# Marginal interactions for Fermi surfaces

I am struggling to understand Polchinski’s derivation (https://arxiv.org/abs/hep-th/9210046) of the conditions for marginality of the 4-fermi operator.

For a scattering process $$(\mathbf{p}_1,\mathbf{p}_2) \to (\mathbf{p}_3,\mathbf{p}_4)$$, the delta function Imposing momentum conservation can be written as

$$\delta^d(\mathbf{p}_1 + \mathbf{p}_2 - \mathbf{p}_3 - \mathbf{p}_4) = \delta^d(\delta \mathbf{k}_3 + \delta \mathbf{k}_4 + \delta \mathbf{l}_3+\delta \mathbf{l}_4)$$

where we have expressed the momenta as $$\mathbf{p} = \mathbf{k} + \mathbf{l}$$ where $$\mathbf{k}$$ is the orthogonal projection of $$\mathbf{p}$$ to the Fermi surface (unique assuming convex Fermi surface) and $$\mathbf{l}$$ is orthogonal to the Fermi surface.

For an inversion-symmetric Fermi surface, it is clear that the momentum changes $$\delta\mathbf{k}_3$$ and $$\delta \mathbf{k}_4$$ are parallel to each other for $$\mathbf{p}_1 = \pm \mathbf{p}_2$$. Polchinski seems to be claiming on page 19 that for these initial momentum configurations, in the scaling limit $$\mathbf{l} \to s\mathbf{l}$$, the delta function scales as $$s^{-1}$$. I do not follow his logic here. He claims that one component of $$\delta\mathbf{k}_3+\delta\mathbf{k}_4$$ vanishes, but in which coordinate system? Is there a coordinate-free explanation for this phenomenon?

I am even more baffled by this statement in footnote 7 on the bottom of page 20:

“One would seem to find the same enhancement for $$\mathbf{p}_1= +\mathbf{p}_2$$ In that case, however, the delta-function is degenerate only at one point on the Fermi surface, so that second order terms in $$\delta \mathbf{k}$$ are nonzero and the enhancement is only by $$s^{-1/2}$$.”

What second-order terms is he referring to?

Edit: Here is an attempt to rationalize the first statement. Clearly $$\mathbf{p}=\mathbf{p}'$$ implies $$\mathbf{k}=\mathbf{k}'$$ and $$\mathbf{l}=\mathbf{l}'$$ assuming convex Fermi surface. For an inversion-symmetric Fermi surface we further have that $$\mathbf{p}=-\mathbf{p}'$$ implies $$\mathbf{k}=-\mathbf{k}'$$ and $$\mathbf{l}=-\mathbf{l}'$$.

Now consider the following subsets of phase space

$$S = \{ (\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3,\mathbf{p}_4) \in \mathbb{R}^{4d} : \mathbf{p}_1 = -\mathbf{p}_2 \}$$ $$S' = \{ (\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3,\mathbf{p}_4) \in \mathbb{R}^{4d} : \mathbf{p}_1 = -\mathbf{p}_2, \; \mathbf{k}_3 = -\mathbf{k}_4 \}$$ $$S'' = \{ (\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3,\mathbf{p}_4) \in \mathbb{R}^{4d} : \mathbf{p}_1 = -\mathbf{p}_2, \; \mathbf{p}_3 = -\mathbf{p}_4 \}$$

By inversion symmetry $$\mathbf{p}_3 = -\mathbf{p}_4 \implies \mathbf{k}_3 = -\mathbf{k}_4, \; \mathbf{l}_3 = -\mathbf{l}_4$$ and thus $$S'' \subseteq S' \subseteq S$$.

Configurations in $$S''$$ conserve momentum so the delta function is divergent there. For a generic $$(\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3,\mathbf{p}_4) \in S'$$, however, we have

$$\delta\mathbf{k}_3 = \mathbf{k}_3 - \mathbf{k}_1 = -\mathbf{k}_4 + \mathbf{k}_2 = -\delta\mathbf{k}_4$$ and thus $$\delta^d(\mathbf{p}_1 + \mathbf{p}_2 - \mathbf{p}_3 - \mathbf{p}_4) = \delta^d(\delta \mathbf{l}_3 + \delta \mathbf{l}_4)$$ so we recover the desired scaling of the delta function.

So it appears what is left to understand is the case $$\mathbf{p}_1=\mathbf{p}_2$$ in footnote 7 on page 20.