Let $\mathfrak g$ be a Lie algebra. The Chevalley-Eilenberg complex is defined to be $\wedge^\* \mathfrak g$ with differential $d\colon \wedge^\* \mathfrak g\to \wedge^{\*-1}\mathfrak g$ defined by $$d(a\_1\wedge\cdots \wedge a\_k)=\sum\_{i,j}(-1)^{i+j-1}[a\_i,a\_j] a\_1\wedge \cdots\wedge\hat{a\_i}\wedge\cdots\wedge\hat{a}\_j\wedge\cdots\wedge a\_k.$$ The differential $d$ is not a derivation with respect to the exterior product $\wedge$, but the deviation from being a derivation is a binary operation which defines a graded Lie algebra structure on $\wedge^\* \mathfrak g$: If $\underline{a},\underline{b}\in\wedge^\*\mathfrak g$, let
$$[\underline{a},\underline{b}]\_{s}=d(\underline{a}\wedge\underline{b})-d\underline{a}\wedge b+\underline{a}\wedge d\underline{b}$$
(I'm omitting some signs.) This bracket operation vanishes once you take homology, since if $d\underline{a}=d\underline{b}=0$ then it is obvious that $d(\underline{a}\wedge\underline{b})=[\underline{a},\underline{b}]\_s$. However, I was talking to Jim Stasheff several years ago, and he mentioned that the Schouten bracket doesn't necessarily vanish on Lie algebra homology if there are coefficients in a nontrivial $\mathfrak g$ module, $M$. However, I don't know what the definition of the Schouten bracket is in this case. The Chevalley-Eilenberg complex is easy enough to understand: $\wedge^\*\mathfrak g\otimes M$, where the differential includes terms where the $a\_i$ act on $M$, but the obvious generalization of the above construction fails since two elements of $M$ somehow need to get combined into one element. So my basic question is how you define a Schouten bracket on the Chevalley-Eilenberg complex with coefficients in a nontrivial $\mathfrak g$-module?



What is the Schouten bracket for the Chevalley-Eilenberg complex with coefficients in a nontrivial module?
Theo Johnson-Freyd:


A truly terrible way to get at this bracket is as follows. If $g$ acts on $M$, then it also acts on the dual space $M^\*$, which you should think of as a geometric space, and so there is a map $g\to\Gamma(TM^\*)$ (sections of tangent bundle). The Schouten bracket on $\wedge^\* g\otimes M$ is the pullback of said bracket on $\wedge^\*\Gamma(TM^\*)$ to $g$ , and restricted to those sections that are linear in the base $M^\*$ . As I said, this is a terrible way to get at this bracket.


(JC: I'm trying to clear the unanswered question backlog.)


