Surface measure $\mathrm{d}S$ of $M$ corresponding to parametrisation

Question

I'm reading these notes, and there's a part I don't understand on page 72 (76 in the PDF), right after the equation marked with two stars.

Note that $\sqrt{1+|Dg(y')|^2}dy'$ is the surface measure $\mathrm{d}S$ of $M$ corresponding to the parametrisation $y'\mapsto(g(y'),y')$ of $M$

A reference is then given, but I can only find it in German and can't understand it.

$y'$ is some point in $\mathbb{R}^{n-1}$, $M$ is a hypersurface in an open subset of $\mathbb{R}^n$. $g$ is a function from $\mathbb{R}^{n-1}$ to $\mathbb{R}$.

Answer 1

This hypersurface $S$ is parametrized by $$y'\mapsto (y',g(y'))\in{\mathbb R}^n\qquad(y'\in{\mathbb R}^{n-1})\ ,$$ hence is considered as graph of the scalar function $g$ defined on ${\mathbb R}^{n-1}$.

If $S$ were a hyperplane $y_n=a_1y_1+a_2y_2+\ldots a_{n-1}y_{n-1}$ then the surface measure on $S$ would be ${1\over\cos\phi}{\rm d}(y')$, where ${\rm d}(y')$ is the $(n-1)$-dimensional euclidean (or Lebesgue) measure on the coordinate hyperplane $y_n=0$, and $\phi$ is the angle between the normal of $S$ and the $y_n$-axis. Now $${1\over\cos\phi}={1\cdot\sqrt{1+a_1^2+a_2^2+\ldots+a_{n-1}^2}\over\langle{\bf e}_n\cdot(-a_1,-a_2,\ldots,-a_{n-1},1)\rangle}=\sqrt{1+a_1^2+a_2^2+\ldots+a_{n-1}^2}=\sqrt{1+|{\bf a}|^2}\ .$$

In the case at hand the tangent plane $T_p$ to the curved surface $S$ at some point $p\in S$ is given by $$T_p:\quad y_n-p_n=\nabla g(p')\cdot (y'-p')\ .$$ The local factor ${1\over\cos\phi}$ therefore is given by $\sqrt{1+|\nabla g(y')|^2}$, as indicated in your source.

Answer 2

Another way to derive the result beside Christian Blatter's excellent answer is to calculate the volume form on the Riemannian manifold.

We'll use $y=(y^1,y^2,\dots,y^n)\in \mathbb{R}^{n-1}$ instead of $y'$ to denote the coordinates. Since we'll reserve $g$ for the metric, we'll use $f$ for the function instead. Let $M=\{(y,f(y))\mid y\in\mathbb{R}^{n-1}\}$ be the graph of $f.$ We'll regard $M$ as an embedded submanifold of $\mathbb{R}^n,$ with metric induced by the canonical metric on $\mathbb{R}^n.$ The surface measure $dS$ is (by abuse of language) just the volume form on $M.$

Note that $M$ is naturally diffeomorphic to $\mathbb{R}^{n-1}.$ By abuse of notation, we'll write this diffeomorphism as $f.$ Now, we'll calculate the pushforward of $\partial_{y^i}$ under $f,$ which will form the global coordinate vector field. Let $\gamma_i=(y^1,\dots,y^{i-1},y^i+t,y^{i+1},\dots,y^{n-1}).$ Note that $(\gamma_i)_\ast(\frac{d}{dt})|_{t=0}=\partial_{y^i},$ so we have $$f_\ast\partial_{y^i}=f_\ast\circ(\gamma_i)_\ast(\frac{d}{dt})|_{t=0}=(f\circ\gamma)_\ast(\frac{d}{dt})|_{t=0}=\frac{d}{dt}(f\circ\gamma)|_{t=0}=e_i+f_{y^i}e_n,$$ where $e_i$ is the $i$-th unit coordinate vector in Euclidean space, and lower indices indicate partial differentiation.

Thus, we have $g(f_\ast\partial_{y^i},f_\ast\partial_{y^j})=\delta_{ij}+f_{y^i}f_{y^j}.$ Using the matrix determinant lemma, we can prove that the determinant of the matrix $[g(f_\ast\partial_{y^i},f_\ast\partial_{y^j})]$ is precisely ${1+|Df|^2}.$ Thus, we have $$ \begin{align} \int_MdS=&\int_{\mathbb{R}^{n-1}}f^\ast dS\\ =&\int_{\mathbb{R}^{n-1}}f^\ast( \sqrt{|g|}\ (f_\ast\partial_{y^1})^\flat\wedge\cdots\wedge (f_\ast\partial_{y^{n-1}})^\flat)\\ =&\int_{\mathbb{R}^{n-1}}(\sqrt{|g|}\circ f )\ f^\ast(f_\ast\partial_{y^1})^\flat\wedge\cdots\wedge f^\ast (f_\ast\partial_{y^{n-1}})^\flat\\ =&\int_{\mathbb{R}^{n-1}}(\sqrt{|g|}\circ f )\ dy^1\wedge\cdots\wedge dy^{n-1}\\ =&\int_{\mathbb{R}^{n-1}}\sqrt{1+|Df|^2}\ dy^1\wedge\cdots\wedge dy^{n-1}\\ \end{align}. $$

Answer 3

From my point of view, the only way one should think to that expression is using area formula (Evans Gariepy, Measure theory and fine properties of functions, in which there is an entire chapter devoted to the proof), since first of all, you can always embed any manifold in $\mathbb{R}^n$ by Whitney theorem, and moreover area formula is way more general since it applies to Lipschitz surfaces.

Hovever always keep in mind that this statement comes from the fact that rectifiable sets, when zoomed enough look like planes. Thus, first of all you prove the statement for affine subspaces, and then you proceed by approximation.

Proofs given above do not quite explain why the formula should be true. Check it out in Evans' book.