The quantum Serre sums vanish in the shuffle algebra, and the opposite Serre ideal annihilates the shuffle half

Put $N=1-a_{ij}$, $x=[v_i]$, $y=[v_j]$, $t=q_i^{-2}$, and $B_{m,r}(t)=0$ outside $0\le r\le m$. Let $W_m=[v_i\mathrm{\mid }\cdots \mathrm{\mid }{v}_i]$ (with $m$ letters), and let $w_p$ be the word with $p$ copies of $v_i$, then $v_j$, then $N-p$ copies of $v_i$ (omitting a zero-length block). Repeatedly shuffling one $x$ into a block of $m$ equal $x$-letters gives $x^{\ast m}=A_m{W}_m$, where $A_m=q_i^{-m(m-1)\mathrm{/}2}[m{]}_i!$: the insertion weights sum to $1+q_i^{-2}+\cdots +q_i^{-2m}=q_i^{-m}[m+1{]}_i$. In a three-block shuffle $W_a\ast y\ast W_b$, with $a+b=N$, fix $w_p$, and let $u$ (respectively $v$) count the $b$-block letters before (respectively after) $y$, so $u+v=b$. The prefix interleavings contribute $B_{p,u}(t)$, the suffix interleavings contribute $B_{N-p,v}(t)$, the $u$ letters before $y$ cross $y$ and the $N-p-v$ letters after it, and each of those $u$ letters crosses those $N-p-v$ suffix $x$-letters; therefore ${\mathrm{coeff}}_{w_p}(W_a\ast y\ast W_b)={\sum }_{u+v=b}{B}_{p,u}(t)B_{N-p,v}(t)q_i^{-a_{ij}(u+N-p-v)-2u(N-p-v)}$.

The pairing is diagonal in the tensor-word bases: for tensor words $u=[v_{i_1}\mathrm{\mid }\cdots \mathrm{\mid }{v}_{i_k}]$ and $v={\xi }_{j_1}\cdots {\xi }_{j_l}$ the defining formula gives $⟨u,v⟩={\delta }_{kl}{\delta }_{i_1{j}_1}\cdots {\delta }_{i_k{j}_k}{\mathrm{\hslash }}^{-k}{\prod }_t{d}_{i_t}^{-1}$, and ${\prod }_t{d}_{i_t}^{-1}$ depends only on $u$. Consequently, for $x={\sum }_w{x}_w[w]$ and tensor words $y,y^{\mathrm{\prime }}$ with letter-sequences $u,u^{\mathrm{\prime }}$, both sides of $⟨x,y{y}^{\mathrm{\prime }}⟩={\sum }_{k\ge 0}⟨x_{\le k},y⟩⟨x_{>k},y^{\mathrm{\prime }}⟩$ equal $x_{u{u}^{\mathrm{\prime }}}{\mathrm{\hslash }}^{-(\mathrm{\mid }u\mathrm{\mid }+\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid })}\left({\prod }_{s\in u}{d}_s^{-1}\right)\left({\prod }_{s\in u^{\mathrm{\prime }}}{d}_s^{-1}\right)$: the left side because only the word $u{u}^{\mathrm{\prime }}$ has nonzero pairing with $y{y}^{\mathrm{\prime }}$, the right side because only the cut $k=\mathrm{\mid }u\mathrm{\mid }$ contributes, the product in $T(V^{\ast })$ being concatenation and the cut coproduct of the word $[u]$ being $\mathrm{\Delta }([u])={\sum }_k[u_{\le k}]\otimes [u_{>k}]$. Extending by bilinearity, this adjunction rule holds for all $x\in \mathrm{S}\mathrm{h}(V)$ and all $y,y^{\mathrm{\prime }}\in T(V^{\ast })$.

For homogeneous one-letter words $[x_1],\dots ,[x_m]\in V$ put $w(a,b):=⟨\mathrm{deg}{x}_a,\mathrm{deg}{x}_b⟩$. Then $[x_1]\ast \cdots \ast [x_m]={\sum }_{\sigma \in S_m}{q}^{-{\sum }_{a<b,\sigma (a)>\sigma (b)}w(a,b)}[x_{{\sigma }^{-1}(1)}\mathrm{\mid }\cdots \mathrm{\mid }{x}_{{\sigma }^{-1}(m)}]$. Here $\sigma$ maps each original letter position to its output position, consistently with [F5]. For $m=1$ this is the identity. If it holds for $m-1$, write the product as $([x_1]\ast \cdots \ast [x_{m-1}])\ast [x_m]$; inserting $[x_m]$ into the $k$-th position of a word $[x_{{\sigma }^{-1}(1)}\mathrm{\mid }\cdots \mathrm{\mid }{x}_{{\sigma }^{-1}(m-1)}]$ multiplies its coefficient by $q^{-{\sum }_{a<m,\sigma (a)\ge k}w(a,m)}$, exactly the sum of the weights of the pairs $(a,m)$ inverted by the resulting permutation, so the block formula reproduces the displayed expansion; the map $(\sigma ,k)\mapsto$ the resulting permutation is a bijection onto $S_m$.

In the coefficient of $w_p$ in ${\sum }_{s=0}^N(-1{)}^s{C}_{N,s}{x}^{\ast (N-s)}\ast y\ast x^{\ast s}$, set $Q=N-p$, write $u+v=s$, and use $A_{N-s}{A}_s{C}_{N,s}=q_i^{-((N-s)(N-s-1)+s(s-1))\mathrm{/}2}[N{]}_i!$ and $-a_{ij}=N-1$. The coefficient simplifies to $[N{]}_i!q_i^{-N(N-1)\mathrm{/}2+(N-1)Q}({\sum }_{u=0}^p(-1{)}^u{B}_{p,u}(t)q_i^{u(2p-1)-u^2})({\sum }_{v=0}^Q(-1{)}^v{B}_{Q,v}(t)q_i^{v-v^2})$. The quotient definition gives $B_{m,r}(t)=B_{m-1,r}(t)+t^{m-r}{B}_{m-1,r-1}(t)$; induction then yields ${\sum }_{r=0}^m(-1{)}^r{t}^{r(r-1)\mathrm{/}2}{B}_{m,r}(t)z^r={\prod }_{j=0}^{m-1}(1-z{t}^j)$. Thus the first factor in parentheses is ${\prod }_{j=0}^{p-1}(1-q_i^{2p-2-2j})$ and vanishes when $p>0$; the second is ${\prod }_{j=0}^{Q-1}(1-q_i^{-2j})$ and vanishes when $Q>0$. Since $p+Q=N\ge 1$, at least one factor vanishes for every $p=0,\dots ,N$. These words exhaust the color degree $N{ϵ}_i+{ϵ}_j$, proving part (i).

Fix $i\mathrm{\ne }j$ and $N=1-a_{ij}$, and put $w(i,i)=2d_i$, $w(i,j)=w(j,i)=d_i{a}_{ij}$; these are the values $⟨\mathrm{deg}{v}_a,\mathrm{deg}{v}_b⟩$, and $w$ is symmetric by [F1]. Let $z:={\sum }_{p=0}^N(-1{)}^p{C}_{N,p}{\xi }_i^p{\xi }_j{\xi }_i^{N-p}$; since $C_{N,N-p}=C_{N,p}$, the element $z$ differs from the printed Serre generator ${\sum }_s(-1{)}^s{C}_{N,s}{\xi }_i^{N-s}{\xi }_j{\xi }_i^s$ by the global sign $(-1{)}^N$, so the two generate the same two-sided ideal. For $t\in \{0,\dots ,N\}$ let ${\zeta }_t:=[v_i{]}^{\ast t}\ast [v_j]\ast [v_i{]}^{\ast (N-t)}$, and for $p\in \{0,\dots ,N\}$ let $w_p$ be the tensor word with $p$ letters $v_i$, then $v_j$, then $N-p$ letters $v_i$. By [F6] and steps 1.2 and 1.3, $⟨{\zeta }_t,z⟩={\mathrm{\hslash }}^{-(N+1)}{d}_i^{-N}{d}_j^{-1}{\sum }_{p=0}^N(-1{)}^p{C}_{N,p}M(t,p)$, where $M(t,p):={\sum }_{\sigma \in S_{N+1},\sigma (t+1)=p+1}{q}^{-{\sum }_{a<b,\sigma (a)>\sigma (b)}W(a,b)}$ and $W(a,b)$ is the weight $w$ of the letters at positions $a,b$ of the sequence $(i^t,j,i^{N-t})$. Then $M(t,p)=M(p,t)$: the map $\sigma \mapsto {\sigma }^{-1}$ is a bijection from the permutations with $\sigma (t+1)=p+1$ onto those with $\tau (p+1)=t+1$, and it preserves the exponent, because the inversion pairs of $\sigma$ correspond to those of $\tau ={\sigma }^{-1}$ with the same pair of letters: if $c=\sigma (b)<d=\sigma (a)$ for an inverted pair $a<b$, then $\tau (c)=b>a=\tau (d)$, and the weight of the pair $\{a,b\}$ equals the weight of the letter pair at the positions $d,c$ of the transposed sequence $(i^p,j,i^{N-p})$, since $\sigma$ carries the position $t+1$ of $j$ onto the position $p+1$, so the letter at position $\sigma (x)$ of the transposed sequence is the letter at position $x$ of the original sequence for every $x$.

The cut coproduct is an algebra map from $\mathrm{S}\mathrm{h}(V)$ to its tensor square with product $(u\otimes v)(u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }})=q^{-⟨\mathrm{deg}{u}^{\mathrm{\prime }},\mathrm{deg}v⟩}(u\ast u^{\mathrm{\prime }})\otimes (v\ast v^{\mathrm{\prime }})$. Indeed a shuffle followed by a cut consists uniquely of shuffles of the two prefixes and two suffixes; its cross-cut inversions contribute $q^{-⟨\mathrm{deg}{u}_{-},\mathrm{deg}{v}_{+}⟩}$, exactly the scalar in this tensor product. On a letter the coproduct is primitive, so it sends every product of letters into $⟨V⟩{\otimes }_R⟨V⟩$. Write this finite sum as $\mathrm{\Delta }(x)={\sum }_a{x}_a^{\mathrm{\prime }}\otimes x_a^{\mathrm{\prime }\mathrm{\prime }}$ with $x_a^{\mathrm{\prime }},x_a^{\mathrm{\prime }\mathrm{\prime }}\in ⟨V⟩$; individual prefixes of ambient tensor words need not themselves belong to the generated half. If $y$ annihilates $⟨V⟩$, step 1.2 gives $⟨x,yu⟩={\sum }_a⟨x_a^{\mathrm{\prime }},y⟩⟨x_a^{\mathrm{\prime }\mathrm{\prime }},u⟩=0$ and likewise $⟨x,uy⟩=0$. Thus $⟨V{⟩}^{\perp }$ is a two-sided ideal.

By part (i) the element $\zeta :={\sum }_{s=0}^N(-1{)}^s{C}_{N,s}{\zeta }_{N-s}$ is zero in $⟨V⟩\subseteq \mathrm{S}\mathrm{h}(V)$, so all of its tensor-word coefficients vanish. By step 1.3 the coefficient of $w_p$ in ${\zeta }_{N-s}$ is $M(N-s,p)$, hence ${\sum }_s(-1{)}^s{C}_{N,s}M(N-s,p)=0$ for every $p$. Substituting $t=N-s$ and using $C_{N,N-t}=C_{N,t}$ (so that $(-1{)}^{N-t}$ is the global factor $(-1{)}^N$) gives ${\sum }_t(-1{)}^t{C}_{N,t}M(t,p)=0$ for every $p$; by the symmetry of step 2.2 this is ${\sum }_p(-1{)}^p{C}_{N,p}M(t,p)=0$ for every $t$. Hence step 2.2 gives $⟨{\zeta }_t,z⟩=0$ for every $t$.

A shuffle product $[x_1]\ast \cdots \ast [x_m]$ of one-letter words is homogeneous of color degree $\mathrm{deg}{x}_1+\cdots +\mathrm{deg}{x}_m$, and $⟨V⟩$ is generated by the one-letter words, so every $x\in ⟨V⟩$ is a finite sum of such products. Unless the letters are exactly $N$ copies of $v_i$ and one copy of $v_j$, their color degree differs from that of $z$, so the diagonal pairing vanishes. In the remaining case the letter weights agree with the two-letter computation of step 2.2, and step 3.1 gives $⟨[x_1]\ast \cdots \ast [x_m],z⟩=0$. By linearity $⟨x,z⟩=0$ for every $x\in ⟨V⟩$, that is, $z\in ⟨V{⟩}^{\perp }$. Consequently every Serre generator lies in the two-sided ideal $⟨V{⟩}^{\perp }$ of step 2.3, and hence so does the ideal $J_{-}$ they generate: $⟨⟨V⟩,J_{-}⟩=0$. This is the stated annihilation and the asserted descent of the wordwise pairing.