Given an elliptic curve $E$, it seems that $max(ord(Aut(E)))$ over the prime 2 is 24, and $(max(ord(Aut(E)))$ over the prime 3 is 12.

The endomorphism algebra of an elliptic curve over $\overline{\mathbb{F}}_p$ is, after tensoring with $\mathbb{Q}$, either an imaginary quadratic field or the unique division algebra $D_p$ which is non-split at $p$ and infinity. So, in order to compute the maximal order of the automorphism algebra, we just compute the units of $D_p$, and then count them.

Here's my question: Given a Shimura variety, how does one compute the maximal order of its automorphism group?

For example, the variety $\mathbb{C}^{\times 3}/ \mathbb{Z}[\zeta_7]^{\times 3}$ where the embedding is given by $\sigma_1 \times \sigma_2 \times \sigma_3: \mathbb{Z}[\zeta_7]^{\times 3} \to \mathbb{C}^{\times 3}$. Here, we look at its automorphisms over $\mathbb{F}_p$ where $p$ is $2$ or $4$ mod $7$.

I have heard that Shimura varieties have finite dimensional automorphism groups (analogous to the fact that $Aut(A, p)$ is finite, where $p$ is a polarization of the abelian variety $A$). What is the automorphism group of this Shimura variety?

  • $\begingroup$ Your question is potentially an interesting one, but as formulated it is not well defined. What Shimura variety are you referring to here? You said the words "Shimura variety" but you didn't mention any one in specific. Are you thinking of this abelian 3-fold as a "point" on a certain Siegel modular 3-fold? A unitary Shimura variety? With level structure and polarization? There are many possibilities... $\endgroup$ – Ari Shnidman Jul 27 '17 at 19:41

Summary: The automorphism group of this variety at least contains $Z/2 \oplus Z/6 \simeq \mathbb{Z}[\zeta_7]^\times$,and is isomorphic to it if the variety is nondegenerate.

More generally, I learned today that given $A$ of the form $\mathbb{C}^{\phi(n)/2}/\mathbb{Z}[\zeta_n]$ (with the appropriate choice of embedding); then $Aut(A)$ is indeed simply $\mathbb{Z}[\zeta_n]^\times$ (and $End(A)$ is $\mathbb{Z}[\zeta_n]$.

To convince oneself that this is at the very least a subgroup of the endomorphism group, recall that endomorphisms of $\mathbb{C}^g/\Lambda$ are maps which preserve the lattice $\Lambda$ (or at least map the lattice to a scaled version of itself, such as $\Lambda \mapsto 2\Lambda$. For example, for any $x \in \mathbb{Z}[\zeta_n]$, we can multiply component-wise every point, taking the lattice to an $x$ scaled version of itself (where we consider $\zeta_n \in \mathbb{C}$ as $e^{2 \pi i/n}$):

$$\mathbb{C}^g \to \mathbb{C}^g$$

$$(p_1, ..., p_g) \mapsto (\sigma_1(x)p_1, ..., \sigma_g(x)p_g)$$

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  • $\begingroup$ There seem to be many misconceptions in this answer. The group of units of the ring of integers in a degree 6 cyclotomic field is a fin. gen. abelian group of rank 2 (in particular, infinite). Perhaps you wanted to choose a polarization as well, to make the automorphism group finite? Also, the abelian variety you write down really does depend on the CM type (a choice of 3 complex embeddings, which you haven't specified). For general n, it can certainly happen that the abelian variety is not simple and hence there's no chance that the endomorphism rind is a domain. $\endgroup$ – Ari Shnidman Jul 27 '17 at 19:42
  • $\begingroup$ Oh, sorry, it looks like I missed your parenthetical remark about appropriate choice of embedding. Do you have a reference for the fact that there is always a primitive CM type for the cyclotomic field, for every n? $\endgroup$ – Ari Shnidman Jul 27 '17 at 21:46

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