# Volume of submanifold as integral of delta-function

Let $$M$$ be an $$n-m$$ dimensional sub-manifold of $$\mathbf R^n$$ defined by the following set of equations: $$$$f_1(\vec x)=0, \\ \vdots \\ f_m(\vec x)=0,$$$$ (where $$\vec x$$ are coordinates in $$\mathbf R^n$$). Is it true that the volume of $$M$$ is $$$$\int d^nx\sqrt{\det (JJ^T)}\prod_{i=1}^m\delta(f_i(\vec x))$$$$ where $$\delta()$$ is the Dirac-Delta and where $$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$$?

• 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 $0$. Sep 21 at 17:34
• @VladimirZolotov No $\det(J J^T)=1$ because $(J J^T)_{ij}=\sum_\mu J_{i\mu}J_{j\mu}$ and $J_{i\mu}\equiv J_{1\mu}=\delta_{1\mu}$. Sep 21 at 18:51
• That integral does not transform appropriately. If you scale $f$, the integral changes, but the volume of the manifold does not. Sep 22 at 7:44
• @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. Sep 22 at 12:44
• 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. Sep 22 at 15:43

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 $$\omega(v_1,\ldots,v_{n-m}) = \frac{\Omega(\nabla f^1,\ldots,\nabla f^m,v_1,\ldots,v_{n-m})}{\det(J)^{1/2}},$$ 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 $$\omega = \frac{f_x\,\mathrm{d}y-f_y\,\mathrm{d}x}{\sqrt{{f_x}^2+{f_y}^2}}\,.$$ There is, of course, no explicit formula for $$\int_M\omega$$ in terms of $$f$$, even when $$f$$ is a polynomial.