(crystalline cohomology version's) Tate's conjecture for K3 surfaces

Question

Let $X$ be a K3 over $\overline{\mathbb{F}_p}$. The (crystalline version's) Tate conjecture predicts:

$c_1: Pic(X)\otimes\mathbb{Q}_p\rightarrow H^2_{crys}(X/W)^{\Phi=p}\otimes\mathbb{Q}_p$

is an isomorphism.

If $\lambda=1-\frac{1}{h}$ is the smallest slope of $H^2_{crys}(X/W)$, then $H^2_{crys}(X/W)$ must be isogenous to $H_{\lambda}\oplus H_1^{22-2h} \oplus H_{2-\lambda}$, where $H_\alpha$ is the F-crystal that $\Phi^t=p^s$ for $\alpha=s/t$. The $\Phi=p$ parts of the latter F-crystal apparently has rank $22-2h$, so the $\Phi=p$ part of $H^2_{crys}(X/W)$ must also has rank $22-2h$? Which means the above conjecture is equivalent to $\rho=22-2h$ always hold true? But the later one is not true in general, as a general K3 in char $p$ should have Picard number $2$, which is usually less than $22-2h$.

Where was I wrong?

Answer 1

This is related to a phenomenon called 'hypersymmetry' (or the lack of it). The term comes from Chai and Oort, who explored this for abelian varieties. The essential point is that the category of F-isocrystals over the algebraic closure of a finite field is not the filtered colimit of the corresponding categories over its finite subfields.

For the simplest example, take a product of two non-isogenous ordinary elliptic curves $A = E_1\times E_2$ over $\overline{\mathbb{F}}_p$. You can get such a pair by choosing two Weil $q$-numbers of weight $1$ generating distinct quadratic extensions of $\mathbb{Q}$. Then we have (with the Tate twist on the left hand side)

$H^2_{crys}(A/W(\overline{\mathbb{F}}_p)(1) = \mathrm{Hom}(H^1_{crys}(E_1/W(\overline{\mathbb{F}}_p),H^1_{crys}(E_2/W(\overline{\mathbb{F}}_p))\oplus H^2_{crys}(E_1/W(\overline{\mathbb{F}}_p))(1) \oplus H^2_{crys}(E_2/W(\overline{\mathbb{F}}_p) )(1)$

Since both elliptic curves are ordinary, they have isogenous (as $F$-crystals) crystalline cohomology groups over $\overline{\mathbb{F}}_p$. In particular, if you take $\varphi$-invariants, you will get a 4-dimensional $\mathbb{Q}_p$-vector space.

However, if you repeat this over any finite field containing the $j$-invariants of $E_1$ and $E_2$, then the crystalline Tate conjecture says that the $\varphi$-invariants will only be 2-dimensional over $\mathbb{Q}_p$, since the first Hom space should have no $\varphi$-invariant elements.

If you take the Kummer K3 surface associated with $A$, you will find a K3 surface that has the same feature: The $\varphi$-invariants in $H^2_{crys}$ over $W(\overline{\mathbb{F}}_p)$ will have two extra ‘transcendental’ dimensions that are not visible over any $W(\mathbb{F}_q)$.

Answer 2

I see where is the problem.

First of all, the right statement of Tate conjecture is for $X$ over $\mathbb{F}_q$ (not over $\overline{\mathbb{F}_q}$!), and Tate conjecture predicts:

$c_1: Pic(X)\otimes\mathbb{Q}_p\rightarrow H^2_{crys}(X/W(\mathbb{F}_q))^{\Phi=p}\otimes\mathbb{Q}_p$ is an isomorphism.

Secondly, Dieudonn$\acute{e}$-Manin classification only works over algebraically closed fields like $\overline{\mathbb{F}_q}$, doesn't work over $\mathbb{F}_q$, and when we talk about slopes of an F-crystal $H$ over $\mathbb{F}_q$, we actually mean the slopes of $H\otimes W(\overline{\mathbb{F}_q})$.

Now assume the K3 crystal $H^2_{crys}(X/W(\mathbb{F}_q))$ has slopes $1-\frac{1}{h},1,1+\frac{1}{h}$, and assume it's non-supersingular.

The fact is, although the rank of $(H^2_{crys}(X/W(\mathbb{F}_q))\otimes W(\overline{\mathbb{F}_q}))^{\Phi=p}$ is indeed $22-2h$, the rank of $H^2_{crys}(X/W(\mathbb{F}_q))^{\Phi=p}$ can be smaller:

Here is an easy example of F-crystal over $\mathbb{F}_9$ of slope 1 rank 1 but $\Phi=p$ space is trivial:

Since $W(\mathbb{F}_9)\cong \mathbb{Z}_9$ there is an 8-th roots-of-unity $\zeta_8$ in $W(\mathbb{F}_9)$, let $H$ be the rank 1 $W(\mathbb{F}_9)$-module where $\Phi$ maps $1$ to $\zeta_8p$.

Then this F-crystal maps any $x\in W(\mathbb{F}_9)$ to $\zeta_8p\sigma(x)$, where $\sigma\in Gal(\mathbb{Q}_9/\mathbb{Q}_3)$ is the only non-trivial (Frobenius) map.

Then no element in $\mathbb{Z}_9$ satisfy $\zeta_8p\sigma(x)=px$. (This is simply because the multiplicative subgroup of $\mathbb{Q}_9^{\times}$ generated by elements of the form $\sigma(x)/x$ is actually $\mu_4\times(\frac{4+3i}{5})^{\mathbb{Z}_3}$, the group of 4th-roots of unities-this is an easy computation.)

But after base change to $W(\overline{\mathbb{F}_9})$, suppose $\zeta_{16}$ is a square root of $\zeta_8$, let $x=\zeta_{16}^7$, then $\Phi(x)=px$ which means $H$ indeed has slope $1$.