4
$\begingroup$

Is it true that every involution $\sigma$ (i.e., $\sigma^2=identity$) of an Enriques surface $X$ acts trivially on $K_X^{\otimes 2}$ i.e., for any $\omega\in K_X^{\otimes 2}$ we have $\sigma^* \omega=\omega$, where by $K_X^{\otimes 2}$ we mean the tensor 2 of the conanical bundle of $X$.

$\endgroup$

2 Answers 2

9
$\begingroup$

Let me try an argument different from Christian's: $\sigma$ does not act freely as $\chi(\mathcal O_X)=1$ and hence not divisible by $2$. At a fixed point $x$, $\sigma$ acts by $\pm1$ on the fibre of $\omega_X$ and hence acts by $1$ on the fibre of $\omega_X^{\otimes2}$. It also acts by a scalar on a global non-zero section of $\omega_X^{\otimes2}$ but as that section is non-zero at $x$ this scalar must be $1$.

Addendum: It seems that it is essential that there are fixed points. If we look at a bielliptic example; $E\times F$ the product of two elliptic curves with $\tau$ acting by an automorphism of order $4$ on $E$ (assumed to have one) and translation by an element of order $4$ on $F$ then if we divide by $\tau^2$ we have that $\tau$ induces an involution which acts by multiplication by $-1$ on global sections of $\omega^{\otimes 2}$.

$\endgroup$
6
  • $\begingroup$ nice! I was hoping for such a more elegant/elementary argument! $\endgroup$ Sep 1, 2011 at 4:31
  • $\begingroup$ @Torsten and Chritian: Thanks for you answer. I am only a student so let me ask some dum questions: 1.I was wondering if what do you mean by $\chi(\mathcal{O}_X)$? 2.Torsten: Do you mean $\tilde{\sigma}$ in Christians notation by $\sigma$? 3. Why $\tilde{\sigma}^2 \in <\tau>$? I know that $<\tilde{\sigma}^2>=<\tau>\times<\sigma>$ Torsten: Can you please explain your answer specifically? Thanks $\endgroup$
    – user13559
    Sep 5, 2011 at 16:28
  • $\begingroup$ No, I am using your notation and don't use the K3 double cover at all. $\endgroup$ Sep 5, 2011 at 16:30
  • $\begingroup$ Edit my last commment: I was meaning $<\tilde{\sigma}>$ instead of $<\tilde{\sigma}^2>$ $\endgroup$
    – user13559
    Sep 5, 2011 at 16:33
  • $\begingroup$ Thanks and what is $\chi(\mathcal{O}_X)$. You please just tell the name and I will search. Thanks $\endgroup$
    – user13559
    Sep 5, 2011 at 17:23
5
$\begingroup$

Yes! However, as Rita, Torsten and Ru pointed out, my first ideas were too simple-minded. Although this makes the remarks by them somewhat unreadable, let me give the corrected answer:

So, let $X$ be a complex Enriques surface and $\sigma$ an involution. Let us denote by $\tilde{X}\to X$ the associated K3-cover, and let $\tau$ be the associated involution, i.e., $X=\tilde{X}/\langle \tau\rangle$.

Now, the automorphism group of $X$ in terms of $\tilde{X}$ is $$ {\rm Aut}(X)={\rm Aut}(\tilde{X},\tau) := {} ( \psi\in{\rm Aut}(\tilde{X})| \psi\tau=\tau\psi ) / \langle\tau\rangle $$ (typesetting the usual brackets does not seem to work?!). In particular, $\sigma$ lifts to an automorphism $\tilde{\sigma}$ of $\tilde{X}$.

Clearly, we have $\tilde{\sigma}^2\in\langle\tau\rangle$. I claim that $\tilde{\sigma}^2={\rm id}$. For otherwise, we would have $\tilde{\sigma}^2=\tau$, and $\tilde{\sigma}$ would be an automorphism of order $4$. In this case, since $\tau$ acts freely on $\tilde{X}$, the same would be true for $\tilde{\sigma}$. However, a K3 surface cannot possess a fixed-point free automorphism of order $4$: the quotient $S$ of $X$ by this automorphism would satisfy $\chi({{\mathcal O}_S})=1/2$, which is absurd. Thus, $\tilde{\sigma}$ is an involution on $\tilde{X}$.

Being an involution, $\tilde{\sigma}$ acts as $\pm{\rm id}$ on the $1$-dimensional vectorspace $H^0(\omega_{\tilde{X}})$. Since $H^0(\omega_{\tilde{X}})^{\otimes2}\to H^0(\omega_{\tilde{X}}^{\otimes 2})$ is onto, we conclude that $\tilde{\sigma}$ acts trivially on global sections of $\omega_{\tilde{X}}^{\otimes2}$.

Now, $\tilde{\sigma}$ induces $\sigma$ on $X$, and global sections of $\omega_X^{\otimes2}$ pull back to global sections of $\omega_{\tilde{X}}^{\otimes2}$. Since $\tilde{\sigma}$ acts trivially on these, we conclude that $\sigma$ acts trivially on global sections of $\omega_X^{\otimes2}$.

It is less obvious, but still true, that automorphisms of order $3$ and $5$ also act trivially on global sections of $\omega_X^{\otimes2}$. Mukai and Ohashi exploit this in their recent analysis of automorphisms of Enriques surfaces.

$\endgroup$
12
  • $\begingroup$ Thanks Christian, 1. If this is true then what is the order of a $\sigma$ that is acting anti semi symplectically? i.e., $\sigma(\omega)=-\omega$. 2. How can you be sure that these local sections glue together to make such global section? Doesnt it mean that $m_\alpha$ will glue together too in this way? $\endgroup$
    – user13559
    Aug 31, 2011 at 16:14
  • $\begingroup$ I don't understand the second paragraph. Finding such an $m_\alpha$ should mean giving a section over $U_\alpha$ of the canonical double cover. However, that is not possible on a Zariski open $U_\alpha$ as then the canonical double cover would be trivial. $\endgroup$ Aug 31, 2011 at 17:09
  • $\begingroup$ Hmm. I really thought there should be an easy argument, but I'm convinced that my arguments are too simple minded. Before editing further, what about the following? Let $\tilde{X}\to X$ be the associated K3 cover, and denote by $\tau$ the associated involution on $\tilde{X}$. Then, ${\rm Aut}(X)$ is isomorphic to ${\rm Aut}(\tilde{X},\tau)$, where this latter group is $\{ \varphi\in{\rm Aut}(X), \varphi \tau=\tau\varphi \}$ modulo $\tau$. In particular, the involution $\sigma$ lifts to an automorphism $\tilde{\sigma}$ of $\tilde{X}$. $\endgroup$ Aug 31, 2011 at 18:08
  • 1
    $\begingroup$ Now, if we had $\tilde{\sigma}^2=\tau$, then this would give rise to a free $\mathbb{Z}/4\mathbb{Z}$-action on the K3 surface $\tilde{X}$, which is absurd. Thus, $\tilde{\sigma}$ is an involution on $\tilde{X}$. Since every section of $\omega_{\tilde{X}}^{\otimes2}$ arises as square of a section of $\omega_{\tilde{X}}$, this implies that $\tilde{\sigma}$ acts trivially on global sections of $\omega_{\tilde{X}}^{\otimes2}$. Now, every global section of $\omega_X^{\otimes2}$ pulls back to a global section of $\omega_{\tilde{X}}^{\otimes2}$ and by compatibility of the actions, $\endgroup$ Aug 31, 2011 at 18:11
  • 1
    $\begingroup$ I've never thought about this, but you might want to look for a $\mathbb{Z}/4\mathbb{Z}$-action $\psi$ on $\tilde{X}$ such that $\psi\tau=\tau\psi$ (then, it will descend to $X$), and such that $\psi$ acts on global sections of $\omega_{\tilde{X}}$ via multiplication by $\sqrt{-1}$. $\endgroup$ Aug 31, 2011 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.