Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

Questions tagged [lie-superalgebras]

The tag has no usage guidance.

1
vote
0answers
20 views

Coordinate superalgebra

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 ...
0
votes
1answer
98 views

super commutation of matrices

Let $M_{p|q}(\mathbb{C}) = M_{p|q}(\mathbb{C})_0 \oplus M_{p|q}(\mathbb{C})_1$ be the super algebra of all $(p+q) \times (p+q)$ matrices. Let $A, B \in M_{p|q}(\mathbb{C})$ (not necessarily ...
2
votes
0answers
64 views

General commutators of derivations of the exterior algebra

Let $M$ be a smooth manifold and let $\Omega(M)$ be the exterior algebra of smooth differential forms over $M$. The $\mathbb R$-linear map $D:\Omega(M)\rightarrow\Omega(M)$ is a derivation of the ...
2
votes
0answers
25 views

Action of quadratic Casimir on cochain/cohomology spaces

Let $\mathfrak g$ be a semisimple Lie algebra and let $M$ be a nontrivial simple $\mathfrak g$-module. Then quadratic Casimir of $\mathfrak g$ acts on $M$ as multiplication by a nonzero scalar $c_M$. ...
5
votes
1answer
57 views

Is Whitehead lemma true for super Lie algebras?

Classical Whitehead lemma states that if $\mathfrak g$ is a finite-dimensional complex Lie algebra and $M$ is a finite-dimensional $\mathfrak g$-module, then first cohomology group $H^1(\mathfrak g, M)...
0
votes
1answer
25 views

Semidirect product of Lie superalgebras

I am studying Lie superalgebrs and need the concept of semidirect product for the case of Lie superalgebras. May you please give me some information on that.
1
vote
0answers
50 views

How to obtain all classical simple Lie superalgebras of rank 2?

I'm reading Musson's book on "Lie superalgebras and Enveloping Algebras" as well Kac' Advances paper on this subject. In both references the rank 2 classical simple Lie superalgebras just appear at ...
2
votes
0answers
30 views

What is contragredient about contragrediant Lie superalgebras

Among the Lie superalgebras there is the class of contragredient Lie superalgebras. Roughly speaking these are those Lie superalgebras that can be defined with a matrix $a_{ij}$ and commutation ...
1
vote
0answers
55 views

Fundamental representations of Lie superalgebras

I have the following question concerning the topic of highest weight representations of a Lie superalgebra $\mathfrak g$. In the case of standard Lie algebras, there exists the concept of fundamental ...
4
votes
0answers
53 views

Finite-dimensionality of modules over simple Lie superalgebras

Let $\mathfrak{g}=\mathfrak{g}_0\oplus\mathfrak{g}_1$ be a simple Lie superalgebra, and for simplicity assume $\mathfrak{g}=\mathfrak{osp}(m|2n)$, so that $\mathfrak{g}_0=\mathfrak{so}(m)\times\...
1
vote
2answers
76 views

What is the commutator in $gl(m|n)$?

For Lie algebra $gl(m)$, the commutator is \begin{align} [E_{ij}, E_{kl}] = \delta_{jk}E_{il} - \delta_{li}E_{kj}. \end{align} What is the commutator $[E_{ij}, E_{kl}]$ in Lie superalgebra $gl(m|n)$? ...