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?



(crystalline cohomology version's) Tate's conjecture for K3 surfaces
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 $W(\mathbb{F}\_q)$.


