The Lorentz exponential: exp(sK) is proper orthochronous, and boosts are exponentials of generators

Fix $K\in \mathfrak{g}$ and $s\in \mathbb{R}$. By [L4], $V$ is a Banach space for the frame-coordinate norm and every linear map $V\to V$, in particular $K$, is a bounded operator; hence $\mathcal{B}(V)=\mathcal{L}(V,V)$ and [L5] applies with $A=K$: the series $E(t)={\sum }_{n\ge 0}{t}^n{K}^n\mathrm{/}n!$ converges absolutely in operator norm and uniformly on compact intervals, $E(t)$ satisfies $\parallel E(t)\parallel \le e^{\mathrm{\mid }t\mathrm{\mid }\parallel K\parallel }$, $E(0)=I$, $E(t+u)=E(t)E(u)$, $t\mapsto E(t)$ is ${\mathcal{C}}^{\mathrm{\infty }}$ with $E^{\mathrm{\prime }}(t)=KE(t)=E(t)K$, and the difference quotients at $0$ converge to $K$. Writing $E(s)=\exp (sK)$ for $t=s$ this is exactly clause (i), including $K={\exp }^{\mathrm{\prime }}(0)$.

Let $e={\sum }_{i=1}^3{\widehat{e}}_i{e}_i\in S$ with $\delta (e,e)=1$. Since each $M_{0i}$ lies in the linear subspace $\mathfrak{g}$ by [L1], $N_e=-{\sum }_{i=1}^3{\widehat{e}}_i{M}_{0i}$ lies in $\mathfrak{g}$. From $(M_{0i}{)}^{\mu }{}_{\nu }={\delta }_0^{\mu }{\eta }_{i\nu }-{\delta }_i^{\mu }{\eta }_{0\nu }={\delta }_0^{\mu }{\delta }_{i\nu }+{\delta }_i^{\mu }{\delta }_{0\nu }$ (using ${\eta }_{i\nu }={\delta }_{i\nu }$ for spatial indices and ${\eta }_{0\nu }=-{\delta }_{0\nu }$) one reads $M_{0i}{e}_0=e_i$ and $M_{0i}{e}_j={\delta }_{ij}{e}_0$ for $j=1,2,3$, since the components vanish unless $\mu \in \{0,i\}$ and $\nu \in \{0,j\}$. Hence $N_e{e}_0=-{\sum }_i{\widehat{e}}_i{e}_i=-e$ and $N_e{e}_j=-{\sum }_i{\widehat{e}}_i{\delta }_{ij}{e}_0=-{\widehat{e}}_j{e}_0$, so $N_{e}e=-{\sum }_j{\widehat{e}}_j^2{e}_0=-e_0$ because $\delta (e,e)={\sum }_j{\widehat{e}}_j^2=1$. Every $v\in V$ decomposes as $v=v^0{e}_0+se+y$ with $v^0,s\in \mathbb{R}$ and $y\in S$, $\delta (e,y)=0$: indeed [L8] gives $v=v^0{e}_0+w$ with $w\in S$, and $w=\delta (e,w)e+y$ with $y:=w-\delta (e,w)e$ satisfies $\delta (e,y)=\delta (e,w)-\delta (e,w)\delta (e,e)=0$. For such $v$, linearity gives $N_{e}v=v^0(-e)+s(-e_0)+0=-v^{0}e-s{e}_0$ and $N_e^{2}v=N_e(-v^{0}e-s{e}_0)=v^0{e}_0+se$; in particular $N_{e}y=0$ and $N_e^2=P$, where $P$ is the projection onto $\mathrm{span}(e_0,e)$ along $\{y\in S:\delta (e,y)=0\}$. This last set is the $\eta$-orthogonal complement of $\mathrm{span}(e_0,e)$, because $\eta (v,v^{\mathrm{\prime }})=0$ for all $v^{\mathrm{\prime }}\in \mathrm{span}(e_0,e)$ holds exactly for $v\in e_0^{\perp }=S$ with $\delta (e,v)=0$; and in the ordered basis $(e_0,e)$ the map $N_e$ has the matrix $\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$.

Let $M(t)$ be the matrix of $\exp (tK)$ in the ordered basis $(e_0,e_1,e_2,e_3)$ and $M_K$ the matrix of $K$, so that $M_K^T{\eta }_0+{\eta }_0{M}_K=0$ by [L1]. By step 1.1, $M$ is ${\mathcal{C}}^{\mathrm{\infty }}$ with $M^{\mathrm{\prime }}(t)=M_{K}M(t)$ (the matrix of $KE(t)$ is the product of the matrices) by [L7]. Define $F(t):=M(t{)}^T{\eta }_{0}M(t)$; its entries are finite sums of products of ${\mathcal{C}}^{\mathrm{\infty }}$ functions, so $F$ is differentiable and the product rule of [L9] gives $F^{\mathrm{\prime }}(t)=M^{\mathrm{\prime }}(t{)}^T{\eta }_{0}M(t)+M(t{)}^T{\eta }_0{M}^{\mathrm{\prime }}(t)=M(t{)}^T{M}_K^T{\eta }_{0}M(t)+M(t{)}^T{\eta }_0{M}_{K}M(t)=M(t{)}^T(M_K^T{\eta }_0+{\eta }_0{M}_K)M(t)=0$. Since $F(0)=I^T{\eta }_{0}I={\eta }_0$ and each entry has vanishing derivative on $\mathbb{R}$, [L9] makes each entry constant, so $F(t)={\eta }_0$ for all $t$ and all $s$. Thus for all $u,v\in V$ with coordinate columns $x,y$ one has $\eta (\exp (sK)u,\exp (sK)v)=x^{T}M(s{)}^T{\eta }_{0}M(s)y=x^T{\eta }_{0}y=\eta (u,v)$, so $\exp (sK)$ is a linear Lorentz map.

Since $N_e^2=P$ with $P^2=P$ by step 1.2, the powers of $N_e$ are $N_e^{2m}=P$ for $m\ge 1$, $N_e^{2m+1}=N_e$ for $m\ge 0$. The exponential series of the bounded operator $r{N}_e$ converges absolutely in operator norm by [L5]; splitting its partial sums into even and odd parts, each part is a convergent series of scalars times $P$ respectively $N_e$, and $\exp (r{N}_e)=I+({\sum }_{m\ge 1}{r}^{2m}\mathrm{/}(2m)!)P+({\sum }_{m\ge 0}{r}^{2m+1}\mathrm{/}(2m+1)!)N_e=I+(\cosh (r)-1)P+\sinh (r)N_e$ with the hyperbolic functions of [L11].

By step 2.1, $\exp (sK)$ is a Lorentz map, so [L3] gives $\det \exp (sK)=\pm 1$, the determinant being that of $M(s)$ by [L7]. Applying the group law of step 1.1 at $t=u=s\mathrm{/}2$ gives $\exp (sK)=\exp ((s\mathrm{/}2)K{)}^2$, so by [L7] $M(s)=M(s\mathrm{/}2)M(s\mathrm{/}2)$ and $\det M(s)=\det (M(s\mathrm{/}2){)}^2=(\pm 1{)}^2=1$, using [L3] once more for $\det M(s\mathrm{/}2)=\pm 1$. Hence $\exp (sK)$ is proper.

By [L2] and [L7], $C^{+}$ is open in the coordinate-norm topology and $e_0\in C^{+}$ with coordinates $(1,0,0,0)$ and $\parallel e_0\parallel =1$; choose $\rho >0$ with $\{v:\parallel v-e_0\parallel <\rho \}\subseteq C^{+}$. By step 1.1 the map $t\mapsto \exp (tK)$ is continuous at $0$ with $\exp (0)=I$, so there is $\delta >0$ with $\parallel \exp (tK)-I\parallel <\rho \mathrm{/}2$ whenever $\mathrm{\mid }t\mathrm{\mid }<\delta$. For such $t$, [L6] gives $\parallel \exp (tK)e_0-e_0\parallel =\parallel (\exp (tK)-I)e_0\parallel \le \parallel \exp (tK)-I\parallel \parallel e_0\parallel <\rho \mathrm{/}2<\rho$, so $\exp (tK)e_0\in C^{+}$. By step 2.1 the map $\exp (tK)$ is a Lorentz map, so by [L3] its image of $C^{+}$ is one of the two components $C^{+},C^{-}$; since it contains $\exp (tK)e_0\in C^{+}$, it equals $C^{+}$. Hence $\exp (tK)$ is orthochronous for every $\mathrm{\mid }t\mathrm{\mid }<\delta$.

Evaluate the formula of step 2.2 on the spanning set of step 1.2: using $N_e{e}_0=-e$, $N_{e}e=-e_0$, $N_{e}y=0$ and $Py=0$ for $\delta (e,y)=0$, one gets $\exp (r{N}_e)e_0=e_0+(\cosh r-1)e_0-\sinh re=\cosh r{e}_0-\sinh re$, $\exp (r{N}_e)e=\cosh re-\sinh r{e}_0$, and $\exp (r{N}_e)y=y$. By [L11], $B(\tanh (r)e)$ acts on $(e_0,e)$ by the matrix $\left(\begin{array}{cc}\cosh r& -\sinh r\\ -\sinh r& \cosh r\end{array}\right)$ and fixes every $y\in S$ with $\delta (e,y)=0$; both maps are linear and agree on the spanning set $\{e_0,e\}\cup \{y\in S:\delta (e,y)=0\}$ of $V$, so they are equal: $\exp (r{N}_e)=B(\tanh (r)e)$.

Let $s\in \mathbb{R}$. For $s=0$ the map $\exp (0)=I$ is orthochronous. Let $s\mathrm{\ne }0$ and choose $n\ge 1$ with $1\mathrm{/}n<\delta \mathrm{/}\mathrm{\mid }s\mathrm{\mid }$ by [L10], so $t:=s\mathrm{/}n$ satisfies $\mathrm{\mid }t\mathrm{\mid }<\delta$ and $\exp (tK)$ is orthochronous by step 3.2. If $s>0$ then $s=nt$, and the group law of step 1.1 gives $\exp (sK)=\exp (tK{)}^n$, a finite product of proper orthochronous Lorentz maps, hence proper orthochronous by [L3]. If $s<0$ then $-s>0$, so $\exp (-sK)$ is proper orthochronous by the previous sentence; the group law at $t=s$, $u=-s$ gives $\exp (sK)=\exp (-sK{)}^{-1}$, and the inverse of a proper orthochronous Lorentz map is proper orthochronous by [L3]. Hence $\exp (sK)$ is orthochronous for every $s\in \mathbb{R}$.

Let $\beta \in (-1,1)$ and let $e$ be a unit spatial vector. By [L11] there is a unique $r\in \mathbb{R}$ with $\beta =\tanh (r)$; step 3.3 then gives $B(\beta e)=B(\tanh (r)e)=\exp (r{N}_e)$, and for $\beta =0$ one has $B(0)={\mathrm{id}}_V=\exp (0)$ by the definition of the boost. Writing $\mathit{\beta }=\mathrm{\mid }\mathit{\beta }\mathrm{\mid }u$ with a unit spatial vector $u$ when $\mathit{\beta }\mathrm{\ne }0$, this exhibits every passive boost as the exponential of a generator in $\mathfrak{g}$. Finally, [L5] applied to the single operator $N_e$ and the group law of step 1.1 give $\exp (r{N}_e)\exp (s{N}_e)=\exp ((r+s)N_e)$ for all $r,s\in \mathbb{R}$.

Clause (i) is step 1.1, together with the derivative statement $K={\exp }^{\mathrm{\prime }}(0)$ noted there. Clause (ii) is steps 2.1, 3.1, 3.2 and 4.1: step 2.1 makes $\exp (sK)$ a Lorentz map, step 3.1 makes it proper, steps 3.2 and 4.1 make it orthochronous; combined with the ${\mathcal{C}}^{\mathrm{\infty }}$ property of step 1.1 this is the curve statement in $S{O}^{+}(V,\eta )$, and every $K\in \mathfrak{g}$ arises as ${\exp }^{\mathrm{\prime }}(0)$ by step 1.1. Clause (iii) is steps 1.2, 2.2, 3.3 and 4.2. Nothing proved here asserts the converse direction (that every element of $S{O}^{+}(V,\eta )$ is an exponential of an element of $\mathfrak{g}$), which is exactly the disclaimer in the statement.