# Frobenius eigenvalues of abelian variety

Let $A$ be an abelian variety over a finite field $\mathbb{F}_q$ and $x_i$ the Frobenius eigenvalues on $H^1$. Does $x_i \mapsto q/x_i$ permute the $x_i$, and why? It should follow from Poincare duality.

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This sounds like homework. If so, you should ask it on math.stackexchange.com. That said, try the Riemann hypothesis instead of Poincare duality. – Will Sawin Jul 25 '12 at 14:17
maybe PD with polarization. – shenghao Jul 25 '12 at 18:03

In this case, you can deduce without the Weil conjectures that $x_{i} \mapsto \displaystyle \frac{q}{x_{i}}$ permutes the eigenvalues. Let $f$ denote the Frobenius isogeny on $A$ and let $f^{\vee}$ be its dual isogeny. Then, we have $$f^{\vee}\circ f = q$$ which shows that if $x_{i}$ is an eigenvalue of $f$, then $\displaystyle \frac{q}{x_{i}}$ will be an eigenvalue of $f^{\vee}$. So, it suffices to show that the characteristic polynomials of $f$ and $f^{\vee}$ are the same. To see this, let $P_{f}$ and $P_{f^{\vee}}$ be the respective characteristic polynomials. Using Chapter IV Theorem 4 of Mumford's Abelian Varieties, we have for $n \in \mathbb{Z}$, $$deg(n_{A}-f) = P_{f} (n)$$ Then, observe that $(n_{A} - f)^{\vee} = n_{A} - f^{\vee}$. Since the degree of an endomorphism is the same as the degree of its dual, we obtain $$P_{f} =deg(n_{A}-f) = deg(n_{A} - f^{\vee}) = P_{f^{\vee}}$$ as desired.