Geodesic curvature change under conformal metrics

Question

Suppose that $\sigma_0$ is a fixed metric on a compact riemannian 2-manifold $M$ with boundary $\partial M$. Let $\sigma=\rho \sigma_{0}$, where $\rho=e^{2\varphi}$ with $\varphi \in C^{\infty}(M)$, be the family of metrics conformal to $\sigma_0$.

Show that $$k=e^{-\varphi}\left(k_0+\partial_n\varphi \right)$$ where $k$ and $k_0$ are the geodesic curvatures of a curve (in $\partial M$) in the $\sigma$ and $\sigma_0$ metrics, respectively, while $\partial_n$ is the right pointing normal derivative (the outer normal derivative for $\partial M$) with respect the metric $\sigma_0$.

A possible way would be: I should use that $$\bar{\nabla_{X}}^{Y}=\nabla_{X}^{Y} + (X \varphi)Y + (Y\varphi)X-\sigma_0(X,Y)\nabla_{0} \varphi.$$

$$N_0(\gamma(t)=e^{\varphi}N(\gamma(t))$$

$$k(\gamma(t))= \left<\bar\nabla_{\gamma^{'}(t)}^{\gamma^{'}(t)}, N(\gamma(t)\right>$$

$$\partial_n \varphi=\left<\nabla_0 \varphi, N\right> \quad in \quad \partial M.$$

where $\bar{\nabla}, \nabla$ are the Levi-Civita connections of $\sigma$ and $\sigma_0$, and $N$ is a unique smooth outward-pointing unit normal vector field along $\partial M$, with respect $\sigma$, and $\{\gamma\} \subset \partial M$ is a regular differentiable curve parametrized by arc length.

The relation between this fiedls with respect $\sigma$ and $\sigma_0$ metrics are $N_0(\gamma(t))=-\dfrac{e^{2\varphi} \nabla f}{e^{\varphi} \left|\nabla f\right|}=e^{\varphi}N(\gamma({t}))$, because $$(\nabla f)^{i}=\sum _{j=1}^{2} \sigma^{ij} \dfrac{\partial f}{\partial x^{j}}=e^{-2\varphi}(\nabla_0 f)^{i}$$ where $f$ is a boundary defining function.

Now I should do $$e^{-\varphi(\gamma(t)}(k_0(\gamma(t)) + \partial_n \varphi(\gamma(t)))= e^{-\varphi}\left(\left<\nabla_{\gamma^{'}(t)}^{\gamma^{'}(t)}, N_0(\gamma(t))\right>_0)+ \left<\nabla_0 \varphi(\gamma(t)), N_0(\gamma(t))\right>_0\right)=e^{-\varphi}\left<\bar \nabla_{\gamma^{'}(t)}^{\gamma^{'}(t)} + 2 \nabla_0 \varphi (\gamma(t)), N_0(\gamma(t))\right>_0$$, in this point I don't know what I can do. Is there anyone who can help me?

Answer 1

I am not sure if it is right but following is my calculation. First, the geodesic curvature is defined to be $$k_g=g(\nabla_{v}v,N)$$ where $v,\ N$ respectively are the unit tangent vector and inner unit normal with respect to $g$. Now if $\tilde{g}=e^{2\varphi}g$, we choose $\dot{\gamma}(t)$ (as you denoted if you think it as parametrization) such that $\tilde{g}(\dot{\gamma}(t),\dot{\gamma}(t))=1$, so we have $g(e^{\varphi}\dot{\gamma}(t),e^{\varphi}\dot{\gamma}(t))=1.$ And we choose $g(N,N)=1$, so $\tilde{g}(e^{-\varphi}N,e^{-\varphi}N)=1$. (Basically I just chose tangent and inner unit normal vectors with respect to two metrics). Thus \begin{eqnarray*} k_{\tilde{g}}&=&\tilde{g}(\tilde{\nabla}_{\dot{\gamma}}\dot{\gamma},e^{-\varphi}N)\\ &=& e^{2\varphi}g(\nabla_{\dot{\gamma}}\dot{\gamma}+2g(\dot{\gamma},\nabla \varphi)\dot{\gamma}-g(\dot{\gamma},\dot{\gamma})\nabla \varphi, e^{-\varphi}N)\\ &=&e^{2\varphi}g(\nabla_{\dot{\gamma}}\dot{\gamma}-e^{-2\varphi}\nabla \varphi, e^{-\varphi}N)\\ &=& e^{\varphi}g(\nabla_{\dot{\gamma}}\dot{\gamma},N)-e^{-\varphi}g(\nabla \varphi,N)\\ &=& e^{\varphi}g(e^{-2\varphi}\nabla_{e^{\varphi}\dot{\gamma}}(e^{\varphi}\dot{\gamma})-e^{-\varphi}g(\dot{\gamma},\dot{\gamma})\dot{\gamma},N)-e^{-\varphi}g(\nabla \varphi,N)\\ &=& e^{\varphi}g(e^{-2\varphi}\nabla_{e^{\varphi}\dot{\gamma}}(e^{\varphi}\dot{\gamma}),N)-e^{-\varphi}g(\nabla \varphi,N)\\ &=& e^{-\varphi}k_g-e^{-\varphi}\frac{\partial \varphi}{\partial N} \end{eqnarray*} So if you want outer normal the last part becomes plus.