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  Orthosymplectic group, matrix representations

+ 4 like - 0 dislike

We have the orthosymplectic $osp(n,m|2k)$. The bosonic part is $so(n,m)\times sp(2k)$. The lie algebra generators are given in eg


where the group is described as a supermatrix acting on super space. For the groups $so(n,m)$ it is easy to go from the lie algebra to the group by exponentiating, these group elements will just be Lorentz transformations in a signature $(n,m)$ space time. Also the same can be done for $sp(2k)$. However, when including the anticommuting generators, is it possible to go from the lie algebra generators to the group by exponentiating? Is there a supermatrix representation of this group?

This post imported from StackExchange MathOverflow at 2015-03-25 13:39 (UTC), posted by SE-user user2133437

asked Jan 11, 2015 in Mathematics by user2133437 (55 points) [ revision history ]
edited Mar 25, 2015 by dimension10
There can be odd generators with odd parameters the result is an even element of the superalgebra where the exponential is defined as usual. These conventions are used, for example, to calculate the usual super Poincare transformations on $\mathbb{R}^{4|4}$. The odd SUSY generator is paired with a fermionic parameter in that calculation. One nice place to read about this would be Ideas and Methods of Supersymmetry and Supergravity: Or a Walk Through Superspace by Kuzenko and Buchbinder.

This post imported from StackExchange MathOverflow at 2015-03-25 13:39 (UTC), posted by SE-user James S. Cook

One doesn't get a group but a supergroup, which is rather a sort of algebra. See, e.g., here or here.

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