# Ngô Quốc Anh

## October 18, 2018

### Jacobian determinant of diffeomorphisms measures the quotient of area of small balls

Filed under: Uncategorized — Ngô Quốc Anh @ 23:50

This post concerns a widely mentioned feature of the Jacobian determinant of diffeomorphisms whose proof is not easy to find. The precise statement of the result is as follows:

Geometric meaning of the Jacobian determinant: Let $U \subset \mathbf R^n$ be open and $\phi : U \to \phi (U)$ be a diffeomorphism in $\mathbf R^n$. Fix a point $a \in U$. Then $\displaystyle |\det J_\phi (a) | = \lim_{r \searrow 0} \frac{{\rm vol}(\phi(B(a,r)))}{{\rm vol}(B(a,r))},$

where $B(a, r)$ denotes the open ball in $\mathbf R^n$ centered at $a$ with radius $r$.

As a remark and to be more exact, we require $\phi$ to be a $C^1$-diffeomorphism. Before proving the above result, it is worth noting that it is true for linear maps, whose proof is not hard. One way to realize this is to make use of the change of variable formula for multiple integrals. The proof presented here is inspired by the proof of Lemma 5.1.12 in this book.

We now proceed with the proof whose proof is divide into a few steps.

Step 1. First we use $\| \cdot \|$ to denote a norm on $\mathbf R^n$. Clearly, because $\phi$ is a $C^1$-diffeomorphism we can write $\displaystyle \phi (x) = \phi (a) + J_\phi (a) \cdot (x- a) + \vec \varepsilon (x) \| x- a\|,$

where the error vector-valued function $\vec \varepsilon$ enjoys the following properties $\displaystyle \| \vec \varepsilon (x) \| \to 0$

as $x \to a$.

Step 2. Applying the linear map represented by the matrix $J_\phi^{-1} (a)$ to get $\displaystyle J_\phi^{-1} (a) \cdot \phi (x) = J_\phi^{-1} (a) \cdot \phi (a) + x- a + \| x- a\| J_\phi^{-1} (a) (\vec \varepsilon (x) ).$

Thus $\displaystyle \big \| J_\phi^{-1} (a) \cdot \phi (x) - J_\phi^{-1} (a) \cdot \phi (a) \big\| = x- a + \| x- a\| J_\phi^{-1} (a) (\vec \varepsilon (x) ),$

which, by the triangle inequality, yields $\displaystyle \big \| J_\phi^{-1} (a) \cdot \phi (x) - J_\phi^{-1} (a) \cdot \phi (a) \big\| \leqslant \big( 1 + J_\phi^{-1} (a) (\vec \varepsilon (x) ) \big) \| x- a\| .$

Since the linear map $J_\phi^{-1} (a) x \mapsto J_\phi^{-1} (a) \cdot x$ is continuous, there is a constant $\alpha > 0$ such that $\displaystyle \| J_\phi^{-1} (a) \cdot x \| \leqslant\alpha \| x\|$

for any $x \in \mathbf R^n$. Using this, we deduce that $\displaystyle \big \| J_\phi^{-1} (a) \cdot \phi (x) - J_\phi^{-1} (a) \cdot \phi (a) \big\| \leqslant \big( 1 + \alpha \| \vec \varepsilon (x) \| \big) \| x- a\| .$

Let $t \in (0,1)$ be arbitrary but fixed. We also let $r_0 >0$ sufficiently small in such a way that $\displaystyle \alpha \| \vec \varepsilon (x) \| < t$

for any $x \in B(a, r_0)$. From this we obtain the first fundamental estimate $\displaystyle J_\phi^{-1} (a) \cdot \phi \big( B(a,r) \big)\subset B\big( J_\phi^{-1} (a) \cdot \phi (a), (1+t)r \big)$

for any $r. In the next part of the proof, we aim to find a small ball centered at $J_\phi^{-1} (a) \cdot \phi (a)$ with smaller radius but compatible with $(1+t)r$. To this purpose, we crucially exploit the homemorphism of $\phi$. The idea of such an argument comes from Mr. Tiến.

Step 3. Going back to the linear approximation for $\phi (x)$ we get $\displaystyle \big\| J_\phi^{-1} (a) \cdot \phi (x) - J_\phi^{-1} (a) \cdot \phi (a) \big\| = \big\| x- a + \| x- a\| J_\phi^{-1} (a) (\vec \varepsilon (x) ) \big\| ,$

which implies $\displaystyle \big\| J_\phi^{-1} (a) \cdot \phi (x) - J_\phi^{-1} (a) \cdot \phi (a) \big\| \geqslant \big(1-\| J_\phi^{-1} (a) (\vec \varepsilon (x) ) \|\big) \|x-a\|.$

Observe that $\displaystyle 1-\| J_\phi^{-1} (a) (\vec \varepsilon (x) ) \| \geqslant 1 - \alpha \|\vec \varepsilon (x)\| \geqslant 1-t$

so long as $x \in B(a,r)$ with $r \leqslant r_0$. Thus we have just obtained the second fundamental estimate $\displaystyle \big\| J_\phi^{-1} (a) \cdot \phi (x) - J_\phi^{-1} (a) \cdot \phi (a) \big\| \geqslant (1-t) \|x-a\|$

for all $x \in B(a,r)$ with $r \leqslant r_0$.

Step 4. Since $B(a, r_0)$ is open, $\phi$ is diffeomorphism (then homeomorphism), and $J_\phi^{-1} (a)$ is linear, the set $\displaystyle J_\phi^{-1} (a) \cdot \phi \big( B(a, r_0) \big)$

is open, which also contains the point $J_\phi^{-1} (a) \cdot \phi (a)$. We denote this set by $V$, namely, $V = J_\phi^{-1} (a) \cdot \phi \big( B(a, r_0) \big)$. Clearly, there is $\delta >0$ sufficiently small such that $\displaystyle B\big( J_\phi^{-1} (a) \cdot \phi (a), \delta \big) \subset V.$

From now on, we let $\displaystyle r < \min\left\{ r_0, \frac \delta{1-t}\right\}$

and observe that $\displaystyle B\big( J_\phi^{-1} (a) \cdot \phi (a), (1-t)r \big) \subset V.$

It is now clear to see that $\displaystyle B\big( J_\phi^{-1} (a) \cdot \phi (a), (1-t)r \big) \subset J_\phi^{-1} (a) \cdot \phi \big( B(a,r)\big).$

Indeed, let $y \in B\big( J_\phi^{-1} (a) \cdot \phi (a), (1-t)r \big)$ be arbitrary. There is some $x \in B(a, r_0)$ such that $\displaystyle J_\phi^{-1} (a) \cdot \phi (x) = y.$

Clearly, making use of the second fundamental estimate above gives $\displaystyle \| x -a \| \leqslant \frac {\big\| J_\phi^{-1} (a) \cdot \phi (x) - J_\phi^{-1} (a) \cdot \phi (a) \big\|}{1-t}=\frac {\big\| y- J_\phi^{-1} (a) \cdot \phi (a) \big\|}{1-t}

which gives the desired result.

Step 5. So far, we have already shown that $\displaystyle B\big( J_\phi^{-1} (a) \cdot \phi (a), (1-t)r \big) \subset J_\phi^{-1} (a) \cdot \phi \big( B(a,r)\big) \subset B\big( J_\phi^{-1} (a) \cdot \phi (a), (1+t)r \big).$

From this we deduce that $\displaystyle \Omega_n (1-t)^nr^n \leqslant {\rm vol} \big( J_\phi^{-1} (a) \cdot \phi \big( B(a,r)\big) \big) \leqslant \Omega_n (1+t)^nr^n$

for some dimensional constant $\Omega_n$. It is now easy to verify that $\displaystyle (1-t)^n \leqslant \frac 1{|\det J_\phi (a)|}\frac{{\rm vol} \big( \phi \big( B(a,r)\big) \big)}{{\rm vol} \big(B(a,r)\big)} \leqslant (1+t)^n.$

First sending $r$ down to zero, then sending $t$ down to 0 we obtain the desired identity.

## 1 Comment »

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