Questions tagged [lie-superalgebras]

The tag has no usage guidance.

Filter by
Sorted by
Tagged with
1 vote
1 answer
261 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 \...
user avatar
  • 5,885
9 votes
0 answers
170 views

Super-extensions of the Poincaré Lie algebra

For $(\mathfrak{g},[-,-])$ an ordinary Lie algebra let me say that a super-extension of it (maybe not the best terminology) is a super-Lie algebra $(\mathfrak{s}, [-,-]_{\mathfrak{s}})$ whose bosonic ...
user avatar
3 votes
0 answers
136 views

Classical Yang-Baxter equation for Lie algebras and Lie superalgebras

The classical Yang-Baxter equation is \begin{align} [r_{12}, r_{13}] + [r_{12}, r_{23}] + [r_{13}, r_{23}] = 0. \quad (1) \end{align} What are the differences between this equation in the case of Lie ...
user avatar
  • 5,885
0 votes
1 answer
90 views

References request: vector representations of Lie superalgebras

Are there some references of fundamental representations of Lie superalgebras (in particular for the Lie superalgebra $sl(m|n)$? Thank you very much.
user avatar
  • 5,885
0 votes
2 answers
634 views

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, ...
user avatar
  • 5,885
4 votes
1 answer
130 views

Solution of the Yang-Baxter equation associated to the $U_q[osp(2n+2|2m)^{(2)}]$ Lie superalgebra

I have a solution (a $R$ matrix) of the Yang-Baxter equation, \begin{equation} R_{12}(x_{1})R_{13}(x_{1}x_{2})R_{23}(x_{2})=R_{23}(x_{2})R_{13}(x_{1}x_{2})R_{12}(x_{1}) \end{equation} that probably ...
user avatar
3 votes
0 answers
125 views

Irreducible representations in BGG category $\mathcal{O}$ over (finitely) direct sum of general linear Lie superalgebra

Let $\mathfrak{g} = \oplus_i^k\mathfrak{gl}(m_i|n_i)$ be a direct sum of general linear Lie superalgebras $\mathfrak{gl}(m_i|n_i)$'s with the Cartan subalgebra $\mathfrak{h} = \oplus_i^k \mathfrak{h}...
user avatar
  • 159
4 votes
0 answers
150 views

Mirror Symmetry for Flag Supermanifolds

I recently learned the following two things, and I wish to know how to make them reconciled. (1) As far as I understand, the flag manifolds serve as a tractable class of examples for the very ...
user avatar
21 votes
1 answer
763 views

Why, in terms of quantum groups, does the knot determinant appear as an evaluation of both the Jones and Alexander polynomials?

The Jones polynomial can be computed from the representation theory of $\mathcal{U}_q(\mathfrak{sl}(2))$. The Alexander polynomial has an analogous description in terms of the representation theory of ...
user avatar
  • 1,444
1 vote
1 answer
137 views

Non-graded representations over Lie superalgebra $\mathfrak{gl}(m,n)$

I have the following questions: Let $m,n$ be positive integers. Consider representations over the general linear Lie super-algebra $\mathfrak{gl}(m,n)$. Namely, modules over the associative algebra $U(...
user avatar
  • 55
3 votes
0 answers
586 views

Orthosymplectic group, matrix representations

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 http://cds.cern.ch/record/524737/files/0110257.pdf$ where the group ...
user avatar
5 votes
1 answer
270 views

homomorphism of Lie superalgebras

In the book Shun-Jen Cheng, Weiqiang Wang Dualities and Representations of Lie Superalgebrasm. One founds the following definition(Definition 1.3): Let $\mathfrak{g}$ and $\mathfrak{g'}$ be Lie ...
user avatar
  • 81
5 votes
1 answer
613 views

Kauffman's state model for the Alexander polynomial, via representation theory

I've been reading Oleg Viro's paper on "quantum relatives of the Alexander polynomial" (arXiv:math/0204290), which, among other more general things, derives state-sum formulas for the Alexander ...
user avatar
  • 1,444
3 votes
1 answer
273 views

Linear independence in (graded) Lie algebras

I asked a mixed-up version of this question earlier. The Lie algebras I have in mind are the homotopy Lie algebras of wedges of finitely many spheres (in dimensions greater than $1$). Thus each ...
user avatar
  • 12.2k
0 votes
1 answer
317 views

Definition of the supertrace in superalgebra representations

Let us consider a matrix superalgebra $A$ with generators satisfying $[L_a,L_b]=i L_c f^c{}_{ab}.$ The generators are matrices on which supertrace is defined bu the usual trace on the bosonic part ...
user avatar
12 votes
1 answer
617 views

What are the simple Lie superalgebras of type E?

Background Simple finite dimensional Lie superalgebras over $\Bbb C$ have been classified. There are the Cartan type superalgebras (algebras of purely odd vector fields), two strange families P(n) ...
user avatar
1 vote
1 answer
386 views

Lie superalgebra in two dimensions

The standard formulation of two dimensional $N=(2,2)$ and $N=(0,2)$ supersymmetry algebras in physics is an explicit one; I am not aware of the underlying abstract Lie superalgebras. Does anyone know ...
user avatar
  • 248
1 vote
0 answers
84 views

Finite dimensional consistently graded Lie superalgebras of depth greater than 2

Victor Kac, in the paper "Classification of infinite-dimensional simple linearly compact Lie superalgebras", http://www.mat.univie.ac.at/~esiprpr/esi605.pdf writes at the beginning of section 5 (p....
user avatar
5 votes
0 answers
332 views

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 ...
user avatar
  • 3,209
1 vote
0 answers
336 views

Do all finite $W$-superalgebras have 1-dimensional representations?

Premet proved the famous KW-conjecture in modular Lie algebra. After, Premet introduced the finite $W$-algebra $U(g, e)$. Also, Premet proposed the conjecture every algebra $U(g, e)$ admits a $1$-...
user avatar
  • 63
9 votes
1 answer
1k views

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 (...
user avatar
5 votes
2 answers
361 views

Building Lie-like algebras from modules over semisimple Lie algebras

Here is a construction of a very broad class of "Lie-like" algebras, and I want to know more about them. Here is the main definition: Suppose $\mathfrak{g}$ is a complex semsimple Lie algebra and $\...
user avatar
10 votes
1 answer
382 views

Is there much theory of superalgebras acting on manifolds by alternating polyvector fields?

Usual story: vector fields on $M$, with their Lie bracket, form a Lie algebra. We can consider "actions" of some other Lie algebra ${\mathfrak g}$ on $M$ by looking at Lie homomorphisms ${\mathfrak g}\...
user avatar
12 votes
5 answers
1k views

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 ...
user avatar

1
2