Area-preserving map of punctured disk to itself

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:

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

• Let's forget about rotation. Consider circular ring, bounded by circumferences of radii $r_1$ and $r_2$. Part of it of radial angle $\alpha$ has area $\alpha \cdot(r_2^{2} - r_1^{2})$.After the mapping, it still has radial angle $\alpha$, but radii are now $\sqrt{1-r_1^{2}}$ and $\sqrt{1-r_2^{2}}$, so the area stays the same. But using such "blocks" we can approximate any adequate subset. – Rybin Dmitry Aug 13 at 10:24
• Added to the above comment.1) the area-preserving change of variables $x=\sqrt 2 r^{1/2} \cos \theta,y=\sqrt 2 r^{1/2} \sin \theta$ transforms this to a question about rectangles which is very transparent. 2) in a certain sense your formula presents (almost) the only such transformation which leaves the polar coordinate system invariant. The precise statement (which is relevant in cartography) is too delicate to state here but for the plane it means that up to dilations of the coordinate axes there is only one such which leaves the cartesian ones invariant. – user131781 Aug 13 at 14:33