Algebraic/Categorical motivation for the Chevalley Eilenberg Complex

Question

Is there a purely algebraic or categorical way to introduce the Chevalley-Eilenberg complex in the definition of Lie algebra cohomology?

In group cohomology, for example, the bar resolution of a group is equal to the chain complex associated to the nerve of the group when considered as a category; I wonder if something like this is also possible for the Chevalley-Eilenberg complex?

I know that the Chevalley-Eilenberg complex arises as the subcomplex of left-invariant differential forms on a Lie group, but apart from this geometric origin, its definition doesn't seem very natural to me, so it would be nice to have some algebraic or categorical construction as well.

Answer 1

Lie algebras are algebras over an operad, usually denoted $\mathscr{L}\mathit{ie}$. This is a quadratic operad, which happens to be Koszul. It therefore comes with a prefered cohomology theory (there is an analogue of Hochschild cohomology of algebras over a Koszul operad) which is defined in terms of a certain canonical complex —unsurprisingl called the Koszul complex. If you work out the details in this general construction, you obtain the Chevalley-Eilenberg resolution.

Alternatively, if $\mathfrak{g}$ is a Lie algebra, we can view $U(\mathfrak g)$, its enveloping algebra, as a PBW deformation of the symmetric algebra $S(\mathfrak g)$. The latter is just a polynomial ring, so we have a nice resolution for it, the Koszul complex, and there is a more or less canonical way of deforming that resolution so that it becomes a resolution for the PBW deformation. Again, working out the details rapidly shows that the deformed complex is the Chevalley-Eilenberg complex.

Finally (I haven't really checked this, but it should be true :) ) if you look at $U(\mathfrak g)$ as presented by picking a basis $B=\{X_i\}$ for $\mathbb g$ and dividing the free algebra it generates by the ideal generated by the relations $X_iX_j-X_jX_i-[X_i,X_j]$, as usual, you can construct the so called Annick resolution. Picking a sensible order for monomials in the free algebra (so that standard monomials are precisely the elements of the PBW basis of $U(\mathfrak g)$ constructed from some total ordering of $B$), this is the Chevalley-Eilenberg complex again.

These three procedures (which are of course closely interrelated!) construct the resolution you want as a special case of a general procedure. History, of course, goes in the other direction.

Answer 2

Both (Lie) group and Lie algebra cohomology are essentially part of a more general procedure. Namely, we take an abelian category $C$ with enough projectives or enough injectives, take a (say, left) exact functor $F$ from $C$ to abelian groups (or modules over a commutative ring) and compute the (right) derived functors of $F$ using projective or injective resolutions.

For example, if $\mathfrak{g}$ is a Lie algerba over a field $k$, we can take the category of $U(\mathfrak{g})$-modules as $C$ and $$M\mapsto\mathrm{Hom}_{\mathfrak{g}}(k,M)$$

as $F$ (here $k$ is a trivial $\mathfrak{g}$-module). Notice that this takes $M$ to the set of all elements annihilated by any element of $\mathfrak{g}$; this is not a $\mathfrak{g}$-module, only a $k$-module, so the target category is the category of $k$-vector spaces.

In the category of $U(\mathfrak{g})$-modules there are enough projectives and enough injectives, so in principle to compute the Exts from $k$ to $M$ we can use either an injective resolution of $M$ or a projective resolution of $k$ as $U(\mathfrak{g})$-modules. I've never seen anyone considering injective $U(\mathfrak{g})$-modules, probably because they are quite messy. So most of the time people go for the second option and construct a projective resolution of $k$. One of the ways to choose such a resolution is the "standard" resolution with

$$C_q=U(\mathfrak{g}\otimes\Lambda^q(\mathfrak{g})$$ but any other resolution would do, e.g. the bar-resolution (which is "larger" then Chevalley-Eilenberg, but more general, it exists for arbitrary augmented algebras). Applying $\mathrm{Hom}_{\mathfrak{g}}(\bullet,M)$ to the standard resolution we get the Chevalley-Eilenberg complex.

All the above holds for group cohomology as well. We have to replace $U(\mathfrak{g})$ by the group ring (or algebra) of a group $G$. The only difference is that there is no analogue of the Chevalley-Eilenberg complex, so one has to use the bar resolution. Probably, Van Est cohomology of a topological group can also be described in this way.

Answer 3

Consider the universal enveloping algebra $U(\mathfrak g)$ of your Lie algebra $\mathfrak g$. Since it is a Hopf algebra, then you can construct a filtered simplicial cocommutative coalgebra $A_\bullet$:

• $A_i=U(\mathfrak g)^{\otimes i}$

• face maps are given by applying the product

• degeneracy maps are given by applying the unit

The $E_1$ term of the associated spectral sequence is precisely the Chevalley-Eilenberg chain complex*.

In other words, the Chevalley-Eilenberg complex of $\mathfrak g$ is a by-product of the Bar complex of $U(\mathfrak g)$. And the Bar complex of an augmented unital algebra $A$ arises as the chain complex associated to the simplicial set $Nerve(A)$ (where I view $A$ as a linear category with one object).

• this is another way of saying what Mariano Suárez-Alvarez says in his answer.