Proof. Write $ P=s_{i_1}\cdots s_{i_r} $ and let $x_k=s_{i_1}\cdots s_{i_k}$ , with $x_0=e$ . Since P is reduced,

$$ \begin{align*} \ell_W(x_k)-\ell_W(x_{k-1})=1 \quad\text{for every } k. \end{align*} $$

Suppose first that $i_k<n$ . Right multiplication by $\iota (s_{i_k})=t_{i_k}t_{\sigma (i_k)}$ exchanges two positions in the first half and two positions in the second half of a symmetric permutation. Hence,

$$ \begin{align*} \mathrm{ex}(\iota(x_k))=\mathrm{ex}(\iota(x_{k-1})). \end{align*} $$

By (3.2),

$$ \begin{align*} \ell_{\widetilde W}(\iota(x_k))-\ell_{\widetilde W}(\iota(x_{k-1}))=2. \end{align*} $$

The two appended simple reflections commute and multiplication by either one changes length by $1$ or $-1$ . Thus, both letters in the block $t_{i_k}t_{\sigma (i_k)}$ increase length.

Now, suppose that $i_k=n$ . Then,

$$ \begin{align*} \iota(x_k)=\iota(x_{k-1})t_n. \end{align*} $$

The entries in positions n and $n+1$ of a symmetric permutation sum to $2n+1$ , so exactly one of them is larger than n. Consequently,

$$ \begin{align*} \Delta\ \mathrm{ex}:=\mathrm{ex}(\iota(x_k))-\mathrm{ex}(\iota(x_{k-1}))\in\{-1,1\}. \end{align*} $$

Since $t_n$ is a simple reflection,

$$ \begin{align*} \Delta\ell:=\ell_{\widetilde W}(\iota(x_k))-\ell_{\widetilde W}(\iota(x_{k-1}))\in\{-1,1\}. \end{align*} $$

Taking the difference of (3.2) at $x_k$ and $x_{k-1}$ gives

$$ \begin{align*} 2=\Delta\ell+\Delta\ \mathrm{ex}. \end{align*} $$

Therefore, $\Delta \ell =\Delta\ \mathrm {ex}=1$ , and the appended letter $t_n$ increases length. Every letter of $\widehat P$ therefore increases the length of the preceding unfolded prefix, which proves statement (i).

For statement (ii), the element $\iota (w_0(I))$ belongs to $\widetilde W_{\,\widehat I}$ . Since

$$ \begin{align*} \mathcal D_L^W(w_0(I))=I, \end{align*} $$

(3.3) gives

$$ \begin{align*} \mathcal D_L^{\widetilde W}(\iota(w_0(I)))=\widehat I. \end{align*} $$

The unique element of the finite Coxeter group $\widetilde W_{\,\widehat I}$ having every simple reflection in $\widehat I$ as a left descent is its longest element. Hence, $\iota (w_0(I))=w_0(\widehat I)$ . The final assertion follows from statement (i).

Proof. If $a=1$ , then $J=\{t_1,\ldots ,t_m\}$ is the full set of simple reflections. Consequently, the assumption $\mathcal D_L(v)\cap J=\varnothing $ forces $v=e$ , so P is empty and the conclusion is immediate. We may therefore assume that $a>1$ .

Write $ P=t_{i_1}\cdots t_{i_\ell }=v. $ Remove the letters from the left by setting

$$ \begin{align*} v^{(0)}=v, \quad v^{(r)}=t_{i_r}v^{(r-1)}=t_{i_{r+1}}\cdots t_{i_\ell} \quad(1\leq r\leq\ell). \end{align*} $$

Each $t_{i_r}$ is a left descent of $v^{(r-1)}$ and $v^{(\ell )}=e$ .

The assumption $\mathcal D_L(v)\cap J=\varnothing $ is equivalent to

$$ \begin{align*} v^{-1}(a)<v^{-1}(a+1)<\cdots<v^{-1}(m+1), \end{align*} $$

so $\eta _a(v)=0$ . Left multiplication by $t_j$ with $j\leq a-2$ only interchanges values outside the terminal interval and therefore leaves $\eta _a$ unchanged. As long as $t_{a-1}$ has not occurred, no letter $t_j$ with $j\geq a$ can be removed: such a letter would have to be a left descent, while $\eta _a=0$ means that the terminal values remain in increasing order.

If $t_{a-1}$ never occurs, then no letter from J occurs and the result follows. Suppose that $t_{a-1}$ occurs once. Immediately before that step, the values

$$ \begin{align*} a,a+1,\ldots,m+1 \end{align*} $$

are still in increasing order. The reflection $t_{a-1}$ interchanges the positions of the values $a-1$ and a. Among the terminal values, only a can thereby move out of its increasing position. Hence, at most, the pairs

$$ \begin{align*} (a,a+1),\ldots,(a,m+1) \end{align*} $$

can become inversions and after the boundary crossing, we have $\eta _a\leq m-a+1$ .

There is no second boundary crossing. Every later removed letter $t_j$ with $j\leq a-2$ leaves $\eta _a$ unchanged. Every later removed letter $t_j$ with $j\geq a$ is a left descent and interchanges the inverted consecutive values j and $j+1$ ; it decreases $\eta _a$ by exactly one. Since $\eta _a$ is nonnegative, the number of letters from J is at most its value immediately after the unique boundary crossing, which is at most $m-a+1$ .

Proof. Put $ u=w_0(I)w. $ Choose the factorised I-witness from Subsection 2.3:

$$ \begin{align*} R=R_IQ, \quad R_I\in\mathrm{Red}(w_0(I)), \quad Q\in\mathrm{Red}(u). \end{align*} $$

Let $m_i=m_{s_i}(Q)$ . By Proposition 3.1,

$$ \begin{align*} \widehat R=\widehat R_I\widehat Q \end{align*} $$

is a reduced word for $\iota (w)$ and $\widehat R_I$ is a reduced word for $w_0(\widehat I)$ . It remains to verify the witness bounds for the residual word $\widehat Q$ .

First, consider a node $j\notin \widehat I$ . It is either the image of a node $i\notin I$ or its mirror under $\sigma $ . In either case, its number of occurrences in $\widehat Q$ is $m_i\leq 1$ by (2.2).

Let $C=[a,b]$ be a connected component of I with $b<n$ . It unfolds into the two components $ [a,b] $ and $[2n-b,2n-a] $ of $\widehat I$ , each having $|C|$ nodes. By (2.3), the number of residual letters of $\widehat Q$ in either component is

$$ \begin{align*} \sum_{i=a}^{b}m_i\leq|C|. \end{align*} $$

It remains to consider a component containing the special node n. Write it as

$$ \begin{align*} C=[a,n], \quad k=|C|=n-a+1. \end{align*} $$

It unfolds into the single type A component

$$ \begin{align*} \widehat C=[a,2n-a], \quad |\widehat C|=2k-1. \end{align*} $$

Set

$$ \begin{align*} M=\sum_{i=a}^{n}m_i. \end{align*} $$

By (2.3), $M\leq k$ . The number of letters of $\widehat Q$ in $\widehat C$ is

(4.1) $$ \begin{align} 2\sum_{i=a}^{n-1}m_i+m_n=2M-m_n. \end{align} $$

If $m_n\geq 1$ , then

$$ \begin{align*} 2M-m_n\leq2k-1=|\widehat C|. \end{align*} $$

Suppose now that $m_n=0$ . Then, Q uses only $s_1,\ldots ,s_{n-1}$ , so u belongs to the standard parabolic subgroup $ W_{\{s_1,\ldots ,s_{n-1}\}}\cong A_{n-1}. $ The word Q is reduced in this parabolic subgroup. Equation (2.1) gives

$$ \begin{align*} \mathcal D_L(u)\cap\{s_a,\ldots,s_{n-1}\}=\varnothing. \end{align*} $$

If $a>1$ , then $s_{a-1}\notin I$ because $[a,n]$ is a connected component of I and, therefore, $m_{a-1}\leq 1$ by (2.2). If $a=1$ , there is no boundary generator. Applying Lemma 4.1 in the type $A_{n-1}$ parabolic subgroup yields

$$ \begin{align*} \sum_{i=a}^{n-1}m_i\leq n-a=k-1. \end{align*} $$

Together with (4.1), this gives

$$ \begin{align*} 2M-m_n=2\sum_{i=a}^{n-1}m_i\leq2k-2<2k-1=|\widehat C|. \end{align*} $$

Thus, for every connected component D of $\widehat I$ , the total number of residual letters belonging to D is at most $|D|$ .

Finally, a reduced word for $w_0(\widehat I)$ contains exactly $\ell (w_0(D))$ letters from each connected component D of $\widehat I$ . Adding the residual bounds just proved shows that $\widehat R$ satisfies both witness conditions.

Proof of Theorem 1.2.

We first prove statement (i). Suppose that $u=w_0(I)w$ is a standard Coxeter element. By the left parabolic decomposition of w with respect to $W_I$ , the assumption $I\subseteq \mathcal D_L(w)$ implies that the factorisation $w=w_0(I)u$ is length additive. Choose $R_I\in \mathrm {Red}(w_0(I))$ and a reduced word Q for u with pairwise distinct letters. Then, $R_IQ$ is an I-witness, so the standard Coxeter condition implies the reduced word condition in every finite Weyl group.

For the converse, it is enough to work in type $C_n$ . Assume that w satisfies the reduced word condition with respect to I, put $ u=w_0(I)w $ and choose the factorised witness $R=R_IQ$ from Subsection 2.3. Proposition 4.2 shows that $\iota (w)$ satisfies the reduced word condition with respect to $\widehat I$ . By the type A equivalence [Reference Gao, Hodges and Yong8, Theorem 1.3], $ w_0(\widehat I)\iota (w) $ is a standard Coxeter element of a standard parabolic subgroup of $A_{2n-1}$ . Proposition 3.1 gives

$$ \begin{align*} w_0(\widehat I)\iota(w) = \iota(w_0(I))\iota(w) = \iota(u). \end{align*} $$

Thus, $\iota (u)$ is a standard Coxeter element.

Again by Proposition 3.1, $\widehat Q$ is a reduced word for $\iota (u)$ . Every reduced word of a standard Coxeter element has pairwise distinct letters. If Q contained two occurrences of $s_i$ for some $i<n$ , then $\widehat Q$ would contain two occurrences of both $t_i$ and $t_{2n-i}$ . If Q contained two occurrences of $s_n$ , then $\widehat Q$ would contain two occurrences of $t_n$ . Both conclusions are impossible. Therefore, the letters of Q are pairwise distinct and ${u=w_0(I)w}$ is a standard Coxeter element. The Coxeter systems of types $B_n$ and $C_n$ have the same Coxeter matrix, so statement (i) also holds in type $B_n$ .

We now prove statement (ii). Combining statement (i) with the all-type geometric classification [Reference Gao, Hodges and Yong9, Theorem 1.3] proves the original criterion in types $B_n$ and $C_n$ . Type $A_n$ follows from [Reference Gao, Hodges and Yong8, Theorem 1.3], and type $G_2$ follows from the rank two theorem of Hodges and Yong [Reference Hodges and Yong13, Theorem 2.10] together with the all-type classification.

The two conditions differ in type $D_4$ . With the numbering used in [Reference Gao, Hodges and Yong9, Example 2.3],

$$ \begin{align*} w=s_3s_2s_3s_4s_2s_1s_2, \quad I=\{s_2,s_3\}, \end{align*} $$

satisfies the original reduced word condition, whereas

$$ \begin{align*} w_0(I)w=s_4s_2s_1s_2 \end{align*} $$

is not a standard Coxeter element. This failure passes to every finite Coxeter group containing the example as a standard parabolic subgroup: lengths, reduced words, left descents and $w_0(I)$ are unchanged, and the support of $w_0(I)w$ remains inside the parabolic subgroup. Every type $D_n$ with $n\geq 4$ and each of $E_6,E_7,E_8$ contains a standard parabolic subgroup of type $D_4$ . Finally, the two conditions also differ in type $F_4$ [Reference Gao, Hodges and Yong8, Section 2]. These are all the remaining irreducible finite Weyl types.