bio  website  math.princeton.edu/~wsawin 

location  Princeton, NJ  
age  21  
visits  member for  3 years, 3 months 
seen  39 secs ago  
stats  profile views  20,014 
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

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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

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How is the shape of $A+R+T$?
Do extensions of algebras really form a group? 
2d

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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 
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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 preimage of the real points in the complex upperhalf 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 
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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 primeto$p$ 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 
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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 
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Numbers with all Ndigit prefixes divisible by N
My question is related: mathoverflow.net/questions/126911/… 
Dec 17 
answered  Vector fields whose divergence are proper maps 
Dec 12 
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Nonunique 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  Nonunique splittings of homotopy idempotents 
Dec 12 
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Nonunique 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 
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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 
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Nonunique 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 
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Nonunique 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 2homotopy 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 
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Nonunique 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 2homotopy between them? And then you furthermore have a cube of ways to get from $f^4$ to $f$, and the aforementioned 2homotopy gives you the faces of the cube, and you demand a 3homotopy filling the cube, and so on. 
Dec 10 
revised 
Homogeneous spaces that are homotopy tori
added 94 characters in body 