Let $M$ be a (complete) Riemannian manifold with boundary $\partial M$ and with $\mu$ being the Riemannian measure on $M$. Given a compact set $K$ in $M$ and $\Omega$ being a precompact open set in $M$ with $K\subset \Omega$, let $$\text{cap}(K, \Omega)=\inf\int_\Omega |\nabla u|^2d\mu,$$ where the infimum is taken over all $u\in C_{0}^{\infty}(\Omega)$ such that $u\equiv 1$ on $K$, we say that $K$ has locally positive capacity if there exists some precompact open set $\Omega$ with $K\subset \Omega$ and $\text{cap}(K, \Omega)>0$.
My question:
Let $S$ be a non-empty compact smooth hypersurface such that
(1) $S\cap \partial M= \emptyset$ or
(2) $S\subset \partial M$.
Does $S$ have locally positive capacity then? I know that if $K\subset M\setminus \partial M$ is compact and the closure of a non-empty open set, then $K$ has locally positive capacity, but how about $S$?
Thanks in advance!
Best Answer
Here's a sketch of one way of showing that embedded hypersurfaces have locally positive capacity:
It's usefule to first show that $\{0\}\subset\mathbb{R}$ has locally positive capacity. Let $\partial_x$ and $dx$ be the standard unit vector field and unit $1$-form on $\mathbb{R}$ respectively. We can choose a precompact interval $(-\epsilon,\epsilon)$ and an arbitrary function $f\in C^\infty_0((-\epsilon,\epsilon))$ and apply the Cauchy-Schwarz inequality w.r.t. the $L^2$ iner product: $$ 2\epsilon\int_{-\epsilon}^\epsilon(\partial_x f)^2dx\ge\left(\int_{-\epsilon}^\epsilon|\partial_xf|dx\right)^2 $$ This, along with the inequality $\int_a^b|\partial_xf|dx\ge|f(b)-f(a)|$ gives $$ \int_{-\epsilon}^\epsilon(\partial_x f)^2dx\ge\frac{2}{\epsilon} $$ Which gives the desired lower bound.
For a hypersurface $S\subset\operatorname{int}(M)$, we can start with the case where $S$ has a global unit normal vector field $n:S\to TM$. In this case we can define a map $\varphi:S\times(-\epsilon,\epsilon)$ by $$ \varphi(p,x)=\gamma_{n(p)}(x) $$ Where $\gamma_{n(p)}$ is the unique geodesic with initial velocity $\dot{\gamma}(0)=n(p)$. This map is an embedding for sufficiently small $\epsilon$. Let $\widetilde{g}$ be the metric on $M$ and $g$ be the induced metric on $S$. We can define two metrics on $S\times(-\epsilon,\epsilon)$: The pullback metric $\varphi^*\widetilde{g}$ and the product metric $g\oplus dx^2$. Since these agree on $S\times\{0\}$ we can ensure that $2dV_{\varphi^*\widetilde{g}}\ge dV_{g\oplus dx^2}$ by shrinking $\epsilon$ as needed Further, $\partial_x$, the standard vector field tangent to the $\mathbb{R}$-factor, is a unit vector field in both metrics. Now, letting $\Omega=\operatorname{im}(\varphi)$ and $f\in C^\infty_0(\Omega)$, we have $$ \int_\Omega \|\nabla f\|^2dV_{\widetilde{g}}=\int_{S\times(-\epsilon,\epsilon)}\|\nabla f\|^2dV_{\varphi^*\widetilde{g}}\ge\int_{S\times(-\epsilon,\epsilon)}(\partial_x f)^2dV_{\varphi^*\widetilde{g}}\ge\frac{1}{2}\int_{S\times(-\epsilon,\epsilon)}(\partial_xf)^2dV_{g\oplus dx^2}=\int_SFdV_g \\ \text{where}\ \ \ F(p)=\int_{(-\epsilon,\epsilon)}(\partial_x(f|_{\{p\}\times\mathbb{R}}))^2dx\ge\frac{2}{\epsilon} \\ \implies\int_\Omega\|\nabla f\|^2dV_{\widetilde{g}}\ge\frac{1}{\epsilon}\operatorname{vol}(S) $$ If there is no global unit normal vector, the normal bundle $NM$ can be used in place of $S\times\mathbb{R}$, and the computation is essentially the same, since each fiber of $NM$ can be canonically identified with $\mathbb{R}$ up to sign. An almost indetical argument also works for $S\subseteq\partial M$, using $[0,\infty)$ in place of $\mathbb{R}$, $[0,\epsilon)$ in place of $(-\epsilon,\epsilon)$, and using the inward-pointing unit normal.