# Morphisms for good reduction are maps respecting filtration

So, let $A,A'/K$ be abelian varieties where $K$ is a $p$-adic local field with residue field $k$. Suppose further that they have good reduction with models $\mathscr{A},\mathscr{A}'/\mathcal{O}_K$. I think it follows from Serre-Tate theory that

$$\text{Hom}(A,A')=\left\{g\in\text{Hom}(\mathscr{A}_k,\mathscr{A}'_k):g^\ast\text{ preserves Hodge filtration}\right\}$$

where

$$g^\ast:H^1_\text{crys}(\mathscr{A}'_k/W(k))\to H^1_\text{crys}(\mathscr{A}/W(k))$$

is the obvious map.

In particular, we get a factorization

$$\text{Hom}(A,A')\hookrightarrow \text{Hom}(\mathscr{A}_k,\mathscr{A}'_k)\hookrightarrow \text{Hom}_{F,V}(H^1_\text{crys}(\mathscr{A}'/W(k)),H^1_\text{crys}(\mathscr{A}_k/W(k))$$

Now, it's the famous result of Tate that the induced map

$$\text{Hom}(\mathscr{A}_k,\mathscr{A}'_k)\otimes_\mathbb{Z}\mathbb{Z}_p\to \text{Hom}_{F,V}(H^1_\text{crys}(\mathscr{A}'/W(k)),H^1_\text{crys}(\mathscr{A}_k/W(k)))$$

is an isomorphism. My hope is then that the injection

$$\text{Hom}(A,A')\otimes_\mathbb{Z}\mathbb{Z}_p \hookrightarrow \text{Hom}_{F,V}(H^1_\text{crys}(\mathscr{A}'_k/W(k)),H^1_\text{crys}(\mathscr{A}_k/W(k)))$$

has image precisely those $F,V$-commuting maps

$$H^1_\text{crys}(\mathscr{A}'_k/W(k))\to H^1_\text{crys}(\mathscr{A}/W(k))$$

that preserve the Hodge filtration.

But, this is not obvious to me. The examples I've computed seem to work. This is also the desired analogue of what happens over $\mathbb{C}$ where

$$\text{Hom}(A,A')=\left\{g\in\text{Hom}(H^1_\text{sing}((A')^\text{an},\mathbb{Z}),H^1_\text{sing}(A^\text{an},\mathbb{Z}):g_\mathbb{C}\text{ preserves Hodge filtration}\right\}$$

which, again, is motivation but not proof.

I think that via the fully faithfulness of $D_\text{crys}$ that this may be related to an $\ell=p$ version of Tate's conjecture for $p$-adic fields. Specifically, let's restrict to the case when $K=\mathbb{Q}_p$. Then, the fully faithfulness of $D_\text{crys}$ should say that

$$\text{Hom}_{\mathbb{Q}_p[G_{\mathbb{Q}_p}]}(H^1_{\acute{e}\text{t}}(A'_{\overline{K}},\mathbb{Q}_p),H^1_{\acute{e}\text{t}}(A_{\overline{K}},\mathbb{Q}_p))\xrightarrow{\approx}\text{Hom}_{F,V,\text{Fil}}(H^1_\text{crys}(\mathscr{A}'_{\mathbb{F}_p}/\mathbb{Q}_p),H^1_\text{crys}(\mathscr{A}_{\mathbb{F}_p}/\mathbb{Q}_p))$$

and thus it seems as though asking whether

$$\text{Hom}(A,A')\otimes_\mathbb{Z}\mathbb{Q}_p\to \text{Hom}_{F,V,\text{Fil}}(H^1_\text{crys}(\mathscr{A}'_{\mathbb{F}_p}/\mathbb{Z}_p),H^1_\text{crys}(\mathscr{A}_{\mathbb{F}_p}/\mathbb{Z}_p))\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$$

is an isomorphism amounts essentially to whether

$$\text{Hom}(A,A')\otimes_\mathbb{Z}\mathbb{Q}_p\to \text{Hom}_{\mathbb{Q}_p[G_{\mathbb{Q}_p}]}(H^1_{\acute{e}\text{t}}(A'_{\overline{K}},\mathbb{Q}_p),H^1_{\acute{e}\text{t}}(A_{\overline{K}},\mathbb{Q}_p))$$

is an isomorphism which, of course, is something like Tate's isogeny conjecture for $\ell=p$.

Any help is greatly appreciated!

EDIT: Assuming that I've correctly applied the fully faithfuless of $D_\text{crys}$ above, and my question does imply Tate's isogeny theorem for $\ell=p$ for $p$-adic fields, then this question (as well, perhaps, as the comments below) give counterexamples to the claim.

So, could someone just verify that my deduction of Tate's isogeny theorem from an affirmative answer to my original question is correct? Thanks!

EDIT EDIT: I am now confident that I didn't err in my transformation of the original question into Tate's isogeny conjecture. So, unless someone would like to give an answer to the question adding something else of interest, I'll just answer my question as a community wiki. Thanks!

• No. Let $A=A'$ be the canonical lift of an ordinary abelian variety $A_0$ over the perfect residue field $k=\mathbf{F}_p$, so ${\rm{H}}^1(A_{\overline{K}}, \mathbf{Q}_p)=V_p(A)^{\vee}=M^{\vee} \oplus M(-1)$ where $M$ is the unramified Galois module corresponding to the etale part of $A_0[p^{\infty}]$. The $\mathbf{Q}_p$-algebra of Galois-equivariant linear endomorphisms of the cohomology is $R \times R^{\rm{opp}}$ for the ring $R$ of Galois-equivariant linear endomorphisms of $M$. But ${\rm{End}}(A) \otimes \mathbf{Q}_p$ often has no nontrivial idempotents (often is $\mathbf{Q}_p$). – nfdc23 Apr 29 '16 at 4:38
• @nfdc23 Hello nfdc23, thanks for your comment. A few questions. First, it seems as though you're answering the $\ell=p$ Tate's isogeny theorem part of my question—you agree then that this is an equivalent formulation of the original statement? Second, I'd appreciate if you could expand upon your comment. Why does the generic fiber of the canonical lift have this cohomology necessarily? I looked in some articles (e.g. Katz) and couldn't find anything. – Alex Youcis Apr 29 '16 at 4:53
• @nfdc23 Also, unless I'm being terribly silly (and I might) won't the canonical lift's endomorphism algebra contain the lift of Frobenius—how can it then be $\mathbb{Q}_p$? Finally, do you know conditions on $A$ that will make this true? – Alex Youcis Apr 29 '16 at 4:54
• Yes, I'm addressing the end of your question, and the decomposition expresses that $A_0[p^{\infty}]$ is (uniquely) the direct product of its etale and connected parts (the residue field is perfect) which are dual to each other, and by definition the canonical lift has $p$-divisible group that is the unique direct product lift of these two factors (related by duality). Hence, the $p$-adic Tate module of the generic fiber is such a product. At the end I was wrong; the $p$-part of Tate's isogeny theorem over finite fields gives equality (i.e., affirmative answer!) for ordinary canonical lifts. – nfdc23 Apr 29 '16 at 13:47
• In particular, the upshot is that one has go to "non-canonical" lifts to see the failure of equality in the ordinary case (even for elliptic curves, as in the link you found). My error was that when writing the end of my first comment I was thinking about general perfect residue fields, over which $A_0$ may be non-CM (even with endomorphism ring $\mathbf{Z}$), but of course over finite fields that cannot happen. – nfdc23 Apr 29 '16 at 13:54

The original question, as stated, has a negative answer. Namely, it is not true that the induced map

$$\text{Hom}(A,A')\otimes\mathbb{Z}_p\to \text{Hom}_{F,V,\text{Fil}}(H^1_{\text{crys}}(\mathscr{A}'_k/W(k)),H^1_\text{crys}(\mathscr{A}_k/W(k)))\qquad (1)$$

is an isomorphism.

For simplicity let $K=\mathbb{Q}_p$ (so $k=\mathbb{F}_p$). By the fully faithfulness of the $D_\text{crys}$ functor we know that

$$\text{Hom}(H^1_{\acute{e}\text{t}}(A',\mathbb{Q}_p),H^1_{\acute{e}\text{t}}(A,\mathbb{Q}_p))\to \text{Hom}(D_\text{crys}(H^1_{\acute{e}\text{t}}(A',\mathbb{Q}_p)),D_{\text{crys}}(H^1_{\acute{e}\text{t}}(A,\mathbb{Q}_p)))$$

is an isomorphism but

$$D_\text{crys}(H^1_{\acute{e}\text{t}}(A',\mathbb{Q}_p))=H^1_{\text{crys}}(\mathscr{A}'_k/K)$$

and similarly for $A$. Thus, if $(1)$ were true then the map

$$\text{Hom}(A,A')\otimes\mathbb{Z}_p\to \text{Hom}(H^1_{\acute{e}\text{t}}(A',\mathbb{Q}_p),H^1_{\acute{e}\text{t}}(A,\mathbb{Q}_p))$$

and thus the map

$$\text{Hom}(A,A')\otimes\mathbb{Z}_p\to\text{Hom}(T_p A,T_p A')$$

is a isomorphism. But, this is Tate's isogeny conjecture for $\mathbb{Q}_p$ which is false. See, for example, this post.

NB: If anyone has anything interesting to add, I would be more than happy to hear it/accept it as an answer!