User dmitri nikshych - MathOverflow

Answer by Dmitri Nikshych for Are there two groups which are categorically Morita equivalent but only one of which is simple

2010-01-05T06:11:14Z

Hi Noah,

Categorically Morita equivalent groups were studied by Deepak Naidu in arXiv:math/0605530. He obtained there a complete description of Morita equivalent groups. It is also shown that simple groups are categorically Morita rigid.

Best, Dmitri

Outer automorphisms of Borel subgroup

2013-03-05T20:07:13Z

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?

Answer by Dmitri Nikshych for Semisimple Hopf algebras with commutative character ring

2010-05-29T20:28:04Z

Sebastian,

No, it does not follow.

In this paper (Example 6.14) 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.

Answer by Dmitri Nikshych for Monoidal structures on von Neumann algebras

2010-05-28T20:48:33Z

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 Lance Barnett, Free Product Von Neumann Algebras of Type III. I wonder if this helps.

Answer by Dmitri Nikshych for Non-symmetric Braiding on finite group Representation Categories

2010-05-28T16:11:25Z

Eric,

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.

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).

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 $S_3$.

Comment by Dmitri Nikshych

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).

Comment by Dmitri Nikshych

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 deepblue.lib.umich.edu/handle/2027.42/30473 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)

Comment by Dmitri Nikshych

2013-03-05T21:17:00Z

Hi Jon! nice to see you here.

Comment by Dmitri Nikshych

2010-05-28T16:49:31Z

Victor is right. I will edit the above answer.