# Image of a half space in $\mathbb{R}^n\setminus \{\vec{0}\}$ under the inversion $x \mapsto \frac{x}{|x|^2}$

Let $n \ge 2$, and consider the inversion $\Phi\colon \mathbb{R}^n \setminus \{\vec{0}\} \to \mathbb{R}^n \setminus \{\vec{0}\}$ given by $x \mapsto \frac{x}{|x|^2}$. For $a > 0$, define the half space $H_a^+ = \{(x_1, \dots, x_{n-1}, x_n) \in \mathbb{R}^n | x_n > a\}$. Prove that $\Phi(H_a^+) = B((0, \dots, 0, \frac{1}{2a}), \frac{1}{2a})$. That is, we want to show that $\Phi(H_a^+)$ equals the $n$-dimensonal ball of radius $\frac{1}{2a}$, centered at the point $(0, \dots, 0, \frac{1}{2a}) \in \mathbb{R}^n$.

This question has appeared on an old PDE qualifying exam. In the course, we covered several topics, including the wave and heat equations, harmonic functions, the Dirichlet problem, and eigenfunctions of the Laplacian. I'm wondering if there is some technique that's related to these concepts and can be used to solve this problem.

One observation I have made is that the set $\{(0, \dots,0, r) \in \mathbb{R}^n| r > a\} \subseteq H_a^+$ is mapped onto $\{(0, \dots, 0, \frac{1}{r}) \in \mathbb{R}^n| r> a\} \subseteq B((0, \dots, 0, \frac{1}{2a}), \frac{1}{2a})$. I know there is much more to be done, but this is the best I have so far! Hints or solutions are greatly appreciated.

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This is not a PDE problem; it is a (Euclidean) geometry problem which is relevant to PDE because it helps to move between boundary value problems on a half-space and on a ball.

Let $O$ be the origin $(0,\dots,0)$, and $A=(0,\dots,a^{-1})$. Consider a point $B$ on the plane $x_n=a$. Note that $|OB|=\dfrac{a}{\cos \angle BOA}$. Let $B^*$ be the image of $B$ under inversion; by definition it lies on the half-line from $O$ through $B$. From $|OB^*|=|OB|^{-1} = a^{-1}\cos\angle BOA = |OA|\,\cos\angle B^*OA$ we conclude that $B^*OA$ is a right triangle with hypotenuse $OA$. Thus, $B^*$ lies on the sphere with diameter $OA$.

So far we saw that the boundary of half-space is mapped onto the sphere. Since the inversion is a homeomorphism of $\mathbb R^n\setminus\{O\}$ onto itself, it remains to check whether the halfspace goes inside or outside the sphere. But this you already did.

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Thank you Pavel! – JZShapiro Dec 16 '12 at 14:16

About 5 years ago, I wrote a short paper on the geometry of the inversion map. I will use a diagram and the argument from the section that deals with inversion of the plane here.

$\hspace{36mm}$

Note that $\triangle PCB$ is similar to $\triangle PBD$. That means $$\frac{PC}{PB}=\frac{PB}{PD}$$ With a little bit of extapolation, this shows that the inversion map maps the plane to a sphere which passes through the point of projection.

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