0
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ 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

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.