=0pt =0

Proof of #1.

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Proof of #1.

Boundary states of a bulk gapped ground state in 2-d quantum spin systems

Yoshiko Ogata
Graduate School of Mathematical Sciences

The University of Tokyo, Komaba, Tokyo, 153-8914, Japan.

We introduce a natural mathematical definition of
boundary states of a bulk gapped ground state,  in the operator algebraic framework
of $2$-d quantum spin systems.
With approximate Haag duality at the boundary,
we derive a $C^*$-tensor category $\tilde \caM$ out of such boundary state.
Under a non-triviality condition of the braiding in the bulk,
we show that the Drinfeld center (with an asymptotic constraint) of
$\tilde \caM$ is equivalent to the bulk braided $C^*$-tensor category
derived in [15].

§ INTRODUCTION

Since the discovery of Kitaev's beautiful model [8], the classification problem of topologically ordered gapped
systems has attracted
a lot of attention.
Ingenious models were made  [10]
a lot of categorical arguments were carried out.
On the other hand,  Naaijkens carried out an analysis of
Kitaev quantum double  models, from the operator algebraic/ algebraic quantum field theoretic (DHR theory)
point of view.
The advantage of this direction is that DHR theory allows us to
treat non-solvable models, hence it allows us to think of the classification.
In[15], we derived braided $C^*$-tensor categories out of
gapped ground states on two-dimensional quantum spin systems
satisfying the approximate Haag duality.
There, following the DHR theory [3],[4],[5],[6],[11]
our objects are representations satisfying the superselection criterions.
In the same paper, we showed that this braided $C^*$-tensor category is an
invariant of the classification of gapped ground state phases.
This can be regarded as an analytical realization of the statement
that anyons are invariant of topological phases.
In particular, the existence of anyons means the ground state is
long-range entangled [14], as was shown in [7]
for a class of commuting Hamiltonians.

Bravyi and Kitaev introduced two gapped boundaries of the toric code model [2].
In [9], Kitaev and Kong studied the model from the categorical point of view.
There, Kitaev and Kong conjectured that the bulk theory is the Drinfeld center of the
gapped boundary theory, and “physically proved” it
for all 2+1D Levin-Wen models.
Recently,  Wallick analyzed the corresponding theory
in the toric code, in terms of superselection sectors, following the corresponding bulk analysis
by Naaijkens [13][12].
There, they showed the bulk-boundary correspondence above.

Although it is a broadly studied subject, as far as we know, there is no mathematical formulation to specify what a boundary state of a bulk gapped ground state means, in the operator algebraic framework.
In this paper, we suggest a natural mathematical definition of it. See Definition <ref>.
Physically, a pure state $\obd$
on the half-plane is a boundary state of a bulk gapped ground state $\obk$ if
they are macroscopically the same away from the boundary.
We analyze such a boundary state $\obd$ in this paper.
We derive a $C^*$-tensor category $\tilde \caM$
for $\obd$
under the boundary version of approximate Haag duality.

We then construct a Drinfeld center $\zam$
of $\tilde\caM$, with some asymptotic constraint (Definition <ref>).
We show that
$\zam$ is a braided $C^*$-tensor category equivalent to the bulk braided $C^*$-tensor category
derived in [15],
if the braiding is nontrivial in a certain sense (Assumption <ref> and Remark <ref>).
This gives proof for the Kitaev and Kong's conjecture, in our setting.

§.§ Setting

Throughout this paper, we fix some $2\le d\in\nan$.
We denote the algebra of $d\times d$ matrices by $\Mat_{d}$.
For each $z\in\bbZ^2$,  let $\caA_{\{z\}}$ be an isomorphic copy of $\Mat_{d}$, and for any finite subset $\Lambda\subset\bbZ^2$, we set $\caA_{\Lambda} = \bigotimes_{z\in\Lambda}\caA_{\{z\}}$.

For finite $\Lambda$, the algebra $\caA_{\Lambda} $ can be regarded as the set of all bounded operators acting on
the Hilbert space $\bigotimes_{z\in\Lambda}{\bbC}^{d}$.
We use this identification freely.
If $\Lambda_1\subset\Lambda_2$, the algebra $\caA_{\Lambda_1}$ is naturally embedded in $\caA_{\Lambda_2}$ by tensoring its elements with the identity.
For an infinite subset $\Gamma\subset \bbZ^{2}$,
is given as the inductive limit of the algebras $\caA_{\Lambda}$ with $\Lambda$, finite subsets of $\Gamma$.
We call $\caA_{\Gamma}$ the quantum spin system on $\Gamma$.

For a subset $\Gamma_1$ of $\Gamma\subset\bbZ^{2}$,
the algebra $\caA_{\Gamma_1}$ can be regarded as a subalgebra of $\caA_{\Gamma}$.
For $\Gamma\subset \bbR^2$, with a bit of abuse of notation, we write $\caA_{\Gamma}$
to denote $\caA_{\Gamma\cap \bbZ^2}$.
Also, $\Gamma^c$ denotes the complement of $\Gamma$ in $\bbR^2$.
We denote the algebra of local elements in $\Gamma$ by $\caA_{\Gamma, \rm{loc}}$,
i.e. $\caA_{\Gamma, \rm{loc}}=\cup_{\Lambda\Subset\Gamma}\caA_\Lambda$.
The upper half-plane $\bbR\times [0,\infty)$ is denoted by $\hu$,
and the lower half-plane $\bbR\times (-\infty,0)$ is denoted by $\hd$.
each $\bm a\in \bbR^2$, $\theta\in\bbR$ and $\varphi\in (0,\pi)$,
we set

\begin{align*}
\Lambda_{\bm a, \theta,\varphi}
=&\lmk \bm a+\left\{
t\bm e_{\beta}\mid t>0,\quad \beta\in (\theta-\varphi,\theta+\varphi)\right\}\rmk.
\end{align*}

where  $\bm e_{\beta}=(\cos\beta,\sin\beta)$.
We call a set of this form a cone, and denote by $\Cbk$ the set of all cones.
We set $\arg\Lambda_{\bm a, \theta,\varphi}:=(\theta-\varphi,\theta+\varphi)$ and $|\arg\Lambda_{\bm a, \theta,\varphi}|=2\varphi$.
We also set $\bm e_\Lambda:=\bm e_{\theta}$ for $\Lambda=\Lambda_{\bm a, \theta,\varphi}$.
For $\varepsilon>0$ and $\Lambda=\Lambda_{\bm a, \theta,\varphi}$, $\Lambda_\varepsilon$
denotes $\Lambda_\varepsilon=\Lambda_{\bm a, \theta,\varphi+\varepsilon}$.
We consider the following sets of cones:

\begin{align}
\begin{split}
\Lambda_{\bm a,\theta,\varphi}\mid \bm a\in \bbR^2,\;
\right\},\\
\Lambda_{(a,0),\varphi,\varphi}\mid a\in\bbR,\; 0<\varphi<\frac\pi 2
\right\},\\
\hu\cap \lmk \Lambda_r\rmk^c\mid \Lambda_r\in\Crbd
\right\}=\left\{
\overline{\Lambda_{(a,0),\frac\pi 2+\varphi,\frac\pi 2-\varphi}}\mid a\in\bbR,\; 0<\varphi<\frac\pi 2
\right\},\\
\Lambda\mid \arg\Lambda\cap\bbA_{[\theta-\varphi,\theta+\varphi]}=\emptyset
\right\}.
\end{split}
\end{align}

See Figures <ref>-<ref>.
Note that they are upward-filtering sets with respect to the inclusion relation.
We denote by $\pc$ the set of all pairs $(\lz,\lzr)\in \CUbk\times \Crbd$
with $\lz\subset \lzr$.

Let $\obk$ be a pure state on $\abk$ with a GNS triple
$(\hbk, \pbk,\Omega_{\mopbk} )$.
Let $\obd$ be a pure state on $\abd$ with a GNS triple
$(\hbd, \pbd,\Omega_{\mopbd} )$.
Now we introduce our definition of boundary states of a bulk state.

We say that $\obd$ is a boundary state of the bulk state $\obk$ if
$\pbd\vert_{\caA_{\lhu}}$ and $\pbk\vert_{\caA_{\lhu}}$ are quasi-equivalent
for any $\Lambda\in\CUbk$.

Recall that in quantum spin systems, two quasi-equivalent states can be approximately represented by
to each other's local perturbations.
From this point of view, this definition is telling us that  $\obd$ is a boundary state of the bulk state $\obk$ if they are macroscopically the same away from the boundary.
Namely, if we restrict these states to the regions of the form Figure <ref> or Figure <ref>,
then they look macroscopically the same.

We show this property for ground states of frustration-free Hamiltonians satisfying “strict cone
LTQO”. See section <ref>.
Although it is not the main target of this paper, we take this definition also for the case that the boundary state has a current at the edge.

[domain=-4:4, scale=0.4]

[fill=lightgray, opacity=.5, very thick] (-3.5,5)–(-1,1)–(1.5,5);

[dashed] (-5,0)–(5,0);

$\Lambda\cap\hu$ for $\ld\in \CUbk$

[domain=-4:4, scale=0.4]

[fill=lightgray, opacity=.5, very thick] (-3.5,5)–(-2,0)–(1,0)–(2.5,5);

[dashed] (-5,0)–(5,0);

$\Lambda\cap\hu$ for $\ld\in \CUbk$

[domain=-4:4, scale=0.4]

[fill=lightgray, opacity=.5, very thick] (-5,0)–(1.5,0)–(-5,5);

[dashed] (-5,0)–(5,0);

$\lm l\in \Clbd$

[domain=-4:4, scale=0.4]

[fill=gray, opacity=.5, very thick] (5,0)–(-1.5,0)–(5,5);

[dashed] (-5,0)–(5,0);

$\lm r\in \Crbd$

We set

\begin{align}
\begin{split}
&\caF_{\mopbk}^{U(0)}:=\cup_{\Lambda\in\CUbk}\pbk\lmk\caA_{\Lambda \cap \hu }\rmk'',\quad
\caF_{\mopbk}^U:=\overline{\caF_{\mopbk}^{U(0)}}^n,\\
&\caF_{\mopbd}^{U(0)}:=\cup_{\Lambda\in\CUbk}\pbd\lmk\caA_{\Lambda \cap \hu}\rmk'',\quad
\caF_{\mopbd}^U:=\overline{\caF_{\mopbd}^{U(0)}}^n,\\
\caG:=\overline{\caG^{(0)}}^n,\\
&\caB_l^{(0)}:=\cup_{\Lambda_l\in \Clbd}\pbd\lmk\caA_{\Lambda_l}\rmk'',\quad
\caB_l:=\overline{\caB_l^{(0)}}^n,\\
\overline{\cup_{\Lambda\in\caC_{(\theta,\varphi)} }\pbk(\caA_{\Lambda})'' }^n.
\end{split}
\end{align}

where ${\overline{\cdot}}^n$ indicates the norm closure.
Note that $\caG=\caB_{(\frac{3\pi}2,\frac\pi 2)}$ in the notation of [15].
Because $\CUbk,\Crbd,\Clbd,\caC_{(\theta,\varphi)}$ are upward-filtering sets,
$\caF^U_{\mopbk}$, $\caF^U_{\mopbd}$, $\caG$, $\caB_l$,
$\caB_{(\theta,\varphi)}$ are $C^*$-algebras.

For example, to construct $\caF_{\mopbd}^{U(0)}$, we collect all the von Neumann algebras
$\pbd\lmk \caA_{\ld \cap\hu}\rmk''$ corresponding to the areas
of the shape Figure <ref> or Figuare <ref>.
Then take the norm closure and obtain $\fbd$.
We regard this $C^*$-algebra $\fbd$, the $C^*$-algebra
corresponding to operations using only the bulk.
To construct $\caB_l^{(0)}$, we collect all the von Neumann algebras
$\pbd\lmk\caA_{\lm l}\rmk''$ corresponding to the areas
of the shape Figure <ref>.
Then take the norm closure and obtain $\fbd$.
We regard this $C^*$-algebra $\bl$, the $C^*$-algebra
corresponding to operations using the left boundary.
By definition, we can easily see

\begin{align}
\pbk\lmk \abd\rmk\subset \fbk,\quad \pbd\lmk \abd\rmk\subset \fbd,\quad \pbk\lmk \abk\rmk\subset \caG.
\end{align}

For a representation $\rho$ of $\abk$ on $\hbk$ and $\Lambda\in \Cbk$,
we define a set of unitaries

\begin{align}
\Vbk{\rho}{\Lambda}:=\left\{
V_{\rho\Lambda}\in\caU\lmk \caH_{\mopbk}\rmk
\mid \left.\Ad\lmk V_{\rho\Lambda}\rmk\circ\rho\right\vert_{\caA_{\Lambda^c}}
=\left.\pbk \right\vert_{\caA_{\Lambda^c}}
\right\}.
\end{align}

We denote by $\Obk$ the set of all representations $\rho$ of $\abk$ on $\hbk$,
which has a non-empty $\Vbk{\rho}{\Lambda}$ for any $\Lambda\in \Cbk$.
We denote by $\OUbk$ the set of all representations $\rho$ of $\abk$ on $\hbk$,
which has a non-empty $\Vbk{\rho}{\Lambda}$ for any $\Lambda\in \CUbk$.
Note that $\Obk\subset\OUbk$.
For $\Lambda_0\in \CUbk$, we denote by $\OUbkl$
the set of all $\rho\in \OUbk$ such that $\rho\vert_{\caA_{\Lambda_0^c}}=\pbk\vert_{\caA_{\Lambda_0^c}}$.

For a representation $\rho$ of $\abd$ on $\hbd$ and $\Lambda_r\in \Crbd$,
we define sets of unitaries

\begin{align}
\begin{split}
\left\{
V_{\rho\Lambda_r}\in\caU\lmk \bl\rmk
\mid \left.\Ad\lmk V_{\rho\Lambda_r}\rmk\circ\rho\right\vert_{\caA_{\Lambda_r^c\cap\hu}}
=\left.\pbk \right\vert_{\caA_{\Lambda_r^c\cap\hu}}
\right\}\\
\left\{
V_{\rho\Lambda_r}\in\caU\lmk \fbd\rmk
\mid \left.\Ad\lmk V_{\rho\Lambda_r}\rmk\circ\rho\right\vert_{\caA_{\Lambda_r^c\cap\hu}}
=\left.\pbk \right\vert_{\caA_{\Lambda_r^c\cap\hu}}
\right\}.
\end{split}
\end{align}

We denote by $\Orbd$ the set of all representations $\rho$ of $\abd$ on $\hbd$,
which has a non-empty $\Vbd{\rho}{\Lambda_r}$ for any $\Lambda_r\in \Crbd$.
We denote by $\OrUbd$ the set of all representations $\rho$ of $\abd$ on $\hbd$,
which has a non-empty $\VUbd{\rho}{\Lambda_r}$ for any $\Lambda_r\in \Crbd$.
Note that $\OrUbd\subset \Orbd$.
Physically, objects in $\OrUbd$ (resp. $\Orbd$) correspond to low energy excitations
which can be localized into the area of the form Figure <ref>,
with operations using only the bulk (resp. operations using the left boundary).

For $\Lambda_{r0}\in \Crbd$, we denote by $\Orbdl$
(resp. $\OrUbdl$)
the set of all $\rho\in \Orbd$ (resp. $\rho\in \OrUbd$) such that $\rho\vert_{\caA_{\Lambda_{r0}^c}}=\pbd\vert_{\caA_{\Lambda_{r0}^c}}$.

For a representation $\rho$ of $\abd$ on $\hbd$ and $\Lambda\in \CUbk$,
we define sets of unitaries

\begin{align}
\begin{split}
\left\{
V_{{\rho}\Lambda}\in\caU\lmk \fbd\rmk
\mid \left.\Ad\lmk V_{\rho\Lambda}\rmk\circ\rho\right\vert_{\caA_{\Lambda^c\cap\hu}}
=\left.\pbd \right\vert_{\caA_{\Lambda^c\cap\hu}}
\right\}.
\end{split}
\end{align}

We denote by $\Obu$ the set of all representations $\rho$ of $\abd$ on $\hbd$,
which has a non-empty $\Vbu{\rho}{\Lambda}$ for any $\Lambda\in \CUbk$.
Physically, objects in $\Obul$ correspond to low energy excitations
which can be localized into the area of the form Figure <ref> and Figuare <ref>,
with operations using only the bulk.
For $\lz\in \CUbk$, we denote by
$\Obul$ the set of all $\rho\in \Obu$ such that
For any $\lm r\in\Crbd$, there exists a $\ld\in \CUbk$ such that
$(\ld,\lm r)\in\pc$.
Then we have $\Vbu\rho\ld\subset\Vbd\rho{\lm r}$,
and we have $\Obul\subset \Orbdl$.

For each $\rho,\sigma\in \OUbk$,
the set of intertwiners between them are denoted by

\begin{align}
\left\{
R\in \caB(\hbk) \mid R\rho(A)=\sigma(A)R,\;\; \text{for all}\;\; A\;\in \abk
\right\}.
\end{align}

For any $\rho,\sigma\in \Orbd$, we set

\begin{align}
\begin{split}
\left\{
R\in \bl \mid R\rho(A)=\sigma(A)R,\;\; \text{for all}\;\; A\;\in \abd
\right\},\\
\left\{
R\in \fbd \mid R\rho(A)=\sigma(A)R,\;\; \text{for all}\;\; A\;\in \abd
\right\}.
\end{split}
\end{align}

Note that $(\rho,\sigma)_U\subset (\rho,\sigma)_l$.

For representations $\rho,\sigma$ of $\abd$ on $\hbd$, we write  $\rho\simeq_l\sigma$
(resp. $\rho\simeq_U\sigma$)
if there exists a unitary $U\in\bl$ (resp. $U\in\fbk$ ) such that $\rho=\Ad U\sigma$.
It is important to note that $\simeq_l$ is coarser than $\simeq_U$.
Physically, we understand $\rho\simeq_U\sigma$ means we can map $\rho$ to $\sigma$
using only “bulk operation”.
On the other hand, $\rho\simeq_l\sigma$ can require “truely boundary operation” to map
$\rho$ to $\sigma$.

§.§ Assumptions

Let us consider Setting <ref>.
We introduce our assumptions in this section.
First of all, we consider the situation that $\obd$ is a boundary state of $\obk$.

The state $\obd$ is a boundary state of $\obk$.
Namely, for any $\Lambda\in\CUbk$, representations
$\pbd\vert_{\caA_{\lhu}}$ and $\pbk\vert_{\caA_{\lhu}}$ are quasi-equivalent.

In order to carry out DHR-theory, we need to assume approximate Haag duality.
We introduced the following approximate  Haag duality in [15].

Consider the setting in subsection <ref>.

For any $\varphi\in (0,2\pi)$ and
$\varepsilon>0$ with
there is some $R^{\mopbk}_{\varphi,\varepsilon}>0$ and decreasing
functions $f^{\mopbk}_{\varphi,\varepsilon,\delta}(t)$, $\delta>0$
on $\bbR_{\ge 0}$
with $\lim_{t\to\infty}f^{\mopbk}_{\varphi,\varepsilon,\delta}(t)=0$
such that

for any cone $\Lambda$ with $|\arg\Lambda|=\varphi$, there is a unitary
$U^{\mopbk}_{\Lambda,\varepsilon}\in \caU(\hbk)$

\begin{align}\label{lem7p}
\pbk\lmk\caA_{\Lambda^c}\rmk'\subset
\Ad\lmk U^{\mopbk}_{\Lambda,\varepsilon}\rmk\lmk
\pbk\lmk \caA_{\lmk \Lambda-R^{\mopbk}_{\varphi,\varepsilon}\bm e_\Lambda\rmk_\varepsilon}\rmk''
\rmk,
\end{align}

for any $\delta>0$ and $t\ge 0$, there is a unitary
$\tilde U^{\mopbk}_{\Lambda,\varepsilon,\delta,t}\in \pbk\lmk \caA_{\Lambda_{\varepsilon+\delta}-t\bm e_{\Lambda}}\rmk''$

\begin{align}\label{uappro}
\lV
U^{\mopbk}_{\Lambda,\varepsilon}-\tilde U^{\mopbk}_{\Lambda,\varepsilon,\delta,t}
\rV\le f^{\mopbk}_{\varphi,\varepsilon,\delta}(t).
\end{align}

The following Lemma is immediate from the definition.

Consider the setting in subsection <ref> and assume Assumption <ref>.
Then for any $\Lambda\in \CUbk$, we have

The following two assumptions are boundary version of the approximate Haag duality.

For any $\varphi\in (0,\frac\pi 2)$, $\varepsilon>0$ with $2\varphi+2\varepsilon<\pi$,
there is some $R^{(r)\mopbd}_{\varphi,\varepsilon}>0$ and decreasing
functions $f^{(r)\mopbd}_{\varphi,\varepsilon,\delta}(t)$, $\delta>0$
on $\bbR_{\ge 0}$
with $\lim_{t\to\infty}f^{(r)\mopbd}_{\varphi,\varepsilon,\delta}(t)=0$
such that

for any cone $\Lambda_r=\lr a \varphi\in \Crbd$, there is a unitary
$U^{(r)\mopbd}_{\Lambda_r,\varepsilon}\in \caU(\hbd)$

\begin{align}\label{lem7p}
\pbd\lmk\caA_{\lrhu}\rmk'\subset
\Ad\lmk U^{(r)\mopbd}_{\Lambda_r,\varepsilon}\rmk\lmk
\pbd\lmk \caA_{ \lr{a-R^{(r)\mopbd}_{\varphi,\varepsilon}}{\varphi+\varepsilon}}\rmk''
\rmk,
\end{align}

for any $\delta>0$ and $t\ge 0$, there is a unitary
$\tilde U^{(r)\mopbd}_{\Lambda_r,\varepsilon,\delta,t}\in \pbd\lmk
\caA_{
\lr{a-t}{\varphi+\varepsilon+\delta}
}\rmk''\cap \fbd$

\begin{align}\label{uappro}
\lV
U^{(r)\mopbd}_{\Lambda_r,\varepsilon}-\tilde U^{(r)\mopbd}_{\Lambda_r,\varepsilon,\delta,t}
\rV\le f_{\varphi,\varepsilon,\delta}(t).
\end{align}

Note in particular, we have $U^{(r)\mopbd}_{\Lambda_r,\varepsilon}\in \fbd$.

For any $\varphi\in (0,\frac\pi 2)$, $\varepsilon>0$ with $2\varphi+2\varepsilon<\pi$,
there is some $R^{(l)\mopbd}_{\varphi,\varepsilon}>0$ and decreasing
functions $f^{(l)\mopbd}_{\varphi,\varepsilon,\delta}(t)$, $\delta>0$
on $\bbR_{\ge 0}$
with $\lim_{t\to\infty}f^{(l)\mopbd}_{\varphi,\varepsilon,\delta}(t)=0$
such that

for any cone $\Lambda_l=\lef a \varphi\in \Clbd$, there is a unitary
$U^{(l)\mopbd}_{\Lambda_l,\varepsilon}\in \caU(\hbd)$

\begin{align}\label{lem7p}
\pbd\lmk\caA_{(\lm l)^c\cap \hu}\rmk'\subset
\Ad\lmk U^{(l)\mopbd}_{\Lambda_l,\varepsilon}\rmk\lmk
\pbd\lmk \caA_{ \lef{\lmk a+R^{(l)\mopbd}_{\varphi,\varepsilon}\rmk}{(\varphi+\varepsilon)}}\rmk''
\rmk,
\end{align}

for any $\delta>0$ and $t\ge 0$, there is a unitary
$\tilde U^{(l)\mopbd}_{\Lambda_l,\varepsilon,\delta,t}\in \pbd\lmk
\caA_{
\lef{a+t}{(\varphi+\varepsilon+\delta)}
}\rmk''\cap \fbd$

\begin{align}\label{uappro}
\lV
U^{(l)\mopbd}_{\Lambda_l,\varepsilon}-\tilde U^{(l)\mopbd}_{\Lambda_l,\varepsilon,\delta,t}
\rV\le f^{(l)}_{\varphi,\varepsilon,\delta}(t).
\end{align}

Note in particular, we have $U^{(l)\mopbd}_{\Lambda,\varepsilon}\in \fbd$.

In order to have direct sum and subobjects, we need the following assumption.

For any $\Lambda\in \CUbk$, $\pbk(\caA_{\Lambda})''$ is an infinite factor.

The boundary version is the following.

For any $\Lambda\in \CUbk$, $\pbd(\caA_{\Lambda\cap\hu})''$ is an infinite factor.

From Assumption <ref>, we can derive infiniteness of von Neumann algebras of larger regions.

Assume Assumption <ref>.
Then for any $\lm l\in \Clbd$, $\pbd(\caA_{\lm{l}})''$ is an infinite factor.

Because $\obd$ is pure, $\pbd(\caA_{\lm{l}})''$ is a factor  for any $\lm l\in \Clbd$.
For any $\lm l\in \Clbd$, there exists a $\ld\in \CUbk$ such that
$\ld\cap\hu\subset\lm l$, hence $\pbd(\caA_{\Lambda\cap\hu})''\subset \pbd(\caA_{\lm{l}})''$.
Therefore, if Assumption <ref> holds, then $\pbd(\caA_{\lm{l}})''$ is an infinite factor.

We also note the following.

If Assumption <ref> and Assumption <ref> hold,
then Assumption <ref> holds.

This is because
$\Lambda\setminus \lmk \Lambda\cap \hu\rmk$ is finite
for any $\ld\in\CUbk$.

Combining this and Lemma 5.3 Lemma 5.5 of [15], for boundary states of gapped ground states, Assumption <ref>
and Assumption <ref> hold automatically.
Suppose that $\obk$ is a gapped ground state in the sense it satisfies
the gap inequality (5.1)of [15], and $\obd$ is a boundary state of $\obk$.
Then Assumption <ref> and Assumption <ref> hold.

The following assumption means that in $\obk$, any quasi-particles
in the upper half plane can freely move to the lower half plane.

In this paper, we consider such a situation.

§.§ Results

We consider the setting in subsection <ref>.
In [15], we showed the following.
Consider the setting in subsection <ref>.
Under Assumption <ref> and Assumption <ref>,
$\Obkl$ with their morphisms $(\rho,\sigma)$,
$\rho,\sigma\in \Obkl$
forms a braided $C^*$-tensor category $\Cabkl$.

Analogous to this theorem, we can derive various $C^*$-tensor categories
from our boundary state $\obd$. We can consider different categories
depending on our choices of objects (representations of $\caA_{\hu}$) and
morphisms (intertwiners between representations).
Assumption <ref> we can show that
all of the following categories are $C^*$-tensor categories
(Theorem <ref>, Theorem <ref>, Theorem <ref>).

* The category $\Carbdl$ with objects $\Orbdl$
and morphisms $(\rho,\sigma)_l$ between objects $\rho,\sigma\in \Orbdl$.

* The category $\CarUbdl$ with objects $\OrUbdl$
and morphisms $(\rho,\sigma)_U$ between objects
$\rho,\sigma\in \OrUbdl$.

* The category $\Cabul$ with objects $\Obul$
and morphisms $(\rho,\sigma)_U$ between objects $\rho,\sigma\in \Obul$.

For each $\rho\in \OrUbdl$ we can define
a family of natural isomorphisms
$\hb - \rho:\Obul\to (-\otimes \rho,\rho\otimes -)_U$, satisfying the half-braiding equation
(Proposition <ref> Lemma <ref> Lemma <ref> Lemma <ref>),
under Assumption <ref> Assumption <ref>.
This $\hb -\rho$  satisfies some asymptotic property (Lemma <ref>).
When we restrict this to $\Obul$,
$\Cabul$ becomes braided $C^*$-tensor category (Theorem <ref>).
Using the condition that $\obd$ is a boundary of $\obk$ (Assumption <ref>),
we can show the following. See Corollary <ref> for a more detailed statement.

The braided $C^*$-tensor category $\Cabul$ and the bulk braided $C^*$-tensor category
$\Cabkl$ are equivalent.

Hence at the boundary, we have something corresponding to the bulk theory.
However, $\Cabul$ is not preferable as a boundary theory.
This is because the morphisms are too limited there, i.e.,
we allow ourselves to use only $\fbd$.
In particular, for toric code,
with rough boundary, $e$ particle is not equivalent to the vacuum
with respect to
$\simeq_U$, while
it is equivalent to the vacuum
with respect to
Therefore, we prefer to use $\bl$ as morphisms.
Hence in section <ref> we introduce a category  $\caM$ with object $\Obul$,
and morphisms between objects $\rho,\sigma\in \Obul$
defined by
The category $\caM$ is a tensor category.
However, it is not a $C^*$-tensor category, in the sense that we cannot
construct a subobject in general.
In order to overcome this difficulty, we introduce the idempotent completion $\tilde\caM$
of $\caM$.
Then we obtain the following (see Proposition <ref>)

This $\tilde\caM$ is a $C^*$-tensor category

We regard this $\tilde \caM$ as our boundary theory.
For this $C^*$-tensor category $\tilde\caM$,
we introduce  the Drinfeld center with an asymptotic constraint
$\caZ_a(\tilde \caM)$ of $\tilde \caM$.
This is the same as the usual Drinfeld center  (Definition <ref>), except that the half-braidings
are required to satisfy the asymptotic property satisfied by
our $\iota^{\llz}$.
Now we assume that the braiding of $\Cabkl$ is non-trivial in the following sense :
if two quasi-particles in the bulk have the asymptotically unitarily equivalent braiding, then they
are equivalent. For more precise condition, see Assumption <ref> and Remark <ref>.
Then we obtain the following.

The Drinfeld center with the asymptotic constraint
$\zam$ is a braided $C^*$-tensor category equivalent to the bulk braided $C^*$-tensor category
derived in [15].

§ BOUNDARY THEORIES

First we extend each $\rho\in\Orbdl$ to an endomorphism of $\bl$.

Consider the setting in subsection <ref>.
Then for any $\rho\in\Orbd$, $\lzr\in\Crbd$ and $\Vrl\rho\lzr\in \Vbd\rho\lzr$,
there exists a unique $*$-homomorphism $\Tbdv\rho\lzr: \bl\to\caB(\hbd)$
such that

$\Tbdv\rho\lzr\pbd=\Ad\lmk \Vrl\rho\lzr\rmk\rho$,

$\Tbdv\rho\lzr$ is $\sigma$-weak continuous on
$\pbd(\caA_{\Lambda_l})''$ for all $\Lambda_l\in\Clbd$.

It satisfies $\Tbdv\rho\lzr(\bl)\subset \bl$ and defines an endomorphism
$\Tbdv\rho\lzr : \bl\to \bl$.
This endomorphism also satisfies

\begin{align}
\left.\Tbdv\rho\lzr\right\vert_{\pbd\lmk\caA_{(\lzr)^c\cap\hu}\rmk''}
\end{align}

If furthermore $\rho\in\OrUbd$ and $\Vrl\rho\lzr\in \VUbd\rho\lzr$
then we have $\Tbdv\rho\lzr(\fbd)\subset \fbd$.

Each $\lm l\in\Clbd$ can be written as $\lm l=(\lm r)^c\cap \hu$
with some $\lm r=\lr a \varphi\in\Crbd$.

\[
\kappa_{\lm l}:=\left\{
\Gamma_r\mid\Gamma_r=\lr b{\varphi_1},\;
a<b,\; 0<\varphi_1<\varphi
\right\}
\subset\Crbd.
\]

Then, as in Lemma 2.11 of [15],
for any $\rho\in\Orbd$, $\lzr\in\Crbd$ and $\Vrl\rho\lzr\in \Vbd\rho\lzr$,

\begin{align}
T_\rho^{(0)}(x):=\Ad\lmk\Vrl\rho{\lzr}\Vrl\rho{K_{\lm l}}^*\rmk(x),\quad\text{if}\;
x\in \pbd\lmk\caA_{\lm l }\rmk'',\;\lm l\in\Clbd
\end{align}

defines a isometric $*$-homomorphism $T_\rho^{(0)}: \bl^{(0)}\to\bl$,
independent of the choice of $K_{\lm l}$, $\Vrl\rho{K_{\lm l}}$, $\lm l$.
As in Lemma 2.13 of [15], this $T_\rho^{(0)}$ extends to
the $\Tbdv\rho\lzr$ with the desired property.
The last statement is trivial from the definition above.

We extend the equivalence relations $\simeq_l$,  $\simeq_U$
to endomorphisms of $\bl$ in an obvious way.

By the same proof as that of [15] Lemma 2.14,
for any $\lm {r1},\lm{r2}\in\Crbd$, $\rho\in \Orbd$, $\Vrl{\rho}{\lm {ri}}\in \Vbd\rho{\lm {ri}}$, $i=1,2$,
we have
$\Tbdv\rho{{\lm {r2}}}=\Ad\lmk\Vrl{\rho}{\lm {r2}}\Vrl{\rho}{\lm {r}}^*\rmk\Tbdv\rho{{\lm {r1}}}$.
For $\lmr i\in\Crbd$ $\rho_i\in \Orbd$ (resp. $\rho_i\in \OrUbd$), $\Vrl{\rho_i}{\lmr i}\in \Vbd{\rho_i}{\lmr i}$
(resp. $\Vrl{\rho_i}{\lmr i}\in \VUbd{\rho_i}{\lmr i}$), $i=1,2$
with $\rho_1\simeq_l\rho_2$
(resp. $\rho_1\simeq_U\rho_2$) we have
$\Tbdv{\rho_1}{\lmr 1}\simeq_l\Tbdv{\rho_2}{\lmr 2}$
(resp. $\Tbdv{\rho_1}{\lmr 1}\simeq_U\Tbdv{\rho_2}{\lmr 2}$).

Consider the setting in subsection <ref>, and let $\lzr\in \Crbd$.
Then for any $\rho,\sigma\in \Orbdl$, we have

\begin{align}\label{iwate}
\rho\circ_{\lzr} \sigma:=
\Tbd\rho{\lzr}\unit\;\Tbd\sigma{\lzr}\unit\;\pbd\in\Orbdl,
\end{align}

and this defines a composition $\circ_{\lzr} : \Orbdl\times \Orbdl \to \Orbdl$.
This composition $\circ_{\lzr}$ satisfies the following properties.

For $\rho,\sigma\in \Orbdl$, we have $\Tbd{\rho\circ_{\lzr} \sigma}\lzr\unit

For $\rho,\sigma,\gamma\in \Orbdl$,
$\lmk \rho\circ_{\lzr} \sigma\rmk\circ_{\lzr} \gamma=\rho\circ_{\lzr}\lmk \sigma\circ_{\lzr} \gamma\rmk$.

If $\rho,\sigma\in \OrUbdl$, then $\rho\circ_{\lzr} \sigma\in\OrUbdl $.

(iv) If $\llz\in\pc$ and $\rho,\sigma\in \Obul$, then
${\rho\circ_{\lzr} \sigma}\in \Obul$.

This follows from the same proof as that of Lemma 3.3, 3.6, 3.7 of [15],
and definition of $\Obul$,$\OrUbdl$.
The proof of (<ref>), (i)(ii) are the same as that of Lemma 3.3, 3.6, 3.7 of [15].
(iii) follows because we have

\begin{align}
\Tbdv\rho{\lm r }\Tbdv\sigma{\lm r }\pbd=\Ad\lmk\Vrl{\rho}{\lm r}\rmk\Tbd\rho{{\lzr}}\unit
\Ad\lmk\Vrl{\sigma}{\lm r}\rmk\Tbd\sigma{{\lzr}}\unit\pbd
=\Ad\lmk\Vrl{\rho}{\lm r}\Tbd\rho{{\lzr}}\unit\lmk \Vrl{\sigma}{\lm r}\rmk\rmk
\Tbd\rho{{\lzr}}\unit
\Tbd\sigma{{\lzr}}\unit\pbd,
\end{align}

and in the setting of (iii), may take $\Vrl{\rho}{\lm r}\Tbd\rho{{\lzr}}\unit\lmk \Vrl{\sigma}{\lm r}\rmk\in \fbd$.
In the setting of (iv), for each $\ld\in\CUbk$, fix $\Vrl\rho\ld\in\Vbu\rho\ld$
and $\Vrl\sigma\ld\in\Vbu\sigma\ld$.
Then $\Vrl\rho\ld\Tbd\rho\lzr\unit \lmk \Vrl\sigma\ld\rmk\in \fbd$ and

\begin{align}
\left. \Ad\lmk \Vrl\rho\ld\Tbd\rho\lzr\unit \lmk \Vrl\sigma\ld\rmk \rmk
\Tbd\rho\lzr\unit\Tbd\sigma\lzr\unit\pbd\right\vert_{\caA_{\ld^c\cap\hu}}
\end{align}

Hence ${\rho\circ_{\lzr} \sigma}\in \Obul$.

Let $\rho_i,\sigma_i\in \Orbd$ (resp $\rho_i,\sigma_i\in \OrUbd$),
$\lm {ri},\lm {ri}'\in \Crbd$, $\Vrl{\rho_i}{\lm{ri}}\in \Vbd{\rho_i}{\lm{ri}}$,
$\Vrl{\sigma_i}{\lm{ri}'}\in \Vbd{\sigma_i}{\lm{ri}'}$
(resp $\Vrl{\rho_i}{\lm{ri}}\in \VUbd{\rho_i}{\lm{ri}}$,
$\Vrl{\sigma_i}{\lm{ri}'}\in \VUbd{\sigma_i}{\lm{ri}'}$), for $i=1,2$.
If $\rho_1\simeq_l\rho_2$ and $\sigma_1\simeq_l\sigma_2$
(resp. $\rho_1\simeq_U\rho_2$ and $\sigma_1\simeq_U\sigma_2$)
then we have
$\Tbdv{\rho_1}{\lm {r1}}\Tbdv{\sigma_1}{\lm {r1}'}\simeq_l \Tbdv{\rho_2}{\lm {2r}}\Tbdv{\sigma_2}{\lm {2r}'}$
(resp. $\Tbdv{\rho_1}{\lm {r1}}\Tbdv{\sigma_1}{\lm {r1}'}\simeq_U\Tbdv{\rho_2}{\lm {2r}}\Tbdv{\sigma_2}{\lm {2r}'}$), by the same proof as Lemma 3.4 of [15].

Consider the setting in subsection <ref>.
For any $\rho_i,\rho_i'\in \Orbd$,
$\lm {ri},\lm {ri}'\in \Crbd$, $\Vrl{\rho_i}{\lm{ri}}\in \Vbd{\rho_i}{\lm{ri}}$,
$\Vrl{\rho_i'}{\lm{ri}}\in \Vbd{\rho_i'}{\lm{ri}}$
and $R_i\in \lmk\Tbdv{\rho_i}{\lm {ri}}, \Tbdv{\rho_i'}{\lm {ri}'}\rmk_l$
for $i=1,2$,
we have

\begin{align}
R_1\otimes R_2:=R_1\Tbdv{\rho_1}{\lm {r1}}\lmk R_2\rmk
\in \lmk
\Tbdv{\rho_1}{\lm {r1}}\Tbdv{\rho_2}{\lm {r2}},
\Tbdv{\rho_1'}{\lm {r1}'}\Tbdv{\rho_2'}{\lm {r2}'}
\rmk_l,
\end{align}

and this defines the tensor product of intertwiners.
This tensor product satisfies following.

(i)$(R_1\otimes R_2)^*=R_1^*\otimes R_2^*$.

(ii) With
$\rho_i''\in \Orbd$,
$\lm {ri}''\in \Crbd$,
$\Vrl{\rho_i''}{\lm{ri}''}\in \Vbd{\rho_i''}{\lm{ri}''}$
and $R_i'\in \lmk\Tbdv{\rho_i'}{\lm {ri}'}, \Tbdv{\rho_i''}{\lm {ri}''}\rmk_l$
for $i=1,2$,
we have
$R'_1R_1\otimes R_2'R_2=(R_1'\otimes R_2')(R_1\otimes R_2)$.

(iii) If $\rho_i,\rho_i'\in \OrUbd$,
$\Vrl{\rho_i}{\lm{ri}}\in \VUbd{\rho_i}{\lm{ri}}$,
$\Vrl{\rho_i'}{\lm{ri}}\in \VUbd{\rho_i'}{\lm{ri}}$
and $R_i\in \lmk\Tbdv{\rho_i}{\lm {r1}}, \Tbdv{\rho_i'}{\lm {r1}'}\rmk_U$,
then $R_1\otimes R_2\in\fbd$.

The proof is identical to that of Lemma 4.3,4.5 of [15].

Next we introduce direct sum to $\Orbdl$.

Consider the setting in subsection <ref>.
Let $\llz\in \pc$.
Assume Assumption <ref>.
Then for any $\rho,\sigma\in \Orbdl$, there exist
$\gamma\in \Orbdl$, isometries
$u\in (\rho,\gamma)_l$, $v\in (\sigma,\gamma)_l$
such that $uu^*+vv^*=\unit$.
In particular, $\gamma=\Ad u \circ\rho+\Ad v \circ \sigma$.
If $\rho,\sigma\in \OrUbdl$, then
$\gamma$ and $u,v$ can be chosen so that
$\gamma\in \OrUbdl$,
$u\in (\rho,\gamma)_U$, $v\in (\sigma,\gamma)_U$.
If $\rho,\sigma\in \Obul$, then
$\gamma$ and $u,v$ can be chosen so that
$\gamma\in \Obul$,
$u\in (\rho,\gamma)_U$, $v\in (\sigma,\gamma)_U$.

For each $\ld\in\CUbk$, with $\Lambda\subset\hu$,
$ \pbd(\caA_{\ld})''$ is properly infinite by Assumption <ref>.
Therefore, there exist

$u_{\ld},v_{\ld}\in  \pbd(\caA_{{\ld}})''\subset\fbd$ such that

For each $\lm r\in \Crbd$, choose $\lm {\lm r}\in \CUbk$ with $(\lm {\lm r},\lm r)\in \pc$.

Set $\gamma:=\Ad(u_{\lz})\rho+\Ad(v_{\lz})\sigma$, $u:=u_{\lz}$, $v:=v_{\lz}$ and
$W_{{{\lm r}}}:=u_{{{\ld}_{\lm r}}}\Vrl{\rho}{{{\lm r}}}u_{\lz}^*+v_{{{\ld}_{\lm r}}}\Vrl{\sigma}{{{\lm r}}}v_{\lz}^*$,
with $\Vrl{\rho}{{{\lm r}}}\in \Vbd\rho{{{\lm r}}}$, $\Vrl{\sigma}{{{\lm r}}}\in \Vbd\sigma{{{\lm r}}}$.
Then as in the proof of 5.7 [15], we have
$\gamma\in \Orbdl$, $u\in (\rho,\gamma)_U$, $v\in (\sigma,\gamma)_U$
and $W_{\lm r}\in \Vbd\rho{\lm r}$.

Suppose $\rho,\sigma\in \Obul$.
For any $\ld\in\CUbk$, choose $\tilde\Lambda_\ld\in \CUbk$ with
$X_\ld:=u_{\tilde\Lambda_\ld}\Vrl\rho\ld u_{\lz}^*+v_{\tilde\Lambda_\ld}\Vrl\sigma\ld v_{\lz}^*$
with $\Vrl\rho\ld\in \Vbu\rho\ld$, $\Vrl\sigma\ld\in \Vbu\sigma\ld$.
Then we have $X_\ld\in\Vbu\gamma\ld$ and

This defines the direct sum.
Note that it is unique up to $\simeq_l$
(resp. $\simeq_U$) because
for other $\tilde \gamma\in \Orbdl$
(resp. $\tilde \gamma\in \OrUbdl$)
$\tilde u\in (\rho,\tilde \gamma)_l$, $\tilde v\in (\sigma,\tilde \gamma)_l$,
(resp. $\tilde u\in (\rho,\tilde \gamma)_U$, $\tilde v\in (\sigma,\tilde \gamma)_U$), we have
$\tilde u u^*+\tilde v v^*\in (\gamma,\tilde\gamma)_l$
(resp. $\tilde u u^*+\tilde v v^*\in (\gamma,\tilde\gamma)_U$).

Next we prove the existence of subobject.
First we prepare several Lemmas.

Consider the setting in subsection <ref>.
Then for any $\lm 1,\lm 2\subset \hu$,
$\pbd\lmk \caA_{\lm 1}\rmk ''\cap\pbd\lmk \caA_{(\lm 2)^c\cap\hu}\rmk'$
is a factor.

Note that $\caM:=\pbd\lmk \caA_{\lm 1}\rmk ''\cap\pbd\lmk \caA_{(\lm 2)^c\cap\hu }\rmk'$,
includes $\pbd\lmk\caA_{\lm 1\cap\lm 2}\rmk''$.
Therefore, we have

\begin{align}
\caM\cap\caM'
\subset
\pbd\lmk \caA_{(\lm 2)^c\cap\hu}\rmk '\cap\pbd\lmk \caA_{(\lm 1)^c\cap\hu }\rmk'\cap \pbd\lmk\caA_{\lm 2\cap\lm 1}\rmk'
\end{align}

because $\pbd$ is irreducible.

Consider the setting in subsection <ref> and assume Assumption <ref>.
Let $\lm{r1},\lm{r2}\in\Crbd$, $D\in \CUbk$ with
$D\cap\lm {r1}=\emptyset$ and $D,\lm {r1}\subset\lm{r2}$.
Then for any projection $p\in \pbd\lmk \caA_{(\lm {r1})^c\cap\hu}\rmk'\cap\bl$,
there exists an isometry
$w\in \pbd\lmk \caA_{(\lm {r2})^c\cap\hu}\rmk'\cap\bl$
such that $ww^*=p$.

We apply Lemma 5.10 of [15].
Let $\delta>0$ be the number given in  Lemma 5.10 of [15].
We also use Lemma <ref>, Lemma <ref> and Lemma <ref>.
Consider the number $\delta_3(\delta_2(\delta))$ given for
$\delta>0$ with the functions
$\delta_2$, $\delta_3$ given in Lemma <ref>, Lemma <ref>.

Let $p\in \pbd\lmk \caA_{(\lm {r1})^c\cap\hu}\rmk'\cap\bl$ be a projection.
Then, because $p\in\bl$, there exists $\lm l\in \Clbd$,
and self-adjoint $x\in \pbd(\caA_{\lm l})''$ such that

\rV\le \delta_3(\delta_2(\delta))$.

Taking larger cone if necessary, we may assume
$(\lm{r1})^c\cap \hu\subset\lm l$ and that  there exists $\Lambda\in\CUbk$
such that
\begin{align}\label{shizuoka}
\Lambda\subset\lm l\cap\lm {r1}\subset \lm l \cap\lm {r2}.
\end{align}

Applying Lemma <ref> with replacing
$\lm 1$, $\lm 2$, $\Gamma$, $(\caH,\pi)$, $x$, $y$, $\varepsilon$
$\lm {r1}$, $\lm l$, $\hu$, $(\hbd,\pbd)$, $p$, $x$, $\delta_3(\delta_2(\delta))$
respectively, we obtain self-adjoint
$z\in \pbd\lmk\caA_{(\lm {r1})^c\cap\hu}\rmk'\cap\pbd(\caA_{\lm l})''$
such that $\lV p-z\rV\le \delta_3(\delta_2(\delta))$.

Now apply Lemma <ref> with
$\caH$, $\caA$, $p$, $x$
replaced by $\hbd$, $\pbd\lmk\caA_{(\lm {r1})^c\cap\hu}\rmk'\cap\pbd(\caA_{\lm l})''$,
$p$, $z$ respectively.
Then from Lemma <ref>, we obtain
a projection $q\in \pbd\lmk\caA_{(\lm {r1})^c\cap\hu}\rmk'\cap\pbd(\caA_{\lm l})''$
such that $\lV q-p\rV<\delta_2(\delta)$.

Next we apply Lemma <ref> with
$\caA$, $p$, $q$ replaced by
$ \pbd\lmk\caA_{(\lm {r1})^c\cap\hu}\rmk'\cap\bl$, $p$, $q$ respectively.
Then from Lemma <ref>, we obtain a unitary
$u\in \pbd\lmk\caA_{(\lm {r1})^c\cap\hu}\rmk'\cap\bl$
such that $p=uqu^*$, $\lV u-\unit\rV<\delta$.

From Lemma <ref>, $\caM:=\pbd\lmk\caA_{(\lm {r2})^c\cap\hu}\rmk'\cap\pbd(\caA_{\lm l})''$
is a factor.
For the cone $\Lambda\in\CUbk$ satisfying (<ref>),
$\pbd(\caA_{\Lambda})''$ is an infinite factor from Assumption (<ref>).
Because $\caM$ includes $\pbd(\caA_{\Lambda})''$, it means
$\caM$ is an infinite factor.

Now we apply Lemma 5.10 of [15], with
$\caH$, $\caN$, $\caM$, $p$, $u$
replaced by $\hbd$, $\pbd(\caA_{D})''$, $\caM$, $p$, $u^*$.
Then from Lemma 5.10 of [15], we have
$q\sim \unit$ in $\caM$.
Namely, there exists an isometry $v\in\caM$
such that

Set $w:=uv\in \pbd\lmk\caA_{(\lm {r2})^c\cap\hu}\rmk'\cap\bl$.
Then this $w$ is isometry with
$ww^*=uqu^*=p$, proving the claim.

Now we are ready to introduce subobjects.

Consider the setting in subsection <ref> and assume Assumption <ref>.
Let $\llz\in\pc$.
Then for any $\rho\in\Orbdl$ and a projection $p\in(\rho,\rho)_l$, there exists $\gamma\in\Orbdl$
and an isometry $v\in (\gamma,\rho)_l$
such that $vv^*=p$.
In particular, $\gamma=\Ad v^*\rho$.

For any $\lm r\in\Crbd$, fix $\Vrl\rho{\lm r}\in\Vbd\rho{\lm r}$ and set
$p_{\lm r}:=\Ad(\Vrl\rho{\lm r})(p)$.
Then as in the proof of Lemma 5.8 [15],
$p_{\lm r}$ is a projection in $\pbd\lmk\caA_{\lc{r}}\rmk'\cap\bl$.

For each $\lm r\in\Crbd$, fix $\Gamma_{\lm r}\in\Crbd$, $D_{\lm r}\in \CUbk$
such that $D_{\lm r}\cap \Gamma_{\lm r}=\emptyset$ and
$D_{\lm r},\Gamma_{\lm r}\subset\lm r$.
By Lemma <ref>, for the projection
$p_{{\Gamma_{\lm r}}}\in \pbd\lmk\caA_{(\Gamma_{\lm r})^c\cap\hu}\rmk'\cap\bl$, there exists
an isometry $w_{\lm r}\in \pbd\lmk\caA_{\lc{r}}\rmk'\cap\bl$
such that $w_{\lm r}w_{\lm r}^*=p_{\Gamma_{\lm r}}$.
We set $\gamma:=\Ad\lmk  w_{\lzr}^* \Vrl\rho{\Gamma_{\lzr}}\rmk\circ\rho$,
$X_{\lm r}:=w_{\lm r}^*\Vrl\rho{\Gamma_{\lm r}}\Vrl{\rho}{\Gamma_{\lzr}}^* w_{\lzr}$,
$v:=\Vrl{\rho}{\Gamma_{\lzr}}^* w_{\lzr}$.
Then, as in the proof of Lemma 5.8 [15],
we get
$\gamma\in\Orbdl$ $X_{\lm r}\in \Vbd\gamma{\lm r}$
and $v\in (\gamma,\rho)_l$ is an isometry
such that $vv^*=p$.

This defines the subobjects.
Note that it is unique up to $\simeq_l$ because
for other $\tilde \gamma\in\Orbd$,
$\tilde v\in (\gamma,\rho)_l$, $v^*\tilde v\in(\tilde\gamma,\gamma)_l$.
Hence we obtain the following.

Consider the setting in subsection <ref> and assume
Assumption <ref>.
Let $\lzr\in\Crbd$.
Let $\Carbdl$ be the category with objects $\Orbdl$ and morphisms
$(\rho,\sigma)_l$ between each $\rho,\sigma\in\Orbdl$.
Then $\Carbdl$ is the $C^*$-tensor category with respect to the
tensor structure $\otimes$,

\begin{align}\label{osaka}
\begin{split}
&\rho\otimes\sigma:=\rho\circ_{\lm {r0}}\sigma,\quad \rho,\sigma\in \Orbdl,\\
&R\otimes S:=R\Tbd\rho{\lzr}\unit(S),\quad
R\in (\rho,\rho')_l,\quad S\in (\sigma,\sigma')_l,\quad
\rho,\rho',\sigma,\sigma'\in\Orbdl.
\end{split}
\end{align}

The representation $\pbd\in \Orbdl$ is the tensor unit,
with $\unit$ as the associativity isomorphisms as well as
left and right multiplication by the tensor unit $\pbd$.

Recall the definition of the $C^*$-tensor category given in Definition 2.1.1 <cit.>.
Clearly, for any $\rho,\sigma\in \Orbdl$,
$(\rho,\sigma)_l$ is a Banach space
and composition as operators on $\hbd$

\begin{align}
(\sigma,\gamma)_l\times(\rho,\sigma)_l\ni (S,T)\mapsto ST\in(\rho,\gamma)_l
\end{align}

is a well defined bilinear map
satisfying $\lV ST\rV\le \lV S\rV\lV T\rV$.
The adjoint as operators on $\hbd$ gives a map

\begin{align}
(\rho,\sigma)_l\ni T\mapsto T^*\in (\sigma,\rho)_l
\end{align}

satisfying $T^{**}=T$, $\lV T^*T\rV=\lV T\rV^2$, $T^*T\ge 0$.
The formula (<ref>) gives
$\rho\otimes\sigma\in\Orbdl$ by Lemma <ref>, and
$R\otimes S\in (\rho\otimes\sigma,\rho'\otimes \sigma')$
by Lemma <ref>.
By Lemma <ref>,
$\unit : (\rho\otimes\sigma)\otimes\gamma\mapsto \rho\otimes(\sigma\otimes\gamma $
is an isomorphism
satisfying the naturality

\begin{align}
\begin{split}
(R\otimes S)\otimes T
=R\Tbd\rho{\lzr}\unit(S)\Tbd{\rho\otimes \sigma}{\lzr} \unit(T)
=R\Tbd\rho{\lzr}\unit(S)\Tbd{\rho}{\lzr} \unit\Tbd{\sigma}{\lzr}(T)\\
=R\Tbd\rho{\lzr}\unit\lmk S\Tbd{\sigma}{\lzr}(T)\rmk
=R\otimes\lmk S\otimes T\rmk
\end{split}
\end{align}

for $R\in (\rho,\rho')_l$, $S\in(\sigma,\sigma')_l$, $T\in (\gamma,\gamma')_l$,
using Lemma <ref>.
The unit $\unit$ on $\hbd$ also
gives natural isomorphisms
$\unit : \pbd\otimes\rho\to\rho$, $\unit :\rho\otimes\pbd\to \rho$
corresponding to the left and right multiplication by $\pbd$.

By Lemma <ref>, we have $(S\otimes T)^*=S^*\otimes T^*$.
By Lemma <ref>, the direct sums exist and by Lemma <ref>,
subobjects exist.
Because $\pbd$ is irreducible, $\pbd$ is simple.

Analogous to Lemma <ref>, we have the following Lemma.

Consider the setting in subsection <ref> and assume Assumption <ref>.
Let $\lm{1},\lm{2}\in\CUbk$, $D\in \CUbk$ with
$D\cap\lm {1}=\emptyset$ and $D,\lm {1}\subset\lm{2}\cap\hu $.
Then for any projection $p\in \pbd\lmk \caA_{(\lm {1})^c\cap\hu}\rmk'\cap\fbd$,
there exists an isometry
$w\in \pbd\lmk \caA_{(\lm {2})^c\cap\hu}\rmk'\cap\fbd$
such that $ww^*=p$.

Remove at next version

We apply Lemma 5.10 of [15].
Let $\delta>0$ be the number given in  Lemma 5.10 of [15].
We also use Lemma <ref>, Lemma <ref>.
Consider the number $\delta_3(\delta_2(\delta))$ given for
$\delta>0$(the number given in  Lemma 5.10 of [15]) with the functions
$\delta_2$, $\delta_3$ given in Lemma <ref>, Lemma <ref>.

Let $p\in \pbd\lmk \caA_{(\lm {1})^c\cap\hu}\rmk'\cap\fbd$ be a projection.
Then, because $p\in\fbd$, there exists $\lm {}\in \CUbk$,
and self-adjoint $x\in \pbd(\caA_{\Lambda\cap\hu})''$ such that

\rV\le \delta_3(\delta_2(\delta))$.
We may assume $D, \lm 2\subset\Lambda$.

Apply Lemma <ref> with
$\lm 1$, $\lm 2$, $\Gamma$, $(\caH,\pi)$,
replaced by
$\lm {1}$, $\ld\cap\hu$, $\hu$, $(\hbd,\pbd)$
Then we obtain a selfadjoint
$z\in \pbd\lmk\caA_{(\lm {1})^c\cap\hu}\rmk'\cap\pbd(\caA_{\lm {}\cap\hu })''$
such that $\lV p-z\rV\le \delta_3(\delta_2(\delta))$.

Now apply Lemma <ref> with
$\caH$, $\caA$, $p$, $x$
replaced by $\hbd$, $\pbd\lmk\caA_{(\lm {1})^c\cap\hu}\rmk'\cap\pbd(\caA_{\lm {}\cap\hu})''$,
$p$, $z$ respectively.
Then from Lemma <ref>, we obtain
a projection $q\in \pbd\lmk\caA_{(\lm {1})^c\cap\hu}\rmk'\cap\pbd(\caA_{\lm {}\cap\hu})''$
such that $\lV q-p\rV<\delta_2(\delta)$.

Note that $p, q\in \pbd\lmk\caA_{(\lm {1})^c\cap\hu}\rmk'\cap\fbd$.
Next we apply Lemma <ref> with
$\caA$, $p$, $q$ replaced by
$ \pbd\lmk\caA_{(\lm {1})^c\cap\hu}\rmk'\cap\fbd$, $p$, $q$ respectively.
Then from Lemma <ref>, we obtain a unitary
$u\in \pbd\lmk\caA_{(\lm {1})^c\cap\hu}\rmk'\cap\fbd$
such that $p=uqu^*$, $\lV u-\unit\rV<\delta$.

From Lemma <ref>, $\caM:=\pbd\lmk\caA_{(\lm {2})^c\cap\hu}\rmk'\cap\pbd(\caA_{\lm {}\cap\hu })''$
is a factor.
Because $\caM$ includes $\pbd(\caA_{\Lambda_2\cap\hu})''$, which is an infinite factor from Assumption <ref>,
$\caM$ is an infinite factor.

Now we apply Lemma 5.10 of [15], with
$\caH$, $\caN$, $\caM$, $p$, $u$
replaced by $\hbd$, $\pbd(\caA_{D})''$, $\caM$, $p$, $u^*$.
Then from Lemma 5.10 of [15], we have
$q\sim \unit$ in $\caM$.
Namely, there exists an isometry $v\in\caM$
such that

Set $w:=uv\in \pbd\lmk\caA_{(\lm {2})^c\cap\hu}\rmk'\cap\fbd$.
Then this $w$ is isometry with
$ww^*=uqu^*=p$, proving the claim.

Consider the setting in subsection <ref> and assume Assumption <ref>.
Let $\llz\in\pc$.
Then for any $\rho\in\Obul$ and a projection $p\in(\rho,\rho)_U$, there exists $\gamma\in\Obul$
and an isometry $v\in (\gamma,\rho)_U$
such that $vv^*=p$.
In particular, $\gamma=\Ad v^*\rho$.

For each $\ld \in\CUbk$, fix $\Vrl\rho{\ld}\in \Vbu\rho{\ld}$ and set
$p_{\ld }:=\Ad(\Vrl\rho{\ld})(p)$.
Then as in the proof of Lemma 5.8 [15],
$p_{\ld}$ is a projection in $\pbd\lmk\caA_{\ld^c\cap\hu }\rmk'\cap\fbd$.

For each $\ld\in \CUbk$, fix some $\Gamma_\ld,D_\ld\in \CUbk$
such that $\Gamma_\ld,D_\ld\subset \ld\cap \hu$ and $\Gamma_\ld\cap D_\ld=\emptyset$.
Applying Lemma <ref>, with
$\lm{1},\lm{2}$, $D$, $p$ with
replaced by $\Gamma_\ld$, $\ld$, $D_\ld$, $p_{\Gamma_\ld}$,
there exists an isometry
$w_\Lambda\in \pbd\lmk \caA_{(\ld)^c\cap\hu}\rmk'\cap\fbd$
such that $w_\ld w_\ld^*=p_{\Gamma_\ld}$.

We set $\gamma:=\Ad\lmk  w_{\lz}^* \Vrl\rho{\Gamma_{\lz}}\rmk\circ\rho$,
$X_{\ld}:=w_{\ld}^*\Vrl\rho{\Gamma_{\ld}}\Vrl{\rho}{\Gamma_{\lz}}^* w_{\lz}$,
$v:=\Vrl{\rho}{\Gamma_{\lz}}^* w_{\lz}$.
Then, as in the proof of Lemma 5.8 [15],
we get
$\gamma\in\Obul$, $X_{\ld}\in \Vbu\gamma{\ld}$
and $v\in (\gamma,\rho)_U$ is an isometry
such that $vv^*=p$.

Consider the setting in subsection <ref> and assume
Assumption <ref>.
Let $\llz\in\pc$.
Let $\Cabul$ be the category with objects $\Obul$ and morphisms
$(\rho,\sigma)_U$ between each $\rho,\sigma\in\Obul$.
Then $\Cabul$ is a $C^*$-tensor subcategory of $\Carbdl$ in the following sense:

(i) $\Cabul$ is closed under $*$, composition of morphisms, and tensor.

(ii) Direct sum $\gamma$, $u\in (\rho,\gamma)_l$,
$v\in (\sigma,\gamma)_l$ of $\rho,\sigma\in \Obul$ can be taken so that
$\gamma\in \Obul$, $u\in (\rho,\gamma)_U$,
$v\in (\sigma,\gamma)_U$.

(iii) For any $\rho\in \Obul$ and projection $p\in (\rho,\rho)_U$,
subobject $\gamma$, $v\in (\gamma,\rho)_l$ can be taken so that $\gamma\in \Obul$
and $v\in (\gamma,\rho)_U$ .

For any $\rho,\sigma\in \Obul$, $(\rho,\sigma)_U\subset (\rho,\sigma)_l$, and (i) is
(iv) of Lemma <ref>.
(ii) is proven in Lemma <ref>.(iii) is Lemma <ref>.

The same proof gives the following.

Consider the setting in subsection <ref> and assume
Assumption <ref>.
Then the same statement as Theorem <ref>  holds
when we replace $\Cabul$ by $\CarUbdl$,
the category with objects $\OrUbdl$ and morphisms
$(\rho,\sigma)_U$ for each $\rho,\sigma\in \OrUbdl$.

§ BRAIDING ON $\OBUL$

In this section, we show that for $\Obul$, we may assign braiding.
We use the following notation.

For cones $\{\lm {1\alpha}\}\subset\CUbk$
(resp. $\{\lm {l2\alpha}\}\subset\Clbd$) and $\{\lm {r2\beta}\}\subset \Crbd$,
we write $\{\lm {1\alpha}\}\leftarrow_r\{\lm {r2\beta}\}$ (resp $\{\lm {l2\alpha}\}\leftarrow_r\{\lm {r2\beta}\}$) if
there exists an $\varepsilon>0$ such that
$\arg(\lm {1\alpha})_\varepsilon\cap \arg(\lm {r2\beta})_\varepsilon=\emptyset$
(resp. $\arg(\lm {l2\alpha})_\varepsilon\cap \arg(\lm {r2\beta})_\varepsilon=\emptyset$
) for all $\alpha,\beta$.
We also write $\lm 1\leftarrow_r\lm {r2}$ if $\{\lm 1\}\leftarrow_r\{\lm {r2}\}$.
For $\lm {1}=\Lambda_{\bm a,\theta,\varphi}\in\CUbk$,
$\lm {r2}\in \Crbd$, $\lm {l2}\in \Clbd$ and $t\in \bbR$, we set
$\lm 1(t):=\lm 1+t\bm e_\theta$,
$\lm {r2}(t):=\lm{r2}+t\bm e_{0}$, $\lm {l2}(t):=\lm{l2}+t\bm e_{0}$, where $\bm e_0:=(1,0)$.

We first investigate the asymptotic behaviors of our endomorphisms.

Consider the setting in subsection <ref>. Assume Assumption <ref>.
Let $\lm {l1}\in\Clbd$ and $\lm {r2}\in \Crbd$,
with $\lm {l1}\leftarrow_r\lm {r2}$.
Then for any $\sigma\in \Orbd$ and $\Vrl\sigma{\lm {r2}(t)}\in \Vbd\sigma{\lm {r2}(t)}$, $t\ge 0$, we have

\begin{align}
\lim_{t\to\infty}
\lV\left.
\lmk
\Tbdv\sigma{\lm {r2}(t)}-\id
\rmk\right\vert_{\pbd\lmk\caA_{(\lm {l1})^c\cap\hu} \rmk'\cap \bl}
\rV=0,
\end{align}

\begin{align}
\lim_{t\to\infty}
\sup_{\sigma\in O^{r}_{\mathop{\mathrm{bd},\lm {r2}(t)}}}
\lV\left.
\lmk
\Tbd\sigma{\lm {r2}(t)}\unit-\id
\rmk\right\vert_{\pbd\lmk\caA_{(\lm {l1})^c\cap\hu} \rmk'\cap \bl}
\rV=0.
\end{align}

Let $\lm {l1}=\lef a \varphi\in \Clbd$.
Fix $\varepsilon>0$ such that $\arg(\lm {l1})_{2\varepsilon}\cap \arg(\lm {r2})_{2\varepsilon}
Then there exists a $t_1\in\bbR_+$ such that
$\lmk \lef {a+s} {(\varphi+2\varepsilon)} \rmk\cap \lm {r2}(t)=\emptyset$ for all $t,s\in\bbR_+$ with
$t-s\ge t_1$.

By Assumption <ref>, for any $s\ge R^{(l)\mopbd}_{\varphi,\varepsilon}$,
we have

\begin{align}
{\pbd\lmk\caA_{(\lm {l1})^c\cap\hu} \rmk'}\subset_{2 f^{(l)}_{\varphi,\varepsilon,\varepsilon}(s)} \pbd\lmk
\caA_{
\lef{a+s}{(\varphi+2\varepsilon)}
\end{align}

By Lemma <ref>, if $\lmk \lef {a+s} {(\varphi+2\varepsilon)} \rmk\cap \lm {r2}(t)=\emptyset$ ,
(hence if $t-s\ge t_1$)
we have

\begin{align}\left.
\lmk
\Tbdv\sigma{\lm {r2}(t)}-\id
\rmk
\right\vert_{\pbd\lmk\caA_{ \lef{a+s}{(\varphi+2\varepsilon)} }\rmk''}=0.
\end{align}

Therefore, if $t-s\ge t_1$, we have

\begin{align}
\lV
\left.
\lmk
\Tbdv\sigma{\lm {r2}(t)}-\id
\rmk
\right\vert_{{\pbd\lmk\caA_{(\lm {l1})^c\cap\hu} \rmk'}\cap\bl}\rV
\le 4f^{(l)}_{\varphi,\varepsilon,\varepsilon}(s).
\end{align}

This completes the proof of the first equality.
The proof of the second one is the same.

The following Lemma gives the support of $\Tbd\rho{\lzr}{\Vrl\rho\lz}$
for $\rho\in\Obu$

Consider the setting in subsection <ref>.
Let $\lz\in\CUbk$, $\lzr\in\Crbd$, $\lm {l0}\in \Clbd$ satisfying
$\hu\cap\lz\subset \lzr\cap \lm {l0}$.
If $\rho\in\Obu$ and $\Vrl\rho\lz\in \Vbu\rho\lz\subset \VUbd\rho{\lzr}$ then

\begin{align}\label{fukushima}\left.
\Tbd\rho{\lzr}{\Vrl\rho\lz}\right\vert_{\pbd\lmk \caA_{(\lm {l0})^c\cap\hu}\rmk''\cap\fbd}=\id,
\end{align}

\begin{align}\label{tochigi}
\left.
\Tbd\rho{\lzr}{\Vrl\rho\lz}\pbd\right\vert_{ \caA_{\lmk (\lm {l0})^c\cup (\lm {r0})^c\rmk\cap\hu}}
\end{align}

By interchanging the role of left and right in Lemma <ref>,
we can define an endomorphism ${T_\rho^{\mathrm{(r)} \lm {l0}\Vrl{{\rho}}{\lz}}}$ of
$\caB_r:=\overline{\cup_{\Lambda_r\in \Crbd}\pbd\lmk\caA_{(\Lambda_r)^c\cap \hu}\rmk''}^n$
which is $\sigma$-weak continuous on any
$\pbd(\caA_{{(\lm {l})^c\cap \hu}})''$, $\lm{l}\in\Clbd$, and

\begin{align}\label{yamagata}
{T_\rho^{\mathrm{(r)} \lm {l0}\Vrl{{\rho}}{\lz}}}\pbd=\Ad\lmk \Vrl\rho\lz\rmk\rho=\Tbd\rho{\lzr}{\Vrl\rho\lz}\pbd
\end{align}

\begin{align}\label{miyagi}
\left.{T_\rho^{\mathrm{(r)} \lm {l0}\Vrl{{\rho}}{\lz}}}\right\vert_{\pbd\lmk \caA_{(\lm {l0})^c\cap\hu}\rmk''}=\id.
\end{align}
By the $\sigma$-weak continuity and
(<ref>), we have
${T_\rho^{\mathrm{(r)} \lm {l0}\Vrl{{\rho}}{\lz}}}\vert_{\fbd}=\Tbd\rho{\lzr}{\Vrl\rho\lz}\vert_{\fbd}$.
Combining this with (<ref>),
we obtain (<ref>).

By Lemma <ref>, we have $
\Tbd\rho{\lzr}{\Vrl\rho\lz}\vert_{\pbd\lmk \caA_{\lmk (\lm {r0})^c\rmk\cap\hu}\rmk}=\id
$. We also have $
\Tbd\rho{\lzr}{\Vrl\rho\lz}\vert_{\pbd\lmk \caA_{\lmk (\lm {l0})^c\rmk\cap\hu}\rmk}=\id
$  from (<ref>).
Hence we obtain (<ref>).

Consider the setting in subsection <ref>, and assume Assumption <ref>.
Let $(\lm 1,\lm {r1})\in \pc$, $\lm {r2}\in \Crbd$
with $\lm 1\leftarrow_r \lm {r2}$.
Let $\rho\in \Obu$.
Then the following hold.

For any $\Vrl \rho{\lm 1}\in \Vbu\rho{\lm 1}$,

\begin{align}
\lim_{t\to\infty }
\lV\left.
\lmk \Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1}}-\id\rmk
\right\vert_{\pbd\lmk \caA_{\lmk \lm {r2}(t)\rmk^c\cap\hu}\rmk'\cap\fbd}
\rV=0.
\end{align}

For any
$\Vrl\rho{\lm 1(t)}\in \Vbu\rho{\lm 1(t)}$, $t\ge 0$
we have

\begin{align}
\lim_{t\to\infty}
\lV
\left.
\lmk \Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t)}}-\id\rmk
\right\vert_{{\pbd\lmk \caA_{\lmk \lm {r2}\rmk^c\cap\hu}\rmk'\cap\fbd}}
\rV=0.
\end{align}

The cone $\lm {r2}$ is of the form $\lm {r2}=\lr a\varphi$.
Because $\lm 1\leftarrow_r \lm {r2}$, there exists $\varepsilon>0$ such that
$\arg \lmk \lm 1\rmk_\varepsilon\cap \arg\lmk \lm{r2}\rmk_\varepsilon=\emptyset$.

Fix $b\in \bbR$ large enough so that
$\lm l:=\lmk\lr  {b} {(\varphi+\varepsilon)}\rmk^c\cap \hu\in\Clbd$ satisfies
$\Lambda_1\subset \lm l\cap\lm {r1}$.
We apply Lemma <ref> replacing
$\lz$, $\lm {r0}$, $\lm {l0}$, $\rho$, $\Vrl\rho{\lz}$ by
$\lm 1$, $\lm {r1}$, $\lm l$, $\rho$,
$\Vrl\rho{\lm 1}\in\Vbu\rho{\lm 1}$ respectively.
Then we obtain

\begin{align}\label{taiwan}
\left.\Tbd\rho{\lm {r1}}{\Vrl\rho{\lm1}}\right\vert_{\pbd\lmk \caA_{(\lm {l})^c\cap\hu}\rmk''\cap\fbd}=\id.
\end{align}

By Assumption <ref>, for any $t\ge 2R^{(r)\mathop{\mathrm {bd}}}_{\varphi,\frac \varepsilon 2}$,
there exists $W_t\in \caU(\fbd)$ such that

\begin{align}\label{ibaraki}
&\pbd\lmk \caA_{\lmk \lm {r2}(t)\rmk^c\cap\hu}\rmk'
\subset
\Ad\lmk W_t\rmk\lmk
\pbd\lmk
\caA_{\lr {a+\frac t2}{\varphi+\varepsilon}}
\rmk''
\rmk,\\
\rV\le {f_{\varphi, \frac \varepsilon 2,\frac \varepsilon 2 }\lmk \frac t2\rmk}.
\end{align}

If $t\ge \max\{2(b-a),2R^{(r)\mathop{\mathrm {bd}}}_{\varphi,\frac \varepsilon 2}\}$, then
$\lr {(a+\frac t2)}{\varphi+\varepsilon}\subset \lr {b}{\varphi+\varepsilon}=(\lm l)^c\cap\hu$,
and combining with (<ref>), we have

\begin{align}
\begin{split}
\pbd\lmk \caA_{\lmk \lm {r2}(t)\rmk^c\cap\hu}\rmk'
\subset
\Ad\lmk W_t\rmk
\lmk
\pbd\lmk
\caA_{(\lm l)^c\cap\hu}
\rmk''
\rmk.
\end{split}
\end{align}

Now for any $t\ge \max\{2(b-a),2R^{\mathop{\mathrm {bd}}}_{\varphi,\frac \varepsilon 2}\}$
and  $x\in {\pbd\lmk \caA_{\lmk \lm {r2}(t)\rmk^c\cap\hu}\rmk'\cap\fbd}$,
we have $\Ad(W_t^*)(x)\in \pbd\lmk
\caA_{(\lm l)^c\cap\hu}
\rmk''\cap\fbd
$ and

\begin{align}\label{gunma}
\begin{split}
&\lV\lmk \Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1}}-\id\rmk(x)\rV
\le
\lV\lmk \Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1}}-\id\rmk\lmk \Ad(W_t^*)(x)\rmk\rV
+2\lV x-\Ad(W_t^*)(x)\rV<4f_{\varphi, \frac \varepsilon 2,\frac \varepsilon 2 }\lmk \frac t2\rmk\lV x\rV,
\end{split}
\end{align}

using (<ref>)
This proves (i).

(ii)For each $s\ge R^{\mathop{\mathrm {bd}}}_{\varphi,\frac \varepsilon 2}$,
by Assumption <ref>, there exists $\tilde W_s\in\caU(\fbd)$ such that

\begin{align}
&\pbd\lmk \caA_{\lmk \lm {r2}\rmk^c\cap\hu}\rmk'
\subset
\Ad\lmk\tilde W_s\rmk
\lmk
\pbd\lmk
\caA_{\lr {a-s}{\varphi+\varepsilon}}
\rmk''
\rmk,\\
\tilde W_s-\unit
\rV\le {f_{\varphi, \frac \varepsilon 2,\frac \varepsilon 2 }(s)}
\end{align}

For each $s\ge 0$, there exists a $t_0(s)\ge 0$ such that
$\lm 1(t)\cap {\lr {a-s}{\varphi+\varepsilon}}=\emptyset$ for all $t\ge t_0(s)$.

Applying Lemma <ref> replacing
$\lz$, $\lm {r0}$, $\lm {l0}$, $\rho$, $\Vrl\rho{\lz}$ by
$\lm 1(t)$, $\lm {r1}$, $\lmk \lr {a-s}{\varphi+\varepsilon}\rmk^c\cap\hu$, $\rho$,
$\Vrl\rho{\lm 1(t)}\in\Vbu\rho{\lm 1(t)}$ respectively, for $t\ge t_0(s)$
and obtain

\begin{align}
\begin{split}
\left.\Tbd{\rho}{\lm {r1}}{\Vrl \rho{\lm 1(t)}}\right\vert_{\pbd\lmk
\caA_{{\lr {a-s}{\varphi+\varepsilon}}}\rmk''\cap\fbd}
\end{split}
\end{align}

Now for any $x\in {{\pbd\lmk \caA_{\lmk \lm {r2}\rmk^c\cap\hu}\rmk'\cap\fbd}}
$s\ge R^{(r)\mathop{\mathrm {bd}}}_{\varphi,\frac \varepsilon 2}$ and
$t\ge t_0(s)$
and  we have $\Ad(\tilde W_s^*)(x)\in \pbd\lmk
\caA_{\lr {a-s}{\varphi+\varepsilon}}
\rmk''\cap\fbd
$ and

\begin{align}\label{nagasaki}
\begin{split}
&\lV \lmk \Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t)}}-\id\rmk(x)\rV\\
\lV \lmk \Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t)}}-\id\rmk
\lmk \Ad(\tilde W_s^*)(x)\rmk\rV
+2\lV \Ad(\tilde W_s^*)(x)-x\rV
\le {4f_{\varphi, \frac \varepsilon 2,\frac \varepsilon 2 }(s)}\lV x\rV
\end{split}
\end{align}

Hence for any $s\ge R^{(r)\mathop{\mathrm {bd}}}_{\varphi,\frac \varepsilon 2}$ and
$t\ge t_0(s)$,
we have
$\lV \lmk \Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t)}}-\id\rmk\vert_{{{\pbd\lmk \caA_{\lmk \lm {r2}\rmk^c\cap\hu}\rmk'\cap\fbd}}}\rV
\le {4f_{\varphi, \frac \varepsilon 2,\frac \varepsilon 2 }(s)}$,
and this proves (ii).

Consider the setting in subsection <ref>.
Assume Assumption <ref> and Assumption <ref>.

Let $\rho,\rho'\in \Obu$, $\sigma,\sigma'\in \OrUbd$.
Let $\lm 1,\lm 1'\in \CUbk$, $\lm {r2},\lm {r2}'\in\Crbd$
with $\{\lm 1,\lm 1'\}\leftarrow_r \{\lm {r2},\lm {r2}'\}$.
Let $\lm {r1}, \lm {r1}'\in\Crbd$ with
$(\lm {1}, \lm {r1})\in \pc$, $(\lm {1}', \lm {r1}')\in \pc$.
Let $\Vrl\rho{\lm 1(t_1)}\in \Vbu\rho{\lm 1(t_1)}\subset \Vbd\rho{\lm {r1}}$,
$\Vrl{\rho'}{{\lm 1}'(t_1')}\in \Vbu{\rho'}{{\lm 1}'(t_1')}\subset \Vbd\rho{\lm {r1}'}$,
$\Vrl\sigma{\lm {r2}(t_2)}\in \VUbd\sigma{\lm {r2}(t_2)}$,
$\Vrl{\sigma'}{\lm {r2}'(t_2')}\in \VUbd{\sigma'}{\lm {r2}'(t_2)}$,
for $t_1,t_1',t_2,t_2'\ge 0$.
Then for any

\begin{align}
\begin{split}
S_1^{(t_1,t_1')}\in \lmk
\Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t_1)}},
\Tbd{\rho'}{\lm {r1}'}{\Vrl{\rho'}{\lm 1'(t_1')}}
\rmk_U,\\
S_2^{(t_2,t_2')}\in \lmk
\Tbd\sigma{\lm {r2}}{\Vrl\sigma{\lm {r2}(t_2)}},
\Tbd{\sigma'}{\lm {r2}'}{\Vrl{\sigma'}{\lm {r2}'(t_2')}}
\rmk_U
\end{split}
\end{align}

with $\lV S_1^{(t_1,t_1')}\rV, \lV S_2^{(t_2,t_2')}\rV\le 1$,
we have

\begin{align}
\lV
S_1^{(t_1,t_1')}\otimes S_2^{(t_2,t_2')}-
S_2^{(t_2,t_2')}\otimes S_1^{(t_1,t_1')}
\rV\to 0,\quad t_1,t_2,t_1',t_2'\to\infty.
\end{align}

By the assumption $\{\lm 1,\lm 1'\}\leftarrow_r \{\lm {r2},\lm {r2}'\}$,
there exist an $\varepsilon>0$ and ${\hlm {1}}\in\CUbk$, ${\hlm {r2}}\in \Crbd$

such that

\begin{align}
\begin{split}
&\arg\lmk\lm 1\rmk_\varepsilon \cup \arg\lmk\lm 1'\rmk_\varepsilon\subset \arg{\hlm 1},\quad
\arg\lmk\lm {r2}\rmk_\varepsilon\cup \arg\lmk\lm {r2}'\rmk_\varepsilon\subset \arg{\hlm {2r}},\\
&(\hlm 1)_{\varepsilon}\cap(\hlm {r2})_\varepsilon=\emptyset,\quad
\arg (\hlm 1)_{\varepsilon}\cap\arg (\hlm {r2})_\varepsilon=\emptyset, \\
&{\hlm 1}\subset \hu,\quad |\arg\hlm 1|+4\varepsilon<2\pi.
\end{split}
\end{align}

By Lemma A.2 of [15], for any $t\ge 0$ fixed, there exists $t_0(t)\ge 0$
such that

\begin{align}
\lm 1(t_1), \lm 1'(t_1')\subset \hlm 1(t),\quad
t_1,t_1'\ge t_0(t).
\end{align}

Similarly, for $t\ge 0$ fixed, there exists $\tilde t_0(t)\ge 0$ such that

\begin{align}
\lm {r2}(t_2), \lm {r2}'({t_2'})\subset \hlm{r2}(t),\quad
t_2,t_2'\ge\tilde t_0(t).
\end{align}

By Lemma <ref> (<ref>), for any $t\ge 0$, $t_1,t_1'\ge t_0(t)$,
$\Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t_1)}}\pbd$
and $\Tbd{\rho'}{\lm {r1}'}{\Vrl{\rho'}{\lm 1'(t_1')}}\pbd$
are $\pbd$ on $\caA_{(\hlm 1(t))^c\cap \hu}$.
as in Lemma 4.2 of [15], we have

\begin{align}\label{chiba}
\begin{split}
S_1^{(t_1,t_1')}\in \lmk
\Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t_1)}},
\Tbd{\rho'}{\lm {r1}'}{\Vrl{\rho'}{\lm 1'(t_1')}}
\rmk_U
\subset
\pbd\lmk\caA_{(\hlm 1(t))^c\cap \hu}\rmk'\cap\fbd
\end{split}
\end{align}

for any $t\ge 0$, $t_1,t_1'\ge t_0(t)$.
Similarly, we have

\begin{align}\label{kanagawa}
S_2^{(t_2,t_2')}\in \lmk
\Tbd\sigma{\lm {r2}}{\Vrl\sigma{\lm {r2}(t_2)}},
\Tbd{\sigma'}{\lm {r2}'}{\Vrl{\sigma'}{\lm {r2}'(t_2')}}
\rmk_U\subset \pbd\lmk\caA_{(\hlm {r2}(t))^c\cap \hu}\rmk'\cap\fbd,
\end{align}

for any $t\ge 0$, $ t_2,t_2'\ge\tilde t_0(t)$.
As in the proof of Lemma <ref> (<ref>),
by Assumption <ref>
there exists $T_0\ge 0$ such that for any $t\ge T_0$
we have

\begin{align}
\begin{split}
&\pbd\lmk \caA_{\lmk \hlm {r2}(t)\rmk^c\cap\hu}\rmk'
\subset
\Ad\lmk W_t\rmk\lmk
\pbd\lmk
\caA_{\lmk \lmk \hlm{r2}\rmk_\varepsilon\cap\hu\rmk+\frac t 2\bm e_1}
\rmk''
\rmk,
\end{split}
\end{align}

with some $W_t\in\caU(\fbd)$ such that $\lim_{t\to\infty}\lV
\rV=0$.
Note that
$\lmk \lmk \hlm{r2}\rmk_\varepsilon\cap\hu\rmk+\frac t 2\bm e_1\subset (\hlm 1(t))^c\cap \hu$
for any $t\ge 0$.
Therefore, from (<ref>), we have

\begin{align}
\begin{split}
& \lV
\left[S_1^{(t_1,t_1')}, S_2^{(t_2,t_2')}\right]
\rV
% =\lV
% \left[S_1^{(t_1,t_1')}, W_t\cdot \Ad (W_t^*)\lmk S_2^{(t_2,t_2')}\rmk \cdot W_t^*\right]
% \rV\\
\le 4\lV W_t-\unit\rV
\end{split}
\end{align}

for any $t\ge 0$, $t_1,t_1'\ge t_0(t)$ and $ t_2,t_2'\ge\tilde t_0(t)$.
By Lemma <ref> and (<ref>), we have

\begin{align}
\begin{split}
\lim_{t_1,t_1',t_2\to\infty} \lV \lmk\Tbdv\sigma{\lm {r2}(t_2)}-\id\rmk\lmk S_1^{(t_1,t_1')}\rmk\rV
\end{split}
\end{align}

By Lemma <ref> and (<ref>), we have

\begin{align}
\begin{split}\lim_{t_2,t_2',t_1\to\infty}
\lV
\lmk \Tbd{\rho}{\lm {r1}}{\Vrl \rho{\lm 1(t_1)}}-\id\rmk
\lmk S_2^{(t_2,t_2')}\rmk
\rV=0.
\end{split}
\end{align}

Combining them we obtain

\begin{align}
\begin{split}
\lV
S_1^{(t_1,t_1')}\otimes S_2^{(t_2,t_2')}-
S_2^{(t_2,t_2')}\otimes S_1^{(t_1,t_1')}
\rV\\
\lV S_1^{(t_1,t_1')}
\Tbd{\rho}{\lm {r1}}{\Vrl \rho{\lm 1(t_1)}}\lmk S_2^{(t_2,t_2')}\rmk
-S_2^{(t_2,t_2')}\Tbdv\sigma{\lm {r2}(t_2)}\lmk S_1^{(t_1,t_1')} \rmk
\rV\\
&=\lim_{t_1,t_1',t_2,t_2'\to\infty} \lV
\left[S_1^{(t_1,t_1')}, S_2^{(t_2,t_2')}\right]
\rV=0.
\end{split}
\end{align}

The statement of Lemma <ref> holds even when
we replace the condition $\{\lm 1,\lm 1'\}\leftarrow_r \{\lm {r2},\lm {r2}'\}$,
by $\lm 1\leftarrow_r \lm {r2}$, $\lm 1'\leftarrow_r \lm {r2}'$.

Note that analogous to Lemma 4.9 of [15], there are sequences of cones $\{\lm 1^{(i)}\}_{i=0}^3\subset\CUbk$,
$\{\lm {r2}^{(i)}\}_{i=0}^3\subset \Crbd$ such that
$\{\lm 1^{(i)},\lm 1^{(i+1)}\}\leftarrow_r \{ \lm {r2}^{(i)}, \lm {r2}^{(i+1)}\}$, $i=0,\ldots,2$
and $\lm 1^{(0)}=\lm 1$, $\lm {r2}^{(0)}=\lm {2r}$,
$\lm 1^{(3)}=\Lambda_{\bm 0, \frac\pi 2,\frac\pi{64}}$,
$\lm {r2}^{(3)}=\lr 0 {\frac\pi{64}}$.
Then the rest is the same as Lemma 4.10 of [15].

Fix some $\tlm {r1}\in \Crbd$ including all $\lm 1^{(i)}$.

With $\Vrl{\rho}{\lm 1^{(i)}(t_1^{(i)})}\in \Vbu{\rho}{\lm 1^{(i)}(t_1^{(i)})}$,
$\Vrl{\sigma}{\lm {r2}^{(i)}(t_2^{(i)})}\in \VUbd \sigma {\lm {r2}^{(i)}}$,
we set

\begin{align}
W_{\rho}^{(i)}\lmk t_1^{(i)}, t_1^{(i+1)}\rmk:=
\Vrl{\rho}{\lm 1^{(i+1)}(t_1^{(i+1)})}\lmk \Vrl{\rho}{\lm 1^{(i)}(t_1^{(i)})}\rmk^*
\in \lmk
\Tbd\rho {\tlm {r1}}{\Vrl{\rho}{\lm 1^{(i)}(t_1^{(i)})}},
\Tbd\rho {\tlm {r1}}{\Vrl{\rho}{\lm 1^{(i+1)}(t_1^{(i+1)})}}
\rmk_U,
\end{align}

\begin{align}
\tilde W_{\sigma}^{(i)}\lmk t_2^{(i)}, t_2^{(i+1)}\rmk:=
\Vrl{\sigma}{\lm {r2}^{(i+1)}(t_2^{(i+1)})}\lmk \Vrl{\sigma}{\lm {r2}^{(i)}(t_2^{(i)})}\rmk^*
\in \lmk
\Tbdv\sigma {{\lm {r2}^{(i)}(t_2^{(i)})}},
\Tbdv\sigma {{\lm {r2}^{(i+1)}(t_2^{(i+1)})}}
\rmk_U.
\end{align}

Then by Lemma <ref>, we have

\begin{align}\label{thailand}
\begin{split}
\lim_{t_1^{(i)}, t_2^{(i)}, t_1^{(i+1)}, t_2^{(i+1)}\to\infty}
\lV
W_{\rho}^{(i)}\lmk t_1^{(i)}, t_1^{(i+1)}\rmk\otimes
\tilde W_{\sigma}^{(i)}\lmk t_2^{(i)}, t_2^{(i+1)}\rmk
-\tilde W_{\sigma}^{(i)}\lmk t_2^{(i)}, t_2^{(i+1)}\rmk\otimes W_{\rho}^{(i)}\lmk t_1^{(i)}, t_1^{(i+1)}\rmk
\rV=0.
\end{split}
\end{align}

We carry out exactly the same argument to $\lm 1'$, $\lm {r2}'$, and obtain the $'$-version of the objects above.

We set

\begin{align}\label{tokyo}
\begin{split}
&R_{\bm t}:=
{W'}_{\rho'}^{(2)}\lmk {t_1'}^{(2)}, {t_1'}^{(3)}\rmk\cdots{W'}_{\rho'}^{(0)}\lmk {t_1'}^{(0)}, {t_1'}^{(1)}\rmk
\lmk W_{\rho}^{(0)}\lmk t_1^{(0)}, t_1^{(1)}\rmk \rmk^*\cdots \lmk W_{\rho}^{(2)}\lmk t_1^{(2)}, t_1^{(3)}\rmk\rmk^*\\
\lmk \Tbd\rho{\tlm {r1}}{\Vrl \rho {\Lambda_{\bm 0, \frac\pi 2,\frac\pi{64}}(t_1^{(3)})}}, \Tbd{\rho'}{\tlm {r1}'}{\Vrl {\rho'} {\Lambda_{\bm 0, \frac\pi 2,\frac\pi{64}}({t_1'}^{(3)})}}\rmk,\\
&\tilde R_{\bm t}:=
{\tilde {W'}}_{\sigma'}^{(2)}\lmk {t_2'}^{(2)}, {t_2'}^{(3)}\rmk\cdots{\tilde{W'}}_{\sigma'}^{(0)}\lmk {t_2'}^{(0)}, {t_2'}^{(1)}\rmk
\lmk W_{\sigma}^{(0)}\lmk t_2^{(0)}, t_2^{(1)}\rmk \rmk^*\cdots \lmk W_{\sigma}^{(2)}\lmk t_2^{(2)}, t_2^{(3)}\rmk\rmk^*\\
\lmk \Tbdv\sigma{{\lr 0 {\frac\pi{64}}(t_2^{(3)})}}, \Tbdv{\sigma'}{{\lr 0 {\frac\pi{64}}({t_2'}^{(3)})}}\rmk_U
\end{split}
\end{align}

Then by Lemma <ref>, we have

\begin{align}
\lim_{\bm t\to\infty}
\lV
R_{\bm t}\otimes \tilde R_{\bm t}-\tilde R_{\bm t}\otimes R_{\bm t}
\rV=0.
\end{align}

As in Lemma 4.10 of [15], combining this and (<ref>),
we obtain $\lV
S_1^{(t_1,t_1')}\otimes S_2^{(t_2,t_2')}-
S_2^{(t_2,t_2')}\otimes S_1^{(t_1,t_1')}
\rV\to 0$.

Consider the setting in subsection <ref> and assume Assumption <ref>, <ref>.
Let $(\lz,\lzr)\in \pc$, and  $\rho\in \Obul$, $\sigma\in \OrUbdl$.
Then the norm limit

\begin{align}\label{halfdef}
\begin{split}
\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk
\lmk \Vrl\sigma{\lm {r2}(t_2)}\otimes \Vrl\rho{\lm 1(t_1)}\rmk^*
\lmk \Vrl\rho{\lm 1(t_1)}\otimes  \Vrl\sigma{\lm {r2}(t_2)}\rmk\in\fbd
\end{split}
\end{align}

exists for any
$\lm 1\in \CUbk$, $\lm {r2}\in \Crbd$
with $\lm 1\leftarrow_r \lm{r2}$,
and $\Vrl\rho{\lm 1(t_1)}\in \Vbu\rho{\lm 1(t_1)}$, $t_1\ge 0$,
$\Vrl \sigma{\lm {r2}(t_2)}\in \VUbd\sigma{\lm {r2}(t_2)}$, $t_2\ge 0$,
and its value is independent of the choice of
$\lm 1\in \CUbk$, $\lm {r2}\in \Crbd$
with $\lm 1\leftarrow_r \lm{r2}$,
and $\Vrl\rho{\lm 1(t_1)}\in \Vbu\rho{\lm 1(t_1)}$, $t_1\ge 0$,
$\Vrl \sigma{\lm {r2}(t_2)}\in \VUbd\sigma{\lm {r2}(t_2)}$, $t_2\ge 0$.
Furthermore, we have

\begin{align}\label{niigata}
\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk=
\lim_{t_2\to\infty}\lmk \Vrl \sigma{\lm {r2}(t_2)}\rmk^*\Tbd\rho{\lzr}\unit
\lmk\Vrl \sigma{\lm {r2}(t_2)}\rmk.
\end{align}

for any
$\lm {r2}\in \Crbd$ with $\lz\leftarrow_r \lm {r2}$
$\Vrl \sigma{\lm {r2}(t_2)}\in \VUbd\sigma{\lm {r2}(t_2)}$, $t_2\ge 0$.

From (<ref>), what we are doing here can be understood as follows.
Recall that $\Vrl \sigma{\lm {r2}(t_2)}$ sends $\sigma$ particle from
$\lzr$ to “right infinity " $\lm {r2}(t_2)$, as $t_2\to \infty$.
When we do so, our $\sigma$ particle may cross the string creating the $\rho$-particle.
What we are measuring with $\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk$
is the braiding which is caused at the cross.

For any $\lm 1,\lm 1'\in \CUbk$, $\lm {r2},\lm {r2}'\in \Crbd$
with $\lm 1\leftarrow_r \lm{r2}$,  $\lm 1'\leftarrow_r \lm{r2}'$.
Let $(\lm 1,\lm{r1}), (\lm 1',\lm{r1}')\in\pc$.
Let $\Vrl\rho{\lm 1(t_1)}\in \Vbu\rho{\lm 1(t_1)}$,
$\Vrl \sigma{\lm {r2}(t_2)}\in \VUbd\sigma{\lm {r2}(t_2)}$.
Then we have

\begin{align}\begin{split}
&\lmk \Vrl\sigma{\lm {r2}'(t_2')}\otimes \Vrl\rho{\lm 1'(t_1')}\rmk^*
\lmk \Vrl\rho{\lm 1'(t_1')}\otimes  \Vrl\sigma{\lm {r2}'(t_2')}\rmk\\
& =\lmk \Vrl\sigma{\lm {r2}(t_2)}\otimes \Vrl\rho{\lm 1(t_1)}\rmk^*\\
\Vrl\sigma{\lm {r2}'(t_2')}\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk^*
\otimes
\Vrl\rho{\lm 1'(t_1')}\Vrl\rho{\lm 1(t_1)}^*
\rmk^*\\
& \lmk
\Vrl\rho{\lm 1'(t_1')}\lmk \Vrl\rho{\lm 1(t_1)}\rmk^*\otimes
\Vrl\sigma{\lm {r2}'(t_2')}\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk^*\rmk
\\
& \lmk \Vrl\rho{\lm 1(t_1)}\otimes  \Vrl\sigma{\lm {r2}(t_2)}\rmk
\end{split}
\end{align}

By Lemma <ref>, we have

\begin{align}
\begin{split}
\lmk
\Vrl\sigma{\lm {r2}'(t_2')}\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk^*
\otimes
\Vrl\rho{\lm 1'(t_1')}\Vrl\rho{\lm 1(t_1)}^*
\rmk^*\lmk
\Vrl\rho{\lm 1'(t_1')}\lmk \Vrl\rho{\lm 1(t_1)}\rmk^*\otimes
\Vrl\sigma{\lm {r2}'(t_2')}\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk^*\rmk
\rV\\
&\quad\to 0,\quad\quad\quad t_1,t_1',t_2,t_2'\to\infty.
\end{split}
\end{align}

Hence the norm limit (<ref>) exists and it is independent of the choice of
$\lm 1\in \CUbk$ etc.
The first part is identical to the proof of Lemma 4.11 [15].
Because $ \Vrl\sigma{\lm {r2}(t_2)}$ etc belong to $\fbd$, by Lemma <ref>,
$\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk$ belongs to $\fbd$.

For any
$\lm {r2}\in \Crbd$  with $\lz\leftarrow_r \lm {r2}$
$\Vrl \sigma{\lm {r2}(t_2)}\in \Vbd\sigma{\lm {r2}(t_2)}$, $t_2\ge 0$,
set $\lm 1:=\lz \in \CUbk$, and take any $\Vrl\rho{\lm 1(t_1)}\in \Vbu\rho{\lm 1(t_1)}$, $t_1\ge 0$.
Then we have

\begin{align}
\begin{split}
\lmk \Vrl\sigma{\lm {r2}(t_2)}\otimes \Vrl\rho{\lm 1(t_1)}\rmk^*
\lmk \Vrl\rho{\lm 1(t_1)}\otimes  \Vrl\sigma{\lm {r2}(t_2)}\rmk
=\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk^*\Tbdv \sigma {\lm {r2}(t_2)}\lmk \Vrl\rho{\lm 1(t_1)}^*\rmk
\Vrl\rho{\lm 1(t_1)}\Tbd\rho{\lz}\unit\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk.
\end{split}
\end{align}

Because $\lm 1:=\lz $, we have
$ \Vrl\rho{\lm 1(t_1)}\in
\pbd\lmk \caA_{\lz^c\cap\hu }\rmk'\cap \fbd$.
Because of $\lz\leftarrow_r \lm {r2}$, from Lemma <ref>, we have
$\Tbdv \sigma {\lm {r2}(t_2)}\lmk \Vrl\rho{\lm 1(t_1)}^*\rmk
\Vrl\rho{\lm 1(t_1)}\to \unit$ , $t_2\to 0$ in norm, uniformly in $t_1\ge 0$.
Hence we obtain (<ref>).

The proof of the following Lemma  is identical to that of Lemma 4.13 [15].

Consider the setting in subsection <ref> and assume Assumption <ref>, <ref>.
Let $\rho\in \Obul$, $\sigma\in \OrUbdl$ and $(\lz,\lzr)\in \pc$.
Then we have $\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk\in \lmk \Tbd \rho{\lzr}{\unit}\Tbd\sigma{\lzr}\unit, \Tbd\sigma{\lzr}\unit \Tbd \rho{\lzr}{\unit}\rmk_U$.

Let $(\lm 1,\lm{r1})\in\pc$.
For any $A\in \aloch$, we have

\begin{align}
\begin{split}
&\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk \Tbd \rho{\lzr}{\unit}\Tbd\sigma{\lzr}\unit\pbd(A)\\
\lmk \Vrl\sigma{\lm {r2}(t_2)}\otimes \Vrl\rho{\lm 1(t_1)}\rmk^*
\Tbd \rho{\lm {r1}}{\lm 1(t_1)}\Tbdv\sigma{\lm {r2}(t_2)}\pbd(A)
\lmk \Vrl\rho{\lm 1(t_1)}\otimes  \Vrl\sigma{\lm {r2}(t_2)}\rmk\\
\lmk \Vrl\sigma{\lm {r2}(t_2)}\otimes \Vrl\rho{\lm 1(t_1)}\rmk^*
\pbd(A)
\lmk \Vrl\rho{\lm 1(t_1)}\otimes  \Vrl\sigma{\lm {r2}(t_2)}\rmk\\
&= \lim_{t_1,t_2\to\infty}
\lmk \Vrl\sigma{\lm {r2}(t_2)}\otimes \Vrl\rho{\lm 1(t_1)}\rmk^*
\Tbdv\sigma{\lm {r2}(t_2)}\Tbd \rho{\lm {r1}}{\lm 1(t_1)}\pbd(A)
\lmk \Vrl\rho{\lm 1(t_1)}\otimes  \Vrl\sigma{\lm {r2}(t_2)}\rmk\\
&=\Tbd\sigma{\lzr}\unit \Tbd \rho{\lzr}{\unit}\pbd(A)\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk,
\end{split}
\end{align}

because  $A\in\caA_{(\lm 1(t_1)^c\cap (\lm {r2}(t_2))^c}$ eventually.
By the $\sigma$-weak continuity of $\Tbd \rho{\lzr}{\unit}\Tbd\sigma{\lzr}\unit=\Tbd {\rho\circ_{\lzr}\sigma}{\lzr}{\unit}$, etc
(Lemma <ref>, <ref>), this proves the claim.

The proof of the following Lemma is identical to that of Lemma 4.14 [15],
using Lemma <ref>.

Consider the setting in subsection <ref> and assume Assumption <ref>, <ref>.
Let $(\lz,\lzr)\in \pc$ and $\rho\in \Obul$, $\sigma,\sigma'\in \OrUbdl$.
Then for any
$S\in \lmk\Tbd\sigma{\lzr}\unit, \Tbd{\sigma'}{\lzr}\unit\rmk_U $,
we have

\begin{align}
\iota^{(\lz,\lzr)}\lmk\rho: \sigma'\rmk\lmk \unit_\rho\otimes S\rmk
=\lmk S\otimes \unit_\rho \rmk\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk.
\end{align}

Note that $S$ is required to be in $\fbd$, not $\bl$.

For any
$\lm 1\in \CUbk$, $\lm {r2}\in \Crbd$
with $\lm 1\leftarrow_r \lm{r2}$,

\begin{align}
\begin{split}
\iota^{(\lz,\lzr)}\lmk\rho: \sigma'\rmk\lmk \unit_\rho\otimes S\rmk
-\lmk S\otimes \unit_\rho\rmk\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk
\rV\\
\lV
\begin{gathered}
\lmk \Vrl{\sigma'}{\lm {r2}(t_2)}\otimes \Vrl{\rho}{\lm 1(t_1)}\rmk^*
\lmk \Vrl{\rho}{\lm 1(t_1)}\otimes  \Vrl{\sigma'}{\lm {r2}(t_2)}\rmk
\lmk \unit_\rho\otimes S\rmk
-\lmk S\otimes \unit_\rho\rmk
\lmk \Vrl\sigma{\lm {r2}(t_2)}\otimes \Vrl\rho{\lm 1(t_1)}\rmk^*
\lmk \Vrl\rho{\lm 1(t_1)}\otimes  \Vrl\sigma{\lm {r2}(t_2)}\rmk
\end{gathered}
\rV\\
\lV
\begin{gathered}
\unit_{\Tbd\rho{\lm {r1}}{\lm 1(t_1)}}
\otimes
\Vrl{\sigma'}{\lm {r2}(t_2)} S \lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk^*
\Vrl{\sigma'}{\lm {r2}(t_2)}S\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk^*
\otimes
\unit_{\Tbd\rho{\lm {r1}}{\lm 1(t_1)}}
\end{gathered}
\rV\\
\end{split}
\end{align}

by Lemma <ref>.

Consider the setting in subsection <ref> and assume Assumption <ref>, <ref>.
Let $(\lz,\lzr)\in \pc$ and $\rho,\rho'\in \Obul$, $\sigma\in \OrUbdl$.
Then for any $R\in \lmk \Tbd\rho{\lzr}\unit, \Tbd{\rho'}{\lzr}\unit\rmk_l$,
we have

\begin{align}
\iota^{(\lz,\lzr)}\lmk\rho': \sigma\rmk\lmk R\otimes \unit_\sigma\rmk
=\lmk \unit_\sigma \otimes R\rmk\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk.
\end{align}

For any
$\lm 1\in \CUbk$, $\lm {r2}\in \Crbd$
with $\lm 1\leftarrow_r \lm{r2}$,

\begin{align}
\begin{split}
\iota^{(\lz,\lzr)}\lmk\rho': \sigma\rmk\lmk R\otimes \unit_\sigma\rmk
-\lmk \unit_\sigma\otimes R\rmk\iota^{(\lz,\lzr)}\lmk\rho: \sigma\rmk
\rV\\
\lV
\begin{gathered}
\lmk \Vrl{\sigma}{\lm {r2}(t_2)}\otimes \Vrl{\rho'}{\lm 1(t_1)}\rmk^*
\lmk \Vrl{\rho'}{\lm 1(t_1)}\otimes  \Vrl{\sigma}{\lm {r2}(t_2)}\rmk
\lmk R\otimes \unit_\sigma\rmk
-\lmk \unit_\sigma\otimes R\rmk
\lmk \Vrl\sigma{\lm {r2}(t_2)}\otimes \Vrl\rho{\lm 1(t_1)}\rmk^*
\lmk \Vrl\rho{\lm 1(t_1)}\otimes  \Vrl\sigma{\lm {r2}(t_2)}\rmk
\end{gathered}
\rV\\
\lV
\begin{gathered}
\Vrl{\rho'}{\lm 1(t_1)} R\lmk \Vrl\rho{\lm 1(t_1)}\rmk^*
\otimes
\Vrl{\sigma}{\lm {r2}(t_2)} \unit_\sigma \lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk^*
\Vrl{\sigma}{\lm {r2}(t_2)}\unit_\sigma\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk^*
\otimes
\Vrl{\rho'}{\lm 1(t_1)} R\Vrl\rho{\lm 1(t_1)}^*
\end{gathered}
\rV\\
\lim_{t_1,t_2\to\infty}
\lV
\begin{gathered}
\Vrl{\rho'}{\lm 1(t_1)} R\lmk \Vrl\rho{\lm 1(t_1)}\rmk^*
\Tbdv{\sigma}{\lm {r2}(t_2)}\lmk
\Vrl{\rho'}{\lm 1(t_1)} R\Vrl\rho{\lm 1(t_1)}^*
\rmk
\end{gathered}
\rV\\
\end{split}
\end{align}

At the last equality, we used the fact

\begin{align}
\begin{split}
\Vrl{\rho'}{\lm 1(t_1)} R\Vrl\rho{\lm 1(t_1)}^*\in
\pbd\lmk
\caA_{{{\lm 1}(t_1)}^c\cap \hu}
\rmk'\cap\bl
\end{split}
\end{align}

and Lemma <ref>.

Furthermore, we have the following.

Consider the setting in subsection <ref> and assume Assumption <ref>, <ref>.
Let $(\lz,\lzr)\in \pc$ and $\rho\in \Obul$, $\sigma,\gamma\in \OrUbdl$.
Then we have

\begin{align}
\begin{split}
\hb{\rho}{\sigma\circ_{\lzr}\gamma}
=\lmk \unit_{\sigma}\otimes \hb\rho\gamma\rmk
\lmk \hb\rho\sigma\otimes\unit_\gamma\rmk.
\end{split}
\end{align}

The proof is the same as [15] Lemma 4.17.

Fix some $(\lm 1,\lm {r1})\in \pc$, $\lm {r2}\in \Crbd$
so that $\lm 1\leftarrow_r\lzr$,
$\lz\leftarrow_r\lm{r2}$, $\lm{r2}\subset \lzr$.
Then we have $\lm 1\leftarrow_r \lm{r2}$, and from
Proposition <ref>,

\begin{align}\label{ishikawa}
\begin{split}
&\lmk \unit_{\sigma}\otimes \hb\rho\gamma\rmk
\lmk \hb\rho\sigma\otimes\unit_\gamma\rmk\\
\Tbd{\sigma}{\lzr}\unit \lmk
\Tbd{\gamma}{\lzr}\unit \lmk\Vrl\rho{\lm 1(t_1)}^*\rmk
\Vrl\gamma{\lm {r2}(t_2)}^*\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit\lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\rmk\\
&\quad \quad \Tbd{\sigma}{\lzr}\unit \lmk\Vrl\rho{\lm 1(t_1)}^*\rmk
\Vrl\sigma{\lm {r2}(t_2)}^*\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk\\
\Tbd{\sigma}{\lzr}\unit \lmk
\Tbd{\gamma}{\lzr}\unit \lmk\Vrl\rho{\lm 1(t_1)}^*\rmk
\Vrl\gamma{\lm {r2}(t_2)}^*
\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit\lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\Vrl\rho{\lm 1(t_1)}^*\rmk
\Vrl\sigma{\lm {r2}(t_2)}^*\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk\\
\Tbd{\sigma}{\lzr}\unit \lmk
\Tbd{\gamma}{\lzr}\unit \lmk\Vrl\rho{\lm 1(t_1)}^*\rmk
\Vrl\gamma{\lm {r2}(t_2)}^*
\Tbd\rho{\lm{r1}}{\Vrl\rho{\lm 1(t_1)}}\lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\rmk
\Vrl\sigma{\lm {r2}(t_2)}^*\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk\\
\Tbd{\sigma}{\lzr}\unit \lmk
\Tbd{\gamma}{\lzr}\unit \lmk\Vrl\rho{\lm 1(t_1)}^*\rmk
\Vrl\gamma{\lm {r2}(t_2)}^*
\rmk
\Vrl\sigma{\lm {r2}(t_2)}^*\\
&\quad\quad\quad\Tbdv\sigma{\lm {r2}(t_2)}\Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t_1)}}\lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk.
\end{split}
\end{align}

Here, because $\lm{r2}\subset \lzr$, we have

\begin{align}
\begin{split}
\Vrl\gamma{\lm {r2}(t_2)},\;
\Tbdv\sigma{\lm {r2}(t_2)}\lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk\in
\pbd\lmk
\caA_{\lrhuz}
\rmk'\cap\fbd.
\end{split}
\end{align}

Therefore, by Lemma <ref>, we have

\begin{align}
\begin{split}
\lim_{t_1,t_2\to\infty}
\lV
\lmk
\Tbdv\sigma{\lm {r2}(t_2)}\Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t_1)}}
-\Tbd\rho{\lm {r1}}{\Vrl\rho{\lm 1(t_1)}}\Tbdv\sigma{\lm {r2}(t_2)}
\rmk
\lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\rV=0.
\end{split}
\end{align}

Substituting this to (<ref>), we obtain

\begin{align}
\begin{split}
&\lmk \unit_{\sigma}\otimes \hb\rho\gamma\rmk
\lmk \hb\rho\sigma\otimes\unit_\gamma\rmk\\
\Tbd{\sigma}{\lzr}\unit \lmk
\Tbd{\gamma}{\lzr}\unit \lmk\Vrl\rho{\lm 1(t_1)}^*\rmk
\Vrl\gamma{\lm {r2}(t_2)}^*
\rmk
\Vrl\sigma{\lm {r2}(t_2)}^*\\
&\quad\quad\quad\Tbd\rho{\lm{r1}}{\Vrl\rho{\lm 1(t_1)}}\Tbdv\sigma{\lm {r2}(t_2)}\lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk\\
\Tbd{\sigma}{\lzr}\unit \lmk
\Tbd{\gamma}{\lzr}\unit \lmk\Vrl\rho{\lm 1(t_1)}^*\rmk
\rmk
\Tbd{\sigma}{\lzr}\unit \lmk
\Vrl\gamma{\lm {r2}(t_2)}^*
\rmk
\Vrl\sigma{\lm {r2}(t_2)}^*\\
\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit
\lmk
\Tbdv\sigma{\lm {r2}(t_2)}\lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk
\rmk\\
\Tbd{\sigma}{\lzr}\unit \lmk
\Tbd{\gamma}{\lzr}\unit \lmk\Vrl\rho{\lm 1(t_1)}^*\rmk
\rmk
\lmk
\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk
\Tbd\sigma{\lzr}\unit \lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\rmk^*\\
\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit
\lmk\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk
\Tbd\sigma{\lzr}\unit \lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\rmk\\
\lmk\lmk  \lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk
\Tbd\sigma{\lzr}\unit \lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk\rmk
\otimes
\lmk\Vrl\rho{\lm 1(t_1)}
\rmk\rmk^*
\\
\Vrl\rho{\lm 1(t_1)}
\Tbd\rho{\lzr}\unit
\lmk\lmk \Vrl\sigma{\lm {r2}(t_2)}\rmk
\otimes  \lmk \Vrl\gamma{\lm {r2}(t_2)}\rmk
\rmk\\
\end{split}
\end{align}

Let $\llz\in\pc$.
Consider the setting in subsection <ref>, and
assume Assumption <ref>,Assumption <ref>.
Then for any $\rho,\sigma\in \Obul$ and $\gamma\in \OrUbdl$, we have

\begin{align}
\begin{split}
\hb { \rho}{\gamma}\otimes
\unit_{\Tbd{\sigma}\lzr\unit}
\rmk
\lmk
\unit_{\Tbd{\rho}\lzr\unit}
\otimes
\hb { \sigma}{\gamma}
\rmk
\end{split}
\end{align}

For any $\lm {r2}\in \Crbd$ with $\lz\leftarrow_r \lm{r2}$ and  $\Vrl\gamma{\lm {r2}(t)}\in \VUbd\gamma{{\lm {r2}(t)}}$,
$t\ge 0$
from Proposition <ref>,
we have

\begin{align}
\begin{split}
\hb {\rho}{\gamma}\otimes
\unit_{\Tbd{\sigma}\lzr\unit}
\rmk
\lmk
\unit_{\Tbd{\rho}\lzr\unit}
\otimes
\hb {\sigma}{\gamma}
\rmk\\
\lim_{t\to\infty} \Vrl\gamma{\lm {r2}(t)}^*
\Tbd{\rho}{\lzr}\unit \lmk \Vrl\gamma{\lm {r2}(t)}\rmk
\lmk
\Vrl\gamma{\lm {r2}(t)}^*
\Tbd{\sigma}{\lzr}\unit \lmk \Vrl\gamma{\lm {r2}(t)}\rmk
\rmk\\
\end{split}
\end{align}

Hence if we restrict our $\iota^{\llz}$ to $\Obul$, then it becomes a braiding, and
we obtain the following.

Consider the setting in subsection <ref> and assume
Assumption <ref>, Assuumption <ref>,
Assumption <ref>.
Let $\llz\in\pc$.
Then $\Cabul$ is braided with braiding

\begin{align}
\begin{split}
\epsilon_{\Cabul}(\rho:\sigma)
:=\hb\rho\sigma,\quad\rho,\sigma\in \Obul.
\end{split}
\end{align}

Our $\hb-\rho$ satisfies the following asymptotic property.

Consider the setting in subsection <ref> and assume Assumption <ref>, <ref>.
Let $(\lz,\lzr)\in \pc$ and
$\lm 1,\lm 3\in \CUbk$ with $\lm 1\leftarrow_r \lm 3$, $\lm 1,\lm3\subset\lz$.
Then for any $\rho\in \Obul$ and $\Vrl\rho{\lm3}\in \Vbu\rho{\lm 3}$
we have

\begin{align}
\begin{split}
\lim_{t\to\infty}\sup_{\sigma\in \Obun{\lm 1(t)}}
\lV
\hb\sigma \rho -\Vrl\rho{\lm3}^* \Tbd\sigma{\lzr}\unit\lmk \Vrl\rho{\lm3}\rmk
\rV=0.
\end{split}
\end{align}

Note in particular

\begin{align}
\begin{split}
\lim_{t\to\infty}\sup_{\sigma\in \Obun{\lm 1(t)}}
\lV
{\Vrl\rho{\lm3}'}^* \Tbd\sigma{\lzr}\unit\lmk {\Vrl\rho{\lm3}'}\rmk-\Vrl\rho{\lm3}^* \Tbd\sigma{\lzr}\unit\lmk \Vrl\rho{\lm3}\rmk
\rV
\end{split}
\end{align}

for any other $\Vrl\rho{\lm3}'\in \Vbu\rho{\lm 3}$.

If $\rho\in \Obun{\lm3}$, we may take $\Vrl\rho{\lm3}^* \Tbd\sigma{\lzr}\unit\lmk \Vrl\rho{\lm3}\rmk=\unit$,
and we have $\lim_{t\to\infty}\sup_{\sigma\in \Obun{\lm 1(t)}}
\lV
\hb\sigma \rho -\unit
\rV=0$.
In our definition of $\hb\sigma \rho$, (roughly) we are moving the $\rho$ particle
to right infinity, and measuring the braiding at the cross with $\sigma$'s string operator.
If $\rho$ is on the right of $\sigma$ from the beginning (for example , if $\rho\in \Obun{\lm3}$), then, there will be no
crossing. Therefore $\hb\sigma \rho\approx \unit$. That is what this Lemma is telling us.

For $\rho\in \Obul$ and $\Vrl\rho{\lm3}\in \Vbu\rho{\lm 3}$, we set

\begin{align}
\tilde\rho:=\Ad\lmk \Vrl\rho{\lm3}\rmk\rho\in \Obun{\lm3}\subset\Obul.
\end{align}

Then, because $\Vrl\rho{\lm3}\in (\rho,\tilde\rho)_U$,
by Lemma <ref>,
we have

\begin{align}\label{usagi}
\begin{split}
\hb \sigma{\tilde\rho}\lmk \unit_\sigma\otimes \Vrl\rho{\lm3} \rmk
=\lmk \Vrl\rho{\lm3}\otimes \unit_\sigma\rmk
\hb \sigma\rho,
\end{split}
\end{align}

for each $\sigma\in \Obun{\lm 1(t)}$.
Let $\lm{r2},\lm{r4}\in \Crbd$ with
$\lz\leftarrow_r \lm{r2}$,
$\lm 1\leftarrow \lm{r4}$,
$\lm 3\subset \lm{r4}$.
Then for any $\Vrl{\tilde\rho}{\lm {r2}(t_2)}\in \VUbd{\tilde\rho}{\lm {r2}(t_2)}$,
$t_2\ge 0$, we have
$\Vrl{\tilde\rho}{\lm {r2}(t_2)}\in \pbd\lmk \caA_{\lm {r4}^c\cap\hu}\rmk'\cap\fbd$
for $t_2\ge 0$ large enough.
Hence we  have

\begin{align}
\begin{split}
&\sup_{\sigma\in \Obun{\lm 1(t)}}
\lV
\hb \sigma {\tilde\rho}-\unit
\rV
\sup_{\sigma\in \Obun{\lm 1(t)}}
\lim_{t_2\to \infty}
\lV
\Vrl{\tilde\rho}{\lm {r2}(t_2)}^*
\lmk
\Tbd\sigma{\lm {r0}}\unit-\id
\rmk\lmk \Vrl{\tilde\rho}{\lm {r2}(t_2)}\rmk
\rV\\
\sup_{\sigma\in \Obun{\lm 1(t)}}
\lV
\lmk\left.
\Tbd\sigma{\lm {r0}}\unit-\id
\rmk\right\vert_{\pbd\lmk \caA_{\lm {r4}^c\cap\hu}\rmk'\cap\fbd}
\rV.
\end{split}
\end{align}

By the same argument as Lemma <ref>, the last line
goes to $0$ as $t\to\infty$.
Combining this and (<ref>), we obtain the result.

§ A BULK-TO-BOUNDARY MAP

Recall from Lemma 2.13 of [15], for any
$\theta\in\bbR$, $\varphi\in (0,\pi)$, $\rho\in \Obk$, $\lz\in C_{(\theta,\varphi)}$,
$\Vrl{\rho}{\lz}\in \Vbk{\rho}{\lz}$,
we may define an endomorphism $T_{\rho}^{(\theta,\varphi)\lz \Vrl{\rho}{\lz}}$
on $\caB_{(\theta,\varphi)}$.
We use this symbol for the rest of this paper.
In this section, we construct a map from bulk to boundary under the Assumption <ref>.
Assumption <ref> gives a $*$-isomorphism between $\fbk$ and $\fbd$.

Consider the setting in subsection <ref>. Assume Assumption <ref>.
Then there exists a $*$-isomorphism
$\tau: \fbk\to \fbd$ such that

\begin{align}
\left.\tau\pbk\right\vert_{\abd}=\pbd.
\end{align}

By Assumption <ref>, for each $\Lambda\in\CUbk$, there exists
a $*$-isomorphism $\tau_\Lambda : \pbk\lmk {\caA_{\lhu}}\rmk''\to \pbd\lmk {\caA_{\lhu}}\rmk''$
such that

\begin{align}\label{hokkaido}
\tau_\Lambda\pbk(A)=\pbd(A),\quad\text{ for all} \;A\in {\caA_{\lhu}}.
\end{align}

From (<ref>) and $\sigma$-weak continuity of $\tau_\Lambda$s, for any $\Lambda_1,\Lambda_2\in\CUbk$
with $\Lambda_1\subset\Lambda_2$, we have $\tau_{\Lambda_2}\vert_{\pbk\lmk {\caA_{\Lambda_1}}\rmk''}
=\tau_{\Lambda_1}$. By the upward-filtering property of $\CUbk$, this means
that we can define an isometric $*$-homomorphism
$\tau_0$ from $\caF_{\mopbk}^{U(0)}$ onto $\caF_{\mopbd}^{U(0)}$

\begin{align}
\tau_0(x)=\tau_\Lambda(x), \quad \text{if}\; \;x\in \pbk\lmk {\caA_{\lhu}}\rmk'',\quad \Lambda\in\CUbk.
\end{align}

Because $\tau_0$ is an isometry, we can extend it to an isometric $*$-homomorphism
$\tau: \fbk\to \fbd$.
It is onto because $\tau(\fbk)\supset \tau_0(\caF_{\mopbk}^{U(0)})=\caF_{\mopbd}^{U(0)}$
and $\tau(\fbk)$ is closed.
Hence $\tau$ is a $*$-isomorphism.
For any $A\in \aloch$, there exists $\Lambda\in \CUbk$ such that $A\in \caA_{\lhu}$.
Hence we have

\begin{align}
\tau\pbk(A)=\tau_0\pbk\lmk A\rmk=\pbd(A).
\end{align}

As this holds for any $A\in \aloch$, the same is true for any $\abd$.

Consider the setting in subsection <ref>.
We have

\begin{align}\label{hyogo}
\gu\cap \pbk\lmk \caA_{\hd}\rmk'\subset\fbk.
\end{align}

For any $x\in \gu\cap \pbk\lmk \caA_{\hd}\rmk'$, by the definition of $\gu$,
there exists a sequence of $\Lambda_n\in \CUbk$ and $x_n\in \pbk\lmk\caA_{\Lambda_n}\rmk''$, $n\in\bbN$
such that $\lim_{n\to\infty}\lV x-x_n\rV=0$.
Set $S_n:=\Lambda_n\cap\hd$. Note that $S_n$ is a finite set, because $\Lambda_n\in \CUbk$.
We apply Lemma <ref> replacing
$\Gamma$, $S$, $(\caH,\pi)$ with $\Lambda_n$, $S_n$, $(\hbk,\pbk\vert_{\caA_{\Lambda_n}})$.
Then we obtain projections of norm $1$ $\caE_{S_n}: \caB(\hbk)\to \pbk(\caA_{S_n})'$
such that $\caE_{S_n}\lmk \pbk\lmk {\caA_{\Lambda_n}} \rmk''\rmk=
\pbk \lmk {\caA_{\Lambda_n\setminus S_n}} \rmk''=\pbk \lmk {\caA_{\Lambda_n\cap H_U}} \rmk''
\subset \fbk$.
From this, for $x_n\in \pbk\lmk\caA_{\Lambda_n}\rmk''$, we have
$\caE_{S_n}(x_n)\in \fbk$.
Because $x\in  \pbk\lmk \caA_{\hd}\rmk'\subset \pbk \lmk \caA_{S_n} \rmk'$,
we have $\caE_{S_n}(x)=x$.

\begin{align}
\lV
\rV=\lV \caE_{S_n}\lmk x-x_n\rmk\rV\le \lV x-x_n\rV\to 0.
\end{align}

Because $\caE_{S_n}(x_n)\in \fbk$, this means $x\in \fbk$.

Note that $\fbk\subset \caB_{(\frac {3\pi}2,\frac\pi 2)}=\caG$.
As in [15], we may define an endomorphism
$\Tbkv\rho {\Lambda_0}$ on $\caG$
for each $\rho\in \OUbkl$.
The restriction of this to $\fbk$ gives us an endomorphism on $\fbk$.

Consider the setting in subsection <ref>.
Assume Assumption <ref>.
Then for any $\Lambda_0\in \CUbk$ with $\Lambda_0\subset H_U$,
$\rho\in \Obk$, $\Vrl\rho{\Lambda_0}\in\Vbk \rho {\Lambda_0}$,
we have

\begin{align}
\Tbkv\rho{\Lambda_0}\lmk\fbk\rmk\subset \fbk.
\end{align}

We first claim $\Tbkv\rho{\Lambda_0}\lmk\pbk\lmk\caA_{\lhu}\rmk''\rmk\subset \pbk\lmk\caA_{\hd}\rmk'$
for any $\Lambda\in \CUbk$.
To see this, note that for any $A\in\caA_{\hd}\subset\caA_{\Lambda_0^c}$, we have
$\Tbkv\rho{\Lambda_0}\lmk \pbk(A)\rmk=\pbk(A)$ by Lemma 2.13 of[15].
Therefore, for any
$x\in \pbk\lmk\caA_{\lhu}\rmk''$, and $A\in \caA_{\hd}$ we have

\begin{align}\label{shika}
\left[
\pbk(A), \Tbkv\rho{\Lambda_0}(x)
\right]
= \Tbkv\rho{\Lambda_0}
\lmk
\left[
\pbk(A),x
\right]
\rmk=0,
\end{align}

proving the claim.

Next, we claim that for any $\Lambda\in\CUbk$,
\Tbkv\rho{\Lambda_0}\lmk \pbk\lmk \caA_{\lhu}\rmk''\rmk\subset\gu$.
To see this, note from the upward-filtering property of $\CUbk$,
for $\Lambda,\Lambda_0\in\CUbk$, there exists a $\Lambda_1$
such that $\Lambda,\Lambda_0\subset \Lambda_1$.
Then for any
$A\in \caA_{(\Lambda_1)^c}\subset \caA_{(\Lambda_0)^c}$,
we have $\Tbkv\rho{\Lambda_0}\lmk \pbk(A)\rmk=\pbk(A)$,
and the same argument as in (<ref>) implies
$\Tbkv\rho\lz\lmk \pbk(\caA_{\lhu})''\rmk\subset \pbk\lmk\caA_{\Lambda_1^c}\rmk'\subset \gu$, proving the claim.
Here we used Lemma <ref>.

Combining these with Lemma <ref>, for any $\Lambda\in\CUbk$,
we obtain

\begin{align}
\Tbkv\rho{\Lambda_0}\lmk \pbk\lmk \caA_{\lhu}\rmk''\rmk\subset\gu\cap \pbk\lmk\caA_{\hd}\rmk'
\subset\fbk.
\end{align}

By the definition of $\fbk$, we obtain $ \Tbkv\rho{\Lambda_0}\lmk \fbk\rmk\subset\fbk$.

Consider the setting in subsection <ref>.
Assume Assumption <ref>.Then for any $\rho,\sigma\in \Obk$,
$\lm1,\lm2\in\CUbk$ with $\lm1,\lm2\subset H_U$, we have
$\lmk\Tbkv\rho{\lm 1},\Tbkv\sigma{\lm 2}\rmk\subset\fbk$.

Because $\Tbkv\rho{\lm1}$ and $\Tbkv\rho{\lm2}$ are identity on $\pbk\lmk\caA_{(\Lambda_1\cup\Lambda_2)^c}\rmk$
we have $\lmk\Tbkv\rho{\lm1},\Tbkv\sigma{\lm2}\rmk\subset \pbk\lmk\caA_{(\Lambda_1\cup\Lambda_2)^c}\rmk'$.

From the upward filtering property of $\CUbk$ and Lemma <ref>,
we obtain $\pbk\lmk\caA_{(\Lambda_1\cup\Lambda_2)^c}\rmk'\subset \fbk$.

Consider the setting in subsection <ref>.
Assume Assumption <ref> and Assumption <ref>.
Let $(\lz,\lzr)\in \pc$, and $\tau : \fbk\to \fbd$ the $*$-isomorphism given in
Lemma <ref>.
Then the following hold.

For any $\rho\in \OUbkl$,

\begin{align}\label{aomori}
\end{align}

belongs to ${\Obul}$
and defines a map $ F_0^{\llz}: \OUbkl\to {\Obul}$.

(ii)For any $\rho\in \OUbkl$, $(\Lambda,\Lambda_r)\in \pc$ and $\Vrl\rho\Lambda\in \Vbk\rho\Lambda$,
$\tau(\Vrl\rho\Lambda)$ belongs to $\Vbu{F_0^{\llz}(\rho)}{\Lambda}$.

For any $\rho,\sigma\in \OUbkl$ and $R\in (\rho,\sigma)$,

\begin{align}\label{kyoto}
\begin{split}
F_1^{\llz}(R):=\tau(R)\in \lmk F_0^{\llz}(\rho), F_0^{\llz}(\sigma)\rmk_U
\end{split}
\end{align}

and defines a map $ F_1^{\llz}: (\rho,\sigma)\to \lmk F_0^{\llz}(\rho), F_0^{\llz}(\sigma)\rmk_U$.

For any $\rho\in \OUbkl$, $(\lm 1,\lm {r1})\in \pc$ and
$\Vrl\rho{\lm 1}\in \Vbk\rho{\lm {1}}$,

\begin{align}\label{akita}
\left. \Tbd{ F_0^{\llz}(\rho)}{\lm {r1}}{\tau(\Vrl\rho{\lm 1})}\right\vert_{\fbd}
=\left.\tau\;\Tbkv\rho{\lm 1}\;\tau^{-1}\right\vert_{\fbd}.
\end{align}

For any $\rho\in \OUbkl$,
the composition (<ref>) makes sense because
$\pbk(\abd)\subset \fbk$, Lemma <ref>, Lemma <ref>.
For any $\rho\in \OUbkl$, $(\Lambda,\Lambda_r)\in \pc$ and $\Vrl\rho\Lambda\in \Vbk\rho\Lambda$, by Lemma 2.14 of [15] and Lemma <ref> we have $\Vrl\rho\Lambda\in
\lmk\Tbk\rho\lz\unit,\Tbkv\rho\Lambda\rmk\subset\fbk$.
Hence $\tau(\Vrl\rho\Lambda)\in\fbd$ is well-defined, and
by definition,

\begin{align}
\begin{split}
&\left.\Ad\lmk \tau(\Vrl\rho\Lambda)\rmk F_0^{\llz}(\rho)\right\vert_{\caA_{\lhuc}}
=\left.\Ad\lmk \tau(\Vrl\rho\Lambda)\rmk  \tau\;\Tbk\rho\lz\unit\;\tau^{-1}\pbd\right\vert_{\caA_{\lhuc}}
=\left.\Ad\lmk \tau(\Vrl\rho\Lambda)\rmk  \tau\;\Tbk\rho\lz\unit\;\pbk\right\vert_{\caA_{\lhuc}}\\
&=\left.\Ad\lmk \tau(\Vrl\rho\Lambda)\rmk  \tau\;\rho\right\vert_{\caA_{\lhuc}}
\end{split}
\end{align}

In particular, if we take $\Lambda=\lz$ and $\Vrl\rho\lz=\unit$,
we see that $\left.F_0^{\llz}(\rho)\right\vert_{\caA_{\lhucz}}=\left.\pbd\right\vert_{\caA_{\lhucz}}$.
This implies  (i) and (ii).
(Note that for any $\ld_1\in\CUbk$, there exists a $\ld\in \CUbk$ such that
$\ld\subset \ld_1\cap\hu$.)
For any $\rho,\sigma\in \OUbkl$ and $R\in (\rho,\sigma)$,
$R\in \lmk\Tbk\rho\lz\unit, \Tbk\sigma\lz\unit\rmk\subset \fbk$ by Lemma <ref>
and $\tau(R)\in \fbd$ well-defined.
It is then straight forward to check that $\tau(R)\in \lmk F_0^{\llz}(\rho), F_0^{\llz}(\sigma)\rmk_U$,
proving (iii).
To show (iv), note that $ \left.  \Tbd{ F_0^{\llz}(\rho)}{\lm {r1}}{\tau(\Vrl\rho{\lm 1})}\right\vert_{\fbd}$
is $\sigma$-weak continuous on $\pbd\lmk\caA_{\lhu}\rmk''$ for any $\Lambda\in\CUbk$,
because  for any $\Lambda\in\CUbk$, there is some $\Lambda_l\in \Clbd$ such that $\lhu\subset \Lambda_l$.
(Recall Lemma <ref>.)
Therefore, $\tau^{-1}\; \left. \Tbd{ F_0^{\llz}(\rho)}{\lm {r1}}{\tau(\Vrl\rho{\lm 1})}\right\vert_{\fbd}\;\tau$ is $\sigma$-weak continuous on
$\pbk\lmk\caA_{\lhu}\rmk''$ for any $\Lambda\in\CUbk$.
On the other hand, for any $A\in \abd$,

\begin{align}\begin{split}
\tau^{-1}\; \left. \Tbd{ F_0^{\llz}(\rho)}{\lm {r1}}{\tau(\Vrl\rho{\lm 1})}\right\vert_{\fbd}\;\tau\pbk(A)=
\tau^{-1}\; \left. \Tbd{ F_0^{\llz}(\rho)}{\lm {r1}}{\tau(\Vrl\rho{\lm 1})}\right\vert_{\fbd}\;\pbd(A)=
\tau^{-1}\;\Ad\lmk \tau(\Vrl\rho{\lm 1})\rmk
=\tau^{-1}\;\Ad\lmk \tau(\Vrl\rho{\lm 1})\rmk
\tau\;\Tbk\rho\lz\unit\;\tau^{-1}\pbd(A)
=\;\Ad\lmk \Vrl\rho{\lm 1}\rmk
\;\Tbk\rho\lz\unit\;\pbk(A)
\Tbkv\rho{\lm 1}\;\pbk(A).
\end{split}
\end{align}

This proves (iv), as in the proof of Lemma 2.14 [15], via the $\sigma$weak continuity above.

Consider the setting in subsection <ref>.
Assume Assumption <ref>, Assumption <ref>, Assumption <ref> and Assumption <ref>.
Let $\epsilon_-^{(\lz)}(\rho:\sigma)$ be the braiding defined in Definition 4.12 [15] for
$\rho,\sigma\in \Obk\subset\OUbk$.
Then we have $F_1^{\llz}\lmk\epsilon_-^{(\lz)}(\rho:\sigma)\rmk=\hb{F_0^{\llz}(\rho)}{F_0^{\llz}(\sigma)}$.

Let $\lm 1,\lm 2\in \CUbk$ with $\lm 1,\lm 2\subset \hu$
such that $\lm 1\leftarrow_{(\frac{3\pi}2,\frac\pi 2)}\lm 2$,
$\lm 1\perp_{(\frac{3\pi}2,\frac\pi 2)}\lm 2$ (see Definition 2.6 Definition 4.8 of [15]).
Let $\Vrl\rho{\lm 1(t_1)}\in \Vbk\rho{\lm 1(t_1)}$, $t_1\ge 0$,
$\Vrl\sigma{\lm 2(t_2)}\in \Vbk\sigma{\lm 2(t_2)}$, $t_2\ge 0$.
By the Definition 4.12 [15],
we have

\begin{align}
\epsilon_-^{(\lz)}(\rho:\sigma)
={\mathop{\mathrm {norm}}}-\lim_{t_1,t_2\to\infty}
\lmk\Vrl\sigma{\lm 2(t_2)}\Tbk\sigma\lz\unit\lmk \Vrl\rho{\lm 1(t_1)}\rmk  \rmk^*
\Vrl\rho{\lm 1(t_1)} \Tbk\rho\lz\unit\lmk \Vrl\sigma{\lm 2(t_2)}\rmk.
\end{align}

Applying $\tau$ to this equation, from Lemma <ref>, we have

\begin{align}\begin{split}
&\tau\lmk \epsilon_-^{(\lz)}(\rho:\sigma)\rmk
={\mathop{\mathrm {norm}}}-\lim_{t_1,t_2\to\infty}
\lmk\tau\lmk \Vrl\sigma{\lm 2(t_2)}\rmk \Tbd{F_0^{\llz}(\sigma)}\lzr\unit
\lmk \tau\lmk \Vrl\rho{\lm 1(t_1)}\rmk \rmk  \rmk^*
\tau\lmk \Vrl\rho{\lm 1(t_1)} \rmk\Tbd{F_0^{\llz}(\rho)}\lzr\unit\lmk \tau\lmk \Vrl\sigma{\lm 2(t_2)}\rmk\rmk.
\end{split}
\end{align}

Let $\lm {r2}\in \Crbd$ with $\lm 1\leftarrow \lm {r2}$, $\arg\lm 2\subset \arg\lm{r2}$,
and let $\lm {r1}\in \Crbd$ with $(\lm 1,\lm {r1})\in\pc$.
Then by Lemma A.2 of [15], for any $t\ge 0$, there exists $t_0(t)\ge 0$
such that
$\lm 2(t_2)\subset \lm {r2}(t)$ for all $t_2\ge t_0(t)$, with $\lim_{t\to\infty} t_0(t)=\infty$.
From Lemma <ref>, for such $t_2\ge t_0(t)$,
$\tau\lmk \Vrl\sigma{\lm 2(t_2)}\rmk\in \Vbu{F_0^{\llz}(\sigma)}{\lm {2}(t_2)}
\subset\VUbd{F_0^{\llz}(\sigma)}{\lm{r2}(t)}$.
Hence we have

\begin{align}\begin{split}
&\tau\lmk \epsilon_-^{(\lz)}(\rho:\sigma)\rmk
={\mathop{\mathrm {norm}}}-\lim_{t_1,t\to\infty}
\lmk\tau\lmk \Vrl\sigma{\lm 2(t_0(t))}\rmk \Tbd{F_0^{\llz}(\sigma)}\lzr\unit
\lmk \tau\lmk \Vrl\rho{\lm 1(t_1)}\rmk \rmk  \rmk^*
\tau\lmk \Vrl\rho{\lm 1(t_1)} \rmk\Tbd{F_0^{\llz}(\rho)}\lzr\unit\lmk \tau\lmk \Vrl\sigma{\lm 2(t_0(t))}\rmk\rmk\\
\end{align}

In [15], we showed that $\Obkl$ with their morphisms $(\rho,\sigma)$,
$\rho,\sigma\in \Obkl$
forms a braided $C^*$-tensor category $\Cabkl$.

Hence we have a functor between this bulk theory and our boundary theory.

Let $\llz\in\pc$.
Consider the setting in subsection <ref>, and
assume Assumption <ref>, Assumption <ref>, Assumption <ref>,
Assumption <ref>, Assumption <ref>, and
Assumption <ref>.
(Note from Lemma <ref>, Assumption <ref> automatically holds.)
Then $F^{\llz} : \Cabkl\to \Cabul$ given by
$F^{\llz}_0$ (<ref>), $F^{\llz}_1$ (<ref>)
gives a fully faithful unitary tensor functor.

Furthermore, it is a central functor with central structure

\begin{align}\label{nara}
\hb{F_0^{\llz}(\rho)}{\sigma} : F_0^{\llz}(\rho)\otimes \sigma
\to \sigma\otimes F_0^{\llz}(\rho),\quad
\rho\in\Obkl, \sigma\in\Obul.
\end{align}

The map $F^{\llz}$ is a functor
because any $\rho,\sigma,\gamma\in \OUbkl$, $R\in (\rho,\sigma)$, $S\in (\sigma,\gamma)$,
we clearly have
$F^{\llz}_1(SR)=F^{\llz}_1(S)F^{\llz}_1(R)$ and $F^{\llz}_1(\unit)=\unit$.
It is linear on morphism.

Note that $F^{\llz}_0(\pbk)=\pbd$, because $\Tbk{\pbk}{\lz}\unit=\id$.
Then $\unit : \pbd\to F^{\llz}_0(\pbk)=\pbd$ gives an isomorphism.

We also see that
$\unit : F^{\llz}_0(\rho)\otimes F^{\llz}_0(\sigma)\to F^{\llz}_0(\rho\otimes\sigma)$
give natural isomorphisms.
In fact, note that from (<ref>), Lemma <ref>, and Lemma 3.6 of [15],

\begin{align}\label{hirosaki}
\begin{split}
&F^{\llz}_0(\rho)\otimes F^{\llz}_0(\sigma)
=\tau\Tbk\rho{\lz}\unit \Tbk\sigma{\lz}\unit\tau^{-1}\pbd\\
&=\tau\Tbk{\rho\otimes \sigma}{\lz}\unit \tau^{-1}\pbd
\end{split}
\end{align}

Furthermore,  let $\rho,\rho',\sigma,\sigma'\in \OUbkl$
and $R\in (\rho,\rho')$, $S\in (\sigma,\sigma')$.
Then, from (<ref>),

\begin{align}
\begin{split}
&F^{\llz}_1(R)\otimes F^{\llz}_1(S)
=F^{\llz}_1(R)\Tbd{F^{\llz}_0(\rho)}{\lz}\unit \lmk  F^{\llz}_1(S)\rmk
=\tau(R)\Tbd{F^{\llz}_0(\rho)}{\lz}\unit\lmk  \tau (S)\rmk\\
=\tau\lmk R\Tbk\rho{\lz}\unit\lmk S\rmk\rmk
R\otimes S
\rmk,
\end{split}
\end{align}

proving the naturality.
These isomorphisms trivially satisfies the diagrams given in Definition 2.1.3 <cit.>.
Hence $F^{\llz}$ is a tensor functor.
Clearly, it is unitary.
It is a central functor from Lemma <ref>, Lemma <ref>,  Lemma <ref>, Lemma <ref>,
Lemma <ref>.

§ A BOUNDARY TO BULK MAP

In this section, we construct an inverse of the map constructed in the previous section.

Consider the setting in subsection <ref>, and assume Assumption <ref>.
Then for any $\rho\in\OrUbd$, there exists a unique $*$-homomorphism
$G(\rho) : \abk\to \caB(\hbk)$ such that

\begin{align}
=\tau^{-1}\rho(A),\quad A\in\abd,\label{nagano}\\
G(\rho)(B)=\pbk(B),\quad B\in \caA_{\hd}\label{gifu}.
\end{align}

If furthermore $\rho\in \Obu$, then
$G(\rho)\in \OUbk$.
In particular, for $\llz\in\pc$, $\rho\in \Obul$, we have
$G(\rho)\in \OUbkl$, and $F_0^{{\llz}}\lmk G(\rho)\rmk=\rho$.
For $\hat\rho\in \OUbkl$, we have
The map preserves tensor i.e., $G(\rho\otimes\sigma)=G(\rho)\otimes G(\sigma)$
for any $\rho,\sigma\in \Obul$.

Let $\llz\in\pc$ and $\rho\in\OrUbd$, $\Vrl\rho{\lzr}\in \VUbd\rho{\lzr}$.
Note that for any $A\in\abd$,

$$\rho(A)=\Ad\lmk \Vrl{\rho}\lzr^*\rmk\Tbdv\rho{\lzr}\pbd(A)\in\fbd.$$

Therefore, $\tau^{-1}\rho(A)$ is well-defined.
Hence there are representations
$G_1(\rho)$( resp.$G_2(\rho)$) of $\abd$ (resp $\caA_{\hd}$)
on $\hbk$, given by the right hand side of
(<ref>)( resp. (<ref>)). Because $G_1(\rho)(\abd)\subset\fbd$
and $\fbd$ commutes with $\pbk(\caA_{\hd})$
by definition, the range of $G_1(\rho)$ and $G_2(\rho)$ commutes.
Because $\abd$, $\caA_{\hd}$ are nuclear,
there exists a unique representation $G(\rho) : \abk \to \caB(\hbk)$
such that $G(\rho)\vert_{\abd}=G_1(\rho)$, $G(\rho)\vert_{\caA_{\hd}}=G_2(\rho)$.
(Proposition 4.7 IV of [16].)

Now suppose that $\rho\in \Obu$.
For any $\lm {}\in \CUbk$, take $\Vrl{\rho}{\lm{}}\in \Vbu\rho{\lm {}}\subset\fbd$.
Then for any $A\in \caA_{\lhuc}$,

\begin{align}
\begin{split}
&\Ad\lmk\tau^{-1}\lmk \Vrl{\rho}{\lm{}}\rmk\rmk G(\rho)(A)
=\Ad\lmk\tau^{-1}\lmk \Vrl{\rho}{\lm{}}\rmk\rmk\lmk \tau^{-1}\rho(A)\rmk
=\tau^{-1}\lmk \Ad\lmk\lmk \Vrl{\rho}{\lm{}}\rmk\lmk \rho(A)\rmk \rmk\rmk
\end{split}
\end{align}

For any $B\in \caA_{\hd}$, we have

\begin{align}
\Ad\lmk\tau^{-1}\lmk \Vrl{\rho}{\lm{}}\rmk\rmk G(\rho)(B)
=\Ad\lmk\tau^{-1}\lmk \Vrl{\rho}{\lm{}}\rmk\rmk\lmk \pbk(B)\rmk
\end{align}

because $ \pbk(B)\in{\fbk}'$.
Combining these, we obtain

\begin{align}
\left.\Ad\lmk\tau^{-1}\lmk \Vrl{\rho}{\lm{}}\rmk\rmk G(\rho)\right\vert_{\caA_{\Lambda^c}}
\end{align}

for any $\Lambda\in\CUbk$, proving $G(\rho)\in \OUbk$.
If $\rho\in \Obul$, then we may take $ \Vrl{\rho}{\lz}=\unit$ above.
Hence $G(\rho)\in \OUbkl$.
For any $A\in\hu$, we have

\begin{align}
\begin{split}
F_0^{\llz}\lmk G(\rho)\rmk(A)
=\tau \Tbk{G(\rho)}\lz\unit\tau^{-1}\pbd(A)
=\tau G(\rho)(A)
\end{split}
\end{align}

Hence we have $F_0^{\llz}\lmk G(\rho)\rmk=\rho$.

For $\hat\rho\in \OUbkl$, we have

\begin{align}
\begin{split}
\lmk
\rmk
=\tau^{-1}F_0^{\llz}\lmk \hat\rho\rmk(A)
=\tau^{-1}\tau\Tbk{\hat\rho}\lz\unit \tau^{-1}\pbd(A)
=\Tbk{\hat\rho}\lz\unit \pbk(A)
=\hat\rho(A),\quad A\in \abd,\\
=\pbk(B)=\hat\rho(B),\quad B\in\caA_{\hd}.
\end{split}
\end{align}

we have

For any $\rho,\sigma\in \Obul$,
from above and Theorem <ref>, we have

\begin{align}
\begin{split}
=F_0^{\llz}G(\rho)\otimes F_0^{\llz}G(\sigma)
=F_0^{\llz}\lmk G(\rho)\otimes G(\sigma)\rmk.
\end{split}
\end{align}

Applying $G$ to this, we have

\begin{align}
\begin{split}
=GF_0^{\llz}\lmk G(\rho)\otimes G(\sigma)\rmk
=G(\rho)\otimes G(\sigma).
\end{split}
\end{align}

This proves the last statement.

The information of $\rho\in\Obul$ is not enough
to recover the full theory of the bulk.
Namely, it cannot guarantee in general if $G(\rho)$ satisfies the superselection criterion
of the bulk.
However, if we combine it with other half boundary information, we can do that.
To do so, note that we can re-do all the things we have done for upper half-plane
to lower half-plane, just reflect things with respect to the $x$-axis.
We denote such lower half-plane version(reflected with respect to the $x$-axis version), by adding $(-)$ symbol.
We prepare $\obdm$ a pure state on $\caA_{\hd}$ and
its GNS representation $(\hbdm,\pbdm)$.
The set of cones $\CUbk$ is reflected to $\CUbkm$
algebra $\fbd$, $\fbk$ is reflected to $\fbdm$, $\fbkm$
the equivalence relation $\simeq_U$
is reflected to $\simeq_{U(-)}$, and so on.

Assumption <ref>$(-)$, Assumption <ref>$(-)$,
Assumption <ref>$(-)$, etc, are
Assumption <ref>, Assumption <ref>,
Assumption <ref> reflected with respect to
And assuming such $(-)$-versions of assumptions,
we obtain braided $C^*$-tensor category $\Obulm$,
boundary to bulk map $G_{(-)}$, and $\tau_{(-)}$, corresponding to
$\tau$ in Lemma <ref>.

Consider the setting in subsection <ref>, and
assume Assumption <ref>, Assumption <ref>, Assumption <ref>,
Assumption <ref>, Assumption <ref>, Assumption <ref>, and their $(-)$-versions.
Let $\lz\in\CUbk$, $\lzm\in \CUbkm$.
We denote by
$\Ototl$ the set of all $\rho\in \Obul$
which allows some $\rho_-\in \Obulm$ such that
$G(\rho)\simeq G_{(-)}(\rho_-)$.

For the rest of this section, we consider the setting of Definition <ref>
and fix some $\varphi_i\in (0,\frac\pi 2)$ such that
$\lz,\lzm\in C_{(0,\varphi_1)}$.
The $\rho_-$ is determined essentially uniquely.

For each $\rho\in\Ototl$, $\rho_-\in \Obulm$
satisfying $G(\rho)\simeq G_{(-)}(\rho_-)$ are determined uniquely up to

If $G(\rho)\simeq G_{(-)}(\rho_-)\simeq G_{(-)}(\rho'_-)$, then
$(-)$-version of Lemma <ref>, Lemma <ref> imply

Elements in $\Ototl$ gives bulk elements in $\Obkl$.

In the setting of Definition <ref>, for any
$\rho\in \Ototl$, we have $G(\rho)\in\Obkl$.

From this property, we may regard $\rho_-$
as dual object of $\rho$.

By the definition, there are unitaries
$W_\rho$ on $\hbk$ and
$\rho_-\in \Obulm$
such that
\Ad(W_\rho)G_{(-)}(\rho_-)=G(\rho)
For any cone $\ld$ in $\bbZ^2$,
either (i) $\lm 1\in\CUbk$ with
$\lm 1\subset\ld$
or (ii) $\lm 2\in \CUbkm$ with
$\lm 2\subset\ld$ exists.
If (i) occurs,
choose $\Vrl\rho{\lm 1}\in \Vbu\rho{\lm 1}\subset\fbd$
and we have

\begin{align}
\begin{split}
\left.\Ad\lmk \tau^{-1}\lmk \Vrl\rho{\lm 1}\rmk\rmk G(\rho)\right\vert_{\caA_{\ld^c}}=
\left. \pbk\right\vert_{\caA_{\ld^c}}.
\end{split}
\end{align}

If (ii) occures, choose $\Vrl{\rho_-}{\lm 2}\in \Vbum{\rho_-}{\lm 2}\subset\fbdm$
and we have

\begin{align}
\begin{split}
\left.\Ad\lmk \tau_{(-)}^{-1}\lmk \Vrl{\rho_-}{\lm 2}\rmk W_\rho^*\rmk G(\rho)\right\vert_{\caA_{\ld^c}}=
\left. \pbk\right\vert_{\caA_{\ld^c}}.
\end{split}
\end{align}

Hence we have $G(\rho)\in\Obkl$.

First, $\Otot$ is closed under tensor.

In the setting of Definition <ref>, for any
$\rho,\sigma\in\Ototl$, we have

Let $\rho,\sigma\in\Otot$.
By the definition, there are unitaries
$W_\rho,W_\sigma$ on $\hbk$ and
$\rho_-,\sigma_-\in \Obulm$
such that

\begin{align}\label{morioka}
\begin{split}
\Ad(W_\rho)G_{(-)}(\rho_-)=G(\rho),\quad
\Ad(W_\sigma)G_{(-)}(\sigma_-)=G(\sigma)
\end{split}
\end{align}

Note that $W_\sigma\in \pbk\lmk\caA_{\lmk \lz\cup\lzm\rmk^c}\rmk'\subset \caB_{(0,\varphi_1)}$,
because $G(\sigma)$ and $G_{(-)}(\sigma_-)$ are $\pbk$ on $\caA_{\lmk \lz\cup\lzm\rmk^c}$.

From Lemma <ref> and its $(-)$ version,
$G(\rho), G(\sigma), G_{(-)}(\rho_-), G_{(-)}(\sigma_-)\in\Obkl$.
By Lemma 2.14 of [15]

${\caB_{\lmk \frac{3\pi+\varphi_1}{2},\frac{\varphi_1+\pi}2\rmk}}
\subset \caB_{(\frac{3\pi}2,\frac\pi 2)}\cap\caB_{(0,\varphi_1)}$,

we have

\begin{align}
\begin{split}
=\left. T_{\hat\rho}^{(0,\varphi_1)\lz\unit}\right\vert_{\caB_{\lmk \frac{3\pi+\varphi_1}{2},\frac{\varphi_1+\pi}2\rmk}}
=\left.\Tbk{\hat\rho}\lz\unit\right\vert_{\caB_{\lmk \frac{3\pi+\varphi_1}{2},\frac{\varphi_1+\pi}2\rmk}},
\end{split}
\end{align}

for any $\hat\rho\in\Obkl$.
Hence for $G(\rho),G(\sigma)\in \Obkl$, we have

\begin{align}
\begin{split}
G(\rho)\otimes G(\sigma)
\end{split}
\end{align}

\begin{align}
\begin{split}
\end{split}
\end{align}

and $T_{G(\rho)}^{(0,\varphi_1)\lz\unit}$,
are $\sigma$-weak continuous
on $\pbk(\caA_{\Lambda})''$ for $\Lambda\in C_{(0,\varphi_1)}$,
we have $T_{G(\rho)}^{(0,\varphi_1)\lz\unit}=\Ad(W_\rho)T_{G_{(-)}(\rho_-)}^{(0,\varphi_1)\lzm\unit}$.
Similarly we have  $T_{G(\sigma)}^{(0,\varphi_1)\lz\unit}=\Ad(W_\sigma)T_{G_{(-)}(\sigma_-)}^{(0,\varphi_1)\lzm\unit}$.

From those facts and  Lemma <ref> and their $(-)$ version, we have

\begin{align}
\begin{split}
&G(\rho\otimes\sigma)=G(\rho)\otimes G(\sigma)
\Ad(W_\sigma)T_{G_{(-)}(\sigma_-)}^{(0,\varphi_1)\lzm\unit}\pbk\\
\Ad\lmk W_\rho T_{G_{(-)}(\rho_-)}^{(0,\varphi_1)\lzm\unit}\lmk W_\sigma \rmk \rmk
=\Ad\lmk W_\rho T_{G_{(-)}(\rho_-)}^{(0,\varphi_1)\lzm\unit}\lmk W_\sigma \rmk \rmk
G_{(-)}(\rho_-)\otimes G_{(-)}(\sigma_-)\\
&=\Ad\lmk W_\rho T_{G_{(-)}(\rho_-)}^{(0,\varphi_1)\lzm\unit}\lmk W_\sigma \rmk \rmk
G_{(-)}(\rho_-\otimes \sigma_-).
\end{split}
\end{align}

Hence we have $\rho\otimes\sigma\in\Ototl$.

Next $\Otot$ is closed under direct sum

In the setting of Definition <ref>, for any
$\gamma\in\Obul$, $u\in (\rho,\gamma)_U$, $v\in (\sigma,\gamma)_U$
isometry with $uu^*+vv^*=\unit$, we have

Let $W_\rho,W_\sigma$ unitaries on $\hbk$ and
$\rho_-,\sigma_-\in \Obulm$ satisfying (<ref>).
For $\rho_-,\sigma_-\in\Obulm$, there are
$\gamma_-\in\Obulm$, $u_-\in (\rho,\gamma)_{U(-1)}$, $v_-\in (\sigma,\gamma_{U(-)}$
isometry with $u_-u_-^*+v_-v_-^*=\unit$.
It is straightforward to check

\begin{align}
\begin{split}
G(\gamma)=\Ad\tau^{-1}(u) G(\rho)+\Ad\tau^{-1}(v) G(\sigma),\quad
G_{(-)}(\gamma_-)=\Ad\tau_{(-)}^{-1}(u_-) G_{(-)}(\rho_-)+\Ad\tau_{(-)}^{-1}(v_-) G_{(-)}(\sigma_-)
\end{split}
\end{align}

\begin{align}
\begin{split}
\tau^{-1}(u)\cdot  W_\rho\cdot  \tau_{(-)}^{-1}(u_-^*)
+\tau^{-1}(v)\cdot  W_\sigma\cdot  \tau_{(-)}^{-1}(v_-^*)\in\caU(\hbk),
\end{split}
\end{align}

we have

\begin{align}
\begin{split}
\end{split}
\end{align}

Hence we obtain $\gamma\in \Ototl$.

Finally we show that $\Ototl$ is closed under taking subobject.
In the setting of Definition <ref>,
let $\rho\in\Ototl$, $p\in (\rho,\rho)_U$ a projection,
$\gamma\in\Obul$, $v\in (\gamma,\rho)_U$ isometry with $vv^*=p$.
Then we have $\gamma\in \Ototl$

It is straightforward to check $G(\gamma)=\Ad\lmk\tau^{-1}(v^*)\rmk G(\rho)$.
There are unitaries
$W_\rho$ on $\hbk$ and
$\rho_-\in \Obulm$
such that
\Ad(W_\rho)G_{(-)}(\rho_-)=G(\rho)
We claim $\Ad W_\rho^*\lmk \tau^{-1}(p)\rmk\in G_{(-)}(\rho_-)(\abk)'$.
In fact for any $A\in\abk$, we have

\begin{align}
\begin{split}
&\Ad W_\rho^*\lmk \tau^{-1}(p)\rmk\cdot G_{(-)}(\rho_-)(A)
=W_\rho^*\tau^{-1}(p) W_\rho \cdot G_{(-)}(\rho_-)(A) \cdot W_\rho^* W_\rho
=W_\rho^*\tau^{-1}(p) G(\rho)(A) W_\rho\\
&=W_\rho^* G(\rho)(A)\tau^{-1}(p)  W_\rho
=G_{(-)}(\rho_-)(A) W_\rho^*\tau^{-1}(p)W_\rho.
\end{split}
\end{align}

Because $G_{(-)}(\rho_-)\vert_{\caA_{\lzm^c}}=\pbk\vert_{\caA_{\lzm^c}}$,
we have

\begin{align}
\begin{split}
\Ad W_\rho^*\lmk \tau^{-1}(p)\rmk\in
G_{(-)}(\rho_-)(\abk)'\subset G_{(-)}(\rho_-)\lmk \caA_{\lzm^c}\rmk'
=\pbk\lmk {\caA_{\lzm^c}}\rmk'\subset \fbkm,
\end{split}
\end{align}

and $\tau_{(-)}\lmk \Ad W_\rho^*\lmk \tau^{-1}(p)\rmk\rmk$
is a well-defined projection in $\fbdm$
and $\tau_{(-)}\lmk \Ad W_\rho^*\lmk \tau^{-1}(p)\rmk\rmk\in (\rho_-,\rho_-)_{U(-)}$.
For this projection, because $\rho_-\in\Obulm$, there exist
$\gamma_-\in \Obulm$, isometry $v_-\in (\gamma_-,\rho_-)_{U(-)}$
such that $v_-v_-^*=\tau_{(-)}\lmk \Ad W_\rho^*\lmk \tau^{-1}(p)\rmk\rmk$.

\tau^{-1}(v^*) W_\rho \tau_{(-)}^{-1}(v_-)
It is a unitary on $\hbk$, and we have

\begin{align}
\begin{split}
=\Ad\lmk\tau^{-1}(v^*)\rmk G(\rho)
=\Ad\lmk \tau^{-1}(v^*) W_\rho \tau_{(-)}^{-1}(v_-v_-^*)\rmk G_{(-)}(\rho_-)\\
&=\Ad\lmk \tau^{-1}(v^*) W_\rho \tau_{(-)}^{-1}(v_-)\rmk
\Ad\lmk\tau_{(-)}^{-1}(v_-^*) \rmk G_{(-)}(\rho_-)
=\Ad(W_\gamma) G_{(-)}(\gamma_-).
\end{split}
\end{align}

Hence we obtain $\gamma\in\Ototl$.

Hence we obtain the following.

The functor $F^{\llz} $ in Theorem <ref> is a braided monoidal equivalence
between braided $C^*$-tensor categories.

The functor $F^{\llz}$ is also essentially surjective from Lemma <ref>.
Hence it is an equivalence.

Recall the definition of center $\caZ(\caC)$
of a strict tensor category $\caC$ (Definition XIII.4.1 of <cit.>).
Here, we introduce a category $\caZ(\Cabul)$
which requires the asymptotic behavior (<ref>).

Consider the setting in subsection <ref>, and
assume Assumption <ref>, Assumption <ref>, Assumption <ref>,
Assumption <ref>, Assumption <ref>, Assumption <ref>.
Let $\lz\in\CUbk$.
An object of $\caZ(\Cabul)$ is a pair
$(\rho, \hat\iota(\rho : -))$ where $\rho\in \Obul$
and $\hat\iota(\rho : -)$ is a family of natural isomorphisms

\begin{align}
\hat\iota(\rho : \sigma) :
\rho\otimes\sigma \to\sigma\otimes\rho,
\end{align}

defined for all objects $\sigma\in\Obul$ such that

\begin{align}\label{shimane}
\lim_{t\to\infty}
\sup_{\sigma\in O^{rU}_{\mathop{\mathrm{bd},\lm {r2}(t)}}}
\lV
\hat\iota(\rho :\sigma)-\unit
\rV=0,
\end{align}
for any $\lm {r2}\in\Crbd$ with $\lz\leftarrow_r\lm {r2}$, $\lm {r2}\subset \lzr$
and for all objects
$\sigma,\gamma\in \Obul$ we have

\begin{align}\label{kouchi}
\hat\iota\lmk {\rho}: {\sigma\otimes \gamma}\rmk
=\lmk \unit_{\sigma}\otimes\hat\iota\lmk \rho: \gamma\rmk\rmk\cdot
\lmk \hat\iota\lmk \rho :\sigma\rmk\otimes\unit_\gamma\rmk.
\end{align}

The morphism from $(\rho, \hat\iota(\rho : -))$ to $(\sigma, \hat\iota(\sigma : -))$
is a morphism $R: \rho\to\sigma$ in $\Cabul$
such that for each object $\gamma \in \Cabul$, we have

\begin{align}
\begin{split}
\hat\iota(\sigma: \gamma)
\lmk R\otimes \unit_\gamma\rmk
=\lmk \unit_\gamma\otimes R\rmk\hat\iota(\rho:\gamma)
\end{split}
\end{align}

By Lemma <ref>, we have $\hat\iota(\rho:\sigma)=\hb{\rho}\sigma$.
From this and the naturality of $\hb{\rho}\sigma$, Lemma <ref>,
the set of all morphisms in  $\caZ(\Cabul)$ from $(\rho, \hat\iota(\rho : -))$ to $(\sigma, \hat\iota(\sigma : -))$
is $(\rho,\sigma)_U$.

As in Theorem XIII.4.2 of <cit.>, $\caZ(\Cabul)$
is a braided $C^*$-tensor category with
the tensor product

\begin{align}
(\rho, \hat\iota(\rho : -))\otimes(\sigma, \hat\iota(\sigma : -))
\rho\otimes\sigma, \hat\iota(\rho\otimes\sigma: -)
\rmk.
\end{align}

Following Theorem XIII.4.2 of <cit.>, we define the morphism $ \hat\iota(\rho\otimes\sigma: -)$

\begin{align}\label{yamaguchi}
\begin{split}
\hat\iota(\rho\otimes\sigma: \gamma)
:=\lmk\hat\iota(\rho : \gamma)\otimes\unit_\sigma\rmk
\lmk \unit_\rho\otimes \hat\iota(\sigma,\gamma)\rmk,
\end{split}
\end{align}

but because of Lemma <ref>, it turns out
this is equal to $\hb{\rho\otimes\sigma}{-}$,
the unique family of morphism
satisfying the condition in Definition <ref>
for $\rho\otimes\sigma$.
For morphism $R$ from
$(\rho, \hat\iota(\rho : -))$ to $(\rho', \hat\iota(\rho' : -))$
and  morphism $S$ from
$(\sigma, \hat\iota(\sigma : -))$ to $(\sigma', \hat\iota(\sigma' : -))$,
the tensor product of $R$ and $S$ in $\caZ(\Cabul)$
by the tensor product $R\otimes S$ in $\Cabul$.
By Lemma <ref> and Lemma <ref>,
this $R\otimes S$ is in fact a morphism
$(\rho, \hat\iota(\rho : -))\otimes (\sigma, \hat\iota(\sigma : -))$ to
$(\rho', \hat\iota(\rho' : -))\otimes (\sigma', \hat\iota(\sigma' : -))$.
With $\hat\iota(\pbd : \sigma)=\unit$, $\sigma\in \Obul$,
$(\pbd,\hat\iota (\pbd,-))$ is the tensor unit
of $\Cabul$ and
$\unit :  (\pbd, \hat\iota(\pbd : -))\otimes (\rho, \hat\iota(\rho : -))\to (\rho, \hat\iota(\rho:-))$,
$\unit :  (\rho, \hat\iota(\rho : -))\otimes (\pbd, \hat\iota(\pbd : -))\to (\rho, \hat\iota(\rho:-))$
give left and right multiplication of the unit.
Note that the tensor unit is irreducible.
Furthermore, $\unit : \lmk (\rho, \hat\iota(\rho : -))\otimes (\sigma, \hat\iota(\sigma : -)\rmk
\otimes
(\gamma, \hat\iota(\gamma : -))
\to (\rho, \hat\iota(\rho : -))\otimes
\lmk (\sigma, \hat\iota(\sigma : -))\otimes
(\gamma, \hat\iota(\gamma : -))\rmk$ is
the associativity isomorphism.

Next we consider direct sum.
For objects $(\rho, \hat\iota(\rho : -)), (\sigma, \hat\iota(\sigma : -))$
of $\caZ\lmk\Cabul\rmk$,
let $\gamma\in \Obul$, $u\in (\rho,\gamma)_U$, $v\in (\sigma,\gamma)_U$ isometry
with $uu^*+vv^*=\unit$.
Then, for any $\zeta\in \Obul$, set

\begin{align}
\hat\iota(\gamma:\zeta):=
\Tbd\zeta{\lzr}\unit(u)\hat\iota(\rho,\zeta) u^*
+\Tbd\zeta{\lzr}\unit(v)\hat\iota(\sigma,\zeta) v^*
\in \caU(\fbd).
\end{align}

Then for any $A\in\abd$, we have

\begin{align}
\begin{split}
&\hat\iota(\gamma,\zeta)\cdot \gamma\otimes\zeta(A)
\Tbd\zeta{\lzr}\unit(u)\hat\iota(\rho,\zeta) u^*
+\Tbd\zeta{\lzr}\unit(v)\hat\iota(\sigma,\zeta) v^*
\rmk \Tbd\gamma{\lzr}\unit\Tbd\zeta{\lzr}\unit\pbd(A)\\
\Tbd\zeta{\lzr}\unit(u)\hat\iota(\rho,\zeta)   \lmk \Tbd\rho{\lzr}\unit\Tbd\zeta{\lzr}\unit\pbd(A)\rmk
+\Tbd\zeta{\lzr}\unit(v)\hat\iota(\sigma,\zeta) \lmk \Tbd\sigma{\lzr}\unit\Tbd\zeta{\lzr}\unit\pbd(A)\rmk
&=\Tbd\zeta{\lzr}\unit(u) \cdot \Tbd\zeta{\lzr}\unit\Tbd\rho{\lzr}\unit\pbd(A)
\hat\iota(\rho,\zeta)
+\Tbd\zeta{\lzr}\unit(v)\cdot \Tbd\zeta{\lzr}\unit\Tbd\sigma{\lzr}\unit\pbd(A)
\hat\iota(\sigma,\zeta)
&=\Tbd\zeta{\lzr}\unit\lmk u\Tbd\rho{\lzr}\unit \pbd(A)\rmk
\hat\iota(\rho,\zeta)
+\Tbd\zeta{\lzr}\unit\lmk v\Tbd\sigma{\lzr}\unit \pbd(A)\rmk
\hat\iota(\sigma,\zeta)
&=\Tbd\zeta{\lzr}\unit\lmk \Tbd\gamma{\lzr}\unit \pbd(A)\cdot u\rmk
\hat\iota(\rho,\zeta)
+\Tbd\zeta{\lzr}\unit\lmk \Tbd\gamma{\lzr}\unit \pbd(A)\cdot v\rmk
\hat\iota(\sigma,\zeta)
&=\lmk \zeta\otimes\gamma(A) \rmk\hat\iota(\gamma,\zeta).
\end{split}
\end{align}

Hence we have $\hat\iota(\gamma,\zeta)\in (\gamma\otimes\zeta,\zeta\otimes\gamma)$.
It is also natural. In fact, for any $\zeta,\zeta'\in \Obul$ and $S\in (\zeta,\zeta')_U$,
we have

\begin{align}
\begin{split}
\lmk
\Tbd\zeta{\lzr}\unit(u)\hat\iota(\rho,\zeta) u^*
+\Tbd\zeta{\lzr}\unit(v)\hat\iota(\sigma,\zeta) v^*
\rmk\\
&=\Tbd{\zeta'}{\lzr}\unit(u) S\hat\iota(\rho,\zeta) u^*
+\Tbd{\zeta'}{\lzr}\unit(v) S\hat\iota(\sigma,\zeta) v^*
=\Tbd{\zeta'}{\lzr}\unit(u) \hat\iota(\rho,\zeta') \Tbd\rho{\lzr}\unit(S)u^*
+\Tbd{\zeta'}{\lzr}\unit(v) \hat\iota(\sigma,\zeta') \Tbd\sigma{\lzr}\unit(S)v^*\\
&=\Tbd{\zeta'}{\lzr}\unit(u) \hat\iota(\rho,\zeta') u^*\Tbd\gamma{\lzr}\unit(S)
+\Tbd{\zeta'}{\lzr}\unit(v) \hat\iota(\sigma,\zeta')v^*\Tbd\gamma{\lzr}\unit(S)\\
=\hat\iota(\gamma,\zeta')\lmk\unit_\gamma\otimes S\rmk.
\end{split}
\end{align}

Furthermore, for any $\zeta,\zeta'\in \Obul$,

\begin{align}
\begin{split}
&\lmk\unit_\zeta\otimes \hat\iota(\gamma,\zeta')\rmk
\lmk\hat\iota(\gamma,\zeta)\otimes\unit_{\zeta'}\rmk\\
\lmk
\Tbd{\zeta'}{\lzr}\unit(u)\hat\iota(\rho,\zeta') u^*
+\Tbd{\zeta'}{\lzr}\unit(v)\hat\iota(\sigma,\zeta') v^*
\rmk
\lmk
\Tbd\zeta{\lzr}\unit(u)\hat\iota(\rho,\zeta) u^*
+\Tbd\zeta{\lzr}\unit(v)\hat\iota(\sigma,\zeta) v^*
\rmk\\
\Tbd{\zeta}{\lzr}\unit\lmk \Tbd{\zeta'}{\lzr}\unit(u)\hat\iota(\rho,\zeta') u^* u\rmk\cdot \hat\iota(\rho,\zeta) u^*
+\Tbd{\zeta}{\lzr}\unit\lmk \Tbd{\zeta'}{\lzr}\unit(v)\hat\iota(\sigma,\zeta') v^* v\rmk\cdot \hat\iota(\sigma,\zeta) v^*\\
\Tbd{\zeta}{\lzr}\unit\Tbd{\zeta'}{\lzr}\unit(u)\cdot
\Tbd{\zeta}{\lzr}\unit \lmk \hat\iota(\rho,\zeta')\rmk\cdot \hat\iota(\rho,\zeta) u^*
\Tbd{\zeta}{\lzr}\unit\lmk \hat\iota(\sigma,\zeta'\rmk) \cdot \hat\iota(\sigma,\zeta) v^*\\
\Tbd{\zeta}{\lzr}\unit\Tbd{\zeta'}{\lzr}\unit(u)\cdot
\lmk
\unit_{\zeta}\otimes \hat\iota(\rho,\zeta')
\rmk
\lmk  \hat\iota(\rho,\zeta) \otimes \unit_{\zeta'}\rmk\cdot u^*
\lmk
\unit_\zeta\otimes \hat\iota(\sigma,\zeta')
\rmk
\lmk
\hat\iota(\sigma,\zeta)\otimes \unit_{\zeta'}
\rmk v^*\\
&=\Tbd{\zeta\otimes\zeta'}{\lzr}\unit (u)
\hat\iota\lmk \rho : \zeta\otimes\zeta'\rmk u^*
\hat\iota\lmk\sigma:\zeta\otimes\zeta'\rmk v^*\\
&=\hat\iota\lmk \gamma: \zeta\otimes\zeta'\rmk.
\end{split}
\end{align}

To check the asymptotic behavior (<ref>),
let $\lm {r2}\in\Crbd$ with $\lz\leftarrow_r\lm {r2}$, $\lm {r2}\subset \lzr$.
Note that $u,v\in \pbd\lmk\caA_{\lhucz}\rmk'\cap\fbd$.
Therefore, Lemma <ref> implies

\begin{align}
\begin{split}
\lim_{t\to \infty}\sup_{\sigma\in O^{rU}_{\mathop{\mathrm{bd},\lm {r2}(t)}}}
\lV \Tbd\zeta{\lzr}\unit(u)-u\rV=0,\quad
\lim_{t\to \infty}\sup_{\sigma\in O^{rU}_{\mathop{\mathrm{bd},\lm {r2}(t)}}}
\lV \Tbd\zeta{\lzr}\unit(v)-v\rV=0.
\end{split}
\end{align}

Combining this with

\begin{align}
\begin{split}
\lim_{t\to\infty}
\sup_{\zeta\in O^{rU}_{\mathop{\mathrm{bd},\lm {r2}(t)}}}
\lV
\hat\iota(\rho :\zeta)-\unit
\rV=
\lim_{t\to\infty}
\sup_{\zeta\in O^{rU}_{\mathop{\mathrm{bd},\lm {r2}(t)}}}
\lV
\hat\iota(\sigma :\zeta)-\unit
\rV
\end{split}
\end{align}
we obtain

\begin{align}
\begin{split}
\lim_{t\to\infty}
\sup_{\zeta\in O^{rU}_{\mathop{\mathrm{bd},\lm {r2}(t)}}}
\lV
\hat\iota(\gamma :\zeta)-\unit
\rV=0.
\end{split}
\end{align}

Hence we conclude
$(\gamma,\hat\iota(\gamma : -))$ is an object of $\caZ\lmk\Cabul\rmk$,
with a morphism $u$ from $(\rho,\hat\iota(\rho : -))$ to $(\gamma,\hat\iota(\gamma : -))$
and a morphism $v$ from $(\sigma,\hat\iota(\sigma : -))$ to $(\gamma,\hat\iota(\gamma : -))$
such that $uu^*+vv^*=\unit$.

Next we consider subobjects.
Let $(\rho,\hat\iota(\rho : -))$ be an object of $\caZ\lmk\Cabul\rmk$
and $p$ a projection in the endomorphism of $(\rho,\hat\iota(\rho : -))$.
Let $\gamma\in\Obul$ and $v\in (\gamma,\rho)_U$ isometry
such that $vv^*=p$.
We then set $\hat\iota\lmk\gamma : -\rmk$ as

\begin{align}
\hat\iota\lmk\gamma : \zeta\rmk:=
\Tbd\zeta{\lzr}\unit(v^*)\hat\iota(\rho :\zeta) v.
\end{align}

This is in fact a half-braiding.
Using the naturality of $\hat\iota(\rho :\zeta)$, we have

\begin{align}
\begin{split}
=\Tbd\zeta{\lzr}\unit(v^*)\hat\iota(\rho :\zeta) v
v^*\hat\iota(\rho :\zeta)^*\Tbd\zeta{\lzr}\unit(v)\\
&=\Tbd\zeta{\lzr}\unit(v^*)\hat\iota(\rho :\zeta) p \hat\iota(\rho :\zeta)^*\Tbd\zeta{\lzr}\unit(v)
=\Tbd\zeta{\lzr}\unit(v^*)\Tbd\zeta\lzr\unit (p)\hat\iota(\rho :\zeta) \hat\iota(\rho :\zeta)^*\Tbd\zeta{\lzr}\unit(v)=\unit
\end{split}
\end{align}

\begin{align}
\begin{split}
=v^*\hat\iota(\rho :\zeta)^*\Tbd\zeta{\lzr}\unit(v)\Tbd\zeta{\lzr}\unit(v^*)\hat\iota(\rho :\zeta) v\\
&=v^* \hat\iota(\rho :\zeta)^*\Tbd\zeta{\lzr}\unit(p)\hat\iota(\rho :\zeta) v
=v^* \hat\iota(\rho :\zeta)^*\hat\iota(\rho :\zeta) pv=\unit.
\end{split}
\end{align}

Hence we have $\hat\iota\lmk
\gamma : \zeta
\rmk\in\caU(\fbd)$.
For any $A\in\abd$,
we have

\begin{align}
\begin{split}
&\hat\iota\lmk\gamma : \zeta\rmk\lmk\gamma\otimes \zeta\rmk(A)
=\Tbd\zeta{\lzr}\unit(v^*)\hat\iota(\rho :\zeta) v\Tbd\gamma\lzr\unit\Tbd\zeta{\lzr}\unit\pbd(A)\\
&=\Tbd\zeta{\lzr}\unit(v^*)\hat\iota(\rho :\zeta) \Tbd\rho\lzr\unit\Tbd\zeta{\lzr}\unit\pbd(A) v
=\Tbd\zeta{\lzr}\unit(v^*)\Tbd\zeta{\lzr}\unit \Tbd\rho\lzr\unit\pbd(A)\hat\iota(\rho :\zeta)  v\\
&=\Tbd\zeta{\lzr}\unit \lmk \Tbd\gamma\lzr\unit\pbd(A) v^*\rmk\hat\iota(\rho :\zeta)  v\\
&=(\zeta\otimes\gamma)(A) \hat\iota\lmk\gamma : \zeta\rmk.
\end{split}
\end{align}

For any $\zeta,\zeta'\in \Obul$, we have

\begin{align}
\begin{split}
\lmk \hat\iota(\gamma: \zeta)\otimes\unit_{\zeta'}\rmk
=\Tbd\zeta\lzr\unit\lmk  \Tbd{\zeta'}{\lzr}\unit(v^*)\hat\iota(\rho :\zeta') v\rmk
\Tbd\zeta{\lzr}\unit(v^*)\hat\iota(\rho :\zeta) v\\
&=\Tbd\zeta\lzr\unit\lmk  \Tbd{\zeta'}{\lzr}\unit(v^*)\hat\iota(\rho :\zeta') p\rmk
=\Tbd\zeta\lzr\unit\lmk  \Tbd{\zeta'}{\lzr}\unit(v^* p)\hat\iota(\rho :\zeta') \rmk
\hat\iota(\rho :\zeta) v\\
\hat\iota\lmk\rho:\zeta\otimes \zeta'\rmk v
\end{split}
\end{align}

Furthermore, as in the direct sum case,
we have $\lim_{t\to\infty}
\sup_{\zeta\in O^{rU}_{\mathop{\mathrm{bd},\lm {r2}(t)}}}
\lV
\hat\iota(\gamma :\zeta)-\unit
\rV=0$.

\begin{align}
\begin{split}
\epsilon_{\zc}\lmk
(\rho,\hat\iota(\rho:-)), (\sigma,\hat\iota(\sigma:-))
\rmk
\end{split}
\end{align}

from (<ref>) and (<ref>),
this defines a braiding of $\caZ(\Cabul)$.

Let $\llz\in\pc$.
Consider the setting in subsection <ref>, and assume Assumption <ref>,
Assumption <ref>, Assumption <ref>, Assumption <ref>, Assumption <ref>.
Let $\hfc : \CaUbkl\to \caZ\lmk\Cabul\rmk$ be a functor
defined by

\begin{align}
\begin{split}
\hfc(\rho):=
\lmk
\ffc_0(\rho), \hb{ \ffc_0(\rho)} {-}
\rmk
\end{split}
\end{align}

for $\rho\in \OUbkl$,

\begin{align}
\begin{split}
\hfc(R):=\ffc_1(R)=\tau(R)\in
\lmk\hfc(\rho),\hfc(\sigma)\rmk
\end{split}
\end{align}

for any $\rho,\sigma\in \OUbkl$ and $R\in (\rho,\sigma)$.

In fact $\rho\in \OUbkl$,
$\hfc(\rho)$ is an object of $\zc$because of
Lemma <ref>, <ref>,<ref>, <ref>, <ref>, Proposition <ref>.
For any $\rho,\sigma\in \OUbkl$ and $R\in (\rho,\sigma)$,
$\hfc(R)=\tau(R)\in (\ffc_0(\rho),\ffc_0(\sigma))_U$
is in the morphism of $\zc$ between objects
\ffc_0(\rho), \hb{ \ffc_0(\rho)} {-}
\rmk $
\ffc_0(\sigma), \hb{ \ffc_0(\sigma)} {-}
\rmk $.
Clearly $\hfc$ is a functor which is linear on morphisms.
Note that

\begin{align}
\begin{split}
\unit : (\pbd,\hat\iota(\pbd: -))\to
\ffc_0(\pbk),
\hb{\ffc_0(\pbk)}{-}
\rmk\\
&=\lmk\pbd,\hat\iota(\pbd :-)\rmk
\end{split}
\end{align}

is an isomorphism.
Furthermore, by Lemma <ref>,

\begin{align}
\begin{split}
=\lmk \ffc_0(\rho)\otimes\ffc_0(\sigma),
\hi{\ffc_0(\rho)\otimes\ffc_0(\sigma)}{-}
\rmk\\
&=\lmk \ffc_0(\rho\otimes\sigma),
\hi{\ffc_0(\rho\otimes \sigma)}{-}
\rmk
= \hfc(\rho\otimes\sigma),
\end{split}
\end{align}

and its endomorphisms are $(\ffc_0(\rho)\otimes \ffc(\sigma),\ffc(\rho\otimes\sigma))_U$
and $\unit : \hfc(\rho)\otimes\hfc(\sigma)\to \hfc(\rho\otimes\sigma)$
is natural, from the proof of Lemma <ref>.
Hence $\hfc$ is a unitary tensor functor.

Next we show that $\hfc$ is equivalence.
In order to do so, it suffices to show that $\hfc$ is fully faithful and surjective.
First we show that $\hfc$ is fully faithful  i.e.,
for any $\rho,\sigma\in\OUbkl$,

\begin{align}
\begin{split}
\hfc :(\rho,\sigma)\ni R\mapsto \tau(R)\in \lmk \hfc(\rho),\hfc(\sigma)\rmk
=\lmk\ffc_0(\rho), \ffc_0(\sigma)\rmk_U
\end{split}
\end{align}

is an isomorphism.
Note that it is injective and linear.
To show that it is also surjective, let $\tilde R$
be an arbitrary element from $\lmk\ffc_0(\rho), \ffc_0(\sigma)\rmk_U \subset\fbd$.
Then $\tau^{-1}\lmk \tilde R\rmk$

\begin{align}
\begin{split}
&\tau^{-1}\lmk \tilde R\rmk\rho(A)
=\tau^{-1}\lmk \tilde R \tau\Tbk\rho\lz\unit\tau^{-1}\pbd(A)  \rmk
=\tau^{-1}\lmk \tilde R\lmk \ffc_0(\rho)\pbd(A)\rmk  \rmk \\
&=\tau^{-1}\lmk\lmk\ffc_0(\sigma)\pbd(A)  \rmk\tilde R \rmk
=\Tbk\sigma\lz\unit\pbk(A)\tau^{-1}(\tilde R)
=\sigma(A)\tau^{-1}(\tilde R),
\end{split}
\end{align}

for any $A\in \abd$.
Hence $\tau^{-1}(\tilde R)\in (\rho,\sigma)$, and we have

\begin{align}
\hfc\lmk \tau^{-1}(\tilde R)\rmk
=\tau\tau^{-1}(\tilde R))=\tilde R.
\end{align}

This proves the surjectivity.

Next we show that $\hfc$ is surjective.
Let  $(\rho,\hi \rho -)$ be an arbitrary object in $\zc$.
By Lemma <ref>, there exists
there exists a $G(\rho)\in \OUbk$ such that
$\ffc_0\lmk {G(\rho)}\rmk =\rho$.
Then we have
\hfc(G(\rho))
\rho,\hb\rho -
\rmk

Finally, $\hfc$ is braided because from Lemma <ref>,

\begin{align}
\begin{split}
&\epsilon_{\zc}\lmk\hfc(\rho), \hfc(\sigma)\rmk\\
\epsilon\lmk
\lmk \ffc_0(\rho),\hi{\ffc_0(\rho)}\rmk), \lmk\ffc_0(\sigma),\hi{\ffc_0(\sigma}-\rmk
\rmk\\
&=\ffc_1\lmk \epsilon_+(\rho,\sigma)\rmk
=\hfc\lmk \epsilon_+(\rho,\sigma)\rmk.
\end{split}
\end{align}

§ BULK-BOUNDARY CORRESPONDENCE

In this section, we construct a tensor category $\caM$
by enriching $\Cabul$ with morphisms from $\bl$.
By taking the idempotent completion of $\caM$,
we obtain a braided $C^*$-tensor category $\tilde \caM$.
We introduce the Drinfeld center with an asymptotic constraint
$\caZ_a(\tilde \caM)$ of $\tilde \caM$.
This is the same as the usual Drinfeld center, except that the half-braidings
are required to satisfy the asymptotic property we saw in Lemma <ref>.

Note that in Corollary <ref>, we did not assume
the gap condition on the boundary.
We now introduce a condition which we expect to be related to the gap condition, which is
Assumption <ref>.
It can be interpreted as a kind of non-triviality of the braiding.
Under this condition,
$\caZ_a(\tilde \caM)$ gets a braided $C^*$-tensor category which is
equivalent to the bulk category $\CaUbk$.

Consider the setting in subsection <ref>.
Let $\llz\in\pc$.
Let $\caM$ be a category with object $\Obj\caM:=\Obul$,
and morphisms between objects $\rho,\sigma\in \Obj\caM:=\Obul$
defined by
Then $\caM$ is a strict tensor category with respect to the tensor $\otimes$
of $\Cabul$.

That  $\caM$ is a strict tensor category
follows from the fact that $\Cabul$ is closed under tensor
and $\Cabul$ is a subcategory of $\Carbdl$.
The tensor category $\caM$ is not necessarily a $C^*$-tensor category.
By taking the idempotent completion, we obtain a $C^*$-tensor category

Let $\llz\in\pc$.
Consider the setting in subsection <ref>, and

Assumption <ref>.
Let $\tilde \caM$ be a category with objects

\begin{align}
\begin{split}
\Obj\tilde\caM:=
\left\{
\rho\in \Obj\caM=\Obul,\quad p\in (\rho,\rho)_l\;\; \text{projection}
\right\},
\end{split}
\end{align}

\begin{align}
\begin{split}
\Mor_{\tilde\caM}\lmk (\rho,p), (\sigma,q)\rmk:=
\left\{
R\in (\rho,\sigma)_l\mid
\right\}\quad
\text{for} \quad (\rho,p),(\sigma,q)\in \Obj\tilde\caM.
\end{split}
\end{align}

The category $\tilde\caM$ is a $C^*$-category with

\begin{align}\label{suika}
\begin{split}
\id_{(\rho,p)}:=p,\quad \text{for each}\quad (\rho,p)\in \Obj\tilde\caM
\end{split}
\end{align}

composition of morphisms

\begin{align}\label{kaki}
\begin{split}
\Mor_{\tilde\caM}\lmk (\sigma,q), (\gamma,r)\rmk\times
\Mor_{\tilde\caM}\lmk (\rho,p), (\sigma,q)\rmk\ni (S,R)
\mapsto  SR\in  \Mor_{\tilde\caM}\lmk (\rho,p), (\gamma,r)\rmk,
\end{split}
\end{align}

and an antilinear contravariant functor $* : \tilde\caM\to\tilde\caM$
given by the adjoint $*$ of $\caB(\hbd)$
for morphisms

\begin{align}\label{melon}
\begin{split}
\Mor_{\tilde\caM}\lmk (\rho,p), (\sigma,q)\rmk\ni T\to
T^*\in \Mor_{\tilde\caM}\lmk (\sigma,q),(\rho,p)\rmk.
\end{split}
\end{align}

Furthermore, $\tilde\caM$ is a $C^*$-tensor category
with the tensor

\begin{align}\label{ringo}
\begin{split}
(\rho,p)\otimes_{\tilde M}(\sigma,q):=
(\rho\otimes\sigma,p\otimes q),\quad
\text{for}\quad(\rho,p),(\sigma,q)\in \Obj\tilde\caM
\end{split}
\end{align}

\begin{align}\label{nashi}
\begin{split}
&R\otimes_{\tilde M}S
:=R\otimes S\in \Mor_{\tilde\caM}\lmk (\rho,p)\otimes_{\tilde M}(\sigma,q),
(\rho',p')\otimes_{\tilde M}(\sigma',q')\rmk\\
\text{for}\quad
R\in \Mor_{\tilde\caM}\lmk (\rho,p),  (\rho',p')\rmk,\quad
S\in \Mor_{\tilde\caM}\lmk (\sigma,q), (\sigma',q')\rmk.
\end{split}
\end{align}

The tensor unit is $(\pbd,\unit)$.

Clearly, each $\Mor_{\tilde\caM}\lmk (\rho,p), (\sigma,q)\rmk$ is a Banach space
with the norm inherited from $\caB(\hbd)$.
In (<ref>), we have $SR\in  \Mor_{\tilde\caM}\lmk (\rho,p), (\gamma,r)\rmk$
because $SR\in (\rho,\gamma)_l$ and $rSR=SR=SRp$.
The map (<ref>) is bilinear and we have $\lV SR\rV\le \lV S\rV \lV R\rV$.
The associativity of the composition follows from that of $\caB(\hbd)$.
In (<ref>), we have $T^*\in \Mor_{\tilde\caM}\lmk (\sigma,q),(\rho,p)\rmk$
because $T^*\in (\sigma,\rho)_l$ and $pT^*=T^*=T^*q$.
That $T^{**}=T$, $\lV T^*T\rV=\lV T\rV^2$
$T^*T\ge 0$ follows from the corresponding properties of $\caB(\hbd)$.
(<ref>) gives the identity morphism :
clearly, for each object $(\rho,p)\in \Obj\tilde\caM$, we have
$\id_{(\rho,p)}:=p\in \Mor_{\tilde\caM}\lmk (\rho,p),  (\rho,p)\rmk$
because $p\in (\rho,\rho)_l$ and $pp=p=pp$.
Furthermore, for any
$R\in \Mor_{\tilde\caM}\lmk (\rho,p), (\sigma,q)\rmk$
we have $R\circ\id_{(\rho,p)}=Rp=R=qR=\id_{(\sigma,q)}\circ R$.
Hence, $\tilde\caM$ is a $C^*$-tensor category.

The formula  in (<ref>)
gives an object of $\tilde\caM$  because
$\rho\otimes\sigma\in \Obj\caM=\Obul$
and $p\otimes q\in (\rho\otimes\sigma,\rho\otimes\sigma)_l$

\begin{align}
\begin{split}
(p\otimes q)^*=p^*\otimes q^*=p\otimes q,\quad
(p\otimes q)^2=p^2\otimes q^2=p\otimes q,
\end{split}
\end{align}

by Lemma <ref>.
The formula in (<ref>) is well-defined because
$R\otimes_{\tilde M}S
:=R\otimes S\in (\rho\otimes\sigma,\rho'\otimes \sigma')_l$

\begin{align}
\begin{split}
(p'\otimes q')(R\otimes S)
=p'R\otimes q'S=R\otimes S
=Rp\otimes Sq=(R\otimes S)(p\otimes q),
\end{split}
\end{align}

by Lemma <ref>.
If we further have
$R'\in \Mor_{\tilde\caM}\lmk (\rho',p'),  (\rho'',p'')\rmk$,
$S'\in \Mor_{\tilde\caM}\lmk (\sigma',q'), (\sigma'',q'')\rmk$,
then we have

\begin{align}\label{aka}
\begin{split}
&\lmk R'\otimes_{\tilde M}S'\rmk\circ\lmk R\otimes_{\tilde M}S\rmk
=\lmk R'\otimes S'\rmk\lmk R\otimes S\rmk
=R'R\otimes S'S
=R'R\otimes_{\tilde M} S'S,\\
&\lmk R\otimes_{\tilde\caM} S\rmk^*
=\lmk R\otimes S\rmk^*=R^*\otimes S^*=R^*\otimes_{\tilde\caM} S^*,
\end{split}
\end{align}

by Lemma <ref>.
We also have

\begin{align}\label{ao}
\begin{split}
\id_{(\rho,p)\otimes_{\tilde M}(\sigma,q)}=\id_{(\rho\otimes\sigma,p\otimes q)}
=p\otimes q=
p\otimes_{\tilde \caM} q=\id_{(\rho,p)}\otimes_{\tilde \caM}\id_{(\sigma,\rho)}.
\end{split}
\end{align}

As the associativity morphism, we take

\begin{align}
\begin{split}
&\alpha_{(\rho,p), (\sigma,q),(\gamma,r)}
:=(p\otimes q)\otimes r\\
\Mor_{\tilde \caM} \lmk
\lmk (\rho,p)\otimes_{\tilde\caM}(\sigma,q)\rmk\otimes_{\tilde\caM}
\lmk (\sigma,q)\otimes_{\tilde\caM}
\Mor_{\tilde \caM} \lmk
\lmk \rho\otimes\sigma\otimes\gamma, p\otimes q\otimes r\rmk, \lmk \rho\otimes\sigma\otimes\gamma, p\otimes q\otimes r\rmk
\rmk.
\end{split}
\end{align}

In fact, we have

\begin{align}
\begin{split}
\alpha_{(\rho,p), (\sigma,q),(\gamma,r)}
:=p\otimes q\otimes r
\in (\rho\otimes\sigma\otimes\gamma, \rho\otimes\sigma\otimes\gamma)_l
\end{split}
\end{align}

\begin{align}
\begin{split}
&\lmk p\otimes q\otimes r\rmk \alpha_{(\rho,p), (\sigma,q),(\gamma,r)}
=\lmk p\otimes q\otimes r\rmk^2
=\lmk p\otimes q\otimes r\rmk=\alpha_{(\rho,p), (\sigma,q),(\gamma,r)},\\
&\alpha_{(\rho,p), (\sigma,q),(\gamma,r)}\lmk p\otimes q\otimes r\rmk
=\lmk p\otimes q\otimes r\rmk^2
=\lmk p\otimes q\otimes r\rmk=\alpha_{(\rho,p), (\sigma,q),(\gamma,r)}.
\end{split}
\end{align}

In fact it is an isomorphism because $\id_{\lmk \rho\otimes\sigma\otimes\gamma, p\otimes q\otimes r\rmk}=p\otimes q\otimes r$.
The naturality of $\alpha$ and the Pentagon axiom follows from
the definition and Lemma <ref>.
The object $(\pbd,\unit)$ is the tensor unit with
left/right unit constraints

\begin{align}
\begin{split}
&l_{(\rho,p)}:=p \in
\Mor_{\tilde\caM}\lmk (\pbd,\unit)\otimes_{\tilde\caM} (\rho,p),
(\rho,p)\rmk= \Mor_{\tilde\caM}\lmk(\rho, p),
&r_{(\rho,p)}:=p \in\Mor_{\tilde\caM}\lmk
(\rho,p) \otimes_{\tilde\caM} (\pbd,\unit), (\rho, p)\rmk= \Mor_{\tilde\caM}\lmk(\rho, p),
\end{split}
\end{align}

Their naturalities follow from the definition of morphisms and identity morphisms.The triangle axiom
follows by

\begin{align}
\begin{split}
\lmk \id_{(\rho,p)}\otimes_{\tilde\caM}l_{(\sigma,q)}\rmk
\alpha_{(p,\rho), (\pbd,\unit),(\sigma,q)}
=\lmk p\otimes q\rmk\lmk p\otimes\unit\otimes q\rmk
=\lmk p\otimes q\rmk
=r_{(\rho,p)}\otimes_{\tilde\caM} \id_{(\sigma,q)}.
\end{split}
\end{align}

Now we construct a direct sum for
$(\rho,p), (\sigma,q)\in \Obj\tilde\caM$.
By Lemma <ref>, there exist $\gamma\in \Obj\caM=\Obul$
and isometries $u\in (\rho,\gamma)_U$, $v\in (\sigma,\gamma)_U$
such that $uu^*+vv^*=\unit$.
We set $r:=upu^*+vqv^*$, which is a projection in $(\gamma,\gamma)_l$.
Hence we have $(\gamma, r)\in \Obj\tilde\caM$.

We have $up\in \Mor_{\tilde\caM} \lmk (\rho, p), (\gamma,r)\rmk$
because $up\in (\rho,\gamma)_l$ and $rup=up=upp$.
It is an isometry because $pu^*up=p=\id_{(\rho,p)}$.
Similarly, we have  $vq\in \Mor_{\tilde\caM} \lmk (\sigma, q), (\gamma,r)\rmk$
and it is an isometry.
We also have $upu^*+vqv^*
Hence we obtain the direct sum.

Next we introduce a subobject of $(\rho,p)\in \Obj\tilde\caM$
with respect to a projection $q\in \mm\lmk (\rho,p), (\rho,p)\rmk$.
We have $(\rho,q)\in \om$ because $\rho\in \Obj(\caM)$,
$q\in (\rho,\rho)_l$ a projection.
Because  $q\in \mm\lmk (\rho,p), (\rho,p)\rmk$, we have
Note also that $q\in \mm\lmk (\rho,q), (\rho,p)\rmk$