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How does one compute the following integral

$$\int _Vf(\pmb{r})\delta (g(\pmb{r}))d\pmb{r}$$

if $g: \mathbb{R}^n \to \mathbb{R}^m$. In this question and the Wikipedia page on the Dirac delta function an answer is given for $m=1$ as

$$\int _Vf(\pmb{r})\delta (g(\pmb{r}))d\pmb{r}=\int _{g^{-1}(0)}\frac{f(\pmb{r})}{|\nabla g(\pmb{r})|}d\sigma\ .$$

So my question concerns the case wheere $m>1$. Is there also some good reference for citing the corresponding result?

Igor
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  • I'm not aware of any definition of the Dirac Delta which doesn't have the real numbers as it's domain. Under what circumstance are you encountering this? – Spencer Jan 03 '18 at 17:29
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    @Spencer. On $U \subseteq \mathbb{R}^n$ the Dirac distribution is still defined by $\langle \delta(\mathbf{r}), \phi(\mathbf{r}) \rangle = \phi(\mathbf{0})$ for $\phi \in C_c^\infty(U).$ – md2perpe Jan 03 '18 at 17:33
  • Oh jeez I posted that before I had any coffee this morning. Of course that definition exists, thank you. – Spencer Jan 03 '18 at 18:08
  • A guess after having having a quick look on it is that $\nabla g(\pmb{r})$ should just be replaced with the Jacobian. – md2perpe Jan 03 '18 at 18:50
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    I think it should involve the determinant of the Jacobian. The determinant is however only defined for square matrices. In case of m<n, my guess would be that $\nabla g(\pmb{r})$ would be replaced by $\sqrt{ \det(J_g(\pmb{r},)J_g(\pmb{r})^\top) }$. For now this is however not more than just a guess. – Igor Jan 03 '18 at 20:04

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Pretend for now that $\delta$ is a bounded measurable function. Consider the measure $\mu$ with $d\mu/dr=f$ (here dr denotes the corresponding Lebesgue measure). By the change of variables formula, $$\int \delta\circ gd\mu=\int\delta d(g_*\mu)$$ where $(g_*\mu)(U)=\mu(g^{-1}U)$. By definition of the delta function, the right hand side is equal to ${\frac {d(g_*\mu)}{dr}}(0)$, i.e. the Radon-Nikodym derivative evaluated at the origin.

Using the Smooth Coarea formula, it's easy to check that

https://en.wikipedia.org/wiki/Smooth_coarea_formula

$${\frac {d(g_*\mu)}{dr}}(y)=\int_{g^{-1}y} {\frac {d\mu/dr}{\sqrt{det(dg(dg)^T)}}}dN$$ where $dN$ is the induced Riemannian measure on the preimage.

You also probably want g to be a submersion (i.e. differential is always surjective).

Simon Segert
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    The Wikipedia formulation applies only to cases where $m\leq n$ (there, the roles of $m$ and $n$ are opposite as in my question). In the meantime I found a -in some sense- more general statement of the coarea formula (without calling it smooth there). It is discussed in Krantz & Parks, Geometric Measure Theory, Theorem 5.3.9. Therefore, I consider my question as answered. Thanks! – Igor Jan 04 '18 at 00:43
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    No problem! If I'm not mistaken, the theorem you cite also seems to have a similar restriction (namely assuming that the dimension of the domain exceeds that of the target). – Simon Segert Jan 04 '18 at 02:20
  • You're right. It does. How would one address the general case? – Igor Jan 04 '18 at 05:43