Coxeter Polyhedral Gluings and Intrinsic Metrics — Examples

(The cell $H$ and its face poset.) The six points $v_0,\dots ,v_5$ lie in $H$: for $v_1=(\frac{1}{2},\frac{\sqrt{3}}{2})$ one has $\mathrm{\mid }{x}_2\mathrm{\mid }=\frac{\sqrt{3}}{2}$ and $\sqrt{3}{x}_1+x_2=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\sqrt{3}$ with $\sqrt{3}{x}_1-x_2=0$, and the remaining five cases follow by the same substitution, the values used being $(\sqrt{3}{)}^2=3$ and $\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{3}=\sqrt{3}$ [F7]. The six equalities among the defining inequalities are the three pairs of parallel lines $x_2=\pm \frac{\sqrt{3}}{2}$, $\sqrt{3}{x}_1+x_2=\pm \sqrt{3}$, $\sqrt{3}{x}_1-x_2=\pm \sqrt{3}$; each of the twelve pairs of equalities from different pairs determines a unique point, namely $v_1$ for $x_2=\frac{\sqrt{3}}{2}$ with $\sqrt{3}{x}_1+x_2=\sqrt{3}$, $v_2$ for $x_2=\frac{\sqrt{3}}{2}$ with $-\sqrt{3}{x}_1+x_2=\sqrt{3}$, $v_0$ for $\sqrt{3}{x}_1+x_2=\sqrt{3}$ with $\sqrt{3}{x}_1-x_2=\sqrt{3}$, and symmetrically $v_3,v_4,v_5$ for the three remaining admissible cases, while in the other six cases one of the remaining inequalities fails (for instance $x_2=\frac{\sqrt{3}}{2}$ with $\sqrt{3}{x}_1+x_2=-\sqrt{3}$ forces $x_1=-\frac{3}{2}$, and then $\sqrt{3}{x}_1-x_2=-2\sqrt{3}<-\sqrt{3}$ violates the inequality). Intersections of three or more equalities are contained in one of these, so by [F1] the nonempty faces of $H$ are exactly the cell $H$ itself, the six sides $[v_0,v_1],[v_1,v_2],[v_2,v_3],[v_3,v_4],[v_4,v_5],[v_5,v_0],$ and the six vertices $v_0,\dots ,v_5$, and $H$ is $2$-dimensional because it contains the non-collinear points $v_0,v_1,v_3$. Consecutive vertices satisfy $\parallel v_k-v_{k+1}{\parallel }_2^2=1$ in each of the six cases (for $(v_0,v_1)$ and $(v_4,v_5)$ it is $\frac{1}{4}+\frac{3}{4}=1$, for $(v_1,v_2)$ it is $1$, and for $(v_2,v_3)$, $(v_3,v_4)$ and $(v_5,v_0)$ it is $\frac{1}{4}+\frac{3}{4}=1$ $($the first coordinate step $\frac{1}{2}$ and the second $\frac{\sqrt{3}}{2}$, with $(\frac{\sqrt{3}}{2}{)}^2=\frac{3}{4}))$, so the sides have length $1$; also $\parallel v_k{\parallel }_2^2=1$ for all six $k$. The centre $o:=(0,0)$ is the barycentre $\frac{1}{6}(v_0+\cdots +v_5)$ of $H$: the first coordinates $1,\frac{1}{2},-\frac{1}{2},-1,-\frac{1}{2},\frac{1}{2}$ sum to $0$ and so do the second coordinates $0,\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},0,-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$, and the midpoint of the side with endpoints $u,w$ is $m=\frac{1}{2}(u+w)$.

(The gluing, (H1)-(H3) and clauses (i), (ii).) The shape $P$ is the full face poset of $H$, including the empty face; its cells are $C_p=p$ for every nonempty face, with identity inclusions as face isometries. Each principal down-set is the face poset of its face, meets are intersections of faces, and the cocycle condition holds for inclusions. The quotient identifies each face copy with its subset in $H$, so the intersection condition holds and the map ${\iota }_H\text{\hspace{0.17em}⁣}:H\to X$ is a bijection; a subset of $X$ is open exactly when its trace on ${\iota }_H(H)=X$ is relatively open, so ${\iota }_H$ is a homeomorphism. $X$ is path-connected: for $x,y\in H$ and $t\in [0,1]$ the point $(1-t)x+ty$ lies in $H$ because the six inequalities are affine, $(1-t)\mathrm{\ell }(x)+t\mathrm{\ell }(y)\le \max \{\mathrm{\ell }(x),\mathrm{\ell }(y)\}\le c$ for each defining inequality $\mathrm{\ell }\le c$ [F5]; the segment path $t\mapsto (1-t)x+ty$ is $\parallel x-y{\parallel }_2$-Lipschitz for $d_2$, since $\parallel ((1-t)x+ty)-((1-s)x+sy){\parallel }_2=\mathrm{\mid }t-s\mathrm{\mid }\parallel x-y{\parallel }_2$ [F4], hence continuous [F5]; composing with ${\iota }_H$ gives a path in $X$, so $X$ is path-connected and therefore connected [F5]. There are thirteen nonempty faces and three shapes (points, unit intervals and $H$), so (H2) local finiteness and (H3) finite shapes hold, and $D=\dim H=2$; by [F3] the chain metric $d$ is a metric inducing the weak topology and $(X,d)$ is proper and complete. Clause (i): for $x,y\in H$ the one-step chain shows $d(x,y)\le \parallel x-y{\parallel }_2$ [F3], while every chain has steps inside $H$ and length the sum of Euclidean distances of its steps, which is at least $\parallel x-y{\parallel }_2$ by the triangle inequality for $d_2$ [F4]; taking the infimum, $d(x,y)=\parallel x-y{\parallel }_2$ for all $x,y\in H$. Clause (ii): hence $d$ is the Euclidean metric, the straight segment $[x,y]\subseteq H$ (which lies in $H$ by convexity, verified in the first sentence) has the unit-speed parametrization $\gamma (t)=x+(t\mathrm{/}R)(y-x)$ on $[0,R]$ when $R=\parallel x-y{\parallel }_2>0$, satisfying $d(\gamma (s),\gamma (t))=\mathrm{\mid }s-t\mathrm{\mid }$; for $x=y$ use the constant map on $[0,0]$. Thus it is a geodesic segment from $x$ to $y$ (Geodesics and geodesic metric spaces), and $(H,d)$ is complete by [F3].

(Clause (iii): the barycentric triangulation.) By [F2] the order complex $K$ of the face poset of $H$ gives a compatible barycentric triangulation of $H$ whose maximal simplices are the maximal chains $v<e<H$, twelve in number, one for every incident vertex-edge pair; the associated triangle has vertices $b_v=v$, $b_e=m$ (the midpoint of the side $e$) and $b_H=o$, where $m$ and $o$ are as in step 1.1. Let $e=[u,w]$ be a side, $v$ one of its endpoints and $m=\frac{1}{2}(u+w)$ its midpoint. Then $\parallel u{\parallel }_2=\parallel w{\parallel }_2=1$ and $\parallel u-w{\parallel }_2=1$ by step 1.1, so $⟨u,w⟩=\frac{1}{2}(\parallel u{\parallel }_2^2+\parallel w{\parallel }_2^2-\parallel u-w{\parallel }_2^2)=\frac{1}{2}$ by [F9], and with $v-m=\pm \frac{1}{2}(u-w)$ and $o-m=-\frac{1}{2}(u+w)$ one gets $⟨v-m,o-m⟩=\mp \frac{1}{4}(\parallel u{\parallel }_2^2-\parallel w{\parallel }_2^2)=0$: the triangle has a right angle at $m$, its legs are $\parallel v-m{\parallel }_2=\frac{1}{2}\parallel u-w{\parallel }_2=\frac{1}{2}$ and $\parallel o-m{\parallel }_2=\frac{1}{2}\parallel u+w{\parallel }_2=\frac{1}{2}\sqrt{2+2⟨u,w⟩}=\frac{\sqrt{3}}{2}$, and its hypotenuse is $\parallel v-o{\parallel }_2=\parallel v{\parallel }_2=1$, consistently with $\frac{1}{4}+\frac{3}{4}=1$ [F7]. Since $m$ lies on the line $vm$ and, for every $p$ on that line, $o-p=(o-m)+(m-p)$ is an orthogonal decomposition [F9], $m$ is the point of the line closest to $o$ and the perpendicular height over the base $vm$ is $\parallel o-m{\parallel }_2=\frac{\sqrt{3}}{2}$, so the triangle content is $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}$ by [F6]. The three barycentric-coordinate slopes of this triangle are the reciprocals of the distances from $v$, $m$, $o$ to the opposite sidelines; by the base-height form of [F6] these distances are $\frac{2\mathrm{cont}}{\parallel m-o{\parallel }_2}=\frac{1}{2}$, $\frac{2\mathrm{cont}}{\parallel v-o{\parallel }_2}=\frac{\sqrt{3}}{4}$ and $\frac{2\mathrm{cont}}{\parallel v-m{\parallel }_2}=\frac{\sqrt{3}}{2}$ respectively, so the slopes are $2$, $\frac{4}{\sqrt{3}}$ and $\frac{2}{\sqrt{3}}$. Every one of the twelve triangles arises this way, from a side and one of its endpoints, so all twelve have these three slopes; the lower-dimensional simplices of $K$ are the chains $v<H$, $e<H$, $v<e$ with barycentres $v,o$; $m,o$ and $v,m$ and slopes $1$, $\frac{2}{\sqrt{3}}$ and $2$. Singleton simplices have constant coordinates with slope $0$. Hence the maximum slope over all simplices is $\frac{4}{\sqrt{3}}$, and $L=\max \{1,\frac{4}{\sqrt{3}}\}=\frac{4}{\sqrt{3}}$ because $\sqrt{3}<2$ gives $\frac{4}{\sqrt{3}}>2>1$ [F7].

(The $A_2$ orbit and its Cayley graph.) Put $c:=\sqrt{3}\mathrm{/}2$ and define linear maps $s(x,y):=(x\mathrm{/}2+cy,cx-y\mathrm{/}2)$ and $t(x,y):=(x\mathrm{/}2-cy,-cx-y\mathrm{/}2)$. Direct multiplication gives $s^2=t^2=1$, $r:=st$ with $r(x,y)=(-x\mathrm{/}2-cy,cx-y\mathrm{/}2)$, and $r^3=1$; their matrices are orthogonal by [F9]. The fixed lines of $s,t$ are respectively $y=x\mathrm{/}\sqrt{3}$ and $y=-x\mathrm{/}\sqrt{3}$, so they are reflections in the two walls of the sector $x\ge 0$, $\mathrm{\mid }y\mathrm{\mid }\le x\mathrm{/}\sqrt{3}$. The point $v_0=(1,0)$ is interior to this sector and at distance $1\mathrm{/}2$ from each wall, by the perpendicular-distance formula [F6]. In the group presentation $⟨s,t\mid s^2=t^2=(st{)}^3=1⟩$, any word first reduces to an alternating word; the relation gives $stst=ts$ and $tsts=st$ and then $sts=tst$, leaving at most the six words $1,s,t,st,ts,sts$. The six matrices represented by these words send $v_0$ respectively to $v_0,v_1,v_5,v_2,v_4,v_3$, all distinct. Thus the reflection group has exactly six elements and exactly this presentation, the rank-two Coxeter presentation of type $A_2$. Its generic orbit is precisely the six vertices. By step 2.2 the triangles cover $H$, and every triangle vertex is a polygon vertex, a midpoint of two such vertices or their average $o$; hence $H$ is their convex hull. This verifies directly the orbit-hull definition of its Coxeter cell. Label a group element $w$ by the vertex $w{v}_0$. Its right-generator neighbors $ws{v}_0=w{v}_1$ and $wt{v}_0=w{v}_5$ are the two neighbors of $w{v}_0$ in the hexagon: orthogonal maps in this group permute the six vertices and preserve their distances, and the only vertices at Euclidean distance $1$ from $v_0$ are $v_1,v_5$ by the displayed coordinates, hence the same holds at $w{v}_0$. This proves that the Cayley graph for $s,t$ is exactly the hexagonal $1$-skeleton.

(Clause (iv): the vertex distances.) By step 2.1 clause (i), $d(v_0,v_k)=\parallel v_0-v_k{\parallel }_2$. For $k=1$: $\parallel v_0-v_1{\parallel }_2=\parallel (\frac{1}{2},-\frac{\sqrt{3}}{2}){\parallel }_2=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$ [F7]. For $k=2$: $v_0-v_2=(\frac{3}{2},-\frac{\sqrt{3}}{2})$, so $\parallel v_0-v_2{\parallel }_2^2=\frac{9}{4}+\frac{3}{4}=3$ and $\parallel v_0-v_2{\parallel }_2=\sqrt{3}$. For $k=3$: $v_0-v_3=(2,0)$, so $\parallel v_0-v_3{\parallel }_2=2$ [F7].

(Clause (iii): the star radius.) Substituting $D=2$ and $L=\frac{4}{\sqrt{3}}$ into [F2] gives $\delta =\frac{1}{2L(D+1)}=\frac{1}{2\cdot \frac{4}{\sqrt{3}}\cdot 3}=\frac{\sqrt{3}}{24},$ using $(\sqrt{3}{)}^2=3$ and $\frac{1}{24\mathrm{/}\sqrt{3}}=\frac{\sqrt{3}}{24}$ [F7].

(Clause (iv): the graph distance and the comparison.) The hexagonal $1$-skeleton is the cycle $v_0{v}_1{v}_2{v}_3{v}_4{v}_5{v}_0$ on six vertices, a connected simple graph, so it carries the graph path metric $d_{\mathrm{g}\mathrm{r}}$ [F8]; for $k\in \{1,2,3\}$ the boundary path $v_0,v_1,\dots ,v_k$ has length $k$, so $d_{\mathrm{g}\mathrm{r}}(v_0,v_k)\le k$, while a path of $r$ edges consists of index increments $+1$ or $-1$ modulo $6$, whose integer sum $j$ satisfies $j\equiv k(\mathrm{m}\mathrm{o}\mathrm{d}6)$ and $\mathrm{\mid }j\mathrm{\mid }\le r$. The minimum of $\mathrm{\mid }j\mathrm{\mid }$ among integers congruent to $k$ is $\min (k,6-k)=k$ for $k=1,2,3$, hence $r\ge k$; thus $d_{\mathrm{g}\mathrm{r}}(v_0,v_k)=k$. Comparing with step 3.2: $\sqrt{3}\mathrm{\ne }2$ and $2\mathrm{\ne }3$ [F7], so the chain metric and the graph metric already differ at the vertex pairs $(v_0,v_2)$ and $(v_0,v_3)$, and in particular the graph metric of the $1$-skeleton is not the restriction of the cell metric; the boundary path has length $3$ between $v_0$ and $v_3$, strictly more than $d(v_0,v_3)=2$.

(The gluing and (H1)-(H3).) Every principal down-set of $P$ is finite and isomorphic to the face poset of the corresponding cell: $P_{\le u}=\{\mathrm{\varnothing },u\}$ for the one-point cell $C_u$, and $P_{\le e}=\{\mathrm{\varnothing },a,b,e\}$ for the edge $e=\{a,b\}$, which is the face poset of the interval $C_e$ with its two vertices. Meets are $u\wedge e=u$ for an incident pair, $e\wedge f=u$ for distinct edges sharing the vertex $u$, and $\mathrm{\varnothing }$ for disjoint edges, distinct vertices or nonincident vertex-edge pairs; also $p\wedge p=p$ and $p\wedge \mathrm{\varnothing }=\mathrm{\varnothing }$. The face isometries satisfy the cocycle condition vacuously (the only nonempty strict chains are $u<e$), the assignment $p\mapsto F_{p,e}$ is the poset isomorphism from $\{a,b,e\}$ onto the nonempty faces of $C_e$, and the intersection condition holds: the cell images are injective and two of them meet exactly in the image of the meet, which for distinct edges sharing a vertex is that vertex and otherwise is empty. Hence $X_{\mathrm{\Gamma }}$ is an isometric polyhedral gluing. Local finiteness (H2): a point of ${\iota }_e(C_e)$ that is not a vertex point lies in that cell alone, and a vertex point lies in its point cell and exactly the cells of the finitely many edges at $u$. Finite shapes (H3): every cell is a point or an interval of one of the finitely many lengths $\{{\mathrm{\ell }}_e:e\in E\}$. Connectedness (H1): order the edges $e_1,\dots ,e_m$ so that each $e_i$ with $i\ge 2$ shares a vertex with ${\bigcup }_{j<i}{e}_j$, which is possible because $\mathrm{\Gamma }$ is connected; each cell inclusion is continuous by the weak-open trace criterion of [F1], so each ${\iota }_{e_i}(C_{e_i})$ is connected [F6], and $Y_i:={\bigcup }_{j\le i}{\iota }_{e_j}(C_{e_j})$ is connected by induction on $i$, since $Y_{i-1}$ and ${\iota }_{e_i}(C_{e_i})$ are connected and meet in the shared vertex [F6]; finally $X_{\mathrm{\Gamma }}={\bigcup }_i{\iota }_{e_i}(C_{e_i})$, because every vertex of $\mathrm{\Gamma }$ is an endpoint of an edge (an isolated vertex would contradict connectedness of $\mathrm{\Gamma }$ with $E\mathrm{\ne }\mathrm{\varnothing }$), so its point cell is a face of an edge cell and its image is already covered. Thus (H1)-(H3) hold, and [F2] gives that $d$ is a metric inducing the weak topology and that $(X_{\mathrm{\Gamma }},d)$ is proper and complete.

(Cutting a leaf edge: the induction claim.) Let $Z$ be the interval realization of a finite tree ${\mathrm{\Gamma }}_0$ with $m_0\ge 1$ edges and positive lengths, with chain metric $d_0$. We prove: there is a function ${\lambda }_0$ on pairs of points of $Z$, defined recursively below, such that (a) for all $x,y$ there is a chain from $x$ to $y$ of length ${\lambda }_0(x,y)$ and every chain from $x$ to $y$ has length at least ${\lambda }_0(x,y)$; (b) for vertices $u,w$ of ${\mathrm{\Gamma }}_0$, ${\lambda }_0(u,w)$ is the sum of the lengths of the edges of the unique reduced path from $u$ to $w$; (c) exactly one geodesic segment joins any two points of $Z$. Base: if $m_0=1$ with edge $e$, then $Z={\iota }_e(C_e)$, all points have coordinates in $[0,{\mathrm{\ell }}_e]$ and the chains are sequences of such coordinates; a chain of $k$ steps has length ${\sum }_i\mathrm{\mid }{t}_{i-1}-t_i\mathrm{\mid }\ge \mathrm{\mid }{t}_x-t_y\mathrm{\mid }$ [F4], with equality for the one-step chain, so (a) holds with ${\lambda }_0(x,y):=\mathrm{\mid }{t}_x-t_y\mathrm{\mid }$; (b) is the case of two vertices, where ${\lambda }_0$ is ${\mathrm{\ell }}_e$ for the two distinct vertices; and for (c), an isometric map $\gamma$ with $\gamma (0)=x$, $\gamma (L)=y$, $L=\mathrm{\mid }x-y\mathrm{\mid }$, satisfies $\mathrm{\mid }\gamma (s)-x\mathrm{\mid }=s$ and $\mathrm{\mid }\gamma (s)-y\mathrm{\mid }=L-s$ at every $s$ [F3], and in the real interval the unique point with these distances is $x$ if $L=0$, and otherwise the one at fraction $s\mathrm{/}L$ from $x$ to $y$ [F4], so $\gamma$ is unique. Step: let $m_0\ge 2$; take a simple path of maximal length in ${\mathrm{\Gamma }}_0$. An endpoint $v$ has no neighbour outside this path (otherwise it extends), and no neighbour on the path except its next vertex $w$ (otherwise there is a cycle); hence $v$ is a leaf. Removing $v$ and $e=\{v,w\}$ leaves a connected acyclic graph ${\mathrm{\Gamma }}_1$ with $m_0-1\ge 1$ edges, since paths between remaining vertices cannot pass through the leaf; the induction hypothesis supplies ${\lambda }_1$ and (a)-(c) for $Z_1:=X_{{\mathrm{\Gamma }}_1}$ with its own chain metric $d_1$; we will prove that $d_1=d_0{\mathrm{\mid }}_{Z_1\times Z_1}$. Label the cell $C_e$ so that $w$ corresponds to ${\mathrm{\ell }}_e$, and write $t_x$ for the coordinate in $[0,{\mathrm{\ell }}_e]$ of $x\in {\iota }_e(C_e)$. Define ${\lambda }_0(x,y):=\mathrm{\mid }{t}_x-t_y\mathrm{\mid }$ if $x,y\in {\iota }_e(C_e)$; ${\lambda }_0(x,y):=({\mathrm{\ell }}_e-t_x)+{\lambda }_1(w,y)$ if $x\in {\iota }_e(C_e)$, $y\in Z_1$; use the symmetric cross formula when $x\in Z_1$, $y\in {\iota }_e(C_e)$, and put ${\lambda }_0(x,y):={\lambda }_1(x,y)$ if $x,y\in Z_1$. The two formulas agree for $x=w\in {\iota }_e(C_e)\cap Z_1$, so ${\lambda }_0$ is well defined on all pairs.

(The induction step, continued.) $Z={\iota }_e(C_e)\cup Z_1$ and ${\iota }_e(C_e)\cap Z_1={\iota }_w(C_w)=\{w\}$: the cell $C_v$ is a face of $C_e$ alone (as $e$ is the only edge at the leaf $v$), and every cell of ${\mathrm{\Gamma }}_1$ meets $C_e$ in the image of the common face, which by the intersection condition is a face of the vertex cell $C_w$, hence contained in $\{w\}$. A chain passing from the leaf interval to $Z_1$ must visit $w$: at its first step leaving that interval the common cell is a cell of ${\mathrm{\Gamma }}_1$, so the departing point lies in both pieces. Reversing a chain gives the same assertion for entry. In a chain with endpoints in $Z_1$, every maximal excursion into the leaf interval therefore starts and ends at $w$; delete these excursions. The remaining chain is in $Z_1$ and has no greater length, so every original chain has length at least $d_1(x,y)={\lambda }_1(x,y)$, and an attaining chain in $Z_1$ gives the reverse bound. Thus $d_0=d_1$ on $Z_1\times Z_1$. For endpoints in the leaf interval, chains staying there have length at least $\mathrm{\mid }{t}_x-t_y\mathrm{\mid }$ by [F4]. A chain leaving it has an initial portion from $x$ to $w$ and a final portion from $w$ to $y$, both in the interval, of total length at least $({\mathrm{\ell }}_e-t_x)+({\mathrm{\ell }}_e-t_y)\ge \mathrm{\mid }{t}_x-t_y\mathrm{\mid }$; all intervening lengths are nonnegative. The one-step chain attains $\mathrm{\mid }{t}_x-t_y\mathrm{\mid }$, hence $d_0(x,y)=\mathrm{\mid }{t}_x-t_y\mathrm{\mid }$ and $d_0(x,w)={\mathrm{\ell }}_e-t_x$. For $x$ in the leaf interval and $y\in Z_1$, split at the first visit to $w$ preceding departure; the initial portion has length at least ${\mathrm{\ell }}_e-t_x$, and the rest has length at least $d_0(w,y)=d_1(w,y)={\lambda }_1(w,y)$. Concatenating the interval segment with an attaining $Z_1$ chain gives equality. The symmetric case follows by reversing chains. This proves (a) and $d_0={\lambda }_0$ without presupposing any restriction identity. For (b): the unique reduced path from the leaf $v$ to a vertex $y$ of ${\mathrm{\Gamma }}_1$ begins with the edge $e$, so the weighted path length from $v$ to $y$ is ${\mathrm{\ell }}_e$ plus the weighted path length from $w$ to $y$, and the remaining vertex pairs lie in ${\mathrm{\Gamma }}_1$; this is exactly the recursion defining ${\lambda }_0$.

(The interval between two points, and geodesic uniqueness.) With the notation of steps 1.2 and 2.1, put $I(x,y):=\{z\in Z:d_0(x,z)+d_0(z,y)=d_0(x,y)\}$. We show by the same induction that $I(x,y)$ is a set on which $z\mapsto d_0(x,z)$ is injective. Base $m_0=1$: all points have coordinates in the interval and $\mathrm{\mid }x-z\mathrm{\mid }+\mathrm{\mid }z-y\mathrm{\mid }=\mathrm{\mid }x-y\mathrm{\mid }$ holds exactly for $z$ between $x$ and $y$ [F4], so $I(x,y)$ is the coordinate segment between $t_x$ and $t_y$, on which $z\mapsto d_0(x,z)=\mathrm{\mid }{t}_x-t_z\mathrm{\mid }$ is injective. Step, case $x,y\in Z_1$: if $z\in {\iota }_e(C_e)$, then by the additivity of step 2.1, $d_0(x,z)+d_0(z,y)=(d_0(x,w)+d_0(w,z))+(d_0(z,w)+d_0(w,y))\ge d_0(x,w)+d_0(w,y)\ge d_0(x,y)$, with equality only if $z=w$ (which is a point of $Z_1$) and $d_0(x,w)+d_0(w,y)=d_0(x,y)$; thus $I(x,y)$ is computed in $Z_1$, where the induction hypothesis applies. Step, case $x,y\in {\iota }_e(C_e)$: for $z\in {\iota }_e(C_e)$ the sum is $\mathrm{\mid }{t}_x-t_z\mathrm{\mid }+\mathrm{\mid }{t}_z-t_y\mathrm{\mid }$, equal to $\mathrm{\mid }{t}_x-t_y\mathrm{\mid }$ exactly for $t_z$ between $t_x$ and $t_y$ [F4]; for $z\in Z_1$ with $z\mathrm{\ne }w$ the sum is $(d_0(x,w)+d_0(w,z))+(d_0(w,z)+d_0(w,y))>d_0(x,w)+d_0(w,y)\ge \mathrm{\mid }{t}_x-t_y\mathrm{\mid }$ by step 2.1; and $z=w$ is a point of ${\iota }_e(C_e)$ with coordinate ${\mathrm{\ell }}_e$. Hence $I(x,y)$ is the set of points of ${\iota }_e(C_e)$ whose coordinate lies between $t_x$ and $t_y$, on which $z\mapsto d_0(x,z)=\mathrm{\mid }{t}_x-t_z\mathrm{\mid }$ is injective. Step, case $x\in {\iota }_e(C_e)$, $y\in Z_1$: for $z\in {\iota }_e(C_e)$ the sum is $d_0(x,z)+d_0(z,w)+d_0(w,y)\ge d_0(x,w)+d_0(w,y)=d_0(x,y)$, with equality exactly when $d_0(x,z)+d_0(z,w)=d_0(x,w)$, that is, when $t_z$ lies between $t_x$ and ${\mathrm{\ell }}_e$; for $z\in Z_1$ the sum is $d_0(x,w)+d_0(w,z)+d_0(z,y)$, equal to $d_0(x,y)$ exactly when $d_0(w,z)+d_0(z,y)=d_0(w,y)$, that is, when $z\in I_{Z_1}(w,y)$. On the first part $z\mapsto d_0(x,z)=\mathrm{\mid }{t}_x-t_z\mathrm{\mid }$ is injective, on the second it is $d_0(x,w)+d_0(w,z)$, injective by the induction hypothesis, the distance values on the first part lie in $[0,d_0(x,w)]$ and those on the second lie in $[d_0(x,w),d_0(x,y)]$, with the shared boundary value attained only at $w$. Thus injectivity holds across the two parts as well. For the opposite ordering $x\in Z_1$, $y\in {\iota }_e(C_e)$, apply this case to $I(y,x)=I(x,y)$: injectivity of $z\mapsto d_0(y,z)$ and the identity $d_0(x,z)=d_0(x,y)-d_0(y,z)$ on $I(x,y)$ give injectivity from $x$ as well. Consequently, if $\gamma$ is a geodesic segment from $x$ to $y$ with $R=d_0(x,y)$, then for every $s$ the point $\gamma (s)$ satisfies $d_0(x,\gamma (s))+d_0(\gamma (s),y)=s+(R-s)=R$, so $\gamma (s)\in I(x,y)$ and $d_0(x,\gamma (s))=s$; injectivity of $z\mapsto d_0(x,z)$ on $I(x,y)$ determines $\gamma (s)$ uniquely for every $s$. Hence the geodesic segment is unique, proving (c); existence follows recursively: use the straight unit-speed interval segment for two points in the leaf interval, the inductively supplied geodesic for two points in $Z_1$, and their concatenation at $w$ for the cross case. Step 2.1 gives the isometry equality also for parameter pairs on opposite sides of $w$; when $x=y$ use the constant map on $[0,0]$.