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Let $X$ be a complex algebraic variety, and let $\tilde X\to X$ be its universal cover. Suppose that there exists a Kahler-Einstein metric on $\tilde X$. Note that $\pi_1(X) \subset Aut(\tilde X)$.

Question. Assume $X$ is compact. Is $\pi_1(X)$ actually a subgroup of the group of isometries of $\tilde X$ with respect to the Kahler-Einstein metric?

I suspect it is, but I don't know enough about analytic geometry to be completely sure.

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    $\begingroup$ 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. $\endgroup$
    – Henri
    Commented Aug 24, 2013 at 19:56
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    $\begingroup$ @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. $\endgroup$
    – Ben McKay
    Commented Aug 24, 2013 at 20:43
  • $\begingroup$ @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 $\pi_1(X)$). $\endgroup$
    – Henri
    Commented Aug 24, 2013 at 23:01
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    $\begingroup$ 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 $\mathbb C$. $\endgroup$
    – Henri
    Commented Aug 24, 2013 at 23:41
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    $\begingroup$ 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 $\mathrm{Aut}(\tilde X) = \mathrm{Isom}(\tilde X, \omega_{\rm KE})$. $\endgroup$
    – Henri
    Commented Aug 25, 2013 at 9:26

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The answer to your question is no without further assumptions.

Here are three counterexamples (one for each curvature's sign):

  • Positive curvature: take $X=\mathbb C/(\mathbb Z \oplus i \mathbb Z)$ the standard torus, and $\tilde X = \mathbb C$. If you endow $\mathbb C$ with the restriction of the Fubini-Study on $\mathbb P^ 1$ (under any embedding $\mathbb C \hookrightarrow \mathbb P^1$), then you get a metric with constant positive curvature on $\mathbb C$ which is clearly not invariant under the natural action of $\mathbb Z \oplus i \mathbb Z$ on $\mathbb C$.
  • Zero curvature: Take as before the torus, and choose on $\mathbb C$ a metric $e^u |dz|^2$ where $u$ is harmonic but not invariant under the action of $\mathbb Z \oplus i \mathbb Z$ (for example $u=\mathrm{Re}(z)$).
  • Negative curvature: Take $X$ to be any compact complex curve with genus $g\geqslant 2$. And put on the unit disk in $\mathbb C$ the restriction of Poincaré metric of the disk of radius $2$. Clearly it won't be invariant under the fundamental group of $X$ (at least if $X$ is well chosen).

However, we can still say something in the last case. Indeed, if you assume that the Kähler-Einstein metric on $\tilde X$ is complete, then Yau's maximum principle shows that it is the unique such metric. Hence any automorphism of $\tilde X$ preserves this metric, so that $\pi_1(X)$ acts by isometries with respect to this metric.

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