Proposition: A visible fourth-order boundary

On this evaluation,

Also $[{\kappa }^4]{\mathcal{R}}_{\mathrm{ref}}=0$ on this evaluation. Thus for $\lambda >0$ the displayed fourth-order coefficient of the chosen polynomial lift is nonzero.

Proof. Substitution in (eq:B) gives (eq:positiveB). A nonzero smooth $a$ is nonzero on an open set, which cannot be contained in the hyperplane $x^1=0$. The integral is therefore strictly positive.

When $h=k=0$, the order-three reference action contains a gravitational term of degree five in $r$ and a scalar term with at least three metric factors. The gauge-fixing action has no order-three term in this centered split. Consequently $I_3$ and all its first derivatives vanish at $r=0$. Moreover $H_2=0$ in that sector, so $(S_2,S_2)=0$. Equation (eq:R4) at $\lambda =0$ proves the claim.