# Questions tagged [lie-superalgebras]

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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 \...
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### 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 ...
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### 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 ...
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### 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.
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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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### 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 ...
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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$where the group ... 5 votes 1 answer 270 views ### homomorphism of Lie superalgebras In the book Shun-Jen Cheng, Weiqiang Wang Dualities and Representations of Lie Superalgebrasm. One founds the following definition(Definition 1.3): Let$\mathfrak{g}$and$\mathfrak{g'}$be Lie ... 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 ... 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 ... 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 ... 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) ... 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 ... 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.... 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 ... 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$1$-... 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 (... 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$\...
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}\...