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438146
bio website math.berkeley.edu/~theojf
location Berkeley
age 29
visits member for 4 years, 11 months
seen 7 hours ago

I am a recent graduate from UC Berkeley. Starting Fall 2013, I am a Boas Assistant Professor at Northwestern University. My research is broadly centered on quantum field theory — my interests include category theory, representation theory, homological algebra, algebraic topology, Poisson geometry, and theoretical physics.


2d
awarded  Enlightened
2d
awarded  Nice Answer
Sep
8
comment Bi-dual of vector spaces
As @YemonChoi correctly says, without further assumptions you cannot prove the map to be surjective, because it's not true in general. Thus the fact alone that vector spaces form a category is not enough to prove injectivity, since otherwise the proof would apply to the opposite category to prove surjectivity. More damningly, the map can fail to be injective for modules over a ring, so somehow your category needs to "know" that it's over a field to prove injectivity.
Sep
6
comment Recognize this strange expression from linear algebra?
No offense taken. Sorry about misstating the end.
Sep
4
comment When is Rep(U_q(g)) invariant under q -> -q and why?
I don't know if the conventions match yours, but Temperley–Lieb with the circle valued at $\delta = -q^2 - q^{-2}$ is invariant us a monoidal category under $q \mapsto -q$ (of course, since the monoidal category only depends on $q$ via $\delta$), and even as a braided category (via the natural isomorphism that acts by $-1$ on the defining object), but not as a pivotal category.
Sep
4
revised Recognize this strange expression from linear algebra?
admitted an error.
Sep
4
revised Recognize this strange expression from linear algebra?
corrected a 0 to a 2
Sep
3
awarded  Enlightened
Sep
3
awarded  Nice Answer
Sep
3
answered Recognize this strange expression from linear algebra?
Sep
2
comment Recognize this strange expression from linear algebra?
@SteveHuntsman Maybe $g_{kl}$ in place of $c_{kl}$?
Sep
2
comment Are linear algebraic groups rigid?
I don't really know the general definition of "reductive", but is this $G$ reductive "over $k[t]$"? There are many examples of reductive groups degenerating to solvable groups, of which this is a particularly nice one.
Aug
12
comment Hyperfinite type II_1 factor as the Clifford algebra
I am far from an expert. My impression is that "the hyperfinite II_1 factor" has lots of automorphisms, and different manifestations of it, although isomorphic, often are not canonically isomorphic. The classification of factors only says that a given one is isomorphic to the hyperfinite II_1 factor, and doesn't tell you that they are really "the same" in any meaningful way. But I repeat: I am far from an expert.
Aug
9
comment What's the state of affairs concerning the identification between quantum group reps at root of unity, and positive energy affine Lie algebra reps?
Great question! I have nothing useful to contribute towards an answer.
Aug
2
awarded  Popular Question
Aug
2
comment String diagrams for bimodules over noncommutative algebras?
@DavidRoberts: Sorry for the slow response. Yes, of course Joyal--Street --- I clearly wasn't thinking straight, and went "Street was one of the names ... what's the name that goes with Street?"
Jul
28
answered String diagrams for bimodules over noncommutative algebras?
Jul
22
comment Removing an article from arxiv
Never use that journal.
Jul
22
comment When does a monoidal functor between ribbon categories preserve cups and caps, but not necessarily braidings?
Right. The strong monoidal functor from supervector spaces to $(\mathbb Z/2)$-modules (with the usual symmetric structure) does not preserve quantum dimension. And any two (right, say) duals are canonically isomorphic, but that isomorphism often is not the identity for some looks-convenient coordinates.
Jul
14
comment Are there isomeasure simplices?
I assume you are familiar with Schanuel's excellent paper "What is the length of a potato?", but your notion of isomeasure reminded me of it, so in the off chance you don't know that paper, I thought I'd mention it.