Coxeter words are commutation-connected; the Euler and skew forms depend only on the Coxeter element

Base of clause (1): for $j=0$ the empty word is reduced and $S(w_0)=\mathrm{\varnothing }$, where $w_0:=1$.

Induction hypothesis of clause (1): let $s_1\cdots s_n$ be a Coxeter word, so that $s_1,\dots ,s_n$ are pairwise distinct with $\{s_1,\dots ,s_n\}=S$, and let $0\le j<n$ with $w_j:=s_1\cdots s_j$ reduced and $S(w_j)=\{s_1,\dots ,s_j\}$.

One direction of the single-pair equivalence: let $u=r_1\cdots r_m$ be reduced, with $t\mathrm{\notin }S(u)$, and suppose every $r_i$ commutes with $t$. For each $i$, the homomorphism property [F3] and reflection conjugation [F18] give $r_{\rho (r_i)e_t}=\rho (r_i)r_{e_t}\rho (r_i{)}^{-1}=r_{e_t}$. By the reflection formula [F2], the $(-1)$-eigenspace of $r_a$ is $\mathbb{R}a$ whenever $B(a,a)\mathrm{\ne }0$; [F4] gives $B(\rho (r_i)e_t,\rho (r_i)e_t)=1$, and [F2] gives $B(e_t,e_t)=1$. Equality of the reflections therefore implies $\rho (r_i)e_t=\pm e_t$. The value $-e_t$ is impossible because $\rho (r_i)e_t=e_t-2B(e_t,e_{r_i})e_{r_i}$ and the distinct basis vectors $e_t,e_{r_i}$ are linearly independent. Thus each $\rho (r_i)$ fixes $e_t$, and so does $\rho (u)$.

Converse setup: assume $\rho (u)e_t=e_t$ and argue by induction on $m=\mathrm{\ell }(u)$. The base $m=0$ is immediate. For $m\ge 1$, write $u=r_1{u}^{\mathrm{\prime }}$, where $u^{\mathrm{\prime }}=r_2\cdots r_m$ is reduced. Since $t\mathrm{\notin }S(u^{\mathrm{\prime }})$ by $S(u^{\mathrm{\prime }})\subseteq S(u)$ from [F8], and $\rho (r_1)=r_{e_{r_1}}$ with $B(e_{r_1},e_{r_1})=1$, the reflection formula [F2] gives $\rho (r_1{)}^{-1}=\rho (r_1)$. The positive-span result [F5] therefore gives $\rho (u^{\mathrm{\prime }})e_t=\rho (r_1)e_t=e_t+{\sum }_{i=2}^m{a}_i{\gamma }_i^{\mathrm{\prime }}$, where $a_i\ge 0$ and ${\gamma }_i^{\mathrm{\prime }}=\rho (r_2\cdots r_{i-1})e_{r_i}$. The reflection formula also gives $\rho (r_1)e_t=e_t+\delta e_{r_1}$ with $\delta =-2B(e_t,e_{r_1})\ge 0$, because $r_1\mathrm{\ne }t$ and the off-diagonal entries of $B$ are nonpositive; thus ${\sum }_i{a}_i{\gamma }_i^{\mathrm{\prime }}=\delta e_{r_1}$. Each ${\gamma }_i^{\mathrm{\prime }}$ is a positive root by [F7] and belongs to ${\mathrm{\Phi }}_{S(u^{\mathrm{\prime }})}$ by [F13], so [F12] gives nonnegative simple coordinates and [F13] gives zero $e_t$-coordinate.

Exclude $\delta >0$: the equality in step 1.4 would then have nonzero right side, so some $a_i>0$, and coordinatewise nonnegativity forces each such ${\gamma }_i^{\mathrm{\prime }}$ to lie on the positive $e_{r_1}$-ray. By [F4] and [F2], $B({\gamma }_i^{\mathrm{\prime }},{\gamma }_i^{\mathrm{\prime }})=B(e_{r_1},e_{r_1})=1$, so if ${\gamma }_i^{\mathrm{\prime }}=\lambda e_{r_1}$ with $\lambda >0$, then ${\lambda }^2=1$ and ${\gamma }_i^{\mathrm{\prime }}=e_{r_1}$. But [F7] puts ${\gamma }_i^{\mathrm{\prime }}$ in $N((u^{\mathrm{\prime }}{)}^{-1})$, so [F10] gives $r_1\in D_L(u^{\mathrm{\prime }})$ and [F9] gives $\mathrm{\ell }(r_1{u}^{\mathrm{\prime }})<\mathrm{\ell }(u^{\mathrm{\prime }})$, contradicting that $u=r_1{u}^{\mathrm{\prime }}$ is reduced. Thus $\delta =0$.

Step of clause (1): under the hypothesis of step 1.2 we have $s_{j+1}\mathrm{\notin }S(w_j)$, so F5 gives $\rho (w_j)e_{s_{j+1}}\in {\mathrm{\Phi }}_{+}$ and then F6 gives $\mathrm{\ell }(w_j{s}_{j+1})=\mathrm{\ell }(w_j)+1$; hence $w_{j+1}$ is reduced, and since $s_1\cdots s_{j+1}$ is a reduced expression of it, F8 gives $S(w_{j+1})=\{s_1,\dots ,s_{j+1}\}$.

Finish the converse by induction: since $\delta =0$, step 1.4 gives ${\sum }_i{a}_i{\gamma }_i^{\mathrm{\prime }}=0$; each ${\gamma }_i^{\mathrm{\prime }}$ is a nonzero positive root, so all $a_i=0$ and $\rho (u^{\mathrm{\prime }})e_t=e_t$. Induction shows every letter of $u^{\mathrm{\prime }}$ commutes with $t$. Step 1.4 also gives $\rho (r_1)e_t=e_t$; using the homomorphism property [F3], the isometry [F4], reflection conjugation [F18] and faithfulness [F17] yields $r_{1}t{r}_1^{-1}=t$, so $r_1$ commutes with $t$ as well.

Clause (1) follows from steps 1.1 and 2.2 by induction on $j$: for the value $c=s_1\cdots s_n$ we get $\mathrm{\ell }(c)=n$ and $S(c)=S$, so every Coxeter word is reduced; conversely a reduced expression of $c$ is a word of length $n=\mathrm{\ell }(c)$ whose letters lie in $S(c)=S$, hence it uses every element of $S$ exactly once and is a Coxeter word.

Clause (5): for each $j$, applying F5 to the pair $(w_{j-1},s_j)$, whose hypothesis $s_j\mathrm{\notin }S(w_{j-1})=\{s_1,\dots ,s_{j-1}\}$ holds by step 3.2 and the pairwise distinctness of the letters, gives ${\beta }_j=e_{s_j}+{\sum }_{i<j}{a}_{ij}{e}_{s_i}$ with $a_{ij}\ge 0$ and support in $\{s_1,\dots ,s_j\}$. The matrix whose columns are $({\beta }_1,\dots ,{\beta }_n)$ in the basis $(e_{s_1},\dots ,e_{s_n})$ is thus upper unitriangular with diagonal entries $1$ and so invertible; hence ${\beta }_1,\dots ,{\beta }_n$ is a basis of $V$.

Clause (2), left descents: for $1\le k\le n$, the descent/inversion criterion F10 gives $s_k\in D_L(c)⟺e_{s_k}\in N(c^{-1})$, and the prefix formula [F7] gives $N(c^{-1})=\{{\beta }_1,\dots ,{\beta }_n\}$. By step 4.1 and linear independence of the basis $(e_s{)}_{s\in S}$, $e_{s_k}={\beta }_j$ forces $j=k$ and $a_{ik}=0$ for all $i<k$, that is, ${\beta }_k=e_{s_k}$. Conversely ${\beta }_k=e_{s_k}$ gives $e_{s_k}\in N(c^{-1})$ and hence $s_k\in D_L(c)$. By the descent definition [F9] and steps 1.3 and 3.1 applied to $u=s_1\cdots s_{k-1}$, the identity $\rho (s_1\cdots s_{k-1})e_{s_k}=e_{s_k}$ holds exactly when $s_k$ commutes with $s_1,\dots ,s_{k-1}$. This gives the formula for $D_L(c)$; if $k<j$ and $s_k,s_j\in D_L(c)$, that formula shows $s_j$ commutes with the earlier letter $s_k$, so the elements of $D_L(c)$ pairwise commute.

Clause (2), right descents and initial letters: apply step 5.1 to $c^{-1}$, whose reduced words are the reverses of the reduced words of $c$ by [F16]; the descent definition [F9] gives $D_R(c)=D_L(c^{-1})$, hence $D_R(c)=\{s_k:s_k\text{ commutes with }{s}_{k+1},\dots ,s_n\}$. If $s_k\in D_L(c)$, then $\mathrm{\ell }(s_{k}c)=\mathrm{\ell }(c)-1$; the exchange condition [F11] gives $s_{k}c=s_1\cdots \widehat{s_i}\cdots s_n$ for some $i$. Since $s_k^2=1$ by [F14], $c=s_k(s_{k}c)$ is a length-$n$ word for $c$, so it is reduced and starts with $s_k$. The right-handed exchange condition gives symmetrically that each $s_k\in D_R(c)$ is the last letter of some reduced expression of $c$.

Clause (3), commutation connectivity, by induction on $n=\mathrm{\mid }S\mathrm{\mid }$: let $u=s_1\cdots s_n$ and $v=t_1\cdots t_n$ be reduced Coxeter words for $c$ (for $n\le 1$ there is only one such word). If $s_1=t_1$, then $s_{1}c=s_2\cdots s_n$ by $s_1^2=1$ in [F14]; the tails $s_2\cdots s_n$ and $t_2\cdots t_n$ are reduced Coxeter words for this element in $W_{S\setminus \{s_1\}}$. By [F15], induction connects them by adjacent commuting transpositions. If $s_1\mathrm{\ne }{t}_1$, let $s=s_1$. Both $s$ and $t_1$ lie in $D_L(c)$, because their left products with $c$ have length $n-1$; step 5.1 shows they commute. Write $s=t_j$ with $j\ge 2$. Step 5.1 applied to $v$ shows $s=t_j$ commutes with $t_1,\dots ,t_{j-1}$, so adjacent commuting swaps move $s$ to the front, giving $s{v}^{\mathrm{\prime }}$ with $v^{\mathrm{\prime }}=t_1\cdots t_{j-1}{t}_{j+1}\cdots t_n$. This remains a reduced word for $c$, and $v^{\mathrm{\prime }}$ is a reduced Coxeter word for $sc\in W_{S\setminus \{s\}}$, using $s^2=1$ in [F14]; induction in that parabolic connects the tails $s_2\cdots s_n$ and $v^{\mathrm{\prime }}$. This proves commutation connectivity. Each such swap preserves the relative order of every noncommuting pair, so that relative order is determined by $c$. Conversely, suppose two reduced Coxeter words have the same relative order for every noncommuting pair. Move the first letter of $v$ leftward in $u$: every letter it crosses has the opposite relative order and therefore must commute with it. Once their first letters agree, repeat on the tails; the two words are connected by adjacent commuting swaps.

Clause (4): by step 6.2 any two reduced Coxeter words for $c$ are connected by adjacent swaps of commuting letters. If $s,t$ commute, step 1.3 gives $\rho (s)e_t=e_t$; the reflection formula [F2] then forces $B(e_t,e_s)=0$, hence $K(e_s,e_t)=0$ by [F1]. If $w^{\mathrm{\prime }}$ is obtained from $w=x_1\cdots x_n$ by swapping the adjacent letters $x_p=s$, $x_{p+1}=t$, every entry $E(e_a,e_b)$ in [F1] depends only on the relative order of $a,b$, and the swap changes that order only for the pair $\{a,b\}=\{s,t\}$. For this pair, both entries $E(e_s,e_t),E(e_t,e_s)$ are zero before and after the swap because $K(e_s,e_t)=0$; every other entry is unchanged. Thus $E_c$ is unchanged by each swap and is independent of the reduced Coxeter word, as is ${\omega }_c=E_c-E_c^{\mathsf{T}}$. Finally $E_c+E_c^{\mathsf{T}}=K$ for each word: for $a\mathrm{\ne }b$, exactly one of the two Euler entries is $K(e_a,e_b)$ and the other is $0$; on the diagonal their sum is $1+1=2=K(e_a,e_a)$ by [F1] and [F2].

Steps 3.2 and 6.2 discharge the length and rank inductions; together with steps 4.1, 5.1, 6.1, and 7.1 they establish clauses (1)–(5). No Choice is used.