Such integrals are usually defined when the submanifold $\mathcal{N}\subset \mathbb{R}^n$ is given (on a neighborhood $V$ of a point $p$) by an immersion $\iota: \mathbb{R}^k\supset U \longrightarrow V \subset \mathbb{R}^n$ or a graph $\Gamma_f :=\left\lbrace (\mathbf{x},f(\mathbf{x})),\ \mathbf{x}\in U\right\rbrace$ of a function $f: U \longrightarrow W \subset \mathbb{R}^{n-k}$ which are such that
$$V\cap \mathcal{N} = \iota(U) = \Gamma_f$$
The parametrization as a graph can in fact be simply be understood as an immersion $\iota:\mathbb{x}\longmapsto (\mathbf{x},f(\mathbf{x}))$. The measure is then defined with the Euclidean structure of $\left( \mathbb{R}^n, \langle\, \cdot \, , \, \cdot \, \rangle \right) $ by
$$ d\sigma (\mathbb{x})= \sqrt{\det\left(\left\langle \partial_i \iota, \partial_j \iota \right\rangle_{i,j=1}^{k} \right) }\, d\mathbf{x} \tag{M}\label{M}$$
(The matrix that appears is known as Gram matrix $G_{d\iota}$ (of the Jacobian $d\iota$), it is a $k\times k$ matrix, related to $d\iota$ by $G_{d\iota}={}^t(d\iota)\cdot d\iota $ when the components of each $\partial_i\iota \in \mathbb{R}^n$ are given in an orthonormal basis. $d\iota$ is an $n\times k$ matrix whose determinant makes no sense. $G_{d\iota}$ can be understood as its square and its determinant makes sense.) This is mentionned for example in here, here or here; and for books the best references I found are 1) "Cours d'analyse - Théorie des distributions et analyse de Fourier" (2001) by Jean-Michel Bony (A.6) p.229 but I would recommend stating from p.224 onward; 2) "Calcul différentiel et intégral" (2001), by François Laudenbach p.194 forward. He turns the immersion into a diffeomorphism, really interesting how he obtains the usual formula p.196; 3) "Analysis 3" by Otto Forster, Chap. 14. He gives an interesting formula for the determinant.
(I plan to make explicit the link with induced $k$-form as explained here one day) but what I want to do now is to define the same measure (\ref{M}) but expressed with the parametrization of $\mathcal{N}$ as the $0$-level set of a submersion $\Phi: \mathbb{R}^n \to \mathbb{R}^{n-k}$ (for simplicity let me stop saying that the map is defined from open subsets to an open subset of the respective spaces. Normally one should say $\forall\ p\in \mathcal{N},\ \exists\ V\subseteq \mathbb{R}^n$ neighborhood of $p$ (in $\mathbb{R}^n$) and a submersion $\Phi:V\to W$ such that $V\cap \mathcal{N}= \Phi^{-1}(\{\mathbf{0}\}),\ \mathbb{0}\in \mathbb{R}^{n-k}$). By relabelling the components in the source space, one can assume that (hypothesis of submersion)
$$ d\Phi =\left( \frac{\partial \Phi^i}{\partial x_j}\right)_{\genfrac{}{}{0pt}{1}{1\leq i\leq n-k}{1\leq j\leq n}} = \begin{pmatrix} d_{\mathbb{x}'}\Phi & d_{\mathbb{x}''}\Phi \end{pmatrix}\quad \text{with}\ d_{\mathbb{x}''}\Phi \ \text{invertible},\ \mathbf{x}'=(x_1,\cdots, x_k),\ \mathbf{x}''=(x_{k+1},\cdots, x_n) $$
(Again, by laziness I don't write there exists an appropriate open subset) For $\mathbb{x}'$ in some open subset, $\exists !\ \mathbb{x}''$ in some other open s.t. $\mathbb{x}:=(\mathbb{x}',\mathbb{x}'') \in \Phi^{-1}(\{\mathbf{0}\}) = V\cap \mathcal{N}$. This is the graph theorem, namely that locally (in the neighborhood $V$), $\mathcal{N}$ is the graph of a function $f$, i.e. $\mathbb{x}''= f(\mathbb{x}')$ (characterizes the $\mathbb{x}\in \mathcal{N}$). Moreover, one has
$$\Phi(\mathbf{x}',f(\mathbf{x}'))=\mathbf{0} \quad \Longrightarrow \quad d_{\mathbb{x}'}\Phi(\mathbf{x}',f(\mathbf{x}')) = 0 = d_{\mathbb{x}'}\Phi + d_{\mathbb{x}''}\Phi \circ d f \quad \Longrightarrow \quad df = -\left(d_{\mathbb{x}''}\Phi\right)^{-1} \circ d_{\mathbb{x}'}\Phi \tag{D}\label{D}$$
It suffices now to take $\iota(\mathbb{x}'):=\big(\mathbf{x}',f(\mathbf{x}')\big) $ plug it into (\ref{M}) and use (\ref{D}) to re-express everything in terms of the submersion $\Phi$. Assuming that one has an orthonormal basis of $\mathbb{R}^n$ (in which one write the components of the jacobian of $\mathbb{x}' \longmapsto (\mathbb{x}',f(\mathbb{x}'))$, one obtains $$ d\sigma(\mathbf{x}') = \sqrt{\det \left(\mathrm{Id}_k + {}^t(df)\cdot df \right)}\, d\mathbf{x}' = \sqrt{\det \left(\mathrm{Id}_k + G_{\left(d_{\mathbb{x}''}\Phi\right)^{-1} \circ d_{\mathbb{x}'}\Phi} \right)}\, d\mathbf{x}' \tag{F}\label{F}$$
Question: in an attempt to generalize definition 5.5.3 p.240 of "A Course on Integration Theory" (2014) by Nicolas Lerner, I'm trying to show that (\ref{F}) coincides with
$$d\sigma(\mathbf{x}') \overset{?}{=}\frac{\sqrt{\det\left(\left\langle \nabla \Phi^i, \nabla \Phi^j \right\rangle_{i,j=1}^{n-k} \right) }}{\det \left(d_{\mathbb{x}''}\Phi\right)}\, d\mathbf{x}' \tag{R}\label{R}$$
Comments:
- formally the definition of an integral on a submanifold with a submersion uses a kind of change of variable formula for distributions, cf. (2.1) p. 83 "Éléments de distributions et d’équations aux dérivées partielles" ( 2002) by Claude Zuily: $$\delta^{(d-k)}\left( \Phi(\mathbf{x}) \right)\, d\mathbf{x}= \delta^{(d-k)} \left( \mathbf{u}\right)\, \left\lvert \det \left( d \tilde{\Phi}\right)^{-1} \left( \mathbf{x}',\mathbf{u}\right) \right\rvert\, d \mathbf{x}' \, d \mathbf{u} = \frac{1}{\left\lvert\det \left( d \tilde{\Phi}(\mathbf{x})\right) \right\rvert}\, d\mathbf{x}' \tag{C}\label{C}$$ where $$\tilde{\Phi}(\mathbf{x}',\mathbf{x}'')=\left(\mathbf{x}',\Phi(\mathbf{x}) \right)= \left(\mathbf{x}',\mathbf{u} \right)\; \Rightarrow\; d\tilde{\Phi}= \begin{pmatrix} \mathrm{Id}_{k} & 0 \\ d_{\mathbf{x}'}\Phi & d_{\mathbf{x}''}\Phi \end{pmatrix}\; \Rightarrow\; \det d\tilde{\Phi} = \det d_{\mathbf{x}''}\Phi $$
- The Gram matrix here is different $$ \left(\left\langle \nabla \Phi^i, \nabla \Phi^j \right\rangle_{i,j=1}^{n-k} \right) =G_{{}^td\Phi} = (d\Phi)\cdot {}^t(d\Phi)$$ Indeed if $M$ is an $n\times k$ matrix with $k< n$ then $G_M:={}^t M\cdot M$ may be invertible (may have a non trivial determinant), but $G_{{}^tM}$, even if it is perfectly defined, cannot be invertible, so its determinant is $0$....
What I have done so far: $$G_{{}^{t}d \Phi} =\left( d \Phi \right) \cdot {}^{t} \left( d \Phi\right) = \begin{pmatrix} d_{\mathbf{x}'}\Phi & d_{\mathbf{x}''}\Phi \end{pmatrix} \cdot \begin{pmatrix} {}^{t}d_{\mathbf{x}'}\Phi \\ {}^{t}d_{\mathbf{x}''}\Phi \end{pmatrix} = G_ {{}^t d_{\mathbf{x}'}\Phi} + G_{{}^t d_{\mathbf{x}''}\Phi} $$ Let us examine the behavior of its determinant under the congruence relation $M \cong {}^{t}P M P$, here with $P:=\left({}^t d_{\mathbf{x}''}\Phi \right)^{-1}$)
$$\frac{\det \left(G_{{}^{t}d \Phi} \right) }{\det \left(d_{\mathbb{x}''}\Phi\right)^2}= \det \left({}^{t}P\cdot G_{{}^{t}d \Phi}\cdot P \right) = \det \left( G_ {{}^t d_{\mathbf{x}'}\Phi\circ \left({}^t d_{\mathbf{x}''}\Phi \right)^{-1}} + \mathrm{Id}_{n-k} \right) \tag{P}\label{P}$$ Hence it remains to show its square root is indeed equal to (\ref{F})
