The world around us appears to have three independent directions, namely East/West, North/South, and up/down. Special relativity tells us to include time; now we have a fourth “direction”, namely toward the future or past.

It is customary to label these directions with time first, so introduce coordinates $\{t,x,y,z\}$. A spacetime vector is therefore a vector with four components, such as \begin{equation} v = \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} \end{equation} which is also called a 4-vector.

The inner product on four-dimensional Minkowski space, the arena for special relativity, is given by (4) of §6.5, where now \begin{equation} g = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \end{equation} so that \begin{equation} |v|^2 = v^Tgv = -t^2 + x^2 + y^2 + z ^2 \end{equation} which is also called the squared interval between the event $(t,x,y,z)$ and the origin.

The Lorentz group in $3+1$ spacetime dimensions is given by \begin{equation} \SO(3,1) = \{ M\in\RR^{4\times4} : M^T g M = g, \det M = 1 \} \end{equation} As with $\SO(4)$, it is sufficient to give generators for $\SO(3,1)$, namely generalized rotations in the ${4\choose2}=6$ independent coordinate planes. As before, we have the rotations \begin{align} R_{yz} &= R_{yz}(\alpha) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos\alpha & -\sin\alpha \\ 0 & 0 & \sin\alpha & \cos\alpha \end{pmatrix}\\ R_{zx} &= R_{zx}(\alpha) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \sin\alpha & 0 & \cos\alpha \\ 0 & 0 & 1 & 0 \\ 0 & \cos\alpha & 0 & -\sin\alpha \end{pmatrix}\\ R_{xy} &= R_{xy}(\alpha) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos\alpha & -\sin\alpha & 0 \\ 0 & \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \end{align} corresponding to rotations in the $yz$, $zx$, and $xy$ planes, respectively. But we also have the Lorentz transformations given by \begin{align} R_{tx} &= R_{tx}(\alpha) = \begin{pmatrix} \cosh\alpha & \sinh\alpha & 0 & 0 \\ \sinh\alpha & \cosh\alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\\ R_{ty} &= R_{ty}(\alpha) = \begin{pmatrix} \cosh\alpha & 0 & \sinh\alpha & 0 \\ 0 & 1 & 0 & 0 \\ \sinh\alpha & 0 & \cosh\alpha & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\\ R_{tz} &= R_{tz}(\alpha) = \begin{pmatrix} \cosh\alpha & 0 & 0 & \sinh\alpha \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \sinh\alpha & 0 & 0 & \cosh\alpha \\ \end{pmatrix} \end{align} corresponding to hyperbolic rotations in the $tx$, $ty$, and $tz$ planes, respectively, which are also called boosts in the $x$, $y$, and $z$ directions, respectively. Any element of the Lorentz group can be obtained as a product of these six generators (with suitable parameters). 1)

As was the case in $\SO(4)$, not every element of $\SO(3,1)$ corresponds to a rotation or boost in a single plane. But there are also elements of $\SO(3,1)$ that do not correspond to rotations or boosts in any plane.

Both the rotation $R_{xy}$ and the boosts $R_{tx}$ and $R_{ty}$ leave the $z$ direction fixed; this rotation also fixes a timelike direction ($t$), whereas these boosts also fix a spacelike direction ($y$ and $x$, respectively). But consider the matrix \begin{equation} N = N(\alpha) = \begin{pmatrix} 1+\alpha^2/2 & \alpha & -\alpha^2/2 & 0 \\ \alpha & 1 & -\alpha & 0 \\ \alpha^2/2 & \alpha & 1-\alpha^2/2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \end{equation} which also fixes the $z$ direction. What else does this transformation leave fixed? Direct computation shows that $N$ fixes the null direction $y=t$; $N$ is a null rotation.