Let S be a hyperbolic surface, and let $\gamma\subset S$ be a simple, closed geodesic. I have been wondering about the following problem:
For $\epsilon>0$ sufficiently small, is it true that $\partial B_{\epsilon}(\gamma)$ is a union of closed geodesics?
My intuition is yes. My thinking is that one should choose $\epsilon$ small enough that $B_{\epsilon}(\gamma)$ is a tubular neighborhood foliated by geodesics passing perpendicularly through $\gamma$. Then hopefully the foliation would give us a parametrization of the two components of $\partial B_{\epsilon}(\gamma)$ that would satisfy the geodesics equation.
Another idea might would be to show that orthogonal projection provides an isometry from each components of $\partial B_{\epsilon}(\gamma)$ to $\gamma$.
Thoughts? Thank you for any input!
Hyperbolic geometry is quite different from Euclidean geometry in this regard! You'll need to adjust your intuition.
For $\epsilon > 0$ sufficiently small, $B_\epsilon(\gamma)$ is an annulus, and neither of the two circles comprising $\partial B_\epsilon(\gamma)$ is a simple closed geodesic (this is assuming $\gamma$ preserves orientation; otherwise $B_\epsilon(\gamma)$ is a Möbius band and its one circle is not a simple closed geodesic).
Instead, at each point $p \in \partial B_\epsilon(\gamma)$ there is a nonzero curvature vector, that vector points towards the inside of $B_\epsilon(\gamma)$, and its magnitude is a constant independent of the point $p$. The value of that constant is a function $\kappa(\epsilon)$ that approaches $0$ as $\epsilon \to 0$; one can work out a precise formula for $\kappa(\epsilon)$ although I do not know it off the top of my head.
And regarding your "other idea", one can work out quite precisely what happens under the orthogonal projection map $\partial B_\epsilon(\gamma) \mapsto \gamma$. This formula I do know: rather than that projection map being an isometry, its derivative decreases the length of each tangent vector by a constant factor equal to $\cosh(\epsilon)$.