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Let $A$ be a principally polarized abelian surface, and $\Gamma \subset A$ a finite subgroup. Denote by $\Theta$ a symmetric theta divisor in $A$. Is it always true that for any $g_1, g_2 \in \Gamma$ the intersection $\Theta \cap \Theta_{g_1} \cap \Theta_{g_2}$ is empty? (Here, $\Theta_g$ denotes the translate of $\Theta$ by $g$).

If this is true, could you explain why? If not, is the locus of principally polarized abelian surfaces for which the above claim fails a divisor in the moduli space?

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    $\begingroup$ You should specify that $g_1$ and $g_2$ are distinct and not equal to $0$. $\endgroup$
    – Joe Silverman
    Commented Dec 15, 2013 at 18:24

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This is always false. Suppose your $A$ is irreducible, i.e. is the Jacobian of a genus 2 curve $C$. Let $p,q,r$ be Weierstrass points of $C$. Take $\Theta =C-p$, $g_1=p-q$ and $g_2=p-r$. Then $\Theta \cap \Theta_{g_1} \cap \Theta_{g_2}$ contains $0$.

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  • $\begingroup$ Thank you for your answer. In your counterexample, both $g_1$ and $g_2$ have order $2$. Are there similar counterexamples in the case where $\Gamma$ contains no element of order $2$? $\endgroup$
    – ginevra86
    Commented Dec 15, 2013 at 15:55
  • $\begingroup$ Similar, no. The point is that torsion points of order $>2$ are much harder to write down. I would be tempted to guess that such counterexamples do not exist for a generic curve, but that would probably be very hard to prove. $\endgroup$
    – abx
    Commented Dec 15, 2013 at 16:07
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perhaps you could also look at Proposition 3.3 in http://www.ams.org/journals/tran/2003-355-08/S0002-9947-03-03303-8/S0002-9947-03-03303-8.pdf

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    $\begingroup$ How is it relevant? $\endgroup$
    – abx
    Commented Dec 15, 2013 at 17:24
  • $\begingroup$ the relevance is the degree of self intersection. The intersection of three algebraically equivalent divisors on a surface is expected to be zero, if they are general enough. using translates, it is subtle and this is well-know problem in diophantine geometry, see works of daniel Bertrand. $\endgroup$
    – john
    Commented Dec 15, 2013 at 17:42

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