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bio website math.berkeley.edu/~theojf
location Berkeley
age 29
visits member for 5 years, 6 months
seen 6 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.


1d
awarded  Popular Question
2d
comment Are lax functor categories into a cartesian closed 2-category cartesian closed?
Well, I misread, and didn't see you wanted cartesian closed categories, rather than just cartesian categories. But I think via a "functor of points" approach the same coordinate-full calculation works.
2d
answered Are lax functor categories into a cartesian closed 2-category cartesian closed?
Apr
21
awarded  Nice Answer
Apr
21
comment Example s.t. the unbased loop-space is not $\Omega X \times X$
The end result being that X = figure eight provides the example @Jens is looking for.
Apr
17
accepted Given a map of classifying spaces, can the target be described as a groupoid quotient of the source mod some action of some (co)kernel?
Apr
15
awarded  Nice Answer
Apr
14
comment Given a map of classifying spaces, can the target be described as a groupoid quotient of the source mod some action of some (co)kernel?
Oh, these formulae certainly make sense for (derived?) algebraic stacks. Do you know if they are known to be "correct" in the algebrogeometric world? E.g. I really do want to work with the algebraic groups G,H qua schemes, and not just work with the topological groups of $\mathbb C$-points.
Apr
14
comment Given a map of classifying spaces, can the target be described as a groupoid quotient of the source mod some action of some (co)kernel?
Awesome. I'll have to think a bit to make sure I understand this formula. Is it clear that this double coset groupoid is groupal, so that I can take B of it? Or on the right did you just mean the space corresponding to the groupoid $G // (H \times H)$?
Apr
14
asked Given a map of classifying spaces, can the target be described as a groupoid quotient of the source mod some action of some (co)kernel?
Apr
7
comment What's the cardinality of a higher category?
By calling this number "Euler characteristic", you also learn what to do (in some cases) when the homotopy groups are infinite or there are infinitely many of them (if you work with $\infty$-groupoids aka homotopy types).
Mar
27
comment In set theory, is there a name for a function which maps the empty set to zero and all the others to one?
As @Tim says, this function is quite important, and shows up a lot. I had no trouble understanding your question. $0^{0^{|S|}}$ is a horrible name, although a great formula. I would call your function "$\pi_{-1}$", since "$k$-truncation" in the homotopy type theory sense is "$\pi_k$", the $k$th fundamental groupoid. (For _pointed_ spaces, by "$\pi_k$" you might mean the $k$th fundamental _group_. But for unpointed spaces, $\pi_0 = $ "connected components", and $\pi_1$ should mean the fundamental groupoid.)
Mar
20
comment Are $(\infty,1)$-categories $A_\infty$ categories?
@Dmitri Awesome, I'll take a look.
Mar
20
comment Commutation of tensor products with inverse limits in a specific case
That's the one.
Mar
19
comment Commutation of tensor products with inverse limits in a specific case
As user74230 suggested in an answer below, I assume you mean $\otimes = \otimes_R$? Then there is a paper by Goodearl --- I am traveling and don't remember a more precise reference --- that studies more generally the map $M \otimes_R \prod_i N_i \to \prod_i(M\otimes N_i)$. My memory is that it is always injective when $R$ is Noetherian, but at that level of generality injectivity can fail when $R$ is not Neotherian. Actually, I think the failure is witnessed by modules that are isomorphic to $R^X$ for some $X$.
Mar
4
answered Establishing Duality in Tannakian Categories
Mar
4
awarded  Enlightened
Mar
4
awarded  Nice Answer
Mar
2
awarded  Notable Question
Feb
24
awarded  Nice Answer