bio | website | sbseminar.wordpress.com |
---|---|---|
location | Bloomington, Indiana | |
age | 34 | |
visits | member for | 5 years, 2 months |
seen | 1 hour ago | |
stats | profile views | 6,205 |
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 |