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user12832
user12832
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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? I thinkIt should follow from Poincare duality only gives us this with $q$ replaced by $g^g$, $g$ the dimension of $A$.

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? I think Poincare duality only gives us this with $q$ replaced by $g^g$, $g$ the dimension of $A$.

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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user12832
user12832
  • 417
  • 2
  • 12

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? I think Poincare duality only gives us this with $q$ replaced by $g^g$, $g$ the dimension of $A$.