Lorentz spinor double cover ($SL(2,\mathbb{C})⟶S{O}^{+}(1,3)$)

Identify Minkowski vectors with Hermitian matrices $X=x^{0}I+x^a{\sigma }_a$. The action $X\mapsto SX{S}^{†}$ preserves $\det X=(x^0{)}^2-|\mathbf{x}{|}^2$. It gives a surjective homomorphism from $SL(2,\mathbb{C})$ to the proper orthochronous Lorentz group with kernel $\{I,-I\}$. Polar decomposition retracts the covering group onto SU(2), making it simply connected. This cover permits half-integer spin group representations that do not descend to the Lorentz group itself.

A real four-vector corresponds to a two-by-two Hermitian matrix with determinant equal to its Minkowski squared norm. The action $X\mapsto NX{N}^{†}$ for $N\in SL(2,\mathbb{C})$ preserves that determinant and the future cone. Expanding with $\frac{1}{2}\mathrm{Tr}({\bar{\sigma }}^{\mu }{\sigma }_{\nu })={\delta }^{\mu }{}_{\nu }$ yields ${\Lambda }^{\mu }{}_{\nu }(N)=\frac{1}{2}\mathrm{Tr}({\bar{\sigma }}^{\mu }N{\sigma }_{\nu }{N}^{†})$. The map has kernel $\{\pm I\}$ and image $S{O}^{+}(1,3)$.

A nonzero Hermitian rank-one positive semidefinite matrix has zero determinant and positive trace, representing a future-directed null vector. It factors as $\chi {\chi }^{†}$, with an irrelevant spinor phase. A past-directed null vector instead requires a minus sign; a positive outer product alone does not cover both time directions.

With $X_{12}=x+iy$, a spinor whose components have phases $e^{-i\varphi /2}$ and $e^{i\varphi /2}$ has $x+iy\propto e^{-i\varphi }$, so its physical azimuth is $-\varphi$. The usual positive azimuth for those phases uses $X_{12}=x-iy$. Both conventions are valid if used consistently.

The real matrix space defined by $C+C^{\ast }=\mathrm{tr}(C)I$ consists exactly of $C=aI+i\mathbf{v}\cdot \mathbf{\sigma }$, with $a$ and the three components of $\mathbf{v}$ real. The Pauli matrices have $\mathrm{tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$, giving the displayed real symmetric bilinear form of signature $(3,1)$. The real isomorphism $C\mapsto aI-\mathbf{v}\cdot \mathbf{\sigma }$ to Hermitian matrices preserves this form when the latter have the polarized negative-determinant form $[\mathrm{tr}(AB)-\mathrm{tr}A\mathrm{tr}B]/2$.