Tetrads, Lorentz connections and spin interfaces

Write the columns of $A$ as $u,v\in {\mathbb{C}}^2$. Since $\det A=1$, $u$ and $v$ are linearly independent. Put $r_{11}:=\mathrm{\parallel }u\mathrm{\parallel }>0$, $q_1:=u\mathrm{/}{r}_{11}$, $\alpha :=q_1^{†}v$, $w:=v-\alpha q_1$, and $r_{22}:=\mathrm{\parallel }w\mathrm{\parallel }>0$. Then $q_1^{†}w=0$ and $q_2:=w\mathrm{/}{r}_{22}$ makes $Q:=(q_1,q_2)$ unitary; with $R:=\left(\begin{array}{cc}{r}_{11}& \alpha \\ 0& r_{22}\end{array}\right)$ we have $A=QR$. By [F1]–[F2], $1=\mathrm{\mid }\det A\mathrm{\mid }=\mathrm{\mid }\det Q\mathrm{\mid }{r}_{11}{r}_{22}=r_{11}{r}_{22}$, and $\det A=\det Q{r}_{11}{r}_{22}=1$ then gives $\det Q=1$. Therefore $Q\in SU(2)$ and $R=\left(\begin{array}{cc}r& z\\ 0& r^{-1}\end{array}\right)$ for $r>0$ and $z\in \mathbb{C}$.

If the first column of $Q$ is $(a,c{)}^{\mathsf{T}}$, its unit orthogonal vectors are $z(-\bar{c},\bar{a}{)}^{\mathsf{T}}$ with $\mathrm{\mid }z\mathrm{\mid }=1$. The determinant of the matrix with these columns is $z(\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2)=z$; $\det Q=1$ therefore forces $z=1$. Thus $Q=\left(\begin{array}{cc}a& -\bar{c}\\ c& \bar{a}\end{array}\right)$ and $\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2=1$. This identifies $SU(2)$ with the unit sphere $S^3\subset {\mathbb{R}}^4$. Under this identification use the standard Euclidean inner product $\cdot$ and norm $\parallel \cdot {\parallel }_2$ on ${\mathbb{R}}^4$. The sphere is path-connected: non-antipodal points are joined by normalizing their straight segment. For antipodal points $x,-x\in S^3$, at least one of the first two standard basis vectors $e_1,e_2$ is not parallel to $x$; choose such an $e_j$ and set $w:=\frac{e_j-(e_j\cdot x)x}{\parallel e_j-(e_j\cdot x)x{\parallel }_2}\mathrm{.}$ The denominator is nonzero because $e_j$ is not parallel to $x$; hence $w\in S^3$ and $w\cdot x=0$. The normalized straight segments from $x$ to $w$ and from $w$ to $-x$ give a path. Hence there is a continuous path $Q_t$ in $SU(2)$ from $I$ to $Q$.

For $R=\left(\begin{array}{cc}r& z\\ 0& r^{-1}\end{array}\right)$, the path $R_t:=\left(\begin{array}{cc}{r}^t& tz\\ 0& r^{-t}\end{array}\right)$ has continuous entries because $r^{\pm t}=\exp (\pm t\log r)$ for the fixed $r>0$, and the real exponential is continuous. It has determinant one for every $t\in [0,1]$, with $R_0=I$ and $R_1=R$. Thus $A_t:=Q_t{R}_t$ is a continuous path in $SL(2,\mathbb{C})$ from $I$ to $A=QR$. Since $A$ was arbitrary, $SL(2,\mathbb{C})$ is path-connected.

For $u\in {\mathbb{R}}^3$, set $\gamma :=\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}$ and $S:=I_3+\frac{u{u}^{\mathsf{T}}}{1+\gamma }$. Define $B(u):=\left(\begin{array}{cc}\gamma & u^{\mathsf{T}}\\ u& S\end{array}\right)$. The identities $\mathrm{\mid }u{\mathrm{\mid }}^2={\gamma }^2-1$ and $S^2=I_3+(\frac{2}{1+\gamma }+\frac{\mathrm{\mid }u{\mathrm{\mid }}^2}{(1+\gamma {)}^2})u{u}^{\mathsf{T}}=I_3+u{u}^{\mathsf{T}}$ give $B(u{)}^{\mathsf{T}}\eta B(u)=\eta$: the time-time entry is $-{\gamma }^2+\mathrm{\mid }u{\mathrm{\mid }}^2=-1$, the time-space block is $-\gamma u^{\mathsf{T}}+u^{\mathsf{T}}S=0$, and the spatial block is $-u{u}^{\mathsf{T}}+S^2=I_3$. Also $B(u)e_0=(\gamma ,u)$ is future timelike. For $t\in [0,1]$, the entries of $B(tu)$ are continuous: $\sqrt{1+t^2\mathrm{\mid }u{\mathrm{\mid }}^2}$ is continuous and its denominator $1+\sqrt{1+t^2\mathrm{\mid }u{\mathrm{\mid }}^2}$ is positive. Thus the path $t\mapsto B(tu)$ consists of Lorentz maps and starts at the identity, so its determinant, which is always $\pm 1$, stays $+1$. Each invertible Lorentz map is a homeomorphism of the timelike set and therefore maps connected components onto connected components. Since $B(tu)e_0$ lies in the future component, $B(tu)$ preserves that component by [F2]. Thus $B(u)\in S{O}^{+}(1,3)$.

For $\mathrm{\Lambda }\in S{O}^{+}(1,3)$, write $\mathrm{\Lambda }{e}_0=(\gamma ,u)$. Preservation of $\eta$ and the future cone gives $\gamma =\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}>0$, so the matrix in step 1.1 satisfies $B(u)e_0=\mathrm{\Lambda }{e}_0$. Then $R:=B(u{)}^{-1}\mathrm{\Lambda }$ is a proper orthochronous Lorentz map fixing $e_0$. Its first column is $e_0$, and preservation of $\eta$ makes its first row $e_0^{\mathsf{T}}$; therefore $R=\mathrm{diag}(1,R_3)$ with $R_3^{\mathsf{T}}{R}_3=I_3$ and $\det R_3=1$.

The real cubic characteristic polynomial of $R_3$ has a real root, and every complex eigenvalue $\lambda$ of an orthogonal matrix has $\mathrm{\mid }\lambda \mathrm{\mid }=1$: on the complexification, $\mathrm{\parallel }{R}_{3}v{\mathrm{\parallel }}^2=v^{†}{R}_3^{\mathsf{T}}{R}_{3}v=\mathrm{\parallel }v{\mathrm{\parallel }}^2$ for an eigenvector $v\mathrm{\ne }0$. In particular every real eigenvalue is $+1$ or $-1$. If there were no $+1$ eigenvalue, then either all three eigenvalues would be $-1$, giving determinant $-1$, or the other two would be a conjugate pair with product $1$, again giving determinant $-1$. Both contradict $\det R_3=1$. Choose a unit eigenvector $n$ for eigenvalue $+1$. Its orthogonal plane is invariant; the restriction to that plane is orthogonal with determinant $1$, hence is a two-dimensional rotation by an angle $\theta$. Choose a unit vector $m\perp n$ and then a unit vector $p\perp n,m$ so $(n,m,p)$ is positively oriented; such $m,p$ are obtained by finite orthogonalization in the two-dimensional plane. In this basis the matrix is $\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \theta & -\sin \theta \\ 0& \sin \theta & \cos \theta \end{array}\right)$. Let $C=(n,m,p)$ be the orthogonal change-of-basis matrix and denote the displayed matrix by $D(\theta )$. Then $R_3=CD(\theta )C^{\mathsf{T}}$, so $R_3(s):=CD(s\theta )C^{\mathsf{T}}$ for $s\in [0,1]$ is a continuous path in $SO(3)$ from $I_3$ to $R_3$.

Use the path $B(2tu)$ for $0\le t\le \frac{1}{2}$, followed by $B(u)\mathrm{diag}(1,R_3(2t-1))$ for $\frac{1}{2}\le t\le 1$, where $R_3(s)$ is the path from step 3.1. This is a continuous path in $S{O}^{+}(1,3)$ from $I_4$ to $\mathrm{\Lambda }=B(u)R$. Hence $S{O}^{+}(1,3)$ is path-connected.

In $O(1,3)$, the determinant takes only the values $+1,-1$, so the subset with determinant $+1$ is both open and closed. The continuous map $L\mapsto L{e}_0$ takes $O(1,3)$ into the timelike set; by [F2], the preimage of the future timelike component is also both open and closed. Their intersection is $S{O}^{+}(1,3)$ and contains $I_4$, so the connected component of $I_4$ in $O(1,3)$ lies in $S{O}^{+}(1,3)$. Step 4.1 joins every member of $S{O}^{+}(1,3)$ to $I_4$, giving the reverse inclusion.

For each point $p$, use [F1] to choose a neighborhood $U$ and a smooth future-directed tetrad $E$ there. Take any coordinate chart around $p$. Relative to its coordinate determinant frame, the smooth orientation ray [F2] has locally constant sign; shrink around $p$ to keep that sign fixed, and reverse one coordinate if the sign is negative. This gives a local oriented coordinate frame. In it, the determinant of the component matrix of $E$ is a smooth nonzero real function; shrink $U$ to a connected neighborhood of $p$, so its sign is constant. If that sign is negative, replace $E_1$ by $-E_1$. This preserves orthonormality and the future direction of $E_0$ and makes $E$ positively oriented by [F2]. Thus positively oriented future tetrads exist locally.

Let $u,v$ lie in a common fiber of $P_g^{+}(M)$. Since $u$ is an isomorphism, $A:=u^{-1}v$ is the unique real matrix with $v=u\circ A$. The isometry condition gives $A^{\mathsf{T}}\eta A=\eta$. Positivity of the two frame orientations gives $\det A>0$, while the Lorentz identity gives $\det A=\pm 1$, hence $\det A=1$. Because $u(e_0)$ and $v(e_0)$ are future timelike, $A{e}_0$ is future timelike; and the invertible Lorentz map $A$ is a homeomorphism of the timelike set. It therefore maps each connected component onto a component, and the image of $e_0$ shows that it preserves the future component. Thus $A\in S{O}^{+}(1,3)$. This proves that the right action is transitive on each fiber; uniqueness of $A$ proves freeness.

For a local positively oriented future tetrad section $E$ on $U$, define ${\mathrm{\Psi }}_E:U\times S{O}^{+}(1,3)\to (P_g^{+}(M)){\mathrm{\mid }}_U$ by ${\mathrm{\Psi }}_E(p,A)=E(p)\circ A$. Step 1.2 makes this map bijective. Its set-theoretic inverse sends a frame $u$ over $p$ to $(p,E(p{)}^{-1}\circ u)$. We construct the topology and smooth structure using these bijections, rather than assuming smoothness of functions on the total space.

If $E^{\mathrm{\prime }}$ is another such section on an overlap, write $E^{\mathrm{\prime }}(p)=E(p)\circ A(p)$. Its coefficients are $A^I{}_J(p)=e_p^I(E_J^{\mathrm{\prime }}(p))$, hence are smooth functions of the base point: both $e$ and $E^{\mathrm{\prime }}$ are smooth fields on $M$. Thus $A$ is a smooth map into the embedded matrix Lie group [F4], and step 1.2 places every value in $S{O}^{+}(1,3)$. The transition between the charts is $(p,B)\mapsto (p,A(p)B)$, which is smooth; its inverse $(p,B)\mapsto (p,A(p{)}^{-1}B)$ is smooth because inversion is smooth in the Lie group of [F4]. Declare a subset of the total space open exactly when its preimage under every ${\mathrm{\Psi }}_E$ is open. The overlap maps just computed are diffeomorphisms on open product domains, so each ${\mathrm{\Psi }}_E$ is a homeomorphism onto an open subset; combined with base and group coordinate charts, they give a compatible smooth atlas. In this atlas the right action is $(p,B,C)\mapsto (p,BC)$, hence smooth, and the projection is the product projection.

The charts of step 2.1 and smooth transitions from step 3.1 give $P_g^{+}(M)$ local products $U\times S{O}^{+}(1,3)$. Different base points separate by the Hausdorff base [F6], and distinct frames over one point separate in the Hausdorff matrix group [F4], so the total space is Hausdorff. For second countability, let $\mathcal{B}$ be a countable base of $M$; for each $B\in \mathcal{B}$ contained in at least one local-section neighborhood, use [A1] to select one pair consisting of such a neighborhood and a positively oriented future tetrad on it. The selected countable family of neighborhoods covers $M$, and countable bases of each selected $U\times S{O}^{+}(1,3)$ give a countable base of the total space by [F4,F6]. Thus the local charts define a smooth manifold structure. Freeness and fiberwise transitivity were proved in step 1.2; by [F5], this is a smooth right principal $S{O}^{+}(1,3)$-bundle. The construction supplies local sections only, so no global tetrad is asserted.

Every Hermitian matrix has a unique form $X(x)$, since its two real diagonal entries determine $x^0,x^3$ and its complex off-diagonal entry determines $x^1,x^2$. Direct evaluation gives $X(x)=\left(\begin{array}{cc}{x}^0+x^3& x^1-i{x}^2\\ x^1+i{x}^2& x^0-x^3\end{array}\right),\det X(x)=(x^0{)}^2-(x^1{)}^2-(x^2{)}^2-(x^3{)}^2\mathrm{.}$ Near its identity, the set $SL(2,\mathbb{C})$ is a smooth real six-dimensional manifold: where the upper-left entry $a$ is nonzero, writing a matrix as $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the equation $ad-bc=1$ uniquely gives $d=(1+bc)\mathrm{/}a$, so $(a,b,c)$ are six real local coordinates. Left translates of this chart cover the group and have smooth transitions because left multiplication is a smooth invertible matrix map. The group product and inverse are smooth matrix operations, so [F8] makes $SL(2,\mathbb{C})$ a Lie group.

For $A\in SL(2,\mathbb{C})$, multiplicativity and the conjugate-transpose determinant identity in [F1] give $\det (AX{A}^{†})=\det X$. Thus the action defines a real linear Lorentz map $\rho (A)$; composition satisfies $\rho (AB)=\rho (A)\rho (B)$, and the coordinate entries are quadratic polynomials in the entries of $A$, so $\rho$ is smooth. Given any $A$, choose a path $A_t$ in $SL(2,\mathbb{C})$ from $I$ to $A$ by [F2]. Then $t\mapsto \rho (A_t)$ is a path of Lorentz maps from the identity to $\rho (A)$. By [F4], its endpoint is in $S{O}^{+}(1,3)$.

A tangent matrix at $I$ to $SL(2,\mathbb{C})$ has trace zero: differentiating $ad-bc=1$ at $I$ gives $Z_{11}+Z_{22}=0$. Conversely, for any trace-zero $Z=\left(\begin{array}{cc}z& b\\ c& -z\end{array}\right)$, the curve $A(t)=\left(\begin{array}{cc}1+tz& tb\\ tc& (1+t^{2}bc)\mathrm{/}(1+tz)\end{array}\right)$ has determinant one near $0$ and derivative $Z$ at $0$. Hence the real tangent space has dimension six. Differentiating the action gives $d{\rho }_I(Z)X=ZX+X{Z}^{†}\mathrm{.}$ Since [F1] shows that the action preserves $-\det X$, differentiation places $d{\rho }_I(Z)$ in the tangent space of the Lorentz group. If it vanishes for every Hermitian $X$, then $X=I$ gives $Z^{†}=-Z$, and the remaining equations give $ZX=XZ$ for every Hermitian $X$. Commuting with $\mathrm{diag}(1,0)$ makes $Z$ diagonal; commuting with ${\sigma }_1$ makes its diagonal entries equal. Its trace is zero, so $Z=0$. Thus $d{\rho }_I$ is injective. Every tangent $K$ to $S{O}^{+}(1,3)$ at $I$ is the derivative of a smooth curve $\mathrm{\Lambda }(t)$ in that group. Differentiating $\mathrm{\Lambda }(t{)}^{\mathsf{T}}\eta \mathrm{\Lambda }(t)=\eta$ at $t=0$ gives $K^{\mathsf{T}}\eta +\eta K=0$, so by [F6] the target tangent lies in the six-dimensional eta-skew space. By [F5] its own dimension is six, hence it equals that space. The injective differential between the six-dimensional source and target tangent spaces is therefore an isomorphism.

Apply [F7] in smooth coordinate charts at the identity, using step 3.1: after shrinking a neighborhood $U$ of $I$ in $SL(2,\mathbb{C})$, $\rho$ maps $U$ diffeomorphically onto a neighborhood $V$ of the identity in $S{O}^{+}(1,3)$. Since $\rho$ is a homomorphism, its image is a subgroup containing $V$ and hence is open. Every coset of this subgroup is open; its complement is a union of open cosets, so the image is also closed. The target is connected by [F4], and the image is nonempty, so it is all of $S{O}^{+}(1,3)$.

If $A$ lies in the kernel, then $AX{A}^{†}=X$ for every Hermitian $X$. Setting $X=I$ gives $A{A}^{†}=I$, so $A$ is unitary and the same equations imply $AX=XA$. Commuting with $\mathrm{diag}(1,0)$ and ${\sigma }_1$ as in step 3.1 forces $A=zI$; $\det A=1$ gives $z^2=1$. Both $I$ and $-I$ act trivially, hence the kernel is exactly $\{I,-I\}$. Shrink $U$ from step 4.1 so that $U\cap (-U)=\mathrm{\varnothing }$. If $\rho (B)\in V$, let $C\in U$ be the unique point with $\rho (C)=\rho (B)$; then $B{C}^{-1}$ is in the kernel, so $B=C$ or $B=-C$. Therefore ${\rho }^{-1}(V)=U\bigsqcup (-U)$ and both pieces map diffeomorphically to $V$. Translating this neighborhood by any target element gives evenly covered neighborhoods everywhere, proving that $\rho$ is a two-sheeted covering.

For smooth vector fields $X,Y$ on $U$, evaluating exterior derivative and the wedge product gives $T^I(X,Y)=X(e^I(Y))-Y(e^I(X))-e^I([X,Y])+{\omega }^I{}_J(X)e^J(Y)-{\omega }^I{}_J(Y)e^J(X)$. Since $Y=e^J(Y)E_J$, the connection definition gives $e^I({\mathrm{\nabla }}_{X}Y)=X(e^I(Y))+{\omega }^I{}_J(X)e^J(Y)$; the corresponding formula with $X,Y$ exchanged shows $T^I(X,Y)=e^I({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y])$. Thus all coframe torsion forms vanish exactly when the induced affine-connection torsion vanishes.

Write $\alpha :=d\mathrm{\Lambda }{\mathrm{\Lambda }}^{-1}$ and $\bar{\omega }:=\mathrm{\Lambda }\omega {\mathrm{\Lambda }}^{-1}$, so ${\omega }^{\mathrm{\prime }}=\bar{\omega }-\alpha$. By [F3] applied componentwise and [F4], differentiating $\mathrm{\Lambda }{\mathrm{\Lambda }}^{-1}=I$ gives $d{\mathrm{\Lambda }}^{-1}=-{\mathrm{\Lambda }}^{-1}d\mathrm{\Lambda }{\mathrm{\Lambda }}^{-1}$ and then $d\alpha =d^2\mathrm{\Lambda }{\mathrm{\Lambda }}^{-1}-d\mathrm{\Lambda }\wedge d{\mathrm{\Lambda }}^{-1}=\alpha \wedge \alpha \mathrm{.}$ The same graded product rule gives $d\bar{\omega }=d\mathrm{\Lambda }\wedge \omega {\mathrm{\Lambda }}^{-1}+\mathrm{\Lambda }d\omega {\mathrm{\Lambda }}^{-1}-\mathrm{\Lambda }\omega \wedge d{\mathrm{\Lambda }}^{-1}=\alpha \wedge \bar{\omega }+\mathrm{\Lambda }(d\omega ){\mathrm{\Lambda }}^{-1}+\bar{\omega }\wedge \alpha \mathrm{.}$ Expanding $d{\omega }^{\mathrm{\prime }}+{\omega }^{\mathrm{\prime }}\wedge {\omega }^{\mathrm{\prime }}$ and substituting these identities cancels the $\alpha$ terms and leaves $\mathrm{\Lambda }(d\omega +\omega \wedge \omega ){\mathrm{\Lambda }}^{-1}$, hence $F^{\mathrm{\prime }}=\mathrm{\Lambda }F{\mathrm{\Lambda }}^{-1}$. Also [F3] gives $d{e}^{\mathrm{\prime }}=d\mathrm{\Lambda }\wedge e+\mathrm{\Lambda }de$ and ${\omega }^{\mathrm{\prime }}\wedge e^{\mathrm{\prime }}=\mathrm{\Lambda }\omega \wedge e-d\mathrm{\Lambda }\wedge e$, hence $T^{\mathrm{\prime }}=\mathrm{\Lambda }T$.

Restrict the unique Levi-Civita connection supplied by [F2] to $U$. Its connection coefficients in $E$ are smooth; metric compatibility and the constant frame metric ${\eta }_{IJ}$ give ${\omega }_{IJ}=-{\omega }_{JI}$. Its tangent torsion is zero, so step 1.1 makes every $T^I$ zero. This proves existence of an $\eta$-skew solution.

Conversely, an $\eta$-skew $\omega$ induces a metric-compatible connection. For $V=V^I{E}_I$ and $W=W^J{E}_J$, the connection terms in $g({\mathrm{\nabla }}_{X}V,W)+g(V,{\mathrm{\nabla }}_{X}W)$ are ${\omega }_{KJ}{V}^J{W}^K+{\omega }_{IJ}{V}^I{W}^J=0$ after interchanging the summed indices in the first term and using ${\omega }_{IJ}=-{\omega }_{JI}$; the remaining terms equal $X(g(V,W))$. If also $T^I=0$, step 1.1 gives zero tangent torsion. By uniqueness in [F2], this connection is the Levi-Civita connection. Hence the solution from step 2.1 is unique, and the displayed torsion equation is equivalent to torsion-freeness.

Steps 1.1–3.1 establish the torsion equivalence and unique Levi-Civita identification; step 1.2 establishes both transformation laws.