The inversion formula $\mathrm{\mid }N(w)\mathrm{\mid }=\mathrm{\ell }(w)$, the root-reflection dictionary and strong exchange

Set-up. Fix a reduced expression $w=s_1\cdots s_n$ of an element $w\in W$, write $w_j:=s_1\cdots s_j$ and $r_i:=w_{i-1}{s}_i{w}_{i-1}^{-1}\in T$ for the prefix reflections, and note $\rho (w_j)=\rho (s_1)\cdots \rho (s_j)$ by [F6] and $\rho (r_i)=\rho (w_{i-1}{s}_i{w}_{i-1}^{-1})=r_{\rho (w_{i-1})e_{s_i}}$.

The inversion formula (2). We prove by induction on $m$ the assertion: for every $x$ with $\mathrm{\ell }(x)=m$ and every reduced expression $x=s_1\cdots s_m$ one has $N(x)=\{\rho (s_{i+1}\cdots s_m{)}^{-1}{e}_{s_i}:1\le i\le m\}$ with pairwise distinct elements, and $\mathrm{\mid }N(x)\mathrm{\mid }=m$. For $m=0$ this is $N(1)=\mathrm{\varnothing }$ by [F5]. For the step let $x=u{s}_m$ with $u:=s_1\cdots s_{m-1}$, so $\mathrm{\ell }(u)=m-1$ and $\mathrm{\ell }(u{s}_m)=m>\mathrm{\ell }(u)$; the first case of the step recursion of [F5] gives $N(x)=\{e_{s_m}\}\bigsqcup s_{m}N(u)$, the union being disjoint with $s_{m}N(u)\subseteq {\mathrm{\Phi }}_{+}\setminus \{e_{s_m}\}$. By induction $N(u)=\{\rho (s_{i+1}\cdots s_{m-1}{)}^{-1}{e}_{s_i}:1\le i\le m-1\}$ with pairwise distinct elements; since $\rho (s_m)\rho (s_{i+1}\cdots s_{m-1}{)}^{-1}=\rho (s_m{s}_{m-1}\cdots s_{i+1})=\rho (s_{i+1}\cdots s_m{)}^{-1}$ for $i\le m-1$ by [F6], one has $s_{m}N(u)=\{\rho (s_{i+1}\cdots s_m{)}^{-1}{e}_{s_i}:1\le i\le m-1\}$, and the term $i=m$ with the empty product contributes $e_{s_m}$. Hence $N(x)=\{\rho (s_{i+1}\cdots s_m{)}^{-1}{e}_{s_i}:1\le i\le m\}$ with pairwise distinct elements, and $\mathrm{\mid }N(x)\mathrm{\mid }=1+\mathrm{\mid }N(u)\mathrm{\mid }=m$. Applying the same formula to the reversed reduced expression $x^{-1}=s_m\cdots s_1$ and using $\rho (s_{i-1}\cdots s_1{)}^{-1}=\rho (s_1\cdots s_{i-1})$ gives $N(x^{-1})=\{\rho (s_1\cdots s_{i-1})e_{s_i}:1\le i\le m\}$, again with pairwise distinct elements. The induction principle [F8] gives (2) for every element.

The dictionary (i) and (ii). Let $\alpha \in \mathrm{\Phi }$, written as $\alpha =\rho (w)e_s=\rho (w^{\mathrm{\prime }})e_{s^{\mathrm{\prime }}}$. Then $\rho (ws{w}^{-1})=r_{\rho (w)e_s}=r_{\alpha }=r_{\rho (w^{\mathrm{\prime }})e_{s^{\mathrm{\prime }}}}=\rho (w^{\mathrm{\prime }}{s}^{\mathrm{\prime }}{w}^{\mathrm{\prime }-1})$ by [F1], and injectivity of $\rho$ from [F2] gives $ws{w}^{-1}=w^{\mathrm{\prime }}{s}^{\mathrm{\prime }}{w}^{\mathrm{\prime }-1}$; so $t_{\alpha }$ is independent of the representation, which is (i). For (ii), $\rho (t_{\alpha })=\rho (ws{w}^{-1})=r_{\rho (w)e_s}=r_{\alpha }$; and writing $\alpha =\rho (x)e_s$ one has $\rho (w)\alpha =\rho (wx)e_s$, so $t_{\rho (w)\alpha }=(wx)s(wx{)}^{-1}=w(xs{x}^{-1})w^{-1}=w{t}_{\alpha }{w}^{-1}$.

The shorteners are exactly the prefix reflections. Keep the reduced expression of 1.1 and let $r\in T$. If $r=r_i$ is a prefix reflection of the word, then $r_{i}w=w_{i-1}{s}_i{w}_{i-1}^{-1}{w}_{i-1}{s}_i{s}_{i+1}\cdots s_n=w_{i-1}{s}_{i+1}\cdots s_n=s_1\cdots \widehat{s_i}\cdots s_n$, so $\mathrm{\ell }(r_{i}w)<n=\mathrm{\ell }(w)$; and by [F4] the set $\{r_1,\dots ,r_n\}$ equals $\mathrm{\Phi }(w)=\{r:\eta (r,w)=-1\}$ and its members are pairwise distinct. Conversely let $r\in T$ with $\mathrm{\ell }(rw)<\mathrm{\ell }(w)$ and suppose $\eta (r,w)=+1$. The formula of [F4] implies the cocycle identity $\eta (x,uv)=\eta (x,u)\eta (u^{-1}xu,v)$ for all $x\in T$, $u,v\in W$: indeed $(\epsilon ,x)\cdot (uv)=((\epsilon ,x)\cdot u)\cdot v$, and comparing first coordinates of the displayed formula gives the identity. With $x=r$, $u=r$, $v=w$ this gives $\eta (r,rw)=\eta (r,r)\eta (r,w)=\eta (r,r)$; also $\eta (r,1)=1$, since the action of $1$ is the identity. We claim $\eta (r,r)=-1$. Write $r=us{u}^{-1}$ with $s\in S$, $u\in W$, so that $u^{-1}ru=s$; applying the action formula of [F4] successively along the word $us{u}^{-1}$ to $(\epsilon ,r)$ gives $(\epsilon ,r)\cdot u=(\epsilon \eta (r,u),u^{-1}ru)=(\epsilon \eta (r,u),s)$, then $U_s(\epsilon \eta (r,u),s)=(-\epsilon \eta (r,u),s)$ by the definition of $U_s$ in [F4], and then $(\epsilon ,r)\cdot (us{u}^{-1})=(-\epsilon \eta (r,u)\eta (s,u^{-1}),r)$. Comparing with $(\epsilon ,r)\cdot r=(\epsilon \eta (r,r),r)$ from [F4] yields $\eta (r,r)=-\eta (r,u)\eta (s,u^{-1})$. Applying the cocycle identity with $x=s$, $u=u^{-1}$, $v=u$ gives $\eta (s,1)=\eta (s,u^{-1})\eta (us{u}^{-1},u)$, that is $1=\eta (s,u^{-1})\eta (r,u)$; hence $\eta (r,r)=-\eta (r,u{)}^2=-1$ under the supposition $\eta (r,w)=+1$. Therefore $\eta (r,rw)=-1$, so $r\in \mathrm{\Phi }(rw)=\{x\in T:\eta (x,rw)=-1\}$; by [F4] applied to the shorter element $rw$, the element $r$ is one of the prefix reflections of a reduced expression of $rw$, and the first paragraph of this step applied to that element gives $\mathrm{\ell }(r\cdot rw)<\mathrm{\ell }(rw)$, that is $\mathrm{\ell }(w)<\mathrm{\ell }(rw)$, contradicting $\mathrm{\ell }(rw)<\mathrm{\ell }(w)$. Hence $\eta (r,w)=-1$ and $r\in \mathrm{\Phi }(w)=\{r_1,\dots ,r_n\}$. Thus $\{r\in T:\mathrm{\ell }(rw)<\mathrm{\ell }(w)\}=\{r_1,\dots ,r_n\}$.

The dictionary (iii) and (iv). Let $\alpha ,\beta \in \mathrm{\Phi }$. Since $\rho (s)e_s=-e_s$ by [F6], the root $-\alpha$ has the representation $-\alpha =-\rho (w)e_s=\rho (ws)e_s$, so $t_{-\alpha }=(ws)s(ws{)}^{-1}=ws{w}^{-1}=t_{\alpha }$. If $t_{\alpha }=t_{\beta }$, then $r_{\alpha }=\rho (t_{\alpha })=\rho (t_{\beta })=r_{\beta }$ by 1.3 (ii); by [F3] and $B(\alpha ,\alpha )=1$ of [F1] the $(-1)$-eigenspace of $r_{\alpha }$ is $\mathbb{R}\alpha$, and that of $r_{\beta }$ is $\mathbb{R}\beta$, so $\mathbb{R}\alpha =\mathbb{R}\beta$ and $\beta =\lambda \alpha$ with $\lambda \mathrm{\ne }0$; then $1=B(\beta ,\beta )={\lambda }^{2}B(\alpha ,\alpha )={\lambda }^2$, so $\lambda =\pm 1$. Conversely $t_{\alpha }=t_{\pm \alpha }$ by the first part. Hence $t_{\alpha }=t_{\beta }$ if and only if $\beta =\pm \alpha$, which is (iii). For (iv), the map $[\alpha ]\mapsto t_{\alpha }$ on classes $\{\pm \alpha \}$ is well defined by (i) and (iii) and injective by (iii); it is surjective because every element of $T$ is of the form $ws{w}^{-1}=t_{\rho (w)e_s}$ by 1.1. Hence it is a bijection. Since $\mathrm{\Phi }={\mathrm{\Phi }}_{+}\bigsqcup {\mathrm{\Phi }}_{-}$ with ${\mathrm{\Phi }}_{-}=-{\mathrm{\Phi }}_{+}$ by [F6], every class $\{\alpha ,-\alpha \}$ contains exactly one positive root; composing $\alpha \mapsto [\alpha ]$ with the class bijection $[\alpha ]\mapsto t_{\alpha }$ gives a bijection ${\mathrm{\Phi }}_{+}\to T$, $\alpha \mapsto t_{\alpha }$.

Strong exchange (3). Let $w=s_1\cdots s_n$ be a reduced expression and $t\in T$ with $\mathrm{\ell }(tw)<\mathrm{\ell }(w)$. By 1.4 the shorteners of $w$ are exactly the prefix reflections $r_1,\dots ,r_n$, which are pairwise distinct; hence $t=r_i$ for a unique $i\in \{1,\dots ,n\}$, and $tw=r_{i}w=s_1\cdots \widehat{s_i}\cdots s_n$ by the computation in 1.4. For the root clause let $\alpha \in {\mathrm{\Phi }}_{+}$ satisfy $t=t_{\alpha }$. By 1.1, $r_i=w_{i-1}{s}_i{w}_{i-1}^{-1}=t_{\rho (w_{i-1})e_{s_i}}$; since $\mathrm{\ell }(w_{i-1}{s}_i)=i>\mathrm{\ell }(w_{i-1})$, the criterion [F7] gives $\rho (w_{i-1})e_{s_i}\in {\mathrm{\Phi }}_{+}$. Both $\alpha$ and $\rho (w_{i-1})e_{s_i}$ are positive roots with the same $t$-image $t$, so (iii) from 2.1 gives $\alpha =\rho (w_{i-1})e_{s_i}=\rho (s_1\cdots s_{i-1})e_{s_i}$. Finally, the prefix formula of (2) from 1.2 shows that this root lies in $N(w^{-1})$. This proves (3), while (1) is 1.3 with 2.1 and (2) is 1.2.