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If I have a smooth surface $M$ (2D embedded in 3D), under what conditions I can assure that there exists a finite collection of charts $\{U_i, \phi_i \}_{i}$, with $\phi_i : U_i \to M$, such that its directional derivatives have always the same magnitud and that they are orthogonal set of coordinates?, i.e.

$$ |\partial_x \phi_i| = |\partial_y \phi_i| \\ \partial_x \phi_i \cdot \partial_y \phi_i = 0 $$

Trivially flat surfaces have this property, and I suspect that is the general case, but I'm not really sure.

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    $\begingroup$ Isn’t this a purely local thing? You just pull back the Riemannian metric on $M$ to a chart (which we can assume is a disc), and choose an orthonormal basis for the tangent space at the origin. This induces an orthonormal frame on the disc via geodesic flow, and integral curves for these two vector fields provide a coordinate system with this property. I am not the most schooled in Riemannian stuff so please correct me if I am wrong. $\endgroup$
    – Bma
    Commented Apr 7, 2023 at 19:08
  • $\begingroup$ I'm confuse on the integral curves part. Given that you have the vector field $e_1(x,y), e_2(x,y)$ generated by the exponential map, how do you generate the parametrization? $\endgroup$
    – user3646987
    Commented Apr 7, 2023 at 19:33
  • $\begingroup$ You travel along the geodesics tangent to $e_1$ and $e_2$— that’s your coordinate system $\endgroup$
    – Bma
    Commented Apr 7, 2023 at 20:07

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You can always do this, but it's not as simple as using some kind of ODE (such a flow of vector fields) to construct such charts.

First, assume that your surface is connected and simply-connected. Then it's either compact, in which case it's a $2$-sphere, or it's diffeomorphic to $\mathbb{R}^2$. In the compact, case, the uniformization theorem implies that it's conformally equivalent to the standard $2$-sphere, i.e., there is a smooth diffeomorphism $f:S^2\to M$ such that the $f$-pullback of the induced metric on $M$ is a multiple of the standard metric on $S^2$. Now composing $f$ with the inverses of the usual stereographic projections (which are conformal) of the sphere to the plane from the north and south poles gives you two coordinate patches on $M$ that are conformal. In the non-compact case, the uniformization theorem says that there is a single smooth diffeomorphism $\phi:U\to M$ that is conformal, i.e., the $\phi$-pullback of the induced metric on $M$ is a multiple of the standard metric on $U\subset\mathbb{R}^2$. Here, by the Riemann mapping theorem, we can either take $U$ to be $\mathbb{R}^2$ or to be the unit disk.

To cover the general case, you only need to know that $M$ has a finite open cover by simply-connected open sets. This is a standard fact about 2-dimensional surfaces.

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  • $\begingroup$ I see— I ignored the finite assumption $\endgroup$
    – Bma
    Commented Apr 8, 2023 at 18:21

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