Exact realization of rational signed-relation spaces

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Theorem — proved

Let $V\leq\mathbb Q^n$ be a rational subspace containing none of

$$e_i,\quad e_i-e_j,\quad e_i+e_j\qquad(i\ne j). \tag{1}$$

Let $\mathcal H(V)$ be the hypergraph on $[n]$ whose edges are the supports of nonzero vectors in $V\cap\{-1,0,1\}^n$. There are distinct positive integers $a_1,\ldots,a_n$ whose signed-relation hypergraph is exactly $\mathcal H(V)$.

Proof

In the quotient $W=\mathbb Q^n/V$, put $v_i=e_i+V$. Consider the finite set of all nonzero vectors of the form $\sum_i\epsilon_i v_i$, $\epsilon_i\in\{-1,0,1\}$. Choose a rational linear functional $\ell:W\to\mathbb Q$ which is nonzero on each of these vectors. Such a functional exists: the excluded functionals form finitely many proper hyperplanes over the infinite field $\mathbb Q$. Multiplying $\ell$ by a common positive denominator gives integers $b_i=\ell(v_i)$, and

$$\sum_i\epsilon_i b_i=0\quad\Longleftrightarrow\quad (\epsilon_i)_i\in V \qquad(\epsilon_i\in\{-1,0,1\}). \tag{2}$$

By (1), all $b_i$ are nonzero and $b_i\ne\pm b_j$ for distinct indices. Set $a_i=|b_i|$ and $\sigma_i=\operatorname{sgn}(b_i)$. Then $\sum_i\epsilon_i a_i=0$ if and only if $(\sigma_i\epsilon_i)_i\in V$. The diagonal sign change is a bijection of the ternary cube preserving every support, proving the theorem.

Fully explicit finite projection

Represent the $v_i$ as integer columns in $\mathbb Z^d$, after clearing denominators. Let $M$ bound all absolute entries and choose an integer $B>nM+1$. The map

$$L(x_0,\ldots,x_{d-1})=\sum_{j=0}^{d-1}x_jB^j$$

is nonzero on every nonzero signed sum of columns. Indeed, if its highest nonzero coordinate is $s$, the leading term has absolute value at least $B^s$, while all lower terms together have absolute value at most $nM(B^s-1)/(B-1)<B^s$. Taking absolute values as above produces the positive integer realization. This is a finite encoding, not an infinite bounded-base encoding.

Signed closure is the essential difficulty

If $\mathcal R\subseteq\{-1,0,1\}^n$ is a list of planted relations, every realization necessarily also satisfies every vector in

$$\operatorname{span}_{\mathbb Q}(\mathcal R)\cap\{-1,0,1\}^n.$$

Conversely, if that rational span obeys (1), the theorem realizes exactly this closure, up to signs. Thus an abstract high-chromatic hypergraph is not enough: one must control the ternary vectors in the entire rational span of its signed edges, not only the prescribed edges.

For any rational $V$, replacing it by $V'=\operatorname{span}_{\mathbb Q}(V\cap\{-1,0,1\}^n)$ leaves its ternary intersection unchanged: $V'\subseteq V$ gives one inclusion and the definition gives the other. A search may therefore restrict to spaces generated by their signed relations.

Consequence for E0774

It suffices to construct spaces $V_t$ obeying (1) such that $\mathcal H(V_t)$ has one uniform hereditary independence ratio and unbounded chromatic number. This is an exact alternative formulation of the finite-block obligation, not a construction satisfying it.