Exact limits of three moment flattening mechanisms

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The critical moment family has uniform hereditary full-dissociation extraction. Every fixed palette also forces a palette-dependent-height relation. The missing conclusion remains an actual monochromatic ternary relation. The three results here delimit particular ways of connecting these facts; they do not exclude color-adaptive, nonpointwise constructions.

1. Full tensors of critical moment configurations lose extraction

Let $E_i=E_{d_i,k_i}$, where $1\leq k_i\leq d_i/2$, and put

$$n_i=2^{d_i},\quad r_i=\sum_{j\leq k_i}\binom{d_i}j,\quad D_i=(d_i-k_i)\binom{d_i}{k_i},\quad N=\prod_i n_i,\quad R=\prod_i r_i.$$

For the full tensor $A=\bigotimes_{i=1}^s E_i$, its largest dissociated subset satisfies the exact finite bound

$$\boxed{\alpha(A)\leq \frac12\sum_{i=1}^s D_i\prod_{j\ne i}r_j+c_0R,\qquad c_0=\tfrac12\log_2(2\pi e/3).} \tag{1}$$

In particular its hereditary extraction constant is at most the right side divided by $N$.

Proof. Identify the columns with points of a Boolean cube of dimension $d=\sum_i d_i$. The retained Walsh rows have indices $S=(S_i)$ with $|S_i|\leq k_i$; this is a downset, though not a total-degree ball. The monomial and Walsh rows are related by an invertible triangular change, preserving dissociation. For a dissociated subset $F$ of size $M$, take independent fair bits $\xi_x$, $x\in F$, and form

$$Y_S=\sum_{x\in F}\xi_x\chi_S(x).$$

The vector $Y$ has entropy $M$ bits, because subset sums are distinct. Order rows by total degree. Expanding each Walsh character into squarefree monomials shows that, conditional on earlier rows, $Y_S$ lies in one residue class modulo $2^{|S|}$. All proper subindices occur earlier in this downset; their Walsh sums determine the needed monomial sums by the triangular expansion.

For an integer variable on a mesh of step $h$, add independent uniform interval jitter after dividing by $h$. Gaussian maximal entropy gives the bound $\frac12\log_2(2\pi e[\operatorname{Var}/h^2+1/12])$. Apply this conditionally, use concavity of logarithm, and use total variance to bound the averaged conditional variance by $\operatorname{Var}Y_S=M/4$. Thus

$$H(Y_S\mid\text{earlier rows}) \leq\tfrac12\log_2\!\left(2\pi e [M/4^{|S|+1}+1/12]\right).$$

Since $M\leq2^d$ and $2|S|\leq d$, the bracket is at most

$$2^{d-2|S|}/4+1/12\leq2^{d-2|S|}/3.$$

The contribution is consequently at most $d/2-|S|+c_0$. Summing and using

$$\sum_{j\leq k_i}(d_i/2-j)\binom{d_i}j=D_i/2$$

proves (1). The full family itself suffices for the claimed upper bound on hereditary extraction. Cube-flip transitivity also gives $\eta(A)=\alpha(A)/N$, but is not needed for the exclusion. $\square$

For two critical factors with both dimensions tending to infinity,

$$\eta(E_1\otimes E_2) \leq\frac{D_1}{2n_1}\frac{r_2}{n_2} +\frac{D_2}{2n_2}\frac{r_1}{n_1} +c_0\frac{r_1r_2}{n_1n_2}\longrightarrow0. \tag{2}$$

Indeed $D_i/(2n_i)\to1/(2\sqrt{2\pi})$ and $r_i/n_i\to0$. No comparison of the two dimension-growth rates is required. For $s$ identical factors,

$$\eta(E_{d,k}^{\otimes s})\leq s\frac D{2n}(r/n)^{s-1}+c_0(r/n)^s. \tag{3}$$

Thus every fixed $s\geq2$ loses uniform extraction along the critical sequence. This is an actual exclusion, not merely the observation that rectangular tensors differ from total-degree moments. It does not apply to sparse tensor subsets or to the distinct family $E_{\sum d_i,k_{\sum d_i}}$.

The analogous simplicial and mixed full tensors

These two additional bounds are author-recorded. For simplicial blocks $S_{n_i,k_i}$, write $N_i=\binom{n_i}{k_i}$, $M_i=\binom{n_i}{k_i-1}$, and $L_i=n_i-k_i+1$. Every row in their full tensor has $L_1L_2$ nonzero entries, of magnitude one. Fair subset-sum entropy and the one-coordinate integer Gaussian bound give

$$\eta(S_1\otimes S_2)\leq \frac{k_1k_2}{2L_1L_2} \log_2\!\left(2\pi e[L_1L_2/4+1/12]\right). \tag{3a}$$

Indeed each row variance is at most $L_1L_2/4$; sum the entropy bounds over its $M_1M_2$ rows and divide by $N_1N_2$. No row independence is assumed. At fixed critical constants $n_i/k_i\sim C_i\log n_i$, (3a) tends to zero when both $n_i$ grow, of order $1/\log n_1+1/\log n_2$.

For a moment block $E_{d,k}$, write $n,r,D$ as above. For a simplicial block write $L_s=n_s-k_s+1$. The same proof, now conditioning the moment rows at each simplicial row, gives

$$\eta(E_{d,k}\otimes S_{n_s,k_s})\leq \frac{k_s}{L_s}\left\{\frac D{2n} +\frac rn\left[\tfrac12\log_2 L_s+c_0\right]\right\}. \tag{3b}$$

The conditional mesh is $2^{|S|}$, and the unconditional row variance is at most $nL_s/4$; the derivation of (1) thus adds $\frac12\log_2L_s$ to each moment-row entropy charge. At the two critical scales, the first term in (3b) vanishes with $n_s$, and the second vanishes with $d$, since $(k_s/L_s)\log_2L_s$ stays bounded. Thus this mixed full tensor also loses extraction whenever both critical indices tend to infinity.

2. A one-dimensional height-two space with no ternary vector

For $1\leq\ell\leq d-2$, put $a=\ell-1$, $b=d-\ell-2$, and

$$C=\{u\in\{0,1\}^3:|u|\ne2\}\times\{0\}^{a}\times\{0,1\}^{b}.$$

The real multilinear polynomials of degree at most $\ell$ supported on $C$ form exactly the line generated by

$$P(u,z,w)=(2-u_1-u_2-u_3)\prod_{j=1}^{a}(1-z_j). \tag{4}$$

To see this, support on $z=0$ forces any function to equal $\prod_j(1-z_j)Q(u,w)$, by uniqueness of multilinear interpolation. The degree of this product is $a+\deg Q$, so $Q$ is affine. Vanishing at every weight-two $u$, for all $w$, first kills all free-coordinate coefficients. The three remaining equations on the affine coefficients force $Q$ to be a multiple of $2-u_1-u_2-u_3$.

The values of $P$ include $2,1,-1$. Thus it has integer height two, but no nonzero scalar multiple takes only the values $0,\pm1$. Parity multiplication identifies degree-$\ell$ functions with relations for moment order $d-\ell-1$, as in the existing moment construction. Consequently a real annihilator, even an integer one of height two, does not automatically give a ternary annihilator on the same support.

This applies at the exact critical degree $\ell=d-k_d-1$, eventually. The support density is $5/2^{\ell+2}$, tending to zero. It does not construct a bounded proper palette or contradict an arbitrary-color theorem for large color classes.

3. Recoverable pointwise ternarization incurs full original degree

Let $d/2\leq\ell<d$, and choose proper subsets $A,B\subset[d]$ with $|A|=|B|=\ell$ and $A\cup B=[d]$. Put

$$p(x)=x_A+x_B\in\{0,1,2\}.$$

Suppose $Y$ is finite and an encoder $\Phi:\{0,1,2\}\times Y\to\{0,\pm1\}$ obeys

$$\Phi(0,y)=0,\qquad \sum_{y\in Y}w_y\Phi(h,y)=h \quad(h=0,1,2) \tag{5}$$

for fixed real weights $w_y$. For some $y$,

$$b_y:=\Phi(2,y)-2\Phi(1,y)\ne0.$$

Otherwise ternarity would force both $\Phi(1,y)$ and $\Phi(2,y)$ to vanish for every $y$, contradicting (5). On the Boolean cube,

$$\Phi(p(x),y)=\Phi(1,y)(x_A+x_B)+b_yx_{[d]}. \tag{6}$$

Thus some fiber has degree exactly $d$. If $Y=\{0,1\}^t$, the total multilinear degree of $F(x,y)=\Phi(p(x),y)$ is at least $d$, since restriction cannot increase degree. For unweighted recovery, $\sum_y b_y=0$, so its nonzero coefficient function $b(y)$ is nonconstant and the total degree is at least $d+1$.

In particular $t=o(d)$ ancillary coordinates cannot give a universal encoder of this form within a critical half-degree bound on $d+t$ variables. The theorem concerns pointwise, zero-preserving, linearly recoverable encoders. It does not prohibit palette-dependent or nonpointwise operations.

Checks

All three proofs are finite algebra and entropy arguments, with no computational search, and are author-recorded. Statements (4)--(6) were checked against a separate derivation, and a separate reviewer checked the rectangular tensor calculation, including its mesh conditioning, constants, distinct columns, and asymptotic quantifiers. These are internal reviews only.