Questions tagged [lie-superalgebras]
19 questions
5
votes
1 answer
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 $H^1(\mathfrak g, M)$ (defined for example as cohomology of the…
Blazej
- 3,200
5
votes
0 answers
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 relations of the generators $H_i,E_i$ and $F_i$ which…
Buzz Lee
- 51
4
votes
0 answers
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 $\mathfrak{g}_0=\mathfrak{so}(m)\times\mathfrak{sp}(2n)$. Take…
freeRmodule
- 1,892
3
votes
0 answers
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 exterior algebra of degree $r\in\mathbb Z$ if it maps…
Bence Racskó
- 7,919
2
votes
0 answers
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 $c_M$. We also have representation $\widehat \rho$ of…
Blazej
- 3,200
2
votes
0 answers
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 some point. There is no word about how to obtain…
P. Wegener
- 121
2
votes
0 answers
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 representations $\Lambda_i$ with $i=1,\cdots…
Stefano
- 21
1
vote
0 answers
On real and imaginary roots in Kac–Moody superalgebras (following Serganova)
I am reading the paper “Kac–Moody superalgebras and integrability” by Serganova (2011), which is highly relevant to my ongoing research. Properly understanding real roots of a Kac–Moody superalgebra $\mathfrak{g}$ (see Definition 4.1 of the paper)…
Irfan
- 322
1
vote
0 answers
Relation between higher and lower weight ans simplicity
I'm studying the article of Kac "Lie Superalgebras" (1974) and in several times he use the fact that if $L$ is a semisimple Lie Algebra, $V$ its faithful, irreducible and finite-dimensional module with $\lambda$ highest weight and $\mu$ lowest…
1
vote
1 answer
Simple quotient of Verma Module $ M_\mathfrak{b}(\lambda) $
I'm starting to look into Lie Superalgebras and understand a bit more about simple f.d. modules.
Right now I'm studying fairly simple examples to try and get a feel for it.
I'm trying to find the simple modules of $ \mathfrak{gl}(1|1) $ and I…
MrP
- 41
1
vote
0 answers
Lie superalgebra sl(m|n)
I know sl(m|n) is the A-series Lie superalgebra by Kac
But in some literature, people also use the terminology su(m|n), which seems to be subclass of A-series.
I am wondering how these two are related in general? For example, representeation theory…
jtkw
- 109
1
vote
0 answers
Question about a specific functor in Lie superalgebras
Given any superassociative algebra, a Lie superalgebra with a Lie superbracket $[a,b]=ab- (-1)^{|a||b|} ba$ is constructed and satisfies
$[a,b]=-(-1)^{|a||b|}[b,a]$
$[a,[b,c]] = [[a,b],c] + (−1)^{|a||b}[b,[a,c]].$
Let we have a parity reversion…
Nil
- 1,336
1
vote
0 answers
Is the exterior algebra a universal enveloping superalgebra?
Let $V$ be a vector space and $\Lambda(V)$ the exterior algebra of $V$. Is $\Lambda(V)$ the universal enveloping superalgebra of some Lie superalgebra? If so, which?
Bipolar Minds
- 1,602
1
vote
0 answers
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 weight. It follows from the Borel-Weil-Bott theorem…
NongAm
- 427
1
vote
1 answer
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 homogeneous), If anybody can help me with finding the…
GA316
- 4,522