Let \(C/\mathbf F_q\) be a smooth projective geometrically connected curve. Center the numerator of its zeta function into an even entire function \(X_C\), and write
$$ L_C(z)=\frac{X_C'(z)}{X_C(z)}. $$
For positive \(\lambda,\mu\), define
$$ Q_C(\lambda,\mu) =4\frac{\mu L_C(\lambda)-\lambda L_C(\mu)} {\mu^2-\lambda^2}, $$
with the diagonal value taken confluently.
Then for every choice of positive nodes \(\lambda_1,\ldots,\lambda_n\), the matrix
$$ [Q_C(\lambda_i,\lambda_j)]_{i,j=1}^n $$
is positive semidefinite. This holds at every matrix order.
The proof has the exact shape one keeps wanting in zeta problems: a geometric polarization forces spectral purity, purity produces imaginary zero frequencies, and the logarithmic derivative becomes a Gram sum.
Rosati positivity forces Frobenius onto one circle#
Let \(J=\operatorname{Jac}(C)\), and let \(\pi\) be its \(q\)-power Frobenius endomorphism. The Rosati involution from a polarization satisfies
$$ \pi^\dagger\pi=[q]_J. $$
Rosati positivity implies that \(\mathbb Q[\pi]\) is a product of fields, stable under \(\dagger\), with no nilpotents. After tensoring with \(\mathbb R\), the involution is ordinary conjugation on every complex factor. Therefore every embedding obeys
$$ \sigma(\pi^\dagger)=\overline{\sigma(\pi)} $$
and hence
$$ |\sigma(\pi)|^2=q. $$
Every Frobenius root has modulus \(\sqrt q\). This is the spectral purity step, and it comes from an actual positive involution rather than from a functional equation alone.
Purity creates imaginary frequency towers#
Normalize the Frobenius roots to the unit circle. Reciprocal pairing makes \(X_C\) even, while the finite-field logarithm makes it periodic in the imaginary direction.
The zeros therefore lie on explicit towers \(z=i\gamma\), including the exceptional angles \(0\) and \(\pi\) with their correct even multiplicities. One gets a paired product
$$ X_C(z) =Cz^{m_0} \prod_{\gamma>0} \left(1+\frac{z^2}{\gamma^2}\right)^{m_\gamma}. $$
Put
$$ H_C(x) =\frac{X_C'(\sqrt x)} {\sqrt x\,X_C(\sqrt x)}. $$
Logarithmic differentiation gives the Stieltjes sum
$$ H_C(x) =\frac{m_0}{x} +2\sum_{\gamma>0}\frac{m_\gamma}{x+\gamma^2}. $$
Its negative divided difference is
$$ M_C(x,y) =\frac{m_0}{xy} +2\sum_{\gamma>0} \frac{m_\gamma} {(x+\gamma^2)(y+\gamma^2)}. $$
Every summand is rank-one positive. Thus \(M_C\) is a Gram kernel at every order.
Finally, with \(x_i=\lambda_i^2\),
$$ Q_C(\lambda_i,\lambda_j) =4\lambda_i\lambda_jM_C(x_i,x_j). $$
This is a positive diagonal congruence of the Gram matrix. All-order positivity follows immediately.
The counterfeit explains what the polarization contributed#
A reciprocal polynomial with positive formal closed-point exponents can still have roots of unequal normalized modulus. One explicit quadratic counterfeit has the right-looking reciprocity and Euler data but produces off-axis centered zeros and a negative order-two Loewner determinant.
So the functional equation and positive arithmetic coefficients are not secretly doing the job. Rosati positivity is the load-bearing input that forces purity.
The theorem is confined to curves over finite fields. It does not construct a number-field polarization, and it gives no Riemann-zeta consequence. That boundary is not disappointing; it identifies the missing object with unusual precision. Over finite fields, the entire chain exists and closes. Over the integers, the analogous positive global polarization is exactly what we do not have.
Notebook references: C-0016, O-0065