Conformal Groups

Section 8.7 Conformal Groups

Consider the vector

\begin{equation} V = \begin{pmatrix} T \\ X \\ Y \\ Z \\ P \\ Q \end{pmatrix}\tag{8.7.1} \end{equation}

and assume the norm is given by

\begin{equation} |V|^2 = -T^2 + X^2 + Y^2 + Z^2 + P^2 - Q^2\tag{8.7.2} \end{equation}

so that \(V\in\RR^6\) with signature \((4,2)\text{.}\) Assume further that \(V\) is null, that is, that \(|V|=0\text{,}\) and set

\begin{equation} v = \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} = \frac{1}{P+Q} \begin{pmatrix} T \\ X \\ Y \\ Z \end{pmatrix} .\tag{8.7.3} \end{equation}

How does \(\SO(4,2)\) act on \(v\text{?}\)

The subgroup \(\SO(3,1)\subset\SO(4,2)\) acts as usual on \(T\text{,}\) \(X\text{,}\) \(Y\text{,}\) \(Z\text{,}\) but leaves \(P\) and \(Q\) invariant; thus, \(\SO(3,1)\) also acts as usual on \(v\text{.}\) The boost in the \((P,Q)\)-plane takes \(P+Q\) to

\begin{equation} (P\cosh\alpha+Q\sinh\alpha) + (Q\cosh\alpha+P\sinh\alpha) = (P+Q)\,e^\alpha\tag{8.7.4} \end{equation}

and thus takes \(v\) to a multiple of itself; this transformation is called a dilation. But what about the remaining 8 elements of \(\SO(4,2)\text{,}\) which mix up \((T,X,Y,Z)\) with \((P,Q)\text{?}\)

Consider for example the rotation \(R_X\) in the \((X,P)\)-plane, and the boost \(B_X\) in the \((X,Q)\)-plane. We have

\begin{equation} \begin{aligned} R_X: (X,P) \amp\longmapsto (X\cos\alpha+P\sin\alpha,P\cos\alpha-X\sin\alpha) , \\ B_X: (X,Q) \amp\longmapsto (X\cosh\alpha+Q\sinh\alpha,Q\cosh\alpha-X\sinh\alpha)\end{aligned}\tag{8.7.5} \end{equation}

from which the corresponding Lie algebra elements \(r_X\text{,}\) \(b_X\) are easily seen to be

\begin{equation} \begin{aligned} r_x: (X,P,Q) \amp\longmapsto (P,-X,0) , \\ b_x: (X,P,Q) \amp\longmapsto (Q,0,X) ,\end{aligned}\tag{8.7.6} \end{equation}

or in matrix form

\begin{equation} r_x = \begin{pmatrix} 0 \amp 1 \amp 0 \\ -1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ \end{pmatrix} , \qquad b_x = \begin{pmatrix} 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \\ \end{pmatrix} .\tag{8.7.7} \end{equation}

If we do both transformations at once, we obtain the null rotations

\begin{equation} r_x \pm b_x = \begin{pmatrix} 0 \amp 1 \amp 1 \\ -1 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \\ \end{pmatrix}\tag{8.7.8} \end{equation}

which have the interesting property that their cube is zero. Thus, the corresponding group element is easy to obtain using a power series; we have

\begin{equation} \exp\bigl( (r_x \pm b_x) \alpha\bigr) = \begin{pmatrix} 1 \amp \alpha \amp \pm\alpha \\ -\alpha \amp 1-\frac12\alpha^2 \amp \mp\frac12\alpha^2 \\ \pm\alpha \amp \pm\frac12\alpha^2 \amp 1+\frac12\alpha^2 \\ \end{pmatrix}\tag{8.7.9} \end{equation}

or equivalently

\begin{equation} \begin{aligned} X \amp\longmapsto X + (P\pm Q)\,\alpha , \\ P \amp\longmapsto P - X\,\alpha - \frac12(P\pm Q)\,\alpha^2 , \\ Q \amp\longmapsto Q \pm X\,\alpha \pm \frac12(P\pm Q)\,\alpha^2 ,\end{aligned}\tag{8.7.10} \end{equation}

which can be combined to yield

\begin{equation} \begin{aligned} P \pm Q \amp\longmapsto P\pm Q , \\ P \mp Q \amp\longmapsto P\mp Q - 2X\,\alpha - (P\pm Q)\,\alpha^2 .\end{aligned}\tag{8.7.11} \end{equation}

These null rotations thus leave one of the null directions \(P\pm Q\) invariant; hence the name.

So what does the null rotation generated by \(r_x+b_x\) do to \(v\text{?}\) We have

\begin{equation} x = \frac{X}{P+Q} \longmapsto \frac{X+(P+Q)\,\alpha}{P+Q} = x + \alpha ,\tag{8.7.12} \end{equation}

with \(y\text{,}\) \(z\text{,}\) \(t\) held fixed; this is a translation in the \(x\)-direction. Translations in the \(y\text{,}\) \(z\text{,}\) and \(t\) directions can be constructed similarly.

What about \(r_x-b_x\text{?}\) Now we have

\begin{equation} x = \frac{X}{P+Q} \longmapsto \frac{X+(P-Q)\,\alpha}{P+Q-2X\,\alpha-(P-Q)\,\alpha^2} = \frac{x+s\,\alpha}{1-2x\,\alpha-s\,\alpha^2}\tag{8.7.13} \end{equation}

where

\begin{equation} s = \frac{P-Q}{P+Q} = \frac{P^2-Q^2}{(P+Q)^2} .\tag{8.7.14} \end{equation}

Since we are assuming \(|V|=0\text{,}\) we can replace \(P^2-Q^2\) by \(T^2-X^2-Y^2-Z^2\text{,}\) so that

\begin{equation} s = -|v|^2\tag{8.7.15} \end{equation}

and therefore

\begin{equation} x \longmapsto \frac{x-|v|^2\,\alpha}{1-2x\,\alpha+|v|^2\,\alpha^2} .\tag{8.7.16} \end{equation}

A conformal translation of \(v\) along \(a\) is defined by

\begin{equation} v \longmapsto \left( v^{-1}+a \right)^{-1} = \frac{v+a|v|^2}{1+2\langle v,a\rangle+|a|^2|v|^2}\tag{8.7.17} \end{equation}

where

\begin{equation} v^{-1} = \frac{v}{|v|^2}\tag{8.7.18} \end{equation}

for \(|v|\ne0\text{.}\) ¹ Comparing (8.7.17) with (8.7.16) shows that the latter is just the conformal translation of \(v\) in the (negative) \(x\)-direction. Conformal translations in the \(y\text{,}\) \(z\text{,}\) and \(t\) directions can be constructed similarly.

In summary, there is a nonlinear action of \(\SO(4,2)\) on vectors \(v\) in \(3+1\)-dimensional Minkowski space, which is associated with transformations that preserve the inner product up to scale. Such transformations are known as conformal transformations, and \(\SO(4,2)\) is referred to as the conformal group of \(3+1\)-dimensional Minkowski space. More generally, we refer to \(\SO(4,2)\) as the conformalization of \(\SO(3,1)\text{;}\) the same construction can be applied to any orthogonal group. Conformalization adds two new degrees of freedom to the representation, thus adding an internal symmetry \(\SO(1,1)\) (the dilation), together with one translation and one conformal translation (the two sets of null rotations) for each existing degree of freedom.

Although inversion is not defined for \(|v|=0\text{,}\) conformal translation (8.7.17) is still well-defined for null vectors (and the zero vector). Conformal translations take null vectors to multiples of themselves, and preserve the (spatial) angles between them.