Integer graph and hard-kernel reductions for moment theta mass

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The fixed-scale extraction/cancellation target below is obtained from full semistability and the weak Gaussian bound. The graph and puncturing identities and their countermodel retain their exact scopes. The growing-modulus reduction is an additional equivalent formulation, not an unresolved extraction prerequisite.

This page proves an exact version of the low-rank reduction. It does not itself prove the constant-scale theta estimate required by the finite extraction criterion; that estimate comes from the stable-lattice bridge.

1. The explicit integer graph

Let 0<=k<d, n=2^d, r=sum_(j<=k)binom(d,j), m=n-r. Identify cube points with subsets of [d]. Let P={B:|B|<=k} and S={T:|T|>k}. The moment matrix A has entries A_(R,T)=1_(R subset T), |R|<=k. Its square submatrix U on P is the Boolean zeta matrix, with determinant one in cardinality order and inverse

$$(U^{-1})_{B,R}=(-1)^{|R|-|B|}1_{B\subseteq R}.$$

Put C=-U^{-1}A_S, an integer r-by-m matrix. The relation lattice K, in coordinate order (S,P), is exactly

$$K=\{(z,Cz):z\in\mathbb Z^m\}. \tag{1}$$

Truncated binomial inversion gives its explicit entries:

$$C_{B,T}= \begin{cases} (-1)^{k-|B|+1}\binom{|T|-|B|-1}{k-|B|},&B\subseteq T,\\ 0,&B\not\subseteq T. \end{cases} \tag{2}$$

Indeed, the relevant sum in -U^{-1}A_S is -sum_(j=0)^(k-|B|)(-1)^j binom(|T|-|B|,j), giving (2).

The graph basis has Gram matrix Q=I_m+C^T C. The projected coordinate basis {P_W e_T:T in S} of Lambda=P_W Z^n is dual to it: its pairing with (e_U,C_U) is delta_(T,U). Therefore its Gram matrix is exactly

$$G=Q^{-1}=I_m-C^T(I_r+CC^T)^{-1}C. \tag{3}$$

Thus the low-rank contraction is an identity, not a heuristic model. For theta_L(s)=sum_(z in L)exp(-pi s^2||z||^2), (1) also gives

$$\vartheta_K(s)=\vartheta_{\mathbb Z}(s)^m \mathbb E e^{-\pi s^2\|CZ\|^2}, \tag{4}$$

where the coordinates of Z are independent with probabilities P(Z_i=j)=exp(-pi s^2 j^2)/theta_Z(s). The missing extraction estimate is a genuinely exponential suppression in (4); dropping the expectation gives no such conclusion.

2. Remove the graph penalty without changing limiting free energy

Let H=ker_Z C, viewed in R^m. For every fixed s>0,

$$\boxed{\vartheta_H(s)\leq\vartheta_K(s) \leq\vartheta_{\mathbb Z}(s)^r\vartheta_H(s).} \tag{5}$$

The lower bound uses (z,0) for z in H. For the upper bound, group (1) by y=Cz. Any nonempty fiber is z_y+H. Decompose z_y into its orthogonal components parallel and perpendicular to span H. The perpendicular component only decreases Gaussian mass. The Gaussian mass of a translate of H within its span is at most theta_H(s): the Fourier expansion of its periodization has nonnegative Gaussian coefficients, each multiplied by a unit phase under translation, so the real sum is largest at zero. Consequently each fiber has mass at most theta_H(s). Summing the remaining factor exp(-pi s^2||y||^2) over y in im C, and then over all Z^r, proves (5). This argument also covers the zero-rank kernel.

In particular, whenever r=o(n),

$$0\leq\frac1n\log\vartheta_K(s) -\frac1n\log\vartheta_H(s) \leq\frac rn\log\vartheta_{\mathbb Z}(s)=o(1). \tag{6}$$

The sufficient strict fixed-scale extraction inequality transfers between K and H with an arbitrarily small loss in its positive constant. This reduces the analytic target; it does not establish it.

3. Surjectivity and determinant retention

Assume 2k<d. The high-weight set P'={T:|T|>=d-k} has r points and is disjoint from P. Complementing all variables acts on the degree-at-most-k monomials by an integral triangular matrix with diagonal entries plus or minus one, and maps P to P'. Hence A_(P') is unimodular, as is the corresponding square submatrix -U^{-1}A_(P') of C. In particular C maps Z^m onto Z^r, and rank H=m-r=n-2r.

The primitive-kernel determinant formula gives

$$\det H=\sqrt{\det(CC^T)},\qquad \det K=\sqrt{\det(I_r+CC^T)}. \tag{7}$$

For completeness, extend an integer basis of H by integer lifts of the standard basis of Z^r under C. They form a unimodular basis of Z^m. Projecting its last r vectors orthogonally off span H gives a volume of 1/det H. The map C on this perpendicular space has volume factor sqrt(det(CC^T)) and sends those projected vectors to the standard basis. This proves the first identity; the second follows from the graph Gram matrix and det(I+C^TC)=det(I+CC^T).

Put D=(d-k)binom(d,k) and s_0=2^(D/(2m)), so det K=s_0^m. The previously proved full semistability applies to H embedded as (H,0) in K, and (7) gives the upper bound:

$$s_0^{m-r}\leq\det H\leq s_0^m. \tag{8}$$

Equivalently,

$$0\leq\log\det K-\log\det H\leq r\log s_0, \qquad \det(I_r+(CC^T)^{-1})\leq s_0^{2r}. \tag{9}$$

At the critical boundary r=o(n) and s_0 converges to a constant greater than one. Thus K and the hard kernel H have the same limiting normalized logarithmic determinant as well as the same fixed-scale theta free energy. The operation preserves the exponential determinant scale. It leaves an analytic theta estimate to be supplied separately, rather than reducing it to rank alone.

4. General coordinate-puncturing theorem

The preservation phenomenon is not specific to a unimodular pivot choice. Let L<=Z^N be any lattice of rank M, let J be a coordinate set, and put L_0={x in L:x_J=0} and q=rank(proj_J L). Then

$$\vartheta_{L_0}(s)\leq\vartheta_L(s) \leq\vartheta_{\mathbb Z}(s)^q\vartheta_{L_0}(s), \qquad \det L_0\leq\det L. \tag{10}$$

For theta, the same Gaussian fiber proof first gives a factor theta_(proj_J L)(s). Choose q coordinate positions giving an injective projection on span(proj_J L). Projecting onto them decreases norms and maps lattice points injectively into Z^q, proving the displayed factor.

For determinants, L_0 is primitive in L. Its orthogonal quotient has covolume det L/det L_0 and maps injectively and contractively under proj_J onto the integral lattice proj_J L. The latter has covolume at least one, because its squared covolume is a positive integer Gram determinant. Thus det L/det L_0>=1, as required.

If L is semistable and det L=s_0^M, its nonzero-rank sublattice L_0 also satisfies det L_0>=s_0^(M-q); the same inequality holds with determinant one at rank zero. In particular, for q=o(N), M=N-o(N), and bounded log s_0, any such puncturing preserves normalized log-theta and logdet up to o(1). It subsumes any number of successive puncturings whose total rank loss is o(N).

This also identifies a limitation precisely. When s_0 tends to a fixed constant, the normalized covolume of the surviving lattice lies between s_0 and s_0^(M/(M-q)), hence still tends to that same constant. Puncturing alone cannot raise it to the growing determinant scale of the existing reverse-Minkowski coloring test. Any useful further argument must exploit additional structure of the surviving lattice.

No finite computation or external theorem is needed for (1)--(10). The graph formulas, affine-class theorem and puncturing argument are author-recorded; separate reviewers checked the first two. No external review is claimed.

5. Why full sublattice stability still matters

There is an explicit countermodel to dropping that hypothesis from the remaining analytic problem. Fix 0<c<log(sqrt 2), let m>=3 and M_m=max(1,floor(exp(cm)/sqrt m)), and take the one-row integer matrix C_m=(1,M_m,...,M_m). Its graph lattice has normalized determinant tending to exp(c)>1, and C_m is surjective. The Gram contraction has rank one and largest diagonal contraction at most 1/(m-1) for all sufficiently large m. Nevertheless, for every fixed s>0,

$$\frac1m\log\vartheta_{\operatorname{graph}C_m}(s) \longrightarrow\log\vartheta_{\mathbb Z}(s). \tag{11}$$

The upper bound drops the graph penalty. For the lower bound use its hard-kernel vectors with first coordinate zero and the other m-1 coordinates summing to zero. Their Gaussian mass is theta_Z(s)^(m-1) times P(Z_1+...+Z_(m-1)=0), for the fixed-scale integer Gaussians in (4). This probability is at least c_s/sqrt m. For an even number of terms, split into equal halves and apply Cauchy--Schwarz to the convolution at zero, restricting the half-sum to an interval of length O_s(sqrt m) whose probability is at least 3/4 by Chebyshev. For an odd number, additionally prescribe one Gaussian to be zero. The polynomial loss proves (11).

Thus the extraction target's limiting left side would be log(s theta_Z(s))=log theta_Z(1/s)>0, by one-dimensional Gaussian Poisson summation. Low rank, surjectivity, and small individual diagonal contractions do not establish the needed negative value.

This is not a counterexample to a full-semistability bridge. The same hard kernel contains a rank-(m-2) root lattice of determinant sqrt(m-1), whose normalized covolume tends to one. It violates full semistability while the graph's shortest nonzero vector has length sqrt 2>exp(c). In particular, checking only short single vectors would also miss the obstruction. The actual moment lattice does have full semistability; the stable-lattice bridge is how it enters the fixed-scale bound.