Two cyclotomic shears give a periodic integrable based quantum sl-two chain

The two denominators and the weights. One has $p=q^{-1}{\mathrm{\Phi }}_4$ and $h=q^{-2}{\mathrm{\Phi }}_8$, so [F2] makes them nonassociate prime elements of $A$. Both are bar fixed, and $1\mathrm{/}p=q\mathrm{/}(1+q^2)$ and $1\mathrm{/}h=q^2\mathrm{/}(1+q^4)$ belong to $q{A}_0$ and, by bar invariance, to $q^{-1}{A}_{\mathrm{\infty }}$. The source highest weight $n_j$ is divisible by ${\mathrm{\ell }}_j$, and $n_{j+1}=n_j+2r_j$ for both parities. Hence every displayed target has the same weight as its source and lies in the target string.

All the modular Gaussian coefficients. For $\mathrm{\ell }\in \{2,4\}$ reduce modulo $q^{-\mathrm{\ell }\mathrm{/}2}{\mathrm{\Phi }}_{2\mathrm{\ell }}(q)$, which is the corresponding $p_j$. Then $z=q^2$ has order $\mathrm{\ell }$ and $q^{\mathrm{\ell }}=-1$. The generating identity ${\prod }_{a=0}^{N-1}(1+z^{a}X)={\sum }_t{z}^{t(t-1)\mathrm{/}2}{\left(\genfrac{}{}{0px}{}{N}{t}\right)}_z{X}^t$ follows by multiplication and the Gaussian recurrence; each full block of length $\mathrm{\ell }$ is $1-(-X{)}^{\mathrm{\ell }}$. It follows that a Gaussian coefficient with lower remainder larger than upper remainder modulo $\mathrm{\ell }$ vanishes. For $r=\mathrm{\ell }-1$ and $t\ge 0$, the coefficient ${\left(\genfrac{}{}{0px}{}{r+t}{t}\right)}_q$ vanishes unless $t=v\mathrm{\ell }$; in that case its unsymmetrized Gaussian value is $1$ and its symmetrizing factor is $q^{-v\mathrm{\ell }(\mathrm{\ell }-1)}=(-1{)}^v$, since $\mathrm{\ell }$ is even. This gives exactly the coefficients of $T_j$ modulo $p_j$. All reductions are in the integral domain $A\mathrm{/}(p_j)$, so they are divisibilities in $A$.

Full divided-power intertwining modulo each denominator. Fix $j$, put $n=n_j=a{\mathrm{\ell }}_j$, $r={\mathrm{\ell }}_j-1$ and $m=n+2r$, and take $v=u_{j+1,r}$ in $V_A(m)\mathrm{/}(p_j)$. Its weight is $n$. For $1\le t\le r$, the coefficient of $E^{(t)}v$ is ${\left(\genfrac{}{}{0px}{}{(a+1){\mathrm{\ell }}_j-1+t}{t}\right)}_q$, with upper remainder $t-1$ and lower remainder $t$, so it vanishes by step 2.1; larger $t$ give zero by the endpoint. Also $F^{(t)}v$ vanishes for every $t>n$: until $t=(a+1){\mathrm{\ell }}_j-1$ its nonzero remainder forces vanishing by step 2.1, and larger $t$ exceed the target endpoint. The assignment $u_{j,t}\mapsto F^{(t)}v$ therefore intertwines all divided $F$ actions by Gaussian multiplication. Divided $E$ reordering, using the annihilation of all positive divided $E$ powers on the weight-$n$ vector $v$, gives exactly ${\left(\genfrac{}{}{0px}{}{n-t+s}{s}\right)}_q{F}^{(t-s)}v$ for $E^{(s)}{F}^{(t)}v$, including zero beyond the source endpoints. Toral elements act by matched weight scalars. Step 2.1 identifies this assignment with $T_j$ modulo $p_j$, proving $[a,T_j]V_A(n_j)\subseteq p_j{V}_A(n_{j+1})$ for every $a\in U_A$.

Algebraic invertibility and integral cross terms. Every $T_j$ is supported on even positions and has odd target positions. Thus $T_{j+1}{T}_j=0$ exactly, and all other twofold products vanish by their summand indices. Every column is finite, so $S^2=0$ and $C^{-1}=1-S$ on the algebraic direct sum. For $a\in U_A$, expansion gives $C^{-1}aC=a+[a,S]-SaS$. The commutator summands are integral by step 3.1. The only cross terms are $T_{j+1}a{T}_j\mathrm{/}(p_{j+1}{p}_j)$. Their numerators are divisible by $p_j$, using $T_{j+1}(a{T}_j-T_{j}a)$, and by $p_{j+1}$, using $(T_{j+1}a-a{T}_{j+1})T_j$. Adjacent denominators are the two distinct primes of step 1.1, so their product divides every matrix entry by [F2]. The conjugated operator therefore preserves the standard free integral span, proving stability of the $B$-span. No nonadjacent repeated denominator creates a cross term.

The compatible basis and crystal charts. The inverse in step 4.1 proves that $B$ is a field basis with free integral span; its homogeneity is step 1.1. The coefficients $(-1{)}^t\mathrm{/}{p}_j$ are bar fixed, so the standard module bar fixes $B$ and remains compatible with the algebra action. The module is integrable because every vector occupies finitely many finite standard strings. By step 1.1, $C$ and its inverse preserve both regular charts and are the identity on both residues, so the charts and crystal vertices coincide with the standard direct-sum ones. Their Kashiwara operators are exactly the finite shifts in [F1]. Thus all four conditions of [F3] hold.

Balancedness, finite versions and exact column agreement. In the common basis $B$, the three coefficient rings are $A,A_0,A_{\mathrm{\infty }}$; their simultaneous intersection is $\mathbb{Z}$ and $A\cap A_0=\mathbb{Z}[q]$, since a Laurent polynomial regular at both ends is constant. Coefficientwise reduction therefore gives the integral balanced isomorphism, with finite support throughout. Omitting shears at indices at least $J$ preserves every preceding argument and gives the stated finite version. A standard action $a$ preserves each constituent summand; $C$ can raise the constituent index by at most one and $C^{-1}$ by at most one. Thus the raw column of $C^{-1}aC$ at index $j$ has indices at most $j+2$ and uses only shears with index at most $j+1$, all present when $J\ge j+2$; its finite and infinite computations agree. Finally the signs at positions $0,2$ in the first string give exactly the two stated seed formulas and factor weights. Every vector, map and index is specified explicitly, so no choice is used.