If $D_r = \{v\in \mathbb{R}^2 : 0 \lt |v| \lt r\}$, consider the map $f_r: D_r \to D_r$ given by:

$$f_r(x,y) = \frac{\sqrt{r^2-x^2-y^2}}{\sqrt{x^2+y^2}}\left(-y,x\right)$$

Geometrically, $f_r(v) \cdot v =0$ and $|v|^2 + |f_r(v)|^2=r^2$, i.e. $f_r$ rotates $v$ by $\frac{\pi}{2}$ and rescales it to its “Pythagorean conjugate” wrt $r$. It is easy to see that $f_r(f_r(v))=-v$.

While it is not hard to show by an explicit computation that the determinant of the Jacobian for $f_r$ is $-1$, and hence $f_r$ is area-preserving, my question is whether there is a more direct or conceptual route to that conclusion. I don't think the mere fact that $f_r(f_r(v))=-v$ suffices, since this only tells us that the product of the Jacobian at two different points, $v$ and $f_r(v)$, is minus the identity matrix.

This function doesn't satisfy the Cauchy-Riemann equations, so it does not correspond to a holomorphic function of a complex variable. But I am curious as to whether it has been given a label in the literature; it seems like a kind of “conjugate” that might have shown up in a range of contexts in geometry or analysis.

The map $f_r$ can be used to show that the volume of the unit ball in $\mathbb{R}^{2n}$ is equal to $\frac{\pi^n}{n!}$, by means of the following argument. If we pick $n$ points $p_i$ uniformly in the square in $\mathbb{R}^2$ of side length 2, and compute the probability $P_1$ that:

$$|p_n| \le |p_{n-1}| \le \dots \le |p_1| \le 1$$

we have:

$$P_1 = \frac{\pi^n}{4^n n!}$$

since all $n$ points must lie in the unit disk, and there will be equal probabilities for each of the $n!$ permutations to have descending magnitudes.

We can also pick $n$ points $q_i$ uniformly from the same square and ask whether:

$$|q_1|^2+|q_2|^2+\dots +|q_n|^2 \le 1$$

i.e. whether the $n$ points taken together give the coordinates of a point in the unit ball in $\mathbb{R}^{2n}$. The probability for this will be:

$$P_2 = \frac{V(B^{2n})}{4^n}$$

If the $q_i$ correspond to a point in the unit ball, we can construct $n$ points:

$$p'_1 = f_1(q_1)$$ $$|p'_1|^2 = 1 - |q_1|^2$$ $$p'_2 = f_{|p'_1|}(q_2)$$ $$|p'_2|^2 = 1 - |q_1|^2 - |q_2|^2$$ $$p'_3 = f_{|p'_2|}(q_3)$$ $$\dots$$

These $p'_i$ will then satisfy:

$$|p'_n| \le |p'_{n-1}| \le \dots \le |p'_1| \le 1$$

Since each $f_r$ is area-preserving, the probability of arriving at these $p'_i$ will be the same as that of picking the $q_i$ from which they were computed. But the probability of the $q_i$ giving us a point in the unit ball in $\mathbb{R}^{2n}$ must then be the same as $P_1$, since the $p'_i$ satisfy the relevant condition. So:

$$P_1=P_2$$

and so:

$$V(B^{2n}) = \frac{\pi^n}{n!}$$