The A-tilde 1 infinity edge, separated from finite dihedral families and from the 4-edge

The matrix in the statement gives $B(x_s{e}_s+x_t{e}_t,x_s{e}_s+x_t{e}_t)=(x_s-x_t{)}^2$. Also $B(x,e_s)=x_s-x_t$ and $B(x,e_t)=x_t-x_s$, so the radical is exactly $\mathbb{R}(e_s+e_t)$. Thus $B$ is positive semidefinite of corank one; the diagram is connected, so the pair is of affine form type by [F4].

In the ordered basis $(e_s,e_t)$, direct substitution in [F1] gives $[r_{e_s}]=\left(\begin{array}{cc}-1& 2\\ 0& 1\end{array}\right),[r_{e_t}]=\left(\begin{array}{cc}1& 0\\ 2& -1\end{array}\right)\mathrm{.}$ Both square to the identity. Their product is $[r_{e_s}{r}_{e_t}]=\left(\begin{array}{cc}3& -2\\ 2& -1\end{array}\right)=I+N,N=\left(\begin{array}{cc}2& -2\\ 2& -2\end{array}\right),N\mathrm{\ne }0,N^2=\left(\begin{array}{cc}2\cdot 2+(-2)\cdot 2& 2\cdot (-2)+(-2)\cdot (-2)\\ 2\cdot 2+(-2)\cdot 2& 2\cdot (-2)+(-2)\cdot (-2)\end{array}\right)=0.$ Since $m(s,t)=\mathrm{\infty }$, the presentation in [F3] has no relation beyond the two involutions, so the assignment $s\mapsto r_{e_s}$, $t\mapsto r_{e_t}$ defines a representation of $W$. For every integer $k$, $(I+N{)}^k=I+kN$ (use $(I+N{)}^{-1}=I-N$ for $k<0$), which is never the identity when $k\mathrm{\ne }0$. Thus the product is a nonidentity unipotent of infinite order and $W$ is infinite.

Here ${\delta }_s={\delta }_t=1$. The slice condition is $\phi (e_s)+\phi (e_t)=1$, so with $\alpha =\phi (e_s)$ its points are exactly $(\alpha ,1-\alpha )$ for $\alpha \in \mathbb{R}$. The vertices from [F5] are $v_s=(1,0)$ and $v_t=(0,1)$; the alcove inequalities give $0<\alpha <1$, and its closure is the Euclidean segment $[v_t,v_s]$. The two walls are its endpoints $\alpha =0$ and $\alpha =1$.

For the left dual action, $(w\cdot \phi )(v)=\phi (\rho (w{)}^{-1}v)$. Since each reflection is involutory, evaluating at $e_s$ gives $(s\cdot \phi )(e_s)=\phi (-e_s)=-\alpha$. Also $r_{e_t}(e_s)=e_s+2e_t$ by [F1], so $(t\cdot \phi )(e_s)=\alpha +2(1-\alpha )=2-\alpha$. Therefore $s\cdot \alpha =-\alpha$ and $t\cdot \alpha =2-\alpha$. Under the left-action convention, $(st)\cdot \alpha =s\cdot (t\cdot \alpha )=\alpha -2$, while $(ts)\cdot \alpha =t\cdot (s\cdot \alpha )=\alpha +2$.

If $m=m(s,t)<\mathrm{\infty }$, then $m\ge 2$ and $0<\pi \mathrm{/}m\le \pi \mathrm{/}2$. By [F7]-[F8] and $\cos 0=1$ from [F12], $0\le c:=\cos (\pi \mathrm{/}m)<1$. Therefore $\det \left(\begin{array}{cc}1& -c\\ -c& 1\end{array}\right)=1-c^2={\sin }^2(\pi \mathrm{/}m)>0,$ where the identity is [F9] and positivity follows from $0\le c<1$. Moreover the quadratic form is $(x_s-c{x}_t{)}^2+(1-c^2)x_t^2$, positive for every nonzero $(x_s,x_t)$; thus it has no radical. From [F3], $ts=(st{)}^{-1}$ and every word reduces to an alternating word, hence to $(st{)}^k$ or $(st{)}^{k}s$; the finite relator $(st{)}^m=1$ reduces $k$ modulo $m$, leaving at most $2m$ elements. At $m=2$ the generators commute; the presentation maps onto $A_1\times A_1$ by sending them to the two factors, and the normal-form bound gives at most four elements, so $W\cong A_1\times A_1$. For $m\ge 3$, let $X=\mathbb{Z}\mathrm{/}m\mathbb{Z}$ and define permutations $\sigma (j)=-j$ and $\tau (j)=1-j$. They are involutions and $(\sigma \tau )(j)=j-1$, which has exact order $m$. The $m$ maps $(\sigma \tau {)}^k$ are distinct translations, and the $m$ maps $(\sigma \tau {)}^k\sigma$ are distinct reflections. No reflection is a translation: equality of $j\mapsto -j-k$ and $j\mapsto j+\mathrm{\ell }$ at $j=0,1$ would imply $m\mid 2$, contrary to $m\ge 3$. By [F3], these permutations define a homomorphic image of $W$ with $2m$ elements. Together with the upper bound, this proves that $W$ is the finite dihedral group $I_2(m)$ of order $2m$, with the standard Coxeter notation [F17].

Compositions of $\alpha \mapsto -\alpha$ and $\alpha \mapsto 2-\alpha$ are precisely the maps $\alpha \mapsto \pm \alpha +2k$ for $k\in \mathbb{Z}$: the product $(ts{)}^k$ is translation by $2k$, and composing it with $s$ gives every orientation-reversing map of this form. The images of $(0,1)$ are all integer intervals: translations by $2k$ give $(2k,2k+1)$ and composing those translations with $\alpha \mapsto 2-\alpha$ gives $(2k+1,2k+2)$. A positive-orientation map stabilizes $(0,1)$ only when $k=0$; a negative-orientation map sends it to $(2k-1,2k)$, which cannot equal $(0,1)$ for integral $k$. Hence the action is simply transitive on these alcoves. This is the rank-one affine reflection group $W_a(A_1)$; its translation subgroup is $2\mathbb{Z}$ and its linear part is $\{\pm 1\}$, hence $W_a(A_1)\cong 2\mathbb{Z}⋊\{\pm 1\}$ as in Xiong's rank-one affine Weyl group example.

For $m(s,t)=4$, the finite diagram has one label-$4$ edge, whereas ${\tilde{A}}_1$ has one label-$\mathrm{\infty }$ edge by [F6]; these labelled graphs are distinct. Xiong's low-rank coincidence [F16] matches the two names $B_2$ and $C_2$ for this finite type. Their rank-two affine extensions have one additional label-$4$ edge by [F10], giving the three-vertex path $(4,4)={\tilde{B}}_2={\tilde{C}}_2$, not ${\tilde{A}}_1$.

The matrix entry $B(e_s,e_t)=-1$ for $m(s,t)=\mathrm{\infty }$ is the defining convention in [F1]. For any $\epsilon >0$, [F20] gives $\epsilon \mathrm{/}\pi >0$, so apply [F19] to choose $N\ge 1$ with $1\mathrm{/}N<\epsilon \mathrm{/}\pi$. If $m\ge N$, then $0<N\mathrm{/}m\le 1$, so $0<\pi \mathrm{/}m\le \pi \mathrm{/}N<\epsilon$. By the Lipschitz bound [F18] and $\cos 0=1$ [F12], $\mathrm{\mid }\cos (\pi \mathrm{/}m)-1\mathrm{\mid }\le \pi \mathrm{/}m<\epsilon$; hence ${\lim }_{m\to \mathrm{\infty }}\cos (\pi \mathrm{/}m)=1$, confirming that the matrix convention is the numerical limit. The infinite label still imposes no finite relator by [F3]. By [F15], Davis's cited Euclidean simplex criterion assumes every Coxeter label is finite, so the rank-one $\mathrm{\infty }$-edge case is established directly here. No choice principle is used.