Root colorings, transversals, and symmetry

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The prime-tensor extraction theorem proves the local property for all roots of unity. This page makes the remaining global coloring problem more concrete. None of the reductions below resolves it.

Local extraction, fractional coloring, and the precise source gap

Let $\Psi(n)$ be the largest dissociated cardinality in $\mu_n$. For any $F\subseteq\mu_n$ and a maximum dissociated set $I\subseteq\mu_n$, rotation preserves dissociation and gives

$$\frac1n\sum_{g\in\mu_n}|F\cap gI| =\frac{|F|\Psi(n)}n.$$

Thus the hereditary dissociation constant of $\mu_n$ is exactly $\Psi(n)/n$: averaging proves the lower bound and taking $F=\mu_n$ proves the upper bound. The weighted version is identical, replacing cardinalities by sums of nonnegative vertex weights. Consequently the fractional chromatic number of its signed-relation hypergraph is $n/\Psi(n)$. For a direct proof, give each of the $n$ indexed rotates of $I$ weight $1/\Psi(n)$, so every vertex is covered with weight one; conversely summing the fractional covering constraints and using the maximum class size proves the matching lower bound.

It follows that $\inf_n\Psi(n)/n>0$ is equivalent to proportional dissociation of all roots. The prime extraction theorem proves this local condition. The target is still integral chromatic growth; bounded fractional chromatic number is not its substitute.

The primary 2006–2007 working-group contribution of Myriam Déchamps explicitly separates these issues: Question 36 concerns $n/\Psi(n)$, Question 38 asks finite quasi-independent covering of all roots, and Question 39 asks Sidonicity. The definitions and statements were checked on pp. 19–20 of the full primary document. This is a historical source check, not a claim about the present literature status or publication priority of the result. It supplies no proof of the missing coloring assertion.

Relation to the odd-prime integer-block candidate

Let $Q=\sup_k\chi(E_k)$ for the odd-prime grids of the main argument. Restriction gives $Q\leq\chi(\mu_\infty)$. Conversely, if $Q<\infty$, every odd squarefree root set is a slice of a prefix grid and has a proper $Q$-coloring. An even squarefree root set is the disjoint union of an odd root set and its negative, so has a $2Q$-coloring using separate palettes. The already proved prime-power coset separation extends that bound to every $\mu_n$. A diagonal subsequence of colorings of $\mu_{j!}$ stabilizes the color of each root, yielding a global proper coloring; every relation has finite support. Hence

$$Q\leq\chi(\mu_\infty)\leq2Q\qquad(Q<\infty).$$

Thus finite colorability of all roots and bounded coloring of the odd-prime grids are equivalent. Full roots are used only for this reformulation: their antipodal pairs are not passed directly to positive-integer encoding.

The new profile-compression theorem applies to arbitrary putative colorings, without symmetry assumptions. It forces a summable reciprocal loss when symbols having identical global color profiles are merged. This excludes fixed-order bounded-state colorings, but it permits all coordinate profiles to be distinct and hence does not give a contradiction.

Equivalence with a dissociated transversal — proved

Write $G=\mu_\infty$ for all roots of unity, with dissociation always referring to addition in $\mathbb C$. The following are equivalent:

1. $G$ has a finite dissociated partition.
2. For some prime $\ell$, there is a dissociated set containing exactly one representative of every multiplicative $\mu_\ell$-coset in $G$.

A proper $q$-coloring gives such a transversal for every prime $\ell>q$. The cosets have finite size $\ell$; their quotient is infinite.

Proof

Fix $\ell>q$, let $T$ consist of roots of order coprime to $\ell$, and put $K=\mathbb Q(T)$. Restrict the coloring to classes $C_1,\ldots,C_q$ of $T$. Choose distinct $a_c\in\mu_\ell$. They are linearly independent over $K$: for every finite conductor $m$ coprime to $\ell$, the cyclotomic degree formula gives $[\mathbb Q(\zeta_{m\ell}):\mathbb Q(\zeta_m)]=\ell-1$, and any $\ell-1$ distinct $\ell$-th roots form a rational basis. Every hypothetical dependence over $K$ lies at such a finite stage.

Therefore $D_0=\bigcup_c a_c C_c$ is dissociated: a signed relation splits first by these independent coefficients and then by each color class. It meets each $\mu_\ell$-coset in $\mu_\ell T$ exactly once.

Choose representatives $r_\alpha\in\mu_{\ell^\infty}$ modulo $\mu_\ell$. The spaces $r_\alpha K(\zeta_\ell)$ have direct sum. For any finite collection, pass to a common $\ell^j$ level and a finite coprime conductor $m$. There are $\ell^{j-1}$ cosets, and the degree over $\mathbb Q(\zeta_{m\ell})$ is exactly $\ell^{j-1}$; their spanning representatives are consequently a basis.

Thus $D=\bigcup_\alpha r_\alpha D_0$ is dissociated and is the desired transversal in the entire $G$. This step is necessary: simply choosing an unused prime does not handle its unbounded prime-power levels.

Conversely, a dissociated transversal gives the partition $G=\bigsqcup_{a\in\mu_\ell}aD$, whose $\ell$ classes are dissociated. This proves the equivalence.

Exact remaining gap: no argument here excludes dissociated $\mu_\ell$-transversals for arbitrarily large primes $\ell$. The checked $r(\mu_{105})=52$ gives only the finite bound $\chi(\mu_{210})\geq5$: antipodal reorientation proves $r(\mu_{210})=52$. It is not an unbounded obstruction.

Exact condition on selector labels

For the coprime-order group $T$ above and any map $f:T\to\mathbb Z/\ell\mathbb Z$, put $C_a=f^{-1}(a)$ and $D_f=\{\zeta_\ell^{f(t)}t:t\in T\}$. Then $D_f$ is dissociated exactly when all $C_a$ are dissociated and

$$\bigcap_{a=0}^{\ell-1}\Sigma(C_a)=\{0\}. \tag{*}$$

Indeed a selector relation is $\sum_a\zeta_\ell^a u_a=0$, where $u_a\in\Sigma(C_a)\subset K$. The sole $K$-linear relation among these $\ell$ roots forces all $u_a$ equal. A common nonzero value gives a nontrivial selector relation; a common zero value is harmless exactly when every class is dissociated. Every representation has finite support, so this criterion also applies to the infinite classes. It is the same coefficient-sensitive intersection condition as the finite tensor layer lemma, not just a condition excluding monochromatic regular polygons.

Regular-polygon cosets alone cannot force unbounded coloring. Write $x\in\mathbb Q/\mathbb Z$ in its finite-support primary decomposition $x=\sum_p x_p$, with $x_p\in[0,1)$, and color by the parity of the number of coordinates in $[1/p,2/p)$. In every order-$p$ coset exactly one point has that $p$-flag and the other flags are fixed. Every nontrivial finite subgroup coset therefore has both colors. This is not a dissociated coloring: the rest of the signed relations remain relevant.

Equivariant partitions require exactly one color per divisor

If $n$ is squarefree with $k$ prime factors, the least number of dissociated classes in a partition of $\mu_n$ permuted by its Galois group is exactly $2^k$.

For the upper bound, partition by multiplicative order. The primitive $d$-th roots, for squarefree $d$, form the tensor product of the primitive-prime bases and are rationally independent.

For the lower bound write the Galois group as $\mathcal G=\prod_{p\mid n}\mathcal G_p$, acting also on the palette. Fix a transitive palette orbit $O$, and let $a$ be the product of primes whose coordinate group acts nontrivially on $O$. If primitive $d$-th roots map to $O$, then $a\mid d$. Indeed the stabilizer of a primitive $d$-th root includes every coordinate outside $d$. Those coordinates fix its color and hence all of $O$, since $\mathcal G$ is abelian and acts transitively on $O$.

Write $d=ab$. All coordinate groups in $b$ act trivially on $O$, so the color map on primitive $d$-th roots is independent of the primitive-$b$ coordinate. Its remaining equivariant map on primitive $a$-th roots is determined by one color in $O$.

If more than $|O|$ orders map to $O$, two distinct orders $ab,ac$ give the same such map. Fix a primitive $a$-th root $u$. The two disjoint sets $u\operatorname{Prim}(b)$ and $u\operatorname{Prim}(c)$ are monochromatic in the same color, and their sums are $\mu(b)u$, $\mu(c)u$, with both Möbius values equal to $\pm1$. Explicitly, the forbidden signed relation is

$$\mu(c)\sum_{x\in u\operatorname{Prim}(b)}x -\mu(b)\sum_{y\in u\operatorname{Prim}(c)}y=0.$$

Thus at most $|O|$ orders feed this orbit. Summing over palette orbits shows that at least the $2^k$ divisors' worth of colors are required.

In particular any finite dissociated partition of all roots, if it exists, must have infinite Galois orbit. Otherwise its common refinement with its finitely many Galois translates would be a finite equivariant dissociated partition, contradicting the finite theorem for arbitrarily large $k$.

There is no justification for assuming a general coloring is equivariant or has finite orbit. Compactness or averaging alone does not give that assumption. Therefore this exponential bound is not a lower bound for unrestricted dissociated coloring.