# Isomorphism of the $\ell$-adic Tate module of an elliptic curve with CM

Let $$E$$ be an elliptic curve over $$K$$ (totally real number field) with complex multiplication by the field $$L$$. Let $$\psi$$ be the Grössencharacter associated to $$E$$, assume that $$\psi$$ of type $$(-r,0)$$ (i.e., the restriction of $$\psi$$ to the archimedian part $$\mathbb{C}^\times$$ of the idele group of $$K$$ has the form $$z\mapsto z^{-r}$$). Set $$V_{\ell}(\psi):= (H^{1}(E(\mathbb{C}),\mathbb{Q})^{\otimes r}\otimes_K L)\otimes_{K\otimes\mathbb{Q}_\ell}L_{\tilde{\ell}}$$ where $$\otimes r$$ is taken over $$K$$.

My question is: How to prove that the $$\ell$$-adic Tate module $$V_\ell(E)$$ is isomorphic to $$V_{\ell}(\psi)\oplus\imath V_{\ell}(\psi)$$ as representations of $$G_K$$, where $$\imath\in G_K$$ is the complex conjugation. (References would be appreciated).

• What do you mean by the Grössencharacter associated to $E$? Jun 24 '19 at 3:29
• @WillSawin, I meant the Hecke character attached to $E$ via the main theorem of complex multiplication (see Silverman II, Theorems 9.1 and 9.2). Jun 24 '19 at 11:04
• @WillSawin, I am missing something? Jun 24 '19 at 11:14
• Doesn't this claim about the Tate module follow from the main theorem of complex multiplication? Jun 24 '19 at 12:17
• @WillSawin, Actually, I don't know. Do you have any guesses about how we can prove that using the main theorem of CM? Jun 24 '19 at 12:32

Edit: Following Will Sawin's comments, I realized that my first answer contained many mistakes. I try to emend my answer, writing $$G_\mathbb{Q}=\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$$ and $$G_K=\operatorname{Gal}(\overline{\mathbb{Q}}/K)$$ throughout.

First of all, your $$r$$ is actually equal to $$1$$: indeed, if you look at Theorem II.9.1 in Silverman's Advanced Topics in the Arithmetic of Elliptic Curves, you see that what he calls $$\alpha_{E/L}$$ is a character which is trivial when restricted to infinite idèles (use property (i) there), and therefore the property in Theorem II.9.2 defining the Grössencharakter, namely $$\psi_{E/L}(x)=\alpha_{E/L}(x)N^L_K(x^{-1})_\infty$$, shows that $$\psi_{E/L}$$ is of infinite type $$(-1,0)$$. Secondly, I think that you have a sign-problem: you should have the same sign in the infinity-type (negative) and in the tensor power showing up in your equation.

Secondly, you say in the comments that you took the above equation from Kato's 2004 Astérisque paper $$p$$-adic Hodge theory and values of zeta functions of modular forms but beware that in his setting $$K$$ is imaginary quadratic (see the beginning of §15) and that he is working with a classical modular form (of some weight $$k$$): it is certainly a good idea to try to grasp what he's doing in case $$k=2$$, but this corresponds to an elliptic curve $$E$$ precisely when the curve is defined over $$\mathbb{Q}$$, so that $$K$$ is an imaginary quadratic field of class number $$1$$ (assuming that $$E$$ has complex multiplication by the full ring of integers). A final remark about Kato's paper: although on page 160 he says that $$$$as representation of $$G_\mathbb{Q}$$ there is an isomorphism $$V_\ell(E)\cong V_{\ell}(\psi)\oplus \sigma V_\ell(\psi)"$$ he does not mean that the above is a direct sum of representations. Indeed, in his notation three lines below his statement, if you pick $$(x,0)\in V_\ell(E)$$ and $$\gamma=\iota\tau\notin G_K$$, you get $$\gamma(x,0)=(0,\gamma x)$$, showing that the direct sum is not a sum of representation. Indeed, as Will Sawin observed, the Galois representation $$V_\ell(E)$$ is irreducible.

I consider then an elliptic curve $$E/\mathbb{Q}$$ with CM by $$\mathcal{O}_K$$. I write $$E_{/K}$$ for its base-change to $$K$$ and I write $$V_\ell(E),V_\ell(E_{/K})$$ for the respective Tate modules. They are somehow different creatures: the first is endowed with an action of $$G_\mathbb{Q}$$ while the second has an action of the subgroup $$G_K$$ and is, moreover, a free $$\mathbb{Q}_\ell\otimes K=K_\ell$$-module of rank $$1$$, as discussed for instance in Tate module of CM elliptic curves: this is not true for $$V_\ell(E)$$, because the complex multiplication is not defined over $$\mathbb{Q}$$ and, in particular, it does not commute with the Galois action of the full group $$G_\mathbb{Q}$$. Let's analyze these two representations separately:

1. As said, $$V_\ell(E_{/K})$$ is of rank $$1$$ over $$K_\ell$$ and I claim that it is isomorphic to $$K_\ell(\psi)$$, which is the rank-$$1$$-module $$K_\ell$$ endowed with an action of $$G_K$$ given by $$g\cdot v=\psi(g)v$$, seeing $$\psi(g)\in K^\times\subseteq K_\ell^\times$$. This is the commutativity of the diagram in Theorem 9.1 (ii) of Silverman's Advanced topics..., together with the definition of $$\psi$$ in Theorem 9.2 ibid.. As a by-product, you get an isomorphism $$K_\ell(\psi)\cong H^1(E(\mathbb{C}),\mathbb{Q}_\ell)^{-1}$$ because on the one hand $$H^1(E(\mathbb{C}),\mathbb{Q}_\ell)^{-1}\cong H^1_\mathrm{et}(E_{/K},\mathbb{Q}_\ell)^{-1}\cong V_\ell(E_{/K})$$ (this is well-explained here Etale cohomology and l-adic Tate modules, upon realizing that $$(\ast\otimes -1)$$ means taking the dual''); and, on the other, because $$H^1(E(\mathbb{C}),\mathbb{Q}_\ell)^{-1}=\mathbb{Q}_\ell\otimes_{\mathbb{Q}}H^1(E(\mathbb{C}),\mathbb{Q})^{-1}=\mathbb{Q}_\ell\otimes_{\mathbb{Z}}H_1(E(\mathbb{C}),\mathbb{Z})=\mathbb{Q}_\ell\otimes \mathcal{O}_K$$ where the last isomorphism is the map $$\gamma\mapsto \int{_\gamma} {\omega_E}$$ (in general, the image of this map is a lattice $$\Lambda$$ such that $$E(\mathbb{C})=\mathbb{C}/\Lambda$$, so in this case $$\Lambda=\mathcal{O}_K$$).

2. The tensor product $$V_\ell(E)\otimes_{\mathbb{Q}_\ell}K_\ell$$ has rank $$2$$ over $$K_\ell$$ and $$G_\mathbb{Q}$$ acts on it on the first factor. It turns out that there is an isomorphism of $$G_\mathbb{Q}$$-representations $$V_\ell(E)\otimes_{\mathbb{Q}_\ell}K_\ell\cong \operatorname{Ind}^K_\mathbb{Q}\big(K_\ell(\psi)\big)$$ as proven (in the general setting of CM cuspforms) in Ribet's 1977 paper Galois representations attached to eigenforms with Nebentypus, Theorem 4.5. You can retrace the line of Ribet's argument (who was inspired by the classical proof for elliptic curves) by doing exercices 2.29-2.32 in Silverman's Advanced topics...: the main point is to compare the traces of both representations and then to invoke a general result saying that two $$2$$-dimensional irreducible representations sharing the same trace are isomorphic.

Now, by restricting the $$G_\mathbb{Q}$$-representation constructed above to the subgroup $$G_K$$ you don't find $$V_\ell(E_{/K})$$ (which is of rank $$1$$ over $$K_\ell$$), but you rather find $$\operatorname{Res}_{G_K}\big(V_\ell(E)\otimes K_\ell\big)\cong\operatorname{Res}_{G_K}\operatorname{Ind}^K_\mathbb{Q}\big(K_\ell(\psi)\big)$$ and you can apply Proposition 22 of Serre's Représentations linéaires des groupes finis to deduce that this latter object is $$K_\ell(\psi)\oplus \sigma K_\ell(\psi)$$, in turn isomorphic to $$\Big(H^1(E(\mathbb{C}),\mathbb{Q}_\ell)^{-1}\Big)\oplus \sigma \Big(H^1(E(\mathbb{C}),\mathbb{Q}_\ell)^{-1}\Big)$$ by point 1. above.

• Thank you very much for your answer. Jun 24 '19 at 16:24
• This answer is wrong or something is very strange. $H^1 ( E_{\overline{K}}, \mathbb Q)$ has no interesting Galois action, so it can't be isomorphic to the dual of the Tate module. Jul 1 '19 at 23:50
• The Tate module only splits as a sum of two modules when $E$ is CM, CM is crucial here. Jul 1 '19 at 23:51
• @WillSawin You're certainly right, I will write something more reasonable as soon as I have time. Thank you. Jul 2 '19 at 15:48
• This is much better. Indeed the key point is to use the main theorem of CM. In your equation in part 1, I don't think you mean to view $H^1(E(\mathbb C), \mathbb Q_\ell)$ as a $\mathbb Q_\ell$-module and then tensoring with $K_\ell$, as that produces a rank $2$ $K_\ell$-module. Instead I think you want to view it as a $K_\ell$-module directly by the action of $K$, and dualize there. This also affects the last line. In part 2, it doesn't make sense to tensor over $\mathbb Q_\ell$ with $K$ . - you want to tensor over $\mathbb Q$. Jul 5 '19 at 17:15