 Solution

Under Fermi-surface renormalization-group scaling, tangential momentum $k$ is fixed whileIn frequency space the quadratic action is invariant whenequivalently, in the time representation $t\mapsto s^{-1}t$ and $\psi (t,k,\ell )\mapsto s^{-1/2}\psi (t,k,\ell )$. The free Fermi liquid action is therefore marginal.

The deformation $\mu (k){\psi }^{†}\psi$ has no factor of frequency or normal momentum, so it is relevant. It shifts the zero of the quasiparticle energy from $\ell =0$ to $\ell =\mu (k)/v_F(k)$ and hence deforms the Fermi surface.

A generic quartic interaction is irrelevant because momentum conservation fixes a normal component and leaves an additional positive power of $s$. It can be marginal only when all four momenta remain on the Fermi surface while satisfyingFor a smooth generic Fermi surface, the robust possibilities are forward or exchange scattering and the opposite-momentum BCS channel ${\mathbf{k}}_2=-{\mathbf{k}}_1$, ${\mathbf{k}}_4=-{\mathbf{k}}_3$.