Which commutative algebras admit a nonzero Poisson bracket? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:54:23Z http://mathoverflow.net/feeds/question/72028 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72028/which-commutative-algebras-admit-a-nonzero-poisson-bracket Which commutative algebras admit a nonzero Poisson bracket? Qiaochu Yuan 2011-08-03T19:34:16Z 2011-08-03T19:53:27Z <p>Let $A$ be a commutative algebra, not necessarily unital, over a field $k$ (of characteristic not equal to $2$, or even equal to $0$, if it helps). A <strong>second-order formal deformation</strong> of $A$ is a $k[h]/h^3$-bilinear associative product $\star$ on $A[h]/h^3$ such that quotienting by $h$, we obtain the original product on $A$. Writing such a product as</p> <p>$$a \star b = ab + h m_1(a, b) + h^2 m_2(a, b), a, b \in A$$</p> <p>it's not hard to verify that $\{ a, b \} = m_1(a, b) - m_1(b, a)$ is a <strong>Poisson bracket</strong> on $A$, that is, a Lie bracket satisfying the Leibniz rule $\{ a, bc \} = \{ a, b \} c + b \{ a, c \}$. Given a nonzero Poisson bracket on $A$, it is interesting to ask whether we can find a formal deformation (replace $k[h]/h^3$ with $k[[h]]$) which gives rise to it as above ("deformation quantization"). </p> <p>But of course we can't ask this question until we have a nonzero Poisson bracket in the first place. So:</p> <blockquote> <p>Which commutative algebras admit a nonzero Poisson bracket? </p> </blockquote> <p>If there is no reasonable description in general feel free to restrict to the finitely-generated case or smooth functions on manifolds etc.</p> <p>What I know: any polynomial algebra in $2$ or more variables admits a nonzero Poisson bracket (take the symmetric algebra on a nonabelian Lie algebra). Any nonzero Poisson bracket gives a nonzero element of the alternating part of the second Hochschild cohomology $H^2(A, A)$, so if this group is trivial then no such brackets exist. I doubt this implication can be reversed in general, but I don't know a counterexample. If you do, I have a <a href="http://math.stackexchange.com/questions/55258/does-every-poisson-bracket-on-a-commutative-algebra-come-from-a-second-order-defo" rel="nofollow">math.SE question</a> you should answer!</p> http://mathoverflow.net/questions/72028/which-commutative-algebras-admit-a-nonzero-poisson-bracket/72029#72029 Answer by Stefan Waldmann for Which commutative algebras admit a nonzero Poisson bracket? Stefan Waldmann 2011-08-03T19:48:32Z 2011-08-03T19:48:32Z <p>In general, I think the question is too broard to expect some reasonable answer. But there are many examples and constructions of Poisson brackets which might be interesting for you.</p> <p>1.) Whenever you have nonzero commuting derivations on your algebra you can build a Poisson bracket out of them. Indeed, if $D_1, \ldots, D_n, E_1, \ldots, E_n$ are commuting derivations then $P = \sum_i D_i \otimes E_i$ gives a Poisson bracket via $$\lbrace a, b\rbrace = \mu \circ (P - P \circ \tau) (a \otimes b)$$ where $\tau$ is the canonical flip and $\mu\colon A \otimes A \longrightarrow A$ is the multiplication. This example allows for an immediate deformation quantization, the star product quantizing it is $$a \star b = \mu \circ \exp(\hbar P)(a \otimes b),$$ a formula going back at least to Gerstenhaber himself (I think in his famous deformation of algebras papers, Nr III).</p> <p>Now this is not sooo special as it may seem at the first sight. In particular, the following nice geometric construction shows that on a smooth manifold you always have nontrivial Poisson brackets:</p> <p>On $\mathbb{R}^d$ there are $d$ vector fields with support inside the compact ball of radius $1$, which coincide with the coordinate vector fields $\partial_1, \ldots, \partial_d$ inside the ball of radius $1/2$, and which commute everywhere. Out of them you can build a Poisson tensor having maximal rank (either $d$ or $d-1$) inside the smaller ball but with compact support. Thus you can implant it to any other manifold using this as a chart :) The existence of such vector fields can be shown by various constructions. Either you can shrink $\mathbb{R}^d$ with a suitable diffeo into the ball and take the images of the coordinate vector fields (I learned this one from Alan Weinstein) or you can play around with two (!) suitable bump functions...</p> <p>2.) Related to this but more sophisticated: there are universal deformation formulas for certain (Lie) groups. So whenever you have an action of such a group on an algebra by automorphisms, one can induce a deformation quantization and hence in particular a Poisson bracket on the algebra. It depends on the action how nontrivial the bracket will be. Part 1.) is a special case for the group $\mathbb{R}^d$.</p> <p>OK, there are several more constructions, but this might already be interesting for you.</p> http://mathoverflow.net/questions/72028/which-commutative-algebras-admit-a-nonzero-poisson-bracket/72031#72031 Answer by Jan Weidner for Which commutative algebras admit a nonzero Poisson bracket? Jan Weidner 2011-08-03T19:53:27Z 2011-08-03T19:53:27Z <p>Smooth functions on a manifold always admit lots of Poisson structures. Indeed, one can construct Poisson structures on $\mathbb R^n$ with support in $(-1,1)^n$ and any prescribed value $\pi^{ij}e_i \wedge e_j$ of the Poisson-Bivector at the origin. This is done in two steps: At first one constructs $n$ commuting vector fields $X_i$ with support in $(-1,1)^n$. Then on just defines the $\pi:=\pi^{ij} X_i\wedge X_j$.</p>