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



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> 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?
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Curve with given Frobenius polynomial
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).


