bio  website  math.princeton.edu/~wsawin 

location  Princeton, NJ  
age  21  
visits  member for  3 years, 8 months 
seen  1 hour ago  
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I am a graduate student at Princeton studying arithmetic algebraic geometry.
1h

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How many traces are there on TemperleyLieb, FussCatalan, IwahoriHecke, BirmanWenzlMurakamiKauffman, … 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

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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

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Lifting points of étale group scheme
use the fact that etale implies formally etale. 
May 19 
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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 welldefined everywhere hence has no monodromy. 
May 19 
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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 
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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 
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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 
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Nonstandard 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 
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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 
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Nonstandard Gauss sums
This is a finite field hypergeometric sum. There is no simple formula, but it's bounded by $2\sqrt{p}$. 
May 16 
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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 
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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 finitebyabelian. 
May 16 
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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 