# Dirac delta uder coordinate transformation

I am having some issues understanding Dirac's delta function/distribution behaviour under change of coordinates. There is a statement, if $(x_1,\ldots,x_n)$ are cartesian coordinates and $y_1,\ldots,y_n$ are some general coordinates in Euclidean $\mathbb{E}^n$ space, then: $$x_1=x_1(y_1,\ldots,y_n),\\ \vdots \\ x_n=x_n(y_1,\ldots,y_n)$$ and the Jacobian: $$J(y_1,\ldots,y_n) = \begin{vmatrix} \frac{\partial x_1}{\partial y_1} & \cdots & \frac{\partial x_1}{\partial y_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial x_n}{\partial y_1} & \cdots & \frac{\partial x_n}{\partial y_n} \end{vmatrix}$$

define the transformation. Provided that $J(y_1^{(P)},\ldots,y_n^{(P)})\ne0$ for $(y_1^{(P)},\ldots,y_n^{(P)})$ coordinates of some point $P$ coresponding to cartesian coordinates $(x_1^{(P)},\ldots,x_n^{(P)})$, then:

$$\delta(x_1-x_1^{(P)})\cdots\delta(x_n-x_n^{(P)}) = \frac{1}{|J(y_1^{(P)},\ldots,y_n^{(P)})|} \delta(y_1-x_1^{(P)})\cdots\delta(x_n-y_n^{(P)})$$.

I wonder how to show this fact? If I consider $\delta$ in the original idea by Dirac, then:

$$f(x_1^{(P)},\ldots,x_n^{(P)})=\int_{\Omega}\delta(x_1-x_1^{(P)})\cdots\delta(x_n-x_n^{(P)}) f(x_1,\ldots,x_n)\,{\rm d}x_1\ldots{\rm d}x_n=\\= \int\ldots\int\frac{\delta(y_1-y_1^{(P)})\cdots\delta(y_n-y_n^{(P)})}{|J(y_1^{(P)},\ldots,y_n^{(P)})|} f(y_1,\ldots,y_n) |J(y_1^{(P)},\ldots,y_n^{(P)})|\,{\rm d}y_1\ldots{\rm d}y_n=\\ = \begin{cases} f(y_1^{(P)},\ldots,y_n^{(P)}) & \quad \text{all integration bounds contain the point P}\\ 0 & \quad \text{Any one of the integration bounds does not contain the point P} \end{cases}$$

And almost all seems fine but I am not sure if such proof is sufficient for this interpretation.

When I consider $\delta$ as a distribution, I do not really understand how to show that this holds not even whether it is true. I remember that for a distribution, one can define a multiplication with a $C^\infty$ function, but I still cannot see how to show that the above holds, since in terms of distributions $\delta$ is defined only as a continuous linear functional that does $<\delta,\phi>=\phi(x)$ to any $C^\infty$ test function with compact support in some set (there is really no integration involved).

• Hint: The space of distributions comes equipped with a topology in which you can approximate the $\delta$ distribution by integration against smooth functions. See what happens if you do that. – Chris Janjigian Aug 31 '18 at 15:11
• I suggest you take $n=1$ first, and see what happens. All the ideas of the general proof are already present in that simpler case, but you will not have to deal with cluttered expressions. – Antoine Aug 31 '18 at 15:29

Suppose that we have some diffeomorphism/parameterization $x = p(y)$. If $u$ is a continuous function and $\phi$ is a test function, then we have $$\left< u \circ p, \phi \right> = \int u(p(y)) \, \phi(y) \, dy = \int u(x) \, \phi(p^{-1}(x)) \, \left|\det\frac{\partial y}{\partial x}\right| \, dx = \left< \left|\det\frac{\partial y}{\partial x}\right| u, \phi \circ p^{-1} \right> ,$$ where $\frac{\partial y}{\partial x}$ is the Jabobian matrix. This result can a bit sloppy be written $$u(y) = \left|\det\frac{\partial y}{\partial x}\right| u(x).$$
We then define $u \circ p$ for a distribution $u$ by the above result.
Note that $$\left| \det\frac{\partial y}{\partial x} \right| = \left| \det \left( \left( \frac{\partial x}{\partial y} \right)^{-1} \right) \right| = \left| \left( \det\frac{\partial x}{\partial y} \right)^{-1} \right| = \left| \det\frac{\partial x}{\partial y} \right|^{-1} = \left| \det J \right|^{-1} ,$$ where $J$ is as in your post.