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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!

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    What does the right-hand side mean? – Randall Apr 04 '23 at 17:20
  • @Randall The RHS of the expression that defines the differential? The bracket is the bracket of the dg Lie algebra, there are some signs that come from the degrees of elements. Does this help? I tried to edit to make it clearer. –  Apr 04 '23 at 17:33
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    Oh, I meant the right-hand side of your iso. Is $H^{\bullet}(L)$ naturally a dgla? – Randall Apr 04 '23 at 17:40
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    @Randall Sorry! Like I said, I have no expertise in this area, but I think the answer is yes. A quick search seems to confirm this is the case, with zero differential and the induced bracket. –  Apr 04 '23 at 17:48
  • @Randall I'm reading more into this and maybe that's not always well-defined. However, I am comfortable with assuming that is the case, that there exists an induced bracket. –  Apr 04 '23 at 17:58
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    Why are you expecting this to be true? Notice that $H(C(L))$ has zero differential, while $C(H(L))$ has a non-zero differential unless $L$ is Abelian, so they cannot be isomorphic in general...! – Pedro Apr 05 '23 at 03:27
  • @Pedro Thank you! Can you clarify why $H(C(L))$ has zero differential in general? I'm not necessarily expecting it to be true. I want to understand if it could be true in some cases, because I saw a talk where it really seemed like something like this was implicitly used, by that I mean, computing the homology before applying the Chevalley-Eilenberg functor, which I find weird, but I don't understand this material well enough to make sense of it. This is a great answer, because $H(L)$ might in fact be abelian. –  Apr 05 '23 at 08:34
  • @flower_i Once you take homology you use up the differential, so again I do not see why you are expecting any differential at all on the homology of a complex. – Pedro Apr 06 '23 at 02:56
  • @Pedro Right, that makes sense. Thank you! –  Apr 06 '23 at 08:24

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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.

Pedro
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