Manuscript X-I-ST@1 · X-I-ST.tex · lemma · 9745fd40

Lemma: Relative reference symmetry

A finite reference action through third order › Reference master identity · lines 912–942 · proof included or linked

X-I-ST@1#lem:referenceCME

The natural reference action satisfies $s I = 0$ as a local variational class. Consequently

\frac{1}{2} (S_{ref}^{[3]}, S_{ref}^{[3]}) = 0 (mod κ^{4}) .

Proof. Naturality of the curvature and scalar densities gives

s E [q] = κ \partial_{μ} (c^{μ} E [q]), s L_{q} = κ \partial_{μ} (c^{μ} L_{q}) .

The background subtraction is $s$ -inert. Also $s (s Ψ_{q ˉ}) = 0$ . Thus $s I = 0$ modulo divergences.

For the formal reference Hamiltonian $I + H_{s}$ ,

\frac{1}{2} (I + H_{s}, I + H_{s}) = s I + \frac{1}{2} (H_{s}, H_{s}) = 0.

Extracting the first four nonnegative coefficients gives the asserted finite-order identity.

Displayed formulas in this passage

unlabeled · line 917
$\frac{1}{2} (S_{ref}^{[3]}, S_{ref}^{[3]}) = 0 (mod κ^{4}) .$
unlabeled · line 925
$s E [q] = κ \partial_{μ} (c^{μ} E [q]), s L_{q} = κ \partial_{μ} (c^{μ} L_{q}) .$
unlabeled · line 936
$\frac{1}{2} (I + H_{s}, I + H_{s}) = s I + \frac{1}{2} (H_{s}, H_{s}) = 0.$