The interior of the Tits cone, finite parabolic stabilizers, and local finiteness

The empty-rank case and the parabolic span. If $S=\mathrm{\varnothing }$, then $W=\{1\}$, $V^{\ast }=C=U=U^{\circ }=\{0\}$, there is one chamber and no walls, and all five clauses hold with radius $1$; hence assume $S\mathrm{\ne }\mathrm{\varnothing }$. Let $I\subseteq S$, put $V_I:=\mathrm{span}\{e_s:s\in I\}$ and let $f\in C$ with $S(f)=I$. The finite words in $I$ form a subgroup containing $I$, and any subgroup containing $I$ contains every such word; therefore their values are exactly $W_I$. For a generator $t\in I$ the reflection formula gives $r_{t}v=v-2B(v,e_t)e_t$ and $r_t{e}_s-e_s=-2B(e_s,e_t)e_t\in V_I$ for every $s\in S$; since $r_t$ is an involution, this shows by induction on a product of generators — no finiteness of $W_I$ is used — that $\rho (u)$ preserves $V_I$ and that $\rho (u{)}^{-1}{e}_s-e_s\in V_I$ for every $u\in W_I$ and every $s\in S$. In particular $\rho (u{)}^{-1}{e}_s\in V_I$ for $s\in I$, and $f$ vanishes on $V_I$ because $S(f)=I$.

An invariant inner product on $V_I^{\ast }$. Assume $W_I$ finite and $I\mathrm{\ne }\mathrm{\varnothing }$. On the finite-dimensional dual $V_I^{\ast }$ consider the action $(u\cdot \phi )(v)=\phi (\rho (u{)}^{-1}v)$ and the positive definite form $⟨\phi ,\psi {⟩}_0:={\sum }_{s\in I}\phi (e_s)\psi (e_s)$ (positive definite because a functional on $V_I$ vanishing on the basis $(e_s{)}_{s\in I}$ is zero). Since $W_I$ is a finite group, the average $⟨\phi ,\psi ⟩:=\frac{1}{\mathrm{\mid }{W}_I\mathrm{\mid }}{\sum }_{u\in W_I}⟨u\cdot \phi ,u\cdot \psi {⟩}_0$ is well defined, bilinear and symmetric; it is positive definite because for $\phi \mathrm{\ne }0$ every summand is $\ge 0$ and the summand with $u=1$ equals ${\sum }_{s\in I}\phi (e_s{)}^2>0$. Hence $⟨\cdot ,\cdot ⟩$ is an inner product on $V_I^{\ast }$, and it is $W_I$-invariant: substituting $u^{\mathrm{\prime }}=ut$ in the sum shows $⟨t\cdot \phi ,t\cdot \psi ⟩=⟨\phi ,\psi ⟩$ for every $t\in W_I$.

Reflections of this inner product. For $t\in I$ the functional $\phi \mapsto \phi (e_t)$ on $V_I^{\ast }$ is nonzero (evaluate at any $\phi$ with $\phi (e_t)\mathrm{\ne }0$), so by Riesz representation there is $z_t\in V_I^{\ast }$ with $⟨\phi ,z_t⟩=\phi (e_t)$ for all $\phi$; then $z_t\mathrm{\ne }0$, and $z_t$ is orthogonal to the hyperplane $H_t:=\{\phi :\phi (e_t)=0\}$. The action of $t$ is an involution fixing $H_t$ pointwise, because $t\cdot \phi =\phi$ holds exactly when $\phi$ vanishes on $(\rho (t)-\mathrm{i}\mathrm{d})V_I=\mathbb{R}{e}_t$. As $t$ is an isometry of $⟨\cdot ,\cdot ⟩$, for all $\psi$ one has $⟨t\cdot \psi ,z_t⟩=⟨t^2\cdot \psi ,t\cdot z_t⟩=⟨\psi ,t\cdot z_t⟩$, while also $⟨t\cdot \psi ,z_t⟩=(t\cdot \psi )(e_t)=\psi (r_t{e}_t)=\psi (-e_t)=-⟨\psi ,z_t⟩$. Hence $⟨\psi ,t\cdot z_t⟩=-⟨\psi ,z_t⟩$ for all $\psi$, and positive definiteness gives $t\cdot z_t=-z_t\mathrm{\ne }{z}_t$. Every $\psi$ decomposes as $\psi =(\psi -\frac{⟨\psi ,z_t⟩}{\mathrm{\parallel }{z}_t{\mathrm{\parallel }}^2}{z}_t)+\frac{⟨\psi ,z_t⟩}{\mathrm{\parallel }{z}_t{\mathrm{\parallel }}^2}{z}_t$ with the first summand orthogonal to $z_t$, and $t$ agrees with $R(\psi ):=\psi -\frac{2⟨\psi ,z_t⟩}{\mathrm{\parallel }{z}_t{\mathrm{\parallel }}^2}{z}_t$ on $z_t^{\perp }$ and on $z_t$; hence $t\cdot \psi =\psi -2\frac{\psi (e_t)}{\mathrm{\parallel }{z}_t{\mathrm{\parallel }}^2}{z}_t(\psi \in V_I^{\ast })\mathrm{.}$

The finite-parabolic lemma. If $I=\mathrm{\varnothing }$, then $V_I^{\ast }=\{0\}$ and $u=1$ gives the conclusion; assume $I\mathrm{\ne }\mathrm{\varnothing }$. Let $\phi \in V_I^{\ast }$ and let $\gamma \in V_I^{\ast }$ be the functional $\gamma (v):={\sum }_{s\in I}{c}_s$ for $v={\sum }_{s\in I}{c}_s{e}_s$; then $\gamma (e_t)=1$ for every $t\in I$ and $\gamma =0$ when $I=\mathrm{\varnothing }$. Choose $u\in W_I$ maximizing the real number $⟨u\cdot \phi ,\gamma ⟩$ over the nonempty finite set $\{⟨v\cdot \phi ,\gamma ⟩:v\in W_I\}$. If $(u\cdot \phi )(e_t)<0$ for some $t\in I$, then step 1.3 applied to $\psi :=u\cdot \phi$ gives $⟨t\cdot (u\cdot \phi ),\gamma ⟩=⟨u\cdot \phi ,\gamma ⟩-\frac{2(u\cdot \phi )(e_t)}{\mathrm{\parallel }{z}_t{\mathrm{\parallel }}^2}⟨z_t,\gamma ⟩=⟨u\cdot \phi ,\gamma ⟩-\frac{2(u\cdot \phi )(e_t)}{\mathrm{\parallel }{z}_t{\mathrm{\parallel }}^2}>⟨u\cdot \phi ,\gamma ⟩,$ using $⟨z_t,\gamma ⟩=\gamma (e_t)=1$; this contradicts maximality, so $(u\cdot \phi )(e_s)\ge 0$ for every $s\in I$. Thus every element of $V_I^{\ast }$ has a $W_I$-translate in the closed chamber of $V_I^{\ast }$.

The infinite-parabolic case (4)(i): points outside $U$ arbitrarily close to $f$. Let $f\in C$ with $I=S(f)$ and $W_I$ infinite. If lengths were bounded on $W_I$ by some $N$, every element of $W_I$ would have a reduced expression of length at most $N$, hence would be the image of one of the finitely many words in $S$ of length at most $N$, and $W_I$ would be finite; so $\mathrm{\ell }$ is unbounded on $W_I$. Every element of $W_I$ has a reduced expression with letters in $I$: writing it as a product of elements of $I$ and repeatedly deleting two letters until no further deletion shortens the word produces a reduced expression whose letters are among the original ones, hence lie in $I$. For such a reduced expression $u=s_1\cdots s_n$ with $s_i\in I$, the suffix formula exhibits every element of $N(u)$ as $\rho (v{)}^{-1}{e}_s$ with $v\in W_I$, $s\in I$; since $\rho (v)$ preserves $V_I$, these roots lie in ${\mathrm{\Phi }}_{+}\cap V_I$. As $\mathrm{\mid }N(u)\mathrm{\mid }=\mathrm{\ell }(u)$ is unbounded on $W_I$, the set ${\mathrm{\Phi }}_{+}\cap V_I$ is infinite. Now let ${\delta }_I:={\sum }_{s\in I}{f}_s$ for the dual basis functionals $f_s$, and put $g_t:=f-t{\delta }_I$ for $t>0$. Every $\alpha \in {\mathrm{\Phi }}_{+}\cap V_I$ is a nonnegative combination $\alpha ={\sum }_{s\in I}{c}_s{e}_s$, so $f(\alpha )={\sum }_s{c}_{s}f(e_s)=0$ and ${\delta }_I(\alpha )={\sum }_s{c}_s>0$, whence $g_t(\alpha )=-t{\delta }_I(\alpha )<0$; therefore $\mathrm{Neg}(g_t)$ is infinite and $g_t\mathrm{\notin }U$ by the criterion of finite negativity. Since $g_t-f=-t{\delta }_I$ has coordinate $-t$ on $I$ and $0$ outside $I$, one has $d(g_t,f)=t$, and $t$ can be taken arbitrarily small; hence every neighborhood of $f$ contains points outside $U$, and $f\mathrm{\notin }{U}^{\circ }$. This proves the reverse direction of (1) and clause (4)(i).

The finite case of (1): a neighborhood of $f$ lies in $U$. Keep the notation of step 1.1. Applying the lemma of step 2.1 to $\phi :=g{\mathrm{\mid }}_{V_I}$ for an arbitrary $g\in V^{\ast }$ gives $u\in W_I$ with $(u\cdot (g{\mathrm{\mid }}_{V_I}))(e_s)\ge 0$ for all $s\in I$; since $\rho (u{)}^{-1}{e}_s\in V_I$ for $s\in I$, this is $(u\cdot g)(e_s)\ge 0$ for $s\in I$. For $s\mathrm{\notin }I$ write $\rho (u{)}^{-1}{e}_s=e_s+v_u$ with $v_u\in V_I$; then $(u\cdot g)(e_s)=f(e_s)+(g-f)(e_s)+(g-f)(v_u)$, because $f$ vanishes on $V_I$. If $I\mathrm{\ne }S$, put $\delta :={\min }_{s\mathrm{\notin }I}f(e_s)>0$ and $M:=\max \{\mathrm{\parallel }\rho (u{)}^{-1}{e}_s-e_s{\mathrm{\parallel }}_1:u\in W_I,s\mathrm{\notin }I\}<\mathrm{\infty }$ (a maximum over the finite set $W_I\times (S\setminus I)$ in the coordinates of [F16]), and let $\epsilon :=\delta \mathrm{/}(1+M)$; if $I=S$ put $\epsilon :=1$ and skip the estimate outside $I$. Every $g$ with $d(f,g)<\epsilon$ then satisfies $(u\cdot g)(e_s)\ge f(e_s)-\mathrm{\parallel }g-f{\mathrm{\parallel }}_{\mathrm{\infty }}-\mathrm{\parallel }g-f{\mathrm{\parallel }}_{\mathrm{\infty }}M>\delta -\epsilon (1+M)=0$ for $s\mathrm{\notin }I$, by [F16], and $\ge 0$ for $s\in I$; that is $u\cdot g\in C$ and $g\in u^{-1}C\subseteq U$. Hence the ball $\{g:d(f,g)<\epsilon \}$ lies in $U$, so $f\in U^{\circ }$; this is the finite case of (1).

The boundary meets infinitely many chambers, (4)(ii). Keep $f$ and $I$ from step 2.2. For every $u\in W_I$ the chamber $uC$ contains $f$, because $u\in W_{S(f)}$ fixes $f$; distinct $u$ give distinct chambers: if $uC=u^{\mathrm{\prime }}C$, then with $L:=u^{\mathrm{\prime }-1}u$ one has $L(C)=C$, and for any point $x\in C^{\circ }$, which is nonempty by [F21] and lies in $C$, the points $x$ and $L(x)$ are in $C$ and $L\cdot x=L(x)$, so the collision theorem gives $L\in W_{S(x)}=W_{\mathrm{\varnothing }}=\{1\}$ and $u=u^{\mathrm{\prime }}$. Since $W_I$ is infinite, the infinitely many distinct chambers $uC$ all contain $f$, so every neighborhood of $f$ meets infinitely many distinct chambers. Together with step 2.2 this is (4).

Chambers and walls near $f$, the case $w=1$. Keep $\epsilon$ from step 3.1 and let $vC$ be a chamber meeting the ball $\{g:d(f,g)<\epsilon \}$ at a point $g$; by step 3.1 there is $u\in W_I$ with $g\in uC$, so $u^{-1}\cdot g\in C$ and $v^{-1}\cdot g\in C$. Apply the collision theorem to the pair $(u^{-1}\cdot g,v^{-1}\cdot g)$ and the element $v^{-1}u$: since $(v^{-1}u)\cdot (u^{-1}\cdot g)=v^{-1}\cdot g$, we get $v^{-1}u\in W_{S(u^{-1}\cdot g)}$. For $s\mathrm{\notin }I$ one has $(u^{-1}\cdot g)(e_s)=g(\rho (u)e_s)=f(e_s)+(g-f)(e_s)+(g-f)(\rho (u)e_s-e_s)>0$, because $\mathrm{\parallel }\rho (u)e_s-e_s{\mathrm{\parallel }}_1\le M$ (the map $u\mapsto u^{-1}$ permutes $W_I$) and $d(f,g)<\epsilon$, by [F16]; hence $S(u^{-1}\cdot g)\subseteq I$ and $v^{-1}u\in W_I$, so $v\in u{W}_I\subseteq W_I$ and $vC$ is one of the chambers $u^{\mathrm{\prime }}C$ with $u^{\mathrm{\prime }}\in W_I$. If instead a wall $v{H}_{e_r}$ meets the ball, put $\alpha :=\rho (v)e_r\in \mathrm{\Phi }$, so that $v{H}_{e_r}=H_{\alpha }$; for every $u\in W_I$ the intersection $H_{\alpha }\cap u{C}^{\circ }$ is empty, because a point $u\cdot x$ with $x\in C^{\circ }$ has $(u\cdot x)(\alpha )=x(\rho (u{)}^{-1}\alpha )\mathrm{\ne }0$ as $\rho (u{)}^{-1}\alpha$ is a root and $x$ is strictly signed on roots. Any point $y\in H_{\alpha }$ in the ball lies in some $uC$ with $u\in W_I$ by the covering of step 3.1, and $y\mathrm{\notin }u{C}^{\circ }$, so $y$ is a boundary point of the closed polyhedral cone $uC=\{f:f(\rho (u)e_s)\ge 0\text{ for all }s\}$ and therefore lies in some wall $u{H}_{e_s}$. Thus the nonempty open subset $H_{\alpha }\cap \{g:d(f,g)<\epsilon \}$ of the hyperplane $H_{\alpha }$ is covered by the finitely many subspaces $H_{\alpha }\cap u{H}_{e_s}$ ($u\in W_I$, $s\in S$), each of which is either $H_{\alpha }$ or a proper subspace; a finite union of proper subspaces of a real vector space cannot contain a nonempty open set (given a point of the open set outside the first $m-1$ subspaces, the affine line through it in a direction outside the last subspace meets each remaining subspace in at most one parameter, so some nearby parameter lies in the open set but in no subspace), so $H_{\alpha }=u{H}_{e_s}$ for some $u\in W_I$ and $s\in S$, and the wall is one of the walls $u{H}_{e_s}$ of the chambers $uC$, $u\in W_I$.

Local finiteness at every point of $U^{\circ }$: (3)(i)-(iii). Let $f\in U^{\circ }$ and write $f=w\cdot f_0$ with $f_0\in C$, and put $I:=S(f_0)$; this is possible because $U={\bigcup }_{w}wC$. The map $L(x):=w\cdot x$ is linear by [F20] and is a bijection with inverse $L^{-1}(x)=w^{-1}\cdot x$; both are given in the coordinates $(e_s)$ by real matrices $(m_{st})$ and $(m_{st}^{\mathrm{\prime }})$, so $d(Lx,Ly)\le Cd(x,y)$ and $d(L^{-1}x,L^{-1}y)\le C^{\mathrm{\prime }}d(x,y)$ with $C:={\max }_s{\sum }_t\mathrm{\mid }{m}_{st}\mathrm{\mid }$ and $C^{\mathrm{\prime }}:={\max }_s{\sum }_t\mathrm{\mid }{m}_{st}^{\mathrm{\prime }}\mathrm{\mid }$ (if $S=\mathrm{\varnothing }$ then $V^{\ast }=\{0\}$ and there is nothing to prove), by [F16]; in particular $C>0$ when $S\mathrm{\ne }\mathrm{\varnothing }$. Since $f\in U^{\circ }$ there is ${\epsilon }^{\mathrm{\prime }}>0$ with $B(f,{\epsilon }^{\mathrm{\prime }})\subseteq U$; then $L(B(f_0,{\epsilon }^{\mathrm{\prime }}\mathrm{/}C))\subseteq B(f,{\epsilon }^{\mathrm{\prime }})\subseteq U$, so $B(f_0,{\epsilon }^{\mathrm{\prime }}\mathrm{/}C)\subseteq L^{-1}(U)=U$ because $L(U)=U$ by [F1]; hence $f_0\in U^{\circ }$, and since $f_0\in C$, the contrapositive of step 2.2 gives that $W_I$ is finite. By steps 3.1 and 4.1 applied to $f_0$ there is ${\epsilon }_0>0$ such that the ball about $f_0$ of radius ${\epsilon }_0$ is covered by the chambers $uC$ ($u\in W_I$), every chamber meeting it is one of them, and every wall meeting it is one of the walls $u{H}_{e_s}$ ($u\in W_I$, $s\in S$). Hence $L(B(f_0,{\epsilon }_0))\supseteq B(f,\epsilon )$ for $\epsilon :={\epsilon }_0\mathrm{/}{C}^{\mathrm{\prime }}$: for $y\in B(f,\epsilon )$ the point $x:=L^{-1}(y)$ satisfies $d(x,f_0)=d(L^{-1}y,L^{-1}f)\le C^{\mathrm{\prime }}d(y,f)<{\epsilon }_0$. So the ball about $f$ is contained in ${\bigcup }_{u\in W_I}wuC$, since $L(uC)=wuC$ for every $u$ by [F20]. If a chamber $A=vC$ meets $B(f,\epsilon )$ at $y$, then $L^{-1}(A)=w^{-1}vC$ meets $B(f_0,{\epsilon }_0)$ at $x$, so $w^{-1}vC=uC$ for some $u\in W_I$ by the $w=1$ case of step 4.1, that is $A=wuC$; and if a wall $A=v{H}_{e_s}$ meets $B(f,\epsilon )$, then $L^{-1}(A)=(w^{-1}v)H_{e_s}$ meets $B(f_0,{\epsilon }_0)$, so $w^{-1}v{H}_{e_s}=u{H}_{e_r}$ for some $u\in W_I$, $r\in S$ by step 4.1, that is $A=(wu)H_{e_r}$. Thus every chamber meeting the ball about $f$ is one of the chambers $wuC$ and every wall meeting it is one of the walls $wu{H}_{e_s}$ with $u\in W_I$, $s\in S$. This is (3) for general $f$.

Compact subsets. Let $K\subseteq U^{\circ }$ be compact and let $\mathcal{B}$ be the family of all balls $B(x,r)$ with $x\in K$, $r>0$, that meet only finitely many chambers and walls. Step 5.1 shows that $\mathcal{B}$ covers $K$, without selecting a radius at each point. These balls are open: for $y\in B(x,r)$, the triangle inequality gives $B(y,r-d(x,y))\subseteq B(x,r)$. By [F18], finitely many members of $\mathcal{B}$ cover $K$ (none if $K=\mathrm{\varnothing }$). Every chamber or wall meeting $K$ meets one of these balls, so only finitely many chambers and walls meet $K$.

Clause (2) and the vertex (5). First, $U^{\circ }$ is $W$-invariant: for each $w$, the map $L_w(x)=w\cdot x$ is a linear bijection by [F20], and as in step 5.1 its two coordinate matrices give $d(L_{w}x,L_{w}y)\le C_{w}d(x,y)$ and $d(L_w^{-1}x,L_w^{-1}y)\le C_w^{\mathrm{\prime }}d(x,y)$ by [F16], so $L_w$ is a homeomorphism; since $L_w(U)=U$ by [F1], the image $L_w(U^{\circ })$ is open and contained in $U$, hence $L_w(U^{\circ })\subseteq U^{\circ }$, and applying the same to $L_w^{-1}$ gives $L_w(U^{\circ })=U^{\circ }$. Now if $f\in U^{\circ }$ then $f=w\cdot g$ with $g\in C$, and $g\in U^{\circ }$ by this invariance, so the criterion (1), whose two directions are steps 2.2 and 3.1, makes the group $W_{S(g)}$ finite and $g\in C^f$, whence $f\in w\cdot C^f$; conversely every point of $w\cdot C^f$ lies in $U^{\circ }$ by the finite case of (1) applied to its $w$-translate in $C$ and the $W$-invariance of $U^{\circ }$. Since $C^f=\bigcup \{C_I:I\subseteq S,W_I\text{ finite}\}$ and $f\in C_I$ means $S(f)=I$, this is the identification $C^f=\{f\in C:W_{S(f)}\text{ finite}\}$ and hence $U^{\circ }={\bigcup }_{w\in W}w\cdot C^f$. Applying (1) to $f=0$, whose zero set is $S(0)=S$, gives $0\in U^{\circ }$ if and only if $W_S=W$ is finite; this is (5).