Coefficient-growth flattening beyond the Hall condition

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Standing

This is a positive extension of Hall/determinant flattening. It permits differing coefficient heights and does not require a Hall-tight subfamily. It still requires the product in question to have one actual color. The presently available focusing argument supplies no coefficient growth bound meeting the new criterion, so the main coloring theorem remains unproved.

All relations have unrestricted finite support. Coefficient height is not the fixed-order notion of freeness.

1. An exact sufficient inequality

For $1\le i\le r$, let $X_i$ be finite sets of roots in pairwise coprime conductor groups, with nonzero relations

$$\sum_{x\in X_i}c_{i,x}x=0, \qquad c_{i,x}\in\mathbb Z\setminus\{0\}.$$

Assume $X_1\cdots X_r$ is monochromatic. Put

$$I_i=\{|c_{i,x}|:x\in X_i\},\quad m_i=|I_i|.$$

If some $m_i=1$, dividing that relation by its common magnitude gives a ternary relation; fix arbitrary points in the other factors. Otherwise all $m_i\ge2$. Write

$$b_i=\min I_i,\quad d_i=m_i-1,\quad N=\prod_i m_i,\quad D=\prod_i d_i,$$

and, for $t\in\prod_i(I_i\setminus\{b_i\})$, put

$$W_t=\prod_i(b_i+t_i).$$

Theorem. There is a nonzero monochromatic ternary relation if

$$2^N>\prod_t(1+W_t). \tag{1}$$

In particular, the simpler sufficient condition is

$$\boxed{ \prod_i\frac{m_i}{m_i-1} \ge 1+\sum_i\frac1{m_i-1} \sum_{a\in I_i\setminus\{b_i\}}\log_2(b_i+a). } \tag{2}$$

The weak inequality in (2) is intentional.

Proof

Group the given relation as

$$w_{i,a}=\sum_{|c_{i,x}|=a}\operatorname{sign}(c_{i,x})x, \qquad \sum_{a\in I_i}a w_{i,a}=0.$$

In $\mathbb Z^{d_i}$, with coordinates indexed by $I_i\setminus\{b_i\}$, define

$$u_{i,a}=b_i e_a\quad(a\ne b_i),\qquad u_{i,b_i}=-\sum_{a\ne b_i}a e_a.$$

The rational linear map $e_a\mapsto w_{i,a}/b_i$ sends every $u_{i,a}$ to $w_{i,a}$, including the distinguished column.

Consider the $N$ indexed integer tensor columns

$$u_{1,a_1}\otimes\cdots\otimes u_{r,a_r}\in\mathbb Z^D.$$

If they have no nonzero indexed ternary relation, their $2^N$ subset sums are distinct. At coordinate $t$, the sum of the absolute values of all column entries is exactly $W_t$: in factor $i$ the two nonzero contributions have absolute values $b_i,t_i$, and these sums multiply under tensoring. The subset sums therefore occupy an integer interval of width $W_t$ in that coordinate. Consequently

$$2^N\le\prod_t(1+W_t).$$

Its contrapositive proves the existence of a nonzero indexed ternary relation under (1).

Apply the tensor product of the displayed rational maps, followed by multiplication in the compositum. The resulting relation is between the products $\prod_i w_{i,a_i}$. Expand their signed root sums. Different magnitude tuples use disjoint, nonempty Cartesian blocks of the $X_i$. Multiplication on root tuples is injective because the conductor groups form a direct product. Thus a nonzero formal coefficient survives, and all expanded coefficients are $0,\pm1$. The support lies in the specified monochromatic product. Extra linear dependencies among the grouped sums do not affect this support argument.

Finally $W_t>1$, and

$$\log_2(1+W_t)<1+\log_2W_t.$$

Summing over the $D$ coordinates gives

$$\frac1D\sum_t\log_2(1+W_t) <1+\sum_i\frac1{d_i}\sum_{a\ne b_i}\log_2(b_i+a).$$

Thus (2) implies the strict inequality (1). This proves the theorem.

2. Two magnitude groups and an example beyond Hall

If $I_i=\{a_i,b_i\}$, the universal columns are, up to sign, the indexed scalar products choosing one of $a_i,b_i$ at each factor. Every ternary relation on that indexed scalar family lifts by the proof above. In this case (1) becomes

$$2^{2^r}>1+\prod_i(a_i+b_i). \tag{3}$$

This is a sufficient, not a necessary, test.

For example, magnitude pairs $\{1,2\},\{1,3\}$ satisfy $16>1+3\cdot4=13$. More directly, from

$$w_{1,1}+2w_{1,2}=0,\qquad w_{2,1}+3w_{2,3}=0$$

one obtains

$$w_{1,2}w_{2,3}-w_{1,1}w_{2,3}+w_{1,2}w_{2,1}=0.$$

Its three Cartesian blocks are disjoint. The Hall test fails: the one-row unions have size two and the two-row union has size three. Rescaling the two original rows cannot make their magnitude pairs coincide, because their ratios are two and three.

This example is a conditional algebraic example: no assertion is made here that an arbitrary coloring supplies actual root factors with these two magnitude pairs.

3. A necessary growth rate for dissociated products

Put $h_i=\max I_i$. Since $b_i+a\le2h_i-1$, a dissociated full product necessarily satisfies

$$\sum_{i=1}^r\log_2(2h_i-1) >\prod_{i=1}^r\frac{m_i}{m_i-1}-1. \tag{4}$$

If all $m_i\le M$, this implies

$$\sum_{i=1}^r\log_2(2h_i-1)> \left(\frac M{M-1}\right)^r-1,$$

and hence

$$\max_{i\le r}h_i> \frac{1+2^{((M/(M-1))^r-1)/r}}2. \tag{5}$$

Only the maximum height among the first $r$ factors is forced to escape at this scale, not every individual height.

This broadens the target for adaptive focusing: one can now seek monochromatic products violating (4), instead of insisting on one common height or a Hall-tight subfamily. No such growth control has been proved. The present full-profile palette $q^{|F|}$ allows much faster height growth, and the only automatic magnitude-count bound is $m_i\le h_i$.

4. Exact escape and actual roots

In the universal scalar model, take

$$I_i=\{1,H_i\},\qquad H_i=2^{2^{i-1}}.$$

The indexed scalar products, up to signs, are exactly $\{2^a:0\le a<2^r\}$, a dissociated set. The identity

$$\prod_{i=1}^r(1+H_i)=2^{2^r}-1$$

attains equality in the exact subset-counting bound. Thus a counting proof cannot replace the strict inequality in (1) by a weak one.

The scalar calculation alone is not an actual-root construction. The separate root-circuit construction realizes this escape with conductor-disjoint root circuits and exactly these magnitude pairs. The stronger recurrent-root construction also shows that infinite rational coloring number in every tail does not itself repair the missing coefficient-growth control.

5. Different actual colors on partially monochromatic blocks

The full product need not have one color for the following extension. Keep the relations and universal integer columns $u_\gamma$ from Section 1. Let $\Gamma$ be any collection of magnitude tuples whose Cartesian blocks are individually monochromatic in a putative proper dissociated $q$-coloring. Different blocks may have different colors. Set

$$K=|\Gamma|,\qquad W_t(\Gamma)=\sum_{\gamma\in\Gamma}|(u_\gamma)_t|.$$

Then necessarily

$$\boxed{2^K\le\prod_t\left(1+\frac{W_t(\Gamma)}q\right)^q.} \tag{6}$$

A strict violation forces a monochromatic ternary relation supported only on the good blocks.

Indeed split $\Gamma$ into actual block-color classes $\Gamma_c$, and put $W_{c,t}=\sum_{\gamma\in\Gamma_c}|(u_\gamma)_t|$. Each indexed integer family $(u_\gamma)_{\gamma\in\Gamma_c}$ must be dissociated: the proof of Section 1 would lift a ternary dependence to disjoint root blocks of color $c$. Consequently

$$2^{|\Gamma_c|}\le\prod_t(1+W_{c,t}).$$

Multiply these inequalities and apply arithmetic--geometric mean to the $q$ factors at each coordinate, using $\sum_cW_{c,t}=W_t(\Gamma)$. This proves (6), including empty classes and zero widths. For all $N$ blocks, $W_t(\Gamma)=\prod_i(b_i+t_i)$.

For the constant-ratio-two circuits $2w_i=\sum_{b\in B_i}\epsilon_b b$, there is a sharper elementary statement. Index the binary product blocks by the coordinates $J$ using the full leaf set rather than singleton $\{w_i\}$. The signed sum on that block is

$$2^{|J|}\prod_iw_i.$$

Two distinct good blocks of the same Boolean rank and the same actual color therefore give a nonzero ternary relation by subtraction. Their supports are disjoint and multiplication is injective. Hence every rank has at most $q$ good blocks, and

$$K\le q(r+1). \tag{7}$$

These inequalities broaden the actual-color target beyond a monochromatic rectangle. They do not produce good blocks: a coloring that splits every local leaf set can leave just the all-pivot singleton good. No lower bound on $K$ contradicting (6) or (7) has been forced for arbitrary colorings.

The theorem, coefficient-safe expansion, strictness, growth consequences, and partial-block extension received independent internal review. No numerical experiment was used. None establishes the required counterexample.