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.


1 Answer 1


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.

  • $\begingroup$ 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?) $\endgroup$
    – Will Chen
    Apr 14, 2015 at 19:00
  • $\begingroup$ @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 $\mathbb Q$. $\endgroup$
    – Will Sawin
    Apr 14, 2015 at 19:04
  • $\begingroup$ @WillSawin: You really mean at the end of your answer that the endomorphism ring is a maximal order in such a quaternion algebra (this resting on work of Tate or Deuring) and computing the units of that maximal order (which is unique up to conjugation in these specific cases since such elliptic curves in characteristics 2 and 3 are unique in their isogeny classes). $\endgroup$
    – user74230
    Apr 15, 2015 at 3:01
  • $\begingroup$ @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. $\endgroup$
    – Will Chen
    Apr 17, 2015 at 3:36
  • 2
    $\begingroup$ @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. $\endgroup$
    – Will Sawin
    Apr 17, 2015 at 12:42

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