Curve with given Frobenius polynomial

Question

Does there exist a prime $p$ and a smooth genus 2 curve $C / \mathbf{F}_p$ such that the characteristic polynomial of Frobenius on the Tate module of $J(C)$ is given by $(T^2 - p)^2$?

More generally, for a curve of arbitrary genus, it possible that both $\sqrt{p}$ and $-\sqrt{p}$ can occur as eigenvalues of the Frobenius?

(This is essentially an idle question, which came up when I was giving an expository talk on the Weil conjectures; I wanted to state the functional equation in the form "we can write the char poly of Frobenius as $\prod_{i=1}^{2g} (T - \lambda_i)$ where $\lambda_{2g+1-i} = p / \lambda_i$", and later I realised that this wouldn't work if both $\sqrt{p}$ and $-\sqrt{p}$ showed up as eigenvalues.)

EDIT. As David Speyer points out, I made a mistake in formulating my original question; the roots of $(T^2 - p)^2$ can still be arranged in this way. Perhaps a better question is

Does there exist a curve over $\mathbf{F}_{p^2}$, of any genus, for which both $p$ and $-p$ occur to odd multiplicity as Frobenius eigenvalues?

Answer 1

In its action on the Tate module, $\operatorname{Frob}_q$ is an element of $GSP_{2g}(\mathbb Q_\ell)$ whose action on the symplectic form is multiplication by $q$. This is due to the Weil pairing, or Poincare duality for etale cohomology.

Every such matrix has eigenvalues $\lambda_1, \dots, \lambda_{2g}$ whose with $\lambda_i \lambda_{2g-i} = q$. This is because the maximal torus in $GSP_{2g}$ has eigenvalues of that form, and every semisimple element is conjugate to an element of the maximal torus (so take the semisimplification of Frobenius, or prove that it is semisimple).

Answer 2

After thinking about this a bit more, I realised that no such example exists. Let $q = p^{2}$ (or $p^{2f}$ for any integer $f$), let $C$ be a curve over $\mathbf{F}_q$, and let $J = J(C)$. The Frobenius must be compatible with the Weil pairing $T_\ell J \times T_\ell J \to \mathbf{Z}_\ell(1)$, and Frobenius acts as multiplication by $q$ on the latter.

Since the Weil pairing is a perfect skew-symmetric pairing, this implies that the determinant of the Frobenius on $T_\ell J$ is $q^g$, where $g$ is the genus of $C$. (This follows from the same argument as is used classically to show that a symplectic matrix must have determinant $+1$.)

However, all the eigenvalues of Frobenius on $T_\ell J$ are either non-real, or are equal to $\pm \sqrt{q}$. The non-real eigenvalues occur in conjugate pairs, each of which has product $q$. So if $-\sqrt{q}$ occurs to odd multiplicity, the determinant would have to be $< 0$, a contradiction.

Answer 3

For genus $1$ any supersingular elliptic curve will have trace of Frobenius $0$, and so have minimal polynomial of Frobenius $T^2-p$. If you permit an abelian surface, not just a Jacobian, then the product of this elliptic curve with itself will have the desired minimal polynomial of Frobenius.

For genus $2$ hyperelliptic curves a supersingular hyperelliptic curve is one whose Jacobian is isogenous to a product of $2$ supersingular elliptic curves, and they are known to exist. See Choie, Jeong and Lee "Supersingular Hyperelliptic Curves of Genus 2 over Finite Fields".