The noncrossing interval of a dihedral group: a five-reflection claw for I2(5) and its complement

(The dihedral group and its reflections.) The defining relations give $c^m=1$, $t=sc$, and $scs=c^{-1}$. The order of $c$ is exactly $m$: if $m\ge 3$, let $R(i)=i+1$ and $J(i)=-i$ on $\mathbb{Z}\mathrm{/}m\mathbb{Z}$. Since $J^2=1$ and $JRJ=R^{-1}$, the assignment $s\mapsto J$, $t\mapsto JR$ satisfies $s^2=t^2=1$ and $st\mapsto R$, so [F1] gives a homomorphism with the image of $c$ of order $m$. If $m=2$, map $s,t$ to the independent coordinate flips of $(\mathbb{Z}\mathrm{/}2\mathbb{Z}{)}^2$; these are commuting involutions, satisfy the defining relations, and their product has order $2$. Since $c^m=1$ in $W$, in both cases $c$ has exact order $m$. Now $W=⟨s,c⟩$ and $s{c}^k=c^{-k}s$, so every word reduces to $c^k$ or $c^{k}s$, with $k$ taken modulo $m$. These at most $2m$ forms are distinct: the $c^k$ are distinct by the exact order just proved, the $c^{k}s$ are distinct by cancellation, and the two families are separated by the homomorphism $\epsilon :W\to \{1,-1\}$ with $\epsilon (s)=\epsilon (t)=-1$, which exists by [F1] because both simple generators map to $-1$ and $st$ maps to $1$. Thus $\mathrm{\mid }W\mathrm{\mid }=2m$. The reflection set is exactly $T=\{c^{k}s:0\le k<m\}$. Every conjugate of $s$ or $t$ has $\epsilon =-1$ and is therefore in this list. Conversely, for every integer $j$, $c^{j}s{c}^{-j}=c^{2j}s$ and, since $t=sc$, $c^{j}t{c}^{-j}=c^{2j-1}s$. The exponents $2j$ and $2j-1$ cover all residues modulo $m$, so every $c^{k}s$ is a conjugate of a simple reflection. In particular $\mathrm{\mid }T\mathrm{\mid }=m$.

(The Coxeter form and its plane action.) Put $\theta =\pi \mathrm{/}m$ and $q=\cos \theta$. For $x=a{e}_s+b{e}_t$, [F3] and [F5] give $B(x,x)=a^2-2qab+b^2=(a-qb{)}^2+{\sin }^2\theta b^2$. Since $0<\theta \le \pi \mathrm{/}2<\pi$, [F5] gives $\sin \theta >0$, so this is positive for every nonzero $(a,b)$. In orthonormal coordinates take $e_s=(1,0)$ and $e_t=(-\cos \theta ,\sin \theta )$. The reflection formula $R(v)=I-2v{v}^{\mathsf{T}}$ and [F5] give $\rho (c)=\rho (s)\rho (t)=\left(\begin{array}{cc}\cos (2\theta )& -\sin (2\theta )\\ \sin (2\theta )& \cos (2\theta )\end{array}\right)\mathrm{.}$ Thus $c$ rotates this plane through $2\pi \mathrm{/}m$. The order calculation in step 1.1 proves finite type, including $m=2$, where $s,t$ commute and the diagram consists of two isolated vertices.

(Reflection lengths.) Every $r\in T$ is nonidentity and is itself a reflection, so ${\mathrm{\ell }}_T(r)=1$. Each nonidentity rotation $c^k$ is not in $T$ by the sign $\epsilon$, and $c^k=(c^{k}s)s$ is a product of two reflections; hence ${\mathrm{\ell }}_T(c^k)=2$. In particular $c$ has length $2$.

(The interval below $c$.) The identity and $c$ lie below $c$. For $r_k=c^{k}s\in T$, $r_k$ is a conjugate of an involutory simple reflection, so $r_k^{-1}=r_k$, and $r_k^{-1}c=r_{k}c=c^{k}sc=c^{k-1}s=r_{k-1}\in T,$ so ${\mathrm{\ell }}_T(r_k)+{\mathrm{\ell }}_T(r_k^{-1}c)=2={\mathrm{\ell }}_T(c)$ and every reflection lies below $c$. If $c^k$ is a rotation other than $1$ or $c$, then $c^{1-k}$ is also a nonidentity rotation, so ${\mathrm{\ell }}_T(c^k)+{\mathrm{\ell }}_T((c^k{)}^{-1}c)=2+2=4\mathrm{\ne }2$. These are all group elements by step 1.1, proving the interval formula. When $m=2$ there are no rotations other than $1,c$, so the same argument covers that case.

(Lattice operations.) The interval in step 3.1 has bottom $1$, top $c$, and $m$ distinct reflections of equal length $1$ between them. Distinct reflections are incomparable by [F2], since each has the same length and a strict absolute-order comparison would require positive length increase. Thus two distinct reflections have only $1$ as common lower bound and only $c$ as common upper bound; operations with $1$ and $c$ are forced by their bottom/top roles. This proves the displayed lattice operations directly and verifies the finite-type lattice conclusion in this example.

(Kreweras action on reflections.) By [F4], $K$ is an order-reversing bijection; its explicit action is $K(1)=c$, $K(c)=1$, and $K(r_k)=r_{k-1}$ by step 3.1. It therefore cycles through all $m$ reflections. Direct multiplication gives $K^2(w)=c^{-1}wc$ and $c^{-1}{r}_{k}c=c^{k-2}s=r_{k-2}$. By [F6] and step 1.2, this conjugation rotates each reflection axis through $-2\pi \mathrm{/}m$ in the displayed orientation (an angle of magnitude $2\pi \mathrm{/}m$); for $m=2$, a rotation through $\pi$ fixes every unoriented axis. Iterating $k\mapsto k-2$ returns to $k$ exactly when $m$ divides $2j$. The least positive such $j$ is $m$ for odd $m$ and $m\mathrm{/}2$ for even $m$. Hence $K^2$ has one cycle for odd $m$, two cycles for even $m$, and is the identity on $T$ when $m=2$.