A couple of weeks ago, my student Yan Mary He presented a nice proof of Liouville’s theorem to me during our weekly meeting. The proof was the one from Benedetti-Petronio’s Lectures on Hyperbolic Geometry, which in my book gets lots of points for giving careful and complete details, and being self-contained and therefore accessible to beginning graduate students. Liouville’s Theorem is the fact that a conformal map between open subsets of Euclidean space of dimension at least 3 are Mobius transformations — i.e. they look locally like the restriction of a composition of Euclidean similarities and inversions on round spheres. This implies that the image of a piece of a plane or round sphere is a piece of a plane or round sphere, a highly rigid constraint. This sort of rigidity is in stark contrast to the case of conformal maps in dimension 2: any holomorphic (or antiholomorphic) map between open regions in the complex plane is a conformal map (and conversely). The proof given in Benedetti-Petronio is certainly clear and readable, and gives all the details; but Mary and I were a bit unsatisfied that it did not really provide any geometric insight into the meaning of the theorem. So the purpose of this blog post is to give a short sketch of a proof of Liouville’s theorem which is more geometric, and perhaps easier to remember.

Remember that a *conformal map* is one which infinitesimally takes round spheres to round spheres. That is, it is *angle preserving*, at least infinitesimally. In particular, it is smooth. So let’s think about a conformal map between open regions in Euclidean 3-space (for concreteness). The image of a flat plane P is a smooth surface f(P). Pick a point p in P and look at its image f(p). Infinitesimal round circles around p in P get taken to infinitesimal round circles around f(p) in f(P). And straight lines perpendicular to P get taken to smooth curves perpendicular to f(P). If you take a smooth surface S in Euclidean 3-space, and a small round circle in S, and push the circle off S in the perpendicular direction, some directions will be distorted more than others (typically): the infinitesimal circle gets distorted to an infinitesimal ellipse, whose major and minor axes are the directions of principal curvature on the surface S. But these ellipses are the conformal image of small round circles in the domain, and therefore should also be (almost) round. In other words: the principal curvatures at each point of f(P) should be *equal*. A point on a surface where the principal curvatures are equal is called an *umbilical point*, and a surface on which every point is umbilical is called *totally umbilical*.

It is a classical fact, proved by Meusnier in 1785, that an umbilical surface in Euclidean space is locally a piece of a plane or sphere. One way to see this is as follows. Let G denote the Gauss map, so that the condition of being umbilical at a point says exactly that dG is a multiple of the identity at that point (note: we are using here in the usual way the canonical identification between the tangent space to the surface and the tangent space to the round sphere at the image of the Gauss map at each point to think of dG as a map from the tangent space to itself). So if a surface is totally umbilical, there is some function f so that dG is equal to f times the identity at each point. Let’s denote by X a local chart on the surface giving rise to local coordinates u and v. So the definition of f says in this notation that and . But then

Since u and v are local coordinates, their tangent vectors and are independent, and therefore . This means that is (locally) constant. But this means that the surface osculates a sphere (or plane) of *fixed* curvature to first order at every point, and therefore (by developing this sphere along a path in the surface) the center of this osculating sphere is fixed and the surface agrees (locally) with the sphere (or plane). Incidentally, Gauss was only 8 years old in 1785, so whatever Meusnier’s proof was, he could not have mentioned the Gauss map by name. Does any reader know Meusnier’s argument?

Once we know that a conformal map takes subsets of planes and round spheres to subsets of planes and round spheres, we can intersect these planes and spheres with perpendicular planes and spheres to see that it takes straight segments and arcs of round circles to straight segments and arcs of round circles. From this it is easy to deduce Liouville’s theorem.

By the way, I strongly suspect that the connection between totally umbilical surfaces and conformal maps is classical and well-known, and for all I know this was how Liouville thought of his theorem in the first place.

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October 31, 2013 at 4:18 pm

Ian AgolI think there might be a synthetic way to see that umbilic spheres are round. Consider an umbilic sphere whose curvature is not constant. As you point out, the Gauss map is conformal. In particular, it should be convex, since the Jacobian is non-zero everywhere, so the sign of the curvature can’t change. Take a point whose curvature is minimal; then then osculating sphere should lie inside of the umbilic sphere (I think one may prove this by considering the cut locus in the interior of the sphere). Now, take a point in the interior of the umbilic sphere, but outside of the minimal osculating sphere. Do an inversion through this point, turning the umbilic sphere inside out. Then the osculating sphere goes to the outside of the inverted umbilic sphere, and therefore the inversion is not convex, a contradiction. There’s some details that need to be checked, but I think this might work.

November 1, 2013 at 5:30 am

Danny CalegariThis sounds like an argument to me.

November 1, 2013 at 9:55 am

Ian AgolA correction: I meant a point whose curvature is maximal, or radius of curvature is minimal.

Also, to see the point about the osculating sphere being in the interior: take the radius r tubular neighborhood about the umbilic sphere, for r less than the minimal radius of curvature. Then the interior boundary will also be umbilic (since shrinking the osculating sphere diameter by r will preserve the 2nd order contact), and will become singular at the minimal radius of curvature. At this point, the first point of the cut locus will appear at the focal point of the minimal curvature radius, so a sphere centered at this point will be embedded.

January 29, 2014 at 8:03 am

Ludwig Bach@Ian Agol: Your idea is interesting. There’s however a simpler proof I found in G. Valiron, “The classical differential geometry of curves and surfaces” (that book is, by the way, a treasure of elegance). It goes like this (slightly altered):

Take a point p on the surface, and its normal N(p). Take a plane P containing N(P), and cut the surface with it: you’ll get a curve C. Since every point is umbilic, C is a line of curvature. This means that, all along the curve C, the normals to the surface all belong to the plane P. This implies that any two of them intersect (possibly at infinity). By choosing a different plane P and repeating, you can cover any pair of points on the surface, so that: any pair of normals to the surface intersect (possibly at infinity).

From here it’s easy. Just use the fact that if n lines in projective space intersect pairwise, they either belong to a plane or have a common intersection. This common intersection is the center of our sphere, QED.

January 29, 2014 at 8:05 am

Ludwig BachCorrection: N(P) should be N(p)