Spherical inversion in terms of special conformal transformation

Question

I want to consider conformal maps on suitable compactifications of $\mathbb{R}^{n}$. I know that a special conformal transformation: $$x_i\mapsto\frac{x_i-x^{2}b_i}{1-2b\cdot x+b^{2}x^{2}}$$ can be written as a composition of a spherical inversion, a translation (by $b$) and another inversion about the same circle. Since the conformal group is generated by special conformal transformations, translations, dilations and rotations, it should be possible to write the spherical inversion: $$x_i\mapsto\frac{x_i}{x^{2}}$$ as a composition of these maps. I can't, however, think of such a representation. How can it be obtained?

Answer 1

The special conformal transformations as well as the translations, dilations and rotations are all continuously connected to the identity. This means that they contain parameters such that at some particular value the trasformation becomes trivial. For example, for $b=0$ the special conformal trasformation you write is simply $x_i\mapsto x_i$. The inversion map $$x_i \mapsto \frac{x_i}{x^2},$$ on the other hand, is not connected to the identity.

We can also note that an inversion changes the orientation of the space, while the other conformal transformations preserve the orientation. In $D$ (Euclidean) dimensions the special conformal transformations, translations, dilations and rotations together form the Lie group $SO(D+1,1)$. This is what, at least in physics, is normally referred to as "the conformal group". If we also allow for inversions this group is extended to $O(D+1,1)$.

Answer 2

I am talking about Euclidian signature. It is true that you cannot obtain $$x_i \mapsto \frac{x_i}{x^2}$$ from special conformal transformations, dilatations, translations and rotations because, as Olof has noted, this map changes the orientations, whereas the listed transformations preserve it. However, what you can obtain is e.g. $$ x_i \mapsto R_{1ik}\frac{x_k}{x^2}, $$ where $R_1$ is the reflection 'along' first axis. In odd-dimensional spaces this allows you to get e.g. $$ x_i\mapsto -\frac{x_i}{x^2}, $$ which is pretty close to your goal. To show this, consider the following composition of special conf.transformation and translation: $$ x_i \mapsto \frac{x_i-b_i x^2}{1-2b\cdot x+b^2x^2}+\frac{b_i}{b^2}=\frac{x_i+\frac{1-2b\cdot x}{b^2}b_i}{b^2x^2\left(1+\frac{1-2b\cdot x}{b^2x^2}\right)}. $$ Now, add dilatation with factor $b^2$ to obtain $$ x_i\mapsto\frac{x_i+\frac{1-2b\cdot x}{b^2}b_i}{x^2\left(1+\frac{1-2b\cdot x}{b^2x^2}\right)}\xrightarrow[b\to\infty]{}\frac{x_i-2x^{1}_i}{x^2}=R_{1ik}\frac{x_k}{x^2}, $$ where the limit $b\to\infty$ means that $b$ goes to infinity along the 1st axis, and $x^1_i$ is the vector projection of $x_i$ onto the 1st axis.

You can have a look at this video. It is for $D=2$, and the above mentioned limit is demonstrated at 2:00. You can notice that result differs from inversion by a reflection.

Answer 3

Edit. I just realized that I didn't answer your question; I'll leave this up in case it helps answer your real question...

Let use denote the operations of inversion and translation by $$ I(x) = \frac{x}{x^2}, \qquad T_b(x) = x+b $$ Then notice what happens when we perform the following composite transformation: \begin{align} I\circ T_{-b}\circ I (x) &=I(T_{-b}(I(x))) = I(T_{-b}(x/x^2)) = I(x/x^2 - b) \\ &= \left(\frac{x}{x^2}-b\right)^{-2}\left(\frac{x}{x^2}-b\right) \\ &= \frac{x^4}{x^2-2b\cdot xx^2+b^2x^4}\frac{x-bx^2}{x^2} \\ &= \frac{x-x^2b}{1-2b\cdot x+b^2x^2} \end{align} But this is precisely the special conformal transformation we wanted. So we see that a special conformal transformation can be written as an inversion followed by a translation, followed by an inversion.