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Suppose $(M,g)$ is a 2-dimenensional compact Riemannian manifold with boundary $S$. Furthermore, assume that $S$ is totally geodesic. Now consider the conformal change of metric $\widetilde{g}=e^{\phi}g$ for some smooth function $\phi$. In order for $\widetilde{g}$ to satisfy the condition that $S$ is still totally geodesic, then it is necessary that $\partial_N \phi=0$. Is this condition sufficient?

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Yes, note that under the conformal change of metric $\tilde{g}=e^{2u}g$, the geodesic curvature changes as follows: (see p.1050 in here for example) $$\frac{\partial u}{\partial\nu}+k_g=k_{\tilde{g}}e^u$$ where $\frac{\partial u}{\partial\nu}$ is the normal derivative of $u$. Therefore, if $S$ is geodesic with respect to $g$ (i.e. $k_g=0$) and $\frac{\partial u}{\partial\nu}=0$, we have $k_{\tilde{g}}=0$, i.e $S$ is geodesic with respect to $\tilde{g}$.

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