adjacent lists. Cost=1.
Ex:
m=3, n=4 :
Legal:
Swap((0,0)(1,1))
Swap((1,0)(2,3)) (circularly adjacent)
Illegal:
Swap((0,0)(0,1)) (same list)
Swap((0,0)(2,1)) (lists are not adjacent)
Swap((0,0)(1,0)) (indices are not circularly adjacent (they're the same)
Swap((0,0)(1,2)) (indices are not circularly adjacent)
2. Circularly shifting one of the lists (Cost=number of shifts)
Your algorithm must efficiently calculate minimum cost required to manipulate the lists such that the resulting lists are all rotations of each other (meaning the puzzle can be fully solved from this state using only rotation moves) i.e.:
[[0, 1, 2, 3] [[2, 1, 0, 3]
[3, 0, 1, 2] and [0, 3, 2, 1]
[1, 2, 3, 0]] [3, 2, 1, 0]]
...are both valid final states.
Instead of lists, you may use any data structure(s) of your choice to represent the puzzle, so long as the cost of simulating a valid move (sliding or rotating) on the puzzle with this representation is O(n*m). The setup cost of initializing this data structure can be disregarded.
A winning solution will compute the cost in the lowest asymptotic runtime in terms of m and n. Execution time will be assessed as a tie breaker.
|
https://codegolf.stackexchange.com/questions/182647/minimum-cost-of-solving-the-eni-puzzle
|
[
"puzzle-solver",
"fastest-algorithm",
"sliding-puzzle"
] | 6 | 2019-04-03T19:53:28 |
[
"@Jonah Because The costs I've outlined here are not the actual number of moves required to make equivalent moves on the puzzle. \"Swapping\" moves have a higher cost. To solve the actual puzzle, you first reach the state I described, or a state that is only one sliding move away. Because my specifications don't allow direct sliding moves (swapping equal indices) like the puzzle does for the blank space, I don't need to consider this latter case. This reduces the problem so that only the two move types I described need to be considered and the concept of a moving \"blank space\" can be ignored.",
"@JonathanAllan Yes. I've edited this to be more clear.",
"why not define the winning state as the actual state where they're all lined up?",
"I edited the tags. Thanks!",
"Hi and welcome to PPCG :) I think you do not need fastest-code or code-challenge since you only actually specify the winning criterion as the algorithmic speed. Further, I think you should specify if you mean the asymptotic worst case, or average case (best case is not a good idea, for reasons that are hopefully obvious). For future use, we have a sandbox to post challenges to work out issues before you post to the main site. Good luck!"
] | 5 |
Technology
| 0 |
3
|
crypto
|
Share Conversion between Different Finite Fields
|
Let us have any linear secret sharing scheme (LSSS) that works on some field $Z_{p}$, where p is some prime or a power of a prime e.g., Shamir Secret Sharing, Additive secret Sharing. The problem at hand is simple, for any secret shared value in $Z_{p}$, is it possible to convert it (and its shares) to elements on $Z_{q}$, where q could be either greater or smaller than p? Note that **I do not need** conversion from $Z_{2^n}$ to $Z_{2}$.
I am aware of the results on [PRSS](http://link.springer.com/chapter/10.1007%2F978-3-540-30576-7_19), and well as in this [2008](https://eprint.iacr.org/2008/221.pdf) work, however the former does not address field conversion, and the latter, have explanations that are not clear nor it offers perfect security.
Any help is more than welcome!
|
https://crypto.stackexchange.com/questions/47554/share-conversion-between-different-finite-fields
|
[
"secret-sharing",
"multiparty-computation",
"finite-field",
"garbled-circuits",
"function-evaluation"
] | 11 | 2017-05-19T06:58:27 |
[
"@Dragos... Btw.... Are you from Bristol's group? that Dragos?",
"I see..Even converting shares from $Z_p \\rightarrow Z_2^{r}$ is a hard problem - at least I have encountered it several times and don't know how to solve it efficiently other than bit decompose and simulate the binary circuit in $Z_p$.",
"@Dragos. My inputs are bounded by the conversion target, if I do any transformation to a smaller field, let say $Z_{2}$ I would loose information (or be forced to use bit decomposition which I also want to avoid). This is why I cannot use $Z_{2}$, as I pointed out. Any idea is welcome!",
"For me, arithmetic means $Z_p$. You can use the method above by first converting from $Z_p \\rightarrow Z_2$ and then from $Z_2 \\rightarrow Z_q$.",
"@Dragos, as I pointed out on the explanation of the problem I 'm not interested on $Z_{2}$ to $Z_{2^r}$ conversions. Thanks anyways!",
"There is a way to do share conversion from binary to arithmetic and vice-versa using this paper: thomaschneider.de/papers/DSZ15.pdf"
] | 6 |
Technology
| 0 |
4
|
crypto
|
RSA key such that pi deciphers to your name per RSA-OAEP
|
Can you efficiently construct an RSA public/private key pair with $8k$-bit public modulus such that $C=\left\lfloor\pi\,2^{8k-2}\right\rfloor$ deciphers per RSA-OAEP to your name as a bytestring in ASCII or UTF-8?
The decryption must be per RSAES-OAEP of [PKCS#1v2.2](http://mpqs.free.fr/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp_EMC_Corporation_Public-Key_Cryptography_Standards_\(PKCS\).pdf#page=21), SHA-256 hash, no label, MGF1 with SHA-256, and sucessful. The key pair must conform¹ to [section 3](http://mpqs.free.fr/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp_EMC_Corporation_Public-Key_Cryptography_Standards_\(PKCS\).pdf#page=6). It could come as $n$, $e$, $d$, or as a PEM text encoding of the private key per the format of [section A.1.2](http://mpqs.free.fr/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp_EMC_Corporation_Public-Key_Cryptography_Standards_\(PKCS\).pdf#page=41) acceptable by an [ASN.1 decoder](https://lapo.it/asn1js/). Kudos if it passes² OpenSSL's [`RSA_check_key`](https://gitlab.com/gitlab-org/build/omnibus-mirror/openssl/blob/706457b7bda7fdbab426b8dce83b318908339da4/crypto/rsa/rsa_chk.c), more kudos if $8k\ge1024$, the higher the better.
$C$ is the first $k$ bytes of [these](http://mpqs.free.fr/C.txt) (in hex) for $8k$ up to $2^{16}$ bits.
* * *
Update: [this](https://crypto.stackexchange.com/a/80322/555) answers in the affirmative, for whatever usual $k$, with $p$ and $q$ equal-size primes, and $N$ hard to factor. The next frontier is a low $e$, say in $[3,2^{256})$, ideally $65537$.
Further stunts are possible, like having $C$ or $2$ the signature of anything per any usual RSA signature scheme, even deterministic and/or with message recovery.
As a consequence: verifying a signature of a file against Alice's RSA public key (including, certified and used for other purposes) is proof neither that the file is unchanged since signature, nor that Alice was the signer.
Example where that matters: the signature of a file is on a trusted site, the file is not. Alice manages to get her public key (which she uses normally) in the verifier's keystore (with the key identifier of the legitimate signer, if there is such thing in the signature). The verifier, who just uses crypto as a magic tool, sees that the signature verifies, and (rightly) trusts the secure site to hold only trusted signatures of trusted files. There is a chance that a file alteration by Alice goes uncaught, or/and that she can wrongly convince the verifier that she made the signature at the date notarized by the trusted site.
Update 2: Some consequences of the same possibility are discussed in Dennis Jackson, Cas Cremers, Katriel Cohn-Gordon and Ralf Sasse's [_Seems Legit: Automated Analysis of Subtle Attacks on Protocols that Use Signatures_](https://eprint.iacr.org/2019/779), originally in [proceedings of CCS 2019](https://doi.org/10.1145/3319535.3339813).
* * *
¹ That is: public modulus $n$ must be the product of $u\ge2$ distinct odd primes, odd public exponent $e$ with $3\le e<n$, private exponent $d$ with $0<d<n$ and $e\,d\equiv1\pmod{r-1}$ for each prime $r$ divisor of $n$.
² For keys otherwise per the above, it limits $u$ (including $u=2$ for $n$ less than 1024-bit, $u\le3$ for $n$ less than 4096-bit, see [`rsa_multip_cap`](https://gitlab.com/gitlab-org/build/omnibus-mirror/openssl/-/blob/706457b7bda7fdbab426b8dce83b318908339da4/crypto/rsa/rsa_mp.c#L100)). It also requires the full factorization of $n$ to be given, as well as reduced exponents and multiplicative inverses.
|
https://crypto.stackexchange.com/questions/80211/rsa-key-such-that-pi-deciphers-to-your-name-per-rsa-oaep
|
[
"rsa",
"oaep"
] | 13 | 2020-04-23T04:22:27 |
[] | 0 |
Technology
| 0 |
5
|
crypto
|
Finding $x$ such that $g^x\bmod p<p/k$?
|
In a Schnorr group as used for DSA, of prime modulus $p$, prime order $q$, generator $g$ (with $p/g$ small), how can we efficiently exhibit an $x$ with $0<x<q$ such that $g^x\bmod p<p/k$, for sizable $k$ but $k\ll\sqrt q$ ?
I see that for small $k$, it is enough to try incremental values of $x$ until finding an $x$ that fits, with expected cost $O(k)$ modular multiplications in $\mathbb Z_p^*$. Is there better?
Assume $p$ is an $l$-bit prime; $q$ is an $n$-bit prime with $q$ dividing $p-1$; and $g=h^{(p-1)/q}$ for some (random?) $h$ in $\mathbb Z_p^*$. For concrete values, assume $l\approx1024$, $n\approx160$, $k\approx2^{64}$.
Update: I'm expecting the problem to be hard, and thus that the more difficult problem of exhibiting $x$ with $g^x\bmod p$ in a range of width $p/k$ must be hard.
|
https://crypto.stackexchange.com/questions/48503/finding-x-such-that-gx-bmod-pp-k
|
[
"discrete-logarithm",
"group-theory"
] | 19 | 2017-06-20T08:57:11 |
[
"Can you please provide me with some real values of variables in question instead of me attempting with random values.",
"@pintor: no, I have made no progress nor got more feedback than the above comments.",
"@fgrieu, did you get any more updates on the problem? I'm also curious",
"This seems to reduce to a discrete logarithm problem in the multi-target setting, with a potentially very high target number depending on $k$. For a $k\\approx \\sqrt{p}$ index calculus seems like the best choice. Once you have one solution, getting more is relatively fast. Unsure if there's something better.",
"@SamuelNeves: yes; the only constraint on $g^x\\bmod p$ is that it is small (about $\\log_2(k)$-bit smaller than $p$ is).",
"Do you get to choose what the $g^x \\mod p$ value is?",
"@puzzlepalace: thank you for finding a hole; plugged it.",
"Choose $x = 0$ ? Runs in $\\Theta(1)$ ;)"
] | 8 |
Technology
| 0 |
6
|
crypto
|
Time-memory tradeoffs in Shor's algorithm
|
Can a quantum computer with insufficient qubits to factor an integer of a given size make _any_ progress in factoring it? For example, what if a quantum computer is only one qubit short of what is necessary to attack a specific integer? Is it capable of making any progress in factoring it, or would it be just as useless as a 2 qubit machine? Assume qubits refer to logical qubits in a general-purpose cryptanalytic quantum computer with sufficient quantum error correction to solve the problem of decoherence.
Given that qubits can be thought of as the memory for quantum computers, I am essentially asking if time-memory tradeoffs are possible with Shor's algorithm when too few qubits are available.
|
https://crypto.stackexchange.com/questions/67910/time-memory-tradeoffs-in-shors-algorithm
|
[
"factoring",
"quantum-cryptanalysis",
"shors-algorithm"
] | 8 | 2019-03-09T22:20:47 |
[
"Interesting question. One problem certainly could occur when you have less qubits than the number of the period $r$ ($a^r \\bmod N$). This would probably mess up the inverse quantum fourier transform which amplifies the correct period (quantum wave interference), thus rendering the whole computation impractical. In that case you can't find $r$.",
"Quantum computers, like classical ones, needs a minimum amount of memory to calculate any given pair of algorithm and input. This is why Turing completeness assumes infinite memory. Even quantum computers can only evaluate the the possibilities that fits in the qubit memory it has. Every algorithm and input has different limits, but it always exists. Exceeding the minimum necessary memory allows you to add further optimizations to attempt to improve performance. However, sometimes you can break the algorithm into smaller individual steps that each need less memory.",
"This answer from Quantum Computing.SE gives some interesting references in the third paragraph."
] | 3 |
Technology
| 0 |
7
|
crypto
|
Does the bias in RC4 drop asymptotically further in the keystream?
|
It's well-known that the RC4 keystream has significant biases that become less prominent later in the keystream. The most severe bias is in the second byte, which has a 128-1 bias towards zero. Other biases remain, and it's typically recommended to drop between 768 and 3072 bytes of the keystream.
Will dropping one more byte _always_ reduce the bias, or is there a point when the bias is as low as it is going to get and it won't get lower as the keystream goes on? It doesn't matter if the bias is negligible.
Note that I am only talking about short-term biases, not long-term ones which RC4-dropN can't solve.
**Does an _N_ exist that results in minimum bias in RC4-dropN?**
|
https://crypto.stackexchange.com/questions/72353/does-the-bias-in-rc4-drop-asymptotically-further-in-the-keystream
|
[
"stream-cipher",
"rc4",
"distinguisher"
] | 7 | 2019-08-03T01:44:25 |
[
"@MaartenBodewes I have not yet, but I also haven't done much research into it. There's a lot of academic text on the (in)security of RC4 which I should familiarize myself with.",
"Hey Forest. This question popped up again. Unfortunately I don't know the answer, but maybe you've found one in the mean time?",
"@kodlu The paper I referenced in my first comment is rc4nomore.com/vanhoef-usenix2015.pdf.",
"please point to the right paper, there are a few with these authors. also, proving convergence rates in random processes is notoriously hard, in general.",
"@IlmariKaronen I am asking whether a minimum level of bias can be reached in a finite number of steps.",
"If I understand your last paragraph right, you seem to actually be asking whether the asymptotic long-term level of bias is actually attained within a finite number of steps, or whether the asymptote is only approached but never reached. Is that correct?",
"I'm reading the Vanhoef and Piessens paper from 2015 to see if I can find the answer myself. If this is a particularly stupid question (it probably is) and the answer turns out to be obvious, I'll delete or self-answer it."
] | 7 |
Technology
| 0 |
8
|
cstheory
|
Partial circulant matrices: Is there a non-zero vector $v\in \{-1,0,1\}^n$ such that $Mv=0$?
|
The following question arose as a side product of some work I have been part of recently. An $m$ by $n$ $(0,1)$-matrix $M$ is partial circulant if it can be formed by taking the first $m$ rows of a [circulant matrix](http://en.wikipedia.org/wiki/Circulant_matrix) and all its entries are either $0$ or $1$. We assume that $m\leq n$.
Given such a partial circulant matrix $M$, what is the computational complexity of the following question?
> Is there a non-zero vector $v\in \\{-1,0,1\\}^n$ such that $Mv=0$?
The problem is clearly in NP but my guess is that it is not NP-hard (unlike [this related question](https://cstheory.stackexchange.com/questions/20277/decide-whether-a-matrixs-kernel-contains-any-non-zero-vector-all-of-whose-entri)). I have not, however, found a poly time solution.
|
https://cstheory.stackexchange.com/questions/12060/partial-circulant-matrices-is-there-a-non-zero-vector-v-in-1-0-1-n-such
|
[
"cc.complexity-theory",
"ds.algorithms",
"matrices"
] | 20 | 2014-11-04T06:46:47 |
[
"The circulant case was in fact considered at mathoverflow.net/questions/168474/… . However, it seems that there is fatal bug in one of the answers and the other answer is not a complete solution.",
"would it be more natural to study the circulant case 1st vs the partial circulant? also (idea) the other problem answers are formulated in terms of subset sum problem, how would this problem be formulated that way (ie as a constrained or special case of subset sum problem)? note there are many subset sum variants studied in the literature that might be close..."
] | 2 |
Science
| 1 |
9
|
cstheory
|
Are monotone Boolean functions in P well-approximated by monotone polynomial-size circuits?
|
**Question 1:** Is it true that for every polynomial $p(n)$ and $\epsilon >0$ there is a polynomial $q(n)$ such that every monotone Boolean function on $n$ variables that can be expressed by a Boolean circuit of size $p(n)$ can be $\epsilon$-approximated by a _monotone_ Boolean circuit of size at most $q(n)$.
A function $f$ is $\epsilon$-approximated by a function $g$ if they agree on at least $(1-\epsilon)$ fraction of inputs.
We can ask a slightly more general question. Let $\mu_p$ denote the Bernoulli measure on $\\{0,1\\}^n$ where every variable is '1' with probability $p$. For a (non-constant) monotone function $f$ let $p_c(f)$ be the value such that $\mu_{p_c}(\\{x: f(x)=1\\})=1/2$.
**Question 2:** Is it true that for every polynomial $p(n)$ and $\epsilon >0$ there is a polynomial $q(n)$ such that every monotone Boolean function $f$ on $n$ variables, there is a Boolean function $g$ that can be expressed by a monotone Boolean circuit of size $q(n)$ such that for $p=p_c(f)$, $\mu_p(\\{x: f(x) \ne g(x) \\}) \le \epsilon$.
Of course, the instinctive thought is that the answers for both these question is **NO**. Is it known? Razborov famously proved that matching cannot be expressed by a monotone Boolean circuit of polynomial size. But matching can be approximated in its critical probability by the property that there are no isolated vertices and this property admits a monotone (low depth) circuit.
I asked the analog questions (Problems 1,2,3) for $AC^0$ and $TC^0$ in [this blog post](https://gilkalai.wordpress.com/2010/02/10/noise-stability-and-threshold-circuits/).
|
https://cstheory.stackexchange.com/questions/31473/are-monotone-boolean-functions-in-p-well-approximated-by-monotone-polynomial-siz
|
[
"cc.complexity-theory",
"circuit-complexity",
"pr.probability"
] | 18 | 2015-05-12T23:58:06 |
[
"Dear Gil, now I see the point: I just wrongly interpreted your question! We indeed have that every monotone circuit, which needs only to coincide with b-Clique on an extremely small, but special set of inputs, must be large. You, however, do not specify this subset of \"hard\" inputs, only its ratio $1-\\epsilon$. This is a very interesting question. B.t.w. that result (on clique approximation) is Thm. 9.26 in my book.",
"\"Every monotone circuit separating graphs consisting of complete a-partite graphs with a > 32 from graphs consisting of b-cliques for a < b < n/32 must have size exponential in min{a,n/b}^{1/4}.\"This is interesting (reference?) but I am not sure it is relevent - namely I don't see that my notion of $\\epsilon$-approximation is relevant to the result you mention.",
"Dear Gil, I interpreted your Question 1 as \"if a monotone boolean function $f$ cannot be $\\epsilon$-approximated by a small monotone circuit, then $f$ has no small non-monotone circuits\" (with \"small\" meaning \"of polynomial size\"). If the interpretation is correct, the function I mentioned only weakly approximates b-Clique function, and still requires large monotone circuits.",
"Dear Stasy, I don't see how the results you mention refer to Questions 1 and 2. (As I said, I also expect a 'no' answer..)",
"Gill, what about the fact that every monotone circuit separating graphs consisting of complete a-partite graphs with a > 32 from graphs consisting of b-cliques for a < b < n/32 must have size exponential in min{a,n/b}^{1/4}. The approximation density seems to be here enough (is it?) to say that your conjecture would definitely imply P!=NP.",
"Thanks Gil, this seems like a good reason to update my prior to \"NO\".",
"Dear Andras, Alas, counting arguments are too weak for things of this kind. A YES answer will tell you that an argument that (1) some monotone NP-complete problem cannot be approximated by a polynomial size monotone circuit in the critical probability, automatically implies that (2) it cannot be approximated by a polynomial size circuit in the critical probability. (2) is a stronger statement than $NP \\ne P$ but I dont regard (1) as hopeless. (You may think that the answer is YES but proving it is hopeless, but I find it easier to think that the answer is NO and its not beyond reach.)",
"Why is NO the instinctive response? Monotone functions have all sorts of nice properties, so a priori this doesn't seem all that unlikely to me. Are you perhaps thinking of a counting argument?"
] | 8 |
Science
| 0 |
10
|
cstheory
|
Weighted Hamming distance
|
Basically my question is, what kind of geometry do we get if we use a "weighted" Hamming distance. This is not necessarily Theoretical Computer Science but I think similar things come up sometimes, for instance in randomness extraction.
Define:
$d(x,y)=$ the Hamming distance between binary strings $x$ and $y$ of length $n$, $=$ the cardinality of $\\{k: x(k)\ne y(k)\\}$.
For a set of strings $A$,
$d(x,A)=\min \\{ d(x,y): y\in A\\}$.
The _$r$ -fold boundary of_ $A\subseteq \\{0,1\\}^n$ is
$$\\{x\in\\{0,1\\}^n: 0 < d(x,A)\le r\\}.$$
Balls centered at $0$ are given by
$$ B(p)=\\{x: d(x,0)\le p\\}, $$ where $0$ is the string of $n$ many zeroes.
A _Hamming-sphere_ is a set $H$ with $B(p)\subseteq H\subseteq B(p+1)$. (So it's more like a ball than a sphere, but this is the standard terminology...)
Now, Harper in 1966 showed that for each $k$, $n$, $r$, one can find a Hamming-sphere that has minimal $r$-fold boundary among sets of cardinality $k$ in $\\\\{0,1\\\\}^n$. So a ball is a set having minimal boundary -- just like in Euclidean space.
The cardinality of $B(p)$ is ${n\choose 0}+\cdots {n\choose p}$.
The $r$-fold boundary of $B(p)$ is just the set $B(p+r)\setminus B(p)$, which then has cardinality ${n\choose p+1}+\cdots+{n\choose p+r}$.
So far, so good. But now suppose we replace $d$ by a different metric $D$: first let $d_j(x,y)$ be the Hamming distance between the prefixes of $x$ and $y$ of length $j$, and then $$ D(x,y)=\max_{j\le n}\ \frac{d_j(x,y)}{f(j) }$$ where $0\le f(j)\le j$. (For example we could have $f(j)=\sqrt{j}$, or $f(j)=j/\log j$.)
This is supposed to make $D(x,y)$ small if the differences of $x$ and $y$ do not clump together at small values of $j$.
# Questions
> Is the minimum $r$-fold boundary (under $D$) realized by a $D$-ball?
>
> Is there a better definition of $D$?
>
> Under the metric $D$, what's the minimum size of the $r$-fold boundary of a subset of $\\\\{0,1\\\\}^n$ having cardinality $k$? (A reasonable lower bound would be nice.)
(Cross-[posted](https://mathoverflow.net/questions/38221/geometry-in-a-hamming-box) on MathOverflow).
|
https://cstheory.stackexchange.com/questions/2637/weighted-hamming-distance
|
[
"co.combinatorics",
"randomness"
] | 20 | 2010-11-01T22:06:48 |
[] | 0 |
Science
| 1 |
11
|
cstheory
|
To what extent MSO = WS1S, when adding relations?
|
[This question has been asked on MathOverflow with no luck a month ago.]
Let me first clarify my definitions. For a word $w \in \Sigma^*$, with $\Sigma =\\{a_1, \ldots, a_n\\}$, I define two structures:
$\mathbb{N}(w) = \langle \mathbb{N}, <, Q_{a_1}, \ldots, Q_{a_n} \rangle$,
and the more usual _word model_ :
$\mathbb{N}^r(w) = \langle \\{0, \ldots, |w|-1\\}, <, Q_{a_1}, \ldots, Q_{a_n} \rangle$,
where $Q_{a_i} = \\{p \mid w_p = a_i\\}$.
Then WS1S is the set of second order formulas with models of the form $\mathbb{N}(w)$, with order, and for which second order quantification is limited to finite subsets of the domain. MSO is the set of monadic second order formulas with models of the form $\mathbb{N}^r(w)$, with order.
The usual proof that REG = WS1S proves at the same time that MSO = WS1S. My question is then, for which first or second order relations can we keep this to be true?
For instance, if we add a unary predicate $E(X)$ which says that a (monadic) second order variable contains an even number of objects, we add no power, as $E(X)$ is expressible as "there exists $X_1$ and $X_2$ that partition $X$, in such a way that if an element is in $X_i$ the next one in $X$ is in $X_j$, $i \neq j$, and the first element of $X$ is in $X_1$ and the last is in $X_2$."
Now, if we add a predicate $|X| < |Y|$, then WS1S becomes undecidable (see Klaedtke & Ruess, 10.1.1.7.3029), while MSO stays trivially decidable.
Thank you.
* * *
**Edit:** As a side question, ... is this question of interest? I mean, I'm no expert in the field, so I'm not sure this question is relevant.
|
https://cstheory.stackexchange.com/questions/15/to-what-extent-mso-ws1s-when-adding-relations
|
[
"lo.logic",
"automata-theory",
"descriptive-complexity"
] | 19 | 2010-08-16T13:32:53 |
[
"Your model $\\mathbb{N}(w)$ is very unusual : what is the need of considering a finite word on an infinite underlying structure ? If you forget the letter predicates (which do not change the problem a lot), you are in fact comparing MSO on finite linear orders, and WS1S on the order $\\omega$. This is strange, because it is more intuitive to compare different logic formalisms on the same structure.",
"What are your reasons to think this question is uninteresting? From a practical point of view, it is useful to know when you can restrict yourself to finite models, isn't it? For the record, I received no comment on MO.",
"Was there no joy on MO because they couldn't figure it out, or because they found your question uninteresting? (I'm guessing the latter)",
"also see the greasemonkey hack mentioned here: meta.cstheory.stackexchange.com/questions/3/latex-math-support",
"Ain't we just a click away from it?",
"We do need Latex math support..."
] | 6 |
Science
| 0 |
12
|
cstheory
|
Descriptive complexity of communication complexity classes
|
It is well known that some major complexity classes, like P or NP, admit a full logical characterization (e.g NP = existential 2nd order logic by Fagin's theorem). On the other hand, one can also define complexity classes in communication complexity (where P = problems solvable with poly(logN) communication etc. - see [Complexity classes in communication complexity theory](http://dl.acm.org/citation.cfm?id=1382962) for more).
My question is - are any descriptive complexity characterizations known for communication complexity classes (or do any such results from standard complexity classes transfer to communication setting easily)?
|
https://cstheory.stackexchange.com/questions/9041/descriptive-complexity-of-communication-complexity-classes
|
[
"cc.complexity-theory",
"lo.logic",
"communication-complexity"
] | 18 | 2011-11-19T14:15:36 |
[
"You may want to limit the computational power of the parties, which is not done in the usual communication complexity models."
] | 1 |
Science
| 0 |
13
|
cstheory
|
Does $EXP\neq ZPP$ imply sub-exponential simulation of BPP or NP?
|
By simulation I mean in the Impaglazzio-Widgerson [IW98] sense, i.e. sub-exponential deterministic simulation which appears correct i.o to every efficient adversary.
I think this is a proof: if $EXP\neq BPP$ then from [IW98] we get that BPP has such a simulation. Otherwise we have that $EXP=BPP$, which implies $RP=NP$ (because $NP \subseteq BPP$) and $EXP \in PH$. Now if $NP=RP=ZPP$ we have that $PH$ collapses to $ZPP$ and as a result $EXP$, but this cannot happen because of the assumption, so $RP\neq ZPP$ and this by the Kabanets paper "Easiness assumptions and hardness tests: trading time for zero error" implies that RP has such a simulation and as a result also NP.
This sounds like it could be a useful gap theorem. Does anyone know if it appears anywhere?
|
https://cstheory.stackexchange.com/questions/430/does-exp-neq-zpp-imply-sub-exponential-simulation-of-bpp-or-np
|
[
"cc.complexity-theory",
"reference-request",
"complexity-classes",
"conditional-results"
] | 29 | 2010-08-23T13:21:04 |
[
"ZPP is closed under complement, and contained in NP. So, $\\mathrm{NP\\subseteq ZPP}$ implies NP = ZPP = coNP, hence PH = NP = ZPP.",
"Why does NP in ZPP imply PH in ZPP?",
"(Just for reference): Identical question on MO: mathoverflow.net/questions/35945. Maybe someone finds the comments there inspiring.",
"Very interesting. I haven't seen this argument before, but I'm not by any means an expert in these things..."
] | 4 |
Science
| 1 |
14
|
cstheory
|
Interesting PCP characterization of classes smaller than P?
|
The PCP theorem, $\mathsf{NP} = \mathsf{PCP}(\mathsf{log}\, n, 1)$, involves probabilistically checkable proofs with polynomial time verifiers, so the smallest class that can be characterized in this way (that is, $\mathsf{PCP}(0, 0)$) must be $\mathsf{P}$. There are also PCP characterizations of larger complexity classes (for example, $\mathsf{NEXP} = \mathsf{PCP}(\mathsf{poly}, \mathsf{poly})$), also using polynomial time verifiers.
Can we achieve an interesting (that is, not immediately following from the definitions) PCP characterization of smaller complexity classes by restricting the time or space used by the verifier? For example, by using a logarithmic space verifier, or an $\mathsf{NC}$ circuit verifier?
|
https://cstheory.stackexchange.com/questions/12060/interesting-pcp-characterization-of-classes-smaller-than-p
|
[
"cc.complexity-theory",
"complexity-classes",
"pcp"
] | 20 | 2012-07-18T11:54:42 |
[
"Your comment is contradictory to your question where you count “PCP(0, 0) = P” as a PCP characterization of P.",
"I suppose I meant are there any characterizations that don't follow immediately from the definition, in the same way that PCP(log n, 1) is non-obvious characterization of NP.",
"Why not? If you restrict the space of the verifier to log, then PCP(0, 0) obviously becomes equal to L. And this is a PCP characterization of L if you count PCP(0, 0)=P as a PCP characterization of P. I cannot see the point of the question."
] | 3 |
Science
| 1 |
15
|
cstheory
|
Is Hankelability NP-hard?
|
I asked this question on [SO](https://stackoverflow.com/questions/29484864/an-algorithm-to-detect-permutations-of-hankel-matrices) on April 7 and added a bounty which has now expired but no poly time solution has been found yet.
I am trying to write code to detect if a matrix is a permutation of a Hankel matrix. Here is the spec.
**Input:** An n by n matrix M whose entries are 1 or 0.
**Output:** A permutation of the rows and columns of M so that M is a [Hankel matrix](http://en.m.wikipedia.org/wiki/Hankel_matrix) if that is possible. A Hankel matrix has constant skew-diagonals (positive sloping diagonals).
When I say a permutation, I mean we can apply one permutation to the order of the rows and a possibly different one to the columns.
A very nice $O(n^2)$ time algorithm [is known](https://stackoverflow.com/questions/20704900/determine-if-some-row-permutation-of-a-matrix-is-toeplitz) for this problem if we only allow permutation of the order of rows.
Peter de Rivaz pointed out this [paper](http://www.mat.ucsb.edu/~g.legrady/academic/courses/15w259/d/re_orderableMatrix.pdf) as a possible route to proving NP-hardness but I haven't managed to get that to work.
|
https://cstheory.stackexchange.com/questions/31174/is-hankelability-np-hard
|
[
"cc.complexity-theory"
] | 28 | 2015-04-16T10:44:55 |
[
"Cross-posted now to mathoverflow.net/questions/204294/is-hankelability-np-hard",
"@IgorShinkar I mean first some permutation on the order of the rows and then a possibly different permutation on the order of the columns, then stop.",
"Do you mean apply first some permutation on the rows and then some permutation on the columns? or do you allow more permutations?",
"@Kaveh Thanks for the question. The permutation applied to the order of the columns can be different from the one applied to the order of the rows.",
"Do you mean that the same permutation is applied to rows and columns? (my got feeling says this is in P, if a matrix is Hankelable with different diagonal values there aren't that many possibilities for the permutation.)",
"Just a guess: maybe this problem is GI-complete?"
] | 6 |
Science
| 1 |
16
|
cstheory
|
Model-checking for three-variable logics and restricted structures
|
Denote the $k$-variable fragment of logic $L$ by $L^{(k)}$. The model-checking problem for a logic $L$ with respect to a class of structures $C$, denoted $MC(L,C)$, is the decision problem
> $MC(L,C)$
> _Input:_ formula $\phi$ of $L$, structure $S$ from $C$
> _Question:_ does $S$ satisfy $\phi$?
Is there a logic $L$, with associated class of structures $C$, and a subclass $D$ of $C$, such that
1\. $MC(L^{(3)},D)$ is "easy",
2\. $MC(L,D)$ is "hard", and
3\. $MC(L^{(3)},C)$ is "hard"?
With "easy" vs. "hard" I mean decidable vs. undecidable, PTIME vs. NP-hard, LogCFL vs. P-complete, or similar dichotomies.
# Motivation
For many logics the three-variable fragment is enough to express "hard" properties (this is true for MSO and FO over ordered structures). So results about undecidability often carry across from the logic to its three-variable fragment. There is also often a way to reduce the arity of predicates used in formulas efficiently (for instance, SAT can be reduced to 3SAT with a linear increase in formula size). Reducing arity facilitates expressing some formulas with fewer variables.
On the other hand, there exist logics which are "easy" for some classes of structures yet "hard" in general. For instance, the model-checking problem for MSO is "easy" (linear time) on unordered graphs satisfying the property "has treewidth less than $k$" for any fixed $k$, but "hard" (NP-complete) on unordered graphs in general.
I am asking whether there are any known examples where the restriction to 3 variables interacts with the property defining the subclass $D$ in an essential way to yield tractability, yet where neither of these restrictions on their own is enough.
Apologies for the vagueness of the question: I am trying to leave this general because I can't think of any examples. As soon as I restrict $D$ enough to ensure condition 1 holds, condition 2 seems to break. I suspect there are some unnatural classes which would work, but ideally $D$ should be "nice", at least hereditary and closed under isomorphism. As a further non-example, model-checking for FO is PSPACE-complete while model-checking for $FO^{(3)}$ is PTIME-complete; here the restriction to a finite number of variables is already enough to ensure an "easy" model-checking problem and condition 3 fails.
Definitions for completeness: FO is [first-order logic](http://en.wikipedia.org/wiki/First-order_logic), MSO is monadic [second-order logic](http://en.wikipedia.org/wiki/Second-order_logic), where quantification over sets of individuals is allowed, as well as quantification over individuals.
|
https://cstheory.stackexchange.com/questions/2637/model-checking-for-three-variable-logics-and-restricted-structures
|
[
"cc.complexity-theory",
"co.combinatorics",
"lo.logic",
"descriptive-complexity"
] | 20 | 2010-11-03T11:47:19 |
[
"@MichaëlCadilhac alas, no.",
"Have you made any progress on that question, András?"
] | 2 |
Science
| 1 |
17
|
cstheory
|
In an $m$ by $n$ Boolean matrix, can you find a square block whose four corners are ones in $O(m \cdot n)$ time?
|
**Decision Problem**
Input: An $m$ by $n$ Boolean matrix $M$.
Decision Question: Does there exist a square block within $M$ such that upper-left corner entry == upper-right corner entry == lower-left corner entry == lower-right corner entry == 1? That is, all four corners of the square block are 1's.
**Cubic Time Solution**
We know that this problem can be solved in $O(m \cdot n \cdot min\\{m,n\\})$ time. The approach involves scanning through the matrix row by row. For each row, for each pair of 1's in that row, we check whether those 1's form an edge of a square block whose four corners are 1's.
**Our Question**
What is the time complexity of this problem? Is there a quadratic time solution? In particular, can we solve this decision problem in $O(m \cdot n)$ time?
**Extra Background**
1. If you're just looking for a rectangular block whose four corners are ones, this can be solved in $O(m \cdot n)$ time. Many variations to this problem can be solved in $O(m \cdot n)$ time as well. I co-authored a paper on this subject.
2. The paper ["Finding squares and rectangles in sets of points"](https://doi.org/10.1007/BF01931281) investigates a related problem from computational geometry where you're given a set of points in a 2D plane and you want to know if there are four points that form an axis-parallel square.
Written together with Eevvoor.
|
https://cstheory.stackexchange.com/questions/47588/in-an-m-by-n-boolean-matrix-can-you-find-a-square-block-whose-four-corners
|
[
"ds.algorithms",
"matrices",
"computational-geometry",
"search-problem",
"boolean-matrix"
] | 18 | 2020-09-18T11:04:38 |
[
"It feels somehow like an FFT would help, although I haven't been able to puzzle out details. You're looking for something translation-invariant, and working in frequency space often helps with that sort of thing. Tossing out this comment in case someone else can spot something.",
"@MichaelWehar: ok, let me know if you make some progress!",
"@MarzioDeBiasi It turns out that my reduction doesn't quite work. I'm seeing if I can fix it.",
"@MarzioDeBiasi I guess that the right isosceles triangle problem is reducible to triangle finding. We can convert the matrix to a graph where vertices are rows, columns, and diagonals. Also, entries with 1's are associated with edges. But, can we do any better than matrix multiplication time?",
"@MarzioDeBiasi Please feel welcome to share these works! I think they could be pretty helpful. :)",
"@michaelwehar my idea of right triangle is a square with three 1s on the corners (no matter where the missing one is) but there is a reduction from mine to yours, where the missing one is in the bottom right (just rotate 90 and replicate four times). BTW there are some works both on monochromatic square and triangle free grids (somewhat related if you interpret \"colors\" with small monochromatic (sub)boxes with 1s on the diagonal)",
"@MarzioDeBiasi I thought about your right isosceles triangle problem. Just to confirm, this problem is asking for a square block where upper-left corner entry == upper-right corner entry == lower-left corner entry == 1, but we don't care whether lower-right corner entry is 0 or 1? I think that I have some potentially promising ideas for this problem. I'm going to try to formalize the ideas and I hope to share in a few days. Thanks again!",
"Even if you restrict the search to right isosceles triangles, it seems impossible to have an $O(mn)$ algorithm",
"@DavidEppstein I wasn't quite sure how many 1's there could be without forcing there to be a square block so this is very helpful! User whosyourjay showed a construction related to affine planes on how many 1's there could be without forcing there to be a rectangular block whose corners are 1's. If I recall correctly, the near-tight bound was something like $m \\cdot n^{1/2}$ (it had degree 3/2).",
"This is more dead end than answer, but: there exist sets of $mn^{1-o(1)}$ nonzeros with no square, formed by filling diagonals according to a near-linear-size progression-free set (en.wikipedia.org/wiki/Salem%E2%80%93Spencer_set). So an algorithm that answers yes if dense enough to force a square and switches to the naive O(nonzeros times min(m,n)) otherwise can't be strongly subcubic.",
"@JoshuaGrochow Yes, when searching for a rectangular block, the problem is related to finding a four cycle in a graph which is solvable in quadratic time. However, when searching for a square block, I don't know what can be done. Maybe there is a way to reduce triangle finding to this problem.",
"Seems a little similar to finding a triangle in a graph. If you get a subcubic reduction from that problem that'd be the end of the story for now.",
"@daniello Yes! That's a good point. When the input matrix is rectangular, we can reduce it to the case where the input matrix is square. Thank you. :)",
"You can assume wlog that $m \\leq n$ (otherwise transpose) and that $n \\leq 2m$ since otherwise you can cover the matrix by $2n/m$ matrices of size $m \\times 2m$ that cover all square blocks. So wlog the input matrix is square $n$ by $n$."
] | 14 |
Science
| 0 |
18
|
cstheory
|
Problem unsolvable in $2^{o(n)}$ on inputs with $n$ bits, assuming ETH?
|
If we assume the Exponential-Time Hypothesis, then there is no $2^{o(n)}$ algorithm for $n$-variable 3-SAT, and many other natural problems, such as 3-COLORING on graphs with $n$ vertices. Notice though that, in general, encoding the input for $n$-variable 3-SAT or $n$-vertex 3-COLORING takes something like $O(n\log n)$ bits. For example, to describe a sparse graph as input to 3-COLORING, for each edge we would have to list its endpoints. So the lower bound is not exponential in the length of the input. Therefore, my question is the following:
Is there a problem for which no $2^{o(n)}$ algorithm exists for inputs of length $n$ bits (assuming ETH)?
Ideally, the problem would be in NP (no cheating with succinct NEXP-hard problems!) and be reasonably natural, but I won't be picky.
Let me also note that after digging around I found that there are efficient ways to encode planar graphs with $O(n)$ bits. So, if one could find a problem that takes time exponential in the number of vertices even for planar graphs, the question would be settled. However, because planar graphs have treewidth $O(\sqrt{n})$, most natural problems have sub-exponential algorithms in this case.
|
https://cstheory.stackexchange.com/questions/16148/problem-unsolvable-in-2on-on-inputs-with-n-bits-assuming-eth
|
[
"cc.complexity-theory",
"sat",
"planar-graphs",
"succinct"
] | 48 | 2013-01-19T08:24:59 |
[
"I believe that it is an open question if we can have a lower bound $2^{\\Omega(n)}$ under ETH, where $n$ is the bit size. We know that proving lower bound $\\Omega(c^n)$ under SETH would disprove SETH.",
"How about the following problem, for some large constant $c>0$? Given the encoding of a non-deterministic TM $M$ and binary string $w$, does $M(w)$ have an accepting computation of length at most $|w|^c$? This should work if anything does... (?)",
"@AviTal This paper comes to mind arxiv.org/abs/cs/0102005 . The usual keywords for this line of research seem to be \"compact\" and \"succinct\" encodings or representations. (succinct is stronger in that it is optimal up to an additive term instead of a multiplicative factor for compact).",
"Would disordered lattice tiling work? Without disorder the problem is NEXP-complete [arXiv:0905.2419] but with disorder (e.g., a vertex-dependent constraint on the tiles) the input would need to be $O(n)$. This is reasonably natural: it would arise in physics as the zero-temperature partition function of a disordered vertex model.",
"Can you point to a reference where I can find efficient ways to encode planar graphs with O(n) bits? (I assume that n is the number of vertices). If so, there are NP-Complete problems for planar graphs and maybe the problem can also be presented in O(n) bits using these methods. Thanks.",
"It is probably not possible to compress $n$ variable SAT instances to $O(n \\log n)$ as stated in the question (Dell, van Malkebeek, \"Satisfibility Allows No Nontrivial Sparsification Unless The Polynomial-Time Hierarchy Collapses\"). However, 3-SAT has no $2^{o(m)}$ algorithm where $m$ is the number of clauses, as shown by Impagliazzo and Paturi (assuming ETH). Since the hard cases have $O(m)$ variables we have $b=O(m \\log m)$, so $m=\\Omega(b/\\log b)$, so 3-SAT has no $2^{o(b/\\log b)}$ algorithm where $b$ is the input size.",
"A problem in NP for which our present algorithmic knowledge is consistent with such a ETH result is edge chromatic number in a dense graph. AFAIK, we do not yet know of any $2^{o(|E|)}$ time algorithms...",
"If you are fine with dropping the condition that the problem is in NP, then you can use any undecidable problem and you do not even need the exponential time hypothesis. So probably you should be picky."
] | 8 |
Science
| 1 |
19
|
cstheory
|
Sylver Coinage Game
|
A game in which the players alternately name positive integers that are not sums of previously named integers (with repetitions being allowed). The person who names 1 (so ending the game) is the loser.
The question is: If player 1 names ‘16’, and both players play optimally thereafter, then who wins?
It has been known that if player 1 name "5n," then player 1 wins. If player name "5n+2", the result is known(maybe palyer 2 wins, but I haven't found the source yet). 16 is the minimum number of "5n+1".
I guess that this problem is PSPACE-hard, but I haven't proved it yet.
|
https://cstheory.stackexchange.com/questions/22112/sylver-coinage-game
|
[
"co.combinatorics",
"combinatorial-game-theory"
] | 18 | 2014-04-14T04:03:06 |
[
"@NealYoung The algorithm in Winning Ways only gives whether an opening move is winning or not -- in accordance with Conway's comment, it relies on a theorem saying finitely moves of the form 2^a 3^b win, and I imagine (Conway having told me of that fact before) that this is what he meant when he said that an algorithm exists. If so, he was being unclear; however, this means that Jeffe's original comment was still correct.",
"Here's an updated link for Jeffε's comment (as of May 2017). Although I read Conway's comment as suggesting the problem is decidable: \"Is there an algorithm for Sylver Coinage that tells you, by looking over all (possibly infinite) options, what the status of the a position is and what if any are the winning moves? The answer is yes (see Winning Ways) but I do not know what the algorithm is.\"",
"@PyRulez just writing down that regex is exponential time. However, there's an easy polynomial time test for whether you have to name 1 as your next move (and lose): it's true iff both 2 and 3 have already been named (and 1 hasn't). For if they have, then no higher numbers are available, and if they haven't, then one of those two numbers can be named next.",
"@SureshVenkat It is efficient. For example, if the current coins are $4$, $5$, and $7$, you effectively have a regex of (aaaa)*(aaaaa)*(aaaaaaa)*, which can be matched against efficiently.",
"I am considering using the construction of sum-free set, for example, the set of odd numbers is a sum-free subset of the integers, and the set $\\{N/2+1, ..., N\\}$ forms a large sum-free subset of the set $\\{1,...,N\\}$ ($N$ even). Fermat's Last Theorem is the statement that the set of all nonzero nth powers is a sum-free subset of the integers for $n > 2$.However, some basic problems of sum-free set are still unknown. How many sum-free subsets of $\\{1, ..., N\\}$ are there, for an integer $N$? Ben Green has shown that the answer is $O(2^{N/2})$. Can this help?",
"Is it even efficient (i.e P time) to check if someone has lost ? Seems like the \"you lost\" test is an unbounded knapsack problem.",
"Apparently it's open whether Sylver Coinage is decidable. Cool. And at least as of 2002, even Conway didn't know who wins from 16."
] | 7 |
Science
| 0 |
20
|
cstheory
|
Complexity of the homomorphism problem parameterized by treewidth
|
The _homomorphism problem_ $\text{Hom}(\mathcal{G}, \mathcal{H})$ for two classes $\mathcal{G}$ and $\mathcal{H}$ of graphs is defined as follows:
> **Input:** a graph $G$ in $\mathcal{G}$, a graph $H$ in $\mathcal{H}$
>
> **Output:** decide if there is a homomorphism from $G$ to $H$, i.e., a mapping $h$ from the vertices of $G$ to those of $H$ such that, for any edge $\\{x, y\\}$ of $G$, $\\{h(x), h(y)\\}$ is an edge of $H$.
For each $k \in \mathbb{N}$, I will call $\mathcal{T}_k$ the class of the graphs of [treewidth](http://en.wikipedia.org/wiki/Treewidth) at most $k$. I'm interested in the problem $\text{Hom}(\mathcal{T}_k, \mathcal{T}_k)$, which I see as a _parameterized_ problem (by the treewidth bound $k$). My question is: **what is the complexity of this parameterized problem?** Is it known to be FPT? or is it W[1]-hard?
Here are some things that I found about the $\text{Hom}$ problem, but which do not help me answer the question. (I write $-$ for the class of all graphs.)
* <http://www.sciencedirect.com/science/article/pii/009589569090132J>: If $\mathcal{H}$ is bipartite then $\text{Hom}(-, \mathcal{H})$ is in PTIME, otherwise it is NP-complete, but of course the NP-hardness relies on allowing arbitrary $G$.
* [http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.86.9013&rep=rep1&type=pdf](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.86.9013&rep=rep1&type=pdf): If the treewidth of $\mathcal{G}$ (modulo homomorphic equivalence) is bounded by a constant then $\text{Hom}(\mathcal{G}, -)$ is in PTIME (and otherwise it isn't, assuming FPT != W[1]). Hence, in particular my problem $\text{Hom}(\mathcal{T}_k, \mathcal{T}_k)$ is in PTIME for fixed $k$, but this doesn't tell me what is the dependency on the parameter.
* From Flum and Grohe's book _Parameterized Complexity Theory_ , Corollary 13.17: The problem $\text{Hom}(\mathcal{T}_k, -)$ is FPT when parameterized by the _size_ of $G$ (but I am parameterizing by the treewidth)
* <http://users.uoa.gr/~sedthilk/papers/homo.pdf>, Corollary 3.2: When fixing a specific graph $H$, the problem $\text{Hom}(\mathcal{T}_k, \\{H\\})$, parameterized by k, is FPT (this even holds for more complicated counting variants), but I do not want to restrict to fixed $H$.
|
https://cstheory.stackexchange.com/questions/34877/complexity-of-the-homomorphism-problem-parameterized-by-treewidth
|
[
"graph-algorithms",
"parameterized-complexity",
"treewidth",
"fixed-parameter-tractable",
"homomorphism"
] | 18 | 2016-06-02T03:37:34 |
[
"This question is still open, but one remark: there is an FPT algorithm parameterized by treewidth for the graph isomorphism problem, here: epubs.siam.org/doi/abs/10.1137/… (Daniel Lokshtanov, Marcin Pilipczuk, Michał Pilipczuk, and Saket Saurabh, \"Fixed-Parameter Tractable Canonization and Isomorphism Test for Graphs of Bounded Treewidth\", SICOMP.) As far as I know, unfortunately, this does not say anything about the homomorphism problem."
] | 1 |
Science
| 0 |
21
|
cstheory
|
Is Node Multiway Cut NP-complete on planar graphs when all terminals lie on the outer face?
|
I am interested in the following problem.
**Node Multiway Cut on Planar Graphs with terminals on the outer face**
* Instance: A plane graph G, and integer k, and a set $S \subseteq V(G)$ of terminals which are all incident on the outer face of G.
* Question: Is there a set of vertices $X \subseteq V(G)$ of size at most $k$ such that all vertices of $S \setminus X$ belong to different connected components of $G - X$?
Informally, the problem asks whether all terminals can be separated with k vertex deletions, knowing that all terminals lie on the outer face. If we replace vertex deletions by edge deletions, then this is known to be possible in polynomial time by considering multiway cuts in the planar dual, and relating them to certain kind of Steiner trees ([Efficient Algorithms for k-Terminal Cuts on Planar Graphs](https://doi.org/10.1007/3-540-45678-3_29)). Since the edge-deletion version reduces to the vertex-deletion version, the edge-deletion version seems easier; no reduction in the reverse direction is known.
The question is: is Node Multiway Cut on Planar Graphs with terminals on the outer face NP-complete? The reason I am interested in this problem is that it would provide me with a starting point for another complexity lower bound if this is indeed NP-complete.
|
https://cstheory.stackexchange.com/questions/8969/is-node-multiway-cut-np-complete-on-planar-graphs-when-all-terminals-lie-on-the
|
[
"cc.complexity-theory",
"np-hardness",
"planar-graphs"
] | 17 | 2011-11-14T09:06:13 |
[
"I know for sure that the algorithm I mentioned works for edge weighted Steiner tree, but one should be able to adapt it to work for vertices.",
"I think that this is equivalent to solving vertex weighted Steiner tree where all terminals on the outer face. The very loose idea would be something like: consider the boundary path between every pair of consecutive terminals you want to separate; connect all the vertices on that path to a new vertex, which becomes a terminal for the Steiner tree problem. Now all these Steiner terminals will all lie on the same face. This should be solvable in polynomial time using the algorithm of Erickson et al. (jstor.org/pss/3689922)."
] | 2 |
Science
| 0 |
22
|
cstheory
|
Can short-distance connectivity be harder than connectivity?
|
Has anybody seen the following (or similar) question being considered:
> Can it be **easier** to determine the presence/absence of $s$-$t$ paths than to determine the presence/absence of _short_ $s$-$t$ paths?
A bit more formally, the _distance_ -$k$ _connectivity_ problem STCON(n,k) is, given a subgraph of a complete undirected graph $K_n$ on $[n]=\\{1,\ldots,n\\}$, to decide whether there is a path from node $s=1$ to node $t=n$ of length $\leq k$. By the _length_ , I mean the number of inner nodes (just for notational convenience). The _connectivity_ problem STCON(n) corresponds to the case when $k=n-2$: is there any $s$-$t$ path at all? These problems are monotone boolean functions of $\binom{n}{2}$ variables $x_{i,j}$ corresponding to the edges of $K_n$.
It is well known, that $O(kn^2)$ fanin-$2$ AND and OR gates are enough to compute STCON(n,k). The desired monotone circuit is given by [Bellman-Ford](http://en.wikipedia.org/wiki/Bellman%E2%80%93Ford_algorithm): first compute $F_j^{(0)}=x_{s,j}$ for all nodes $j$, and then use the recursion $F_j^{(l+1)}=\bigvee_{i\in[n]} F_i^{(l)}\land x_{i,j}$. Then $F_j^{(l)}$ accepts a given graph iff it has an $s$-$j$ path of length $\leq l$. Hence, $F_n^{(k)}=$STCON(n,k).
> **Question 1:** Can STCON(n) have more than constant times **smaller** monotone circuits than STCON(n,k), for some $k\ll n$?
A boolean function $f(x_1,\ldots,x_n)$ is a _monotone projection_ of a boolean function $g(y_1,\ldots,y_m)$, if the function $f$ can be obtained from $g$ by replacing each $y$-variable by $0$ or $1$ or by some $x$-variable. Say, we trivially have that every THRESHOLD(n,k) is a monotone projection of MAJORITY(2n).
By taking $k$ copies of the input graph for STCON(n,k), one can easily show that STCON(n,k) is a monotone projection of STCON(kn). Thus, STCON(kn) cannot have smaller circuits than STCON(n,k). Can the factor "k" here be replaced by some constant?
> **Question 2:** Is STCON(n,k) a monotone projection of STCON(m) for some $m < kn$?
**ADDDED** (Apr.7 ): Among related results I know are the following:
1. _STCON and monotone**NC$^1$**_ : if a boolean function $f$ has a monotone formula of size $s$, then $f$ is a monotone projection of STCON(m) for $m=O(s)$; this is easy: just view formulas as parallel-sequential networks. In particular, this implies that _every_ monotone boolean function of $n$ variables _is_ a monotone projection of STCON(m) for some $m\leq n2^n$ (take a monotone DNF).
2. _STCON itself is not in monotone**NC$^1$**_ : every monotone circuit for STCON(n,k) requires depth about $(\log k)(\log n)$ (independent on circuit size!). This was shown by [Karchmer and Wigderson](http://www.math.ias.edu/~avi/PUBLICATIONS/MYPAPERS/KW90/KW90.pdf). As observed by [Grigni and Sipser](http://www.mathcs.emory.edu/~mic/papers/4.ps), this holds even when AND gates are allowed to have unbounded fanin! If, however, OR gates can have large fanin and AND gates have fanin-$2$, then depth $2\log k$ and size $O(n^3\log k)$ is already enough: repeated squaring of the adjacency matrix $\log k$ times.
3. _$s$-$t$ Connectivity vs. Connectivity_ : If $f=x_1x_2\lor x_3x_4\lor\cdots\lor x_{n-1}x_n$ is a monotone projection of CONN(m), then $m\geq 2^n+1$. This was proved by [Skyum and Valiant](http://dl.acm.org/citation.cfm?id=3158&CFID=497574256&CFTOKEN=86163804). Here CONN is a "sibling" of STCON: accepts a graph iff it is connected. Note that this particular function $f$ _is_ a monotone projection of STCON(m) even for $m=n$: take a bunch of $n/2$ parallel paths of length two.
4. _Formulas vs. Circuits_ : If the depth is restricted to $o(\log n)$ then STCON(n,k) requires even non-monotone(!) unbounded fanin _formulas_ of size $n^{\Omega(\log k)}$ for all $k\leq \log\log n$. This was proved by [Rossman](http://eccc.hpi-web.de/report/2013/169/download), and separated for the first time bounded-depth circuits from formulas, because for such values of $k$, repeated squaring gives the desired small (even monotone) _circuit_ of depth $2\log k\leq 2\log\log\log n=o(\log n)$.
5. _Directed vs. Undirected Connectivity_ : The directed reachability problem DSTCON(n) (where input graphs are directed) is a monotone projection of STCON(m) only if $m=n^{\Theta(\log n)}$. This was proved by [Potechin](http://eccc.hpi-web.de/report/2009/142/revision/1/download), and separated monotone versions of **L** and **NL**.
Does anybody know some results more tightly related to the STCON(n,k) vs. STCON(n) question itself?
* * *
**CORRECTION** (Apr 9): Unfortunately, just taking $k$ copies of the input graph $G$ for STCON(n,k), only shows that STCON(n,k) is a monotone projection of **D** STCON(kn), the **directed** (even acyclic) version of STCON. The difficult direction is to ensure that the modified graph $H=H_G$ has no s-t paths, when $G$ _has_ s-t paths, but all they are longer than $k$. If $H$ is allowed to be directed, this is easy to ensure by arranging the copies of $G$ into $k$ layers, and drawing _directed_ edges between layers according to $G$. But if the edges $H$ are required to be _undirected_ , this simple layering trick won't work. So, I must refine my Question 2 as (cf. with the 5-th result by Potechin above):
> **Question 2':** Is STCON(n,k) a monotone projection of **D** STCON(m) for some $m < kn$?
Both answers would be interesting. If YES, this would show that layering is not the best way to "count". If NO, this would show that every monotone switching network for STCON(n,k) requires $kn$ nodes. (By the Bellman-Ford algorithm - which itself is just the "layering trick" in disguise - $kn$ nodes are also enough.)
|
https://cstheory.stackexchange.com/questions/31005/can-short-distance-connectivity-be-harder-than-connectivity
|
[
"cc.complexity-theory",
"graph-algorithms",
"circuit-complexity",
"lower-bounds",
"dynamic-programming"
] | 17 | 2015-04-03T10:00:23 |
[] | 0 |
Science
| 0 |
23
|
cstheory
|
Sequences with sublogarithmic concat and approximate split
|
Is there a data structure for representing sequences that supports the operations:
* **concat** takes two sequences of size $m$ and $n$ and produces a new sequence of size $m+n$ by joining them in time $o(\lg \min(n,m/n))$ (or $o(\lg \min(m,n/m))$ if $n>m$).
* For some constants $c$ and $N$, **approximate split** takes a sequence of size $n > N$ and divides it into two contiguous non-overlapping sequences of sizes $p$ and $q$ such that $p+q = n$ and $1/c < p/q < c$ in time $o(\lg n)$
* **singleton** takes a single element and produces a sequence containing only that element in $O(1)$ time.
* **linearize** takes a sequence of size $n$ and produces an array of size $n$ containing the elements of the sequence in time $O(n)$.
I'm really looking for worst-case or expected bounds on the time, but amortized bounds would also be interesting. Results in the word RAM would be interesting as well.
Here are some results I am aware of. In this table, **sloppy split** is like approximate split, but the bounds are $1/c < (\lg p)/(\lg q) < c$.
structure | concat | approx split | sloppy split
==================================================================================
doubly-linked lists | O(1) | O(n) | *
arrays | O(n+m) | O(1) | *
weight balanced trees | O(lg (n+m)) | O(1) | *
treaps and skip lists | O(lg min(n,m)) ** | O(1) ** | *,**
functional finger trees | O(min(lg lg m/n, lg lg n) | O(lg n) | O(1)
*: same as approx split
**: expected
By functional finger trees I mean "[Purely functional representations of catenable sorted lists](http://www.math.tau.ac.il/~haimk/papers/loglog23.ps)" by Kaplan and Tarjan.
|
https://cstheory.stackexchange.com/questions/5964/sequences-with-sublogarithmic-concat-and-approximate-split
|
[
"ds.data-structures"
] | 17 | 2011-04-08T22:01:48 |
[] | 0 |
Science
| 0 |
24
|
cstheory
|
Linear-time algorithm to test if clique number equals degeneracy bound?
|
Given a connected simple graph $G=(V,E)$, let $d$ denote its [degeneracy](https://en.wikipedia.org/wiki/Degeneracy_\(graph_theory\)) and let $\omega$ denote the size of a [maximum clique](https://en.wikipedia.org/wiki/Clique_problem).
A well-known bound on the clique number is $\omega\le d+1$, which is helpful when solving the maximum clique problem or when [enumerating all maximal cliques](https://dl.acm.org/doi/abs/10.1145/2543629?casa_token=fxzBWpiA1ZsAAAAA:CrMNWZ-0GU9DU3L0-UE43AniI2H-2gvXWFzx341kOOA8MvUU2QnCVAQUxcpoQQGtfriWot8YZGi9TA).
How fast can one test whether this bound is tight, $\omega=d+1$? An algorithm that runs in time $O(dm)=O(m^{1.5})$ is [known](https://pubsonline.informs.org/doi/10.1287/opre.2019.1970).
Is there a faster algorithm, say, running in linear time?
Or, is $O(m^{1.5})$ best possible under something like [SETH](https://en.wikipedia.org/wiki/Exponential_time_hypothesis)?
|
https://cstheory.stackexchange.com/questions/47947/linear-time-algorithm-to-test-if-clique-number-equals-degeneracy-bound
|
[
"graph-algorithms",
"parameterized-complexity",
"clique",
"fixed-parameter-tractable"
] | 17 | 2020-11-30T14:49:56 |
[
"Maybe you are aware, but this problem recently appeared in codeforces codeforces.com/contest/1439/problem/B. It seems nobody in the discussion mentions anything better than $O(m^{1.5})$.",
"Um...it looks like their article was just published and they list your question as Open Problem 1 (minus the possible connection to SETH)...",
"Ah, I see now. (After your comment, I thought I could answer your Q, but then looked at the article and saw I just rediscovered their algorithm in the case $p=0$.) I wonder if you don't even need SETH - it seems like maybe it's possible to show that if you could do better than $O((m+n) + d^2 n)$ (which is what they have) then either you'd be able to compute $k$-cores faster than linear (not possible by query complexity) or test whether a $d$-vertex graph is a clique in $o(d^2)$ (also not possible by query complexity). I don't quite see a reduction, but maybe one can do it w/o SETH...",
"The decomposition is linear time, but I don’t think that solves the problem. It works, for example, for G=C_3 or G=C_4 but fails for their disjoint union (or for their disjoint union with an added edge between them).",
"Can't you just compute the k-core decomposition (which is quasilinear time IIRC), and then just check if the last core is a clique?"
] | 5 |
Science
| 0 |
25
|
cstheory
|
Is it possible to boost the error probability of a Consensus protocol over dynamic network?
|
Consider the binary consensus problem in a synchronous setting over dynamic network (thus, there are $n$ nodes, and some of them are connected by edges that may change round to round). Given a randomized $\delta$-error Monte Carlo protocol for Consensus in this setting, is it possible to use this protocol as a black box to obtain another Consensus protocol with smaller error?
Note that for problems with deterministic outcome (such as Sum), this can be done by repeatedly executing the protocol and taking the majority as the answer. However I am unable to figure out how for Consensus, so I wonder whether this is actually impossible to do.
\--- Details ---
**Dynamic Network.** There are $n$ nodes on a dynamic network over synchronous setting (thus the protocol will proceed in rounds). Each node has a unique ID of size $\log(n)$. There are edges connecting pairs of nodes, and this set of edges may change from round to round (but remains fixed throughout any particular round). There may be additional restrictions to the dynamic network that allows the problem to be solved by some protocol (such as having a fixed diameter), but it is irrelevant to our discussion.
The only information known to each node is their unique ID. They do not know $n$ or the topology of the network. In particular, they do not know who their neighbors are in the current round (neighbors are nodes that are adjacent to them with an edge). The only way to figure out the neighbors is after receiving messages from them.
**How nodes communicate.** The only method of communication between nodes is by broadcasting messages under the $CONGEST$ model (D. Peleg, "Distributed Computing, A Locality-Sensitive Approach"). More specifically, in each round, every node may broadcast an $O(\log n)$ sized message. Then, every node will receive ALL messages that were broadcasted by all of its neighbors in that particular round in an arbitrary order. In particular, nodes may attach their IDs onto their messages, since their IDs are of size $O(\log n)$.
**Binary Consensus.** The binary consensus problem over this network is as follows. Each node has a binary input ($0$ or $1$), and each must decide either $0$ or $1$ under the the validity, termination, and agreement for consensus with $\delta$ error:
* validity means that for $z \in \\{0, 1\\}$, if all nodes have $z$ as its initial input, then the only possible decision is $z$.
* termination means that each node must eventually decide.
* agreement means that all nodes must have the same decision.
* The error rate means that under worst case input and worst case network, the probability that the protocol does not satisfy any of the three conditions is at most $\delta$ over average coin flips of the protocol's coin.
**Question.** For a particular set of dynamic networks, suppose we have a lower bound on the average number of rounds incurred by any randomized $\delta$-error protocol $P$ for solving binary consensus under this settings such that $\delta < \frac{1}{3}$, under worst case dynamic network from this set, worst case nodes' inputs, and on average coin flips of the protocol. Does this bound automatically hold for any $\delta'$-error protocol for $\delta < \delta' < \frac{1}{3}$?.
|
https://cstheory.stackexchange.com/questions/32442/is-it-possible-to-boost-the-error-probability-of-a-consensus-protocol-over-dynam
|
[
"ds.algorithms",
"reductions",
"dc.distributed-comp"
] | 16 | 2015-08-23T23:24:43 |
[
"@Peter Also this question is more of a whether it is possible to obtain a better protocol using a black box protocol. For example, in Sum, each node can run the protocol multiple times (either in parallel or in a row) and takes a majority. Using Chernoff's bound, for any $0 < \\delta' < \\delta < \\frac{1}{3}$, one can obtain protocol with $\\delta'$ error from a $\\delta$ error protocol for Sum by executing the protocol a constant amount of time and taking a majority as the answer.",
"@Peter No unless your protocol is specifically designed to be able to do that. The only way for the nodes to communicate is through broadcasting and receiving messages. E.g., the lack of message from a particular node may mean something. There are no \"magical\" communication channels otherwise.",
"Do you assume that, when an instance of your blackbox consensus algorithm fails, all nodes are aware that the consensus algorithm has failed? This seems to depend on the precise properties of the dynamic network that you're assuming...",
"suggest try to fit it into the following framework. The Consensus Problem in Unreliable Distributed Systems (A Brief Survey) (2000) / Fisher. there are some impossibility results known & could your idea possibly violate any?",
"@vzn No, I'm not really familiar with this topic. Thus I thought that this question is well-known, and I was hoping for a quick answer here. Seems like this is not well-known then. I wasn't aware of the TCS stackexchange before, I'll try to ask the admin to cross post this question there.",
"are you familiar with Paxos, \"a family of protocols for solving consensus in a network of unreliable processors\"? does it not match the requirements for some reason? there are many (standard) consensus protocols, can you give an example of a \"randomized δ-error protocol for Consensus\"?",
"it appears research level to me! the whole question seems to be built on a research framework! seriously consider Theoretical Computer Science! can you define \"randomized δ-error protocol\"? try also Theoretical Computer Science Chat"
] | 7 |
Science
| 0 |
26
|
cstheory
|
complexity of checking if a subspace is a Euclidean section of L1
|
If $X$ is a linear subspace of ${\mathbb R}^n$, $X$ is high-dimensional, and for every $x\in X$ we have
$(1-\epsilon) \sqrt n ||x||_2 \leq ||x||_1 \leq \sqrt n ||x||_2$
for some small $\epsilon >0$, then we say that $X$ is an almost-Euclidean section of $\ell_1^n$, and (the matrix whose image is) X is useful in compressed sensing. A random subspace works excellently, and there is a huge research program devoted to the **explicit construction** of such spaces.
Is it known what is the complexity of approximating the "Euclidan-sectionness" of $X$? That is, given a subspace $X$, say presented via a basis, consider the problem of finding the unit (in $\ell_2$ norm) vector in $X$ of smallest $\ell_1$ norm.
> What is the complexity of this problem? Are hardness of approximation results known?
Apart from specific applications, these seem to be interesting problems. Is it known what is the complexity of finding the vector of **maximum** $\ell_1$ norm among the unit vectors of a given subspace?
|
https://cstheory.stackexchange.com/questions/4992/complexity-of-checking-if-a-subspace-is-a-euclidean-section-of-l1
|
[
"cc.complexity-theory",
"cg.comp-geom",
"norms",
"compressed-sensing"
] | 17 | 2011-02-17T18:27:44 |
[
"This is not an answer to your question. But in the vein of computational problems arising from compressed sensing applications, Koiran and Zouzias have a recent paper on checking whether a matrix satisfies the restricted isoperimetry property (RIP) and related problems.",
"Thanks Sariel, I had inverted the places of $\\ell_1$ and $\\ell_2$. (If you want $\\ell_2$ in the middle, then it should be $||x||_1/\\sqrt n \\leq ||x||_2 \\leq (1+\\epsilon)||x||_1 / \\sqrt n$, because it's always $||x||_1 \\leq \\sqrt n ||x||_2$.)",
"The above inequality is slightly wrong. You probably mean that $(1-\\varepsilon)||x||_1 /\\sqrt{n} \\leq ||x||_2 \\leq ||x||_1/\\sqrt{n}$. Indeed, for any $x$, $||x||_1 \\geq ||x||_2$."
] | 3 |
Science
| 0 |
27
|
cstheory
|
Looking for an operator on polynomials
|
I have a small, self-contained, math question, whose motivation is from theoretical computer science (specifically, list decoding of algebraic codes, derivative/multiplicity codes, etc). I wonder whether someone might have an idea.
I'm looking for an operator T that can be applied to m-variate polynomials over a finite field. When applied to a polynomial, the operator should yield a polynomial of not much higher degree. The operator should satisfy the following property: for every m-variate polynomial $F$, and m univariate non-constant polynomials $g_1,...,g_m$, if you know $g_1,...,g_m$ and $F(g_1(t),...,g_m(t))$ for a parameter t, then you can also compute $TF(g_1(t),...,g_m(t))$. The operator should be non-trivial, in the sense that for a fixing $x_1,..,x_m$, the value $F(x_1,...,x_m)$ does not determine the value of $TF(x_1,...,x_m)$.
For $m=1$, the derivative operator gives exactly that: By the chain rule $(F(g(t)))' = F'(g(t))\cdot g'(t)$, which implies that $F'(g(t)) = (F(g(t)))'/g'(t)$. My question is whether there is an operator that works for general m.
|
https://cstheory.stackexchange.com/questions/10881/looking-for-an-operator-on-polynomials
|
[
"it.information-theory",
"coding-theory",
"algebra",
"polynomials"
] | 16 | 2012-03-28T08:17:46 |
[
"Honestly, I asked this question so long ago, I only barely remember what I needed it for :-)... I believe that the issue is that multiplying by a fixed polynomial is \"trivial\" in the sense that to compute the multiplication at a point, you only need to know the value of F at the point (and nothing about F's inner-workings).",
"Dana, I may be missing something, what about $TF(x_1,...,x_n) = T(x_1,...,x_n) P(x_1, ..., x_n)$, where $P$ is some multivariate polynomial?",
"F(g(t))' and g'(t) are formal polynomials in t. You can divide the formal polynomials.",
"Suppose $g(t) = t^2$. You still can't evaluate $F'$ at 0? (I understand that you want point-wise evaluation)",
"If g' is identically zero then you can't expect anything (I'll edit the question to specify the g_i's are not identically zero).",
"Your example in one dimensional case is not flawless. What if $g'(t)=0$ ?",
"Suppose you take $TF(g_1(t),..,g_m(t)) = \\sum_{i=1}^{m} g_i'(t) dF(x_1,...,x_m)/dx_i$ (known as the total derivative). How you do compute $dF(x_1,...,x_m)/dx_i$ from $F(g_1(t),...,g_m(t))$?",
"Why total derivative wouldn't fit your requirements?"
] | 8 |
Science
| 0 |
28
|
cstheory
|
When does adding edges decrease the cover time of a graph?
|
When first learning about random walks on a graph $G$, one may have an intuitive feeling that adding edges to $G$ will decrease its cover time $C(G)$. However, this is not the case. The [path graph](https://en.wikipedia.org/wiki/Path_graph) $P_n$ has cover time $C(P_n) = \Theta(n^2)$ while the [($\frac{n}{2}$, $\frac{n}{2}$)-lollipop graph](http://mathworld.wolfram.com/LollipopGraph.html) $L_{\frac{n}{2}, \frac{n}{2}}$ has cover time $C(L_{\frac{n}{2}, \frac{n}{2}}) = \Theta(n^3)$. Eventually adding edges does decrease the cover time because the complete graph $K_n$ has cover time $C(K_n) = \Theta(n \log n)$ by analogy to the [coupon collector problem](https://en.wikipedia.org/wiki/Coupon_collector%27s_problem).
Under what assumptions though would the supposed intuition hold true? Specifically, what about [vertex transitivity](https://en.wikipedia.org/wiki/Vertex-transitive_graph)?
> Let $G$ be a vertex-transitive graph with $n$ vertices. Consider any vertex-transitive graph $G^+$ obtained from $G$ by adding edges (i.e. both $G$ and $G^+$ are vertex transitive and $G$ is a spanning subgraph of $G^+$).
>
> Is it the case that $C(G^+) \le C(G)$?
|
https://cstheory.stackexchange.com/questions/33071/when-does-adding-edges-decrease-the-cover-time-of-a-graph
|
[
"graph-theory",
"pr.probability"
] | 16 | 2015-11-12T18:35:47 |
[
"My bad, I confused spanning with induced.",
"@chazisop Those subgraphs are not spanning.",
"Wouldn't the complete graph $K_{n}$ and any of its subgraphs $K_{k}$, $k < n$ be a counterexample for general vertex transitivity?"
] | 3 |
Science
| 0 |
29
|
cstheory
|
Complexity of approximating the range of a matrix
|
Given an $m$ by $n$ matrix $M$ with $m \leq n$ and elements from $\\{-1,1\\}$, let us define:
$$S_M = |\\{Mx : x \in \\{-1,1\\}^n\\}|.$$
I believe that it is NP-hard to compute $S_M$ exactly, by applying the reductions from [Decide whether a matrix's kernel contains any non-zero vector all of whose entries are -1, 0, or 1](https://cstheory.stackexchange.com/questions/20277/decide-whether-a-matrixs-kernel-contains-any-non-zero-vector-all-of-whose-entri) to the following decision problem: does $S_M = 2^n$?
> Is it possible to approximate $S_M$ to within a constant factor in polynomial time? If not, what is the best one can do in polynomial time?
[Cross-posted to <https://mathoverflow.net/questions/229852/complexity-of-approximating-the-size-of-the-range-of-a-matrix> ]
|
https://cstheory.stackexchange.com/questions/33676/complexity-of-approximating-the-range-of-a-matrix
|
[
"cc.complexity-theory",
"approximation-algorithms",
"linear-algebra"
] | 15 | 2016-01-28T02:00:10 |
[
"(1) If there is a poly-time $2^{n^{1-\\epsilon}}$-approximation for any $\\epsilon>0$, then there is a poly-time $(1+1/q(n))$-approximation for any polynomial $q$. To see this, given $M$, consider the block-diagonal matrix $M(k)$ formed by $k$ independent copies of $M$ along the diagonal, so $S_M=(S_{M(k)})^{1/k}$. Take $k=(n q(n))^{1/\\delta}$. If $x$ is a $2^{(nk)^{1-\\epsilon}}$-approximation of $S_{M(k)}$, then $x^{1/k}$ is a $(1+1/q(n))$-approximation of $S(M)$. (2) One has $2^k\\le S_M\\le 2^n$ where $k$ is the rank of $M$, but this doesn't help as $S_M$ can be large even when $k=m=1$."
] | 1 |
Science
| 0 |
30
|
cstheory
|
Intersecting Complexity Classes with Advice
|
In [on hiding information from an oracle](http://linkinghub.elsevier.com/retrieve/pii/0022000089900184), the authors (Abadi, Feigenbaum, and Kilian) wrote:
> $(\mathsf{NP/poly} \cap \mathsf{co\text-NP}{/poly})$ ... is **not known** to be equal to $(\mathsf{NP} ∩ \mathsf{co\text-NP}){/poly}$.
They highlighted that in the conference paper, they mistook the two classes. Apparently, the latter is a subset of the former, but we don't know if the containment is strict.
Assuming $X$ and $Y$ are complexity classes, and $F$ is a set of functions specifying the length of the advice strings, are there recent results comparing $(X_{/F} \cap Y_{/F})$ and $(X \cap Y)_{/F}$, resolving issues like the one pointed above?
|
https://cstheory.stackexchange.com/questions/5839/intersecting-complexity-classes-with-advice
|
[
"cc.complexity-theory",
"reference-request",
"complexity-classes",
"advice-and-nonuniformity"
] | 15 | 2011-04-03T03:17:43 |
[
"@HenryYuen: I think that a Language $L$ is in $(NP \\cap coNP)/poly$ iff there is a language $K$ in $NP \\cap coNP$ and $a_i$ advice of polynomial length s.t. $x \\in L$ iff $(x, a_{|x|}) \\in K$.",
"@HenryYuen: Oh, I got your point. Strangely, the definition of (NP ∩ coNP)/poly in Complexity Zoo (qwiki.stanford.edu/index.php/Complexity_Zoo:N#npiconppoly) does not mention machines with advice, but \"languages\" with advice. However, quoting from An Oracle's Builder Toolkit: \"NP ∩ coNP is defined by the enumeration $M_1,M_2,\\ldots$, where $M_i$ with $i = \\langle i_1,i_2 \\rangle$ is the pair of non-deterministic oracle Turing machines $N_{i_1}$ and $N_{i_2}$ such that both machines run in time $n^i$...\" (Read page 27 for more info)",
"The reason I ask about the computational model is because I don't know what (NP ∩ coNP)/poly means. NP/poly is the set of languages that are accepted by NP machines with polynomial amount of advice. Presumably, (NP ∩ coNP)/poly is the set of languages accepted by (NP ∩ coNP) machines with polynomial advice -- or is there another definition that isn't machine-definition-dependent?",
"@HenryYuen: I actually don't know if there's an underlying model of computation, but the definitions of complexity classes are clear, as in the case of ZPP. Besides being a very interesting problem, is there any significance (to my question) if NP ∩ coNP does (or does NOT) admit a computation model?",
"If you take the definition that ZPP = RP ∩ coRP, that doesn't immediately give you a model of computation that accepts only ZPP languages: it only says that every language in ZPP has both an RP machine and a coRP machine (which is well defined). You have to prove that ZPP is the class of languages that admit Las Vegas algorithms. Similarly -- NP ∩ coNP defines a set of languages that have both NP and coNP machines, but is there a model of computation that accepts precisely NP ∩ coNP?",
"@HenryYuen: Maybe the definition of ZPP helps: ZP = RP ∩ CoRP. See en.wikipedia.org/wiki/ZPP_(complexity)#Intersection_definition.",
"What is the definition of (NP ∩ coNP)/poly? What is a NP ∩ coNP machine?"
] | 7 |
Science
| 0 |
31
|
cstheory
|
Intermediate problems between PSPACE and EXPTIME
|
Intermediate problems between P and NP are quite famous, and are sometimes considered as complexity classes by themselves.
Do you know of any problem that is known to be PSPACE-hard and in EXPTIME, and resisting all efforts to be proved complete for one of these classes ?
Lifted (succinct) versions of problems between LOGSPACE and PTIME are accepted, but even more interesting would be problems that are not of this form.
Here are some I found, this area of complexity theory seems to be the realm of games:
* Phutball: <https://arxiv.org/abs/0804.1777>
* Go (with Chinese rules): D. Lichtenstein and M. Sipser, Go is polynomial-space hard
|
https://cstheory.stackexchange.com/questions/38393/intermediate-problems-between-pspace-and-exptime
|
[
"cc.complexity-theory",
"complexity-classes",
"exp-time-algorithms",
"pspace"
] | 17 | 2017-06-09T06:50:47 |
[
"I think the more correct way of stating the question is asking for candidate problems in ExpTime - PSpace - ExpTime-hard. Note that e.g. in the case of P vs. NP we don't know if Factoring or GI is P-hard."
] | 1 |
Science
| 0 |
32
|
cstheory
|
Mutual information vs. Product sets
|
Suppose we have two _dependent_ random variables $X$ and $Y$, each of which is uniform over $\\{0,1\\}^n$, such that their mutual information $I(X;Y)$ is small, say, at most $\sqrt{n}$. Does this imply that there exist large sets $\mathcal{X}, \mathcal{Y} \subset \\{0,1\\}^n$ such that the product set $\mathcal{X} \times \mathcal{Y}$ is contained in the support of the distribution $(X,Y)$?
A variant of this question can be phrased in graph-theoretic terms: Suppose we have a sufficiently dense bipartite graph G. Does this imply that $G$ contains a large complete sub-graph $G'$? (i.e., $G'$ is required to be a complete bipartite graph with bi-partition $(A,B)$ such that $A$ and $B$ are relatively dense in the corresponding parts of $G$)
|
https://cstheory.stackexchange.com/questions/10225/mutual-information-vs-product-sets
|
[
"co.combinatorics",
"it.information-theory"
] | 15 | 2012-02-15T17:49:06 |
[
"If $(X,Y)$ is a product of iid trials $(X_i,Y_i)$ then I am inclined to say yes.",
"If my calculation is correct, a standard use of the probabilistic method shows that for sufficiently large N, there exists a bipartite graph on N+N vertices with at least (1/2)N^2 edges that does not contain K_{k,k} as a subgraph, where k = floor(2.001 lg N). (“lg” denotes the logarithm to base two.)",
"I'm not an expert, but perhaps this paper on Ramsey theory applied to bipartite graphs can help you: math.mit.edu/~fox/paper-density-theorems-final.pdf"
] | 3 |
Science
| 0 |
33
|
cstheory
|
Computability of a "weird" set
|
The starting point of this question is the observation that the smallest positive integers $a,b,c$ satisfying
$$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4$$
are [absurdly high](https://plus.google.com/+johncbaez999/posts/Pr8LgYYxvbM). This leads to the following general question: Is the set $C\subseteq {\mathbb N}$ defined by $$ C = \left\\{n\in\mathbb{N}\setminus\\{0\\}: \big(\exists a,b,c \in\mathbb{N}\setminus\\{0\\}\big):\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n\right\\}$$ computable?
|
https://cstheory.stackexchange.com/questions/39383/computability-of-a-weird-set
|
[
"computability"
] | 14 | 2017-10-26T07:49:40 |
[
"Cross-posted from mathoverflow.net/questions/278747/is-this-set-computable",
"I'm not an expert, but if you simplify the equation you get a cubic diophantine equation in three variables, and I think it's an open problem if universality can be achieved in such a setting. In every case, quadratic diophantine equations in two variables ($ax^2 + by + c = 0$) are already NP-complete (see NP-complete decision problems for quadratic polynomials) ... so I'm not surprised if switching to cubic+three vars leads to very hard instances (outside NP)."
] | 2 |
Science
| 0 |
34
|
cstheory
|
Can a Penrose tile cellular automaton be Turing-complete?
|
This question was based on an incorrect premise ... see Colin's comment below. Forget it.
This was inspired by the discussion on [this Math Overflow question](https://mathoverflow.net/questions/45378/undecidability-in-conways-game-of-life/45382). First, I need to define our terms.
In a Penrose tiling, there are generally several different legal patterns of tiles which will cover a given region. A Penrose tile cellular automaton should be some rule that deterministically takes a Penrose tiling $T$ to a Penrose tiling $T'$, where each tile in $T'$ is deterministically given by looking at the tiles in a radius $r$ around the corresponding spot in $T$. You should be able to implement these by having a set of rules of the form: when a region of shape $S$ is covered by a pattern $X$ of tiles, replace it with a pattern $Y$ of tiles. In order to do this deterministically, you either have to make sure no two of your replaceable regions can overlap or have some priority ranking on these replacement rules. But certainly you can ensure this, and define lots of different Penrose tiling cellular automata.
What does it mean for a Penrose tiling celluar automaton to be Turing-complete. Here's what I think it should mean (although I won't insist on it in the answers): the halting problem should be reducible to the behavior of such an automaton. Let's take some canonical Penrose tiling $T_0$ that provides constraints on our starting position. We start with an input $G$ to a universal Turing machine, and we'd like to decide whether the machine halts on input $G$. I would like some computable map from $G$ to a starting Penrose tiling $S$ such that $S$ differs from $T_0$ on only a finite set of tiles. We then run the Penrose tiling cellular automaton, starting with $S$, and wait for some finite canonical configuration $C$ at position $0$ in the Penrose tiling. I want configuration $C$ to appear at position $0$ if and only if the universal Turing machine halts on input $G$.
The strange thing about Penrose tiling cellular automata is that every finite legal configuration of tiles appears infinitely often in every Penrose tiling of the plane. Thus, our universal Penrose tiling is not only computing the behavior of our universal Turing machine on input $G$, it's also simultaneously carrying out the first steps of every other possible computation. This makes me doubt whether Penrose tiling cellular automata could be Turing complete. On the other hand, if they're not Turing complete, what is the class of computations that Penrose tiling cellular automata can perform?
|
https://cstheory.stackexchange.com/questions/2883/can-a-penrose-tile-cellular-automaton-be-turing-complete
|
[
"computability",
"machine-models",
"cellular-automata"
] | 15 | 2010-11-11T06:14:56 |
[
"Yes, that is true for tilings formed via the deflation process. But an arbitrary planar tiling might have a \"rift\" along a line in the plane, along which the higher level tiles do not match up (like a tiling of the plane by squares in which the upper half plane is shifted right by a fraction of a unit). This is certainly possible with Robinson tiles and Ammann tiles, and I would expect it also for the various Penrose tiles although I haven't checked. If rifts can occur, local patches along the rift will appear infinitely often in the same tiling but needn't appear in another tiling.",
"@Matt: What I should have said is that every finite pattern that appears once in an infinite Penrose tiling appears infinitely often in every Penrose tiling.",
"If every finite pattern would occur, you would not be able to \"get stuck\" when attempting to tile the plane (because your finite pattern, whatever it is so far, would be guaranteed to be part of a valid tiling of the plane). But in fact, you do need to know what you are doing in order to not get stuck, regardless of which Penrose tile set you are talking about. Only very occasionally do you really have a choice of what tile to place (in the sense that either tile would be ok for creating a full tiling of the plane), approximately once per inflation scale level.",
"Yes, it's 14 years later, but I can't help noticing that something is wrong on the internet (like the xkcd cartoon). It is not correct that every finite pattern occurs in a Penrose tiling.",
"@Peter: This question intrigued me because it seems related to one of the main open problems in tile self-assembly, which is \"the power of multiple nucleation.\" If the Penrose tiling rule were somehow limited to radiate out one tile at a stage from the origin, instead of everywhere being tileable at once, I think the model would be very close to self-assembly models. The open questions there are complexity-related, not computability, e.g.: is there a finite shape such a restricted Penrose tiling could achieve that a GCA could not achieve with only polynomial complexity increase?",
"@Colin, Actually, I was wrong about this. Forget the question. Maybe I should delete it.",
"Can you give an example of a region which may be legally covered by different patterns of tiles? I'm not familiar with the subject, but there is a unique way to break up a triangle into smaller triangles in \"up-down\" generation - see for example Prop 6.1 of Quasicrystals and geometry.",
"If someone reduced the Turing Machine Immortality Problem to a Penrose CA, would that be progress? (TMIP = given a TM with an input tape with infinitely many nonblank symbols on it, does it halt?) That problem is undecidable for TMs.",
"I don't think the existence of every finite pattern is any evidence against Turing completeness - in the game of life it seems plausible that there is a (finite) initial state that runs every turing machine G and puts a \"halt\" marker at some position f(G) if and only if G halts.",
"I hope it's clearer now. I want the rules to be deterministic and local, but exactly how they're formulated is less imporant.",
"typo: I meant \"$\\delta \\in S$\"",
"Okay, second round. There are Generalized Cellular Automata, where each cell $x$ transitions based on a \"stencil\" $\\delta_x$, which is a neighborhood of tiles around $x$, and the size and shape of these neighborhoods may vary from cell $x$ to cell $y$. Is a Penrose CA a generalization of GCA's, where now each $x$ has a stencil set $S$ (perhaps common to all cells) and some $\\delta \\subseteq S$ can be applied to $x$ and its surrounding cells? This then leads to multiple possible tilings of the plane for a single start state. Does every (\"nice\"?) start state converge to a stable tiling?",
"The replacement rules are indeed connected templates of finite size. The problem with your periodic \"halting\" state is that it can't happen. You always can find any specific finite pattern in a Penrose tiling, so if any Penrose tiling CA has a starting state that is eventually periodic, then any starting state for that Penrose tiling CA is periodic with that period.",
"I am confused by your definition, sorry. Is a Penrose CA just like a normal (2D) CA, except that the replacement rules are connected templates of finite size, instead of a local transition function that is applied cell by cell? So there's an initial configuration at time step 0 that specifies the input, and the connected templates specify the program? One way small models of computation specify \"halting\" is to repeat a special state infinitely often. So maybe if the simulated TM halts the Penrose CA keeps recreating a certain pattern, and if the TM runs forever, anything goes."
] | 14 |
Science
| 0 |
35
|
cstheory
|
NP-Hardness of 4-cycle packing problem in complete bipartite digraph?
|
A directed complete bipartite graph is a bipartite graph where there is exactly one directed edge between any two vertices from its two different parts. In other words, it's an orientation of a complete bipartite graph.
Given a directed complete bipartite graph, we are asked to find the largest set of edge-disjoint 4-cycles. Note that these 4-cycles can share node(s).
Is this problem NP-Hard? I tried to reduce from 2P2N-3SAT to this problem, but failed. On the other hand, I found no polynomial algorithm could solve it optimally.
|
https://cstheory.stackexchange.com/questions/45998/np-hardness-of-4-cycle-packing-problem-in-complete-bipartite-digraph
|
[
"graph-theory",
"np-hardness"
] | 14 | 2019-12-10T00:40:50 |
[] | 0 |
Science
| 0 |
36
|
cstheory
|
Is there a P-complete language X such that succinct-X is in P?
|
I came across a paper called "A Note on Succinct Representation of Graphs". It seems that in the discussion section they claim that for any problem $X$ that is $\mathrm{P}$-hard under projections, $\mbox{succinct-X}$ is $\mathrm{EXP}$-hard.
This made me wonder if the theorem fails for more complicated kinds of reductions. This led me to the following.
> **My Question**
>
> Are there any problems $X$ such that $X$ is $\mathrm{P}$-complete under logspace reductions and $\mbox{succinct-X}$ is in $\mathrm{P}$?
**Additional Questions:**
(1) If this is an open problem, then what are the implications if such an $X$ existed?
(2) If this is an open problem, is it still open is we weaken the requirement to $\mbox{succinct-X}$ in $PH$?
(3) It seems that such an $X$ would also satisfy that $X$ is $\mathrm{P}$-hard under logspace reductions, but not $\mathrm{P}$-hard under projections. Are such problems known to exist?
(4) I might be mistaken, but I think that I have a construction for such an $X$ assuming that $NP = L$. Would this be interesting?
**Note:** I'm still learning about projections so please let me know if I made any mistakes. Thank you!
|
https://cstheory.stackexchange.com/questions/42185/is-there-a-p-complete-language-x-such-that-succinct-x-is-in-p
|
[
"cc.complexity-theory",
"complexity-classes",
"reductions",
"succinct"
] | 14 | 2019-01-09T22:34:19 |
[
"Werent p-projections used in Valiants' original completeness and algebra paper?",
"@EmilJeřábek Yes, I noticed this too. :)",
"Seeing that projections are a restricted class of polylogtime reductions, let me add that what I wrote above about logtime reductions also applies to polylogtime reductions.",
"@EmilJeřábek Thank you for the clarification! I really appreciate it.",
"Yes, that's it. (Note that logtime Turing machines are defined so that they receive input by means of a query tape that requests individual bits instead of the usual input tape, hence they readily take input represented by a circuit.) For pspace-succinct-$X$, one can take a variant of succinct-$X$ where the input is not represented by an ordinary Boolean circuit, but by a quantified Boolean formula (or circuit).",
"@MichaëlCadilhac Thanks Michaël! I appreciate the helpful suggestions. :)",
"@user3483902 Thank you very much for the good questions! I'm still learning about projections so I think it's best if I refer you to a reference that includes a definition for time bounded projections: theoryofcomputing.org/articles/v013a004",
"@EmilJeřábek In regards to the second part of your comment, did you mean the following? If $X$ is $P$-complete under logspace reductions, it doesn't necessarily mean that $\\mbox{succinct-X}$ is $EXP$-hard. But, $\\mbox{pspace-succinct-X}$ would be $EXP$-hard where $\\mbox{pspace-succinct}$ is a more powerful notion of succinctness (that we haven't formally defined) that offers a higher compression ratio in some cases.",
"@EmilJeřábek Now, although $x$ has length $n$ and $pad$-$x$ has length $2^n$, $pad$-$x$ can be encoded as a circuit of size $n$. Further, because $r$ is in dlogtime, it can be represented by a circuit of size $n^2$ on inputs of size $2^n$. And, it seems that we can somehow combine these two circuits to get a circuit $c$ of size roughly $n^3$ that encodes $\\mbox{r(pad-x)}$ so that $x \\in L$ iff $c \\in \\mbox{succinct-X}$.",
"@EmilJeřábek Thanks for the helpful comment! So if I understand the first part correctly, the proof would go something like this? Given a language $X$ that is $P$-complete under dlogtime reductions and a language $L$ in $DTIME(2^n)$. Consider padding $L$ to get a language $pad$-$L$. Since $pad$-$L$ is in $P$, there is a dlogtime reduction $r$ from $pad$-$L$ to $X$. We have that for every input string $x$, $x \\in L$ iff $pad$-$x \\in pad$-$L$ iff $\\mbox{r(pad-x)} \\in X$.",
"A simple padding argument shows that if $X$ is P-complete under dlogtime reductions, then succinct-$X$ is EXP-complete (in fact, still under dlogtime reductions). I thought at first this would generalize to less restrictive notions of reductions so that, e.g., P-completeness of $X$ under logspace reductions would imply EXP-completess of succinct-$X$ under PSPACE reductions, but it does not seem to work that way; rather, it gives the EXP-completeness under polytime (or even more efficient) reductions of \"PSPACE-succinct\" $X$, if you get what I mean by that.",
"what is a \"projection\" , does it pertain to reductions only - reductions with logspace requirements have this property, or something that imposes succintness to the representation?",
"My 2¢ here: Question 3 certainly looks like a prerequisite, so it may be worth asking separately. If this doesn't get an answer, you may want to contact P-hardness specialists such as P. McKenzie."
] | 13 |
Science
| 0 |
37
|
cstheory
|
Fourier spectrum of the parity of two monotone Boolean functions
|
This is a question that I've been pondering, on and off, for a while, and unsuccessfully. I'd be very interested in any insight regarding this conjecture. (Or rather, these conjectures.)
Recall that, given a Boolean function $f\colon \\{-1,1\\}^n \to \\{-1,1\\}$, the Kahn—Kalai—Linial theorem states that $$\max_{i\in[n]} \operatorname{Inf}_i f \geq \operatorname{Var}[f]\cdot \Omega\\!\left(\frac{\log n}{n}\right) \tag{1} $$ from which we get that, for any _monotone_ Boolean function $f\colon \\{-1,1\\}^n \to \\{-1,1\\}$, $$\max_{i\in[n]} \widehat{f}(i) \geq \operatorname{Var}[f]\cdot \Omega\\!\left(\frac{\log n}{n}\right) = (1-\widehat{f}(0)^2)\cdot \Omega\\!\left(\frac{\log n}{n}\right) \tag{2} $$ (writing $\widehat{f}(i)$ for $\widehat{f}(\\{i\\})$, $i\in\\{0,\dots,n\\}$). Moreover, (2) is tight, as shown by considering the $\textsf{Tribes}_n$ function. In particular, this implies that $$ W^{(0)}[f]+W^{(1)}[f] = \sum_{i=0}^n \widehat{f}(i)^2 = \Omega\\!\left(\frac{\log^2 n}{n^2}\right) \tag{3} $$ for any monotone Boolean function $f\colon \\{-1,1\\}^n \to \\{-1,1\\}$, where $W^{(k)}[f] \stackrel{\rm def}{=} \sum_{S: \lvert S\rvert =k} \widehat{f}(S)^2$.
Now, consider _two_ monotone Boolean functions $f,g\colon \\{-1,1\\}^n \to \\{-1,1\\}$, and let $h\stackrel{\rm def}{=}fg$ be their parity. It is easy to see that we could have $W^{(0)}[h]+W^{(1)}[h]=0$, e.g. by considering $f,g$ to be two different dictator functions (but in that very specific case, $W^{(2)}[h]=1$). _But must there be some non-negligible Fourier mass on the first 3 Fourier levels, then?_
> What can we say about $W^{(0)}[h]+W^{(1)}[h]+W^{(2)}[h] = \sum_{S: \lvert S\rvert \leq 2} \widehat{h}(S)^2$?
**Conjecture 1.** For any two monotone Boolean functions $f,g\colon \\{-1,1\\}^n \to \\{-1,1\\}$, one must have $W^{(0)}[fg]+W^{(1)}[fg]+W^{(2)}[fg] > 0$.
Actually, I'd be actually inclined to believe the following stronger statement:
**Conjecture 2.** For any two monotone Boolean functions $f,g\colon \\{-1,1\\}^n \to \\{-1,1\\}$, one must have $W^{(0)}[fg]+W^{(1)}[fg]+W^{(2)}[fg] \geq \frac{1}{\operatorname{poly}(n)}$.
_Side note:_ this would have implications about _weak learning_ of the class of "2-monotone" functions, which are exactly those functions obtained as parity or anti-parity of $2$ monotone functions. (See e.g. [BCOST15].) But more importantly, this is a very simple-looking question, that has been nagging at me for way too long.
* * *
[BCOST15] Eric Blais, Clément L. Canonne, Igor Carboni Oliveira, Rocco A. Servedio, Li-Yang Tan. _Learning Circuits with few Negations._ APPROX-RANDOM 2015: 512-527
|
https://cstheory.stackexchange.com/questions/37722/fourier-spectrum-of-the-parity-of-two-monotone-boolean-functions
|
[
"boolean-functions",
"fourier-analysis",
"monotone"
] | 14 | 2017-03-07T06:55:43 |
[
"@daniello As far as I can tell, not much, at least in terms of blackbox interpretation of the results. The question above would have applications to weak learning (learning to advantage $1/2+1/\\mathrm{poly}(n)$), while the paper you link shows lower bound on testing (and a relaxation, parameterized testing). As far as I am aware, there is no direct connection between the two (the general connection is \"(strong) learning implies testing\").",
"Can you shed some light on what arxiv.org/pdf/1705.04205.pdf means for this question?"
] | 2 |
Science
| 0 |
38
|
cstheory
|
Is it possible to find the median with a linear size sorting network?
|
Is there a sorting network that makes only $O(n)$ comparisons and finds the median?
The AKS sorting network sorts with $O(\log n)$ parallel steps, but here I am only interested in the number of comparisons. The median of medians algorithm finds the median with $O(n)$ comparisons, but it cannot be implemented as a sorting network.
Remark. In fact, in a recent work, we needed a version that is less powerful than pairwise comparison and resembles sorting networks. In our model one could input two elements, $a$ and $b$, and the output was either "a" or "b", such that the output is at least as big as the other number. (In case of equality, either one of them, and we are interested in worst case complexity.) In this variant we could prove that there is a solution with $O(n)$ comparisons. Of course I am also interested if anyone knows anything about this model ever being studied.
|
https://cstheory.stackexchange.com/questions/27765/is-it-possible-to-find-the-median-with-a-linear-size-sorting-network
|
[
"ds.algorithms",
"sorting",
"sorting-network",
"selection"
] | 14 | 2014-12-09T06:40:04 |
[
"Actually the comparator circuit model is a bit different, since you are allowed to repeat inputs and their negations. What you are looking for is a comparator network for majority.",
"@Yuval Your result seems to be the only hit on google for \"comparator circuit complexity\" and even there I cannot see the definition as the minimum number of gates needed, but I suppose you mean the same thing as I do.",
"Equivalently, you are asking what is the comparator circuit complexity of (Boolean) majority."
] | 3 |
Science
| 0 |
39
|
cstheory
|
Exponential-time factorization of polynomials
|
Let an _explicit_ field be a field for which equality is decidable (in some standard model of computation). I am interested in the factorization of univariate polynomials over an explicit field.
It is known that over some explicit fields, testing irreducibility of polynomials is undecidable [1]. Though, over many fields, the irreducible factorization of polynomials is known to be computable in polynomial time, either deterministically (over $\mathbb Q$ [2] or any number field) or probabilistically (over finite fields [3]). Here the size of the input is the size of the _dense representation_ of the polynomial, that is the list of all its coefficients.
> Is there an explicit field for which
>
> * either the best known algorithm has an exponential running time,
> * or (better) there exists an exponential lower bound for irreducibility testing?
>
**Quick clarification:** There may be very different _explicit fields_ , depending on the complexity (for instance) on the equality test or of the standard operations (addition, multiplication). Actually, I am not aware of any result about polynomial factorization which is not a polynomial time upper bound, or undecidability. So I am interested in any kind of different result (exponential time is just an example). Conditional lower bounds are of interest too.
**References.**
[1] A. Fröhlich et J. C. Shepherdson (1955): [On the factorisation of polynomials in a finite number of steps](http://dx.doi.org/10.1007/BF01180640).
[2] A. K. Lenstra, H. W. Lenstra Jr., L. Lovász (1982): [Factoring polynomials with rational coefficients](http://dx.doi.org/10.1007/BF01457454).
[3] E.R. Berlekamp (1967): [Factoring Polynomials Over Finite Fields](http://www3.alcatel-lucent.com/bstj/vol46-1967/articles/bstj46-8-1853.pdf).
|
https://cstheory.stackexchange.com/questions/18502/exponential-time-factorization-of-polynomials
|
[
"algebraic-complexity",
"polynomials",
"exp-time-algorithms"
] | 14 | 2013-07-30T01:12:45 |
[
"Thanks @JoshuaGrochow! I found the paper incredibly easy to read, even though my German lessons are quite old now...",
"Depending on your definition of \"explicit field\", the result you cite as [1] was also proven by van der Waerden much earlier (1930). Amazingly, he proved this before the notion of computability had been formally defined! See: ams.org/mathscinet-getitem?mr=1512605",
"I was aware while writing my question that some this question should be addressed. Actually, I am interested in both cases. I would be very much interested by the implication to ETH you mention. In other words, I am also interested in conditional lower bounds.",
"What do you mean by \"explicit field\" ? Are the operations is the field given by oracles or by algorithms ? I am asking because if the field's operations are given by efficient algorithms proving an exponential lower bound for factoring seems to imply of proof of ETH."
] | 4 |
Science
| 0 |
40
|
cstheory
|
Pseudorandom functions in ACC^0?
|
In the lower bound result by Ryan Williams (Non-uniform $\mathsf{ACC}$ circuit lower bounds), there is a mention of "little evidence that Pseudorandom function generators exist in $\mathsf{ACC}^0$. Is there any development in this regard that might be of interest? (Even old results implying something along these lines will be great.)
|
https://cstheory.stackexchange.com/questions/10220/pseudorandom-functions-in-acc0
|
[
"cc.complexity-theory",
"reference-request",
"cr.crypto-security",
"circuit-complexity",
"natural-proofs"
] | 14 | 2012-02-15T01:26:37 |
[
"The following paper doesn't answer your question but does show somewhat simpler candidate PRFs ccs.neu.edu/home/viola/papers/spn.pdf"
] | 1 |
Science
| 0 |
41
|
cstheory
|
DPLL and Lovász Local Lemma
|
Let $\varphi$ be a CNF formula. Suppose that each of $\varphi$'s clauses consist of exactly $t$ literals (and, moreover, all literals within one particular clause correspond to different variables). It is well known that if every clause has less than $2^t / e$ clauses that share variables with it, then $\varphi$ is satisfiable (let us call such formulae _easy_). Satisfiability can be proved easily using [Lovász local lemma](http://en.wikipedia.org/wiki/Lov%C3%A1sz_local_lemma). Moreover, using [a recent result](http://arxiv.org/abs/0903.0544) by Moser and Tardos one can show that one of the satisfying assignments can be found in polynomial expected time using the following very simple procedure:
* Pick a random assignment
* While there exists an unsatisfied clause, resample all its variables.
On the other hand, most of modern SAT solvers are [DPLL](http://en.wikipedia.org/wiki/DPLL_algorithm)-based. It means that they try to find a satisfying assignment using brute force with two simple prunings:
* If a formula contains clause with one literal, then we can fix it.
* If one variable occurs in a formula only with (or without) negation, then we can fix it.
**The question: is it true that a version of DPLL that splits on random variables finds a satisfying assignment of any easy formula in polynomial expected time?**
|
https://cstheory.stackexchange.com/questions/7720/dpll-and-lov%c3%a1sz-local-lemma
|
[
"cc.complexity-theory",
"ds.algorithms",
"sat"
] | 14 | 2011-08-13T02:32:29 |
[
"@ilyaraz Still hoping you'll post an official answer to this question.",
"@Stasys Actually, Dmitriy Itsykson (who you probably know) convinced me that the answer is negative. I'll elaborate on it soon. Argument is more or less the following: we pick a hard unsatisfiable formula and pad it in order to make it easy.",
"I think the answer to your question is (apparently) \"No\". Hirsch (2000) exhibited a simple CNF with exactly one satisfying assignment such that no \"randomized DPLL\" (randomized local search algorithm) can find it in expected polynomial time. See my writeup thi.informatik.uni-frankfurt.de/~jukna/BFC-book/sat-chapter.pdf User: friend, password: catchthecat."
] | 3 |
Science
| 0 |
42
|
cstheory
|
Which monotone DNFs are evasive?
|
A Boolean function $\phi$ on variables $X$ is _[evasive](https://en.wikipedia.org/wiki/Evasive_Boolean_function)_ if every decision tree for $\phi$ has height $|X|$. In other words, for any strategy that picks variables of $X$ and asks for their value, an adversary can answer the queries such that the strategy needs to ask about the value of all variables of $X$ to know the value to which $\phi$ evaluates.
A _monotone DNF_ is a disjunction of conjunction of variables, representing a monotone Boolean function on the set $X$ of variables that it uses.
**Which monotone DNFs are evasive?** Is there a characterization of them? What is the complexity, given a monotone DNF, of determining whether it is evasive?
Examples of evasive monotone DNFs include single-clause DNFs, e.g., $x \land y \land z$, where we must query the value of every variable (and the adversary can answer $1$) until the last one which determines the value of the formula. For another example, take $(x \land y) \lor (y \land z)$: if I ask about $x$ or $z$ then the opponent answers $0$ and I'm left with evaluating the other clause, if I ask about $y$ the opponent answers $1$ and I must evaluate $x \lor z$ which requires me to query both variables.
However, not all monotone DNFs are evasive. For instance, for $(s \land t) \lor (t \land u) \lor (u \land v)$, I can query variable $t$. If it is false, then I know that clauses $s\land t$ and $t \land u$ are false, so I can evaluate the function just knowing $u$ and $v$ (I don't need to query $s$). If it is true, then I have to evaluate $s \lor u \lor (u \land v)$, and as the second clause subsumes the third, I just need to know $s$ and $u$ (I don't need to query $v$).
Playing a bit with the problem in the case of graph DNFs (with two variables per clause), it looks like for a path $(x_0 \land x_1) \lor \cdots \lor (x_{n-1} \land x_{n})$ with $n>0$ the function is not evasive iff $n$ is a multiple of 3, and this seems to generalize to trees if there is some kind of tree structure on nodes that are 3 steps apart. For general graphs, I have no idea. I have looked for related work about evasiveness and decision tree complexity for monotone CNFs and DNFs but I didn't find anything relevant.
On graph DNFs, the problem can be seen as a kind of game on the graph of the formula. You pick a vertex $v$ (= variable) on the graph, and the opponent can choose between:
* removing $v$ and all its incident edges (= $v$ evaluates to $0$, all clauses involving it are removed), or
* removing the open ball of radius 2 centered on $v$, i.e., $v$, its adjacent edges, the neighbors of $v$, and their adjacent edges (= $v$ evaluates to $1$, all clauses involving $v$ become singleton clauses involving $v$'s neighbors, you will need to ask about each of these neighbors because each has a singleton clause involving it, but these singleton clauses subsume all other clauses involving the neighbors so you get to remove them).
The question is about characterizing and recognizing the graphs $G$ for which you have a strategy that guarantees that you win, where "winning" means "reaching a graph with an isolated vertex" (= a variable that you do not need to ask about).
|
https://cstheory.stackexchange.com/questions/44278/which-monotone-dnfs-are-evasive
|
[
"np-hardness",
"boolean-functions"
] | 13 | 2019-07-16T13:28:04 |
[
"Since the Alexander dual of the corresponding simplicial complex consists of all the independent sets of $G$, a necessary condition for evasiveness is that the independence polynomial of $G$ vanishes at $-1$.",
"Possibly relevant references for the lazy: A survey of evasiveness: Lower bounds on the decision-tree complexity of boolean functions, Lecture Notes on Evasiveness of Graph Properties"
] | 2 |
Science
| 0 |
43
|
cstheory
|
Question on Products of Graphs
|
Let $S_{n}(G)$, $C_{n}(G)$, $T_{n}(G)$ be the $n$-fold Strong Product, Cartesian Product and Tensor Product of a graph $G$ on $|V|$ vertices.
Let the chromatic number ($\chi(G)$) and the independence number ($\alpha(G)$) of $G$ be known through a (possibly exponential time) algorithm. Is it known that the calculation of $\chi(S_{n}(G))$, $\chi(C_{n}(G))$, $\chi(T_{n}(G))$,$\alpha(S_{n}(G))$, $\alpha(C_{n}(G))$ and/or $\alpha(T_{n}(G))$ require exponential in $n$ complexity for any $n$ say even when restricted to some special graphs?
|
https://cstheory.stackexchange.com/questions/7522/question-on-products-of-graphs
|
[
"cc.complexity-theory",
"graph-theory",
"graph-colouring"
] | 14 | 2011-07-26T18:45:56 |
[
"@Andrew: The first one",
"By $n$-fold strong product do you mean $G\\boxtimes G\\boxtimes \\ldots \\boxtimes G$, or $G\\boxtimes K_n$?",
"Here is Wikipedia's article on Hedetniemi's conjecture.",
"Hi Hsien: Actually you answered part of the question. If you assume the conjecture, then we already have an $o(0)$ algorithm for those cases.",
"Hi Hsien: You are correct.",
"You mean other than the trivial case like $\\chi(C_n(G)) = \\chi(G)$, or $\\chi(T_n(G)) = \\chi(G)$ assuming Hedetniemi's Conjecture, right?"
] | 6 |
Science
| 0 |
44
|
cstheory
|
Deterministic context-free languages that can be represented as the word problem of a group
|
Consider a group $G$. We call $G$ virtually free is it contains a free subgroup of finite index.
If $G$ is finitely generated by some set $X \subseteq G$ one can consider the _word problem_ $W\\!P(G)$ that is the formal language consisting of all words over the alphabet $X \cup X^{-1}$ that evaluate to the unit in $G$.
It is a famous result by Mueller and Schupp that the $W\\!P(G)$ is context-free if and only if $G$ is virtually free.
It is also known that if $W\\!P(G)$ is context-free, it is deterministic context-free.
My question is: Do we know more about how the class of deterministic context-free languages that can be represented as the word problem of some group, is contained in the class of all (deterministic) context-free languages?
|
https://cstheory.stackexchange.com/questions/39954/deterministic-context-free-languages-that-can-be-represented-as-the-word-problem
|
[
"fl.formal-languages",
"gr.group-theory"
] | 13 | 2018-01-11T04:49:31 |
[
"I'm not sure if you're still interested in this, but it is known that such languages are closed under deletion and have universal prefix. This is true for all word problems, not just DCFL. There's a proof here: D. W. Parkes and R. M. Thomas. Groups with context-free reduced word problem. Communications in Algebra, 30:3143–3156, 2002.",
"@MichaëlCadilhac Thank you. This answer of yours is also very interesting.",
"Maybe the class of languages DLOGTIME-reducible to some WP(G) would be more robust.",
"For that above question, I believe $\\{a^nb^m \\mid m < n\\}$ is not the WP of some group. I guess you'd be quite interested in Greibach's hardest language, and her notion of jump PDAs.",
"No not in particular; but do you have a DCFL in mind that is not the WP of some group? For example: The WP of any free group is a DCFL and is solvable in logspace (Lipton, Zalcstein, 1977). Maybe there are more connections between WP of groups and DCFL solvable in logspace. But again, nothing in particular.",
"Are you just asking whether there exists a DCFL that is not the WP of some group? Or something more general?"
] | 6 |
Science
| 0 |
45
|
cstheory
|
Complexity to compute the eigenvalue signs of the adjacency matrix
|
Let $A$ be the $n\times n$ adjacency matrix of a (non-bipartite) graph. Assume that we are given the amplitudes of its eigenvalues, i.e., $|\lambda_1|=a_1,\ldots, |\lambda_n|=a_n$, and we would like to calculate their signs. Is there a faster way of computing the signs of these eigenvalues, other than recomputing the eigenvalues themselves?
|
https://cstheory.stackexchange.com/questions/16789/complexity-to-compute-the-eigenvalue-signs-of-the-adjacency-matrix
|
[
"cc.complexity-theory",
"ds.algorithms",
"linear-algebra",
"spectral-graph-theory"
] | 13 | 2013-03-07T12:38:48 |
[
"Thanks, that's exactly the point. I was wondering if the knowledge of the amplitudes made the problem any simpler. I think that this could have interesting implications in problems where eigenvalues are well approximated by the squared degrees of the graph, etc.",
"Nice question! For comparison, the best known running time for computing the eigenvalues themselves (upto additive error 1/4) is $O(n^3)$ (cstheory.stackexchange.com/a/2810/15)"
] | 2 |
Science
| 0 |
46
|
cstheory
|
Name and references for balanced variant of the long code?
|
The [long code](https://en.wikipedia.org/wiki/Long_code_%28mathematics%29) (arising in PCP theory etc) is an encoding of a set of $k$ values, using binary strings of length $2^k$ (double exponential in the number of bits needed to specify a value), with one coordinate position in the encoding for each of the $2^k$ different binary functions on the input values. Instead, I want to consider a shorter but still doubly exponential code, with one coordinate position for each of the $\tbinom{k}{k/2}/2$ partitions of the $k$ values into two equal-sized subsets (assuming $k$ is even). Each partition can be thought of as corresponding to two different binary functions (the functions whose inverse images form the given partition) and we choose arbitrarily which one of these two functions to use.
The minimum distance of the code (and in fact the distance between all pairs of codewords) is $\tbinom{k-2}{(k-2)/2}$, smaller than the length of the code by only a factor of $\tfrac{k}{2k-1}>\tfrac{1}{2}$. For instance for $k=8$ we get a code of length 35 with minimum distance 20, which <http://www.codetables.de/> tells me matches the [Griesmer bound](https://en.wikipedia.org/wiki/Griesmer_bound) for binary linear codes of the same size (rank 3 or size 8).
I am interested in this for reasons only vaguely related to coding theory, but it seems likely to me that the coding theorists have already studied this code. Unfortunately, I don't know of a good way of looking up coding schemas by the formulas for their parameters. If it has been previously studied, what is it called? And what are some references to it?
(BTW, the property I actually care about, which is easy enough to prove using Sperner's theorem, is that this is the longest possible binary code in which each two positions are independent of each other, in the sense that all four possible combinations of symbols in the two positions are realized by codewords. There appears to be something in the literature called a Sperner code but it's transposed from this — each codeword has equally many zeros and ones, rather than the property of the code I'm looking at, where each bit position has equally many codewords for which that position is a 0 or 1.)
|
https://cstheory.stackexchange.com/questions/19602/name-and-references-for-balanced-variant-of-the-long-code
|
[
"reference-request",
"coding-theory"
] | 13 | 2013-10-31T16:11:44 |
[
"Ok, if it wasn't in the literature before, it is now. See section 6 of arxiv.org/abs/1303.1136 — this code gives the smallest set of vertices to delete from a hypercube graph in order to make all remaining hypercube subgraphs have at most 1/8 of the number of vertices of the original graph."
] | 1 |
Science
| 0 |
47
|
cstheory
|
Generalizing limit-colimit coincidence to Scott-continuous adjunctions: any uses?
|
In Abramsky and Jung's 1994 handbook chapter on denotational semantics, after proving that the limit and colimit of expanding sequences exist and coincide, they have the following to say about generalizing this theorem:
> We can generalize Theorem 3.3.7 [the limit-colimit coincidence of expanding sequences in DCPO] [...] we can use general Scott-continuous adjunctions instead of e-p-pairs. [...] By the passage from embeddings to, no longer injective, lower adjoints, we allow domains not only to grow but also to shrink as we move on in the index set. Thus points, which at some stage looked different, may at a later stage be recognized to be the same. [...] For the main text, we must remind ourselves that this generalization has so far not found any application in semantics.
It's been over fifteen years since this chapter was written. Have there been any applications of this generalization in semantics since then?
|
https://cstheory.stackexchange.com/questions/2528/generalizing-limit-colimit-coincidence-to-scott-continuous-adjunctions-any-uses
|
[
"pl.programming-languages",
"semantics",
"denotational-semantics",
"domain-theory"
] | 13 | 2010-10-28T10:07:39 |
[] | 0 |
Science
| 0 |
48
|
cstheory
|
The best known upper bound for two-way probabilistic finite automata with one-counter
|
It is known that the class of languages recognized by two-way deterministic finite automata with one-counter (2D1CAs) is a proper subset of $ \mathsf{L} $ (deterministic log-space): A 2D1CA can run at most $ O(n^2) $ steps if it halts on a given input, and there is a language in $ \mathsf{L} $, i.e. \begin{equation} \mathtt{DG} = \lbrace w_0 \$ w_1 \$ \cdots \$ w_k \mid k \geq 1, w_{i \in \lbrace 0, \ldots, k \rbrace} \in \lbrace a,b\rbrace^* , \mbox{ and } w_0=w_i \mbox{ for some } 1 \leq i \leq k \rbrace, \end{equation} which cannot be recognized by any 2D1CA [(Duris and Galil, 1982)](http://www.sciencedirect.com/science/article/pii/0304397582900871).
What happens if such an automaton can use a coin?
More precisely, what is the best known upper bound for the class of languages recognized by two-way **probabilistic** finite automata with one-counter (2P1CAs) with bounded error?
|
https://cstheory.stackexchange.com/questions/11080/the-best-known-upper-bound-for-two-way-probabilistic-finite-automata-with-one-co
|
[
"cc.complexity-theory",
"reference-request",
"automata-theory",
"upper-bounds"
] | 14 | 2012-04-13T13:36:31 |
[] | 0 |
Science
| 0 |
49
|
history
|
Are there any references to entombed animals in ancient India?
|
The 13th century Hindu philosopher Arulnandi Shivacharya wrote a work called the Shiva Jnana Siddhiyar, which among other things contains a refutation of Buddhist philosophy. In [this excerpt](https://i.sstatic.net/KxNh7.jpg), various Buddhist theories of what the ultimate cause of the body is are refuted:
> If you say bodies are formed from the mixture of four elements, then these cannot unite as their natures are opposed to each other. **If you say they are formed by the union of blood and semen, then account for toads being found in the heart of rocks, and worms in the heart of trees.** If you say the real cause is good and bad Karma, then these, being opposed, cannot join and form bodies. If food is the cause, then the food which in youth develops the body is not capable of preventing decay in old age. If intelligence is the cause, then that which is formless Chaitanya cannot assume Achaitanya (non-intelligent) form. If you assert that bodies are formed from nothing, then we could cull flowers from the sky.
My question is about the part in bold, which is arguing that intercourse cannot be the ultimate cause of the body, because there are toads which are found in the center of solid rock, and they couldn't have been produced by intercourse since there's no way that toads could have gone inside a solid rock and reproduced.
This statement struck me as rather odd, but then I found out from the Wikipedia article [Entombed animal](https://en.wikipedia.org/wiki/Entombed_animal) that claims of toads being found in solid rock are surprisingly common across many cultures:
> Entombed animals are animals reportedly found alive after being encased in solid rock, or coal or wood, for an indeterminate amount of time. The accounts usually involve frogs or toads. The reputed phenomenon, sometimes called "toad in the hole", has been dismissed by mainstream science, but has remained a topic of interest to Fortean researchers.... References to entombed animals have appeared in the writings of William of Newburgh, J. G. Wood, Ambroise Paré, Robert Plot, André Marie Constant Duméril, John Wesley, and others. Even Charles Dickens mentioned the phenomenon in his journal All the Year Round. **According to the Fortean Times, about 210 entombed animal cases have been described in Europe, North America, Africa, Australia, and New Zealand since the fifteenth century.**
Conspicuously absent from that list of places is Asia. So my question is, are there any references in ancient India to entombed animals? The work I quoted above was written in the 13th century. But I'm wondering whether there are earlier references to this phenomenon. Arulnandi Shivachrya refers to this phenomenon as if it's common knowledge, so I'm guessing older sources would mention it as well.
|
https://history.stackexchange.com/questions/34768/are-there-any-references-to-entombed-animals-in-ancient-india
|
[
"india",
"science",
"animals",
"ancient-india",
"folklore"
] | 16 | 2017-01-05T23:06:24 |
[
"THe proposed edit seems to replace a question with an answer. I'm confused.",
"Unless someone can say what's available in the article I linked to above, there's not much else that can be said—at least the last time I researched this, I spent a fair bit of time and didn't uncover anything useful, but it's so very hard to prove a negative.",
"This, though not readily available online, might be useful. Didn't find anything else that would refer to this with a quick search.",
"It reminds me Schrödinger's cat..."
] | 4 |
Culture & Recreation
| 0 |
50
|
history
|
What was the first overland road from Sweden to Finland?
|
The [Swedish post road](https://en.wikipedia.org/wiki/King%27s_Road_\(Finland\)) from Norway, through Sweden, used the Åland archipelago to pass into Sweden, and this is easily found (evidence of) in the south of Finland to the present day. **When (and where) was the first overland route constructed overland from Sweden into (Swedish) Finland?**
The only (poor) evidence I have for roads existing in the north is by the [War of 1808–9](https://en.wikipedia.org/wiki/Finnish_War) where Russian forces were planning to advance overland into Sweden (along with an army group advancing across the Gulf of Bothnia). One of the WP article's references does say "In addition, several new good roads had been built into Finland greatly reducing the earlier dependency on naval support for any large operation in Finland." but it doesn't specify where these roads were.
I looked through all articles on [Swedish](https://en.wikipedia.org/wiki/Category:Roads_in_Sweden) and Finnish road networks on the English Wikipedia, and the most I found was a reference to a ['Finnmark path'](https://en.wikipedia.org/wiki/Arctic_Ocean_Highway) which was meant to have gone from Finnish Lapland to Finnmark in the 16th century. The [Finnish WP article](https://fi.wikipedia.org/wiki/J%C3%A4%C3%A4merentie) for the same page _does not_ mention the Finnmark path at all, and I couldn't find anything else on a road of that name.
I understand—from the comments—that the term "road" can be meaningless without further definition for a period much longer than a few centuries ago. For clarity, I'm defining road as purpose-built (or purpose-developed) and used regionally for that purpose, such as the post road mentioned above. This would mean hunting tracks that slowly developed don't count, while a merchant-led endeavour to expand (and maintain) the tracks between two townships would.
|
https://history.stackexchange.com/questions/62286/what-was-the-first-overland-road-from-sweden-to-finland
|
[
"transportation",
"sweden",
"finland"
] | 6 | 2020-12-21T03:13:26 |
[
"Carl Linnaeus's 1732 trip to Lapland was recorded in a fascinating book. Of course he was not interested in the speediest or easiest route, so much as the most interesting, but it does give first-hand detailed account of his travels & travel conditions. Wikipedia summary: \"Linnaeus travelled clockwise around the coast of the Gulf of Bothnia, making major inland incursions from Umeå, Luleå and Tornio. He returned from his six-month-long, over 2,000 kilometres (1,200 mi) expedition in October, having gathered and observed many plants, birds and rocks\".",
"No. It was a national road net (for low values of \"road\"). There is a map from 1742, but I cannot access it here. Whatever. There were \"roads\" a long time ago, between 1809 and 1917 it was a hostile border, and then after Finland became independent they opened a railroad.",
"@TomasBy: That's not what I wrote. \"local roads\" means those used to get around town, and its immediate vicinity. Emphasis on immediate vicinity. Magnus Ericson's law basically requires that towns build and maintain a local urban road network for commerce.",
"@PieterGeerkens are you suggesting they were not actually connected? Each man just had a small bit of private road in isolation?",
"@TomasBy: Read the next sentence, these are local roads only: \"... stipulate that the land-owning farmers are responsible for each road lot, bridge, ice road or ferry. The length of the road lot depends on the size of the agricultural property.\" It would be insanity to require small isolated villages otherwise quite content to build and maintain long-haul access rods they don't themselves use or need.",
"@PieterGeerkens history of roads in Sweden: \"The national law of Magnus Eriksson (1350) decrees that each landholder is responsible for a section of road, a bridge, an ice road, or a ferry.\".",
"Torneå was founded in the early 17th c. I assume there were roads to it from both east and west."
] | 7 |
Culture & Recreation
| 0 |
51
|
history
|
Which misdeeds of F. Reineke was J.B.B. de Lesseps "obliged to suppress"?
|
F. Reineke was commandant of Kamchatka from about 1780 to 1784. According to Витер's _История символов как история административного деления государства_ he was responsible for moving the territorial administration from Bolsheretesk to Nizhnekamchatsk. According to J.B.B. de Lesseps, Reineke resigned in 1784 ["for reasons which I am obliged to suppress"](https://archive.org/details/travelsinkamtsch01less/page/140).
It was common for Siberian administrators to be venal. According to Forsyth's _A history of the peoples of Siberia_ , the very first governor of Siberia, Matvei Gagarin, was hanged for fraud, nepotism, and selling offices; another governor in Irkutsk was relieved from office for gross corruption and replaced with his second, who was "vicious" and "ruthless".
**Was Reineke guilty of the same old corruption and exploitation, or something worse?** I'm skeptical that de Lesseps, a Frenchman, had any reason to suppress a story of Russian corruption.
|
https://history.stackexchange.com/questions/55995/which-misdeeds-of-f-reineke-was-j-b-b-de-lesseps-obliged-to-suppress
|
[
"russia",
"18th-century",
"government",
"crime",
"kamchatka"
] | 6 | 2019-12-31T20:47:10 |
[
"One more thing: According to this source, in 1786 F. Reineke moves from Kamchatka to Irkutsk and in 1798 retires from the civil service. All this suggests that his record in Kamchatka was just fine. There was some controversy related to his subordinates, as well as some misdeeds in the Russian postal service in Kamchatka during Reineke's tenure. Maybe this is what Lesseps alludes to.",
"The original says \"деятельный\"; \"Active\" might be a better translation. You can also try to get the book \"Правители Камчатки : 1700-2007\" via an interlibrary loan and look there for further information.",
"@MoisheKohan Very interesting. As Reineke was a public servant, I wonder if \"entrepreneurial\" was a backhanded compliment.",
"For what it is worth, A. S. Sgibnev in his detailed \"History of main events in Kamchatka, 1650 - 1856\", describes F. Reineke as \"honest and entrepreneurial.\"",
"Gagarin was not hanged (just) for corruption, Russian Wikipedia mentions that he attempted to form separate Siberian kingdom : ru.wikipedia.org/wiki/…",
"Thanks, Mark, for your help along the way!",
"Consistently high quality questions. Thank you for your contributions to the site."
] | 7 |
Culture & Recreation
| 0 |
52
|
history
|
Who owned the ship Ensayo, and what were they doing near Baja California in 1842?
|
According to Plummer's _The Shogun's Reluctant Ambassadors_ , in 1842, sea drifters from the _Eijū-maru_ were picked up and by the _Ensayo_ , a "Spanish pirate ship" with a Philippine crew. It was "carrying illegal goods between Spain or Mexico and Manila," and "[b]ecause it was on a 'wanted list', it could not enter any harbors freely." The Japanese men were treated badly and forced ashore on a beach in Baja California.
I found the description confusing. Of Pacific ports, only the Philippines were still Spanish. The Manila galleon system had been over for decades. During Mexico's first twenty years of independence, as far as I know, it didn't trade with East Asia. The Spanish were a minor presence in the Pacific even when they controlled a lot of its coastline.
Who owned _Ensayo_ and what was it up to?
|
https://history.stackexchange.com/questions/47476/who-owned-the-ship-ensayo-and-what-were-they-doing-near-baja-california-in-1842
|
[
"19th-century",
"spain",
"age-of-sail",
"philippines",
"piracy"
] | 6 | 2018-08-05T23:12:12 |
[
"@AlbertoYagos: I concur with T.E.D.'s view.",
"I had \"Vida en México de trece náufragos japoneses\" in my hands, but forgot to look for this point!!",
"@AlbertoYagos - I can't always speak for all our voters, but I for one would consider what you have there acceptable, with perhaps a GoodReads link to the book in question, and an admission that you are working from memory.",
"@T.E.D. I don't have access to either of those books and I can't quote more. I just have that reference. That is why it's a comment and not an answer.",
"@AlbertoYagos - Any way you could see your way to expanding that to an answer? It probably wouldn't take much, as it seems like you already put in enough research for it to be one. You ought to reap the reward for that work, and Aaron may eventually even want to accept it.",
"According to Del tratado al tratado. 120 años de relaciones entre Japón y México by Aurelio Asiain, 2008 (p. 46), the Ensayo didn't commit piracy but smuggling. She smuggled goods from and to Mexico. Adiain's source seems to be Sano, Yoshikazu, Vida en México de trece náufragos japoneses, 1842."
] | 6 |
Culture & Recreation
| 0 |
53
|
history
|
What would a criminal defense attorney charge in late 19th century America?
|
What would a criminal defense attorney charge on the [American frontier](https://en.wikipedia.org/wiki/American_frontier#The_Postbellum_West) in a murder trial in 1885? I am specifcally interested in what it would have cost three cowpokes to mount a murder defence in Rapid City, Dakota Territory 1885. I know they were defended by Colonel William Parker who was a successful lawyer and politician in the Territory at the time, but so far I have found no clue what his fees were.
|
https://history.stackexchange.com/questions/68240/what-would-a-criminal-defense-attorney-charge-in-late-19th-century-america
|
[
"united-states",
"19th-century",
"law",
"old-west"
] | 5 | 2022-02-04T11:16:22 |
[
"Would also significantly help if you documented your preliminary research. I imagine that the answer is highly variable - urban setting vs rural, proximity of a a railroad, the presence of skilled attorney vs having to recruit one from elsewhere, Of course, the cost will range from zero (self representation) to what the market will bear.",
"Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer."
] | 2 |
Culture & Recreation
| 0 |
54
|
linguistics
|
In which non-Sinitic languages do negative clauses retain older constituent order in SVC-derived complex predicates?
|
Many complex predicates are historically derived from serial verb constructions. This is not only true of the Sinitic family. For example, in Saramaccan (Byrne 1987, as cited in Givón 2009):
(1) a bi-fefi di-wosu kaba.
he TNS-paint the-house finish
He finished painting the house.
However, the Chinese constructions are more innovative in that the object has been moved to the end of the sentence, for the most part. Yet in most dialects of modern Chinese except Mandarin, complex predicates also sometimes preserve the older V + O + C order similar to the Saramaccan example in (1). This is most common in negative clauses. (The C is what is traditionally termed a _complement_ in Chinese linguistics, but it's more like a _satellite_ in Talmy's sense, and is developed from the 'lighter' verb of an asymmetric serial verb construction). An example from Cantonese is (2):
(2) nei wan ngo m dou
you find me NEG arrive
You cannot find me.
My question: I know negative clauses are more conservative in general, but are there specific cases outside the Sinitic branch where object placement in complex predicates is more conservative in negative clauses? Thanks!
Givón, T. (2009). _The genesis of syntactic complexity: Diachrony, ontogeny, neuro-cognition, evolution_. John Benjamins Publishing.
|
https://linguistics.stackexchange.com/questions/21555/in-which-non-sinitic-languages-do-negative-clauses-retain-older-constituent-orde
|
[
"syntax",
"historical-linguistics",
"word-order",
"negation"
] | 7 | 2017-04-06T01:26:05 |
[] | 0 |
Science
| 0 |
55
|
linguistics
|
Are there any languages where the first person cannot be an object?
|
In some languages, nouns low on the animacy hierarchy, particularly inanimates cannot surface as A, and if a situation arises where they are underlyingly A, some reparative strategy such as a passive must be used. Some theorise that this was the case in early PIE, and a friend of mine mentioned Algonquian and Japhung Qiangic in the conversation that lead me to consider this:
Are there any languages where a similar rule is found at the other end of the animacy hierarchy, such that the 1st person is defective and cannot be O, and an underlying 1st person O requires a syntactic derivation to be used, moving the first person either to S (a passive) or oblique status (an antipassive)?
[Chukchi](https://en.wikipedia.org/wiki/Chukchi_language) seems to come close, as it has a verbal prefix _(i)ne- ~ (e)na-_ , which Comrie(1979) analyses as a "de-ergativising prefix" which in some cases acts like an antipassive is required with 1st person singular Os when the A is not 3rd person plural (in which case a 1sg.S/O suffix _is_ used), however it's not quite a full antipassive as in these cases, as while there is a change to verbal person agreement (absolutive forms are used to mark underlying A) there is not change to nominal case marking, and furthermore, in the "present II"-tense it's used with a much larger set of A/P combinations. Further examples of borderline cases like Chukchi would be fine, but what I'm really interested in is if there are any more clear-cut examples where the construction involved is clearly a syntactic derivation.
* * *
Comrie, B. 1979: _Degrees of Ergativity: Some Chukchee evidence_ , pp. 219-40 in Plank, F. 1979 (ed.): _[Ergativity - Towards a Theory of Grammatical Relations](https://drive.google.com/file/d/1f3eXlLDI2qC3yeMG1YStsmtH5RSXzHLE/view)_ , Academic Press
|
https://linguistics.stackexchange.com/questions/27161/are-there-any-languages-where-the-first-person-cannot-be-an-object
|
[
"syntax",
"grammar",
"list-of-languages",
"grammatical-object",
"person"
] | 7 | 2018-02-16T13:02:04 |
[
"Inanimate objects not being able to be agents makes some sense (if your language doesn't use a lot of metaphor), but why would any language prohibit 1st person objects? I can imagine this as a kind of taboo, but not as an actual ungrammaticality."
] | 1 |
Science
| 0 |
56
|
mathoverflow
|
2, 3, and 4 (a possible fixed point result ?)
|
The question below is related to the classical [Browder-Goehde-Kirk fixed point theorem](https://www.researchgate.net/publication/38323269_An_elementary_proof_of_the_fixed_point_theorem_of_Browder_and_Kirk).
Let $K$ be the closed unit ball of $\ell^{2}$, and let $T:K\rightarrow K$ be a mapping such that $$\Vert Tx-Ty\Vert _{\ell^{4}}\leq\Vert x-y\Vert _{\ell^{3}}$$ for all $x,y\in K$.
Is it true that $T$ has fixed points?
|
https://mathoverflow.net/questions/18264/2-3-and-4-a-possible-fixed-point-result
|
[
"fa.functional-analysis",
"open-problems",
"banach-spaces"
] | 79 | 2010-03-15T06:26:45 |
[
"$FP(p,p,\\infty)$ is false for any $p$. We can let $(Tx)_i = x_{i-1}$ for $i<1$ and $(Tx)_0 =(1/2)( 1 - \\sum_i x_i^2)$. Then $||Tx||_{\\ell_2}^2 = (1/4)(1-||x||_{\\ell_2}^2)^2+ x_{\\ell_2}^2 \\leq 1$ since $x_{\\ell_2}\\leq 1$ and, because each coordinate is a $1$-Lipschitz function in the $\\ell_2$ norm, the whole thing is a $1$-Lipschitz transformation from $\\ell^\\infty$ to $\\ell^2$.",
"As far as I understood, the BGK (Browder-Göhde/Göbel-Kirk) fixed point theorem states that every non-expansive self-mapping on a non-empty, closed and convex subset of a uniformly convex Banach space has a fixed point.",
"What are the related known results?",
"Any progress on this?",
"Would someone please state the classical Browder-Goehde-Kirk fixed point theorem?",
"@Fabrizio Polo I'm not claiming that. Just that, e.g., $FP(2,4,3)$ holds, via BGK. Please read carefully my comment.",
"@Ady: Since your question is whether or not $FP(2,3,4)$ holds, I'm confused. You seem to be claiming that BGK resolves your question.",
"BGK $\\Longrightarrow$ FP$(p,q,r)$ for 1 < p $\\leq$ r $\\leq q< \\infty $ , e.g.",
"For fun, I started considering variants of this question but made no real progress on them either. Let ${\\bf FP}(p,q,r)$ be the same statement with $2,3,4$ replaced by $p,q,r$. The BGK fixed point theorem gives us ${\\bf FP}(p,p,p)$ for $1 < p < \\infty.$ For what other values of $p,q,r$ can you prove or disprove this statement? Do you know of a counterexample to $FP(1,\\infty, \\infty)$?",
"Why did this question get a downvote? Seems interesting to me.",
"I don't think it does. I just meant to make sure other readers could easily tell what you were asking; I didn't downvote and I hope I didn't encourage anyone else to.",
"If that deserves a \"-1\", it's cool to me. ",
"It would be easier to read if you just wrote $\\|Tx−Ty\\|_4 \\le \\|x−y\\|_3$. It took me a minute to get the point of your question since I could barely see the difference between the supersubscripts"
] | 13 |
Science
| 1 |
57
|
mathoverflow
|
Grothendieck's Period Conjecture and the missing p-adic Hodge Theories
|
Singular cohomology and algebraic de Rham cohomology are both functors from the category of smooth projective algebraic varieties over $\mathbb Q$ to $\mathbb Q$-vectors spaces. They come with the extra structure of Weil cohomology theories - grading, cup product, orientation map, and cycle class map. We can consider the space of all invertible natural transformations from singular cohomology to algebraic de Rham cohomology that respect this structure. For purely formal reasons, this is an affine scheme. The coordinates are the matrix coefficients of the natural transformation for each algebraic variety, and the relations are given by the various commutative diagrams that must hold. Conjecturally, this scheme is a torsor for the motivic Galois group of $\mathbb Q$.
de Rham's theorem, together with GAGA, gives us an isomorphism between these two functors when tensored with $\mathbb C$. In other words, it gives us a $\mathbb C$-point of this affine scheme.
Grothendieck's period conjecture says that this point is a generic point - so the ring of periods in $\mathbb C$ is equal to the whole ring of functions on this scheme.
* * *
On the other hand, consider the $p$-adic analogue. $p$-adic etale cohomology and algebraic De Rham cohomology are both functors from the category of smooth projective algebraic varieties over $\mathbb Q_p$ to $\mathbb Q_p$-vector spaces. They are Weil cohomology theories.
We can consider the space of all isomorphisms between these two functors as an affine scheme over $\mathbb Q_p$.
$p$-adic Hodge theory gives a $B_{dR}$-valued point of this scheme.
However, it is not the generic point. The reason is that the isomorphism described by $p$-adic Hodge theory naturally factors through the category of $\operatorname{Gal}(\overline{\mathbb Q}_p|\mathbb Q_p)$-representations. Hence it lies in a closed subscheme that is a torsor under, not the full motivic Galois group, but the Zariski closure of $\operatorname{Gal}(\overline{\mathbb Q}_p|\mathbb Q_p)$ inside it. This group is much smaller because the Tate conjecture fails over $\mathbb Q_p$, and a map of Tannakian categories that is not a full functor corresponds to a map of Tannakian groups that is not surjective.
* * *
So the analogue of the period conjecture fails badly over $\mathbb Q_p$. This leads me to my question: Grothendieck's period conjecture essentially gives us a description of the scheme of isomorphisms betweeen singular and de Rham cohomology - it's the spectrum of the period ring. However, we do not have a similar picture of the scheme of isomorphisms between $p$-adic etale cohomology and de Rham cohomology. The period ring only gives us one point.
> Can we explicitly describe any other points of this scheme of isomorphisms?
|
https://mathoverflow.net/questions/200083/grothendiecks-period-conjecture-and-the-missing-p-adic-hodge-theories
|
[
"nt.number-theory",
"tannakian-category",
"p-adic-hodge-theory"
] | 61 | 2015-03-15T14:18:19 |
[
"I should correct myself: to obtain that \"over a finite field mixed motives are pure\" we must assume that such mixed motives exist (which follows from the Tate conjecture) and that they have a weight filtration respected by Frobenius. Then for each mixed motive $M$ appropriate polynomials in Frobenius are idempotent on $M$ which split it as a direct sum of pure motives. I am not sure how far we are from proving the Tate conjecture for varieties over finite fields; i imagine it's not considered out of reach. And i'm not sure how useful motives over finite fields are apart as source of intuition.",
"Thank you @WillSawin for this excellent question. I think that for the usual equal characteristic $0$ Grothendieck conjecture one must consider arbitrary mixed motives which complicates things. So for varieties over p-adic fields shouldn't one also consider motives of nonprojective varieties ? It seems that p-adic étale and similar cohomologies can get complicated for those. Im not familiar enough with the field. We have Voevodsky's $DM_{gm}^{eff}(Q_p,Q)$ but no t-structure i believe. Also have you thought about varieties over a finite field where mixed motives with $Q$-coefficients are pure ?",
"@AknazarKazhymurat Thanks! I have figured nothing out.",
"representation is abnormally small (compared to the full motivic group) and try to work from there (one can find examples among abelian varieties). Another question is whether you should restrict to isomorphism respecting the filtration on both sides (as I think you should).",
"Will Sawin, first of all, thank you for this interesting (but hard) question. My impression is that part of the difficulty stems from the fact that your analogy is perhaps not so close: a closer analogue might be the collection of all comparison theorems between étale cohomology for each $p$ and de Rham cohomology for a variety over $\\mathbb Q$ (rather than the single comparison theorem for a variety over $\\mathbb Q_{p}$. In that case, one recovers I believe the full motivic group. As for your precise question, I guess one can take a variety over $\\mathbb Q$, select a $p$ such that the Galois",
"To get an isomorphism between algebraic de Rham cohomology and singular cohomology one only needs de Rham's theorem and Serre's GAGA, so no Hodge theory is required.",
"@Olivier I meant to write $B_{dR}$ instead of $\\overline{\\mathbb C}_p$. Given that, do I need to use a different scheme?"
] | 7 |
Science
| 1 |
58
|
mathoverflow
|
Does every triangle-free graph with maximum degree at most 6 have a 5-colouring?
|
## A very specific case of Reed's Conjecture
Reed's $\omega$,$\Delta$, $\chi$ conjecture proposes that every graph has $\chi \leq \lceil \tfrac 12(\Delta+1+\omega)\rceil$. Here $\chi$ is the chromatic number, $\Delta$ is the maximum degree, and $\omega$ is the clique number.
When restricted to triangle-free graphs, the equivalent question is, Does every triangle-free graph have chromatic number $\leq \frac \Delta 2 +2$?
This is known for $\Delta\leq 4$. In general for triangle-free graphs, $\chi \leq O(\Delta/\log \Delta)$, so the conjecture is also true for very large $\Delta$.
How about $\Delta=5$? $\Delta=6$? Because of parity, $\Delta=6$ is the easier of these two cases (and actually easily implies the $\Delta=5$ case. Can anyone prove it?
Kostochka proved that every triangle-free graph has $\chi \leq \frac 2 3 \Delta +2$. He also proved that $\chi\leq \frac \Delta 2 +2$ for graphs of sufficiently large girth depending on $\Delta$. Can anyone prove it for girth $\geq 5$? $4$?
This would at least provide some hope for proving Reed's Conjecture for triangle-free graphs.
> Does every triangle-free graph with $\Delta\leq 6$ have $\chi \leq 5$? What about every graph with girth at least five?
|
https://mathoverflow.net/questions/37923/does-every-triangle-free-graph-with-maximum-degree-at-most-6-have-a-5-colouring
|
[
"graph-theory",
"graph-colorings",
"co.combinatorics"
] | 55 | 2010-09-06T12:58:11 |
[
"Another possible tangential attack is to establish the conjecture for the \"warmth\" $w$ of the graph, because Brightwell and Winkler proved that $w \\le \\chi$.",
"Good question; it is known to hold, without the round-up. Reed never published the result in a paper but it's in Graph Colouring and the Probabilistic Method (with Molloy), in the chapter on hard-core distributions (Chapter 23 I think). (It wouldn't let me comment twice in a row, so I deleted the old comment to add: ) A proof of a stronger result, noted by McDiarmid, appears in Section 2.2 of my thesis, which is here: columbia.edu/~ak3074/papers/phdthesis.pdf",
"This is not addressing your question, but I wonder if it is known whether Reed's conjecture holds when $\\chi$ is replaced by $\\chi_f$, the fractional chromatic number?"
] | 3 |
Science
| 1 |
59
|
mathoverflow
|
Class function counting solutions of equation in finite group: when is it a virtual character?
|
Let $w=w(x_1,\dots,x_n)$ be a word in a free group of rank $n$. Let $G$ be a finite group. Then we may define a class function $f=f_w$ of $G$ by
$$ f_w(g) = |\\{ (x_1,\dots, x_n)\in G^n\mid w(x_1,\dots,x_n)=g \\}|. $$ The question is:
> Can we characterize the words $w$ for which $f_w$ is a virtual character for any group $G$?
(A virtual character is a difference of two characters, sometimes also called generalized character.) That $f_w$ is a virtual character of $G$ is equivalent to
$$ (f_w,\chi) = \frac{1}{|G|} \sum_{x_1, \dots,x_n\in G} \chi(w(x_1, \dots,x_n)) \in \mathbb{Z} $$ for all $\chi \in \operatorname{Irr}(G)$.
Some examples and non-examples:
1. $n=1$, $w= x^k$. Then $(f_w,\chi)\in \mathbb{Z}$ is the $k$-th Frobenius-Schur-indicator.
2. $w=[x,y]=x^{-1}y^{-1}xy$. Then $(f_w,\chi)= \frac{|G|}{\chi(1)}$, so $f_w$ is in fact a character.
3. For $w=x^{-k}y^{-1}x^ly$, it can be shown that $(f_w,\chi)= \frac{|G|}{\chi(1)} a$ for some integer $a$.
4. $f_w$ is a virtual character for words of the form $w= x_1^{e_1}x_2^{e_2} \dots x_n^{e_n}$.
5. For $w=[x^2,y^2]$ and $G= SL(2,7)$, the class function $f_w$ is not a generalized character.
It is always the case that $(f_w, \chi)$ is an algebraic integer in $\mathbb{Q}(\chi)$. This follows from an old result of Solomon saying $f_w(g)$ is divisible by $|C_G(g)|$ when the number of variables is $>1$. (See also [this paper](http://de.arxiv.org/abs/1105.6066v1) by Rodriguez Villegas and Gordon, where my question is discussed very briefly.) So for rational-valued groups, $f_w$ is a virtual character for any word. (In fact, it suffices that $G$ is a normal subgroup of a rational-valued group.) In view of 4., the same is true for abelian groups, and I don't know examples of solvable groups where $f_w$ is not a virtual character.
Another remark is that $f_w$ does not change if we replace $w$ by its image under an automorphism of the free group in question. Using this, one can of course find rather complicated looking words for which $f_w$ is trivial, that is, $f_w(g)\equiv |G|^{n-1}$.
Added later:
The proofs that $(f_w,\chi)\in \mathbb{Z}$ in 2.-4. above basically use the same "trick". I'm also interested in other examples where perhaps another idea of proof is used than in the examples 1.-4., or "more conceptual" explanations why $f_w$ sometimes is a (virtual) character. (I admit that this is somewhat vague.)
A related question is when $f_w$ is a character, not only a virtual character. For $G$ the symmetric group, this is discussed in Exercise 7.69k of "Enumerative Combinatorics 2" by Richard Stanley. (Note that in this case, $f_w$ is always a virtual character.)
|
https://mathoverflow.net/questions/105854/class-function-counting-solutions-of-equation-in-finite-group-when-is-it-a-virt
|
[
"gr.group-theory",
"rt.representation-theory",
"finite-groups",
"characters"
] | 54 | 2012-08-29T08:38:56 |
[
"To add to HJRW's comment a function is a $\\mathbb Q$-linear combination of characters iff it is constant on elements generating conjugate subgroups. In your set up because it is an algebraic integer combination of characters you just need it to be a $\\mathbb Q$-linear combination to be a virtual character. So at the end of the day you need that if $m$ is relatively prime to the order of $G$ then the number of solutions for $g$ and $g^m$ are the same. For the symmetric group this happens always because $g$ and $g^m$ are conjugate. The same is true for any group with all characters rational.",
"This question is considered (although I don't think answered definitively) in this papr of Amit and Vishne: u.math.biu.ac.il/~vishne/publications/S0219498811004690.pdf ."
] | 2 |
Science
| 1 |
60
|
mathoverflow
|
How many algebraic closures can a field have?
|
Assuming the axiom of choice given a field $F$, there is an algebraic extension $\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is algebraically closed, then $K\cong\overline F$ via an isomorphism which fixes $F$. We can therefore say that $\overline F$ is _the_ algebraic closure of $F$.
(In fact, much less than the axiom of choice is necessary.)
Without the axiom of choice, it is consistent that some fields do not have an algebraic closure. It is consistent that $\Bbb Q$ has two non-isomorphic algebraically closed algebraic extensions.
It therefore makes sense to ask: Suppose there are two non-isomorphic algebraically closed algebraic extensions. Is there a third? Are there infinitely many? Are there Dedekind-infinitely many?
> What is provable from $\sf ZF$ about the spectrum of algebraically closed algebraic extensions of an arbitrary field? What about the rational numbers?
|
https://mathoverflow.net/questions/325756/how-many-algebraic-closures-can-a-field-have
|
[
"set-theory",
"ac.commutative-algebra",
"lo.logic",
"axiom-of-choice",
"field-extensions"
] | 54 | 2019-03-19T02:41:22 |
[
"@DavidRoberts: Should, yes. One day...",
"Shouldn't you add an answer about this now, pointing to your own paper?",
"@AsafKaragila thanks!",
"@Holo: arxiv.org/abs/1911.09285",
"@Holo: The one I am putting in arxiv later today. Check Monday's Replacements.",
"@AsafKaragila hi, may I ask which paper of yours implies this result? I would like to read it",
"@Noah: Some months ago I was talking with another MO user about this question, and I realised that one of my recent papers implies that it is consistent that every field has a proper class of pairwise non-isomorphic algebraic closures, unless it is real-closed or algebraically closed. It's still not clear whether or not we can have some of them having the same cardinality. That's a great little idea.",
"@Noah: That's a good question. I don't know. It is true that the standard axiomatization of ACF is complete.",
"What happens to the model theoretic results about algebraically closed fields without choice? Can you have uncountable nonisomorphic ACFs with the same characteristic and cardinality without choice?",
"@WillSawin My intuition says yes, but I’m not sure — perhaps field properties like being formally real or having a particular valuation could have some impact on the answer. Hopefully looking at the arguments in the paper Asaf linked above can sharpen it somewhat.",
"@AlecRhea The answer seems pretty likely to be \"proper class many\", right?",
"@Gerhard: In the case of the rationals, I guess? There is a canonically constructed closure which is countable. But it is consistent that there is an algebraically closed algebraic extension which is not countable. In fact, it does not have a real-closed subfield.",
"Does the consistency of two closures follow because there are two sets for which no bijection exists and each set is contained by one of the closures? In which case, I suspect there are more than two closures. Gerhard \"It's Not About The Algebra?\" Paseman, 2019.03.19.",
"It seems we can ask a similar question even with choice (but without global choice) about proper class sized fields like the field of fractions of the Grothendieck ring of the ordinals, since the poset/chains we want to Zorn are proper classes. It might be interesting to see how many algebraic closures a proper class sized field can have in the presence of choice without global choice.",
"@Matt: Not much, I expect. But maybe with time someone will find this interesting enough to pursue from a research point of view.",
"Oh my misunderstanding. As an outsider to this, I'm excited to see what comes out of this question. Cheers.",
"@Matt: Not really. That tells us that for the rational numbers there is a canonical algebraically closed algebraic extension, so the spectrum is non-empty. But it's pretty far from anything I'm asking.",
"Is this related? math.stackexchange.com/questions/114978/…",
"Much appreciated Asaf; it seems I take choice for granted all too often, I thought these facts were more intrinsic to field theory.",
"Alec, it appears in Hodges' Läuchli's algebraic closure of $Q$. Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289–297. MR422022.",
"Very cool question, could you provide a reference/proof hint for the comment about the consistency $\\mathbb{Q}$ having more than one algebraic closure without choice?",
"Not fully sure about the tags, though."
] | 22 |
Science
| 1 |
61
|
mathoverflow
|
Atiyah's paper on complex structures on $S^6$
|
M. Atiyah has posted a preprint on arXiv on the non-existence of complex structure on the sphere $S^6$.
<https://arxiv.org/abs/1610.09366>
It relies on the topological $K$-theory $KR$ and in particular on the forgetful map from topological complex $K$-theory to $KR$.
Question: what are the nice references to learn about $KR$?
**Edit :** Thank you very much for the comments and suggestions, M. Atiyah's paper "K-theory and reality." Quart. J. Math., Oxford (2), 17 (1966),367-86 is a fantastic paper. However I have more questions.
Question 1: how to build the morphism $$KSp(\mathbb{R}^6)\rightarrow K^{7,1}(pt)?$$
Question 2: why do we use $\mathbb{R}^{7,1}$?
In fact I do not understand the sentence "The 6-sphere then appears naturally as the base of the light-cone". And why it suffices to look at this particular model of $S^6$.
|
https://mathoverflow.net/questions/253577/atiyahs-paper-on-complex-structures-on-s6
|
[
"dg.differential-geometry",
"at.algebraic-topology",
"complex-geometry",
"kt.k-theory-and-homology",
"index-theory"
] | 52 | 2016-10-31T13:01:46 |
[
"I think question should be edited to remove any implication that it is asking about possible correctness of Atiyah's paper. If the title were to say something about your actual question on KR, that would be good.",
"Regarding the map you wish to understand, I think that the KSp should be seen as coming from one specific KR. Btw web.archive.org/web/20230113145507/https://www.math.umd.edu/… should be rather useful, but I haven't been able to figure out the dimension correctly from there. Rosenberg uses a different grading to Atiyah.",
"For your final question, the section titled \"Ambient Space Construction\" on this blog entry of mine is related. Basically it is a way to fix one conformal structure on $\\mathbb{S}^6$.",
"@RyanBudney I suspect that the sentence \"integers goes to 0\" means integrable ACS goes to 0 in KR, while non-integrable ones goes to 1. It seems that he was using this map to separate integrable and non-integrable ACSs, it cannot be done if it is trivial. I feel that this might be the most crucial step, but do not have a clear idea what's going on...",
"Karoubi briefly treats KR-theory in \"K-Theory: An Introduction\" (page 177). The book is a good reference for K-Theory in general.",
"I think basics of $KR$ theory is easy to learn from [1]. What is puzzling is the claim on p.5 of arxiv.org/abs/1610.09366 that \"Because of the Atiyah-Singer theory, the linear algebra acquires a topological meaning, and that is embodied in $KR$ theory.\"",
"And the involution on the base space would be trivial.",
"@DenisNardin You go from $E$ to $E \\oplus \\overline{E}$; your involution is the obvious one that swaps factors.",
"@RyanBudney Can you explain a little bit more how to treat a complex bundle as a real bundle? You need to throw in a C_2-action and I really don't see how to do that (this is a bit embarassing because I should know KR very well).",
"I think it's just a straight-forward computation, isn't it? I've only scanned the paper and this is my first time thinking about $KR$, but I suspect what's going on is that the generator for complex k-theory is the bundle with Euler class $1$. In $KR$ theory this is twice the generator, because the generator is one where the fixed-point set of the involution is a Moebius bundle. I haven't thought about this in detail but that's what I suspect is going on.",
"Yes of course, but do you know why it is trivial?",
"The integers he's talking about is 6-dimensional complex k-theory of $S^6$, i.e. this group is infinite cyclic.",
"This sentence is on p4 of Atiyah's preprint.",
"The forgetful map appears to be that you can treat a complex bundle as a real bundle, that gives your map from complex k-theory to this \"KR\" variant. Where is the sentence you are quoting?",
"I am reading [1], it is beautifully written. However as I am a complete amateur, I would like to understand the relationships between topological complex K-theory and KR-theory. In particular this sentence: \"There are natural forgetful maps from complex K-theory to KR-theory and in dimension 6 the integers go to 0\".",
"It looks like you implicitly do not accept Atiyah's reference [1] in his paper. What isn't \"nice\" about that?"
] | 16 |
Science
| 0 |
62
|
mathoverflow
|
Concerning proofs from the axiom of choice that ℝ³ admits surprising geometrical decompositions: Can we prove there is no Borel decomposition?
|
This question follows up on a [comment](https://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines#comment238532_92919) I made on Joseph O'Rourke's recent [question](https://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines), one of several questions here on mathoverflow concerning surprising geometric partitions of space using the axiom of choice.
1. Joseph O'Rourke: [Filling $\mathbb{R}^3$ with skew lines](https://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines)
2. Zarathustra: [Is it possible to partition $\mathbb{R}^3$ into unit circles?](https://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles)
3. Benoît Kloeckner: [Is it still impossible to partition the plane into Jordan curves without choice?](https://mathoverflow.net/questions/21327/is-it-still-impossible-to-partition-the-plane-into-jordan-curves-without-choice)
The answers to these questions show that the axiom of choice implies, by rather easy and flexible arguments, that space admits several surprising geometrical decompositions. To give a few examples:
* $\mathbb{R}^3$ is the union of disjoint circles, all with the same radius. Proof: Well-order the points in type $\frak{c}$. At stage $\alpha$, consider the $\alpha^\text{th}$ point. If it is not yet covered, we may find a unit circle through this point that avoids the fewer-than-$\mathfrak{c}$-many previously chosen circles. QED
* $\mathbb{R}^3$ is the union of disjoint circles, with all different radii. Proof: at stage $\alpha$, find a circle of a previously unused radius that avoids the fewer-than-$\mathfrak{c}$-many previously chosen circles. QED
* $\mathbb{R}^3$ is the union of disjoint skew lines, each with a different direction. Proof: at stage $\alpha$, put a line through the $\alpha^\text{th}$ point with a different direction than any line used previously. QED (Edit: it appears that one might achieve this one constructively by using the $z$-axis and a nested collection of [hyperboloids](http://en.wikipedia.org/wiki/Ruled_surface), which unless I am mistaken would give lines of different directions.)
* $\mathbb{R}^3$ is the union of disjoint lines, each of which pierces the unit ball.
* And so on.
The excellent article _Jonsson, M.; Wästlund, J._ , [**Partitions of $\mathbb{R}^3$ into curves**](http://dx.doi.org/10.7146/math.scand.a-13850), Math. Scand. 83, No. 2, 192–204 (1998). [ZBL0951.52018](https://zbmath.org/?q=an:0951.52018). [JStor](https://www.jstor.org/stable/24493132), proves among other things that $\mathbb{R}^3$ can be partitioned into unlinked unit circles, either all with the same radius or all with different radii, and they have a very general theorem about partitioning $\mathbb{R}^3$ into isometric copies of any family of continuum many real algebraic curves.
So the general situation is that the axiom of choice constructions are both easy and flexible and not entirely ungeometrical.
Meanwhile, there are also a few concrete constructions, which do not use the axiom of choice. For example, Szulkin (see theorem 1.1 in [the Jonsson, Wästlund article](http://dx.doi.org/10.7146/math.scand.a-13850)) shows that $\mathbb{R}^3$ is the union of disjoint circles by a completely explicit method. And there are others. But my question is not about these cases where there is a concrete construction. Rather, my question is:
**Question 1.** Can we prove, in any of the cases where there is no concrete construction, that there is none?
For example, taking Borel as the measure of "explicit", the question would be: can we prove, for any of the kinds of decompositions I mention above or similar ones, that there is no Borel such decomposition? This would mean a decomposition of $\mathbb{R}^3$ such that the relation of "being on the same circle" would be a Borel subset of $\mathbb{R}^6$, and in this case, the function that maps a point to the nature of the circle on which it was to be found (the center, radius, and so on) would also be a Borel function.
A closely related question would be:
**Question 2.** Can we prove for any of the decomposition types that it requires the axiom of choice?
For example, perhaps there is an argument that in one of the standard models of $\text{ZF}+\neg\text{AC}$ that there is no such kind of decomposition.
My only idea for an attack on question 1 is this: if there were a Borel decomposition of $\mathbb{R}^3$ into circles or lines of a certain kind, then the assertion that this particular Borel definition was such a partition would be $\Pi^1_2$ and hence absolute to all forcing extensions. So the same Borel definition would work to define such a partition even in forcing extensions of the universe having additional reals. Furthermore, the ground model partitions would be a part of that new partition. And this would seem to be a very delicate geometric matter to fit the new reals into the partition in between the ground model objects. But I don't know how to push an argument through.
|
https://mathoverflow.net/questions/93601/concerning-proofs-from-the-axiom-of-choice-that-%e2%84%9d%c2%b3-admits-surprising-geometrical
|
[
"lo.logic",
"set-theory",
"axiom-of-choice",
"mg.metric-geometry",
"descriptive-set-theory"
] | 51 | 2012-04-09T13:32:24 |
[
"@LSpice Thanks!",
"Sounds like a good idea. Perhaps one could make such a kind of argument work with determinacy or $\\text{AD}_{\\mathbb{R}}$?",
"I really like these questions and every so often I go back and think about them. For me the obvious strategy to attempt is analytic - under some convenient assumption like \"all sets of reals are measurable\", show that such a decomposition cannot exist by some argument that involves, e.g. the probability distribution of the angle of the unique line through a random point. However I've been unable to make anything like this work. Is there a philosophical reason why this sort of arugment cannot be the right way to go?",
"Joel, you might be interested in this paper of Zoltan Vidnyansky: arxiv.org/pdf/1209.4267.pdf. In it, he provides a \"black box\" theorem that allows one to deduce, under the assumption $V=L$, that sets like these, which are typically produced by transfinite recursion in the way you describe, can be coanalytic. His construction is still by transfinite recursion, but he shows how to do it very carefully so that the end result is coanalytic. This does not really answer either of your main questions, but it seems relevant so I thought I should make you aware of it!",
"Good morning, Paul! Yes, of course full AC is not necessary, since as you say a well-ordering of the reals is sufficient. My question was whether one can prove that ZF (or ZF+DC), if consistent, is not able to prove the existence of decompositions. I like your selector idea, and perhaps that is a good strategy.",
"Hi Joel. The second question is whether some form (or consequence) of the Axiom of Choice not following from ZF (or ZF + DC$_{\\mathbb{R}}$) follows from the existence of one of these decompositions? For instance, a selector for some Borel equivalence relation, or even a set of reals without the Baire property? It seems that you already know that full AC is not necessary, since the constructions all work from the existence of a wellordering of the reals. Is that right? Perhaps the existence of a selector for some equivalence relation is sufficient to carry out the constructions.",
"Thanks! I'm really looking forward to hearing the outcome for the constructive case---please post an answer when you arrive at something. I would be interested to know the connection between the constructive case and the case of continuous partitions in classical logic, which could of course be viewed as a stricter form of the Borel partitions I asked about. What I expect is that one could hope to rule out continuous partitions in many of the cases, but ruling out Borel partitions seems mysterious to me.",
"The constructive version is currently discussed at groups.google.com/group/constructivenews/browse_thread/thread/…",
"@Joel: The trouble, constructively, is in showing that each sphere intersects the circles in exactly two points. When the points of intersection pass from one circle to another, as we vary the radius of the spehere, there is a discontinuity (the points jump suddenly). This is a \"model-theoretic\" way of seeing that something isn't right constructively.",
"Joseph, thanks! These lines are skew in the sense that any two of them have different directions, but not skew in the sense of your question, which I know how to achieve only by using AC. ",
"@Joel: Your idea to use hyperboloids to construct skew lines is nice!",
"Asaf, since it seems difficult to make any proof at all, it would be interesting to see that ZF alone, without DC, cannot give a certain kind of construction. So your idea of considering a model where the reals are a countable union of countable sets is a good one, but I don't see how to make it work. In that model, the two questions become in effect the same, since every set is Borel there.",
"And no, I don't think we know that we can't do all these things constructively. Perhaps there are nice explicit decompositions of each type. When there is, then we can prove that there is just by exhibiting it. But when there isn't, I don't see how we could prove that there isn't, and this is what my question is about.",
"Andrej, I was using the term \"construction\" loosely, in classical logic, but could you explain why the argument isn't constructive in your sense? You have a circle in the $xy$ plane with center at $(4k+1,0,0)$ and a sphere of radius $r$ at the origin. So you are intersecting two circles; you can solve for the intersection points, and they vary step-wise continuously with $r$. (I am guesing that this \"step-wise\" step is the issue for constructive logic...)",
"By the way, theorem 1.1 in Jonsson & Wästlund article is not constructive as far as I can tell. There is a problem with \"each sphere intersects one of the circles in exactly two points\".",
"Do we know that these things can't happen constructively?",
"Shooting from the hip, what happens when the real numbers are a countable union of countable sets? Should you require DC as well, as it ensures that most mathematics behaves normally, and at least ensures that there are more sets than Borel sets?"
] | 17 |
Science
| 1 |
63
|
mathoverflow
|
Enriched Categories: Ideals/Submodules and algebraic geometry
|
While working through Atiyah/MacDonald for my final exams I realized the following:
The category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals
$$I\cdot J:=\langle\\{ab\mid a\in I, b\in J\\}\rangle$$
The internal Hom is given by the ideal quotient; let's denote it $[I, J]:=(J:I)=\lbrace a\in A\mid aI\subset J\rbrace$. We have
$$IJ\subset K \Leftrightarrow I\subset [J,K].$$
Furthermore the category(poset) of submodules $U(M)$ of an A-Module $M$ is enriched over the ideal category via
$$[U,V]:=\lbrace a\in A\mid aU\subset V\rbrace.$$
Where does this enrichment come from? More specifically: As in algebraic geometry we are dealing with the set of prime ideals ( wich is more or less the subcategory
$$\mathrm{Int^{op}}/A\subset\mathrm{CRing^{op}}/A$$
of ring-maps $A\to B$ with $B$ an integral domain), I'd like to see the category of Ideals as some "flattened" version of $\mathrm{CRing^{op}}/A$ by taking the kernel. Or more generally i'd like to see like to see it as a "flattened" version of the category of modules (by taking the annihilator?). So:
**Question:** Is it possible to build the enriched categories $I(A)$ and $U(M)$ out of $\mathrm{CRing^{op}}/A$ and $\mathrm{Mod}_A/M$ in a nice way? So: Can we for example write $I\cdot J$ down as a kernel of some $A\to B$? Or $\mathrm{Ann}(M)\cdot \mathrm{Ann}(N)$ just by using $N,M$ and constructions in/on $\mathrm{Mod}_A$?
|
https://mathoverflow.net/questions/76443/enriched-categories-ideals-submodules-and-algebraic-geometry
|
[
"ac.commutative-algebra",
"ag.algebraic-geometry",
"ct.category-theory",
"monoidal-categories",
"enriched-category-theory"
] | 48 | 2011-09-26T11:24:59 |
[
"\"The category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals\" That's how I'll explain commutative algebra to my son!",
"Very interesting question. $I(A)$ is equivalent to the full subcategory of the comma-category which consists of regular epimorphisms $A \\to ?$. I've already asked here (mathoverflow.net/questions/69037/…) how we can describe the ideal product under this equivalence."
] | 2 |
Science
| 0 |
64
|
mathoverflow
|
Extending a line-arrangement so that the bounded components of its complement are triangles
|
Given a finite collection of lines $L_1,\dots,L_m$ in ${\bf{R}}^2$, let $R_1,\dots,R_n$ be the connected components of ${\bf{R}}^2 \setminus (L_1 \cup \dots \cup L_m)$, and say that $\\{L_1,\dots,L_m\\}$ is _triangulating away from infinity_ iff every $R_j$ that is bounded is triangular.
Can every finite line-arrangement be augmented (by adding more lines) to yield one that is triangulating away from infinity?
(The old and new arrangements need not be generic; they may have parallel lines and/or multiple intersections.)
Some background: A slightly different (and, in my view, less natural) question was raised by me about twenty years ago; it has never been answered either. That problem first appeared in Richard K. Guy and Richard J. Nowakowski’s American Mathematical Monthly column on open problems (see “Bite-Sized Combinatorial Geometry Problems”, AMM, Vol. 103, No. 4, Apr. 1996, p. 342 and “Monthly Unsolved Problems”, AMM, Vol. 104, No. 10, Dec. 1997, pp. 969-970). Stan Wagon invited the readers of his on-line column to completely solve a one-parameter class of problems of this kind in which three lines symmetrically divide an equilateral triangle; see [Link](https://web.archive.org/web/20171230114418/http://mathforum.org/wagon/spring96/p812.html) .
A related problem that might be simpler to solve in the negative is a version in which we not only require that all bounded components have 3 sides but also require that all unbounded components have 2 or 3 sides. If nobody can solve my original problem affirmatively or negatively, a negative solution to this problem would deserve the bounty.
(Historical aside: Many years ago I proposed the original version of this problem to John Conway. He said “Let me think about that,” and as the curiosity-bug bit into his brain, he began to draw pictures, make observations, and formulate conjectures. But he was no fool; he could see that I was deliberately trying to entice him into working on the problem, and he was too proud to want to be seen as one who is so easily seduced. He shuddered as if shaking off an unpleasant memory and said “You know, I don’t have to work on just ANY damned problem!”)
|
https://mathoverflow.net/questions/155196/extending-a-line-arrangement-so-that-the-bounded-components-of-its-complement-ar
|
[
"discrete-geometry",
"plane-geometry",
"hyperplane-arrangements"
] | 47 | 2014-01-20T11:45:57 |
[
"the dual version of Silvester-Caley thm in a stronger version shows that a simplicial argangment has to satisfy an optimal incidency condition that cannot hold if one move some lines in a certain neigbourhood while some other will remain fixed. In another hand a set of lines construced with the condition of my first comment should not change (up to homeomorphism) if we move the lines that are constructed thanks to the 5th line according to any choice of this 5th line in a fine neigbourhood, while a set of line including the first four one will remain fixed. (it's the heuristic argmt)",
"The weaker condition (unbounded are 2 or 3 side) is stronger than the condition in the projective plane because if two unbounded component have both 3 sides, and correspound to a unique component in the projective plane, then this projective component will have 4 sides",
"@JamesPropp : would you be ok with a negative answer in the projective plane (it would still be weaker than the weaker condition) I think it should not be difficult to show that if we start from 4 projective lines in general position and we add one more line that can't be constructed (with the ruler) from the first 4 lines, then it won't be possible to augment in a simplicial arrangement. Maybe it is not strong enought to be in an answer eventhought I thing the general and wealker case can be deduced from the argument that I think woukd work in the more simple case of the projective plane",
"There are more simplicial pseudoline arrangements known than simplicial line arrangements; see Berman's \"Symmetric Simplicial Pseudoline Arrangements\", combinatorics.org/ojs/index.php/eljc/article/view/737 — but that doesn't make the problem trivial, or even easier, for that variation.",
"@LSpice : I had no idea that had happened. The point was a MathJax usage correction.",
"Another partial solution: Put a vertical line through every intersection. Then every bounded region is either a triangle or a trapezoid, because all vertices must lie on those vertical lines.",
"Leaving aside the fact that Yoav is thinking of affine space rather than projective space, the truth is even more embarrassing than his comment would suggest: if I'm thinking clearly today (as opposed to yesterday), any generic 3-line arrangement in projective space IS a triangulation! So I retract my proposal. To replace it by a more credible proposal, I'd have to acquaint myself with the prior literature that David Eppstein is implicitly citing.",
"Isn't it the case that a generic 3-line arrangement can be extended to a simplicial arrangement by adding at each intersection of two lines the parallel to the third line?",
"If anyone can show that a generic triple of lines in the projective plane does not belong to any simplicial arrangement (which ought to be true, given what David Eppstein tells us), I would be inclined to award the bounty to that person.",
"In case you're unaware (since I didn't see any mention of it in the question): these things are more commonly studied in the projective plane, where there is no distinction between bounded and unbounded. An arrangement of lines or pseudolines in the projective plane for which all faces are three-sided is called a simplicial arrangement. There are only a few infinite families of them known, so certainly it is not known that everything else can be augmented to become simplicial.",
"It would be also interesting to find whether the answer to the question is affirmative if we replace lines by pseudolines (that is --- unbounded curves or olylines satisfying the property that every two of them intersect at exactly one point; here we omit parallel lines).",
"@James: I would add combinatorial-geometry and/or discrete-geometry.",
"I have modified the question again, this time to suggest a related question whose solution would win the bounty if the original question proves to be too hard.",
"@domotorp: Are there other appropriate tags I should add?",
"@James, regarding bountyhunting: I think you might have received low attention due to the appended tags that are not followed by many. Offering a bounty certainly draws more attention, as there are only a few questions with a bounty.",
"Don't take $A$ to be on one of the old lines. Any 3 new lines all pass through $A$, so 3 new lines cannot all be sides of a convex region. Thus, any region $D$ lies in a wedge between two consecutive new lines, call them $L$ and $L'$. Then $\\partial D \\setminus (\\partial D \\cap (L \\cup L'))$ consists of two paths, call them $p$ and $q$. (Or, if $A$ is at a corner of $D$, one path.) If $p$ contains more than one old line, then it has an internal vertex $v$. But there would be a new line through $v$, contradicting that $L$ and $L'$ are consecutive. So $p$ is a single segment, and likewise $q$.",
"I like Ilya Bogdanov's observation, but I don't follow his proof. Why couldn't there be a five-sided face whose sides go \"old\", \"new\", \"old\", \"new\", \"new\"? Or a six-sided face whose sides go \"old\", \"new\", \"old\", \"new\", \"old\", \"new\"? Clearly I'm missing something. Also, in the case where a line is both \"old\" AND \"new\" (if say $A$ is one of the original intersection points), should we interpret \"old\" as \"non-new\" in Ilya's argument?",
"It is easy to \"quadrilaterate away\" just by choosing a point $A$ and drawing all lines connecting $A$ with all intersection points of the original lines. Each resulting region will have at most four sides because it cannot contain two consecutive sides on \"old\" lines.",
"@AdamP.Goucher I am confused. If a line crosses a square at two adjacent sides, this leaves a pentagon behind. I can also get a pentagon by having two lines cross inside a square, and both exit on parallel sides.",
"I've revised the question to include reference to an earlier, similar problem, though I doubt that this will be of much use to anyone, except as an indication that the problem is subtler than it appears at first. (Though I'm hoping that both problems eventually turn out to be simple after all, when approached in the right way.)",
"Re Gerhard Paseman's comment, I think I've learned something from this experience about modifying questions in MathOverflow. It's not enough to revise the wording of a question in response to people's queries; it's important to make it clear to everyone that one has done so! I'll try to remember to do this in the future.",
"Re Adam Goucher's suggestion, I find the phrase \"all lines that intersect the square pass through a single point\" obscure. Adam, can you clarify this for me?",
"Re Sam Hopkins' first question, I suspect that the problem becomes easier (and has a negative answer) in higher dimensions, but I have no counterexamples to propose.",
"Observe that it is possible to extend an arrangement to one that is `quadrilateralating away from infinity' by superimposing a sufficiently large square mesh (two perpendicular families of parallel lines), where the squares are sufficiently small that all lines that intersect the square pass through a single point.",
"I have delayed in responding because I found the question incomplete. In particular, I wanted Joseph's question to be acknowledged. Now that I notice the acknowledgment, I am inclined to believe the answer is no, as one may set up a sequence of \"problem quadrilaterals\", where simple attempts to triangulate them produce more \"problem quads\". Further, if there is a delay of about a day in responding to comments, you might expect twice that in responding to responses. Personally, I disagree about the \"bounty bias\". Gerhard \"Does Like Thinking For Rewards\" Paseman, 2014.01.22",
"@JamesPropp: Do you have a counterexample for the analogous statement (involving simplices, of course) in higher dimensions?",
"Well, now there's a bounty as extra incentive.",
"I find it strange that this post has received no answers and so few comments. I wonder if the whole bounty-system is distorting the priorities of the MathOverflow community, by causing MathOverflow contributors to withhold information about solved problems (or to hold off on tackling unsolved problems) until they can get a bounty for their contribution.",
"How generic is the original/new arrangement? Do you admit multiple points? Parallel lines?",
"May I ask: What does it mean to \"extend\" a line arrangement?"
] | 30 |
Science
| 0 |
65
|
mathoverflow
|
Homotopy type of TOP(4)/PL(4)
|
It is known (e.g. the Kirby-Siebenmann book) that $\mathrm{TOP}(n)/\mathrm{PL}(n)\simeq K({\mathbb Z}/2,3)$ for $n>4$. I believe it is also known (Freedman-Quinn) that $\mathrm{TOP}(4)/\mathrm{PL}(4)\to K({\mathbb Z}/2,3)$ is 5-connected. Is it known whether $\mathrm{TOP}(4)/\mathrm{PL}(4)$ is not equivalent to $K({\mathbb Z}/2,3)$?
**Edit:** The following lists the relevant definitions.
* The topological group $\mathrm{TOP}(n)$ is the group of self-homeomorphisms of ${\mathbb R}^n$ with the compact-open topology.
* The topological group $\mathrm{PL}(n)=|\mathrm{PL}_S(n)|$ is defined as the geometric realization of the simplicial group $\mathrm{PL}_S(n)$. The $k$-simplices of $\mathrm{PL}_S(n)$ are the piecewise linear homeomorphisms $\Delta^k\times{\mathbb R}^n\to\Delta^k\times{\mathbb R}^n$ which commute with the projection onto $\Delta^k$.
* With the above definitions, there exists a canonical map of topological groups $\mathrm{PL}(n)\to\mathrm{TOP}(n)$. Then the space $\mathrm{TOP}(n)/\mathrm{PL}(n)$ is defined as the homotopy fibre of the induced map $B\mathrm{PL}(n)\to B\mathrm{TOP}(n)$. It is _not_ actually a quotient of a group by a subgroup.
* Here is a way of recovering the homotopy type of $\mathrm{TOP}(n)/\mathrm{PL}(n)$ as an actual quotient. Let $\mathrm{TOP}_S(n)$ be the singular complex of $\mathrm{TOP}(n)$: $\mathrm{TOP}_S(n)$ is the simplicial set whose $k$-simplices are continuous maps $\Delta^k\to\mathrm{TOP}(n)$; these are in canonical bijection with the homeomorphisms $\Delta^k\times{\mathbb R}^n\to\Delta^k\times{\mathbb R}^n$ commuting with the projection onto $\Delta^k$. Hence we obtain an inclusion of simplicial groups $\mathrm{PL}_S(n)\hookrightarrow\mathrm{TOP}_S(n)$, which induces by adjunction the previous map of topological groups $\mathrm{PL}(n)\to\mathrm{TOP}(n)$. The space $\mathrm{TOP}(n)/\mathrm{PL}(n)$ is weakly homotopy equivalent to the geometric realization of the simplicial set $\mathrm{TOP}_S(n)/\mathrm{PL}_S(n)$ (which is levelwise given by taking cosets).
|
https://mathoverflow.net/questions/106971/homotopy-type-of-top4-pl4
|
[
"at.algebraic-topology",
"gt.geometric-topology",
"homotopy-theory"
] | 45 | 2012-09-11T19:38:57 |
[
"@Ricardo: yes, that's what I meant.",
"@Peter: Thank you for remarking that $\\mathrm{PL}(n)$ is not defined as a subgroup of $\\mathrm{TOP}(n)$. In fact, by $\\mathrm{TOP}(n)/\\mathrm{PL}(n)$ I meant the homotopy fibre of the map $B \\mathrm{PL}(n)\\to B \\mathrm{TOP}(n)$, which I believe is the usual meaning. I edited the question to clarify that. @John: Are you by any chance referring to Michael Weiss' recent sequence of articles titled \"Smooth maps to the plane and Pontryagin classes\"?",
"@Peter: Michael Weiss has made some recent progress on understanding $\\text{Top}(n)$.",
"PL(n) needs a different definition; subspace of Top(n) won't do; see for example 1.10 in Madsen and Milgram, \"The classifying spaces for surgery and cobordism of manifolds\". It is BTop(n) and BPL(n) that classify the appropriately defined bundles. As far as I know the question asked is still open. In general, BTop(n) and cognate spaces are not well understood; even their cohomology is not known, although the cohomology when one goes to the limit and considers BTop and BPL is completely understood.",
"$\\text{TOP}(n)$ is in fact the group of self-homeomorphisms of ${\\mathbb R}^n$, as Qfwfq states.",
"@Qfwfq : This is not really my area, but I'm pretty sure that $TOP(n)$ is suppposed to classify topological microbundles on a space. This means that it is something like the classifying space for something like the group of germs of homeomorphisms of $\\mathbb{R}^n$ (or maybe the pseudogroup of homeomorphisms between open sets in $\\mathbb{R}^n$). Similarly for $PL(n)$. I don't have it at hand, but there are proper definitions in a book by Madsen and Milgram.",
"@Steven Landsburg: From books.google.it/… I gather $TOP(n)$ is the group of self-homeomorphisms of $\\mathbb{R}^n$ and $PL(n)$ the subgroup of piecewise linear ones.",
"It was unknown 20 years ago, when this question was proposed as Conjecture 3.10 in 'Differential Topology, Foliations, and Group Actions' by Paul A. Schweitzer. Link: tinyurl.com/cddp8oq ",
"What is $TOP(n)$?"
] | 9 |
Science
| 0 |
66
|
mathoverflow
|
A kaleidoscopic coloring of the plane
|
> **Problem.** Is there a partition $\mathbb R^2=A\sqcup B$ of the Euclidean plane into two Lebesgue measurable sets such that for any disk $D$ of the unit radius we get $\lambda(A\cap D)=\lambda(B\cap D)=\frac12\lambda(D)$?
(I.V.Protasov called such partitions _kaleidoscopic_).
Observe that for the $\ell_1$\- or $\ell_\infty$-norms on the plane such partitions exist: just take a suitable chessboard coloring.
The problem can be reformulated in terms of convolutions: _Is there a measurable function $f:\mathbb R^2\to\\{1,-1\\}$ such that its convolution with the characteristic function $\chi_D$ of the unit disk $D$ is identically zero?_
(The problem was posed 08.11.2015 by T.Banakh and I.Protasov on [page 19](http://www.math.lviv.ua/szkocka/viewpage.php?vol=0&page=19) of [Volume 0](http://www.math.lviv.ua/szkocka/viewbook.php?vol=0) of the [Lviv Scottish Book](http://www.math.lviv.ua/szkocka)).
|
https://mathoverflow.net/questions/219860/a-kaleidoscopic-coloring-of-the-plane
|
[
"real-analysis",
"mg.metric-geometry",
"measure-theory",
"harmonic-analysis",
"geometric-measure-theory"
] | 44 | 2015-10-02T11:55:32 |
[
"@AntonPetrunin No, I do not know such a figure. It seems that for any triangle a kaleidoscopic coloring does exist.",
"Do you know any plane figure (maybe nonconvex) without kaleidoscopic colorings? (wrt parallel translations)",
"I seem to see a 3-color kaleidoscope for the perfect hexagonal ball.",
"I'll record the basic observations in the Fourier domain: Let $f$ be the stated function, let $\\hat{f}$ be its Fourier transform and let $\\hat{\\chi_D}$ be the Fourier transform of $\\chi_D$. Then $\\hat{f} \\hat{\\chi_D}$ is orthogonal to every character of $\\mathbb{R}^2$. In my naive nonanalyst way, I think this means $\\hat{f}$ is supported on the zeroes of $\\hat{\\chi_D}$. According to Mathematica, $\\hat{\\chi_D}(k_1, k_2) = 2 \\pi \\tfrac{J_1(k)}{k}$ where $J_1$ is the first Bessel function and $k = \\sqrt{k_1^2+k_2^2}$. Unfortunately, $J_1$ has infinitely many zeroes.",
"About the tag scottish-book, see the meta discussion: meta.mathoverflow.net/questions/3948/tag-referring-to-a-book/…"
] | 5 |
Science
| 0 |
67
|
mathoverflow
|
What does the theta divisor of a number field know about its arithmetic?
|
This question is about a remark made by van der Geer and Schoof in their beautiful article "Effectivity of Arakelov divisors and the theta divisor of a number field" (from '98) ([link](http://arxiv.org/pdf/math/9802121v3)).
Let me first of all recall that a curve $C$ over a function field can be recovered from its Jacobian $\operatorname{Pic}(C)$ and its theta divisor $\Theta \subset \operatorname{Pic}^{g-1}(C)$ ($g$ the genus of $C$).
In their article van der Geer and Schoof construct analogues of the Jacobian and the theta divisor in the case of number fields.
Namely they consider the Arakelov-Picard group $\operatorname{Pic}(F)$ of a number field $F$ and define its theta divisor as restriction of the function $h^0$ on $\operatorname{Pic}(F)$ given by $$h^0(D) = \log\left(\sum_{f \in I} e^{-\pi \|f\|^2_D}\right),$$ where $I$ is a lattice associated with the Arakelov divisor $D$, to the subspace $\operatorname{Pic}^{(d)}(F)$ of Arakelov divisors of degree $d$. Here $d$ is a suitable analogue of the genus of a curve.
Now they say that "it should be possible to reconstruct the arithmetic of the number field $F$ from $\operatorname{Pic}^{(d)}(F)$ together with $h^0$".
My question is now of course which parts of the arithmetic of $F$ are known to be recovered from their analogy? For example I would be very interested in the question if the units of $F$ can be recovered somehow.
|
https://mathoverflow.net/questions/47920/what-does-the-theta-divisor-of-a-number-field-know-about-its-arithmetic
|
[
"nt.number-theory",
"arakelov-theory",
"theta-functions",
"reference-request"
] | 41 | 2010-12-01T08:17:15 |
[
"Since I think it is useful information to have around, I summarize from memory from the deleted answer and subsequent discussion: the above mentioned paper is the thesis of R. Groenewegen \"Vector bundles and geometry of numbers\", more specifically its second part \"Torelli for number fields\", which explores the problem mentioned in the question, obtaining interesting results related to it, yet not answering the question itself. (I hope this is an accurate and neutral summary of the situation.) The document itself is available eg here: math.leidenuniv.nl/en/theses/13 ",
"Anyway, the purpose is solved. You came to know about the paper I have kept . I am happy. I am deleting my answer. Why to welcome troubles unnecessarily. Let peace prevail. "
] | 2 |
Science
| 0 |
68
|
mathoverflow
|
Homology of $\mathrm{PGL}_2(F)$
|
**Update:** As mentioned below, the answer to the original question is a strong No. However, the case of $\pi_4$ remains, and actually I think that this one would follow from Suslin's conjecture on injectivity of $H_i(\mathrm{GL}_n(F))\to H_i(\mathrm{GL}_{n+1}(F))$ up to torsion (in the case $i=4$, $n=3$) together with some results of Dupont (specifically exact sequence (15.10) in Dupont's book "Scissors congruences, group homology and characteristic classes") and Goncharov's conjecture on motivic cohomology in terms of Bloch groups in weights $2$ and $3$. Concretely:
> Question. Let $F$ be any field. Consider the inclusion $F^\times\to \mathrm{PGL}_2(F)$ of the diagonal matrices. Is the induced map $$ \bigwedge^4 F^\times_{\mathbb Q} = H_4(F^\times,\mathbb Q)\to H_4(\mathrm{PGL}_2(F),\mathbb Q)$$ surjective?
(Explanation: In Dupont's (15.10), the kernel of the last map is precisely $H^2(F,\mathbb Q(3))$ under Goncharov's conjecture. Under Suslin's conjecture, $H_4(\mathrm{PGL}_3(F),\mathbb Q)\to H_4(\mathrm{PGL}_4(F),\mathbb Q)$ is injective, and the target is stable. Computing in terms of the homology of the $K$-theory space, one sees that the contributions to $H_4(\mathrm{PGL}_4(F),\mathbb Q)$ are $\mathrm{Sym}^2 K_2^M(F)_{\mathbb Q}$ and $H^2(F,\mathbb Q(3))$ and $K_4^M(F)_{\mathbb Q}$. The last term $K_4^M(F)_{\mathbb Q}$ is precisely the cokernel of the map $H_4(\mathrm{PGL}_3(F),\mathbb Q)\to H_4(\mathrm{PGL}_4(F),\mathbb Q)$ by Suslin's theorem, so $H_4(\mathrm{PGL}_3(F),\mathbb Q)$ should have contributions from $H^2(F,\mathbb Q(3))$ and $\mathrm{Sym}^2 K_2^M(F)_\mathbb{Q}$. The latter admits a surjection from $\mathrm{Sym}^2(\bigwedge^2 F^\times)_{\mathbb{Q}} = H_4((F^\times)^2\rtimes \Sigma_3)$, which is precisely the decomposable part of $H_4(\mathrm{PGL}_3(F),\mathbb Q)$ in Dupont's work. Looking back at (15.10), this shows that the first term should be zero, which leads to the question. There is some confusion between $\mathrm{PSL}$ and $\mathrm{PGL}$ in what I said, but at least assuming that all elements of $F$ have $6$-th roots, the argument should work.)
* * *
Here is a hopelessly naive question. Please point me to the relevant literature!
Let $F$ be any field. The Cartan subgroup of $\mathrm{PGL}_2(F)$ is $F^\times\rtimes \Sigma_2$. Let $X_2(F)$ be the homotopy cofiber of $B(F^\times\rtimes\Sigma_2)\to B\mathrm{PGL}_2(F)$, so there is a long exact sequence $$ \ldots\to H_i(F^\times\rtimes\Sigma_2)\to H_i(\mathrm{PGL}_2(F))\to H_i(X_2(F))\to \ldots $$ (I'm not sure if I should write $H_i(G)$ or $H_i(BG)$...). If I am not making a stupid mistake, then the known computations of the homology of $\mathrm{GL}_2(F)$ going up to the first unstable group $H_3(\mathrm{GL}_2(F))$ can be summarized by saying that $H_i(X_2(F))=0$ for $i=1,2$, and $H_3(X_2(F))=\mathfrak p(F)$ is the pre-Bloch group of $F$. (The pre-Bloch group is the quotient of $\mathbb Z[F^\times]$ by the ``$5$-term relations''.) (I might be off by some $2$-torsion.) By Hurewicz (and the check that $\pi_1 X_2(F)=0$), the same holds true for the homotopy groups of $X_2(F)$.
As far as I could find, little is known about the homology of $\mathrm{GL}_2(F)$ or $\mathrm{PGL}_2(F)$ in degrees $>3$. The rational structure could be determined by the following very naive question.
> Question. Are the homotopy groups $\pi_i X_2(F)$ bounded torsion for $i\neq 3$?
I might even expect the implicit bound to be independent of $F$ (but of course depend on $i$; the order of magnitude should be $i!$).
The only evidence I have is that a back-of-the-envelope calculation seems to suggest that this holds true for finite fields. In that case the order of $\mathrm{PGL}_2(\mathbb F_q)$ is the product of $q-1$, $q$ and $q+1$. Localized at $q-1$, the homology agrees with the homology of the Cartan; localized at $q$, it is only in degrees larger than $q$ (which is OK as the bound on torsion may depend on $i$), and localized at $q+1$, the homology seems to agree with the homology of a $K(\mathbb Z/(q+1),3)$ (up to bounded torsion in each degree), which is roughly $\mathbb Z/(q+1)$ in all degrees $\equiv 3\mod 4$ (and is bounded torsion in other degrees). The pre-Bloch group $\mathfrak p(\mathbb F_q)$ is also $\mathbb Z/(q+1)$ up to $2$-torsion, so things add up.
The case of $F=\mathbb Q$ might be amenable to computations, but I was unable to do those. Are there any relevant results about $H_4(\mathrm{GL}_2(F))$ that might shed light on $\pi_4 X_2(F)$?
A final remark: I expect that it is critical that $F$ is a field. The similar statement should be false already for discrete valuation rings, or $\mathbb Z[\tfrac 1n]$ (which is why the case of $F=\mathbb Q$ is not so easy to compute; I presume one would try to first compute for all $\mathbb Z[\tfrac 1n]$ and then pass to a colimit, and the desired structure should only appear in the colimit).
**Update:** As mentioned by Matthias Wendt in the comments, a special case of the work of Borel-Yang says that $H_i(\mathrm{PGL}_2(\mathbb Q),\mathbb Q)=0$ for $i>0$. Inserting this into the above long exact sequence, one immediately sees that $X_2(\mathbb Q)$ is in fact rather complicated, and for example has nontorsion $\pi_5$. Thus, the answer to the question is a strong No. Thanks for the help in sorting this out!
|
https://mathoverflow.net/questions/298336/homology-of-mathrmpgl-2f
|
[
"gr.group-theory",
"algebraic-groups",
"kt.k-theory-and-homology",
"group-cohomology",
"algebraic-k-theory"
] | 39 | 2018-04-20T02:58:29 |
[
"@Matthias Wendt: Thanks a lot for the Borel-Yang reference! Let's not worry right now about the exact shape of the torsion; I'd be happy enough with the statement rationally (but then my \"evidence\" would be vacuous). About homotopy vs. homology: One reason is that the homotopy of the K-theory space is usually simpler than the homology of the K-theory space (just the primitives). Maybe the better answer is that I have some roundabout argument that led me to believe in the question, but it's not worth spelling this out while there is no nontrivial evidence.",
"For the function field case, Harder's results show that rational cohomology of $SL_2(F)$ is torsion. I don't know of any results bounding torsion there. The homology of $X_2(F)$ is the equivariant homology of a complex of configurations of points on $\\mathbb{P}^1$. The exponential bounds for torsion (in cohomology) could follow if we knew that all torsion comes from symmetry groups of configurations. But I guess we don't know that. Finally: why should we assume that it's easier to study homotopy of $X_2(F)$ instead of its homology?",
"According to Borel and Yang, the rational cohomology of $SL_2(F)$ with $F$ a number field is the tensor product: for every real place an exterior algebra generated by an Euler class in degree 2, for every complex place an exterior algebra generated by a class in the Bloch group. So it's reasonable to assume that in this case $X_2(F)$ is rationally a product of 3-spheres. I haven't checked the details, but that would be partial support for assuming that the higher homotopy groups (above degree 3) of $X_2(F)$ are torsion.",
"@S. carmeli: I don't think the cohomological dimension of $F$ should be relevant. In fact, there is no \"Galois descent\" (even approximate, like maybe in degrees bigger than the cohomological dimension or so (like for $K$-theory)) for $X_2(F)$. In fact, for $F=\\overline{\\mathbb F}_p$, the Bloch group is $0$, and all homotopy groups of $X_2(F)$ are bounded torsion. Thus, there is no hope to recover $X_2(\\mathbb F_p)$ from $X_2(\\overline{\\mathbb F}_p)$.",
"Don't you want to assume a bound on the cohomological dimension of $F$? I think that his is sort of important in the case of a finite field, since etale homotopically the space X(F) over the algebraic closure is very close to an Eilenberg-Mclane space and then taking fixed points don't take high homotopies to far down, but this sort of reasoning breaks in high cohomological dimension.",
"Related to your final remark, K. Knudson has computed the homology of PGL_2(A) where A is the coordinate ring of an elliptic curve, using an action of this group on a certain Bruhat-Tits tree, see \"On the K-theory of elliptic curves\". Of course the result involves the homology of PGL_2 of the base field k, so is computable if k is finite.",
"@Matthias Wendt: \"Bounded torsion\" means \"bounded exponent\". I would expect something like \"$\\pi_i X_2(F)$ is killed by $i!$ for $i\\neq 3$\". I would imagine that the torsion is of the same cardinality as $F$.",
"@Will Sawin: This is basically Quillen's classical computation.",
"I think for all that's known now we can't even rule out the possibility that $H_i(X_2(F))$ has cardinality of $F$ for any algebraically closed field $F$. Does \"bounded torsion\" mean bound on size or bound on exponent?",
"How does one do the computations in the $PGL_2(\\mathbb F_q)$ case?"
] | 10 |
Science
| 0 |
69
|
mathoverflow
|
Correspondence between eigenvalue distributions of random unitary and random orthogonal matrices
|
In the course of a physics problem ([arXiv:1206.6687](http://arxiv.org/abs/1206.6687)), I stumbled on a curious correspondence between the eigenvalue distributions of the matrix product $U\bar{U}$, with $U$ a random unitary matrix and $\bar{U}$ its complex conjugate, on the one hand, and the random orthogonal matrix $O$ with determinant $-1$, on the other hand. The random matrices $U$ and $O$ are uniformly distributed with respect to the Haar measure on the unitary group $U(N)$ and the orthogonal group $O(N+1)$.
One eigenvalue of $O$ is fixed at $-1$ (to ensure that det $O=-1$). The other $N$ eigenvalues have a distribution $p_O$ which was [known](http://en.wikipedia.org/wiki/Circular_ensemble) since Girko (1985). We calculated the distribution $p_{U\bar{U}}$ of the eigenvalues of $U\bar{U}$ (which we did not find in the literature _\--- has anyone seen it before?_). We discovered to our surprise that $p_{U\bar{U}}=p_O$. This holds for both $N$ even and odd (in the latter case both $U\bar{U}$ and $O$ have an eigenvalue fixed at $+1$).
**Question:** Is there a more direct route to arrive at this identity between the two eigenvalue distributions, without going through a separate calculation of each one? (You can find two such separate calculations in the Appendix of arXiv:1206.6687, but this seems a rather unsatisfactory way of understanding the correspondence.)
* * *
Some intuition for what is going on: for both $U\bar{U}$ and $O$ the eigenvalues different from $\pm 1$ come in complex conjugate pairs $e^{\pm i\theta}$. The matrix $O$ has an unpaired eigenvalue at $-1$, which repels $\theta$ from $\pi$. The matrix $U\bar{U}$ cannot have an unpaired eigenvalue at $-1$ by construction and somehow this leads to a repulsion of $\theta$ from $\pi$ which is mathematically equivalent to what happens for the matrix $O$.
For example, when $N=2$ the eigenvalue $e^{i\theta}$ has the same distribution $P(\theta)=(2\pi)^{-1}(1+\cos\theta)$ for both $U\bar{U}$ and $O$. For $N=3$ the distribution is $P(\theta)=\pi^{-1}(1-\cos^2\theta)$, in addition to an eigenvalue fixed at $+1$, again the same for $U\bar{U}$ and $O$. The correspondence continues for larger $N$, when factors $(\cos\theta_k-\cos\theta_l)^2$ appear in both distributions.
|
https://mathoverflow.net/questions/105512/correspondence-between-eigenvalue-distributions-of-random-unitary-and-random-ort
|
[
"pr.probability",
"random-matrices",
"haar-measure",
"mp.mathematical-physics",
"linear-algebra"
] | 37 | 2012-08-26T03:42:59 |
[
"@student -- the first statement [$U^TU\\in{\\rm COE}(n)$] is correct, but the second statement is not, I think. The eigenvalue distribution of $U\\bar{U}$, with $U\\in{\\rm CUE}(n)$ is not the eigenvalue distribution of ${\\rm CRE}(n)$, it's the distribution of the eigenvalues $\\neq -1$ in ${\\rm CRE}(n+1)$. So there is this increment by one of the dimensionality. Additionally, note that the inverse of $U\\bar{U}$ is not equal to its transpose, so it's not an element of ${\\rm O}(n)$ (as it should be, for the CRE).",
"so we have that $U^TU = e^{i\\bar{H}} e^{iH} = e^{2\\operatorname{Re} H + [H, \\bar{H}]/2 + \\ldots}$ is COE$(n)$ and apparently $U\\bar{U} = e^{iH} e^{-i\\bar{H}} = e^{-2\\operatorname{Im} H + [H, \\bar{H}]/2 + \\ldots}$ is CRE$(n)$ ...",
"@CarloBeenakker I will try to write a proof when time permits.",
"@Henry.L --- Thank you for your feedback. Can you explain what you mean by \"after you have figured out this case\" ? How do I find the distribution of $\\Lambda$ and show that it is equal to the distribution of the eigenvalues of orthogonal matrices?",
"@CarloBeenakker I think that you can assume without loss of generality that $U\\bar{U}=\\Lambda$ is a diagonal matrix first whose eigenvalues must come in pairs. After you figured out this case, we can multiply both sides with another unitary random matrix $V$ and its Hermitian conjugate $V^*$ to bend the distribution in a linear manner. I do not think complex conjugate provide anything more than a real structure with double dimension? Or did I understand your description wrongly?",
"@Z254R: as far as I know, the Weingarten formula gives averages of polynomials in the matrix elements of $U$ or $O$; that does not seem a viable route to find the distribution of the eigenvalues of $U\\bar{U}$, nor to show that the eigenvalues of $U\\bar{U}$ and $O$ have the same distribution.",
"@Carlo: thanks for the clarification. Time for me to go and read a book on this subject before further commenting :-) (I think I made all my $U\\bar{U}$ related errors so far because of \"wrongly\" simulating a random unitary matrix. I was simulating random symmetric unitaries, not arbitrary ones). I still believe that there should be a slick solution to your problem!",
"@Suvrit, $U\\bar{U}$ is not a symmetric matrix, meaning it is not equal to its transpose; the probability distribution of unitary symmetric matrices is indeed well known, it is the circular orthogonal ensemble (COE) of random matrix theory; this is an unfortunate name, because it can create a confusion with the ensemble of random orthogonal matrices, but these are entirely different objects.",
"Is the observation that $U\\bar{U}$ is a unitary symmetric matrix (and every symmetric unitary matrix $S$ can be written as such a product) of value to you here (I say that because such matrices seem to be used when studying orthogonal ensembles...)",
"indeed, the eigenvalue distributions of the classical compact groups are related, notably $p_O$ for $N$ even is also the eigenvalue distribution of the compact symplectic group; the matrices of the form $U\\bar{U}$ do not form a group, so this does not seem to help much; note also that the correspondence between $p_{U\\bar{U}}$ and $p_O$ is not an asymptotic large-$N$ result (like the correspondence to a normal distribution): it holds exactly for any finite $N$.",
"There is an identity for the eigenvalue distributions in this paper: web.williams.edu/go/math/sjmiller/public_html/ntandrmt/handouts/… (see page 6) Maybe this would help. It is also a known fact that the orthogonal group is similar to a standard normal distribution: www-stat.stanford.edu/~cgates/PERSI/papers/random_matrices.pdf (see page 56). Thus it is possible to solve equality of these two distributions to obtain the answer to your question. ",
"argh...i made stupid calculation error!",
"well, $U\\bar{U}$ is not itself orthogonal (its inverse is not equal to its transpose); I can construct an orthogonal matrix $O$ with the same eigenvalues as $U\\bar{U}$, but this orthogonal matrix is not uniformly distributed, so that does not seem to help much."
] | 13 |
Science
| 0 |
70
|
mathoverflow
|
Is there any positive integer sequence $c_{n+1}=\frac{c_n(c_n+n+d)}n$?
|
In a [recent answer](https://mathoverflow.net/a/282154/17581) Max Alekseyev provided two recurrences of the form mentioned in the title which stay integer for a long time. However, they eventually fail.
> **QUESTION** Is there any (**added:** strictly increasing) sequence of positive integers $c_1,c_2,c_3,\dots$ satisfying the relation $$ c_{n+1}=\frac{c_n(c_n+n+d)}n $$ for for all $n\geq 1$ (and some integer constant $d$)?
(As Sam Hopkins notes, it would be also very interesting if the indices of the sequence start from some $k>1$ rather than from 1.)
|
https://mathoverflow.net/questions/282159/is-there-any-positive-integer-sequence-c-n1-fracc-nc-nndn
|
[
"nt.number-theory",
"integer-sequences",
"recurrences"
] | 37 | 2017-09-27T08:35:39 |
[
"@WillSawin: phrased this way, the transformation sort of looks like those found in the theory of cluster algebras, so maybe if there is a positive answer to the question it could be found via the \"Laurent phenomenon.\"",
"If we ignore the division by n, a necessary condition for the integrality of the original sequence is for the modified $c_{p+1}$ to divide $p$ for all primes $p$. Might it be possible to show that given any $c_1$ and $d$, that there must exist a prime $p$ such that this is not the case?",
"A note that you may want to allow the sequence to start at $c_k$ for some $k>1$ because otherwise the motivating example is not really of this form.",
"We can express this as a time-invariant dynamical system by adding an additional variable $n$, i.e. $(x,y) \\to (x+1, y(y+x+d)/x)$. For each prime $p$, this naturally gives an algebraic dynamical system on $\\mathbb Q_p \\times \\mathbb Z_p$. One wants to know if there is a $d$ such that all these systems stay inside $\\mathbb Z_p \\times \\mathbb Z_p$ forever. Maybe techniques of $p$-adic algebraic dynamics would be helpful here?"
] | 4 |
Science
| 0 |
71
|
mathoverflow
|
What is the three-dimensional hyperbolic volume of a four-manifold?
|
Every smooth closed orientable 4-manifold may be constructed via a handle decomposition. Before asking a couple of questions, I recall some well-known facts about handle-decompositions of 4-manifolds.
We can of course order the handles according to their index. Handles of index 0 and 1 form a connected 4-dimensional handlebody, whose boundary is a closed 3-manifold, diffeomorphic to a connected sum $\\#_g(S^2\times S^1)$ of some $g$ copies of $S^2\times S^1$ (if $g=0$ we get $S^3$). Handles of index 2 are attached to some framed link $L\subset \\#_g(S^2\times S^1)$. Since 3- and 4-handles also form a 1-dimensional handlebody, after the attaching of the 2-handles we must necessarily obtain a 4-manifold whose boundary is again diffeomorphic to $\\#_h(S^2\times S^1)$, for some $h$ which is not necessarily equal to $g$. The new $\\#_h(S^2\times S^1)$ is obtained by _surgery_ along the framed link $L$.
Therefore, in some sense, constructing closed orientable 4-manifolds reduces to constructing framed links $L\subset \\#_g(S^2\times S^1)$ that produce some $\\#_h(S^2\times S^1)$ via surgery. Let us define the _3-dimensional hyperbolic volume_ of a handle decomposition as the Gromov norm of $(S^2\times S^1) \setminus L$ (which is in turn the sum of the hyperbolic volumes of its pieces according to geometrization, whence the name). Let us then define the _3-dimensional hyperbolic volume_ of a closed orientable 4-manifold as the infimum of all the hyperbolic 3-dimensional volumes among all its handle decompositions. The infimum is actually a minimum because the set of 3-dimensional hyperbolic volumes is well-ordered.
The general question is:
> What can we say about the 3-dimensional hyperbolic volume of a closed 4-manifold?
A more specific one:
> Which closed 4-manifolds have zero 3-dimensional hyperbolic volume?
which is equivalent to the following:
> Which closed 4-manifolds admit a handle decomposition such that $\\#_g(S^2\times S^1)\setminus L$ is a graph manifold?
Complex projetive plane belongs to this class, and also many doubles of 2-handlebodies: if you take any link $L\subset S^2\times S^1$ whose complement is a graph manifold, you can attach 2-handles to it, and then make the double of the resulting bounded 4-manifold. The resulting double has volume zero.
Finally, we have the following very specific question:
> Is there a 4-manifold with positive 3-dimensional hyperbolic volume?
I would expect that most (all?) aspherical 4-manifolds have positive volume, and maybe also many simply connected ones, but I don't know the answer.
|
https://mathoverflow.net/questions/69819/what-is-the-three-dimensional-hyperbolic-volume-of-a-four-manifold
|
[
"gt.geometric-topology",
"4-manifolds",
"3-manifolds",
"knot-theory"
] | 37 | 2011-07-08T12:14:14 |
[
"I think this might be related to this question: mathoverflow.net/q/19974/1345 In particular, if the 3-dimensional volume is zero, I suspect that the manifold admits an F-structure, and the simplicial volume is zero (and likely minimal volume =0). The point is that the circle action on a Seifert manifold M extends over dehn fillings and attaching 2-handles along MxI. If all of the level sets of a Morse function on a 4-manifold are graph manifolds (and the complements of the 2-handles attaching maps), then I think that these local circle actions may be assembled to get an F-structure.",
"This question sounds a bit similar to what is written in the last lines of a recent article of Gromov and Guth : arxiv.org/abs/1103.3423"
] | 2 |
Science
| 0 |
72
|
mathoverflow
|
Two-convexity ⇒ Lefschetz?
|
Assume that
* $\Omega$ is an open simply connected set in $\mathbb R^n$
* _(two-convexity)_ if 3 faces of a 3-simplex belong to $\Omega$ then whole simplex in $\Omega$.
Is it true that any component of intersection of $\Omega$ with any 2-plane is simply connected?
**Comments:**
* Note that an open connected set $\Omega\subset\mathbb R^n$ is convex (in the usual sense) if together with two sides of triangle $\Omega$ contains whole triangle.
* If the boundary of $\Omega$ is a smooth hypersurface then the answer is YES. [The above property implies that at most one principle curvature of the boundary is negative. Then the statement follows easily from a Morse-type argument; see Gromov's "Sign and geometric meaning of curvature", Lefschetz theorem in Section $\tfrac12$. In fact this argument, shows that in this case any component of intersection of $\Omega$ with any affine subspace (not necessary 2-dimensional) is simply connected.]
* The answer is YES if $n=3$. [In this case I can mimic the Morse-type argument.]
* The statement would follow if one could approximate any $\Omega$ by two-convex domains with smooth boundary. BUT the example below shows that such approximation does not exist for $n\ge 4$.
* I was surprised that my question has a lot in common with [this question](https://mathoverflow.net/questions/29373/minimal-length-embeddings-of-braids-into-r3-with-fixed-endpoints/)
**Example:** We will construct two-convex simply connected open set $\Omega$ in $\mathbb R^4$ such that intersection $L_{t_0}$ of $\Omega$ with some hyperplane is not simply connected. [This NOT a counterexample, but it shows that it is impossible to prove it using smoothing, as indicated in the comments.]
Set $$\Pi=\\{\,(x,y,z,t)\in\mathbb R^4\mid\,y< x^2\\}.$$ Let $\Pi'$ be the image of $\Pi$ under a generic rotation of $\mathbb R^4$, say $(x,y,z,t)\mapsto(z,t,x,y)$.
Note that $\Pi$ is open and two-convex. Therefore $\Omega=\Pi\cap \Pi'$ is also open two-convex set. One can choose coordinates so that $\Omega$ is an epigraph for a function $f\colon\mathbb R^3\to\mathbb R$ like $$f=\max\\{\alpha_1-\beta_1^2,\alpha_2-\beta_2^2 \\},$$ where $\alpha_i$ and $\beta_i$ are linear functions. In particular $\Omega$ is contactable.
Let $L_{t_0}$ be the intersection of $\Omega$ with hyperplane $t=t_0$; it is a complement of two convex parabolic cylinders in general position. If these cylidners have a point of intersection then $\pi_1 L_{t_0}=\mathbb Z$.
[In particular, the function $f$ can not be approximated by smooth functions which Hessian has at most one negative eigenvalue value at all points.]
|
https://mathoverflow.net/questions/55788/two-convexity-%e2%87%92-lefschetz
|
[
"convexity",
"convex-geometry",
"morse-theory",
"tag-removed"
] | 36 | 2011-02-17T14:12:07 |
[
"@WillSawin, I guess your question is about last par, starting with \"Let $L_{t_0}$...\". We are in hypereplane, i.e., in $\\mathbb R^3$. Imagine that your parabolic cylinders have exactly one point of intersection. Then the complement is homotoipically equivalent to $\\mathbb{S}^1$.",
"I don't understand the example. I'm visualizing pairs parabolic cylinders in general position and all of them have simply-connected complements. Could you describe a specific hyperplane that does the job?",
"Here's an article of mine where (a slight variant of) higher convexity is used: staff.science.uu.nl/~henri105/PDF/advgometry.pdf",
"For an open set $\\Omega$, why is [ if 3 faces of a 3-simplex belong to $\\Omega$ then whole simplex in $\\Omega$ ] not the same as convex? "
] | 4 |
Science
| 0 |
73
|
mathoverflow
|
Orthogonal vectors with entries from $\{-1,0,1\}$
|
Let $\mathbf{1}$ be the all-ones vector, and suppose $\mathbf{1}, \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_{n-1}} \in \\{-1,0,1\\}^n$ are mutually orthogonal non-zero vectors. Does it follow that $n \in \\{1,2\\} \cup \\{4k : k \in \mathbb{N}\\}$?
A few notes are in order:
1. If we do not allow "$0$" entries in the vectors then this is a well-known necessary condition for the existence of an $n \times n$ [Hadamard matrix](https://en.wikipedia.org/wiki/Hadamard_matrix), so the answer to my question is "yes" if we replace $\\{-1,0,1\\}$ with$\\{-1,1\\}$.
2. With some computer help, I have shown that there do not exist collections of vectors with these properties when $n = 3$, $5$, or $6$, so the answer to my question is "yes" when $n \leq 6$.
**Edit (May 19, 2023):** In the comments, Max Alekseyev has shown that the answer is "yes" when $n \leq 12$, and Ilya Bogdanov has shown that the answer is "yes" when $n$ is an odd prime.
|
https://mathoverflow.net/questions/446995/orthogonal-vectors-with-entries-from-1-0-1
|
[
"co.combinatorics",
"linear-algebra",
"matrices"
] | 35 | 2023-05-17T12:10:00 |
[
"I've computationally verified the conjecture $n=15$ and together with Ilya's and Mikhail's arguments that makes it proved for all $n\\leq 20$. For $n=21$, I can tell for now that if a counterexample exists, it should have at least 6 vectors with exactly 14 nonzero components.",
"@MaxAlekseyev Same argument that establishes that Hadamard matrices are of order $1$, $2$, or $4k$. If there are three mutually orthogonal ±1-vectors $\\mathbf{1}$, $v_1$, $v_2$, then the number of indices $i$ where $v_{1i} = s_1$, $v_{2i} = s_2$ is the same for any of the four combinations of $s_1, s_2 \\in \\{+1, -1\\}$.",
"@MikhailTikhomirov: What does \"invoke Hadamard\" stand for?",
"Of course, $2p^k$ ditto.",
"Further, $n = 2p$ for odd prime $p$ is also a no-go. $1 = 1/2p + \\ldots$ requires at least one more $1/2p$, and their sum $1 / p$ requires an extra (even) denominator divisible by $p$ (thus, another $2p$). We now have three full rows, which is enough to invoke Hadamard.",
"@IlyaBogdanov's argument also seems to apply when $n = p^k$ for $p$ odd prime. For the diagonal element we have $1 = 1/p^k + \\ldots$, which has to involve at least $p - 1$ more $1/p^k$'s. But two odd-sized full rows are already a contradiction.",
"@kodlu Yes, even many such weighing matrixes exist. But the all-1 column changes a lot.",
"@IlyaBogdanov - You're quite right, thank you.",
"@NathanielJohnston The life seems more complicated for me. E.g., it could happen that the sum involves $1/15$, $2\\times 1/6$ and $6\\times 1/10$ (,the other denominators should be even!). However, this may fasten algorithms searching such matrices.",
"@IlyaBogdanov - If you agree with my reasoning, please feel free to write this up as an answer and I will accept it. Your denominators-cancelling argument is what blew this open. Thank you!",
"@IlyaBogdanov - And a similar argument works to rule out the $n \\equiv 2\\pmod{4}$ cases, thus proving the entire thing. Suppose $n = 2m$ for some odd number $m$. Then (again, by your denominators-cancelling argument) $A$ must have at least $2$ columns with all entries non-zero (one of which is the all-ones column). If it had $3$ or more such columns we get the Hadamard contradiction, so assume it has exactly $2$. The remaining columns each have their number of non-zero entries each equal to factors of $m$, which are all odd. Such columns cannot be orthogonal to $\\mathbf{1}$. Contradiction.",
"Very interesting. If the all 1 vector is removed there are solutions with odd lengths.",
"@IlyaBogdanov - Very nice, thank you! I think your argument can be extended to all odd numbers, not just odd primes. For example, if $n = 15$ then (by your denominators-cancelling argument) there must be at least 3 columns of $A$ that have all $15$ entries with no $0$s (since then you can get $1/15 + 1/15 + 1/15 = 1/5$ to reduce the denominator). But the standard proof of the Hadamard necessary condition says that you can't have $3$ orthogonal columns of $\\pm 1$ entries unless $n \\in \\{1,2\\} \\cup \\{4k\\}$.",
"@NathanielJohnston Each diagonal element of $BB^T$ is a sum of $n$ njmbers each of them is either 0 or of the form $1/k$ with $k\\leq n$. At least one summand is $1/n$; if there are less than $n$ such, then $n$ in the denominators cannot cancel.",
"@Ilya - Can you expand upon the statement \"But the diagonal elements in $BB^T$ may equal $1$ only if $D = nI$\"? I expect that this is the step that is using the hypothesis of $n$ being an odd prime, but I'm having a hard time seeing how.",
"As a first step, $n$ cannot be an odd prime. This is because the matrix $A$ composed from those columns should be such that $D=A^TA$ is diagonal, so $B=AD^{-1/2}$ should be orthogonal. But the diagonal element s in $BB^T$ may equal $1$ only I’d $D=nI$ which is impossible by Hadamard. It seems that one may investigate $BB^T$ or, equivalently, the relation $AD^{-1}A^T=I$ further…",
"@domotorp - Only when there's a Hadamard matrix.",
"Do you have a construction for such a matrix for every $n$ that is divisible by 4, or only for those $n$ for which there exists a Hadamard matrix?",
"The question seems related to weighing matrices except that the all-ones vector kind of messes things up.",
"@PeterMueller: I've also checked with cliquer and it confirmed no solutions for $n=11$ within a couple of hours. So, the \"yes\" answer holds for all $n\\leq 12$.",
"Maybe linear algebra modulo $4$ is relevant, as these vectors miss $2 \\bmod 4$? There is also a natural combinatorial interpretation -- with each $\\mathbf{v}_i$ associate disjoint subsets $A_i^+$, $A^-_i$ of $\\{1,2,\\ldots,n\\}$ defined by $A^{\\pm}_i=\\{j: (\\mathbf{v}_{i})_j=\\pm 1\\}$. Then $|A^+_i|=|A^-_i|$ by orthogonality of $\\mathbf{v}_i$ with $\\mathbf{1}$. The rest of the orthogonality conditions are encoded by $|A^+_i \\cap A^+_j| + |A^-_i \\cap A^-_j| = |A^+_i \\cap A^-_j| + |A^-_i \\cap A^+_j|$ for $i \\neq j$.",
"@MaxAlekseyev ... and so did I using cliquer. Presently running $n=11$ which might take a while ...",
"I confirm the affirmative answer for all $n\\leq 10$ with ILP.",
"Somewhat relevant: mathoverflow.net/q/418009",
"@Peter: Thanks, I do need that condition so I'll add that in. (I'm reminded of a time when I asked my linear algebra students to produce some orthogonal vectors satisfying certain conditions, and lots of them got easy marks since I forgot to specify \"non-zero\"...)",
"@NathanielJohnston Which condition prevents all $\\mathbf{v_i}$ to be the $0$-vectors?",
"@Amir: If you just require linear independence then you can always (for all $n$) find such a set of vectors. Furthermore, if you just require linear independence then whenever $n$ is even you can find such a set of vectors with entries in $\\{-1,1\\}$.",
"Interesting! Would the answer change if we replace the \"orthogonality\" requirement with \"independence\"?"
] | 28 |
Science
| 0 |
74
|
mathoverflow
|
Is there a Mathieu groupoid M_31?
|
I have read something which said that the large amount of common structure between the simple groups $SL(3,3)$ and $M_{11}$ indicated to Conway the possibility that the Mathieu groupoid $M_{13}$ might exist.
Indeed, it exists, and the group of permutations of the other points generated by paths in which the hole starts and ends at the same point is the Mathieu group $M_{12}$.
It is natural to ask whether or not something analogous exists for the Mathieu group $M_{24}$. The corresponding linear group seems to be $GL(5,2)$.
$2^{4}:A_{8}$ is a vector (or hyperplane) stabilizer in $GL(5,2)$ and an octad stabilizer in $M_{24}$. $2^{6}:(GL(2,2)\times GL(3,2))$ is likewise the stabilizer of a 2-dimensional (or 3-dimensional) space in $GL(5,2)$, and the stabilizer of a 'trio' (to use the term from SPLaG) in $M_{24}$.
The natural setting, therefore, in which to look for something analogous to $M_{13}$ is $(\mathbb{Z}/(2))^{5} \backslash 0$. (This can also be regarded as $4$-dimensional projective space over $\mathbb{Z}/(2)$, but I will here think of it as $5$-dimensional space over $\mathbb{Z}/(2)$ with the origin removed.)
Then, to have $24$ points left over when the 'holes' are put in place, it would be natural to have a 'missing $3$-dimensional space' for a hole.
A move should consist of moving the $3$-dimensional 'hole' to some other $3$-dimensional subspace (minus the origin) of $(\mathbb{Z}/(2))^{5}$. The intersection of these spaces can have dimension $1$ or $2$, and there should be different rules for moving the hole to another location depending on the dimension of its intersection with the new location.
Who has tried this before? Are there any papers on it?
|
https://mathoverflow.net/questions/70680/is-there-a-mathieu-groupoid-m-31
|
[
"gr.group-theory",
"finite-groups",
"reference-request"
] | 34 | 2011-07-18T14:48:41 |
[
"It's good to know. 31 vs. 13 for M12."
] | 1 |
Science
| 0 |
75
|
mathoverflow
|
Subalgebras of von Neumann algebras
|
In the late 70s, Cuntz and Behncke had a paper
H. Behncke and J. Cuntz, _Local Completeness of Operator Algebras_ , Proceedings of the American Mathematical Society, Vol. 62, No. 1 (Jan., 1977), pp. 95- 100
the about the following question. Let $A$ be a $C^\star$-algebra and let $B \subset A$ be some $*$-subalgebra. Let $B$ be dense and such that every maximal abelian subalgebra of $B$ is norm-closed. Is it true that $B=A$? They proved this in various cases.
I want to ask a related question in the von-Neumann-setting. More precisely: Let $A$ be a von Neumann algebra and $B \subset A$ be some ultra-weakly dense $\star$-subalgebra such every MASA of $B$ is ultra-weakly closed in $A$. Is it true that $B=A$? A result of Gert Pedersen implies that once $B$ is a $C^\star$-algebra, then $B=A$. Hence, the two questions are closely related. Taking the work of Behncke-Cuntz and Pedersen together, it is known that $B=A$ if $A$ has no $II_1$-part.
> **Question:** Is it true for every von Neumann algebra?
or a little more modest
> **Question:** Is it true for the hyperfinite $II_1$-factor?
Another strenghtening of the the assumption (which could help) would be the following:
> **Question:** Let $A$ be a $II_1$-factor and $B \subset A$ be a ultra-weakly dense $\star$-subalgebra such that for every hyperfinite subalgebra $R \subset A$ one has that $R \cap B$ is ultra-weakly closed in $A$. Is it true that $B=A$?
|
https://mathoverflow.net/questions/34692/subalgebras-of-von-neumann-algebras
|
[
"oa.operator-algebras",
"fa.functional-analysis"
] | 33 | 2010-08-05T14:40:37 |
[
"If I understand correctly, you need to add a density hypothesis in your question?"
] | 1 |
Science
| 0 |
76
|
mathoverflow
|
Isometric embeddings of finite subsets of $\ell_2$ into infinite-dimensional Banach spaces
|
Question: Does there exist a finite subset $F$ of $\ell_2$ and an infinite-dimensional Banach space $X$ such that $F$ does not admit an isometric embedding into $X$?
There are some results of the type: each finite ultrametrics admits an isomeric embedding into any infinite dimensional Banach space. See Shkarin, S. A. Isometric embedding of finite ultrametric spaces in Banach spaces. Topology Appl. 142 (2004), no. 1-3, 13-17 for this, and also the papers: Dekster, B. V. Simplexes with prescribed edge lengths in Minkowski and Banach spaces. Acta Math. Hungar. 86 (2000), no. 4, 343-358; Faver, Timothy; Kochalski, Katelynn; Murugan, Mathav Kishore; Verheggen, Heidi; Wesson, Elizabeth; Weston, Anthony Roundness properties of ultrametric spaces. Glasg. Math. J. 56 (2014), no. 3, 519-535.
There is a related question on MathOverflow: [Almost isometric embeddability implies isometric embeddability](https://mathoverflow.net/questions/209435/almost-isometric-embeddability-implies-isometric-embeddability)
Added on 9/29/2016: Related results were obtained by James Kilbane in <https://arxiv.org/abs/1609.08971>
Added on 4/3/2017: In another [recent paper](https://arxiv.org/pdf/1704.00319.pdf) James Kilbane proved that the set of possible counterexamples (if they exist) is small in a certain sense.
|
https://mathoverflow.net/questions/221181/isometric-embeddings-of-finite-subsets-of-ell-2-into-infinite-dimensional-ban
|
[
"mg.metric-geometry",
"banach-spaces",
"metric-embeddings"
] | 33 | 2015-10-17T17:44:32 |
[
"The last result I know in this direction is in Petrov, F. V.; Stolyarov, D. M.; Zatitskiy, P. B. On embeddings of finite metric spaces in ln∞. Mathematika 56 (2010), no. 1, 135–139",
"Is much known when $X$ is $\\ell^n_{\\infty}$ for $n<\\#F$?"
] | 2 |
Science
| 0 |
77
|
mathoverflow
|
Peano Arithmetic and the Field of Rationals
|
In 1949 Julia Robinson showed the undecidability of the first order theory of the field of rationals by demonstrating that the set of natural numbers $\Bbb{N}$ is _first order definable_ in $(\Bbb{Q}, +, \cdot$).
It is not hard to see that Robinson's result can be reformulated in the following symmetric form.
**Theorem A.** _The structures_ ($\Bbb{N}, +, \cdot$) _and_ $(\Bbb{Q}, +, \cdot$) _are bi-interpretable._
The following generalization of Theorem A is considered folkore (I am not aware of a published reference).
**Theorem B.** _If $(M, +, \cdot)$ is a model of $PA$ (Peano arithmetic), then the field of rationals $\Bbb{Q}^M$ of $(M, +, \cdot)$ is bi-interpretable with $(M, +, \cdot )$._
Let $EFA$ denote the _exponential function arithmetic_ fragment of $PA$, a fragment also known as $I\Delta_{0}+exp$.
Based on _a posteriori_ evidence _classical_ theorems of Number Theory do not require the full power of $PA$ since they can be already verified in $EFA$ (indeed Harvey Friedman has conjectured that even FLT can be verified in $EFA$, with a proof that would be very different from Wiles').
This suggests that in Theorem B one should be able to weaken $PA$ to $EFA$, hence my question:
**Question**. _Is there a published reference for the strengthening of Theorem B, where $PA$ is weakened to $EFA$?_
P.S. The following paper provides an excellent expository account of Robinson's theorem (and related results).
D. Flath and S. Wagon, _How to Pick Out the Integers in the Rationals: An Application of Logic to Number Theory_ , American Mathematical Monthly, Nov. 1991.
|
https://mathoverflow.net/questions/65664/peano-arithmetic-and-the-field-of-rationals
|
[
"lo.logic",
"nt.number-theory"
] | 32 | 2011-05-21T11:01:13 |
[
"@Samuel Ried: No I have not (sorry for the tardy reply, I have been \"away\" for a long while).",
"@AliEnayat: Have you found an answer to your question yet? I would be very interested to know.",
"SJR: The answer to your first question is positive, i.e., nothing fancier is going on. Regarding the second one: the arguments of Robinson's paper should go through in $EFA$ (based on $EFA$'s track record in handling \"elementary\" number theory); my question is whether anyone has actually shown - in a publishd source - that this is indeed the case.",
"Ali: Suppose $M\\models PA$. To interpret $M$ in $\\mathbb{Q}^M$ does Julia Robinson's formula with the operations of $\\mathbb{Q}^M$ do the job, or is there something fancier going on? Is your question about how much of the argument in Robinson's paper goes through in models of $I\\Delta_0+exp$?"
] | 4 |
Science
| 0 |
78
|
mathoverflow
|
$f\circ f=g$ revisited
|
This may be related to [solving $f(f(x))=g(x)$](https://mathoverflow.net/questions/17614). Let $C(\mathbb{R})$ be the linear space of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$, and let $\mathcal{S}:=\\{g\in C(\mathbb{R}) ; \exists\; f\in C(\mathbb{R})$ s.t. $f\circ f=g \\}$ . Is there some infinite dimensional (or, at least, bidimensional) linear subspace of $C(\mathbb{R})$ contained in $\mathcal{S}$ ?
P.S. As a remark, there is a [maybe] interesting connection between [How to solve $f(f(x)) = \cos(x)$?](https://mathoverflow.net/questions/17605) and [Borsuk pairs of Banach spaces ](https://mathoverflow.net/questions/11675). Namely, let $E$ be the closed subspace of $C[-1,1]$ consisting of all even functions, and let $K$ be the closed unit ball of $E$. Then the continuous mapping $\Psi:$ $K$ $\rightarrow$ $E$ expressed by $\Psi(f)$ $:=$ $f\circ f$ $+$ $\left(\left\Vert f\right\Vert _{\infty}-1\right)\cdot\cos$
is odd on $\partial K$, and has no zeros in $K$.
|
https://mathoverflow.net/questions/18882/f-circ-f-g-revisited
|
[
"ca.classical-analysis-and-odes",
"fractional-iteration"
] | 32 | 2010-03-20T14:35:10 |
[
"In the previous comment, the full stop was taken as a part of the url - however, the intention was clearly to link to this post: solving $f(f(x))=g(x)$.",
"Unfortunately, not even two vectors in that cone generate a subspace in ${\\mathcal S}$ since ${\\mathcal S}$ contains no decreasing functions (see mathoverflow.net/questions/17614.) Any solution to this problem is a linear space containing no injective functions.",
"All increasing bijections are ok, so there is an infinite-dimensional cone. ",
"ad least there is a 2-dimensional cone (a space which is closed under linear combinations with positive coefficients) given by the space of all nondecreasing linear maps."
] | 4 |
Science
| 0 |
79
|
mathoverflow
|
The central insight in the proof of the existence of a class of Kervaire invariant one in dimension 126
|
I understand from [a helpful earlier MO question](https://mathoverflow.net/questions/101031/kervaire-invariant-why-dimension-126-especially-difficult) that the techniques leading to the celebrated resolution of the Kervaire invariant one problem in the other candidate dimensions yield no insight on dimension 126, to the extent that until recently there was not consensus as to which way it was likely to go.
For a few weeks now, I've been aware of [a manuscript of Minami](https://arxiv.org/pdf/2106.14604) which, if I'm reading correctly, asserts the existence of such an element in dimension 126 as a consequence of Wang and Xu's 2017 paper on the stable 61-stem. Presumably they would not have failed to remark on the resolution of the Kervaire invariant one problem if such was readily visible from what they had done, so it would seem that there is something that is not _obviously_ a consequence of it, but which can nevertheless be extracted in eight pages if one does somehow know what to look for. It's common that when an old conjecture like this is resolved, it is as an application of some newly discovered techique or heretofore-ungleaned insight into the inner workings of the encompassing theory.
**What is the new insight here?**
The work on the stable stems is forbiddingly technical work I am unlikely to ever grasp the full details of, but it's still possible sometimes to get the flavor of the ideas involved, and that's what I would hope for here.
|
https://mathoverflow.net/questions/397905/the-central-insight-in-the-proof-of-the-existence-of-a-class-of-kervaire-invaria
|
[
"at.algebraic-topology",
"homotopy-theory",
"stable-homotopy"
] | 31 | 2021-07-19T16:48:31 |
[
"Probably you aware by now but just in case, there is also this preprint \"On the Last Kervaire Invariant Problem\" by Weinan Lin, Guozhen Wang, and Zhouli Xu (arxiv.org/abs/2412.10879) which also claims to resolve the final case of dimension 126 in the affirmative.",
"You're most welcome. \"Suitable for MathOverflow\" is among my favorite genres of MathOverflow question!",
"Thank you for asking this question in a way that is suitable for MO! (For those wondering: it's asking about mathematical content/insight, not correctness of a preprint)"
] | 3 |
Science
| 0 |
80
|
mathoverflow
|
On the definition of regular (non-noetherian, commutative) rings
|
All rings are commutative with unit. A ring $R$ is called _regular_ if it satisfies
**(Reg)** Every finitely generated ideal of $R$ has finite projective dimension.
Clearly this gives the usual definition if $R$ is noetherian. In the non-noetherian world, valuation rings are regular (this is the reason why I got interested), as are polynomial rings in arbitrary sets of variables over a field.
I am wondering about the reason for this definition, as opposed in particular to:
**(Reg')** Every finitely presented $R$-module has finite projective dimension.
Clearly (Reg') implies (Reg), and they are equivalent if $R$ is _coherent_ (every finitely generated ideal is finitely presented).
To me, (Reg') looks much more natural than (Reg), and from my limited knowledge of the literature it seems that the association "coherent+regular" is quite popular. So, my question is
> What are examples of regular rings not satisfying (Reg')?
Since my post is really about motivating the definition, I am of course more interested in "natural" examples (not just constructed for the purpose).
On the other hand, all examples are welcome since I don't know any.
|
https://mathoverflow.net/questions/327031/on-the-definition-of-regular-non-noetherian-commutative-rings
|
[
"ag.algebraic-geometry",
"ac.commutative-algebra",
"homological-algebra",
"regular-rings",
"valuation-rings"
] | 31 | 2019-04-03T01:38:58 |
[
"If I understand correctly, to show that regular Noetherian is regular in this sense, you have to use the fact that for Noetherian rings being proj dimension <n at a point implies it in a neighborhood?",
"@KestutisCesnavicius: Thanks for the reference. Both properties are stable under any localization, and are Zariski-local on the spectrum. I doubt that they are local in the stronger sense of localizing at primes, but this should be true if the ring is assumed coherent, which is the case for valuation rings and finitely generated flat algebras over them. See Knaf, J. of Alg. Appl. 7 (2008) #5, 575-591.",
"Are these notions local on $R$? Perhaps it would be better to impose these conditions on the local rings of $R$? Anyway, I just wanted to mention that for a valuation ring $V$, not only $V$ itself is regular (in the stronger sense) but also so is every finite type, flat $V$-algebra $R$ with regular $R$-fibers (e.g., $R$ could be $V$-smooth), at least if one takes the local definition I have alluded to. This is a result of Gabber--Ramero, Prop. 11.4.1 in arxiv.org/abs/math/0409584"
] | 3 |
Science
| 0 |
81
|
mathoverflow
|
Todd class as an Euler class
|
Let $X$ be a relatively nice scheme or topological space.
In various physics papers I've come accross, the Todd class $\text{Td}(T_X)$ is viewed as the Euler class of the normal bundle to $X\to LX$. Here $LX$ is the loop space of $X$. They usually call it $e_{reg}$, the regularised Euler class.
**Question:** Is there a way of making this notion of regularised Euler class rigorous?
Whatever the correct definition is, it should work for
* finite dimension vector bundles, where it reduces to usual notion of Euler class,
* loop spaces, where it reduces to the Todd class.
|
https://mathoverflow.net/questions/379590/todd-class-as-an-euler-class
|
[
"ag.algebraic-geometry",
"at.algebraic-topology",
"loop-spaces"
] | 31 | 2020-12-23T03:29:19 |
[
"You can define $i_*$ for any closed embedding between smooths actually, and $i^*i_*=e$ is still true.",
"@Pulcinella The two questions are indeed related: in finite dimensions in the equivariant setting, after localization both $e(N_i)$ and $i^*$ become invertible, so that one can define $i_*$ from the formula $i^*i_*=e(N_i)$. This is another of the reasons inducing me to think that the equivariant setting is the correct one to handle the Euler class of the normal bundle to $i\\colon X\\to LX$. For Gysin I admit I have never thought to that: it is a very good question!",
"Also I'd be curious whether you think it's necessary to work equivariantly to get an answer. The Atiyah paper worked non equivariantly, but a lot of their intermediate algebra makes it look like equivariant setup is the one to use.",
"@domenicofiorenza That's a good question! It was part of what I was getting at originally. A question I might have for you is if you know whether 1. this regularised Euler class might fit into a Gysin sequence, or 2. if it satisfies a formula like $i^*i_*=e(N_i)$ for $i:X\\to LX$. Obviously $i_*$ is undefined at the moment, but ignore that for now...",
"@Pulcinella Well, in the finite dimensional case that is the definition of equivariant Chern classes. In the infinite dimensional case the finite dimensional formulas become infinite products and one has to make sense of them. So the real question, I guess, is what singles out the zeta-regularization among the possible ways of giving a meaning to those infinite products? Surely zeta-regularization has good properties (see, e.g., the paper by Quine,Heydari and Song), but I do not know how to completely avoid an element of choice making it absolutely canonical.",
"@domenicofiorenza I've had a look and I'm a bit worried by the fact that your top chern classes (of the normal bundle of $X$ inside $LX$) are defined by hand like in Atiyah's original paper, rather than follow from some more general notion of euler class, from which the formal Weierstrass-regularisation type formulas also fall out.",
"I'd say Sections 2 and 6. In Section 6 you'll have to replace every occurrence of $\\mathbb{C}/\\Lambda$ with $\\mathbb{R}/\\Lambda$ or ask Mattia for a copy of his thesis, where you can find the computation spelled out in detail",
"@domenicofiorenza Thanks! Could you point to the most relevant section(s)?",
"Always unpleasant to mention a paper one has contributed to, but something very close to an answer to the original question can be found in arxiv.org/pdf/2106.14945.pdf . There, only the Witten class case is treated in detail as we considered Atiyah's treatment in \"Circular symmetry and stationary-phase approximation\" to be complete for the Todd class. Yet, the details on the Todd class case in the spirit of our paper can be found in Mattia Coloma's thesis.",
"You might also want to look at the closely related literature deriving the Witten and A^ genus from [rigorous mathematical] quantum field theory - see Costello on the Witten genus arxiv.org/abs/1006.5422, and Grady-Gwilliam and Grady on the A^ genus arxiv.org/abs/1110.3533 and arxiv.org/abs/1211.6816",
"There's a version of this question in derived algebraic geometry (explaining the Todd genus and Grothendieck-Riemann-Roch via derived versions of loop spaces) that's well understood - the idea is due to a visionary but hard to understand paper of Markarian arxiv.org/abs/math/0610553, a related story is explained in arxiv.org/abs/1305.7175 and the final derivation of the Todd genus is due to Kondyrev and Prikhodko arxiv.org/abs/1906.00172 --- see also closely related arxiv.org/abs/1511.03589 and arxiv.org/abs/1804.00879.",
"I suspect this might still be unknown: see Atiyah's paper \"Circular symmetry and stationary-phase approximation\", where it's apparent that these sorts of questions are still not rigorously solved (and Brylinski's more recent book on loop spaces where he repeats this).",
"@kiran Of course. This comes up a lot in papers of Rongmin Lu, e.g. \"Regularized Equivariant Euler Classesand Gamma Functions\", section 6 digital.library.adelaide.edu.au/dspace/bitstream/2440/50479/8/…",
"Could you include a reference or give a little more context for this \"regularized euler class\" ?"
] | 14 |
Science
| 0 |
82
|
mathoverflow
|
When are two C*-algebras isomorphic as Banach spaces?
|
We may consider each $C^*$-algebra as a Banach space (by forgetting the multiplication and adjoint). I wonder how drastic this step is, i.e., which properties of the $C^*$-algebra are reflected by its Banach space structure.
Clearly, finite-dimensionality and other cardinality properties can be detected. Thus, to avoid these cases, let us restrict to separable, infinite-dimensional $C^*$-algebras.
For commutative $C^*$-algebras the problem has been solved by Milutin. In particular, $C(X)$ and $C(Y)$ are isomorphic as Banach spaces for any uncountable, compact, metric spaces $X$ and $Y$.
In the paper
Hamana. **On linear topological properties of some $C^*$-algebras** , Tohoku Math. J., II. Ser. 29, 157-163 (1977).
it is shown that the Banach space structure also reflects if all irreducible representations of the $C^*$-algebra are finite-dimensional.
Let us ask what happens at the 'opposite' end of the scale:
> Question: Are all simple, separable, infinite-dimensional $C^*$-algebras isomorphic as Banach spaces?
|
https://mathoverflow.net/questions/156621/when-are-two-c-algebras-isomorphic-as-banach-spaces
|
[
"fa.functional-analysis",
"oa.operator-algebras"
] | 31 | 2014-02-03T15:01:12 |
[
"This has been answered many times in the comments but is the pair $c_0$ (the null sequences) and the algebra of continuous functions on the closed unit interval not a simple counter example to your initial question?",
"It is an open problem whether $C^∗_r(F_2)$ is Banach isomorphic to a nuclear $C^*$-algebra (say the CAR algebra $B$ as a model). It is known that (1) the dual of $B$ has the bounded approximation property (because it's AFD), (2) $B$ has a Schauder basis (Junge--Nielsen--Ruan--Xu, Adv. Math. 2004), and (3) $B\\cong B\\otimes_\\alpha B$ for some reasonable tensor product (in this case the spatial tensor product). None of these properties is known for $C^∗_r(F_2)$.",
"@HannesThiel, so maybe you could organize all this long conversation into the single answer?",
"@Tomek: Thank you for pointing out that reference, I was not aware of it. It deals exaclty with the topic I asked about.",
"As Taka pointed out, among the simple, separable, non-type I C*-algebras, all the nuclear ones are isomorphic as Banach spaces (even much more is true) but beyond the nuclear case this is no longer the case. This certainly answers my question, thanks.",
"Hannes, this ma.utexas.edu/users/rosenthl/pdf-papers/93.pdf survey article of Rosenthal gives a nice overview of the Banach/operator-space structure of C*-algebras.",
"@BillJohnson: HRS proved that none of Type II and III noncommutative $L^1$ spaces Banach embed into a type I noncommutative $L^1$ space. If $A$ is a type I $C^*$-algebra, then $A^*$ is a predual of type I von Neumann algebra.",
"Thanks, Nik. Taka, can you give a lazy old man a hint as to why the HRS classification theorem yields that Type I is invariant under Banach space isomorphisms?",
"By Szankowski (Acta Math 1981), there is a separable (simple, unital, etc.) $C^*$-algebra which does not have the approximation property. Such a $C^*$-algebra cannot be Banach isomorphic to a nuclear one.",
"A complement to Caleb's remark: Being type I is invariant under a Banach space isomorphism (at least in the separable case). This follows from Haagerup--Rosenthal--Sukochev's classification of noncommutative $L^1$-spaces up to Banach isomorphism.",
"@Caleb: Thanks for pointing out the result of Kirchberg. This clarifies the picture a lot. The question now seems to be whether the Banach space structure can detect nuclearity of the C*-algebra.",
"@Qiaochu: Thanks for clarifying. Here, two Banach spaces E and F are called isomorphic if there exists a bounded linear map from E to F that is bijective. (The inverse map will automatically be bounded, too). I do not want to assume isometric isomorphism.",
"Hi Hannes, Kirchberg showed in his subalgebras of CAR algebra paper that all separable nuclear, non Type I C*-algebras are isomorphic as operator spaces (much stronger than just isomorphic as Banach spaces). I think Christensen and others also did some things in this direction for von Neumann algebras: those results are in Pisier's operator space book, but my copy isn't with me.",
"@Bill: the \"simplest\" example of a simple C*-algebra (hard not to make puns here) is the algebra $K(H)$ of compact operators on a separable Hilbert space. The irrational rotation algebras $A_\\theta \\subset B(L^2({\\bf T}))$ generated by multiplication by $e^{2\\pi i t}$ and rotation by the angle $2\\pi\\theta$ for irrational $\\theta$ are also simple. The CAR algebra is obtained by embedding $M_{2^n}$ in $M_{2^{n+1}}$ via the map $A \\mapsto \\left[\\matrix{A&0\\cr 0&A}\\right]$ and completing the union $\\bigcup_{n=1}^\\infty M_{2^n}$; it is also simple.",
"I assume \"isomorphic as Banach spaces\" means linear homeomorphism. It's well known that there are non-isomorphic C*-algberas that are isometric as Banach spaces. (Google \"C*-algebra not isomorphic to its opposite algebra.) The correct statement is: A and B are isomorphic as C*-algebras if and only if they are completely isometric as operator spaces.",
"What are some simple and complicated examples of simple separable $C^*$-algebras?",
"Is an isomorphism of Banach spaces a bounded linear map with bounded linear inverse or an isometric isomorphism?"
] | 17 |
Science
| 0 |
83
|
mathoverflow
|
"Three great cocycles" in Complex Analysis as cohomology generators
|
In his [lecture notes](http://www.math.harvard.edu/~ctm/home/text/class/harvard/213a/00/html/course/course.pdf), C. McMullen discusses "the three great cocycles" in Complex Analysis: the derivative $$f\mapsto\log f',$$ the non-linearity $$f\mapsto (\log f')'dz$$ and the Schwarzian derivative $$f\mapsto \left(\frac{f'''}{f'}-\frac32\left(\frac{f''}{f'}\right)^2\right)(dz)^2$$ I would like to understand what "cocycle" means here precisely, and in which sense these maps are "the only" cocycles.
For diffeomorphisms of the circle, this is very clearly explained, e. g., in the [book](http://www.math.psu.edu/tabachni/Books/BookPro.pdf) by Ovsienko and Tabachnikov. Namely, $\mathrm{Diff}(S^1)$ is a group acting naturally on each of the spaces $F_\lambda(S^1)$ of tensor densities of degree $\lambda$. The corresponding (group) 1-cohomology is trivial unless $\lambda$ equals $0$, $1$, or $2$, where it is one-dimensional and generated by the above cocycles.
Question: is there an equally simply formulated statement in the context of complex variable, e. g. treating the above cocycles as generators of a group (or perhaps groupoid?) cohomology? This [MO answer](https://mathoverflow.net/a/38133/56624) suggests that the group is $\mathrm{Aut}$(meromorphic functions), but, at least naively, this seems to coincide with a subgroup of the Mobius group.
**Update:** Let me restate the question in simpler terms.The maps above satisfy the cocycle equation $$ D(g\circ f)(z)=(f'(z))^\lambda D(g)(f(z))+D(f(z)), $$ for $\lambda=0,1,2$, respectively. For a tensor density $\psi$, one can define the coboundary $$ D_\psi:f\mapsto (f')^\lambda(\psi\circ f)-\psi, $$ satisfying the same equation. So, the question is: what are the explicit conditions guaranteeing that the three maps above are the only, up to coboundaries, solutions to the cocycle equation?
|
https://mathoverflow.net/questions/231823/three-great-cocycles-in-complex-analysis-as-cohomology-generators
|
[
"reference-request",
"cv.complex-variables",
"group-cohomology"
] | 31 | 2016-02-22T05:57:23 |
[
"I'm confused by the notation, especially $D$ and $\\circle$.",
"See perhaps \"Explicit formulae for cocycles of holomorphic vector fields with values in $ \\lambda$ densities\" by Wagemann (arxiv.org/pdf/math-ph/0003035.pdf).",
"@WillSawin, well, in the case of the circle the space of tensor densities of degree $\\Lambda$ is just the vector space of smooth functions on the circle, endowed with the diffeomorphism group action $g\\mapsto (f')^\\lambda \\cdot g\\circ f$. In the complex variable case, it's a part of my question what this space should be...",
"What is a tensor density?"
] | 4 |
Science
| 0 |
84
|
mathoverflow
|
When is a compact topological 4-manifold a CW complex?
|
Freedman's $E_8$-manifold is nontriangulable, as proved on page (xvi) of the Akbulut-McCarthy 1990 Princeton Mathematical Notes "Casson's invariant for oriented homology 3-spheres". Kirby showed that a compact 4-manifold has a handlebody structure if and only if it is smoothable: [1](http://www.manifoldatlas.him.uni-bonn.de/index.php/Questions_about_surgery_theory) and [2](https://mathoverflow.net/questions/7921/failure-of-smoothing-theory-for-topological-4-manifolds/38739#38739). When is a compact topological 4-manifold a CW complex?
|
https://mathoverflow.net/questions/73428/when-is-a-compact-topological-4-manifold-a-cw-complex
|
[
"at.algebraic-topology",
"open-problems",
"4-manifolds",
"cw-complexes",
"topological-manifolds"
] | 31 | 2011-08-22T11:32:18 |
[
"I have good reason to believe that it is an open question! Apologies - I hadn't seen the earlier posting mathoverflow.net/questions/36838/… ",
"I take it this is an open problem? Small note, but your question was effectively asked in another form, here: mathoverflow.net/questions/36838/… I usually think of CW-complexes as being a tool for describing homotopy-types rather than homeomorphism types, so my answer was to a weaker question than the one asked. "
] | 2 |
Science
| 0 |
85
|
mathoverflow
|
Cohomology of symmetric groups and the integers mod 12
|
When $n \ge 4$, the third homology group $H_3(S_n,\mathbb{Z})$ of the symmetric group $S_n$ contains $\mathbb{Z}_{12}$ as a summand. Using the universal coefficient theorem we get $\mathbb{Z}_{12}$ as a summand of the cohomology group $H^3(S_n,\mathbb{Z}_{12})$.
**What's a nice 'formula' for a $\mathbb{Z}_{12}$-valued 3-cocycle on $S_n$ that generates this summand of the third cohomology?**
In more concrete terms, I'm looking for a nontrivial recipe to get an element $c(g,g',g'') \in \mathbb{Z}_{12}$ from three elements of $S_n$, such that
$$ c(g',g'',g''') - c(gg',g'',g''') + c(g,g'g'',g''') - c(g,g',g''g''') + c(g,g',g'') = 0 $$
for all $g,g',g'',g''' \in S_n$. Here "nontrivial" means that for all nonzero $\alpha \in \mathbb{Z}_{12}$, we do not have
$$ \alpha c(g,g',g'') = f(g',g'') - f(gg',g'') + f(g,g'g'') - f(g,g') $$
for some $f \colon S_n \times S_n \to \mathbb{Z}_{12}$.
|
https://mathoverflow.net/questions/314626/cohomology-of-symmetric-groups-and-the-integers-mod-12
|
[
"gr.group-theory",
"group-cohomology"
] | 30 | 2018-11-05T12:17:19 |
[
"@DevSinha - can you explain that class and how it works in more detail, or point me toward an explanation? I wanted a \"formula\", but an explicit geometrical description like this would also be nice.",
"@PhillippLampe - another typo! I'll fix it, thanks.",
"@JohnBaez Why do the second and the third summand in the cocycle condition have the same sign? Should the sign not alternate?",
"From a number theoretic perspective, the third homology group of the general linear group is related to algebraic $K$-theory, namely $K_3$ (this seems relevant because $S_n$ embeds into $\\mathrm{GL}_n$). In general the definition of $K$-group is highly non-constructive but $K_3$ is related to the Bloch group which is very explicit and some people have studied its torsion. The group $K_3(\\mathbb{Z})=K_3(\\mathbb{Q})$ is cyclic of order 48. By conjectures of Quillen-Lichtenbaum and Bloch-Kato the size of $(K_3)_\\mathrm{tors}$ is linked with $\\zeta'(-1)=-1/12$.",
"@JohnBaez, I can tell you about the geometry of this cocycle, using Fox-Neuwirth cochains in the unordered configuration space model for $BS_n$: this class is represented by having four points which share a coordinate, and three points which all share a coordinate, with two sharing another coordinate. There is probably a prettier subvariety which represents this, but I don't have any techniques/ideas to get at those.",
"@JohnBaez -- I guess you've seen the formulae for 3-cocycles on dihedral groups in the math-phys literature? It doesn't directly help, since you assume $n \\geq 4$... but it gives some idea of what might be possible.",
"It would be equivalent (and also interesting) to demonstrate a $\\Bbb Z/12$-valued invariant for bordism classes of oriented 3-manifolds equipped with $n$-fold covering space, $n \\geq 4$. It's not so clear to me that this is any easier, though.",
"This doesn’t really address your question, but I once saw Fernando Rodriguez-Villegas give a talk called “All 12s are the same.” The theme was exactly your motivation here - that of describing deep connections that underlie various occurrences of the number 12 throughout mathematics. I don’t recall if this occurrence was discussed, but he might be a good guy to hit up as a curator of 12s.",
"@YCor Yes thank you, I guess I constructed an element of $H_3(S_n)$ coming from the generator of $H_3(A_n)$ but maybe I am confused here too... At any rate I did not think at all about how to move to $H^3$",
"@მამუკაჯიბლაძე $H_3$ is not cyclic in this case, so I'm confused what you want to call a generator in $H^3(S_n,Z/12Z)$. You maybe mean an element of order 12.",
"On the afterthought - I think I had (at least) one thing imprecise: $S_5$ embeds in $O(3)$, and its pullback to $Pin(3)$ is actually a double cover of $S_5$, so you get a 3-cycle for that double cover, which must then be descended back to $S_5$ via transfer/induction/Gysin/whateveritscalled...",
"@JohnBaez Yes I think one might also obtain a generator of $H_3(A_n)$ this way. I said a little bit about it in an answer here",
"Our comments crossed paths. Okay, so your idea (in the case $S_5$) is based on embedding $S_5$ as a subgroup of $\\mathrm{Pin}(3)$. This is closely related to Epa and Gantner's method embedding of the double cover of $A_n$ in $\\mathrm{Spin}(n)$.",
"I didn't see that recipe in there. I like it but I don't quite understand it. It seems there should be a direct way to see how 3-simplices in the triangulation you describe give 3-chains in some chain complex that computes the homology of $S_5$. But I'm not seeing it.",
"Hmmm I don't actually see clearly a proof that for larger $n$ the cycle I describe will continue to be a generator - in fact that it will not even bound. In detail one should do this: realise $S_5$ as a subgroup in $Pin(3)$; for larger $n$, find an $m$ such that there is an embedding of $Pin(3)$ into $Pin(m)$ that (1) induces isomorphism on $H_3$ and (2) carries $S_5$ into a subgroup of a $S_n$ subgroup. While this will give what is needed, what I don't see is whether you can choose such a thing in such a way that $S_5$ will land in a subgroup as indicated in my previous comment...",
"And I think that answer linked by @QiaochuYuan contains also a recipe for an explicit 3-cycle representing a generator in $H_3$. Take the Cayley graph for some presentation of $S_5$ by generators and relations - e. g. the Coxeter one. It gives a polyhedral subdivision of the 3-sphere (for the Coxeter presentation it consists of all (hyper)faces of a permutahedron). Triangulate it, and take the sum of all obtained 3-simplices (with signs according to the orientation). For larger $n$, just choose any $S_5$ subgroup on a 5-element subset of $\\{1,...,n\\}$",
"Yes, it needs that summand. Fixed, thanks!",
"I believe the last (non-coboundary) condition needs one more summand, $-f(g,g')$",
"@MinseonShin - yes, I'll fix that. Thanks!",
"Should the domain of $f$ be two copies of $S_{n}$?",
"@QiaochuYuan - that's a great explanation of what's secretly motivating my question! However, I'm looking for something maximally concrete: a 'formula', as explicit as possible, that computes an element of $\\mathbb{Z}_{12}$ from 3 permutations, obeying the conditions I listed. If you have to use conformal nets to get that element, okay... but I'm hoping there's a formula that combinatorists or finite group theorists could more easily enjoy. Sort of like the sign of a permutation, only a lot subtler.",
"Here's a thing I wrote which is maybe relevant to your question: mathoverflow.net/a/196130/290 In the frame of that answer you are looking for something like a suitably nontrivial action of $S_n$ on an object in a suitably nice 3-category, and you can try to get such a thing by taking powers of an invertible object in a symmetric monoidal 3-category (for example, as suggested in that answer, conformal nets). The action of $S_n$ on such powers factors through the $3$-truncation of the sphere spectrum which naturally introduces a $\\mathbb{Z}_{24}$.",
"OK. Note that this is equivalent to independently asking for (a) an explicit 3-cocycle $c_4$ valued in $Z/4Z$, such that $2c_4$ is not a coboundary; (b) an explicit 3-cocycle $c_3$ valued in $Z/3Z$, which is not a coboundary. [The solutions $c_3$ and $2c_4$ are unique modulo coboundaries, but not $c_4$.]",
"@ChrisGerig - an explicit 4-term exact sequence would be a big step in the right direction and in some ways even better than the cocycle.",
"Since I'm trying to get to the bottom of why the numbers 12 and 24 show up in many contexts in mathematics, I greatly prefer a $\\mathbb{Z}_{12}$-valued cocycle and suspect there will be a good answer to my question, even if technically speaking it involves some noncanonical choices. However, beggars can't be choosers! So, a $\\mathbb{Z}_6$-valued cocycle would also be okay.",
"The homology is isomorphic to $C_{12}\\oplus C_2(\\oplus C_2)$ for $n\\ge 4$ (without the second $\\oplus C_2$ for $n=4,5$ groupprops.subwiki.org/wiki/…) but there is no reason that the $C_{12}$ summand to be canonical a priori. If you're looking for a nice formula it's likely that you don't want non-canonical choices and hence have to accept these few $C_2$ terms (on the other hand, modding out the 2-torsion should give something canonical, in $C_6$).",
"Would you accept the following rephrasal? The generator of your summand in $H^3(S_n,\\mathbb{Z}/12)$ corresponds to a 4-term exact sequence (a crossed module extension) $0\\to \\mathbb{Z}/12\\to N\\to E\\to S_n\\to 0$, from which you can write down an explicit 3-cocycle using a choice of cross-section $S_n\\to E$. I am not sure if this is due to MacLane (could check later)."
] | 27 |
Science
| 0 |
86
|
mathoverflow
|
Is there an Ehrhart polynomial for Gaussian integers
|
Let $N$ be a positive integer and let $P \subset \mathbb{C}$ be a polygon whose vertices are of the form $(a_1+b_1 i)/N$, $(a_2+b_2 i)/N$, ..., $(a_r+b_r i)/N$, with $a_j + b_j i$ being various Gaussian integers. For any Gaussian integer $u+vi$, let $h_P(u+vi)$ be the number of lattice point in $(u+vi) \cdot P$. I would like to say that $h_P(u+vi)$ is in some sense "quasi-polynomial modulo $N$". To emphasize: **The reason that this does not follow from the ordinary Ehrhart polynomial theorem is that I am multiplying by a Gaussian integer, not just an ordinary integer.**
The case $N=1$ can be studied using [Pick's theorem](http://en.wikipedia.org/wiki/Pick%27s_theorem). The area of $P$ transforms simply under multiplication by Gaussian integers, the number of lattice points on the boundary is only a little messier. In particular, it is easy to show that there is some non-zero Gaussian integer $D$ such that, as long as $GCD(D, u+vi)=1$ and $GCD(u,v)=1$, then $h_P(u+vi) = (u^2+v^2) \mathrm{Area}(P) + \mbox{constant}$.
A result that would satisfy me is that there is some non-zero non-zero Gaussian integer $D$ such that, as long as $GCD(D, u+vi)=1$ and $GCD(u,v)=1$, then $$h_p(u,v) = (u^2+v^2) \mathrm{Area}(P) + a(u,v) u + b(u,v) v + c(u,v)$$ where $a$, $b$ and $c$ are periodic modulo $N$ in each input.
Motivation: This is enough to give an elementary deduction that the [quartic residue](http://en.wikipedia.org/wiki/Quartic_reciprocity#Quartic_residue_character) symbol $\left[ \frac{z}{u+vi} \right]$ is periodic as a function of $u+vi$.
Motivation behind the motivation: There are a series of papers: [Habicht](http://www.ams.org/mathscinet-getitem?mr=113867), [Kubota](http://www.ams.org/mathscinet-getitem?mr=921585), [Hill](http://www.ams.org/mathscinet-getitem?mr=1324545) which promise to prove $m$-th power reciprocity laws roughly by counting lattice points. I can't follow any of them, so I thought I would true to reverse engineer what might be true in order to make the $\mathbb{Z}[i]$ proof come out.
For this reason, I would also be interested in statements for other number fields, but I don't have a precise question worked out in that setting.
|
https://mathoverflow.net/questions/165884/is-there-an-ehrhart-polynomial-for-gaussian-integers
|
[
"nt.number-theory",
"mg.metric-geometry",
"lattices"
] | 30 | 2014-05-11T18:46:47 |
[
"Dear David. Sorry for the late question, but, did you eventually found some place (in addition to the articles you mention) where this is explained/worked, hopefully for other number fields?",
"Yeah, right, I didn't think straight. Sorry for noise.",
"I do not see this. A rotation can change the lattice in a quite messy way. What am I missing?",
"naively, at least if $|a+ib|\\in\\mathbb{Z}$, this should follow from the classical case; indeed, by multiplying by $a+ib$ you apply to your $\\mathbb{C}$-plane the composition of a rotation and the scalar matrix $C=|a+ib|I$, and so the integer points would behave in the same way as by scaling with $C$ alone.",
"@DimaPasechnik Because I'm multiplying by Gaussian integers, not ordinary integers. Edited to clarify.",
"I was under impression that the quasiperiodicity of Ehrhart polynomial holds for any lattice $\\Lambda$ in $\\mathbb{R}^d$ and a polytope with vertices in $\\Lambda$. How does your setting differ?"
] | 6 |
Science
| 0 |
87
|
mathoverflow
|
Nontrivial tangent bundle that is diffeomorphic to the trivial bundle
|
Is there an example of a smooth $n$-manifold $M$ whose tangent bundle is nontrivial as a bundle but is nonetheless (abstractly) diffeomorphic to the trivial bundle $M \times \mathbb{R}^n$?
(This question was inspired by [Trivial fiber bundle](https://mathoverflow.net/questions/58685/trivial-fiber-bundle).)
|
https://mathoverflow.net/questions/58793/nontrivial-tangent-bundle-that-is-diffeomorphic-to-the-trivial-bundle
|
[
"dg.differential-geometry",
"differential-topology"
] | 30 | 2011-03-17T16:25:39 |
[
"$T^*G\\cong G\\times \\mathfrak g^*\\cong G\\times \\mathbb R^{n}$",
"@Dylan and @Somnath: Tangent bundles over spheres never give such examples. Moreover, in general, if the homotopy-equivalence $M\\to M$ induced by a diffeomorphism $TM\\to M\\times {\\mathbb R}^n$ is homotopic to a diffeomorphism $M\\to M$, then $TM$ is trivial. See \"Diffeomorphism of total spaces and equivalence of bundles\" by De Sapio and Waldschap. This still leaves open the case when $M$ is a homotopy-sphere though. ",
"@Dylan - That was the example I was thinking of too! It is clear that $TS^5$ is not the trivial rank $5$ bundle over $S^5$. It only has $3$ linearly independent nowhere vanishing vector fields (for lack of anything elementary one can use Adams' vector fields on spheres result). Although I don't know how to do it myself, if you can prove (extend your sphere bundle result to vector bundles) your claim, we're done! ",
"I think I can show that the unit tangent bundle over $S^5$ is diffeomorphic to the trivial one... but I'm having trouble otherwise, since most of the diffeomorphisms I can think of rely on theorems about compact spaces. ",
"@David, is no reason for the diffeomorphism $TM\\to M\\times \\mathbb R^n$ to cover a diffemorphism of $M$. @Faisal, the answer to your question is almost certainly yes, but I do not see an explicit example at the moment, cf. my answer to mathoverflow.net/questions/58685/trivial-fiber-bundle.",
"A trivial observation: such an isomorphism would cover an automorphism of $M$. A possibly unhelpful observation: One could try to put a Riemannian metric on $M$ and see what constraints this gives in the light of my first remark."
] | 6 |
Science
| 0 |
88
|
mathoverflow
|
Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$?
|
> Is there a field $F$ such that $F \cong F(X,Y)$ as fields, but $F \not \cong F(X)$ as fields?
I know only an example of a field $F$ such that $F$ isomorphic to $F(x,y)$ : this is something like $F=k(x_0,x_1,\dots)$. But in this case we have $F \cong F(x)$.
This is related to [this](https://mathoverflow.net/questions/225896) question. This was previously [asked](https://math.stackexchange.com/questions/2000333) on MSE but after 2 months and a bounty, no answer was provided.
Thank you very much!
|
https://mathoverflow.net/questions/258647/is-there-a-field-f-which-is-isomorphic-to-fx-y-but-not-to-fx
|
[
"ra.rings-and-algebras",
"examples",
"fields"
] | 30 | 2017-01-03T03:25:32 |
[
"Oops, I forgot to mention that I've posted a related question here: mathoverflow.net/questions/259117/…",
"I have an example where I know $F\\cong F(X,Y)$ but I don't know whether $F\\cong F(X)$. There are torsion-free abelian groups $G$ such that $G\\cong G\\oplus\\mathbb{Z}\\oplus\\mathbb{Z}$ but $G\\not\\cong G\\oplus\\mathbb{Z}$ (see mathoverflow.net/questions/218113/… for example). Let $\\mathbb{Q}(G)$ be the field of fractions of the group algebra $\\mathbb{Q}[G]$. Then $\\mathbb{Q}(G)\\cong\\mathbb{Q}(G)(X,Y)$, but I don't see why it should be isomorphic to $\\mathbb{Q}(G)(X)$.",
"Related on MSE: math.stackexchange.com/questions/388291"
] | 3 |
Science
| 0 |
89
|
mathoverflow
|
The field of fractions of the rational group algebra of a torsion free abelian group
|
Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions.
> Are there non-isomorphic torsion free abelian groups $G$ and $H$ such that $\mathbb{Q}(G)\cong\mathbb{Q}(H)$? If so, are there easy examples?
This is motivated by [this question](https://mathoverflow.net/questions/258647/is-there-a-field-f-which-is-isomorphic-to-fx-y-but-not-to-fx). There are [examples](https://mathoverflow.net/questions/218113/a-is-isomorphic-to-a-oplus-mathbbz2-but-not-to-a-oplus-mathbbz) of torsion free abelian groups $G$ where $G\cong G\oplus\mathbb{Z}\oplus\mathbb{Z}$, but $G\not\cong G\oplus\mathbb{Z}$. Since $\mathbb{Q}(G\oplus\mathbb{Z}\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X,Y)$ and $\mathbb{Q}(G\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X)$, a negative answer to my question would give a positive answer to the motivating question.
I've found a little about this construction in the literature, but not much. For example, in
_Gervasio G. Bastos and T. M. Viswanathan_ , MR 942063 [**Torsion-free abelian groups, valuations and twisted group rings**](http://dx.doi.org/10.4153/CMB-1988-021-x), _Canad. Math. Bull._ **31** (1988), no. 2, 139--146.
it is proved that $\mathbb{Q}(G)$ is a purely transcendental extension of $\mathbb{Q}$ if and only if $G$ is free abelian.
Any other relevant references would be welcome. Also, there are obvious variants: e.g., using a different field, such as $\mathbb{F}_2$ instead of $\mathbb{Q}$.
|
https://mathoverflow.net/questions/259117/the-field-of-fractions-of-the-rational-group-algebra-of-a-torsion-free-abelian-g
|
[
"gr.group-theory",
"ac.commutative-algebra",
"fields",
"abelian-groups"
] | 29 | 2017-01-08T13:30:07 |
[
"This question could be related.",
"Lovely question Jeremy. I only wish I could say something useful about it!",
"Just a remark: if $G$ has no primitive element (in the sense that for every $g\\in G$ there exists $n\\ge 2$ such that $n^{-1}g\\in G$) then I guess that $\\pm G$ is characterized within $\\mathbb{Q}(G)^*$ as those elements with roots of arbitrarily high order. (This holds if $G$ is a $\\mathbb{Z}[1/p]$-module, or if $G$ is noncyclic of rank 1.) Since $G$ is isomorphic to $(\\pm G,\\cdot)$ modulo its torsion subgroup, it follows that $\\mathbb{Q}(G)$ recognizes $G$ in this case."
] | 3 |
Science
| 0 |
90
|
mathoverflow
|
Linking formulas by Euler, Pólya, Nekrasov-Okounkov
|
Consider the formal product $$F(t,x,z):=\prod_{j=0}^{\infty}(1-tx^j)^{z-1}.$$ (a) If $z=2$ then on the one hand we get Euler's $$F(t,x,2)=\sum_{n\geq0}\frac{(-1)^nx^{\binom{n}2}}{(x;x)_n}t^n,$$ on the other we get Pólya's formula (the "cycle index decomposition", see Stanley's EC2, p.19) $$F(t,x,2)=\sum_{n\geq0}Z(S_n,(1-x)^{-1},\dots,(1-x^n)^{-1})t^n.$$ (b) If $t=x$ then we get Nekrasov-Okounkov's $$F(x,x,z)=\sum_{n\geq0}x^n\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\left(1-\frac{z}{h_{\square}^2}\right);$$ where $h_{\square}$ is the hook-length (for notations and references you may see _[Theorems, problems and conjectures](http://dauns.math.tulane.edu/%7Etamdeberhan/conjectures.html)_).
> **My Question still waiting for an answer.** Is there a unifying combinatorial right-hand side in $$\prod_{j\geq0}(1-tx^j)^{z-1}=?$$
**Remark 1.** One possibility is perhaps a refinement in line with (b) involving hook-lengths.
**Remark 2.** Even special cases, such as evaluation for specific numerical values of $t$ and/or $z$, would be interesting.
**Remark 3.** Yet another direction is to find a hook-length expansion for $F(t,x,2)$. This could be fascinating because the result will connect the cycle index polynomial with hooks.
|
https://mathoverflow.net/questions/251068/linking-formulas-by-euler-p%c3%b3lya-nekrasov-okounkov
|
[
"nt.number-theory",
"co.combinatorics",
"rt.representation-theory",
"power-series",
"gauge-theory"
] | 29 | 2016-09-29T21:18:19 |
[
"Neither one follows from the other.",
"never mind -- NO formula generalizes the cycle index",
"does Nekrasov-Okounkov formula really fall out of cycle index for permutation groups??"
] | 3 |
Science
| 0 |
91
|
mathoverflow
|
Non-linear expanders?
|
Recall that a family of graphs (indexed by an infinite set, such as the primes, say) is called an _expander family_ if there is a $\delta>0$ such that, on every graph in the family, the discrete Laplacian (or the adjacency matrix) has spectral gap $|\lambda_0-\lambda_1|\geq \delta$. (Assume all graphs in the family have the same degree (= valency) $d$.)
If, for every $p$, we specify some maps from $\mathbb{Z}/p\mathbb{Z}$ to itself, then we are defining a graph with vertex set $\mathbb{Z}/p\mathbb{Z}$ for each $p$: a vertex is adjacent to the vertices it is mapped to.
If we consider the maps $x\to x+1$, $x\to 3x$ (say), then we do not get expanders. On the other hand, if we take $x\to x+1$, $x\to x^{-1}$, then, as is widely known, we do get an expander family.
Consider the maps
$x\mapsto x+1$, $x\mapsto x^3$.
Do they (and their inverses, if you wish) give an expander family? (Let $p$ range only over primes $p\equiv 2 \mod 3$, so as to keep the map $x\mapsto x^3$ injective.)
What about the maps $x\mapsto x+1$, $x\mapsto 3 x$, $x\mapsto x^3$? Do they, taken together, give an expander family?
(Is there a way to relate these maps to the action of a linear group? Are there examples of sets of not neessarily linear maps giving rise to expander families?)
UPDATE: What if these were shown _not_ to be expanders? What would be some interesting consequences?
|
https://mathoverflow.net/questions/168426/non-linear-expanders
|
[
"nt.number-theory",
"gr.group-theory",
"graph-theory"
] | 29 | 2014-05-28T07:17:01 |
[
"Can you say anything on the structure of those $A\\subset{\\mathbb F}_p$ with $|(3A)\\setminus(A+1)|<c|A|$ (or smaller than $|A|^\\delta$ at least)? If so, it might be possible to use this information to show that such $A$ cannot be stable under the map $x\\mapsto x^3$.",
"Yes, as I said it should do something like expand $|A|$ to $|A| + |A|^\\delta$ for some small $\\delta$.",
"That would give only a very, very weak expansion result, wouldn't it?",
"I was thinking something more along the lines of Proposition 4.3 of my paper with Ben: arxiv.org/abs/0709.3562 , which should show enough equidistribution in the function $x \\mapsto x^3$ on medium-sized intervals in ${\\bf Z}/p{\\bf Z}$ (maybe cutting out the interval of size $p^{1/3}$ or so around the origin, and maybe removing other \"major arcs\" also) to show that the image of that interval has small intersection with other medium-sized intervals.",
"I can see that $|A\\cup (A+1)|\\leq (1+\\delta) |A|$ implies that $\\sum_{x\\in A} e(\\xi x)$ is small for $\\epsilon p\\leq \\xi \\leq (1-\\epsilon) p$ (say), and that one can get from that (by Weyl differencing) that $\\sum_{x\\in A} e(\\xi x^2)$ must be small for all $\\xi$ in a range. But how would one show from that that $|A\\cup A^3|$ is large?",
"Actually, thinking about it more, Bourgain-Konyagin (which has to do with how arithmetic progressions and geometric progressions interact) is more relevant for the $x \\mapsto x+1, x \\mapsto 3x$ case. For $x \\mapsto x+1, x \\mapsto x^3$, it is indeed Weyl differencing which is the key: to show that if A is basically the union of a small number of intervals, then $A^{\\cdot 3}$ does not behave like the union of a small number of intervals.",
"It looks like Bourgain-Konyagin type exponential sum estimates may be able to prove a very weak expansion result of the form $|A \\cup (A+1) \\cup A^{\\cdot 3} | \\geq |A| + |A|^\\delta$ when $p^\\varepsilon \\leq |A| \\leq p^{1-\\varepsilon}$, where $A^{\\cdot 3} = \\{ a^3: a \\in A \\}$, which would almost give a non-trivial diameter bound, but the improvement is slight and one also would have to treat the case of very small or very large A somehow. [Maybe a Weyl differencing argument would work?]",
"Agreed - yes, these are my reasons for asking! But what about diameter bounds? I actually want expansion, but that might already be somewhat interesting.",
"... proving expansion rigorously, though, would probably require substantial new ideas since there does not appear to be any natural group involved other than the free group. But one could imagine that sum-product theorems could at least give some non-trivial bounds on diameter and mixing time, even if they fall short of true expansion.",
"As for the last question, as you well know, it was proven long ago by Pinsker that any pair of random permutations will form an expander on $\\{1,\\dots,n\\}$ with probability $1-o(1)$. Absent any obvious nilpotent or solvable structure (as with the $x \\mapsto x+1, x \\mapsto 3x$ case) I would conjecture that $x \\mapsto x+1, x \\mapsto x^3$ behaves like two random maps for the purposes of expansion, since there do not appear to be any local relations between these maps (in contrast to the former case in which we have $3(x+1) = 3x + 1 + 1 + 1$)."
] | 10 |
Science
| 0 |
92
|
mathoverflow
|
Can every 3-dimensional convex body be trapped in a tetrahedral cage?
|
Can every 3-dimensional convex body be trapped in a tetrahedral cage?
Although the question is fairly unambiguous, I give all relevant definitions:
$\bullet$ A subset $C$ of $\mathbb{R}^n$ is an $n$-dimensional _**convex body**_ if $C$ is convex, compact, and has non-empty interior.
$\bullet$ A _**polyhedral cage**_ $P^{(1)}$ in $\mathbb{R}^3$ is the union of all edges (_i.e._ , the 1-skeleton) of a convex $3$-dimensional polyhedron $P$. In particular, a _**tetrahedral cage**_ is the union of the six edges of some tetrahedron.
$\bullet$ A convex $3$-dimensional body $C$ is _**trapped**_ by the tetrahedral cage $T^{(1)}$, that is, by the $1$-skeleton of the tetrahedron $T$, if the cage is fixed (motionless), and if for every continuous rigid motion (rotations allowed) $f_t(C);\ 0\le t\le 1$, either $f_t(C)$ intersects $T$ for every $t\in [0,1]$ or $T^{(1)}$ contains an interior point of $f_t(C)$ for some $t_0\in [0,1]$.
In other words, $C$ cannot be moved arbitrarily far from $T$ while, during the entire motion, avoiding the cage's bars penetrating $C$'s interior.
$\bullet$ Remark. In dimension $n>3$, one can consider analogous questions, with a variety of types of a simplicial cage, by taking the $i$-skeleton of a convex $n$-dimensional simplex, with $1\le i\le n-2$.
A trivial example: A ball is trapped in the cage consisting of the $1$-skeleton of the regular tetrahedron edge-tangent to the ball. Also, obviously, every convex body can be trapped in some polyhedral cage.
For an interested reader, I suggest a few of somewhat less trivial exercises: each of the following convex bodies can be trapped in a tetrahedral cage: the cube, the circular cylinder of any (finite) height, the circular cone, the regular tetrahedron.
|
https://mathoverflow.net/questions/379560/can-every-3-dimensional-convex-body-be-trapped-in-a-tetrahedral-cage
|
[
"mg.metric-geometry",
"convex-geometry"
] | 28 | 2020-12-22T20:17:05 |
[
"@Anton The half-circle, for example, and many other, related.",
"In 2-dimensional case, one can ask which convex figures can be trapped by three nails. Rectangle cannot be trapped, but I do not see other examples.",
"@DanRomik Romik : Yes, it is quite a bit different.",
"The paper How to cage an egg by Oded Schramm (Inventiones Math. 107 (1992), 543-560) comes to mind as potentially relevant, although the question he answers is a bit different than yours.",
"@JosephO'Rourke: Yes. More precisely, some of the traps for the cone also trap the spindle, no matter how thin it is.",
"Does your trap for a cone also trap a thin spindle? Say, two tall, base-to-base cones?"
] | 6 |
Science
| 0 |
93
|
mathoverflow
|
Mathieu group $M_{23}$ as an algebraic group via additive polynomials
|
An elegant description of the Mathieu group $M_{23}$ is the following: Let $C$ be the multiplicative subgroup of order $23$ in the field $F=\mathbb F_{2^{11}}$ with $2^{11}$ elements. Then $M_{23}$ is the group of additive maps of $F$ to itself which permute the set $C$. The restriction to $C$ then is the natural action of $M_{23}$ of degree $23$.
Any additive map of $F$ to itself has the form $x\mapsto\sum_{i=0}^{10}a_ix^{2^i}$ for $a_i\in F$. The condition that such maps preserve $C$ can be expressed easily as an equivalent set of polynomial conditions on the coefficients $a_0,\dots,a_{10}$. For instance, for a variable $t$, compare coefficients of $$\prod_{c\in C}(t-\sum_{i=0}^{10}a_ic^{2^i})=t^{23}-1.$$ In addition, we need the equations $a_i^{2^{11}}=a_i$ to ensure that the $a_i$ are in $F$.
However, the polynomials we obtain this way (or some other way, there are somewhat smarter possibilities) are extremely complicated.
Of course, any finite subgroup of a linear group is an algebraic set, that is a solution of a system of polynomial equations. And in general such a system will be messy.
Here, I somehow expect that there should be a simple system of polynomials discribing $M_{23}$ as an algebraic set. The obvious attempt would be to first take a system as above, and then hope that a Groebner basis looks better. After some naive attempts it seems that the systems are too complex for Magma or Singular to compute the Groebner bases.
Before trying to refine the approach, here is my question: Has anyone seen this version to describe $M_{23}$?
|
https://mathoverflow.net/questions/264513/mathieu-group-m-23-as-an-algebraic-group-via-additive-polynomials
|
[
"ag.algebraic-geometry",
"gr.group-theory",
"polynomials",
"finite-groups"
] | 28 | 2017-03-13T06:48:14 |
[
"@DaveBenson Indeed, I had also tried something along this idea. The upper bound of the Galois group of $Y^{23}+XY^3+1$ is achieved by showing that this polynomial divides an additive polynomial of degree $2^{11}$. So actually this additive polynomial has Galois group $M_{23}$, and its groups acts on an $11$-dimensional $\\mathbb F_2$-space.",
"This is a very old question, but I can't help adding the following comment, that may or may not be relevant. I wonder whether Abhyankar's paper, \"Mathieu group coverings in characteristic two\" is relevant here? He demonstrates that there are very innocuous looking degree 23 trinomials over ${\\mathbb F}_2(X)$ with Galois group $M_{23}$. His example is $Y^{23}+XY^3+1$.",
"The additive subgroup generated by $C$ is a subfield, is not reduced to $F_2$, and hence equals $F_{11}$. Hence the fixator of $C$ in $GL_{11}$ is reduced to the trivial subgroup. Hence the stabilizer $G$ of $C$ in $GL_{11}$ is finite (as an algebraic group). Since $|C|=23$ and the dimension is 11, there exists a 3-element subset $F$ of $C$ such that $C-F$ generates $F_{11}$ additively. Hence a 3-cycle on $F$ cannot belong to $G(K)$ for any extension $K$ of $F_2$. So the (faithful) action of $G(K)$ on $C$ is reduced to $M_{23}$, and hence $G(K)$ is reduced to $G(F_2)=M_{23}$ for every $K$.",
"If I understand correctly, you don't know the answer (whether $G$ has positive dimension, where $G\\subset\\mathrm{GL}_{11}$ is the stabilizer of $C$).",
"@YCor: I tried to answer that in my first comment above.",
"Why are you not happy with just defining the $G$ as the subgroup $\\mathrm{GL}_{11}$ that preserves $C$? then $G(\\mathbf{F}_2)$ is reduced to the Mathieu group. No need to incorporate the conditions that entries are in $\\mathbf{F}_2$: this is part of the definition of $G(\\mathbf{F}_2)$. If $G$ is defined as I just did, I don't even know if it's positive-dimensional.",
"one natural way to produce an explicit set of equations is to identify a polynomial $p$ left invariant by $M_{23}$, but not by any overgroup of it in GL(11,2), and write down equations for the matrix entries saying that $p$ is invariant.",
"@DimaPasechnik Yes, I know that. But I believe that has nothing to do with my question.",
"It is known how to express generators of $M_{23}$ as a permutation group in terms of rational univariate maps over $\\mathbb{F}_{23}$. (Not sure about the original reference, I saw it in a book by D. Gorenstein ams.org/mathscinet-getitem?mr=698782). More precisely, it gives such representation for $M_{24}$, as the maps in PSL(2,23) and one extra binomial of relatively high degree.",
"@DimaPasechnik Right, one could try some smaller toy examples. The group $PSL(3,2)$ maybe is not that interesting though, because it equals $GL(3,2)$. So all that we polynomially encode would be bijectivity of the linear maps.",
"@JimHumphreys I hope to use it for certain combinatorial questions about $M_{23}$, for instance in the study of sharply transitive sets of permutations. Such question have a polynomial formulation, so if the $M_{23}$ has a nice polynomial formulation too, I have some hope.",
"E.g. one may ask for such a description for PSL(3,2): I suppose you have to replace 11 by 3 and 23 by 7, i.e. $|C|=7$ and $F=\\mathbb{F}_{2^3}$, and consider PSL(3,2) as the group of additive maps of $F$ preserving $C$. (Disclaimer---I didn't check all the details here :-)).",
"the natural object preserved by the group here is dual of the perfect binary Golay code en.wikipedia.org/wiki/Binary_Golay_code One might ask for similar descriptions of smaller codes to begin with.",
"This looks like a reasonable (though maybe quite difficult) question. But I wonder what consequences such a description would have?"
] | 14 |
Science
| 0 |
94
|
mathoverflow
|
Derivative of Class number of real quadratic fields
|
Let $\Delta$ be a fundamental quadratic discriminant, set $N = |\Delta|$, and define the Fekete polynomials $$ F_N(X) = \sum_{a=1}^N \Big(\frac{\Delta}a\Big) X^a. $$ Define $$ f_N(X) = \frac{F_N(X)}{X^N-1}. $$ Canceling common factors it can be shown that $$ f_N(X) = \frac{H_N(X)}{\Phi_N(X)}, $$ where $\Phi_N$ is the $N$-th cyclotomic polynomial and $H$ has integral coefficients. (These definitions seem rather ad hoc, but in fact the objects above are all quite "natural".)
If $\Delta < 0$, then $f_N(1) = \frac{2h}w$, where $h$ is the class number of $\Delta$ and $w$ the number of roots of unity in the corresponding quadratic number fields. In fact we get $$ f_N(1) = - \frac{\sum_{a=1}^{N-1} a \Big( \frac{\Delta}a\Big)}N . $$
If $\Delta > 0$, then $f_N(1) = 0$, which suggests looking at the first derivative. Here we find that $f_N'(1)$ is always even, so we set $f_N'(1) = - 2 h$ and get $$ h = \frac{\sum_{a=1}^{N-1} \frac{a(a-1)}2 \Big( \frac{\Delta}a\Big)}N . $$ Here is a small table:
$ $ $ \begin{array} {c|ccccccccccc} N & 12 & 13 & 17 & 21 & 24 & 28 & 29 & 33 & 37 & 41 & 44 \\\ \hline h & 1 & 1 & 2 & 2 & 3 & 4 & 3 & 6 & 5 & 8 & 7 \end{array} $
It is quite remarkable that $h$ seems to possess, for positive discriminants, properties analogous to class numbers. In particular, $h$ is odd if and only if $\Delta = p > 5$ is a prime $\equiv 5 \bmod 8$, and if $\Delta$ is divisible by $n$ distinct primes, then $h$ is divisible by $2^{n-2}$ etc.
In the case $\Delta < 0$, Hasse and Bergstrom have shown how to extract the "genus factor" from the analytic class number formula, but the necessary calculations were horrible. Thus if we want to avoid going through similar technicalities in the case $\Delta > 0$ then we would need a natural interpretation of $h$ as the order of an analogue of a "class group".
So I guess my question is whether there is such a mysterious group (connected to real quadratic number fields) with order $h$. Even a proof that $h \ne 0$ would be welcome, since in the case of negative discriminants this is equivalent to the nonvanishing of Dirichlet's L-function at $s = 1$.
**Edit.** As Rene Schoof has pointed out, the numbers in question are essentially generalized Bernoulli numbers $B_{2,\chi}$ for (quadratic) Dirichlet characters with conductor $N$. These numbers are defined by $$ B_{2,\chi} = \sum_{a=1}^N \chi(a) B_2(a/N), $$ where $B_2(X) = X^2 - X + \frac16$ is the second Bernoulli polynomial. The constant term $\frac 16$, in which this expression differs from the one I have used above, is irrelevant since $\sum \chi(a) = 0$.
These numbers $B_{2,\chi}$ show up as factors of eigenspaces of cuspidal divisor class groups as studied in the book by Kubert-Lang (more exactly, the $p$-part of the group is reflected in the divisibility by $p$ of the Bernoulli numbers). The Birch-Tate-Lichtenbaum conjecture also predicts that the $B_{2,\chi}$ are factors of $K_2({\mathcal O})$, where ${\mathcal O}$ is the ring of integers of the maximal real subfield of the $N$-th roots of unity (Kubert-Lang, p. 151).
Although these numbers have such a prominent place in number theory, their elementary properties apparently have not been investigated (I've looked into Washington's and Lang's books on cyclotomic number fields). Numerical experiments suggest e.g. the following:
Let $\chi$ be an even Dirichlet character with conductor $N$. Then $h_2(N) = \frac{N}4 B_{2,\chi}$ is an integer for all $N > 8$.
Moreover, $h_2(N)$ is odd if and only if
1. $N = p$ for primes $p \not \equiv 1 \bmod 8$;
2. $N = 4p$ for primes $p \equiv 3 \bmod 8$;
3. $N = 8p$ for odd primes $p \not \equiv \pm 1 \bmod 8$.
In absence of any kind of genus theory for cuspidal divisor class groups it seems that the only option for proving such results is working directly with the definition of these numbers. As mentioned above, this becomes technical for integers with many prime factors.
There are similar results for $h_3(N) = \frac{N^2}6 B_{3,\chi}$, and there can be no doubt that most of this generalizes in some form to all generalized Bernoulli numbers $B_{n,\chi}$.
This whole set of question came out of my approach to quadratic reciprocity in Chapter 7 of my notes on [Pell conics](http://www.rzuser.uni-heidelberg.de/~hb3/pell.html).
|
https://mathoverflow.net/questions/114434/derivative-of-class-number-of-real-quadratic-fields
|
[
"nt.number-theory",
"algebraic-number-theory"
] | 28 | 2012-11-25T10:14:45 |
[
"Hi! I think these are generalized Bernoulli numbers $B_{2,\\chi}$ associated the quadratic character $\\chi$ of conductor $\\Delta$. Apart from a power of $2$ perhaps, they are equal to the order of the $\\chi$-part of the cuspidal divisor class group of the modular curve $X_1(\\Delta)$. See the book by Kubert-Lang for more details."
] | 1 |
Science
| 0 |
95
|
mathoverflow
|
derived category of equivariant coherent sheaves and fixed points
|
The K-group $K^T(X)$ of $T$(torus)-equivariant coherent sheaves on a variety $X$ is isomorphic to $K^T(X^T)$, that of the fixed point locus via the inclusion homomorphism, when we tensor the quotient field of the representation ring $R(T)$ of $T$.
Is there a similar result for $D_T(Coh X)$ and $D_T(Coh X^T)$, derived categories of $T$-equivariant coherent sheaves ? I do not know even how to formulate `the quotient field of $R(T)$' in the derived category case.
|
https://mathoverflow.net/questions/48407/derived-category-of-equivariant-coherent-sheaves-and-fixed-points
|
[
"kt.k-theory-and-homology",
"ag.algebraic-geometry"
] | 28 | 2010-12-05T16:09:13 |
[
"@Ben, you're absolutely right, somehow I missed the salient word \"coherent\" in the question. But even for constructible sheaves, is there a localization statement at the level of derived categories?",
"It doesn't seem like that to me at all. Is there anything about coherent sheaves in GKM?",
"This seems like exactly the sort of thing that ought to be one of the main results in the paper of Goresky-Kottwitz-MacPherson. Unfortunately, I haven't been able to extract the kind of statement you're asking for; it would be very interesting if someone can give a precise answer. They do give a localization-type theorem for an element of the equivariant derived category in (7.8), but only after taking cohomology."
] | 3 |
Science
| 0 |
96
|
mathoverflow
|
Computational complexity of topological K-theory
|
I am a novice with K-theory trying to understand what is and what is not possible.
Given a finite simplicial complex $X$, there of course elementary ways to quickly compute the cohomology of $X$ with field coefficients. However, it is known that computing the rational homotopy groups of $X$ is at least $NP$-hard (in fact, at least $\\# P$-hard), simply because it is possible to reduce NP-hard problems to the computations of some rational homotopy groups.
For any of: (1) Real K-theory, (2) Complex K-theory, (3) p-adically completed K-theory, is there an algorithm to compute $K^0$ of a finite simplicial complex? Is this algorithm polynomial-time?
If there is no known algorithm, is there at least evidence that any theoretical algorithm, should it exist, must be at least $NP$-hard? In other words, is there any way to reduce an $NP$-hard problem to the calculation of some $K$ group?
|
https://mathoverflow.net/questions/125745/computational-complexity-of-topological-k-theory
|
[
"kt.k-theory-and-homology",
"at.algebraic-topology",
"computational-complexity"
] | 28 | 2013-03-27T10:15:43 |
[
"Just thought I'd add a citation for the fact that rational homotopy groups of simply-connected spaces are NP hard here. In particular, this is not just some $\\pi_1$ issue.",
"Link: jointmathematicsmeetings.org/meetings/national/jmm2012/…",
"Have you looked at Gunnar Carlsson's stuff? They have this software for computing homology; maybe it could be made to work with K-theory, or maybe they've addressed the computational complexity and that could help you bound the complexity of K-theory. Here is a link: comptop.stanford.edu. In a similar vein, Rob Ghrist has done work on algorithms for cohomology theory, but I'm not sure if he's addressed computational complexity or not. Last thought: at the JMM last year there was a special session about K-theory and electrical engineering. Maybe those guys have looked into this question",
"Dominique Arlettaz has a result regarding the order of differentials in the Atiyah-Hirzebruch spectral sequence, see \"The order of the differentials in the Atiyah-Hirzebruch spectral sequence\". I would imagine that one might be able to make certain conclusions if you assumed enough about the torsion of the homology of the space. However, on rereading your question I realize this is not so much what you were interested in. Thought I'd mention the paper of Arlettaz just in case though.",
"I have thought about this occasionally. There is some reason to think that there might be a tractable combinatorial construction of a minimal Kan complex of homotopy type $\\Omega^\\infty(KU/2)$. From that it should be possible to determine the computational complexity of $(KU/2)^*(X)$. Odd primes might be possible as well. I have various kinds of calculations related to this, but no real conclusion.",
"Probably not. The first differential is easy (for KU), it's determined by a cohomology operation, but I think the higher ones are algorithmically intractable and would describe higher-order cohomology operations. Maybe one should try to describe vector bundles on finite simplicial complexes combinatorially instead? ",
"Yes, but only as a black-box. I do not know an algorithm to compute differentials or solve extension problems. Is this possible?",
"Are you familiar with the Atiyah-Hirzebruch SS?"
] | 8 |
Science
| 0 |
97
|
mathoverflow
|
Finite-dimensional subalgebras of $C^\star$-algebras
|
Let $A$ be a unital $C^\star$-algebra and let $a_1,\dots,a_n$ be a finite list of normal elements in $A$ which (together with their adjoints) generate a norm-dense $\star$-subalgebra $B \subset A$. Clearly, if $A$ is finite-dimensional, then every element in $A$ (and hence $B$) has finite spectrum. I am asking for the converse.
> **Question:** Assume that each element of $B$ has finite spectrum. Is it true that $A$ has to be finite-dimensional?
The existence of finitely-generated infinite torsion groups shows that this might be a highly non-trivial problem. In this case one would consider the reduced group $C^\star$-algebra and note that all monomials in the generators of the group and their inverses (which are equal to the adjoints of the generators) would have finite spectrum. However, the generated algebra would still be infinite-dimensional. I do not know of any simpler way to come up with such an example. In this case it is conceivable that the random-walk operator associated with the generating set (which is an element in the real group ring) has infinite spectrum, even though I did not prove this.
Maybe there is also need to consider the spectra of elements in matrices over $B$ (which of course follow to be finite if $A$ is finite-dimensional.) In view of this, I am not only asking for an answer to the question but also for the right question (or a better one) if the answer to the original question is negative.
A stronger assumption would be to assume that $A$ itself consists only of elements with finite spectrum. This case seems much easier to approach and the answer seems to be positive. In fact every infinite-dimensional $C^\star$-algebra should contain an element with infinite spectrum.
Just to get started, a more concrete instance of the question above is:
> **Question:** Let $p_1, p_2$ and $p_3$ be three projections in a $C^*$-algebra with the property that every (non-commutative) polynomial in $p_1, p_2$ and $p_3$ has finite spectrum. Is it true that the projections generate a finite-dimensional algebra?
|
https://mathoverflow.net/questions/35207/finite-dimensional-subalgebras-of-c-star-algebras
|
[
"oa.operator-algebras",
"c-star-algebras",
"von-neumann-algebras",
"open-problems"
] | 28 | 2010-08-11T02:49:03 |
[
"Every finitely generated *-algebra is generated by a finite list of normal elements (the real and imaginary parts of its generators) so the assumption that the $a_i$ are normal is superfluous.",
"@AliTaghavi: Yes, in this case, the spectrum of $A$ is equal to the joint spectrum of $a_1,\\dots,a_n$ (which is contained in the product of the spectra of $a_1,\\dots,a_n$).",
"Is the answer to the original qustion obvious for commutative C* algebras?",
"Let $G$ be a finitely-generated Group, $H$ a Hilbert space, $h : G \\mapsto \\mathbb{B}(H)$ a unitary representation of $G$ such that $h(G)$ is linear independent and $B = span(h(G))$. Then G is finite. For a simple proof see Proposition 2 in ON THE SPECTRUM OF THE SUM OF GENERATORS OF A FINITELY GENERATED GROUP, II, matwbn.icm.edu.pl/ksiazki/cm/cm65/cm65111.pdf",
"To \"In fact every infinite-dimensional C⋆-algebra should contain an element with infinite spectrum.\" : For a proof see ALAN W. TULLO, CONDITIONS ON BANACH ALGEBRAS WHICH IMPLY FINITE DIMENSIONALITY, journals.cambridge.org/action/… ",
"Could be; I do not know. I was just talking with Joachim Cuntz and George Elliott about the problem, but they did not mention Richard Kadison. There is a related problem to find an element with infinite spectrum in an infinite dimensional $C^\\ast$-algebra, which is a lot easier. Where did you hear that Kadison asked the question?",
"@Andreas, Why don't you go ahead and ask this as a separate question. I was thinking about this as an approach to your question. One next step (which would be much harder) would to say if the spectra of these elements isn't the solution of a polynomial equation, then one of these elements has an infinite spectrum. I have no idea how to prove this, or even if it's true, but it seems to work for a few examples. ",
"I do not know. Beautiful question! Do you want to (or may I) ask this as a separate question?",
"Suppose you have a C$^*$ algebra generated by $g_1$, $g_2$ and $g_3$. Consider the elements $x_1g_1+x_2g_2+x_3g_3$, where the $x_i$ are variables. If the spectra of these elements are the solutions of a polynomial equation in $x_1$, $x_2$ and $x_3$, is the algebra finite dimensional? ",
"I have just learned that Choi-Elliott proved in (ams.org/mathscinet-getitem?mr=1081290) that the set of self-adjoint elements with finite spectrum in certain rotation algebras are dense in the set of all self-adjoint elements. (Interestingly, this result is not explicit, in the sense that there is no concrete $\\theta$ for which it happens.) However, from what I understand, it seems unclear whether there is a finitely generated dense sub-algebra such that each element has finite spectrum."
] | 10 |
Science
| 0 |
98
|
mathoverflow
|
Can one divide by the cardinal of an amorphous set?
|
This question arose in a discussion with Peter Doyle.
It is provable in ZF that one can divide by any positive finite cardinal $k$: if $X \times \\{1,\ldots,k\\} \simeq Y \times \\{1,\ldots,k\\}$ then $X \simeq Y,$ where I use $\simeq$ to denote the existence of a bijection between two sets.
Assuming the Axiom of Choice, the positive finite cardinals are precisely the cardinals that one can divide by since $1\cdot\mathfrak{m} = 2\cdot\mathfrak{m}$ for any infinite cardinal $\mathfrak{m}$. The ultimate question is whether this is actually provable in ZF:
> If $\mathfrak{m}$ is an infinite cardinal, must there be two cardinals $\mathfrak{p} \neq \mathfrak{q}$ such that $\mathfrak{p}\cdot\mathfrak{m} = \mathfrak{q}\cdot\mathfrak{m}$?
As a candidate for a counterexample, I thought it might be possible to divide by the cardinal of a (strongly) amorphous set. Specifically, in the First Fraenkel model:
> Is it true that if $X \times A \simeq Y \times A$ then $X \simeq Y$, where $A$ is the set of atoms?
|
https://mathoverflow.net/questions/235798/can-one-divide-by-the-cardinal-of-an-amorphous-set
|
[
"lo.logic",
"set-theory",
"axiom-of-choice"
] | 28 | 2016-04-09T15:05:50 |
[
"@Lorenzo: A hint would imply that this is easy... it's not! Peter Doyle has written up division by three and division by four - math.dartmouth.edu/~doyle",
"Can someone give me a hint on how to prove that in $\\text{ZF}$ we can divide by finite cardinals?",
"I like this question!",
"There is Lovasz's unique factorization and unique roots for finite structures, and related results. It might be prudent (and useful if not tangential for this question) to see in which logical domains his arguments can be repeated. Gerhard \"Forward Engineering For Reverse Mathematics\" Paseman, 2016.04.09."
] | 4 |
Science
| 0 |
99
|
mathoverflow
|
Unital $C^{*}$ algebras whose all elements have path connected spectrum
|
A unital $C^{*}$ algebra is called a "Path connected algebra" if the spectrum of all its elements is a path connected subset of $\mathbb{C}$.
> **What is an example of a non commutative path connected $C^*$ algebra? What is an example of a simple path connected $C^{*}$ algebra? In particular does $C^{*}_{red} F_{2}$ satisfiy this property?**
>
> As another question: Is the tensor product of two path connected algebra, a path connected algebra?(For spatial norm).
|
https://mathoverflow.net/questions/195900/unital-c-algebras-whose-all-elements-have-path-connected-spectrum
|
[
"fa.functional-analysis",
"oa.operator-algebras",
"operator-theory",
"sp.spectral-theory",
"c-star-algebras"
] | 28 | 2015-02-06T16:24:12 |
[
"@JohannesHahn Thanks for your comment. As you pointed out a minimal dynamical system gives a simple C* algebra. But what about the spectrum of its elements(Path connected spectrum)?",
"@HannesThiel Thank you very much for your comment and your attention. The motivation for the second part of my question, about the reduced $C^*$ algebra, is obvious. Regarding the first part: the motivation is that the product topology of two path connected space is a path connected space. But according to your comment, one can ask that: Is the tensor product of two idempotent less C* algebras, an idempotent less C* algebra?",
"@AliTaghavi What is your motivation to consider path connected and not just connected spectra?",
"Example of a connected C∗ algebra : See B.E. Blackadar. A simple unital projectionless C∗-algebra. J. Operator Theory, 5:63–71, 1981. By the comments of Hannes Thiel and Sam Evington this algebra is an example of a connected C* algebra. But is it also path connected (remark by Ali Taghavi) ?",
"Hannes' observation can be improved: all elements of a C*-algebra with no non-trivial projections have connected spectrum. Indeed given an element with disconnected spectrum the holomorphic functional calculus gives a non-trivial idempotent. This will be similar to some self-adjoint idempotent (see Blackadar's K-Theory book) which will be non-trivial.",
"@HannesThiel Thank you for the comment. So my second question leads to the following: What is an example of a projectionless simple $C^{*}$ algebra for which the spectrum of all elements are path connected.",
"The spectrum of a nontrivial projection (i.e., nonzero and not the unit) is disconnected. Thus, a unital C*-algebra with the property you consider cannot contain any nontrivial projections. Conversely, if a C*-algebra contains a normal element with disconnected spectrum, then it contains a nontrivial projection. Thus, in a unital, simple C*-algebras without nontrivial projections (such as the Jiang-Su algebra), at least the spectrum of every normal element is connected.",
"@JohannesHahn Thanks for your comment. If I am not mistaken, this algebra is simple iff the action of $G$ on $X$ does not have a nontrivial compact invariant set. Now how you controle the spectrum of elements?",
"Well $\\mathbb{C}$ is simple. But I assume you want something less trivial. A stab in the dark: One could imagine putting some group action on $C(X)$ such that none of its non-trivial ideals are $G$-invariant and hope that $C(X)\\rtimes G$ is simple.",
"@JohannesHahn yes I mean without closed two sided ideal.",
"Oh you meant simple as in simple not as in easy...",
"@JohannesHahn But $C(X)$ is not simple.",
"An easy example: $C(X)$ for some path connected compactum $X$ is path connected in this sense because the spectrum of $f\\in C(X)$ coincides with the range of $f$."
] | 13 |
Science
| 0 |
100
|
mathoverflow
|
A question on simultaneous conjugation of permutations
|
Given $a,b\in S_n$ such that their commutator has at least $n-4$ fixed points, is there an element $z\in S_n$ such that $a^z=a^{-1}$, and $b^z=b^{-1}$? Here $a^z:=z^{-1}az$.
Magma says that the answer is yes up to $n\leq 11$ (computed by M. Zieve). The answer is also yes if $a$ and $b$ commute via the argument below. It would be great to have an answer in case a,b do not commute, even under the assumption that $\langle a,b\rangle$ contains the alternating group $A_n$. If $\langle a,b\rangle$ contains $A_n$ with $n\geq 4$, then any $z$ as above is unique.
The commutative case: If $a^b=a$, the orbits (and cycles) of $a$ are permuted by $b$. Say $b$ permutes under conjugation cyclicly the orbits $A_1,A_2,\ldots,A_r$ of $a$. Let $a_i\in S_n$ be the cycle induced by $a$ on $A_i$, for $i=1,\ldots,r$. Choose $c\in S_n$ to be element of order $r$ such that $b=c$ on $A_i$ for $i=1,...,r-1$. (This forces $a_r^c=a_1$). Since $b=c$ outside $A_1$ and $a_1^{bc^{-1}}=a_1$, we have $b=a_1^e c$ for some integer $e$.
Choose $v$ to be an involution on $A_1$ such that $a_1^v=a_1^{-1}$ on $A_1$, and trivial outside $A_1$. Choose $w$ to be the involution in the symmetric group on $1,\ldots,r$ such that $(1,\ldots,r)^w=(r,\ldots,1)$ and $1^w=2$. Define $u$ to be a permutation such that if $(i,j)$, $i\leq j$, is an orbit of $w$, then $\alpha^u = \alpha^{c^{j-i}}$ for every $\alpha\in A_i$, $i=1,\ldots,r$. This way $c^u =c^{-1}$. Let $z:=vv^c\cdots v^{c^{r-1}}u$, so that $c^z=c^{-1}$, and $a^z=a^{-1}$. (Moreover, $a_i^z=a_{i^w}^{-1}$ for $i=1,\ldots,r$). It follows that $b^z=a_{1^w}^{-e}c^{-1} = a_2^{-e}c^{-1}=c^{-1}a_1^{-e} = b^{-1}$.
Some might consider this proof (in the commutative case) to be simpler in the language of wreath products.
|
https://mathoverflow.net/questions/267191/a-question-on-simultaneous-conjugation-of-permutations
|
[
"co.combinatorics",
"gr.group-theory",
"permutations",
"algebraic-combinatorics"
] | 27 | 2017-04-14T04:11:08 |
[
"For the record: This is proved in Theorem 4.9 of arXiv:1810.08971v4 by Junyao Pan. (I have not checked the proof.)",
"Consider the polygonal mesh with vertices $\\{1,\\dots,n\\}$, directed edges $i\\to a(i)$ and $i\\to b(i)$ for each $i$, and for each (possibly trivial) cycle $(k\\;\\cdots\\;k+m-1)$ of $[a,b]$ a face with boundary $[a,b]^m$ starting at $k$. The genus depends on the cycle type of $[a,b]$. Lifting to the universal cover gives a tessellation of the Euclidean or hyperbolic plane. It suffices to find a symmetry of the plane sending each $a$ (resp $b$) edge to an $a$ (resp $b$) edge with opposite orientation. In the commutative case this is realized by a $180^\\circ$ rotation around any square face.",
"Have you tried using Rutilo Moreno, Luis Manuel Rivera, Blocks in cycles and $k$-commuting permutations, SpringerPlus, December 2016, 5:1949? (A preprint also appears as arXiv:1306.5708.) Corollary 1 characterizes when the commutator of two permutations has at least $n-k$ fixed points (the authors say that these two permutations $k$-commute).",
"@SamHopkins I don't think so. I think a reason why many fixed points matter is as follows: Suppose $a^{-1}b^{-1}ab=x$. If $z$ inverts $a$ and $b$, then $x^z=a^{-z}b^{-z}a^zb^z=aba^{-1}b^{-1}$, so $x^{zab}=x$. Thus the $z$ we are looking for is in the coset $C_G(x)b^{-1}a^{-1}$. Note that $C_G(x)$ is $S_{n-k}$ times a subgroup of $S_k$, where $k\\le4$. I believe the reason for the existence of $z$ is the hugeness of $C_G(x)$.",
"Totally naive comment, but would the 4 vs 5 thing have anything to do with when A_n is simple?",
"Note also that it is very easy to find examples in which $[a,b]$ has $n-5$ fixed points and there is no element $z$ inverting both $a$ and $b$.",
"After doing lots of random experiments I am starting to believe that this might be true. If so, then I am afraid that it might not be easy to prove!"
] | 7 |
Science
| 0 |
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