A sufficient antipodal-lift formulation and its global gap

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This is a new sufficient formulation, not a proof of Erdős 963. It uses a weaker requirement than Boolean lifting: no integrality or Boolean coordinates are required. Antipodality is not claimed to follow from lattice emptiness. Its global existence assertion remains unproved.

Let $P=\{p_1,\ldots,p_n\}$ be distinct positive reals, set $p_0=0$, and put

$$H=\{u\in\mathbb R^{n+1}:\sum_i u_i=0,\ \sum_i u_i p_i=0\}.$$

For $K\subseteq H$, project the vertices of the standard affine $n$-simplex modulo $K$. The resulting configuration $X_K$ has affine dimension $n-\dim K$, and the prescribed heights $p_i$ descend to an affine functional on it.

Exact criterion

A finite configuration is antipodal if every pair of distinct points lies on two distinct parallel supporting hyperplanes of its convex hull. The quotient $X_K$ is antipodal, with all its labeled points distinct, if and only if

$$\boxed{\max_i u_i-\min_i u_i\le\tfrac12\|u\|_1 \quad\text{for every }u\in K.} \tag{1}$$

For a fixed pair $i,j$, a normalized supporting functional is a vector $y\in K^\perp$ satisfying

$$y_i=1,\quad y_j=-1,\quad |y_\ell|\le1\quad(\ell\ne i,j).$$

The image of the box of remaining coordinates in $K^*$ is compact and convex. The elementary separating-hyperplane criterion for membership in that image gives precisely

$$|u_i-u_j|\le\sum_{\ell\ne i,j}|u_\ell|\quad(u\in K). \tag{2}$$

For a nonzero balanced vector $u$, write $M=\max u_i>0$ and $m=\min u_i<0$. Applying (2) at these two coordinates yields $2(M-m)\le\|u\|_1$. Conversely, for arbitrary $i,j$,

$$|u_i-u_j|+|u_i|+|u_j|\le2(M-m),$$

so (1) implies (2). This proves the criterion. In particular, no vector $e_i-e_j$ can belong to an admissible $K$, which also ensures distinct quotient points.

Every individual ordinary ternary relation gives an admissible line in $H$. Suppose it has $a$ positive and $b$ negative coefficients on $P$, and augment it by $u_0=b-a$. Positivity forces $a,b\ge1$. If $a=b$, distinctness excludes $a=b=1$, so its coordinate range is $2\le a=\|u\|_1/2$. If $a>b$, its range is $a-b+1\le a=\|u\|_1/2$; reversal handles $b>a$. This proves the claim for the whole line by homogeneity.

Why a global lift would prove the target

An antipodal set $X$ of affine dimension $r$ has at most $2^r$ points. Here is the full volume argument. Let $Q=\operatorname{conv}X$ in its affine span. Each homothet $(x+Q)/2$, $x\in X$, lies in $Q$. Their relative interiors are pairwise disjoint. Otherwise, for distinct $x,y$ there would be $u,v\in\operatorname{relint}Q$ with $x+u=y+v$. A functional attaining its maximum at $x$ and minimum at $y$ gives

$$f(x)-f(y)=f(v)-f(u)<\max_Q f-\min_Q f,$$

a contradiction. Comparing $r$-dimensional volumes proves $|X|2^{-r}\operatorname{vol}Q\le\operatorname{vol}Q$.

Consequently, the following unproved assertion suffices:

$$\text{There is }K\subseteq H\text{ satisfying (1) with } \dim K\ge n-d(P). \tag{3}$$

It would give

$$|P|+1=|X_K|\le2^{n-\dim K}\le2^{d(P)}.$$

No implication from maximum dissociation to (3) has been established. Individual admissible ternary lines cannot simply be summed.

An exact two-relation test

For two augmented ternary relations $u,v$, their span satisfies (1) if and only if (1) holds for the following five vectors, ignoring zero vectors:

$$u,\qquad v,\qquad u+v,\qquad u-v,\qquad v_0u-u_0v. \tag{4}$$

Only coordinate zero can have coefficient outside $\{-1,0,1\}$. Thus coordinate signs of $su+tv$ can change only on the lines

$$s=0,\quad t=0,\quad s=t,\quad s=-t,\quad u_0s+v_0t=0.$$

On each sector of this arrangement, $\|su+tv\|_1$ is linear. Each inequality $(su+tv)_i-(su+tv)_j\le\|su+tv\|_1/2$ is consequently linear there, so it suffices to check the bounding rays. Their images, up to sign, are exactly the vectors in (4). Necessity is immediate. This is a finite compatibility test, not an extension theorem.

Minimal ternary circuits do not justify greedy projection

There is a projection-minimal antipodal lift of dimension greater than ordinary dissociation dimension even when its kernel is generated entirely by support-minimal ternary relations.

Take

$$P=\{1,2,3,4,5,6,7,8,10,13,17,22,28\}.$$

Its sum is $126$. Any seven-element subset has its $128$ subset sums in $\{0,\ldots,126\}$, and hence is dependent. Thus $d(P)\le6$. Also $\{1,2,4,8,17\}$ is dissociated, since the first four elements have subset sums $0,\ldots,15$, disjoint from their translate by $17$. In particular this set is not a counterexample to the cardinality bound.

Order seven leaves as $(7,1,2,3,4,5,6)$, lift them to $e_1,\ldots,e_7$, and lift the six successive prefix sums

$$8,10,13,17,22,28$$

to the corresponding prefix indicator vectors. Together with zero this is a subset of $\{0,1\}^7$, hence antipodal. Its heights are given by the leaf weights. Its six independent defining relations are

$$7+1=8,\quad8+2=10,\quad10+3=13,\quad13+4=17, \quad17+5=22,\quad22+6=28.$$

Each is a support-minimal ternary relation, because every pair of distinct positive numbers is dissociated. The six relations generate the full affine kernel, which has dimension $13-7=6$. Every intermediate span satisfies (1), since it is contained in the final admissible kernel.

Nevertheless this lift admits no lower-dimensional antipodal projection retaining all labeled points distinctly, even if the height requirement is dropped. It contains

$$0,e_1,\ldots,e_7,\mathbf1.$$

For any such distinctness-preserving antipodal image under a linear map $T$, normalize a supporting functional for $0,T(e_i)$ to have values $0,1$. All displayed image points have values in $[0,1]$. Since $T(\mathbf1)=\sum_jT(e_j)$, nonnegativity forces this functional to vanish on every $T(e_j)$ with $j\ne i$. The seven such functionals show that $T(e_1),\ldots,T(e_7)$ are linearly independent. Thus the projection still has dimension at least seven, greater than $d(P)$.

There is no contradiction to the existence assertion (3): a different five-dimensional cube lift is already supplied by the binary digits of the integers in $P\subseteq[1,31]$. The obstruction proves that earlier relations may have to be replaced, not just supplemented, in a global construction.

Status

The exact norm criterion, the sufficient implication, the two-relation test, and the greedy obstruction are proved. Existence of a globally chosen $K$ as in (3) is the precise remaining gap. No new universal bound on $d(P)$ or on $f(n)$ follows from this note.