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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.

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2 Answers 2

up vote 5 down vote accepted

I suggest to look at my paper - P. E. Pushkar', “Generalization of the Chekanov Theorem. Diameters of Immersed Manifolds and Wave Fronts” Local and global problems of singularity theory, Collection of papers dedicated to the 60th anniversary of academician Vladimir Igorevich Arnold, Tr. Mat. Inst. Steklova, 221, Nauka, Moscow, 1998, 289–304

You can find it here - http://wenku.baidu.com/view/0514da4d852458fb770b56c6.html?from=related

Diameters are double normals!

In particular, there is an estimate in the paper - number of double normals of generic immersed submanifold $M^n$ of the Euclidean space is at least $(B^2-B)/2+nB/2$. Here $B$ is $\dim H_*(M,Z_2)$. This estimation is exact for product of spheres, oriented surfaces.

There is a proof (by Maxim Kazaryan) of my estimate for the case of embeddings, which use only square-function - http://www.mi.ras.ru/~kazarian/papers/homology05.pdf in the section 16. Unfortunately it is in Russian.

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I know your work, but it is not what I'm looking for. Diameters are not oriented double normals. The round sphere, for example has no oriented double normals. –  alvarezpaiva Feb 5 '12 at 15:55
    
It seems to me that my passing diameters are exactly your oriented double normals (page 14 in the paper) for a hypersurface. –  Petya Feb 5 '12 at 16:06
    
"passing diameters" : Great! I'll take a look. –  alvarezpaiva Feb 5 '12 at 16:23
    
Yes, passing diameters, that's the thing! Great! If I understand your theorem, any hypersurface that is not a homology sphere will have oriented double normals. Is that right? –  alvarezpaiva Feb 5 '12 at 16:32
    
It seems to me I could not deduce that statement from results. With some restrictions - yes (a hypersurfase should be a result of regular homotopy without passing self-tangencies of a boundary of neibourhood of a submanifold of codimencion at least 2). It was long time ago! –  Petya Feb 5 '12 at 19:15

A proof in the embedded case.

In the embedded case the torus $T$ bounds a domain $D\subset \mathbb{R}^3$. Any unit vector $\vec{u}\in\mathbb{R}^3$ defines a linear map

$$ \mathbb{R}^3\ni x\mapsto (\vec{u},\vec{x})\in\mathbb{R}. $$

We denote by $f_{\vec{u}}$ its restriction to the torus $T$. For generic $\vec{u}$ the function $f_{\vec{u}}$ is Morse. Thus it will have at least 4 critical points. The critical set $C(\vec{u})$ of this function can be given an alternate descritiption. Consider the Gauss map

$$ \mathcal{G}: \partial D\to S^2 $$

given by the outer normal. Then $\vec{u}\in S^2$ and

$$ C(\vec{u})= \mathcal{G}^{-1}\lbrace\vec{u}\rbrace \; \cup \; \mathcal{G}^{-1}\lbrace-\vec{u}\rbrace =: C_+(\vec{u})\cup C_-(\vec{u}) . $$

From the Gauss-Bonnet theorem we deduce that the both sets $C_\pm(\vec{u})$ have even cardinalities. Thus, it suffices to show that they are both nonempty. Note that the minimma of $f_{\vec{u}}$ belong to $C_-(\vec{u})$ while the maxima of $f_{\vec{u}}$ belong to $C_+(\vec{u})$.

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Thanks Liviu. I may be a bit thick today, but I can't see why the critical points in $C_+(u)$ (or $C_{-}(u)$) will be aligned in such a way that the line joining them will have $u$ or $-u$ as direction. Could you please be a bit more explicit on this point? –  alvarezpaiva Feb 5 '12 at 14:27
    
Ooops! I misread your problem. –  Liviu Nicolaescu Feb 5 '12 at 15:07

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