Timeline for Do elements of the fundamental group give rise to isometries
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 28, 2013 at 9:54 | answer | added | Henri | timeline score: 6 | |
| Aug 25, 2013 at 9:26 | comment | added | Henri | Not exactly; I am just saying that if $\tilde X$ admits a complete Kähler-Einstein metric with negative curvature, then it is unique by Yau's maximum principle, and therefore $\mathrm{Aut}(\tilde X) = \mathrm{Isom}(\tilde X, \omega_{\rm KE})$. | |
| Aug 25, 2013 at 9:01 | comment | added | Leertje | @Henri Great! If I understand correctly, if $K_X$ is ample, then $X$ is KE and thus also $\tilde X$, right? And in this particular case $\pi_1(X)$ is a subgroup of Isom$(\tilde X)$. | |
| Aug 25, 2013 at 0:07 | comment | added | Henri | @Leertje: I'll wait a bit to see if someone has got a better approach and if not I'll write an answer. | |
| Aug 24, 2013 at 23:54 | history | edited | Leertje | CC BY-SA 3.0 |
added 6 characters in body
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| Aug 24, 2013 at 23:48 | comment | added | Leertje | @Henri: could you make your comments into an answer? They have been very helpful already! Thank you. | |
| Aug 24, 2013 at 23:41 | comment | added | Henri | Two more remarks: - in the negatively curved case, it is enough to assume that the KE metric on $\tilde X$ is complete to ensure its uniqueness by Yau's maximum principle - in the Ricci flat case, the answer to the question is $\textbf{no}$ : take $X= \mathbb C/(\mathbb Z \oplus i \mathbb Z)$, and any metric $e^u |dz|^2$ where $u$ is harmonic and not invariant under the natural action of $\mathbb Z \oplus i \mathbb Z$ on $\mathbb C$. | |
| Aug 24, 2013 at 23:01 | comment | added | Henri | @BenMcKay: you're right, I was too quick, thanks for noticing. I even assumed compactness which doesn't seem to be part of the hypothesis. But if $\tilde X$ is compact, then the argument I gave for negative Ricci still works (the KE metric will be preserved by any element of $\mathrm{Aut}(\tilde X)$ hence by $\pi_1(X)$). | |
| Aug 24, 2013 at 20:43 | comment | added | Ben McKay | @Henri: only if the metric is on $X$. But the metric is on $\tilde{X}$. Maybe $X$ isn't KE. If the KE metric has positive Ricci, then the argument works, but what about negative or zero Ricci. | |
| Aug 24, 2013 at 19:56 | comment | added | Henri | A first observation: as X is Kähler-Einstein, $K_X$ is either ample, antiample or numerically trivial. If $K_X>0$, then the KE metric is unique hence is fixed by any biholomorphism. If X is Fano, then it is simply connected, and your statement is trivially checked. Remains the case of a Calabi Yau variety. | |
| Aug 24, 2013 at 19:19 | review | Close votes | |||
| Aug 24, 2013 at 20:39 | |||||
| Aug 24, 2013 at 18:10 | history | asked | Leertje | CC BY-SA 3.0 |