13,301 reputation
139101
bio website sbseminar.wordpress.com
location Bloomington, Indiana
age 34
visits member for 5 years, 2 months
seen 1 hour ago
Assistant Professor at Indiana University, working on tensor categories and their relationships to operator algebras and topology.

Nov
10
comment In Algebraic Compact Quantum Groups, is an Irreducible Corepresentation equivalent to its Conjugate?
No problem. For that the easiest example is the trivial corep which is always selfdual.
Nov
6
comment In Algebraic Compact Quantum Groups, is an Irreducible Corepresentation equivalent to its Conjugate?
@JpMcCarthy: Ok, I rewrote it with more detail.
Nov
6
revised In Algebraic Compact Quantum Groups, is an Irreducible Corepresentation equivalent to its Conjugate?
More detail
Nov
5
answered In Algebraic Compact Quantum Groups, is an Irreducible Corepresentation equivalent to its Conjugate?
Oct
16
awarded  Necromancer
Oct
8
answered No basis change in a fusion ring allowed?
Sep
30
awarded  Explainer
Sep
29
awarded  Yearling
Sep
17
comment Does an equivalence of fusion categories depend on choice of simple objects within isomorphism classes?
p.s. Let's talk sometime this week. I started looking at our draft again today.
Sep
17
answered Does an equivalence of fusion categories depend on choice of simple objects within isomorphism classes?
Sep
16
comment Square roots of elements in a finite group and representation theory
@FriederLadisch: I think the explanation for why this happens often, is that you do get that multiplicity-free summands of tensor products behave as predicted. So if you have enough multiplicity free summands in your tensor products then you'll get a FS grading.
Sep
4
comment When is Rep(U_q(g)) invariant under q -> -q and why?
With the "usual" conventions the dimension of the 2d rep of $U_q(\mathfrak{sl}_2)$ is $q+q^{−1}$. Your formula has two changes. There's a variable $s$ with $s^L=q$ where $L$ is the index of the root lattice in the weight lattice (so $L=2$ for $\mathfrak{sl}_2$). You need this $s$ to write down the braiding. Your q is s, which explains the appearance of $q^2$ in your formula. The minus sign is coming because TL is not the "usual" pivotal structure (since it's real instead of quaternionic). (The latter point is not important since changing piv. str. won't affect whether Rep is symmetric.)
Sep
4
revised When is Rep(U_q(g)) invariant under q -> -q and why?
added 14 characters in body
Sep
4
asked When is Rep(U_q(g)) invariant under q -> -q and why?
Aug
30
revised Does the notion of a “coherent state” exist in TQFTs? (ETQFTs?)
added 7 characters in body
Aug
10
awarded  Nice Answer
Jul
16
comment How weird can Modular Tensor Categories be over non-algebraically closed fields?
There's a bunch of great stuff related to Andre's comment in this short paper of Greg's.
Jul
16
comment How weird can Modular Tensor Categories be over non-algebraically closed fields?
Modularity should be thought of as being as far from symmetry as possible. In the symmetric case the S-matrix has rank 1, and in the modular case it has full rank. On the other hand, Rep(G) is contained in its center which is modular, so @DavidSpeyer's example still works.
Jul
6
comment When is/isn't the monoidal unit compact projective?
Very minor point, for A-mod-A projectivity is separability of A which in enough generality is stronger than semisimplicity (though they agree for algebras over a perfect field).
Jul
6
answered Rank vanishing in tensor categories