MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $g$?

share|cite|improve this question
up vote 12 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer
"deformations of algebraic structures are always governed by a cohomology theory" --- cohomology theory in what sense? I thought deformations are supposed to be governed by a dg Lie algebra. – Kevin H. Lin Jan 17 '10 at 0:26
I think we are both saying the same thing. The complex computing the cohomology groups where the infinitesimal deformations and the obstructions live has the structure of a graded Lie algebra, hence dually a differential graded algebra. This is explained in Nijenhuis and Richardson. – José Figueroa-O'Farrill Jan 17 '10 at 1:02
is there written somewhere the construction of the Chevalley-Eilenberg complex in the dg setting? sure I could grind it out, but a reference? – Jim Stasheff Apr 14 '15 at 20:09

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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.