Q:

Do elements of the fundamental group give rise to isometries

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.

A:

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. 

