Recurrent root sets do not supply a uniform height

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Standing

Let $G=\bigoplus_{p\text{ odd}}\mu_p$, with additive relations interpreted in $\mathbb C$. Write $\chi_h$ for coloring by height-$h$ independent classes, and $\chi_{\mathbb Q}$ for coloring by rationally independent classes.

This page proves that even rationally rich, product-recurrent root sets need not have a relation at one uniform height. It gives an exact stronger target: excluding height-$h$ independent piecewise-syndetic sets for one fixed $h$ is equivalent to the premise of fixed-height focusing. That exclusion remains unproved. No finite coloring of $G$ is constructed here.

1. A height-safe tensor extension

Let $B\subset\mathbb Z^d$ be height-$h$ independent, with all column entries in $\{0,\pm1\}$. Let $C$ be height-$h|B|$ independent. Then the indexed tensor product $B\otimes C$ is height-$h$ independent.

Indeed, for a relation with coefficients $a_{b,c}\in[-h,h]$, each coordinate $u$ of $B$ gives a $C$-relation

$$\sum_{c\in C}\left(\sum_{b\in B}a_{b,c}b_u\right)c=0$$

with coefficient magnitudes at most $h|B|$. Every bracket therefore vanishes. For each $c$, the equations say $\sum_ba_{b,c}b=0$; height-$h$ independence of $B$ finishes the proof.

More generally, $|B|$ can be replaced by the maximum absolute row sum of the integer column matrix of $B$. We use the simpler bound below. For roots on disjoint conductor blocks, multiplication identifies the tensor product faithfully with the actual root products. Standard prime-grid columns have the required $0,\pm1$ entries.

2. Fixed-height extraction with arbitrarily large rank defect

The established root extraction theorem makes $G$ Sidon in the additive, discrete, torsion-free group $\mathbb C$. The arXiv v1 copy of Hare--Yang was checked after its digest: Proposition 2, Lemma 3, and Theorem 2 in Section 3 give, for every fixed integer $H$, one $\delta_H>0$ such that every finite root set contains a height-$H$ independent subset of relative size at least $\delta_H$. The identity excluded by that theorem is additive $0$, not root $1$.

On any prescribed conductor tail, full grids have arbitrarily large cardinality-to-rank ratio

$$\frac N d=\prod_{p\in J}\frac p{p-1}.$$

For example, take all permitted odd primes through a cutoff. These finite products dominate the harmonic sum over odd integers up to that cutoff and coprime to the fixed excluded conductor, and those sums diverge. Thus for any fixed $H,L$ one may choose a finite fresh grid $E$ with $\delta_H|E|/\operatorname{rank}_{\mathbb Q}E\ge L$. An extracted set $C\subset E$ then satisfies

$$C\text{ is height-}H\text{ independent},\qquad \frac{|C|}{\operatorname{rank}_{\mathbb Q}C}\ge L. \tag{1}$$

Multiplying $C$ by the inverse of one of its roots arranges $1\in C$. This preserves all relations, stays in the same conductor group, and permutes the full grid's standard $0,\pm1$ columns.

In particular, no fixed-height lower bound of the form $\operatorname{rank}_{\mathbb Q}C\ge\epsilon_H|C|$ holds for actual root subsets. This is a consequence of the checked local extraction theorem, not a new coloring theorem.

3. Recurrent tails with infinite rational coloring number

Set $B_0=\{1\}$. Recursively choose disjoint finite conductor blocks and sets $C_i$ satisfying

$$1\in C_i,\qquad H_i=i|B_{i-1}|, \qquad C_i\text{ height-}H_i\text{ independent}, \qquad \frac{|C_i|}{\operatorname{rank}_{\mathbb Q}C_i}\ge i+1, \tag{2}$$

and put $B_i=C_1\cdots C_i$. Existence follows from (1) at each finite stage. These choices can, if desired, be made by the first successful finite subset in a fixed enumeration; the extraction theorem guarantees termination. No search was performed here.

Define

$$D_j=\bigcup_{n\ge j}C_j\cdots C_n. \tag{3}$$

Then:

1. $D_j$ is height-$j$ independent, so $D_1$ is dissociated.
2. $\chi_{\mathbb Q}(D_j)=\infty$ for every $j$.
3. The sets $D_j$ are nested and eventually lie in every fixed conductor tail.
4. For $x\in D_j$, sufficiently late $k$ satisfies $xD_k\subset D_j$.

For (1), induct along a finite tail product. Its prefix has at most $|B_{i-1}|$ elements, and $j|C_j\cdots C_{i-1}|\le i|B_{i-1}|=H_i$. Apply the tensor lemma at height $j$, and then use finiteness of every relation's support to pass to the union. For (2), each $C_i\subset D_j$ for $i\ge j$, since all factors contain $1$; any rationally independent class in $C_i$ has at most its rational rank, so its coloring number is at least $i+1$. Assertions (3)--(4) follow directly from disjoint conductor blocks and disjoint late factor positions.

Thus arbitrary-depth monochromatic products of rationally dependent factors can all lie in one dissociated set whose every tail has infinite rational coloring number. The missing issue is not just choosing rationally dependent profile classes. Each $D_j$ still has some rational dependence, but the least relation heights have no common finite bound.

4. An ordinary idempotent still does not fix the quantifier

For completeness, the preceding construction supports an idempotent ultrafilter $\mathcal U$ on $G$ such that

$$D_j\in\mathcal U\quad\text{for every }j, \qquad \chi_{\mathbb Q}(A)=\infty\quad(A\in\mathcal U). \tag{4}$$

This is not an assertion of minimality in the ideal sense.

Here is a self-contained semigroup argument. Let $\mathcal I=\{A\subset G:\chi_{\mathbb Q}(A)<\infty\}$. This is a translation-invariant ideal: restriction and finite unions preserve finite colorability. The requirements of containing all $D_j$ and all complements of members of $\mathcal I$ have the finite intersection property. Otherwise a $D_j$ would be covered by a finite union of finitely colorable sets, contradicting Section 3. Ultrafilter compactness therefore makes the following a nonempty compact set:

$$K=\{\mathcal U:D_j\in\mathcal U\ (j\ge1),\quad \mathcal U\cap\mathcal I=\varnothing\}.$$

Define ultrafilter multiplication by

$$A\in\mathcal U\mathcal V \iff\{x:x^{-1}A\in\mathcal V\}\in\mathcal U.$$

It is associative by expanding the definition; right multiplication by a fixed ultrafilter is continuous, since the inverse image of the basic clopen set $\{\mathcal U:A\in\mathcal U\}$ is another such set. For $\mathcal U,\mathcal V\in K$, tail recurrence shows that the return set of $D_j$ contains $D_j$, so $D_j\in\mathcal U\mathcal V$. For $A\in\mathcal I$, no translate $x^{-1}A$ belongs to $\mathcal V$, so its return set is empty. Hence $K$ is a compact subsemigroup.

Choose a minimal nonempty compact subsemigroup $M\subset K$, using the intersection property of chains of compact sets. For $a\in M$, the set $Ma$ is a nonempty compact subsemigroup, and therefore is $M$. The nonempty set $\{x\in M:xa=a\}$ is also a compact subsemigroup, so it is $M$. In particular $a^2=a$. This proves (4); the ultrafilter is nonprincipal because finite sets belong to $\mathcal I$.

It contains the dissociated set $D_1$, and for every $h$ it contains the height-$h$ independent set $D_h$. Thus idempotence and even infinite rational coloring number in every member cannot replace the common height in the earlier idempotent focusing proposition.

5. The exact additional global-color condition

A set $T\subset G$ is thick if every finite $E\subset G$ has a translate $Et\subset T$. A set $A$ is piecewise syndetic if $FA$ is thick for some finite $F\subset G$.

Proposition. For every fixed finite $h$,

$$\boxed{\chi_h(G)<\infty \quad\Longleftrightarrow\quad G\text{ has a height-}h\text{ independent piecewise-syndetic subset}.} \tag{5}$$

For the forward direction, first prove that a cell in every finite partition of $G$ is piecewise syndetic. Suppose $A\cup B$ is piecewise syndetic, with $F(A\cup B)$ thick, and $A$ is not piecewise syndetic. Since $FA$ is not thick, choose finite nonempty $K$ such that $Kx\not\subset FA$ for every $x$. Given finite $E$, take $t$ with $KEt\subset FA\cup FB$. For every $e\in E$, some $k\in K$ has $ket\notin FA$, so $ket\in FB$. Therefore $Et\subset K^{-1}FB$. The fixed finite set $K^{-1}F$ witnesses that $B$ is piecewise syndetic. Induction, starting from $G$, proves the partition assertion; apply it to a finite height-$h$ coloring.

Conversely, suppose $A$ is height-$h$ independent and $FA$ is thick. For any finite $E\subset G$, choose $t$ with $Et\subset FA$. Assign each $x\in E$ an $f\in F$ with $f^{-1}xt\in A$. Within a color, multiplication by $f^{-1}t$ preserves all additive relations and is injective. This is a proper height-$h$ coloring with at most $|F|$ colors. Finite-color compactness gives a coloring of $G$. This proves (5).

The sets $D_j$ from Section 3 are not piecewise syndetic. Each $C_i$ is a proper subset of its full conductor group $E_i$, since the full group has a ternary prime-polygon relation. Given finite $F$, choose a later block disjoint from all conductor primes in $F$. Projection of $FD_1$ to that block is precisely $C_i\ne E_i$. No translate of the finite subgroup $E_i$ can be contained in $FD_1$, since its projection would be all of $E_i$. Thus $FD_1$ is not thick. Its subsets $D_j$ cannot be piecewise syndetic either.

Syndetic and uniformly recurrent normalization

The same existence question can use a syndetic independent set, meaning one with $FB=G$ for finite $F$, without enlarging a given thick-cover witness. Suppose $A$ is height-$h$ independent and $FA$ is thick. Exhaust $G$ by increasing finite $E_n$, choose $t_n$ with $E_nt_n\subset FA$, and take a coordinatewise convergent subsequence of $t_n^{-1}A$ in $\{0,1\}^G$. Its limit $B$ is height-$h$ independent: every forbidden relation has finite support. Also $FB=G$: for each $x$, the covering disjunction over the fixed finite set $F$ persists under pointwise convergence. Thus (5) is equivalent to existence of a syndetic height-$h$ independent set.

Let $H=\langle F\rangle$, a finite multiplicative root subgroup. Since $HB=G$, choosing one point of $B$ in each $H$-coset gives an independent global transversal. This agrees with the earlier selector formulation; it does not restrict the tail color labels beyond that formulation.

There is also a minimal-shift normalization. The orbit closure of $B$ consists of independent sets $C$ satisfying the same $FC=G$: both conditions are closed, and $G$ is abelian. Choose a minimal nonempty closed invariant subsystem by compactness. Every point $C$ in it is uniformly recurrent. Explicitly, for a nonempty clopen pattern neighborhood $U$, minimality makes its inverse shifts cover the subsystem; a finite subcover makes the occurrence set of $U$ syndetic. Taking $U=\{D:X\subset D\}$ for finite $X\subset C$ shows that

$$\bigcap_{x\in X}x^{-1}C$$

is syndetic.

The covering witness for this last intersection depends on the entire pattern $X$. No bound on that dependence is proved. Fixed initial coverage therefore does not keep coefficient heights or magnitude-group counts under control during focusing. In particular this normalization does not upgrade the ordinary idempotent in Section 4 to a minimal-ideal idempotent containing its constructed $D_1$.

Combining (5) with fixed-height focusing gives a precise sufficient proof target: for one fixed $h$, prove that every piecewise-syndetic subset of $G$ has a height-$h$ relation. This is an equivalent reformulation of the unresolved premise, not a proof of it. Existing bounds allow the height to depend on the finite thick-cover size.

Positive patterns themselves force cover growth

The growth of return-cover witnesses is not merely caused by intersecting with a conductor tail. Let $B\ne\varnothing$ be height-$h$ independent for $h\ge1$, and put $R_X=\{t:Xt\subset B\}$ for a finite $X\subset B$.

First, $Bk\subset B$ implies $k=1$. Finite order turns that inclusion into equality. A nonidentity stabilizer then contains an element of prime order, whose orbit on any point of $B$ is a forbidden ternary prime-polygon relation.

Consequently, given any nonempty finite $F\subset G$, there is a pattern $X\subset B$ with $|X|\le|F|$ such that $FR_X\ne G$. Choose $t\notin F$. For each $f\in F$, the preceding fact supplies $b_f\in B$ with $b_f f^{-1}t\notin B$. Taking $X=\{b_f:f\in F\}$ excludes $t$ from $FR_X$. Thus even the initial covering set need not persist through finitely many positive-pattern constraints.

There is also an arbitrarily large lower bound on covering cardinality. Suppose $FB=G$, and choose a finite subgroup $K\supset F$. Translate $B$ so that $X=B\cap K$ has maximal size among all its $K$-coset sections. Then

$$|X|\ge\lceil |K|/|F|\rceil, \qquad R_X\text{ meets each }K\text{-coset at most once}. \tag{6}$$

The first claim follows by covering $K$ with the sets $f(B\cap K)$. For a return $r$, maximality gives $B\cap Kr=Xr$. Two different returns in the same coset would therefore yield a nontrivial stabilizer of the independent set $X$, which the preceding argument excludes.

If $ER_X=G$ for a finite $E$, choose a finite subgroup $H\supset E\cup K$. Then $H=E(R_X\cap H)$ and $|R_X\cap H|\le |H|/|K|$, proving

$$|E|\ge |K|. \tag{7}$$

If the maximizing translation was $B'=Ba^{-1}$, transport its pattern back as $Xa\subset B$; commutativity gives $R_{Xa}(B)=R_X(B')$. Hence this conclusion concerns patterns inside one original $B$. If $B$ is uniformly recurrent, all these positive-pattern returns are syndetic, but their minimum cover sizes are unbounded. This proves no coefficient-height growth or coloring theorem; it specifies why minimal recurrence alone does not fix that gap. Both pattern statements received independent internal review.

The source hypotheses, tensor lemma, tail construction, ultrafilter argument, and piecewise-syndetic equivalence received independent internal checks. No numerical experiment was used.