*Given an embedded two-torus in three-dimensional Euclidean space, paint the inside of the torus red and the outside blue. Show that there is an oriented line in ${\mathbb R}^3$ that cuts the torus perpendicularly in (at least) two points at which it crosses from red to blue.*

This is true (I'll say why in a minute), but I'd like to know if there is a simple proof using standard critical point theory.

Here is a proof that works even if the torus is immersed:

Consider the space of geodesics of ${\mathbb R}^3$ which is well-known to be symplectomorphic to the cotangent of the two-sphere. The congruence of oriented lines normal to the immersed torus is an immersed exact Lagrangian manifold in the space of geodesics. The *oriented double normal* we're looking for is just a double (multiple) point for this immersed Lagrangian. In other words, we would like to know that this immersion cannot be an embedding. The result now follows from a theorem of Claude Viterbo (JDG **47** (1997) 120-168).

**Theorem (Viterbo).** *There is no exact Lagrangian embedding of the two-torus in the cotangent space of the two sphere*.

This theorem and the arguments I gave before settle the problem not only in Euclidean space, but also in hyperbolic space, three-dimensional Hadamard manifolds, three-dimensional normed spaces with smooth, quadratically-convex spheres and, in general, it works for any three-dimensional Finsler manifold whose space of geodesics is symplectomorphic to the cotangent of the two-sphere.

Isn't there some simpler argument that works in ${\mathbb R}^n$ $(n > 3)$ and yields **something** like: if an compact oriented manifold is immersed as a hypersurface in ${\mathbb R}^n$, then it either admits an oriented double normal or it is homeomorphic/diffeomorphic to a sphere ?

**Relation to the standard double-normal problem.** In the standard (non-oriented) double-normal problem, everything reduces to considering the critical points of the distance-squared function defined on the symmetric product of the immersed manifold with itself. What
I can't see is whether there is some minimax procedure that constructs critical values (and points) that correspond to oriented double normals.