Ordered roots and the mu-dot-root matrix in I2(5)

Put $\theta =\pi \mathrm{/}5$ and $u=\cos \theta >0$, since $\theta$ is acute. As $3\theta =\pi -2\theta$, [F10] gives $4u^3-3u=-(2u^2-1)$, or $(u+1)(4u^2-2u-1)=0$. Thus $2u=(1+\sqrt{5})\mathrm{/}2=\phi$, verifying the stated exact value. The Gram matrix is $G=\left(\begin{array}{cc}1& -\phi \mathrm{/}2\\ -\phi \mathrm{/}2& 1\end{array}\right)$ with determinant $1-{\phi }^2\mathrm{/}4=(3-\phi )\mathrm{/}4>0$, so it is positive definite. By [F2], $C_V=R_1{R}_2$, which has exact order $5$ by [F1]; [F9] gives $c^5=1$, so $h=\mathrm{ord}(c)=5$. From $c=s_1{s}_2$ and [F9] we have $s_2=s_{1}c$ and $s_{1}c{s}_1=c^{-1}$, so every group word has one of the ten normal forms $c^q$ or $c^q{s}_1$, $0\le q<5$. The five rotations are distinct by the order of $c$, as are the five elements $c^q{s}_1$; their determinants under $\rho$ differ, so the two lists are disjoint. Thus $\mathrm{\mid }W\mathrm{\mid }=10$. In the ordered basis $({\alpha }_1,{\alpha }_2)$, the reflection matrices give $R_1=\left(\begin{array}{cc}-1& \phi \\ 0& 1\end{array}\right)$ and $R_2=\left(\begin{array}{cc}1& 0\\ \phi & -1\end{array}\right)$, hence $C_V=R_1{R}_2=\left(\begin{array}{cc}\phi & -\phi \\ \phi & -1\end{array}\right)$.

Direct application of the reflection formula gives ${\rho }_1=(1,0)$ and ${\rho }_2=(\phi ,1)$ in simple-root coordinates. Multiplication by $C_V$ gives ${\rho }_3=(\phi ,\phi )$, ${\rho }_4=(1,\phi )$, and ${\rho }_5=(0,1)$; a second application gives ${\rho }_6=(-1,0)=-{\rho }_1$ and ${\rho }_7=(-\phi ,-1)=-{\rho }_2$. Since ${\rho }_{i+2}=C_V{\rho }_i$, induction on each parity gives ${\rho }_{i+5}=-{\rho }_i$ for every $i\ge 1$. The first five vectors are distinct, and [F5] identifies ${\mathrm{\Phi }}_{+}=\{{\rho }_1,\dots ,{\rho }_5\}$, so they exhaust the positive roots. Their squared norms are $Q(a,b):=a^2+b^2-\phi ab$: $Q(1,0)=Q(\phi ,1)=Q(\phi ,\phi )=Q(1,\phi )=Q(0,1)=1$ using [F7]. This proves (i).

The inverse Gram matrix is $G^{-1}=\frac{4}{3-\phi }\left(\begin{array}{cc}1& \phi \mathrm{/}2\\ \phi \mathrm{/}2& 1\end{array}\right)$, so its columns give the displayed ${\beta }_1,{\beta }_2$. From ${\mu }_{i+2}=C_V{\mu }_i$ in [F3] and $(C_V-\mathrm{i}\mathrm{d}){\mu }_i=-2{\rho }_i$, which follows from [F5] and the definition of $\mu$ in [F3], we obtain ${\mu }_{i+2}={\mu }_i-2{\rho }_i$; taking $i=1,2,3$ gives the three displayed recursions.

For coefficient vectors $(a,b)$ and $(u,v)$, the pairing is $B((a,b),(u,v))=au+bv-\frac{\phi }{2}(av+bu)$. The columns of the root-coordinate matrix and the $\mu$-coordinate matrix are respectively $R=\left(\begin{array}{ccccc}1& \phi & \phi & 1& 0\\ 0& 1& \phi & \phi & 1\end{array}\right)$ and $M=\left(\begin{array}{ccccc}\frac{4}{d}& \frac{2\phi }{d}& \frac{4}{d}-2& \frac{2\phi }{d}-2\phi & \frac{4}{d}-2-2\phi \\ \frac{2\phi }{d}& \frac{4}{d}& \frac{2\phi }{d}& \frac{4}{d}-2& \frac{2\phi }{d}-2\phi \end{array}\right)$, with $d:=3-\phi$. Therefore the desired dot-product matrix is $M^{\mathsf{T}}GR$, which, using [F7], is the displayed matrix. For example, its $(3,1)$ entry is $({\beta }_1-2{\alpha }_1)\cdot {\alpha }_1=1-2=-1$, and its $(4,3)$ entry is $({\beta }_2-2{\rho }_2)\cdot {\rho }_3=\phi -2(\phi \mathrm{/}2)=0$. The displayed entries give the stated signs and the one-step subdiagonal zeros. Specifically, these signs agree with F8(b),(d), and for $n=2$ the zero immediately below the diagonal is F8(c) with $t=1$. Finally ${\mu }_{i+2}=C_V{\mu }_i$ and ${\rho }_{j+2}=C_V{\rho }_j$, so ${\mu }_{i+2}\cdot {\rho }_{j+2}={\mu }_i\cdot {\rho }_j$ by $B$-invariance of $C_V$. This proves (ii)-(iii).

The odd-$h$ clause of [F5] gives $w_0=c^{(5-1)\mathrm{/}2}a=c^2{s}_1=(s_1{s}_2{)}^2{s}_1$ and asserts the alternating word $(s_1{s}_2{)}^2{s}_1$ is reduced of Coxeter length $5$ with prefix roots ${\rho }_1,\dots ,{\rho }_5$. It is the unique longest element by [F6]. The coordinate matrix $A=2(I-B)$ is $\left(\begin{array}{cc}0& \phi \\ \phi & 0\end{array}\right)$, so $A(1,1)=\phi (1,1)$; by [F7] and the definition of $\phi$, $\lambda =\phi =2\cos (\pi \mathrm{/}5)$. Here $J=\{s_1\}$ and $K=\{s_2\}$, so the theorem's $u=p_1{\alpha }_1$ and $v=p_2{\alpha }_2$ span $V$ since $p_1,p_2>0$; hence its Coxeter plane $P=\mathrm{span}(u,v)$ is $V$. No Choice is used.