The quadratic
$$ P(z)=1+2z+\frac{101}{100}z^2 $$
has roots
$$ \boxed{ z=\frac{-100\pm10i}{101}. } $$
They are plainly not real. Now complete and normalize it by
$$ A(z)=\frac{8(101z^2+200z+100)}{2501}. $$
$$ T(z)=\frac{2(404z+901)}{2501}. $$
Every coefficient of both polynomials is positive, and the exact boundary condition is
$$ A\left(\frac14\right)=\frac12. $$
Nevertheless every nonzero completed fixed-column Casoratian orientation in the reciprocal-Newton hierarchy is positive:
$$ \mathcal C_1=\frac{2299}{800}, \qquad \mathcal C_2=\frac{505399}{80000}, $$
$$ \mathcal C_3=\frac{70475679}{8000000}, \qquad \mathcal C_4=\frac{21354130737}{100000000}. $$
Every later coefficient vanishes by rank two.
So this is a positive completed symbol with the right modular boundary value, a positive primitive quotient, and a nonreal conjugate root pair that passes every nonzero orientation available to the completed hierarchy.
The reason is one extra monomial. I keep staring at it.
First halve the determinant#
The completed-boundary problem begins with
$$ A(z)=b+(z-\alpha)T(z), $$
and two reciprocal fields
$$ \rho(z)=\frac{zA'(z)}{A(z)} =\sum_{k\geq1}\rho_kz^k. $$
$$ V(z)=\frac{T(z)}{A(z)} =\sum_{k\geq0}v_kz^k. $$
The original fixed-column Lambert filter produces a depth-\(j\) determinant whose entries are complicated Laurent coefficients. Its highest slow-scale term looks like an unpleasant all-depth Pluecker polynomial.
There is a much cleaner coordinate system.
Split the asymptotic rows by parity. A completed-coefficient Toeplitz gauge turns every even row into a monomial pivot, so those rows can be eliminated exactly. If
$$ p=\left\lfloor\frac j2\right\rfloor, \qquad q=\left\lfloor\frac{j-1}{2}\right\rfloor, $$
define
$$ M_{r,s} =-\rho_{p+1+s-r} +\alpha^{r+1}v_{p+1+s}. $$
Then the remaining leading coefficient is
$$ \boxed{ (-1)^{p(p+1)/2} \det[M_{r,s}]_{r,s=0}^{q}. } $$
The opaque depth-\(j\) object has become a determinant of roughly half the size.
If the modular \(V\)-term is deleted, checkerboarding and row reversal turn the odd and even depth blocks into Newton Hankel fields
$$ [S_{1+r+s}] \qquad\text{and}\qquad [S_{2+r+s}], $$
where \(S_k\) are the power sums of the reciprocal zeros of \(A\). The even-depth block is the orientation-squared Newton field.
That reduction is lovely, but it is not a sign theorem. The \(\alpha^{r+1}v_{p+1+s}\) term is rank one and determinant-scale significant. For the actual theta head, its determinant ratio against the uncharged block grows from about \(1.0031\) at depth one to \(9.66\times10^7\) at depth thirty. Rank one does not mean negligible. Here it means surgical.
The discriminant moves#
Take a general positive quadratic
$$ P(z)=1+\sigma z+\pi z^2 $$
and normalize
$$ A(z)=b\frac{P(z)}{P(\alpha)}, \qquad T(z)=\frac{A(z)-b}{z-\alpha}. $$
Then
$$ T(z)=\frac{b}{P(\alpha)} \left(\sigma+\alpha\pi+\pi z\right), $$
so \(A\) and \(T\) have positive coefficients whenever \(\alpha,b,\sigma,\pi>0\), regardless of whether \(P\) is real-rooted.
The terminal completed orientations factor as
$$ \mathcal C_3 =12\pi P(\alpha) \left(\sigma^2-4\pi+\alpha\pi\sigma\right), $$
$$ \mathcal C_4 =288\pi^2P(\alpha) \left(\sigma^2-4\pi+\alpha\pi\sigma\right). $$
Ordinary quadratic root reality is controlled by
$$ \Delta=\sigma^2-4\pi. $$
The completed hierarchy sees instead
$$ \boxed{ \Delta_\alpha =\sigma^2-4\pi+\alpha\pi\sigma. } $$
That last positive term opens a real chamber between \(\Delta=0\) and \(\Delta_\alpha=0\). Quadratics inside it have nonreal roots but the completed terminal orientations remain positive.
For the concrete choice
$$ \alpha=\frac14, \qquad \sigma=2, \qquad \pi=\frac{101}{100}, $$
the ordinary discriminant is
$$ \Delta=-\frac1{25}, $$
while the completed chamber value is
$$ \Delta_\alpha =-\frac1{25} +\frac14\cdot2\cdot\frac{101}{100} =\frac{93}{200}. $$
The conjugate pair is not sitting on a delicate numerical boundary. It lies strictly inside the false-positive chamber created by the modular completion.
What this kills, and what it does not#
The completed cross-minor hierarchy is not a finite Hermite criterion. Nonnegative orientations alone cannot imply root reality, even after imposing positive completed coefficients, positive primitive coefficients, and the exact boundary value \(A(\alpha)=b\).
This does not rule out a strict infinite-rank theorem for the actual theta function. The quadratic counterfeit has rank two, so every coefficient after depth four vanishes rather than remaining strictly positive. An all-shift hierarchy, the uncompleted orientation-squared Newton field, or an additional reciprocal channel could still recover a genuine root signature.
But the cheap converse is gone. Once the completed determinant is reduced to its clean reciprocal-Newton core, the little modular term does not merely perturb the answer. It moves the boundary of the property being detected.
That is a wonderfully rude failure mode.
Notebook references: K-0155, O-0301