Let $\mathfrak g$ be a Lie algebra. The ChevalleyEilenberg complex is defined to be $\wedge^* \mathfrak g$ with differential $d\colon \wedge^* \mathfrak g\to \wedge^{*1}\mathfrak g$ defined by $$d(a_1\wedge\cdots \wedge a_k)=\sum_{i,j}(1)^{i+j1}[a_i,a_j] a_1\wedge \cdots\wedge\hat{a_i}\wedge\cdots\wedge\hat{a}_j\wedge\cdots\wedge a_k.$$ 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 ChevalleyEilenberg 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 ChevalleyEilenberg complex with coefficients in a nontrivial $\mathfrak g$module?

$\begingroup$ The analog thing for associative algebras, which is the Gerstenhaber bracket, is not defined for notregular values (which is the analogue of trivial values) $\endgroup$ – Mariano SuárezÁlvarez Oct 4 '10 at 20:06

$\begingroup$ By "values" I mean "coefficients"... Sometime ago I decided that homology takes coefficients in the module, and cohomology takes values in the module :) $\endgroup$ – Mariano SuárezÁlvarez Oct 4 '10 at 20:06

1$\begingroup$ A truly terrible way to get at this bracket is as follows. If $\mathfrak g$ 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 $\mathfrak g \to \Gamma(\rm TM^*)$ (sections of tangent bundle). The Schouten bracket on $\wedge^\bullet\mathfrak g\otimes M$ is the pullback of said bracket on $\wedge^\bullet\Gamma(\rm TM^*)$ to $\mathfrak g$, 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. $\endgroup$ – Theo JohnsonFreyd Oct 5 '10 at 2:46
Theo JohnsonFreyd:
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.)