Quantum Current and Holographic Categorical Symmetry

Tian Lan1$\star$
and Jing-Ren Zhou1

Department of Physics, The Chinese University of Hong Kong, Shatin, New Territories, Hong Kong, China

${}^\star$<EMAIL_ADDRESS>
August 27, 2024

§ ABSTRACT

We establish the formulation for quantum current. Given a symmetry group $G$, let $\mathcal{C}:=\mathop\mathrm{Rep} G$ be its representation category. Physically, symmetry charges are objects of $\mathcal{C}$ and symmetric operators are morphisms in $\mathcal{C}$. The addition of charges is given by the tensor product of representations. For any symmetric operator $O$ crossing two subsystems, the exact symmetry charge transported by $O$ can be extracted. The quantum current is  defined as symmetric operators that can transport symmetry charges over an arbitrary long distance. A quantum current exactly corresponds to an object in the Drinfeld center $Z_1(\mathcal{C})$. The condition for quantum currents to be superconducting is also specified, which corresponds to condensation of anyons in one higher dimension. To express the local conservation, the internal hom must be used to compute the charge difference, and the framework of enriched category is inevitable. To illustrate these ideas, we develop a rigorous scheme of renormalization in one-dimensional lattice systems and analyse the fixed-point models. It is proved that in the fixed-point models, superconducting quantum currents form a Lagrangian algebra in $Z_1(\mathcal{C})$ and the boundary-bulk correspondence is verified in the enriched setting. Overall, the quantum current provides a natural physical interpretation to the holographic categorical symmetry.

§ INTRODUCTION

The electric charge is a conserved quantity. Classically, we think that the electric charge is a continuous quantity and we talk about the charge density $\rho$. The global charge $Q=\int \rho$
is conserved. If we divide the whole system into two parts $\BA$ and $\BB$, and denote $Q_\BA=\int_\BA \rho$ and $Q_\BB=\int_\BB \rho$, then

\begin{align}
0=\Delta Q=\Delta Q_\BA+ \Delta Q_\BB  \implies \Delta Q_\BA =-\Delta Q_\BB .
\end{align}

The change of charge in one subsystem must compensate that in the other. When the charge in $\BA$ increases, there must be charge flowing from $\BB$ to $\BA$, i.e., a current. Again one can consider the current density $\bl j$, and we have the famous differential equation for local conservation of charge

\begin{align}
\frac{\partial \rho}{\partial t}+\nabla\cdot \bl j=0.
\end{align}

However, we know in reality that electric charge is discrete instead of continuous. The above treatment, including the density of charge or current as well as the differential equation, is only an effective approximation at a macroscopic or statistical level.

The case becomes even worse when we consider other symmetries. The electric charge is just the conserved quantity of global $U(1)$ symmetry, taking value in integers (with appropriate unit). The addition of electric charges is the usual addition of integers. We may switch to, for example, angular momentum, which is the conserved quantity of the rotation symmetry, $SU(2)$. The quantum angular momentum takes value in, no longer ordinary numbers, but the representations of the symmetry group $SU(2)$. The addition of angular momentum is also more complicated than the addition of numbers. For example, the total angular momentum of two spin $1/2$'s together is the “direct sum”  of a spin $0$ singlet and a spin $1$ triplet: $1/2\otimes 1/2=0\oplus 1$.

In this paper we give a serious treatment to the local conservation of quantum symmetry charge, which seems long missing in the literature, although the mathematics involved is quite fundamental. The essential technical step is to realize that in a quantum system with symmetry, the symmetric operators are just the morphisms in the tensor (or fusion when the group is finite) category of symmetry charges, or equivalently, symmetric (or invariant) tensor networks. With this in mind, we show that for any symmetric operator crossing two subsystems, a symmetry charge can be extracted which we interpret as the symmetry charge transported by the operator. Moreover, we define the notion of quantum current, which is the collection of symmetric operators that can transport a symmetry charge over a long distance.

It can be seen directly from our definition that a quantum current is an object in the Drinfeld center of the fusion category of symmetry charges.
Our work is a generalization of the Noether theorem: when the symmetry, which can be discrete, is given, the possible current is fully determined, by computing the Drinfeld center. The local conservation is now written as

\begin{equation}
\Hom_\cC(Q\ot X,Y)\cong \Hom_{Z_1(\cC)}((Q,\beta),[X,Y]_{Z_1(\cC)}), \end{equation}

where $\cC$ is the category of symmetry charges and $Z_1(\cC)$ is the category of quantum currents. The functor on the left hand side $\Hom(-\ot X,Y)$ compute the ways how $X$ can be converted to $Y$.  The internal hom $[X,Y]_{Z_1(\cC)}$ is an object in $Z_1(\cC)$ that represents the functor $\Hom(-\ot X,Y)$. Physically, $[X,Y]_{Z_1(\cC)}$ is the quantum current providing the universal answer for the ways how $X$ can be converted to $Y$. We have to use the internal hom to compute the quantum current, including the charge difference and how the charge flows, for discrete symmetry charges.

We like to comment on the difference between our formulation and the traditional notion of current in quantum mechanics or  current operator in quantum field theory. The current carried by a charged quantum particle is traditionally defined as the charge times the probability current. Based on such notion, one can only conclude that the expectation value of the charge is locally conserved. Similarly, a current operator in quantum field theory is such an operator that its expectation value (correlation function) satisfies a local conservation condition. We consider these traditional notions of current as only “semi-classical”, in that (1) the local conservation is only satisfied on average, at a macroscopic or statistical level; (2) they can not be used to deal with the discrete or quantized charge transport in a single quantum mechanical process; and (3) they usually require continuous space-time and continuous symmetry. There are recent works <cit.> extending the “semi-classical” current to lattice systems. Our formulation, on the contrary, is truly quantum: it can apply to discrete space and discrete symmetry, and can be used to analyse quantized charge transport exactly instead of on average.

On the other hand, quantum currents can be identified with excitations in a topological order in one higher dimension. In fact, our work provides a concrete physical interpretation to the categorical symmetry<cit.>.

The term categorical symmetry was first proposed in Ref. <cit.>, which aims at providing an invariant shared by gapped phases before and after a phase transition. For example, the 1+1D transverse field Ising model has two gapped phases, $\Z_2$ symmetric and $\Z_2$ symmetry breaking: One exhibits the $\Z_2$ symmetry while the other exhibits the dual $\Z_2$ symmetry. Putting the two $\Z_2$ together, one says that there is a categorical symmetry of 2+1D $\Z_2$ topological order (or toric code model <cit.>), which has both $\Z_2$ charges and $\Z_2$ fluxes. From a purely mathematical point of view, this categorical symmetry is the Drinfeld center of the fusion category $\Rep \Z_2$ of symmetry charges. Because of the boundary-bulk correspondence of topological ordered phases <cit.>, one can say, virtually, that categorical symmetry is the topological order in one dimension higher. Because of this relation, in this paper we call such invariant (topological order in one dimension higher) as the holographic categorical symmetry, in order to avoid possible confusion with e.g., fusion category symmetry <cit.> and other similar notions which may also be referred to as categorical symmetry.[In a more general sense, any generalized version of symmetry relating to category theory may be called categorical symmetry.]

However, the physical meaning of holographic categorical symmetry remains mysterious.
There are even other names for the same notion, such as symmetry TFT <cit.> and topological symmetry <cit.>.
Ref. <cit.> proposed to view holographic categorical symmetry as the transparent patch operators. Based on the idea of topological Wick rotation <cit.>, Refs. <cit.> proposed to view holographic categorical symmetry (or the background category of the enriched category description<cit.> of gapped phase) as the sectors of non-local symmetric operators.
Moreover, in the context of topological order, one can condense the excitations in the bulk topological order to obtain a boundary theory, or fuse defects in the bulk with the  boundary theory. Thus, by topological Wick rotation, the algebras and defects in the holographic categorical symmetry (corresponding to the topological order in the bulk)  should play an important role in classifying phases and phase transitions (corresponding to the boundary theory). Such point of view has been emphasized in Refs. <cit.>.

Our formulation of quantum currents clarifies the confusion around various concepts regarding holographic categorical symmetry. We give a rigorous definition, Definition <ref>, for what operators qualify as quantum currents. We also give a definition, Definition <ref>, for what quantum currents are superconducting. These definitions are tested in concrete one dimensional fixed-point lattice models. We prove that the superconducting quantum currents in the fixed-point model indeed form a Lagrangian algebra in the Drinfeld center, corresponding to a maximal anyon condensation in one higher dimension, which also determines the fixed-point defects (or excitations) in the fixed-point model. Therefore, we established the correspondence between quantum currents and the holographic categorical symmetry; holographic categorical symmetry is about the transport property of the physical system. We also want to emphasize that the quantum currents can be measured, observed and understood, in the same dimension of the physical system; no fictional one dimensional higher bulk is required.

The paper is organized as follows: in Section <ref> we give some necessary preliminary notions and fix our notations; in Section <ref> we explain basic techniques on how to manipulate symmetric operators, viewing them as graphs or tensor networks; in Section <ref> we motivate the definition for quantum current and introduce some related notions such as the charge transported by symmetric operators and the condensation of quantum currents;
in Section <ref> we give a rigorous treatment to the renormalization process in 1+1D lattice models; finally in Section <ref> we show that in the fixed-point lattice models superconducting quantum currents form a Lagrangian algebra and give a series of examples. We summarize main contents in this paper in Table <ref>.

[c] |m.28|m.36|m.28|
Physical quantity
Mathematical description
Graphical representation

Symmetry charge/
Hilbert space on
local region $\BK$
$(\cH_\BK,\rho^\BK)\in \cC:=\Rep G$
\itk{\filldraw (0,1.3) -- node[right]{$\cH_\BK$} (0,0);}$

General symmetric
operator/intertwiner     $\Hom(\cH_{\BK_1},\cH_{\BK_2})$ consists of
$f\in \Hom_\Ve(\cH_{\BK_1},\cH_{\BK_2})$
such that $\forall g\in G$,
$\rho^{\BK_2}_g  f (\rho^{\BK_1}_g)^{-1}=f$     [baseline=(current bounding box.center),>=Triangle[open,width=0pt 20],yscale=0.6][rectangle, draw] (l) at (0,0) $f$;
(0,-2) – node[right]$\cH_{\BK_1}$ (l) – node[right]$\cH_{\BK_2}$ (0,2);

Symmetric operator on
bipartite system $\BA$ and $\BB$     $O\in \Hom(\cH_\BA\ot \cH_\BB,\cH_\BA\ot\cH_\BB)$
[rectangle, minimum width=4em, draw] (f) at (0,0) $O$;
(v1) at (-0.5,1) $\cH_\BA$;
(vn) at (0.5,1) $\cH_\BB$;
(w1) at (-0.5,-1) $\cH_\BA$;
(wm) at (0.5,-1) $\cH_\BB$;
(v1) – (v1 |- f.north)
(vn) – (vn |- f.north)
(w1) – (w1 |- f.south)
(wm) – (wm |- f.south);

Transformation on $O$
to extract transported
symmetry charge     $O\overset{\epsilon}{\to}\bar O=\bar r \bar l\overset{\epsilon^{-1}}{\to} O=rl$
(Definition <ref>)     $\bar O=$[baseline=(current bounding box.center),>=Triangle[open,width=0pt 20],scale=0.8]
[rectangle, draw] (e) at (0,-1) $\bar l$;
[rectangle, draw] (m) at (0,0) $\bar r$;
(1,1)node[above]$\cH_\BB^*$ – (m)– (-1,1)node[above]$\cH_\BB$;
(e) – (0,-0.5)node[right]$\Ima \bar O$–(m);
(-1,-2)node[below]$\cH_\BA^*$ – (e) – (1,-2)node[below]$\cH_\BA$;

Transported symmetry
charge from region
$\BA$ to $\BB$     $\curr{\BA}{\BB}=\Ima \bar O$      $O=$
[rectangle, draw] (l) at (0,1) $l$;
[rectangle, draw] (r) at (2,2) $r$;
(l)–node[above]$\curr{\BA}{\BB}$ (r);

Quantum current     $(Q,\beta_{Q,-})\in Z_1(\cC)$
(Definition <ref>)

Simple objects
in $Z_1(\Rep G)$: $(C_x,\tau)$
(Appendix <ref>)
[rectangle,draw] (l) at (-2,-1) $l$;
[rectangle,draw] (m) at (0,0) $\beta_{Q,\cH_\BM}$;
[rectangle,draw] (r) at (2,1) $r$;
(-2,-2)node[below]$\cH_s$ – (l) – (-2,2)node[above]$\cH_s$
(0,-2)node[below]$\cH_\BM$ – (m) – (0,2)node[above]$\cH_\BM$
(2,-2)node[below]$\cH_t$ – (r) – (2,2)node[above]$\cH_t$
(l)–node[above]$Q$ (m) – node[above]$Q$ (r);

Translation invariant
fixed-point model
$(\Z,A,-m^\dag m)$     Hilbert space: $\cH_i=A,\forall i\in\Z$,
where $(A,m,\eta)$ is a
Frobenius algebra in $\cC$

Hamiltonian: $H=-\sum_i (m^\dag m)_i$

Ground state subspace: $A$      $(m^\dag m)_i$=
(1,1)node[right]$A$ – (0,0)– (-1,1)node[left]$A$;
(0,-1)node[below]$m$ – (0,-0.5)node[right]$A$  –(0,0)node[above]$m^\dag$ ;
(-1,-2)node[left]$A$ – (0,-1) – (1,-2)node[right]$A$;

Half-infinite chain model
with boundary condition
$(\N,A,-m^\dag m, M,-\rho^\dag\rho)$     Boundary Hilbert space:
$\cH_0=M$, where
$(M,\rho)$ is a right $A$-module in $\cC$

$H=-(\rho^\dag\rho)_0-\sum_{i\geq 1} (m^\dag m)_i$

Ground state subspace: $M$

Boundary change: $A$-module map

All boundary conditions: $\cC_A$    $(\rho^\dag\rho)_0$=
(-1,-1)node[left]$\rho$ – (0,-2)node[right]$A$;
(-1,1)node[left]$M$ – (-1,-2)node[left]$M$;
(0,1)node[right]$A$ – (-1,0)node[left]$\rho^\dag$;

Finite chain model with
two-side boundary
$(\BL=\{0,...,J\},A,-m^\dag m,$
$ M,-\rho^\dag\rho,N,-\lambda^\dag\lambda)$     Boundary Hilbert spaces:
$\cH_0=M$, $\cH_J=N$, where
$(M,\rho)$ is a right $A$-module and
$(N,\lambda)$ is a left $A$-module in $\cC$

Hamiltonian: $H=-(\rho^\dag\rho)_0$
$-(\lambda^\dag\lambda)_{J-1}-\sum_{1\leq i\leq J-2} (m^\dag m)_i$

Ground state subspace:
$\Ima P=M\otr[A] N$     $P$=
(1,1)node[right]$N$ – (1,-2)node[right]$N$;
(-1,1)node[left]$M$ – (-1,-2)node[left]$M$;
(-1,0)node[left]$\rho$ – (0,-.5)node[below]$A$ – (1,-1)node[right]$\lambda^\dag$;

Fixed-point defects/
Excitations     Defect Hilbert space:
$A$-$A'$-bimodule $B$ in $\cC$

Defect change: $A$-$A'$-bimodule map

Ground state subspace
of two defects ${}_{A''}B'_A$ and $_A B_{A'}$:
$\Ima P'=B'\otr[A]B$

All defects: $_A\cC_{A'}\cong\Fun_\cC(\cC_A,\cC_A')$

Excitations: $_A\cC_A\cong$
$\Fun_\cC(\cC_A,\cC_A)^{\text{rev}}\cong Z_1(\cC)_{[A,A]}$    $P'$=
(1,1)node[right]$B$ – (1,-2)node[right]$B$;
(-1,1)node[left]$B'$ – (-1,-2)node[left]$B'$;
(-1,0)node[left]$\rho'$ – (0,-.5)node[below]$A$ – (1,-1)node[right]$\lambda^\dag$;

quantum currents     Internal hom $[A,A]\in Z_1(\cC)$,
which is also a Lagrangian
algebra in $Z_1(\cC)$

categorical symmetry    $Z_0(^{\cC}\cC_A)\cong ^{Z_1(\cC)}{\text{Fun}_{\cC}(\cC_A,\cC_A)}$

Universal model
$ -(\iota_A\ot \iota_A) m^\dag m (\iota_A^\dag\ot\iota_A^\dag))$
for $\cC=\Rep G$
with trivial $\omega_2$     Frobenius algebras in $\Rep G$
are classified by $(H\subset G,\omega_2)$,
where $\omega_2\in H^2(H,U(1))$

For trivial $\omega_2$, $A:=\Fun(G/H)$,
$\iota_A: A\to \Fun(G)$
(Subsection <ref>)

Realizing 1+1D spontaneous
symmetry breaking phases
with unbroken subgroup $H$
for different choices of $H$     [baseline=(current bounding box.center),>=Triangle[open,width=0pt 20],yscale=0.8]
[draw] (e1) at (-1,-1) $\iota_A^\dag$;
[draw](e2) at (1,-1) $\iota_A^\dag$;
[draw](g) at (-1,2) $\iota_A$;
[draw](h) at (1,2) $\iota_A$;
(-1,-2)node[below]$\Fun(G)$ – (e1) – (-.5,-.5)node[above]$A$ – (0,0)node[below]$m$ – (0,.5)node[right]$A$ – (0,1)node[above]$m^\dag$ – (-.5,1.5)node[below]$A$ – (g) – (-1,3)node[above]$\Fun(G)$
(1,-2)node[below]$\Fun(G)$ – (e2) – (.5,-.5)node[above]$A$ – (0,0) – (0,1) – (.5,1.5)node[below]$A$ – (h) – (1,3)node[above]$\Fun(G)$;

Universal model
for $\cC=\Rep G$
with nontrivial $\omega_2$    For nontrivial $\omega_2$, $A$ is given by
Example <ref>

Realizing generic 1+1D
$G$-symmetric phases labeled by
$(H,\omega_2)$ for different choices
of $H$ and $\omega_2$
($H$ is the unbroken subgroup
and $\omega_2$ labels symmetry protected
topological order under $H$)

A summary of contents.

§ PRELIMINARIES AND NOTATIONS

We first review the group representation category and fix our notation for graphical calculus.

Let $G$ be a compact group. The group representation category is the functor category $\Fun(BG, \Ve)=:\Rep G$, where $BG$ is the category with one object $\bullet$ and $BG(\bullet,\bullet)=G$, and $\Ve$ is the category of finite dimensional vector spaces over $\C$. Spelling it out,

* An object in $\Rep G$  is a pair $(V,\rho)$, called a group representation, where $V$ is a vector space and $\rho: G\to GL(V)$ is a group homomorphism, i.e. there are $g$-indexed invertible linear maps on $V$, denoted by $\rho_g$, such that
\begin{equation} \rho_{gh}=\rho_g\rho_h. \end{equation}
$\rho_g$ is referred to as the group action or symmetry action. An object in $\Rep G$ is physically referred to as a ($G$-)symmetry charge.

We may use simply $V$ to denote a representation $(V,\rho)$, and write the group action $\rho_g(a)$ as $g\triangleright a$ or $ga$, when the explicit form of $\rho$ is not important for the discussion.

* A morphism between two representations $(V,\rho)$ and $(W,\tau)$ is a linear map $f:V\to W$ that commutes with group actions, $f\rho_g=\tau_g f$ for all $g\in G$. The space of morphisms from $(V,\rho)$ to $(W,\tau)$ is denoted by $\Hom((V,\rho),(W,\tau))$. In particular, the endomorphism space $\Hom((V,\rho),(V,\rho))$ is exactly the subspace of symmetric operators on $V$, which satisfy $\rho_g f\rho_{g^{-1}}=f$. The following terms

* morphism in $\Rep G$,

* intertwiner, intertwining operator

* invariant tensor,

* symmetric tensor,[In this paper “symmetric” always mean invariant under group action, and never mean invariant under permutation of tensor indices.]

* symmetric operator,

will be used interchangeably.

Graphically, an object is represented by a line and a morphism is represented by a node between two lines

\begin{align}
f:(V,\rho)\to(W,\tau) & =
\itk{
\filldraw (0,-2) -- node[right]{$(V,\rho)$} (l) -- node[right]{$(W,\tau)$} (0,2);
\node[rectangle, draw] (l) at (0,0) {$f$};
\end{align}

Composition of morphisms is done from bottom to top

\begin{align}
gf = \itk{
\node[rectangle, draw] (l) at (0,1) {$f$};
\node[rectangle, draw] (r) at (0,2) {$g$};
\filldraw (0,0) -- (l) -- (r) -- (0,3);
}\quad .
\end{align}

The representation category enjoys additional nice structures, one of which is the tensor product. Given two representations $(V,\rho)$ and $(W,\tau)$, their tensor product

\begin{equation} (V,\rho)\ot(W,\tau):=(V\ot W, \rho\ot \tau) \end{equation}
is again a representation with action given by

\begin{align}
(\rho\ot\tau)_g=\rho_g\ot \tau_g.
\label{tensor}
\end{align}

Tensor product is graphically represented by juxtaposing lines

\begin{align}
(V,\rho)\ot (W,\tau) = \itk{
\draw (0,0) -- node[left]{$(V,\rho)$} (0,2);
\draw (1,0) -- node[right]{$(W,\tau)$} (1,2);
\end{align}

There is always the trivial representation $\one:=(\C,\id)$, $\id_g=\id_\C$ for all $g\in G$. $\one$ is the unit of tensor product. For any representation $(V,\rho)$, there is a dual representation $(V^*,\rho^\star)$[Dual group action and dual object, dual morphism are distinct. We denote dual group action of $\rho$ as $\rho^\star$, and dual object of $V$, dual morphism of $f$ as $V^*$, $f^*$ respectively.] where the underlying vector space is the dual vector space $V^*:=\Hom_\Ve(V,\C)$[When confusion is possible, subscripts are added to clarify $\Hom(-,-)$ in different categories.], and the group action is $\rho^\star_g:=-\circ\rho_{g^{-1}}=(\rho_{g^{-1}})^*$.  More concretely

\begin{align}
\rho^\star_g: V^*&\to V^*,\nonumber\\
\varphi&\mapsto \varphi\rho_{g^{-1}},
\label{dualrep}
\end{align}

graphically represented as

\begin{align}
\itk{
\filldraw (0,-1.5) -- node[right]{$V^*$} (r) -- node[right]{$V^*$} (0,1.5);
\node[rectangle, draw] (r) at (0,0) {$\rho^\star_g$};
\itk{
\filldraw (0,-1.5) -- node[right]{$V^*$} (r) -- node[right]{$V^*$} (0,1.5);
\node[rectangle, draw] (r) at (0,0) {$(\rho_{g^{-1}})^*$};
\itk{
\draw (-1,-.5) -- (l) -- (-1,.5);
\node[right] at (-1,-.65) {$V$};
\node[right] at (-1,.65) {$V$};
\node[rectangle, draw] (l) at (-1,0) {$\rho_{g^{-1}}$};
\draw (-1,1.5 |-1,.5) arc(0:180:0.5);
\draw (-2,.5) -- (-2,-1.5)node[left]{$V^*$};
\draw (-1,.5 |-1,-.5) arc(180:360:0.5);
\draw (0,-.5) -- (0,1.5)node[right]{$V^*$};
\end{align}

If one choose a basis of $V$ and the corresponding dual basis of $V^*$, then the matrix representation of $g$ on $V^*$ is the transpose of the matrix representation of $g^{-1}$ on $V$. We can check that the above defined $(V^*,\rho^\star)$ is indeed the dual object of $(V,\rho)$ in $\Rep G$: The pairing between $V^*$ and $V$

\begin{align}
V^*\ot  V&\to \C,\nonumber\\
\varphi\ot v&\mapsto \varphi(v)
\end{align}

is symmetric:
\begin{equation}
\rho^\star_g \varphi(\rho_g (v))=\varphi\circ \rho_{g^{-1}} \circ \rho_g(v)= \varphi(v).
\end{equation}
The copairing

\begin{align}
\C&\to V\ot V^*,\nonumber\\
1&\mapsto \sum_a a\ot \delta_a,
\end{align}

where $\{a\}$ is a basis of $V$ and $\{\delta_a\}$ the corresponding dual basis $\delta_a(b)=\delta_{a,b}$, is also symmetric
\begin{equation}\sum_a \rho_g(a)\ot \rho_g^\star (\delta_a)=\sum_a \rho_g(a)\ot (\delta_a\circ \rho_{g^{-1}})=\sum_a \rho_g(a)\ot \delta_{\rho_g(a)} =\sum_a a\ot \delta_a.\end{equation}

Therefore, the pairing and copairing both remain as morphisms in $\Rep G$ and exhibit $(V^*,\rho^\star)$ as the dual object of $(V,\rho)$.

Another structure is the direct sum. Given two representations $(V,\rho)$ and $(W,\tau)$, their direct sum $(V,\rho)\oplus (W,\tau):=(V\oplus W, \rho \oplus \tau)$ is again a representation with action given by

\begin{align}
(\rho\oplus\tau)_g=\rho_g\oplus\tau_g\in \End_\Ve(V)\oplus \End_\Ve(W)\subset\End_\Ve(V\oplus W,V\oplus W).
\end{align}

An isomorphism is a morphism invertible under composition. Two representations are isomorphic $V\cong W$ when there is an isomorphism between them; in other words, $V$ and $W$ differ by a basis change which commutes with group actions. A nonzero representation is irreducible (or simple) if it is not isomorphic to a direct sum of two nonzero representations. For a compact group $G$, all finite dimensional representations are completely reducible, i.e. isomorphic to a direct sum of irreducible representations. For reader's convenience, we review the categorical formulation of direct sum here

In a category whose hom-sets form abelian groups and composition is bilinear,
the direct sum of two objects $A,B$, if exists, is an object $Y$ together with two pairs of morphisms $p_A:Y\to A,$ $q_A:A\to Y,$ $ p_B:Y\to B,$ $q_B:B\to Y$ satisfying

\begin{align}
p_Aq_A=\id_A, \quad &p_Bq_B=\id_B,\nonumber\\
p_Aq_B=0, \quad &p_Bq_A=0,\nonumber\\
\end{align}

We refer to $p_A,p_B$ as projections[A projection may also refer to an operator that squares to itself. $q_Ap_A$ and $q_Bp_B$ are projections in this sense. In this paper we use both meanings of projection and the precise meaning should be clear from the context. As the two meanings are closely related and both standard, we avoid inventing a new name.] and $q_A,q_B$ as embeddings.

Such an object $Y$ is simultaneously a product and coproduct of $A$ and $B$, and by the universal property of limit, is unique up to unique isomorphism. Thus it is fine to talk about the direct sum and denote it by $A\oplus B$.

$\Rep G$ is also a unitary category with unitary structure given by the usual Hermitian conjugate.

In a unitary category such as $\Rep G$, it is always possible to choose $q_i=p_i^\dag$.

Now suppose that

\begin{align}
V\ot W\cong \oplus_i X_i,
\end{align}

where $X_i$ are irreps (irreducible representations). In more elementary words, the above means that after a change of basis of $V\ot W$, the group actions all become block-diagonal. We depict the projection map $p_i$ from $V\ot W$ to $X_i$ by

\begin{align}
p_i: V\ot W\to X_i=
\itk{
\draw[-<] (0,2)node[right]{$X_i$}-- (0,1);
\draw (1,0)node[right]{$W$} -- (0,1) -- (-1,0)node[left]{$V$} ;
\label{projection}
\end{align}

and the embedding map by

\begin{align}
q_i=p_i^\dagger: X_i\to V\ot W=
\itk{
\draw (1,1)node[right]{$W$} -- (0,0) -- (-1,1)node[left]{$V$};
\draw[-<] (0,-1)node[right]{$X_i$}  --(0,0) ;
\label{embedding}
\end{align}

The normalization is taken to be

\begin{align}
p_j p_i^\dagger=\delta_{ij}\id_{X_i},\quad \sum_{i}p_i^\dagger p_i=\id_{V\otimes W},
\end{align}

so that $p_i,p_i^\dag$ exhibit $V\ot W$ as a direct sum of $X_i$'s.

Two fundamental constructions will be useful later and we fix the notation here:

Given a set $S$, the vector space of finite formal linear combinations of elements in $S$, is called the free vector space on $S$, and denoted by $C(S)$.

Given a subset $T$ of a vector space $V$, the subspace of all linear combinations of vectors in $T$, is called the space spanned by $T$, and denoted by $\langle T\rangle$.

$S$ is automatically a basis of $C(S)$. $\langle T\rangle$ is the smallest subspace of $V$ containing $T$, and $T$ is not necessarily a basis of $\langle T\rangle$.

The following notion is also useful in later analysis:

Given two finite dimensional Hilbert spaces $V$ and $W$, a linear map $U:V\to W$ is called isometric if

\begin{equation}
\forall v_1,v_2\in V, \ \la v_1|v_2 \ra
=\la Uv_1|Uv_2 \ra,\end{equation}

which is equivalent to $U^{\dagger}U=\id_V$. On the other hand, $U:V\to W$ is called partially isometric if any one of the following equivalent conditions holds (denote by $\ker U^\perp$ the orthogonal complement of $\ker U$ in $V$ and by $\Ima U$ the image of $U$):

* The restriction $U: \ker U^\perp \to W$ is isometric;

* The restriction $U^\dag: \Ima U\to V$ is isometric;

* The restriction $U: \ker U^\perp \to \Ima U$ is unitary;

* The restriction $U^\dag:\Ima U \to \ker U^\perp$ is unitary;

* $U^\dag U$ is a projection, $(U^\dag U)^2=U^\dag U$;

* $U U^\dag$ is a projection, $(U U^\dag)^2=U U^\dag$.

For technical simplicity, in this paper we mainly work with the example $\Rep G$ whose objects and morphisms have underlying vector spaces and linear maps that can be calculated concretely. We would like to emphasize that the graphical calculus techniques here and below remain valid even if we consider a more general unitary fusion category (UFC). For a general fusion category, we continue to interpret objects as symmetry charges and morphisms as symmetric operators; however, we lose access to underlying vectors and linear maps, unless we are willing to deal with (weak) Hopf algebras and their (co-)modules.

Throughout this paper, it is understood that $\cC=\Rep G$, but when we write $\cC$, we are making statements applicable to general unitary fusion categories, and when we write $\Rep G$ we are making statements specific to $\Rep G$. Note that we also need to assume that $G$ is finite for $\Rep G$ to be a unitary fusion category.

$\cC$ is local <cit.> or anomaly-free if there exists a fiber functor $\cC\to \Ve$, in which case one can reconstruct $\cC$ as the representation category of a Hopf algebra, and consider such Hopf algebra as the global symmetry of the system. The fiber functor $\cC\to\Ve$ allows us to work still with vector spaces and linear maps. When there does not exist a fiber functor, the fusion category is a representation category of a weak Hopf algebra and describes an anomalous symmetry. It is anomalous in the sense that the local tensor product structure of the lattice system is no longer the usual one. Given two representations over a weak Hopf algebra, the usual vector space tensor product of them is no longer a representation. Instead, one needs to take the relative tensor product over a subalgebra of the weak Hopf algebra.

Let $(\cA,\ot,\alpha)$ be a monoidal category, its Drinfeld center $Z_1(\cA)$ is the braided monoidal category defined as follows:

* An object of $Z_1(\cA)$ is a pair $(X,\beta_{X,-})$ where $X\in \cA$ is an object in $\cA$ and $\beta_{X,V}:X\ot V\to  V\ot X$ is a collection of isomorphisms for each $V\in \cA$ such that

* $\beta_{X,-}$ is natural: for any $f:V\to W$,

\begin{align}\label{Drinfeld1}(f\ot \id_X)\beta_{X,V}=\beta_{X,W}(\id_X\ot f).\end{align}

* $\beta_{X,-}$ satisfies the hexagon equation
\begin{equation}\label{Drinfeld2}
\begin{tikzcd}
& X\ot(V\ot W)
\rar{\beta_{X,V\ot W}}
& (V\ot W)\ot X
\drar{\alpha_{V,W,X}}
(X\ot V)\ot W
\urar{\alpha_{X,V,W}}
\drar{\beta_{X,V}\ot\id_W}
&&& V\ot (W\ot X)
\\ & (V\ot X)\ot W
\rar{\alpha_{V,X,W}}
& V\ot (X\ot W)
\urar{\id_V\ot\beta_{X,W}}
\end{tikzcd}
\end{equation}

* A morphism $g$ from $(X,\beta_{X,-})$ to $(Y,\beta_{Y,-})$[We follow the usual convention and abuse the same notation $\beta_{X,-}$ and $\beta_{Y,-}$ for different pairs $(X,\beta_{X,-})$ and $(Y,\beta_{Y,-})$. The reader is reminded that the pairs should be understood as a whole; the half-braiding $\beta_{X,-}$ depends on the pair $(X,\beta)$ instead of only the object $X$. Indeed, the same object $X$ may be equipped with different half-braidings to form different objects in the Drinfeld center.] is a morphism $g:X\to Y$ in $\cA$ satisfying $\beta_{Y,V}(g\ot\id_V)=(\id_V\ot g)\beta_{X,V}$ for any $V\in\cA$.

* The tensor product of $(X,\beta_{X,-})$ and $(Y,\beta_{Y,-})$ is given by $(X\ot Y,\beta_{X\ot Y,-})$ where
\begin{equation} \beta_{X\ot Y,V}=\alpha_{V,X,Y}(\beta_{X,V}\ot\id_Y)\alpha^{-1}_{X,V,Y}(\id_X\ot \beta_{Y,V})\alpha_{X,Y,V}. \end{equation}

* The braiding is $c_{(X,\beta_{X,-}),(Y,\beta_{Y,-})}=\beta_{X,Y}$.

[Permutation group $S_3$]

$S_3$ is the simplest non-Abelian group with 6 elements:

\begin{equation} S_3=\{1,a,ab,ab^2,b,b^2|a^2=1,b^3=1,aba=b^2\}, \end{equation}

where $a,b$ are called two generators of $S_3$. $S_3$ has three irreducible representations $\lambda_0,\lambda_1,\tau$ listed below:

$1$     $a$     $ab=b^2 a$     $ab^2=ba$     $b$     $b^2$
$\lambda_0$     1     1     1     1     1     1
$\lambda_1$     1     $-1$     $-1$     $-1$     1     1
$\Lambda$     $\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}$     $\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}$     $\begin{pmatrix}
0 & \omega^* \\
\omega & 0
\end{pmatrix}$     $\begin{pmatrix}
0 & \omega \\
\omega^* & 0
\end{pmatrix}$     $\begin{pmatrix}
\omega & 0 \\
0 & \omega^*
\end{pmatrix}$     $\begin{pmatrix}
\omega^* & 0 \\
0 & \omega
\end{pmatrix}$

where $\omega=e^{2\pi i/3}$. Let $\irr(\cC)$ denote for the set of isomorphism classes of simple objects in semisimple category $\cC$. We have $\irr(\Rep S_3)=\{\lambda_0,\lambda_1,\Lambda\}$. The dual representation $\Lambda^*$ of $\Lambda$ is listed as

$1$     $a$     $ab=b^2 a$     $ab^2=ba$     $b$     $b^2$
$\Lambda^*$     $\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}$     $\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}$     $\begin{pmatrix}
0 & \omega \\
\omega^* & 0
\end{pmatrix}$     $\begin{pmatrix}
0 & \omega^* \\
\omega & 0
\end{pmatrix}$     $\begin{pmatrix}
\omega^* & 0 \\
0 & \omega
\end{pmatrix}$     $\begin{pmatrix}
\omega & 0 \\
0 & \omega^*
\end{pmatrix}$

Denote the basis of $\Lambda$ as $\{\bm 0,\bm 1\}$. Correspondingly $\Lambda^*$ has dual basis $\{\delta_{\bm 0},\delta_{\bm 1}\}$. Then $\Lambda^*$ is isomorphic to $\Lambda$ through the isomorphism $\bm 0 \mapsto \delta_{\bm 1},
\bm 1 \mapsto \delta_{\bm 0}$.

We also list the data in $Z_1(\Rep S_3)$ (See the calculation of $Z_1(\Rep G)$ for general $G$ in Appendix <ref>):

Centralizer subgroups     $N_1\cong S_3$     $N_a=\{1,a\} \cong N_{ab}=\{1,ab\}$
$ \cong N_{ab^2}=\{1,ab^2\} \cong \mathbb{Z}_2$     $N_b=N_{b^2}$
$=\{1,b,b^2\} \cong \mathbb{Z}_3$
Conjugacy classes      $C_1\cong G/{N_1}\cong \{1\}$     $C_a\cong C_{ab}\cong C_{ab^2}$
$\cong G/{N_a}\cong \{a,ab,ab^2\}$     $C_b \cong C_{b^2}\cong G/{N_b}$
$\cong \{b,b^2\}$
Representation of
centralizer subgroup     $\Rep S_3$     $\Rep \mathbb{Z}_2$
$\irr(\Rep\Z_2)=\{+,-\}$     $\Rep \mathbb{Z}_3$

Therefore, there are 8 simple objects in $Z_1(\Rep S_3)$ labelled as:

\begin{align}
\irr(Z_1(\Rep S_3))=\{
\end{align}

[Quaternion group $Q_8$]

$Q_8$ is a non-Abelian group with 8 elements:

\begin{equation} Q_8=\{\pm 1,\pm i, \pm j,\pm k|i^2=j^2=k^2=ijk=-1,ikj=1\}. \end{equation}

$Q_8$ has five irreducible representations $\gamma_0,\gamma_1,\gamma_2,\gamma_3,\Gamma$ listed below:

$\pm 1$     $\pm i$     $\pm j$     $\pm k$
$\gamma_0$     1     1     1     1
$\gamma_1$     1     1     $-1$     $-1$
$\gamma_2$     1     $-1$     1     $-1$
$\gamma_3$     1     $-1$     $-1$     1
$\Gamma$     $\pm I$     $\mp i\sigma_x$     $\mp i\sigma_y$     $\mp i\sigma_z$

where $\sigma_x,\sigma_y,\sigma_z$ are Pauli matrices. The dual representation $\Gamma^*$ of $\Gamma$ is listed as

$\pm 1$     $\pm i$     $\pm j$     $\pm k$
$\Gamma^*$     $\pm I$     $\pm i\sigma_x$     $\mp i\sigma_y$     $\pm i\sigma_z$

Denote the basis of $\Gamma$ as $\{|0\ra,|1\ra\}$. Correspondingly $\Gamma^*$ has dual basis $\{\delta_{|0\ra},\delta_{|1\ra}\}$. Then $\Gamma^*$ is isomorphic to $ \Gamma$ through the isomorphism $|0\ra \mapsto \delta_{|1\ra},
|1\ra \mapsto -\delta_{|0\ra}$.

[Special unitary group $SU(2)$]
$SU(2)$ is a non-Abelian Lie group with infinite elements:

\begin{equation} SU(2)=\left\{\begin{pmatrix}\alpha & -\ov\beta \\\beta & \ov{\alpha} \end{pmatrix}\bigg|\alpha,\beta\in\C,|\alpha|^2+|\beta|^2=1\right\}, \end{equation}

where $\ov\alpha$ denotes the complex conjugate of $\alpha$.
$SU(2)$ has infinite numbers of irreps labelled by $l$, which are non-negative integers and half-integers. The dimension of irrep labelled by $l$ is $2l+1$. We list generators for irreps $l=0,1/2,1$ below:

$J_x$     $J_y$     $J_z$
$l=0$     0     0     0
$l=\frac{1}{2}$     $\frac12\sigma_x$     $\frac12\sigma_y$     $\frac12\sigma_z$
$l=1$     $\begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}$     $\begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}$     $\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

For each $l$, group elements in $SU(2)$ represented by $l$ are $\{\ee^{\ii\theta (n_xJ_x+n_yJ_y +n_z J_z)} |\theta,n_x,n_y,n_z\in \R,n_x^2+n_y^2+n_z^2=1\}$. The dual representation $l=\frac{1}{2}^*$ of $l=\frac1{2}$ is listed as

$J_x$     $J_y$     $J_z$
$l=\frac{1}{2}^*$     $-\frac12\sigma_x$     $\frac12\sigma_y$     $-\frac12\sigma_z$

Denote the basis of $l=\frac{1}{2}$ as $\{|\frac12,\frac12\ra,|\frac12,-\frac12\ra\}$. Correspondingly $l=\frac12^*$ has dual basis $\{\delta_{|\frac12,\frac12\ra},\delta_{|\frac12,-\frac12\ra}\}$. Then $\frac12^*$ is isomorphic to $ \frac12$ through the isomorphism $|\frac12,\frac12\ra \mapsto \delta_{|\frac12,-\frac12\ra},
|\frac12,-\frac12\ra \mapsto -\delta_{|\frac12,\frac12\ra}$.

Fix a set $\BL$ (whose elements are viewed as lattice sites) and a group $G$. A quantum system with onsite symmetry $G$, on $\BL$, or just a system for short, consists of the following data

* For each subset $\BK\subset \BL$, there is a Hilbert space $\cH_\BK$ which carries a group representation $(\cH_\BK,\rho^\BK)$;

* A Hermitian operator (the total Hamiltonian) $H$ on $\cH_\BL$,

such that

* For any two disjoint subsets $\BK_1$ and $\BK_2$, the representation associated to their disjoint union is the tensor product of those associated to $\BK_1$ and $\BK_2$

\begin{equation} (\cH_{\BK_1\coprod  \BK_2},\rho^{ \BK_1\coprod         \BK_2})=(\cH_{\BK_1},\rho^{ \BK_1})\otimes(\cH_{\BK_2},\rho^{\BK_2}); \end{equation}

* The total Hamiltonian has the form

\begin{equation} H=\sum_{\BK\subset \BL} H_\BK, \end{equation}

where $H_ \BK=\tilde H_\BK\ot \id_{\cH_{\BL\backslash \BK}}$ and $\tilde H_\BK$ is a symmetric operator on $\cH_\BK$, i.e., $ \rho_g^\BK  \tilde H_\BK (\rho_g^\BK)^{-1}=\tilde H_\BK,$ $\forall g\in G.$

We denote such a quantum system by $(\BL,\cH_\BK,H)$.

The empty set $\emptyset\subset \BL$ is necessarily associated with the trivial group representation. For the singleton subset $\{i\}\subset \BL$ ($i$ is just a lattice site), we will simply write the associated representation as $(\cH_i,\rho^i)$. It is clear that

\begin{align}
\cH_\BK&=\otr[i\in \BK] \cH_i,\\
\rho^\BK_g&=\otr[i\in \BK]\rho^i_g.
\end{align}

An operator on $\cH_\BL$ of the form $O_\BK=\tilde O_\BK\ot \id_{\cH_{\BL\backslash \BK}}$ is called supported on $\BK$. When no confusion arises, we will abuse notation and do not distinguish $O_\BK$ from $\tilde O_\BK$.

§ SYMMETRIC OPERATORS AS GRAPHS

In this section we set up the formulation that any symmetric operator in a system with onsite symmetry $G$ can be represented by graphical calculus in $\Rep G$. In the language of tensor network, any symmetric operator can be represented by a $G$-invariant tensor network.

Each local Hilbert space can be decomposed to irreducible representations.

A charge decomposition consists of the following choices:

* A representative $X_a$ for each isomorphic class $a$ of irreps, $a\in \irr(\Rep G)$.

* For each local Hilbert space or more generally each representation $V$, the projection maps to and the embedding maps from the representative irreps: $p_V^{a;n}:V\to X_a, q^V_{a;n}:X_a\to V$ where $n$ goes from 1 to the multiplicity of $X_a$ in $V$.

* In particular, for each tensor product of two representative irreps, the projection maps to and the embedding maps from the representative irreps:
\begin{equation} p_{ab}^{c;n}:X_a\ot X_b \to X_c, q_{c;n}^{ab}:X_c\to X_a\ot X_b. \end{equation}

Tree graphs decorated by $X_a$ and $p,q$'s in a charge decomposition serve as bases of intertwiner spaces.
We may draw any graph in $\Rep G$ and interpret it as an intertwiner between several representations. More specifically, let $V_1,\dots,V_n$ and $W_1,\dots,W_m$ be representations, an intertwiner in $f\in \Hom(V_1\ot\dots\ot V_n, W_1\ot\dots\ot W_m)$ can be represented by the graph:

\begin{equation} f=
\itk{
\node[rectangle, minimum width=12em, draw] (f) at (0,0) {$f$};
\node (v1) at (-2,2) {$W_1$};
\node (vn) at (2,2) {$W_m$};
\node (w1) at (-2,-2) {$V_1$};
\node (wm) at (2,-2) {$V_n$};
\draw
(v1) -- (v1 |- f.north)
(vn) -- (vn |- f.north)
(w1) -- (w1 |- f.south)
(wm) -- (wm |- f.south);
\draw[dash pattern= on 3pt off 12pt, very thick] (-1.88,1)--(1.88,1) (-1.88,-1)--(1.88,-1);
\end{equation}

With a chosen charge decomposition, the intertwiner above can be expanded in terms of “basis” graphs. As an illustrative example, consider the intertwiner space $\Hom(V_1\ot V_2,W_1\ot W_2)$ with four external legs $V_1, V_2, W_1, W_2$. One first decompose all external legs to irreps, and then fuse the irreps one by one in a chosen order. One choice of basis graphs is

\begin{equation}
\itk{
\node (v1) at (-2,-2) {$V_1$};
\node (v2) at (2,-2) {$V_2$};
\node (w1) at (-2,2) {$W_1$};
\node (w2) at (2,2) {$W_2$};
\draw[->] (v1)--(-2,-1)node[left]{$p_{V_1}^{a;i}$};
\draw[->](v2)--(2,-1)node[right]{$p_{V_2}^{b;j}$};
\draw[->](w1)--(-2,1)node[left]{$q^{W_1}_{d;k}$};
\draw[->](w2)--(2,1)node[right]{$q^{W_2}_{e;l}$};
\draw (-2,-1)--node[below]{$X_a$} (0,-0.5)
(2,-1)--node[below]{$X_b$} (0,-0.5)
(-2,1)--node[above]{$X_d$} (0,0.5)
(2,1)--node[above]{$X_e$} (0,0.5);
\draw[-<] (0,0)node[right]{$X_t$}--(0,0.5)node[above]{$q_{t;n}^{de}$};
\draw[-<] (0,0)--(0,-0.5)node[below]{$p^{t;m}_{ab}$};
\end{equation}

where $a,b,t,d,e,i,j,k,l,m,n$ runs over all possible values. Another choice is

\begin{equation}
\itk{
\node (v1) at (-2,-2.5) {$V_1$};
\node (v2) at (2,-2.5) {$V_2$};
\node (w1) at (-2,2.5) {$W_1$};
\node (w2) at (2,2.5) {$W_2$};
\draw[->] (v1)--(-2,-1.5)node[left]{$p_{V_1}^{a;i}$};
\draw[->](v2)--(2,-1.5)node[right]{$p_{V_2}^{b;j}$};
\draw[->](w1)--(-2,1.5)node[left]{$q^{W_1}_{d;k}$};
\draw[->](w2)--(2,1.5)node[right]{$q^{W_2}_{e;l}$};
\draw[-<] (-2,-1.5)--node[right]{$X_a$} (-2,-0.5) ;
\draw[-<] (2,1.5)--node[left]{$X_e$} (2,0.5);
\draw(2,-1.5)--node[left]{$X_b$} (2,0.5)node[right]{$p^{e;m}_{cb}$};
\draw(-2,1.5)--node[right]{$X_d$} (-2,-0.5)node[left]{$q_{a;n}^{dc}$};
\draw (-2,-0.5)--node[above]{$X_c$}(2,0.5);
\end{equation}

where $a,b,c,d,e,i,j,k,l,m,n$ runs over all possible values. The only remaining third choice is left as an exercise for the reader.

In general, a basis of the intertwiner space $\Hom(V_1\ot\dots\ot V_n, W_1\ot\dots\ot W_m)$ can be obtained by decorating a tree graph, which is bivalent between external and internal legs and trivalent between internal legs:

* The external legs are fixed and decorated by $V_1,\dots,V_n,W_1,\dots,W_m$;

* The internal legs are decorated by representative irreps;

* The vertices are decorated by projection and embedding maps from a chosen charge decomposition.

Different orders to fuse internal legs lead to different tree graphs and thus different sets of basis intertwiners. Basis intertwiners on different tree graphs are related to each other by the F-moves.

After fixing basis graphs, every symmetric operator has a unique linear expansion and concrete calculation is possible. However, for the purpose to read out the total symmetry charge being transported by symmetric operators, we are going to use a slightly more effective method to rewrite the symmetric operators.

Consider an intertwiner $f\in \Hom(V,W)$. An image decomposition $(\bar l,\bar r)$ of $f$ is a factorization of $f$ through its image

\begin{equation}\label{eq.imdecomp}
f=V\xrightarrow{\bar l}\Ima f \xrightarrow{\bar r} W,
\end{equation}

where $\Ima f$ is the image of $f$ (Definition <ref>). Graphically,

\begin{align}
\itk{
\node[rectangle, draw] (a) at (0,1.5) {$f$};
\filldraw (0,0)node[right]{$V$} -- (a) -- (0,3)node[right]{$W$};
=\quad \itk{
\filldraw (0,0)node[right]{$V$} -- (l) -- (0,1.8)node[right]{$\Ima f$} -- (r) -- (0,3.6)node[right]{$W$};
\node[rectangle, draw] (l) at (0,1) {$\bar l$};
\node[rectangle, draw] (r) at (0,2.6) {$\bar r$};
} .
\end{align}

In more concrete words, we can think the image decomposition as a singular value decomposition performed on symmetric tensors, which we call symmetric singular value decomposition (SSVD). Explicitly, there are two steps in an SSVD of $f$:

* Direct sum decomposition of $V$ and $W$. Suppose the direct sum decomposition of $V,W$ is given by the projection maps $p_V^{a;n},p_W^{a;m},n=1,\dots,n_a,m=1,\dots,m_a$, then $f$ can be represented by a block diagonal matrix, whose blocks correspond to irreps:

\begin{align}
M^a_{mn}:=p_W^{a;m}f(p_V^{a;n})^\dagger,\quad f=\sum_{a,mn} (p_W^{a;m})^\dag M^a_{mn} p_V^{a;n}.
\label{eq.Mamn}
\end{align}

* The usual SVD performed on the $m_a\times n_a$ matrices $M^a_{mn}$ for all representative irreps $X_a$, leads to a SSVD of $f$:

\begin{align}
\underset{a}{\bigoplus}M^a=\underset{a}{\bigoplus} (w^a)^\dag\Sigma^a v^a,
\end{align}

where $v^a$ and $w^a$ are $n_a\times n_a$ and $m_a\times m_a$ unitary matrices, and $\Sigma^a$ is a $m_a\times n_a$ rectangular diagonal matrix with non-negative real entries.
The number of non-zero singular values of $M^a$ tells us how many copies of $X_a$ are in $\Ima f$. Graphically, for each isomorphism class $a$ of irreps,

\begin{align}
\itk{
\node[rectangle, draw] (m) at (0,2) {$M^a_{mn}$};
\draw[-<] (m)-- (0,1)node[right]{$X_a$} -- (0,.5);
\draw (0,.5)node[left]{$ p^{a;n}_V$} -- (0,0)node[right]{$V$};
\draw[-<] (m) -- (0,3)node[right]{$X_a$} -- (0,3.5);
\draw (0,3.5)node[left]{$ (p^{a;m}_W)^\dag$} -- (0,4)node[right]{$W$};
\itk{
\node[rectangle, draw] (m) at (0,2) {$\Sigma^a_{kk}$};
\draw[-<] (m)-- (0,1)node[right]{$X_a$} -- (0,.5);
\draw (0,.5)node[left]{$v^a_{kn} p^{a;n}_V$} -- (0,0)node[right]{$V$};
\draw[-<] (m) -- (0,3)node[right]{$X_a$} -- (0,3.5);
\draw (0,3.5)node[left]{$\ov{w^a_{km}} (p^{a;m}_W)^\dag$} -- (0,4)node[right]{$W$};
\end{align}

And therefore,

\begin{align}f=\sum_{a,mn}
\itk{
\node[rectangle, draw] (m) at (0,2) {$M^a_{mn}$};
\draw[-<] (m)-- (0,1)node[right]{$X_a$} -- (0,.5);
\draw (0,.5)node[left]{$ p^{a;n}_V$} -- (0,0)node[right]{$V$};
\draw[-<] (m) -- (0,3)node[right]{$X_a$} -- (0,3.5);
\draw (0,3.5)node[left]{$ (p^{a;m}_W)^\dag$} -- (0,4)node[right]{$W$};
\itk{
\node[rectangle, draw] (m) at (0,2) {$\Sigma^a_{kk}$};
\draw[-<] (m)-- (0,1)node[right]{$X_a$} -- (0,.5);
\draw (0,.5)node[left]{$v^{a;k}_V$} -- (0,0)node[right]{$V$};
\draw[-<] (m) -- (0,3)node[right]{$X_a$} -- (0,3.5);
\draw (0,3.5)node[left]{$(w^{a;k}_W)^\dag$} -- (0,4)node[right]{$W$};
\end{align}

where $v^{a;k}_V:=\sum_n v^a_{kn} p^{a;n}_V $ and $w^{a;k}_W:=\sum_m w^a_{km} p^{a;m}_W$ are another two sets of projections of the direct sum decomposition of $V,W$.

Here it is easy to see that the ambiguity of $\bar l$ and $\bar r$ arises from choices of block diagonal unitary matrices in the SSVD and how one separate the non-zero singular values of $M^a$.

In the factorization $f=\bar r\bar l$, $\bar r$ is a monomorphism. And in any abelian category (in this paper all categories used are moreover semisimple), $\bar l$ is an epimorphism, proof of which is omitted.

The pair $(\bar l,\bar r)$ is clearly not unique. Such ambiguity is related to the sectors of quantum currents and morphisms between quantum currents. We will come to this point later.

We introduce a trial-and-error method to compute the image, which is more straightforward (if succeed). Given two representations $(V,\rho), (W,\tau)$, note that we can endow the operator space $\Hom_\Ve(V,W)$ with a natural group action by post-composing $\tau_g$. For an intertwiner $f:V\to W$, we can then consider the cyclic sub-representation of $\Hom_\Ve(V,W)$ generated by $f$, $C(G)f:=\la \tau_g f,g \in G\ra$. Pick any $v\in V$, there is an intertwiner:

\begin{align}
\xi_v: C(G)f &\to W ,
\nonumber\\
O&\mapsto O(v).
\end{align}

$O\in C(G)f$ has the form $O=\sum_{h\in G} c_h \tau_h f=\sum_{h\in G}c_h f\rho_h$, with which it is easy to check that $\xi_v$ is symmetric.
If $C(G)f$ happens to be isomorphic to $\Ima f$ for some $v$, the computation is completed. For this method to work, the necessary and sufficient conditions are

* $\Ima f \cong C(G)f$ is a cyclic sub-representation of $W$. $v$ should be chosen as a preimage of a cyclic vector in $\Ima f$. In this case $\xi_v$ automatically maps onto $\Ima f$.

* $\xi_v$ must also be injective, which means that $\ker \xi_v=0$, i.e., $\xi_v(O)=O(v)=0$ implies that $O$ is the zero operator $O(w)=0,\ \forall w\in V$.

There are simple cases that these conditions are satisfied, for example, $V$ is the regular representation $C(G)$ of Abelian group $G$, where one can take $v=e$ the identity element of $G$.
However, in practice, the second condition is difficult to check, and also we can not know in advance whether $\Ima f$ is a cyclic representation or not. Such method can only be used with trial-and-error.

Let $G=S_3$ (Example <ref>). Consider the direct sum decomposition
$\Lambda\ot \Lambda \cong \Lambda\oplus \lambda_0\oplus \lambda_1$. We take the basis of $\Lambda$ to be $\{\bm 0,\bm 1\}$, and the tensor basis of $\Lambda\ot \Lambda$ to be $\{\bm 0\bm 0,\bm 0\bm 1,\bm 1\bm 0,\bm 1\bm 1\}$ (similarly for later examples involving 2-dimensional representations). Then the basis change between $\Lambda\ot \Lambda$ and $\Lambda\oplus \lambda_0\oplus \lambda_1$ is

\begin{gather}
\begin{pmatrix}
1 &0&0&0\\
0 &0&0&1
\end{pmatrix},\
\begin{pmatrix}
0 &\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0
\end{pmatrix},\
\begin{pmatrix}
0 &\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0
\end{pmatrix}.
\end{gather}

Consider an intertwiner $f\in \text{Hom}(\Lambda\ot \Lambda,\Lambda\ot\Lambda)$ taking the following form:

\begin{align}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\underbrace{
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
0 & 1 & 0 & 0
\end{pmatrix}}_{\makecell{\small\text{Changing basis from }\\\small\Lambda\oplus \lambda_0\oplus \lambda_1\text{ to }\Lambda\ot\Lambda
\underbrace{
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}}_{\makecell{\small M^\Lambda\oplus M^{\lambda_0}\oplus M^{\lambda_1}}}
\underbrace{
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0
\end{pmatrix}}_{\makecell{\small\text{Changing basis from }\\\small\Lambda\ot\Lambda\text{ to }\Lambda\oplus \lambda_0\oplus \lambda_1
\nonumber\\
\begin{pmatrix}
p^\Lambda_{\Lambda\ot \Lambda}\\
p^{\lambda_0}_{\Lambda\ot \Lambda}\\
p^{\lambda_1}_{\Lambda\ot \Lambda}
\end{pmatrix}^\dag
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
p^\Lambda_{\Lambda\ot \Lambda}\\
p^{\lambda_0}_{\Lambda\ot \Lambda}\\
p^{\lambda_1}_{\Lambda\ot \Lambda}
\end{pmatrix}
\end{align}

where the similarity transformation can be thought as changing to the basis with fixed symmetry charge, i.e., changing $f\in\End(\Lambda\ot\Lambda)$ to $\underset{a}{\bigoplus}M^a\in \End(\Lambda\oplus \lambda_0\oplus \lambda_1)$. Since $M^\Lambda=1$, there is only one non-zero singular value in $M^{\Lambda}$,[Note that the matrix $M^\Lambda$ is not indexed by the basis of $\Lambda$. By Definition (<ref>), $M^\Lambda$ is a $1\times 1$ matrix.] we see $\Ima f\cong \Lambda$. Graphically,

\begin{align}
f=\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.9]
\node[rectangle, draw] (e) at (0,-1) {$\bar l$};
\node[rectangle, draw] (m) at (0,0) {$\bar r$};
\draw (1,1)node[right]{$\Lambda$} -- (m)-- (-1,1)node[left]{$\Lambda$};
\draw (e) -- (0,-0.5)node[right]{$\Lambda$}--(m);
\draw (-1,-2)node[left]{$\Lambda$} -- (e) -- (1,-2)node[right]{$\Lambda$};
\end{tikzpicture},
\end{align}

where by requiring $\bar l$ and $\bar r$ to be intertwiners, the image decomposition is solved as
\begin{equation}
\bar l=\begin{pmatrix}
0 & 0 & 0 & c_1\\
c_1 & 0& 0& 0
\end{pmatrix},\quad \bar r=\begin{pmatrix}
0 & c_2\\
0& 0\\
c_2 &0
\end{pmatrix},\end{equation}
where $c_1,c_2$ are non-zero complex numbers such that $c_1 c_2=1$. A different choice of $(\bar l,\bar r)$ correspond to changing $c_1$ and $c_2$ to $c'_1$ and $c'_2$ such that $c'_1 c'_2=1$. Here since $\dim \Hom(\Lambda\ot \Lambda,\Lambda)=1$, choices of $(\bar l,\bar r)$ only differ by a scalar.

Similarly, a general intertwiner $f\in \text{Hom}(\Lambda\ot \Lambda,\Lambda\ot\Lambda)$ reads

\begin{align}
&=c_1 (
+c_2 (
p_{\Lambda\ot\Lambda}^{\lambda_0}+c_3 (
\begin{pmatrix}
p^\Lambda_{\Lambda\ot \Lambda}\\
p^{\lambda_0}_{\Lambda\ot \Lambda}\\
p^{\lambda_1}_{\Lambda\ot \Lambda}
\end{pmatrix}^\dag
\begin{pmatrix}
c_1 & 0 & 0 & 0 \\
0 & c_1 & 0 & 0 \\
0 & 0 & c_2 & 0 \\
0 & 0 & 0 & c_3
\end{pmatrix}
\begin{pmatrix}
p^\Lambda_{\Lambda\ot \Lambda}\\
p^{\lambda_0}_{\Lambda\ot \Lambda}\\
p^{\lambda_1}_{\Lambda\ot \Lambda}
\end{pmatrix}\nonumber\\
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
0 & 1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
c_1 & 0 & 0 & 0 \\
0 & c_1 & 0 & 0 \\
0 & 0 & c_2 & 0 \\
0 & 0 & 0 & c_3
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0
\end{pmatrix}.
\nonumber\\&=
\begin{pmatrix}
c_1 & 0 & 0 & 0 \\
0 & \frac{c_2+c_3}{{2}} & \frac{c_2-c_3}{{2}} & 0 \\
0 & \frac{c_2-c_3}{{2}} & \frac{c_2+c_3}{{2}} & 0 \\
0 & 0 & 0 & c_1
\end{pmatrix}
\end{align}

Let $G=Q_8$ (Example <ref>). Consider the direct sum decomposition $\Gamma^*\ot \Gamma \cong \Gamma\ot \Gamma^*$ $\cong\gamma_0\oplus \gamma_1\oplus\gamma_2\oplus\gamma_3$, and the basis change between them is[Denote the basis of $\Gamma$ to be $\{|0\ra,|1\ra\}$ and the dual basis of $\Gamma^*$ to be $\{\delta_{|0\ra},\delta_{|1\ra}\}$. If we choose the isomorphism between $\Gamma^*\ot \Gamma$ and $\Gamma\ot\Gamma^*$ to be $\delta_{|0\ra} |0\ra \mapsto |0\ra\delta_{|0\ra}$, $\delta_{|0\ra} |1\ra\mapsto |0\ra\delta_{|1\ra}$, $\delta_{|1\ra} |0\ra\mapsto |1\ra\delta_{|0\ra}$, $\delta_{|1\ra} |1\ra\mapsto |1\ra\delta_{|1\ra}$, the matrix transformations changing their basis to $\gamma_0\oplus \gamma_1\oplus\gamma_2\oplus\gamma_3$ happen to be the same.]

\begin{gather}
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}}
\end{pmatrix},\
\begin{pmatrix}
0 &\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0
\end{pmatrix},\nonumber\\
\begin{pmatrix}
0 &\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0
\end{pmatrix},\
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}}
\end{pmatrix};\\
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}}
\end{pmatrix},\
\begin{pmatrix}
0 &\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0
\end{pmatrix},\nonumber\\
\begin{pmatrix}
0 &\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0
\end{pmatrix},\
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}}
\end{pmatrix}.
\end{gather}

Consider an intertwiner $f\in \text{Hom}(\Gamma^*\ot \Gamma,\Gamma\ot\Gamma^*)$ taking the following form:

\begin{align}
0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} \\
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}}
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} \\
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}}
\end{pmatrix}
\nonumber\\
\begin{pmatrix}
\end{pmatrix}^\dag
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
\end{pmatrix}
\end{align}

where the similarity transformation can be thought as changing to the basis with fixed symmetry charge in $\text{Hom}(\gamma_0\oplus \gamma_1\oplus\gamma_2\oplus\gamma_3,\gamma_0\oplus \gamma_1\oplus\gamma_2\oplus\gamma_3)$. Since there is only one non-zero singular value in $M^{\gamma_1}$, we see $\Ima f\cong \gamma_1$. Graphically,

\begin{align}
f=\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.9]
\node[rectangle, draw] (e) at (0,-1) {$\bar l$};
\node[rectangle, draw] (m) at (0,0) {$\bar r$};
\draw (1,1)node[right]{$\Gamma^*$} -- (m)-- (-1,1)node[left]{$\Gamma$};
\draw (e) -- (0,-0.5)node[right]{$\gamma_1$}  --(m);
\draw (-1,-2)node[left]{$\Gamma^*$} -- (e) -- (1,-2)node[right]{$\Gamma$};
\end{tikzpicture},
\end{align}

where the image decomposition is solved as
\begin{equation}
\bar l=c_1 \begin{pmatrix}
0 & 1 & 1 & 0
\end{pmatrix},\quad \bar r=c_2\begin{pmatrix}
\end{pmatrix},\end{equation}
where $c_1,c_2$ are non-zero complex numbers such that $c_1 c_2=1$.

§ QUANTUM CURRENT

In this section we motivate the definition of quantum current. We would like to define a quantum current as a collection of symmetric operators that can transport symmetry charge all over the space. This physical idea will be put into precise mathematical definition later. For concreteness we will first focus on lattice system in one spatial dimension with onsite symmetry. We will show that a quantum current carries two important quantities: one is the symmetry charge being transported, and the other is the half-braiding that determines how the current extends over the space without leaving things behind along its path. Together, a quantum current is identified with an object in the Drinfeld center $Z_1(\cC)$.

§.§ Every symmetric operator carries a symmetry charge

To begin with, let's consider a bipartite system,

\begin{align}
\cH= \cH_\BA\ot \cH_\BB,
\end{align}

with symmetry actions $\rho^i_g\in GL(\cH_i), i=\BA,\BB$ and $\rho_g=\rho^\BA_g\ot \rho^\BB_g=:\rho^\BA_g\rho^\BB_g$ where we abuse the notation $\rho^\BA_g=\rho^\BA_g\ot \id_\BB$, $\rho^\BB_g=\id_\BA\ot \rho^\BB_g$. A symmetric operator acting on the total space, is by definition an operator $O\in\Hom_\Ve(\cH,\cH)$ that commutes with symmetry actions

\begin{align}
\rho_g  O \rho_g^{-1}=O, \quad \forall g\in G.
\label{symop}
\end{align}

In general, $O$ does not commute with “partial” group actions $\rho^\BA$ or $\rho^\BB$. Indeed, it can transport symmetry charge between $\BA$ and $\BB$. We are tempted to represent $O$ by a trivalent tree graph

\begin{align}
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\cH_\BA$}--(l)--(0,3)node[above]{$\cH_\BA$};
\draw (2,0)node[below]{$\cH_\BB$}--(r)--(2,3)node[above]{$\cH_\BB$};
\draw (l)--node[above]{$X$} (r);
}. \label{eq.tcX}
\end{align}

Then we interpret $X$ as the symmetry charge transported by $O$ from subsystem $\BA$ to subsystem $\BB$ and the intertwiners $l,r$ as describing how the charge $X$ leaves $\BA$ and arrives as $\BB$.
However, a large enough representation $X$ can always do the job to represent $O$. We need to find the smallest $X$.

Note that there is a natural isomorphism
$\epsilon$ by “bending legs”:

\begin{align}
\Hom(\cH_\BA\ot \cH_\BB,\cH_\BA\ot\cH_\BB)& \overset{\epsilon}{\cong} \Hom(\cH_\BA^*\ot \cH_\BA,\cH_\BB\ot \cH_\BB^*)
\nonumber\\
\itk{
\node[rectangle, minimum width=4em, draw] (f) at (0,0) {$O$};
\node (v1) at (-0.5,1) {$\cH_\BA$};
\node (vn) at (0.5,1) {$\cH_\BB$};
\node (w1) at (-0.5,-1) {$\cH_\BA$};
\node (wm) at (0.5,-1) {$\cH_\BB$};
\draw
(v1) -- (v1 |- f.north)
(vn) -- (vn |- f.north)
(w1) -- (w1 |- f.south)
(wm) -- (wm |- f.south);
}& \overset{\epsilon}{\mapsto}
\itk{
\node[rectangle, minimum width=4em, draw] (f) at (0,0) {$O$};
\node (vn) at (0.5,1) {$\cH_\BB$};
\node (w1) at (-0.5,-1) {$\cH_\BA$};
\node (w1d) at (-1.5,-1) {$\cH_\BA^*$};
\node (vnd) at (1.5,1) {$\cH_\BB^*$};
\draw
(vn) -- (vn |- f.north)
(vn |- f.south) arc(180:360:0.5) -- (vnd)
(w1) -- (w1 |- f.south)
(w1 |- f.north) arc(0:180:0.5) -- (w1d);
}=: \bar O. \label{eq.bending}
\end{align}

The smallest $X$ is nothing but $\Ima \bar O$.

If we choose bases of $\cH_\BA$ and $\cH_\BB$, $O$ can be represented by a tensor with four indices: $O_{ab}^{a'b'}$. Under the isomorphism $\epsilon$, while choosing the corresponding dual bases of $\cH_\BA^*$ and $\cH_\BB^*$, $\bar O$ can be represented by the tensor $\bar O_{a'a}^{b'b}=O_{ab}^{a'b'}$. In short, bending the legs on the graph corresponds to rotating the positions of indices of the tensors (in the corresponding bases and dual bases).
Readers familiar with tensor network may find that the above is just doing SSVD for the tensor $O$ while viewing two $\cH_\BA$ legs $a',a$ as input and two $\cH_\BB$ legs $b',b$ as output.

A more physical way to understand the above is to note that the dual group representation is the anti symmetry charge. Thus, $\cH_\BA^* \ot \cH_\BA$ may be thought as calculating the initial charge minus the final charge in $\BA$, i.e., the charge decrease in $\BA$, while $\cH_\BB\ot \cH_\BB^*$ may be thought as calculating the final charge minus the initial charge in $\BB$, i.e., the charge increase in $\BB$. The result $\Ima \bar O$ is the transported charge. We conclude the above discussion in the following definition.

Given a symmetric operator $O$ acting on a bipartite system $\cH_\BA\ot \cH_\BB$, the symmetry charge transported by $O$ from $\BA$ to $\BB$ is $\Ima \bar O$, denoted by $\curr{\BA}{\BB}:=\Ima \bar O$, where $\bar O$ is given by the “bending leg” isomorphism (<ref>).

Now we have

\begin{align}
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\cH_\BA$}--(l)--(0,3)node[above]{$\cH_\BA$};
\draw (2,0)node[below]{$\cH_\BB$}--(r)--(2,3)node[above]{$\cH_\BB$};
\draw (l)--node[above]{$\curr{\BA}{\BB}$} (r);
\label{eq.tc}
\end{align}

where $O=(\id_{\cH_\BA}\ot r)(l\ot \id_{\cH_\BB})$, and $(l,r)$ is a pair of intertwiners coming from the image decomposition $(\bar l,\bar r)$ of $\bar O$. Explicitly, given a symmetric operator $O\in\Hom (\cH_\BA\ot \cH_\BB,\cH_\BA\ot \cH_\BB)$, the symmetry charge transported by $O$ is extracted by the following transformations:

\begin{align}
\label{process}
\itk{
\node[rectangle, minimum width=4em, draw] (f) at (0,0) {$O$};
\node (v1) at (-0.5,1) {$\cH_\BA$};
\node (vn) at (0.5,1) {$\cH_\BB$};
\node (w1) at (-0.5,-1) {$\cH_\BA$};
\node (wm) at (0.5,-1) {$\cH_\BB$};
\draw
(v1) -- (v1 |- f.north)
(vn) -- (vn |- f.north)
(w1) -- (w1 |- f.south)
(wm) -- (wm |- f.south);}
\itk{
\node[rectangle, minimum width=4em, draw] (f) at (0,0) {$O$};
\node (vn) at (0.5,1) {$\cH_\BB$};
\node (w1) at (-0.5,-1) {$\cH_\BA$};
\node (w1d) at (-1.5,-1) {$\cH_\BA^*$};
\node (vnd) at (1.5,1) {$\cH_\BB^*$};
\draw
(vn) -- (vn |- f.north)
(vn |- f.south) arc(180:360:0.5) -- (vnd)
(w1) -- (w1 |- f.south)
(w1 |- f.north) arc(0:180:0.5) -- (w1d);}
=:\bar O
\overset{\bar O=\bar r\bar l}{=\joinrel=}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.8]
\node[rectangle, draw] (e) at (0,-1.2) {$\bar l$};
\node[rectangle, draw] (m) at (0,0.2) {$\bar r$};
\draw (1,1)node[above]{$\cH_\BB^*$} -- (m)-- (-1,1)node[above]{$\cH_\BB$};
\draw (e) -- (0,-0.5)node[right]{$\Ima \bar O=:\curr{\BA}{\BB}$}--(m);
\draw (-1,-2)node[below]{$\cH_\BA^*$} -- (e) -- (1,-2)node[below]{$\cH_\BA$};
\end{tikzpicture}
\overset{\epsilon^{-1}}{\mapsto}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.8]
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\cH_\BA$}--(l)--(0,3)node[above]{$\cH_\BA$};
\draw (2,0)node[below]{$\cH_\BB$}--(r)--(2,3)node[above]{$\cH_\BB$};
\draw (l)--node[above]{$\curr{\BA}{\BB}$} (r);
\end{tikzpicture},
\end{align}

where $\epsilon$ is the natural isomorphism (<ref>), and there are further natural isomorphism
$\Hom(\cH_\BA,\cH_\BA\ot \curr{\BA}{\BB}) \cong$ $ \Hom(\cH_\BA^*\ot \cH_\BA,\curr{\BA}{\BB})$ giving a bijection between $\bar l$ and $l$, and
$\Hom(\curr{\BA}{\BB}\ot\cH_\BB,\cH_\BB) \cong \Hom(\curr{\BA}{\BB},\cH_\BB\ot\cH_\BB^*)$ giving a bijection between $\bar r$ and $r$. These three steps are

* Natural isomorphism $\epsilon$: We apply the rotation of symmetric tensor $\bar O_{a'a}^{b'b}=O_{ab}^{a'b'}$;

* Image decomposition $\bar O=\bar r\bar l$: SSVD in Definition <ref>;

* Natural isomorphism $\epsilon^{-1}$: We apply the inverse rotation of tensors $l_{b}^{ac}=\bar l_{ab}^{c}$ and $r_{ca}^{b}=\bar r_{c}^{ba}$.

The symmetry charge $\curr{\BB}{\BA}$ transported by $O$ from $\BB$ to $\BA$ is similarly defined, and one have $\curr{\BB}{\BA}=(\curr{\BA}{\BB})^*$; the total symmetry charge is conserved.

Let $G=S_3$ (Example <ref>). Consider a symmetric operator $O\in\Hom(\Lambda\ot\Lambda,\Lambda\ot\Lambda)$ taking the following form:

\begin{align}
\itk{
\node[rectangle, minimum width=4em, draw] (f) at (0,0) {$O$};
\node (v1) at (-0.5,1) {$\Lambda$};
\node (vn) at (0.5,1) {$\Lambda$};
\node (w1) at (-0.5,-1) {$\Lambda$};
\node (wm) at (0.5,-1) {$\Lambda$};
\draw
(v1) -- (v1 |- f.north)
(vn) -- (vn |- f.north)
(w1) -- (w1 |- f.south)
(wm) -- (wm |- f.south);}=\frac{1}{2}(\sigma^\BA_x\sigma^\BB_x+\sigma^\BA_y \sigma^\BB_y)=
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}.
\end{align}

We go through the process in Diagram (<ref>):

* After rotation of symmetric tensor, we have $\bar O^{01}_{10}=O^{10}_{01}=1$ and $\bar O^{10}_{01}=O^{01}_{10}=1$.

* Consider the direct sum decomposition
$\Lambda^*\ot \Lambda \cong\Lambda\ot \Lambda^*\cong \Lambda\oplus \lambda_0\oplus \lambda_1$ and the basis change between them is

\begin{gather}
\begin{pmatrix}
0 &1&0&0\\
0 &0&1&0
\end{pmatrix},\
\begin{pmatrix}
\frac{1}{\sqrt{2}}&0&0&\frac{1}{\sqrt{2}}
\end{pmatrix},\
\begin{pmatrix}
\frac{1}{\sqrt{2}}&0&0&-\frac{1}{\sqrt{2}}
\end{pmatrix};\nonumber\\
\begin{pmatrix}
0 &0&1&0\\
0 &1&0&0
\end{pmatrix},\
\begin{pmatrix}
\frac{1}{\sqrt{2}}&0&0&\frac{1}{\sqrt{2}}
\end{pmatrix},\
\begin{pmatrix}
\frac{1}{\sqrt{2}}&0&0&-\frac{1}{\sqrt{2}}
\end{pmatrix}.
\end{gather}

Then we have

\begin{align}
\bar O&=\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
\frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}}
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}}
\end{pmatrix}
\nonumber\\
\begin{pmatrix}
\end{pmatrix}^\dag
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
\end{pmatrix}
\end{align}

where we find $\curr{\BA}{\BB}=\Ima \bar O=\Lambda$. And we solve $\bar l$ and $\bar r$ as the following form:

\begin{align}
\bar O=\underbrace{c_1\begin{pmatrix}
0 & 0 \\ 0 & 1 \\ 1 & 0 \\ 0 & 0
\end{pmatrix}}_{\bar r}
\underbrace{c_2\begin{pmatrix}
0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0
\end{pmatrix}}_{\bar l}
=\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.9]
\node[rectangle, draw] (e) at (0,-1) {$\bar l$};
\node[rectangle, draw] (m) at (0,0) {$\bar r$};
\draw (1,1)node[right]{$\Lambda^*$} -- (m)-- (-1,1)node[left]{$\Lambda$};
\draw (e) -- (0,-0.5)node[right]{$\Lambda$}  --(m);
\draw (-1,-2)node[left]{$\Lambda^*$} -- (e) -- (1,-2)node[right]{$\Lambda$};
\end{tikzpicture},
\end{align}

where $c_1$ and $c_2$ are non-zero complex numbers such that $c_1 c_2=1$.

* After inverse rotation of symmetric tensors, we find $l^{00}_1=\bar l^0_{01}=1$, $l^{11}_0=\bar l^1_{10}=1$, $r^0_{11}=\bar r^{01}_1=1$, $r^1_{00}=\bar r^{10}_0=1$, i.e.,

\begin{align}
\underbrace{
\begin{pmatrix}
I\ot c_1\begin{pmatrix}
0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0
\end{pmatrix}\end{pmatrix}}_{\id_{\Lambda}\ot r}
\underbrace{
\begin{pmatrix}
0 & 1 \\ 0 & 0 \\ 0 & 0 \\ 1 & 0
\end{pmatrix}\ot I
\end{pmatrix}}_{l\ot \id_{\Lambda}}
= \itk{
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\Lambda$}--(l)--(0,3)node[above]{$\Lambda$};
\draw (2,0)node[below]{$\Lambda$}--(r)--(2,3)node[above]{$\Lambda$};
\draw (l)--node[above]{$\Lambda$} (r);
\end{align}

Let $G=Q_8$ (Example <ref>). Consider a symmetric operator $O\in \text{Hom}(\Gamma\ot \Gamma,\Gamma\ot\Gamma)$ taking the following form:

\begin{align}
\itk{
\node[rectangle, minimum width=4em, draw] (f) at (0,0) {$O$};
\node (v1) at (-0.5,1) {$\Gamma$};
\node (vn) at (0.5,1) {$\Gamma$};
\node (w1) at (-0.5,-1) {$\Gamma$};
\node (wm) at (0.5,-1) {$\Gamma$};
\draw
(v1) -- (v1 |- f.north)
(vn) -- (vn |- f.north)
(w1) -- (w1 |- f.south)
(wm) -- (wm |- f.south);}&=\sigma^\BA_x\ \sigma^\BB_x=
\begin{pmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix}
\nonumber\\
\underbrace{
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}}_{\bar O}
\underbrace{c_1\begin{pmatrix}
0 & 1 & 1 & 0
\end{pmatrix}}_{\bar r}
\underbrace{c_2\begin{pmatrix}
\end{pmatrix}}_{\bar l}
=\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.9]
\node[rectangle, draw] (e) at (0,-1) {$\bar l$};
\node[rectangle, draw] (m) at (0,0) {$\bar r$};
\draw (1,1)node[right]{$\Gamma^*$} -- (m)-- (-1,1)node[left]{$\Gamma$};
\draw (e) -- (0,-0.5)node[right]{$\gamma_1$}  --(m);
\draw (-1,-2)node[left]{$\Gamma^*$} -- (e) -- (1,-2)node[right]{$\Gamma$};
\end{tikzpicture}
\nonumber\\
\underbrace{
\begin{pmatrix}
I\ot c_1 \begin{pmatrix}
0 & 1 \\ 1 & 0
\end{pmatrix}\end{pmatrix}}_{\id_{\Gamma}\ot r}
\underbrace{
\begin{pmatrix}
0 & 1 \\ 1 & 0
\end{pmatrix}\ot I
\end{pmatrix}}_{l\ot \id_{\Gamma}}=
\itk{
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\Gamma$}--(l)--(0,3)node[above]{$\Gamma$};
\draw (2,0)node[below]{$\Gamma$}--(r)--(2,3)node[above]{$\Gamma$};
\draw (l)--node[above]{$\gamma_1$} (r);
\end{align}

where we go through the process in Diagram (<ref>) and apply the result in Example <ref> (recall that $\curr{\BA}{\BB}=\Ima\bar O=\gamma_1$, $c_1 c_2=1$).

Let $G=SU(2)$ (Example <ref>). Suppose that each subsystem $\BA,\BB$ carries a spin $\frac{1}{2}$. Consider a symmetric operator $O\in\Hom(\frac{1}{2}\ot\frac{1}{2},\frac{1}{2}\ot\frac{1}{2})$ in the form of Heisenberg interaction:

\begin{align}
\itk{
\node[rectangle, minimum width=4em, draw] (f) at (0,0) {$O$};
\node (v1) at (-0.5,1) {$\frac12$};
\node (vn) at (0.5,1) {$\frac12$};
\node (w1) at (-0.5,-1) {$\frac12$};
\node (wm) at (0.5,-1) {$\frac12$};
\draw
(v1) -- (v1 |- f.north)
(vn) -- (vn |- f.north)
(w1) -- (w1 |- f.south)
(wm) -- (wm |- f.south);}=\frac13\vec{\sigma}^\BA \cdot\vec {\sigma}^\BB=\frac13
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & -1 & 2 & 0 \\
0 & 2 & -1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}.
\end{align}

We go through the first two steps in Diagram (<ref>):

* After rotation of symmetric tensor, we have $\bar O^{00}_{00}=\bar O^{11}_{11}=\frac13$, $\bar O^{11}_{00}=\bar O^{00}_{11}=-\frac13$ and $\bar O^{01}_{10}=\bar O^{10}_{01}=\frac23$.

* Consider the direct sum decomposition
$\frac12^*\ot \frac12 \cong\frac12\ot \frac12^*\cong 1\oplus 0$, and the basis change between them is

\begin{gather}
\begin{pmatrix}
\frac1{\sqrt{2}}&0&0&-\frac1{\sqrt{2}}\\
\end{pmatrix},\
\begin{pmatrix}
\end{pmatrix};\nonumber\\
\frac1{\sqrt{2}}&0&0&-\frac1{\sqrt{2}}\\
\end{pmatrix},\
\begin{pmatrix}
\frac{1}{\sqrt{2}}&0&0&\frac{1}{\sqrt{2}}
\end{pmatrix}.
\end{gather}

Then we have

\begin{align}
\bar O&=\frac13\begin{pmatrix}
1 & 0 & 0 & -1 \\
0 & 0 & 2 & 0 \\
0 & 2 & 0 & 0 \\
-1 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
0 & \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
-1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}
\end{pmatrix}\frac13
\begin{pmatrix}
2 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 & -1 & 0 \\
\frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}} \\
0 & 1 & 0 & 0 \\
-\frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}}
\end{pmatrix}
\nonumber\\
\begin{pmatrix}
\end{pmatrix}^\dag
\frac13\begin{pmatrix}
2 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
\end{pmatrix}
\end{align}

where we find $\curr{\BA}{\BB}=\Ima \bar O=1$.

At this stage, we find that $O$ transports an angular momentum of spin 1.

\begin{align}
\frac13\vec{\sigma}^\BA \cdot\vec {\sigma}^\BB = \itk{
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\frac{1}{2}$}--(l)--(0,3)node[above]{$\frac{1}{2}$};
\draw (2,0)node[below]{$\frac{1}{2}$}--(r)--(2,3)node[above]{$\frac{1}{2}$};
\draw (l)--node[above]{$1$} (r);
\end{align}
Denote the basis of spin $\frac12$ as $\{|\frac12,\frac12\ra,|\frac12,-\frac12\ra\}$ and the basis of spin $1$ as $\{|1,1\ra,|1,0\ra,|1,-1\ra\}$. One possible choice of $l,r$ can be obtained from the Clebsch-Gordan coefficients:

\begin{align}
l\left|\frac12,\frac12\right\ra &=
\sqrt{\frac23} \left|\frac12,-\frac12\right\ra |1,1\ra
l\left|\frac12,-\frac12\right\ra &= \sqrt{\frac13} \left|\frac12,-\frac12\right\ra |1,0\ra
r|1,1\ra \left|\frac12,-\frac12\right\ra =& \sqrt{\frac23}
\left|\frac12,\frac12\right\ra,\
r|1,0\ra \left|\frac12,\frac12\right\ra = -\sqrt{\frac13}
\left|\frac12,\frac12\right\ra,\nonumber\\
r|1,0\ra \left|\frac12,-\frac12\right\ra =& \sqrt{\frac13}
\left|\frac12,-\frac12\right\ra,\
r|1,-1\ra \left|\frac12,\frac12\right\ra = -\sqrt{\frac23}
\left|\frac12,-\frac12\right\ra.
\end{align}

Other choices of $l,r$ are again only up to a scalar.

When the conditions in Remark <ref> are satisfied, we can simply define the space $\curr{\BA}{\BB}$ to be the operator space spanned by $\rho^\BB_g O \rho^\BB_{g^{-1}}, g\in G$:

\begin{align}
\curr{\BA}{\BB}=\langle\rho^\BB_g O \rho^\BB_{g^{-1}}, g\in G\rangle= \{X\in \Hom(\cH,\cH)|X=\sum_{g\in G} a_g \rho^\BB_g O \rho^\BB_{g^{-1}},a_g\in \C\}.
\end{align}

In other words, $\curr{\BA}{\BB}$ is the closure of $O$ under the “partial” conjugation group actions on operators.
Note that

* $\rho^\BB_g O \rho^\BB_{g^{-1}}=\rho^\BA_{g^{-1}}O\rho^\BA_{g}.$

* $\rho^\BB_g O \rho^\BB_{g^{-1}}$ is in general no longer symmetric,
\begin{align}
\rho_h \rho^\BB_g O \rho^\BB_{g^{-1}} \rho_h^{-1}=\rho^\BB_{hgh^{-1}} O (\rho^\BB_{hgh^{-1}})^{-1}.
\end{align}

* $\rho^\BB_g O \rho^\BB_{g^{-1}}$ are in general not linearly independent for $g\in G$.

* The group action on $\curr{\BA}{\BB}$ is the “partial” conjugation by $\rho^\BB$, with respect to which the tensors $l,r$ in Eq. (<ref>) are intertwiners.
More precisely, one can associate the group representation $\tau^\BB:G\to GL(\curr{\BA}{\BB})$ to $\curr{\BA}{\BB}$ where

\begin{align}
\tau^\BB_h(\rho^\BB_g O \rho^\BB_{g^{-1}})=\rho^\BB_h \rho^\BB_g O \rho^\BB_{g^{-1}}\rho^\BB_{h^{-1}}=\rho^\BB_{hg} O \rho^\BB_{(hg)^{-1}}.
\end{align}

In other words, as a group representation, $\curr{\BA}{\BB}$ is understood as $(\langle\rho^\BB_g O \rho^\BB_{g^{-1}}, g\in G\rangle,\tau^\BB)$.
But, it is easy to see that one can also associate this operator space with the group representation $\tau^\BA:G\to GL(\curr{\BA}{\BB})$ where

\begin{align}
\tau^\BA_h(\rho^\BB_g O \rho^\BB_{g^{-1}})=\rho^\BA_h \rho^\BB_g O \rho^\BB_{g^{-1}}\rho^\BA_{h^{-1}}=\rho^\BA_{hg^{-1}} O \rho^\BA_{h^{-1}g}=\rho^\BB_{gh^{-1}}O\rho^\BB_{g^{-1}h}.
\end{align}

It is easy to see that $(\langle\rho^\BB_g O \rho^\BB_{g^{-1}}, g\in G\rangle,\tau^\BA)=\curr{\BB}{\BA}$ represents the symmetry charge transported by $O$ from $\BB$ to $\BA$.
It is clear that $\curr{\BB}{\BA}$ and $\curr{\BA}{\BB}$ are dual to each other; the total symmetry charge of $\BA,\BB$ together is conserved.

When the conditions in Remark <ref> are satisfied by $\bar O$,

\begin{align}
\curr{\BA}{\BB}:=\Ima \bar O=C(G)\bar O=\la \rho_g^\BB \ot (\rho^\BB)_g^\star \bar O, g\in G\ra.
\end{align}

Then graphically,

\begin{align}
\itk{
\node[right] (m) at (0,0.5) {$\Ima \bar O$};
\node(e1) at (-1,-1) {$\cH_\BA^*$};
\node(e2) at (1,-1) {$\cH_\BA$};
\node[draw](g) at (-1,2) {$\rho_g^\BB$};
\node[draw](h) at (1,2) {$(\rho^\BB)_g^\star$};
\node[draw](my) at (0,0) {$\bar l$};
\node[draw](dy) at (0,1) {$\bar r$};
\draw (-1,3)node[above]{$\cH_\BB$} -- (g);
\draw (1,3)node[above]{$\cH_\BB^*$} -- (h);
\draw
(e1)-- (my) -- (dy) --(g)
(e2)-- (my) -- (dy) --(h);}
\itk{
\node[right] (m) at (0,0.5) {$\Ima \bar O$};
\node(e1) at (-1,-1) {$\cH_\BA^*$};
\node(e2) at (1,-1) {$\cH_\BA$};
\node[draw](g) at (-1,2) {$\rho_g^\BB$};
\node[draw](my) at (0,0) {$\bar l$};
\node[draw](dy) at (0,1) {$\bar r$};
\draw (-1,3)node[above]{$\cH_\BB$} -- (g);
\draw (.5,2.5)node[left]{$\cH_\BB^*$} -- (0.5,1.5);
\draw
(e1)-- (my) -- (dy) --(g)
(e2)-- (my) -- (dy) --(.5,1.5);
\draw (.5,2.5) arc(180:0:0.5);
\node[draw](z) at (1.5,2) {$\rho_{g^{-1}}^\BB$};
\draw (1.5,1.5) arc(180:360:0.5);
\draw (2.5,3)node[above]{$\cH_\BB^*$} -- (2.5,1.5);}
\overset{\epsilon^{-1}}{\mapsto}
\itk{
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\node[draw](y) at (2,2.9) {$\rho_g^\BB$};
\draw (l)--node[above]{$\curr{\BA}{\BB}$} (r);
\node[draw](x) at (2,.5) {$\rho_{g^{-1}}^\BB$};
\draw (0,-.5)node[below]{$\cH_\BA$}--(l)--(0,3.5)node[above]{$\cH_\BA$};
\draw (2,-.5)node[below]{$\cH_\BB$}--(x)--(r)--(y)--(2,3.5)node[above]{$\cH_\BB$};
\end{align}

whereby $\curr{\BA}{\BB}=\langle\rho^\BB_g O \rho^\BB_{g^{-1}}, g\in G\rangle\cong \langle\rho^\BB_{g^{-1}} O \rho^\BB_g, g\in G\rangle=\langle\rho^\BA_g O \rho^\BA_{g^{-1}}, g\in G\rangle$.

For the method in Remark <ref> to work, the image of $\bar O$, i.e., the transported symmetry charge, is necessarily a cyclic representation.
In Proposition <ref>, given an arbitrary symmetric operator $O\in\End(\cH)$, we actually do not know in advance whether the transported symmetry charge $\curr{\BA}{\BB}$ is a cyclic representation or not.
It is just a convenient method that we can try at first. If $\curr{\BA}{\BB}$ is not cyclic, we will fail to obtain the correct result, i.e., there is no solution for $l$ and $r$ in Eq. (<ref>) if we wrongly put $X=(\langle\rho^\BB_g O \rho^\BB_{g^{-1}}, g\in G\rangle,\tau^\BB)$. In this situation, we should extract the transported symmetry charge by the original method in Definition <ref>.

The regular representation and all irreducible representations of $G$ are cyclic representations of $G$.

Let $G=\Z_2:=\{1,\zeta\}$. $\irr(\Rep \Z_2)=\{(\alpha_+,\rho^+),(\alpha_-,\rho^-)\}$, where $\alpha_+$ is the trivial representation and $\alpha_-$ is the sign representation. Consider the regular representation on a qubit $R:=\  C(|1\ra,|\zeta\ra)$, where we identify $|1\ra$ with the logical 0 and $|\zeta\ra$ with the logical 1, and write the nontrivial action $\rho_\zeta=\ \sigma_x$. Then we have  $R\cong\alpha_+\oplus \alpha_-$ where $\alpha_+\cong \la |+\ra:=\frac1{\sqrt2}(|1\ra+|\zeta\ra)\ra$ and $\alpha_-\cong \la |-\ra:=\frac1{\sqrt2}(|1\ra-|\zeta\ra)\ra$. Consider two regions $\BA$ and $\BB$ both carrying the regular representation $R$, and a symmetric operator $O=\ \sigma^\BA_z \sigma^\BB_z\in$ $\Hom(R\ot R,R\ot R)$. Let us try the method in Proposition <ref> to extract the symmetry charge transported by $O$:
\begin{equation} \curr{\BA}{\BB}=(\langle \sigma^\BA_z \sigma^\BB_z,-\sigma^\BA_z\sigma^\BB_z\rangle,\tau^\BB)= (\ \la\sigma^\BA_z \sigma^\BB_z\ra,\tau^\BB)\cong (\alpha_-,\rho^-). \end{equation}

We find no contradiction in solving $l,r$. Therefore,

\begin{align}
\sigma^\BA_z\ \sigma^\BB_z = \itk{
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\alpha_+\oplus \alpha_-$}--(l)--(0,3)node[above]{$\alpha_+\oplus \alpha_-$};
\draw (2,0)node[below]{$\alpha_+\oplus \alpha_-$}--(r)--(2,3)node[above]{$\alpha_+\oplus \alpha_-$};
\draw (l)--node[above]{$\alpha_-$} (r);
\end{align}

where one possible choice of $l,r$ is

\begin{align}
l|1\ra=|1\ra|-\ra,\ l|\zeta\ra=-|\zeta\ra|-\ra;\nonumber\\
\ r|-\ra|1\ra=|1\ra,\ r|-\ra|\zeta\ra=-|\zeta\ra,
\end{align}

or in more compact tensor notation:
\begin{equation}
l_{|1\ra}^{|1\ra|-\ra}= - l_{|\zeta\ra}^{|\zeta\ra|-\ra}=\ r_{|-\ra|1\ra}^{|1\ra}=-r_{|-\ra|\zeta\ra}^{|\zeta\ra}=1,\end{equation}
with other entries being zero. Other choices of $l$ and $r$ are only up to a scalar.

This result is physically easy to understand: the operator $\sigma_z$ flips the $\Z_2$ charge, and $O=\sigma_z^\BA\sigma_z^\BB$ flips the $\Z_2$ charges of $\BA$ and $\BB$ together. Therefore, $O$ indeed moves an odd $\Z_2$ charge from $\BA$ to $\BB$.

Let $G=S_3$ (Example <ref>).

\begin{align}
\sigma^\BA_z \sigma^\BB_z = \itk{
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\Lambda$}--(l)--(0,3)node[above]{$\Lambda$};
\draw (2,0)node[below]{$\Lambda$}--(r)--(2,3)node[above]{$\Lambda$};
\draw (l)--node[above]{$\lambda_1$} (r);
\end{align}

Denote the basis of $\Lambda$ as $\{\bm 0,\bm 1\}$ and the basis of $\lambda_1$ as $\{\bm o\}$. One possible choice of $l$, $r$ is

\begin{align}
l(\bm 0)=\bm 0\bm o,\  l(\bm 1)=-\bm 1\bm o;\nonumber\\
r(\bm e\bm 0)=\bm 0,\  r(\bm e\bm 1)=-\bm 1.
\end{align}

Here since the intertwiner spaces $\Hom(\Lambda\ot \lambda_1,\Lambda), \Hom(\lambda_1\ot\Lambda,\Lambda)$ are all of dimension 1, other choices of $l$ and $r$ are only up to a scalar.

Let $G=Q_8$ (Example <ref>).

\begin{align}
\sigma^\BA_y \sigma^\BB_y = \itk{
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\Gamma$}--(l)--(0,3)node[above]{$\Gamma$};
\draw (2,0)node[below]{$\Gamma$}--(r)--(2,3)node[above]{$\Gamma$};
\draw (l)--node[above]{$\gamma_2$} (r);
\end{align}

Denote the basis of $\Gamma$ as $\{|0\ra,|1\ra\}$ and the basis of $\gamma_2$ as $\{|x_2 \ra\}$. One possible choice of $l$, $r$ is

\begin{align}
l|0\ra=|1\ra |x_2 \ra,\  l|1\ra=-|0\ra |x_2 \ra;\nonumber\\
r|x_2 \ra |0\ra=- |1\ra,\  r|x_2 \ra |1\ra=|0\ra.
\end{align}

Other choices of $l$ and $r$ are only up to a scalar. And

\begin{align}
\sigma^\BA_z\sigma^\BB_z = \itk{
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\Gamma$}--(l)--(0,3)node[above]{$\Gamma$};
\draw (2,0)node[below]{$\Gamma$}--(r)--(2,3)node[above]{$\Gamma$};
\draw (l)--node[above]{$\gamma_3$} (r);
\end{align}

Denote the basis of $\gamma_3$ as $\{|x_3 \ra\}$. One possible choice of $l$, $r$ is

\begin{align}
l|0\ra=|0\ra |x_3 \ra,\  l|1\ra=-|1\ra |x_3 \ra;\nonumber\\
r|x_3 \ra |0\ra= |0\ra,\  r|x_3 \ra |1\ra=-|1\ra.
\end{align}

Other choices of $l$ and $r$ are only up to a scalar.

§.§ Symmetry charge flows via half-braiding

Now consider a tripartite subsystem

\begin{align}
\cH_{\{s\}\coprod \BM\coprod \{t\}}=\cH_s\ot \cH_\BM\ot\cH_t.
\end{align}

For symmetric operator $O$ acting on this subsystem, we can similarly extract the symmetry charge transported from $s\BM:={\{s\}}\cup \BM$ to $t$, or from $s$ to $\BM t:=\BM\cup \{t\}$. By repeatedly performing the SSVD process in Diagram (<ref>), $O$ can be represented by the following symmetric tensor

\begin{equation}
\node[rectangle,draw] (l) at (-2,-1) {$l$};
\node[rectangle,draw] (m) at (0,0) {$h$};
\node[rectangle,draw] (r) at (2,1) {$r$};
\draw (-2,-2)node[below]{$\cH_s$} -- (l) -- (-2,2)node[above]{$\cH_s$}
(0,-2)node[below]{$\cH_\BM$} -- (m) -- (0,2)node[above]{$\cH_\BM$}
(2,-2)node[below]{$\cH_t$} -- (r) -- (2,2)node[above]{$\cH_t$}
(l)--node[above]{$\curr{s}{\BM t}$} (m) -- node[above]{$\curr{s\BM}{t}$} (r);
} .\label{eq.generalsymop}
\end{equation}

With such graphical representation, we can read out two transported charges $\curr{s}{\BM t}$ and $\curr{s\BM}{t}$, together with a symmetric tensor $h$ telling us how the charge flows through $\BM$. Instead of such general form, here we would like to focus on the most simple form of charge transport.
Suppose that we want the operator $O$ to only transport charge from $s$ to $t$ with no other effect on the intermediate region $\BM$, and moreover, the intermediate region $\BM$ can be arbitrary; in other words, we want to study how a constant symmetry charge can be freely transported all over the space.
We arrive at the following definition of quantum current:

In a lattice system with onsite symmetry, $(\BL,\cH_\BK)$, (the Hamiltonian can be arbitrary), a quantum current $(Q,\beta)$ is the collection of symmetric operators of the following form

\begin{equation}
\itk{
\node[rectangle,draw] (l) at (-2,-1) {$l$};
\node[rectangle,draw] (m) at (0,0) {$\beta_{Q,\cH_\BM}$};
\node[rectangle,draw] (r) at (2,1) {$r$};
\draw (-2,-2)node[below]{$\cH_s$} -- (l) -- (-2,2)node[above]{$\cH_s$}
(0,-2)node[below]{$\cH_\BM$} -- (m) -- (0,2)node[above]{$\cH_\BM$}
(2,-2)node[below]{$\cH_t$} -- (r) -- (2,2)node[above]{$\cH_t$}
(l)--node[above]{$Q$} (m) -- node[above]{$Q$} (r);
\end{equation}

* $s,t\in \BL$ are called the source and target sites[To emphasize the locality of the charge transport, we use two sites instead of two regions as the source and target of the quantum current. Under renormalization, $s$ or $t$ can be the combination of several neighbouring sites in the original lattice.];

* $\BM\subset \BL$ is an arbitrary intermediate subregion, $\cH_\BM=\otr[i\in \BM] \cH_i$;

* $Q$ is the symmetry charge transported from $s$ to $t$;

* $l,r$ are arbitrary intertwiners in $\Hom(\cH_s, \cH_s\ot Q)$ and $\Hom(Q\ot \cH_t,\cH_t)$ respectively, and called source and target intertwiners;

* $\beta$ is a collection of invertible intertwiners $\{\beta_{Q,V}:Q\ot V\to V\ot Q\}$ for any $V\in \cC$,

\begin{equation}
\beta_{Q,V} = \itk{
\node (l) at (-2,-1) {};
\node[] (m) at (0,0) {};
\node (r) at (2,1) {};
\draw
(0,-2)node[below]{$V$} -- (m)node[below right]{$\beta$} -- (0,2)node[above]{$V$}
(l)--node[above]{$Q$} (0,0) -- (r);
\end{equation}

where for simplicity we omitted the subscript of $\beta$ in the graph which can be unambiguously read out from the decorations on the legs. We also draw the $\beta$ node intuitively as a crossing-over.
$\beta$ must be compatible with the arbitrary choice of $\BM$, which turns out to be the following conditions:

* $\beta$ commutes with local symmetric operators (naturality):
for any $f:V\to W$ in $\cC$

\begin{equation}\label{eq.qca}
\itk{
\node (l) at (-2,-1) {};
\node[] (m) at (0,0) {};
\node (r) at (2,1) {};
\node[rectangle,draw] (f) at (0,1) {$f$};
\draw
(0,-2)node[below]{$V$} -- (m)node[below right]{$\beta$} --(f)-- (0,2)node[above]{$W$}
(l)--node[above]{$Q$} (0,0) --  (r);
} =
\itk{
\node (l) at (-2,-1) {};
\node[] (m) at (0,0) {};
\node (r) at (2,1) {};
\node[rectangle,draw] (f) at (0,-1) {$f$};
\draw
(0,-2)node[below]{$V$} --(f)-- (m)node[below right]{$\beta$} -- (0,2)node[above]{$W$}
(l)--node[above]{$Q$} (0,0) -- (r);
} .
\end{equation}

* $\beta$ is compatible with any bipartition of the intermediate region: for any $V,W\in \cC$,

\begin{equation}\label{eq.qcb}
\itk{
\node (l) at (-2,-1) {};
\node[] (m) at (0,0) {};
\node (r) at (2,1) {};
\draw
(0,-2)node[below]{$V\ot W$} -- (m)node[below right]{$\beta$} -- (0,2)node[above]{$V\ot W$}
(l)--node[above]{$Q$} (0,0) -- (r);
\itk{
\node (l) at (-2,-1) {};
\node[] (m1) at (-1,-0.5) {};
\node[] (m2) at (1,0.5) {};
\node (r) at (2,1) {};
\draw
(-1,-2)node[below]{$V$} -- (m1)node[below right]{$\beta$} -- (-1,2)node[above]{$V$}
(1,-2)node[below]{$W$} -- (m2) node[below right]{$\beta$}-- (1,2)node[above]{$W$}
(l)node[left]{$Q$} --  (r);
} .
\end{equation}

Reducing to classical systems, the constant current in 1+1D consists of only one datum $Q$. In the quantum setting, we see that although the transported charge $Q$ is a constant, we still need the datum $\beta$ to describe how quantum mechanically the charge $Q$ flows through the local Hilbert spaces. Moreover, even if $Q$ is the same, different ways of flowing $\beta$ should be regarded as different quantum currents $(Q,\beta)$. Such $\beta$ is exactly the half-braiding. The above two conditions are exactly Eq. (<ref>) and (<ref>). Due to these conditions, not all charge $Q$ can flow over an arbitrarily long distance.
Our convention for $\beta$ is also justified.

It is then clear that a quantum current is associated with the pair $(Q,\beta_{Q,-})$ which is an object in the Drinfeld center $Z_1(\cC)$. Conversely, given an object $(X,\beta_{X,-})\in Z_1(\cC)$ we can construct quantum current operators by pasting $\beta_{X,-}$ in the intermediate sites and fusing $X$ to the source and target sites with some choice of source and target intertwiners.

Similar notions as transparent patch operator<cit.> and unconfined (non-local) symmetric operator<cit.> are defined elsewhere.
Our definition here contrast with them in two aspects: (1) our definition is model independent; it only cares about the symmetry, or group representations, but not specific Hamiltonians. The condition (<ref>), which is mathematically the naturality of half-braiding (<ref>), requires a quantum current to commute with any possible local symmetric operators (including local Hamiltonians) that does not overlap with the source and the target.  (2) We point out the condition (<ref>), which is mathematically the hexagon equation for half-braiding (<ref>), and physically the requirement for a quantum current to be able to consistently extend over an arbitrary long distance. This condition is overlooked in previous works <cit.>.

A symmetric operator $O\in (Q,\beta)$ with a fixed choice of $s,\BM,t,l,r$ is called a realization of the quantum current $(Q,\beta)$.

Let $s,\BM,t,l$ be fixed and $r$ vary, all such realizations form a vector space, denoted by $(Q,\beta)^{s,\BM,t}_{l,-}$. This operator space admits a natural two-sided action of the local symmetric operator algebra at $t$, $\End(\cH_t):=\Hom(\cH_t,\cH_t)$, by pre and post composition of operators. In other words, $(Q,\beta)^{s,\BM,t}_{l,-}$ is an $\End({\cH_t})$-bimodule.
Similarly, $(Q,\beta)^{s,\BM,t}_{-,r}$ is an $\End({\cH_s})$-bimodule.

However, if one lets both $l,r$ vary, the set of operators $(Q,\beta)^{s,\BM,t}_{-,-}$ with fixed $s,\BM,t$ is, in general, not even closed under addition[For a quantum current $(Q,\beta)$, given two different realizations of it $(Q,\beta)^{s,\BM,t}_{r,l}$ and $(Q,\beta)^{s,\BM,t}_{r',l'}$ as two different symmetric operators, the addition of them may not still be in the set $(Q,\beta)^{s,\BM,t}_{-,-}$. This is what we mean by not closed under addition.]. This phenomenon is related to quantum entanglement.

Let $G=\Z_2=\{1,\zeta\}$ and $\irr(Z_1(\Rep \Z_2))=\{\mathbbm{1},m,e,\psi\}$, where we borrow the labelling of anyons in the well-known toric code model. Consider two qubits $\BA$ and $\BB$ both carrying the regular $\Z_2$ representation $R$. Denoting the simple $\Z_2$ charge by $(\alpha_+,\rho^+)$ and $(\alpha_-,\rho^-)$ and the charge basis of each qubit by $|+\ra,|-\ra$, we have $R\cong \alpha_+\oplus\alpha_-=C(|+\ra,|-\ra)$ (we note that the basis we choose here is different from that in Example <ref>). We take $\beta$ to be trivial in this example ($\mathbbm{1}$ and $e$ in $Z_1(\Rep \Z_2)$ have trivial $\beta$) and thus omit the intermediate region $\BM$. Let $P_i|j\ra=\delta_{ij}|j\ra, i,j=+,-$ be the projection operator to the fixed charge. Then
\begin{equation}
P_+\ot P_+=\begin{pmatrix}
1 & 0  \\
0 & 0 \\
\end{pmatrix}\otimes \begin{pmatrix}
1 & 0  \\
0 & 0 \\
\end{pmatrix},\quad
P_-\ot P_-=\begin{pmatrix}
0 & 0  \\
0 & 1 \\
\end{pmatrix}\otimes \begin{pmatrix}
0 & 0  \\
0 & 1 \\
\end{pmatrix}.\end{equation}

By Proposition <ref> and $\rho_\zeta=\sigma_z$ in basis $\{|+\ra,|-\ra\}$, $P_+\ot P_+$ and $P_-\ot P_-$ both transport charge $\alpha_+$. However, if one adds them up,
\begin{equation}
O:=P_+\ot P_+ +P_-\ot P_-=\begin{pmatrix}
1 & 0  \\
0 & 0 \\
\end{pmatrix}\otimes \begin{pmatrix}
1 & 0  \\
0 & 0 \\
\end{pmatrix}+\begin{pmatrix}
0 & 0  \\
0 & 1 \\
\end{pmatrix}\otimes \begin{pmatrix}
0 & 0  \\
0 & 1 \\
\end{pmatrix}.\end{equation}

We extract the transported charge by $O$ by going through the process in Diagram (<ref>). We have $R^*\otimes R\cong R\otimes R^*\cong \alpha_+\oplus\alpha_+\oplus\alpha_-\oplus\alpha_-$[Denote the dual basis of $R^*$ as $\{\delta_{|+\ra},\delta_{|-\ra}\}$. $R$ is isomorphic to $R^*$ through the isomorphism $|+\ra\mapsto\delta_{|+\ra}$, $|-\ra\mapsto\delta_{|-\ra}$.], and the basis change between them is

\begin{gather}
p_{R^*\otimes R}^{\alpha_+;1}=
\begin{pmatrix}
1 & 0 & 0 & 0
\end{pmatrix},\
p_{R^*\otimes R}^{\alpha_-;1}=
\begin{pmatrix}
0 & 1 & 0 & 0
\end{pmatrix},\nonumber\\
p_{R^*\otimes R}^{\alpha_-;2}=
\begin{pmatrix}
0 &0&1&0
\end{pmatrix},\
p_{R^*\otimes R}^{\alpha_+;2}=
\begin{pmatrix}
0 & 0 & 0 & 1
\end{pmatrix};\nonumber\\
p_{R\otimes R^*}^{\alpha_+;1}=
\begin{pmatrix}
1 & 0 & 0 & 0
\end{pmatrix},\
p_{R\otimes R^*}^{\alpha_-;1}=
\begin{pmatrix}
0 & 1 & 0 & 0
\end{pmatrix},\nonumber\\
p_{R\otimes R^*}^{\alpha_-;2}=
\begin{pmatrix}
0 &0&1&0
\end{pmatrix},\
p_{R\otimes R^*}^{\alpha_+;2}=
\begin{pmatrix}
0 & 0 & 0 & 1
\end{pmatrix}.
\end{gather}

Then we have

\begin{align}
\bar O&=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\nonumber\\
&        =
\begin{pmatrix}
p_{R\otimes R^*}^{\alpha_+;1}\\
p_{R\otimes R^*}^{\alpha_-;1}\\
p_{R\otimes R^*}^{\alpha_-;2}\\
p_{R\otimes R^*}^{\alpha_+;2}
\end{pmatrix}^\dag
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
p_{R^*\otimes R}^{\alpha_+;1}\\
p_{R^*\otimes R}^{\alpha_-;1}\\
p_{R^*\otimes R}^{\alpha_-;2}\\
p_{R^*\otimes R}^{\alpha_+;2}
\end{pmatrix}=(
p_{R\otimes R^*}^{\alpha_+;1}
p_{R^*\otimes R}^{\alpha_+;1}+(
p_{R\otimes R^*}^{\alpha_+;2}
p_{R^*\otimes R}^{\alpha_+;2},
\end{align}

where we see the transported charge by $O$ is $\curr{\BA}{\BB}=\alpha_+\oplus\alpha_+$. Graphically,

\begin{align}
O = \itk{
\node[rectangle, draw] (l) at (0,1) {$l$};
\node[rectangle, draw] (r) at (2,2) {$r$};
\draw (0,0)node[below]{$\alpha_+\oplus \alpha_-$}--(l)--(0,3)node[above]{$\alpha_+\oplus \alpha_-$};
\draw (2,0)node[below]{$\alpha_+\oplus \alpha_-$}--(r)--(2,3)node[above]{$\alpha_+\oplus \alpha_-$};
\draw (l)--node[above]{$\alpha_+\oplus \alpha_+$} (r);
\end{align}

The explicit forms of $l,r$ are easy to compute and we omit them.

Therefore, we conclude that symmetric operators $P_+\ot P_+$, $P_-\ot P_-$ transport the symmetry charge $\alpha_+$, different from $\alpha_+\oplus\alpha_+$ transported by their addition, i.e., $(\alpha_+,\beta)^{\BA,\BM,\BB}_{-,-}$ is not closed under addition.

Because of this subtlety, we introduce a finer notion, that instead of allowing arbitrary source and target intertwiners, we restrict to bimodules of $\End(\cH_s)\ot \End(\cH_t)$.
Fix $s,\BM,t$ to obtain a set of realizations of $(Q,\beta)$.
Pick subspaces
\begin{equation} \mathfrak L\subset\Hom(\cH_s,\cH_s\ot Q),\quad \mathfrak R\subset \Hom(Q\ot\cH_t,\cH_t). \end{equation}
We call the set of realizations with $s,\BM,t,l\in \mathfrak L, r\in \mathfrak R$, a sector of $(Q,\beta)$, denoted by $(Q,\beta)^{s,\BM,t}_{\mathfrak L,\mathfrak R}$, if $(Q,\beta)^{s,\BM,t}_{\mathfrak L,\mathfrak R}$ forms a vector space, and moreover an $\End(\cH_s)\ot \End(\cH_t)$-bimodule. We call a sector simple if this bimodule is simple.

Suppose a Hamiltonian $H=\sum_\BK H_\BK$ is given.
A non-zero realization $O\in (Q,\beta)$, with non-empty $\BM$, is called superconducting if $OH=HO$. A quantum current $(Q,\beta)$ is called superconducting if it has a realization that is superconducting.

A superconducting quantum current can transport charges over a long distance without costing any energy.
In the earlier version of this work, we used the terminology that the quantum current is condensed instead of superconducting, because of the correspondence with anyon condensation in topological orders in one higher dimension. The terminology “superconducting” fits better with the idea of quantum current.

Alternatively, a realization $O\in (Q,\beta)^{s,\BM,t}_{-,-}$ automatically commutes with all terms $H_\BK$, $\BK\subset \BM$, i.e. terms whose support is within $\BM$. Therefore, $O$ can only cost energy, or create/annihilate excitations around $s$ or $t$. We can also consider whether a simple sector $(Q,\beta)^{s,\BM,t}_{\mathfrak L,\mathfrak R}$ is superconducting. Suppose that $O',O\in (Q,\beta)^{s,\BM,t}_{\mathfrak L,\mathfrak R}$ and that $O'$ is superconducting. By simpleness, the difference between $O$ and $O'$ must be local symmetric operators on $s$ and $t$. Thus, we know that excitations created/annihilated by $O$ around $s$ or $t$ must be of the trivial type. Therefore, a simple sector is superconducting if its realizations create/annihilate only excitations of the trivial type.

For non-simple quantum current, e.g., $(Q,\beta)=(Q_1,\beta_1)\oplus (Q_2,\beta_2)$, by definition, if any of $(Q_1,\beta_1)$ and $(Q_2,\beta_2)$ is superconducting, we will say $(Q,\beta)$ is superconducting. This definition is more natural than requiring that all realizations are superconducting, as can be seen in Section <ref> and further explained in Remark <ref>. There, we will also elaborate more on the deep connection between
the superconducting of quantum currents and the emergent symmetry of gapped quantum phases.

§.§ Morphisms between quantum currents

We will show that a morphism in the Drinfeld center defines a reasonable transformation, or morphism, between quantum currents.

A morphism in the Drinfeld center is an intertwiner that commutes with the half-braiding. Suppose we have a realization of quantum current with charge $Q$ and half-braiding $\beta_{Q,-}$. Pick a morphism $f:(Q,\beta_{Q,-})\to (X,\beta_{X,-})$ in the Drinfeld center. Inserting $f$ in the graph we obtain a new realization

\begin{equation}
\itk{
\node[rectangle,draw] (l) at (-2,-1) {$l$};
\node[] (m) at (0,0) {};
\node[rectangle,draw] (f) at (1,0.5) {$f$};
\node[rectangle,draw] (r) at (2,1) {$r'$};
\draw (-2,-2)node[below]{$\cH_s$} -- (l) -- (-2,2)node[above]{$\cH_s$}
(0,-2)node[below]{$\cH_\BM$} -- (m)node[below right]{$\beta$} -- (0,2)node[above]{$\cH_\BM$}
(2,-2)node[below]{$\cH_t$} -- (r) -- (2,2)node[above]{$\cH_t$}
(l)--node[above]{$Q$} (0,0) --  (f) --node[above]{$X$} (r);
} =
\itk{
\node[rectangle,draw] (l) at (-2,-1) {$l$};
\node[] (m) at (0,0) {};
\node[rectangle,draw] (f) at (-1,-0.5) {$f$};
\node[rectangle,draw] (r) at (2,1) {$r'$};
\draw (-2,-2)node[below]{$\cH_s$} -- (l) -- (-2,2)node[above]{$\cH_s$}
(0,-2)node[below]{$\cH_\BM$} -- (m)node[below right]{$\beta$} -- (0,2)node[above]{$\cH_\BM$}
(2,-2)node[below]{$\cH_t$} -- (r) -- (2,2)node[above]{$\cH_t$}
(l)-- node[below]{$Q$} (f) -- (0,0) --node[above]{$X$}  (r);
\end{equation}

which is associated with charge $X$ and half-braiding $\beta_{X,-}$. We also see that the morphism $f$ in the Drinfeld center can change the intertwiners $l,r$ on the source and target sites. The entire process of dragging $f$ from $t$ to $s$ thus defines a map between $(Q,\beta)^{s,\BM,t}_{-,r'(f\ot \id_{\cH_t})}$ and $(X,\beta)^{s,\BM,t}_{-,r'}$. Clearly, such map, which happens on the $Q$ or $X$ leg, commutes with the action of local symmetric operators on $s$, which happens on the $\cH_s$ legs. In other words, $f$ defines an $\End(\cH_s)$-bimodule map. Meanwhile, the map $r'\mapsto r'(f\ot \id_{\cH_t})$ also commutes with action of $\End(\cH_t)$.

It is desired to have a notion of morphisms between quantum currents, purely from the point of view of symmetric tensors, such that the category of quantum currents is equivalent to the Drinfeld center. We are unclear about such a definition for the moment. The above discussion suggests to define the bimodule maps over local operator algebras as the morphisms between quantum currents, which is the point of view taken in <cit.>. It should work for half-infinitely long quantum currents, but not work well for finitely long quantum currents.

The morphisms between $(Q,\beta_{Q,-})$ and $(X,\beta_{X,-})$ in the Drinfeld center are in bijection with the $\End(\cH_s)\ot\End(\cH_t)$-bimodule maps between the sectors $(Q,\beta)^{s,\BM,t}_{\mathfrak L,\mathfrak R}$ and $(X,\beta)^{s,\BM,t}_{\mathfrak L',\mathfrak R'}$ for large enough $\mathfrak L,\mathfrak R,\mathfrak L',\mathfrak R'$ and $\cH_\BM$.

§ RENORMALIZATION OF 1+1D LATTICE SYSTEM WITH SYMMETRY

Physically, the renormalization fixed-points represent phases of matter and are thus of great interest.
In this section we try to give a rigorous treatment of the renormalization process of 1+1D lattice model. Based on this, we give a general analysis of 1+1D gapped lattice model at fixed-point.

§.§ Renormalization fixed-point and Frobenius algebra

Recall Definition <ref>. For a 1D lattice, without losing generality, we may think the set $\BL$ of lattice sites as a linearly ordered set.

Given two 1+1D quantum systems with symmetry $G$, $(\BL, \cH_\BK, H)$ and $(\BL', \cH_{\BK'}, H')$:

* A lattice renormalization is an order preserving map $f:\BL\to \BL'$.

* A Hilbert space renormalization is a pair $(f,\{U_i\})$, a lattice renormalization $f$ with a collection of intertwiners indexed by $\BL'$, $\{U_i:\cH_{f^{-1}(i)}\to \cH_i\}_{i\in \BL'}$ such that $U_i$ are partial isometries (recall Definition <ref>).
$(f,\{U_i\})$ is called proper when $f$ is surjective. Graphically for example,

\begin{equation}
\begin{tikzcd}
{} & \bullet & \bullet & {} \\
\\
\\
\\
\bullet & \bullet & \bullet & \bullet & \bullet
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{\cH_{f^{-1}(i_1)}}}"', draw=none, from=5-1, to=5-2]
\arrow["{\cH_{i_1}}", draw=none, from=1-1, to=1-3]
\arrow["{\cH_{i_2}}", draw=none, from=1-2, to=1-4]
\arrow[from=5-1, to=1-2]
\arrow["{U_{i_1}}", from=5-2, to=1-2]
\arrow[from=5-3, to=1-3]
\arrow["{U_{i_2}}"{description}, from=5-4, to=1-3]
\arrow[from=5-5, to=1-3]
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{\cH_{f^{-1}(i_2)}}}"', draw=none, from=5-3, to=5-5]
\end{tikzcd},
\end{equation}

where we use $\bullet$ to depict a lattice site, and $f:\BL\to \BL'$, $i_1,i_2\subset \BL'$, and $f^{-1}(i_1),f^{-1}(i_2)\subset \BL$.

* Let $U=\otimes_{i\in \BL'} U_i$. When the following condition holds,

\begin{align}
U H U^\dag =UU^\dag(H'+\Delta E)UU^\dag, \label{eq.effH}
\end{align}

where $\Delta E$ is some energy zero point shift, we call the pair $(f,\{U_i\})$  a (system) renormalization,
and meanwhile call $(\BL',\cH_{\BK'},H')$ the renormalized system of $(\BL,\cH_\BK,H)$ after $(f,\{U_i\})$.

For a proper Hilbert space renormalization $(f,\{U_i\})$, the image of $U_i$ should be viewed as the low energy degrees of freedom after renormalization, which justifies our definition: $U_i^\dag U_i$ and $ U_iU_i^\dag$ are projections onto the low energy subspace, and the inner product is preserved within the low energy subspace. Equation (<ref>) just says that the restrictions of $H'$ and $H$ in the low energy subspace are equal, and serve as the effective Hamiltonian. The energy shift can be equivalently moved to the left-hand side

\begin{align}
U(H-\Delta E)U^\dag =UU^\dag H' UU^\dag,
\end{align}

When $f$ is not surjective, for $f^{-1}(i)=\emptyset$, $U_i:\cH_\emptyset=\C\to \cH_i$ just fixes a state in $\cH_i$. In other words, the non-proper Hilbert space renormalization adds sites in fixed states to the original system.

There is a natural composition of Hilbert space renormalizations

\begin{align}
\label{eq.renorm}
(g,\{ W_i\})\circ(f,\{U_j\}):= (gf, \{W_i(\otimes_{j\in g^{-1}(i)}U_j)\}).
%\{ \cH_{(fg)^{-1}(i)}=\otimes_{j\in g^{-1}(i)}\cH_{f^{-1}(j)}\xrightarrow{\otimes_{j\in g^{-1}(i)}U_j} \otimes_{j\in g^{-1}(i)} \cH_{j}=\cH_{g^{-1}(i)}\xrightarrow{W_i}\cH_i\}).
\end{align}

Graphically for example,

\begin{equation}
\begin{tikzcd}
{} & \bullet & \bullet & {} \\
\\
{} & \bullet & \bullet & \bullet \\
\\
\bullet & \bullet & \bullet & \bullet & \bullet
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{\cH_{f^{-1}(j_1)}}}"', draw=none, from=5-1, to=5-2]
\arrow["{\cH_{i_1}}", draw=none, from=1-1, to=1-3]
\arrow["{\cH_{i_2}}", draw=none, from=1-2, to=1-4]
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{\cH_{f^{-1}(j_3)}}}"', draw=none, from=5-4, to=5-5]
\arrow["{\cH_{f^{-1}(j_2)}}"', draw=none, from=5-2, to=5-4]
\arrow["{U_{j_3}}"', from=5-4, to=3-4]
\arrow[from=5-5, to=3-4]
\arrow[from=3-4, to=1-3]
\arrow["{U_{j_2}}", from=5-3, to=3-3]
\arrow["{W_{i_2}}"', from=3-3, to=1-3]
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{\cH_{g^{-1}(i_2)}}}"', draw=none, from=3-3, to=3-4]
\arrow[from=5-1, to=3-2]
\arrow["{U_{j_1}}", from=5-2, to=3-2]
\arrow["{W_{i_1}}", from=3-2, to=1-2]
\arrow["{\ \ \ \ \ \ \ \ \cH_{g^{-1}(i_1)}}"', draw=none, from=3-1, to=3-3]
\end{tikzcd}
\begin{tikzcd}
{} & \bullet & \bullet & {} \\
\\
\\
\\
\bullet & \bullet & \bullet & \bullet & \bullet
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{\cH_{f^{-1}g^{-1}(i_1)}}}"', draw=none, from=5-1, to=5-2]
\arrow["{\cH_{i_1}}", draw=none, from=1-1, to=1-3]
\arrow["{\cH_{i_2}}", draw=none, from=1-2, to=1-4]
\arrow[from=5-1, to=1-2]
\arrow["{W_{i_1}U_{j_1}}", from=5-2, to=1-2]
\arrow["{W_{i_2}(U_{j_2}\otimes U_{j_3})}"', from=5-3, to=1-3]
\arrow[from=5-4, to=1-3]
\arrow[from=5-5, to=1-3]
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{\cH_{f^{-1}g^{-1}(i_2)}}}"', draw=none, from=5-3, to=5-5]
\end{tikzcd},
\end{equation}

where $f:\BL\to \BL'$, $g:\BL'\to \BL''$, $i_1,i_2\subset \BL''$, $g^{-1}(i_1),g^{-1}(i_1)\subset \BL'$ and $f^{-1}(j_1),f^{-1}(j_2),f^{-1}(j_3)\subset \BL$.

We can speak of whether a system $(\BL,\cH_\BK,H)$ is invariant with respect to a renormalization $(f,\{U_i\})$. However, in practice we expect to learn from renormalization some information of quantum phases, which are quantum systems at thermodynamic limit (infinite system size) whose interactions are local. To this end, we focus on the special case $\BL=\Z$, i.e. infinite chain.

A quantum system $(\Z,\cH_\BK, H=\sum_\BK H_\BK)$ is called local if all terms $H_\BK$ are of finite support.

The quantum phases are also almost uniform with (at-least emergent) translation invariance.

A local quantum system $(\Z,\cH_\BK,H)$ is called translation invariant if $\cH_{i}=A$ for all $i\in \Z$ and $H=\sum_i P_i$ where $P_i$ is supported on $\{i,...,i+n-1\}$ and has the form $P_i=P\ot \id$ for some $P\in\End(A^{\ot n})$. We may simply denote a translation invariant system by $(\Z,A,P)$.

For translation invariant systems, we hope to study the renormalizations that only depends how many sites are combined into one, but not on positions.

Let $\{ m_{n}:A^{\otimes n}\to A\}_{n\in \N}$ be partial isometries. A Hilbert space renormalization $(f,\{U_i\})$, between the translation invariant local systems $(\Z,A,P)$ and $(\Z,A,P')$, is called (locally) generated by $\{m_{n}\}$ if for any $i\in \Z$, $f^{-1}(i)$ is finite, and $U_i=m_{|f^{-1}(i)|}$.

Graphically, for example we sketch a Hilbert space renormalization generated by $\{m_0,m_1,m_2\}$ on the following local patch,

\begin{equation}\label{eq.genem}
\itk{
\draw (0,1) circle (2pt);
\draw (0,1.1) node[left]{$m_0$} -- (0,2) node[above]{$A$};
\draw (1.5,0) node[below]{$A$} -- (1.5,1) node[left]{$m_1$} -- (1.5,2) node[above]{$A$};
\draw (3,0) node[below]{$A$} -- (3.5,1) node[left]{$m_2$} -- (3.5,2) node[above]{$A$};
\draw (4,0) node[below]{$A$} -- (3.5,1);
\begin{tikzcd}
{} & \bullet & \bullet & \bullet & {} \\
\\
{} & \circ & \bullet & \bullet & \bullet & {}
\arrow["{U_1}", from=3-3, to=1-3]
\arrow["{\cH_0=A}", draw=none, from=1-1, to=1-3]
\arrow["{U_2}"', from=3-4, to=1-4]
\arrow[from=3-5, to=1-4]
\arrow["{\cH_1=A}", draw=none, from=1-2, to=1-4]
\arrow["{\cH_2=A}", draw=none, from=1-3, to=1-5]
\arrow["{U_0}", from=3-2, to=1-2]
\arrow["{\mathbb{C}}"', draw=none, from=3-1, to=3-3]
\arrow["{\cH_{f^{-1}(1)}=A}"', draw=none, from=3-2, to=3-4]
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{\cH_{f^{-1}(2)}=A\ot A}}"', draw=none, from=3-3, to=3-6],
\end{tikzcd}
\end{equation}

where we use $\circ$ to depict the empty subset of the lattice whose associated Hilbert space is $\C$.

Now we are ready to define renormalization fixed-point.

We call $\{m_{n}\}$ a fixed-point local generator (of renormalization), if $m_1=\id_A$ and for any two $(f,\{U_j\})$ and $(g,\{W_i\})$ generated by $\{m_{n}\}$, their composition $(gf, W_i(\otimes_{j\in g^{-1}(i)}U_j))$ is still generated by $\{m_{n}\}$.

Given a fixed-point local generator $\{m_{n}\}$, $m:=m_2$ is an associative binary operation on $A$, and $\eta:=m_0$ is the unit of $m$. $(A,m,\eta)$ is an isometric associative unital algebra, $(A,m^\dag,\eta^\dag)$ is an associative unital coalgebra, and $(A,m,\eta,m^\dag,\eta^\dag)$ is a Frobenius algebra. Note that here algebra all means algebra object in the unitary fusion category $\cC$.

Recall the composition of Hilbert space renormalization in Eq. (<ref>), we have the following equations on renormalizations:

\begin{equation}
\begin{tikzcd}
& {\underset{\bullet}{A}} \\
{\overset{\bullet}{A}} & {\overset{\bullet}{A}} \\
{\overset{\circ}{\C}} & {\overset{\bullet}{A}}
\arrow[from=2-2, to=1-2]
\arrow[from=3-2, to=2-2]
\arrow[from=2-1, to=1-2]
\arrow[from=3-1, to=2-1]
\end{tikzcd}
\ \ \ \ =\ \ \ \
\begin{tikzcd}
{\underset{\bullet}{A}} \\
{\textcolor{white}{\overset{\bullet}{A}}} \\
\arrow[from=3-1, to=1-1]
\end{tikzcd}
\ \ \ \ =\ \ \ \
\begin{tikzcd}
{\underset{\bullet}{A}} \\
{\overset{\bullet}{A}} & {\overset{\bullet}{A}} \\
{\overset{\bullet}{A}} & {\overset{\circ}{\C}}
\arrow[from=3-2, to=2-2]
\arrow[from=3-1, to=2-1]
\arrow[from=2-1, to=1-1]
\arrow[from=2-2, to=1-1]
\end{tikzcd},
\end{equation}

\begin{equation}
\begin{tikzcd}
& {\underset{\bullet}{A}} \\
{\overset{\bullet}{A}} & {\overset{\bullet}{A}} \\
{\overset{\bullet}{A}} & {\overset{\bullet}{A}} & {\overset{\bullet}{A}}
\arrow[from=3-1, to=2-1]
\arrow[from=2-1, to=1-2]
\arrow[from=3-2, to=2-2]
\arrow[from=2-2, to=1-2]
\arrow[from=3-3, to=2-2]
\end{tikzcd}
\begin{tikzcd}
& {\underset{\bullet}{A}} \\
& {\textcolor{white}{\overset{\bullet}{A}}} \\
{\overset{\bullet}{A}} & {\overset{\bullet}{A}} & {\overset{\bullet}{A}}
\arrow[from=3-1, to=1-2]
\arrow[from=3-2, to=1-2]
\arrow[from=3-3, to=1-2]
\end{tikzcd}
\begin{tikzcd}
& {\underset{\bullet}{A}} \\
& {\overset{\bullet}{A}} & {\overset{\bullet}{A}} \\
{\overset{\bullet}{A}} & {\overset{\bullet}{A}} & {\overset{\bullet}{A}}
\arrow[from=3-2, to=2-2]
\arrow[from=2-2, to=1-2]
\arrow[from=2-3, to=1-2]
\arrow[from=3-3, to=2-3]
\arrow[from=3-1, to=2-2]
\end{tikzcd}.
\end{equation}

The above renormalizations are generated by $\{m_0,m_1,m_2\}$ as shown in Diagram (<ref>). Therefore, we obtain

\begin{gather}
\label{Eq.algunit} m_2(m_0\ot m_1)=m_1=\id_A= m_2(m_1\ot m_0),\\
m_2(m_1\ot m_2)=m_3= m_2(m_2\ot m_1),
\end{gather}

which means that $(A,m=m_2,\eta=m_0)$ is an unital associative algebra. Taking Hermitian conjugate one has that $(A,m^\dag,\eta^\dag)$ is a unital associative coalgebra. By definition $m_n$ are partial isometries, while unitality implies that $m$ is an epimorphism[$f_1m=f_2m\implies f_1m(\id_A\ot \eta)=f_2m(\id_A\ot \eta)\implies f_1=f_2$] and $\Ima m=A$, thus

\begin{equation}\label{eq.mmdag}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (0,2)node[below]{$m$} --(0,3)node[above]{$A$} ;
\draw (-1,1)node[left]{$A$} -- (0,2) -- (1,1)node[right]{$A$};
\draw (1,1)node[right]{$A$} -- (0,0)-- (-1,1)node[left]{$A$};
\draw (0,-1)node[below]{$A$} -- (0,0)node[above]{$m^\dag$} ;\end{tikzpicture}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (0,3)node[above]{$A$} -- (0,-1)node[below]{$A$} ;\end{tikzpicture},
\end{equation}

i.e., $mm^\dag=\id_A$. To show the Frobenius condition

\begin{equation}\label{eq.Fro}
\itk{
\draw (1,1)node[right]{$A$} -- (1,-2)node[right]{$A$};
\draw (-1,1)node[left]{$A$} -- (-1,-2)node[left]{$A$};
\draw (1,0)node[right]{$m$} -- (-1,-1)node[left]{$m^\dag$};}
\itk{
\draw (1,1)node[right]{$A$} -- (0,0)-- (-1,1)node[left]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$}  --(0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[left]{$A$} -- (0,-1) -- (1,-2)node[right]{$A$};}
\itk{
\draw (1,1)node[right]{$A$} -- (1,-2)node[right]{$A$};
\draw (-1,1)node[left]{$A$} -- (-1,-2)node[left]{$A$};
\draw (-1,0)node[left]{$m$} -- (1,-1)node[right]{$m^\dag$};},
\end{equation}

i.e., $ (\id_A\ot m)(m^\dag\ot \id_A)=m^\dag m=(m\ot \id_A)(\id_A \ot m^\dag)$, we introduce an auxiliary operator $\alpha=(\id_A\ot m)(m^\dag\ot \id_A)-m^\dag m$ and prove $\alpha=0$. Direct calculation shows that
\begin{equation} \alpha^\dag \alpha=(m\ot \id_A)(\id_A \ot m^\dag)(\id_A\ot m)(m^\dag\ot \id_A)-m^\dag m,  \end{equation}
\begin{equation}
0=(m\ot \id_A)(\id_A\ot \alpha^\dag\alpha)(m^\dag\ot\id_A)=((\id_A\ot \alpha)(m^\dag\ot\id_A))^\dag (\id_A\ot \alpha)(m^\dag\ot\id_A).\end{equation}

Then by positive definiteness we conclude $(\id_A\ot \alpha)(m^\dag\ot\id_A)=0$ and thus
\begin{equation} \alpha=(\eta^\dag\ot \id_{A\ot A})(\id_A\ot \alpha)(m^\dag\ot\id_A)=0. \end{equation}

For $n\geq 3$, $m_n$ is just the $n$-ary multiplication, generated by (any sequence of) the binary multiplication $m=m_2$:

\begin{align}
m_n&=m(m\ot\id_A)(m\ot\id_{A\ot A})\cdots(m\ot \id_{A^{\ot (n-2)}})
\nonumber\\
&=m(m\ot\id_A)(m\ot\id_{A\ot A})\cdots(\id_A\ot m\ot \id_{A^{\ot (n-3)}})
\nonumber\\&=\cdots
\nonumber\\&=m(\id_A\ot m)(\id_{A\ot A}\ot m)\cdots(\id_{A^{\ot (n-2)}}\ot m).
\end{align}
Therefore, for any $n\geq 1$, $m_n$ is an epimorphism and $\Ima m_n=A$. In particular, we have $m_n m_n^\dag=\id_A$.

$(\Z,A,P)$ is called a (translation invariant local) fixed-point lattice model with respect to a fixed-point local generator $\{m_{n}\}$ if any proper Hilbert space renormalization generated by $\{m_n\}$ is a system renormalization between $(\Z,A,P)$ and itself. Spelling it out (recall Definition <ref> and Corollary <ref>), $(\Z,A,P)$ is invariant under any renormalization $(f:\Z\to \Z,\{U_i=m_{|f^{-1}(i)|}\})$, where $f$ is surjective, $UU^\dag=\ot_i U_iU^\dag_i=\id$ and Eq. (<ref>) becomes

\begin{align}
U H U^\dag =H+\Delta E.
\end{align}

In the following a Frobenius algebra is always assumed to be isometric.

For readers familiar with the abstract notions, we have established a connection between the renormalization process in 1+1D and the operad theory. We believe that renormalization fixed-points always have certain (weakly) associative algebraic structures.

§.§ Commuting projector Hamiltonian and ground states

Given a Frobenius algebra $(A,m,\eta)$ in $\cC$,  $(\Z,A,-m^\dag m)$ is a fixed-point lattice model. Moreover, $m^\dag m$ is a projector and commutes with each other when acting on different sites, i.e., $(\Z,A,-m^\dag m)$ is a commuting projector fixed-point lattice model[The commuting-projector model is automatically gapped.].

From the Frobenius condition (<ref>) and the associativity of $m^\dag$, we have

\begin{equation}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (0,0) -- (-1,1) -- (-1,3)node[above]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$} -- (0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[below]{$A$} -- (0,-1) -- (1,-2)node[below]{$A$};
\draw (2,3)node[above]{$A$} -- (1,2) -- (0,3)node[above]{$A$};
\draw (1,1)node[below]{$m$} -- (1,1.5)node[right]{$A$} -- (1,2)node[above]{$m^\dag$} ;
\draw (0,0) -- (.5,.5) -- (1,1) -- (2,0) -- (2,-2)node[below]{$A$};
\end{tikzpicture}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (-1,-2)node[below]{$A$} --  (-1,3)node[above]{$A$};
\draw (1,-2)node[below]{$A$} --  (1,3)node[above]{$A$};
\draw (3,-2)node[below]{$A$} --  (3,3)node[above]{$A$};
\draw (-1,0)node[left]{$m$} -- (1,-1)node[right]{$m^\dag$};
\draw (3,2)node[right]{$m$} -- (1,1)node[left]{$m^\dag$};
\end{tikzpicture}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (-1,-2)node[below]{$A$} --  (-1,3)node[above]{$A$};
\draw (1,-2)node[below]{$A$} --  (1,3)node[above]{$A$};
\draw (3,-2)node[below]{$A$} --  (3,3)node[above]{$A$};
\draw (-1,2)node[left]{$m$} -- (1,1)node[right]{$m^\dag$};
\draw (3,0)node[right]{$m$} -- (1,-1)node[left]{$m^\dag$};
\end{tikzpicture}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (1,3)node[above]{$A$} -- (0,2) -- (-1,3)node[above]{$A$};
\draw (0,1)node[below]{$m$} -- (0,1.5)node[right]{$A$} -- (0,2)node[above]{$m^\dag$} ;
\draw (-1,-2)node[below]{$A$} -- (-1,0) -- (0,1) -- (1,0);
\draw (2,3)node[above]{$A$} -- (2,1) -- (1,0) -- (0,1);
\draw (1,-1)node[below]{$m$} -- (1,-.5)node[right]{$A$} -- (1,0)node[above]{$m^\dag$} ;
\draw (0,-2)node[below]{$A$} -- (.5,-1.5) -- (1,-1) -- (2,-2)node[below]{$A$};
\end{tikzpicture},
\end{equation}

i.e., $(\id_A\ot m^\dag m)(m^\dag m\ot\id_A)=(m^\dag m\ot\id_A)(\id_A\ot m^\dag m)$, which implies $m^\dag m$ on neighboring sites commutes. And from Eq. (<ref>), we have

\begin{equation}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (1,4)node[right]{$A$} -- (0,3)-- (-1,4)node[left]{$A$};
\draw (0,2)node[below]{$m$} -- (0,2.5)node[right]{$A$}  --(0,3)node[above]{$m^\dag$} ;
\draw (-1,1)node[left]{$A$} -- (0,2) -- (1,1)node[right]{$A$};
\draw (1,1)node[right]{$A$} -- (0,0)-- (-1,1)node[left]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$}  --(0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[left]{$A$} -- (0,-1) -- (1,-2)node[right]{$A$};
\end{tikzpicture}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (1,1)node[right]{$A$} -- (0,0)-- (-1,1)node[left]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$}  --(0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[left]{$A$} -- (0,-1) -- (1,-2)node[right]{$A$};
\end{tikzpicture},
\end{equation}

i.e., $m^\dag m m^\dag m=m^\dag m$, which implies that $m^\dag m$ is a projector.

Next, we show $(\Z,A,-m^\dag m)$ is a fixed-point model. Recall Definitions <ref> and <ref>. Let $j\in\Z$ be a site. Consider a special lattice renormalization

\begin{equation} f_{j,n} (i)=\begin{cases}
i, i<j,\\
j, j\leq i<j+n,\\
i-(n-1), i\geq j+n.
\end{cases} \end{equation}

$f^{-1}_{j,n} (j)=\{j,j+1,...,j+n-1\}$, $U_{j}=m_n$ and $U_{i}=\id_A,\ \forall i\neq j$. $(\Z,A,-m^\dag m)$ is a fixed-point model with respect to fixed-point local generators $\{m_n\}$ if for any $j\in\Z$ and $n\geq 1$ we have

\begin{equation}\label{proof.fixed}
m_n H m_n^\dag= H+\Delta E,
\end{equation}

where the Hamiltonian is $H=-\sum_i (m^\dag m)_i$.
We check for

* $n=1$. Obviously Eq. (<ref>) is satisfied for any $j\in\Z$ and $\Delta E=0$.

* $n=2$ and $m=m_2$. We compute each term $(m^\dag m)_i\ot\id$ in the Hamiltonian wrapped by $m_2, m_2^\dag$[We distinguish $m$ and $m_2$ here in order to emphasize $m^\dag m$ is a term in Hamiltonian while $m_2$ is a Hilbert space renormalization local generator.]. For term $i=j-1$, by Frobenius condition (<ref>),

\begin{equation}
m_2 (m^\dag m)_{j-1} m_2^\dag=
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (1,1)node[right]{$A$} -- (0,0)-- (-1,1)node[left]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$}  --(0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[left]{$A$} -- (0,-1) -- (1,-2)node[right]{$A$};
\draw (3,-2) -- (3,1);
\draw (1,-2)node[right]{$A$} -- (2,-3)node[above]{$m_2^\dag$}-- (3,-2)node[right]{$A$};
\draw (2,-3) -- (2,-4)node[right]{$A$};
\draw (1,1)node[right]{$A$} -- (2,2)node[below]{$m_2$}-- (3,1)node[right]{$A$};
\draw (2,2) -- (2,3)node[right]{$A$};
\end{tikzpicture}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (1,1)node[right]{$A$} -- (0,0)-- (-1,1)node[left]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$}  --(0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[left]{$A$} -- (0,-1) -- (1,-2)node[right]{$A$};
\end{tikzpicture}=(m^\dag m)_{j-1}.
\end{equation}

For term $i=j$,

\begin{equation}
m_2 (m^\dag m)_{j} m_2^\dag=
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (0,-3) -- (0,-4)node[right]{$A$};
\draw (1,-2)node[right]{$A$} -- (0,-3)node[above]{$m_2^\dag$}-- (-1,-2)node[left]{$A$};
\draw (0,2)node[below]{$m_2$} -- (0,3)node[right]{$A$};
\draw (-1,1)node[left]{$A$} -- (0,2) -- (1,1)node[right]{$A$};
\draw (1,1)node[right]{$A$} -- (0,0)-- (-1,1)node[left]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$}  --(0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[left]{$A$} -- (0,-1) -- (1,-2)node[right]{$A$};
\end{tikzpicture}=\quad
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (0,1)node[right]{$A$} -- (0,-2)node[right]{$A$};
\end{tikzpicture}=(\id_A)_{j}.
\end{equation}

For term $i=j+1$,

\begin{equation}
m_2 (m^\dag m)_{j+1} m_2^\dag=
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (1,1)node[right]{$A$} -- (0,0)-- (-1,1)node[left]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$}  --(0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[left]{$A$} -- (0,-1) -- (1,-2)node[right]{$A$};
\draw (-3,-2) -- (-3,1);
\draw (-1,-2)node[left]{$A$} -- (-2,-3)node[above]{$m_2^\dag$}-- (-3,-2)node[left]{$A$};
\draw (-2,-3) -- (-2,-4)node[left]{$A$};
\draw (-1,1)node[left]{$A$} -- (-2,2)node[below]{$m_2$}-- (-3,1)node[left]{$A$};
\draw (-2,2) -- (-2,3)node[left]{$A$};
\end{tikzpicture}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (1,1)node[right]{$A$} -- (0,0)-- (-1,1)node[left]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$}  --(0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[left]{$A$} -- (0,-1) -- (1,-2)node[right]{$A$};
\end{tikzpicture}=(m^\dag m)_{j}.
\end{equation}

All other terms obviously commute with $m_2$. Therefore, $m_2 H m_2^\dag=H-1$.

* $n\geq 3$. By similar calculations, we have $m_n H m_n^\dag=H-(n-1)$.

Based on this example we explain why in Definition <ref> we only require the fixed-point to be invariant under proper renormalizations. Consider the non-proper renormalization $m_0$, which adds an extra site (in the state of the unit of the algebra) to the model. We may try to include more interactions, such as $m_n^\dag m_n$. Straightforward calculation shows that the interaction $m^\dag_n m_n$ overlapping with this extra site gets renormalized to the interaction $m^\dag_{n-1} m_{n-1}$. Moreover, to preserve translation invariance, we necessarily needs to add extra sites by non-proper renormalizations in a translation invariant way. Such  fixed-point in a stronger sense is possible to be defined, but greatly complicates the analysis. We thus focus on the nearest neighbour interactions which is at fixed-point in a weaker sense (only invariant under proper renormalizations).

The ground state subspace (of total Hilbert space) of the commuting projector fixed-point lattice model $(\Z,A,-m^\dag m)$ (an infinite chain model with translation invariance) is $A$.

The Hamiltonian is $H=-\sum_i (m^\dag m)_i$, where $(m^\dag m)_i$ is supported on $\{i,i+1\}$ and has the form $(m^\dag m)_i=m^\dag m\ot\id
$ for $m^\dag m\in\End(A\ot A)$.
Since all terms of $H$ are commuting projections, the ground states of $H$ are given by the common eigenstates of $(m^\dag m)_i$ for all $i$ with eigenvalue $+1$, i.e., the ground state subspace is

\begin{align}
\{ |\psi\ra\in \cH|\forall i,(m^\dag m)_{i}|\psi\ra=|\psi\ra\}=\{ |\psi\ra\in \cH|\prod_i (m^\dag m)_{i}|\psi\ra=|\psi\ra\}
=\Ima \prod_i (m^\dag m)_{i},
\end{align}

where we note that the equation holds as eigenvalues of a projector can only be $0$ or $+1$. Graphically, from the Frobenius condition (<ref>), we have

\begin{equation}
\prod_i (m^\dag m)_{i}= \begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (0,0) -- (-1,1) -- (-1,3.5)node[left]{$...$} --(-1,5)node[above]{$A$};
\draw (0,-1)node[below]{$m$} -- (0,-0.5)node[right]{$A$} node[left]{$...\quad$} -- (0,0)node[above]{$m^\dag$} ;
\draw (-1,-2)node[below]{$A$} -- (0,-1) -- (1,-2)node[below]{$A$};
\draw (2,3) -- (1,2) -- (0,3) -- (0,5)node[above]{$A$};
\draw (1,1)node[below]{$m$} -- (1,1.5)node[right]{$A$} -- (1,2)node[above]{$m^\dag$} ;
\draw (0,0) -- (.5,.5) -- (1,1) -- (2,0) -- (2,-2)node[below]{$A$};
\draw (3,5)node[above]{$A$} -- (2,4) -- (1,5)node[above]{$A$};
\draw (2,3)node[below]{$m$} -- (2,3.5)node[right]{$A$} node[right]{$\quad ...$} -- (2,4)node[above]{$m^\dag$} ;
\draw (1,2) -- (1.5,2.5) -- (2,3) -- (3,2) --(3,-0.5)node[right]{$...$} -- (3,-2)node[below]{$A$};
\end{tikzpicture}
\begin{tikzpicture}[baseline=(current bounding box.center),>={Triangle[open,width=0pt 20]},scale=0.6]
\draw (-1,-2)node[below]{$A$} -- (0,-1) -- (1,-2)node[below]{$A$};
\draw (0,-1)node[below]{$m$} node[left]{$...\quad$} -- (1,0) -- (3,-2)node[below]{$A$};
\draw (1,0)node[below]{$m$} -- (2,1) --(4,-1)node[right]{$\quad ...$}-- (5,-2)node[below]{$A$};
\draw (2,1)node[below]{$m$} -- (2,1.5)node[right]{$A$} -- (2,2)node[above]{$m^\dag$};
\draw (3,3)node[above]{$m^\dag$} -- (2,2) -- (0,4)node[left]{$...\quad$}-- (-1,5)node[above]{$A$};
\draw (4,4)node[above]{$m^\dag$} node[right]{$\quad ...$} -- (3,3) -- (1,5)node[above]{$A$};
\draw (5,5)node[above]{$A$} -- (4,4) -- (3,5)node[above]{$A$};
\end{tikzpicture},
\end{equation}

where we see $\Ima \prod_i (m^\dag m)_{i}=A$ (we can check it satisfies Definition <ref>).

§.§ Fixed-point boundary conditions

Now we extend the above discussion to include boundaries of 1D lattice. Without losing generality, let's consider the special case that the lattice is $\BL=\N$.

A system $(\N, \cH_\BK, H)$ is called a translation invariant local system with boundary condition $(M,Q)$ if

* $\cH_0=M$, $\cH_i=A$ for all $i\geq 1$;

* $H=Q+\sum_{i\geq 1}P_i$ where $P_i$ supported in $\{ i,\dots,i+n-1\}$, $P_i=P\ot \id$ for some $P\in \End(A^{\ot n})$, and $0$ is in the finite support of $Q$.
We may simply denote this system by $(\N,A,P,M,Q)$.

Let $\{\rho_n:M\ot A^{\otimes n}\to M, m_{n}:A^{\otimes n}\to A\}_{n\in \N}$ be partial isometries. A Hilbert space renormalization $(f,\{U_i\})$, between the systems $(\N,A,P,M,Q)$ and $(\N,A,P',M,Q')$, is called (locally) generated by $\{\rho_n, m_{n}\}$ if for any $i\in\N$, $f^{-1}(i)$ is finite, $f^{-1}(0)$ is nonempty, $U_0=\rho_{|f^{-1}(0)|-1},$ and $U_i=m_{|f^{-1}(i)|}$ for $i\geq 1$.

Graphically, for example we sketch a Hilbert space renormalization generated by $\{\rho_1,m_2\}$ on the following local patch,

\begin{equation}
\itk{
\draw (0,0) node[below]{$M$} -- (0,1) node[left]{$\rho_1$} -- (0,2) node[above]{$M$};
\draw (1,0) node[below]{$A$} -- (0,1);
\draw (2,0) node[below]{$A$} -- (2.5,1) node[left]{$m_2$} -- (2.5,2) node[above]{$A$};
\draw (3,0) node[below]{$A$} -- (2.5,1);
\begin{tikzcd}
{} & \bullet & {} & \bullet & {} \\
\\
{} & \bullet & \bullet & \bullet & \bullet & {}
\arrow["{U_1}"', from=3-4, to=1-4]
\arrow[from=3-5, to=1-4]
\arrow["{U_0}"', from=3-2, to=1-2]
\arrow[draw=none, from=3-1, to=3-3]
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{A\otimes A}}"', draw=none, from=3-3, to=3-6]
\arrow[from=3-3, to=1-2]
\arrow["{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{M\otimes A}}"', draw=none, from=3-1, to=3-4]
\arrow["{\cH_0=M}"', draw=none, from=1-3, to=1-1]
\arrow["{\cH_1=A}", draw=none, from=1-3, to=1-5].
\end{tikzcd}
\end{equation}

We call $\{\rho_n, m_{n}\}$ a fixed-point local generator (with boundary), if
\begin{equation} \rho_0=\id_M,\quad m_1=\id_A \end{equation}
and for any two $(f,\{U_j\})$ and $(g,\{W_i\})$ generated by $\{\rho_n,m_{n}\}$, their composition
is still generated by $\{\rho_n, m_{n}\}$. $(\N,A,P,M,Q)$ is called a (translation invariant local) fixed-point lattice model with respect to a fixed-point local generator $\{\rho_n,m_{n}\}$ if any proper Hilbert space renormalization generated by $\{\rho_n, m_n\}$ is a system renormalization between $(\N,A,P,M,Q)$ and itself.

Given a fixed-point local generator $\{\rho_n, m_{n}\}$ and $(A,m:=m_2,\eta:=m_0)$ is a Frobenius algebra as before, then $(M,\rho:=\rho_1)$ is a right $A$-module (Definition <ref>).

Similar to the proof of Theorem <ref>, by composition of Hilbert space renormalization, we have $\rho(\rho\ot\id_A)=\rho(\id_M\ot m)$ and $\rho(\id_M\ot \eta)=\id_M$.