Lower bound on length of closed curve

Question

Let $\Sigma = \{ (x,y,z)\in \mathbb{R}^3 \colon -z^2+x^2+y^2 =1 \}$ be the one-sheeted hyperboloid, so there is a unique closed geodesic, which is of length $2\pi$, given by the intersection of $\Sigma$ with the plane $\{z=0\}$. Fix a point $p=(x_0,y_0,z_0)\in \Sigma$ with $z_0>0$. Does there exist an $\epsilon>0$ such that every smooth, homotopically non-trivial, closed curve $\gamma \colon [0,2\pi] \to \Sigma$ with $\gamma(0)=p$ has length $l(\gamma)\geq 2\pi+\epsilon$?

This seems to me to be intuitively the case, but I do not know how to prove it. Can anyone offer a proof or any suggestions towards a proof? Many thanks!

Answer 1

I would use following arguments:

• The curve $\gamma$ lies outside the right cylinder of axis $Oz$ and radius equal to one.
• The length of $\gamma$ is larger or equal to the one of its projection $\delta$ on the $Oxy$ plane. This is a general result: the length of a curve is larger or equal to the one of its orthogonal projection on any plane.
• $\delta$ is included in the region of the plane $Oxy$ minus the disk $D$ centered on $O$ with radius $1$ (consequence of the first argument).
• Using the Isoperimetric inequality you get that the length of $\delta$ is greater of equal to $2\pi$. This is therefore also the case for the length of $\gamma$.

You have to use an extra argument to get that $l(\gamma) \ge 2\pi +\epsilon$ with $\epsilon >0$ by using the isoperimetric inequality and the fact that the projection of $p$ on $Oxy$ is outside the disk $D$