The dual action, the faces, and the rank-two chamber tiling

The dual action is an action by linear maps. For $w\in W$ and $f\in V^{\ast }$ the formula $(w\cdot f)(v)=f(\rho (w{)}^{-1}v)$ defines an element of $V^{\ast }$ because a composite of linear maps is linear, and $f\mapsto w\cdot f$ is linear because evaluation at $v$ is linear in $f$. Moreover $\mathrm{i}\mathrm{d}\cdot f=f$, since $\rho (\mathrm{i}\mathrm{d})=\mathrm{i}\mathrm{d}$, and $w_1\cdot (w_2\cdot f)=f\circ \rho (w_2{)}^{-1}\circ \rho (w_1{)}^{-1}=f\circ \rho (w_1{w}_2{)}^{-1}=(w_1{w}_2)\cdot f,$ using $\rho (w_1{w}_2)=\rho (w_1)\rho (w_2)$ and hence $\rho (w_1{w}_2{)}^{-1}=\rho (w_2{)}^{-1}\rho (w_1{)}^{-1}$.

Generators on the dual plane. The formulas in [F1] show $r_s(P)=P=r_t(P)$, since each map preserves $P$ and is its own inverse. Hence every $g\in W_{s,t}=⟨r_s,r_t⟩$ and its inverse preserve $P$. The restriction $g\mapsto g{\mathrm{\mid }}_P$ is a homomorphism into $\mathrm{G}\mathrm{L}(P)$, so $(g\cdot \phi )(v)=\phi (g^{-1}v)$ is well-defined on $P^{\ast }$ and is a left action: $(gh)\cdot \phi =\phi \circ h^{-1}\circ g^{-1}=g\cdot (h\cdot \phi )$, and the identity acts trivially. Here $s\cdot \phi$ and $t\cdot \phi$ denote the actions of $r_s$ and $r_t$ respectively. The dual basis vectors satisfy $(s\cdot f_s)(e_s)=f_s(\rho (s{)}^{-1}{e}_s)=f_s(-e_s)=-1$ and $(s\cdot f_s)(e_t)=f_s(\rho (s)e_t)=f_s(e_t+2c{e}_s)=2c$, while $(s\cdot f_t)(e_s)=f_t(-e_s)=0$ and $(s\cdot f_t)(e_t)=f_t(e_t+2c{e}_s)=1$; hence in coordinates $s\cdot (y_s{f}_s+y_t{f}_t)=(-y_s)f_s+(2c{y}_s+y_t)f_t$, that is $\rho (s{)}^{\ast }(y_s,y_t)=(-y_s,2c{y}_s+y_t)\mathrm{.}$ The same computation with $s$ replaced by $t$ gives $\rho (t{)}^{\ast }(y_s,y_t)=(y_s+2c{y}_t,-y_t)$. Both displayed maps are involutions: applying the first twice returns $(y_s,y_t)$, and applying the second twice returns $(y_s,y_t)$. The first fixes the line $\{\phi :\phi (e_s)=0\}=\{y_s=0\}$ pointwise and the second fixes $\{\phi :\phi (e_t)=0\}=\{y_t=0\}$ pointwise, since these are exactly the vectors whose coordinate in question vanishes. For the functional $\delta (\phi )=\phi (e_s+e_t)=y_s+y_t$ one has $\delta (s\cdot \phi )=(2c-1)y_s+y_t$ and $\delta (t\cdot \phi )=y_s+(2c-1)y_t$, so each generator preserves $\delta$ exactly when $c=1$, i.e. when $m(s,t)=\mathrm{\infty }$; in that case $W_{s,t}$ preserves $\delta$ because it is generated by $r_s$ and $r_t$.

Faces. Let $I\subseteq S$ and put $f_I:={\sum }_{s\mathrm{\notin }I}{f}_s\in V^{\ast }$. By the dual-basis property $f_I(e_s)=1$ for $s\mathrm{\notin }I$ and $f_I(e_s)=0$ for $s\in I$; hence $f_I\in C_I$ and every face is nonempty. In particular $I=\mathrm{\varnothing }$ gives ${\sum }_{s\in S}{f}_s\in C^{\circ }$, so the interior is nonempty, and $I=S$ gives $C_S=\{f:f(e_s)=0\text{ for all }s\in S\}=\{0\}$, since a functional vanishing on the basis $(e_s)$ of $V$ is zero. For a root $\alpha$, $B(\alpha ,\alpha )=1$ makes $\alpha \mathrm{\ne }0$, so some coordinate $\alpha (s)\mathrm{\ne }0$ and $f_s(\alpha )=\alpha (s)\mathrm{\ne }0$. Thus ${\mathrm{e}\mathrm{v}}_{\alpha }:V^{\ast }\to \mathbb{R}$ is a nonzero linear map and $H_{\alpha }$ is a hyperplane by [F9].

The finite rank-two model. Assume $m<\mathrm{\infty }$ and put $\theta :=\pi \mathrm{/}m$ and $c=\cos \theta$. Then $0<\theta \le \pi \mathrm{/}2$, $\sin \theta \mathrm{\ne }0$ and $1-c^2={\sin }^2\theta$. On $P$ put $u:=e_s$ and $v:=(e_t+c{e}_s)\mathrm{/}\sin \theta$. Then $B(u,u)=B(e_s,e_s)=1$, $B(u,v)=(B(e_s,e_t)+cB(e_s,e_s))\mathrm{/}\sin \theta =(-c+c)\mathrm{/}\sin \theta =0$ and $B(v,v)=(B(e_t,e_t)+2cB(e_s,e_t)+c^{2}B(e_s,e_s))\mathrm{/}{\sin }^2\theta =(1-2c^2+c^2)\mathrm{/}{\sin }^2\theta =(1-c^2)\mathrm{/}{\sin }^2\theta =1$: thus $(u,v)$ is a $B{\mathrm{\mid }}_P$-orthonormal basis of $P$ and $e_s=u$, $e_t=-cu+\sin \theta v$. Since $r_s$ fixes $\ker B(-,e_s)\cap P=\mathbb{R}v$ pointwise and sends $u=e_s$ to $-u$, in the coordinates $(x_u,x_v)$ of the basis $(u,v)$ it is $r_s(x_u,x_v)=(-x_u,x_v)$; since $\ker B(-,e_t)\cap P$ is spanned by $(\sin \theta ,c)$ and $r_t(e_t)=-e_t$, the map $r_t$ is the reflection in the line $\mathbb{R}(\sin \theta ,c)$. Consequently the composite $A:=r_s{r}_t$ satisfies $A(u)=r_s(e_s+2c{e}_t)=r_s((1-2c^2)u+2c\sin \theta v)=(2c^2-1)u+2c\sin \theta v=(\cos 2\theta )u+(\sin 2\theta )v$ and $A(v)=-\sin 2\theta u+\cos 2\theta v$, so in the orthonormal coordinates $A(x_u,x_v)=(x_u\cos 2\theta -x_v\sin 2\theta ,x_u\sin 2\theta +x_v\cos 2\theta ),$ the rotation through $2\theta$, whose order is exactly $m$.

The rank-two roots. Assume $m<\mathrm{\infty }$ and keep the orthonormal coordinates $(u,v)$ of 1.4, in which $A$ is the rotation through $2\theta$. Then $A^k{e}_s=(\cos 2k\theta )u+(\sin 2k\theta )v$ for every $k\in \mathbb{Z}$, and $r_s{A}^k{e}_s=(-\cos 2k\theta )u+(\sin 2k\theta )v=(\cos (m-2k)\theta )u+(\sin (m-2k)\theta )v$ because $\pi =m\theta$ and $\cos (\pi -\alpha )=-\cos \alpha$, $\sin (\pi -\alpha )=\sin \alpha$. Moreover $e_t=-\cos \theta u+\sin \theta v=(\cos (m-1)\theta )u+(\sin (m-1)\theta )v$, and the same two formulas applied to $e_t$ give $A^k{e}_t$ and $r_s{A}^k{e}_t$ at angles $(m-1+2k)\theta$ and $(1-2k)\theta$. As $k$ varies over $\mathbb{Z}$ the angles $2k\theta$ are exactly the even multiples of $\theta$ and the angles $(1-2k)\theta$ are exactly the odd multiples, so every element of $S:=\{(\cos j\theta )u+(\sin j\theta )v:j=0,1,\dots ,2m-1\}$ is one of the four displayed families and $S\subseteq {\mathrm{\Phi }}_P$; conversely $S$ contains $e_s$ and $e_t$ and is stable under $A$ (which adds $2\theta$ to an angle) and under $r_s$ (which sends an angle $\phi$ to $\pi -\phi =(m\theta -\phi )$), hence stable under $W_{s,t}=⟨\rho (s),\rho (t)⟩=⟨\rho (s),A⟩$, so ${\mathrm{\Phi }}_P\subseteq S$. Therefore ${\mathrm{\Phi }}_P=W_{s,t}\{e_s,e_t\}=\{(\cos j\theta )u+(\sin j\theta )v:j=0,1,\dots ,2m-1\},$ a set of exactly $2m$ unit vectors, and ${\mathrm{\Phi }}_P$ is stable under $\beta \mapsto -\beta$.

The infinite case. Assume $m=\mathrm{\infty }$, so $c=1$ and by 1.2 the generators act by $s\cdot (y_s,y_t)=(-y_s,2y_s+y_t)$ and $t\cdot (y_s,y_t)=(y_s+2y_t,-y_t)$ and preserve $\delta =y_s+y_t$. On the affine line $\{\delta =1\}$ parametrized by $\tau =y_s$ these maps read $\tau \mapsto -\tau$ and $\tau \mapsto 2-\tau$, the two reflections of the line about $0$ and $1$. Every word in the two involutions rewrites to $A^k$ or $A^k{r}_s$ with $k\in \mathbb{Z}$, where $A=r_s{r}_t$; their actions on this affine line are respectively $\tau \mapsto \tau -2k$ and $\tau \mapsto -\tau -2k$. These are pairwise distinct, so restriction to $P$ is faithful on $W_{s,t}$. The interval images are $A^k[0,1]=[-2k,1-2k]$ and $A^k{r}_s[0,1]=[-1-2k,-2k]$, so they are exactly $[j,j+1]$ for $j\in \mathbb{Z}$, with distinct elements giving distinct intervals. These cover the affine line by [F8] and have pairwise disjoint interiors. Since $C_P$ is the cone on $[0,1]$ in the half-plane $\{\delta \ge 0\}$ and the action is linear, the chamber $w{C}_P$ is the cone on the interval corresponding to $w$; these cones have pairwise disjoint interiors and their union is $\{\phi :\delta (\phi )>0\}\cup \{0\}$, because every point with $\delta >0$ is a positive multiple of a unique point of $\{\delta =1\}$ lying in a unit interval, while every point of a cone on a unit interval has $\delta \ge 0$ and only $0$ has $\delta =0$. By [F1] and the normal forms above, the roots are precisely the four families $A^k{e}_s=(1+2k,2k)$, $A^k{e}_t=(-2k,1-2k)$, $A^k{r}_s{e}_s=(-1-2k,-2k)$ and $A^k{r}_s{e}_t=(2+2k,1+2k)$ for $k\in \mathbb{Z}$. Equivalently ${\mathrm{\Phi }}_P=\{(j+1,j),(j,j+1):j\in \mathbb{Z}\}$, since the first and fourth families cover the even and odd values of $j$ in the first pair, and the second and third do so in the second pair. On $\{\delta =1\}$ the equation $\phi (a{e}_s+b{e}_t)=0$ is $(a-b)y_s+b=0$, so the two root families give wall traces $-j$ and $j+1$, exactly the integers. If two chamber intervals are $[j,j+1]$ and $[l,l+1]$ with $j<l$, the wall whose trace is $j+1$ places their interiors in opposite open half-planes, because its defining linear form on $\{\delta >0\}$ has the sign of $y_s\mathrm{/}\delta -(j+1)$ up to a fixed nonzero factor.

The rank-two walls. Assume $m<\mathrm{\infty }$. The map $\mathrm{♭}:P\to P^{\ast }$, $x\mapsto B(x,\cdot )$, is an isomorphism: in the orthonormal basis $(u,v)$ of 1.4 its inverse sends $\phi$ to $\phi (u)u+\phi (v)v$. Moreover it intertwines the two actions, since $(g\cdot \mathrm{♭}(x))(z)=B(x,g^{-1}z)=B(gx,z)=\mathrm{♭}(gx)(z)$ for $g\in W_{s,t}$. For $\beta \in P$ one has $\mathrm{♭}({\beta }^{\perp })=H_{\beta }\cap P^{\ast }$, where ${\beta }^{\perp }=\{x\in P:B(x,\beta )=0\}$: indeed $\mathrm{♭}(x)(\beta )=B(x,\beta )$ vanishes exactly when $x\in {\beta }^{\perp }$. By 2.1 each $\beta \in {\mathrm{\Phi }}_P$ is a unit vector at an angle $j\theta$, so ${\beta }^{\perp }$ is the line at angle $j\theta +\pi \mathrm{/}2$, and the lines $H_{\beta }\cap P^{\ast }$ depend only on the pair $\{\beta ,-\beta \}$: there are exactly $m$ of them, at angles $\pi \mathrm{/}2+j\theta$, $j=0,\dots ,m-1$, equally spaced by $\theta$. The chamber $C_P=\{\phi \in P^{\ast }:\phi (e_s)\ge 0,\phi (e_t)\ge 0\}$ corresponds under $\mathrm{♭}$ to $\{x\in P:B(x,e_s)\ge 0,B(x,e_t)\ge 0\}$, whose boundary lines are $e_s^{\perp }$ (the line at angle $\pi \mathrm{/}2$) and $e_t^{\perp }$ (the line at angle $(m-1)\theta +\pi \mathrm{/}2$, i.e. $\pi \mathrm{/}2-\theta$): so $C_P=\mathrm{♭}$ of the closed sector between these two consecutive walls, of angle $\theta$, and no root hyperplane line meets $C_P^{\circ }$ because the $m$ walls are equally spaced by $\theta$ and contain both boundary lines.

The finite chamber tiling. Assume $m<\mathrm{\infty }$. Put $a:=r_s$, $b:=r_t$, and $A:=ab$. Since $a^2=b^2=1$ and $aAa=A^{-1}$, every word in $a,b$ rewrites to $A^k$ or $A^{k}a$. The finite relation $A^m=1$ therefore gives at most $2m$ elements, represented by $0\le k<m$. Their restrictions to $P$ are all distinct: the rotations $A^k{\mathrm{\mid }}_P$ are distinct by their exact order $m$, the maps $A^{k}a{\mathrm{\mid }}_P$ are distinct after cancellation of $a{\mathrm{\mid }}_P$, and a rotation cannot equal a reflection because their determinants on $P$ are $1$ and $-1$. Thus restriction to $P$ is faithful on this matrix subgroup, $\mathrm{\mid }{W}_{s,t}\mathrm{\mid }=2m$, and it is the dihedral group of order $2m$. Its contragredient action on $P^{\ast }$ has the same distinct elements. The action maps the set of $m$ walls to itself, because $g\cdot (H_{\beta }\cap P^{\ast })=H_{g\beta }\cap P^{\ast }$ for $g\in W_{s,t}$ and ${\mathrm{\Phi }}_P$ is $W_{s,t}$-stable, so it permutes the $2m$ closed sectors; it maps the sector $C_P$ to the sectors obtained by successive rotations through $2\theta$ and by the reflection $\rho (s)$, which realizes all $2m$ of them: under ${\mathrm{♭}}^{-1}$, $C_P$ has angular interval $[\pi \mathrm{/}2-\theta ,\pi \mathrm{/}2]$ and $r_s{C}_P$ has interval $[\pi \mathrm{/}2,\pi \mathrm{/}2+\theta ]$; their $A^k$ images add $2k\theta$, giving all consecutive sectors modulo $2\pi$. Thus the orbit of $C_P$ is exactly the set of closed sectors cut out by the $m$ root hyperplanes. Since $\mathrm{\mid }{W}_{s,t}\mathrm{\mid }=2m$ equals the number of sectors, the action on the orbit is simply transitive, and the sectors have pairwise disjoint interiors and union $P^{\ast }$ as the $2m$ sectors of $m$ distinct lines through the origin. Finally, a sign choice for each of the $m$ wall forms defines an intersection of open half-planes. If nonempty it is convex, by linearity of those forms, and so lies in a single sector: the segment joining any two of its points meets no wall. Thus distinct sector interiors have different signs for some wall form, which supplies a root hyperplane separating their entire interiors.