bio  website  sbseminar.wordpress.com 

location  Bloomington, Indiana  
age  35  
visits  member for  5 years, 9 months 
seen  15 hours ago  
stats  profile views  6,659 
Assistant Professor at Indiana University, working on tensor categories and their relationships to operator algebras and topology.
2h

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 
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Are the twoside 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 twoside 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?). 
May 6 
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Intuition behind the definition of quantum groups
I knew this story for SL(2) from Kassel's book, but I had been under the impression that there wasn't such a nice story for higher SL(n). But you seem to be saying there is such a story. Where can I read about it? 
May 6 
awarded  Nice Answer 
Mar 31 
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Nilpotence of the stable Hopf map via framed cobordism
The 12 seems to big to be easily geometrically explained, but it's enough to just show that the 3torus is cobordant to $M \coprod M$ for some 3manifold M. 
Mar 31 
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Nilpotence of the stable Hopf map via framed cobordism
@QiaochuYuan: You're right, I forgot that there were primes other than 2. (Or rather, I was looking at the picture for just the 2part.) 
Mar 31 
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Nilpotence of the stable Hopf map via framed cobordism
A closely related fact, which quickly implies this one, is that the 3torus is cobordant to a disjoint union of four copies of the 3sphere (with its unit quaternion framing). Since that's just 3dimensional it might be easier to see. 
Mar 31 
awarded  Nice Question 