36,954 reputation
258135
bio website math.princeton.edu/~wsawin
location Princeton, NJ
age 21
visits member for 3 years, 11 months
seen 55 mins ago

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?(non-Alphabetical)
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 $u(1-u)$ 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 space-filling 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 $2^{n-1}-k$ 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.algebraic-topology
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 finite-dimensional 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 non-quasicompact morphisms from a non-reduced 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 $\binom{n}{k}= \binom{n}{n-k}$, and then it counts the number of subsets of subsets of $[2^n]$ of size $2^{n-1}$ in a transparent manner - by counting, for each $k$, the number of subsets with $k$ elements in the first half and $2^{n-1}-k$ elements in the second half.