bio | website | math.berkeley.edu/~theojf |
---|---|---|
location | Berkeley | |
age | 29 | |
visits | member for | 5 years, 5 months |
seen | Mar 29 at 15:07 | |
stats | profile views | 14,964 |
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. |