surface curvature

Question

I would like to proof the existence or the non-existence of a finite surface which has 2 different radius of curvature $R_1$ and $R_2$ that are:

• constant on the whole surface

• finite

• different each other

I am working on the theory of thin shells and the use of surfaces where curvature is constant greatly simplify the equations. I am looking for the most general case to perform a test. Unfortunately I can not figure out such surface. Is it possible or not?

Answer 1

If by "radii of curvature" you mean principal curvatures, then you're asking about isoparametric [sic, not isoperimetric] surfaces. According to this survey article (see Theorem 4 and Lemma 5), the only examples are cylinders, planes, and spheres. (If memory serves, this is a local result, i.e., not a consequence of completeness.)

Answer 2

The only such closed, compact surface is the torus. A (distorted) hyperbolic space or the cylinder can give examples of noncompact surfaces. A topological disk could carry the required metric.

Suppose that a closed, compact, orientable surface $S$ has two constant, distinct, finite principal curvatures $R_1, R_2$. Then the principal curvature directions give two distinct, everywhere orthogonal, nonvanishing vector fields on $S$, or a nonvanishing quadratic differential (the shape operator). So $S$ cannot be a sphere by the Hairy Ball Theorem. Nor can $S$ have genus greater than 1, by Riemann-Roch. Gauss-Bonnet implies that $2\pi \chi =AR_1R_2$. If $S$ is the torus, it can be equipped with a metric such that $R_1$ is any positive real number and $R_2=0$. (A cylinder with identified ends.)

I am less sure of noncompact or nonclosed surfaces, but some would work. If $S$ is topologically a disk, it could be a hyperbolic space with the metric adjusted so that $x$ curves faster than $y$. The annulus could have a cylindrical metric with $R_1>0$ and $R_2=0$. By deleting small disks you can always make surfaces with arbirarily many boundary components.