Theorem: Corrected positivity ledger

Under Assumption ass:product, the following hold.

1. The Friedrichs extension $A_{F,\beta }$ of $-{\Delta }_q$ is nonnegative. Its form domain is the closure of compactly supported smooth sections in

   $${\Vert u\Vert }^2+{\Vert {\nabla }_{\beta }u\Vert }^2.$$

2. In the logarithmic mode, ${\mathsf{L}}_L$ has exactly one transverse negative eigenvalue

   $$-{{\kappa }_L}^2,{\kappa }_L=\frac{2e^{-{\gamma }_E}}{L},$$

   with radial eigenfunction $K_0({\kappa }_{L}r)$.

3. In a mode $0<\nu <1$, ${\mathsf{E}}_{\lambda }$ has a transverse negative eigenvalue exactly when $\lambda <0$. It is unique in that mode and is $-{\kappa }^2$, where

   $$\lambda =\frac{\Gamma (-\nu )}{\Gamma (\nu )}{\left(\frac{\kappa }{2}\right)}^{2\nu }.$$

   For $\lambda \in [0,\infty ]$ the transverse extension is nonnegative.

4. Every transverse ledger extension has essential spectrum $[0,\infty )$, with only the finitely many negative eigenvalues just described below it. Let

   $${\kappa }_{\ast }=\max (\{0\}\cup \{{\kappa }_j:-{{\kappa }_j}^2\text{ is a transverse negative eigenvalue}\}).$$

   Then the full product operator satisfies

   $$\mathrm{spec}(B)={\mathrm{spec}}_{\mathrm{ess}}(B)=[m^2-{\kappa }_{\ast }^2,\infty ),B\ge 0⟺m\ge {\kappa }_{\ast }.$$

   The full operator has no $L^2(N)$ point spectrum.

Proof. The first assertion is the Friedrichs construction applied to the nonnegative covariant Dirichlet form ReedSimonII.

At a negative radial energy $-{\kappa }^2$, square integrability at infinity forces the solution to be a multiple of $K_{\nu }(\kappa r)$. For $\nu =0$,

if and only if $L=2e^{-{\gamma }_E}/\kappa$. This gives one and only one negative eigenvalue for each finite $L$.

For $0<\nu <1$,

The ratio of the two boundary coefficients is (eq:kappa). Since $\Gamma (-\nu )<0$, its right-hand side is a bijection from $\kappa >0$ onto $\lambda <0$.

The Friedrichs radial spectrum is $[0,\infty )$, as follows from the Bessel spectral transform. Changing finitely many endpoint conditions gives a finite-rank transverse resolvent difference, so it does not change the transverse essential spectrum. The preceding square-integrable solutions account for all negative eigenvalues. There is no transverse zero eigenfunction: the solution permitted at zero cannot also be square integrable at infinity. These facts also follow from the explicit half-line analysis in DerezinskiRichard.

Fourier transformation in $z$ turns $B$ into the direct integral of $B_{\perp }+{k_z}^2+m^2$. Each transverse bound state therefore generates a continuous band $[m^2-{{\kappa }_j}^2,\infty )$, not a full-space bound eigenfunction. The transverse continuum generates $[m^2,\infty )$. Their union is (eq:full-spectrum). For any fixed spectral value, the equation ${k_z}^2+\lambda +m^2=E$ holds on a Lebesgue-null set of $k_z$ for each transverse spectral value $\lambda$; the direct-integral spectral representation therefore has no full-space eigenvectors.