Let $d\in\mathbb N$, $k\in\{1,\ldots,d\}$ and $\Omega$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary. Assume, for simplicity, that $\Omega$ is described by a single chart, i.e. there is a $C^1$-diffeomorphism from $\Omega$ onto an open subset $U$ of $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$.

In the definition of the surface measure, we need to consider the Gram matrix$^1$ $$G_{\phi^{-1}}(u):=\left|{\rm D}\phi^{-1}(u)\right|^2\;\;\;\text{for }u\in U$$ associated to $\phi^{-1}$ and the square-root $$\sqrt{g_{\phi^{-1}}(u)}=\det\left|{\rm D}\phi^{-1}(u)\right|\;\;\;\text{for all }u\in U\tag1$$ of the Gram determinant $$g_{\phi^{-1}}:=\det G_{\phi^{-1}}.$$

Now we know that $$\tilde\phi:=\pi\circ\left.\phi\right|_{\partial\Omega},$$ where $\pi$ is the canonical projection of $\mathbb R^k$ onto $\mathbb R^{k-1}$ with $\pi(\partial\mathbb H^k)=\mathbb R^{k-1}$, is the chart for the manifold boundary $$\partial\Omega=\{x\in\Omega:\phi(x)\in\partial\mathbb H^k\}\tag2.$$

Question: Can we find an expression for $g_{\tilde\phi^{-1}}$ which does only involve $\phi$?

It might be useful to note that $${\rm D}\tilde\phi^{-1}(\tilde u)={\rm D}\phi^{-1}(\iota(\tilde u))\circ\iota\;\;\;\text{for all }\tilde u\in\tilde U\tag3,$$ where $\iota$ is the canonical embedding of $\mathbb R^{k-1}$ into $\mathbb R^k$ with $\iota\mathbb R^{k-1}=\partial\mathbb H^k:=\mathbb R^{k-1}\times\{0\}$ and $\tilde U:=\tilde\phi(\partial\Omega)$. Noting that $\iota^\ast=\pi$, we obtain $$G_{\tilde\phi^{-1}}(\tilde u)=\pi\circ G_{\phi^{-1}}(\iota(\tilde u))\circ\iota\;\;\;\text{for all }\tilde u\in\tilde U.$$

Assuming that $k=d$, I was able to derive $$\sqrt{g_{\tilde\phi^{-1}}(\tilde u)}=\left|\det{\rm D}\phi^{-1}(\iota(\tilde u))\right|\left\|\left({\rm D}\phi^{-1}(\iota(\tilde u))^{-1}\right)^\ast e_d\right\|\;\;\;\text{for all }\tilde u\in\tilde U\tag5,$$ where $(e_1,\ldots,e_k)$ denotes the standard basis of $\mathbb R^k$.

$^1$ If $A$ is any matrix, then $|A|:=\sqrt{A^\ast A}$.


1 Answer 1


I think I've figured it out, but I'd really appreciate if someone could verify my solution:

We will need the following basic facts about the determinant of a matrix $A\in\mathbb R^{m\times n}$, where $m,n\in\mathbb N$:

  1. If $m=n$ and $i,j\in\{1,\ldots,n\}$, then $$(-1)^{i+j}\det A^{ij}=\det A(A^{-1})^\ast_{ij}\tag6,$$ where $A^{ij}$ denotes the submatrix obtained from $A$ by delting the $i$th row and $j$th column.
  2. If $m=n$, then $$\sqrt{\det A}=\det\sqrt A\tag7.$$
  3. $$\det|A|=\sqrt{\det{A^\ast A}}\tag8.$$
  4. If $m=n$, then $$\det|A|=\left|\det A\right|\tag9.$$

Now let $\tilde u\in\tilde U$, i.e. $\tilde u=\tilde\phi(x)=\pi(\phi(x))$ for some $x\in\partial\Omega$ (and since $\left.\iota\circ\pi\right|_{\partial\mathbb H^k}=\operatorname{id}_{\partial\mathbb H^k}$ and $\phi(x)\in\partial\mathbb H^k$, we have $\iota(\tilde u)=\phi(x)$).

Let $$A:={\rm D}\phi^{-1}(\iota(\tilde u))=T_x(\phi)^{-1}\tag{10},$$ where $T_x(\phi):T_x\:\Omega\to\mathbb R^k$ denotes the pushforward of $\phi$ at $x$, and $$\tilde A:={\rm D}\tilde\phi^{-1}(\tilde u)=A\circ\iota\tag{11}.$$ Since $$\tilde A^\ast\tilde A=B^{kk},$$ we obtain $$\det{\tilde A^\ast\tilde A}=\det B(B^{-1})^\ast_kk\tag{12}$$ from $(6)$. Now, $$B=(T_x(\phi)^\ast)^{-1}T_x(\phi)^{-1}\tag{13}$$ and hence $$(B^{-1})^\ast=B^{-1}=T_x(\phi)T_x(\phi)^\ast\tag{14}.$$ So, $$g_{\tilde\phi^{-1}}(\tilde u)=\det B\left\|T_x(\phi)^\ast e_k\right\|^2=g_{\phi^{-1}}(\iota(\tilde u))\left\|T_{\phi^{-1}(\iota(\tilde u))}(\phi)^\ast e_k\right\|^2\tag{15}.$$ This can be reformulated as $$\sqrt{g_{\tilde\phi^{-1}}(\tilde u)}=\det\left|{\rm D}\phi^{-1}(\iota(\tilde u))\right|\left\|T_{\phi^{-1}(\iota(\tilde u))}(\phi)^\ast e_k\right\|\tag{16}.$$ In the special case $k=d$, we know $$\det\left|{\rm D}\phi^{-1}(\iota(\tilde u))\right|=\left|\det{\rm D}\phi^{-1}(\iota(\tilde u))\right|$$ from $(9)$.


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