Timeline for Are automorphisms of abelian varieties detected by the formal group?
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6 events
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| Oct 22, 2014 at 3:03 | comment | added | Lennart Meier | @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? | |
| Oct 6, 2014 at 6:53 | comment | added | jacob | 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? | |
| Jan 4, 2014 at 22:46 | comment | added | Damian Rössler | 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 | |
| Jan 4, 2014 at 22:31 | comment | added | Lennart Meier | Thanks! I am the one who missed something. Can one use it to answer also the second question? | |
| Jan 4, 2014 at 16:18 | comment | added | Damian Rössler | 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). | |
| Jan 3, 2014 at 16:27 | history | asked | Lennart Meier | CC BY-SA 3.0 |