The Lorentzian spin double cover from Hermitian matrices

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.