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5 votes
1 answer
893 views

Definition of orthosymplectic supergroups

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 definitio …
Jianrong Li's user avatar
  • 6,201
3 votes
1 answer
163 views

Reference request: coordinate ring of $OSP(2p|n)$

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 …
Jianrong Li's user avatar
  • 6,201
1 vote
1 answer
395 views

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 \ …
Jianrong Li's user avatar
  • 6,201
1 vote
1 answer
189 views

Construct super Poisson brackets on the coordinate rings of Lie super groups

On line 7 of page 61 of the book a guide to quantum groups, a Poisson bracket is defined on $\mathbb{C}[GL_n]$ for every classical $r$-matrix as follows. Let $V$ be a vector space with a basis $v_1, …
Jianrong Li's user avatar
  • 6,201
0 votes
Accepted

How to prove a bracket is super anti-commutative?

This question was solved by Vladimir Dotsenko in the comments of the question.
Jianrong Li's user avatar
  • 6,201
0 votes
1 answer
185 views

How to prove a bracket is super anti-commutative?

On page 12 of the paper, there is a formula about super Poisson bracket on a Lie super group $G$: \begin{align} \{\phi, \psi\} = \sum_{\mu, \nu} (-1)^{|\phi||\nu|} r^{\mu \nu} ( R_{\mu} \phi R_{\nu} \ …
Jianrong Li's user avatar
  • 6,201