Let $M$ be an $nm$ dimensional submanifold of $\mathbf R^n$ defined by the following set of equations: \begin{equation} f_1(\vec x)=0, \\ \vdots \\ f_m(\vec x)=0, \end{equation} (where $\vec x$ are coordinates in $\mathbf R^n$). Is it true that the volume of $M$ is \begin{equation} \int d^nx\sqrt{\det (JJ^T)}\prod_{i=1}^m\delta(f_i(\vec x)) \end{equation} where $\delta()$ is the DiracDelta and where $J$ is the rectangular matrix $J_{i\mu}\equiv \frac{\partial f_i(\vec x)}{\partial x^\mu}$ with $i=1,...,m$ and $\mu=1,...,n$?

$\begingroup$ What if we only have one $f_1 = x_1$ but $n$ is big? It looks like $\det(JJ^T) = 0$ so the whole integral is $0$. $\endgroup$– Vladimir ZolotovSep 21 at 17:34

$\begingroup$ @VladimirZolotov No $\det(J J^T)=1$ because $(J J^T)_{ij}=\sum_\mu J_{i\mu}J_{j\mu}$ and $J_{i\mu}\equiv J_{1\mu}=\delta_{1\mu}$. $\endgroup$– dennisSep 21 at 18:51

3$\begingroup$ That integral does not transform appropriately. If you scale $f$, the integral changes, but the volume of the manifold does not. $\endgroup$– Ryan BudneySep 22 at 7:44

$\begingroup$ @RyanBudney Actually I don't think that's true. If each $f_i$ is scaled by a constant $k_i$, then $\prod_i \delta(f_i)\to \prod_i \frac{\delta(f_i)}{k_i}$ and $\det(JJ^T)\to\det(JJ^T)\prod_i k_i^2$ such that all $k_i$ cancel. $\endgroup$– dennisSep 22 at 12:44

$\begingroup$ What is your definition of "volume" of the manifold $M$? I think if you are taking the induced Riemann metric from being a submanifold of Euclidean space, this integral won't agree. Your integral it something like the dual of the Thom class, so you have a normalization problem. $\endgroup$– Ryan BudneySep 22 at 15:43
1 Answer
One should distinguish between the volume of the submanifold (a number that might be infinite) and the volume form, an exterior differential form $\omega$ of degree $n{}m$ on the (presumed regular) 0 level set $M = f^{1}(0)\subset \mathbb{R}^n$ of the mapping $f:\mathbb{R}^n\to\mathbb{R}^m$.
The formula for $\omega$ is easy to write down: If $f = (f^1,\ldots,f^m)$ and we set $J^{ij} = J^{ji} = \nabla f^i\cdot\nabla f^j$, then $$ \omega(v_1,\ldots,v_{nm}) = \frac{\Omega(\nabla f^1,\ldots,\nabla f^m,v_1,\ldots,v_{nm})}{\det(J)^{1/2}}, $$ where $\Omega = \mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ is the volume form on $\mathbb{R}^n$. Regularity is equivalent to the condition that $\det(J)$ be nonvanishing on $M=f^{1}(0)$. This assumes, of course, that $M$ is given the orientation for which $\omega$ is a positive $n{}m$ form.
For example, when $n=2$ and $m=1$, one finds that $$ \omega = \frac{f_x\,\mathrm{d}yf_y\,\mathrm{d}x}{\sqrt{{f_x}^2+{f_y}^2}}\,. $$ There is, of course, no explicit formula for $\int_M\omega$ in terms of $f$, even when $f$ is a polynomial.