Assumption 2.1. Unless otherwise noted, we always assume that Coxeter groups are irreducible and their Coxeter matrices have the signature $(n-1,1)$ , where $n$ is the rank.

Proof. By Remark 3.1, $K$ contains $x$ with $x\not =s_{\unicode[STIX]{x1D6FC}_{0}}\cdot x$ for some $\unicode[STIX]{x1D6FC}_{0}\in \unicode[STIX]{x1D6E5}$ .

Fix $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E5}$ arbitrarily. Since we assume that $W$ is irreducible, the Coxeter graph associated with $W$ is connected (cf. [Reference Humphreys7, Section 2.2]). Hence there is a path from $\unicode[STIX]{x1D6FC}_{0}$ to $\unicode[STIX]{x1D6FC}$ in the Coxeter graph, that is to say, there is a sequence of simple roots $(\unicode[STIX]{x1D6FC}_{0},\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{k})$ such that $\unicode[STIX]{x1D6FC}_{k}=\unicode[STIX]{x1D6FC}$ and $B(\unicode[STIX]{x1D6FC}_{i-1},\unicode[STIX]{x1D6FC}_{i})\not =0$ for $i=1,\ldots ,k$ , where $k$ is some positive integer.

For $i=0$ , by the above discussions, there is a point $x_{0}\in K$ such that $x_{0}$ is not fixed by $s_{\unicode[STIX]{x1D6FC}_{0}}$ , i.e., $x_{0}\not =s_{\unicode[STIX]{x1D6FC}_{0}}\cdot x_{0}$ . For $i=1$ , we have $B(\unicode[STIX]{x1D6FC}_{0},\unicode[STIX]{x1D6FC}_{1})\not =0$ . On the other hand, since

$$\begin{eqnarray}\displaystyle B(s_{\unicode[STIX]{x1D6FC}_{0}}\cdot x_{0},\unicode[STIX]{x1D6FC}_{1}) & = & \displaystyle \frac{B(x_{0}-2B(x_{0},\unicode[STIX]{x1D6FC}_{0})\unicode[STIX]{x1D6FC}_{0},\unicode[STIX]{x1D6FC}_{1})}{|s_{\unicode[STIX]{x1D6FC}_{0}}(x_{0})|_{1}}\nonumber\\ \displaystyle & = & \displaystyle \frac{B(x_{0},\unicode[STIX]{x1D6FC}_{1})-2B(x_{0},\unicode[STIX]{x1D6FC}_{0})B(\unicode[STIX]{x1D6FC}_{0},\unicode[STIX]{x1D6FC}_{1})}{|s_{\unicode[STIX]{x1D6FC}_{0}}(x_{0})|_{1}},\nonumber\end{eqnarray}$$

if $B(x_{0},\unicode[STIX]{x1D6FC}_{1})=0$ , then $B(s_{\unicode[STIX]{x1D6FC}_{0}}\cdot x_{0},\unicode[STIX]{x1D6FC}_{1})\not =0$ because $B(x_{0},\unicode[STIX]{x1D6FC}_{0})\not =0$ and $B(\unicode[STIX]{x1D6FC}_{0},\unicode[STIX]{x1D6FC}_{1})\not =0$ . Hence either $B(x_{0},\unicode[STIX]{x1D6FC}_{1})$ or $B(s_{\unicode[STIX]{x1D6FC}_{0}}\cdot x_{0},\unicode[STIX]{x1D6FC}_{1})$ is nonzero. This means that either $x_{0}$ or $s_{\unicode[STIX]{x1D6FC}_{0}}\cdot x_{0}$ is not fixed by $s_{\unicode[STIX]{x1D6FC}_{1}}$ . Let $x_{1}$ be such point. Here we remark that since $x_{0}\in K$ and $K$ is $W$ -invariant, we know that $s_{\unicode[STIX]{x1D6FC}_{0}}\cdot x_{0}\in K$ , so $x_{1}$ belongs to $K$ . Similarly, we obtain that either $x_{1}$ or $s_{\unicode[STIX]{x1D6FC}_{1}}\cdot x_{1}$ is not fixed by $s_{\unicode[STIX]{x1D6FC}_{2}}$ . Let $x_{2}$ be such point. By repeating this consideration, we eventually obtain $x_{k}\in K$ such that $x_{k}$ is not fixed by $s_{\unicode[STIX]{x1D6FC}}$ , as required.◻

Proof. Let $K\subset \widehat{Q}$ be a $W$ -invariant closed subset. We may show that $E\subset K$ .

Let

$$\begin{eqnarray}K^{\prime }=\left(\mathop{\bigcup }_{x,y\in K,x\not =y}L(x,y)\bigg\backslash\hspace{0.0pt}W(V_{0})\right)\setminus \widehat{Q_{-}}.\end{eqnarray}$$

Thus $K^{\prime }$ is also a $W$ -invariant set. In what follows, we claim the inclusion

$$\begin{eqnarray}\overline{K^{\prime }}\cap \widehat{Q}\subset K,\end{eqnarray}$$

where $\overline{K^{\prime }}$ is the closure of $K^{\prime }$ with respect to the Euclidean topology on $V$ . Suppose, on the contrary, that $\overline{K^{\prime }}\cap \widehat{Q}\not \subset K$ , i.e., there is $p\in \overline{K^{\prime }}\cap \widehat{Q}$ such that $p\in \widehat{Q}\setminus K$ . Then there is an open neighborhood of $p$ in $\widehat{Q}$ such that $U\subset \widehat{Q}\setminus K$ because $K$ is closed. Take a sequence $\{p_{i}\}$ in $K^{\prime }$ converging to $p$ . By the definition of $K^{\prime }$ , there are $x_{i}$ and $y_{i}$ in $K$ with $x_{i}\not =y_{i}$ such that $p_{i}\in L(x_{i},y_{i})$ for each $i$ . We fix such $x_{i}$ and $y_{i}$ and assume that $x_{i}$ is visible from $p_{i}$ and $y_{i}$ is not. Let $v_{i}=\frac{1}{2}p_{i}+\frac{1}{2}x_{i}$ . Then we have $B(v_{i},v_{i})=\frac{1}{4}(B(p_{i},p_{i})+2B(p_{i},x_{i}))$ . Since $p_{i}$ converges to $p\in \widehat{Q}$ , $B(p_{i},p_{i})$ goes to 0. However, since $x_{i}$ never goes to $p$ , we have $B(p_{i},p_{i})\leqslant |B(p_{i},x_{i})|$ for sufficiently large $i$ . (Otherwise, by taking a subsequence $\{p_{j_{k}}\}$ of $\{p_{i}\}$ with $B(p_{j_{k}},p_{j_{k}})>|B(p_{j_{k}},x_{j_{k}})|$ for each $k$ , one has $0\leqslant |B(p_{j_{k}},x_{j_{k}})|<B(p_{j_{k}},p_{j_{k}})\rightarrow 0$ , which means that $x_{j_{k}}$ goes to $p$ , a contradiction.) Moreover, since $x_{i}$ never goes to $p$ again, one has $B(p_{i},x_{i})<0$ for sufficiently large $i$ by Proposition 2.8. Hence, for sufficiently large $i$ , we have

$$\begin{eqnarray}\displaystyle B(v_{i},v_{i}) & = & \displaystyle \frac{1}{4}(B(p_{i},p_{i})+2B(p_{i},x_{i}))\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{1}{4}(|B(p_{i},x_{i})|+2B(p_{i},x_{i}))\leqslant \frac{B(p_{i},x_{i})}{4}\leqslant 0.\nonumber\end{eqnarray}$$

On the one hand, since $x_{i}$ is visible from $p_{i}$ , each point in $L[x_{i},p_{i}]\setminus \{x_{i},p_{i}\}$ does not belong to $\widehat{Q}\cup \widehat{Q_{-}}$ . In particular, $v_{i}\not \in \widehat{Q}\cup \widehat{Q_{-}}$ . Thus, we have $B(v_{i},v_{i})>0$ for every $i$ , a contradiction. Therefore, we conclude the desired inclusion $\overline{K^{\prime }}\cap \widehat{Q}\subset K$ .

For each $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E5}$ , when we take $x\in K$ with $x\not =s_{\unicode[STIX]{x1D6FC}}\cdot x$ , $L(x,s_{\unicode[STIX]{x1D6FC}}\cdot x)$ intersects with $\unicode[STIX]{x1D6FC}$ . Since $x$ and $s_{\unicode[STIX]{x1D6FC}}\cdot x$ belong to $K$ , $\unicode[STIX]{x1D6FC}$ belongs to $K^{\prime }$ . By Lemma 3.2, we can take such an element of $K$ for every $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E5}$ . Hence $\unicode[STIX]{x1D6E5}\subset K^{\prime }$ . Since $K^{\prime }$ is $W$ -invariant, we also have $W\cdot \unicode[STIX]{x1D6E5}=\widehat{\unicode[STIX]{x1D6F7}}\subset K^{\prime }$ . Thus, the accumulation points of $W\cdot \unicode[STIX]{x1D6E5}$ , which is nothing but $E$ , should be contained in $\overline{K^{\prime }}\cap \widehat{Q}\subset K$ . Therefore, we have $E\subset K$ .◻

Proof of Theorem 1.3.

For any $x\in E$ , it is obvious that $\overline{W\cdot x}\subset E$ . Moreover, since $\overline{W\cdot x}$ is $W$ -invariant closed set, from Proposition 3.3, we also have $E\subset \overline{W\cdot x}$ , as desired.◻

Proof. Fix $k\in \mathbb{Z}$ arbitrarily. By the remark above, we have a constant $C_{k}\geqslant 1$ , which depends only on $k$ , so that for each $n\in \mathbb{N}$ ,

$$\begin{eqnarray}{C_{k}}^{-1}||w^{n}(x)|_{1}|\leqslant ||w^{n+k}(x)|_{1}|\leqslant C_{k}||w^{n}(x)|_{1}|.\end{eqnarray}$$

Then we see that

$$\begin{eqnarray}\displaystyle |B(w^{n_{i}}\cdot x,w^{n_{i}+k}\cdot x)| & = & \displaystyle \frac{|B(w^{n_{i}}(x),w^{n_{i}+k}(x))|}{||w^{n_{i}}(x)|_{1}|\cdot ||w^{n_{i}+k}(x)|_{1}|}\nonumber\\ \displaystyle & = & \displaystyle \frac{|B(x,w^{k}(x))|}{||w^{n_{i}}(x)|_{1}|\cdot ||w^{n_{i}+k}(x)|_{1}|}\leqslant C_{k}\frac{|B(x,w^{k}(x))|}{||w^{n_{i}}(x)|_{1}|^{2}}\rightarrow 0,\nonumber\end{eqnarray}$$

as $n_{i}\rightarrow \infty$ . By Proposition 2.8, we have the conclusion.◻

Proof. We see this by calculating directly;

$$\begin{eqnarray}\displaystyle B(p,p^{\prime })=B(w(p),w(p^{\prime }))=\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}^{\prime }B(p,p^{\prime }). & & \displaystyle \nonumber\end{eqnarray}$$

Since $p$ and $p^{\prime }$ are distinct and residing on $\widehat{Q}$ , we have $B(p,p^{\prime })\not =0$ by Proposition 2.8. Hence $\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}^{\prime }=1$ , as required.◻

Proof. Since $\widehat{Q}$ is compact, there exists a converging subsequence $(w^{n_{i}}\cdot x)_{n_{i}}$ of $(w^{n}\cdot x)_{n}$ . Let $y$ be the convergent point of the sequence above.

Notice that $w$ has two eigenvectors in $\widehat{Q}$ when $\unicode[STIX]{x1D706}\not =\pm 1$ . In such case, we denote the other eigenvector by $p^{\prime }$ and the corresponding eigenvalue by $\unicode[STIX]{x1D706}^{\prime }$ . By Lemma 4.2, $\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}^{\prime }=1$ must be satisfied.

We consider an eigenvalue $\unicode[STIX]{x1D706}^{\prime }$ with $|\unicode[STIX]{x1D706}^{\prime }|<1$ and the corresponding eigenvector $p^{\prime }$ . Then we have

$$\begin{eqnarray}\displaystyle |B(w^{n_{i}}\cdot x,p^{\prime })|=\frac{|B(w^{n_{i}}(x),p^{\prime })|}{||w^{n_{i}}(x)|_{1}|}=\frac{|B(x,w^{-n_{i}}(p^{\prime }))|}{||w^{n_{i}}(x)|_{1}|}=\frac{|\unicode[STIX]{x1D706}^{n_{i}}|\cdot |B(x,p^{\prime })|}{||w^{n_{i}}(x)|_{1}|} & & \displaystyle \nonumber\end{eqnarray}$$

and $|B(w^{n_{i}}\cdot x,p^{\prime })|\rightarrow |B(y,p^{\prime })|$ as $n_{i}\rightarrow \infty$ . Since $|\unicode[STIX]{x1D706}^{n_{i}}|\rightarrow \infty$ , we have $||w^{n_{i}}(x)|_{1}|\rightarrow \infty$ . Now there exists a constant $C\geqslant 1$ so that

$$\begin{eqnarray}C^{-1}||x|_{1}|\leqslant ||w(x)|_{1}|\leqslant C||x|_{1}|,\end{eqnarray}$$

which is independent of $x$ . Therefore,

$$\begin{eqnarray}\displaystyle & & \displaystyle |B(w^{n_{i}}\cdot x,w\cdot y)|=\frac{|B(w^{n_{i}}(x),w(y))|}{||w^{n_{i}}(x)|_{1}|\cdot ||w(y)|_{1}|}=\frac{|B(w^{n_{i}-1}(x),y)|}{||w^{n_{i}}(x)|_{1}|\cdot ||w(y)|_{1}|}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \,\frac{C|B(w^{n_{i}-1}(x),y)|}{||w^{n_{i}-1}(x)|_{1}|\cdot ||w(y)|_{1}|}=\frac{C}{||w(y)|_{1}|}|B(w^{n_{i}-1}\cdot x,y)|\rightarrow 0,\nonumber\end{eqnarray}$$

as $n_{i}\rightarrow \infty$ by Lemma 4.1. This implies that $y=w\cdot y$ . Since we can apply this argument for each converging subsequence of $(w^{n}\cdot x)_{n}$ and its convergent point, we can deduce that the convergent point, say, $y$ , is fixed by $w$ . By Lemma 4.2, we have the following two possibilities:

1. (1) $w$ has only one fixed point in $\widehat{Q}$ ;

2. (2) $w$ has two fixed points $p,p^{\prime }$ in $\widehat{Q}$ .

If (1), then it is obvious that $y=p$ . If (2), since $|\unicode[STIX]{x1D706}^{-1}|<1$ and $||w^{n_{i}}(x)|_{1}|\rightarrow \infty$ as $n_{i}\rightarrow \infty$ , one has

$$\begin{eqnarray}\displaystyle |B(w^{n_{i}}\cdot x,p)| & = & \displaystyle \frac{|B(w^{n_{i}}(x),p)|}{||w^{n_{i}}(x)|_{1}|}=\frac{|B(x,w^{-n_{i}}(p))|}{||w^{n_{i}}(x)|_{1}|}\nonumber\\ \displaystyle & = & \displaystyle \frac{|\unicode[STIX]{x1D706}^{-n_{i}}|\cdot |B(x,p)|}{||w^{n_{i}}(x)|_{1}|}\rightarrow 0.\nonumber\end{eqnarray}$$

In both cases (1) and (2), all converging subsequences of $(w^{n}\cdot x)_{n}$ converge to the same point. Thus we have the conclusion.◻

Proof. When $B(\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FF}^{\prime })=-1$ , let $\unicode[STIX]{x1D702}=\frac{1}{2}\unicode[STIX]{x1D6FF}+\frac{1}{2}\unicode[STIX]{x1D6FF}^{\prime }$ . Then it is easy to see that $\widehat{\unicode[STIX]{x1D702}}=\lim _{n\rightarrow \infty }(s_{\unicode[STIX]{x1D6FF}}s_{\unicode[STIX]{x1D6FF}^{\prime }})^{n}\cdot \unicode[STIX]{x1D70C}\in E$ . Moreover, we have $B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FF}})=B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FF}^{\prime }})=0$ . Thus $\widehat{\unicode[STIX]{x1D702}}$ is visible from $\widehat{\unicode[STIX]{x1D6FF}}$ and $\widehat{\unicode[STIX]{x1D6FF}^{\prime }}$ by Lemma 2.5(i). In addition, by Remark 3.1, $\widehat{\unicode[STIX]{x1D702}}$ is not visible from $\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ . Let $A_{\unicode[STIX]{x1D6FF}^{\prime \prime }}=\{\widehat{\unicode[STIX]{x1D702}}\}$ . Then $A_{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ satisfies the desired properties.

Assume $B(\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FF}^{\prime })>-1$ . Then the order of $s_{\unicode[STIX]{x1D6FF}}s_{\unicode[STIX]{x1D6FF}^{\prime }}$ is finite. Let $m$ be a positive integer such that $(s_{\unicode[STIX]{x1D6FC}}s_{\unicode[STIX]{x1D6FD}})^{m}=1$ . Thus the order of the parabolic subgroup $W^{\prime }$ generated by $s_{\unicode[STIX]{x1D6FC}}$ and $s_{\unicode[STIX]{x1D6FD}}$ is $2m$ . Let $T=W^{\prime }\cdot \unicode[STIX]{x1D70C}$ and let $T_{\unicode[STIX]{x1D6FC}}\subset T$ (resp. $T_{\unicode[STIX]{x1D6FD}}\subset T$ ) be the set of the points in $T$ which are visible from $\unicode[STIX]{x1D6FC}$ (resp. $\unicode[STIX]{x1D6FD}$ ). We prove that there is $\widehat{\unicode[STIX]{x1D702}}\in \widehat{Q}$ such that $\{\widehat{\unicode[STIX]{x1D702}}\}$ satisfies the required property.

Suppose that there does not exist $\widehat{\unicode[STIX]{x1D702}}$ , i.e., $T_{\unicode[STIX]{x1D6FC}}\cap T_{\unicode[STIX]{x1D6FD}}=\emptyset$ . Then $T_{\unicode[STIX]{x1D6FC}}$ and $T_{\unicode[STIX]{x1D6FD}}$ have the same cardinality. In fact, $s_{\unicode[STIX]{x1D6FD}}:T_{\unicode[STIX]{x1D6FC}}\rightarrow T_{\unicode[STIX]{x1D6FD}}$ is well defined by $T_{\unicode[STIX]{x1D6FC}}\cap T_{\unicode[STIX]{x1D6FD}}=\emptyset$ and Lemma 2.5(ii). Moreover, this is injective. Thus, one has $|T_{\unicode[STIX]{x1D6FC}}|\leqslant |T_{\unicode[STIX]{x1D6FD}}|$ . Similarly, $|T_{\unicode[STIX]{x1D6FD}}|\leqslant |T_{\unicode[STIX]{x1D6FC}}|$ . Hence $|T_{\unicode[STIX]{x1D6FC}}|=|T_{\unicode[STIX]{x1D6FD}}|$ , denoted by $m^{\prime }$ . Moreover, since $s_{\unicode[STIX]{x1D6FC}}:T\setminus T_{\unicode[STIX]{x1D6FC}}\rightarrow T_{\unicode[STIX]{x1D6FC}}$ is injective, one has $|T_{\unicode[STIX]{x1D6FC}}|\geqslant |T\setminus T_{\unicode[STIX]{x1D6FC}}|=|T|-|T_{\unicode[STIX]{x1D6FC}}|$ . Thus, $|T|\leqslant 2|T_{\unicode[STIX]{x1D6FC}}|=|T_{\unicode[STIX]{x1D6FC}}|+|T_{\unicode[STIX]{x1D6FD}}|\leqslant |T|$ . Hence $T\setminus (T_{\unicode[STIX]{x1D6FC}}\cup T_{\unicode[STIX]{x1D6FD}})=\emptyset$ . We write $T_{\unicode[STIX]{x1D6FC}}=\{\unicode[STIX]{x1D70C}_{1},\ldots ,\unicode[STIX]{x1D70C}_{m^{\prime }}\}$ and $T_{\unicode[STIX]{x1D6FD}}=\{\unicode[STIX]{x1D70C}_{1}^{\prime },\ldots ,\unicode[STIX]{x1D70C}_{m^{\prime }}^{\prime }\}$ . Observe that $s_{\unicode[STIX]{x1D6FC}}$ and $s_{\unicode[STIX]{x1D6FD}}$ act on $\{1,\ldots ,m^{\prime }\}$ as permutations $\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}},\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FD}}:\{1,\ldots ,m^{\prime }\}\rightarrow \{1,\ldots ,m^{\prime }\}$ so that $s_{\unicode[STIX]{x1D6FD}}\cdot T_{\unicode[STIX]{x1D6FC}}=\{\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}(1)}^{\prime },\ldots ,\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}(m^{\prime })}^{\prime }\}$ and $s_{\unicode[STIX]{x1D6FC}}\cdot T_{\unicode[STIX]{x1D6FD}}=\{\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FD}}(1)},\ldots ,\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FD}}(m^{\prime })}\}$ . In particular, we recognize that each image of the points in $T_{\unicode[STIX]{x1D6FD}}$ by $s_{\unicode[STIX]{x1D6FC}}$ must be visible from $\unicode[STIX]{x1D6FC}$ and vice versa. Moreover, the permutation $\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}}:=\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FD}}$ has order $m^{\prime }$ . By Proposition 2.13(i), each $B$ -reflection extends the length of visible curves. Thus we see that

$$\begin{eqnarray}\displaystyle d_{B}(\unicode[STIX]{x1D70C}_{i},\unicode[STIX]{x1D70C}_{j}) & {\geqslant} & \displaystyle d_{B}(\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}}(i)},\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}}(j)})\nonumber\\ \displaystyle & {\geqslant} & \displaystyle d_{B}(\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}}^{2}(i)},\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}}^{2}(j)})\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \cdots \nonumber\\ \displaystyle & {\geqslant} & \displaystyle d_{B}(\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}}^{m^{\prime }}(i)},\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}}^{m^{\prime }}(j)})=d_{B}(\unicode[STIX]{x1D70C}_{i},\unicode[STIX]{x1D70C}_{j}),\nonumber\end{eqnarray}$$

for any $i,j\in \{1,2,\ldots ,m^{\prime }\}$ . Thus all the equalities of these inequalities must be satisfied. Since $\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}}(i)}=s_{\unicode[STIX]{x1D6FC}}s_{\unicode[STIX]{x1D6FD}}\cdot \unicode[STIX]{x1D70C}_{i}$ , $d_{B}(\unicode[STIX]{x1D70C}_{i},\unicode[STIX]{x1D70C}_{j})=d_{B}(s_{\unicode[STIX]{x1D6FD}}\cdot \unicode[STIX]{x1D70C}_{i},s_{\unicode[STIX]{x1D6FD}}\cdot \unicode[STIX]{x1D70C}_{j})$ also holds. From Proposition 2.13(iii), $d_{B}(\unicode[STIX]{x1D70C}_{i},\unicode[STIX]{x1D70C}_{j})=d_{B}(s_{\unicode[STIX]{x1D6FD}}\cdot \unicode[STIX]{x1D70C}_{i},s_{\unicode[STIX]{x1D6FD}}\cdot \unicode[STIX]{x1D70C}_{j})$ implies that $\unicode[STIX]{x1D70C}_{i}$ and $\unicode[STIX]{x1D70C}_{j}$ should belong to $\unicode[STIX]{x2202}V_{\unicode[STIX]{x1D6FD}}\cap T\subset T_{\unicode[STIX]{x1D6FD}}$ if $\unicode[STIX]{x1D70C}_{i}\not =\unicode[STIX]{x1D70C}_{j}$ , while $\unicode[STIX]{x1D70C}_{i}$ and $\unicode[STIX]{x1D70C}_{j}$ belong to $T_{\unicode[STIX]{x1D6FC}}$ . However, since $T_{\unicode[STIX]{x1D6FC}}\cap T_{\unicode[STIX]{x1D6FD}}=\emptyset$ , this cannot happen. Hence, $\unicode[STIX]{x1D70C}_{i}=\unicode[STIX]{x1D70C}_{j}$ . In particular, each of $T_{\unicode[STIX]{x1D6FC}}$ and $T_{\unicode[STIX]{x1D6FD}}$ consists of one element. Let $T_{\unicode[STIX]{x1D6FC}}=\{\unicode[STIX]{x1D70C}_{1}\}$ and $T_{\unicode[STIX]{x1D6FD}}=\{\unicode[STIX]{x1D70C}_{1}^{\prime }\}$ . Then four points $\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D70C}_{1},\unicode[STIX]{x1D70C}_{1}^{\prime }$ and $\unicode[STIX]{x1D6FD}$ should lie in the same line $L(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})$ . On the other hand, since $W^{\prime }$ is finite, $L(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})$ does not intersect with $\widehat{Q}$ (see Remark 5.1), a contradiction. Hence $T_{\unicode[STIX]{x1D6FC}}\cap T_{\unicode[STIX]{x1D6FD}}$ is not empty. This means that $\widehat{\unicode[STIX]{x1D702}}$ exists in $W^{\prime }\cdot \unicode[STIX]{x1D70C}$ . Thus $\widehat{\unicode[STIX]{x1D702}}\in E$ .

Let $\widehat{\unicode[STIX]{x1D702}}=x_{\unicode[STIX]{x1D6FC}}\widehat{\unicode[STIX]{x1D6FC}}+x_{\unicode[STIX]{x1D6FD}}\widehat{\unicode[STIX]{x1D6FD}}+x_{\unicode[STIX]{x1D6FE}}\widehat{\unicode[STIX]{x1D6FE}}$ . Since $\widehat{\unicode[STIX]{x1D702}}\in E\subset \text{conv}(\widehat{\unicode[STIX]{x1D6E5}})$ , we have $0\leqslant x_{\unicode[STIX]{x1D6FC}},x_{\unicode[STIX]{x1D6FD}},x_{\unicode[STIX]{x1D6FE}}\leqslant 1$ and $x_{\unicode[STIX]{x1D6FC}}+x_{\unicode[STIX]{x1D6FD}}+x_{\unicode[STIX]{x1D6FE}}=1$ . Since

$$\begin{eqnarray}\displaystyle & \displaystyle B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D702}})=x_{\unicode[STIX]{x1D6FC}}B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FC}})+x_{\unicode[STIX]{x1D6FD}}B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FD}})+x_{\unicode[STIX]{x1D6FE}}B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FE}})=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FC}})\geqslant 0\quad \text{and}\quad B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FD}})\geqslant 0, & \displaystyle \nonumber\end{eqnarray}$$

we have $B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FE}})<0$ or $x_{\unicode[STIX]{x1D6FC}}B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FC}})=x_{\unicode[STIX]{x1D6FD}}B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FD}})=x_{\unicode[STIX]{x1D6FE}}B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FE}})=0$ . Suppose that the latter case happens. Since $\widehat{\unicode[STIX]{x1D702}}\not =\widehat{\unicode[STIX]{x1D6FE}}$ , (a) both of $x_{\unicode[STIX]{x1D6FC}}$ and $x_{\unicode[STIX]{x1D6FD}}$ are positive, or (b) either $x_{\unicode[STIX]{x1D6FC}}$ or $x_{\unicode[STIX]{x1D6FD}}$ is 0 and the other is positive.

1. (a) Suppose that both of $x_{\unicode[STIX]{x1D6FC}}$ and $x_{\unicode[STIX]{x1D6FD}}$ are positive. If $x_{\unicode[STIX]{x1D6FE}}>0$ , then we have $B(\widehat{\unicode[STIX]{x1D702}},\unicode[STIX]{x1D6FC})=B(\widehat{\unicode[STIX]{x1D702}},\unicode[STIX]{x1D6FD})=B(\widehat{\unicode[STIX]{x1D702}},\unicode[STIX]{x1D6FE})=0$ , a contradiction to Remark 3.1. If $x_{\unicode[STIX]{x1D6FE}}=0$ , then $\widehat{\unicode[STIX]{x1D702}}\in E\cap \text{conv}(\{\widehat{\unicode[STIX]{x1D6FC}},\widehat{\unicode[STIX]{x1D6FD}}\})$ , a contradiction to the finiteness of $W^{\prime }$ .

2. (b) Suppose that either $x_{\unicode[STIX]{x1D6FC}}$ or $x_{\unicode[STIX]{x1D6FD}}$ is 0 and the other is positive, say, $x_{\unicode[STIX]{x1D6FC}}>0$ and $x_{\unicode[STIX]{x1D6FD}}=0$ . Then $x_{\unicode[STIX]{x1D6FE}}>0$ because of $\widehat{\unicode[STIX]{x1D702}}\not =\widehat{\unicode[STIX]{x1D6FC}}$ . Thus $\widehat{\unicode[STIX]{x1D702}}\in \text{conv}(\{\widehat{\unicode[STIX]{x1D6FC}},\widehat{\unicode[STIX]{x1D6FE}}\})$ . Since

   $$\begin{eqnarray}B(\widehat{\unicode[STIX]{x1D702}},\unicode[STIX]{x1D6FD})=x_{\unicode[STIX]{x1D6FC}}B(\widehat{\unicode[STIX]{x1D6FC}},\unicode[STIX]{x1D6FD})+x_{\unicode[STIX]{x1D6FE}}B(\widehat{\unicode[STIX]{x1D6FE}},\unicode[STIX]{x1D6FD})\geqslant 0,\end{eqnarray}$$
   we have $B(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})=B(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FD})=0$ . This is a contradiction to our assumption “irreducible”. ◻

Proof. The line joining a point in $A_{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ and $\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ always crosses $\widehat{Q}\setminus A_{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ . This means that $\text{int}(A_{\unicode[STIX]{x1D6FF}^{\prime \prime }})$ is not visible from $\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ , where $\text{int}(\cdot )$ denotes the interior relative to $\widehat{Q}$ . By Remark 5.1, it follows that both two end points of $A_{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ are contained in $E$ such that one end point of $A_{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ is visible from $\widehat{\unicode[STIX]{x1D6FF}}$ and the other is visible from $\widehat{\unicode[STIX]{x1D6FF}^{\prime }}$ . Moreover, each of both two end points $\widehat{\unicode[STIX]{x1D702}}$ is contained in $\text{conv}(\{\widehat{\unicode[STIX]{x1D6FF}},\widehat{\unicode[STIX]{x1D6FF}^{\prime }}\})$ , so it is written like $\widehat{\unicode[STIX]{x1D702}}=r\widehat{\unicode[STIX]{x1D6FF}}+(1-r)\widehat{\unicode[STIX]{x1D6FF}^{\prime }}$ , where $0<r<1$ . Then it follows that $B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }})=rB(\widehat{\unicode[STIX]{x1D6FF}},\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }})+(1-r)B(\widehat{\unicode[STIX]{x1D6FF}^{\prime }},\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }})\leqslant 0$ . Our assumption “irreducible” says that either $B(\widehat{\unicode[STIX]{x1D6FF}},\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }})$ or $B(\widehat{\unicode[STIX]{x1D6FF}^{\prime }},\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }})$ is negative. Thus we obtain $B(\widehat{\unicode[STIX]{x1D702}},\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }})<0$ , i.e., $\widehat{\unicode[STIX]{x1D702}}$ is not visible from $\widehat{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ by Lemma 2.5(i). Hence $A_{\unicode[STIX]{x1D6FF}^{\prime \prime }}$ satisfies the required property.◻

Proof of Theorem 1.2 in the case of rank 3.

Fix three sets $A_{\unicode[STIX]{x1D6FC}},A_{\unicode[STIX]{x1D6FD}}$ and $A_{\unicode[STIX]{x1D6FE}}$ in the statements of Lemmas 5.2 and 5.3. For our proof, we introduce the following notation.

• ∙ Let $C_{\unicode[STIX]{x1D6FE}}\subset \widehat{Q}$ be the connected closed set joining end points of $A_{\unicode[STIX]{x1D6FC}}$ and $A_{\unicode[STIX]{x1D6FD}}$ satisfying $C_{\unicode[STIX]{x1D6FE}}\subset V_{\unicode[STIX]{x1D6FE}}$ . Such a set $C_{\unicode[STIX]{x1D6FE}}$ is uniquely determined. Similarly, we also define $C_{\unicode[STIX]{x1D6FC}}$ and $C_{\unicode[STIX]{x1D6FD}}$ . Note that the end points of each $C_{\unicode[STIX]{x1D6FF}}$ ( $\unicode[STIX]{x1D6FF}\in \unicode[STIX]{x1D6E5}$ ) are some points in $E$ .

Note that we have the equality

(3) $$\begin{eqnarray}\displaystyle \widehat{Q}=A_{\unicode[STIX]{x1D6FC}}\sqcup A_{\unicode[STIX]{x1D6FD}}\sqcup A_{\unicode[STIX]{x1D6FE}}\sqcup \text{int}(C_{\unicode[STIX]{x1D6FC}})\sqcup \text{int}(C_{\unicode[STIX]{x1D6FD}})\sqcup \text{int}(C_{\unicode[STIX]{x1D6FE}}). & & \displaystyle\end{eqnarray}$$

Suppose, on the contrary, that $E\subsetneq \widehat{Q}\setminus (\bigcup _{i=1}^{m}W\cdot D_{i})$ . Since $E$ is closed, there exists a connected open set $U_{1}\subset (\widehat{Q}\setminus (\bigcup _{i=1}^{m}W\cdot D_{i}))\setminus E$ . Note that the boundary of $U_{1}$ consists of two points. By taking $U_{1}$ as a maximal one, we may assume that $\overline{U_{1}}\cap E$ is nonempty and consists of two points $a_{1},b_{1}$ which are the end points of $\overline{U_{1}}$ . Since the end points of $C_{\unicode[STIX]{x1D6FC}},C_{\unicode[STIX]{x1D6FD}}$ and $C_{\unicode[STIX]{x1D6FE}}$ are contained in $E$ , $\overline{U_{1}}\subset C_{\unicode[STIX]{x1D6FC}}$ or $\overline{U_{1}}\subset C_{\unicode[STIX]{x1D6FD}}$ or $\overline{U_{1}}\subset C_{\unicode[STIX]{x1D6FE}}$ occurs. Moreover, by $C_{\unicode[STIX]{x1D6FF}}\subset V_{\unicode[STIX]{x1D6FF}}$ for $\unicode[STIX]{x1D6FF}\in \{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}\}$ and Proposition 2.13(i), $\overline{U_{1}}$ is a geodesic.

Let, say, $\overline{U_{1}}\subset C_{\unicode[STIX]{x1D6FE}}$ . Let $U_{2}=s_{\unicode[STIX]{x1D6FE}}\cdot U_{1}$ and let $a_{2}$ and $b_{2}$ be the end points of $U_{2}$ . Then $U_{2}\cap D_{i}=\emptyset$ for any $i$ by definition of $U_{2}=s_{\unicode[STIX]{x1D6FE}}\cdot U_{1}$ . Namely, $U_{2}\cap A_{\unicode[STIX]{x1D6FF}}=\emptyset$ for any $\unicode[STIX]{x1D6FF}\in \{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}\}$ . By (3) together with $U_{2}=s_{\unicode[STIX]{x1D6FE}}\cdot U_{1}\not \subset C_{\unicode[STIX]{x1D6FE}}\subset V_{\unicode[STIX]{x1D6FE}}$ , $U_{2}\subset \text{int}(C_{\unicode[STIX]{x1D6FC}})$ or $U_{2}\subset \text{int}(C_{\unicode[STIX]{x1D6FD}})$ occurs. Moreover, $\overline{U_{2}}\cap E=\{a_{2},b_{2}\}$ . From Proposition 2.13(i) and (iii), we notice that $\ell _{B}(U_{1})<\ell _{B}(U_{2})$ .

Similarly, for each $n\geqslant 1$ , if $U_{n}\subset \text{int}(C_{\unicode[STIX]{x1D6FF}})$ for some $\unicode[STIX]{x1D6FF}\in \{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}\}$ , then let $U_{n+1}=s_{\unicode[STIX]{x1D6FF}}\cdot U_{n}$ and let $a_{n+1}$ and $b_{n+1}$ be the end points of $\overline{U_{n+1}}$ . Moreover, we also have $\ell _{B}(U_{n})<\ell _{B}(U_{n+1})$ . In particular, $U_{i}\not =U_{j}$ for any $i$ and $j$ with $i\not =j$ . In addition, $\overline{U_{n}}\cap E=\{a_{n},b_{n}\}$ , where $a_{n}$ and $b_{n}$ are the end points of $\overline{U_{n}}$ . If $U_{i}\cap U_{j}\not =\emptyset$ for some $i$ and $j$ with $i\not =j$ , then an end point of $U_{i}$ belongs to $U_{j}$ or an end point of $U_{j}$ belongs to $U_{i}$ . Since each end point of $U_{i}$ and $U_{j}$ is an element of $E$ , we obtain $U_{i}\cap E\not =\emptyset$ or $U_{j}\cap E\not =\emptyset$ , a contradiction. Hence $U_{i}\cap U_{j}=\emptyset$ for all $i$ and $j$ with $i\not =j$ .

Now it follows that one of $C_{\unicode[STIX]{x1D6FC}}$ , $C_{\unicode[STIX]{x1D6FD}}$ and $C_{\unicode[STIX]{x1D6FE}}$ , say, $C_{\unicode[STIX]{x1D6FE}}$ , contains infinitely many open sets $U_{n}$ . Let $U_{i_{k}}\subset \text{int}(C_{\unicode[STIX]{x1D6FE}})$ for $k\geqslant 1$ , where $i_{1}<i_{2}<i_{3}<\cdots \,$ . Since $U_{i_{k}}\cap U_{i_{k^{\prime }}}=\emptyset$ for any $k\not =k^{\prime }$ , the disjoint union $\bigsqcup _{k=1}^{\infty }U_{i_{k}}$ is contained in $\text{int}(C_{\unicode[STIX]{x1D6FE}})$ . On the one hand, we have $\ell _{B}(C_{\unicode[STIX]{x1D6FE}})<\infty$ . On the other hand, since we have $0<\ell _{B}(U_{i_{1}})<\ell _{B}(U_{i_{2}})<\cdots \,$ , we obtain that $\sum _{k=1}^{\infty }\ell _{B}(U_{i_{k}})=\infty$ , a contradiction.

Therefore, we conclude that $E=\widehat{Q}\setminus \bigcup _{i=1}^{m}W\cdot D_{i}$ , as required.◻