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T. Amdeberhan
T. Amdeberhan
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How to proofprove a bracket is super anti-commutative?

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Jianrong Li
Jianrong Li
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On page 12 of the paper, there is a formula about super Poisson bracket on a Lie super group $G$: \begin{align} \{\phi, \psi\} = \sum_{\mu, \nu} (-1)^{|\phi||\nu|} r^{\mu \nu} ( R_{\mu} \phi R_{\nu} \psi - L_{\mu} \phi L_{\nu} \psi ), \end{align} where $\phi, \psi$ are functions on $G$.

I am trying to verify that the bracket satisfies \begin{align} \{\psi, \phi\} = - (-1)^{|\psi| |\phi|} \{\phi, \psi\}. \end{align}

We have \begin{align} \{\psi, \phi\} & = \sum_{\mu, \nu} (-1)^{|\psi||\nu|} r^{\mu \nu} ( R_{\mu} \psi R_{\nu} \phi - L_{\mu} \psi L_{\nu} \phi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} r^{\nu \mu} ( R_{\nu} \psi R_{\mu} \phi - L_{\nu} \psi L_{\mu} \phi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} r^{\nu \mu} ( -(-1)^{|R_{\nu}\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi + (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} (-1)^{|\mu||\nu|} r^{ \mu \nu} ( (-1)^{|R_{\nu}\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi - (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ). \end{align} Therefore \begin{align} & (-1)^{|\psi||\phi|}\{\psi, \phi\} \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\phi|} (-1)^{|\psi||\mu|} (-1)^{|\mu||\nu|} r^{ \mu \nu} ( (-1)^{|R_{\nu }\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi - (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ). \end{align} Therefore we need to show that \begin{align} & (-1)^{|\psi||\phi| + |\psi||\mu| + |\mu||\nu| + |R_{\nu }\psi | | R_{\mu} \phi| + |\psi| |\phi| } = 1, \\ & (-1)^{|\psi||\phi| + |\psi||\mu| + |\mu||\nu| + |L_{\nu }\psi | | L_{\mu} \phi| + |\psi| |\phi| } = 1. \end{align}\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} In general, do we have the above identities? How to prove them? Thank you very much.

On page 12 of the paper, there is a formula about super Poisson bracket on a Lie super group $G$: \begin{align} \{\phi, \psi\} = \sum_{\mu, \nu} (-1)^{|\phi||\nu|} r^{\mu \nu} ( R_{\mu} \phi R_{\nu} \psi - L_{\mu} \phi L_{\nu} \psi ), \end{align} where $\phi, \psi$ are functions on $G$.

I am trying to verify that the bracket satisfies \begin{align} \{\psi, \phi\} = - (-1)^{|\psi| |\phi|} \{\phi, \psi\}. \end{align}

We have \begin{align} \{\psi, \phi\} & = \sum_{\mu, \nu} (-1)^{|\psi||\nu|} r^{\mu \nu} ( R_{\mu} \psi R_{\nu} \phi - L_{\mu} \psi L_{\nu} \phi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} r^{\nu \mu} ( R_{\nu} \psi R_{\mu} \phi - L_{\nu} \psi L_{\mu} \phi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} r^{\nu \mu} ( -(-1)^{|R_{\nu}\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi + (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} (-1)^{|\mu||\nu|} r^{ \mu \nu} ( (-1)^{|R_{\nu}\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi - (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ). \end{align} Therefore \begin{align} & (-1)^{|\psi||\phi|}\{\psi, \phi\} \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\phi|} (-1)^{|\psi||\mu|} (-1)^{|\mu||\nu|} r^{ \mu \nu} ( (-1)^{|R_{\nu }\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi - (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ). \end{align} Therefore we need to show that \begin{align} & (-1)^{|\psi||\phi| + |\psi||\mu| + |\mu||\nu| + |R_{\nu }\psi | | R_{\mu} \phi| + |\psi| |\phi| } = 1, \\ & (-1)^{|\psi||\phi| + |\psi||\mu| + |\mu||\nu| + |L_{\nu }\psi | | L_{\mu} \phi| + |\psi| |\phi| } = 1. \end{align} In general, do we have the above identities? How to prove them? Thank you very much.

On page 12 of the paper, there is a formula about super Poisson bracket on a Lie super group $G$: \begin{align} \{\phi, \psi\} = \sum_{\mu, \nu} (-1)^{|\phi||\nu|} r^{\mu \nu} ( R_{\mu} \phi R_{\nu} \psi - L_{\mu} \phi L_{\nu} \psi ), \end{align} where $\phi, \psi$ are functions on $G$.

I am trying to verify that the bracket satisfies \begin{align} \{\psi, \phi\} = - (-1)^{|\psi| |\phi|} \{\phi, \psi\}. \end{align}

We have \begin{align} \{\psi, \phi\} & = \sum_{\mu, \nu} (-1)^{|\psi||\nu|} r^{\mu \nu} ( R_{\mu} \psi R_{\nu} \phi - L_{\mu} \psi L_{\nu} \phi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} r^{\nu \mu} ( R_{\nu} \psi R_{\mu} \phi - L_{\nu} \psi L_{\mu} \phi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} r^{\nu \mu} ( -(-1)^{|R_{\nu}\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi + (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} (-1)^{|\mu||\nu|} r^{ \mu \nu} ( (-1)^{|R_{\nu}\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi - (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ). \end{align} Therefore \begin{align} & (-1)^{|\psi||\phi|}\{\psi, \phi\} \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\phi|} (-1)^{|\psi||\mu|} (-1)^{|\mu||\nu|} r^{ \mu \nu} ( (-1)^{|R_{\nu }\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi - (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ). \end{align} Therefore we need to show 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} In general, do we have the above identities? How to prove them? Thank you very much.

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Jianrong Li
Jianrong Li
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How to proof a bracket is super anti-commutative?

On page 12 of the paper, there is a formula about super Poisson bracket on a Lie super group $G$: \begin{align} \{\phi, \psi\} = \sum_{\mu, \nu} (-1)^{|\phi||\nu|} r^{\mu \nu} ( R_{\mu} \phi R_{\nu} \psi - L_{\mu} \phi L_{\nu} \psi ), \end{align} where $\phi, \psi$ are functions on $G$.

I am trying to verify that the bracket satisfies \begin{align} \{\psi, \phi\} = - (-1)^{|\psi| |\phi|} \{\phi, \psi\}. \end{align}

We have \begin{align} \{\psi, \phi\} & = \sum_{\mu, \nu} (-1)^{|\psi||\nu|} r^{\mu \nu} ( R_{\mu} \psi R_{\nu} \phi - L_{\mu} \psi L_{\nu} \phi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} r^{\nu \mu} ( R_{\nu} \psi R_{\mu} \phi - L_{\nu} \psi L_{\mu} \phi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} r^{\nu \mu} ( -(-1)^{|R_{\nu}\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi + (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ) \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\mu|} (-1)^{|\mu||\nu|} r^{ \mu \nu} ( (-1)^{|R_{\nu}\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi - (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ). \end{align} Therefore \begin{align} & (-1)^{|\psi||\phi|}\{\psi, \phi\} \\ & = \sum_{\mu, \nu} (-1)^{|\psi||\phi|} (-1)^{|\psi||\mu|} (-1)^{|\mu||\nu|} r^{ \mu \nu} ( (-1)^{|R_{\nu }\psi | | R_{\mu} \phi|} R_{\mu} \phi R_{\nu} \psi - (-1)^{|L_{\nu} \psi| |L_{\mu} \phi|} L_{\mu} \phi L_{\nu} \psi ). \end{align} Therefore we need to show that \begin{align} & (-1)^{|\psi||\phi| + |\psi||\mu| + |\mu||\nu| + |R_{\nu }\psi | | R_{\mu} \phi| + |\psi| |\phi| } = 1, \\ & (-1)^{|\psi||\phi| + |\psi||\mu| + |\mu||\nu| + |L_{\nu }\psi | | L_{\mu} \phi| + |\psi| |\phi| } = 1. \end{align} In general, do we have the above identities? How to prove them? Thank you very much.