We defined conformal and isometric maps for surfaces $f,g: \Omega \subset \mathbb{R}^2 \rightarrow S \subset \mathbb{R}^3$.

Under a reparametrization of $f$ I understand a diffeomorphism $\Phi : M \subset \mathbb{R}^2 \rightarrow \Omega$ such that our surface is given by $f \circ \Phi$.

So $f$ is conformal to $g$ if there is a positive function $\lambda$ such that $Df^T Df = \lambda Dg^T Dg$ and isometric if $\lambda =1$.

Now, I was wondering how I can find out if there is a reparametrization of $f$ and $g$ sucht that they are locally/globally conformal/isometric?

We had one theorem heading in this direction that was like: If the two surfaces that have the same Gauß curvature locally, then they are locally isometric.

But what about the global case and what about conformal reparametrizations?

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    $\begingroup$ In this and your other question, it looks like you've confused smooth with Riemannian embeddings. Every surface in $\mathbb{R}^3$ might be locally a smooth embedding, but there's no definition of "Riemannian surface in $\mathbb{R}^3$" that requires the surface to be a local Riemannian embedding. It is false that any two surfaces are locally isometric. (The sphere and the plane, for example, are not locally isometric; this is why you can't draw accurate flat maps of the Earth.) $\endgroup$
    – mollyerin
    Jan 23, 2015 at 3:34
  • $\begingroup$ It is very much unclear what your question (or questions) is. You may want to separate questions about isometric and conformal immersions. On conformal side, every simply-connected surface in $R^3$ admits a conformal parameterization by $R^2$ or the open unit disk (this is Riemann's theorem). $\endgroup$ Jan 25, 2015 at 1:49
  • $\begingroup$ On the isometric side, a necessary (but far from sufficient) condition for an isometric parameterization is that $S$ has zero Gaussian curvature. I do not think you can find a (non-tautological) necessary and sufficient condition since given a flat (zero curvature) incomplete metric on an open disk, there is no easy way to tell if it embeds isometrically into $R^2$. If this is what you wanted to know, I can post a detailed answer. $\endgroup$ Jan 25, 2015 at 1:50
  • $\begingroup$ @studiosus thank you for your comment. my question was: given two surfaces in $\mathbb{R}^3$, how can I decide if I can reparametrize one of them in such a way that they are isometric or conformal to each other. $\endgroup$
    – user66906
    Jan 25, 2015 at 2:38
  • $\begingroup$ Isometric/conformal class of a surface is independent of a parameterization. In general, determining if two surfaces are isometric or conformal to each other is an impossible task. If you have two simply-connected surfaces, things are simpler since there are exactly 3 conformal classes of such surfaces. $\endgroup$ Jan 25, 2015 at 4:27

1 Answer 1


As I said in my comment, isometric or conformal type of a surface in $R^3$ is independent of a parameterization.

This is an answer in the conformal setting (isometric setting is hopeless). You need:

Uniformization Theorem: Every simply-connected Riemann surface is conformal either to the unit disk, or to the complex plane or to the 2-sphere.

In view of this theorem, if you have two simply-connected surfaces $S_1, S_2$ in $R^3$, equipped with the induced Riemannian metric, then they have one of the above three conformal types. To show that they are conformal to each other, you need a bit of luck. For instance, if each surface appears as a graph of a smooth bounded function over a bounded ( simply-connected) domain, then one can prove that they are conformal to the unit disk and, hence, to each other.

You can read more about uniformization and the 3 types of simply-connected Riemann surfaces (and how to tell them apart!) in the book by Ahlfors and Sario "Riemann surfaces".


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