# Invertibility of Condensation Defects and
Symmetries of 2 + 1d QFTs

Matthew Buican1 and Rajath Radhakrishnan2

###### Abstract

We characterize discrete (anti-)unitary symmetries and their non-invertible
generalizations in $2+1$d topological quantum field theories (TQFTs) through
their actions on line operators and fusion spaces. We explain all possible
sources of non-invertibility that can arise in this context. Our approach
gives a simple $2+1$d proof that non-invertible generalizations of unitary
symmetries exist if and only if a bosonic TQFT contains condensable bosonic
line operators (i.e., these non-invertible symmetries are necessarily “non-
intrinsic”). Moving beyond unitary symmetries and their non-invertible
cousins, we define a non-invertible generalization of time-reversal symmetries
and derive various properties of TQFTs with such symmetries. Finally, using
recent results on 2-categories, we extend our results to corresponding
statements in $2+1$d quantum field theories that are not necessarily
topological.

Contents

toc

##  1 Introduction

The modern perspective on symmetries in quantum field theory (QFT) is that
they correspond to sectors of topological defects of various dimensions [1,
2]. Fully understanding the symmetry of a given QFT then amounts to finding
all topological operators in the theory and their correlations functions.
Thinking in this way leads down a path beyond groups to the world of
(non-)invertible symmetries, which are conjecturally characterized by higher-
fusion categories [3, 4, 5].

In spite of this understanding, even classifying the set of topological
defects in a QFT can be difficult. However, some general structural properties
of symmetries can be deduced from studying consistency conditions in higher-
fusion categories.111For certain non-invertible symmetries, some of the
associativity relations in the higher-fusion category have been worked out
[6]. For the purposes of this paper, we will be interested in the rich case of
$2+1$d QFTs and their discrete symmetries. Mathematically, the relevant
structures are fusion 2-categories, which capture the properties of
topological surface and line operators [7, 8, 9].

In this setting, it is interesting to understand how the different sub-sectors
of fusion 2-categories interact with and constrain each other.222Understanding
how more general sectors of QFTs constrain each other is of course also
interesting, although typically much harder (e.g., see [10] for a relation
between non-topological local operators and topological surfaces in certain
broad classes of theories). To that end, the authors of [11] found an elegant
and inspirational result: if a $2+1$d QFT does not contain any topological
line operators, then all topological surfaces are invertible.

The argument is simple and ingenious: consider a topological surface operator,
$S$, on a cylinder and shrink the cylinder to get a line operators, $L_{S}$
(see Fig. 1).

$\xrightarrow{\text{Shrink }S}$$S$$L_{S}$ Fig. 1: Shrinking a topological
surface operator, $S$, to get a topological line operator, $L_{S}$.

In general, $L_{S}$ decomposes into a sum of simple lines.333A simple line
operator is a line that hosts only a single local topological operator on it.
Now, consider the junction of $L_{S}$ with the trivial line operator (see Fig.
2).

$L_{S}$$1$$\alpha$ Fig. 2: A junction of $L_{S}$ with the trivial line.

The junction is labelled by a point operator, $\alpha\in\text{Hom}(L_{S},1)$,
where the dimension of $\text{Hom}(L_{S},1)$ determines the number of copies
of the trivial line in $L_{S}$. Replacing $L_{S}$ with the surface operator,
$S$, we get Fig. 3.

$S$$1$$S$$1$$\alpha$$\alpha$$\xrightarrow{\text{Flatten }S}$ Fig. 3: The
distinct junctions of $L_{S}$ with the trivial line operator imply distinct
local operators on $S$.

Upon flattening $S$, we find that each $\alpha$ at the junction of $L_{S}$
with the trivial line implies a distinct local operator on $S$. This argument
shows that if we shrink a simple surface operator,444A simple surface operator
is a surface which hosts only a single local topological operator on it. then
$L_{S}$ contains only one copy of the trivial line operator.

Now, suppose a 2+1D QFT has no non-trivial topological line operators.
Shrinking a simple surface operator, $S$, of this QFT, we get the trivial
line. This implies that $S$ is invertible. Indeed, suppose

$S\times\bar{S}=S_{1}+S_{2}+\cdots~{},$ (1.1)

for some surface operators, $S_{1}$, $S_{2}$, $\cdots$. Shrinking each of
these surfaces, we get the fusion rule

$L_{S}\times L_{\bar{S}}=L_{S_{1}}+L_{S_{2}}+\cdots~{}.$ (1.2)

Since the QFT only has the trivial line operator, we get

$1\times 1=1+1+\cdots~{},$ (1.3)

which is inconsistent.

What about the converse of the above theorem? Does the existence of
topological line operators guarantee that there are non-invertible surface
operators? No. For example, $SU(2)_{1}$ Chern-Simons theory has a non-trivial
line operator. However, this TQFT does not have any invertible or non-
invertible surface operators. This discussion motivates us to study the
following question in $2+1$d QFT:

Question: What properties of a theory’s line operators determine whether there
are non-invertible surface operators?

Except for a discussion of non-genuine lines in Appendix Appendix C. Action on
twisted sectors, we will only consider genuine line operators of a 2+1d QFT
and symmetries implemented by topological surfaces acting on them. In
particular, we will only consider surface operators which act faithfully on
the line operators and their fusion spaces. In other words, we assume that a
surface operator which acts trivially on the topological line operators and
its fusion spaces is the trivial surface operator.555The non-invertible
symmetries of 2+1D TQFTs can all be implemented by condensation defects [8,
12]. As we will discuss in section 5.2.1, condensation defects satisfy this
assumption.

Another important aspect of symmetries in QFT is their use in constraining the
space of QFTs. Important examples of these ideas include $c$-theorems in
different dimensions and their constraints on renormalization group flows,
studies of metrics on conformal manifolds, and the constraints that
topological modular forms place on more general deformations of 2d QFTs.

In many of these cases, the resulting relations between different theories can
be understood in terms of interfaces of different kinds: RG interfaces (e.g.,
see [13, 14, 15]), Janus interfaces (e.g., see [16, 17]), and more general
domain walls. Since we will be particularly interested in theories that can be
related to each other through various topological manipulations like discrete
gauging and condensation, the interfaces that are relevant to us are gapped
domain walls.

Such gapped domain walls give, in a more mathematical language, theories with
symmetries characterized by Morita equivalent (higher-)fusion categories. In
the case of fusion 2-categories, Décoppet showed that, up to Morita
equivalence, any such category can be written in the following factorized form
[18]

$\text{2Vec}_{G}^{\omega}~{}\boxtimes\text{ Mod}(\mathcal{C})~{},$ (1.4)

where the first factor denotes a set of invertible surface operators with
fusion group $G$ and anomaly $\omega\in H^{4}(G,U(1))$, and the second factor
denotes the line operators of a modular tensor category (MTC) [19, 20],
$\mathcal{C}$, and the surfaces formed by all possible higher gaugings of
these lines. In other words, $\text{ Mod}(\mathcal{C})$ denotes a (non-spin)
$2+1$d topological quantum field theory (TQFT). Therefore, to answer our above
question, it suffices to answer it in the context of $2+1$d TQFTs.

As a result, we embark on a study of the action of surfaces on lines and
fusion spaces (i.e., trivalent junctions of lines) in different classes of
$2+1$d TQFTs.666This approach can be thought of as a more “active” version of
the more “passive” approach of understanding which lines can end on the
boundary of a folded theory (or, in a more mathematical language, of
understanding Lagrangian algebras of the folded theory) [21, 22, 8]. As a
simple starting point, we analyze $2+1$d non-spin Abelian TQFTs, where the
action of surfaces on lines alone leads to a theorem on invertibility of
surface operators. These are theories in which all line operators are
invertible (i.e., their fusion is described by an Abelian group, $A$). As is
well known, Abelian TQFTs are classified by the topological spins of their
lines and $A$. In other words, these theories are described by their modular
data. As we will see, modularity is then sufficiently powerful to lead to the
following result:

Theorem: In a $2+1$d Abelian TQFT (see table 1 for details), non-invertible
surfaces exist if and only if there is a non-anomalous 1-form symmetry.

We then turn our attention to non-Abelian TQFTs. In this case, the modular
data does not determine the theory. Moreover, as we will see, the action of
surfaces on lines does not suffice to give a generalization of Theorem 1. To
make progress on such a generalization, we need to understand more precisely
how a non-invertible symmetry acts on fusion spaces. To that end, we prove a
lemma showing that fusion spaces are not new sources of non-invertibility: if
a symmetry has an invertible action on lines, then it has an invertible action
on fusion spaces. With this lemma in hand we prove that:

Theorem: $2+1$d TQFTs (see table 1 for details) without bosons only have
invertible symmetries.

The above theorem generalizes one direction of our result in the Abelian case.
To find a partial generalization of the converse, it turns out we need to
delve deeper into the structure of the “higher gauging” [23, 12, 24] that
leads to the surfaces in the ${\rm Mod}(\mathcal{C})$ factor of the fusion
2-category in (1.4).777In studying condensation surface operators, there are
two extreme cases where the classification becomes easier. One is when the the
category of lines, $\mathcal{C}$, is symmetric. In other words, the
$\mathcal{S}$-matrix of braidings of the lines is trivial. In particular, when
the line operators are all bosonic, $\mathcal{C}$ is equivalent to the
representation category of a finite group, $\mathcal{C}\simeq\text{Rep}(G)$
[25]. The condensation surfaces obtained in this case were classified in terms
of higher representation theory in [26, 27, 28]. Various properties, like the
fusion rules of these surfaces, can be determined explicitly. In particular,
condensation surface operators obtained from higher-gauging line operators
described by a Rep$(G)$ category always include non-invertible surface
operators for any non-trivial group $G$. In a gauge theory, the fusion rules
of these surface operators contain crucial information about the gauge group
[29]. In deriving our results on TQFTs, we study the opposite case where the
$\mathcal{S}$-matrix of the line operators is non-degenerate. In so doing we
find the following general constraint:

Theorem: A $2+1$d TQFT (see table 1 for details) has a non-invertible symmetry
if and only if it contains condensable bosonic line operators.

The above theorems follow from the existence of certain “universal” non-
invertible surfaces (i.e., surfaces that exist whenever there is a non-
invertible symmetry).888These surface operators are a generalization of the
universal surface operators found in [26, 27]. However, given theories
typically have a much more complicated zoology of non-invertible surfaces that
includes these universal surfaces as a subsector. We give an abstract
characterization of this zoology in terms of special Frobenius Algebras.
Moreover, in the case of a condensation defect arising from gauging Abelian
lines (even in a non-Abelian TQFT), we give a precise characterization of
invertibility of the surface operator in terms of the data of the Abelian
lines and their embedding in the larger theory.

Using the above results, combined with a simple pictorial derivation of the
result of [18] that leads to the factorization in (1.4), we arrive at the
following theorem:

Theorem: A $2+1$d QFT (see table 1 for details) has a non-invertible symmetry
if and only if it contains condensable bosonic line operators.

We conclude our discussion of non-invertible symmetries by rephrasing our
results in terms of “non-intrinsic” non-invertible symmetries [30, 31].

Terminology | Meaning
---|---
Abelian TQFT | non-spin Abelian TQFT
TQFT | non-spin TQFT
QFT | QFT without a transparent fermionic line
unitary symmetry | Invertible symmetry that preserves spin
anti-unitary / time-reversal symmetry | Invertible symmetry that conjugates spin
non-invertible symmetry | Non-invertible generalization of a unitary symmetry
non-invertible time-reversal symmetry | Non-invertible generalization of a time-reversal symmetry
line operator | genuine line operator
MTC | unitary MTC
Table 1: The terminology employed in this paper. We adopt a shorthand so as
not to overload definitions, theorems, and discussions with too many
adjectives. In particular, note that all TQFTs we study in this paper are non-
spin (i.e., they do not require a spin structure to be well-defined).

The final topic we tackle in this paper is to give a definition of a non-
invertible generalization of anti-unitary or “time-reversal” symmetries (e.g.,
see [32] for discussion in the context of Maxwell theory and QCD). We point
out various peculiar features of these symmetries, including the fact that it
is not always possible to distinguish them from standard non-invertible
symmetries (see table 1 for an explanation of our terminology) via their
actions on lines, even when the theory contains lines that are genuinely
anyonic (i.e., not bosonic or fermionic).

The structure of the paper is as follows. In the next section, we start with a
review of what is meant by a (non-)invertible surface. After that, we proceed
to a discussion of symmetries of abelian TQFTs in Sec. 3. We define a non-
invertible symmetry of an abelian TQFT, derive various consequences, and prove
a theorem on the existence of non-invertible symmetries in such theories. We
move on to define non-invertible symmetries of non-Abelian TQFTs in Sec. 4,
and we find a partial generalization of our theorem in Abelian theories. We
then discuss higher-gauging and condensation surfaces in Sec. 5. We use this
machinery to prove a general theorem on the existence of non-invertible
symmetries in $2+1$d TQFT. We then discuss how our results fit in to a general
theorem on $2+1$d QFT in Sec. 6. In Sec. 7 we move on to non-invertible time-
reversal symmetries and describe certain peculiar features of these
symmetries. After concluding with some open questions, we include various
appendices that show how our “active” definitions of symmetries relate to more
standard “passive” definitions in terms of Lagrangian algebras in folded
theories and also explain some basic aspects of how our work extends to non-
genuine lines.

##  2 Basics of surface fusion and invertibility

In this section we give a basic idea of what it means for a topological
surface operator to be invertible, and we sketch out some facts about the
action of (non-)invertible surface operators on lines. In subsequent sections,
we give more precise definitions of surfaces and derive various consequences.

A surface operator, $S$, is said to be invertible if there exists another
surface operator, $\bar{S}$, such that

$S\times\bar{S}=\bar{S}\times S=\mathds{1}~{},$ (2.1)

where $\mathds{1}$ is the trivial surface operator. The invertible surface
operators form a group under fusion.

To prove this statement, note that, if $S_{1}$ and $S_{2}$ are invertible
surface operators, then $S_{1}\times S_{2}\neq 0$. Indeed, suppose this fusion
was zero, then multiplying both sides on the right by $\bar{S}_{2}$ implies
that $S_{1}$ is the zero operator, which is not consistent with invertibility
of $S_{1}$. Given that $S_{1}\times S_{2}\neq 0$, this product is itself an
invertible surface operator with inverse $\bar{S}_{2}\times\bar{S}_{1}$.
Therefore, the invertible surface operators are closed under fusion. Assuming
that the fusion is associative, we have an associative multiplication on the
surface operators with $\mathds{1}$ being the identity. This means that the
invertible surface operators form a monoid. The inverse of $S$ is unique
because otherwise

$S\times\bar{S}=\mathds{1}=S\times\bar{S}^{\prime}~{},$ (2.2)

implies that $\bar{S}=\bar{S}^{\prime}$. Therefore, the invertible surface
operators indeed form a group.

Since $\mathds{1}$ acts trivially on all line operators, it is clear that the
action of $S$ and $\bar{S}$ on any line operator cannot be zero. In fact, the
action of an invertible surface operator on a simple line operator should
result in a single outcome. To understand this, suppose we have instead

$S\cdot p=p_{1}+p_{2}~{},$ (2.3)

for some simple lines, $p_{1}$ and $p_{2}$. Then, we find

$\bar{S}\times S\cdot p=\bar{S}\cdot p_{1}+\bar{S}\cdot p_{2}~{}.$ (2.4)

Since the action of $\bar{S}$ on a line operator cannot be trivial, the R.H.S
of this expression has at least two line operators. This contradicts the
trivial action of $\mathds{1}$ on $p$. Therefore, the action of an invertible
surface operator on a simple line operator should result in a single outcome.
Equation (2.1) implies that the action of $S$ and $\bar{S}$ should be
invertible functions on the set of simple line operators: permutations.

Next let us suppose $S$ is a non-invertible surface, and let us further assume
that the action of surface operators on line operators is faithful. It is easy
to then show that the action of $S$ on the simple lines cannot be a
permutation. To the contrary, assume that the action of $S$ on the simple
lines is a permutation. Let $N$ be the order of this permutation. Then, the
action of $S^{N}$ on the line operators is trivial. If the action of the
surface operators on the lines is faithful, then

$S^{N}=\mathds{1}~{},$ (2.5)

and we have

$S\times\bar{S}=\mathds{1}~{},$ (2.6)

where $\bar{S}:=S^{N-1}$. This result contradicts the assumption that $S$ is
non-invertible. As we will see in section 4, surfaces in non-Abelian TQFTs can
act trivially on lines but still act non-trivially on fusion spaces. However,
we will show that such surfaces (including the hypothetical $S^{N}$ above) are
necessarily invertible and so our above discussion can be appropriately
modified to show that non-invertible surfaces cannot act via
permutations.999Since we are dealing with surface operators which from a
fusion 2-category, there must be a finite $N$ such that $\mathds{1}\in S^{N}$.
In particular, a surface which acts trivially on the lines and non-trivially
on the fusion spaces also has a finite order, and therefore the above argument
holds for such surface operators.

In the next section we give a precise definition of a symmetry in an Abelian
TQFT and derive various consequences.

##  3 (Non-)invertible symmetries of Abelian TQFTs

In this section we discuss basic properties of Abelian TQFTs (recall from
table 1 that we will only consider non-spin theories) before moving on to a
definition of topological surfaces in these theories. We then derive some
consequences of our definition, culminating in theorem 1.

The line operators of an abelian TQFT, $\mathcal{T}$, have fusion rules given
by an abelian group, $A$. The correlation functions of the line operators are
then fixed by the choice of a non-degenerate quadratic form on $A$ called the
topological spin [33, 34, 35]

$\theta:A\to U(1)~{}.$ (3.1)

The non-degeneracy of the quadratic form is, by definition, the invertiblity
of the matrix

$\mathcal{S}_{a,b}:=\frac{1}{\sqrt{|A|}}\frac{\theta_{a\times
b}}{\theta_{a}\theta_{b}}~{},$ (3.2)

where $|A|$ is the order of $A$. The group, $A$, along with the topological
spin, $\theta$, define a pointed MTC, $\mathcal{C}$ [20]. In fact, since we
can reconstruct $A$ from $\mathcal{S}$ via the Verlinde formula, we can
equivalently think of the MTC as being defined by the modular data,
$(\mathcal{S},\theta)$. We will use the notation $\mathcal{T}$ and
$\mathcal{C}$ interchangeably to describe a TQFT.

In the next subsection, we review the definition of a unitary invertible
symmetry of an abelian TQFT and interpret it in terms of the action of a
surface operator on the line operators. Then we will define non-invertible
generalizations of these symmetries (which we will simply call “non-invertible
symmetries”) and study their properties.

###  3.1 Characterizing unitary symmetries of Abelian TQFTs and their non-
invertible generalizations

A unitary 0-form symmetry of an abelian TQFT must preserve the fusion rules of
the lines. Therefore, it is a permutation, $\sigma$, of the line operators
that is an automorphism of the fusion group, $A$. This permutation should also
preserve the correlation functions of line operators. Since the correlation
functions are completely determined by the topological spins, we require

$\theta_{\sigma(a)}=\theta_{a}~{}.$ (3.3)

Using (3.2), it follows that $\mathcal{S}$ is also preserved under the
symmetry action

$\mathcal{S}_{\sigma(a)\sigma(b)}=\mathcal{S}_{ab}~{}.$ (3.4)

Therefore, $\sigma$, viewed as an $|A|\times|A|$ permutation matrix, commutes
with $\mathcal{S}$ and $\mathcal{T}$, where

$\mathcal{T}_{ab}:=\delta_{a,b}\theta_{b}~{}.$ (3.5)

The group of unitary invertible symmetries of abelian TQFTs have been
classified for many types of abelian TQFTs [36, 37].

A unitary 0-form symmetry is implemented by an invertible surface operator,
$S_{\sigma}$. Let us depict the conditions described above diagrammatically.
If we have

$S_{\sigma}\cdot a=b~{},$ (3.6)

then the line operators $a$ and $b$ can form a junction on the surface
operator, $S_{\sigma}$, as in Fig. 4.

$S_{\sigma}$$a$$b$ Fig. 4: The symmetry action of a surface on a line
operator.

$S_{\sigma}$ implements a symmetry if it preserves twists and if the condition
in Fig. 5 is satisfied.

$c$$d$$a$$b$$S_{\sigma}$$b\times d$$a\ \times\ c$$\Longrightarrow$$S_{\sigma}$
Fig. 5: A necessary condition for the $S_{\sigma}$ action to be a symmetry.

This condition is natural, since we can bring the two lines, $a$ and $c$, on
the right arbitrarily close to each other and fuse them to form $a\times c$.
Since the junction on the surface is topological, this maneuver brings the
lines, $b$ and $d$, arbitrarily close to each other, which then fuse to give
the outcome $b\times d$.

To describe non-invertible symmetries, let us relax the assumption that $S$
should act as a permutation (see the discussion in Sec. 2). In this case, $S$
can be a function on the set of line operators to itself which has a non-
trivial kernel. In other words, defining

${\rm Ker}(S):=\left\\{a\in\mathcal{T}|S\cdot a=1\right\\}~{},$ (3.7)

we could have $1\neq a\in{\rm Ker}(S)$ (here $1$ is the trivial line).
Moreover, we could have $S\cdot b=0$ for some line operator $b\neq 1$. We will
denote this “null” set

${\rm Null}(S):=\left\\{a\in\mathcal{T}|S\cdot a=0\right\\}~{}.$ (3.8)

Lines in ${\rm Null}(S)$ cannot form a junction on the surface operator with
any other line operator.101010Note that lines in ${\rm Null}(S)$ may be mapped
by $S$ to non-genuine / twisted sector lines (i.e., to lines that are attached
to surfaces). Otherwise, moving $S$ through certain non-vanishing
configurations of lines leads to discontinuities in the space of the
corresponding correlation functions. Although it will be irrelevant for our
discussion of genuine lines in the main text (we will briefly discuss certain
non-genuine lines in App. Appendix C. Action on twisted sectors.), it would be
interesting to explore these ideas further and understand if such
discontinuities are allowed or forbidden. Note that ${\rm Ker}(S)$ and ${\rm
Null}(S)$ are distinct spaces and that

${\rm Ker}(S)\cap{\rm Null}(S)=\emptyset~{}.$ (3.9)

More generally, $S\cdot a$ is of the form

$S\cdot a=\sum_{b}n_{S\cdot a}^{b}b~{},\ \ \ n_{S\cdot
a}^{b}\in\mathbb{N}~{}.$ (3.10)

If $n_{S\cdot a}^{b}\neq 0$, we get Fig. 6.

$S$$a$$b$$\alpha$ Fig. 6: If $b\in S\cdot a$, then $a$ and $b$ can form a
junction on $S$. Here $\alpha$ is a local operator that lives at the junction
and belongs to a vector space, $V_{S\cdot a}^{b}$, with dimension $n_{S\cdot
a}^{b}$.

Let us repeat the analysis in Fig. 5 but now keeping track of the local
operators, $\alpha$, at the junctions. We get Fig. 7.

$c$$a$$S$$b\times d$$a\ \times\
c$$\Longrightarrow$$\beta$$\alpha$$\gamma$$d$$b$$S$ Fig. 7: A necessary
condition for the $S$ action to be a non-invertible symmetry.

For every choice of $\alpha\in V_{S\cdot a}^{b}$ and $\beta\in V_{S\cdot
c}^{d}$, there should be a $\gamma\in V_{S\cdot(a\times b)}^{b\times d}$. This
logic implies that the following inequality should be satisfied

$n_{S\cdot a}^{b}n_{S\cdot c}^{d}\leq n_{S\cdot(a\times c)}^{b\times d}~{}.$
(3.11)

From this definition, it is possible to show that, for any unitary symmetry
(including the trivial one), we have $|A|^{2}(|A|-1)$ strict inequalities and
$|A|^{4}-|A|^{3}+|A|^{2}$ equalities. Interestingly, this statement still
holds for the non-invertible generalizations we will discuss in this section.
Also, we must have $n_{S\cdot 1}^{1}\neq 0$, since the identity line forms a
junction with itself on the surface. We will only consider simple surface
operators (i.e., surfaces that do not have any non-trivial local operators on
them). A generic non-simple surface operator is a direct sum of simple
surfaces. Therefore, we have $n_{S\cdot 1}^{1}=1$.

To find non-invertible generalizations of unitary symmetries (hereafter
referred to as “non-invertible symmetries”), we require that
$\theta_{a}=\theta_{b}$ if $n_{S\cdot a}^{b}\neq 0$. This demand is equivalent
to requiring that $n_{S\cdot a}^{b}$ commutes with $\mathcal{T}$ as a matrix.
For unitary symmetries, this implied that components of the
$\mathcal{S}$-matrix related by the symmetry had to be equal.

On the other hand, non-invertible symmetries can (and as we will see below do)
act by mapping some simple line operators to direct sums of simple lines.
Moreover, since the modular data defines an Abelian TQFT, the $n_{S}$ matrix
should also commute with $\mathcal{S}$,111111This fact also follows from the
sandwich picture of Fig. 17, where the TQFT sits in a slab bounded by surfaces
supporting 2d chiral and anti-chiral CFT modes. Our statements then amount to
the modular invariance of the CFT torus partition function. and the unitary
condition generalizes to

$\sum_{i}n_{S\cdot a}^{i}\mathcal{S}_{ib}=\sum_{j}\mathcal{S}_{aj}n_{S\cdot
j}^{b}\ \Rightarrow\ [n_{S},\mathcal{S}]=0~{},$ (3.12)

or equivalently

$n_{S\cdot a}^{b}=\sum_{i,j}\mathcal{S}_{ai}n_{S\cdot
i}^{j}\mathcal{S}_{jb}^{-1}~{}.$ (3.13)

In particular, choosing $a$ and $b$ in the above equation to be the trivial
line operator, we get

$\sum_{i,j}n_{S\cdot i}^{j}=|A|~{}.$ (3.14)

As a consequence of this discussion, we define a (non-invertible) symmetry of
an abelian TQFT as follows:

Definition: A (non-invertible) symmetry of an abelian TQFT is a map, $S$, on
the ring of line operators such that

  1. 1.

If $n_{S\cdot a}^{b}\neq 0$, then $\theta_{a}=\theta_{b}$.

  2. 2.

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

  3. 3.

$\sum_{a,b\in A}n_{S\cdot a}^{b}=|A|$.

Condition 1 in the definition above ensures that $n_{S\cdot a}^{b}$ commutes
with $\mathcal{T}$, and condition 3 ensures that $n_{S\cdot a}^{b}$ commutes
with $\mathcal{S}$ (for example, see [38, Theorem 3.4]). In appendix Appendix
A. Relating our definitions to Lagrangian subalgebras in folded theories, we
show that this definition is equivalent to the definition of boundary
conditions in terms of Lagrangian subgroups in the folded theory.

Let us derive some consequences of this definition. To that end:

If $S$ implements a (non-invertible) symmetry of an abelian TQFT,
$\mathcal{T}$, then the integers $n_{S\cdot a}^{b}\in\left\\{0,1\right\\}$.

To understand this statement, suppose $n_{S\cdot a}^{b}$ is non-zero. Using
condition (2), we get

$n_{S\cdot a}^{b}n_{S\cdot\bar{a}}^{\bar{b}}\leq n_{S\cdot 1}^{1}=1~{},$
(3.15)

where $\bar{a}$ is the inverse of $a\in A$. We have the following
possibilities: $n_{S\cdot\bar{a}}^{\bar{b}}=0$ or $n_{S\cdot
a}^{b}=n_{S\cdot\bar{a}}^{\bar{b}}=1$. Let us show that
$n_{S\cdot\bar{a}}^{\bar{b}}$ cannot be zero. Since $n_{S\cdot a}^{b}$ is non-
zero, we have

$n_{S\cdot a}^{b}n_{S\cdot a}^{b}\leq n_{S\cdot a^{2}}^{b^{2}}\implies
n_{S\cdot a^{2}}^{b^{2}}\neq 0~{}.$ (3.16)

Similarly, we can show that $n_{S\cdot a^{k}}^{b^{k}}\neq 0$ for any $k$. Now,
let $k^{\prime}=\text{lcm}(O_{a},O_{b})-1$, where $O_{a}$ is the order of
$a\in A$. Then $n_{S\cdot
a^{k^{\prime}}}^{b^{k^{\prime}}}=n_{S\cdot\bar{a}}^{\bar{b}}\neq 0$. $\square$

Let us derive some more properties of non-invertible symmetries of abelian
TQFTs. First, we claim:

Under the action of a non-invertible symmetry on an abelian TQFT, there is at
least one line operator which is mapped to a non-simple line.

To prove this assertion, suppose $S$ acts on the line operators to give a
unique non-zero outcome so that $n_{S\cdot a}^{b}=n_{S\cdot c}^{d}=1$.
Condition 2 simplifies to

$n_{S\cdot a}^{b}n_{S\cdot c}^{d}=n_{S\cdot(a\times c)}^{b\times d}~{},$
(3.17)

where the integers $n_{S\cdot a}^{b}\in\\{0,1\\}$. In other words,

$S\cdot(a\times b)=S\cdot a\times S\cdot b~{}.$ (3.18)

That is, $S$ is an endomorphism of the fusion group, $A$. Now, for $S$ to be
non-invertible, it has to map some non-trivial line operator to the trivial
line (it has to be a non-injective endomorphism on the group of simple lines
$A$). In other words, we require:

For a non-invertible symmetry, $S$, ${\rm Ker}(S)$ is non-trivial. In other
words, there exists a non-trivial $\ell\in{\rm Ker}(S)$.

Now, from condition 1, we know that all lines in Ker$(S)$ must be bosons. We
have (using (3.18))

$S\cdot a=S\cdot(e\times a)~{},\ \forall a\in\mathcal{T}\ {\rm and}\ \forall
e\in{\rm Ker}(S)~{}.$ (3.19)

Condition 1 implies

$\theta_{a}=\theta_{e\times a}\implies S_{ea}=\frac{\theta_{e\times
a}}{\theta_{e}\theta_{a}}=1~{},\ \forall a\in\mathcal{T}~{},$ (3.20)

which is inconsistent with modularity of the TQFT. Therefore, a non-invertible
symmetry cannot act on line operators to give a unique outcome. There is at
least one line operator on which $S$ acts to give a sum of simple lines as
claimed. $\square$

This result in turn implies that, to satisfy condition 3, the surface operator
must send at least one line operator to $0$:

For a non-invertible symmetry, $S$, ${\rm Null}(S)$ is non-trivial. In other
words, there exists non-zero $n\in{\rm Null}(S)$.

In other words, there is at least one line operator which cannot form a
junction on the surface with any line operator. This should be contrasted with
a non-invertible symmetry action of line operators on other lines operators
through linking in $2+1$d or line operators on local operators in $1+1$d,
where the non-invertibility of the lines does not necessarily imply that the
result of the symmetry action is zero (for example, consider the Fibonacci
category).121212Note that in $2+1$d TQFTs described by weakly integral modular
categories, the non-invertible symmetries implemented by line operators always
set some line to zero. This statement follows from the existence of zeroes in
the modular $S$ matrix [39].

Given the universality of the trivial line (i.e., it is always present), we
next argue for the following useful statement:

The number of simple line operators that can appear on the R.H.S. of the non-
invertible symmetry action is completely determined by the symmetry action on
the trivial line.

To understand this claim, consider the action of $S$ on the trivial line. The
trivial line can form a junction with itself on the surface operator.
Therefore, we have

$S\cdot 1=1+\sum_{\ell}n_{S\cdot 1}^{\ell}~{}\ell~{}.$ (3.21)

If $n_{S\cdot 1}^{\ell}\neq 0$, then, from condition 1, $\ell$ is a boson. Let
us show that the integers $n_{S\cdot 1}^{\ell}$ cannot all be zero. From our
discussion above, we know that a surface operator, $S$, implementing a non-
invertible symmetry acts on some line operator, $a$, to give more than one
outcome. Suppose we have

$S\cdot a=\sum_{i=1}^{N}b_{i}~{},\ N>1~{}.$ (3.22)

In particular, $n_{S\cdot a}^{b_{i}}=n_{S\cdot a}^{b_{j}}\neq 0$ with $i\neq
j$. Then, using (3.17), we have

$n_{S\cdot a}^{b_{i}}n_{S\cdot\bar{a}}^{\bar{b}_{j}}=n_{S\cdot
1}^{b_{i}\times\bar{b}_{j}}\neq 0~{}.$ (3.23)

Therefore, the action of $S$ on the trivial line contains the outcome
$b_{i}\times\bar{b}_{j}$. If $i\neq j$, then $b_{i}\times\bar{b}_{j}$ is a
non-trivial line operator. This shows that $b_{i}\times\bar{b}_{j}$ for all
$j\neq i\in\\{1,...,N\\}$ should be one of the outcomes in the fusion $S\cdot
1$. Therefore:

A symmetry of an abelian TQFT is non-invertible if and only if its action on
the trivial line operator results in more than one outcome.

In particular, if $S\cdot 1$ is a sum of $N$ line operators, then the action
of $S$ on any other line operator (if non-trivial) also results in $N$
outcomes. If $b\in S\cdot a$, then $b\times l_{i}\in S\cdot a$ where

$S\cdot 1=1+\sum_{i=1}^{N-1}l_{i}~{}.$ (3.24)

This proves our assertion. $\square$

Moreover, our discussion implies the following:

If an abelian TQFT does not have any non-trivial bosons, then it does not have
any non-invertible symmetries.

Let us now find a more precise characterization of $N$. To that end, it is
easy to see that the larger the number of simple lines, $N$, appearing on the
R.H.S. of (3.24), the larger $|{\rm Null}(S)|$ is (i.e., the larger the number
of simple lines that cannot form a junction on $S$). To derive this statement,
let $S\cdot a=0$, and take $n:=|A|-|{\rm Null}(A)|$ be the number of simple
lines on which $S$ acts non-trivially. Clearly $n\geq 1$, since $S$ acts non-
trivially on $1$. In fact $n\geq 2$, because, if $n=1$, then, to satisfy
condition 3, $S\cdot 1$ should be the sum of all anyons in the TQFT. This
scenario requires all anyons in the TQFT to be bosons, which is inconsistent
with modularity. Moreover, we know that on each of the $n$ simple lines, $S$
acts to give $N$ outcomes. From condition 3, we get

$Nn=N(|A|-|{\rm Null}(A)|)=|A|~{}.$ (3.25)

Then, the action of $S$ on $|A|-n=|{\rm Null}(A)|$ simple line operators is
zero. If $N=1$, then $n=|A|$ and we have an invertible symmetry (clearly
$|{\rm Null}(A)|=0$ in the invertible case).

If there are non-trivial bosons in the TQFT, is there always a non-invertible
symmetry? Yes! Let $B$ be a non-anomalous bosonic 1-form symmetry of
$\mathcal{T}$.131313Given a boson, $b$, in an Abelian TQFT, $b^{n}$ is also a
boson for all $n\in\left\\{0,1,\cdots O_{b}\right\\}$ (where $O_{b}$ is the
order of $b\in A$), since $\theta$ is a quadratic form. This
$\mathbb{Z}_{O_{b}}$ 1-form symmetry is then non-anomalous since $\mathcal{S}$
restricted to this subspace is completely degenerate. Then we can define the
non-invertible symmetry implemented by a surface, $S_{B}$, which acts on the
anyons as follows

$S_{B}\cdot a=\begin{cases}B\text{-orbit of }a&\text{ if a braids trivially
with }B~{},\\\ 0&\text{ if a braids non-trivially with }B~{}.\end{cases}$
(3.26)

As we will see in Section 5, a typical theory has a much larger set of non-
invertible symmetries. However, the symmetries described in (3.26) are
universal in the sense that they are present in all theories with non-
invertible symmetries.141414Let us verify that (3.26) indeed gives a symmetry
according to our definition. To that end, the fact that $n_{S\cdot a}^{c}\neq
0$ if and only if $a$ braids trivially with $B$, and $c=\beta\times a$ for
some (bosonic) $\beta\in B$ means that $\theta_{a}=\theta_{c}$ (i.e.,
condition 1 of our definition is satisfied). Now, if $a$ and $c$ braid
trivially with $B$, then so too does $a\times c$. Moreover, if $b$ and $d$ are
in the $B$-orbits of $a$ and $c$ respectively, then $b\times d$ is in the
$B$-orbit of $a\times c$. As a result, condition 2 is also satisfied. A result
of Müger [40] implies that $|B|\cdot|\mathcal{C}_{B}|=|A|$, where
$\mathcal{C}_{B}$ consists of the lines that braid trivially with $B$. As a
result, condition 3 is also satisfied.

We have therefore arrived at the following result:

Theorem 1: An abelian TQFT has non-invertible symmetries if and only if it has
a non-trivial non-anomalous bosonic 1-form symmetry.151515This is a stricter
version of the result in [11] described in the introduction. We have shown
that the fusion 2-category of modules over a pointed modular category
$\mathcal{C}$, Mod$(\mathcal{C})$, is group-like if and only if $\mathcal{C}$
has no non-trivial objects on which the non-degenerate quadratic form,
$\theta$, vanishes.

Abelian TQFTs without non-invertible symmetries are precisely those which do
not contain any bosonic line operators. These correspond to anisotropic
pointed categories. Anisotropic pointed categories have been completely
classified and are [41]:

  1. 1.

$A_{p}$ TQFT $\cong SU(p)_{1}$ Chern-Simons theory or its Galois conjugate
depending on the odd prime $p$.

  2. 2.

$B_{p}$ TQFT $\cong$ Galois conjugate of $A_{p}$.

  3. 3.

$A_{p}\boxtimes A_{p}=B_{p}\boxtimes B_{p}$ TQFT.

  4. 4.

Semion and $\overline{\text{Semion}}$.

  5. 5.

Semion $\boxtimes$ Semion and
$\overline{\text{Semion}}\boxtimes\overline{\text{Semion}}$.

  6. 6.

3-Fermion Model $\cong$ Spin$(8)_{1}$ Chern-Simons theory.

  7. 7.

$SU(4)_{1}$ Chern-Simons theory and Galois conjugates.

  8. 8.

$SU(4)_{1}$ Chern-Simons theory $~{}\boxtimes$ Semion, $SU(4)_{1}$ Chern-
Simons theory $\boxtimes~{}\overline{\text{Semion}}$ and Galois conjugates.

Here, Semion is $U(1)_{2}$ Chern-Simons theory, and $p$ is an odd
prime.161616A subset of these abelian TQFTs also play a crucial role in the
classification of TQFTs invariant under Galois conjugations [42]. All abelian
TQFTs with non-invertible symmetry come from gauging a 0-form symmetry of an
anisotropic abelian TQFT. Note that this 0-form symmetry must act trivially on
the line operators. If not, the theory obtained after gauging will be a non-
abelian TQFT.

Another way to present the above results is to say that non-invertible
symmetries of Abelian TQFTs are emergent or “non-intrinsic” in the language of
[30, 31]. This characterization follows because non-invertible symmetries
arise from performing a topological operation (0-form gauging) on an input
theory that lacks non-invertible symmetries (in this case, an anisotropic
Abelian TQFT).

Finally, we conclude by noting that the presence of a non-anomalous $2+1$d
1-form symmetry is the condition for an orbifold to produce a 2d “code” CFT in
the construction of [43] (see other related constructions in [44, 45, 46, 47,
48, 49, 50]). This is a CFT living on the boundary of a slab of non-spin
Abelian TQFT (as in Fig. 17) related to a quantum code by the quantum
stabilizer code / CFT map of [43]. Therefore, from Theorem 1, we see that
theories with non-invertible symmetries give rise to orbifold 2d code CFTs in
the language of [43] (see also [51] for an appearance of non-invertible
symmetries in the context of stabilizer codes).171717The discussion in [43]
applies to $\mathbb{Z}_{2}^{k}$ 1-form symmetry, but this discussion can be
generalized. It would be interesting to explore and generalize this connection
further.

In the next section, we will shift our attention to non-Abelian TQFTs.
However, we will revisit Abelian TQFTs and their topological surfaces from the
perspective of higher-gauging in Sec. 5. This perspective will allow us to
better understand the action of more general topological surfaces on the line
operators.

##  4 (Non-)invertible symmetries of non-Abelian TQFTs

In this section, our goal is to begin generalizing the results of the previous
section to non-Abelian TQFTs (again, recall from table 1 that we only study
non-spin theories). To that end, consider a non-abelian TQFT, $\mathcal{T}$,
whose lines are described by some unitary MTC, $\mathcal{C}$. To characterise
non-invertible symmetries of $\mathcal{T}$, we generalize Fig. 7 to non-
abelian TQFTs as in Fig. 8.

$c$$a$$S$$e$$f$$\Longrightarrow$$\beta$$\alpha$$\gamma$$d$$b$$S$ Fig. 8: A
necessary condition for $S$ to generate a non-invertible symmetry.

In this latter figure, $e\in b\times d$ and $f\in a\times c$. Here $\alpha$
and $\beta$ belong to the vector spaces $V_{S\cdot a}^{b}$ and $V_{S\cdot
c}^{d}$ respectively. There are a total of $n_{S\cdot a}^{b}n_{S\cdot c}^{d}$
possible such junctions. For each of these junctions, there should be some
$e\in b\times d$, $f\in a\times c$, and $\gamma\in V_{Sf}^{e}$ such that the
diagram on the R.H.S. is allowed. Therefore, we get the constraint

$n_{S\cdot a}^{b}n_{S\cdot c}^{d}\leq\sum_{e,f}N_{ac}^{f}N_{bd}^{e}n_{S\cdot
f}^{e}~{}.$ (4.1)

Moreover, if dim$(V_{S\cdot f}^{e})\neq 0$, there must be a morphism between
junction vector spaces $V_{S\cdot a}^{b}\otimes V_{S\cdot c}^{d}$ and
$\bigoplus_{f,e}n_{S\cdot f}^{e}~{}V_{S\cdot f}^{e}$. In an explicit choice of
basis, we get the constraint in Fig. 9.

$c$$a$$S$$e$$f$$\alpha$$\beta$$\gamma$$d$$b$$S$$a$$c$$d$$b$$\mu$$\nu$$=\sum\limits_{f,e,\mu,\nu,\gamma}\left[\tilde{F}_{(a,b),(c,d),(f,e)}\right]_{(\alpha,\beta)}^{(\mu,\nu,\gamma)}$
Fig. 9: A morphism between junction vector spaces.

Considering the action of a surface operator on three line operators and using
the associativity of the fusion rules, we get a constraint on the $\tilde{F}$
matrices given by Fig. 10.181818See [52, 53] for a discussion of these
constraints in the context of module categories.

$S$$S$$S$$S$$S$$\tilde{F}$$\tilde{F}$$\tilde{F}$$F,F$ $\tilde{F}$ Fig. 10:
Constraint on $\tilde{F}$ matrices from associativity of fusion. On the right
bottom of the diagram, the $F,F$ action indicates the $F$-move on either side
of the surface operator.

In addition, the action of the surface operator must be compatible with the
braiding of line operators. This reasoning leads to the constraint in Fig. 11,
which leads to the equation

$c$$d$$a$$b$$S$$c$$d$$a$$b$$S$$=$ Fig. 11: Compatibility of $S$ with
braiding.

$R_{bd}^{e}\tilde{F}_{(c,d),(a,b),(f,e)}(R_{ac}^{f})^{-1}=\tilde{F}_{(a,b),(c,d),(f,e)}~{}.$
(4.2)

Following this discussion, we can define a (non-invertible) symmetry of a non-
abelian TQFT as follows:

Definition: A (non-invertible) symmetry of a non-abelian TQFT, $\mathcal{T}$,
with MTC, $\mathcal{C}$, is a map, $S$, on the ring of line operators such
that:

  1. 1.

The $\tilde{F}$ matrices defined in Fig. 9 satisfying the conditions in Figs.
10 and 11 exist.

  2. 2.

$n_{S\cdot a}^{b}n_{S\cdot c}^{d}\leq\sum_{e,f}N_{ac}^{f}N_{bd}^{e}n_{S\cdot
f}^{e}$ and $n_{S\cdot 1}^{1}=1$.

  3. 3.

$\sum_{a,b\in\mathcal{C}}n_{S\cdot a}^{b}d_{a}d_{b}=\text{dim}(\mathcal{C})$.

In appendix Appendix A. Relating our definitions to Lagrangian subalgebras in
folded theories, we show that this definition is equivalent to the definition
of boundary conditions in terms of Lagrangian algebras in
$\mathcal{T}\boxtimes\bar{\mathcal{T}}$.

Note that the last two conditions above involve gauge-invariant quantities
(i.e., quantities independent of fusion spaces), while the first condition
involves $\tilde{F}$ and $R$, which are gauge-dependent. The final condition
can be derived, as in the case of the corresponding Abelian condition 3, from
commutation of $n_{S}$ with the $\mathcal{S}$ matrix (see the discussion
around (3.12)).

Furthermore, note that the last two conditions are necessary but not
sufficient to have a non-invertible symmetry [54, 55]. For example, in [54]
Davydov shows that, while the choice $n_{S\cdot a}^{b}=\delta_{a,b}$ is always
physical/consistent (since it holds for the identity interface), the choice
$n_{S\cdot a}^{b}=\delta_{a,\bar{a}}$ is not always consistent. In other
words, charge conjugation is not always a symmetry of a TQFT, even though it
is a symmetry of the modular data and hence of the fusion rules (see also
[56]). It is therefore not a surprise that $n_{S\cdot
a}^{b}=\delta_{a,\bar{a}}$ satisfies the last two conditions in our
definition.191919It would be very interesting to obtain a set of constraints
using only gauge-invariant quantities to characterise a non-invertible
symmetry.

In addition, note that, for a given choice of integers, $n_{S\cdot a}^{b}$,
satisfying conditions 2 and 3, we may have distinct $\tilde{F}$ solutions
satisfying 1 [57][58, Remark 3.9]. Such surface operators have identical
action on the line operators and correspond to distinct isospectral boundary
CFTs in the configuration in Fig. 17. For example, in [57] Davydov finds two
isospectral CFTs with the charge-conjugation modular invariant. This result
implies that, in the corresponding bulk TQFT, there is an invertible surface
operator which acts trivially on all line operators, but acts non-trivially on
the fusion spaces.

This discussion motivates us to more explicitly consider the action of
symmetries on fusion spaces. To that end, take a junction of three line
operators, $a$, $b$, and $c$. We can move a (generically non-invertible)
surface operator operator across the junction as in Fig. 12 to get a matrix,
$U_{S}(a,b,c,f)_{e,d}$, acting on the fusion space, $V_{ed}^{f}$.

$S$$f$$c$$\gamma$$a$$b$$\nu$$S$$f$$a$$b$$\nu^{\prime}$$d$$e$$=\
\sum\limits_{e,d,\nu^{\prime}}[U_{S}(a,b,c,f)_{e,d}]_{\nu\nu^{\prime}}$ Fig.
12: Action of $S$ on fusion spaces.

At the level of fusion spaces, this action can be written as

$U_{S}(a,b,c):\bigoplus_{d,e,f}N_{de}^{f}~{}n_{Sa}^{d}~{}n_{Sb}^{e}~{}n_{Sc}^{f}~{}V_{de}^{f}\rightarrow
V_{ab}^{c}~{},$ (4.3)

where $U_{S}(a,b,c)$ is a block-diagonal matrix, with the block
$U_{S}(a,b,c,f)_{d,e}$ acting on $V_{de}^{f}$. Note that in general $U_{S}$ is
a rectangular matrix, and is therefore non-invertible.

The collection of matrices $\\{U_{S}(a,b,c)\\}$ cannot be chosen arbitrarily.
They must satisfy consistency conditions with the $\tilde{F}$ matrices.
Writing these full consistency conditions and deriving their consequences are
beyond the scope of this paper (this route would amount to a non-invertible
generalization of [59]).202020In principle, the full set of consistency
conditions is already encoded in the definition of the 2-category of module
categories over a modular tensor category. However, it would be illuminating
to reduce the consistency conditions to a set of polynomial equations to
enable explicit calculations. However, the discussion in [57] motivates us to
ask the following simpler question: if a surface acts invertibly on lines, can
it act non-invertibly on fusion spaces?

To answer this question, we first return to Fig. 9 and specialize to the case
that the surface acts invertibly (i.e., as a permutation) on the lines. We get
Fig. 13, where we have simplified the notation for $\tilde{F}$.

$c$$a$$S$$\sigma(f)$$f$$\sigma(c)$$\sigma(a)$$S$$a$$c$$\sigma(c)$$\sigma(a)$$\mu$$\nu$$=\sum\limits_{\mu,\nu,f}\left[\tilde{F}^{\sigma}_{a,c,f}\right]^{(\mu,\nu)}$
Fig. 13: A surface, $S$, with a permutation action on lines.

Let the action on $S$ on the fusion space, $V_{ab}^{c}$, be given by the
matrix

$U_{S}(a,b,c)_{\mu,\nu}~{},$ (4.4)

where $\mu,\nu$ runs over an orthonormal basis for $V_{ab}^{c}$. Then, moving
the surface on the R.H.S. of Fig. 13 over the fusion space, we get Fig. 14.

$c$$a$$S$$\sigma(f)$$\sigma(c)$$\sigma(a)$$S$$a$$c$$\sigma(c)$$\sigma(a)$$\mu$$\nu^{\prime}$$=\sum\limits_{\mu,\nu,f}\left[\tilde{F}^{\sigma}_{a,c,f}\right]^{(\mu,\nu)}\sum\limits_{\nu^{\prime}}U_{S}(a,c,f)_{\nu\nu^{\prime}}$
Fig. 14: The action of $S$ on fusion spaces.

Using the relation in Fig. 15 on the L.H.S. of Fig. 14 we get Fig. 16. Since
the sum in the L.H.S and R.H.S of this diagram is over the same basis, we can
equate the components to get the equation

$\sqrt{\frac{d_{f}}{d_{a}d_{c}}}\delta_{\nu^{\prime},\mu^{\dagger}}=\sum_{\nu^{\prime}}[\tilde{F}^{\sigma}_{a,c,f}]^{\mu,\nu}~{}U_{S}(a,c,f)_{\nu\nu^{\prime}}~{}.$
(4.5)

$\sigma(c)$$\sigma(a)$$f$$\sigma(a)$$\sigma(c)$$\nu$$\sigma(c)$$\sigma(a)$$\mu$$=\sum\limits_{\mu,\nu,f}\sqrt{\frac{d_{f}}{d_{a}d_{c}}}\
\delta_{\nu,\mu^{\dagger}}$ Fig. 15: The completeness relation for lines.
Note that $d_{\sigma(\ell)}=d_{\ell}$ by commutativity with $\mathcal{S}$.
$S$$a$$c$$\mu$$\nu^{\prime}$$=\sum\limits_{\mu,\nu,f}\left[\tilde{F}_{a,c,f}^{\sigma}\right]^{(\mu,\nu)}\sum\limits_{\nu^{\prime}}U_{S}(a,c,f)_{\nu\nu^{\prime}}$$S$$\sigma(f)$$\sigma(c)$$\sigma(a)$$\mu$$\nu^{\prime}$$\sum\limits_{\mu,\nu^{\prime}f}\sqrt{\frac{d_{f}}{d_{a}d_{c}}}\
\delta_{\nu^{\prime},\mu^{\dagger}}$$c$$a$$\sigma(c)$$\sigma(a)$$\sigma(f)$
Fig. 16: Using completeness relation for lines on the L.H.S of Fig. 14.

The L.H.S. of (4.5) is an invertible matrix and the R.H.S. is a product of two
square matrices. Therefore, both $\tilde{F}$ and $U_{S}$ have to be invertible
matrices. In particular, we see that:

Lemma 2: If a surface acts invertibly on lines, then it also acts invertibly
on fusion spaces.

As a result, the invertible ambiguity found by Davydov in [57] does not
generalize to a non-invertible ambiguity. Said differently, an invertible
symmetry and a non-invertible symmetry cannot differ only by their action on
fusion spaces.

Given these results, let us prove a partial generalization of theorem 1 to
unitary non-Abelian TQFTs. To that end, suppose we have a TQFT without any
bosonic line operators. Since the action of a surface operator preserves
topological twists, it cannot map a non-trivial line operator in such a theory
to the trivial line operator. By the same logic, a surface cannot map the
trivial line operator to a non-trivial line. Therefore, we have

$n_{S\cdot 1}^{b}=n_{S\cdot b}^{1}=\delta_{b,1}~{}.$ (4.6)

Next, since $n_{S\cdot a}^{b}$ commutes with the modular $S$-matrix of the
TQFT (it describes a modular invariant in the boundary CFT), we have

$\sum_{c}\mathcal{S}_{ac}n_{S\cdot c}^{b}=\sum_{c}n_{S\cdot
a}^{c}\mathcal{S}_{cb}~{},$ (4.7)

where $\mathcal{S}$ is the modular S-matrix. Choosing $b=1$ and using (4.6),
we get

$d_{a}=\sum_{c}n_{S\cdot a}^{c}d_{c}~{},$ (4.8)

where $d_{a}$ are the quantum dimensions of the line operators. This logic
shows that, when the TQFT does not have any bosons, the action of the surface
operator preserves the quantum dimensions. On the other hand, choosing $a=1$
and using (4.6), we get

$\sum_{c}n_{S\cdot c}^{b}d_{c}=d_{b}~{}.$ (4.9)

In a unitary TQFT, the quantum dimensions satisfy $d_{a}\geq 1$ for all line
operators $a$. Also, note that $n_{S\cdot a}^{b}$ are non-negative integers.
Suppose $n_{S\cdot e}^{f}\neq 0$, then using (4.8) for $a=e$ we find that
$d_{f}\leq d_{e}$. Also, using (4.9) for $b=f$, we have $d_{f}\geq d_{e}$.
Therefore, $d_{e}=d_{f}$ and the only consistent solution is

$n_{S\cdot a}^{b}=\delta_{b,\sigma(a)}~{},$ (4.10)

for some permutation, $\sigma$, of the lines. That is, in a unitary TQFT
without bosons, all symmetries act as permutations on the line operators. In
the boundary CFT, the corresponding statement is that, after the maximal
extension of the chiral algebra, the partition function is always given by a
permutation modular invariant [60, 61].

Therefore, using lemma 2, we arrive at:

Theorem 3: $2+1$d TQFTs without bosons only have invertible symmetries.

In the Abelian case, we know from theorem 1 that the converse also holds.
However, the converse does not hold more generally. Indeed, there are non-
abelian TQFTs with bosonic line operators that do not have any non-invertible
symmetries. To understand this statement, let us assume that a TQFT,
$\mathcal{T}$, has a non-invertible symmetry implemented by some surface
operator, $S$. From the argument leading to theorem 3, we know that either $S$
should map a non-trivial line operator the trivial line operator, or $S$
should map the trivial line operator to a non-trivial line operator. That is,

$n_{S\cdot 1}^{a}\neq 0\text{ or }n_{S\cdot a}^{1}\neq 0~{},$ (4.11)

for some non-trivial line operator, $a$. Therefore, the TQFT, $\mathcal{T}$,
has a bosonic line operator, $a$. More precisely, let $\mathcal{A}_{L}$ be the
set of bosonic line operators such that $n_{S\cdot a}^{1}\neq 0$, and
similarly define $\mathcal{A}_{R}$ to be the set of bosonic line operators
such that $n_{S\cdot 1}^{a}\neq 0$. The authors of [53] showed that
$\mathcal{A}_{L}$ and $\mathcal{A}_{R}$ admit the structure of a commutative
Frobenius algebra. In other words, $\mathcal{A}_{L}$ and $\mathcal{A}_{R}$ are
bosons that can be condensed.212121As we will discuss below, the bosonic lines
in $\mathcal{A}_{L}/\mathcal{A}_{R}$ can be condensed on a surface to define a
surface operator. The action of the surface operator on the line operators
correspond to a type 1 parent of the original partition function in the
boundary CFT. Type 1 partition functions are sums of squares of linear
combinations of characters [52]. Therefore, a necessary condition for a TQFT
to have a non-invertible symmetry is that it must contain a bosonic line
operator which can be condensed.

For example, the Fibonacci TQFT does not contain any bosons. Therefore, it is
clear that this theory does not have any non-invertible symmetries. If we
stack five copies of Fibonacci TQFT, then this theory does contain a boson.
However, the authors of [62, 63] showed that this boson cannot be condensed.
Therefore, this TQFT, despite having a boson, does not have any non-invertible
symmetries. More generally, $SO(3)_{k}$ Chern-Simon theory at odd $k$, or any
of its powers do not have non-invertible symmetries [64].222222A modular
tensor category which does not contain condensable bosons is called completely
anisotropic. This is a stronger condition than anisotropy, which only requires
the TQFT does not contain a Rep$(G)$ subcategory for some non-trivial group
$G$.

To find a converse of theorem 3, we will need to introduce aspects of higher
gauging in the next section. In other words, we will need to understand some
details of how to construct topological surfaces from lines.

However, before moving on to the resolution of surfaces in terms of lines, let
us consider an Abelian TQFT with pointed MTC, $\mathcal{C}$, and show that our
general non-Abelian symmetry definition reduces to our Abelian one. Then the
conditions 2 and 3 reduce to corresponding conditions in the Abelian case.
Moreover, condition 1 leads to (4.2) which, in turn, leads (for $b=d$ and
$a=c$) to $\theta_{a}=\theta_{b}$, which reproduces the first condition in the
Abelian case.

##  5 Resolving surfaces into lines: higher gauging and applications

In this section we build on the general results of the previous two sections
by specializing to the case of condensation surfaces. As alluded to above,
these are defects that are constructed by summing over lines (higher gauging)
on a two-dimensional submanifold of spacetime.

After reviewing basic properties of higher gauging in Abelian TQFTs, we
generalize the “universal” non-invertible surfaces we constructed in (3.26)
and consider arbitrary condensation surfaces. We explicitly construct the
action of these general surfaces on the lines of Abelian TQFTs and show how to
reconstruct the higher-gauging data from the action on lines.

Then, we move on to condensation surfaces in non-Abelian TQFTs and give a
construction of a non-Abelian analog of the condensation surface in (3.26).
This result leads to a generalization of theorem 1 to non-Abelian TQFTs.

Finally, we conclude this section by considering arbitrary condensation
surfaces built out of Abelian lines in general (not necessarily Abelian)
TQFTs. We characterize precisely when such surfaces are (non-)invertible.

###  5.1 Abelian TQFTs

As in section 3, we consider a non-spin $2+1$d TQFT with invertible
topological line operators forming an abelian group, $A$, under fusion. The
full data that describes the 1-form symmetry is the group, $A$, along with the
fusion and braiding phases: $F(a,b,c)$ and $R(a,b)$ respectively. This data
forms a pointed braided fusion category, $\mathcal{C}$.

Consider the higher gauging of a 1-form symmetry subgroup, $G\leq A$, on a
2-dimensional submanifold, $\Sigma$ [12]. For this procedure to be consistent,
the 1-anomaly of $G$ should vanish. In other words, $F(g,h,k)$ restricted to
$G$ should be trivial in $H^{3}(G,U(1))$. Therefore, there exist phases,
$\tau_{G}(a,b):G\times G\to\mathds{C}$, such that

$F(g,h,k)=\frac{\tau_{G}(h,k)\tau_{G}(g,h+k)}{\tau_{G}(g+h,k)\tau_{G}(g,h)}~{},~{}~{}g,h,k\in
G~{}.$ (5.1)

This condition ensures that the gauging of the $G$-lines on $\Sigma$ does not
depend on a choice of a triangulation of $\Sigma$. While higher gauging $G$ on
$\Sigma$, we have the freedom to add an SPT phase (a.k.a. discrete torsion)
labelled by a 2-cocycle, $\sigma\in H^{2}(G,U(1))$. Let us label the
condensation surface operator obtained in this way as $S(\Sigma,G,\sigma)$.

As a concrete example, consider the higher gauging of a subgroup,
$\mathds{Z}_{2}\leq A$, on $\Sigma$ [12]. The line operator, $a$, generating
this subgroup can be higher-gauged if and only if $\theta_{a}=\pm 1$. This
condition is equivalent to the existence of a gauge in which the
$\mathbb{Z}_{2}$ restriction of the $F$ matrices satisfy
$F|_{\mathds{Z}_{2}}=1$. We can then define

$S(\Sigma,\mathds{Z}_{2},1):=\frac{1}{\sqrt{|H_{1}(\Sigma,\mathds{Z}_{2})}|}\sum_{\gamma\in
H_{1}(\Sigma,\mathds{Z}_{2})}a(\gamma)~{}.$ (5.2)

This condensation surface has the following fusion rule depending on
$\theta_{a}$:

$\displaystyle\theta_{a}=1$ $\displaystyle:$ $\displaystyle
S(\Sigma,\mathds{Z}_{2},1)\times
S(\Sigma,\mathds{Z}_{2},1)=\sqrt{|H_{1}(\Sigma,\mathds{Z}_{2})|}~{}S(\Sigma,\mathds{Z}_{2},1)~{},$
(5.3) $\displaystyle\theta_{a}=-1$ $\displaystyle:$ $\displaystyle
S(\Sigma,\mathds{Z}_{2},1)\times S(\Sigma,\mathds{Z}_{2},1)=\mathds{1}~{},$
(5.4)

where $\mathds{1}$ is the trivial surface operator. Whether the condensation
defect, $S(\Sigma,\mathds{Z}_{2},1)$, acts non-trivially on $\mathcal{C}$ or
not depends crucially on how the line operator, $a$, is embedded in
$\mathcal{C}$. For example, if $\mathcal{C}$ is $\mathds{Z}_{2}$ discrete
gauge theory, $S(\Sigma,\mathds{Z}_{2},1)$ implements electric-magnetic
duality. On the other hand, if $\mathcal{C}$ is the Ising category, then
$S(\Sigma,\mathds{Z}_{2},1)$ acts trivially.

Note that the fusion rule for $S(\Sigma,G,\sigma)$ depends on the 2-manifold,
$\Sigma$. For example, in the case of $S(\Sigma,\mathds{Z}_{2},1)$, we get the
fusion rule [12]

$S(\Sigma,\mathds{Z}_{2},1)\times
S(\Sigma,\mathds{Z}_{2},1)=\sqrt{|H_{1}(\Sigma,\mathds{Z}_{2})|}~{}S(\Sigma,\mathds{Z}_{2},1)~{}.$
(5.5)

In particular, for $\Sigma=\mathds{T}^{2}$, we find

$S(\mathds{T}^{2},\mathds{Z}_{2},1)\times
S(\mathds{T}^{2},\mathds{Z}_{2},1)=2~{}S(\mathds{T}^{2},\mathds{Z}_{2},1)~{}.$
(5.6)

From now on, we will fix $\Sigma=\mathds{T}^{2}$ and write the surface
operator as $S(G,\sigma)$ without explicitly mentioning the dependence on the
2-manifold. We will be mainly interested in distinguishing invertible and non-
invertible surface operators, and so there is no loss of generality in
choosing $\Sigma=\mathds{T}^{2}$. Indeed, we have

$H_{1}(\Sigma,M)\cong M^{2g}~{},$ (5.7)

where $\Sigma$ is a compact genus $g$ surface, and $M$ is an abelian group.
Therefore, if the fusion rules are non-invertible over the torus, they are
non-invertible on any higher genus surface.

####  5.1.1 Action on line operators

To describe the action of the condensation defect on line operators, we must
consider the direction normal to the surface as well (since the surface acts
on lines that pierce it) and therefore the braiding data of the lines
naturally enters.232323Since the surface itself is two dimensional, the
obstruction to constructing it only depends on the fusion data (i.e., the $F$
matrices). Moreover, even though the 2d SPT does not affect the 1-anomaly of
the lines, it naturally enters the action of the surface on lines (and
therefore also the surface fusion rules). In particular, it turns out to be
useful to define the following phase

$\Xi_{[G,\sigma]}(g,h):=R(g,h)\frac{\tau_{G}(g,h)}{\tau_{G}(h,g)}\frac{\sigma(g,h)}{\sigma(h,g)}~{},$
(5.8)

which explicitly contains braiding and SPT data. Indeed, we will show that the
action of the condensation defect, $S(G,\sigma)$, on a line operator, $p$, is
given by

$S(G,\sigma)\cdot p=\sum_{g\in G}\Bigg{[}\frac{1}{|G|}\sum_{h\in
G}\frac{\theta_{h\cdot
p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)\Bigg{]}g\cdot p~{}.$ (5.9)

From this expression, note that $g\cdot p\in S(G,\sigma)\cdot p$ if

$\frac{1}{|G|}\sum_{h\in G}\frac{\theta_{h\cdot
p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)\neq 0~{}.$ (5.10)

Since this sum is over $|G|$ phases, we have

$\frac{1}{|G|}\sum_{h\in G}\frac{\theta_{h\cdot
p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)\in\\{0,1\\}~{}.$ (5.11)

For this sum to be equal to $1$, $p$ should satisfy

$\frac{\theta_{h\cdot
p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)=1~{},~{}\forall~{}h\in G~{}.$
(5.12)

Now, let us understand how the expression (5.9) for the action of the surface
operator on the line operators can be derived. Consider a CFT $\mathcal{T}$
with charge conjugation partition function

$Z_{\mathcal{T}}=\sum_{p}\chi_{p}\bar{\chi}_{\bar{p}}~{},$ (5.13)

where $p$ labels the representations of some chiral algebra, $V$. We will
choose a chiral algebra such that Rep$(V)\simeq\mathcal{C}$ as modular tensor
categories. Then, this CFT has a 0-form symmetry implemented by Verlinde lines
labelled by $a\in A$. Gauging a subgroup, $G\leq A$, with discrete torsion,
$\sigma\in H^{2}(G,U(1))$, results in a CFT, $\mathcal{T}/(G,\sigma)$, with
partition function [65]

$Z_{\mathcal{T}/(G,\sigma)}=\sum_{g\in G}\Bigg{[}\frac{1}{|G|}\sum_{h\in
G}\frac{\theta_{h\cdot
p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)\Bigg{]}\chi_{p}\bar{\chi}_{\overline{g\cdot
p}}~{}.$ (5.14)

$a$$b$$O_{(a,b)}$$S(G,\sigma)$$\Sigma_{1}$$\Sigma_{2}$$I$ Fig. 17: The
pairing of 2d CFT left and right movers on $\Sigma_{1}$ and $\Sigma_{2}$
respectively is specified by an abelian TQFT on $X\simeq\Sigma_{1}\times I$
with a surface operator, $S(G,\sigma)$, inserted between [21, 66]. Different
$S(G,\sigma)$ lead to different modular invariants.

Consider the 3d-2d picture where the bulk theory is the TQFT described by the
modular tensor category $\mathcal{C}$, and the CFT left and right movers are
on $\Sigma_{1}$ and $\Sigma_{2}$ respectively (see Fig. 17). The partition
function of the CFT is described by how Wilson lines end on the surface
operator $S(G,\sigma)$ of the bulk TQFT. More precisely, if the Wilson lines
$p$ and $g\cdot p$ form a junction on the surface operator, then the CFT
contains the primary operator $O_{(p,g\cdot p)}$. Conversely, if the CFT
contains a primary operator $O_{(p,g\cdot p)}$, then the Wilson lines with
labels $p$ and $g\cdot p$ can form a junction on the surface operator
$S(G,\sigma)$. This logic shows that the action of $S(G,\sigma)$ on the Wilson
line, $p$, contains the outcome $g\cdot p$ (possibly among others). Therefore,
using the general expression in equation (5.14) for the partition function, we
get the action of the surface operator on Wilson lines in equation (5.9).

Note that, in an Abelian TQFT, all non-trivial surface operators act non-
trivially on the line operators. Indeed, suppose we have some condensation
defect, $S(G,\sigma)$, which acts trivially on all line operators. Then, using
(5.12), we get

$\frac{\theta_{h\cdot p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)=1~{},\
\forall\ h\in G~{},\ \forall p\in\mathcal{C}~{},$ (5.15)

if and only if $g$ is the trivial line operator. This logic implies

$\frac{\theta_{h\cdot p}}{\theta_{h}\theta_{p}}=|A|\mathcal{S}_{h,p}=1~{},\
\forall h\in G~{},\ \forall p\in\mathcal{C}~{}.$ (5.16)

To be consistent with modularity, $h$ must be the trivial line operator.
Therefore, if $S(G,\sigma)$ acts trivially on all line operators, then $G$
must be the trivial subgroup. Note that $S(\mathds{Z}_{1},1)$ is the trivial
surface operator, and this is the only such surface operator which acts
trivially on all line operators.

Therefore, non-trivial surface operators of an abelian TQFT are completely
characterised by their action on the line operators. As we discussed in
section 4, this is in contrast with non-abelian TQFTs where surface operators
can act non-trivially on fusion spaces but trivially on the lines [57].

####  5.1.2 Reconstructing the higher-gauged symmetry and SPT

Given the construction of condensation defects, $S(G,\sigma)$, in the previous
subsection, we would like to understand if we can reconstruct $G\leq A$ and
$\sigma\in H^{2}(G,U(1))$ from the symmetry action. Recall from (5.9) that the
action of this surface operator on the lines is given by

$S(G,\sigma)\cdot p=\sum_{g\in G}\Bigg{[}\frac{1}{|G|}\sum_{h\in
G}\frac{\theta_{h\cdot
p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)\Bigg{]}g\cdot p~{}.$ (5.17)

Therefore, the elements of $G$ as a set are given by

$G=\\{(S(G,\sigma)\cdot p)\cdot\bar{p}~{}|~{}p\in\mathcal{C}\\}~{}.$ (5.18)

Let us verify that fusions of the form $(S(G,\sigma)\cdot p)\cdot\bar{p}$ give
us all line operators, $g$, that are condensed. In other words, we have to
show that for every $g$ in the group being condensed, there is at least one
line operator, say $p$, such that $S(G,\sigma)\cdot p=g\cdot p$. We know that
the line operators $p$ and $g\cdot p$ are related by a symmetry action if and
only if the condition (5.12) is satisfied. Let us show that, for a given $g$,
there is at least one line operator, $p$, satisfying this condition.

From the modularity of the S-matrix, we know that all 1-dimensional
representations of the fusion rules of line operators are given by

$\lambda_{x}(a)=\frac{\mathcal{S}_{ax}}{\mathcal{S}_{1x}}~{}.$ (5.19)

where $x,a\in A$. For a given line operator $x$, the function
$\lambda_{x}(a)$, when restricted to the abelian subgroup of line operators
$G$, must be a character of $G$. Moreover,

$\lambda_{x}(a)=\lambda_{x^{\prime}}(a)~{},$ (5.20)

when $x^{\prime}=x\cdot e$, and $e\in Z_{A}(G)$. Here, $Z_{A}(G)$ is the set
of line operators in $A$ which braid trivially with all line operators in $G$.
Therefore, the number of distinct characters is

$\frac{|A|}{|Z_{A}(G)|}=|G|~{},$ (5.21)

where we have used the result that $|Z_{A}(G)|=\frac{|A|}{|G|}$ [40].
Therefore, restricting $\lambda_{x}(a)$ to the subgroup $G$ gives us all
characters of the group $G$. Now, the constraint (5.12) can be written in
terms of the S-matrix as

$\lambda_{p}(h)=\frac{\mathcal{S}_{hp}}{\mathcal{S}_{1p}}=(\Xi_{[G,\sigma]}(g,h))^{-1},~{}\forall~{}h\in
G~{}.$ (5.22)

$\Xi_{[G,\sigma]}^{-1}$ is a bicharacter (and in particular a character) of
the group $G$. Since the functions $\lambda_{p}(h)$ for $p\in A$ exhaust all
characters of $G$, there must be some $p$ such that the above equation is
satisfied. This finishes the proof that for a given $g$ there is always some
$p$ such that $S(G,\sigma)\cdot p=g\cdot p$.

Also, if $g\cdot p\in S\cdot p$, then, using (5.12), we get

$\frac{\sigma(g,h)}{\sigma(h,g)}=\frac{\theta_{h}\theta_{p}}{\theta_{h\cdot
p}}\frac{\tau_{G}(h,g)}{\tau_{G}(g,h)}\frac{1}{R(g,h)}~{}.$ (5.23)

As a result, the choice of discrete torsion is also fixed by the symmetry
action.242424In a non-Abelian TQFT with a symmetry implemented by a surface
operator, $S(G,\sigma)$, obtained from higher-gauging invertible line
operators, $G$ can be fixed by choosing some invertible line operator
$g\in\mathcal{C}$ such that  $S(G,\sigma)\cdot p\ni g\cdot p~{},$ (5.24) for
each $p\in\mathcal{C}$. The discrete torsion, $\sigma$, can then be fixed
using (5.23). Note that there may not be a unique set of $G$ and $\sigma$
satisfying these constraints. Indeed, distinct algebra objects in non-abelian
TQFTs can be Morita equivalent. For example, consider the surface operator,
$S$, that acts trivially on the Ising model. We have  $S\cdot
1=1~{},~{}S\cdot\psi=\psi~{},~{}S\cdot\sigma=\sigma~{}.$ (5.25) We can choose
the group $G$ to be either the trivial group or a $\mathds{Z}_{2}$ group
generated by $\psi$. Indeed the algebra objects  $1\text{ and }1+\psi~{},$
(5.26) are Morita equivalent in the Ising category. This discussion
illustrates that the embedding of the $G$-lines in the larger TQFT is
important for understanding the action of the corresponding surface.

Next, we consider a set of examples to illustrate the above points. The first
example has non-trivial invertible and non-invertible surfaces, while the
second and third examples only have non-trivial non-invertible and invertible
surfaces respectively. Finally, in Appendix Appendix B. Determining all
symmetries in further Abelian examples, we consider two infinite classes of
theories and determine all their surface operators. In all the examples, we
find that non-invertible surface operators are always of the form $S_{\rm
inv}\times S_{B}$, where $S_{B}$ is a universal surface operator described in
(3.26), and $S_{\rm inv}$ is an invertible surface operator. It will be
interesting to understand if this statement is true for non-invertible surface
operators of an arbitrary Abelian TQFT.

####  5.1.3 Example 1: Toric code $\cong$ Spin$(16)_{1}$ Chern-Simons theory
$\cong$ $\mathbb{Z}_{2}$ discrete gauge theory

Consider the Toric code (also known as Spin$(16)_{1}$ Chern-Simons Theory or
$\mathbb{Z}_{2}$ discrete gauge theory) with lines operators $1$, $e$, $m$,
and $\psi$ corresponding to topological spins

$\theta_{1}=1,~{}\theta_{e}=1,~{}\theta_{m}=1,~{}\theta_{\psi}=1~{}.$ (5.27)

Let us use the characterisation of symmetries in terms of their action on line
operators to determine the symmetries of this TQFT. Let us consider the action
of the symmetry on the trivial line operator. $S\cdot 1$ cannot be a sum of
more than two outcomes. This follows from applying (3.25)

$Nn=4~{},$ (5.28)

where $N$ is the number of outcomes for lines that are not annihilated by $S$,
and $n$ is the number of lines that are not annihilated by $S$. Recall that
$n\geq 2$ from modularity. Therefore, $N$ has to be $2$ or $1$. Let us
consider these cases in turn: $N=1$. In this case, the trivial line operator
is fixed by the symmetry action. Therefore, the symmetry is invertible.
Invertible symmetries of the Toric code are the trivial symmetry and the
electric-magnetic duality (which exchanges $e$ and $m$ while fixing $1$ and
$\psi$). $N=2$. In this case, we have

$S\cdot 1=1+\ell~{}.$ (5.29)

Preservation of topological spins implies that $\ell$ is $e$ or $m$. Also, if
$S\cdot\psi$ is non-trivial, then it should be a sum of two fermionic lines,
but $\psi$ is the only fermion. Therefore, $S\cdot\psi=0$.

Now, let us consider the case with $\ell=e$ in more detail (the case of
$\ell=m$ will be related to it by fusion with the electric-magnetic duality
surface). If the action of $S$ on $e$ is non-trivial, it should be a sum of
two line operators. Note that $m\not\in S\cdot e$ since then $\psi\in S\cdot
e$, which is inconsistent with preserving spins. Therefore, we have two
options. $S\cdot e=1+e$ or $S\cdot e=0$. [${\bf(a)}$] $S\cdot e=1+e$. To
satisfy (3.25) we must have $S\cdot m=0$. This is the universal $S_{B}$
symmetry for $B=\\{1,e\\}$ described in (3.26). [${\bf(b)}$] $S\cdot e=0$. In
this case, to satisfy (3.25), $S\cdot m$ must be non-trivial. The only
consistent choice is $S\cdot m=1+e$. Including the surfaces related to the two
non-invertible symmetries described above via electric-magnetic duality (i.e.,
via $e\leftrightarrow m$), we get $4$ non-invertible symmetries and $2$
invertible symmetries (including the trivial one). This result agrees with the
known classification of symmetries in terms of condensation defects [12].

Next, let us use the discussion in section 5.1.2 to find the lines that need
to be condensed to get surface operators implementing the symmetries above.
For the electric-magnetic duality symmetry discussed above, using (5.18) we
find that the lines to be condensed are

$1~{},\ \psi~{},$ (5.30)

without any discrete torsion, as
$H^{2}(\mathds{Z}_{2},U(1))\cong\mathds{Z}_{1}$.

Now, let us consider the symmetry

$S\cdot 1=S\cdot e=1+e~{},~{}S\cdot m=S\cdot\psi=0~{}.$ (5.31)

Using (5.18), we find that the lines to be condensed are

$1~{},\ e~{}.$ (5.32)

Similarly, for the symmetry

$S\cdot 1=S\cdot m=1+m~{},~{}S\cdot e=S\cdot\psi=0~{},$ (5.33)

the lines to be condensed are

$1~{},\ m~{}.$ (5.34)

Now, let us consider the symmetry

$S\cdot 1=1+e~{},~{}S\cdot m=1+e~{},~{}S\cdot e=S\cdot\psi=0~{}.$ (5.35)

Using (5.18), we find that the lines to be condensed are

$1~{},\ e~{},\ m~{},\ \psi~{}.$ (5.36)

Since the 1-form symmetry to be condensed is
$\mathds{Z}_{2}\times\mathds{Z}_{2}$, we have a choice of discrete torsion:
$H^{2}(\mathds{Z}_{2}\times\mathds{Z}_{2},U(1))\cong\mathds{Z}_{2}$. To
understand which discrete torsion corresponds to the above surface, we can
choose a basis where the $F$ matrices are all trivial and the R-matrices given
by [67]

$\displaystyle
R_{e,e}^{1}=R_{m,m}^{1}=R_{\psi,m}^{e}=R_{e,m}^{\psi}=R_{e,\psi}^{m}=1~{},$
(5.37) $\displaystyle
R_{\psi,\psi}^{1}=R_{m,e}^{\psi}=R_{m,\psi}^{e}=R_{\psi,e}^{m}=-1~{}.$ (5.38)

For the Toric code, (5.23) simplifies to

$\frac{\sigma(g,h)}{\sigma(h,g)}=\frac{1}{R(h,p)R(p,h)R(h,g)}~{},$ (5.39)

and we get

$\frac{\sigma(e,\psi)}{\sigma(\psi,e)}=-1~{},~{}\frac{\sigma(e,m)}{\sigma(m,e)}=-1~{},~{}\frac{\sigma(m,\psi)}{\sigma(\psi,m)}=-1~{}.$
(5.40)

Clearly, we have to choose the non-trivial discrete torsion in
$H^{2}(\mathds{Z}_{2}\times\mathds{Z}_{2},U(1))$.

Finally, let us consider the symmetry

$S\cdot 1=1+m~{},~{}S\cdot e=1+m~{},~{}S\cdot m=S\cdot\psi=0~{}.$ (5.41)

Using (5.18) we find that the lines to be condensed are

$1~{},\ e~{},\ m~{},\ \psi~{}.$ (5.42)

Using

$\frac{\sigma(g,h)}{\sigma(h,g)}=\frac{1}{R(h,p)R(p,h)R(h,g)}~{},$ (5.43)

we get

$\frac{\sigma(g,h)}{\sigma(h,g)}=1~{}\forall g,h\in\\{1,e,m,\psi\\}~{}.$
(5.44)

Therefore, this symmetry is obtained from higher-gauging the
$\mathds{Z}_{2}\times\mathds{Z}_{2}$ 1-form symmetry without discrete torsion.

The set of condensable lines found from symmetry actions in this section
agrees with the discussion in [12, Section 6.3] up to a relabelling of the
line operators, $e\leftrightarrow m$.

####  5.1.4 Example 2: Double semion Model $\cong$ twisted $\mathbb{Z}_{2}$
discrete gauge theory

Consider the double semion model (also known as the twisted $\mathbb{Z}_{2}$
discrete gauge theory) with line operators $1$, $s$, $\bar{s}$, and $d$ and
topological spins

$\theta_{1}=1~{},~{}\theta_{s}=i~{},~{}\theta_{\bar{s}}=-i~{},~{}\theta_{d}=1~{}.$
(5.45)

To determine the symmetries, let us consider the action of the symmetry on the
trivial line operator. As in the previous example, $S\cdot 1$ cannot be a sum
of more than two outcomes, since (3.25) implies

$Nn=4~{}.$ (5.46)

We again know that $n\geq 2$. Therefore, $N=1,2$. $N=1$. In this case, the
trivial line operator is fixed by the symmetry action. Therefore, the symmetry
is invertible. The Double semion model does not have any non-trivial unitary
invertible symmetries. $N=2$. In this case we have

$S\cdot 1=1+d~{},~{}S\cdot d=1+d~{},~{}S\cdot s=S\cdot\bar{s}=0~{}.$ (5.47)

This is the universal $S_{B}$ symmetry, for $B=\\{1,d\\}$, described in
(3.26). From (5.18), $B$ is precisely the 1-form symmetry group that needs to
be higher-gauged to get the surface, $S_{B}$.

####  5.1.5 Example 3: 3-Fermion Model $\cong$ Spin$(8)_{1}$ Chern-Simons
Theory

As a final example, consider the 3-Fermion model with lines operators $1$,
$\psi_{1}$, $\psi_{2}$, $\psi_{3}$ and topological twists

$\theta_{1}=1~{},~{}\theta_{\psi_{1}}=\theta_{\psi_{2}}=\theta_{\psi_{3}}=-1~{}.$
(5.48)

All line operators in this TQFT are higher-gaugeable. A priori one might
imagine that there are non-invertible symmetries. However, from theorem 1, we
know that this model cannot have any non-invertible symmetries since it lacks
a non-anomalous 1-form symmetry. Therefore, the 3-fermion models only has
invertible symmetries. Any permutation of the three fermionic lines is a
symmetry of this theory. Therefore, the invertible symmetry group is $S_{3}$.

Consider the order-two symmetry which acts as

$S\cdot\psi_{1}=\psi_{2}~{},~{}S\cdot\psi_{2}=\psi_{1}~{},$ (5.49)

and leaves the other lines invariant. Using (5.18) we find that the surface
operator which implements this symmetry is obtained from higher-gauging a
$\mathds{Z}_{2}$ 1-form symmetry generated by the line operator $\psi_{3}$. In
general, condensing the line operator $\psi_{i}$ gives the order two symmetry
which fixes $\psi_{i}$ and permutes the other two fermionic lines. Finally,
the surfaces implementing the two order three symmetries are obtained from
higher-gauging the full $\mathds{Z}_{2}\times\mathds{Z}_{2}$ 1-form symmetry,
with and without discrete torsion.

###  5.2 Non-Abelian TQFTs

In this section, we introduce basic aspects of higher gauging in the context
of non-Abelian TQFTs. Our goal is to give a partial converse of theorem 3 and,
in so doing, find a non-Abelian generalization of theorem 1.

To that end, recall from the discussion in section 4 that a necessary
condition to have a non-invertible symmetry is that the TQFT in question,
$\mathcal{T}$, must have condensible bosons living in commutative Frobenius
algebras $\mathcal{A}_{L}$ or $\mathcal{A}_{R}$ [53]. These algebras consist
of bosons satisfying $n_{S\cdot a}^{1}\neq 0$ and $n_{S\cdot 1}^{a}\neq 0$
respectively.

Since the bosons are condensible, we can gauge them in a three-dimensional
slab of spacetime. Then, as shown in Fig. 18, we can compress this slab to
form a non-Abelian generalization of the universal surfaces described in
(3.26). All that remains to show is that such surfaces are always non-
invertible.

$S_{B}{}$$\xrightarrow{\text{Shrink}}$$B$ Fig. 18: We can gauge the
commutative algebra object, $B$, in a 2+1d slab and shrink the slab to get the
surface operator, $S_{B}$.

More explicitly, suppose a TQFT contains a condensable algebra, $B$, and that
we compress a slab of $B$ condensate to form a surface operator, $S_{B}$. Let
$\mathcal{C}/B$ be the MTC obtained from condensing $B$. Under the
condensation, a line operator, $a\in\mathcal{C}$, becomes a superposition of
line operators in $\mathcal{C}/B$, and we have [63]

$a\to\sum_{t\in\mathcal{C}/B}n_{a}^{t}~{}t~{},$ (5.50)

where $n_{a}^{t}$ are non-negative integers. Using this condensation, we can
define $S_{B}$, whose action on the line operators in $\mathcal{C}$ is given
by

$n_{S_{B}a}^{b}=\sum_{i\in\mathcal{C}/B}n_{a}^{i}n_{b}^{i}~{}.$ (5.51)

Note that, as promised, this is a generalization of the universal surface
operator, $S_{B}$, defined in (3.26) for abelian TQFTs.

Let us show that the action of the surface $S_{B}$ satisfies

$S\cdot 1=\sum_{m\in B}m~{},$ (5.52)

where $1$ is the trivial line operator. To understand this claim, note that
$S_{B}$ is made up of a network of the algebra object, $B$. A line operator,
$b$, is in the output of the $S_{B}$-action on a line $a$ if $a$, $B$, and $b$
form a junction as in Fig. 19.

$S_{B}{}$$a$$b$$B$$\alpha$ Fig. 19: Action of $S_{B}$ on a line.

However, not all junctions, $\alpha\in$Hom$(B\times a,b)$, are consistent with
changing the triangulation of $B$ lines on the surface. Indeed, only those
junctions that satisfy the condition in Fig. 20 are allowed [53].252525We
thank I. Runkel for a discussion of this point.

$a$$b$$B$$b$$B$$a$$=$$\alpha$$\alpha$ Fig. 20: A consistency condition on
$\alpha\in\text{Hom}(B\times a,b)$ from a change of triangulation.

Consider the case when $a$ is the trivial line operator. Then, the L.H.S. of
the equality in Fig. 20 can be simplified as in Fig. 21.

$1$$b$$B$$=$$\alpha$$b$$B$$1$$\alpha$$=$$b$$B$$1$$\alpha$ Fig. 21:
Consistency of the $S_{B}$ action on the trivial line. In the first equality,
we have used the fact that $B$ is a commutative algebra. Therefore, $B$ lines
can be braided freely. In the second equality, we use the fact that bubbles in
a network of $B$ lines can be removed, since $B$ is a Frobenius algebra.

Therefore, if $B$ is a commutative Frobenius algebra, any morphism,
$\alpha\in\text{Hom}(B\times a,b)$, satisfies the constraint in Fig. 20. As a
result, any element $m\in B$ is a consistent outcome for the action of $S_{B}$
on the trivial line operator. Hence, for any condensable algebra, $B$, the
surface, $S_{B}$, obtained from higher-gauging $B$ implements a non-invertible
symmetry. In summary, we have the following result:

Theorem 4: A $2+1$d TQFT has a non-invertible symmetry if and only if it
contains condensable bosonic line operators.

This is our promised generalization of the Abelian theorem 1 to general (non-
spin) non-Abelian TQFTs. In particular, we have learned that:

All non-invertible symmetries in non-spin TQFTs are emergent: they come from
sequentially gauging symmetries starting from a theory with no condensable
bosons. In modern language, non-invertible symmetries in these theories are
non-intrinsic.

As in the Abelian case, non-Abelian TQFTs typically have a plethora of non-
universal surface operators. Indeed, given a modular tensor category,
$\mathcal{C}$, let $\mathcal{A}$ be a special Frobenius algebra in
$\mathcal{C}$. The lines forming $\mathcal{A}$ can be higher-gauged on a
surface to get an operator, $S(\mathcal{A})$ [12]. The action of this surface
operator can then be deduced from the partition function of the boundary RCFT
defined by $\mathcal{A}$. Note that, unlike the universal case described
above, unless $\mathcal{A}$ is commutative, the surface cannot be “expanded”
into a slab.262626In general, we can imagine trying to extend the surface into
a slab by summing over the condensable algebras $\mathcal{A}_{L}$ and
$\mathcal{A}_{R}$ in a three-dimensional slab ending on $S$. Consider
performing this operation with $\mathcal{A}_{L}$ “to the right of $S$” (where
“right” can be defined from the configuration in Fig. 17). This extension
involves summing over insertions of the identity line ending on the defect and
producing a three-dimensional network of the object $\mathcal{A}_{L}$ upon the
action of $S$. When $\mathcal{A}_{L}=\mathcal{A}$, this maneuver amounts to
expanding a universal surface into a slab (i.e., the inverse procedure of Fig.
18). On the other hand, when $\mathcal{A}\neq\mathcal{A}_{L}$, this process
instead amounts to defining a transition of the non-invertible surface to a
new non-trivial surface.

Assuming that the fusion rules of $\mathcal{C}$ are multiplicity-free, and
that the algebra object $\mathcal{A}$ contains at most one copy of any given
simple line operator, the action of the surface operator is given by [53]

$S(\mathcal{A})\cdot a=\sum_{b\in\mathcal{C}}n_{S\cdot a}^{b}b~{},$ (5.53)

where

$n_{S(\mathcal{A})\cdot
a}^{b}=\sum_{i,j,k\in\mathcal{A}}m_{jk}^{i}\Delta_{i}^{kj}\sum_{c\in\mathcal{C}}(F_{kjb}^{\bar{a}})^{-1}_{ic}R_{kj}^{i}\frac{\theta_{c}}{\theta_{b}}(F_{kjb}^{\bar{a}})_{ci}~{}.$
(5.54)

In this latter expression, $m_{jk}^{i}$ and $\Delta_{i}^{kj}$ specify the
multiplication and co-multiplication of $\mathcal{A}$ respectively. For
example, all special Frobenius algebras in an MTC describing a discrete gauge
theory have been classified in [68]. Plugging in the data of these algebras
into the formula above (or its generalization for the case of theories with
multiplicity in the fusion rule), we can compute the action of all surface
operators on the lines.

Our definition of non-invertible symmetries was directly in terms of the
consistent ways in which a surface operator can act on the line operators and
fusion spaces. Given a set of integers, $n_{Sa}^{b}$, satisfying this
definition and an action of $S$ on the fusion spaces, it would be interesting
to derive the full data of the algebra $\mathcal{A}$ (the mathematical
machinery in [69, 38] will likely be of use). We leave this generalization of
section 5.1.2 for future work.

####  5.2.1 Faithfullness of condensation defects

Condensation defects obtained from higher-gauging a non-Morita trivial algebra
object act faithfully on the untwisted sector of the TQFT. To understand this
statement, note that condensation defects of a TQFT with MTC, $\mathcal{C}$,
are in one-to-one correspondence with Lagrangian algebras of
$\mathcal{C}\boxtimes\bar{\mathcal{C}}$ [22, 12]. The condensation defects
which act trivially on all line operators correspond to Lagrangian algebras
obtained from soft braided tensor auto-equivalences of $\mathcal{C}$ [57]. In
particular, the only soft braided tensor auto-equivalence which acts trivially
on the fusion spaces is the trivial one. This symmetry corresponds to the
canonical Lagrangian algebra in $\mathcal{C}\boxtimes\bar{\mathcal{C}}$.
Therefore, the only condensation defect that acts trivially on both the line
operators and fusion spaces is the trivial surface operator.

####  5.2.2 Non-invertibility from $\mathcal{A}$

From lemma 2 in section 4, we know that the invertibility of a surface
operator can be determined by studying its action on line operators. Let
$\mathcal{A}$ be a special Frobenius algebra. When is the surface operator $S$
obtained from higher-gauging $\mathcal{A}$ on a surface non-invertible?

To answer this question, let us review the definition of left and right center
of the algebra $\mathcal{A}$ [53]. The left center, $C_{L}(\mathcal{A})$ of
$A$, is the maximal sub-object of $\mathcal{A}$ such that

$m_{a,i}^{b}~{}R_{i,a}^{b}=m_{i,a}^{b}~{},$ (5.55)

where $m$ is the multiplication in $\mathcal{A}$, and $R_{i,a}^{b}$ is the
braiding of the lines $a\in\mathcal{A}$ and $i\in C_{L}(\mathcal{A})$.
Graphically, we have Fig. 22.

$\mathcal{A}$$C_{L}(\mathcal{A})$$\mathcal{A}$$C_{L}(\mathcal{A})$$\mathcal{A}$$\mathcal{A}$$=$
Fig. 22: Left center of $\mathcal{A}$.

Similarly, the right center, $C_{R}(\mathcal{A})$, is defined as the maximal
sub-object of $\mathcal{A}$ such that

$m_{j,a}^{b}~{}R_{a,j}^{b}=m_{a,j}^{b}~{},$ (5.56)

where $a\in\mathcal{A}$ and $j\in C_{R}(\mathcal{A})$. Graphically, we have
Fig. 23.

$\mathcal{A}$$C_{R}(\mathcal{A})$$\mathcal{A}$$C_{R}(\mathcal{A})$$\mathcal{A}$$\mathcal{A}$$=$
Fig. 23: Right center of $\mathcal{A}$.

Now, we will show that if $\mathcal{A}$ has a non-trivial left center, then

$n_{S\cdot i}^{1}=1~{},\ \forall i\in C_{L}(\mathcal{A})~{}.$ (5.57)

In other words, $C_{L}(\mathcal{A})$ is precisely the $\mathcal{A}_{L}$
algebra we discussed in the previous part of section 5.2. To see this, note
that $n_{S\cdot a}^{1}\neq 0$, if and only if the line $a$ can end on the
surface operator $S$. For the end-point to be consistent with a change of
triangulation of the network of $\mathcal{A}$ defects on the surface, it
should satisfy Fig. 24. This figure does indeed hold, since $i$ belongs to the
left center of $\mathcal{A}$, and therefore the $i$ and $\mathcal{A}$ lines
can be unbraided freely.

$i$$\mathcal{A}$$\mathcal{A}$$i$$=$$1$$1$ Fig. 24: Condition for
$n_{Si}^{1}\neq 0$.

Similarly, we can show that if the algebra $\mathcal{A}$ has a non-trivial
right center, then

$n_{S\cdot 1}^{a}=1~{},\ \forall a\in C_{R}(\mathcal{A})~{}.$ (5.58)

Clearly, $C_{R}(\mathcal{A})$ is just the $\mathcal{A}_{R}$ algebra we
discussed before.

In conclusion, if the algebra $\mathcal{A}$ has a non-trivial left or right
center, then the corresponding surface operator acts non-invertibly on the
line operators. See [53, Section 5.5] for a detailed proof of this in the
context of extending chiral algebras of the boundary CFT.

In the following section, we will derive an explicit necessary and sufficient
criterion to determine invertibility of surface operators obtained from
higher-gauging an invertible 1-form symmetry in both Abelian and non-Abelian
TQFTs.

####  5.2.3 A precise condition for invertibility of condensation defects from
Abelian lines

Property | Determined by
---|---
Condensability |  $F$ matrices of $\mathcal{A}$.
Fusion Rules |  $R$ matrices of $\mathcal{C}$ and $\sigma$
Action on lines |  $\mathcal{A}\hookrightarrow\mathcal{C}$ and $\sigma$
Table 2: This table summarizes the specific data in $\mathcal{C}$ that
determines various properties of condensation defects. Here, $\sigma\in
H^{2}(\mathcal{A},U(1))$ is the discrete torsion.

Recall the notation for a condensation defect, $S(G,\sigma)$, obtained from
higher-gauging a 1-form symmetry, $G$, with discrete torsion, $\sigma$
introduced in Sec. 5.1. The important data is summarized in table 2. From Sec.
2, we know that $S(G,\sigma)$ implements an invertible symmetry if and only if
its action on the lines is a permutation. Therefore, the action of the surface
operator on a line, $p$, should result in a unique outcome. In the boundary
CFT, this means that the CFT partition function should be a permutation
modular invariant.

For an invertible action, using (5.9) and (5.12), for a given $p$ the
condition

$\frac{\theta_{h\cdot p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)=1~{},\
\forall~{}h\in G~{},$ (5.59)

should not be satisfied for all but one value of $g\in G$. Moreover, we should
not have

$S(G,\sigma)\cdot p=S(G,\sigma)\cdot q~{},$ (5.60)

for $p\neq q$.

Note that the last two equations equations along with (5.9) and the results in
[65] upon which this discussion is based apply to any non-spin TQFT (including
non-abelian ones) with a $G$ invertible 1-form symmetry subgroup. We are
therefore ready to prove the following lemma:

Lemma 5: In a $2+1$d TQFT with invertible 1-form symmetry subgroup $G$, the
condensation defect, $S(G,\sigma)$, is invertible if and only if
$\Xi_{[G,\sigma]}(g,h)$ is a non-degenerate bicharacter.

Proof: If: Assume that the condensation defect, $S(G,\sigma)$, is invertible,
and suppose the bicharacter, $\Xi_{(G,\sigma)}(g,h)$, is degenerate. That is,
there exists $g$ and $\ell$ such that272727Since $\Xi_{[G,\sigma]}(g,h)$ is a
bicharacter, for a fixed $g$ (or fixed $h$) it must be a linear character of
the group $G$. As a matrix, a non-degenerate $\Xi_{[G,\sigma]}$ has distinct
characters of the group $G$ as its rows and columns. A degenerate
$\Xi_{[G,\sigma]}$ has some repeated rows or columns. Since row rank and
column rank agree, we will only consider the case of repeated columns.

$\Xi_{(G,\sigma)}(g,k)=\Xi_{(G,\sigma)}(\ell,k)~{},\ \forall~{}k\in G~{}.$
(5.61)

This expression can be rearranged to get

$\displaystyle\Xi_{(G,\sigma)}(g,k)\Xi^{-1}_{(G,\sigma)}(\ell,k)$
$\displaystyle=$ $\displaystyle\Xi_{(G,\sigma)}(g,k)\Xi_{(G,\sigma)}(-\ell,k)$
(5.62) $\displaystyle=$
$\displaystyle\Xi_{(G,\sigma)}(g-\ell,k)=\Xi_{(G,\sigma)}(u,k)=1~{},\
\forall~{}k\in G~{}.$ (5.63)

We have used the bicharacter property of $\Xi_{(G,\sigma)}$ and defined
$u:=g-\ell$ to be some non-identity element of $G$. This argument implies that
the action of $S(G,\sigma)$ on the trivial line, $1$, of the MTC,
$\mathcal{C}$, gives the outcome

$1+u\in S(G,\sigma)\cdot 1~{}.$ (5.64)

This result contradicts the assumption that $S(G,\sigma)$ is invertible.
Hence, for an invertible action, the bicharacter, $\Xi_{(G,\sigma)}(g,h)$, is
non-degenerate and satisfies

$\Xi_{[G,\sigma]}(g,k)=\Xi_{[G,\sigma]}(l,k)~{},\ \forall~{}k\in G\implies
g=l~{}.$ (5.65)

Only if: Assume that $\Xi_{[G,\sigma]}(g,h)$ is non-degenerate. Then, for a
given $p\in\mathcal{C}$, there is always some $g\in G$ which satisfies

$\frac{\theta_{h\cdot p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)=1~{},\
\forall~{}h\in G~{}.$ (5.66)

To see this, let us rewrite this equation in terms of the S-matrix

$\frac{\mathcal{S}_{hp}}{\mathcal{S}_{1p}}=(\Xi_{[G,\sigma]}(g,h))^{-1}~{},\
\forall~{}h\in G~{}.$ (5.67)

Recall that $\lambda_{p}(h):=\frac{\mathcal{S}_{hp}}{\mathcal{S}_{1p}}$ is a
character of the group $G$. By assumption, $\Xi_{[G,\sigma]}(g,h)$ is non-
degenerate which implies that the rows of $\Xi_{[G,\sigma]}$ exhaust all
characters of $G$. Therefore, there exists some $g$ such that the above
equation is satisfied. Moreover, this $g$ must be unique. If this were not
true, then the bicharacter, $\Xi_{[G,\sigma]}(g,k)$, would not be non-
degenerate. Indeed, suppose (5.66) held for distinct $g,\ell\in G$, then

$\Xi_{[G,\sigma]}(g,k)=\Xi_{[G,\sigma]}(\ell,k)~{},\ \forall~{}k\in G~{},$
(5.68)

contradicting non-degeneracy of $\Xi_{[G,\sigma]}$. Moreover, the non-
degeneracy of $\Xi_{[G,\sigma]}(g,h)$ implies that

$S(G,\sigma)\cdot p\neq S(G,\sigma)\cdot q~{},$ (5.69)

for any $p,q\in\mathcal{C}$ with $p\neq q$. To see this statement holds,
suppose we have

$S(G,\sigma)\cdot p=S(G,\sigma)\cdot q~{}.$ (5.70)

Let $S(G,\sigma)\cdot p=g\cdot p$ and $S(G,\sigma)\cdot q=\ell\cdot q$ for
some $g,\ell\in G$. We will show that the non-degeneracy of $\Xi_{[G,\sigma]}$
implies that $p=q$. Note that $g,\ell$ are invertible line operators while
$p,q$ are non-invertible in general. From the assumptions, we have

$g\cdot p=\ell\cdot q\implies p=\bar{g}\cdot\ell\cdot q~{},$ (5.71)

and

$\frac{\theta_{h\cdot p}}{\theta_{h}\theta_{p}}\Xi_{[G,\sigma]}(g,h)=1~{},\
\forall~{}h\in G~{},~{}~{}~{}\frac{\theta_{h\cdot
q}}{\theta_{h}\theta_{q}}\Xi_{[G,\sigma]}(\ell,h)=1~{},\ \forall~{}h\in G~{}.$
(5.72)

Now, let us write the topological twists in terms of $\mathcal{S}$-matrix
elements. We have

$\frac{\theta_{h\cdot p}}{\theta_{h}\theta_{p}}=\frac{D}{d_{h\cdot
p}}\mathcal{S}_{h,p}=\frac{D}{d_{h\cdot\bar{g}\cdot\ell\cdot
q}}\mathcal{S}_{h,\bar{g}\cdot\ell\cdot
q}=\frac{D}{d_{h\cdot\bar{g}\cdot\ell\cdot
q}}\frac{\mathcal{S}_{h,\bar{g}\cdot\ell}\mathcal{S}_{h,q}}{\mathcal{S}_{h,1}}=\frac{1}{d_{\bar{g}\cdot\ell}}\frac{\mathcal{S}_{h,\bar{g}\cdot\ell}}{\mathcal{S}_{h,1}}\frac{D}{d_{h\cdot
q}}\mathcal{S}_{h,q}=\frac{1}{d_{\bar{g}\cdot\ell}}\frac{\mathcal{S}_{h,\bar{g}\cdot\ell}}{\mathcal{S}_{h,1}}\frac{\theta_{h\cdot
q}}{\theta_{h}\theta_{q}}~{},$ (5.73)

where $D$ is the total quantum dimension, $1$ is the trivial line operator,
and, in writing the third equality, we have used

$\frac{\mathcal{S}_{ax}}{\mathcal{S}_{1x}}\frac{\mathcal{S}_{bx}}{\mathcal{S}_{1x}}=\sum_{c}N_{ab}^{c}\frac{\mathcal{S}_{cx}}{\mathcal{S}_{1x}}~{},$
(5.74)

for $a=\bar{g}\cdot\ell$, $b=q$, and $x=h$. We can substitute the expression
obtained above into the first equation in (5.72) to get

$\frac{1}{d_{\bar{g}\cdot\ell}}\frac{S_{h,\bar{g}\cdot\ell}}{\mathcal{S}_{h,1}}\frac{\theta_{h\cdot
q}}{\theta_{h}\theta_{q}}\Xi_{[G,\sigma]}(g,h)=1~{},\ \forall~{}h\in G~{}.$
(5.75)

Using the second equation in (5.72) and rearranging the terms, we get

$\Xi_{[G,\sigma]}(l,h)=\frac{1}{d_{\bar{g}\cdot
l}}\frac{\mathcal{S}_{h,\bar{g}\cdot
l}}{\mathcal{S}_{h,1}}\Xi_{[G,\sigma]}(g,h)~{},\ \forall~{}h\in G~{}.$ (5.76)

Note that invertibility of $\bar{g}\cdot\ell$ implies $d_{\bar{g}\cdot\ell}=1$
and that therefore $\mathcal{S}$ in the expression above can be written as

$\frac{\mathcal{S}_{h,\bar{g}\cdot\ell}}{\mathcal{S}_{h,1}}=\Xi_{[G,\sigma]}(h,\bar{g}\cdot\ell)\Xi_{[G,\sigma]}(\bar{g}\cdot\ell,h)~{}.$
(5.77)

This statement follows from the definition of $\Xi$ given in (5.8). Therefore,
we get

$\Xi_{[G,\sigma]}(\ell,h)=\Xi_{[G,\sigma]}(h,\bar{g}\cdot\ell)\Xi_{[G,\sigma]}(\bar{g}\cdot\ell,h)\Xi_{[G,\sigma]}(g,h)=\Xi_{[G,\sigma]}(h,\bar{g}\cdot\ell)\Xi_{[G,\sigma]}(\ell,h)~{},\
\forall~{}h\in G~{},$ (5.78)

where, in the last equality, we used the bicharacter property of
$\Xi_{[G,\sigma]}$. This argument implies

$\Xi_{[G,\sigma]}(h,\bar{g}\cdot\ell)=1~{},~{}~{}~{}\forall~{}h\in G~{}.$
(5.79)

Since $\Xi_{[G,\sigma]}$ is non-degenerate, we get
$\bar{g}\cdot\ell=1\implies\ell=g$. Then, from $p=\bar{g}\cdot\ell\cdot q$ we
have $p=q$. $\square$

For a non-invertible surface operator, $S(G,\sigma)$, let $U$ be the set of
elements in $G$ such that any $u\in U$ satisfies

$\Xi_{[G,\sigma]}(u,k)=1~{}\forall~{}k\in G~{}.$ (5.80)

From the bicharacter property of $\Xi_{[G,\sigma]}$, $U$ is in fact a subgroup
of $G$. The order of $U$ can be seen as the “amount of non-invertibility” of
the surface operator. In particular, the action of $S(G,\sigma)$ on the
trivial line operator is given by

$S(G,\sigma)\cdot 1=\sum_{u\in U}u~{}.$ (5.81)

Note that, in general, the line operators being condensed cannot all end on
the surface operator. For example, if the condensation of a non-trivial group
$G$ of lines results in an invertible surface operator, then only the trivial
line operator can end on this surface.

For example, let us consider the case of higher-gauging a $\mathds{Z}_{2}$
1-form symmetry generated by a line $a$. The expression (5.8) for $\Xi$
reduces to

$\Xi_{[\mathds{Z}_{2},1]}(g,h)=R(g,h)~{},$ (5.82)

where we have used $\tau_{G}(g,h)=1$ since there is a gauge in which the $F$
matrix is trivial for the $\mathds{Z}_{2}$ to be higher-gaugeable. Moreover,
there is no non-trivial choice of discrete torsion in this case. For the
invertibility of the condensation defect, we want $\Xi_{[\mathds{Z}_{2},1]}$
to be non-degenerate.

$\Xi_{[\mathds{Z}_{2},1]}(a,a)\neq 1\implies R(a,a)\neq 1~{}.$ (5.83)

For an order two line operator $a$, $R(a,a)=\pm 1$. Therefore, for the
invertibility of the condensation defect we require the line operator being
condensed to be fermionic. This agrees with the discussion in [12].

##  6 From TQFTs to more general QFTs

In this section, our goal is to first give a simple pictorial “derivation” of
the result from [18] discussed around (1.4). Our reason for doing this is to
more precisely show how our above results (in particular theorem 4; in the
final part of this section we will also discuss lemma 5) generalize to non-
topological $2+1$-dimensional QFTs.

As discussed in the introduction, it is generally interesting to understand
relations between different QFTs and to encode these relations in domain walls
between the theories being compared. Here we will consider gapped interfaces
between QFTs related by “topological manipulations.” More specifically,
suppose we have a QFT, $\mathcal{Q}$, with a non-invertible symmetry,
$\mathcal{C}_{\mathcal{Q}}$. If we perform a topological manipulation, $\phi$,
on $\mathcal{Q}$ to obtain another QFT, $\mathcal{Q}^{\prime}$, such that
$\phi(\mathcal{C}_{\mathcal{Q}})$ is an invertible symmetry of
$\mathcal{Q}^{\prime}$, then $\mathcal{C}_{\mathcal{Q}}$ is called a “non-
intrinsically non-invertible symmetry” [30, 31]. This statement amounts to
being able to construct gapped domain walls between the two theories hosting
$\mathcal{C}_{\mathcal{Q}}$ and $\mathcal{C}_{\mathcal{Q}^{\prime}}$.

We will look at the structure of non-invertible symmetries in a 2+1D QFT and
show that non-invertible symmetries implemented by surface operators are non-
intrinsic in 2+1D QFTs without transparent fermionic line operators.

Before explaining this statement further, let us briefly recapitulate our
results leading up to theorem 4. First, recall that in 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 described by a strongly anisotropic modular tensor
category

$\mathcal{T}_{1}\xrightarrow{\text{condense
}B_{1}}\mathcal{T}_{2}\xrightarrow{\text{condense
}B_{2}}\mathcal{T}_{3}\cdots\xrightarrow{\text{condense
}B_{n-1}}\mathcal{T}_{n}~{}.$ (6.1)

From our results, we know that $\mathcal{T}_{n}$ only has invertible surface
operators. Therefore, all non-invertible surface operators in
$\mathcal{T}_{1}$ arise from starting from $\mathcal{T}$, and sequentially
gauging symmetries (implemented by surface operators) to move in the reverse
direction in the sequence of TQFTs (6.1). This logic shows that there is a
topological manipulation on the TQFT $\mathcal{T}_{1}$ such that all non-
invertible surface operators become invertible. Therefore, non-invertible
symmetries implemented by surface operaors in a 2+1D TQFT are non-intrinsic.

Now, consider the set of topological surface and line operators in a 2+1D QFT,
$\mathcal{Q}$. In general, these operators may form an infinite family. We
will focus on the subset of topological operators which form a fusion
2-category. The set of topological line operators form a braided fusion
category, $\mathcal{C}_{\mathcal{Q}}$. We will assume that there are no
fermionic line operators which braided trivially with all the other line
operators. Therefore, the set of line operators which braid trivially with all
lines in $\mathcal{C}_{\mathcal{Q}}$ are bosonic. Moreover, the pictorial
equalities in Fig. 25 show that transparent lines are closed under fusion.

$b$$a$$c$$b$$a$$c$$b$$a$$c$$b$$a$$c$$\frac{S_{xc}}{S_{1c}}$$=$$=$$=$$x$$x$
Fig. 25: Suppose the line operator $a$ and $b$ are transparent. Then, $c$ is
also transparent.

As a result, they form a symmetric fusion category, Rep$(K)$, for some finite
group, $K$ [25]. Condensing Rep$(K)$ gives a QFT, $\mathcal{Q}^{\prime}$, with
no transparent line operators. [70, Theorem 2]. In other words,
$\mathcal{C}_{\mathcal{Q}^{\prime}}$ is an MTC. Let us denote the full fusion
2-category of surfaces and lines in $\mathcal{Q}^{\prime}$ as
$\mathcal{C}_{\mathcal{Q}^{\prime}}^{(2)}$.

One can show that it is possible to perform further topological manipulations
on $\mathcal{Q}^{\prime}$ to get a QFT, $\mathcal{Q}^{\prime\prime}$, such
that, as alluded to in the introduction, the fusion 2-category of symmetries
of $\mathcal{Q}^{\prime\prime}$ is of the form [18]

$\text{2Vec}_{G}^{\omega}\boxtimes\text{
Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})~{},$ (6.2)

for some finite group, $G$, and 4-cocycle $\omega\in Z^{4}(G,U(1))$. Here,
$\text{2Vec}_{G}^{\omega}$ denotes a set of invertible surface operators which
form the group $G$ under fusion with anomaly $\omega$. The second factor,
$\text{ Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})$, denotes the line operators
which form the MTC, $\mathcal{C}_{\mathcal{Q}^{\prime}}$, and their
condensation defects.

Mathematically, the above result is the statement that any fusion 2-category
is Morita equivalent to a fusion 2-category of the form (6.2). The argument in
[18] is as follows. We start by recalling the fact that the MTC,
$\mathcal{C}_{\mathcal{Q}^{\prime}}\boxtimes\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}}$,
admits a canonical gapped boundary. Including surface operators, the resulting
structure can be written in terms of fusion 2-categories as the statement that
$\text{Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})\boxtimes\text{Mod}(\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}})$
admits a gapped boundary. In other words, there is a gapped interface between
$\text{Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})\boxtimes\text{Mod}(\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}})$
and 2Vec (see Fig. 26).

2Vec$\text{Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})\boxtimes\text{Mod}(\overline{\mathcal{C}}_{\mathcal{Q}^{\prime}})$
Fig. 26: Gapped interface between 2Vec and
$\text{Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})\boxtimes\text{Mod}(\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}})$.

Stacking the 2-category, $\mathcal{C}_{\mathcal{Q}^{\prime}}^{(2)}$, on both
sides of the interface, we find that there is a gapped interface between
$\mathcal{B}:=\text{Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})\boxtimes\text{Mod}(\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}})\boxtimes\mathcal{C}_{\mathcal{Q}^{\prime}}^{(2)}$
and $\mathcal{C}_{\mathcal{Q}^{\prime}}^{(2)}$ (see Fig. 27).

$\mathcal{C}_{\mathcal{Q}^{\prime}}^{(2)}$$\mathcal{B}$ Fig. 27: Gapped
interface between $\mathcal{C}_{\mathcal{Q}^{\prime}}^{(2)}$ and
$\mathcal{B}:=\text{Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})\boxtimes\text{Mod}(\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}})\boxtimes\mathcal{C}_{\mathcal{Q}^{\prime}}^{(2)}$

The line operators in
$\mathcal{B}:=\text{Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})\boxtimes\text{Mod}(\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}})\boxtimes\mathcal{C}_{\mathcal{Q}^{\prime}}^{(2)}$
form the MTC
$\mathcal{C}_{\mathcal{Q}^{\prime}}\boxtimes\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}}\boxtimes\mathcal{C}_{\mathcal{Q}^{\prime}}$.
Condensing the canonical Lagrangian algebra in
$\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}}\boxtimes\mathcal{C}_{\mathcal{Q}^{\prime}}$
we reduce the fusion 2-category
$\text{Mod}(\bar{\mathcal{C}}_{\mathcal{Q}^{\prime}})\boxtimes\mathcal{C}_{\mathcal{Q}^{\prime}}^{(2)}$
to a fusion 2-category with only surface operators. As explained pictorially
in the introduction, such a fusion 2-category must necessarily have only
invertible surface operators and is of the form 2Vec${}_{G}^{\omega}$ for some
finite group $G$ and 4-cocycle $\omega$ [11]. Therefore, under this
condensation, the fusion 2-category $\mathcal{B}$ becomes
$\text{Mod}(\mathcal{C}_{{\mathcal{Q}^{\prime}}})\boxtimes\text{2Vec}_{G}^{\omega}$
(see Fig. 28).

$\mathcal{\mathcal{C}}_{\mathcal{Q}^{\prime}}^{(2)}$$\mathcal{B}$$\text{2Vec}_{G}^{\alpha}\boxtimes\text{Mod}(\mathcal{\mathcal{C}}_{\mathcal{Q}^{\prime}})$
Fig. 28: A set of line operators in $\mathcal{B}$ can be gauged to get a
gapped interface between it and
$\text{2Vec}_{G}^{\alpha}\boxtimes\text{Mod}(\mathcal{\mathcal{C}}_{\mathcal{Q}^{\prime}})$

This completes the proof in [18].

The only source of non-invertible surface operators in
$\text{2Vec}_{G}^{\alpha}\boxtimes\text{Mod}(\mathcal{\mathcal{C}}_{\mathcal{Q}^{\prime}})$
are the condensation defects denoted by objects in the fusion 2-category
$\text{Mod}(\mathcal{\mathcal{C}}_{\mathcal{Q}^{\prime}})$. Note that
$\text{Mod}(\mathcal{\mathcal{C}}_{\mathcal{Q}^{\prime}})$ describes the
surface and line operators of a 2+1D TQFT. The fusion 2-category is completely
determined by the MTC $\mathcal{C}_{\mathcal{Q}^{\prime}}$. From our theorem
4, we know that there are non-invertible condensation defects if and only if
there are condensable bosonic line operators in
$\mathcal{C}_{\mathcal{Q}^{\prime}}$. Therefore, we know that sequentially
condensing all the condensable line operators in
$\mathcal{C}_{\mathcal{Q}^{\prime}}$ gives a strongly anisotropic MTC, say
$\mathcal{C}$, which does not contain any condensable bosons. Using this
discussion, we can extend Fig. 28 to get Fig. 29.

In summary, we have shown that, given the symmetries of a 2+1D QFT described
by a fusion 2-category, there is a sequence of topological manipulations
leading to a QFT which has only invertible topological surface
operators.282828Note that to invert the sequence of topological manipulations
(gauging line operators) in Fig. 29 we may have to keep track of dual
symmetries implemented by surfaces which act trivially on the topological
lines and their fusion spaces. For example, gauging the boson in the Toric
code results in the trivial TQFT. To invert this gauging procedure we have to
note that the resulting trivial TQFT is in fact a $\mathds{Z}_{2}$-SPT.
Therefore, we have arrived at the following result:

Theorem 6: In a 2+1d QFT without transparent fermionic lines, non-invertible
symmetries implemented by non-invertible surface operators are non-intrinsic.

$\mathcal{C}_{\mathcal{Q}^{\prime}}$$\mathcal{B}$$\text{2Vec}_{G}^{\alpha}\boxtimes\text{Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})$$\text{2Vec}_{G}^{\alpha}\boxtimes\text{Mod}(\mathcal{C})$
Fig. 29: Condensing all condensable line operators in the fusion 2-category
$\text{2Vec}_{G}^{\alpha}\boxtimes\text{Mod}(\mathcal{C}_{\mathcal{Q}^{\prime}})$
leads to a fusion 2-category $\text{2Vec}_{G}^{\alpha}\boxtimes
Mod(\mathcal{C})$ which does not contain any non-invertible surface operators
where $\mathcal{C}$ is some strongly-anisotropic modular tensor category.

###  6.1 Generalizing lemma 5 from TQFT to QFT

In a similar spirit to our generalization of theorem 4 from TQFT to more
general QFTs and theorem 5 in the previous subsection, let us discuss how to
extend lemma 5 in Sec. 5.2.3 to any $2+1$d QFT that has an invertible 1-form
symmetry.

To that end, let $\mathcal{C}$ be the braided fusion category of line
operators in a $2+1$D TQFT. The condensation defects that can be constructed
by higher-gauging these line operators are described by module categories of
$\mathcal{C}$. Consider a set of invertible line operators which form a group,
$A$, under fusion. In this case, $\mathcal{C}\cong\text{Vec}_{A}^{\omega}$ for
some $\omega\in H^{3}(A,U(1))$ (we have suppressed the braiding). A module
category, $\mathcal{M}(G,\sigma)$ over $\mathcal{C}$, is labelled by a
subgroup $G\leq A$ such that $\omega|_{G}$ is trivial in cohomology, and
$\sigma\in H^{2}(G,U(1))$. In the notation used in other sections of this
paper, we have $\mathcal{M}(G,\sigma)=S(G,\sigma)$. In [71], the authors study
fusion rules of module categories of $\text{Vec}_{A}^{\omega}$ and their
invertibility under fusion. Rephrasing Proposition 5.5 in [71] in our
language, we get the following statement noting that this proposition holds
even with a degenerate $\mathcal{S}$ matrix

Theorem 7: In any $2+1$d QFT with abelian 1-form symmetry subgroup, $G$, the
condensation defect $S(G,\sigma)$ is invertible if and only if the phase
$\Xi_{[G,\sigma]}(g,h)$ is non-degenerate.

##  7 Non-invertible time-reversal symmetries of TQFTs

In previous sections, we considered non-invertible generalizations of unitary
symmetries (which we will continue to refer to as “non-invertible
symmetries”). In this section, we consider non-invertible generalizations of
anti-unitary / time-reversal symmetries, which we will colloquially call “non-
invertible time-reversal symmetries.”

As we have seen, unitary symmetries give rise to automorphisms of the fusion
group that preserve the topological spins. On the other hand, anti-unitary /
time-reversal symmetries give rise to automorphisms of the fusion group that
conjugate the topological spins

$\theta_{S_{U}\cdot a}=\theta_{a}~{},\ \ \ \theta_{S_{T}\cdot
a}=\theta^{*}_{a}~{},$ (7.1)

where the “$U$” subscript stands for unitary, and the “$T$” subscript stands
for “time reversal.”

At the level of an Abelian MTC, we require at least one line with complex
topological spin, $\theta_{a}\neq\theta_{a}^{*}$, in order to have an anti-
unitary symmetry (otherwise all the data of the MTC can be gauge-transformed
to be real, and the symmetry is unitary). Toric code / $\mathbb{Z}_{2}$
Dijkgraaf-Witten theory (see section 5.1.3) is an example of a theory with a
real MTC that lacks a time-reversal symmetry, while double semion / twisted
$\mathbb{Z}_{2}$ Dijkgraaf-Witten theory (see section 5.1.4) has an MTC with
complex topological spins and a time-reversal symmetry. In the non-Abelian
case, we can a priori imagine that there are non-trivial time reversal
symmetries even if all spins are real (since the spins do not determine the
MTC).292929On the other hand, up to rank eleven, there are no non-Abelian
unitary MTCs consisting solely of bosons and fermions (i.e., all examples with
real spins up to this rank are Abelian MTCs) [72]. It would be interesting to
understand if this statement is true more generally.

When we have a complex topological spin for some line, $a$, then (7.1) implies
that unitary and time-reversal symmetries have different actions on $a$. In
particular

$\exists\ b\ {\rm s.t.}\ n_{S_{U}\cdot a}^{b}\neq n_{S_{T}\cdot a}^{b}~{}.$
(7.2)

In other words, it is always possible to distinguish a unitary and time-
reversal symmetry of an MTC via symmetry actions on the complex-spin
topological lines. On the other hand, we claim that:

Non-invertible time-reversal symmetries of $2+1$d TQFTs and non-invertible
symmetries (i.e., non-invertible generalizations of unitary symmetries) can
have identical actions on lines, even when the topological spins of those
lines are complex.

One way to understand this claim is to recall from our discussion in section
3.1 that, in an Abelian TQFT, a non-invertible symmetry, $S$, has a non-
trivial space of lines it annihilates (see (3.8)): ${\rm
Null}(S)\neq\emptyset$. Now, let us suppose that $S$ annihilates all lines,
$a_{i}$, with complex spin, $\theta_{a_{i}}\neq\theta_{a_{i}}^{*}$. If the
theory has an invertible time-reversal symmetry, $S_{T}$, then the product

$S\times S_{T}=S_{T^{\prime}}~{},$ (7.3)

is a non-invertible time-reversal symmetry.303030Consider a configuration in
which the $S$ and $S_{T}$ defects are close together but have not fused. This
configuration gives rise to a time-reversal action on the theory and therefore
so to does the fusion product. However, it is clear that ${\rm
Null}(S_{T^{\prime}})={\rm Null}(S)$ and that, unlike the invertible case in
(7.2), here

$n_{S_{T^{\prime}}\cdot a}^{b}=n_{S\cdot a}^{b}~{},\ \forall\
a,b\in\mathcal{C}~{},$ (7.4)

even though $S_{T^{\prime}}$ is a time-reversal symmetry, and $S$ is not.

As we will explain in our example subsection below, the phenomenon discussed
in (7.4) manifests itself in the simplest theory with time reversal: the
double semion. In this theory there is a universal non-invertible surface,
$S_{B}$, playing the role of $S$ and an invertible time reversal symmetry,
$S_{T}$, exchanging the semion and anti-semion. Moreover, $S_{B}$ annihilates
the complex spin lines (i.e., the semion and anti-semion lines).

We can explicitly describe this scenario more generally in Abelian TQFT. The
conditions for the existence of an $S_{B}$ surface and an $S_{T}$ surface are:

  1. 1.

$\theta_{b}=1$ for some non-trivial $b\in\mathcal{C}$.

  2. 2.

$\sum_{a\in\mathcal{C}}\theta_{a}^{n}\in\mathds{R}$ for all positive integer
$n$ [73].

We then have a non-invertible time-reversal symmetry, $S_{T^{\prime}}$, via
(7.3). If all lines, $a$, with $\theta_{a}\neq\theta_{a}^{*}$ braid non-
trivially with some of the bosons, $B$, we are condensing, then we are
precisely in the situation of (7.4), where the non-invertible time reversal
symmetry is indistinguishable from the non-invertible symmetry. On the other
hand, if a line, $a$, with complex spin braids trivially with $B$, then we can
distinguish the non-invertible time reversal symmetry from the non-invertible
symmetry via the action on lines.

Our goal in the next subsection will be to give a generalization of the
definition of (non-invertible) symmetries of 2+1D TQFTs to the time reversal
case. After this general discussion, we give a set of explicit examples. All
the non-invertible time reversal symmetries in these theories are of the form
$S_{T}\times S_{B}$.313131It will be interesting to understand if all non-
invertible time-reversal symmetries are of this form.

###  7.1 A general definition of a (non-invertible) time reversal symmetry

To motivate our general definition of (non-invertible) time-reversal
symmetries, we first return to condition 2 above for the existence of an
invertible time reversal symmetry. This condition implies that

$\frac{\sum_{(a,b)\in\mathcal{C}\times\mathcal{C}}\theta_{(a,b)}^{n}}{|\sum_{(a,b)\in\mathcal{C}\times\mathcal{C}}\theta_{(a,b)}^{n}|}=\frac{\sum_{a,b\in\mathcal{C}}\theta_{a}^{n}\theta_{b}^{n}}{|\sum_{a,b\in\mathcal{C}}\theta_{a}^{n}\theta_{b}^{n}|}=1~{},$
(7.5)

for all $n$, which, in turn, implies that $\mathcal{C}\boxtimes\mathcal{C}$
admits a gapped boundary [74]. Therefore, the TQFT described by the category
$\mathcal{C}\boxtimes\mathcal{C}$ admits a gapped boundary if $\mathcal{C}$
has a time-reversal symmetry.

Generalizing the discussion in the case of Abelian TQFTs, we find that a
general TQFT admits a time-reversal symmetry only if there is a gapped domain
wall between itself and the complex-conjugated TQFT that arises from unfolding
the $\mathcal{C}\boxtimes\mathcal{C}$ theory with gapped boundary. In terms of
MTCs, this is a domain wall between $\mathcal{C}$ and $\bar{\mathcal{C}}$.

We can generalize the definition of non-invertible symmetries to accommodate
time-reversal as follows. We have a surface operator implementing the time-
reversal symmetry with $\tilde{F}$ as defined in Fig. 9. However, now the
$\tilde{F}$ are anti-linear maps. The constraints on $\tilde{F}$ from
associativity of fusion is as in Fig. 30.

$S$$S$$S$$S$$S$$\tilde{F}$$\tilde{F}$$\tilde{F}$$F^{*},F$ $\tilde{F}$ Fig.
30: The constraints on $\tilde{F}$ matrices from associativity of fusion. On
the right bottom of the diagram, the $F^{*},F$ action indicates the $F$-move
on either side of the surface operator. Note that, unlike in the case of
unitary symmetry and its non-invertible generalization, in this case we use
the complex conjugate $F$-move on the L.H.S. of the surface operator since the
TQFT on the L.H.S. is the time-reversal of the TQFT on the R.H.S. of the
surface.

Moreover, the constraint on $\tilde{F}$ from braiding is

$(R_{bd}^{e})^{*}\tilde{F}_{(c,d),(a,b),(f,e)}(R_{ac}^{f})^{-1}=\tilde{F}_{(a,b),(c,d),(f,e)}~{},$
(7.6)

where $S\cdot a\ni b$ and $S\cdot c\ni d$. With these conditions in place, we
can define a time-reversal symmetry as follows:

Definition: A (non-invertible) time-reversal symmetry of a non-Abelian TQFT,
$\mathcal{T}$, with MTC, $\mathcal{C}$, is a map, $S$, on the ring of line
operators such that

  1. 1.

The $\tilde{F}$ matrices defined in Fig. 9 satisfying the conditions in Fig.
30 and (7.6) exist.

  2. 2.

$n_{Sa}^{b}n_{Sc}^{d}\leq\sum_{e,f}N_{ac}^{f}N_{bd}^{e}n_{Sf}^{e}$ and
$n_{S1}^{1}=1$.

  3. 3.

$\sum_{a,b\in\mathcal{C}}n_{Sa}^{b}d_{a}d_{b}=\text{dim}(\mathcal{C})$.

The action of the non-invertible time-reversal symmetry on the fusion spaces
is given by an anti-linear map $U_{T}(a,b,c,f)_{e,d}$, acting on the fusion
space, $V_{ed}^{f}$. At the level of fusion spaces, this action can be written
as

$U_{T}(a,b,c):\bigoplus_{d,e,f}N_{de}^{f}~{}n_{Sa}^{d}~{}n_{Sb}^{e}~{}n_{Sc}^{f}~{}V_{de}^{f}\rightarrow
V_{ab}^{c}~{},$ (7.7)

where $U_{T}(a,b,c)$ is a block-diagonal matrix, with the block
$U_{T}(a,b,c,f)_{d,e}$ acting on $V_{de}^{f}$. Note that in general $U_{T}$ is
a rectangular matrix, and is therefore non-invertible.

When the time-reversal symmetry is invertible, then the constraints on
$\tilde{F}$ discussed above imply that $F_{abc}^{d}$ and $R_{ab}^{c}$ are
related by symmetry action on the fusion spaces to $(F_{S\cdot a,S\cdot
b,S\cdot c}^{S\cdot d})^{*}$ and $(R_{S\cdot a,S\cdot b}^{S\cdot c})^{*}$
respectively. This statement agrees with the discussion of invertible time-
reversal symmetries in [59].323232Non-invertible time-reversal symmetries in
QFTs including Maxwell theory and massive QCD were first discussed in [32].

Many results that we proved for non-invertible generalizations of unitary
symmetries also hold for non-invertible time-reversal symemtries. For example,
if a TQFT admits a non-invertible time-reversal symmetry, then it necessarily
contains bosonic line operators. For abelian TQFTs, this can be seen by noting
that the argument we used to prove the corresponding statement for abelian
TQFTs in section 3 (specifically around (3.18)) also applies to non-invertible
time-reversal symmetries.

Consider a (non-abelian) TQFT without bosonic line operators. Let $S_{T}$ be a
non-invertible time-reversal symmetry with the action on the line operators
specified by the non-negative integers $n_{S_{T}\cdot a}^{b}$. Since there are
no bosonic line operators, we have

$n_{S_{T}\cdot 1}^{b}=n_{S_{T}\cdot b}^{1}=\delta_{b,1}~{}.$ (7.8)

Also, note that the matrix of integers $n_{S_{T}\cdot a}^{b}$ satisfies
[75][55, Theorem 3.3]

$\sum_{c}\mathcal{S}_{ac}n_{S_{T}\cdot c}^{b}=\sum_{c}n_{S_{T}\cdot
a}^{c}\bar{\mathcal{S}}_{cb}~{},$ (7.9)

where $\bar{\mathcal{S}}_{cb}$ is the complex-conjugate of the S-matrix.
Choosing $b=1$, using (7.8) and noting that quantum dimensions of line
operators are real numbers, we get

$d_{a}=\sum_{c}n_{S\cdot a}^{c}d_{c}~{},$ (7.10)

where $d_{a}$ are the quantum dimensions of the line operators. On the other
hand, choosing $a=1$ and using (7.8), we get

$\sum_{c}n_{S\cdot c}^{b}d_{c}=d_{b}~{}.$ (7.11)

In a unitary TQFT, the quantum dimensions satisfy $d_{a}\geq 1$ for all line
operators $a$. Therefore, the constraints (7.10) and (7.11) imply that

$n_{S\cdot a}^{b}=\delta_{b,\sigma(a)}~{},$ (7.12)

for some permutation, $\sigma$, of the lines. That is, in a unitary TQFT
without bosons, (non-invertible) time-reversal symmetries act as permutations
on the line operators. In fact, if a time-reversal symmetry acts as
permutations on the line operators, then its action on fusion spaces must be
invertible. This statement follows from noting that the corresponding argument
for non-invertible symmetries around (4.5) can be readily generalized for non-
invertile time-reversal symmetries using the anti-linear maps $\tilde{F}$ and
$U_{T}$.

In summary, if the TQFT does not contain any bosonic line operators, it cannot
have any non-invertible time-reversal symmetries. In the Abelian case, the
existence of bosonic line operators is enough to guarantee the existence of a
non-invertible generalization of a unitary symmetry and so, if an Abelian TQFT
has a non-invertible time-reversal symmetry, then that TQFT also has a non-
invertible symmetry.333333This is a necessary but not sufficient condition to
prove that all time reversal symmetries in Abelian TQFTs are of the form
$S_{B}\times S_{T}$.

Non-invertible symmetries have a description in terms of Lagrangian algebras
in the $\mathcal{C}\boxtimes\bar{\mathcal{C}}$ MTC (see Appendix Appendix A.
Relating our definitions to Lagrangian subalgebras in folded theories). For
time-reversal symmetries, the analogous structure is a Lagrangian subalgebra
in $\mathcal{C}\boxtimes\mathcal{C}$. In other words, if $\mathcal{C}$ has a
time-reversal symmetry, then $\mathcal{C}\boxtimes\mathcal{C}$ has a non-
anomalous symmetry implemented by topological line operators, which when
gauged gives the trivial TQFT. That is, $\mathcal{C}\boxtimes\mathcal{C}$ must
admit a Lagrangian algebra. Note that while
$\mathcal{C}\boxtimes\bar{\mathcal{C}}$ always has at least one Lagrangian
algebra, $\mathcal{C}\boxtimes\mathcal{C}$ may not have any. This statement
corresponds to the fact that the any TQFT has the trivial (unitary) symmetry
while it may not have any time-reversal symmetries.

Note that even if $\mathcal{C}\boxtimes\mathcal{C}$ has a Lagrangian algebra
object, it does not imply that the $\mathcal{C}$ is time-reversal invariant.
For example, $SU(9)_{1}$ MTC does not have a non-trivial time-reversal
symmetry even though $SU(9)_{1}\boxtimes SU(9)_{1}$ has a Lagrangian subgroup.

###  7.2 Example 1: Double semion model $\cong$ twisted $\mathbb{Z}_{2}$
discrete gauge theory

Consider the double semion model with line operators $1$, $s$, $\bar{s}$, $d$
and topological spins

$\theta_{1}=1,~{}\theta_{s}=i,~{}\theta_{\bar{s}}=-i,~{}\theta_{d}=1~{}.$
(7.13)

This theory has an invertible time-reversal symmetry implemented by a surface,
$S_{T}$, satisfying

$S_{T}\cdot 1=1~{},\ S_{T}\cdot d=d,~{}S_{T}\cdot s=\bar{s}~{}.$ (7.14)

$S_{T}$ corresponds to the Lagrangian subgroup

$\left\\{(1,1)~{},\ (d,d)~{},\ (s,\bar{s})~{},\
(\bar{s},s)\right\\}\in\mathcal{C}\boxtimes\mathcal{C}~{},$ (7.15)

where $\mathcal{C}$ is the MTC of the double semion model.

Next, recall that, in section 5.1.4, we found a non-invertible symmetry in
this TQFT implemented by the surface operator, $S_{B}$. We showed that that
this surface operator acts as

$S_{B}\cdot 1=S_{B}\cdot d=1+d~{},~{}S_{B}\cdot s=S_{B}\cdot\bar{s}=0~{}.$
(7.16)

From this action, and the action of $S_{T}$, we can construct the symmetry

$S_{T^{\prime}}:=S_{B}\times S_{T}~{}.$ (7.17)

Note that $S_{T^{\prime}}$ acts on lines in the same way as $S_{B}$

$S_{T^{\prime}}\cdot 1=S_{T^{\prime}}\cdot d=1+d~{},\ S_{T^{\prime}}\cdot
s=S_{T^{\prime}}\cdot\bar{s}=0~{}.$ (7.18)

This is the simplest example of the general phenomenon discussed in the
introduction to this section: at the level of the action on line operators,
non-invertible symmetries and their time-reversal counterparts need not be
distinguishable.

Let us look at the action of $S_{T^{\prime}}$ on the fusion spaces. This is
determined by the action of $S_{T}$ and $S_{B}$ on the fusion spaces. $S_{T}$
acts on the fusion spaces as

$U_{T}(a,b):V_{ab}^{c}\to V_{S_{T}(a),S_{T}(b)}^{S_{T}(c)}~{}.$ (7.19)

We can choose a gauge in which $F(a,b,c)=1$ $,~{}a,b,c\in\mathcal{C}$ [67].
Also, $1,d$ are bosons. Therefore, it is consistent to choose

$U_{T}(a,b,c)=1~{}\forall~{}a,b,c~{}\in\\{1,d\\}~{}.$ (7.20)

Now, let us consider the $S_{B}$ symmetry. Since both $F$ and $R$ matrices of
the line operators $1,d$ are trivial, we can choose

$\tilde{F}^{S_{B}}_{a,b,c}=1~{},$ (7.21)

where $\tilde{F}^{S_{B}}$ is only defined when $a,b,c\in\\{1,d\\}$. Since the
action of $S_{B}$ on $s$ and $\bar{s}$ is zero, the action of $S_{B}$ on
fusion spaces involving these line operators is ill-defined. Using (4.5) and
the choice of $\tilde{F}^{S_{B}}$ above, the action of $S_{B}$ on fusion
spaces $V_{ab}^{c},~{}a,b,c\in\\{1.d\\}$ is given by

$U_{S_{B}}(a,b,c)=1~{}.$ (7.22)

The action of the symmetry $S_{T^{\prime}}$ on the fusion spaces is given by

$U_{T^{\prime}}(a,b,c)=U_{T}(a,b,c)U_{S_{B}}(a,b,c)=1~{}\forall~{}a,b,c\in\\{1,d\\}~{},$
(7.23)

and the action on other fusion spaces is ill-defined. Therefore, we conclude
that the data $\\{n_{S_{B}\cdot a}^{b},U_{S_{B}}(a,b,c)\\}$ and
$\\{n_{S_{T^{\prime}}\cdot a}^{b},U_{T^{\prime}}(a,b,c)\\}$ is identical.
Therefore, this data cannot be used to distinguish between the symmetries
$S_{T^{\prime}}$ and $S_{B}$.

We expect that the above ambiguity can be resolved by studying the action of
$S_{T^{\prime}}$ on non-genuine line operators. In particular, we expect that
once we include twisted sectors, the action on $S_{T^{\prime}}$ on the lines
$s,\bar{s}$ will map them to non-genuine lines. Therefore, in this description
the action of $S_{T^{\prime}}$ on, say $V_{ss}^{1}$, will map it to a fusion
space for non-genuine lines. This extra data could then be used to distinguish
between $S_{T^{\prime}}$ and $S_{B}$ symmetries. We leave this study for
future work.

###  7.3 Example 2: $B_{5}$ TQFT (Galois conjugate of $SU(5)_{1}$ Chern-Simons
Theory)

$U(1)_{n}$ Chern-Simons theories for even $n$ do not have any invertible time-
reversal symmetries [36]. On the other hand, abelian TQFTs with an odd number
of line operators often have time-reversal symmtries. The simplest example is
the $B_{5}$ prime abelian TQFT (which is a Galois conjugate of $SU(5)_{1}$
Chern-Simons theory). It has anyons that we label as $\ell_{i}$ for
$i=0,\cdots,4$, with topological spins

$\theta_{\ell_{0}}=1~{},~{}\theta_{\ell_{1}}=\theta_{\ell_{4}}=e^{\frac{2\pi
i}{5}}~{},~{}\theta_{\ell_{2}}=\theta_{\ell_{3}}=e^{-\frac{2\pi i}{5}}~{}.$
(7.24)

Since there are no non-trivial bosons, we know from our general discussion
that this theory only has invertible time reversal symmetries.

We clearly have the invertible time-reversal symmetry that acts as

$S_{T}\cdot\ell_{0}=\ell_{0}~{},~{}S_{T}\cdot\ell_{1}=\ell_{2}~{},~{}S_{T}\cdot\ell_{2}=\ell_{4}~{},~{}S_{T}\cdot\ell_{4}=\ell_{3}~{},~{}S_{T}\cdot\ell_{3}=\ell_{1}~{},$
(7.25)

but there are other time reversal symmetries as well since we have the fusion

$S_{T}\times S_{T}=S_{C}~{},$ (7.26)

where $S_{C}$ is the surface implementing the charge conjugation symmetry.
Therefore, we also have

$S_{T}^{3}\neq\mathds{1}~{}.$ (7.27)

The non-anomalous 1-form symmetry in $B_{5}\boxtimes B_{5}$ corresponding to
$S_{T}$ is a non-anomalous $\mathbb{Z}_{5}$ 1-form symmetry

$\left\\{(\ell_{0},\ell_{0})~{},(\ell_{1},\ell_{2})~{},(\ell_{2},\ell_{4})~{},(\ell_{4},\ell_{3})~{},(\ell_{3},\ell_{1})\right\\}\in