Existence and uniqueness of adapted coordinates, and the affine structure of simultaneous slices

For every $v\in V$ the vector $v-\tau (v)u$ lies in $K$, because $\tau (v-\tau (v)u)=\tau (v)-\tau (v)\tau (u)=\tau (v)-\tau (v)=0$ by linearity [L2]. Hence $v=\tau (v)u+k$ with $k:=v-\tau (v)u\in K$, so $V=\mathbb{R}u+K$.

Suppose $v=tu+k=t^{\mathrm{\prime }}u+k^{\mathrm{\prime }}$ with $t,t^{\mathrm{\prime }}\in \mathbb{R}$ and $k,k^{\mathrm{\prime }}\in K$. Applying $\tau$ gives $t=t^{\mathrm{\prime }}$ since $\tau (u)=1$ and $\tau (k)=\tau (k^{\mathrm{\prime }})=0$; then $k=v-tu=k^{\mathrm{\prime }}$. Hence every $v$ has exactly one decomposition $v=tu+k$, and $t=\tau (v)$.

Fix $t\in \mathbb{R}$. The event $o+tu$ lies in $A_t$, because $\tau ((o+tu)-o)=t\tau (u)=t$, so $A_t$ is nonempty. For $k\in K$ and $p\in A_t$ one has $\tau ((p+k)-o)=\tau (p-o)+\tau (k)=t$, so $p+k\in A_t$; the action of $K$ on $A_t$ so defined is associative, and it is free and transitive because displacements in $A_t$ are displacements in $A$ and lie in $K$: if $p,q\in A_t$ then $\tau (q-p)=\tau (q-o)-\tau (p-o)=0$, so $q-p\in K$ and $p+(q-p)=q$ with the displacement unique in $V$ and hence in $K$ [L1, L2]. Thus $A_t$ is an affine space modelled on $K$.

Let $p\in A$ and put $v:=p-o$, the unique displacement vector of [L1]. By steps 1.1 and 2.1 there are unique $t\in \mathbb{R}$ and $k\in K$ with $v=tu+k$; by [L3] there is a unique $x\in {\mathbb{R}}^3$ with $k={\sum }_i{x}_i{e}_i$. Hence $(t,x)$ is the unique pair with $p=o+tu+{\sum }_i{x}_i{e}_i$, and the map $\iota$ is well defined. The assignment $(t,x)\mapsto o+tu+{\sum }_i{x}_i{e}_i$ is a left and right inverse of $\iota$ by the defining property of the action [L1]; hence $\iota$ is a bijection onto $\mathbb{R}\times {\mathbb{R}}^3$. This proves clause 1.

For $p,q\in A_t$ the coordinate expressions $p-o=t_{p}u+{\sum }_i{x}_i(p)e_i$ and $q-o=t_{q}u+{\sum }_i{x}_i(q)e_i$ have $t_p=t_q=t$ by definition of $A_t$, so subtracting the two displays and using [L1] gives $p-q={\sum }_i(x_i(p)-x_i(q))e_i$. By [L3], $g(p-q,p-q)={\sum }_i(x_i(p)-x_i(q){)}^2$, so $d_g(p,q)=\mathrm{\parallel }x(p)-x(q){\mathrm{\parallel }}_2$. This proves clause 2.

Choose $o\in A$, possible because $A$ is nonempty, and choose $u\in V$ with $\tau (u)=1$, possible because $\tau$ is surjective. By [L4], choose an ordered $g$-orthonormal basis $(f_1,f_2,f_3)$ of $K$. If it belongs to the fixed orientation class, put $(e_1,e_2,e_3)=(f_1,f_2,f_3)$; otherwise put $(e_1,e_2,e_3)=(f_2,f_1,f_3)$. The interchange preserves the basis property and all identities $g(e_i,e_j)={\delta }_{ij}$, and its determinant is $-1$ by [L5], so it changes the orientation class. Since there are exactly two classes, the resulting basis is positively oriented. Thus $(o,u,(e_1,e_2,e_3))$ is an adapted chart, proving clause 3. These are finitely many existential choices, not a selection from an arbitrary family.