Category: Differential geometry

Dupin’s theorem in general dimension

Let d \geq 3. We say that two smooth foliations F, G of codimension 1 in \mathbb R^d are orthogonal if for any p \in \mathbb R^d, the leaves X, Y of F, G passing through p are orthogonal at p.

Dupin’s theorem. Let \vec F = (F_1, \dots, F_d) be a d-tuple of mutually orthogonal foliations of codimension 1 in \mathbb R^d. Let p \in \mathbb R^d, let X_i denote the leaf of F_i passing through p, and let
Y_i := \bigcap_{j \neq i} X_j. Then Y_i is tangent to a principal curvature vector of X_j at p whenever i \neq j.

This theorem is well-known in dimension d = 3, but it was brought to my attention by Hyunwoo Kwon that there does not seem to be a good reference if d \geq 4. This appears to be for superficial reasons: the usual proof is based on properties of the cross product which clearly only hold when d = 3. I call this a superficial reason, because we can easily replace the cross product with the wedge product, and then the argument will work in general. In this blog post, I will record the proof which works for any dimension d \geq 3. (Actually, it works in case d \leq 2 as well, but Dupin’s theorem is vacuously true in that case.)

We first check that the conclusion of the theorem makes sense. In fact, Y_i is the intersection of d - 1 smooth hypersurfaces, each of which are mutually orthogonal and therefore are transverse. So Y_i is a smooth curve, and its tangent space at p is a line.

Near p, we can find a coordinate system \vec \xi = (\xi^1, \dots, \xi^d) such that the level sets of \xi^i are the leaves of F_i. Let g be the euclidean metric tensor written in the coordinates \vec \xi. Orthogonality of F_i and F_j means that \langle d\xi^i, d\xi^j\rangle = 0. Therefore the cometric

\displaystyle g^{ij} = \frac{\partial \xi^i}{\partial x^\mu} \frac{\partial \xi^j}{\partial x^\nu} \delta^{\mu \nu}

is diagonal; therefore the same holds for the metric

\displaystyle g_{ij} = \frac{\partial x^\mu}{\partial \xi^i} \frac{\partial x^\nu}{\partial \xi^j} \delta_{\mu \nu}.

Let (i, j, k) be a distinct triple. Differentiating the equation g_{ij} = 0, we obtain

\displaystyle 0 = \frac{\partial^2 x^\mu}{\partial \xi^i \partial \xi^k} \frac{\partial x^\nu}{\partial \xi^j} \delta_{\mu \nu} + \frac{\partial x^\mu}{\partial \xi^i} \frac{\partial^2 x^\nu}{\partial \xi^j \partial \xi^k} \delta_{\mu \nu}.

Thus if we set

\displaystyle a_{ijk} := \frac{\partial^2 x^\mu}{\partial \xi^i \partial \xi^j} \frac{\partial x^\nu}{\partial \xi^k} \delta_{\mu \nu}

then we have the system of equations a_{ijk} + a_{ikj} = 0, a_{jik} + a_{jki} = 0, a_{kij} + a_{kji} = 0, a_{ijk} - a_{jik} = 0, a_{ikj} - a_{kij} = 0, a_{jki} - a_{kji} = 0. Here, the first three equations arise from the formula for the differentiated metric and the last three originate from symmetry of second partial derivatives. The only solution of this system of this equation is a_{ijk} = 0 for every distinct triple (i, j, k), or in other words

\displaystyle 0 = \frac{\partial^2 x^\mu}{\partial \xi^i \partial \xi^j} \frac{\partial x^\nu}{\partial \xi^k} \delta_{\mu \nu}.

Combining this equation with the equation g_{ij} = 0 and the formula for the metric, we see that every vector in the list

\displaystyle L := \left(\frac{\partial^2 \vec x}{\partial \xi^i \partial \xi^j}, \frac{\partial \vec x}{\partial \xi^1}, \dots, \widehat{\frac{\partial \vec x}{\partial \xi^k}}, \dots, \frac{\partial \vec x}{\partial \xi^d}\right)

(where the roof means to remove that entry from the list) is orthogonal to \partial \vec x/\partial \xi^k. Therefore every vector in L is contained in a vector space of dimension d - 1, but the length of L is d, so L is linearly dependent.

Let A_{(k)} be the second fundamental form of X_k, written in the coordinate system

\displaystyle \vec \xi_k := (\xi^1, \dots, \widehat{\xi^k}, \dots, \xi^d).

Then A_{(k)ij} is equal to a scalar times the Hodge star of the wedge product of all of the vectors in L. Since the wedge product of L vanishes, so does A_{(k)ij}, so A_{(k)} is diagonal. Since the metric h_{(k)} on X_k is given by h_{(k)ij} = g_{ij}, h_{(k)} is also diagonal in the coordinate system \vec \xi_k.

It follows that \partial_{\xi_i} is a principal curvature vector of X_k whenever i \neq k. But \partial_{\xi_i} is normal to X_i, so it is tangent to the curve Y_i, as desired.