Proof. A root $\gamma $ satisfies $\gamma \in \Phi ^+$ and $s_i \gamma \in \Phi ^-$ if and only if $\gamma =\beta _{s_i}=\alpha _i$ . Also, by the definition of separators,

$$\begin{align*}\mathrm{Sep}(s_i u) \setminus \mathrm{Sep}(u)=\{\beta \in \mathrm{R}({\mathscr{L}}): s_i u \beta \in \Phi^- \text{ and } u\beta \in \Phi^+\}.\end{align*}$$

This set is $\{u^{-1} \alpha _i\}$ if $u^{-1} \alpha _i \in \mathrm {R}({\mathscr {L}})$ and is $\varnothing $ otherwise. The lemma now follows from the definition of $\tau _i$ (Theorem 1.2).

Proof. Fix some $w \in {\mathscr {L}}$ . For every $\beta \in \mathrm {Sep}(u)$ , we have $w\beta \in \Phi ^+$ and $u\beta \in \Phi ^-$ . Writing $u\beta =(uw^{-1})(w\beta )=(uw^{-1})\beta _{r_{w\beta }}$ , we see that $r_{w\beta }$ is a right inversion of $uw^{-1}$ . It follows that ${|\mathrm {Sep}(u)|\leq |\mathrm {Inv}(uw^{-1})|=\ell (uw^{-1})<\infty }$ .

Proof. We proceed by induction on $M = \ell (w)$ . The base case $M = 0$ is trivial, so assume that $M> 0$ . Write $\mathsf {w} = i_M\cdots i_1$ . Let $\mathsf {w}' = i_{M-1}\cdots i_{1}$ be the word obtained by deleting the leftmost letter of $\mathsf {w}$ , and let $w' = s_{i_M} w \in W$ be the element of W represented by the word $\mathsf {w}'$ . Then $\tau _{\mathsf w}(u) = \tau _{i_M}(\tau _{\mathsf w'}(u))$ .

First, suppose that $\tau _{\mathsf w}(u) = s_{i_M}\tau _{\mathsf w'}(u)$ . Then $(w')^{-1} \tau _{\mathsf {w}'}(u) = w^{-1} \tau _{\mathsf {w}}(u)$ . Hence, by Theorem 3.1 and the inductive hypothesis, we have

$$\begin{align*}\mathrm{Sep}(\tau_{\mathsf{w}}(u)) \subseteq \mathrm{Sep}(\tau_{\mathsf{w}'}(u)) \subseteq \mathrm{Sep}((w')^{-1} \tau_{\mathsf{w}'}(u)) = \mathrm{Sep}(w^{-1} \tau_{\mathsf{w}}(u)),\end{align*}$$

as desired.

Next, suppose that $\tau _{\mathsf w}(u) = \tau _{\mathsf w'}(u)$ . Let $v = \tau _{\mathsf w}(u) = \tau _{\mathsf w'}(u)$ . We wish to prove that

$$\begin{align*}{\mathrm{Sep}(v) \subseteq \mathrm{Sep}(w^{-1}v)}.\end{align*}$$

Choose any $\beta \in \mathrm {Sep}(v)$ ; we will show that $\beta \in \mathrm {Sep}(w^{-1} v)$ (i.e., that $w^{-1} v \beta \in \Phi ^-$ ). Recall that

(3.1) $$ \begin{align} s_{i_M} v \beta = v\beta - 2B(v \beta, \alpha_{i_M}) \alpha_{i_M}. \end{align} $$

We now consider two cases.

Case 1. Assume $B(v\beta , \alpha _{i_M}) \geq 0$ . Then applying $(w')^{-1}$ to each side of (3.1) yields

$$\begin{align*}w^{-1} v \beta = (w')^{-1} v \beta - 2B(v\beta, \alpha_{i_M}) (w')^{-1} \alpha_{i_M}.\end{align*}$$

By the inductive hypothesis, we have $\beta \in \mathrm {Sep}((w')^{-1}v)$ , so $(w')^{-1} v \beta \in \Phi ^-$ . Moreover, $s_{i_M}$ is not a right inversion of $(w')^{-1}$ , so by (2.4), we have $(w')^{-1} \alpha _{i_M} \in \Phi ^+$ . It follows that $w^{-1} v \beta \in \Phi ^-$ , as desired.

Case 2. Assume $B(v\beta , \alpha _{i_M}) < 0$ . We will prove that $v^{-1} s_{i_M} v \beta \in \mathrm {Sep}((w')^{-1} v)$ ; this will imply that

$$\begin{align*}w^{-1} v \beta = (w')^{-1} v (v^{-1} s_{i_M} v \beta) \in \Phi^-,\end{align*}$$

as desired. Our inductive hypothesis tells us that $\mathrm {Sep}(v)\subseteq \mathrm {Sep}((w')^{-1} v)$ , so it suffices to show that $v^{-1} s_{i_M} v \beta \in \mathrm {Sep}(v)$ ; that is, we want to show that $v^{-1} s_{i_M} v \beta \in \mathrm {R}({\mathscr {L}})$ and $v(v^{-1} s_{i_M} v) \beta \in \Phi ^-$ .

Applying $v^{-1}$ to each side of (3.1) yields

$$\begin{align*}v^{-1} s_{i_M} v \beta = \beta - 2B(v\beta, \alpha_{i_M}) v^{-1} \alpha_{i_M}.\end{align*}$$

We know that $\beta \in \mathrm {R}({\mathscr {L}})$ . Since $\tau _{\mathsf w}(u) = \tau _{\mathsf w'}(u)=v$ , we also have $v^{-1} \alpha _{i_M} \in \mathrm {R}({\mathscr {L}})$ . Because $\mathrm {R}({\mathscr {L}})$ is closed, this implies that $v^{-1}s_{i_M} v \beta \in \mathrm {R}({\mathscr {L}})$ . Since $-\beta \not \in \mathrm {R}({\mathscr {L}})$ , we also know that $-\beta \neq v^{-1}\alpha _{i_M}$ , so $v\beta \neq -\alpha _{i_M}$ . Since $v\beta \in \Phi ^-\setminus \{-\alpha _{i_M}\}$ , we also conclude that $v(v^{-1} s_{i_M} v) \beta = s_{i_M} v \beta \in \Phi ^-$ .

Proof. Fix an arbitrary $v \in \mathrm {Str}(R)$ . Each element $u\in \mathrm {Str}(R)$ is uniquely determined by the set of hyperplanes in the Coxeter arrangement $\mathcal H_W$ that separate $\mathbb {B} v$ from $\mathbb {B} u$ . Each such hyperplane must be of the form $\mathrm {H}_\beta $ for some root $\beta $ in the finite set $\Phi \setminus (\mathrm {R}({\mathscr {L}})\cup (-\mathrm {R}({\mathscr {L}})))$ .

Proof. It is clear that (1) implies (2) and that (3) implies (2). Therefore, it is enough to show that (2) implies both (1) and (3). Write $c = s_{i_n} \cdots s_{i_1}$ , where $i_1, \ldots , i_n$ is an ordering of I.

Assume (2); we will prove (1). Let ${\mathscr {L}} \subseteq W$ be a (possibly infinite) convex set, and let $u_0, u_1, u_2, \ldots $ be a periodic billiards trajectory (defined with respect to the convex set ${\mathscr {L}}$ and the ordering $i_1, \ldots , i_n$ of I). We wish to show that $u_0 \in {\mathscr {L}}$ . Our strategy will be to construct a finite convex subset ${\mathscr {L}}' \subseteq {\mathscr {L}}$ such that the billiards trajectory remains the same if the convex set ${\mathscr {L}}$ is replaced by ${\mathscr {L}}'$ . It will then follow from (2) that $u_0 \in {\mathscr {L}}'$ and, consequently, that $u_0 \in {\mathscr {L}}$ .

Choose a positive integer K such that the period of $u_0, u_1, u_2, \ldots $ divides $Kn$ . For each $j \in [Kn]$ such that $u_j = s_{i_j}u_{j-1}$ , we have $u_{j-1}^{-1}\alpha _{i_j} \not \in \mathrm {R}({\mathscr {L}})$ by the definition of the noninvertible Bender–Knuth toggles, so we may choose an element

$$\begin{align*}w_j \in {\mathscr{L}} \setminus H^+_{u_{j-1}^{-1}\alpha_{i_j}}.\end{align*}$$

Let ${\mathscr {L}}'$ be the convex hull of $\{w_j : j \in [Kn],\, u_j = s_{i_j}u_{j-1}\}$ . Then ${\mathscr {L}}'$ is a finite convex subset of ${\mathscr {L}}$ (see Theorem 1.1).

For $i \in I$ , let $\tau _i'$ denote the noninvertible Bender–Knuth toggle defined with respect to the convex set ${\mathscr {L}}'$ . Let $u^{\prime }_0 = u_0$ , and define a billiards trajectory $u^{\prime }_0, u^{\prime }_1, u^{\prime }_2, \ldots $ using the convex set ${\mathscr {L}}'$ and the ordering $i_1, \ldots , i_n$ of I. (That is, define this sequence via the recurrence relation $u^{\prime }_j = \tau ^{\prime }_{i_j}(u^{\prime }_{j-1})$ .) We claim that $u_j = u^{\prime }_{j}$ for all nonnegative integers j. It suffices to prove that $u_j = u^{\prime }_j$ for all $j \in [Kn]$ . To do so, we use induction on j.

Suppose first that $u_j = u_{j-1}$ . Then $u_{j-1}^{-1}\alpha _{i_j} \in \mathrm {R}({\mathscr {L}})$ . Since ${\mathscr {L}}' \subseteq {\mathscr {L}}$ , this implies that ${u_{j-1}^{-1}\alpha _{i_j} \in \mathrm {R}({\mathscr {L}}')}$ , so

$$\begin{align*}u^{\prime}_j = \tau^{\prime}_{i_j}(u^{\prime}_{j-1}) = \tau^{\prime}_{i_j}(u_{j-1}) = u_{j-1} = u_j.\end{align*}$$

Next, suppose that $u_j = s_{i_j} u_{j-1}$ . Then

$$\begin{align*}w_j \in {\mathscr{L}}' \setminus H^+_{u_{j-1}^{-1}\alpha_{i_j}},\end{align*}$$

so $u_{j-1}^{-1}\alpha _{i_j} \not \in \mathrm {R}({\mathscr {L}}')$ . Therefore,

$$\begin{align*}u^{\prime}_j = \tau^{\prime}_{i_j}(u^{\prime}_{j-1}) = \tau^{\prime}_{i_j}(u_{j-1}) = s_{i_j}u_{j-1} = u_{j}.\end{align*}$$

This proves our claim that $u_j = u^{\prime }_{j}$ for all nonnegative integers j. By (2), we have $u_0 \in {\mathscr {L}}'\subseteq {\mathscr {L}}$ , as desired.

Now, still assuming (2), we will prove (3). Let ${\mathscr {L}} \subseteq W$ be a finite convex set, and let $u_0, u_1, u_2, \ldots \in W$ be a billiards trajectory (defined with respect to the convex set ${\mathscr {L}}$ and the ordering $i_1, \ldots , i_n$ of I). We wish to show that the billiards trajectory eventually reaches ${\mathscr {L}}$ .

By Theorem 3.1, we have

$$\begin{align*}\mathrm{Sep}(u_0) \supseteq \mathrm{Sep}(u_1) \supseteq \mathrm{Sep}(u_2) \supseteq \cdots.\end{align*}$$

By Theorem 3.3, the sequence $\mathrm {Sep}(u_0), \mathrm {Sep}(u_1), \mathrm {Sep}(u_2), \ldots $ is eventually constant; equivalently, there exists a single stratum $\mathrm {Str}(R)$ that contains $u_k$ for all sufficiently large positive integers k. The stratum $\mathrm {Str}(R)$ is finite by Theorem 3.6, so the billiards trajectory $u_0, u_1, u_2, \ldots $ is eventually periodic. According to (2), we have $u_k \in {\mathscr {L}}$ for all sufficiently large positive integers k. This proves (3).

Proof. Let $v \in {\mathscr {L}}$ be arbitrary. Choose some $w \in \mathrm {Str}(R)$ that minimizes the length $\ell (vw^{-1})$ . Since $w\neq v$ , there exists $i \in I$ such that $\ell (v w^{-1} s_i) = \ell (vw^{-1}) - 1$ . We claim that $-w^{-1}\alpha _i$ is a transmitting root. To prove this, it is enough to show that $-w^{-1}\alpha _i \in \mathrm {R}({\mathscr {L}})$ .

By the minimality of $\ell (vw^{-1})$ , we have that $s_iw\not \in \mathrm {Str}(R)$ , so $\mathrm {Sep}(s_i w) \neq \mathrm {Sep}(w)$ . That is, there exists a root $\beta \in \mathrm {R}({\mathscr {L}})$ such that $s_i w \beta $ and $w \beta $ have different signs. It follows that $\beta = \pm w^{-1} \alpha _i$ . Since $\ell (vw^{-1}s_i)<\ell (vw^{-1})$ , we see that $s_i$ is a right inversion of $vw^{-1}$ , so $vw^{-1}\alpha _i \in \Phi ^-$ . Since $v\beta \in \Phi ^+$ , we cannot have $\beta = w^{-1} \alpha _i$ , so $\beta =-w^{-1}\alpha _i$ . Therefore, $-w^{-1}\alpha _i \in \mathrm {R}({\mathscr {L}})$ , as desired.

Proof. Let us rephrase the hypotheses of the lemma as follows. We have that $u \in H^+_{u^{-1}\gamma }$ , that $\beta \neq -u^{-1} \gamma $ , and that

$$\begin{align*}H_{u^{-1}\gamma}^+ \cap H^+_{\beta} \cap H_{u^{-1}\alpha_i}^+ = \varnothing.\end{align*}$$

See Figure 5 for an illustration of this scenario.

Figure 5 An illustration of the proof of Theorem 3.9, drawn using the positive projectivization of the Tits cone and Coxeter arrangement of $\widetilde G_2$ , whose Coxeter graph is . The assumption that $\tau _i(u)=u$ implies that ${{\mathscr {L}}\subseteq H^+_{\beta }\cap H_{u^{-1}\alpha _i}^+\subseteq H_{u^{-1}\gamma }^-}$ , which ends up contradicting the hypothesis that $\beta $ is a transmitting root of the stratum containing u.

Assume for the sake of contradiction that $\tau _i(u) = u$ . By the definition of a transmitting root, there exist $w \in \mathrm {Str}(R)$ and $i' \in I$ such that $-w \beta = \alpha _{i'}$ . We will show that

$$\begin{align*}s_{i'}w \in H_{u^{-1}\gamma}^+ \cap H^+_{\beta} \cap H_{u^{-1}\alpha_i}^+,\end{align*}$$

which will contradict the fact that $H_{u^{-1}\gamma }^+ \cap H^+_{\beta } \cap H_{u^{-1}\alpha _i}^+ = \varnothing $ .

First, we will show that $s_{i'} w \in H_{u^{-1}\gamma }^+$ . Since $\tau _i(u) = u$ , we have $\beta , u^{-1} \alpha _i \in \mathrm {R}({\mathscr {L}})$ , so

$$\begin{align*}{\mathscr{L}} \subseteq H^+_\beta \cap H_{u^{-1}\alpha_i}^+ \subseteq H_{-u^{-1}\gamma}^+.\end{align*}$$

Thus, ${-u^{-1}\gamma \in \mathrm {R}({\mathscr {L}})}$ . We also have $u(-u^{-1}\gamma ) = -\gamma \in \Phi ^-$ , so $-u^{-1}\gamma \in \mathrm {Sep}(u) = R = \mathrm {Sep}(w)$ . Therefore, $w(-u^{-1}\gamma ) \in \Phi ^-$ , or equivalently, $wu^{-1}\gamma \in \Phi ^+$ . Since $\gamma \neq -u\beta $ , we have $wu^{-1}\gamma \neq \alpha _{i'}$ . It follows that $s_{i'} w u^{-1} \gamma \in \Phi ^+$ as well. Hence, $s_{i'}w \in H_{u^{-1}\gamma }^+$ .

Second, we have $s_{i'} w \in H_{\beta }^{+}$ since $s_{i'} w \beta = s_{i'}(-\alpha _{i'}) = \alpha _{i'} \in \Phi ^+$ .

Finally, we will show that $s_{i'} w \in H_{u^{-1}\alpha _i}^+$ . We have $u^{-1}\alpha _i \in \mathrm {R}({\mathscr {L}})$ (since $\tau _i(u)=u$ ) and $u(u^{-1}\alpha _i) = \alpha _i \in \Phi ^+$ , so $u^{-1}\alpha _i \not \in \mathrm {Sep}(u) = R = \mathrm {Sep}(w)$ . Therefore, $wu^{-1}\alpha _i \in \Phi ^+$ . Since $w^{-1} \alpha _{i'} = -\beta \not \in \mathrm {R}({\mathscr {L}})$ , we have $w^{-1} \alpha _{i'} \neq u^{-1}\alpha _i$ , so $wu^{-1}\alpha _i \neq \alpha _{i'}$ . It follows that $s_{i'}wu^{-1}\alpha _i \in \Phi ^+$ , so $s_{i'}w \in H_{u^{-1}\alpha _i}^+$ .

Proof. First, suppose that $u\beta $ and $\alpha _i$ are linearly dependent. Since $\beta \in \mathrm {Sep}(u)$ , this implies that $-u\beta = \alpha _i$ . Since $\beta \in \mathrm {R}({\mathscr {L}})$ , we have $u^{-1}\alpha _i = -\beta \not \in \mathrm {R}({\mathscr {L}})$ , so $\tau _i(u) = s_i u$ by the definition of $\tau _i$ .

Next, suppose that $u\beta $ and $\alpha _i$ are linearly independent. Let $\gamma = - (au \beta + a' \alpha _i) \in \Phi $ . We have $- u \beta \in \Phi ^+$ , so when $\gamma $ is written in the basis of simple roots, all the coefficients are nonnegative except possibly the coefficient of $\alpha _i$ . But $\gamma \neq -\alpha _i$ , so $\gamma \in \Phi ^+$ . Moreover, $\gamma \neq -u \beta $ , and

$$\begin{align*}H_\gamma^+ \cap H^+_{u\beta} \cap H_{\alpha_i}^+ = \{w \in W : \{-awu\beta - a'w\alpha_i,\, wu\beta,\, w\alpha_i\} \subseteq \Phi^+\} = \varnothing.\end{align*}$$

By Theorem 3.9, we have $\tau _i(u) = s_iu$ .

Proof. We have $s_i u \beta = u \beta - 2B(u\beta , \alpha _i) \alpha _i \in \Phi $ . Now take $a = 1$ and $a' = -2B(u\beta , \alpha _i)$ in Theorem 3.10.

Proof. We have $-u \beta \in \Phi ^+$ because $\beta \in R = \mathrm {Sep}(u)$ . Now, let $\gamma \in \Phi ^+$ , and suppose that

(3.2) $$ \begin{align}-u \beta \mathrel{\mathsf{dom}} \gamma.\end{align} $$

We must show that $\gamma = -u\beta $ .

We know by (3.2) that $-\gamma \mathrel {\mathsf {dom}} u \beta $ . Applying $u^{-1}$ to each side yields $-u^{-1} \gamma \mathrel {\mathsf {dom}} \beta $ . Then ${{\mathscr {L}} \subseteq H^+_{\beta } \subseteq H^+_{-u^{-1}\gamma }}$ , so $-u^{-1}\gamma \in \mathrm {R}({\mathscr {L}})$ . Moreover, $u (-u^{-1}\gamma ) = -\gamma \in \Phi ^-$ , so ${-u^{-1}\gamma \in \mathrm {Sep}(u) = R}$ .

By the definition of a transmitting root, there exists $w \in \mathrm {Str}(R)$ such that $-w\beta $ is a simple root. Since $-u^{-1} \gamma \in R = \mathrm {Sep}(w)$ , we have that $wu^{-1}\gamma \in \Phi ^+$ . Applying $wu^{-1}$ to each side of (3.2) yields $-w\beta \mathrel {\mathsf {dom}} wu^{-1}\gamma $ . Since $-w\beta $ is simple, it is small, so $wu^{-1}\gamma = -w\beta $ . Therefore, $\gamma = -u \beta $ , as desired.

Proof. According to [Reference Develin, Macauley and Reiner10, Theorem 1.15], the acyclic orientations $\mathrm {ao}(c)$ and $\mathrm {ao}(c')$ are flip equivalent. Hence, we may assume that $\mathrm {ao}(c')$ is obtained from $\mathrm {ao}(c)$ via a single flip from a source to a sink. That is, there is a reduced word $i_ni_{n-1}\cdots i_1$ for c such that $i_{n-1}\cdots i_1i_{n}$ is a reduced word for $c'$ .

Fix an arbitrary nonempty convex set ${\mathscr {L}}\subseteq W$ . Suppose c is futuristic. Let $u\in W$ be a periodic point of $\mathrm {Pro}_{c'}$ . We have that $\mathrm {Pro}_c \circ \tau _{i_n} = \tau _{i_n} \circ \mathrm {Pro}_{c'}$ , so $\tau _{i_n}(u)$ must be a periodic point of $\mathrm {Pro}_c$ . Since c is futuristic, this implies that $\tau _{i_n}(u)\in {\mathscr {L}}$ . Because u is a periodic point of $\mathrm {Pro}_{c'}$ , it can be obtained by applying a sequence of noninvertible Bender–Knuth toggles to $\tau _{i_n}(u)$ . Hence, $u\in {\mathscr {L}}$ . This proves that $c'$ is futuristic.

We have shown that if c is futuristic, then $c'$ is futuristic. A similar argument proves the other direction.

Proof. Assume that c is futuristic; we will show that $c'$ is futuristic. Let ${\mathscr {L}} \subseteq W_J$ be a nonempty convex set; we will show that every periodic point of $\mathrm {Pro}_{c'}$ lies in ${\mathscr {L}}$ .

Note that $W_J$ is precisely the set of elements of W with no right inversions in $W \setminus W_J$ , so $W_J$ is a convex subset of W, with $\mathrm {R}(W_J)=\Phi ^+ \setminus \Phi _J$ . Now, write $\mathrm {R}_J({\mathscr {L}})=\{\beta \in \Phi _J:{\mathscr {L}}\subseteq H_\beta ^+\}$ . We have

$$\begin{align*}\mathrm{R}({\mathscr{L}}) \cap \Phi_J=\mathrm{R}_J({\mathscr{L}}),\end{align*}$$

and the inclusion ${\mathscr {L}} \subseteq W_J$ gives $\Phi ^+ \setminus \Phi _J \subseteq \mathrm {R}({\mathscr {L}})$ . At the same time, we have $w\beta \in \Phi ^-$ for all $w\in {\mathscr {L}}$ and $\beta \in \Phi ^-\setminus \Phi _J$ , so $\mathrm {R}({\mathscr {L}})$ is disjoint from $\Phi ^-\setminus \Phi _J$ . Putting this all together yields

(3.3) $$ \begin{align} \mathrm{R}({\mathscr{L}})=\mathrm{R}_J({\mathscr{L}})\cup(\Phi^+\setminus\Phi_J). \end{align} $$

Moreover, ${\mathscr {L}}$ is the intersection of the half-spaces $H_\beta ^+$ for $\beta \in \mathrm {R}_J({\mathscr {L}}) \cup (\Phi ^+ \setminus \Phi _J)$ , so ${\mathscr {L}}$ is convex as a subset of W.

For $i\in I$ , let $\tau _i\colon W\to W$ denote the noninvertible Bender–Knuth toggle on W that is defined with respect to ${\mathscr {L}}$ . For $i\in J$ , let $\tau ^J_{i}\colon W_J\to W_J$ denote the noninvertible Bender–Knuth toggle on $W_J$ that is defined with respect to ${\mathscr {L}}$ .

We claim that for all $i \in J$ , the restriction of $\tau _i$ to $W_J$ is $\tau _i^J$ . Moreover, we claim that for all $i \in I \setminus J$ , the restriction of $\tau _i$ to $W_J$ is the identity function on $W_J$ . Let $i \in I$ and $u \in W_J$ . Then $\tau _i(u) = u$ if and only if $u^{-1} \alpha _{i} \in \mathrm {R}({\mathscr {L}})$ . By (3.3), this holds if and only if $u^{-1}\alpha _i \in \mathrm {R}_J({\mathscr {L}})$ or $i \not \in J$ ; this proves both claims.

It follows that the restriction of $\mathrm {Pro}_c$ to $W_J$ is $\mathrm {Pro}_{c'}$ . Since c is futuristic, every periodic point of $\mathrm {Pro}_{c}$ lies in ${\mathscr {L}}$ . Hence, every periodic point of $\mathrm {Pro}_{c'}$ lies in ${\mathscr {L}}$ , as desired.

Proof. By passing to the parabolic subgroup $W_{{\mathfrak {o}} \cup {\mathfrak {o}}'}$ , we may assume that ${\mathfrak {o}} \cup {\mathfrak {o}}' = I$ .

It is immediate from the definitions that $m({\mathfrak {o}},{\mathfrak {o}}')=2$ if $\deg _{{\mathfrak {o}}'}({\mathfrak {o}})=0$ .

Suppose that $\deg _{{\mathfrak {o}}'}({\mathfrak {o}})=1$ . The irreducible factors of W are pairwise isomorphic Coxeter groups, and the Coxeter graph of each irreducible factor is a star graph whose edges all have the same label. To compute the order of ${\prod }_{i\in {{\mathfrak {o}}}}s_i{\prod }_{i'\in {\mathfrak {o}}'}s_{i'}$ in W, we compute the order after projecting to each irreducible factor, and we then take the least common multiple of the orders obtained this way (which will all be equal because the irreducible factors are isomorphic). We can check directly that this order is a if ${\mathrm {lab}}_{\mathfrak {o}}(\mathfrak {o}')=\{a\}$ , is $4$ if $\mathrm {lab}_{{\mathfrak {o}}}({\mathfrak {o}}')=\{3,3\}$ , is $6$ if $\mathrm {lab}_{{\mathfrak {o}}}({\mathfrak {o}}')=\{3,3,3\}$ , and is $\infty $ otherwise.

Finally, suppose that ${\deg }_{{\mathfrak {o}}'}(\mathfrak {o}){\geq } 2$ . Let $\mathfrak {o}=\{i_{1},\ldots ,i_{k}\}$ and ${\mathfrak {o}'}=\{i_{1}',\ldots ,i_{k'}'\}$ , and let $b{\geq } 1$ be an integer. Upon inspection, we find that it is is impossible to apply a sequence of commutation moves to the word $(i_{1}\cdots i_{k}i_{1}'\cdots i_{k'}')^{b}$ to obtain a word to which we can apply a nil move or a braid move that is not a commutation move. By Matsumoto’s (Theorem 2.1), this implies that $(i_{1}\cdots i_{k}i_{1}'\cdots i_{k'}')^{b}$ is a reduced word. Hence, ${\prod }_{i\in {\mathfrak {o}}}s_i{\prod }_{i'\in {\mathfrak {o}}'}s_{i'}$ does not have order b. As b was arbitrary, we conclude that $m(\mathfrak {o},{\mathfrak {o}}')=\infty $ .

Proof. Fix , and assume without loss of generality that ${{\deg }}_{{\mathfrak {o}}'}({\mathfrak {o}})\leq {{\deg }}_{{\mathfrak {o}}}({\mathfrak {o}}')$ . The proof follows from Theorem 3.17 and (3.5). For example, if ${{\deg }}_{{\mathfrak {o}}'}({\mathfrak {o}})=1$ and $\mathrm {lab}_{{\mathfrak {o}}}({\mathfrak {o}}')=\{3,3,3\}$ , then

and $m({\mathfrak {o}},{\mathfrak {o}}')=6$ by Theorem 3.17. If ${\deg }_{{\mathfrak {o}}'}({\mathfrak {o}})\geq 2$ , then

and $m({\mathfrak {o}},{\mathfrak {o}}')=\infty $ by Theorem 3.17. Similar arguments handle the other cases.

Proof. According to (3.7), we have $v\alpha _{{\mathfrak {o}}}=\sqrt {|\mathfrak {o}|}\omega (\iota (v)\alpha _i)$ , so the proof follows from the fact that $\omega $ maps each simple root of W to a positive scalar multiple of a simple root of .

Proof. Fix and . We wish to prove that $\iota (\tau _{\mathfrak {o}}(x))=\overline {\tau }_{\mathfrak {o}}(\iota (x))$ . To do so, it suffices to show that for each $i\in \mathfrak {o}$ , we have $\tau _{\mathfrak {o}}(x)=x$ if and only if $\tau _i(\iota (x))=\iota (x)$ .

Fix $i\in {\mathfrak {o}}$ . We have $\tau _{\mathfrak {o}}(x)=x$ if and only if for all . According to Theorem 3.19, this holds if and only if $\iota (w)(\iota (x)^{-1}\alpha _i)\in \Phi ^+$ for all . Because ${\mathscr {L}}$ is the convex hull of , it follows that $\tau _{\mathfrak {o}}(x)=x$ if and only if $z(\iota (x)^{-1}\alpha _i)\in \Phi ^+$ for all $z\in {\mathscr {L}}$ . This, in turn, holds if and only if $\tau _i(\iota (x))=\iota (x)$ .

Proof. Preserve the notation from above. We prove the contrapositive. Suppose that c is not futuristic. Then there exist a nonempty convex set and an element such that $\mathrm {Pro}_c^K(u)=u$ for some integer $K\geq 1$ . Let ${\mathscr {L}}$ be the convex hull in W of . It follows from Theorem 3.20 that $\iota \circ \mathrm {Pro}_c=\mathrm {Pro}_{\iota (c)}\circ \iota $ . Hence, $\mathrm {Pro}_{\iota (c)}^K(\iota (u))=\iota (\mathrm {Pro}_c^K(u))=\iota (u)$ .

We need to show that $\iota (u)\not \in {\mathscr {L}}$ . Because , there exists such that and for all . We can write $\beta =y\alpha _{{\mathfrak {o}}}$ for some and . Choose $i\in \mathfrak {o}$ . For each , we can use Theorem 3.19 to see that if and only if $\iota (v)(\iota (y)\alpha _i)\in \Phi ^+$ . It follows that $\iota (u)(\iota (y)\alpha _i)\in \Phi ^-$ and that $\iota (w)(\iota (y)\alpha _i)\in \Phi ^+$ for all . Since ${\mathscr {L}}$ is the convex hull of , we deduce that $\iota (u)\not \in {\mathscr {L}}$ .

Proof. It may help to keep the following geometric intuition in mind. As in Figure 1, we can think of the Bender–Knuth billiards process as taking place within the Tits cone ${\mathbb {B} W}$ , or, equivalently, in the positive projectivization $\mathbb P({\mathbb {B} W})=({\mathbb {B} W}\setminus \{0\})/\mathbb {R}_{>0}$ , which has the structure of an affine Euclidean space. By Theorem 3.11, the billiards trajectory can never reflect off of a one-way mirror that forms an obtuse angle with the hyperplane $\mathrm {H}_\beta $ . Thus, if the billiards trajectory is ever moving toward the hyperplane $\mathrm {H}_\beta $ , it can never be redirected to move away from $\mathrm {H}_\beta $ . Since the billiards trajectory is periodic, the only possibility is that it always moves parallel to $\mathrm {H}_\beta $ and hits only one-way mirrors that are orthogonal to $\mathrm {H}_\beta $ . In other words, if $u_j = u_{j-1}$ , then $B(u_{j-1}\beta , \alpha _{i_j}) = 0$ .

Of course, the billiards trajectory $u_0, u_1, u_2, \ldots $ is really a sequence of elements of W rather than an actual piecewise linear path inside $\mathbb P({\mathbb {B} W})$ . (But see Section 11 and Theorem 12.4.) So, in order to turn our geometric intuition into a rigorous proof, we must formalize the notions of ‘moving toward’ and ‘moving away from’ $\mathrm {H}_\beta $ . We can do so as follows.

Consider the Coxeter element $c=s_{i_n}\cdots s_{i_1}$ . There is an integer $K\geq 1$ such that ${\mathrm {Pro}_c^K(u_0)=u_0}$ and . For $j \geq 1$ , define

$$\begin{align*}\lambda_j = \begin{cases} 2B(u_{j-1}\beta, \alpha_{i_j}) & \text{if }u_j = u_{j-1}; \\ 0 & \text{if }u_j = s_{i_j} u_{j-1}. \end{cases} \end{align*}$$

By Theorem 3.11, we have $\lambda _j \geq 0$ for all j, and we wish to prove that $\lambda _j = 0$ for all j. Observe that we have defined $\lambda _j$ in such a way that

(5.1) $$ \begin{align}u_j \beta = s_{i_j} u_{j-1}\beta + \lambda_j \alpha_{i_j}.\end{align} $$

For $j \geq 0$ , define

$$\begin{align*}c_j = s_{i_{j+n}} \cdots s_{i_{j+2}} s_{i_{j+1}}\in W\quad\text{and}\quad \gamma_j = c_j^K u_j \beta - u_j \beta \in V.\end{align*}$$

One may think of $c_j^K u_j = s_{i_{j+Kn}} \cdots s_{i_{j+2}}s_{i_{j+1}} u_j$ as the location that the billiards trajectory would reach if it were to start at $u_j$ and then progress $Kn$ steps, in the absence of any one-way mirrors. Roughly, $\gamma _j$ measures the angle between the hyperplane $\mathrm {H}_\beta $ and the ‘limiting direction’ of the billiards trajectory.

Note that for all integers $j \geq 0$ and all vectors $\epsilon \in V$ , we have

$$\begin{align*}\pi(c_j^K \epsilon - \epsilon) = \pi(c_j^K) \pi(\epsilon) - \pi(\epsilon).\end{align*}$$

Since $c_j$ is conjugate to c and we have , we must also have . It follows that $\pi (c_j^K \epsilon - \epsilon ) = 0$ , so $c_j^K \epsilon - \epsilon \in V^W$ . Taking $\epsilon = u_{j} \beta $ , we find that $\gamma _j \in V^W$ for all j. Taking $\epsilon = \alpha _{i_j}$ yields that $c_j^K \alpha _{i_j} - \alpha _{i_j} \in V^W$ for all j.