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Jared Weinstein's user avatar
Jared Weinstein's user avatar
Jared Weinstein's user avatar
Jared Weinstein
  • Member for 15 years, 2 months
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Density results for equality of Galois/automorphic representations
Here's a silly comment: In the n = 1 case, you can't have epsilon any smaller than 1/2. If chi is the quotient between two l-adic Galois characters under comparison, then the density of primes p for which chi(p) = 1 is 1/[G_Q^ab : ker(chi)]. If chi is not injective then this density is no more than 1/2. Certainly the bound 1/2 can be attained by choosing chi to be a nontrivial quadratic character. I think the question becomes significantly more interesting when n = 2! (If you restrict to irreducible Galois representations, can you attain epsilon = 1/2?)
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Locally profinite fields ?
Point well taken. By no means is it necessarily a bad phrase. I only meant that there could be some ambiguity in how the phrase is interpreted (whereas there is much precedent for "locally compact group").
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Which p-adic numbers are also algebraic?
Victor: Pete was only saying the two fields were abstractly isomorphic -- he did not claim the isomorphism preserved any topology. I've also used that isomorphism before. I would describe its use as "violent."
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Does $\operatorname{Con}\sf(ZF)$ imply $\operatorname{Con}\sf(ZF + \operatorname{Aut}{\bf C = Z/\mathrm 2Z})$?
So many prompt and interesting responses: This is why I love MO. I agree that Pete's notes give a proof that in ZF, Aut C only has Z/2Z as a nontrivial finite subgroup. But I still don't see any reason why Aut C couldn't be something terrifying like a semidirect product of Z by Z/2Z. Any comments on this?
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Convergence of a sequence of continuable Dirichlet series
The question requires f(s) to be entire, which is excluded if V is trivial -- I've thought of this counterexample already. Can anything be said if V is nontrivial? Still, this example hints at a negative answer to the question, since zeta(s) is regular at 0 whereas the f_n(s) do not converge. So, +1.
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What's the "best" proof of quadratic reciprocity?
Keith: Showing that 2/p doesn't belong to Z[zeta_p] follows from the Q-linear independence of 1, zeta, ... , zeta^(p-2), correct? That's not so hard to get across if they've had linear algebra. But in practice, when I gave this proof I wasn't at all "honest" in the sense you use above. In particular, I never gave a full proof that the pth cyclotomic polynomial was irreducible. In fact, a lot of what I did in that class was very impressionistic and left off details that were either relegated to office hours or simply not treated. It would have made you sick to watch, I'm sure!
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Cohomology of rigid-analytic spaces
Thank you -- I hadn't thought about the functoriality of semistable models. It appears from your comment that at the very least, $X$ admits a semistable model whose reduction contains something birational to the reduction of $Z$, correct? In which case the cohomology of $\overline{Z}$ ought to appear in the cohomology of $X$, which is what I wanted to know.
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