bio  website  math.princeton.edu/~wsawin 

location  Princeton, NJ  
age  21  
visits  member for  3 years, 11 months 
seen  55 mins ago  
stats  profile views  24,061 
I am a graduate student at Princeton studying arithmetic algebraic geometry.
1d

comment 
Curvature of a finite metric space
Every finite metric space embeds into $\mathbb R^n$. Maybe try a notion of combined curvature and dimension? 
1d

reviewed  Close Eigenvalues of a random matrix 
1d

reviewed  Leave Open A compact Alexandrov space with curvature bounded below has curvature bouneded above? 
1d

reviewed  Close floating point representation via the perspective of TTE/computable analysis 
1d

reviewed  Leave Open Indecomposable decomposition for a commutative ring 
1d

comment 
Indecomposable decomposition for a commutative ring
Noetherian is a sufficient condition, no? 
2d

awarded  Nice Answer 
2d

reviewed  Close Encyclopedia of Mathematics?(nonAlphabetical) 
2d

reviewed  Leave Open Optimisation of betting strategy 
2d

comment 
What is the fewest number of points you must delete from $\mathbb{R}^3$ to make it not simply connected?
@JoelDavidHamkins The fibers of that map are certainly countable. Now add $u(1u)$ times that direction times a variable. Each removed point can only remove countably many values of $u$, and you win. 
2d

comment 
What is the fewest number of points you must delete from $\mathbb{R}^3$ to make it not simply connected?
@JoelDavidHamkins Sort of. I don't think you can get nonintersecting surfaces. Take a spacefilling curve, and try to contract it in two ways that don't intersect. But you could ask for a weaker condition  like that each point is only contained in countably many surfaces. I think you can achieve that by a variant of my construction. Take an analytic disc with boundary your curve. If it contains uncountably many lines, they form a single analytic family, so you can always choose a direction such that the disc contains no lines in that direction. Project onto the perpendicular direction. 
2d

answered  What is the fewest number of points you must delete from $\mathbb{R}^3$ to make it not simply connected? 
Aug
30 
comment 
Radius of convergence of Taylor expansion of $z \mapsto (1  z \cdot a)^{1}$
I don't understand these terms and how they relate to the title. 
Aug
30 
comment 
Number of linearly bisected subsets in finite vector space $F_2^n$
@user50982 The only reason to work with $2^{n1}k$ is to give a bijective proof of the first formula. I'm sure there's multiple ways to obtain it. And yes, I mean subspaces. 
Aug
30 
awarded  at.algebraictopology 
Aug
30 
comment 
Is $\mathbb{R}^3 \setminus \mathbb{Q}^3$ simply connected?
@JoelDavidHamkins I think so. It might be interesting to understand this number more fully. One should be able to replace the space of homotopies with a finitedimensional space and understand exactly what sort of spaces are being removed to get a better lower bound. 
Aug
29 
comment 
Motivic fundamental group of the moduli space of curves?
@BenWieland To show a monodromy group is large, I can work with a particular family of curves and show that has large monodromy. To show the monodromy of a representation is large, I just need to show that the invariants of tensor powers of this representation are small. This is controlled by expected value of powers of the trace of Frobenius acting on the representation of a random point on that family of curves over a finite field. This type of thing can often be controlled with standard methods from probability theory. 
Aug
29 
answered  Vanishing natural transformation and strong generator 
Aug
29 
comment 
definition of “immersion” of schemes (without open or closed)
Indeed, the opposite order is the same as the usual order, except in the case of nonquasicompact morphisms from a nonreduced scheme. stacks.math.columbia.edu/tag/01QV stacks.math.columbia.edu/tag/03DQ The opposite order appears in Hartshorne and, if I remember correctly, causes some strife by making some of the exercises unnecessarily difficult. 
Aug
29 
comment 
Number of linearly bisected subsets in finite vector space $F_2^n$
@kodlu Sure. Using $\binom{n}{k}= \binom{n}{nk}$, and then it counts the number of subsets of subsets of $[2^n]$ of size $2^{n1}$ in a transparent manner  by counting, for each $k$, the number of subsets with $k$ elements in the first half and $2^{n1}k$ elements in the second half. 