bio | website | sbseminar.wordpress.com |
---|---|---|
location | Bloomington, Indiana | |
age | 35 | |
visits | member for | 5 years, 11 months |
seen | 15 hours ago | |
stats | profile views | 6,732 |
Assistant Professor at Indiana University, working on tensor categories and their relationships to operator algebras and topology.
Aug
26 |
awarded | Popular Question |
Aug
11 |
awarded | Popular Question |
Aug
2 |
awarded | Nice Answer |
Jul
22 |
answered | Symmetries of module categories over the category of representations of quantum $sl(2)$ |
Jul
22 |
comment |
Symmetries of module categories over the category of representations of quantum $sl(2)$
If your category is Vec(G) for a finite group G, then any cohomology class $H^2(G,\mathbb{C}^\times)$ will give you a tensor autoequivalence of Vec(G) whose underlying functor is trivial. (You just use the cohomology class to define the natural tranformation $1_{gh} = \mathrm{id}(1_g \otimes 1_h) \rightarrow \mathrm{id}(1_g \otimes 1_h) = 1_{gh}$.) |
Jul
22 |
comment |
Symmetries of module categories over the category of representations of quantum $sl(2)$
You need to be a little careful, just counting the automorphisms isn't enough as you might have that some of them act trivially on the Grothendieck group. |
Jul
7 |
awarded | Good Question |
Jun
5 |
comment |
What is $\infty^6$?
Doesn't it just mean the same (imprecise) thing as "There are six degrees of freedom"? |
Jun
5 |
awarded | Good Answer |
May
28 |
comment |
Is the (hyperfinite) TLJ subfactor unique at fixed index (if it exists)?
The proof has never appeared. |
May
22 |
comment |
Distinct 2D RCFTs with the same underlying MTC
More simply, any even unimodular lattice gives a CFT whose MTC is trivial. |
May
17 |
comment |
For what $G$ is $Rep(D(S_3))_{ad}$ Grothendieck equivalent to $Rep(G)$?
I don't know of any references that are particularly better than Serre. |
May
16 |
answered | For what $G$ is $Rep(D(S_3))_{ad}$ Grothendieck equivalent to $Rep(G)$? |
May
7 |
comment |
Is the space of immersions of $S^n$ into $\mathbb R^{n+1}$ simply connected?
I think there's a fibration $\Omega^n(SO(n+1))\rightarrow X \rightarrow SO(n+1)$, where X is the space of unbased maps from $S^n$ to $SO(n+1)$. So you can look at the long exact sequence attached to this fibration to try to compute $\pi_1(X)$. Since $\pi_2(SO(n+1))$ vanishes, this only has five nontrivial terms. $\star \rightarrow \pi_{n+1}(SO(n+1)) \rightarrow \pi_1(X) \rightarrow \mathbb{Z}_2 \rightarrow \pi_n(SO(n+1)) \rightarrow \pi_0(X) \rightarrow \star$. By Kervaire's table for $\pi_{n+1}(SO(n+1))$ you basically have the answer up to knowing how $\pi_1$ acts on $\pi_n$ for $SO(n+1)$ |
May
7 |
comment |
Is the space of immersions of $S^n$ into $\mathbb R^{n+1}$ simply connected?
Wait, doesn't the very next Johannson table (summarizing work of Kervaire) give the answer for larger n? $\mathbb{Z}_2^3, \mathbb{Z}_2^2, \mathbb{Z} \oplus \mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}_2^2, \mathbb{Z}_2, \mathbb{Z}_4, \mathbb{Z}$ depending on $n$ mod 8? |
May
7 |
comment |
Is the space of immersions of $S^n$ into $\mathbb R^{n+1}$ simply connected?
For n large, this is only two steps away from the stable range, so one could hope to get roughly the right answer by looking at the fiber sequence for SO twice. This question suggests that it's possible to calculate $\pi_n(\mathrm{SO}(n+1))$ exactly using it once. Using it again it seems to me you can get the answer up to at worst a factor of 2. |
May
7 |
comment |
Are the two-side TLJ subfactors maximal?
What you should have in mind here as an example of a similar flavor is $\mathbb{C} \subset M_2(\mathbb{C})$ sitting as the scalar matrices. The only intermediate $C^*$ algebra is the diagonal matrices, which has nontrivial center. This example looks exactly the same, except with gradings around that don't actually do anything important. |
May
7 |
answered | Are the two-side TLJ subfactors maximal? |
May
7 |
answered | Is the (hyperfinite) TLJ subfactor unique at fixed index (if it exists)? |
May
7 |
comment |
Is the (hyperfinite) TLJ subfactor unique at fixed index (if it exists)?
Do you mean subfactors of the hyperfinite? Otherwise what do you mean by unique, as you could just change the factor? If you do mean of the hyperfinite, then existence is open, so what you mean by uniqueness is again unclear (unique, if it exists?). |