Manuscript X-E@1 · X-E.tex · lemma · 293fae6a

Lemma: Quadratic part, derived

E3: the quadratic operator, its inverse, and the residual algebra › The quadratic part of the Einstein–Hilbert density · lines 983–1008 · proof included or linked

X-E@1#lem:quadratic

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Up to a total derivative,

\int - q R [η + κh] = - \frac{1}{2} κ^{2} \int h_{μν} G_{lin}^{μν} [h] + O (κ^{3}), S_{gr} = S_{FP} [h] + κ (2 S_{FP} (h, k) + S_{3} [h]) + O (κ^{2}),

S_{FP} [h] = \frac{1}{4} \int h_{μν} G_{lin}^{μν} [h], \frac{δ S _{FP}}{δ h _{μν}} = \frac{1}{2} G_{lin}^{μν} [h],

where $S_{FP} (h, k)$ is the polarization and $S_{3} [h] = - \frac{1}{2} \int [- q R]_{(3)} [h]$ the cubic vertex.

Proof. Let $φ (s) = \int - q R [η + s h]$ for compactly supported $h$ (the density statement follows). By (eq:palatini), $φ^{'} (s) = - \int - q G^{μν} [η + s h] h_{μν}$ , so $φ^{'} (0) = 0$ and $φ^{''} (0) = - \int G_{lin}^{μν} [h] h_{μν}$ . The Euler derivative follows from the symmetry of $G_{lin}$ , which is the symmetry of the second variation.

Displayed formulas in this passage

unlabeled · line 986
$\int - q R [η + κh] = - \frac{1}{2} κ^{2} \int h_{μν} G_{lin}^{μν} [h] + O (κ^{3}), S_{gr} = S_{FP} [h] + κ (2 S_{FP} (h, k) + S_{3} [h]) + O (κ^{2}),$
unlabeled · line 992
$S_{FP} [h] = \frac{1}{4} \int h_{μν} G_{lin}^{μν} [h], \frac{δ S _{FP}}{δ h _{μν}} = \frac{1}{2} G_{lin}^{μν} [h],$

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section: The quadratic part of the Einstein–Hilbert density