The hexagonal $A_2$ cell: Euclidean cell metric versus graph distance

(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 $d(v_0,v_3)=2$.