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Write $S=H/T$ and take $n\ge 2$. The unit-length constraint prevents an arbitrary multiplicative field rescaling, so the running coupling is found by renormalizing the coefficient of $|\nabla \sigma {|}^2$. We compute that coefficient in a slowly rotating planar background, which is enough because $O(n)$ symmetry makes the two-derivative stiffness isotropic on the sphere.

Let $n_0=(\cos \theta ,\sin \theta ,0,\dots )$, $e_1=(-\sin \theta ,\cos \theta ,0,\dots )$ and $e_a$ for $2\le a\le n-1$ the remaining constant orthonormal tangent vectors. Parameterize the fast tangent fluctuations byThis satisfies ${\sigma }^2=1$ exactly. Put ${\Omega }_{\mu }={\partial }_{\mu }\theta$, and choose it locally constant and small compared with the shell momenta. Since ${\partial }_{\mu }{n}_0=e_1{\Omega }_{\mu }$ and ${\partial }_{\mu }{e}_1=-n_0{\Omega }_{\mu }$, direct differentiation givesAt quadratic order in the fast fields, the cross term is $2{\Omega }_{\mu }{\partial }_{\mu }{\pi }_1$, whose integral is zero for constant ${\Omega }_{\mu }$. The term $(\partial q{)}^2$ is quartic in the fast fields. Remaining cross terms are cubic and have zero Gaussian expectation; their effects on the two-derivative stiffness start at higher loop order. Field-measure corrections independent of the background contribute normalization terms and do not alter the leading gradient coefficient in this calculation.

The quadratic fast action is $(2g{)}^{-1}\int |\nabla \pi {|}^2$. In the shell $\Lambda /b<|p|<\Lambda$ its connected correlation function is $⟨{\pi }_a(p){\pi }_b(-p)⟩=g{\delta }_{ab}/p^2$ with the momentum delta function understood. Thus $⟨{\pi }_a(x{)}^2⟩=g{I}_D$, wherewith the limit $I_2(b)=(2\pi {)}^{-1}\log b$. Here $S_D=2{\pi }^{D/2}/\Gamma (D/2)$ is the angular measure of the unit $(D-1)$-sphere. Averaging the background-gradient coefficient gives the one-loop stiffness of a spherical nonlinear sigma model:The coefficient is $n-2$, not $n-1$: the fluctuation tangent to the rotation plane does not reduce the gradient cost, because its contribution cancels between $q^2$ and ${\pi }_1^2$. All other $n-2$ transverse components do reduce it.

After integrating out the shell, put $x=b{x}^{\prime }$ to restore the cutoff. The constrained field keeps unit norm, so $g^{\prime }=b^{-ϵ}{g}_{\mathrm{eff}}$, or at this orderFor $b=e^{d\ell }$, $I_D=(S_D/(2\pi {)}^D){\Lambda }^{ϵ}d\ell +O(d{\ell }^2)$. Therefore the renormalization-group beta function isThe displayed terms are the requested one-loop flow. They are a weak-coupling expansion; the discrete $n=1$ target has no tangent modes and must instead be treated by domain walls, as in the first part.