Exact limits of the new convex, topological, and spectral approaches

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None of these examples is a counterexample to Erdős 963. Coherence and the new geometric proposal are defined in the convex-fiber note.

Dissociation is weaker than coherence

The set $D=(3,5,6,7)$ has the distinct subset sums

$$0,3,5,6,7,8,9,10,11,12,13,14,15,16,18,21.$$

But in its weight-$21$ fiber,

$$\mathbf1=\tfrac13\bigl((0,0,0,3)+(1,0,3,0)+(2,3,0,0)\bigr).$$

Thus $\mathbf1$ is not a vertex. A dissociated cube excludes two-point midpoint decompositions, not convex decompositions with three or more terms. One cannot replace maximum dissociation by convex extremality without proving a stronger extraction theorem.

Maximum coherent sets need not ternary-span

The set $P=(2,5,8,9)$ has the distinct subset sums

$$0,2,5,7,8,9,10,11,13,14,15,16,17,19,22,24.$$

Nevertheless,

$$\mathbf1=\tfrac13\bigl((3,0,0,2)+(0,3,0,1)+(0,0,3,0)\bigr)$$

in its weight-$24$ fiber. Every triple is coherent by the proved triple lemma, so $c(P)=3<d(P)=4$. Since $P$ is dissociated, the omitted element lies outside the ternary span of every maximum coherent triple. Integer generation, proved in the main note, is strictly weaker than ternary spanning.

Largest-element exchange fails even for coherent sets

Let $D=(4,7,8,9)$ and append $12$. The weight-$28$ fiber of $D$, apart from $\mathbf1$, consists of

$$(0,4,0,0),(1,0,3,0),(3,0,2,0),(3,1,0,1),(5,0,1,0),(7,0,0,0).$$

This list follows by considering the coefficient of $9$, which can only be zero or one. The cost $(1,0,0,-3)$ is $-2$ at $\mathbf1$ and nonnegative at all six alternatives. Hence $D$ is coherent.

Every four-set containing $12$ is noncoherent. With coordinates in the indicated order, the following are exact convex decompositions within the appropriate total-sum fiber:

Their fiber weights are $31,32,33,36$, respectively. Heredity also excludes coherence of the full five-set. Its coherent dimension is four, but no maximum coherent subset contains its largest element.

Deleting the largest coherent element need not give modular dissociation

Take the coherent $C=\{4,7,8,9\}$ and put

$$D=3C\cup\{28\}=\{12,21,24,27,28\}.$$

In the weight-$112$ fiber, the coefficient of $28$ is congruent to one modulo three, hence is at least one. Minimizing that coefficient and then exposing the old $C$-fiber proves coherence of $D$. But

$$12+21+24+27=84=3\cdot28.$$

Thus $D\setminus\{28\}$ is not dissociated modulo $28$.

Canonical image lattices do not supply an essential inequality system

For $P=\{1,2,3\}$ and maximum dissociated $D=\{1,2\}$, the unique ternary representation matrix has rows

$$V=\begin{pmatrix}1&0\\0&1\\1&1\end{pmatrix}.$$

The other maximum choices give $e_1,e_2,-e_1+e_2$, up to row order. All these matrices are totally unimodular.

For any system $Vz\le u$ with the origin feasible, replace $u$ by its integer floor without changing the feasible integer points. The resulting polyhedron is integral. Hence the origin is an integer-hull vertex exactly when it is an ordinary vertex. Two independent tight rows witness this, so the third row is inessential.

Consequently no maximum $D$, canonical ternary lifts, or right-hand side makes all three rows essential to origin vertexhood. The associated-prime construction proposed in the main note is false in this form, not merely unproved. Noncanonical integer lifts or a different polyhedral construction would be needed.

Ordinary cube topology discards essential arithmetic information

For actual positive weights, every nonextreme level section of the real cube is a convex polytope of dimension $n-1$, hence a ball. Its ordinary homotopy type does not distinguish generic weights from weights with many subset-sum collisions.

Even strict coordinate monotonicity, antipodal symmetry, and distinct singleton values together are insufficient without affine additivity. For arbitrary $n\ge3$, let $m(S)=\max S$, with $m(\varnothing)=0$, and choose $0<\varepsilon<1/(4n)$. On cube vertices define

$$f(S)=|S|+\varepsilon\bigl(m(S)-m(S^c)+n\bigr).$$

This is strictly increasing along edges, satisfies $f(\varnothing)=0$ and $f(S)+f(S^c)=n+2\varepsilon n$, and has distinct singleton values. Its perturbation range is less than $1/2$, so different cardinality layers have different values.

Every four-coordinate face has a collision. If its free coordinates are $i_1<i_2<i_3<i_4$ and its fixed base is $B$, the two subsets

$$B\cup\{i_4,i_1\},\qquad B\cup\{i_4,i_2\}$$

have equal sizes and maxima, and their complements have equal maxima because $i_3$ remains outside both. The face on the last three coordinates is vertex-injective: the values of $m(S)-m(S^c)$ within its singleton layer are $-2,-1,1$, and within its pair layer are $-1,1,2$. Thus the largest injective coordinate-face dimension is exactly three for arbitrarily large $n$. The multilinear extension retains monotonicity and the antipodal identity, but it is not affine. This rules out arguments using only those weakened geometric features.

Spectral mass on maximum subset-sum cubes is also insufficient

Let $P=\{3,4,5,6\}$, with $d(P)=3$. Every triple is dissociated, whereas $3+6=4+5$. For a maximum triple $D$, set $S=\Sigma(D)$ and take any nonzero spectral factor

$$Q(t)=\sum_{s\in S}c_s e^{ist},\qquad K(t)=\frac{|Q(t)|^2}{\sum_s|c_s|^2}.$$

Let $A_D$ join vertices of $S$ whose absolute difference lies in $P$. Then

$$2\sum_{p\in P}\operatorname{Re}\widehat K(p) =\frac{c^*A_Dc}{\|c\|^2}<4. \tag{1}$$

Indeed, each graph is connected, has maximum degree four, and has a vertex of degree three. These claims follow from the following subset-sum and degree lists, in the same order:

Connectivity also follows from the canonical basis-color cube edges. The strict inequality in (1) follows from

$$4\|c\|^2-c^*A_Dc =\sum_{\{s,t\}\in E}|c_s-c_t|^2 +\sum_s(4-\deg s)|c_s|^2>0.$$

Equality would force $c$ to be constant on the connected graph and zero at a deficient-degree vertex, hence zero everywhere.

Thus arbitrary spectral factors on maximum cubes, even after convex mixing, cannot give the proposed aggregate Fourier domination $2\sum_{p\in P}\operatorname{Re}\widehat K(p)\ge|P|$. This only excludes that particular positive spectral-counting method.

Coherent odd quadruples can have more than four smaller blocked extensions

Let $D=(15,17,27,33)$. Its pair sums are distinct, and its three smallest elements sum to more than its largest, so it is dissociated. By the proved odd-four-set lemma it is coherent. All six smaller odd numbers

$$p\in\{1,5,7,9,11,13\}$$

give noncoherent extensions. Four are ruled out by

$$1=33-15-17,\quad5=15+17-27,\quad 9=15+27-33,\quad11=17+27-33.$$

For $p=7,13$, the five-sets remain dissociated: their pair sums are distinct, and their three smallest entries sum to more than their largest. Nevertheless they have the exact fiber decompositions

$$\mathbf1= \frac{(2,0,5,0,0)+2(0,3,0,2,0)+(4,0,1,2,0)+2(0,0,0,0,3)}6$$

for the weight-$99$ fiber of $(7,15,17,27,33)$, and

$$\mathbf1= \frac{(0,0,3,2,0)+(0,3,0,1,1)+(3,0,0,0,2)}3$$

for the weight-$105$ fiber of $(13,15,17,27,33)$.

Thus blocked smaller odd extensions cannot be counted just by the four triples of a coherent four-set, or bounded by its cardinality. This does not refute the global coherent bound. Indeed the ten-element union in this example contains the coherent set $\{1,5,7,9,27\}$: its first four entries form a dissociated odd quadruple, and $27$ exceeds their sum, so maximizing the coefficient of $27$ in the total-sum fiber reduces to coherence of that quadruple.

Minimum-element exchange also fails for coherence

Let

$$P=\{33,36,37,38,41,45\},\qquad D=P\setminus\{33\}.$$

Then $c(P)=5$, but no maximum coherent subset contains $33$.

In the coordinate order $(36,37,38,41,45)$, the weight-$197$ fiber of $D$ consists of $\mathbf1$ and

$$(0,0,4,0,1),\quad(0,3,0,1,1),\quad (0,2,0,3,0),\quad(1,0,1,3,0).$$

For exhaustiveness, every solution has degree five because $4\cdot45<197<6\cdot36$. Taking deficits from $45$ reduces its enumeration to the weight-$28$ fiber on $(4,7,8,9)$ already listed above, restricted to representations of degree at most five. The cost $(-3,0,0,1,0)$ equals $-2$ at $\mathbf1$ and is nonnegative at the other four points, so $D$ is coherent.

Four five-subsets containing $33$ have a ternary relation from

$$33+41=36+38,\qquad 33+45=37+41.$$

The remaining set $E=(33,36,37,38,45)$ has the decomposition

$$\mathbf1=\frac14\bigl[ (3,0,0,0,2)+(0,4,0,0,1) +(1,0,3,0,1)+(0,0,1,4,0) \bigr].$$

Every point displayed has weight $189=\sum E$, so $E$ is noncoherent. Heredity also excludes coherence of the full $P$. Thus neither extremal-element exchange principle can justify an iteration of the one-step empty-lift construction.

A coherent vertex need not admit any modular pivot

The coherent set

$$D=(9,20,42,44,49,63)$$

has no pivot $p\in D$ for which $D\setminus\{p\}$ is dissociated modulo $p$. Moreover its all-ones point is coordinatewise maximal among vertices of all its integer fibers. Thus squarefree vertices cannot in general be reduced to an asymptotic group-relaxation branch.

To prove coherence, its total weight is $227$. Minimize the cost $12u_9+8u_{20}+5u_{44}$. Modulo seven, a feasible point of cost at most $25$ must have

$$(u_9,u_{20},u_{44})=(0,0,5),\ (0,1,2),\ \text{or }(1,1,1).$$

This follows directly by taking $u_9=0,1,2$ in $2u_9+6u_{20}+2u_{44}\equiv3\pmod7$ and the cost bound. The residual weights, divided by seven, are $1,17,22$, to be represented using $6,7,9$. The first two are impossible and the last has the unique representation $6+7+9$. Hence $\mathbf1$ is the unique cost minimizer.

Every proposed pivot has a modular collision:

$$\begin{aligned} 2\cdot9&=20+42-44,&2\cdot20&=49-9,\\ 2\cdot42&=44+49-9,&3\cdot44&=20+49+63,\\ 2\cdot49&=44+63-9,&2\cdot63&=42+44+49-9. \end{aligned}$$

The coefficient-two relations make $\mathbf1+e_p$ a midpoint nonvertex for every $p\ne44$. For the remaining coordinate, in the displayed ordering of $D$,

$$\mathbf1+e_{44}=\tfrac13\bigl[ (1,0,1,5,0,0)+(0,2,2,0,3,0)+(2,1,0,1,0,3)\bigr].$$

All three terms have weight $271$. Translating these decompositions by nonnegative integer vectors excludes every vertex properly dominating $\mathbf1$ coordinatewise. This is an obstruction to a pivot-based proof, not to the coherent cardinality conjecture.