How can I tell whether a Poisson structure is symplectic "algebraically"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:01:27Z http://mathoverflow.net/feeds/question/19989 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19989/how-can-i-tell-whether-a-poisson-structure-is-symplectic-algebraically How can I tell whether a Poisson structure is symplectic "algebraically"? Theo Johnson-Freyd 2010-03-31T19:26:44Z 2010-04-01T03:22:09Z <p>My primary motivation for asking this question comes from the discussion taking place in the comments to <a href="http://mathoverflow.net/questions/19932/" rel="nofollow">What is a symplectic form intuitively?</a>.</p> <p>Let $M$ be a smooth finite-dimensional manifold, and <code>$A = \cal C^\infty(M)$</code> its algebra of smooth functions. A <em>derivation</em> on $A$ is a linear map <code>$\{\}: A \to A$</code> such that <code>$\{fg\} = f\{g\} + \{f\}g$</code> (multiplication in $A$). Recall that all derivations factor through the de Rham differential, and so: <strong>Theorem:</strong> Derivations are the same as vector fields.</p> <p>A <em>biderivation</em> is a linear map <code>$\{,\}: A\otimes A \to A$</code> such that <code>$\{f,-\}$</code> and <code>$\{-,f\}$</code> are derivations for each $f\in A$. By the same argument as above, biderivations are the same as sections of the tensor square bundle <code>${\rm T}^{\otimes 2}M$</code>. <em>Antisymmetric</em> biderivations are the same as sections of the exterior square bundle <code>${\rm T}^{\wedge 2}M$</code>. A <em>Poisson structure</em> is an antisymmetric biderivation such that <code>$\{,\}$</code> satisfies the Jacobi identity.</p> <p>Recall that sections of <code>${\rm T}^{\otimes 2}M$</code> are the same as vector-bundle maps <code>${\rm T}^*M \to {\rm T}M$</code>. A <em>symplectic structure</em> on $M$ is a Poisson structure such that the corresponding bundle map is an isomorphism. Then its inverse map makes sense as an antisymmetric section of <code>${\rm T^*}^{\otimes 2}M$</code>, i.e. a differential 2-form, and the Jacobi identity translates into this 2-form being closed. So this definition agrees with the one you may be used to of "closed nondegenerate 2-form".</p> <blockquote> <p><strong>Question:</strong> Is there a "purely algebraic" way to test whether a Poisson structure is symplectic? I.e. a way that refers only to the algebra $A$ and not the manifold $M$?</p> </blockquote> <p>For example, it is necessary but not sufficient that <code>$\{f,-\} = 0$</code> implies that $f$ be locally constant, where I guess "locally constant" means "in the kernel of every derivation". The easiest way that I know to see that it is necessary is to use Darboux theorem to make $f$ locally a coordinate wherever its derivative doesn't vanish; it is not sufficient because, for example, the rank of the Poisson structure can drop at points.</p> <p>Please add tags as you see fit.</p> http://mathoverflow.net/questions/19989/how-can-i-tell-whether-a-poisson-structure-is-symplectic-algebraically/20023#20023 Answer by Ben Webster for How can I tell whether a Poisson structure is symplectic "algebraically"? Ben Webster 2010-04-01T01:17:52Z 2010-04-01T01:17:52Z <p>The moral answer is "Yes. A Poisson structure is symplectic if and only if the algebra has no interesting Poisson ideals."</p> <p>The idea is this: a Poisson ideal is one which is closed under the operation of Poisson bracket with every function (not to be confused with a coisotropic ideal, which one closed under taking the bracket of two elements in the ideal). Obviously, the vanishing set of a Poisson ideal is a Poisson submanifold, so if you're symplectic, you'd better not have any whose vanishing set is a submanifold (not empty or the whole space). And if you have a Poisson submanifold, than the ideal vanishing on the submanifold is certainly Poisson. </p> <p>So basically up to nebulous concerns about how the Nullstellensatz works for manifolds, that's the right answer.</p> http://mathoverflow.net/questions/19989/how-can-i-tell-whether-a-poisson-structure-is-symplectic-algebraically/20032#20032 Answer by Mariano Suárez-Alvarez for How can I tell whether a Poisson structure is symplectic "algebraically"? Mariano Suárez-Alvarez 2010-04-01T03:22:09Z 2010-04-01T03:22:09Z <p>In the purely algebraic setting, Daniel Farkas proved in his beautiful paper [Farkas, Daniel R. Characterizations of Poisson algebras. Comm. Algebra 23 (1995), no. 12, 4669--4686. <a href="http://www.ams.org/mathscinet-getitem?mr=MR1352562" rel="nofollow">MR1352562</a>] that a Poisson-simple linear Poisson algebra over an algebraically closed field is a regular symplectic domain, a partial converse of the much easier fact that a commutative regular affine domain which is symplectic is Poisson-simple. There are examples of non symplectic Poisson-simple polynomial algebras, though.</p>