Row symmetrizers and signed column symmetrizers vanish outside dominance

The intersection matrix. For $d\in S_n$ put $a_{ij}(d):=\mathrm{\mid }{B}_i\cap d(C_j)\mathrm{\mid }$, the number of positions of the $\nu$-block $C_j$ that $d$ sends into the $\alpha$-block $B_i$. Then ${\sum }_j{a}_{ij}(d)=\mathrm{\mid }{B}_i\mathrm{\mid }={\alpha }_i$ and ${\sum }_i{a}_{ij}(d)=\mathrm{\mid }{C}_j\mathrm{\mid }={\nu }_j$. For $w\in S_{\alpha }$ and $v\in S_{\nu }$ one has $w^{-1}(B_i)=B_i$ and $v(C_j)=C_j$, so $a_{ij}(wdv)=\mathrm{\mid }{B}_i\cap wdv(C_j)\mathrm{\mid }=\mathrm{\mid }{w}^{-1}(B_i)\cap dv(C_j)\mathrm{\mid }=\mathrm{\mid }{B}_i\cap d(C_j)\mathrm{\mid }=a_{ij}(d)$; hence the matrix $a(d)$ is constant on the double coset $S_{\alpha }d{S}_{\nu }$.

The eigenvector identities. By [F2], $x_{\alpha }{T}_w=T_w{x}_{\alpha }=q^{\mathrm{\ell }(w)}{x}_{\alpha }$ for all $w\in S_{\alpha }$. For a generator $s_i\in S_{\nu }$ split $S_{\nu }$ into $A=\{u:\mathrm{\ell }(u{s}_i)=\mathrm{\ell }(u)+1\}$ and its complement $B$; right multiplication by $s_i$ bijects $A$ onto $B$, and the multiplication rule gives $y_{\nu }{T}_i={\sum }_{u\in A}(-q{)}^{-\mathrm{\ell }(u)}{T}_{u{s}_i}+{\sum }_{u\in B}(-q{)}^{-\mathrm{\ell }(u)}((q-1)T_u+q{T}_{u{s}_i})=(-q+q-1){\sum }_{v\in B}(-q{)}^{-\mathrm{\ell }(v)}{T}_v-{\sum }_{v\in A}(-q{)}^{-\mathrm{\ell }(v)}{T}_v=-y_{\nu }\mathrm{.}$ Hence $y_{\nu }{T}_v=(-1{)}^{\mathrm{\ell }(v)}{y}_{\nu }$ for every $v\in S_{\nu }$ by induction along reduced words, and since $y_{\nu }^{\ast }=y_{\nu }$ one gets $T_v{y}_{\nu }=(y_{\nu }{T}_{v^{-1}}{)}^{\ast }=(-1{)}^{\mathrm{\ell }(v)}{y}_{\nu }$.

Sorted representatives exist and are unique. For a matrix $a=(a_{ij})$ with row sums ${\alpha }_i$ and column sums ${\nu }_j$, process positions $x=1,\dots ,n$ increasingly. In each block $C_j$, request target blocks $B_i$ in increasing order, requesting $B_i$ exactly $a_{ij}$ times, and send each position to the smallest unused member of its requested block. Before the $r$-th request for $B_i$ within $C_j$, exactly ${\sum }_{j^{\mathrm{\prime }}<j}{a}_{i{j}^{\mathrm{\prime }}}+r-1<{\sum }_{j^{\mathrm{\prime }}\le j}{a}_{i{j}^{\mathrm{\prime }}}\le {\alpha }_i$ members have been used. Thus the rule defines a bijection $d_0$, increasing on each $C_j$, with $d_0^{-1}$ increasing on each $B_i$, and $a(d_0)=a$. These two monotonicities also force the rule: the counts determine the increasing target-block list in each $C_j$, and inverse monotonicity forces successive requests for $B_i$ to take its members in increasing order. Hence $d_0$ is unique with this matrix and these monotonicities.

The dominance inequality. If $a_{ij}\le 1$ for all $i,j$, then for every $k$ one has ${\sum }_{i\le k}{\alpha }_i={\sum }_{i\le k}{\sum }_j{a}_{ij}={\sum }_j{\sum }_{i\le k}{a}_{ij}\le {\sum }_j\min (k,{\nu }_j)$; since $\min (k,{\nu }_j)$ counts the pairs $(i,j)$ with $i\le k$ and $i\le {\nu }_j$, this equals ${\sum }_{i\le k}{\nu }_i^{\mathrm{\prime }}$, where ${\nu }_i^{\mathrm{\prime }}$ is the $i$-th column height of $\nu$. Hence if $\alpha$ is not dominated by ${\nu }^{\mathrm{\prime }}$, so that ${\sum }_{i\le k}{\alpha }_i>{\sum }_{i\le k}{\nu }_i^{\mathrm{\prime }}$ for some $k$, then every matrix with row sums $\alpha$ and column sums $\nu$ has an entry at least $2$.

Reduction by descents. Starting with $g\in S_n$, if $g^{-1}$ is not increasing on an $\alpha$-block, choose an adjacent pair $i,i+1$ in that block with $g^{-1}(i)>g^{-1}(i+1)$ and put $h=s_{i}g$. Then $\mathrm{\ell }(h)=\mathrm{\ell }(g)-1$, $T_g=T_i{T}_h$, and $x_{\alpha }{T}_g{y}_{\nu }=q{x}_{\alpha }{T}_h{y}_{\nu }$. If $g$ is not increasing on a $\nu$-block, choose adjacent positions $j,j+1$ there with $g(j)>g(j+1)$ and put $h=g{s}_j$. Then $T_g=T_h{T}_j$ and $x_{\alpha }{T}_g{y}_{\nu }=-x_{\alpha }{T}_h{y}_{\nu }$. These length criteria follow by counting inversions, also on the inverse permutation. Each move stays in the same double coset and preserves its matrix by step 1.1. Length strictly decreases, so the process ends at the unique sorted representative $d_0$ of step 1.3. Consequently $x_{\alpha }{T}_g{y}_{\nu }$ is a unit multiple of $x_{\alpha }{T}_{d_0}{y}_{\nu }$, which proves the required vanishing equivalence without assuming $T_{wdv}=T_w{T}_d{T}_v$ for non-additive products.

A repeated intersection gives adjacent transpositions. Let $d_0$ be the sorted representative with matrix $a$, and suppose $a_{ij}\ge 2$. Put $X:=C_j\cap d_0^{-1}(B_i)$, the positions of $C_j$ sent into $B_i$, and $Y:=d_0(X)=B_i\cap d_0(C_j)$; then $\mathrm{\mid }X\mathrm{\mid }=\mathrm{\mid }Y\mathrm{\mid }=a_{ij}\ge 2$. Both are intervals: if $x_0<x<x_1$ with $x_0,x_1\in X$ and $x\in C_j$, then $d_0(x_0)<d_0(x)<d_0(x_1)$ by monotonicity of $d_0$ on $C_j$ and the two outer values lie in the interval $B_i$, so $d_0(x)\in B_i$ and $x\in X$; and if $p<r<q$ with $p,q\in Y$ and $r\in B_i$, then $d_0^{-1}(p)<d_0^{-1}(r)<d_0^{-1}(q)$ by monotonicity of $d_0^{-1}$ on $B_i$ and the outer preimages lie in $C_j$, so $r\in Y$. The increasing bijection $d_0:X\to Y$ between finite intervals of the same size is the shift $x\mapsto x+(\min Y-\min X)$. Writing $a:=\min X$ we have $a,a+1\in C_j$ and $d_0(a),d_0(a)+1\in B_i$ with $d_0(a+1)=d_0(a)+1$. Then $t:=s_a\in S_{\nu }$ and $s:=s_{d_0(a)}\in S_{\alpha }$ satisfy $s{d}_0=d_{0}t$: both sides agree at $a$ and $a+1$ and fix all other positions.

The factorization and vanishing. Let $s\in S_{\alpha }$, $t\in S_{\nu }$ be simple with $s{d}_0=d_{0}t$ as in step 2.2. If $\mathrm{\ell }(s{d}_0)=\mathrm{\ell }(d_0)+1$ then $T_s{T}_{d_0}=T_{s{d}_0}=T_{d_{0}t}=T_{d_0}{T}_t$ by [F1], and if $\mathrm{\ell }(s{d}_0)=\mathrm{\ell }(d_0)-1$ then both sides equal $(q-1)T_{d_0}+q{T}_{s{d}_0}$; in both cases $T_s{T}_{d_0}=T_{d_0}{T}_t$. Split $S_{\alpha }$ into right cosets $\{u,us\}$ of the subgroup $\{1,s\}$ with minimal representatives $u$, and $S_{\nu }$ into left cosets $\{v,tv\}$ of $\{1,t\}$ with minimal representatives $v$; then $x_{\alpha }=X(1+T_s)$ with $X:={\sum }_u{T}_u$, because $T_u{T}_s=T_{us}$, and $y_{\nu }=(1-q^{-1}{T}_t)Y$ with $Y:={\sum }_v(-q{)}^{-\mathrm{\ell }(v)}{T}_v$, because $T_{tv}=T_t{T}_v$ and $(-q{)}^{-\mathrm{\ell }(tv)}=(-q{)}^{-1}(-q{)}^{-\mathrm{\ell }(v)}=-q^{-1}(-q{)}^{-\mathrm{\ell }(v)}$. Therefore $x_{\alpha }{T}_{d_0}{y}_{\nu }=X(1+T_s)T_{d_0}(1-q^{-1}{T}_t)Y=X{T}_{d_0}(1+T_t)(1-q^{-1}{T}_t)Y=0,$ since $(1+T_t)(1-q^{-1}{T}_t)=1-q^{-1}{T}_t+T_t-q^{-1}{T}_t^2=1+(1-q^{-1})T_t-q^{-1}((q-1)T_t+q)=0$. No division or cancellation of $q+1$ occurs.

All terms of such a double coset vanish. Let $C$ be a double coset whose matrix $a(C)$ has an entry at least $2$, and let $d_0$ be its sorted representative. Then $x_{\alpha }{T}_{d_0}{y}_{\nu }=0$ by steps 2.2 and 3.1, so $x_{\alpha }{T}_d{y}_{\nu }=0$ for every $d\in C$ by step 2.1.

Suppose $\alpha$ is not dominated by ${\nu }^{\mathrm{\prime }}$. By step 1.4 every double coset matrix has an entry at least $2$, so by step 4.1 every $x_{\alpha }{T}_d{y}_{\nu }$ with $d\in S_n$ vanishes; since the $T_d$ are an $R$-basis of $H$ by [F1], $x_{\alpha }H{y}_{\nu }=0$. All structure constants lie in $\mathbb{Z}[q^{\pm 1}]$, no factor $q+1$ is inverted, and no choice principle is used; for $n=0,1$ the groups are trivial and the claim is immediate.