Uniform extraction from axis lines with linearly growing sides

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Theorem and scope

For integers $d_i\geq3$, put

$$X_t=\prod_{i=1}^t[d_i],\qquad N_t=|X_t|,$$

and let $L_t\subset\mathbb Z^{X_t}$ consist of the indicator vectors of all full axis-parallel lines. Thus an $i$-line has $d_i$ entries equal to one. Different lines give distinct, nonzero, non-antipodal vectors.

Write $\rho(L_t)$ for the positive-real Sidon parameter: for every complex coefficient family $(c_\ell)_{\ell\in L_t}$,

$$\max_{z\in\mathbb T^{X_t}} \operatorname{Re}\sum_{\ell\in L_t}c_\ell\prod_{x\in\ell}z_x \ \geq\ \rho(L_t)\sum_{\ell\in L_t}|c_\ell|. \tag{1}$$

The direct finite estimate is

$$\boxed{\displaystyle \rho(L_t)\geq \min_{1\leq i\leq t} \prod_{j=i+1}^t\cos(\pi/d_j)^{d_i}.} \tag{2}$$

In particular, if

$$M=\sup_{i\geq1}d_i\sum_{j>i}d_j^{-2}<\infty, \tag{3}$$

then every $L_t$ has positive-real Sidon parameter at least $2^{-9M}$, and there is one $\delta>0$ such that every subset of every $L_t$ contains a dissociated subset of relative size at least $\delta$.

This includes the dense linear schedule $d_i=i+2$, with $M\leq1$. A sharper elementary estimate below gives $\rho(L_t)\geq\exp(-\pi^2/2)$. More generally, for $d_i=a+b(i-1)$, integers $a\geq3,b\geq1$, one has $M\leq1/b$ and $\rho(L_t)\geq\exp(-\pi^2/(2b))$.

This is a new direct estimate for the line subfamily. It does not assert proportional dissociation of the full simplex tensors at a linear schedule, which the full-tensor entropy obstruction rules out. It also does not prove that the coloring numbers $\chi_1(L_t)$ are unbounded. The proof below is self-contained until its final application of the supplied Pisier theorem. Independent internal review checked the finite induction, constants, embeddings, and exact Pisier hypotheses; no defect was found.

1. A target-independent root-polygon rounding lemma

For every integer $d\geq3$ and $u\in\mathbb T$, there is a finite random variable $Y\in\mathbb T$ satisfying

$$Y^d=u\quad\hbox{surely},\qquad \mathbb EY=a_d:=\cos(\pi/d)>0. \tag{4}$$

Indeed the $d$-th roots of $u$ are the vertices of a rotated regular $d$-gon. Its convex hull contains the closed disk centered at zero of radius $\cos(\pi/d)$. The positive real number $a_d$ therefore is a convex combination of those vertices. Use the combination's weights as probabilities.

The distribution may depend on $u$, but its expectation does not. Independent copies of (4), with possibly different prescribed targets, therefore satisfy

$$\mathbb E\prod_{x\in F}Y_x=a_d^{|F|} \tag{5}$$

for every finite set $F$ of distinct indices. No uniform choice of the whole distribution, and no measurable choice in the target, is required: only finitely many targets occur in a given step.

2. Directional induction

For a polynomial on $L_t$, denote its total coefficient magnitude in direction $i$ by $W_i$. We prove the stronger directional assertion

$$\max_z\operatorname{Re}P(z) \geq\sum_{i=1}^t r_{i,t}W_i, \qquad r_{i,t}=\prod_{j=i+1}^t a_{d_j}^{d_i}. \tag{6}$$

For $t=1$, the family has a single line, whose character can be assigned any unit value, so $r_{1,1}=1$ works.

Suppose the assertion holds through $t-1$, and put $d=d_t$. Decompose the point variables into layers $z^{(s)}=(z_{x,s})_{x\in X_{t-1}}$, $s\in[d]$. An arbitrary polynomial on $L_t$ has the form

$$P(z)=\sum_{s=1}^d P_s(z^{(s)}) +\sum_{x\in X_{t-1}}b_x\prod_{s=1}^d z_{x,s}, \tag{7}$$

where each $P_s$ uses precisely the old line directions.

In each $P_s$, multiply every direction-$i$ coefficient by $a_d^{d_i}$, obtaining $\widetilde P_s$. Choose the layers $z^{(s)}$ independently using the induction hypothesis for these rescaled polynomials. Their total real value satisfies

$$\sum_s\operatorname{Re}\widetilde P_s(z^{(s)}) \geq\sum_{i<t}r_{i,t-1}a_d^{d_i}W_i. \tag{8}$$

All layer choices are now fixed.

For each $x$, if $b_x\ne0$, prescribe

$$u_x= \frac{\overline b_x}{|b_x|} \left(\prod_s z_{x,s}\right)^{-1}; \tag{9}$$

if $b_x=0$, choose any $u_x\in\mathbb T$. Choose independent random variables $Y_x$ from (4), with targets $u_x$, and replace every layer coordinate by

$$z'_{x,s}=z_{x,s}Y_x. \tag{10}$$

Every new-axis term then equals $|b_x|$ deterministically, since $Y_x^d=u_x$. An old $i$-line character, in any layer, is multiplied by the product of $Y_x$ over its $d_i$ distinct base points. By (5), its expected multiplier is exactly $a_d^{d_i}$. Consequently

$$\begin{aligned} \mathbb E\operatorname{Re}P(z') &=\sum_s\operatorname{Re}\widetilde P_s(z^{(s)})+W_t\\ &\geq\sum_{i<t}r_{i,t-1}a_d^{d_i}W_i+W_t. \end{aligned} \tag{11}$$

The random choice has finite support, so one outcome attains at least this expectation. This proves (6), with $r_{i,t}=r_{i,t-1}a_{d_t}^{d_i}$ and $r_{t,t}=1$, hence (2).

Rescaling before optimization is essential: no sign or phase of an old individual term is assumed favorable. The target-dependent distributions are selected only after the old layers are fixed. Independence is required across base points, not across layers; the same $Y_x$ is deliberately used in all layers so that (9) is solved exactly.

3. Uniform constants and dense schedules

For $d\geq3$,

$$-\log\cos(\pi/d)\leq\frac{9\log2}{d^2}. \tag{12}$$

One elementary proof observes that $F(x)=-\log\cos x\,/x^2$ is nondecreasing on $(0,\pi/3]$. The sign of its derivative is the sign of $h(x)=x\tan x+2\log\cos x$. Here $h(0)=0$, and

$$h'(x)=x\sec^2x-\tan x =\frac{x-\sin x\cos x}{\cos^2x}\geq0.$$

Thus $F(x)\leq F(\pi/3)=9\log2/\pi^2$, giving (12). Combining (12) with (3) gives

$$r_{i,t}\geq \exp\left(-9\log2\,d_i\sum_{j>i}d_j^{-2}\right) \geq2^{-9M}. \tag{13}$$

For the arithmetic schedule $d_i=a+b(i-1)$, monotonicity and an integral comparison give

$$\sum_{j>i}d_j^{-2} \leq\int_0^\infty(d_i+bu)^{-2}\,du =\frac1{bd_i}. \tag{14}$$

For these increasing arithmetic schedules, improve (12) to

$$-\log\cos(\pi/n)\leq\frac{\pi^2}{2n(n-1)},\qquad n\geq4. \tag{14a}$$

For $n\geq5$, set $u=\pi^2/(2n^2)$. The elementary inequalities $\cos(\pi/n)\geq1-u$ and $-\log(1-u)\leq u/(1-u)$ show that the left side of (14a) is at most $\pi^2/(2n^2-\pi^2)$, which is at most its right side because $\pi^2<10\leq2n$. For $n=4$, the left side is $(\log2)/2<3/8<\pi^2/24$, using $\log2<3/4$ and $\pi>3$. Every future side $d_j$, $j>i$, is at least four, and

$$\begin{aligned} d_i\sum_{j=i+1}^t-\log\cos(\pi/d_j) &\leq\frac{\pi^2d_i}{2} \sum_{j=i+1}^t\frac1{d_jd_{j-1}}\\ &=\frac{\pi^2d_i}{2b} \left(\frac1{d_i}-\frac1{d_t}\right) \leq\frac{\pi^2}{2b}. \end{aligned} \tag{14b}$$

Here $d_j-1\geq d_{j-1}$, so (14a) justifies the first inequality. Consequently:

• $d_i=i+2$ has $\rho(L_t)\geq\exp(-\pi^2/2)$.
• $d_i=10i$ has $\rho(L_t)\geq\exp(-\pi^2/20)$.
• A dyadic block schedule in which side $2^m$ is repeated $2^m$ times, for $m\geq2$, has $M\leq2$: from any point in its block, the remaining same-block contribution to $d_i\sum_{j>i}d_j^{-2}$ is less than one, and all later blocks contribute exactly one. Hence it has the uniform bound $\rho(L_t)\geq2^{-18}$.

The number of columns is

$$|L_t|=N_t\sum_{i\leq t}\frac1{d_i}, \tag{15}$$

so the arithmetic and displayed repeated-dyadic schedules have an unbounded line-to-point ratio. Equation (15) alone is not a chromatic lower bound: dissociated classes may have rational dependencies.

4. Independent additive-loss version

There is also a simpler, weaker directional estimate

$$\rho(L_t)\geq \min_{i\leq t}\left(1- \frac{\pi^2d_i}{2}\sum_{j=i+1}^t d_j^{-2}\right). \tag{16}$$

For each prescribed target $u_x=e^{i\theta_x}$, choose a finite random real lift $X_x$ with $e^{iX_x}=u_x$, mean zero, and $\mathbb EX_x^2\leq\pi^2$: take the two adjacent lifts surrounding zero with their mean-zero probabilities. Use independent lifts across base points. Replace all $z_{x,s}$ by $z_{x,s}e^{iX_x/d}$. New terms again align exactly, while the phase correction on an old $i$-line has mean zero and variance at most $\pi^2d_i/d^2$. The inequality

$$\operatorname{Re}(we^{iv}) \geq\operatorname{Re}w-v\operatorname{Im}w-\tfrac12|w|v^2$$

therefore bounds its expected loss by $\pi^2d_i|w|/(2d^2)$. This time optimize the unscaled old layer polynomials and induct with an additive loss for each direction, yielding (16). For $d_i=10i$, it gives the independent explicit bound $\rho(L_t)\geq1-\pi^2/20>1/2$.

The multiplicative root-polygon argument is stronger because its expectation is exactly real and target-independent; unlike the accumulated additive estimate, it remains positive for $d_i=i+2$.

5. One full-dissociation extraction constant

Embed $H_t=\mathbb Z^{X_t}$ into $H_{t+1}$ by $e_x\mapsto e_{(x,1)}$. This is a primitive direct-summand embedding, and takes every old axis line to a line in the fixed new-coordinate slice. Consequently the $L_t$ are nested in the direct-limit free abelian group $H$. Every character of $H_t$ extends to $H$.

Under (3), every finite polynomial supported on $L=\bigcup_tL_t\subset H$ therefore satisfies the ordinary Sidon estimate with constant at most $2^{9M}$, by (13). Apply Pisier's Theorem 2 once to this one infinite set in the dual of the compact abelian group $\widehat H$. It gives one integer $k$ such that every finite subset of $L$ has a dissociated subset of size at least its cardinality divided by $k$. Set $\delta=1/k$.

The exact theorem statement was checked in the supplied full source, Theorem 2 on reading-copy page 3, after reading its digest. The source does not provide an explicit dependence of $k$ on the Sidon constant, so no numerical extraction constant is asserted. The analytic constants above are explicit.

The carry-free integer realization and separated-block assembly already available in the wiki apply to these finite $0,1$-column families. They do not settle the remaining obligation

$$\sup_t\chi_1(L_t)=\infty$$

for any of these schedules. No completion claim follows from this extraction theorem alone.