Proof sketch for the dissociated covering attack

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Target

Problem 774 is open: is every proportionately dissociated set $A\subset\mathbb N_{>0}$ a finite union of dissociated sets? For a finite configuration $E$ of nonzero vectors write $r(F)$ for the largest dissociated subset of $F$, $\eta(E)=\min_{\varnothing\ne F\subseteq E}r(F)/|F|$ for the hereditary extraction constant, and $\chi_1(E)$ for the least number of dissociated classes covering $E$; dissociated means no nonzero relation with coefficients in $\{-1,0,1\}$. By the inherited finite-block theorem and rational realization, both author-recorded and elementary, the problem is equivalent to deciding

$$\text{(F)}\qquad\exists\,\delta>0\ \forall q\ \exists\text{ finite }E_q:\ \eta(E_q)\ge\delta,\ \chi_1(E_q)>q,$$

whose truth is a negative answer, against the positive alternative $\forall\delta>0\ \exists q(\delta)\ \forall E:\ \eta(E)\ge\delta\Rightarrow \chi_1(E)\le q(\delta)$.

Held links

• Repeated simplexes. L271, tier 0 proves semistability of the full relation lattice of $S_q^{\otimes k}$ for $q$ a power of two. Together with the inherited stable-theta and finite phase-entropy results at their author-recorded standing, it gives one hereditary extraction constant at $k=\lfloor q(\log\log q+\log\log\log q+c)\rfloor$, for any one fixed real $c$. No coloring bound for this family is assumed.
• Uniform extraction for the odd-root grids. All roots of unity are proportionately dissociated with one constant: L21, tier 2, via the analytic Sidon bound L20, tier 2. The transfer to the integer blocks of the inherited main argument is author-recorded.
• The pivot. By that argument, (F) follows from $\sup_k\chi_1(\mu_{N_k})=\infty$ for $N_k=3\cdot5\cdots p_k$, its equation (4). Conversely, if the supremum is finite, the root-transversal equivalence (author-recorded) makes all roots of unity a finite union of dissociated sets, a positive answer to Déchamps's Question 38, and closes the prime-tensor route.
• Layer criterion and separation parameter (tensor coloring, author-recorded): $\chi_1(\mu_{Mp})\le q$ exactly when $p\ge\tau_q(\mu_M)$. Root coverage (tensor holes, author-recorded): $\tau_q\ge\lceil1/\alpha_q\rceil$ and the nonjump recurrence.
• Arcs. L123 (tier 0): $q$ arcs are proper exactly when $qD(N)\ge N$; L124 (open) would bound every grid by arcs.
• Symmetry. Coordinatewise permutations of $\prod\mathbb Z/p_i$ are exactly the permutations of $\mu_N$ preserving quasi-independence (Ramsey and Graham, Theorem 1.2.1, a source theorem consumed as stated).

The current line: repeated factors with uniform extraction

The repeated-simplex argument changes the tensor schedule. Identify the common side $q=2^b$ with a Boolean block. The relation lattice consists of integer functions whose Walsh frequencies miss at least one whole block. Boolean spectral compression and a monotone coupling of the bit measure conditioned on an empty block prove semistability of this lattice, with normalized covolume

$$\sigma_{q,k}=q^{k(q-1)^{k-1}/(2(q^k-(q-1)^k))}.$$

At $k=\lfloor q(\log\log q+\log\log\log q+c)\rfloor$ this tends to $\exp(e^{-c}/2)>1$. The stable-theta and phase-entropy interfaces then give uniform extraction by fully dissociated subsets, including inside every finite subset by transitivity. The finite target would follow if the actual coloring numbers at one fixed $c$ were unbounded. This is the exact remaining obligation for this family.

The entropy upper bound in the same note shows that taking $k/q-\log\log q-\log\log\log q\to+\infty$ makes extraction decay to zero. That parameter change cannot be used in an amplification. L269, tier 0 proves an exact obstruction to scalar linear colorings: over $\mathbb F_q$, an affine hyperplane with all normal coefficients nonzero is dissociated exactly when $k<q$. At and above $q$ it contains an explicit flat trade. This rules out those colorings and their coordinate-permuted versions, but leaves arbitrary palettes open. Indeed, at $k=q\ge7$, three rationally independent classes suffice, although all scalar linear fiber colorings fail. The research note also proves that Boolean affine fibers of codimension $t$ are dependent when $k\ge2^t$, and that every Boolean quadratic two-coloring fails when $k\ge q+1$. These statements exclude those templates only; multiple quadratic output bits and arbitrary palettes are not covered.

The finite claim L211, tier 0 supplies a separate exact gap: thirteen positive integers have $\eta=7/13$, $\chi_f=37/19<2$, and $\chi_1=3$. Thus even weighted half-extraction does not imply a two-color partition. This is a finite obstruction, not an unbounded family. Every point deletion is two-colorable, so the inherited critical-core lemma gives $\tau_3\le6$ and prevents using this seed alone for arbitrarily cheap successive simplex jumps.

The odd-root alternative

The twisted-layer note builds proper colorings of each grid from twisted copies of a proper coloring of the previous one. Its Lemmas 1 to 3, Corollary 4, Lemma 5, Proposition 6 and Lemmas 7 to 11 are author-recorded proofs: coordinate permutations act on signed spans; random twists kill a root common value with exact probability $\kappa$ per layer, $\kappa$ the root-span fraction of the class; robust layering preserves $\kappa$ exactly, so the root-coverage parameter is constant along robust steps; survival bounds for values touching few layers transfer from one grid to the next, on average over the twists; the exact first moment of common values over random twists diverges along the odd-prime grids (Lemma 7); and certificates that use only empty layers or pairs of layers with dissociated union never certify a set beyond the rank (Lemma 8); and values whose survival sets are covered by few translates of a small set of twists are excluded together, without a union bound (Lemma 9); the only coordinate permutations preserving the arc class of three arcs are the identity and a reflection (Lemma 10); and twisted layerings whose twists avoid certain coincidences have classes with no coordinate symmetry at all (Lemma 11).

The roots lose their layer-count race with three colors at every prime: they need about $\log N/\log3$ layers and the next prime supplies about $\log N$. Lemma 7 shows that this is the only part of the first-moment count that holds. Beyond the roots the expected number of common values is astronomically large, carried by twist families of tiny probability, so no union bound over values can prove a twisted layering proper. Whether three colors suffice for every odd squarefree order is the open claim L210, whose proof would close the prime-tensor route and whose refutation needs one order requiring four classes; it is a conjecture supported by the root race, the rank bound through seven primes and small-order data. The exact finite data: arc classes below the rank threshold span no outside root through $255255$, signed pairs of roots survive twists with probability at most $2/15$ against $\kappa^2=1/9$ at $105$ and $1155$, and two colors suffice at $165$ and $195$ although two arcs fail there.

Constraints on any counterexample

These are proved here, author-recorded, and constrain any family meeting (F), tensor or not.

A. Rank density must vanish over every field. Let $E$ consist of integer vectors (clear a common denominator) and let $\mathbb K$ be $\mathbb Q$ or $\mathbb F_\ell$ for a prime $\ell$. A subset whose images in $\mathbb K^d$ are linearly independent is dissociated, since a relation with coefficients $\pm1$ reduces to a nontrivial $\mathbb K$-linear relation. When no element of $E$ maps to zero, Edmonds' matroid partition theorem therefore gives $\chi_1(E)\le\max_{F\subseteq E}\lceil|F|/\mathrm{rank}_{\mathbb K}F\rceil$. A family meeting (F) must have $\max_F|F|/\mathrm{rank}_{\mathbb K}F\to\infty$ over every such field: no field sees its relations as sparse, which is why kernel counting over a finite field, as in Lewko's bounded-torsion theorem, cannot transfer directly.

B. The window bound. For a finite $E\subset\mathbb N_{>0}$, $\chi_1(E)\le\max_x|E\cap[x/2,x]|$. Color the elements in increasing order, giving $x$ the least color whose current largest element $m$ satisfies $x>2m$ (or which is unused). Each class has consecutive ratios above two, so each element exceeds the sum of the smaller ones in its class and the class is dissociated. The colors blocked for $x$ have largest elements in $[x/2,x)$, distinct elements of $E$, so $x$ receives a color of index at most $|E\cap[x/2,x]|$. If moreover $\eta(E)\ge\delta$, a dissociated $D\subseteq[x/2,x]$ has $2^{|D|}$ distinct subset sums in $[0,|D|x]$, so $|E\cap[x/2,x]|\le(\log_2x+\log_2\log_2x+O(1))/\delta$. A counterexample needs windows $[x/2,x]$ holding unboundedly many elements.

B'. Balanced windows. If $E\subseteq(X-Y,X]$ and $2|E|Y\le X$, every relation of $E$ has equally many coefficients $+1$ and $-1$: writing $x=X-y_x$, a relation gives $(P-Q)X=\sum\epsilon_xy_x$, whose right side is smaller than $X$ in absolute value, with $P$ and $Q$ the numbers of plus and minus signs. Inside a narrow window, dissociation is the absence of two distinct equal-size subsets with equal sums.

C. No line-forcing in tensor powers. If nonzero $a_1,\ldots,a_m\in\mathbb Q^r$ are pairwise non-parallel, the tensor powers $a_t^{\otimes w}$ are linearly independent for every $w\ge m-1$. A dependence $\sum_tc_ta_t^{\otimes w}=0$ would give $\sum_tc_tP(a_t)=0$ for every form $P$ of degree $w$. Choose a linear form $\lambda$ nonzero at every $a_t$; the points $a_t/\lambda(a_t)$ are distinct on the hyperplane $\lambda=1$, where forms of degree $w$ restrict to all polynomials of degree at most $w$, and $m$ distinct points are separated by polynomials of degree $m-1$, so for each $t_0$ some $P$ gives $c_{t_0}\lambda(a_{t_0})^w=0$. Hence in a tensor power of one atom of $m$ pairwise non-parallel vectors, every combinatorial line with at least $m-1$ wildcard coordinates is independent, and Hales–Jewett, Graham–Rothschild or parameter-word theorems, which force lines of uncontrolled wildcard length, cannot force a monochromatic relation there.

D. Disjointness forcing collapses. Topological lower bounds for chromatic numbers of Kneser-type hypergraphs force $r$ pairwise disjoint monochromatic members of a family of $k$-sets in $[n]$, with one vector $v_A$ for each member $A$. Turning that into a monochromatic relation by the same mechanism needs a relation $\sum_i\pm v_{A_i}=0$ for every pairwise disjoint $r$-tuple. Fix a disjoint $(r-1)$-tuple $B_1,\ldots,B_{r-1}$; every member $A$ avoiding it then satisfies $v_A=\sum_is_iv_{B_i}$ for a sign vector $s\in\{\pm1\}^{r-1}$. If more than $2^{r-1}$ members avoid the tuple, which holds once $\binom{n-(r-1)k}k>2^{r-1}$, two distinct members get the same signs and $v_A=v_{A'}$, a collapse that a configuration of distinct vectors excludes. If $n\le rk$, coloring each member by whether it contains the point $1$ leaves no monochromatic disjoint $r$-tuple. Disjointness hypergraphs therefore carry this forcing at most in the narrow range between these two conditions.

E. Every family needs an integral gap. Write $\chi_f(E)$ for the fractional dissociated covering number, the least total weight of dissociated sets covering every element with weight at least one. If a group of permutations of $E$, each mapping dissociated sets to dissociated sets, acts transitively on $E$, then the images of one largest dissociated set under the group cover every element equally often, so $\chi_f(E)=|E|/r(E)=1/\eta(E)$. The odd-root grids and every simplex tensor (coordinate permutations, Lemma 1 of the twisted-layer note), the critical Boolean moments (coordinate flips) and the simplicial boundaries (signed vertex permutations) of the inherited main argument are transitive in this sense. Along any of them a counterexample needs $\chi_1/\chi_f\to\infty$ with $\chi_f\le1/\delta$: a pure gap between the integral and fractional covering numbers. More generally, Bourgain's weighted extraction theorem gives an absolute $c_0>0$ with $1/\eta(E)\le\chi_f(E)\le1/(c_0\eta(E))$ for every finite $E$, with no symmetry hypothesis. The fractional-covering note states the precise source interface and proves the linear-programming consequence. Thus (F) is equivalent to bounded fractional covering with unbounded integral covering, even for asymmetric families. The inherited Grow–Whicher computation separates exact reciprocal extraction from fractional covering at one small set, $\chi_f=27/13$ against $\eta=1/2$, but an unbounded fractional separation at a fixed extraction constant is ruled out by Bourgain's theorem.

Provisional premises

None of the conclusions above uses an unproved premise. The race model rests on its two stated assumptions; Lemma 7 of the note proves its first assumption false for values of large complexity, and the model is used here only for the roots, where it is exact, and for values of small complexity.

Gaps

1. (F) itself. No family is known with uniform extraction and unbounded $\chi_1$. For the odd-root grids, a proof of (4) would have to defeat robust twisted layerings at infinitely many primes for every palette (Corollary 4 of the note); since the roots lose their race, that needs tightness or common values of complexity at least two on typical twist families.
2. Repeated-simplex coloring. At one fixed $c$, prove unbounded $\chi_1(S_q^{\otimes k(q)})$ as $q$ ranges over powers of two, or construct a bounded palette for this family. Uniform extraction is established in the repeated-simplex argument. The obstructions to scalar and Boolean linear fibers and to quadratic two-colorings do not cover arbitrary colorings, and moving above the critical scale loses extraction.
3. Small complexity in the odd-root route. For classes produced by robust twisted layerings, bound survival probabilities, with the orbit counts, strongly enough that the sum of the probabilities of all unwanted common values of complexity at most $s_0$ is small, uniformly in the conductor. A bound $\Pr_g[\hat gF\in\Sigma(C)]\le\kappa^{c\,s(F)}$ with an unspecified $c>0$ does not by itself suffice. Under the crude count $(2M)^s$ and $p\sim\log M$ layers one needs $c\log(1/\kappa)>1$; sharper orbit counts can improve that requirement. No such exponential survival bound holds for all complexities: it fails on $\Sigma(C)$ beyond $\log|G_M|/(c\log(1/\kappa))$ (Lemma 7). Proposition 6 transfers such bounds only on average and only for values touching at most $p-\tau_q$ layers.
4. Rigidity. Outside an event of small probability, $p$ independent uniform twists of one class have no common value of complexity above $s_0$. Quantitative small-complexity bounds and this rigidity bound, with their error probabilities summing to less than one for all colors and the required robustness tests, would prove the open claim L210. Rigidity needs a structural argument, since Lemma 7 rules out a union bound over values, and Lemma 8 shows that it concerns many layers at once. Lemma 9 of the note (author-recorded) reduces this rigidity gap to a property of a single class: apart from few orbits of short or thin values, every value of large complexity must have its survival set, the twists that carry it into the class's signed span, covered by fewer than $p$ translates (for layerings that keep each color on copies of one class) of a set $\mathcal N$ of twists with $\binom p2|\mathcal N^{-1}\mathcal N|$ small against $|G_M|$. Then two of the $p$ twists must fall in one translate, an event of small probability, and no union bound over values is needed. The set $\mathcal N$ must contain the cyclotomic affine maps, since values on a short initial arc survive many Galois automorphisms. A value with a globally unique signed representation cannot survive all layers once common roots have been excluded: its support would have to lie in every twisted class. The open case therefore concerns values that admit different representations across the layers. The difference of two such representations can have coefficients in $\{-2,-1,0,1,2\}$, so it must not automatically be treated as a ternary relation. Natural survival of the whole class is Lemma 10 of the note: the arc class has exactly two coordinate symmetries, the identity and a reflection. At the conductor 105, exact survival sets over all twists show values on at least 20 points of the arc class clustered in fewer than 11 affine cosets, the next prime, while values on 10 to 17 points can be spread or survive through other representations, within the range the first moment handles there. The classes of a robust twisted layering at 105 are more rigid still: the whole-class sums survive only the identity, and sums on 20 points at most two twists; Lemma 11 shows that the whole-class part of this rigidity propagates to every level of the induction. Values with many representations at large conductors are the open case.
5. Other relation sources. Constraints A to E leave the inherited simplicial and moment families open, but by E they need a gap between integral and fractional covering numbers. Unlike the tensor grids they have no layer structure with many layers per step, so the race model does not apply to them.

Next moves

• Seek a flat relation under an arbitrary fixed palette in the repeated power-of-two simplex family at one fixed critical offset $c$. Its lattice and extraction inputs are available, and the stated linear and quadratic palettes are excluded. A successful obstruction must use more than a monochromatic affine fiber; bounded proper colorings would instead retire this candidate family.
• Attack the single-class clustering property of Lemma 9 for arc classes and their twisted layerings: extend Lemma 10 from the whole class to large subsets of it, bound how many pairwise far twisted copies of a class can share a large set, and control the values that change representation through relations on two or more grid lines, which a class meeting every line in at most $\lceil p/q\rceil$ points forces and which at 105 are crosses and stars in the directions 3, 5 and 7.
• Test the small-complexity bound further: survival of values of complexity three for arc classes, $\tau_2(\mu_{165})$ (which decides $\chi_1(\mu_{3\cdot5\cdot11\cdot13})$), and $\kappa$ of arc classes at $4849845$.
• Treat the bounded coloring of all roots of unity as a target in its own right, and move the negative attack to relation sources without many-layer product structure.