Riemann surface with negative curvature

Question

Let $(M,g)$ be a noncompact Riemann surface with Gaussian curvature $K<0$. Can we show that the universal cover of $M$ must be conformal to the disk?

From the uniformization theorem, we know that the universal cover must be conformal to either the plane or the disk. How can we distinguish them by curvature?

Answer 1

If $M$ is not compact or if you allow $K\to 0$ there is no reason.

Consider a conformal metric $e^{\varphi (z)} dz d \bar z$ on $\bf C$. Then $K= -2 e^{-\varphi } {\partial ^2 \varphi \over \partial z \partial \bar z}$.

Take for instance $\varphi (z)= z.\bar z = r^2$. Then $K=-4 e^{-r^2}<0$.