Rank-one skew kernels in a numerical quantum Serre algebra

First define $D,D^{\mathrm{\prime }}$ on the free algebra by the displayed word rules. Write $e=e_i$. Both operators lower the number of letters $e$ in a word by one, so are locally nilpotent there. For each Serre polynomial on letters $e_j,e_k$, put $r=1-a_{jk}$. If $i$ is distinct from $j,k$, its image under $D$ is zero. If $i=j$, set $a=a_{jk}=1-r$. The two terms in $D(e^{r-s}{e}_k{e}^s)$ have respective coefficients $q^{r-s-1}[r-s{]}_q$ and $q^{2(r-s)+a+s-1}[s{]}_q$ on the words with the appropriate one fewer $e$. On a fixed resulting word $e^{r-s-1}{e}_k{e}^s$, the contributions from indices $s$ and $s+1$ cancel: their common $q$-power is $q^{r-s-1}$ and their remaining coefficients cancel by [F1]. If $i=k$, removing its single letter gives $e_j^r$ times ${\sum }_s(-1{)}^s{\left[\genfrac{}{}{0px}{}{r}{s}\right]}_q{q}^{a_{ij}(r-s)}$. Symmetry gives $a_{ij}=a_{kj}=a_{jk}=1-r$, so this sum is zero by [F1]. Thus $D$ kills every generating Serre polynomial. Its skew product rule then preserves their two-sided ideal, so it descends to $S$. Reversing word order interchanges $D,D^{\mathrm{\prime }}$ and preserves that ideal, since reversal multiplies each Serre polynomial by $(-1{)}^r$. Hence $D^{\mathrm{\prime }}$ also descends. The degree-lowering argument proves local nilpotence on $S$.

The operators commute. To check this without any presentation comparison, induct on word length in the free algebra. For homogeneous words $x,y$ of $i$-weights $m_x,m_y$, both $D{D}^{\mathrm{\prime }}(xy)$ and $D^{\mathrm{\prime }}D(xy)$ expand to $q^{m_y}D{D}^{\mathrm{\prime }}(x)y+D(x)D^{\mathrm{\prime }}(y)+q^{m_x+m_y-2}{D}^{\mathrm{\prime }}(x)D(y)+q^{m_x}xD{D}^{\mathrm{\prime }}(y)$, using the inductive commutation on $x,y$. They agree on generators, so agree on all words and on $S$. In particular $D^{\mathrm{\prime }}$ preserves $B_i$. The skew product rule shows that $B_i$ is a graded subalgebra.

For homogeneous $b\in B_i$ of $i$-weight $m$, $D(eb)=b$ and $D(be)=q^{m}b$, so $R(b)=eb-q^{-m}be$ belongs to $B_i$. For homogeneous $b,c\in B_i$ one has $R(bc)=R(b)c+q^{-m_b}bR(c)$. Moreover direct application of the second skew rule gives $D^{\mathrm{\prime }}(R(b))=(q^m-q^{-m})b+R(D^{\mathrm{\prime }}(b))$, since $D^{\mathrm{\prime }}(b)$ has weight $m-2$. With $F=-D^{\mathrm{\prime }}\mathrm{/}(q-q^{-1})$, this is $RF(b)-FR(b)=[m{]}_{q}b$. The grading gives $KR{K}^{-1}=q^{2}R$ and $KF{K}^{-1}=q^{-2}F$, and $K$ is invertible. Thus these operators satisfy every defining relation of $U_q({\mathfrak{s}\mathfrak{l}}_2)$ on $B_i$. Their product rules are $R(bc)=R(b)c+K^{-1}(b)R(c)$, $F(bc)=F(b)K(c)+bF(c)$, and $K(bc)=K(b)K(c)$.

For $j\mathrm{\ne }i$ put $N=-a_{ij}$. Iterating the definition of $R$ gives $R^r(e_j)={\sum }_{s=0}^r(-1{)}^s{q}^{-s(a_{ij}+r-1)}{\left[\genfrac{}{}{0px}{}{r}{s}\right]}_q{e}^{r-s}{e}_j{e}^s\mathrm{.}$ Indeed the case $r=0$ is immediate, and passing from $r$ to $r+1$ uses $R(R^r{e}_j)=e{R}^r{e}_j-q^{-a_{ij}-2r}(R^r{e}_j)e$ and the quantum-binomial recurrence. At $r=N+1$ the extra $q$-factor is $1$, so the numerical Serre relation is exactly $R^{N+1}(e_j)=0$. Since $F(e_j)=0$ and its weight is $-N$, the commutator in step 3.1 gives by induction $F(R^r{e}_j)=[r{]}_q[N-r+1{]}_q{R}^{r-1}{e}_j$ for $r\ge 1$. The needed scalar identity is ${\sum }_{t=0}^{r-1}[m+2t{]}_q=[r{]}_q[m+r-1{]}_q$, obtained by summing the defining geometric series. Therefore $V_j=\mathrm{span}\{R^r(e_j):0\le r\le N\}$ is a finite-dimensional invariant subspace with weights $-N,-N+2,\dots ,N$.

Let $B_0$ be the subalgebra generated by all these $V_j$, $j\mathrm{\ne }i$. It is contained in $B_i$ and stable under $R$, by its product rule and the terminating strings of step 4.1. The normal-order identity $be=q^{m}eb-q^{m}R(b)$ for homogeneous $b\in B_0$ shows that ${\sum }_{n\ge 0}{e}^n{B}_0$ is stable under right multiplication by $e$. It is also stable under right multiplication by every $e_j$ with $j\mathrm{\ne }i$, since these lie in $B_0$. It contains $1$, hence every word in the generators and all of $S$. Thus $S={\sum }_{n\ge 0}{e}^n{B}_0$.

This sum is direct. For $b\in B_i$, iterating $D$ gives $D(e^{n}b)=q^{n-1}[n{]}_q{e}^{n-1}b$ and $D^n(e^{n}b)=q^{n(n-1)\mathrm{/}2}[n{]}_q!b$. If ${\sum }_{n=0}^d{e}^n{b}_n=0$ with $b_n\in B_i$ and $b_d\mathrm{\ne }0$, application of $D^d$ gives $q^{d(d-1)\mathrm{/}2}[d{]}_q!b_d=0$, impossible by [F1]. Consequently the expression in step 5.1 is unique. Applying $D$ to that expression and using the directness just proved shows that an element in $\ker D$ has $b_n=0$ for every $n>0$. Hence $B_i=B_0$, and rescaling $e^n$ by the nonzero $[n{]}_q!$ gives the asserted divided-power direct sum.

The action on $B_i$ is integrable. Each $V_j$ in step 4.1 is finite-dimensional with a finite set of integer weights, so its raising and lowering operators are nilpotent. The product formulas of step 3.1 make multiplication $V_{j_1}\otimes \cdots \otimes V_{j_t}\to B_i$ equivariant for the coproducts $\mathrm{\Delta }(R)=R\otimes 1+K^{-1}\otimes R$, $\mathrm{\Delta }(F)=F\otimes K+1\otimes F$, $\mathrm{\Delta }(K)=K\otimes K$. These coproduct formulas satisfy the same rank-one relations, as is verified directly using $KR=q^{2}RK$ and $KF=q^{-2}FK$. Each finite tensor product has only finitely many weights; $R$ increases weight by two and $F$ decreases it by two, so both are nilpotent on it. Every element of $B_i=B_0$ is a finite sum of products of root-string generators, hence lies in a finite sum of such invariant images. Thus $R,F$ are locally nilpotent and $K$ acts with eigenvalues $q^m$, $m\in \mathbb{Z}$. No complete reducibility theorem for rank-one representations is used.

Finally let $0\mathrm{\ne }z\in S_{\beta }$ satisfy $D(z)=D^{\mathrm{\prime }}(z)=0$, and put $m=m(\beta )$. Then $z\in B_i$ and $Fz=0$. By step 7.1 there is an integer $n\ge 0$ with $R^{n}z\mathrm{\ne }0$ and $R^{n+1}z=0$. The same commutator induction as in step 4.1 yields $F{R}^{r}z=-[r{]}_q[m+r-1{]}_q{R}^{r-1}z$. At $r=n+1$ it follows that $[n+1{]}_q[m+n{]}_q{R}^{n}z=0$. Both $[n+1{]}_q$ and $R^{n}z$ are nonzero, so $[m+n{]}_q=0$. Since $q>0$ and $q\mathrm{\ne }1$, a symmetric quantum integer $[r{]}_q$ vanishes only when $r=0$. Thus $m=-n\le 0$, as required.