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The displayed matrices are the three Pauli matrices, in the order ${\sigma }_3,{\sigma }_1,{\sigma }_2$. They satisfy $\mathrm{tr}{A}_i=0$ and $\mathrm{tr}(A_i{A}_j)=2{\delta }_{ij}$. Every Hermitian matrix has a unique expression $aI+{\sum }_i{v}_i{A}_i$ with all four coefficients real. Therefore they form a basis together with $I$.

The preceding trace identities give $b(I,I)=-1$, $b(I,A_i)=0$, and $b(A_i,A_j)={\delta }_{ij}$. The Gram matrix is thusConsequently rank $=4$ and inertia signature $(3,1)$. If signature is defined as the number of positive squares minus the number of negative squares, its value is $2$. This is the Hermitian matrix representation of Minkowski four-vectors, with the negative-determinant sign convention.

Identify a Minkowski vector $(t,x,y,z)$ with the Hermitian matrix representation of Minkowski four-vectorsFor $A\in SL(2,\mathbb{C})$, the real-linear action $X\mapsto AX{A}^{†}$ 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 $AX{A}^{†}=X$ 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:The quotient does not include the disconnected time-reversing component of $SO(3,1)$.