21,508 reputation
450168
bio website math.berkeley.edu/~theojf
location Berkeley
age 30
visits member for 5 years, 9 months
seen 2 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.


2h
comment Is the $\infty$-category of presentable $\infty$-categories presentable?
... are well-studied in some areas of computer science, because they allow for coinductive, rather than inductive, reasoning. A more down-to-earth reason not to be afraid of Cantor's paradox is the well-known theorem that the homotopy category of topological spaces is not concretizable (because every non-empty object has a proper class, not a set, of subobjects).
2h
comment Is the $\infty$-category of presentable $\infty$-categories presentable?
I disagree: in principle the answer should be yes. Indeed, $Pr^L$ contains all limits and colimits (the strict version of this statement is due to Greg Bird in his unpublished '76 thesis, if my memory is correct), and feels like it is generated under colimits by some basic building blocks. The reason Cantor's paradox does not apply is that we are in the homotopical world, not the set-theoretic world. Indeed, there are good homotopical models (although now I am speaking outside my expertise) in which all functors, and in particular the "power set" functor, have fixed points. These worlds ...
8h
awarded  Popular Question
Jul
1
awarded  Nice Question
Jul
1
accepted Does projective imply flat?
Jun
30
comment Does projective imply flat?
@EricWofsey Ah, good point.
Jun
30
comment Does projective imply flat?
Well, grumble. And this gives an example where I don't even have flat resolutions, since by the same logic $(N,1)$ fails to be flat for every $N$. (Tensoring with $(M,0)$ and then projecting onto the $\mathcal B$ factor still gives the functor $F$.)
Jun
30
comment Does projective imply flat?
@AlexDegtyarev Do you mind spelling out your abstract nonsense? Eric Wofsey in an answer below seems to provide a counterexample.
Jun
30
comment Does projective imply flat?
@AlexDegtyarev The thing I know how to do is to use the fact that projective implies flat to conclude that Tor groups can be computed by projectively resolving only one of the two variables. Please explain what you have in mind? All the categories I care about have enough projectives, so I don't mind assuming that as an axiom.
Jun
30
comment Does projective imply flat?
@FernandoMuro Yes, all the categories I care about have enough injectives, so I'm happy to add that as an axiom. If you have a proof available, I'll be happy to accept it as an answer.
Jun
30
comment Intuition behind the definition of quantum groups
... the "quantum plane" above I imagine as having spectrum $\mathbb R$ (or maybe $\mathbb C$). Then again, coordinates satisfying $XY = qYX$ have funny behavior along the axes... I guess the main difference is that $xp - px = i\hbar$ is natural if you have some translation-invariance, and for QM on a line we do want $x \mapsto x+x_0$ to be a symmetry. Whereas $XY = qYX$ is natural if you want an action by scaling --- if you're trying to do "quantum linear algebra".
Jun
30
comment Intuition behind the definition of quantum groups
@AndreKornell Good question. The word "plane" is, I hope, self-explanatory --- this is some kind of 2-dimensional linear space. Probably this use of "quantum" is nothing better than an abbreviation for "noncommutative". In the quantum mechanics of a particle on the line, the phase space is a plane with coordinates $x,p$, but those satisfy $xp - px = i\hbar$. Setting $q = e^{i\hbar}$, $X = e^x$ and $Y = e^Y$ does give coordinates satisfying $XY = qYX$ (or perhaps I'm off by a sign), but those should have spectrum $\mathbb R_+$ (or maybe $\mathbb C^\times$), whereas the coordinates on ...
Jun
30
asked Does projective imply flat?
Jun
28
awarded  Nice Question
Jun
23
revised Why does the bitxor function appear in Nim?
error in an equation
Jun
23
comment Why does the bitxor function appear in Nim?
@Halbort: Yes; I will correct it.
Jun
23
comment Why does the bitxor function appear in Nim?
Huh. I started writing this last night, but only posted it this morning, and only then saw that Will Sawin has given essentially the same argument.
Jun
23
answered Why does the bitxor function appear in Nim?
Jun
22
comment Is there a symmetric monoidal 2-category “SuperDuperVect”?
@MikeShulman I meant only that I thought I remember at least some discussion of symmetric monoidal bicategories in one of your papers, perhaps on double categories?
Jun
21
comment Lurie's Endomorphism Space vs. Endomorphisms
In other examples I've thought about, to produce the map $A \to Map_{Top}(M,M)$ qua algebras requires more than just the action $A \times M \to M$, but also the associativity data of that action. Does that work here?