Proof. This is a consequence of [Reference Devillers, Mühlherr and Van MaldeghemDMV12, Lemma $18$ ] and [Reference Abramenko and BrownAB08, Lemma $5.107$ ].

Convention 2.3. For the rest of this paper, we let $(W, S)$ be a Coxeter system of finite rank and we define $\Phi := \Phi (W, S)$ (respectively $\Phi _+ := \Phi (W, S)_+$ and $\Phi _- := \Phi (W, S)_-$ ). Moreover, we assume that $m_{st} \in \{2, 3, 4, 6, 8, \infty \}$ for all $s\neq t \in S$ .

Proof. This is a consequence of [Reference Caprace and MühlherrCM05, Lemma $2.3$ ] and its proof.

Proof. The implication (i) $\Rightarrow $ (ii) follows exactly as in [Reference Abramenko and BrownAB08, Proposition $3.165$ ]. Now, suppose item (ii) and assume that there exists $R \in \partial ^2 \alpha \cap \partial ^2 \beta $ . As R is finite, there exists $k \in \mathbb {N}$ such that $(r_{\alpha } r_{\beta })^k$ fixes a chamber in R, that is, ${(r_{\alpha } r_{\beta })^k w = (r_{\alpha } r_{\beta })^k(w) = w}$ for some $w\in R$ . However, this implies $(r_{\alpha } r_{\beta })^k = 1$ . As $o(r_{\alpha } r_{\beta }) = \infty $ , we obtain a contradiction. Now, suppose that none of $\{ \alpha , \beta \}, \{ -\alpha , \beta \}$ is nested. Then, we have $\alpha \not \subseteq \beta , \beta \not \subseteq \alpha $ and $(-\alpha ) \not \subseteq \beta , \beta \not \subseteq (-\alpha )$ . This implies that none of $\alpha \cap (-\beta ), \beta \cap (-\alpha ), (-\alpha ) \cap (-\beta ), \beta \cap \alpha $ is the empty set. Arguing as in the proof of [Reference WeissWei09, Proposition $29.24$ ] and using Lemma 2.5, there exists $R \in \partial ^2 \alpha \cap \partial ^2 \beta $ and we are done. The second part of (a) follows from the first part of (a) and [Reference Abramenko and BrownAB08, Lemma $8.42(c)$ ].

To show assertion (b), we note that by assertion (a), there exists $R \in \partial ^2 \alpha \cap \partial ^2 \beta $ . We deduce $\emptyset \neq R \cap \alpha \cap \beta \subseteq \gamma $ and $\emptyset \neq R \cap (-\alpha ) \cap (-\beta ) \subseteq (-\gamma )$ . If follows from Lemma 2.4 that $R \in \partial ^2 \gamma $ . In particular, $R \in \partial ^2 \alpha \cap \partial ^2 \beta \cap \partial ^2 \gamma $ . We deduce $o(r_{\alpha } r_{\gamma }) < \infty $ and $o(r_{\beta } r_{\gamma }) < \infty $ from assertion (a).

Proof. Let $R \in \partial ^2 \alpha \cap \partial ^2 \beta \cap \partial ^2 \gamma $ be a residue, let $\delta \in \{ \beta , \gamma \}$ and let $T \in \partial ^2 \alpha \cap \partial ^2 \delta $ . It suffices to show that $T \in \partial ^2 \alpha \cap \partial ^2 \beta \cap \partial ^2 \gamma $ . Using Lemma 2.7, we deduce that R and T are parallel. Then, Lemma 2.5 implies that a reflection of $\Sigma (W, S)$ stabilizes R if and only if it stabilizes T. As $r_{\alpha }, r_{\beta }, r_{\gamma }$ stabilize R, they also stabilize T and assertion (a) follows. Before we show assertion (b), we prove the following claim:

Claim: $\gamma \notin (\alpha , \beta ) \Rightarrow \alpha \cap \beta \cap (-\gamma ) \cap R \neq \emptyset $ .

We have $\alpha \cap \beta \not \subseteq \gamma $ or $(-\alpha ) \cap (-\beta ) \not \subseteq (-\gamma )$ . In the first case, we have $\alpha \cap \beta \cap (-\gamma ) \neq ~\emptyset $ and, as roots are convex, the claim follows from [Reference Abramenko and BrownAB08, Lemma $5.45$ ]. Thus, we can assume $(-\alpha ) \cap (-\beta ) \not \subseteq (-\gamma )$ . As roots are convex, [Reference Abramenko and BrownAB08, Lemma $5.45$ ] implies $(-\alpha ) \cap (-\beta ) \cap \gamma \cap R \neq \emptyset $ . Let x be contained in this set and let $y \in R$ be opposite to x in R. Then, $y \in \alpha \cap \beta \cap (-\gamma ) \cap R$ and the claim follows.

We are now in the position to prove assertion (b). We assume $(\alpha , \beta ) \cap \{ \gamma , -\gamma \} = \emptyset = (-\alpha , \beta ) \cap \{ \gamma , -\gamma \}$ . By the above, we deduce the following:

$$ \begin{align*} &x \in \alpha \cap \beta \cap (-\gamma) \cap R \quad\text{and}\quad x' \in \alpha \cap \beta \cap \gamma \cap R, \\ &y \in (-\alpha) \cap \beta \cap (-\gamma) \cap R \quad\text{and}\quad y' \in (-\alpha) \cap \beta \cap \gamma \cap R. \end{align*} $$

As residues and roots are convex, there exist $P, Q \in \partial \gamma $ such that $P \subseteq \alpha \cap \beta \cap R$ and $Q \subseteq (-\alpha ) \cap \beta \cap R$ . As $P \subseteq \alpha $ and $Q \subseteq (-\alpha )$ , we have $P \neq Q$ and Lemma 2.2 implies that there exist $p\in P, q\in Q$ that are opposite in R. Using [Reference WeissWei03, Proposition $5.4$ ], every chamber in R lies on a minimal gallery from p to q. As roots are convex and $p, q \in \beta $ , we infer $R \subseteq \beta $ , which is a contradiction to $R \in \partial ^2 \beta $ .

Proof. Note that by Lemma 2.8 and [Reference Abramenko and BrownAB08, Lemma $8.42(3)$ ], the pair $\{ -\alpha , \gamma \}$ is prenilpotent. Thus, we have to show that $(-\alpha ) \cap \gamma \subseteq \beta $ and $\alpha \cap (-\gamma ) \subseteq (-\beta )$ . As $\gamma \in (\alpha , \beta )$ , we have $\alpha \cap \beta \subseteq \gamma $ and $(-\alpha ) \cap (-\beta ) \subseteq (-\gamma )$ . In particular, we have $(-\gamma ) \subseteq (-\alpha ) \cup (-\beta )$ and $\gamma \subseteq \alpha \cup \beta $ . We compute

$$ \begin{align*} (-\alpha) \cap \gamma &\subseteq (-\alpha) \cap (\alpha \cup \beta) \subseteq (-\alpha) \cap \beta \subseteq \beta, \\ \alpha \cap (-\gamma) &\subseteq \alpha \cap ((-\alpha) \cup (-\beta)) \subseteq \alpha \cap (-\beta) \subseteq (-\beta).\\[-3pc] \end{align*} $$

Proof. Assertion (a) is [Reference BischofBis22, Proposition $2.3$ ] and assertion (b) is a consequence of the classification in [Reference FeliksonFel98] (see Figure $8$ in § 5.1 in loc. cit.).

Proof. Let $i \in \{1, 2\}$ and assume that $o(r_{\alpha _3} r_{\beta }) < \infty $ . Then, $\{ r_{\alpha _i}, r_{\beta }, r_{\alpha _3} \}$ is a reflection triangle by Lemmas 2.8(b) and 2.9(a). We apply Remark 2.11 to determine the triangle that induces $\{ r_{\alpha _1}, r_{\beta }, r_{\alpha _3} \}$ . We note that $\alpha _2 \in (-\alpha _1, \beta )$ and $\alpha _1 \in (-\alpha _2, \beta )$ hold by Lemma 2.10.

There exist $R \in \partial ^2 \alpha _1 \cap \partial ^2 \alpha _2 = \partial ^2 \alpha _1 \cap \partial ^2 \beta $ with $R \subseteq \alpha _3$ , and $R' \in \partial ^2 \alpha _1 \cap \partial ^2 \alpha _3$ with $R' \subseteq \alpha _2$ . Thus, $\emptyset \neq R' \cap \alpha _1 \subseteq \alpha _1 \cap \alpha _2 \subseteq \beta $ and Lemma 2.4 yields $R \subseteq \beta $ . This implies that there exists $\varepsilon \in \{+, -\}$ such that $\{ \varepsilon \alpha _1, \beta , \alpha _3 \}$ is a triangle. If $\varepsilon = +$ , then $(-\alpha _1, \beta ) = \emptyset $ by Lemma 2.12. However, this is a contradiction to the fact that $\alpha _2 \in (-\alpha _1, \beta )$ . Thus, we have $\varepsilon = -$ . We show that $\{ \beta , \alpha _2, \alpha _3 \}$ is a triangle.

There exists $R_3 \in \partial ^2 \alpha _1 \cap \partial ^2 \alpha _2 = \partial ^2 \beta \cap \partial ^2 \alpha _2$ with $R_3 \subseteq \alpha _3$ . There exists $R_1 \in \partial ^2 \alpha _2 \cap \partial ^2 \alpha _3$ with $R_1 \subseteq \alpha _1$ . In particular, $\emptyset \neq R_1 \cap \alpha _2 \subseteq \alpha _1 \cap \alpha _2 \subseteq \beta $ . Now, as $\{ -\alpha _1, \beta , \alpha _3 \}$ is a triangle, there exists $T \in \partial ^2 \beta \cap \partial ^2 \alpha _3$ with $T \subseteq (-\alpha _1)$ . Thus, ${\emptyset \neq \beta \cap T \subseteq \beta \cap (-\alpha _1) \subseteq \alpha _2}$ . We infer that $\{ \beta , \alpha _2, \alpha _3 \}$ is a triangle and hence $(-\alpha _2, \beta ) = \emptyset $ by Lemma 2.12. However, this is again a contradiction. We conclude $o(r_{\alpha _3} r_{\beta }) = \infty $ .

Note that $\emptyset \neq R_3 \cap (-\beta ) \subseteq \alpha _3$ and $\emptyset \neq R_1 \cap \alpha _2 \cap (-\alpha _3) \subseteq \alpha _1 \cap \alpha _2 \subseteq \beta $ . In particular, $\{-\alpha _3, \beta \}$ is a prenilpotent pair and by Lemma 2.8, it is nested. We have ${\emptyset \neq R_3 \cap \beta \subseteq \alpha _3}$ . However, this implies $\beta \not \subseteq (-\alpha _3)$ . We deduce $(-\alpha _3) \subseteq \beta $ . This finishes the proof of assertion (a).

Let $\{x, y\} \in \partial \alpha _3$ such that $\bigcap _{i=1}^3 \alpha _i = \{y\}$ (see Lemma 2.12) and let $R \in \partial ^2 \alpha _1 \cap \partial ^2 \alpha _2$ . By Lemma 2.7(b), R is unique and hence $y\in R$ . Let $d\in R$ be opposite to y in R and let $(c_0 = x, c_1 = y, \ldots , c_n = d)$ be a minimal gallery. Then, $c_i \in R$ for each $1 \leq i \leq n$ . Let $(\beta _1, \ldots , \beta _n)$ be the sequence of roots crossed by $(c_0, \ldots , c_n)$ . Then, $\beta _1 = -\alpha _3$ and $o(r_{\beta _i} r_{\beta }) < \infty $ for each $2 \leq i \leq n$ by Lemma 2.8. Assume $(-\alpha _3, \beta ) \neq \emptyset $ . Then, [Reference Abramenko and BrownAB08, Lemma $3.69$ ] implies that for each $\gamma \in (-\alpha _3, \beta )$ , there exists $2 \leq i \leq n-1$ with $\gamma = \beta _i$ . As $\gamma \subsetneq \beta $ , this is a contradiction and hence, $(-\alpha _3, \beta ) = \emptyset $ .

Proof. The first part is [Reference BischofBis25a, Corollary $3.3$ ] and the second part is a consequence of the first.

Proof. We show the hypothesis by induction on $n-i$ . If $n-i = 0$ , then there is nothing to show. Thus, we suppose $n-i>0$ . Let $J_i$ be the type of $R_i$ and let $w := {\mathrm {proj}}_{R_i} 1_W = {\mathrm {proj}}_{P_{i-1}} 1_W \in P_{i-1}$ . As $P_{i-1} \neq P_i$ are contained and opposite in $R_i$ by Lemma 2.5, there exists $w' \in P_i$ such that w and $w'$ are opposite in $R_i$ , that is, ${w' = wr_{J_i}}$ . For $t\in S$ with $J_i \cap J_{i+1} = \{t\}$ , we have $R_i \cap R_{i+1} = P_i = \mathcal {P}_t(w') = \mathcal {P}_t(wr_{J_i})$ . As ${w = {\mathrm {proj}}_{R_i} 1_W}$ , we deduce $\ell ( {\mathrm {proj}}_{P_i} 1_W ) = \ell (w (r_{J_i} t)) \geq \ell (w) +2$ . Using Lemma 2.14, we infer $\ell (( {\mathrm {proj}}_{P_i} 1_W)s) = \ell ( {\mathrm {proj}}_{P_i} 1_W) +1$ for $s \in J_{i+1} \backslash J_i$ and hence $ {\mathrm {proj}}_{R_{i+1}} 1_W = {\mathrm {proj}}_{P_i} 1_W$ . Using induction, the claim follows.

Proof. This is a consequence of Lemma 2.14.

Proof. We use the canonical linear representation of $(W, S)$ (see [Reference Abramenko and BrownAB08, Ch. $2.5$ ]). Let $V := \mathbb {R}^S$ be the vector space over $\mathbb {R}$ with standard basis $(e_s)_{s\in S}$ and let $(\cdot , \cdot )$ be the symmetric bilinear form on V given by

$$ \begin{align*} (e_s, e_t) := -\cos\bigg( \frac{\pi}{m_{st}} \bigg) = \begin{cases} 1 & \text{if } s=t, \\ -\dfrac{\sqrt{2}}{2} & \text{otherwise}. \end{cases} \end{align*} $$

Then, W acts on V via $\sigma : W \to \mathrm {GL}(V), s \mapsto ( \sigma _s: V \to V, x \mapsto x - 2(x, e_s) e_s )$ and $(\cdot , \cdot )$ is invariant under this action. Let $\alpha $ and $\beta $ be as in the statement. Without loss of generality, we can assume $\alpha = \alpha _r$ and $\beta \in \{ rs\alpha _t, rst\alpha _p \}$ . At first, we consider the case $\beta = rs\alpha _t$ . We compute

$$ \begin{align*} (e_r, \sigma(rs)(e_t)) &= ( e_r, \sigma_r( \sigma_s(e_t) ) ) = ( \sigma_r(e_r), \sigma_s(e_t) ) = (-e_r, e_t + \sqrt{2} e_s) = \dfrac{\sqrt{2}}{2} +1>1. \end{align*} $$

Now, we assume $\beta = rst\alpha _p$ and we compute

$$ \begin{align*} (e_r, \sigma(rst)(e_p)) &= (e_r, \sigma_r(\sigma_s(\sigma_t(e_p)))) \\ &= ( \sigma_s(-e_r), \sigma_t(e_p) ) \\ &= (-e_r -2(-e_r, e_s)e_s, e_p -2(e_p, e_t)e_t) \\ &= -(e_r, e_p) + 2(e_p, e_t)(e_r, e_t) + 2(e_r, e_s)(e_s, e_p) -4(e_r, e_s)(e_p, e_t)(e_s, e_t) \\ &= -(e_r, e_p) +1 +1 +\sqrt{2}>1. \end{align*} $$

Using [Reference Abramenko and BrownAB08, Lemma $2.77$ ], we obtain that $o(r_{\alpha } r_{\beta }) = \infty $ . As $\{ \alpha , \beta \}$ is a prenilpotent pair, Lemma 2.8 yields that $\{ \alpha , \beta \}$ is a pair of nested roots and hence $\alpha \subsetneq \beta $ .

Convention 2.19. For the rest of this section, we assume that $(W, S)$ is $2$ -spherical and $m_{st} \geq 4$ for all $s\neq t \in S$ .

Proof. Suppose $R_1 \neq R$ and $\ell (s_1 \cdots s_{k-1}r) = k$ . Then, $ {\mathrm {proj}}_{R_1} c_0 = {\mathrm {proj}}_{P_0} c_0$ and the claim follows from Lemma 2.15. Now, we suppose $n>1$ . Assume that $R_1 = R$ . Then, Lemma 2.14 implies $ {\mathrm {proj}}_{R_2} 1_W = {\mathrm {proj}}_{P_1} 1_W$ and the claim follows from Lemma 2.15. Now, we suppose $R_1 \neq R$ . If $\ell (s_1 \cdots s_{k-1}r) = k$ , the claim follows from assertion (a). Thus, we can assume that $\ell (s_1 \cdots s_{k-1}r) = k-2$ . Then, Lemma 2.14 yields $k-1 = i+1$ , where $c_i = {\mathrm {proj}}_R c_0$ . Define $d := {\mathrm {proj}}_{R_1} c_0$ and replace G by a minimal gallery $(d_0 = c_0, \ldots , d, c_{k-1}, c_k)$ . Now, we are in the case $R_1 = R$ and the claim follows.

Proof. We have $\ell (1_W, {\mathrm {proj}}_P 1_W) = k-1$ . Thus, it suffices to show $\ell (1_W, {\mathrm {proj}}_Q 1_W)>k-1$ . For $n=1$ , we obtain $\ell (1_W, {\mathrm {proj}}_Q 1_W) \geq k$ . Now, we assume $n>1$ . Then, we have $ {\mathrm {proj}}_{R_n} 1_W = {\mathrm {proj}}_{P_{n-1}} 1_W$ by Lemma 2.21. Since $Q \subseteq R_n$ , we deduce $\ell (1_W, {\mathrm {proj}}_Q 1_W) \geq \ell (1_W, {\mathrm {proj}}_{R_n} 1_W) = \ell (1_W, {\mathrm {proj}}_{P_{n-1}} 1_W)$ and the claim follows by induction.

Proof. We define

$$ \begin{align*} d := \begin{cases} {\mathrm{proj}}_{P_0} c_0 & \text{if } R \neq R_1 \text{ and } \ell(s_1 \cdots s_{k-1}r) = k, \\ {\mathrm{proj}}_{P_1} c_0 & \text{otherwise}. \end{cases} \end{align*} $$

We first show that $\ell (c_0, {\mathrm {proj}}_Q c_0) = \ell (c_0, {\mathrm {proj}}_{R_{\alpha , Q}} c_0) + \ell ( {\mathrm {proj}}_{R_{\alpha , Q}} c_0, d) + \ell ( d, {\mathrm {proj}}_Q c_0 )$ . By definition, we have $R_{\alpha , Q} = R_{\alpha , P_i}$ for all $1 \leq i \leq n$ . We prove the hypothesis by induction on n. Suppose first $n=1$ and that one of the following holds:

• • $R = R_1$ ;

• • $R\neq R_1$ and $\ell (s_1 \cdots s_{k-1}r) = k-2$ .

Then, $Q = P_1 \subseteq R_{\alpha , Q}, d = {\mathrm {proj}}_Q c_0$ and the claim follows. We prove the case $R \neq R_1$ and $\ell (s_1 \cdots s_{k-1}r) = k$ together with the case $n>1$ simultaneously. Lemma 2.21 provides in both cases $ {\mathrm {proj}}_{R_n} c_0 = {\mathrm {proj}}_{P_{n-1}} c_0$ . If $n>1$ , we have $R_{\alpha , Q} = R_{\alpha , P_{n-1}}$ ; if $n=1$ , we have $P_{n-1} = P_0 \subseteq R_{\alpha , Q}$ and $d = {\mathrm {proj}}_{P_{n-1}} c_0$ . This is used in the third equation below. We compute the following:

$$ \begin{align*} \ell(c_0, {\mathrm{proj}}_Q c_0) &= \ell(c_0, {\mathrm{proj}}_{R_n} c_0) + \ell({\mathrm{proj}}_{R_n}c_0, {\mathrm{proj}}_Q c_0) \\ &= \ell(c_0, {\mathrm{proj}}_{P_{n-1}} c_0) + \ell({\mathrm{proj}}_{P_{n-1}} c_0, {\mathrm{proj}}_Q c_0) \\ &= \ell(c_0, {\mathrm{proj}}_{R_{\alpha, Q}} c_0) + \ell({\mathrm{proj}}_{R_{\alpha, Q}} c_0, d) + \ell(d, {\mathrm{proj}}_{P_{n-1}} c_0) \\ & \qquad + \ell({\mathrm{proj}}_{P_{n-1}} c_0, {\mathrm{proj}}_Q c_0) \\ &\geq \ell(c_0, {\mathrm{proj}}_{R_{\alpha, Q}} c_0) + \ell({\mathrm{proj}}_{R_{\alpha, Q}} c_0, d) + \ell(d, {\mathrm{proj}}_Q c_0) \\ &\geq \ell(c_0, {\mathrm{proj}}_Q c_0). \end{align*} $$

Thus, concatenating a minimal gallery from $c_0$ to $ {\mathrm {proj}}_{R_{\alpha , Q}} c_0$ , a minimal gallery from $ {\mathrm {proj}}_{R_{\alpha , Q}} c_0$ to d and a minimal gallery from d to $ {\mathrm {proj}}_Q c_0$ yields a minimal gallery from $c_0$ to $ {\mathrm {proj}}_Q c_0$ . Using Lemma 2.6, there exists a minimal gallery from d to $ {\mathrm {proj}}_Q c_0$ such that every chamber of this gallery is contained in $\mathcal {C}(\partial ^2 \alpha )$ and for two adjacent chambers, there exists a residue in $\partial ^2 \alpha $ containing both. Since $R_{\alpha , Q} \in \{R, R_1\} \subseteq \partial ^2 \alpha $ and, as $R_{\alpha , Q}$ is convex, each chamber of a minimal gallery from $ {\mathrm {proj}}_{R_{\alpha , Q}} c_0$ to d is contained in $R_{\alpha , Q}$ ; the claim follows.

Proof. Recall that $\alpha = \alpha _G$ . As $R \in \partial ^2 \alpha $ , we have $\alpha \neq \pm \beta $ . By Lemma 2.8(a), there exists $C \in \partial ^2 \alpha \cap \partial ^2 \beta $ . Then, there exists a panel $Q' \in \partial \alpha $ that is contained in C. We let $ {\mathrm {proj}}_{Q'} c_0 \neq y \in Q'$ . Let $P_i, R_i$ be as before (with $P_n = Q'$ ), let $G' := (c_0, \ldots , c_{k-1})$ and let $G" := (c_0, \ldots , c_k, c_{k+1})$ be the minimal gallery of type $(s_1, \ldots , s_k, s_{k-1})$ . Let E be a minimal gallery from $c_0$ to y as in Lemma 2.23. We can extend this minimal gallery (if necessary) to a minimal gallery from $c_0$ to $e \in C$ , where $\ell (e) = {\ell ( {\mathrm {proj}}_C c_0) +4}$ . Let $Q" \in \partial \beta $ be a panel contained in C and let $ {\mathrm {proj}}_{Q"} c_0 \neq y' \in Q"$ . Take a minimal gallery $H = (d_0 = c_0, \ldots , d_{m-2} = {\mathrm {proj}}_{R_{\beta , Q"}} c_0, \ldots , d_q := {\mathrm {proj}}_{Q"} d_0, d_{q+1} := y')$ as in Lemma 2.23. Then, $m = k_{\beta } \leq k$ . As before, we can extend H (if necessary) to a minimal gallery from $d_0$ to e. Note that $R \neq C$ by assumption, and since $R \in \partial ^2 \alpha _{G'} \cap \partial ^2 \alpha _{G"}, R \notin \partial ^2 \beta $ , we have $\alpha _{G'} \neq \pm \beta \neq \alpha _{G"}$ .

(i) Assume that $R = R_1$ . Since $R \in \partial ^2 \alpha _{G"} \cap \partial ^2 \alpha , C \in \partial ^2 \alpha $ and $\alpha _{G"} \neq \pm \beta $ , Lemma 2.7(b) implies $C \notin \partial ^2 \alpha _{G"}$ and hence the gallery H has to cross the wall $\partial \alpha _{G"}$ . Assume that $(d_0, \ldots , d_{m-2})$ crosses the wall $\partial \alpha _{G"}$ . Let $1 \leq j \leq m-2$ be such that $\{ d_{j-1}, d_j \} \in \partial \alpha _{G"}$ . Then, $k = k_{\alpha _{G"}} \leq j \leq m-2 \leq k-2$ , which is a contradiction. Thus, the gallery $(d_0, \ldots , d_{m-2})$ does not cross the wall $\partial \alpha _{G"}$ and hence $(d_{m-1}, \ldots , d_{q+1})$ has to cross the wall $\partial \alpha _{G"}$ . Let $m \leq j \leq q+1$ be such that $\{ d_{j-1}, d_j \} \in \partial \alpha _{G"}$ . By Lemma 2.23, there exists $L \in \partial ^2 \beta $ such that $\{ d_{j-1}, d_j \} \subseteq L$ . Then, $L \in \partial ^2 \beta \cap \partial ^2 \alpha _{G"}$ and hence $o(r_{\alpha _{G"}} r_{\beta }) < \infty $ . As $\partial ^2 \alpha \cap \partial ^2 \alpha _{G"} = \{R\} \neq \{C\} = \partial ^2 \alpha \cap \partial ^2 \beta $ (see Lemma 2.7(b)), we have $\partial ^2 \alpha \cap \partial ^2 \beta \cap \partial ^2 \alpha _{G"} = \emptyset $ and hence $\{ r_{\alpha }, r_{\alpha _{G"}}, r_{\beta } \}$ is a reflection triangle. As $ {\mathrm {proj}}_R c_0 \in R \cap \beta \neq \emptyset $ and $R \notin \partial ^2 \beta $ , we deduce $R \subseteq \beta $ . As $e \in C \cap (-\alpha _{G"}) \neq \emptyset $ and ${C \notin \partial ^2 \alpha _{G"}}$ , we deduce $C \subseteq (-\alpha _{G"})$ . As $L \in \partial ^2 \alpha _{G"} \cap \partial ^2 \beta , \{ d_{j-1}, d_j \} \subseteq L \cap \alpha $ and $L \notin \partial ^2 \alpha $ , we deduce $L \subseteq \alpha $ . Thus, $T := \{ \alpha , -\alpha _{G"}, \beta \}$ is a triangle. For $d \in W$ with $\delta (c_{k-2}, d) = s_k s_{k-1}$ , we have $d\in \bigcap _{\gamma \in T} \gamma $ and Lemma 2.12(b) implies $\bigcap _{\gamma \in T} \gamma = \{ d \}$ . If $\ell (s_1 \cdots s_{k-2}s_kr) = k$ , then $k_{\beta } = k+1$ . Thus, $\ell (s_1 \cdots s_{k-2}s_kr) = k-2$ and part (b) follows.

(ii) Assume that $R \neq R_1$ . Since $R \in \partial ^2 \alpha _{G'} \cap \partial ^2 \alpha $ and $R \neq C \in \partial ^2 \alpha $ , Lemma 2.7(b) implies $C \notin \partial ^2 \alpha _{G'}$ and hence H has to cross the wall $\partial \alpha _{G'}$ . Suppose that $(d_0, \ldots , d_{m-2})$ does not cross the wall $\partial ^2 \alpha _{G'}$ . Replacing $\alpha _{G"}$ by $\alpha _{G'}$ in part (i), we obtain that $T := \{ \alpha , -\alpha _{G'}, \beta \}$ is a triangle. Using Lemma 2.12(b), we have $\bigcap _{\gamma \in T} \gamma = \{ c_{k-1} \}$ and hence assertion (a) follows. Now, we suppose that $(d_0, \ldots , d_{m-2})$ crosses the wall $\partial \alpha _{G'}$ and let $1 \leq j \leq m-2$ be such that $P' := \{ d_{j-1}, d_j \} \in \partial \alpha _{G'}$ . Note that $1 \leq m-2 \leq k-2$ and hence $k \geq 3$ . Let Z be the $\{ s_{k-1}, r \}$ -residue containing $c_{k-2}$ . Then, $\alpha _{G'}$ is not a simple root of Z and hence $k_{\alpha _{G'}} \in \{ k-2, k-1 \}$ . This implies $k-2 \leq k_{\alpha _{G'}} \leq j \leq m-2 \leq k-2$ . Lemma 2.22 implies $P' = P_{\alpha _{G'}}$ and hence $P'$ is contained in Z. Moreover, we have $j = m-2$ and $R_{\beta , Q"} = R_{\{r, s_k\}}(d_j)$ . Both nonsimple roots of $R_{\beta , Q"}$ contain $-\alpha $ by Lemma 2.18. As one of them is equal to $\beta $ , we have a contradiction.

Proof. One can easily check that the conditions in part (ii) must be satisfied if $\mathcal {M}$ is a commutator blueprint. Thus, we can assume part (ii). It suffices to show $\vert U_w \vert = 2^{\ell (w)}$ . We show this by induction. For $\ell (w) \leq 1$ , there is nothing to show. Thus, we can assume $\ell (w)>1$ . Let $G = (c_0, \ldots , c_{k+1}) \in \mathrm {Min}(w)$ and let $(\alpha _1, \ldots , \alpha _{k+1})$ be the sequence of roots crossed by G. We define $H := (c_0, \ldots , c_k)$ . Using induction, we have $\vert U_H \vert = 2^k$ . We show that $U_G \cong U_H \rtimes _{\varphi } \mathbb {Z}_2$ , where $\varphi \in {\mathrm {Aut}}(U_H)$ acts on $U_H$ as $u_{\alpha _{k+1}}$ . We consider the mapping

$$ \begin{align*} u_{\alpha_{k+1}}: \{ u_{\alpha} \mid c_k \notin \alpha \} \to U_H, u_{\alpha} \mapsto u_{\alpha} \prod\limits_{\gamma \in M_{\alpha, \alpha_{k+1}}^G} u_{\gamma}. \end{align*} $$

By Axioms (C2) and (C3), $u_{\alpha _{k+1}}$ extends to an endomorphism on $U_H$ . Condition (C1) guarantees $u_{\alpha _{k+1}}^2 = {\mathrm {id}}$ and hence $u_{\alpha _{k+1}} \in {\mathrm {Aut}}(U_H)$ . Thus, we can define the semi-direct product $U_H \rtimes _{u_{\alpha _{k+1}}} \mathbb {Z}_2$ . As $U_w \cong U_G \cong U_H \rtimes _{u_{\alpha _{k+1}}} \mathbb {Z}_2$ , the claim follows.

Proof. Suppose $x, y, z \in G$ and let $x_1, \ldots , x_k, y_1, \ldots , y_l, z_1, \ldots , z_m \in X$ be such that $x = x_1 \cdots x_k, y = y_1 \cdots y_l, z = z_1 \cdots z_m$ . We show $[x, [y, z]] = 1$ . We first assume $l = 1 = m$ . Induction on k yields $[x, [y, z]] = [xx_k^{-1}, [y, z]]^{x_k} [x_k, [y, z]] = 1$ . Now, assume $l=1$ . Induction on m implies $[x, [y, z]] = [x, [y, z_m] [y, zz_m^{-1}]^{z_m}] = [x^{(z_m^{-1})}, [y, zz_m^{-1}]]^{z_m} [x, [y, z_m]]^{[y, zz_m^{-1}]^{z_m}} = 1$ . Now, induction on l yields

$$ \begin{align*} [x, [y, z]] = [x, [yy_l^{-1}, z]^{y_l} [y_l, z]] = [x, [y_l, z]] [x^{(y_l^{-1})}, [yy_l^{-1}, z]]^{y_l[y_l, z]} = 1.\\[-34pt] \end{align*} $$

Proof. Let $f_1, \ldots , f_r, h_1, \ldots , h_m \in X$ such that $g = f_1 \cdots f_r, h = h_1 \cdots h_m$ . We show the claim by induction on $r+m$ . If $r+m \in \{0, 1\}$ , the claim follows directly. Thus, we assume $r+m \geq 2$ . Again, if $0 \in \{r, m\}$ , the claim follows directly. Thus, we can assume $r, m \geq 1$ . Using the nilpotency class, we obtain

$$ \begin{align*} [g, h]^2 &= ( [g, h_m] [g, hh_m^{-1}]^{h_m} )^2 \\ &= ( [gf_r^{-1}, h_m]^{f_r} [f_r, h_m] [g, hh_m^{-1}] )^2 \\ &= [gf_r^{-1}, h_m]^2 [f_r, h_m]^2 [g, hh_m^{-1}]^2. \end{align*} $$

Using the nilpotency class and the fact that $[f_r, h_m]^2 = [f_r, [f_r, h_m]]$ holds, the claim follows by induction.

Proof. Suppose that $\mathcal {M}$ is a commutator blueprint of type $(W, S)$ and that the groups $U_w$ are nilpotent of class at most $2$ . Then, $\mathcal {M}$ is a pre-commutator blueprint of type $(W, S)$ by Lemma 3.3 and it is not hard to see that $\mathcal {M}$ satisfies items ( $2$ -n $1$ ), ( $2$ -n $2$ ) and ( $2$ -n $3$ ) (see Lemma 3.7).

Now, we suppose that $\mathcal {M}$ satisfies the conditions in part (ii). We apply Lemma 3.5. Let $G = (c_0, \ldots , c_{k+1}) \in \mathrm {Min}(w)$ be a minimal gallery, let $(\alpha _1, \ldots , \alpha _{k+1})$ be the sequence of roots crossed by G and let $H := (c_0, \ldots , c_k)$ . Note that condition (C1) holds by items ( $2$ -n $3$ ) (applied with $j=k+1$ ) and ( $2$ -n $2$ ). Moreover, condition (C2) follows from items ( $2$ -n $1$ ) and ( $2$ -n $2$ ). Using items ( $2$ -n $3$ ) and ( $2$ -n $1$ ), one can show that condition (C3) also holds. Now, Lemma 3.5 implies that $\mathcal {M}$ is a commutator blueprint. It is left to show that the group $U_w \cong U_G$ is nilpotent of class at most $2$ . We prove this by induction on $\ell (w)$ . The claim is obvious for $\ell (w) \leq 1$ . Thus, we can assume $\ell (w)>1$ . Using item ( $2$ -n $3$ ), we compute for all $1 \leq i < j \leq k+1$ :

$$ \begin{align*} [ [ u_{\alpha_i}, u_{\alpha_j} ], u_{\alpha_{k+1}} ] &= [ u_{\alpha_i}, u_{\alpha_j} ]^{-1} \prod\limits_{\gamma \in M_{\alpha_i, \alpha_j}^G} u_{\alpha_{k+1}}^{-1} u_{\gamma} u_{\alpha_{k+1}} \\ &= \bigg( \prod\limits_{\gamma \in M_{\alpha_i, \alpha_j}^G} u_{\gamma} \bigg)^{-1} \cdot \prod\limits_{\gamma \in M_{\alpha_i, \alpha_j}^G} \bigg( u_{\gamma} \prod\limits_{\delta \in M_{\gamma, \alpha_{k+1}}^G} u_{\delta} \bigg) = 1. \end{align*} $$

Now, the claim follows from Lemma 3.6.

Convention 4.1. In this subsection, we assume that $(W, S)$ is of universal type (that is, $m_{st} = \infty $ for all $s\neq t \in S$ ) and has rank at least $2$ .

Proof. We abbreviate $\mathcal {M} := \mathcal {M}(K, \mathcal {J}, s, t)$ and $M_{\alpha , \beta }^G := M(K, \mathcal {J}, s, t)_{\alpha , \beta }^G$ for all $(G, \alpha , \beta ) \in \mathcal {I}$ . By definition, $\mathcal {M}$ satisfies Axioms (CB $1$ ) and (CB $2$ ), and by Remark 4.2, it also satisfies Axiom (PCB). Hence, $\mathcal {M}$ is a pre-commutator blueprint. We apply Theorem 3.8. Thus, we suppose $(G, \alpha , \beta ) \in \mathcal {I}$ with $M_{\alpha , \beta }^G \neq \emptyset $ . Then, $\alpha , \beta $ are roots of type s and every $\gamma \in M_{\alpha , \beta }^G$ is a root of type t. Now, it is straight forward to verify that $\mathcal {M}$ satisfies the conditions in Theorem 3.8(ii). We infer that $\mathcal {M}$ is a commutator blueprint and the groups $U_w$ are nilpotent of class at most $2$ . Moreover, $\mathcal {M}$ is Weyl-invariant, as $M_{\alpha , \beta }^G$ does only depend on the existence of a suitable gallery H and not on G.

Proof. We abbreviate $M_{\alpha , \beta }^G := M( s_0, s_1 )_{\alpha , \beta }^G$ for all $(G, \alpha , \beta ) \in \mathcal {I}$ . Similarly as in Theorem 4.6, $\mathcal {M}( s_0, s_1 )$ is a pre-commutator blueprint and we apply Theorem 3.8. Let $G = (c_0, \ldots , c_{k+1}) \in \mathrm {Min}$ be a minimal gallery and let $(\alpha _1, \ldots , \alpha _{k+1})$ be the sequence of roots crossed by G. We first show the following.

Claim. If $\alpha _i$ has type $s_1$ and $\alpha _{i+1}$ has type $s_0$ for some $1 \leq i \leq k$ , then $u_{\alpha _i} u_{\alpha _{i+1}} \in Z( U_G )$ .

Let $1 \leq j \leq k+1$ . We have to show that $u_j$ commutes with $u_{\alpha _i} u_{\alpha _{i+1}}$ . If $j \in \{i, i+1\}$ , the claim follows. We next suppose $j<i$ . However, then it follows by construction that $M_{\alpha _j, \alpha _i}^G = M_{\alpha _j, \alpha _{i+1}}^G$ . This follows for $j>i+1$ similarly.

Now, the conditions in Theorem 3.8(ii) follow essentially from the claim. Hence, $\mathcal {M}( s_0, s_1 )$ is a commutator blueprint and the groups $U_w$ are nilpotent of class at most $2$ . As $M_{\alpha , \beta }^G$ only depends on the existence of a suitable minimal gallery between $\alpha $ and $\beta $ , it is also Weyl-invariant.

Proof. We abbreviate $M_{\sigma , \rho }^H := M_{\mathrm {nil}}(n)_{\sigma , \rho }^H$ for all $(H, \sigma , \rho ) \in \mathcal {I}$ . We first show assertion (a). We can assume that $M_{\alpha , \delta }^G = \{ \gamma \}$ . By definition, there exists an $(i, j, j, k)$ -gallery between $\alpha $ and $\delta $ for some $i, j, k \in \mathbb {N}$ with $i < j < k \leq n$ , and this gallery has type $(c_j, c_{j+1}, \ldots , c_{k-1}, c_k)$ . Note that

$$ \begin{align*} c_k = (s, r_{1, k}, s, r_{2, k}, \ldots, s, r_{i-1, k}, \textbf{s}, r_{i, k}, \ldots, s, r_{j-1, k}, s) \end{align*} $$

and $\gamma $ is the root of type $\textbf {s}$ crossed by $c_k$ directly before $r_{i, k}$ .

Assume that $M_{\gamma , \beta }^G \neq \emptyset $ . Then, there would exist an $(i', j', j', k')$ -gallery between $\gamma $ and $\beta $ for some $i', j', k' \in \mathbb {N}$ with $i' < j' < k' \leq n$ . Because the type of G after crossing $\gamma $ is $r_{i, k}$ , we know that $i' = i$ and $j' = k$ . However, then we have an $(i, j, j', k')$ -gallery between $\alpha $ and $\beta $ by Remark 4.11, which is a contradiction. This finishes the proof of the claim.

We now show assertion (b). Again, we can assume ${M_{\delta , \beta }^G = \{ \gamma \}}$ . By definition, there exists an $(i, j, j, k)$ -gallery between $\delta $ and $\beta $ for some $i, j, k \in \mathbb {N}$ with $i<j<k \leq n$ , and this gallery has type $(c_j, \ldots , c_k)$ . Note that

$$ \begin{align*} &c_{k-1} = (\ldots, s, r_{k-2, k-1}) \quad\text{and} \quad c_k = (s, r_{1, k}, \ldots, s, r_{j-1, k}, s). \end{align*} $$

Assume that $M_{\alpha , \gamma }^G \neq \emptyset $ . Then, there would exist an $(i', j', j', k')$ -gallery between $\alpha $ and $\gamma $ for some $i', j', k' \in \mathbb {N}$ with $i' < j' < k' \leq n$ . We distinguish the following cases.

1. (i) $i=1$ . Then, $\gamma $ is the first root crossed by $c_k$ and the type of G just before crossing $\gamma $ is $r_{k-2, k-1}$ . This implies that $j' = 1$ and $k' = k$ , which is a contradiction.

2. (ii) $i>1$ . In this case, the type of G just before crossing $\gamma $ is $r_{i-1, k}$ . This implies $j' = i$ and $k' = k$ . However, then we have an $(i', j', j, k)$ -gallery between $\alpha $ and $\beta $ , which is a contradiction.

Proof. We abbreviate $\mathcal {M} := \mathcal {M}_{\mathrm {nil}}(n)$ and $M_{\alpha , \beta }^G := M_{\mathrm {nil}}(n)_{\alpha , \beta }^G$ for all $(G, \alpha , \beta ) \in \mathcal {I}$ . By definition, $\mathcal {M}$ satisfies Axioms (CB $1$ ) and (CB $2$ ), and by Remark 4.2 it also satisfies Axiom (PCB). Hence, $\mathcal {M}$ is a pre-commutator blueprint. We apply Lemma 3.5. Suppose $G = (d_1, \ldots , d_p) \in \mathrm {Min}$ , let $\beta \in \Phi (G)$ be the last root that is crossed by G and let $\alpha \in \Phi (G) \backslash \{ \beta \}$ . If $M_{\alpha , \beta }^G = \emptyset $ , then Axioms (C1) and (C2) are satisfied. Otherwise, there exists an $(i, j, j, k)$ -gallery between $\alpha $ and $\beta $ for some $i, j, k \in \mathbb {N}$ with ${1 \leq i < j < k \leq n}$ and we have $M_{\alpha , \beta }^G = \{ \gamma \}$ . Then, $M_{\alpha , \gamma }^G = \emptyset = M_{\gamma , \beta }^G$ holds by definition (see Remark 4.11). Thus, Axioms (C1) and (C2) are satisfied and it remains to check Axiom (C3).

Let $\alpha , \delta \in \Phi (G) \backslash \{ \beta \}$ with $\alpha \leq _G \beta $ . If there is an $(i_1, j_1, i_2, j_2)$ -gallery between $\alpha $ and $\beta $ for some $i_1, j_1, i_2, j_2 \in \mathbb {N}$ , then Axiom (C3) follows essentially from the fact that Axiom (C3) holds for upper triangular matrices. Thus, we can assume that there does not exist any $(i_1, j_1, i_2, j_2)$ -gallery between $\alpha $ and $\beta $ for all $i_1, j_1, i_2, j_2 \in \mathbb {N}$ . In particular, $M_{\alpha , \beta }^G = \emptyset $ and we have to check that the following is a relation in $U_{(d_0, \ldots , d_{p-1})}$ :

$$ \begin{align*} \bigg[ u_{\alpha}, u_{\delta} \prod\limits_{\gamma \in M_{\delta, \beta}^G} u_{\gamma} \bigg] = \prod\limits_{\gamma \in M_{\alpha, \delta}^G} \bigg( u_{\gamma} \prod\limits_{\omega \in M_{\gamma, \beta}^G} u_{\omega} \bigg). \end{align*} $$

Suppose $M_{\alpha , \delta }^G \neq \emptyset $ and let $\gamma \in M_{\alpha , \delta }^G$ . Then, Lemma 4.12 implies $M_{\gamma , \beta }^G = \emptyset $ and the right-hand side of the previous equation is equal to $\prod \nolimits _{\gamma \in M_{\alpha , \delta }^G} u_{\gamma }$ in both cases $M_{\alpha , \delta }^G = \emptyset $ and $M_{\alpha , \delta }^G \neq \emptyset $ .

If $M_{\delta , \beta }^G \kern1.5pt{=}\kern1.5pt \emptyset $ , then Axiom (C3) holds and we are done. Thus, we can assume ${M_{\delta , \beta }^G \kern1.3pt{=}\kern1.3pt \{ \gamma \}}$ . Using Remark 4.11, we deduce $M_{\delta , \gamma }^G = \emptyset $ . Moreover, Lemma 4.12 yields $M_{\alpha , \gamma }^G = \emptyset $ . This implies that the following is a relation in $U_{(d_0, \ldots , d_{p-1})}$ and hence Axiom (C3) holds:

$$ \begin{align*} \bigg[ u_{\alpha}, u_{\delta} \prod\limits_{\omega \in M_{\delta, \beta}^G} u_{\omega} \bigg] = [ u_{\alpha}, u_{\delta} u_{\gamma} ] = [ u_{\alpha}, u_{\gamma} ] [ u_{\alpha}, u_{\delta} ]^{u_{\gamma}} = [u_{\alpha}, u_{\delta}] = \prod\limits_{\omega \in M_{\alpha, \delta}^G} u_{\omega}. \end{align*} $$

Now, Lemma 3.5 yields that $\mathcal {M}$ is a commutator blueprint. The Weyl-invariance follows from the fact that $M_{\alpha , \beta }^G$ does not depend on G, but only on the existence of a suitable minimal gallery between $\alpha $ and $\beta $ .

We show parts (i) and (ii). Let $n \in \mathbb {N}_{\geq 3}$ and $G \in \mathrm {Min}(w)$ be a $(1, 2, n-1, n)$ -gallery. Then, $U_w \cong U_G$ is not nilpotent of class at most $n-2$ . This proves part (ii) and the second part of part (i). Thus, it is left to show that for every $w\in W$ , the group $U_w$ is nilpotent of class at most $n-1$ .

Let $w\in W$ . We prove the claim by induction on $\ell (w)$ . If $\ell (w) \leq 1$ , the claim holds. Thus, we can assume $\ell (w)>1$ . Let $G = (c_0, \ldots , c_p) \in \mathrm {Min}(w)$ , let $(\alpha _1, \ldots , \alpha _p)$ be the sequence of roots crossed by G and let $H := (c_0, \ldots , c_{p-1})$ . Using induction, the group $U_H$ is nilpotent of class at most $n-1$ . If $M_{\alpha _i, \alpha _p}^G = \emptyset $ for all $1 \leq i \leq p-1$ , then ${U_w \cong U_H \times \mathbb {Z}_2}$ and $U_w$ is again nilpotent of class at most $n-1$ . Thus, we can assume $M_{\alpha _i, \alpha _p}^G \neq \emptyset $ for some $1 \leq i \leq p-1$ and hence, there exists a $(j, k, k, l)$ -gallery between $\alpha _i$ and $\alpha _p$ for some $j, k, l \in \mathbb {N}$ with $1 \leq j<k<l\leq n$ . Then, the type of G ends with

$$ \begin{align*} (\ldots, s, r_{1, l}, \ldots, s, r_{k-1, l}, s). \end{align*} $$

Let $1 \leq i' \leq i$ be minimal with the property that there exists an $(i_1, j_1, i_2, j_2)$ -gallery between $\alpha _{i'}$ and $\alpha _p$ for some $i_1, j_1, i_2, j_2 \in \mathbb {N}$ with $1\leq i_1 < j_1, 1 \leq i_2 < j_2$ and $j_1 < j_2$ . Note that $M_{\alpha _{i'}, \alpha _p}^G = \emptyset $ is possible. Comparing the type of G and the type of the $(i_1, j_1, i_2, j_2)$ -gallery, we deduce $i_2 = k$ and $j_2 = l \leq n$ .