Here $\delta_1$ is the inclusion map which maps $S^1$ to the boundary of the 2-dimensional disk $D^2$; $\gamma_2$ is the map that compactifies the boundary of $D^2$ to single point; $f_2$ is the map in notknow and $f_1$ is the map obtained by applying $\beta$ to $f_2$, i.e., $\beta[f_2]=f_1$. By knowing whether $f_1$ is a non-trivial map representing $(1,1)$ in demap we can deduce whether $f_2$ is a non-trivial soliton configuration. In the commutative diagram $\lambda$ is the homotopy lift of $f_2 \circ \gamma_2$. The fact that such a lift exists is because $D^2$ is homeomorphic to a two dimensional cube, i.e., a square, hence by the homotopy lifting property $\lambda$ exists.

Because the image of $\gamma_2 \, \circ \, \delta_1$ is a point, so does the image of $ f_2 \,\circ \gamma_2 \, \circ  \, \delta_1 = p \, \circ\, \lambda \,\circ\, \delta_1$. This implies $(\lambda \,\circ\, \delta_1)[S^1]$ is  in the kernel of the map $O(n)\xrightarrow{p}\frac{O(n)}{O(\frac{n}{2}) \times O(\frac{n}{2})}$. Since $(\lambda \,\circ\, \delta_1)[S^1]$ is projected to a point in the base space, it must be contained entirely in a single fiber
$O\left({n\over 2}\right)\times O\left({n\over 2}\right).$ Therefore the sought-after $f_1$ is given by

$$f_1=\lambda \,\circ\, \delta_1.$$
The above arguments allows us take the map $f_2$ as input and produce the map $f_1$ as output, i.e., we have constructed $\beta$.

In the following we apply the construction discussed above to the following proposed soliton configuration[It is illuminating to compare $ Q_{\rm sol}^{\mathbb{R}}$ with $Q_{\rm sol}^\mathbb{C}$ in gsolt, namely,
$Q_{\rm sol}^{\mathbb{R}}=\Re [Q_{sol}^{\mathbb{C}}]\otimes I + \Im[Q_{sol}^{\mathbb{C}}]\otimes E$.]

\begin{align}
\label{propsol}
f_2:(\theta,\phi)\ra Q_{\rm sol}^{\mathbb{R}}=& \left( n_1 XI + n_2 EE + n_3 ZI \right) \oplus Z \oplus Z ...
\end{align}

Because $Q_{\rm sol}^{\mathbb{R}}$ is a real symmetric matrix, it can be diagonalized by orthogonal transformation

\begin{align*}Q_{\rm sol}^{\mathbb{R}}=W \cdot \left[ \begin{pmatrix}
1	&	0	& 0	&	0	\\
0	& 	1	&	0	&	0	\\
0	&	0	&	-1	&	0	\\
0	&	0	&	0	&	-1
\end{pmatrix} \oplus Z \oplus Z... \right]  \cdot W^T\end{align*}

\begin{align*}
W(\theta,\phi)=& \begin{pmatrix}
\cos\frac{\theta}{2} \cos\phi	&	-\cos\frac{\theta}{2} \sin\phi	& -\sin\frac{\theta}{2}	&	0	\\
\cos\frac{\theta}{2} \sin\phi	&	\cos\frac{\theta}{2} \cos\phi	&	0	&	-\sin\frac{\theta}{2}	\\
\sin\frac{\theta}{2}			&	0								&	\cos\frac{\theta}{2} \cos\phi	&	\cos\frac{\theta}{2} \sin\phi	\\
0								&	\sin\frac{\theta}{2}			&	-\cos\frac{\theta}{2} \sin\phi	&	\cos\frac{\theta}{2} \cos\phi
\end{pmatrix} \oplus I ...
\end{align*}

Naively, one might think $W(\theta,\phi)$ is a mapping between $S^2$ and $O(n)$. However, this is not true. To see it, let's inspect $W(0,\phi)$ and $W(\pi,\phi)$,

\begin{align*}
W (0,\phi)=& \begin{pmatrix}
\cos\phi	&	-\sin\phi	& 0	&	0	\\
\sin\phi	&	\cos\phi	&	0	&0	\\
0			&	0								&	\cos\phi	&	\sin\phi	\\
0								&	0			&	-\sin\phi	&	 \cos\phi
\end{pmatrix} \oplus I ...,
\end{align*}

\begin{align*}
W (\pi,\phi)=& \begin{pmatrix}
0	&	0& -1	&	0	\\
0	&0	&	0	&	-1\\
1		&	0								&0	&0	\\
0								&	1			&0&	0
\end{pmatrix} \oplus I ...
\end{align*}

The fact that $W(0,\phi)$ depends on $\phi$ and $W(\pi,\phi)$ does not implies that we should view $W$ as a mapping between $D^2$ and $O(n)$, where $\theta=0$ corresponds to the boundary  while $\theta=\pi$ corresponds to the center of $D^2$ (i.e., the radius of $D^2$ is $\pi-\theta$). In fact, $W$ is the map $\lambda$ in the commutative diagram, namely,
Hence the map $f_1$ is given by

\begin{align}
\label{dedf1} f_1(\phi)=
W (0,\phi)=& \begin{pmatrix}
\cos\phi	&	-\sin\phi	& 0	&	0	\\
\sin\phi	&	\cos\phi	&	0	&0	\\
0			&	0								&	\cos\phi	&	\sin\phi	\\
0								&	0			&	-\sin\phi	&	 \cos\phi
\end{pmatrix} \oplus I ...\\=&\left\{\begin{pmatrix}
\cos\phi	&	-\sin\phi\\
\sin\phi	&	\cos\phi\end{pmatrix}\oplus 1...\right\}\oplus \left\{\begin{pmatrix}
\cos\phi	&	\sin\phi\\
-\sin\phi	&	\cos\phi\end{pmatrix}\oplus 1...\right\}. \notag
\end{align}

To summarize, given the map $f_2$ in propsol, we have obtained the  map $f_1$  in the commutative diagram via dedf1.

Now we are ready to determine whether propsol is a topological non-trivial soliton configuration. It is known that the following map from $S^1$ to $O(n/2)$

$$\tilde{f}_1(\phi)= \left\{\begin{pmatrix}
\cos\phi	&	\mp\sin\phi\\
\pm\sin\phi	&	\cos\phi\end{pmatrix}\oplus 1...\right\}$$
is a representative of the generator of
Thus the mapping class of $f_1$ in dedf1 is the $(1,1)$ element of $\pi_1(O(n/2)\times O(n/2))=\mathbb{Z}_2\times\mathbb{Z}_2$. It follows that  $f_2$ in propsol is a representative of the generator of
$\pi_2(\frac{O(n)}{O(n/2)\times O(n/2)})=\mathbb{Z}_2$, i.e., it is a soliton configuration.

§.§.§ The Berry phase of a self-rotating $\mathbb{Z}_2$ soliton

To calculate the Berry's phase due to a $2 \pi$ self-rotation of the soliton in propsol, we rotate the soliton configuration to produce $Q^\mathbb{R}(\theta,\phi,\tau)$ in the same way in the appendix <ref>. After all dust settles, we end up with

\begin{align*}
&Q^\mathbb{R}(\theta,\phi,\tau) = \Re [Q^{\mathbb{C}}(\theta,\phi,\tau) ]\otimes I + \Im[Q^{\mathbb{C}}(\theta,\phi,\tau) ]\otimes E= \\
\begin{pmatrix}
\cos \theta					&	0							&		\sin\theta \cos (\phi+\tau)	&		-\sin\theta \sin (\phi+\tau)		\\
0							&	\cos\theta					&		\sin\theta \sin (\phi+\tau)	&		\sin\theta \cos (\phi+\tau)			\\
\sin\theta \cos(\phi+\tau)	&	\sin\theta \sin(\phi+\tau)	&		-\cos\theta					&		0	\\
-\sin\theta \sin(\phi+\tau)	&	\sin\theta \cos(\phi+\tau)	&		0							&	-\cos\theta
\end{pmatrix} \oplus Z \oplus Z ...
\end{align*}

Similarly, we can extend the space-time to one extra dimension by defining
$$\tilde{Q}^\mathbb{R}(\theta,\phi,\t,u):=\Re [\tilde{Q}^{\mathbb{C}}(\theta,\phi,\t,u)]\otimes I + \Im[\tilde{Q}^{\mathbb{C}}(\theta,\phi,\t,u)]\otimes E.$$
(It's easy to check it suffices all the properties we want for extension). Plugging the extended $\tilde{Q}_2^\mathbb{R}$ into the WZW term, we find

\begin{align*}
W_{WZW}[Q^\mathbb{R}] =& 2 \pi i \left( -  \frac{1}{512 \pi^2}      \int\limits_{\mathcal{B}} \tr \left[  \tilde{Q}^\mathbb{R}   \, (d\tilde{Q}^\mathbb{R})^4 \right] \right)
= i \pi.
\end{align*}

Hence the soliton is again a fermion.

§.§ Complex class in $(3+1)$-D

The mass manifold in this situation is $U(n)$. $\pi_3$ (relevant to the existence of soliton) and the $\pi_5$ (relevant to the existence of the WZW term) are both stabilized for $n\ge 3$.  In the following, we shall restrict ourselves to such conditions.

The fact a unit soliton in this mass manifold is a fermion has already been discussed in [22]. We briefly repeat the argument here for completeness. To construct a static soliton we start from $n=2$ (as far as $\pi_3$ is concerned, it stabilizes for $n\ge 2$ with $\pi_3(U(n))=\mathbb{Z}$.)  Thus the $n=2$ unit soliton is just the degree one map of $S^3 \rightarrow
SU(2)\sim S^3~~$[Note that $\pi_3(U(n))=\mathbb{Z}$ originates from the $SU(n)$ part of $U(n)$. Among other things, it means that we can limit ourselves to the $SU(n)$ WZW term for the Berry phase calculation.]. We can choose the unit soliton configuration to be

\begin{align*}
Q_{\rm sol}^\mathbb{C}(\vec{\Omega}) = \begin{pmatrix}
\Omega_0 + i \Omega_3		&		i(\Omega_1 - i \Omega_2)		\\
i(\Omega_1 + i \Omega_2)	&		\Omega_0 - i \Omega_3			\\
\end{pmatrix}
\end{align*}

$$\Omega_0^2 +\Omega_1^2+\Omega_2^2+\Omega_3^2=1$$
are the coordinate on $S^3$. For $n\ge 3$  one can write the unit soliton as

\begin{align*}
Q_{\rm sol}^\mathbb{C}(\vec{\Omega})=\begin{pmatrix}
\Omega_0 + i \Omega_3		&		i(\Omega_1 - i \Omega_2)		\\
i(\Omega_1 + i \Omega_2)	&		\Omega_0 - i \Omega_3			\\
\end{pmatrix}\oplus 1\oplus 1...
\end{align*}

Next, we rotate the unit soliton in, say, the $\Omega_1$-$\Omega_2$ plane by $2 \pi$. The time-dependent soliton configuration can be written as

\begin{align*}
Q^\mathbb{C}(\vec{\Omega},\t)= &\left[\begin{pmatrix}
e^{-i\frac{\tau}{2}}			&	0	&	0		\\
0	&	e^{+i\frac{\tau}{2}}		&	0		\\
0			&	0			&	1
\end{pmatrix}\oplus 1...\right] \cdot Q_{\rm sol}^\mathbb{C}(\vec{\Omega}) \cdot
\left[\begin{pmatrix}
e^{+i\frac{\tau}{2}}			&	0	&	0		\\
0	&	e^{-i\frac{\tau}{2}}		&	0		\\
0			&	0			&	1
\end{pmatrix}\oplus 1 ..\right] \\
=&\left[ \begin{pmatrix}
1			&	0	&	0		\\
0			&	e^{+i \tau}		&	0		\\
0			&	0			&	e^{-i \tau}
\end{pmatrix}\oplus 1..\right] \cdot  Q_{\rm sol}^\mathbb{C}(\vec{\Omega})   \cdot
\left[\begin{pmatrix}
1			&	0	&	0		\\
0	&	e^{-i \tau}		&	0		\\
0			&	0			&	e^{+i \tau}
\end{pmatrix}\oplus 1..\right]
\end{align*}

where $\tau \in [0,2 \pi]= S^1$ is the time parameter. One can extend the configuration to $\mathcal{B} = S^3 \times D^2$, where $D^2$ is the two-dimensional disk with radius $u \in [0,\pi]$, by

\begin{align}
\label{3drtsol}
\tilde{Q}^\mathbb{C}(\vec{\Omega},\t,u)=& \left[\begin{pmatrix}
1			&	0	&	0		\\
0			&	\sin u \, e^{+i \tau}		&	-\cos u		\\
0			&	\cos u			&	\sin u \, e^{-i \tau}
\end{pmatrix}\oplus 1...\right] \cdot  Q_{sol}^\mathbb{C}(\vec{\Omega}) \cdot
\left[\begin{pmatrix}
1			&	0	&	0		\\
0	&	\sin u \, e^{-i \tau}		&	\cos u		\\
0			&	\cos u			&	\sin u \, e^{+i \tau}
\end{pmatrix}\oplus 1..\right]
\end{align}

Plugging 3drtsol into the WZW term gives

\begin{align*}
W_{WZW}[\tilde{Q}^\mathbb{C}] = &2 \pi i \left(  \frac{i}{480\pi^3}    \int\limits_{\mathcal{B}}  \,\tr \Big[   \left(\tilde{Q}^{\mathbb{C}\dagger}   d \tilde{Q}^\mathbb{C} \right)^5    \Big] \right)= i \pi.
\end{align*}

Hence the unit soliton is a fermion.

§.§ Real class in $(3+1)$-D

The mass manifold is $U(n)/O(n)$. Here $\pi_3$ (relevant to the existence of soliton) and $\pi_5$ (relevant to the existence of the WZW term) are both stabilized for $n\ge 6$. To write down the degree one soliton in $U(n)/O(n)$, let's first look at the fibration O(n)  U(n)  U(n)/O(n).Here the projection $p$ is defined by u  u_S=u^T ·u, where u∈U(n),  u_S ∈U(n)/O(n).After the homotopy groups are stabilized, the long exact sequence associated with ufib is given by

\begin{align*}
... & \pi_4(U(n)/O(n)) \ra \pi_3(O(n))\xrightarrow{i_*} \pi_3(U(n)) \xrightarrow{p_*}\pi_3 (U(n)/O(n))\ra\pi_2(O(n))...\\
... &~~~~~~~~~~~0~~~~~~~~\ra~~~~~~\mathbb{Z}~~~~\xrightarrow{i_*}~~~~~\mathbb{Z}~~~~~\xrightarrow{p_*}~~~~~~~~~~~\mathbb{Z}_2~~~~~~~\ra ~~~~~~0~~~~~...
\end{align*}

This implies  that we can construct the unit soliton in $U(n)/O(n)$ by taking a unit soliton in $U(n)$, namely $Q_{\rm sol}^\mathbb{C}$ in appendix <ref>, and perform the projection map in uprojj, namely,
$$ Q_{\rm sol}^\mathbb{R}=\left(Q_{\rm sol}^\mathbb{C}\right)^T \cdot Q_{\rm sol}^\mathbb{C}.$$

The time dependent $Q^\mathbb{R}(\vec{\Omega},\tau)$ can be constructed by the similar projection of $Q^\mathbb{C}(\vec{\Omega},\tau)$, i.e.,

$$Q^\mathbb{R}(\vec{\Omega},\tau)=\left(Q^\mathbb{C}(\vec{\Omega},\tau)\right)^T\cdot Q^\mathbb{C}(\vec{\Omega},\tau).$$

The extended $\tilde{Q}^\mathbb{R}$  can also be constructed by the same projection

$$\tilde{Q}^\mathbb{R}(\vec{\Omega},\tau,u)=\left(\tilde{Q}^\mathbb{C}(\vec{\Omega},\tau,u)\right)^T\cdot \tilde{Q}^\mathbb{C}(\vec{\Omega},\tau,u).$$
The result  $\tilde{Q}^\mathbb{R}(\vec{\Omega},\tau,u)$ can be substituted into the WZW term to obtain

\begin{align*}
W_{WZW}[Q^\mathbb{R}] =&  2 \pi i \left(  \frac{i}{960\pi^3}    \int\limits_{\mathcal{B}}   \, \tr \Big[   \left(\tilde{Q}^{\mathbb{R}\dagger}   d \tilde{Q}^\mathbb{R} \right)^5    \Big] \right)= i \pi
\end{align*}

Therefore  the unit soliton is again a fermion.

§ BOSONIZATION FOR SMALL FLAVOR NUMBER

In this appendix we discuss the bosonization in cases when $n$, the number of flavors, is less than the value necessary to stabilize $\pi_{D+1}({\rm mass~manifold})$, or the WZW term.

In some cases, although the homotopy group $\pi_{D+1}({\rm mass~manifold})$ is not yet stabilized, it already contains $\mathbb{Z}$ as a subgroup. For instance, for real class in $(3+1)$-D, $\pi_5(U(3)/O(3)) = \mathbb{Z} \times \mathbb{Z}_2$. The nonlinear sigma model derived from fermion integration in appendix <ref> contains the level-$1$ WZW term, which is $2 \pi$ times the topological invariant of the $\mathbb{Z}$ part of $\pi_5$. This is also true for $n=6,8$ of the non-charge-conserved cases in $(2+1)$-D. In these cases the story is unchanged.

In other cases  $\pi_{D+1}({\rm mass~manifold})$  is a finite abelian group, e.g., $\mathbb{Z}_2$, or even $0$.  This requires a case-by-case study. Here, instead of attempting at studying all possible cases, we shall focus on the case that is relevant to the applications in section <ref> of the main text, namely, the case of $n=2$ complex class in $(2+1)$-D (which is relevant to the discussions in subsection <ref> of the main text).

§.§ Complex class in $2+1$ D with $n=2$

The mass manifold is
$$\frac{U(2)}{U(1)\times U(1)} = S^2$$

\begin{align*}
\pi_4( S^2) =  0 ,~~{\rm but}~~\pi_3(S^2)=\mathbb{Z}.
\end{align*}

The generator of $\pi_3(S^2)$ is called Hopf map [58]. The question at hand is whether this signifies the presence of a topological term in the nonlinear sigma model. A similar situation occurs for, e.g., the non-linear sigma model describing the anti-ferromagnetic spin chains in $(1+1)$-D. There, the mass manifold is $S^2$ and $\pi_3(S^2)=0$ but $\pi_2(S^2)=\mathbb{Z}$. In the nonlinear sigma model, there is a topological term associated with the $\pi_2$ in the non-linear sigma model, the $\theta$ term, which is responsible for the difference between the integer and half-integer spin chains [59, 60, 40].

To answer the question posed above, the derivation in appendix <ref> is not applicable. This is because the Hopf term (or the $\theta$ term) is invariant under arbitrary infinitesimal deformation of $Q^{\mathbb{C}}$.

§.§ Mass manifold enlargement

One way to proceed is to enlarge the mass manifold (or the target space of the order parameter). The idea [61] is as follows. If two order parameter configurations cannot be deformed into each other, as in the case where configurations correspond to different elements of $\pi_D$ (in this case $\pi_3$), we can enlarge the mass manifold so that after the enlargement, one configuration can be continuously deformed to the other. One can then compute the Berry phase difference caused by infinitesimal order parameter variation using the method explained in <ref>, and integrate the result. However, it is important to note that enlarging the mass manifold requires adding extra fermion flavors. It is important to make sure that the initial and the final order parameters couple to the added fermion flavors in a trivial way (i.e., in the added flavor space, the order parameters are the same constant) so that the Berry phase difference is originated from the original fermions. Finally, one accounts for the Berry phase by picking the coefficient in front of the $\pi_D$ (here $\pi_3$) topological invariant.

Using this technique, Abanov [51] enlarged $\frac{U(2)}{U(1)\times U(1)} = S^2$ to
$$\frac{U(l+1)}{U(l)\times U(1)} = \mathbb{CP}^l$$
and showed that the nonlinear sigma model from the $n=2$ fermion integration contains a $\theta=\pi$ Hopf term. In the following we will choose an alternative enlargement, namely,
$$\frac{U(2)}{U(1)\times U(1)} = S^2\ra \frac{U(4)}{U(2)\times U(2)}.$$
We will show that the result is consistent with that of Abanov. Because of the Hopf term, the unit soliton has fermion statistics [52].

§.§ The derivation of the Hopf term

The Hopf map is a map from $S^3$ with coordinate $(\Omega_0,\Omega_1,\Omega_2,\Omega_3 )$ where $\sum_{i=0}^3\Omega_i^3=1$ to
$S^2$ with coordinate $(n_1,n_2,n_3)$ where $\sum_{i=1}^3n_i^2=1.$ More explicitly,

Ω⃗ n⃗= z^†σ^a z
where    z:=
Ω_0+ i Ω_1
Ω_2 + i Ω_3  .
The $Q^{\mathbb{C}}$ of the non-linear sigma model is given by

\begin{align}
\label{qzeig}
Q^\mathbb{C}	(\vec{\Omega}) = \sum_{a=1}^3n^a(\vec{\Omega})\cdot  \sigma^a
= 2 \, z(\vec{\Omega}) \, z(\vec{\Omega})^\dagger - I_{2},
\end{align}
where $z$ is given by zexpr. In writing down the 2nd equality we have used the identity
$$\sum_{a=1}^3\sigma^a_{ij} \sigma^a_{kl} = 2\delta_{il} \delta_{jk} - \delta_{ij} \delta_{kl}. $$
In qzeig the $2\times 2$ matrix $Q^\mathbb{C}$ has eigenvalues $\pm 1$, and $z$ is the eigenvector associated with eigenvalue $+1$.

In the following, we enlarge the order parameter so that $Q^\mathbb{C}$ can be deformed to $\sigma_z$. To do so, we add two additional fermion flavors and enlarge the mass manifold to
$$\frac{U(4)}{U(2)\times U(2)}.$$

In the enlarged space the order parameter is given by

=  Q^ℂ(Ω⃗)    0
0      -Z       =  2   z(Ω⃗)   z(Ω⃗)^†- I_2     0
0      -Z
where $z(\vec{\Omega})$ is given by zexpr. Here the fermions associated with extra flavors couple to the mass term $Y\otimes (-Z)$ (see table <ref>). Although the $Q^{\mathbb{C}}$ given by qzeig cannot be deformed to a constant configuration in the space $\frac{U(2)}{U(1)\times U(1)} = S^2$ (i.e., within the first $2 \times 2$ block) because

it is possible to deform the $4 \times 4$ ${Q'}^{\mathbb{C}}$ to a constant matrix.  This is because
$$\pi_3\left( \frac{U(4)}{U(2)\times U(2)}\right)=0.$$

Now we explicitly construct such a deformation. First we rewrite extendedHopf as

\begin{align*}
= \begin{pmatrix} 2 \, z'(\vec{\Omega}) \, {z'}(\vec{\Omega})^{\dagger} - I_3 & 0  \\  0  & +1 \end{pmatrix} \\
&\text{where }  z'(\vec{\Omega}) := \begin{pmatrix} \Omega_0+ i \Omega_1 \\ \Omega_2 + i \Omega_3 \\0 \end{pmatrix}.
\end{align*}

We then write down a continuous deformation, as a function of $u$, as follows
=  2   z̃^'(Ω⃗,u)     z̃^'†(Ω⃗,u)   - I_3     0
0      +1
z̃^'(Ω⃗,u)  :=

-cosu ( Ω_0+ i Ω_1 ), -cosu ( Ω_2 + i Ω_3 ) , sinu ^T    for  u∈[π/2, π]

cosu, 0, sinu ^T   for  u∈[0,π/2).
This extends the configuration from the space-time $\mathcal{M}=S^3$ at $u=\pi$ to a four dimensional disk $\mathcal{B}=D^4$ with $u\in[0,\pi]$. Here $\partial{\mathcal{B}} = \mathcal{M}$ and with $u$ as the radial direction of $D^4$. For $u=\pi$ qwext reduces to
extendedHopf, while for $u=0$
$${\tilde{Q'}}^\mathbb{C}(\vec{\Omega},u=0)= \begin{pmatrix} 1&0&0&0\\
0&-1&0&0\\0&0&-1&0\\ 0&0&0&+1 \end{pmatrix}=Z\oplus (-Z).$$
Therefore at $u=0$ and $u=\pi$ the fermions associated with the added flavors couples to exactly the same mass term $Y\otimes (-Z)$ according to table  <ref>.

qwext constitutes an extension we need to define the WZW term (which is stabilized at $n=4$).
Now we can plug qwext into the WZW term of the $\frac{U(4)}{U(2)\times U(2)}$ non-linear sigma model. When all dust settles we obtain

\begin{align*}
W_{\rm WZW}[\tilde{Q'}^{\mathbb{C}}] = & -  \frac{2\pi i}{256 \pi^2}      \int\limits_{\mathcal{B}} \tr\left[  \tilde{Q'}^{\mathbb{C}}   \, (d \tilde{Q'}^{\mathbb{C}})^4 \right]=i \pi.\\
\end{align*}

This result agrees with that of  Ref.[51] and suggests the existence of a $\theta=\pi$ Hopf term.

§.§ Gauging small $n$ non-linear sigma models

In appendix <ref>, we have shown how to gauge the nonlinear sigma models. Recall that the non-trivial gauge coupling terms all originate from the WZW term. For small $n$, the WZW term does not exist. One might think we need to re-derive the gauging procedure. Fortunately, we can use the mass manifold enlargement idea discussed in the preceding subsection to derive the  coupling between $Q^{\mathbb{C},\mathbb{R}}$ and the gauge field. Without going into details we (1) add additional fermion flavors until the WZW term is stabilized. (2) Proceed as usual to gauge the continuous symmetries. (3) Restrict $Q^{\mathbb{C},\mathbb{R}}$ to the proper sub-mass manifold and the gauge group to the proper subgroup (so that the gauge field does not couple to the added fermion flavors). Following this procedure, we gauged the small $n$ nonlinear sigma model following the same try-and-error method.

As an example, we shall write down the charge-U(1) gauged nonlinear sigma model for $n=2$ in the $(2+1)$-D complex class. As shown in appendix <ref>, the bosonized theory the $S^2$ nonlinear sigma model with the $\theta=\pi$ Hopf term. Plugging $Q^\mathbb{C} =n^a \sigma^a$ into the gauge coupling part in gaugedWZW2C, we arrive at

\begin{align}
W[\boldsymbol{\beta} ,A] = {1\over \lambda_3} \int\limits_{\mathcal{M}} d^3 x \, (\partial_\mu \hat{n})^2 + i\pi H(\hat{n})  +  \int\limits_{\mathcal{M}} d^3x \left[ i A_\mu \left( \frac{1}{8\pi} \epsilon^{abc} \epsilon^{\mu\nu\rho} n^a  \partial_\nu  n^b  \partial_\rho n^c \right) \right], \notag
\end{align}

where the last term makes the $S^2$ solitons carry $U(1)$ charge. For $Q^\mathbb{C}$ in the $l=0$ and $l=2$ component of te mass manifold, we have a constant configuration $Q^\mathbb{C} = \pm I \in \frac{U(2)}{U(2) \times U(0)}$. Plugging it into gaugedWZW2C, we get

\begin{align}
\pm \frac{i}{4\pi}  \int\limits_{\mathcal{M}} d^3 x \, \epsilon^{\mu\nu\rho} A_\mu \partial_\nu A_\rho,
\end{align}

which gives $\sigma_{xy}=\pm 1$.

§ MASSLESS FERMIONS AS THE BOUNDARY OF BULK TOPOLOGICAL INSULATORS/SUPERCONDUCTORS

The idea behind our bosonization is to fluctuate the bosonic order parameters ($Q^{\mathbb{C}}$ or  $Q^{\mathbb{R}}$) to restore the full emergent symmetries of the massless fermion theory.  These order parameters are chosen so that  when they are static, any $Q^{\mathbb{C},\mathbb{R}}(\v x)$ configuration will fully gap out the fermions.
As shown in appendix <ref>, a static $Q^{\mathbb{C},\mathbb{R}}(\v x)$ configuration breaks at least some of the emergent symmetries. Conversely, if the full emergent symmetries are unbroken the fermion spectrum should remain gapless. Putting it succinctly, the emergent symmetries protect the gapless fermions.

The above situation reminds us of the boundary gapless modes of SPTs. Therefore, it is natural to suspect that each of the gapless fermion theories can be realized at the boundary of certain emergent-symmetry-protected SPT.
In this appendix, we show that this is indeed the case. Moreover, we shall construct the bulk SPT explicitly.

§.§ The $\mathbb{Z}$ classification

As discussed in Ref.[6], the classification of free fermionic SPTs can be determined by checking how many copies of  the boundary theory can be “stacked” together before a symmetry allowed mass term emerges. For example, a $\mathbb{Z}_N$ classification implies, after stacking $N$ copies of the massless fermion theory, a mass term can be found without breaking any of the protecting symmetry (here the emergent symmetries). In the following, we show that the emergent-symmetry-protected SPT has  $\mathbb{Z}$ classification.

For the sake of generality, we shall use the the Majorana fermion representation, even for complex classes.  $N$ copies of the boundary theory is described by the gamma matrices and the matrices that execute symmetry transformations,

\begin{align*}
\Gamma_i^{(N)} 	=& \G_i \otimes I_N,~~i=1,...,d \\
T^{(N)} 		=& t \otimes I_N \\
U^{(N)}			=& u \otimes I_N
\end{align*}

Here $t$ and $u$ are orthogonal matrices obeying $\{t,\G_i\}=[u,\G_i]=0$. The symbol $t$ and $u$ stand for anti-unitary and unitary, respectively. It is important to note that $t$ and $u$ represent the complete set of anti-unitary and unitary  transformation matrices, from the product of which all symmetry matrices can be constructed. In addition, $\G_i, t, u$  stand for the gamma and symmetry matrices for one copy of the massless fermion theory.

Existence of a mass term for the stacked massless fermion theory, implies that there exist a matrix $M^{(N)}$ that anti-commutes with all of the gamma matrices. The general form of $M^{(N)}$ is

$$M^{(N)}=m_s\otimes A_N+m_a\otimes S_N$$
where $m_s,S_N$ and $m_a, A_N$ are symmetric and anti-symmetric matrices, respectively.  Since  $\G_i^{(N)}=\G_i \otimes I_N,  ~T^{(N)} = t \otimes I_N, ~U^{(N)}= u \otimes I_N$  it follows that

\begin{align*}
\{m_{s,a} ,\G_i \} = \{m_{s,a},t\}=[m_{s,a},u]=0
\end{align*}

If such a $m_a\ne 0$ exists, we can use it as the mass term for the original massless fermion theory. This contradicts the statement that under the protection of emergent symmetry there is no mass term. Thus $m_a=0$ and $M^{(N)}$ reduces to M^(N)=m_s⊗A_N.

On the other hand, $m_{s}$ can then be used to construct an anti-unitary symmetry. By our assumption, such anti-unitary symmetry matrix $m_s$ must be the product $t$'s and $u$'s.  Thus the matrix

\begin{align*}
T^{(N)\prime} =m_{s} \otimes I_N
\end{align*}

is an anti-unitary symmetry matrix of the stacked fermion theory. However such $T^{(N)\prime}$ commutes with  MNA which is a contradiction.
(Recall that in Majorana fermion representation, a mass matrix must anti-commute with all anti-unitary symmetry matrices.) Therefore ${M}^{(N)}$ can not exist for any $N$. Consequently, the classification of the massless fermion theory must be $\mathbb{Z}$.

§.§ Construction of the bulk SPT

To construct the bulk SPT, we follow the “holographic construction” in Ref.[62]. In the following, we just summarize the results.

For a massless fermion theory, with gamma matrices $\{\G_i | \, i=1,2,...,d \}$, anti-unitary symmetry $t$, and unitary symmetries $\{u\}$, we can construct the bulk matrices,

\begin{align*}
\G_i^{\rm(bk)}= &
\begin{cases}
\G_i \otimes Z 				&\text{ for } i=1,...,d \\
I_{\rm dim(\G_i)} \otimes X &\text{ for } i=d+1
\end{cases}\\
T^{\rm(bk)}= & t \otimes Z \\
U^{\rm(bk)} = & u \otimes I
\end{align*}

Here the label $\rm(bk)$ is for distinguishing the bulk from the boundary matrices. In [62], it's shown that as long as the boundary massless fermion is irreducible, and  $t$,$u$ prohibit any mass term,  then there is single allowed bulk mass term which respects all the symmetries[Here the irreducibility means the gamma and the symmetry matrices cannot be  simultaneously block-diagonalized non-trivially. The fact that this is true for our case is because the inclusion of the full emergent symmetries. (Proof omitted.)]. Such mass term is given by

\begin{align*}
M^{\rm(bk)} = I_{\rm dim(\G_i)} \otimes Y.
\end{align*}

The above mass term can be used to regularize and gap out the fermion in the bulk. In Wilson's regularization, the SPT (single-particle) Hamiltonian in momentum space is given by

\begin{align*}
h^{\rm(bk)}(\v k) = \sum\limits_{i=1}^{d+1} \sin k_i \, \Gamma_i^{\rm(bk)} + \left(d+1 + m_B - \sum\limits_{i=1}^{d+1} \cos k_i \right) M^{\rm(bk)}
\end{align*}

When $m_B<0$, and when the lattice is cut open in the $(d+1)$th direction (actually the gapless boundary modes exist when the cut is along any direction), the boundary low energy theory is that of the original massless fermions.

§ THE DECOUPLING OF THE CHARGE-SU(2) GAUGE FIELD FROM THE LOW ENERGY NON-LINEAR SIGMA MODEL AFTER CONFINEMENT

In this appendix, we  show that the charge-SU(2) gauge field is not coupled to o5. To recap, the charge-SU(2) singlet $Q^\mathbb{R}$ is given by
Q^ℝ=n_i N_i  where      N_i = (YXY, IYY, YZY, IIX, IIZ ) .
Following appendix <ref> after gauging the charge-SU(2) symmetry, the $\frac{O(8)}{O(4) \times O(4)}$ nonlinear sigma model with $k=1$ WZW term becomes

\begin{align}
W[Q^\mathbb{R}, a] =&\frac{1}{4 \lambda_3} \int\limits_{\mathcal{M}} d^3 x \, \tr\Big[\Big(\p_\mu Q^\mathbb{R}+i [a_\mu,Q^\mathbb{R}]\Big)^2\Big]
-  \frac{ 2 \pi i }{512\pi^2}   \Big\{   \int\limits_{\mathcal{B}}  \tr \Big[\tilde{Q}^\mathbb{R}    \,\left( d \tilde{Q}^\mathbb{R} \right)^4 \Big] \notag\\
+& 8 \int\limits_{\mathcal{M}} \tr \Big[ i a Q^\mathbb{R} (dQ^\mathbb{R})^2 - (a Q^\mathbb{R})^2 dQ^\mathbb{R} - \frac{i}{3} (a Q^\mathbb{R})^3 + i a^3 Q^\mathbb{R} -a Q^\mathbb{R} f - a f Q^\mathbb{R} \Big]
\Big\}.
\label{gzero}
\end{align}

Since all $N_i$ commute with the charge-SU(2) group, it follows that $Q^\mathbb{R}=n_i N_i$ commutes with charge-SU(2) gauge field $a$. Hence the gauge coupling term in the stiffness term vanishes.

To show this is also true for the gauged WZW term part, we shall take the
$$\tr\left[a Q^\mathbb{R} (dQ^\mathbb{R})^2 \right]$$
term in gzero as an example. Plugging in $Q^\mathbb{R}=n_i N_i$, we obtain

$$  \tr\left[a Q^\mathbb{R} (dQ^\mathbb{R})^2 \right] = \sum_{i,j,k}  \tr[a \, N_i N_j N_k] n_i \, dn_j \, dn_k.$$

In the following we shall prove that each term in the sum vanishes, i.e.,
$$\tr[a \, N_i \, N_j \,  N_k]=0~~~~\forall (i,j,k).$$

To achieve that we insert a $N_l^2=1$ where $l\ne i,j,k$ into the trace and leave it invariant, i.e.,

$$\tr[a N_i N_j N_k]=\tr[N_l^2 a  N_i N_j N_k].$$

Due to the commutivity between $N_l$ and $a$ and the anti-commutivity between $N_l$ and each of the $N_{i,j,k}$, we can move one $N_l$ all the way to the right end and use the cyclic property of trace to put it back to the front

$$\tr[N_l^2  a N_i N_j N_k]=-\tr[N_l a  N_i N_j N_k N_l]=-\tr[N_l N_l a  N_i N_j N_k ]=-\tr[ a  N_i N_j N_k ].$$

$$\tr[ a  N_i N_j N_k ]=-\tr[ a  N_i N_j N_k ]\Rightarrow \tr[ a  N_i N_j N_k ]=0.$$

This proof can be applied to all gauge coupling terms in gzero because there is an odd number of $Q^\mathbb{R}$s for every term that  couples to the gauge field. Therefore the charge-SU(2) gauge field is not coupled to $Q^\mathbb{R}=n_i N_i$.

§ THE WZW TERM IN THE  $(3+1)$-D REAL CLASS NON-LINEAR SIGMA MODEL

In this section, we will show that upon the decomposition in  U(1)_S5 of subsection <ref>, namely,
$$Q^\mathbb{R}(x)= e^{i\theta(x)} \cG_S(x),$$
the contribution of the WZW term is solely from the $\cG_S(x)$ part, i.e.,

\begin{align*}
\tr\left[ \left( Q^{\mathbb{R}\dagger} dQ^{\mathbb{R}} \right)^5 \right] = \tr\left[ \left( \cG_S^\dagger d\cG_S \right)^5 \right]
\end{align*}

First, note that one can at most choose $d\theta$ once in the expansion of $\tr\left[ \left( Q^{\mathbb{R}\dagger} dQ^{\mathbb{R}} \right)^5 \right]= \tr\left[(\cG_S^\dagger d\cG_S + i d\theta)^5\right]$, otherwise the differential form vanishes because $(d\theta)^2=0$. The only term that can possibly survive other than $\tr\left[(\cG_S^\dagger d \cG_S )^5\right]$ is then of the form

\begin{align*}
\tr\left[d \theta \left( \cG_S^\dagger d\cG_S \right)^4 \right]
= d \theta  \left( \cG_S^\dagger d\cG_S \right)^a \left( \cG_S^\dagger d\cG_S \right)^b \left( \cG_S^\dagger d\cG_S \right)^c \left( \cG_S^\dagger d\cG_S \right)^d \tr\left[  t^a t^b t^c t^d\right]
\end{align*}

Here $\{ t^a\}$ are the complete basis for the generators of $SU(n)$ in the fundamental representation (note that $\cG_S$ are the symmetric special unitary matrices, which are special kind of unitary matrices). In the following, we will show that for every term from the trace $\tr\left[  t^a t^b t^c t^d\right]$, it is at least symmetric with respect to two of the  indices in $a,b,c,d$. If so, because the scalar valued one forms $\left( \cG_S^\dagger d\cG_S \right)^a$ anti-commute with each others $\tr\left[d \theta \left( \cG_S^\dagger d\cG_S \right)^4 \right]$ vanishes.

We shall choose the conventions

\begin{align}
\label{tatbOrtho}
\tr \left[ t^a t^a \right] = \frac{1}{2} \delta_{ab} \\
\label{tatbComm}
[t^a , t^b] = i f_{abc} t^c
\end{align}

where $f_{abc}$ is the structure constant for $SU(n)$. Here the Einstein summation convention is used.  $f_{abc}$ is real and totally anti-symmetric. We shall also define

\begin{align}
\label{dabc}
d_{abc} = 2 \, \tr\left[ \{t^a,t^b \}t^c \right]
\end{align}

It can be shown simply that due to the cyclic property of trace and the hermiticity of $t^a$, $d_{abc}$ is real and totally symmetric with respect to $a,b,c$.

As a pre-step, we would calculate $t^a  t^b$. Because the identity matrix $I_n$ together with $\{ t^a \}$ form a complete basis for all $n\times n$ complex matrices, we can decompose $t^a t^b$ in terms of them. The coefficients can be calculated making use of tatbOrtho, tatbComm, and dabc,

\begin{align}
\label{tatb}
t^a t^b  =& \frac{1}{n}\tr \left[ t^a t^b\right]I_n +  \tr\left[\left(\{t^a,t^b \} + [t^a,t^b]\right)t^c \right] t^c\notag \\
=& \frac{1}{2} \left[ \frac{1}{n} \delta_{ab} I_n + (d_{abc} + i f_{abc}) t^c\right].
\end{align}

The equation above implies

\begin{align}
\label{tatbAntiComm}
\{t^a,t^b \} = \frac{1}{n} \delta_{ab} I_n + d_{abc} t^c
\end{align}

For later usage, we will derive another formula for the product of two $f_{abc}$s. By direct expansion, one can prove the following identity

\begin{align*}
[t^a , [ t^b, t^c]] = \{ \{t^a,t^b\},t^c\} - \{ \{t^a,t^c\},t^b\}.
\end{align*}

Using of tatbComm and tatbAntiComm twice in the equation above, we  get

\begin{align}
\label{ff}
f_{abe} f_{cde} = \frac{2}{n} \left(\delta_{ac} \delta_{bd} - \delta_{ad} \delta_{bc} \right) + \left(d_{ace} d_{bde} - d_{ade} d_{bce} \right)
\end{align}

Now we can calculate $\tr \left[ t^a t^b t^c t^d\right]$  by applying tatb twice and carrying out the trace. After some algebra and the help of ff, we get

\begin{align*}
\tr \left[ t^a t^b t^c t^d\right]=& \frac{1}{4} \tr \left[\left( \frac{1}{n} \delta_{ab} I_n + (d_{abc} + i f_{abe}) t^e\right) \left( \frac{1}{n} \delta_{cd} I_n + (d_{cdf} + i f_{cdf}) t^f\right)\right]\\
=& \frac{1}{4} \Big[
\thead{
\frac{1}{n} (\delta_{ab} \delta_{cd} - \delta_{ac} \delta_{bd} + \delta_{ad} \delta_{bc} ) \\
+\frac{1}{2} (d_{abe} d_{cde} - d_{ace} d_{bde} + d_{ade}d_{bce} )\\
+\frac{i}{2} ( f_{abe} d_{cde} + f_{cde} d_{abe} )
\Big]
\end{align*}

By the symmetry properties of $\delta_{ab}$ and $d_{abc}$, every term is least symmetric with respect to two indices. For example, $f_{abe} d_{cde}$ is symmetric with respect to $c,d$. This concludes our proof.

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