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In a paper of A. Weinstein on the geometry of Poisson manifolds, he relates the formal linearization around a zero, p, of the Poisson bivector to extensions of the Lie algebra induced by the bivector on the tangent space over p.

I wanted to know if this is part of a big picture, possibly relating deformations of Lie brackets to some extensions of Lie algebras.

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up vote 5 down vote accepted

In fact the picture is extremely simple and works indeed for any type of algebra as follows :

Let $\mu$ be a Lie algebra on a vector space $V$, $c$ a two-cocycle $c\in CE^2(V;M)$ (Chevalley Eilenberg cohomology) where $M$ is a module over the Lie algebra. The extension of $\mu$ by $c$ is nothing else than a deformation of $\mu$, but in the space of Lie algebras on the vector space $V\oplus M$. The deformed Lie algebra has a bracket $\mu'$ given by $\mu'(v+m,v'+m')=(\mu(v,v'),v.m'-v'.m+c(v,v'))$. One can easily check that the Jacobi condition for the deformed algebra $\mu'$ is equivalent to the data of the Jacobi condition for $\mu$, the module structure of $M$ and the cocycle condition for $c$. From this point of view one can also view $\mu'$ as the semi-direct product of $\mu$ and $M$ when the cocycle $c$ is null.

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There are big pictures that I'll let others describe. Here's a little picture which cogeneralizes the Weinstein remark. (To "cogeneralize" is to make more specific, rather than less.)

Recall that a Lie bialgebra is a vector space $\mathfrak g$ with a "Lie bracket" $\mathfrak g^{\wedge 2} \to \mathfrak g$ satisfying Jacobi, a "Lie cobracket" $\mathfrak g \to \mathfrak g^{\wedge 2}$ satisfying Jacobi, and such that the two structures satisfy a compatibility condition which has lots of equivalent formulations: one of them is that the cobracket is a 1-cochain for the Chevalley-Eilenberg complex of $\mathfrak g$ with values in $\mathfrak g^{\wedge 2}$ (diagonal adjoint action).

Then the first result you prove about these things is: Any such structure defines (and is equivalent to) an "extension" (although it's not a short exact sequence), called the double of $\mathfrak g$. As a vector space, the double is the sum $\mathfrak g \oplus \mathfrak g^\ast$ (where $\mathfrak g^\ast$ is the dual vector space, and is a Lie algebra by turning around the Lie cobracket), and indeed each of the summands $\mathfrak g,\mathfrak g^\ast$ inside the double is a Lie subalgebra. The two terms do interact: they interact in the unique way making the canonical pairing $(\mathfrak g \oplus \mathfrak g^\ast)^{\otimes 2} \to \mathbb k$ ad-invariant. For various equivalent descriptions, and if you want to see this all in pictures, I have a short expository note on Lie bialgebras at .

Anyway, why is this a cogeneralization of what Alan's doing? There is a generalization of Lie algebra to Lie algebroid, which I can define if you like, but I would assume that it's in Alan's paper, and one example of a Lie algebroid is that the tangent bundle of a manifold has a canonical Lie algebroid structure. A Lie algebroid structure on the cotangent bundle is precisely the same as a Poisson bivector. So a Poisson manifold is (almost) an example of a "Lie bialgebroid", because the tangent bundle is both an algebroid and a coalgebroid. I say "almost", because a priori there is no cocycle condition. But the linearization of a Poisson structure near a zero thereof I think should satisfy a cocycle condition --- I haven't worked out the details, so take this paragraph with a grain of salt.

Anyway, having not read this paper, I'm not sure if I've answered the question you asked, or a related one.

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$\otimes 2$ should be an exponent, right? – darij grinberg Apr 5 '11 at 15:32
@darij: yes, fixed. – Theo Johnson-Freyd Apr 22 '11 at 3:16

Hinich says in "Deformation theory and lie algebra homology", that Grothendieck says that "to each deformation problem we can assign a sheaf of Lie algebras over X; the sheaf of infinitesimal automorphisms".

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@cdm80: Thank you! but where does the Lie algebra extension show up? Is it possible to associate such an extension to that sheaf of infinitesimal automorphisms? I will give a look to that Hinich's paper. – Feri Apr 5 '11 at 2:50

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