$\displaystyle\sum_{l,\eta}\sum_{\mathbf{Q}_{l\eta}}\sum_{\mathbf{Q}^{\prime}_{-l\eta}}(b_{-\eta\mathbf{K}_{D}+\mathbf{k}-\mathbf{k}^{\prime}-\mathbf{Q}_{l\eta}+\mathbf{Q}^{\prime}_{-l\eta},l,A_{1}}+b_{\eta\mathbf{K}_{D}-\mathbf{k}+\mathbf{k}^{\prime}+\mathbf{Q}_{l\eta}-\mathbf{Q}^{\prime}_{-l\eta},l,A_{1}}^{\dagger})\sum_{\alpha\beta}u^{n\star}_{\mathbf{k},\mathbf{Q}_{l\eta},\alpha,\eta}\sigma^{x}_{\alpha\beta}u^{n^{\prime}}_{\mathbf{k}^{\prime},\mathbf{Q}^{\prime}_{-l\eta},\beta,-\eta}.$
(162)

By defining

$\mathbf{Q}_{l\eta}-\mathbf{Q}^{\prime}_{-l\eta}=\mathbf{Q}^{\prime\prime}_{-l\eta},$
(163)

the Inter-valley Hamiltonian becomes

$\displaystyle H_{inter-
vall}^{op}\approx\frac{\gamma_{3}}{\sqrt{2N_{G}{\sqrt{M(\alpha_{1}+\alpha_{2}+\alpha_{3})}}}}\sum_{\mathbf{k},\mathbf{k}^{\prime}\in
MBZ}\sum_{n,n^{\prime}=1,2}\sum_{\eta,s}\gamma_{\mathbf{k},n,\eta,s}^{\dagger}\gamma_{\mathbf{k}^{\prime},n^{\prime},-\eta,s}$
$\displaystyle\sum_{l}\sum_{\mathbf{Q}^{\prime\prime}_{-l\eta}}(b_{-\eta\mathbf{K}_{D}+\mathbf{k}-\mathbf{k}^{\prime}-\mathbf{Q}^{\prime\prime}_{-l\eta},l,A_{1}}+b_{\eta\mathbf{K}_{D}-\mathbf{k}+\mathbf{k}^{\prime}+\mathbf{Q}^{\prime\prime}_{-l\eta},l,A_{1}}^{\dagger})\sum_{\mathbf{Q}_{l\eta}}\sum_{\alpha\beta}u^{n\star}_{\mathbf{k},\mathbf{Q}_{l\eta},\alpha,\eta}\sigma^{x}_{\alpha\beta}u^{n^{\prime}}_{\mathbf{k}^{\prime},\mathbf{Q}_{l\eta}-\mathbf{Q}^{\prime\prime}_{-l\eta},\beta,-\eta}.$
(164)

###  VI.2 Phonon-Mediated Electron-Electron Interaction for the Flat Bands

We rewrite

$\displaystyle H_{inter-
vall}^{op}\approx\frac{1}{\sqrt{N_{G}}}\sum_{\mathbf{k},\mathbf{k}^{\prime}\in
MBZ}\sum_{n,n^{\prime}=1,2}\sum_{\eta,s}\sum_{l=\pm}\sum_{\mathbf{Q}_{-l\eta}}\times$
$\displaystyle\times
G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
nn^{\prime}l}\gamma_{\mathbf{k},n,\eta,s}^{\dagger}\gamma_{\mathbf{k}^{\prime},n^{\prime},-\eta,s}(b_{-\eta\mathbf{K}_{D}+\mathbf{k}-\mathbf{k}^{\prime}-\mathbf{Q}_{-l\eta},l,A_{1}}+b_{\eta\mathbf{K}_{D}-\mathbf{k}+\mathbf{k}^{\prime}+\mathbf{Q}_{-l\eta},l,A_{1}}^{\dagger})$
$\displaystyle G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
nn^{\prime}l}=\frac{\gamma_{3}}{\sqrt{2M\omega_{A_{1}}}}\sum_{\mathbf{Q}^{\prime}_{l\eta}}\sum_{\alpha\beta}u^{n\star}_{\mathbf{k},\mathbf{Q}^{\prime}_{l\eta},\alpha,\eta}\sigma^{x}_{\alpha\beta}u^{n^{\prime}}_{\mathbf{k}^{\prime},\mathbf{Q}^{\prime}_{l\eta}-\mathbf{Q}_{-l\eta},\beta,-\eta}$
$\displaystyle\omega_{A1}=\sqrt{(\alpha_{1}+\alpha_{2}+\alpha_{3})/M}$ (165)

In the above expression, $\mathbf{Q}_{\pm l\eta}=\pm
l\eta\mathbf{q}_{2}+\mathbf{G}_{M}$ and the summation is made over the
$\mathbf{G}_{M}$.

With the electron Hamiltonian

$H_{el}=\sum_{m,\mathbf{k}\in
MBZ,\eta,s}E_{m\mathbf{k}}\gamma_{\mathbf{k}m\eta
s}^{\dagger}\gamma_{\mathbf{k}m\eta s}$ (166)

and phonon Hamiltonian for the $A_{1}$ mode

$H_{ph}=\sum_{\mathbf{q},l}\omega_{A1}b_{\mathbf{q}l,A_{1}}^{\dagger}b_{\mathbf{q}l,A_{1}}$
(167)

we apply the transformation $S$

$-H_{inter-vall}^{op}=[H_{el}+H_{ph},S]$ (168)

and obtain

$\displaystyle S=\frac{1}{\sqrt{N_{G}}}\sum_{\mathbf{k},\mathbf{k}^{\prime}\in
MBZ}\sum_{n,n^{\prime}=1,2}\sum_{\eta,s}\sum_{l=\pm}\sum_{\mathbf{Q}_{-l\eta}}\times$
$\displaystyle\gamma_{\mathbf{k},n,\eta,s}^{\dagger}\gamma_{\mathbf{k}^{\prime},n^{\prime},-\eta,s}(A_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
nn^{\prime}l}b_{-\eta\mathbf{K}_{D}+\mathbf{k}-\mathbf{k}^{\prime}-\mathbf{Q}_{-l\eta},l,A_{1}}-A_{\mathbf{k}^{\prime},\mathbf{k},-\mathbf{Q}_{-l\eta}}^{-\eta
n^{\prime}nl\star}b_{\eta\mathbf{K}_{D}-\mathbf{k}+\mathbf{k}^{\prime}+\mathbf{Q}_{-l\eta},l,A_{1}}^{\dagger})$
$\displaystyle A_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
nn^{\prime}l}=-\frac{\mathbf{G}_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
nn^{\prime}l}}{E_{n\mathbf{k}}-E_{n^{\prime}\mathbf{k}^{\prime}}-\omega_{A_{1}}}$
(169)

One should note that it is important to keep $-\mathbf{Q}_{-l\eta}$ in
$A_{k^{\prime},k,-\mathbf{Q}_{-l\eta}}^{-\eta n^{\prime}nl\star}$ and not
rewrite it as $\mathbf{Q}_{l\eta}$ since
$-\mathbf{Q}_{-l\eta}=l\eta\mathbf{q}_{2}-\mathbf{G}_{M}$ whereas
$\mathbf{Q}_{l\eta}=l\eta\mathbf{q}_{2}+\mathbf{G}_{M}$. They hence differ by
a Moiré reciprocal lattice vector, but since this vector is what we sum over,
it is important to keep the correct form.

We then obtain the phonon-mediated electron-electron interaction (we now
expand explicitly the $\mathbf{Q}_{l\eta}$ and perform the summation over
$\mathbf{G}_{M}$)

$\displaystyle H_{el-
el}=\frac{1}{N_{G}}\sum_{\mathbf{k}\mathbf{k}^{\prime},\mathbf{k}_{1}\mathbf{k}_{1}^{\prime},\mathbf{G}_{M},l,s,s_{1}}\frac{G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta
nn^{\prime}l}G_{\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{-\eta
n_{1}n_{1}^{\prime}l}\omega_{A_{1}}}{(E_{n_{1}\mathbf{k}_{1}}-E_{n_{1}^{\prime}\mathbf{k}_{1}^{\prime}})^{2}-\omega_{A_{1}}^{2}}\delta_{\mathbf{k}-\mathbf{k}^{\prime},-\mathbf{k}_{1}+\mathbf{k}_{1}^{\prime}}\gamma_{\mathbf{k},n,\eta,s}^{\dagger}\gamma_{\mathbf{k}_{1},n_{1},-\eta,s_{1}}^{\dagger}\gamma_{\mathbf{k}_{1}^{\prime},n_{1}^{\prime},\eta,s_{1}}\gamma_{\mathbf{k}^{\prime},n^{\prime},-\eta,s}$

Since the optical phonon frequency $\omega_{A1}$ is much larger than the
bandwidth of the flat bands in TBG, we neglect the latter. In the Cooper
channel
$\mathbf{k}_{1}=-\mathbf{k},\mathbf{k}_{1}^{\prime}=-\mathbf{k}^{\prime}$, we
have

$\displaystyle H_{el-
el}=-\frac{1}{N_{G}}\frac{1}{\omega_{A_{1}}}\sum_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{G}_{M},l,s,s_{1}}G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta
nn^{\prime}l}G_{-\mathbf{k},-\mathbf{k}^{\prime},l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{-\eta
n_{1}n_{1}^{\prime}l}\gamma_{\mathbf{k},n,\eta,s}^{\dagger}\gamma_{-\mathbf{k},n_{1},-\eta,s_{1}}^{\dagger}\gamma_{-\mathbf{k}^{\prime},n_{1}^{\prime},\eta,s_{1}}\gamma_{\mathbf{k}^{\prime},n^{\prime},-\eta,s}.$
(171)

If one wants to include more bands in the system, one can extend the indices
$n,n^{\prime}$, although due to the larger dispersion of the remote bands, one
might need to keep the kinetic energies.

It should be noted as the summation over $\mathbf{k},\mathbf{k}^{\prime}$ is
within the MBZ, so we should split the factor $N_{G}$ (which represents the
total number of atomic unit cells) into two parts, $N_{G}=N_{M}\times N_{0}$,
with $N_{M}$ the total number of Moiré unit cell and $N_{0}$ the number of
atomic unit cells in one Moiré unit cell. Consequently, we can rewrite the
above Hamiltonian as

$\displaystyle H_{el-
el}=-\frac{1}{N_{M}}\sum_{\mathbf{k},\mathbf{k}^{\prime},s,s_{1}}V^{\eta
nn^{\prime},n_{1}n_{1}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};-\mathbf{k},-\mathbf{k}^{\prime}}\gamma_{\mathbf{k},n,\eta,s}^{\dagger}\gamma_{-\mathbf{k},n_{1},-\eta,s_{1}}^{\dagger}\gamma_{-\mathbf{k}^{\prime},n_{1}^{\prime},\eta,s_{1}}\gamma_{\mathbf{k}^{\prime},n^{\prime},-\eta,s}$
(172)

where

$V^{\eta
nn^{\prime},n_{1}n_{1}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=\frac{1}{N_{0}}\frac{1}{\omega_{A_{1}}}\sum_{\mathbf{G}_{M}}\sum_{l}G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta
nn^{\prime},l}G_{\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{-\eta
n_{1}n_{1}^{\prime},l}$ (173)

represents the effective attractive electron-electron interaction mediated by
phonons in TBG.

###  VI.3 Gauge Fixing for Eigen-state Basis and Chern Band Basis

####  VI.3.1 Eigen-state Basis

The eigen-energy basis of the bands in the TBG is given by

$\gamma_{\mathbf{k},n,\eta,s}^{\dagger}=\sum_{\mathbf{Q}\alpha}u_{\mathbf{Q}\alpha;n\eta}\left(\mathbf{k}\right)c_{\mathbf{k},\mathbf{Q},\eta,\alpha
s}^{\dagger}\ ,$ (174)

where $u_{\mathbf{Q}\alpha;n\eta}(\mathbf{k})$ is the eigenstate of energy
band $n$ of the first quantized single-particle Hamiltonian
$h_{\mathbf{Q},\mathbf{Q}^{\prime}}^{\left(\eta\right)}\left(\mathbf{k}\right)$
in valley $\eta$

$\sum_{\mathbf{Q}^{\prime},\beta}[h_{\mathbf{Q},\mathbf{Q}^{\prime}}^{\left(\eta\right)}\left(\mathbf{k}\right)]_{\alpha\beta}u_{\mathbf{Q}^{\prime}\beta;n\eta}(\mathbf{k})=\epsilon_{n,\eta}(\mathbf{k})u_{\mathbf{Q}\alpha;n\eta}(\mathbf{k})\
,$ (175)

where $\epsilon_{n,\eta}(\mathbf{k})$ is the single-particle energy of the
eigen-state $u_{\mathbf{Q}\alpha;n\eta}(\mathbf{k})$. In each valley and spin,
we use integers $n>0$ to label the $n$-th conduction band, and use integer
$n<0$ to label the $|n|$-th valence band (thus $n\neq 0$). The lowest
conduction and valence bands in each valley-spin flavor is thus labeled by
$n=\pm 1$.

The BM Hamiltonian becomes

$\hat{H}_{0}=\sum_{\mathbf{k}}\sum_{n\eta
s}\epsilon_{n,\eta}(\mathbf{k})\gamma_{\mathbf{k}n\eta
s}^{\dagger}\gamma_{\mathbf{k}n\eta s}.$ (176)

Bloch periodicity and periodic gauge require

$c_{\mathbf{k}+\mathbf{b}_{Mi},\mathbf{Q},\eta\alpha
s}^{\dagger}=c_{\mathbf{k},\mathbf{Q}-\mathbf{b}_{Mi},\eta\alpha
s}^{\dagger}\implies
u_{\mathbf{Q}\alpha;n\eta}\left(\mathbf{k}+\mathbf{b}_{Mi}\right)=u_{\mathbf{Q}-\mathbf{b}_{Mi},\alpha;n\eta}\left(\mathbf{k}\right)\
\implies\gamma_{\mathbf{k}+\mathbf{b}_{Mi},n\eta
s}^{\dagger}=\gamma_{\mathbf{k}n\eta s}^{\dagger}$ (177)

and the $\hat{C}_{2z}$ and $\hat{P}$ symmetries require

$\hat{C}_{2z}:\;\;\epsilon_{n,\eta}(\mathbf{k})=\epsilon_{n,-\eta}(-\mathbf{k})\
,\qquad\hat{P}:\;\;\epsilon_{n,\eta}(\mathbf{k})=-\epsilon_{-n,\eta}(-\mathbf{k})\
.$ (178)

####  VI.3.2 Gauge Fixing for Eigen-state Basis

We fix the gauge for the eigen-state basis of energy bands
$\gamma_{\mathbf{k}n\eta s}^{\dagger}$ in Eq. (174) for the symmetry operators
$\hat{C}_{2z},\hat{T}$ and $\hat{P}$. We denote the wave function
$u_{\mathbf{Q}\alpha;n\eta}(\mathbf{k})$ as a column vector
$u_{n\eta}(\mathbf{k})$ in the space of indices $\\{\mathbf{Q},\alpha\\}$. A
representation matrix $D(\hat{g})$ of an operation $\hat{g}$ acts on a wave
function $u_{n\eta^{\prime}}(\mathbf{k})$, we denote the resulting wave
function in the valley $\eta$ for short as
$[D(g)]_{\eta\eta^{\prime}}u_{n\eta^{\prime}}(\mathbf{k})$, the components of
which are given by
$\sum_{\mathbf{Q}^{\prime}\beta\eta^{\prime}}[D(g)]_{\mathbf{Q}\alpha\eta,\mathbf{Q}^{\prime}\beta\eta^{\prime}}u_{\mathbf{Q}^{\prime}\beta;n\eta^{\prime}}(\mathbf{k})$.
For the symmetries $\hat{C}_{2z},\hat{T}$ and $\hat{P}$, the sewing matrices
$B^{\hat{g}}(\mathbf{k})$ in the band and valley space are defined by

$[D(\hat{C}_{2z})]_{\eta\eta^{\prime}}u_{n\eta^{\prime}}(\mathbf{k})=\sum_{m}[B^{C_{2z}}(\mathbf{k})]_{m\eta,n\eta^{\prime}}u_{m\eta}(-\mathbf{k})\
,\;\;\;[D(\hat{T})]_{\eta\eta^{\prime}}u_{n\eta^{\prime}}^{*}(\mathbf{k})=\sum_{m}[B^{T}(\mathbf{k})]_{m\eta,n\eta^{\prime}}u_{m\eta}(-\mathbf{k})\
,$ (179)
$[D(\hat{P})]_{\eta\eta^{\prime}}u_{n\eta^{\prime}}(\mathbf{k})=\sum_{m}[B^{P}(\mathbf{k})]_{m\eta,n\eta^{\prime}}u_{m\eta}(-\mathbf{k})\
.$ (180)

For non-degenerate wave function $u_{n\eta^{\prime}}(\mathbf{k})$ in the
valley $\eta^{\prime}$, since $\hat{C}_{2z}$ and $\hat{T}$ commute with the
$\hat{H}_{0}$ and flips the valley $\eta$, while $\hat{P}$ anti-commutes with
$\hat{H}_{0}$ and preserves the valley $\eta$, we generically have

$\begin{split}&[B^{C_{2z}}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=\delta_{\eta,-\eta^{\prime}}\delta_{m,n}e^{i\varphi^{C_{2z}}_{n,\eta^{\prime}}(\mathbf{k})}\
,\quad[B^{T}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=\delta_{\eta,-\eta^{\prime}}\delta_{m,n}e^{i\varphi^{T}_{n,\eta^{\prime}}(\mathbf{k})}\
,\\\
&[B^{P}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=\delta_{\eta,\eta^{\prime}}\delta_{-m,n}e^{i\varphi^{P}_{n,\eta^{\prime}}(\mathbf{k})}\
.\end{split}$ (181)

Accordingly the energy band fermion operators transform as

$\hat{g}\gamma_{\mathbf{k},n,\eta^{\prime},s}^{\dagger}\hat{g}^{-1}=\sum_{m\eta}[B^{g}(\mathbf{k})]_{m\eta,n\eta^{\prime}}\gamma_{\hat{g}\mathbf{k},m,\eta,s}^{\dagger}\
.$ (182)

Since these three symmetries satisfy the relations

$C_{2z}^{2}=1\ ,\ T^{2}=1\ ,\ P^{2}=-1\ ,\ \\{P,C_{2z}\\}=0\ ,\ \\{P,T\\}=0\
,\ [C_{2z},T]=0\ ,$ (183)

With the above notations, the symmetries $\hat{C}_{2z},\hat{T}$ and $\hat{P}$
allows us to define

$\begin{split}&B^{C_{2z}}(-\mathbf{k})B^{C_{2z}}(\mathbf{k})=B^{T}(-\mathbf{k})B^{T*}(\mathbf{k})=-B^{P}(-\mathbf{k})B^{P}(\mathbf{k})=I\
,\qquad
B^{P}(-\mathbf{k})B^{C_{2z}}(\mathbf{k})=-B^{C_{2z}}(-\mathbf{k})B^{P}(\mathbf{k})\
,\\\ &\qquad\qquad
B^{P}(-\mathbf{k})B^{T}(\mathbf{k})=-B^{T}(-\mathbf{k})B^{P*}(\mathbf{k})\
,\qquad
B^{T}(-\mathbf{k})B^{C_{2z}*}(\mathbf{k})=B^{C_{2z}}(-\mathbf{k})B^{T}(\mathbf{k})\
,\end{split}$ (184)

where $B^{g*}(\mathbf{k})$ stands for the complex conjugation of matrix
$B^{g}(\mathbf{k})$, and $I$ is the identity matrix in the $n,\eta$ space. We
have two independent symmetry operations $\hat{C}_{2z}\hat{T}$ (anti-unitary)
and $\hat{C}_{2z}\hat{P}$ (unitary), which do not change $\mathbf{k}$. Their
sewing matrices are defined by

$[D(\hat{C}_{2z})D(\hat{T})]_{\eta\eta^{\prime}}u_{n\eta^{\prime}}^{*}(\mathbf{k})=\sum_{m}[B^{C_{2z}T}(\mathbf{k})]_{m\eta,n\eta^{\prime}}u_{m\eta}(\mathbf{k})\
,$ (185)
$[D(\hat{P})D(\hat{C}_{2z})]_{\eta\eta^{\prime}}u_{n\eta^{\prime}}(\mathbf{k})=\sum_{m}[B^{C_{2z}P}(\mathbf{k})]_{m\eta,n\eta^{\prime}}u_{m\eta}(\mathbf{k})\
.$ (186)

For non-degenerate eigenstates at momentum $\mathbf{k}$ (non-degenerate within
one valley), we have

$[B^{C_{2z}T}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=\delta_{\eta,\eta^{\prime}}\delta_{m,n}e^{i\varphi^{C_{2z}T}_{n,\eta^{\prime}}(\mathbf{k})}\
,\qquad[B^{C_{2z}P}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=\delta_{-\eta,\eta^{\prime}}\delta_{-m,n}e^{i\varphi^{C_{2z}P}_{n,\eta^{\prime}}(\mathbf{k})}\
,$ (187)

where by definition we have
$\varphi^{C_{2z}T}_{n,\eta^{\prime}}(\mathbf{k})=\varphi^{T}_{n,\eta^{\prime}}(\mathbf{k})+\varphi^{C_{2z}}_{n,-\eta^{\prime}}(-\mathbf{k})$,
and
$\varphi^{C_{2z}P}_{n,\eta^{\prime}}(\mathbf{k})=\varphi^{C_{2z}}_{n,\eta^{\prime}}(\mathbf{k})+\varphi^{P}_{n,-\eta^{\prime}}(-\mathbf{k})$.
The sewing matrices of $C_{2z}T$ and $C_{2z}P$ are subject to the constraint
that

$(C_{2z}T)^{2}=(C_{2z}P)^{2}=1\ ,\qquad[C_{2z}T,C_{2z}P]=0\ ,$ (188)

and thus they satisfy

$B^{C_{2z}T}(\mathbf{k})B^{C_{2z}T*}(\mathbf{k})=[B^{C_{2z}P}(\mathbf{k})]^{2}=I\
,\qquad
B^{C_{2z}P}(\mathbf{k})B^{C_{2z}T}(\mathbf{k})=B^{C_{2z}T}(\mathbf{k})B^{C_{2z}P*}(\mathbf{k})\
.$ (189)

We will now fixe the gauge of the wave functions and sewing matrices of the
$\mathbf{k}$ preserving symmetry operations $C_{2z}T$ and $C_{2z}P$. By Eqs.
(187) and (189), we make the $\mathbf{k}$-independent choices for the sewing
matrices

$[B^{C_{2z}T}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=\delta_{\eta,\eta^{\prime}}\delta_{m,n}\
,\qquad[B^{C_{2z}P}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=-\text{sgn}(n)\eta^{\prime}\delta_{-\eta,\eta^{\prime}}\delta_{-m,n}\
.$ (190)

Accordingly, the symmetry actions on the band basis fermion operators are
given by

$(C_{2z}T)\gamma_{\mathbf{k},n,\eta,s}^{\dagger}(C_{2z}T)^{-1}=\gamma_{\mathbf{k},n,\eta,s}^{\dagger}\
,\qquad(C_{2z}P)\gamma_{\mathbf{k},n,\eta,s}^{\dagger}(C_{2z}P)^{-1}=-\text{sgn}(n)\eta\gamma_{\mathbf{k},-n,-\eta,s}^{\dagger}\
.$ (191)

This does not yet fix the entire phases of the energy basis at momentum
$\mathbf{k}$, since the sewing matrices in Eq. (190) are invariant under the
unitary transformation of wave functions
$u_{n\eta}(\mathbf{k})\rightarrow\text{sgn}(n)\eta u_{n\eta}(\mathbf{k})$ at
each individual $\mathbf{k}$. To further fix this gauge freedom for different
$\mathbf{k}\in\text{MBZ}$, we start by choosing a momentum
$\mathbf{k}=\mathbf{k}_{0}$ where the eigenstates within one valley are non-
degenerate, and choose a gauge fixing of the eigen-state basis at
$\mathbf{k}_{0}$ satisfying Eq. (190). We then fix the band basis of bands
$\pm n$ at other $\mathbf{k}\neq\mathbf{k}_{0}$ by requiring

$f_{n,\eta}(\mathbf{k}+\mathbf{q},\mathbf{k})=\left|u^{\dagger}_{n,\eta}(\mathbf{k}+\mathbf{q})u_{n,\eta}(\mathbf{k})-u^{\dagger}_{-n,\eta}(\mathbf{k}+\mathbf{q})u_{-n,\eta}(\mathbf{k})\right|$
(192)

to be a continuous function of $\mathbf{k}$ and $\mathbf{q}$, and satisfy

$\lim_{\mathbf{q}\rightarrow\mathbf{0}}f_{n,\eta}(\mathbf{k}+\mathbf{q},\mathbf{k})=0$
(193)

for all $\mathbf{k}$. This fixes the relative sign between wave functions
$u_{n,\eta}(\mathbf{k})$ and $u_{-n,\eta}(\mathbf{k})$ in a way that is
continuous in $\mathbf{k}$. We do not require the wave function
$u_{n,\eta}(\mathbf{k})$ itself to be globally continuous in $\mathbf{k}$ of
the entire MBZ; locally $u_{n,\eta}(\mathbf{k})$ can always be chosen to be
continuous in $\mathbf{k}$, provided $u_{n,\eta}(\mathbf{k})$ is non-
degenerate at momentum $\mathbf{k}$.

Within the space of each pair of PH symmetric bands with band indices $n=\pm
n_{\text{B}}$, if we use $\zeta^{a}$ and $\tau^{a}$ ($a=0,x,y,z$) to denote
the identity and Pauli matrices in the energy band $n=\pm n_{\text{B}}$ space
and the valley space:

$B^{C_{2z}T}(\mathbf{k})=\zeta^{0}\tau^{0}\ ,\qquad
B^{C_{2z}P}(\mathbf{k})=-\zeta^{y}\tau^{y}\ .$ (194)

For $n_{\text{B}}=1$ (i.e., within the lowest conduction and valence bands
$n=\pm 1$ per spin per valley) when $\mathbf{k}$ is at $\mathbf{K}_{M}$ or
$\mathbf{K}_{M}^{\prime}$ point of the MBZ, the bands $n=+1$ and $n=-1$ are
degenerate. In this case, we can still choose the eigenstate basis at
$\mathbf{K}_{M}$ or $\mathbf{K}_{M}^{\prime}$ point such that Eqs. (194) and
(193) are satisfied.

We further fix the relative gauge between wave functions at momenta
$\mathbf{k}$ and $-\mathbf{k}$ by fixing the sewing matrices of $\hat{C}_{2z}$
and $\hat{P}$. For $\mathbf{k}$ not at the $P$-invariant momenta, which are
$\Gamma_{M}$ and the three equivalent $M_{M}$ in TBG, we have

$B^{C_{2z}}(\mathbf{k})=\zeta^{0}\tau^{x}\ ,\qquad
B^{T}(\mathbf{k})=\zeta^{0}\tau^{x}\ ,\qquad
B^{P}(\mathbf{k})=-i\zeta^{y}\tau^{z}\ .$ (195)

which are consistent with Eq. (194). With the continous gauge condition, the
sewing matrix $B^{P}(\mathbf{k})$ must have additional minus signs, i.e.,
$B^{P}(\mathbf{k})=i\zeta^{y}\tau^{z}$, at an odd (even) number of the four
$P$-invariant momenta if the the two bands $n=\pm n_{B}$ have an odd (even)
topological winding number protected by $C_{2z}T$; and at the other odd (even)
$P$-invariant momenta $B^{P}(\mathbf{k})$ are $-i\zeta^{y}\tau^{z}$, same as
those at generic momenta. Accordingly, the sewing matrices
$B^{C_{2z}}(\mathbf{k})$ and $B^{T}(\mathbf{k})$ also have the additional
minus at momenta where $B^{P}(\mathbf{k})$ has the minus sign. In this work,
we choose $B^{P}(\mathbf{k}_{\Gamma_{M}})=-i\zeta^{y}\tau^{z}$ and
$B^{P}(\mathbf{k}_{M_{M}})=i\zeta^{y}\tau^{z}$.

####  VI.3.3 Gauge Fixing for Chern band basis

After the gauge fixing in Eqs. (194) and (193), we define a new irrep basis
$\gamma^{(n_{\text{B}}){\dagger}}_{\mathbf{k},e_{Y},\eta,s}$

$\gamma^{(n_{\text{B}}){\dagger}}_{\mathbf{k},e_{Y},\eta,s}=\frac{\gamma^{\dagger}_{\mathbf{k},n_{\text{B}},\eta,s}+ie_{Y}\gamma^{\dagger}_{\mathbf{k},-n_{\text{B}},\eta,s}}{\sqrt{2}}\
,\quad(e_{Y}=\pm 1).$ (196)

For $n_{\text{B}}=1$, we call them the Chern band basis within the lowest two
bands
$\gamma^{(1){\dagger}}_{\mathbf{k},e_{Y},\eta,s}=\gamma^{\dagger}_{\mathbf{k},e_{Y},\eta,s}$,
as given in Eq. (196), where $e_{Y}=\pm 1$.

The basis $\gamma^{(n_{\text{B}}){\dagger}}_{\mathbf{k},e_{Y},\eta,s}$ defines
a band with well-defined Berry curvature; for fixed $e_{Y},\eta,s$, Eq.(196)
gives a band with Chern number

$C^{n_{\text{B}}}_{e_{Y},\eta,s}=e_{Y}e_{2,n_{\text{B}}}\ ,$ (197)

where $e_{2,n_{\text{B}}}\in\mathbb{Z}$ is the Wilson loop winding number of
the two bands $n=\pm n_{\text{B}}$, provided the pair of bands $n=\pm
n_{\text{B}}$ are disconnected with other bands. The wave functions of the
Chern basis in Eq. (196) are given by

$u_{e_{Y},n_{\text{B}},\eta}^{\prime}(\mathbf{k})=\frac{u_{+n_{\text{B}},\eta}(\mathbf{k})+ie_{Y}u_{-n_{\text{B}},\eta}(\mathbf{k})}{\sqrt{2}}\
.$ (198)

Due to the continuous gauge condition, we see that
$\lim_{\mathbf{q}\rightarrow\mathbf{0}}u^{\dagger}_{+n_{\text{B}},\eta}(\mathbf{k}+\mathbf{q})u_{+n_{\text{B}},\eta}(\mathbf{k})=\lim_{\mathbf{q}\rightarrow\mathbf{0}}u^{\dagger}_{-n_{\text{B}},\eta}(\mathbf{k}+\mathbf{q})u_{-n_{\text{B}},\eta}(\mathbf{k})$.
Therefore, the Chern band wave functions satisfy the continuous condition

$\lim_{\mathbf{q}\to
0}|u_{e_{Y},n_{\text{B}},\eta}^{\prime{\dagger}}(\mathbf{k}+\mathbf{q})u_{e_{Y}^{\prime},n_{\text{B}},\eta}^{\prime}(\mathbf{k})|=\frac{1}{2}\lim_{\mathbf{q}\to
0}|u^{\dagger}_{+n_{\text{B}},\eta}(\mathbf{k}+\mathbf{q})u_{+n_{\text{B}},\eta}(\mathbf{k})+e_{Y}e_{Y}^{\prime}u^{\dagger}_{-n_{\text{B}},\eta}(\mathbf{k}+\mathbf{q})u_{-n_{\text{B}},\eta}(\mathbf{k})|=\delta_{e_{Y},e_{Y}^{\prime}},$
(199)

This allows us to define a continuous Berry curvature for the Chern band wave
function $u_{e_{Y},n_{\text{B}},\eta}^{\prime}(\mathbf{k})$.

We first focus in the valley $\eta=+$. The sewing matrix for $C_{2z}T$
restricted in the valley $\eta=+$ is given by
$B^{C_{2z}T}(\mathbf{k})=\zeta^{0}$ (see Eq. (194)). Under this gauge,
according to [73], the non-abelian Berry’s connection
$[\mathbf{A}(\mathbf{k})]_{mn}=iu^{\dagger}_{m,+}(\mathbf{k})\partial_{\mathbf{k}}u_{n,+}(\mathbf{k})$
will take the form

$\mathbf{A}(\mathbf{k})=\begin{pmatrix}0&i\mathbf{a}(\mathbf{k})\\\
-i\mathbf{a}(\mathbf{k})&0\end{pmatrix}.$ (200)

The sign of wave functions $u_{n,+}(\mathbf{k})$ is fixed in such a way that
$\mathbf{a}(\mathbf{k})$ is globally continuous in the MBZ excluding the Dirac
nodes between the two bands $\pm n_{\text{B}}$ (recall that we assume the
bands $\pm n_{\text{B}}$ are disconnected from other bands, and thus there can
be Dirac nodes between them only if $n_{B}=1$), which is always possible [73].
In particular, this way of sign fixing is consistent with Eq. (193), since the
vanishing of the diagonal Berry’s connection requires $\lim_{\mathbf{q}\to
0}|u^{\dagger}_{m,\eta}(\mathbf{k}+\mathbf{q})u_{n,\eta}(\mathbf{k})|=\delta_{m,n}$.

It is known that the Wilson loop winding number of two bands isolated from
other bands is given by the Euler class [73]:

$e_{2,n_{\text{B}}}=\frac{1}{2\pi}\sum_{i}\oint_{\partial
D_{i}}d\mathbf{k}\cdot\mathbf{a}(\mathbf{k})=\frac{1}{2\pi}\int_{\mathrm{MBZ}-\sum_{i}D_{i}}d^{2}\mathbf{k}\
\Omega(\mathbf{k})\ ,$ (201)

where $D_{i}$ is a sufficiently small region containing the $i$th Dirac point
in the BZ, and
$\Omega(\mathbf{k})=\nabla_{\mathbf{k}}\times\mathbf{a}(\mathbf{k})$.

With Eq. (200), we can derive the Berry connection of the irrep band basis
$\gamma^{\dagger}_{\mathbf{k},e_{Y},+,s}$ at $\mathbf{k}$ away from Dirac
points as

$\displaystyle\mathbf{A}^{\prime}_{e_{Y}}(\mathbf{k})=iu^{\prime{\dagger}}_{e_{Y},n_{\text{B}},+}(\mathbf{k})\partial_{\mathbf{k}}u_{e_{Y},n_{\text{B}},+}^{\prime}(\mathbf{k})$
$\displaystyle=\frac{i}{2}[u^{\dagger}_{+n_{\text{B}},+}(\mathbf{k})\partial_{\mathbf{k}}u_{+n_{\text{B}},+}(\mathbf{k})+ie_{Y}u^{\dagger}_{+n_{\text{B}},+}(\mathbf{k})\partial_{\mathbf{k}}u_{-n_{\text{B}},+}(\mathbf{k})-ie_{Y}u^{\dagger}_{-n_{\text{B}},+}(\mathbf{k})\partial_{\mathbf{k}}u_{+n_{\text{B}},+}(\mathbf{k})+u^{\dagger}_{-n_{\text{B}},+}(\mathbf{k})\partial_{\mathbf{k}}u_{-n_{\text{B}},+}(\mathbf{k})]$
$\displaystyle=e_{Y}\mathbf{a}(\mathbf{k})\ .$ (202)

Furthermore, the Berry curvature can be shown to be non-divergent at the Dirac
points between the two bands $n=\pm n_{\text{B}}$ (see proof in [58]). If
$n_{\text{B}}>1$, there are no Dirac points between bands $n=\pm
n_{\text{B}}$). Therefore, by Eq. (201), we find the irrep band basis
$\gamma^{(n_{\text{B}})\dagger}_{\mathbf{k},e_{Y},+,s}$ carries a Chern number
given by Eq. (197). We also note that the $\hat{C}_{2z}$ symmetry maps the
irrep band basis $\gamma^{(n_{\text{B}})\dagger}_{\mathbf{k},e_{Y},+,s}$ into
$\gamma^{(n_{\text{B}})\dagger}_{-\mathbf{k},e_{Y},-,s}$ (see Eq. (195)).
Since $\hat{C}_{2z}$ does not change the Chern number, we conclude that the
Chern number of the Chern band basis
$\gamma^{(n_{\text{B}})\dagger}_{\mathbf{k},e_{Y},\eta,s}$ in the MBZ is
simply given by Eq. (197).

In particular, for the lowest two bands $n_{\text{B}}=1$, the bands are
topological and carry a winding number $e_{2}=1$ [71, 74, 73]. Therefore, for
the Chern band basis (the irrep basis with $n_{\text{B}}=1$)
$\gamma^{\dagger}_{-\mathbf{k},e_{Y},-,s}$, we have Chern number

$C_{e_{Y},\eta,s}=e_{Y}\ ,$ (203)

thus the name “Chern band basis” within the lowest two bands (see [58] for a
more careful treatment at the Dirac points at CNP, which does not change the
conclusion).

Now we show that if $e_{2,n_{B}}$ is odd, then the sign of the sewing matrix
$B^{P}(\mathbf{k})$ _must_ be $\mathbf{k}$-dependent: for $\eta=+$,
$B^{P}(\mathbf{k})$ can be chosen as $-i\zeta^{y}$ at all the momenta except
one or three of the $P$-invariant momenta, where $B^{P}(\mathbf{k})$ must be
$i\zeta^{y}$. To see this, we assume
$B^{P}(\mathbf{k})=-i\chi(\mathbf{k})\zeta^{y}$, where $\chi(\mathbf{k})=\pm
1$, and transform it into the Chern band basis (198). We obtain

$B^{P\prime}_{e_{Y},e_{Y}^{\prime}}(\mathbf{k})=u^{\prime\dagger}_{e_{Y},n_{B},+}(-\mathbf{k})D(P)u^{\prime}_{e_{Y},n_{B},+}(\mathbf{k})=-i\chi(\mathbf{k})e_{Y}\delta_{e_{Y},e_{Y}^{\prime}}.$
(204)

Therefore, $\hat{P}$ leaves each branch of the Chern band basis, which has the
Chern numbers $e_{2,n_{B}}e_{Y}$, invariant. $i\hat{P}$ can be equivalently
thought as an inversion symmetry for each Chern band since it squares to 1 and
changes $\mathbf{k}$ to $-\mathbf{k}$. The “inversion” eigenvalues of the
Chern band $e_{Y}$ are given by $\chi(\mathbf{k})e_{Y}$ for $\mathbf{k}$ being
the $P$-invariant momentum. Due to the relation between Chern number and
inversion eigenvalues, we have

$(-1)^{C^{n_{\text{B}}}_{e_{Y},\eta,s}}=\prod_{K}\chi(K),$ (205)

where $K$ indexes the four $P$-invariant momenta. Therefore, the right hand
side must be -1 (1) if $C^{n_{\text{B}}}_{e_{Y},\eta,s}$ is odd (even),
implying $\chi(K)=-1$ at one or three (zero, two, or four) of the four
$P$-invariant momenta. The sign of $B^{P}(\mathbf{k})$ in the other valley
$\eta=-$ can be obtained from the constraint of $B^{C_{2z}P}(\mathbf{k})$ and
$B^{P}(\mathbf{k})$.

####  VI.3.4 The chiral symmetry at $w_{0}=0$

At $w_{0}=0$, the single-particle Hamiltonian of TBG acquires an additional
unitary chiral symmetry $\hat{C}$, which satisfies the anti-commutation
relation with the full single-particle Hamiltonian $\hat{H}_{0}$ in Eq. (141)

$\\{\hat{C},\hat{H}_{0}\\}=0;\;\;\;\;\hat{C}c^{\dagger}_{\mathbf{k},\mathbf{Q},\eta,\alpha,s}\hat{C}^{-1}=\sum_{\mathbf{Q}^{\prime}\eta^{\prime}\beta}[D(\hat{C})]_{\mathbf{Q}^{\prime}\eta^{\prime}\beta,\mathbf{Q}\eta\alpha}c^{\dagger}_{\mathbf{k},\mathbf{Q}^{\prime},\eta^{\prime},\beta,s}\
,$ (206)

with the representation matrix

$\quad[D(\hat{C})]_{\mathbf{Q}^{\prime}\eta^{\prime}\beta,\mathbf{Q}\eta\alpha}=\delta_{\mathbf{Q}^{\prime},\mathbf{Q}}\delta_{\eta^{\prime},\eta}(\sigma_{z})_{\beta,\alpha}\
.$ (207)

Note that $\hat{C}$ preserves the electron momentum $\mathbf{k}$. Since
$\hat{C}$ flips the single-particle Hamiltonian $\hat{H}_{0}$, it is not a
commuting symmetry of TBG, but only reflects a relation between the positive
and negative energy spectra. The transformation $\hat{C}$ satisfies

$\hat{C}^{2}=1,\qquad\\{\hat{C},\hat{C}_{2z}\\}=0,\qquad[\hat{C},\hat{T}]=0,\qquad[\hat{C},\hat{P}]=0,\qquad\\{\hat{C},\hat{C}_{2z}T\\}=0,\qquad\\{\hat{C},\hat{C}_{2z}\hat{P}\\}=0.$
(208)

When transformed into the energy band basis, the chiral symmetry $\hat{C}$
implies

$\epsilon_{n,\eta}(\mathbf{k})=-\epsilon_{-n,\eta}(\mathbf{k}),\;\;\;[D(\hat{C})]_{\eta\eta^{\prime}}u_{n\eta^{\prime}}(\mathbf{k})=\sum_{m}[B^{C}(\mathbf{k})]_{m\eta,n\eta^{\prime}}u_{m\eta}(\mathbf{k})\
,$ (209)

where

$[B^{C}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=\delta_{\eta,\eta^{\prime}}\delta_{-m,n}e^{i\varphi^{C}_{n,\eta^{\prime}}(\mathbf{k})};\;\;\;\hat{C}\gamma_{\mathbf{k},n,\eta^{\prime},s}^{\dagger}\hat{C}^{-1}=\sum_{m\eta}[B^{C}(\mathbf{k})]_{m\eta,n\eta^{\prime}}\gamma_{\mathbf{k},m,\eta,s}^{\dagger}\
.$ (210)

By the relations $\\{\hat{C},\hat{C}_{2z}T\\}=\\{\hat{C},\hat{C}_{2z}P\\}=0$,
the sewing matrix of $C$ satisfies

$B^{C}(\mathbf{k})B^{C_{2z}T}(\mathbf{k})=-B^{C_{2z}T}(\mathbf{k})B^{C*}(\mathbf{k})\
,\qquad
B^{C}(\mathbf{k})B^{C_{2z}P}(\mathbf{k})=-B^{C_{2z}P}(\mathbf{k})B^{C}(\mathbf{k})\
.$ (211)

Under the gauge fixing of Eq. (194), we have
$B^{C_{2z}T}(\mathbf{k})=\zeta^{0}\tau^{0}$, and
$B^{C_{2z}P}(\mathbf{k})=-\zeta^{y}\tau^{y}$. The only
$\mathbf{k}$-independent gauge for sewing matrix of $C$ in consistency with
Eqs. (210) and (211) within each pair of bands $n=\pm n_{\text{B}}$ is then
(up to a global minus sign)

$B^{C}(\mathbf{k})=\zeta^{y}\tau^{z}\ .$ (212)

In particular, this $\mathbf{k}$-independent gauge fixing (212) of $C$
automatically ensures the continuous gauge fixing condition (193), which is
crucial for defining the irrep band basis in Eq. (196). To see this, note that
Eq. (212) tells us that $u_{-n,\eta}(\mathbf{k})=i\text{sgn}(n)\eta
u_{n,\eta}(\mathbf{k})$ for band $n=\pm n_{\text{B}}$, and thus we have

$f_{n,\eta}(\mathbf{k}+\mathbf{q},\mathbf{k})=\left|u^{\dagger}_{n,\eta}(\mathbf{k}+\mathbf{q})u_{n,\eta}(\mathbf{k})-u^{\dagger}_{-n,\eta}(\mathbf{k}+\mathbf{q})u_{-n,\eta}(\mathbf{k})\right|=\left|u^{\dagger}_{n,\eta}(\mathbf{k}+\mathbf{q})u_{n,\eta}(\mathbf{k})[1-\text{sgn}(n)^{2}\eta^{2}]\right|=0$
(213)

for any $\mathbf{k}$ and $\mathbf{q}$, satisfying Eq. (193).

We also note that this gauge fixing of $C$ is consistent with the gauge
fixings of both $C_{2z}$ and $P$ separately in Eq. (195). Basically, the
relations $\\{\hat{C},\hat{C}_{2z}\\}=0$ and $[\hat{C},\hat{P}]=0$ requires

$B^{C}(-\mathbf{k})B^{C_{2z}}(\mathbf{k})=-B^{C_{2z}}(\mathbf{k})B^{C}(\mathbf{k})\
,\qquad
B^{C}(-\mathbf{k})B^{P}(\mathbf{k})=B^{P}(\mathbf{k})B^{C}(\mathbf{k})\ ,$
(214)

which is satisfied by Eq. (212)

###  VI.4 Discrete Symmetry Analysis of the Intra-Layer Inter-valley Electron-
Phonon Interaction and the phonon-induced electron-electron interaction

We transform the phonon-mediated electron-electron interaction $H_{el-el}$
into the Chern band basis and the Hamiltonian is written as

$\displaystyle H_{el-
el}=-\frac{1}{N_{G}\omega_{A_{1}}}\sum_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},\mathbf{G}_{M},l,s,s_{1}}G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta
e_{Y}e_{Y}^{\prime}l}G_{\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{-\eta
e_{Y_{1}}e_{Y_{1}}^{\prime}l}\delta_{\mathbf{k}-\mathbf{k}^{\prime},-\mathbf{k}_{1}+\mathbf{k}_{1}^{\prime}}\gamma_{\mathbf{k}e_{Y}\eta
s}^{\dagger}\gamma_{\mathbf{k}_{1}e_{Y_{1}}-\eta
s_{1}}^{\dagger}\gamma_{\mathbf{k}_{1}^{\prime},e_{Y_{1}}^{\prime},\eta,s_{1}}\gamma_{\mathbf{k}^{\prime},e_{Y}^{\prime},-\eta,s}$

with

$G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}e_{Y}^{\prime}l}=\frac{\gamma_{3}}{\sqrt{2M\omega_{A_{1}}}}\sum_{\mathbf{Q}^{\prime}_{l\eta}}\sum_{\alpha\beta}u^{e_{Y}\star}_{\mathbf{k},\mathbf{Q}^{\prime}_{l\eta},\alpha,\eta}\sigma^{x}_{\alpha\beta}u^{e_{Y}^{\prime}}_{\mathbf{k}^{\prime},\mathbf{Q}^{\prime}_{l\eta}-\mathbf{Q}_{-l\eta},\beta,-\eta}.$
(216)

Here we consider the chiral limit with $w_{0}=0$ and will discuss the
constraint on the form of
$G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{l\eta}}^{\eta
e_{Y}e_{Y}^{\prime}l}$, as well as $V^{\eta
e_{Y}e_{Y}^{\prime},e_{Y1}e_{Y1}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}$,
from different discrete symmetries.

  * •

Chiral Symmetry $\hat{C}$:

$\displaystyle[D(\hat{C})]_{\eta\eta^{\prime}}u_{n\eta^{\prime}}(\mathbf{k})=\sum_{m}[B^{C}(\mathbf{k})]_{m\eta,n\eta^{\prime}}u_{m\eta}(\mathbf{k}),\;\;B^{C}(\mathbf{k})=\zeta^{y}\tau^{z},$
$\displaystyle[B^{C}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=-im\delta_{m,-n}\eta\delta_{\eta\eta^{\prime}},\;\;\quad[D(\hat{C})]_{\mathbf{Q}^{\prime}\eta^{\prime}\beta,\mathbf{Q}\eta\alpha}=\delta_{\mathbf{Q}^{\prime},\mathbf{Q}}\delta_{\eta^{\prime},\eta}(\sigma_{z})_{\beta,\alpha}=\delta_{\mathbf{Q}^{\prime},\mathbf{Q}}\delta_{\eta^{\prime},\eta}(-1)^{\alpha+1}\delta_{\beta,\alpha}$
(217)

which leads to

$(-1)^{\alpha+1}u_{\mathbf{k},\mathbf{Q},\alpha,n,\eta}=in\eta
u_{\mathbf{k},\mathbf{Q},\alpha,-n,\eta}\implies\boxed{u_{\mathbf{k},\mathbf{Q},\alpha,n,\eta}=-in\eta(-1)^{\alpha}u_{\mathbf{k},\mathbf{Q},\alpha,-n,\eta}}$
(218)

in the eigen-state basis. On the Chern band basis, we have

$\displaystyle
u_{\mathbf{k},\mathbf{Q},\alpha,e_{Y},\eta}=\frac{u_{\mathbf{k},\mathbf{Q},\alpha,+,\eta}+ie_{Y}u_{\mathbf{k},\mathbf{Q},\alpha,-,\eta}}{\sqrt{2}}=\frac{-i\eta(-1)^{\alpha}u_{\mathbf{k},\mathbf{Q},\alpha,-,\eta}+ie_{Y}(-i)(-1)\eta(-1)^{\alpha}u_{\mathbf{k},\mathbf{Q},\alpha,+,\eta}}{\sqrt{2}}$
$\displaystyle=\frac{-i\eta(-1)^{\alpha}u_{\mathbf{k},\mathbf{Q},\alpha,-,\eta}-e_{Y}\eta(-1)^{\alpha}u_{\mathbf{k},\mathbf{Q},\alpha,+,\eta}}{\sqrt{2}}=-e_{Y}\eta(-1)^{\alpha}\frac{ie_{Y}u_{\mathbf{k},\mathbf{Q},\alpha,-,\eta}+u_{\mathbf{k},\mathbf{Q},\alpha,+,\eta}}{\sqrt{2}}=-e_{Y}\eta(-1)^{\alpha}u_{\mathbf{k},\mathbf{Q},\alpha,e_{Y},\eta}.$
(219)

This shows that, depending on the valley $\eta$ and Chern number $e_{Y}$, the
Chern band basis is fully sub-lattice polarized (in the chiral limit) and that
the Chern number for $u_{\mathbf{k}\mathbf{Q}\alpha e_{Y}\eta}$ is
$e_{Y}=\eta(-1)^{\alpha+1}$. The sewing matrix in the Chern band basis can be
obtained from

$B^{g}_{e_{Y_{1}}\eta;e_{Y}\eta^{\prime}}=\sum_{\mathbf{Q},\alpha}u^{\star}_{\hat{g}\mathbf{k},\mathbf{Q},\alpha,e_{Y_{1}},\eta}\sum_{\mathbf{Q}^{\prime}\alpha^{\prime}}[D(\hat{g})]_{\mathbf{Q}\eta\alpha;\mathbf{Q}^{\prime}\eta^{\prime}\alpha^{\prime}}u_{\mathbf{k}\mathbf{Q}^{\prime}\alpha^{\prime}e_{Y}\eta^{\prime}}$
(220)

which leads to

$B^{C}_{e_{Y_{1}}\eta;e_{Y}\eta^{\prime}}=e_{Y}\eta\delta_{e_{Y_{1}},e_{Y}}\delta_{\eta\eta^{\prime}}.$
(221)

From the above constraint equation for the wave function, the constraint
equation for $G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{l\eta}}^{\eta
e_{Y}e_{Y}^{\prime}l}$ is given by

$\displaystyle G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}e_{Y}^{\prime}l}=e_{Y}e_{Y}^{\prime}G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}e_{Y}^{\prime}l}$ $\displaystyle\implies
G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}e_{Y}^{\prime}l}=\delta_{e_{Y},e_{Y}^{\prime}}G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}e_{Y}l}=:\delta_{e_{Y},e_{Y}^{\prime}}G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}l}$ (222)

We only keep one $e_{Y}$ index in the last equality. This greatly simplifies
the form of the electron-electron interaction, which becomes

$\displaystyle H_{el-
el}=-\frac{1}{N_{G}\omega_{A_{1}}}\sum_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},\mathbf{G}_{M},l,s,s_{1},e_{Y},e_{Y}^{\prime}}G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta
e_{Y}l}G_{\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{-\eta
e_{Y1}l}\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}\gamma_{\mathbf{k}_{1},e_{Y_{1}},-\eta
s_{1}}^{\dagger}\gamma_{\mathbf{k}_{1}^{\prime},e_{Y_{1}},\eta,s_{1}}\gamma_{\mathbf{k}^{\prime},e_{Y},-\eta,s}$
(223)

with

$G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}l}=\frac{\gamma_{3}}{\sqrt{2M\omega_{A_{1}}}}\sum_{\mathbf{Q}^{\prime}_{l\eta}}\sum_{\alpha\beta}u^{e_{Y}\star}_{\mathbf{k},\mathbf{Q}^{\prime}_{l\eta},\alpha,\eta}\sigma^{x}_{\alpha\beta}u^{e_{Y}}_{\mathbf{k}^{\prime},\mathbf{Q}^{\prime}_{l\eta}-\mathbf{Q}_{-l\eta},\beta,-\eta}.$
(224)

  * •

$\hat{C}_{2z}\hat{P}$:

$\displaystyle[D(C_{2z}P)]_{\eta\eta^{\prime}}u_{n\eta^{\prime}}(\mathbf{k})=\sum_{m}[B^{C_{2z}P}(\mathbf{k})]_{m\eta,n\eta^{\prime}}u_{m\eta}(\mathbf{k}),\;\;\;B^{C_{2z}P}(\mathbf{k})=-\zeta^{y}\tau^{y}$
$\displaystyle[B^{C_{2z}P}(\mathbf{k})]_{m\eta,n\eta^{\prime}}=m\eta\delta_{m,-n}\delta_{\eta,-\eta^{\prime}},\;[D(\hat{C}_{2z}\hat{P})]_{\mathbf{Q}\alpha\eta;\mathbf{Q}^{\prime}\beta\eta^{\prime}}=\delta_{\mathbf{Q},\mathbf{Q}^{\prime}}\zeta_{\mathbf{Q}^{\prime}}\sigma^{x}_{\alpha\beta}\delta_{\eta,-\eta^{\prime}}$
$\displaystyle\implies\delta_{\eta,-\eta^{\prime}}\zeta_{\mathbf{Q}}\sigma^{x}_{\alpha\beta}u_{\mathbf{k},\mathbf{Q},\beta,n\eta^{\prime}}=m\eta\delta_{m,-n}\delta_{\eta,-\eta^{\prime}}u_{\mathbf{k}\mathbf{Q}\alpha
m\eta}\implies$
$\displaystyle\boxed{u_{\mathbf{k},\mathbf{Q},\alpha,m\eta}=m\eta\zeta_{Q}u_{\mathbf{k},\mathbf{Q},\bar{\alpha},-m,-\eta}}$
(225)

On the Chern band basis, this gives

$\displaystyle
u_{\mathbf{k},\mathbf{Q},\alpha,e_{Y},\eta}=-ie_{Y}\eta\zeta_{Q}u_{\mathbf{k},\mathbf{Q},\bar{\alpha},e_{Y},-{\eta}}$
(226)

and thus

$B^{C_{2z}P}_{e_{Y_{1}}\eta;e_{Y}\eta^{\prime}}=-ie_{Y}\eta^{\prime}\delta_{e_{Y_{1}},e_{Y}}\delta_{\eta,-\eta^{\prime}}.$
(227)

$\hat{C}_{2z}\hat{P}$ can lead to the constraint

$\displaystyle G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}l}=G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{-\eta
e_{Y}-l}$ (228)

where we have used
$\zeta_{\mathbf{Q}^{\prime}_{l\eta}}=l\eta;\;\;\zeta_{\mathbf{Q}^{\prime}_{l\eta}-\mathbf{Q}_{-l\eta}}=\zeta_{\mathbf{Q}^{\prime\prime}_{-l\eta}}=-l\eta$.

  * •

$\hat{C}_{2z}\hat{T}$:

$u_{\mathbf{k},\mathbf{Q},\beta,n,\eta}=\sigma^{x}_{\beta\alpha}u_{\mathbf{k},\mathbf{Q},\alpha,n,\eta}^{\star}\implies
u_{\mathbf{k},\mathbf{Q},\alpha,e_{Y},\eta}=u_{\mathbf{k},\mathbf{Q},\bar{\alpha},-e_{Y}\eta}^{\star}$
(229)

and the sewing matrix is

$B^{C_{2z}T}_{e_{Y_{1}}\eta;e_{Y}\eta^{\prime}}=\delta_{e_{Y_{1}},-e_{Y}}\delta_{\eta\eta^{\prime}}$
(230)

This now gives the following transformation property

$G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta
e_{Y}l}=(G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta-
e_{Y}l})^{\star}.$ (231)

  * •

Particle-hole symmetry $\hat{P}$: we next deal with the particle-hole
symmetry, of which the gauge fixing cannot be done with a constant matrix over
the entire MBZ as discussed in Sec. VI.3.2 and VI.3.3. We write the sewing
matrix for $\hat{P}$ as

$B^{P}(\mathbf{k})=e^{i\phi_{\mathbf{k}}}i\zeta^{y}\tau^{z},$ (232)

where $e^{i\phi_{\Gamma_{M}}}=-1$, $e^{i\phi_{M_{M}}}=1$ and
$e^{i\phi_{\mathbf{k}}}=-1$ for a generic $\mathbf{k}$. With

$\sum_{\mathbf{Q}^{\prime}\alpha^{\prime}}[D(\hat{P})]_{\mathbf{Q}\eta\alpha;\mathbf{Q}^{\prime}\eta^{\prime}\alpha^{\prime}}u_{\mathbf{k}\mathbf{Q}^{\prime}\alpha^{\prime}n\eta^{\prime}}=\sum_{m}B^{P}_{m\eta;n\eta^{\prime}}u_{-\mathbf{k},\mathbf{Q},\alpha,m,\eta}$
(233)

and
$B^{P}_{m\eta;n\eta^{\prime}}=e^{i\phi_{\mathbf{k}}}m\eta\delta_{m,-n}\delta_{\eta,\eta^{\prime}}$,
we obtain

$\zeta_{\mathbf{Q}}u_{\mathbf{k},\mathbf{Q},\alpha
n\eta^{\prime}}=e^{i\phi_{\mathbf{k}}}m\eta\delta_{m,-n}\delta_{\eta,\eta^{\prime}}u_{-\mathbf{k},-\mathbf{Q},\alpha
m\eta}\implies
u_{\mathbf{k},\mathbf{Q},\alpha-m\eta}=e^{i\phi_{\mathbf{k}}}\zeta_{\mathbf{Q}}m\eta
u_{-\mathbf{k},-\mathbf{Q},\alpha m\eta}.$ (234)

on the eigen-state basis and

$\displaystyle
u_{\mathbf{k},\mathbf{Q},\alpha,e_{Y},\eta}=ie_{Y}e^{i\phi_{\mathbf{k}}}\zeta_{\mathbf{Q}}\eta
u_{-\mathbf{k},-\mathbf{Q},{\alpha},e_{Y},{\eta}}$ (235)

on the Chern band basis. The corresponding sewing matrix on the Chern band
basis reads

$B^{P}_{e_{Y_{1}}\eta;e_{Y}\eta^{\prime}}=ie_{Y}\eta
e^{i\phi_{\mathbf{k}}}\delta_{e_{Y_{1}},e_{Y}}\delta_{\eta\eta^{\prime}},$
(236)

which leads to the constraint

$\displaystyle
G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}=-l\eta\mathbf{q}_{2}+\mathbf{G}}^{\eta
e_{Y}l}=e^{i(\phi_{\mathbf{k}^{\prime}}-\phi_{\mathbf{k}})}G_{-\mathbf{k},-\mathbf{k}^{\prime},\mathbf{Q}_{l\eta}=l\eta\mathbf{q}_{2}-\mathbf{G}}^{\eta
e_{Y}-l}.$ (237)

  * •

$\hat{C}_{3z}$: The sewing matrix $B^{C_{3z}}(\mathbf{k})$ satisfies the
following conditions (we write the $\mathbf{k}$ dependence on the sewing
matrices of $\hat{C}_{2z}\hat{T}$ and $\hat{C}_{2z}P$ for completeness,
although we know these matrices have been chosen $\mathbf{k}$-independent):

$\displaystyle
Unitarity:\;\;\;B^{C_{3z}}(\mathbf{k})B^{C_{3z}\dagger}(\mathbf{k})=B^{C_{3z}\dagger}(\mathbf{k})B^{C_{3z}}(\mathbf{k})=\zeta_{0}\tau_{0}$
$\displaystyle\hat{C}_{3z}^{3}=1:\;\;\;B^{C_{3z}}(\hat{C}_{3}^{2}\mathbf{k})B^{C_{3z}}(\hat{C}_{3}\mathbf{k})B^{C_{3z}}(\mathbf{k})=\zeta_{0}\tau_{0}$
$\displaystyle\hat{C}_{2z}\hat{T}:\;\;\;B^{C_{3z}}(\mathbf{k})B^{C_{2z}T}(\mathbf{k})=B^{C_{2z}T}(\hat{C}_{3z}\mathbf{k})B^{C_{3z}\star}(\mathbf{k})$
$\displaystyle\hat{C}_{2z}\hat{P}:\;\;\;B^{C_{3z}}(\mathbf{k})B^{C_{2z}P}(\mathbf{k})=B^{C_{2z}P}(\hat{C}_{3z}\mathbf{k})B^{C_{3z}}(\mathbf{k})$
(238)

It is easiest to solve the constraint from $\hat{C}_{2z}\hat{T}$ first, since
its sewing matrix is identity. Hence, the matrix elements of
$B^{C_{3z}}(\mathbf{k})$ are all real. We also know $\hat{C}_{3z}$ does not
change the valley and hence

$B^{C_{3z}}(\mathbf{k})_{m\eta;n\eta^{\prime}}=A_{mn}^{\eta}\delta_{\eta,\eta^{\prime}}$
(239)

with the real matrix $A_{mn}^{\eta}$ to be determined. The unitarity condition
gives $AA^{T}=1$ and hence $A$ is a rotation matrix,

$[B^{C_{3z}}(\mathbf{k})]_{m\eta;n\eta^{\prime}}=\delta_{\eta\eta^{\prime}}\begin{pmatrix}\cos\theta^{\eta}(\mathbf{k})&-\sin\theta^{\eta}(\mathbf{k})\\\
\sin\theta^{\eta}(\mathbf{k})&\cos\theta^{\eta}(\mathbf{k})\end{pmatrix}$
(240)

Since
$[B^{C_{2z}P}(\mathbf{k})]_{m\eta;n\eta^{\prime}}=m\eta\delta_{m,-n}\delta_{\eta,-\eta^{\prime}}$,
the condition becomes

$\displaystyle A^{-\eta}_{r,-n}(-n)=rA^{\eta}_{-r,n}\implies
A^{-\eta}_{1,1}=A^{\eta}_{2,2},\;\;A^{-\eta}_{1,2}=-A^{\eta}_{2,1}$ (241)

where $n,r=\pm$ translates into $r=1,2$ in the matrix. This condition implies
the $C_{3z}$ sewing matrix is the same at either of the valleys, namely

$B^{C_{3z}}(\mathbf{k})_{m\eta;n\eta^{\prime}}=\delta_{\eta\eta^{\prime}}\begin{pmatrix}\cos\theta(\mathbf{k})&-\sin\theta(\mathbf{k})\\\
\sin\theta(\mathbf{k})&\cos\theta(\mathbf{k})\end{pmatrix}$ (242)

Finally $\hat{C}_{3z}^{3}=1$ implies

$\theta(\mathbf{k})+\theta(\hat{C}_{3z}\mathbf{k})+\theta(\hat{C}_{3z}^{2}\mathbf{k})=2\pi
n,\;\;\;n\in Z$ (243)

where $n$ is an integer. On the eigen-state basis, this reads

$e^{i\eta\frac{2\pi}{3}(-1)^{\alpha+1}}u_{\mathbf{k},\mathbf{Q},\alpha
n\eta}=\sum_{m=1,2}A_{mn}u_{\hat{C}_{3}\mathbf{k},\hat{C}_{3}\mathbf{Q},\alpha
m\eta}$ (244)

while on the Chern band basis, this then gives

$u_{\hat{C}_{3}\mathbf{k},\hat{C}_{3}\mathbf{Q},\alpha
e_{Y}\eta}=e^{ie_{Y}\theta(\mathbf{k})}e^{i\eta\frac{2\pi}{3}(-1)^{\alpha+1}}u_{\mathbf{k},\mathbf{Q},\alpha
e_{Y}\eta},$ (245)

from which we find the sewing matrix of $\hat{C}_{3z}$ on the Chern basis is
diagonal

$B^{C_{3z}}_{e_{Y_{1}}\eta,e_{Y}\eta^{\prime}}(\mathbf{k})=e^{-ie_{Y}\theta(\mathbf{k})}\delta_{e_{Y},e_{Y_{1}}}\delta_{\eta\eta^{\prime}}.$
(246)

This can be understood from the fact that in the chiral limit, there exists a
relation between the Chern number and the sublattice/valley polarization,
$e_{Y}=\eta(-1)^{\alpha+1}$.

We can see that $\theta(\mathbf{k})$ cannot be constant over the whole MBZ. At
the $\hat{C}_{3z}$-invariant points, i.e. $\mathbf{k}=\Gamma_{M}$,
$\mathbf{k}=\mathbf{K}_{M}$, or $\mathbf{k}=\mathbf{K}_{M}^{\prime}$,
$e^{-ie_{Y}\theta(\mathbf{k})}$ is the $\hat{C}_{3z}$-eigenvalue, and there
exists a relation between the Chern number and $\hat{C}_{3z}$-eigenvalues,

$e^{i\frac{2\pi}{3}C}=B(\Gamma_{M})B(\mathbf{K}_{M})B(\mathbf{K}_{M}^{\prime})=e^{-ie_{Y}(\theta(\Gamma_{M})+\theta(\mathbf{K}_{M})+\theta(\mathbf{K}_{M}^{\prime}))},$
(247)

in which $C$ is the Chern number and it should take $C=e_{Y}$ for the Chern
band basis. The $\theta$ value at $\Gamma_{M}$ can be obtained from the
relation (243),

$\theta(\Gamma_{M})+\theta(\Gamma_{M})+\theta(\Gamma_{M})=2\pi n,\;\;\;n\in
Z\implies\theta(\Gamma_{M})=2\pi/3\mod 2\pi.$ (248)

So if we assume $\theta$ is a constant over the whole MBZ, we will find
$e^{-ie_{Y}(\theta(\Gamma_{M})+\theta(\mathbf{K}_{M})+\theta(\mathbf{K}_{M}^{\prime}))}=1$,
which would mean zero Chern number (mod 3) and not $e_{Y}=1$ as it is. From
the TBG, we know the $\hat{C}_{3z}$-eigenvalues of the flat bands is $1$ at
$\Gamma$, and $\exp(\pm 2\pi i/3)$ at $\mathbf{K}_{M}$,
$\mathbf{K}_{M}^{\prime}$ (for the degenerate point). When we separate the
bands into Chern band basis, the $\hat{C}_{3z}$-eigenvalue at $\mathbf{K}_{M}$
of the Chern band and that at $\mathbf{K}_{M}^{\prime}$ have to be the same
(if they were opposite, and with eigenvalue $1$ at $\Gamma_{M}$, we would get
zero Chern number). We hence deduce

$\theta(\Gamma_{M})=0\mod 2\pi
n;\;\;\;\theta(\mathbf{K}_{M})=\frac{2\pi}{3}\mod
2\pi;\;\;\;\theta(\mathbf{K}_{M}^{\prime})=\frac{2\pi}{3}\mod 2\pi$ (249)

which gives

$e^{i\frac{2\pi}{3}C}=B(\Gamma_{M})B(K_{M})B(K_{M}^{\prime})=e^{-ie_{Y}(\theta(\Gamma_{M})+\theta(K_{M})+\theta(K_{M}^{\prime}))}=e^{i\frac{2\pi}{3}e_{Y}},$
(250)

leading to $C=e_{Y}$ as advertised.

The $\hat{C}_{3z}$ symmetry gives a constraint on the $G$ function,

$G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}l}=e^{ie_{Y}(\theta(\mathbf{k})-\theta(\mathbf{k}^{\prime}))}G_{\hat{C}_{3z}\mathbf{k},\hat{C}_{3z}\mathbf{k}^{\prime},\hat{C}_{3z}\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}l}$ (251)

  * •

$\hat{C}_{2x}$: The sewing matrix for $\hat{C}_{2x}$ takes the form

$B^{C_{2x}}_{m\eta;n\eta^{\prime}}(\mathbf{k})=A^{\eta}_{mn}\delta_{\eta\eta^{\prime}}$
(252)

and we need to again determine the matrix $A^{\eta}_{mn}$. For
$\hat{C}_{2z}\hat{T}$,

$\hat{C}_{2z}\hat{T}:\;\;\;B^{C_{2x}}(\mathbf{k})B^{C_{2z}T}(\mathbf{k})=B^{C_{2z}T}(\hat{C}_{2x}\mathbf{k})B^{C_{2x}\star}(\mathbf{k})\implies
A=A^{\star}$ (253)

and the unitarity condition requires $AA^{T}=1$. Furthermore, for the chiral
symmetry $\hat{C}$,

$\displaystyle\\{C,C_{2x}\\}=0:\;\;B^{C_{2x}}(\mathbf{k})B^{C}(\mathbf{k})=-B^{C}(\hat{C}_{2x}\mathbf{k})B^{C_{2x}}(\mathbf{k})\implies
A^{\eta}(\mathbf{k})=\cos[\alpha^{\eta}(\mathbf{k})]\zeta_{z}+\sin[\alpha^{\eta}(\mathbf{k})]\zeta_{x}$
(254)

with $\zeta$ being in $n=1,2$ band space and $\alpha^{\eta}$ being an angle
(not to be confused for the orbital $\alpha$). For $\hat{C}_{2z}\hat{P}$,

$\displaystyle\\{\hat{C}_{2z}\hat{P},\hat{C}_{2x}\\}=0:\;\;B^{C_{2x}}(\mathbf{k})B^{C_{2z}P}(\mathbf{k})=-B^{C_{2z}P}(\hat{C}_{2x}\mathbf{k})B^{C_{2x}}(\mathbf{k}),$
$\displaystyle
A^{-\eta}_{-m,-n}=-mnA^{\eta}_{m,n}\implies\alpha^{\eta}(\mathbf{k})=\alpha^{-\eta}(\mathbf{k})=\alpha(\mathbf{k}).$
(255)

For the $\hat{C}_{3z}$ rotation,

$\displaystyle
B^{C_{2x}}(\hat{C}_{3z}^{2}\mathbf{k})B^{C_{3z}}(\hat{C}_{3z}\mathbf{k})B^{C_{3z}}(\mathbf{k})=B^{C_{3z}}(\hat{C}_{2x}\mathbf{k})B^{C_{2x}}(\mathbf{k})\implies$
$\displaystyle\alpha_{\hat{C}_{3z}^{2}\mathbf{k}}-\alpha_{\mathbf{k}}-\theta_{\hat{C}_{3z}\mathbf{k}}-\theta_{\mathbf{k}}-\theta_{\hat{C}_{2x}\mathbf{k}}=0\mod
2\pi$ (256)

Putting all the above together, one can find the transformation for the Chern
band basis

$u_{\mathbf{k},\mathbf{Q},\alpha
e_{Y}\eta}=e^{ie_{Y}\alpha(\mathbf{k})}u_{\hat{C}_{2x}\mathbf{k},\hat{C}_{2x}\mathbf{Q},\bar{\alpha},-e_{Y},\eta}$
(257)

from which we see that

$\alpha(\mathbf{k})-\alpha(\hat{C}_{2x}\mathbf{k})=0\mod 2\pi.$ (258)

The sewing matrix is then

$B^{C_{2x}}_{e_{Y_{1}}\eta;e_{Y}\eta^{\prime}}=\delta_{e_{Y_{1}},-e_{Y}}e^{ie_{Y}\alpha(\mathbf{k})}\delta_{\eta\eta^{\prime}}.$
(259)

$\hat{C}_{2x}$ imposes the following constraint

$\displaystyle
G_{\mathbf{k},\mathbf{k}^{\prime},Q_{-l\eta}=-l\eta\mathbf{q}_{2}+\mathbf{G}}^{\eta,e_{Y},l}=e^{ie_{Y}(\alpha(\mathbf{k}^{\prime})-\alpha(\mathbf{k}))}G_{\hat{C}_{2x}\mathbf{k},\hat{C}_{2x}\mathbf{k}^{\prime},\hat{C}_{2x}\mathbf{Q}_{-l\eta}=\hat{C}_{2x}(-l\eta\mathbf{q}_{2}+\mathbf{G})}^{\eta,-e_{Y},-l}$
(260)

  * •

Hermiticity: From the definition of
$G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta
nn^{\prime}l}$, one can obtain the identity

$G_{\mathbf{k}^{\prime},\mathbf{k},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta,n^{\prime}nl\star}=G_{\mathbf{k},\mathbf{k}^{\prime},l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{-\eta,nn^{\prime}l}$
(261)

which is required for the hermiticity condition $H_{el-el}^{\dagger}=H_{el-
el}$.

We now define the $F$-function

$F^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=\frac{1}{N_{0}\omega_{A_{1}}}\sum_{\mathbf{G}_{M},l}G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta,e_{Y},l}G_{\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{-\eta,e_{Y}^{\prime},l}$
(262)

so that the phonon-mediated electron-electron interaction takes the form

$H_{el-
el}=-\frac{1}{N_{M}}\sum_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},s,s_{1},e_{Y},e_{Y}^{\prime}}F^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}\gamma_{\mathbf{k}_{1},e_{Y_{1}},-\eta,s_{1}}^{\dagger}\gamma_{\mathbf{k}_{1}^{\prime},e_{Y_{1}},\eta,s_{1}}\gamma_{\mathbf{k}^{\prime},e_{Y},-\eta,s}$
(263)

We need to keep in mind that this term actually came from neglecting the one-
body term in the phonon-mediated electron-electron interaction, which is

$H_{el-
el}=-\frac{1}{N_{M}}\sum_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},s,s_{1},e_{Y},e_{Y}^{\prime}}F^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}\gamma_{\mathbf{k}^{\prime},e_{Y},-\eta,s}\gamma_{\mathbf{k}_{1},e_{Y_{1}},-\eta,s_{1}}^{\dagger}\gamma_{\mathbf{k}_{1}^{\prime},e_{Y_{1}},\eta,s_{1}}$
(264)

We can apply all the symmetry operators to derive the constraint equations for
the $F$-function, which are summarized as follows.

$\displaystyle\hat{C}_{2z}\hat{P}:\boxed{F^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=F^{-\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}}$
$\displaystyle\hat{C}_{2z}\hat{T}:\boxed{F^{\eta
e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=F^{\eta,-e_{Y},-e_{Y}^{\prime}\star}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}}$
$\displaystyle
Hermiticiy:\boxed{F^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=F^{\eta,e_{Y}^{\prime},e_{Y}\star}_{\mathbf{k}_{1}^{\prime},\mathbf{k}_{1};\mathbf{k}^{\prime},\mathbf{k}}}$
$\displaystyle
Reshuffling:\boxed{F^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=F^{-\eta,e_{Y}^{\prime},e_{Y}}_{\mathbf{k}_{1},\mathbf{k}_{1}^{\prime};\mathbf{k},\mathbf{k}^{\prime}}}$
$\displaystyle\hat{P}:\boxed{F^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=e^{i(\phi_{\mathbf{k}^{\prime}}-\phi_{\mathbf{k}}+\phi_{\mathbf{k}_{1}^{\prime}}-\phi_{\mathbf{k}_{1}})}F^{\eta,e_{Y},e_{Y}^{\prime}}_{-\mathbf{k},-\mathbf{k}^{\prime};-\mathbf{k}_{1},-\mathbf{k}_{1}^{\prime}}}$
$\displaystyle\hat{C}_{3z}:\;\;\boxed{F^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=e^{ie_{Y}(\theta(\mathbf{k})-\theta(\mathbf{k}^{\prime}))+ie_{Y}^{\prime}(\theta(\mathbf{k}_{1})-\theta(\mathbf{k}_{1}^{\prime}))}F^{\eta,e_{Y},e_{Y}^{\prime}}_{\hat{C}_{3z}\mathbf{k},\hat{C}_{3z}\mathbf{k}^{\prime};\hat{C}_{3z}\mathbf{k}_{1},\hat{C}_{3z}\mathbf{k}_{1}^{\prime}}}$
$\displaystyle\hat{C}_{2x}:\;\;\boxed{F^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=e^{-ie_{Y}(\alpha(\mathbf{k})-\alpha(\mathbf{k}^{\prime}))-ie_{Y}^{\prime}(\alpha(\mathbf{k}_{1})-\alpha(\mathbf{k}_{1}^{\prime}))}F^{\eta,-e_{Y},-e_{Y}^{\prime}}_{\hat{C}_{2x}\mathbf{k},\hat{C}_{2x}\mathbf{k}^{\prime};\hat{C}_{2x}\mathbf{k}_{1},\hat{C}_{2x}\mathbf{k}_{1}^{\prime}}}$
(265)

Since we have already derived all the symmetry constraint for the
$G$-function, all the symmetry constraint on the $F$-function can be derived
from the definition of $F$-function (Eq. 262). Here we only show an example of
reshuffling symmetry.

$\displaystyle F^{\eta
e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}=\sum_{\mathbf{G}_{M}}\sum_{l}G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta
e_{Y}l}G_{\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{-\eta
e_{Y}^{\prime}l}$ $\displaystyle F^{-\eta
e_{Y}^{\prime},e_{Y}}_{\mathbf{k}_{1}\mathbf{k}_{1}^{\prime};\mathbf{k},\mathbf{k}^{\prime}}=\sum_{\mathbf{G}_{M}}\sum_{l}G_{\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{-\eta
e_{Y}^{\prime}l}G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{\eta
e_{Y}l}=F^{\eta
e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}.$
(266)

The above symmetry relations relate different components of the $F$-function
and let’s next count the number of independent parameters. Out of $2\times
16=32$ complex numbers for each set of
$\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}$
momenta, we are left with $2$ complex numbers
$F^{+++}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}$,
$F^{++-}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}$
out of which we can obtain the entire set of values.

Next we consider the phonon-mediated electron-electron interaction in the
Cooper channel (similar to Eq. 172 but in the Chern band basis),

$\displaystyle H_{el-
el}=-\frac{1}{N_{M}}\sum_{\mathbf{k},\mathbf{k}^{\prime},s,s_{1},e_{Y},e_{Y}^{\prime}}V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}\gamma_{\mathbf{k}e_{Y}\eta
s}^{\dagger}\gamma_{-\mathbf{k}e_{Y_{1}}-\eta
s_{1}}^{\dagger}\gamma_{-\mathbf{k}^{\prime},e_{Y_{1}},\eta,s_{1}}\gamma_{\mathbf{k}^{\prime}e_{Y},-\eta,s}$
(267)

with

$\displaystyle
V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=F^{\eta
e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};-\mathbf{k},-\mathbf{k}^{\prime}}.$
(268)

From the above constraint equations for the $F$-function, we can easily
extract the corresponding relations for
$V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}$, which are
listed as follows.

$\displaystyle\hat{C}_{2z}\hat{P}:\boxed{V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=V^{-\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}}$
$\displaystyle\hat{C}_{2z}\hat{T}:\boxed{V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=V^{\eta,-e_{Y},-e_{Y}^{\prime}\star}_{\mathbf{k},\mathbf{k}^{\prime}}}$
$\displaystyle
Hermiticiy:\boxed{V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=V^{\eta,e_{Y}^{\prime},e_{Y}\star}_{-\mathbf{k}^{\prime},-\mathbf{k}}}$
$\displaystyle
Reshuffling:\boxed{V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=V^{-\eta,e_{Y}^{\prime},e_{Y}}_{-\mathbf{k},-\mathbf{k}^{\prime}}}$
$\displaystyle\hat{P}:\boxed{V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=V^{\eta,e_{Y},e_{Y}^{\prime}}_{-\mathbf{k},-\mathbf{k}^{\prime}}}$
$\displaystyle\hat{C}_{3z}:\;\;\boxed{V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=e^{ie_{Y}(\theta(\mathbf{k})-\theta(\mathbf{k}^{\prime}))+ie_{Y}^{\prime}(\theta(-\mathbf{k})-\theta(-\mathbf{k}^{\prime}))}V^{\eta,e_{Y},e_{Y}^{\prime}}_{\hat{C}_{3z}\mathbf{k},\hat{C}_{3z}\mathbf{k}^{\prime}}}$
$\displaystyle\hat{C}_{2x}:\;\;\boxed{V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=e^{-ie_{Y}(\alpha(\mathbf{k})-\alpha(\mathbf{k}^{\prime}))-ie_{Y}^{\prime}(\alpha(-\mathbf{k})-\alpha(-\mathbf{k}^{\prime}))}V^{\eta,-e_{Y},-e_{Y}^{\prime}}_{\hat{C}_{2x}\mathbf{k},\hat{C}_{2x}\mathbf{k}^{\prime}}}$
(269)

We now have the constraint equations for $\hat{C}_{2z}\hat{P}$ and
$\hat{C}_{2z}\hat{T}$ for the $V$ potential at the same momentum. In addition,
the combination of reshuffling and particle-hole symmetry also give one more
constraint equation that does not change the momentum,

$\displaystyle\text{Reshuffling and
}\hat{P}:\boxed{V^{\eta,e_{Y}e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=V^{-\eta,e_{Y}^{\prime}e_{Y}}_{\mathbf{k},\mathbf{k}^{\prime}}}.$
(270)

Putting the above results together, we can explicitly show the following
constraints for different components of the $V$ potential at the same momentum

$\displaystyle
V^{+++}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{-++}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{+--\star}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{---\star}_{\mathbf{k}\mathbf{k}^{\prime}}$
$\displaystyle
V^{++-}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{--+}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{+-+}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{-+-}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{+-+\star}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{-+-\star}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{++-\star}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{--+\star}_{\mathbf{k}\mathbf{k}^{\prime}}$
(271)

We notice that there are more constraints for the case of $e_{Y}\neq
e^{\prime}_{Y}$, provided by the combination of Reshuffling and $P$. This
extra constraint is trivial for the case of $e_{Y}=e^{\prime}_{Y}$ since the
constraint from Reshuffling and $P$ is the identical to that from $C_{2z}P$.
Hence fundamentally, in the chiral limit, the phonon-mediated electron-
electron interaction at every $\mathbf{k},\mathbf{k}^{\prime}$ depends only on
one complex parameter $V^{+++}_{\mathbf{k}\mathbf{k}^{\prime}}$ and one _real_
parameter $V^{++-}_{\mathbf{k}\mathbf{k}^{\prime}}$.

For the above symmetry constraint results, we have two more comments, as
discussed below.

  * •

While we see the change of $e^{i\phi_{\mathbf{k}}}$ in the sewing matrix of
$\hat{P}$ between $k=\Gamma_{M}$ and $k=M_{M}$, this does not help to
determine the sign changes of the $V$ potential. The reason is that, for the
Cooper channel
($\mathbf{k}_{1}=-\mathbf{k},\mathbf{k}_{1}^{\prime}=-\mathbf{k}^{\prime}$),
we have

$F^{\eta
e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};-\mathbf{k},-\mathbf{k}^{\prime}}=e^{i(\phi_{\mathbf{k}^{\prime}}-\phi_{\mathbf{k}}+\phi_{-\mathbf{k}^{\prime}}-\phi_{-\mathbf{k}})}F^{\eta
e_{Y},e_{Y}^{\prime}}_{-\mathbf{k},-\mathbf{k}^{\prime};\mathbf{k},\mathbf{k}^{\prime}}$
(272)

We can also find a relation between $\phi_{\mathbf{k}}$ and
$\phi_{-\mathbf{k}}$

$\displaystyle
u_{\mathbf{k},\mathbf{Q},\alpha,e_{Y},\eta}=ie_{Y}e^{i\phi_{\mathbf{k}}}\zeta_{\mathbf{Q}}\eta
u_{-\mathbf{k},-\mathbf{Q},\bar{\alpha},e_{Y},{\eta}}\implies-
ie_{Y}e^{-i\phi_{\mathbf{k}}}\eta=\sum_{\mathbf{Q},\alpha}u_{\mathbf{k},\mathbf{Q},\alpha,e_{Y},\eta}^{\star}\zeta_{\mathbf{Q}}u_{-\mathbf{k},-\mathbf{Q},\bar{\alpha},e_{Y},{\eta}}\implies$
$\displaystyle-
ie_{Y}e^{-i\phi_{-\mathbf{k}}}\eta=-ie_{Y}e^{i\phi_{\mathbf{k}}}\eta\implies
e^{i(\phi_{\mathbf{k}}+\phi_{-\mathbf{k}})}=1$ (273)

Thus, we have

$F^{\eta
e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime};-\mathbf{k},-\mathbf{k}^{\prime}}=F^{\eta
e_{Y},e_{Y}^{\prime}}_{-\mathbf{k},-\mathbf{k}^{\prime};\mathbf{k},\mathbf{k}^{\prime}},$
(274)

which corresponds to the constraint equation for $V$ potential from $\hat{P}$.
Hence, the derivation here show that no sign change occurs because of the
$\hat{P}$ sewing matrix.

  * •

We now analyze the constraint from $\hat{C}_{3z}$. First we put
$\mathbf{k}^{\prime}=0$ and we have

$V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},0}=e^{ie_{Y}(\theta(\mathbf{k})-2\theta(0))+ie_{Y}^{\prime}\theta(-\mathbf{k})}V^{\eta,e_{Y},e_{Y}^{\prime}}_{\hat{C}_{3}\mathbf{k},0}$
(275)

Now notice that if we put $\mathbf{k}=\mathbf{K}_{M}$ then
$-\mathbf{k}=\mathbf{K}_{M}^{\prime}$ but
$\theta(\mathbf{K}_{M})=\theta(\mathbf{K}_{M}^{\prime})=2\pi/3\mod 2\pi$, and
we have ($\theta_{0}=0\mod 2\pi$)

$V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{K}_{M},0}=e^{i(e_{Y}+e_{Y}^{\prime})\frac{2\pi}{3}}V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{K}_{M},0}$
(276)

(one notes that
$V_{\mathbf{k}+\mathbf{G},\mathbf{k}^{\prime}}=V_{\mathbf{k},\mathbf{k}^{\prime}+\mathbf{G}}=V_{\mathbf{k},\mathbf{k}^{\prime}}$).
Hence, there are nodes for the channel with $e_{Y}=e_{Y}^{\prime}$, but not
for $e_{Y}=-e_{Y}^{\prime}$, as shown below.

$e_{Y}=e_{Y}^{\prime}:\;\;V^{\eta,e_{Y},e_{Y}}_{\mathbf{K}_{M},0}=e^{ie_{Y}\frac{4\pi}{3}}V^{\eta,e_{Y},e_{Y}}_{\mathbf{K}_{M},0}\implies
V^{\eta,e_{Y},e_{Y}}_{\mathbf{K}_{M},0}=0.$ (277)

###  VI.5 Continuous Symmetry Analysis of the Intra-layer and Inter-valley
Electron-Phonon Interaction and the phonon-induced electron-electron
interaction

It is known that the projected Coulomb interaction Hamiltonian into the flat
bands, in the chiral limit, enjoys a large $U(4)\times U(4)$ symmetry with
generators

$S^{ab}_{e_{Y}}=\sum_{\mathbf{k}\eta
s\eta^{\prime}s^{\prime}}\gamma^{\dagger}_{\mathbf{k}e_{Y}\eta
s}\tau^{a}_{\eta\eta^{\prime}}s^{b}_{ss^{\prime}}\gamma_{\mathbf{k}e_{Y}\eta^{\prime}s^{\prime}},\;\;\;\
e_{Y}=\pm 1,\;a,b=0,1,2,3$ (278)

However, this enlarged symmetry will be broken by the electron-phonon
coupling. To see that, we calculate $[H_{el-el},S^{ab}_{e_{Y_{2}}}]$ and with
some tedious calculation, we find

$\displaystyle[H_{el-el},S^{ab}_{e_{Y_{2}}}]=2F^{\eta
e_{Y},e_{Y_{2}}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}\gamma_{\mathbf{k}e_{Y}\eta
s}^{\dagger}\gamma_{\mathbf{k}^{\prime}e_{Y}-\eta
s}(\gamma_{\mathbf{k}_{1}e_{Y_{2}}-\eta
s_{2}}^{\dagger}\tau^{a}_{\eta,\eta_{2}}s^{b}_{s_{2}s_{1}}\gamma_{\mathbf{k}_{1}^{\prime}e_{Y_{2}}\eta_{2}s_{1}}-\gamma_{\mathbf{k}_{1}e_{Y_{2}}\eta_{2}s_{2}}^{\dagger}\tau^{a}_{\eta_{2},-\eta}s^{b}_{s_{2}s_{1}}\gamma_{\mathbf{k}_{1}^{\prime}e_{Y_{2}}\eta
s_{1}})+$ $\displaystyle+\text{one-body terms},$ (279)

from which one can see that the two body terms only vanish for

$a=0,\;\;\forall b\implies U_{s}(2)_{e_{Y}=1}\times U_{s}(2)_{e_{Y}=-1}$ (280)

where $s$ means spin. The electron-phonon coupling breaks the $U(4)\times
U(4)$ symmetry in the chiral flat-band limit to a $U_{s}(2)_{e_{Y}=1}\times
U_{s}(2)_{e_{Y}=-1}$ spin $U(2)$ symmetry in each Chern sector, with the
generators

$S^{b}_{e_{Y}}=\sum_{\mathbf{k}\eta s}\gamma^{\dagger}_{\mathbf{k}e_{Y}\eta
s}s^{b}_{ss^{\prime}}\gamma_{\mathbf{k}e_{Y}\eta s^{\prime}},\;\;\;\ e_{Y}=\pm
1,\;b=0,1,2,3$ (281)

This symmetry is broken away from the chiral flat-band limit, leaving behind
just the total $U(2)$ spin symmetry.

We also have $a=3,b=0$ (Valley $U(1)$ charge _per Chern sector_ ) for which
$\tau^{a}_{\eta\eta^{\prime}}=\eta\delta_{\eta\eta^{\prime}}$. In this case,
one can show

$\displaystyle[H_{el-
el},S^{30}_{e_{Y_{2}}}]=-\frac{2}{N_{M}}\sum_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{k}_{1},\mathbf{k}_{1}^{\prime},s,s_{1},e_{Y}}\eta$
$\displaystyle[F^{\eta
e_{Y},e_{Y_{2}}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}\gamma_{\mathbf{k},e_{Y},\eta
s}^{\dagger}\gamma_{\mathbf{k}^{\prime},e_{Y},-\eta
s}\gamma_{\mathbf{k}_{1},e_{Y_{2}},-\eta,s_{1}}^{\dagger}\gamma_{\mathbf{k}_{1}^{\prime},e_{Y_{2}},\eta,s_{1}}-F^{\eta
e_{Y_{2}},e_{Y}}_{\mathbf{k},\mathbf{k}^{\prime};\mathbf{k}_{1},\mathbf{k}_{1}^{\prime}}\gamma_{\mathbf{k},e_{Y_{2}},\eta
s}^{\dagger}\gamma_{\mathbf{k}^{\prime},e_{Y_{2}},-\eta,s}\gamma_{\mathbf{k}_{1},e_{Y},-\eta,s_{1}}^{\dagger}\gamma_{\mathbf{k}_{1}^{\prime},e_{Y},\eta,s_{1}}.$
(282)

And we can see that Valley $U(1)$ charge per Chern band sector is _not_
conserved. However, the _total (summed over both chern sectors)_ $U(1)$ Valley
charge

$Q_{v}=\sum_{\mathbf{k}e_{Y}\eta s}\gamma^{\dagger}_{\mathbf{k}e_{Y}\eta
s}\tau^{z}_{\eta\eta^{\prime}}\gamma_{\mathbf{k}e_{Y}\eta s}$ (283)

is a good quantum number; it remains so away from the chiral flat-band limit.

###  VI.6 Numerical results for the phonon-induced electron-electron
interaction

In this section, we will discuss our numerical results for the electron-phonon
interaction and the phonon-induced electron-electron interaction in TBG and
compare the results with the symmetry analysis of $V^{\eta
e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}$ in the previous section.

Our numerical calculation is based on the expression (137) for electron-phonon
coupling $g^{s}_{\alpha\beta}(\mathbf{p},\mathbf{p}^{\prime})$ in one single
layer and then project it into the flat band basis in TBG as

$\displaystyle G_{\mathbf{k},\mathbf{k}^{\prime},Q_{-l\eta}}^{\eta
nn^{\prime}ls}=\sum_{\mathbf{Q}^{\prime}_{l\eta},\alpha\beta}u^{n\star}_{\mathbf{k},\mathbf{Q}^{\prime}_{l\eta},\alpha,\eta}g^{s,l}_{\alpha\beta}(\mathbf{p},\mathbf{p}^{\prime})u^{n^{\prime}}_{\mathbf{k}^{\prime},\mathbf{Q}^{\prime}_{l\eta}-\mathbf{Q}_{-l\eta},\beta,-\eta},$
(284)

with $\mathbf{p}=\mathbf{k}+\eta\mathbf{K}_{D}-\mathbf{Q}^{\prime}_{l\eta}$
and
$\mathbf{p}^{\prime}=\mathbf{k}^{\prime}-\eta\mathbf{K}_{D}-\mathbf{Q}^{\prime}_{l\eta}+\mathbf{Q}_{-l\eta}$
and $\mathbf{k},\mathbf{k}^{\prime}$ is chosen within the Moiré Brillouin
zone. The corresponding inter-valley electron-phonon interaction is then given
by

$\displaystyle H_{inter-
vall}=\frac{1}{\sqrt{N_{G}}}\sum_{\mathbf{k},\mathbf{k}^{\prime}\in
MBZ}\sum_{n,n^{\prime}=1,2}\sum_{\eta,s}\sum_{l=\pm}\sum_{Q_{-l\eta}}G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
nn^{\prime}ls}\gamma_{\mathbf{k},n,\eta,s}^{\dagger}\gamma_{\mathbf{k}^{\prime},n^{\prime},-\eta,s}$
$\displaystyle(b_{-\eta\mathbf{K}_{D}+\mathbf{k}-\mathbf{k}^{\prime}-\mathbf{Q}_{-l\eta},l,A_{1}}+b_{\eta\mathbf{K}_{D}-\mathbf{k}+\mathbf{k}^{\prime}+\mathbf{Q}_{-l\eta},l,A_{1}}^{\dagger})$
(285)

Since the Moiré Hamiltonian is expanded around $\mathbf{K}_{D}$, namely
$|\mathbf{K}_{D}|\gg k,k^{\prime},|\mathbf{Q}^{\prime}_{l\eta}|$ and
$g^{s}_{\alpha\beta}(\mathbf{p},\mathbf{p}^{\prime})$ is a smooth function of
$\mathbf{p}$ and $\mathbf{p}^{\prime}$, we can approximate
$g^{s}_{\alpha\beta}(\mathbf{p},\mathbf{p}^{\prime})$ by
$g^{s}_{\alpha\beta}(\eta\mathbf{K}_{D},-\eta\mathbf{K}_{D})$ in our numerical
calculations. We test this approximation by directly evaluating
$g^{s}_{\alpha\beta}(\mathbf{p},\mathbf{p}^{\prime})$ and find that the
derivation of $g^{s}_{\alpha\beta}(\mathbf{p},\mathbf{p}^{\prime})$ from
$g^{s}_{\alpha\beta}(\eta\mathbf{K}_{D},-\eta\mathbf{K}_{D})$ is only within
5% at the relevant momentum scale of Moiré reciprocal lattice vectors. For
$g^{s}_{\alpha\beta}(\eta\mathbf{K}_{D},-\eta\mathbf{K}_{D})$, only the phonon
mode $s=A1$ give non-zero contribution, so we only keep $s=A1$ below and drop
the s-index in calculating coupling strength

$\displaystyle G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
nn^{\prime}l}=\sum_{\mathbf{Q}^{\prime}_{l\eta},\alpha\beta}u^{n\star}_{\mathbf{k},\mathbf{Q}^{\prime}_{l\eta},\alpha,\eta}g^{l}_{\alpha\beta}(\eta\mathbf{K}_{D},-\eta\mathbf{K}_{D})u^{n^{\prime}}_{\mathbf{k}^{\prime},\mathbf{Q}^{\prime}_{l\eta}-\mathbf{Q}_{-l\eta},\beta,-\eta}.$
(286)

A similar expression of electron-phonon coupling parameter for the Chern band
basis is given by

$\displaystyle G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}e_{Y}^{\prime}l}=\sum_{\mathbf{Q}^{\prime}_{l\eta},\alpha\beta}u^{\star}_{\mathbf{k},\mathbf{Q}^{\prime}_{l\eta},\alpha,e_{Y},\eta}g^{l}_{\alpha\beta}(\eta\mathbf{K}_{D},-\eta\mathbf{K}_{D})u_{\mathbf{k}^{\prime},\mathbf{Q}^{\prime}_{l\eta}-\mathbf{Q}_{-l\eta},\beta,e_{Y}^{\prime},-\eta}.$
(287)

The corresponding phonon-induced electron-electron interaction in the Cooper
pair channels is given by

$H_{el-
el}=-\frac{1}{N_{M}}\sum_{\mathbf{k},\mathbf{k}^{\prime},s_{1},s_{2},e_{Y},e_{Y}^{\prime}}V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}\gamma_{\mathbf{k},e_{Y},\eta,s_{1}}^{\dagger}\gamma_{-\mathbf{k},e_{Y}^{\prime},-\eta,s_{2}}^{\dagger}\gamma_{-\mathbf{k}^{\prime},e_{Y}^{\prime},\eta,s_{2}}\gamma_{\mathbf{k}^{\prime},e_{Y},-\eta,s_{1}}$
(288)

where

$V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=\frac{1}{N_{0}}\frac{1}{\omega_{A_{1}}}\sum_{\mathbf{G}_{M}}\sum_{l}G_{\mathbf{k},\mathbf{k}^{\prime},-l\eta\mathbf{q}_{2}+\mathbf{G}_{M}}^{\eta
e_{Y}l}G_{-\mathbf{k},-\mathbf{k}^{\prime},l\eta\mathbf{q}_{2}-\mathbf{G}_{M}}^{-\eta
e_{Y}^{\prime}l}.$ (289)

Here we have already added the required factors $N_{0}$ and $N_{M}$ and
divided the expression by $\omega_{A_{1}}$ so
$V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}$ has the
correct energy unit.

In the above expression, the phonon momentum is
$\mathbf{q}=-\eta\mathbf{K}_{D}+\mathbf{k}-\mathbf{k}^{\prime}-\mathbf{Q}_{-l\eta}$
and the e-ph coupling parameter
$G_{\mathbf{k},\mathbf{k}^{\prime},\mathbf{Q}_{-l\eta}}^{\eta
e_{Y}e_{Y}^{\prime}l}$ depends on the Moiré momentum points
$\mathbf{Q}_{-l\eta}$. In Fig. 11 and 12, we plot $|G|$ as a function of
different values of $|\mathbf{Q}_{-l\eta}|$ for the non-chiral and chiral
limit, separately, on the eigen-state basis. In Fig. 13, we plot $|G|$ as a
function of different values of $|\mathbf{Q}_{-l\eta}|$ in the chiral limit on
the Chern-band basis. The corresponding lattice of $\mathbf{Q}_{-l\eta}$ and
different shells are shown in Fig. 10(b). One can see that while $|G|$
generally decays away for large $|G|$’s in both cases, the decay rate is
faster for the chiral limit. Furthermore, we notice that in the chiral limit
on the Chern-band basis (Fig.13), $|G^{e_{Y}e_{Y}^{\prime}}|$ is zero if
$e_{Y}\neq e_{Y}^{\prime}$, which has been derived in our analytical results,
see Eq. • ‣ VI.4. The detailed form of $G$ function is quite complicated and
thus, we, instead, show
$V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}$ as a function
of $\mathbf{k}^{\prime}$ in Fig. 14 for
$\mathbf{k}=\Gamma_{M},\mathbf{K}_{M},M_{M}$. Several features are noticed:
(1) Intra-Chern-band channel ($e_{Y}=e_{Y}^{\prime}=\pm$) shows nodes at
$\mathbf{k}=\Gamma_{M},\mathbf{k}^{\prime}=\mathbf{K}_{M}$, which is
consistent with Eq. (277) from the symmetry analysis. (2) We notice that the
interaction for the inter-Chern-band channels keeps its value over the whole
MBZ while that for the intra-Chern-band channels drops quickly near the MBZ
boundary due to the presence of the nodes.

Figure 10:  (a) The band dispersion at $\theta=1.06^{\circ}$ with
$w_{0}/w_{1}=0.8$. Other parameters are the same as in Fig. 9. (b) The Moiré
BZ and the $\mathcal{Q}_{\pm}$ lattice with the first three shells (dashed
lines). The momentum $k_{x}$ and $k_{y}$ is in the unit of
$k_{\theta}=2|K_{D}|\sin\left(\frac{\theta}{2}\right)$ with
$\theta=1.06^{\circ}$.  Figure 11:  The parameter
$G_{\mathbf{k},\mathbf{k}^{\prime},Q}^{\eta nml}$ as a function of $|Q|$ for
the projection of the electron-phonon interaction for TBG with
$w_{0}/w_{1}=0.8$. $|Q|$ in the unit of
$k_{\theta}=2|K_{D}|\sin\left(\frac{\theta}{2}\right)$. We choose
$\mathbf{k}={\bf\Gamma}_{M}$ and
$\mathbf{k}^{\prime}={\bf\Gamma}_{M},\mathbf{K}_{M},{\bf M}_{M}$ for (a), (b)
and (c), respectively. The plot is on the eigen-state basis with
$(n,m)=(-,-),(-,+),(+,-),(+,+)$ for red circles, blue stars, black plus signs,
and green crosses, respectively.  Figure 12:  The parameter
$G_{\mathbf{k},\mathbf{k}^{\prime},Q}^{\eta nml}$ as a function of $|Q|$ for
the projection of the electron-phonon interaction for TBG in the chiral limit
($w_{0}=0$). $|Q|$ in the unit of
$k_{\theta}=2|K_{D}|\sin\left(\frac{\theta}{2}\right)$. We choose
$\mathbf{k}={\bf\Gamma}_{M}$ and
$\mathbf{k}^{\prime}={\bf\Gamma}_{M},\mathbf{K}_{M},{\bf M}_{M}$ for (a), (b)
and (c), respectively. The plot is on the eigen-state basis with
$(n,m)=(-,-),(-,+),(+,-),(+,+)$ for red circles, blue stars, black plus signs,
and green crosses, respectively.  Figure 13:  The parameter
$G_{\mathbf{k},\mathbf{k}^{\prime},Q}^{\eta e_{Y}e_{Y}^{\prime}l}$ as a
function of $|Q|$ for the projection of the electron-phonon interaction for
TBG on the Chern-band basis in the chiral limit ($w_{0}=0$). $|Q|$ in the unit
of $k_{\theta}=2|K_{D}|\sin\left(\frac{\theta}{2}\right)$. We choose
$\mathbf{k}={\bf\Gamma}_{M}$ and
$\mathbf{k}^{\prime}={\bf\Gamma}_{M},\mathbf{K}_{M},{\bf M}_{M}$ for (a), (b)
and (c), respectively. The plot is on the Chern-band basis with
$(e_{Y},e_{Y}^{\prime})=(-,-),(-,+),(+,-),(+,+)$ for red circles, blue stars,
black plus signs, and green crosses, respectively.  Figure 14:
$V_{\mathbf{k},\mathbf{k}^{\prime}}^{+e_{Y},e^{\prime}_{Y}}$ as a function of
$\mathbf{k}^{\prime}$ for a fixed $\mathbf{k}$. Here
$V_{\mathbf{k},\mathbf{k}^{\prime}}^{+e_{Y},e^{\prime}_{Y}}$ in unit of eV and
$\mathbf{k}^{\prime}$ in the unit of
$k_{\theta}=2|K_{D}|\sin\left(\frac{\theta}{2}\right)$. (a) and (b) are for
$\mathbf{k}={\bf\Gamma}_{M}$; (c) and (d) are for $\mathbf{k}={\bf M}_{M}$;
(e) and (f) are for $\mathbf{k}=\mathbf{K}_{M}$.

###  VI.7 Symmetry Classification of Irreducible Pairing Channels

Since the phonon energy and the electron-phonon coupling are strong, the eige-
state basis might not be the best basis to work in here. We hence work in the
Chern band basis, for which the symmetry transformation properties are
summarized as follows.

$\displaystyle\hat{g}\gamma^{\dagger}_{\mathbf{k}e_{Y}\eta^{\prime}s}\hat{g}^{-1}=\sum_{e_{Y_{1}}\eta}B^{g}_{\eta
e_{Y_{1}};e_{Y}\eta^{\prime}}\gamma^{\dagger}_{\hat{g}\mathbf{k},e_{Y_{1}}\eta
s}$
$\displaystyle\hat{C}\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}\hat{C}^{-1}=e_{Y}\eta\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}$
$\displaystyle\hat{C}_{2z}\hat{P}\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}(\hat{C}_{2z}\hat{P})^{-1}=ie_{Y}\eta\gamma_{\mathbf{k},e_{Y},-\eta,s}^{\dagger}$
$\displaystyle\hat{C}_{2z}\hat{T}\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}(\hat{C}_{2z}\hat{T})^{-1}=\gamma_{\mathbf{k},-e_{Y},\eta,s}^{\dagger}$
$\displaystyle\hat{P}\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}\hat{P}^{-1}=ie_{Y}\eta
e^{i\phi_{\mathbf{k}}}\gamma_{-\mathbf{k},e_{Y},\eta,s}^{\dagger}$
$\displaystyle\hat{C}_{3z}\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}\hat{C}_{3z}^{-1}=e^{-ie_{Y}\theta(\mathbf{k})}\gamma_{\hat{C}_{3z}\mathbf{k},e_{Y},\eta,s}^{\dagger}$
$\displaystyle\hat{C}_{2x}\gamma_{\mathbf{k},e_{Y},\eta,s}^{\dagger}\hat{C}_{2x}^{-1}=e^{ie_{Y}\alpha(\mathbf{k})}\gamma_{\hat{C}_{2x}\mathbf{k},-e_{Y},\eta,s}^{\dagger}$
(290)

####  VI.7.1 Discrete Symmetry analysis of the Gap Function

The mean-field Bogoliubov-de Gennes (BdG) Hamiltonian for the inter-valley
pairing can be written as

$H_{\Delta}=\sum_{\mathbf{k},e_{Y_{1}},e_{Y_{2}},s_{1},s_{2},\eta}\gamma_{\mathbf{k},e_{Y_{1}},\eta,s_{1}}^{\dagger}\Delta^{\eta}_{\mathbf{k};e_{Y_{1}}s_{1},e_{Y_{2}}s_{2}}\gamma_{-\mathbf{k},e_{Y_{2}},-\eta,s_{2}}^{\dagger}+h.c.$
(291)

First we notice the reshuffling symmetry of the gap function,

$\Delta^{\eta}_{\mathbf{k};e_{Y_{1}}s_{1},e_{Y_{2}}s_{2}}=-\Delta^{-\eta}_{-\mathbf{k};e_{Y_{2}}s_{2},e_{Y_{1}}s_{1}},$
(292)

which can be obtained by interchanging two $\gamma^{\dagger}$ operators and
then relabelling the indices.

Because of the $SU(2)$ spin symmetry of the system, We can define

$\Delta^{\eta}_{\mathbf{k};e_{Y_{1}}s_{1},e_{Y_{2}}s_{2}}=\Delta^{\eta}_{\mathbf{k};e_{Y_{1}},e_{Y_{2}}}{\mathcal{S}}_{s_{1},s_{2}}.$
(293)

Where ${\mathcal{S}}$ is the spin matrix, which takes the form

$\text{Singlet:
}{\mathcal{S}}_{s_{1}s_{2}}=(is_{y})_{s_{1}s_{2}};\;\;\;\text{Triplet:
}{\mathcal{S}}_{s_{1}s_{2}}=((\mathbf{d}\cdot{\bf s})is_{y})_{s_{1}s_{2}};$
(294)

where $\mathbf{d}$ is the three component d-vector for triplet pairing. We
note that

$\text{Singlet:
}{\mathcal{S}}_{s_{1}s_{2}}=-{\mathcal{S}}_{s_{2}s_{1}};\;\;\;\text{Triplet:
}{\mathcal{S}}_{s_{1}s_{2}}={\mathcal{S}}_{s_{2}s_{1}};$ (295)

Rep Symbol |  Basis Functions |  Dim |  $C_{2z}$ |  $C_{2x}$ |  $C_{3z}$
---|---|---|---|---|---
$A_{1}$ |  1, or $x^{2}+y^{2}$, or $z^{2}$ |  1 |  1 |  1 |  1
$A_{2}$ |  z |  1 |  1 |  -1 |  1
$B_{1}$ |  $y(3x^{2}-y^{2})$ |  1 |  -1 |  -1 |  1
$B_{2}$ |  $x(x^{2}-3y^{2})$ |  1 |  -1 |  1 |  1
$E_{1}$ |  $(x,y)$ |  2 |  $-\sigma_{0}$ |  $\sigma_{z}$ |  $\exp(-i2\pi\sigma_{y}/3)$
$E_{2}$ |  ($\frac{x^{2}-y^{2}}{2},xy$) |  2 |  $\sigma_{0}$ |  $\sigma_{z}$ |  $\exp(i2\pi\sigma_{y}/3)$
Table 4:  Irreducible pairing channels.

We now write down the basis function of the irreducible representation (irrep)
of the $C_{6v}$ group, which is the symmetry group of the TBG Hamiltonian with
two valleys, in the table 4. The pairing Hamiltonian $H_{\Delta}$ can
transform as a representation of the $C_{6v}$ group. Thus, based on the table
4, we next consider the symmetry classification of the irreducible pairing
channels [75, 76].

Let’s first consider (spinless) $\hat{C}_{2z}$ symmetry, which generally plays
the role of inversion and allows us to identify the parity of different
pairing channels. For $\hat{C}_{2z}$, the electron operator is transformed as

$\hat{C}_{2z}\gamma^{\dagger}_{\mathbf{k},e_{Y},\eta
s}\hat{C}_{2z}^{-1}=e^{i\phi_{\mathbf{k}}}\gamma^{\dagger}_{-\mathbf{k},e_{Y},-\eta
s}$ (296)

From the table 4, We expect the pairing Hamiltonian $H_{\Delta}$ should
transform as certain representations under $C_{2z}$, with the eigenvalue
$\lambda_{C_{2z}}=\pm 1$,

$\hat{C}_{2z}H_{\Delta}\hat{C}_{2z}^{-1}=\lambda_{C_{2z}}H_{\Delta}.$ (297)

In the case of $\lambda_{C_{2z}}=1$, the Hamiltonian is invariant under
$\hat{C}_{2z}$. For the case of $\lambda_{C_{2z}}=-1$, the Hamiltonian is not
invariant under $\hat{C}_{2z}$, as the gap term changes sign. However, it is
invariant under $\hat{C}_{2z}$ followed by a gauge transformation
$c\rightarrow ic$, i.e. a projective representation. Hence, the spectrum is
still $\hat{C}_{2z}$ symmetric. With
$\exp(i\phi_{\mathbf{k}}+i\phi_{-\mathbf{k}})=1$, the gap function then
follows the relation

$\displaystyle\Delta^{\eta}_{\mathbf{k};e_{1}s_{1},e_{2}s_{2}}=\lambda_{C_{2z}}\Delta^{-\eta}_{-\mathbf{k};e_{1}s_{1},e_{2}s_{2}}.$
(298)

Together with the reshuffling symmetry (292), we have

$\displaystyle\Delta^{\eta}_{\mathbf{k};e_{1},s_{1},e_{2},s_{2}}=-\lambda_{C_{2z}}\Delta^{\eta}_{\mathbf{k};e_{2},s_{2},e_{1},s_{1}},$
(299)

which leads to

$\text{Singlet:
}\Delta^{\eta}_{\mathbf{k};e_{1},e_{2}}=\lambda_{C_{2z}}\Delta^{\eta}_{\mathbf{k};e_{2},e_{1}};\;\;\;\text{Triplet:
}\Delta^{\eta}_{\mathbf{k};e_{1},e_{2}}=-\lambda_{C_{2z}}\Delta^{\eta}_{\mathbf{k};e_{2},e_{1}}.$
(300)

Next let’s consider the (spinless) $\hat{C}_{3z}$ symmetry. Since the $1$-D
irrep of $C_{6v}$ have only eigenvalue $1$ for $\hat{C}_{3z}$, the
transformation of the pairing Hamiltonian is given by

$\hat{C}_{3z}H_{\Delta}\hat{C}_{3z}^{-1}=H_{\Delta},$ (301)

which leads to

$\Delta^{\eta}_{\mathbf{k};e_{1},e_{2}}=e^{-i(e_{1}\theta_{\mathbf{k}}+e_{2}\theta_{-\mathbf{k}})}\Delta^{\eta}_{\hat{C}_{3z}^{-1}\mathbf{k};e_{1},e_{2}}$
(302)

irrespective of the triplet and singlet paring channels. At the
$\mathbf{K}_{M}$ or $\mathbf{K}_{M}^{\prime}$ points, this gap then becomes

$\Delta^{\eta}_{\mathbf{K}_{M};e_{1},e_{2}}=e^{-i(e_{1}\theta_{\mathbf{K}_{M}}+e_{2}\theta_{-\mathbf{K}_{M}})}\Delta^{\eta}_{\mathbf{K}_{M};e_{1},e_{2}}=e^{-i(e_{1}+e_{2})\theta_{\mathbf{K}_{M}}}\Delta^{\eta}_{\mathbf{K}_{M},e_{1},e_{2}}$
(303)

where we have used $\theta_{\mathbf{K}_{M}}=\theta_{-\mathbf{K}_{M}}=2\pi/3$
for the TBG flat bands, as we previously showed. This means that any $1D$
irrep will have, for the pairing between the bands with equal Chern number
(called ”intra-Chern-band pairing” below),

$\Delta_{\mathbf{K}_{M};e_{Y},e_{Y}}=0$ (304)

No such condition exists for pairing between opposite Chern number bands
$e_{Y_{1}}=-e_{Y_{2}}$ (called ”inter-Chern-band pairing” below). Hence we
have the following theorem: _The gap function of the 1D irrep pairing channel
between two bands with the equal Chern number $\pm 1$ (intra-Chern-band
pairing) of the $\hat{C}_{6v}$ symmetry group has nodes at $\mathbf{K}_{M}$,
$\mathbf{K}_{M}^{\prime}$._

Based on the above discussion, we can summarize the 1D irrep pairing channels
as the following.

  * •

$A_{1},A_{2}$ irrep channels:

$\text{Singlet:
}\Delta^{\eta}_{\mathbf{k};e_{1},e_{2}}=\Delta^{\eta}_{\mathbf{k};e_{2},e_{1}};\;\;\;\text{Triplet:
}\Delta^{\eta}_{\mathbf{k};e_{1},e_{2}}=-\Delta^{\eta}_{\mathbf{k};e_{2},e_{1}};\;\;\;\Delta^{\eta}_{\mathbf{k};e_{1},e_{2}}=e^{-i(e_{1}\theta_{\mathbf{k}}+e_{2}\theta_{-\mathbf{k}})}\Delta^{\eta}_{\hat{C}_{3z}^{-1}k,e_{1},e_{2}}.$
(305)

For these irrep channels, the singlet pairing between the bands with the same
Chern number has a node at $\mathbf{K}_{M}$, while the triplet pairing between
the bands with the same Chern number vanishes identically as
$\lambda_{C_{2z}}=1$ for the $A_{1},A_{2}$ pairing channels.

  * •

$B_{1},B_{2}$ irrep channels:

$\text{Singlet:
}\Delta^{\eta}_{\mathbf{k};e_{1},e_{2}}=-\Delta^{\eta}_{\mathbf{k};e_{2},e_{1}};\;\;\;\text{Triplet:
}\Delta^{\eta}_{\mathbf{k};e_{1},e_{2}}=\Delta^{\eta}_{\mathbf{k};e_{2},e_{1}};\;\;\;\Delta^{\eta}_{\mathbf{k};e_{1},e_{2}}=e^{-i(e_{1}\theta_{\mathbf{k}}+e_{2}\theta_{-\mathbf{k}})}\Delta_{\hat{C}_{3z}^{-1}\mathbf{k};e_{1},e_{2}}$
(306)

For these representations, the triplet pairing between same Chern number bands
has a zero at $K_{M}$, while the singlet pairing between the same Chern
numbers vanishes identically as $\lambda_{C_{2z}}=-1$ for the $B_{1},B_{2}$
pairing channels.

We now move to the $2D$ irrep pairing channels, which necessarily require two
pairing terms $(H_{\Delta_{1}},H_{\Delta_{2}})$ and two different gaps
$\Delta_{1;\mathbf{k};e_{1},e_{2}},\Delta_{2;\mathbf{k};e_{1},e_{2}}$. The
pairing Hamiltonian is written as

$\displaystyle
H_{\Delta}=H_{\Delta_{1}}+H_{\Delta_{2}}+H^{\dagger}_{\Delta_{1}}+H^{\dagger}_{\Delta_{2}}$
(307) $\displaystyle
H_{\Delta_{1}}=\sum_{\mathbf{k},e_{Y_{1}},e_{Y_{2}},\eta}\gamma_{\mathbf{k},e_{Y_{1}},\eta}^{\dagger}\Delta^{\eta}_{1;\mathbf{k};e_{Y_{1}},e_{Y_{2}}}{\mathcal{S}}\gamma_{-\mathbf{k},e_{Y_{2}},-\eta}^{\dagger}$
$\displaystyle
H_{\Delta_{2}}=\sum_{\mathbf{k},e_{Y_{1}},e_{Y_{2}},\eta}\gamma_{\mathbf{k},e_{Y_{1}},\eta}^{\dagger}\Delta^{\eta}_{2;\mathbf{k};e_{Y_{1}},e_{Y_{2}}}{\mathcal{S}}\gamma_{-\mathbf{k},e_{Y_{2}},-\eta}^{\dagger}.$

Here we have suppressed the spin index and all the spin information is
included in the ${\mathcal{S}}$ function (singlet or triplet). According to
the table 4, the $\hat{C}_{2z}$ symmetry should give the constraints

$\hat{C}_{2z}H_{\Delta_{1}}\hat{C}_{2z}^{-1}=\lambda_{C_{2z}}H_{\Delta_{1}},\;\;\hat{C}_{2z}H_{\Delta_{2}}\hat{C}_{2z}^{-1}=\lambda_{C_{2z}}H_{\Delta_{2}};\;\;\;\Delta^{\eta}_{1;\mathbf{k};e_{1},e_{2}}=\pm\lambda_{C_{2z}}\Delta^{\eta}_{1;\mathbf{k};e_{2},e_{1}};\;\;\Delta^{\eta}_{2;\mathbf{k};e_{1},e_{2}}=\pm\lambda_{C_{2z}}\Delta^{\eta}_{2;\mathbf{k};e_{2},e_{1}}$
(308)

with $+$ for singlet pairing and $-$ for triplet pairing (and
$\lambda_{C_{2z}}=-(+)$ for $E_{1}(E_{2})$ irrep channels, respectively). One
notices that $(H_{\Delta_{1}}$ and $H_{\Delta_{2}})$ have to be both singlet
or both triplet. The actual order parameter can be a combination of
$aH_{\Delta_{1}}+bH_{\Delta_{2}}$, where $a$ and $b$ can be complex numbers.
This breaks $\hat{C}_{3z}$ symmetry and is hence nematic. Another possibility
would be a time-reversal breaking order parameter, which can respect
$\hat{C}_{3z}$. We now treat each of the two $2D$ irrep channels separately as
follows.

  * •

$E_{2}$ irrep channels: We have

$\hat{C}_{3z}H_{\Delta_{\alpha}}\hat{C}_{3z}^{-1}=H_{\Delta_{\beta}}(e^{-i\frac{2\pi}{3}\sigma_{y}})_{\beta\alpha}\implies
e^{-i(e_{1}\theta_{\mathbf{k}}+e_{2}\theta_{-\mathbf{k}})}\Delta^{\eta}_{\alpha;\hat{C}_{3z}^{-1}\mathbf{k};e_{1},e_{2}}=\Delta^{\eta}_{\beta;\mathbf{k};e_{1},e_{2}}(e^{-i\frac{2\pi}{3}\sigma_{y}})_{\beta\alpha},$
(309)

where $\alpha,\beta=1,2$ indices represent the different pairing channels and
$\sigma$ is the corresponding Pauli matrix. For $\mathbf{k}=\Gamma_{M}$ with
$\theta_{\Gamma_{M}}=0$, the equation above becomes

$\Delta^{\eta}_{2;\Gamma_{M};e_{1},e_{2}}=\sqrt{3}\Delta^{\eta}_{1;\Gamma_{M};e_{1},e_{2}};\;\;\Delta^{\eta}_{1;\Gamma_{M};e_{1},e_{2}}=-\sqrt{3}\Delta^{\eta}_{2;\Gamma_{M};e_{1},e_{2}}\implies\Delta^{\eta}_{1;\Gamma_{M};e_{1},e_{2}}=\Delta^{\eta}_{2;\Gamma_{M};e_{1},e_{2}}=0$
(310)

Hence the nematic $2D$ irrep $E_{2}$ channel has a zero in the gap function at
the $\Gamma_{M}$ point for both the same and different Chern number pairing.
This is different from the $1D$ irrep pairing channels. At $\mathbf{K}_{M}$,
we have

$\Delta^{\eta}_{1;\mathbf{K}_{M};e_{1},-e_{1}}=\Delta^{\eta}_{2;\mathbf{K}_{M};e_{1},-e_{1}}=0;\;\;\;\;ie_{1}\Delta^{\eta}_{1;\mathbf{K}_{M};e_{1},e_{1}}=\Delta^{\eta}_{2;\mathbf{K}_{M};e_{1},e_{1}}$
(311)

Hence the pairing between the bands with opposite Chern number is zero at
$\mathbf{K}_{M}$, while the pairing between the bands with the same Chern
number is nonzero and satisfies the constraint above. From $\hat{C}_{2z}$, the
$\lambda_{C_{2z}}=1$ for the $E_{2}$ representation and we have

$\Delta^{\eta}_{1;\mathbf{k}e_{1},e_{2}}=\pm\Delta_{1;\mathbf{k}e_{2},e_{1}};\;\;\Delta_{2;\mathbf{k};e_{1},e_{2}}=\pm\Delta_{2;\mathbf{k};e_{2},e_{1}},$
(312)

where, again, we have $\pm$ for singlet/triplet. We can then see that for same
chern number pairing, there cannot be a triplet component.

  * •

$E_{1}$ irrep channels: We have

$\hat{C}_{3z}H_{\Delta_{\alpha}}\hat{C}_{3z}^{-1}=H_{\Delta_{\beta}}(e^{i\frac{2\pi}{3}\sigma_{y}})_{\beta\alpha}\implies
e^{-i(e_{1}\theta_{\mathbf{k}}+e_{2}\theta_{-\mathbf{k}})}\Delta^{\eta}_{\alpha;\hat{C}_{3z}^{-1}\mathbf{k};e_{1},e_{2}}=\Delta^{\eta}_{\beta;\mathbf{k};e_{1},e_{2}}(e^{i\frac{2\pi}{3}\sigma_{y}})_{\beta\alpha}.$
(313)

For $\mathbf{k}=\Gamma_{M}$ and $\theta_{\Gamma_{M}}=0$, the equation above
becomes

$\Delta^{\eta}_{2;\Gamma_{M};e_{1},e_{2}}=-\sqrt{3}\Delta^{\eta}_{1;\Gamma_{M};e_{1},e_{2}};\;\;\Delta^{\eta}_{1;\Gamma_{M};e_{1},e_{2}}=\sqrt{3}\Delta^{\eta}_{2;\Gamma_{M};e_{1},e_{2}}\implies\Delta^{\eta}_{1;\Gamma_{M};e_{1},e_{2}}=\Delta^{\eta}_{2;\Gamma_{M};e_{1},e_{2}}=0.$
(314)

Hence, similar as the $E_{2}$ irrep case, the nematic $2D$ irrep $E_{1}$
channel also has a zero in the gap function at the $\Gamma_{M}$ point for both
the same and different Chern number pairing. At $K_{M}$ we have

$\Delta^{\eta}_{1;\mathbf{K}_{M};e_{1},-e_{1}}=\Delta^{\eta}_{2;\mathbf{K}_{M};e_{1},-e_{1}}=0;\;\;\;\;ie_{1}\Delta_{1;\mathbf{K}_{M};e_{1},e_{1}}=-\Delta_{2;\mathbf{K}_{M};e_{1},e_{1}}$
(315)

Hence, the pairing between the bands with different Chern numbers is zero at
$\mathbf{K}_{M}$ while the pairing between the bands with the same Chern
number is nonzero and satisfies the constraint above. From $\hat{C}_{2z}$, the
$\lambda_{C_{2z}}=-1$ for the $E_{2}$ representation and we have

$\Delta^{\eta}_{1;\mathbf{k};e_{1},e_{2}}=\mp\Delta^{\eta}_{1;\mathbf{k};e_{2},e_{1}};\;\;\Delta^{\eta}_{2;\mathbf{k};e_{1},e_{2}}=\mp\Delta^{\eta}_{2;\mathbf{k};e_{2},e_{1}}$
(316)

with $\mp$ for singlet/triplet. We can then see that for the pairing between
the bands with the same Chern number, there cannot be a singlet component.

Finally, let’s discuss the constraint due to the presence of time reversal
(TR) symmetry. It should be noted that TR may be spontaneously broken by the
gap function, so the constraint derived below can only be applied to the TR
invariant pairing. We can combine the spinless $\hat{C}_{2z}$ with the
_spinful_ TR $\hat{T}$ (both are independent symmetries of the TBG system),
which gives

$\hat{C}_{2z}T\gamma^{\dagger}_{\mathbf{k},e_{Y},\eta,s}(\hat{C}_{2z}\hat{T})^{-1}=\sum_{s^{\prime}}\gamma^{\dagger}_{\mathbf{k},-e_{Y},\eta,s^{\prime}}(i{\mathcal{S}}_{y})_{s^{\prime}s}.$
(317)

Applying this symmetry to the pairing Hamiltonian (291) implies that

$\epsilon_{s_{1}^{\prime}s_{1}}(\Delta^{\eta}_{\mathbf{k};-e_{Y_{1}}s_{1},-e_{Y_{2}}s_{2}})^{\star}\epsilon_{s_{2}^{\prime}s_{2}}=\Delta_{k;e_{Y_{1}}s_{1}^{\prime},e_{Y_{2}}s_{2}^{\prime}},$
(318)

where $\epsilon_{s_{1}^{\prime}s_{1}}$ is the Levi-Civita symbol. Hence, for
spin singlet and triplet, we have two transformations for the gap function,

$\displaystyle\text{Singlet:
}\epsilon_{s_{1}^{\prime}s_{1}}(\Delta^{\eta}_{\mathbf{k};-e_{Y_{1}},-e_{Y_{2}}})^{\star}\epsilon_{s_{1},s_{2}}\epsilon_{s_{2}^{\prime}s_{2}}=\Delta^{\eta}_{\mathbf{k};e_{Y_{1}},e_{Y_{2}}}\epsilon_{s_{1}^{\prime},s_{2}^{\prime}}\implies\boxed{(\Delta^{\eta}_{\mathbf{k};-e_{Y_{1}},-e_{Y_{2}}})^{\star}=\Delta_{\mathbf{k};e_{Y_{1}},e_{Y_{2}}}}$
$\displaystyle\text{Triplet:
}(\Delta^{\eta}_{\mathbf{k};-e_{Y_{1}},-e_{Y_{2}}})^{\star}(\mathbf{d}\cdot\ss)^{\star}_{s_{1}s_{2}}=\Delta^{\eta}_{\mathbf{k};e_{Y_{1}},e_{Y_{2}}}(\mathbf{d}\cdot\ss)_{\bar{s}_{1}\bar{s}_{2}}(-1)^{s_{1}+s_{2}}\implies\boxed{(\Delta^{\eta}_{\mathbf{k};-e_{Y_{1}},-e_{Y_{2}}})^{\star}=-\Delta^{\eta}_{\mathbf{k};e_{Y_{1}},e_{Y_{2}}}}$
(319)

where $s=1,2$ are $\uparrow,\downarrow$ while $\bar{1}=2$ and vice versa.

####  VI.7.2 Continuous Symmetry Analysis of Irreducible Pairing Channels

Now let’s consider the constraint from the continuous symmetry in the chiral
limit on the phonon-mediated electron-electron interaction in Eq. (288) when
taking the mean field decomposition for different pairing channels.

In the chiral limit, we have an $SU(2)$ for each chern number sector and we
now separate the two $SU(2)$ operators per chern number sector into

$S^{+}_{e_{Y}}=\sum_{\mathbf{k}\eta}\gamma^{\dagger}_{\mathbf{k},e_{Y},\eta,\uparrow}\gamma_{\mathbf{k},e_{Y},\eta,\downarrow},\;\;\;\;S^{-}_{e_{Y}}=\sum_{\mathbf{k}\eta}\gamma^{\dagger}_{\mathbf{k},e_{Y},\eta,\downarrow}\gamma_{\mathbf{k},e_{Y},\eta,\uparrow},$
(320)

which can be further linearly combined to form the total (tot) spin and
relative (rel) spin raising and lowering operators,

$\displaystyle
S^{+}_{tot}=\frac{1}{\sqrt{2}}(S^{+}_{+}+S^{+}_{-}),\;\;\;S^{-}_{tot}=\frac{1}{\sqrt{2}}(S^{-}_{+}+S^{-}_{-})$
$\displaystyle
S^{+}_{rel}=\frac{1}{\sqrt{2}}(S^{+}_{+}-S^{+}_{-}),\;\;\;S^{-}_{rel}=\frac{1}{\sqrt{2}}(S^{-}_{+}-S^{-}_{-}).$
(321)

We first separate the gap function based on the total spin - singlet and
triplet components. This will be the same even away from the chiral limit.

We now define the two-fermion pairing operators

$\Pi^{\dagger}_{\mathbf{k},\eta;e_{1}s_{1};e_{2},s_{2}}=\gamma_{\mathbf{k},e_{1},\eta
s_{1}}^{\dagger}\gamma^{\dagger}_{-\mathbf{k},e_{2},-\eta s_{2}}$ (322)

and decompose them into singlet and triplet channels,

$\Pi^{\dagger}_{\mathbf{k},\eta,e_{1}s_{1};e_{2},s_{2}}=\sum_{S=0,1;M=-S\ldots
S}C_{s_{1}s_{2}}^{SM}\Pi^{\dagger}_{\mathbf{k},\eta;e_{1},e_{2},S,M}$ (323)

where $C_{s_{1}s_{2}}^{SM}$ are the Clebsh-Godron Coefficients. The operator
form of the $S,M$ pairing channel is then given by

$\displaystyle\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},0,0}=\frac{1}{\sqrt{2}}(\gamma_{\mathbf{k},e_{1},+,\uparrow}^{\dagger}\gamma_{-\mathbf{k},e_{2},-,\downarrow}^{\dagger}-\gamma_{\mathbf{k},e_{1},+,\downarrow}^{\dagger}\gamma_{-\mathbf{k},e_{2},-,\uparrow}^{\dagger})$
$\displaystyle\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},1,0}=\frac{1}{\sqrt{2}}(\gamma_{\mathbf{k},e_{1},+\uparrow}^{\dagger}\gamma_{-\mathbf{k},e_{2},-\downarrow}^{\dagger}+\gamma_{\mathbf{k},e_{1},+,\downarrow}^{\dagger}\gamma_{-\mathbf{k},e_{2},-,\uparrow}^{\dagger})$
$\displaystyle\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},1,1}=\gamma^{\dagger}_{\mathbf{k},e_{1},+,\uparrow}\gamma^{\dagger}_{-\mathbf{k},e_{2},-,\uparrow}$
$\displaystyle\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},1,-1}=\gamma^{\dagger}_{\mathbf{k},e_{1},+,\downarrow}\gamma^{\dagger}_{-\mathbf{k},e_{2},-,\downarrow}.$
(324)

Besides the usual rotations within spin $1$ by the raising and lowering total
spin operators

$\displaystyle
S_{tot}^{-}\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},1,1}\left|0\right\rangle=\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},1,0}\left|0\right\rangle,\;\;\;(S_{tot}^{-})^{2}\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},1,1}\left|0\right\rangle=\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},1,-1}\left|0\right\rangle,$
(325)

we also have, for $e_{1}=e_{2}$,

$\displaystyle
S_{rel}^{-}\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{1},1,1}\left|0\right\rangle=e_{1}\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},1,0}\left|0\right\rangle;\;\;\;\;(S_{rel}^{-})^{2}\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{1},1,1}\left|0\right\rangle=\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},1,-1}\left|0\right\rangle$
(326)

As such, we can see that the action of $S_{rel}^{-}$ and $S_{tot}^{-}$ on the
pairing states with the same Chern number is identical (albeit some signs).

_However_ , for $e_{1}\neq e_{2}$, we have that the singlet and the triplet
states are related, due to the relative (rel) $SU(2)$ we find:

$\displaystyle
S_{rel}^{-}\Pi^{\dagger}_{\mathbf{k},+;e_{1},-e_{1},1,1}\left|0\right\rangle=-e_{1}\Pi^{\dagger}_{\mathbf{k},+;e_{1},-e_{1},0,0}\left|0\right\rangle;\;\;\;\;$
$\displaystyle
S_{rel}^{-}\Pi^{\dagger}_{\mathbf{k},+;e_{1},-e_{1},0,0}\left|0\right\rangle=e_{1}\Pi^{\dagger}_{\mathbf{k},+;e_{1},-e_{1},1,-1}\left|0\right\rangle;\;\;\;\;$
(327)

with similar relations for $\Pi^{\dagger}_{\mathbf{k},-;e_{1},-e_{1},S,M}$ and
hence the triplet pairing is actually related to the singlet pairing of the
inter-Chern-band pairing, which implies the _inter-Chern-band singlet and
triplet pairings to be degenerate_ , in the chiral flat-band limit; while the
intra-Chern-band triplet and singlet pairings do not need this degeneracy.

We now can rewrite the Hamiltonian (288) into the form

$H_{el-
el}=-\frac{1}{N_{M}}\sum_{\mathbf{k},\mathbf{k}^{\prime}}\sum_{\eta,e_{1},e_{2},s_{1},s_{2}}V_{\mathbf{k},\mathbf{k}^{\prime}}^{\eta
e_{1}e_{2}}\sum_{S=0,1}\sum_{M=-S\ldots
S}\Pi^{\dagger}_{\mathbf{k},\eta;e_{1},e_{2},S,M}\Pi_{\mathbf{k}^{\prime},-\eta;e_{1},e_{2},S,M}.$
(328)

With

$\Pi^{\dagger}_{\mathbf{k},\eta
e_{1}s_{1};e_{2},s_{2}}=-\Pi^{\dagger}_{-\mathbf{k},-\eta;e_{2}s_{2};e_{1},s_{1}}$
(329)

we have

$\Pi^{\dagger}_{\mathbf{k},\eta;e_{1},e_{2},0,0}=\Pi^{\dagger}_{-\mathbf{k},-\eta;e_{2},e_{1},0,0};\;\;\;\Pi^{\dagger}_{\mathbf{k},\eta;e_{1},e_{2},1,M}=-\Pi^{\dagger}_{-\mathbf{k},-\eta;e_{2},e_{1},1,M},\;\;M=-1,0,1$
(330)

Using the reshuffling property, we find that

$H_{el-
el}=-\frac{2}{N_{M}}\sum_{\mathbf{k},\mathbf{k}^{\prime}}\sum_{e_{1},e_{2},s_{1},s_{2}}V_{\mathbf{k},\mathbf{k}^{\prime}}^{+e_{1}e_{2}}\sum_{S=0,1}\sum_{M=-S\ldots
S}\Pi^{\dagger}_{\mathbf{k},+;e_{1},e_{2},S,M}\Pi_{\mathbf{k}^{\prime},-;e_{1},e_{2},S,M}$
(331)

###  VI.8 Gap equation and Possible Pairing Channels for the Flat Bands

####  VI.8.1 Linearized gap equation

In this section, we will first drive the linearized gap equation for the TBG
Hamiltonian and discuss the properties of the linearized gap equation.

Now let us consider the general interaction

$H_{el-
el}=-\frac{1}{N_{M}}\sum_{\mathbf{k}\mathbf{k}^{\prime}}\sum_{\alpha\beta\mu\nu}V_{\alpha,\beta,\mu,\nu}(\mathbf{k},\mathbf{k}^{\prime})\gamma_{\mathbf{k},\alpha
s_{1}}^{\dagger}\gamma_{-\mathbf{k},\beta
s_{2}}^{\dagger}\gamma_{-\mathbf{k}^{\prime}\mu
s_{2}}\gamma_{\mathbf{k}^{\prime}\nu s_{1}}$ (332)

with composite index
$\alpha=e_{1},\eta;\beta=e_{2},-\eta;\mu=e_{2},\eta;\nu=e_{1},-\eta$, and the
interaction function $V$ is defined as

$V_{\alpha,\beta,\mu,\nu}(\mathbf{k},\mathbf{k}^{\prime})=V_{e_{1},\eta;e_{2},-\eta;e_{2},\eta;e_{1},-\eta}(\mathbf{k},\mathbf{k}^{\prime})=V_{\mathbf{k},\mathbf{k}^{\prime}}^{\eta
e_{1}e_{2}}.$ (333)

Next we define the gap function as

$\Delta^{\eta}_{\mathbf{k};e_{1}s_{1},e_{2}s_{2}}=-\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}V_{\mathbf{k}\mathbf{k}^{\prime}}^{\eta
e_{1}e_{2}}\langle\gamma_{-\mathbf{k}^{\prime}e_{2}\eta
s_{2}}\gamma_{\mathbf{k}^{\prime}e_{1}-\eta s_{1}}\rangle,$ (334)

or

$\Delta_{\mathbf{k};\alpha s_{1},\beta
s_{2}}=-\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}V_{\alpha\beta\mu\nu}(\mathbf{k},\mathbf{k}^{\prime})\langle\gamma_{-\mathbf{k}^{\prime}\mu
s_{2}}\gamma_{\mathbf{k}^{\prime}\nu s_{1}}\rangle,$ (335)

in terms of the notation $\alpha,\beta,\mu,\nu$. With the above form of the
gap function, we can perform the mean field decomposition of the interaction
Hamiltonian,

$H_{el-
el}=\frac{1}{N_{M}}\sum_{\mathbf{k}}\sum_{\alpha\beta\mu\nu}\left(\Delta_{\mathbf{k};\alpha
s_{1},\beta s_{2}}\gamma^{\dagger}_{\mathbf{k},\alpha
s_{1}}\gamma^{\dagger}_{-\mathbf{k},\beta s_{2}}+\Delta^{*}_{\mathbf{k};\alpha
s_{1},\beta s_{2}}\gamma_{-\mathbf{k},\beta s_{2}}\gamma_{\mathbf{k},\alpha
s_{1}}-\Delta_{\mathbf{k};\alpha s_{1},\beta
s_{2}}\langle\gamma^{\dagger}_{\mathbf{k},\alpha
s_{1}}\gamma^{\dagger}_{-\mathbf{k},\beta s_{2}}\rangle\right).$ (336)

We introduce the anomalous Green’s function in the imaginary time $\tau$ as

${\mathcal{F}}_{\nu s_{1},\mu s_{2}}(\mathbf{k},\tau)=\langle
T_{\tau}\gamma_{\mathbf{k}\nu s_{1}}(\tau)\gamma_{-\mathbf{k}\mu
s_{2}}(0)\rangle,$ (337)

where $\langle...\rangle$ represents the thermal average, and its Fourier
transform

${\mathcal{F}}_{\nu s_{1},\mu
s_{2}}(\mathbf{k},\tau)=\frac{1}{\beta}\sum_{i\omega_{n}}{\mathcal{F}}_{\nu
s_{1},\mu s_{2}}(\mathbf{k},i\omega_{n})e^{-i\omega_{n}\tau},$ (338)

where $\beta=1/k_{B}T$ and $i\omega_{n}=(2n+1)\pi/\beta$ is the Matsubara
frequency.

The gap function can be related to the anomalous Green’s function by

$\Delta_{\mathbf{k};\alpha s_{1},\beta
s_{2}}=\sum_{\mathbf{k}^{\prime},\mu\nu}V_{\alpha\beta\mu\nu}(\mathbf{k}\mathbf{k}^{\prime}){\mathcal{F}}_{\nu
s_{1},\mu
s_{2}}(\mathbf{k}^{\prime},0)=\frac{1}{\beta}\sum_{\mathbf{k}^{\prime},i\omega_{n},\mu\nu}V_{\alpha\beta\mu\nu}(\mathbf{k}\mathbf{k}^{\prime}){\mathcal{F}}_{\nu
s_{1},\mu s_{2}}(\mathbf{k}^{\prime},i\omega_{n}).$ (339)

With the above definition, we can derive the equation of motion for the
anomalous Green’s function, given by

$\frac{d{\mathcal{F}}_{\nu s_{1},\mu
s_{2}}(\mathbf{k},\tau)}{d\tau}=-\sum_{\mu^{\prime}}h_{\mathbf{k},\nu\mu^{\prime}}{\mathcal{F}}_{\mu^{\prime}s_{1},\mu
s_{2}}(\mathbf{k},\tau)-2\sum_{\beta s_{1}^{\prime}}\Delta_{\mathbf{k},\nu
s_{1},\beta s_{1}^{\prime}}G_{\mu s_{2},\beta
s_{1}^{\prime}}(-\mathbf{k},-\tau),$ (340)

where $h_{\mathbf{k},\nu\mu^{\prime}}$ is the single-particle Hamiltonian

$H_{0}=\sum_{\mathbf{k},\alpha\beta,s}h_{\mathbf{k},\alpha\beta}\gamma^{\dagger}_{\mathbf{k},\alpha,s}\gamma_{\mathbf{k},\beta
s}$ (341)

and the Green’s function $G$ is defined as

$G_{\alpha s_{1},\beta s_{2}}(\mathbf{k},\tau)=-\langle
T_{\tau}\gamma_{\mathbf{k},\alpha
s_{1}}(\tau)\gamma^{\dagger}_{\mathbf{k},\beta s_{2}}(0)\rangle$ (342)

and its Fourier transform is

$G_{\alpha s_{1},\beta
s_{2}}(\mathbf{k},\tau)=\frac{1}{\beta}\sum_{i\omega_{n}}G_{\alpha s_{1},\beta
s_{2}}(\mathbf{k},i\omega_{n})e^{-i\omega_{n}\tau}.$ (343)

In the frequency space, the equation of motion for the anomalous Green’s
function then reads

$\sum_{\mu^{\prime}}(i\omega_{n}\delta_{\nu\mu^{\prime}}-h_{\mathbf{k},\nu\mu^{\prime}}){\mathcal{F}}_{\mu^{\prime}s_{1},\mu
s_{2}}(\mathbf{k},i\omega_{n})=2\sum_{\mu^{\prime}s_{1}^{\prime}}\Delta_{\mathbf{k},\nu
s_{1},\mu^{\prime}s_{1}^{\prime}}G_{\mu
s_{2},\mu^{\prime}s_{1}^{\prime}}(-\mathbf{k},-i\omega_{n}).$ (344)

Next we can make the approximation of replacing the full Green’s function $G$
by the zero-order Green’s function $G_{0}$, in which the thermal average is
only for the single-particle Hamiltonian $H_{0}$. From the equation of motion,
$G_{0}$ can be solved analytically and given by

$G_{0;\alpha s_{1},\beta
s_{2}}(\mathbf{k},i\omega_{n})=\delta_{s_{1}s_{2}}[(i\omega_{n}-h_{\mathbf{k}})^{-1}]_{\alpha\beta}$
(345)

Thus, up to the lowest order in the gap function $\Delta$, we find the
anomalous Green’s function is given by

${\mathcal{F}}_{\nu s_{1},\mu
s_{2}}(\mathbf{k},i\omega_{n})=2\sum_{\alpha\beta,s_{1}^{\prime}s_{2}^{\prime}}G_{0;\nu
s_{1},\alpha s_{1}^{\prime}}(\mathbf{k},i\omega_{n})\Delta_{\mathbf{k},\alpha
s_{1}^{\prime},\beta^{\prime}s_{2}^{\prime}}G_{0;\mu
s_{2},\beta^{\prime}s_{2}^{\prime}}(-\mathbf{k},-i\omega_{n}).$ (346)

Substituting the above form of ${\mathcal{F}}$ into Eq. (339), we obtain the
linearized gap equation

$\Delta_{\alpha s_{1}\beta s_{2}}(\mathbf{k})=\frac{2}{\beta
N_{M}}\sum_{i\omega_{n},\mathbf{k}^{\prime},\mu,\nu}V_{\alpha,\beta,\mu,\nu}(\mathbf{k},\mathbf{k}^{\prime})[G_{0}(\mathbf{k}^{\prime},i\omega_{n})\Delta(\mathbf{k}^{\prime})G_{0}^{T}(-\mathbf{k}^{\prime},-i\omega_{n})]_{\nu
s_{1},\mu s_{2}},$ (347)

It should be noted that in the Chern-band basis, the single-particle Green’s
function is not diagonal, and the explicit form of the linearized gap equation
on the Chern-band basis is written as

$\Delta_{e_{1}s_{1},e_{2}s_{2};\mathbf{k}}^{\eta}=\frac{2}{\beta
N_{M}}\sum_{i\omega_{n},\mathbf{k}^{\prime},e_{1}^{\prime},e_{2}^{\prime}}V^{\eta,e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}G_{0,e_{1}e_{1}^{\prime}}^{-\eta}(\mathbf{k}^{\prime},i\omega_{n})\Delta^{-\eta}_{e_{1}^{\prime}s_{1},e_{2}^{\prime}s_{2};\mathbf{k}^{\prime}}G^{\eta}_{0,e_{2}e_{2}^{\prime}}(-\mathbf{k}^{\prime},-i\omega_{n}),$
(348)

It should be noted that the left hand side of the above equation is for gap
function $\Delta_{e_{1}s_{1},e_{2}s_{2};\mathbf{k}}^{\eta}=\Delta_{e_{1}\eta
s_{1},e_{2}-\eta s_{2}}$, which has the opposite $\eta$ value from
$\Delta^{-\eta}_{e_{1}^{\prime}s_{1},e_{2}^{\prime}s_{2};\mathbf{k}^{\prime}}=\Delta_{e_{1}^{\prime}-\eta
s_{1},e_{2}^{\prime}\eta s_{2}}$ on the right hand side of the equation. On
the Chern-band basis, the single particle Hamiltonian reads

$h_{\eta,\mathbf{k}}=(d_{0,\eta,\mathbf{k}}-\mu)\zeta^{0}+d_{x,\eta,\mathbf{k}}\zeta^{x},$
(349)

with

$d_{0,\eta,\mathbf{k}}=(\epsilon_{+,\eta,\mathbf{k}}+\epsilon_{-,\eta,\mathbf{k}})/2;\;\;d_{x,\eta,\mathbf{k}}=(\epsilon_{+,\eta,\mathbf{k}}-\epsilon_{-,\eta,\mathbf{k}})/2,$
(350)

and $\epsilon_{n,\eta,\mathbf{k}}$ are the single-particle eigen-energies of
the BM model $\hat{H}_{0}$ so they should be real. Time reversal symmetry
requires

$\epsilon_{n,\eta,\mathbf{k}}=\epsilon_{n,-\eta,-\mathbf{k}}\implies
d_{0,-\eta,-\mathbf{k}}=d_{0,\eta,\mathbf{k}};\;d_{x,-\eta,-\mathbf{k}}=d_{x,\eta,\mathbf{k}}\implies
h_{-\eta,-\mathbf{k}}=h_{\eta,\mathbf{k}}.$ (351)

The corresponding single-particle Green function is given by

$G_{0}^{\eta}(\mathbf{k},i\omega_{n})=(i\omega_{n}-h_{\eta,\mathbf{k}})^{-1}=\sum_{m}\frac{P_{m,\mathbf{k},\eta}}{i\omega_{n}-\xi_{m,\eta,\mathbf{k}}},$
(352)

where $m=\pm$,
$\xi_{m,\eta,\mathbf{k}}=d_{0,\eta,\mathbf{k}}-\mu+m|d_{x,\eta,\mathbf{k}}|$
and the projection operator is defined as

$P_{m,\mathbf{k},\eta}=\frac{1}{2}\left(1+m\frac{d_{x,\eta,\mathbf{k}}}{|d_{x,\eta,\mathbf{k}}|}\zeta^{x}\right).$
(353)

With such form of the single-particle Green’s function, the linearized gap
equation can be written as

$\Delta_{e_{1}s_{1},e_{2}s_{2};\mathbf{k}}^{\eta}=\frac{2}{N_{M}}\sum_{\mathbf{k}^{\prime},m_{1}m_{2}}V^{\eta,e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}\left[P_{m_{1},\mathbf{k}^{\prime},-\eta}\Delta^{-\eta}_{s_{1}s_{2},\mathbf{k}^{\prime}}P^{T}_{m_{2},-\mathbf{k}^{\prime},\eta}\right]_{e_{1}e_{2}}T^{\eta}_{m_{1}m_{2}\mathbf{k}^{\prime}},$
(354)

where the Mastubara frequency summation can be performed

$T^{\eta}_{m_{1}m_{2}\mathbf{k}^{\prime}}=\frac{1}{\beta}\sum_{i\omega_{n}}\frac{1}{i\omega_{n}-\xi_{m_{1},-\eta,\mathbf{k}^{\prime}}}\frac{1}{-i\omega_{n}-\xi_{m_{2},\eta,-\mathbf{k}^{\prime}}}=-\frac{n_{F}(\xi_{m_{1},-\eta,\mathbf{k}^{\prime}})-n_{F}(-\xi_{m_{2},\eta,-\mathbf{k}^{\prime}})}{\xi_{m_{1},-\eta,\mathbf{k}^{\prime}}+\xi_{m_{2},\eta,-\mathbf{k}^{\prime}}}.$
(355)

In the flat band limit, we have
$\xi_{\mathbf{k},e_{1}\eta}=\xi_{\mathbf{k}}\ll k_{B}T$, and up to the lowest
order term, we find

$T^{\eta}_{m_{1}m_{2}\mathbf{k}^{\prime}}\approx\frac{\beta}{4},$ (356)

which is independent of $m_{1},m_{2},\mathbf{k}^{\prime},\eta$. In this limit,
the gap equation is simplified to

$\Delta_{e_{1}s_{1},e_{2}s_{2};\mathbf{k}}^{\eta}=\frac{\beta}{2N_{M}}\sum_{\mathbf{k}^{\prime},m_{1}m_{2}}V^{\eta,e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}\left[P_{m_{1},\mathbf{k}^{\prime},-\eta}\Delta^{-\eta}_{s_{1}s_{2},\mathbf{k}^{\prime}}P^{T}_{m_{2},-\mathbf{k}^{\prime},\eta}\right]_{e_{1}e_{2}}.$
(357)

Furthermore, one can show that

$\sum_{m_{1}m_{2}}P_{m_{1},\mathbf{k}^{\prime},-\eta}\Delta^{-\eta}_{s_{1}s_{2},\mathbf{k}^{\prime}}P^{T}_{m_{2},-\mathbf{k}^{\prime},\eta}=\Delta^{-\eta}_{s_{1}s_{2},\mathbf{k}^{\prime}},$
(358)

and thus in the flat band limit, the linearized gap equation is simplified as

$2k_{B}T\Delta^{\eta}_{\mathbf{k};e_{1}s_{1},e_{2}s_{2}}=\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}V^{\eta
e_{1}e_{2}}_{\mathbf{k},\mathbf{k}^{\prime}}\Delta^{-\eta}_{\mathbf{k}^{\prime};e_{1}s_{1},e_{2}s_{2}},$
(359)

where we denote
$\Delta^{\eta}_{\mathbf{k};e_{1}s_{1},e_{2}s_{2}}=\Delta_{e_{1}\eta
s_{1},e_{2}-\eta s_{2}}(\mathbf{k})$. The above equation can be written in an
eigen-problem form

$\displaystyle
2k_{B}T\left(\begin{array}[]{c}\Delta^{\eta}_{\mathbf{k};e_{1}s_{1},e_{2}s_{2}}\\\
\Delta^{-\eta}_{\mathbf{k};e_{1}s_{1},e_{2}s_{2}}\end{array}\right)=\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}\mathcal{V}^{\eta,e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}\left(\begin{array}[]{c}\Delta^{\eta}_{\mathbf{k}^{\prime};e_{1}s_{1},e_{2}s_{2}}\\\
\Delta^{-\eta}_{\mathbf{k}^{\prime};e_{1}s_{1},e_{2}s_{2}}\end{array}\right),$
(364)

where

$\displaystyle\mathcal{V}^{\eta,e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}=\left(\begin{array}[]{cc}0&V^{\eta
e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}\\\ V^{-\eta
e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}&0\end{array}\right)$ (367)

needs to be diagonalized and possesses a off-diagonal form.

We can decompose the gap function into the orbital part and spin part. Let’s
denote the spin part as ${\mathcal{S}}^{SM}_{s_{1}s_{2}}$, where $S=0,1$ and
$M=-S,...,S$. Explicitly, ${\mathcal{S}}^{00}=is_{y}$ for spin singlet and
${\mathcal{S}}^{1,\pm 1}=\frac{i}{\sqrt{2}}(s_{x}\pm
is_{y})s_{y}=\frac{1}{\sqrt{2}}(-s_{z}\mp s_{0}),{\mathcal{S}}^{1,0}=s_{x}$
for spin triplet, where $s_{0}$ is the identity matrix and $s_{x,y,z}$ are
three Pauli matrices. Then we can decompose
$\Delta^{\eta}_{\mathbf{k};e_{1}s_{1},e_{2}s_{2}}=\sum_{S,M}\Delta^{\eta,SM}_{\mathbf{k};e_{1}e_{2}}{\mathcal{S}}^{SM}_{s_{1}s_{2}}$.
Since $V^{\eta e_{1}e_{2}}_{kk^{\prime}}$ is independent of spin and
$Tr(({\mathcal{S}}^{SM})^{\dagger}{\mathcal{S}}^{S^{\prime}M^{\prime}})=2\delta_{SS^{\prime}}\delta_{MM^{\prime}}$,
the eigen-equation for the gap function is

$\displaystyle
2k_{B}T\left(\begin{array}[]{c}\Delta^{\eta,SM}_{\mathbf{k};e_{1}e_{2}}\\\
\Delta^{-\eta,SM}_{\mathbf{k};e_{1}e_{2}}\end{array}\right)=\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}\mathcal{V}^{\eta,e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}\left(\begin{array}[]{c}\Delta^{\eta,SM}_{\mathbf{k}^{\prime};e_{1}e_{2}}\\\
\Delta^{-\eta,SM}_{\mathbf{k}^{\prime};e_{1}e_{2}}\end{array}\right)$ (372)

for any $S,M$.

Since we only have two independent $V^{\eta
e_{1}e_{2}}_{\mathbf{k},\mathbf{k}^{\prime}}$ for the fix
$\mathbf{k},\mathbf{k}^{\prime}$, namely one complex
$V^{+++}_{\mathbf{k},\mathbf{k}^{\prime}}$ and one real
$V^{++-}_{\mathbf{k},\mathbf{k}^{\prime}}$, we can explicitly write down two
independent equations explicitly, namely,

$\displaystyle 2k_{B}T\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k};++}\\\
\Delta^{-,SM}_{\mathbf{k};++}\end{array}\right)=\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}\mathcal{V}^{+,++}_{\mathbf{k},\mathbf{k}^{\prime}}\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k}^{\prime};++}\\\
\Delta^{-,SM}_{\mathbf{k}^{\prime};++}\end{array}\right),\;\;\mathcal{V}^{+,++}_{\mathbf{k},\mathbf{k}^{\prime}}=\left(\begin{array}[]{cc}0&V^{+++}_{\mathbf{k}\mathbf{k}^{\prime}}\\\
V^{-++}_{\mathbf{k}\mathbf{k}^{\prime}}&0\end{array}\right),$ (379)

and

$\displaystyle 2k_{B}T\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k};+-}\\\
\Delta^{-,SM}_{\mathbf{k};+-}\end{array}\right)=\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}\mathcal{V}^{+,+-}_{\mathbf{k},\mathbf{k}^{\prime}}\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k}^{\prime};+-}\\\
\Delta^{-,SM}_{\mathbf{k}^{\prime};+-}\end{array}\right),\;\;\mathcal{V}^{+,+-}_{\mathbf{k},\mathbf{k}^{\prime}}=\left(\begin{array}[]{cc}0&V^{++-}_{\mathbf{k}\mathbf{k}^{\prime}}\\\
V^{-+-}_{\mathbf{k}\mathbf{k}^{\prime}}&0\end{array}\right).$ (386)

With the hermitian and reshuffling conditions, one can show
$V^{\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}}=(V^{-\eta,e_{Y},e_{Y}^{\prime}}_{\mathbf{k}^{\prime},\mathbf{k}})^{*}$,
or more explicitly,
$V^{-++}_{\mathbf{k}\mathbf{k}^{\prime}}=(V^{+++}_{\mathbf{k}^{\prime}\mathbf{k}})^{*}$
and
$V^{-+-}_{\mathbf{k}\mathbf{k}^{\prime}}=(V^{++-}_{\mathbf{k}^{\prime}\mathbf{k}})^{*}$,
so both equations are the eigen-equations for hermitian matrices
$\mathcal{V}^{+,e_{1}e_{2}}_{\mathbf{k},\mathbf{k}^{\prime}}$, which can be
solved numerically through diagonalization process.

Now let us discuss several properties of the gap function
$\Delta^{\eta,SM}_{\mathbf{k};e_{1}e_{2}}$. Although some results discussed
below have been derived in Sec. VI.7.1 from the symmetry view of the gap
function, we will show similar results can be obtained from the linearized gap
equation (372). With the reshuffling symmetry Eq. (292),
${\mathcal{S}}^{00}_{s_{1}s_{2}}=-{\mathcal{S}}^{00}_{s_{2}s_{1}}$ and
${\mathcal{S}}^{1M}_{s_{1}s_{2}}={\mathcal{S}}^{1M}_{s_{2}s_{1}}$, we have

$\Delta^{\eta,00}_{\mathbf{k};e_{1}e_{2}}=\Delta^{-\eta,00}_{-\mathbf{k};e_{2}e_{1}}$
(387)

for spin singlet and

$\Delta^{\eta,1M}_{\mathbf{k};e_{1}e_{2}}=-\Delta^{-\eta,1M}_{-\mathbf{k};e_{2}e_{1}}$
(388)

for spin triplet.

As well known for normal superconductors, when there is inversion symmetry,
the orbital part of the gap function has definite parity, and we only have
spin-singlet even-parity or spin-triplet odd-parity gap functions. There are
two differences for TBG: (1) There is no inversion symmetry for TBG and thus
the spinless $C_{2z}$ symmetry plays the role of inversion; and (2) Due to the
existence of Chern-band index, spin-singlet odd-parity or triplet even-parity
states can also exist. Below we will discuss these two points in details.

Let us first consider the $\hat{C}_{2z}$ symmetry, which requires $V^{\eta
e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}=V^{-\eta
e_{1}e_{2}}_{-\mathbf{k}-\mathbf{k}^{\prime}}$. Thus, the gap equation (359)
can be transformed as

$\displaystyle
2k_{B}T\Delta^{\eta,SM}_{\mathbf{k};e_{1}e_{2}}=\sum_{\mathbf{k}^{\prime}}V^{\eta
e_{1}e_{2}}_{\mathbf{k},\mathbf{k}^{\prime}}\Delta^{-\eta,SM}_{\mathbf{k}^{\prime};e_{1}e_{2}}\rightarrow
2k_{B}T\Delta^{-\eta,SM}_{-\mathbf{k};e_{1}e_{2}}=\sum_{\mathbf{k}^{\prime}}V^{-\eta
e_{1}e_{2}}_{-\mathbf{k},-\mathbf{k}^{\prime}}\Delta^{\eta,SM}_{-\mathbf{k}^{\prime};e_{1}e_{2}}$
$\displaystyle\rightarrow
2k_{B}T\Delta^{-\eta,SM}_{-\mathbf{k};e_{1}e_{2}}=\sum_{\mathbf{k}^{\prime}}V^{\eta
e_{1}e_{2}}_{\mathbf{k},\mathbf{k}^{\prime}}\Delta^{\eta,SM}_{-\mathbf{k}^{\prime};e_{1}e_{2}}$
(389)

In the above derivation, we replace
$\mathbf{k}\rightarrow-\mathbf{k},\mathbf{k}^{\prime}\rightarrow-\mathbf{k}^{\prime},\eta\rightarrow-\eta$
in the first step and use $C_{2z}$ symmetry in the second step. We can write
the above equation into an eigen-equation form

$\displaystyle
2k_{B}T\left(\begin{array}[]{c}\Delta^{-\eta,SM}_{-\mathbf{k};e_{1}e_{2}}\\\
\Delta^{\eta,SM}_{-\mathbf{k};e_{1}e_{2}}\end{array}\right)=\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}\mathcal{V}^{\eta,e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}\left(\begin{array}[]{c}\Delta^{-\eta,SM}_{-\mathbf{k}^{\prime};e_{1}e_{2}}\\\
\Delta^{\eta,SM}_{-\mathbf{k}^{\prime};e_{1}e_{2}}\end{array}\right),$ (394)

which can be compared with Eq. (372), and one can see that

$\displaystyle\Delta^{-\eta,SM}_{-\mathbf{k};e_{1}e_{2}}=\lambda_{C_{2z}}\Delta^{\eta,SM}_{\mathbf{k};e_{1}e_{2}}$
(395)

with $|\lambda_{C_{2z}}|=1$ for any $S,M$. If the above procedure is performed
twice, we need to go back to the original equation, so
$\lambda_{C_{2z}}^{2}=1$, which means $\lambda_{C_{2z}}=\pm 1$. This can be
understood as the definite parity $\pm 1$ for the gap function with respect to
$\hat{C}_{2z}$ symmetry.

We combine the anti-symmetry condition (Eq. (387) and (388)) with the definite
parity of the orbital part Eq. (395) and obtain

$\Delta^{\eta,00}_{\mathbf{k};e_{1}e_{2}}=\lambda_{C_{2z}}\Delta^{\eta,00}_{\mathbf{k};e_{2}e_{1}}$
(396)

for spin singlet and

$\Delta^{\eta,1M}_{\mathbf{k};e_{1}e_{2}}=-\lambda_{C_{2z}}\Delta^{\eta,1M}_{\mathbf{k};e_{2}e_{1}}$
(397)

for spin triplet.

Next let us consider the role of Chern-band index $e_{1,2}$ in the gap
function. We decompose the gap function into

$\Delta^{\eta}_{\mathbf{k};e_{1}s_{1},e_{2}s_{2}}=\sum_{\mu,SM}\Delta^{\eta,SM}_{\mathbf{k};\mu}(\zeta^{\mu})_{e_{1}e_{2}}({\mathcal{S}}^{SM})_{s_{1}s_{2}},$
(398)

where $\mu=0,x,y,z$ and $\zeta^{\mu}$ are the identity and Pauli matrices for
the Chern-band index. The total gap function
$\Delta^{\eta}_{\mathbf{k};e_{1}s_{1},e_{2}s_{2}}$ needs to be anti-symmetric.
For intra-band channels, $\zeta^{\mu}$ should be $\zeta^{0}$ or $\zeta^{z}$,
both of which are symmetric matrices. So to make the total gap function being
anti-symmetric, the singlet channel $\Delta^{\eta,00}_{\mathbf{k};\mu}$ can
only be even-parity ($\lambda_{C_{2z}}=+$) and the triplet channel
$\Delta^{\eta,1M}_{\mathbf{k};\mu}$ can only be odd-parity
($\lambda_{C_{2z}}=-$) for intra-Chern-band pairings. This is consistent with
the standard spin-singlet even-parity or spin-triplet odd-parity pairings.

For inter-Chern-band channels, $\zeta^{\mu}$ can be symmetric $\zeta^{x}$ or
anti-symmetric $\zeta^{y}$. To make the total gap function anti-symmetric, the
singlet channel $\Delta^{\eta,00}_{\mathbf{k};\mu}$ can either be even-parity
for $\mu=x$ as $\zeta^{x}$ is symmetric, or odd-parity for $\mu=y$ as
$\zeta^{y}$ is anti-symmetric. Similarly, the triplet channel
$\Delta^{\eta,1M}_{\mathbf{k};\mu}$ can be either even-parity for $\mu=y$ or
odd-parity for $\mu=x$. Next we will decompose the linearized gap equation
(359) into the $\zeta^{x}$ and $\zeta^{y}$ channels and show that they have
the same form. We substitute the decomposition (398) into Eq. (359) and with
the orthonormal relation $Tr(\zeta^{\mu}\zeta^{\nu})=2\delta_{\mu\nu}$, we
find

$4k_{B}T\Delta^{\eta,SM}_{\mathbf{k};\nu}=\frac{1}{N_{M}}\sum_{\mathbf{k},\mu}\left(\sum_{e_{1}e_{2}}V^{\eta
e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}(\zeta^{\mu})_{e_{1}e_{2}}(\zeta^{\nu})_{e_{2}e_{1}}\right)\Delta^{\eta,SM}_{\mathbf{k}^{\prime},\mu}.$
(399)

For the inter-Chern-band channels, we take $\mu,\nu$ to be $x,y$ and find

$\sum_{e_{1}e_{2}}V^{\eta
e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}(\zeta^{\mu})_{e_{1}e_{2}}(\zeta^{\mu})_{e_{2}e_{1}}=V^{\eta+-}_{\mathbf{k}\mathbf{k}^{\prime}}+V^{\eta-+}_{\mathbf{k}\mathbf{k}^{\prime}}=2V^{\eta+-}_{\mathbf{k}\mathbf{k}^{\prime}},\;\mu=x,y$
(400) $\sum_{e_{1}e_{2}}V^{\eta
e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}(\zeta^{x})_{e_{1}e_{2}}(\zeta^{y})_{e_{2}e_{1}}=\sum_{e_{1}e_{2}}V^{\eta
e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}\delta_{e_{1},-e_{2}}(-i)\epsilon_{e_{2}e_{1}}=(-i)(-V^{\eta+-}_{\mathbf{k}\mathbf{k}^{\prime}}+V^{\eta-+}_{\mathbf{k}\mathbf{k}^{\prime}})=0$
(401)

and

$\sum_{e_{1}e_{2}}V^{\eta
e_{1}e_{2}}_{\mathbf{k}\mathbf{k}^{\prime}}(\zeta^{y})_{e_{1}e_{2}}(\zeta^{x})_{e_{2}e_{1}}=0,$
(402)

where we have used $V^{\eta+-}_{kk^{\prime}}=V^{\eta-+}_{kk^{\prime}}$ from
the constraints of the $V$ function in Eqs. (VI.4). With the above relations,
we obtain

$2k_{B}T\Delta^{\eta,SM}_{\mathbf{k};\mu}=\frac{1}{N_{M}}\sum_{\mathbf{k},\mu}V^{\eta+-}_{\mathbf{k}\mathbf{k}^{\prime}}\Delta^{\eta,SM}_{\mathbf{k}^{\prime},\mu},\;\;\mu=x,y.$
(403)

Thus, $\Delta^{\eta,SM}_{\mathbf{k};x}$ and $\Delta^{\eta,SM}_{\mathbf{k};y}$
satisfy the same linearized gap equation. The largest eigen-value of the above
gap equation, which determines the $T_{c}$ for the inter-Chern-band channel,
has a certain $C_{2z}$-parity, say even-parity, and then both singlet channel
$\Delta^{\eta,00}_{\mathbf{k};\mu=x}$ and triplet channel
$\Delta^{\eta,1M}_{\mathbf{k};\mu=y}$ can exist and are degenerate with the
same $T_{c}$. This is consistent with the conclusion that singlet and triplet
channels are degenerate for the inter-Chern-band pairings in the chiral flat-
band limit from the continuous symmetry analysis in Sec. VI.7.2.

####  VI.8.2 Intra-Chern-band channels from linearized gap equation

Now let us discuss the intra-Chern-band channels. Besides Eq. (379), we have
another linearized gap equation for
$V^{\pm,--}_{\mathbf{k},\mathbf{k}^{\prime}}$,

$\displaystyle 2k_{B}T\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k};--}\\\
\Delta^{-,SM}_{\mathbf{k};--}\end{array}\right)=\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}\left(\begin{array}[]{cc}0&V^{+,--}_{\mathbf{k}\mathbf{k}^{\prime}}\\\
V^{-,--}_{\mathbf{k}\mathbf{k}^{\prime}}&0\end{array}\right)\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k}^{\prime};--}\\\
\Delta^{-,SM}_{\mathbf{k}^{\prime};--}.\end{array}\right),$ (410)

which is directly related to Eq. (379) by symmetry. Since $C_{2z}T$ requires
$V^{\eta,e_{Y},e^{\prime}_{Y}}_{\mathbf{k},\mathbf{k}^{\prime}}=(V^{\eta,-e_{Y},-e_{Y}^{\prime}}_{\mathbf{k},\mathbf{k}^{\prime}})^{*}$,
the matrix $\mathcal{V}^{+,--}_{\mathbf{k}\mathbf{k}^{\prime}}$ is just the
complex conjugate of that in Eq. (379). Therefore, we expect (1) The eigen-
equations (379) and (410) give the same $T_{c}$, and (2) the corresponding
eigen-vectors are related by
$\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k};--}\\\
\Delta^{-,SM}_{\mathbf{k};--}\end{array}\right)=\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k};++}\\\
\Delta^{-,SM}_{\mathbf{k};++}.\end{array}\right)^{*}$ up to an arbitrary phase
factor, which can always be chosen to be $1$. For the convenience of the
notation, we define

$\displaystyle\Psi^{SM}_{\mathbf{k},e_{Y}e^{\prime}_{Y}}=\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k};e_{Y}e^{\prime}_{Y}}\\\
\Delta^{-,SM}_{\mathbf{k};e_{Y}e^{\prime}_{Y}}\end{array}\right)$ (413)

so the above relation can be simplified as

$\displaystyle\Psi^{SM}_{\mathbf{k},--}=(\Psi^{SM}_{\mathbf{k},++})^{*}.$
(414)

Before we show the numerical results, we first consider some simplified
version for the analytical results. Let us consider the phonon-mediated el-el
interaction with the following form for the intra-Chern-band channel [46, 64]

$\displaystyle
V^{\eta,e_{Y},e_{Y}}_{\mathbf{k},\mathbf{k}^{\prime}}=U^{*}_{0,e_{Y},\mathbf{k}}U_{0,e_{Y},\mathbf{k}^{\prime}}+U^{*}_{1,e_{Y},\mathbf{k}}U_{1,e_{Y},\mathbf{k}^{\prime}};\;U_{0,e_{Y},\mathbf{k}}=\frac{\sqrt{A}b^{2}}{k^{2}+b^{2}};\;U_{1,e_{Y},\mathbf{k}}=\frac{\sqrt{V_{0}}}{k^{2}+b^{2}}k_{e_{Y}}^{2};\;\;k_{e_{Y}}=k_{x}+ie_{Y}k_{y},$
(415)

where $e_{Y}=\pm$ and $A$, $V_{0}$ and $b$ are material dependent parameters
with their values $A=0.44meV$, $V_{0}=0.4meV$ and $b=0.185k_{\theta}$ obtained
from the heavy fermion model for TBG. We note that $U_{0}$ is independent of
$e_{Y}$ index (for s-wave) but the form of $U_{1}$ critically depends on
$e_{Y}$ index (for d-wave). Both $U_{0}$ and $U_{1}$ are independent of the
valley $\eta$ index. The corresponding linearized gap equation for intra-
Chern-band channel is written as

$\displaystyle
2k_{B}T\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k},e_{Y}e_{Y}}\\\
\Delta^{-,SM}_{\mathbf{k},e_{Y}e_{Y}}\end{array}\right)=\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime},\alpha}U_{\alpha,e_{Y},\mathbf{k}^{\prime}}U^{*}_{\alpha,e_{Y},\mathbf{k}}\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k}^{\prime},e_{Y}e_{Y}}\\\
\Delta^{-,SM}_{\mathbf{k}^{\prime},e_{Y}e_{Y}}\end{array}\right),$ (422)

where $\alpha=0,1$. The above decomposition form of the interaction allows us
to define

$\tilde{\Delta}^{\eta,SM}_{\alpha,e_{Y}}=\frac{1}{N_{M}}\sum_{\mathbf{k}^{\prime}}U_{\alpha,e_{Y},\mathbf{k}^{\prime}}\Delta^{\eta,SM}_{\mathbf{k}^{\prime},e_{Y}e_{Y}}$
(423)

for the intra-Chern-band channel and simplify the gap equation as

$\displaystyle
2k_{B}T\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k},e_{Y}e_{Y}}\\\
\Delta^{-,SM}_{\mathbf{k},e_{Y}e_{Y}}\end{array}\right)=\sum_{\alpha}U^{*}_{\alpha,e_{Y},\mathbf{k}}\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)\left(\begin{array}[]{c}\tilde{\Delta}^{+,SM}_{\alpha,e_{Y}}\\\
\tilde{\Delta}^{-,SM}_{\alpha,e_{Y}}\end{array}\right).$ (430)

Now we can introduce

$\tilde{V}_{\beta\alpha,e_{Y}}=\frac{1}{N_{M}}\sum_{\mathbf{k}}U_{\beta,e_{Y},\mathbf{k}}U^{*}_{\alpha,e_{Y},\mathbf{k}},$
(431)

and transform the gap equation into an eigen-problem

$\displaystyle
2k_{B}T\left(\begin{array}[]{c}\tilde{\Delta}^{+,SM}_{\beta,e_{Y}}\\\
\tilde{\Delta}^{-,SM}_{\beta,e_{Y}}\end{array}\right)=\sum_{\alpha}\tilde{V}_{\beta\alpha,e_{Y}}\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)\left(\begin{array}[]{c}\tilde{\Delta}^{+,SM}_{\alpha,e_{Y}}\\\
\tilde{\Delta}^{-,SM}_{\alpha,e_{Y}}\end{array}\right).$ (438)

Crucially, we find that

$\tilde{V}_{01,-e_{Y}}=\tilde{V}_{10,e_{Y}}=\frac{1}{N_{M}}\sum_{\mathbf{k}}\frac{\sqrt{AV_{0}}b^{2}k_{e_{Y}}^{2}}{(k^{2}+b^{2})^{2}}=0$
(439)

after performing the angular integral of the momentum. Thus, two channels
(s-wave and d-wave) are decoupled,
$\tilde{V}_{\beta\alpha,e_{Y}}=\tilde{V}_{\alpha\alpha,e_{Y}}\delta_{\alpha\beta}$.
Furthermore, we find

$\tilde{V}_{00,e_{Y}}=\frac{1}{N_{M}}\sum_{\mathbf{k}}\frac{Ab^{4}}{(k^{2}+b^{2})^{2}},\;\;\tilde{V}_{11,e_{Y}}=\frac{1}{N_{M}}\sum_{\mathbf{k}}\frac{V_{0}k^{4}}{(k^{2}+b^{2})^{2}},$
(440)

both of which are independent of the parameter $e_{Y}$, so we drop the index
$e_{Y}$ for $\tilde{V}_{\alpha\alpha}$ ($\alpha=0,1$) below.

Now the gap equation becomes

$\displaystyle
2k_{B}T\left(\begin{array}[]{c}\tilde{\Delta}^{+,SM}_{\alpha,e_{Y}}\\\
\tilde{\Delta}^{-,SM}_{\alpha,e_{Y}}\end{array}\right)=\left(\begin{array}[]{cc}0&\tilde{V}_{\alpha\alpha}\\\
\tilde{V}_{\alpha\alpha}&0\end{array}\right)\left(\begin{array}[]{c}\tilde{\Delta}^{+,SM}_{\alpha,e_{Y}}\\\
\tilde{\Delta}^{-,SM}_{\alpha,e_{Y}}\end{array}\right),$ (447)

with $e_{Y}=\pm$ and $\alpha=0,1$ for s-wave and d-wave channels, separately.
This eigen-problem can be easily solved with two eigen-solutions
$\pm\frac{\tilde{V}_{\alpha\alpha}}{2}$, and we only take the positive value
one, namely

$k_{B}T_{c,\alpha}=\frac{\tilde{V}_{\alpha\alpha}}{2}.$ (448)

To get an estimate of $T_{c}$ in each channel, We can change the momentum
summation in Eq. (440) to the momentum integral with the cut-off $\sim
k_{\theta}$ and obtain

$\tilde{V}_{00}=\frac{S_{M}}{2\pi}\int_{0}^{k_{\theta}}kdk\frac{Ab^{4}}{(k^{2}+b^{2})^{2}},\;\;\tilde{V}_{11}=\frac{S_{M}}{2\pi}\int_{0}^{k_{\theta}}kdk\frac{V_{0}k^{4}}{(k^{2}+b^{2})^{2}}$
(449)

where $S_{M}=\frac{\sqrt{3}}{2}a_{M}^{2}$, $a_{M}$ is the Moire lattice
constant

Numerical integral over the momentum gives $\tilde{V}_{00}\approx 0.018meV$
and $\tilde{V}_{11}\approx 0.387meV$, which give rise to $k_{B}T_{c,0}\approx
0.009meV$ for s-wave channel and $k_{B}T_{c,1}\approx 0.193meV$ for the d-wave
channel. Based on this estimate, we conclude that the d-wave pairing channel
will win and give the critical temperature

$k_{B}T_{c}^{intra}=\frac{\tilde{V}_{11}}{2}\approx 0.193meV$ (450)

for the intra-Chern-band channel.

The corresponding normalized eigen-vector is
$(\tilde{\Delta}^{+,SM}_{+},\tilde{\Delta}^{-,SM}_{+})=\frac{1}{\sqrt{2}}(1,1)$.
With the normalized eigen-vector and Eq. (430), we have

$\displaystyle\Psi^{SM}_{\mathbf{k},++}=\left(\begin{array}[]{c}\Delta^{+,SM}_{\mathbf{k},++}\\\
\Delta^{-,SM}_{\mathbf{k},++}\end{array}\right)=\frac{\sqrt{V_{0}}}{\sqrt{2}\tilde{V}_{11}(k^{2}+b^{2})}(k_{x}-ik_{y})^{2}\left(\begin{array}[]{c}1\\\
1\end{array}\right).$ (455)

We also consider the gap equation (410), which should given the same $T_{c}$,
and the corresponding eigen-state is given by

$\displaystyle\Psi^{SM}_{\mathbf{k},--}=(\Psi^{SM}_{\mathbf{k},++})^{*}=\frac{\sqrt{V_{0}}}{\sqrt{2}\tilde{V}_{11}(k^{2}+b^{2})}(k_{x}+ik_{y})^{2}\left(\begin{array}[]{c}1\\\
1\end{array}\right).$ (458)

according to Eq. (414). It should be noted that both
$\Psi^{SM}_{\mathbf{k},++}$ and $\Psi^{SM}_{\mathbf{k},--}$ have even parity,
$\lambda_{C_{2z}}=1$, so they can only form spin-singlet pairing ($S,M=0,0$ in
the above gap function).

We emphasize that $\Psi^{SM}_{\mathbf{k},++}$ and $\Psi^{SM}_{\mathbf{k},--}$
are two independent degenerate pairing channels with the same $T_{c}$, and
they together form a 2D $E_{2}$ irreducible representation. To see that, we
need to consider the operator form of the gap function and define

$\displaystyle
H_{\Delta_{e_{Y}}}=\frac{1}{N_{M}}\sum_{\mathbf{k},s_{1},s_{2},\eta}\gamma_{\mathbf{k},e_{Y},\eta,s_{1}}^{\dagger}\Delta^{\eta,00}_{\mathbf{k};e_{Y}e_{Y}}(is_{y})_{s_{1}s_{2}}\gamma_{-\mathbf{k},e_{Y},-\eta,s_{2}}^{\dagger}$
$\displaystyle=\frac{\sqrt{2V_{0}}}{\tilde{V}_{0}}\sum_{\mathbf{k},s_{1},s_{2}}\frac{k_{-e_{Y}}^{2}}{k^{2}+b^{2}}\gamma_{\mathbf{k},e_{Y},+,s_{1}}^{\dagger}(is_{y})_{s_{1}s_{2}}\gamma_{-\mathbf{k},e_{Y},-,s_{2}}^{\dagger}$
(459)

where $e_{Y}=\pm$. Here $H_{\Delta_{e_{Y}}}$ only involves the creation
operator terms, so the full gap Hamiltonian should be
$H_{\Delta_{e_{Y}}}+H^{\dagger}_{\Delta_{e_{Y}}}$ to keep the Hamiltonian
hermitian. Now we want to check the transformation property of
$H_{\Delta_{e_{Y}}}$ under $C_{3z}$ and $C_{2x}$ (spinless). Since we are
dealing with the continuous model around $\mathbf{k}=0$ (excluding Moiré BZ
boundary), we should be able to choose the phase factors
$\theta(\mathbf{k})=\alpha(\mathbf{k})=0$ for $C_{3z}$ and $C_{2x}$ in Eq.
(VI.7), so $C_{3z}\gamma_{\mathbf{k},e_{Y}\eta
s}^{\dagger}C_{3z}^{-1}=\gamma_{C_{3}\mathbf{k},e_{Y}\eta s}^{\dagger}$ and
$C_{2x}\gamma_{\mathbf{k},e_{Y}\eta
s}^{\dagger}C_{2x}^{-1}=\gamma_{C_{2x}\mathbf{k},-e_{Y}\eta s}^{\dagger}$.
Then, one can show

$C_{3z}H_{\Delta_{e_{Y}}}C_{3z}^{-1}=e^{ie_{Y}\frac{2\pi}{3}}H_{\Delta_{e_{Y}}};C_{2x}H_{\Delta_{e_{Y}}}C_{2x}^{-1}=H_{\Delta_{-e_{Y}}}.$
(460)

We further define

$H_{\Delta_{1}}=H_{\Delta_{+}}+H_{\Delta_{-}};H_{\Delta_{2}}=i(H_{\Delta_{+}}-H_{\Delta_{-}})$
(461)

and on this basis $(H_{\Delta_{1}},H_{\Delta_{2}})$, we have

$C_{3z}H_{\Delta_{\alpha}}C_{3z}^{-1}=\sum_{\beta}H_{\Delta_{\beta}}(e^{-i\frac{2\pi}{3}\sigma_{y}})_{\beta\alpha};C_{2x}H_{\Delta_{\alpha}}C_{2x}^{-1}=\sum_{\beta}H_{\Delta_{\beta}}(\sigma_{z})_{\beta\alpha},$
(462)

where $\sigma$ is the Pauli matrix for two components of pairings. Since
$(H_{\Delta_{1}},H_{\Delta_{2}})$ is even under $C_{2z}$,
$\lambda_{C_{2z}}=1$, we conclude $(H_{\Delta_{1}},H_{\Delta_{2}})$ belongs to
the 2D $E_{2}$ irrep by comparing with the irrep table (Tab. 4). It should be
emphasized that $H_{\Delta_{\alpha}}$ ($\alpha=1,2$) cannot be written in the
$k_{x}^{2}-k_{y}^{2}$ or $k_{x}k_{y}$ form because of the Chern-band index.
For example, the explicit form of $H_{\Delta_{1}}$ is

$\displaystyle H_{\Delta_{1}}=H_{\Delta_{+}}+H_{\Delta_{-}}$
$\displaystyle=\frac{\sqrt{2V_{0}}}{\tilde{V}_{0}N_{M}}\sum_{\mathbf{k},s_{1},s_{2}}\frac{1}{k^{2}+b^{2}}(is_{y})_{s_{1}s_{2}}\left(k_{-}^{2}\gamma_{\mathbf{k},+,+,s_{1}}^{\dagger}\gamma_{-\mathbf{k},+,-,s_{2}}^{\dagger}+k_{+}^{2}\gamma_{\mathbf{k},-,+,s_{1}}^{\dagger}\gamma_{-\mathbf{k},-,-,s_{2}}^{\dagger}\right),$
(463)

and it is clear that we cannot make the summation over $k_{-}^{2}$ and
$k_{+}^{2}$ to get $k_{x}^{2}-k_{y}^{2}$ form.

The above analytical solutions fit well with our numerical results as
discussed below. The largest three eigen-values of the linearized gap
equations (Eqs. 379 and 386) are shown in Tab. 5. The largest eigen-value of
the intra-Chern-band channel correspond to $k_{B}T_{c}=0.16meV$, which is
quite close to the value of $k_{B}T_{c}\approx 0.19meV$ obtained from the
effective interaction discussed above.

Intra-Chern-band channel | 0.16 meV | 0.04 meV | 0.0002 meV
---|---|---|---
Inter-Chern-band channel | 0.21 meV | 0.008 meV | 0.0079 meV
Table 5:  $k_{B}T$ for the largest three eigen-values of the linearized gap
equations (Eqs. 379 and 386). The largest one determines $k_{B}T_{c}$.