most recent 30 from http://mathoverflow.net 2013-05-23T13:52:49Z http://mathoverflow.net/feeds/question/41055 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41055/what-is-the-schouten-bracket-for-the-chevalley-eilenberg-complex-with-coefficient Jim Conant 2010-10-04T19:53:48Z 2011-02-24T11:56:15Z

<p>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?</p>

http://mathoverflow.net/questions/41055/what-is-the-schouten-bracket-for-the-chevalley-eilenberg-complex-with-coefficient/56515#56515 Jim Conant 2011-02-24T11:56:15Z 2011-02-24T11:56:15Z

<p>Theo Johnson-Freyd:</p> <p>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 <code>$M^*$</code>, which you should think of as a geometric space, and so there is a map <code>$g\to\Gamma(TM^*)$</code> (sections of tangent bundle). The Schouten bracket on <code>$\wedge^* g\otimes M$</code> is the pullback of said bracket on <code>$\wedge^*\Gamma(TM^*)$</code> 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.</p> <p>(JC: I'm trying to clear the unanswered question backlog.)</p>