Hodge-Tate proliferation

Let {C/\mathbf{Q}_{p}} be a complete algebraically closed extension, and let {A/C} be an abelian variety, with p-adic Tate module {T_{p}A} and dual abelian variety {A^{\ast}}. There are then at least three natural candidates for a “Hodge-Tate map”

\displaystyle \mathrm{HT}_{A}:T_{p}A\rightarrow\omega_{A^{\ast}}.

To wit:

1. (Scholze) For any smooth proper rigid space {X/C}, we have the Hodge-Tate spectral sequence

\displaystyle E_{2}^{i,j}=H^{i}(X,\Omega_{X/C}^{j})(-j)\Rightarrow H_{\mathrm{et}}^{i}(X,\mathbf{Q}_{p})\otimes C.

Taking {X=A}, we get canonical identifications {H^{1}(A,\mathcal{O}_{A})=\mathrm{Lie}A^{\ast}} and {H_{\mathrm{et}}^{1}(A,\mathbf{Q}_{p})\otimes C=\mathrm{Hom}_{\mathbf{Z}_{p}}(T_{p}A,C)}, so the edge map {E_{2}^{1,0}\rightarrow H^{1}} gives a map {\mathrm{Lie}A^{\ast}\rightarrow\mathrm{Hom}_{\mathbf{Z}_{p}}(T_{p}A,C)}, and we may define {\mathrm{HT}_{A}} as its {C}-dual.

2. (Tate, Fargues) Supposing {G/\mathcal{O}_{C}} is a p-divisible group, there is a very direct definition of a Hodge-Tate map for {G}: writing {G^{D}} for the Cartier dual, we have

\displaystyle \begin{array}{rcl} T_{p}G & \cong & \mathrm{Hom}_{\mathrm{pdiv}/\mathcal{O}_{C}}(G^{D},\mu_{p^{\infty}})\\ x & \mapsto & \lambda_{x},\end{array}

and we define

\displaystyle \begin{array}{rcl} \mathrm{HT}_{G}:T_{p}G & \rightarrow & \omega_{G^{D}}\\ x & \mapsto & \lambda_{x}^{\ast}\frac{dT}{T}.\end{array}

For {A} with good reduction, we can apply this right away to {G=A[p^{\infty}]} – but this really only works over {\mathcal{O}_{C}}, since {\mu_{p^{n}}} doesn’t have any differentials over {C}! To push this through in general, let {\mathcal{A}^{\ast}} be the formal semiabelian scheme over {\mathcal{O}_{C}} obtained by completing the Neron model of {A^{\ast}} along the identity component of its special fiber. The rigid generic fiber of {\mathcal{A}^{\ast}} is naturally a rigid analytic subgroup of (the rigid space associated with) {A^{\ast}}, and in particular we get a canonical inclusion {\mathcal{A}^{\ast}[p^{n}](\mathcal{O}_{C})\subset A^{\ast}[p^{n}](C)}. This dualizes and compiles into a surjection {\tau:T_{p}A\rightarrow T_{p}\mathcal{A}^{\ast D}}. Now {\mathcal{A}^{\ast}[p^{\infty}]} is an honest p-divisible group over {\mathcal{O}_{C}}, so the discussion above gives a map {\mathrm{HT}':T_{p}\mathcal{A}^{\ast D}\rightarrow\omega_{\mathcal{A}^{\ast}}}, and finally {\omega_{\mathcal{A}^{\ast}}} is naturally an {\mathcal{O}_{C}}-lattice in {\omega_{A^{\ast}}}, and we may define {\mathrm{HT}_{A}} as the composite map

\displaystyle T_{p}A\overset{\mathrm{HT}'\circ\tau}{\rightarrow}\omega_{\mathcal{A}^{\ast}}\overset{\mathrm{incl}}{\rightarrow}\omega_{A^{\ast}}.

3. (Coleman, Hodge-Tate periods and p-adic abelian integrals) Given an element {x=(0,x_{1},\dots,x_{n},\dots)\in T_{p}A} with {x_{n}\in A[p^{n}](C)}, choose {D_{n}} a divisor on {A^{\ast}} corresponding to {x_{n}\in A=\mathrm{Pic}^{0}(A^{\ast})}. Since {p^{n}x_{n}=0}, we may choose rational functions {f_{n}} on {A^{\ast}} such that {(f_{n})=p^{n}D_{n}}. The limit

\displaystyle \lim_{n\rightarrow\infty}\mathrm{dlog}f_{n}

exists and defines an element of {H^{0}(A^{\ast},\Omega_{A^{\ast}/C}^{1})\cong\omega_{A^{\ast}}} depending only on {x} (the nonholomorphic bits of the individual {\mathrm{dlog}f_{n}}‘s go to zero p-adically).

In practice, definitions 1 and 2 are both pretty useful. Definition 1 varies well in families, and e.g. plays a crucial role in Scholze’s torsion paper. Definition 2 captures more delicate integral behavior and simultaneously allows one to import information from the p-divisible groups universe; for example, the proof of Theorem D in this post uses definition 2 crucially. Fortunately, Scholze proves the equivalence of these two definitions in his CDM article (and it requires a pretty ridiculous diagram chase!). Coleman’s definition is a lot crazier, and I don’t think it’s been proven to agree with the others in general. To quote Coleman’s paper, “It is also possible to make sense of the limit…when {A} has bad reduction; however, we have not established its connection with Hodge-Tate.” Anyone up for it?

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