# Immersed figure-8 annuli and a strong isomorphism conjecture

Bowen Shi  Department of Physics, University of California at San Diego, La
Jolla, CA 92093, USA

###### Abstract

Immersion (i.e., local embedding) is relevant to the physics of topologically
ordered phases through entanglement bootstrap. An annulus can immerse in a
disk or a sphere as a “figure-8”, which cannot be smoothly deformed to an
embedded annulus. We investigate a simple problem: is there an Abelian state
on the figure-8 annulus? We show that if the answer is affirmative, a strong
sense of isomorphism must hold: two homeomorphic immersed regions must have
isomorphic information convex sets, even if they cannot smoothly deform to
each other on the background physical system. We explain why to care about
strong isomorphism and give proof in the context of Abelian anyon theory. We
further discuss a connection between immersed annuli and anyon transportation
in the presence of topological defects. In appendices, we discuss related
problems in broader contexts.

##  I Introduction

It is fascinating that many universal properties of a many-body system, such
as topologically ordered phases, symmetry-protected phases, and quantum
critical points, can be extracted from a single wave function [1, 2, 3, 4, 5,
6, 7, 8, 9, 10, 11]. A further surprise is that in recent years, we have seen
a hope to derive [12, 13] these rules from conditions on a state: entanglement
bootstrap [14, 15, 16, 17] is a framework that aims to do this. In
entanglement bootstrap, we start with a single wave function (or reference
state) on a topologically trivial region, e.g., a ball or a sphere. We impose
some conditions on the wave function as the starting point and derive
(bootstrap) laws of the emergent theory from there. On the way to deriving the
emergent physical laws, we identify information-theoretic concepts (forms of
many-body entanglement), such as information convex sets and the modular
commutator. The goal of entanglement bootstrap has similarities and
distinctions with quantum field theory, bootstrap in other physical contexts,
renormalization group, categorical theory, and solvable models. Some of these
aspects have been discussed in recent works on this subject.

In this work, we investigate a simple aspect of entanglement bootstrap, the
role of immersion. As we shall explain, immersion, i.e., local embedding of a
topological space to another, is natural from the internal theoretical
structure of the entanglement bootstrap. (Here, we are mainly interested in
the case that a topological space is immersed in a background of the same
space dimension. See the immersed “figure-8” annulus in Fig. 1 for an
example.) The importance of immersion is noticed only gradually: see [17] for
a state-of-the-art discussion and [18] for an early application with the
terminology “immersion” unnoticed. We suspect that the full extent of the
benefits we can reap from immersion remains largely hidden beneath the
surface.

To motivate the usage of immersed regions, let us first recall why embedded
regions are of interest. For systems with anyons, it is natural to consider
states locally indistinguishable from the ground state. On closed manifolds,
there can be multiple locally indistinguishable ground states. Similarly, it
is meaningful to think of such states reduced to subsystems. In entanglement
bootstrap, this intuition is made precise with the concept “information convex
set” $\Sigma(\Omega)$. It is a set of density matrices on $\Omega$, locally
indistinguishable from the reference state. Originally, $\Omega$ is considered
to be a subsystem, i.e., a region embedded in the physical system of interest.
Smooth deformations of $\Omega$ must preserve the structure of information
convex sets, according to the isomorphism theorem [14]. Thus, only the
“topology class” of $\Omega$ is of interest.

Immersed regions are a broader class of regions. They are locally embedded but
need not be globally embedded. Because of this, states on immersed regions can
use multiple copies of local Hilbert spaces. For instance, states on the
figure-8 annulus in Fig. 1, use the Hilbert space of the overlapping region
(the region covered by two sheets) twice. Local embedding is sufficient for an
information convex set to be defined [16] because we only needed local
consistency of state in the information convex set with the reference state.
Similarly, information convex sets are shown to be isomorphic under smooth
deformation of the immersed region. To our knowledge, the topological notion
of immersion rarely appeared in physics before. A notable exception is a
sequence of connections observed by Hastings [19] and with Freedman [20] on
invertible phases and the classification of quantum cellular automata.

Figure 1: An immersed “figure-8” annulus ($X_{8}$) on a background
2-dimensional physical system on which we have a reference state $\sigma$. The
background is often taken to be a ball or a sphere. Entanglement bootstrap
axioms A0 and A1 are imposed on bounded radius balls such as the ball (yellow)
illustrated. Entropy combinations $\Delta(B,C)$ and $\Delta(B,C,D)$ are
defined in Eq. (1).

What insight can we gain by considering immersed regions? One powerful
application, in entanglement bootstrap of arbitrary dimensions [17], is the
ability to construct very rich types of immersed punctured manifolds in an
$n$-dimensional sphere. For instance, a punctured torus cannot embed in a
2-sphere, but it can immerse in it. With the trick of healing the puncture, we
obtain states on diverse types of closed manifolds, knowing only a state on a
ball or a sphere! The ability to construct states on closed manifolds from
immersion secretly agrees with that of a non-immersion approach mentioned by
Kitaev in the context of invertible phases.111See [21] at around 1 hour and 35
minutes, where the class of manifolds is referred to as normally framed
manifolds.

Despite the progress mentioned above, the study of immersed regions in
entanglement bootstrap is still in its early stages. Some topological
manifolds can immerse in topologically distinct ways, even in a sphere. A
simple example is the immersed “figure-8” annulus, which cannot be smoothly
deformed to an embedded annulus. One natural question is whether two
homeomorphic regions $\Omega,\Omega^{\prime}$ immersed in the $n$-dimensional
sphere in inequivalent classes, still have isomorphic information convex sets
(Conjecture 6 and 14), which we call the strong isomorphism conjecture. One
may also wonder if smooth deformation (and also “tunneling” Sec. III.1)
between different immersed regions can give interesting automorphism of
information convex sets. These two problems are closely related, as we shall
explain, and both are open problems.

In the main body of this work, we shall focus on the simplest 2-dimensional
(2D) context. The setup of the problem is recalled in Sec. II. We note that
figure-8 $X_{8}$ represents the only nontrivial class of immersed annulus on a
sphere (Table 1). We explain that the strong isomorphism conjecture for 2D
boils down to a simple question about the figure-8 annulus: “is there an
Abelian extreme point in $\Sigma(X_{8})$?” See Sec. III, where the intuition
is that an Abelian sector on figure-8 implies that tunneling a figure-8 (Sec.
III.1) not only changes the immersion class but also acts as an isomorphism
between information convex sets. Such tunneling operations are powerful enough
to relate homeomorphic immersed surfaces in different classes.

Is it true that the figure-8 annulus always detects an Abelian sector? This
problem appears to be nontrivial, and we provide a full solution only for
Abelian theories (Sec. IV), i.e., when all the anyon types detected by the
embedded annulus are Abelian. The more general case is left as a conjecture.

A crucial intuition we used in the proof is that anyons transported along the
figure-8 annulus cannot be permuted. We discuss a thought experiment
“transportation experiment” (Sec. V), that broadens the scope of this idea and
applies it to physical contexts with topological defects. It relates the
property of immersed annulus that thickens the transportation loop to the
question of if anyons are permuted after the transportation. Many ideas and
questions generalize to higher dimensions. In the appendix, we discuss a few
tools useful in thinking about such generalizations.

##  II Background

This section is a review of some necessary background of entanglement
bootstrap. We shall focus on the concept of immersion regions and related
facts. In particular, we explain why there are precisely two classes of
immersed annuli in a sphere; see Table 1.

###  II.1 Setup and axioms

We consider entanglement bootstrap setup in two spatial dimensions (2D). We
consider a reference state $\sigma$ on a 2D surface. Unless stated otherwise,
we consider a reference state on the sphere $\mathbf{S}^{2}$. (We used bold
font to emphasize that a reference state is defined on the sphere.) In this
case, $\sigma=|\psi\rangle\langle\psi|$.222Starting with a ball or a sphere is
equivalent due to the “completion trick” [16]. We shall take $\mathbf{S}^{2}$
to be explicit. The important thing is that we do not need to a state on a
topologically nontrivial manifold. We assume that the total Hilbert space is a
tensor product of on-site Hilbert spaces from sites (or cells) of a coarse-
grained lattice. The local Hilbert space for each coarse-grained site is
finite-dimensional.

The axioms A0 and A1 are imposed on small balls containing a few coarse-
grained lattice sites. The regions are described in Fig. 1, where

$\Delta(B,C,D)\equiv
S_{BC}+S_{CD}-S_{B}-S_{D},\quad\textrm{and}\quad\Delta(B,C)\equiv\Delta(B,C,\emptyset).$
(1)

Here $S_{A}|_{\rho}$ means $S(\rho_{A})=-{\rm Tr}(\rho_{A}\ln\rho_{A})$ is the
von Neumann entropy. The strong subadditivity [22] implies useful inequalities
of similar (but diverse) forms. A particularly useful one to keep in mind is
that, for any $\rho_{ABCD}$, $\Delta(B,C,D)_{\rho}\geq I(A:C|B)_{\rho}\geq 0$.
Here $I(A:C|B)\equiv S_{AB}+S_{BC}-S_{B}-S_{ABC}$ is the conditional mutual
information.

######  Remark.

The axioms A0 and A1 stated that the entropy conditions hold precisely. This
can be a pretty strong requirement even though whole classes of solvable
models satisfy these conditions precisely [23]. Nevertheless, there is
evidence that precisely satisfied axioms are too strong to accommodate chiral
topological orders.333At least, this should be true if the local Hilbert space
on each site is finite-dimensional. One piece of evidence comes from an
unpublished work with Daniel Ranard, Ting-Chun Lin, and Isaac Kim, where a
commuting projector parent Hamiltonian is identified. From the physical
standing point, the right way to appreciate these axioms is that we expect
them to hold more accurately in larger systems. The first appearance of
entropy in physics is in thermodynamics, where the prediction works best for
large systems. We draw the parallel and argue that entanglement bootstrap
predictions are also relevant for systems with approximate axioms. In fact,
for chiral phases, a certain formula initially argued used precise area law
[8] found confirmation with another theoretical approach [24] in the range of
validity of the latter.

The axioms are imposed on an intermediate length scale, which is much smaller
than the total system size and larger than the correlation length and
(microscopic) lattice spacing. This can be viewed as one instance of starting
from the “middle” [25] rather than the “bottom”, where entanglement bootstrap
may be considered as one of those most quantum cases. Indeed, suppose the two
axioms are satisfied on some intermediate scale. In that case, the same
conditions hold on all larger length scales, and they define a renormalization
group fixed point in this sense.

######  Remark.

It is a conjecture that if the violations of the axioms are small enough,
coarse-grain the lattice further will result in more accurate axioms. These
are “threshold conjectures”; see Chapter 11 of [26] for a version and an
explanation of why a threshold $0<\delta<\ln 2$ of violation of A1 is
consistent with current examples from spurious topological entanglement
entropy, non-Abelian anyons, domain walls, and defects.

Although we are mainly interested in systems where the axioms hold, some
concepts and insights developed in entanglement bootstrap, such as the idea of
considering quantum states on immersed regions, may be useful in broader
contexts as well.

###  II.2 Immersed annuli and its classification

Immersed regions are locally embedded regions. For each immersed region, an
information convex set [14, 16] can be defined. An information convex set is a
set of density matrices on a region $\Omega$ that can locally extend and are
locally identical to the reference state $\sigma$. Any locally embedded region
has a “good enough” set of balls to check such consistency conditions, and
thus, an information convex set $\Sigma(\Omega)$ is defined. Embedded regions,
also called “subsystems,” are special cases of immersed regions.(We shall
denote “$\Omega$ is immersed in $Y$” by $\Omega\looparrowright Y$.)

The isomorphism theorem says if two immersed regions
($\Omega,\Omega^{\prime}\looparrowright\mathbf{S}^{2}$) can be smoothly444In
entanglement bootstrap, we have a coarse-grained lattice rather than a smooth
manifold, in which context “smooth deformation” always means applying
elementary steps; see Section 3.3 of [14]. deformed to each other, the
information convex sets associated with them are isomorphic. We denote this as

$\Omega\overset{\mathbf{S}^{2}}{\sim}\Omega^{\prime}\quad\Rightarrow\quad\Sigma(\Omega)\cong\Sigma(\Omega^{\prime}).$
(2)

By isomorphism “$\cong$”, we mean the existence of an invertible map
$\Phi:\Sigma(\Omega)\to\Sigma(\Omega^{\prime})$ that preserves both the
entropy difference and the fidelity between any pair of elements. Generally,
for two immersed regions $\Omega,\Omega^{\prime}\looparrowright Y$, we say
$\Omega\overset{Y}{\sim}\Omega^{\prime}$ if two regions can be smoothly
deformed to each other on $Y$.555An alternative way to say two regions can be
deformed to one another smoothly through immersed configurations is to say
there is a regular homotopy, a term adopted from differential topology. More
precisely, regular homotopy is about maps, and we have modulo-out
automorphisms. This convention can similarly be traced back to math
literature, e.g. “immersed surface” defined in [27].

We say two regions can be smoothly deformed to each other if
$\Omega,\Omega^{\prime}\looparrowright Y$ and $\Omega$ can deform to
$\Omega^{\prime}$ on the background $Y$ by a sequence of elementary steps.

region’s name | embedded annulus | figure-8
---|---|---
symbol | $X$ | $X_{8}$
region |  |
turning number | $1$ | $0$
superselection sectors | ${\mathcal{C}}=\\{1,a,b,\cdots\\}$ | ${\mathcal{C}}_{8}=\\{\mu,\nu,\lambda,\cdots\\}$
Table 1: The two topological classes of immersed annuli in a sphere. One is
represented by the embedded annulus $X$, and another is represented by the
nontrivially immersed “figure-8” $X_{8}$. Labels of superselection sectors, in
the last line, follow from the simplex theorem reviewed in Section II.3.

Below, we shall be interested in immersed annuli. As the isomorphism theorem
implies, only the topological class is of interest. We explain below that
there are only two topological classes of immersed annuli, denoted by $X$ and
$X_{8}$; see Table 1. (We adopt two alternative ways to draw $X_{8}$. One is
as in Fig. 1, where the overlap part is shown as two layers, one on top and it
covers another layer. The second way is as Table 1, where the overlap part is
transparent so that we see both layers. We note that the “top and bottom” in
the first way is introduced for drawing purposes only, and this is not part of
the data to specify immersion. We shall frequently use the first approach as
it simplifies drawing.) We shall be interested in knowing if
$\Sigma(X)\cong\Sigma(X_{8})$, which is not answered by the isomorphism
theorem.

According to Table 1, figure-8 $X_{8}$ is the only nontrivial immersion type
on a sphere. This fact follows from the following argument. First, on the disk
($B^{2}$), each immersed annulus is the thickening of some curve in $B^{2}$,
where the curve has no sharp corners. Therefore, they are classified by the
turning number (modulo the sign). Two regions with different turning numbers
cannot be converted to each other, as is well-known [28]. Some allowed and not
allowed deformations on $B^{2}$ are illustrated as follows:

(3)

Here, each thick curve represents an immersed annulus, $\overset{B^{2}}{\sim}$
means two immersed regions can be smoothly deformed to each other on the
background $B^{2}$, whereas ${\not\sim}$ means impossible to do such
deformation. On the sphere, there is an extra deformation that lets the
annulus sweep over the entire sphere! This process can change the turning
number by an even integer and thus further reduces the classification:

(4)

Consequently, we end up with two equivalence classes of immersed annuli. These
are $X$ and $X_{8}$, as described in Table 1.

The central nontrivial claim of the isomorphism theorem is the _existence_ and
uniqueness of a density matrix on another region that shares some property of
an existing density matrix. This comes from a powerful “merging technique”,
which enters the proof of the isomorphism theorem. The merging technique works
even if we modify the topology of the region. In a later section, we shall
review the merging technique when we use it in important ways.

###  II.3 Superselection sectors and fusion spaces

Information convex sets detect superselection sectors and fusion spaces. In
2D, an embedded annulus $X$ detects the anyon types
${\mathcal{C}}=\\{1,a,b,\cdots\\}$, and here $1$ denotes the vacuum sector.
The 2-hole disk detects the fusion spaces. These follow from the structure
theorems: simplex theorem and Hilbert space theorem in the original paper
[14]. The simplex theorem states that the information convex set $\Sigma(X)$
is a simplex:

$\Sigma(X)=\left\\{\rho_{X}\left|\rho_{X}=\sum_{a\in{\mathcal{C}}}p_{a}\,\rho^{a}_{X}\right.\right\\},$
(5)

where the extreme points $\rho^{a}_{X}$ (which are density matrices) are
mutually orthogonal, that is, the fidelity between $\rho^{a}_{X}$ and
$\rho^{b}_{X}$ is zero when $a\neq b$. $\\{p_{a}\\}$ is a probability
distribution, and the set of extreme point labels ${\mathcal{C}}$ is finite.

######  Remark.

More generally, the simplex theorem works for any sectorizable region.
Sectorizable is a topological condition that says $S$ contains disjoint
subsets $S_{1}$ and $S_{2}$, each of which can deform back to $S$ by a
sequence of elementary extensions. In 2D, immersed annuli and disks are the
only sectorizable regions. A Möbius strip embedded in $RP^{2}$ is not
sectorizable, even though it is the thickening of some loop.

Of importance is the fact that the simplex theorem works equally well for
immersed regions, e.g., figure-8 $X_{8}$. The vacuum state is, however,
defined _only_ for embedded annulus:

  * •

${\mathcal{C}}=\\{1,a,b,\cdots\\}$ is the superselection sectors defined from
embedded annulus $X$. The vacuum “1” is associated with a special extreme
point, the reference state $\rho^{1}_{X}\equiv\sigma_{X}$.

  * •

${\mathcal{C}}_{8}=\\{\mu,\nu,\lambda,\cdots\\}$ is the set of superselection
sectors defined from the figure-8 $X_{8}$. We do not assume the existence of a
vacuum.666The existence of the vacuum may be a logical consequence of the
axioms. In this work, we only attempt to solve a simpler problem.

Here are a few words about antiparticles. ${\mathcal{C}}$ has a natural notion
of antiparticle, that is, for each $a\in{\mathcal{C}}$, there is a unique
$\bar{a}\in{\mathcal{C}}$ such that $a\times\bar{a}=1+\cdots$; one way to
define antiparticle is to deform the annulus $X$ back to itself in a
nontrivial way on the sphere. It is a further interesting question on whether
${\mathcal{C}}_{8}$ has a natural definition of antiparticle, too. It seems
natural to consider the following automorphism of $\Sigma(X_{8})$:

(6)

A more careful (and broader) discussion is postponed in Appendix B.

A general immersed region ($Y$) can detect a fusion space $\mathbb{V}_{I}(Y)$.
Let $\partial Y$ be the thickened boundary of $Y$. In general, $\partial Y$
can have multiple connected components, and each component is an annulus,
either an embedded annulus (like $X$) or a nontrivially immersed annulus (like
$X_{8}$). The label $I$ is then a collection of ordered labels from
${\mathcal{C}}$ and ${\mathcal{C}}_{8}$. Two instances are shown in Fig. 2,
where $I=(a,b,c)$ for Fig. 2(a) and $I=(\mu,a,\nu)$ for Fig. 2(b). Once we
specify the label $I$, we get a convex subset $\Sigma_{I}(Y)$ isomorphic to
the state space (i.e., the space of density matrices) of a finite-dimensional
fusion space $\mathbb{V}_{I}$:

$\Sigma_{I}(Y)\cong{\mathcal{S}}(\mathbb{V}_{I}).$ (7)

This is the main content of the Hilbert space theorem. Furthermore, the
thickened boundary $\partial Y$ is sectorizable, and it is useful to denote
the set of superselection sectors as ${\mathcal{C}}_{\partial Y}$ and note
$I\in{\mathcal{C}}_{\partial Y}$.

Figure 2: Examples of fusion spaces. (a) A region that detects
$\mathbb{V}_{ab}^{c}$, and (b) a region that detects $\mathbb{V}_{\mu
a}^{\nu}$. Here, $a,b,c\in{\mathcal{C}}$ and $\mu,\nu\in{\mathcal{C}}_{8}$.

###  II.4 Quantum dimensions

We give a definition of quantum dimension for superselection sectors detected
by immersed annuli. (A generalization into arbitrary spatial dimensions is
presented in Appendix A.) Formally, the definition is as follows:

######  Definition 1 (Quantum dimension).

Let $Z$ be an immersed annulus. Let $Z=BCD$, where $BD=\partial Z$ is the
thickened boundary and $C=Z\setminus\partial Z$ is the interior; $B$ and $D$
each is a union of two disks, as illustrated in Fig. 3. For any extreme point
$\rho^{h}_{Z}\in\Sigma(Z)$,

$d_{h}\equiv\exp{\left(\frac{\Delta(B,C,D)_{\rho^{h}}}{4}\right)}$ (8)

is the quantum dimension of the superselection sector $h$.

Figure 3: The partition of an immersed annulus $Z=BCD$ used in defining
quantum dimension (Definition 1). We first map the immersed annulus to a
topological annulus by some homeomorphism indicated by the arrow. We then
partition the topological annulus as illustrated.

Immediately following this definition, we have

  1. 1.

$d_{h}\geq 1$ for any $h\in{\mathcal{C}}_{Z}$, where ${\mathcal{C}}_{Z}$ is
the set of superselection sectors characterized by $\Sigma(Z)$. This follows
immediately from strong subadditivity $\Delta(B,C,D)\geq 0$. Note that
${\mathcal{C}}_{Z}={\mathcal{C}}$ if $Z\overset{\mathbf{S}^{2}}{\sim}X$ and
${\mathcal{C}}_{Z}={\mathcal{C}}_{8}$ if
$Z\overset{\mathbf{S}^{2}}{\sim}X_{8}$.

  2. 2.

Importantly, Definition 1 does not require the knowledge (or existence) of a
vacuum state on $Z$. On the other hand, for the embedded annulus $X$, where
the vacuum is defined, Definition 1 is equivalent to another definition
$d_{a}\equiv\exp{(\frac{S(\rho^{a}_{X})-S(\rho^{1}_{X})}{2})}$.

  3. 3.

Definition 1 works even if we relax the topology of the region that defines
the reference state. This general setup allows topological defects and is
relevant to some discussion later in Section V and in Appendix B.

######  Definition 2 (Abelian sector).

A superselection sector $h$ of an immersed annulus is Abelian if $d_{h}$ in
Definition 1 gives $d_{h}=1$.

######  Definition 3 (Abelian theory).

A 2D theory is Abelian if $d_{a}=1,\forall a\in{\mathcal{C}}$.

Note that the definition of Abelian theory only puts a requirement on embedded
annuli. For various reasons, we will need the following lemma.

######  Lemma 4 (Abelian sector criterion).

Let $\rho_{Z}\in\Sigma(Z)$, where $Z$ is an immersed annulus, the following
are equivalent:

  * 0.

$\rho_{Z}$ is an extreme point of $\Sigma(Z)$ associated with an Abelian
sector.

  * 1.

$\Delta(B,C,D)_{\rho}=0$ for the partition in Fig. 3.

  * 2.

$\Delta(B,C,D)_{\rho}=0$ for the partition in Fig. 4(a).

  * 3.

$\Delta(B,C,D)_{\rho}=0$ for the partition in Fig. 4(b).

  * 4.

$I(A:C|B)_{\rho}=0$ for the partition in Fig. 4(c).

According to Lemma 4, any one of the conditions 1, 2, 3, and 4 can be used to
determine if a state in $\Sigma(Z)$ is an extreme point associated with
Abelian superselection sector.

Figure 4:  As with Fig. 3, we mapped the immersed annulus to the topological
annulus by some (fixed) homeomorphism. Three distinct partitions of the
annulus are shown in (a), (b), and (c).

###### Proof.

First, suppose that $\rho_{Z}$ is an extreme point of $\Sigma(Z)$. According
to the definition of quantum dimension, $\Delta(B,C,D)_{\rho}=0$ for the
partition in Fig. 3. It follows from Lemma 12 in Appendix A that any one of
conditions 1, 2, 3, 4 implies the rest of them. (Lemma 12 in the appendix is a
more general statement about the equivalence of definitions for quantum
dimension, works beyond 2D setups.) Second, it is easy to see that none of 1,
2, 3, 4 is true if $\rho_{Z}=\sum_{h}p_{h}\,\rho^{h}_{Z}$ is a non-extreme
point. (For non-extreme points, the probability distribution $\\{p_{h}\\}$ has
$p_{h}<1$ for any $h\in{\mathcal{C}}_{Z}$.) This is because a non-extreme
point has contribution $\sum_{h}p_{h}\ln\left(\frac{1}{p_{h}}\right)$ added to
$\Delta$s and $I$ in conditions 1, 2, 3, 4. This contribution is always
positive for non-extreme points. These two considerations derive Lemma 4. ∎

##  III Fun with figure-8

We have learned that the “figure-8” region $X_{8}$ represents the only
nontrivial class of immersed annulus in the sphere (Table 1). It is arguably
the simplest nontrivial immersed region and is fun to play with. The set of
superselection sectors detected by $X_{8}$ are
${\mathcal{C}}_{8}=\\{\mu,\nu,\cdots\\}$. Throughout this section, we assume a
reference state $\sigma=|\psi\rangle\langle\psi|$ on a sphere
$\mathbf{S}^{2}$, and the axioms are satisfied everywhere on the reference
state. Also, recall that $X$ represents the embedded annulus, which can detect
the anyon types ${\mathcal{C}}=\\{1,a,b,\cdots\\}$.

The seemingly simple figure-8 region ($X_{8}$) already has interesting
puzzles. The simplest one is on the existence of an Abelian sector in
${\mathcal{C}}_{8}$:

######  Conjecture 5 (Abelian sector on figure-8).

$\exists\mu\in{\mathcal{C}}_{8}$, such that $d_{\mu}=1$.

We emphasize that the conjecture is made in the context where the reference
state is defined on sphere $\mathbf{S}^{2}$. An alternative version of the
conjecture requires the reference state $\sigma$ defined on a disk. These two
versions are equivalent due to the ability to “complete” the disk into a
sphere. One may wonder if it is possible to relax the support of the reference
state further. This is impossible. If the reference state is defined on an
annulus, there may not be an Abelian sector, even for an embedded annulus;
this is due to topological defects, see Section V.1. We shall prove Conjecture
5 for Abelian theory in Section IV.

Another natural question about figure-8 is whether

$\Sigma(X_{8})\cong\Sigma(X).$ (9)

The weaker statement $|{\mathcal{C}}_{8}|=|{\mathcal{C}}|$, i.e., the number
of extreme points of $\Sigma(X_{8})$ equals the number of extreme points of
$\Sigma(X)$ is true; we will discuss this in Section V. If the two information
convex sets $\Sigma(X_{8})$ and $\Sigma(X)$ are isomorphic in the most natural
manner, we expect an invertible map between them that preserves the quantum
dimensions. Then, in addition to $|{\mathcal{C}}|=|{\mathcal{C}}_{8}|$, we
should have an Abelian sector of ${\mathcal{C}}_{8}$ that is mapped to the
vacuum $1\in{\mathcal{C}}$. This is one way to motivate Conjecture 5. (It is a
further question whether there is a “canonical vacuum” on
${\mathcal{C}}_{8}$.)

An intriguing and nontrivial observation, which we shall make later in this
section, is that Conjecture 5 not only implies the seemingly stronger
$\Sigma(X_{8})\cong\Sigma(X)$ but also implies a very powerful “strong
isomorphism”:

######  Conjecture 6 (2D strong isomorphism conjecture).

$\Omega,\Omega^{\prime}\looparrowright\mathbf{S}^{2}\textrm{ are
homeomorphic}\quad\Rightarrow\quad\Sigma(\Omega)\cong\Sigma(\Omega^{\prime}).$
(10)

Importantly, we only required that $\Omega$ and $\Omega^{\prime}$ are
homeomorphic, and this is strictly weaker than
$\Omega\overset{\mathbf{S}^{2}}{\sim}\Omega^{\prime}$. (The weaker statement:
$\Omega\overset{\mathbf{S}^{2}}{\sim}\Omega^{\prime}$ implies
$\Sigma(\Omega)\cong\Sigma(\Omega^{\prime})$ is a known fact, from isomorphism
theorem.) This motivates the name “strong isomorphism conjecture”. As is
evident, Eq. (9) is implied by the strong isomorphism conjecture. It is
straightforward to generalize Conjecture 6 to arbitrary spatial dimensions;
see Appendix C. As a reminder, $\cong$ means there exists an isomorphism. It
does not imply a “canonical” isomorphism. $\cong$ preserves the entropy
difference and fidelity between two elements.

###  III.1 Tunneling trick

We introduce a tunneling trick. The essential observation is that tunneling a
figure-8 through a strip can add a twist to the strip. If the “figure-8”
region carries an Abelian sector, this process generates an isomorphism
between information convex sets associated with homeomorphic topological
regions in possibly distinct regular homotopy classes. We first explain the
tunneling process at a topological level. After that, we discuss the details
of entanglement bootstrap and explain what tunneling does on the quantum state
(density matrices) and why an Abelian sector in ${\mathcal{C}}_{8}$ plays a
role.

Figure 5: The topological detail of tunneling a figure-8 $X_{8}$ through a
strip. The strip can be thought of as a “local piece” (in fact, a 1-handle) of
some larger immersed region.

At the topological level, the tunneling process contains a few steps. This is
illustrated in Fig. 5. In the first step, we attach $X_{8}$ to the strip by a
small disk on the right. In the second and third steps, we deform the region
such that $X_{8}$ disappears from the right and appears at the left of the
strip, still attached. In the fourth step, we detach $X_{8}$. By the end of
the tunneling process, a twist is added to the strip. It is instructive to
compare the tunneling of $X_{8}$ with the tunneling of an embedded annulus
$X$:

(11)

Tunneling an embedded annulus does not add any twist to the strip. The
difference in the turning number between $X_{8}$ and $X$ matters, and thus
$X_{8}$ is of importance here.

What does the tunneling process in Fig. 5 mean in entanglement bootstrap? It
represents a transformation of the quantum state (density matrix) on the
region containing the strip. Let the region containing the original strip be
$A$ and the region containing the strip after tunneling be $A^{\prime}$, (see
Fig. 6). Suppose, the state on “figure-8” $C$ is an extreme point
$\rho^{\mu}_{C}$, with $\mu\in{\mathcal{C}}_{8}$. The transformation, as we
shall discuss, must be a quantum channel that depends on $\mu$:

$\Phi(\mu):\quad\Sigma(A)\to\Sigma(A^{\prime}).$ (12)

The quantum channel $\Phi(\mu)$ can be written as a product of three quantum
channels

$\Phi(\mu)=\Gamma^{\rm det}\circ\Phi^{\rm deform}_{ABC\to
A^{\prime}B^{\prime}C^{\prime}}\circ\Gamma^{\rm att},$ (13)

where the attaching map $\Gamma^{\rm
att}:\Sigma(A)\to\Sigma_{[\rho_{C}^{\mu}]}(ABC)$ is a merging process, as we
explain in the next paragraph. $\Sigma_{[\rho_{C}^{\mu}]}(ABC)$ is the subset
of $\Sigma(ABC)$ with the constraint that the state reduces to $C$ gives
$\rho^{\mu}_{C}$. The deformation map $\Phi^{\rm deform}_{ABC\to
A^{\prime}B^{\prime}C^{\prime}}:\Sigma_{[\rho_{C}^{\mu}]}(ABC)\to\Sigma_{[\rho_{C^{\prime}}^{\mu}]}(A^{\prime}B^{\prime}C^{\prime})$
is associated with the deformation (steps 2 and 3 of Fig. 5). This channel is
an isomorphism (by the isomorphism theorem). The detaching map $\Gamma^{\rm
det}:\Sigma_{[\rho_{C^{\prime}}^{\mu}]}(A^{\prime}B^{\prime}C^{\prime})\to\Sigma(A^{\prime})$
is the partial trace ${\rm Tr}_{B^{\prime}C^{\prime}}$. In summary, the three
quantum channels are

$\displaystyle\Gamma^{\rm att}:$
$\displaystyle\quad\Sigma(A)\to\Sigma_{[\rho_{C}^{\mu}]}(ABC),\quad\qquad\qquad\,\textrm{step
1 of Fig.~{}\ref{fig:tunneling-main}}$ (14) $\displaystyle\Phi^{\rm
deform}_{ABC\to A^{\prime}B^{\prime}C^{\prime}}:$
$\displaystyle\quad\Sigma_{[\rho_{C}^{\mu}]}(ABC)\to\Sigma_{[\rho_{C^{\prime}}^{\mu}]}(A^{\prime}B^{\prime}C^{\prime}),\qquad\textrm{step
2 and 3 of Fig.~{}\ref{fig:tunneling-main}}$ $\displaystyle\Gamma^{\rm det}:$
$\displaystyle\quad\Sigma_{[\rho_{C^{\prime}}^{\mu}]}(A^{\prime}B^{\prime}C^{\prime})\to\Sigma(A^{\prime}),\qquad\qquad\,\textrm{step
4 of Fig.~{}\ref{fig:tunneling-main}}$

In fact, any one of them is a sequence of Petz maps [29] and partial traces.

Figure 6: Analyzing the attachment (detachment) of an immersed figure-8
annulus $C$ ($C^{\prime}$). Only part of $A$, $A^{\prime}$ are shown. (a) A
partition for merging. (b) and (c) have special usage when the state on
“figure-8” is an Abelian extreme point.

The attachment map $\Gamma^{\rm att}$ is defined as follows. Consider an
arbitrary state $\lambda_{A}\in\Sigma(A)$ and an arbitrary extreme point
$\rho_{C}^{\mu}\in\Sigma(C)$, with $\mu\in{\mathcal{C}}_{8}$; the partitions
are those in Fig. 6(a). First, we extend $\lambda_{A}$ to
$\lambda_{AB}\in\Sigma(AB)$ and extend $\rho^{\mu}_{C}$ to
$\rho^{\mu}_{BC}\in\Sigma(BC)$. Such an extension is unique because we enlarge
the regions smoothly. $B=B_{1}B_{2}$. The two states obey the conditions for
merging:

$I(A:B_{2}|B_{1})_{\lambda}=0,\quad I(C:B_{1}|B_{2})_{\rho^{\mu}}=0,\quad{\rm
Tr}_{A}\,\lambda_{AB}={\rm Tr}_{C}\,\rho^{\mu}_{BC}.$ (15)

According to the merging lemma [30, 14], there is a unique state $\tau_{ABC}$
satisfying (i) $\tau_{ABC}$ reduces to $\lambda_{AB}$ on $AB$ and reduces to
$\rho^{\mu}_{BC}$ on $BC$, (ii) $I(A:C|B)_{\tau}=0$. Then, by the merging
theorem [14, 15], the state $\tau_{ABC}$ is in the information convex set
$\Sigma(ABC)$, and in fact it is in the convex subset
$\Sigma_{[\rho^{\mu}_{C}]}(ABC)$. The attachment map $\Gamma^{\rm att}$ is
defined such that $\tau_{ABC}=\Gamma^{\rm att}(\lambda_{A})$. Importantly,
$\Gamma^{\rm att}$ is a sequence of Petz maps acting in a local neighborhood
of the strip. It depends on $\mu$ through merging, and it is independent of
$\lambda_{A}$.

The discussion of the attachment map so far is for general
$\mu\in{\mathcal{C}}_{8}$. If $\mu$ is non-Abelian, the map $\Gamma^{\rm
att}:\Sigma(A)\to\Sigma_{[\rho^{\mu}_{C}]}(ABC)$ may not be an isomorphism.
(The same is true already for tunneling an embedded annulus in the context of
Eq. (11).) The important thing, as we shall explain, is that if
$\mu\in{\mathcal{C}}_{8}$ is Abelian, $\Gamma^{\rm att}$ and $\Gamma^{\rm
det}$ are isomorphisms. The special property of Abelian sectors relevant to
$\Gamma^{\rm att}$ is that for any Abelian $\mu\in{\mathcal{C}}_{8}$,

$I(A:C|B)_{\alpha}=0,\quad\forall\alpha\in\Sigma_{[\rho^{\mu}_{C}]}(ABC).$
(16)

The derivation of Eq. (16) is as follows. We first extend the state
$\alpha_{ABC}\in\Sigma_{[\rho^{\mu}_{C}]}(ABC)$ to a state in $\Sigma(ABCD)$,
as in Fig. 6(b). If $\mu$ is Abelian, by Lemma 4, we have
$\Delta(B,C,D)_{\rho^{\mu}}=0$. Then by strong subadditivity,
$I(A:C|B)_{\alpha}\leq\Delta(B,C,D)_{\rho^{\mu}}=0$. Because of its quantum
Markov state structure, $\alpha_{ABC}$ can be recovered from $\alpha_{AB}$ by
a quantum channel independent of $\alpha_{AB}$ (but depend on $\rho^{\mu}$).
The fact that this $\alpha$-independent quantum channel reverses the partial
trace ${\rm Tr}_{C}$ implies that $\Gamma^{\rm att}$ is an isomorphism. For
the same reason, the detachment map $\Gamma^{\rm det}$ is an isomorphism too.
(The state exists by merging. The reversibility follows from
$I(A^{\prime}:C^{\prime}|B^{\prime})=0$.)

In summary, tunneling a figure-8 region (Fig. 5) induces a quantum channel of
the form Eq. (12). The channel $\Phi(\mu)$ is an isomorphism if
$\mu\in{\mathcal{C}}_{8}$ is Abelian.

As an application, we see that $\Sigma(X_{8})\cong\Sigma(X)$ if
$\exists\mu\in{\mathcal{C}}_{8}$ such that $d_{\mu}=1$, (that is if Conjecture
5 holds). This is because we can tunnel a figure-8 through $X$ and turn $X$
into a figure-8 $X_{8}$. This result is summarized as follows:

(17)

Another way to illustrate this result uses the idea of handle decomposition;
see e.g., Chapter 4 of [31]. An annulus is the union of a 0-handle with a
1-handle. The difference between the embedded annulus $X$ and the immersed
“figure-8” $X_{8}$ is the absence or existence of a twist on the 1-handle. We
thus redraw (17) as:

(18)

Here, a 0-handle is a disk (the round part of the figure), and a 1-handle is
the strip (thick lines in the figure). Both 0-handles and 1-handles are
topological balls. The difference is the way they attach. This way of
illustration will be convenient in the following.

###  III.2 Abelian sector on “figure-8” implies 2D strong isomorphism

Now, we are in the position to explain why if the figure-8 annulus has an
Abelian sector (i.e. if Conjecture 5 is true), then the strong isomorphism
conjecture (Conjecture 6) holds. This uses the tunneling trick discussed in
Section III.1. We explain why the trick is general enough.

In the remainder of the section, we assume the existence of
$\mu\in{\mathcal{C}}_{8}$, such that $d_{\mu}=1$. The proof of strong
isomorphism of an arbitrary region reduces to the study of connected regions
pretty straightforwardly. Thus, we only consider connected regions
$\Omega\looparrowright\mathbf{S}^{2}$ below. Because $\Omega$ is connected and
is immersed in $\mathbf{S}^{2}$, it is either the sphere itself or it is an
orientable surface with boundaries. Nothing needs to be proved for the case
$\Omega$ is the sphere itself, and thus we consider the latter case. An
orientable surface with boundaries can be written as a connected sum of $k\geq
0$ tori777The $k=0$ case, in our convention, is a sphere with $m$ holes.
($\\#k\,T^{2}$) with $m\geq 1$ holes; see, e.g. [32]. Thus, such an $\Omega$
is a _boundary connected sum_ of $k$ punctured tori $T^{2}\setminus B^{2}$ and
$m-1$ annuli.

Figure 7: Tunneling trick applied to a connected region immersed in the
sphere.

Objects participating in the boundary connected sum can slide around each
other since they are connected by 1-handles. (Permutation by sliding works as
deformations of immersed regions.) It follows that we can smoothly deform
$\Omega$ on $\mathbf{S}^{2}$ such that, it looks like the left figure of Fig.
7. It is a 0-handle with $2k+m-1$ 1-handles attached in the order (around the
0-handle) indicated in the figure. Some of the 1-handles have twists. Upon the
removal of possible twists, $\Omega$ becomes $\Omega^{\star}$ in the right
figure of Fig. 7. In $\Omega^{\star}$, the 1-handles form two groups: $(m-1)$
of them are separated from others (dark brown), and there are $k$ pairs of
1-handles (dark blue), the two 1-handles in each pair “cross” each other.
Importantly, in both figures in Fig. 7, we can arrange the 1-handles in the
same order, circling around the 0-handle (the chunky gray disk), thanks to the
ability to slide the components around each other.

Thus, for any connected $\Omega\looparrowright\mathbf{S}^{2}$, we can find a
“standard” immersed region $\Omega^{\star}\looparrowright\mathbf{S}^{2}$ in
the homeomorphism class of $\Omega$ such that we can convert $\Omega$ to
$\Omega^{\star}$ by the tunneling trick. Then, suppose there is an Abelian
sector on “figure-8”, the tunneling trick implies the isomorphism

$\Sigma(\Omega)\cong\Sigma(\Omega^{\star}).$ (19)

Then, consider another region $\Omega^{\prime}\looparrowright\mathbf{S}^{2}$
in the homeomorphism class of $\Omega$, we can do the same trick and show
$\Sigma(\Omega^{\prime})\cong\Sigma(\Omega^{\star})$. It follows that
$\Sigma(\Omega)\cong\Sigma(\Omega^{\prime})$. This completes the argument.

##  IV Proofs in the context of 2D Abelian theory

We can derive a concrete result (Proposition 7) for Abelian theories in 2D. It
implies that Conjecture 5 and Conjecture 6 are true for Abelian theories.

######  Proposition 7.

If $d_{a}=1,\forall a\in{\mathcal{C}}$, then
$d_{\mu}=1,\forall\mu\in{\mathcal{C}}_{8}$.

The proof uses Fig. 8, where the key idea is that anyon transported along the
“figure-8” does not change its type. Same with Section III, we would like to
think of the figure-8 region in Fig. 8 to be embedded in either a ball or a
sphere. However, it turns out that the argument only needs a certain property
of the neighborhood of the figure-8 region. This is advantageous, and the same
idea works in the transportation experiment in the next section.

Figure 8: Operations on an extreme point $\rho^{\mu}_{X_{8}}\in\Sigma(X_{8})$,
where $X_{8}$ is an immersed figure-8 and $\mu\in{\mathcal{C}}_{8}$. These are
useful in proving Proposition 7, where we require $d_{a}=1,\forall
a\in{\mathcal{C}}$. The end result of the analysis is a useful condition
$I(A:E|BCD)_{\rho^{\mu}}=0$ on the figure we redraw on the right; only part of
the annuli $A$ and $E$ are shown.

###### Proof.

We consider a sequence of operations applying to a state on “figure-8” $X_{8}$
carrying an arbitrary $\mu\in{\mathcal{C}}_{8}$; see Fig. 8. We denote this
extreme point of $\Sigma(X_{8})$ as $\rho^{\mu}_{X_{8}}$. We draw $X_{8}$ in
such a way that part of the “figure-8” annulus is blown up. The region is like
a “watch with a twisted belt”. We shall explain the steps in Fig. 8 and the
fact that they imply a useful condition Eq. (23) on the $\rho^{\mu}_{X_{8}}$.
Below are the details.

In Step 1, we take a partial trace to create a hole in the interior of
$X_{8}$. We call the resulting region $X_{8}^{\circ}$. The hole carries the
vacuum sector $1\in{\mathcal{C}}$. The two boundaries of the original figure-8
shape carry identical superselection sectors $\mu\in{\mathcal{C}}_{8}$. In
Step 2, we merge a disk ($YZ$) to the hole. The resulting region
$X_{8}^{\circ}YZ$ is in a state

$\tilde{\rho}^{\mu}_{X_{8}^{\circ}YZ}=\frac{1}{|{\mathcal{C}}|}\sum_{a\in{\mathcal{C}}}\rho^{a\bar{a}\mu\mu}_{X_{8}^{\circ}YZ},\qquad
I(X_{8}^{\circ}:Z|Y)_{\tilde{\rho}^{\mu}}=0,$ (20)

where the state $\tilde{\rho}^{\mu}_{X_{8}^{\circ}YZ}$ is an element in the
information convex set $\Sigma(X_{8}^{\circ}YZ)$ and it is the indicated
convex combination of extreme points $\rho^{a\bar{a}\mu\mu}_{X_{8}^{\circ}YZ}$
of $\Sigma(X_{8}^{\circ}YZ)$; each state
$\rho^{a\bar{a}\mu\mu}_{X_{8}^{\circ}YZ}$ has the superselection sectors
indicated on respective boundaries in the figure (after step 2).

In step 3, we smoothly deform the region “figure-8-with-2-holes” back to
itself. The deformation is such that the hole with anyon sector $a$ is
transported along the figure-8 by almost a complete loop (initially moving
upward), and the hole that carries $\bar{a}$ is also transported upwards
slightly. Thus regions $Y$ and $Z$ are stretched during step 3 and become thin
(and long) immersed strips $Y^{\prime}$ and $Z^{\prime}$. After step 3,
$\bar{a}$ is in the upper hole and $a$ is in the lower hole of
“figure-8-with-2-holes”.

Although it might sound an obvious comment, we emphasize that transporting an
anyon along the immersed figure-8 (embedded in a ball or a sphere) cannot
change anyon type.888This is a special instance of Lemma 4.3 of [14], which
says no matter how we transport an annulus within a fixed large disk (keep the
annulus embedded during the process) and come back to its original position,
anyons cannot be permuted. It is because of this important property that we
can conclude the argument. Suppose $X_{8}^{\circ}$, $Y$ and $Z$ are deformed
into ${X_{8}^{\circ}}^{\prime}$, $Y^{\prime}$ and $Z^{\prime}$ respectively
after step 3. Then the state after step 3 is

$\tilde{\rho}^{\prime\mu}_{{X_{8}^{\circ}}^{\prime}Y^{\prime}Z^{\prime}}=\frac{1}{|{\mathcal{C}}|}\sum_{a\in{\mathcal{C}}}\rho^{a\bar{a}\mu\mu}_{{X_{8}^{\circ}}^{\prime}Y^{\prime}Z^{\prime}},\qquad
I(X^{\prime}:Z^{\prime}|Y^{\prime})_{\tilde{\rho}^{\prime\mu}}=0.$ (21)

In fact,
$\rho^{a\bar{a}\mu\mu}_{{X_{8}^{\circ}}^{\prime}Y^{\prime}Z^{\prime}}=\rho^{a\bar{a}\mu\mu}_{{X_{8}^{\circ}}YZ}$
because we deformed the ”figure-8-with-2-holes” back to itself. (The
uniqueness of the state with labeling $a\bar{a}\mu\mu$ follows from the fact
that $a,\bar{a}\in{\mathcal{C}}$ are Abelian.) Thus
$\tilde{\rho}^{\prime\mu}=\tilde{\rho}^{\mu}$. The vanishing of conditional
mutual information in (21) follows from that in (20) and the fact that smooth
deformation of the regions preserves the conditional mutual information.

In Step 4, we trace out the horizontal strip, and this effectively brings $a$
and $\bar{a}$ together. According to the Abelian fusion rule
$a\times\bar{a}=1$ (derived in entanglement bootstrap in a self-contained way)
the fusion outcome must be in the vacuum sector $1\in{\mathcal{C}}$. The
resulting state on $X_{8}^{\circ}=ABCDE$ is nothing but the reduced density
matrix of $\rho^{\mu}_{X_{8}}$! (Step 5 is intended to be a dummy step
indicating that it is possible to fill in the hole and get a state
$\rho^{\mu}_{X_{8}}$. This is precisely the content of the previous sentence.)
Let us relabel the regions as the figure we redraw on the right, i.e.,
${X_{8}^{\circ}}^{\prime}\supset AE$, $Y^{\prime}=BD$ and $Z^{\prime}=C$. By
the strong subadditivity

$I(AE:C|BD)_{\rho^{\mu}}\leq
I({X_{8}^{\circ}}^{\prime}:Z^{\prime}|Y^{\prime})_{\tilde{\rho}^{\prime\mu}}=0.$
(22)

We used the fact that $\tilde{\rho}^{\prime\mu}$ is identical to
$\rho^{\mu}_{X_{8}^{\circ}}$ after we reduced it to $X_{8}^{\circ}$.

Below, we show $I(AE:C|BD)_{\rho^{\mu}}=0$ implies a useful condition

$I(A:E|BCD)_{\rho^{\mu}}=0.$ (23)

The steps are as follows.

$\displaystyle 0$ $\displaystyle=I(AE:C|BD)_{\rho^{\mu}}$ (24)
$\displaystyle=S_{ABDE}+S_{BCD}-S_{BD}-S_{ABCDE}$
$\displaystyle=(S_{AB}+S_{DE})+(S_{BC}+S_{CD}-S_{C})-(S_{B}+S_{D})-S_{ABCDE}$
$\displaystyle=(S_{AB}+S_{BC}-S_{B})+(S_{DE}+S_{CD}-S_{D})-S_{C}-S_{ABCDE}$
$\displaystyle=S_{ABC}+S_{CDE}-S_{C}-S_{ABCDE}$
$\displaystyle=I(AB:DE|C)_{\rho^{\mu}}$ $\displaystyle\geq
I(A:E|BCD)_{\rho^{\mu}}.$

In the third line, we use the fact that the extreme point $\rho^{\mu}_{X_{8}}$
factorizes on some region pairs such as $(AB,DE)$ and $(B,D)$ and the quantum
Markov state condition $I(B:D|C)=0$. In the fifth line, we used $I(A:C|B)=0$
and $I(C:E|D)=0$. The last line follows from strong subadditivity. Thus, Eq.
(23) holds. Then, by Lemma 4, item 4 $\Rightarrow$ item 0, we conclude that
any $\mu\in{\mathcal{C}}_{8}$ is Abelian. ∎

The key ingredient of the above proof is that transporting an anyon along the
figure-8 does not change its type; here, we assume that the figure-8 region is
immersed in a ball or a sphere. We emphasize again that the crucial input is
the anyon transportation property in the neighborhood of the immersed region,
and it is not essential to assume the immersion is in a ball or sphere if we
know this property. This observation is useful in Section V.

The special property of Abelian theory we make use of is the deterministic
fusion outcome $a\times\bar{a}=1$. Because of this, the proof idea does not
generalize easily to non-Abelian theories. One may wonder if a careful
analysis of the fusion spaces involving multiple sectors in ${\mathcal{C}}$
and ${\mathcal{C}}_{8}$ can be helpful in the general case. A complementary
idea is to take full advantage of the fact that the figure-8 annulus is
immersed in a sphere. We leave the general case as an open problem.

##  V Transportation experiment

We discuss a thought experiment, “transportation experiment,” which reveals
the physical connection between immersed annuli with topological defects. The
entanglement bootstrap setup in this section is relaxed compared with that in
Sections III and IV. Let us consider 2D for simplicity. We allow the reference
state to have “defect points” where the axioms may be violated.
(Alternatively, we consider a reference state defined on a manifold $N$ with
nontrivial topology, e.g., a $k$-hole disk.) The setup is illustrated in Fig.
9.

Figure 9: The entanglement bootstrap setup of Section V and the main idea of
the “transportation experiment”. The black dots in the figure are places we do
not impose the axioms. The loop ${\mathcal{L}}$ along which we transport the
“test annulus” (blue) is away from these black dots. The thickening of
${\mathcal{L}}$ denoted by $X_{{\mathcal{L}}}$ (green) is an immersed annulus.

We want to transport an anyon along a closed loop ${\mathcal{L}}$ and back. In
entanglement bootstrap, this is done by transporting a “test annulus” $X$,
which is a sufficiently small embedded annulus that detects the anyon type by
the information convex set $\Sigma(X)$. If we transport the test annulus $X$
along closed loop ${\mathcal{L}}$ and back to its original position, we have
an automorphism of the information convex set determined by the loop
${\mathcal{L}}$:999Accurately speaking, the automorphism is determined by a
loop ${\mathcal{L}}$ with a chosen orientation. We may denote an oriented loop
as $\vec{\mathcal{L}}$. However, we shall use ${\mathcal{L}}$ for simplicity.
Quantities of interest below, such as the subset of anyons not permuted by the
transportation do not depend on the orientation. (For a discussion of
automorphisms of information convex sets in broader setups, see Appendix B.)

$\Phi({\mathcal{L}}):\Sigma(X)\to\Sigma(X).$ (25)

In general, the automorphism $\Phi({\mathcal{L}})$ can permute extreme points
of $\Sigma(X)$, and thus, it permutes anyon type. In other words, for each
${\mathcal{L}}$, we have an automorphism of labels
$\varphi_{\mathcal{L}}:{\mathcal{C}}\to{\mathcal{C}}$. The automorphism
depends only on the topological class of ${\mathcal{L}}$, i.e., smoothly
deforming ${\mathcal{L}}$ without passing the defect points preserves $\Phi$
and $\varphi$. Explicit examples come from topological defects, which we
recall in Section V.1. Throughout this section, we consider loops
${\mathcal{L}}$ whose thickening ($X_{\mathcal{L}}$) is an immersed
annulus.101010This excludes some loops on nonorientable manifolds. For
instance, the noncontractible loop on $RP^{2}$ can be thickened to an Möbius
strip but not an annulus. We define

${\mathcal{I}}({\mathcal{L}})\equiv\left\\{a\in{\mathcal{C}}|\,a=\varphi_{\mathcal{L}}(a)\right\\}$
(26)

as the subset of superselection sectors in ${\mathcal{C}}$ that are preserved
under the automorphism. Therefore,
$|{\mathcal{I}}({\mathcal{L}})|=|{\mathcal{C}}|$ if and only if no anyons are
permuted. We let ${\mathcal{C}}_{\mathcal{L}}=\\{\mu,\nu,\cdots\\}$ be the set
of superselection sectors associated with the extreme points of
$\Sigma(X_{\mathcal{L}})$.

######  Definition 8 (parallel).

Transportation along ${\mathcal{L}}$ is parallel if
$|{\mathcal{I}}({\mathcal{L}})|=|{\mathcal{C}}|$.

In this section, we ask the following questions.

$\textrm{Question:}\,\,\qquad|{\mathcal{I}}({\mathcal{L}})|=|{\mathcal{C}}|\quad\overset{?}{\Longleftrightarrow}\quad\exists\mu\in{\mathcal{C}}_{\mathcal{L}}\,\,\textrm{
s.t. }d_{\mu}=1.$ (27)

We explain why the “$\Leftarrow$” direction is a relatively simple fact
(Proposition 10). We conjecture that the statement is true in the
“$\Rightarrow$” direction (Conjecture 11). We explain the fact that
“$\Rightarrow$” is true for Abelian theories, adopting the discussion in
Section IV to this context. Note that Conjecture 11 implies the existence of
an Abelian sector on the figure-8 annulus as a special case; thus, it also
implies the strong isomorphism conjecture in 2D. Another result is Proposition
9, which states
$|{\mathcal{C}}_{\mathcal{L}}|=|{\mathcal{I}}({\mathcal{L}})|$. It implies
$|{\mathcal{C}}_{8}|=|{\mathcal{C}}|$ as a special case.

###  V.1 Topological defects: an example

A well-known example of topological defect is the one that permutes $e$ and
$m$ in the toric code model [33]. The anyon types of the toric code are
${\mathcal{C}}=\\{1,e,m,f\\}$. Their quantum dimensions are
$d_{1}=d_{e}=d_{m}=d_{f}=1$. Topological defects in this model are non-Abelian
${\mathcal{C}}^{\prime}=\\{\sigma_{+},\sigma_{-}\\}$ with
$d_{\sigma_{+}}=d_{\sigma_{-}}=\sqrt{2}$. Anyons $e$ and $m$ are permuted when
they are transported around the defect (either $\sigma_{+}$ or $\sigma_{-}$).
While defects are most familiar in the context where global symmetries are
present [34], it is understood that the phenomenon of permuting anyons does
not go away under arbitrary local perturbations, not preserving any symmetry.

Figure 10: Information convex set is affected by the presence of a defect
(black dot). The illustration is for the toric code with a defect that
permutes $e$ and $m$.

The presence of a defect affects information convex sets. This is illustrated
in Fig. 10 for the toric code example. The annulus $X^{\prime}$ surrounds the
defect, and its information convex set $\Sigma(X^{\prime})$ has two extreme
points. The quantum dimension for each extreme point, defined according to
Definition 1, agrees with $d_{\sigma_{+}}=d_{\sigma_{-}}=\sqrt{2}$. Thus, no
Abelian sector exists on annulus $X^{\prime}$. If we consider an embedded
annulus $X$ not surrounding any defect, then the information convex set
$\Sigma(X)$ is a simplex with four extreme points. These extreme points
correspond to Abelian sectors $1$, $e$, $m$, and $f$. The model is solvable,
and explicit computation111111Similar computation can be done in some non-
Abelian models, such as quantum double using “minimum diagram” technique [16,
35]. can be done to verify these statements. Putting aside the details,
examples like this are nontrivial contexts for question (27).

###  V.2 Constraints of transportation

Below, we analyze the constraints of the transportation problem (Fig. 9). The
entanglement bootstrap setup and notations have been described at the
beginning of the section. As a useful fact, we note
$1\in{\mathcal{I}}({\mathcal{L}})$. This is because the test annulus $X$
remains embedded in the entire process, and we can reversibly fill in the hole
at any step if it carries the vacuum sector $1\in{\mathcal{C}}$. We start with
a general statement:

######  Proposition 9.

$|{\mathcal{C}}_{\mathcal{L}}|=|{\mathcal{I}}({\mathcal{L}})|$. In particular,
if the transportation is parallel,
$|{\mathcal{C}}_{\mathcal{L}}|=|{\mathcal{C}}|$.

As a corollary, $|{\mathcal{C}}|=|{\mathcal{C}}_{8}|$. This is because
transporting an anyon along a figure-8 immersed in a ball or sphere cannot
change the anyon type.

Here is a sketch of the key idea of proof of Prop. 9. The trick is to attach
the test annulus to another annulus by a strip. After transporting the test
annulus, the strip will be stretched along loop ${\mathcal{L}}$. After gluing
the test annulus with the other annulus (by merging with another annulus), we
get a punctured torus ${\mathcal{W}}_{\mathcal{L}}$. This process is
illustrated in Fig. 11. The main part of the proof is to analyze
$\Sigma({\mathcal{W}}_{\mathcal{L}})$, or more conveniently $\Sigma(T^{2})$,
where the torus $T^{2}$ has a reference state obtained by an analog of “vacuum
block completion” (see Example 4.27 of [17]).

###### Proof.

Consider the process illustrated in Fig. 11. We start with a disk away from
the defect points (black dots) and remove two small balls from it. Let the
resulting 2-hole disk be $Y$. Consider a state labeled by $a,1,a$ on the three
boundaries $\rho^{a1a}_{Y}\in\Sigma(Y)$. It must be an extreme point because
the corresponding fusion space is 1 dimensional.

In step 1, we move the annulus that carries $a\in{\mathcal{C}}$ (the test
annulus we choose) upwards so that the region $Y$ is deformed into an immersed
region $Y^{\prime}$. One of the boundaries of $Y^{\prime}$ is now labeled by
$\hat{1}$, indicating that it is the vacuum sector deformed onto an immersed
annulus, so we cannot compare the state directly to the reference state. A
fact useful later is that $\hat{1}$ is Abelian. In step 2, we transport the
test annulus further along the closed loop ${\mathcal{L}}$ and call the
resulting region $Y_{\mathcal{L}}$, as illustrated in Fig. 11(c).

Figure 11: Construction of an extreme point in
$\Sigma_{\hat{1}}({\mathcal{W}}_{\mathcal{L}})$, useful in the proof of Prop.
9.

If $a\in{\mathcal{I}}({\mathcal{L}})$, that is, $a$ is invariant under the
transportation along ${\mathcal{L}}$, we do step 3 of Fig. 11, which glues
annulus $X$ with the state on $Y_{\mathcal{L}}$ to a punctured torus
${\mathcal{W}}_{\mathcal{L}}$. In terms of quantum states, we merge two
quantum states, which are extreme points of $\Sigma(X)$ and
$\Sigma(Y_{\mathcal{L}})$ (thickening a certain region before merging if
needed) with the superselection sectors labeled in Fig. 11(c). Here,
$a\in{\mathcal{I}}({\mathcal{L}})$ is required so that the two states match on
their overlap regions. For each valid choice of $a$, we get an extreme point
of $\Sigma_{\hat{1}}({\mathcal{W}}_{\mathcal{L}})$. Note that
$1\in{\mathcal{I}}({\mathcal{L}})$, and therefore
$\Sigma_{\hat{1}}({\mathcal{W}}_{\mathcal{L}})$ is nonempty. Note that the
punctured torus ${\mathcal{W}}_{\mathcal{L}}$ contains annuli
$X_{\mathcal{L}}$ and an embedded test annulus $X$.

Next, we “complete” ${\mathcal{W}}_{\mathcal{L}}$ into a torus $T^{2}$ with a
valid reference state. This step is optional, but it is both natural and
convenient. The details are as follows. Choose $a=1$, and run the process in
Fig. 11 to obtain an extreme point
$\rho^{|1_{X}\rangle}_{{\mathcal{W}}_{\mathcal{L}}}\in\Sigma_{\hat{1}}({\mathcal{W}}_{\mathcal{L}})$.
By the completion trick (Lemma 4.4 of [17]), we first thicken
${\mathcal{W}}_{\mathcal{L}}$ and then heal the puncture. By doing this, we
obtain a pure state $|1_{X}\rangle$ on a torus $T^{2}$. By construction,
$T^{2}\supset{\mathcal{W}}_{\mathcal{L}}$, and because $\hat{1}$ on the
thickened boundary of ${\mathcal{W}}_{{\mathcal{L}}}$ is Abelian,
$|1_{X}\rangle$ is a valid reference state: it satisfies A0 and A1 everywhere.
(The notation $|1_{X}\rangle$ is a reminder that if we reduce it to $X$, we
get the reference state $\sigma_{X}$.)

Having explained the construction, we explain various constraints on the
information convex sets, which leads to the conclusion. First, by applying the
associativity to step 3 (combining $X$ and $Y_{\mathcal{L}}$ into
${\mathcal{W}}_{\mathcal{L}}$), we have
$\dim\mathbb{V}_{1}(\mathcal{W}_{\mathcal{L}})=|{\mathcal{I}}({\mathcal{L}})|$,
this is by counting the number of ways to match the boundary conditions in
Fig. 11. If we take the reference state $|1_{X}\rangle$ on the “completed”
torus $T^{2}$, we further have
$\dim\mathbb{V}(T^{2})=|{\mathcal{I}}({\mathcal{L}})|$. This equality is easy
to understand from the point of view that $T^{2}=X\bar{X}$, where $\bar{X}$ is
the complement of $X$ on $T^{2}$. Both $X$ and $\bar{X}$ are annuli and
sectorizable. Applying the associativity to the gluing of the two annuli, we
have

$\dim\mathbb{V}(T^{2})=\sum_{a,b\,\in\,{\mathcal{C}}}\delta_{a,b}\,\delta_{a,\varphi_{\mathcal{L}}(b)}=|{\mathcal{I}}({\mathcal{L}})|,$
(28)

where $\delta_{a,b}$ comes from the matching of sector on one shared boundary
of the two annuli, and $\delta_{a,\varphi_{\mathcal{L}}(b)}$ comes from the
other shared boundary. The appearance of $\varphi_{\mathcal{L}}$ is due to the
transportation property.

Similarly, we cut the torus along another direction into two annuli,
$T^{2}=X_{\mathcal{L}}\bar{X}_{\mathcal{L}}$. Not only are $X_{\mathcal{L}}$
and $\bar{X}_{\mathcal{L}}$ both sectorizable, but they must also have
isomorphism information convex sets
$\Sigma(X_{\mathcal{L}})\cong\Sigma(\bar{X}_{\mathcal{L}})$. This is because
$X_{\mathcal{L}}$ deforms to $\bar{X}_{\mathcal{L}}$ smoothly on $T^{2}$. So
they both characterize ${\mathcal{C}}_{\mathcal{L}}$. From a computation
parallel to Eq. (28), we see

$\dim\mathbb{V}(T^{2})\leq|{\mathcal{C}}_{\mathcal{L}}|.$ (29)

The possibility to have “$<$” is associated with the possibility that only a
subset of labels in ${\mathcal{C}}_{\mathcal{L}}$ survive the matching and
appear on the torus. In particular, if “$<$” applies, we cannot obtain the
maximum-entropy state in $\Sigma(X_{\mathcal{L}})$ by a partial trace from any
state in $\Sigma(T^{2})$.

Below, we show “$=$” in Eq. (29) must be realized. This is done by explicitly
finding a state in $\Sigma(T^{2})$ that reduces to the maximum-entropy state
of $\Sigma(X_{\mathcal{L}})$. The state we identify is precisely
$|1_{X}\rangle$. (The logic below is familiar. See, e.g., Lemma 5.7 of [17].)
Recall that $|1_{X}\rangle$ reduces to the test annulus $X$ gives the vacuum
state $\rho^{1}_{X}\equiv\sigma_{X}$, an Abelian extreme point. Now apply the
definition of quantum dimension (8) to the partition of $X=BCD$ as Fig. 3,
with the requirement that $B=\partial X\cap X_{\mathcal{L}}$. By strong
subadditivity,

$0=\Delta(B,C,D)_{|1_{X}\rangle}\geq I(\bar{X}:C|B)_{|1_{X}\rangle}\geq
I(A:C_{-}|B)_{|1_{X}\rangle},$ (30)

for any $A\subset\bar{X}$ and $C_{-}\subset C$. The specific choice
$A=X_{\mathcal{L}}\setminus X$ and
$C_{-}=X_{\mathcal{L}}\cap(X\setminus\partial X)$ is what we want because
$ABC_{-}$ is now a Levin-Wen partition [23] of the immersed annulus
$X_{\mathcal{L}}$ (namely, $A$ and $C_{-}$ are two disks, and they are
separated by $B$). $I(A:C_{-}|B)_{|1_{X}\rangle}=0$ for the Levin-Wen
partition [2] of $X_{\mathcal{L}}$ implies that $|1_{X}\rangle$ reduces to
$X_{\mathcal{L}}$ is the maximum-entropy state of $\Sigma(X_{\mathcal{L}})$.
This implies “$=$” applies to Eq. (29). ∎

######  Proposition 10.

If there exists $\mu\in{\mathcal{C}}_{\mathcal{L}}$ with $d_{\mu}=1$, the
transportation along ${\mathcal{L}}$ is parallel.

###### Proof.

An annulus is a disk with a puncture, homeomorphically speaking. If the state
on $X_{\mathcal{L}}$ is an extreme point, we can use the completion trick
(Lemma 4.4 of [17]) to heal a puncture to obtain a state on the disk. The
resulting state satisfies the axioms on the entire disk if the extreme point
is associated with an Abelian sector ($\mu\in{\mathcal{C}}_{\mathcal{L}}$ with
$d_{\mu}=1$). Let this state be $\tilde{\sigma}$. It is known that
transporting a test annulus on a disk cannot permute anyon types, supposing
that the disk has a reference state. (This is by Lemma 4.3 of [14]). As the
loop $X_{\mathcal{L}}$ is identified as part of the disk for which we have a
valid reference state $\tilde{\sigma}$, the transportation of testing annulus
on $X_{\mathcal{L}}$ cannot permute anyons. Thus, the transportation along
${\mathcal{L}}$ is parallel. ∎

######  Conjecture 11.

If the transportation along ${\mathcal{L}}$ is parallel, there exists
$\mu\in{\mathcal{C}}_{\mathcal{L}}$ with $d_{\mu}=1$.

###### Proof of conjecture 11 for Abelian theories.

For Abelian anyon theory (that is $d_{a}=1$ for $\forall a\in{\mathcal{C}}$),
the conjecture holds. The proof is that of Proposition 7, with a few minor and
easy-to-state modifications. We only need to replace the immersed figure-8
annulus in Fig. 8 with the immersed annulus $X_{\mathcal{L}}$ in Fig. 9. As we
noted, every step works here because nowhere in the proof of Proposition 7 did
we use the condition that the annulus is immersed in a sphere. All we needed
was the fact that a test annulus carrying $\forall a\in{\mathcal{C}}$
transported along the immersed annulus remains in the same superselection
sector $a$. For Abelian theory, what we prove is $d_{\mu}=1$ for any
$\mu\in{\mathcal{C}}_{\mathcal{L}}$. ∎

We do not know a proof of the general case. This conjecture implies Conjecture
5, and thus, it also implies the 2-dimensional strong isomorphism conjecture
(Conjecture 6).

##  VI Discussion

In this work, we started by asking a simple question: “is there an Abelian
superselection sector detected by the immersed figure-8 annulus?” The question
is relevant to gapped topologically ordered systems in 2D, which support
anyons and is formulated in the framework of entanglement bootstrap. We gave
an affirmative answer to this question for Abelian anyon theories but leave
the general case as a conjecture (Conjecture 5). We explained the fact that if
there is an Abelian sector on figure-8, a certain strong isomorphism
(Conjecture 6) of information convex sets must be true: if two homeomorphic
regions $\Omega$ and $\Omega^{\prime}$ are immersed in the sphere or a disk,
the information convex sets $\Sigma(\Omega)$ and $\Sigma(\Omega^{\prime})$
must be isomorphic. Importantly, it does not matter if $\Omega$ can be
smoothly deformed to $\Omega^{\prime}$ on the sphere (or disk), a situation
where the established isomorphism theorem does not apply.

If the conjecture of the existence of an Abelian sector on “figure-8” annulus
turns out false, a counterexample (e.g., a solvable model) would be extremely
interesting. This also means the existence (absence) of Abelian sectors on
immersed figure-8 will be a nontrivial (and exotic) criterion for classifying
wave functions with entanglement area law as well as topologically ordered
phases.

Suppose the existence of an Abelian sector on figure-8 (Conjecture 5) can be
proved. Strong isomorphism will be a fact. While strong isomorphism is
powerful, it seems to imply that we cannot learn much by looking at a
topological space immersed differently. Let me argue this is far from the
case. If strong isomorphism holds, a couple of intriguing questions can be
pursued. In particular, it means tunneling processes (Section III.1 and Fig.
5) can generate explicit isomorphisms not allowed by smooth deformation. Thus,
automorphisms of information convex sets will not only come from topologically
nontrivial ways to smoothly deform an immersed region back to itself but also
from ways that utilize tunneling in intermediate steps. We expect these
diverse automorphisms to be informative. A natural question is: “Can we use
these automorphisms (dancing of quantum states) to extract all the data of the
modular tensor category underlying an anyon model?”121212An optimism to this
possibility was found independent by Kyle Kawagoe who went to my chalk talk at
math department Ohio State University. If the answer is affirmative, it will
manifestly use a single quantum state and this will be complementary to the
“microscopic approach” [36]. Can the mathematical structure coming out of
immersion be comparable with the mapping class group? The analogous questions
in higher dimensions can be asked in parallel, but are more open.

A stronger conjecture we postponed until now is the existence of a canonical
(Abelian) vacuum state on the immersed figure-8 annulus. If this is true, the
process in Fig. 15 of the appendix may well be on the right track in
extracting the full set of anyon topological spins. This is a meaningful
problem that deserves an effort elsewhere. Proving the existence of a
canonical vacuum on figure-8 remains an open problem, even for Abelian
theories, in which context we proved Conjecture 5.

We further discussed the relevance of immersed annuli in contexts with
topological defects. The central question we asked was Eq. (27), relating
“parallel” transportation to the existence of an Abelian sector on the
immersed annulus that thickens the transportation loop. The question has two
directions, and we provided an affirmative answer to one direction. The other
direction is a conjecture that implies our main conjecture about the existence
of an Abelian state on figure-8 (Conjecture 5) as well as strong isomorphism.

Although this work focuses on 2D, we expect that the insights gained are
useful in higher dimensions as well. Questions stirring in the right direction
could be: what are generalizations of the tunneling trick (Section III.1 and
Fig. 5) and transportation experiments (Section V and Fig. 9) in higher
dimensions? We developed some tools and discussed a few further open problems
for higher dimensions in the appendices.

Finally, we ask whether considering quantum states on regions immersed in a
physical system can be useful in other physical contexts. Immersed regions
utilize the Hilbert space more efficiently by using local pieces more than
once, and they can be made larger and topologically (or geometrically) more
interesting than the original physical system available to us. Smooth
deformation of immersion region, a property natural in systems with
entanglement area law, will be absent in broader contexts such as gapless
systems. Nevertheless, many other benefits of immersion may persist.

## Acknowledgments

After settling the Abelian case, I made multiple attempts on the general proof
of Conjecture 5 without progress. These years gave me the time to contemplate
the potential implications of this conjecture, especially strong isomorphism
and tunneling. Conversations with many helped me sharpen these thoughts,
including Anuj Apte, Yu-An Chen, Tarun Grover, Isaac Kim, Xiang Li, Ting-Chun
Lin, Hanqing Liu, Daniel Ranard, Hao-Yu Sun, Chong Wang, Xueda Wen, and
Carolyn Zhang. I thank John McGreevy and Jin-Long Huang for collaborating on
projects that taught me immersion-related topology. I thank Michael Levin for
a conversation about topological spin during a UQM conference at Caltech in
2022 and an intriguing discussion with John McGreevy on a train to that
conference. I thank my classmate Yang Fan for sharing the video “outside in”
[28] with me many years ago. I thank Xinyao Zhao for providing a selection of
colors, some of which I chose for some figures. I thank Ilya Gruzberg, Kyle
Kawagoe, Yuan-Ming Lu, David Penneys, and Stuart Raby for related
conversations during my visit to Ohio State University near the completion of
this work. This work was supported by the Simons Collaboration on Ultra-
Quantum Matter, a grant from the Simons Foundation (652264).

##  Appendix A Equivalent definitions of quantum dimension

Immersed sphere shells, in the context of $n$-dimensional entanglement
bootstrap, are sectorizable regions. We derive an equivalence relation on the
definition of quantum dimensions (Lemma 12) for immersed sphere shells in
$n$-dimensions, where $n\geq 2$ and is arbitrary. It applies to immersed
annuli as a special case ($n=2$).

Because the setup is general, we introduce some diagrammatic notation for the
partitions. Let $Z=S^{n-1}\times\mathbb{I}$ be a sphere shell in $n$ space
dimensions, where $S^{n-1}$ is a $(n-1)$-sphere and $\mathbb{I}$ is an
interval. In Fig. 12, the horizontal direction indicates the direction of the
interval $\mathbb{I}$. Therefore, each vertical strip that connects the top
and the bottom represents a sphere shell. If a strip is cut in half (top and
bottom), each piece is a half-sphere shell (topologically a ball). We denote
the superselection sectors detected by $\Sigma(Z)$ as ${\mathcal{C}}_{Z}$.

Figure 12: Partitions of a sphere shell $Z=S^{n-1}\times\mathbb{I}$. The
horizontal direction is the direction of the interval $\mathbb{I}$.

######  Lemma 12.

The following are equivalent definitions of quantum dimension $d_{h}$ of
$h\in{\mathcal{C}}_{Z}$, where $Z$ is an immersed sphere shell in
$n$-dimension.

  1. (a)

$\Delta(B,C,D)=2\ln d_{h}$ for the partition in Fig. 12(a).

  2. (b)

$\Delta(B,C,D)=2\ln d_{h}$ for the partition in Fig. 12(b).

  3. (c)

$\Delta(B,C,D)=4\ln d_{h}$ for the partition in Fig. 12(c).

  4. (d)

$I(A:C|B)=2\ln d_{h}$ for the partition in Fig. 12(d).

######  Remark.

We shall see in the proof that we do not need to deform the annulus.
Therefore, Lemma 12 holds not only for sphere shells immersed in a sphere but
also applicable to contexts allowing topological defects (i.e., when the
reference state is on a manifold with a nontrivial topology).

###### Proof.

First, we consider the partition in Fig. 13(a), where $Z=ABCDEF$. By the
strong subadditivity, $\Delta(B,C,DE)_{\rho^{h}}\geq I(A:C|B)_{\rho^{h}}$.
Furthermore, because $\rho_{Z}^{h}$ is an extreme point of $\Sigma(Z)$, it
factorizes as $\rho_{BCDEF}^{h}=\rho_{BCDE}^{h}\otimes\rho_{F}^{h}$, which
implies

$\Delta(B,C,DE)_{\rho^{h}}=\Delta(B,C,DEF)_{\rho^{h}}.$ (31)

The region $Z$, as a sectorizable region, satisfies $\Delta(\partial
Z,Z\setminus\partial Z)_{\rho^{h}}=0$, the so-called “extreme point
criterion”. It implies

$\displaystyle S_{CDEF}$ $\displaystyle=S_{Z}+S_{AB},$ (32) $\displaystyle
S_{DEF}$ $\displaystyle=S_{Z}+S_{ABC},$

where the first line follows letting $\partial Z$ be $CDEF$. The second line
uses $0=\Delta(EF,ABCD)\geq\Delta(DEF,ABC)$.

Figure 13: Partitions used in the proof of Lemma 12. In (a) and (b), the
convention is the same as that in Fig. 12. (a’) illustrates the partition of
(a) in the concrete case that $Z=X_{8}$ is the “figure-8”.

Therefore, combining (31) and (32), we get

$\displaystyle\Delta(B,C,DE)_{\rho^{h}}=$
$\displaystyle\Delta(B,C,DEF)_{\rho^{h}}$ (33) $\displaystyle=$
$\displaystyle\left(S_{BC}-S_{B}+S_{CDEF}-S_{DEF}\right)_{\rho^{h}}$
$\displaystyle=$
$\displaystyle\left(S_{BC}-S_{B}+S_{AB}-S_{ABC}\right)_{\rho^{h}}$
$\displaystyle=$ $\displaystyle I(A:C|B)_{\rho^{h}}.$

Thus, among the statements in Lemma 12, (a) and (d) are equivalent. Similarly,
(b) and (d) are equivalent. Therefore, the three statements (a), (b), and (d)
in Lemma 12 are equivalent.

The next step is establishing the equivalence between statement (c) and the
rest. For this purpose, we consider Fig. 13(b), where $Z=BCDD^{\prime}E$. We
will show:

$\Delta(B,CD^{\prime}E,D)_{\rho^{h}}=\Delta(B,C,DD^{\prime})_{\rho^{h}}.$ (34)

Then, because the extreme point $\rho^{h}_{Z}$ factorizes as a tensor product
on the left of $E$ and right of $E$, we have
$\Delta(B,CD^{\prime}E,D)_{\rho^{h}}=\Delta_{L}+\Delta_{R}$. $\Delta_{L}$ and
$\Delta_{R}$ give contributions identical to conditions (a) and (b) of Lemma
12. Thus, the claimed result holds.

The rest of the proof is the justification of (34). In fact, this is a special
case of the decoupling lemma (Lemma D.1 of [16]). Nonetheless, we spell out
the details so readers do not need to read the more general decoupling lemma.

All regions below are those in Fig. 13(b). Because $\rho^{h}_{Z}$ is an
extreme point, it satisfies

$\displaystyle S_{BCD^{\prime}E}$
$\displaystyle=S_{BC}-S_{C}+S_{CD^{\prime}E}$ (35) $\displaystyle
S_{CDD^{\prime}E}$
$\displaystyle=S_{CDD^{\prime}}-S_{D^{\prime}}+S_{D^{\prime}E}$ $\displaystyle
S_{D}+S_{D^{\prime}}$ $\displaystyle=S_{DD^{\prime}}$ $\displaystyle 0$
$\displaystyle=S_{CD^{\prime}E}-S_{C}+S_{D^{\prime}E}.$

The first and second lines follow from the vanishing of conditional mutual
information, bounded by the extreme point criterion: (1st line)
$0=\Delta(C,D^{\prime}E)_{\rho^{h}}\geq I(B:D^{\prime}E|C)_{\rho^{h}}$, (2nd
line) $0=\Delta(D^{\prime},E)_{\rho^{h}}\geq I(CD:E|D^{\prime})_{\rho^{h}}$.
The third line is because any extreme point is factorized into a tensor
product along disjoint shells arranged in the interval $\mathbb{I}$ direction.
The fourth line is $\Delta(C,D^{\prime}E)_{\rho^{h}}=0$, also by the extreme
point criterion. Therefore,

$\displaystyle\Delta(B,CD^{\prime}E,D)_{\rho^{h}}$
$\displaystyle=S_{BCD^{\prime}E}+S_{CDD^{\prime}E}-S_{B}-S_{D}$ (36)
$\displaystyle=(S_{BC}-S_{C}+S_{CD^{\prime}E})+(S_{D^{\prime}E}-S_{D^{\prime}}+S_{CDD^{\prime}})-S_{B}-S_{D}$
$\displaystyle=S_{BC}+(S_{CD^{\prime}E}-S_{C}+S_{D^{\prime}E})+S_{CDD^{\prime}}-S_{B}-(S_{D^{\prime}}+S_{D})$
$\displaystyle=S_{BC}+S_{CDD^{\prime}}-S_{B}-S_{DD^{\prime}}$
$\displaystyle=\Delta(B,C,DD^{\prime})_{\rho^{h}}.$

The brackets in the middle steps are replacements according to Eq. (35). This
completes the proof. ∎

Lemma 12 implies that $d_{h}\geq 1$, for any superselection sector $h$
detected by a sphere shell. Such superselection sectors are associated with
point excitations. It is worth noting that, in $n\geq 3$, not every type of
superselection sector is associated with point excitations, and many of them
are not detected by sphere shells. The general definition of quantum dimension
for immersed sectorizable regions is currently lacking. Nonetheless, a
generalization is known for regions participating in a pairing [17].

##  Appendix B Automorphism of information convex sets: tools and examples

Automorphism (i.e., self-isomorphism) of information convex sets plays an
important role in entanglement bootstrap. The simplest application is the
definition of antiparticles, but its implication is much broader. In this
appendix, we talk about two things related to automorphisms of information
convex sets.

In Section B.1, we introduce a “immersion complex” $\mathfrak{M}(\omega,N)$
(Definition 13). It is a simplicial complex whose points represent immersed
regions. It is an attempt to provide a compact mathematical way to describe
two things: (i) topological classes of immersion ($\omega$ into $N$) by
counting the connected component ($\pi_{0}$) of $\mathfrak{M}$, and (ii) the
topological classes of deformation of an immersed region
$\Omega\looparrowright N$ back itself (keeping the intermediate configurations
immersed), as the fundamental group ($\pi_{1}$) of a connected component of
$\mathfrak{M}$. In Section B.2, we utilize tunneling (Section III.1) to make
automorphisms not available otherwise.

###  B.1 Math related to immersion and deformation classes

Consider the immersion of an $n$-dimensional space with boundary into another,
$\Omega\looparrowright N$. Let the homeomorphism class of $\Omega$ be
$\omega$. In our convention, $\Omega$ knows all the data of the immersed
region while $\omega$ forgets everything but the homeomorphism class. (We
intend to first describe the mathematical problem and postpone the physical
relevance to automorphisms of information convex sets and examples in the
latter part of the section.)

Let $\mathcal{R}(\omega,N)=\\{\Omega_{1}(\omega),\Omega_{2}(\omega),\cdots\\}$
be the set of representatives of equivalence classes of immersion of $\omega$
into $N$. That is

  1. 1.

$\Omega_{i}(\omega)\looparrowright N$ for any $i$;

  2. 2.

$\Omega_{i}(\omega)\overset{N}{\not\sim}\Omega_{j}(\omega)$ if $i\neq j$;

  3. 3.

$\Omega_{i}(\omega)$ is homeomorphic to $\omega$ for any $i$;

  4. 4.

any $\Omega\looparrowright N$ homeomorphic to $\omega$ must satisfy
$\Omega\overset{N}{\sim}\Omega_{i}(\omega)$ for some $i$.

Here, $Y\overset{N}{\sim}Z$ means two immersed regions $Y,Z\looparrowright N$
can be smoothly deformed to each other on $N$, where the intermediate
configurations are immersed. In particular, if there is no way to immerse
$\omega$ to $N$, $\mathcal{R}(\omega,N)=\emptyset$. Now we are ready to
introduce the “immersion complex”.

######  Definition 13 (Immersion complex).

Let $\omega$ and $N$ be two $n$-dimensional manifolds, possibly with
boundaries. We define the immersion complex $\mathfrak{M}(\omega,N)$ as

$\mathfrak{M}(\omega,N)\equiv\coprod_{\Omega\in\mathcal{R}(\omega,N)}\mathfrak{m}(\Omega,N).$
(37)

Here, $\coprod$ means disjoint union, and $\mathfrak{m}(\Omega,N)$, for any
$\Omega\looparrowright N$, is the simplicial 2-complex such that

  1. 1.

$\Omega^{\prime}\looparrowright N$ represents a point (0-simplex) of
$\mathfrak{m}(\Omega,{N})$ if and only if
$\Omega^{\prime}\overset{N}{\sim}\Omega$.

  2. 2.

Two distinct points $\Omega,\Omega^{\prime}\in\mathfrak{m}(\Omega,{N})$ are
connected by a 1-simplex131313A 1-simplex is a line segment connecting two
points. One may alternatively call it a 1-cell. of $\mathfrak{m}(\Omega,{N})$
if and only if $\Omega\cap\Omega^{\prime}$ can be converted to $\Omega$ and
$\Omega^{\prime}$ by two sequences of extensions (one for each).

  3. 3.

Three distinct points
$\Omega^{\prime},\Omega^{\prime\prime},\Omega^{\prime\prime\prime}\in\mathfrak{m}(\Omega,{N})$
are connected by a triangle (2-simplex) of $\mathfrak{m}(\Omega,{N})$ if and
only if both of the following holds

     * •

$\Omega\cap\Omega^{\prime}\cap\Omega^{\prime\prime}$ can be converted to
$\Omega\cap\Omega^{\prime}$, $\Omega^{\prime}\cap\Omega^{\prime\prime}$ and
$\Omega\cap\Omega^{\prime\prime}$ by three sequences of extensions (one for
each).

     * •

Points in each of the three pairs $(\Omega,\Omega^{\prime})$,
$(\Omega^{\prime},\Omega^{\prime\prime})$ and $(\Omega,\Omega^{\prime\prime})$
are connected by a 1-simplex of $\mathfrak{m}(\Omega,{N})$, according to rule
2 above.

######  Remark.

We emphasize that the definition above serves as an attempt to formulate the
mathematical problem, but we do not claim it has the desired mathematical
rigor. Even if it does, we do not know if it is the most convenient to work
with. In practice, it is likely to be useful to take some cellular
decomposition of manifold $N$ and only consider only “large but finite-sized”
immersed regions $\Omega$. By doing so, the number of 1-simplex connecting a
given point will be finite. (This setup is also similar to entanglement
bootstrap, where the manifold $N$ will be interpreted as the manifold on which
we define the reference state, and thus $N$ is equipped with a coarse-grained
lattice.)

By definition $\mathfrak{m}(\Omega,N)=\mathfrak{m}(\Omega^{\prime},N)$ if
$\Omega^{\prime}\overset{N}{\sim}\Omega$. Furthermore,
$\mathfrak{m}(\Omega,N)$ must be connected. Therefore,
$\mathfrak{M}(\omega,N)$ is the disjoint union of these connected components.
Also by definition, for any triangle of $\mathfrak{m}(\Omega,N)$, its three
edges must be included in $\mathfrak{m}(\Omega,N)$, and for any 1-simplex of
$\mathfrak{m}(\Omega,N)$ its two endpoints must be included in
$\mathfrak{m}(\Omega,N)$. This means both $\mathfrak{m}(\Omega,N)$ and
$\mathfrak{M}(\omega,N)$ are simplicial 2-complexes.

######  Remark.

One may ask if it is meaningful to add higher dimensional $k$-simplexes
($k\geq 3$) instead of stopping at $k=2$. The relevance to physics we can
think of so far (as we explain shortly) depends on $\pi_{0}$ and $\pi_{1}$.
Adding higher dimensional simplexes will not change them. Nonetheless, it
remains an interesting question if higher homotopy (or homology) groups of
$\mathfrak{M}$ can provide further information.141414We thank John McGreevy
for this question.

####  B.1.1 physical relevance of $\mathfrak{M}$ through $\pi_{0}$ and
$\pi_{1}$

It is easy to explain why $\pi_{0}$ of $\mathfrak{M}(\omega,N)$ is of physical
relevance. Consider the entanglement bootstrap setup where $N$ is the manifold
on which we define our reference state $\sigma_{N}$. The $n$-dimensional
version of axioms A0 and A1 are satisfied everywhere on $N$. Then
$\pi_{0}(\mathfrak{M}(\omega,N))$, i.e., the number of connected components of
$\mathfrak{M}(\omega,N)$, is the number of inequivalent immersions of the
topological space $\omega$ in $N$. For instance,
$\pi_{0}(\mathfrak{M}(\textrm{annulus},S^{2}))$ is $\mathbb{Z}_{2}$ (the
cyclic group of order 2). We discussed this in the main text without using
this notation. In broader contexts, $\pi_{0}(\mathfrak{M})$ is the set of
regular homotopy class of immersed regions, which has been a topic of study in
math literature; see, e.g., [37, 38, 27]. Within each connected component,
$\mathfrak{m}(\Omega,N)$, $\Omega\in\mathcal{R}(\omega,N)$, we can smoothly
deform the region and the isomorphism theorem implies that such deformation
preserves the information convex set. Configurations in two connected
components are harder to relate, although tunneling (Section III.1) provides
one way to generate isomorphisms between such configurations (when a certain
Abelian sector exists).

The fundamental group of the connected components of $\mathfrak{M}$, i.e.,
$\pi_{1}(\mathfrak{m}(\Omega,N))$, by (37) is relevant to the study of
automorphism of information convex sets:

  * •

Immersed regions up to small deformations are represented by nearby points
connected by 1-simplex in $\mathfrak{m}(\Omega,N)$. For each oriented loop
$\vec{\gamma}$ in $\mathfrak{m}(\Omega,N)$ that starts at $\Omega$ and ends at
$\Omega$, we have an associated automorphism:

$\Phi(\vec{\gamma}):\quad\Sigma(\Omega)\to\Sigma(\Omega).$ (38)

The reason (38) makes sense is that each oriented 1-simplex connecting two
adjacent points $\Omega$ and $\Omega^{\prime}$, defines the following
isomorphism $\Sigma(\Omega)\to\Sigma(\Omega^{\prime})$: we first reduce states
in $\Sigma(\Omega)$ to $\Omega\cap\Omega^{\prime}$ by elementary steps of
restrictions, and then we extend the support of these states to
$\Omega^{\prime}$ by elementary steps of extensions. These steps are
isomorphisms between information convex sets. After we apply these
isomorphisms on $\vec{\gamma}$, in a sequence given by the orientation, we
generate a unique automorphism (38).

  * •

Triangles (2-simplex) are the places where the loop can deform
($\vec{\gamma}\to\vec{\gamma}^{\prime}$) without affecting the automorphism.
We illustrate the deformation as

(39)

where only part of the loops $\vec{\gamma}$ and $\vec{\gamma}^{\prime}$ in the
neighborhood of the triangle are shown. The fact that
$\Phi(\vec{\gamma})=\Phi(\vec{\gamma}^{\prime})$ can be checked from
(Definition 13) and the discussion below Eq. (38).

We therefore argue that $\pi_{1}(\mathfrak{m}(\Omega,N))$ is the information
that determines the nontrivial classes of deformation of immersed region
$\Omega$ back to itself in the background manifold $N$. Automorphisms of
$\Sigma(\Omega)$ coming from smooth deformations can be different _only if_
two closed loops $\vec{\gamma}$ and $\vec{\gamma}^{\prime}$ (both pass
$\Omega$) represent two different elements of
$\pi_{1}(\mathfrak{m}(\Omega,N))$. In particular, if the loop $\vec{\gamma}$
can shrink to a point on $\mathfrak{m}(\Omega,N)$, the associated automorphism
$\Phi(\vec{\gamma})$ must be the (trivial) identity map.

####  B.1.2 Examples

We give a few simple examples to illustrate the immersion complex (Definition
13) and discuss its physical relevance. Recall

$\mathfrak{M}(\omega,N)\equiv\coprod_{\Omega\in\mathcal{R}(\omega,N)}\mathfrak{m}(\Omega,N),$
(40)

where $\mathcal{R}(\omega,N)$ is the set of representatives of immersed
regions homeomorphic to $\omega$, immersed in $N$. Our first example is about
annuli immersed in a disk. Let $\omega$ be the annulus and $N=B^{2}$ be the
disk.

$\mathfrak{M}(\textrm{annulus},B^{2})=\coprod_{j=0}^{\infty}\mathfrak{m}(X(j),B^{2}),$
(41)

where $X(j)$ for $j\geq 0$ represents an immersed annulus with turning number
$j$. In particular, we can let $X(0)=X_{8}$ be the immersed “figure-8” and let
$X(1)=X$ be the embedded annulus. We expect $\mathfrak{m}(X,B^{2})$ to be
simply connected and $\pi_{1}(\mathfrak{m}(X,B^{2}))$ is trivial.151515If we
deform $X$ back to itself, keeping the intermediate configurations embedded,
the deformation must correspond to a loop in $\mathfrak{m}(X,B^{2})$ that can
shrink to a point. The argument is precisely that used to prove Lemma 4.3 of
[14]. However, for the deformations allowing immersed intermediate
configurations, the argument does not apply. What is
$\pi_{1}(\mathfrak{m}(X_{8},B^{2}))$? From the intuition around Eq. (6), we
conjecture that $\pi_{1}(\mathfrak{m}(X_{8},B^{2}))=\mathbb{Z}_{2}$. If this
is true, the definition of anti-sector on $X_{8}$, discussed near Eq. (6) is
justified.

The second example is about annuli immersed in a sphere.

$\mathfrak{M}(\textrm{annulus},S^{2})=\mathfrak{m}(X,S^{2})\coprod\mathfrak{m}(X_{8},S^{2}),$
(42)

where $X$ and $X_{8}$ are the only two classes of immersed annuli in the
sphere (as discussed in Table 1). Thus $\mathfrak{M}(\textrm{annulus},S^{2})$
has two connected components. What are the fundamental groups $\pi_{1}$ of the
connected components? $\pi_{1}(\mathfrak{m}(X,S^{2}))$ should contain
$\mathbb{Z}_{2}$ as a subgroup. The related observation is that $X$ can turn
inside out on the sphere and the fact that such process can turn an anyon
$a\in{\mathcal{C}}$ into anti-anyon $\bar{a}\in{\mathcal{C}}$, which is, in
general, a different sector. (For a related discussion, see Appendix G of
[14].) We conjecture that $\pi_{1}(\mathfrak{m}(X_{8},S^{2}))$ is
$\mathbb{Z}_{2}$.

One simple example in 3-dimensional space is

$\mathfrak{M}(\textrm{sphere shell},B^{3})=\mathfrak{m}(Z,B^{3}),$ (43)

where $Z$, on the right-hand side, represents the embedded sphere shell in the
3-dimensional ball $B^{3}$. Eq. (43) follows from the fact that there is a
unique way to immerse a sphere $S^{2}$ in $B^{3}$ (or $R^{3}$), up to regular
homotopy; this is an influential early result by Smale [37], which is also the
topic of video [28]. It seems plausible that
$\pi_{1}(\mathfrak{m}(Z,B^{3}))=\mathbb{Z}_{2}$, but we do not know proofs. It
is further interesting to consider

$\mathfrak{M}(\textrm{genus-{\hbox{g}}
handlebodies},B^{3})=\mathfrak{m}(G_{g},B^{3}),$ (44)

where $G_{g}$ is an embedded genus-$g$ handlebody and $g$ is a positive
integer. As is indicated in Eq. (44), $\mathfrak{M}$ is connected in this
example. Genus-$g$ handlebodies detect the “graph excitations” studied in [16,
17], where for $g=1$ these graph excitations are flux loops. In general,
$\pi_{1}(\mathfrak{m}(G_{g},B^{3}))$ can have pretty nontrivial structure even
though $G_{g}$ are sectorizable regions.

Figure 14: A deformation process, which corresponds to an oriented loop in
$\mathfrak{m}(X,X_{+})$. Here, $X$ is an annulus (thick black loop in the top
figure) embedded in a wider annulus $X_{+}$ (gray).

Our last example is for 2-dimensional regions. Consider an annulus $X$
embedded in an annulus $N=X_{+}$ that thickens $X$, as shown in Fig. 14. We
ask

$\pi_{1}(\mathfrak{m}(X,X_{+}))=?$ (45)

Note that, $\pi_{1}(\mathfrak{m}(X,X_{+}))$ can be nontrivial even if
$\mathfrak{m}(X,B^{2})$ is trivial. Intuitively, in the process described in
Fig. 14, a pair of twists are created locally and then transported around the
annulus before they annihilate again. (The process in Fig. 14 is inspired by
Fig. 8 of [38] as well as Fig. 7 of [20].) Can this process represent a
nontrivial element of $\pi_{1}(\mathfrak{m}(X,X_{+}))$? We do not know the
answer. If this process indeed corresponds to a non-contractible loop in
$\mathfrak{m}(X,X_{+})$, it may have nontrivial physical consequences in
contexts with topological defects. Recall that references state on annulus
$X_{+}$ may come from a system with a defect line that passes through $X_{+}$
as discussed in Section V.

In summary, the problem we try to formulate in this section is meant to be a
pure topology problem. This problem is carefully separated from the study of
quantum states on such regions. Nonetheless, we believe this problem is a
crucial mathematical problem towards a clear understanding of the automorphism
of information convex sets. We wonder if the answers to these questions (or
ones that are close analog) have appeared in math literature. The question on
$\pi_{0}(\mathfrak{M}(\omega,N))$ is essentially the classification of
immersion up to regular homotopy. Some related results can be looked
up161616Even in this case, we do not always find the results we want. We are
most interested in manifolds with boundaries (such as punctured manifolds)
immersed in a manifold with _the same_ dimension. Sometimes, a math result we
spent time searching for came out fresh [39]. in existing math literature,
e.g., [37, 38]. It is unclear to us if the fundamental group
$\pi_{1}(\mathfrak{m}(\Omega,N))$, e.g., for examples listed above, has been
studied in math literature. (That is why we only provided guesses to most
answers!) In any case, entanglement bootstrap provides (adds) a motivation for
the study of such mathematical objects.

###  B.2 Automorphisms utilizing tunneling

Tunneling trick (Section III.1) provides extra isomorphism of information
convex sets by “large changes” of the region, such as adding or removing a
twist on a 1-handle. This is in contrast with the smooth deformation built
from elementary steps. Tunneling is not described by lines in
$\mathfrak{m}(\Omega,N)$. The tunneling trick, as described in the main text,
works for the specific context of 2D. Below, we broaden the meaning of
tunneling and use it to refer to any “large change” of an immersed region
(together with the associated information convex set).

By allowing both smooth deformations and tunneling, we can obtain more diverse
classes of automorphisms of $\Sigma(\Omega)$. We will not discuss the entire
scope of how this might work, acknowledging that it is an interesting
mathematical problem. Instead, we describe a nontrivial example in 2D.

Figure 15: Automorphism of the information convex set of an immersed punctured
torus, where tunneling is used. The punctured torus is visualized as an
annulus (horizontal, gray) with a 1-handle (black) connecting both boundaries
of the annulus.

Let $\mathcal{W}$ be a punctured torus immersed in a disk $B^{2}$, as shown in
the left figure of Fig. 15. If there is an Abelian sector on figure-8 region,
$\mu\in{\mathcal{C}}_{8}$, then we can design an automorphism of
$\Sigma(\mathcal{W})$ utilizing the tunneling trick in one of the steps; see
the process shown in Fig. 15. The entire process takes the immersed punctured
torus $\mathcal{W}$ back to itself. Steps 1 and 2 are smooth
deformations.171717As a side note, I described steps 1 and 2 in a math student
seminar in 2020 [40]; see the first backup slide, which was my attempt to ask
about an unknown-to-me terminology. Later, my collaborators and I learned in
the projects [16, 17] that the term I looked for was immersion and regular
homotopy. Now step 3 is added, which gives a reason to ask more questions.
Step 3 is a tunneling. Thus, the automorphism cannot be associated with a
closed loop in $\mathfrak{m}(\mathcal{W},B^{2})$. Is there any physical
significance of this process in Fig. 15? We anticipate that the answer is yes,
relating this to the problem of defining topological spins from a state, but
we postpone the discussion to future work.

##  Appendix C Strong isomorphism conjecture in arbitrary dimensions

Here is the statement of strong isomorphism conjecture for an arbitrary space
dimension $n$, generalizing $n=2$ case (Conjecture 6) in the main text. We
denote the $n$-dimensional sphere on which we define the reference state as
$\mathbf{S}^{n}$. Axioms A0 and A1 are satisfied everywhere.

######  Conjecture 14.

Suppose two immersed regions
$\Omega,\Omega^{\prime}\looparrowright\mathbf{S}^{n}$ are homeomorphic, then

$\Sigma(\Omega)\cong\Sigma(\Omega^{\prime}).$ (46)

Suppose this conjecture is true, then we have similar beliefs (as 2D,
discussed in the discussion section) that the explicit isomorphisms that come
up during the proof can be informative and useful in extracting universal
information.

Besides, the strong isomorphism conjecture (Conjecture 14) seems to resonate
with another basic question: “What is the definition of quantum dimension for
an arbitrary immersed sectorizable region $S\looparrowright\mathbf{S}^{n}$?”
This question is open for dimensions $n\geq 3$. Many (if not all) nontrivially
immersed $S$ have an embedded version
$S^{\star}\hookrightarrow\mathbf{S}^{n}$, where $S^{\star}$ is in the
homeomorphism class of $S$. Since $S^{\star}$ is embedded, the quantum
dimension of a superselection sector $I\in{\mathcal{C}}_{S^{\star}}$
(characterized by $\Sigma(S^{\star})$) can be defined from the entropy
difference of an extreme point $\rho^{I}$ with the vacuum $\sigma$:

$d_{I}\equiv\exp{\left(\frac{S(\rho_{S^{\star}}^{I})-S(\sigma_{S^{\star}})}{2}\right)},\quad
I\in{\mathcal{C}}_{S^{\star}}.$ (47)

While Eq. (47) is familiar in early works, we are still interested in knowing
if $d_{I}\geq 1$ in general. Nontrivial checks from 3D and gapped domain walls
are consistent with $d_{I}\geq 1$, but the general proof for an arbitrary $n$
is lacking. This puzzle can be restated as a question about the vacuum state:
“Does the vacuum state have the absolute minimum entropy among states in
$\Sigma(\Omega)$, for any $\Omega\hookrightarrow\mathbf{S}^{n}$?” We would
like to conjecture the affirmative direction.

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