Root normals inside the moved space, factorizations into reflections, and independent normals

Let $w\mathrm{\ne }1$. First $\dim M(w)>0$: otherwise $M(\rho (w))=0$, so $\rho (w)-\mathrm{i}\mathrm{d}=0$ and $\rho (w)=\mathrm{i}\mathrm{d}$, whence $w=1$ by [F4], a contradiction. Next, a root $\alpha$ with $F(w)\subseteq H_{\alpha }$ exists. If $F(w)=0$ then F6 gives $M(w)=F(w{)}^{\perp }=V\mathrm{\ne }0$, so $S\mathrm{\ne }\mathrm{\varnothing }$ and $\alpha :=e_s$ for any $s\in S$ is a root with $F(w)=0\subseteq H_{\alpha }$. Suppose now that $F(w)\mathrm{\ne }0$; let ${\mathrm{\Phi }}^{\mathrm{\prime }}\subseteq \mathrm{\Phi }$ be the finite set of roots with $F(w)\subseteq ̸H_{\alpha }$, so that each $H_{\alpha }\cap F(w)$ with $\alpha \in {\mathrm{\Phi }}^{\mathrm{\prime }}$ is a proper subspace of the finite-dimensional real space $F(w)$; the finite family consisting of those subspaces and $\{0\}$ consists of proper subspaces of $F(w)$, since $F(w)\mathrm{\ne }0$; by [F2] its union does not cover $F(w)$, so there is $0\mathrm{\ne }x\in F(w)$ with $x\mathrm{\notin }{H}_{\alpha }$ for every $\alpha \in {\mathrm{\Phi }}^{\mathrm{\prime }}$, including when ${\mathrm{\Phi }}^{\mathrm{\prime }}=\mathrm{\varnothing }$; for every root $\beta$ the implication $x\in H_{\beta }\Rightarrow F(w)\subseteq H_{\beta }$ holds, and $x$ is fixed by $w$, so $w\in {\mathrm{Stab}}_W(x)$. By [F1] there are $w_0\in W$ and $I⊊S$ with $x\in w_0{C}_I$ and ${\mathrm{Stab}}_W(x)=w_0{W}_I{w}_0^{-1}$; since $w\mathrm{\ne }1$ lies in this stabiliser, $I\mathrm{\ne }\mathrm{\varnothing }$, so for $s\in I$ one has $x\in w_0{C}_I\subseteq w_0{H}_{e_s}=H_{\rho (w_0)e_s}$ by [F1], the equality following from the $B$-invariance of $\rho (w_0)$ [F5]; the implication above with $\beta :=\rho (w_0)e_s$ gives $F(w)\subseteq H_{\rho (w_0)e_s}$ for the root $\rho (w_0)e_s\in \mathrm{\Phi }$, so in this case a root with the required property exists as well. Finally, if $F(w)\subseteq H_{\alpha }$, then $B(f,\alpha )=0$ for all $f\in F(w)$, so $\alpha \in F(w{)}^{\perp }=M(w)$ by F6.

Let $t\in T$ and let $\alpha \in \mathrm{\Phi }$ satisfy $\rho (t)=r_{\alpha }$, as supplied by [F3]. Then $r_{\alpha }(v)-v=-2B(v,\alpha )\alpha$ for every $v$ by [F5], so $M(t)=M(r_{\alpha })=\mathbb{R}\alpha$ and $\dim M(t)=1$; also $\rho (t{)}^{-1}=\rho (t)=\rho (t^{-1})$ because $t^2=1$.

For $X,Y\in \mathrm{O}(V)$ one has $\dim M(XY)\le \dim M(X)+\dim M(Y)$ by F6.

For $t_1,\dots ,t_m\in T$, the telescoping identity is $\rho (t_1\cdots t_m)-\mathrm{i}\mathrm{d}={\sum }_{i=1}^m\rho (t_1\cdots t_{i-1})(\rho (t_i)-\mathrm{i}\mathrm{d})$: each summand is $\rho (t_1\cdots t_i)-\rho (t_1\cdots t_{i-1})$, by the homomorphism property of $\rho$. Thus $M(t_1\cdots t_m)\subseteq {\sum }_{i=1}^m\rho (t_1\cdots t_{i-1})M(t_i)$.

Linear-algebra tool. Let $Z$ be a finite-dimensional vector space spanned by vectors $w_1,\dots ,w_m$. Then $\dim Z\le m$: by [F8] $Z$ has a basis $B$ with $\mathrm{\mid }B\mathrm{\mid }=\dim Z$, and $B$ is linearly independent while $Z$ is spanned by $w_1,\dots ,w_m$, so [F9] gives $\dim Z\le m$. If moreover $\dim Z=m$, then $w_1,\dots ,w_m$ are linearly independent: otherwise a nontrivial relation expresses some $w_j$ as a combination of the remaining $m-1$ vectors by [F10], those remaining vectors still span $Z$ by [F10], and [F9] would give $\dim Z\le m-1$, a contradiction.

A product of $k$ elements of $T$ has moved dimension at most $k$: by step 1.2 each factor has moved dimension $1$, and step 1.3 applied $k$ times bounds the moved dimension of the product by the sum $k$.

Let $\alpha \in \mathrm{\Phi }$ satisfy $F(w)\subseteq H_{\alpha }$ and $\alpha \in M(w)$, as produced by step 1.1. Then $\mathbb{R}\alpha$ is a line contained in $M(\rho (w))$, and by F6 and [F3] the unique reflection of $\mathrm{O}(V)$ with moved space $\mathbb{R}\alpha$ is $\mathrm{i}\mathrm{d}-2{\mathrm{\Pi }}_{\mathbb{R}\alpha }=r_{\alpha }=\rho (t_{\alpha })$; hence $\rho (t_{\alpha })=(\rho (w){)}_{\mathbb{R}\alpha }$ in the notation of [F6]. Applying F6 with $A=\rho (w)$ and $U=\mathbb{R}\alpha$ gives $\rho (t_{\alpha }){\le }_{\mathrm{O}}\rho (w)$ and $\dim M(w)=\dim \mathbb{R}\alpha +\dim M(\rho (t_{\alpha }{)}^{-1}\rho (w))=1+\dim M(t_{\alpha }w)$, the last equality because $\rho (t_{\alpha }{)}^{-1}\rho (w)=\rho (t_{\alpha }^{-1}w)=\rho (t_{\alpha }w)$ by steps 1.1 and 1.2.

Let $t_1,\dots ,t_m\in T$ and put $v_i:=\rho (t_1\cdots t_{i-1}){\alpha }_i$ and $Z:=\mathrm{span}\{v_1,\dots ,v_m\}$. Steps 1.2 and 1.4 give $M(t_1\cdots t_m)\subseteq Z$. Since $Z$ is spanned by these $m$ vectors, step 1.5 gives $\dim Z\le m$; [F8] applied to the subspace $M(t_1\cdots t_m)$ of $Z$ yields $\dim M(t_1\cdots t_m)\le \dim Z\le m$.

Carter's formula and factorization: every $w\in W$ satisfies ${\mathrm{\ell }}_T(w)=\dim M(w)$, and $w$ is a product of exactly $\dim M(w)$ elements of $T$. If $w=1$, then $\dim M(w)=0$ and the empty product represents $1$, giving both assertions. If $w\mathrm{\ne }1$ with $\dim M(w)=k\ge 1$, step 2.2 supplies $t:=t_{\alpha }\in T$ with $\dim M(tw)=k-1$; by induction on $k$ (applied to $tw$, whose moved dimension is $k-1$) there are $t_2,\dots ,t_k\in T$ with $tw=t_2\cdots t_k$, so $w=t\cdot (tw)=t{t}_2\cdots t_k$ is a product of $k$ elements of $T$. For the reverse inequality let $w=s_1\cdots s_j$ with $s_1,\dots ,s_j\in T$; then $k=\dim M(w)=\dim M(\rho (s_1)\cdots \rho (s_j))\le j$ by step 2.1, so no shorter product of elements of $T$ represents $w$ and ${\mathrm{\ell }}_T(w)=k$ by [F7].

Suppose $\dim M(t_1\cdots t_m)=m$ and put $v_i:=\rho (t_1\cdots t_{i-1}){\alpha }_i$ as in step 2.3. Step 2.3 gives $M(t_1\cdots t_m)\subseteq Z$ with $m=\dim M(t_1\cdots t_m)\le \dim Z\le m$, so $\dim Z=m$ and $M(t_1\cdots t_m)=Z$ by [F8]. Thus the $v_i$ span $M(t_1\cdots t_m)$, and by step 1.5 the vectors $v_1,\dots ,v_m$ are linearly independent and hence form a basis of $M(t_1\cdots t_m)$: this is the independence and spanning assertion of (3). Finally, if $t_1\cdots t_m$ is a shortest reflection factorization of its product $u:=t_1\cdots t_m$, then $m={\mathrm{\ell }}_T(u)=\dim M(u)$ by step 3.1, so the hypothesis $\dim M(t_1\cdots t_m)=m$ holds and the same conclusion applies. This proves (1), (2) and (3).