Timeline for Poisson algebras as deformations vs. Poisson algebras in algebraic topology
Current License: CC BY-SA 4.0
8 events
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| Sep 28, 2021 at 22:19 | history | edited | John Baez | CC BY-SA 4.0 |
changed "invoque" to "invoke"
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| Sep 13, 2011 at 1:56 | vote | accept | Qiaochu Yuan | ||
| Sep 12, 2011 at 22:05 | comment | added | Qiaochu Yuan | @Theo: thanks! Alright, so it looks like the answer is something like "this is true in a more meaningful sense than the sense in which you wanted it to be true," which I guess I can get behind. | |
| Sep 12, 2011 at 17:25 | comment | added | Theo Johnson-Freyd | @Qiaochu: I said it badly in my answer, but the Pois^d bracket absolutely comes from a "commutator" in the E_n algebra. You just have to modify what you mean by "commutator". The correct definition of "the commutator" in an n-algebra is the 2-ary operation corresponding to the fundamental class of the (n-1)-sphere (this sphere is the space of 2-ary operations). When n=1, the 0-sphere consists of two points, and the fundamental class is the one that signs the two points differently. | |
| Sep 12, 2011 at 16:01 | comment | added | Qiaochu Yuan | I'm not familiar enough with any of this material that my opinion is trustworthy, but I think Ben Wieland's comment on David Ben-Zvi's answer still applies: it sounds like this abstract machinery produces a Poisson bracket that doesn't come from the (super?) commutator on an $\mathbb{E}_n$-algebra, but that comes from some kind of special commutator already present in the $\mathbb{E}_n$-structure. Is this accurate? $A = C(\Omega^d(X), \mathbb{Q})$ comes with some kind of (super?) commutator which passes to $\text{gr}(A)$. Is this the $\text{Pois}^d$ bracket? | |
| Sep 12, 2011 at 7:40 | history | edited | DamienC | CC BY-SA 3.0 |
remark about Costello-Gwilliam work added
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| Sep 12, 2011 at 7:33 | history | edited | DamienC | CC BY-SA 3.0 |
added 6 characters in body
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| Sep 12, 2011 at 7:27 | history | answered | DamienC | CC BY-SA 3.0 |