Symmetry TFTs for Non-invertible Defects

Abstract

Given any symmetry acting on a d-dimensional quantum field theory, there is an associated $(d+1)$-dimensional topological field theory known as the Symmetry TFT (SymTFT). The SymTFT is useful for decoupling the universal quantities of quantum field theories, such as their generalized global symmetries and ’t Hooft anomalies, from their dynamics. In this work, we explore the SymTFT for theories with Kramers-Wannier-like duality symmetry in both $(1+1)$d and $(3+1)$d quantum field theories. After constructing the SymTFT, we use it to reproduce the non-invertible fusion rules of duality defects, and along the way we generalize the concept of duality defects to higher duality defects. We also apply the SymTFT to the problem of distinguishing intrinsically versus non-intrinsically non-invertible duality defects in $(1+1)$d.

Appendices

Correlation Functions of k-dimensional Operators in $(2k+1)$d

In this appendix, we provide derivations for Eqs. (3.9), (3.10), and (6.8); namely we derive the linking of k-dimensional operators in $(2k+1)$-dimensional ${\mathbb {Z}}_N$ gauge theory, as well as their commutation relations in 2k-dimensions. We begin with the action for ${\mathbb {Z}}_N$ gauge theory in $(2k+1)$ dimensions,

$$\begin{aligned} S= \frac{2\pi }{N}\int _{X_{2k+1}} b^{(k)} \cup \delta a^{(k)}~. \end{aligned}$$

(A.1)

This theory has $N^2$ k-dimensional operators given by

$$\begin{aligned} L_{(e,m)}(M_k) = \exp \left( \frac{2\pi i}{N} \oint _{M_k} e a^{(k)} \right) \exp \left( \frac{2\pi i}{N} \oint _{M_k} m b^{(k)}\right) , \quad (e,m)\in {\mathbb {Z}}_N\times {\mathbb {Z}}_N \end{aligned}$$

(A.2)

with $L_{(1,0)}$ and $L_{(0,1)}$ together generating a ${\mathbb {Z}}_N^{(k)}\times {\mathbb {Z}}_N^{(k)}$ k-form symmetry. When $k=1$, we obtain a standard ${\mathbb {Z}}_N$ gauge theory in $(2+1)$d, as discussed in Sect. 3. When $k=2$, we obtain a ${\mathbb {Z}}_N^{(1)}$ gauge theory in $(4+1)$d, as discussed in Sect. 6.²⁴

We consider the mutual braiding between two operators labeled by (e, m) and $(e',m')$. To do so, we evaluate the correlation functions of these two operators on manifolds $M_k$ and $M_k'$ which form a Hopf link. In practice, this means that we have insertions in the action of the form

$$\begin{aligned} S= & {} {2 \pi \over N} \int _{X_{2k+1}} b^{(k)} \delta a^{(k)} + {2 \pi \over N} \int _{M_k} (e a^{(k)} + m b^{(k)} ) + {2 \pi \over N} \int _{M_k'} (e' a^{(k)} + m' b^{(k)})\nonumber \\= & {} {2 \pi \over N} \int _{X_{2k+1}} b^{(k)} \delta a^{(k)} + {2 \pi \over N} \int _{X_{2k+1}} ( e \omega _{M_k}+ e' \omega _{M_k'}) a^{(k)} + {2 \pi \over N} \int _{X_{2k+1}} ( m \omega _{M_k}+ m' \omega _{M_k'}) b^{(k)},~\nonumber \\ \end{aligned}$$

(A.3)

where $\omega _{M_k}$ is the Poincaré dual of the k-cycle $M_k$ with respect to $X_{2k+1}$, and hence is a cocycle of degree $k+1$. Integrating out the field $b^{(k)}$ enforces

$$\begin{aligned} \delta a^{(k)} = -(m \omega _{M_k}+ m' \omega _{M_k'})~. \end{aligned}$$

(A.4)

Defining $V_{k+1}$ such that $\partial V_{k+1} = m M_k + m' M_k'$, we have $a^{(k)} =- \textrm{PD}(V_{k+1})$. Plugging this back into the action then gives contribution

$$\begin{aligned} -{2 \pi \over N} \int _{X_{2k+1}} ( e \omega _{M_k}+ e' \omega _{M_k'}) \cup \textrm{PD}(V_{k+1}) = - {2 \pi \over N} \int _{X_{2k+1}} \textrm{PD}( (e M_k+ e' M_k') \cap V_{k+1}) ~.\nonumber \\ \end{aligned}$$

(A.5)

In general, the intersection pairing $M_k \cap V_{d-k}$ between $M_k \in H_k(X_{d}, {\mathbb {Z}})$ and $V_{d-k} \in H_{d-k}(X_d, {\mathbb {Z}})$ satisfies

$$\begin{aligned} M_k\cap V_{d-k} = (-1)^{k(d-k)} V_{d-k} \cap M_k~, \end{aligned}$$

(A.6)

and thus in the current case we may write

$$\begin{aligned} -{2 \pi \over N} \int _{X_{2k+1}} \left[ e \,\textrm{PD}( M_k\cap V_{k+1}) + e'\, \textrm{PD}( V_{k+1} \cap M_k' )\right] ~. \end{aligned}$$

(A.7)

It is convenient to formally rewrite $V_{k+1}=m \partial ^{-1}M_k + m' \partial ^{-1}M_k'$ by moving the boundary operator $\partial$ to the right hand side in the definition below (A.4). Then $\textrm{PD}( M_k\cap V_{k+1})= \textrm{PD}( M_k\cap \partial ^{-1} M_k') \equiv \text {link}(M_k, M_k')$. Moreover, by using integration by parts, we have $\textrm{PD}( V_{k+1} \cap M_k' ) = \textrm{PD}( \partial ^{-1}M_{k} \cap M_k' ) = (-1)^{k+1}\textrm{PD}( M_k\cap \partial ^{-1} M_k') \equiv (-1)^{k+1}\text {link}(M_k, M_k')$. Substituting these into (A.7) gives

$$\begin{aligned} -{2 \pi \over N} (e m' + (-1)^{k+1}m e')\, \textrm{link}(M_k, M_k')~. \end{aligned}$$

(A.8)

We thus conclude that the braiding between two k-dimensional operators is

$$\begin{aligned} \langle L_{(e,m)}(M_k) L_{(e',m')}(M_k')...\rangle = \exp \left( - \frac{2\pi i}{N} (em' + (-1)^{k+1} me') \text {link}(M_k, M_k')\right) \langle ...\rangle \nonumber \\ \end{aligned}$$

(A.9)

We next derive the equal time correlation function between generic dyonic lines (e, m) and $(e',m')$, given for $k=1$ and $k=2$ in (3.10) and (6.10). We start with the commutation relation

$$\begin{aligned} L_{(1,0)}(M_k) L_{(0,1)}(M_k') = \exp \left( -{2 \pi i \over N} \langle M_k, M_k'\rangle \right) L_{(0,1)}(M_k') L_{(1,0)}(M_k)\quad \end{aligned}$$

(A.10)

which can be obtained, for example, by canonical quantization.²⁵ The pairing $\langle M_k, M_k'\rangle$ is the intersection pairing between two k-manifolds in 2k-dimensions. Moving the phase to the left side and relabeling $M_k\leftrightarrow M_k'$, we have

$$\begin{aligned} \begin{aligned} L_{(0,1)}(M_k) L_{(1,0)}(M_k')&= \exp \left( {2 \pi i \over N} \langle M_k', M_k\rangle \right) L_{(1,0)}(M_k') L_{(0,1)}(M_k)\\&=\exp \left( {2 \pi i \over N}(-1)^k \langle M_k, M_k'\rangle \right) L_{(1,0)}(M_k') L_{(0,1)}(M_k)\\ \end{aligned}\nonumber \\ \end{aligned}$$

(A.11)

where in the second line we used ${\langle {M_k,M_k'}\rangle }= (-1)^k {\langle {M_k',M_k}\rangle }$.²⁶ Using (A.10) and (A.11) and the definition of the dyonic line (3.7), we may then determine the equal time commutation relation between dyonic lines,

$$\begin{aligned} L_{(e,m)}(M_k) L_{(e',m')}(M_k')= & {} \left( L_{(1,0)}(M_k)\right) ^{\otimes e} \left( L_{(0,1)}(M_k)\right) ^{\otimes m} \nonumber \\{} & {} \quad \left( L_{(1,0)}(M_k')\right) ^{\otimes e'} \left( L_{(0,1)}(M_k')\right) ^{\otimes m'} \nonumber \\{} & {} =e^{-\frac{2\pi i}{N}(e m' +(-1)^{k+1} m e'){\langle {M_k, M_k'}\rangle }} \left( L_{(1,0)}(M_k')\right) ^{\otimes e'} \left( L_{(0,1)}(M_k')\right) ^{\otimes m'}\nonumber \\{} & {} \quad \left( L_{(1,0)}(M_k)\right) ^{\otimes e} \left( L_{(0,1)}(M_k)\right) ^{\otimes m} \nonumber \\{} & {} = e^{-\frac{2\pi i}{N}(e m' +(-1)^{k+1} m e'){\langle {M_k, M_k'}\rangle }} L_{(e',m')}(M_k') L_{(e,m)}(M_k) \end{aligned}$$

(A.12)

which reproduces (3.10) and (6.10). Note that the phase from the equal time commutation relation in (A.12) coincides with the phase from the linking in (A.9), with the intersection pairing number being replaced by the linking number.

Measuring EM Charges of Defect Junctions

Fig. 33

The ${\mathbb {Z}}_2^{\textrm{EM}}$ charge of the trivalent junction denoted by a triangle can be measured by enclosing the junction with the surface $D_{\textrm{EM}}$. This surface is a condensate of the algebra anyon ${{\mathcal {A}}}$, and may be decomposed as in the middle image. The remaining discs of $D_{\textrm{EM}}$ can be shrunk to give orange and purple junctions, which will be studied in the text

The goal of this appendix is to compute the ${\mathbb {Z}}_2^{\textrm{EM}}$ charge of the junctions involving $L_{(N/2,N/2)}$ and $\Sigma _{(e)}$. Recall that there are various such junctions depending on the angle between the line and the surface anchored on $\Sigma _{(e)}$, c.f. Fig. 15, and we will focus here on the charge of the triangle junction. The ${\mathbb {Z}}_2^{\textrm{EM}}$ charge of this junction can be measured by enclosing the junction with the surface $D_{\textrm{EM}}$, as shown in the left of Fig. 33. This surface is a condensate of the algebra anyon ${{\mathcal {A}}}$, and we may simplify the configuration to the one in the middle of Fig. 33. This leaves us with discs of $D_{\textrm{EM}}$ surrounding each of the outgoing lines, which may in general involve complicated intersections of ${{\mathcal {A}}}$ with the external lines. Instead of studying the details of these discs, we may instead shrink them to point-like junctions as shown in the right of Fig. 33. Our goal will now be to understand the junctions appearing here, which will allow us to evaluate the configuration and obtain the charge. Throughout we will neglect real number normalization factors, since these will in any case cancel out to give the final ${\mathbb {Z}}_2$-valued charge.

To begin, we consider the 4-valent junction between $L_{(e,m)}$ and the algebra object ${{\mathcal {A}}}$. In this appendix, we will always draw invertible $L_{(e,m)}$ lines in blue, ${{\mathcal {A}}}$ lines in red, and the junction between the two as a purple dot. The junction between $L_{(e,m)}$ and ${{\mathcal {A}}}$ is the one encountered when e.g. $L_{(e,m)}$ pierces the ${\mathbb {Z}}_2^{\textrm{EM}}$ surface as in Fig. 34. We note that, in Fig. 34, the configuration should not depend on precisely where along the surface $L_{(e,m)}$ intersects. As such, we obtain the consistency condition on the junction shown in Fig. 35. Note that we have introduced an associative (co)multiplication junction $\mu$, defined in Fig. 36, which every algebra object ${{\mathcal {A}}}$ is automatically equipped with.

Fig. 34

The intersection of $L_{(m,e)}$ and $D_{\textrm{EM}}$, with the latter resolved into a mesh of ${\mathcal {A}}$

Fig. 35

Consistency condition involving the purple junction between $L_{(e,m)}$ and ${{\mathcal {A}}}$, and the trivalent junction $\mu$ between three ${{\mathcal {A}}}$ lines

Fig. 36

Definition of the trivalent junction $\mu$ between three ${{\mathcal {A}}}$ lines

On general grounds, the junction between $L_{(e,m)}$ and ${{\mathcal {A}}}$ must be of the form shown in Fig. 37, with $\alpha _{e,m}(p)$ a series of undetermined constants. In fact, imposing the consistency condition in Fig. 35 is sufficient to fix these constants. This may be shown by expanding both sides of Fig. 35 and making use of Figs. 36 and 37. Once expanded, the right-hand side is given as in Fig. 38, whereas the left-hand side is given in the first line of Fig. 39. One may rearrange the configuration in Fig. 39 via a series of F-moves to get the second line in Fig. 39, and upon using the half-braid and more F-moves one may put it in the form shown in the last line of Fig. 39. This configuration may now be compared to the one in Fig. 38. Equating the two gives

$$\begin{aligned} \alpha _{e,m}(p+q) = e^{{2 \pi i \over N}m q}\alpha _{e,m}(p)~, \end{aligned}$$

(B.1)

and choosing $\alpha _{e,m}(0)=1$ (which is simply a choice of convention) we derive that

$$\begin{aligned} \alpha _{e,m}(p) =e^{{2 \pi i \over N}m p}~. \end{aligned}$$

(B.2)

This completely specifies the junction in Fig. 34.

Fig. 37

The general form of the junction between $L_{(e,m)}$ and ${{\mathcal {A}}}$, with $\alpha _{e,m}(p)$ a series of to-be-determined constants

Fig. 38

The right-hand side of Fig. 35, expanded out using Fig. 37

Fig. 39

The left-hand side of Fig. 35, expanded out using Fig. 37. A series of F-moves and a half-braid give the configuration in the last line. Equating this with Fig. 39 fixes the constants $\alpha _{e,m}(p)$

Having understood the junction between $L_{(e,m)}$ and ${{\mathcal {A}}}$, we next study the junctions between $L_{(e,m)}$ and $\Sigma _{(e')}$. As mentioned in the main text, this junction depends on the angle between $L_{(e,m)}$ and the EM duality defect $D_{\textrm{EM}}$ which is anchored on $\Sigma _{(e')}$. This is illustrated in Fig. 15. There we have highlighted two particular cases: first, the case in which $\theta = 0^+$, which we denote by a square; and second, the case in which $\theta = \pi$, which we denote by a triangle. These junctions are subject to consistency conditions illustrated in Fig. 16, which physically correspond to the statement that the junction should not depend on where along $\Sigma _{(e')}$ the line $L_{(e,m)}$ is anchored. We will now solve these consistency conditions, beginning with the triangle junction.

Fig. 40

The definition of the triangle junction, given in terms of a series of undetermined coefficients $\beta _{e,m,e'}(p)$. These coefficients may be fixed by imposing the consistency condition in Fig. 16

Fig. 41

The left-hand side of the consistency condition in Fig. 16. Comparing this to Fig. 40 fixes the undetermined coefficients $\beta _{e,m,e'}(p)$

The general form of the triangle junction is given in Fig. 40, which depends on a series of undetermined coefficients $\beta _{e,m,e'}(p)$. We may solve for these coefficients by imposing the constraint in Fig. 16. The right-hand side of the equation for the triangle junction is precisely the configuration in Fig. 40, whereas the left-hand side is as shown in the first line of Fig. 41 (we use the fact that the junctions $\mu _L$ and $\mu ^\vee _L$ do not involve any phases in their expansions). A series of F-moves and shrinking of a loop gives the second line in Fig. 41. From this we obtain the constraint

$$\begin{aligned} \beta _{e,m,e'}(p+q) = \beta _{e,m,e'}(p)~, \end{aligned}$$

(B.3)

and choosing the convention $\beta _{e,m,e'}(0)=1$, we see that all phases can be taken to be trivial,

$$\begin{aligned} \beta _{e,m,e'}(p)=1~. \end{aligned}$$

(B.4)

This completely fixes the triangle junction.

Fig. 42

The definition of the square junction, given in terms of a series of undetermined coefficients $\gamma _{e,m,e'}(p)$

Fig. 43

The left-hand side of the consistency condition in Fig. 16. Comparing this to Fig. 40 fixes the undetermined coefficients $\beta _{e,m,e'}(p)$

Having fixed the triangle junction, we may now fix the square junction in an analogous way. The most general form of the expansion is given in Fig. 42, with the factors $\gamma _{e,m,e'}(p)$ to be determined. On the other hand, the left-hand side of the consistency condition in Fig. 16 is as shown in Fig. 43. A series of F-moves and braids gives the final line of Fig. 43, from which we read off a constraint on $\gamma _{e,m,e'}(p)$. Choosing conventions such that $\gamma _{e,m,e'}(0)=1$, the solution is

$$\begin{aligned} \gamma _{e,m,e'}(p) = e^{ {2 \pi i \over N} e p}~. \end{aligned}$$

(B.5)

This completely fixes the square junction. We note that, by the braiding properties of the lines $L_{(e,m)}$ and $L_{(e'+n,-n)}$, the square junction obtained here is equivalent to a half-braided triangle junction, as shown in Fig. 17. This was used in Sect. 3.5 of the main text, but will not be needed here.

We have now successfully understood the purple circle and triangle junctions appearing in Fig. 33. We have also understood the purple square junction, which does not appear in Fig. 33 explicitly but which will be important for its computation. All that remains is to evaluate the orange circle junctions, i.e. the four-fold intersection of ${{\mathcal {A}}}$ and $\Sigma _{(e)}$.

Fig. 44

Definition of the orange junctions between ${{\mathcal {A}}}$ and $\Sigma _{(e)}$ in terms of the square and triangle junctions. For simplicity we only define the orange junctions in pairs, as shown. The coefficients $G_e(p,q)$ may be determined by imposing the consistency condition in Fig. 45

Fig. 45

A consistency condition on the orange junctions. This follows from the fact that the ${\mathbb {Z}}_2^{\text {EM}}$ charge of the trivial junction between $\Sigma _{(e)}$, $\Sigma _{(e)}$, and $L_{(0,0)}$ is trivial

In fact, instead of computing the four-fold junction itself, for our purposes it suffices to compute a pair of orange junction, as shown in the left-hand side of Fig. 44. This in general admits an expansion in terms of square and triangle junctions as in the right-hand side of Fig. 44. Here $G_e(p,q)$ are a series of undertermined constants, which are subject to the consistency condition shown in Fig. 45. Physically, what this condition says is that any point on $\Sigma _{(e)}$ is ${\mathbb {Z}}_2^{\textrm{EM}}$ invariant.

By using the definitions of the square and triangle junctions obtained above, it is straightforward to derive an expression for $G_e(p,q)$ from the consistency condition in Fig. 45. We will only give the result,

$$\begin{aligned} G_e(p,q) = e^{-{2\pi i \over N} (pq-p^2)}\, \delta _{p+q+e \text { mod } 2}~, \end{aligned}$$

(B.6)

where the delta function is included to enforce that the line in the interior of the bubble is the EM dual of the incoming line (since it passes $D_{\textrm{EM}}$ in between). This is the final piece of data needed to compute the charge of the junction in Fig. 33.

We finally return to the original calculation of interest, namely the computation of the charge of the junction between $L_{(N/2,N/2)}$ and $\Sigma _{(e)}$. The computation is illustrated in Fig. 46. In words, we begin by using the definition of a pair of orange circle junctions given in Fig. 44, together with the definition of the purple circle junction given in Fig. 37. This gives a factor of

$$\begin{aligned} G_e(p,q) \,\alpha _{{N\over 2},{N\over 2}}(-q) = e^{-{2 \pi i \over N}(pq-p^2)} e^{-\pi i q} \delta _{p+q+e \text { mod } 2}~. \end{aligned}$$

(B.7)

We may then use the definition of the square and triangle junctions given in Figs. 40 and 42, which gives a factor of

$$\begin{aligned} \gamma _{p,-p,e}(r+q)\times \gamma _{-p,p,e}\left( r+p+{N\over 2}\right) = e^{{2\pi i \over N}p(r+q)}e^{-{2 \pi i \over N} p (r + p + N/2)}~. \end{aligned}$$

(B.8)

Fig. 46

The final computation of the junction charge. Details are given in the text

Noting further that the line $L_{(e+r+q+p,-r-q-p)}$ in the interior of the bubble must be the EM dual of the incoming line $L_{(e+r,-r)}$ (since it passes $D_{\textrm{EM}}$ in between) forces us to restrict to triplets r, p, q satisfying

$$\begin{aligned} e+r+q+p = -r\,\,\,\textrm{mod}\,\,N~, \end{aligned}$$

(B.9)

or in other words $p+q = -e-2r$ mod N. Thus in total we produce a phase of

$$\begin{aligned} e^{-{2 \pi i \over N}(pq-p^2)} e^{-\pi i q}\times e^{{2\pi i \over N}p(r+q)}e^{-{2 \pi i \over N} p (r + p + N/2)} = e^{-\pi i (p+q)} = (-1)^e~, \end{aligned}$$

(B.10)

which is the final result for the charge of the junction. In summary, for even e the junction in Fig. 33 is ${\mathbb {Z}}_2^{\text {EM}}$ even. On the other hand, for odd e the junction is ${\mathbb {Z}}_2^{\text {EM}}$ odd, and hence there are an odd number of K lines terminating on the junction.

Gauge Invariance of Twisted ${\mathbb {Z}}_N$ Gauge Theory

In this appendix we provide a bit more detail on the proof of gauge invariance of the action (4.26). We first note that $\delta _C {\textbf{a}}$ is gauge invariant under (4.25). This means that the gauge variation of (4.27) is

$$\begin{aligned} \begin{aligned} (\delta _C {\textbf{g}}^T)_{ij} K^{C_{ij}+1} (\delta _C {\textbf{a}})_{jkl}&= \left( {\textbf{g}}^T_{j} K^{C_{ij}}- {\textbf{g}}^T_{i}\right) K^{C_{ij}+1} (\delta _C {\textbf{a}})_{jkl} \\&= \left( {\textbf{g}}^T_{j} - {\textbf{g}}^T_{i} K^{C_{ij}} \right) K (\delta _C {\textbf{a}})_{jkl}~. \end{aligned} \end{aligned}$$

(C.1)

We further use

$$\begin{aligned} (\delta _C \delta _C {\textbf{a}})_{ijkl} = K^{C_{ij}}(\delta _C {\textbf{a}})_{jkl} - (\delta _C {\textbf{a}})_{ikl} + (\delta _C {\textbf{a}})_{ijl} - (\delta _C {\textbf{a}})_{ijk}=0 \end{aligned}$$

(C.2)

to replace $K^{C_{ij}} (\delta _C {\textbf{a}})_{jkl}$ by $(\delta _C {\textbf{a}})_{ikl} - (\delta _C {\textbf{a}})_{ijl} + (\delta _C {\textbf{a}})_{ijk}$ in (C.1). The second line in (C.1) becomes

$$\begin{aligned} {\textbf{g}}^T_{j} K (\delta _C {\textbf{a}})_{jkl} - {\textbf{g}}^T_{i} K (\delta _C {\textbf{a}})_{ikl} + {\textbf{g}}^T_{i} K (\delta _C {\textbf{a}})_{ijl} - {\textbf{g}}^T_{i} K (\delta _C {\textbf{a}})_{ijk} \equiv \left( \delta ({\textbf{g}}^T K \delta _C {\textbf{a}})\right) _{ijkl}\nonumber \\ \end{aligned}$$

(C.3)

which is a total derivative. This shows that (4.26) is gauge invariant.

The invariance of (4.26) under background gauge transformation (4.28) is also straightforward. Under (4.28), the term (4.27) becomes

$$\begin{aligned} ({\textbf{a}}_{ij})^T K^{\gamma _i} K^{C_{ij}+ \gamma _j - \gamma _i +1} \left( K^{C_{jk}+ \gamma _k-\gamma _j} K^{-\gamma _k}{\textbf{a}}_{kl}- K^{-\gamma _j}{\textbf{a}}_{jl} + K^{-\gamma _j} {\textbf{a}}_{jk}\right) \end{aligned}$$

(C.4)

and all of the $\gamma _i$ manifestly cancel.

Fusion Rules of the $(2+1)$d SymTFT of $TY({\mathbb {Z}}_N)$ from Modular S Matrices

In this appendix, we collect the data of the Drinfeld center ${{\mathcal {Z}}}(TY({\mathbb {Z}}_N))$ of $TY({\mathbb {Z}}_N)$ obtained in the mathematics literature [64, 65]. In particular, we record the explicit form of the modular S matrices, which can be used to rederive the fusion rules of Sect. 4.1 via the Verlinde formula.

1.1 Objects and Modular S Matrices

In the notation of [64], the objects in the category ${{\mathcal {Z}}}(TY({\mathbb {Z}}_N))$ include

• 1. 2N invertible lines: $X_{g,i}, g\in {\mathbb {Z}}_N, i\in {\mathbb {Z}}_2$.

• 2. $N(N-1)/2$ non-invertible lines of quantum dimension 2: $Y_{[g,h]}, g,h\in {\mathbb {Z}}_N, g<h$.

• 3. 2N non-invertible lines of quantum dimension $\sqrt{N}$: $Z_{g,i}, g\in {\mathbb {Z}}_N, i\in {\mathbb {Z}}_2$.

They are in one-to-one correspondence with $L_{(e)}^{q}, L_{[e,m]}, \Sigma _{(e)}^{q}$ in the main text. In this notation, the modular S matrices are given by²⁷

$$\begin{aligned} \begin{aligned}&S_{X_{g,i}, X_{h,j}}=\frac{1}{2N}e^{- \frac{4\pi i}{N} gh}~,\\&S_{X_{g,i}, Z_{h,j}}= S_{Z_{h,j}, X_{g,i}} = \frac{(-1)^i}{2 \sqrt{N}} e^{- \frac{2\pi i}{N} gh}~,\\&S_{X_{g,i}, Y_{[h,k]}}= S_{ Y_{[h,k]}, X_{g,i}}= \frac{1}{N} e^{-\frac{2\pi i}{N} g (h+k)}~,\\&S_{Z_{g,i}, Y_{[h,k]}}= S_{ Y_{[h,k]}, Z_{g,i}} = 0~,\\&S_{Z_{g,i},Z_{h,j}}= \frac{(-1)^{i+j}}{2N} \omega _g \omega _h \sum _{k=0}^{N-1} e^{\frac{2\pi i}{N}(k-g-h)k}~,\\&S_{Y_{[g,h]},Y_{[g',h']}}=\frac{1}{N} \left( e^{-\frac{2\pi i}{N}(gh'+hg')}+ e^{-\frac{2\pi i}{N}(gg' + hh')}\right) ~, \end{aligned} \end{aligned}$$

(D.1)

where

$$\begin{aligned} \omega _g= \left( \frac{1}{\sqrt{N}} \sum _{h=0}^{N-1} e^{-\frac{2\pi i}{N}gh}\gamma (h)\right) ^{1/2}, \quad \gamma (h) ={\left\{ \begin{array}{ll} (-1)^h e^{-\frac{\pi i}{N} h^2}, &{} N\in 2{\mathbb {Z}}+1\\ e^{-\frac{\pi i}{N} h^2}, &{} N\in 2{\mathbb {Z}}\end{array}\right. } \end{aligned}$$

(D.2)

For completeness, we also record the spins of the lines here,

$$\begin{aligned} \theta _{X_{g,i}} = e^{-{2 \pi i \over N} g^2}~, \hspace{0.5 in} \theta _{Y_{[g,h]}}=e^{-{2 \pi i \over N} gh}~, \hspace{0.5 in} \theta _{Z_{g,i}} = (-1)^i\, \omega ^*_g~. \end{aligned}$$

(D.3)

The Verlinde formula [84] enables us to derive the fusion rules between $W_1, W_2, W_3\in \{X_{g,i}, Y_{[g,h]}, Z_{g,i}\}$ from the modular S matrices,

$$\begin{aligned} N_{W_1, W_2}^{W_3} = \sum _{W} \frac{S_{W_1, W} S_{W_2, W} S^*_{W_3, W}}{S_{X_{0,0}, W}}~. \end{aligned}$$

(D.4)

As a consistency check, we will now apply these formulae to small values of N in order to reproduce the fusion rules found in the main text.

1.2 Fusion Rules for Small N

$N=2$: The fusion rule between invertible lines is $X_{g,i} \otimes X_{h,j} = X_{g+h, i+j}$, and we therefore reproduce the results in the text upon identifying $X_{g,i} \leftrightarrow {\widehat{L}}_{(g)}^{(-1)^i}$. The fusion rules involving non-invertible lines are more interesting. The S-matrix elements and Verlinde formula give rise to,

$$\begin{aligned} \begin{aligned}&X_{1,i} \otimes Z_{0,j} = Z_{0, i+j+1}~, \hspace{1cm} X_{1,i} \otimes Z_{1,j} = Z_{1, i+j}~,\\&Z_{g,i}\otimes Z_{g,j} = X_{0,i+j} \oplus X_{1,i+j+g+1}~. \end{aligned} \end{aligned}$$

(D.5)

Comparing with the fusion rule (4.9), we find the identification $Z_{1,j}\leftrightarrow \widehat{\Sigma }_{(0)}^{(-1)^j}, Z_{0,j}\leftrightarrow \widehat{\Sigma }_{(1)}^{(-1)^j}$. As there is only one line with quantum dimension 2, we also have the identification $Y_{[0,1]}\leftrightarrow {\widehat{L}}_{[0,1]}$. Indeed, the following fusion rules

$$\begin{aligned} X_{g,i} \otimes Y_{[0,1]}= Y_{[0,1]}~, \hspace{1cm} Z_{0,i}\otimes Z_{1,j} = Y_{[0,1]}~, \end{aligned}$$

(D.6)

match with those observed in Sect. 4.1.4.

$N=3$: The fusion rules between invertible lines are again $X_{g,i} \otimes X_{h,j} = X_{g+h, i+j}$, and we therefore match the results in the text upon identify $X_{g,i} \leftrightarrow {\widehat{L}}_{(g)}^{(-1)^i}$. The fusion rules involving non-invertible lines are more interesting. The ones involving quantum dimension $\sqrt{3}$ lines are

$$\begin{aligned} \begin{aligned}&X_{0,i}\otimes Z_{0,j}= Z_{0,i+j}~, \quad X_{1,i}\otimes Z_{0,j}= Z_{2,i+j+1}~, \quad X_{2,i}\otimes Z_{0,j}= Z_{1,i+j+1}~. \end{aligned} \end{aligned}$$

(D.7)

Comparing with the fusion rule (4.4), we find the identification $Z_{0,j}\leftrightarrow \widehat{\Sigma }_{0}^{(-1)^j}$, $Z_{1,j}\leftrightarrow \widehat{\Sigma }_{(1)}^{-(-1)^j}$, and $Z_{2,j}\leftrightarrow \widehat{\Sigma }_{(2)}^{-(-1)^j}$. Other fusion rules between X and Z can be derived by starting with (D.7) and fusing additional invertible lines on both sides. The fusion rule between X and Y is

$$\begin{aligned} X_{g,i}\otimes Y_{[g',h']}= Y_{[g+g', g+h']} \end{aligned}$$

(D.8)

which leads to the idenfication $Y_{[g,h]}\leftrightarrow {\widehat{L}}_{[g,h]}$. Other fusion rules can be similarly worked out, and we find that they match with those from Sect. 4.1 upon using the above identifications.

$N=4$: The fusion rules between invertible lines are again $X_{g,i} \otimes X_{h,j} = X_{g+h, i+j}$, and we therefore identify $X_{g,i} \leftrightarrow {\widehat{L}}_{(g)}^{(-1)^i}$ once again. The fusion between X and Z lines are

$$\begin{aligned} \begin{aligned} X_{1,i}\otimes Z_{0,j}= Z_{2,i+j}~, \quad X_{1,i}\otimes Z_{1,j}= Z_{3,i+j+1}~, \quad X_{2,i}\otimes Z_{g,j}= Z_{g,i+j+g}~. \end{aligned} \end{aligned}$$

(D.9)

Note that upon fusing $X_{h,j}$, the $Z_{g,i}$’s for even g form a closed orbit, while those for odd g form another closed orbit. We therefore identify $Z_{0,j}\leftrightarrow \widehat{\Sigma }_{(0)}^{(-1)^{j}}$, $Z_{1,j}\leftrightarrow \widehat{\Sigma }_{(1)}^{(-1)^{j}}$, $Z_{2,j}\leftrightarrow \widehat{\Sigma }_{(2)}^{(-1)^{j}}$, and $Z_{3,j}\leftrightarrow \widehat{\Sigma }_{(3)}^{-(-1)^{j}}$. The fusion rule between X and Y is

$$\begin{aligned} X_{g,i}\otimes Y_{[g',h']}= Y_{[g+g', g+h']} \end{aligned}$$

(D.10)

which leads to the idenfication $Y_{[g,h]}\leftrightarrow {\widehat{L}}_{[g,h]}$. Other fusion rules can be similarly worked out, and we find that they match with those from Sect. 4.1 upon using the above identifications.

Condensation Defects for the EM Exchange Symmetry in $(4+1)$d

In this appendix, we discuss the four-dimensional defect $D_{\text {EM}}$ generating the ${\mathbb {Z}}_4^{\text {EM}}$ (or ${\mathbb {Z}}_2^{\text {EM}}$ for $N=2$) symmetry in $(4+1)$d ${\mathbb {Z}}_N^{(1)}$ gauge theory. Since the defect is different depending on the value of N, we must discuss three separate cases.

1.1 Odd N

For N odd the operator $D_{\text {EM}}$ is given by (6.16), which we reproduce here for convenience,

$$\begin{aligned} D_{\text {EM}}(M_4)= & {} \frac{|H^0(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^2} \sum _{\sigma , \sigma '\in H_2(M_4, {\mathbb {Z}}_N)} \nonumber \\{} & {} \quad \exp \left( \frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle })\right) S_{(1,-1)}(\sigma ) S_{(1,1)}(\sigma ')~. \end{aligned}$$

(E.1)

From (6.13), the expected fusion rule would be $S_{(e,m)}(\tau ) D_{\text {EM}}(M_4) = D_{\text {EM}}(M_4) S_{(-m,e)}(\tau )\exp (-\frac{2\pi i em}{N}{\langle {\tau ,\tau }\rangle })$. We will check this now, together with the invertibility of the defect.

1.1.1 Fusion rule $S_{(e,m)} D_{\textrm{EM}}= D_{\textrm{EM}} S_{(-m,e)} e^{-2\pi i em/N {\langle {\tau ,\tau }\rangle }}$

We begin by verifying the fusion rule above. We have

$$\begin{aligned} \begin{aligned}&S_{(e,m)}(\tau ) D_{\text {EM}}(M_4)\\ {}&\quad = \frac{|H^0(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^2} \sum _{\sigma , \sigma '\in H_2(M_4, {\mathbb {Z}}_N)} \exp \left( \frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle })\right) S_{(e,m)}(\tau ) S_{(1,-1)}(\sigma ) S_{(1,1)}(\sigma ') \\&\quad =\frac{|H^0(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^2} \sum _{\sigma , \sigma '\in H_2(M_4, {\mathbb {Z}}_N)} \exp \left( \frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle }+ (e+m){\langle {\tau , \sigma }\rangle }+ (m-e){\langle {\tau ,\sigma '}\rangle })\right) \\&\qquad \times S_{(1,-1)}(\sigma ) S_{(1,1)}(\sigma ')S_{(e,m)}(\tau )~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.2)

Now we use the fusion rule (6.9) to rewrite $S_{(e,m)}(\tau )$ as $e^{2\pi i m^2/N {\langle {\tau ,\tau }\rangle }}S_{(e,-e)}(\tau ) S_{(m,m)}(\tau ) S_{(-m,e)}(\tau )$. The above expression then becomes

$$\begin{aligned} \begin{aligned} S_{(e,m)}(\tau ) D_{\text {EM}}(M_4)&=\frac{|H^0(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^2} \sum _{\sigma , \sigma '\in H_2(M_4, {\mathbb {Z}}_N)}\\ {}&\quad \exp \bigg (\frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle }+ (e+m){\langle {\tau , \sigma }\rangle }+ (m-e){\langle {\tau ,\sigma '}\rangle }\\&\quad + m^2 {\langle {\tau ,\tau }\rangle })\bigg ) S_{(1,-1)}(\sigma ) S_{(1,1)}(\sigma ')S_{(e,-e)}(\tau ) S_{(m,m)}(\tau ) S_{(-m,e)}(\tau )~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.3)

Note that $(e,-e)=e(1,-1), (m,m)=m(1,1)$, and hence one can switch the order of the surface operators and combine the terms using the quantum torus algebra (6.11). The result is

$$\begin{aligned} \begin{aligned}&S_{(e,m)}(\tau ) D_{\text {EM}}(M_4)\\&\quad =\frac{|H^0(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^2} \sum _{\sigma , \sigma '\in H_2(M_4, {\mathbb {Z}}_N)} \exp \bigg (\frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle }+ m{\langle {\tau , \sigma }\rangle }+ (e+2m){\langle {\tau ,\sigma '}\rangle }\\&\qquad + m^2 {\langle {\tau ,\tau }\rangle })\bigg ) S_{(1,-1)}(\sigma +e \tau ) S_{(1,1)}(\sigma '+m\tau ) S_{(-m,e)}(\tau )~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.4)

We finally make a change of variables $\sigma \rightarrow \sigma -e\tau$, $\sigma '\rightarrow \sigma '-m\tau$, upon which the above expression then becomes

$$\begin{aligned} \begin{aligned}&S_{(e,m)}(\tau ) D_{\text {EM}}(M_4)\\&\quad =\frac{|H^0(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^2} \sum _{\sigma , \sigma '\in H_2(M_4, {\mathbb {Z}}_N)} \exp \bigg (\frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle } -em {\langle {\tau ,\tau }\rangle } )\bigg ) \\&\qquad \times S_{(1,-1)}(\sigma ) S_{(1,1)}(\sigma ') S_{(-m,e)}(\tau ) \\&\quad = D_{\text {EM}}(M_4)S_{(-m,e)}(\tau ) \exp \left( -\frac{2\pi i em}{N}{\langle {\tau ,\tau }\rangle }\right) ~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.5)

Using the commutation relation (6.10), the surface operator in the last step can finally be re-expressed as

$$\begin{aligned} S_{(-m,e)}(\tau ) \exp \left( -\frac{2\pi i em}{N}{\langle {\tau ,\tau }\rangle }\right)= & {} S_{(-m,0)}(\tau ) S_{(0,e)}(\tau ) \exp \left( -\frac{2\pi i em}{N}{\langle {\tau ,\tau }\rangle }\right) \nonumber \\= & {} S_{(0,e)}(\tau ) S_{(-m,0)}(\tau )~. \end{aligned}$$

(E.6)

Thus $D_{\text {EM}}$ maps $S_{(e,0)}$ to $S_{(0,e)}$ and $S_{(0,m)}$ to $S_{(-m,0)}$, exactly as expected in (6.13).

1.1.2 Charge conjugation operator $C=D_{\text {EM}}^2$

Before checking $D_{\text {EM}}^4=1$, we first compute $C=D_{\text {EM}}^2$, which should be the charge conjugation operator.²⁸ To this effect, we have

$$\begin{aligned} \begin{aligned} D_{\text {EM}}(M_4)^2&= \frac{|H^0(M_4, {\mathbb {Z}}_N)|^4}{|H^1(M_4, {\mathbb {Z}}_N)|^4} \sum _{\sigma , \sigma ',\tau ,\tau '\in H_2(M_4, {\mathbb {Z}}_N)}\\&\quad \quad \exp \left( \frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle }+ {\langle {\tau ', \tau '}\rangle } + {\langle {\tau , \tau '}\rangle })\right) \\&\qquad \times S_{(1,-1)}(\sigma ) S_{(1,1)}(\sigma ')S_{(1,-1)}(\tau ) S_{(1,1)}(\tau ')~. \end{aligned} \end{aligned}$$

(E.7)

To simplify the above expression, we switch the order of surface operators, combine the terms with the same charge using the quantum torus algebra, and make a change of variables. The final expression is

$$\begin{aligned} \begin{aligned} C(M_4)=D_{\text {EM}}(M_4)^2&= \frac{|H^0(M_4, {\mathbb {Z}}_N)|^4}{|H^1(M_4, {\mathbb {Z}}_N)|^4} \sum _{\sigma , \sigma ',\tau ,\tau '\in H_2(M_4, {\mathbb {Z}}_N)}\\&\qquad \exp \bigg (\frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle }+ {\langle {\tau ', \tau '}\rangle }\\&\quad + {\langle {\tau , \tau }\rangle } - {\langle {\sigma ',\tau '}\rangle }- {\langle {\sigma ,\tau '}\rangle }+ {\langle {\sigma ',\tau }\rangle }-{\langle {\sigma ,\tau }\rangle } )\bigg ) S_{(1,-1)}(\sigma ) S_{(1,1)}(\sigma ')~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.8)

Since $\tau , \tau '$ only appear in the exponent and not in the surface operators, we can perform the sum over them to simplify the expression. We first sum over $\tau$. The relevant part is

$$\begin{aligned} \sum _{\tau \in H_2(M_4, {\mathbb {Z}}_N)} \exp \left( \frac{2\pi i}{N} ({\langle {\tau , \tau }\rangle }+ {\langle {\sigma '-\sigma , \tau }\rangle })\right) ~. \end{aligned}$$

(E.9)

To complete the square, it is useful to introduce $\rho$ satisfying $2\rho =\sigma '-\sigma$. Note that $\rho$ exists since the equality is defined modulo $N\lambda$ where $\lambda$ is the generator of $H_2(M_4, {\mathbb {Z}}_N)$ and N is odd. Hence (E.9) can be simplified to

$$\begin{aligned} \sqrt{|H_2(M_4, {\mathbb {Z}}_N)|} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N] \exp \left( - \frac{2\pi i}{N} {\langle {\rho , \rho }\rangle }\right) ~, \end{aligned}$$

(E.10)

where

$$\begin{aligned} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]:= \frac{1}{\sqrt{|H_2(M_4, {\mathbb {Z}}_N)|}}\sum _{\tau \in H_2(M_4, {\mathbb {Z}}_N)} \exp \left( \frac{2\pi i}{N} {\langle {\tau , \tau }\rangle }\right) \end{aligned}$$

(E.11)

defines an invertible TQFT, which trivializes on a spin manifold. Similarly, we can sum over $\tau '$ by introducing $2\omega =\sigma +\sigma '$, which gives

$$\begin{aligned} \sqrt{|H_2(M_4, {\mathbb {Z}}_N)|} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N] \exp \left( - \frac{2\pi i}{N} {\langle {\omega , \omega }\rangle }\right) ~. \end{aligned}$$

(E.12)

The charge conjugation defect thus becomes

$$\begin{aligned} \begin{aligned} C(M_4)&= \frac{|H^0(M_4, {\mathbb {Z}}_N)|^4 |H_2(M_4, {\mathbb {Z}}_N)|}{|H^1(M_4, {\mathbb {Z}}_N)|^4} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^2 \sum _{\sigma , \sigma '\in H_2(M_4, {\mathbb {Z}}_N)} \exp \bigg (\frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } \\ {}&\hspace{1cm} +{\langle {\sigma ,\sigma '}\rangle }-{\langle {\rho ,\rho }\rangle }-{\langle {\omega ,\omega }\rangle } )\bigg ) S_{(1,-1)}(\sigma ) S_{(1,1)}(\sigma ')~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.13)

To further simplify the expression, it is useful to replace $\sigma ,\sigma '$ in terms of $\rho ,\omega$ via

$$\begin{aligned} \sigma =\omega -\rho , \hspace{1cm} \sigma '= \omega + \rho ~. \end{aligned}$$

(E.14)

Summing over $\sigma ,\sigma '$ is equivalent to summing over $\rho ,\omega$. Hence the expression for the charge conjugation defect simplifies to

$$\begin{aligned} C(M_4)= & {} \frac{|H^0(M_4, {\mathbb {Z}}_N)|^4 |H_2(M_4, {\mathbb {Z}}_N)|}{|H^1(M_4, {\mathbb {Z}}_N)|^4} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^2 \sum _{\rho , \omega \in H_2(M_4, {\mathbb {Z}}_N)} \nonumber \\{} & {} \qquad \exp \bigg (\frac{2\pi i}{N} (-{\langle {\rho ,\rho }\rangle }+{\langle {\omega ,\omega }\rangle }+ 2 {\langle {\rho ,\omega }\rangle })\bigg ) \nonumber \\{} & {} \quad \times S_{(1,-1)}(\omega -\rho ) S_{(1,1)}(\omega +\rho )\nonumber \\{} & {} = \frac{|H^0(M_4, {\mathbb {Z}}_N)|^4 |H_2(M_4, {\mathbb {Z}}_N)|}{|H^1(M_4, {\mathbb {Z}}_N)|^4} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^2 \sum _{\rho , \omega \in H_2(M_4, {\mathbb {Z}}_N)}\nonumber \\{} & {} \qquad \exp \bigg (-\frac{4\pi i}{N} {\langle {\rho , \omega }\rangle }\bigg ) S_{(0,2)}(\rho ) S_{(2,0)}(\omega )~. \end{aligned}$$

(E.15)

1.1.3 Invertibility of $D_{\text {EM}}$

We finally check the invertibility of $D_{\text {EM}}$ by computing $D_{\text {EM}}^4$, and conforming it is the identity, up to a local counterterm. Since we have already computed $C=D_{\text {EM}}^2$, we only need to compute $C^2$ as follows.

$$\begin{aligned} C(M_4)^2= & {} \frac{|H^0(M_4, {\mathbb {Z}}_N)|^8 |H_2(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^8} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^4 \sum _{\sigma ,\sigma ',\tau ,\tau '\in H_2(M_4, {\mathbb {Z}}_N)} \nonumber \\{} & {} \qquad \exp \bigg (-\frac{4\pi i}{N} ({\langle {\sigma , \sigma '}\rangle }+ {\langle {\tau ,\tau ')}\rangle }\bigg )\nonumber \\{} & {} \quad \times S_{(0,2)}(\sigma ) S_{(2,0)}(\sigma ')S_{(0,2)}(\tau ) S_{(2,0)}(\tau ')\nonumber \\= & {} \frac{|H^0(M_4, {\mathbb {Z}}_N)|^8 |H_2(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^8} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^4 \sum _{\sigma ,\sigma ',\tau ,\tau '\in H_2(M_4, {\mathbb {Z}}_N)}\nonumber \\{} & {} \qquad \exp \bigg (\frac{2\pi i}{N} (-2{\langle {\sigma , \sigma '}\rangle }-2 {\langle {\tau ,\tau ')}\rangle }\nonumber \\{} & {} \quad -4{\langle {\sigma ',\tau }\rangle }\bigg ) S_{(0,2)}(\sigma +\tau ) S_{(2,0)}(\sigma '+\tau ')\nonumber \\= & {} \frac{|H^0(M_4, {\mathbb {Z}}_N)|^8 |H_2(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^8} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^4 \sum _{\sigma ,\sigma ',\tau ,\tau '\in H_2(M_4, {\mathbb {Z}}_N)}\nonumber \\{} & {} \qquad \exp \bigg (\frac{2\pi i}{N} (-2{\langle {\sigma , \sigma '}\rangle }+2 {\langle {\sigma ,\tau ')}\rangle }\nonumber \\{} & {} \quad -2{\langle {\sigma ',\tau }\rangle }\bigg ) S_{(0,2)}(\sigma ) S_{(2,0)}(\sigma ')\nonumber \\= & {} \frac{|H^0(M_4, {\mathbb {Z}}_N)|^8 |H_2(M_4, {\mathbb {Z}}_N)|^4}{|H^1(M_4, {\mathbb {Z}}_N)|^8} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^4 = \chi [M_4,{\mathbb {Z}}_N]^4 {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^{4}. \nonumber \\ \end{aligned}$$

(E.16)

From the second to the last line, we have combined the surface operators with the same charge, made a change of variables $\sigma \rightarrow \sigma -\tau , \sigma '\rightarrow \sigma '-\tau '$, integrated out $\tau$ and $\tau '$ which enforced $\sigma ,\sigma '$ to be trivial 2-cycles in $H_2(M_4, {\mathbb {Z}}_N)$, and finally used the definition of the 4d Euler counterterm in (5.3). Note that $\chi [M_4,{\mathbb {Z}}_N]$ is a local counterterm and ${{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]$ is a phase, which proves that $C(M_4)^2= D_{\text {EM}}(M_4)^4$ is an invertible operator.

1.1.4 Summary of algebra of co-dimension one

We now summarize the algebra involving the $D_{\text {EM}}$ defect. We may define the orientation reversal of $D_{\text {EM}}$ as $\overline{D}_{\text {EM}}$ via

$$\begin{aligned} \overline{D}_{\text {EM}}(M_4)= & {} \chi [M_4,{\mathbb {Z}}_N]^{-2} \frac{|H^0(M_4, {\mathbb {Z}}_N)|^2}{|H^1(M_4, {\mathbb {Z}}_N)|^2} \sum _{\sigma , \sigma '\in H_2(M_4, {\mathbb {Z}}_N)} \nonumber \\{} & {} \quad \exp \left( -\frac{2\pi i}{N} ({\langle {\sigma , \sigma }\rangle } + {\langle {\sigma , \sigma '}\rangle })\right) S_{(1,1)}(\sigma ') S_{(1,-1)}(\sigma ) \end{aligned}$$

(E.17)

where motivated by (5.6) the Euler counterterm is included. This also renders $\overline{D}_{\text {EM}}\times D_{\text {EM}}=1$. The fusion rules are then

$$\begin{aligned} \begin{aligned}&C(M_4)= D_{\text {EM}}(M_4)^2~,\\&D_{\text {EM}}(M_4)^4=\chi [M_4, {\mathbb {Z}}_N]^4 {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^4~,\\&\overline{D}_{\text {EM}}(M_4)\times D_{\text {EM}}(M_4)=1~,\\&\overline{D}_{\text {EM}}(M_4)= \chi [M_4,{\mathbb {Z}}_N]^{-4} {{\mathcal {Z}}}_{Y}[M_4, {\mathbb {Z}}_N]^{-4} C(M_4) D_{\text {EM}}(M_4)^2~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.18)

1.2 $N=2$

When $N=2$, there is only one type of condensate, and the EM symmetry is ${\mathbb {Z}}_2^{\text {EM}}$. The topological defect for ${\mathbb {Z}}_2^{\text {EM}}$ is

$$\begin{aligned} \begin{aligned} D_{\text {EM}}(M_4) = \frac{|H^0(M_4,{\mathbb {Z}}_2)|}{|H^1(M_4,{\mathbb {Z}}_2)|} \sum _{\sigma \in H_2(M_4,{\mathbb {Z}}_2)} S_{(1,0)}(\sigma ) S_{(0,1)}(\sigma +[w_2^{TM}])~, \end{aligned} \end{aligned}$$

(E.19)

where $[w_2^{TM}]$ is the Poincaré dual of the second Stiefel-Whitney class of the tangent bundle of the spacetime manifold $w_2^{TM}$.²⁹ To confirm that this is the correct result, we check the commutation relations with the surface operators $S_{(e,m)}(\tau )$, as well as that $D_{\text {EM}}(M_4)^2=1$ up to Euler counterterm.

1.2.1 Fusion with $S_{(e,m)}$

We begin by computing

$$\begin{aligned} \begin{aligned}&S_{(e,m)}(\tau ) D_{\text {EM}}(M_4)\\&= \frac{|H^0(M_4,{\mathbb {Z}}_2)|}{|H^1(M_4,{\mathbb {Z}}_2)|} \sum _{\sigma \in H_2(M_4,{\mathbb {Z}}_2)} S_{(e,m)}(\tau ) S_{(1,0)}(\sigma ) S_{(0,1)}(\sigma +[w_2^{TM}])\\&= \frac{|H^0(M_4,{\mathbb {Z}}_2)|}{|H^1(M_4,{\mathbb {Z}}_2)|} \sum _{\sigma \in H_2(M_4,{\mathbb {Z}}_2)} \exp \left( i \pi (m {\langle {\tau ,\sigma }\rangle }+ e {\langle {\tau , \sigma +[w_2^{TM}]}\rangle })\right) \\&\quad S_{(1,0)}(\sigma ) S_{(0,1)}(\sigma +[w_2^{TM}])S_{(e,m)}(\tau )~.\\ \end{aligned}\nonumber \\ \end{aligned}$$

(E.20)

Using $S_{(e,m)}(\tau )= S_{(1,0)}((e-m)\tau )S_{(0,1)}((m-e)\tau ) S_{(m,e)}(\tau )e^{i\pi m(m-e){\langle {\tau ,\tau }\rangle }}$ and combining the surface operators with the same charge, the above expression simplifies to

$$\begin{aligned} \begin{aligned}&S_{(e,m)}(\tau ) D_{\text {EM}}(M_4)\\&= \frac{|H^0(M_4,{\mathbb {Z}}_2)|}{|H^1(M_4,{\mathbb {Z}}_2)|} \sum _{\sigma \in H_2(M_4,{\mathbb {Z}}_2)} \exp \Big (i \pi (m {\langle {\tau ,\sigma }\rangle }+ e {\langle {\tau , \sigma +[w_2^{TM}]}\rangle }+m(m-e){\langle {\tau ,\tau }\rangle }\\ {}&\hspace{0.5cm}+(e-m){\langle {\tau ,\sigma +[w_2^{TM}]}\rangle })\Big ) S_{(1,0)}(\sigma +(e-m)\tau ) S_{(0,1)}(\sigma +[w_2^{TM}]+(e-m)\tau )S_{(m,e)}(\tau )~.\\ \end{aligned}\nonumber \\ \end{aligned}$$

(E.21)

We further make a change of variable $\sigma \rightarrow \sigma -(e-m)\tau$, and use ${\langle {\tau ,\tau }\rangle }={\langle {\tau ,w_2^{TM}}\rangle } \textrm{mod}\,\, 2$, upon which the expression simplifies to

$$\begin{aligned}{} & {} S_{(e,m)}(\tau ) D_{\text {EM}}(M_4)\nonumber \\{} & {} \quad = \frac{|H^0(M_4,{\mathbb {Z}}_2)|}{|H^1(M_4,{\mathbb {Z}}_2)|} \sum _{\sigma \in H_2(M_4,{\mathbb {Z}}_2)} \exp \left( i \pi em {\langle {\tau ,\tau }\rangle }\right) S_{(1,0)}(\sigma ) S_{(0,1)}(\sigma +[w_2^{TM}])S_{(m,e)}(\tau )\nonumber \\{} & {} \quad =D_{\text {EM}}(M_4) S_{(m,e)}(\tau ) \exp \left( i \pi em {\langle {\tau ,\tau }\rangle }\right) \end{aligned}$$

(E.22)

as expected.³⁰

1.2.2 Invertibility of $D_{\textrm{EM}}(M_4)$

We next compute

$$\begin{aligned} D_{\text {EM}}(M_4)^2= & {} \frac{|H^0(M_4,{\mathbb {Z}}_2)|^2}{|H^1(M_4,{\mathbb {Z}}_2)|^2} \sum _{\sigma ,\tau } S_{(1,0)}(\sigma ) S_{(0,1)}(\sigma +[w_2^{TM}]) S_{(1,0)}(\tau ) S_{(0,1)}(\tau +[w_2^{TM}]) \nonumber \\= & {} \frac{|H^0(M_4,{\mathbb {Z}}_2)|^2}{|H^1(M_4,{\mathbb {Z}}_2)|^2} \sum _{\sigma ,\tau } S_{(1,0)}(\sigma +\tau ) S_{(0,1)}(\sigma +\tau ) \exp \left( i\pi {\langle {\sigma +[w_2^{TM}], \tau }\rangle }\right) \nonumber \\= & {} \frac{|H^0(M_4,{\mathbb {Z}}_2)|^2}{|H^1(M_4,{\mathbb {Z}}_2)|^2} \sum _{\sigma ,\tau } S_{(1,1)}(\sigma ) \exp \left( i\pi {\langle {\sigma +[w_2^{TM}]-\tau , \tau }\rangle }\right) ~. \end{aligned}$$

(E.23)

Using ${\langle {\tau ,\tau }\rangle }= {\langle {[w_2^{TM}],\tau }\rangle }\,\,\textrm{mod}\,\,2$, this expression can be simplified to

$$\begin{aligned} \begin{aligned} D_{\text {EM}}(M_4)^2&=\frac{|H^0(M_4,{\mathbb {Z}}_2)|^2}{|H^1(M_4,{\mathbb {Z}}_2)|^2} \sum _{\sigma ,\tau } S_{(1,1)}(\sigma ) \exp \left( i\pi {\langle {\sigma , \tau }\rangle }\right) \\ {}&= \frac{|H^0(M_4,{\mathbb {Z}}_2)|^2 |H_2(M_4,{\mathbb {Z}}_2)|}{|H^1(M_4,{\mathbb {Z}}_2)|^2} = \chi [M_4,{\mathbb {Z}}_2]~. \end{aligned} \end{aligned}$$

(E.24)

We can also define the orientation reversal as

$$\begin{aligned} \overline{D}_{\text {EM}}= \chi [M_4, {\mathbb {Z}}_2]^{-1} D_{\text {EM}} \end{aligned}$$

(E.25)

upon which (E.24) can be rewritten as

$$\begin{aligned} D_{\text {EM}}(M_4) \times \overline{D}_{\text {EM}}(M_4) = 1 \end{aligned}$$

(E.26)

Hence $D_{\text {EM}}(M_4)$ is invertible, and generates a ${\mathbb {Z}}_2$ symmetry.

1.3 Even N and $N\ge 4$:

The symmetry defect in this case is almost identical to that for odd N, but with some minor modifications. In particular, we now have

$$\begin{aligned} D_{\text {EM}}(M_4)= & {} \frac{|H^0(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)|}{|H^1(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2))|} \sum _{(\sigma , \sigma ')\in H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)} \nonumber \\{} & {} \qquad \exp \left( \frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle })\right) \nonumber \\{} & {} \quad \times S_{(1,-1)}(\sigma )S_{(0,1)}\left( \frac{N}{2}[w_2^{TM}]\right) S_{(1,1)}(\sigma ')~. \end{aligned}$$

(E.27)

As in the case of $N=2$, for generic even N, we are allowed to turn on the background field $w_2^{TM}$.

1.3.1 Gauge invariance

As explained in Sect. 6.2.2, summing over $H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)$ requires that the defect should be invariant under the gauge transformation $\sigma \rightarrow \sigma +\frac{N}{2}\lambda , \sigma '\rightarrow \sigma '+\frac{N}{2}\lambda$, which follows from the identification $S_{(N/2,N/2)}(\sigma )= S_{(N/2,-N/2)}(\sigma )$. Indeed, it is straightforward to check that (E.27) is invariant under this gauge transformation, due to the proper coupling to the background field $[w_2^{TM}]$. This means that summing over the elements $(\sigma ,\sigma ')$ in $H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)$ amounts to summing over $\sigma$ and $\sigma '$ in $H_2(M_4,{\mathbb {Z}}_N)$ separately, but only over half of the total domain.³¹

The ${\mathbb {Z}}_4^{\text {EM}}$ defect (E.27) differs from that for the odd N case by coupling to the background field $[w_2^{TM}]$. If we do not couple to the background field in (E.27) and use (E.1) instead, then under the gauge transformation $\sigma \rightarrow \sigma +\frac{N}{2}\lambda , \sigma '\rightarrow \sigma '+\frac{N}{2}\lambda$ we would find that $D_{\text {EM}}$ transforms as $D_{\text {EM}}\rightarrow (-1)^{\frac{N}{2}{\langle {\lambda ,\lambda }\rangle }} D_{\text {EM}} = (-1)^{\frac{N}{2}{\langle {\lambda ,[w_2^{TM}]}\rangle }}D_{\text {EM}}$, which is not invariant and renders the sum over $H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)$ ill-defined. The coupling to $[w_2^{TM}]$ in (E.27) precisely compensates this non-invariance.

1.3.2 Fusing with $S_{(e,m)}(\tau )$

To justify that (E.27) is the correct condensation defect of ${\mathbb {Z}}_4^{\text {EM}}$, we may check that the commutation relation with the surface operators is of the desired form (6.13),

$$\begin{aligned} S_{(e,m)}(\tau ) D_{\text {EM}}(M_4) = D_{\text {EM}}(M_4) S_{(-m,e)}(\tau ) \exp \left( -\frac{2\pi i e m }{N} {\langle {\tau ,\tau }\rangle }\right) ~. \end{aligned}$$

(E.28)

The proof is identical to the case for odd N, and will not be reproduced here.

1.3.3 Two useful summation formulas

Before discussing the charge conjugation operator and verifying the invertibility of $D_{\text {EM}}$, it is useful to discuss a summation formula which will be used later on. We start with the discrete Fourier transformation,

$$\begin{aligned} \frac{1}{\sqrt{N}} \sum _{x\in {\mathbb {Z}}_N} \exp \left( \frac{2\pi i}{2N} x^2 + \frac{2\pi i}{N} x p \right) = \exp \left( \frac{\pi i}{4}- \frac{\pi i p^2}{N}\right) \end{aligned}$$

(E.29)

and

$$\begin{aligned} \begin{aligned}&\frac{1}{N} \sum _{x,y\in {\mathbb {Z}}_N, x+y\in 2{\mathbb {Z}}} \exp \left( \frac{2\pi i}{2N} (x^2+y^2)\right) = {\left\{ \begin{array}{ll} i, &{} N\in 4{\mathbb {Z}}\\ 0, &{} N\in 4{\mathbb {Z}}+2 \end{array}\right. }\\&\frac{1}{N} \sum _{x,y\in {\mathbb {Z}}_N, x+y\in 2{\mathbb {Z}}+1} \exp \left( \frac{2\pi i}{2N} (x^2+y^2)\right) = {\left\{ \begin{array}{ll} 0, &{} N\in 4{\mathbb {Z}}\\ i, &{} N\in 4{\mathbb {Z}}+2 \end{array}\right. } \end{aligned} \end{aligned}$$

(E.30)

First Formula We now prove the first of two summation formulas of interest. To motivate the first, we begin by considering the function

$$\begin{aligned} f(a,b)= \frac{1}{N}\sum _{\begin{array}{c} x\in {\mathbb {Z}}_N, y\in {\mathbb {Z}}_N\\ x+y\in 2{\mathbb {Z}} \end{array}} \exp \left( \frac{2\pi i}{N} \bigg ( \frac{x^2+y^2}{2}- ax - by\bigg )\right) \end{aligned}$$

(E.31)

where N is an even number, and $a,b\in {\mathbb {Z}}_N$. Note that f(a, b) is invariant under shifting a, b by N/2 simultaneously, i.e.

$$\begin{aligned} f(a,b)= f\left( a+\frac{N}{2}, b+\frac{N}{2}\right) \end{aligned}$$

(E.32)

which is guaranteed by $x+y\in 2{\mathbb {Z}}$ in the range of sum. To perform the sum, we complete the square and make a change of variable. Then f(a, b) simplifies to

$$\begin{aligned} f(a,b)= \exp \left( - \frac{2\pi i}{2N} (a^2+b^2)\right) \frac{1}{N}\sum _{\begin{array}{c} x\in {\mathbb {Z}}_N, y\in {\mathbb {Z}}_N\\ x+y+a+b\in 2{\mathbb {Z}} \end{array}} \exp \left( \frac{2\pi i}{2N} (x^2+y^2)\right) ~. \end{aligned}$$

(E.33)

The sum is invariant under the shift $(a,b)\rightarrow (a+\frac{N}{2}, b+\frac{N}{2})$, but the phase in front is not. The phase transforms as

$$\begin{aligned} \exp \left( - \frac{2\pi i}{2N} (a^2+b^2)\right) \rightarrow e^{- i \pi (a+b) - i \pi N/2}\exp \left( - \frac{2\pi i}{2N} (a^2+b^2)\right) ~. \end{aligned}$$

(E.34)

On the other hand, f(a, b) satisfies (E.32), and hence $f(a,b)=0$ whenever the phase $e^{- i \pi (a+b) - i \pi N/2}\ne 1$. In particular, this implies that $f(a,b)=0$ when $(N,a+b)\in (4{\mathbb {Z}}, 2{\mathbb {Z}})$ and $(N,a+b)\in (4{\mathbb {Z}}+2, 2{\mathbb {Z}}+1)$. Further evaluating the summation (E.33) via (E.30), we obtain

$$\begin{aligned} f(a,b)= \exp \left( -\frac{2\pi i}{2N} (a^2+b^2) + \frac{\pi i}{2}\right) \delta _{a+b+N/2\,\,\textrm{mod}\,\,2}~. \end{aligned}$$

(E.35)

Comparing this to (E.29), this is the generalized discrete Fourier transformation on a space with constraints.

We would like to generalize (E.35) so that the variables are the elements in $H_2(M_4,{\mathbb {Z}}_N)$, and the product is replaced by the intersection pairing or the Poincaré dual of the Pontryagin square. Concretely, we are interested in evaluating the generalized discrete Fourier transformation

$$\begin{aligned} \frac{1}{|H_2(M_4,{\mathbb {Z}}_N)|}\sum _{\begin{array}{c} \tau ,\tau '\in H_2(M_4,{\mathbb {Z}}_N),\\ \tau +\tau '= 2\eta \in H_2(M_4,{\mathbb {Z}}_N) \end{array}} \exp \left( \frac{2\pi i}{2N} ({{\mathcal {P}}}([\tau ]) + {{\mathcal {P}}}([\tau '])) -\frac{2\pi i}{N} ({\langle {a,\sigma }\rangle } + {\langle {b,\tau }\rangle })\right) ~,\nonumber \\ \end{aligned}$$

(E.36)

where $a,b\in H_2(M_4,{\mathbb {Z}}_N)$, and $[\tau ]$ is the Poincaré dual of $\tau$. The condition $\tau +\tau '= 2\eta \in H_2(M_4,{\mathbb {Z}}_N)$ simply means that $\tau +\tau '$ is an even element in $H_2(M_4,{\mathbb {Z}}_N)$, i.e. $\tau +\tau '$ is trivial when regarded as an element in $H_2(M_4,{\mathbb {Z}}_2)$. The sum (E.36) is invariant under the gauge transformation

$$\begin{aligned} a\rightarrow a+ \frac{N}{2}\lambda , \hspace{1cm} b\rightarrow b+ \frac{N}{2}\lambda ~, \end{aligned}$$

(E.37)

and hence (a, b) can be regarded as an element in $H_2(M_4,({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)$. Following the same discussion which lead to (E.35), we can complete the square in (E.36), and find

$$\begin{aligned} \begin{aligned}&\exp \left( -\frac{2\pi i}{2N}({{\mathcal {P}}}([a])+ {{\mathcal {P}}}([b]))\right) \frac{1}{|H_2(M_4,{\mathbb {Z}}_N)|}\\&\quad \sum _{\begin{array}{c} \tau ,\tau '\in H_2(M_4,{\mathbb {Z}}_N),\\ \tau +\tau '+a+b= 2\eta \in H_2(M_4,{\mathbb {Z}}_N) \end{array}} \exp \left( \frac{2\pi i}{2N} ({{\mathcal {P}}}([\tau ]) + {{\mathcal {P}}}([\tau '])) \right) ~. \end{aligned} \end{aligned}$$

(E.38)

The sum is gauge invariant, and defines an invertible TQFT ${{\mathcal {Z}}}_Y(M_4)$ as follows,

$$\begin{aligned} \begin{aligned}&\frac{1}{|H_2(M_4,{\mathbb {Z}}_N)|}\sum _{\begin{array}{c} \tau ,\tau '\in H_2(M_4,{\mathbb {Z}}_N),\\ \tau +\tau '= 2\eta \in H_2(M_4,{\mathbb {Z}}_N) \end{array}}\\&\quad \exp \left( \frac{2\pi i}{2N} ({{\mathcal {P}}}([\tau ]) + {{\mathcal {P}}}([\tau '])) \right) = {\left\{ \begin{array}{ll} {{\mathcal {Z}}}_{Y}[M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}]~, &{} N\in 4{\mathbb {Z}}\\ 0, &{} N\in 4{\mathbb {Z}}+2 \end{array}\right. } \\&\frac{1}{|H_2(M_4,{\mathbb {Z}}_N)|}\sum _{\begin{array}{c} \tau ,\tau '\in H_2(M_4,{\mathbb {Z}}_N),\\ \tau +\tau '+\xi = 2\eta \in H_2(M_4,{\mathbb {Z}}_N) \end{array}}\\&\quad \exp \left( \frac{2\pi i}{2N} ({{\mathcal {P}}}([\tau ]) + {{\mathcal {P}}}([\tau '])) \right) = {\left\{ \begin{array}{ll} 0, &{} N\in 4{\mathbb {Z}}\\ {{\mathcal {Z}}}_{Y}[M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}]~, &{} N\in 4{\mathbb {Z}}+2 \end{array}\right. } \\ \end{aligned}\nonumber \\ \end{aligned}$$

(E.39)

where $\xi$ is an generator of $H_2(M_4,{\mathbb {Z}}_N)$, and the sum is independent of the choice of $\xi$. In particular, we can choose $\xi =[w_2^{TM}]\mod 2$. The above formula is a generalization of (E.30) and is similar to ${{\mathcal {Z}}}_Y$ for the odd N case, c.f. (E.11). But the phase in (E.38) is not gauge invariant. This enforces that (E.36) vanishes for certain (N, a, b). To find the vanishing condition, we perform the gauge transformation

$$\begin{aligned}{} & {} \exp \left( -\frac{2\pi i}{2N}({{\mathcal {P}}}([a])+ {{\mathcal {P}}}([b]))\right) \rightarrow e^{-\pi i {\langle {a+b,\lambda }\rangle } - \pi i \frac{N}{2} {\langle {w_2^{TM}, \lambda }\rangle }} \nonumber \\{} & {} \quad \exp \left( -\frac{2\pi i}{2N}({{\mathcal {P}}}([a])+ {{\mathcal {P}}}([b]))\right) ~. \end{aligned}$$

(E.40)

Because $\lambda$ is arbitrary, we require $a+b+\frac{N}{2}w_2^{TM}$ to be a trivial element in $H_2(M_4,{\mathbb {Z}}_2)$ for the sum (E.36) to be nontrivial. In summary, the sum (E.36) equals

$$\begin{aligned} \begin{aligned} \exp \left( -\frac{2\pi i}{2N}({{\mathcal {P}}}([a])+ {{\mathcal {P}}}([b]))\right) {{\mathcal {Z}}}_{Y}\left[ M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}\right] \delta _{a+b+\frac{N}{2}w_2^{TM}\,\,\textrm{mod}\,\,2}~. \end{aligned} \end{aligned}$$

(E.41)

Second Formula We further consider the summation

$$\begin{aligned} \frac{1}{N^2/4} \sum _{\begin{array}{c} (x,y)\in ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2,\\ x+y=0\mod 2 \end{array}} \exp \left( \frac{2\pi i}{N} (-x a + y b)\right) \delta _{a+b\mod 2} = \delta _{a} \delta _{b}~. \end{aligned}$$

(E.42)

Upgrading the summation variables to elements in homology, we find

$$\begin{aligned}{} & {} \frac{|H_2(M_4,{\mathbb {Z}}_2)|^2}{|H_2(M_4,{\mathbb {Z}}_N)|^2} \sum _{\begin{array}{c} (\tau ,\tau ')\in H_2(M_4,({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)\\ \tau +\tau '=0\mod 2 \end{array}} \nonumber \\{} & {} \quad \exp \left( \frac{2\pi i}{N} (-{\langle {\tau ', \sigma }\rangle }+ {\langle {\tau , \sigma '}\rangle })\right) \delta _{\sigma +\sigma '\mod 2} = \delta _{\sigma }\delta _{\sigma '} ~. \end{aligned}$$

(E.43)

1.3.4 Charge conjugation $C= D_{\text {EM}}^2$

We now return to the evaluation of powers of $D_{\text {EM}}$. Taking the square of (E.27), we obtain

$$\begin{aligned} \begin{aligned} C(M_4)&= D_{\text {EM}}(M_4)^2 \\&= \frac{|H^0(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)|^2}{|H^1(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2))|^2} \sum _{\begin{array}{c} (\sigma , \sigma ')\in \\ H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2) \end{array}} \\&\qquad \exp \left( \frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle }+ \frac{N}{2}{\langle {[w_2^{TM}], \sigma '}\rangle })\right) \\ {}&\quad \times S_{(1,-1)}(\sigma ) S_{(1,1)}(\sigma ') \sum _{\begin{array}{c} (\tau , \tau ')\in \\ H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2) \end{array}} \\&\qquad \exp \bigg ( \frac{2\pi i}{N} ({\langle {\tau ',\tau '}\rangle }+ {\langle {\tau ,\tau }\rangle }- {\langle {\sigma +\sigma '+\frac{N}{2}[w_2^{TM}], \tau '}\rangle } \\ {}&\quad - {\langle {\sigma -\sigma '+\frac{N}{2}[w_2^{TM}], \tau }\rangle })\bigg )~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.44)

Summing over $(\tau ,\tau ')\in H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2))$ can be reorganized as summing over ${\widetilde{\tau }}:=\tau '+\tau$ and ${\widetilde{\tau }}':=\tau '-\tau$ in $H_2(M_4, {\mathbb {Z}}_N)$ respectively, but with the constraint ${\widetilde{\tau }}\pm {\widetilde{\tau }}'$ being an even element in $H_2(M_4,{\mathbb {Z}}_N)$, or equivalently a trivial element in $H_2(M_4,{\mathbb {Z}}_2)$. This gives

$$\begin{aligned} \begin{aligned}&\sum _{\begin{array}{c} (\tau , \tau ')\in \\ H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2) \end{array}} \exp \Bigg ( \frac{2\pi i}{N}\bigg ( {\langle {\tau ',\tau '}\rangle }+ {\langle {\tau ,\tau }\rangle }\\&\qquad - {\langle {\sigma +\sigma '+\frac{N}{2}[w_2^{TM}], \tau '}\rangle } - {\langle {\sigma -\sigma '+\frac{N}{2}[w_2^{TM}], \tau }\rangle }\bigg )\Bigg )\\&\quad =\sum _{\begin{array}{c} {\widetilde{\tau }}, {\widetilde{\tau }}'\in H_2(M_4, {\mathbb {Z}}_N)\\ {\widetilde{\tau }}+{\widetilde{\tau }}'=2\eta \in H_2(M_4,{\mathbb {Z}}_N) \end{array}}\exp \Bigg (\frac{2\pi i}{2N} ({{\mathcal {P}}}([{\widetilde{\tau }}])+ {{\mathcal {P}}}([{\widetilde{\tau }}'])) -\frac{2\pi i}{N}({\langle {\sigma +\frac{N}{2}[w_2^{TM}], {\widetilde{\tau }}}\rangle }+{\langle {\sigma ', {\widetilde{\tau }}'}\rangle })\Bigg )~. \\ \end{aligned}\nonumber \\ \end{aligned}$$

(E.45)

In particular, the gauge invariance under $(\sigma ,\sigma ')\rightarrow (\sigma +\frac{N}{2}\lambda , \sigma '+ \frac{N}{2}\lambda )$ is manifest for the first line, and can be seen in the second line from the constraint in the summation domain ${\widetilde{\tau }}+{\widetilde{\tau }}'=2\eta \in H_2(M_4,{\mathbb {Z}}_N)$. The last expression is of precisely the form (E.36), where $(a,b)= (\sigma +\frac{N}{2}[w_2^{TM}], \sigma ')$, and hence we can apply the result (E.41) here. The sum simplifies to

$$\begin{aligned} |H_2(M_4,{\mathbb {Z}}_N)| {{\mathcal {Z}}}_{Y}[M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}] \exp \left( - \frac{2\pi i}{2N} ({{\mathcal {P}}}([\sigma ]+\frac{N}{2}w_2^{TM})+ {{\mathcal {P}}}([\sigma ']))\right) \delta _{\sigma +\sigma '\,\,\textrm{mod}\,\,2}~.\nonumber \\ \end{aligned}$$

(E.46)

Substituting the above calculations into the expression of $C(M_4)$, we get

$$\begin{aligned} \begin{aligned} C(M_4)&= \frac{|H^0(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)|^2 |H_2(M_4,{\mathbb {Z}}_N)|}{|H^1(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2))|^2} {{\mathcal {Z}}}_{Y}[M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}] \\ {}&\quad \times \sum _{\begin{array}{c} (\sigma , \sigma ')\in H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2),\\ \sigma +\sigma ' =0 \in H_2(M_4,{\mathbb {Z}}_2) \end{array}} \exp \bigg (\frac{2\pi i}{N} ({\langle {\sigma ', \sigma '}\rangle } + {\langle {\sigma , \sigma '}\rangle }+ \frac{N}{2}{\langle {[w_2^{TM}], \sigma '}\rangle } \\ {}&\quad -\frac{1}{2} {{\mathcal {P}}}([\sigma ]+ \frac{N}{2}w_2^{TM})-\frac{1}{2} {{\mathcal {P}}}([\sigma ']))\bigg ) S_{(1,-1)}(\sigma )S_{(1,1)}(\sigma ') ~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.47)

1.3.5 Invertibility of $D_{\text {EM}}$

We finally show that $D_{\text {EM}}^4 = C^2$ is the identity up to an Euler counterterm and an invertible phase ${{\mathcal {Z}}}_{Y}$. Taking the square of (E.47), we find

$$\begin{aligned} \begin{aligned} D_{\text {EM}}(M_4)^4&= C(M_4)^2\\&= \frac{|H^0(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)|^4 |H_2(M_4,{\mathbb {Z}}_N)|^2}{|H^1(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2))|^4} {{\mathcal {Z}}}_{Y}[M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}]^2 \\&\quad \times \sum _{\begin{array}{c} (\sigma , \sigma '),(\tau ,\tau ')\in \\ H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2),\\ \sigma +\sigma ', \tau +\tau '=0 \in \\ H_2(M_4,{\mathbb {Z}}_2) \end{array}}\\ {}&\quad \exp \bigg (\frac{2\pi i}{N} (\frac{1}{2}{{\mathcal {P}}}([\sigma '])- \frac{1}{2}{{\mathcal {P}}}([\sigma ])+ {\langle {\sigma ,\sigma '}\rangle }- \frac{N}{2}{\langle {\sigma -\sigma ',[w_2^{TM}]}\rangle }\\ {}&\quad - \frac{N^2}{4}{\langle {[w_2^{TM}],[w_2^{TM}]}\rangle }-{\langle {\tau ',\sigma }\rangle }+ {\langle {\tau ,\sigma '}\rangle })\bigg ) S_{(1,-1)}(\sigma )S_{(1,1)}(\sigma ') ~. \end{aligned} \end{aligned}$$

(E.48)

We further sum over $\tau ,\tau '$ by applying the formula (E.43), which constrains $\sigma ,\sigma '$ to be trivial provided $\sigma +\sigma '=0\mod 2$. Hence the above expression simplifies to

$$\begin{aligned} \begin{aligned} D_{\text {EM}}(M_4)^4&= C(M_4)^2\\&= \frac{|H^0(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)|^4 |H_2(M_4,{\mathbb {Z}}_N)|^2}{|H^1(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2))|^4} {{\mathcal {Z}}}_{Y}[M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}]^2 \frac{|H_2(M_4,{\mathbb {Z}}_N)|^2}{|H_2(M_4,{\mathbb {Z}}_2)|^2}\\&=\frac{|H^0(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)|^4 |H^2(M_4,({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)|^2}{|H^0(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)|^4}{{\mathcal {Z}}}_{Y}[M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}]^2 \\&= \chi \left[ M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}\right] ^2 {{\mathcal {Z}}}_{Y}\left[ M_4, \frac{{\mathbb {Z}}_N\times {\mathbb {Z}}_N}{{\mathbb {Z}}_2}\right] ^2~. \end{aligned}\nonumber \\ \end{aligned}$$

(E.49)

1.3.6 Summary of algebra of co-dimension one

We close by summarizing the algebra involving the $D_{\text {EM}}$ defect. Noting that the orientation reversal of $D_{\text {EM}}$ is

$$\begin{aligned} \begin{aligned} \overline{D}_{\text {EM}}(M_4)&= \chi [M_4,({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2]^{-1} \frac{|H^0(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)|}{|H^1(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2))|}\\&\quad \times \sum _{(\sigma , \sigma ')\in H_2(M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2)} \exp \left( -\frac{2\pi i}{N} {\langle {\sigma , \sigma '}\rangle }\right) \\&\qquad S_{(1,1)}(\sigma ') S_{(0,-1)}(\sigma +\frac{N}{2}[w_2^{TM}]) S_{(1,0)}(\sigma )~, \end{aligned}\nonumber \\ \end{aligned}$$

(E.50)

we have

$$\begin{aligned}{} & {} C(M_4)= D_{\text {EM}}(M_4)^2~,\nonumber \\{} & {} D_{\text {EM}}(M_4)^4= \chi [M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2]^2 {{\mathcal {Z}}}_{Y}[M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2]^2~,\nonumber \\{} & {} \overline{D}_{\text {EM}}(M_4)\times D_{\text {EM}}(M_4)=1~,\nonumber \\{} & {} \overline{D}_{\text {EM}}(M_4)=\chi [M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2]^{-2} {{\mathcal {Z}}}_{Y}[M_4, ({\mathbb {Z}}_N\times {\mathbb {Z}}_N)/{\mathbb {Z}}_2]^{-2} C(M_4) D_{\text {EM}}(M_4)^{2}.\qquad \quad \end{aligned}$$

(E.51)

More on ${{\widehat{S}}}_{\{N/2,0\}}$

Given an operator ${\mathcal {O}}$ invariant under a symmetry G, upon gauging G we may get a series of operators $\widehat{{\mathcal {O}}}$ transforming in representations of the quantum dual ${{\widehat{G}}}$. In this final appendix, we discuss how to assign representations of subgroups of the quantum symmetry ${{\widehat{G}}}$ to operators $\widehat{{\mathcal {O}}}$ descending from ${\mathcal {O}}$ that are invariant under only a subgroup $H \subset G$. We will focus only on the case of interest to us in this paper, namely $G= {\mathbb {Z}}_4^{\textrm{EM}}$, $H = {\mathbb {Z}}_2^{\textrm{EM}}$, and the operator ${{\widehat{O}}} = {{\widehat{S}}}_{\{N/2,0\}}$.

Recall from the discussion in Sect. 7.1.1 that, unlike the surface ${{\widehat{S}}}_{(N/2,N/2)}$ which is invariant under ${\mathbb {Z}}_4^{\textrm{EM}}$, or the surface ${{\widehat{S}}}_{[e,m]}$ whose constituents are not invariant under any subgroup of ${\mathbb {Z}}_4^{\textrm{EM}}$, the constituents of ${{\widehat{S}}}_{\{N/2,0\}}$ are invariant under a ${\mathbb {Z}}_2^{\textrm{EM}}$ subgroup of ${\mathbb {Z}}_4^{\textrm{EM}}$. That is, under $(e,m) \rightarrow (-e,-m)$ we see that both $S_{(N/2,0)}$ and $S_{(0,N/2)}$ are left unchanged. This means that ${{\widehat{S}}}_{[e,m]}$ can be assigned a representation of the quantum ${\widehat{{\mathbb {Z}}}}_2^{\textrm{EM}}\subset {\widehat{{\mathbb {Z}}}}_4^{\textrm{EM}}$, and hence that the identity line in ${{\widehat{S}}}_{[e,m]}$ carries a two-fold index, i.e. ${\mathfrak {I}}_1^p({{\widehat{S}}}_{[e,m]})$ for $p = 0,1$.

Physically, the statement is that the bare identity line ${\mathfrak {I}}_1^0({{\widehat{S}}}_{\{N/2,0\}})$ cannot absorb a single copy of K, and hence when stacked with K gives a distinct line ${\mathfrak {I}}_1^1({{\widehat{S}}}_{\{N/2,0\}})$. However, it can absorb $K^2$, and hence there are no other distinct lines generated in this way. To see how these statements arise, let us focus on the global fusion. We first consider coincident loops of ${\mathfrak {I}}_1^0({{\widehat{S}}}_{\{N/2,0\}})$ and K, and ask whether this configuration can be distinguished from a loop of ${\mathfrak {I}}_1^0({{\widehat{S}}}_{\{N/2,0\}})$ in isolation (c.f. Fig. 24 and the discussion surrounding it). A loop of K gives a non-trivial contribution to correlation functions if and only if it links with n units of EM flux, with n non-zero modulo 4. If $n=1,3$, then K gives a non-trivial result, but the loop of ${\mathfrak {I}}_1^0({{\widehat{S}}}_{\{N/2,0\}})$ evaluates to zero since the constituent lines ${\mathfrak {I}}_1^0(S_{(N/2,0)})$ and ${\mathfrak {I}}_1^0(S_{(0,N/2)})$ (which are not lines in the EM-gauged theory, but which do exist in the pregauged theory) are not invariant under the primitive generator of ${\mathbb {Z}}_4^{\textrm{EM}}$. Hence in the presence of $n=1,3$ units of EM flux, one cannot distinguish the configurations with ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ stacked with K from the configuration with ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ in isolation. However, if we consider the case of $n=2$ units of EM flux, then the loop of K is again non-trivial, but now ${\mathfrak {I}}_1^0(S_{(0,N/2)})$ does not evaluate to zero since the constituent lines ${\mathfrak {I}}_1^0(S_{(N/2,0)})$ and ${\mathfrak {I}}_1^0(S_{(0,N/2)})$ are invariant under ${\mathbb {Z}}_2^{\textrm{EM}} \subset {\mathbb {Z}}_4^{\textrm{EM}}$. Thus by considering the configuration with $n=2$ units of EM flux, one can distinguish between the configurations with ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ stacked with K and with ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ in isolation. In other words, ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ cannot absorb K.

We may now ask whether ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ can absorb $K^2$. As before, we consider coincident loops of ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ and $K^2$, and ask whether this configuration can be distinguished from a loop of ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ in isolation. The main difference from before is that the loop of $K^2$ gives a non-trivial result only in the presence of an odd number of units of EM flux. However, as we have argued before, in the presence of an odd number of units of EM flux, the loop of ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ evaluates to zero. Hence ${\mathfrak {I}}_1^0(S_{\{N/2,0\}})$ can absorb $K^2$.