3
$\begingroup$

Here is the new version of the question which is more explicit. The older version is below.

I am looking for complex projective varieties (in dimensions $2$ and higher) admitting a fixed-point free holomorphic involution that sit as a hypersurface in variety with lots of rational curves (say $\mathbb{P}^n$). Fermat quartic surface in $\mathbb{P}^3$ is an example. What are the examples and restrictions? e.g., for which $d\geq 3$, there is a hypersurface of degree $d$ in $\mathbb{P}^n$ admitting such an involution?

In complex dimension 1, (some) Riemann surfaces of odd genus admit a holomorphic involution without fixed-point. In complex dimension 2, abelian surfaces and K3 surfaces are the first examples that come to mind (for admitting a holomorphic involution without fixed-point). There are possibly other example that are elliptic fibrations over a curve.

I am looking for a larger pool of such varieties in complex dimensions 2 and 3.

  • What are the restrictions imposed by the existence of a holomorphic involution without fixed-point? (For instance, rationally connected varieties do not admit such an involution)

  • Are all examples in dim 2 and 3 abelian or K3 fibered varieties?

Improve this question
$\endgroup$
6
  • 6
    $\begingroup$ Given one such variety $X$, for every variety $Y$, the product $X\times Y$ is another such variety. Starting from $X$ a curve of genus $g>0$, you can produce examples in every dimension $n$, e.g., $X\times \mathbb{P}^{n-1}$. I wrote a bit more about this once in a comment to an MO post by user abx . . . $\endgroup$
    – Jason Starr
    Commented Jul 11, 2022 at 18:33
  • 2
    $\begingroup$ Many compact ball quotients admit fixed-point free holomorphic involutions. $\endgroup$
    – Moishe Kohan
    Commented Jul 11, 2022 at 18:59
  • 2
    $\begingroup$ In the opposite direction --- if $Y$ is a variety such that $\pi_1(Y)$ has a surjection onto $\mathbb{Z}/2$, the corresponding etale double covering $X$ of $Y$ has such an involution. $\endgroup$
    – Sasha
    Commented Jul 11, 2022 at 20:44
  • 1
    $\begingroup$ Thanks, Jason. That's helpful. Is there a way to tell which hypersurfaces in $\mathbb{P}^n$ have such an involution? (or better said, for what $d\geq 0$, there is a degree $d$ hypersurface in $\mathbb{P}^n$ admitting such an involution. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Jul 12, 2022 at 14:16
  • 1
    $\begingroup$ "Is there a way to tell which hypersurfaces in $\mathbb{P}^n$ have such an involution?" For all $n\geq 4$, there is no such hypersurface. For $n\geq 4$, the restriction map on Picard groups from $\mathbb{P}^n$ to the hypersurface is an isomorphism. Thus, every automorphism extends to $\mathbb{P}^n$. The fixed locus of the involution of $\mathbb{P}^n$ equals two linear subspaces, at least one of which has dimension $>1$. The intersection of this linear subspace with the hypersurface is nonempty. $\endgroup$
    – Jason Starr
    Commented Jul 13, 2022 at 11:24

2 Answers 2

Reset to default
7
$\begingroup$

For $n\geq 4$, a smooth hypersurface $X\subset\mathbb{P}^n$ never admits a fixed point free involution. By the Lefschetz theorem $\operatorname{Pic}(X) $ is cyclic, so any automorphism of $X$ preserves $\mathscr{O}_X(1)$, hence is induced by an automorphism of $\mathbb{P}^n$. Now an involution of $\mathbb{P}^n$ has two fixed subspaces of dimension $p$ and $q$ with $p+q=n-1$, so one (at least) of these subspaces intersect $X$.

The same result holds for $n=3$ and $\deg (X)\neq 4$. In this case $\mathscr{O}_X(1)$ is the unique line bundle $L$ on $X$ such that $L^{d-4}=K_X$, hence it is again preserved by any automorphism of $X$, and the same argument applies.

Improve this answer
$\endgroup$
3
  • $\begingroup$ This is concrete, thanks. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Jul 13, 2022 at 13:41
  • 1
    $\begingroup$ I am just adding some additional information here. Even if you consider actions of finite cyclic groups of arbitrary order $r>1$ on smooth complete intersections of codimension $c$, then the same kind of argument shows that there must be fixed points if $n+1\geq r(c+1)$. A similar combinatorial argument shows there are fixed points if $d\leq n$. For (irreducible) hyper-Kaehler manifolds, there can only be a free action if the dimension of the hyper-Kaehler is congruent to $-2$ module $2r$. $\endgroup$
    – Jason Starr
    Commented Jul 13, 2022 at 18:19
  • 1
    $\begingroup$ Actually the optimal inequality above is $n+1 > rc$ (a bit better than the inequality I wrote down). If you choose $n+1$ equal to $rc$ and you choose each degree equal to $r$ (so that the sum of the degrees $d$ equals $n+1$), then there are complete intersections with no fixed points. $\endgroup$
    – Jason Starr
    Commented Jul 14, 2022 at 10:32
4
$\begingroup$

The general recipe for constructing such varieties is the following.

Start with your favorite smooth variety $Y$ such that $\operatorname{Pic}^0(Y) \neq 0$ (for instance, this condition is automatically satisfied if $H^0(Y, \, \Omega_Y^1) \neq 0$), and choose a non-zero, $2$-torsion divisor $\mathcal{L}$ in $\operatorname{Pic}^0(Y)$. Correspondingly, there is an étale double cover $$\pi \colon X \to Y, \quad \pi_*\mathcal{O}_X= \mathcal{O}_Y \oplus \mathcal{L},$$ and the generator of the deck transformations of $\pi$ is a fixed-point free holomorphic involution on $X$.

The Kodaira dimension can only increase under this procedure, namely $\operatorname{kod}(X) \geq \operatorname{kod}(Y)$. In particular, if we start with $Y$ a variety of general type (for example, a product of curves of genus $\geq 2$), it follows that $X$ is of general type, too. Thus, the answer to your second question is no.

Improve this answer
$\endgroup$
9
  • 1
    $\begingroup$ Honestly, I do not understand your comment. This is the general recipe, so all the examples (including Jason's) must arise in this way. I also do not understand the "equally hard": for instance, take any irregular surface of general type (there is plenty of examples, for instance, product of curves with genus $\geq 2$, or complete intersections in Abelian varieties), choose any two-torsion divisor and you have your new variety endowed with the holomorphic involution. This is completely explicit. $\endgroup$
    – Francesco Polizzi
    Commented Jul 12, 2022 at 16:10
  • 1
    $\begingroup$ Furthermore, it seems that you are ignoring the fact that I answered (negatively) your second question. Anyway, it does not matter... $\endgroup$
    – Francesco Polizzi
    Commented Jul 12, 2022 at 16:12
  • 1
    $\begingroup$ Ok. But, in your question, you write "I am looking for a larger pool of such varieties in complex dimensions 2 and 3.", and then you ask "(2) Are all examples in dim 2 and 3 abelian or K3 fibered varieties?". Indeed, my answer provides lots of examples in every dimension (since there are plenty of irregular varieties in every dimension) and answers negatively question (2). If you wanted to know specifically about hypersurfaces in projective space, in my opinion you should have asked it explicitly. Please note that I do not want to make any controversy, I am just explaining my point. $\endgroup$
    – Francesco Polizzi
    Commented Jul 12, 2022 at 18:56
  • 1
    $\begingroup$ By the way, I do not remember now whether there exists an Enriques involution over a K3 which is a quartic hypersurface of $\mathbb{P}^3$. $\endgroup$
    – Francesco Polizzi
    Commented Jul 12, 2022 at 19:14
  • 1
    $\begingroup$ @FrancescoPolizzi. There do exist some such Enriques involutions on quartic hypersurfaces. But none of these Enriques involutions are induced by (regular) automorphisms of the projective space. There is a nice model of all such (complex) K3 surfaces as $(2,2,2)$ complete intersections in $\mathbb{P}^5$ that are (setwise) fixed by an involution of $\mathbb{P}^5$ with fixed locus a disjoint union of two $2$-planes (both disjoint from the K3 surface). I used this model in one article with Graber, Harris, and Mazur (and a similar model in another article I wrote about char $p$). $\endgroup$
    – Jason Starr
    Commented Jul 13, 2022 at 18:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .