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255126
bio website math.princeton.edu/~wsawin
location Princeton, NJ
age 21
visits member for 3 years, 8 months
seen 1 hour ago

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


1h
comment How many traces are there on Temperley-Lieb, Fuss-Catalan, Iwahori-Hecke, Birman-Wenzl-Murakami-Kauffman, … algebras?
@მამუკაჯიბლაძე and of course the "traces" are just the traces of the irreducible representations of $G$, which are equinumerous with the conjugacy classes.
2d
comment Lifting points of étale group scheme
Every object in your question matches up with something in the definition of formally etale. Using Wikipedia's notation, you want to take $C= S/m^{i+1}S$, $B=G$, $J= m^i S$. Then the statement it gives is precisely the statement you ask for. The key point is that $m^i$ is nilpotent in $S/m^{i+1}$.
2d
comment Lifting points of étale group scheme
use the fact that etale implies formally etale.
May
19
comment Interpretation of the monomorphism $H^2(\pi_1(X),\mathbb{Z}) \rightarrow H^2(X,\mathbb{Z})$
BTW, it's clear that you obtain a central extension. Clearly it's a normal subgroup, and the conjugation action of $\pi_1(X)$ on $\pi_1(S^1)$ is just the monodromy of the circle bundle, which is trivial because the cohomology class goes in the direction of the circle action, which is well-defined everywhere hence has no monodromy.
May
19
comment Interpretation of the monomorphism $H^2(\pi_1(X),\mathbb{Z}) \rightarrow H^2(X,\mathbb{Z})$
ACL's method does secretly give you the correct map. Because the correct map is injective but not surjective, it has an inverse that's a partial function, and I'm pretty sure that's exactly ACL's map. ACL's map is only defined when the long exact sequence map $\pi_2(X) \to \pi_1(S^1)=\mathbb Z$ vanishes. This map is determined using the map $\pi_2(X) \to H_2(X)$ and the pairing $H_2(X) \times H^2(X,\mathbb Z) \to \mathbb Z$.I'm pretty sure it vanishes exactly when the line bundle comes from a central extension, and he computes the correct central extension.
May
19
comment What is the cokernel of the map $H^2\big(\pi_1(X), \mathbb Z\big) \longrightarrow H^2(X,\mathbb Z).$
$H_2(\overline{X}, \mathbb Z)=Hom(\pi_2(X),\mathbb Z)$. So we get a map $H^2(X,\mathbb Z) \to Hom(\pi_2(X),mathbb Z)$. Presumably this is the "obvious" map. Then you can describe this as the group of functions on spheres in $X$ that come from cohomology classes. I'm not sure if that's simpler.
May
19
revised What is the cokernel of the map $H^2\big(\pi_1(X), \mathbb Z\big) \longrightarrow H^2(X,\mathbb Z).$
This is about manifolds, homotopy groups, and cohomology, hence is algebric topology.
May
18
comment Congruence properties of $x_1^6+x_2^6+x_3^6+x_4^6+x_5^6 = z^6$?
I can strengthen your "it is unlikely that" statement. Saying that $p$ divides $x^6+y^6+z^6+w^6$ for some coprime $x,z,t,w$ is equivalent to saying that the algebraic surface in $\mathbb P^3$ with equation $x^6+y^6+z^6+w^6$ has at least one $\mathbb F_p$ points. This is a smooth surface, so by the Weil conjectures its number of points is $p^2+a_p+1$ with $|a_p|< 106p$. (This can also be proved more directly using Gauss sums). Hence it has at least one point for $p>106$. Combined with your calculations, we see it has at least one point for all $p\neq 7, 31$.
May
17
comment Non-standard Gauss sums
@Liss It's an upper bound on the absolute value. For a reference you can go back to Weil: "On Some Exponential Sums". Example 1 after equation (5) on the bottom of page 206 is the bound, where $\mathfrak d=\{0,-1\}$ so $R_{\mathfrak d}(t) = t(t+1)$, $\chi$ is the quadratic character, and $\psi(x) = \omega_p^{lx}$.
May
17
awarded  Enlightened
May
17
awarded  Good Question
May
17
awarded  Nice Answer
May
17
answered Strings with no long runs from proper subalphabets
May
16
comment how to evaluate the following double summation to infinity without using integration method?
In your notation the area of a circle is $\pi r$. The sum you get with the aporoximation can be computed exactly by summation by parts again.
May
16
comment Non-standard Gauss sums
This is a finite field hypergeometric sum. There is no simple formula, but it's bounded by $2\sqrt{p}$.
May
16
comment Strings with no long runs from proper subalphabets
There is some recurrence that has $k$ choose $b$ terms for $N_{n,k,b}$ with fixed $b,k$ and varying $n$. You choose the distance from the most recent appearance of each of the $b$ letters to the end of the string, and consider what happens when you add one more letter. So there is some kind of asymptotic...
May
16
revised How bad can $\pi_1$ of a linear group orbit be?
deleted 122 characters in body
May
16
comment How bad can $\pi_1$ of a linear group orbit be?
@YCor Then I think my amended answer, together with your answer, shows that it is actually finite-by-abelian.
May
16
comment How bad can $\pi_1$ of a linear group orbit be?
@FrancoisZiegler Indeed.
May
16
revised How bad can $\pi_1$ of a linear group orbit be?
added 379 characters in body