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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

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    $\begingroup$ While this is not a complete answer to your question, you might be interested in looking up the classical integral theorems, the most general of which is Stokes theorem. $\endgroup$ Commented Feb 26, 2019 at 10:47
  • $\begingroup$ Loosely speaking, integrating over a lower dimensional object in a higher dimensional space is not something trivial because if you were to use the classical volume element, then your lower dimensional object would have volume zero and thus your integral is automatically zero. Therefore you need to integrate using so-called surface elements corresponding to the dimension of the object you want to integrate over. This is usually done using forms, but in the case of $\mathbb{R}^3$ Stokes theorem tells you how you can translate a surface integral back to a volume integral and then compute it. $\endgroup$ Commented Apr 10, 2019 at 10:36
  • $\begingroup$ I just used 3 dimensions for illustration. My space is higher dimensional. $\endgroup$ Commented Apr 12, 2019 at 9:15

2 Answers 2

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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.

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    $\begingroup$ Notice that for a square matrix A $\det A^tA = (\det A)^2$. In your calculation you want to reach the det of the right submatrix and avoid the zero row/columns. You get rid of that extra stuff you do not want by again looking at $\phi ' ^t \phi$. Notice how this also makes it a square matrix so its determinant is defined! Therefore, you have "proved" what the non-equal dimensional Jacobians should be! $\endgroup$ Commented Nov 8, 2021 at 15:11
  • $\begingroup$ I'm very eager to know if you ever answered this question, as I have the same one. $\endgroup$ Commented Mar 20, 2023 at 1:21
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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.

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  • $\begingroup$ what is a textbook for this formula $\endgroup$ Commented Oct 14 at 14:21

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