Tetrads, Lorentz connections and spin interfaces — model 2

Put $c=\cosh r$ and $s=\sinh r$. Multiplying the upper-left $2\times 2$ block gives $B(r{)}^{\mathsf{T}}\eta B(r)=\eta$, because $c^2-s^2=1$ and the off-diagonal time-space terms cancel. The determinant of that block is $1$, the remaining block is identity, and $B(r)e_0=(c,s,0,0)$ is future timelike since $c>\mathrm{\mid }s\mathrm{\mid }$. Hence $B(r)\in S{O}^{+}(1,3)$. For the Euclidean form, however, $\delta (B(r)e_0,B(r)e_0)=c^2+s^2>1=\delta (e_0,e_0)$ for $r\mathrm{\ne }0$, so $B(r)\mathrm{\notin }SO(4)$. Thus $B(r)$ is not in the image of any map with codomain $SO(4)$.

Write $A(r)=\cosh (r\mathrm{/}2)I+\sinh (r\mathrm{/}2){\sigma }_1$. Its eigenvalues are $e^{r\mathrm{/}2}$ and $e^{-r\mathrm{/}2}$, so $\det A(r)=1$. Pauli multiplication gives $AI{A}^{†}=\cosh rI+\sinh r{\sigma }_1$, $A{\sigma }_1{A}^{†}=\sinh rI+\cosh r{\sigma }_1$, and $A{\sigma }_j{A}^{†}={\sigma }_j$ for $j=2,3$; thus its Hermitian action is exactly $B(r)$. By [F4] it is the Lorentzian spin lift, consistent with the target group in [F3].

On the Cartesian tetrad $e^I=d{x}^I$, $d{e}^I=0$, so $\omega =0$ gives zero torsion and is $\eta$-skew. By [F2] it induces the unique Levi-Civita connection. The boost $B(r)$ is constant on spacetime, so $e^{\mathrm{\prime }}=B(r)e$ also has $d{e}^{\mathrm{\prime }}=0$ and its transformed connection is ${\omega }^{\mathrm{\prime }}=0$ by [F1]; the torsion remains zero. The local matrix identity from [F5] uses the same $\eta$; it does not identify this operator as one induced by a group representation. Thus the Euclidean orthogonal comparison group does not supply this Lorentzian interface.