Let $q$ be a power of a prime $p$. Deligne's paper "*Variétés abéliennes ordinaires sur un corps fini*" seems to describe an equivalence of categories between


1. ordinary abelian varieties over a finite field $\mathbf F\_q$,
2. complex abelian varieties equipped with an endomorphism $\pi$ which is a $q$-Weil number.


Is that true? The paper actually states the equivalence of categories between (1) and a category of free Z-modules. It sketches the relation with complex abelian varieties, but does it remain an equivalence of categories?


This question has been discussed a bit here:
[Canonical lifts from $\mathbb F\_q$ and CM-theory](https://mathoverflow.net/questions/162987/canonical-lifts-from-mathbb-f-q-and-cm-theory). The answer seems to be "yes". I still ask this question because of the following, which looks like a contradiction to me:


Fix a $q$-Weil number $\pi$, let $K = \mathbf Q(\pi)$, and $\mathcal O\_K$ the maximal order in $K$. Then, $\mathrm{Cl}(K)$ acts freely and transitively on the set of isomorphism classes of complex abelian varieties with endomorphism ring $\mathcal O\_K$. On the other hand, this action of $\mathrm{Cl}(K)$ on the abelian varieties over $\mathbf F\_q$ with endomorphism ring $\mathcal O\_K$ is still free but not necessarily transitive (Waterhouse, *Abelian varieties over finite fields*, Theorem 5.3: the number of orbits is $2^s$ where $s$ is the number of primes factors of $p$ in $K\_0$ staying prime in $K$, and $K\_0$ is the real subfield of $K$).


Did I misinterpret Deligne's paper? Or is this $s$ actually always 0?



Ordinary abelian varieties over a finite field
Because the abelian variety is ordinary, $\pi$ has the property that it's $p$-adic valuation at every place is either $0$ or $1$ (this follows easily from condition (IV) on the first page of Deligne's paper.) That means no prime $p'$of $\mathbb Q(\pi)$ lying over $p$ is equal to its complex conjugate, because if the $p'$-adic valuation of $\pi$ is $1$ then the $\overline{p}'$-adic valuation of $\pi$ is $0$, and vice versa.


On the other hand, if a prime of $K\_0$ remained a prime of $K$ then it would equal its complex conjugate.


Hence $s=0$.


