Timeline for Elliptic factors in the Jacobian and zeta function
Current License: CC BY-SA 4.0
11 events
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| Oct 13, 2020 at 12:01 | comment | added | Ari Shnidman | @DavidESpeyer nice, this seems to give a counterexample for abelian varieties. There is definitely more to check for the Jacobian issue-- e.g. if $E$ is not CM then I think this exact construction doesn't work since I believe $E^3$ has only reducible principal polarizations (even over an algebraic closure), so no twist of it can have an irreducible principal polarization. But maybe with some extra level structure or CM this can be modified to work... | |
| Oct 11, 2020 at 21:35 | comment | added | David E Speyer | (The key property of $A_4 \subset GL_3$ I am using is that every group element has a $1$-eigenvalue.) I have no idea whether such a thing can occur as the Jacobian of a genus 3 curve, and I also am not confident enough in my number theory to be sure that this works the way I think it will. | |
| Oct 11, 2020 at 21:34 | comment | added | David E Speyer | Here is an idea. Let $E$ be an elliptic curve over $\mathbb{Q}$. Then $A_4 \subset GL_3(\mathbb{Z}) \subseteq \mathrm{Aut}(E^{\oplus 3})$ (where $A_4$ is the alternating group on $4$-elements. Choose a map $\rho : \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to A_4$ (in other words, an $A_4$ extension of $\mathbb{Q}$). It seems to me that we should be able to twist $E^{\oplus 3}$ by $\rho$ and get a $3$-dimensional abelian variety $X$ over $\mathbb{Q}$. such that $X_p$ maps to $E_p$ for all $p$ but where $X$ doesn't map to $E$ over $\mathbb{Q}$. | |
| Oct 11, 2020 at 21:24 | comment | added | David E Speyer | Unfortunately, this is pretty far from the kind of subgroups of $GL_n(\mathbb{Q}_{\ell})$ that come from abelian varieties. | |
| Oct 11, 2020 at 21:23 | comment | added | David E Speyer | Okay, actually, I can answer this one: No. Let $n$ be odd and let $G$ be the alternating group $A_{n+1}$, acting on $\mathbb{Q}_{\ell}^n$ by the standard $n$-dimensional representation of $A_{n+1}$. For $g \in A_{n+1}$, the multiplicity of $1$ as an eigenvalue of $g$ is $(\mbox{number of cycles of $g$}) - 1$. Since $n$ is odd, every element of $A_{n+1}$ has at least $2$ cycles, so every element of $G$ has $1$ as an eigenvalue, and hence every characteristic polynomial is divisible by $\lambda - 1$. Yet $\mathbb{Q}_{\ell}^n$ has no invariant line. | |
| Oct 11, 2020 at 21:14 | comment | added | David E Speyer | Here is a pure representation theory question which seems related, simpler, and hard: Let $G$ be a closed compact subgroup of $GL_n(\mathbb{Q}_{\ell})$. Suppose that, for every $g \in G$, the characteristic polynomial of $g$ has a degree $k$ factor (in $\mathbb{Q}_{\ell}[t]$). Does this imply that $\mathbb{Q}_{\ell}^{\oplus n}$ has a $G$-invariant $k$-dimensional subspace? | |
| Oct 11, 2020 at 19:56 | answer | added | Ari Shnidman | timeline score: 5 | |
| S Oct 11, 2020 at 11:27 | history | suggested | gmvh |
Added top-level tag
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| Oct 11, 2020 at 11:08 | review | Suggested edits | |||
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| Oct 11, 2020 at 0:46 | comment | added | Jackson Morrow | For the first question, I believe the elliptic factor will appear in the factorization of $Z_p$, namely $Z_p$ won't be irreducible. I'm not sure about the second part, but I don't think so. One good thing to look up would be the Honda--Tate theorem. | |
| Oct 10, 2020 at 23:29 | history | asked | T. Combot | CC BY-SA 4.0 |