Cartesian restrictions cannot be extracted from almost full input mass

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Status and reason for considering this question

The propositions below are proved. They concern every fixed diagonal form of degree $r$ in $r$ variables, $r\ge3$. They do not settle either target question about $S_r$.

The bounded-fiber criterion would give positive density from a positive fraction of input tuples lying in bounded fibers. Those tuples need not form a Cartesian product. The distinction matters for a new possible method: Croot, Mao, Pohoata, Sheffer, and Yip's weighted entropic sieve, Section 4 assumes independent coordinate laws in Proposition 4.3. Its Theorem 4.1 assumes Cartesian input sets and a uniform bound on every output fiber; the proof uses these hypotheses separately in equations (12)--(14). Theorem 1.7 bounds a $B_3[g]$ subset of the first $N$ cubes by $O(g^{1/9}N\exp(-c\sqrt{\log N}/\log\log N))$. These statements and the relevant proof passages were checked in the full primary text. None asserts bounded typical multiplicity for all triples, or a density estimate for their values.

The next construction shows, within the exact diagonal-form setting, why one cannot automatically replace those hypotheses by high input mass or even by a vanishing relative-entropy deficit.

An almost full set with small Cartesian rectangles

Fix $r\ge3$ and nonzero integers $a_1,\ldots,a_r$, and put

$$F(\mathbf x)=\sum_{i=1}^r a_i x_i^r, \qquad g=\gcd(r,2),\qquad Y=\frac{\log N}{10}.$$

For sufficiently large $N$, let

$$\mathcal P_N=\{p\in[Y,2Y]:p\equiv-1\pmod{2r},\quad p\nmid a_1\cdots a_r\}, \quad \ell_N=|\mathcal P_N|, \quad \Delta_N=\prod_{p\in\mathcal P_N}p.$$

The prime number theorem in this fixed progression gives

$$\ell_N\sim\frac{\log N}{10\varphi(2r)\log\log N}, \qquad \log\Delta_N\sim\frac{Y}{\varphi(2r)}. \tag{1}$$

In particular, $\Delta_N\le N^{1/5}$ eventually. Define the output-dependent input restriction

$$\Gamma_N=\{\mathbf x\in[N]^r: p\nmid F(\mathbf x)\text{ for all }p\in\mathcal P_N\}.$$

Theorem. One has

$$|\Gamma_N|=\left(1-O_r\left(\frac1{\log\log N}\right)\right)N^r. \tag{2}$$

Nevertheless, every Cartesian rectangle $B_1\times\cdots\times B_r\subseteq\Gamma_N$, with $B_i\subseteq[N]$, satisfies

$$\prod_{i=1}^r|B_i| \le(1+o_r(1))N^r(g/r)^{\ell_N}. \tag{3}$$

If $B^r\subseteq\Gamma_N$, the stronger common-set conclusion is

$$|B|\le(1+o_r(1))N(g/r)^{\ell_N}. \tag{4}$$

Thus (3) saves $\exp(-c_r\log N/\log\log N)$ in volume, and (4) saves such a factor in the root-set size, for some $c_r>0$.

Proof. For each selected prime, $\gcd(r,p-1)=\gcd(r,2)=g$. The power map on $\mathbb F_p$ therefore has fibers of size at most $g$, including its fiber at zero. After fixing $x_2,\ldots,x_r$, the congruence $F=0\pmod p$ allows at most $g$ residue classes for $x_1$. Hence

$$\#\{\mathbf x\in[N]^r:p\mid F(\mathbf x)\} \le g(N/p+1)N^{r-1}.$$

The union bound, (1), and $p\ge Y$ prove (2), since $\sum_{p\in\mathcal P_N}p^{-1}=O_r(1/\log Y)$.

For a nonempty rectangle as in the theorem, write $s_{i,p}=|B_i\bmod p|$ and $T_{i,p}=\{a_i b^r:b\in B_i\bmod p\}$. The sum $T_{1,p}+\cdots+T_{r,p}$ omits zero. Indeed, any choice of these residues has independent representatives in the corresponding $B_i$, and a zero sum would contradict the restriction defining $\Gamma_N$. Iterated Cauchy--Davenport therefore gives

$$\sum_i|T_{i,p}|\le p+r-2.$$

Since $|T_{i,p}|\ge s_{i,p}/g$,

$$\sum_i s_{i,p}\le g(p+r-2). \tag{5}$$

Assign each $p$ to an index $i$ for which $s_{i,p}\le g(p+r-2)/r$. Let $\mathcal P_i$ be the resulting partition and $\Delta_i=\prod_{p\in\mathcal P_i}p$, allowing $\Delta_i=1$. CRT and interval counting imply

$$|B_i|\le (N+\Delta_i) \prod_{p\in\mathcal P_i}\frac{g(p+r-2)}{rp}.$$

Multiply over $i$. Every $\Delta_i\le\Delta_N\le N^{1/5}$, and

$$\prod_{p\in\mathcal P_N}\left(1+\frac{r-2}{p}\right)=1+o_r(1).$$

This proves (3). Empty rectangles are trivial. If every $B_i=B$, then all the $s_{i,p}$ are equal, so (5) gives the same bound for $|B\bmod p|$ at every prime. Applying CRT once proves (4). Finally $g<r$, and (1) gives the stated exponential savings. $\square$

Vanishing relative entropy is not sufficient for the weighted step

Specialize to $F=x^3+2y^3+4z^3$. The preceding prime progression is $p\equiv5\pmod6$, equivalently $p\equiv2\pmod3$ for the large primes being used. Let $U_N$ be uniform on $[N]^3$, and let $\mu_N=U_N(\,\cdot\mid\Gamma_N)$. If $\varepsilon_N=1-|\Gamma_N|/N^3$, then exactly

$$D(\mu_N\Vert U_N)=-\log(1-\varepsilon_N) =O(1/\log\log N)=o(1). \tag{6}$$

Here $D$ is relative entropy, with natural logarithms. The identity

$$D(\mu_N\Vert U_N) =\left(\sum_{i=1}^3H((\mu_N)_i)-H(\mu_N)\right) +\sum_{i=1}^3D((\mu_N)_i\Vert\operatorname{Unif}[N]) \tag{7}$$

shows that the total correlation and all marginal entropy deficits also tend to zero; $H$ denotes Shannon entropy with the same logarithm.

Define nonnegative weights

$$m_p(\mathbf x)=p\,1_{p\mid F(\mathbf x)},\qquad K_N(\mathbf x)=\prod_{p\in\mathcal P_N}(1+m_p(\mathbf x)).$$

Cubing permutes $\mathbb F_p$, so $F=0\pmod p$ has exactly $p^2$ solutions. Fixing any one coordinate leaves precisely $p$ solutions in the other two. Thus $m_p$ has mean one and each one-coordinate marginal mean is one. CRT gives

$$\Delta_N^{-3}\sum_{\mathbf x\bmod\Delta_N}K_N(\mathbf x)=2^{\ell_N}.$$

Every residue vector has $(N/\Delta_N)^3(1+O(\Delta_N/N))$ representatives in the box, uniformly. Positivity of the weights permits summing these relative errors, yielding

$$\mathbb E_{U_N}K_N=(1+O(\Delta_N/N))2^{\ell_N}, \qquad \mathbb E_{\mu_N}K_N=1. \tag{8}$$

The latter identity holds because every factor is one on $\Gamma_N$.

Consequently a product-law lower bound cannot automatically extend to correlated laws in the form

$$\mathbb E_\mu K_N\ge 2^{\ell_N}\exp(-C D(\mu\Vert U_N))$$

with fixed $C$: (6) and (8) contradict it. By (7), replacing the entropy penalty by marginal deficits plus total correlation does not repair this assertion. This example tests such an extension; it does not contradict a theorem stated for independent coordinates.

Consequence for the density problem

A set of bounded-fiber input tuples of size $cN^r$, if obtained, could be intersected with $\Gamma_N$ at a cost of only $o(N^r)$ tuples. Thus the positive-density criterion remains viable on these non-Cartesian restrictions. Conversely, (3)--(4) show that high input mass alone cannot supply the Cartesian hypothesis required by the cited bounded-multiplicity sieve. No estimate for the actual integer fibers of $F$, and no new infinitude case, follows from this construction.