Weight subsets with equal root sums are unique

If $w\mathrm{\ne }1$, choose a reduced word $w=s_{i_1}\cdots s_{i_k}$ with $k=\mathrm{\ell }(w)>0$. The product $s_{i_1}w=s_{i_2}\cdots s_{i_k}$ is represented by a word of length $k-1$, so $\mathrm{\ell }(s_{i_1}w)\le k-1=\mathrm{\ell }(w)-1$; by [F1] the length changes by exactly one, so $\mathrm{\ell }(s_{i_1}w)=\mathrm{\ell }(w)-1$ and, with $\alpha :={\alpha }_{i_1}$, the criterion of [F1] gives $w^{-1}\alpha <0$. Write $w=s_{\alpha }{w}^{\mathrm{\prime }}$ with $w^{\mathrm{\prime }}=s_{\alpha }w$, so $\mathrm{\ell }(w)=\mathrm{\ell }(w^{\mathrm{\prime }})+1$ and $\alpha \in {\mathrm{\Pi }}_w$.

For $\alpha$ as in step 1.1 one has $s_{\alpha }({\mathrm{\Phi }}^{+}\setminus \{\alpha \})={\mathrm{\Phi }}^{+}\setminus \{\alpha \}$ by [F2] and $\alpha \mathrm{\notin }{w}^{\mathrm{\prime }}({\mathrm{\Phi }}^{-})$ because $(w^{\mathrm{\prime }}{)}^{-1}\alpha =w^{-1}{s}_{\alpha }\alpha =-w^{-1}\alpha >0$. Hence $s_{\alpha }({\mathrm{\Pi }}_{w^{\mathrm{\prime }}})=s_{\alpha }({\mathrm{\Phi }}^{+})\cap s_{\alpha }(w^{\mathrm{\prime }}({\mathrm{\Phi }}^{-}))=(({\mathrm{\Phi }}^{+}\setminus \{\alpha \})\cup \{-\alpha \})\cap w({\mathrm{\Phi }}^{-})=({\mathrm{\Phi }}^{+}\setminus \{\alpha \})\cap w({\mathrm{\Phi }}^{-})={\mathrm{\Pi }}_w\setminus \{\alpha \}$, so ${\mathrm{\Pi }}_w=s_{\alpha }({\mathrm{\Pi }}_{w^{\mathrm{\prime }}})\cup \{\alpha \}$ is a disjoint union.

Induction on $\mathrm{\ell }(w)$ proves the three identities simultaneously. For $w=1$ all three sides vanish. For $w\mathrm{\ne }1$, apply step 2.1 and the induction hypothesis to $w^{\mathrm{\prime }}$: $\mathrm{\#}{\mathrm{\Pi }}_w=\mathrm{\#}{\mathrm{\Pi }}_{w^{\mathrm{\prime }}}+1=\mathrm{\ell }(w^{\mathrm{\prime }})+1=\mathrm{\ell }(w)$; similarly $\rho -w\rho =(\rho -s_{\alpha }\rho )+s_{\alpha }(\rho -w^{\mathrm{\prime }}\rho )=\alpha +s_{\alpha }{\sum }_{\beta \in {\mathrm{\Pi }}_{w^{\mathrm{\prime }}}}\beta =\alpha +{\sum }_{\beta \in {\mathrm{\Pi }}_{w^{\mathrm{\prime }}}}{s}_{\alpha }\beta ={\sum }_{\gamma \in {\mathrm{\Pi }}_w}\gamma$, using $s_{\alpha }\rho =\rho -\alpha$; and therefore $w\circ 0=w\rho -\rho =-{\sum }_{\alpha \in {\mathrm{\Pi }}_w}\alpha$.

It remains to prove uniqueness. Let $S\subseteq {\mathrm{\Phi }}^{+}$ with ${\sum }_{\alpha \in S}\alpha ={\sum }_{\alpha \in {\mathrm{\Pi }}_w}\alpha$. We induct on $\mathrm{\ell }(w)$. For $w=1$, ${\mathrm{\Pi }}_1=\mathrm{\varnothing }$ and the assumed sum of $S$ is zero. Every positive root has nonnegative simple-root coefficients with at least one positive coefficient, so a nonempty set of positive roots has a sum with at least one positive coefficient and cannot sum to zero. Thus $S=\mathrm{\varnothing }={\mathrm{\Pi }}_1$. For $w\mathrm{\ne }1$ use the element $\alpha$ of step 1.1. If $\alpha \in S$, put $S^{\mathrm{\prime }}=s_{\alpha }(S\setminus \{\alpha \})$. Then $S^{\mathrm{\prime }}\subseteq {\mathrm{\Phi }}^{+}$ by [F2], and using step 3.1 for $w^{\mathrm{\prime }}$ and linearity of $s_{\alpha }$ one computes ${\sum }_{\beta \in S^{\mathrm{\prime }}}\beta =s_{\alpha }({\sum }_{\beta \in S}\beta -\alpha )=s_{\alpha }(\rho -w\rho -\alpha )=\rho -w^{\mathrm{\prime }}\rho ={\sum }_{\beta \in {\mathrm{\Pi }}_{w^{\mathrm{\prime }}}}\beta$. By induction $S^{\mathrm{\prime }}={\mathrm{\Pi }}_{w^{\mathrm{\prime }}}$, hence $S=s_{\alpha }({\mathrm{\Pi }}_{w^{\mathrm{\prime }}})\cup \{\alpha \}={\mathrm{\Pi }}_w$.

If $\alpha \mathrm{\notin }S$, then $S\subseteq {\mathrm{\Phi }}^{+}\setminus \{\alpha \}$, so $s_{\alpha }(S)\subseteq {\mathrm{\Phi }}^{+}\setminus \{\alpha \}$ and $s_{\alpha }(S)\cup \{\alpha \}\subseteq {\mathrm{\Phi }}^{+}$ is a disjoint union with ${\sum }_{\beta \in s_{\alpha }(S)\cup \{\alpha \}}\beta =s_{\alpha }(\rho -w\rho )+\alpha =\rho -w^{\mathrm{\prime }}\rho ={\sum }_{\beta \in {\mathrm{\Pi }}_{w^{\mathrm{\prime }}}}\beta$. By induction $s_{\alpha }(S)\cup \{\alpha \}={\mathrm{\Pi }}_{w^{\mathrm{\prime }}}$, but $\alpha \mathrm{\notin }{\mathrm{\Pi }}_{w^{\mathrm{\prime }}}$ because $(w^{\mathrm{\prime }}{)}^{-1}\alpha =-w^{-1}\alpha >0$ by step 1.1. This contradiction rules out $\alpha \mathrm{\notin }S$, so the previous case applies and $S={\mathrm{\Pi }}_w$ always.