Suppose $M\_{g}$ is the mapping torus $\Sigma\_{g} \times [0, 1]/ (x, 0) \equiv (\tau x, 1)$, where $\Sigma\_{g}$ is the hyperbolic space with genus $g,$ and $\tau : \Sigma\_{g} \to \Sigma\_{g}$ is an isometry map. 


Now suppose $\gamma$ is a simple closed geodesic in $M\_{g}$, could we find a tubular neighborhood of $\gamma$ in $M\_{g}$ and with radius $\varepsilon$, $\varepsilon$ is small enough, such that the boundary torus of this tubular neighborhood has flat metric?



The Tubular Neighborhood of a Closed Geodesic
No, it is flat only in case if the geodesic is horisontal; i.e., locally we have $\gamma(t)=(x,t)$.


In general, the torus will have an $\mathbb S^1$-invariant metric which is not flat.


In the simplest case of vertical geodesic the metric tensor in coordinates $(t,\theta)$
will look like this
$$
\begin{pmatrix}
1&0
\\\
0&\cosh(\sin\theta\cdot\epsilon)
\end{pmatrix}
$$
(or something like this). 


