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 $X$.


2 Answers 2


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 $\omega^{\otimes 2}$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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