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Here is a problem in contact topology that was suggested by Petya's answer to this mathoverflow question of mine.

Let $S^* \mathbb{R}^n$ be the space of cooriented contact elements of $\mathbb{R}^n$. I will think of contact elements as pairs of point-cooriented hyperplane, where the point lies on the hyperplane. It will then make sense to say that two elements of $S^* \mathbb{R}^n$ are parallel.

Question. Let $i: S^{n-1} \rightarrow S^* \mathbb{R}^n$ be a Legendrian embedding that is Legendrian isotopic to the manifold of all cooriented hyperplanes passing through a point. Does there necessarily exist a point $x \in S^{n-1}$ such that the contact elements $i(x)$ and $i(-x)$ are parallel?

Remark. Because Petya's proof is a simple application of critical point theory for the support function and the support function is just a simple instance of a generating function for a Legendrian submanifold, I'm guessing off the top of my head that his proof extends. The condition of being Legendrian isotopic to the manifold of all cooriented hyperplanes passing through a point garantees the existence of a generating function quadratic at infinity.

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This is true for mere topological reasons: By a result due to G. Hirsch, every self-map of a sphere of odd degree sends a pair of antipodal points to a pair of antipodal points. See e.g. Theorem 6.2 in §9 of [A. Granas & J. Dugundji; Fixed Point Theory, Springer (2003)]

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  • $\begingroup$ The map takes the point $x$ in the sphere to the normal unit vector to the contact element $i(x)$, right? Yeap, that works! Thanks. $\endgroup$ – alvarezpaiva Sep 5 '16 at 4:14

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