The finite-negativity criterion, the reduction step, and convexity of the Tits cone

Let $f\in U$ and let $w\in W$ satisfy $g:=w\cdot f\in C$; such a $w$ exists because $U={\bigcup }_{u\in W}uC$. For $\alpha \in \mathrm{Neg}(f)$, the dual action gives $g(\rho (w)\alpha )=f(\alpha )<0$. Since $g$ is nonnegative on $V_{+}$ and $\rho (w)\alpha$ is a signed root, this forces $\rho (w)\alpha \in {\mathrm{\Phi }}_{-}$, hence $\alpha \in N(w)$. Thus $\mathrm{Neg}(f)\subseteq N(w)$ and $\mathrm{\mid }\mathrm{Neg}(f)\mathrm{\mid }\le \mathrm{\mid }N(w)\mathrm{\mid }=\mathrm{\ell }(w)$, proving the forward implication of (1) and all of (5).

For every $f\in V^{\ast }$ one has $\mathrm{Neg}(f)=\mathrm{\varnothing }$ if and only if $f\in C$. If $f\in C$ and $\alpha \in {\mathrm{\Phi }}_{+}$, write $\alpha ={\sum }_s{c}_s{e}_s$ with $c_s\ge 0$; then $f(\alpha )={\sum }_s{c}_{s}f(e_s)\ge 0$, so no positive root is negative at $f$ and $\mathrm{Neg}(f)=\mathrm{\varnothing }$. Conversely, if $\mathrm{Neg}(f)=\mathrm{\varnothing }$ then $f(e_s)\ge 0$ for every $s$, because $e_s\in {\mathrm{\Phi }}_{+}$ and $f(e_s)<0$ would put $e_s$ into $\mathrm{Neg}(f)$; hence $f\in C$. This is clause (2).

Let $N:=\mathrm{Neg}(f)$ be finite and nonempty, with $n:=\mathrm{\mid }N\mathrm{\mid }$, and let $f\mathrm{\notin }C$, so that $N\mathrm{\ne }\mathrm{\varnothing }$ by step 1.2. Since $N\subseteq V_{+}\setminus \{0\}$, an element $\alpha ={\sum }_s{c}_s{e}_s$ of $N$ has all coefficients $\ge 0$ and some coefficient $>0$, and the strict inequality $f(\alpha )={\sum }_s{c}_{s}f(e_s)<0$ forces $f(e_s)<0$ for some $s\in S$; for that $s$ one has $e_s\in {\mathrm{\Phi }}_{+}$ and $e_s\in N$. This is the existence clause of (3).

Keep $N$, $f$ and $s$ from step 2.1, so that $e_s\in N$ and $f(e_s)<0$, and let $\beta \in {\mathrm{\Phi }}_{+}$. Because $\rho (s)=r_s$ is an involution, the dual action gives $(s\cdot f)(\beta )=f(r_s\beta )$. For $\beta =e_s$ this is $f(r_s{e}_s)=f(-e_s)=-f(e_s)>0$, so $e_s\mathrm{\notin }\mathrm{Neg}(s\cdot f)$. For $\beta \mathrm{\ne }{e}_s$ the reflection $r_s$ permutes ${\mathrm{\Phi }}_{+}\setminus \{e_s\}$, so $r_s\beta \in {\mathrm{\Phi }}_{+}\setminus \{e_s\}$, and $\beta \in \mathrm{Neg}(s\cdot f)$ if and only if $f(r_s\beta )<0$, that is if and only if $r_s\beta \in N\setminus \{e_s\}$; since $\beta \mapsto r_s\beta$ is a bijection of ${\mathrm{\Phi }}_{+}\setminus \{e_s\}$ onto itself, this gives $\mathrm{Neg}(s\cdot f)=r_s(N\setminus \{e_s\})$ and $\mathrm{\mid }\mathrm{Neg}(s\cdot f)\mathrm{\mid }=\mathrm{\mid }N\setminus \{e_s\}\mathrm{\mid }=n-1$. This is the reduction identity of (3).

Claim: every $f\in V^{\ast }$ with $\mathrm{Neg}(f)$ finite lies in $U$. Induct on the natural number $n=\mathrm{\mid }\mathrm{Neg}(f)\mathrm{\mid }$. For $n=0$ step 1.2 gives $f\in C\subseteq U$. For $n\ge 1$ one has $f\mathrm{\notin }C$ by step 1.2, so step 2.1 supplies $s\in S$ with $e_s\in N$, and step 3.1 gives $\mathrm{\mid }\mathrm{Neg}(s\cdot f)\mathrm{\mid }=n-1$, so by the induction hypothesis $s\cdot f\in U$, say $s\cdot f=u\cdot g$ with $g\in C$; then $f=s\cdot (s\cdot f)=(su)\cdot g\in U$ because $w^{\mathrm{\prime }}U=U$ for every $w^{\mathrm{\prime }}\in W$. Iterating the reduction step decreases the size of the negative-root set by exactly one each time and stops precisely when that set is empty, which by step 1.2 is exactly when the current element lies in $C$; hence after exactly $n=\mathrm{\mid }\mathrm{Neg}(f)\mathrm{\mid }$ steps one reaches $w\in W$ with $w\cdot f\in C$. This proves the converse direction of (1) and the iteration clause of (3).

For $\lambda >0$ one has $\mathrm{Neg}(\lambda f)=\mathrm{Neg}(f)$, since $(\lambda f)(\alpha )=\lambda f(\alpha )$ and $\lambda >0$; also $\mathrm{Neg}(0)=\mathrm{\varnothing }$, because $0(\alpha )=0$ is not $<0$. For $t\in [0,1]$, if $\alpha \in \mathrm{Neg}((1-t)f+tg)$ then $(1-t)f(\alpha )+tg(\alpha )<0$, which forces $f(\alpha )<0$ or $g(\alpha )<0$, since both coefficients are $\ge 0$; hence $\mathrm{Neg}((1-t)f+tg)\subseteq \mathrm{Neg}(f)\cup \mathrm{Neg}(g)$. By the criterion of steps 1.1 and 4.1, every $f\in U$ satisfies $\lambda f\in U$ for all $\lambda \ge 0$ (the case $\lambda =0$ is $0\in C\subseteq U$), and $(1-t)f+tg\in U$ for all $f,g\in U$ and $t\in [0,1]$, endpoints $t=0$ and $t=1$ included. This is (4).

No Choice is used: the induction is finite, and each reduction instantiates a single simple reflection with negative coordinate. Steps 1.1–5.1 establish all five clauses.