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I found two versions of definitions of orthosymplectic supergroups. It seems that they are not equivalent. I don't know which version of the definition is standard.

The first version of the definition is in the paper. The orthosymplectic supergroup $OSP(2p|n)$ is defined as follows.

Let $A = A_0 \oplus A_1$ be a supercommutative superalgebra, where elements in $A_0$ are even and elements in $A_1$ are odd, and \begin{align} A_i A_j \subset A_{i+j \ (\text{mod} \ 2)}, \quad i, j \in \{0, 1\}. \end{align}

The general linear Lie supergroup $GL(m|n)$ is defined by \begin{align} & GL(m|n) = \left\{ \left( \begin{array}{c|c} X_{11} & Y_{12} \\ \hline Y_{21} & X_{22} \end{array} \right) \right\}, \\ & X_{11}=(x_{ij})_{i,j=1,\ldots,m}, X_{22}=(x_{ij})_{i,j=m+1,\ldots,m+n}, \\ & Y_{12}=(y_{ij})_{\substack{ i=1,\ldots,m \\ j = m+1, \ldots, m+n}}, Y_{21}=(y_{ij})_{\substack{i=m+1,\ldots,m+n \\ j = 1, \ldots, m}}, \end{align} where $x_{ij} \in A_0$ and $y_{ij} \in A_1$ and $\det(X_{11})\neq 0$, $\det(X_{22}) \neq 0$.

We have \begin{align} x_{ij} x_{kl} = x_{kl} x_{ij} \\ y_{ij} y_{kl} = - y_{kl} y_{ij} \\ y_{ij} x_{kl} = - x_{kl} y_{ij}. \end{align}

The Lie supergroup $OSP(m|n)$ is defined by \begin{align} & OSP(m|n) = \{ M \in GL(m|n): m=2p, M^{\text{st}}HM = H \}, \end{align} where \begin{align} & H = \left( \begin{array}{c|c} Q & 0 \\ \hline 0 & I_n \end{array} \right), \\ & Q = \left( \begin{array}{c|c} 0 & I_p \\ \hline -I_p & 0 \end{array} \right), \end{align} $I_n$ is the identity matrix of order $n$, and for \begin{align} M = \left( \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right), \end{align} \begin{align} M^{\text{st}} = \left( \begin{array}{c|c} A^T & -C^T \\ \hline B^T & D^T \end{array} \right). \end{align}

The second version of the definition is in the paper. The definition is in Section 3.1. It is similar as the first definition above. But the matrix $H$ is different. It is defined as follows (formulas (3.2), (3.3), it is denoted by $\mathfrak{J}_{2n|2m+1}, \mathfrak{J}_{2n|2m}$). \begin{align} \mathfrak{J}_{2n|2m+1} = \left( \begin{array}{c|c|c|c|c} 0 & I_n & 0 & 0 & 0 \\ \hline -I_n & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & I_m & 0 \\ \hline 0 & 0 & I_m & 0 & 0 \\ \hline 0 & 0 & 0 & 0 & 1 \end{array} \right) \end{align} \begin{align} \mathfrak{J}_{2n|2m} = \left( \begin{array}{c|c|c|c} 0 & I_n & 0 & 0 \\ \hline -I_n & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & I_m \\ \hline 0 & 0 & I_m & 0 \end{array} \right) \end{align}

Which definition is standard? Thank you very much.

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1 Answer 1

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They are both standard. It depends on your choice of non-degenerate supersymmetric bilinear form, which doesn't matter. It seems that your definition is for $SPO(2p|n)$. SPO means symplectic-orthogonal, while OSP means othogional-symplectic. Be careful with non-degenerate bilinear form: $H=diag(Q,I_n)$ in the case of $SPO(2p|n)$, and $H=diag(I_n,Q)$ in the case of $OSP(n|2p)$.

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  • $\begingroup$ thank you very much. The lower part of the matrix $H$ in both cases are different: one is diagonal matrix and the other is \begin{align} \left( \begin{array}{c|c} 0 & I_m \\ \hline I_m & 0 \end{array} \right), \end{align} or \begin{align} \left( \begin{array}{c|c|c} 0 & I_m & 0 \\ \hline I_m & 0 & 0 \\ \hline 0 & 0 & 1 \end{array} \right), \end{align} Are the coordinate ring of $SPO(2p|n)$ of the definitions in both cases the same? $\endgroup$ Mar 17, 2017 at 6:49
  • $\begingroup$ They are not the same, because you have different bilinear forms. But it doesn't matter. Before you define the SPO, you should consider which kind of bilinear form may simplify your notation or problem. $\endgroup$
    – User
    Mar 18, 2017 at 1:31

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