# Gram determinant of a boundary chart of a submanifold

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 $$|A|:=\sqrt{A^\ast A}$$.

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 $$(9)$$.