Timeline for Why is $\Pi_r^L$ a non-degenerate Poisson structure on $G\ $?
Current License: CC BY-SA 4.0
8 events
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| Jan 18, 2023 at 18:38 | comment | added | Anil Bagchi. | @PavelSafronov$:$ Nice point. Another thing is that let $G$ be a Lie group with Lie algebra $\mathfrak {g}.$ Let $\omega$ be a left-invariant non-degenerate closed $2$-form on $G.$ Then how to show that $\omega (e)$ is a $2$-cocycle on $\mathfrak {g}$ i.e. $$\omega (e) ([a,b], c) + \omega(e) ([b,c], a) + \omega (e) ([c,a], b) = 0$$ for all $a,b,c \in \mathfrak {g},$ where $[\cdot, \cdot]$ is the Lie bracket on $\mathfrak {g}.$ Thanks for your time. | |
| Jan 18, 2023 at 9:37 | comment | added | Pavel Safronov | As you've probably already understood, $\Pi^L_r$ is non-degenerate iff it is non-degenerate at the unit (by left invariance) iff $r$ is non-degenerate. | |
| Jan 17, 2023 at 15:06 | comment | added | Anil Bagchi. | @PavelSafronov$:$ If $r = \sum\limits_{s,t} r^{st} X_s \otimes X_t$ then $r$ is non-degenerate if and only if the matrix $(r^{st})$ is invertible. | |
| Jan 17, 2023 at 14:36 | comment | added | Anil Bagchi. | @PavelSafronov$:$ Sorry, I forgot to mention that $r$ is itself non-degenerate i.e. we assume that the map $\xi \mapsto (\xi \otimes \text {id}) (r)$ is an isomorphism from $\mathfrak {g}^{\ast} \longrightarrow \mathfrak {g}$ which clearly discards the case $r = 0.$ | |
| Jan 17, 2023 at 14:30 | history | edited | Anil Bagchi. | CC BY-SA 4.0 |
added 14 characters in body
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| Jan 17, 2023 at 14:05 | comment | added | Pavel Safronov | $r=0$ seems to satisfy your assumptions, but clearly $\Pi^L_r=0$ is not non-degenerate. | |
| Jan 17, 2023 at 12:33 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
Typo in the title corrected
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| Jan 17, 2023 at 10:29 | history | asked | Anil Bagchi. | CC BY-SA 4.0 |