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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}

Are there some references which describe the coordinate ring of $OSP(2p|n)$ explicitly? In the case of $OSP(2|1)$, its coordinate ring is described in the paper, page 15, equation (3).

Thank you very much.

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  • $\begingroup$ Look at Section 3 on page 7 in arxiv.org/pdf/0706.0348.pdf by Bin Shu and Weiqiang Wang, which may be helpful. $\endgroup$ – Mee Seong Im Mar 11 '17 at 5:01
  • $\begingroup$ @MeeSeongIm, thank you very much. It seems that the matrix $H$ ( (59) of the paper ) is different from the matrix (3.2) in the paper. I am confused. $\endgroup$ – Jianrong Li Mar 11 '17 at 8:09
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Since the orthosymplectic supergroup $G=\mathrm{OSP}(m|n)$ is affine algebraic (indeed, this is a Chevalley supergroup), you can apply results in Masuoka's paper ``Harish-Chandra pairs for algebraic affine supergroup schemes over an arbitrary field'' (see Proposition 4.5 in https://arxiv.org/abs/1111.2387).

For an affine algebraic supergroup scheme $G$ over a field, we get the following data: (1) the (ordinary) affine algebraic group scheme $G_{\mathrm{res}}$ associated with $G$, and (2) the Lie superalgebra $\mathrm{Lie}(G)$ of $G$. The answer of your question is that there is an isomorphism of superalgebras from the coordinate ring $\mathcal{O}(G)$ of $G$ to the superalgebra $\mathcal{O}(G_{\mathrm{res}}) \otimes \bigwedge (\mathrm{Lie}(G)_1)^*$, where $\square^*$ denotes the linear dual of $\square$ and $\bigwedge(\square)$ denotes the exterior algebra over $\square$.

For general Chevalley supergroups, you can find a lot of results in "Chevalley supergroups" by R.~Fioresi and F.~Gavarini (https://arxiv.org/abs/0808.0785) and in "Algebraic supergroups and Harish-Chandra pairs over a commutative ring" by A.~Masuoka and T.~Shibata (Section 6 in https://arxiv.org/abs/1304.0531).

Good luck!

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