The row and column Specht conventions in rank three

Shape (3). For the one-row shape, $x_{(3)}={\sum }_{w\in S_3}{T}_w$ and pairing $u$ and $u{s}_i$ with $\mathrm{\ell }(u{s}_i)=\mathrm{\ell }(u)+1$ gives $(T_u+T_{u{s}_i})T_i=q(T_u+T_{u{s}_i})$, hence $x_{(3)}{T}_i=q{x}_{(3)}$; since $z_{(3)}=x_{(3)}{T}_{w_{(3)}}{y}_{(1^3)}=x_{(3)}$ (here $w_{(3)}=1$ and $y_{(1^3)}=1$) the module $P((3))$ is the free rank-one module on $x_{(3)}$ with action $q$. Squaring gives $x_{(3)}^2={\sum }_{v,w}{T}_v{T}_w={\sum }_v{T}_v{x}_{(3)}=\left({\sum }_v{q}^{\mathrm{\ell }(v)}\right)x_{(3)}=p_3(q)x_{(3)}$, because the length distribution in $S_3$ is $(1,2,2,1)$ and $1+2q+2q^2+q^3=(1+q)(1+q+q^2)$; hence the Murphy cell has form value $G_{(3)}(c_{(3)},c_{(3)})=p_3(q)$ and action $q$, and $\beta (x_{(3)},x_{(3)})=1$ by [F1].

Shape (1^3). For the one-column shape, $x_{(1^3)}=1$ and $P((1^3))=y_{(3)}H$ with $y_{(3)}={\sum }_{w\in S_3}(-q{)}^{-\mathrm{\ell }(w)}{T}_w$. For each pair $u,u{s}_i$ with $\mathrm{\ell }(u{s}_i)=\mathrm{\ell }(u)+1$, its contribution is $a(T_u-q^{-1}{T}_{u{s}_i})$, where $a=(-q{)}^{-\mathrm{\ell }(u)}$. The quadratic relation gives $(T_u-q^{-1}{T}_{u{s}_i})T_i=T_{u{s}_i}-q^{-1}((q-1)T_{u{s}_i}+q{T}_u)=-(T_u-q^{-1}{T}_{u{s}_i})$. Summing the pairs gives $y_{(3)}{T}_i=-y_{(3)}$ for $i=1,2$, the sign representation of the rank-one module. The distinguished basis of $M((1^3))$ is the standard basis and $\beta$ is diagonal with entries $q^{\mathrm{\ell }(w)}$, so $\beta (y_{(3)},y_{(3)})={\sum }_w{q}^{-2\mathrm{\ell }(w)}{q}^{\mathrm{\ell }(w)}={\sum }_w{q}^{-\mathrm{\ell }(w)}=q^{-3}{p}_3(q)$, the last equality by $w\mapsto w_{0}w$. Modulo the higher ideal $I^{⊳(1^3)}$, which contains $1+T_i$ because $1+T_1=x_{(2,1)}$ is a higher Murphy element and conjugation by the invertible braid word $T_1{T}_2$ sends $T_1$ to $T_2$, placing $1+T_2$ in the same two-sided ideal, the class $c_{(1^3)}$ satisfies $T_i=-1$ and $G_{(1^3)}(c_{(1^3)},c_{(1^3)})=P_{(1^3)}(q)=1$ by [F2].

The comparison maps. By [F3] applied to $\lambda =(3)$, where ${\lambda }^{\mathrm{\prime }}=(1^3)$ and $\mathrm{\ell }(w_0(S_{(1^3)}))=\mathrm{\ell }(w_{(3)})=0$, the isomorphism ${\mathrm{\Phi }}_{(3)}$ has unit $u_{(3)}=1$ and the identity $G_{(1^3)}^{\theta }({\mathrm{\Phi }}_{(3)}{x}_{(3)},{\mathrm{\Phi }}_{(3)}{x}_{(3)})=\beta (x_{(3)},x_{(3)})=1$ is consistent because both sides are the value of the respective rank-one forms; explicitly ${\mathrm{\Phi }}_{(3)}(x_{(3)})=c_{(1^3)}$ by [F3]. Applying it to $\lambda =(1^3)$, where ${\lambda }^{\mathrm{\prime }}=(3)$ and $\mathrm{\ell }(w_0(S_{(3)}))=3$, $\mathrm{\ell }(w_{(1^3)})=0$, gives $u_{(1^3)}=q^3$, and the identity $G_{(3)}(c_{(3)},c_{(3)})=q^3{q}^{-3}{p}_3(q)=p_3(q)$ is consistent; explicitly ${\mathrm{\Phi }}_{(1^3)}(y_{(3)})=c_{(3)}$. In the target $C_{\mathrm{M}}((1^3){)}^{\theta }$ the generator acts by the ordinary action of $\theta (T_i)=(q-1)-T_i$, hence by $(q-1)-(-1)=q$ for every unit parameter; at $q=1$ this is $\theta (T_i)=-T_i$ and scalar $+1$; in the target $C_{\mathrm{M}}((3){)}^{\theta }$ it acts by the ordinary action of $(q-1)-T_i$ on the class acting by $q$, hence by $(q-1)-q=-1$; these match the source actions $q$ on $P((3))$ and $-1$ on $P((1^3))$ computed in steps 1.1 and 1.2.

Specialization and conclusion. At $q=1$ over $\mathbb{Q}$ the four form values become $1$, $1^{-3}{p}_3(1)=6$, $6$ and $1$, and $\theta$ is the sign automorphism exchanging the trivial and sign modules, in agreement with [F4]; the module actions match as in step 2.1. Every displayed identity is an identity in $A=\mathbb{Z}[q^{\pm 1}]$ between Laurent polynomials, so all of them specialize along any ring map $A\to R$ with $q\mapsto q\in R^{\times }$. Only the rank-one computations, the trace-free form evaluations and the comparison theorem were used; no choice principle is used.