 Solution

Ignoring interactions, every field has dimension $(d-2)/2$, and every Laplacian contributes two derivatives. The operator ${\varphi }^n({\nabla }^2\varphi {)}^m$ therefore has dimensionBecause the integrated interaction $\int d^{d}xg\mathcal{O}$ is dimensionless, the coupling has scaling dimensionIt is a relevant coupling when ${\Delta }_g>0$, a marginal coupling when ${\Delta }_g=0$, and an irrelevant coupling when ${\Delta }_g<0$.