TITLE: Is there the longest geodesic?
QUESTION [9 upvotes]: Given a closed 2-surface $M$ together with a Riemannian metric $g$.
We pick a free homotopy class $\gamma \in \pi_1(M)$ and consider the set $C(\gamma)$ of all closed geodesics homotopic to $\gamma$.
Of course, the set $C(\gamma)$ may be infinite. The length of a geodesic loop gives a function on $C(\gamma)$. My question is whether this function is bounded (in terms of $g$)?

REPLY [6 votes]: No such bound $C(\gamma)$ exists, even when $S$ is the two-sphere and even assuming that all geodesics considered are simple.  Here is the example (which generalises to surfaces with genus).
Suppose that $S$ is the two-sphere.  Pick four open disks $(D_i)_{i = 1}^4$ whose closures are closed disks and which are pairwise disjoint. (For example, use small round disks with respect to the usual round metric on $S$.)  Let $P = S - \bigcup_i D_i$; so $P$ is a "four-holed sphere".  We equip $P$ with a hyperbolic metric $g_P$ where all boundary components are geodesic.  We now claim the following:

$(P, g_P)$ has infinitely many closed simple geodesics (produced, for example, by a "braiding" construction).
All but four of these (namely, the curves of $\partial P$) are disjoint from $\partial P$.
In any infinite collection of these geodesics, their lengths are unbounded.

We now choose any riemannian metric $g_S$ on $S$ that extends $g_P$.  Note that all of the previous geodesics (except perhaps the components of $\partial P$) remain geodesic with respect to the metric $g_S$.  Since all of those geodesics are null-homotopic in $S$, we are done.
Morally: The riemannian surface $(S, g_S)$ has four "mushrooms" (the disks $D_i$).  The geodesics used above wander around the surface avoiding the tops of the mushrooms.