Take the 2-minute tour ×

A chain complex of vector spaces $X_k$ is a sequence of linear mappings

$\dots \overset{d_{k-1}}{\longrightarrow} X_k \overset{d_{k}}{\longrightarrow} X_{k+1} \overset{d_{k+1}}{\longrightarrow} \dots$

such that $d_{k+1} \circ d_k = 0$. Here $k \in \mathbb N$ runs through the natural numbers. I have held the view that a chain complex $(X_k,d_k)$ is actually just a way to write down a pair of direct sums $(X,d) = ( \bigoplus X_k, \bigoplus d_k )$ that satisfy an additional property (namely, the graduation and the chain property).

However, the notions of tensor products that apply naturally to each of these points of view appear incompatible. Let $(X_k,d_k)$ and $(Y_l,D_l)$ be chain complexes. Then we define

$(X \otimes Y )_{p} := \bigoplus\limits_{k+l=p} X_k \otimes Y_l$

The tensor product of $d = \bigoplus d_k$ and $D = \bigoplus D_l$ in the sense of linear mappings is then defined by

$d \otimes D : X_k \otimes Y_l \rightarrow X_{k+1} \otimes Y_{l+1}$,

$( d \otimes D ) ( v \otimes w ) = ( d_k v \otimes D_l w )$

On the other hand, in the case of tensor products of chain complexes, we rather define

$d \otimes D : X_k \otimes Y_l \rightarrow X_{k+1} \otimes Y_{l} + X_{k} \otimes Y_{l+1}$,

$( d \otimes D ) ( v \otimes w ) = d_k v \otimes w + (-1)^k v \otimes D_l w$

The apparent incompatibility of these definitions contradicts my (likely simplified) view that chain complexes are only special notation for pairs of vector spaces and linear mappings. I expect this to become even more confusing as more structures are regarded (e.g., if the complex "is" a graded algebra).

Question: Is there an intuitive point of view, that there relates these two products? This is not discussed in literature (MacLane, Homology; Lang, Algebra).

share|improve this question
At least on my computer, the term $(X \otimes Y ){p} := \oplus{k+l=p} X_k \otimes Y_l$ does not seem to render as expected. If others see this as well, I am sorry about that. –  shuhalo Apr 7 '12 at 21:16
Briefly one should think of the differential as an element of a super-Lie algebra (so for starters think about how a Lie algebra acts on tensor products of its representations). There are MO questions elaborating on this but I can't find them at the moment. –  Qiaochu Yuan Apr 7 '12 at 21:22
Your "naive" definition of tensor product doesn't yield an object that you would interpret as a chain complex, since the new differentials are changing the grading by 2. Since the even and odd degrees are not related by any maps, this object could be interpreted as a pair of "even" and "odd" complexes, not a single complex. –  Jack Huizenga Apr 7 '12 at 21:30
Martin: the Markdown used on this forum regularly gets in LaTeX's way. Use backticks (``) to enclose "endangered" LaTeX formulas (these are usually the formulas with many underscroes). –  darij grinberg Apr 7 '12 at 21:54
This operator $x\otimes y\mapsto dx\otimes y+(-1)^{|x|}x\otimes Dy$ is never called $d\otimes D$; it's called some kind of $d$ again. Try thinking of this as a Leibniz rule (product rule). Boundary maps in chain complexes are a lot like derivations. –  Tom Goodwillie Apr 8 '12 at 1:38

3 Answers 3

Even at the risk of using slightly too heavy armory, I would like to shortly explain, how I would view the situation in a light, that naturally produces and "explains" the observations collected above:

  • @Martin: Tthe natural extension on tensor products
  • @TomGoodwillie: The "skew-Leibniz"-rule one has to use
  • @JackHuizenga: the even-odd grading associated to it
  • @Qiaochu: The mysterious "super-Lie-algebra"
  • and in general the "additional" $d^2=0$ condition

First, being an algebraicist I "almost never" believe in maps $V\rightarrow V$. I rather always expect an entire algebra $H$ of operators acting on $V,W,...$ (e.g. universal enveloping of Lie algebras instead of Lie algebras or only single elements themselves). Furthermore as in this example, it sould be clearifies, how $H$ also acts on tensor products, the unit object $V=k$ and dual spaces .

Pretty much (and this is not entirely correct, but in a certain sense and with restrictions one can provo this, e.g. Etinghof's theorem) the above amount to $H$ is a Hopf algebra having additional structures:

  • A coproduct $\Delta: H\rightarrow H\otimes H$ that tells you how to act on $V\otimes W$: Split up as $\Delta$ says, then act on each space $V,W$ as you already knew.
  • A counit $\epsilon: H\rightarrow k$ telling you, with which scalar we act on the unit object $V=k$
  • An antipode $S:H\rightarrow H$ that has to be used on an element $h\in H$ before letting it act in the argument of a linear form $f\in V^*$ (e.g. to get combined action again "the right way around")
  • Such that... well pretty much of what you expect is true: You get again actions at all ($\Delta,\epsilon$ be algebra maps) and the so-defined action respects the re-bracking-iso of triple tensor products ("coassociativity"), the isomorphism $V\otimes k\cong V$ ("counitality") and the evaluation $V^*\otimes V\rightarrow k$ and dual basis ("antipode condition"). Respect means here, the action on the left side (e.g. in the second case via $(id\otimes \epsilon)\Delta$) matches the action on the right...or the isomorphism entertwines these two actions, or is a module homomorphisms or how you'd like to put it.

Take as examples (and if more interested read e.g. Susan Montgomory's book!):

  • A group ring $k[G]$ with group elements $\Delta(g)=g\otimes g$, $\epsilon(g)=1$ and $S(g)=g^{-1}$, meaning just copy the element and act on each side of the tensor product as you wish (and inverse on "contragradiant" representations), as usual for group representation.
  • A universal Lie-algebra enveloping $U(\ell)$ with Lie elements $\Delta(x)=1\otimes x+x\otimes 1$, $\epsilon(x)=0$ and $S(x)=-x$. As we expect: Lie algebras act on tensor products via Leibniz ;-)
  • A quantum group such as the often discussed $U_q(\ell)$ or many exotic more discovered by Schneider/Andruskiewitsch ("Classifying pointed Hopf algebras with abelian coradical").
  • Finally the Taft algebra for OUR purposes: $H_{Taft}=\langle g,x\rangle$ with $g^2=1$, $x^2=0$ and $gx=-xg$. Then $\Delta(g)=g\otimes g$ but now $\Delta(x)=g\otimes x+x\otimes 1$ a skew-LieElement/-derivation/-primitive.

It's the RadfordBiproduct/Majid/Bosonization of $k[\mathbb{Z}_2]$ with the BradiedHopfAlgebra/NicholsAlgebra $k[x]/(x^2)$, where the braiding is induced naturally by the sign-graduation and turns out to be the fermionic/Kozul-braiding $x^a\otimes x^b\rightarrow (-1)^{ab}x^b\otimes x^a$. For related more general cases see below... The Taft algebra has dimension only $4$ and somewhat the smallest nontrivial example - the noncommutativity of $g$ and $x$ has to exactly match skew-Leibniz-coproduct and the "premature" truncation $x^2=0$. Generally, Hopf algebras are pretty "picky" about their structures fitting together ;-) (ANSWER TO BELOW: if we take the subalgebra generated just by one such differential operator $x^2=0$, and require the graduation action to be faithful, and the Nichols algebra to be "indecomposable", we uniquely get the Taft algebra)

What does this have to do with the question?

Well, I find it natural to thing of a chain complex $(X_k,d_k)$ as a vector-space $V=\oplus_k X_k$ with an action of $H_{Taft}$ by $x=\oplus_k d_k$ acts differential ($x^2=0$!), $g$ as the odd/even operator $g.a=(-1)^{|a|}a$ (which you may want to refine to a $\mathbb{Z}$-graduation), both exactly anticommuting by the graduation shift.

And now the tensor product is right and "has-to-be" - satisfied with that?

$$x.(a\otimes b):=(x.a)\otimes (1.b)+(g.a)\otimes (x.b)=(x.a)\otimes b+(-1)^{|a|}a\otimes (x.b)$$

I would like to close with some remarks on the "super-Lie-algebras" and the general case:

The classification I mentioned roughly uses, that you may split a (pointed graded) Hopf algebra into (the "Radford-Biproduct") of a groupring (the ""coradical") and a "braided" Hopf algebra (i.e. in a category with a braiding induced by the group's conjugacy action and the groupelement-skewness). The generic case is that the braided Hopf algebra is like the universal enveloping of a Lie algebra, but in these braided "Nichols algebras" we may have truncations. When and how is pretty tricky and open especially over nonabelian groups, while over abelian groups, they're nowadays well understood by the works of Heckenberger. You can image them (and the theory of root systems carries overs!!) as Lie algebras in a braided sense, and this allows specific but sometimes much more exotic Dynkin diagrams - in the easiest case of group $\mathbb{Z_2},\mathbb{Z_3}$ these are e.g. in physics called Super- and Color-Lie-algebras, but of course there are much more interesting cases ;-) ;-)

I you have to have more, you might want to check out my article "Nichols algebras" on Wikipedia ;-) ;-) and run over "our" Nichols algebra $k[x]/(x^2)$ and braiding underlaying the Taft algebra.

share|improve this answer

It seems to me that your question is actually about something that is clearer in a more basic context than chain complexes. Fix a field $k$ and let's work instead in the category $C$ of pairs $(V, T)$ where $V$ is a vector space and $T : V \to V$ is a linear transformation. This category, as a $k$-linear category, admits the following two descriptions:

  • It is the category of $k$-linear representations of the monoid $\mathbb{Z}_{\ge 0}$.
  • It is the category of $k$-linear representations of the one-dimensional Lie algebra over $k$.

These two descriptions lead to two different notions of tensor product (that is, two monoidal structures): for one, the tensor product $V_1 \otimes V_2$ is acted on by $T_1 \otimes T_2$, but for the other, the tensor product $V_1 \otimes V_2$ is acted on by $$T_1 \otimes I + I \otimes T_2.$$

(This is necessary in order for the action to exponentiate to the correct action of the corresponding Lie group when $k = \mathbb{R}$, for example.) Abstractly, both descriptions above say that $C$ is equivalent to the category of modules over $k[x]$, the first description via the construction of the monoid algebra and the second description via the construction of the universal enveloping algebra. However, the natural notions of tensor product in each setting is different: both the monoid algebra and universal enveloping algebra constructions actually spit out bialgebras, and the comultiplication on each bialgebra gives rise to the tensor product. The first construction spits out $k[x]$ with the comultiplication $$x \mapsto x \otimes x$$

whereas the second construction spits out $k[x]$ with the comultiplication $$x \mapsto x \otimes 1 + 1 \otimes x.$$

So what you've observed is that the category of chain complexes is morally closer to the second construction than the first. Why this is the case is something that I don't have a completely satisfactory answer for, but in any case the point I want to make is that the second comultiplication is just as natural as the first from the appropriate perspective.

Algebro-geometrically, comultiplications on $k[x]$ correspond to algebraic monoidal structures on the affine line $\mathbb{A}^1$. The first construction comes from multiplication and the second construction comes from addition.

share|improve this answer
The above answer doesn't explain the sign; if you're curious about that I think you should ask a separate question, although there are some MO questions that already cover similar background. –  Qiaochu Yuan Apr 9 '12 at 4:49
"Why this is the case" I'm sure is nothing more than because the differential should drop the grading by 1 and not 2. –  Chris Gerig Apr 9 '12 at 5:06
This answer goes in a very similar direction than my answer above; and I think the sign and the shift do arrise if the algebra and coalgebra are treated together...if you consider the $x\otimes x$ you're actually tliking about the groupring $k[\mathbb{Z}]$, i.e. $x$ has to be invertible. If you want on the other hand $x^2=0$ you MUST have the sign as I explained above - except of course you're in characteristic 2 where $k[x]/(x^2)$ IS the proper universal Lie enveloping :-) –  Simon Lentner Apr 9 '12 at 17:17

This is just a naive comment on the second to last paragraph in Qiaochu's answer:


Surely the motivation comes from thinking of the associated double complex and the desire to 'embed' both complexes into the tensor product complex? From this point of view the choice $x \mapsto x \otimes 1 + 1 \otimes x$ is very natural.

The signs in the formula are then intepreted as arising from the braiding i.e. Given complexes $(C,\partial)$ and $(D,\delta)$, we evaluate $\partial \otimes 1 + 1 \otimes \delta$ on tensors $\sum_i c_i \otimes d_i$ via the pairing

$$\mathrm{End}(C) \otimes \mathrm{End}(D) \otimes C \otimes D \xrightarrow{\Psi} \mathrm{End}_k(C) \otimes C \otimes \mathrm{End}(D) \otimes D \to C \otimes D $$

Where $\Psi$ is the braiding morphism. We see that taking $\Psi$ to be the plain tensor flip doesn't work and then modify this so that on homogeneous elements $c$ and $d$ we have $\Psi\colon c \otimes d \mapsto (-1)^{|c||d|} d \otimes c$ which does work.

To me this is main unsatisfactory part of the story: why the Koszul braiding? The only explanation I have seen so far is: 'It works! Just accept it.' I guess if I could see that the Koszul braiding was the only braiding that worked then I would be happy but I can't see that off the top of my head...

share|improve this answer
If you requre $x^2=0$ (!) the ONLY possible braiding is $x\otimes \rightarrow -x\otimes x$. Surely this 1) could come (and does!) from a more complex grading operator than $\mathbb{Z}_2$ (e.g. $\mathbb{Z}$) BUT with the same quotient! 2) there could be additional differentials $y,z,...$ with very complicated interaction, e.g. at a $S_3,S_4,S_5$-grading BUT the part generated only by $x$ would again look like I showed. With this conditions ($x^2=0$, faithful action, "indecomposable") the TAFT algebra is UNIQUE! –  Simon Lentner Apr 12 '12 at 13:05
Thanks for this Simon - I think I understand now. We just consider the braiding on a one dimensional vector space where we see that there is but one choice and then uniqueness of the Koszul braiding follows from there via naturality. It seems that the difference between how you are stating things (with the action of the Taft algebra providing the signs) vs. the signs coming from a braiding is somewhat like the Taft Algebra acting as some kind of 'bosonization' a la Majid. –  Jake Apr 12 '12 at 13:41
YEAH, absolutely! The Taft algebra IS exactly the bosonization/RadfordBiproduct/MajidConstruction (all equivalent) of the groupring $k[\mathbb{Z}_2]$ (which provides the sign-graduation and with it a unique braiding) and the braided Hopf Algebra thereover (esp. Nichols Algebra) $k[x]/(x^2)$, which in turn in it's existance requires exactly the prescribed braiding / sign-graduation –  Simon Lentner Apr 12 '12 at 22:48

Your Answer


