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Let $A$ be an abelian variety and and $\sigma$ an automorphism of $A$. Suppose $f:A\rightarrow P^1$ is a morphism. Is it true that $\sigma$ descends to an automorphism of $P^1$? I seem to remember reading this somewhere (or something like it). Does anybody have any references?

thank you!

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Let E be an elliptic curve and let x,y be points on it. Then the divisor

[x] + [-x] + [y] - 2[2y] - [-3y]

is principal, so is div(f) for some function f:E -> P^1; this function sends x and y to the same point of P^1 (namely, 0) but sends -x and -y to different points (unless you were very unlucky in your choices) so [-1] can't descend along f.

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By the way, the point of this answer is 1) to answer your question, and 2) to make the point that when you are wondering "does every X satisfy Y" you should try the simplest possible X first! – JSE Dec 1 '09 at 3:52

Something is wrong with this statement: your map to $\mathbb{P}^{1}$ is given by a pencil in a linear system $|L|$ on $A$. Since the vector space dimension of the linear system is at least two it follows that there is only a finite group of translations on $A$ that will preserve the isomorphism class of $L$. So if $\sigma$ is not in this finite group $\sigma^{*}L \neq L$ and so $\sigma$ does not descend to an automorphism of $\mathbb{P}^{1}$.

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Regarding this, Tony, there's a notational question of whether translations count for the poster as automorphisms of the abelian variety -- for me, Aut(A) consists of automorphisms of the variety preserving the identity. – JSE Dec 1 '09 at 3:50
JSE's comment is well-taken (as is his example). Nevertheless, TP's example was the first one that leapt to my mind as well: the fact that for an (ample, basepoint free) line bundle L on an abelian variety A, the group of translations on A commuting with the corresponding morphism to projective space is the finite group Ker Phi_L: A -> A^{\vee} is of vital importance in the geometry of abelian varieties, e.g. in Mumford's definition of theta groups. – Pete L. Clark Dec 1 '09 at 4:22
Ah! I see. I misunderstood the question and thought that $\sigma$ can be any automorphism of $A$ as a variety, not as a group. – Tony Pantev Dec 1 '09 at 5:13

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