Do Chevalley-Eilenberg homology functor and taking the cohomology commute?

Question

I've stumbled upon some ideas from homological algebra that I'm trying to piece together from a talk I heard. I don't have much background in this area, so I'm not sure if this is a reasonable thing to expect. Consider the Chevalley-Eilenberg homology as a functor $C_\bullet$ from dg Lie algebras to cochain complexes that to a dg Lie algebra $L$ associates the vector space $C_\bullet(L) = \bigwedge^\bullet(L) = \text{Sym}(L[1])$ with differential given by $$ \mathrm{d}(x_1 \wedge \dots \wedge x_n) = \sum_{i < j} \pm [x_i, x_j] \wedge x_1 \wedge \dots \hat{x_i} \dots \hat{x_j}\dots \wedge x_n$$ Does taking the cohomology commute with $C_\bullet$, that is, is there some reason why we would have $H^\bullet (C_\bullet(L)) \cong C_\bullet(H^\bullet(L))$? Even if this is not true, I would appreciate some reference as to be able to understand (and check) if it is true in some particular case, as I don't know how I would go about proving this. Thank you for your time!

Answer 1

Consider the case in which $L$ is a plain Lie algebra concentrated in degree zero (with zero differential). That is, consider the case when $L= L^0$ so that $d = 0$ since $L^1 = 0$, so the codomain of the only possible non-trivial differential is already zero.

In this case, $H(L) = H^0(L) = L$ so that you are asking whether $C(L)$ and its homology $H(C(L))$ are isomorphic. As soon as the differential of $C(L)$ is non-trivial (i.e. as soon as $L$ is not Abelian) then simply by dimension considerations $C(L)$ will be (locally, in each degree) necessarily larger than $H(C(L))$, so it is impossible for these two things to be isomorphic.

The reason is simple: homology is a quotient of a submodule, so the dimension will fall as soon as $d\neq 0$, i.e. as soon as $\ker d$ is not the ambient space.