The Lorentz group preserves “lengths”, that is, it preserves the squared interval. We can ask instead what transformations preserve “angles”.

In Euclidean space, we can define angles in terms of the dot product, that is, the angle $\theta$ between two vectors $\vv$ and $\ww$ is given by \begin{equation} \cos\theta = \frac{\vv\cdot\ww}{|\vv|\,|\ww|} \end{equation} Any transformation that preserves length will preserve angles. But if we rescale $\vv$ and $\ww$, the angle between them still doesn't change.

This formula for the angle between two vectors relies on the Euclidean signature, but we can use the notion of scale-invariance to generalize the notion of “angle-preserving” transformations. A conformal transformation is one which preserves the inner product up to scale. Clearly, all orthogonal transformations are also conformal transformations. And if we multiply all vectors by an arbitrary factor, that is if we “stretch” or “dilate” our space by a given amount, then we have simply multiplied all inner products by a corresponding factor. That is, \begin{equation} (\lambda v)^T (\lambda w) = \lambda^2 (v^T w) \end{equation} The transformation $v\longmapsto\lambda v$ is called a dilation; dilations are another type of conformal transformations.

Conformal transformations should be thought of as acting on the endpoints of two vectors. If we shift every point in space by the same amount in a given direction, inner products don't change at all; translations are also conformal transformations.

There is one additional type of conformal transformation, known as a conformal translation. Let $v\in V$ be non-null, so that $|v|\ne0$. We can then define the inverse of $v$ via \begin{equation} v^{-1} = \frac{v}{|v|^2} \end{equation} since the inner product of $v$ with $v^{-1}$ is 1. Taking the inverse of a vector “inverts” it through the unit circle, and we use this to define a conformal translation by a constant vector $a$ to be the transformation \begin{equation} v \longmapsto \left( v^{-1}+\alpha \right)^{-1} = \frac{v+\alpha|v|^2}{1+2\langle v,\alpha\rangle+|\alpha|^2|v|^2} \end{equation} If you see a resemblance between this transformation and the null translations in the previous section, that's not a coincidence.

Conformal translations are clearly not linear. Nonetheless, conformal transformations can be identified with orthogonal groups. We consider here only the conformal group of Minkowski space, that is, we start with $\SO(3,1)$ and obtain $\SO(4,2)$, but a similar construction can be used in other cases.

Consider the vector \begin{equation} V = \begin{pmatrix} T \\ X \\ Y \\ Z \\ P \\ Q \end{pmatrix} \end{equation} and define \begin{equation} v = \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} = \frac{1}{P+Q} \begin{pmatrix} T \\ X \\ Y \\ Z \end{pmatrix} \end{equation} Then we claim that $\SO(4,2)$ acting as usual on $V$ induces conformal transformations acting on $v$.

We assume that $T$ and $Q$ are the timelike coordinates, that is, that \begin{equation} |V|^2 = -T^2 + X^2 + Y^2 + Z^2 + P^2 - Q^2 \end{equation} and we further assume that $V$ (but not $v$!) is null, that is, that \begin{equation} |V|^2 = 0 \end{equation} Then $\SO(3,1)\subset\SO(4,2)$ acts as usual on all of $T$, $X$, $Y$, $Z$, but leaves $P$ and $Q$ alone. But this means that $\SO(3,1)$ acts as usual on $v$, since the factor $P+Q$ goes along for the ride. Thus, Lorentz transformations on $v$ are contained in $\SO(4,2)$.

Consider now the boost $R_{PQ}\in\SO(4,2)$, which leaves $T$, $X$, $Y$, and $Z$ alone, but takes $P+Q$ to \begin{equation} (P\cosh\alpha+Q\sinh\alpha) + (Q\cosh\alpha+P\sinh\alpha) = (P+Q)\,e^\alpha \end{equation} Thus, $P+Q$ goes to a multiple of itself, and therefore so does $v$; this is the dilation on $v$.

We have accounted for seven of the ${6\choose2}=15$ elements of $\SO(4,2)$; the eight remaining transformations are best understood in terms of null rotations. Consider for example \begin{equation} T_\pm = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & \pm\alpha & \alpha \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & \mp\alpha & 0 & 0 & 1-\alpha^2/2 & \mp\alpha^2/2 \\ 0 & \alpha & 0 & 0 & \pm\alpha^2/2 & 1+\alpha^2/2 \end{pmatrix} \end{equation} Direct computation shows that $T_+$ induces a translation, and $T_-$ a conformal translation, both in the $x$ direction; similar constructions yield the remaining translations and conformal translations.

We have shown that $\SO(4,2)$, the orthogonal group acting on $V$, is also the conformal group when acting on $v$, that is, the conformal group of $3+1$-dimensional Minkowski space is precisely $\SO(4,2)$.