1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.