How can I show that inversion is continuously connected to a reflection?

Question

From Ex 3.1 in the TASI lectures on the conformal bootstrap: http://arxiv.org/abs/1602.07982 the problem is the inversion map (with Euclidean signature) $$ I\colon x^\mu \mapsto \frac{x^\mu}{x^2} $$ and a reflection map $$ R\colon x^0\mapsto -x^0 \\ x^i\mapsto x^i $$ are ``continuously connected''. I think this means to show that the two maps are homotopic $R\sim I$ which is valid since $I$ is continuous in the 1-point compactification $\mathbb{R}^n\cup \{\infty\}$. Equivalently I chould show that $R-I \sim f_c$ where $f_c$ is some constant map. I don't think the compactified space is contractible (its an $n$-sphere, right? ) so I can't use $f_c\sim 1$.

I don't know enough topology to make an argument why such a homotopy should exist, and my first guess of an explicit homotopy by convex combination fails due to a discontinuity at the origin $$ F(x^\mu,t)=(1-t)R+tI $$

I can see intuitively how this connection might be plausible. Since the reflection creases the unit sphere about its equator. Somehow this crease may allow the interior of the sphere to swap with the exterior yielding the inversion map. But I don't know how to make this idea precise.

Answer 1

Recall that the (global) conformal group is given by

$${\rm Conf}(p,q)~\cong~O(p\!+\!1,q\!+\!1)/\{\pm {\bf 1} \},\tag{1} $$

cf. e.g. this Phys.SE post. Using the embedding $\imath: \mathbb{R}^{p,q}\hookrightarrow \overline{\mathbb{R}^{p,q}}$ into the conformal compactifification $\overline{\mathbb{R}^{p,q}}$, one may show after a short calculation that the inversion map $I:\mathbb{R}^{p,q}\to \mathbb{R}^{p,q}$, given by

$$I(x)~:=~\frac{x}{\eta^{p,q}(x,x)} ,\tag{2}$$

is represented by the $O(p\!+\!1,q\!+\!1)$ matrix

$$ I~=~{\rm diag}(-1,1,1,\ldots, 1) . \tag{3}$$

Secondly, the reflection $R:\mathbb{R}^{p,q}\to \mathbb{R}^{p,q}$ of the first coordinate, given by

$$ R(x)~:=~(-x^1,x^2,\ldots,x^{p+q}),\tag{4}$$

is represented by the $O(p\!+\!1,q\!+\!1)$ matrix

$$ R~=~{\rm diag}(1,-1,1,\ldots, 1) . \tag{5}$$

Therefore the composition $IR$ is represented by the $O(p\!+\!1,q\!+\!1)$ matrix

$$ IR~=~{\rm diag}(-1,-1,1,\ldots, 1) . \tag{6}$$

We can now reformulate OP's question as follows.

Question: When does the composition $IR$ belong to the connected component ${\rm Conf}_0(p,q)$ that contains the identity element?

Answer: One may show that this happens precisely

• if $p>0$, or

• if $p=0$ and $q$ is odd,

by using the fact that the indefinite orthogonal group $O(p\!+\!1,q\!+\!1)$ has four connected components, and methods sketched in the above mentioned Phys.SE post.

Example: In the 2D Euclidean case $\mathbb{R}^{2,0}\cong \mathbb{C}$ with $p=2$ and $q=0$, the inversion (2) is

$$ I(z)~=~\frac{1}{\bar{z}}, \tag{7}$$

the reflection (4) is minus complex conjugation

$$ R(z)~=~-\bar{z},\tag{8} $$

and the composition $IR$ is the Möbius transformation

$$IR(z) ~=~-\frac{1}{z},\tag{9}$$

which is represented by the $SL(2,\mathbb{C})$ matrix

$$ \begin{pmatrix} 0 &\pm 1 \cr \mp 1 &0 \end{pmatrix}. \tag{10}$$ The Möbius transformation (9) - (10) belongs to the connected component

$${\rm Conf}_0(2,0)~\cong~ SO^+(3,1) ~\cong~SL(2,\mathbb{C}) /\{\pm {\bf 1} \},\tag{11}$$

that contains the identity element, i.e. the restricted Lorentz group.

Answer 2

Conceptually, the idea is that planes and spheres are equivalent from the point of conformal geometry. Conformal transformations map {planes, spheres} to {planes, spheres}, and in fact do this transitively -- any object in the set {planes, spheres} can be obtained from any other by a conformal transformation.

Inversion is a reflection against a sphere, and you just have to conjugate it by a transformation which turns this sphere into a plane to get a reflection. This latter transformation can be connected to a trivial one continuously by moving the center of the sphere to infinity while keeping a point on the boundary fixed.

More precisely, suppose that $T_1$ maps the unit sphere to a plane $x_1=const$. Then $$ R_1=T_1IT_{1}^{-1} $$ is a reflection if $I$ is the standard inversion wrt the unit sphere. If we have a homotopy $T_t$, $t\in[0,1]$ between $T_1$ and $T_0=\mathrm{id}$, then we have a homotopy between $R_1$ a reflection and $R_0=I$ the standard inversion.

In order to find such a homotopy consider the special conformal transformation, which is given by $$ K(a)=IP(a)I= x_\mu\mapsto \frac{x_\mu+a_\mu x^2}{1+a^2x^2+2(a\cdot x)} $$ for $P(a)$ the translation by $a$. The idea is that the first inversion maps the unit sphere to itself, and then we can translate by $a=(1,0,0,\ldots)$ to map the point $-a$ of the unit sphere to the origin, which will then be mapped to infinity by the second inversion. A sphere with a point at infinity is a plane. It is easy to see that it will be given by the equation $x_1=1/2$ by considering where does the point $a$ of the sphere map, together with the symmetry of the construct.

We thus can take $T_t=K(ta)$, since $K(0)=\mathrm{id}$. It is somewhat messy to work out the explicity formulas, but the concept should be clear.

Answer 3

Thinking of the sphere $S^n$ as the one-point compactification of $\mathbb{R}^n$, we can consider the stereographic projection from the plane defined by $x^0 = 0$ to the unit sphere $\{x\in \mathbb{R}^n\,:\,|x| = 1\}$. This map is actually defined on $\mathbb{R}^n\cup\{\infty\}$, it takes the point $\infty$ to the north pole of the unit sphere. Moreover, these two subsets are exactly the fixed point sets of $R$ and $I$, respectively. Both maps behave as reflections over these fixed point sets, as you hinted at. Using the geometric picture of the stereographic projection (the map pulls the point in the plane along a line connecting that point to the north pole of the unit sphere until it first hits the sphere), we see we can continuously pass from the identity map to the projection map by moving continuously along these lines. You should be able to write down the relevant formulas without too much trouble. We may as well parameterize this process by $t\in [0,1]$ so that the image of the plane at $t = 0$ is the plane itself and at time $t = 1$ is the unit sphere. Denote by $S_t$ the image of this map at time $t$. Defining a family of maps of $S^n$ to be the reflection across $S_t$ at time $t$ gives a continuous path from $R$ to $I$ in the space of maps from $S^n$ to itself.

Depending on the specifics of the problem, you may need to check whether this path of maps is contained in the conformal group (i.e. for each $t\in [0,1]$ whether the reflection across $S_t$ is conformal).