# Is Nijenhuis–Richardson bracket a BV bracket?

Let $g$ be a finite dimensional Lie algebra, and let me denote $A=(\bigwedge g^* \otimes g, d)$ the Chevalley-Eilenberg complex that calculates cohomology of the Lie algebra with coefficients in the adjoint representation. On the complex $A$ there is Nijenhuis–Richardson bracket and the resulting DGLA could be used to describe deformations of the given algebra.

Suppose that $g$ is unimodular i.e. determinant of the adjoint representation is trivial representation. In such situation there are isomorphisms $$\bigwedge^i g^* \otimes g \cong \bigwedge^{n-i} g \otimes g,$$ thus differential in the complex for homology of the lie algebra induces one more map on the complex (this map goes in the opposite direction on $A$). I want to mimic Ginzburg's construction of the Batalin-Vilkovisky algebra structure on the Hochshild cohomology complex of a CY algebra and such induced map should play a role of BV differential, and unimodularity condition plays role of the CY condition. But one important ingredient is missing in this situation, namely there is no associative algebra structure on $A$.

I wounder is it possible to introduce some multiplication on $A$ such that Nijenhuis–Richardson bracket become a BV bracket?

There is one silly answer to your question, and you are probably aware of it: if you use as coefficients $S(g)$, the symmetric algebra of $g$, and not just $g$, everything will work wonderfully. Alas I have never seen anything smaller in this context.
• It looks like in such situation PBW isomorphism reduces question to the Hochshild cohomology of $U(g)$ that is CY if $g$ is unimodular, so it is really just an example of Ginzburg construction. But I don't understand deformation theoretic meaning of the Chevalley-Eilenberg complex for $S(g)$, is it some sort of "extended deformations"? – Sasha Pavlov Sep 28 '13 at 1:05