Let $V$ be a finite dimensional vector space and $S(V)$ the corresponding symmetric algebra. Suppose that we have a Poisson bracket $\lambda = \{,\}: S(V) \otimes S(V) \to S(V)$. Let $V^*$ be the dual vector space of $V$. Is there a natural Poisson bracket on $S(V^*)$ which is related to the Poisson bracket $\lambda$ on $S(V)$? Thank you very much.

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    $\begingroup$ If $\lambda$ is quadratic, i.e. $\lambda$ arises from a map $b: \Lambda^2(V) \to S^2(V)$, then its dual identifies with a map $b^*:S^2(V^*) \to \Lambda^2(V^*)$ (away from characteristic two) which gives a graded Poisson structure on the exterior algebra $\Lambda(V^*)$ rather than on $S(V^*)$ (and the two Poisson enveloping algebras are Koszul duals). $\endgroup$
    – M T
    Sep 25, 2015 at 9:50

1 Answer 1


Under the natural interpretation of the term "natural", the answer is negative. For example, consider a linear bracket corresponding to a Lie algebra that does not admit an invariant bilinear form (e.g., 3-dimensional Heisenberg Lie algebra). The dual vector space has no "natural" Poisson bracket.

The answer is positive in the important special case of nondegenerate Poisson bracket, because in this case $V$ and its dual are canonically isomorphic by means of the corresponding symplectic structure.


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