B_{5}\boxtimes B_{5}~{}.$ (7.28)

On the other hand, the $S_{T}^{3}$ time reversal symmetry corresponds to the
distinct non-anomalous $\mathds{Z}_{5}$ 1-form symmetry

$\left\\{(\ell_{0},\ell_{0})~{},~{}(\ell_{2},\ell_{1})~{},~{}(\ell_{4},\ell_{2})~{},~{}(\ell_{3},\ell_{4})~{},~{}(\ell_{1},\ell_{3})\right\\}\in
B_{5}\boxtimes B_{5}~{}.$ (7.29)

Note that $B_{5}\boxtimes B_{5}$ is dual to the $\mathds{Z}_{5}$ Dijkgraaf-
Witten theory. The non-anomalous $\mathds{Z}_{5}$ subgroups found above
correspond to the two Rep$(\mathds{Z}_{5})$ subcategories in the
$\mathds{Z}_{5}$ gauge theory.

###  7.4 Example 3: $B_{25}$ TQFT (Galois conjugate of $SU(25)_{1}$ Chern-
Simons theory)

The $B_{25}$ TQFT has line operators, $\ell_{a}$, labelled by integers $0\leq
a\leq 24$ with topological spins $e^{\frac{2\pi ia^{2}}{25}}$. This theory has
all the cast of characters we have been introduced to above: a unitary
symmetry, anti-unitary symmetries, and non-invertible generalizations of both.

To understand this statement, note that we have the order four time-reversal
symmetry given by

$S_{T}\cdot\ell_{a}=\ell_{7a\text{ mod }25}~{}.$ (7.30)

Moreover, it is easy to see that

$S_{T}\times S_{T}=S_{C}$ (7.31)

where $S_{C}$ is the charge conjugation surface operator. Therefore, there is
also an $S_{T}^{3}$ invertible time-reversal symmetry.

Additionally, since we have non-trivial bosons, we have a non-invertible
symmetry implemented by the surface operator, $S_{B}$, where
$B=\\{\ell_{0},~{}\ell_{5},~{}\ell_{10},~{}\ell_{15},~{}\ell_{20}\\}$. We have

$S_{B}\cdot\ell_{a}=\ell_{0}+\ell_{5}+\ell_{10}+\ell_{15}+\ell_{20}~{},$
(7.32)

when $a\in\\{0,~{}5,~{}10,~{}15,~{}20\\}$, and zero otherwise. It is clear
that the surface operator

$S_{B}\times S_{T}=S_{T^{\prime}}$ (7.33)

has the same action on the lines as $S_{B}$. Note that the $B_{25}\boxtimes
B_{25}$ TQFT is dual to the $\mathds{Z}_{25}$ Dijkgraaf-Witten theory. The
$\mathds{Z}_{25}$ gauge theory has three Rep$(\mathds{Z}_{25})$ subgroups
coming from the three normal subgroups of $\mathds{Z}_{25}$ [76]. These non-
anomalous 1-form symmetries correspond to the three surface operators,
$S_{T}$, $S_{T}^{3}$, and $S_{B}$. This example shows that for a TQFT
described by an MTC $\mathcal{C}$, some of the symmetries specified by
Lagrangian subgroups of $\mathcal{C}\boxtimes\mathcal{C}$ correspond to
symmetries that are not of time-reversal type.

###  7.5 Example 4: $B_{125}$ TQFT (Galois conjugate of $SU(125)_{1}$ Chern-
Simons Theory)

Finally, let us look at an example where the non-invertible time-reversal
symmetry action can be observed at the level of the action on the line
operators. The $B_{125}$ TQFT has a time-reversal symmetry given by

$S_{T}\cdot\ell_{a}=\ell_{57a\text{ mod }125}~{},$ (7.34)

where $0\leq a\leq 124$. We have the fusion rules $S_{T}\times S_{T}=S_{C}$
where $S_{C}$ is the charge conjugation surface operator. We also have the
non-invertible symmetry $S_{B}$ where
$B=\\{\ell_{0},\ell_{25},\ell_{50},\ell_{75},\ell_{100}\\}$. The surface
operator, $S_{B}\times S_{T}$, is a non-invertible time-reversal symmetry. Its
action on the line operators is given by

$(S_{B}\times S_{T})\cdot\ell_{a}=B\text{-orbit of }\ell_{57a\text{ mod
}125}~{},$ (7.35)

if $\ell_{57a\text{ mod }125}$ braids trivially with $B$ and zero otherwise.
For example, we have

$(S_{B}\times
S_{T})\cdot\ell_{5}=\ell_{35}+\ell_{60}+\ell_{85}+\ell_{110}+\ell_{15}~{}.$
(7.36)

Note that $\theta_{\ell_{5}}=e^{\frac{2\pi i}{5}}$ and
$\theta_{\ell_{35}}=\theta_{\ell_{60}}=\theta_{\ell_{85}}=\theta_{\ell_{110}}=\theta_{\ell_{15}}=e^{\frac{2\pi
i4}{5}}$.

##  8 Conclusion

In this paper, we characterised non-invertible symmetries of $2+1$d TQFTs in
terms of their action on line operators and fusion spaces. We learned that in
these theories, non-invertible symmetries appear if and only if there are
condensable bosons (in the case of Abelian TQFTs, this condition simplifies to
the mere existence of bosons since all bosons are condensable in this case).
As a result, all non-invertible symmetries are emergent or non-intrinsic in
the sense that they appear via gauging 0-form symmetries in theories that have
only invertible symmetries.

Indeed, given a TQFT, $\mathcal{T}_{1}$, with a set of condensable bosons,
$B_{1}$, we can condense $B_{1}$ to get a TQFT, $\mathcal{T}_{2}$. If
$\mathcal{T}_{2}$ has condensable bosons, $B_{2}$, we can further condense
these bosons to get a new TQFT, $\mathcal{T}_{3}$. Continuing this way, we end
up with a TQFT, $\mathcal{T}_{n}$, which does not contain any condensable
bosons. In other words, $\mathcal{T}_{n}$ is discribed by a strongly-
anisotropic modular tensor category.

$\mathcal{T}_{1}\xrightarrow{\text{condense
}B_{1}}\mathcal{T}_{2}\xrightarrow{\text{condense
}B_{2}}\mathcal{T}_{3}\cdots\xrightarrow{\text{condense
}B_{n-1}}\mathcal{T}_{n}~{}.$ (8.1)

Our results show that $\mathcal{T}_{n}$ only has invertible surface operators.
Therefore, all non-invertible surface operators in the TQFT $\mathcal{T}_{1}$
arise starting from $\mathcal{T}_{n}$, and then sequentially gauging
symmetries (implemented by surface operators) to move in the reverse direction
in the sequence of TQFTs (8.1).

Using [18], we argued that our arguments above can be used to show that non-
invertible symmetries in 2+1d QFTs are non-intrinsic as well.

We reviewed the construction of surface operators by higher-gauging line
operators and related it to the classification of surface operators in terms
of surface action on the TQFT. In particular, we found an explicit criterion
to determine the 1-form symmetry group and the discrete torsion data of the
condensation defect from the action of the surface operator on the lines.
Moreover, we derived an explicit criterion to determine the invertibility of a
surface operator obtained from higher-gauging an invertible 1-form symmetry.

We reviewed the definition of time-reversal symmetry of a $2+1$d TQFT, and
generalized it to include non-invertible time-reversal symmetries. We argued
that non-invertible time-reversal symmetries of a TQFT described by an MTC,
$\mathcal{C}$, can be classified by studying the Lagrangian algebras in the
category $\mathcal{C}\boxtimes\mathcal{C}$. We analyzed various explicit
examples of Abelian TQFTs admitting invertible and non-invertible time-
reversal symmetries and used our definition to understand the curious fact
that non-invertible time reversal symmetries can sometimes act on lines in the
same way as standard non-invertible symmetries, even when there are lines with
complex topological spin.

The results in this paper point to several questions to further understand
non-invertible symmetries in 2+1d:

  * •

SymTFTs are powerful tools to study possible topological manipulations of a
QFT [77, 78, 30, 31]. As discussed above, any $2+1$d QFT with a non-invertible
symmetry can be related to a QFT with only invertible surface operators
through sequential gaugings. Therefore, the SymTFT of any $2+1$d QFT can be
related to that of a QFT with only invertible surface operators. It will be
very interesting to explore this direction further, and, in particular, to
understand how taking the Drinfeld center of (1.4) relates to constructing the
SymTFT.

  * •

It will be interesting to understand gauging 0-form symmetries at the level of
surface operators of a TQFT. This will make explicit how gauging 0-form
symmetries of strongly anisotropic TQFTs lead to TQFTs with non-invertible
symmetries. Better understanding this gauging may yield interesting insights
into the classification of topological phases in $2+1$d.

  * •

Motivated by the action on non-invertible symmetries in 1+1D, for example in
the Ising model, one would expect that the genuine line operators that are set
to zero by the non-invertible symmetry action are actually mapped to a
twisted-sector line. It will be very interesting to show this explicitly, and
more generally, to work out the action of the non-invertible symmetries
discussed in this paper on the non-genuine lines bounding surface operators.

  * •

Many results discussed in this paper can be applied to spin TQFTs. It will be
useful to systematically study possible condensations in this case.

  * •

We have seen glimmers that Galois theory has a role to play in our discussion.
For example, some of the anisotropic TQFTs that play a role in the questions
above also play a role in constructing Galois invariant theories. Moreover,
time reversal symmetries are clearly related to Galois actions. It is tempting
to try to understand this interplay even more generally.

Acknowledgments

We would like to thank Anuj Apte, Mahesh Balasubramanian, Francesco Benini,
Lakshya Bhardwaj, Mathew Bullimore, Yichul Choi, Christian Copetti, Iñaki
García Etxebarria, David Jordan, Zohar Komargodski, Kantaro Ohmori, Adrian
Padellaro, Pavel Putrov, Shu-Heng Shao, Ingo Runkel and Yunqin Zheng for
discussions and comments. M. B. would like to thank SISSA, Nordita, and Durham
University where part of this work was completed. R. R. would like to thank
Durham University, International Centre for Mathematical Sciences,
International Centre for Theoretical Sciences, Nordita, Queen Mary University
of London, Simons Centre for Geometry and Physics and SwissMap Research
Station for hospitality during various stages of this project. Some of the
results in this paper were first announced at Durham University. M. B. is
supported by the Royal Society under the grant, “Relations, Transformations,
and Emergence in Quantum Field Theory” and by STFC under the grant,
“Amplitudes, Strings and Duality.” No new data were generated or analyzed in
this study.

## Appendix A. Relating our definitions to Lagrangian subalgebras in folded
theories

In this appendix, we explain how to relate our “active” definition of
symmetries on lines and fusion spaces to a more “passive” definition in terms
of boundaries and Lagrangian algebras in folded theories. We start with the
case of Abelian TQFTs (see Sec. 3) and then move on to non-Abelian TQFTs (see
Sec. 4).

###  A.1 The Abelian case

Let us show that our definition of (non-invertible) symmetries of Abelian
TQFTs is equivalent to the classification of (non-invertible) symmetries in
terms of Lagrangian subgroups of the fusion group of the folded TQFT,
$\mathcal{T}\times\bar{\mathcal{T}}$. If the original theory, $\mathcal{T}$,
has fusion group, $A$, the folded theory has fusion group, $A\times A$. Using
$n_{S\cdot a}^{b},n_{S\cdot c}^{d}\in\\{0,1\\}$, for non-zero $n_{S\cdot
a}^{b}$ and $n_{S\cdot c}^{d}$ we can write condition 2 as

$n_{S\cdot a}^{b}n_{S\cdot c}^{d}=n_{S\cdot(a\times c)}^{b\times d}~{}.$ (A.1)

Given a set of non-zero integers, $n_{S\cdot a}^{b}$, satisfying (A.1) along
with conditions 1 and 3, the map

$n_{S\cdot a}^{b}\to(a,\bar{b})\in\mathcal{T}\times\bar{\mathcal{T}}~{},$
(A.2)

defines a Lagrangian subgroup of the fusion group $A\times A$ of
$\mathcal{T}\times\bar{\mathcal{T}}$. Conversely, given a Lagrangian subgroup,
$L<A\times A$, we can define a non-invertible surface, $S_{L}$, whose action
on the line operators are determined by the integers

$n_{S\cdot a}^{b}=\begin{cases}1&\text{ if }(a,\bar{b})\in L~{},\\\ 0&\text{
otherwise }~{}.\end{cases}$ (A.3)

Therefore, our definition of non-invertible symmetries of $\mathcal{T}$ is
equivalent to the classification of (non-invertible) symmetries in terms of
Lagrangian subgroups of the fusion group, $A\times A$, of the folded TQFT,
$\mathcal{T}\times\bar{\mathcal{T}}$.

###  A.2 The non-Abelian case

Let us show that our definition of (non-invertible) symmetries in non-Abelian
TQFTs is equivalent to the description of surface operators in terms of
Lagrangian algebras of $\mathcal{C}\times\bar{\mathcal{C}}$. To that end,
suppose we have a map, $S$, satisfying the conditions in our non-Abelian
definition. If $n_{S\cdot a}^{b}$ is non-zero, consider the algebra object

$L:=\sum_{a,b\in\mathcal{C}}n_{Sa}^{b}(a,\bar{b})\in\mathcal{C}\times\bar{\mathcal{C}}~{}.$
(A.4)

Conditions 1, 2, and 3 imply that we can define a Lagrangian algebra structure
on $L$. Indeed, using the “folding trick” on Fig. 9, we can convert the
surface operator, $S$, into a boundary condition. $\tilde{F}$ is then related
to the multiplication in the Lagrangian algebra, $L$, describing this boundary
condition [58][79, Appendix D].343434We refer the reader to [80, Appendix A]
for a review of these constraints and the relation of $\tilde{F}$ to
multiplication in the algebra in the context of 2d QFTs and their boundary
conditions. The constraint in Fig. 5 ensures that $\tilde{F}$ defines an
associative algebra structure on $L$. Moreover, the constraint in Fig. 11
ensures that this algebra is commutative. Using condition 3, we have

$\text{dim}(L)=\sum_{a,b\in\mathcal{C}}n_{Sa}^{b}d^{2}_{(a,\bar{b})}=\sum_{a,b\in\mathcal{C}}n_{Sa}^{b}d_{a}^{2}d_{b}^{2}=\text{dim}(\mathcal{C})=\sqrt{\text{dim}(\mathcal{C}\times\bar{\mathcal{C}})}~{}.$
(A.5)

Also, since $n_{S1}^{1}=1$, the trivial line,
$(1,1)\in\mathcal{C}\times\bar{\mathcal{C}}$, is an element of $L$ with
multiplicity 1. Therefore, $L$ is a connected commutative algebra. Corollary
3.8 in [58] states that a connected commutative algebra, say
$V=\sum_{v}n_{v}v$, in a unitary MTC, $\mathcal{D}$, with
$\text{dim}(V)^{2}=\text{dim}(\mathcal{D})$ is a Lagrangian algebra if and
only if the following inequality holds for all $v,w\in\mathcal{D}$

$n_{v}n_{w}\leq\sum_{u}N_{vw}^{u}n_{u}~{}.$ (A.6)

Using this theorem and condition 2, we see that the algebra, $L$, defined
above using the symmetry action, $S$, is indeed a Lagrangian algebra.

Conversely, given a Lagrangian algebra

$L=\sum_{(a,\bar{b})\in\mathcal{C}\times\bar{\mathcal{C}}}n_{(a,\bar{b})}(a,\bar{b})~{},$
(A.7)

in $\mathcal{C}\times\bar{\mathcal{C}}$, we can define a non-invertible
symmetry of the non-abelian TQFT with MTC, $\mathcal{C}$, as

$n_{S\cdot a}^{b}:=n_{(a,\bar{b})}~{}.$ (A.8)

From the Lagrangian property of $L$, the integers $n_{Sa}^{b}$ satisfy the
definition of a non-invertible symmetry.

## Appendix B. Determining all symmetries in further Abelian examples

In this section we determine all symmetries of several infinite classes of
Abelian TQFTs. All non-invertible symmetries in these theories are of the form
$S_{B}\times S_{U}$, where $S_{B}$ is a universal symmetry described in (3.26)
and $S_{U}$ is an (invertible) unitary symmetry.

###  B.1 $B_{p^{r}}$ prime abelian TQFTs for odd prime $p$

Consider the $B_{p^{r}}$ family of prime abelian TQFTs [37] with fusion rules
given by the group, $\mathds{Z}_{p^{r}}$, and topological twists,
$\theta_{\ell_{a}}=e^{\frac{2\pi ia^{2}}{p^{r}}}$, where we have labelled the
line operators as $\ell_{a},a\in\\{0,1,\dots,p^{r}\\}$. Before discussing the
general case, let us explicitly consider some theories with low rank:

  * •

Let us first consider the case $r=1$. We get a TQFT with $\mathds{Z}_{p}$
fusion rules which does not have any non-trivial bosons. Therefore, this TQFT
does not have any non-invertible symmetries. There is an invertible surface
operator, $S_{C}$, which implements the charge conjugation symmetry.

  * •

Now, let us consider $r=2$. We get a TQFT with lines
$\ell_{0},\ell_{1},\cdots,\ell_{p^{2}-1}$. We have the non-anomalous 1-form
symmetry $B\cong\mathds{Z}_{p}=\\{\ell_{0},\ell_{p}\\}$. We get a universal
non-invertible surface operator, $S_{B}$, of the type (3.26). We also have an
invertible charge conjugation surface operator, $S_{C}$. We have the fusion
rule $S_{C}\times S_{B}=S_{B}$.

  * •

Now, let us consider $r=3$. We get a TQFT with lines
$\ell_{0},\ell_{1},\cdots,\ell_{p^{3}-1}$. We have the non-anomalous 1-form
symmetry $B\cong\mathds{Z}_{p}=\langle\ell_{p^{2}}\rangle$. We get a non-
invertible surface operator, $S_{B}$, of the type (3.26). We have an
invertible charge conjugation surface operator, $S_{C}$. We also have the non-
invertible surface operator $S_{C}\times S_{B}$.

  * •

Now, let us consider $r=4$. We get a TQFT with lines
$\ell_{0},\ell_{1},\dots,\ell_{p^{4}-1}$. We have the non-anomalous 1-form
symmetry $B\cong\mathds{Z}_{p^{2}}=\langle\ell_{p^{2}}\rangle$. We get a non-
invertible surface operator, $S_{B}$, of the type (3.26). We also get a non-
invertible surface operator, $S_{K}$, for
$K=\langle\ell_{p^{3}}\rangle\cong\mathds{Z}_{p}<B$. We also have an
invertible charge conjugation surface operator, $S_{C}$. We have $S_{C}\times
S_{B}=S_{B}$, while $S_{C}\times S_{K}$ is a new surface operator.

For general $r$, from the expression for the topological spin of the line
operators

$\theta_{\ell_{a}}=e^{\frac{2\pi ia^{2}}{p^{r}}}~{},$ (B.1)

it is clear that the line operator, $\ell_{p^{\frac{r+1}{2}}}$, for odd $r$
and the line operator, $\ell_{p^{\frac{r}{2}}}$, for even $r$ are bosons.
These line operators generate a $B\cong\mathds{Z}_{p^{\frac{r+1}{2}}}$ for odd
$r$ and $B\cong\mathds{Z}_{p^{\frac{r}{2}}}$ non-anomalous 1-form symmetry for
even $r$. We get the non-invertible surface operator $S_{B}$ of type (3.26).
In fact, we get non-invertible surface operators for every subgroup $K<B$. We
also have the invertible charge conjugation surface operator, $S_{C}$. Fusing
the two types of surface operators we get new non-invertible surfaces
$S_{C}\times S_{K}$. Suppose

$S_{C}\times S_{K}=S_{K}~{}.$ (B.2)

For this to be true, the $K$-orbit of $\ell_{a}$ must be equal to the
$K$-orbit of $\bar{\ell}_{a}$ for all $\ell_{a}$ that braids trivially with
$K$. In other words,

$\ell_{a}+\ell_{a}\in K~{},$ (B.3)

for all $\ell_{a}$ that braid trivially with $K$.

Suppose $r$ is even. Consider the group
$K=\langle\ell_{p^{\frac{r}{2}+q}}\rangle$ for some $0\leq q\leq\frac{r}{2}$.
For different values of $q$, we get all the subgroups of $B$. The line
operators which braid trivially with $K$ satisfy

$e^{2\pi i\frac{ap^{\frac{r}{2}+q}}{p^{r}}}=1\implies a=0\text{ mod
}p^{\frac{r}{2}-q}~{}.$ (B.4)

Imposing the condition (B.3) we get

$l_{2p^{\frac{r}{2}-q}}\in K~{}.$ (B.5)

Since $l_{p^{\frac{r}{2}+q}}$ is a generator of $K$, there should be an
integer $k$ such that

$kp^{\frac{r}{2}+q}=2p^{\frac{r}{2}-q}\implies k=2p^{-2q}~{}.$ (B.6)

Here $k$ is an integer only for $q=0$. Therefore, we get the fusion rule

$S_{C}\times S_{B}=S_{B}~{}.$ (B.7)

For all proper subgroups $K<B$, we have

$S_{C}\times S_{K}\neq S_{K}~{}.$ (B.8)

Therefore, in total, we get $r+1$ surfaces. This counting agrees with the
classification of surface operators of the $B_{p^{r}}$ TQFT in terms of
condensation defects. Indeed, any subgroup of the $\mathds{Z}_{p^{r}}$ fusion
group can be higher gauged. There are $r+1$ distinct subgroups leading to
$r+1$ surface operators. The exact map between the surface operators discussed
above and condensation defects can be explicitly worked out using the
discussion in section 5.1.2.

Suppose $r$ is odd. Consider the group
$K=\langle\ell_{p^{\frac{r+1}{2}+q}}\rangle$ for some $0\leq
q\leq\frac{r+1}{2}$. For different values of $q$, we get all the subgroups of
$B$. The line operators which braid trivially with $K$ satisfy

$e^{2\pi i\frac{p^{\frac{r+1}{2}+q}a}{p^{r}}}=1\implies a=0\text{ mod
}p^{\frac{r-1}{2}-q}~{}.$ (B.9)

Imposing the condition (B.3) we get

$\ell_{2p^{\frac{r-1}{2}-q}}\in K~{}.$ (B.10)

Since $\ell_{p^{\frac{r+1}{2}+q}}$ is a generator of $K$, there should be an
integer $k$ such that

$kp^{\frac{r+1}{2}+q}=2p^{\frac{r-1}{2}-q}\implies k=2p^{-2q-1}~{}.$ (B.11)

Therefore, there is no integer solution for $k$ for any $q$. This argument
shows that when $r$ is odd, we have $S_{C}\times S_{K}\neq S_{K}$ for all $K$.
Again, we find that there are $r+1$ surface operators in total, which agrees
with the classification of surface operators in terms of condensation defects.

$A_{p^{r}}$ TQFTs is another class of prime abelian TQFTs which have the same
fusion group as $B_{2^{r}}$ TQFTs but different topological spins. The
classification of surface operators of $A_{p^{r}}$ is identical to that of
$B_{p^{r}}$ since these two classes of TQFTs are related by a Galois
conjugation [42].

###  B.2 $A_{2^{r}}$ prime abelian TQFTs $\cong U(1)_{2^{r}}$ Chern-Simons
Theory

Next, consider the $A_{2^{r}}$ family of prime abelian TQFTs with fusion rules
given by the group, $\mathds{Z}_{2^{r}}$, and topological spins,
$\theta_{\ell_{a}}=e^{\frac{2\pi ia^{2}}{2^{r+1}}}$. Before discussing the
general case, let us explicitly consider some theories with low rank:

  * •

Let us first consider the case $r=1$. We get the Semion model, which does not
have any non-trivial bosons. Therefore, this TQFT does not have any non-
invertible symmetries.

  * •

Now, let us consider $r=2$. We get a TQFT with lines
$\ell_{0},\ell_{1},\ell_{2},\ell_{3}$ and topological twists
$\theta_{\ell_{0}}=1,\theta_{\ell_{1}}=e^{\frac{2\pi
i}{8}},\theta_{\ell_{2}}=-1,\theta_{\ell_{3}}=e^{\frac{2\pi i}{8}}$. Once
again, there are no bosons, so this theory does not have any non-invertible
symmetries. However, there is clearly an invertible symmetries which permutes
$\ell_{1}\leftrightarrow\ell_{3}$. This invertible surface operator can be
obtained from higher-gauging the fermionic line operator, $\ell_{2}$.

  * •

Now, let us consider $r=3$. We get the line operators
$\ell_{0},\ell_{1},\ell_{2},\ell_{3},\ell_{4},\ell_{5},\ell_{6},\ell_{7}$ and
topological spins $\theta_{\ell_{0}}=1,\theta_{\ell_{1}}=e^{\frac{2\pi
i}{16}},\theta_{\ell_{2}}=e^{\frac{2\pi i}{4}},\theta_{\ell_{3}}=e^{\frac{2\pi
i9}{16}},\theta_{\ell_{4}}=1,\theta_{\ell_{5}}=e^{\frac{2\pi
i9}{16}},\theta_{\ell_{6}}=e^{\frac{2\pi
i}{4}},\theta_{\ell_{7}}=e^{\frac{2\pi i1}{16}}$. We have a boson which forms
the non-anomalous 1-form symmetry group $B=\\{\ell_{0},\ell_{4}\\}$. We have a
universal non-invertible symmetry of the type (3.26) implemented by the
surface $S_{B}$. The action of this surface on the line operators is

$S_{B}\cdot\ell_{0}=\ell_{0}+\ell_{4},~{}S_{B}\cdot\ell_{2}=\ell_{2}+\ell_{6},~{}S_{B}\cdot\ell_{4}=\ell_{0}+\ell_{4},~{}S_{B}\cdot\ell_{6}=\ell_{2}+\ell_{6}~{},$
(B.12)

and the action on $\ell_{1},\ell_{3},\ell_{5},\ell_{7}$ are all zero. We also
have the charge conjugation symmetry implemented by an invertible surface, say
$S_{C}$. Moreover, $S_{B}\times S_{C}=S_{B}$. This exhausts all non-invertible
symmetries of this TQFT.

  * •

Let $r=4$. In this case, we have the bosonic line operators
$B=\\{\ell_{0},\ell_{8},\ell_{16},\ell_{24}\\}\cong\mathds{Z}_{4}$. We get the
non-invertible surface operators, $S_{B}$, as well as $S_{K}$, where
$K=\\{\ell_{0},\ell_{16}\\}\subset B$. Moreover, we have the charge
conjugation surface operator $S_{C}$. By fusing these surface operators, we
get the non-invertible surface $S_{B}\times S_{C}=S_{C}\times S_{B}=S_{B}$ and
$S_{K}\times S_{C}=S_{C}\times S_{K}$. In total, we get four non-trivial
surface operators.

Now let us consider the case of general $r$. To identity the non-invertible
symmetries, we need to find the structure of bosons in the theory. Suppose $n$
is an integer such that $r=2n+1$ or $r=2n+2$. From the expression for the
topological spin

$\theta_{\ell_{a}}=e^{\frac{2\pi ia^{2}}{2^{r+1}}}~{},$ (B.13)

it is clear that the line operator $\ell_{2^{r-n}}$ is a boson that generates
a non-anomalous $B\cong\mathds{Z}_{2^{n}}$ subgroup of the full 1-form
symmetry group. Let $S_{B}$ be the non-invertible surface operator defined by
$B$. There are also non-invertible surface operators $S_{K}$ defined by a
subgroup $K<B$. In total, we get $n$ such non-invertible surface operators.
Also, we have the charge conjugation invertible surface operator, $S_{C}$. All
other non-invertible surface operators of this TQFT are obtained from fusing
$S_{K}$ and $S_{C}$. We have,

$(S_{C}\times S_{K})\cdot\ell_{a}=S_{C}\cdot(\text{$K$-orbit of
}\ell_{a})=\text{ $K$-orbit of }\bar{\ell}_{a}=(S_{K}\times
S_{C})\cdot\ell_{a}~{},$ (B.14)

if $\ell_{a}$ braids trivially with $K$ and zero otherwise. Suppose we have

$S_{C}\times S_{K}=S_{C}\times S_{K^{\prime}}~{}.$ (B.15)

Then the $K$-orbit of $\ell_{a}$ must be equal to the $K^{\prime}$-orbit of
$\ell_{a}$, which is possible only if $K=K^{\prime}$. Now, suppose

$S_{C}\times S_{K}=S_{K}~{}.$ (B.16)

For this equality to hold, the $K$-orbit of $\ell_{a}$ must be equal to the
$K$-orbit of $\bar{\ell}_{a}$ for all $\ell_{a}$ that braids trivially with
$K$. In other words,

$\ell_{a}+\ell_{a}\in K~{},$ (B.17)

for all $\ell_{a}$ that braids trivially with $K$. We have
$B=\langle\ell_{2^{r-n}}\rangle$. Then the line operators which braid with $B$
satisfy

$e^{2\pi i\frac{2^{r-n}a}{2^{r}}}=1\implies a=0\text{ mod }2^{n}~{}.$ (B.18)

Imposing the condition (B.17) we get

$\ell_{2^{n+1}}\in B~{}.$ (B.19)

Since $\ell_{2^{r-n}}$ is a generator of $K$, there should be an integer, $k$,
such that

$k2^{(r-n)}=2^{n+1}\implies k=2^{2n+1-r}~{}.$ (B.20)

Here, $r$ is either $2n+1$ or $2n+2$. In the former case $k=1$, and in the
later case we don’t have an integer solution for $k$. Therefore, if $r=2n+1$
we have the fusion rules

$S_{C}\times S_{B}=S_{B}~{},$ (B.21)

and, if $r=2n+2$, then $S_{C}\times S_{B}$ is a new non-invertible surface
operator.

Now let us consider the group $K=\langle\ell_{2^{r-n+q}}\rangle$ for some
$0\leq q\leq n$. For different values of $q$, we get all the subgroups of $B$.
The line operators which braid trivially with $K$ satisfy

$e^{2\pi i\frac{2^{r-n+q}a}{2^{r}}}=1\implies a=0\text{ mod }2^{n-q}~{}.$
(B.22)

Imposing the condition (B.17) we get

$\ell_{2^{n-q+1}}\in K~{}.$ (B.23)

Since $\ell_{2^{r-n-q}}$ is a generator of $K$, there should be an integer $k$
such that

$k2^{(r-n+q)}=2^{n-q+1}\implies k=2^{2n+1-r-2q}~{}.$ (B.24)

For $r=2n+1$ we get $k={-2q}$, which is an integer only for $q=0$. For
$r=2n+2$, we get $k=2^{-1-2q}$ which is not an integer for any $q$. Therefore,
the fusions

$S_{C}\times S_{K}\neq S_{K}~{},$ (B.25)

for $q>0$ and any value of $r$.

$A_{2^{r}}$ TQFTs can be realized as $U(1)_{2^{r}}$ Chern-Simons theory. The
surface operators in this Chern-Simons theory were studied in [35, 12]. Our
classification of surface operators above agrees with the classification in
these papers. There are other prime abelian TQFTs labelled as
$B_{2^{r}},C_{2^{r}}$ and $D_{2^{r}}$ TQFTs which have the same fusion group
as $A_{2^{r}}$ but different topological spins [37]. The classification of
surface operators in these latter theories are identical to that in
$A_{2^{r}}$ since these theories are related by Galois conjugation [42].

Note that in the examples studied so far, the full set of surface operators
can be constructed by first finding the invertible symmetries, then finding
the $S_{B}$ type surface operators defined in (3.26) for every non-anomalous
1-form symmetry group $B$ of the TQFT and then studying their fusions. It
would be interesting to understand if this is a general property of symmetries
in Abelian TQFT.

## Appendix C. Action on twisted sectors

Throughout this paper we have been concerned with the action of symmetries on
genuine lines. That is to say, we only considered the symmetry action of
surfaces on lines that are not attached to surfaces. In this appendix, we
begin to generalize this study (leaving a more complete picture for future
work).

Indeed, the action of a surface operator on the genuine line operators of a
QFT can be extended to an action on the non-genuine line operators that live
at the end of surface operators (a.k.a. “twisted sector” lines). Similar to
the various conditions the action of a surface operator on genuine lines
satisfy, we get various constraints on the surface action on twisted sectors.

$S$$a^{(S_{1})}$$S_{1}$$S_{2}$$b^{(S_{2})}$ Fig. 31: Action of a surface
operator on a non-genuine line.

Let $a^{(S_{1})}$ be a non-genuine line at the end of surface operators
$S_{1}$. Then the action of a surface operator, $S$, on $a^{(S_{1})}$ is given
by

$S\cdot a^{(S_{1})}=\sum_{S_{2}}\sum_{b^{(S_{2})}}n_{S\cdot
a^{(S_{1})}}^{b^{(S_{2})}}~{}b^{(S_{2})}~{},$ (C.1)

where the fusion coefficient $n_{S\cdot a^{(S_{1})}}^{b^{(S_{2})}}\neq 0$ if
Fig. 31 is consistent. For $n_{S\cdot a^{(S_{1})}}^{b^{(S_{2})}}$ to be non-
zero, there are various constraints on $S_{2}$ and $b^{(S_{2})}$ that must be
satisfied. First of all, there must a non-trivial topological junction between
the surface operators

$S_{1}\times S\text{ and }S\times S_{2}~{}.$ (C.2)

In other words,

$\text{Hom}(S_{1}\times S,S\times S_{2})\neq 0~{}.$ (C.3)

In particular, if the surface operators are all invertible, then we get the
constraint

$S_{2}=S^{-1}\times S_{1}\times S~{}.$ (C.4)

That is, $S_{2}$ and $S_{1}$ should be related by conjugation. Also, we can
generalize Fig. 8 to get Fig. 32.

$S$$a^{(S_{1})}$$S_{1}$$S_{2}$$b^{(S_{2})}$$S_{3}$$S_{4}$$a^{(S_{2})}$$b^{(S_{4})}$$\Longrightarrow$$S$$a^{(S_{i})}$$S_{i}$$S_{j}$$b^{(S_{j})}$$\sum\limits_{S_{i}\in
S_{1}\times S_{3},\ S_{j}\in S_{2}\times S_{4}}\
\sum\limits_{a^{(S_{i})},a^{(S_{j})}}\
N_{a^{(S_{1})}a^{(S_{2})}}^{a^{(S_{i})}}N_{b^{(S_{1})}b^{(S_{2})}}^{b^{(S_{i})}}$
Fig. 32: Condition on $S$-action from fusion of non-genuine lines

From this diagram, we get the constraint

$n_{Sa^{(S_{1})}}^{b^{(S_{2})}}n_{Sa^{(S_{3})}}^{b^{(S_{4})}}~{}\leq\sum\limits_{S_{i}\in
S_{1}\times S_{3},\ S_{j}\in S_{2}\times S_{4}}\
\sum\limits_{a^{(S_{i})},a^{(S_{j})}}\
N_{a^{(S_{1})}a^{(S_{2})}}^{a^{(S_{i})}}N_{b^{(S_{1})}b^{(S_{2})}}^{b^{(S_{i})}}~{}~{}n_{Sa^{(S_{i})}}^{b^{(S_{j})}}~{}.$
(C.5)

In the rest of this appendix, we will focus on some examples of abelian TQFTs
and study the action of invertible line operators on the non-genuine line
operators.

###  C.1 Semion $\times$ Semion TQFT

Consider the Semion $\times$ Semion TQFT with line operators
$1,s,\tilde{s},s\tilde{s}$ with topological twists $1,i,i,-1$, respectively.
This theory has a $\mathds{Z}_{2}$ symmetry implemented by a surface operator,
$S$, which acts as

$S\cdot s=\tilde{s}~{}.$ (C.6)

Note that $S$ has two topological boundary conditions, and let $a_{1}^{(S)}$
and $a_{2}^{(S)}$ be the non-genuine line operators labelling these
boundaries. We have the fusion rules

$a_{2,1}^{(S)}=s\times a_{1,2}^{(S)}=\tilde{s}\times a_{1,2}^{(S)}~{},$ (C.7)

and

$a_{1,2}^{(S)}\times a_{1,2}^{(S)}=1+s\tilde{s},~{}~{}a_{1,2}^{(S)}\times
a_{2,1}^{(S)}=s+\tilde{s}~{}.$ (C.8)

The action of the surface operator $S$ on $a_{1,2}^{(S)}$ must fix them. To
understand this statement, suppose we have

$S\cdot a_{1}^{(S)}=a_{2}^{(S)}~{}.$ (C.9)

Then we get Fig. 33

$S$$a_{1}^{(S)}$$S$$S$$a_{2}^{(S)}$$\Longrightarrow$$a_{2}^{(S)}$$a_{1}^{(S)}$$S$$\mathds{1}$
Fig. 33: Action of $S$ on $a_{1,2}^{(S)}$

whose R.H.S. is obtained from “flattening” the L.H.S. In other words, by
fusing two patches of the $S$ surface on the left diagram on either side of
the surface $S$ in the middle we get the diagram on the right. In the right
diagram we have the non-genuine lines, $a_{1,2}^{(S)}$, forming a junction at
the end of $S$. This situation is not consistent since $a_{1}^{(S)}$ and
$a_{2}^{(S)}$ are distinct simple boundary conditions of $S$. Therefore, $S$
must act trivially on the two non-genuine lines, $a_{1,2}^{(S)}$.

More generally, we can consider the action of $S$ on non-genuine lines
$a^{(\bar{S})}$ bounding the inverse surface $\bar{S}$, and use the argument
in Fig. 33 to show that the action of $S$ on any non-genuine line $a^{(S)}$ is
trivial.

###  C.2 3-fermion Model

Consider the 3-fermion model with line operators
$1,\psi_{1},\psi_{2},\psi_{3}$ and topological twists
$\theta_{1}=1,\theta_{\psi_{i}}=-1$. This TQFT has an $S_{3}$ 0-form symmetry
which permutes the three fermions. Consider the surface operator $S_{(12)}$
which permutes $\psi_{1}$ and $\psi_{2}$. We have two non-genuine line
operators $a_{1,2}^{(S_{(12)})}$ with fusion rules

$a_{2,1}^{(S_{(12)})}=\psi_{1}\times a_{1,2}^{(S_{(12)})}=\psi_{2}\times
a_{1,2}^{(S_{(12)})}~{},$ (C.10)

and

$a_{1,2}^{(S_{(12)})}\times
a_{1,2}^{(S_{(12)})}=1+\psi_{3},~{}~{}a_{1,2}^{(S_{(12)})}\times
a_{2,1}^{(S_{(12)})}=\psi_{1}+\psi_{3}~{}.$ (C.11)

Similarly, we have the surface operators $S_{(13)},S_{(23)}$ and corresponding
non-genuine lines. Now, let us consider the surface operators
$S_{(123)},S_{(132)}$ which permute all three fermionic lines. These surface
operators have a single non-genuine line at their boundary $b^{(S_{(123)})}$,
$c^{(S_{(132)})}$.

The action of the surface operators on these non-genuine lines is mostly fixed
by the fact the output and input surface operators must be conjugated by the
surface operator being acted with. Moreover, $S$ acts trivially on a non-
genuine line at its own boundary.

Let us consider the action of the surface operator $S_{(123)}$ on
$a_{1,2}^{(S_{(12)})}$. We know that

$S_{(123)}\cdot a_{1}^{(S_{(12)})}=n_{S\cdot
a_{1}^{(S_{(12)})}}^{a_{1}^{S_{(23)}}}~{}a_{1}^{(S_{(23)})}+n_{S\cdot
a_{1}^{(S_{(12)})}}^{a_{2}^{S_{(23)}}}~{}a_{2}^{(S_{(23)})}~{}.$ (C.12)

Since the surface operator $S_{(123)}$ is invertible, we want the action of
$S_{(123)}$ to preserve all gauge-invariant topological data associated with
the non-genuine lines. From the fusion rules of non-genuine lines described
above we know that both $a_{1,2}^{(S_{(12)})}$ and $a_{1,2}^{(S_{(13)})}$ have
quantum dimension $\sqrt{2}$. Therefore, to preserve quantum dimensions only
one among $n_{S\cdot a_{1}^{(S_{(12)})}}^{a_{1}^{S_{(23)}}}$ and $n_{S\cdot
a_{1}^{(S_{(12)})}}^{a_{2}^{S_{(23)}}}$ must be non-zero. To exactly determine
the outcome, we must compare the topological data of $a_{a_{1}}^{(S_{(12)})}$
with $a_{a_{1}}^{(S_{(23)})}$ and $a_{a_{2}}^{(S_{(23)})}$. For a non-genuine
line operator $a$, the following combination of the $G$-crossed braided
$R$-matrix is gauge invariant

$\frac{\sum_{\mu}[R_{aa}^{c}]_{\mu\mu}}{\sum_{\nu}[R_{aa}^{c^{\prime}}]_{\nu\nu}}~{},$
(C.13)

where $\mu,\nu$ run over the basis of the relevant fusion spaces [59]. Using
[59, Section XI M], we find

$\frac{R_{{a_{1}}^{(S_{(12)})}{a_{1}}^{(S_{(12)})}}^{1}}{R_{{a_{1}}^{(S_{(12)})}{a_{1}}^{(S_{(12)})}}^{\psi_{3}}}=e^{\frac{i\pi}{2}},~{}\frac{R_{{a_{1}}^{(S_{(23)})}{a_{1}}^{(S_{(23)})}}^{1}}{R_{{a_{1}}^{(S_{(23)})}{a_{1}}^{(S_{(23)})}}^{\psi_{1}}}=e^{\frac{i\pi}{2}},~{}\frac{R_{{a_{2}}^{(S_{(23)})}{a_{2}}^{(S_{(23)})}}^{1}}{R_{{a_{2}}^{(S_{(23)})}{a_{2}}^{(S_{(23)})}}^{\psi_{1}}}=e^{\frac{-i\pi}{2}}~{}.$
(C.14)

Since the action of $S_{(123)}$ must preserve gauge-invariant topological
data, we find

$S_{(123)}\cdot a_{1}^{(S_{(12)})}=a_{1}^{(S_{(23)})}~{}.$ (C.15)

More generally, the action of an invertible surface operator on non-genuine
lines must be such that

$\frac{\sum_{\mu}[R_{S\cdot a,S\cdot a}^{S\cdot
c}]_{\mu\mu}}{\sum_{\nu}[R_{S\cdot a,S\cdot a}^{S\cdot
c^{\prime}}]_{\nu\nu}}=\frac{\sum_{\mu}[R_{aa}^{c}]_{\mu\mu}}{\sum_{\nu}[R_{aa}^{c^{\prime}}]_{\nu\nu}}~{},$
(C.16)

for all $c,c^{\prime}$. The equations (C.3),(C.5) and (C.16) are necessary
conditions to be satisfied for consistent action of invertible surface
operators on the genuine and non-genuine line operators. It will be
interesting to find a necessary and sufficient set of conditions that
determine the action of surface operators on non-genuine lines. Note that the
conditions (C.3),(C.5) and (C.16) are sufficient to determine the action of
invertible symmetries on the twisted sectors in the 3-fermion model. For
invertible surfaces, this is related to determining which higher
representations of the symmetry group are realized in the TQFT [81, 82]. For
non-invertible surface operators, this would amount to finding which
representations of higher Tube algebra are realized in these TQFTs [83, 84].
We leave this study for future work.

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