# Uncountably many non-isomorphic Tate modules

Do there exist uncountably many abelian surfaces with good reduction over $$\mathbb{Q}_p$$ with pairwise non-isomorphic rational $$p$$-adic Tate modules?

If we took $$l$$-adic Tate modules there would be countably many at most by a result of Kisin and the fact that the moduli of ppav is of finite type.

This answer would suggest that there is only countably many $$p$$-adic Tate modules for elliptic curves with good reduction.

Is it at least true that there are uncountably many non-isomorphic crystalline representations of $$\mathrm{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$$ coming from geometry?

• Given an abelian variety $A$ over a finite field $k$ its lifts to abelian schemes over $W(k)$ are in bijection with rank $g$ saturated submodules of $H^1_{cris}(A/W(k))$ that reduce mod $p$ to the 1st step of the Hodge filtration and are isotropic for the cup product so they are parametrized by (the $W(k)[1/p]$-points of) a residue disk in the isotropic Grassmanian. The isomorphism classes of Galois representations are then parametrized by the quotient of this disk by $Aut_{\varphi} H^1_{cris}(A)[1/p]$. – SashaP Aug 9 '20 at 20:30
• If the dimension of this algebraic group is less than $g(g+1)/2$ then the quotient has uncountably many points. This happens e.g. if $A$ is the product of $4$ pairwise non-isogenous elliptic curves, for in that case the automorphisms of a $\varphi$-module is an $8$-dimensional group and $g(g+1)/2=10$. Maybe this happens in smaller dimensions as well because the action on the Grassmanian is not free, but I'm not sure. – SashaP Aug 9 '20 at 20:30
• *4 pairwise non-isogenous ordinary elliptic curves – SashaP Aug 9 '20 at 20:47
• @MikhailBondarko Sorry, I forgot to mention that we should pick a polarization on $A$ of degree prime to $p$ (let's assume that $A$ has such, if it doesn't then the question of existence of algebraizable deformations becomes subtler), it induces a perfect pairing on $H^1_{cris}$, and the claim is that lifts of the Hodge filtration isotropic for this pairing give lifts of the abelian variety that admit a lift of the polarizing line bundle. – SashaP Aug 12 '20 at 22:01
• Deformations of abelian schemes are unobstructed and one can use that to show that any lift of the Hodge filtration is realized by a lift of $A$ to a formal abelian scheme (for a reference, see e.g. Messing's "Crystals associated to Barsotti-Tate group", though there the approach is slightly different). The chosen polarization lifts to a given formal lift if and only if the lift of the Hodge filtration is isotropic. this is because the obstruction to lifting a line bundle onto $\mathfrak{A}$ is the image of its crystalline Chern class in the group $H^2(\mathfrak{A},\mathcal{O})$. – SashaP Aug 12 '20 at 22:01