Theorem: Single-mode corrections and shifted branches

Let $n$ be deficient with $q=n+\beta \ne 0$ and $\nu =|q|/\alpha <1$. Take ${\mathsf{E}}_0$ in that mode and Friedrichs in the others. Then

For one complex field,

Let $P_X(\beta )$ be any Friedrichs polynomial formula for $X=F,j_{\phi },T^{\mu }{}_{\nu }$, regarded as a polynomial rather than periodically reduced. Then

If both modes are deficient and both are irregular,

Proof. For $c=c_0(\nu )$, the radial heat kernel is obtained by replacing $I_{\nu }$ by $I_{-\nu }$. The Laplace transform used in Proposition prop:modes remains valid because $-\nu >-1$. Their difference gives (eq:E0kernel).

Let $c=\cosh \eta$. Near $c=1$,

This proves regularity directly and supplies its second jet. Coincidence gives $\delta F=\nu /(4{\pi }^2\alpha r^2)$; the angular derivative gives $\delta j_{\phi }=-2eq\delta F$. The time-split second derivative, conservation, and the improvement identity give the displayed stress components.

For the branch identity set

and fix their displayed logarithms. Define the entire function of the branch parameter

On $0<b<1$, this is the physical Friedrichs angular sum. Irregularizing mode $0$ replaces $x^{\beta }$ by $y^{-\beta }$; the result is $\mathcal{S}(\beta +1)$. Irregularizing mode $-1$ similarly gives $\mathcal{S}(\beta -1)$. The coincidence jets of this entire branch give the stated polynomial identities. Independent mode corrections add.