The natural half-shift kernel in this problem has a \(2\times2\) minor equal to
$$ -\frac{d+1}{d}(d!)^2. $$
Negative. In every dimension. Right at the front.
And yet, after the same half-shift, every admissible cup coordinate indexed by a partition with at most four positive parts has nonnegative coefficients. I keep staring at those two facts beside each other. The positivity is not inherited from a positive kernel. It appears only after the determinant and the inverse cup transform have been allowed to interact.
That is a much fussier mechanism, and a much cuter one.
The array and the half-shift#
Start with the flagged rectangular array
$$ A_{r,j}(a) =(r+1)h_{2j-r-1}(a,a+1,\ldots,a+r+1), $$
where \(h_k\) is a complete homogeneous symmetric polynomial. A partition \(\lambda\) selects a maximal minor \(H_\lambda(a)\). The allowed partitions are exactly those satisfying
$$ \lambda_i\le d+1-i. $$
These minors are then expanded in the noncrossing-cup basis. Concretely, there is a signed lower-unitriangular incidence matrix \(W_d\), determined by which subsets cut every cup exactly once, and the gauged cup coordinates \(\alpha_\lambda\) are defined by
$$ W_d\alpha=S_d(H_K)_K. $$
So this is already a composition of two alternating objects: a determinant and an inverse signed incidence transform. Neither one looks coefficientwise positive on its own.
Now make the half-shift
$$ b=a-\frac12, \qquad Y=2a+d=2b+d+1. $$
The principal minor suddenly factors into positive linear pieces:
$$
P_d(b)=H_\varnothing
=d!\prod_{0\le p That factorization is the first hint that \(b\), rather than \(a\), is the honest coordinate. The theorem is: If \(\lambda\) is admissible and has at most four positive parts, then
\(\alpha_\lambda(b+\tfrac12)\in\mathbb Q_{\ge0}[b]\). For four positive parts this means $$
d\ge n\ge m\ge \ell\ge r\ge1,
\qquad
m\le d-1,\ \ell\le d-2,\ r\le d-3.
$$ Every coefficient is nonnegative, uniformly in the rank \(d\). The half-shift factors the flagged array through a Pascal-type matrix: $$
A\!\left(b+\frac12\right)=F P^{\mathrm{Pas}}_d(b),
\qquad
P^{\mathrm{Pas}}_d(b)_{n,j}=\binom{2j}{n}b^{2j-n}.
$$ This is exactly where one wants a planar network. If the constant kernel \(F\), or the relevant compound kernel built from it, were totally nonnegative, then minors would come with positive path families and the whole theorem might collapse into one luminous picture. It does not. The leading \(2\times2\) minor is the negative number above. Therefore the natural half-shift kernel is not totally nonnegative, and a positive planar-network explanation for that kernel cannot exist. I like this obstruction because it is tiny and rude. It does not merely say that nobody found the network. It points to the two-by-two square where the network fantasy dies. The replacement proof keeps the determinant and inverse cup transform together. First, divide each minor by the principal product and clear its natural rising-factorial content. For four rows the normalized amplitude is $$
Z_d(n,m,\ell,r)
=\frac{(Y)_n(Y-1)_m(Y-2)_\ell(Y-3)_r}{P_d}
H_{(n,m,\ell,r)}.
$$ The important property is not ordinary monotonicity. For each coordinate define a weighted backward difference $$
\mathcal D_iF(p)
=(2b+d+p_i-i)F(p-e_i)-F(p).
$$ Every nonempty mixed difference \(\mathcal D_S Z_d\), for all fifteen subsets \(S\subseteq\{1,2,3,4\}\), is coefficientwise nonnegative on every admissible Boolean cube. These weighted amplitude differences are precisely shaped to absorb the content factors that appear when neighboring partition parts move. Next comes the inverse cup transform. Fixing the top part \(n\), invert the lower three-coordinate cup matrix to obtain \(\gamma^{(n)}_{m,\ell,r}\). Universal inverse-transform formulas rewrite its rows using values, first differences, and mixed differences of the normalized amplitudes. Because those ingredients are coefficientwise nonnegative, one gets both $$
\gamma^{(n)}_{m,\ell,r}\ge_{\mathrm{coeff}}0
$$ and the adjacent-top contraction $$
\gamma^{(n-1)}_{m,\ell,r}
-\gamma^{(n)}_{m,\ell,r}
\ge_{\mathrm{coeff}}0.
$$ That contraction is the little engine. It is what remains when an odd step is paired with the preceding even step. Cup recurrences are clean in the strict interior and then acquire correction terms when parts collide. The top diagonal \((k,k,\ell,r)\) and top subdiagonal \((k+1,k,\ell,r)\) are not cosmetic boundary cases; they are where a collision-free recurrence would quietly become false. The exact boundary rows express those coordinates as positive sums of lower cup coordinates plus signed lower-triple transforms. Parity pairing converts the signed terms into the positive adjacent-top contractions above. One remaining all-equal resonance is handled by an exact inverse-transform identity. Once the diagonal and subdiagonal are initialized, the strict-top recurrence propagates positivity upward, with every collision correction already a positive unit-coefficient sum of lower coordinates. The two-row theorem supplies the first complete layer. The three-row argument adds fixed-leg inverse transforms, coefficientwise contractions, and a separate bottom-one recurrence. The four-row proof then repeats the architecture one level higher, but only after auditing the new collision geometry instead of pretending it is the same recurrence with another index attached. That exactness matters. The theorem stops at four positive parts. It does not prove five-row positivity, arbitrary-row positivity, total nonnegativity of the complete cup kernel, or a planar-network model for the failed half-shift kernel. And it has no implication for the Riemann hypothesis. Four rows is the boundary proved here. The negative \(2\times2\) minor stays negative. The final cup coordinates stay positive anyway. Notebook references: C-0141, C-0132, C-0133, C-0112
The obvious proof is unavailable#
Positivity after composition#
Collisions are part of the proof#