A proved one-step empty lattice lift

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For a positive integer set $P=\{a_1,\ldots,a_n\}$, recall that coherence means that $\mathbf1$ is a vertex of

$$\operatorname{conv}\{u\in\mathbb Z_{\ge0}^n:a\cdot u=\sum_i a_i\}.$$

Let $c(P)$ be the largest coherent-subset size. Coherence implies dissociation, so $c(P)\le d(P)$.

An empty lattice polytope has no lattice points other than its vertices. The following is a proved construction, not a construction in dimension $c(P)$ for general $P$.

One-step quotient theorem

Theorem. If $P$ is noncoherent, there is an empty lattice polytope of dimension $n-1$ with $n+1$ vertices, together with an integer linear functional taking the values $0,a_1,\ldots,a_n$ on those vertices. The polytope has a unimodular triangulation that is a star at the height-zero vertex, and is normal.

A relation with one unit-coefficient side

Take a nontrivial convex decomposition

$$\mathbf1=\sum_\nu \theta_\nu u^{(\nu)},\qquad \theta_\nu>0,\quad \sum_\nu\theta_\nu=1,$$

in the integer fiber, with all $u^{(\nu)}\ne\mathbf1$. Put $r^{(\nu)}=u^{(\nu)}-\mathbf1$. Each entry is at least $-1$, and $a\cdot r^{(\nu)}=0$.

There exists one such $r$ with $\sum_i r_i\le0$ for which the affine relation

$$v=(v_0,v_1,\ldots,v_n) =\left(-\sum_i r_i,r_1,\ldots,r_n\right) \tag{1}$$

has at least two positive and two negative entries. Every negative entry is $-1$.

To prove this, first suppose some participating $u$ has degree less than $n$. Then $v_0>0$, and positivity of $a$ forces $r$ to have a positive entry too. If $r$ had only one negative entry, its total positive coefficient sum would be less than one, which is impossible in a nonzero relation on positive weights. Thus both sides have at least two entries.

Otherwise every participating point has degree at least $n$; their average degree is $n$, so all have degree exactly $n$. If every associated $r$ had only one positive entry, say $r_j=m$, its negative entries would be $-1$ on a set $J$ of size $m$. The case $m=1$ would equate two distinct weights, so $m\ge2$, and

$$ma_j=\sum_{i\in J}a_i.$$

Strict convexity of the square function gives

$$\sum_i a_i^2r_i=ma_j^2-\sum_{i\in J}a_i^2<0.$$

This contradicts the fact that the weighted average of these $r$ is zero. Hence some $r$ has at least two positive entries; degree equality then gives at least two negative entries as well. This proves the claim. In particular, $r$ is primitive because it has entries equal to $-1$.

The quotient polytope

Let

$$\Lambda=\mathbb Z^n/\mathbb Zr,\qquad Q=\operatorname{conv}(\bar0,\bar e_1,\ldots,\bar e_n) \subset\Lambda\otimes\mathbb R.$$

Primitivity makes $\Lambda$ a free lattice of rank $n-1$. The height $x\mapsto a\cdot x$ descends to $\Lambda$.

The entire space of affine relations on the $n+1$ displayed points is spanned by (1). As its positive and negative supports both have size at least two, no displayed point lies in the convex hull of the others and no two coincide. Thus all $n+1$ points are vertices.

Let $J=\{j:v_j=-1\}$. The simplices obtained by deleting one vertex $j\in J$ form a unimodular triangulation of $Q$. Indeed, for any barycentric vector $\lambda\ge0$ with $\sum\lambda_i=1$, replace it by

$$\lambda+t v,\qquad t=\min_{j\in J}\lambda_j.$$

Its entries remain nonnegative, its sum remains one, and at least one negative-side coordinate becomes zero. This proves coverage.

For the intersection assertion, suppose $\lambda_j=0$, $\mu_k=0$, and $\mu-\lambda=t v$ represent a point in the two simplices. At coordinate $j$ nonnegativity gives $t\le0$; at coordinate $k$ it gives $t\ge0$. Hence $t=0$, and both omitted coordinates vanish. Their intersection is exactly the common face.

Each simplex is affinely independent and unimodular. The coefficient $v_j=-1$ expresses the omitted vertex as an integral affine combination of the retained ones, so the retained vertices generate the entire affine lattice. Every lattice point of $Q$ lies in a unimodular simplex and is consequently one of its vertices. This proves emptiness. The same triangulation proves normality: a lattice point in $mQ$ has nonnegative integer barycentric coefficients of total $m$ in one of the dilated unimodular simplices. Also $v_0\ge0$, so no simplex deletes the height-zero vertex; the triangulation is a star there.

Relation to the target and exact gap

An empty lattice polytope in dimension $k$ has at most $2^k$ vertices: two vertices of the same parity would have a lattice midpoint that is not a vertex. Therefore an empty lattice lift of the heights $\{0\}\cup P$ in dimension at most $c(P)$ would establish

$$|P|\le2^{c(P)}-1\le2^{d(P)}-1.$$

The theorem supplies such a lift when $c(P)=n-1$; the standard simplex supplies one when $c(P)=n$. Arbitrary greedy iteration is false. The exact example in the global lift note has $c(P)=3$ and a valid normal, star-triangulated first quotient in dimension four which admits no further empty, vertex-preserving height projection. A different first quotient does reach dimension three. Thus compatible choices, or replacement of earlier relations, are essential.

The substantive next question is whether a minimum-dimensional empty lift (or one with a unimodular triangulation) must have dimension at most $c(P)$. This is an additional unproved statement, not an established duality. “Minimum-dimensional” here must mean a global minimum, not minimality under further projections of a fixed lift. The same new note strengthens the one-step lift to a Boolean-coordinate lift, but does not prove that this construction can be iterated. The failed minimum- and maximum-element exchange principles in the obstruction note do not supply it.

An exact criterion for an empty lift in dimension three

Let $S\subset\mathbb Z$ contain zero, with $5\le|S|\le8$. For a block $B$ of at least two heights, write $g(B)=\gcd\{b-b':b,b'\in B\}$. There is an empty three-dimensional lattice lift of these heights if and only if $S$ has a partition $B_0\sqcup B_1$ with the following properties:

• Each block has at most four elements.
• In a four-element block $b_1<b_2<b_3<b_4$, $b_1+b_4=b_2+b_3$.
• If both blocks have at least three elements, their $g$-values agree.
• If one block has two elements and the other at least three, the latter's $g$-value divides the former's difference.

A singleton imposes no gcd condition. These are all cases, since there are at least five heights.

For necessity use Howe's width-one theorem, in the source passage already checked in the global lift note. All vertices occupy two consecutive lattice planes. An empty triangle is unimodular, and an empty quadrilateral is a unimodular parallelogram. Hence a block of at least three vertices generates the entire plane difference lattice; its height differences generate the image of that lattice under the height functional. Both such blocks therefore have the same gcd. A two-point block is a primitive segment, whose height difference is a multiple of that image generator. The parallelogram gives the displayed pair-sum identity.

For sufficiency label the blocks so that $0\in B_0$, take the generator $g$ from a block of at least three elements, and put the plane height functional equal to $g x$ on $\mathbb Z^2$. For a triangle, its two normalized differences are coprime, so Bézout's identity makes them the first coordinates of a unimodular basis. For a four-block use the two differences from its smallest element to its middle elements; their sum is the difference to the largest. The same construction gives a unimodular parallelogram. A segment of normalized height difference $t$ uses primitive direction $(t,1)$. Put the blocks in layers $z=0,1$, and choose the third height coefficient to match their offsets. Every lattice point in the resulting convex hull lies in one of the two layers, where emptiness is already proved. The layer with at least three vertices and the other nonempty layer give dimension three. Finally translate the height-zero vertex to the origin.

This criterion controls emptiness, not normality. It supplies a finite arithmetic test in dimension three, but no proof that $c(P)=3$ always forces the required partition of $\{0\}\cup P$.