Uniformly Sidon tensor-simplex blocks

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Standing and precise result

Proved here, but not an E0774 counterexample. Put

$$q_i=\left\lceil1000i\log_2(i+1)\right\rceil, \qquad S_q=\{e_1,\ldots,e_{q-1},-e_1-\cdots-e_{q-1}\},$$

and define the distinct integer-vector configuration

$$E_k=S_{q_1}\otimes\cdots\otimes S_{q_k} \subseteq\mathbb Z^{d_k},\qquad d_k=\prod_{i\leq k}(q_i-1),\quad N_k=|E_k|=\prod_{i\leq k}q_i. \tag{1}$$

There is one $\delta>0$ such that every subset of every $E_k$ contains a dissociated subset of relative size at least $\delta$. Moreover $d_k/N_k\to0$, so their rational-linearly-independent coloring numbers are unbounded. The latter assertion is not a lower bound for dissociated coloring. The unproved obligation is

$$\sup_k\chi_{\mathrm{diss}}(E_k)=\infty. \tag{2}$$

The proof of the extraction statement uses the checked full-source Pisier Theorem 2 only at its last step. The new analytic estimate is proved below. Its lattice-rounding and analytic steps are author-recorded and were checked by a separate reviewer; this is not a claim of external review or publication priority.

1. The real Sidon estimate to prove

For a finite frequency family $E$ in a free abelian group with dual torus $G$, write $\rho(E)$ for the largest number such that

$$\max_{g\in G}\operatorname{Re}\sum_{v\in E}c_v\,v(g) \geq\rho(E)\sum_{v\in E}|c_v| \tag{3}$$

for all complex coefficients. Here $v(g)$ denotes the unit-modulus character value. A positive lower bound for $\rho$ gives the usual Sidon constant at most $1/\rho$, since the supremum of absolute value is at least the supremum of real part. The initial configuration is the free singleton $E_0=\{1\in\mathbb Z\}$, on the dual torus $\mathbb T$, for which $\rho=1$. This is the character $z\mapsto z$, not the constant character. We use the empty product convention $Q_0=\pi^2$.

We shall prove uniformly in $k$ that

$$\rho(E_k)>1-\frac1{90000}. \tag{4}$$

2. Integer lattice of phase lifts

For side sizes $q_1,\ldots,q_k$, let $I=\prod_i[q_i]$, $N=|I|$, and let $W\subseteq\mathbb R^I$ be the space of arrays whose sum on every coordinate line is zero. Set $L=W\cap\mathbb Z^I$.

The integer vectors $b_j=e_j-e_q$, $1\leq j<q$, form a basis of the one-factor zero-sum lattice. Their tensor products form an integer basis of $L$. Indeed, array entries with every coordinate less than its last value are arbitrary integers, and the line-sum equations uniquely determine all remaining entries by signed sums of those entries. In particular

$$L=\bigotimes_i A_{q_i-1},\qquad \dim W=\prod_i(q_i-1).$$

The character-value torus of (1) is precisely $\exp(2\pi iW)\subseteq\mathbb T^I$. In these normalized coordinates, the map $x\mapsto\exp(2\pi ix)$ has period lattice $L$. Equivalently, choose the phases on the interior grid arbitrarily and extend by reciprocal products on boundary lines. This also proves that this torus is connected. In radian coordinates the map is $X\mapsto\exp(iX)$, and its period lattice is $2\pi L$. The random variable $x$ in Section 3 is normalized; $X$ in (7) and Section 4 is measured in radians.

3. Randomized lattice rounding — proved

For every $x\in W$ there is a finite random lattice point $Z\in L$ such that $\mathbb E(x-Z)=0$ and, for every coordinate $a\in I$,

$$\mathbb E(x_a-Z_a)^2 \leq\frac14\prod_{i=1}^k \frac{q_i-1+H_{q_i-1}}{q_i}, \qquad H_n=\sum_{j=1}^n\frac1j. \tag{5}$$

Here is the full rounding argument. Take an ordered integer basis $b_1,\ldots,b_d$ of $L$, with orthogonal Gram–Schmidt vectors $u_1,\ldots,u_d$. Starting with residual $x$, process indices in descending order. If its coefficient along $u_i$ is $\alpha_i$, round $\alpha_i$ to one of its adjacent integers, with probabilities chosen so that the rounded integer has conditional expectation $\alpha_i$, and subtract that integer times $b_i$. At termination the residual is

$$e=x-Z=\sum_i\eta_i u_i.$$

Conditional on the choices at larger indices, $\mathbb E\eta_i=0$ and $\mathbb E\eta_i^2\leq1/4$. For $i<j$, $\eta_j$ is determined by those larger-index choices, so $\mathbb E\eta_i\eta_j=0$. Therefore

$$\mathbb E e_a^2\leq\frac14\sum_i u_i(a)^2. \tag{6}$$

The same conditional expectations also give $\mathbb E e=0$. Both assertions hold for every fixed target and every chosen basis, so the mean-zero assertion survives the basis randomization below.

For the anchored basis of $A_{q-1}$, the Gram–Schmidt squared lengths are $2,3/2,\ldots,q/(q-1)$: its leading $j$-by-$j$ Gram determinant is $j+1$, because the Gram matrix is $I+\mathbf1\mathbf1^T$. Thus their sum is $q-1+H_{q-1}$. In lexicographic order the tensor basis has tensor-product Gram–Schmidt vectors (by triangularity and orthogonality). Its total squared Gram–Schmidt length is consequently

$$T=\prod_i(q_i-1+H_{q_i-1}).$$

Finally choose a uniform independent permutation of the $q_i$ coordinate values in every factor and use that permuted lattice basis for rounding. These permutations preserve $L$, act transitively on $I$, and preserve $T$. Averaging (6) gives $T/(4N)$ in every coordinate, which is (5). The target $x$ stays fixed: (6) is a target-independent bound for each chosen basis. There is no assumption that $x$ is permutation invariant.

Multiplying the residual by $2\pi$ gives a random lift $X$ of any prescribed torus element such that

$$\mathbb E X=0,\qquad \mathbb E X_a^2\leq Q_k, \qquad Q_k=\pi^2\prod_{i\leq k} \frac{q_i-1+H_{q_i-1}}{q_i},\quad Q_0=\pi^2. \tag{7}$$

4. Tensor extension inequality — proved

Write a polynomial on $E\otimes S_q$ as $q$ column polynomials $f_1,\ldots,f_q$ on the base torus $G$. Their character tuples are exactly $(g_1,\ldots,g_q)$ satisfying $\prod_jg_j=1$.

First maximize $\operatorname{Re}f_j$ independently, at points $g_j$. Their total real value is at least $\rho(E)\sum_{v,j}|c_{v,j}|$. Choose a random real lift $X$ of $(\prod_jg_j)^{-1}$ obeying (7), and replace each $g_j$ by $g_j\exp(iX/q)$. The new tuple has product one and is therefore admissible for the tensor polynomial.

For a single complex term $z$,

$$\operatorname{Re}(ze^{it})\geq \operatorname{Re}z-t\operatorname{Im}z-\tfrac12|z|t^2.$$

Since $\mathbb E X_a=0$, each first-order term vanishes in expectation. Taking expectations of the second-order bound and summing the columns proves

$$\rho(E_{k+1})\geq\rho(E_k)-\frac{Q_k}{2q_{k+1}^2}. \tag{8}$$

All random choices have finite support; an outcome attains at least the expected value. No simultaneous phase interpolation or tensor multiplicativity of ordinary Sidon constants is assumed.

The earlier stationarity proof is also valid: at each column maximum its derivative vanishes in every tangent direction $X\in W$. The mean-zero proof records a stronger property of the rounding lemma and does not require that calculation. A separate adversarial check covered the lattice basis, the target-dependent quantifiers, and the cyclotomic transfer and found no defect. This is internal checking, not external review.

5. Explicit summable loss

Let $a_i=(q_i-1+H_{q_i-1})/q_i$. With the side sizes in (1),

$$q_i\geq1000i\log_2(i+1),\qquad q_i\leq2000i^2,$$

and $H_{q_i-1}-1\leq\log q_i$. All unmarked logarithms here are natural. Consequently

$$\log a_i\leq\frac{\log q_i}{q_i} \leq\frac{\log8000}{1000i}<\frac1{100i}.$$

It follows that $Q_{i-1}\leq\pi^2e^{1/100}i^{1/100}$. Hence

$$\sum_{i\geq1}\frac{Q_{i-1}}{2q_i^2} \leq\frac{\pi^2e^{1/100}}{2\cdot10^6} \sum_{i\geq1}i^{-199/100} <\frac1{90000}. \tag{9}$$

For explicit elementary numerical bounds use $\pi^2<10$, $e^{1/100}\leq100/99$, and $\sum_i i^{-199/100}\leq1+\int_1^\infty x^{-199/100}dx=199/99$. Summing (8) from $\rho(E_0)=1$ proves (4).

6. One full dissociation constant, not a sequence of constants

Embed the ambient free abelian group of $E_k$ into that of $E_{k+1}$ by $v\mapsto v\otimes e_1$. This is a primitive direct-summand embedding, and it identifies $E_k$ with a slice of $E_{k+1}$. In the direct-limit free abelian group, let $E=\bigcup_kE_k$. Every finite polynomial on $E$ lies in some $E_k$, and every character of that finite-stage group extends to the limit group. By (4), $E$ is Sidon with constant at most $90000/89999$.

Pisier's Theorem 2, checked in the held reading copy on page 3, applies once to this single set $E$. It supplies one $\delta>0$ for dissociated extraction from every finite subset of $E$, hence from every subset of every block $E_k$. This avoids assuming any unprovided quantitative dependence in Pisier's announcement.

This last implication uses Pisier's stated theorem as a literature theorem; the supplied announcement refers elsewhere for its proof. No numerical value of $\delta$ is obtained in these notes.

The blocks have no zero or equal-up-to-sign columns: each factor has that property for $q_i\geq3$, and equality of nonzero pure tensors forces proportionality in every factor. The finite realization lemma therefore turns each block into distinct positive integers, preserving its entire signed support hypergraph and the common extraction constant.

Explicitly, order the $d_k$ coordinates, put $B_k=N_k+2$, and take

$$F_k=\left\{\left|\sum_{j=0}^{d_k-1}v_jB_k^j\right|:v\in E_k\right\}. \tag{10}$$

Every column entry is $0$ or $\pm1$, so a signed sum has coordinate absolute values at most $N_k<B_k-1$. The leading nonzero digit therefore cannot be canceled. This verifies the exact finite integer encoding directly. It preserves dissociated subsets and their coloring number, up to coordinate sign changes. We do not claim that this scalar encoding preserves the displayed numerical real-Sidon constant.

7. What is still missing

Since $q_i\leq2000i\log_2(i+1)$, $\sum_i1/q_i=\infty$. Therefore

$$\frac{d_k}{N_k}=\prod_{i\leq k}(1-1/q_i)\longrightarrow0.$$

Any partition into rationally independent classes needs at least $N_k/d_k$ classes. But dissociated classes can be rationally dependent; the saturated recurrence example emphasizes this distinction. No argument yet proves (2), and the constructed integer blocks are not claimed to have unbounded dissociated coloring number.

The next task is specifically to prove (2) for this tensor family, or to modify the family while retaining the summable bound (9) and introducing a verified unbounded dissociated coloring obstruction.