Parity fibers and the rank-one odd obstruction

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The parity-fiber argument improves the proved rank-defect threshold from two to three. It also isolates a necessary condition on a least-dimensional counterexample to the stronger odd-slice bound. It does not settle the unrestricted target.

A parity-prescribed integer realization

Let $L$ be a free abelian group of finite rank, $X\subset L$ finite, and $f:L\to\mathbb R$ a homomorphism positive on $X$. Suppose $f$ preserves exactly the zero ternary relations on $X$, and let $\chi:L\to\mathbb F_2$ be a character with $\chi(x)=1$ for $x\in X$. Then $X$ has a positive odd integer realization preserving all ternary relations and distinctness.

Choose a basis $b_1,\ldots,b_r$ of $L$. For large $t>0$, choose integers $m_i\equiv\chi(b_i)\pmod2$ with $\lvert m_i-tf(b_i)\rvert\le1$, and let the integer functional $\ell_t$ send $b_i$ to $m_i$. On the finite family of ternary combinations of $X$, the error $\ell_t-tf$ is bounded independently of $t$. Thus all nonzero combinations retain their signs when $t$ is sufficiently large; zero combinations remain zero. The prescribed parity makes all $\ell_t(x)$ odd. Positivity and distinctness follow from the one- and two-term combinations.

If the supplied positive functional kills some nonzero ternary combinations, first perturb it within the positive cone to avoid the finitely many corresponding proper hyperplanes. This realizes the relations of the lattice configuration, rather than those additional relations of the original functional.

The fiber inequalities

Let $P\subset L$, $L=\langle P\rangle_{\mathbb Z}$, have rank $r$, and suppose a positive real evaluation preserves all ternary relations on $P$. Put $d=d(P)$. Let

$$F(k)=\max\{|O|:O\subset\mathbb Z_{>0}\text{ odd},\ d(O)\le k\}, \quad G(k)=\max\{|Q|:Q\subset\mathbb R_{>0},\ d(Q)\le k\},$$

with $G(0)=0$. The elementary ternary-span bound makes these maxima finite. Then

$$|P|\le(2^r-1)F(d-r+1)+G(d-r). \tag{1}$$

If a character $\lambda:L\to\mathbb F_2$ equals one on every element of $P$, then

$$|P|\le2^{r-1}F(d-r+1). \tag{2}$$

To prove this, partition $P$ into fibers $P_v$ modulo $2L$. The occurring residues span $L/2L\cong\mathbb F_2^r$. For $v\ne0$, extend $v$ to a basis using $r-1$ occurring residues, represented by $b_1,\ldots,b_{r-1}\in P$. Any dissociated subset $B\subseteq P_v$ remains dissociated after adjoining these representatives. Indeed, a ternary relation reduced modulo $2L$ forces every coefficient on the $b_i$ to vanish, and then dissociation of $B$ applies. Therefore

$$d(P_v)\le d-r+1\quad(v\ne0).$$

Adjoining a full residue basis to a dissociated subset of $P_0$ similarly gives $d(P_0)\le d-r$.

For each nonzero $v$, choose a character taking value one on $v$. The realization lemma shows that $|P_v|\le F(d-r+1)$; also $|P_0|\le G(d-r)$. There are $2^r-1$ nonzero residues, proving (1). Under the hypothesis of (2), only the $2^{r-1}$ residues with $\lambda(v)=1$ can occur.

Application to intrinsic rank

Use the universal rational lift from [[research/erdos_963/corank_one|the intrinsic-rank note]], and let $L$ be its generated integer lattice. Its rank is the intrinsic ternary-relation rank $\rho$. The positive evaluation preserves all ternary relations, so (1) applies. For an original positive odd integer set, its evaluation modulo two gives the character required for (2).

The established small cases in [[research/erdos_963/proof_sketch|the proof sketch]] give

$$F(k)\le2^{k-1}\quad(1\le k\le4),\qquad G(k)\le2^k-1\quad(0\le k\le3).$$

Consequently

$$\boxed{\rho\ge d(P)-3\ \Longrightarrow\ |P|\le2^{d(P)}-1.} \tag{3}$$

For positive odd sets under the same rank hypothesis,

$$\boxed{|P|\le2^{d(P)-1}.} \tag{4}$$

Indeed, put $t=d-\rho\le3$ in (1) to obtain $(2^\rho-1)2^t+(2^t-1)=2^d-1$; (2) gives (4).

The short-relation obstruction to odd induction

Suppose $d$ is the least dimension for which the stronger odd bound fails, and let

$$O\subset\mathbb Z_{>0}\text{ be odd},\qquad d(O)=d,\qquad |O|>2^{d-1}.$$

Let $R\subseteq\mathbb Z^O$ be generated by all genuine zero ternary relations supported on at most $d+1$ coordinates, and put

$$H=\mathbb Z^O/R,\qquad s=\dim_{\mathbb F_2}(H/2H).$$

Then

$$\boxed{s=1.} \tag{5}$$

The original integer evaluation descends to $H$. Modulo two it gives a character $\lambda$ taking value one on every coordinate column, so $s\ge1$. If $s\ge2$, choose a character $\chi$ different from both zero and $\lambda$. Its values on the generating columns are not constant. Let $O_0,O_1$ be the two nonempty color classes.

Both classes have dimension at most $d-1$. If $O_0$ contained a dissociated $d$-subset, adjoin an element of $O_1$. A dependence on these $d+1$ original elements belongs to the defining relations of $H$. Applying $\chi$ forces the coefficient of the adjoined element to vanish, contradicting dissociation. For a $d$-subset of $O_1$, use $\chi+\lambda$ instead.

By minimality of $d$, each color class has at most $2^{d-2}$ elements. Their union cannot be $O$, a contradiction proving (5).

Thus the supports of the short relations span the full even-weight hyperplane in $\mathbb F_2^O$. Select $|O|-1$ independent rows modulo two. Their integer relation matrix has rank $|O|-1$, with an odd maximal minor, and annihilates the original weight vector. Its rational row span is therefore the entire rational relation space of $O$. In particular, the rational universal lift imposing only these short relations already has rank one.

The short relations cannot all be balanced: otherwise both the all-ones vector and the distinct original weights would lie in the rational kernel, contradicting nullity one.

A restriction on the common difference divisor

Normalize the odd weights to have greatest common divisor one, and put

$$h=\gcd\{x-y:x,y\in O\}.$$

Then $h$ is even, and

$$h\le d-1. \tag{6}$$

Indeed, every weight is congruent to a fixed unit modulo $h$, so the coefficient sum of every integer relation is divisible by $h$. A nonbalanced short relation exists by the preceding paragraph. Positivity requires both signs; hence its nonzero coefficient sum has absolute value at most $d-1$, proving (6).

More precisely, some nonbalanced short relation has coefficient sum of (2)-adic valuation exactly $v_2(h)$. Let $K$ be the full integer kernel of the primitive weight vector. The short-relation lattice $R$ has odd index in $K$: its rows span $K/2K$, since they span the even-weight hyperplane and $K$ is a primitive corank-one lattice. The coefficient-sum map $\sigma:K\to\mathbb Z$ has image $h\mathbb Z$. For the reverse inclusion in this assertion, prescribe $\sum_i z_i=h$ and eliminate the last variable. The equation $\sum_i x_i z_i=0$ then becomes

$$\sum_{i<n}(x_i-x_n)z_i=-x_nh,$$

which is soluble in integers by the definition of $h$. The image $\sigma(R)$ has odd index in $\sigma(K)$, as it is a quotient of $K/R$. Its generator therefore has valuation $v_2(h)$. The image is generated by the coefficient sums of the short relations, so at least one of those generators has exactly this valuation.

This rules out an excessively fine common congruence class in a least-dimensional odd counterexample, but does not exclude the typical case $h=2$.

These restrictions concern a least-dimensional failure of the strong odd bound. They do not, without further odd-bound hypotheses, prove that a least-dimensional failure of the ordinary target has intrinsic rank one.

Remaining gap

No argument is known here for a rank-one short-relation presentation whose supports span the entire even-weight binary hyperplane. In this case every nonconstant binary coloring is crossed oddly by some short relation; the two-color induction above cannot be applied. Positivity, distinctness, and this binary condition have not been shown to force a dissociated $(d+1)$-subset, nor has a violating odd set been found.

Alternative-representation generation is false

The existing seven-point example

$$O=\{3,15,31,33,35,41,55\}$$

refutes the proposed generation reduction for every choice of maximum dissociated $D$. Every five-subset is maximum, and every representation of either omitted element on $D$ has full support. It is unique: subtracting two full-support representations and dividing by two would give a nonzero ternary relation on $D$. Thus differences of alternative representations generate zero.

Nevertheless the seven displayed six-term relations in the cited note have coefficient matrix congruent modulo two to $I+J$ of order seven. Its kernel over $\mathbb F_2$ consists exactly of the constant vectors, so it has rank six. The actual weight vector annihilates the integer matrix; its rational rank is therefore exactly six, and the intrinsic rank of $O$ is one.

For example, on $D=\{3,15,31,33,35\}$ the two unique outside representations are

$$41=-3+15+31+33-35,\qquad 55=3+15-31+33+35.$$

Keeping these two equations while deforming $D$ does not preserve the other short relations. This is a counterexample to the auxiliary generation assertion, not to the logarithmic bound.

A further restriction in dimension five

The extended unit-congruence theorem proves $|O|\le16$ whenever $d(O)\le5$ and $O$ lies in one odd class modulo four. Consequently a primitive least-dimensional failure of the strong odd bound with $d=5$ has

$$\gcd\{x-y:x,y\in O\}=2.$$

Indeed, (6) makes this positive even divisor at most four, and the new theorem excludes four. The case of common difference divisor two remains unresolved.