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I am considering $K3$ surfaces in $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$ with an automorphism that preserves an ample divisor class.

For an automorphism $\rho$ of a $K3$ surface, let ${\rm Pic}^\rho(S) = \{ D \in {\rm Pic}(S) \mid \rho^*(D) = D \}$.

Let me state the question:

For a given primitive ample divisor class $H$ of $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$, does there exist a pair $(S, \rho)$ of a smooth $K3$ surface $S$ that is an anti-canonical section of $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1 $ and an automorpism $\rho$ of $S$ such that $${\rm Pic}^\rho(S) = \left \langle H|_S \right \rangle?$$

According to Arnaud BEAUVILLE (https://arxiv.org/pdf/math/0211313), a $20-3=17$ dimensional family of $K3$ surfaces lie inside $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1 $. I do not see any obstacles which prevent such a pair from existing.

I guess that the Torelli theorem may help.

Does anyone have an answer?

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    $\begingroup$ What kind of automorphism? There is an obvious cyclic automorphism of order $3$ of $\mathbb{P}^1\times\mathbb{P}^1\times \mathbb{P}^1$ with rank $1$ invariant subgroup of the Picard group., It looks like the usual Noether-Lefschetz argument will give that a sufficiently general $(2,2,2)$-hypersurface that is stabilized by the automorphism will have the same Picard group (and will have $6$ fixed points). However, if your automorphism is free on the K3 surface, obviously this forces the K3 surface to have "invariant Picard rank" equal to $10$. $\endgroup$
    – Jason Starr
    Commented May 1, 2022 at 10:07
  • $\begingroup$ @Jason star, no other requirements are put on, hence you answered the question! Being excited by such a quick answer, I changed the question a bit, which I would like to know the answer originally. $\endgroup$
    – Basics
    Commented May 1, 2022 at 10:26
  • $\begingroup$ You added a hypothesis about a specified primitive ample divisor class. For most choices of primitive ample divisor class, the answer is "no". The normal bundle of the surface in $\mathbb{P}^1\times \mathbb{P}^1\times \mathbb{P}^1$ is already one $\rho$-invariant divisor class. If you specify another $\rho$-invariant divisor class that is not a (rational) multiple of the class of the normal bundle, then necessarily the "invariant Picard rank" is at least $2$. $\endgroup$
    – Jason Starr
    Commented May 1, 2022 at 15:01
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    $\begingroup$ Why is the normal bundle $\rho$-invariant? Are you assuming that $\rho$ comes from an automorphism of $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$? $\endgroup$
    – Basics
    Commented May 1, 2022 at 15:36
  • $\begingroup$ Yes, I was assuming that the automorphism comes from $\mathbb{P}^1\times \mathbb{P}^1\times \mathbb{P}^1$. If the automorphism does not arise in this way, then there is no reason that the normal bundle should be $\rho$-invariant. $\endgroup$
    – Jason Starr
    Commented May 1, 2022 at 16:26

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In the generic case you'll have $\text{NS}(S)\otimes\mathbb R\cong\mathbb R^3$ and you can compute the $3$-by-$3$ matrix for the action of the three involutions $i_1,i_2,i_3$ on, say, a basis consisting of the three pullbacks of a point via the projections $\pi_1,\pi_2,\pi_3:S\to\mathbb P^1$. My recollection is that the eigen-divisors of infinite-order automorphsims such as $i_1\circ i_2$ are nef, but not ample, so they'll lie on the boundary of the ample cone. In other words, you may be able to get a nef eigen-divisor in $\text{NS}(S)\otimes\mathbb R$, but it seems unlikely you can get an ample eigen-divisor, at least when the picard number is $3$.

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  • $\begingroup$ I also expect that such a K3 surface $S$ is not a generic one. I simply would like to know whether such a K3 surface exists, regardless of generic or not. $\endgroup$
    – Basics
    Commented May 1, 2022 at 12:00
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    $\begingroup$ @user69559 Fair enough, but you said that there is a 17-dimensional family of such surfaces (up to isomorphism) and that you "do not see any obstacles which prevent such a pair from existing," so I was noting that picard number 3 is probably an obstacle, so you'll need to look within the (countably many?) subfamilies of dimension 16 having picard number ${}\ge4$ for an example. $\endgroup$
    – Joe Silverman
    Commented May 1, 2022 at 15:19

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