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Let $\mathfrak g$ be a Lie algebra (if it matters, right now I only care about finite-dimensional Lie algebras in characteristic $0$, although I'm never opposed to hearing about more general cases). Recall that it determines a differential graded algebra ("the complex that computes Lie algebra cohomology"), with $k$th component $\wedge^k \mathfrak g^*$ and differential determined by the bracket. It is a complex by the Jacobi identity. Thinking in terms of supergeometry, I will call this dga $\mathcal C^\infty([-1]\mathfrak g)$.

Moreover, if $\mathfrak h \to \mathfrak g$ is a homomorphism of Lie algebras, then we get a homomorphism of dgas $\mathcal C^\infty([-1]\mathfrak g)\to\mathcal C^\infty([-1]\mathfrak h)$, so that $\mathcal C^\infty\circ [-1]$ is a contravariant functor. The dga is a complete invariant: the functor is full and faithful.

Suppose that $\mathfrak h,\mathfrak g$ are two Lie algebras and $f: \mathfrak h \to \mathfrak g$ a Lie algebra homomorphism. Suppose furthermore that the corresponding map $f^*: \mathcal C^\infty([-1]\mathfrak g)\to\mathcal C^\infty([-1]\mathfrak h)$ is a quasi-isomorphism, i.e. it induces an isomorphism on cohomology. Does it follow that $f$ is an isomorphism?

When I ask it this way, it sounds strongly like the answer should be "no": almost never is cohomology a complete invariant. For example, the cohomology in degree $1$ sees only the abelianizations of $\mathfrak g,\mathfrak h$. But on the other hand, research I'm doing on Lie algebroids suggests that the similar statement with "algebra" replaced by "algebroid" throughout should be true. I don't see a direct proof even in the "algebra" case, but I feel like there should be either a trivial counterexample or an easy argument in favor. In either case, though, and maybe because it's late at night, I'm stuck.

Which is all to say that secretly I care about algebroids, so if any of y'all know a good reference for the problem at that generality, please send it my way. But I will happily accept an answer just for algebras if one is provided.

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I like to think of $C^\infty([-1]g)$ as being the Lie-algebra version of the (nerve of the) 1-object Lie groupoid associated to the 1-connected Lie group G integrating g. This is of course related to the classifying space BG. Probably your f corresponds to a homotopy equivalence of H and G, but I can't make any of my intuitions (such as they are) rigorous.. –  David Roberts Sep 20 '10 at 9:44
If you replace the Lie algebra cohomology with the Lie algebra bar homology then a theorem like this may hold. Any thoughts? –  James Griffin Sep 21 '10 at 10:09

4 Answers 4

up vote 3 down vote accepted

Here are some comments to your answer that I hope will be helpful (it's still sufficiently confusing that I might make some mistakes). Given an $L_{\infty}$ algebra, we can define the "Koszul dual" Chevalley $\it{chains}$ $C_*(L)$(for several reasons it's more natural to think of the Chevalley chains rather than cochains). The ordinary notion of a quasi-isomorphism of $L_{\infty}$ is a morphism of Chevalley complexes whose first Taylor coefficient induces a quasi-isomorphism of complexes. We could ask what happens if we declare two Lie algebras weakly equivalent if there is a quism of Chevalley complexes? We have seen that this notion can differ from the standard one. Another example is when $g_1=sl_2(\mathbb{C})$ and $g_2=\mathbb{C}[2]$ an abelian Lie algebra concentrated in degree 2. From this point of view, your notion is a very natural relaxing of the usual notion of quasi-isomorphism --- it's the notion of weak equivalence transported from Koszul duality. It's worth noting that the two notions will coincide if your dg-Lie algebra is concentrated in positive degree (or more generally with some nilpotency hypotheses on the action of the degree zero piece $g^0$ on your dg lie algebra.) This explains Trial's answer above.

It's useful to have the rational homotopy picture in mind. Given a 1-connected CW complex Xwhich is finite in each degree, you can construct a Lie model L for X whose $H_{*}(L)=\pi_{*}(\Omega(M)) $ with Whitehead bracket. One gets chains on your space by taking $C_{*}(L)$. To go back, you have something known as the cobar construction which spits out $C_{*}(\Omega(M))$, which is U(L), the universal enveloping algebra of L. Again, things start to break down in the non-simply connected case if you don't assume some hypothesis about the action of $\pi_1$ on the higher homotopy groups. From this point of view, the comment at the end of the last paragraph is a reflection of the fact that in these instances, when you have an isomorphism on homology groups, you have an isomorphism on homotopy groups.

Finally when thinking about the idea that the more relaxed notion gives you a derived equivalence, you must define the derived category properly. Let's think about the abelian case. The most classical case of Koszul duality says that given a finite dimensional graded vector space, $D^{+}(SV) \cong D^{+}(\wedge(V^*))$. Here we are considering the derived category of modules which are bounded below. It's pretty easy to see that this cannot be extended an equivalence of unbounded derived categories.

The right notion which makes the whole thing work for unbounded derived categories can be found in Positselski's works and is called the coderived category. Let g be a Lie algebra over $\mathbb{C}$. Then the equivalence between the derived category of modules over U(g) and the coderived category of co-modules over it's Chevalley complex $C_{∗}(g)$ in which $M \to C_{∗}(g,M)$. Under the assumption g is finite, we can consider these co-modules to be modules over C*(g)(now looking at cochains again) and look at the corresponding localization of the category of dg-modules over C*(g). What we see is that we are localizing at a smaller set of weak equivalences than quasi-isomorphisms.

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This is awesome, thanks! But I'm excited and confused in one place. Namely, given a Lie algebra $L$ (satisfying...), I construct some space $X$, which is roughly "$\mathrm BL$", or "derived $\mathrm BL$", with $\operatorname{Chains}(X) = \operatorname{CE Chains}(L)$. Fine. Then I think you said I should take loops $\Omega X$, and lo and behold, $\operatorname{Chains}(\Omega X) = \mathrm U(L)$. This shouldn't be surprising, because $\mathrm U(L)$ is a cocommutative coalgebra, as is (should be) $\operatorname{Chains}(Y)$, and also because I expect $\Omega\mathrm BG$ to be (continued) –  Theo Johnson-Freyd Mar 21 '11 at 14:46
(continuation) roughly the same as $G$, and $\mathrm U(L)$ is a good model for the "formal group" integrating $L$. Except, I actually expect $\Omega\mathrm BG = G/G^{\rm ad}$, the adjoint groupoid. Or, no, $\operatorname{Maps}(S^1 \to \mathrm BG = G/G^{\rm ad}$. In this setting, I'm confused whether there's a difference between $\Omega(-)$ and $\operatorname{Maps}(S^1,-)$. This isn't really about your question, more just me expressing a confusion I have. –  Theo Johnson-Freyd Mar 21 '11 at 14:55
Ah, nevermind, I've unconfused myself. A homotopy of maps from $S^1$ is allowed to rotate the basepoint around the circle, and this is precisely what generates the adjoint action. A homotopy of based loops is not, and so I just expect $\Omega\mathrm B G = G$. Nevermind all my mumbling. –  Theo Johnson-Freyd Mar 21 '11 at 14:57

It seems to me that if $\frak{g}$ is a 1-dimensional Lie algebra and $\frak{h}$ is a nonabelian 2-dimensional Lie algebra then a nontrivial map $\frak{h}\to\frak{g}$ will induce an isomorphism on (trivial coefficients) cohomology.

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Yes, thanks. Now that it's <strike>morning</strike>afternoon, I also thought of that example. Thanks! –  Theo Johnson-Freyd Sep 20 '10 at 20:35

If $\mathfrak{g}$ and $\mathfrak{h}$ are say nilpotent and finite dimensional then the map $\mathfrak{g}\to\mathfrak{h}$ is an isomorphism. It follows from uniqueness of minimal Sullivan model, in this case $\bigwedge\mathfrak{g}^* $ (there should be a simpler reason, but I don't see it). As noted above by Tom Goodwillie, it is not true for solvable Lie algebras.

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I think what's going on more deeply is the following. I will describe it here for two reasons: (1) so that experts might correct errors, and (2) so that someone who wanders here via Google might get more data.

There is something called "Koszul duality" — or rather, there are many such things — and I don't understand any of them. But some version of it connects dg Lie algebras, or rather $L_\infty$ algebras, to dg commutative algebras, or rather $E_\infty$ algebras. In one direction, the map from $L_\infty$ algebras to $E_\infty$ algebras is essentially what I have described as "Lie algebra cohomology": in particular, every Lie algebra is an $L_\infty$ algebra supported only in degree $0$, and every dg commutative algebra is an $E_\infty$ algebra in which all higher associators vanishing, and the map I have described is the part of Koszul duality restricted to these particular cases.

Then the word "duality" means that somehow there should be an inverse to this map, and the word $_\infty$ means "only care up to quasi-isomorphism". Of course, as we've seen, there is not an inverse if the source is "dg Lie algebras up to quasi-iso" and the target is "dg commutative algebras up to quasi-iso": the one- and non-abelian-two- dimensional Lie algebras are not quasi-iso as dg Lie algebras, but their chain complexes are quasi-iso as dg commutative algebras. (Perhaps the problem is that I don't understand the words "$E_\infty$" well enough — are the chain complexes for the one- and two-dimensional Lie algebras the same $A_\infty$ algebra?)

So where's the duality? At least one answer is (as far as I can tell — I am very not an expert) that if two Lie algebras have quasi-isomorphic cohomology, then their derived representation theories are the same. Here the word "derived" means "consider chain complexes in the category, and force quasi-isomorphisms to be isomorphisms." In particular, you force exact sequences to be zero, and so in particular you force extensions to split.

Well, for the example above, you can simply dive in and work out enough of the representation theory. The representation theory of the one-dimensional Lie algebra goes by the name "undergraduate linear algebra": it is the theory of a vector space along with a matrix. The indecomposable representations (provided that I work in an algebraically closed setting, and I am already in characteristic $0$, so let's call the ground field $\mathbb C$) are Jordan blocks: matrices that look like $$ \begin{pmatrix} \lambda & 1 & & \\ & \lambda & \ddots & \\ & & \ddots & 1 \\ & & & \lambda \end{pmatrix} $$ There are nontrivial extensions whenever $\lambda = \lambda$. But there is a unique irreducible representation for each $\lambda \in \mathbb C$, and it is one-dimensional, and this is all of them. The full representation theory of the two-dimensional Lie algebra is much more complicated, but again the irreducibles are all one-dimensional and parameterized by $\lambda \in \mathbb C$. Indeed, the abelianization map form the two-dimensional algebra to the one-dimensional algebra induces the obvious bijection on irreducible representations. This is essentially a proof (where "essentially" means "modulo everything I don't understand") that the derived categories of representations are the same. Or maybe it's not a proof: better is that it is a symptom of the equivalence of the derived representation theory.

If I am correct in the story I have said, then it should be that quasi-isos between semisimple Lie algebras are necessarily isos: there just aren't very many interesting chain complexes in a semisimple category. But there is enough I don't understand that I would not say this with any certainty.

If you see this post and know more than I do, please explain further in the comments. If you see a way to extract from this "answer" a question that we can post and attract more attention, that would also be awesome!

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(1) Almost, but $C^*(\mathfrak{g};k)=RHom_g(k,k)$ only knows about the part of the derived category generated by the trivial module. So it's not an equivalence between the whole derived categories, but only that part. (1b) I don't think your analysis of Tom's example is correct. (2) Also if the category of representations of a semi-simple Lie algebra really were semi-simple, there would be no cohomology at all. The cohomology records the existence of interesting chain complexes, but these chain complexes are made of infinite-dimensional modules, hence the failure of semi-simplicity. –  Ben Wieland Mar 21 '11 at 23:31
(3) A quasi-isomorphism of $A_\infty$-algebras yields an equivalence of derived categories. The category of modules over a Lie algebra has the further structure of being symmetric monoidal (and this preserves the subcategory generated by the unit object, also known as the trivial representation). This is recorded by the $E_\infty$-structure on the algebra. So an equivalence of $E_\infty$-algebras is equivalent to a symmetric monoidal equivalence. –  Ben Wieland Mar 21 '11 at 23:35

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