Periodic cyclotomic tensor corrections escape every finite cut

Finite cuts and their correction. In a finite cut every evaluated degree is bounded by the finite divided strings, and [F2] is a finite sum of integral divided actions. Thus ${\mathrm{\Psi }}_J$ is an integral semilinear involution. Degree zero is the identity, while degree $r>0$ raises first weight by $2r$. Finite first-weight induction constructs the corrected seed: at the next weight its residual is anti-bar by involutivity, so [F3] supplies its unique positive-polynomial correction. There are only finitely many raw vectors and weights in the cut, hence this process terminates. If two normalized fixed lifts differed, the coefficients at the least first weight of their difference would be in $q\mathbb{Z}[q]$ and equal their own bars, hence zero. This proves existence and uniqueness without an infinite completion.

Projection coherence, including the cutoff. Let $P_J$ keep raw tensor coordinates with both constituent indices at most $J$ and map them to the corresponding cut raw basis. To calculate a raw bar column, expand by $C\otimes C$, apply [F2], and reexpress by $C^{-1}\otimes C^{-1}$. All nonidentity expansion and reexpression terms increase a constituent index, and the middle operator preserves both indices. A term using the omitted outgoing shear at index $J$ is outside the retained rectangle and cannot return. An input outside that rectangle likewise cannot return. The retained raw columns are consequently exactly the cut columns, giving $P_J{\mathrm{\Psi }}_L={\mathrm{\Psi }}_J{P}_J$ for $L\ge J$ and also $P_J\mathrm{\Psi }={\mathrm{\Psi }}_J{P}_J$ for the algebraic full module. Since $P_J$ preserves the seed, weights and polynomial correction coefficients, step 1.1 gives $P_J{g}^{(L)}=g^{(J)}$. This proves endpoint independence at the actual deleted-last-shear boundary.

Four states and three transfer bands. Work at total weight zero, and use first position $a$ and second bottom-distance $d$. The seed has $(a,d)=(0,0)$. On $0\le a,d\le 3$, even shears send offsets $0\mapsto 1$ and $2\mapsto 3$, odd shears send only $0\mapsto 3$, and inverse shears have the same supports. Degree $r$ decreases both offsets by $r$. Thus all reached raw coordinates remain in this envelope and $r\le 3$. At first weight $2+2t$, both factors satisfy $n_j\mathrm{/}2-a=t+1$ or $n_l\mathrm{/}2-d=t+1$. For $t=4k+s$, $s=0,1,2,3$, the two possibilities are respectively $((2k,0),(2k+1,1))$, $((2k+1,0),(2k+2,3))$, $((2k+2,2),(2k+3,3))$, and $((2k+2,1),(2k+3,2))$. Order their tensor pairs $LL,LH,HL,HH$. Let $B_{t,r}$ be the four-by-four raw coefficient matrix from layer $t$ to $t+r$, obtained by the three finite operations of step 2.1. Then $B_{t,0}=I$ and there are no bands beyond $r=3$. The coefficient family $f_0=(1,0,0,0{)}^T$, $f_t=\mathcal{P}({\sum }_{r=1}^{\min (3,t)}{B}_{t-r,r}{\bar{f}}_{t-r})$ is well defined layer by layer: each partial sum is a finite algebraic vector, its least unsolved residual is anti-bar, and the integral full bar makes that residual Laurent. No infinite vector is asserted to belong to the module. The symmetry of the shear signs and the two Gaussian factors in [F2] preserves $f_t=(P,Q,-Q,R{)}^T$.

The scalar passed between four-layer cycles. At layer $4k+1$, standard constituent $2k+2$ first appears, at offsets $(3,3)$, with coefficient $c_k=-(P-2hQ-h^{2}R)\mathrm{/}{h}^2$. This follows by expanding the four raw states: the lower constituent is odd, its first top shear has coefficient $1\mathrm{/}h$, and its second bottom shear has coefficient $-1\mathrm{/}h$. Earlier layers cannot enter constituent $2k+2$, because odd shears require offset zero and that first occurs at this layer. At this point a future residual with an old first constituent is impossible above its top weight; a mixed new/old constituent term has an old second bottom or an old first top, so its positive quasi-$R$ terms vanish. The only remaining future residual is therefore $c_k(\mathrm{\Psi }{v}_{2k+2,3;2k+2,3}-v_{2k+2,3;2k+2,3})$. Its initial center coefficient is bar fixed, since no earlier standard coordinate in this diagonal constituent can contribute to its first weight. Consequently the next four-layer correction depends only on $c_k$. Positive-part extraction is linear over $\mathbb{Q}$ (and over $\mathbb{Z}$ on integral coefficients); no linearity over $\mathbb{Q}(q)$ is claimed. In the induction below only the integer scalar $a_k$ is taken outside it. At $k=0$, [F2] gives $f_1=(-q^5+q^3-2q,-q^3+q,q^3-q,q{)}^T$, so $P-2hQ-h^{2}R=-p(h+1)$.

Explicit matrices and integral ghost residual. Let $E_{21}$ be the two-by-two matrix with its only nonzero entry $1$ at $(2,1)$, and put $D_t=(I+{\alpha }_t{E}_{21})\otimes (I-{\alpha }_t{E}_{21})$, with ${\alpha }_t=1\mathrm{/}p,1\mathrm{/}h,-1\mathrm{/}p,0$ according as $t<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>4=0,1,2,3$. This is exactly $C\otimes C$ between the four raw and standard states of step 3.1. For a source factor state $(j,a)$ and a target state $(j,a-r)$, put $\gamma (j,a,r)={\left(\genfrac{}{}{0px}{}{n_j-a+r}{r}\right)}_q$; put the entry zero if that same constituent target is absent. Define $T_{t,r}$ on tensor pairs by $A_r\gamma (j,a,r)\gamma (l,d,r)$ in the matching target pair and zero elsewhere. Then the completely specified four-by-four matrices are $B_{t,r}=D_{t+r}^{-1}{T}_{t,r}{D}_t$. This includes $B_{t,0}=I$. For the diagonal ghost in step 4.1 the new string has $n=8k+10$, and its three positive divided coefficients are $G_r={\left(\genfrac{}{}{0px}{}{8k+7+r}{r}\right)}_q$, $r=1,2,3$. Each $G_r$ is divisible by $h$: the upper and lower remainders modulo $4$ are respectively $(0,1),(1,2),(2,3)$, so the Gaussian block argument proved in the construction gives zero modulo $h$. Likewise $p\mid G_1,G_3$ by their upper/lower remainders modulo $2$. Thus $h^2\mid G_r^2$ for all three degrees, and $p^2{h}^2\mid G_r^2$ for degrees $1,3$. Reexpressing the resulting standard offset $3-r$ uses no shear for $r=2$, and at most the two denominators $p$ for $r=1,3$. Hence the future residual of the normalized rational ghost $-p((2k-1)h-1)\mathrm{/}{h}^2$ has raw Laurent integer coefficients. Subsequent raw integral corrections preserve this property, and involutivity makes every least residual anti-bar. Thus the positive-part recursion used next has genuine Laurent input for every integer $k\ge 0$, despite its initial rational ghost.

Uniform quotient extraction. Put $U_1=(0,0,0,-p((2k-1)h-1)\mathrm{/}{h}^2{)}^T$ at layer $4k+1$, and set $U_i=\mathcal{P}({\sum }_{r=1}^{\min (3,i-1)}{B}_{4k+i-r,r}{\bar{U}}_{i-r})$ for $i=2,3,4,5$. In the matrix formula of step 5.1 replace ${\left(\genfrac{}{}{0px}{}{8k+N}{r}\right)}_q$ by ${\prod }_{j=1}^r(z{q}^{N-j+1}-z^{-1}{q}^{-N+j-1})\mathrm{/}(q^j-q^{-j})$; thus $z=q^{8k}$, and crossing a four-layer period replaces $z$ by $q^{8}z$. Write $k=3m+c$, $c=0,1,2$, and $x=q^{24m}$. Each residual is a finite Laurent polynomial in $x$ with rational $q,m$ coefficients. Polynomial division at infinity and removal of its constant term defines a $\mathbb{Q}(m)$-linear operator on these rational coefficients; on their Laurent sum it is exactly $\mathcal{P}$. Write each nonzero-$x$-power coefficient, after cancelling powers of $q$, as $N\mathrm{/}D$ with $D(0)\mathrm{\ne }0$. For positive $x$-powers the numerator is a polynomial with zero valuation at least $18$, and every negative-power coefficient has infinity degree at most $-18$; the finite bounds and denominator products are calculated below. Accordingly $q^{24am}N\mathrm{/}D$ has zero constant at zero for $a>0,m\ge 0$, and for $a<0$ its infinity degree is negative for all $m\ge 0$, so contributes no polynomial part. If $a>0$ and $D\mid (q^{24}-1{)}^3$, put $u=q^{24}-1$ and $\rho ={\mathrm{rem}}_D(N(1+amu+am(am-1)u^2\mathrm{/}2))$. The binomial expansion of $(1+u{)}^{am}$ is exact modulo $D$, including $m=0$ and $am=1$, because $u^3=0$ there. Thus $\rho ={\mathrm{rem}}_D(q^{24am}N)$, and Euclidean division gives the exact positive part $q^{24am}N\mathrm{/}D-\rho \mathrm{/}D+\rho (0)\mathrm{/}D(0)$: the quotient's constant is $-\rho (0)\mathrm{/}D(0)$, since the dividend vanishes at zero. For $x$-power zero, ordinary numerator/denominator division followed by constant removal applies, even if the coefficient has a pole at zero. These arguments justify extraction term by term, with no specialization at a root and no missing small-$m$ case.

The finite all-parameter identity. The following quotient/remainder identities result from substitution in the explicit matrices of step 5.1 and the extraction rule of step 6.1. Let $H=(q-1)(q+1)$, $J=q^2+1$, $K=q^4+1$, and $L=(q^2-q+1)(q^2+q+1)$. For residual layers $i=2,3,4,5$ respectively, the common denominators after removing powers of $q$ divide $HJ{K}^2$, $H^2{J}^2{K}^2$, $H^3{J}^2{K}^{2}L$, and $H^3{J}^3{K}^{2}L$; the least positive-$x$-power zero valuations are respectively $18+16c,21+16c,22+16c,27+16c$, and the greatest negative-$x$-power infinity degrees are their negatives. All factors $H,J,K,L$ divide $q^{24}-1$, with their distinct cyclotomic factors pairwise relatively prime, and the displayed exponents are at most three; hence each listed denominator divides $(q^{24}-1{)}^3$. To obtain these identities form $T_{t,r}$ from its matching pairs, multiply on the two sides by $D_{t+r}^{-1},D_t$, collect powers of $x$, and for each positive power replace its remainder by the three-term polynomial of step 6.1; collect again after each of the four layers. The scalar $U_{5,LL}-h{U}_{5,LH}+h{U}_{5,HL}-h^2{U}_{5,HH}$ then has the following coefficients of $ph,p$, with all other coefficients zero: for $c=0$, $(-(216m^3+216m^2+66m+6),36m^2+30m+6)$; for $c=1$, $(-(216m^3+432m^2+282m+60),36m^2+54m+20)$; for $c=2$, $(-(216m^3+648m^2+642m+210),36m^2+78m+42)$. These are finite rational-function identities, verified by the indicated polynomial divisions; the formulas above specify every input and operation. Substitution of $k=3m+c$ in the three output pairs gives exactly $-(2k+2)(2k+3)p((2k+1)h-1)$. Since the extraction bounds include $m=0$, this proves the cycle identity for every $k\ge 0$.

Factorial invariant and nonzero endpoint. Apply step 7.1 to step 4.1, using only linearity under the integer multiplier $a_k$. Starting with $a_0=1$, the scalar multiplier satisfies $a_{k+1}=-(2k+2)(2k+3)a_k$, hence $a_k=(-1{)}^k(2k+1)!$ and $P-2hQ-h^{2}R=a_{k}p((2k-1)h-1)$ at every odd layer $4k+1$. Modulo $q^4+1$, $h=0$ and $-p=q^3-q$, so the first coordinate $P$ has remainder $a_k(q^3-q)$. Its divisor is monic and the displayed nonzero remainder has smaller degree. This proves nonvanishing of every odd endpoint in the coefficient family.

Identification with all finite cuts and conclusion. For any fixed $J$, raw coordinate projection of the layerwise family is a finite vector: only finitely many raw states have both indices at most $J$. Step 2.1 shows that each coefficient equation projects to its cut equation; the seed and positive-polynomial corrections project as well. Step 1.1 therefore identifies this vector with $g^{(J)}$. In particular all retained odd endpoints have the formula from step 8.1. The largest odd index at most $J$ is $J$ or $J-1$, giving the claimed escaping support in every cut. The finite projection comparison was proved with the omitted outgoing shear, so no full-formal/cut identification has been assumed. Every parameter, basis and finite calculation is specified, and no choice is used.