Riemannian Green identity, boundary trace, and wave lifting

Let $DM$ be the labelled double. Around each $x\in \mathrm{\partial }M$, [F3] supplies a smooth local extension of $g$ as a covariant two-tensor. Symmetrizing it preserves its restriction to $M$, and continuity of positive definiteness at the seam lets us shrink the chart so that this symmetric extension $g_x$ is positive definite. Compactness of $\mathrm{\partial }M$ gives finitely many smaller seam coordinate balls $B_i$ covering it, with compact closures inside metric-extension charts $U_i$. By [F4], choose nonnegative smooth bumps ${\psi }_i$ equal to one on $\overline{B_i}$ and supported in $U_i$. Put $\sigma ={\sum }_i{\psi }_i$ and $W=\{\sigma >0\}$. By [F4], choose a smooth cutoff $\eta$ equal to one on an open neighborhood $W_0$ of $\mathrm{\partial }M$ and supported in $W$. Define ${\varphi }_i=\eta {\psi }_i\mathrm{/}\sigma$ on $W$ and zero outside; these are smooth and supported in $U_i$, and ${\sum }_i{\varphi }_i=1$ on $W_0$. Extend each ${\varphi }_i{g}_i$ by zero outside $U_i$ and put $\tilde{g}={\sum }_i{\varphi }_i{g}_i$ on $W_0$. This is positive definite as a convex combination, and it equals $g$ on $W_0\cap M$ because every $g_i$ restricts to $g$ there. Thus $\tilde{g}$ is a positive smooth metric on an open neighborhood of $\mathrm{\partial }M$ in the boundaryless double. Apply [F6] to its geodesic ODE. For small $s\ge 0$, set $E(x,s)={\exp }_x(-s{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}(x))$. At $s=0$ its differential sends $(w,a)\in T_x\mathrm{\partial }M\oplus \mathbb{R}$ to $w-a{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}$, an isomorphism, so [F6] gives a local collar chart. The inward normal velocity points into $M$, so these images lie in $M^{\circ }$ for small positive $s$. Compactness gives a uniform collar width and local invertibility on $\mathrm{\partial }M\times [0,ϵ)$. If no positive width made $E$ injective, there would be distinct $(x_k,s_k)$ and $(y_k,r_k)$ with equal images and $s_k,r_k\downarrow 0$. Compactness gives convergent subsequences $x_k\to x$ and $y_k\to y$; continuity and equality of the images imply $x=y$. Both parameter sequences then lie in one local inverse chart at $(x,0)$, contradicting local injectivity. Hence $E$ is a diffeomorphism from $\mathrm{\partial }M\times [0,ϵ)$ onto a collar.

In a coordinate chart write $G=(g_{ij})$, $\rho =\sqrt{\det G}$, and let ${\mathrm{\Gamma }}_{ij}^k$ be the Levi–Civita symbols. Applying the Koszul formula from [F5] to coordinate vector fields (whose brackets vanish) and raising the last index gives ${\mathrm{\Gamma }}_{ij}^k=\frac{1}{2}{g}^{k\mathrm{\ell }}({\mathrm{\partial }}_i{g}_{j\mathrm{\ell }}+{\mathrm{\partial }}_j{g}_{i\mathrm{\ell }}-{\mathrm{\partial }}_{\mathrm{\ell }}{g}_{ij})\mathrm{.}$ By [F5], divergence is the coefficient of the Lie derivative of $\rho \mathrm{\mid }d{x}^1\cdots d{x}^n\mathrm{\mid }$, and the coordinate Lie derivative gives ${\mathcal{L}}_W(\rho \mathrm{\mid }d{x}^1\cdots d{x}^n\mathrm{\mid })={\mathrm{\partial }}_i(\rho W^i)\mathrm{\mid }d{x}^1\cdots d{x}^n\mathrm{\mid },$ so ${\mathrm{div}}_{g}W={\rho }^{-1}{\mathrm{\partial }}_i(\rho W^i)$. The connection trace is ${\mathrm{\partial }}_i{W}^i+{\mathrm{\Gamma }}_{ik}^i{W}^k$. Its Christoffel trace is ${\mathrm{\Gamma }}_{ik}^i=\frac{1}{2}{g}^{i\mathrm{\ell }}({\mathrm{\partial }}_i{g}_{k\mathrm{\ell }}+{\mathrm{\partial }}_k{g}_{i\mathrm{\ell }}-{\mathrm{\partial }}_{\mathrm{\ell }}{g}_{ik})=\frac{1}{2}{g}^{i\mathrm{\ell }}{\mathrm{\partial }}_k{g}_{i\mathrm{\ell }}={\mathrm{\partial }}_k\log \rho \mathrm{.}$ In the middle equality, relabeling $i$ and $\mathrm{\ell }$ makes the first and third terms cancel; the last equality is the determinant differential formula in [F5]. Hence the connection trace is also ${\mathrm{\partial }}_i{W}^i+{\mathrm{\Gamma }}_{ik}^i{W}^k={\rho }^{-1}{\mathrm{\partial }}_i(\rho W^i)\mathrm{.}$ These two formulas establish the required coordinate divergence identity. Use the finite smooth partition of unity from [F4]. For a field supported in an interior chart, integrating each coordinate derivative gives zero. In a boundary chart flattened to $x_n\ge 0$, tangential derivatives integrate to zero because the field is supported away from the chart sides, while the normal derivative integrates to $-\sqrt{\det G}{W}^n$ at $x_n=0$. The induced boundary density and outward conormal satisfy $g(W,{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}})d{S}_g=-\sqrt{\det G}{W}^{n}d{x}^1\cdots d{x}^{n-1}$ there, so summing the chart identities gives ${\int }_M{\mathrm{div}}_{g}Wd{V}_g={\int }_{\mathrm{\partial }M}g(W,{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}})d{S}_g$. With $W=u{\mathrm{\nabla }}_{g}v-v{\mathrm{\nabla }}_{g}u$, the product rule gives ${\mathrm{div}}_{g}W=u{\mathrm{\Delta }}_{g}v-v{\mathrm{\Delta }}_{g}u$, hence ${\int }_M((-{\mathrm{\Delta }}_{g}u)v+u{\mathrm{\Delta }}_{g}v)d{V}_g={\int }_{\mathrm{\partial }M}(u{\mathrm{\partial }}_{{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}}v-v{\mathrm{\partial }}_{{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}}u)d{S}_g\mathrm{.}$

Choose $\chi \in C_c^{\mathrm{\infty }}([0,ϵ))$ equal to $1$ near $0$. Define $H(E(x,s),t)=\chi (s)h(x,t)$ in the collar and set $H=0$ outside it. The time support of $h$ and the normal cutoff make this a smooth function vanishing near $t=0,T$. The collar map and inverse have bounded first derivatives on a smaller compact collar, so the chain rule gives $\mathrm{\parallel }H{\mathrm{\parallel }}_{C^1([0,T]\times M)}\le C\mathrm{\parallel }h{\mathrm{\parallel }}_{C^1(\mathrm{\Sigma })}$.

For each fixed $x$, one-dimensional integration by parts gives ${\int }_0^T(u_{tt}v-u{v}_{tt})dt=[u_{t}v-u{v}_t{]}_0^T$. Integrating over $M$ and adding the spatial identity in step 1.2 yields the stated space-time Green identity.

Since $\mathrm{\mid }{\mathrm{\partial }}_{{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}}u\mathrm{\mid }=\mathrm{\mid }g({\mathrm{\nabla }}_{g}u,{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}})\mathrm{\mid }\le \mathrm{\mid }{\mathrm{\nabla }}_{g}u\mathrm{\mid }$, integration over the finite-measure compact boundary gives $\mathrm{\parallel }{\mathrm{\partial }}_{{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}}u{\mathrm{\parallel }}_{L^2(\mathrm{\Sigma })}^2\le T{\mathrm{area}}_g(\mathrm{\partial }M)\mathrm{\parallel }{\mathrm{\nabla }}_{g}u{\mathrm{\parallel }}_{C^0}^2$. Taking square roots proves the normal-trace estimate.