The Hecke sign automorphism, pullback and contragredient dual

The automorphism and its first properties. Put $\theta (T_i):=(q-1)-T_i$. Since $T_i^{-1}=q^{-1}(T_i-(q-1))$ by [F1], one has $\theta (T_i)=-q{T}_i^{-1}$. The assignment satisfies the Hecke relations: $\theta (T_i{)}^2=(q-1{)}^2-2(q-1)T_i+T_i^2=(q-1{)}^2-2(q-1)T_i+(q-1)T_i+q=(q-1)\theta (T_i)+q$; the braid relation holds because $\theta (T_i)=-q{T}_i^{-1}$ multiplies every generator by the same scalar $-q$, and inverses satisfy the braid relations; the distant commutation holds similarly, and the scalar squares cancel. Since $H$ is presented by these relations, the assignment extends uniquely to an $R$-algebra endomorphism $\theta$ of $H$; applying it twice fixes each generator, so ${\theta }^2={\mathrm{i}\mathrm{d}}_H$ by [F2] and $\theta$ is an involutive automorphism. For a reduced word $w=s_{i_1}\cdots s_{i_{\mathrm{\ell }}}$ one gets $\theta (T_w)=(-q{)}^{\mathrm{\ell }}{T}_{i_1}^{-1}\cdots T_{i_{\mathrm{\ell }}}^{-1}=(-q{)}^{\mathrm{\ell }(w)}{T}_{w^{-1}}^{-1}$, and $\theta$ commutes with $\ast$ because both sides of $\theta (T_w^{\ast })=(-q{)}^{\mathrm{\ell }(w)}{T}_w^{-1}$ and $\theta (T_w{)}^{\ast }=(-q{)}^{\mathrm{\ell }(w)}{T}_{w^{-1}}^{-\ast }=(-q{)}^{\mathrm{\ell }(w)}{T}_w^{-1}$ agree. All generator images have coefficients in $\mathbb{Z}[q^{\pm 1}]$, so $\theta$ is compatible with base change by [F5].

Pullback of modules. If $M$ is a right $H$-module, then $M^{\theta }$ with $m\cdot h:=m\theta (h)$ is a right $H$-module, because $(m\cdot h)\cdot k=m\theta (h)\theta (k)=m\theta (hk)=m\cdot (hk)$. Since $\theta$ is an involutive automorphism, the assignments $M\mapsto M^{\theta }$ on modules and $N\mapsto N$ on underlying $R$-modules are inverse bijections on submodules: a subset $N\subseteq M$ is closed under the old action if and only if it is closed under the new one, since $\theta$ is bijective and $\theta (H)=H$. Hence $(-{)}^{\theta }$ is exact, preserves submodules and quotients, and $(M^{\theta }{)}^{\theta }=M$; it also carries simple modules to simple modules and preserves the length of finite composition series with multiplicities, because the two module structures have exactly the same submodule lattice.

Contragredient modules. For a right $H$-module $M$, put $M^{\circ }:={\mathrm{Hom}}_R(M,R)$ with $(f\cdot h)(m):=f(m{h}^{\ast })$. Then $((f\cdot h)\cdot k)(m)=f(m{k}^{\ast }{h}^{\ast })=f(m(hk{)}^{\ast })$ by [F2], so $f\cdot (hk)=(f\cdot h)\cdot k$ and $M^{\circ }$ is a right $H$-module. If $M\cong R^r$ is finite free, then $M^{\circ }\cong R^r$ by [F4] applied with $N=R$, and the dual-basis identification is natural in $R$, so $(-{)}^{\circ }$ commutes with base change. Finally $(M^{\theta }{)}^{\circ }=(M^{\circ }{)}^{\theta }$: both have underlying module ${\mathrm{Hom}}_R(M,R)$, and for the two actions one computes $(f\cdot h)(m)=f(m\theta (h^{\ast }))$ for the first and $(f\cdot h)(m)=f(m\theta (h{)}^{\ast })=f(m\theta (h^{\ast }))$ for the second, using that $\theta$ commutes with $\ast$ by step 1.1.

The common eigenspace. Let $H_{\lambda }$ be the $R$-span of $\{T_w:w\in S_{\lambda }\}$, the subalgebra generated by the $T_i$ with $s_i\in S_{\lambda }$; it is $\theta$-stable because $\theta (T_i)=(q-1)-T_i\in H_{\lambda }$. The element $\theta (x_{\lambda })$ lies in $H_{\lambda }$ and satisfies $T_i\theta (x_{\lambda })=\theta (\theta (T_i)x_{\lambda })=\theta (((q-1)-T_i)x_{\lambda })=\theta ((q-1)x_{\lambda }-q{x}_{\lambda })=-\theta (x_{\lambda })$ for every generator $T_i$ of $S_{\lambda }$, by [F3]; so it is a common left eigenvector for eigenvalue $-1$. Conversely let $a={\sum }_{w\in S_{\lambda }}{a}_w{T}_w\in H_{\lambda }$ satisfy $T_{i}a=-a$ for all these generators. Comparing in the expansion $T_{i}a={\sum }_w{a}_w{T}_i{T}_w$, using the left multiplication rule of [F1], the coefficient of $T_v$ equals $q{a}_{s_{i}v}$ when $\mathrm{\ell }(s_{i}v)=\mathrm{\ell }(v)+1$ and equals $(q-1)a_v+a_{s_{i}v}$ when $\mathrm{\ell }(s_{i}v)=\mathrm{\ell }(v)-1$, and it must equal $-a_v$; the first case gives $a_{s_{i}v}=-q^{-1}{a}_v$, and the second gives the same relation with the roles of $v$ and $s_{i}v$ exchanged. Hence $a_{s_{i}u}=-q^{-1}{a}_u$ whenever $\mathrm{\ell }(s_{i}u)=\mathrm{\ell }(u)+1$. Every nonidentity $w\in S_{\lambda }$ is a length-increasing product of generators of $S_{\lambda }$: the permutation $w^{-1}$ restricted to some row block $B_r$ is not the identity, hence has an adjacent descent, which exhibits a generator $s_i$ of $S_{\lambda }$ with $\mathrm{\ell }(s_{i}w)=\mathrm{\ell }(w)-1$. Iterating, $a_w=(-q{)}^{-\mathrm{\ell }(w)}{a}_1$ for all $w\in S_{\lambda }$, so $a=a_1{y}_{\lambda }$: the common eigenspace is exactly the free line $R{y}_{\lambda }$, and it contains $\theta (x_{\lambda })$.

The case $q=1$. At $q=1$ the quadratic relation is $T_i^2=1$, so $T_{w^{-1}}^{-1}=T_w$ and step 1.1 gives $\theta (T_w)=(-1{)}^{\mathrm{\ell }(w)}{T}_w=\mathrm{sgn}(w)T_w$. The relations $T_i^2=1$, braid and distant commutation are the Coxeter relations of the adjacent transpositions of $S_n$, so the specialization $H_{R,1}(S_n)$ is the group algebra $R[S_n]$ with $T_w$ the group basis element $w$; the displayed identity then says that $\theta$ is the pullback along the sign automorphism $w\mapsto \mathrm{sgn}(w)w$. For $N=0$, as in the shapes $\mathrm{\varnothing }$ and $(1^n)$, the unit below is $q^0=1$.

The unit. By step 2.2 write $\theta (x_{\lambda })=c{y}_{\lambda }$ with $c\in R$. In the expansion $\theta (T_w)={\prod }_j\theta (T_{i_j})={\prod }_j((q-1)-T_{i_j})$ of a reduced word, every term other than the full product of the $-T_{i_j}$ has fewer than $\mathrm{\ell }(w)$ generator factors and so cannot contribute to the standard basis element of length $N$; hence the coefficient of $T_{w_0(S_{\lambda })}$ in $\theta (x_{\lambda })={\sum }_{w\in S_{\lambda }}\theta (T_w)$ is $(-1{)}^N$, the contribution of $w=w_0(S_{\lambda })$ alone. In $c{y}_{\lambda }$ the coefficient of $T_{w_0(S_{\lambda })}$ is $c(-q{)}^{-N}$, so $c=(-1{)}^N(-q{)}^N=q^N$. Therefore $\theta (x_{\lambda })=q^N{y}_{\lambda }$, and applying the involution $\theta$ and using $\theta (q^N)=q^N$ gives $\theta (y_{\lambda })=q^{-N}{x}_{\lambda }$.

Statements (1), (2), (3) and (4) are established by steps 1.1, 1.2, 2.1 and 3.1, with the $q=1$ clause in step 2.3 and the unit $q^{\mathrm{\ell }(w_0(S_{\lambda }))}$ displayed in step 3.1. Only the presentation of $H$, inversion counts, the standard basis and finite free duality are used; no choice principle is used.