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Let $C$ be a (very) general genus 1 curve embedded in $\mathbb{CP}^1\times \mathbb{CP}^1$ as a (2,2)-divisor.

Each projection defines $C$ as a double cover of $\mathbb{CP}^1$ and induces an involution $\tau_i:C\to C$. Let $G\simeq \mathbb Z/2 * \mathbb Z/2$ be the subgroup of $Aut(C)$ generated by these. Note that $G$ is infinite.

Are there points on $C$ with finite orbit under $G$?

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No. Letting $\sigma$ and $\tau$ denote the two involutions, $\sigma \circ \tau$ is a translation by an element of $\mathrm{Pic}^0(C)$. In general, this translation will not be torsion, so its orbit through any point is infinite. (In fact, if the translation IS torsion, then $G$ is a finite dihedral group, not $\mathbb{Z}/2 \ast \mathbb{Z}/2$.)

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  • $\begingroup$ Thanks for the answer! Do you also know what happens if you replace $C$ with a K3 surface in $(\mathbb{CP}^1)^3$ with the $3$ induced involutions? $\endgroup$
    – gsvr
    Commented Oct 4, 2015 at 19:14
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    $\begingroup$ “I was gratified to be able to answer promptly, and I did. I said I didn’t know.” Mark Twain $\endgroup$
    – David E Speyer
    Commented Oct 4, 2015 at 19:45
  • $\begingroup$ Hehe ok thanks. Well, the K3 is fibered by elliptic curves like that in 3 ways, so it is perhaps likely that also here the points have infinite order. $\endgroup$
    – gsvr
    Commented Oct 4, 2015 at 20:21
  • $\begingroup$ [Edited only to supply the missing ")" at the end.] $\endgroup$
    – Noam D. Elkies
    Commented Oct 5, 2015 at 3:05
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Regarding K3 surfaces in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$, you might find the following two articles relevant. (Also forward trace them as references for further articles.)

  1. Baragar, A. Rational points on K3 surfaces in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$. Math. Ann. 305 (1996), no. 3, 541-558. (MR1397435)

  2. Wang, Lan Rational points and canonical heights on K3-surfaces in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$. Recent developments in the inverse Galois problem (Seattle, WA, 1993), 273-289, Contemp. Math., 186, Amer. Math. Soc., Providence, RI, 1995 (MR1352278)

Addendum Since you mention that you're interested in higher dimensional analogous, I'll mention the following. The involutions on a $(2,2,2)$-surface in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$ are morphisms. But if $n\ge4$, then the involutions on a $(2,2,\ldots,2)$ hypersurface in $\underbrace{\mathbb P^1\times\mathbb P^1\times\cdots\times\mathbb P^1}_{\text{$n$ factors}}$ have non-trivial indeterminacy locus, so some points have orbits that terminate.

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  • $\begingroup$ Thank you! I'm mostly interested the question for K3s and higher dimensional Calabi-Yaus, so these references look useful! $\endgroup$
    – gsvr
    Commented Oct 4, 2015 at 23:28
  • $\begingroup$ A reference for this construction in higher-dimensional settings is Cantat--Oguiso, " Birational automorphism groups and the movable cone theorem for Calabi-Yau manifolds of Wehler type via universal Coxeter groups". In general you only get rational maps, but they have codimension 2 indeterminacy -- such maps are sometimes called "pseudoautomorphisms". $\endgroup$
    – user47305
    Commented Oct 5, 2015 at 15:01
  • $\begingroup$ @Mark Thanks for the reference. But is your point that the indeterminacy locus in that setting is exactly codimension 2. In general, the indeterminacy locus of any rational map between smooth varieties has codimension at least 2. $\endgroup$
    – Joe Silverman
    Commented Oct 5, 2015 at 17:08
  • $\begingroup$ Whoops, sorry. The point is that not only is the indeterminacy locus codim 2 (duh), but that the map is an isomorphism in codimension 1 -- no divisors are contracted in either direction. This means in particular that there is still a functorial pullback on $N^1(X)$ (i.e. $(f^\ast)^n = (f^{\circ n})^\ast$), which you won't get for a general rational map. Thanks! $\endgroup$
    – user47305
    Commented Oct 5, 2015 at 23:14
  • $\begingroup$ @Mark Ah, the map is what the dynamics crowd calls algebraically stable. That's definitely of interest. $\endgroup$
    – Joe Silverman
    Commented Oct 6, 2015 at 1:15

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