Distinct cyclotomic shears give integrable based quantum sl-two chains

Cyclotomic denominators. For $m=2\mathrm{\ell }\ge 6$, primitive $m$th roots occur in inverse pairs and none equals $1$ or $-1$. Thus ${\mathrm{\Phi }}_m$ has even degree $2d$, integer coefficients, constant term $1$, and is reciprocal. Consequently $p=q^{-d}{\mathrm{\Phi }}_m(q)$ belongs to $A$, is bar fixed, and is irreducible up to a unit in the unique-factorization ring $A$. Moreover $1\mathrm{/}p=q^d\mathrm{/}{\mathrm{\Phi }}_m(q)$ is regular at zero with strictly positive valuation; bar invariance gives strictly positive valuation at infinity in the parameter $q^{-1}$. Different values of $\mathrm{\ell }$ give nonassociate irreducibles. Finally $p$ divides $[\mathrm{\ell }{]}_q$, since a primitive $2\mathrm{\ell }$th root is a simple zero of $q^{\mathrm{\ell }}-q^{-\mathrm{\ell }}$ and not a zero of $q-q^{-1}$.

The needed Gaussian congruences. Put $z=q^2$ and reduce modulo ${\mathrm{\Phi }}_{2\mathrm{\ell }}(q)$, so $z$ is a primitive $\mathrm{\ell }$th root and $q^{\mathrm{\ell }}=-1$. The finite identity ${\prod }_{h=0}^{N-1}(1+z^{h}X)={\sum }_{k=0}^N{z}^{k(k-1)\mathrm{/}2}{\left(\genfrac{}{}{0px}{}{N}{k}\right)}_z{X}^k$, proved by multiplying one factor at a time, and the block identity ${\prod }_{h=0}^{\mathrm{\ell }-1}(1+z^{h}X)=1-(-X{)}^{\mathrm{\ell }}$ give the Gaussian digit rule: a coefficient with lower remainder exceeding upper remainder modulo $\mathrm{\ell }$ vanishes; when the remainders agree, the remaining coefficient comes from the ordinary binomial expansion of the full blocks. Since ${\left(\genfrac{}{}{0px}{}{N}{k}\right)}_q=q^{-k(N-k)}{\left(\genfrac{}{}{0px}{}{N}{k}\right)}_{q^2}$, for $r=\mathrm{\ell }-1$ this gives ${\left(\genfrac{}{}{0px}{}{r+k}{k}\right)}_q=0<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p$ unless $k=t\mathrm{\ell }$, and gives ${\left(\genfrac{}{}{0px}{}{r+t\mathrm{\ell }}{t\mathrm{\ell }}\right)}_q=(-1{)}^{t(\mathrm{\ell }-1)}<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p$. Indeed the lower remainder is nonzero exactly in the vanishing cases, while in the other cases the unsymmetrized coefficient is $1$ and the symmetrizing factor is $q^{-t\mathrm{\ell }(\mathrm{\ell }-1)}=(-1{)}^{t(\mathrm{\ell }-1)}$. The argument takes place in the integral domain $A\mathrm{/}(p)$ and proves divisibility in $A$, rather than merely pointwise numerical vanishing.

Modular divided-power homomorphisms. Fix $j$, write $\mathrm{\ell }={\mathrm{\ell }}_j$, $n=j\mathrm{\ell }$, $r=\mathrm{\ell }-1$ and $m=n+2r=n_{j+1}$. In $V_A(m)\mathrm{/}(p_j)$, the vector $v=u_{j+1,r}$ has weight $n$. For $1\le a\le r$, the coefficient of $E^{(a)}v$ is ${\left(\genfrac{}{}{0px}{}{m-r+a}{a}\right)}_q={\left(\genfrac{}{}{0px}{}{(j+1)\mathrm{\ell }-1+a}{a}\right)}_q$, whose upper remainder $a-1$ is smaller than the lower remainder $a$; hence it is zero by step 2.1. For $a>r$ the action is already zero. Also $F^{(k)}v={\left(\genfrac{}{}{0px}{}{r+k}{k}\right)}_q{u}_{j+1,r+k}$ vanishes modulo $p_j$ whenever $k>n$: until $k=(j+1)\mathrm{\ell }-1$ its remainder is nonzero, and for $k\ge (j+1)\mathrm{\ell }$ the target position exceeds $m$. Therefore $u_{j,k}\mapsto F^{(k)}v$ defines a homomorphism of the complete divided strings modulo $p_j$. To check all generators, divided $F$ compatibility is the Gaussian product identity for $F^{(a)}{F}^{(k)}$; divided $E$ compatibility follows by reordering $E^{(a)}{F}^{(k)}$ and using the just-proved annihilation of all positive divided $E$ powers on the weight-$n$ vector $v$, leaving exactly ${\left(\genfrac{}{}{0px}{}{n-k+a}{a}\right)}_q{F}^{(k-a)}v$; the vanished positions $k>n$ respect the source endpoints. Every toral generator and toral quantum binomial has the same weight scalar on matched positions. Step 2.1 identifies this homomorphism with $T_j$ modulo $p_j$, so $(a{T}_j-T_{j}a)V_A(n_j)\subseteq p_j{V}_A(n_{j+1})$ for every $a\in U_A$.

Exact vanishing of consecutive shears. The nonzero target positions of $T_j$ are $k=r_j+t{\mathrm{\ell }}_j=(t+1){\mathrm{\ell }}_j-1$, $0\le t\le j$. Reduced modulo ${\mathrm{\ell }}_{j+1}={\mathrm{\ell }}_j+1$, their residues are ${\mathrm{\ell }}_j-t-1$, ranging from $2$ to ${\mathrm{\ell }}_j-1$. None is zero. Since $T_{j+1}$ is supported precisely on positions divisible by ${\mathrm{\ell }}_{j+1}$, one has $T_{j+1}{T}_j=0$ exactly. Other products $T_i{T}_j$ vanish because their source and target summands do not match. On every vector only finitely many $T_j$ act, so $S$ is algebraic, $S^2=0$, and $C^{-1}=1-S$. The equality $n_{j+1}=n_j+2r_j$ also shows each $T_j$ preserves weights.

Integral action, including the cross terms. Put $L_A={⨁}_j{V}_A(n_j)$. For $a\in U_A$, the standard action preserves each summand of $L_A$, and $C^{-1}aC=a+[a,S]-SaS$ by step 4.1. Step 3.1 makes every column of $[a,T_j]\mathrm{/}{p}_j$ integral. The only potentially nonzero summands of $SaS$ are $T_{j+1}a{T}_j\mathrm{/}(p_{j+1}{p}_j)$. The numerator is divisible by $p_j$, because it equals $T_{j+1}(a{T}_j-T_{j}a)$ using $T_{j+1}{T}_j=0$; it is divisible by $p_{j+1}$, because it also equals $(T_{j+1}a-a{T}_{j+1})T_j$. These are entrywise divisibilities of matrices in standard divided bases. The distinct irreducibles $p_j,p_{j+1}$ are coprime in the unique-factorization ring $A$, so their product divides every entry. Thus all cross terms are integral. For every input column there are at most two shear positions, so all sums remain finite. It follows that $C^{-1}aC$ preserves $L_A$, equivalently $a{M}_A\subseteq M_A$. This checks the full divided-power integral form, not just $E$ and $F$.

The basis and its compatible bar. The inverse from step 4.1 makes $B=C\{u_{j,k}\}$ a field basis and makes its $A$-span free with that basis. Homogeneity was checked there. Each coefficient $(-1{)}^{t({\mathrm{\ell }}_j-1)}\mathrm{/}{p}_j$ is bar fixed by step 1.1, so the standard module bar fixes each $b_{j,k}$; consequently this is exactly the basis-fixing bar of $B$ and retains its compatibility with the $U$ action. Every vector has finitely many summand components, each in a finite divided string, so $E,F$ are locally nilpotent, the chosen weight grading is a direct sum, and $M$ is integrable. No arbitrary selection of vectors or bases is used.

Crystal charts and balancedness. Let $L_0$ and $L_{\mathrm{\infty }}$ be the standard direct-sum regular charts. Step 1.1 shows that both $C$ and $C^{-1}$ preserve these charts and are the identity on their zero and infinity residues: their nonidentity entries are signed $1\mathrm{/}{p}_j$, in $q{A}_0$ and $q^{-1}{A}_{\mathrm{\infty }}$, respectively. Therefore the $B$-generated regular charts equal the standard ones, and the residues of $B$ equal the standard divided-string crystal vertices. Kashiwara stability and the crystal inverse relations follow directly from [F1], so the fourth based axiom holds. In the common basis $B$, the integral lattice and charts have coefficient rings $A,A_0,A_{\mathrm{\infty }}$. Their simultaneous intersection has integer coefficients because a Laurent polynomial regular at both zero and infinity is an integer constant. Also $A\cap A_0=\mathbb{Z}[q]$. Hence the integral balanced map is the coefficientwise isomorphism ${⨁}_B\mathbb{Z}⟶{⨁}_B\mathbb{Z}[q]\mathrm{/}q\mathbb{Z}[q]$. All coefficient families have finite support.

The explicit top action and conclusion. Since $u_{j,0}$ is highest, $E^{(r_j)}{u}_{j,0}=0$. Also $b_{j,0}=u_{j,0}+u_{j+1,r_j}\mathrm{/}{p}_j$, so [F1] gives $E^{(r_j)}{b}_{j,0}=(G_j\mathrm{/}{p}_j)u_{j+1,0}$. Replacing $u_{j+1,0}$ by $b_{j+1,0}-u_{j+2,r_{j+1}}\mathrm{/}{p}_{j+1}$ gives the stated formula, because $r_{j+1}=j+3$ is not divisible by ${\mathrm{\ell }}_{j+2}=j+5$ and thus $u_{j+2,r_{j+1}}=b_{j+2,r_{j+1}}$. Gaussian polynomials are nonzero in $K$, so $G_j\mathrm{\ne }0$; step 5.1 proves that both coefficients are in $A$. Steps 4.1–7.1 establish every claimed basis, action, bar, integrability, crystal and balanced property for this actual algebraic module.