Chamber faces and their stabilizers in $A_2$

The set-up and the six sectors, (i). Put $c=\cos (\pi \mathrm{/}3)$. To derive $c=\frac{1}{2}$, the addition formulas give $\cos 2x={\cos }^{2}x-{\sin }^{2}x$ and $\sin 2x=2\sin x\cos x$, hence $\cos 3x=\cos (2x+x)=\cos 2x\cos x-\sin 2x\sin x=4{\cos }^{3}x-3\cos x$; at $x=\pi \mathrm{/}3$ this reads $-1=\cos \pi =4c^3-3c$, that is $(c+1)(2c-1{)}^2=0$, and $c>0$ because $0<\pi \mathrm{/}3<\pi \mathrm{/}2$ and cosine is strictly decreasing on $[0,\pi ]$ with $\cos (\pi \mathrm{/}2)=0$, while $\pi >0$; hence $c=\frac{1}{2}$. Here $c=\cos (\pi \mathrm{/}3)=\frac{1}{2}$, so $2c=1$ and $B(e_s,e_t)=-\frac{1}{2}$, while $B(e_s,e_s)=B(e_t,e_t)=1$; the reflection formula gives $r_s{e}_s=-e_s$, $r_t{e}_t=-e_t$ and $r_s{e}_t=e_t+2c{e}_s=e_t+e_s$, $r_t{e}_s=e_s+2c{e}_t=e_s+e_t$, so the dual action is $s\cdot f=(-x_s,x_s+x_t)$ and $t\cdot f=(x_s+x_t,-x_t)$. The three displayed lines $H_{e_s}=\{x_s=0\}$, $H_{e_t}=\{x_t=0\}$ and $H_{e_s+e_t}=\{x_s+x_t=0\}$ are root hyperplanes, since $e_s+e_t=r_t{e}_s$ is a root. The rank-two picture with $m=3$ and $P=\mathbb{R}{e}_s+\mathbb{R}{e}_t=V$ states that the $2m=6$ closed sectors cut out by these root lines are exactly the chambers $w{C}_P$ with $w\in W_{s,t}$, that these chambers have pairwise disjoint interiors and tile the plane, and that $W_{s,t}$ acts on them simply transitively. Here $W=⟨s,t⟩$, so $\rho (W)=⟨\rho (s),\rho (t)⟩=W_{s,t}$, and by [F9] the homomorphism $\rho$ is an isomorphism $W\to W_{s,t}$; hence $W$ acts on the same six sectors, the transitivity and freeness of the $W_{s,t}$-action pass to $W$, and the six sectors are exactly the six chambers $wC$ with $\mathrm{\mid }W\mathrm{\mid }=\mathrm{\mid }{W}_{s,t}\mathrm{\mid }=2m=6$.

Wall stabilizers, (ii). The functional $f=(0,1)$ lies in $C$ with $f(e_s)=0$ and $f(e_t)=1$, so $S(f)=\{s\}$ and the stabilizer formula gives ${\mathrm{Stab}}_W(f)=W_{\{s\}}=\{1,s\}$; every point of the open face $C_{\{s\}}=\{x_s=0,x_t>0\}$ has the same zero set $\{s\}$, so the same formula applies to all of them, and symmetrically every point of $C_{\{t\}}=\{x_t=0,x_s>0\}$ has stabilizer $\{1,t\}$.

Interior and vertex stabilizers, (iii). The functional $(1,1)$ lies in $C^{\circ }$ and has empty zero set, so its stabilizer is $W_{\mathrm{\varnothing }}=\{1\}$; the origin has $S(0)=\{s,t\}$, so its stabilizer is $W_S=W$, of order $6$ by step 1.1.

An orbit and the intersection rule, (iv). Using the generator formulas, $s\cdot (0,1)=(0,1)$, $t\cdot (0,1)=(1,-1)$ and $st\cdot (0,1)=(-1,0)$; moreover the three-element set $S_0:=\{(0,1),(1,-1),(-1,0)\}$ is stable under $s$ and $t$, since $s$ fixes $(0,1)$ and interchanges $(1,-1)$ with $(-1,0)$, while $t$ interchanges $(0,1)$ with $(1,-1)$ and fixes $(-1,0)$; hence $W\cdot (0,1)\subseteq S_0$. Conversely $(0,1)$, $t\cdot (0,1)$ and $st\cdot (0,1)$ are three distinct elements of the orbit, so $W\cdot (0,1)=S_0$, of cardinality $3=\mathrm{\mid }W\mathrm{\mid }\mathrm{/}\mathrm{\mid }{\mathrm{Stab}}_W(0,1)\mathrm{\mid }=6\mathrm{/}2$ by steps 1.1 and 1.2, and only $(0,1)$ has both coordinates $\ge 0$, so it is the unique point of the orbit in $C$, in accordance with the collision theorem. Finally, for $f\in C$, membership in $sC$ is equivalent to $s\cdot f=(-x_s,x_s+x_t)\in C$, hence to $x_s=0$; therefore $sC\cap C=\{x_s=0,x_t\ge 0\}$, as asserted.

The whole plane is the Tits cone, (v). Every standard parabolic subgroup of the finite group $W$ is finite, so every point of $C$ has finite $W_{S(f)}$ and the interior criterion gives $C\subseteq U^{\circ }$. Since $U^{\circ }$ is $W$-invariant and the six chambers cover $V^{\ast }$ by step 1.1, one has $V^{\ast }={\bigcup }_{w\in W}wC\subseteq U^{\circ }\subseteq U\subseteq V^{\ast }$, so all three sets are equal.