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Let $A$ be an abelian variety of dimension $g$ over an algebraically closed field $k$.

Assume $k$ has characteristic $p$ and denote by $A(p)$ the $p$-divisible group of dimension $g$ associated with $A$. Then the map $$ End(A) \to End(A(p)) $$ is injective and, if $k = \overline{\mathbb{F}}_p$, even a bijection after $p$-completion (by a theorem of Tate). In particular, every automorphism of $A$ inducing the identity on the p-divisible group is already the identity.

In dimension $1$, i.e. for elliptic curves something significantly stronger is true. Every automorphism of an elliptic curve inducing the identitiy on the formal group is the identity. Indeed, the following stronger statement is true and proven in Notes by Charles Rezk, Proposition 12.2: A choice of coordinate of the formal group up to terms of order 5 is equivalent to the choice of Weierstrass coordinates (and no non-trivial automorphism fixes the Weierstrass coordinates).

Is it true in general that automorphisms of abelian varieties are detected by the formal group in the sense that every automorphism of an abelian variety $A$ inducing the identity on the formal group of $A$ is already the identity?

Note that the corresponding result is clearly true if $k=\mathbb{C}$ as already every automorphism fixing the Lie algebra is the identity.

In some situations, it is possible to assign to an abelian variety $A$ of dimension $g$ a $1$-dimensional instead of a $g$-dimensional formal group/$p$-divisible group. To that purpose assume (as in the statement of a PEL-Shimura moduli problem):

  • A is ($\mathbb{Z}_{(p)}$)-principally polarized.
  • For a quadratic imaginary extension $F$ of $\mathbb{Q}$, we have a morphism $\mathcal{O}_F \to End(A)$ such that complex conjugation is sent to the Rosatti involution.
  • $p$ splits in $F$ so that $\mathcal{O}_F \otimes \mathbb{Z}_p \cong \mathbb{Z}_p\times \mathbb{Z}_p$. Denote a corresponding non-trivial idempotent in $\mathcal{O}_F \otimes \mathbb{Z}_p$ by $e$.
  • eA(p) is $1$-dimensional.

Also here it is true that every automorphism of $A$, fixing the polarization and the endomorphism structure, that induces the identity on $eA(p)$ is already the identity.

Is it true in the situation above that every automorphism of $A$, fixing the polarization and the endomorphism structure, that induces the identity on the formal group of $eA(p)$ is already the identity?

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    $\begingroup$ An automorphism that is the identity on the formal group induces the identity on the local ring at $0$ (because the formal group is defined on the completion of that local ring, and the local ring injects into its completion). Hence it induces the identity on the function field of $A$ and thus it must be the identity (but maybe I missed an obvious point ? Let me know). $\endgroup$
    – Damian Rössler
    Commented Jan 4, 2014 at 16:18
  • $\begingroup$ Thanks! I am the one who missed something. Can one use it to answer also the second question? $\endgroup$
    – Lennart Meier
    Commented Jan 4, 2014 at 22:31
  • $\begingroup$ I am not sure but note that an automorphism of $A$, which fixes the polarization must be of finite order. See for instance Milne's 'Abelian varieties' (in the book edited by Cornell and Silverman), Prop. 17.5 $\endgroup$
    – Damian Rössler
    Commented Jan 4, 2014 at 22:46
  • $\begingroup$ I think the answer to both questions is yes. Consider the endomorpyism $\phi$-1. Since it has a kernel on the fundamental group it must be of degree 0, and hence kills a positive-dimensional abelian subvariety $B$ of $A$, such that $B(p)$ contains $eA(p)$. Moreover, replacing $B$ by $\mathcal{O}_F\cdot B$ we have a splitting up to isogeny $A\sim B+C$ which is complatible with the endomorphism action. I think this contradicts $eA(p)$ being one-dimensional. Is that right? $\endgroup$
    – jacob
    Commented Oct 6, 2014 at 6:53
  • $\begingroup$ @jacob: I agree that $\phi−1$ kills a positive-dimensional abelian subvariety $B$. I do not see that $B(p)$ contains $eA(p)$ as we do not know that $\phi$ induces identity on $A(p)$. I also do not see the contradiction at the end: Why should $eC(p)$ not be zero-dimensional? $\endgroup$
    – Lennart Meier
    Commented Oct 22, 2014 at 3:03

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