The mu-dot-root identities, the cone separation, and the canonical simple systems of the subintervals [1, sigma]

For every $i\ge 1$, [F2] gives $(C-\mathrm{i}\mathrm{d}){\mu }_i=-2{\rho }_i$. Since $C$ is an isometry and $B({\rho }_i,{\rho }_i)=1$, $B({\mu }_i,{\rho }_i)=1$ follows by expanding $B(C{\mu }_i,C{\mu }_i)=B({\mu }_i-2{\rho }_i,{\mu }_i-2{\rho }_i)$. The same identity gives $C{\mu }_i={\mu }_i-2{\rho }_i$, so $r_{{\rho }_i}C{\mu }_i={\mu }_i$ and ${\mu }_i\in F(r_{{\rho }_i}C)$. For $1\le i\le n$, the reflection word $r_{{\rho }_i}C$ cancels to $R_1\cdots R_{i-1}{R}_{i+1}\cdots R_n$, so its reflection length is at most $n-1$; the triangle inequality and ${\mathrm{\ell }}_T(c)=n$ give the lower bound $n-1$. Thus its fixed space is one-dimensional. Since $r_{{\rho }_{i+n}}C=C(r_{{\rho }_i}C)C^{-1}$, the same dimension holds for all $i$. The vector ${\mu }_i$ is nonzero because ${\mu }_i\cdot {\rho }_i=1$, and it is the unique vector in that line with this pairing. This proves (1).

For all $i,j\ge 1$, ${\mu }_i\cdot {\rho }_j=-{\mu }_{j+n}\cdot {\rho }_i$. Indeed, ${\rho }_i=\frac{1}{2}({\mu }_i-C{\mu }_i)$ by [F2], so $-{\mu }_{j+n}\cdot {\rho }_i=-C{\mu }_j\cdot \frac{1}{2}({\mu }_i-C{\mu }_i)=\frac{1}{2}({\mu }_j-C{\mu }_j)\cdot {\mu }_i={\rho }_j\cdot {\mu }_i$, using the isometry of $C$ and [F1]-[F2]. This is (2)(a).

Suppose $1\le i\le j\le nh\mathrm{/}2$. If $i>n$, then $j>n$ and $C$-equivariance gives ${\mu }_i\cdot {\rho }_j={\mu }_{i-n}\cdot {\rho }_{j-n}$. Repeating reduces to a first index at most $n$ while keeping both indices positive and ordered. For $i\le n$, ${\mu }_i={\beta }_i$, and ${\beta }_i\cdot {\rho }_j$ is the coefficient of ${\alpha }_i$ in the positive root ${\rho }_j$; this is nonnegative by [F4]. Thus (2)(b) holds.

Every reflection lies below $c$. Indeed, $M(c)=V$ by [F1]. For a positive root $\alpha$, the orthogonal reflection with moved line $\mathbb{R}\alpha$ is uniquely $r_{\alpha }=\rho (t_{\alpha })$ by [F3],[F6]; the Wall restriction for this line is below $C=\rho (c)$ and has moved space $\mathbb{R}\alpha$. The Wall dimension identity and Carter's formula therefore give $t_{\alpha }{\le }_{T}c$.

We use two elementary consequences of absolute order. First, if $w=r_1\cdots r_m$ is a reduced reflection factorization and $p<q$, then $r_p{r}_q{\le }_{T}w$: distinct root-reflections have product moved space the two-plane spanned by their distinct normal lines, so their product has reflection length $2$ by Carter's formula. Write $w=x{r}_{p}y{r}_{q}z$; then $(r_p{r}_q{)}^{-1}w=r_q{r}_{p}x{r}_{p}y{r}_{q}z=(r_q(r_{p}x{r}_p)y{r}_q)z$, a product of $m-2$ reflections because conjugation preserves reflections. The triangle inequality forces this complement to have length $m-2$. Second, if $u{\le }_{T}v{\le }_{T}w$, then $v^{-1}w{\le }_T{u}^{-1}w$: write $v=ux$, $w=vy$ with additive reflection lengths; then $u^{-1}w=xy$, while $y^{-1}xy$ has the same length as $x$, so $y{\le }_{T}xy$. These facts will be used below.

Put $E=M(\sigma )$ and $F=E^{\perp }$. Then $W_{\sigma }=\{w:M(w)\subseteq E\}$ is the pointwise stabilizer of $F$, because $M(w)=F(w{)}^{\perp }$. If $F\mathrm{\ne }0$, choose $x\in F$ outside $\{0\}$ and every proper subspace $F\cap H_{\alpha }$; this is possible by [F9] and [F12]. By [F10], ${\mathrm{Stab}}_W(x)=g{W}_I{g}^{-1}$ for some $g,I$. Each conjugate simple generator has a root hyperplane containing $x$, hence, by the choice of $x$, containing all of $F$; therefore every generator of this stabilizer fixes $F$ pointwise. Conversely every element fixing $F$ fixes $x$, so $W_{\sigma }=g{W}_I{g}^{-1}$. If $F=0$, take $g=1$ and $I=S$.

To prove (2)(c), first reduce $i$ modulo $n$ using $C$-equivariance, which preserves ${\mu }_{i+t}\cdot {\rho }_i$, and assume $1\le i\le n$. If $i+t\le n$, then ${\mu }_{i+t}={\beta }_{i+t}$ and ${\rho }_i=R_1\cdots R_{i-1}{\alpha }_i$ has no ${\alpha }_{i+t}$ coefficient, so the pairing is zero. If $i+t>n$, put $j=i+t-n$, so $1\le j<i$. By (2)(a), ${\mu }_{i+t}\cdot {\rho }_i={\mu }_{j+n}\cdot {\rho }_i=-{\mu }_i\cdot {\rho }_j=-{\beta }_i\cdot {\rho }_j=0$, since ${\rho }_j$ is supported on ${\alpha }_1,\dots ,{\alpha }_j$. Now if $1\le i<j\le nh\mathrm{/}2$ and $j<i+n$, (2)(d) is (2)(c) with $t=j-i$; if $j\ge i+n$, (2)(a) and (2)(b) give ${\mu }_j\cdot {\rho }_i=-{\mu }_{i+n}\cdot {\rho }_j\le 0$. This proves (2)(c)-(d).

For every $\sigma {\le }_{T}c$, one has $P_{\sigma }={\mathrm{\Phi }}_{+}\cap M(\sigma )$. If $\alpha \in P_{\sigma }$, then $M(t_{\alpha })=\mathbb{R}\alpha \subseteq M(\sigma )$ by [F5]. Conversely, if $\alpha \in {\mathrm{\Phi }}_{+}\cap M(\sigma )$, then $t_{\alpha }{\le }_{T}c$ by step 1.4 and $\sigma {\le }_{T}c$, so moved-space rigidity in [F5] gives $t_{\alpha }{\le }_T\sigma$. Also $P_{\sigma }$ spans $M(\sigma )$: choose a reduced reflection factorization of $\sigma$; its prefix-conjugated normals are a basis of $M(\sigma )$ by [F7], and the reflections with those normals have moved lines in $M(\sigma )$, hence lie below $\sigma$ by the same common-upper-bound argument. Taking the positive sign of each normal puts a spanning set in $P_{\sigma }$.

For any increasing tuple $a_1<\cdots <a_m$ of positive roots, one has ${\mathrm{\ell }}_T(R(a_1)\cdots R(a_m)c)=n-m$ if and only if $\mu (a_i)\cdot a_j=0$ for all $i>j$. If $m=0$, the left side is ${\mathrm{\ell }}_T(c)=n$ by [F1], and the zero-pairing condition is vacuous; hence assume $m\ge 1$. Put $r_i=R(a_i)$. The length equality is equivalent to the reverse product $q=r_m\cdots r_1$ being a reduced element below $c$: its complement is $q^{-1}c=r_1\cdots r_{m}c$. In fact, if ${\mathrm{\ell }}_T(q^{-1}c)=n-m$, then ${\mathrm{\ell }}_T(q)\le m$ and $n={\mathrm{\ell }}_T(c)\le {\mathrm{\ell }}_T(q)+{\mathrm{\ell }}_T(q^{-1}c)\le n$, so both inequalities are equalities; the converse is the defining absolute-order equality. For the forward implication, the pair-product fact in step 1.5 gives $r_i{r}_j{\le }_{T}c$ whenever $i>j$. Since ${\mathrm{\ell }}_T(r_{i}c)=n-1$, this gives $r_j{\le }_T{r}_{i}c$ from ${\mathrm{\ell }}_T(r_j{r}_{i}c)=n-2$; hence $M(r_j)\subseteq M(r_{i}c)=F(r_{i}c{)}^{\perp }$. The vector $\mu (a_i)$ is a nonzero member of the one-dimensional fixed space $F(r_{i}c)$, so $\mu (a_i)\cdot a_j=0$. Conversely, suppose all these pairings vanish. The matrix $[\mu (a_i)\cdot a_j]$ is upper triangular with diagonal $1$ by (1), so both the $a_i$ and the $\mu (a_i)$ are linearly independent; in particular $m\le n$. Induct downward on $i$ to prove $q_i:=r_m\cdots r_i{\le }_{T}c$ with length $m-i+1$. The base $q_m=r_m$ is step 1.4. If $q_{i+1}{\le }_{T}c$, then each factor $r_j$ with $j>i$ satisfies $r_j{\le }_T{q}_{i+1}$: writing a reduced factorization $q_{i+1}=x{r}_{j}y$, the product $r_j{q}_{i+1}=(r_{j}x{r}_j)y$ uses one fewer reflections, and the triangle inequality makes that expression reduced. Put $v=q_{i+1}^{-1}c$. Complement order reversal from step 1.5 gives $v{\le }_T{r}_{j}c$, so $\mu (a_j)\in F(r_{j}c)\subseteq F(v)$ for every $j>i$. These $m-i$ independent vectors form a basis of $F(v)$, whose dimension is $m-i$ by Carter's formula. The assumed zero pairings give $a_i\in F(v{)}^{\perp }=M(v)$. The Wall restriction for the line $\mathbb{R}{a}_i$ gives $r_i{\le }_{T}v$; writing $v=r_{i}z$ then gives $c=q_{i+1}{r}_{i}z$, with lengths adding, so $q_i=q_{i+1}{r}_i{\le }_{T}c$ and ${\mathrm{\ell }}_T(q_i)=m-i+1$. At $i=1$ this proves the claimed equivalence. The same argument applies to every subtuple because the vanishing conditions are inherited.

If ${\rho }_k={\sum }_{r=1}^m{a}_r{\rho }_{i_r}$ with $a_r\ge 0$ and $i_1<\cdots <i_m<k$, then (2)(d) gives ${\mu }_k\cdot {\rho }_{i_r}\le 0$ for every $r$, so $1={\mu }_k\cdot {\rho }_k={\sum }_r{a}_r({\mu }_k\cdot {\rho }_{i_r})\le 0$, a contradiction. This proves (3).

Let ${\mathrm{\Phi }}_I=\{\rho (u){\alpha }_s:u\in W_I,s\in I\}$ be the intrinsic root system from [F14]; its representation is the restriction to $V_I=\mathrm{span}\{{\alpha }_s:s\in I\}$, since the reflection formulas agree. An ambient reflection $t\in T\cap W_I$ has a reduced $I$-expression by [F14]. Apply strong exchange to $tt=1$: it expresses $t$ as a prefix-conjugate of a letter of that expression, so its root normal belongs to $\pm {\mathrm{\Phi }}_I$ by [F3]. Conversely every intrinsic root gives an ambient reflection in $W_I$. Conjugating by $g$, the root-reflection dictionary therefore identifies the roots of $g{W}_I{g}^{-1}$ with precisely the ambient roots whose reflections fix $F$ pointwise, namely $\mathrm{\Phi }\cap E=\pm P_{\sigma }$ by step 2.2. Their span is both $\rho (g)V_I$ and $E$, since $P_{\sigma }$ spans $E$.

Choose $f\in {\mathcal{C}}^{\circ }$ and project it orthogonally to $E$. Its pairing with every root of $P_{\sigma }$ is positive by [F4], so it avoids the subsystem arrangement. Apply [F10] to the finite Coxeter system $(W_I,I)$ and transfer its chambers to $E$ by $\rho (g)$. The chamber containing the projection has inward unit normals $\mathrm{\Delta }=\{\rho (gu){\alpha }_s:s\in I\}$ for some $u\in W_I$. These are a basis of $E$, their reflections are the conjugates of the simple generators of $W_I$, and their Gram matrix is the restricted Coxeter Gram matrix. In particular distinct normals have nonpositive pairings. By the root-sign theorem [F4] applied to this conjugate Coxeter system, its positive roots are exactly the roots positive on the projection, hence $P_{\sigma }$, and every such root is a nonnegative combination of $\mathrm{\Delta }$. This supplies the simple system and the Coxeter presentation used below.

In any such positive root subsystem with simple system $\mathrm{\Gamma }$, its first root in the global order is a member of $\mathrm{\Gamma }$. For if the first root $\eta$ were not simple, write $\eta ={\sum }_{\gamma \in \mathrm{\Gamma }}{b}_{\gamma }\gamma$ with $b_{\gamma }\ge 0$; each $\gamma$ is later than $\eta$, so (2)(d) gives $\mu (\gamma )\cdot \eta \le 0$. Linearity of $\mu$ and (1) would give $1=\mu (\eta )\cdot \eta =\sum b_{\gamma }(\mu (\gamma )\cdot \eta )\le 0$, impossible. Also the last root of $P_{\sigma }$ belongs to $\mathrm{\Delta }$: otherwise it is a nonnegative combination of the earlier simple roots, contrary to step 3.1.

We prove the recursive selection and the factorization $\sigma =R({\delta }_1)\cdots R({\delta }_k)$ by induction on $k$. For $k=1$, $W_{\sigma }$ has rank one, $P_{\sigma }$ has its single positive root ${\delta }_1$, and $\sigma =R({\delta }_1)$. For $k>1$, order the simple roots increasingly; step 4.2 makes the last one ${\delta }_k={\tau }_t$. Put $r=R({\delta }_k)$. Since $r{\le }_T\sigma$, write $\sigma =ru$ with ${\mathrm{\ell }}_T(u)=k-1$; then ${\sigma }^{\mathrm{\prime }}:=\sigma r=rur$ has length $k-1$. Write $c=\sigma v$ with ${\mathrm{\ell }}_T(v)=n-k$. The identity $c={\sigma }^{\mathrm{\prime }}(rv)$ and the bounds ${\mathrm{\ell }}_T(rv)\le n-k+1$ and $n={\mathrm{\ell }}_T(c)\le {\mathrm{\ell }}_T({\sigma }^{\mathrm{\prime }})+{\mathrm{\ell }}_T(rv)$ force ${\mathrm{\ell }}_T(rv)=n-k+1$, so ${\sigma }^{\mathrm{\prime }}{\le }_{T}c$. Also ${\sigma }^{\mathrm{\prime }}$ fixes $F$ pointwise, hence $M({\sigma }^{\mathrm{\prime }})\subseteq E$.

Let $q$ be the global index with ${\delta }_k={\rho }_q$ and put $\nu =C{\mu }_q={\mu }_{q+n}$. For every ${\rho }_j\in P_{\sigma }$ one has $j\le q$ and, by (2)(a) and $C$-equivariance, $\nu \cdot {\rho }_j={\mu }_{q+n}\cdot {\rho }_j=-{\mu }_{j+n}\cdot {\rho }_{q+n}=-{\mu }_j\cdot {\rho }_q\le 0$ by (2)(b); at $j=q$ this is $-1$. Since ${\mu }_q\in F(rc)$, conjugating by $C$ gives $\nu \in F(cr)$; and ${\mathrm{\ell }}_T(cr)=n-1$, so $F(cr)=\mathbb{R}\nu$ and $M(cr)={\nu }^{\perp }$. Moreover ${\sigma }^{\mathrm{\prime }}{\le }_{T}cr$: its complement is ${\sigma }^{\mathrm{\prime }-1}cr=rvr$, of length $n-k$, while ${\mathrm{\ell }}_T(cr)=n-1$; the lengths add to $n-1$. Hence $U:=M({\sigma }^{\mathrm{\prime }})$ is contained in $E\cap {\nu }^{\perp }$, and both have dimension $k-1$ because $\nu \cdot {\delta }_k=-1$. Thus $U=E\cap {\nu }^{\perp }$. The roots $P_{{\sigma }^{\mathrm{\prime }}}={\mathrm{\Phi }}_{+}\cap U$ span $U$ by step 2.2. Each is a nonnegative combination of $\mathrm{\Delta }$ and is orthogonal to $\nu$; since $\nu \cdot {\delta }_k=-1$ and $\nu \cdot {\delta }_j\le 0$ for every $j$, every root in $P_{{\sigma }^{\mathrm{\prime }}}$ is a combination only of the ${\delta }_j$ for which $\nu \cdot {\delta }_j=0$. Those zero-pairing simple roots must span the $(k-1)$-space $U$, so they are exactly ${\delta }_1,\dots ,{\delta }_{k-1}$. They form a simple system for $P_{{\sigma }^{\mathrm{\prime }}}$. Induction gives ${\sigma }^{\mathrm{\prime }}=R({\delta }_1)\cdots R({\delta }_{k-1})$ and proves the reverse selection at each rank. Therefore $\sigma =R({\delta }_1)\cdots R({\delta }_k)$. For each $i$, its prefix $p_i:=R({\delta }_1)\cdots R({\delta }_i)$ and complementary suffix $R({\delta }_{i+1})\cdots R({\delta }_k)$ multiply to $\sigma$ and have at most $i$ and $k-i$ reflections, so both lengths attain these bounds. Thus $M(p_i)$, contained in $\mathrm{span}({\delta }_1,\dots ,{\delta }_i)$, has dimension $i$ and equals that span. Since $\sigma R({\delta }_k)\cdots R({\delta }_{i+1})=p_i$, the induction shows that ${\delta }_i$ is the last root of $P_{\sigma }$ in this subspace.

The other recursive description follows by considering the suffix $q_i:=R({\delta }_i)\cdots R({\delta }_k)$. The factorization in step 6.1 gives ${\mathrm{\ell }}_T(q_i)=k-i+1$: its prefix and suffix factors together multiply to $\sigma$ of length $k$, and their lengths are bounded by their numbers of reflections, whose sum is $k$. Writing $p_i:=R({\delta }_1)\cdots R({\delta }_{i-1})$, the same length equality forces ${\mathrm{\ell }}_T(p_i)=i-1$, so $q_i^{-1}\sigma =q_i^{-1}{p}_i{q}_i$ has length $i-1$ and $q_i{\le }_T\sigma$. If $c=\sigma v$ as above, then $q_i^{-1}c=q_i^{-1}(R({\delta }_1)\cdots R({\delta }_{i-1}))q_{i}v$ has length at most $(i-1)+(n-k)=n-{\mathrm{\ell }}_T(q_i)$; the triangle inequality forces equality, so $q_i{\le }_{T}c$. Its moved space is $\mathrm{span}({\delta }_i,\dots ,{\delta }_k)$, since its factors fix the orthogonal complement of that span and its moved dimension is $k-i+1$. Thus $P_{q_i}=P_{\sigma }\cap M(q_i)$, and $\{{\delta }_i,\dots ,{\delta }_k\}$ is a simple system for this subsystem: a root in the tail span has zero coefficients on ${\delta }_1,\dots ,{\delta }_{i-1}$ in its nonnegative $\mathrm{\Delta }$-expansion. By step 4.2 the first root of $P_{q_i}$ is a simple root of this tail system; its first simple root is ${\delta }_i$. Finally $R({\delta }_{i-1})\cdots R({\delta }_1)\sigma =q_i$ by cancellation, so ${\delta }_i$ is the first root of $P_{\sigma }$ in the moved space in (4)(i), and ${\delta }_1={\tau }_1$. This proves (4)(i).

Put $D_i:=R({\delta }_1)\cdots R({\delta }_{i-1})$ and ${\epsilon }_i:=D_i{\delta }_i$. The prefix $R({\delta }_1)\cdots R({\delta }_i)$ has reflection length $i$ by the factorization of $\sigma$ and the triangle inequality, so this simple-generator word in the Coxeter system of $W_{\sigma }$ is reduced. The root-inversion formula [F13], applied to that parabolic system, gives ${\epsilon }_i\in P_{\sigma }$, hence it is a positive root. Conjugation gives $R({\epsilon }_i)=D_{i}R({\delta }_i)D_i^{-1}$; multiplying these identities and cancelling adjacent inverse prefixes yields $\sigma =R({\epsilon }_k)\cdots R({\epsilon }_1)$. Also $R({\epsilon }_i)D_{i}R({\delta }_{i+1})\cdots R({\delta }_k)=D_{i}R({\delta }_i)R({\delta }_{i+1})\cdots R({\delta }_k)=\sigma$. This proves (4)(ii).

Suppose ${\mathrm{\ell }}_T(\sigma )=2$ and $P_{\sigma }=\{{\tau }_1<\cdots <{\tau }_t\}$. By (4)(i), $\mathrm{\Delta }=\{{\tau }_1,{\tau }_t\}$ and $\sigma =R({\tau }_1)R({\tau }_t)$. Step 4.1 realizes this as a canonical rank-two system. If its exponent is $m$, [F11] gives simple-root angle $\pi -\pi \mathrm{/}m$ and a rotation of magnitude $2\pi \mathrm{/}m$. Write its generators as $r,s$ and $q=rs$. The identities $rqr=q^{-1}$ and $s=rq$ reduce every word to $q^j$ or $q^{j}r$, $0\le j<m$; their distinct actions are the $m$ rotations and $m$ reflections in lines spaced by $\pi \mathrm{/}m$. Every reflection is a conjugate of $r$ or $s$: the conjugates give respectively $q^{2j}r$ and $q^{2j-1}r$. The root-reflection dictionary thus gives $t=m$ positive roots, equally spaced between the simple-root rays. The global order agrees with angular order: a decrease after ${\tau }_r$ would put ${\tau }_{r+1}$ in the cone on earlier roots ${\tau }_1,{\tau }_r$, contrary to step 3.1. A product of reflections has rotation angle twice the difference of its normal angles, modulo $2\pi$; hence the two-reflection factorizations of $\sigma$ are exactly $R({\tau }_1)R({\tau }_t)=R({\tau }_2)R({\tau }_1)=\cdots =R({\tau }_t)R({\tau }_{t-1})$. These exhaust factorizations in $W$: if $\sigma =R(\alpha )R(\beta )$, then $M(\sigma )\subseteq \mathrm{span}(\alpha ,\beta )$, and both spaces have dimension $2$, so the positive representatives of $\alpha ,\beta$ lie in $P_{\sigma }$ by step 2.2. The endpoint angle gives ${\tau }_1\cdot {\tau }_t=-\cos (\pi \mathrm{/}t)\le 0$, while each adjacent angle gives ${\tau }_i\cdot {\tau }_{i+1}=\cos (\pi \mathrm{/}t)\ge 0$.

Fix $i$. The second factorization in (4)(ii) writes $\sigma =R({\epsilon }_i)q_i^{\mathrm{\prime }}$ where $q_i^{\mathrm{\prime }}$ is the product of the other $k-1$ simple reflections in $\mathrm{\Delta }$. Thus ${\mathrm{\ell }}_T(q_i^{\mathrm{\prime }})=k-1$, $R({\epsilon }_i){\le }_T\sigma$, and $q_i^{\mathrm{\prime }}=R({\epsilon }_i)\sigma$. Since $c=\sigma v$ with ${\mathrm{\ell }}_T(v)=n-k$, $R({\epsilon }_i)c=q_i^{\mathrm{\prime }}v$ is reduced, so $q_i^{\mathrm{\prime }}{\le }_{T}R({\epsilon }_i)c$. It follows that $\mu ({\epsilon }_i)\in F(q_i^{\mathrm{\prime }})$. The moved space $M(q_i^{\mathrm{\prime }})$ is exactly $\mathrm{span}\{{\delta }_j:j\mathrm{\ne }i\}$: it is contained there because those are its reflection normals, and both dimensions are $k-1$. Hence $\mu ({\epsilon }_i)\cdot {\delta }_j=0$ for $j\mathrm{\ne }i$. Since ${\epsilon }_i-{\delta }_i$ is a linear combination of ${\delta }_1,\dots ,{\delta }_{i-1}$, (1) gives $1=\mu ({\epsilon }_i)\cdot {\epsilon }_i=\mu ({\epsilon }_i)\cdot {\delta }_i$. Thus the restrictions to $E$ of the functionals $B(\mu ({\epsilon }_i),-)$ form the dual basis to $\mathrm{\Delta }$. For $x={\sum }_j{a}_j{\delta }_j$ with $a_j\ge 0$, $\mu ({\epsilon }_i)\cdot x=a_i$; intersecting the cone with this zero hyperplane and with the unit sphere proves the wall equality, and the supporting hyperplane in $E$ is $E\cap \mu ({\epsilon }_i{)}^{\perp }$. This proves (4)(iii).

If distinct positive roots ${\rho }_i,{\rho }_j$ satisfy $R({\rho }_i)R({\rho }_j){\le }_{T}c$, their product has reflection length $2$ and belongs to the rank-two case of step 8.1. Both roots lie in the corresponding $P_{\sigma }$ by moved-space rigidity. If $i<j$, the only increasing-order two-factorization in step 8.1 is the endpoint factorization, so ${\rho }_i\cdot {\rho }_j\le 0$. If $i>j$, the factorization is one of the adjacent descending pairs, so ${\rho }_i\cdot {\rho }_j\ge 0$. This proves (4)(v).

If $i<j$ and ${\epsilon }_i>{\epsilon }_j$, the factorization $\sigma =R({\epsilon }_k)\cdots R({\epsilon }_1)$ is reduced, and the pair-product fact of step 1.5 gives $R({\epsilon }_j)R({\epsilon }_i){\le }_T\sigma {\le }_{T}c$. The rank-two sign result of step 9.1 gives ${\epsilon }_i\cdot {\epsilon }_j\le 0$. On the other hand, since distinct simple roots have nonpositive pairings by step 4.1, applying $R({\delta }_{j-1}),\dots ,R({\delta }_{i+1})$ successively to ${\delta }_j$ yields a nonnegative combination of ${\delta }_{i+1},\dots ,{\delta }_j$, and ${\epsilon }_i\cdot {\epsilon }_j=-{\delta }_i\cdot (R({\delta }_{i+1})\cdots R({\delta }_{j-1}){\delta }_j)\ge 0$. Thus the pairing is zero, and the two reflections commute. Sorting the ${\epsilon }_i$ into global order uses only swaps of such inverted commuting pairs, so the product is unchanged and $\sigma =R({\theta }_k)\cdots R({\theta }_1)$.