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Let $E$ be a supersingular elliptic curve over $\mathbf{F}_p$, and $H$ its endomorphism algebra $\text{End}(E)\otimes_{\mathbf{Z}}\mathbf{Q}$, a quaternion algebra (non split at $p$ and $\infty$).

For every prime $\ell\neq p$, there is a faithful $\ell$-adic algebra-representation:

$$\rho_{\ell} : H\to \text{End}_{\mathbf{Q}_{\ell}}(V_{\ell}(E))$$

where $V_{\ell}(E) := (\varprojlim_{n\ge 0} E_{\overline{\mathbf{F}}_p}({\overline{\mathbf{F}}_p})[\ell^n])\otimes_{\mathbf{Z}_{\ell}}\mathbf{Q}_{\ell}$ is the rational $\ell$-adic Tate module of $E$.

  • Using the natural isomorphism $H^1(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell}) \simeq \text{Hom}_{\mathbf{Q}_{\ell}}(V_{\ell}(E),\mathbf{Q}_{\ell})$ we have an $H\otimes_{\mathbf{Q}}\mathbf{Q}_{\ell}$-module structure on $H^1(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})$, as is classically known.

  • Now we make a more naive construction. Functoriality of the étale site of $E_{\overline{\mathbf{F}}_p}$ gives that for any endomorphism $f : E\to E$ there is a map $$H^i(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})\to H^i(E_{\overline{\mathbf{F}}_p},f^{-1}\mathbf{Q}_{\ell})$$ and since $\mathbf{Q}_{\ell}$ is constant (here I am being imprecise about the nature of $\mathbf{Q}_{\ell}$, which is not a constant sheaf on the étale site, but the meaning is clear from the context) we also have an isomorphism $H^i(E_{\overline{\mathbf{F}}_p},f^{-1}\mathbf{Q}_{\ell})\simeq H^i(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})$, and we call $$f^* : H^i(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})\to H^i(E_{\overline{\mathbf{F}}_p},f^{-1}\mathbf{Q}_{\ell})\to H^i(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})$$ the composition of the two. In other words, every element $f\in\text{End}(E)$ has an effect $f^*$ on $H^i(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})$. On the other hand, the effect of each element of $\mathbf{Z}\subset\text{End}(E)$ on $H^i(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})$ is invertible, and so the above construction defines, for every element $f\in \text{End}(E)\otimes_{\mathbf{Z}}\mathbf{Q} = H$, an effect $f^*$ on $\ell$-adic cohomology.

Does the construction in the second point, for $i=1$, give an action of $H$ on $H^1(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})$ as an algebra? If so, does this action agree with the one constructed in the first point?

My expectation is that the answer to the first question is no, and hence so is the answer to the second.

(1) The second construction should only define on $H^i(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})$ the structure of a representation of $H^{\times}$, and it should not be possible to upgrade this to an algebra action of $H$.

(2) I expect the problem with the second construction is that, for two elements $f,g\in H$, we may have $(f+g)^*\neq f^*+g^*$, where the first $+$ is in $H$ and the second is in $H^i$.

(3) Existence of the $\ell$-adic Tate-module functor, and the fact that for abelian varieties $A,B$ the map $\text{Hom}_{\rm AV}(A,B)\otimes_{\mathbf{Z}}\mathbf{Q}_{\ell}\to \text{Hom}_{\mathbf{Q}_{\ell}}(V_{\ell}(A),V_{\ell}(B))$ is injective, should be crucial to have an algebra action of $H$ on $H^1(E_{\overline{\mathbf{F}}_p},\mathbf{Q}_{\ell})$, and it feels it should not be possible to construct it just as a consequence of functoriality of the étale site, a much less deep fact. Am I right?

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  • $\begingroup$ How are these constructions different, and why do you expect the second construction to be non-additive? $\endgroup$ – S. Carnahan Dec 29 '18 at 7:28
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The fact that the action of $f+g$ by pullback on $H^1$ is equal to the action of $f$ plus the action of $g$ does not follow, as far as I can see, just from functoriality of the etale site, but it does follow from general properties of etale cohomology, in particular the Kunneth formula.

From the Kunneth formula we get $H^1(E x E) = H^1(E) \times H^1(E)$ with the isomorphism given by pullback along the two maps $(E \times E ) \to E$. From this one can check that the maps $E \to (E x E)$ that send $x$ to $(x,id)$ or $(id,x)$ or $(x,x)$ act in the obvious way on $H^1$ (i.e. the dual of how they act on the Tate module), and from this we can check that the map $(E \times E) \to E$ given by the group law acts in the obvious way on the $H^1$.

Then the map $f+g$ on $H^1$ is just the diagonal map, followed by $f \times g$, followed by the group law and we can check that its action by $H^1$ is exactly what we expect.

This should all follow from the axioms of a Weil cohomology theory and so no additional assumptions are needed.

The injectivity of the Tate module functor seems irrelevant here as one is just asking for an action, not an injective action, and anyways the quaternion algebra is a division algebra and so will have a hard time acting non-injectively.

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