Proper orthochronous Lorentz group

The proper orthochronous Lorentz group is the identity component of the Lorentz group. Its elements have determinant one and preserve the future light cone.

Use the Lorentz spinor double cover $SL(2,\mathbb{C})\to S{O}^{+}(1,3)$. For a lift with distinct eigenvalues $\lambda ,{\lambda }^{-1}$, diagonalize and choose logarithms $z,-z$; this gives a traceless matrix logarithm. For a repeated eigenvalue, choose the sign of the lift so it is $I+N$, where $N^2=0$, and use $\log (I+N)=N$. The two lifts have the same Lorentz image. Exponentiating the differential of the covering map therefore expresses every proper orthochronous transformation as the exponential of a real element of the Lie algebra of the Lorentz group. This argument does not assert that every element of $SL(2,\mathbb{C})$ itself has a traceless logarithm.

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 $S{O}^{+}(1,3)$.

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 $[\mathrm{tr}(AB)-\mathrm{tr}A\mathrm{tr}B]/2$.