Finite inversion sets are recognized by their rank-two initial or final segments

Rank-two setup. By [F10], the finite-type hypothesis gives positive definiteness of $B$. For every root-spanned plane $P$, [F1] identifies the subgroup generated by its root reflections with a finite dihedral group $W_P$, and identifies its positive roots with the angular list ${\beta }_{u_1},\dots ,{\beta }_{u_m}$. Since ${\mathrm{\Phi }}_{+}\to T$ is bijective by [F4], for every positive root $\alpha$ one has $t_{\alpha }\in W_P$ exactly when $\alpha \in {\mathrm{\Phi }}_P^{+}$. No choice of a family of points or subsystems is made.

Global closure of inversion sets. Suppose $\alpha ,\beta \in {\mathrm{\Phi }}_{+}$, $a,b>0$, and $\gamma :=a\alpha +b\beta \in {\mathrm{\Phi }}_{+}$. If $\alpha ,\beta \in N(w)$, then $\rho (w)\alpha ,\rho (w)\beta \in {\mathrm{\Phi }}_{-}\subseteq -V_{+}$, so $\rho (w)\gamma =a\rho (w)\alpha +b\rho (w)\beta$ is a nonzero vector in $-V_{+}$; since it is a root, [F2] gives $\rho (w)\gamma \in {\mathrm{\Phi }}_{-}$ and $\gamma \in N(w)$. If instead $\alpha ,\beta \mathrm{\notin }N(w)$, their images are positive roots by [F2], so $\rho (w)\gamma$ is a nonzero vector in $V_{+}$ and the same sign criterion gives $\rho (w)\gamma \in {\mathrm{\Phi }}_{+}$; hence $\gamma \mathrm{\notin }N(w)$.

Reflection stability, clause (3)(c). Let $e_s\in I$ and put $I^{\mathrm{\prime }}:=s(I\setminus \{e_s\})$. By [F2], $I^{\mathrm{\prime }}\subseteq {\mathrm{\Phi }}_{+}$ and is finite. Fix a noncommutative root plane $P$. If $e_s\mathrm{\notin }P$ and $P\subseteq e_s^{\perp }$, then [F9] shows $\rho (s)=r_{e_s}$ fixes $P$ pointwise, so $I^{\mathrm{\prime }}\cap {\mathrm{\Phi }}_P^{+}=I\cap {\mathrm{\Phi }}_P^{+}$ is a segment. If $e_s\mathrm{\notin }P$ and $P\subseteq ̸e_s^{\perp }$, put $P^{\mathrm{\prime }}:=\rho (s)P$. Then $e_s\mathrm{\notin }{P}^{\mathrm{\prime }}$ because $e_s\in P^{\mathrm{\prime }}$ would imply $\rho (s)e_s=-e_s\in P$ and hence $e_s\in P$. The map $\rho (s)$ bijects ${\mathrm{\Phi }}_{P^{\mathrm{\prime }}}^{+}$ with ${\mathrm{\Phi }}_P^{+}$ and preserves or reverses their angular order; therefore $I^{\mathrm{\prime }}\cap {\mathrm{\Phi }}_P^{+}=\rho (s)(I\cap {\mathrm{\Phi }}_{P^{\mathrm{\prime }}}^{+})$ is a segment by (ii). If $e_s\in P$, it is an extreme positive root of this subsystem: the simple root $e_s$ spans an extreme ray of $V_{+}$, while [F1] puts all positive roots of ${\mathrm{\Phi }}_P$ in the sector generated by the two extreme roots of $P$; if $e_s$ were strictly inside that sector, its unique nonnegative simple-root coordinates would force both extreme roots onto the same ray ${\mathbb{R}}_{>0}{e}_s$, impossible. Orient the angular list so $e_s={\beta }_1$. The reflection $\rho (s)$ reverses the angular order and permutes the positive roots other than $e_s$, so its order-reversing bijection sends ${\beta }_j$ to ${\beta }_{m+2-j}$ for $2\le j\le m$. Since $I\cap {\mathrm{\Phi }}_P^{+}$ is a segment containing ${\beta }_1$, it is $\{{\beta }_1,\dots ,{\beta }_q\}$; deleting ${\beta }_1$ and reflecting gives the final segment $\{{\beta }_{m+2-q},\dots ,{\beta }_m\}$, with the empty case when $q=1$. If $e_s={\beta }_m$, reverse the angular order and obtain the initial-segment counterpart. Thus $I^{\mathrm{\prime }}$ satisfies (ii).

Rank-two closure, clause (3)(a). Fix $P$ and write ${\beta }_i:={\beta }_{u_i}$. If $m=2$, the subsystem has only its two orthogonal positive root rays; a positive combination of two distinct roots on these rays is not a root in the subsystem, and a root on either ray has unit norm, so closure is immediate. If $m>2$, condition (ii) makes $I\cap {\mathrm{\Phi }}_P^{+}$ an initial or final segment or empty, and its complement within ${\mathrm{\Phi }}_P^{+}$ is also a segment of one of these forms. The positive roots lie in a pointed sector of angle less than $\pi$ by step 1.1. If distinct roots ${\beta }_i,{\beta }_j$ with $i<j$ are given, every root direction strictly between their rays is a positive combination of them: in angular coordinates ${\theta }_i<{\theta }_k<{\theta }_j$, the unit vector on ray ${\theta }_k$ equals $\frac{\sin ({\theta }_j-{\theta }_k)}{\sin ({\theta }_j-{\theta }_i)}{\beta }_i+\frac{\sin ({\theta }_k-{\theta }_i)}{\sin ({\theta }_j-{\theta }_i)}{\beta }_j$, whose coefficients are positive. Thus a positive-root combination that is a root lies between its two distinct input rays, or is the same root when the inputs are proportional. Each initial or final segment contains every listed root between two of its members, so both the segment and its complement are closed as claimed.

Inversion sets satisfy the rank-two condition, (1)(i)$\Rightarrow$(ii). Let $I=N(w)$ and fix a noncommutative $W_P$. If ${\beta }_i,{\beta }_j\in I\cap {\mathrm{\Phi }}_P^{+}$ with $i<j$, every intermediate ${\beta }_k$ is a positive combination of these roots, so lies in $I$ by step 1.2. Thus the intersection is empty or a consecutive block $\{{\beta }_p,\dots ,{\beta }_q\}$. If $p>1$ and $q<m$, then ${\beta }_{p-1},{\beta }_{q+1}\mathrm{\notin }I$ and ${\beta }_p$ is a positive combination of them, contradicting the complement closure of step 1.2. Therefore $p=1$ or $q=m$, which proves (ii).

A nonempty set satisfying (ii) contains a simple root, clause (3)(b). Suppose to the contrary that $I\mathrm{\ne }\mathrm{\varnothing }$ has no simple root. Choose $\alpha ={\sum }_{t\in S}{a}_t{e}_t\in I$ of minimum height $\mathrm{ht}(\alpha ):={\sum }_t{a}_t$, where $a_t\ge 0$ are the unique simple-root coordinates. Then $\alpha$ is not simple. Since $1=B(\alpha ,\alpha )={\sum }_t{a}_{t}B(\alpha ,e_t)$ by [F3], there is $s$ with $a_s>0$ and $B(\alpha ,e_s)>0$. Put $\beta :=\rho (s)\alpha =\alpha -2B(\alpha ,e_s)e_s$ by [F9]. By [F2], $\beta \in {\mathrm{\Phi }}_{+}\setminus \{e_s\}$, and its height is strictly less than that of $\alpha$, so $\beta \mathrm{\notin }I$ by minimality; also $e_s\mathrm{\notin }I$. The roots $\alpha ,e_s$ are distinct and nonproportional, and $P_0:=\mathrm{span}(\alpha ,e_s)$ is a root plane. Invariance of $B$ and $\rho (s)e_s=-e_s$ give $B(\beta ,e_s)=-B(\alpha ,e_s)<0$. Thus the two distinct reflections $t_{\beta }$ and $s=t_{e_s}$ do not commute: in the positive-definite plane by step 1.1 their normal lines are neither equal nor orthogonal, and distinct orthogonal reflections commute only when their normal lines are perpendicular. Hence $W_{P_0}$ is noncommutative. Since $\alpha =\beta +2B(\alpha ,e_s)e_s$ is a positive combination of two roots in the complement of $I$ in ${\mathrm{\Phi }}_{P_0}^{+}$, clause (3)(a) gives $\alpha \mathrm{\notin }I$, a contradiction. Hence $I$ contains a simple root.

Reflection sequences of reduced words, forward direction of (2). Let $r_1\cdots r_k$ be reduced, put $w_j:=r_1\cdots r_j$, and let $t_i$ be its reflection sequence. If $k=0$, the empty sequence is the reflection sequence of the empty reduced word. For $k>0$, [F5] gives the roots of $N(w_j^{-1})$ as the prefix roots; by [F4], the reflection for the root $\rho (r_1\cdots r_{i-1})e_{r_i}$ is $r_1\cdots r_{i-1}{r}_i{r}_{i-1}\cdots r_1=t_i$. Fix $W_P$. By (1), each prefix intersection with ${\mathrm{\Phi }}_P^{+}$ is empty, an initial segment, or a final segment. These intersections are nested and each step adds at most one root. A nested chain of initial/final segments can change sides only at the full set; consequently its added roots are $u_1,u_2,\dots$ from the first endpoint or $u_m,u_{m-1},\dots$ from the other. The reflection subsequence in $W_P$ is therefore initial or final in the stated endpoint-inward convention.

Recognition in the reverse direction, base and induction. We prove (ii)$\Rightarrow$(i) by induction on $\mathrm{\mid }I\mathrm{\mid }$. If $I=\mathrm{\varnothing }$, then $I=N(1)$. If $I\mathrm{\ne }\mathrm{\varnothing }$, step 3.1 gives a simple root $e_s\in I$. Define $I^{\mathrm{\prime }}:=\rho (s)(I\setminus \{e_s\})$. By step 1.3, $I^{\mathrm{\prime }}$ satisfies (ii), and [F2] shows $\rho (s)$ bijects ${\mathrm{\Phi }}_{+}\setminus \{e_s\}$ with itself, so $\mathrm{\mid }{I}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }I\mathrm{\mid }-1$. The induction hypothesis supplies $v\in W$ with $N(v)=I^{\mathrm{\prime }}$.

Reconstructing the element. One has $e_s\mathrm{\notin }{I}^{\mathrm{\prime }}$: if $e_s=\rho (s)\alpha$ for $\alpha \in I\setminus \{e_s\}$, then $\alpha =\rho (s)e_s=-e_s$, impossible for a positive root. Hence $e_s\mathrm{\notin }N(v)$, so $\rho (v)e_s\in {\mathrm{\Phi }}_{+}$ by [F2] and $\mathrm{\ell }(vs)>\mathrm{\ell }(v)$ by [F6]. For any positive root $\gamma \mathrm{\ne }{e}_s$, $\rho (s)\gamma$ is positive by [F2], and $\gamma \in N(vs)$ exactly when $\rho (v)\rho (s)\gamma \in {\mathrm{\Phi }}_{-}$, which holds exactly when $\rho (s)\gamma \in N(v)$. Since $e_s\mathrm{\notin }N(v)$ and $\rho (s)$ permutes ${\mathrm{\Phi }}_{+}\setminus \{e_s\}$, this gives $\rho (s)N(v)=N(vs)\setminus \{e_s\}$. Also $\rho (vs)e_s=-\rho (v)e_s\in {\mathrm{\Phi }}_{-}$, so $e_s\in N(vs)$. Therefore $N(vs)=\{e_s\}\bigsqcup \rho (s)N(v)=\{e_s\}\bigsqcup (I\setminus \{e_s\})=I$. This proves (i) and discharges the induction.

Prefix recognition for a candidate sequence. Conversely suppose distinct reflections $t_1,\dots ,t_k$ satisfy the rank-two subsequence condition, and let ${\alpha }_i\in {\mathrm{\Phi }}_{+}$ be the unique root with $t_i=t_{{\alpha }_i}$ by [F4]. For $0\le j\le k$, put $A_j:=\{{\alpha }_1,\dots ,{\alpha }_j\}$. For every $W_P$, the subsequence in $W_P$ among the first $j$ reflections is a prefix of the full subsequence; by the endpoint-inward convention, it is again initial or final, so $A_j$ satisfies (ii). By (1), each $A_j$ is the inversion set of some $v_j\in W$. These finitely many witnesses can be selected by finite induction on $j$, which is finite choice only and does not use AC. Take $v_0=1$; [F5] gives $\mathrm{\ell }(v_j)=\mathrm{\mid }{A}_j\mathrm{\mid }=j$.

Build the reduced word. Since $A_j\subseteq A_{j+1}$, the weak-order criterion [F7] gives $v_j^{-1}{\le }_R{v}_{j+1}^{-1}$; their lengths differ by one, so [F8] gives a simple generator $s_{j+1}$ with $v_{j+1}^{-1}=v_j^{-1}{s}_{j+1}$. Thus $v_k^{-1}=s_1\cdots s_k$ is reduced. By [F5], the reflection sequence of each prefix $s_1\cdots s_j$ corresponds to the roots in $N(v_j)=A_j$, and [F4] identifies those roots' reflections with the prefix reflections. Taking successive set differences shows its $j$th reflection is $t_j$, so the given sequence is the reflection sequence of this reduced word. This proves (2).

Bijection and Choice. Surjectivity follows from steps 2.2, 4.1 and 5.1. If $N(u)=N(v)$, then the inversion-set criterion in [F7] gives $u^{-1}{\le }_R{v}^{-1}$ and $v^{-1}{\le }_R{u}^{-1}$; antisymmetry gives $u=v$. Thus $w\mapsto N(w)$ is injective, and [F5] ensures every $N(w)$ is finite. No Axiom of Choice is used: the inductions are on finite sets or words, and every witness is a single existential instantiation for the fixed object under consideration; the finite sequence of representatives in step 6.1 uses only finite choice, provable by induction, and no arbitrary family of choices is formed.