# The remaining prime bands and the cost of completion **The sufficient twin margin remains OPEN.** The pilot is now controlled by [prime-dispersion.md](prime-dispersion.md), which consumes the reductions here and uses a different second moment. This note narrows the arithmetic and audits a particular theorem interface. It does not reduce the remaining exponent deficit or control another full rectangle. The reductions below are written derivations; the finite checks are separate and do not establish asymptotic rates. ## 1. Remove the small distinguished primes without separating their averages Keep x dyadic, D=floor(x^(277/1000)), E=floor(x^(467/1000)), I=(D,2D], J=(E,2E], V=floor(x^(6/25)), Z=floor(x^(1/20)). For any prime set B define the squarefree coefficient $$ Q_{I,B}(\ell)=\mu(\ell)\sum_{p\in B,\ p\mid\ell,\ \ell/p\in I}\log p. \tag{1} $$ It is bounded in modulus by log l. Equivalently it is minus the sum of mu(d)log p over dp=l, d in I, p in B, p not dividing d. These are the Q coefficients of [coefficient-structure.md](coefficient-structure.md). Set V_0=floor(x^(47/200)), Z_0=floor(x^(9/200)), and split each coefficient into primes below its new cutoff and primes above it, denoted Q_- and Q_+. The exact product split is $$ Q_{I,\le V}Q_{J,\le Z} =Q_{I,-}Q_{J,\le Z}+Q_{I,+}Q_{J,-}+Q_{I,+}Q_{J,+}. \tag{2} $$ No prime-by-prime triangle inequality is taken. The first two products have expanded upper support exponents (0.512,0.517) and (0.517,0.512). In [endpoint-pairing.md](endpoint-pairing.md) the sufficient exponent is $$ \frac3{20}+\frac7{10}(a+b)+\frac14\max(a,b) =\frac{19991}{20000}<1. \tag{3} $$ Its endpoint proof applies to any separate logarithmically bounded coefficients with these supports. Take tau=1/10000: (3)+3tau<1. Thus both discarded **endpoint** pieces have a power saving, including all dyadic boxes and both gcd branches. No independent density estimate is needed for this step, since the coefficients multiply Delta itself. Let E_+ be the remaining exact QQ discrepancy. Then for every fixed H, $$ R_{I,J}(x)=E_+(x)+O_H(x/\log^H x),\qquad E_+=\sum_{p\in(V_0,V],\ q\in(Z_0,Z]}\log p\log q \sum_{d\in I,e\in J\atop p\nmid d,\ q\nmid e} \mu(d)\mu(e)\Delta_{dp,eq}(x). \tag{4} $$ Here p,q are prime. The other sectors and the total density were already controlled in coefficient-structure.md. A separate high-band density estimate is also available from that note's excluded-prime argument, since restricting the outer prime sums only decreases their absolute harmonic cost. At finite x the discrepancy still includes its actual density subtraction. ## 2. A low-frequency bound uniform in the expanded divisors The preceding note bounded h<=x^0.03 on the top expanded boxes only. That qualification matters. This section supplies one cutoff valid on **every** box of the pilot. Write P=MN, Q=M+N, u=P/x for reduced divisor sizes M,N and let positive h run in a dyadic block of size A. The paired polynomial bound is, apart from a fixed arbitrarily small power loss, $$ (1+A/u)^{1/2}\min(1,A/u) \left[P^{17/20}Q^{1/4}A^{-3/20}+P^{7/8}Q^{1/8}\right]. \tag{5} $$ Put r=P/(Ax). If r>=1, the two terms are bounded by x^(17/20)A^(7/10)Q^(1/4)r^(-3/20) and x^(7/8)A^(7/8)Q^(1/8)r^(-1/8), up to constants. If r<=1, the corresponding extra powers of r are 7/20 and 3/8. In both cases they are at most one. Consequently (5) is bounded by $$ x^{17/20}A^{7/10}Q^{1/4} +x^{7/8}A^{7/8}Q^{1/8}. \tag{6} $$ Since Q<2h eventually. The latter follows from 0.0342<0.045, including fixed constants and floors by taking x sufficiently large. It is not asserted on the saved finite prefixes. Thus both primes are odd and coprime to (2/g)h. The top-box first deficit is still 61/20000 before strict losses. Nothing in (8) is an improvement of that exponent, a statement about the proportion of residual mass removed, or a uniform product cutoff. ## 3. Exact completion in the right prime modulus Here is an explicit interface to ordinary Kloosterman sums. Fix a reduced-divisor box and a compatible g in {1,2}. Since p,q are odd, write the original divisors as gd and ge, so the expanded divisors are gdp, geq and (dp,eq)=1. Keep the weights mu(gd)mu(ge); they enforce the squarefree parity restrictions, including odd d,e when g=2. Put theta=2/g and z_0=x/2, with z in [x/2,x]. Define I_g=(D/g,2D/g] and J_g=(E/g,2E/g], and also impose the relevant box conditions dp~M, eq~N. Compatibility requires (e,pq)=1. For fixed p,e,q satisfying this, put e_c(v)=exp(2 pi i v/c) and $$ \begin{split} b_h&=-\theta h\,\overline{pe}\pmod q,\\ \Phi_h(d)&=\mathrm e\left(\frac{hz_0}{gdepq}\right) -\mathrm e\left(\frac{hz}{gdepq}\right),\\ F_h(d)&=\mu(gd)1_{d\in I_g,\ dp\sim M}1_{(d,pe)=1} \mathrm e_e\left(-\theta h\,\overline{pq}\,\overline d\right) \Phi_h(d),\\ \widehat F_h(t)&=\sum_d F_h(d)\mathrm e_q(-td). \end{split} \tag{9} $$ The inverses in e_e are modulo e; the inverse in b_h is modulo q. There is deliberately no restriction q not dividing d inside F_h. It will be supplied by the completed kernel. For compatible d, the elementary CRT phase split is $$ \mathrm e\left(-\theta h\frac{\overline{dp}}{eq}\right) =\mathrm e_e\left(-\theta h\,\overline{pq}\,\overline d\right) \mathrm e_q\left(-\theta h\,\overline{pe}\,\overline d\right). \tag{10} $$ Define the unnormalized classical sum S(t,b;q)=sum_{u mod q,u!=0} e_q(tu+b inv(u)). Finite Fourier inversion gives the exact identity $$ \sum_{d,\ q\nmid d}F_h(d)\mathrm e_q(b_h\overline d) =\frac1q\sum_{t\bmod q}S(t,b_h;q)\widehat F_h(t). \tag{11} $$ Indeed q^(-1)sum_t e_q(t(u-d)) is 1 precisely for u=d mod q. In particular it gives zero when q divides d. This verifies the sign, the normalization 1/q and the coprimality condition without a theorem import. For the original positive-h polynomial, (11) is multiplied by c_h=-W(h/(T+1))/(2 pi i h), then summed over h,p,q,e with the outer weight mu(ge)log p log q. Negative h give its complex conjugate, so twice the real part is taken. The e interval and eq~N cut remain. This identity keeps all prime averages: no absolute value has yet been taken over p,q,e or h. It also keeps the true endpoint perturbation. Two obstacles prevent treating this as an ordinary separate-coefficient bilinear form in h and t. The transform has **all** t modulo q, and F_h depends on h through both its modulo-e phase and Phi_h. It also depends on p,e,q and the actual box. No short dual support or factorization hat F_h(t)=alpha_h beta_t has been established. Even with Phi_h replaced by 1, the modulo-e phase causes this issue. An exact small example is g=1, D=4, p=17, e=11, q=13, h=1,2. The nonzero d values in (4,8] are 5,6,7, distinct modulo q. The ratios F_2(d)/F_1(d) are e_11(4), e_11(7), e_11(6), respectively. They differ, so these two residue vectors and their invertible discrete Fourier transforms are not proportional. The validator also checks the actual endpoint-dependent transform on explicit examples. This is a failure of an automatic separation identity, not an asymptotic claim about effective low-rank approximation. ## 4. The full dual range has an exact operator-norm obstruction This is a precise closure for arbitrary coefficients, not for (9). Let q be prime, lambda a unit modulo q, and H a set of r distinct nonzero residues with 2<=r<=q-1. Define the r by q matrix K_{h,t}=S(t,lambda h;q). Orthogonality gives $$ \sum_{t\bmod q}K_{h_1,t}\overline{K_{h_2,t}} =q\sum_{u\ne0}\mathrm e_q(\lambda(h_1-h_2)\overline u) =\begin{cases}q(q-1),&h_1=h_2,\\-q,&h_1\ne h_2.\end{cases} \tag{12} $$ Thus K K*=q^2 I-q J. On the subspace orthogonal to the all-ones vector its eigenvalue is exactly q^2, and its other eigenvalue is q(q-r). Consequently ||K||_(2 to 2)=q exactly. The kernel normalized by 1/q in (11) has norm exactly 1: it has no uniform power saving. More specifically, every coefficient vector v satisfies $$ \|K^*v\|_2^2=q^2\|v\|_2^2-q\left|\sum_h v_h\right|^2 \ge q^2(1-r/q)\|v\|_2^2. \tag{12a} $$ Here r<=T=o(q). Thus even fixing the actual harmonic weights cannot supply a fixed-power saving against arbitrary full-range dual coefficients merely through this operator estimate. Deleting t=0 does not repair this. Since S(0,lambda h;q)=-1, the new Gram matrix is q^2 I-(q+1)J, with the same largest eigenvalue q^2. For example the row vector (1,-1,0,...) attains it. A bound q^(1-eta) times the two coefficient norms, for fixed eta>0 and all arbitrary full-range vectors, is therefore false. In (8), h