*Open Problem:** Does every triangular billiards admit a periodic orbit.md?
Answer (1775): Yes for acute triangles, there is at least one periodic orbit:
For arbitrary triangles: unknown!
Let \(M\) be a Hamiltonian.md.
For regular value) r\in {\mathbf{R}} of the Hamiltonian, $H^{-1}(r \(r/in /RR\) of the Hamiltonian, \(H/inv(r) \subset M\) is a submanifold \(Y\subset M\) with a smooth vector field \(X_H\) called a regular level set.
Question: Does \(X_H\) have a closed orbit on every regular level set? What conditions do you need to guarantee the existence of a closed orbit?
Turns out not to depend on \(H\), and only on the hypersurface \(Y\). The existence of a closed orbit is equivalent to the existence of a closed embedded curve \(\gamma\) that is everywhere tangent to \(\ker( {\left.{{\omega}} \right|_{{Y}} } )\).
Question: When is such a curve guaranteed to exist?
Theorem (Weinstein, 1972): If \(Y\) is convex.
Theorem (Rabinowitz): If \(Y\) is “star-shaped” (exists a point \(p\) that can “see” all points via straight lines).
Theorem (1987): Every contact-type hypersurface in the periodic orbit.md.
Conjecture (Weinstein, 1978): Let \((M, \xi)\) be a closed (compact) periodic orbit.md.
Theorem (Weinstein, Dimension 3, Overtwisted. 1993): Let \((M, \lambda, \xi)\) be a closed overtwisted contact structure.md. Then the periodic orbit.md.