28,644 reputation
243105
bio website math.princeton.edu/~wsawin
location Princeton, NJ
age 21
visits member for 3 years, 3 months
seen 39 secs ago

I am a graduate student at Princeton studying arithmetic algebraic geometry.


1h
answered Number of elements of “$\mathrm{SL}_n(\mathbb{F}_p^\times)$” mod $p$
2d
comment Moments of random special unitary matrices
You can view it as the trace of a $k$-cycle in $S_k$ acting on the invariant subspace of $V^{\otimes k}$.
2d
comment How is the shape of $A+R+T$?
Do extensions of algebras really form a group?
2d
comment The uniform boundedness of rational torsion for traceless abelian surfaces over a function field
this kind of work might be helpful. I think it essentially says that you can't obtain unbounded torsion with bounded gonality from covers of a single curve: math.polytechnique.fr/~cadoret/Gonality.pdf
Dec
22
comment The letters of the word “ART”
@AliTaghavi Aren't such $X$ closed subsets of $\mathbb R^2$, hence locally compact?
Dec
21
answered The letters of the word “ART”
Dec
21
answered Can the pre-image of the real points in the complex upper-half plane of a modular elliptic curve under the modular parametrization be identified?
Dec
19
answered abelian $\ell$-adic representations in $\widehat{SL(2,Z)}$
Dec
17
comment Etale fundamental group of a curve in characteristic $p$
I think $\pi_1$ is not known for any hyperbolic curve in any reasonable sense. For instance, just knowing for each prime-to-$p$ cover the $p$-rank of its Jacobian seems very hard - it is not at all obvious that there is a usable finite description of this data.
Dec
17
comment Etale fundamental group of a curve in characteristic $p$
Take $G_1$ an infinitely generated free $p$-group and $G_2$ the product of $G_1$ with $\mathbb Z/p$. Then they have the same finite quotients but are distinct. This is not so great an example, because group cohomology distinguishes them.
Dec
17
comment Numbers with all N-digit prefixes divisible by N
My question is related: mathoverflow.net/questions/126911/…
Dec
17
answered Vector fields whose divergence are proper maps
Dec
12
comment Non-unique splittings of homotopy idempotents
@MikeShulman What part of the argument are you most skeptical of? Perhaps precisely identifying the problem here will help prove or disprove the claim in general.
Dec
12
answered Non-unique splittings of homotopy idempotents
Dec
12
comment Non-unique splittings of homotopy idempotents
I think it's more complicated even than that: two of nontrivial faces are not the $2$-homotopy, but rather the $2$-homotopy composed with $f$. In my case $f$ coming first gives the same thing but afterwards is trivial. Maybe the simplical perspective is better? I think my idea of a constant function must be wrong, because any homotopy splitting of it must be contractible which makes all the maps and homotopies unique.
Dec
10
comment Canonical form of cubic curves over general fields
There is no canonical form for cubic curves because the moduli space of cubic curves is not a scheme.
Dec
10
comment Non-unique splittings of homotopy idempotents
because opposite faces cancel. The only problem is if the nontrivial element of $\pi_2(X)$ pulls back to a trivial element of $\pi_2( Map(X,X))$, which I don't know how to rule out.
Dec
10
comment Non-unique splittings of homotopy idempotents
If this is right, I think you might be able to construct two different coherentizations of the homotopy $X \to pt \to X$ for some space $X$, maybe $X = \mathbb CP^\infty$. All the functions involved in the homotopy will be constant functions. Start with the obvious homotopies, but choose the 2-homotopy between the two homotopies $f^3 \to f$ to be a sphere representing a nontrivial element of $\pi_2(X)$. Then you can glue on a $3$-cell, $4$-cell, etc. to ensure the higher homotopies exist. The only cell that can trivialize that element of $\pi_2(X)$ is the $3$-cell, and I think it doesn't
Dec
10
comment Non-unique splittings of homotopy idempotents
Let me see if I understand what a fully coherent idempotent is: You have a homotopy $f \circ f \sim f$, which gives you two homotopies $ f\circ f \circ f \sim f\circ f \sim f$, and then you demand a 2-homotopy between them? And then you furthermore have a cube of ways to get from $f^4$ to $f$, and the aforementioned 2-homotopy gives you the faces of the cube, and you demand a 3-homotopy filling the cube, and so on.
Dec
10
revised Homogeneous spaces that are homotopy tori
added 94 characters in body