Uniformization theorem and conformally equivalent metrics on the disk.

I am studying the Uniformization theorem, and in particular I would like to understand the link between the complex version https://en.wikipedia.org/wiki/Uniformization_theorem:

Every simply connected 1-complex manifold is biholomorphic to the unit disk, complex plane or to the Riemann sphere

and the real version:

Every Riemannian 2 manifold has a conformally equivalent metric of constant Gaussian curvature.

I understand that a complex structure is equivalent to a conformal structure on an oriented smooth surface but I am stuck with the example of the disk $\mathbb{D}\subseteq \mathbb{C}$. Take the unit disk $\mathbb{D}$ you can view it as the restriction of the complex plane $\mathbb{C}$ with his complex structure, this complex structure give us the conformal structure that contains the hyperbolic metric of the disk. Althought if we take the induced subspace metric we get the flat metric on the disk. Now I am wondering, does this imply the two metrics are conformally equivalent?

If yes, how does the Uniformization theorem in the real case imply the one in the complex case, if there are two conformally equivalent metric on the disk that have different curvature?

Solution 1:

I found the answer to this question some times ago, now I will post just for completeness, I doubt anyone will ever wonder. The real version of the theorem state more: there exist a complete metric of constant curvature. The two metric on the disk are constant curvature but only hyperbolic is also complete. Also from Poincare Hopf theorem we get that if there is a complete contant curvature metric of negative curvature then the space is isometric to the hyperbolic space, so the real version of the theorem is indeed equivalent to the complex version.