Periodic cyclotomic shear tensor corrections escape every finite cut

Finite normalized correction. In raw coordinates the bar is $(C^{-1}\otimes C^{-1})\mathrm{\Theta }(C\otimes C)$ followed by scalar bar. Its coefficients are Laurent integral: each term $a_r{E}^{(r)}\otimes F^{(r)}$ acts integrally by [F1], and $a_r\in \mathbb{Z}[q,q^{-1}]$. Its diagonal is $1$ and every other term strictly raises the first weight, since $C$ preserves weights and every $r>0$ term does so. In a finite cut there are finitely many raw weight-zero vectors. Successively remove the lowest remaining first-weight error: involutivity makes its scalar coefficient $x$ satisfy $x+\bar{x}=0$, so its positive Laurent part is the unique coefficient in $q\mathbb{Z}[q]$ whose difference from its bar is $x$. This finite induction gives existence and uniqueness of $G^{(J)}$. All its nonseed weights have first weight at least $2$. Every map in the raw bar raises constituent indices; projecting away indices exceeding $J$ commutes with the bar computed in a larger cut. Uniqueness therefore proves agreement of the projected normalized lifts and of $f_j^{(J)}$.

A uniformly bounded set of positions. Write an ordinary weight-zero tensor vector as $u_{j,a}\otimes u_{l,n_l-d}$; then $n_j-n_l=2(a-d)$. The seed has $a=d=0$. Both $C$ and $C^{-1}$ preserve $a\le 3$ and $d\le 3$: on an even link the only eligible positions or endpoint distances in this range are $0,2$, which become $1,3$; on an odd link only $0$ is eligible and it becomes $3$. In endpoint distances the same assertion follows from $n_j\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{\mathrm{\ell }}_j)$ and $n_{j+1}=n_j+2({\mathrm{\ell }}_j-1)$. Every term of $\mathrm{\Theta }$ lowers both $a$ and $d$. Thus the raw bar and the correction induction preserve these bounds. Nonadjacent string lengths differ by at least $8$, whereas $\mathrm{\mid }2(a-d)\mathrm{\mid }\le 6$, so only equal or adjacent constituent indices occur.

Ordinary normalized invariant coordinates. In a constituent pair of lengths $m,n$, put $D_{a,d}=\mathrm{\Delta }(F^{(a)})(u_0\otimes u_{n-d})$ if $a\ge d$, and $D_{a,d}=\mathrm{\Delta }(E^{(d)})(u_a\otimes u_n)$ if $a<d$. Their leading ordinary vector is $u_a\otimes u_{n-d}$. Their term with first position $a-t$ has respectively coefficient $q^{t(m-a+t)}{\left(\genfrac{}{}{0px}{}{n-d+t}{t}\right)}_q$ or $q^{t(n-d+t)}{\left(\genfrac{}{}{0px}{}{m-a+t}{t}\right)}_q$, and is omitted if an endpoint is exceeded. These vectors are bar fixed by [F2], because their starting tensors have highest first factor or lowest second factor. The lowest exponent of either nonleading coefficient is respectively $t(a-d+t)$ or $t(d-a+t)$, strictly positive. Hence their transition and inverse are unitriangular matrices over $\mathbb{Z}[q]$. Expand the ordinary-coordinate vector $(C\otimes C)G^{(J)}$ in these $D$ vectors, with coefficients $z_{j,l,a}$. They are bar fixed and regular at $q=0$; except for the original seed coefficient their residues are zero. In pair $(j,l)$ their denominator divides $p_{j-1}{p}_{l-1}$, with missing index $-1$ interpreted as denominator $1$, because these are the only incoming shear denominators and the inverse $D$ matrix is Laurent integral.

Parity and the bounded rational states. Put ${\delta }_j={\mathrm{deg}}_q{p}_j(\mathrm{m}\mathrm{o}\mathrm{d}2)$ and define ${\tau }_0=0$, ${\tau }_{j+1}={\tau }_j+j+{\delta }_j(\mathrm{m}\mathrm{o}\mathrm{d}2)$. Assign tuple $(j,a,l,n_l-d)$ the parity $ϵ=a(1+j+l)+{\tau }_j+{\tau }_l+jl(\mathrm{m}\mathrm{o}\mathrm{d}2)$. Its coefficient in every vector under consideration contains only powers of $q$ of that parity. Indeed $n_j\mathrm{/}2\equiv j+1(\mathrm{m}\mathrm{o}\mathrm{d}2)$; a first shear has even source position and a second shear has even source position $n_l-d$, hence $a\equiv j+l(\mathrm{m}\mathrm{o}\mathrm{d}2)$ in the latter case. In both cases the change of $ϵ$ is ${\delta }_j$ or ${\delta }_l$, exactly the parity of the shear coefficient. A term of $\mathrm{\Theta }$ with index $r$ has coefficient parity $r(1+j+l)$, exactly the change from $a$ to $a-r$. The seed has parity zero, positive Laurent extraction preserves parity, and the $D$ coefficients have parity $t(1+j+l)$. In particular diagonal $z_{j,j,a}$ has parity $j+a$. A bar-fixed Laurent polynomial is a polynomial in $p=q+q^{-1}$, by pairing its opposite exponents and the recurrence $q^s+q^{-s}=p(q^{s-1}+q^{1-s})-(q^{s-2}+q^{2-s})$. Multiplying $z$ by its bar-fixed denominator therefore gives a polynomial in $p$; zero residue at both ends bounds its degree strictly below that of the denominator. Thus on an odd diagonal $z_3=c_3\mathrm{/}{p}^2,z_2=A_2\mathrm{/}p,z_1=c_1\mathrm{/}{p}^2,z_0=A_0\mathrm{/}p$; on an even diagonal $z_3=up\mathrm{/}{h}^2+vp\mathrm{/}h$, and initially $z_2,z_0$ have the form $A\mathrm{/}h+B\mathrm{/}{h}^2$, while $z_1$ has the form $Ap\mathrm{/}h+Bp\mathrm{/}{h}^2$. The constants may be treated over $\mathbb{Q}$.

The scalar calculations used below. On a diagonal of length $m$ write $H_{a,t}=q^{t(m-a+t)}{\left(\genfrac{}{}{0px}{}{m-a+t}{t}\right)}_q$; this is the coefficient from $D_a$ to first position $a-t$. All following evaluations follow by substituting in $[s]=(q^s-q^{-s})\mathrm{/}(q-q^{-1})$ and ${\left(\genfrac{}{}{0px}{}{N}{t}\right)}_q={\prod }_{v=0}^{t-1}[N-v]\mathrm{/}[t-v]$. Modulo $h$, one has $p^2=2$, $q^4=-1$, $[8s]\mathrm{/}h=-2sp$. For $m=8k+10$ these give $H_{3,1}\mathrm{/}h=-2(k+1)p$, $H_{2,2}=-1$, $H_{1,1}=q^{2}p$, $H_{3,3}\mathrm{/}h=2(k+1)q^{2}p$, and $H_{3,2}$ is divisible by $h$. For $m=8k+4$, all endpoint coefficients $H_{a,a}$ with $1\le a\le 3$ vanish modulo $h$. Modulo $p$, one has $q^2=-1$; for $m=8k+4$ the evaluations are $H_{3,1}\mathrm{/}p=-(4k+1)$, $H_{3,2}=4k+1$, $H_{1,1}\mathrm{/}p=-(4k+2)$, $H_{2,2}=4k+2$, and $H_{3,3}\mathrm{/}p=-(4k+2)(4k+1)$. For $m=8k+2$ one has $H_{2,2}=4k+1$. Quotients here are taken only when the numerator is divisible by the indicated denominator, and evaluations are in $\mathbb{Q}[q,q^{-1}]\mathrm{/}(q^4+1)$ or $\mathbb{Q}[q,q^{-1}]\mathrm{/}(q^2+1)$ understood as the corresponding residue fields.

Base odd diagonal. For $j=1$, $m=4$, the first-weight bound excludes positions $2,3$, so only $z_1=c\mathrm{/}{p}^2,z_0=A\mathrm{/}p$ occur. The double pole in the ordinary component at position $1$ comes only from shearing both seed factors; its sign is negative, hence $c=-1$. The ordinary endpoint has no incoming first shear and no admissible backward mixed tensor with first position $0$, so it equals its Laurent integral raw coefficient. Since $H_{1,1}\mathrm{/}p=-2$ modulo $p$, cancellation of its simple pole forces $A=2c=-2$. Thus the odd scalar starts at $c_0=-1$, with endpoint residue $-2\mathrm{/}p=-p=q^3-q$ modulo $h$.

Odd to even: the double pole. Suppose the odd diagonal $j=2k+1$, $m=8k+4$, has $z_3=z_1=c\mathrm{/}{p}^2$, $z_2=(4k+1)c\mathrm{/}p$ and $z_0=(4k+2)c\mathrm{/}p$; for $k=0$ use the base state with positions $2,3$ absent. Its endpoint residue modulo $h$ is $(2k+1)cp$, because all the other endpoint coefficients vanish there. On the next even diagonal, of length $8k+10$, the only double $h$ pole at first position $3$ is the product of the two incoming shears of the previous endpoint. The first sign is positive and the second is $(-1{)}^{(8k+4)\mathrm{/}4}=-1$. Mixed incoming terms have only a single pole. Therefore the numerator $up$ of $z_3=up\mathrm{/}{h}^2+vp\mathrm{/}h$ is the negative of that endpoint residue, and $u=-(2k+1)c$.

Odd to even: the simple poles. On this even diagonal, the ordinary components at first positions $0,1,2$ equal their integral raw components: an incoming first shear lands only at position $3$, and a backward adjacent pair has first position at least $3$ by the weight equation. In the component at position $2$, $H_{3,1}$ is divisible by $h$, so there is no double pole; the rational-state bounds force $z_2=C\mathrm{/}h$. Cancellation of its simple pole gives $C=4(k+1)u$. Similarly $H_{3,2}$ and $H_{3,3}$ are divisible by $h$, forcing $z_1=Bp\mathrm{/}h$ and $z_0=A\mathrm{/}h$. The endpoint pole equation modulo $h$ is $A-C+(2B+C)q^2=0$, by step 5.1. The elements $1,q^2$ are rationally independent in that residue field, so $A=C$ and $B=-C\mathrm{/}2$. This proves $z_2=z_0=C\mathrm{/}h$, $z_1=-Cp\mathrm{/}(2h)$, with $C=-4(k+1)(2k+1)c$; no value of $v$ is needed. Its endpoint modulo $p$ is $-(2k+3)C$, since the odd-position states vanish there and $H_{2,2}=4k+5$ for length $8k+10$.

Even to odd: the two double poles. Let the preceding even diagonal have length $8k+2$, with the state just proved and scalar $C$, and consider the next odd diagonal of length $m=8k+4$, with $k\ge 1$. At first position $3$, the double $p$ pole comes only from both shears of the preceding raw vector at position $2$. Its two signs are negative and positive. That raw coefficient evaluated modulo $p$ is $C\mathrm{/}h=-C\mathrm{/}2$: its ordinary component is integral by step 8.1, and the $z_3$ contribution vanishes because it has a factor $p$. Thus $c_3=C\mathrm{/}2=:c^{\mathrm{\prime }}$. At first position $1$, the double pole comes from both shears of the preceding endpoint, again with negative total sign. That endpoint equals $-(2k+1)C$ modulo $p$. On the current diagonal its double-pole numerator is $c_1+(4k+1)c^{\mathrm{\prime }}$, using $H_{3,2}=4k+1$. Equating it to $(2k+1)C=(4k+2)c^{\mathrm{\prime }}$ forces $c_1=c^{\mathrm{\prime }}$. Single-pole mixed terms cannot alter either double-pole equation.

Even to odd: the simple poles. The ordinary component at first position $2$ is integral. An incoming first shear lands at positions $1,3$; an incoming second shear would have to start in the backward mixed pair at second position $m-3$, which is odd and is therefore killed by the even link. Hence $z_2+c^{\mathrm{\prime }}{H}_{3,1}\mathrm{/}{p}^2$ has no pole, giving $A_2=(4k+1)c^{\mathrm{\prime }}$. The ordinary endpoint is also integral: no first shear lands at position $0$, and the backward mixed pair requires first position at least $1$. Its pole equation is $A_0-(4k+2)c^{\mathrm{\prime }}+(4k+2)A_2-(4k+2)(4k+1)c^{\mathrm{\prime }}=0$, so $A_0=(4k+2)c^{\mathrm{\prime }}$. Consequently the odd state has exactly the form used in step 7.1. Combining steps 8.1 and 9.1 gives the recurrence $c_{k+1}=-(2k+2)(2k+1)c_k$, and therefore $c_k=(-1{)}^{k+1}(2k)!$.

Nonzero endpoint residues in every cut. The odd endpoint residue is $(2k+1)c_{k}p=(-1{)}^k(2k+1)!(q^3-q)$ modulo $h$. The following even scalar is $C=(-1{)}^{k}2(2k+2)!$, and its endpoint residue is $-(2k+3)C=(-1{)}^{k+1}2(2k+3)!$ modulo $p$. Multiplication by powers of $q$ identifies these ideals with $(q^4+1)$ and $(q^2+1)$ respectively. These residues are nonzero in their characteristic-zero residue fields, so every asserted endpoint coefficient is nonzero. The computation used only incoming shears and thus applies to every containing finite cut; step 1.1 also gives their coherence. Finite arithmetic, explicit projections and finite induction used no choice. This proves the stated escape of normalized corrected lifts.