I am slightly confused about the interchangeability of the terms isometry and biholomorphism. This confusion is rooted in the following statement, which I have seen more than once in some discussions on equivalent versions of the Uniformization Theorem.

Suppose $(M,g)$ is a complete, simply connected, $2$-dimensional Riemannian manifold with constant curvature. By the Killing-Hopf theorem, it is isometric to either $\mathbb{R}^2$, $S^2$, or $\mathbb{H}^2$. Identify $M$ as a simply connected Riemann surface and identify $\mathbb{R}^2$, $S^2$, and $\mathbb{H}^2$ as, respectively, $\mathbb{C}$, $\hat{\mathbb{C}}$, and $D_1$. Then $M$ is biholomorphically equivalent to $\mathbb{C}$, $\hat{\mathbb{C}}$, or $D_1$.

To me, an isometry between two Riemannian manifolds $(M,g)$ and $(M',g')$ is a diffeomorphism $f:M\longrightarrow M'$ such that $g=f^*g'$. Thus, it preserves angles and can be described as a conformal mapping.

On the other hand, given two Riemann surfaces $M$ and $N$, a biholomorphism is a bijective map $f:M \longrightarrow N$ such that both $f$ and $f^{-1}$ are holomorphic.

So, in the context of the italicized statement, does it really follow that the isometry turns into a biholomorphism just because we switched from Riemannian manifolds to Riemann surfaces, and because both isometries and biholomorphisms preserve angles? Or is there something more subtle going on here?


1 Answer 1


The italicized statement is quite trivial once you know about the unit disk model of the hyperbolic plane; in contrast, the uniformization theorem is highly nontrivial. As for your question in the last paragraph: Each oriented Riemannian 2-manifold $M$ has a canonical complex structure making it a Riemann surface $X=X(M)$; under this correspondence $M\mapsto X(M)$, every orientation preserving isometry is biholomorphic. In other words, you have a functor from the category of oriented Riemannian surfaces (where morphisms are orientation preserving local isometries) to the category of Riemann surfaces (where morphisms are holomorphic maps).

Every simply connected surface is orientable, but this is not the same as to say that every simply connected surface is oriented. (The concepts oriented and orientable are frequently conflated but that's an abuse of language.)


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