# how do automorphisms of elliptic curves act on the Tate module?

Let $E/k$ be an elliptic curve over some algebraically closed field $k$ of characteristic $p\ge 0$. It's known that $Aut(E)$ acts faithfully on the Tate module $T_\ell(E)$ ($\ell\ne p$) with determinant 1. Is there a complete description of the actions of $Aut(E)$ on $T_\ell(E)$? Ie, for any elliptic curve as above, can we describe the subgroup $Aut(E)\subset SL_2(\mathbb{Z}_\ell)\subset Aut(T_\ell(E))$ up to conjugacy?

In characteristic 0 the action can be computed analytically and we find that the "extra automorphisms" $i,\rho$ of orders 4,6 (corresponding to $j$-invariant 1728,0) essentially act via conjugates of

$$M_i =\begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}\qquad M_\rho = \begin{bmatrix}1 & 1 \\ -1 & 0\end{bmatrix}$$

Reducing mod $p$ one finds that the reductions $\overline{i},\overline{\rho}$ act on the $\ell$-power torsion in the same way, and so the matrices are the same.

Thus, my question reduces to: If $char(k) = 2$ or 3, and $j = 0\equiv 1728$, then we have automorphism groups of order either 12 (characteristic 3), or 24 (characteristic 2).

In this case what is the subgroup $Aut(E)\subset SL_2(\mathbb{Z}_\ell)$? We certainly get both the automorphisms $\overline{i},\overline{\rho}$ with matrices $M_i,M_\rho$, but there isn't necessarily a single choice of basis for $T_\ell(E)$ such that $\overline{i},\overline{\rho}$ have matrices $M_i,M_\rho$ respectively. Also, in characteristic 2, there are additional automorphisms which aren't in the group generated by $\overline{i},\overline{\rho}$.

I feel like this must have been done somewhere, but I can't find any references for this.

How many subgroups of order $12$ and $24$ are there in $SL_2(\mathbb Z_l)$? By the classification of finite subgroups of $SO(3)$ there are just two of each, one abelian and one non-abelian. It is easy to see that the abelian one cannot appear because the characteristic polynomial of each element should be integral. The non-abelian ones of order $12$ and $24$ are the degree $2$ central extensions of $S_3$ and $A_4$ respectively.
A more abstract computation would be observing that the endomorphism algebra is a quaternion algebra ramified at $p$ and $\infty$ and computing the group of units of this algebra.
• By the characteristic polynomial being integral, do you mean having coefficients in $\mathbb{Z}$? Can you explain why that must be true (and how it is related to the group being abelian?) Apr 14, 2015 at 19:00
• @oxeimon Yes, that's what I mean. It's because the determinant of the endomorphism of an abelian variety acting on the Tate module is its degree, so the characteristic polynomial which can be expressed as a determinant takes integral values. I guess that only shows the coefficients are rational, not integral. Regardless, the abelian groups are in fact cyclic - the cyclic groups of order $12$ and $24$. Their generators have eigenvalues that are primitive $12$th and $24$th roots of unity. These eigenvalues are not roots of degree $2$ polynomials over $\mathbb Q$. Apr 14, 2015 at 19:04
• @WillSawin So this is really weird. I feel like since the automorphisms have to act on $T_\ell(E)$ for every $\ell\ne 2$ (or 3), they really should act via matrices in $SL(2,\mathbb{Z}$. So suppose $E_0$ is the elliptic curve over $\overline{\mathbb{F}_2}$ with $j$-invavriant $0\equiv 1728$ with 24 automorphisms. You can lift this to characteristic 0 (say, over the maximal unramified ext of $\mathbb{Z}_2$) to either $j$-invariant 0 or $j$-invariant 1728, where you find cyclic automorphism groups of order 6,4 respectively. Apr 17, 2015 at 3:36
• @oxeimon Your idea that they should act via matrices in $SL(2,\mathbb Z)$ is reasonable but not quite correct. What's true is that they should act via matrices in an algebraic group $G$ over $\mathbb Z$ such that $G \otimes \mathbb Z_\ell = SL_2(\mathbb Z_\ell)$ for every $\ell$ but $2$ or $3$. $SL_2$ satisfies this, but the group of norm $1$ elements in a maximal order of a quaternion algebra ramified at $2$ or $3$ also satisfies this. Apr 17, 2015 at 12:42