## Update on the short Reeb orbit conjecture

In an earlier post, I stated a conjecture that every contact three-manifold has a Reeb orbit with an upper bound on its symplectic action in terms of the volume of the contact manifold. There hasn’t been much progress on this conjecture, so now I would like to state a slightly stronger conjecture (the constant is improved) and discuss some examples.

To state the conjecture, let $(Y,\lambda)$ be a closed contact three-manifold. Define the volume $vol(Y,\lambda) = \int_Y \lambda\wedge d\lambda$. If $\gamma$ is a Reeb orbit, define its symplectic action by $A(\gamma) = \int_\gamma\lambda$.

Conjecture 1 (“Short Reeb orbit conjecture”). If $(Y,\lambda)$ is a closed contact three-manifold, then there exists a Reeb orbit $\gamma$ with $A(\gamma)^2\le vol(Y,\lambda)$.

Example 1: $S^1$-invariant contact forms. Let $Y$ be an $S^1$-bundle over a surface $\Sigma$ with Euler class $e>0$, and let $\lambda$ be a connection 1-form on $Y$ with nowhere vanishing curvature $\omega$, regarded as a real 2-form on $\Sigma$. Let $\pi:Y\to\Sigma$ denote the projection and let $f:\Sigma\to {\mathbb R}$ be a positive function. Then $(\pi^*f)\lambda$ is a contact form on $Y$. I claim that the conjecture holds for this contact form, with equality if and only if $e=1$ and $f$ is constant.

To see this, observe first that $vol(Y,\lambda) = 2\pi\int_\Sigma f^2\omega$. In particular, if $f_0$ denotes the minimum value of $f$, then $vol(Y,\lambda) \ge 4\pi^2f_0^2e$. On the other hand, if $p_0\in\Sigma$ satisfies $f(p_0)=f_0$, then $\pi^{-1}(p_0)$ is a Reeb orbit $\gamma$ with $A(\gamma) = 2\pi f_0$. The claim follows immediately.

A special case of this example is where $Y$ is a compact star-shaped hypersurface in ${\mathbb R}^4$ which is $S^1$-invariant, where $S^1$ acts on ${\mathbb C}^2$ as multiplication by scalars of absolute value $1$.

Example 2: geodesics in $S^2$Let $g$ be any Riemannian metric on $S^2$ and let $Y$ be the unit cotangent bundle of $S^2$ with its canonical contact form $\lambda$, so that Reeb orbits correspond to closed geodesics, and symplectic action corresponds to length of geodesics. Then $vol(Y,\lambda) = 2\pi Area(S^2,g)$. The conjecture then implies that there is a closed geodesic whose length $\ell$ satisfies $\ell^2\le 2\pi Area(S^2,g)$.

For example, if $g$ is the standard round metric, then a great circle has $\ell=2\pi$ and so the desired inequality is $(2\pi)^2\le 2\pi(4\pi)$, which holds with a factor of $2$ to spare. However there is a famous example in systolic geometry where the minimum of $\ell^2/Area(S^2,g)$ is greater than that of $S^2$. The example is obtained by taking two equilateral triangles in the plane, say of side length 1, and gluing their edges together, to obtain $S^2$ with a metric which is flat except for three cone singularities. Smooth these. The area is then approximately $\sqrt{3}/2$, while the length of the shortest geodesic (which goes from a vertex of the triangle to the center of the opposite side and back) is approximately $\sqrt{3}$.  The desired inequality is now $(\sqrt{3})^2\le 2\pi(\sqrt{3}/2)$, which still holds, but with less than a factor of $2$ to spare. And this is possibly a little worrisome, because of:

Example 3: even geodesics in $S^2$. As I learned from discussions with Vinicius Gripp, the conjecture actually says a little more about geodesics in $S^2$. Continuing with the notation from Example 2, observe that $\pi_1(Y)={\mathbb Z}/2$. We can pull back the contact form $\lambda$ to the double cover of $Y$, which is $S^3$, to obtain a contact form with twice the volume. Also, there are two different kinds of geodesics on $S^2$: let us call a geodesic “even” if it corresponds to a homotopically trivial loop in $Y$, and “odd” if it corresponds to a homotopically nontrivial loop in $Y$. Then Reeb orbits in $Y$ correspond to even geodesics (including even degree covers of odd geodesics). So Conjecture 1 above implies the following:

Conjecture 2. For any Riemannian metric $g$ on $S^2$, there exists an even geodesic of length $\ell$ with $\ell^2 \le 4\pi Area (S^2,g)$.

This conjecture is now sharp for the standard metric on $S^2$. The reason is that a great circle is an odd geodesic, so the shortest even geodesic is a double cover of this which has length $4\pi$.

In general, you can see whether a geodesic is even or odd by removing a point from $S^2$ which is not on the geodesic to get a plane curve and seeing whether the rotation number of the plane curve is even or odd. Also, the parity of the rotation number of the plane curve is opposite the parity of the number of double points of the curve.

Now what about the example with two triangles glued together? Here the shortest geodesic is again odd. Its double cover is even, but it has length $2\sqrt{3}$, so the desired inequality here is $(2\sqrt{3})^2\le 4\pi(\sqrt{3}/2)$, which fails. Does this mean that we have a counterexample to Conjecture 1?

No. There is another, slightly longer geodesic which is even. This geodesic consists of six segments going between the centers of the sides of the triangles and has length $3$. It is an even geodesic because it has an odd number (three) of double points. Here the desired inequality is $(3)^2\le 4\pi(\sqrt{3}/2)$, which holds.

Conclusion. We don’t yet have any counterexample to Conjecture 1, although I think one should try to make one using open books and see what happens. As for Conjecture 2, there is lots of literature on systolic geometry (concerning the existence of a closed geodesics which are short with respect to the Riemannian volume), although I don’t know if this specific conjecture about “even” geodesics has been considered.