A point outside the Tits cone with infinite stabilizer

Block decomposition, (i). For a cross pair the Coxeter entry is $2$, so the corresponding off-diagonal form entry is $-\cos (\pi \mathrm{/}2)=0$, and the form is block diagonal with blocks $\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$ on $\{e_{s_1},e_{s_2}\}$ and on $\{e_{s_3},e_{s_4}\}$. For a generator $s$ of one block and a basis vector $e_j$ of the other block, the reflection formula gives $r_s{e}_j=e_j-2B(e_j,e_s)e_s=e_j$, so each generator acts trivially on the other block. Put $V_i:=\mathrm{span}\{e_j:j\in S_i\}$; the same computation shows that every $\rho (s)$ preserves each $V_i$ and that $\rho$ decomposes as $\rho (w_1,w_2)={\rho }_1(w_1)\oplus {\rho }_2(w_2)$, once $W$ is identified with $W_1\times W_2$: cross generators $s\in S_1$, $t\in S_2$ satisfy $(st{)}^2=1$, hence $st=t^{-1}{s}^{-1}=ts$ because both are involutions, so the blocks commute. The assignment $s_1\mapsto (s_1,1),s_2\mapsto (s_2,1),s_3\mapsto (1,s_3),s_4\mapsto (1,s_4)$ respects the relators (within-block relators hold factorwise and cross pairs satisfy $((s_i,1)(1,s_j){)}^2=(s_i^2,s_j^2)=1$), so by the universal property of [F7] it induces a homomorphism $\mathrm{\Phi }:W\to W_1\times W_2$, which is inverse to the multiplication map $W_1\times W_2\to W$ because both composites are homomorphisms agreeing with the identity on generators. By the intrinsic parabolic presentation each $W_i$ is moreover the Coxeter group presented by the restricted rank-two matrix $m{\mathrm{\mid }}_{S_i}$ (Support, intrinsic parabolic presentations, minimal coset representatives and length additivity, with the type-A identification (2)), so the rank-two facts of [F1] apply to $W_1$ and $W_2$. The dual action is therefore componentwise, and $C=\{f:f(e_j)\ge 0\text{ for all }j\}=\{f_1:f_1(e_{s_1}),f_1(e_{s_2})\ge 0\}\times \{f_2:f_2(e_{s_3}),f_2(e_{s_4})\ge 0\}=C_1\times C_2$, whence $U={\bigcup }_{(w_1,w_2)}{w}_1{C}_1\times w_2{C}_2=U_1\times U_2$. This is (i).

The vector, (ii). By clause (iii) of the rank-two example, $U_1=\{f_1:{\mathrm{\Delta }}_1(f_1)>0\}\cup \{0\}$. For $f=(-1,0,1,-1)$ one has ${\mathrm{\Delta }}_1(f_1)=-1<0$, so $f_1\mathrm{\notin }{U}_1$; by step 1.1 $U=U_1\times U_2$, hence $f\mathrm{\notin }U$.

The stabilizer and the contrast, (iii). For the first factor, use the rank-two formulas with $(s,t)=(s_1,s_2)$ read off from the infinite dihedral example: $s_1:(x_1,x_2)\mapsto (-x_1,2x_1+x_2)$ and $s_2:(x_1,x_2)\mapsto (x_1+2x_2,-x_2)$, so $s_2$ fixes the line $x_2=0$ pointwise, and in particular fixes $f_1=(-1,0)$; writing $u_1:=s_1{s}_2$ one has $u_1^k\cdot x=x+2k{\mathrm{\Delta }}_1(x)(-1,1)$, so $u_1^k\cdot f_1=(-1+2k,-2k)$ equals $f_1$ only for $k=0$, while $u_1^k{s}_1\cdot f_1=(1+2k,-2-2k)$ equals $(-1,0)$ only for $k=-1$, which is the element $u_1^{-1}{s}_1=s_2$; since $W_1=\{u_1^k\}\cup \{u_1^k{s}_1\}$, this gives ${\mathrm{Stab}}_{W_1}(f_1)=\{1,s_2\}$. For the second factor $f_2=(1,-1)$ has ${\mathrm{\Delta }}_2(f_2)=0$ and $f_2\mathrm{\ne }0$, so the rank-two stabilizer computation gives ${\mathrm{Stab}}_{W_2}(f_2)=⟨s_3{s}_4⟩$, which is infinite cyclic because $\rho (s_3{s}_4)=r_{s_3}{r}_{s_4}$ has infinite order. Since the action is componentwise by step 1.1, an element $(w_1,w_2)$ fixes $f$ exactly when $w_1$ fixes $f_1$ and $w_2$ fixes $f_2$, so ${\mathrm{Stab}}_W(f)=\{1,s_2\}\times ⟨s_3{s}_4⟩$; this subgroup is infinite (it contains $(1,(s_3{s}_4{)}^k)$ for $k\mathrm{\ne }0$), hence it cannot be a finite parabolic subgroup of $W$. Moreover $f$ lies outside the closure: $U\subseteq \{{\mathrm{\Delta }}_1\ge 0,{\mathrm{\Delta }}_2\ge 0\}$ and this set is closed, so $\bar{U}\subseteq \{{\mathrm{\Delta }}_1\ge 0,{\mathrm{\Delta }}_2\ge 0\}$, and conversely every point $(x_1,x_2,x_3,x_4)$ of it is approached by $(x_1+\epsilon ,x_2+\epsilon ,x_3+\epsilon ,x_4+\epsilon )$ as $\epsilon >0$ tends to zero; these approximants have both ${\mathrm{\Delta }}_i>0$ and lie in $U$; since ${\mathrm{\Delta }}_1(f_1)=-1<0$, the functional $f$ is strictly outside $\bar{U}$. For contrast, in the rank-two example the infinite stabilizer $⟨st⟩$ occurs exactly on the boundary line $\mathrm{\Delta }=0$ with the origin removed, which lies outside $U$ but in its closure, and every point with $\mathrm{\Delta }\mathrm{\ne }0$ has stabilizer of order at most $2$: no $u^k$ with $k\mathrm{\ne }0$ fixes it, and $u^{k}s\cdot f=f$ reduces to the single equation $x_s=-k\mathrm{\Delta }$, which determines at most one $k$, the corresponding element $u^{k}s$ being an involution. This is (iii).