On line 7 of page 61 of the book a guide to quantum groups, a Poisson bracket is defined on $\mathbb{C}[GL_n]$ for every classical $r$-matrix as follows.

Let $V$ be a vector space with a basis $v_1, \ldots, v_n$, and $r \in \mathfrak{g} \otimes \mathfrak{g}$ is a skew-symmetric classical $r$-matrix. Suppose that \begin{align} r(v_i \otimes v_j) = \sum_{k,l} r_{ij}^{(kl)} v_k \otimes v_l, \end{align} for some $r_{ij}^{(kl)} \in \mathbb{C}$.

Define \begin{align} \{c_{i j}, c_{kl}\} = \sum_{s,s'} r_{s',s}^{(j l)} c_{ks} c_{i s'} - \sum_{s, s' } r_{ik}^{(s' s)} c_{s' j} c_{s l}, \ i,j,k,l = 1,2, \ldots, n. \quad (1) \end{align} Here $c_{ij}$ are natural coordinates on $\mathbb{C}[GL_n]$. By using classical $r$-matrix \begin{align} \sum_{s} (r_{sj}^{(s_3s_4)} r_{ik}^{(ss_2)} - r_{sk}^{(s_3s_2)} r_{ij}^{(ss_4)} + r_{is}^{(s_3s_4)} r_{jk}^{(ss_2)} - r_{sk}^{(s_4s_2)} r_{ij}^{(s_3s)} + r_{is}^{(s_3s_2)} r_{jk}^{(s_4s)} - r_{js}^{(s_4s_2)} r_{ik}^{(s_3s)}) =0 \end{align} and \begin{align} r_{ij}^{(kl)} = - r_{ji}^{(lk)}, \end{align} we can show that the bracket defined by (1) is Poisson. Is it possible to modify (1) such that it is a super Poisson bracket on $\mathbb{C}[GL(m|n)]$? Thank you very much.


1 Answer 1


I think that it is done in the paper of Andruskiewitsch "Lie superbialgebras and Poisson-Lie supergroups", Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 63 (1993), 147-163, http://link.springer.com/article/10.1007/BF02941339 (the key result is Proposition 3 in that paper).

  • $\begingroup$ 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. $\endgroup$ Feb 7, 2017 at 14:50
  • $\begingroup$ @JianrongLi it would help to know what exactly you tried. Maybe you can include it in that new question you posted. $\endgroup$ Feb 7, 2017 at 15:08
  • $\begingroup$ 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} $\endgroup$ Feb 7, 2017 at 15:24
  • $\begingroup$ @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. $\endgroup$ Feb 7, 2017 at 15:31
  • $\begingroup$ thank you very much. Yes, your answer solved my problem. $\endgroup$ Feb 7, 2017 at 15:36

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