Exploring the Connes–Consani framework through symmetry, art, and computation
Operator: $(Af)(x) = \int_{-1}^{1} \frac{\sin 2\pi(x-y)}{\pi(x-y)} f(y)\,dy$, bandwidth $c=2\pi$, order $m=0$.
| $n$ | $\lambda_n$ | $\lambda_{n+1}/\lambda_n$ |
|---|
Asymptotic: $\lambda(n) \sim (4n+1)^{-2n-1/2} (e\pi)^{2n+1/2}$.
| $n$ | $A(n)$ (asymptotic) | $\lambda_n$ | Ratio $\lambda_n/A(n)$ |
|---|
Test function: $g(x) = e^{-\pi x^2}$ restricted to $[2^{-1/2}, 2^{1/2}]$.
$Q = \sum_{n=0}^{9} \lambda_n |\langle \xi_n, g\rangle|^2$
$W_\infty(g) = \int_1^2 (g(\rho)+g(1/\rho))\log\rho\,\frac{d\rho}{\rho} - 2g(1)\log(2\pi)$ for the same Gaussian.
Conclusion: $Q > 0$ but $W_\infty(g) < 0$ → Weil positivity fails for the cut‑off Gaussian (inadmissible test function).
Design your own test function $g(x)$ and see if it satisfies Weil positivity at $\lambda=\sqrt{2}$.
$Q$ (quadratic form) = —
$W_\infty$ (Weil distribution) = —
The plot shows $g(x)$ (blue) and its mirror $g(1/x)$ (red) inside the support interval.