Bipartite Coxeter Elements and Ordered Root Complexes

Write $\mathrm{\parallel }x{\mathrm{\parallel }}_0^2:={\sum }_{s\in S}{x}_s^2$ for the standard coordinate norm and set $q(x):=x^{\mathsf{T}}Ax$. If $m(s,t)=\mathrm{\infty }$ for distinct $s,t$, then $B(e_s+e_t,e_s+e_t)=0$, contradicting positive definiteness; hence every off-diagonal Coxeter exponent is finite. Therefore $A_{ss}=0$ and $A_{st}=-2B(e_s,e_t)=2\cos (\pi \mathrm{/}m(s,t))\ge 0$ for $s\mathrm{\ne }t$, with $A_{st}>0$ exactly on the edges of the connected diagram. Also $q(x)=2\mathrm{\parallel }x{\mathrm{\parallel }}_0^2-2B(x,x)<2\mathrm{\parallel }x{\mathrm{\parallel }}_0^2$ for every nonzero $x$.

For $1\le i\le n$, the first $n$ prefix vectors satisfy ${\rho }_i={\alpha }_i$ for $i\le r$ and ${\rho }_i=\rho (a){\alpha }_i={\alpha }_i+{\sum }_{s\in J}{A}_{si}{\alpha }_s$ for $i>r$: the earlier reflections in the same color class fix ${\alpha }_i$, while each reflection in the other class adds its $A_{si}{\alpha }_s$ and fixes the other roots of that class. Hence the coordinate matrix of $({\rho }_1,\dots ,{\rho }_n)$ relative to the ordered basis $({\alpha }_1,\dots ,{\alpha }_n)$ is block upper triangular with identity diagonal blocks, so these vectors form a basis. The cyclic definitions give ${\rho }_{i+n}=C_V{\rho }_i$ and ${\mu }_{i+n}=C_V{\mu }_i$ for every $i\ge 1$, because each block of $n$ reflections has product $c$. Each ${\rho }_i$ is a root, being the image of a simple root by an element of $W$.

The unit sphere for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_0$ is compact, so $q$ attains a maximum $\lambda$ there; normalizing $e_s+e_t$ for an edge $\{s,t\}$ gives a positive value of $q$, so $\lambda >0$, and step 1.1 at a maximizer gives $\lambda <2$. Let $p$ be a unit maximizer. For $y$ with $⟨p,y{⟩}_0=0$, maximality at $(p+ty)\mathrm{/}(1+t^2\mathrm{\parallel }y{\mathrm{\parallel }}_0^2{)}^{1\mathrm{/}2}$ gives $2t⟨Ap,y{⟩}_0+t^2(q(y)-\lambda \mathrm{\parallel }y{\mathrm{\parallel }}_0^2)\le 0$ for every real $t$. If $⟨Ap,y{⟩}_0\mathrm{\ne }0$, a sufficiently small $t$ of the same sign makes the linear term dominate the quadratic term, a contradiction. Thus $Ap$ is orthogonal to every $y\perp p$, so it is a multiple of $p$; pairing with $p$ gives $Ap=\lambda p$.

For $1\le i\le n$, duality gives $B({\beta }_i,{\alpha }_j)=0$ for every $j\mathrm{\ne }i$, so $R_j{\beta }_i={\beta }_i$ for every $j\mathrm{\ne }i$ and in particular ${\mu }_i=R_1\cdots R_{i-1}{\beta }_i={\beta }_i$. Also $R_i{\beta }_i={\beta }_i-2{\alpha }_i$. Thus $C_V{\beta }_i=R_1\cdots R_n{\beta }_i=R_1\cdots R_{i-1}({\beta }_i-2{\alpha }_i)={\mu }_i-2{\rho }_i={\beta }_i-2{\rho }_i$, so $(C_V-{\mathrm{i}\mathrm{d}}_V){\beta }_i=-2{\rho }_i$.

Since every entry of $A$ is nonnegative, $q(\mathrm{\mid }p\mathrm{\mid })\ge q(p)=\lambda$, while $\mathrm{\parallel }\mathrm{\mid }p\mathrm{\mid }{\mathrm{\parallel }}_0=1$, so $\mathrm{\mid }p\mathrm{\mid }$ is also a maximizer and satisfies $A\mathrm{\mid }p\mathrm{\mid }=\lambda \mathrm{\mid }p\mathrm{\mid }$ by step 2.1. Put $r:=\mathrm{\mid }p\mathrm{\mid }$. If $r_s>0$ and $A_{st}>0$, then $\lambda r_t={\sum }_u{A}_{tu}{r}_u\ge A_{ts}{r}_s>0$; since $\lambda >0$ and the edge graph of $A$ is connected, positivity propagates to every coordinate. Thus we may take $p\in {\mathbb{R}}_{>0}^S$, proving the existence claim in (1).

The vectors ${\beta }_1,\dots ,{\beta }_n$ form a basis by the construction in the preceding definition, and ${\rho }_1,\dots ,{\rho }_n$ form a basis by 1.2; thus 2.2 shows that $C_V-{\mathrm{i}\mathrm{d}}_V$ maps one basis to the other up to the nonzero scalar $-2$, so it is invertible. If $i=mn+j$ with $m\ge 0$ and $1\le j\le n$, then the cyclic recursions in F1 and the identity in 2.2 give $(C_V-{\mathrm{i}\mathrm{d}}_V){\mu }_i=C_V^m(C_V-{\mathrm{i}\mathrm{d}}_V){\mu }_j=-2C_V^m{\rho }_j=-2{\rho }_i$. Applying the conditional map of the preceding definition now yields $\mu ({\rho }_i)={\mu }_i$ for all $i\ge 1$, proving (4) and the definition's well-definedness justification.

Put $d^{\mathrm{\prime }}:={\sum }_{t\in K}{p}_t^2$. Since distinct simple roots in either color class are orthogonal, $B(u,u)=d$ and $B(v,v)=d^{\mathrm{\prime }}$. Multiplying the eigenvector equations ${\sum }_{t\in K}{A}_{st}{p}_t=\lambda p_s$ by $p_s$ and summing over $s\in J$ gives ${\sum }_{s\in J,t\in K}{p}_s{A}_{st}{p}_t=\lambda d$; summing the equations indexed by $K$ gives the same left side equal to $\lambda d^{\mathrm{\prime }}$, hence $d=d^{\mathrm{\prime }}>0$. Finally $B(u,v)=-\frac{1}{2}{\sum }_{s\in J,t\in K}{p}_s{A}_{st}{p}_t=-\lambda d\mathrm{/}2$. The Gram matrix on $(u,v)$ is $d\left(\begin{array}{cc}1& -\lambda \mathrm{/}2\\ -\lambda \mathrm{/}2& 1\end{array}\right)$ with determinant $d^2(1-{\lambda }^2\mathrm{/}4)>0$, so $u,v$ are independent and $P$ is a plane.

For $s\in J$, $R_s$ fixes ${\alpha }_{s^{\mathrm{\prime }}}$ for $s^{\mathrm{\prime }}\in J\setminus \{s\}$ and negates ${\alpha }_s$; the corresponding statement holds in $K$. For $t\in K$, the commuting reflections in $a$ give $\rho (a){\alpha }_t={\alpha }_t+{\sum }_{s\in J}{A}_{st}{\alpha }_s$, and for $s\in J$, $\rho (b){\alpha }_s={\alpha }_s+{\sum }_{t\in K}{A}_{ts}{\alpha }_t$. Summing with coefficients $p_t$ and $p_s$ and using $Ap=\lambda p$ yields $\rho (a)u=-u$, $\rho (a)v=v+\lambda u$, $\rho (b)v=-v$, and $\rho (b)u=u+\lambda v$. Thus $P$ is invariant under $\rho (a),\rho (b)$ and $C_V=\rho (a)\rho (b)$.

In the basis $(u,v)$, $\rho (a)=\left(\begin{array}{cc}-1& \lambda \\ 0& 1\end{array}\right)$, $\rho (b)=\left(\begin{array}{cc}1& 0\\ \lambda & -1\end{array}\right)$, and $C_V{\mathrm{\mid }}_P=\left(\begin{array}{cc}{\lambda }^2-1& -\lambda \\ \lambda & -1\end{array}\right)$. The vectors $f_1:=u\mathrm{/}\sqrt{d}$ and $f_2:=(v+\frac{\lambda }{2}u)\mathrm{/}\sqrt{d(1-{\lambda }^2\mathrm{/}4)}$ are $B$-orthonormal, and direct substitution gives $C_V{f}_1=({\lambda }^2\mathrm{/}2-1)f_1+\lambda \sqrt{1-{\lambda }^2\mathrm{/}4}{f}_2$. With $\theta :=\arccos (\lambda \mathrm{/}2)\in (0,\pi \mathrm{/}2)$ these coefficients are $\cos (2\theta )$ and $\sin (2\theta )$; since $C_V$ preserves $B$ and has determinant $1$ on $P$, it is the rotation through $2\theta$.

For $x=\xi u+\eta v\in P$, one has $B(x,{\alpha }_s)=p_s(\xi -\frac{\lambda }{2}\eta )$ for $s\in J$ and $B(x,{\alpha }_t)=p_t(\eta -\frac{\lambda }{2}\xi )$ for $t\in K$. Thus $\mathcal{C}\cap P$ is given by $\xi \ge \frac{\lambda }{2}\eta$ and $\eta \ge \frac{\lambda }{2}\xi$; because $0<\lambda \mathrm{/}2<1$, these inequalities imply $\xi ,\eta \ge 0$, and their strict versions place the relative interior in ${\mathcal{C}}^{\circ }$. Its boundary rays are $x_J:=\lambda u+2v$ (the $J$ simple-root hyperplanes) and $x_K:=2u+\lambda v$ (the $K$ simple-root hyperplanes). Their squared $B$-norms are both $d(4-{\lambda }^2)$ and $B(x_J,x_K)=\lambda d(2-{\lambda }^2\mathrm{/}2)$, so the sector angle has cosine $\lambda \mathrm{/}2$ and equals $\theta$. The restrictions of $\rho (a)$ and $\rho (b)$ are reflections in the lines $\mathbb{R}{x}_J$ and $\mathbb{R}{x}_K$: they fix those lines, are involutions, and are nonidentity on $P$ by 4.1.

Let $m_0$ be the order of $C_V{\mathrm{\mid }}_P$. Since it rotates through $2\theta \in (0,\pi )$, $m_0\ge 3$ and $2\theta =2\pi j\mathrm{/}{m}_0$ for an integer $j$ with $1\le j<m_0\mathrm{/}2$ and $\gcd (j,m_0)=1$. The subgroup $D:=⟨\rho (a){\mathrm{\mid }}_P,\rho (b){\mathrm{\mid }}_P⟩$ consists of the $m_0$ distinct rotations $(C_V{\mathrm{\mid }}_P{)}^k$ and the $m_0$ distinct reflections $\rho (a){\mathrm{\mid }}_P(C_V{\mathrm{\mid }}_P{)}^k$, so $\mathrm{\mid }D\mathrm{\mid }=2m_0$. Distinct elements of $D$ have lifts $\tilde{g}\in ⟨a,b⟩$ with distinct restrictions to $P$, hence are distinct elements of $W$; by F4 their chamber interiors are distinct and disjoint. Thus the orbit of $S_0:=\mathcal{C}\cap P$ consists of $2m_0$ distinct sectors of angle $\theta$, each of the form $\rho (\tilde{g})\mathcal{C}\cap P$, with pairwise disjoint relative interiors. Hence $2m_0\theta \le 2\pi$, so $j=1$ and $\theta =\pi \mathrm{/}{m}_0$; equality shows the sectors cover $P$. As $m_0\mid h$ by F1, the element $c^{m_0}$ fixes every point of $P$, in particular a point of ${\mathcal{C}}^{\circ }\cap P$; its stabiliser is trivial, so $c^{m_0}=1$ and $h\mid m_0$. Thus $m_0=h$ and $\lambda =2\cos (\pi \mathrm{/}h)$, completing (1) and the angle claims of (2).

For every root $\alpha$, the trace $H_{\alpha }\cap P$ is a line: if $P\subseteq H_{\alpha }$, the nonidentity reflection $t_{\alpha }$ fixes a point of ${\mathcal{C}}^{\circ }\cap P$, contradicting its trivial stabiliser; otherwise the nonzero linear functional $x\mapsto B(x,\alpha )$ restricts to a nonzero functional on the two-plane $P$, whose kernel is a line. Each such line is a root-hyperplane trace and cannot meet the relative interior of any sector $\rho (\tilde{g})S_0$, because that relative interior lies in the chamber interior $\rho (\tilde{g}){\mathcal{C}}^{\circ }$, which is disjoint from the arrangement. Conversely the two boundary lines of $S_0$ are traces of simple-root hyperplanes by 7.1, and their images under $D$ are root-hyperplane traces because $\rho (W)$ permutes roots. These images are all the sector boundary lines: the $2h$ sectors from 8.1 tile $P$ and have $2h$ distinct boundary rays, while every root trace must be one of those lines because it cannot meet a sector interior. Opposite rays define the same line, so the traces are exactly $h$ distinct lines. The connected components of the complement of these lines are the sector interiors. Each component is connected and the ambient chamber interiors form a disjoint open cover of it, so it lies in one chamber interior. Conversely, if a full-dimensional chamber section contained points from two components, choose them in its relative interior; their segment lies in that relative interior by convexity and crosses a root trace, contradicting disjointness of chamber interiors from the arrangement. Hence the full-dimensional chamber sections are exactly the $2h$ sectors.

The ray ${\mathbb{R}}_{>0}{x}_J$ lies in the relative interior of the face ${\mathcal{C}}_J$, so its stabiliser is $W_J$; similarly the stabiliser of ${\mathbb{R}}_{>0}{x}_K$ is $W_K$. Every element of $W_J$ reduces to a product of a subset $I\subseteq J$ because its simple generators are commuting involutions; the subset products are distinct because their $\rho$-images have different signs on the independent simple roots and $\rho$ is faithful by [F2]. The image of the product for $I$ acts by $-1$ on $\mathrm{span}\{{\alpha }_s:s\in I\}$ and by $+1$ on its $B$-orthogonal complement, so its $(-1)$-eigenspace has dimension $\mathrm{\mid }I\mathrm{\mid }$; therefore it is a reflection exactly when $\mathrm{\mid }I\mathrm{\mid }=1$, and the reflections in $W_J$ are precisely its $\mathrm{\mid }J\mathrm{\mid }=r$ simple generators. The same reasoning gives exactly $\mathrm{\mid }K\mathrm{\mid }=n-r$ reflections in $W_K$. The $2h$ distinct boundary rays from 8.1 alternate between the two types, so there are $h$ of each; conjugating the stabiliser count around them gives $hr+h(n-r)=hn$ incidences between rays and ambient root hyperplanes. Each hyperplane contributes exactly two incidences because its trace is a line. Hence the arrangement has $hn\mathrm{/}2$ hyperplanes, so $\mathrm{\mid }T\mathrm{\mid }=hn\mathrm{/}2$ by the root-reflection correspondence of [F7].

If $h$ is odd, a semicircle from a point of ${\mathcal{C}}^{\circ }\cap P$ to its antipode crosses each trace line once; since every root-hyperplane trace is one of these lines, it meets each ambient root hyperplane exactly once. The boundary-ray types alternate, so one type occurs $(h+1)\mathrm{/}2$ times and the other $(h-1)\mathrm{/}2$ times. Counting the hyperplanes met using 10.1 gives $hn\mathrm{/}2=((h+1)\mathrm{/}2)r+((h-1)\mathrm{/}2)(n-r)$ if the more frequent rays have type $J$, and the same equation with $r$ and $n-r$ interchanged if they have type $K$. Either equation gives $r=n\mathrm{/}2$. Thus $n$ is even and $\mathrm{\mid }J\mathrm{\mid }=\mathrm{\mid }K\mathrm{\mid }=n\mathrm{/}2$.

Choose the direction around the circle so that the first boundary ray of $S_0$ is $x_J$. The rays then occur in the cyclic order $p_{2k}=C_V^k{\mathbb{R}}_{>0}{x}_J$ and $p_{2k+1}=C_V^k\rho (a){\mathbb{R}}_{>0}{x}_K$ for $k=0,\dots ,h-1$: $\rho (a)$ reflects across the $J$ boundary, while $C_V$ advances by two sector angles. The hyperplanes through $p_{2k}$ are exactly $H_{C_V^k{\alpha }_s}$ for $s=s_1,\dots ,s_r$, and those through $p_{2k+1}$ are exactly $H_{C_V^k\rho (a){\alpha }_t}$ for $t=s_{r+1},\dots ,s_n$, by the stabiliser counts and the description of the reflections in $W_J,W_K$ from 10.1. These are precisely the blocks $H_{{\rho }_{kn+1}},\dots ,H_{{\rho }_{kn+r}}$ and $H_{{\rho }_{kn+r+1}},\dots ,H_{{\rho }_{(k+1)n}}$ by the cyclic prefix definition in 1.2. The first $h$ rays carry $hn\mathrm{/}2$ hyperplanes: this is $hn\mathrm{/}2$ from $h\mathrm{/}2$ complete color pairs when $h$ is even, and $n(h-1)\mathrm{/}2+r=hn\mathrm{/}2$ when $h$ is odd by 11.1. The opposite semicircle carries the other $hn\mathrm{/}2$ hyperplanes. Each half-turn list is pairwise distinct because every trace line meets that semicircle only once, and the listed hyperplanes in each boundary block are distinct simple-root images.

Fix $k\le hn\mathrm{/}2$. The first half-turn list of hyperplanes in 12.1 is pairwise distinct, so ${\rho }_k\mathrm{\ne }\pm {\rho }_j$ for every $j<k$. For $1\le j\le k$, put $y_j:=R_{j-1}\cdots R_1{\rho }_k$, with $y_1={\rho }_k$ and $y_k={\alpha }_k$; for $j<k$, $y_j\mathrm{\ne }\pm {\alpha }_j$, since otherwise applying $R_1\cdots R_{j-1}$ would give ${\rho }_k=\pm {\rho }_j$. Each $R_j$ preserves the sign of every root other than $\pm {\alpha }_j$ by [F3], so $y_j$ and $y_{j+1}=R_j{y}_j$ have the same sign. Since $y_k={\alpha }_k\in {\mathrm{\Phi }}_{+}$, descending induction gives ${\rho }_k\in {\mathrm{\Phi }}_{+}$ for every $k\le hn\mathrm{/}2$.

Let $q:=hn\mathrm{/}2$. The word $w=c^{h\mathrm{/}2}$ for even $h$, or $w=c^{(h-1)\mathrm{/}2}a$ for odd $h$, has exactly $q$ letters; in the odd case this uses $r=n\mathrm{/}2$ from 11.1. Its prefix roots are ${\rho }_1,\dots ,{\rho }_q$ by the cyclic definition, and these are positive by 13.1. At each step the root-length criterion therefore increases the prefix length by one, so the word is reduced and $\mathrm{\ell }(w)=q=\mathrm{\ell }(w_0)$. By uniqueness of the longest element, $w=w_0$, proving the formulas for $w_0$ in (3).