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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?

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$\gamma$ is a simple closed curve. – Lizhi Chen Oct 30 '11 at 3:18

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).

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