Plane subsystems, their canonical generators, and the angular order of their roots

By F6, $B$ is positive definite. Fact [F7] identifies $V$ with $V^{\ast }$ by $b(v)=B(v,\cdot )$ and gives $b(\rho (w)v)=w\cdot b(v)$; it also gives the vector descriptions of $C,C^{\circ },H_{\alpha }$. Thus the chamber tiling and stabilizer statements F8, F9 apply in this model. Put $S(v):=\{s\in S:B(v,e_s)=0\}$.

Finite-union base cases: if $N=0$, the empty union misses $0$ in every vector space; if $N=1$, a proper subspace $V_1⊊W$ misses a point of $W$ by definition.

Induction hypothesis of the finite-union lemma: for $N\ge 2$, assume that for every real vector space $W$ and every family of $N-1$ proper subspaces the union is not all of $W$.

Step of the finite-union lemma, $N\ge 2$: let $V_1,\dots ,V_N$ be proper subspaces of $W$. If $V_N\subseteq {\bigcup }_{i<N}{V}_i$, then ${\bigcup }_{i\le N}{V}_i={\bigcup }_{i<N}{V}_i\mathrm{\ne }W$ by step 1.3. Otherwise choose $u\in W\setminus {\bigcup }_{i<N}{V}_i$ (nonempty by step 1.3) and $w\in W\setminus V_N$ (nonempty since $V_N$ is proper), and form the line $L:=\{u+\lambda w:\lambda \in \mathbb{R}\}$; each $V_i$ meets $L$ in at most one point, because two distinct points of $L$ in $V_i$ give $w\in V_i$ and then $u\in V_i$. Hence at most $N$ values of $\lambda$ are excluded, and since $\mathbb{R}$ is infinite some $\lambda$ has $u+\lambda w\mathrm{\notin }{\bigcup }_{i\le N}{V}_i$; this proves the lemma for $N$, and every selection made is a single existential instantiation from a set already known to be nonempty, so no choice principle is used.

Orbit and stabilizer: by F8 the orbit $Wx$ meets $C$ in exactly one point $y$ (for $x=0$ one has $y=0$, and $0\in C$ by F11); fix $w\in W$ with $x=w\cdot y$. Then ${\mathrm{Stab}}_W(x)=w{\mathrm{Stab}}_W(y)w^{-1}=w{W}_{S(y)}{w}^{-1}$ by F9, where $S(y)=\{s:B(y,e_s)=0\}$. Hence $W^{\mathrm{\prime }}={\mathrm{Stab}}_W(x)$ is a conjugate of the standard parabolic $W_{S(y)}$.

Existence of $x$ and clause (5): if $n=2$, then $P=V$ and the family indexed by $\mathrm{\Phi }\setminus P$ is empty, so take $x=0$. If $n>2$, some simple root is outside $P$ because the simple roots span $V$, hence the finite family $P^{\perp }\cap H_{\alpha }$, $\alpha \in \mathrm{\Phi }\setminus P$, is nonempty. Each member is a proper subspace of $P^{\perp }$: $P^{\perp }\subseteq H_{\alpha }$ would mean $B(y,\alpha )=0$ for all $y\in P^{\perp }$, i.e. $\alpha \in (P^{\perp }{)}^{\perp }=P$, contrary to $\alpha \mathrm{\notin }P$, since $B$ is positive definite. If there is one such subspace, step 1.2 supplies a point outside it; if there are at least two, step 2.1 supplies a point outside their union. This gives $x\in P^{\perp }$ with $B(x,\alpha )\mathrm{\ne }0$ for every root $\alpha \in \mathrm{\Phi }\setminus P$. This proves the existence asserted in the statement and shows that no Choice is used (clause (5)).

Roots of $W^{\mathrm{\prime }}$: for $\alpha \in \mathrm{\Phi }$ one has $t_{\alpha }\in W^{\mathrm{\prime }}$ if and only if $\alpha \in P$. If $\alpha \in \mathrm{\Phi }\cap P$, then $B(x,\alpha )=0$ because $x\in P^{\perp }$, so $r_{\alpha }(x)=x-\frac{2B(x,\alpha )}{B(\alpha ,\alpha )}\alpha =x$ by [F1], and $\rho (t_{\alpha })=r_{\alpha }$ by F5, hence $t_{\alpha }\in {\mathrm{Stab}}_W(x)=W^{\mathrm{\prime }}$. Conversely, if $t_{\alpha }\in W^{\mathrm{\prime }}$, then the same two formulas give $x=r_{\alpha }(x)=x-\frac{2B(x,\alpha )}{B(\alpha ,\alpha )}\alpha$, hence $B(x,\alpha )=0$ (as $\alpha \mathrm{\ne }0$), and the defining property of $x$ from step 3.1 forces $\alpha \in P$. In particular $\{\alpha \in \mathrm{\Phi }:t_{\alpha }\in W^{\mathrm{\prime }}\}=\mathrm{\Phi }\cap P$, and this proves the second display of clause (1).

Rank and span: put $J:=S(y)$ and ${\mathrm{\Phi }}_J:=\mathrm{\Phi }\cap V_J$. By F5 and F10, $t_{\alpha }\in w{W}_J{w}^{-1}⟺w^{-1}{t}_{\alpha }w\in W_J⟺t_{\rho (w{)}^{-1}\alpha }\in W_J⟺\rho (w{)}^{-1}\alpha \in \mathrm{\Phi }\cap V_J⟺\alpha \in \rho (w)(\mathrm{\Phi }\cap V_J)$; the third equivalence also uses $t_{-\gamma }=t_{\gamma }$ when $\gamma$ is negative. Thus the roots of the conjugate parabolic $w{W}_J{w}^{-1}$ are $\rho (w)(\mathrm{\Phi }\cap V_J)$. Comparing with step 4.1 gives $\rho (w)(\mathrm{\Phi }\cap V_J)=\mathrm{\Phi }\cap P$; since $e_s\in {\mathrm{\Phi }}_J$ for $s\in J$, the set ${\mathrm{\Phi }}_J$ spans $V_J$, and invertibility of $\rho (w)$ shows $\dim V_J=\dim \mathrm{span}(\mathrm{\Phi }\cap P)=\dim P=2$. Hence $\mathrm{\mid }J\mathrm{\mid }=\dim V_J=2$ because $(e_s{)}_{s\in J}$ is a basis of $V_J$, and $\mathrm{\Phi }\cap P$ spans $P$ because it equals the image of ${\mathrm{\Phi }}_J$. Thus $W^{\mathrm{\prime }}$ is a conjugate of a standard parabolic of rank two.

The finite dihedral model: $W_J=⟨s:s\in J⟩=⟨t_{e_s}:s\in J⟩$, and conjugating the generating set by $w$ gives $W^{\mathrm{\prime }}=⟨w{t}_{e_s}{w}^{-1}:s\in J⟩=⟨t_{\rho (w)e_s}:s\in J⟩$ by [F2, F5]; each $\rho (w)e_s$ lies in $\rho (w){\mathrm{\Phi }}_J=\mathrm{\Phi }\cap P=:R$, so $W^{\mathrm{\prime }}$ is contained in $⟨t_{\alpha }:\alpha \in R⟩$. Conversely every $t_{\alpha }$ for $\alpha \in R$ belongs to $W^{\mathrm{\prime }}$ by step 4.1, proving equality. Moreover $R={\mathrm{\Phi }}_P^{+}\bigsqcup (-{\mathrm{\Phi }}_P^{+})$ with ${\mathrm{\Phi }}_P^{+}:=\mathrm{\Phi }\cap P\cap V_{+}$ by [F4], and $R$ is finite by [F6].

Faithful plane action: $W^{\mathrm{\prime }}=w{W}_J{w}^{-1}$ preserves $P=\rho (w)V_J$, because $W_J$ preserves $V_J$ by F10. Since $B$ is positive definite, $V=V_J\oplus V_J^{\perp }$. Every generator $s\in J$ fixes $V_J^{\perp }$ pointwise by the reflection formula [F1] and $\rho (s)=r_{e_s}$ [F2]; hence every $v\in W_J$ fixes $V_J^{\perp }$ pointwise. If $u=wv{w}^{-1}$ acts trivially on $P$, then $\rho (v)$ acts trivially on $V_J=\rho (w{)}^{-1}P$ and on $V_J^{\perp }$, so it is the identity on $V$ and $v=1$ by [F12]. Thus $G:=\rho (W^{\mathrm{\prime }}){\mathrm{\mid }}_P$ is faithful.

Orthogonal plane action: $G$ is finite by [F6] and is contained in $O(P)$ because $W^{\mathrm{\prime }}$ preserves $P$ and is generated by the B-isometric reflections from step 6.1 and [F3]. Each $t_{\alpha }$ with $\alpha \in R$ acts as a nontrivial orthogonal reflection on $P$: [F5] gives $B(\alpha ,\alpha )=1$, so its normal line lies in $P$ and its restriction fixes the one-dimensional orthogonal line and negates $\alpha$.

Identify plane reflections with group reflections: take $g\in G$ with determinant $-1$ and let $u\in W^{\mathrm{\prime }}$ be its unique preimage under the faithful action of step 7.1. Every determinant-$-1$ map in $O(P)$ is a reflection, since its eigenvalues are $1$ and $-1$. By step 6.1, $W^{\mathrm{\prime }}$ is generated by $t_{\alpha }$ with $\alpha \in R\subseteq P$; each $\rho (t_{\alpha })=r_{\alpha }$ fixes $P^{\perp }$ pointwise by [F1]. By positive definiteness [F7], $V=P\oplus P^{\perp }$, so $\rho (u)$ is the reflection $g$ on $P$ and the identity on $P^{\perp }$, hence has fixed hyperplane $M$. If $M$ is not a root hyperplane, each $M\cap H_{\beta }$ is a proper subspace of $M$; the arrangement is finite by [F7]. Applying the finite-union lemma from step 2.1 inside $M$ gives $z\in M$ outside every root hyperplane. Choose $v\in W$ with $y:=v^{-1}\cdot z\in C$ using F8. The arrangement is $W$-invariant, so $y$ also avoids every root hyperplane; because $y\in C$, this makes $y\in C^{\circ }$, and F9 gives ${\mathrm{Stab}}_W(y)=\{1\}$. Since $z=v\cdot y$, its stabilizer is conjugate to the trivial stabilizer of $y$, contradicting $u\mathrm{\ne }1$ and $z\in M=\mathrm{Fix}(\rho (u))$. Therefore $M=H_{\alpha }$ for some root $\alpha$. The unique $B$-orthogonal reflection with fixed hyperplane $H_{\alpha }$ is $r_{\alpha }$, so [F5] gives $\rho (u)=r_{\alpha }=\rho (t_{\alpha })$ and faithfulness [F12] yields $u=t_{\alpha }\in T$; step 4.1 forces $\alpha \in P$. Conversely every $t_{\alpha }$ with $\alpha \in \mathrm{\Phi }\cap P$ lies in $W^{\mathrm{\prime }}$ by step 4.1 and acts as a reflection on $P$. Hence the determinant-$-1$ elements of $G$ correspond exactly to $T\cap W^{\mathrm{\prime }}$.

Finite orthogonal plane groups: $R$ spans $P$, so choose two nonproportional roots in $R$. Their reflections restrict to distinct reflections of $G$ by steps 6.1 and 8.1, and their product is a nonidentity rotation and the rotation subgroup $H:=G\cap SO(P)$ is nontrivial. The determinant maps $G$ onto $\{\pm 1\}$, with kernel $H$, so $\mathrm{\mid }G\mathrm{\mid }=2\mathrm{\mid }H\mathrm{\mid }$. Let $\theta$ be the least positive rotation angle in the finite group $H$. For any angle $\phi \in (0,2\pi )$ of an element of $H$, division by $\theta$ gives $\phi =d\theta +\delta$ with $0\le \delta <\theta$; the rotation of angle $\delta$ is in $H$, so minimality forces $\delta =0$. Dividing $2\pi$ by $\theta$ likewise gives $2\pi =N\theta +\delta$ with $0\le \delta <\theta$; the inverse of the rotation through $N\theta$ has angle $\delta$, so again $\delta =0$. Thus $h$, the rotation through $\theta$, has exact order $N$ and every element of $H$ is a power of $h$, so $N=\mathrm{\mid }H\mathrm{\mid }=:k\ge 2$ and $\theta =2\pi \mathrm{/}k$. The coset $Ht$ for any reflection $t\in G$ consists of all $k$ orientation-reversing orthogonal maps, each a reflection in a line of $P$. If ${\theta }_0$ is the angle of a unit normal to the reflection line of $t$, then the unit normal to $h^{j}t$ has angle ${\theta }_0+j\pi \mathrm{/}k$; including both orientations gives $2k$ equally spaced normal directions.

Angular order and count: let $k:=\mathrm{\mid }H\mathrm{\mid }=\mathrm{\mid }T\cap W^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Phi }}_{+}\cap P\mathrm{\mid }$ by steps 9.1 and 8.2 and the positive-root/reflection bijection [F5]. Put $C^{\mathrm{\prime }}:=\mathrm{cone}(R\cap V_{+})$; since $R\cap V_{+}$ spans $P$ by steps 5.1 and 6.1, it is a pointed, finitely generated full-dimensional cone in the plane and has exactly two extreme rays, each containing a generator $r_1,r_2\in R\cap V_{+}$. Its positive roots lie in the closed angular sector between those rays, and a root in that sector is positive, so this sector contains exactly the $k$ positive roots counted above. The normal lines are spaced by $\pi \mathrm{/}k$ by step 9.1; therefore these $k$ roots occupy consecutive directions, and the sector has angle $(k-1)\pi \mathrm{/}k$. The unit normals $r_1,r_2$ therefore have angle $(k-1)\pi \mathrm{/}k$, so the product of their linear reflections $r_{r_1}{r}_{r_2}$ is a rotation through $2\pi \mathrm{/}k$, of exact order $k$. Since $\rho (ab)=r_{r_1}{r}_{r_2}$ and $\rho$ is injective by [F12], $m:=\mathrm{ord}(ab)=k$.

The alternating list: put $a:=t_{r_1}$, $b:=t_{r_2}$ and $q:=ab$. Step 10.1 gives $m=k$ and the angle between the unit roots $r_1,r_2$ as $\pi -\pi \mathrm{/}m$, whence $B(r_1,r_2)=-\cos (\pi \mathrm{/}m)$. The conjugate formulas in the Statement show each $u_j$ is a reflection in $W^{\mathrm{\prime }}$. Their positive roots satisfy ${\beta }_{u_1}=r_1$ and ${\beta }_{u_2}=\rho (a)r_2=r_{r_1}(r_2)=r_2+2\cos (\pi \mathrm{/}m)r_1$, which has angle $\pi \mathrm{/}m$ from $r_1$. For $2\le j<m$, the alternating-word identity $u_{j+1}=q{u}_{j-1}{q}^{-1}$ and the root-conjugation identity [F5] give the root $\rho (q){\beta }_{u_{j-1}}$ for $u_{j+1}$; its angle is $j\pi \mathrm{/}m<\pi$, so it is the positive root ${\beta }_{u_{j+1}}$. By step 10.1, $\rho (q)$ rotates through $2\pi \mathrm{/}m$. Starting from $u_1,u_2$, induction now gives ${\beta }_{u_j}$ at angle $(j-1)\pi \mathrm{/}m$ from $r_1$ for all $1\le j\le m$. These are the $m=k$ consecutive positive roots; at $j=m$ the vector is the unit root on the ray of $r_2$, hence equals $r_2$ and [F5] gives $u_m=b$. The conjugate formulas show odd-indexed $u_j$ are conjugate to $a$ and even-indexed $u_j$ to $b$.

Clauses (2) and (3): by steps 8.2 and 11.1, the $2k$ roots of $R$ are $\{\pm {\beta }_{u_1},\dots ,\pm {\beta }_{u_m}\}$ with $m=k$ and distinct positive roots, so this is all of $\mathrm{\Phi }\cap P$ and its positive roots are ordered from the ray $r_1$ to $r_2$ between its two extreme rays. By step 6.1, $W^{\mathrm{\prime }}$ is generated by its reflections $T\cap W^{\mathrm{\prime }}$; steps 8.2 and 11.1 together with [F5] identify that set with the alternating elements $u_j$, each a word in $a,b$, so $W^{\mathrm{\prime }}=⟨a,b⟩$. The involutions $a,b$ with product of order $m$ give a surjection from the dihedral group of order $2m$ onto $W^{\mathrm{\prime }}$, and $\mathrm{\mid }{W}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }=2k=2m$ by steps 7.1, 9.1 and 10.1; hence this surjection is an isomorphism. Every reflection is conjugate to $a$ or $b$ by step 11.1. For the canonical-system characterization stated in (3), it remains to check positive spanning and extremality. The roots in ${\mathrm{\Phi }}_P^{+}$ lie in the cone generated by the extreme roots $r_1,r_2$, so condition (i) holds. Extremality gives condition (ii): if $r_i$ were a nonnegative combination of other positive subsystem roots, every nonzero summand would have to lie on its extreme ray; since every root has norm one, the only positive root on that ray is $r_i$ itself, a contradiction. Thus $\{r_1,r_2\}$ is the canonical system of $W^{\mathrm{\prime }}$.

Clause (4) and the reversal: a $2$-dimensional subspace $Q\subseteq P$ equals $P$, so $\mathrm{\Phi }\cap Q=\mathrm{\Phi }\cap P$; therefore the extreme rays, rank-two subsystem, reflection subgroup, and angular order constructed in $Q$ are the same as those already established in steps 10.1--12.1 and 6.1. If the extreme rays are exchanged, let $r_1^{\mathrm{\prime }}:=r_2$ and $r_2^{\mathrm{\prime }}:=r_1$. Repeating the calculation of step 11.1 with the rays exchanged (so the product is $ba=(ab{)}^{-1}$) gives the new alternating roots at angle $(j-1)\pi \mathrm{/}m$ from $r_2$, hence at angle $(m-j)\pi \mathrm{/}m$ from $r_1$; this is the angle of ${\beta }_{u_{m+1-j}}$. The root-reflection bijection [F5] then gives $u_j^{\mathrm{\prime }}=u_{m+1-j}$, so the index order reverses.

The finite-union induction is discharged by steps 1.2, 1.3 and 2.1, and the alternating-root induction by step 11.1; together with steps 3.1--10.1, 12.1, and 13.1 these establish clauses (1)--(5). The proof uses no Choice.