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144108
bio website sbseminar.wordpress.com
location Bloomington, Indiana
age 35
visits member for 5 years, 11 months
seen 15 hours ago
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?).