In the context of $R$ gravity, the perturbation expansion appears as:
$$ S=\int \left( \partial \tilde{h} \partial \tilde{h} + X \tilde{h} \partial \tilde{h} \partial \tilde{h} + ... \right) d^4 x $$
To argue the non-renormalizability of this theory using the power counting argument, I've considered: $[\partial]=1$, $[h]=1$, and $[X]=[-1]$. Given these dimensions, all the terms in the perturbation expansion have a net dimension of 4, which with the factor $[d^4x]=-4$, renders the action dimensionless. Furthermore, because $[X]<0$, higher energy terms are enhanced rather than suppressed, leading to the conclusion that the theory is not power-counting renormalizable. Is my understanding correct?
For $ a R^2 $ gravity (where $[a]=-2$ and $[R^2]=4$), the perturbation expansion is:
$$ S=\int \left( ( \partial \tilde{h} \partial \tilde{h} )^2 + 2\sqrt{X} \tilde{h}(\partial \tilde{h} \partial {h})^2 + ... \right) d^4 x $$
A particular paper (RG) I'm referencing (see equation 51 on page 10) posits that $[X] = 0$, which would imply the renormalizability of $aR^2 $ gravity.
I'm struggling with applying the power counting argument here. If we maintain $[h]=1$ and $[\partial]=1$ as in the $R$ gravity scenario, then $[( \partial \tilde{h} \partial \tilde{h} )^2]$ ends up being 8, which does not make the action dimensionless. My subsequent deduction would be that $[\sqrt{X}]$ must cancel out the dimension of $[h]$, resulting in $[\sqrt{X}]=-1$. The only resolution I see is if the metric perturbation $h$ in $ aR^2 $ gravity does not retain the same dimensions as in $ R $ gravity. Could $ h $ in $ aR^2 $ be dimensionless, implying $[X]=0$? Is there a justification for $ h $ having different dimensions across the two theories, even though both $ h $'s represent metric perturbations?