# complete metric on a Riemann Surface

I'm reading a book and found a sentence I don't understand:

Every Riemann surface $S$ supports an essentially unique complete metric of constant curvature $1$, $0$ or $-1$. Every point of $S$ has a neighborhood $U$ which is isometrically isomorphic to an open set in the Riemann sphere ($\mathbb{P}$), the complex plane ($\mathbb{C}$) or the open unit disc ($\mathbb{D}$).

Speciffically my doubt is: What does "isometrically isomorphic" means in this context? I searched in the web and found something about vector spaces, but the book has not mentioned vector spaces so far so i'm confused.

• This simply means isometric, that is, there exists a diffeomorphism preserving Riemannian metrics – Moishe Kohan Oct 17 '13 at 14:58
• Where can I get more information about the result mentioned in the sentence? – antony almeida Oct 17 '13 at 15:08
• "isometrically isomorphic" only makes sense when the metric is specified on both sides. On ${\mathbb P}$ the usual spherical metric is meant, on ${\mathbb C}$ the euclidean metric, and on ${\mathbb D}$ the metric $ds={|dz|\over 1-|z|^2}$. – Christian Blatter Oct 17 '13 at 15:10
• almeida: For references, google "uniformization theorem". Your book is also missing that the constant curvature metric is conformal with respect to the given complex structure. – Moishe Kohan Oct 17 '13 at 15:41