Roots, inversions and chamber images in $I_2(5)$, $A_2$ and infinite dihedral type

The case $A_2$: the constant. Put $x:=\pi \mathrm{/}3$, $q:=\cos x$ and $r:=\sin x$. The addition formulas give $\cos 2x={\cos }^{2}x-{\sin }^{2}x=2q^2-1$ and $\sin 2x=2qr$, hence $\cos 3x=\cos (2x+x)=(2q^2-1)q-(2qr)r=2q^3-q-2q(1-q^2)=4q^3-3q$; since $3x=\pi$ and $\cos \pi =-1$, we get $4q^3-3q+1=0$, that is $(q+1)(2q-1{)}^2=0$. Now $0<\pi \mathrm{/}3<\pi \mathrm{/}2<\pi$ and cosine is strictly decreasing on $[0,\pi ]$ with $\cos 0=1$ and $\cos (\pi \mathrm{/}2)=0$, so $0<q<1$; hence $q\mathrm{\ne }-1$ and $q=1\mathrm{/}2$. Thus in this case $c=1\mathrm{/}2$, and the reflection formulas of [F1] give $\rho (s)e_t=e_t+e_s$ and $\rho (t)e_s=e_s+e_t$.

The case $I_2(5)$: the chambers. By [F6] the ten chambers $u_k{C}^{\circ }$ $(0\le k<10)$ are exactly the ten sectors cut out by the five root lines. By [F5], $(st{)}^5=1$ together with $s^2=t^2=1$ gives $s{u}_k=u_{1-k}$ and $t{u}_k=u_{-1-k}$ with indices modulo $10$; combined with $\mathrm{\ell }(u_k)=\min (k,10-k)$ this gives $\mathrm{\ell }(s{u}_k)=\mathrm{\ell }(u_k)-1$ exactly for $1\le k\le 5$, and $\mathrm{\ell }(t{u}_k)=\mathrm{\ell }(u_k)-1$ exactly for $5\le k\le 9$. By the shortening equivalence of [F6], the chambers on the negative side of $H_{e_s}$ are exactly $u_1{C}^{\circ },\dots ,u_5{C}^{\circ }$ and those on the negative side of $H_{e_t}$ exactly $u_5{C}^{\circ },\dots ,u_9{C}^{\circ }$.

The case $I_2(\mathrm{\infty })$: the invariant functional direction. Here $c=1$, so $B(u,u)=B(e_s,e_s)+2B(e_s,e_t)+B(e_t,e_t)=1-2+1=0$, $B(u,e_s)=B(e_s,e_s)+B(e_t,e_s)=1-1=0$ and $B(u,e_t)=B(e_s,e_t)+B(e_t,e_t)=-1+1=0$, where $u:=e_s+e_t$; by bilinearity $B(u,\cdot )=0$ on $\mathbb{R}{e}_s+\mathbb{R}{e}_t$. Moreover $\rho (s)u=\rho (s)e_s+\rho (s)e_t=-e_s+(e_t+2e_s)=u$ and $\rho (t)u=(e_s+2e_t)-e_t=u$, so $\rho (w)u=u$ for every $w\in W$, because $W$ is generated by $s,t$ and $\rho$ is a homomorphism.

The case $A_2$: the positive roots. By 1.1, the three vectors $e_s=\rho (1)e_s$, $e_t=\rho (1)e_t$ and $e_s+e_t=\rho (t)e_s$ are roots; they lie in $V_{+}\setminus \{0\}$, hence in ${\mathrm{\Phi }}_{+}$, and they are pairwise distinct as functions on $\{s,t\}$, with values $(1,0)$, $(0,1)$, $(1,1)$. By [F6] the six chambers are cut out by the three root hyperplanes $H_{\beta }$, $\beta \in \mathrm{\Phi }$; the map $\alpha \mapsto H_{\alpha }$ from ${\mathrm{\Phi }}_{+}$ onto these lines is injective, because two roots on one line are proportional and both have $B$-norm one by [F2], forcing the proportionality factor to be $\pm 1$, and it is surjective because every $H_{\beta }$ equals $H_{-\beta }$ with exactly one of $\pm \beta$ in ${\mathrm{\Phi }}_{+}$. Hence $\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }=3$ and ${\mathrm{\Phi }}_{+}=\{e_s,e_t,e_s+e_t\}$.

The case $A_2$: the chambers. Write $y_s:=f(e_s)$ and $y_t:=f(e_t)$ for $f\in V^{\ast }$, so that $f$ is described by the pair $(y_s,y_t)$. By [F6], $f\in w{C}^{\circ }$ if and only if $f(\rho (w)e_s)>0$ and $f(\rho (w)e_t)>0$. Using the products $\rho (s)e_s=-e_s$, $\rho (s)e_t=e_s+e_t$, $\rho (t)e_t=-e_t$, $\rho (t)e_s=e_s+e_t$ from [F1] and 1.1, together with $\rho (st)e_s=e_t$, $\rho (st)e_t=-e_s-e_t$, $\rho (ts)e_s=-e_s-e_t$, $\rho (ts)e_t=e_s$, $\rho (sts)e_s=-e_t$, $\rho (sts)e_t=-e_s$ (computed from $\rho (uv)=\rho (u)\rho (v)$), this gives $C^{\circ }=\{y_s>0,y_t>0\}$, $s{C}^{\circ }=\{y_s<0,y_s+y_t>0\}$, $st{C}^{\circ }=\{y_t>0,y_s+y_t<0\}$, $sts{C}^{\circ }=\{y_s<0,y_t<0\}=-C^{\circ }$, $ts{C}^{\circ }=\{y_s>0,y_s+y_t<0\}$ and $t{C}^{\circ }=\{y_t<0,y_s+y_t>0\}$. On each of these six sets the signs of $(y_s,y_t,y_s+y_t)$ are the six distinct patterns $(+,+,+)$, $(-,+,+)$, $(-,+,-)$, $(-,-,-)$, $(+,-,-)$, $(+,-,+)$; these are exactly the six combinations not excluded by one of the three equations $y_s=0$, $y_t=0$, $y_s+y_t=0$, so the six chambers are the six sectors, in the stated cyclic order. By [F5], $s^2=t^2=1$ and $(st{)}^3=1$ give $s(st{)}^j=(ts{)}^{j}s$ and $t(st{)}^j=(ts{)}^{j}t$ for all $j\ge 0$ and $(ts{)}^j=(st{)}^{-j}$, whence $s{u}_k=u_{1-k}$ and $t{u}_k=u_{-1-k}$ with indices read modulo $6$; combined with $\mathrm{\ell }(u_k)=\min (k,6-k)$ this gives $\mathrm{\ell }(s{u}_k)=\mathrm{\ell }(u_k)-1$ exactly for $k\in \{1,2,3\}$ and $\mathrm{\ell }(t{u}_k)=\mathrm{\ell }(u_k)-1$ exactly for $k\in \{3,4,5\}$. By the shortening equivalence of [F6], $H_{e_s}$ separates $C^{\circ }$ from $u_1{C}^{\circ },u_2{C}^{\circ },u_3{C}^{\circ }$ and $H_{e_t}$ from $u_3{C}^{\circ },u_4{C}^{\circ },u_5{C}^{\circ }$.

The case $I_2(5)$: the constant. Put $c:=\cos (\pi \mathrm{/}5)$ (the constant of the form at $m=5$) and $q:=\cos (2\pi \mathrm{/}5)$. By 1.1, $\cos (\pi \mathrm{/}3)=1\mathrm{/}2$; since $0<\pi \mathrm{/}5<\pi \mathrm{/}3<\pi$ and cosine is strictly decreasing on $[0,\pi ]$, we have $c>1\mathrm{/}2>0$. The double-angle formula gives $q=2c^2-1$, and since $4\pi \mathrm{/}5=\pi -\pi \mathrm{/}5$ with $\cos (\pi -x)=\cos ((-x)+\pi )=-\cos (-x)=-\cos x$, we get $\cos (4\pi \mathrm{/}5)=-c$; on the other hand $\cos (4\pi \mathrm{/}5)=2q^2-1=2(2c^2-1{)}^2-1$. Hence $2(2c^2-1{)}^2-1+c=0$, that is $8c^4-8c^2+c+1=0$, and because $8c^4-8c^2+c+1=(2c-1)(c+1)(4c^2-2c-1)$ we obtain $4c^2-2c-1=0$, as $c>1\mathrm{/}2$ and $c\mathrm{\ne }-1$. Finally $4X^2-2X-1=4(X-\frac{1+\sqrt{5}}{4})(X-\frac{1-\sqrt{5}}{4})$ by expansion, so $c$ is one of the two numbers $\frac{1\pm \sqrt{5}}{4}$; since $\sqrt{5}>1$ gives $\frac{1-\sqrt{5}}{4}<0<c$, we conclude $c=\frac{1+\sqrt{5}}{4}=\varphi \mathrm{/}2$ with $\varphi :=\frac{1+\sqrt{5}}{2}$, and then $2c=\varphi$ and $4c^2-1=2c+1-1=2c=\varphi$.

The case $I_2(\mathrm{\infty })$: the positive roots. Put $\mathrm{\Psi }:=\{e_s+ku:k\ge 0\}\cup \{e_t+ku:k\ge 0\}$. First, $\mathrm{\Psi }\subseteq {\mathrm{\Phi }}_{+}$: clearly $e_s,e_t\in \mathrm{\Psi }\cap {\mathrm{\Phi }}_{+}$, and for $k\ge 1$ the identities $e_s+ku=\rho (s)(e_t+(k-1)u)$ and $e_t+ku=\rho (t)(e_s+(k-1)u)$ hold by 1.3, so induction on $k$ shows that every element of $\mathrm{\Psi }$ is a root; every element of $\mathrm{\Psi }$ has nonnegative coefficients in $e_s,e_t$, so $\mathrm{\Psi }\subseteq \mathrm{\Phi }\cap V_{+}\setminus \{0\}={\mathrm{\Phi }}_{+}$. Conversely, $\mathrm{\Psi }\cup (-\mathrm{\Psi })$ is stable under $\rho (s)$ and $\rho (t)$: by 1.3 one has $\rho (s)(e_s+ku)=-e_s+ku$, which is $e_t+(k-1)u$ for $k\ge 1$ and $-e_s$ for $k=0$; $\rho (s)(e_t+ku)=(e_t+2e_s)+ku=e_s+(k+1)u$; $\rho (t)(e_t+ku)=-e_t+ku$, which is $e_s+(k-1)u$ for $k\ge 1$ and $-e_t$ for $k=0$; and $\rho (t)(e_s+ku)=(e_s+2e_t)+ku=e_t+(k+1)u$. Since $W$ is generated by $s,t$ and $\rho$ is a homomorphism, $\mathrm{\Psi }\cup (-\mathrm{\Psi })$ is stable under every $\rho (w)$; it contains $e_s,e_t$, so it contains the orbit $\mathrm{\Phi }$. Hence $\mathrm{\Phi }\subseteq \mathrm{\Psi }\cup (-\mathrm{\Psi })$, and an element of ${\mathrm{\Phi }}_{+}$ lies in $\mathrm{\Psi }$: it lies in $\mathrm{\Psi }\cup (-\mathrm{\Psi })$, and no element of $-\mathrm{\Psi }$ has all coefficients nonnegative, while the elements of $V_{+}$ do. Therefore ${\mathrm{\Phi }}_{+}=\mathrm{\Psi }$; finally the two families $e_s+ku$ and $e_t+ku$ are disjoint and each is injective in $k$, because $u\mathrm{\ne }0$ while evaluating $e_s+ku=e_s+k^{\mathrm{\prime }}u$ at $s$ gives $k=k^{\mathrm{\prime }}$, and $e_s+ku=e_t+k^{\mathrm{\prime }}u$ would give $1+k=k^{\mathrm{\prime }}$ at $s$ and $k=1+k^{\mathrm{\prime }}$ at $t$, hence $k=k+2$, a contradiction; so ${\mathrm{\Phi }}_{+}$ is infinite.

The case $I_2(\mathrm{\infty })$: the inversion sets. By [F5] the $u_k$ are pairwise distinct with $\mathrm{\ell }(u_k)=k$, and $u_{2k}=(st{)}^k$, $u_{2k+1}=(st{)}^{k}s$. We prove by induction on $k\ge 0$ the two formulas $N((st{)}^k)=\{ju+e_t:0\le j\le 2k-1\}$ and $N((st{)}^{k}s)=\{ju+e_s:0\le j\le 2k\}$. For $k=0$ the first is $N(1)=\mathrm{\varnothing }$ and the second is $N(s)=\{e_s\}$. Assume the second formula for some $k\ge 0$. Since $\mathrm{\ell }((st{)}^{k+1})=2k+2>2k+1=\mathrm{\ell }((st{)}^{k}s)$, the recursion of [F3] gives $N((st{)}^{k+1})=\{e_t\}\bigsqcup \rho (t)N((st{)}^{k}s)$; by 1.3, $\rho (t)e_s=e_s+2e_t=u+e_t$, so $\rho (t)\{ju+e_s:0\le j\le 2k\}=\{ju+u+e_t:0\le j\le 2k\}=\{(j+1)u+e_t:0\le j\le 2k\}$ and therefore $N((st{)}^{k+1})=\{ju+e_t:0\le j\le 2k+1\}$, which is the first formula with $k+1$. Since $\mathrm{\ell }((st{)}^{k+1}s)=2k+3>2k+2=\mathrm{\ell }((st{)}^{k+1})$, again, but with $\rho (s)e_t=e_t+2e_s=u+e_s$ from 1.3, $N((st{)}^{k+1}s)=\{e_s\}\bigsqcup \rho (s)\{ju+e_t:0\le j\le 2k+1\}=\{ju+e_s:0\le j\le 2k+2\}$, the second formula with $k+1$. So both formulas hold for all $k\ge 0$. Their elements are pairwise distinct because $u\mathrm{\ne }0$, so the two sets have cardinalities $2k$ and $2k+1$, equal to $\mathrm{\ell }((st{)}^k)=2k$ and $\mathrm{\ell }((st{)}^{k}s)=2k+1$ by [F5], in agreement with [F4].

The case $A_2$: the inversion sets. By [F3] and 1.1, and since $\mathrm{\ell }(u_0)=0<\mathrm{\ell }(u_1)=1<\mathrm{\ell }(u_2)=2<\mathrm{\ell }(u_3)=3$ by [F5], one has $N(s)=\{e_s\}\bigsqcup \rho (s)N(1)=\{e_s\}$, $N(t)=\{e_t\}$, $N(st)=\{e_t\}\bigsqcup \rho (t)N(s)=\{e_t,e_s+e_t\}$, $N(ts)=\{e_s\}\bigsqcup \rho (s)N(t)=\{e_s,e_s+e_t\}$ and $N(sts)=\{e_s\}\bigsqcup \rho (s)N(st)=\{e_s,e_s+e_t,\rho (s)(e_s+e_t)\}=\{e_s,e_s+e_t,e_t\}={\mathrm{\Phi }}_{+}$, where we used $\rho (s)e_t=e_s+e_t$ and $\rho (s)(e_s+e_t)=\rho (s)e_s+\rho (s)e_t=-e_s+(e_s+e_t)=e_t$; also $N(1)=\mathrm{\varnothing }$. The cardinalities $0,1,1,2,2,3$ equal $\mathrm{\ell }(1),\mathrm{\ell }(s),\mathrm{\ell }(t),\mathrm{\ell }(st),\mathrm{\ell }(ts),\mathrm{\ell }(sts)$ by [F5], as asserted.

The case $I_2(5)$: the positive roots. By 2.3 and [F1], $\rho (s)e_t=e_t+\varphi e_s$ and $\rho (t)e_s=e_s+\varphi e_t$, and using $\rho (uv)=\rho (u)\rho (v)$ and ${\varphi }^2=\varphi +1$ one computes $\rho (st)e_s=\rho (s)(e_s+\varphi e_t)=-e_s+\varphi (e_t+\varphi e_s)=({\varphi }^2-1)e_s+\varphi e_t=\varphi (e_s+e_t)$. Hence the five vectors $e_s=\rho (1)e_s$, $e_t=\rho (1)e_t$, $e_t+\varphi e_s=\rho (s)e_t$, $e_s+\varphi e_t=\rho (t)e_s$ and $\varphi (e_s+e_t)=\rho (st)e_s$ are roots; each lies in $V_{+}\setminus \{0\}$, so all five lie in ${\mathrm{\Phi }}_{+}$, and they are pairwise distinct as functions on $\{s,t\}$, with values $(1,0)$, $(0,1)$, $(\varphi ,1)$, $(1,\varphi )$, $(\varphi ,\varphi )$. By [F6] the ten chambers are cut out by the five root hyperplanes $H_{\beta }$, $\beta \in \mathrm{\Phi }$, and as in 2.1 the map $\alpha \mapsto H_{\alpha }$ is a bijection from ${\mathrm{\Phi }}_{+}$ onto these lines, so $\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }=5$ and the five listed vectors exhaust ${\mathrm{\Phi }}_{+}$; each of them has $B$-norm one because it is a root, by [F2].

The case $I_2(5)$: the inversion sets $N(u_k)$, $1\le k\le 5$. By [F5], $\mathrm{\ell }(u_k)=k$ for $1\le k\le 5$, so each step below is a shortening-free step of the recursion of [F3] and the union is disjoint. Starting from $N(1)=\mathrm{\varnothing }$ and using the reflection values $\rho (t)e_s=e_s+\varphi e_t$, $\rho (s)e_t=e_t+\varphi e_s$, $\rho (s)(e_s+\varphi e_t)=\varphi (e_s+e_t)$, $\rho (t)(e_t+\varphi e_s)=\varphi (e_s+e_t)$, $\rho (t)\varphi (e_s+e_t)=\varphi e_s+e_t$ and $\rho (s)\varphi (e_s+e_t)=e_s+\varphi e_t$ (all from [F1], $\rho (uv)=\rho (u)\rho (v)$ and ${\varphi }^2=\varphi +1$) one computes $N(u_1)=\{e_s\}$, $N(u_2)=\{e_t\}\bigsqcup \rho (t)N(u_1)=\{e_t,e_s+\varphi e_t\}$, $N(u_3)=\{e_s\}\bigsqcup \rho (s)N(u_2)=\{e_s,e_t+\varphi e_s,\varphi (e_s+e_t)\}$, $N(u_4)=\{e_t\}\bigsqcup \rho (t)N(u_3)=\{e_t,e_s+\varphi e_t,\varphi (e_s+e_t),e_t+\varphi e_s\}$ and $N(u_5)=\{e_s\}\bigsqcup \rho (s)N(u_4)=\{e_s,e_t+\varphi e_s,\varphi (e_s+e_t),e_s+\varphi e_t,e_t\}={\mathrm{\Phi }}_{+}$, the last equality by 3.2. This gives the displayed sets $N(u_1),\dots ,N(u_5)$, of cardinalities $k=\mathrm{\mid }N(u_k)\mathrm{\mid }=\mathrm{\ell }(u_k)$ as required by [F4].