A source–sink move in A3: transporting the Euler and skew forms by an initial letter

The word $c^{\mathrm{\prime }}=s_{1}c{s}_1$ evaluates to $s_1{s}_1{s}_2{s}_3{s}_1=s_2{s}_3{s}_1$ using $s_1^2=1$ [F7], and it uses each of $s_1,s_2,s_3$ once, so $c^{\mathrm{\prime }}$ is a Coxeter word and, by [F2], a reduced Coxeter word for the conjugate Coxeter element $s_{1}c{s}_1$; in it $s_1$ is the final letter. This is the source-sink move.

Compute $E_{c^{\mathrm{\prime }}}$ in the ordered basis $(q_1,q_2,q_3)=(e_{s_2},e_{s_3},e_{s_1})$ from [F1] and the entries $K=2B$ of [F3]: $K(e_{s_2},e_{s_3})=K(e_{s_3},e_{s_2})=-1$, $K(e_{s_1},e_{s_2})=K(e_{s_2},e_{s_1})=-1$, $K(e_{s_1},e_{s_3})=K(e_{s_3},e_{s_1})=0$. Since $s_2$ precedes $s_3$, which precedes $s_1$ in the word $c^{\mathrm{\prime }}$, the triangular rule gives $E_{c^{\mathrm{\prime }}}(q_1,q_1)=E_{c^{\mathrm{\prime }}}(q_2,q_2)=E_{c^{\mathrm{\prime }}}(q_3,q_3)=1$, $E_{c^{\mathrm{\prime }}}(q_2,q_1)=K(e_{s_3},e_{s_2})=-1$, $E_{c^{\mathrm{\prime }}}(q_3,q_1)=K(e_{s_1},e_{s_2})=-1$, $E_{c^{\mathrm{\prime }}}(q_3,q_2)=K(e_{s_1},e_{s_3})=0$, and the three upper-triangular entries $E_{c^{\mathrm{\prime }}}(q_1,q_2)=E_{c^{\mathrm{\prime }}}(q_1,q_3)=E_{c^{\mathrm{\prime }}}(q_2,q_3)=0$. This is the displayed matrix; subtracting its transpose gives the displayed ${\omega }_{c^{\mathrm{\prime }}}$. In the fixed basis $(e_{s_1},e_{s_2},e_{s_3})$ the read-off entries are $E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_2})=E_{c^{\mathrm{\prime }}}(q_3,q_1)=-1$, $E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_3})=E_{c^{\mathrm{\prime }}}(q_3,q_2)=0$, $E_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_3})=E_{c^{\mathrm{\prime }}}(q_1,q_2)=0$, while ${\omega }_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_2})=E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_2})-E_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_1})=-1-0=-1$, ${\omega }_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_3})=E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_3})-E_{c^{\mathrm{\prime }}}(e_{s_3},e_{s_1})=0$, and ${\omega }_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_3})=E_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_3})-E_{c^{\mathrm{\prime }}}(e_{s_3},e_{s_2})=0-(-1)=1$.

The reflection $\rho (s_1)=r_{e_{s_1}}$ acts on the simple basis by [F3] and [F4]: $\rho (s_1)e_{s_1}=-e_{s_1}$, $\rho (s_1)e_{s_2}=e_{s_2}-2B(e_{s_2},e_{s_1})e_{s_1}=e_{s_1}+e_{s_2}$, and $\rho (s_1)e_{s_3}=e_{s_3}-2B(e_{s_3},e_{s_1})e_{s_1}=e_{s_3}$.

Check $E_{c^{\mathrm{\prime }}}(\rho (s_1)e_q,\rho (s_1)e_{q^{\mathrm{\prime }}})=E_c(e_q,e_{q^{\mathrm{\prime }}})$ on the nine simple pairs, using bilinearity, step 1.3, the entries of step 1.2 and the values of [F5]. Pair $(q,q^{\mathrm{\prime }})=(s_1,s_2)$: $E_{c^{\mathrm{\prime }}}(-e_{s_1},e_{s_1}+e_{s_2})=-E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_1})-E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_2})=-1+1=0=E_c(e_{s_1},e_{s_2})$. Pair $(s_1,s_1)$: $E_{c^{\mathrm{\prime }}}(-e_{s_1},-e_{s_1})=E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_1})=1=E_c(e_{s_1},e_{s_1})$. Pair $(s_1,s_3)$: $E_{c^{\mathrm{\prime }}}(-e_{s_1},e_{s_3})=-E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_3})=0=E_c(e_{s_1},e_{s_3})$. Pair $(s_2,s_1)$: $E_{c^{\mathrm{\prime }}}(e_{s_1}+e_{s_2},-e_{s_1})=-E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_1})-E_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_1})=-1-0=-1=E_c(e_{s_2},e_{s_1})$. Pair $(s_2,s_2)$: $E_{c^{\mathrm{\prime }}}(e_{s_1}+e_{s_2},e_{s_1}+e_{s_2})=E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_1})+E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_2})+E_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_1})+E_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_2})=1-1+0+1=1=E_c(e_{s_2},e_{s_2})$. Pair $(s_2,s_3)$: $E_{c^{\mathrm{\prime }}}(e_{s_1}+e_{s_2},e_{s_3})=E_{c^{\mathrm{\prime }}}(e_{s_1},e_{s_3})+E_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_3})=0+0=0=E_c(e_{s_2},e_{s_3})$. Pair $(s_3,s_1)$: $E_{c^{\mathrm{\prime }}}(e_{s_3},-e_{s_1})=-E_{c^{\mathrm{\prime }}}(e_{s_3},e_{s_1})=0=E_c(e_{s_3},e_{s_1})$. Pair $(s_3,s_2)$: $E_{c^{\mathrm{\prime }}}(e_{s_3},e_{s_1}+e_{s_2})=E_{c^{\mathrm{\prime }}}(e_{s_3},e_{s_1})+E_{c^{\mathrm{\prime }}}(e_{s_3},e_{s_2})=0-1=-1=E_c(e_{s_3},e_{s_2})$. Pair $(s_3,s_3)$: $E_{c^{\mathrm{\prime }}}(e_{s_3},e_{s_3})=1=E_c(e_{s_3},e_{s_3})$. All nine pairs agree, so by bilinearity the identity holds on all of $V$.

Transport of $\omega$: since ${\omega }_{c^{\mathrm{\prime }}}=E_{c^{\mathrm{\prime }}}-E_{c^{\mathrm{\prime }}}^{\mathsf{T}}$ and ${\omega }_c=E_c-E_c^{\mathsf{T}}$ by [F1], and since $\rho (s_1)$ is linear and preserves the pairing of arguments' roles, step 2.1 gives ${\omega }_{c^{\mathrm{\prime }}}(\rho (s_1)\beta ,\rho (s_1){\beta }^{\mathrm{\prime }})=E_{c^{\mathrm{\prime }}}(\rho (s_1)\beta ,\rho (s_1){\beta }^{\mathrm{\prime }})-E_{c^{\mathrm{\prime }}}(\rho (s_1){\beta }^{\mathrm{\prime }},\rho (s_1)\beta )=E_c(\beta ,{\beta }^{\mathrm{\prime }})-E_c({\beta }^{\mathrm{\prime }},\beta )={\omega }_c(\beta ,{\beta }^{\mathrm{\prime }})$ for all basis vectors, hence for all vectors by bilinearity. This proves (i) by direct computation, illustrating the general identity [F6].

Clause (ii): from [F5], ${\omega }_c(e_{s_1},e_{s_2})=1$, and step 1.2 gives ${\omega }_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_1})=1$. In $c=s_1{s}_2{s}_3$ the noncommuting pair $s_1,s_2$ occurs in the order $s_1$ before $s_2$, so the oriented entry is ${\omega }_c(e_{s_1},e_{s_2})=-K(e_{s_1},e_{s_2})=1$; in $c^{\mathrm{\prime }}=s_2{s}_3{s}_1$ the same pair occurs in the reversed order $s_2$ before $s_1$, and the oriented entry is ${\omega }_{c^{\mathrm{\prime }}}(e_{s_2},e_{s_1})=1$, positive in the reversed order. Moreover step 3.1 with $\beta =e_{s_1},{\beta }^{\mathrm{\prime }}=e_{s_2}$ exhibits the transported identity ${\omega }_{c^{\mathrm{\prime }}}(\rho (s_1)e_{s_1},\rho (s_1)e_{s_2})={\omega }_c(e_{s_1},e_{s_2})$: the new form is the old form read after applying $\rho (s_1)$ to the two canonical roots, so the forms are conjugate, while the sign on the fixed ordered canonical roots is reversed. This is the sign convention used in the rank-two alignment definition.

Conclusion: steps 1.1-1.3, 2.1, 3.1 and 4.1 verify (i) and (ii) by exact matrix and basis-pair computations. Every witness is a fixed basis vector or a fixed word, so no Choice is used.