Questions tagged [lie-superalgebras]
The lie-superalgebras tag has no usage guidance.
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Is there a definition of analogue Weyl group for Lie super algebra?
I heard from some people working in Lie super algebra that there was no proper definition for Weyl group of Lie super algebra. I do not know Lie super algebra at all. But When I searched on Google, I ...
9
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I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt
Question: I am talking about the proof given on pages 50-52 of Pierre Deligne, Pavel Etingof, Daniel S. Freed, Lisa C. Jeffrey, David Kazhdan, John W. Morgan, David R. Morrison, and Edward Witten (...
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Serre relations for Lie Superalgebras
Finite dimensional complex simple Lie algebras are classified using Cartan matrices. One of the main ingredients is Serre Relations. Lets call this Cartan-Killing theory.
I have the following ...
5
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Is the SUSY Algebra isomorphic for all Kähler Manifolds?
For a Kähler manifold, the graded algebra generated by $\partial,\overline{\partial},\partial^*,\overline{\partial}^\ast$, the Lefschetz operator, and the dual Lefschetz operator, is called the ...
2
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Coordinate ring of an equivariant embedding of a homogeneous projective variety
Lie algebra: Let $G$ be a semisimple, simply connected linear algebraic group with a fixed Borel subgroup $B$. Let $P$ be a parabolic subgroup containing $B$. Let $\lambda$ be a dominant integral ...
2
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Two definitions of the super Jacobi identity
In this paper, page 149, the super Jacobi identity is given by
\begin{align}
J(x, y,z) := (-1)^{|x||z|}[[x, y],z] +(-1)^{|z||y|}[[z,x], y]+(-1)^{|y||x|}[[y,z],x] = 0.
\end{align}
But in this article, ...
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Super version of Poisson brackets of tensor products
Let $A$ be a Poisson super algebra ($A$ is a super algebra and $A$ satisfies super Jacobi identity, super commutativity, super Leibniz rule).
Super version of the product of two tensor products is
\...