Link edge lengths versus dihedral mirror angles in type I_2(m)

Interior angle. Put $a=\pi \mathrm{/}(2m)$. At $v_0=(1,0)$, the addition and double-angle formulas give $v_1-v_0=2\sin a(-\sin a,\cos a)$ and $v_{2m-1}-v_0=2\sin a(-\sin a,-\cos a)$. Since $\sin a>0$, the unit incident edge directions are $u_{+}=(-\sin a,\cos a)$ and $u_{-}=(-\sin a,-\cos a)$; their inner product is ${\sin }^{2}a-{\cos }^{2}a=-\cos (2a)=\cos (\pi -\pi \mathrm{/}m)$. Their angular distance is therefore $\pi -\pi \mathrm{/}m$ by [F2]. The directions bound the inward wedge at the vertex, so this is the interior angle. Rotation by $j\pi \mathrm{/}m$ preserves inner products and carries the configuration to $v_j$, giving the same angle at every vertex.

Inward normals. At $v_0=(1,0)$ the edge $[v_0,v_1]$ has midpoint direction $(\cos \frac{\pi }{2m},\sin \frac{\pi }{2m})$ and the edge $[v_{2m-1},v_0]$ has midpoint direction $(\cos \frac{\pi }{2m},-\sin \frac{\pi }{2m})$; the outward unit normals are these midpoint directions, and the inward unit normals are their negatives $n_{+}=(-\cos \frac{\pi }{2m},-\sin \frac{\pi }{2m})$ and $n_{-}=(-\cos \frac{\pi }{2m},\sin \frac{\pi }{2m})$. Then $⟨n_{+},n_{-}⟩={\cos }^2\frac{\pi }{2m}-{\sin }^2\frac{\pi }{2m}=\cos \frac{\pi }{m}$ by [F3], so $d_{\mathrm{a}\mathrm{n}\mathrm{g}}(n_{+},n_{-})=\arccos (\cos \frac{\pi }{m})=\frac{\pi }{m}$, since $0<\frac{\pi }{m}\le \frac{\pi }{2}$ and $\cos$ is injective on $[0,\pi ]$.

The mirror lines. With $c:=\cos (\pi \mathrm{/}m)$ and $v_s:=c{e}_s+e_t$, $v_t:=e_s+c{e}_t$, evaluating the bilinear form [F4] gives $B(v_s,e_s)=c-c=0$ and $B(v_t,e_t)=-c+c=0$, so $\mathbb{R}{v}_s\subseteq H_s$ and $\mathbb{R}{v}_t\subseteq H_t$; since $B{\mathrm{\mid }}_P$ is positive definite by [F5] and $v_s,v_t\mathrm{\ne }0$, each of the subspaces $\{v\in P:B(v,e_s)=0\}$ and $\{v\in P:B(v,e_t)=0\}$ is a line, so these inclusions are equalities. Moreover $B(v_s,v_s)=c^2-2c^2+1=1-c^2$, $B(v_t,v_t)=1-c^2$ and $B(v_s,v_t)=2c-c^3-c=c(1-c^2)$, so the ratio of the Statement is $c(1-c^2)\mathrm{/}(1-c^2)=c$, the denominator being positive because $c=\cos (\pi \mathrm{/}m)<1$ for $m\ge 2$.

The link at a vertex. At $v_0=(1,0)$ the two facets through $v_0$ are the edges $[v_0,v_1]$ and $[v_{2m-1},v_0]$, with edge directions $u_{+}:=\frac{v_1-v_0}{\mid v_1-v_0\mid }$ and $u_{-}:=\frac{v_{2m-1}-v_0}{\mid v_{2m-1}-v_0\mid }$; the angle between $u_{+}$ and $u_{-}$ is the interior angle $\pi -\pi \mathrm{/}m$ of step 1.1. By [F1] the tangent cone is $\{\xi :⟨\xi ,n_{+}⟩\ge 0,⟨\xi ,n_{-}⟩\ge 0\}$ for the inward normals $n_{\pm }$ of the two edges. Its boundary lines are $\mathbb{R}{u}_{+}$ and $\mathbb{R}{u}_{-}$, since $u_{\pm }$ is the unit direction along the facet whose inward normal is $n_{\pm }$, so the cone is the intersection of the two closed half-planes bounded by these lines that contain $u_{+}$ and $u_{-}$; that intersection is exactly the wedge $\{a{u}_{+}+b{u}_{-}:a,b\ge 0\}$, whose unit section is the arc from $u_{+}$ to $u_{-}$. The angular distance of its endpoints is $\arccos ⟨u_{+},u_{-}⟩=\pi -\pi \mathrm{/}m$.

The mirror angle. The angle $\theta$ of the nonzero vectors $v_s,v_t$ in the positive definite plane $(P,B)$ is the number in $[0,\pi ]$ with $\cos \theta =B(v_s,v_t)\mathrm{/}(\mid v_s{\mid }_B\mid v_t{\mid }_B)$, where $\mid v{\mid }_B=\sqrt{B(v,v)}$; by step 1.3 this cosine is $c\in [0,1]$, so $\theta \in [0,\pi \mathrm{/}2]$ and $\theta =\arccos c=\pi \mathrm{/}m$ by [F2]. The mirror lines $H_s,H_t$ therefore meet at angle $\pi \mathrm{/}m$.

The product. Let $A=r_s{r}_t$. By [F5] $A$ is a $B$-preserving linear map of the positive definite plane $(P,B)$ with $\det A=1$ and $\mathrm{tr}A=2\cos (2\pi \mathrm{/}m)$. Choose a $B$-orthonormal basis $(u,w)$ of $P$ by [F6] and let $M=\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)$ be the matrix of $A$ in it; $B$-preservation and $\det M=1$ give $M^{T}M=I$, so $M^{-1}=\left(\begin{array}{cc}s& -q\\ -r& p\end{array}\right)=M^T=\left(\begin{array}{cc}p& r\\ q& s\end{array}\right)$, whence $s=p$ and $r=-q$; thus $M=\left(\begin{array}{cc}p& q\\ -q& p\end{array}\right)$ with $p^2+q^2=1$ and $2p=\mathrm{tr}M=2\cos (2\pi \mathrm{/}m)$. Hence $p=\cos (2\pi \mathrm{/}m)$ and $q^2=1-p^2={\sin }^2(2\pi \mathrm{/}m)$ by [F6], so $q=\pm \sin (2\pi \mathrm{/}m)$ for $m\ge 3$, where $0<2\pi \mathrm{/}m<\pi$ makes $\sin (2\pi \mathrm{/}m)>0$, and $q=0$ for $m=2$. In every $B$-orthonormal basis, therefore, $A$ acts by the rotation of angle $2\pi \mathrm{/}m$ or of angle $-2\pi \mathrm{/}m$: the product $r_s{r}_t$ is a rotation of $(P,B)$ through $2\pi \mathrm{/}m$ up to the choice of orientation.

The link Gram matrix. By step 2.1 the vertex link is the arc with endpoints $u_{+},u_{-}$; its Gram matrix as a spherical $1$-simplex is the matrix with diagonal entries $1$ and off-diagonal entries $\cos (\pi -\pi \mathrm{/}m)$, and $\cos (\pi -\pi \mathrm{/}m)=-\cos (\pi \mathrm{/}m)=-c=B(e_s,e_t)$ by [F3] and [F4], so it equals the Gram matrix of $B{\mathrm{\mid }}_P$ displayed in the Statement. It is positive definite since $1-c^2>0$ and the diagonal entries are $1$. Hence for $m\ge 3$ the link edge angular distance $\pi -\pi \mathrm{/}m\in (\pi \mathrm{/}2,\pi )$ and the mirror angle $\pi \mathrm{/}m\in (0,\pi \mathrm{/}2]$ are distinct and supplementary, and they are equal to $\pi \mathrm{/}2$ only in the square case $m=2$; in either case they must not be interchanged.