User dmitri nikshych - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:21:22Z http://mathoverflow.net/feeds/user/3011 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3309/are-there-two-groups-which-are-categorically-morita-equivalent-but-only-one-of-wh/10790#10790 Answer by Dmitri Nikshych for Are there two groups which are categorically Morita equivalent but only one of which is simple Dmitri Nikshych 2010-01-05T06:11:14Z 2013-05-03T05:24:53Z <p>Hi Noah,</p> <p>Categorically Morita equivalent groups were studied by Deepak Naidu in <a href="http://arxiv.org/abs/math/0605530" rel="nofollow">arXiv:math/0605530</a>. He obtained there a complete description of Morita equivalent groups. It is also shown that simple groups are categorically Morita rigid.</p> <p>Best, Dmitri</p> http://mathoverflow.net/questions/123661/outer-automorphisms-of-borel-subgroup Outer automorphisms of Borel subgroup Dmitri Nikshych 2013-03-05T20:07:13Z 2013-03-06T05:59:26Z <p>Let $B$ be the group of upper triangular $n$-by-$n$ matrices of determinant $1$ over a finite field $F$. What is the group of outer automorphisms of $B$? More generally, what are outer automorphisms of the Borel subgroup of a finite Chevalley group?</p> http://mathoverflow.net/questions/26363/semisimple-hopf-algebras-with-commutative-character-ring/26381#26381 Answer by Dmitri Nikshych for Semisimple Hopf algebras with commutative character ring Dmitri Nikshych 2010-05-29T20:28:04Z 2010-05-29T20:28:04Z <p>Sebastian,</p> <p>No, it does not follow.</p> <p>In <a href="http://lanl.arxiv.org/abs/0704.0195" rel="nofollow">this paper (Example 6.14)</a> we proved that if a Tambara-Yamagami fusion category admits a braiding then its dimension is a power of 2. Note that a Tambara-Yamagami category has a commutative Grothendieck ring. Hopf algebras whose representation category is of Tambara-Yamagami type are classified by Tambara (Representations of tensor categories with fusion rules of self-duality for abelian groups, Isr. J. Math. 118 (2000), 29-60). For example, there is a Hopf algebra $A = k^9 \oplus M_3(k)$ (so-called Kac-Paljutkin algebra) with commutative chracater ring and $Rep(A)$ admitting no braiding. </p> http://mathoverflow.net/questions/23601/monoidal-structures-on-von-neumann-algebras/26308#26308 Answer by Dmitri Nikshych for Monoidal structures on von Neumann algebras Dmitri Nikshych 2010-05-28T20:48:33Z 2010-05-28T20:48:33Z <p>There is a construction of free product of von Neumann algebras due to Voiculescu. It is very popular among operator algebraists. Definitions can be found, e.g., in <a href="http://www.jstor.org/stable/2160912" rel="nofollow">Lance Barnett, Free Product Von Neumann Algebras of Type III</a>. I wonder if this helps.</p> http://mathoverflow.net/questions/26268/non-symmetric-braiding-on-finite-group-representation-categories/26280#26280 Answer by Dmitri Nikshych for Non-symmetric Braiding on finite group Representation Categories Dmitri Nikshych 2010-05-28T16:11:25Z 2010-05-28T16:54:39Z <p>Eric,</p> <p>The answer is no for $Rep(A_5)$ and yes for $Rep(S_4)$, thanks to Victor Ostrik's observation. For a braided category $C$ let $C'$ denote its Mueger center, i.e., the subcategory of objects $Y$ in $C$ such that the square of braiding of $Y$ with any $X$ in $C$ is identity. So $C$ is symmetric if $C=C'$ and $C$ is non-degenerate (or modular) if $C'$ is trivial.</p> <p>Note that $C:=Rep(A_5)$ is simple, i.e., it has no non-trivial proper fusion subcategories. Now if $C$ has a non-symmetric braiding then $C' \neq C$ is a proper subcategory. So $C'$ is trivial, i.e., $C$ with the above braiding is non-degenerate (modular). This cannot happen (e.g., $C$ has a simple object of dimension 5, but in a modular category the square of dimension of any object divides dimension of the category, thanks to the result of Etingof-Gelaki).</p> <p>For $D:= Rep(S_4)$ there is a non-symmetric braiding with $D'=Rep(S_3)$, namely the equivariantization of a pointed category $Vec_{Z/2Z\oplus Z/2Z}$ with respect to an action of $S_3$.</p> http://mathoverflow.net/questions/123661/outer-automorphisms-of-borel-subgroup/123702#123702 Comment by Dmitri Nikshych Dmitri Nikshych 2013-03-06T20:09:33Z 2013-03-06T20:09:33Z Thanks, Aakumadula ! Probably one can deduce that $\theta$ (modulo an inner automorphism) respects root subgroups by using that normal abelian subgroups of $B$ are in bijection with positive roots (at least for the type $A_n$). In particular, maximal normal abelian subgroups of $B$ correspond to simple roots. The automorphism $\sigma'$ of $B$ given by composition of taking transpose w.r.t (1n) -- (n1) diagonal and taking the inverse. Does this coincide with your $\sigma$? (it corresponds to the symmetry of the $A_n$ Dynkin diagram). http://mathoverflow.net/questions/123661/outer-automorphisms-of-borel-subgroup/123668#123668 Comment by Dmitri Nikshych Dmitri Nikshych 2013-03-05T21:37:46Z 2013-03-05T21:37:46Z Thanks! Sure $B$ coincides with its normalizer in $GL(n, F)$. But it appears that there are non-trivial outer automorphisms of $B$ , e.g., coming from automorphisms of $F$. Also, in this paper <a href="http://deepblue.lib.umich.edu/handle/2027.42/30473" rel="nofollow">deepblue.lib.umich.edu/handle/2027.42/30473</a> the group $Out(B)$ is computed when $F$ is the field of $2$ elements. It is surprisingly large (in this case, of course, $B$ is the group of unipotent upper-triangular matrices) http://mathoverflow.net/questions/123661/outer-automorphisms-of-borel-subgroup Comment by Dmitri Nikshych Dmitri Nikshych 2013-03-05T21:17:00Z 2013-03-05T21:17:00Z Hi Jon! nice to see you here. http://mathoverflow.net/questions/26268/non-symmetric-braiding-on-finite-group-representation-categories/26280#26280 Comment by Dmitri Nikshych Dmitri Nikshych 2010-05-28T16:49:31Z 2010-05-28T16:49:31Z Victor is right. I will edit the above answer.