The Wall form and line restrictions of a plane rotation, and the necessity of a common upper bound

By [F1] the matrix of $A$ in the basis $(e_s,e_t)$ has determinant $1$ and trace $2\cos (2\pi \mathrm{/}m)$, and $A$ has order $m$; the trace differs from $\mathrm{tr}\mathrm{i}\mathrm{d}=2$ because $1-\cos (2\pi \mathrm{/}m)=2{\sin }^2(\pi \mathrm{/}m)>0$ by [F6], as $\pi \mathrm{/}m\in (0,\pi \mathrm{/}3]\subset (0,2]$ gives $\sin (\pi \mathrm{/}m)>0$.

Put $u:=(e_s+e_t)\mathrm{/}\sqrt{2-2c}$ and $\tilde{w}:=(e_t-e_s)\mathrm{/}\sqrt{2+2c}$; the square roots are nonzero because $2-2c=4{\sin }^2(\pi \mathrm{/}(2m))>0$ and $2+2c=4{\cos }^2(\pi \mathrm{/}(2m))>0$, since $0<\pi \mathrm{/}(2m)\le \pi \mathrm{/}6<2$ and $\cos$ has no zero in $[0,\pi \mathrm{/}2)$ by [F6]. A direct computation with [F1] gives $B(u,u)=B(\tilde{w},\tilde{w})=1$ and $B(u,\tilde{w})=0$, so $(u,\tilde{w})$ is an oriented orthonormal basis, and computing $Au=\cos \theta u+\sin \theta \tilde{w}$, $A\tilde{w}=-\sin \theta u+\cos \theta \tilde{w}$ with $\theta :=2\pi \mathrm{/}m$: indeed $B(Au,u)=2c^2-1=\cos \theta$, $B(Au,\tilde{w})=2c\sin (\pi \mathrm{/}m)=\sin \theta$, $B(A\tilde{w},u)=-\sin \theta$ and $B(A\tilde{w},\tilde{w})=2c^2-1=\cos \theta$, using $1-c^2={\sin }^2(\pi \mathrm{/}m)$ and the double-angle identities of [F6]. Hence in these coordinates $A$ is the rotation by $\theta =2\pi \mathrm{/}m$, and $1$ is not an eigenvalue of $A$: the matrix of $A-\mathrm{i}\mathrm{d}$ has determinant $(\cos \theta -1{)}^2+{\sin }^2\theta =2-2\cos \theta =4{\sin }^2(\theta \mathrm{/}2)\mathrm{\ne }0$, since $\theta \mathrm{/}2=\pi \mathrm{/}m\in (0,2]$. Therefore $F(A)=\ker (A-\mathrm{i}\mathrm{d})=0$ and $M(A)=F(A{)}^{\perp }=V$ by F3. Moreover $S_A=(A-\mathrm{i}\mathrm{d}{)}^{-1}$, computed from the displayed rotation matrix, is $\frac{1}{2}\left(\begin{array}{cc}-1& \sin \theta \mathrm{/}(1-\cos \theta )\\ -\sin \theta \mathrm{/}(1-\cos \theta )& -1\end{array}\right)=-\frac{1}{2}\mathrm{i}\mathrm{d}-\frac{1}{2}\cot (\theta \mathrm{/}2)J$ with $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, because $\sin \theta \mathrm{/}(1-\cos \theta )=\cot (\theta \mathrm{/}2)$ by the double-angle identities; under the identification of the oriented plane with $\mathbb{C}$, the matrix $J$ is multiplication by $i$, so $S_A$ is multiplication by $-\frac{1}{2}-\frac{i}{2}\cot (\theta \mathrm{/}2)$, and $(e^{i\theta }-1)(-\frac{1}{2}-\frac{i}{2}\cot (\theta \mathrm{/}2))=1$ by the displayed identity $e^{i\theta }=\cos \theta +i\sin \theta$ together with ${\sin }^2\theta +{\cos }^2\theta =1$ and $\sin \theta \mathrm{/}(1-\cos \theta )=\cot (\theta \mathrm{/}2)$, so $S_A$ is multiplication by $1\mathrm{/}(e^{i\theta }-1)$. The identity ${\chi }_A(u,v)+{\chi }_A(v,u)=-B(u,v)$ with symmetric part $-\frac{1}{2}B$ is F3.

Part (iii). Take $m=4$, so $\theta =\pi \mathrm{/}2$ and the rotation matrix of step 1.2 gives $A^2=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=-{\mathrm{i}\mathrm{d}}_V$; hence $M(w_0)=\mathrm{im}(-2\mathrm{i}\mathrm{d})=V=M(A)$ for $w_0:=A^2$, so $M(w_0)\subseteq M(A)$. By [F4] and step 1.2, ${\mathrm{\ell }}_T(A)=\dim M(A)=2$ and ${\mathrm{\ell }}_T(w_0)=\dim M(w_0)=2$; also $w_0^{-1}A=A^{-2}A=A^{-1}$ and $A^{-1}{w}_0=A^{-1}{A}^2=A$, and $A^{-1}=A^3$ has moved space $V$ because $F(A^{-1})=F(A)=0$, so ${\mathrm{\ell }}_T(w_0^{-1}A)={\mathrm{\ell }}_T(A^{-1}{w}_0)={\mathrm{\ell }}_T(A)=2$. Hence ${\mathrm{\ell }}_T(A)=2\mathrm{\ne }4={\mathrm{\ell }}_T(w_0)+{\mathrm{\ell }}_T(w_0^{-1}A)$ and ${\mathrm{\ell }}_T(w_0)=2\mathrm{\ne }4={\mathrm{\ell }}_T(A)+{\mathrm{\ell }}_T(A^{-1}{w}_0)$, so $w_0\mathrm{̸}{\le }_{T}A$ and $A\mathrm{̸}{\le }_T{w}_0$ by [F7]. If some $\delta \in W$ satisfied $A{\le }_T\delta$ and $w_0{\le }_T\delta$, then $2={\mathrm{\ell }}_T(A)\le {\mathrm{\ell }}_T(\delta )$ and ${\mathrm{\ell }}_T(\delta )=\dim M(\delta )\le \dim V=2$ by [F4], so [F7] gives ${\mathrm{\ell }}_T(\delta )={\mathrm{\ell }}_T(A)+{\mathrm{\ell }}_T(A^{-1}\delta )=2$ and hence ${\mathrm{\ell }}_T(A^{-1}\delta )=0$; then $A^{-1}\delta =1$, since ${\mathrm{\ell }}_T(x)=\dim M(x)=0$ forces $\rho (x)=\mathrm{i}\mathrm{d}$ and $x=1$ by [F5], so $\delta =A$, contradicting $w_0\mathrm{̸}{\le }_{T}A$; thus $A$ and $w_0$ have no common upper bound. Finally $w_0=A^2$ is the unique longest element. Since $sAs=A^{-1}$ and $t=A^{-1}s$, every word reduces to $A^k$ or $A^{k}s$ with $0\le k\le 3$. The four rotations are distinct by the order of $A$; the four $A^{k}s$ are distinct by cancellation, and the two lists cannot overlap: overlap would give $s=A^j$, whereas $s$ has moved dimension one by [F3] and [F5], and $A^j$ has moved dimension zero for $j=0$ and two for $j=1,2,3$ by the displayed rotation matrices. These eight elements are $1,s,t,st,ts,sts,tst,stst$, because $As=sts$, $A^{3}s=t$, and $A^{2}s=tst$ (the relation $A^4=1$ gives $stst=tsts$). A word of length at most three reduces, by cancelling adjacent equal generators, to one of the first seven words; these are distinct from $A^2=stst$. Thus $\mathrm{\ell }(A^2)=4$, and every other element has length at most three.

Part (ii), first half. Let $L\subseteq V$ be a line; since $M(A)=V$ by step 1.2, $L\subseteq M(A)$, so the operator $H_L={\mathrm{\Pi }}_L{S}_A{\mathrm{\Pi }}_L$ of F3 is defined. As $L$ is one-dimensional, $H_L$ is multiplication by a real scalar $\lambda$, and the identity $H_L+H_L^{\ast }=-{\mathrm{i}\mathrm{d}}_L$ of F3 gives $2\lambda =-1$, so $\lambda =-\frac{1}{2}$; hence $A_L=\mathrm{i}\mathrm{d}+H_L^{-1}$ acts on $L$ as $-{\mathrm{i}\mathrm{d}}_L$ and on $L^{\perp }$ as the identity, that is $A_L=\mathrm{i}\mathrm{d}-2{\mathrm{\Pi }}_L$. By F3 $M(A_L)=L$, and $A_L{\le }_{\mathrm{O}}A$ by F3. The assignment $L\mapsto A_L$ is a bijection from the lines of $V$ onto the reflections $B{\le }_{\mathrm{O}}A$: F3 gives a bijection $U\mapsto A_U$ from the subspaces of $M(A)=V$ onto $\{B:B{\le }_{\mathrm{O}}A\}$ whose inverse is $B\mapsto M(B)$ and which satisfies $M(A_U)=U$, and the one-dimensional subspaces of $V$ are the lines, while the elements $B{\le }_{\mathrm{O}}A$ with $\dim M(B)=1$ are exactly the reflections $B{\le }_{\mathrm{O}}A$ (the elements $B=\mathrm{i}\mathrm{d}$ and $B=A$ correspond to $U=0$ and $U=V$ and are not reflections, as $\dim M(A)=2$).

$T$ has exactly $m$ elements. Since $A=st$ one has $sA=t$ and $A^{-1}=ts$, and the set $W_0:=\{A^j,A^{j}s:0\le j<m\}$ contains $1$, $A$, $s=A^{0}s$ and $t=A^{m-1}s$ and is closed under multiplication and inversion, since $s{A}^{j}s=A^{-j}$ for every $j$: it is a subgroup of $W$ containing $s$ and $t$, so $W_0=W$. Conjugation by the elements of $W_0$ gives $A^{k}s{A}^{-k}=A^{2k}s$, $A^{k}t{A}^{-k}=A^{2k-1}s$ and $(A^{k}s)t(A^{k}s{)}^{-1}=A^{2k+1}s$, so every element of $T$ lies in $\{A^{j}s:0\le j<m\}$, while conversely $A^{2k}s=A^{k}s{A}^{-k}$ and $A^{2k-1}s=A^{k}t{A}^{-k}$ are conjugates of $s$ and $t$; hence $T=\{A^{j}s:0\le j<m\}$, and these $m$ elements are distinct because $A$ has order $m$ by [F1]. By the bijection $\{\pm \alpha :\alpha \in \mathrm{\Phi }\}\to T$ of [F5] the plane has exactly $m$ root lines, and the lines $\mathbb{R}(e_s+a{e}_t)$ for $a\in \mathbb{R}$ are pairwise distinct, so there are infinitely many lines and some are not root lines.

Part (ii), second half. If $L=\mathbb{R}\beta$ for a root $\beta \in \mathrm{\Phi }$, then by [F5] the reflection $\mathrm{i}\mathrm{d}-2{\mathrm{\Pi }}_L$ with normal $\beta$ equals $r_{\beta }=\rho (t_{\beta })\in \rho (W)$; by step 2.2 this operator is $A_L$, so $A_L\in \rho (W)$. Conversely suppose $A_L\in \rho (W)$, say $A_L=\rho (w)$; then $\dim M(w)=\dim M(A_L)=1$ by step 2.2, so the nonzero moved space $M(w)=L$ contains a root $\beta$ by [F5], and $L=\mathbb{R}\beta$ is a root line. Hence $A_L$ lies in $\rho (W)$ exactly when $L$ is one of the $m$ root lines, and for every other line $L$ the operator $A_L$ is an orthogonal reflection with moved space $L$ such that $A_L\mathrm{\notin }\rho (W)$.

Collecting the verified claims: (i) is steps 1.1 and 1.2; (ii) is steps 2.2, 2.3 and 3.1; (iii) is step 2.1. In particular the converse implication $M(\alpha )\subseteq M(\beta )\Rightarrow \alpha {\le }_T\beta$ of the rigidity theorem fails in $W=I_2(4)$ when no common upper bound is available: $\alpha :=w_0$ and $\beta :=A$ satisfy $M(\alpha )=V=M(\beta )$ by step 1.2, while $w_0\mathrm{̸}{\le }_{T}A$ and $A$ and $w_0$ have no common upper bound in $W$, both by step 2.1.