Geometric meaning of geometric inversion $f(x)=\frac{x}{|x|^2}$.

Question

First the geometric inversion map $f:\mathbb{R}^n\setminus \{0\}\rightarrow \mathbb{R}^n$ is defined by $$f(x)=\frac{x}{|x|^2}=:X.$$ The one of its properties is following:

For any $c\in\mathbb{R}^n, c\not=0$, the sphere $|x-c|=|c|$ is mapped to the hyperplane $2X\cdot c=1$.

I can't understand what it means intuitively. As I know, $2X\cdot c=1$ means the set of $X$ with length of $1/2$ when projection maps $X$ to $c$. However, I failed to find some connection with sphere $|x-c|=|c|$. Can I get any help?

Answer 1

See properties of a circle inversion on a plane – it projects each circle passing through the projection's centre to a line.

In your notation, $c$ is a center of a given sphere. Then $d=2c$ is an origin's antipode on that sphere.
The image of $d$ in the projection is $f(d) = \frac d{|d|^2}.$

Let $a$ be any point of the sphere (except $O$). The triangle $\triangle Oad$ is right, because it's inscribed in a circle and its side $Od$ is a diameter of the circle. Then the angle at $O$ has a cosine $K = \frac{|a|}{|d|} = \frac{|a|}{|2c|}.$

The image of $a$ is $f(a) = \frac a{|a|^2}$ and its modulus is $1/|a|.$

The scalar product of $f(a)$ and $f(d)$ is then $$|f(a)|\cdot|f(d)|\cdot K = \frac 1{|a|}\cdot \frac 1{|d|} \cdot \frac{|a|}{|d|} = \frac 1{|d|^2} = |f(d)|\cdot |f(d)| = \mathrm{const.}$$ equal a scalar product of $f(d)$ with itself.

Which implies $f(a)$ lies on a hyperplane passing through $f(d)$ such that $f(d)$ is an orthogonal projection of $O$ on that plane.

Added

The same can be shown geometrically, with no vectors and their scalar products.

First, note $f$ essentially just scales its argument, so for each $q\ne O$ points $O, q, f(q)$ are collinear.
Second, $|f(q)| = 1/|q|.$

Let $e=f(d)$ and $b=f(a).$

Then triangles $\triangle aOd$ and $\triangle eOb$ are similar: from the collinearity they share the angle at $O$, and from the inversion of lengths they share the sides' ratio: $\text {length}(Ob) : \text {length}(Oe) =\text {length}(Od) : \text {length}(Oa)$.

As a result, the angle at $e$ equals the respective angle at $a$, hence it's right. This implies the set of all $b$, that is an image of the sphere in $f$, is a (hyper)plane normal to the $Ocde$ line at $e$.

Answer 2

Let $$\mathbb{S}^{n - 1} = \{ x \in \mathbb{R}^n : \lVert x \rVert = 1 \}$$ be the unit sphere in $\mathbb{R}^n$. Now, identify points on the punctured space $\mathbb{R}^n \setminus \{ 0 \}$ with pairs $(d, r)$ where $d \in \mathbb{S}^{n - 1}$ and $r$ is a positive number, by taking $(d, r)$ to the scalar product $r d$.

In this way of looking at the punctured plane, your map $f$ has the form $$f(d, r) = f(d, 1 / r).$$ I would say that it "wants to swap the origin with infinity." There's also a topological way of justifying this "swapping", which maybe you can imagine.

Answer 3

No one seems to have said this, so I'll say it. There is a very nice geometric interpretation when $|c|>1/2$. Spheres passing through the origin (let's call such a sphere $S^{n-1}_c$) get sent to hyperplanes (say $H^{n-1}_c$). Moreover (here's the bit no one's said), the unit sphere about the origin, $\mathbb S^{n-1}$, is fixed pointwise. Now, the intersection $S^{n-1}_c\cap\mathbb S^{n-1}$ is a sphere one dimension down $S^{n-2}_c$ on the surface of $\mathbb S^{n-1}$. In fact, $S^{n-1}_c$ is the only sphere passing through the origin and intersecting $\mathbb S^{n-1}$ this way. Since $S^{n-1}_c$ is fixed, $H^{n-1}_c$ must be the unique hyperplane containing $S^{n-2}_c$. This approach can be worked out when $|c|=1/2$, but then $H^{n-1}_c$ becomes the tangent hyperplane to $\mathbb S^{n-1}$ at the single point of intersection of $S^{n-1}_c$ and $\mathbb S^{n-1}$.

This is very easy to visualise in 2D (at least, I think it is)!