20,462 reputation
449158
bio website math.berkeley.edu/~theojf
location Berkeley
age 29
visits member for 5 years, 5 months
seen Mar 29 at 15:07

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.


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
Feb
23
comment Free Loop-Space Recognition Principle
Being a bit cavalier with hom-tensor adjunctions, I guess this is asking: give $G = S^1$ the trivial left $G$ action; then when is $\mathrm{maps}_G(G,X) \simeq X$ as $G$-spaces, where the $G$-action on $\mathrm{maps}_G(G,X)$ is via the usual (right) $G$-action on $G$. I am reminded of the following characterization: a monoid $M$ is a group iff the diagonal $M$ action on $M \times M$ is isomorphic (via some probably-non-identity map on $M \times M$) to the action which acts just on the left factor and trivially on the right.
Feb
23
comment Free Loop-Space Recognition Principle
Great question. Some thoughts: (1) the free loop space of $X$ carries a $\mathrm{Diff}(S^1) \simeq S^1 \rtimes \mathbb Z/2$ action by rotating (and reflecting) the circle, and (2) the target, I think, can be reconstructed as the homotopy fixed points of this action. If I were to drop the orientation reversal, you would be asking: given a homotopy $S^1$-space $X$, when is $X = \mathrm{maps}(S^1,\mathrm{maps}_{S^1}(*,X))$?
Feb
15
accepted In a closed monoidal abelian category, are the compact projectives a monoidal subcategory?
Feb
15
comment In a closed monoidal abelian category, are the compact projectives a monoidal subcategory?
Of course. I should have thought of that. I had spent a while playing with cases where $(0,\mathbb Z)\otimes (0,\mathbb Z) \cong (\mathbb Z,M)$, in which an associator puts very strong restrictions on $M$. Here I guess there might be some choice for the associator for general $M$, but really only one choice for, say, $M = \mathbb Z / (n)$.
Feb
14
asked In a closed monoidal abelian category, are the compact projectives a monoidal subcategory?
Feb
14
accepted When is/isn't the monoidal unit compact projective?
Feb
11
comment Do levelwise quasi-isomorphisms of bicomplexes induce a quasi-isomorphism between the total complexes?
It seems you should be able to package Cone(f) into some big bicomplex, and spectral sequences should compare it to Cone(Tot^\prod f) and Cone(Tot^\oplus f).
Feb
9
comment Which real Pin groups agree?
Credit where it's due: Nige asked me to mention that he learned about the isomorphism $Pin(4,0) \cong Pin(0,4)$, and about the error in my notes, from his student Yumi Boote.
Feb
8
comment Why does inconstructibility of $\sqrt[3]{2}$ imply impossibility of cube doubling?
I would have thought a "compass and straight edge construction in $\mathbb R^3$" meant that you could draw a straight line between any two known points and you could draw a sphere with center any known point and passing through any other known point. Then you may intersect such drawings, and you "know" any isolated point of intersection.
Feb
3
awarded  Popular Question
Feb
1
comment Are all smooth functions composites of 0-, 1-, and 2-ary functions?
My strong recollection is that you get all functions $\mathbb R^n \to \mathbb R$ by composing functions $\mathbb R \to \mathbb R$ and addition $+ : \mathbb R^2 \to \mathbb R$. My recollection is that this is true in both the continuous and smooth cases. But Qiaochu's comment makes me worried that perhaps I'm remembering Arnold's solution to Hilbert's problem for continuous functions, and perhaps my recollections are wrong for the smooth category.