19,943 reputation
445152
bio website math.berkeley.edu/~theojf
location Berkeley
age 29
visits member for 5 years, 2 months
seen 3 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
comment Lurie's approach to the bar-cobar adjunction
I think in "where $\mathcal C$ is the $\infty$-category of Kan complexes" you mean $\mathcal C$ to be $\mathcal S$.
Dec
15
comment Does the $(\mathbb Z/2)$-graded isomorphism $E_n \cong E_{n+2}$ have any nice properties?
@DavidTreumann Good question! Here's my understanding. There is a semidirect product $O(n) \ltimes E_n$ called "framed $E_n$", and its formality makes it isomorphic to what you get by adjoining to $P_n$ a generator that acts as primitive of the bracket in the Hochschild cohomology coming from the commutative multiplication. The isomorphism $P_n \cong P_{n+2}$ is compatible with this extra generator. So I think that there is a sense in which the isomorphism $E_n \cong E_{n+2}$ is equivariant for some sort of $O$ action, but I'd have to think a while to unpack the details.
Dec
14
comment Does the $(\mathbb Z/2)$-graded isomorphism $E_n \cong E_{n+2}$ have any nice properties?
@DavidTreumann Is there a vector space call Q^n, or an interesting connected space of them? Often I feel like there's only one, since when I write Q^n, I usually mean that I have distinguished a basis. But then I don't know if I've distinguished a basis, or an ordered basis; for the former, perhaps there's BS_n of them. And of course often there's BGL(n) of them. Anyway, surely the space of E_n operads is a B(Aut E_n), and surely there's an action of O_n by changing the framing, but I would not expect it to be everything. Over Q, isn't Aut E_2 the Grothendieck-Teichmuller group?
Dec
14
comment Does the $(\mathbb Z/2)$-graded isomorphism $E_n \cong E_{n+2}$ have any nice properties?
@QiaochuYuan: Interesting. As far as I know, formality requires dividing by arbitrary integers, so I don't expect an isomorphism between $E_n \wedge KU$ and $P_n \wedge KU$, at least not if my understanding of "$\wedge KU$" is correct. But either of the following would be awesome: (1) formality over $KU$; (2) interesting statements over $KU$, but no formality, hence interesting statements that don't require formality.
Dec
13
comment Does the $(\mathbb Z/2)$-graded isomorphism $E_n \cong E_{n+2}$ have any nice properties?
In physics, I've been told that $\mu$ is a "mass", or perhaps "mass squared", but I don't claim to understand why. I've also been told that $\mu = \hbar$, a claim that I understand even less.
Dec
13
comment Does the $(\mathbb Z/2)$-graded isomorphism $E_n \cong E_{n+2}$ have any nice properties?
A warm-up question: I don't really know what $\mathbb Q[\mu,\mu^{-1}]$ "means" in terms of derived geometry, but I'd love to. It's something like the functions on $\mathrm B \mathbb G_m$, or $\mathrm B^2 \mathbb G_m$, or something. I mean, $\mathrm{Spec}(\mathbb Q[\mu,\mu^{-1}])$ is a "(derived) abelian group scheme" that integrates the line in degree $2$, thought of as an abelian Lie algebra, and that line is something like $\mathrm B^2$(usual 1-dimensional Lie algebra). What is $\mathbb Q[\mu,\mu^{-1}]$ really?
Dec
13
asked Does the $(\mathbb Z/2)$-graded isomorphism $E_n \cong E_{n+2}$ have any nice properties?
Dec
1
comment The resolution of which conjecture/problem would advance Mathematics the most?
@user74230 Good to know! As I hope was clear from my comment, I am not an expert.
Nov
30
comment The resolution of which conjecture/problem would advance Mathematics the most?
My memory is that in algebraic geometry, Grothendieck hoped for a program to develop quite a lot of machinery in order to prove some outstanding (Weil?) conjectures. When Deligne proved them without all the machinery Grothendieck hoped would be developed, Grothendieck, according to the story, was quite upset: the point wasn't the result, but the rest of the theory. I bring this up to support the idea that "open conjectures that would advance mathematics" might make such an advance because they seem to need dramatically new machinery or ideas, which would then be available for other uses.
Nov
28
comment Lie bialgebras cohomology
I agree with Gabriel's comments, but want to add one remark. There are different ways to model many-to-many operations: properads and props (which are essentially equivalent, by a deep result of Vallette's) allow compositions of arbitrary "genus" (or "loop order"); an alternate version, called "dioperads", uses only "tree-level" compositions. This in general makes a huge difference: the answers to homological-type questions can be very difference. I think that for Lie bialgebras, you happen to get the same answers in the different settings, but for other types of bialgebras you often do not.
Nov
24
awarded  Nice Answer
Nov
20
awarded  Notable Question
Nov
10
comment Does the linear automorphism group determine the vector space?
... well formed. Indeed, the problem is that there is no functorial construction of $V$ from $\mathrm{GL}(V)$, so that, in particular, bundles of groups isomorphic to $\mathrm{GL}(V)$ do not lift to bundles of vector spaces isomorphic to $V$. (The first example I know of is a bundle whose base space is the suspension of $\mathbb R \mathbb P^2$.)
Nov
10
comment Does the linear automorphism group determine the vector space?
@LSpice Incidentally, I was led astray by your discussion of the underlying set $X$. Normally a "vector space structure" consists of particular maps --- it makes sense, then, to ask whether two given vector space structures on the same set are equal. The Erlangen program, at best, is about actually giving a manifold some structure in this strict sense. Your question, as I now understand, was whether the isomorphism type of the group $\mathrm{GL}(V)$ determines the isomorphism type of the vector space $V$. This is within a family of common questions, but I would argue is not really ...
Nov
9
awarded  Notable Question
Nov
8
comment Does the linear automorphism group determine the vector space?
@LSpice: Oh, I'm sorry --- I misunderstood the question. My apologies.
Nov
8
awarded  Good Answer
Nov
7
answered Does the linear automorphism group determine the vector space?
Nov
7
comment Does the linear automorphism group determine the vector space?
This does not answer the question posed, which is whether the vector space structure (in the sense that we teach in undergraduate linear algebra) on the underlying set of $V$ can be recovered from how the group $\mathrm{GL}(V)$ sits inside the group of all permutations of the underlying set.
Nov
6
comment Counterintuitive consequences of the Hahn-Banach theorem
@WillieWong I am reasonably well-educated in set theory and related questions, and am on record as being sick of AC questions on MO. But I was not aware of this "standard reference". I now am, and I am better for the knowledge.