A moved-space intersection in $A_3$ that is not the meet

Put $c:=\cos (\pi \mathrm{/}3)$. By [F7], $0<\pi \mathrm{/}3<\gamma <2$ gives $c>\cos \gamma =0$, while $2c^2-1=\cos (2\pi \mathrm{/}3)=-c$, so $(2c-1)(c+1)=0$ and $c=\frac{1}{2}$; also $\cos (\pi \mathrm{/}2)=0$ by [F7]. Put $v_i:=\frac{1}{\sqrt{2}}(e_i-e_{i+1})\in \mathbf{H}$ for $i=1,2,3$. Then $⟨v_i,v_i⟩=1$, $⟨v_i,v_{i+1}⟩=-\frac{1}{2}$ and $⟨v_i,v_j⟩=0$ whenever $\mathrm{\mid }i-j\mathrm{\mid }\ge 2$, which by [F6] and [F7] matches $B(e_{s_i},e_{s_j})=-\cos (\pi \mathrm{/}m(s_i,s_j))$; the $v_i$ are linearly independent, since the coordinates of $a{v}_1+b{v}_2+c{v}_3$ are $(a,-a+b,-b+c,-c)\mathrm{/}\sqrt{2}$, and every $x\in \mathbf{H}$ equals $\sqrt{2}(x_1{v}_1+(x_1+x_2)v_2+(x_1+x_2+x_3)v_3)$, so they form a basis of the three-dimensional space $\mathbf{H}$ and the linear map $\mathrm{\Theta }:V\to \mathbf{H}$ with $\mathrm{\Theta }(e_{s_i})=v_i$ is a linear isometry onto $\mathbf{H}$. For each $i$ the permutation $(ii+1)$ preserves $\mathbf{H}$, fixes ${v_i}^{\perp }\cap \mathbf{H}$ pointwise and sends $v_i$ to $-v_i$, so it acts on $\mathbf{H}$ as an orthogonal involution with moved space $\mathbb{R}{v}_i$; the image $\mathrm{\Theta }\rho (s_i){\mathrm{\Theta }}^{-1}$ is an orthogonal involution with the same moved space $\mathbb{R}{v}_i$ by [F5], and by the uniqueness in [F5] the two are equal. Since $\phi$ is an isomorphism [F1] and $S$ generates $W$, the two homomorphisms $\mathrm{\Theta }\rho {\mathrm{\Theta }}^{-1}$ and $\sigma \mapsto \sigma {\mathrm{\mid }}_{\mathbf{H}}$ from $W$ to the orthogonal group of $\mathbf{H}$ agree on $S$, hence everywhere: $\mathrm{\Theta }\rho (w){\mathrm{\Theta }}^{-1}=\phi (w){\mathrm{\mid }}_{\mathbf{H}}$ for all $w\in W$. In particular $\mathrm{\Theta }\mathrm{\Phi }=\{\sigma v_i:\sigma \in S_4,i\le 3\}=\{\pm \frac{1}{\sqrt{2}}(e_p-e_q):p\mathrm{\ne }q\}$, because $\phi$ is onto and the transpositions of $S_4$ are the images of the conjugate reflections.

Under the identification of step 1.1, $F(w)={\mathrm{\Theta }}^{-1}\{x\in \mathbf{H}:\phi (w)x=x\}$ and ${\mathrm{\ell }}_T(w)=\dim V-\dim F(w)=3-\dim \{x\in \mathbf{H}:\phi (w)x=x\}$ by [F2]. For $\gamma ={\phi }^{-1}(1234)$ the fixed space in ${\mathbb{R}}^4$ of the $4$-cycle $(1234)$ is $\mathbb{R}(e_1+e_2+e_3+e_4)$, which meets $\mathbf{H}$ in $0$, so ${\mathrm{\ell }}_T(\gamma )=3$. For $\alpha ={\phi }^{-1}((12)(34))$ the fixed space in ${\mathbb{R}}^4$ is $\{x:x_1=x_2,x_3=x_4\}$, whose intersection with $\mathbf{H}$ is $\mathbb{R}(e_1+e_2-e_3-e_4)$, of dimension $1$; hence ${\mathrm{\ell }}_T(\alpha )=2$, and the same computation with $x_1=x_4$, $x_2=x_3$ gives fixed space $\mathbb{R}(e_1-e_2-e_3+e_4)\cap \mathbf{H}$ of dimension $1$ and ${\mathrm{\ell }}_T(\beta )=2$. Finally $\alpha \gamma ={\phi }^{-1}(24)$ and $\beta \gamma ={\phi }^{-1}(13)$ by the multiplication convention $(\sigma \tau )(x)=\sigma (\tau (x))$ applied in $S_4$: for instance $\alpha \gamma$ sends $1\mapsto 1$, $2\mapsto 4$, $3\mapsto 3$, $4\mapsto 2$, so it is the transposition $(24)$; and for a transposition $(pq)$ the fixed space in ${\mathbb{R}}^4$ has dimension $3$ (inside $\mathbf{H}$ its coordinates satisfy $x_p=x_q=a$ and $2a+x_r+x_s=0$ for the remaining indices $r,s$, leaving two free parameters), so ${\mathrm{\ell }}_T({\phi }^{-1}(pq))=3-2=1$.

Since $F(\alpha )$ is the line $\mathbb{R}(e_1+e_2-e_3-e_4)$ in the model of step 1.1 and $\mathrm{\Theta }$ is an isometry, $M(\alpha )=F(\alpha {)}^{\perp }$ is, inside $\mathbf{H}$, the orthogonal complement of $e_1+e_2-e_3-e_4$, namely $\{x\in \mathbf{H}:x_1+x_2-x_3-x_4=0\}=\{x\in \mathbf{H}:x_2=-x_1,x_4=-x_3\}$; likewise $F(\beta )=\mathbb{R}(e_1-e_2-e_3+e_4)$ gives $M(\beta )=\{x\in \mathbf{H}:x_1-x_2-x_3+x_4=0\}=\{x\in \mathbf{H}:x_2=-x_3,x_4=-x_1\}$. Intersecting the two sets gives $x_2=-x_1$, $x_3=-x_2=x_1$, $x_4=-x_1$ (the sum condition is then automatic), so $M(\alpha )\cap M(\beta )=\mathbb{R}(e_1-e_2+e_3-e_4)$ is a line.

Since ${\alpha }^{-1}\gamma =\alpha \gamma ={\phi }^{-1}(24)$ because $\alpha$ is an involution, step 2.1 gives ${\mathrm{\ell }}_T(\gamma )=3=2+1={\mathrm{\ell }}_T(\alpha )+{\mathrm{\ell }}_T({\alpha }^{-1}\gamma )$, so $\alpha {\le }_T\gamma$ by the definition of ${\le }_T$ in [F6], and the analogous computation with $\beta \gamma ={\phi }^{-1}(13)$ gives $\beta {\le }_T\gamma$.

By step 1.1 the roots of $A_3$ in the model are the twelve vectors $\pm \frac{1}{\sqrt{2}}(e_p-e_q)$, $p\mathrm{\ne }q$, and none of these is a real multiple of $e_1-e_2+e_3-e_4$, so the line $\mathbb{R}(e_1-e_2+e_3-e_4)$ of step 2.2 contains no root; hence no $w\in W$ has $M(w)=\mathbb{R}(e_1-e_2+e_3-e_4)$, because a nonzero moved space contains a root by [F3].

Let $\tau \in W$ be a common lower bound of $\alpha$ and $\beta$, so $M(\tau )\subseteq M(\alpha )\cap M(\beta )=\mathbb{R}(e_1-e_2+e_3-e_4)$ by [F4] and step 2.2. If $M(\tau )\mathrm{\ne }0$, then $M(\tau )$ equals that line and $\tau$ is an element with moved space the line, contradicting step 3.2; hence $M(\tau )=0$ and $\tau =1$ by [F2]; since $1{\le }_T\alpha ,\beta$, the greatest common lower bound of $\alpha$ and $\beta$ is $1$, and $M(1)=0$ is strictly smaller than the line $M(\alpha )\cap M(\beta )$. This verifies (i), (ii) and (iii).