Integration by substitution in different dimensions

Question

I want to solve a specific integral, by using substitution. As it is too specific to describe my situation and probably also not of general interest, let me give a toy example.

Let $\overline{\Omega} \subseteq \mathbb{R}^3$ and $\Omega \subseteq \mathbb{R}^2$ be two domains of different dimension. Note that the intrinsic dimension of $\overline{\Omega}$ is two, although the ambient space is three dimensional. Furthermore, I know a bijective map $\varphi: \overline{\Omega} \rightarrow \Omega$. We may assume that all partial derivatives of $\varphi$ exist. I want to solve the following integral as follows:

$$\int_{x\in \Omega} f(x)dx = \int_{x\in \varphi(\overline{\Omega})} f(x)dx = \int_{y\in \overline{\Omega}} f(\varphi(y)) \ ? ? ? \ dy.$$

Where I have written the three question marks, there should be a dependence on $\varphi$. According to Wikipedia, if $\varphi$ would be a function from $\mathbb{R}^n$ to $\mathbb{R}^n$, I should take the absolute value of the determinant of the Jacobian. (https://en.wikipedia.org/wiki/Integration_by_substitution)

But $\varphi'$ is not a square matrix. So something else needs to be done. I feel there should be a general theorem that one can look up if one knows integrals better.

Question 1: What should be at the three question marks?

Question 2: Can someone give me a citeable source?

many thanks Till

Answer 1

Discussing this with a colleague, we found a possible solution and a reason it cannot work the way that I described it.

What we can do is to replace $\Omega$ by $\Omega_n = \Omega \times {0} \subseteq \mathbb{R}^{3}$. In turn, we modify $\varphi_n(y) = (\varphi(y),0)$. At last $f_n(x,0) := f(x)$. Now $\varphi_n'$ is a square matrix.

It is easy to see that

$$ \int_{x\in \Omega}f(x)dx =\int_{y\in \Omega_n}f_n(y)dy. $$

Now, we can apply integration by substitution according to Wikipedia. We get $$ \int_{y\in \Omega_n}f_n(y)dy = \int_{z\in \overline{\Omega}}f_n(\varphi_n(z)) \ |det(\varphi_n'(z))| dz. $$

While this looks great at first it reveals a bigger problem: $det(\varphi_n'(z )) = 0$ This is because the last column is $0$.

So I am in need of a new approach.

Answer 2

So the big inside seems to be that one takes $\sqrt{\det{J J^t}}$ as the stretching factor in the integral.

• $\det$ denotes the determinant.

• $J$ denotes the Jacobian of the function $\varphi$ as above.

This is called the metric tensor. Unfortunately, I have not found a reliable source for it so far. However, one can derive it with fairly elementary methods, as follows.

Let $\varphi : \Omega \rightarrow \overline{\Omega}$. We want to know by how much this mapping is "stretching". Let us look at the Jacobian $J$ of $\varphi$. $J(x)$ describes intuitively the tangent plane at each point $\varphi(x)$. Furthermore, the columns of $J(x)$ form a full-dimensional box $B$ in the tangent space $T$. The amount of stretching at $x$ corresponds to the volume of this box. If $J$ would be a square matrix, the determinant would tell us this volume, but if $\Omega$ and $\overline{\Omega}$ live in ambient spaces of different dimension, then $J$ will not be a square matrix. Naively, we compute an orthonormal basis $O$ of the tangent space $T$. Then we describe $J$ in terms of $O$. After the basis transformation, $J'$ is a square matrix and, as $O$ is orthonormal, $\det J'$ gives us the correct volume. :)

Now, the important insight, using linear algebra, is that $\sqrt{\det{J J^t}}$ gives the same volume.