Tits Cones, Chambers, and Parabolic Stabilizers

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.

Walls are root hyperplanes. For $w\in W$, $s\in S$ and $f\in V^{\ast }$ one has $f\in w{H}_{e_s}$ if and only if $(w^{-1}\cdot f)(e_s)=0$, and by the dual action this is $f(\rho (w)e_s)=0$, that is $f\in H_{\rho (w)e_s}$; hence $w{H}_{e_s}=H_{\rho (w)e_s}$. As $\mathrm{\Phi }=\{\rho (w)e_s:w\in W,s\in S\}$, the walls $w{H}_{e_s}$ of the chambers are exactly the root hyperplanes $H_{\alpha }$ with $\alpha \in \mathrm{\Phi }$. This is (1).

The side rule. Let $w\in W$, $s\in S$, $x\in C^{\circ }$ and put $\beta :=\rho (w{)}^{-1}{e}_s$, so that $(w\cdot x)(e_s)=x(\beta )$. By the signed root system, either $\beta \in {\mathrm{\Phi }}_{+}$ and then $x(\beta )>0$, or $\beta \in {\mathrm{\Phi }}_{-}$ and then $x(\beta )<0$, so the sign of $(w\cdot x)(e_s)$ is the same for every $x\in C^{\circ }$. By the root-length criterion applied to $w^{-1}$, $\beta =\rho (w^{-1})e_s\in {\mathrm{\Phi }}_{+}$ if and only if $\mathrm{\ell }(w^{-1}s)>\mathrm{\ell }(w^{-1})$, and $\mathrm{\ell }(w^{-1}s)=\mathrm{\ell }(sw)$, $\mathrm{\ell }(w^{-1})=\mathrm{\ell }(w)$ because lengths are inversion-invariant; hence $w{C}^{\circ }\subseteq \{f:f(e_s)>0\}$ if and only if $\mathrm{\ell }(sw)>\mathrm{\ell }(w)$, and $w{C}^{\circ }\subseteq \{f:f(e_s)<0\}$ if and only if $\mathrm{\ell }(sw)<\mathrm{\ell }(w)$. Parity $\mathrm{\ell }(sw)=\mathrm{\ell }(w)\pm 1$ makes the two alternatives exclusive and exhaustive. If $\mathrm{\ell }(sw)<\mathrm{\ell }(w)$ and $x\in C$, then $\beta \in {\mathrm{\Phi }}_{-}$, so $\beta =-b$ with $b\in {\mathrm{\Phi }}_{+}\setminus \{0\}$ and $(w\cdot x)(e_s)=x(-b)=-x(b)\le 0$ because $x\ge 0$ on $V_{+}$ and ${\mathrm{\Phi }}_{+}\subseteq V_{+}\setminus \{0\}$; thus $wC\subseteq \{f:f(e_s)\le 0\}$, while $C\subseteq \{f:f(e_s)\ge 0\}$ by definition. This is (2).

Collision, claim and base case. Claim: if $f,g\in C$, $w\in W$ and $w\cdot f=g$, then $f=g$ and $w\in W_{S(f)}$. Proceed by induction on $k:=\mathrm{\ell }(w)$. For $k=0$ one has $w=1$, so $f=g$ and $1\in W_{S(f)}$.

Collision, induction step. Let $k\ge 1$ and suppose the claim known for all elements of length $k-1$. A reduced expression of $w$ begins with some $s\in S$ and has length $k$, so $w=s{w}^{\mathrm{\prime }}$ with $\mathrm{\ell }(w^{\mathrm{\prime }})=k-1$ and $\mathrm{\ell }(sw)=k-1<\mathrm{\ell }(w)$. By step 1.2 the chambers $C$ and $wC$ lie on opposite sides of the wall $H_{e_s}$: $C\subseteq \{f:f(e_s)\ge 0\}$ and $wC\subseteq \{f:f(e_s)\le 0\}$. Since $g=w\cdot f$ lies in both, $g(e_s)=0$. Then $s\cdot g=g$: for $t\mathrm{\ne }s$ the reflection formula gives $r_s{e}_t=e_t-2B(e_t,e_s)e_s$, so $(s\cdot g)(e_t)=g(r_s{e}_t)=g(e_t)-2B(e_t,e_s)g(e_s)=g(e_t)$ using $g(e_s)=0$; and $(s\cdot g)(e_s)=g(r_s{e}_s)=g(-e_s)=0=g(e_s)$. Two linear functionals agreeing on the basis $(e_s{)}_{s\in S}$ are equal, so $s\cdot g=g$ and hence $g=s\cdot g=s\cdot (w\cdot f)=(sw)\cdot f=w^{\mathrm{\prime }}\cdot f$. The induction hypothesis applied to $w^{\mathrm{\prime }}$ and the pair $(f,g)$ gives $f=g$ and $w^{\mathrm{\prime }}\in W_{S(f)}$. Since $g(e_s)=0$ and $f=g$, also $s\in S(f)$; therefore $w=s{w}^{\mathrm{\prime }}\in W_{S(f)}$ because $W_{S(f)}$ is the subgroup generated by $S(f)$. This completes (3).

Point stabilizers. Let $f\in C$. If $s\in S(f)$, the computation of step 2.1 with $g=f$ gives $s\cdot f=f$, so every element of the subgroup $W_{S(f)}$ generated by these $s$ fixes $f$; hence $W_{S(f)}\subseteq {\mathrm{Stab}}_W(f)$. Conversely, if $w\cdot f=f$ with $f\in C$, then (3) applied to the pair $(f,f)$ gives $w\in W_{S(f)}$; hence ${\mathrm{Stab}}_W(f)=W_{S(f)}$. For the second formula, ${\mathrm{Stab}}_W(w\cdot g)=w{\mathrm{Stab}}_W(g)w^{-1}$ holds in any group action: an element $x$ fixes $g$ if and only if $wx{w}^{-1}$ fixes $w\cdot g$. Given $f\in U$ and $w$ with $w^{-1}\cdot f\in C$, applying the first formula to $g:=w^{-1}\cdot f$ gives ${\mathrm{Stab}}_W(f)=w{W}_{S(w^{-1}\cdot f)}{w}^{-1}$. This is (4).

The intersection rule. Let $f\in C$ and $w\in W$. If $f\in wC\cap C$, then $w^{-1}\cdot f\in C$ and (3) applied to the element $w^{-1}$ and the pair $(f,w^{-1}\cdot f)$ gives $f=w^{-1}\cdot f$ and $w^{-1}\in W_{S(f)}$, that is $w\in W_{S(f)}$; conversely, if $f\in C$ and $w\in W_{S(f)}$, then $w\cdot f=f$ by step 3.1, so $f\in wC\cap C$. This proves $wC\cap C=\{f\in C:w\in W_{S(f)}\}$. For the second description, note that $f\in {\bar{C}}_T$ means $f\in C$ and $f(e_s)=0$ for all $s\in T$, so $T\subseteq S(f)$; hence if $f\in {\bar{C}}_T$ and $w\in W_T$ then $w\in W_{S(f)}$, and conversely for $f\in C$ with $w\in W_{S(f)}$ one has $f\in {\bar{C}}_{S(f)}$ with $w\in W_{S(f)}$. Therefore $wC\cap C=\{f\in C:w\in W_{S(f)}\}={\bigcup }_{T\subseteq S,w\in W_T}{\bar{C}}_T$. This is (5).

The strict fundamental domain. Every $f\in U$ lies in some chamber $wC$, so $w^{-1}\cdot f\in C$: every $W$-orbit contained in $U$ meets $C$. If $x,y\in C$ lie in one orbit, say $y=w\cdot x$, then (3) gives $x=y$. For the disjointness of open chambers, if $z\in w{C}^{\circ }\cap v{C}^{\circ }$ put $f:=w^{-1}\cdot z$ and $g:=v^{-1}\cdot z$, so that $f,g\in C^{\circ }$ and $g=(v^{-1}w)\cdot f$; (3) applied to the element $v^{-1}w$ gives $f=g$ and $v^{-1}w\in W_{S(f)}=W_{\mathrm{\varnothing }}=\{1\}$, so $v^{-1}w=1$ and $w=v$. The chambers meeting in a face are described by step 4.1. This is (6).