| Full text | |
| Author(s): |
Total Authors: 2
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| Affiliation: | [1] Univ Sao Paulo, Dept Matemat, Caixa Postal 66281, BR-05315970 Sao Paulo - Brazil
[2] Omsk State Polytech Univ, Mira 11, Omsk 644050 - Russia
Total Affiliations: 2
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| Document type: | Journal article |
| Source: | Journal of Algebra; v. 452, p. 448-473, APR 15 2016. |
| Web of Science Citations: | 4 |
| Abstract | |
It is well known that if the ground field K has characteristic zero and G is a connected algebraic group, defined over K, then the Lie algebra Lie(G') of the commutant G' of G coincides with the commutant Lie(G)' of Lie(G). We show that this result is no longer true in the category of algebraic supergroups. We also construct a reductive supergroup H = X G, where X and G are connected, reduced and abelian supergroups, such that X-u not equal 1 and (H-ev)(u) is non-trivial connected (super)group. Quasi-reductive supergroups have been introduced in {[}10]. We prove that a supergroup H is quasi-reductive if and only if the largest even (super)subgroup of the solvable radical R(H) is a torus, (H) over tilde = H/R(H) contains a normal supersubgroup U, which is quasi-isomorphic to a direct product of normal supersubgroups U-i, and (H) over tilde /U is a triangulizable supergroup with odd unipotent radical. Moreover, for every i, Lie(U-i) = U-i circle times Sym(n(i)) are such that either n(i) = 0 and U-i is a classical simple Lie superalgebra, or n(i) = 1 and U-i is a simple Lie algebra. (C) 2015 Published by Elsevier Inc. (AU) | |