Lemma: Retained first variation and free normalization

Let $\mathcal{E}[q]=J_q{G}^{\mu \nu }[q]$, regarded as an $E^{\vee }$-valued density. Then

The quadratic transverse-traceless spatial action is

In particular its kinetic sign is positive.

Proof. The Palatini identity $\delta R_{\mu \nu }={\nabla }_{\rho }\delta {{\Gamma }^{\rho }}_{\nu \mu }-{\nabla }_{\nu }\delta {{\Gamma }^{\rho }}_{\rho \mu }$, together with $\delta J_q=\frac{1}{2}{J}_q{q}^{\mu \nu }\delta q_{\mu \nu }$ and $\delta q^{\mu \nu }=-q^{\mu \alpha }{q}^{\nu \beta }\delta q_{\alpha \beta }$, gives the first formula. Differentiating it once at $\eta$ gives the quadratic action and its symmetric Hessian. On trace-free, divergence-free spatial fields $G_{\mathrm{lin}}[l]=-\frac{1}{2}\square l$; integration by parts gives the last display.