Timeline for Varieties with the same number of $\mathbb{F}_p$-points for all but finitely many primes
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13 events
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| Sep 20, 2021 at 6:57 | comment | added | alpoge | btw @lkx your question about ab. surfaces was definitely interesting! I just wanted to make sure my argument was clear. Thm. 1.1 of DiPippo-Howe’s arxiv.org/pdf/math/9803097.pdf shows that the number of isog. classes of g-dim’l ab. var.s over $\mathbb{F}_q$ grows like $q^{g(g+1)/4}$ as $q\to\infty$. On the other hand the num. of pt.s on such an ab. var. is asymptotic to $q^g$. So for $g\geq 3$ there are plenty of pairs of $g$-dim’l ab. var.s over large $\mathbb{F}_p$ which aren’t isog. but have the same num. of $\mathbb{F}_p$-pt.s! There’s a ton more to be said, I’m just not an expert. | |
| Sep 20, 2021 at 5:38 | answer | added | lkx | timeline score: 4 | |
| Sep 20, 2021 at 5:28 | comment | added | alpoge | nope!! Brauer-Nesbitt is a statement of the form "two semisimple rep.s are isomorphic iff they have the same trace functions". we're literally directly talking about trace functions here, you're asking does the trace function of that virtual rep. vanish on these particular elements? and im saying yes it does, because it's continuous and vanishes on a dense subset (namely the Frobenii) so vanishes everywhere. does that make sense? you're not asking for an iso. of rep.s in your first q (though i guess you could be talking about my comment on your second q? maybe im confused, if so lemme know) | |
| Sep 20, 2021 at 5:25 | comment | added | lkx | Oh you must be referring to Brauer–Nesbitt theorem? That's what I was missing. | |
| Sep 20, 2021 at 5:15 | comment | added | alpoge | you're not giving me the same Frobenius trace for only one prime, you're giving it to me for all but finitely many primes aka a dense subset of the absolute Galois group. im deducing that the trace of $\mathrm{Frob}_p^n$ for $X$ agrees with that of $Y$ by limiting to the element $\mathrm{Frob}_p^n$ with a sequence of $\mathrm{Frob}_P$'s with $P$ some other primes (which i can do by Chebotarev). does that make sense? again im not saying that if $X$ and $Y$ just have the same trace of the single element $\mathrm{Frob}_p$ then they automatically have the same char. poly. of $\mathrm{Frob}_p$ | |
| Sep 20, 2021 at 5:07 | comment | added | lkx | But we only know the trace so when there is many eigenvalues it should fail right? There probably should be a counterexample with surfaces. Are there non-isogenous abelian surfaces over a finite field with the same number of points? | |
| Sep 20, 2021 at 4:12 | comment | added | alpoge | btw I just saw your edit aka second question and I suspect it's unknown (again, not my expertise) because it's not known if the rep.s are semisimple in general. That said we can at least read off the char. poly. of $\mathrm{Frob}_p$ on $H^i$ by purity ($p$ large and via your hypotheses of smoothness and properness) and so I think it at least follows in your situation that the semisimplifications of the $H^i$'s agree as Galois modules. | |
| Sep 20, 2021 at 4:05 | history | edited | lkx | CC BY-SA 4.0 |
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| Sep 20, 2021 at 3:51 | comment | added | alpoge | The way it goes is: the tr. of all but fin. many $\mathrm{Frob}_p$'s on said virtual rep. is $0$. The trace function is a cont. function on the abs. Gal. gp. valued in $\mathbb{Q}_\ell$. By Chebotarev that set of Frobenii is dense. So the fun. is $0$ on the whole gp., hence $0$ on $\mathrm{Frob}_p^n$ for all $n$. I prefer to think of it via mod-$\ell^N$ rep.s: given $p$ and $n$, for all $N$ the action of $\mathrm{Frob}_p^n$ on the mod-$\ell^N$ rep. matches the action of some $\mathrm{Frob}_q$ by usual Chebotarev for finite Gal. ext.s, so the trace is $0$ mod $\ell^N$, now take $N\to\infty$. | |
| Sep 20, 2021 at 3:36 | comment | added | lkx | The trace does not determine the entire characteristic polynomial does it? | |
| Sep 19, 2021 at 21:18 | comment | added | alpoge | Chebotarev applied to the virtual representation $\sum_i (-1)^i \left(H_c^i(X_{/\overline{\mathbb{Q}}}, \mathbb{Q}_\ell) - H_c^i(Y_{/\overline{\mathbb{Q}}}, \mathbb{Q}_\ell)\right)$ (but lemme know if I’ve said anything stupid! I only think about curves admittedly…). | |
| Sep 19, 2021 at 19:12 | history | edited | YCor |
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| Sep 19, 2021 at 17:37 | history | asked | lkx | CC BY-SA 4.0 |