# Supplemental Material for “UV/IR Mixing in Marginal Fermi Liquids”

Weicheng Ye  Perimeter Institute for Theoretical Physics, Waterloo, Ontario,
Canada N2L 2Y5  Department of Physics and Astronomy, University of Waterloo,
Waterloo, Ontario, Canada N2L 3G1  Sung-Sik Lee  Perimeter Institute for
Theoretical Physics, Waterloo, Ontario, Canada N2L 2Y5  Department of Physics
and Astronomy, McMaster University, Hamilton, Ontario, Canada L8S 4M1  Liujun
Zou  Perimeter Institute for Theoretical Physics, Waterloo, Ontario, Canada
N2L 2Y5

###### Contents

  1. I Framework for Renormalization Group Analysis
  2. II Summary of the RG Results
  3. III Feynman Diagram Calculations
    1. III.1 Second Leading Order Result of Self-Energy
    2. III.2 Third Leading Order Result of Fermion Self-Energy
    3. III.3 Third leading Order Result of Boson Self-Energy
      1. III.3.1 General Strategies of Calculation and Assumptions
      2. III.3.2 All Diagrams at the Third Leading Order
      3. III.3.3 Benz Diagram
      4. III.3.4 3-String Diagram

This Supplemental Material contains the framework for the renormalization
group (RG) analysis (Section I), a summary of the RG results (Section II), and
the details of the Feynman diagram claculations (Section III).

##  I Framework for Renormalization Group Analysis

In this section, we present the framework for our field-theoretic RG analysis.
In terms of the renormalized fields and couplings, the action of the patch
theory for $N$ Fermi surfaces (FS) coupled to a $U(1)$ gauge field is written
as

$\displaystyle S=S_{\psi}+S_{\text{int}}+S_{a},$ (1)

where

$\displaystyle\begin{split}S_{\psi}&=\int[dk]\sum_{i,p}\psi^{\dagger}_{ip}(k)\left[-iZ_{1}k_{\tau}+Z_{2}(pk_{x}+k_{y}^{2})\right]\psi_{ip}(k),\\\
S_{\text{int}}&=\int[dk_{1}][dk_{2}]\sum_{i,p}Z_{2}\lambda_{p}a(-k_{1})\psi^{\dagger}_{ip}(k_{2})\psi_{ip}(k_{1}+k_{2}),\\\
S_{a}&=\int[dk]Z_{3}\frac{N}{2e^{2}\mu^{\epsilon}}|k_{y}|^{1+\epsilon}a(-k)a(k)\\\
\end{split}$ (2)

with $[dk]\equiv\frac{dk_{\tau}dk_{x}dk_{y}}{(2\pi)^{3}}$, $\lambda_{\pm}=\pm
1$, $\mu$ a renormalization scale with the same dimension as $k_{y}$, and
$Z_{1,2,3}$ some renormalization factors. In the above, the rotational
symmetry of the patch theory is invoked for the absence of the relative
renormalization between $k_{x}$ and $k_{y}^{2}$, and the Ward identity is used
to guarantee that the multiplicative renormalization of $S_{\rm int}$ is the
same as that of $(pk_{x}+k_{y}^{2})$ part of the fermion kinetic term
Metlitski and Sachdev (2010). For latter convenience, define
$\alpha\equiv\gamma e^{2}$ with $\gamma\equiv 1/(4\pi)$ as a proxy for the
coupling constant $e^{2}$. The renormalized fields, couplings and momentum are
related to the bare ones via

$\displaystyle\begin{split}&k_{\tau B}=\frac{Z_{1}}{Z_{2}}k_{\tau}\equiv
Z_{\tau}k_{\tau},\quad k_{xB}=k_{x},\quad k_{yB}=k_{y},\\\
&\psi_{ipB}(k_{B})=\sqrt{\frac{Z_{2}}{Z_{\tau}}}\psi_{ip}(k)\equiv\sqrt{Z_{\psi}}\psi_{ip}(k),\quad
a_{B}(k_{B})=\frac{1}{Z_{\tau}}a(k)\equiv\sqrt{Z_{a}}a(k),\\\
&\alpha_{B}\equiv\gamma e_{B}^{2}=\frac{1}{Z_{3}Z_{\tau}}\gamma
e^{2}\mu^{\epsilon}\equiv Z_{\alpha}\alpha\mu^{\epsilon}\end{split}$ (3)

where the subscript $B$ stands for bare. $Z_{\tau}$ is introduced to account
for a nontrivial dynamical exponent, $Z_{\psi}$ and $Z_{a}$ are introduced to
account for anomalous dimensions of the fermion and gauge fields,
respectively, and $Z_{\alpha}$ is introduced to account for the flow of
$\alpha$. At the tree level, the fields and couplings have the following
scaling dimensions in momentum space:

$\displaystyle\begin{split}&[k_{\tau}]_{0}=2,\quad[k_{x}]_{0}=2,\quad[k_{y}]_{0}=1,\\\
&[\psi_{ip}(k)]_{0}=-\frac{7}{2},\quad[a(k)]_{0}=-3,\quad[e^{2}]_{0}=0.\end{split}$
(4)

The $Z$ factors are chosen so that the following renormalization conditions
are satisfied:

$\displaystyle\begin{split}\frac{\partial\Sigma(k_{\tau}=\mu^{2},k_{x}=k_{y}=0)}{\partial
k_{\tau}}&=0,\\\ \Sigma(k_{y}=\mu,k_{\tau}=k_{x}=0)&=0,\\\
\Pi(k_{y}=\mu,k_{\tau}=k_{x}=0)&=0,\end{split}$ (5)

where $\Sigma$ and $\Pi$ are the 1-particle irreducible self-energy of the
renormalized fermionic and gauge fields, respectively, and $\mu$ is the
renormalization scale. Below we focus on the case with $\epsilon=0$. In this
case, it turns out that physical observables have divergence as $\Lambda_{y}$,
the cutoff of $k_{y}$, goes to infinity. Notice that $\Lambda_{y}$ measures
the abundance of gapless degrees of freedom (DOF) on the FS, so it is not a
cutoff as in a usual field theory, which separates high-energy DOF from low-
energy ones. Accordingly, it is more appropriate to view $\Lambda_{y}$ as a
parameter that is on equal footing as the couplings of the theory.

Denote a correlation function involving $n_{\psi}$ fermionic fields and
$n_{a}$ gauge fields by $G^{(n_{\psi},n_{a})}$, which is generically written
as a function of a set of momenta, $\alpha$, $\mu$ and $\Lambda_{y}$. Then
$G^{(n_{\psi},n_{a})}_{B}\cdot\delta(\\{k_{B}\\})=Z_{\psi}^{\frac{n_{\psi}}{2}}Z_{a}^{\frac{n_{a}}{2}}G^{(n_{\psi},n_{a})}(\\{k\\})\cdot\delta(\\{k\\})$,
where $\\{k\\}$ denotes the momenta this correlation function depends on, and
the $\delta$-functions impose momentum conservation. Due to $k_{\tau
B}=Z_{\tau}k_{\tau}$, $\delta(\\{k\\})=Z_{\tau}\delta(\\{k_{B}\\})$,
$G_{B}^{(n_{\psi},n_{a})}=Z_{\psi}^{\frac{n_{\psi}}{2}}Z_{a}^{\frac{n_{a}}{2}}Z_{\tau}G^{(n_{\psi},n_{a})}(\\{k\\})$.
Because all bare quantities and fields are independent of $\mu$,
$\mu\frac{d}{d\mu}G^{(n_{\psi},n_{a})}_{B}(\\{k_{B}\\})=0$, which leads to a
Callan-Symanzik equation for the renormalized correlation function:

$\displaystyle\left[\mu\frac{\partial}{\partial\mu}+b_{\alpha}\alpha\frac{\partial}{\partial\alpha}+\gamma_{z}k_{\tau}\frac{\partial}{\partial
k_{\tau}}+\frac{n_{\psi}\gamma_{\psi}}{2}+(n_{a}-1)\gamma_{z}\right]G^{(n_{\psi},n_{a})}=0$
(6)

with

$\displaystyle
b_{\alpha}\equiv\frac{\mu}{\alpha}\frac{d\alpha}{d\mu}=-\frac{\mu}{Z_{\alpha}}\frac{dZ_{\alpha}}{d\mu},\quad\gamma_{z}\equiv-\frac{\mu}{Z_{\tau}}\frac{dZ_{\tau}}{d\mu},\quad\gamma_{\psi}\equiv\frac{\mu}{Z_{\psi}}\frac{dZ_{\psi}}{d\mu}.$
(7)

Note that the beta function of $\alpha$ is $\beta(\alpha)=b_{\alpha}\alpha$.

On the other hand, suppose the engineering dimension of $G^{(n_{\psi},n_{a})}$
is $\Delta_{0}$. By dimensional analysis, it must have the form
$G^{(n_{\psi},n_{a})}=\mu^{\Delta_{0}}g\left(\frac{\\{k_{\tau}\\}}{\mu^{2}},\frac{\\{k_{x}\\}}{\mu^{2}},\frac{\\{k_{y}\\}}{\mu},\alpha,\frac{\Lambda_{y}}{\mu}\right)$,
where $g$ is some function. Thus,

$\displaystyle\left(\mu\frac{\partial}{\partial\mu}+2k_{\tau}\frac{\partial}{\partial
k_{\tau}}+2k_{x}\frac{\partial}{\partial k_{x}}+k_{y}\frac{\partial}{\partial
k_{y}}+\Lambda_{y}\frac{\partial}{\partial\Lambda_{y}}-\Delta_{0}\right)G^{(n_{\psi},n_{a})}=0.$
(8)

Combining this equation with Eq. (6) yields another form of the Callan-
Symanzik equation:

$\displaystyle\left[(2-\gamma_{z})k_{\tau}\frac{\partial}{\partial
k_{\tau}}+2k_{x}\frac{\partial}{\partial k_{x}}+k_{y}\frac{\partial}{\partial
k_{y}}+\Lambda_{y}\frac{\partial}{\partial\Lambda_{y}}-b_{\alpha}\alpha\frac{\partial}{\partial\alpha}-\Delta_{0}-\frac{n_{\psi}\gamma_{\psi}}{2}-(n_{a}-1)\gamma_{z}\right]G^{(n_{\psi},n_{a})}=0.$
(9)

At the fixed point, $b_{\alpha}=0$, and the above equation implies the
following scaling structure of the correlation function

$\displaystyle
G^{(n_{\psi},n_{a})}(s^{z}\\{k_{\tau}\\},s^{2}\\{k_{x}\\},s\\{k_{y}\\},s\Lambda_{y})=s^{\Delta}G^{(n_{\psi},n_{a})}(\\{k_{\tau}\\},\\{k_{x}\\},\\{k_{y}\\},\Lambda_{y})$
(10)

with $z\equiv 2-\gamma_{z}$ the dynamical exponent, and
$\Delta\equiv\Delta_{0}+\frac{n_{\psi}\gamma_{\psi}}{2}+(n_{a}-1)\gamma_{z}$
the total scaling dimension of this correlation function. It is clear that
$\gamma_{\psi}$ and $2\gamma_{z}$ can be interpreted as the anomalous
dimensions of the fermionic and gauge fields, respectively. Note that the
anomalous dimension of the gauge field $a(k)$ is entirely from a rescaling of
time (by setting $k_{\tau B}=Z_{\tau}k_{\tau}$), while its real-space
counterpart $a(x)$ has no anomalous dimension due to the Ward identity. Also
note that the renormalized correlation function would not depend on
$\Lambda_{y}$ in the absence of any UV/IR mixing, but one of the central
results of this paper is that there is UV/IR mixing.

Before finishing this section, we describe an equivalent way to analyze the
theory, which is also often used in the literature and differs from the above
one by some convention Lee (2008, 2009); Metlitski and Sachdev (2010); Mross
_et al._ (2010); Zou and Chowdhury (2020a, b). We will mainly use the previous
convention, and readers not interested in this part can skip it. In the
alternative convention, the bare action is written as

$\displaystyle\begin{split}S_{\psi}&=\int[dk]\sum_{i,p}\psi^{\dagger}_{ip}(k)\left(-iZ_{f}Z_{\eta}\eta
k_{\tau}+Z_{f}(pk_{x}+k_{y}^{2})\right)\psi_{ip}(k),\\\
S_{\text{int}}&=\int[dk_{1}][dk_{2}]\sum_{i,p}Z_{f}\lambda_{p}a(-k_{1})\psi^{\dagger}_{ip}(k_{2})\psi_{ip}(k_{1}+k_{2}),\\\
S_{a}&=\int[dk]\frac{1}{Z_{e}}\frac{N}{2e^{2}\mu^{\epsilon}}|k_{y}|^{1+\epsilon}a(-k)a(k),\\\
\end{split}$ (11)

and $[dk]=\frac{dk_{\tau}dk_{x}dk_{y}}{(2\pi)^{3}}$, $\lambda_{\pm}=\pm 1$.
Again, the Ward identity is used to guarantee that the renormalization factor
in $S_{\rm int}$ is $Z_{f}$, the same as that of the $pk_{x}+k_{y}^{2}$ part
of the fermionic kinetic term. The difference between this convention and our
previous one is that in this case $k_{\tau}$ has the same dimension as
$k_{x}$. Instead, another dimensionless parameter $\eta$ is introduced to keep
track of a nontrivial dynamical exponent. Below we will see that, not
surprisingly, these two conventions are equivalent. The bare quantities and
fields are related to the renormalized ones via

$\displaystyle(k_{\tau
B},k_{xB},k_{yB})=(k_{\tau},k_{x},k_{y}),~{}\psi_{ipB}=\sqrt{Z_{f}}\psi_{ip},~{}a_{B}=a,~{}\eta_{B}=Z_{\eta}\eta=1,~{}e_{B}^{2}\equiv
Z_{e}e^{2}.$ (12)

The engineering dimensions are also given by (4) with the extra
$[\eta]_{0}=0$. We also define $\tilde{\alpha}\equiv\frac{\gamma e^{2}}{\eta}$
with $\gamma=1/(4\pi)$. Similar to the previous convention, the $Z$ factors
are determined by demanding the renormalization conditions given in Eq. (5),
but with $\mu^{2}$ changed to $\mu^{2}/\eta$ in the first renormalization
condition:

$\displaystyle\frac{\partial\Sigma(k_{\tau}=\mu^{2}/\eta,k_{x}=k_{y}=0)}{\partial
k_{\tau}}$ $\displaystyle=0.$ (13)

Now consider a correlation function involving $n_{f}$ fermionic fields and
$n_{c}$ gauge fields, denoted by $G^{(n_{f},n_{c})}$. One can show that
$G^{(n_{f},n_{c})}=\eta^{\frac{n_{f}}{2}+n_{c}-1}\mu^{\Delta_{0}}g\left(\frac{\eta\\{k_{\tau}\\}}{\mu^{2}},\frac{\\{k_{x}\\}}{\mu^{2}},\frac{\\{k_{y}\\}}{\mu},\tilde{\alpha},\frac{\Lambda_{y}}{\mu}\right)$,
where $\Delta_{0}$ is the engineering dimension of the correlation function,
and $g$ is the same function as in the previous convention. This implies that
$\eta\frac{\partial}{\partial\eta}G^{(n_{f},n_{c})}=\left(\frac{n_{f}}{2}+n_{c}-1+k_{\tau}\frac{\partial}{\partial
k_{\tau}}-e^{2}\frac{\partial}{\partial e^{2}}\right)G^{(n_{f},n_{c})}$.
Combining this result and analogous derivations as before, similar Callan-
Symanzik equations can be obtained

$\displaystyle\begin{split}\left(\mu\frac{\partial}{\partial\mu}+b_{e}e^{2}\frac{\partial}{\partial
e^{2}}+b_{\eta}\eta\frac{\partial}{\partial\eta}+\frac{n_{f}}{2}\gamma_{f}\right)G^{(n_{f},n_{c})}&=0,\\\
\left[\mu\frac{\partial}{\partial\mu}+(b_{e}-b_{\eta})e^{2}\frac{\partial}{\partial
e^{2}}+b_{\eta}k_{\tau}\frac{\partial}{\partial
k_{\tau}}+\frac{n_{f}}{2}(\gamma_{f}+b_{\eta})+(n_{c}-1)b_{\eta}\right]G^{(n_{f},n_{c})}&=0,\\\
\left[(2-b_{\eta})k_{\tau}\frac{\partial}{\partial
k_{\tau}}+2k_{x}\frac{\partial}{\partial k_{x}}+k_{y}\frac{\partial}{\partial
k_{y}}+\Lambda_{y}\frac{\partial}{\partial\Lambda_{y}}-(b_{e}-b_{\eta})e^{2}\frac{\partial}{\partial
e^{2}}-\Delta_{0}-\frac{n_{f}}{2}(\gamma_{f}+b_{\eta})-(n_{c}-1)b_{\eta}\right]G^{(n_{f},n_{c})}&=0,\end{split}$
(14)

where

$\displaystyle
b_{\eta}\equiv\frac{\mu}{\eta}\frac{d\eta}{d\mu},~{}b_{e}\equiv\frac{\mu}{e^{2}}\frac{de^{2}}{d\mu},~{}\gamma_{f}\equiv\frac{\mu}{Z_{f}}\frac{dZ_{f}}{d\mu}.$
(15)

It is not hard to see that $b_{\eta}$, $b_{e}-b_{\eta}$ and
$\gamma_{f}+b_{\eta}$ should be identified as $\gamma_{z}$, $b_{\alpha}$ and
$\gamma_{\psi}$ in the previous convention, respectively, and the two
conventions ultimately lead to identical results. Note that in this
convention, the beta function for $\tilde{\alpha}$ is
$\beta(\tilde{\alpha})=\left(b_{e}-b_{\eta}\right)\tilde{\alpha}$. In this
paper, we will mainly use the previous convention.

##  II Summary of the RG Results

The $Z$ factors introduced in Eq. (2) are calculated in Sec. III. Here we
summarize the results. We expand $Z$ in powers of $1/N$, and at each order of
$1/N$, we keep results relevant to the calculations of $b_{\alpha}$,
$\gamma_{z}$ and $\gamma_{\psi}$ to the leading order in $\Lambda_{y}/\mu$:

$\displaystyle\begin{split}&Z_{1}=1-\frac{1}{N}\cdot\frac{2\alpha}{\pi}\ln\frac{\Lambda_{y}}{\sqrt{\alpha}\mu}+\frac{1}{N^{2}}\cdot\frac{2\alpha^{3}}{\pi^{2}}\left(S^{a}(\alpha)+T^{a}(\alpha)+2G^{a}(\alpha)-S^{b}(\alpha)-T^{b}(\alpha)-2G^{b}(\alpha)\right)\ln\frac{\Lambda_{y}}{\mu},\\\
&Z_{2}=1+\frac{1}{N^{2}}\cdot\frac{2\alpha^{3}}{\pi^{2}}\left(S^{a}(\alpha)+T^{a}(\alpha)-S^{b}(\alpha)-T^{b}(\alpha)\right)\ln\frac{\Lambda_{y}}{\mu},\\\
&Z_{3}\approx
1-\frac{1}{N}\cdot\frac{2\alpha^{3}}{\pi}F(\alpha)+\frac{1}{N^{2}}\cdot\frac{4\alpha^{4}}{3\pi^{2}}\left(\ln\frac{\Lambda_{y}}{\mu}\right)^{2},\end{split}$
(16)

where the functions
$F(\alpha),G^{a}(\alpha),S^{a}(\alpha),T^{a}(\alpha),G^{b}(\alpha),S^{b}(\alpha),T^{b}(\alpha)$
can be found in Eqs. (33), (47), (48), (50), (55), (56), (57), respectively.
Using these expressions, Eqs. (3) and (7), $b_{\alpha}$, $\gamma_{z}$ and
$\gamma_{\psi}$ are obrained to be

$\displaystyle\begin{split}&b_{\alpha}\approx\frac{1}{N}\cdot\frac{2\alpha}{\pi}-\frac{1}{N^{2}}\cdot\frac{8\alpha^{4}}{3\pi^{2}}\ln\frac{\Lambda_{y}}{\mu},\\\
&\gamma_{z}=-\frac{1}{N}\cdot\frac{2\alpha}{\pi}+\frac{1}{N^{2}}\cdot\left(\frac{4\alpha^{3}}{\pi^{2}}\left(G^{a}(\alpha)-G^{b}(\alpha)\right)-\frac{2\alpha^{2}}{\pi^{2}}\right),\\\
&\gamma_{\psi}=-\frac{1}{N}\cdot\frac{2\alpha}{\pi}+\frac{1}{N^{2}}\cdot\left(\frac{2\alpha^{3}}{\pi^{2}}\left(2G^{a}(\alpha)-S^{a}(\alpha)-T^{a}(\alpha)-2G^{b}(\alpha)+S^{b}(\alpha)+T^{b}(\alpha)\right)-\frac{2\alpha^{2}}{\pi^{2}}\right).\end{split}$
(17)

Here we ignore $\frac{1}{N^{2}}\ln\frac{\Lambda_{y}}{\mu}$ term in $Z_{3}$ and
keep only terms proportional to $\ln\frac{\Lambda_{y}}{\mu}$ at order
$1/N^{2}$ in $b_{\alpha}$. Note that almost all terms in $b_{\alpha}$,
$\gamma_{z}$ and $\gamma_{\psi}$ come from explicit $\mu$ dependence in $Z$
except for the $\frac{1}{N^{2}}\alpha^{2}$ term, which comes from the
$\ln\sqrt{\alpha}$ term in $Z_{1}$.

##  III Feynman Diagram Calculations

In this section, we present the details of the Feynman diagram calculations
leading to the results summarized in Section II. Parts of the relevant
calculations can be found in Refs. Lee (2009); Metlitski and Sachdev (2010);
Mross  _et al._ (2010); Zou and Chowdhury (2020a); Holder and Metzner (2015a,
b). Here we focus on the case where $\epsilon=0$.111As discussed in Ref. Lee
(2009), if we use the second convention, we need to set
$\eta\sim\mathcal{O}(1)$ such that the large-$N$ expansion has no extra
factors of $N$ from $\eta$. We will also assume that external $k_{y}$ to be
positive for self-energies unless mentioned otherwise.

First, we collect results of all the Feynman diagrams contributing to the
self-energy of boson and fermion to the third leading order in the large-$N$
expansion. We will see that all of the divergent diagrams give the traditional
logarithmic divergence except for the Benz diagram and the 3-String diagram,
which give the unconventional double logarithmic divergence. In any case,
$Z_{1,2,3}$ can be expanded in $1/N$ as

$\displaystyle
Z_{i}=1+\frac{1}{N}\delta_{i}^{(1)}+\frac{1}{N^{2}}\delta_{i}^{(2)}\quad
i=1,2,3$ (18)

such that renormalization conditions Eq. (5) are satisfied. The propagator of
$a$, denoted as $\mathcal{D}(k)$, and the propagator of $\psi$, denoted as
$\mathcal{G}(k)$, to the third leading order in $1/N$ are written as

$\displaystyle\begin{split}\mathcal{D}^{-1}&=ND^{-1}-\Pi^{(1)}-\frac{1}{N}\Pi^{(2)}+...\\\
\mathcal{G}^{-1}&=G^{-1}-\frac{1}{N}\Sigma^{(1)}-\frac{1}{N^{2}}\Sigma^{(2)}+...,\end{split}$
(19)

where $D$ is the leading-order propagator for $a$ as in e.g. Ref. Metlitski
and Sachdev (2010)

$\displaystyle
D(k_{\tau},k_{y})=\frac{1}{\gamma\frac{|k_{\tau}|}{|k_{y}|}+\frac{|k_{y}|}{e^{2}}}$
(20)

with $\gamma=1/(4\pi)$, and $G$ is the bare fermion propagator,

$\displaystyle G_{\pm}(k_{\tau},k_{x},k_{y})=\frac{1}{-ik_{\tau}\pm
k_{x}+k_{y}^{2}}.$ (21)

We can also write down a large-$N$ expansion of $b_{\alpha}$, $\gamma_{z}$ and
$\gamma_{\psi}$ to the order of $1/N^{2}$,

$\displaystyle\begin{split}b_{\alpha}&=\frac{1}{N}b_{\alpha}^{(1)}+\frac{1}{N^{2}}b_{\alpha}^{(2)}+...\\\
\gamma_{z}&=\frac{1}{N}\gamma_{z}^{(1)}+\frac{1}{N^{2}}\gamma_{z}^{(2)}+...\\\
\gamma_{\psi}&=\frac{1}{N}\gamma_{\psi}^{(1)}+\frac{1}{N^{2}}\gamma_{\psi}^{(2)}+...\\\
\end{split}$ (22)

which can be calculated through solving the Callan-Symanzik equation Eq. (6)
from the propagator $\mathcal{D},\mathcal{G}$ at the third leading order of
$1/N$, or through Eq. (7).

###  III.1 Second Leading Order Result of Self-Energy

The second leading order result of the fermion self-energy is given by the
standard one-loop diagram shown in Fig. 1.

Figure 1: One-loop diagram for fermion self-energy.

The diagram has been computed in Mross  _et al._ (2010); Zou and Chowdhury
(2020a). Here we write down the result of the diagram at $\epsilon=0$ with
$y$-momentum cutoff $\Lambda_{y}$,

$\displaystyle\Sigma_{1LF}(k)=\int[dl]~{}G_{f}(k-l)D(l)=ik_{\tau}\frac{e^{2}}{2\pi^{2}}\left(\frac{1}{2}+\ln\left(\frac{\Lambda_{y}}{\sqrt{\gamma
e^{2}|k_{\tau}|}}\right)\right).$ (23)

To impose the renormalization conditions in Eq. (5), we take

$\displaystyle\delta_{1}^{(1)}=-\frac{2\alpha}{\pi}\ln\left(\frac{\Lambda_{y}}{\mu\sqrt{\alpha}}\right),\quad\delta_{2}^{(1)}=0$
(24)

with $\alpha\equiv e^{2}/(4\pi)$. 222 We note that in the simple minimal
subtraction scheme, one would take
$\delta^{(1)}=-\frac{2\alpha}{\pi}\ln\frac{\Lambda_{y}}{\mu}$, which amounts
to imposing a different RG condition.

Let us then turn to the second leading order result of the self-energy of
boson, $\Pi^{(1)}$. At this order only the so-called Aslamazov-Larkin (AL)
diagram Aslamazov and Larkin (1996) in Fig. 2 gives non-zero but finite
contribution as argued in Ref. Mross  _et al._ (2010).

Figure 2: Aslamazov-Larkin (AL) diagram for boson self-energy.

The AL diagram reads Metlitski and Sachdev (2010)

$\displaystyle\Pi_{AL}(k)=\frac{1}{2}\int\frac{dl_{\tau}dl_{x}dl_{y}}{(2\pi)^{3}}\Gamma^{3}(k,l,-(k+l))\Gamma^{3}(-k,-l,k+l)D(l)D(k+l)$
(25)

where $\Gamma^{3}$ is the fermion-induced three-boson vertex and receives
contributions from both fermion patches as well as both directions in which
fermions can travel in the loop,

$\displaystyle\Gamma^{3}=\Gamma^{3}_{+}+\Gamma^{3}_{-},$ (26)
$\displaystyle\Gamma^{3}_{s}(l_{1},l_{2},-(l_{1}+l_{2}))=\lambda_{s}^{3}\left(f_{s}(l_{1},l_{2},-(l_{1}+l_{2}))+f_{s}(l_{2},l_{1},-(l_{1}+l_{2}))\right),$
(27) $\displaystyle
f_{s}(l_{1},l_{2},-(l_{1}+l_{2}))=-\int\frac{dp_{\tau}dp_{x}dp_{y}}{(2\pi)^{3}}G_{s}(p)G_{s}(p-l_{1})G_{s}(p+l_{2}).$
(28)

First let us compute $f_{\pm}(k,l,-(k+l))$. We perform the integral in the
sequence of $p_{x}$, $p_{y}$ and $p_{\tau}$ to obtain

$\displaystyle\begin{split}f_{\pm}(k,l,-(k+l))&=\frac{1}{4\pi\left(k_{y}(il_{\tau}\mp
l_{x}-l_{y}^{2})-l_{y}(ik_{\tau}\mp k_{x}+k_{y}^{2})\right)}\\\
&\quad\quad\quad\quad\times\left(k_{\tau}\theta\left(\frac{k_{\tau}}{k_{y}}\right)+l_{\tau}\theta\left(\frac{l_{\tau}}{l_{y}}\right)-(k+l)_{\tau}\theta\left(\frac{(k+l)_{\tau}}{(k+l)_{y}}\right)\right).\end{split}$
(29)

Here $\theta(x)$ is the unit step function. An important property is that if
we take the static limit $k_{\tau}\rightarrow 0$, the fermion-induced three-
boson vertex is non-zero only when the internal boson satisfies,

$\displaystyle-
k_{y}<l_{y}<0,~{}~{}~{}\text{when}~{}k_{y}>0~{}~{}~{}~{}~{}~{}\text{or}~{}~{}~{}~{}~{}~{}0<l_{y}<-k_{y},~{}~{}~{}\text{when}~{}k_{y}<0,$
(30)

i.e. the integration of the internal boson momentum only has nonzero
contribution from a restricted finite region in the static limit. 333This is
the reason why we do not see the double log divergence in lower order
diagrams. In $U(1)\times U(1)$ or $U(2)$ gauge theory Zou and Chowdhury
(2020a), this also leads to the fact that there is a fixed line instead of
several isolated fixed points at the leading order of the RG equation. It is
straightforward to see that the fixed line will in principle be lifted because
of the third leading order results of (boson) self-energy. Note that the
imaginary part of the pole of $l_{x}$ for both $f_{s}(k,l,-(k+l))$ and
$f_{s}(l,k,-(k+l))$ with a fixed $s=\pm$ are at
$s(l_{\tau}-k_{\tau}l_{y}/k_{y})$. Considering the integration of $l_{x}$ in
Eq. (25), we see that $\Pi_{AL}$ is non-zero only when the two fermion loops
come from different patches. After the integration of $l_{x}$, we see that the
$k_{x}$ dependence drops out and we have

$\displaystyle\begin{split}\Pi_{\text{AL}}(k_{\tau},k_{y})&=-\frac{1}{64\pi^{4}k_{y}^{2}}\int
dl_{\tau}dl_{y}D(l)D(k+l)\left(k_{\tau}\theta\left(\frac{k_{\tau}}{k_{y}}\right)+l_{\tau}\theta\left(\frac{l_{\tau}}{l_{y}}\right)-(k+l)_{\tau}\theta\left(\frac{(k+l)_{\tau}}{(k+l)_{y}}\right)\right)^{2}\\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{\left(l_{y}^{2}+k_{y}l_{y}\right)^{2}}{|l_{\tau}-k_{\tau}l_{y}/k_{y}|\left(\left(l_{\tau}-k_{\tau}l_{y}/k_{y}\right)^{2}+\left(l_{y}^{2}+k_{y}l_{y}\right)^{2}\right)}.\\\
\end{split}$ (31)

It is straightforward to see that this integral is convergent from the
kinematic constraint in Eq. (30). Here, we focus on
$\Pi_{AL}(k_{\tau}=0,k_{y})$. Also we temporarily assume $k_{y}>0$, which
indicates that $-k_{y}<l_{y}<0$. After a change of variable with
$l_{\tau}=k_{y}^{2}\Delta$ ($-\infty<\Delta<\infty$) and $l_{y}=k_{y}\delta$
($-1<\delta<0$), the final result is given as follows,

$\displaystyle\Pi_{AL}(k_{\tau}=0,k_{y})=-\frac{|k_{y}|}{e^{2}}\frac{2\alpha^{3}}{\pi}F(\alpha),$
(32)

where

$\displaystyle
F(\alpha)\equiv\int_{0}^{1}d\delta\int_{0}^{\infty}d\Delta\frac{\delta}{\delta^{2}+\alpha\Delta}\frac{1-\delta}{(1-\delta)^{2}+\alpha\Delta}\frac{\Delta\left(\delta^{2}-\delta\right)^{2}}{\Delta^{2}+\left(\delta^{2}-\delta\right)^{2}}.$
(33)

To impose the renormalization conditions in Eq. (5), we take the counterterm
to be

$\displaystyle\delta_{3}^{(1)}=-\frac{2\alpha^{3}}{\pi}F(\alpha)$ (34)

###  III.2 Third Leading Order Result of Fermion Self-Energy

Next, we move to the self-energies at the third leading order of $1/N$.
Because the goal is to compute the beta functions to the order of $1/N^{2}$,
it is sufficient to keep only the divergent parts in the self-energies. We
start from the fermion self-energy. There are four non-zero diagrams as shown
in Fig. 3. The two-loop fermion diagram in Fig. 3(a) is convergent, while the
divergence of the two nested diagrams shown in Fig. 3(b) cancel with each
other. Thus, the divergent part only comes from the three-loop diagram, shown
in Fig. 3(c).

(a) Two-Loop Fermion Diagram.

(b) Nested Diagrams of One-Loop Fermion Diagram with AL diagram and its
corresponding counterterm.

(c) 3-Loop Fermion Diagram.

Figure 3: Diagrams contributing to the third leading order of fermion self-
energy.

In the three-loop diagram, the fermion in the top loop can run in the
clockwise or counterclockwise direction. Moreover, only when the fermion
running in the loop lies in a different $\pm$ patch than the external fermion
Metlitski and Sachdev (2010); Sur and Lee (2014) will the diagram give
logarithmic divergence. Let us assume that the external fermion lies in the
$+$ patch. First, consider the case in which the fermion in the loop travels
in counterclockwise direction. Its contribution reads

$\displaystyle\begin{split}\Sigma_{\text{3LF}}^{a}(k)=-\lambda_{+}^{3}\lambda_{-}^{3}\int\frac{[dl_{1}][dl_{2}][dp]}{(2\pi)^{9}}G_{+}(k-l_{1})G_{+}(k-l_{2})&G_{-}(p)G_{-}(p+l_{1})G_{-}(p+l_{2})\\\
\times D(l_{1})D(l_{1}-l_{2})D(l_{2}).\end{split}$ (35)

Integrating over $l_{1x}$, $l_{2x}$ results in

$\displaystyle\begin{split}\Sigma_{3LF}^{a}&(k)=-\int\frac{dp_{\tau}dp_{x}dp_{y}dl_{1\tau}dl_{1y}dl_{2\tau}dl_{2y}}{(2\pi)^{7}}\frac{\theta(l_{1\tau}-k_{\tau})-\theta(-l_{1\tau}-p_{\tau})}{i(k_{\tau}-p_{\tau}-2l_{1\tau})+2(k_{y}+p_{y})l_{1y}-p_{x}+p_{y}^{2}-k_{x}-k_{y}^{2}}\\\
&\times\frac{\theta(l_{2\tau}-k_{\tau})-\theta(-l_{2\tau}-p_{\tau})}{i(k_{\tau}-p_{\tau}-2l_{2\tau})+2(k_{y}+p_{y})l_{2y}-p_{x}+p_{y}^{2}-k_{x}-k_{y}^{2}}\cdot\frac{1}{-ip_{\tau}-p_{x}+p_{y}^{2}}D(l_{1})D(l_{1}-l_{2})D(l_{2}).\end{split}$
(36)

Integrating over $p_{x}$ and $p_{y}$, we obtain

$\displaystyle\Sigma_{3LF}^{a}(k)=\int$
$\displaystyle\frac{dp_{\tau}dl_{1\tau}dl_{1y}dl_{2\tau}dl_{2y}}{(2\pi)^{5}}\frac{D(l_{1})D(l_{1}-l_{2})D(l_{2})}{2\left((l_{2}-l_{1})_{y}G_{+}(k)^{-1}+2i(l_{2y}l_{1\tau}-l_{1y}l_{2\tau})\right)}T(p_{\tau};l_{1\tau},l_{2\tau};l_{1y},l_{2y};k_{\tau}),$
(37)

where $G_{+}(k)^{-1}$ is just the inverse of bare $+$ patch fermion propagator
and

$\displaystyle\begin{split}T(p_{\tau};l_{1\tau},l_{2\tau};l_{1y},l_{2y};k_{\tau})=\left(\theta(l_{1\tau}-k_{\tau})-\theta(-l_{1\tau}-p_{\tau})\right)\left(\theta(l_{2\tau}-k_{\tau})-\theta(-l_{2\tau}-p_{\tau})\right)\\\
\times\Bigg{[}\left(\theta\left(\frac{2l_{2\tau}-k_{\tau}}{l_{2y}}\right)-\theta\left(\frac{2l_{1\tau}-k_{\tau}}{l_{1y}}\right)\right)\theta(-p_{\tau})+\left(\theta\left(\frac{2l_{1\tau}-k_{\tau}}{l_{1y}}\right)-\theta\left(\frac{l_{2\tau}-l_{1\tau}}{l_{2y}-l_{1y}}\right)\right)\theta\left(k_{\tau}-p_{\tau}-2l_{1\tau}\right)\\\
-\left(\theta\left(\frac{2l_{2\tau}-k_{\tau}}{l_{2y}}\right)-\theta\left(\frac{l_{2\tau}-l_{1\tau}}{l_{2y}-l_{1y}}\right)\right)\theta\left(k_{\tau}-p_{\tau}-2l_{2\tau}\right)\Bigg{]}.\\\
\end{split}$ (38)

Suppose $k_{\tau}>0$. Separating $l_{1y},l_{2y}$ plane into six regions as in
Fig. 4(a) and doing the integration of $p_{\tau}$, we obtain

$\displaystyle\mathcal{T}(l_{1\tau},l_{2\tau};l_{1y},l_{2y};k_{\tau})\equiv\int
dp_{\tau}T(p_{\tau};l_{1\tau},l_{2\tau};l_{1y},l_{2y};k_{\tau})$ (39)

in each region shown in Fig. 4(b) ($k_{\tau}$ is set to be 1 in these graphs).

(a) Our separtation of $l_{1y}$-$l_{2y}$ plane.

(b) Result of $\mathcal{T}$ in each region. The units in the $l_{1\tau}$ and
$l_{2\tau}$ axes are both $k_{\tau}$.

Figure 4: Result of $\mathcal{T}$ given different sections on the
$l_{1y}-l_{2y}$ plane. Different sections of $l_{1y}-l_{2y}$ plane are labeled
in Fig. 4(a) and the result of $\mathcal{T}$ in each region is given in Fig.
4(b) where each graph from left to right then from top down corresponds to the
result in section 1-6 respectively.

Note that

$\displaystyle\mathcal{T}(l_{1\tau},l_{2\tau};-l_{1y},-l_{2y};k_{\tau})=-\mathcal{T}(l_{1\tau},l_{2\tau};l_{1y},l_{2y};k_{\tau})$
(40)

and therefore regions connected by $l_{1y},l_{2y}\Longleftrightarrow-
l_{1y},-l_{2y}$ gives the same result. We can thus restrict to regions with
$l_{1y}>0$. First consider region 1. In order to isolate the divergent terms,
let us do the following change of parameters:

$\displaystyle
l_{2y}=l_{1y}\delta,~{}~{}~{}~{}~{}~{}l_{1\tau}=l_{1y}^{2}\Delta_{1},~{}~{}~{}~{}~{}~{}l_{2\tau}=l_{1y}^{2}\Delta_{2},~{}~{}~{}~{}~{}~{}k_{\tau}=l_{1y}^{2}\Delta_{0}.$
(41)

Moreover, we can write down $\mathcal{T}$ as

$\displaystyle\mathcal{T}(l_{2\tau},l_{1\tau};l_{2y},l_{1y};k_{\tau})=l_{1y}^{2}\mathcal{T}_{0}(\Delta_{1},\Delta_{2};\delta;\Delta_{0}).$
(42)

Then we have

$\displaystyle\begin{split}\Sigma_{3LF}^{a}(k)=&\frac{i\alpha^{3}}{2\pi^{2}}\int
dl_{1y}d\delta
d\Delta_{1}d\Delta_{2}\frac{l_{1y}}{(\Delta_{2}-\delta\Delta_{1})-\frac{1}{2}i(1-\delta)\frac{G_{+}(k)^{-1}}{l_{1y}^{2}}}\mathcal{T}_{0}\\\
&\quad\quad\quad\quad\times\frac{1}{1+\alpha|\Delta_{1}|}\frac{|\delta|}{|\delta|^{2}+\alpha|\Delta_{2}|}\frac{|1-\delta|}{|1-\delta|^{2}+\alpha|\Delta_{1}-\Delta_{2}|}.\end{split}$
(43)

The divergent terms come from the integration region
$l_{1y}\rightarrow\infty$, so let us do the integration of $l_{1y}$ from
$k_{y}$, the dynamically generated IR cutoff, to infinity instead. We can thus
first do a large $l_{1y}$ expansion in this region,

$\displaystyle\Sigma_{3LF}^{a}(k)=\frac{i\alpha^{3}}{2\pi^{2}}\left(\sigma_{1}\int
l_{1y}dl_{1y}+i\sigma_{2}~{}G_{+}(k)^{-1}\int
l_{1y}^{-1}dl_{1y}+\sigma_{3}~{}k_{\tau}\int l_{1y}^{-1}dl_{1y}+...\right),$
(44)

where

$\displaystyle\begin{split}\sigma_{1}&=\int d\delta
d\Delta_{1}d\Delta_{2}\frac{1}{\Delta_{2}-\delta\Delta_{1}}\frac{1}{1+\alpha|\Delta_{1}|}\frac{|\delta|}{|\delta|^{2}+\alpha|\Delta_{2}|}\frac{|1-\delta|}{|1-\delta|^{2}+\alpha|\Delta_{1}-\Delta_{2}|}\mathcal{T}_{0}(\Delta_{1},\Delta_{2};\delta;0),\\\
\sigma_{2}&=\int d\delta
d\Delta_{1}d\Delta_{2}\frac{1-\delta}{2(\Delta_{2}-\delta\Delta_{1})^{2}}\frac{1}{1+\alpha|\Delta_{1}|}\frac{|\delta|}{|\delta|^{2}+\alpha|\Delta_{2}|}\frac{|1-\delta|}{|1-\delta|^{2}+\alpha|\Delta_{1}-\Delta_{2}|}\mathcal{T}_{0}(\Delta_{1},\Delta_{2};\delta;0),\\\
\sigma_{3}&=\frac{d}{d\Delta_{0}}\left(\int d\delta
d\Delta_{1}d\Delta_{2}\frac{1}{\Delta_{2}-\delta\Delta_{1}}\frac{1}{1+\alpha|\Delta_{1}|}\frac{|\delta|}{|\delta|^{2}+\alpha|\Delta_{2}|}\frac{|1-\delta|}{|1-\delta|^{2}+\alpha|\Delta_{1}-\Delta_{2}|}\mathcal{T}_{0}(\Delta_{1},\Delta_{2};\delta;\Delta_{0})\right)\Bigg{|}_{\Delta_{0}=0}.\\\
\end{split}$ (45)

The integration of the … terms is convergent. The term proportional to
$\sigma_{1}$ is polynomially divergent and does not contribute to the
logarithmic divergence, so it should give an unimportant shift of chemical
potential. The term proportional to $\sigma_{2}$ and $\sigma_{3}$ give the
following logarithmically divergent contribution to $\Sigma^{a}_{3LF}(k)$:

$\displaystyle\left(\frac{\alpha^{3}}{2\pi^{2}}S^{a}(\alpha)G_{+}(k)^{-1}-\frac{\alpha^{3}}{\pi^{2}}G^{a}(\alpha)ik_{\tau}\right)\ln\left(\frac{\Lambda_{y}}{k_{y}}\right),$
(46)

where $G^{a}(\alpha)$ and $S^{a}(\alpha)$ are defined as follows

$\displaystyle
G^{a}(\alpha)\equiv\int_{0}^{1}d\delta\int_{0}^{\infty}d\Delta\frac{\delta}{\delta^{2}+\alpha\Delta}\frac{1-\delta}{(1-\delta)^{2}+\alpha\Delta},$
(47) $\displaystyle
S^{a}(\alpha)\equiv\int_{0}^{1}d\delta\int_{0}^{\infty}d\Delta_{1}\int_{0}^{\infty}d\Delta_{2}\frac{(1-\delta)\Delta_{2}}{(\Delta_{2}+\delta\Delta_{1})^{2}}\frac{1}{1+\alpha\Delta_{1}}\frac{\delta}{\delta^{2}+\alpha\Delta_{2}}\frac{1-\delta}{(1-\delta)^{2}+\alpha(\Delta_{1}+\Delta_{2})}.$
(48)

The calculation in the region 2 is completely analogous to the calculation in
the region 1. If 1 and 2 are exchanged in Eqs. (37) and (38), one can see that
the result in region 2 is identical to the result in region 1. For region 3,
it is clear that $\sigma_{3}=0$ while from $\sigma_{2}$ we obtain the
divergent part

$\displaystyle\frac{\alpha^{3}}{\pi^{2}}G_{+}(k)^{-1}T^{a}(\alpha)\ln\left(\frac{\Lambda_{y}}{k_{y}}\right),$
(49)

where

$\displaystyle
T^{a}(\alpha)\equiv\int_{0}^{\infty}d\delta\int_{0}^{\infty}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}\frac{(1+\delta)\Delta_{2}}{(\Delta_{2}+\delta\Delta_{1})^{2}}\frac{1}{1+\alpha\Delta_{1}}\frac{\delta}{\delta^{2}+\alpha\Delta_{2}}\frac{1+\delta}{(1+\delta)^{2}+\alpha(\Delta_{1}-\Delta_{2})}.$
(50)

Adding the results of all regions up, we get the result for the case in which
the fermion travels in counterclockwise direction:

$\displaystyle\Sigma_{3LF}^{a}(k)\rightarrow\frac{2\alpha^{3}}{\pi^{2}}\left(S^{a}(\alpha)+T^{a}(\alpha)\right)G_{+}(k)^{-1}\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)-\frac{4\alpha^{3}}{\pi^{2}}G^{a}(\alpha)ik_{\tau}\ln\left(\frac{\Lambda_{y}}{k_{y}}\right).$
(51)

The calculation of the case in which the fermion travels clockwise is
completely analogous with similar expressions. The diagram reads

$\displaystyle\begin{split}\Sigma_{\text{3LF}}^{b}(k)=-\lambda_{+}^{3}\lambda_{-}^{3}\int\frac{[dl_{1}][dl_{2}][dp]}{(2\pi)^{9}}G_{+}(k-l_{1})G_{+}(k-l_{2})&G_{-}(p)G_{-}(p-l_{1})G_{-}(p-l_{2})\\\
\times D(l_{1})D(l_{1}-l_{2})D(l_{2}).\end{split}$ (52)

Integrate over $l_{1x}$, $l_{2x}$, $p_{x}$ and $p_{y}$ in order, we have

$\displaystyle\Sigma_{3LF}^{b}(k)=-\int$
$\displaystyle\frac{dp_{\tau}dl_{1\tau}dl_{1y}dl_{2\tau}dl_{2y}}{(2\pi)^{5}}\frac{D(l_{1})D(l_{1}-l_{2})D(l_{2})}{2\left[(l_{2}-l_{1})_{y}G_{+}(k)^{-1}+2i\left(l_{2y}l_{1\tau}-l_{1y}l_{2\tau}-i(l_{1y}^{2}l_{2y}-l_{1y}l_{2y}^{2})\right)\right]}T(p_{\tau};l_{1\tau},l_{2\tau};l_{1y},l_{2y};k_{\tau}),$
(53)

where $T(p_{\tau};l_{1\tau},l_{2\tau};l_{1y},l_{2y};k_{\tau})$ is precisely
the same as what was defined in Eq. (38). Following closely the procedure
outlined, we have

$\displaystyle\Sigma_{3LF}^{b}(k)\rightarrow-\frac{2\alpha^{3}}{\pi^{2}}\left(S^{b}(\alpha)+T^{b}(\alpha)\right)G_{+}(k)^{-1}\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)+\frac{4\alpha^{3}}{\pi^{2}}G^{b}(\alpha)ik_{\tau}\ln\left(\frac{\Lambda_{y}}{k_{y}}\right),$
(54)

where $G^{b}(\alpha)$ is defined as

$\displaystyle
G^{b}(\alpha)\equiv\int_{0}^{1}d\delta\int_{0}^{\infty}d\Delta\frac{\delta}{\delta^{2}+\alpha\Delta}\frac{1-\delta}{(1-\delta)^{2}+\alpha\Delta}\frac{\Delta^{2}}{\Delta^{2}+\left(\delta^{2}-\delta\right)^{2}},$
(55)

and $S^{b}(\alpha)$ and $T^{b}(\alpha)$ are defined as

$\displaystyle\begin{split}S^{b}(\alpha)&\equiv\int_{0}^{1}d\delta\int_{0}^{\infty}d\Delta_{1}\int_{0}^{\infty}d\Delta_{2}\frac{(1-\delta)\Delta_{2}\left((\Delta_{2}+\delta\Delta_{1})^{2}-(\delta^{2}-\delta)^{2}\right)}{\left((\Delta_{2}+\delta\Delta_{1})^{2}+(\delta-\delta^{2})^{2}\right)^{2}}\\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{1}{1+\alpha\Delta_{1}}\frac{\delta}{\delta^{2}+\alpha\Delta_{2}}\frac{1-\delta}{(1-\delta)^{2}+\alpha(\Delta_{1}+\Delta_{2})},\end{split}$
(56)
$\displaystyle\begin{split}T^{b}(\alpha)&\equiv\int_{0}^{\infty}d\delta\int_{0}^{\infty}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}\frac{(1+\delta)\Delta_{2}\left((\Delta_{2}+\delta\Delta_{1})^{2}-(\delta^{2}+\delta)^{2}\right)}{\left((\Delta_{2}+\delta\Delta_{1})^{2}+(\delta^{2}+\delta)^{2}\right)^{2}}\\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{1}{1+\alpha\Delta_{1}}\frac{\delta}{\delta^{2}+\alpha\Delta_{2}}\frac{1+\delta}{(1+\delta)^{2}+\alpha(\Delta_{1}-\Delta_{2})}.\end{split}$
(57)

Collecting the results for fermion self-energy at this order, we obtain the
counter terms:

$\displaystyle\begin{split}&\delta_{1}^{(2)}=\frac{2\alpha^{3}}{\pi^{2}}\left(S^{a}(\alpha)+T^{a}(\alpha)+2G^{a}(\alpha)-S^{b}(\alpha)-T^{b}(\alpha)-2G^{b}(\alpha)\right)\ln\frac{\Lambda_{y}}{\mu}\\\
&\delta_{2}^{(2)}=\frac{2\alpha^{3}}{\pi^{2}}\left(S^{a}(\alpha)+T^{a}(\alpha)-S^{b}(\alpha)-T^{b}(\alpha)\right)\ln\frac{\Lambda_{y}}{\mu}\end{split}$
(58)

where
$G^{a}(\alpha),S^{a}(\alpha),T^{a}(\alpha),G^{b}(\alpha),S^{b}(\alpha),T^{b}(\alpha)$
are defined in Eqs. (47), (48), (50), (55), (56), (57) respectively.

Before we continue, let us mention that from the Ward identity Metlitski and
Sachdev (2010), the fermion field strength divergence is the same as the
divergence of fermion-gauge vertex, whose contribution to order $1/N^{2}$
comes solely from Fig. 5 Holder and Metzner (2015b).

Figure 5: Diagram contributing to the divergence of fermion-fermion-boson
vertex at order $1/N^{2}$.

###  III.3 Third leading Order Result of Boson Self-Energy

In this subsection, we evaluate the boson self energy at the third leading
order and calculate $\delta_{3}^{(2)}$, the contribution to $Z_{3}$ at the
order of $1/N^{2}$. Again, it is sufficient to keep only the divergent parts.
It turns out that $\delta_{3}^{(2)}$ is double-logarithmic (double-log)
divergent, i.e.,
$\delta_{3}^{(2)}\propto\left(\ln\frac{\Lambda_{y}}{\mu}\right)^{2}$, with
$\mu$ the renormalization scale. The leading contribution in the limit of
$\mu\ll\Lambda_{y}$ is

$\displaystyle\delta_{3}^{(2)}=\frac{4\alpha^{4}}{3\pi^{2}}\left(\ln\frac{\Lambda_{y}}{\mu}\right)^{2}$
(59)

This result can be obtained by combining Eqs. (95), (116) and the
renormalization condition in Eq. (5). The rest of this subsection is devoted
to the computational details.

####  III.3.1 General Strategies of Calculation and Assumptions

(a) Benz Diagram.

(b) 3-String Diagram.

(c) Feynman diagrams with one fermion loop attached to three boson lines.

(d) Feynman diagrams involving counterterms.

Figure 6: Diagrams contributing to the third leading order of boson self-
energy. We do not repeat all the possible reshuffling of boson lines when
attached to a fermion loop. We also do not include Feynman diagrams with only
one fermion loop, which are deemed to be zero in the static limit
$k_{\tau}\rightarrow 0$ Lee (2009); Metlitski and Sachdev (2010).

Since the higher-loop diagrams are much more difficult to compute explicitly,
we begin by outlining the general strategies we use and spell out assumptions
we make to simplify our calculations.

We draw all diagrams contributing to the third leading order, i.e. order
$1/N$, of boson self-energy (see Fig. 6). It turns out that some diagrams have
internal boson momenta restricted as in Eq. (30). Using the calculation of the
AL diagram and the results in quantum field theory (QFT), we make the
following local divergence assumption : integration with restricted
$y$-momentum is convergent when the Feynman diagram contains no divergent
subdiagrams. Hence, only two sets of diagrams can contribute to double-log
divergence, which we call Benz diagram and 3-String diagram, and denote the
results of these two sets of diagrams by $\Pi_{\rm Benz}(k_{y})$ and $\Pi_{\rm
3String}(k_{y})$, respectively.

For these two sets of diagrams, we can first use the residue theorem to
integrate the $x$-momentum of all internal propagators, as well as frequency
and $y$-momentum of the two fermion loops. Then we are left with the
integration of internal boson frequency and $y$-momentum, denoted as
$l_{i\tau}$ and $l_{iy}$, with $i=1,2$ for two internal boson lines. It turns
out that at $k_{\tau}=0$ these two contributions to the self energy take the
form of $\Pi_{\rm Benz}(k_{y})=\frac{1}{|k_{y}|}\int
dl_{1\tau}dl_{1y}dl_{2\tau}dl_{2y}\mathsf{B}(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})$
and $\Pi_{\rm 3String}(k_{y})=\frac{1}{|k_{y}|}\int
dl_{1\tau}dl_{1y}dl_{2\tau}dl_{2y}\mathsf{S}(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})$,
where $\mathsf{B}$ and $\mathsf{S}$ can be written as

$\displaystyle\begin{split}&\mathsf{B}(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})=\mathsf{B}(0,l_{1\tau},l_{1y},l_{2\tau},l_{2y})+a_{B}(l_{1\tau},l_{1y},l_{2\tau},l_{2y})\cdot
k_{y}+f_{B}(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})\cdot k_{y}^{2},\\\
&\mathsf{S}(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})=\mathsf{S}(0,l_{1\tau},l_{1y},l_{2\tau},l_{2y})+a_{S}(l_{1\tau},l_{1y},l_{2\tau},l_{2y})\cdot
k_{y}+f_{S}(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})\cdot
k_{y}^{2},\end{split}$ (60)

where $a_{B,S}$ is the coefficient of the $k_{y}$-term in the Taylor expansion
of $\mathsf{B},\mathsf{S}(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})$, and
$f_{B,S}(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})$ are some analytic functions
of $k_{y}$ defined by the above equations. It turns out that the integrations
of both $\mathsf{B}(0,l_{1\tau},l_{1y},l_{2\tau},l_{2y})$ and
$\mathsf{S}(0,l_{1\tau},l_{1y},l_{2\tau},l_{2y})$ are polynomially divergent
for the individual diagrams. In the end, they are canceled among all diagrams
related to each other by rearranging the order in which photon lines attach to
the fermion loops, e.g. between the Benz diagram in Fig. 6(a) and its twin
diagram in Fig. 7, and among the six cases of 3-String diagrams shown in Fig.
8 Holder and Metzner (2015a). Furthermore, $\Pi(k)$ is an even function of
$k_{y}$ due to the inversion symmetry of the system, so the second terms
(i.e., the terms proportional to $k_{y}$) in the above equations do not
contribute. Hence we just need to integrate
$f_{B,S}(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})$, with a cutoff
$\Lambda_{y}$ for $l_{1y}$ and $l_{2y}$.

To extract potential double-log divergence in a simpler way, we apply the
following expansion trick similar to the calculation of fermion self-energy.
Since we are interested in the divergent part of the integration as
$\Lambda_{y}\rightarrow\infty$, it is enough to consider the region with
sufficiently large $l_{iy}$ and $l_{i\tau}$. Therefore, we can simply
integrate $f(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})$ at $k_{y}=0$ here, and
impose suitable dynamically generated IR cutoff for the integrals, i.e., the
scale above which $f(0,l_{1\tau},l_{1y},l_{2\tau},l_{2y})$ is comparable with
$f(k_{y},l_{1\tau},l_{1y},l_{2\tau},l_{2y})$. In this way, the coefficient of
the double-log divergent term can be extracted. To check the validity of this
strategy, we compute the Benz diagram in Sec. III.3.3 explicitly and confirm
that it gives the same answer as the one obtained from the method explained
above.

####  III.3.2 All Diagrams at the Third Leading Order

All diagrams contributing to the boson self-energy to the third leading order
in $1/N$ are listed in Fig. 6. For brevity, we do not repeat diagrams related
to the ones listed through rearranging the order in which boson lines are
attached to a fermion loop. These diagrams are related to each other by the
fact that their summation vanishes when external momentum vanishes Holder and
Metzner (2015a). Moreover, since we are mostly concerned with divergence in
front of the term $|k_{y}|$ of boson self-energy, we do not include Feynman
diagrams which have only one fermion loop because they vanish in the static
limit $k_{\tau}\rightarrow 0$ Lee (2009); Metlitski and Sachdev (2010).

Because of the kinetic constraint mentioned in the calculation of AL diagram
in (30) Holder and Metzner (2015b), at the static limit $k_{\tau}=0$, at least
one boson’s $y$-momentum in Fig. 6(c) is bounded by external boson
$y$-momentum $k_{y}$. Hence, we conjecture that diagrams in Fig. 6(c) with no
divergent subdiagrams are convergent. In fact, the first two diagrams have
divergent subdiagrams corresponding to one-loop fermion self-energy diagram as
in Fig. 1, whose divergence is cancelled by the third and fourth diagrams in
Fig. 6(d) involving counterterms.444When we apply the Callan-Symanzik equation
(6) to the third leading order of $1/N$, $\mu\frac{\partial}{\partial\mu}$
acted on these two diagrams involving counterterms is nonzero, yet it is
cancelled by $\gamma_{z}k_{\tau}\frac{\partial}{\partial k_{\tau}}$ acted on
AL diagram. We conjecture that the rest six diagrams are simply finite like
the AL diagram. The first and second diagram in Fig. 6(d) only give divergence
to the Landau damping term, i.e., divergence in the term $|k_{\tau}|/|k_{y}|$,
and cancel with each other due to the Ward identity. The fifth diagram in Fig.
6(d) is convergent.

Finally, the Benz diagram as in Fig. 6(a) is explicitly calculated in Section
III.3.3 while the calculation of the 3-String diagram in Fig. 6(b) is outlined
in Section III.3.4 555See also Ref. Holder and Metzner (2015a) for the
calculation of 3-String diagram with quadratic boson kinetic term, i.e. at
$\epsilon=1$, where some $(\ln\frac{\Lambda_{y}}{k_{y}})^{5}$ divergence is
identified., with their double-log divergence given in Eqs . (95) and (116)
(note that they are identically the same).

####  III.3.3 Benz Diagram

Let us briefly sketch the calculation of the Benz diagram. Before the
calculation, we mention that rearanging the order in which photon propagators
attach to fermion loops results in a “twin” diagram of the Benz diagram (see
Fig. 7), which is simply the nested diagram of the 3-loop fermion self-energy
diagram in Fig. 3(c) together with a single fermion loop. It only gives
divergence to the Landau damping term $|k_{\tau}|/|k_{y}|$ that is canceled by
the second diagram in Fig. 6(d) involving counterterms corresponding to
fermion field strength renormalization. Similarly, the $|k_{\tau}|/|k_{y}|$
divergence of the Benz diagram is canceled by the first diagram in Fig. 6(d)
involving counterterms corresponding to vertex renormalization, which can be
explicity checked.

Figure 7: The twin diagram of the Benz diagram after rearranging the order
photon propagators attach to the fermion loop.

Considering different directions fermions in the loops of Benz diagram can
travel as well as different patches the fermions can belong to, there are
altogether $16$ different contributions from the Benz diagram, which we
temporarily denote as $\Pi_{\text{Benz}}=\sum(\Pi_{\pm,C/A;\pm,C/A})$, where
$\Pi_{+C,-A}$ represents the Benz diagram whose outer fermion loop lies on the
$+$ patch and runs in the clockwise (C) direction, while the inner fermion
loop lies on the $-$ patch and runs in the anti-clockwise (A) direction, etc.
Unless the two fermion loops belong to the opposite patches, the diagram
vanishes Sur and Lee (2014); Lee (2009). Since the photon propagator is
independent of the $x$-component momentum $k_{x}$, it is convenient to take
$k_{x}=0$, and it is then straightforward to see that simultaneously flipping
the patch labels of the two fermion loops does not change the result of the
diagram, i.e., $\Pi_{+C,-A}=\Pi_{-C,+A}$, etc. Furthermore, simultaneously
flipping the directions in which fermions run in the two loops will complex
conjugate the result, i.e., $\Pi_{+A,-C}=\Pi^{*}_{+C,-A}$. Therefore, the sum
of all Benz diagrams can be written as

$\displaystyle 4{\rm Re}(\Pi_{+C,-C}+\Pi_{+C,-A}).$ (61)

Therefore, only two cases need to be considered : we can first set the outer
loop fermion to be in the $+$ patch travelling clockwise, while the inner loop
fermion is set to be in the $-$ patch travelling either clockwise (case $a$)
or anti-clockwise (case $b$), corresponding to $\Pi_{+C,-C}$ and $\Pi_{+C,-A}$
respectively.

The Benz diagram in the two cases read

$\displaystyle\begin{split}\Pi_{\text{Benz}}^{a}(k)&=\lambda_{+}^{5}\lambda_{-}^{3}\int[dl_{1}][dl_{2}][dp][dq]~{}G_{+}(q)G_{+}(q+k)G_{+}(q+k-l_{1})G_{+}(q-l_{1})G_{+}(q-l_{2})\\\
&\quad\quad\quad\quad\times
G_{-}(p)G_{-}(p+l_{1})G_{-}(p+l_{2})D(l_{1})D(l_{2})D(l_{1}-l_{2}),\end{split}$
(62)
$\displaystyle\begin{split}\Pi_{\text{Benz}}^{b}(k)&=\lambda_{+}^{5}\lambda_{-}^{3}\int[dl_{1}][dl_{2}][dp][dq]~{}G_{+}(q)G_{+}(q+k)G_{+}(q+k-l_{1})G_{+}(q-l_{1})G_{+}(q-l_{2})\\\
&\quad\quad\quad\quad\times
G_{-}(p)G_{-}(p-l_{1})G_{-}(p-l_{2})D(l_{1})D(l_{2})D(l_{1}-l_{2}).\end{split}$
(63)

Using the residue theorem to calculate the integration of $l_{1x}$, $l_{2x}$,
$p_{x}$, $q_{x}$, $p_{y}$ and $q_{y}$ in order, and taking the real part of
the expression, we obtain the following expression,

$\displaystyle\Pi_{\text{Benz}}^{\prime}(k)\equiv{\rm
Re}(\Pi_{\text{Benz}}^{a}+\Pi_{\text{Benz}}^{b})=\frac{1}{16(2\pi)^{6}|k_{y}|}\int
dl_{1\tau}dl_{1y}dl_{2\tau}dl_{2y}dp_{\tau}dq_{\tau}\left(\sum_{i=1}^{6}\mathsf{B}_{i}\right)D(l_{1})D(l_{2})D(l_{1}-l_{2}),$
(64)

where $\mathsf{B}_{i}$ are defined as follows

$\displaystyle\begin{split}\mathsf{B}_{1}\equiv&\left(\theta(l_{2\tau}-q_{\tau})-\theta(-l_{2\tau}-p_{\tau})\right)\left(\theta(l_{1\tau}-k_{\tau}-q_{\tau})-\theta(-l_{1\tau}-p_{\tau})\right)\left(\theta(k_{\tau}-2l_{1\tau}-p_{\tau}+q_{\tau})-\theta(-p_{\tau})\right)\\\
&\times\left(\theta(k_{\tau}-2l_{1\tau}+q_{\tau})-\theta(k_{\tau}+q_{\tau})\right)\left(\theta\left(\frac{l_{1\tau}}{l_{1y}}\right)-\theta\left(\frac{-k_{\tau}+2l_{1\tau}-2l_{2\tau}}{2l_{1y}-2l_{2y}}\right)\right)\\\
&\times\left(\theta\left(\frac{k_{\tau}l_{1y}+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{2l_{1y}}\right)-\theta(k_{\tau})\right)\mathcal{B}_{1},\end{split}$
(65)
$\displaystyle\begin{split}\mathsf{B}_{2}\equiv&\left(\theta(l_{2\tau}-q_{\tau})-\theta(-l_{2\tau}-p_{\tau})\right)\left(\theta(l_{1\tau}-k_{\tau}-q_{\tau})-\theta(-l_{1\tau}-p_{\tau})\right)\left(\theta(-2l_{2\tau}-p_{\tau}+q_{\tau})-\theta(-p_{\tau})\right)\\\
&\times\left(\theta(-2l_{2\tau}+q_{\tau})-\theta(k_{\tau}+q_{\tau})\right)\left(\theta\left(\frac{-k_{\tau}+2l_{1\tau}-2l_{2\tau}}{2l_{1y}-2l_{2y}}\right)-\theta\left(\frac{k_{\tau}+2l_{2\tau}}{2l_{2y}}\right)\right)\\\
&\times\left(\theta\left(\frac{k_{\tau}l_{1y}+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{2l_{1y}}\right)-\theta(k_{\tau})\right)\mathcal{B}_{1},\end{split}$
(66)
$\displaystyle\begin{split}\mathsf{B}_{3}\equiv&\left(\theta(l_{2\tau}-q_{\tau})-\theta(-l_{2\tau}-p_{\tau})\right)\left(\theta(l_{1\tau}-k_{\tau}-q_{\tau})-\theta(-l_{1\tau}-p_{\tau})\right)\left(\theta(k_{\tau}-2l_{1\tau}-p_{\tau}+q_{\tau})-\theta(-p_{\tau})\right)\\\
&\times\left(\theta(k_{\tau}-2l_{1\tau}+q_{\tau})-\theta(q_{\tau})\right)\left(\theta\left(\frac{-k_{\tau}+2l_{1\tau}-2l_{2\tau}}{2l_{1y}-2l_{2y}}\right)-\theta\left(\frac{-k_{\tau}+2l_{1\tau}}{2l_{1y}}\right)\right)\\\
&\times\left(\theta\left(\frac{k_{\tau}l_{2y}+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{2l_{2y}}\right)-\theta(k_{\tau})\right)\mathcal{B}_{2},\end{split}$
(67)
$\displaystyle\begin{split}\mathsf{B}_{4}\equiv&\left(\theta(l_{2\tau}-q_{\tau})-\theta(-l_{2\tau}-p_{\tau})\right)\left(\theta(l_{1\tau}-k_{\tau}-q_{\tau})-\theta(-l_{1\tau}-p_{\tau})\right)\left(\theta(-2l_{2\tau}-p_{\tau}+q_{\tau})-\theta(-p_{\tau})\right)\\\
&\times\left(\theta(-2l_{2\tau}+q_{\tau})-\theta(q_{\tau})\right)\left(\theta\left(\frac{l_{2\tau}}{l_{2y}}\right)-\theta\left(\frac{-k_{\tau}+2l_{1\tau}-2l_{2\tau}}{2l_{1y}-2l_{2y}}\right)\right)\\\
&\times\left(\theta\left(\frac{k_{\tau}l_{2y}+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{2l_{2y}}\right)-\theta(k_{\tau})\right)\mathcal{B}_{2},\end{split}$
(68)
$\displaystyle\begin{split}\mathsf{B}_{5}\equiv&\left(\theta(l_{2\tau}-q_{\tau})-\theta(-l_{2\tau}-p_{\tau})\right)\left(\theta(l_{1\tau}-q_{\tau})-\theta(-l_{1\tau}-p_{\tau})\right)\left(\theta(-2l_{1\tau}-p_{\tau}+q_{\tau})-\theta(-p_{\tau})\right)\\\
&\times\left(\theta(-2l_{1\tau}+q_{\tau})-\theta(k_{\tau}+q_{\tau})\right)\left(\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)-\theta\left(\frac{k_{\tau}+2l_{1\tau}}{2l_{1y}}\right)\right)\\\
&\times\left(\theta\left(\frac{k_{\tau}(l_{1y}-l_{2y})+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{2(l_{1y}-l_{2y})}\right)-\theta(k_{\tau})\right)\mathcal{B}_{3},\end{split}$
(69)
$\displaystyle\begin{split}\mathsf{B}_{6}\equiv&\left(\theta(l_{2\tau}-q_{\tau})-\theta(-l_{2\tau}-p_{\tau})\right)\left(\theta(l_{1\tau}-q_{\tau})-\theta(-l_{1\tau}-p_{\tau})\right)\left(\theta(-2l_{2\tau}-p_{\tau}+q_{\tau})-\theta(-p_{\tau})\right)\\\
&\times\left(\theta(-2l_{2\tau}+q_{\tau})-\theta(k_{\tau}+q_{\tau})\right)\left(\theta\left(\frac{k_{\tau}+2l_{2\tau}}{2l_{2y}}\right)-\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\\\
&\times\left(\theta\left(\frac{k_{\tau}(l_{1y}-l_{2y})+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{2(l_{1y}-l_{2y})}\right)-\theta(k_{\tau})\right)\mathcal{B}_{3},\end{split}$
(70)

and $\mathcal{B}_{i}=\mathcal{B}^{a}_{i}-\mathcal{B}^{b}_{i}$ are defined as
follows

$\displaystyle\begin{split}\mathcal{B}^{a}_{1}&\equiv\frac{l_{1y}\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+k_{y}^{2}l_{1y}^{2}l_{2y}(l_{1y}-l_{2y})\right)}{\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+k_{y}^{2}l_{1y}^{2}l_{2y}^{2}\right)\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+k_{y}^{2}l_{1y}^{2}(l_{1y}-l_{2y})^{2}\right)},\\\
\mathcal{B}^{a}_{2}&\equiv\frac{l_{2y}}{(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+k_{y}^{2}l_{1y}^{2}l_{2y}^{2}},\\\
\mathcal{B}^{a}_{3}&\equiv\frac{l_{1y}-l_{2y}}{(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+k_{y}^{2}l_{1y}^{2}(l_{1y}-l_{2y})^{2}},\end{split}$
(71)
$\displaystyle\scriptsize\begin{split}\mathcal{B}^{b}_{1}&\equiv\frac{l_{1y}\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}-\left(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y}+k_{y}l_{1y}l_{2y}\right)\left(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y}-k_{y}l_{1y}(l_{1y}-l_{2y})\right)\right)}{\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y}+k_{y}l_{1y}l_{2y})^{2}\right)\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y}-k_{y}l_{1y}(l_{1y}-l_{2y}))^{2}\right)},\\\
\mathcal{B}^{b}_{2}&\equiv\frac{l_{2y}\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}-(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y})(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y}+k_{y}l_{1y}l_{2y})\right)}{\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y})^{2}\right)\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y}+k_{y}l_{1y}l_{2y})^{2}\right)},\\\
\mathcal{B}^{b}_{3}&\equiv\frac{(l_{1y}-l_{2y})\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}-(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y})(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y}-k_{y}l_{1y}(l_{1y}-l_{2y}))\right)}{\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y})^{2}\right)\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+\left(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y}-k_{y}l_{1y}(l_{1y}-l_{2y})\right)^{2}\right)}.\\\
\end{split}$ (72)

Note that $\mathcal{B}^{a}_{1}=\mathcal{B}^{a}_{2}+\mathcal{B}^{a}_{3}$ and
$\mathcal{B}^{b}_{1}=\mathcal{B}^{b}_{2}+\mathcal{B}^{b}_{3}$.

Setting $k_{\tau}=0$, we obtain

$\displaystyle\begin{split}\sum_{i=1}^{6}\mathsf{B}_{i}=&\left(\hat{T_{1}}\theta\left(\frac{l_{1\tau}}{l_{1y}}\right)-\hat{T_{2}}\theta\left(\frac{l_{2\tau}}{l_{2y}}\right)+(\hat{T_{2}}-\hat{T_{1}})\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\left(\theta(l_{1y})-\theta(l_{2y})\right)\text{sgn}\left(l_{1y}l_{2\tau}-l_{1\tau}l_{2y}\right)\mathcal{B}_{2}\\\
&+\left(l_{1}\rightarrow-l_{1},~{}~{}~{}l_{2}\rightarrow
l_{2}-l_{1}\right),\end{split}$ (73)

where the second term denotes the expression coming from transforming the
first expression accordingly and the simplification comes from the left-right
flipping symmetry of the Benz diagram. Here $\hat{T}_{1,2}$ are defined as
follows

$\displaystyle\begin{split}\hat{T}_{1}\equiv&\left(\theta(l_{2\tau}-q_{\tau})-\theta(-l_{2\tau}-p_{\tau})\right)\left(\theta(l_{1\tau}-q_{\tau})-\theta(-l_{1\tau}-p_{\tau})\right)\\\
&\left(\theta(-2l_{1\tau}-p_{\tau}+q_{\tau})-\theta(-p_{\tau})\right)\left(\theta(-2l_{1\tau}+q_{\tau})-\theta(q_{\tau})\right),\end{split}$
(74)
$\displaystyle\begin{split}\hat{T}_{2}\equiv&\left(\theta(l_{2\tau}-q_{\tau})-\theta(-l_{2\tau}-p_{\tau})\right)\left(\theta(l_{1\tau}-q_{\tau})-\theta(-l_{1\tau}-p_{\tau})\right)\\\
&\left(\theta(-2l_{2\tau}-p_{\tau}+q_{\tau})-\theta(-p_{\tau})\right)\left(\theta(-2l_{2\tau}+q_{\tau})-\theta(q_{\tau})\right).\end{split}$
(75)

Then the first term and the second term should give the same contribution to
$\Pi^{a,b}$, and therefore we focus on the first term. We can now explicitly
see that $\Pi^{a,b}$ are even functions of $k_{y}$, as expected.

Following the procedure outlined in Section III.3.1, we expand
$\mathcal{B}_{2}^{a}$ at small $k_{y}$ and focus on terms proportional to
$k_{y}^{2}$,

$\displaystyle\mathcal{B}^{a}_{2}\rightarrow-\frac{l_{1y}^{2}l_{2y}^{3}}{(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+k_{y}^{2}l_{1y}^{2}l_{2y}^{2}\right)}k_{y}^{2}\xrightarrow{f(k_{y}=0)\cdot
k_{y}^{2}}-\frac{l_{1y}^{2}l_{2y}^{3}}{(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{4}}k_{y}^{2}.$
(76)

We will just use the expression $f(0)k_{y}^{2}$ for most of our calculation,
yet the full expression is important in identifying the dynamically generated
IR cutoff for calculating the coefficient of the double-log. For
$\mathcal{B}_{2}^{b}$, since only single-log is present, we just write down
the expression of $f(0)k_{y}^{2}$ here for simplicity,

$\displaystyle\begin{split}\mathcal{B}^{b}_{2}\rightarrow&l_{1y}^{2}l_{2y}^{3}\Bigg{(}-\frac{1}{\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y})^{2}\right)^{2}}+\frac{8(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y})^{2}}{\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y})^{2}\right)^{3}}\\\
&-\frac{8(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y})^{4}}{\left((l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{2}+(l_{1y}l_{2y}^{2}-l_{1y}^{2}l_{2y})^{2}\right)^{4}}\Bigg{)}k_{y}^{2}.\end{split}$
(77)

Note that the relevant terms in $\sum_{i=1}^{6}\mathsf{B}_{i}$ are invariant
if we simultaneously flip the sign of $l_{1y}$, $l_{2y}$ or $l_{1\tau}$,
$l_{2\tau}$. Therefore, let us restrict to the integration region where
$l_{1y}>0$ as well as $l_{1\tau}>0$. In this regime, the first term of
$\sum_{i=1}^{6}\mathsf{B}_{i}$ is nonzero only when $l_{2y}<0$ and
$l_{2\tau}>0$, and after integrating over $p_{\tau}$ and $q_{\tau}$ it becomes
$\mathcal{B}_{2}\times\min(l_{1\tau},l_{2\tau})^{2}$. Now we are left with the
integral over $l_{1\tau}$ $l_{1y}$, and $l_{2\tau}$ and $l_{2y}$. To perform
this integral, first consider the region where $l_{1\tau}>l_{2\tau}$. In order
to isolate the divergence, we change the integration variables as

$\displaystyle
l_{2y}=-l_{1y}\delta,~{}~{}~{}~{}~{}~{}l_{1\tau}=l_{1y}^{2}\Delta_{1},~{}~{}~{}~{}~{}~{}l_{2\tau}=l_{1y}^{2}\Delta_{2}.$
(78)

After the integration of $\delta$, $\Delta_{1,2}$ as well as $l_{1y}$, we have
the following logarithmic divergent piece of $\Sigma^{a,b}$ coming from the
first term in $\sum_{i=1}^{6}\mathsf{B}_{i}$ in the region $l_{1y}>0$,
$l_{2y}<0$, $l_{1\tau}>l_{2\tau}>0$,

$\displaystyle\frac{|k_{y}|}{e^{2}}\frac{\alpha^{4}}{4\pi^{2}}\left(\int\frac{dl_{1y}}{l_{1y}}\int_{0}^{\infty}d\delta\int_{0}^{\infty}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}~{}\Delta_{2}^{2}\delta^{3}(\omega^{a}-\omega^{b})\frac{1}{1+\alpha\Delta_{1}}\frac{\delta}{\delta^{2}+\alpha\Delta_{2}}\frac{1+\delta}{(1+\delta)^{2}+\alpha(\Delta_{1}-\Delta_{2})}\right)$
(79)

where $\omega^{a}$ and $\omega^{b}$ comes from $\mathcal{B}^{a}_{2}$ and
$\mathcal{B}^{b}_{2}$ respectively,

$\displaystyle\omega^{a}=\frac{1}{(\Delta_{2}+\delta\Delta_{1})^{2}\left((\Delta_{2}+\delta\Delta_{1})^{2}+(k_{y}/l_{1y})^{2}\delta^{2}\right)}\xrightarrow{k_{y}=0}\frac{1}{(\Delta_{2}+\delta\Delta_{1})^{4}},$
(80)
$\displaystyle\omega^{b}=\frac{1}{\left((\delta^{2}+\delta)^{2}+(\Delta_{2}+\delta\Delta_{1})^{2}\right)^{2}}-\frac{8(\delta^{2}+\delta)^{2}}{\left((\delta^{2}+\delta)^{2}+(\Delta_{2}+\delta\Delta_{1})^{2}\right)^{3}}+\frac{8(\delta^{2}+\delta)^{4}}{\left((\delta^{2}+\delta)^{2}+(\Delta_{2}+\delta\Delta_{1})^{2}\right)^{4}}.$
(81)

Next consider the region where $l_{2\tau}>l_{1\tau}$. To compare two regions
let us do the following change of parameters in this region:

$\displaystyle
l_{2y}=-|l_{1y}|,~{}~{}~{}~{}~{}~{}l_{1y}=|l_{1y}|\delta,~{}~{}~{}~{}~{}~{}l_{1\tau}=l_{1y}^{2}\Delta_{2},~{}~{}~{}~{}~{}~{}l_{2\tau}=l_{1y}^{2}\Delta_{1}.$
(82)

The logarithmic divergent piece in this region is then written as follows,

$\displaystyle\frac{|k_{y}|}{e^{2}}\frac{\alpha^{4}}{4\pi^{2}}\left(\int\frac{dl_{1y}}{l_{1y}}\int_{0}^{\infty}d\delta\int_{0}^{\infty}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}~{}\Delta_{2}^{2}\delta^{2}(\omega^{a}-\omega^{b})\frac{1}{1+\alpha\Delta_{1}}\frac{\delta}{\delta^{2}+\alpha\Delta_{2}}\frac{1+\delta}{(1+\delta)^{2}+\alpha(\Delta_{1}-\Delta_{2})}\right)$
(83)

Adding the contribution of all regions up is now straightforward, which is
simply adding Eqs. (79) and (83) before multiplied by 4. The second term in
$\sum_{i=1}^{6}\mathsf{B}_{i}$ gives an extra factor of 2 while different
directions in which fermions travel as well as different fermion patch numbers
give an extra factor of 4. Carefully considering all these factors, the Benz
diagram gives the following log divergent contribution to boson self-energy

$\displaystyle
4\Pi_{\text{Benz}}^{\prime}(k)\rightarrow\frac{|k_{y}|}{e^{2}}\frac{8\alpha^{4}}{\pi^{2}}\int\frac{dl_{1y}}{l_{1y}}(M^{a}(\alpha)-M^{b}(\alpha)),$
(84)

where $M_{a,b}(\alpha)$ are defined as

$\displaystyle
M^{a,b}(\alpha)\equiv\int_{0}^{\infty}d\delta\int_{0}^{\infty}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}~{}\Delta_{2}^{2}\delta^{2}(1+\delta)\omega^{a,b}\frac{1}{1+\alpha\Delta_{1}}\frac{\delta}{\delta^{2}+\alpha\Delta_{2}}\frac{1+\delta}{(1+\delta)^{2}+\alpha(\Delta_{1}-\Delta_{2})}$
(85)

with $\omega^{a,b}$ defined in Eqs. (80) and (81). At $k_{y}=0$, even though
$M^{b}(\alpha)$ is just a regular function of $\alpha$, $M^{a}(\alpha)$ is
not, because of the divergent integration coming from the region where
$\Delta_{1}$ and $\Delta_{2}$ go to zero simultaneously, which can be
translated to the dynamical region where

$\displaystyle|l_{1y}|,|l_{2y}|\gg|k_{y}|,~{}~{}~{}~{}~{}~{}|l_{1\tau}|,|l_{2\tau}|\approx|k_{y}l_{1y}|.$
(86)

Note that $|k_{y}l_{1y}|$ is approximately the on-shell energy of a fermion
with $x$-momentum $l_{1y}^{2}$ and $y$-momentum $k_{y}+l_{1y}$. Therefore the
integral is actually double-log divergent instead of single-log divergent.
Denoting the integrand of $M^{a}(\alpha)$ as $m^{a}(\alpha)$, we separate
$M^{a}(\alpha)$ in the following way

$\displaystyle\begin{split}M^{a}(\alpha)&\equiv\int_{0}^{\infty}d\delta\int_{0}^{\infty}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}~{}m^{a}(\alpha)\\\
&=\left(\int_{0}^{\infty}d\delta\int_{1}^{\infty}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}~{}m^{a}(\alpha)\right)\\\
&+\left(\int_{0}^{\infty}d\delta\int_{0}^{1}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}\left(m^{a}(\alpha)-\frac{\Delta_{2}^{2}\delta}{(\Delta_{2}+\delta\Delta_{1})^{2}\left((\Delta_{2}+\delta\Delta_{1})^{2}+(k_{y}/l_{1y})^{2}\delta^{2}\right)}\right)\right)\\\
&+\Bigg{(}\int_{0}^{\infty}d\delta\int_{0}^{1}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}\frac{\Delta_{2}^{2}\delta}{(\Delta_{2}+\delta\Delta_{1})^{2}\left((\Delta_{2}+\delta\Delta_{1})^{2}+(k_{y}/l_{1y})^{2}\delta^{2}\right)}\Bigg{)}.\\\
\end{split}$ (87)

Then only the third term is IR divergent at $k_{y}=0$, which will be related
to the double-log divergence we are after.

In order to compute the coefficient of the double-log through the expansion
trick sketched in Section III.3.1, we need to first identify the IR cutoff for
$l_{\tau}$ and $l_{y}$. For this, we return to the full expression of the
integrand,

$\displaystyle\int_{0}^{\Lambda_{y}}dl_{1y}\int_{0}^{\Lambda_{y}}dl_{2y}\int_{0}^{l_{1y}^{2}}dl_{1\tau}\int_{0}^{l_{1\tau}}dl_{2\tau}\frac{l_{2\tau}^{2}l_{1y}l_{2y}}{(l_{1y}l_{2\tau}-l_{\tau}l_{2y})^{2}\left((l_{1y}l_{2\tau}-l_{\tau}l_{2y})^{2}+k_{y}^{2}l_{1y}^{2}l_{2y}^{2}\right)}.$
(88)

From Eq. (76) or (88), the crossover scale appears when
$(l_{1y}l_{2\tau}-l_{\tau}l_{2y})^{2}$ is of the same order as
$k_{y}^{2}l_{1y}^{2}l_{2y}^{2}$, and therefore the dynamically generated IR
cutoff for $l_{\tau}$ should be $k_{y}l_{y}$. Again, note that $|k_{y}l_{1y}|$
is approximately the on-shell energy of a fermion with $x$-momentum
$l_{1y}^{2}$ and $y$-momentum $k_{y}+l_{1y}$. Accordingly, we can do the
integration of $\Delta_{1}$ from $k_{y}/l_{1y}$ instead of 0 to 1, which gives
us a log term. This log term is reminiscent of the BCS-log associated to the
real Cooper pair Polchinski (1992); Shankar (1994), which is also tied solely
to the integration of frequency, and hence looks like a virtual BCS-log. Then
the third term after the integration of $\delta,\Delta_{1},\Delta_{2}$ as well
as $l_{1y}$ is

$\displaystyle\int\frac{dl_{1y}}{l_{1y}}M^{a}(\alpha)\rightarrow\frac{1}{6}\int_{k_{y}}^{\Lambda_{y}}\frac{dl_{1y}}{l_{1y}}\int_{k_{y}/l_{1y}}^{1}\frac{d\Delta_{1}}{\Delta_{1}}=\frac{1}{12}\left(\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)\right)^{2}.$
(89)

It is also possible to calculate the coefficient of the relevant double-log
divergent terms explicitly. Here, it is possible to do the explicit
integration. After integrating over $\Delta_{1}$, $\Delta_{2}$ and $\delta$,
the result is a function $f(k_{y}/l_{1y})$, explicitly written out as follows,

$\displaystyle
f(x)=\frac{x\left(\left(x^{2}+3\right)\ln\left(x^{2}+1\right)-2x^{2}\ln(x)-4\right)+4\arctan(x)}{12x^{3}}.$
(90)

and the third term now becomes
$\int\frac{dl_{1y}}{l_{1y}}f\left(\frac{k_{y}}{l_{1y}}\right)$. From the
limiting behavior of $f(x)$ at small and large $x$,

$\displaystyle f(x)=\Bigg{\\{}\begin{array}[]{rcl}-\frac{1}{6}\ln(x)&&x\ll
1\\\ \frac{1}{2}\frac{\ln(x)}{x^{2}}&&x\gg 1\end{array}$ (93)

we see that the integration $\int\frac{dx}{x}f(\frac{1}{x})$ is IR convergent
and UV double-log divergent. Hence the integration of $l_{1y}$ is IR
convergent and UV double-log divergent. This integration can be carried out
explicitly as well, which at large $\Lambda_{y}/k_{y}$ is

$\displaystyle\frac{1}{12}\left(\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)\right)^{2}+\frac{5}{36}\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)+\frac{3\pi^{2}+38}{432}.$
(94)

The double-log term in (94) agrees with Eq. (89) that is obtained from the
simplified scheme. This yields the promised double-log divergent term

$\displaystyle\Pi_{\text{Benz}}(k)\rightarrow\frac{|k_{y}|}{e^{2}}\frac{2\alpha^{4}}{3\pi^{2}}\left(\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)\right)^{2}.$
(95)

####  III.3.4 3-String Diagram

Finally, let us briefly sketch the calculation of the 3-String diagram. For
simplicity, we focus on the case where $k_{\tau}=0$ whenever possible. As in
previous cases, the two fermion loops must belong to different patches to give
rise to UV divergence. Since the photon propagator is independent of the
$x$-component momentum $k_{x}$, we can set the left fermion loop to be in the
$-$ patch without loss of generality. We can also fix that fermion in the left
loop runs in counterclockwise direction and fermion in the right loop runs in
clockwise direction. Then there are 6 possible ways in which the three
internal photon propagators attach to the two loops, shown in Fig. 8. After
simultaneously flipping the directions in which fermions run in the two loops,
diagram $a$ and $b$ do not change, while $c,d$ and $e,f$ transform into each
other respectively. Therefore, the contribution of the 3-String diagram as a
whole becomes

$\displaystyle 2(\Pi^{a}+\Pi^{b})+4\text{Re}(\Pi^{c}+\Pi^{e}),$ (96)

where

Figure 8: All possible 3-String diagrams depending on how three internal
photon propagators attach to the fermion loops. From left to right then from
top down are case $a$, $b$, $c$, $d$, $e$ and $f$ respectively.

$\displaystyle\begin{split}\Pi_{\text{3String}}^{a}(k)&=\lambda_{+}^{4}\lambda_{-}^{4}\int[dl_{1}][dl_{2}][dp][dq]~{}G_{-}(p)G_{-}(p+k)G_{-}(p+l_{1}+k/2)G_{-}(p+l_{2}+k/2)\\\
&\quad\quad\times
G_{+}(q)G_{+}(q-k)G_{+}(q-l_{1}-k/2)G_{+}(q-l_{2}-k/2)D(k/2+l_{1})D(k/2-l_{2})D(l_{1}-l_{2}),\end{split}$
(97)
$\displaystyle\begin{split}\Pi_{\text{3String}}^{b}(k)&=\lambda_{+}^{4}\lambda_{-}^{4}\int[dl_{1}][dl_{2}][dp][dq]~{}G_{-}(p)G_{-}(p+k)G_{-}(p+l_{1}+k/2)G_{-}(p+l_{2}+k/2)\\\
&\quad\quad\times
G_{+}(q)G_{+}(q+k)G_{+}(q+l_{1}+k/2)G_{+}(q+l_{2}+k/2)D(k/2+l_{1})D(k/2-l_{2})D(l_{1}-l_{2}),\end{split}$
(98)
$\displaystyle\begin{split}\Pi_{\text{3String}}^{c}(k)&=\lambda_{+}^{4}\lambda_{-}^{4}\int[dl_{1}][dl_{2}][dp][dq]~{}G_{-}(p)G_{-}(p+k)G_{-}(p+l_{1}+k/2)G_{-}(p+l_{2}+k/2)\\\
&\quad\quad\times
G_{+}(q)G_{+}(q+l_{1}+k/2)G_{+}(q+l_{2}+k/2)G_{+}(q+l_{2}-k/2)D(k/2+l_{1})D(k/2-l_{2})D(l_{1}-l_{2}),\end{split}$
(99)
$\displaystyle\begin{split}\Pi_{\text{3String}}^{e}(k)&=\lambda_{+}^{4}\lambda_{-}^{4}\int[dl_{1}][dl_{2}][dp][dq]~{}G_{-}(p)G_{-}(p+k)G_{-}(p+l_{1}+k/2)G_{-}(p+l_{2}+k/2)\\\
&\quad\quad\times
G_{+}(q)G_{+}(q-l_{1}-k/2)G_{+}(q-l_{2}-k/2)G_{+}(q-l_{2}+k/2)D(k/2+l_{1})D(k/2-l_{2})D(l_{1}-l_{2}).\end{split}$
(100)

Again, we use the residue theorem to calculate the integration of $l_{1x}$,
$l_{2x}$, $p_{x}$, $q_{x}$, $p_{y}$ and $q_{y}$ in order. The expression at
non-zero $k_{\tau}$ is extremely lengthy, but there is some simplification
when $k_{\tau}=0$:

$\displaystyle\Pi_{\text{3String}}^{a,b,c,e}(k_{y})=\frac{{\rm
sgn}(k_{y})}{(2\pi)^{6}|k_{y}|}\int
dl_{1\tau}dl_{1y}dl_{2\tau}dl_{2y}dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{a,b,c,e}+\hat{T}_{2}\mathsf{S}_{2}^{a,b,c,e}\right)D(k/2+l_{1})D(k/2-l_{2})D(l_{2}-l_{1}),$
(101)

where $\hat{T}_{1,2}$ are defined in (74) and (75) and

$\displaystyle\begin{split}\mathsf{S}_{i}^{a,b}\equiv&\left(\theta\left(\frac{2l_{i\tau}}{2l_{iy}-k_{y}}\right)-\theta\left(\frac{2l_{i\tau}}{2l_{iy}+k_{y}}\right)\right)\left(\theta\left(\frac{2l_{i\tau}}{2l_{iy}-k_{y}}\right)\mathcal{S}^{a,b}_{i,-}-\theta\left(\frac{2l_{i\tau}}{2l_{iy}+k_{y}}\right)\mathcal{S}^{a,b}_{i,+}\right)\\\
&+\Bigg{(}\left(\theta\left(\frac{2l_{i\tau}}{2l_{iy}-k_{y}}\right)-\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\theta\left(\frac{k_{y}(l_{1\tau}-l_{2\tau})+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{k_{y}(l_{1y}-l_{2y})}\right)\\\
&-\left(\theta\left(\frac{2l_{i\tau}}{2l_{iy}+k_{y}}\right)-\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\theta\left(\frac{k_{y}(l_{1\tau}-l_{2\tau})-2l_{1y}l_{2\tau}+2l_{1\tau}l_{2y}}{k_{y}(l_{1y}-l_{2y})}\right)\Bigg{)}\mathcal{S}^{a,b}_{0},\end{split}$
(102)
$\displaystyle\begin{split}\mathsf{S}_{1}^{c,e}\equiv&\left(\theta\left(\frac{2l_{1\tau}}{2l_{1y}-k_{y}}\right)-\theta\left(\frac{2l_{1\tau}}{2l_{1y}+k_{y}}\right)\right)\theta\left(\frac{2l_{1\tau}}{2l_{1y}+k_{y}}\right)\mathcal{S}^{c,e}_{1}\\\
&+\left(\theta\left(\frac{2l_{1\tau}}{2l_{1y}+k_{y}}\right)-\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\theta\left(\frac{k_{y}(l_{1\tau}-l_{2\tau})-2l_{1y}l_{2\tau}+2l_{1\tau}l_{2y}}{k_{y}(2l_{1y}+k_{y})}\right)\mathcal{S}^{c,e}_{1}\\\
&-\left(\theta\left(\frac{2l_{1\tau}}{2l_{1y}-k_{y}}\right)-\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\Bigg{(}\theta\left(\frac{k_{y}(l_{1\tau}-l_{2\tau})+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{k_{y}(-2l_{2y}+k_{y})}\right)\mathcal{S}^{c,e}_{2}\\\
&+\theta\left(\frac{k_{y}(l_{1\tau}-l_{2\tau})+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{k_{y}(l_{1y}-l_{2y})}\right)\mathcal{S}^{c,e}_{0}\Bigg{)},\end{split}$
(103)
$\displaystyle\begin{split}\mathsf{S}_{2}^{c,e}\equiv&\left(\theta\left(\frac{2l_{2\tau}}{2l_{2y}+k_{y}}\right)-\theta\left(\frac{2l_{2\tau}}{2l_{2y}-k_{y}}\right)\right)\left(\theta\left(\frac{2l_{2\tau}}{2l_{2y}-k_{y}}\right)\mathcal{S}^{c,e}_{2,-}-\theta\left(\frac{2l_{2\tau}}{2l_{2y}+k_{y}}\right)\mathcal{S}^{c,e}_{2,+}\right)\\\
&+\left(\theta\left(\frac{2l_{2\tau}}{2l_{2y}+k_{y}}\right)-\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\theta\left(\frac{k_{y}(l_{1\tau}-l_{2\tau})-2l_{1y}l_{2\tau}+2l_{1\tau}l_{2y}}{k_{y}(2l_{1y}+k_{y})}\right)\mathcal{S}^{c,e}_{1}\\\
&-\left(\theta\left(\frac{2l_{2\tau}}{2l_{2y}-k_{y}}\right)-\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\Bigg{(}\theta\left(\frac{k_{y}(l_{1\tau}-l_{2\tau})+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{k_{y}(-2l_{2y}+k_{y})}\right)\mathcal{S}^{c,e}_{2}\\\
&+\theta\left(\frac{k_{y}(l_{1\tau}-l_{2\tau})+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}}{k_{y}(l_{1y}-l_{2y})}\right)\mathcal{S}^{c,e}_{0}\Bigg{)},\end{split}$
(104)

where the most important $\mathcal{S}$ factors for case $a$ are defined as
follows,

$\displaystyle\begin{split}\mathcal{S}^{a}_{i,\pm}&\equiv\frac{k_{y}\pm
2l_{iy}}{16k_{y}l_{i\tau}\left(k_{y}(l_{1\tau}-l_{2\tau})\mp
2(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})\right)},\\\
\mathcal{S}^{a}_{0}&\equiv\frac{l_{1y}-l_{2y}}{4\left(k_{y}(l_{1\tau}-l_{2\tau})+2l_{1y}l_{2\tau}-2l_{1\tau}l_{2y}\right)\left(k_{y}(l_{1\tau}-l_{2\tau})-2l_{1y}l_{2\tau}+2l_{1\tau}l_{2y}\right)},\end{split}$
(105)

and for case $b$,

$\displaystyle\begin{split}\mathcal{S}^{b}_{i,\pm}\equiv&(k_{y}\pm
2l_{iy})\Bigg{/}\Big{\\{}2k_{y}\left(4l_{i\tau}-i(k_{y}^{2}-4l_{iy}^{2})\right)\\\
&\times\left(2\left(k_{y}(l_{1\tau}-l_{2\tau})\mp
2\left(l_{1y}l_{2\tau}-l_{1\tau}l_{2y}\right)\right)\pm
i\left(k_{y}^{2}(l_{1y}-l_{2y})+4(l_{1y}^{2}l_{2y}-l_{2y}^{2}l_{1y})\right)+2ik_{y}(l_{1y}^{2}-l_{2y}^{2})\right)\Big{\\}},\\\
\mathcal{S}^{b}_{0}\equiv&(l_{1y}-l_{2y})\Bigg{/}\Big{\\{}\left(2\left(k_{y}(l_{1\tau}-l_{2\tau})+2\left(l_{1y}l_{2\tau}-l_{1\tau}l_{2y}\right)\right)-i\left(k_{y}^{2}(l_{1y}-l_{2y})+4(l_{1y}^{2}l_{2y}-l_{2y}^{2}l_{1y})\right)+2ik_{y}(l_{1y}^{2}-l_{2y}^{2})\right)\\\
&\times\left(2\left(k_{y}(l_{1\tau}-l_{2\tau})-2\left(l_{1y}l_{2\tau}-l_{1\tau}l_{2y}\right)\right)+i\left(k_{y}^{2}(l_{1y}-l_{2y})+4(l_{1y}^{2}l_{2y}-l_{2y}^{2}l_{1y})\right)+2ik_{y}(l_{1y}^{2}-l_{2y}^{2})\right)\Big{\\}},\\\
\end{split}$ (106)

and for case $c$,

$\displaystyle\scriptsize\begin{split}\mathcal{S}_{1}^{c}\equiv&(-k_{y}-2l_{1y})\Bigg{/}\Big{\\{}2\left(k_{y}^{2}l_{1y}-k_{y}^{2}l_{2y}-2ik_{y}l_{1\tau}+2k_{y}l_{1y}^{2}+2ik_{y}l_{2\tau}-2k_{y}l_{2y}^{2}-4il_{1\tau}l_{2y}+4l_{1y}^{2}l_{2y}+4il_{1y}l_{2\tau}-4l_{1y}l_{2y}^{2}\right)\\\
&\times\left(k_{y}^{2}l_{1y}-k_{y}^{2}l_{2y}-2ik_{y}l_{1\tau}+2k_{y}l_{1y}^{2}-4k_{y}l_{1y}l_{2y}-2ik_{y}l_{2\tau}+2k_{y}l_{2y}^{2}+4il_{1\tau}l_{2y}-4l_{1y}^{2}l_{2y}-4il_{1y}l_{2\tau}+4l_{1y}l_{2y}^{2}\right)\Big{\\}},\\\
\mathcal{S}_{2}^{c}\equiv&(-k_{y}+2l_{2y})\Bigg{/}\Big{\\{}2\left(-k_{y}^{2}l_{1y}+k_{y}^{2}l_{2y}+2ik_{y}l_{1\tau}-2k_{y}l_{1y}^{2}+4k_{y}l_{1y}l_{2y}-2ik_{y}l_{2\tau}-2k_{y}l_{2y}^{2}-4il_{1\tau}l_{2y}+4l_{1y}^{2}l_{2y}+4il_{1y}l_{2\tau}-4l_{1y}l_{2y}^{2}\right)\\\
&\times\left(-k_{y}^{2}l_{1y}+k_{y}^{2}l_{2y}+2ik_{y}l_{1\tau}-2k_{y}l_{1y}^{2}+4k_{y}l_{1y}l_{2y}+2ik_{y}l_{2\tau}-2k_{y}l_{2y}^{2}-4il_{1\tau}l_{2y}+4l_{1y}^{2}l_{2y}+4il_{1y}l_{2\tau}-4l_{1y}l_{2y}^{2}\right)\Big{\\}},\\\
\mathcal{S}_{0}^{c}=&\mathcal{S}_{1}^{c}-\mathcal{S}_{2}^{c}\equiv(-l_{1y}+l_{2y})\Bigg{/}\Big{\\{}\left(k_{y}^{2}l_{1y}-k_{y}^{2}l_{2y}-2ik_{y}l_{1\tau}+2k_{y}l_{1y}^{2}+2ik_{y}l_{2\tau}-2k_{y}l_{2y}^{2}-4il_{1\tau}l_{2y}+4l_{1y}^{2}l_{2y}+4il_{1y}l_{2\tau}-4l_{1y}l_{2y}^{2}\right)\\\
&\times\left(k_{y}^{2}l_{1y}-k_{y}^{2}l_{2y}-2ik_{y}l_{1\tau}+2k_{y}l_{1y}^{2}-4k_{y}l_{1y}l_{2y}+2ik_{y}l_{2\tau}+2k_{y}l_{2y}^{2}+4il_{1\tau}l_{2y}-4l_{1y}^{2}l_{2y}-4il_{1y}l_{2\tau}+4l_{1y}l_{2y}^{2}\right)\Big{\\}},\\\
\mathcal{S}^{c}_{2,-}\equiv&i(k_{y}-2l_{2y})\Bigg{/}\Big{\\{}8k_{y}l_{2\tau}\left(-k_{y}^{2}l_{1y}+k_{y}^{2}l_{2y}+2ik_{y}l_{1\tau}-2k_{y}l_{1y}^{2}+4k_{y}l_{1y}l_{2y}+2ik_{y}l_{2\tau}-2k_{y}l_{2y}^{2}-4il_{1\tau}l_{2y}+4l_{1y}^{2}l_{2y}+4il_{1y}l_{2\tau}-4l_{1y}l_{2y}^{2}\right)\Big{\\}},\\\
\mathcal{S}^{c}_{2,+}\equiv&i(k_{y}+2l_{2y})\Bigg{/}\Big{\\{}8k_{y}l_{2\tau}\big{(}-k_{y}^{2}l_{1y}+k_{y}^{2}l_{2y}+2ik_{y}l_{1\tau}-2k_{y}l_{1y}^{2}-2ik_{y}l_{2\tau}+2k_{y}l_{2y}^{2}+4il_{1\tau}l_{2y}-4l_{1y}^{2}l_{2y}-4il_{1y}l_{2\tau}+4l_{1y}l_{2y}^{2}\big{)}\Big{\\}},\\\
\end{split}$ (107)

and, finally, for case $e$,

$\displaystyle\begin{split}\mathcal{S}_{1}^{e}\equiv&i(k_{y}+2l_{1y})\Bigg{/}\Big{\\{}4(k_{y}l_{1\tau}-k_{y}l_{2\tau}+2l_{1\tau}l_{2y}-2l_{1y}l_{2\tau})\\\
&\times\left(2k_{y}^{2}l_{1y}-2k_{y}^{2}l_{2y}+k_{y}^{3}+2ik_{y}l_{1\tau}-4k_{y}l_{1y}l_{2y}+2ik_{y}l_{2\tau}-4il_{1\tau}l_{2y}+4il_{1y}l_{2\tau}\right)\Big{\\}},\\\
\mathcal{S}_{2}^{e}\equiv&(k_{y}-2l_{2y})\Bigg{/}\Big{\\{}4\left(-k_{y}^{2}l_{1y}+k_{y}^{2}l_{2y}-ik_{y}l_{1\tau}+2k_{y}l_{1y}l_{2y}+ik_{y}l_{2\tau}-2k_{y}l_{2y}^{2}+2il_{1\tau}l_{2y}-2il_{1y}l_{2\tau}\right)\\\
&\times\left(2k_{y}^{2}l_{1y}-2k_{y}^{2}l_{2y}+k_{y}^{3}+2ik_{y}l_{1\tau}-4k_{y}l_{1y}l_{2y}+2ik_{y}l_{2\tau}-4il_{1\tau}l_{2y}+4il_{1y}l_{2\tau}\right)\Big{\\}},\\\
\mathcal{S}_{0}^{e}=&\mathcal{S}_{1}^{e}-\mathcal{S}_{2}^{e}\equiv-i(l_{1y}-l_{2y})\Bigg{/}\Big{\\{}4(k_{y}l_{1\tau}-k_{y}l_{2\tau}+2l_{1\tau}l_{2y}-2l_{1y}l_{2\tau})\\\
&\times\left(-k_{y}^{2}l_{1y}+k_{y}^{2}l_{2y}-ik_{y}l_{1\tau}+2k_{y}l_{1y}l_{2y}+ik_{y}l_{2\tau}-2k_{y}l_{2y}^{2}+2il_{1\tau}l_{2y}-2il_{1y}l_{2\tau}\right)\Big{\\}},\\\
\mathcal{S}^{e}_{2,-}\equiv&(-k_{y}+2l_{2y})\Bigg{/}\Big{\\{}2k_{y}\left(k_{y}^{2}+4il_{2\tau}-4l_{2y}^{2}\right)\\\
&\times\left(2k_{y}^{2}l_{1y}-2k_{y}^{2}l_{2y}+k_{y}^{3}+2ik_{y}l_{1\tau}-4k_{y}l_{1y}l_{2y}+2ik_{y}l_{2\tau}-4il_{1\tau}l_{2y}+4il_{1y}l_{2\tau}\right)\Big{\\}},\\\
\mathcal{S}^{e}_{2,+}\equiv&i(k_{y}+2l_{2y})\Bigg{/}\Big{\\{}4k_{y}\left(k_{y}^{2}+4il_{2\tau}-4l_{2y}^{2}\right)\left(k_{y}l_{1\tau}-k_{y}l_{2\tau}-2l_{1y}l_{2\tau}+2l_{1\tau}l_{2y}\right)\Big{\\}}.\end{split}$
(108)

First consider case $a$ and $b$. Because of the theta pre-factors, only
$\mathcal{S}^{a,b}_{0}$ contribute to UV divergence according to the local
divergence conjecture in Section III.3.1. Setting $k_{y}=0$ everywhere in the
theta factors except for an overall $\text{sgn}(k_{y})$, we obtain

$\displaystyle\begin{split}-\hat{T}_{1}\mathsf{S}_{1}^{a,b}+\hat{T}_{2}\mathsf{S}_{2}^{a,b}\approx\left(-\hat{T_{1}}\theta\left(\frac{l_{1\tau}}{l_{1y}}\right)+\hat{T_{2}}\theta\left(\frac{l_{2\tau}}{l_{2y}}\right)+(\hat{T_{1}}-\hat{T_{2}})\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\text{sgn}\left(\frac{l_{1y}l_{2\tau}-l_{1\tau}l_{2y}}{k_{y}(l_{1y}-l_{2y})}\right)\mathcal{S}^{a,b}_{0}.\end{split}$
(109)

Again the real part of this expression is invariant if we simultaneously flip
the sign of $k_{y},l_{1y}$, $l_{2y}$ or $l_{1\tau}$, $l_{2\tau}$. Therefore
let us assume $l_{1\tau}>0$ as well as $l_{1y}>0$. Then we have six regions
where the expression is non-zero,

$\displaystyle\begin{split}\text{region
1:}~{}~{}~{}~{}~{}~{}l_{2\tau}>l_{1\tau}>0,~{}l_{1y}>l_{2y}>0,~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{a,b}+\hat{T}_{2}\mathsf{S}_{2}^{a,b}\right)&=-(l_{2\tau}-l_{1\tau})^{2}\text{sgn}(k_{y})\mathcal{S}_{0}^{a,b},\\\
\text{region
2:}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}l_{2\tau}>l_{1\tau}>0,~{}l_{2y}<0,~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{a,b}+\hat{T}_{2}\mathsf{S}_{2}^{a,b}\right)&=-l_{1\tau}^{2}\text{sgn}(k_{y})\mathcal{S}_{0}^{a,b},\\\
\text{region
3:}~{}~{}~{}~{}~{}~{}l_{1\tau}>l_{2\tau}>0,~{}l_{2y}>l_{1y}>0,~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{a,b}+\hat{T}_{2}\mathsf{S}_{2}^{a,b}\right)&=(l_{2\tau}-l_{1\tau})^{2}\text{sgn}(k_{y})\mathcal{S}_{0}^{a,b},\\\
\text{region
4:}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}l_{1\tau}>l_{2\tau}>0,~{}l_{2y}<0,~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{a,b}+\hat{T}_{2}\mathsf{S}_{2}^{a,b}\right)&=-l_{2\tau}^{2}\text{sgn}(k_{y})\mathcal{S}_{0}^{a,b},\\\
\text{region
5:}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}l_{2\tau}<0,~{}l_{2y}>l_{1y}>0,~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{a,b}+\hat{T}_{2}\mathsf{S}_{2}^{a,b}\right)&=l_{1\tau}^{2}\text{sgn}(k_{y})\mathcal{S}_{0}^{a,b},\\\
\text{region
6:}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}l_{2\tau}<0,~{}l_{1y}>l_{2y}>0,~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{a,b}+\hat{T}_{2}\mathsf{S}_{2}^{a,b}\right)&=-l_{2\tau}^{2}\text{sgn}(k_{y})\mathcal{S}_{0}^{a,b}.\\\
\end{split}$ (110)

After interchanging $l_{1\tau},l_{2\tau}$ as well as $l_{1y},l_{2y}$
simultaneously, we see that the expressions in region 1 and 3, 2 and 4, as
well as 5 and 6 are identical to each other respectively, and therefore we can
focus on region 1, 2 and 5. The analysis of case $c$ and $e$ are
straightforward as well. Because of the theta pre-factors,
$\mathcal{S}_{2,\pm}^{c,e}$ do not contribute to the divergence in the UV.
Setting $k_{y}=0$ everywhere in the theta factors except for an overall
$\text{sgn}(k_{y})$, we obtain

$\displaystyle\begin{split}-\hat{T}_{1}\mathsf{S}_{1}^{c,e}+\hat{T}_{2}\mathsf{S}_{2}^{c,e}\approx&\left(-\hat{T}_{1}\theta\left(\frac{l_{1\tau}}{l_{1y}}\right)+\hat{T}_{2}\theta\left(\frac{l_{2\tau}}{l_{2y}}\right)+(\hat{T}_{1}-\hat{T}_{2})\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\\\
\times&\text{sgn}\left(l_{1y}l_{2\tau}-l_{1\tau}l_{2y}\right)\text{sgn}(k_{y})\left(\theta(-l_{1y})\mathcal{S}_{1}^{c,e}-\theta(-l_{2y})\mathcal{S}_{2}^{c,e}-\theta(l_{1y}-l_{2y})\mathcal{S}_{0}^{c,e}\right).\end{split}$
(111)

Restricting to the regime where $l_{1y}>0$ as well as $l_{1\tau}>0$, we have
four different sections on the $l_{2y}-l_{2\tau}$ plane where the expression
is non-zero,

$\displaystyle\begin{split}\text{region
1:}~{}~{}~{}~{}~{}~{}l_{2\tau}>l_{1\tau}>0,~{}l_{1y}>l_{2y}>0,~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{c,e}+\hat{T}_{2}\mathsf{S}_{2}^{c,e}\right)&=(l_{2\tau}-l_{1\tau})^{2}\text{sgn}(k_{y})\mathcal{S}_{0}^{c,e},\\\
\text{region
2:}~{}~{}~{}~{}~{}~{}l_{2\tau}>l_{1\tau}>0,~{}l_{2y}<0,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{c,e}+\hat{T}_{2}\mathsf{S}_{2}^{c,e}\right)&=l_{1\tau}^{2}\text{sgn}(k_{y})\mathcal{S}_{1}^{c,e},\\\
\text{region
4:}~{}~{}~{}~{}~{}~{}l_{1\tau}>l_{2\tau}>0,~{}l_{2y}<0,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{c,e}+\hat{T}_{2}\mathsf{S}_{2}^{c,e}\right)&=l_{2\tau}^{2}\text{sgn}(k_{y})\mathcal{S}_{1}^{c,e},\\\
\text{region
6:}~{}~{}~{}~{}~{}~{}l_{2\tau}<0,~{}l_{1y}>l_{2y}>0,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\int
dp_{\tau}dq_{\tau}\left(-\hat{T}_{1}\mathsf{S}_{1}^{c,e}+\hat{T}_{2}\mathsf{S}_{2}^{c,e}\right)&=l_{2\tau}^{2}\text{sgn}(k_{y})\mathcal{S}_{0}^{c,e}.\\\
\end{split}$ (112)

The expression for single log divergence is very involved, but it is managable
to calculate the double log divergence as sketched in Section III.3.1. First,
we expand the whole expression in powers of $k_{y}$. As in the calculation of
the Benz diagram, we collect terms at the third order in the expansion, i.e.
terms proportional to $k_{y}^{2}$. Note that $k_{y}$ mainly appears in two
different places, in $\mathcal{S}$ factors as well as in
$D(k/2+l_{1})D(k/2-l_{2})$666$k_{y}$ also appears in theta factors, but there
it only contributes to a shift of integration regions in the IR that we can
ignore for the calculation in the UV.. To simplify expressions further, we are
allowed to set $k_{y}=0$ inside the boson propagators except for the case $c$
and region 2 and 4 of case $e$, where we have to take into account the second
term in the expansion of both $\mathcal{S}^{c,e}_{1}$ and
$D(k/2+l_{1})D(k/2-l_{2})$ at small $k_{y}$, i.e. terms proportional to
$k_{y}$ in two terms respectively 777Expanding $\mathcal{S}$ factors at small
$k_{y}$, symbolically we have $\mathcal{S}\approx
x_{0}+x_{1}k_{y}+x_{2}k_{y}^{2}$. Moreover, $D(k/2+l_{1})D(k/2-l_{2})\approx
d_{0}+d_{1}k_{y}+d_{2}k_{y}^{2}$. We can explicitly check that the integration
of $x_{0}d_{0}$, i.e. the constant term, is cancelled among the six cases. The
same calculation also shows that the integration of $x_{0}d_{2}$ is cancelled
among the six cases. What we now have are $x_{2}d_{0}$ as well as
$x_{1}d_{1}$. We then notice that $x_{1}=0$ expect for $\mathcal{S}^{c}_{0,1}$
and $\mathcal{S}^{e}_{1}$..

For the sake of presentation, let us first identify the double log divergence
coming from the dynamical region (86)888Here we need to distinguish between
integration region 1-6, and dynamical region on the boundary of respective
integration region that contributes mostly to the double log divergence.. We
just need to collect terms at the third order of the expansion that are
proportional to $1/l_{\tau}^{4}$ and contribute to the IR divergence in the
integration of $l_{1\tau},l_{2\tau}$. These terms only come from case $e$,
hence we write down the terms proportional to $k_{y}$ in the series expansion
of $\mathcal{S}^{e}$,

$\displaystyle\begin{split}\mathcal{S}^{e}_{0}\Bigg{|}_{k_{y}~{}\text{linear
term}}=&\left(\frac{l_{2y}^{2}(l_{1y}-l_{2y})^{3}}{16(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{4}}-\frac{(l_{1y}-l_{2y})(l_{1\tau}-l_{2\tau})^{2}}{64(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{4}}\right)k_{y}^{2},\\\
\mathcal{S}^{e}_{1}\Bigg{|}_{k_{y}~{}\text{linear
term}}=&\left(\frac{l_{2y}^{2}l_{1y}^{3}}{16(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{4}}-\frac{l_{1y}(l_{1\tau}^{2}+l_{2\tau}^{2})+2l_{2y}l_{1\tau}l_{2\tau}}{64(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{4}}\right)k_{y}^{2}.\\\
\end{split}$ (113)

We see that $k_{y}$-linear terms of $\mathcal{S}^{e}$ are separated into two
terms $\mathcal{S}^{e,(1)}+\mathcal{S}^{e,(2)}$. The first terms
$\mathcal{S}^{e,(1)}$ of the two expansions are proportional to
$1/l_{\tau}^{4}$ while the second terms $\mathcal{S}^{e,(2)}$ are not, and we
isolate the first terms,

$\displaystyle\begin{split}\mathcal{S}^{e,(1)}_{1}&=\frac{l_{1y}^{3}l_{2y}^{2}}{16(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{4}}k_{y}^{2},\\\
\mathcal{S}^{e,(1)}_{2}&=\frac{l_{2y}^{3}(l_{2y}^{2}-3l_{1y}l_{2y}+3l_{1y}^{2})}{16(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{4}}k_{y}^{2},\\\
\mathcal{S}^{e,(1)}_{0}&=\frac{l_{2y}^{2}(l_{1y}-l_{2y})^{3}}{16(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{4}}k_{y}^{2}.\end{split}$
(114)

Just considering these terms coming from case $e$, we see that, to our
surprise, Eq. (111) becomes identical to Eq. (73) after interchanging
$l_{1\tau},l_{1y}$ with $l_{2\tau},l_{2y}$. Namely, after substituting
$\mathcal{S}^{e}$ in (111) with $\mathcal{S}^{e,(1)}$ and interchanging label
$l_{1\tau},l_{1y}$ and $l_{2\tau},l_{2y}$, we have

$\displaystyle\begin{split}\left(-\hat{T}_{1}\mathsf{S}_{1}^{e}+\hat{T}_{2}\mathsf{S}_{2}^{e}\right)^{(1)}=&\frac{{\rm
sgn}(k_{y})k_{y}^{2}}{16}\left(\hat{T_{1}}\theta\left(\frac{l_{1\tau}}{l_{1y}}\right)-\hat{T_{2}}\theta\left(\frac{l_{2\tau}}{l_{2y}}\right)+(\hat{T_{2}}-\hat{T_{1}})\theta\left(\frac{l_{1\tau}-l_{2\tau}}{l_{1y}-l_{2y}}\right)\right)\\\
&\times\left(-\left(\theta(l_{1y})-\theta(l_{2y})\right)\text{sgn}\left(l_{1y}l_{2\tau}-l_{1\tau}l_{2y}\right)\frac{l_{1y}^{2}l_{2y}^{3}}{(l_{1y}l_{2\tau}-l_{1\tau}l_{2y})^{4}}+\left(l_{1}\rightarrow-
l_{1},~{}~{}~{}l_{2}\rightarrow l_{2}-l_{1}\right)\right),\end{split}$ (115)

which is identical to the expression we obtain after substituting (76) into
(73), except for an extra factor $\frac{{\rm sgn}(k_{y})}{16}$ which can be
accounted for from different factors in front of the whole expression as shown
in (64) and (101). Moreover, both contribute to photon self-energy with an
extra factor of $4$ as shown in (61) and (96). Therefore, we calculate the
double log divergence of the 3-String diagram originating from the same source
as the Benz diagram, i.e. coming from the dynamical region Eq. (86),

$\displaystyle\Pi_{\text{3String}}(k)\rightarrow\frac{|k_{y}|}{e^{2}}\frac{2\alpha^{4}}{3\pi^{2}}\left(\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)\right)^{2}.$
(116)

After subtracting these terms out, now we aim to show that the double log
divergences of the rest of the 3-String diagram cancel among the six cases.
These double-log divergences that are canceled among themselves come from
dynamical regions with large $l_{1\tau}$ or $l_{2\tau}$. The UV divergences
from these high-energy modes can only renormalize local terms in the action.
As a result, both single-log and double-log, which require non-local counter
terms, are expected to be canceled. One interesting conclusion we can draw
from this surprising result is that the double log divergence actually does
not appear at this order in the $\epsilon=0$ version of Ising-nematic
transition in a metal Metlitski and Sachdev (2010), which is very closely
related to the nFL theory we discussed here except
$\lambda_{+}=\lambda_{-}=+1$. Here $\lambda_{+}=\lambda_{-}$ because the
critical Ising-nematic order parameter, which is the analog of the field $a$
discussed here, couples to a fermion bilinear operator whose symmetry form
factor requires $\lambda_{+}=\lambda_{-}$.

Let us do the calculation in region 4 of case $e$ as an example. To unearth
the double-log divergence from the second term in the expansion (113), we do
the following change of parameters:

$\displaystyle
l_{2y}=-l_{1y}\delta,~{}~{}~{}~{}~{}~{}l_{1\tau}=l_{1y}^{2}\Delta_{1},~{}~{}~{}~{}~{}~{}l_{2\tau}=l_{1y}^{2}\Delta_{2},$
(117)

and integrate $|l_{1y}|$ from $k_{y}$ to $\Lambda_{y}$. The logarithmic
divergent piece in this region from the second term is then written as
follows,

$\displaystyle\begin{split}\left(\Pi^{e}(k)\right)^{(2)}\Bigg{|}_{\text{region~{}4}}&\approx\frac{|k_{y}|}{(2\pi)^{6}}\int_{0}^{\infty}dl_{1y}\int_{-\infty}^{0}dl_{2y}\int_{0}^{\infty}dl_{1\tau}\int_{0}^{l_{1\tau}}dl_{2\tau}~{}l_{2\tau}^{2}\mathcal{S}_{1}^{e,(2)}D(k/2+l_{1})D(k/2-l_{2})D(l_{2}-l_{1})\\\
&=-\frac{|k_{y}|}{e^{2}}\frac{\alpha^{4}}{16\pi^{2}}\Bigg{(}\int\frac{dl_{1y}}{l_{1y}}\int_{0}^{\infty}d\delta\int_{0}^{\infty}d\Delta_{1}\int_{0}^{\Delta_{1}}d\Delta_{2}~{}\frac{\Delta_{2}^{2}(\Delta_{1}^{2}+\Delta_{2}^{2}+2\delta\Delta_{1}\Delta_{2})}{(\Delta_{2}+\delta\Delta_{1})^{4}}\\\
&\times\frac{1}{1+\alpha\Delta_{1}}\frac{\delta}{\delta^{2}+\alpha\Delta_{2}}\frac{1+\delta}{(1+\delta)^{2}+\alpha(\Delta_{1}-\Delta_{2})}\Bigg{)}.\end{split}$
(118)

The above integration is IR divergent when $\Delta_{2}$ and $\delta$ go to
zero simultanesouly, which can be translated to the dynamical region where

$\displaystyle|l_{1y}|\gg|k_{y}|,~{}~{}~{}|l_{1\tau}|\gg|k_{y}l_{1y}|,~{}~{}~{}~{}~{}~{}|l_{2y}|\approx|k_{y}|,~{}~{}~{}|l_{2\tau}|\approx|k_{y}l_{1y}|.$
(119)

To simplify the expression further, we change $\Delta_{2}$ to
$\Delta_{2}^{\prime}\delta$, then at small $\delta$ the integrand is
approximately

$\displaystyle\frac{\Delta_{2}^{\prime}\Delta_{1}^{2}}{\alpha(\Delta_{1}+\Delta_{2}^{\prime})^{4}(1+\alpha\Delta_{1})^{2}}\frac{1}{\delta},$
(120)

which gives the following double log divergence in the bracket

$\displaystyle\frac{1}{6\alpha^{2}}\int_{k_{y}}^{\Lambda_{y}}\frac{dl_{1y}}{l_{1y}}\int_{k_{y}/l_{1y}}^{1}\frac{d\delta}{\delta}=\frac{1}{12\alpha^{2}}\left(\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)\right)^{2}.$
(121)

The calculation in the other integration regions and other cases are as
straightforward, and we see that the double-log divergent terms in region 2,
4, 5 and 6 of case $a$ and $b$ are all the same, i.e.

$\displaystyle\Pi^{a,b}(k)\Bigg{|}_{\text{region~{}2,4,5,6}}\rightarrow\frac{|k_{y}|}{e^{2}}\frac{\alpha^{2}}{192\pi^{2}}\left(\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)\right)^{2},$
(122)

while the double log divergent terms in region 4 and 6 of case $c$ and the
second term of $e$ are all the same, i.e.

$\displaystyle\Pi^{c}(k),(\Pi^{e}(k))^{(2)}\Bigg{|}_{\text{region~{}4,6}}\rightarrow-\frac{|k_{y}|}{e^{2}}\frac{\alpha^{2}}{192\pi^{2}}\left(\ln\left(\frac{\Lambda_{y}}{k_{y}}\right)\right)^{2},$
(123)

and in all the other regions there is no additional double log divergence.
Adding them all up we see that the additional double log divergence cancels
among the six cases, as promised.

It is worth pointing out that the calculation of the 3-String diagram relies
on the “local divergence conjecture” we make in Section III.3.1 as well as the
correctness of the expansion trick with correct dynamically generated IR
cutoff. It will be very interesting if a more rigorous calculation of these
coefficients, either analytically or numerically, can be done. We defer this
to future work.

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