Arbitrary tuple tests and the ideal comparison

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This examines Section 4 of the preserved source. The source construction is OpenAI's; the independent examination is prepared for Plasma AI. Its scope includes arbitrary square-integrable tests on ordered label tuples, not merely functions of their products.

Collapse does not discard tuple information

A selected set uses at most $m$ slots in each big group. If its product is $n$, unique factorization determines the multiplicities $a_p$ of its labels. With $k_g$ selected slots in group $g$, its probability is

$$\rho(n)=\frac1n\prod_g \frac{k_g!}{V_g^{k_g}\prod_{p\in\mathcal P_g}a_p!} \le\frac{L^\kappa}{n},$$

where $\kappa$ depends on $m$ and the fixed group bounds but not $J$. For a tuple test $f$ supported where the product lies in $[cU,CU]$, put $a_n=\mathbb E(f\mathbf1_{P=n})$. Fiberwise Cauchy–Schwarz gives

$$|a_n|^2\le\rho(n)\mathbb E(|f|^2\mathbf1_{P=n}),\qquad \sum_n|a_n|^2\le\frac{L^\kappa}{cU}\|f\|_2^2. \tag{I1}$$

Thus a test can distinguish different orderings with the same product; only the product-dependent phase is collapsed. Repeated numerical labels are included by the factorials above.

For integer sequences on intervals of lengths $O(U),O(V)$, Cauchy–Schwarz and a geometric sum reduce the bilinear phase bound to

$$U+\sum_{1\le h\ll V} \min(U,\|\alpha h\|_{\mathbb R/\mathbb Z}^{-1}).$$

If $\alpha=u/r+\beta$, $(u,r)=1$, $|\beta|\le r^{-2}$, split the $h$ range into blocks of diameter at most $r/2$. Distinct phases in a block are at least $1/(2r)$ apart: the rational difference is at least $1/r$, and its perturbation is at most $1/(2r)$. The nearest point to zero costs $O(U)$ and the other distances contribute $O(r\log(2r))$. There are $O(1+V/r)$ blocks. For $r=1$ the trivial estimate suffices. Combining this with (I1) yields normalized bilinear norm

$$O\left(L^\kappa \left(V^{-1}+r^{-1}+\frac{\log(2r)}U+ \frac{r\log(2r)}{UV}\right)^{1/2}\right). \tag{I2}$$

Multiplication by $(P_AP_B)^{i\zeta}$ only changes the two tests by unit-modulus factors. This part has no restriction on $\zeta$.

The concentrated-cell attack

Partition product logarithms into cells of width $h$ and retain those of mass at least $\xi$. Let $\mathcal C_Q$ be the primitive characters of conductor at most $Q$, including conductor one. On a retained cell of mass $\nu$, the comparison kernel is

$$\frac{\mathbf1_{D\ {\rm in\ cell}}\mathbf1_{X\ {\rm in\ cell}}}{\nu} \sum_{\chi\in\mathcal C_Q}\overline{\chi(D)}\chi(X). \tag{I3}$$

There are at most $Q^2$ such characters. Two distinct ones have a nonprincipal quotient on their common modulus, at most $Q^2$. Hold all but one prime label fixed. The product-cell condition intersects its big prime group with an interval. Siegel–Walfisz followed by partial summation on this interval gives an absolute error $O_D(L^{-D})$, with any fixed $D$, since the prime is at least $\exp(L^c)$. Averaging the other labels preserves that error, regardless of how short the interval is.

Consequently, on the conditional probability space of a retained cell, the character Gram matrix has diagonal one and off-diagonal entries $O_D(\xi^{-1}L^{-D})$. Its norm is at most $1+O_D(Q^2\xi^{-1}L^{-D})$, and hence at most two with sufficient precision. The kernel in (I3) is the corresponding frame operator. Passing from conditional probability to the original probability multiplies both squared norms by the same $\nu$; it creates no $\xi^{-1}$ norm loss. Different cells are orthogonal blocks. This proves the norm bound on the full tuple space, including tests concentrated in a retained cell of mass close to $\xi$.

For discarded cells, let $E$ be the event that either selected product falls in one. Its probability is at most $(N_A+N_B)\xi$. The comparison kernel vanishes there. For arbitrary tests on the two independent tuple spaces, even if one is concentrated on $E$,

$$\mathbb E\bigl[\mathbf1_E|f(X_B)g(X_A)|\bigr] \le\mathbb P(E)^{1/2}\|f\|_2\|g\|_2. \tag{I4}$$

The second Cauchy–Schwarz factor is $\mathbb E|f(X_B)g(X_A)|^2=\|f\|_2^2\|g\|_2^2$ by independence. The tempting objection that restriction to a small event need not have small $L^2$ operator norm is correct but does not refute (I4), which is a bilinear $L^1$ estimate on independent spaces. It would be invalid to apply this argument to an unrestricted joint test instead of the separated conditional tests used here.

Uniform comparison at all additive frequencies

On product dyads $U,V$, put $Z=UV$. Dirichlet approximation with maximum denominator $\lfloor Z/S\rfloor$ gives

$$\alpha=u/r+\beta,\quad r\le Z/S,\quad |\beta|\le r^{-2},\quad |\beta|Z\le2S/r. \tag{I5}$$

For $r>Q$, use (I2) on the raw term. Move each comparison kernel onto its own test and use its norm at most two; its support remains within a fixed enlargement of the original product dyad. Apply (I1) and (I2) again. Since $U,V\ge\exp(L^c)$ and $\log(2r)\le L^{d+1}$ eventually, both bounds are at most a constant times

$$L^\kappa\bigl(o(L^{-A})+Q^{-1}+L^{d+1}/S\bigr)^{1/2}. \tag{I6}$$

For $r\le Q$, all selected products are units modulo $r$. Character expansion of $e(uv/r)$ on the unit group has coefficient $\ell^1$ norm at most $\sqrt{\varphi(r)}\le\sqrt Q$, by Parseval and Cauchy–Schwarz. Its imprimitive characters agree on these products with their primitive inducing characters, which are present in $\mathcal C_Q$.

For each such character $\psi$, integrating (I3) against $\psi(D)$ gives

$$\mathbb E_D\mathcal K(D,X)\psi(D) =\psi(X)+O_D(Q^2\xi^{-1}L^{-D}) \tag{I7}$$

pointwise for retained $X$. The inducing character supplies the exact diagonal term, and the other terms are the Gram errors just bounded. Using (I7) on both products and summing the rational character expansion costs $O_D(Q^{5/2}\xi^{-1}L^{-D})$.

The remaining scalar function of the product logarithm is $M(v)=\exp(i\zeta v)e(\beta e^v)$. Its derivative is $O(L^{C_4}+S)$ on the enlarged dyadic range. Each matched pair of logarithms differs by less than $h$, so their two sums differ by less than $2h$. Moreover $\mathbb E_D|\mathcal K(D,X)|\le Q^2$ for fixed retained $X$. The pointwise error of moving $M$ from the free product to the endpoint product is therefore $O(Q^4h(S+L^{C_4}))$. This is valid for complex coefficients and does not use positivity of the kernel.

There are at most $L^{d+1}$ product dyads per selected side. For a desired total error $L^{-A_0}$, take a dyadic target $B_0>A_0+2(d+1)+10$. Writing $Q=L^{q_*}$, $S=L^{s_*}$, $h=L^{-h_*}$, $\xi=L^{-x_*}$, sufficient successive choices are

$$\begin{aligned} q_*,s_*&>2(B_0+\kappa+10)+d+1,\\ h_*&>4q_*+\max(s_*,C_4)+B_0+10,\\ x_*&>2B_0+d+1+h_*+10,\\ D&>B_0+x_*+\tfrac52q_*+10. \end{aligned} \tag{I8}$$

They pay respectively for (I6), the log-phase error, the square root of discarded mass in (I4), and (I7) together with the Gram norm. The absolute Siegel–Walfisz precision is chosen last in this construction. No choice uses $J$, $\alpha$, or the individual value of $\zeta$.

Probe cancellation is an identity followed by an extremal bound

Split big groups into two blocks of at least $c_0T/3$ groups each and designate $m$ probes per group. Average the last $\ell$ slots in both endpoint tests; these coordinates occur in no multiplier. Let the resulting tests be $F',H'$, and decompose $F'$ by the averaging projections on the first block's probes and $H'$ by those on the second block. These are orthogonal decompositions because the coordinate projections commute and are orthogonal.

For a symmetric function on all $M=J+\ell$ coordinates of one group, all projection components of the same cardinality $j$ have equal squared norm. The fraction of these components containing all $m$ designated probes and none of the last $\ell$ slots is

$$\frac{\binom{M-m-\ell}{j-m}}{\binom Mj} =\frac{(j)_m(M-j)_\ell}{(M)_{m+\ell}}. \tag{I9}$$

It is zero for infeasible cardinalities. If $M\ge(m+\ell)(m+\ell-1)$, the denominator is at least $\tfrac12M^{m+\ell}$. Maximizing the remaining ratio in its own coordinate $u=j/M$ gives

$$2u^m(1-u)^\ell \le2\left(\frac m{m+\ell}\right)^m \left(\frac\ell{m+\ell}\right)^\ell \le2\ell^\ell m^{-\ell}=:\theta_m. \tag{I10}$$

Resolve the orthogonal decomposition by the vector of cardinalities across groups. Independent permutations make the retained fraction the product of (I9), even for tests entangled across groups. The full-probe components therefore have norms at most $\theta_m^{s_i/2}$ times the original norms. Choose $m$ with $-(c_0/6)\log\theta_m>E_{\rm id}+3$.

For a component pair, let $U,V$ be its missing probes in the respective blocks. A freed pair $A,B$ survives only when $A\subseteq U$ and $B\subseteq V$: otherwise an unshared input coordinate or unshared target coordinate is mean zero and absent from the multiplier. This uses the construction's asymmetry: the first kernel uses target labels, the second input labels.

Conditioning on the labels outside $A\cup B$ leaves precisely an arbitrary tuple test on $X_B$ and one on $X_A$. They are on independent spaces; the fixed shared product only changes the real additive frequency in (I5). Conditional Cauchy–Schwarz averages the error with the original component norms. When $U,V$ are nonempty,

$$\sum_{\varnothing\ne A\subseteq U,\, \varnothing\ne B\subseteq V}(-1)^{|A|+|B|}=1. \tag{I11}$$

The initial negative sign of each comparison consequently cancels that component pair's raw contribution. The exceptional pairs sum before absolute values to $R(F_{\rm full},H')+R(F'-F_{\rm full},H_{\rm full})$. The raw pairing has norm at most one, so (I10) controls this whole sum without a subset-count loss.

Approximation errors have at most $3^P$ component/free-set choices, where $P\le mC_0T$: a probe is indicated, missing but retained, or freed. Taking $A_0>E_{\rm id}+mC_0\log3+10$ absorbs them. The number of comparison patterns is at most $2^P$, and each coefficient is bounded by $Q^4\xi^{-2}$. Thus the total coefficient exponent is fixed before choosing $J$. A larger fixed $J$ can meet the transfer requirement without changing any kernel precision.

The concentrated-cell, repeated-label, imprimitive-character, and conditional-tuple attacks therefore leave the claimed ideal comparison intact. No finite numerical experiment is used to infer its uniformity.