a compact scalar. There are no theta terms since the fermions are massless
(and we assume $\operatorname{rank}(Q)=N_{c}$ so there are no decoupled
photons). Vector-like theories are recovered by taking the Weyl fermions to
come in pairs of opposite charges $Q_{\alpha,a}=-Q_{\alpha^{\prime},a}$ and
packaging them into Dirac fermions.

The gauge anomaly cancellation conditions are $\mathcal{A}_{abc}=0$ and
$\mathcal{A}_{a}=0$, where

$\displaystyle\operatorname{tr}[U(1)_{a}U(1)_{b}U(1)_{c}]\colon\
\mathcal{A}_{abc}$ $\displaystyle=\sum_{\alpha}Q_{\alpha a}Q_{\alpha
b}Q_{\alpha c}$ (5.2) $\displaystyle\operatorname{tr}[U(1)_{a}]\colon\hskip
65.44142pt\mathcal{A}_{a}$ $\displaystyle=\sum_{\alpha}Q_{\alpha a}$

The condition $\mathcal{A}_{a}=0$ guarantees that the theory can be placed on
a curved space consistently. Alternatively, it guarantees that there is no
$2$-group between the Poincaré symmetry and the $1$-form magnetic symmetry.

We insert a monopole with magnetic charges $m_{a}\in\mathbb{Z}$, defined by
removing a point from spacetime and imposing boundary conditions on the gauge
field, schematically of the form

$A_{a}\sim\frac{1}{2}m_{a}(1-\cos\theta)\mathrm{d}\phi\,.$ (5.3)

If we repeat the same analysis as in section 3, we find that each $4d$ Weyl
fermion gives rise, in its lowest angular momentum mode, to a $2d$ fermion
with the same charge $Q_{\alpha a}$ under $U(1)^{N_{c}}$, chirality
$\gamma_{\alpha}^{\star}=\operatorname{sign}(Qm)_{\alpha}$ and $SU(2)_{J}$
representation of dimension $|(Qm)_{\alpha}|$ (i.e., with spin
$j_{0}=\frac{|(Qm)_{\alpha}|-1}{2}$).

Let us check that the $2d$ ’t Hooft anomalies vanish for all the symmetries
preserved by the monopole; this will allow a boundary condition that is
compatible with such symmetries. For example, the quadratic anomaly for the
gauge group $U(1)^{N_{c}}$ is262626Here $\mu$ denotes the $SU(2)_{J}$ weight
and it takes values $\mu=-j_{0},-j_{0}+1,\dots,j_{0}$, with
$j_{0}=\frac{|(Qm)_{\alpha}|-1}{2}$.

$\displaystyle\operatorname{tr}[U(1)_{a}U(1)_{b}]\colon\sum_{\alpha,\mu}Q_{\alpha
a}Q_{\alpha b}\gamma_{\alpha}^{\star}$
$\displaystyle=\sum_{\alpha}|(Qm)_{\alpha}|Q_{\alpha a}Q_{\alpha
b}\operatorname{sign}(Qm)_{\alpha}$ (5.4)
$\displaystyle=\sum_{\alpha,c}Q_{\alpha c}m_{c}Q_{\alpha a}Q_{\alpha b}$
$\displaystyle=\sum_{c}\mathcal{A}_{abc}m_{c}\,,$

which indeed vanishes if $\mathcal{A}_{abc}=0$. The $2d$ gravitational anomaly
is

$\displaystyle c_{L}-c_{R}\colon\sum_{\alpha,\mu}\gamma_{\alpha}^{\star}$
$\displaystyle=\sum_{\alpha}|(Qm)_{\alpha}|\operatorname{sign}(Qm)_{\alpha}$
(5.5) $\displaystyle=\sum_{\alpha,a}Q_{\alpha a}m_{a}$
$\displaystyle=\sum_{a}\mathcal{A}_{a}m_{a}\,,$

which again vanishes if $\mathcal{A}_{a}=0$.

More generally, a $4d$ mixed anomaly of the form $G$-$G$-$U(1)_{a}$ for some
flavor group $G$ induces, after the reduction to the lowest angular momentum
mode, a $G$-$G$ ’t Hooft anomaly in $2d$. If this anomaly is non-zero, the
symmetry $G$ will not be preserved by the boundary. (However, a subgroup of
$G$ may be preserved.) We therefore learn that the scattering experiment will
preserve all symmetries that do not participate in a $2$-group. This is as
expected, since the monopole acts as a source for the magnetic $1$-form
symmetry participating in the $2$-group.

Finally, let us show that $SU(2)_{J}$ is also anomaly-free, which is a
necessary condition for the existence of spherically-symmetric monopoles.
Recall that a $2d$ $G$-$G$ anomaly is measured by the Dynkin index of the
representation under which the fermions transform; for the $d$-dimensional
representation of $SU(2)$, $T(d)=\frac{1}{12}d(d^{2}-1)$. Therefore, the
anomaly of $SU(2)_{J}$ is

$\displaystyle\sum_{\alpha}T(|(Qm)_{\alpha}|)\gamma_{\alpha}^{\star}$
$\displaystyle=\frac{1}{12}\sum_{\alpha}(Qm)_{\alpha}((Qm)_{\alpha}^{2}-1)$
(5.6)
$\displaystyle=\frac{1}{12}\sum_{a,b,c}\mathcal{A}_{abc}m_{a}m_{b}m_{c}-\frac{1}{12}\sum_{a}\mathcal{A}_{a}m_{a}\,,$

which again vanishes if the linear and cubic gauge anomalies vanish.

The rules of the game are, then, as follows. Given an arbitrary $4d$ abelian
gauge theory, a scattering process in the presence of a monopole of magnetic
charge $m_{a}\in\mathbb{Z}$ reduces, at low energies, to an effective $2d$
boundary problem. Each $4d$ Weyl fermion of charge $Q_{\alpha a}$ leads to a
$2d$ free fermion with the same gauge charge, chirality
$\gamma_{\alpha}^{\star}=\operatorname{sign}(Qm)_{\alpha}$, and $SU(2)_{J}$
spin $j_{0}=\frac{|(Qm)_{\alpha}|-1}{2}$. The boundary condition at the
monopole conserves all the symmetries of the original $4d$ theory that do not
participate in a $2$-group. Then, a typical scattering process consists of
sending a left-moving fermion towards the boundary, which will become a right-
moving excitation with the same quantum numbers; such an outgoing state is,
typically, a twist operator that lives at the end of some topological line. In
$4d$, this is interpreted as outgoing radiation that is attached to a $3d$
topological surface. This surface can be invertible or non-invertible,
depending on the details of the theory.

This overall picture is quite similar to the non-chiral example discussed in
the previous sections. That being said, chiral theories do give rise to some
special features. A simple example can be illustrated via the gauge theory
consisting of a single $U(1)$ gauge field and five fermions with charges
$-1,-5,7,8,-9$. As is easily checked, this theory has no cubic or linear gauge
anomalies. The $2d$ reduction, in the presence of a minimally charged
monopole, consists of three right-moving fermions with charges $-1,-5,-9$, and
two left-moving fermions with charges $7,8$. These fermions transform in the
$SU(2)_{J}$ representations $\boldsymbol{1},\boldsymbol{5},\boldsymbol{9}$ and
$\boldsymbol{7},\boldsymbol{8}$, respectively. Consider now what happens if we
send the left-mover with charge $8$ towards the boundary; this fermion
transforms in the $\boldsymbol{8}$ of $SU(2)_{J}$, which is even-dimensional.
On the other hand, all the right-movers transform in odd-dimensional
representations of $SU(2)_{J}$. Therefore, conservation of $SU(2)_{J}$
requires the topological line itself to carry $SU(2)_{J}$ charge. This is only
possible if the line either does not commute with $SU(2)_{J}$, or is non-
invertible, or it carries fermionic zero-modes (recall that invertible lines
without zero-modes never add charge under semi-simple groups, cf. footnote
19). In any case, we see that the out-states in chiral theories can be more
involved than those of vector-like theories. This feature is actually of some
phenomenological relevance. It turns out that certain monopoles of $SU(5)$ GUT
theories lead to a spectrum of $2d$ fermions that cannot conserve the rotation
symmetry unless we attach a topological line that is as in the chiral example
above. We leave the details to future work.

###  5.1 Additional Open Questions

Finally, we highlight a few open questions.

  1. 1.

The approach we took throughout this paper is that the monopole is infinitely
heavy and has an internal gap (for instance a gap to a dyon state or other
possible internal states). It would be interesting to reconsider the problem
when this is not true, for instance, if there are normalizable zero modes on
the ’t Hooft line. This corresponds in the $2d$ reduction to a non-simple
boundary condition!

  2. 2.

We have assumed throughout the paper that monopoles are represented by simple
boundary conditions. This forced the breaking of $\mathbb{Z}_{nN_{f}}$ to
$\mathbb{Z}_{nN_{f}/\gcd(2,nN_{f})}$ by the boundary condition. Instead, a
non-simple boundary condition can preserve the full axial
$\mathbb{Z}_{nN_{f}}$. It would be interesting to understand how these
different scenarios occur in various UV complete theories. A simple example of
a theory that should lead to non-simple monopole boundary conditions is
$SO(3)$ $\mathcal{N}=1$ supersymmetric Yang-Mills theory (see [93, 94] for
nice discussions of the axial symmetry of this theory).

  3. 3.

Is there a similar story for scattering in the presence of axions [95]? What
about in the $\mathbb{CP}^{1}$ NLSM in $3+1$ dimensions [96]? The
$\mathbb{CP}^{1}$ NLSM in $3+1$ dimensions has a $1$-form symmetry and some
non-invertible surfaces, therefore, it would be interesting to see if they may
play a similar role.

  4. 4.

Our discussion did not determine the precise topological surface that comes
out from the scattering gedanken experiment (only the $1$-form symmetry
anomaly of the FQH state). Can it be determined in some UV completions of
QED4?

  5. 5.

Related to the previous point, how do we excite the non-transparent anyons of
the Hall state? What is their physical significance?

  6. 6.

Another interesting generalization is to use fermions of higher electric
charge. A new feature of such theories is that there is an electric
$\mathbb{Z}^{(1)}_{q}$, that mixes with the rest of the symmetries in a non-
trivial way. The reduced $2d$ theory now is a variant of the Schwinger model
with higher charges, which has its own new features [97, 98, 75].

  7. 7.

We have left the question of determining which scattering states have no
centrifugal barrier to the future. It would be very nice to understand that
along with the condensates that are turned on in the monopole vicinity. In the
language of defect conformal field theory this is equivalent to determining
the space of marginal defect operators and the one point functions of bulk
operators. The modes with no centrifugal barrier that we discussed in the text
for $N_{f}>1$ are tilt operators in the language of defect conformal field
theory.

## Acknowledgement

We thank T.D. Brennan, Y. Choi, G. Cuomo, N. Dorey, K. Ohmori, N. Seiberg, A.
Sever, S-H. Shao, M. Watanabe, S. Yankielowicz and Y. Zheng for useful
discussions. ZK is supported in part by the Simons Foundation grant 488657
(Simons Collaboration on the Non-Perturbative Bootstrap) and the BSF grant no.
2018204. DT is supported by the STFC grant ST/L000385/1, the EPSRC grant
EP/V047655/1 “Chiral Gauge Theories: From Strong Coupling to the Standard
Model”, and a Simons Investigator Award. PBS is supported by WPI Initiative,
MEXT, Japan at Kavli IPMU, the University of Tokyo.

##  Appendix A Some technical computations in $\boldsymbol{2d}$ free fermion
CFT

In this appendix we collect some computations regarding free fermion CFTs.

#### Boundary States for the 3450 Model.

In this subsection we sketch the classification of all conformal boundaries
for the 3450 model, i.e., those preserving only Virasoro and $U(1)$ symmetry.
The basic idea is to perform the coset construction by the $U(1)$, yielding a
Virasoro algebra at $c=1$ that must also be preserved by the boundary. Since
the complete classification of boundary states at $c=1$ is known [99, 100],
this leads to the classification for the 3450 model. The results are as
follows: in addition to the $U(1)\times U(1)^{\prime}$\- and $U(1)\times
U(1)^{\prime\prime}$-preserving states described in the main text, there
exists a small family of pathological states, the so-called Janik states
[100], which have continuous spectrum in the open sector and are thus usually
ruled out. Ignoring these states, this justifies our assumption that an extra
${\mathbb{Z}}_{5}$ is preserved by the boundary.

We start by decomposing the partition functions of the left- and right-movers
into characters under the relevant chiral algebras. In our case, these are a
$U(1)_{5}$ current algebra from the preserved $U(1)$ symmetry, and a commuting
$\text{Vir}_{c=1}$ algebra from the coset construction. These have characters

$\chi_{n}^{U(1)_{5}}(\tau,a)=e^{ina}\frac{q^{n^{2}/50}}{\eta(\tau)}\,,\qquad\chi_{h}^{\text{Vir}_{c=1}}(\tau)=\frac{q^{h}}{\eta(\tau)}$
(A.1)

where $a\sim a+2\pi$ is a $U(1)$ fugacity. In terms of these, the partition
functions are

$\displaystyle\mathcal{Z}_{L}(\tau,a)$
$\displaystyle=\sum_{n_{1},n_{2}\in{\mathbb{Z}}}\chi_{3n_{1}+4n_{2}}^{U(1)_{5}}(\tau,a)\chi_{(4n_{1}-3n_{2})^{2}/50}^{\text{Vir}_{c=1}}(\tau)\,,$
(A.2) $\displaystyle\mathcal{Z}_{R}(\tau,a)$
$\displaystyle=\sum_{\tilde{n}_{1},\tilde{n}_{2}\in{\mathbb{Z}}}\chi_{5\tilde{n}_{1}}^{U(1)_{5}}(\tau,a)\chi_{\tilde{n}_{2}^{2}/2}^{\text{Vir}_{c=1}}(\tau)\,.$

All the characters appearing in $\mathcal{Z}_{L,R}$ are distinct irreducibles,
with the sole exception of $\chi_{0}^{\text{Vir}_{c=1}}(\tau)$, which splits
into irreducibles as

$\chi_{0}^{\text{Vir}_{c=1}}(\tau)=\sum_{J=0}^{\infty}\frac{q^{J^{2}}-q^{(J+1)^{2}}}{\eta(\tau)}\,.$
(A.3)

We wish to find all irreducibles $\mathcal{M}_{L}\in\mathcal{Z}_{L}$ and
$\mathcal{M}_{R}\in\mathcal{Z}_{R}$ such that
$\mathcal{M}_{L}\cong\mathcal{M}_{R}$. Ignoring the reducibility of
$\chi_{0}^{\text{Vir}_{c=1}}$ at first, these are precisely the solutions to

$\displaystyle 3n_{1}+4n_{2}$ $\displaystyle=5\tilde{n}_{1}\,,$ (A.4)
$\displaystyle 4n_{1}-3n_{2}$ $\displaystyle=\pm 5\tilde{n}_{2}\,.$

Each solution gives one Ishibashi state preserving $U(1)$ and Virasoro. (This
is essentially the rigorous version of the argument given in section 2.3 based
on conservation of scaling dimension.) However, when $\tilde{n}_{2}=0$,
reducibility of $\chi_{0}^{\text{Vir}_{c=1}}$ implies that we actually obtain
an infinite set of Ishibashi states labeled by $J\geq 0$.

To write down the Ishibashi states, let us split the solutions to (A.4) into
three subsets: $S_{0}$ and $S_{\pm}$, with $S_{0}$ consisting of the solutions
with $\tilde{n}_{2}=0$, and $S_{\pm}$ consisting of the solutions with
$\tilde{n}_{2}\neq 0$ and the $\pm$ choice of sign in (A.4). Then the
Ishibashi states are

$\|(n_{1},n_{2},\tilde{n}_{1},\tilde{n}_{2})\in
S_{\pm}\rangle\\!\rangle\,,\qquad\|(n_{1},n_{2},\tilde{n}_{1},0)\in
S_{0};J\geq 0\rangle\\!\rangle\,.$ (A.5)

The boundary states preserving an additional $U(1)$ symmetry can be written in
terms of these states as

$\displaystyle\frac{1}{\sqrt{5}}|\mathcal{B}_{\pm}(\theta_{1},\theta_{2})\rangle$
$\displaystyle=\sum_{(n_{1},n_{2},\tilde{n}_{1},\tilde{n}_{2})\in
S_{\pm}}e^{i(\tilde{n}_{1}\theta_{1}+\tilde{n}_{2}\theta_{2})}\|(n_{1},n_{2},\tilde{n}_{1},\tilde{n}_{2})\rangle\\!\rangle\
$ (A.6) $\displaystyle+\sum_{(n_{1},n_{2},\tilde{n}_{1},0)\in
S_{0}}\sum_{J=0}^{\infty}e^{i\tilde{n}_{1}\theta_{1}}(\pm
1)^{J}\|(n_{1},n_{2},\tilde{n}_{1},0);J\rangle\\!\rangle\,,$

where the choice of sign corresponds to whether $U(1)^{\prime}$ or
$U(1)^{\prime\prime}$ is the additional preserved symmetry. But these boundary
states do not quite saturate the available Ishibashi states (A.5); the
remaining states take the form of Janik-like states

$\frac{1}{\sqrt{5}}|\mathcal{B}(\theta_{1},x)\rangle=\sum_{(n_{1},n_{2},\tilde{n}_{1},0)\in
S_{0}}\sum_{J=0}^{\infty}e^{i\tilde{n}_{1}\theta_{1}}P_{J}(x)\|(n_{1},n_{2},\tilde{n}_{1},0);J\rangle\\!\rangle\,.$
(A.7)

where $-1\leq x\leq 1$ is a parameter and $P_{\ell}(x)$ is a Legendre
polynomial. However, as shown in [100], these states always exhibit a
continuous spectrum in the open channel, and so are somewhat pathological. We
learn that the complete set of boundary states consists of
$|\mathcal{B}_{\pm}(\theta_{1},\theta_{2})\rangle$ and
$|\mathcal{B}(\theta_{1},x)\rangle$, with only the former being physically
reasonable, and these preserve an additional $U(1)^{\prime}$ or
$U(1)^{\prime\prime}$, as assumed.

#### Scaling dimensions in the 3450 Model.

Here we show that the scattering process (2.28) conserves scaling dimension:

$n_{1}\psi_{1}+n_{2}\psi_{2}\longrightarrow
T^{n_{1}+3n_{2}}+\tilde{n}_{1}\tilde{\psi}_{1}+\tilde{n}_{2}\tilde{\psi}_{2}$
(A.8)

where

$\displaystyle\tilde{n}_{1}$
$\displaystyle=\tfrac{3}{5}n_{1}+\tfrac{4}{5}n_{2}-\tfrac{1}{2}+\bigl{[}\tfrac{1}{2}+\tfrac{2}{5}n_{1}+\tfrac{1}{5}n_{2}\bigr{]}$
(A.9) $\displaystyle\tilde{n}_{2}$
$\displaystyle=\tfrac{4}{5}n_{1}-\tfrac{3}{5}n_{2}-\tfrac{1}{2}+\bigl{[}\tfrac{1}{2}+\tfrac{1}{5}n_{1}-\tfrac{2}{5}n_{2}\bigr{]}$

The in-state carries dimension $(\frac{1}{2}(n_{1}^{2}+n_{2}^{2}),0)$, and
therefore we expect the out-state to carry dimension
$(0,\frac{1}{2}(n_{1}^{2}+n_{2}^{2}))$. In order to check that this is indeed
correct, we proceed as in section 2.1, that is, we study the twisted Hilbert
space defined by the boundary conditions $\psi(\sigma+2\pi)=e^{2\pi
i\eta}\psi(\sigma)$ (see figure 2). Each fermion contributes a vacuum
dimension of

$\tfrac{c}{24}+\sum_{p\in\mathbb{N}+[\eta]}p=\tfrac{c}{24}+\sum_{p\in\mathbb{N}+[\eta]}p^{-s}\biggr{|}_{s\to-1}=\tfrac{c}{24}+\zeta(s,[\eta])\biggr{|}_{s\to-1}=\tfrac{1}{8}-\tfrac{1}{2}[\eta](1-[\eta])$
(A.10)

where in the last equality we used $c=1$ for each complex fermion. Then,
$|T^{k}\rangle$ carries $h_{\text{vac}}=0$ and

$\tilde{h}_{\text{vac}}=\sum_{\text{right}}\tfrac{1}{8}-\tfrac{1}{2}[\tilde{\eta}](1-[\tilde{\eta}])$
(A.11)

We also need to compute the scaling dimension of the oscillators
$\tilde{\psi}^{\tilde{n}}$. Recall that this composite operator is defined via
a subtracted OPE, cf.

$\tilde{\psi}^{\tilde{n}}:=\prod_{j=0}^{\tilde{n}-1}\tilde{\partial}^{j}\tilde{\psi}$
(A.12)

Each derivative carries dimension $\tilde{h}(\tilde{\partial}^{j})=j$, while
the fermion $\tilde{\psi}$ itself carries dimension
$\tilde{h}(\tilde{\psi})=[\operatorname{sign}(\tilde{n})\tilde{\eta}]$ (recall
that negative $\tilde{n}$ means that we use $\tilde{\psi}^{\dagger}$). Thus,
all in all, the oscillators carry scaling dimension

$\tilde{h}(\tilde{\psi}^{\tilde{n}})=\sum_{j=0}^{|\tilde{n}|-1}[\operatorname{sign}(\tilde{n})\tilde{\eta}]+j\equiv\tfrac{1}{2}\tilde{n}^{2}+\tilde{n}([\tilde{\eta}]-\tfrac{1}{2})$
(A.13)

Finally, using (A.11) and (A.13) (with
$\tilde{\eta}_{i}=\frac{1}{2}+\frac{q_{i}k}{5}$ with $q_{i}=3,4$ for
$\tilde{\psi}_{1},\tilde{\psi}_{2}$, respectively) one can easily check that

$\tilde{h}_{\text{vac}}+\tilde{h}(\tilde{\psi}_{1}^{\tilde{n}_{1}})+\tilde{h}(\tilde{\psi}_{2}^{\tilde{n}_{2}})\equiv\frac{1}{2}(n_{1}^{2}+n_{2}^{2})$
(A.14)

and scaling dimension is conserved, as required.

More generally, if we have $N$ fermions charged under $U(1)^{N}$, then the
general scattering experiment is fixed by charge conservation to
$n_{i}X_{i}\longrightarrow n_{i}\mathcal{R}_{ij}\tilde{X}_{j}$. The condition
that scaling dimension is preserved amounts to
$||n||^{2}=||\mathcal{R}n||^{2}$, which holds if and only if $\mathcal{R}$ is
orthogonal, i.e., if $U(1)^{N}$ is anomaly-free. In the fermionic language,
the conservation of scaling dimension is less manifest but can be checked
using the same method as before.

***

Let us also check the conservation of ${\mathbb{Z}}_{5}$ charge in the process
(2.28). Recall that the fermions have charges $q_{i}=0,0,3,4$, so the in-state
is manifestly neutral. The ${\mathbb{Z}}_{5}$ quantum number of the out-state
is the sum $Q_{\text{vac}}(T^{n_{1}+3n_{2}})+3\tilde{n}_{1}+4\tilde{n}_{2}$.
Since $\tilde{n}_{1},\tilde{n}_{2}$ are given in (2.20), we need only compute
the vacuum charge that the endpoint of the line adds:

$\displaystyle Q_{\text{vac}}(T^{n_{1}+3n_{2}})$
$\displaystyle=\sum_{i}q_{i}\times\bigl{(}\tfrac{1}{2}-\bigl{[}\tfrac{1}{2}+\tfrac{1}{5}q_{i}(n_{1}+3n_{2})\bigr{]}\bigr{)}\gamma_{i}^{\star}$
(A.15)
$\displaystyle=5(n_{1}+3n_{2})-3\left\lfloor\tfrac{1}{2}+\tfrac{3}{5}(n_{1}+3n_{2})\right\rfloor-4\left\lfloor\tfrac{1}{2}+\tfrac{4}{5}(n_{1}+3n_{2})\right\rfloor\,.$

Adding the contribution from the local excitations using (2.20) some
straightforward algebra shows that

$Q_{\text{vac}}(T^{n_{1}+3n_{2}})+3\tilde{n}_{1}+4\tilde{n}_{2}=5n_{1}\equiv
0\mod 5\,.$ (A.16)

We have therefore verified by explicit computation that the scattering (2.28)
preserves ${\mathbb{Z}}_{5}$ charge, as indeed it should since
${\mathbb{Z}}_{5}$ sits inside the $U(1)\times U(1)^{\prime}$ symmetry
preserved by the boundary state.

#### Monopole Scattering.

We now show that the process (3.35),

$\psi_{i,\mu}\longrightarrow T^{-2}+\tilde{\psi}_{i,\mu}$ (A.17)

conserves the $\mathbb{Z}_{nN_{f}}$ symmetry that acts as
$\psi\mapsto\psi,\tilde{\psi}\mapsto e^{2\pi i/nN_{f}}\tilde{\psi}$. In the
bosonic picture,

$e^{iX_{i,\mu}}\longrightarrow
e^{i\tilde{X}_{i,\mu}-\frac{2}{nN_{f}}i\sum_{j,\mu^{\prime}}\tilde{X}_{j,\mu^{\prime}}}$
(A.18)

and conservation of the symmetry amounts to checking that the out-state has no
self-monodromy, which is straightforward.

However, in translating this to the fermionic picture, we must take into
account that there is an ambiguity in the definition of the
${\mathbb{Z}}_{nN_{f}}$ charge $q$. Clearly, shifting $q\mapsto q+nN_{f}$
leads to the same transformation for integer vertex operators, but it does not
in general for fractional ones. In particular, the transformation

$e^{-\tfrac{2}{nN_{f}}i\tilde{X}}\mapsto
e^{-\tfrac{2}{nN_{f}}i\tilde{X}-\tfrac{4\pi iq}{(nN_{f})^{2}}}$ (A.19)

leads to different results for $q=1$ and $q=1+nN_{f}$. In this way $T^{-2}$
can be said to act by $\tilde{X}_{i}\mapsto\tilde{X}_{i}+\tfrac{2\pi
q_{i}}{nN_{f}}$ with $q_{i}=-2$ or $q_{i}=nN_{f}-2$. In the outgoing state
(A.18), we take the charges to be $nN_{f}-2$ for $\tilde{X}_{i,\mu}$ and $-2$
for the rest. It is simple to check that this assignment of charges leaves the
out-state invariant. Furthermore, using (2.26), one can check that this
assignment leads to $Q_{\text{vac}}(T^{-2})=-1$. Hence, the out-state has
overall charge $Q(T^{-2}+\tilde{\psi})=-1+1\equiv 0$, i.e., it is
$\mathbb{Z}_{nN_{f}}$-neutral, just like the in-state.

It is important to stress that the assignment of charges being $nN_{f}-2$ for
$\tilde{X}_{i,\mu}$ and $-2$ for the rest means that the line does not act the
same on all fractional vertex operators $e^{i\alpha\tilde{X}}$, but it still
acts as $\tilde{\psi}\mapsto e^{2\pi i/nN_{f}}\tilde{\psi}$ on the local
fermions. This is an artifact of using abelian bosonization.

***

Let us also check that (A.17) conserves scaling dimension. In the bosonic
picture (A.18) this is manifest, but it is less obvious in the fermionic
picture. Using (A.11) one can check that the endpoint of $T^{-2}$ has weight
$\tilde{h}_{\text{vac}}=\frac{2}{nN_{f}}$ while the fermion itself has scaling
dimension $\tilde{h}(\tilde{\psi})=\frac{1}{2}-\frac{2}{nN_{f}}$. Hence, the
scaling dimension of the out-state $T^{-2}+\tilde{\psi}$ is
$\tilde{h}\equiv\frac{1}{2}$, matching the dimension of the in-state.

##  Appendix B Review of some $\boldsymbol{2d}$ Anomalies

Here we collect some known facts about anomalies in $2d$. We always assume
that $c_{L}=c_{R}$ so that there are no purely gravitational anomalies.

First, given a pair of $U(1)$ symmetries, the perturbative
$U(1)$-$U(1)^{\prime}$ anomaly is

$\sum_{i}Q_{i}Q_{i}^{\prime}\gamma_{i}^{\star}$ (B.1)

where $\gamma^{\star}=\pm 1$ is the chirality matrix, and the sum runs over
all fermions in the theory.

Discrete symmetries are a little bit trickier. This is illustrated by the fact
that [101, 102]

$\Omega_{3}^{\text{spin}}(B\mathbb{Z}_{n})=\begin{cases}\mathbb{Z}_{n}&n\text{
odd}\\\ \mathbb{Z}_{2}\times\mathbb{Z}_{2n}&n/2\text{ even}\\\
\mathbb{Z}_{4n}&n/2\text{ odd}\end{cases}$ (B.2)

which classifies anomalies for $\mathbb{Z}_{n}$ symmetries.

For $n$ odd, all anomalies are induced from the embedding
$\mathbb{Z}_{n}\subset U(1)$, namely, they are the mod $n$ reduction of the
perturbative anomaly above. For example, the $\mathbb{Z}_{n}$-$\mathbb{Z}_{n}$
anomaly is

$\sum_{i}q^{2}_{i}\gamma_{i}^{\star}\mod n\qquad\text{(for $n$ odd)}$ (B.3)

where $q_{i}=0,1,\dots,n-1$ denotes the charge of the fermions (in the sense
that, under $k\in\mathbb{Z}_{n}$, they transform as $\psi_{i}\mapsto e^{2\pi
iq_{i}k/n}\psi_{i}$).

For $n$ even, things are not quite so simple. The mixed
$\mathbb{Z}_{n}$-$U(1)$ anomaly is still given by the mod $n$ reduction of the
perturbative anomaly, but the $\mathbb{Z}_{n}$-$\mathbb{Z}_{n}$ anomaly
requires more work. In this case, the induced anomaly is still meaningful, but
it only detects the $\mathbb{Z}_{n}$ subgroup of $\mathbb{Z}_{2n}$ or
$\mathbb{Z}_{4n}$. It turns out that, for even $n$, the answer is

$\sum_{i}q^{2}_{i}\gamma_{i}^{\star}\mod 2n\qquad\text{(for $n$ even)}$ (B.4)

as we show below.

In order to detect the full non-perturbative anomaly, a standard procedure is
to look at the spin of the operators in the $\mathbb{Z}_{n}$ twisted Hilbert
space [44, 103, 104].272727Of course, here we implicitly assume that such
twisted Hilbert space actually exists in the first place. This is indeed the
case when all fermions are Dirac, but an odd number of Majorana-Weyl fermions
do not admit a well-defined, $\mathbb{Z}_{2}$-graded Hilbert space. Let us
focus on the Ramond twisted Hilbert space, defined by

$\psi_{i}(\sigma+2\pi)=e^{2\pi iq_{i}/n}\psi_{i}(\sigma)$ (B.5)

In an anomaly-free theory, an $n$-fold rotation should act trivially on this
Hilbert space. (We look at the R-sector for simplicity but if we were to look
at the NS sector, the claim would be that the $n$-fold rotation should act as
$(-1)^{n}$). An anomaly for $\mathbb{Z}_{n}$ is then detected by exhibiting an
operator whose spin is not an integral multiple of $1/n$, but rather has a
larger denominator. In other words, given that all creation operators have
integral conformal weight, the anomaly of $\mathbb{Z}_{n}$ is measured by
$nE^{\text{R}}_{0}\mod 1$, where $E_{0}^{\text{R}}$ is the vacuum energy of
the twisted Ramond Hilbert space. (As before, if we were to look at the NS
sector, the anomaly would be $n(E_{0}^{\text{NS}}+1/2)\mod 1$.)

This vacuum energy is easy to compute. Each mode with boundary condition
$\psi\mapsto e^{2\pi i\eta}\psi$ contributes

$\sum_{p\in\mathbb{N}+[\eta]}p=\sum_{p\in\mathbb{N}+[\eta]}p^{-s}\biggr{|}_{s\to-1}=\zeta(s,1-[\eta])\biggr{|}_{s\to-1}\equiv-\frac{1}{12}+\frac{1}{2}[\eta](1-[\eta])$
(B.6)

and therefore the total vacuum energy is

$E^{\text{R}}_{0}=\sum_{i}\bigl{(}-\tfrac{1}{12}+\tfrac{1}{2}\bigl{[}q_{i}/n\bigr{]}\bigl{(}1-\bigl{[}q_{i}/n\bigr{]}\bigr{)}\bigr{)}\gamma_{i}^{\star}$
(B.7)

Given that $[\eta]=\eta+\text{integer}$ and $c_{L}=c_{R}$, this expression can
be simplified into

$nE^{\text{R}}_{0}=\frac{1}{2}\sum_{i}q_{i}(1-q_{i}/n)\gamma_{i}^{\star}\mod
1$ (B.8)

The fractional part of this expression then gives the anomaly of
$\mathbb{Z}_{n}$. It is now a simple exercise to show that

$\frac{1}{2}\sum_{i}q_{i}(1-q_{i}/n)\gamma_{i}^{\star}\equiv 0\mod
1\qquad\Longleftrightarrow\qquad\sum_{i}q_{i}^{2}\gamma_{i}^{\star}\equiv
0\mod\gcd(n,2)n$ (B.9)

in agreement with (B.3) and (B.4).

Let us check a few examples. First off, if both left-movers and right-movers
transform with the same charges, $q_{i}=\tilde{q}_{i}$, the anomaly manifestly
vanishes, even the non-perturbative part. We could in fact have guessed this
since in this case it is possible to write down a $\mathbb{Z}_{n}$-preserving
mass term. Likewise, if they transform with opposite charges,
$q_{i}=-\tilde{q}_{i}$, the anomaly is $nE_{0}^{\text{R}}=\sum_{i}q_{i}\mod
1$, which vanishes for integer charges.

For a more interesting example, consider $N$ fermions and a $\mathbb{Z}_{2}$
symmetry where all left-movers transform as $\psi\to-\psi$ and all right-
movers as $\psi\to+\psi$. Then, the anomaly is

$2E^{\text{R}}_{0}=\frac{1}{2}N(1-1/2)\equiv\frac{1}{4}N\mod 1$ (B.10)

which agrees with the known answer [105, 106, 107] (the general result is that
$\nu$ Majorana fermions have anomaly $\nu/8\mod 1$, and our fermions are
Dirac, so $N=2\nu$).

Finally, take $N$ fermions and a $\mathbb{Z}_{N}$ symmetry where all right-
movers transform with charge $+1$, and all left-movers are neutral. The
anomaly reads

$NE^{\text{R}}_{0}=-\frac{1}{2}N(1-1/N)\equiv\frac{1}{2}(1-N)\mod 1$ (B.11)

This vanishes if $N$ is odd, in agreement with the fact that the perturbative
anomaly vanishes, but it is non-zero for $N$ even, showing that for $N$ even
the symmetry is anomalous. On the other hand, for $N$ even, the
$\mathbb{Z}_{N/2}$ subgroup is anomaly-free (to show this we just replace
$1/N\to 2/N$ in the formula above, with result
$(N/2)E_{0}^{\text{R}}=(2-N)/2$, which indeed vanishes mod $1$). We then learn
that only the $\mathbb{Z}_{2}\subset\mathbb{Z}_{N}$ subgroup suffers from non-
perturbative anomalies, while the anomaly of $\mathbb{Z}_{N/2}$ is determined
by the usual perturbative formula.

##  Appendix C Lagrangian for Hall state

In this appendix we collect some properties of the Hall state TQFT
$\mathcal{A}^{N,p}$ that we use throughout the text. This TQFT is labeled by a
pair of integers $N,p$ such that $\gcd(N,p)=1$ [91]. By definition, the fusion
algebra is group-like and isomorphic to $\mathbb{Z}_{N}$; and, furthermore,
the generator of $\mathbb{Z}_{N}$ carries topological spin
$\frac{p}{2N}\mod\frac{1}{2}$. Equivalently, $\mathcal{A}^{N,p}$ is an abelian
TQFT with $1$-form symmetry $\mathbb{Z}_{N}^{(1)}$ and anomaly $p/N\mod 1$.
Note that we regard $\mathcal{A}^{N,p}$ as a spin TQFT. Moreover, it is often
convenient to stack the theory with some invertible TQFT such that the total
central charge vanishes; this can be accomplished by adding factors of
$U(1)_{\pm 1}$ as necessary (which we always leave implicit). When the central
charge vanishes, the defect $\mathcal{D}_{p/N}$ becomes truly topological;
otherwise, its phase depends on the choice of trivialization of the tangent
bundle of its support [108].

Abstractly, the anyons of the Hall state are labeled by an integer
$s\in\\{0,1,\dots,N-1\\}$ and they satisfy fusion rules $s\times
s^{\prime}=(s+s^{\prime}\mod N)$. They carry spin $h_{s}=\frac{ps^{2}}{2N}$
and braid with the generator $s=1$ with phase $e^{-2\pi isp/N}$. For some
purposes, it may be useful to have an explicit Lagrangian for this TQFT. Being
abelian, such theory can always be written as a $U(1)^{q}$ Chern-Simons theory
for some $q$, namely

$\mathcal{L}=\frac{1}{4\pi}a^{t}K\mathrm{d}a+\frac{1}{2\pi}a^{t}v\,\mathrm{d}A$
(C.1)

where $K$ is a $q\times q$ integral symmetric matrix, and $v$ is a $q\times 1$
vector of integers. On the other hand, $A$ is the electromagnetic field
(thought of as a background field here).

The conductivity $p/N$ can be obtained by formally integrating out $a$ using
its equations of motion

$Ka+Av=0$ (C.2)

and plugging the result back to $\mathcal{L}$, cf.

$\mathcal{L}=-\frac{1}{2\pi}(v^{t}K^{-1}v)A\mathrm{d}A$ (C.3)

from where we read the conductivity as $p/N=v^{t}K^{-1}v$. The line that
generates $\mathbb{Z}_{N}^{(1)}$ can be taken as $W=\exp\bigl{(}iv^{t}\int
a\bigr{)}$, and an arbitrary line can be written as $W^{s}$. The spin of
$W^{s}$ is $\frac{1}{2}(sv)^{t}K^{-1}(sv)\equiv\frac{ps^{2}}{2N}$, as above.
Using (C.2), it is easily checked that $W^{s}$ couples to $A$ with coefficient
$sv^{t}K^{-1}v$, namely this anyon carries electromagnetic charge
$sp/N$.282828Note that the lines $s=0,1,\dots,N-1$ all carry fractional
charge, and only the transparent anyon $W^{N}$ has integral charge. Indeed,
transparent anyons describe the worldlines of microscopic fermions.

Although the explicit form of $K$ is seldom needed, it may be useful to know
that a general expression is known (see e.g. [109]). For example, a state with
conductivity $p/N$ can be constructed by taking the $K$-matrix

$K=\begin{pmatrix}+k_{1}&1&0&0&0&\cdots\\\ 1&-k_{2}&1&0&0&\cdots\\\
0&1&+k_{3}&1&0&\cdots\\\ 0&0&1&-k_{4}&1&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{pmatrix}$ (C.4)

where $k_{i}$ are the coefficients of the continued fraction of $p/N$, namely

$\frac{p}{N}=\frac{1}{k_{1}+\frac{1}{k_{2}+\frac{1}{k_{3}+\cdots}}}$ (C.5)

The charge vector is $v=(1,0,\dots)$.

For example, if $K=(N)$ is $1\times 1$ and $v=1$, we find a fractional
coefficient $1/N$, which means that $U(1)_{N}$ realizes the theory
$\mathcal{A}^{N,1}$. On the other hand, if we take $N$ to be an odd integer
and take

$K=\begin{pmatrix}\frac{N+1}{2}&1\\\ 1&2\end{pmatrix}$ (C.6)

and $v=(1\ 0)$, then the fractional coefficient becomes $2/N$, which means
this $K$ matrix realizes $\mathcal{A}^{N,2}$. As a quick check, this matrix
has Smith normal form

$K=\begin{pmatrix}\frac{N+1}{2}&-1\\\ 1&0\end{pmatrix}\begin{pmatrix}1&0\\\
0&N\end{pmatrix}\begin{pmatrix}1&2\\\ 0&1\end{pmatrix}$ (C.7)

which means that the $1$-form symmetry is $\mathbb{Z}_{N}^{(1)}$. Also, a
generic line can be written as $W=(s,0)$, whose spin is $h_{s}=s^{2}/N$, which
indeed confirms that this has $p=2$.

In conclusion, the Hall state for the topological defect $\mathcal{D}_{2/n}$
that we use in the main text admits a Lagrangian description of the form

$\begin{cases}K=\begin{pmatrix}n/2\end{pmatrix}&n\text{ even}\\\\[8.61108pt]
K=\begin{pmatrix}\frac{n+1}{2}&1\\\ 1&2\end{pmatrix}&n\text{ odd}\end{cases}$
(C.8)

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