Product capacity obstruction

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A proposed extension of the bounded-$\Omega$ argument was to exploit the tree structure of prime-power residue classes, hoping for a bound depending only on the number $\omega(q)$ of distinct prime factors. The obstruction below defeats any such argument based only on all individual divisor-face capacities, even when $\omega(q)\le3$ and the moduli are distinct and primitive.

The weights constructed here are not known to be realizable as actual first-kill harmonic masses. This is not a counterexample to Problem 25, or to its bounded-$\omega$ version for arbitrary residues. The later squarefull-kernel theorem proves natural density for that bounded-$\omega$ singleton version. Its arithmetic kernel truncation and activation estimates go beyond the face capacities used here.

External finite rectangle input

We use the following finite form of Carleson's rectangle counterexample. For every integer $N$ there is a finite family $\mathcal R_N$ of dyadic rectangles in $[0,1)^2$ such that

$$\sum_{R\in\mathcal R_N}|R|=1,\qquad \left|\bigcup_{R\in\mathcal R_N}R\right|\le C_1/N, \tag{1}$$

and, for every dyadic rectangle $F$,

$$\sum_{\substack{R\in\mathcal R_N\\R\subseteq F}}|R| \le C_2|F|. \tag{2}$$

The constants are absolute. This statement and its finite construction were checked on pages 1, 3, and 4 of Blasco--Pott, Carleson's counterexample and a scale of Lorentz-BMO spaces on the bitorus. The transfer to congruences and the distinct-modulus refinement below are separate arguments.

Transfer to two different prime coordinates

For each second-coordinate dyadic interval $J$, choose a triadic interval $J'\subseteq J$ with $|J'|>|J|/6$. To do so, choose a triadic grid length $t$ satisfying $|J|/6<t\le|J|/2$; an interval of length at least $2t$ contains a full grid interval of length $t$.

Replace $R=I\times J$ by $R'=I\times J'$ and give it weight

$$c_R=\frac{|R'|}{24C_2}.$$

Then

$$\sum_Rc_R\ge\frac1{144C_2},\qquad \left|\bigcup_RR'\right|\le C_1/N. \tag{3}$$

The modified rectangles satisfy the exact packing bound

$$\sum_{R'\subseteq I_0\times K}c_R\le|I_0\times K| \tag{4}$$

for every dyadic $I_0$ and triadic $K$.

For verification, put $\ell=|K|$. If $R'\subseteq I_0\times K$, then $I\subseteq I_0$, $J$ meets $K$, and $|J|<6\ell$. Choose a dyadic grid length $h$ with $6\ell\le h<12\ell$, using $h=1$ if this choice exceeds the ambient interval. Every such $J$ is contained in an $h$-grid interval meeting $K$. Since $h\ge\ell$, there are at most two such intervals, with total length at most $24\ell$. Applying (2) to $I_0$ times those intervals bounds the sum of the original areas by $24C_2|I_0\times K|$. Modified areas are no larger, proving (4).

The digit maps identify dyadic intervals with $2$-adic balls and triadic intervals with $3$-adic balls, preserving measure and containment. CRT therefore identifies the modified rectangles with full cylinders $C_i=[a_i]_{q_i}$, $q_i=2^{u_i}3^{v_i}$. Combine any duplicate cylinders by adding their weights. Equation (4) becomes every divisor-face bound

$$\sum_{\substack{i:d\mid q_i\\a_i\equiv b\pmod d}}c_i\le1/d. \tag{5}$$

Faces containing other primes have no descendants. This already gives an obstruction with $\omega(q_i)\le2$, but possibly repeated moduli.

Distinct primitive moduli with at most three prime factors

For each base cylinder $C_i$, choose a finite pool $P_i$ of primes greater than $3$, all pools disjoint, such that

$$1\le\sum_{p\in P_i}1/p\le2.$$

Divergence of the prime reciprocal sum permits this. Replace the base cylinder by

$$D_{i,p}=C_i\cap[0]_p,\qquad w_{i,p}=c_i/(2p)\quad(p\in P_i).$$

The modulus is $q_ip$. Each tag prime occurs in exactly one modulus, so all moduli are distinct and no one divides another. Also

$$\omega(q_ip)\le3,\qquad \mu\left(\bigcup_{i,p}D_{i,p}\right)\le C_1/N, \qquad \sum_{i,p}w_{i,p}\ge\frac1{288C_2}. \tag{6}$$

All face capacities remain true. A face without a tag receives at most the old total weight, since $\sum_{p\in P_i}1/(2p)\le1$. A face with a tag prime has at most one descendant. Equation (5) for the base cylinder itself gives $c_i\le1/q_i$, so its descendant weight is at most $1/(2q_ip)\le1/d$ for every relevant divisor $d\mid q_ip$. A face with two tags has no descendants.

The exact harmonic cutoffs do not fix the obstruction

Let $Q$ be the largest tagged modulus, put $X=Q^3$, and halve all the weights in (6). Every relevant divisor $d$ is at most $Q$, and

$$\frac{H_{\lfloor X/d\rfloor}}{H_X} \ge\frac{\log(X/d)}{1+\log X} \ge\frac{2\log Q}{1+3\log Q}>\frac12 \qquad(Q\ge3).$$

The new weights consequently satisfy

$$\sum_{\substack{i:d\mid q_i\\a_i\equiv b\pmod d}}w_i \le\frac{H_{\lfloor X/d\rfloor}}{dH_X},$$

yet have total mass at least $1/(576C_2)$ and full union measure tending to zero. Thus even the precise harmonic face constraints, together with distinctness, primitivity, and $\omega(q)\le3$, cannot imply a uniform small-footprint bound on the abstract total mass.

The previous unbounded-rank face obstruction is therefore not merely a consequence of increasing the number of prime coordinates. Large exponents already permit a fixed-rank obstruction. Additional information specific to actual first-kill masses is necessary.

Comparable two-prime base moduli cannot be repaired by refinement

There is a direct arithmetic obstruction to one attempted realization of these abstract weights. Fix $C\ge1$. Let $U$ be any finite periodic union, allowing multiple residues, all of whose base moduli are of the form $2^u3^v$ and lie in $[Q,CQ]$. Put $\beta=\mu(U)$. Suppose $B$ is a delayed singleton system with globally distinct moduli, each divisible by a base modulus and with each full cylinder contained in $U$. Then, uniformly in $X$,

$$\frac{|B\cap[1,X]|}{X},\quad \frac{H_B(X)}{H_{\lfloor X\rfloor}} \le\max\left\{2\beta,\, O_C\left(\frac{(\log(2Q))^2}{Q}\right)\right\}. \tag{7}$$

Indeed, there are $O_C(\log(2Q))$ distinct base moduli: for each $u$, the allowed $v$'s occupy an interval of bounded length. Their lcm is at most $(CQ)^2$, and their reciprocal sum is $O_C(\log(2Q)/Q)$. Insert these facts into the established arbitrary divisor-refinement bound, equation (5). The number of base residues does not enter that lemma. This proves (7).

In particular, if such a base family has $\beta\to0$ and $Q\to\infty$, no choice of singleton divisor refinements can realize a fixed positive harmonic mass, even with arbitrary composite tags or repeated use of a tag prime on different base moduli. The hypotheses include the comparable-modulus situation obtained when a two-coordinate rectangle construction has common area before the bounded-factor dyadic-to-triadic replacement. No assertion that every rectangle obstruction admits that normalization is needed here.