Noah Snyder

Bloomington, Indiana

sbseminar.wordpress.com

Age: 35

Assistant Professor at Indiana University, working on tensor categories and their relationships to operator algebras and topology.
10h
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
13
comment Is it common to seek outside letters for the third-year renewal of an assistant professor?
The counterargument is that since these letter writers are a subset of the people who will be asked for tenure letters in a couple years, it shouldn't be a huge increase in total time spent as the time spent writing now will save time then.
May
13
comment Is it common to seek outside letters for the third-year renewal of an assistant professor?
We had it such letters for a while in math at Indiana, but don't have it anymore.
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?).
May
6
comment 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
Apr
30
awarded Nice Answer
Apr
3
awarded Great Answer
Mar
31
comment 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 3-torus is cobordant to $M \coprod M$ for some 3-manifold M.
Mar
31
comment 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 2-part.)
Mar
31
comment Nilpotence of the stable Hopf map via framed cobordism
A closely related fact, which quickly implies this one, is that the 3-torus is cobordant to a disjoint union of four copies of the 3-sphere (with its unit quaternion framing). Since that's just 3-dimensional it might be easier to see.
Mar
31
awarded Nice Question
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