# What is the (Koszul? derived?) interpretation of a pair of Lie algebras with the same cohomology?

There are many words and sentences in mathematics that I basically completely don't understand, including the words "Koszul" and "derived". But rather than ask for a complete description of such words, I will ask about a particular example, and hope that an MOer can spell out concretely what's going on. (Of course, links to good write-ups of Koszul, etc., are always welcome.)

There is a relatively easy theorem, probably due to Chevalley and Eilenberg, with many generalizations due to Koszul:

Let $L$ be a (finite-dimensional) vector space (in characteristic $0$); let $[1] L^*$ denote the dual vector space, "shifted" to a graded vector space supported in degree $1$; and let $([1]L^*)^{\vee \bullet}$ denote the free graded-commutative algebra generated by $[1]L^*$ (I impose the Koszul rule for signs, so that as an algebra $([1]L^*)^{\vee \bullet} = (L^*)^{\wedge\bullet}$ is classically an alternating algebra). Then to give a Lie algebra structure to $L$ is the same as giving to $([1]L^*)^{\vee \bullet}$ a square-zero degree-$1$ derivation.

The construction is (contravariantly) functorial and full and faithful, and so embeds the category of Lie algebras fully-faithfully into the (opposite to the) category of commutative dgas.

Here are two examples. I will call my characteristic-$0$ field $\mathbb R$. The one-dimensional Lie algebra corresponds to the dga $\mathbb R \overset 0 \to \mathbb R \xi$, where $\xi$ is the coordinate function on the one-dimensional vector space, and the algebra is graded-commutative, so $\xi^2 = 0$. The two-dimensional non-commutative Lie algebra corresponds to the dga $\mathbb R \overset 0 \to (\mathbb R \xi \oplus \mathbb R \upsilon) \overset{\xi \mapsto 0, \, \upsilon \mapsto \xi\upsilon}\longrightarrow \mathbb R\xi\upsilon$.

Now, once we're in the land of dgas, it makes sense to talk about their (co?)homology. In the above examples, the cohomology of the one- and two-dimensional Lie algebras are isomorphic as algebras; the isomorphism on cohomology is given in one direction by the "abelianization" map from the two-dimensional Lie algebra to the one-dimensional Lie algebra.

Question: How should I interpret "geometrically" the fact that the cohomologies agree?

An idea that I've heard is that somehow dgas-up-to-? should correspond to some notion of "stack". Now, it is certainly not the case that "point mod one-dimensional" and "point mod two-dimensional" present the same stack. So that's not quite what's going on. Perhaps the problem is that these algebras are not the same as ${\rm A}_\infty$ algebras; we should probably add the word "${\rm A}_\infty$" to the list of words I don't really know. If this is the answer, I hope that some MOer will spell it out.

Another idea that I've heard is that the passage "Lie algebra to dga to cohomology" remembers exactly the "derived category of representations of the Lie algebra". Again, I don't really know what that is, but that's OK. Somehow, a representation of a Lie algebra should be a "quasicoherent sheaf on point mod the Lie algebra", and I know that people like "derived categories of quasicoherent sheaves". So should I understand Koszul duality as remembering not all the data of some "stack-like object" but just some "derived data"? If so, again I hope that some MOer will spell it out pedantically in this example.

-
It gets maybe a bit deeper into fancy-words-territory than you might like, but see David Ben-Zvi's answer to my question mathoverflow.net/questions/385/… –  Kevin H. Lin Oct 1 '10 at 18:16
@Kevin: Thanks for the link. I think that David Ben-Zvi's answer does include a lot that I'm interested in. (And maybe wasn't so fancy-words-full as to be un-useful.) –  Theo Johnson-Freyd Oct 2 '10 at 2:27

I know, basically, two answers to this question. The first one starts from the observation that what you call "a relatively easy theorem" generalizes to infinite-dimensional Lie algebras $L$ by replacing the exterior algebra generated by $L^\ast$ with the exterior coalgebra cogenerated by $L$. Then the construction becomes covariant, but produces a (conilpotent) DG-coalgebra rather than a DG-algebra.

Now it is just a fact that the natural equivalence relation on DG-coalgebras is more delicate than the quasi-isomorphism. There is a certain class of "filtered quasi-isomorphisms" that one is supposed to invert when dealing with (conilpotent) DG-coalgebras. The map of Chevalley-Eilenberg complexes that you describe, when considered as a map of the homological, rather than cohomological, Chevalley-Eilenberg complexes, is not a filtered quasi-isomorphism of DG-coalgebras and cannot be obtained from such using compositions and fractions. This saves the derived Koszul duality as an equivalence between appropriate localizations of the categories of (augmented) DG-algebras and (conilpotent) DG-coalgebras.

The second answer describes what happens if one insists on considering the cohomological Chevalley-Eilenberg complex of a finite-dimensional Lie algebra, viewed as a DG-algebra up to a quasi-isomorphism. One can also apply the derived Koszul duality to it, obtaining a certain DG-coalgebra. This DG-coalgebra is, basically, the k-linear dual to the complete topological DG-algebra of derived adic completion of the enveloping algebra U(L) with respect to its augmentation ideal. So if a morphism of finite-dimensional Lie algebras induces an isomorphism of cohomology, this should be understood to mean that it induces an isomorphism of the derived adic completions of the enveloping algebras at their augmentation ideals. Admittedly, this is not a geometric interpretation.

EDIT: I have been asked in a comment what a filtered quasi-isomorphism of DG-coalgebras is, so let me add an explanation of this to the answer. This notion does not presume a fixed filtration on a DG-coalgebra, but rather a class of admissible filtrations. We consider coaugmented DG-coalgebras, i.e., the basic field $k$ is embedded as a subcoalgebra into a DG-coalgebra $C$ and its image is annihilated by the differential. An increasing filtration $F$ on $C$ is admissible if $F_0C=0$, $F_1C=k$, the components $F_iC$ are preserved by the differential, and the filtration $F$ is compatible with the comultiplication.

Not every coalgebra admits an admissible filtration at all! A coalgebra is called conilpotent if it has an admissible filtration (just as a coalgebra, let us drop the condition of compatibility with the differential for a moment). In this case, it has a certain maximal admissible filtration (which is preserved by the differential automatically). However, filtered quasi-isomorphisms of conilpotent DG-coalgebras are defined in terms of arbitrary admissible filtrations. A morphism of DG-coalgebras is a filtered quasi-isomorphism if admissible filtrations can be chosen on its source and target so that the morphism is compatible with them and is a quasi-isomorphism on all filtration components.

-
This is very interesting. Can you say more about what the filtration on the DC coalgebras is? If I understood it, I might be able to get closer to answering the question that prompted me to ask about this example. –  Theo Johnson-Freyd Oct 2 '10 at 2:21

Is that first construction in Theo's question related to Quillen's construction $M\mapsto R\oplus M \to R$ for $M$ an $R$-module (R a commutative $A$-algebra), which sends an R-module to a nilpotent thickening (also an $A$-algebra) of $R$ with kernel isomorphic to M. –  Harry Gindi Oct 2 '10 at 17:24