Timeline for every involution of an Enriques surface is
Current License: CC BY-SA 3.0
18 events
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| Aug 23, 2017 at 20:09 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Sep 6, 2011 at 2:37 | comment | added | user13559 | @Christian: Sorry to bother you again. Why $\tilde{\sigma}^2\in <\tau>$. I know that $<\tilde{\sigma}>=<\sigma>\times<\tau>$ where by $<>$ I mean the group generated by that automorphism. Thanks | |
| Sep 1, 2011 at 6:53 | comment | added | rita | Christian, I deleted my comments now that you have edited again. | |
| Aug 31, 2011 at 23:41 | history | edited | Christian Liedtke | CC BY-SA 3.0 |
completely rewritten answer
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| Aug 31, 2011 at 20:57 | comment | added | Christian Liedtke | 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 $\sqrt{-1}$. | |
| Aug 31, 2011 at 20:50 | comment | added | user13559 | $\lambda\in {\omega_X} ^{\otimes 2}$ | |
| Aug 31, 2011 at 20:45 | comment | added | user13559 | Thanks, I think I undersatnd now. the last question I will be pleased if you help with is: Lets assume $g(λ)=−λ$ where $λ∈ω_X⊗2$ and g∈Aut(X). Does such g exists and what is the possible order(s) of such g? Thanks | |
| Aug 31, 2011 at 20:20 | comment | added | Christian Liedtke | of course, $\mathbb{Z}/4\mathbb{Z}$ may act on a K3 surface. BUT: this action is supposed to be free (!), and this is impossible: the hypothetical quotient $S$ would have $\chi({\mathcal O}_S)= \chi({\mathcal O}_{\tilde{X}})/4=\frac{1}{2}$... I don't understand the other objection: an involution on $\tilde{X}$ acts on global sections of $\omega_X$ by ${\rm id}$ or by $-{\rm id}$. Since the map $H^0(\omega_{\tilde{X}})^{\otimes 2}\rightarrow H^0(\omega_{\tilde{X}}^{\otimes 2})$ is onto, we conlude that every involution acts trivially on global sections of $\omega_{\tilde{X}}^{\otimes2}$! | |
| Aug 31, 2011 at 19:55 | comment | added | user13559 | @Christian: If you have an involution on $\tilde{X}$ it might not be anti symplectic, i.e., it might not act trivially on $\omega_{\tilde{X}}^{\otimes 2}$. Only if your involution acts by multiplication by $\pm 1$ on $\omega_{\tilde{X}}$ it will act trivially on $\omega_{\tilde{X}}^{\otimes 2}$. Also I dont undersatand why it is absurd to hace a $C_4$ action on a K3 surface!! | |
| Aug 31, 2011 at 18:12 | comment | added | Christian Liedtke | we conclude that $\sigma$ acts trivially on global sections of $\omega_X^{\otimes2}$... Is there no easier way to see this?! | |
| Aug 31, 2011 at 18:11 | comment | added | Christian Liedtke | 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, | |
| Aug 31, 2011 at 18:08 | comment | added | Christian Liedtke | 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 $\tilde{X}$. | |
| Aug 31, 2011 at 17:09 | comment | added | Torsten Ekedahl | 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. | |
| Aug 31, 2011 at 16:14 | comment | added | user13559 | 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? | |
| Aug 31, 2011 at 16:08 | vote | accept | user13559 | ||
| Sep 5, 2011 at 15:50 | |||||
| Aug 31, 2011 at 14:37 | history | edited | Christian Liedtke | CC BY-SA 3.0 |
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| Aug 31, 2011 at 14:24 | history | edited | Christian Liedtke | CC BY-SA 3.0 |
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| Aug 31, 2011 at 2:22 | history | answered | Christian Liedtke | CC BY-SA 3.0 |