Root sign coherence and the action of simple reflections on positive roots

Set-up. Every $f\in C^{\circ }$ satisfies $f(e_s)>0$ for all $s\in S$, so that $C^{\circ }\subseteq B_s$ and $C^{\circ }\subseteq B_s\cap B_t$ for all $s\mathrm{\ne }t$; moreover $C^{\circ }\mathrm{\ne }\mathrm{\varnothing }$. Every root is of the form $\rho (w)e_s$ with $B(\rho (w)e_s,\rho (w)e_s)=1$, hence nonzero, and $-\rho (w)e_s=\rho (ws)e_s$ is again a root, so $\mathrm{\Phi }=-\mathrm{\Phi }$.

The sign criterion, forward direction. Let $v={\sum }_s{\lambda }_s{e}_s\in V_{+}\setminus \{0\}$, so all ${\lambda }_s\ge 0$ and some ${\lambda }_r>0$. For $f\in C^{\circ }$ linearity gives $f(v)={\sum }_s{\lambda }_{s}f(e_s)\ge {\lambda }_{r}f(e_r)>0$, since all summands are nonnegative and the summand at $r$ is positive.

The sign criterion, converse direction. Let $v\mathrm{\notin }{V}_{+}\setminus \{0\}$. If $v=0$ then $f(v)=0$ for every $f$, so assume $v\mathrm{\notin }{V}_{+}$; then some coordinate ${\lambda }_r=v(r)$ is negative. Let $g:=f_r+\epsilon {\sum }_s{f}_s$ with $\epsilon >0$ so small that $\epsilon \mathrm{\mid }{\sum }_{s}v(s)\mathrm{\mid }<\mathrm{\mid }{\lambda }_r\mathrm{\mid }$. Then $g(e_t)={\delta }_{rt}+\epsilon >0$ for every $t\in S$, so $g\in C^{\circ }$, and $g(v)={\lambda }_r+\epsilon {\sum }_{s}v(s)<0$. Hence some element of $C^{\circ }$ evaluates $v$ negatively, and the criterion of (1) holds in both directions.

The equivalence of the two criteria. By 1.2 and 1.3, for every $v\in V$: $v\in V_{+}\setminus \{0\}$ if and only if $f(v)>0$ for every $f\in C^{\circ }$.

Roots have a sign. Fix $w\in W$ and $s\in S$. For $f\in V^{\ast }$ one has $(w^{-1}\cdot f)(e_s)=f(\rho (w^{-1}{)}^{-1}{e}_s)=f(\rho (w)e_s)$; by [F1] applied to the element $w^{-1}$ either $w^{-1}{C}^{\circ }\subseteq B_s$ or $w^{-1}{C}^{\circ }\subseteq s{B}_s$. In the first case $f(\rho (w)e_s)>0$ for every $f\in C^{\circ }$, so $\rho (w)e_s\in V_{+}\setminus \{0\}$ by 2.1 and hence in ${\mathrm{\Phi }}_{+}$; in the second case $f(\rho (w)e_s)<0$ for every $f\in C^{\circ }$, so $-f(\rho (w)e_s)=f(-\rho (w)e_s)>0$ for every $f\in C^{\circ }$, whence $-\rho (w)e_s\in V_{+}\setminus \{0\}$ by 2.1 and $\rho (w)e_s\in -V_{+}\setminus \{0\}$, so it lies in ${\mathrm{\Phi }}_{-}$. The two alternatives are exclusive because $V_{+}\cap (-V_{+})=\{0\}$ while $\rho (w)e_s\mathrm{\ne }0$ by 1.1. Since every root is some $\rho (w)e_s$ and $-\rho (w)e_s=\rho (ws)e_s$ is a root, this gives $\mathrm{\Phi }={\mathrm{\Phi }}_{+}\bigsqcup {\mathrm{\Phi }}_{-}$ and ${\mathrm{\Phi }}_{-}=-{\mathrm{\Phi }}_{+}$; taking $w=1$ gives $e_s=\rho (1)e_s\in V_{+}\setminus \{0\}$ and hence $e_s\in {\mathrm{\Phi }}_{+}$. Reading the two cases displayed above as equivalences with 2.1 gives $\rho (w)e_s\in {\mathrm{\Phi }}_{+}⟺w^{-1}{C}^{\circ }\subseteq B_s$ and $\rho (w)e_s\in {\mathrm{\Phi }}_{-}⟺w^{-1}{C}^{\circ }\subseteq s{B}_s$, and with 2.1 the last sentence of (2) follows.

Simple reflections act on positive roots. Fix $s\in S$. That $r_s{e}_s=-e_s$ is [F2]. Let $\alpha \in {\mathrm{\Phi }}_{+}\setminus \{e_s\}$. Since $r_s=\rho (s)$ and $\mathrm{\Phi }$ is $\rho (W)$-invariant by [F3], $r_s\alpha \in \mathrm{\Phi }$, so by 3.1 either $r_s\alpha \in {\mathrm{\Phi }}_{+}$ or $r_s\alpha \in {\mathrm{\Phi }}_{-}$. Suppose $r_s\alpha \in {\mathrm{\Phi }}_{-}$, that is $r_s\alpha \in -V_{+}\setminus \{0\}$, and write $r_s\alpha =\alpha -2B(\alpha ,e_s)e_s$ by [F2] with $B(e_s,e_s)=1$. In coordinates: $(r_s\alpha )(r)=\alpha (r)$ for $r\mathrm{\ne }s$ and $(r_s\alpha )(s)=\alpha (s)-2B(\alpha ,e_s)$. Since $\alpha \in V_{+}\setminus \{0\}$ all coordinates $\alpha (r)$ are nonnegative and not all vanish, and since $r_s\alpha \in -V_{+}\setminus \{0\}$ all its coordinates are nonpositive; for $r\mathrm{\ne }s$ both statements apply to $\alpha (r)$, so $\alpha (r)=0$ for all $r\mathrm{\ne }s$, that is $\alpha =\alpha (s)e_s$ with $\alpha (s)>0$. Then $1=B(\alpha ,\alpha )=\alpha (s{)}^{2}B(e_s,e_s)=\alpha (s{)}^2$, so $\alpha (s)=1$ and $\alpha =e_s$, contradicting the choice of $\alpha$. Hence $r_s\alpha \in {\mathrm{\Phi }}_{+}$; moreover $r_s\alpha \mathrm{\ne }{e}_s$, because $r_s{e}_s=-e_s$ and $r_s$ is an involution, so $r_s\alpha =e_s$ would give $\alpha =-e_s\mathrm{\notin }{\mathrm{\Phi }}_{+}$. Thus $r_s({\mathrm{\Phi }}_{+}\setminus \{e_s\})\subseteq {\mathrm{\Phi }}_{+}\setminus \{e_s\}$, and applying the involution $r_s$ once more gives equality. Finally $r_s\mathrm{\Phi }=\mathrm{\Phi }$: indeed $r_s=\rho (s)$ maps $\mathrm{\Phi }$ into $\mathrm{\Phi }$ by [F3], and $r_s^2=\mathrm{i}\mathrm{d}$ by [F2], so this map is a bijection of $\mathrm{\Phi }$; consequently $r_s{\mathrm{\Phi }}_{+}=({\mathrm{\Phi }}_{+}\setminus \{e_s\})\cup \{-e_s\}$. This is (3), while (1) is 2.1 and (2) is 3.1.