Is there a definition of analogue Weyl group for Lie super algebra? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:48:38Z http://mathoverflow.net/feeds/question/26304 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26304/is-there-a-definition-of-analogue-weyl-group-for-lie-super-algebra Is there a definition of analogue Weyl group for Lie super algebra? Shizhuo Zhang 2010-05-28T20:02:13Z 2011-08-29T21:57:04Z <p>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 found that it seems there still exists some definitions of Weyl group. </p> <p>I wonder whether there is a well-accepted definition for it. </p> <p>The reason I want to ask this question is that I need Weyl group for Lie super algebra to play with the geometry related to super Lie algebra. </p> <p>Another question is that I heard from some experts in Lie super algebra that there was no well-accepted super geometry related to Lie super algebra. </p> <p>However, It seems that one of the students of Manin, who is Dimitry Leites ever developed supergeometry. </p> http://mathoverflow.net/questions/26304/is-there-a-definition-of-analogue-weyl-group-for-lie-super-algebra/26311#26311 Answer by JosÃ© Figueroa-O'Farrill for Is there a definition of analogue Weyl group for Lie super algebra? JosÃ© Figueroa-O'Farrill 2010-05-28T22:04:48Z 2010-05-28T22:04:48Z <p>The answer to the question in the title is affirmative. In the <a href="http://arxiv.org/abs/hep-th/9607161v1" rel="nofollow">Dictionary of Lie superalgebras</a>, there is an entry on the Weyl group of a classical Lie superalgebra. It is generated by reflections associated to the simple even roots, hence it is the standard Weyl group of the even subalgebra. In addition, they also mention that one can extend the Weyl group by the addition of so-called generalised Weyl transformations associated to the odd roots. They also give a couple of references.</p> <p>As for the geometry associated to Lie superalgebras, there is a notion of <a href="http://en.wikipedia.org/wiki/Supergroup_%28physics%29" rel="nofollow">Lie supergroup</a> (this link is the not-particularly-good wikipedia article), which stands in relation to Lie superalgebras just as their non-super counterparts. Lie supergroups are particular examples of <a href="http://en.wikipedia.org/wiki/Supermanifold" rel="nofollow">supermanifolds</a>, on which there is a substantial literature.</p> http://mathoverflow.net/questions/26304/is-there-a-definition-of-analogue-weyl-group-for-lie-super-algebra/26332#26332 Answer by S. Carnahan for Is there a definition of analogue Weyl group for Lie super algebra? S. Carnahan 2010-05-29T04:42:48Z 2010-05-29T05:16:22Z <p>I'm not an expert in this area, but I'm told that the key phrase in the superalgebra world is "Weyl groupoid" rather than Weyl group. <strike>I did not look at the construction long enough to understand it.</strike> Serganova has a <a href="http://math.berkeley.edu/~serganov/papers/superk.pdf" rel="nofollow">paper</a> describing foundations in a super analogue of the Kac-Moody setting, and you can find a description of the Weyl groupoid there.</p> <p><b>Edit:</b> The situation seems to be the following: For Kac-Moody algebras, there is a unique conjugacy class of Cartan subalgebra (under automorphisms), and the Weyl group acts transitively on systems of simple roots. These properties fail to hold in the superalgebra setting. One instead can form a groupoid whose objects are finite size square matrices $A$ with integer entries (or the Lie superalgebras $g(A)$ obtained by a generators-and-relations construction), and whose morphisms from $A$ to $A'$ are superalgebra isomorphisms $g(A) \to g(A')$ that take a Cartan of $g(A)$ to a Cartan of $g(A')$. The Weyl groupoid of $g(A)$ is then the connected component of $A$ in the larger groupoid.</p> <p>Regarding geometry, I think Penkov has done some work with flag supermanifolds and Borel-Weil-Bott. I don't think there is much debate about the foundations of supermanifold theory, but I guess the geometric representation theory doesn't extend from the even case by rote translation of proofs.</p> http://mathoverflow.net/questions/26304/is-there-a-definition-of-analogue-weyl-group-for-lie-super-algebra/74006#74006 Answer by Sasha Voronov for Is there a definition of analogue Weyl group for Lie super algebra? Sasha Voronov 2011-08-29T21:57:04Z 2011-08-29T21:57:04Z <p>From the point of view of super geometry, Manin and I introduced the notion of a super Weyl group in relation to the geometry of homogeneous superspaces. We constructed Schubert supercells which were labeled by elements of a super Weyl group. I am not sure whether this is a well-accepted definition; it was rather a construction done in an ad hoc way for each classical simple Lie supergroup. The results were announced in </p> <p>Manin, Y. I.; Voronov, A. A. Schubert supercells. Functional Anal. Appl. 18, 329-330 (1985).</p> <p>Voronov, A. A. Relative disposition of the Schubert supervarieties and resolution of their singularities. Functional Anal. Appl. 21, 62-64 (1987).</p> <p>and published in detail in</p> <p>Manin, Y. I.; Voronov, A. A. Supercellular partitions of flag superspaces. Current problems in mathematics. Newest results, USSR Acad. Sci., Moscow. 32, 27-70 (1988). (in Russian). English translation: J. Soviet Math. 51(1), 2083-2108 (1990).</p>