Chevalley Eilenberg complex definitions?

Question

In Weibel's An Introduction to Homological Algebra, the Chevalley-Eilenberg complex of a Lie algebra $g$ is defined as $\Lambda^*(g) \otimes Ug$ where $Ug$ is the universal enveloping algebra of $g$. The differential here has degree -1.

I have been told that the Chevalley-Eilenberg complex for $g$ is $C^*(g) = \text{Sym}(g^*[-1])$, the free graded commutative algebra on the vector space dual of $g$ placed in degree 1. The bracket $[,]$ is a map $\Lambda^2 g \to g$ so its dual $d : = [,]* \colon g^* \to \Lambda^2 g^*$ is a map from $C^1(g) \to C^2(g)$. Since $C^*(g)$ is free, this defines a derivation, also called $d$, from $C^*(g)$ to itself. This derivation satisfies $d^2 = 0$ precisely because $[,]$ satisfies the Jacobi identity.

Finally, Kontsevich and Soibelman in Deformation Theory I leave it as an exercise to construct the Chevalley-Eilenberg complex in analogy to the way that the Hochschild complex is constructed for an associative algebra by considering formal deformations.

The first is a chain complex, the second a cochain complex, and what do either have to do with formal deformations of $g$?

Answer 1

The first complex, from Weibel, is a projective resolution of the trivial $\mathfrak g$-module $k$ as a $\mathcal U(\mathfrak g)$-module; I am sure Weibel says so!

Your second complex is obtained from the first by applying the functor $\hom_{\mathcal U(\mathfrak g)}(\mathord-,k)$, where $k$ is the trivial $\mathfrak g$-module. It therefore computes $\mathrm{Ext}_{\mathcal U(\mathfrak g)}(k,k)$, also known as $H^\bullet(\mathfrak g,k)$, the Lie algebra cohomology of $\mathfrak g$ with trivial coefficients.

The connection with deformation theory is explained at length in Gerstenhaber, Murray; Schack, Samuel D. Algebraic cohomology and deformation theory. Deformation theory of algebras and structures and applications (Il Ciocco, 1986), 11--264, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 247, Kluwer Acad. Publ., Dordrecht, 1988.

In particular neither of your two complexes 'computes' deformations: you need to take the projective resolution $\mathcal U(\mathfrak g)\otimes \Lambda^\bullet \mathfrak g$, apply the functor $\hom_{\mathcal U(\mathfrak g)}(\mathord-,\mathfrak g)$, where $\mathfrak g$ is the adjoint $\mathfrak g$-module, and compute cohomology to get $H^\bullet(\mathfrak g,\mathfrak g)$, the Lie algebra cohomology with coefficients in the adjoint representation. Then $H^2(\mathfrak g,\mathfrak g)$ classifies infinitesimal deformations, $H^3(\mathfrak g,\mathfrak g)$ is the target for obstructions to extending partial deformations, and so on, exactly along the usual yoga of formal deformation theory à la Gerstenhaber.

By the way, the original paper [Chevalley, Claude; Eilenberg, Samuel Cohomology theory of Lie groups and Lie algebras. Trans. Amer. Math. Soc. 63, (1948). 85--124.] serves as an incredibly readable introduction to Lie algebra cohomology.

Answer 2

The other comment revealed most of the story. Let me add some points. First, the Chevalley-Eilenberg complex is defined for the most general case of $H^\bullet(g,M)$ --- cohomology with coefficients in a module M. In the case M=k the trivial module, you get your second complex. In the case $M=g$ the adjoint representation, you get the comples Kontsevich and Soibelman ask to construct.

The first complex is some version of Koszul complex $(A^!)^* \otimes A$ for quadratic algebras: one can think of the Koszul dual of the dg-commutative algebra $\Lambda^\bullet( g^* )$ as the quadratic-linear algebra $U(\mathfrak{g})$. It is acyclic, and gives a resolution of the trivial module by free $U(g)$-modules, so you can use is to compute Ext groups $Ext_{U(g)}^\bullet(k,M)=H^\bullet(g,M)$.

Finally, another way to think of your second complex is as follows. If $g$ is the Lie algebra of a Lie group $G$, one can consider the subcomplex of the de Rham complex consisting of left-invariant differential forms. A left-invariant form is defined by its behaviour at the unit of the group, 1-forms are dual to the tangent space and form $ g^* $ , 2-forms give $\Lambda^2(g^*)$ etc., and you get precisely the CE complex, where the differential is what the de Rham differential induces. If $G$ is compact, it is easy to show that this complex has the same cohomology as the de Rham complex, so we compute the de Rham cohomology of the group in this case!

Answer 3

Just to add to what Mariano has written and focussing only on the deformations aspect, the exercise in Kontsevich-Soibelman hints at the "principle" that deformations of algebraic structures are always governed by a cohomology theory and that if you didn't know which, you would discover it by analysing the conditions defining infinitesimal deformations, trivial infinitesimal deformations and obstructions to integrating infinitesimal deformations.

I think that this is a very good exercise and it helps motivate the classical formulas for the differentials, at least in the case of cohomology with values in the adjoint module.

I would add another reference to the one of Gerstenhaber et al, and that is the older paper by Nijenhuis and Richardson Deformations of Lie algebra structures. In general there are a number of classic papers of Nijenhuis and Richardson on this topic. In particular they define a graded Lie algebra structure on the deformation complex which makes very clear the nature of the obstructions.