Timeline for Which commutative algebras admit a nonzero Poisson bracket?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Aug 16, 2011 at 22:06 | vote | accept | Qiaochu Yuan | ||
| Aug 11, 2011 at 0:40 | comment | added | Qiaochu Yuan | @Daniel: thanks for your comments! I am having trouble figuring out which of the general statements you're making is doing the work, since it seems to me that you produce $\alpha$ out of thin air. I admit I'm not very familiar with the definition of the Gerstenhaber bracket (maybe that's what's doing the work?), so if you had some time to spell some simple things like this out for me in an answer it would be much appreciated. | |
| Aug 10, 2011 at 22:06 | comment | added | Daniel Pomerleano | The algebraic version is slightly more complicated for technical reasons and was only proven a bit later here: arxiv.org/abs/math/0310399 | |
| Aug 10, 2011 at 22:06 | comment | added | Daniel Pomerleano | we can extend our multiplication to second order by m_t=m + tπ+t^2\alpha which will be a solution to the Maurer Cartan Equation in HCH^*(A,A)[t]/t^3. So I don't think you will find any counterexample to your Stack Exchange question because I believe it's true. In the smooth case, where HKR is an isomorphism and hence I'm pretty certain that what I've said up to now is true, (an algebraic version of) Kontsevich's formality theorem tells you that a Poisson structure extends to a deformation of all orders. Have a look, it's wonderful math! arxiv.org/abs/q-alg/9709040 | |
| Aug 10, 2011 at 21:46 | comment | added | Daniel Pomerleano | @ Qiaochu: I am not entirely sure about this(I am mostly familiar with the case of smooth algebras where a lot more can be said) because it seems to me that in char=0, there is always an injective (this is the part I haven't checked) map of lie algebras HKR: \oplus \Wedge^i T_A[1] \to HH^*(A,A)[1](when A is smooth this is an iso) in particular if your deformation class $\pi$ is the image of a bivector field the condition [\pi,\pi]=0 in HH^*(A,A)[1] says exactly that your bivector field is Poisson. The above condition says that on the level of Hochschild chains [\pi,\pi]=d\alpha so.... | |
| Aug 4, 2011 at 12:39 | comment | added | Qiaochu Yuan | @Theo: thanks. I figured that had to be the case but didn't know how to go about constructing a counterexample. | |
| Aug 4, 2011 at 9:09 | comment | added | Theo Johnson-Freyd | (continuation) $\partial_x \wedge (x \partial_y - \partial_z)$. So this cannot satisfy Jacobi. Alternately, check directly that this bivector field is not Poisson. For another example, find your favorite invertible de Rham 2-form that is not closed, and take its inverse. Other examples come from the following observation: Jacobi is a quadratic relation, whereas 2-cocycle is a linear relation. So you do not expect Jacobi(a+b) to hold even if both Jacobi(a) and Jacobi(b) hold. | |
| Aug 4, 2011 at 9:05 | comment | added | Theo Johnson-Freyd | I don't have an account at M.SE, and anyway won't answer that question, only comment on its motivation, so I'll leave the comment here. You ask: does the antisymmetrization of a Hochschild 2-cocycle satisfy Jacobi? The answer is NO. Recall that a bivector field is certainly a 2-cocycle, and the Hamiltonian flows for a Poisson bivector field foliate the space into (even-dimensional) symplectic leaves. But there are non-integrable plane distributions in $\mathbb R^3$, the standard one being $\ker(dz+xdy)$. This distribution is generated by Hamiltonian vector fields for the bivector (continued) | |
| Aug 3, 2011 at 19:53 | answer | added | Jan Weidner | timeline score: 4 | |
| Aug 3, 2011 at 19:48 | answer | added | Stefan Waldmann | timeline score: 5 | |
| Aug 3, 2011 at 19:39 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
added 198 characters in body
|
| Aug 3, 2011 at 19:34 | history | asked | Qiaochu Yuan | CC BY-SA 3.0 |