In the Virasoro algebra, which is generated by $L_n$, one has the obvious subalgebra spanned by $L_{-1}$ ,$L_{1}$ and $L_{0}$ which is isomorphic to the Lie algebra $\mathfrak{sl}(2,\mathbb{R})$.

The Neveu schwarz super virasoro algebra, as defined in http://en.wikipedia.org/wiki/Super_Virasoro_algebra, is generated by $L_n$ and $G_r$ with $r$ half integer. Here we also have a subalgebra if we restrict to $L_0$, $L_1$ and $L_{-1}$ and $G_{\pm\frac{1}{2}}$.

My question is, what is this algebra called? Does it also have a (super)matrix representation, that naturally extends $\mathfrak{sl}(2,\mathbb{R})$?


The (super-)algebra you are referring to is called $\mathfrak{osp}(1,2)$, where osp stands for orthosymplectic. I am not sure about the matrix representation, but a google search about "orthosymplectic superalgebra" will give you plenty of references.

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  • $\begingroup$ Thanks. It seems like this lie algebra can be represented as a supermatrix (see arxiv.org/pdf/1205.0119v3.pdf page 3). What would the associated group be? I guess I have to exponentiate the lie algebra, but I am wondering if there is a simpler definition of the associated group? $\endgroup$ – Jonathan Lindgren Jan 14 '15 at 13:12
  • $\begingroup$ For example when exponentiating sl(2,R) we would obtain SL(2,R) which can be defined as all matrices with unit determinant, and I am looking for something similar for the OSP group. $\endgroup$ – Jonathan Lindgren Jan 14 '15 at 13:36
  • $\begingroup$ Oh, is it just the same condition as for the symplectic group in en.wikipedia.org/wiki/Symplectic_group, but with trace replaced by supertrace and $\Omega$ replaced by g in arxiv.org/pdf/1205.0119v3.pdf ? $\endgroup$ – Jonathan Lindgren Jan 14 '15 at 13:47
  • $\begingroup$ Sorry for writing so many comments, but are you sure that osp(1,2) is the correct algebra? In arxiv.org/pdf/1205.0119v3.pdf, where it is represented as a super matrix, the subalgebra seems to be 2x2 symplectic/orthogonal matrices, NOT 2x2 matrices with unit determinant ? $\endgroup$ – Jonathan Lindgren Jan 14 '15 at 13:53
  • $\begingroup$ Sorry forget my last comment, I missed that SL(2,R)=Sp(2,R), not SO(2,R). $\endgroup$ – Jonathan Lindgren Jan 14 '15 at 14:03

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