The two principal radii of curvature at a point on a surface in three dimensions provide a succinct description of the local geometry of the surface. The principal curvatures determine whether a surface is locally flat, convex, or a saddle (Figure 1). The radii (plural of *radius, *pronounced *ray-dee-eye* or ˈrādēˌī) prove even more dispositive in the theory of convex surfaces. Convex surfaces necessarily have positive radii of curvature everywhere, but much more is true: Knowing that the two radii of curvature at every point on a convex surface in fact determines the entire surface up to translations. But when does the reverse hold?

More precisely, when does a function \(\Lambda\) that maps an outward-pointing normal vector \(u \in \mathbb{S}^2\) to a pair of curvatures \(\lambda_1(u), \lambda_2(u)\) always define a convex surface in \(\mathbb{R}^3\)? This question is closely related to deep problems in geometry, notably the Christoffel–Minkowski problem. In this post, we derive a condition on every curvature function \(\Lambda\) that defines a convex surface that limits the available functions.

To state the result, let’s make up a word.

Definition.A point \(x\) on a convex surface \(S\) anequicurved points of \(S\)if all principal curvatures of \(S\) at \(x\) are equal.

In a neighborhood of an equicurved point \(x\), the surface \(S\) looks spherical up to second order. Indeed, spheres are characterized by the fact that every point is a point of equicurvature. Surprisingly, every sufficiently regular three-dimensional convex surface has at least one point of equicurvature.

Theorem 1.Let \(S\subset \mathbb{R}^3\) be a strictly convex twice-differentiable surface. Then \(S\) has at least one point of equicurvature.

The proof requires a few preliminaries.

Fact 2 [Schneider, Chapter 2.5].Let \(S\) be as above, and let \(h\colon \mathbb{S}^2\to \mathbb{R}\) be the support function of \(S\). Then:1. The support function \(h\) is of class \(C^2\), that is, \(h\) is twice continuously differentiable.

2. The eigenvalues of the Hessian \(\nabla^2 h(u)\) of the support function are the principal radii of curvature of \(S\) in the direction \(u\in \mathbb{S}^2\). (Note that this parameterization uses the outward-pointing normal to determine the radius of curvature, as opposed to the base point.)

3. The eigenvectors of \(\nabla^2 h(u)\) corresponding to nonzero eigenvalues are perpendicular to \(u\).

*Proof of Theorem 1.* For a contradiction, suppose that there is no point of equicurvature on \(S\). By Fact 2(1), that the Hessian map \(\nabla^2 h \colon \mathbb{S}^2 \to \mathbb{R}^{3×3}\) is continuous. Moreover, since \(S\) has no point of equicurvature, \(\nabla^2 h(u)\) has distinct eigenvalues everywhere by Fact 2(2).

This implies that the principal unit eigenvector \(\xi_i(u) \in \mathbb{S}^2\) of \(\nabla^2 h(u)\) is also well defined, continuous, and perpendicular to \(u\) by Fact 2(3). In other words, \(\xi \colon \mathbb{S}^2 \to \mathbb{S}^2\) is a non vanishing continuous map from the unit sphere onto it’s tangent space, which is impossible by the hairy ball theorem.

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Fascinating! We’ve used a classical theorem in differential geometry to find a special point on every smooth, strictly convex surface.

Moreover, the proof doesn’t work for surfaces embedded in even dimensions, and even in odd dimensions, the straightforward extension of the method results in a far less satisfying conclusion (no point of equicurvature, but you must have points where the principal curvatures intertwine).

Of course, the theorem is almost vacuously true in one and two dimensions, so we seem to have stumbled upon a truly “three dimensional” fact.

Note that a friend of mine pointed out that this applies to all \(C^2\) surfaces that are topologically equivalent to \(S^2\), since the principal radii of curvature can be parameterized the same way, and the hairy ball theorem also holds for these surfaces.