Chapter 7: Unitary Groups

### The Geometry of $SU(2,2)$

Just as with the orthogonal groups, we can have unitary groups that preserve an inner product with arbitrary signature. As before, we let $g$ be the diagonal matrix with signature $(p,q)$, and define $$\SU(p,q) = \{ M\in\CC^{(p+q)\times(p+q)} : M^\dagger g M = g, \det M = 1 \}$$ If $q=0$ (or $p=0$), we recover the ordinary unitary groups. As a nontrivial example, we consider $$\SU(2,2) = \{ M\in\CC^{4\times4} : M^\dagger g M = g, \det M = 1 \}$$ where $g$ has signature $(2,2)$.

It is easy to see that $\SO(p,q)$ has precisely $pq$ boosts out of $\frac12(n-1)n$ total elements, where $p+q=n$. A little more work shows that $\SU(p,q)$ has $2pq$ boosts out of $n^2-1$ total elements, where again $p+q=n$. Are there cases where these numbers agree (not necessarily for the same value of $n$)?

Two possibilities are \begin{align} \frac12 (3-1)2 &= 3 = 2^2 -1 \\ \frac12 (6-1)6 &= 15 = 4^2 -1 \end{align} With no boosts, that is, with $q=0$ in both cases, this leads to the identifications \begin{align} \SU(2) &\cong \textrm{Spin}(3) \label{su2so3}\\ \SU(4) &\cong \textrm{Spin}(6) \label{su4so6} \end{align} where we must again replace the orthogonal groups by their double-covers, the spin groups. We have already seen (\ref{su2so3}), but (\ref{su4so6}) is new. Taking other signatures into account, we have the two further identifications \begin{align} \SU(1,1) &\cong \textrm{Spin}(2,1) \\ \SU(2,2) &\cong \textrm{Spin}(4,2) \end{align} so that, locally at least, we can identify $\SU(2,2)$ with $\SO(4,2)$, which we have already studied.

Knowing now that we can identify elements of $\SU(2,2)$ with those of $\SO(4,2)$, we write down generators of $\SU(2,2)$, using the appropriate names from the generators of $\SO(4,2)$.

We know that $\SO(4,2)$ acts on vectors with 6 components, such as $$v = \begin{pmatrix} t \\ x \\ y \\ z \\ p \\ q \end{pmatrix}$$ Can we find a 6-dimensional representation of $\SU(2,2)$? Consider the matrix $$P = \begin{pmatrix} 0 & t+iq & z+ip & x+iy \\ -t-iq & 0 & -x+iy & z-ip \\ -z-ip & x-iy & 0 & t-iq \\ -x-iy & -z+ip & -t+iq & 0 \end{pmatrix}$$ whose determinant is $$\det P = (-t^2+x^2+y^2+z^2+p^2-q^2)^2$$ which is precisely the (square of the) $\SO(4,2)$ (squared) norm of $v$. Since $$\det(MPM^T) = \det P \label{transact}$$ so long as $\det M=\pm1$, elements $M\in\SU(2,2)$ will preserve the determinant of $P$, and are therefore in $\Og(4,2)$; whether they are in $\SO(4,2)$ depends on whether they preserve orientation. Furthermore, $P$ is the most general $4\times4$ antisymmetric complex matrix, and $MPM^T$ is automatically antisymmetric if $P$ is. Thus, all we need to do to show that $\SU(2,2)\subset\SO(4,2)$ is to check orientations, and this also suffices to demonstrate the (local) equivalence between $\SU(2,2)$ and $\SO(4,2)$, since each has 15 generators.

It is sufficient to exhibit 15 generators of $\SU(2,2)$, and check that they preserve orientations. One possible choice of such generators is given by \begin{align} R_{xy} &= \begin{pmatrix} e^{-i\alpha} & 0 & 0 & 0 \\ 0 & e^{i\alpha} & 0 & 0 \\ 0 & 0 & e^{i\alpha} & 0 \\ 0 & 0 & 0 & e^{-i\alpha} \\ \end{pmatrix} \\ R_{yz} &= \begin{pmatrix} \cos\alpha & i\sin\alpha & 0 & 0 \\ i\sin\alpha & \cos\alpha & 0 & 0 \\ 0 & 0 & \cos\alpha & i\sin\alpha \\ 0 & 0 & i\sin\alpha & \cos\alpha \\ \end{pmatrix} \\ R_{zx} &= \begin{pmatrix} \cos\alpha & \sin\alpha & 0 & 0 \\ -\sin\alpha & \cos\alpha & 0 & 0 \\ 0 & 0 & \cos\alpha & -\sin\alpha \\ 0 & 0 & \sin\alpha & \cos\alpha \\ \end{pmatrix} \\ R_{px} &= \begin{pmatrix} \cos\alpha & -i\sin\alpha & 0 & 0 \\ -i\sin\alpha & \cos\alpha & 0 & 0 \\ 0 & 0 & \cos\alpha & i\sin\alpha \\ 0 & 0 & i\sin\alpha & \cos\alpha \\ \end{pmatrix} \\ R_{py} &= \begin{pmatrix} \cos\alpha & \sin\alpha & 0 & 0 \\ -\sin\alpha & \cos\alpha & 0 & 0 \\ 0 & 0 & \cos\alpha & \sin\alpha \\ 0 & 0 & -\sin\alpha & \cos\alpha \\ \end{pmatrix} \\ R_{pz} &= \begin{pmatrix} e^{i\alpha} & 0 & 0 & 0 \\ 0 & e^{-i\alpha} & 0 & 0 \\ 0 & 0 & e^{i\alpha} & 0 \\ 0 & 0 & 0 & e^{-i\alpha} \\ \end{pmatrix} \\ R_{tx} &= \begin{pmatrix} \cosh\alpha & 0 & -\sinh\alpha & 0 \\ 0 & \cosh\alpha & 0 & -\sinh\alpha \\ -\sinh\alpha & 0 & \cosh\alpha & 0 \\ 0 & -\sinh\alpha & 0 & \cosh\alpha \\ \end{pmatrix} \\ R_{ty} &= \begin{pmatrix} \cosh\alpha & 0 & i\sinh\alpha & 0 \\ 0 & \cosh\alpha & 0 & -i\sinh\alpha \\ -i\sinh\alpha & 0 & \cosh\alpha & 0 \\ 0 & i\sinh\alpha & 0 & \cosh\alpha \\ \end{pmatrix} \\ R_{tz} &= \begin{pmatrix} \cosh\alpha & 0 & 0 & \sinh\alpha \\ 0 & \cosh\alpha & -\sinh\alpha & 0 \\ 0 & -\sinh\alpha & \cosh\alpha & 0 \\ \sinh\alpha & 0 & 0 & \cosh\alpha \\ \end{pmatrix} \\ R_{tp} &= \begin{pmatrix} \cosh\alpha & 0 & 0 & -i\sinh\alpha \\ 0 & \cosh\alpha & -i\sinh\alpha & 0 \\ 0 & i\sinh\alpha & \cosh\alpha & 0 \\ i\sinh\alpha & 0 & 0 & \cosh\alpha \\ \end{pmatrix} \\ R_{qx} &= \begin{pmatrix} \cosh\alpha & 0 & -i\sinh\alpha & 0 \\ 0 & \cosh\alpha & 0 & -i\sinh\alpha \\ i\sinh\alpha & 0 & \cosh\alpha & 0 \\ 0 & i\sinh\alpha & 0 & \cosh\alpha \\ \end{pmatrix} \\ R_{qy} &= \begin{pmatrix} \cosh\alpha & 0 & -\sinh\alpha & 0 \\ 0 & \cosh\alpha & 0 & \sinh\alpha \\ -\sinh\alpha & 0 & \cosh\alpha & 0 \\ 0 & \sinh\alpha & 0 & \cosh\alpha \\ \end{pmatrix} \\ R_{qz} &= \begin{pmatrix} \cosh\alpha & 0 & 0 & i\sinh\alpha \\ 0 & \cosh\alpha & -i\sinh\alpha & 0 \\ 0 & i\sinh\alpha & \cosh\alpha & 0 \\ -i\sinh\alpha & 0 & 0 & \cosh\alpha \\ \end{pmatrix} \\ R_{qp} &= \begin{pmatrix} \cosh\alpha & 0 & 0 & \sinh\alpha \\ 0 & \cosh\alpha & \sinh\alpha & 0 \\ 0 & \sinh\alpha & \cosh\alpha & 0 \\ \sinh\alpha & 0 & 0 & \cosh\alpha \\ \end{pmatrix} \\ R_{tq} &= \begin{pmatrix} e^{-i\alpha} & 0 & 0 & 0 \\ 0 & e^{-i\alpha} & 0 & 0 \\ 0 & 0 & e^{i\alpha} & 0 \\ 0 & 0 & 0 & e^{i\alpha} \\ \end{pmatrix} \end{align} and we leave checking the orientation as an exercise. When acting via (\ref{transact}), each transformation $R_{ab}$ corresponds to a rotation or boost in the $ab$ plane through an angle (possibly hyperbolic) of $2\alpha$.