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##  Appendix A Monopole operators and discrete symmetries

In this appendix we build the monopole operators $\Phi^{{\dagger}}$ in the
main text and study how the bare discrete symmetries of $\mathrm{QED}_{2+1}$
act on them. The symmetries in consideration are charge conjugation
$\mathcal{C}$, spatial reflection $\mathcal{R}$ and time-reversal
$\mathcal{T}$, where these symmetries act as in Sec. II.2 on fermions and we
have dropped the subscripts for the rest of the Appendix. The best way to
understand the definition of monopoles is in the large $N_{\\!f}$ limit, where
gauge fluctuations are weak and the states can be described by free fermions
and a Gauss law [29].

Consider $\mathrm{QED}_{2+1}$ with $N_{\\!f}=2n$ Dirac fermions , tlet
$\ket{q}$ be states corresponding to the $2\pi q$-flux inserted with all
negative modes filled. Following Ref. [29], using the operator-state
correspondence these states can be thought of as Dirac fermions living a
sphere with $2\pi q$ magnetic field. There will be $2N$ Dirac zero modes
$a_{\alpha,q}$ that are Lorentz scalars but transform as $\psi_{\alpha}$ under
$\mathrm{SU}(N_{\\!f})_{\mathrm{f}}$. Then the ground states of the free
fermion problem correspond to states

$\displaystyle\ket{q},\quad f^{{\dagger}\,\alpha_{1}}_{\,\,q}\ket{q},\quad
f^{{\dagger}\,\alpha_{1}}_{\,\,q}f^{{\dagger}\,\alpha_{2}}_{\,\,q}\ket{q},\quad\dots,\qquad
f^{{\dagger}\,\alpha_{1}}_{\,\,q}f^{{\dagger}\,\alpha_{2}}_{\,\,q}\dots
f^{{\dagger}\,\alpha_{2n}}_{\,\,q}\ket{q},$ (68)

where $f_{\beta q}\ket{q}=0$ for all $\beta$. It turns out that if require to
quantize the zero modes in a consistent $\mathcal{C}\mathcal{R}$ way, the
gauge invariant state will then correspond to states with $N$ of the zero
modes filled. There are $\binom{2n}{n}$ such states and they transform in the
self-conjugate fundamental representation of $\mathrm{SU}(2n)_{\mathrm{f}}$,
i.e., in the representation whose Young tableaux has one column and $n$
boxes141414Or in terms of weights, its heighest weight correspond to
$\varpi_{n}$.

Define $\mathfrak{M}^{{\dagger}\,[\alpha_{1}\dots\alpha_{n}]}_{q}$ as the
operator associated with the
state151515$\mathfrak{M}_{+1}^{{\dagger}\,[\cdot]}$ is $\mathcal{M}$ in the
main text.

$\displaystyle\ket{\mathfrak{M}^{{\dagger}\,[\alpha_{1}\dots\alpha_{n}]}_{q}}=f^{{\dagger}\alpha_{1}}_{q}\dots
f^{{\dagger}\alpha_{n}}_{q}\ket{q}.$ (69)

We can use the result of Ref. [54] to define a $\mathcal{C}\mathcal{T}$
transformation161616This is $\mathcal{T}$ in the notation of Ref. [54] .

$\displaystyle\mathcal{C}\mathcal{T}\cdot\mathfrak{M}^{{\dagger}\,A}_{q}\cdot\mathcal{C}\mathcal{T}^{-1}=(-1)^{\frac{n(n-1)}{2}}E_{AB}\mathfrak{M}^{{\dagger}\,B}_{q}.$
(70)

where we are using the conventions for the antisymmtric multiindices and
$E_{AB}$ in the main text. This operation satisfy
$\left(\mathcal{C}\mathcal{T}\right)^{2}=(-1)^{nM}$ when acting on monopoles.
The monopole operators $\Phi^{{\dagger}\,A}$ in the main text are the
Lorentzian version of $\mathfrak{M}^{{\dagger}\,A}_{+1}$.

Time reversal $\mathcal{T}$ flips the ${U}(1)_{{top}}$ charge without changing
the $\mathrm{SU}(N_{\\!f})_{\mathrm{f}}$ representation and if we assume
$\mathcal{T}^{2}=1$ on monopoles, we can impose

$\displaystyle\mathcal{T}\cdot\Phi^{{\dagger}\,A}\cdot\left(\mathcal{T}\right)^{-1}=\begin{cases}+E^{AB}\Phi_{\,B}&n=0\mod
2,\\\ +iE^{AB}\Phi_{\,B}&n=1\mod 2.\\\ \end{cases}$ (71)

$\hat{\mathcal{T}}=(-)^{M}\circ\mathcal{T}$ is also a time reversal that
squares to one on monopoles but the only difference is a unitary generated by
a local current so it is enough to only consider $\mathcal{T}$ for our
purposes.

Similarly for $\mathcal{C}\mathcal{T}$ we choose

$\displaystyle\mathcal{C}\mathcal{T}\cdot\Phi^{{\dagger}\,A}\cdot\left(\mathcal{C}\mathcal{T}\right)^{-1}=E_{AB}\Phi^{{\dagger}\,B}.$
(72)

Whatever phase we could have included can be absorbed in a redefinition of the
monopole operator.

Finally, $\mathcal{C}\mathcal{R}$ leaves the gauge flux fixed but acts as
conjugation in $\mathrm{SU}(N_{\\!f})_{\mathrm{f}}$. Assuming that
$(\mathcal{C}\mathcal{R})^{2}$ is a unitary generated by a local current, i.e.
it is a ${U}(1)_{{top}}$ element when restricted to monopoles, we have

$\displaystyle\mathcal{C}\mathcal{R}\cdot\Phi^{{\dagger}\,A}\cdot\left(\mathcal{C}\mathcal{R}\right)^{-1}=\begin{cases}+E_{AB}\Phi^{{\dagger}\,B}&n=0\mod
2,\\\ +iE_{AB}\Phi^{{\dagger}\,B}&n=1\mod 2.\\\ \end{cases}$ (73)

where again there was a phase ambiguity that corresponds to multiplying by
some ${U}(1)_{{top}}$ element so that $(\mathcal{C}\mathcal{R})^{2}=1$ on
monopoles.

Then we find that $\mathcal{C}$ and $\mathcal{R}$ act as

$\displaystyle\mathcal{C}\cdot\Phi^{{\dagger}\,A}\cdot\left(\mathcal{C}\right)^{-1}$
$\displaystyle=\begin{cases}+\Phi_{\,A}&n=0\mod 2,\\\ +i\Phi_{\,A}&n=1\mod
2.\\\ \end{cases}$ (74)
$\displaystyle\mathcal{R}\cdot\Phi^{{\dagger}\,A}\cdot\left(\mathcal{R}\right)^{-1}$
$\displaystyle=(-1)^{n}E^{AB}\Phi_{B}.$ (75)

The weird phases for odd $n$ were introduced to simplify actual calculations.
The calculations are usually easier done by diagonalizing $E_{AB}$ and write
the monopoles in this basis. Note that for $n$ even the representation is
orthogonal and for $n$ odd is simplectic, which is why is better to treat the
cases separately.

###  A.1  $N_{\\!f}=2n=0\mod{4}$

Let’s think of $E_{AB}$ as matrix acting on
$\bigwedge^{\frac{N_{\\!f}}{2}}\mathbb{C}^{N_{\\!f}}\cong\mathbb{C}^{\binom{N_{\\!f}}{N_{\\!f}/2}}$.
It is symmetric, has trace zero and squares to the identity. Then there are
some real orthogonal tensors $\hat{\chi}^{A}_{\rho}$ such that

$\begin{split}\hat{\chi}^{A}_{\rho}E_{AB}\hat{\chi}^{B}_{\sigma}&=\lambda_{\rho}\delta_{\rho\sigma}\\\
\hat{\chi}^{A}_{\rho}\hat{\chi}^{A}_{\sigma}&=\delta_{\rho\sigma}\\\
\hat{\chi}^{A}_{\rho}\hat{\chi}^{B}_{\rho}&=\hat{\delta}_{B}^{A}\\\
\end{split}$ (76)

for some numbers $\lambda_{\rho}=\pm 1$, where half of them are positive and
the other half are negative. We can then define
${\chi}^{A}_{\rho}=\sqrt{\lambda_{\rho}}\hat{\chi}^{A}_{\rho}$ so that

$\begin{split}{\chi}^{A}_{\rho}E_{AB}{\chi}^{B}_{\sigma}=\delta_{\rho\sigma}\\\
{\chi}^{A}_{\rho}\left({\chi}^{A}_{\sigma}\right)^{*}=\delta_{\rho\sigma}\\\
{\chi}^{A}_{\rho}\left({\chi}^{B}_{\rho}\right)^{*}=\hat{\delta}^{A}_{B}\end{split}$
(77)

This allow us to define

$\displaystyle\Phi^{{\dagger}}_{\rho}$
$\displaystyle\equiv\chi_{\rho}^{A}E_{AB}\Phi^{{\dagger},B}$ (78)
$\displaystyle\Phi_{\rho}\equiv\left(\Phi^{{\dagger}}_{\rho}\right)^{{\dagger}}$
$\displaystyle=(\chi_{\rho}^{A})^{*}E_{AB}\Phi_{,B}=(\chi_{\rho}^{A})\Phi_{,A}$
(79)

Then

$\begin{split}\mathcal{T}:\Phi^{{\dagger}}_{\rho}\longrightarrow&(O_{T})_{\rho\sigma}\Phi_{\sigma}\\\
\mathcal{R}:\Phi^{{\dagger}}_{\rho}\longrightarrow&\Phi_{\rho}\\\
\mathcal{C}\mathcal{T}:\Phi^{{\dagger}}_{\rho}\longrightarrow&\Phi^{{\dagger}}_{\rho}\\\
\mathcal{C}:\Phi^{{\dagger}}_{\rho}\longrightarrow&(O_{T})_{\rho\sigma}\Phi_{\sigma}\\\
(O_{T})_{\rho\sigma}=(\chi_{\rho}^{A})^{*}E_{AB}\chi_{\sigma}^{B}&=(\chi_{\rho}^{A}\chi_{\sigma}^{A})^{*}=\lambda_{\rho}\delta_{\rho\sigma}\end{split}$
(80)

where $\lambda_{\rho}$ are the eigenvalues $(\pm 1)$ of $E_{AB}$. If is useful
to note that

$\displaystyle O_{T}^{2}=\mathds{1}\quad\text{and}\quad O_{T}^{*}=O_{T}.$
(81a)

If an element $U\in\mathrm{SU}(N_{\\!f})_{\mathrm{f}}$ acts on Dirac fermions
as $\psi_{\alpha}\longrightarrow U_{a}^{\,\,\beta}\psi_{b}$, then
$\Phi^{({\dagger})}_{\rho}\longrightarrow
U_{\rho\sigma}\Phi^{({\dagger})}_{\sigma}$ with

$\displaystyle
U_{\rho\sigma}=\chi_{\rho}^{[\alpha_{1}\dots\alpha_{n}]}U_{\alpha_{1}}^{\,\,\beta_{1}}\cdots
U_{\alpha_{n}}^{\,\,\beta_{n}}\left(\chi_{\sigma}^{[\beta_{1}\dots\beta_{n}]}\right)^{*}.$
(82a)

We can check that $U^{\top}\cdot U=\mathds{1}$ and $U=U^{*}$. This shows
explicitly that the representation is orthogonal and real. We can also show
that $O_{T}\cdot U\cdot O_{T}$ corresponds to the element
$V\in\mathrm{SU}(N_{\\!f})_{\mathrm{f}}$ that acts on the fermions as
$V_{\alpha}^{\,\,\beta}=(U_{\alpha}^{\,\,\beta})^{*}$.

Sometimes it is useful to go to a basis of real (hermitian) monopoles as

$\displaystyle\Phi^{({\dagger})}_{\rho}$
$\displaystyle=\frac{1}{\sqrt{2}}\left(R_{\rho}\pm iI_{\rho}\right)$ (83a)

where the symmetry breaking is evident once the monopole is condensed

$\begin{split}\mathcal{T}:(R_{\rho},I_{\rho})&\longrightarrow\left(O_{T}\right)_{\rho\sigma}(R_{\sigma},I_{\sigma})\\\
\mathcal{C}\mathcal{T}:(R_{\rho},I_{\rho})&\longrightarrow(R_{\rho},-I_{\rho})\\\
\mathcal{R}:(R_{\rho},I_{\rho})&\longrightarrow(R_{\rho},-I_{\rho})\\\
\mathcal{C}:(R_{\rho},I_{\rho})&\longrightarrow\left(O_{T}\right)_{\rho\sigma}(R_{\sigma},-I_{\sigma})\end{split}$
(84)

It is customary to call $R$ and $I$ the real and imaginary part of the
monopole.

###  A.2  $N_{\\!f}=2\mod{4}$

Let’s think of $E_{AB}$ again as matrix acting on
$\mathbb{C}^{\binom{N_{\\!f}}{N_{\\!f}/2}}$. It is antisymmetric and squares
to minus the identity. Then there are some real orthogonal tensors
${\chi}^{A}_{\rho}$ such that

$\begin{split}{\chi}^{A}_{\rho}E_{AB}{\chi}^{B}_{\sigma}&=J_{\rho\sigma}\\\
{\chi}^{A}_{\rho}{\chi}^{A}_{\sigma}&=\delta_{\rho\sigma}\\\
{\chi}^{A}_{\rho}{\chi}^{B}_{\rho}&={\delta}_{B}^{A}\end{split}$ (85)

where $J_{\rho\sigma}$ is the canonical antisymmetric tensor, that in matrix
form takes the form of antidiagonal matrix and squares to
$-\delta_{\rho\sigma}$. Then define

$\displaystyle\Phi^{{\dagger}}_{\rho}$
$\displaystyle\equiv\chi_{\rho}^{A}E_{AB}\Phi^{{\dagger},B}$ (86)
$\displaystyle\Phi_{\rho}\equiv\left(\Phi^{{\dagger}}_{\rho}\right)^{{\dagger}}$
$\displaystyle=(\chi_{\rho}^{A})^{*}E_{AB}\Phi_{,B}=J_{\rho\sigma}\chi_{\sigma}^{A}\Phi_{,A}$
(87)

so that

$\begin{split}\mathcal{T}:\Phi^{{\dagger}}_{\rho}\longrightarrow&iJ_{\rho\sigma}\Phi_{\sigma}\\\
\mathcal{R}:\Phi^{{\dagger}}_{\rho}\longrightarrow&-J_{\rho\sigma}\Phi_{\sigma}\\\
\mathcal{C}\mathcal{T}:\Phi^{{\dagger}}_{\rho}\longrightarrow&J_{\rho\sigma}\Phi^{{\dagger}}_{\sigma}\end{split}$
(88)

The $\mathrm{SU}(N_{\\!f})_{\mathrm{f}}$ group acts on
$\Phi^{{\dagger}}_{\rho}$ by multiplication by

$\displaystyle
U_{\rho\sigma}=\chi_{\rho}^{A}U_{A}^{\,\,B}(\chi_{\sigma}^{B})^{*}.$ (89)

We now have $U^{\top}\cdot J\cdot U=J$
171717$J_{\alpha\beta}U_{\alpha\rho}U_{\beta\sigma}={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}J_{\alpha\beta}\chi_{\alpha}^{A}\chi_{\beta}^{B}}\chi_{\rho}^{C}\chi_{\sigma}^{D}U_{A}^{\,\,C}U_{B}^{\,\,D}={\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}E^{AB}U_{A}^{\,\,C}U_{B}^{\,\,D}}\chi_{\rho}^{C}\chi_{\sigma}^{D}={\color[rgb]{0,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.5,0}E^{CD}\chi_{\rho}^{C}\chi_{\sigma}^{D}}=J_{\rho\sigma}.$and
$U^{{\dagger}}\cdot U=\mathds{1}$. This shows explicitly that the
representation is symplectic.

In contrast to the previous section, we cannot define a real and imaginary
parts that transform nicely under $\mathrm{SU}(N_{\\!f})_{\mathrm{f}}$.
Instead, consider

$\displaystyle\bm{\alpha}_{\rho}=\begin{pmatrix}J_{\rho\sigma}\Phi_{\sigma}\\\
\Phi^{{\dagger}}_{\rho}\end{pmatrix},$ (90a)

so that $(\bm{\alpha}^{{\dagger}}_{\rho})^{\top}=\varepsilon
J_{\rho\sigma}\bm{\alpha}_{\sigma}$, where $\varepsilon=i\tau^{2}$ and
$\tau^{a}$ are Pauli matrices acting on the components of
$\bm{\alpha}_{\rho}$. Then

$\begin{split}\mathcal{T}:\bm{\alpha}_{\rho}&\longrightarrow
i\tau^{1}\bm{\alpha}_{\rho}\\\
\mathcal{R}:\bm{\alpha}_{\rho}&\longrightarrow\tau^{1}J_{\rho\sigma}\bm{\alpha}_{\sigma}\\\
\mathcal{C}\mathcal{T}:\bm{\alpha}_{\rho}&\longrightarrow
J_{\rho\sigma}\bm{\alpha}_{\sigma}\\\
{U}(1)_{{top}}:\bm{\alpha}_{\rho}&\longrightarrow
e^{-i\theta\tau^{3}}\bm{\alpha}_{\rho}\\\
\mathrm{SU}(N_{\\!f})_{\mathrm{f}}:\bm{\alpha}_{\rho}&\longrightarrow
U_{\rho\sigma}\bm{\alpha}_{\sigma}\end{split}\quad.$ (91)

##  Appendix B Symmetry properties from zero modes – the $N_{f}=8$ case on
honeycomb lattice

Following the discussion in Sec II.4, we find that the $\mathrm{SU}(8)$
representation of the monopoles $\mathbf{70}_{8}$ breaks into
$\mathrm{SU}(4)_{s}\times\mathrm{SU}(2)_{v}$ representations as

$\bf{70}_{8}\longrightarrow(\bf{1}_{4},\bf{5}_{2})\oplus(\bf{15}_{4},\bf{3}_{2})\oplus(\bf{20}_{4},\mathbf{1}_{2})$
(92)

The $\mathrm{SU}(4)_{s}$ representations can be understood by the
identification $\mathrm{SU}(4)/\mathds{Z}_{2}=\mathrm{SO}(6)$.
$\mathbf{1}_{4}$, $\mathbf{15}_{4}$ and $\mathbf{20}_{4}$ are the trace
(scalar), antisymmetric part (adjoint) and traceless symmetric part of an
$\mathrm{SO}(6)$ matrix obtained by taking the tensor product of two vectors.
For the $\mathrm{SU}(2)$ case, we can similarly use the identification
$\mathrm{SU}(2)/\mathds{Z}_{2}=\mathrm{SO}(3)$.

Consider $\mathrm{SU}(4)$ tensors with two indices that transform as an
$\mathrm{SO}(6)$ vector

$(s_{\mathfrak{a}})=\left(\mathds{J}\sigma^{1},\mathds{J}\sigma^{2},\mathds{J}\sigma^{3},i\mathds{J}\tau^{1},i\mathds{J}\tau^{2},i\mathds{J}\tau^{3}\right)$
(93)

where $\mathds{J}=i^{2}\sigma^{2}\tau^{2}$ and $\bar{a}=1,\dots,6$ is an
$\mathrm{SO}(6)$ index. The factor of $i$ in the last three components is
important to make $s_{\mathfrak{a}}$ transform as an $\mathrm{SO}(6)$ vector
under $s_{\mathfrak{a}}\rightarrow
U^{\top}s_{\mathfrak{a}}U=R_{\mathfrak{a}}^{\,\,\mathfrak{b}}s_{\mathfrak{b}}$,
with $R\in\mathrm{SO}(6)$. Note that under the partial action of time reversal
$\mathds{J}^{\top}\cdot
s_{\mathfrak{a}}^{*}\cdot\mathds{J}=-\eta_{\mathfrak{a}}^{\mathcal{T}}s_{\mathfrak{a}}$
where $\eta^{\mathcal{T}}=(+1,+1,+1,-1,-1,-1)$.

Similarly for the $\mathrm{SU}(2)_{v}$ factor, we introduce

$(v_{\mathfrak{i}})=\left(\varepsilon_{\mu}\mu^{1},\varepsilon_{\mu}\mu^{2},\varepsilon_{\mu}\mu^{3}\right)$
(94)

with $\varepsilon_{\mu}=i\mu^{2}$ and $\bar{i}=1,2,3$ is an $\mathrm{SO}(3)$
index. Under partial time reversal $\varepsilon_{\mu}^{\top}\cdot
v_{\mathfrak{i}}^{*}\cdot\varepsilon_{\mu}=-v_{\mathfrak{i}}$.

Next we consider the tensors
$S_{\mathfrak{a}\mathfrak{b}}=s_{\mathfrak{a}}\otimes s_{\mathfrak{b}}$ and
$V_{\mathfrak{i}\mathfrak{j}}=v_{\mathfrak{i}}\otimes v_{\mathfrak{j}}$. The
monopole wave-functions are then obtained by antisymmetrizing the tensor
product of the appropriate components (trace, symmetric or antisymmetric
components) of $S_{\mathfrak{a}\mathfrak{b}}$ and
$V_{\mathfrak{i}\mathfrak{j}}$. We also need to introduce an extra
normalization factors:

$\begin{split}\chi_{\mathfrak{i}\mathfrak{j}}^{(\mathcal{V})}&=\mathbb{c}_{\mathcal{V}}\mathcal{A}\left[S_{\mathfrak{a}\mathfrak{a}}\otimes
V_{\langle\mathfrak{i}\mathfrak{j}\rangle}\right]\\\
\chi_{\mathfrak{a}\mathfrak{b},\mathfrak{i}\mathfrak{j}}^{(\mathcal{N})}&=i\mathbb{c}_{\mathcal{N}}\mathcal{A}\left[S_{[\mathfrak{a}\mathfrak{b}]}\otimes
V_{[\mathfrak{i}\mathfrak{j}]}\right]\\\
\chi_{\mathfrak{a}\mathfrak{b}}^{(\mathcal{Q})}&=\mathbb{c}_{\mathcal{Q}}\mathcal{A}\left[S_{\langle\mathfrak{a}\mathfrak{b}\rangle}\otimes
V_{\mathfrak{i}\mathfrak{i}}\right]\end{split}$ (95)

with $\mathbb{c}_{\mathcal{V}/\mathcal{N}/\mathcal{Q}}$ real numbers. In the
above expression, $[\dots]$ denotes antisymmetrization of indices and
$\langle\dots\rangle$ corresponds to the traceless symmetric combination of
indices. The factor of $i$ in $\chi^{(\mathcal{N})}$ is important to make the
representation explicitly real and to make
$(\chi_{\rho}^{A})^{*}E^{AB}=\chi_{\rho}^{B}$. From here we point out that the
time-reversal matrix becomes
$\left(O_{T}\right)_{\rho\sigma}=\pm\mathcal{T}^{\prime}[\chi_{\rho}^{A}](\chi_{\sigma}^{A})^{*}$
(with sum over A), where $\mathcal{T}^{\prime}$ is the partial action of time-
reversal we discussed previously in the section. We have
$\mathcal{T}^{\prime}[\chi_{\mathfrak{i}\mathfrak{j}}^{(\mathcal{V})}]=\chi_{\mathfrak{i}\mathfrak{j}}^{(\mathcal{V})}$,
$\mathcal{T}^{\prime}[\chi_{\mathfrak{a}\mathfrak{b},\mathfrak{i}\mathfrak{j}}^{(\mathcal{N})}]=-\eta^{\mathcal{T}}_{\mathfrak{a}}\eta^{\mathcal{T}}_{\mathfrak{b}}\chi_{\mathfrak{a}\mathfrak{b},\mathfrak{i}\mathfrak{j}}^{(\mathcal{N})}$
181818The extra minus sign comes from the $i$ in the normalization. and
$\mathcal{T}^{\prime}[\chi_{\mathfrak{a}\mathfrak{b}}^{(\mathcal{Q})}]=\eta^{\mathcal{T}}_{\mathfrak{a}}\eta^{\mathcal{T}}_{\mathfrak{b}}\chi_{\mathfrak{a}\mathfrak{b}}^{(\mathcal{Q})}$.
Lastly, we make a change of labels, $[\mathfrak{a}\mathfrak{b}]\rightarrow A$,
for the $\mathcal{N}$ monopoles to match the notation in the main text. Under
this change of basis
$-\eta^{T}_{\mathfrak{a}}\eta^{T}_{\mathfrak{b}}\rightarrow(O_{\mathcal{T}})_{AA}$.
To see this, note that the adjoint indices that get a minus sign are the ones
that rotate only either $\vec{\sigma}$ or $\vec{\tau}$. This exactly
corresponds to the elements where
$\eta^{\mathcal{T}}_{\mathfrak{a}}=\eta^{\mathcal{T}}_{\mathfrak{b}}$.

As mentioned in the main text, the $\mathfrak{i}$ index transforms as $F_{2}$.
It is then easy to see that $V_{\mathfrak{i}\mathfrak{i}}$ transform as
$A_{1}$ and $V_{[\mathfrak{i}\mathfrak{j}]}$ transforms as $F_{2}$. The
tensors $V_{\langle\mathfrak{i}\mathfrak{j}\rangle}$ go into two different
representations: the off-diagonal terms ($\mathfrak{i}\neq\mathfrak{j}$)
transform still as $F_{2}$ while the diagonal terms transform as $E_{2}$. More
explicitly, if we define

$\displaystyle V_{0}$ $\displaystyle=\frac{V_{11}+V_{22}+V_{33}}{2\sqrt{3}},$
$\displaystyle\quad V_{\mathfrak{i}}$
$\displaystyle=\frac{1}{2\sqrt{2}}\absolutevalue{\varepsilon_{\mathfrak{i}\mathfrak{j}\mathfrak{k}}}V_{\mathfrak{j}\mathfrak{k}},$
(96) $\displaystyle V_{\pm}$ $\displaystyle=\frac{V_{11}+\omega^{\pm
1}V_{22}+\omega^{\pm 2}V_{33}}{2\sqrt{3}},$ $\displaystyle\quad
V_{\mathfrak{i}A}$
$\displaystyle=\frac{1}{2\sqrt{2}}\varepsilon_{\mathfrak{i}\mathfrak{j}\mathfrak{k}}V_{\mathfrak{j}\mathfrak{k}},$

the transformation properties under $\mathrm{SU}(2)_{v}$ and
$C_{6v}^{{}^{\prime\prime\prime}}$ are summarized in Tab. 3. As mentioned in
the main text, it is not necessary to build the explicit linear combinations
that transform under a certain $C_{6v}^{{}^{\prime\prime\prime}}$ irreps. We
can directly obtain the irreps by using the fact that restricting
representations to subgroups commute with tensor product operations of irreps.
Using Tab. 5, we see that as $\mathbf{3}\rightarrow F_{2}$ and
$\mathbf{1}\rightarrow 1$ then

$\begin{split}\mathbf{5}=\left(\bigvee\mathbf{3}\right)\ominus\mathbf{1}\longrightarrow&\left(\bigvee
F_{2}\right)\ominus A_{1}\\\ \longrightarrow&\left(A_{1}\oplus E_{2}\oplus
F_{1}\right)\ominus A_{1}\\\ \longrightarrow&E_{2}\oplus F_{1}\end{split}$
(97)

as we showed using the explicit construction.

$\mathrm{R}_{SU(2)_{\mathrm{v}}}$ | $\mathrm{R}_{C_{6v}^{{}^{\prime\prime\prime}}}$ | Basis | $T_{1}$ | $T_{2}$ | $R_{y}$ | $C_{6h}$
---|---|---|---|---|---|---
$\mathbf{1}$ | $A_{1}$ | $V_{0}$ | $\mathds{I}$ | $\mathds{I}$ | $\mathds{I}$ | $\mathds{I}$
$\mathbf{3}$ | $F_{2}$ | $(V_{{1}A},V_{{2}A},V_{{3}A})^{\top}$ | $\left(\begin{smallmatrix}1&0&0\\\ 0&-1&0\\\ 0&0&-1\\\ \end{smallmatrix}\right)$ |  $\left(\begin{smallmatrix}-1&0&0\\\ 0&-1&0\\\ 0&0&1\\\ \end{smallmatrix}\right)$ | $-\left(\begin{smallmatrix}1&0&0\\\ 0&0&1\\\ 0&1&0\\\ \end{smallmatrix}\right)$ | $\left(\begin{smallmatrix}0&0&1\\\ 1&0&0\\\ 0&1&0\\\ \end{smallmatrix}\right)$
| $F_{1}$ | $(V_{{1}S},V_{{2}S},V_{{3}S})^{\top}$ | $\left(\begin{smallmatrix}1&0&0\\\ 0&-1&0\\\ 0&0&-1\\\ \end{smallmatrix}\right)$ | $\left(\begin{smallmatrix}-1&0&0\\\ 0&-1&0\\\ 0&0&1\\\ \end{smallmatrix}\right)$ | $+\left(\begin{smallmatrix}1&0&0\\\ 0&0&1\\\ 0&1&0\\\ \end{smallmatrix}\right)$ | $\left(\begin{smallmatrix}0&0&1\\\ 1&0&0\\\ 0&1&0\\\ \end{smallmatrix}\right)$
$\mathbf{5}$ | $E_{2}$ | $(V_{+},V_{-})^{\top}$ | $\mathds{I}$ | $\mathds{I}$ | $\left(\begin{smallmatrix}0&1\\\ 1&0\\\ \end{smallmatrix}\right)$ | $\left(\begin{smallmatrix}\omega&0\\\ 0&\omega^{2}\\\ \end{smallmatrix}\right)$

Table 3: Summary of the symmetry properties of the
$V_{\mathfrak{i}\mathfrak{j}}$ tensors under $\mathrm{SU}(2)_{\mathrm{v}}$ and
the extended point group $C_{6v}^{{}^{\prime\prime\prime\prime}}$. We denote
the valley irreps ($\mathrm{R}_{\mathrm{SU}(2)_{\mathrm{v}}}$) by their
dimension. The point group irreps
($\mathrm{R}_{C_{6v}^{{}^{\prime\prime\prime}}}$) are denote by the usual
symbols explained in Sec. C.3.2. The symmetry operation
$g\in\\{T_{1},T_{2},R_{y},C_{6h}\\}$

act on the vectors in the basis column by left multiplication.

##  Appendix C Representation theory of space groups

In this section we spell out some details about the representation theory of
extended point grounds used in the main text, focusing mainly in the extension
of $C_{6v}$ to $C_{6v}^{{}^{\prime\prime\prime}}$. We will only talk about
_linear_ representations as we only attempt to classify the representations of
gauge invariant operators, which should transform linearly under the
symmetries of the system. We restrict to unitary representations and leave
time reversal on its own.

###  C.1 Finite groups

Recall that the representation theory of a finite group boils down to find all
the conjugacy classes and then construct the character table. A character of a
representation $R$ is a function from the group ($G$) to the complex numbers
given by $\chi_{R}(g)=\frac{1}{\dim{R}}\tr(\pi_{R}(g))$ where $\pi_{R}(g)$ is
the matrix that represents the element $g$ in the representation $R$. From the
definition it is clear that if $g^{\prime}=h\cdot g\cdot h^{-1}$, then
$\chi_{R}(g)=\chi_{R}(g^{\prime})$, i.e. the characters are constants on
conjugacy classes. We can define an inner product between characters given by
$(\chi_{1},\chi_{2})=\frac{1}{\absolutevalue{G}}\sum_{g\in
G}\bar{\chi}_{1}(g)\chi_{2}(g)$. It turns out that the characters of
irreducible representations (irreps) are orthogonal under this product and
form a basis for characters. There are as many irreps as conjugacy classes and
a irrep is fully classified once we know the the value of its character on
each conjugacy class. Which is why most of the information of the
representation thoery is summarized on a table which gives a table with a list
of the irreps with the value of its character next to it.

Representations can also be combined to form new representations by taking
direct sums or tensor products. It is useful to have a rule about how to
combine irreps under tensor products as this will specify how to write tensor
products of any representations. Furthermore, we can define symmetric square
$\bigvee R$ and antisymmetric square $\bigwedge R$ of a representation $R$ by
taking the symmetric and antisymmetric elements of the tensor product
$R\otimes R$.

The rules to take the product of representations can be figure out by noting
that $\chi_{R_{1}\otimes R_{2}}(g)=\chi_{R_{1}}(g)\chi_{R_{2}}(g)$ and then
taking inner products with respect to the irreps. For the symmetric and
antisymmetric squares of a representations, we can use $\chi_{\bigvee
R}(g)=\frac{1}{2}\left(\chi_{R}(g)^{2}+\chi_{R}(g^{2})\right)$ and
$\chi_{\bigwedge
R}(g)=\frac{1}{2}\left(\chi_{R}(g)^{2}-\chi_{R}(g^{2})\right)$. The last
relations imply that

$\begin{split}\bigvee(R_{1}\oplus R_{2})&=\bigvee R_{1}\oplus\bigvee
R_{2}\oplus(R_{1}\otimes R_{2}),\\\ \bigwedge(R_{1}\oplus R_{2})&=\bigwedge
R_{1}\oplus\bigwedge R_{2}\oplus(R_{1}\otimes R_{2}),\end{split}$ (98)

and

$\begin{split}\bigvee(R_{1}\otimes R_{2})&=(\bigvee R_{1}\otimes\bigvee
R_{2})\oplus(\bigwedge R_{1}\otimes\bigwedge R_{2}),\\\ \bigwedge(R_{1}\oplus
R_{2})&=(\bigvee R_{1}\otimes\bigwedge R_{2})\oplus(\bigvee
R_{1}\otimes\bigwedge R_{2}).\end{split}$ (99)

###  C.2 Space groups

Often we want to classify how objects transform under the symmetries of the
underlying lattice which is encoded in the space group. The space group we
encounter is often a semdirect prodcut of the point group $G_{\text{pt}}$
(symmetries of the unit cell) and translations ($\cong\mathbb{Z}^{2}$). This
group is _not_ a finite group so we cannot directly apply the techniques of
the previous section. Nevertheless, the action of the space group on the
objects we care about is such that some combinations of translations act
trivially. This allow us to study the representation theory of
$G_{\text{pt}}^{(\Lambda)}=(G_{\text{pt}}\ltimes\mathbb{Z}^{2})/(\Lambda)$
where $\Lambda$ is a set of translations that is closed under composition and
conjugation by the point group. This condition ensures that the quotient is
well defined. Notice that if $\Lambda^{\prime}\subset\Lambda$, then
$G_{\text{pt}}^{(\Lambda^{\prime})}$ is isomorphic to
$G_{\text{pt}}^{(\Lambda)}\ltimes(\Lambda/\Lambda^{\prime})$ and thus the
irreps $G_{\text{pt}}^{(\Lambda)}$ are also irreps of
$G_{\text{pt}}^{(\Lambda^{\prime})}$.

We can understand $G_{\text{pt}}^{(\Lambda)}$ as the point group of a super
unit cell that depends on $\Lambda$. This groups are thus often called the
extended point groups. Another way to understand the group quotient is that we
are imposing relations among the translations we are adding.

###  C.3  The $C_{6v}\ltimes\mathbb{Z}^{2}$ case

The triangular, Kagome and honeycomb lattice share the same space group, the
difference is the location of the sites and the edges of the lattices. The
point group is called $C_{6v}$. As our main interest on the main text is on
the honeycomb lattice, we will have in mind said lattice in the examples
below. We denote the extended point group $C_{6v}^{(\Lambda)}$ by
$C_{6v}^{(\absolutevalue{\Lambda})}$ where $\absolutevalue{\Lambda}$ is the
length of the smallest unit vector of $\Lambda$191919At least for the cases
considered in this appendix there is no ambiguity in this notation..

####  C.3.1  $C_{6v}$

$C_{6v}$ is generated by a reflection ($\sigma$)202020This is $R_{y}$ in the
main text. and a $\pi/3$ rotation ($C_{6}$) around the hexagon center. They
satisfy $\sigma^{2}=C_{6}^{6}=(\sigma C_{6})^{2}=e$, where $e$ is the identity
element. There are $6$ conjugacy classes represented by
$\\{e,C_{6},C_{3}=C^{2}_{6},C_{2}=C^{3}_{6},\sigma,\sigma\cdot C_{6}\\}$. The
character table for the 6 irreps is shown in Tab. 4. The one dimensional
representations are actually representations of the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ subgroup generated by reflections
$\sigma$ and inversion $C_{2}=C_{6}^{3}$. The irreps are generated by $A_{2}$
and $B_{1}$. The two dimensional representations correspond to objects with
lattice angular momentum: $E_{l}$ get a $\exp(\pm\frac{2\pi i}{6})l$ under
$C_{6}$. The product rules are simple: all the one dimensional irreps square
to the trivial irrep ($A_{1}$), while $A_{2}\otimes B_{1}=B_{2}$,
$E_{2}=B_{1}\otimes E_{1}$ and $\bigvee E_{1}=A_{1}\oplus E_{2}$ and
$\bigwedge E_{1}=A_{2}$.

$\begin{array}[]{|c||c|c|c|c|c|c|}\hline\cr\framebox{$\left[g\right]$}&\framebox{$\left[e\right]$}&\framebox{$\left[C_{6}\right]$}&\framebox{$\left[C_{6}^{2}\right]$}&\framebox{$\left[C_{6}^{3}\right]$}&\framebox{$\left[\sigma\right]$}&\framebox{$\left[\sigma\cdot{C_{6}}\right]$}\\\
\hline\cr\hline\cr D_{[g]}&1&2&2&1&3&3\\\ \hline\cr\hline\cr
A_{1}&1&1&1&1&1&1\\\ \hline\cr A_{2}&1&1&1&1&-1&-1\\\ \hline\cr
B_{1}&1&-1&1&-1&1&-1\\\ \hline\cr B_{2}&1&-1&1&-1&-1&1\\\ \hline\cr
E_{1}&2&1&-1&-2&0&0\\\ \hline\cr E_{2}&2&-1&-1&2&0&0\\\ \hline\cr\end{array}$
Table 4: Character table for the point group $C_{6v}$. $[g]$ denote the
conjugacy class and $D_{[g]}$ denote its dimension.

####  C.3.2 $C_{6v}^{(2)}\left(C_{6v}^{{}^{\prime\prime\prime}}\right)$

The group $C_{6v}^{(2)}$ is what we called $C_{6v}^{{}^{\prime\prime\prime}}$
in the main text. The quotient corresponds to
$\Lambda=\left(2\mathbb{Z}\right)^{2}$ which correspond to the constrains
$T_{1}^{2}=T_{2}^{2}=e$. We have four new conjugacy classes represented by
$T_{1}$, $T_{1}\cdot C_{2}$, $\sigma\cdot T_{1}$ and $\sigma\cdot C_{6}\cdot
T_{1}$. The four new irreps are three dimensional and correspond to objects
with momentum at the $\bm{M}$ points (the middle points of the edges of the
hexagonal BZ) with different behavior under $C_{2}$ and $\sigma$ \- they can
be obtained from the other three by tensoring with the one dimensional irreps
of $C_{6v}$:

$\begin{split}F_{2}&=A_{2}\otimes F_{1}\\\ F_{3}&=B_{1}\otimes F_{1}\\\
F_{4}&=B_{2}\otimes F_{1}\end{split}$ (100)

The character table for the new irreps of $C_{6v}^{(1,0)}$ is shown in Tab. 6.
The product rules for the representations can be obtained from Tab. 5 and the
expressions of the other irreps in terms of "fundamental" irreps.

$\begin{array}[]{|c||c|c|c|c|c|c|}\hline\cr R&R\otimes A_{2}&R\otimes
B_{1}&R\otimes E_{1}&R\otimes F_{1}&\bigvee R&\bigwedge R\\\ \hline\cr
A_{2}&A_{1}&B_{2}&E_{1}&F_{2}&A_{1}&\cdot\\\ \hline\cr
B_{1}&B_{2}&A_{1}&E_{2}&F_{3}&A_{1}&\cdot\\\ \hline\cr
E_{1}&E_{1}&E_{2}&A_{1}\oplus A_{2}\oplus E_{2}&F_{3}\oplus F_{4}&A_{1}\oplus
E_{2}&A_{2}\\\ \hline\cr F_{1}&F_{2}&F_{3}&F_{3}\oplus F_{4}&A_{1}\oplus
E_{2}\oplus F_{1}\oplus F_{2}&A_{1}\oplus E_{2}\oplus F_{1}&F_{2}\\\
\hline\cr\end{array}$ Table 5: Rules for the tensor products of
representations of $C_{6v}^{{}^{\prime\prime\prime}}$. The last two columns
denote the symmetric and antisymetric squares of the representations.
$\begin{array}[]{|c||c|c|c|c|c|c|c|c|c|c|}\hline\cr\framebox{$[g]$}&\framebox{$[e]$}&\framebox{$[C_{6}]$}&\framebox{$[C_{6}^{2}]$}&\framebox{$[C_{6}^{3}]$}&\framebox{$[\sigma]$}&\framebox{$[\sigma\cdot{C_{6}}]$}&\framebox{$[T_{1}]$}&\framebox{$[C_{6}^{3}\cdot{T_{1}}]$}&\framebox{$[\sigma\cdot
T_{2}]$}&\framebox{$[\sigma\cdot C_{6}\cdot T_{1}]$}\\\
\hline\cr\hline\cr\framebox{$D_{[g]}$}&1&8&8&1&6&6&3&3&6&6\\\
\hline\cr\hline\cr A_{1}&1&1&1&1&1&1&1&1&1&1\\\ \hline\cr
A_{2}&1&1&1&1&-1&-1&1&1&-1&-1\\\ \hline\cr B_{1}&1&-1&1&-1&1&-1&1&-1&1&-1\\\
\hline\cr B_{2}&1&-1&1&-1&-1&1&1&-1&-1&1\\\ \hline\cr
E_{1}&2&1&-1&-2&0&0&2&-2&0&0\\\ \hline\cr E_{2}&2&-1&-1&2&0&0&2&2&0&0\\\
\hline\cr\hline\cr F_{1}&3&0&0&3&1&1&-1&-1&-1&-1\\\ \hline\cr
F_{2}&3&0&0&3&-1&-1&-1&-1&1&1\\\ \hline\cr F_{3}&3&0&0&-3&1&-1&-1&1&-1&1\\\
\hline\cr F_{4}&3&0&0&-3&-1&1&-1&1&1&-1\\\ \hline\cr\end{array}$ Table 6:
Character table for the extended point group
$C_{6v}^{{}^{\prime\prime\prime}}$. $[g]$ denote the conjugacy class and
$D_{[g]}$ denote its dimension.

##  Appendix D Mean field, Projective symmetry group and low-energy basis for
the $\pi$ flux state on the honeycomb lattice

###  D.1 Mean field details

The mean field Hamiltonian in Eq. 21 in momentum space becomes (we are going
to omit spin in this Appendix and set $t=1$)

$\begin{split}H_{MF}&=\sum_{\bm{k}}f^{{\dagger}}_{\bm{k}+\bm{Q}_{1}A}f_{\bm{k}B}+(e^{i\bm{k}\cdot\bm{a}_{2}}+e^{i\bm{k}\cdot\bm{a}_{3}})f^{{\dagger}}_{\bm{k}A}f_{\bm{k}B}+h.c.\end{split}$
(101)

where
$f_{\bm{k}u}=\frac{1}{\sqrt{N_{cell}}}\sum_{\bm{r}}e^{-i\bm{k}\cdot\bm{r}}f_{\bm{r}u}$
and $u=A,B$. As $H_{MF}$ breaks translation by $\bm{a}_{1}$, it mixes partons
with momenta $\bm{k}$ and $\bm{k}+\bm{Q}_{1}$. Here $\bm{Q}_{i}=\bm{G}_{i}/2$
where $\bm{G}_{i}$ are the reciprocal lattice vectors. It is convinient to
define a multicomponent fermion operator $F_{\bm{k}}$ by
$F_{\bm{k},un_{x}n_{y}}\equiv f_{\bm{k}+n_{x}\bm{Q}_{x}+n_{y}\bm{Q}_{y},u}$
where $n_{x},n_{y}\in\\{0,1\\}$ and restrict $\bm{k}$ to $\frac{1}{2}BZ$ which
is the reduced BZ shown in Fig. 4(a).

We introduce three sets of Pauli matrices. The first set $\nu^{i}$ will act on
the sublattice index $u$ while the last two, $\mu^{i}$ and $\rho^{i}$, act on
the $n_{x}$ and $n_{y}$ indices, respectively. $H_{MF}$ becomes

$\begin{split}H_{MF}&=\sum_{\mathbf{k}\in\frac{1}{2}BZ}F_{\bm{k}}^{{\dagger}}\mathfrak{h}_{\bm{k}}F_{\bm{k}}\\\
\mathfrak{h}_{\bm{k}}&=\nu^{+}\left(\mu^{1}+\rho^{3}e^{ik_{2}}+\mu^{3}\rho^{3}e^{ik_{3}}\right)+\text{h.c.}\end{split}$
(102)

where $k_{i}\equiv\bm{k}\cdot\bm{a}_{i}$. The eigenvalues of
$\mathfrak{h}_{\bm{k}}$ can be easily found by first noting that $\rho^{3}$
commutes with it so all the eigenvalues are at least double degenerated.
Second, note that
$\mathfrak{h}_{\bm{k}}^{2}=3+2\left(\cos(k_{1})\mu^{3}+\cos(k_{2})\rho^{3}\mu^{1}-\sin(k_{3})\nu^{3}\mu^{2}\rho^{3}\right)$.
The last three matrices anticommute with each other which implies that the
eigenvalues are

$\displaystyle
E_{\bm{k},s_{1}s_{2}}=s_{1}\sqrt{3+2s_{2}\sqrt{\cos(k_{1})^{2}+\cos(k_{2})^{2}+\sin(k_{1}-k_{2})^{2}}}\quad,s_{1},s_{2}\in\\{+1,-1\\}.$
(103)

Note that band touchings occur when the argument of the inner square root is
either zero or $(3/2)^{2}$. The first option happens for
$\bm{k}=\frac{\bm{Q}_{1}+\bm{Q}_{2}}{2}\equiv\bar{\bm{\Gamma}}$ and the band
touching occur at $\frac{2\pm 1}{4}$ filling. While the second option happens
for $\bm{k}=\pm\frac{\bm{G}_{1}-\bm{G}_{2}}{12}\equiv\bar{\bm{K}}$ at half
filling.

(a)

(b)

Figure 4:  (a) Contour plot for the first band above half-filling. The blue
curve denote a boundary for the BZ while the green curve is the boundary of
the reduced BZ. The Dirac points at half-filling correspond to the momenta
labeled by $1$ and $2$ while the momentum $3$ corresponds to the Dirac points
at quarter filling. (b) Band structure for the $\pi$-flux mean field
Hamiltonian centered around to the $\Gamma$ the center of the reduced BZ.

####  D.1.1  Quantum Hall mass

In this section we explore the Hall mass for the $\pi$-flux Hamiltonian, i.e.
a hopping pattern that breaks time reversal and reflections but preserve their
composition together with translations and rotations.

The idea here is to introduce imaginary second-nearest hoppings such that
there is a $\pi/2$ half flux around triangles with two nearest neighbor bonds
and one second-nearest neighbor bond. This can be achieved by a term

$H^{\prime}=\frac{i\chi}{2}\sum_{\langle\langle
i,j\rangle\rangle}(-1)^{s_{ij}}f_{i}^{{\dagger}}f_{j}$ (104)

with

$\begin{split}s_{\bm{r},\bm{r}+\bm{a}_{1}}^{(u)}&=u+1,\\\
s_{\bm{r},\bm{r}+\bm{a}_{2}}^{(u)}&=r_{1}+u,\\\
s_{\bm{r},\bm{r}+\bm{a}_{3}}^{(u)}&=r_{1},\end{split}$ (105)

$s_{ij}+s_{ji}=1$ and $\chi$ real. In momentum space it reads

$\begin{split}H^{\prime}&=\chi\sum_{\bm{k}\in\frac{1}{2}BZ}F_{\bm{k}}^{{\dagger}}\mathfrak{p}_{\bm{k}}F_{\bm{k}}\\\
\mathfrak{p}_{\bm{k}}=&\left[\nu^{3}\mu^{3}\sin(k_{1})-\nu^{3}\mu^{1}\rho^{3}\sin(k_{2})+\mu^{2}\rho^{3}\cos(k_{3})\right].\end{split}$
(106)

After some algebra we find that the eigenvalues of
$t\mathfrak{h}_{\bm{k}}+\chi\mathfrak{p}_{\bm{k}}$ are

$\begin{split}E_{\bm{k},s_{1}s_{2}}^{\text{Hall}}&=\sqrt{3}s_{1}\sqrt{t^{2}+\chi^{2}\left(1-3g_{\bm{k}}^{2}\right)+2ts_{2}\sqrt{t^{2}g_{\bm{k}}^{2}+\chi^{2}\left(1-4g_{\bm{k}}^{2}\right)}}\quad,s_{1},s_{2}\in\\{+1,-1\\},\\\
g_{\bm{k}}&=\frac{1}{3}\sqrt{\cos(k_{1})^{2}+\cos(k_{2})^{2}+\sin(k_{3})^{2}}.\end{split}$
(107)

Note that the gap for small $\chi$ at the $\bar{\bm{\Gamma}}$ point is
$2\sqrt{3}|\chi|$ while at the $\bar{\bm{K}}$ point is $\sqrt{3}|\chi|$.

###  D.2 Calculation of the Projective Symmetry Group

From Eq. 21, we read the hopping amplitudes as

$\displaystyle t_{\bm{r},\bm{r}}^{AB}$ $\displaystyle=t_{\bm{r},\bm{r}}^{BA}$
$\displaystyle=e^{i\pi r_{1}}$ $\displaystyle
t_{\bm{r},\bm{r}+\mathbf{a}_{2}}^{AB}$
$\displaystyle=t_{\bm{r},\bm{r}-\mathbf{a}_{2}}^{BA}$ $\displaystyle=1$ (108)
$\displaystyle t_{\bm{r},\bm{r}+\mathbf{a}_{3}}^{AB}$
$\displaystyle=t_{\bm{r},\bm{r}-\mathbf{a}_{3}}^{BA}$ $\displaystyle=1$

with all other hopping terms vanishing. Next we compute the $\mathrm{U}(1)$
gauge rotations $e^{i\phi(\bm{r})}$ associated with the physical symmetries.

####  D.2.1 $T_{1}$

Under translation $T_{1}$ we have

$\begin{split}t_{\bm{r},\bm{r}}^{AB}=e^{i\pi
r_{1}}&\,\,\,\,\longrightarrow\,\,\,\,t_{\bm{r}-\mathbf{a}_{1},\bm{r}-\mathbf{a}_{1}}^{AB}=-e^{i\pi
r_{1}}\\\
t_{\bm{r},\bm{r}+\mathbf{a}_{2}}^{AB}=1&\,\,\,\,\longrightarrow\,\,\,\,t_{\bm{r}-\mathbf{a}_{1},\bm{r}-\mathbf{a}_{1}+\mathbf{a}_{2}}^{AB}=1\\\
t_{\bm{r},\bm{r}+\mathbf{a}_{3}}^{AB}=1&\,\,\,\,\longrightarrow\,\,\,\,t_{\bm{r}-\mathbf{a}_{1},\bm{r}-\mathbf{a}_{1}+\mathbf{a}_{3}}^{AB}=1\end{split}$
(109)

This gives

$\begin{split}\phi(\bm{r},B)-\phi(\bm{r},A)&=\pi\mod 2\pi\\\
\phi(\bm{r}+\mathbf{a}_{2},B)-\phi(\bm{r},A)&=0\mod 2\pi\\\
\phi(\bm{r}+\mathbf{a}_{3},B)-\phi(\bm{r},A)&=0\mod 2\pi\end{split}$ (110)

whose solution is

$\begin{split}\phi(\bm{r},A)&=\pi r_{2}\\\
\phi(\bm{r},B)&=\pi(r_{2}+1)\end{split}.$ (111)

####  D.2.2 $T_{2}$

$T_{2}$ does not require any gauge transformation so

$\displaystyle\phi(\bm{r},u)=0.$ (112)

####  D.2.3 $\sigma$

We consider the reflection $\sigma$ ($\mathcal{R}_{y}$ in the main text) that
acts as $(r_{1},r_{2})\rightarrow(r_{1}+r_{2},-r_{2})$ and $A\leftrightarrow
B$.

$\begin{split}t_{\bm{r},\bm{r}}^{AB}=e^{i\pi
r_{1}}&\,\,\,\,\longrightarrow\,\,\,\,t_{\tilde{\bm{r}},\tilde{\bm{r}}}^{BA}=e^{i\pi(r_{1}+r_{2})}\\\
t_{\bm{r},\bm{r}+\mathbf{a}_{2}}^{AB}=1&\,\,\,\,\longrightarrow\,\,\,\,t_{\tilde{\bm{r}},\tilde{\bm{r}}-\mathbf{a}_{3}}^{BA}=1\\\
t_{\bm{r},\bm{r}+\mathbf{a}_{3}}^{AB}=1&\,\,\,\,\longrightarrow\,\,\,\,t_{\tilde{\bm{r}},\tilde{\bm{r}}-\mathbf{a}_{2}}^{BA}=1\end{split}$
(113)

where $\tilde{\bm{r}}=\sigma(\bm{r})$.

$\begin{split}\phi(\bm{r},B)-\phi(\bm{r},A)&=\pi r_{2}\mod 2\pi\\\
\phi(\bm{r}+\mathbf{a}_{2},B)-\phi(\bm{r},A)&=0\mod 2\pi\\\
\phi(\bm{r}+\mathbf{a}_{3},B)-\phi(\bm{r},A)&=0\mod 2\pi\end{split}$ (114)

whose solution is

$\begin{split}\phi(\bm{r},A)&=\pi\frac{r_{2}(r_{2}+1)}{2}\\\
\phi(\bm{r},B)&=\pi\frac{r_{2}(r_{2}-1)}{2}\end{split}$ (115)

####  D.2.4 $C_{6}$

The $C_{6}$ rotation acts as
$(i_{1},i_{2},A)\rightarrow(-i_{2},i_{1}+i_{2},B)$ and
$(i_{1},i_{2},B)\rightarrow(1-i_{2},i_{1}+i_{2}-1,A)$ and the inverse
$C_{6}^{-1}$ acts as $(i_{1},i_{2},A)\rightarrow(i_{1}+i_{2},1-i_{1},B)$ and
$(i_{1},i_{2},B)\rightarrow(i_{1}+i_{2},-i_{1},A)$.

$\begin{split}t_{\bm{r},\bm{r}}^{AB}=e^{i\pi
r_{1}}&\,\,\,\,\longrightarrow\,\,\,\,t_{\tilde{\bm{r}}+\mathbf{a}_{2},\tilde{\bm{r}}}^{BA}=1\\\
t_{\bm{r},\bm{r}+\mathbf{a}_{2}}^{AB}=1&\,\,\,\,\longrightarrow\,\,\,\,t_{\tilde{\bm{r}}+\mathbf{a}_{2},\tilde{\bm{r}}+\mathbf{a}_{1}}^{BA}=t_{\tilde{\bm{r}}+\mathbf{a}_{2},\tilde{\bm{r}}+\mathbf{a}_{2}-\mathbf{a}_{3}}^{BA}=1\\\
t_{\bm{r},\bm{r}+\mathbf{a}_{3}}^{AB}=1&\,\,\,\,\longrightarrow\,\,\,\,t_{\tilde{\bm{r}}+\mathbf{a}_{2},\tilde{\bm{r}}+\mathbf{a}_{2}}^{BA}=e^{i\pi(r_{1}+r_{2})}\end{split}$
(116)

where $\tilde{\bm{r}}=C_{6}^{-1}(\bm{r})$.

$\begin{split}\phi(\bm{r},B)-\phi(\bm{r},A)&=\pi i_{1}\quad\mod 2\pi\\\
\phi(\bm{r}+\mathbf{a}_{2},B)-\phi(\bm{r},A)&=0\quad\quad\mod 2\pi\\\
\phi(\bm{r}+\mathbf{a}_{3},B)-\phi(\bm{r},A)&=\pi(i_{1}+i_{2})\mod
2\pi\end{split}$ (117)

Whose solution is

$\begin{split}\phi(\bm{r},A)&=\pi\frac{r_{1}(r_{1}+1)}{2}+\pi r_{1}r_{2},\\\
\phi(\bm{r},B)&=\pi\frac{r_{1}(r_{1}-1)}{2}+\pi r_{1}r_{2},\\\
\phi(\bm{r},u)&=\pi\frac{r_{1}(r_{1}+1)}{2}+\pi r_{1}u+\pi
r_{1}r_{2}.\end{split}$ (118)

We can rewrite this in in the enlarged unit cell by writing
$r_{i}=2s_{i}+e_{i}$ where $s_{i}\in\mathbb{Z}$ and $e_{i}=0,1$ so

$\displaystyle\phi(\bm{s},u,e_{1},e_{2})$
$\displaystyle=\pi\left(s_{1}+e_{1}+e_{1}u+e_{1}e_{2}\right)$ (119a)

####  D.2.5 $\mathcal{T}_{0}$

As the hopping amplitudes are real numbers, there is no gauge transformation
needed so we can take

$\displaystyle\phi(\bm{r},u)=0.$ (120)

####  D.2.6 PSG action in momentum space

The previous transformation properties in momentum space read

$\begin{split}T_{1}:F_{\mathbf{k}}&\longrightarrow
e^{ik_{1}}\mu^{3}\rho^{1}\nu^{3}F_{\mathbf{k}}\\\
T_{2}:F_{\mathbf{k}}&\longrightarrow e^{ik_{2}}\rho^{3}F_{\mathbf{k}}\\\
\sigma:F_{\mathbf{k}}&\longrightarrow\nu^{1}\exp(-i\frac{\pi}{2}\frac{1-\rho^{1}}{2}\mu^{3}\nu^{3})F_{\sigma(\mathbf{k})+\frac{\mathbf{Q}_{2}}{2}}\\\
C_{6}:F_{\mathbf{k}}&\longrightarrow\left(\nu^{+}+e^{-ik_{2}}\rho^{3}\nu^{-}\right)\cdot\frac{1+\vec{\rho}\cdot\vec{\mu}}{2}\cdot\exp(i\pi\frac{1-\mu^{1}}{2}\frac{1-\rho^{2}\nu^{3}}{2})F_{C_{6}[\mathbf{k}]+\frac{\mathbf{Q}_{1}}{2}}\\\
\mathcal{T}_{0}:F_{\bm{k}}&\longrightarrow F_{-\bm{k}}.\end{split}$ (121)

###  D.3 Low energy Hamiltonian

We find the low-energy Hamiltonian at quarter-filling by defining
$\hat{\psi}_{\bm{q}}=F_{\bm{\Gamma}^{\prime}+\bm{q}}$ with
$\bm{\Gamma}^{\prime}=\frac{\bm{G}_{1}+\bm{G}_{2}}{4}$ and expand for small
$\bm{q}$

$\begin{split}P^{{\dagger}}\cdot\mathfrak{h}_{\bm{\Gamma}^{\prime}+\bm{q}}\cdot
P=-\sqrt{3}\mathds{1}_{4\times
4}+\frac{1}{\sqrt{2}}\bm{q}\cdot\bm{\alpha}+\mathcal{O}(\bm{q}^{2})\end{split}$
(122)

where $P$ is an $8$ by $4$ matrix whose columns are the eigenvectors of
$\mathfrak{h}_{\bm{\Gamma}^{\prime}}$ with eigenvalue $-\sqrt{3}$.

In particular, we can factor $P=\mathcal{U}\cdot P_{0}$ for $\mathcal{U}$ a
unitary and $P_{0}$ another projector. $\mathcal{U}$ is given by
$\mathcal{U}=\mathcal{U}_{1}\cdot\mathcal{U}_{2}\cdot\mathcal{U}_{3}\cdot\mathcal{U}_{4}$
where

$\begin{split}\mathcal{U}_{1}&=\exp(i\pi\frac{\mathds{1}+\mu^{3}\nu^{3}}{2}\frac{\mathds{1}+\rho^{3}}{2})\\\
\mathcal{U}_{2}&=\exp(-i\frac{\pi}{8}\mu^{2})\\\
\mathcal{U}_{3}&=\exp(-i\frac{\arctan(\sqrt{2})}{2}\mu^{3}\nu^{3})\\\
\mathcal{U}_{4}&=i\rho^{3}e^{i\frac{\pi}{4}\nu^{1}}e^{i\frac{\pi}{3}\mu^{2}}\frac{1+\vec{\rho}\cdot\vec{\mu}}{2}\end{split}$
(123)

so that
$\mathcal{U}^{{\dagger}}\cdot\mathfrak{h}_{\bm{\Gamma}^{\prime}}\cdot\mathcal{U}=-\sqrt{3}\nu^{3}$.
We can then simply take $P_{0}=\frac{1+\nu^{3}}{2}$ which ultimately leads to
$\bm{\alpha}^{1}=+\rho^{1}$ and $\bm{\alpha}^{2}=-\rho^{3}$.

The Dirac fermions are then given by
$\psi_{\bm{q}}=P^{{\dagger}}\hat{\psi}_{\bm{q}}$. The Hamiltonian in Eq. 22
can be reproduced by identifying
$(\gamma^{0},\gamma^{1},\gamma^{2})=(-i\rho^{2},\rho^{3},\rho^{1})$ and
introducing back the spin-orbital indices. The action on the Dirac fermions
are then easily found from the explicit expression of $\psi_{\bm{q}}$ and Eq.
121 evaluated at $\bm{k}=\bm{\Gamma}^{\prime}$. For example,
$\mathcal{T}[\psi_{\bm{q}}]=\mathcal{T}[P^{{\dagger}}F_{\bm{\Gamma}^{\prime}+\bm{q}}]=P^{\top}F_{-\bm{\Gamma}^{\prime}-\bm{q}}=P^{\top}\mu^{1}\rho^{1}F_{\bm{\Gamma}^{\prime}-\bm{q}}$.
One can check that $P^{\top}\mu^{1}\rho^{1}=-\mu^{2}\rho^{2}P^{{\dagger}}$ so
that $\mathcal{T}[\psi_{\bm{q}}]=-i\rho^{2}\gamma^{0}\psi_{-\bm{q}}$ which
matches the result in the main text if we introduce back the spin-orbital
action $\sigma^{2}\tau^{2}$.

##  Appendix E Wannier centers from PSG

###  E.1 Introduction

In this appendix, we use the techniques developed in Refs. [39, 37, 38] to
decompose the band structure of the $\pi$-flux state on the Honeycomb lattice.
We follow the approach used in Ref. [15].

The idea is that in order to find the decomposition of the quantum spin Hall
bands we need to calculate the eigenvalue spectrum of the rotations at high-
symmetry points and then compare it with the spectra of the Wannier bands we
define below. In particular, we consider three rotations: hexagon centered
6-fold rotation $C_{6h}$, hexagon centered 3-fold rotation $C_{3h}=C_{6h}^{2}$
and a bond (between sites 1 ad 6) centered 2-fold rotation $C_{2e}$ 212121We
could have also included site centered 3-fold rotations but they do not give
any new information in the present case..

We will compare the spinon band representation to those of Wannier insulator
centered on site ($\Gamma^{v}$), hexagon centers ($\Gamma^{h}$) and bond
centers $\Gamma^{b}$, respectively. The representation of $\Gamma^{v}$ is
straightforward as these are our original basis. For the other two insulators,
we take them to be equal-amplitude superpositions of spinons at the sites on
the boundary of the plaquette or bond. As we have different choices for the
phases between sites, these insulators will have internal quantum numbers. In
particular, the hexagon-centered Wannier insulators come in six different
flavours $\Gamma_{\eta}^{h}$, with $\eta=\pm 1,\pm 2,\pm 3$, corresponding the
different eigenvalues under $C_{6h}$. Similarly, the bond-centered insulators
come in two flavours $\Gamma_{\lambda}^{b}$ corresponding $\lambda=\pm 1$.
Note that in order to make a meaningful comparison between the eigenvalue
spectra we need to pick a gauge and use it for all the bands.

As we will see below, the Dirac points of the spinon bands occur at the high-
symmetry points. Therefore, it is _fundamental_ that we include the quantum
spin Hall mass to split the spinon band into four Krammers sub-bands with
well-defined eigenvalue spectra. The sub-bands will be denoted by
$\Gamma_{\text{PSG}}^{a}$ with $a=+2,+1,-1,-2$, where $a$ denotes the position
of the band with respect to zero energy. Note that each Krammers band
corresponds to two Krammers-pairs of spinons per enlarged unit cell.

(a)

(b)

(c)

Figure 5:  (a) Enlarged unit cell used in the band topology calculation. Pink
bonds correspond to negative hoppings on those bonds and numbers correspond to
the labeling of sites in the enlarged unit cell. (b) & (c) Choice of gauge
transformation for $C_{6h}$ and $C_{3h}$: the plus or minus signs correspond
to a gauge transformation $f_{i}\rightarrow\pm f_{i}$ for $C_{3h}$ and
$f_{i}\rightarrow\pm if_{i}$ for $C_{6h}$.

###  E.2 Band topology calculation

We shall denote the Wannier insulator on the unit cell labeled by $\bm{r}$ on
the sublattice $a=1,\dots,8$ by $\ket{\bm{r},a}$. Using the gauge choice for
$C_{6h}$ depicted in Fig. 5(b) we find that
$\ket{\bm{r},6}=(-1)^{r_{1}+1}i\ket{R_{\pi/3}\bm{r},1}$,
$\ket{\bm{r},7}=(-1)^{r_{1}}i\ket{R_{\pi/3}\bm{r}+\bm{a}_{3},8}$,
$\ket{\bm{r},8}=(-1)^{r_{1}+1}i\ket{R_{\pi/3}\bm{r}-\bm{a}_{3},7}$ and
$C_{6h}\ket{\bm{r},a}=(-1)^{r_{1}}i\ket{R_{\pi/3}\bm{r},a+1}$, where
$R_{\pi/3}$ is the $\pi/3$ clockwise rotation matrix. Going to momentum space,
we find that $C_{6h}$ sends
$\bm{k}\rightarrow\bm{k}^{\prime}=R_{\pi/3}\bm{k}+\bm{Q}_{1}$ and

$\begin{split}C_{6h}\ket{\bm{k},a}&=+i\ket{\bm{k}^{\prime},a+1}\quad,a=1,\dots,5\\\
C_{6h}\ket{\bm{k},6}&=-i\ket{\bm{k}^{\prime},1}\\\
C_{6h}\ket{\bm{k},7}&=+ip_{3}^{*}\ket{\bm{k}^{\prime},8}\\\
C_{6h}\ket{\bm{k},8}&=-ip_{3}\ket{\bm{k}^{\prime},7}\end{split}$ (124)

where $p_{3}=\exp(i\bm{k}^{\prime}\cdot\bm{a}_{3})$. The invariant momentum is
simply $\bm{k}=\bm{Q}_{1}+\bm{Q}_{2}$ and thus $p_{3}=1$. The eigenvalues are
$[1,\Omega,\Omega^{2},,\Omega^{3},\Omega^{4},\Omega^{5}]\oplus[+1,-1]$, where
$\Omega=e^{\frac{\pi i}{3}}$.

For $C_{3h}$ we find

$\begin{split}C_{3h}\ket{\bm{k},a}&=-\ket{\bm{k}^{\prime},a+2}\quad,a=1,\dots,4\\\
C_{3h}\ket{\bm{k},a}&=+\ket{\bm{k}^{\prime},a-4}\quad,a=5,6\\\
C_{3h}\ket{\bm{k},7}&=-p_{2}\ket{\bm{k}^{\prime},7}\\\
C_{3h}\ket{\bm{k},8}&=-p_{2}^{*}\ket{\bm{k}^{\prime},8}\end{split}$ (125)

where $\bm{k}^{\prime}=R_{2\pi/3}\bm{k}+\bm{Q}_{2}$ and
$p_{2}=\exp(i\bm{k}^{\prime}\cdot\bm{a}_{2})$. The invariant momenta are
$\bm{k}=\bm{Q}_{1}+\bm{Q}_{3}:=\bm{M}$ and
$\bm{k}=\frac{\bm{G}_{1}-\bm{G}_{2}}{6}=\bm{K}=-\bm{K}^{\prime}$. The
eigenvalues in this case are
$[1,\omega,\omega^{2}]\oplus[1,\omega,\omega^{2}]\oplus p_{3}\oplus p_{3}^{*}$
222222Note that for the invariant momenta $-p_{2}=p_{3}^{*}$ with $p_{3}$ as
previously defined.. As for $C_{2e}$ there is no gauge transformation needed,
it is easy to see that the eigenvalues are four pairs of $[+1,-1]$.

We next construct the hexagon centered Wannier insulators. We start by taking
the eigenvectors of $C_{6h}$ with eigenvalue $\Omega^{2\eta}$ with weight on
hexagon A, denoted by $\ket{\text{hex},A}$ (omitting the position dependence).
The other three states are taken by translating the previous state by half
lattice vectors. We shall denote the representation by $\Gamma^{h}_{\eta}$.
From the construction we have
$C_{6h}\ket{\text{hex},A}=\Omega^{2\nu-1}\ket{\text{hex},A}$. Next, one can
check that $C_{6h}^{3}=(-1)^{\eta}$ when acting on $\ket{\text{hex},B/C/D}$
and furthermore with the appropiate phase conventions, the matrix becomes

$(-1)^{\eta}\left(\begin{array}[]{ccc}0&1&0\\\ 0&0&1\\\
1&0&0\end{array}\right)$ (126)

whose eigenvalues are
$(-1)^{\eta}[1,\omega,\omega^{2}]=[\Omega^{\eta},\Omega^{\eta+2},\Omega^{\eta+4}]$.
Similarly, we find
$C_{3h}\ket{\text{hex}_{\eta},A}=\omega^{\eta}\ket{\text{hex}_{\eta},A}$ and
$C_{3h}^{3}=1$ on the other hexagons. Thus the eigenvalues are
$\omega^{\eta}\oplus[1,\omega,\omega^{2}]$. As none of the hexagons are
invariant under $C_{2e}$, the eigenvalues are $[+1,-1]\oplus[+1,-1]$.

Similarly the bond Wannier insulators come in two flavours
$\Gamma^{b}_{\lambda}$ with $\lambda=+1,-1$ which correspond to the equal-
amplitude symmetric or antisymmetric superposition of the onsite insulators at
the boundary of the bonds. By a similar calculation as for the previous bands
we can fill Tab. 7.

| $\Gamma^{v}$ | $\Gamma^{e}_{\lambda}$ | $\Gamma^{h}_{\eta}$ | HSM
---|---|---|---|---
$C_{6h}$ | $[\text{Doublet}]\oplus[\text{Sextet}]$ | $2[\text{Sextet}]$ | $\Omega^{\eta}\oplus[\text{Triplet}_{\eta}]$ | $\bm{M}$
$C_{3h}$ | $p_{3}\oplus p_{3}^{*}\oplus 2[\text{Triplet}_{0}]$ | $4[\text{Triplet}_{0}]$ | $\omega^{\eta}\oplus[\text{Triplet}_{0}]$ | $\bm{M},\bm{K},\bm{K}^{\prime}$
$C_{2e}$ | $4[\text{Doublet}]$ | $\lambda(1\oplus p_{1}\oplus p_{2}\oplus p_{3})\oplus 4[\text{Doublet}]$ | $2[\text{Doublet}]$ | $\bm{M},\bm{\Gamma}$
Table 7:  Space group representation of rotations of the Wannier insulators of
the $\pi$-flux state on the honeycomb lattice at high symmetry points. The
multiplets are denoted by
$[\text{Sextet}]=[1,\Omega^{1},\Omega^{2},\Omega^{3},\Omega^{4},\Omega^{5}]$,
$[\text{Triplet}_{\eta}]=(-)^{\eta}[1,\omega^{1},\omega^{2}]$ and
$[\text{Doublet}]=[+1,-1]$. $p_{i}=e^{i\bm{k}\cdot\bm{a}_{i}}$ is the phase
factor obtained under $T_{i=1,2,3}$ translation. The high-symmetry momenta
(HSM) for each group operations are recorded in the last column. They are
$\bm{M}=\frac{\bm{G}_{1}+\bm{G}_{2}}{2}$,
$\bm{K}=\frac{\bm{G}_{1}-\bm{G}_{2}}{6}$ and $\bm{K}^{\prime}=-\bm{K}$, where
$\bm{G}_{1,2}$ are the reciprocal lattice vectors.

We find that the spectrum of the PSG bands by straightforward diagonalization
of the mean-field Hamiltonian at high-symmetry points and then finding the
eigenvalues of the rotations acting on them. The result is shown in Tab. 8. As
previously mentioned, we need to introduce the quantum spin Hall mass to open
a gap and be able to decompose the resulting bands in terms of the Wannier
insulators. We find that the gap splits the eigenvalues equally between the
bands below and above the previous Dirac points. This can be understood by
remaining that the $C_{6h}$ rotation can be written (in the appropriate basis)
as $\exp(-\frac{\pi}{6}\gamma^{0})$ times some matrix $R_{C_{6h}}$ acting on
valley indices. As the QSH mass gives opposite $\gamma_{0}$ eigenvalues to
spin up and spin down, the eigenvalues come in pairs.

| $\Gamma_{PSG(\bm{M})}$ | $\Gamma_{PSG(\bm{\Gamma})}$ | $\Gamma_{PSG(\bm{K})}$
---|---|---|---
$C_{6h}$ | $\\{1,\Omega^{1},\Omega^{2},\Omega^{3}\\}\oplus\\{1,\Omega^{3},\Omega^{4},\Omega^{5}\\}$ | $-$ | $-$
$C_{3h}$ | $\\{1\oplus[\text{Triplet}_{0}]\\}\oplus\\{1\oplus[\text{Triplet}_{0}]\\}$ | $-$ | $\\{[\omega,\omega^{2}]\\}\oplus\\{1\oplus[\text{Triplet}_{0}]\\}\oplus\\{[\omega,\omega^{2}]\\}$
$C_{2e}$ | $\\{2[\text{Doublet}]\\}\oplus\\{2[\text{Doublet}]\\}$ | $\bigoplus_{n=1}^{4}\\{[\text{Doublet}]\\}$ | $-$
Table 8:  Space group representation of rotations of the spinon bands of the
$\pi$-flux state on the honeycomb lattice at high symmetry points. The
eigenvalues are ordered from higher to lower bands. The multiplets are denoted
by
$[\text{Sextet}]=[1,\Omega^{1},\Omega^{2},\Omega^{3},\Omega^{4},\Omega^{5}]$,
$[\text{Triplet}_{\eta}]=(-)^{\eta}[1,\omega^{1},\omega^{2}]$ and
$[\text{Doublet}]=[+1,-1]$. $p_{i}=e^{i\bm{k}\cdot\bm{a}_{i}}$ is the phase
factor obtained under $T_{i=1,2,3}$ translation. The high-symmetry momenta
(HSM) for each group operations are recorded in the last column. They are
$\bm{M}=\frac{\bm{G}_{1}+\bm{G}_{2}}{2}$,
$\bm{K}=\frac{\bm{G}_{1}-\bm{G}_{2}}{6}$ and $\bm{K}^{\prime}=-\bm{K}$, where
$\bm{G}_{1,2}$ are the reciprocal lattice vectors.

A decomposition for the $PSG$ bands is

$\begin{split}2\Gamma^{+2}_{PSG}&=\Gamma^{v}+\left(\Gamma^{v}-\Gamma^{h}_{\eta=0}-\Gamma^{h}_{\eta=3}\right)+\left(\Gamma^{h}_{\eta=1}+\Gamma^{h}_{\eta=2}-\Gamma^{h}_{\eta=4}-\Gamma^{h}_{\eta=5}\right)\\\
2\Gamma^{+1}_{PSG}&=\Gamma^{v}-\left(\Gamma^{v}-\Gamma^{h}_{\eta=0}-\Gamma^{h}_{\eta=3}\right)+\left(\Gamma^{h}_{\eta=1}+\Gamma^{h}_{\eta=2}-\Gamma^{h}_{\eta=4}-\Gamma^{h}_{\eta=5}\right)\\\
2\Gamma^{-1}_{PSG}&=\Gamma^{v}-\left(\Gamma^{v}-\Gamma^{h}_{\eta=0}-\Gamma^{h}_{\eta=3}\right)-\left(\Gamma^{h}_{\eta=1}+\Gamma^{h}_{\eta=2}-\Gamma^{h}_{\eta=4}-\Gamma^{h}_{\eta=5}\right)\\\
2\Gamma^{-2}_{PSG}&=\Gamma^{v}+\left(\Gamma^{v}-\Gamma^{h}_{\eta=0}-\Gamma^{h}_{\eta=3}\right)-\left(\Gamma^{h}_{\eta=1}+\Gamma^{h}_{\eta=2}-\Gamma^{h}_{\eta=4}-\Gamma^{h}_{\eta=5}\right)\\\
\end{split}$ (127)

This decomposition is unique modulo addition of multiples of
$\Gamma^{h}_{\eta=0}+\Gamma^{h}_{\eta=1}+\Gamma^{h}_{\eta=2}+\Gamma^{h}_{\eta=3}+\Gamma^{h}_{\eta=4}+\Gamma^{h}_{\eta=5}-\Gamma^{b}_{\lambda=+1}-\Gamma^{b}_{\lambda=-1}$
which is irrelevant for the monopole quantum numbers. Notice that under time
reversal $\eta\rightarrow 3-\eta$ because as $C_{6h}$ involves a factor of
$i$, time reversal involves a minus sign ($-1=\Omega^{3}$) on top of complex
conjugation of eigenvalues.

##  Appendix F Numerical determination of the Berry phase

###  F.1 Monopole configuration on zero flux background

We first find the gauge to introduce the $2\pi$ flux on the (0-flux) HC
lattice and the gauge transformation needed to implement the lattice
symmetries (in the $L_{1},L_{2}\rightarrow\infty$). The Hamiltonian as usual
is on a $L_{1}$ by $L_{2}$ lattice:

$H=\sum_{\mathbf{r}\mu}e^{iA_{\mu}(\mathbf{r})}f_{\mathbf{r}A}^{{\dagger}}f_{\mathbf{r}+b_{\mu}B}+h.c.$
(128)

where $b_{1}=\bm{0}$, $b_{2}=\bm{a}_{2}$ and $b_{3}=\bm{a}_{3}$. We introduce
the $2\pi$ flux by choosing

$\begin{split}A_{1}(\mathbf{r})&=\frac{2\pi}{L_{1}L_{2}}(r_{1}+s)\\\
A_{2}(\mathbf{r})&=0\\\
A_{3}(\mathbf{r})&=\frac{2\pi}{L_{2}}(C-r_{2})\delta_{r_{1},0}\\\ \end{split}$
(129)

we have to restrict $0\leq r_{i=1,2}<L$. $s$ and $C$ are constants that will
be constrained in the following.

Notice that under $T_{2}$ there is no gauge transformation needed. For
$T_{1}$, $\delta
A_{\mu}(\mathbf{r})=A_{\mu}(X^{-1}[\mathbf{r}])-A_{\mu}(\mathbf{r})$ with
$X=T_{1}$ so that

$\begin{split}\delta
A_{1}(\mathbf{r})&=A_{1}(\mathbf{r}-\mathbf{a}_{1})-A_{1}(\mathbf{r})=-\frac{2\pi}{L_{1}L_{2}}\left(1-\delta_{r_{1},0}L_{1}\right),\\\
\delta A_{2}(\mathbf{r})&=0,\\\ \delta
A_{3}(\mathbf{r})&=\frac{2\pi(r_{2}-C)}{L_{2}}(\delta_{r_{1},0}-\delta_{r_{1},1}).\end{split}$
(130)

The translated Hamiltonian is related to the initial one by threading Wilson
Loops (WLs) along the $\bm{a}_{2}$ direction with total flux $e^{\frac{2\pi
i}{L_{1}}}$ in addition to some boundary terms in the $\bm{a}_{1}$. In order
to remove this WL, a gauge transformation$f_{u\mathbf{r}}\rightarrow
e^{i\phi_{u}(\mathbf{r})}f_{u\mathbf{r}}$ with

$\begin{split}\phi_{A}(\mathbf{r})&=\frac{2\pi(r_{2}-C)}{L_{2}}\delta_{r_{1},0}\\\
\phi_{B}(\mathbf{r})&=\frac{2\pi(r_{2}-1-C)}{L_{2}}\delta_{r_{1},0}\end{split}$
(131)

is used. The net change in the phases is given by

$\begin{split}\delta
A_{1}(\mathbf{r})+\phi_{B}(\mathbf{r})-\phi_{A}(\mathbf{r})&=-\frac{2\pi}{L_{1}L_{2}}\\\
\delta
A_{2}(\mathbf{r})+\phi_{B}(\mathbf{r}+\mathbf{a}_{2})-\phi_{A}(\mathbf{r})&=0\\\
\delta
A_{3}(\mathbf{r})+\phi_{B}(\mathbf{r}+\mathbf{a}_{3})-\phi_{A}(\mathbf{r})&=0\end{split}$
(132)

Note that the remaining WLs in the $2$ direction have a flux
$\exp(\pm\frac{2\pi i}{L_{1}})$ which goes to zero in the
$L_{1}\rightarrow\infty$ limit. This means that in later limit the $T_{1}$
(projective) symmetry is recovered.

Next, we take $L_{1}=L_{2}=L$ in order to be able to do rotations. Consider

$\begin{split}C_{6}[H]&=\sum_{\bm{r},\mu}e^{iA_{\mu}(\mathbf{r})}f_{BC_{6}[\mathbf{r}]}^{{\dagger}}f_{AC_{6}[\mathbf{r}+b_{\mu}-\mathbf{a}_{2}]}+h.c.\end{split}$
(133)

where $C_{6}[\mathbf{r}]$ is clockwise rotation by $\pi/3$ around the
plaquette to the right of $\bm{r}=(0,0)$. Let
$\tilde{b}_{\mu}=C_{6}[\mathbf{a}_{2}-b_{\mu}]$ ( $\tilde{b}_{1}=b_{3}$,
$\tilde{b}_{2}=b_{1}$ and $\tilde{b}_{3}=b_{2}$). Therefore

$\begin{split}C_{6}[H]&=\sum_{\bm{r},\mu}e^{iA_{\mu}(C_{6}^{-1}[\mathbf{r}])}f_{B\mathbf{r}}^{{\dagger}}f_{A\mathbf{r}-\tilde{b}_{\mu}}+h.c.\\\
&=\sum_{\bm{r},\mu}e^{-iA_{\mu}(C_{6}^{-1}[\mathbf{r}+\tilde{b}_{\mu}])}f_{A\mathbf{r}}^{{\dagger}}f_{B\mathbf{r}+\tilde{b}_{\mu}}+h.c.\end{split}$
(134)

The change in the gauge field is

$\begin{split}\delta A_{1}(\mathbf{r})&=-\frac{2\pi}{L^{2}}(r_{1}+s)\\\ \delta
A_{2}(\mathbf{r})&=-\frac{2\pi}{L}(r_{1}+C)\delta_{r_{1}+r_{2},L-1}\\\ \delta
A_{3}(\mathbf{r})&=-\frac{2\pi}{L^{2}}(r_{1}+\tilde{r}_{2}+s)+\frac{2\pi}{L}(r_{2}-C)\delta_{r_{1},0}\\\
\end{split}$ (135)

where $\tilde{r}_{2}=\textit{mod}(r_{1}+r_{2},L)-r_{1}$.

The field $\delta A_{\mu}$ varies slowly in the bulk but has jumps along the
lines $r_{1}=0$ and $r_{1}+r_{2}=L-1$. In order to eliminate them, we perform
a gauge transformation that imposes $\phi_{B}(r+b_{\mu})-\phi_{A}(r)+\delta
A_{\mu}(\mathbf{r})=0$ away from the jumps. A solution is given by

$\begin{split}\phi_{A}(\mathbf{r})&=-\frac{\pi}{L^{2}}({r_{1}(2\tilde{r}_{2}+r_{1}+1)}+s(2(r_{1}+\tilde{r}_{2})+1)),\\\
\phi_{B}(\mathbf{r})&=-\frac{\pi}{L^{2}}({r_{1}(2\tilde{r}_{2}+r_{1}-1)}+s(2(r_{1}+\tilde{r}_{2})-1)).\end{split}$
(136)

The boundary terms become

$\begin{split}\phi_{B}(L-1,x+1-L)-\phi_{A}(0,x)+\delta
A_{3}(x,0)=\frac{\pi(L-1-2C)}{L}\\\
\phi_{B}(L-1,x+1-L)-\phi_{A}(L+1-x,x)+\delta
A_{2}(L+1-x,x)=\frac{2\pi(s-C)}{L}\end{split}$ (137)

Setting $C=\frac{L-1}{2}$ and $s=-\frac{L+1}{2}$ kills both terms. The $C_{6}$
being discussed is hence preserved. The other rotations can be obtained by
combination with translations which we already know how to implement.

A problem we face now is how to decide the overall phase of the Gauge
transformation in order to fix the Berry phase. To see that this is an issue,
consider a uniform gauge transformation, say $e^{i\theta/L^{2}}$, with
$N=nL^{2}$ occupied fermion modes. The expectation value of the symmetry
operation gets an extra $e^{in\theta}$ phase.

In order to fix the phase, we impose that the total phase
$\exp(i\sum_{\mathbf{r}u}\phi_{u}(\mathbf{r}))$ is $1$. This condition comes
from the requirement that our state is gauge invariant. Consider the sum

$\Theta=\sum_{r_{1},r_{2}=0}^{L-1}\sum_{u=A,B}\phi_{u}(\mathbf{r})=2\pi\frac{(2L+5)(L-1)}{6}.$
(138)

and then redefine $\phi\rightarrow\phi-\frac{\Theta}{2L^{2}}$. The gauge
transformation now reads

$\begin{split}\phi_{A}(\mathbf{r})&=-\frac{\pi}{L^{2}}({r_{1}(2\tilde{r}_{2}+r_{1}+1)}-\frac{L+1}{2}(2(r_{1}+\tilde{r}_{2})+1))-2\pi\frac{(2L+5)(L-1)}{12L^{2}},\\\
\phi_{B}(\mathbf{r})&=-\frac{\pi}{L^{2}}({r_{1}(2\tilde{r}_{2}+r_{1}-1)}-\frac{L+1}{2}(2(r_{1}+\tilde{r}_{2})-1))-2\pi\frac{(2L+5)(L-1)}{12L^{2}}.\end{split}$
(139)

We numerically checked that this gauge transformation gives us the same $\pi$
Berry phase as found in Ref. [15].

###  F.2 Monopole configuration on $\pi$ flux background

The monopole configuration of the $\pi$ flux state is obtained by taking

$H=\sum_{\mathbf{r}\mu}t_{\mathbf{r}\mathbf{s}}^{AB}e^{iA_{\mu}(\mathbf{r})}f_{\mathbf{r}A}^{{\dagger}}f_{\mathbf{r}+b_{\mu}B}+h.c.$
(140)

where $A$ is the same as before (Eq. 129) and $t_{\mathbf{r}\mathbf{s}}^{AB}$
are the same hopping parameters (Eq. 21) used in the main text. It is easy to
see that the symmetries can be implemented by simply combining the gauge
transformation of the previous section with the gauge transformation from Sec.
D.2. We numerically checked that these gauge transformations give consistent
phases with the Berry phases obtained in the main text using the band
calculation.

##  Appendix G Mathematical backgrounds for anomaly calculation

###  G.1 General constructions

In this section, we review math constructions used with cohomology groups.

First, recall that the cup product gives a graded bilinear product between
$(\psi_{i},\phi_{j})\in H^{i}(G)\times H^{j}(G)$ to $H^{i+j}(G)$ denoted as
$\psi_{i}\cup\phi_{j}$. Here $G$ is a Abelian group that in our cases is given
simply $\mathbb{Z}$ or $\mathbb{Z}_{n}$ (integers modulo $n$). Note that given
a group homomorphism $\rho:G\rightarrow G^{\prime}$ which abusing notation we
denote also by $\rho$, at the cohomology level. For example, if we take
$G=\mathbb{Z}$ and $G^{\prime}=\mathbb{Z}_{n}$ then the induced map is modulo
$n$ reduction. Let us denote the group and cohomology maps by $\rho_{n}$.
Similarly, we can take $G=G^{\prime}=\mathbb{Z}\,\text{or}\,\mathbb{Z}_{n}$
and consider the map corresponding to multiplication by some element in $G$.
At the level of cohomology this operation is also multplication by said
element.

Note that we can combine the two previous operations at the group level to
obtain the short exact sequence (SEC)

$0\longrightarrow\mathbb{Z}\overset{n\cdot}{\longrightarrow}\mathbb{Z}\overset{\rho_{n}}{\longrightarrow}\mathbb{Z}_{n}\longrightarrow
0.$

From this SEC, we can obtain a long exact sequence at the cohomology level
that allow as to define the Bockstein homomorphism $\beta_{n}$

$\dots\longrightarrow
H^{i}(\mathbb{Z})\overset{n\cdot}{\longrightarrow}H^{i}(\mathbb{Z})\overset{\rho_{n}}{\longrightarrow}H^{i}(\mathbb{Z}_{n})\overset{\beta_{n}}{\longrightarrow}H^{i+1}(\mathbb{Z})\longrightarrow\dots\quad.$

By looking at the long exact sequence, we see that $\beta_{n}$ can be though
of as the obstruction to lifting a $\mathbb{Z}_{n}$ cochain to a $\mathbb{Z}$
one, which is why sometimes it is denoted by
$\beta_{n}(x)=\frac{1}{n}\differential{x}$.

Following Ref. [50], we define the Pontraygin square operation
$\mathcal{P}:H^{i}(G)\rightarrow H^{2i}(\Gamma(G))$232323Note that there is a
different $\mathcal{P}$ for each $G$ but for ease of reading we omit an
explicit dependence on $G$., where $\Gamma(G)$ is the universal quadratic
group $G$. $\mathcal{P}$ is universal in the sense is the only map
$H^{i}(G)\rightarrow H^{2i}(\Gamma(G))$ such that
$\mathcal{P}(x+y)-\mathcal{P}(x)-\mathcal{P}(y)=2x\cup y$. Note that
$\Gamma(\mathbb{Z}_{2n})=\mathbb{Z}_{4n}$ and
$\Gamma(\mathbb{Z}_{2n-1})=\mathbb{Z}_{2n-1}$. In particular, for
$G=\mathbb{Z}_{2n}$ we define $\mathcal{P}(x)=x\cup
x+x\cup_{1}\differential{x}$ and for $G=\mathbb{Z}_{2n-1}$ we can simply take
$\mathcal{P}(x)=x\cup x$. Here $\cup_{1}$ is a generalized bilinear product of
degree -1 defined in the appendix of Ref. [50].

From now on, we will focus on the case where $G=\mathbb{Z}_{2n}$ as our
anomaly calculations only require these cases.

Note that the property $\mathcal{P}(x+y)-\mathcal{P}(x)-\mathcal{P}(y)=2x\cup
y$ implies that
$\mathcal{P}(\sum_{i=1}^{N}x_{i})=\sum_{i=1}^{N}\mathcal{P}(x_{i})+\sum_{1\leq
i<j\leq N}2x_{i}\cup x_{j}$ which we will frequently use. By the explicit
formula for $\mathcal{P}$ for $G=\mathbb{Z}_{2n}$, if $x=\rho_{2n}\hat{x}$ ,
for some integer class $\hat{x}$, then
$\mathcal{P}(x)=\rho_{4n}\left(\hat{x}\cup\hat{x}\right)$.

We actually need a bit more of relationships between the previously defined
operations which can be found in Ref. [55]. We accept that
$\mathbb{Z}_{0}=\mathbb{Z}$. First of all, let us define
$\mu_{p,q}:H^{*}(\mathbb{Z}_{q})\rightarrow H^{*}(\mathbb{Z}_{p})$ by sending
$x$ to the cohomology class of $\frac{p}{(p,q)}x^{\prime}$ where $x^{\prime}$
is a cochain in cohomology class of $x$. $(p,q)$ is largest common divisor of
$p$ and $q$. From the definition, we have $\mu_{p,q}(a\cdot x)=(a\\!\\!\mod
p)\cdot\mu_{p,q}(x)$ for $a\in\mathbb{Z}_{q}$. Note that $\mu_{p,0}=\rho_{p}$
and , if $r\neq 0$, $\mu_{p,rp}$ is reduction modulo $r$ and $\mu_{rq,q}$ is
multiplication by $r$. For ease of reading, we denote the former operations by
$\rho_{p}$ as in the $r=0$ case. The following relations can be found in
Sections 2 and 4 of Ref. [55] 242424Note also that the $\Delta_{q}$ defined in
Ref. [55] is our $\beta_{q}$ . $(r,s\in\mathbb{Z}-\\{0\\})$:

$\displaystyle\beta_{p}\circ\mu_{p,q}$
$\displaystyle=\frac{q}{(p,q)}\beta_{q}$ (141a)
$\displaystyle\mathcal{P}\circ\mu_{2r,m}$
$\displaystyle=\begin{cases}\frac{2r}{(2r,m)}\mu_{4r,2m}\,\circ\,\mathcal{P}\quad&,m\,\,\text{even}\vspace{0.25cm}\\\
\frac{r}{(r,m)}\,\mu_{4r,\,\,m}\,\circ\,\mathcal{P}\quad&,m\,\,\text{odd}\\\
\end{cases}$ (141b) $\displaystyle\mathcal{P}(s\cdot x)$
$\displaystyle=\begin{cases}s^{\,\,}\cdot\mathcal{P}(x)\quad&,x\in
H^{2i-1}(\mathbb{Z}_{2r})\vspace{0.25cm}\\\
s^{2}\cdot\mathcal{P}(x)\quad&,x\in H^{\,\,2i\,\,}(\mathbb{Z}_{2r})\\\
\end{cases}$ (141c)

In particular, we can set $m=2n$ and $r=2mn$ in the second relation to obtain
$\mathcal{P}\circ\mu_{2mn,2n}=m\mu_{4mn,4n}\circ\mathcal{P}$ which can be
sloppily written as $\mathcal{P}(m\cdot x)=m^{2}\cdot\mathcal{P}(x)\mod 4mn$
for $x\in H^{*}(\mathbb{Z}_{2n})$. For our manipulations, there formulas
simply say that $\mathcal{P}$ behaves as a square and we are allowed to move
integers outside of the argument.

Eq. 3.3 of Ref. [55] tells us that for $x\in H^{*}(\mathbb{Z}_{p})$, $y\in
H^{*}(\mathbb{Z}_{q})$ and if $d=(p,q)$ and
$r=\mathrm{lcd}(p,q)=\frac{pq}{d}$, then
$\mu_{r,p}(x)\cup\mu_{r,q}(y)=\mu_{r,d}\left(x\cup y\right)$ where $x\cup
y=\mu_{d,p}(x)\cup\mu_{d,q}(y)$. We can combine this expression with the
linearity of $\mu$ to obtain $\left(a\cdot x\right)\cup\left(b\cdot
y\right)=ab\cdot\left(x\cup y\right)$ where $ap,bq|r$.

Now we introduce the Steenrod squares
$\mathrm{Sq}^{i}:H^{*}(\mathbb{Z}_{2})\rightarrow H^{*+i}(\mathbb{Z}_{2})$
that can be axiomatically defined, see [56]. These operations satisfy
$\mathrm{Sq}^{i}(x)=x\cup x$ if $x\in H^{i}(\mathbb{Z}_{2})$ and are
$\mathbb{Z}_{2}$ linear. $\mathrm{Sq}^{0}$ is the identity map and
$\mathrm{Sq}^{1}(x\cup y)=x\cup\mathrm{Sq}^{1}(y)+\mathrm{Sq}^{1}(x)\cup y$.
There is an interesting connection between $\mathrm{Sq}^{1}$ and $\beta_{2}$
which can be written as $\mathrm{Sq}^{1}=\rho_{2}\circ\beta_{2}$.

Additionally, on a compact $d$ dimensional manifold $X_{d}$, there exist the
Wu classes $\nu_{i}\in H^{i}(X_{d},\mathbb{Z}_{2})$ that for every $x\in
H^{d-j}(X_{d},\mathbb{Z}_{2})$ satisfy $\int_{X_{d}}\nu_{j}\cup
x=\int_{X_{d}}\mathrm{Sq}^{j}(x)\mod 2$. In particular, for 4-dimensional
closed manifolds $\nu_{1}=w_{1}^{TM}$, $\nu_{2}=w_{2}^{TM}+w_{1}^{TM}\cup
w_{1}^{TM}$, where $w_{i}^{TM}$ are the i-th Stiefel-Whitney classes of the
Tangent bundle of the manifold. For orientable manifolds $w_{1}^{TM}=0$. If
the manifold is additionally spin, $w_{2}^{TM}=0$. We point that if $x\in
H^{1}(X_{2n},\mathbb{Z}_{2})$ then $x^{2m}=\mathrm{Sq}^{1}(x^{2m-1})$.
Therefore, for $X_{2n}$ orientable
$\int_{X_{2n}}x^{2n}=\int_{X_{2n}}x^{2n-1}\cup w_{1}^{TM}=0$.

###  G.2 $\mathrm{O}(N)$ bundles

We start by introducing the characteristic classes of the $\mathrm{O}(N)$
bundles we use: the Stiefel-Whitney (SW) characteristic classes denoted by
$w_{i}\in H^{i}(\mathbb{Z}_{2})$ and the Pontraygin classes $p_{i}\in
H^{4i}(\mathbb{Z})$. As we working in $4$ dimensions, we are only interested
in the first four SW classes and the first Pontraygin class. Note that for
$\mathrm{O}(N)$ bundles, we have $w_{i}=0$ if $i>N$. Furthermore, if our
bundle can be restricted to a $\mathrm{SO}(N)$ bundle, then $w_{1}=0$ and the
Pontraygin class $p_{1}$ is related to the gauge curvature $F$ by

$\displaystyle
p_{1}[\mathrm{SO}(N)]=\frac{1}{8\pi^{2}}\int\Tr_{\mathrm{SO}(N)}[F\wedge F],$
(142)

where the trace is evaluated in the vector representation of $\mathrm{SO}(N)$.

There is a relation between $\mathrm{Sq}^{i}$ and the SW classes $w_{j}$ of a
general $\mathrm{O}(N)$ bundle (see [56], §8):

$\mathrm{Sq}^{j}w_{i}=\sum_{k=0}^{j}\binom{j-i}{k}w_{j-k}\cup w_{i+k}$

where $w_{0}=1$ and $\binom{x}{i}=x(x-1)\dots(x-i+1)/i!$. In particular,
$\mathrm{Sq}^{1}(w_{2n})=w_{2n}\cup w_{1}+w_{2n+1}$ and
$\mathrm{Sq}^{1}(w_{2n+1})=w_{1}\cup w_{2n+1}$.

The following identity is valid (see Theorem C of Ref. [57]):

$p_{1}=\mathcal{P}(w_{2})+2\cdot\left(w_{1}\cup\mathrm{Sq}^{1}\left(w_{2}\right)+w_{4}\right)\mod
4$ (143)

which can be equivalently written as
$p_{1}=\mathcal{P}(w_{2})+2\cdot\left(w^{2}_{1}\cup{w_{2}}+\mathrm{Sq}^{1}\left(w_{3}\right)+w_{4}\right)\mod
4.$

Given an $\mathrm{O}(A)$-bundle $\alpha$ and an $\mathrm{O}(B)$-bundle
$\beta$, we can construct the Whitney sum $\alpha\oplus\beta$ which is a
$\mathrm{O}(A+B)$. Naturally, the classes of $\alpha\oplus\beta$ can be
written in terms of the classes of $\alpha$ and $\beta$ (see e.g. Ref. [58]):

$\displaystyle w_{i}[\alpha\oplus\beta]$
$\displaystyle=\sum_{j=0}^{i}w_{j}[\alpha]\cup w_{i-j}[\beta]$ (144a)
$\displaystyle p_{1}[\alpha\oplus\beta]$
$\displaystyle=p_{1}[\alpha]+p_{1}[\beta]+\beta_{2}(w_{1}[\alpha])\cup\beta_{2}(w_{1}[\beta])$
(144b)

###  G.3 $\mathrm{PSU}(N)$ bundles

In Ref. [59] (and reviewed in Ref. [60]), the author classified
$\mathrm{PSU}(N)(\equiv\mathrm{PU}(N))$ bundles over spaces of dimension up to
4 by assigning two cohomology classes $t_{2}\in H^{2}(X,\mathbb{Z}_{N})$ and
$q_{4}\in H^{4}(X,\mathbb{Z})$ to each principal $\mathrm{PU}(N)$-bundle
$P\rightarrow X$. In our notation, $t_{2}$ corresponds to $u_{2}$ , the second
Stifel-Whitney class of the $\mathrm{PSU}(N)$-bundle and $q_{4}$ is defined as
the second Chern number of the complexified adjoint bundle that ends up being
minus the instanton number of the $\mathrm{PSU}(N)$-bundle252525the relative
minus sign with respect to Ref. [59] is due to the convention of the Lie
algebra. Here we take it to be Hermitian while in the original work they take
a anti-Hermitian Lie algebra.

$\begin{split}q_{4}(P):=&-ch_{2}^{\mathrm{PSU}(N)}(P\otimes\mathbb{C})\\\
=&-\frac{1}{8\pi^{2}}\int_{X}Tr_{\mathrm{Adj}}[{F}\wedge{F}\,]=:-p_{1}[\mathrm{PSU}(N)]\\\
\end{split}$ (145)

where $F$ is the curvature of the $\mathrm{PSU}(N)$ bundle and the trace is
taken over the adjoint representation.

We restrict to the case where $N$ is even from now on. The classes $q_{4}$ and
$t_{2}$ are related by $q_{4}(P)=(N+1)\mathcal{P}\left(t_{2}\right)\mod 2N$ or
in our notation

$p_{1}[\mathrm{PSU}(N)]=(N-1)\mathcal{P}(u_{2})\mod 2N$ (146)

Notice that
$\rho_{2}\mathcal{P}(u_{2})=\rho_{2}\rho_{N}\mathcal{P}(u_{2})=\rho_{2}(u_{2}\cup
u_{2})=\rho_{2}(u_{2})\cup\rho_{2}(u_{2})$, then
$\frac{1}{2}\int_{X_{4}}\mathcal{P}(u_{2})=\int_{X_{4}}\rho_{2}(u_{2})\cup
w_{2}^{TM}=0$ when $X_{4}$ is a closed spin oriented manifold (This matches
Witten’s result [61]).

For $N=2$ the formulas reduce to $\mathrm{PSU}(2)=\mathrm{SO}(3)$ where
$u_{2}=w_{2}$ is the SW class. Eq. 146 tells us then
$p_{1}=\mathcal{P}(w_{2})\mod 4$ which matches with Eq. 143 for $N=3$ and
$w_{1}=0$.

###  G.4 PSU(4) and SO(6) bundles

It is known that $\mathrm{SU}(4)/\mathbb{Z}_{2}=\mathrm{SO}(6)$, where the
$\mathbb{Z}_{2}$ is generated by $-\mathds{1}\in\mathrm{SU}(4)$. If a
$\mathrm{PSU}(4)$-bundle can be lifted to a $\mathrm{SO}(6)$-bundle there must
be a relation between the characteristic classes. Specifically, we can lift
the $\mathrm{PSU}(4)$ structure when $2\cdot u_{2}=0$ or in other words there
exists a class $w_{2}\in H^{2}(\mathbb{Z}_{2})$ such that
$u_{2}=\mu_{4,2}w_{2}=2\cdot w_{2}$. This condition can be understood by
thinking of $u_{2}$ as the obstruction of objects charged under the
$\mathrm{SU}(4)$ center, more precisely, if a object of charge $q$ goes around
a 2-cycle $\gamma$ the object picks a phase $\exp(i\frac{2\pi
q}{4}\int_{\gamma}u_{2})$. Now note that if $u_{2}=2w_{2}$, every object of
even charge does not feel $w_{2}$. The charge two objects precisely
corresponds to $\mathrm{SO}(6)$ representations. Therefore, the gauge
structure can be lifted to a $\mathrm{SO}(6)$ bundle.

In the previous situation $p_{1}[\mathrm{PSU}(4)]=\mathcal{P}(2\cdot
w_{2})=4\cdot\mathcal{P}(w_{2})=4\cdot w_{2}\cup w_{2}\mod 8$ which means
$p_{1}$ is divisible by $4$. The Pontraygin classes are related as
$p_{1}[\mathrm{SO}(6)]=\frac{1}{4}p_{1}[\mathrm{PSU}(4)]$262626See the
definition in terms of $F$ and use the following observation:
$Tr_{\bigwedge^{2}f_{N}}[U]=\frac{\Tr[U]^{2}-\Tr[U^{2}]}{2}$, then by setting
$U=e^{i\alpha_{a}T^{a}}e^{i\beta_{b}T^{b}}$ and expanding to order
$\mathcal{O}(\alpha_{a}\beta_{b})$, we get that
$Tr_{\bigwedge^{2}f_{N}}[T^{a}T^{b}]=(N-2)\Tr_{f_{N}}[T^{a}T^{b}]$. Now notice
that the $\mathrm{SO}(6)$ vector representation corresponds to the
antisymmetric square of the fundamental of
$\mathrm{SU}(4)$($\bigwedge^{2}f_{4}$).. This relation is consistent with the
previous condition by recalling that $p_{1}[\mathrm{SO}(6)]=w_{2}\cup
w_{2}\mod 2$.

##  Appendix H DSL in rectangular representations

###  H.1 Emergent compact $\mathrm{U}(p)_{g}\,\mathrm{QCD}_{3}$

We now extend the parton construction to spin systems where the on-site
$\mathrm{SU}(N)_{\mathrm{s}}$ representations with rectangular Young tableaux
with $n$ rows and $p$ columns. In this cases, we can represent the spin system
generators of $\mathrm{SU}(N)_{\mathrm{s}}$ as

$S^{\alpha}_{\,\,\beta}=\sum_{s=1,\dots,p}f^{{\dagger}\alpha s}f_{\beta s}$
(147)

where $s=1,\dots,p$ and $\alpha=1,\dots,N$. We then put the fermions at $n/N$
filling by gauging the $U(1)$ symmetry. At this point, the physical states are
in the antisymmetric representation $(\varpi_{np})$ of $\mathrm{SU}(Np)$.
Next, we gauge the $\mathrm{SU}(p)$ symmetry corresponding to rotations in the
$p$ index as this extracts the states that are trivial under $\mathrm{SU}(p)$.
In other words, it picks the irreps in the decomposition of $(\varpi_{np})$
under the branching
$\mathrm{SU}(Np)\longrightarrow\mathrm{SU}(N)\times\mathrm{SU}(p)$ that have a
trivial irrep in the second factor. From Eq. 18, we see that the only factor
we keep has a rectangular YT with $n$ columns and $p$ rows in the second
factors and thus have the YT we want in the first factor.

We can consider now $p$ copies of the mean-field Hamiltonians with
$IGG=U(1)_{g}$ and $N_{f}$ low energy Dirac fermions. We then obtain
$\mathrm{U}(p)\,\mathrm{QCD}_{3}$ with $N_{\\!f}$ fundamental fermions.
According to Ref. [12] and references therein, the monopoles are classified by
the topological flux along the Cartan directions of the gauge field, i.e.
along the diagonal $\mathrm{U}(1)^{p}$ subgroup of $\mathrm{U}(p)$, modulo the
action of the Weyl group that in this case is just permutation of the
$\mathrm{U}(1)$’s. The monopole with the lowest scaling dimension corresponds
to inserting a $2\pi$ flux along a single direction of the $\mathrm{U}(p)$
gauge field. The symmetry properties of this monopoles are similar to those of
$\mathrm{U}(1)\,\,\mathrm{QED}_{3}$, the difference being the Berry phase
factors comming from the filled bands.

###  H.2 Emergent $\mathrm{Sp}(p)_{g}\,\mathrm{QCD}_{3}$

Consider now spin $S=\frac{p}{2}$ representations of $\mathrm{SU}(2)_{s}$. In
this section, we will discuss a parton decomposition that has a larger gauge
redundancy than $\mathrm{U}(p)$, namely $\mathrm{Sp}(p)$
($\mathrm{Sp}(1)=\mathrm{SU}(2)$). The enhancement of the gauge redundancy is
a generalization of the $S=\frac{1}{2}$ [62, 63] and $S=1$ [64, 65] cases.

Consider $2p$ fermionic partons $f_{\alpha s}$ where
$\alpha=\uparrow,\downarrow$ and $s=1,\dots,p$. Introduce the parton operators

${F}=\left(\begin{array}[]{cccccc}f_{\uparrow 1}&\dots&f_{\uparrow
p}&-f^{{\dagger}\downarrow 1}&\dots&-f^{{\dagger}\downarrow p}\\\
f_{\downarrow 1}&\dots&f_{\downarrow p}&f^{{\dagger}\uparrow
1}&\dots&f^{{\dagger}\uparrow p}\\\ \end{array}\right),$ (148)

then the spin operators are written as

$\mathbf{S}=\frac{1}{2}f^{{\dagger}\alpha
s}\vec{\sigma}_{\alpha}^{\,\,\,\beta}f_{\beta
s}=\frac{1}{4}\Tr[{F}^{{\dagger}}\vec{\sigma}F].$ (149)

We see that there is an apparent larger $\mathrm{U}(2p)$ redundancy
$F\rightarrow FU^{{\dagger}}$ with $U\in\mathrm{U}(2p)$. Nevertheless, the $F$
operators satisfy a reality condition
$F^{{\dagger}}=\sigma^{2}\,F^{\top}\rho^{2}$ 272727$\rho^{a}$ are Pauli
matrices acting on particle-hole channel., then $U$ must satisfy
$U^{\top}\rho^{2}U=\rho^{2}$ and thus the gauge symmetry is $\mathrm{Sp}(p)$.
If we consider $p$ copies of a mean-field Hamiltonian with
$IGG=\mathrm{Sp}(1)=\mathrm{SU}(2)$, the resulting mean-field Hamiltonian has
$IGG=\mathrm{Sp}(p)$. This happen for the $\pi$-flux state on the square
lattice and zero flux state of the honeycomb lattice.

###  H.3 $\mathrm{Sp}(N)_{s}$ spin systems

Let $(\vartheta_{k})$ be the heighest wight representations of
$\mathrm{Sp}(N)_{s}$ that corresponds to the $k$-fold antisymmetric product of
the defining representations modulo contraction with the invariant
antisymmetric tensor282828Equivalently $(\vartheta_{k})$ is the largest
representation when we restrict the $(\varpi_{k})$ representation of
$\mathrm{SU}(2N)$ to its $\mathrm{Sp}(N)$ subgroup.. For a parton construction
for the case the irreps with heighest weight $p\vartheta_{N}$, we introduce
the parton operator

$F=\left(\begin{array}[]{cccccc}f_{1,1}&\dots&f_{1,p}&-f^{{\dagger}N+1,1}&\dots&-f^{{\dagger}N+1,p}\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\
f_{N,1}&\dots&f_{N,p}&-f^{{\dagger}2N1}&\dots&-f^{{\dagger}2N,p}\\\
f_{N+1,1}&\dots&f_{N+1,p}&f^{{\dagger}1,1}&\dots&-f^{{\dagger}1,p}\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\
f_{2N,1}&\dots&f_{N+1,p}&f^{{\dagger}N,1}&\dots&f^{{\dagger}N,p}\\\
\end{array}\right),$ (150)

where $f_{\alpha,s}$, with $\alpha=1,\dots,2N$ and $s=1,\dots,p$, are
fermionic annhilation operators. We then gauge the $\mathrm{Sp}(p)$ symmetry
that acts as $F\longrightarrow F\cdot U^{\top}$. In order to prove that
gauging this symmetry gives the right Hilbert space, recall that the Hilbert
space whose algebra is generated by $f_{\alpha,s}$ is the sum of the spinor
representations of $\mathrm{Spin}(4pN)$. This is so because we can construct
$4pN$ Majorana operators out of the $2pN$ complex fermions. Then, we need to
consider the branching under
$\mathrm{Spin}(4pN)\longrightarrow\mathrm{Sp}(N)\times\mathrm{Sp}(p)$ and find
the factors that are trivial in the second factor. We have not been able to
find a proof but using SAGE [66], we verified that gauging gives the right
answer for all pairs of $(N,p)$ such that $pN\leq 16$.

We can consider $pN$ copies of a mean-field Hamiltonian for $S=\frac{1}{2}$
systems with $IGG=\mathrm{SU}(2)_{g}$ and $2M$ low energy Dirac fermions. We
expect an emergent $\mathrm{Sp}(p)_{g}\,\mathrm{QCD}_{3}$ with $N_{\\!f}=NM$.
If we instead consider copies of mean-field Hamiltonians with
$IGG=\mathrm{U}(1)_{g}$, we expect to find emergent
$\mathrm{U}(p)_{g}\,\mathrm{QCD}_{3}$ with $N_{\\!f}=2NM$.

This motivate the possibility of a stable $\mathrm{U}(1)$ DSL on the vector
$\mathrm{SO}(5)(\cong\mathrm{Sp}(2)/\mathbb{Z}_{2})$ Heisenberg model on the
triangular lattice as the bare monopoles still carry $\bm{K}$ momenta[15, 16]
that cannot be cancelled by the zero-modes contributions. The monopole quantum
numbers can be easily found by starting from 27 and restricting the
$\mathrm{SO}(6)$ reps to $\mathrm{SO}(5)$ reps:

$\begin{split}\mathbf{20}&\longrightarrow\mathbf{14}\oplus\mathbf{5}\oplus\mathbf{1}\\\
\mathbf{15}&\longrightarrow\mathbf{10}\oplus\mathbf{5}\\\
\mathbf{1}&\longrightarrow\mathbf{1}\end{split}$ (151)

where $\mathbf{1}$ is the scalar, $\mathbf{5}$ is the vector, $\mathbf{10}$ is
the adjoint and $\mathbf{14}$ is the symmetric traceless representation.