Tits Cones, Chambers, and Parabolic Stabilizers — Examples

The dual action and $\mathrm{\Delta }$. For every $f=(x_s,x_t)$ the reflection formula gives $r_s{e}_s=-e_s$, $r_s{e}_t=e_t+2e_s$, $r_t{e}_t=-e_t$ and $r_t{e}_s=e_s+2e_t$, so evaluating the dual action gives $s\cdot f=(-x_s,2x_s+x_t)$ and $t\cdot f=(x_s+2x_t,-x_t)$. Both generators fix the vector $e_s+e_t$: indeed $r_s(e_s+e_t)=-e_s+e_t+2e_s=e_s+e_t$ and symmetrically for $t$; hence $\rho (w)$ and $\rho (w{)}^{-1}$ fix $e_s+e_t$ for every $w$ and $\mathrm{\Delta }(w\cdot f)=(w\cdot f)(e_s+e_t)=f(\rho (w{)}^{-1}(e_s+e_t))=\mathrm{\Delta }(f)$. This is the action and the invariance assertion of (i).

The elements of $W$. In $W$ one has $s^2=t^2=1$, and with $u=st$ also $u^{-1}=ts$, so $t=u^{-1}s$; every product of generators can be rewritten as $u^k$ or $u^{k}s$ by the rules $(u^k)s=u^{k}s$, $(u^k)t=u^{k-1}s$, $(u^{k}s)s=u^k$ and $(u^{k}s)t=u^{k+1}$, and induction on the number of factors gives $W=\{u^k:k\in \mathbb{Z}\}\cup \{u^{k}s:k\in \mathbb{Z}\}$.

Sample membership tests, (iv). For $k\ge 0$ put ${\alpha }_k:=e_t+k(e_s+e_t)=k{e}_s+(k+1)e_t$ and ${\beta }_k:=e_s+k(e_s+e_t)=(k+1)e_s+k{e}_t$. Then ${\alpha }_0=e_t$ and ${\beta }_0=e_s$ lie in ${\mathrm{\Phi }}_{+}$, and $B({\alpha }_k,e_s)=-1$, $B({\beta }_k,e_t)=-1$, so the reflection formula gives $r_s{\alpha }_k={\alpha }_k+2e_s={\beta }_{k+1}$ and $r_t{\beta }_k={\beta }_k+2e_t={\alpha }_{k+1}$. Since $r_s$ permutes ${\mathrm{\Phi }}_{+}\setminus \{e_s\}$ and ${\alpha }_k\mathrm{\ne }{e}_s$ for all $k\ge 0$, while $r_t$ permutes ${\mathrm{\Phi }}_{+}\setminus \{e_t\}$ and ${\beta }_k\mathrm{\ne }{e}_t$ for all $k\ge 1$, mutual induction on $k$ gives ${\alpha }_k,{\beta }_k\in {\mathrm{\Phi }}_{+}$ for all $k\ge 0$. For $f=(1,-1)$ one has $f({\alpha }_k)=k-(k+1)=-1<0$ for every $k$, so $\mathrm{Neg}(f)$ is infinite and $f\mathrm{\notin }U$; for $f^{\mathrm{\prime }}=(-1,0)$ one has $f^{\mathrm{\prime }}({\beta }_k)=-(k+1)<0$ for every $k$, so $f^{\mathrm{\prime }}\mathrm{\notin }U$. Finally $s\cdot (-1,3)=(-(-1),2(-1)+3)=(1,1)\in C$, so $(-1,3)=s\cdot (1,1)\in sC\subseteq U$.

The products $u^k$ and $u^{k}s$. With $u=st$ one has $u\cdot f=s\cdot (t\cdot f)=(-(x_s+2x_t),2(x_s+2x_t)-x_t)=(x_s-2\mathrm{\Delta },x_t+2\mathrm{\Delta })=f-2\mathrm{\Delta }(1,-1)$; since $\mathrm{\Delta }$ is $W$-invariant, iterating gives $u^k\cdot f=(x_s-2k\mathrm{\Delta },x_t+2k\mathrm{\Delta }),u^{k}s\cdot f=(-(x_s+2k\mathrm{\Delta }),x_t+2x_s+2k\mathrm{\Delta })$ for every $k\in \mathbb{Z}$.

The chambers. By step 1.2 the chambers of the system are the translates $u^{k}C$ and $u^{k}sC$ of the closed quadrant $C=\{x_s\ge 0,x_t\ge 0\}$.

The traces on an affine line. Write $a:=x_s\mathrm{/}\mathrm{\Delta }$ on the region $\mathrm{\Delta }>0$, so that $x_t\mathrm{/}\mathrm{\Delta }=1-a$. The computed matrices of step 2.1 identify the chambers $u^{k}C$ and $u^{k}sC$ with the intervals: $f\in u^{k}C$ if and only if $u^{-k}\cdot f\in C$, that is $x_s+2k\mathrm{\Delta }\ge 0$ and $x_t-2k\mathrm{\Delta }\ge 0$, which reads $-2k\le a\le 1-2k$; and $f\in u^{k}sC$ if and only if $-(x_s+2k\mathrm{\Delta })\ge 0$ and $x_t+2x_s+2k\mathrm{\Delta }\ge 0$, which reads $-(2k+1)\le a\le -2k$. Thus on each affine line $\{\mathrm{\Delta }=\text{const}>0\}$ the traces of the chambers are the unit intervals with root traces the integers, in agreement with the rank-two picture, and the chambers are exactly the $u^{k}C$ and $u^{k}sC$. This is (ii).

Stabilizers, (v). Since ${\mathrm{Stab}}_W(f)=W_{S(f)}$ for $f\in C$, one has ${\mathrm{Stab}}_W(0)=W_S=W$ because $S(0)=\{s,t\}$. Let $\mathrm{\Delta }(f)=0$ and $f\mathrm{\ne }0$: step 2.1 gives $u^k\cdot f=f$ for every $k$ and $(u^{k}s)\cdot f=-f\mathrm{\ne }f$ for every $k$, while $W=\{u^k\}\cup \{u^{k}s\}$ by step 1.2; since $u$ has infinite order, because $u^k\cdot (1,0)=(1-2k,2k)\mathrm{\ne }(1,0)$ for $k\mathrm{\ne }0$, the stabilizer ${\mathrm{Stab}}_W(f)=⟨u⟩$ is infinite cyclic. For $f\in C^{\circ }$ the zero set is empty, so ${\mathrm{Stab}}_W(f)=\{1\}$; for the open wall $\{x_s=0,x_t>0\}$ it is $\{s\}$, so the stabilizer is $\{1,s\}$, and symmetrically $\{1,t\}$ for $\{x_t=0,x_s>0\}$. For a general $g=w\cdot f\in U$ with $f\in C$ and $f\mathrm{\ne }0$, the stabilizer formula gives ${\mathrm{Stab}}_W(g)=w{W}_{S(f)}{w}^{-1}$, of order at most $2$: $S(f)\subseteq \{s,t\}$, and $S(f)=\{s,t\}$ would give $f=0$, so $\mathrm{\mid }S(f)\mathrm{\mid }\le 1$. Hence every point of $U\setminus \{0\}$ has stabilizer of order at most $2$. Finally, for $f^{\mathrm{\prime }}=(-1,0)$ the generator $t$ fixes $f^{\mathrm{\prime }}$, and among the remaining elements $u^k\cdot (-1,0)=(-1+2k,-2k)$ equals $(-1,0)$ only for $k=0$, while $u^{k}s\cdot (-1,0)=(1+2k,-2-2k)$ equals $(-1,0)$ only for $k=-1$, which is the element $u^{-1}s=t$ already listed; hence ${\mathrm{Stab}}_W(f^{\mathrm{\prime }})=\{1,t\}$.

The Tits cone, (iii). Since $\mathrm{\Delta }$ is $W$-invariant and $C\subseteq \{\mathrm{\Delta }\ge 0\}$, one has $U\subseteq \{\mathrm{\Delta }\ge 0\}$. If $\mathrm{\Delta }(f)>0$, then $a:=x_s\mathrm{/}\mathrm{\Delta }$ satisfies $x_t\mathrm{/}\mathrm{\Delta }=1-a$, and the intervals $[-2k,1-2k]$ and $[-(2k+1),-2k]$, $k\in \mathbb{Z}$, cover $\mathbb{R}$: these are the unit intervals $[n,n+1]$, $n\in \mathbb{Z}$, in the two parity classes, and [F13] places every real $a$ in one of them; by step 3.1 the functional $f$ lies in some chamber, so $f\in U$. If $\mathrm{\Delta }(f)=0$ and $f\mathrm{\ne }0$, then $u^k\cdot f=f$ for all $k$ while $(u^{k}s)\cdot f=-f\mathrm{\ne }f$ and $W=\{u^k\}\cup \{u^{k}s\}$; such an $f$ lies in $U$ if and only if $f\in C$ or $-f\in C$, and both conditions force $f=0$ because $C\cap \{\mathrm{\Delta }=0\}=\{0\}$. Hence $U=\{\mathrm{\Delta }>0\}\cup \{0\}$.

The interior, the closure and the boundary, (iii). The half-plane $\{\mathrm{\Delta }>0\}$ lies in $U$ and is open: if $\mathrm{\Delta }(f)>0$ and $d(f,g)<\mathrm{\Delta }(f)\mathrm{/}2$, then $\mathrm{\mid }\mathrm{\Delta }(g)-\mathrm{\Delta }(f)\mathrm{\mid }=\mathrm{\mid }(g-f)(e_s)+(g-f)(e_t)\mathrm{\mid }\le 2d(f,g)<\mathrm{\Delta }(f)$, so $\mathrm{\Delta }(g)>0$; hence $\{\mathrm{\Delta }>0\}\subseteq U^{\circ }$. The origin is not an interior point: for $\lambda >0$ the point $(-\lambda ,0)$ has $\mathrm{\Delta }=-\lambda <0$ and so lies outside $U$, while $d((-\lambda ,0),0)=\lambda$ can be made arbitrarily small; hence $0\mathrm{\notin }{U}^{\circ }$, and $U^{\circ }=\{\mathrm{\Delta }>0\}$ by step 4.1. Moreover $\bar{U}=\{\mathrm{\Delta }\ge 0\}$: any $f$ with $\mathrm{\Delta }(f)<0$ has the ball of radius $\mathrm{\mid }\mathrm{\Delta }(f)\mathrm{\mid }\mathrm{/}2$ disjoint from $U$, because $\mathrm{\Delta }(g)<0$ for every $g$ in it, while every $f$ with $\mathrm{\Delta }(f)\ge 0$ is a limit of the points $g_{\lambda }$ with coordinates $(x_s+\lambda ,x_t+\lambda )$, $\lambda >0$, which exist as functionals by [F1] and satisfy $\mathrm{\Delta }(g_{\lambda })=\mathrm{\Delta }(f)+2\lambda >0$, so $g_{\lambda }\in U$ by step 4.1, with $d(g_{\lambda },f)=\lambda$ decreasing to $0$ (and $f\in U$ itself when $\mathrm{\Delta }(f)>0$). Therefore the boundary $\bar{U}\setminus U^{\circ }$ is the line $\{\mathrm{\Delta }=0\}$, of which $U$ contains exactly the point $0$; and $U$ is neither open nor closed, because $0\in U$ is not interior while the nonzero boundary points lie in $\bar{U}\setminus U$.

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 $\{1,t\}$.