Proof. Put \(u'_p=v_p(b_{m_p,p})\). These are nonnegative integers, and the factorization argument above shows that \(S_0=\{p:u'_p>0\}\) is finite. For \(p\in S_0\), take \(u_p=u'_p+1\), and set \[D_0=\prod_{p\in S_0}p^{u_p}, \qquad M=\prod_{p\in S_0}p^{u_pm_p+u'_p}.\] Choose the representative \(-D_0<r\le0\) satisfying \(r\equiv\mathfrak z_p\pmod{p^{u_p}}\) for every \(p\in S_0\), by the Chinese remainder theorem. If \(S_0\) is empty, these definitions give \(D_0=M=1\) and \(r=0\).

For \(p\in S_0\), the congruence defining \(r\) shows that \(w_p=(\mathfrak z_p-r)/D_0\) lies in \(\mathbb Z_p\); for \(p\notin S_0\), the same conclusion follows because \(D_0\) is a \(p\)-adic unit. Define \(g\) initially as the displayed rational polynomial. Its centered expansion is \[g(w_p+X)=\sum_{j=m_p}^k b_{j,p}\frac{D_0^j}{M}X^j.\] If \(p\in S_0\), the coefficient at \(j=m_p\) has valuation \(u'_p+u_pm_p-(u_pm_p+u'_p)=0\). For \(j>m_p\), its valuation is at least \[u_p(j-m_p)-u'_p\ge1.\] If \(p\notin S_0\), both \(D_0\) and \(M\) are \(p\)-adic units, so every coefficient is integral and the coefficient at \(j=m_p\) is again a unit. Thus the centered expansion belongs to \(\mathbb Z_p[X]\) for every prime \(p\). Translation by \(-w_p\in\mathbb Z_p\) shows that \(g\) itself belongs to \(\mathbb Z_p[x]\). Every rational coefficient of \(g\) is therefore integral at every prime and hence is an integer.

The elements \(w_p\) are roots of \(g\); reducing them at the prime-power factors of any modulus and applying the Chinese remainder theorem proves intersectivity. The degree and sign of the leading coefficient are preserved because \(D_0,M>0\). Finally, the defining identity for \(g\) gives \(Mg(n)=h(r+D_0n)\), and \(r>-D_0\) gives \(r+D_0n\ge1\) for \(n\ge1\). ◻

Proof. Only primes dividing \(\lambda(l)\) can occur in denominators of the coefficients of \(h_l\). Fix such a prime, put \(u=v_p(l)>0\), and use the affine coordinate \[Y=\frac{r_l-\mathfrak z_p}{p^u}+\frac{l}{p^u}x \quad\text{over }\mathbb Z_p.\] Its slope is a unit, and expansion at \(\mathfrak z_p\) gives \[ h_l(x)= \left(\frac{\lambda(l)}{p^{u m_p}}\right)^{-1} \sum_{j=m_p}^k b_{j,p}p^{u(j-m_p)}Y^j. \tag{3}\] The prefactor is a \(p\)-adic unit, and the displayed polynomial is integral over \(\mathbb Z_p\). This proves \(h_l\in\mathbb Z[x]\).

For every prime \(p\), the element \((\mathfrak z_p-r_l)/l\) belongs to \(\mathbb Z_p\): when \(p\mid l\), this uses the defining congruence for \(r_l\), and otherwise \(l\) is a unit. It is a root of \(h_l\). Reducing these roots at the prime-power factors of an arbitrary modulus and applying the Chinese remainder theorem proves intersectivity.

Substitution of \(\mathfrak z_{p,l}+X\) into the definition of \(h_l\) gives the centered expansion in (1). Complete multiplicativity gives \(v_p(\lambda(l))=m_pv_p(l)\), so \(l^{m_p}/\lambda(l)\) is a unit in \(\mathbb Z_p\). This proves the exact valuation there, including the preservation of the multiplicity and of the unit condition. The assertion is local at \(p\); the displayed quotient need not be an integer.

The degree and leading coefficient follow from the definition. Expanding \(h(r_l+lx)\) and using \(|r_l|<l\) bounds every coefficient by \(O_h(l^k/\lambda(l))=O_h(a_l)\).

The same root choices give \(r_{lD}\equiv r_l\pmod l\). The representatives of this class in \((-lD,0]\) are exactly \(r_l,r_l-l,\ldots,r_l-(D-1)l\), so \(r_{lD}=r_l-il\) for some \(0\le i<D\). Complete multiplicativity of \(\lambda\) now gives the polynomial identity in (2). The remaining bounds follow from \(1\le m_p\le k\), while \(\lambda(p)=p^{m_p}\) gives the equality of prime supports. Finally, \(Dx-i\ge D-(D-1)=1\) for \(x\ge1\). ◻

Proof. Choose an initial threshold so that the preceding nonconstancy holds and \(p>k\). Under Condition [local:unit-condition], this initial threshold can be chosen using only \(k\). The nonzero polynomial \(F'\bmod p\) has degree at most \(k-1\), which gives the count of regular arguments; taking \(p\ge2(k-1)\) also makes at least half of the arguments regular.

First consider the sum modulo \(p\), and let \(k'\) be the degree of \(F\bmod p\). If \(k'=1\), the sum is zero. If \(k'\ge2\), repeated differencing gives \[\left|\mathbb E_{x\bmod p}e(uF(x)/p)\right|^{2^{k'-1}} \le \mathbb E_{b_1,\ldots,b_{k'-1}\bmod p} \left|\mathbb E_{x\bmod p} e\!\left(\frac{u\,\Delta_{b_1}\cdots\Delta_{b_{k'-1}}F(x)}p\right) \right|,\] where \(\Delta_bF(x)=F(x+b)-F(x)\). To obtain the inequality, square the mean, express it as the mean of multiplicative differences, and iterate the triangle inequality and Cauchy–Schwarz. The reduction modulo \(p\) of \(u\Delta_{b_1}\cdots\Delta_{b_{k'-1}}F\) has degree at most one, with coefficient of \(x\) equal to \(u k'!\) times the leading coefficient of \(F\bmod p\) times \(b_1\cdots b_{k'-1}\). Its character mean vanishes unless one increment is zero. The right-hand side is therefore at most \((k'-1)/p\). Since \(8\delta<2^{1-k}\le2^{1-k'}\) and \(1\le k'\le k\), increasing the threshold by an amount depending only on \(k\) gives the first bound in (4) for \(j=1\). Removing at most \(k-1\) arguments gives \[\left|\mathbb E_{\substack{x\bmod p\\x\text{ regular}}}e(uF(x)/p)\right| \le\frac{p}{p-k+1} \left(\left|\mathbb E_{x\bmod p}e(uF(x)/p)\right|+\frac{k-1}{p}\right).\] The first bound and another increase depending only on \(k\) give the second bound in (4).

For \(j\ge2\), put \(n=\lfloor j/2\rfloor\). Write \(x=x_0+p^{j-n}y\), where \(x_0\) runs modulo \(p^{j-n}\) and \(y\) runs modulo \(p^n\). Taylor expansion modulo \(p^j\) leaves only the linear variation, because \(2(j-n)\ge j\). The average over \(y\) vanishes unless \(F'(x_0)\equiv0\pmod{p^n}\). We claim that these arguments occupy at most \(k-1\) classes modulo \(p^v\), where \(v=\lceil n/(k-1)\rceil\).

Otherwise choose \(k\) integer representatives \(x_1,\ldots,x_k\) in distinct such classes with \(p^n\mid F'(x_i)\). Interpolation of the degree-at-most-\(k-1\) polynomial \(F'\) gives \[F'(X)=\sum_{i=1}^k F'(x_i) \frac{\prod_{j\ne i}(X-x_j)}{\prod_{j\ne i}(x_i-x_j)}.\] Each denominator has valuation at most \((k-1)(v-1)<n\), whereas the numerator polynomial has integral coefficients. Every coefficient of every summand is therefore divisible by \(p\) in \(\mathbb Z_p\). This contradicts \(F'\not\equiv0\pmod p\) and proves the claim. Since \(v\le n\le j-n\), the density of the surviving arguments is at most \[(k-1)p^{-\lceil n/(k-1)\rceil} \le (k-1)p^{-j/(3(k-1))}.\] Here \(n\ge j/3\). Since \(1/(3(k-1))>8\delta\), one fixed increase of the prime threshold absorbs the factor \(k-1\) for every \(j\ge2\). All choices of threshold are uniform in \(l\) and \(u\); fix \(P_1\) large enough to satisfy them all. After the initial nonconstancy condition, every increase used only \(k\). This proves the stated degree-only choice under Condition [local:unit-condition]. ◻

Proof. For a nonconstant polynomial over \(\mathbb Z_p\), consider the ideal generated by its nonconstant coefficients. An affine substitution with unit slope preserves this ideal: substitution gives one inclusion, since a constant cannot contribute to a nonconstant coefficient, and the inverse affine substitution gives the reverse inclusion. Thus it preserves the minimum valuation of the nonconstant coefficients.

Apply this observation to the translation by \(\mathfrak z_{p,l}\). Equation (1) displays a nonconstant coefficient of valuation \(v_p(b_{m_p,p})\), so the minimum for \(F=h_l\) is at most that number, uniformly in \(l\). Differentiation multiplies the coefficient of degree \(j\) by \(j\). Hence some coefficient of \(F'\) has valuation at most \(v_p(b_{m_p,p})+T_{k,p}\).

The matrix \(V_k\) maps the coefficient vector of a polynomial of degree at most \(k-1\) to its values at \(0,1,\ldots,k-1\). Its determinant is \(\prod_{0\le i<j\le k-1}(j-i)=\prod_{j=1}^{k-1}j!\), and its adjugate has integer entries. If every value \(F'(x)\), \(0\le x<k\), had valuation greater than \(S_p\), multiplication by \(V_k^{-1}=\operatorname{adj}(V_k)/ \det V_k\) would give valuation greater than \(S_p-v_p(\det V_k)=v_p(b_{m_p,p})+T_{k,p}\) for every coefficient of \(F'\). This contradicts the preceding bound and proves the lemma. ◻

Proof. Taylor expansion over the integers gives \[F(x_0+p^{E-s}y)=F(x_0)+p^E\bigl(c_1y+pC(y)\bigr), \qquad p\nmid c_1,\quad C\in\mathbb Z[y].\] Indeed the linear coefficient is \(p^EF'(x_0)/p^s\), and every term of degree at least two is divisible by \(p^{E+1}\), since \(E-2s\ge1\). The polynomial \(T(y)=c_1y+pC(y)\) induces a bijection modulo every power of \(p\): its difference at two arguments is their difference times a \(p\)-adic unit, so the induced map on each finite residue ring is injective and hence bijective.

For a uniform argument in the stated class modulo \(p^j\), the variable \(y\) is uniform modulo \(p^{j-E+s}\). Its reduction modulo \(p^{j-E}\) is still uniform, with each residue occurring \(p^s\) times. The bijectivity of \(T\) at this latter modulus proves the asserted distribution of \(F\). ◻

Proof. If \(p<P_1\), repeat any admissible residue \(L\) times. The sum is zero because \(p^{E_p}\mid L\).

If \(p\ge P_1\), let \(R\) be \(h_l(x)\) for a uniform regular argument \(x\bmod p\), and take independent copies \(R_1,\ldots,R_L\). Fourier inversion gives \[p\,\mathbb P(R_1+\cdots+R_L=0) =1+\sum_{u=1}^{p-1}\bigl(\mathbb Ee(uR/p)\bigr)^L.\] By (4), the sum following \(1\) has modulus at most \((p-1)p^{-4\delta L}<1\). The probability is therefore positive, and any realization with sum zero supplies the required residues. ◻

Proof. Reversing a directed edge replaces the one-edge operator by its adjoint, so its norm remains at most \(p^{-4\delta}\). The case \(e_G=0\) is immediate; assume \(e_G\ge1\). For the first assertion, replace the edge weights by \(c_p\) one at a time. In a term containing the difference for an edge \(x\to y\), fix all labels other than \(Y_x,Y_y\). Because there is no other edge on this unordered pair, the remaining factors separate into a function of \(Y_x\) and a function of \(Y_y\), each bounded by one and hence of uniform \(L^2\) norm at most one. The one-edge operator estimate bounds this term by \(p^{-4\delta}\). Summing over the \(e_G\) replacements gives the stated error \(e_Gp^{-4\delta}\).

For the second assertion put \(B=K_G-c_p^{e_G}\). If the two configuration spaces have cardinalities \(N_X,N_Y\), the kernel operator \((T_Bf)(X)=\mathbb E_Y B(X,Y)f(Y)\) has Euclidean matrix \(B(X,Y)/\sqrt{N_XN_Y}\) under the isometries from uniform \(L^2\) to Euclidean space. Thus, with two independent copies of each list, matrix multiplication gives \[\|B\|_{\mathrm{op}}^4 \le \sum_j s_j(B)^4 =\mathbb E\prod_{a,b\in\{0,1\}}B(X_a,Y_b),\] where \(s_j(B)\) are the singular values for the normalized measures. Indeed, the trace of \((T_BT_B^*)^2\) is the uniform average of \(B(X_0,Y_0)\overline{B(X_1,Y_0)}B(X_1,Y_1)\overline{B(X_0,Y_1)}\); here \(B\) is real, so the conjugates can be omitted. This explains why no factor depending on \(N_X,N_Y\) occurs. Expand the four copies of \(B\). A product using any subset of the four corners again comes from a graph of the stated kind: an unordered pair of its formal endpoints determines both the original edge and its corner. A term using \(j\) corners has exactly \(je_G\) edges, so its graph error is at most \(je_Gp^{-4\delta}\). Its constant multiplier has modulus \(c_p^{(4-j)e_G}\le1\). All main terms are \(c_p^{4e_G}\) times the corresponding sign, so they cancel. Consequently \[\|B\|_{\mathrm{op}}^4 \le\sum_{j=0}^4\binom4j je_Gp^{-4\delta} =32e_Gp^{-4\delta},\] because \(\sum_{j=0}^4\binom4j j=4\cdot2^3=32\). Taking fourth roots proves (6). ◻

Proof. The interaction between the two halves is governed by the symmetric matrix \[\mathsf A_{xy}=H^S_{x,\theta y},\qquad x,y\in V_+.\] Fix \(x\in V_+\), and suppose that the cut symbols occur at positions \(i<j\) in its cut block, with gap \(g=j-i\). Swapping the cut symbols gives the diagonal entry \(\mathsf A_{xx}=\Delta^g\). Any other transposition carrying \(x\) across the cut must move the first cut symbol to a position beyond \(j\), or the second to a position before \(i\). Its gap is at least \(g+1\). There are fewer than \(L^2\) such transpositions; transpositions in other blocks do not cross the cut. Consequently the absolute off-diagonal row sum is at most \(L^2\Delta^{g+1}=\Delta^g/10\). Symmetric diagonal dominance gives \[\mathsf A\succeq\kappa\operatorname{Id}, \qquad \kappa=\frac9{10}\Delta^{L-1}>0.\]

For \(j\in[\ell_s]^{V_+}\) define \(u(j)=(\mathbf 1_{\{j_x=i\}})_{x\in V_+,\,i\in[\ell_s]}\). After identifying the minus side with the plus side by reflection, the cross-interaction matrix on half-index assignments is \[C_{j,j'}=\exp\bigl(u(j)^T (\mathsf A\otimes\operatorname{Id})u(j')\bigr).\] Split \(\mathsf A=(\mathsf A-\kappa\operatorname{Id})+ \kappa\operatorname{Id}\). The entrywise exponential associated with the first summand is positive semidefinite: its power-series expansion is a sum of Gram matrices of tensor powers. Its diagonal entries are at least one. The exponential associated with the second summand is a tensor product, over \(x\in V_+\), of \[\mathbf1\mathbf1^T+(\exp(\kappa)-1)\operatorname{Id}\] on \([\ell_s]\). It therefore dominates \(a_*\operatorname{Id}\), where \(a_*=(\exp(\kappa)-1)^{n_*}\). The entrywise product of positive semidefinite matrices is positive semidefinite, as follows directly from their Gram representations. Multiply the lower bound for the second exponential entrywise by the first. The resulting diagonal lower bound is at least \(a_*\operatorname{Id}\) because the first exponential has diagonal entries at least one. Thus \(C\succeq a_*\operatorname{Id}\).

The interactions within the two halves give the same diagonal factor on either side of \(C\), with all entries at least one. The full index matrix retains the lower bound \(a_*\operatorname{Id}\). For fixed lists \(Y\), reflected vertices have the same address because \(b\) is marked. The half-label evaluations therefore form the same vector on both sides of this matrix, even when some field labels coincide. There are \(\ell_s^{n_*}\) half-index assignments. Averaging uniformly on both sides converts the matrix lower bound into \[\mathbb E_{j_+,j_-}R_S(j) J((z_v)_{v\in V_+})J((z_{\theta v})_{v\in V_+}) \ge a_*\ell_s^{-n_*}\, \mathbb E_{j_+}J((z_v)_{v\in V_+})^2.\] Multiplying by \(I_S(Y)\) and averaging \(Y\) proves (7) with \(c_*=a_*\ell_{t+1}^{-n_*}\). ◻

Proof. Reflection identifies the minus lists and their internal edge factors with the plus lists and their internal factors. No index interaction crosses this cut: an interaction changes only a marked block, whereas the cut is determined by the unmarked block \(b\). The two within-half index weights are likewise identified by reflection; write \(R_+\) for the one on the plus half.

Let \(K(Y_+,Y_-)\) be the product of all cross-list edge weights. Then \(Q_S(J)\) is the quadratic form of this kernel on the half function \[g(Y_+)=I_+(Y_+)\mathbb E_{j_+} R_+(j_+)J\bigl((Y_{u_S(v),j_v})_{v\in V_+}\bigr).\] Cauchy–Schwarz, the uniform bound on \(R_+\), and \(I_+^2\le I_+\) give \(\|g\|_2^2\le C\mathcal D_S(J)\). By (6), the kernel \(K\) differs in operator norm by \(O(p^{-\delta})\) from the constant \(c_p^{e_{\rm cross}}\), where \(e_{\rm cross}\) is its number of edges. The constant contributes \(c_p^{e_{\rm cross}}(\mathbb Eg)^2\ge0\). This proves (8). In the terminal case \(I_S=1\), so the cross kernel is exactly one and the entire form is nonnegative. ◻

Proof. Let \(\Pi_+\) be the \(r_*\) words in block \(b\) satisfying the cut order. For each new address \(u\in U_{S'}\), the old plus addresses above it are \((u,q)\) with \(q\in\Pi_+\); here \(q\) is inserted in block \(b\). Concatenate their lists by identifying \[[\ell_{s+1}]\ \cong\ \Pi_+\times[\ell_s], \qquad Y'_{u,(q,i)}=Y_{(u,q),i}.\] This is a bijection of independent uniform list entries, so it changes neither their joint distribution nor the value of an average. Figure 1 separates this relabeling from the index restriction used below.

A new address edge requires \(t+1\) shifted blocks outside \(S'\). Those blocks also certify an old edge for every choice of the two words in block \(b\). Hence every factor of \(I_{S'}(Y')\) occurs, with its direction unchanged, among the factors of \(I_+(Y_+)\). Since all factors belong to \([0,1]\), \[I_+(Y_+)\le I_{S'}(Y').\] When \(S'\) is terminal this inequality follows from \(I_{S'}=1\).

Apply (7) to \(S'\). In its nonnegative right-hand side, restrict the new index at each \(v\in V_+\) to the sublist \(\{v_b\}\times[\ell_s]\). Each restriction has probability \(1/r_*\), independently across the \(n_*\) slots. On this event the labels and the remaining independent indices are exactly those in \(\mathcal D_S(J)\). Its probability is \(r_*^{-n_*}\), and the pointwise weight comparison above now yields (9). ◻

Proof. At a cut in block \(b\), pair each \(S\) with \(b\notin S\), \(|S|\le t\), with \(S\cup\{b\}\). Each component containing \(b\) appears in exactly one pair. Equations (8) and (9) bound the quadratic form of the pair below by \[p^{-|S|\delta/2} \left(p^{-\delta/2}c_*r_*^{-n_*}-Cp^{-\delta}\right) \mathcal D_S(J).\] This is nonnegative once \(p\) is sufficiently large. The only unpaired sets are terminal sets omitting \(b\); their forms are nonnegative by Lemma 10. Thus \(M_{l,p}\) is reflection positive at every cut.

For \(S=\varnothing\), each list has length one and \(R_S=1\). Lemma 8 and \(c_p\ge1/(2k)\) show that \(\nu_{\varnothing}(1)\) is bounded below by a fixed positive constant for all sufficiently large \(p\). All component masses are bounded above, and the number of components is fixed. It follows that \[0<c\le M_{l,p}(1)\le C,\qquad \varepsilon_{l,p}(1)\le C p^{-(t+1)\delta/2}.\] Our choice \((t+1)\delta/2>5\) gives the stated bound \(p^{-4}\) after increasing the prime threshold. Summing the nonleading coefficients also gives total probability \(O(p^{-\delta/2})\) for \(S\ne\varnothing\). Removing the terminal components retains the positive leading mass, so the same estimate holds after normalization to \(\mu_{l,p}^{\circ}\).

Simultaneous translation of all list entries preserves their product law, all differences in \(I_S\), and the index weight \(R_S\). It therefore translates every output label and proves the asserted invariances and marginal uniformity.

Finally, a nonterminal \(S\) leaves at least \(t+1\) blocks unmarked. Consecutive cycle vertices shift by \(c\) in all these blocks, including the edge from \(v_{L-1}\) to \(v_0\). Their addresses are consequently joined by a directed edge. Since its weights include every pair of list entries, any positive-weight label tuple has admissible increment on this edge, regardless of the chosen indices. This holds around the entire cycle in every component retained by \(\mu_{l,p}^{\circ}\). ◻

Proof. Equation (2) states that \(h_l(Dx-i)=\lambda(D)h_{lD}(x)\). Both \(D\) and \(\lambda(D)\) are units modulo \(p\). The argument map \(x\mapsto Dx-i\) is therefore a bijection and, on differentiating this identity, preserves regularity. Counting regular arguments gives \[W_{lD,p}(y)=W_{l,p}(\lambda(D)y).\] Apply (10) to every list entry in a component measure. The independent uniform list law and all index weights are unchanged; each old edge weight becomes exactly the corresponding new edge weight. Thus the map pushes \(\nu_{l,p,S}\) to \(\nu_{lD,p,S}\) for each \(S\). All component masses, and in particular the normalizing masses \(M_{l,p}(1)\) and \(M_{lD,p}(1)\), agree. The two claimed pushforwards follow with no additional factor. ◻

Proof. First normalize the leading component \(\nu_{\varnothing}\), and write \(x=z_v\), \(y=(z_{v'})_{v'\ne v}\). Before normalization its density relative to uniform labels is \(K_1(x,y)K_2(y)\), where \(K_1\) is the product of the \(m\) weights on edges incident to \(v\), and \(K_2\) contains the remaining weights. Put \[A(y)=\mathbb E_xK_1(x,y),\qquad B_f(y)=\mathbb E_x f(x)K_1(x,y).\] The incident edges form a graph between the singleton list \(x\) and the list \(y\). Equation (6), applied to this kernel, gives \[\|A-c_p^m\|_2\ll p^{-\delta},\qquad \|B_f\|_2\ll p^{-\delta}\|f\|_2.\] This includes \(m=0\), when \(A=1\) and \(B_f=0\).

Let \(Z_0=\nu_{\varnothing}(1)\), which is bounded below uniformly. The other-label marginal has density \(Z_0^{-1}K_2(y)A(y)\), and the conditional expectation equals \(B_f(y)/A(y)\) where \(A(y)>0\). Its squared norm is therefore exactly \[\frac1{Z_0}\mathbb E_y K_2(y)\frac{|B_f(y)|^2}{A(y)},\] with the ratio taken as zero if \(A(y)=0\). On \(A\ge c_p^m/2\), the bound on \(B_f\) makes this \(O(p^{-2\delta})\|f\|_2^2\). The complement has uniform probability \(O(p^{-2\delta})\), by the bound on \(A-c_p^m\) and the fixed positive lower bound on \(c_p^m\). Everywhere, weighted Cauchy–Schwarz gives \[\frac{|B_f(y)|^2}{A(y)} \le\mathbb E_x |f(x)|^2K_1(x,y)\le\|f\|_2^2.\] Since \(K_2\le1\), the complement has the same required bound. Thus the squared conditional norm in the normalized leading component is \(O(p^{-2\delta})\|f\|_2^2\).

For either mixture, condition additionally on its component index. The tower property and Jensen’s inequality can only increase the squared conditional norm under this extra conditioning. In each nonleading component the trivial bound is \(\|f\|_2^2\), because its single-slot marginal is uniform. Proposition 12 bounds their total probability by \(O(p^{-\delta/2})\). The squared norm for the mixture is consequently \(O(p^{-\delta/2})\|f\|_2^2\), and the norm is \(O(p^{-\delta/4})\|f\|_2\). Since \(\gamma=\delta/10\), the fixed implicit constant is absorbed by taking \(p\) sufficiently large, proving (11). ◻

Proof. For every prime \(p\ge P_1\), the quantitative local input to the estimates above is \[0\le W_{l,p}\le1,\qquad c_p\ge(2k)^{-1},\qquad \max_{u\ne0}|\widehat W_{l,p}(u)|\le p^{-4\delta}.\] Indeed, with the normalized Fourier convention, for \(u\ne0\) the coefficient \(\widehat W_{l,p}(u)\) is \(c_p\) times the corresponding regular-argument average in (4); also \(c_p\le1\). These statements hold uniformly in \(l\). We extract constants from the preceding proofs using \[e_*:=\binom d2,\qquad R_*:=\exp(e_*),\qquad C_{\mathrm{gr}}:=(32e_*)^{1/4},\qquad Z_*:=\frac12(2k)^{-e_*}.\] They depend only on \(k,L,t\), since \(d=(L!)^{2t+1}\).

For a set of \(s\) marked blocks, the graph on list entries has \(K^{w-s}\ell_s=d/2^s\le d\) formal vertices. Its simplicity was checked above before field labels were assigned, so coincidences between their values do not affect the graph bound. Every graph for \(I_S\), every cross-list graph in Lemma 10, and every incident graph in Lemma 14 consequently has at most \(e_*\) edges. Lemma 8 therefore bounds their graph errors by \(e_*p^{-4\delta}\). For the cross-list and incident graphs, which join two lists, it bounds the operator errors by \(C_{\mathrm{gr}}p^{-\delta}\).

Each index interaction \(H^S_{xy}\) is at most one, so \(R_S\) and every within-half weight \(R_+\) are at most \(R_*\). In particular, \(\nu_S(1)\le R_*\). The proof of Lemma 9 gives the explicit constant \[c_*=(\exp(\kappa)-1)^{n_*}\ell_{t+1}^{-n_*}, \qquad \kappa=\frac9{10}(10L^2)^{-(L-1)}.\] The exact half-index factor \(\ell_s^{-n_*}\) proved there and the exact index-restriction factor \(r_*^{-n_*}\) in Lemma 11 give the domination constant \(c_*r_*^{-n_*}\). In the unmarked proof, Cauchy–Schwarz and \(I_+^2\le I_+\) give \(\|g\|_2^2\le R_*^2\mathcal D_S(J)\). The list bound just obtained thus allows the constant in (8) to be \[C_{\mathrm{def}}:=R_*^2C_{\mathrm{gr}}.\] The pairing proof of Proposition 12 is therefore nonnegative at every cut whenever \[p^{\delta/2}\ge C_{\mathrm{def}}r_*^{n_*}/c_*.\]

Let \(e_0\le e_*\) be the number of edges in the leading component. The graph estimate and the lower bound on \(c_p\) give \[\nu_{\varnothing}(1) \ge c_p^{e_0}-e_0p^{-4\delta} \ge (2k)^{-e_*}-e_*p^{-4\delta}.\] Consequently \(\nu_{\varnothing}(1)\ge Z_*\) once \(e_*p^{-4\delta}\le Z_*\). There are at most \(2^w\) components, each of mass at most \(R_*\). Set \[C_{\mathrm{mass}}:=\frac{2^wR_*}{Z_*},\qquad \alpha_{\mathrm{exc}}:=\frac{(t+1)\delta}{2}>5.\] The normalizing mass \(M_{l,p}(1)\) is at least the leading mass, so \[\eta_{l,p}\le C_{\mathrm{mass}}p^{-\alpha_{\mathrm{exc}}}.\] It is at most \(p^{-4}\) if \(p^{\alpha_{\mathrm{exc}}-4}\ge C_{\mathrm{mass}}\). In either \(\mu_{l,p}\) or \(\mu_{l,p}^{\circ}\) the probability of nonleading components is at most \(C_{\mathrm{mass}}p^{-\delta/2}\). For the latter law, its unnormalized mass after removing the terminal terms still contains \(\nu_{\varnothing}(1)\ge Z_*\); hence the same denominator bound applies without an additional normalization factor. Cycle support and translation invariance in Proposition 12 are exact properties of these components and impose no further threshold.

For the leading component in Lemma 14, the incident graph has \(m\le e_*\) edges. Its list operator error is at most \(C_{\mathrm{gr}}p^{-\delta}\). The split into \(A\ge c_p^m/2\) and its complement in that proof therefore bounds the squared conditional norm by \[Z_0^{-1}C_{\mathrm{gr}}^2 \bigl(2c_p^{-m}+4c_p^{-2m}\bigr)p^{-2\delta}\|f\|_2^2.\] Using \(Z_0\ge Z_*\) and \(c_p^{-m}\le(2k)^{e_*}\) bounds this by \[C_{\mathrm{lead}}p^{-2\delta}\|f\|_2^2, \qquad C_{\mathrm{lead}}:=Z_*^{-1}C_{\mathrm{gr}}^2 \bigl(2(2k)^{e_*}+4(2k)^{2e_*}\bigr).\] The case \(m=0\) already has zero conditional norm. For either mixture, conditioning additionally on the component index and applying Jensen’s inequality bounds its squared conditional norm by \[\bigl(C_{\mathrm{lead}}p^{-2\delta} +C_{\mathrm{mass}}p^{-\delta/2}\bigr)\|f\|_2^2 \le (C_{\mathrm{lead}}+C_{\mathrm{mass}}) p^{-\delta/2}\|f\|_2^2.\] Here the bound in every other component is \(\|f\|_2^2\) because its single-slot marginal is uniform. As \(\delta/2-2\gamma=3\delta/10>0\), the desired squared bound \(p^{-2\gamma}\|f\|_2^2\) follows once \[p^{\delta/2-2\gamma}\ge C_{\mathrm{lead}}+C_{\mathrm{mass}}.\]

Choose an integer \(P>P_1\) large enough that the displayed requirements on \(p\) above hold for every \(p\ge P\), and that \(P^{\gamma/2}\ge3\). This is possible because \(\delta>0\), \(\alpha_{\mathrm{exc}}-4>0\), and \(\delta/2-2\gamma>0\), and all the constants depend only on \(k,L,t\). The product condition in (12) can be included in the same choice: for integer \(P\ge2\), \[\prod_{\substack{p\ge P\\p\ \mathrm{prime}}}(1+p^{-3}) \le \exp\left(\sum_{n\ge P}n^{-3}\right) \le \exp\left(\frac{1}{2(P-1)^2}\right),\] whose last expression tends to one. The covariance proof uses only the exact progression identity and bijections when \(p\nmid D\), so it imposes no further restriction on \(P\). This proves the stated dependence of \(P\). Under Condition [local:unit-condition], Section 2 allows \(P_1,L\) to be chosen from \(k\) and each \(E_p\) from \(k,p\). Also, \(t\) is chosen from \(\delta\), which depends only on \(k\). The product defining \(M_0\) then uses a degree-dependent set of primes and degree-dependent exponents. For a general fixed polynomial, \(M_0\) retains any dependence of the local exponents \(E_p\) as well as that of \(P_1\) and \(L\). ◻

Proof. Across any reflection cut of \(V\), the matrix of the product law is the tensor product of the corresponding prime-law matrices. It is therefore positive semidefinite. To explain its consequence for signed inputs, fix a cut \(V=V_+\sqcup V_-\) and its reflection \(\theta\). Define maps \(\phi_\pm:V\to V_\pm\) that fix the selected half and reflect the other half into it. An assignment of functions is a map \(a:v\mapsto g_v\); write its integral as \(\Lambda(a)\). The two real functions on the plus-half labels that enter the cut bilinear form are \[F(x)=\prod_{v\in V_+}a(v)(x_v), \qquad G(x)=\prod_{v\in V_+}a(\theta v)(x_v).\] Its mixed value is \(\Lambda(a)\), and its two quadratic values are \(\Lambda(a\circ\phi_+)\) and \(\Lambda(a\circ\phi_-)\). Cauchy–Schwarz for the positive semidefinite matrix gives \[ |\Lambda(a)|^2 \le \Lambda(a\circ\phi_+)\Lambda(a\circ\phi_-), \qquad \Lambda(a\circ\phi_\pm)\ge0. \tag{14}\] In particular a constant assignment has nonnegative integral, proving the first assertion of (13).

We next give a finite sequence of fold maps that sends every vertex to one vertex. In one block write \(p_i\) for the position of symbol \(i\) in its permutation word. Folding onto the half where \(i\) precedes \(i+1\) sorts the pair \((p_i,p_{i+1})\), leaving all other entries unchanged. Apply the comparisons \(i=1,\ldots,L-1\) in that order, and repeat this sweep \(L-1\) times. A sweep moves the largest entry to the end. Once the largest entries occupy their final positions, the next sweep leaves them there and places the largest remaining entry immediately before them. Thus every position array becomes \((1,\ldots,L)\). Perform these sweeps in each block. If their fold maps in chronological order are \(\phi_1,\ldots,\phi_m\), then \[\phi_m\circ\cdots\circ\phi_1(v)=v_* \quad\text{for every }v\in V,\] where \(v_*\) has the identity word in every block.

Now fix a finite nonempty list of real functions and let \(M\) be the maximum absolute integral over all assignments from that list, allowing repetition. If \(M>0\), take a maximizing assignment. Each of its folded integrals is nonnegative and at most \(M\), while their product is at least \(M^2\) by (14). Both therefore equal \(M\). Hence either fold preserves maximality. Apply the pullbacks in reverse order, first \(\phi_m\), then \(\phi_{m-1}\), and so on. The final assignment is \(a\circ\phi_m\circ\cdots\circ\phi_1\), which is constant. It follows that \(M\) is a diagonal value. Conversely, constant assignments are included in the maximum. We have proved that the maximum absolute mixed integral equals the largest diagonal value in the list; this is also true when \(M=0\).

For \(\epsilon>0\), apply this conclusion to the list \[\frac{g_v}{(Z_{l,\mathcal P}(g_v)+\epsilon)^{1/d}}, \qquad v\in V.\] Each diagonal value in this list is at most one. Multilinearity gives \(|\mathcal Z_{l,\mathcal P}((g_v)_v)| \le\prod_v(Z_{l,\mathcal P}(g_v)+\epsilon)^{1/d}\). Let \(\epsilon\downarrow0\). This proves the second assertion, including cases where some diagonal values vanish. ◻

Proof. Nonnegativity follows from Proposition 16. Since \(d\) is even, homogeneity of \(Z\) gives absolute homogeneity of its \(d\)-th root. Expanding \(Z(g+h)\) by its slots and applying (13) termwise gives \[Z_{l,\mathcal P}(g+h) \le\sum_{S\subseteq V}\|g\|_{l,\mathcal P}^{|S|} \|h\|_{l,\mathcal P}^{d-|S|} =\bigl(\|g\|_{l,\mathcal P}+\|h\|_{l,\mathcal P}\bigr)^d.\] Taking roots proves the triangle inequality. Putting \(g\) in one slot and 1 in all others in (13) proves mean domination, since the single-slot marginal is uniform and \(Z(1)=1\).

The simultaneous translation invariance of the tuple laws proves that translations are isometries. Let \(G_B\subseteq X_{\mathcal P}\) be the additive subgroup of vectors supported on \(B\). Then \[(\mathbb E_Bg)(z)=\frac1{|G_B|}\sum_{a\in G_B}g(z+a).\] The triangle inequality gives the asserted contraction. Since \(\mathbb E_Bg\) is independent of \(B\), the probability laws at those primes integrate out, proving the equality of the two seminorms. Finally, \(\lambda(D)\) is a unit at each remaining prime, and (10) pushes the product law at \(l\) exactly to that at \(lD\). This proves (15). ◻

Proof. At one prime, let \(\mathfrak m_p\) denote the marginal law of the labels in all slots other than \(v\), and let \[T_p f=\mathbb E_{\rho_p}\bigl(f(z_v)\mid(z_u)_{u\ne v}\bigr).\] This is an operator from uniform \(L^2(\mathbb F_p)\) to \(L^2(\mathfrak m_p)\). It preserves constants. It sends uniform-mean-zero inputs to \(\mathfrak m_p\)-mean-zero outputs, since \(\mathbb E_{\mathfrak m_p}T_pf=\mathbb Ef\). By (11), its norm on the mean-zero subspace is at most \(p^{-\gamma}\).

The product law makes the conditional operator for all primes equal to \(T=\bigotimes_pT_p\): this identity holds first for products of one-prime functions, by independence across primes, and then for every function by linearity. The exact-support \(B\) Fourier space has a mean-zero factor at primes in \(B\) and a constant factor elsewhere. On this space the operator norm is at most \(q_B^{-\gamma}\). Furthermore, the images of two distinct exact-support spaces are orthogonal in \(L^2(\bigotimes_p\mathfrak m_p)\). At a prime in the symmetric difference of the supports, one image factor is constant and the other has mean zero. Their inner product is zero, and tensoring preserves this orthogonality.

Decompose \(g_v=\sum_{q_B>R}g_{v,B}\) by exact supports. Both the domain summands and their images are orthogonal, so \[\|Tg_v\|_2^2 =\sum_{q_B>R}\|Tg_{v,B}\|_2^2 \le\sum_{q_B>R}q_B^{-2\gamma}\|g_{v,B}\|_2^2 \le R^{-2\gamma}\|g_v\|_2^2.\] Condition the original integral on the other slots and apply Cauchy–Schwarz. The remaining factor is bounded by \[\left\|\prod_{u\ne v}g_u(z_u)\right\|_2 \le\prod_{u\ne v} \left(\mathbb E|g_u(z_u)|^{2(d-1)}\right)^{1/(2(d-1))} =\prod_{u\ne v}\|g_u\|_{2(d-1)},\] using ordinary Hölder and the uniform single-slot marginals. This proves (16). If \(\mathcal P\) is empty, the frequency hypothesis forces \(g_v=0\), and the conclusion is immediate. ◻

Proof. The zero function is immediate; otherwise scale so that \(\|f\|_\infty\le1\). Write \(N_I=|I|\), and let \(u\) be the largest cardinality of a support whose prime product is at most \(Q\). We first control conditional sums of squares of the \(f_B\), then their square-function moments. Symmetrization will recover the moment of their sum.

Conditional energy. For \(J\subseteq\mathcal P\) with \(q_J\le Q\), and any fixed coordinate vector \(z_J\), we claim that \[ \sum_{\substack{B\in\mathcal F\\B\supseteq J}} \mathbb E\bigl(|f_B|^2\mid z_J\bigr)\ll4^{|J|}. \tag{18}\] The conditional expectation here averages uniformly over all coordinates outside \(J\). If \(B\supseteq J\), the coefficient of an outside character \(\eta\) of exact support \(B\setminus J\), after fixing \(z_J\), is \[c_J(\eta;z_J) =\frac1{N_I}\sum_{x\in I}f(x)e(-x\eta) \prod_{p\in J}\bigl(p\mathbf 1_{\{x\equiv z_p\pmod p\}}-1\bigr).\] Indeed, the sum over the nonzero frequencies at \(p\) is \(\sum_{a=1}^{p-1}e(a(z_p-x)/p) =p\mathbf 1_{\{x\equiv z_p\pmod p\}}-1\). Different supports \(B\supseteq J\) have disjoint sets of outside characters. Conditional Parseval therefore identifies the left side of (18) with the squared Euclidean norm of this coefficient vector on the union of those character sets.

Expand the product in \(c_J\) over subsets \(J'\subseteq J\). Apart from a sign, the corresponding vector has coordinates \[\frac{q_{J'}}{N_I} \sum_{\substack{x\in I\\x\equiv z_{J'}\pmod{q_{J'}}}} f(x)e(-x\eta).\] The congruence denotes the class determined by the fixed coordinates in \(J'\); for the empty subset it imposes no restriction. Write its progression as \(x=a+q_{J'}n\), with \(n\) in a consecutive interval of length \(N'\). Then \[N'\ge0.9N_I/q_{J'}, \qquad N'q_{J'}/N_I\le1.1.\] In these coordinates the frequencies are \(q_{J'}\eta\), up to constant phases. They remain distinct because their denominators are coprime to \(q_J\). Their denominators are at most \(Q\), so they number at most \(Q^2\) and have pairwise circle distance at least \(Q^{-2}\).

The Gram matrix of these characters in normalized \(L^2\) on the progression has diagonal 1. By geometric summation, every off-diagonal entry has modulus at most \(Q^2/(2N')\). Its operator norm is therefore at most its maximum absolute row sum, which is bounded by \[1+\frac{Q^4}{2N'} \le1+\frac{Q^5}{1.8N_I} \le1+\frac1{18Q}.\] The associated Fourier analysis map has norm squared equal to this Gram-matrix norm. Since \(\|f\|_\infty\le1\) and \(N'q_{J'}/N_I\le1.1\), the displayed coefficient vector has Euclidean norm bounded by an absolute constant. The triangle inequality over the \(2^{|J|}\) subsets \(J'\), followed by squaring, proves (18).

Square-function moments. Set \[S(z)=\left(\sum_{B\in\mathcal F}|f_B(z)|^2\right)^{1/2}.\] Expand \(\mathbb ES^{2r}\) and fix the supports of its first \(r-1\) factors. Their union \(W'\) has size at most \((r-1)u\), and their product depends only on \(z_{W'}\). For the final support \(B\), integration outside \(W'\) leaves exactly \(\mathbb E(|f_B|^2\mid z_J)\), where \(J=B\cap W'\). Every intersection that occurs satisfies \(q_J\le q_B\le Q\). For each fixed intersection \(J\), the sum over supports with \(B\cap W'=J\) is bounded by the larger sum over all \(B\supseteq J\) in (18). There are at most \(2^{(r-1)u}\) possible intersections and each has at most \(u\) elements. Thus \[\mathbb ES^{2r}\le C\,2^{(r-1)u}4^u\mathbb ES^{2r-2}.\] The same argument with \(r=1\) uses \(J=W'=\varnothing\). Iteration gives \[ \|S\|_{2r}\le C_r^{1+u}. \tag{19}\] The overlaps among supports have now been controlled. It remains to bound their sum by this square function with a loss exponential only in \(u\).

Recovering the sum. Fix a support size \(j\ge1\) and color every prime coordinate with one of \(j\) colors. Call a support transversal when it has exactly one prime of each color, and let \(\mathcal T\) be the transversal supports in \(\mathcal F\) for this coloring. Take independent uniform pairs \(X_p^0,X_p^1\) at every prime. The operator \(\operatorname{swap}_p\) interchanges these two copies. Define \[D_B=\prod_{p\in B}(1-\operatorname{swap}_p)f_B(X^0), \qquad B\in\mathcal T.\] Averaging over all second copies leaves \(f_B(X^0)\), since every other term in this expansion has mean zero in at least one copied coordinate. Jensen’s inequality gives \[\left\|\sum_{B\in\mathcal T}f_B\right\|_{2r} \le\left\|\sum_{B\in\mathcal T}D_B\right\|_{L^{2r}(X^0,X^1)}.\] Independent swaps preserve the joint law of the copies and multiply \(D_B\) by the product of the corresponding signs. Thus for independent uniform signs \(\epsilon_p\), indexed by primes, \[\left\|\sum_{B\in\mathcal T}D_B\right\|_{L^{2r}(X^0,X^1)} =\left\|\sum_{B\in\mathcal T}D_B\prod_{p\in B}\epsilon_p \right\|_{L^{2r}(X^0,X^1,\epsilon)}.\]

For completeness, the scalar sign estimate \[\left\|\sum_i a_i\epsilon_i\right\|_{2r} \le C_r\left(\sum_i|a_i|^2\right)^{1/2}\] follows by expanding the even moment. Surviving terms have even multiplicity at every index. For nonnegative coefficients their sum is at most \[\frac{(2r)!}{2^r r!}\left(\sum_i a_i^2\right)^r,\] because \((2m)!\ge2^m m!\). For complex coefficients, take absolute values of the surviving terms in \((\sum_i a_i\epsilon_i)^r(\sum_i\overline{a_i}\epsilon_i)^r\) and use the nonnegative case. Iterating this estimate over the color classes gives, for fixed copies, \[\left\|\sum_{B\in\mathcal T}D_B\prod_{p\in B}\epsilon_p \right\|_{L^{2r}(\epsilon)} \le C_r^j\left(\sum_{B\in\mathcal T}|D_B|^2\right)^{1/2}.\] At each step Minkowski is used in the form \(\| (\sum_i|B_i|^2)^{1/2}\|_{2r} \le(\sum_i\|B_i\|_{2r}^2)^{1/2}\). The signs are indexed by individual primes; transversality ensures that each monomial uses exactly one sign from each color class.

In expanding the swaps in \(D_B\), index its \(2^j\) terms by a choice of copy for each color. For any fixed such choice, made consistently across all supports, the selected prime coordinates have the original product law. The triangle inequality in \(L^{2r}(\ell^2(\mathcal T))\) therefore gives \[\left\|\left(\sum_{B\in\mathcal T}|D_B|^2\right)^{1/2}\right\|_{2r} \le2^j\left\|\left(\sum_{B\in\mathcal T}|f_B|^2\right)^{1/2} \right\|_{2r} \le2^j\|S\|_{2r}.\] Consequently the transversal sum has norm at most \((2C_r)^j\|S\|_{2r}\).

Now color independently and uniformly. A fixed size-\(j\) support is transversal with probability \(j!/j^j\ge e^{-j}\), where here \(e\) is Euler’s number; the inequality follows from \(\log(j!)\ge\int_1^j\log x\,dx\). Thus the full size-\(j\) sum is exactly the coloring average of its transversal sum divided by this probability. Minkowski yields \[\left\|\sum_{\substack{B\in\mathcal F\\|B|=j}}f_B\right\|_{2r} \le(2eC_r)^j\|S\|_{2r}.\] Include the constant term of size zero and sum over \(j\le u\). Together with (19), this bounds the norm of the full sum by \((C'_r)^{1+u}\). Finally, the product of \(u\) distinct primes is at least \((u+1)!\), so \((u+1)!\le Q\). Hence \(u=o(\log Q)\) as \(Q\to\infty\), uniformly in the prime set. For every fixed \(r\) and \(\epsilon>0\), \((C'_r)^{1+u}\le C_{r,\epsilon}Q^\epsilon\). Restoring \(\|f\|_\infty\) proves (17) with all the stated uniformities. ◻

Proof. The specified fiber has uniform probability \(1/q_B\). Its contribution to \(\mathbb E|G|^s\) is \(q_B^{-1}\mathbb E|G(z_B,\cdot)|^s\). Dropping the other nonnegative contributions proves the first assertion. For the last assertion, a character with original exact support \(C\) retains exact support \(C\setminus B\) after specialization. The two cuts therefore arise, before specialization, from the complete-support subfamilies \[\{C\in\mathcal F:q_{C\setminus B}\le H\}, \qquad \{C\in\mathcal F:q_{C\setminus B}>H\}.\] Apply Proposition 19 to these subfamilies and then use the fiber bound. Combining coefficients can cancel frequencies but cannot create any outside the indicated ranges. ◻

Proof. The lift is real and has uniform mean \(|A|/N\). The lower bound follows from Lemma 17. Ordinary Hölder and the uniform single-slot marginals give \[Y_l(N,A)\le \left\|\mathcal L_{[N],\mathcal P_N,Q}\mathbf 1_A\right\|_d^d.\] The integer \(d\) is even. Also \(6\beta<1\), so for all sufficiently large \(N\) one has \(N\ge10Q^6\). Apply (17) with moment order \(d\), choosing its exponent small enough that its \(d\)-th power is at most \(C_\epsilon N^\epsilon\). In the bounded remaining range of \(N\), the lift has a bounded number of coefficients, each of modulus at most one. The probability normalization then supplies a uniform bound independent of \(l\), and enlarging \(C_\epsilon\) finishes the proof. ◻

Proof. Put \(A_l=k!a_l\), \(\theta'=A_l\alpha\), and \(Q_0=\lfloor N^{1-\eta}\rfloor\). Pigeonholing the \(Q_0+1\) points \(0,\theta',\ldots,Q_0\theta'\) into \(Q_0\) equal intervals gives a reduced rational \(a'/q'\) such that \[1\le q'\le Q_0,\qquad \|\theta'-a'/q'\|_{\mathbb R/\mathbb Z}\le\frac1{q'Q_0}\le\frac1{(q')^2}.\] Lifting this approximation through multiplication by \(A_l\) places \(\alpha\) within \(1/(A_lq'Q_0)\) of a rational with denominator at most \(A_lq'\). If \(A_lq'\le N^\eta\), this would place \(\alpha\) in a major arc, since \(Q_0^{-1}\le2N^{-1+\eta}\). Therefore \(A_lq'>N^\eta\), and in particular \[N^{\eta/2}\le q'\le N^{1-\eta},\] using \(A_l\le N^{\eta/8}\).

Choose \(M=\lceil2U\rceil\). Since \(X<2U\), this integer is larger than all the interval lengths under consideration. Also \(U\ge1\), so the absolute comparison \(2U\le M\le2U+1\le3U\) holds. Repeated differencing gives \[ \begin{split} \left|\sum_{m\in I}e\bigl(\alpha F(m)+y'm\bigr)\right|^{2^{k-1}} &\le C_k M^{2^{k-1}-k}\\ &\quad\times\sum_{|b_1|,\ldots,|b_{k-1}|<M} \min\!\left(M,\|\theta'b_1\cdots b_{k-1}\|_{\mathbb R/\mathbb Z}^{-1}\right). \end{split} \tag{28}\] Here the minimum equals \(M\) when the distance is zero. To verify the estimate, square the sum and group pairs by their difference. After taking absolute values, square again and use Cauchy–Schwarz on the increment indices. At each step the inner sum is over an intersection of interval translates, hence an interval. After \(s\) steps the prefactor is \(M^{2^s-s-1}\): its exponent is zero for \(s=1\), and the next step doubles that exponent and adds \(s\). After \(k-1\) steps the phase is linear with coefficient \(k!a_l\alpha b_1\cdots b_{k-1}\), and a geometric sum proves (28). The added linear term contributes only a constant after the first differencing and does not change the estimate. This derivation uses only \(|I|\le M\), so also covers short intervals.

The tuples with a zero increment contribute \(O_k(M^{k-1})\) inside the sum in (28). For the others, group by the absolute product \(m\le M^{k-1}\). For each \(\epsilon>0\) their multiplicity is \(O_{\epsilon,k}(m^\epsilon)\). Indeed, at a prime power \(p^a\) the number of exponent choices is a binomial coefficient of fixed degree in \(a\); it is at most \(p^{a\epsilon}\) for all sufficiently large \(p\). The finitely many smaller primes contribute a bounded extra factor, since a fixed polynomial in \(a\) divided by \(p^{a\epsilon}\) is bounded.

Partition the positive product indices into blocks of \(\lfloor q'/2\rfloor\) consecutive integers. If \(0<b<q'/2\), then \[\|\theta'b\|_{\mathbb R/\mathbb Z} \ge\|a'b/q'\|_{\mathbb R/\mathbb Z}-\frac b{(q')^2} \ge\frac1{2q'}.\] Thus the points in a block have spacing at least \(1/(2q')\) on the circle. Ordering their distances to zero and summing the reciprocal-distance bound gives \(O(M+q'\log(2q'))\) per block. There are \(O(M^{k-1}/q'+1)\) blocks. Dividing (28) by \(M^{2^{k-1}}\), and absorbing the divisor bound and logarithms into \(N^\epsilon\), gives \[C_{\epsilon,k}N^\epsilon \left(\frac1{q'}+\frac1M+\frac{q'}{M^k}\right).\] Here \(M^k\ll N\) and \(M^k\gg N^{1-\eta/8}\). The three terms are bounded respectively by \(N^{-\eta/2}\), \(O(N^{-(1-\eta/8)/k})\), and \(O(N^{-7\eta/8})\). For sufficiently small fixed \(\epsilon\), the displayed expression is \(O(N^{-\eta/4})\). Taking its \(2^{k-1}\)-st root proves (27), because \(\eta/(4\cdot2^{k-1})=5\nu\). ◻

Proof. In a periodic mean we may average at any common multiple of the periods and separate prime-power factors by the Chinese remainder theorem. First consider a prime \(p\mid D\), and abbreviate \(E=E_p\), \(s=s_p\). On the chosen argument class, Section 2 gives \[F(x_p+p^{E-s}z)=F(x_p)+p^E(c_1z+pC(z)),\qquad p\nmid c_1,\] where \(C\) has integer coefficients. The polynomial in parentheses is a bijection modulo every power of \(p\): its difference at two arguments is their difference times a unit. Thus for a local frequency of denominator \(p^j\) with \(j\le E\), the phase is constant at its value on \(r\). This remains true if \(j>E-s\). For \(j>E\), uniform arguments in the restricted class modulo \(p^j\) give uniform \(z\) modulo \(p^{j-E+s}\), hence uniform reduction modulo \(p^{j-E}\). The \(F\)-values are consequently uniform among the lifts of \(r\) modulo \(p^j\), and a primitive character has mean zero. The joint factor from primes dividing \(D\) is therefore \[\chi_D(\alpha_0)= \begin{cases} e(\alpha_Dr),&q(\alpha_D)\mid D,\\ 0,&q(\alpha_D)\nmid D. \end{cases}\]

Put \(q_1=q(\alpha_1)\). If \(u\) has a prime not dividing \(q_1\), its factor \(B_p\) has mean zero independently of all the other factors, so \(A_u=0\). Thus only divisors of \(\operatorname{rad}(q_1)=\prod_{p\mid q_1}p\) contribute. For \(p^j\Vert q_1\), let \(\alpha_p\) be the primitive \(p^j\)-denominator part of \(\alpha_1\) and set \[a_{p,j}=\mathbb E_{m\bmod p^j}e(\alpha_pF(m)),\qquad b_{p,j}=\mathbb E_{m\bmod p^j}B_p(m)e(\alpha_pF(m)).\] The Chinese remainder theorem now expresses the truncated coefficient as \[ c(\alpha_0)=\chi_D(\alpha_0) \sum_{\substack{u\mid\operatorname{rad}(q_1)\\u\le H}} \left(\prod_{\substack{p^j\Vert q_1\\p\mid u}}b_{p,j}\right) \left(\prod_{\substack{p^j\Vert q_1\\p\nmid u}}a_{p,j}\right). \tag{30}\] The mean against \(1+B_p=\psi_p\) is \[a_{p,j}+b_{p,j}= \begin{cases} \Gamma_p(\alpha_p),&j=1,\\ 0,&j\ge2. \end{cases}\] For \(j=1\) this is the defining pushforward of \(\psi_p\). For \(j\ge2\), condition on a regular argument modulo \(p\). Its lifts are mapped uniformly to the lifts of its value, so their primitive character mean vanishes. This holds on each argument class separately, regardless of the size of \(\psi_p\) there.

If \(q_1\le H\), all divisors of \(\operatorname{rad}(q_1)\) occur. Thus (30) factors as \[c(\alpha_0)=\chi_D(\alpha_0) \prod_{p^j\Vert q_1}(a_{p,j}+b_{p,j}).\] For squarefree \(q_1\) this is \(\chi_D(\alpha_0)\Gamma(\alpha_1)\); if any exponent is at least two, it is zero. These are precisely the cases in (26). The empty product, when \(q_1=1\), also gives the required coefficient.

Suppose instead that \(q_1>H\). The model coefficient is then zero. By (4), \(|a_{p,j}|\le p^{-8\delta j}\). For \(j=1\) the preceding identity and (21) give \(|b_{p,1}|\le2p^{-\gamma}\); for \(j\ge2\) the identity gives \(b_{p,j}=-a_{p,j}\), hence again \(|b_{p,j}|\le2p^{-\gamma j}\). Taking absolute values in (30) and dropping the divisor cutoff yields \[\begin{align*} |c(\alpha_0)| &\le\prod_{p^j\Vert q_1}(|a_{p,j}|+|b_{p,j}|)\\ &\le3^{\omega(q_1)}q_1^{-\gamma} \le q_1^{-\gamma/2}<H^{-\gamma/2}. \end{align*}\] Here \(\omega(q_1)\) is the number of distinct prime divisors, and the last inequality before the strict one follows from (12), since every prime of \(q_1\) is at least \(P\). This bound also covers \(\operatorname{rad}(q_1)\le H<q_1\): in that case the full divisor product is present and is actually zero, because some exponent is at least two. If \(q(\alpha_D)\nmid D\), all coefficients vanish already through \(\chi_D(\alpha_0)\). ◻

Proof of Lemma 23. On the minor arcs the two bounds already proved give \[|S_1|\ll N^{-\nu+(5/2)\beta},\qquad |S_0|\ll N^{-\eta+2\beta}.\] On a major arc, the error terms in the two approximations are at most \[O(N^{\eta+4\beta-1}),\qquad O\bigl(N^{2\eta+4\beta-(1-\eta/8)/k}\bigr),\] respectively. Each of these four bounds is \(O(N^{-\beta\gamma/4})=O(H^{-\gamma/2})\) by (20). Indeed \(\nu=100\beta\), \(\eta=1/(100k)\) and \(\beta=\eta/(1000\cdot2^k)\), so the two major-arc exponents are less than \(-0.99\) and \(-0.97/k\), respectively; the minor-arc exponents have still larger margins than required. Lemma 25 and \(|V_N(y)|\le1\) now complete the comparison on the major arcs as well. All thresholds used here are uniform under \(l\le N^\zeta\) and \(D\le N^\beta\). ◻

Proof of Proposition 22. Extend \(J_1,J_2\) by zero to \(\mathbb Z\), and for either kernel define \[\mathcal B_i=\frac1N\sum_{x,n\in\mathbb Z}k_i(n)J_1(x)J_2(x+n).\] Let \(\widetilde J_i(\alpha)=\sum_xJ_i(x)e(-x\alpha)\) denote the counting-measure Fourier transforms. Expanding the characters and integrating imposes \(n+x-y=0\), so \[\mathcal B_i=\frac1N\int_0^1 S_i(\alpha)\widetilde J_1(-\alpha)\widetilde J_2(\alpha)\,d\alpha.\] Cauchy–Schwarz, Plancherel and Lemma 23 give \[|\mathcal B_1-\mathcal B_0| \le\frac{\|S_1-S_0\|_\infty}{N} \|J_1\|_{\ell^2(\mathbb Z)}\|J_2\|_{\ell^2(\mathbb Z)} \ll H^{-\gamma/2},\] because the two counting-measure norms are at most \(\sqrt N\).

Every term in \(\mathcal B_1\) uses a positive shift \(F(m)\) with \(m\in\mathbb N_+\), and hence vanishes by avoidance. Thus \(\mathcal B_1=0\), regardless of the signed weights in (23). For a pair in the supports of \(J_1,J_2\), ordering gives \(1\le y-x\le N-1\) and the residue conditions give \(y-x\equiv r\pmod D\). Both indicators in \(k_0\) are therefore automatic, and \[\begin{align*} \mathcal B_0 &=\frac D{N^2}\sum_{x,y\in[N]}J_1(x)J_2(y) \sum_{\xi\in\Xi}\Gamma(\xi)e(-\xi(y-x))\\ &=D\sum_{\xi\in\Xi}\Gamma(\xi) \widehat{J_1}_{[N]}(-\xi)\widehat{J_2}_{[N]}(\xi). \end{align*}\] This proves (22) with its stated normalization and signs. ◻

Proof. The frequencies contributing to \(\widehat g(\xi)\) have the form \(\xi+\omega\), where \(\omega\) ranges over the characters on \(X_B\) that survive the cutoff. The denominators are coprime, so \(q(\xi+\omega)=q(\xi)q(\omega)\le q(\xi)q_B\le Q\). Consequently every character on \(X_B\) occurs in the specialized coefficient, which equals \[\frac{M_0}{|I|}\sum_{x\in I} \mathbf 1_A(x)\mathbf 1_{\{x\equiv s\pmod{M_0}\}}e(-x\xi) \sum_{\omega\text{ on }X_B}e((b-x)\omega).\] Character orthogonality makes the inner sum \(q_B\) when \(x\equiv b_p\pmod p\) for every \(p\in B\), and zero otherwise. The Chinese remainder theorem now gives (32). ◻

Proof. We first record the scale inequalities. By (2), \(D\le L_D\le D^k\), so \(L_D\ge M_0\ge2\) and \(n_B<N\) for large \(N\). The defining ceiling gives (34). Also \[\begin{gather*} n_B\ge M_0^{-k}N^{1-k\tau}\ge N^{1/2},\\ H\le Q':=n_B^\beta\le Q, \qquad q_BH\le Q. \end{gather*}\] Here \(k\tau<1/2\) and \(\tau<\beta/2\) follow from (20).

By the signed Hölder inequality (13), it is enough to prove the asserted common bound for \(Z_{l,\mathcal S}(g)\), where \(g\) is any one of the specialized inputs. Let \(g_{\le H}\) retain exactly the remaining-coordinate frequencies of denominator at most \(H\). We claim that \[Z_{l,\mathcal S}(g) =Z_{l,\mathcal S}(g_{\le H}) +O(H^{-\gamma}q_BN^{o(1)}).\] To see this, telescope the \(d\) slots and apply (16) to the high-frequency input in each term. The moment estimate (17) is applicable to all the other inputs as well as to that high-frequency piece. Indeed, before specialization, a cut on the remaining denominator retains entire supports \(C\subseteq\mathcal P_N\) according to the condition on \(q_{C\setminus B}\); it never selects only part of the frequencies with a given full support \(C\). Fixing \(B\) costs at most \(q_B^{1/2}\) in the distinguished \(L^2\) norm and \(q_B^{1/[2(d-1)]}\) in each of the other \(d-1\) norms. The total cost is therefore \(q_B\). The original input is bounded by \(M_0\), and \(N\ge10Q^6\) for large \(N\), so (17) supplies the remaining \(N^{o(1)}\) factor. Since \[\frac{\beta\gamma}{2}-\tau-\sigma =\tau\left(500k(d+1)-1-\frac\gamma4\right)>0,\] the claimed error is \(O(N^{-\sigma})\). The denominator cuts are symmetric under negation, so the functions used here are real.

We next express \(g_{\le H}\) as an average of shorter lifts. Let \(a'\pmod D\) be the residue in Lemma 26. For each representative \(j\in[L_D]\) with \(j\equiv a'\pmod D\), set \[c_j=j-L_D,\qquad A'_j=\{x'\in[n_B]:L_Dx'+c_j\in A\}.\] These progressions partition the integers of \([N]\) in the class \(a'\pmod D\): the first positive term is \(j\), and \(n_B\) terms suffice for every such integer. Terms above \(N\) contribute no points to \(A'_j\). Moreover, \(A'_j\) satisfies (31) with \(lD\) in place of \(l\). For if a nonzero difference in \(A'_j-A'_j\) were \(h_{lD}(m)\) with \(m\ge1\), the corresponding difference in \(A-A\) would be \[L_Dh_{lD}(m)=h_l(Dm-i),\qquad 0\le i<D,\] by (2); its argument is positive and its value is nonzero.

Put \(\mathcal H_j=\mathcal L_{[n_B],\mathcal S,H}\mathbf 1_{A'_j}\). Every prime of \(L_D\) divides \(D\), so \(L_D\) is invertible on \(X_{\mathcal S}\) and multiplication by \(L_D\) preserves denominators. Lemma 26, applied using \(q_BH\le Q\), gives the exact identity \[ g_{\le H}(z) =\frac{\Lambda_B}{L_D/D} \sum_{\substack{j\in[L_D]\\j\equiv a'\pmod D}} \mathcal H_j\bigl(L_D^{-1}(z-c_j)\bigr). \tag{35}\] For completeness, the summand on the right has Fourier coefficient \(\widehat{\mathcal H_j}(L_D\xi)e(-c_j\xi)\) at \(\xi\). Since \(\Lambda_B/(L_D/D)=Dn_B/N\), summing these coefficients gives \(D/N\) times the sum over the required original integers, exactly as in (32); both sides have zero coefficients above \(H\).

Under the map \(z\mapsto L_D^{-1}(z-c_j)\), the tuple law for \(l\) pushes forward to the tuple law for \(lD\), by (10) at each remaining prime. There are exactly \(L_D/D\) representatives \(j\). The seminorm triangle inequality applied to (35) therefore yields \[Z_{l,\mathcal S}(g_{\le H}) \le\Lambda_B^d\max_j Z_{lD,\mathcal S}(\mathcal H_j).\] This is where the exact change of tuple laws preserves the leading constant in the comparison.

It remains to replace the cutoff \(H\) on the shorter interval by its natural cutoff \(Q'\). Equations (16) and (17), now on \([n_B]\), bound the change of the diagonal by \(O(H^{-\gamma}N^{o(1)})=O(N^{-\sigma})\). Their interval hypothesis holds because \(6\beta<1\), and hence \(n_B\ge10(Q')^6\) for large \(N\). The extended lift only uses primes in \(\mathcal P_{n_B}\). It is obtained from \(\mathcal L_{[n_B],\mathcal P_{n_B},Q'}\mathbf 1_{A'_j}\) by averaging the coordinates in \(B\cap\mathcal P_{n_B}\) and then viewing the result as independent of the additional coordinates in \(\mathcal S\). The contraction under coordinate averaging from Lemma 17 gives \[Z_{lD,\mathcal S}(\mathcal H_j) \le Y_{lD}(n_B,A'_j)+O(N^{-\sigma}) \le\mathcal Y_{lD}(n_B)+O(N^{-\sigma}).\] Combining these estimates with (34) proves the diagonal bound. Applying (13) to the mixed inputs proves (33): the product of \(d\)th roots of quantities bounded by the same nonnegative upper bound is at most that upper bound. ◻

Proof. At every prime-power factor of \(M_0\), the construction in Section 2 supplies a length-\(L\) sequence of admissible residues whose sum is zero. The Chinese remainder theorem combines them into such a sequence modulo \(M_0\). Fix one sequence. There are \(M_0\) choices of the initial cycle residue; the other cycle residues are then determined, and the zero-sum condition closes the cycle. Its \(L\) vertices are distinct. The remaining \(d-L\) residues can be arbitrary, giving at least \(M_0^{d-L+1}\) good assignments out of \(M_0^d\). ◻

Proof. We may replace each \(\mu_{l,p}-\varepsilon_{l,p}\) by its normalized law \(\mu_{l,p}^{\circ}\): passing back multiplies the integral by \(\prod_{p\in\mathcal P_N}(1-\eta_{l,p})\le1\). We first locate an ordered pair of intervals along the cycle. We then truncate the other inputs and condition on their prime supports; the remaining pair integral will have exactly the multipliers appearing in Proposition 22.

Partition \([N]\) into \(\lfloor N^\tau\rfloor\) consecutive intervals whose lengths differ by at most one. Each \(G_s\) is the average of the interval lifts \(\mathcal L_{I,\mathcal P_N,Q}f_s\) with weights \(|I|/N\). Thus we may average independently over an interval assignment to every slot. These intervals have length \(\gg N^{1-\tau}\), so the moment estimate (17) applies: indeed \(6\beta+\tau<7\beta<1\) by (20). For any interval assignment, ordinary Hölder, uniform single-slot marginals and (17) bound the integral in absolute value by \(N^{o(1)}\). The probability that all \(L\) cycle slots choose the same interval is \[\sum_I\left(\frac{|I|}{N}\right)^L \le\left(\max_I\frac{|I|}{N}\right)^{L-1} \ll N^{-(L-1)\tau}.\] Their contribution is \(O(N^{-\sigma})\), since \((L-1)\tau>\sigma=\gamma\tau/4\).

In every other assignment, some directed cycle edge has its first interval \(I_1\) strictly before its second interval \(I_2\). Otherwise the interval indices would be nonincreasing all the way around the cycle, and therefore constant. Fix such an edge for the current assignment. The constants below are uniform in this choice.

Set \(T=N^\tau\) and truncate every interval lift except the two selected ones to denominators at most \(T\). Telescoping the other slots, (16) and (17) give total error \(O(T^{-\gamma}N^{o(1)})=O(N^{-\sigma})\), because \(\gamma\tau>\sigma\). Expand the truncated inputs into characters. Each coefficient has modulus at most \(M_0\), and there are at most \(T^2\) rational frequencies of denominator at most \(T\). Thus the sum of the absolute coefficients in the resulting expansion is \(O(T^{2(d-2)})\).

Fix one term, let \(B\) be the union of the supports of its \(d-2\) characters, and put \[R=q_B\le T^{d-2},\qquad \mathcal S=\mathcal P_N\setminus B.\] Condition on all tuple labels at every prime in \(B\). The expanded characters become constants of modulus one. By independence between primes, the remaining tuple labels still have law \(\bigotimes_{p\in\mathcal S}\mu_{l,p}^{\circ}\). Let \(g_1,g_2\) denote the two selected interval lifts after specialization at this conditioning. They are functions on \(X_{\mathcal S}\). For the chosen directed edge, let \(\Gamma\) be the increment multiplier defined in Section 5. The remaining pair integral is exactly \[ \sum_{\xi\text{ on }X_{\mathcal S}} \widehat g_1(-\xi)\widehat g_2(\xi)\Gamma(\xi). \tag{37}\] Indeed, simultaneous translation of the two labels kills all pairs of characters whose frequencies do not sum to zero. The surviving pair with frequencies \(-\xi,\xi\) has value \(e(\xi(z_2-z_1))\), whose expectation is \(\Gamma(\xi)\).

For the terms with \(q(\xi)>H\), (21) and Parseval give \[\left|\sum_{q(\xi)>H} \widehat g_1(-\xi)\widehat g_2(\xi)\Gamma(\xi)\right| \le H^{-\gamma}\|g_1\|_2\|g_2\|_2 \ll H^{-\gamma}R N^{o(1)}.\] The last inequality uses the interval moment bound and the cost \(R^{1/2}\) of each specialization in \(L^2\).

We apply the kernel estimate to the remaining terms. Put \(D=M_0R\). The parameter choice gives \[(d-2)\tau<\frac\beta2, \qquad HR\le Q, \qquad D\le N^\beta\] for large \(N\); the first inequality follows directly from \(\tau=\beta\gamma/[1000k(d+1)]\). By Lemma 26, their Fourier coefficient formulas for \(q(\xi)\le H\) restrict the underlying integers to classes \(a'_1,a'_2\pmod D\). The difference \(a'_2-a'_1\) is admissible modulo \(M_0\) by goodness. It is admissible at each prime in \(B\) because every conditioning of positive probability is in the cycle support of the nonexceptional law. It is therefore admissible modulo \(D\).

The remaining frequencies of denominator at most \(H\) are precisely the set \(\Xi\) in Proposition 22. Indeed, their denominators are squarefree and coprime to \(D\). Conversely, a squarefree denominator at most \(H\) and coprime to \(D\) has no prime below \(P\), since all those primes divide \(M_0\), and no prime in \(B\); every prime divisor is at most \(H\le Q\).

Define functions on \([N]\) by \[J_i(x)=\mathbf 1_A(x)\mathbf 1_{I_i}(x) \mathbf 1_{\{x\equiv a'_i\pmod D\}},\qquad i=1,2.\] They are bounded by one and have ordered supports. Every difference from the first support to the second is positive, so a zero polynomial value cannot be such a difference; (31) excludes all nonzero values of \(h_l(\mathbb N_+)\). Thus all hypotheses of Proposition 22 hold at scale \(N\). The Fourier normalizations in (32) are \[\widehat g_i(\xi) =\frac{DN}{|I_i|}\widehat{J_i}_{[N]}(\xi) \qquad(q(\xi)\le H).\] Since (22) has a factor \(D\) outside its absolute value, the low-denominator part of (37) is at most \[C H^{-\gamma/2}D\frac{N^2}{|I_1||I_2|} \ll H^{-\gamma/2}D N^{2\tau}.\]

Both parts of (37) are consequently bounded, uniformly in the conditioning and in the term of the character expansion, by \[O\!\left(N^{-\beta\gamma/4+(d+2)\tau+o(1)}\right).\] Averaging the conditioning costs no factor, and summing the character expansion costs at most \(O(N^{2d\tau})\). The final exponent satisfies \[\frac{\beta\gamma}{4}-(3d+2)\tau-\sigma =\tau\left(250k(d+1)-(3d+2)-\frac\gamma4\right)>0.\] Thus the resulting error is \(O(N^{-\sigma})\). The same-interval contribution and the truncation error already satisfy this bound. Finally, the interval assignments were averaged with probabilities summing to one, so the uniform estimate proves (36). ◻