How to motivate the surface element

Question

$\newcommand{\RR}{\mathbb{R}} \newcommand{\dd}{\mathrm{d}}$ In teaching multivariable integration on sub-manifolds in $\RR^n$, i.e. integrals over $k$-dimensional surfaces $M\subset \RR^n$ you define the integral of $f$ over $M$ as follows: If $\phi:T\to M$ is a paremetrization (i.e. $T\subset\RR^k$ is open, $D\phi$ has full rank), then the integral is $$ \int_M f(x)\dd S(x) = \int_T f(\phi(t))\sqrt{\det G(t)}\dd t $$ where $G$ is the matrix with entries $$ G_{ij}(t) = \langle\partial_i\phi(t),\partial_j\phi(t)\rangle. $$ Once it is clear that $\sqrt{\det G(t)}$ is the $k$-dimensional volume of parallelotope spanned by the tangential vectors $\partial_j\phi(t)$ it seem clear why this should be the surface element.

But my question is:

Is there an intuitive way to see that the Gramian $\det G(t)$ is the squared volume of said parallelotope?

Answer 1

Here is an argument that has convinced (or at least silenced :-)) students in office hours. Making it into a rigorous proof would require more time talking about what volume MEANS than I do.

First, suppose I have $k$ vectors $\vec{v}_1$, $\vec{v}_2$, ..., $\vec{v}_k$ in $\mathbb{R}^k$. Let $A$ be the $k \times k$ matrix with columns $\vec{v}_1$, ..., $\vec{v}_k$. Then the volume of the parallelotope is $$|\det(\vec{v}_1, \ldots, \vec{v}_k)| = |\det A| = \sqrt{\det(A^T A)} = \sqrt{\det(\vec{v}_i \cdot \vec{v}_j)}.$$

The nifty thing is that each entry $\vec{v}_i \cdot \vec{v}_j$ of the $k \times k$ matrix is invariant under rotation of coordinates. So we have expressed the rotationally invariant volume in terms of much simpler rotationally invariant quantities.

If the vector $\vec{v}_1$, ..., $\vec{v}_k$ are now in $\mathbb{R}^n$ for some $n>k$, it is still true that the volume is $\sqrt{\det(\vec{v}_i \cdot \vec{v}_j)}$, since we can just choose orthogonal coordinates in the $k$-plane spanned by the $\vec{v}$'s and ignore the larger ambient space. (On one occasion, I drew two vectors on a notepad, shaded the parallogram and tilted the notepad at various angles, to point out that the area and the dot products don't care how I hold the notepad in space.) And we can still write this as $\sqrt{\det(A^T A)}$.

Answer 2

$\newcommand{\RR}{\mathbb{R}}\renewcommand{\span}{\mathrm{span}} \newcommand{\vol}{\mathrm{vol}}$

Here is my own answer:

Let's assume that you know that the volume of an $n$-dimensional parallelotope $A$ spanned by $n$ vectors $a_1,\dots,a_n\in\RR^n$ is $\vol_n(A) = \det(a_1,\dots,a_n)$.

To get the $k$-dimensional volume $\vol_k(A)$ of the parallelotope $A$ spanned by $a_1,\dots,a_k\in\RR^n$ (still in $\RR^n$!) do the following: Do $QR$-factorization of $A=[a_1,\dots,a_k]$ to get $$ A = QR $$ with $Q\in\RR^{n\times n}$ and $R\in\RR^{n\times k}$ upper triangular. This is nothing else than turning (and mirroring) the parallelotope around by $Q^T$ such that $$ \begin{array}{cl} r_1 = Q^T a_1&\in\span(e_1)\\ r_2 = Q^Ta_2&\in\span(e_1,e_2)\\ \vdots & \vdots\\ r_k=Q^Ta_k&\in \span(e_1,\dots,e_k). \end{array} $$ Since $Q^T$ is an isometry, we have that the $k$-dimensional volume of the parallelotope spanned by $a_1,\dots,a_k$ is the same as the one spanned by the columns of $R$. These lie in $\RR^k$ anyway, and hence, this volume is the determinant of the upper $k\times k$ part of $R$, i.e. the determinant of the diagonal entries of $R$. In other words: We have $$ R = \left[\begin{matrix}R_1\\ 0\end{matrix}\right]. $$ and the volume is $\vol_k(A) = \det(R_1)$.

Finally, we see that $$ \det(A^TA) = \det(R^TQ^TQR) = \det(R^T I R) = \det(R_1^T R_1) = \det(R_1)^2 = \vol_k(A)^2. $$

Answer 3

My proof is basically the same as yours, but perhaps this will be a bit more intuitive.

We want to find the volume of a $k$-parallelepiped spanned by $v_1,v_2,...,v_k$ in $\mathbb{R}^k$. We are armed only with the determinant, which lets us find the (signed) volume of $m$-parallelepipeds in $\mathbb{R}^m$.

The crucial observation is that, for any good definition of volumes, we should have that the "$m$ -volume" of $v_1,v_2,...,v_m \in \mathbb{R}^n$ should be the same as the "$m+1$-volume" of $v_1,v_2,...,v_m,w \in \mathbb{R}^n$ if $w$ is perpendicular to all the $v_j$, and is of length $1$. In other words "Volume equals base times height".

So the volume of $v_1,v_2,...,v_k \in \mathbb{R}^n$ is the same as the volume of $v_1,v_2,...,v_k,w_1,w_2,...,w_{n-k}$, where each $w_i$ is orthogonal to all the other vectors in the list, and is of norm $1$.

So we could just say that the volume is $Det(v_1,v_2,...,v_k,w_1,w_2,...,w_{n-k})$. This is a recipe for how to find the volume, but it suffers some flaws as a definition and as a computational tool of $k$-volumes. As a definition, it is difficult because we would need show that the result is independent of the choice of $w_i$, and because the sign is meaningless (it depends on the $w_k$). As a computational tool, one actually needs to carry out the laborious process of calculating $w_i$ which work. It seems we should be able to find a formula which does not involve actually calculating these auxiliary vectors.

The insight is to realize that the determinant of the transpose is the same as the determinant of the original matrix. So letting $M = (v_1 v_2 ... v_k w_1 ... w_{n-k})$ and $A = (v_1 v_2 ... v_k)$

$\textrm{Vol}^2 = Det(M^\top M)$

But when we actually compute the matrix $M^\top M$, we see that it consists of $A^\top A$ and then ones down rest of the diagonal. So its determinant is just $Det(A^\top A)$.

Thus $\textrm{Vol}^2 = Det(A^\top A)$.