# Elliptic factors in the Jacobian and zeta function

Consider a hyperelliptic curve $$\mathcal{C}$$ over $$\mathbb{Q}$$ and its Jacobian $$J(\mathcal{C})$$. Assume that $$J(\mathcal{C})$$ admits an elliptic factor $$\mathcal{E}$$.

For almost all primes, we can reduce $$\mathcal{C}$$ modulo $$p$$, and consider its zeta function $$Z_p$$. Can the elliptic factor property be seen as a special property of $$Z_p$$?

Conversly, if such property is satisfied for almost all primes, can we prove that a hyperelliptic curve possess an elliptic factor in its Jacobian?

• 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. – Jackson Morrow Oct 11 at 0:46
• 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? – David E Speyer Oct 11 at 21:14
• 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. – David E Speyer Oct 11 at 21:23
• Unfortunately, this is pretty far from the kind of subgroups of $GL_n(\mathbb{Q}_{\ell})$ that come from abelian varieties. – David E Speyer Oct 11 at 21:24
• 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 $\mathbb{Q}$. – David E Speyer Oct 11 at 21:34

If you ask the same question over a general number field $$F$$, then you will find counterexamples by considering "false elliptic curves" i.e. principally polarized abelian surfaces with quaternionic multiplication. These are automatically Jacobians of hyperelliptic curves of genus two. They have the property that for almost all primes their reduction is a product of elliptic curves, but over $$F$$ (and even over its algebraic closure) the Jacobian may very well be simple (hence no elliptic factors).
Over $$\mathbb{Q}$$, you might look at abelian surfaces with "potential" quaternion multiplication, but actually I would guess that these don't give counterexamples-- just a guess. In genus greater than 2, it's even less clear to me.