# Is this lattice in the Tate module of an elliptic curve, coming from complex-analytic uniformization, stable under Frobenius?

Let $$E$$ be an elliptic curve over $$\mathbb{Q}$$, and let $$p$$ and $$\ell$$ be two distinct primes of good reduction. Let $$T_\ell = T_\ell(E) = \varprojlim E[\ell^n](\overline{\mathbb{Q}})$$ be the $$\ell$$-adic Tate module, and let $$F_p \in \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$$ be a Frobenius element at $$p$$. Then $$F_p$$ acts $$\mathbb{Z}_\ell$$-linearly on $$T_\ell$$, and this action depends only up to conjugation on the choice of $$F_p$$. In particular, its characteristic polynomial is well-defined. A basic result is that the coefficients of this characteristic polynomial are integers.

This last fact is usually proved by considering the reduction of $$E$$ modulo $$p$$, which does not change the $$\ell$$-adic Tate module, and using that we can realize the $$F_p$$-action in characteristic $$p$$ as coming from an actual morphism of elliptic curves, namely the Frobenius morphism $$E \to E^{[p]}$$. But I was wondering if it is possible to give a more direct proof, namely by constructing a $$\mathbb{Z}$$-lattice $$\Lambda \subset T_\ell$$ (by which I mean a rank 2 free $$\mathbb{Z}$$-module such that the map $$\Lambda \otimes \mathbb{Z}_\ell \to T_\ell$$ is an isomorphism) which is preserved by $$F_p$$ in the sense that $$F_p(\Lambda) \subset \Lambda$$ (note that one cannot expect equality here since the determinant of $$F_p$$ acting on $$T_\ell$$ is $$p$$). Certainly, if you already know that $$F_p$$ has integral characteristic polynomial, then you can easily construct such lattices: take any $$t \in T_\ell \setminus \ell T_\ell$$ that is not an eigenvector for $$F_p$$, then $$\Lambda = t \mathbb{Z} + F_p(t)\mathbb{Z} \subset T_\ell$$ is an $$F_p$$-invariant lattice. So there should be plenty such lattices. But the goal is to construct an $$F_p$$-invariant lattice without using that we already know that $$F_p$$ has integral characteristic polynomial.

One potential lattice can be constructed as follows. We choose a complex-analytic uniformization $$E(\mathbb{C}) = \mathbb{C}/\Lambda_0$$ for some lattice $$\Lambda_0 \subset \mathbb{C}$$. Then we define a map $$\Lambda_0 \to T_\ell$$ by sending $$\lambda \in \Lambda_0$$ to the sequence $$(\ell^{-1} \lambda, \ell^{-2} \lambda, \ell^{-3}\lambda, \ldots) \in T_\ell$$, which is well-defined because $$\ell^{-n}\lambda \in E(\mathbb{C})[\ell^n] = E(\overline{\mathbb{Q}})[\ell^n]$$. Let $$\Lambda_\ell \subset T_\ell$$ be the image of this map. It is not hard to prove that $$\Lambda_\ell$$ is free of rank 2 and that $$\Lambda_\ell \otimes \mathbb{Z}_\ell \to T_\ell$$ is an isomorphism. Also note that $$\Lambda_\ell$$ does not depend on the choice of the uniformization.

Question: Does $$F_p(\Lambda_\ell) \subset \Lambda_\ell$$ hold?

P.S. I've tried searching for results in this direction in various places, but did not find much. If someone has suggestions for references or keywords to search for, I would be much obliged.

There is a subtle problem with this idea, that causes serious problems. You observed that $$\Lambda_\ell \otimes \mathbb Z_\ell = T_\ell$$ but didn't find any other information for it. There is a reason for that.

Let $$K$$ be the field generated by the coordinates of the $$\ell$$-power torsion points of $$E$$. Given an $$\ell$$-power torsion point defined over $$F$$, to make Frobenius act on it, we need to know its reduction mod $$p$$, so we need to embed $$F$$ into the maximal unramified extension $$\mathbb Q_p^{ur}$$ of $$\mathbb Q_p$$.

Given a point in the homology of $$E_{\mathbb C}$$, to find the corresponding point of $$F$$, we need to express the coordinates as complex numbers, so we need to embed $$F$$ into $$\mathbb C$$.

Are these embeddings canonical? Well, if we define $$F$$ as the field generated by the complex coordinates of $$\ell$$-power torsion points, then the second embedding is canonical but the first isn't. If we define $$F$$ as the field generated by the $$p$$-adic coefficients of torsion points, then the first embedding is canonical but the second isn't. So regardless, there is some ambiguity - we can translate one of our embeddings by an automorphism of $$F$$ and get one that looks equally reasonable.

How bad is that ambiguity? Fixing an automorphism $$\sigma \in \operatorname{Gal}(F/\mathbb Q)$$ of $$F$$, making this change of embeddings corresponds exactly to translating your lattice by the action of $$\sigma$$ on $$T_\ell(E)$$. So the set of lattices we obtain your construction is a $$\operatorname{Gal}(F/\mathbb Q)$$-orbit in $$T_\ell(E)$$.

For $$E$$ generic, we have $$\operatorname{Gal}(F/\mathbb Q) \cong GL_2(\mathbb Z_\ell)$$, so the orbit is quite large. In fact every single lattice $$\Lambda$$ with $$\Lambda \otimes \mathbb Z_\ell = T_\ell$$ lies in this orbit, because we can find a matrix in $$GL_2$$ taking the basis of one such lattice to another. So there is no more information available about these lattices than your initial observation that $$\Lambda \otimes \mathbb Z_\ell = T_\ell$$!

Of course, there are examples of such $$\Lambda$$ stable under $$F$$ and examples not stable under $$F$$.

For any $$E$$ non-CM, the situation is the same, because the Galois group is an open subgroup of $$GL_2(\mathbb Z_\ell)$$ and these act transitively on the set $$GL_2(\mathbb Z_\ell)/GL_2(\mathbb Z)$$ of lattices $$\Lambda$$, since $$GL_2(\mathbb Z)$$ is dense in $$GL_2(\mathbb Z_\ell)$$.

For $$E$$ CM, the situation is different, as the Galois group is much smaller. If $$p$$ is a supersingular prime, then David Speyer's argument shows $$\Lambda_\ell$$ is never stable under Frobenius. Conversely, if $$p$$ is an ordinary prime, then the endomorphism $$V =p /F$$ lifts to an endomorphism of the curve over the CM field and thus an endomorphism of the curve over $$\mathbb Q$$, thus always preserves $$\Lambda_\ell$$, and because its determinant is $$p$$, $$F= p/V$$ necessarily preserves $$\Lambda_\ell$$ as well. So for CM curves, Frobenius preserves this lattice if and only if $$p$$ is ordinary.

• Thanks! You are right, the question is subtly but seriously problematic. I actually mentioned in the question already that the action of $F_p$ is well-determined only up conjugation, but somehow I did not realize that this makes the question of whether $F_p(\Lambda_\ell) \subset \Lambda_\ell$ completely meaningless in most cases. – Marc Paul Sep 22 at 16:05

Any construction along these lines is going to run into an obstruction pointed out by Serre. Consider the elliptic curve $$E = \{ y^2 = x^3+x \}$$ over $$\mathbb{Z}[i]$$, and let $$p$$ be a prime which is $$3 \bmod 4$$. Let $$E/p$$ be the reduction of $$E$$ modulo $$p$$ (which remains prime in $$\mathbb{Z}[i]$$). Then $$E/p$$ has the following endomorphisms:

• The $$p$$-power Frobenius $$F(x,y) = (x^p, y^p)$$ and
• The complex muliplication $$J(x,y) = (-x, iy)$$.

These maps obey $$JF=-FJ$$, $$J^2 = -1$$ and $$F^2 = -p$$.

There do not exist $$2 \times 2$$ integer matrices obeying these relations. (Proof below.) So there is no construction which associates a $$\mathbb{Z}$$-lattice to an elliptic curve and is functorial in charateristic $$p$$. So it is impossible that $$J$$ and $$F$$ both preserve your lattice. I didn't think about this in detail, but it seems much more likely that $$J$$ does than $$F$$.

Proof that there are not integer matrices obeying $$J^2 = -1$$, $$JF = -FJ$$ and $$F^2 = -p$$: Suppose otherwise. Using $$J^2 = -1$$, we can choose bases so that $$J = \left[ \begin{smallmatrix} 0&-1 \\ 1&0 \end{smallmatrix} \right]$$. The equation $$JF=-FJ$$ means that $$F$$ is of the form $$F = \left[ \begin{smallmatrix} a&b \\ b&-a \end{smallmatrix} \right]$$. Then $$F^2 = (a^2+b^2) \mathrm{Id}$$. There is no solution to $$a^2+b^2 = -p$$ in integers.

• $J$ certainly preserves the lattice because it extends to an endomorphism over the complex numbers. So indeed $F$ must not. – Will Sawin Sep 22 at 15:07
• Thanks! This argument is quite famous and I have actually seen this argument before, so I should have known that what I was trying to do had no chance of ever working... – Marc Paul Sep 22 at 16:09