Connectedness of the proper orthochronous Lorentz group

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.