Ordinary Hecke Specht Polytabloids and Signed Murphy Comparison

Length and multiplication criteria for one generator. Let $g\in S_n$ and $1\le i<n$. The one-line list of $g{s}_i$ is obtained from that of $g$ by exchanging the entries $g(i),g(i+1)$ at positions $i,i+1$, and only the relative order of this single pair changes, so $\mathrm{\ell }(g{s}_i)=\mathrm{\ell }(g)+1$ if $g(i)<g(i+1)$ and $\mathrm{\ell }(g{s}_i)=\mathrm{\ell }(g)-1$ if $g(i)>g(i+1)$. Replacing $g$ by $g^{-1}$ gives $\mathrm{\ell }(s_{i}g)=\mathrm{\ell }(g^{-1}{s}_i)$ equal to $\mathrm{\ell }(g)+1$ if and only if $g^{-1}(i)<g^{-1}(i+1)$. Thus [F1] gives $T_g{T}_i=T_{g{s}_i}$ under the right ascent hypothesis $g(i)<g(i+1)$; under the left ascent hypothesis $g^{-1}(i)<g^{-1}(i+1)$, applying [F1] to $g^{-1}$ and then $\ast$ gives $T_i{T}_g=T_{s_{i}g}$.

Order preservation inside rows. If $d\in D_{\lambda }$ and $a<b$ are entries lying in the same row of $t^{\lambda }$, then $d^{-1}(a)<d^{-1}(b)$. Indeed the boxes of $a$ and $b$ lie in that row with the box of $a$ to the left, and $t^{\lambda }\cdot d=d^{-1}\cdot t^{\lambda }$ carries those boxes to the entries $d^{-1}(a),d^{-1}(b)$; since $t^{\lambda }\cdot d$ is row standard, its entries increase along the row.

The coset representatives. The row sets of $t^{\lambda }\cdot g$ are $g^{-1}(B_1),\dots ,g^{-1}(B_k)$ by [F4], so $t^{\lambda }\cdot g$ and $t^{\lambda }\cdot g^{\mathrm{\prime }}$ have the same row sets if and only if $g^{\mathrm{\prime }}{g}^{-1}$ preserves every $B_r$, that is, if and only if $g^{\mathrm{\prime }}\in S_{\lambda }g$. For a given right coset $S_{\lambda }g$, the tableau $t^{\lambda }\cdot g$ has some row sets; filling each row by its entries in increasing order gives a row-standard tableau with the same row sets, and it equals $t^{\lambda }\cdot y$ for some $y\in S_n$ because the action of $S_n$ on tableaux is transitive. That $y$ lies in $S_{\lambda }g$, since it has the same row sets as $t^{\lambda }\cdot g$. Two row-standard tableaux with the same row sets are equal, and the action on tableaux is free, so $y$ is unique in its coset; hence $D_{\lambda }$ is a complete set of right coset representatives of $S_{\lambda }$ in $S_n$.

The parabolic identities. For every $w\in S_{\lambda }$ one has $x_{\lambda }{T}_w=q^{\mathrm{\ell }(w)}{x}_{\lambda }$, $T_w{x}_{\lambda }=q^{\mathrm{\ell }(w)}{x}_{\lambda }$ and $x_{\lambda }^{\ast }=x_{\lambda }$. For a generator $s_i\in S_{\lambda }$ split $S_{\lambda }$ into $A=\{u:\mathrm{\ell }(u{s}_i)>\mathrm{\ell }(u)\}$ and its complement $B$; right multiplication by $s_i$ bijects $A$ onto $B$, and the multiplication rule gives $x_{\lambda }{T}_i={\sum }_{u\in A}{T}_{u{s}_i}+{\sum }_{u\in B}((q-1)T_u+q{T}_{u{s}_i})={\sum }_{v\in B}{T}_v+(q-1){\sum }_{u\in B}{T}_u+q{\sum }_{v\in A}{T}_v=q{x}_{\lambda }\mathrm{.}$ Since every product $u{s}_i$ with $u\in S_{\lambda }$ lies in $S_{\lambda }$, induction on $\mathrm{\ell }(w)=\mathrm{\ell }(v{s}_i)=\mathrm{\ell }(v)+1$ gives $x_{\lambda }{T}_w=(x_{\lambda }{T}_v)T_i=q^{\mathrm{\ell }(v)}q{x}_{\lambda }=q^{\mathrm{\ell }(w)}{x}_{\lambda }$. The star map sends each $T_w$ to $T_{w^{-1}}$ and permutes the sum $x_{\lambda }$, so $x_{\lambda }^{\ast }=x_{\lambda }$; then $(T_w{x}_{\lambda }{)}^{\ast }=x_{\lambda }{T}_{w^{-1}}=q^{\mathrm{\ell }(w^{-1})}{x}_{\lambda }=q^{\mathrm{\ell }(w)}{x}_{\lambda }$ by [F2], and applying $\ast$ again gives $T_w{x}_{\lambda }=q^{\mathrm{\ell }(w)}{x}_{\lambda }$.

Length additivity and multiplicativity. For $w\in S_{\lambda }$ and $d\in D_{\lambda }$ one has $\mathrm{\ell }(wd)=\mathrm{\ell }(w)+\mathrm{\ell }(d)$ and $T_w{T}_d=T_{wd}$. Prove this by induction on $\mathrm{\ell }(w)$. For $w=1$ both claims are immediate. For $\mathrm{\ell }(w)\ge 1$ write $w=s_{i}v$ with $s_i\in S_{\lambda }$ a simple reflection of $S_{\lambda }$ and $\mathrm{\ell }(w)=1+\mathrm{\ell }(v)$; then $\mathrm{\ell }(s_{i}v)=\mathrm{\ell }(v)+1$, so $v^{-1}(i)<v^{-1}(i+1)$ by step 1.1. Because $s_i\in S_{\lambda }$, the labels $i,i+1$ lie in one row block $B_r$, and $v^{-1}\in S_{\lambda }$ preserves every row block, so $v^{-1}(i)$ and $v^{-1}(i+1)$ lie in the same row of $t^{\lambda }$; step 1.2 then gives $(vd{)}^{-1}(i)=d^{-1}(v^{-1}(i))<d^{-1}(v^{-1}(i+1))=(vd{)}^{-1}(i+1)$, whence $\mathrm{\ell }(s_{i}vd)=\mathrm{\ell }(vd)+1$ by step 1.1. The induction hypothesis gives $\mathrm{\ell }(vd)=\mathrm{\ell }(v)+\mathrm{\ell }(d)$, so $\mathrm{\ell }(wd)=\mathrm{\ell }(s_{i}vd)=\mathrm{\ell }(v)+1+\mathrm{\ell }(d)=\mathrm{\ell }(w)+\mathrm{\ell }(d)$. For the basis elements, step 1.1 gives $T_{s_i}{T}_{vd}=T_{s_{i}vd}$ since $\mathrm{\ell }(s_{i}vd)=\mathrm{\ell }(vd)+1$, so $T_w{T}_d=T_{s_i}(T_v{T}_d)=T_{s_i}{T}_{vd}=T_{wd}$ using the induction hypothesis $T_v{T}_d=T_{vd}$.

The shape of the three cases. Let $d\in D_{\lambda }$, $t=t^{\lambda }\cdot d$, $s=t\cdot s_i=s_i\cdot t=t^{\lambda }\cdot (d{s}_i)$. First, $s$ is row standard if and only if $i$ and $i+1$ lie in different rows of $t$: if they lie in the same row then $i$ is left of $i+1$ and the swap creates a descent in that row, while if they lie in different rows then each affected row keeps the same two neighbours of the moved entry, the left neighbour smaller than $i$ and the right neighbour larger than $i+1$, so both inequalities persist. Second, the position of a label $j$ in the row-reading word of $t$ equals $d(j)$, because $t=d^{-1}\cdot t^{\lambda }$; so by step 1.1, $\mathrm{\ell }(d{s}_i)=\mathrm{\ell }(d)+1$ holds exactly when $i$ precedes $i+1$ in $t$, and this happens in particular when $i$ and $i+1$ lie in the same row.

The basis. For $d\in D_{\lambda }$, step 2.1 gives $x_{\lambda }{T}_d={\sum }_{w\in S_{\lambda }}{T}_w{T}_d={\sum }_{w\in S_{\lambda }}{T}_{wd}$, whose support in the standard basis is exactly the right coset $S_{\lambda }d$ with coefficient $1$ at every element. By step 1.3 the cosets $S_{\lambda }d$, $d\in D_{\lambda }$, are pairwise disjoint and partition $S_n$, so the supports of the $x_{\lambda }{T}_d$ are disjoint; since the $T_w$ are $R$-linearly independent by [F1], the family $\{x_{\lambda }{T}_d:d\in D_{\lambda }\}$ is linearly independent. For an arbitrary $g\in S_n$, step 1.3 lets us write $g=wd$ with $w\in S_{\lambda }$, $d\in D_{\lambda }$, and then $x_{\lambda }{T}_g=x_{\lambda }{T}_w{T}_d=q^{\mathrm{\ell }(w)}{x}_{\lambda }{T}_d$ by steps 2.1 and 1.4; since the $T_g$ span $H$ by [F1], the family spans the right ideal $M(\lambda )=x_{\lambda }H$. Hence it is a basis and $M(\lambda )$ is free, which is clause (1).

Case of one row. Suppose $i$ and $i+1$ lie in the same row of $t$. Then $s$ has the same row sets as $t$, so $d{s}_i$ and $d$ lie in the same right coset of $S_{\lambda }$ by step 1.3, say $d{s}_i=wd$ with $w\in S_{\lambda }$. By steps 2.1 and 2.2, $\mathrm{\ell }(d)+1=\mathrm{\ell }(d{s}_i)=\mathrm{\ell }(wd)=\mathrm{\ell }(w)+\mathrm{\ell }(d)$, so $\mathrm{\ell }(w)=1$. The multiplication rule [F1] and step 2.2 give $x_{\lambda }{T}_d{T}_i=x_{\lambda }{T}_{d{s}_i}=x_{\lambda }{T}_{wd}=x_{\lambda }{T}_w{T}_d=q^{\mathrm{\ell }(w)}{x}_{\lambda }{T}_d=q{x}_{\lambda }{T}_d$, using step 1.4. This is the first displayed case.

Case of two rows. Suppose $i$ and $i+1$ lie in different rows of $t$. Then $s$ is row standard by step 2.2, so $d^{\mathrm{\prime }}:=d{s}_i$ lies in $D_{\lambda }$ and $t^{\lambda }\cdot d^{\mathrm{\prime }}=s$; this is the element $d^{\mathrm{\prime }}$ of the statement. The multiplication rule [F1] gives $x_{\lambda }{T}_d{T}_i=x_{\lambda }{T}_{d^{\mathrm{\prime }}}$ if $\mathrm{\ell }(d^{\mathrm{\prime }})=\mathrm{\ell }(d)+1$, and $x_{\lambda }{T}_d{T}_i=(q-1)x_{\lambda }{T}_d+q{x}_{\lambda }{T}_{d^{\mathrm{\prime }}}$ if $\mathrm{\ell }(d^{\mathrm{\prime }})=\mathrm{\ell }(d)-1$. These are the second and third displayed cases, and together with step 3.2 they prove clause (3).

Base change. All constructions above are defined over an arbitrary commutative $R$ with $q\in R^{\times }$ by the formulas of Hecke row permutation modules and ordinary Dipper-James polytabloids, and every identity was obtained from the presentation and from inversion counts, so its coefficients are Laurent polynomials in $q$ with integer coefficients. Given a ring map $R\to R^{\mathrm{\prime }}$ with $q\mapsto q^{\mathrm{\prime }}\in (R^{\mathrm{\prime }}{)}^{\times }$, the base change sends $H_{R,q}(S_n)$ to $H_{R^{\mathrm{\prime }},q^{\mathrm{\prime }}}(S_n)$ and the elements $T_w$, $x_{\lambda }$ to their images $T_w^{\mathrm{\prime }}$, $x_{\lambda }^{\mathrm{\prime }}$; the basis $\{x_{\lambda }{T}_d\}$ maps to the corresponding basis, and the formulas of steps 1.4, 2.1 and 3.1 specialize termwise, so clause (4) follows.

Clauses (1), (2), (3) and (4) are established by steps 3.1, 1.4, 3.2, 4.1 and 4.2. Only inversion counts, the presentation of $H$, the structure of the row groups and the standard basis are used; no choice principle is used.

The form is well defined, symmetric and bilinear. By [F1] the family $\{x_d:d\in D_{\lambda }\}$ is an $R$-basis of $M(\lambda )$, so there is exactly one $R$-bilinear form $\beta$ on $M(\lambda )$ whose Gram matrix in this basis is the diagonal matrix with entries $q^{\mathrm{\ell }(d)}$, $d\in D_{\lambda }$; this form is symmetric because its Gram matrix is diagonal, and it is the permutation form of the statement. For $\lambda =\mathrm{\varnothing }$ the basis is $\{1\}$ and $\beta (1,1)=q^0=1$.

Orbits of one generator. Fix $i$ and abbreviate $x_d:=x_{\lambda }{T}_d$. Partition $D_{\lambda }$ into the set $S$ of those $d$ for which $i,i+1$ lie in the same row of $t^{\lambda }\cdot d$, and the remaining set $D$; for $d\in D$ the element $d^{\ast }:=d{s}_i$ again lies in $D_{\lambda }$ and $(d^{\ast }{)}^{\ast }=d$, by the second and third cases of [F1] applied twice. Thus $D$ is partitioned into two-element orbits $\{u,v\}$ with $v=u{s}_i$ and $\mathrm{\ell }(v)=\mathrm{\ell }(u)+1$, and the action on such an orbit is $x_u{T}_i=x_v$, $x_v{T}_i=q{x}_u+(q-1)x_v$, while $x_d{T}_i=q{x}_d$ for $d\in S$. In particular each $x_d{T}_i$ lies in the span of the orbit of $d$.

Self-adjointness of the generators. In an orbit $\{u,v\}$ with $\mathrm{\ell }(v)=\mathrm{\ell }(u)+1$, the two diagonal pairings are $\beta (x_u{T}_i,x_u)=\beta (x_u,x_u{T}_i)=0$ and $\beta (x_v{T}_i,x_v)=\beta (x_v,x_v{T}_i)=(q-1)q^{\mathrm{\ell }(v)}$. The two off-diagonal pairings both equal $q^{\mathrm{\ell }(v)}=q{q}^{\mathrm{\ell }(u)}$ by the action in step 1.2. A singleton orbit contributes $q{q}^{\mathrm{\ell }(d)}$ on its diagonal, and pairings between distinct orbit spans vanish. Thus $\beta (x{T}_i,y)=\beta (x,y{T}_i)$ on basis vectors and hence, by bilinearity, on all of $M(\lambda )$.

Words, all of $H$, and restriction. For every word $h=T_{i_1}\cdots T_{i_r}$ of generators and all $x,y\in M(\lambda )$ one has $\beta (xh,y)=\beta (x,y{h}^{\ast })$, where $h^{\ast }=T_{i_r}\cdots T_{i_1}$: this follows from step 2.1 by induction on $r$, moving one generator at a time from the first to the second argument. Since the generators span $H$ as an $R$-algebra by [F3], every $h\in H$ is a finite $R$-linear combination of such words, and by [F2] the element $h^{\ast }$ is obtained from the same expression by reversing each word; bilinearity therefore gives $\beta (xh,y)=\beta (x,y{h}^{\ast })$ for all $h\in H$. Finally $P(\lambda )$ is a right ideal contained in $M(\lambda )$ by [F1], so the same identity holds for $x,y\in P(\lambda )$; its restriction to $P(\lambda )$ is the DJ form of the statement.

Base change. Under a ring map $R\to R^{\mathrm{\prime }}$ with $q\mapsto q^{\mathrm{\prime }}\in (R^{\mathrm{\prime }}{)}^{\times }$, the distinguished basis identifies $R^{\mathrm{\prime }}{\otimes }_{R}M(\lambda )$ with $M_{R^{\mathrm{\prime }}}(\lambda )$ and the diagonal weights specialize, identifying the scalar extension of $\beta$ with ${\beta }^{\mathrm{\prime }}$. The natural map $j:R^{\mathrm{\prime }}{\otimes }_{R}P(\lambda )\to M_{R^{\mathrm{\prime }}}(\lambda )$ has image $P_{R^{\mathrm{\prime }}}(\lambda )$: the vectors $z_{\lambda }{T}_w$ span $P(\lambda )$ and their images $z_{\lambda }^{\mathrm{\prime }}{T}_w^{\mathrm{\prime }}$ span $P_{R^{\mathrm{\prime }}}(\lambda )$. For pure tensors, and then by bilinearity, ${\beta }^{\mathrm{\prime }}(j(a\otimes p),j(b\otimes r))=ab\beta (p,r)$. This proves compatibility of the restricted forms along the natural map, without inferring injectivity of a tensor map from right exactness; the later split polytabloid basis establishes that this map is also an isomorphism. Contravariance specializes by the same identity.

The permutation form is well defined, symmetric and bilinear by step 1.1, is $H$-contravariant by step 3.1, and its restriction to $P(\lambda )$ satisfies the same identity; both forms are base-change compatible by step 4.1. Only the distinguished basis, inversion counts and the algebra structure of $H$ are used; no choice principle is used.