= Solution
Identify a Minkowski vector $(t,x,y,z)$ with the <Hermitian matrix representation of Minkowski four-vectors>
$$
X=\begin{pmatrix}t+z&x-iy\\x+iy&t-z\end{pmatrix},\qquad
\det X=t^2-x^2-y^2-z^2.
$$
For $A\in SL(2,\mathbb C)$, the real-linear action $X\mapsto AXA^\dagger$ preserves Hermiticity and the <determinant>, so it preserves the Lorentz quadratic form. It also preserves the cone of positive-definite <Hermitian matrices>, hence the future timelike cone. The group $SL(2,\mathbb C)$ is connected, by <polar decomposition of an invertible complex matrix> into $SU(2)$ and positive determinant-one <Hermitian matrices>. Thus its image lies in the <Proper orthochronous Lorentz group>.
If $AXA^\dagger=X$ for every Hermitian $X$, taking $X=I$ makes $A$ unitary; the remaining equations then say it commutes with every <Hermitian matrix>. Their real span is all <Hermitian matrices> and their complex span is all complex <matrices>, so $A$ is scalar. The <determinant> condition leaves precisely $A=\pm I$. A discrete kernel gives an injective differential, and the real dimensions of both groups are six. The image is therefore open and equals the connected <Lorentz group>:
$$
\boxed{SO_0(3,1)\cong SL(2,\mathbb C)/\{\pm I\}.}
$$
The quotient does not include the disconnected time-reversing component of $SO(3,1)$.
Back to article page
