12. Negative association and the four-corner¶

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In the previous chapter 11 we built the block bridge twin_prime_conjecture_of_blocks: the infinitude of twins follows if at every scale N there is a carrier with a survivor, and the "bad" part is strictly smaller than the carrier. With that, the entire content of the problem was compressed into a single block inequality of cardinalities.

The present chapter takes that inequality and translates it into the language of tallies by side ranks of $6m\pm1$, where it assumes the form of a four-corner relation. We will show that the required direction N_{33}<N_{00} is not a density fact but a symmetry fact: it is forced by the negative association of the two ranks, which in turn is a shadow of the exclusive (per-prime) structure of the two.

Tallies by rank and the matrix $N_{ij}$¶

Fix a block of candidate centers m (indices of the survivor class at scale A).

Definition 12.1 (side ranks). To each center m assign the pair of side ranks $$r_-(m)=\bigl\lvert\{p\le A: p\mid 6m-1\}\bigr\rvert,\qquad r_+(m)=\bigl\lvert\{p\le A: p\mid 6m+1\}\bigr\rvert,\tag{12.1}$$ — the number of small prime divisors of the left and the right side respectively. A center is a twin center if and only if both sides are free of small divisors, that is, r_-=r_+=0 (after the sieve-to-the-root this entails the primality of both sides; see 10).

Note. The rank here counts precisely the small divisors $p\le A$; the completeness of the rank (that zero rank entails primality) is guaranteed by the sides lying below A^2 — this is the same sieve-to-the-root mechanism prime_of_no_small_prime_factor from 10.

Coarsen the ranks to the binary feature "zero / nonzero" and introduce the $2\times2$ tally matrix over the pair $(\,[r_-{>}0]\,,[r_+{>}0]\,)$:

\[ \begin{array}{c|cc} & r_+=0 & r_+>0\\\hline r_-=0 & N_{00} & N_{03}\\ r_->0 & N_{30} & N_{33} \end{array} \]

Here $N_{00}$ is the number of centers with both sides clean (these are exactly the twin centers), $N_{33}$ — the centers whose both sides are "spoiled" by a small divisor, and $N_{03},N_{30}$ are the mixed ones (one side clean, the other not). The indexing 0/3 is a legacy of the rank scale 0..3 in the sources; the digit 3 encodes "nonzero rank".

Our goal in this chapter is to compare the corner tallies: to prove that the diagonal N_{33} (both spoiled) is outweighed by the opposite corner N_{00} (both clean). This is exactly what yields the lower bound on twin centers within a block.

The rank-1 approximation and the sign of the rank-2 correction¶

It is natural to assume that, to first approximation, the spoilage indicators of the left and right sides are independent. If the probabilities $\Pr[r_->0]=\pi_-$ and $\Pr[r_+>0]=\pi_+$ were independent, the tally matrix $N_{ij}$ would have rank 1:

\[ N_{ij}\ \approx\ T\cdot u_i\, v_j,\qquad u=(1-\pi_-,\ \pi_-),\quad v=(1-\pi_+,\ \pi_+), \]

where T is the total number of centers. For a rank-1 matrix all four $2\times2$ minors vanish; in particular, the exact equality holds $$N_{00}\,N_{33}\ =\ N_{03}\,N_{30}.\tag{12.2}$$

The observation: the real tally matrix is not of rank 1 — there is a rank-2 correction, and the whole question is its sign.

Definition 12.2 (four-corner defect). Set $$\boxed{\ \Delta_{\mathrm{fc}}\ :=\ N_{03}\,N_{30}-N_{00}\,N_{33}\ }\tag{12.3}$$ — this is (up to normalization) the determinant of the tally matrix, that is, the amplitude of its rank-2 part. The sign of $\Delta_{\mathrm{fc}}$ is the sign of the covariance of the binary ranks: $$\mathrm{sgn}\,\Delta_{\mathrm{fc}}\ =\ -\,\mathrm{sgn}\,\mathrm{Cov}\big([r_-{>}0],[r_+{>}0]\big).$$

The condition $\Delta_{\mathrm{fc}}\ge0$, that is, $$N_{00}\,N_{33}\ \le\ N_{03}\,N_{30},\tag{12.4}$$ is called the four-corner inequality and is equivalent to the negative association of the ranks: the spoilage of one side makes the spoilage of the other less likely. It is precisely this sign that we need, and precisely this sign that is not accidental.

Why the sign is negative: the exclusive structure of the two¶

Negative association is not postulated here — it flows from the already proven exclusivity of divisors (see 02, no_large_shared_divisor): a prime p>2 cannot divide both sides at once, for $\gcd(6m-1,6m+1)\mid 2$. Hence at every prime the contribution to the two ranks is exclusive: p divides either 6m-1, or 6m+1, or nothing — but never both.

From this the per-prime generating function of the ranks is a product over primes with no cross term:

\[ \prod_{2<p\le A}\big(c_p+a_p\,x+b_p\,y\big),\qquad a_p=b_p\ (\pm\text{ symmetry}),\ \textbf{no term }xy, \]

where x marks the contribution to r_-, y — to r_+, and the absence of xy is a direct consequence of "both at once" being forbidden.

Expanding such a product yields at every prime $$\mathrm{Cov}\big(\mathbf 1_{p\mid-},\mathbf 1_{p\mid+}\big)=0-a_pb_p=-a_pb_p<0,$$ and summation over the primes of the layer preserves the sign: $$\mathrm{Cov}(r_-,r_+)=\sum_{p}(-a_pb_p)<0.$$ The negative covariance of the ranks is exactly the four-corner.

Conclusion. The sign is forced by the exclusive structure of the two, not by the density of primes.

Note. In the CRT model the same inequality is derived purely by symmetry — via Maclaurin's inequality on elementary symmetric polynomials: $R_{\mathrm{CRT}}=20\,e_6/e_3^2\le 20\binom{s}{6}/\binom{s}{3}^2<1$. This confirms that the four-corner is a property of the shape of the product c+ax+by, not of analytic prime-counting estimates.

The algebra of the passage: four-corner + side-corner ⟹ $N_{33}<N_{00}$¶

The four-corner inequality by itself compares the diagonal N_{00}N_{33} with the mixed product N_{03}N_{30}. To obtain the desired N_{33}<N_{00}, a second, lateral ingredient is needed — the side-corner: $$N_{03}\,N_{30}\ \le\ N_{00}^{\,2}.\tag{12.5}$$ It says that the mixed corners are small compared with the clean one: low rank is markedly more likely than high rank (the same survivor recursion $q\to q-2$), so the clean corner N_{00} dominates. Formally we have formalized precisely the algebra of the passage — that these two inequalities together force the conclusion.

The non-strict passage is the lemma:

Theorem 12.3 (N33_le_N00_of_four_corner). Let $N_{00},N_{03},N_{30},N_{33}\in\mathbb N$. If $0<N_{00}$, the four-corner $N_{00}\,N_{33}\le N_{03}\,N_{30}$ and the side-corner $N_{03}\,N_{30}\le N_{00}^{2}$ hold, then $N_{33}\le N_{00}$. 🟢

The proof is elementary: the chain $N_{00}N_{33}\le N_{03}N_{30}\le N_{00}\cdot N_{00}$ gives $N_{00}N_{33}\le N_{00}N_{00}$, whence cancelling the positive factor N_{00} (Nat.le_of_mul_le_mul_left) yields $N_{33}\le N_{00}$.

The strict form — what we actually need (a strict gap B_5>0, that is, more clean centers than fully spoiled ones) — is given by the lemma:

Theorem 12.4 (N33_lt_N00_of_four_corner). Let $N_{00},N_{03},N_{30},N_{33}\in\mathbb N$. If $0<N_{00}$, the strict four-corner $N_{00}\,N_{33}<N_{03}\,N_{30}$ and the side-corner $N_{03}\,N_{30}\le N_{00}^{2}$ hold, then $N_{33}<N_{00}$. 🟢

Here the strict inequality $N_{00}N_{33}<N_{03}N_{30}$ is threaded through the side-corner up to $N_{00}N_{33}<N_{00}N_{00}$, and lt_of_mul_lt_mul_left cancels N_{00} on the left, giving the strict conclusion.

Finally, the $\pm$ symmetry of the sides substantially simplifies the side-corner. Since the left and right sides enter the product symmetrically ($a_p=b_p$), the mixed corners are equal: $N_{03}=N_{30}$. The lemma side_corner_of_le records this simplification:

Theorem 12.5 (side_corner_of_le). Let $N_{00},N_{03},N_{30}\in\mathbb N$. If $N_{03}=N_{30}$ and $N_{03}\le N_{00}$, then the side-corner $N_{03}\,N_{30}\le N_{00}^{2}$ holds. 🟢

Substituting the symmetry reduces the quadratic inequality $N_{03}N_{30}\le N_{00}^2$ to a product of two identical factors $N_{03}\cdot N_{03}\le N_{00}\cdot N_{00}$, each of which is majorized via $N_{03}\le N_{00}$ (Nat.mul_le_mul). Thus the entire lateral node rests on a single linear fact: there are no more mixed centers than purely twin centers.

Note. All three lemmas N33_le_N00_of_four_corner, N33_lt_N00_of_four_corner, side_corner_of_le are over $\mathbb N$, machine-checked, and use no sorry. They formalize the algebra of the passage, not the corner inequalities themselves.

Numerical background and magnitudes¶

Denote the corner ratios $$R_{\mathrm{fc}}=\frac{N_{00}N_{33}}{N_{03}N_{30}},\qquad R_{\mathrm{sc}}=\frac{N_{03}N_{30}}{N_{00}^{2}}.$$ Then the proportion of fully-spoiled to clean factorizes: $$\frac{N_{33}}{N_{00}}=R_{\mathrm{fc}}\cdot R_{\mathrm{sc}}.$$

Numerically (tools/RESULTS_fourcorner.md) the direction holds on all blocks: $R_{\mathrm{fc}}<1$ everywhere, and as the layer grows $R_{\mathrm{fc}}$ rises toward 1 from below ($0.876\to0.977$ over $k=21\dots28$), while the lateral ratio remains steadily small, $R_{\mathrm{sc}}\approx0.014$. Hence $$\frac{N_{33}}{N_{00}}\le(1+o(1))\cdot0.014\ll1,$$ and for the strict N_{33}<N_{00} no four-corner margin is needed — the direction itself, $R_{\mathrm{fc}}\le1$, plus the smallness of $R_{\mathrm{sc}}$ suffice.

Note. That $R_{\mathrm{fc}}\to1$ from below, rather than bouncing off 1, is structural too: the pairwise correlations are small ($a_pb_p\sim p^{-2}$), their sum over the layer $(A,A^2]$ tends to zero, and the ranks become asymptotically independent. Exclusivity forbids crossing the mark 1, so the limit is attained strictly from below. But it is precisely this tightness near 1 that hides the parity problem — see below.

Where the node is open: the conjecture and the closing plan¶

The three formalized lemmas give the passage, but not the corner inequalities themselves for the real tallies. The honest boundary — in the house sense of an honestly named open node (see the glossary) — runs here.

Conjecture 12.6 (the real four-corner). For the real CRT tallies of the survivor block at scale A, the strict $N_{00}N_{33}<N_{03}N_{30}$ and the side-corner $N_{03}N_{30}\le N_{00}^2$ hold. 🟡

Closing plan (distribution-free). The model inequality for the product c+ax+by is provable elementarily (Maclaurin / negative association). One step model→reality remains: to show that the exact CRT tallies coincide with this product to an order that does not flip the sign of the rank-2 correction. Formally this is worked out in 14 via the exact decomposition of the remainder $N_{ij}^{\text{real}}=N_{ij}^{\text{model}}+e_{ij}$; the task is to bound e_{ij} so that $\mathrm{sgn}\Delta_{\mathrm{fc}}$ is preserved.

Section takeaway. We do not pass the reduction off as a proof: so far only the algebra of the passage and the model sign are proven; control of the remainder e_{ij} is an open front, but the candidate is distribution-free, not density-based.

A bridge to the next chapter¶

With that, the content of the problem has once again localized — now into one correlational fact: the negative association of the ranks $(r_-,r_+)$ on the real tallies. We have already seen that its root is the exclusive per-prime structure of the two: the product c+ax+by without the term xy.

In the next chapter 13 we lift this observation to the level of the fractal layer of the engine: we show that the same two organizes the self-similar recursion of survivors upward through the primes ($q\to q-2$) and the finite blocker collapse downward in scale, and that it is precisely this fractal exclusivity that is the common source both of the direction $R_{\mathrm{fc}}\le1$ and of the finiteness of the descent depth (EPMI — the impossibility of a perpetual engine, see the glossary).