Does anyone know how to prove (or have a reference for) the fact that the inversion $$I : \Bbb R^n \setminus \{0\} \rightarrow \Bbb R^n \setminus \{0\}\\ x \mapsto \frac x {\|x\|^2}$$ sends circles to circles or straight lines ?

Since $I = L^{-1} \circ I \circ L$ on $\Bbb R^n \setminus \{0\}$ for any linear invertible isometry $L : \Bbb R^n \rightarrow \Bbb R^n$, we can assume that the circle $C$ we consider is contained in an affine $2$-plane, say $\Bbb R^2 \times \{h\}$ for some $h \in \Bbb R^{n-2}$.

If $h = 0$, the proof is straightforward. However, I don't know how to prove the case $h \neq 0$.

But maybe this is not the way to go to prove this statement.

I saw here https://mathoverflow.net/questions/25251/properties-of-the-n-dimensional-stereographic-projection someone saying something about circles being the intersection of several $(n-1)$-spheres (because it's quite easy to prove that $(n-1)$-spheres are sent to $(n-1)$-spheres or hyperplanes), but I don't really want to use this claim.

  • $\begingroup$ As far as I remember in the book Foundations of Hyperbolic Manifolds by John Ratcliffe your claim is proven in the context of Moebius transformations. I might be wrong though, since it has been a couple of years since I read this book. $\endgroup$ – humanStampedist Jun 7 '18 at 11:20
  • $\begingroup$ Why do you say "intersection of (n-1)-spheres"? It is enough to prove your fact for n=3, since a circle and the origin are contained in a 3-dimensional vector subspace, and the image will also lie there. And for n=3 a circle is intersection of two spheres and spheres map to spheres or planes. I consider this proof to be the most natural one, certainly better than the heap of computations suggested below. $\endgroup$ – liaombro May 5 at 5:30

The inverse of a vector $x$ is defined by

$$\frac1x = \frac{x}{x^2} = \Big(\frac1{x\cdot x}\Big)x$$

If the circle's centre is $c$, its radius is $\lVert a\rVert=\lVert b\rVert$, and its plane is spanned by vectors $a$ and $b$ (with $a\cdot b=0$), then it can be parametrized by

$$x = c + a\cos\theta + b\sin\theta$$

Without loss of generality, $a$ and $b$ can be rotated until $b$ is orthogonal to $c$. So now we have

$$a^2 = b^2,\qquad a\cdot b = b\cdot c = 0$$


$$\frac1x = \frac{c+a\cos\theta+b\sin\theta}{c^2+2c\cdot a\cos\theta+a^2}$$

This equation doesn't look like a circle, but of course it must be. The new circle's centre $c'$ should be the midpoint of the two extremal points at $\theta=0,\;\theta=\pi$ :


$$c' = \frac12\Big(\frac1{c+a} + \frac1{c-a}\Big) = \overset{\text{algebra}}{\cdots} = \frac{(c^2+a^2)c-2(c\cdot a)a}{(c+a)^2(c-a)^2}$$

And the new (directed) radius $a'$ should be half the displacement between these points:

$$a' = \frac12\Big(\frac1{c+a} - \frac1{c-a}\Big) = \cdots = \frac{(c^2+a^2)a-2(a\cdot c)c}{(c+a)^2(c-a)^2}$$

and $b' = \lVert a'\rVert\frac{b}{\lVert b\rVert}$ .

(These are undefined if $c+a=0$ or $c-a=0$, but then the result is a straight line, and the problem is 2-dimensional, which is easily solved.)

If these guesses are correct, then the new point must be in this plane, and have constant radius:

$$\Big(\frac1x - c'\Big)\wedge a'\wedge b' \overset{?}{=} 0,\qquad \Big(\frac1x - c'\Big)^2 \overset{?}{=} a'^2$$

(The wedge product $\wedge$ is a measure of linear independence; any vector $a$ has $a\wedge a=0$.)

To verify these equations, first note that the denominator is

$$(c+a)^2(c-a)^2 = (c^2+a^2+2c\cdot a)(c^2+a^2-2c\cdot a) = (c^2+a^2)^2-4(c\cdot a)^2$$


$$c'^2 = \bigg(\frac{(c^2+a^2)c-2(c\cdot a)a}{(c+a)^2(c-a)^2}\bigg)^2 = \frac{(c^2+a^2)^2c^2-4(c^2+a^2)(c\cdot a)^2+4(c\cdot a)^2a^2}{(c+a)^4(c-a)^4}$$

$$= \frac{\big((c^2+a^2)^2-4(c\cdot a)^2\big)c^2}{(c+a)^4(c-a)^4} = \frac{c^2}{(c+a)^2(c-a)^2}$$

$$a'^2 = \cdots = \frac{a^2}{(c+a)^2(c-a)^2}$$


$$c'\wedge a' = \bigg(\frac{(c^2+a^2)c-2(c\cdot a)a}{(c+a)^2(c-a)^2}\bigg)\wedge\bigg(\frac{(c^2+a^2)a-2(a\cdot c)c}{(c+a)^2(c-a)^2}\bigg)$$

$$= \frac{(c^2+a^2)^2(c\wedge a)-2(c^2+a^2)(c\cdot a)(c\wedge c)-2(c^2+a^2)(c\cdot a)(a\wedge a)+4(c\cdot a)^2(a\wedge c)}{(c+a)^4(c-a)^4}$$

$$= \frac{\big((c^2+a^2)^2-4(c\cdot a)^2\big)(c\wedge a)}{(c+a)^4(c-a)^4} = \frac{c\wedge a}{(c+a)^2(c-a)^2}$$

Now we can calculate the two expressions:

$$\Big(\frac1x - c'\Big)\wedge a' = \frac{c+a\cos\theta+b\sin\theta}{c^2+2c\cdot a\cos\theta+a^2}\wedge\frac{(c^2+a^2)a-2(a\cdot c)c}{(c+a)^2(c-a)^2}-c'\wedge a'$$

$$= \frac{(c^2+a^2)(c\wedge a)-0+0-2(c\cdot a\cos\theta)(a\wedge c)}{(c^2+2c\cdot a\cos\theta+a^2)(c+a)^2(c-a)^2}+\frac{b\sin\theta}{x^2}\wedge a'-c'\wedge a'$$

$$= \frac{(c^2+2c\cdot a\cos\theta+a^2)(c\wedge a)}{(c^2+2c\cdot a\cos\theta+a^2)(c+a)^2(c-a)^2}+\frac{b\sin\theta}{x^2}\wedge a'-c'\wedge a'$$

$$= \frac{c\wedge a}{(c+a)^2(c-a)^2}-c'\wedge a'+\frac{b\sin\theta}{x^2}\wedge a'$$

$$= 0+\frac{b\sin\theta}{x^2}\wedge a'$$

$$\Big(\frac1x - c'\Big)\wedge a'\wedge b' = \frac{b\sin\theta}{x^2}\wedge a'\wedge\frac{b\lVert a'\rVert}{\lVert b\rVert} = 0$$


$$\Big(\frac1x - c'\Big)^2 = \frac1{x^2} - 2\frac{x\cdot c'}{x^2} + c'^2$$

$$= \frac1{x^2}-\frac2{x^2}\frac{(c+a\cos\theta+b\sin\theta)\cdot\big((c^2+a^2)c-2(c\cdot a)a\big)}{(c+a)^2(c-a)^2}+c'^2$$

$$= \frac{(c+a)^2(c-a)^2}{x^2(c+a)^2(c-a)^2}-2\frac{(c^2+a^2)c^2-2(c\cdot a)^2+(c^2+a^2)(c\cdot a\cos\theta)-2a^2(c\cdot a\cos\theta)+0-0}{x^2(c+a)^2(c-a)^2}+c'^2$$

$$= \frac{(c^2+a^2)^2-4(c\cdot a)^2}{x^2(c+a)^2(c-a)^2}-2\frac{(c^2+a^2)c^2-2(c\cdot a)^2+(c^2-a^2)(c\cdot a\cos\theta)}{x^2(c+a)^2(c-a)^2}+c'^2$$

$$= \frac{(c^2+a^2)(c^2+a^2-2c^2)-2(c^2-a^2)(c\cdot a\cos\theta)}{x^2(c+a)^2(c-a)^2}+c'^2$$

$$= \frac{(c^2+a^2+2c\cdot a\cos\theta)(a^2-c^2)}{(c^2+2c\cdot a\cos\theta+a^2)(c+a)^2(c-a)^2}+c'^2$$

$$= \frac{a^2-c^2}{(c+a)^2(c-a)^2}+\frac{c^2}{(c+a)^2(c-a)^2}$$

$$= \frac{a^2}{(c+a)^2(c-a)^2}$$



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.