Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups

Let $(e_0,e_1,e_2,e_3)$ be an ordered $\eta$-orthonormal basis, existing by [L2], let $M$ be the matrix of $L$ in it, and write $v={\sum }_a{v}^a{e}_a$, $w={\sum }_b{w}^b{e}_b$ with coordinate columns $x=[v]$, $y=[w]$. Bilinearity of $\eta$ together with $\eta (e_a,e_b)=\mathrm{diag}(-1,1,1,1{)}_{ab}$ gives $\eta (v,w)=x^T{\eta }_{0}y$, and the coordinate column of $Lv$ is $Mx$ by [L13], so $\eta (Lv,Lw)=(Mx{)}^T{\eta }_0(My)=x^T{M}^T{\eta }_{0}My$; the defining identity of the Given is therefore equivalent to $x^T(M^T{\eta }_{0}M-{\eta }_0)y=0$ for all $x,y\in {\mathbb{R}}^4$, and testing this on the standard basis vectors of ${\mathbb{R}}^4$ shows it is equivalent to $M^T{\eta }_{0}M={\eta }_0$. Conversely, if $M$ is invertible with $M^T{\eta }_{0}M={\eta }_0$, the map $\mathrm{\Phi }\circ M\circ {\mathrm{\Phi }}^{-1}$, where $\mathrm{\Phi }(u):={\sum }_a{u}^a{e}_a$, is linear with matrix $M$ and satisfies $\eta ((\mathrm{\Phi }M{\mathrm{\Phi }}^{-1})v,(\mathrm{\Phi }M{\mathrm{\Phi }}^{-1})w)=x^T{M}^T{\eta }_{0}My=x^T{\eta }_{0}y=\eta (v,w)$, so it is a linear Lorentz map. Both directions hold for an arbitrary ordered $\eta$-orthonormal basis, so the criterion holds in one such basis exactly when it holds in every such basis and does not depend on the basis used.

If $Lv=0$, then for every $w\in V$ the defining identity gives $\eta (v,w)=\eta (Lv,Lw)=\eta (0,Lw)=0$; by nondegeneracy [L1] this forces $v=0$. Hence $\ker L=\{0\}$ and $L$ is injective.

If $L_1,L_2$ are linear Lorentz maps, then $\eta (L_1{L}_{2}v,L_1{L}_{2}w)=\eta (L_{2}v,L_{2}w)=\eta (v,w)$ for all $v,w$, so $L_1{L}_2$ is a linear Lorentz map; the identity satisfies $\eta (v,w)=\eta (v,w)$ and is a linear Lorentz map.

Evaluating the defining identity at $w=v$ gives $\eta (Lv,Lv)=\eta (v,v)$ for every $v\in V$. Hence the sign of $\eta (v,v)$ is unchanged, so a timelike $v$ has timelike image and a spacelike $v$ has spacelike image; and a null $v$ has $\eta (Lv,Lv)=0$.

Let $(o;e_0,e_1,e_2,e_3)$ be an inertial frame and let $v_0,\dots ,v_3\in V$ have coordinate columns $x_0,\dots ,x_3$ in its basis; let $M$ be the matrix of $L$ in that basis. By [L13] the matrix whose $j$-th column is the coordinate column of $L{v}_j$ is $M[x_0\dots x_3]$, so by [L6], multiplicativity [L7] and [L10], $ϵ(L{v}_0,\dots ,L{v}_3)=\det (M[x_0\dots x_3])=\det (M)\det ([x_0\dots x_3])=\det (L)ϵ(v_0,\dots ,v_3)$. Consequently the volume form is preserved when $\det L=1$, and the two sides differ by the sign $-1$ when $\det L=-1$.

Let $F,G$ be affine Lorentz maps with linear parts $L,M$, so that $F(p)=F(p_0)+L(p-p_0)$ and $G(p)=G(q_0)+M(p-q_0)$ for all $p$. Expanding gives $G(F(p))=G(F(p_0))+M(F(p)-F(p_0))=G(F(p_0))+ML(p-p_0)$, so $G\circ F$ is affine with linear part $ML$; the identity map of $A$ is affine with linear part the identity. The linear part of an affine map is unique: from $F(p_0+v)=F(p_0)+Lv$ and $F(p_0+v)=F(p_0)+L^{\mathrm{\prime }}v$ for all $v$ we get $Lv=L^{\mathrm{\prime }}v$ for all $v$, and moreover $F(p+v)-F(p)=F(p_0)+L(p+v-p_0)-F(p_0)-L(p-p_0)=Lv$ for every $p\in A$, so $Lv$ is recovered from $F$ alone.

By step 1.2, $\ker L=\{0\}$, so by [L11] $\dim V=\dim (\mathrm{im}L)$, that is $\mathrm{im}L$ is a linear subspace of $V$ with $\dim (\mathrm{im}L)=\dim V$; [L12] gives $\mathrm{im}L=V$, so $L$ is bijective. Define $L^{-1}:V\to V$ by taking $L^{-1}(u)$ to be the unique $v$ with $Lv=u$; then for $u,w\in V$ and $a\in \mathbb{R}$ we have $L(a{L}^{-1}u+L^{-1}w)=au+w$, so $L^{-1}(au+w)=a{L}^{-1}u+L^{-1}w$ and $L^{-1}$ is linear. Moreover $\eta (L^{-1}u,L^{-1}w)=\eta (L(L^{-1}u),L(L^{-1}w))=\eta (u,w)$, so $L^{-1}$ is a linear Lorentz map. Thus $L$ is a linear isomorphism by [L14].

Let $M$ be the matrix of $L$ in an ordered $\eta$-orthonormal basis; by step 1.1, $M^T{\eta }_{0}M={\eta }_0$. Taking determinants and using [L7], [L8] and [L9], $\det (M{)}^2\det ({\eta }_0)=\det (M^T)\det ({\eta }_0)\det (M)=\det ({\eta }_0)$ with $\det ({\eta }_0)=-1$, hence $\det (M{)}^2=1$; by [L10], $\det L=\det (M)\in \{1,-1\}$.

Choose an ordered $\eta$-orthonormal basis $(b_0,b_1,b_2,b_3)$ with existence from [L2], and replace $b_0$ by $-b_0$ if $b_0\mathrm{\notin }{C}^{+}$, so that $b_0\in C^{+}$; this is legitimate because $b_0$ is timelike by [L2] and $T=C^{+}\bigsqcup C^{-}$ by [L4], so every timelike vector lies in exactly one of the two components, and $(-b_0,b_1,b_2,b_3)$ is again $\eta$-orthonormal. Write $e_0:=b_0$; then by [L3] and [L4] the two components are described in this basis as $C^{+}=\{v:v^0>\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }\}$ and $C^{-}=\{v:v^0<-\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }\}$, since the coordinate set $T_{+}$ is a component containing $e_0\in C^{+}$. For $z\in T$ one has $\mathrm{\mid }{z}^0\mathrm{\mid }>\mathrm{\mid }{z}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }\ge 0$, hence $z^0\mathrm{\ne }0$, and $z\in C^{+}$ exactly when $z^0>0$; since $\eta (z,e_0)=-z^0$ and $\eta$ is bilinear with $\eta (e_0,e_0)=-1$ and $\eta (e_i,e_0)=0$, we have $z\in C^{+}$ exactly when $\eta (z,e_0)<0$. Step 1.4 gives $L(C^{+})\subseteq T$. Suppose $p,q\in C^{+}$ with $Lp\in C^{+}$ and $Lq\in C^{-}$; then $\eta (Lp,e_0)<0<\eta (Lq,e_0)$. For every $t\ge 0$ the coordinate inequality $(p+tq{)}^0=p^0+t{q}^0>\mathrm{\mid }{p}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }+t\mathrm{\mid }{q}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }\ge \mathrm{\mid }{p}_{\mathrm{v}\mathrm{e}\mathrm{c}}+t{q}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }$ shows $p+tq\in C^{+}$, so $L(p+tq)\in T$; but $t:=-\eta (Lp,e_0)\mathrm{/}\eta (Lq,e_0)$ is a positive real with $\eta (L(p+tq),e_0)=\eta (Lp,e_0)+t\eta (Lq,e_0)=0$, contradicting that every $z\in T$ has $\eta (z,e_0)\mathrm{\ne }0$. Thus $L(C^{+})$ cannot meet both $C^{+}$ and $C^{-}$, and since it is a nonempty subset of $T=C^{+}\bigsqcup C^{-}$ it is contained in exactly one of them. The same argument with $C^{-}$ in place of $C^{+}$, using $z\in C^{-}\iff z^0<0\iff \eta (z,e_0)>0$ and the reversed coordinate inequality, shows that $L(C^{-})$ is contained in exactly one of $C^{+},C^{-}$.

By step 1.4 the map $L$ carries every timelike vector to a timelike vector, so $L(T)\subseteq T$. Conversely let $w\in T$; by step 2.1 there is a unique $v\in V$ with $Lv=w$, and $\eta (v,v)=\eta (Lv,Lv)=\eta (w,w)<0$, so $v\in T$ and $w\in L(T)$. Hence $L(T)=T$.

The set $O(V,\eta )$ contains the identity and is closed under composition by step 1.3, and the inverse of every $L\in O(V,\eta )$ is a linear Lorentz map by step 2.1; as composition of maps is associative and the elements of $O(V,\eta )$ are invertible linear maps, $O(V,\eta )$ is a subgroup of $GL(V)$ in the sense of [L14]. For $SO(V,\eta )$: if $\det L=\det L^{\mathrm{\prime }}=1$ then $\det (L{L}^{\mathrm{\prime }})=1$ by [L7] and step 2.2, $\det (\mathrm{id})=1$ by [L9] and [L10], and $\det (L^{-1})\det (L)=\det (L^{-1}L)=\det (\mathrm{id})=1$ gives $\det (L^{-1})=1$, so $SO(V,\eta )$ is a subgroup of $O(V,\eta )$. For $O^{\uparrow }(V,\eta )$: if $L(C^{+})=L^{\mathrm{\prime }}(C^{+})=C^{+}$ then $(L{L}^{\mathrm{\prime }})(C^{+})=L(C^{+})=C^{+}$; the identity fixes $C^{+}$; and if $L(C^{+})=C^{+}$ then $L^{-1}(C^{+})=C^{+}$, because for $y\in C^{+}=L(C^{+})$ step 2.1 writes $y=Lx$ with $x\in C^{+}$, giving $L^{-1}y=x\in C^{+}$ and hence $L^{-1}(C^{+})\subseteq C^{+}$, while for $x\in C^{+}$ we have $Lx\in C^{+}$, so $x=L^{-1}(Lx)\in L^{-1}(C^{+})$ and $C^{+}\subseteq L^{-1}(C^{+})$. Finally $S{O}^{+}(V,\eta )=SO(V,\eta )\cap O^{\uparrow }(V,\eta )$ is an intersection of subgroups of $O(V,\eta )$, hence a subgroup.

If $v$ is null, then $v\mathrm{\ne }0$ and $\eta (v,v)=0$ by [L5], so step 1.4 gives $\eta (Lv,Lv)=0$, while $Lv\mathrm{\ne }0$ because $L$ is injective by step 2.1; hence $Lv$ is null by [L5] again, and $L$ carries the null set into itself. Conversely, if $w$ is null, then by step 2.1 there is a unique $v$ with $Lv=w$; then $\eta (v,v)=\eta (Lv,Lv)=\eta (w,w)=0$ and $v\mathrm{\ne }0$ because $Lv=w\mathrm{\ne }0$, so $v$ is null and $w=Lv$ lies in the image of the null set. Hence $L$ maps the set of null vectors onto itself.

Let $F$ be an affine Lorentz map with linear part $L$ and base event $p_0$, so that $F(p)=F(p_0)+L(p-p_0)$ for all $p$ and $L$ is a linear Lorentz map; by step 2.1 the inverse $L^{-1}$ is a linear Lorentz map, and $G(p):=p_0+L^{-1}(p-F(p_0))$ satisfies $G(F(p))=p_0+L^{-1}(F(p_0)+L(p-p_0)-F(p_0))=p_0+(p-p_0)=p$ and $F(G(p))=F(p_0)+L(p_0+L^{-1}(p-F(p_0))-p_0)=F(p_0)+(p-F(p_0))=p$ for all $p\in A$. Thus $G$ is a two-sided inverse of $F$; by step 1.6 the linear part of $G$ is recovered as $G(p+v)-G(p)=L^{-1}v$ for any $p$, so $G$ is an affine Lorentz map, and $F$ is a bijection of $A$ with inverse $G$.

By step 2.3 there are signs $\epsilon ,\delta \in \{+,-\}$ with $L(C^{+})\subseteq C^{\epsilon }$ and $L(C^{-})\subseteq C^{\delta }$. If $\epsilon =\delta$, then $L(T)=L(C^{+})\cup L(C^{-})\subseteq C^{\epsilon }$, which contradicts $L(T)=T$ from step 3.1 because $T=C^{\epsilon }\bigsqcup C^{\text{other}}$ and both components are nonempty by [L3] and [L4]. Hence $\epsilon \mathrm{\ne }\delta$. If $x\in C^{\epsilon }$, then $x\in T=L(T)=L(C^{+})\cup L(C^{-})$; were $x\in L(C^{-})\subseteq C^{\delta }$ we would have $x\in C^{\epsilon }\cap C^{\delta }=\mathrm{\varnothing }$, so $x\in L(C^{+})$. Hence $C^{\epsilon }\subseteq L(C^{+})\subseteq C^{\epsilon }$, that is $L(C^{+})=C^{\epsilon }$, and the same argument with the roles of the two components exchanged gives $L(C^{-})=C^{\delta }$. In particular exactly one of $L(C^{+})=C^{+}$ and $L(C^{+})=C^{-}$ holds, and $L(C^{+})=C^{+}$ if and only if $L(C^{+})\subseteq C^{+}$: the forward implication is trivial, and conversely if $L(C^{+})\subseteq C^{+}$ then the alternative $L(C^{+})=C^{-}$ is impossible because $C^{-}$ is nonempty and disjoint from $C^{+}$.

The composition of two affine Lorentz maps is affine Lorentz: by step 1.6 its linear part is the composition $ML$ of the linear parts, and $ML$ is a linear Lorentz map by step 1.3. The identity map of $A$ is affine with linear part the identity of $V$, which is a linear Lorentz map. By step 3.4 every affine Lorentz map $F$ possesses a two-sided inverse which is again affine Lorentz, with linear part $L^{-1}$ for the linear part $L$ of $F$. Hence the affine Lorentz maps form a group under composition, and every one of them is a bijection of $A$.

Clause (i) is steps 1.1 and 2.1, clause (ii) is steps 2.1, 2.2 and 3.2, clause (iii) is steps 1.4 and 3.3 together with the volume-form computation of step 1.5, clause (iv) is steps 2.3, 3.1 and 4.1, and clause (v) is step 4.2; this proves every clause of the statement.