Timeline for Construct super Poisson brackets on the coordinate rings of Lie super groups
Current License: CC BY-SA 3.0
10 events
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Feb 7, 2017 at 22:43 | comment | added | Jianrong Li | it's fine. I understand. Thank you very much for helping me about this problem. | |
Feb 7, 2017 at 22:27 | comment | added | Vladimir Dotsenko | @JianrongLi : I do not have time to double-check this now, sorry about that! I hope you will be able to unwrap the definition fully. | |
Feb 7, 2017 at 17:19 | comment | added | Jianrong Li | I tried to do computations in the question. I obtained \begin{align} \{t_{ij}, t_{kl}\} = \sum_{a,b} (-1)^{|t_{ij}||E_{kb}|} r^{ik}_{ab} t_{aj}t_{bl} - (-1)^{|t_{ij}||E_{bl}|} r^{ab}_{jl} t_{ia}t_{kb}. \end{align} Is it correct? | |
Feb 7, 2017 at 15:36 | comment | added | Jianrong Li | thank you very much. Yes, your answer solved my problem. | |
Feb 7, 2017 at 15:31 | comment | added | Vladimir Dotsenko | @JianrongLi Note that $|R_\mu\phi|=|\phi|+|\mu|$, because $R_\mu\phi$ is the result of $\mu$ acting on $\phi$, and action respects the grading. I think that this property, and the same for the right-invariant operators $L_\mu$, completely resolves your concern. | |
Feb 7, 2017 at 15:24 | comment | added | Jianrong Li | thank you very much. First I tried to verify the super anti commutativity. I tried to do the computations in another post. But I don't know how to prove that \begin{align} & (-1)^{|\psi||\phi| + |\psi||\mu| + |\mu||\nu| + |R_{\nu }\psi | | R_{\mu} \phi| + |\phi| |\nu| } = 1, \\ & (-1)^{|\psi||\phi| + |\psi||\mu| + |\mu||\nu| + |L_{\nu }\psi | | L_{\mu} \phi| + |\phi| |\nu| } = 1. \end{align} | |
Feb 7, 2017 at 15:08 | comment | added | Vladimir Dotsenko | @JianrongLi it would help to know what exactly you tried. Maybe you can include it in that new question you posted. | |
Feb 7, 2017 at 14:50 | comment | added | Jianrong Li | thank you very much. Do you know thank you very much for the reference. How to write the formula in "Lie superbialgebras and Poisson-Lie supergroups" using coordinates $c_{ij}$ in the case of $GL(m|n)$? I tried but the formula I get didn't satisfy super commutativity. I asked this question in the post. | |
Feb 7, 2017 at 4:36 | vote | accept | Jianrong Li | ||
Feb 7, 2017 at 1:30 | history | answered | Vladimir Dotsenko | CC BY-SA 3.0 |