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Orbits of group scheme action
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Orbits of group scheme action
@Allen: Even for $\mathbf{G}_m$ acting on itself by translation the same phenomenon occurs. What definition of "the orbit" are you using?
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What does it mean for a Deligne-Mumford stack to have trivial generic stabilizers?
I don't understand why one would expect a condition at generic points should be equivalent to a condition at "each point $p$" (which $p$?), nor what "acting faithfully" is suppose to mean (faithfully on what, and in what sense)? I also don't know anything about the topological context, so I'll bow out here.
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Gross's paper on Heegner points
Given $(E,C)$ you form $E \rightarrow E/C$, and given $f:E \rightarrow E'$ you form $(E, \ker f)$. But if you're trying to do something arithmetic (such as analyze fields of definition of specific analytically-defined points relative to a specific $\mathbf{Q}$-structure) then you need a definition of $Y_0(N)$ that goes beyond the upper half-plane description.
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What does it mean for a Deligne-Mumford stack to have trivial generic stabilizers?
Yes, and it would be true for any etale atlas from a scheme. But isn't it more "geometric" to say that the given stack contains a dense open substack that is an algebraic space?
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What does it mean for a Deligne-Mumford stack to have trivial generic stabilizers?
For any quasi-separated Artin stack, the locus of points with trivial automorphism scheme is Zariski-open and (by a theorem of Artin) is an algebraic space. One can ask that this open subspace be Zariski-dense, and it is equivalent to this open subspace containing all generic points. That is, the automorphism schemes at generic points are trivial if and only if there is a dense open substack that is an algebraic space.
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What are the obstructions to showing that $\zeta$ doesn't vanish on the strip $1- \varepsilon < {\rm Re}(s) \leq 1$
Dear Peter Mueller: The reason I put emphasis on the Grothendieck-Artin approach (which also entails no smoothness hypotheses) and not Dwork's in my comment is because of the context of the question asked: it is the cohomological approach (and not Dwork's) that was essential to Deligne's papers Weil I and especially Weil II, the latter for which the crux of the induction goes beyond "constant coefficient" $\zeta$-functions. I was just trying to indicate a context that includes "GRH cases".
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Double coset formulas for Orthogonal groups [Solved]
How do you know that you have "globally constant" signs when working with a disconnected group?
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On non-split extensions of $\mathrm{SL}_d(q)$
The conjugation action of $G$ on its abelian normal subgroup $E$ factors through a $G/E$-action on $E$, so the given extension structure makes $E$ into an ${\rm{SL}}_d(q)$-module. If you specify this module structure then the set of isomorphism classes of such extensions is in bijection with ${\rm{H}}^2({\rm{SL}}_d(q),E)$, so you're asking to compute degree-2 cohomology for ${\rm{SL}}_d(q)$ acting on $\mathbf{F}_p$-vector spaces; sounds hard! The case $|E| = q^d$ isn't special, but natural cases of interest are $\mathbf{F}_q^d$ with usual action and ${\mathfrak{sl}}_d(q)$ with adjoint action.
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Preimage of a maximal compact open subgroup in the simply connected cover
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What are the obstructions to showing that $\zeta$ doesn't vanish on the strip $1- \varepsilon < {\rm Re}(s) \leq 1$
@Mustafa: For varieties $V$ over $\mathbf{F}_q$, $\zeta(s,V)$ is rational in $q^{-s}$ by Grothendieck-Artin, hence periodic mod $2\pi i/\log(q)$, so one gets an $\epsilon_V > 0$ once non-vanishing on ${\rm{Re}}(s)=1$ is proved, as Deligne did by a representation-theoretic variant of the method of Hadamard and de la vallee Poussin via cohomological interpretation of $\zeta(s,V)$. From that Deligne used further ideas to push an initial abstract $\epsilon_V > 0$ all the way to $1/2$. One dreams to do the same for RH, but there's no idea how to even begin to find $\epsilon > 0$. Big problem! :)
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What are the obstructions to showing that $\zeta$ doesn't vanish on the strip $1- \varepsilon < {\rm Re}(s) \leq 1$
The main difficulty is the lack of an idea. :) All known "zero-free regions" are asymptotic to the line ${\rm{Re}}(s)=1$. If anyone ever had a clue how to even go about finding such an $\epsilon$, it would be a tremendous breakthrough, probably the key insight required to solve GRH.
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Preimage of a maximal compact open subgroup in the simply connected cover
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Preimage of a maximal compact open subgroup in the simply connected cover
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