# Irreducible representations of Leavitt algebras

Roozbeh Hazrat  Hazrat: Centre for Research in Mathematics
Western Sydney University
Australia<EMAIL_ADDRESS>,  Raimund Preusser  Preusser:
Chebyshev Laboratory, St. Petersburg State University, Russia
<EMAIL_ADDRESS>and  Alexander Shchegolev  Shchegolev: Department of
Mathematics, St. Petersburg State University, Russia<EMAIL_ADDRESS>
###### Abstract.

For a weighted graph $E$, we construct representation graphs $F$, and
consequently, $L_{K}(E)$-modules $V_{F}$, where $L_{K}(E)$ is the Leavitt path
algebra associated to $E$, with coefficients in a field $K$. We characterise
representation graphs $F$ such that $V_{F}$ are simple $L_{K}(E)$-modules. We
show that the category of representation graphs of $E$,
$\operatorname{\mathbf{RG}}(E)$, is a disjoint union of subcategories, each of
which contains a unique universal object $T$ which gives an indecomposable
$L_{K}(E)$-module $V_{T}$ and a unique irreducible representation graph $S$,
which gives a simple $L_{K}(E)$-module $V_{S}$.

Specialising to graphs with one vertex and $m$ loops of weight $n$, we
construct irreducible representations for the celebrated Leavitt algebras
$L_{K}(n,m)$. On the other hand, specialising to graphs with weight one, we
recover the simple modules of Leavitt path algebras constructed by Chen via
infinite paths or sinks and give a large class of non-simple indecomposable
modules.

###### Key words and phrases:

Leavitt path algebra, Weighted Leavitt path algebra, Irreducible
representation

###### 2020 Mathematics Subject Classification:

16S88

###### Contents

  1. 1 Introduction
  2. 2 Representation graphs of a given graph
    1. 2.1 Covering and immersions
    2. 2.2 Representation graphs
    3. 2.3 The category of the representation graphs of a given graph
    4. 2.4 Quotients of representation graphs
    5. 2.5 The subcategories $\operatorname{\mathbf{RG}}(E)_{C}$
  3. 3 Weighted Leavitt path algebras and their representations
    1. 3.1 Representations of weighted Leavitt path algebras
    2. 3.2 Irreducible representations of weighted Leavitt path algebras
    3. 3.3 The fullness of the functor $V$
    4. 3.4 Indecomposability of $L_{K}(E)$-modules $V_{F}$
  4. 4 Irreducible representations of Leavitt path algebras
    1. 4.1 Chen simple modules via representation graphs
      1. 4.1.1 The case of $\mathcal{N}_{u}$, where $u$ is a sink
      2. 4.1.2 The case of $V_{[p]}$, where $[p]$ is irrational
      3. 4.1.3 The case of $V_{[p]}$, where $[p]$ is rational
  5. 5 Branching systems of weighted graphs and representations
    1. 5.1 Branching systems on an interval
    2. 5.2 Branching systems and representation graphs
    3. 5.3 Exceptional characteristic
  6. A Representation graphs for quivers with relations
    1. A.1 Irreducible representation graphs of a graph with relations

##  1\. Introduction

In a series of papers [21, 22, 23], William Leavitt studied algebras that are
now denoted by $L_{K}(n,n+k)$ and have been coined Leavitt algebras. Let
$X=(x_{ij})$ and $Y=(y_{ji})$ be $n\times(n+k)$ and $(n+k)\times n$ matrices
consisting of symbols $x_{ij}$ and $y_{ji}$, respectively. Then for a field
$K$, $L_{K}(n,n+k)$ is the unital $K$-algebra generated by all $x_{ij}$ and
$y_{ji}$ subject to the relations $XY=I_{n}$ and $YX=I_{n+k}$. Leavitt
established that these algebras are of type $(n,k)$. Recall that a ring $A$ is
of type $(n,k)$ if $n$ and $k$ are the least positive integers such that
$A^{n}\cong A^{n+k}$ as right $A$-modules. He further showed that
$L_{K}(1,k+1)$ are (purely infinite) simple rings and $L_{K}(n,n+k)$, $n\geq
2$ are domains (see also [11]).

A Leavitt path algebra, first introduced in [1, 8], is a certain quotient of
the path algebra of a directed graph. When the graph consists of one vertex,
$k+1$-loops, and their doubles, its Leavitt path algebra is $L_{K}(1,k+1)$.
The Leavitt path algebras $L_{K}(E)$ associated to the graphs $E$ with
coefficients in the field $K$ in general and the algebras $L_{K}(1,k+1)$ in
particular turned out to be a very rich and interesting class of algebras
whose studies so far have constitute over 150 research papers and counting. A
comprehensive treatment of the subject can be found in the book [2].

There have been a substantial number of papers devoted to (irreducible)
representations of a Leavitt path algebra $L_{K}(E)$, i.e., (simple)
$L_{K}(E)$-modules. Starting with a path algebra, the celebrated theorem of
Gabriel characterises graphs $E$ whose path algebras $KE$ have a finite number
of isomorphism classes of indecomposable representations. For Leavitt path
algebras, Ara and Brustenga [5, 6] studied their finitely presented modules,
proving that the category of finitely presented modules over a Leavitt path
algebra $L_{K}(E)$ is equivalent to a quotient category of the corresponding
category of modules over the path algebra $KE$. A similar statement for graded
modules over a Leavitt path algebra was established by Paul Smith [32],
allowing for making a bridge to study finite dimensional algebras with radical
square zero. On the other hand, by introducing the notion of branching systems
for a graph $E$, Gonçalves, and Royer [14] could construct representations for
$L_{K}(E)$, providing sufficient conditions when these representations are
faithful.

In [10], starting from a one-sided infinite path $p$ in a graph $E$, Chen
considered the $K$-vector space $V_{[p]}$ with basis given by the infinite
paths tail equivalent to $p$. Observing that this vector space is endowed with
a natural action of $L_{K}(E)$, he proved that $V_{[p]}$ is a simple
$L_{K}(E)$-module. A similar construction was also given for paths ending on a
sink vertex.

Numerous work followed, noteworthy the work of Ara-Rangaswamy and Rangaswamy
[30, 31, 9] producing new simple modules associated to infinite emitters and
characterising those algebras which have countably (finitely) many distinct
isomorphism classes of simple modules. Abrams, Mantese and Tonolo [3] studied
the projective resolutions for these simple modules. The recent work of Ánh
and Nam [4] provides another way to describe the so-called Chen and Rangaswamy
simple modules. We note that these simple modules can also be recovered via
the setting of Steinberg algebras (see [7, 25, 33]).

The algebras $L_{K}(n,n+k)$, for $n>1$, $k\geq 0$, can not be obtained via
Leavitt path algebras. For this reason, the weighted Leavitt path algebras
were introduced in [17] which give $L_{K}(n,n+k)$ for a weighted graph with
one vertex and $n+k$ loops of weight $n$ (Example 18). If the weights of all
the edges are $1$ (i.e., the graph is unweighted), then the weighted Leavitt
path algebras reduce to the usual Leavitt path algebras (Example 19).

In this note for a weighted graph $E$ we construct a representation graph $F$,
and consequently a representation $V_{F}$ for the weighted Leavitt path
algebra $L_{K}(E)$. We characterise the representation graphs $F$ such that
$V_{F}$ are irreducible representations, i.e, they are simple
$L_{K}(E)$-modules. A graph $F$ is a representation graph for $E$, if there is
a graph homomorphism $F\rightarrow E$ such that, roughly speaking, each edge
of $E$ ‘uniquely’ lifts to $F$ (Definition 2). Then $V_{F}$ is generated as a
$K$-vector space by vertices of $F$ as the basis and the action of $L_{K}(E)$
on each vertex $v$ is uniquely determined by moving $v$ on the graph $F$ along
the edges determined by the monomials of $L_{K}(E)$.

Specialising to graphs with one vertex and $m$ loops of weight $n$, we can
construct irreducible representations for the classical Leavitt algebras
$L_{K}(n,m)$. As an example, the algebra $L_{K}(2,3)$ can be obtained as a
Leavitt path algebra of a weighted graph with one vertex and two loops of
weight $3$:

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The representation graph $F$ below gives rise to a simple $L(2,3)$-module
$V_{F}$, as follows:

$\textstyle{F:\,\,}$$\textstyle{v_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{3}}$$\textstyle{v_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{e_{2}}$$\scriptstyle{e_{3}}$$\textstyle{v_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{3}}$$\textstyle{v_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{e_{2}}$$\scriptstyle{e_{3}}$$\textstyle{v_{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{3}}$$\textstyle{v_{5}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{e_{2}}$$\scriptstyle{e_{3}}$$\textstyle{v_{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{3}}$$\textstyle{v_{7}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\textstyle{v_{8}}$$\textstyle{v_{9}}$$\textstyle{\dots}$
(1)

Here $V_{F}$ is a $K$-vector space with basis
$\\{v_{i}\\}_{i\in\mathbb{N}_{0}}$ with the action of
$\mathtt{e_{i},e_{i}^{*}}\in L_{K}(2,3)$ on $v_{k}$ defined by
$v_{k}\mathtt{e_{i}}=r(e_{i})$ and $v_{k}\mathtt{e_{i}^{*}}=s(e_{i})$ and
similarly for $\mathtt{f_{i}}$’s. Here $\mathtt{e_{i}}$ and $v_{k}$, uniquely
determine an edge $e_{i}$ in $F$ with $s(e_{i})=v_{k}$ and thus
$v_{k}\mathtt{e_{i}}$ slides $v_{k}$ along the edge $e_{i}$ to $r(e_{i})$.
Therefore $v_{5}\mathtt{e_{3}}=v_{9}$ and $v_{6}\mathtt{f_{2}^{*}}=v_{4}$.

On the other hand, specialising to graphs with weight one, we recover simple
modules of Leavitt path algebras constructed by Chen via infinite paths. Our
approach gives a completely new way to represent simple modules of these
algebras. Besides being more visual, this approach allows for carrying
calculus on these modules with ease.

As an example, for the graph

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(2)

the infinite paths $p=\mathtt{efg\,efg\dots}$ and
$q=\mathtt{efef^{2}ef^{3}\dots}$, give rise to Chen simple $L_{K}(E)$-modules
$V_{[p]}$ and $V_{[q]}$. Using our approach, the representation graphs $F$ and
$G$ below give rise to simple $L_{K}(E)$-modules $V_{F}$ and $V_{G}$ such that
$V_{F}\cong V_{[p]}$ and $V_{G}\cong V_{[q]}$.

|
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(4)

Here $V_{F}$ is the $K$-vector space with basis $\\{v_{i}\mid
i\in\mathbb{N}\\}$ and the action of edges slides the vertices along the graph
$F$ such as $v_{1}\mathtt{efg}=v_{1}$ and $v_{9}\mathtt{egef^{*}}=v_{6}$.

We will next study the functor from the category of representation graphs of
the graph $E$ (see §2.3) to the category of (right) $L_{K}(E)$-modules,
arising from our construction:

$\displaystyle V:\operatorname{\mathbf{RG}}(E)$
$\displaystyle\longrightarrow\operatorname{Mod}L_{K}(E),$
$\displaystyle(F,\phi)$ $\displaystyle\longmapsto V_{F}.$

We show that $\operatorname{\mathbf{RG}}(E)$ can be written as a disjoint
union of certain subcategories, each of which contains a unique universal
representation and a unique irreducible representation of $E$, up to
isomorphism. We show that the unique universal representation $T$ of each of
these subcategories gives an indecomposable $L_{K}(E)$-module $V_{T}$, whereas
the irreducible representation $S$, gives a simple $L_{K}(E)$-module $V_{S}$.

Next in Section 5 we describe branching systems for weighted graphs and show
how the representation graphs give rise to examples of branching systems.
Branching systems for Leavitt path algebras were systematically studied by
Gonçalves and Royer ([14, 13, 12]).

As expounded by Green [16], one can describe the module category of a certain
class of quotient path algebras $A_{K}(E,r):=KE/\langle r\rangle$, where $KE$
is the path algebra of the finite graph $E$ with coefficients in the field $K$
and $r$ is a set of certain relations, via the following equivalence of
categories

$\operatorname{Mod}\,A_{K}(E,r)\longrightarrow\operatorname{Rep}(E,r).$ (5)

Here $\operatorname{Mod}\,A_{K}(E,r)$ is the category of right
$A_{K}(E,r)$-modules and $\operatorname{Rep}(E,r)$ is the category of
representations of the graph $E$ with relations as described in [16]. The
objects of the category $\operatorname{Rep}(E,r)$ consist of placing arbitrary
$K$-vector spaces on the vertices of the graph $E$ and assigning linear
transformations to the edges that satisfy the relations $r$ (see Appendix A).
Since (weighted) Leavitt path algebras are among this class, this gives a
justification of why branching systems would give representations for
(weighted) Leavitt path algebras. However it is not clear how this general
machinery of (5) can be used to systematically describe irreducible or
indecomposable representations of such algebras as the delicate case of
acyclic graphs with no relations, which gives the celebrated Gabriel theorem
of indecomposability, shows.

As such we believe that the notion of representation graphs of this paper
allows us, for the first time, to produce irreducible and indecomposable
representations for a wide class of algebras, such as Leavitt algebras
$L_{K}(n,m)$.

Throughout the paper $K$ denotes a field and $K^{\times}:=K\setminus\\{0\\}$.
By a $K$-algebra we mean an associative (but not necessarily commutative or
unital) $K$-algebra. The semigroup of positive integers is denoted by
$\mathbb{N}$ and the monoid of non-negative integers by $\mathbb{N}_{0}$.

##  2\. Representation graphs of a given graph

A _directed graph_ $E$ is a tuple $(E^{0},E^{1},r,s)$, where $E^{0}$ and
$E^{1}$ are sets and $r,s$ are maps from $E^{1}$ to $E^{0}$. We think of each
$e\in E^{1}$ as an edge pointing from vertex $s(e)$ to vertex $r(e)$. In this
paper all directed graphs are assumed to be row-finite, i.e. no vertex emits
infinitely many edges.

A weighted graph is a directed graph $E$ equipped with a map
$w:E^{1}\rightarrow\mathbb{N}$. If $e\in E^{1}$, then $w(e)$ is called the
weight of $e$. We write $(E,w)$ to emphasise that the graph is weighted.
Throughout we develop our concepts in the setting of weighted graphs. When the
weight map is the constant map $1$, i.e., $w(e)=1$ for any $e\in E^{1}$, we
retrieve the notions in the classical case of directed graphs.

The main notion introduced and studied in this paper in relation with the
theory of Leavitt path algebras, is the notion of a representation graph of a
given weighted graph. The concept is closely related to the theory of covering
and immersions in graph theory [34, 20]. We start by recalling these notions
in the setting of weighted graphs.

###  2.1. Covering and immersions

For weighted graphs $E$ and $G$, a _weighted graph homomorphism_
$\phi:E\rightarrow G$, consists of two maps $\phi^{0}:E^{0}\rightarrow G^{0}$
and $\phi^{1}:E^{1}\rightarrow G^{1}$ such that for any $e\in E^{1}$,
$s(\phi^{1}(e))=\phi^{0}(s(e))$, $r(\phi^{1}(e))=\phi^{0}(r(e))$ and
$w(\phi^{1}(e))=w(e)$. Namely, $\phi$ is a homomorphism of graphs which
preserve the weight. For a vertex $v\in E^{0}$, we define

$w(v):=\max\\{w(e)\mid e\in s^{-1}(v)\\}.$

If $v$ is a sink we set $w(v)=0$.

######  Definition 1.

Let $E$ and $T$ be weighted graphs and $\phi=(\phi^{0},\phi^{1}):T\rightarrow
E$ a homomorphism of weighted graphs.

  1. (1)

The pair $(T,\phi)$ is called an _immersion_ in $E$, if for any $v\in T^{0}$,
the map $\phi:s^{-1}(v)\rightarrow s^{-1}(\phi^{0}(v))$ is injective.

  2. (2)

The pair $(T,\phi)$ is called a _covering_ of $E$, if the following hold:

    1. (i)

The morphism $\phi$ is onto, i.e, $\phi^{0}$ and $\phi^{1}$ are surjective.

    2. (ii)

For any $v\in T^{0}$, the map $\phi^{1}:r^{-1}(v)\rightarrow
r^{-1}(\phi^{0}(v))$ and $\phi:s^{-1}(v)\rightarrow s^{-1}(\phi^{0}(v))$ are
bijective.

Putting it another way, a weighted graph immersion or covering is a classical
immersion or covering which preserves the weights.

Let $(E,w)$ be a weighted graph. The directed graph
$\hat{E}=(\hat{E}^{0},\hat{E}^{1},\hat{s},\hat{r})$, where
$\hat{E}^{0}=E^{0}$, $\hat{E}^{1}:=\\{e_{1},\dots,e_{w(e)}\mid e\in E^{1}\\}$,
$\hat{s}(e_{i})=s(e)$ and $\hat{r}(e_{i})=r(e)$, is called the directed graph
associated to $(E,w)$. If $e_{i}\in\hat{E}^{1}$, then
$\operatorname{tag}(e_{i}):=i$ is called the tag of $e_{i}$ and
$\operatorname{st}(e_{i}):=e$ is called the structure edge of $e_{i}$. There
is a forgetful homomorphism

$\displaystyle\hat{E}$ $\displaystyle\longrightarrow E,$ (6) $\displaystyle u$
$\displaystyle\longmapsto u$ $\displaystyle e_{i}$ $\displaystyle\longmapsto
e$

relating these two graphs.

It is easy to see that if $\phi:T\rightarrow E$ is an immersion or a covering
of weighted graphs, so is the graph homomorphism
$\hat{\phi}:\hat{T}\rightarrow\hat{E}$ defined by $\hat{\phi}(u)=\phi(u)$,
$u\in\hat{T}$, and $\hat{\phi}(e_{i}):=\phi(e)_{i}$, $1\leq i\leq w(e)$.

We use the convention that a (finite) path $p$ in a weighted graph $E$ is
either a single vertex $p=v\in E^{0}$ or a sequence $p=e_{1}e_{2}\cdots e_{n}$
of edges $e_{i}$ in $E$ such that $r(e_{i})=s(e_{i+1})$ for $1\leq i\leq n-1$.
We define $s(p)=s(e_{1})$, and $r(p)=r(e_{n})$. We denote by
$\operatorname{Path}(E)$ the set of all finite paths in $E$. Moreover, if
$u,v\in E^{0}$, then we denote by ${}_{u}\\!\operatorname{Path}(E)$ the set of
all finite paths starting in $u$, by $\operatorname{Path}_{v}(E)$ the set of
all finite paths ending in $v$ and by ${}_{u}\\!\operatorname{Path}_{v}(E)$
the intersection of ${}_{u}\\!\operatorname{Path}(E)$ and
$\operatorname{Path}_{v}(E)$.

Given a weighted graph $E$, we define the _double graph_ $E_{d}$ of $E$ as
weighted graph with $E_{d}^{0}=E^{0}$, $E_{d}^{1}=\\{e,e^{*}\mid e\in
E^{1}\\}$ with $w(e^{*})=w(e)$, where $e^{*}$ has orientation the reverse of
that of its counterpart $e\in E^{1}$ (see [2, p. 6]). Note that for a weighted
graph $E$, one can identify $(\hat{E})_{d}$ and $\widehat{(E_{d})}$.

A path $p=x_{1}\dots x_{n}\in\operatorname{Path}(E_{d})$ is called
_backtracking_ if there is a $1\leq j\leq n-1$ such that $x_{j}x_{j+1}=ee^{*}$
or $x_{j}x_{j+1}=e^{*}e$ for some $e\in E^{1}$. We say $p$ is _reduced_ if it
is not backtracking. We also use the following convention: when we say $p$ is
a _reduced path in $E$_, we mean $p$ is not backtracking path of $E_{d}$. This
is in line with literature in graph theory.

Let $E$ be a connected weighted graph. Fix a base point $v\in E^{0}$. The
_universal covering graph of $E$_ is a directed weighted graph $T=T(E,v)$
constructed as follows: let $T^{0}$ be the set of all reduced path in $E$
starting from $v$, $T^{1}=\\{(a,e)\in T^{0}\times E^{1}\mid r(a)=s(e)\\}$ and
put $s(a,e)=a$, $r(a,e)=ae$. Furthermore for $(a,e)\in T^{1}$ we set
$w(a,e)=w(e)$ and identify $(a,e)^{*}$ with $(ae,e^{*})$. If the weight of the
graph is $1$, we obtain the classical case of universal covering. Similar to
the classical case, one can show that $T$ is a tree and the isomorphism class
of $T=T(E,v)$ is independent of the choice of base point $v$ (see [20]). The
notion of universal covers allows us to easily construct representation graphs
for a given weighted graph (see Lemma 4 and Example 5). We note that if $T$ is
a universal cover of weighted graph $E$, $\hat{T}$ is not necessarily the
universal cover for $\hat{E}$.

For a connected (weighted) graph $E$, we denote by $\pi(E)$ its fundamental
group (which is independent of the choice of base-point). We define a _length
map_

$\displaystyle|\,|:\pi(\hat{E})$ $\displaystyle\longrightarrow\mathbb{Z}^{n},$
(7) $\displaystyle p$ $\displaystyle\longmapsto|p|$

where $n=\max\\{w(e)\mid e\in E^{1}\\}$, as follows: For $\\{e_{1},\dots
e_{w(e)}\mid e\in E^{1}\\}$ and $\\{e_{1}^{*},\dots e_{w(e)}^{*}\mid e\in
E^{1}\\}$ and $v\in E^{0}$, set $|v|=0$, $|e_{i}|=(0,\dots,0,1,0,\dots)$ and
$|e_{i}^{*}|=(0,\dots,0,-1,0,\dots)\in\mathbb{Z}^{n}$, where $1\leq i\leq
w(e)$ and $1$ and $-1$ are in the $i$-th component, respectively. One can
extend this to a well-defined map (7) by counting the length of the path $p$
which is a homomorphism of groups. Note that $|\pi(\hat{E})|=0$ if and only if
for paths $p,q$ in $\hat{E}$ with $s(p)=s(q)$ and $r(p)=r(q)$ we have
$|p|=|q|$. This will be used to give a criterion when a representation graph
gives rise to a graded representation (Theorem 25).

###  2.2. Representation graphs

We are in a position to define the main notion of this paper, namely a
representation graph of a given weighted graph $E$. Roughly, a graph $F$ is a
representation graph of the graph $E$ if “locally” $F$ covers all the
structure edges arriving at a vertex and all the tags emitting from a vertex
in $E$. For the next definition, recall that to a weighted graph $E$ one can
associate a directed graph $\hat{E}$ (see §2.1).

######  Definition 2.

Let $(E,w)$ be a weighted graph. A pair $(F,\phi)$, where
$F=(F^{0},F^{1},s_{F},r_{F})$ is a directed graph and
$\phi=(\phi^{0},\phi^{1}):F\rightarrow\hat{E}$ is a homomorphism of directed
graphs is called a representation graph of $E$, if the following hold:

  1. (1)

For any $v\in F^{0}$ and $1\leq i\leq w(\phi^{0}(v))$, there is precisely one
$f\in s_{F}^{-1}(v)$ such that $\operatorname{tag}(\phi^{1}(f))=~{}i$;

  2. (2)

For any $v\in F^{0}$ and $e\in r^{-1}(\phi^{0}(v))$, there is precisely one
$f\in r_{F}^{-1}(v)$ such that $\operatorname{st}(\phi^{1}(f))=e$.

In the definition above, using (6) we identify vertices of $\hat{E}$ with
those of $E$ when needed. Throughout the paper, we simply denote a
representation graph $(F,\phi)$ of $(E,w)$ by $F$ if there is no cause for
confusion in context. Some examples of representation graphs are given in
Introduction (1) and (3).

Let $E$ be a weighted graph, $\hat{E}$ the directed graph associated to $E$
and $(F,\phi)$ a representation graph for $E$. Let $\hat{E}_{d}$ and $F_{d}$
be the double graphs of $\hat{E}$ and $F$, respectively. Clearly the
homomorphism $\phi:F\rightarrow\hat{E}$ induces a map
$\operatorname{Path}(F_{d})\rightarrow\operatorname{Path}(\hat{E}_{d})$, which
we also denote by $\phi$. We call two vertices $u,v\in F^{0}$ in $F$ connected
if ${}_{u}\\!\operatorname{Path}_{v}(F_{d})\neq\emptyset$. A representation
graph $F$ is called connected if any $u,v\in F^{0}$ are connected. This is
equivalent to say that the graph $F$ is connected as an undirected graph. If
$C$ is a connected component of $F$, then $(C,\phi|_{C})$ is again a
representation graph for $E$.

######  Lemma 3.

Let $E$ be a weighted graph and $(F,\phi)$ a representation of $E$ with the
induced map
$\phi:\operatorname{Path}(F_{d})\rightarrow\operatorname{Path}(\hat{E}_{d})$.
Let $q,q^{\prime}\in\operatorname{Path}(F_{d})$ such that
$\phi(q)=\phi(q^{\prime})$. If $s(q)=s(q^{\prime})$ or $r(q)=r(q^{\prime})$,
then $q=q^{\prime}$.

###### Proof.

First suppose that $r(q)=r(q^{\prime})=v$. If one of the paths $q$ and
$q^{\prime}$ is trivial (i.e., is a vertex), then the other must also be
trivial and we have $q=v=q^{\prime}$ as desired. Assume now that $q$ and
$q^{\prime}$ are not trivial. Then $q=x_{n}\dots x_{1}$ and
$q^{\prime}=y_{n}\dots y_{1}$, for some $n\geq 1$ and
$x_{1},\dots,x_{n},y_{1},\dots,y_{n}\in F_{d}^{1}$. We proceed by induction on
$n$.

Case $n=1$: Suppose $\phi(x_{1})=\phi(y_{1})=e_{i}$ for some $e\in E^{1}$ and
$1\leq i\leq w(e)$. It follows from Definition 2(2) that $x_{1}=y_{1}$ and
hence $q=q^{\prime}$. Suppose now that $\phi(x_{1})=\phi(y_{1})=e_{i}^{*}$ for
some $e\in E^{1}$ and $1\leq i\leq w(e)$. Then it follows from Definition 2(1)
that $x_{1}=y_{1}$ and hence $q=q^{\prime}$.

Case $n\to n+1$: Suppose that $q=x_{n+1}\dots x_{1}$ and
$q^{\prime}=y_{n+1}\dots y_{1}$. By the inductive assumption we have
$x_{i}=y_{i}$ for any $1\leq i\leq n$. It follows that
$s_{F_{d}}(x_{n})=s_{F_{d}}(y_{n})=:u$. Hence
$x_{n+1},y_{n+1}\in\operatorname{Path}_{u}(F_{d})$. Now we can apply the case
$n=1$ and obtain $x_{n+1}=y_{n+1}$.

Now suppose that $s(q)=s(q^{\prime})$. Then $r(q^{*})=r((q^{\prime})^{*})$.
Since clearly
$\phi(q^{*})=\phi(q)^{*}=\phi(q^{\prime})^{*}=\phi((q^{\prime})^{*})$, we
obtain $q^{*}=(q^{\prime})^{*}$. Hence $q=q^{\prime}$. ∎

Since the notion of $\operatorname{Path}(F)$ of a graph $F$ plays a prominent
role, we remark that in the language of category Lemma 3 takes on a familiar
form. Recall that a functor $F:\mathcal{C}\rightarrow\mathcal{D}$ between two
small categories $\mathcal{C}$ and $\mathcal{D}$ is called _star injective_ ,
if the map

$F|_{{\rm St}(x)}:{\rm St}(x)\longrightarrow{\rm St}(F(x)),$

is injective, where

${\rm St}(x)=\\{f:x\rightarrow y\text{~{}a morphism
in~{}}\mathcal{C}\;|\;y\in\mathcal{C}\\},$ (8)

for every object $x\in\mathcal{C}$. Similarly $F$ is called _co-star
injective_ if $F|_{{\rm St}(x)^{\rm op}}$ is injective, where ${\rm
St}(x)^{\rm op}=\\{f:y\rightarrow x\text{~{}a morphism
in~{}}\mathcal{C}\;|\;y\in\mathcal{C}\\}$.

One can consider a graph $F$ as a category with vertices as objects and paths
as morphisms. Then the notion $\operatorname{Path}(F)$ represents the
morphisms of the category $F$. A graph homomorphism $\phi:F\rightarrow G$
gives rise to a functor, called $\phi$ again, $\phi:F\rightarrow G$ between
the categories $F$ and $G$. With this interpretation, Lemma 3 states that the
functor $\phi:F_{d}\rightarrow\hat{E}_{d}$ is star and co-star injective.

The next lemma shows that any representation for a covering graph induces a
representation for the graph as well. This allows us to construct many
representations for a given graph (see Example 5).

######  Lemma 4.

Let $E$ be a weighted graph and $T$ a covering of $E$. Then any representation
graph $F$ of $T$ is a representation graph for $E$. On the other hand, if $F$
is a representation of $E$ and $G$ is a covering of $F$ then $G$ is also a
representation of $E$.

###### Proof.

Let $\psi:T\rightarrow E$ be a covering map and $\phi:F\rightarrow\hat{T}$ a
representation for $T$. It is easy to see that there is a covering of
unweighted graphs $\hat{\psi}:\hat{T}\rightarrow\hat{E}$ which respects the
tags, i.e., if $\psi(t)=e$ then $\hat{\psi}(t_{i})=e_{i}$, for $1\leq i\leq
w(t)=w(e)$. Consider the following diagram, where $\sigma$ is the graph
homomorphism $\hat{\psi}\phi$.

$\textstyle{F\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\scriptstyle{\sigma}$$\textstyle{\hat{T}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\hat{\psi}}$$\textstyle{\hat{E}}$

We check that $(F,\sigma)$ is a representation graph for $E$.

Let $v\in F^{0}$, and $1\leq i\leq w(\sigma^{0}(v))$. Since $\phi$ is a
representation of $T$, by Definition 2(1), there is a unique $f\in
s_{F}^{-1}(v)$ such that $\operatorname{tag}(\phi^{1}(f))=i$. Since
$\hat{\psi}$ is a covering, there is a bijection $s^{-1}(\phi(v))\rightarrow
s^{-1}(\sigma(v))$, preserving tags. Thus $f$ is also the unique edge such
that $\operatorname{tag}(\sigma^{1}(f))=i$.

To see the second condition of a representation graph for the morphism
$\sigma$, let $v\in F^{0}$ and $e\in r^{-1}(\sigma^{0}(v))$. Since $\psi$ is a
covering, we have a bijection $\psi:r^{-1}(\phi(v))\rightarrow
r^{-1}(\sigma(v))$. Thus there is a unique $e^{\prime}\in r^{-1}(\phi(v))$
such that $\psi(e^{\prime})=e$. By Definition 2(2) for the representation
$\phi$, there is precisely one $f\in r_{F}^{-1}(v)$ such that
$\operatorname{st}(\phi^{1}(f))=e^{\prime}$ and consequently
$\operatorname{st}(\hat{\psi}\phi^{1}(f))=e$.

The second part of the lemma is easy and we leave it to the reader. ∎

######  Example 5.

Let $E$ be a weighted graph with one vertex, three loops of weight three.

$\textstyle{E:}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathtt{e_{1},e_{2},e_{3}}}$$\scriptstyle{\mathtt{f_{1},f_{2},f_{3}}}$$\scriptstyle{\mathtt{g_{1},g_{2},g_{3}}}$

The universal cover of the weighted graph $E$ resembles the Cayley graph of
the free group $\mathbb{F}_{3}$ with three generators (note that in
comparison, the universal cover of $\hat{E}$ is the Cayley graph of the free
group $\mathbb{F}_{9}$). Below we show the part of the universal cover $T$
generated by $\mathtt{e},\mathtt{f}$ and $\mathtt{g}$. For each node, choosing
$e_{1},f_{2},g_{3}$ to emit and the same arrows to arrive at the node, they
satisfy Conditions (1) and (2) of Definition 2 of the representation graphs
and thus give a representation graph $F$ for $T$ and consequently one for $E$.

$\mathtt{e}$$\mathtt{e}$$\mathtt{f}$$\mathtt{g}$$\mathtt{f}$$\mathtt{f}$$\mathtt{g}$$\mathtt{e}$$\mathtt{g}$$\mathtt{g}$$\mathtt{f}$$\mathtt{e}$

$f_{2}$$e_{1}$$g_{3}$•$f_{2}$$e_{1}$$g_{3}$•

We will see in Section 3 that any representation graph such as $F$ in Example
5 will give a module $V_{F}$ for the weighted Leavitt path algebra $L_{K}(E)$.
However $V_{F}$ in this example is not a simple module. Next we characterise
those representations that the associated module is simple.

######  Definition 6.

Let $E$ be a weighted graph and $(F,\phi)$ a representation graph for $E$ with
the induced map
$\phi:\operatorname{Path}(F_{d})\rightarrow\operatorname{Path}(\hat{E}_{d})$.
We say $F$ is an irreducible representation for $E$ if the following
equivalent conditions hold.

  1. (1)

The graph $F$ is connected and

$\phi(\operatorname{Path}_{u}(F_{d}))\neq\phi(\operatorname{Path}_{v}(F_{d})),\text{
for any }u\neq v\in F^{0}.$

  2. (2)

The graph $F$ is connected and

$\phi({}_{u}\\!\operatorname{Path}(F_{d}))\neq\phi({}_{v}\\!\operatorname{Path}(F_{d})),\text{
for any }u\neq v\in F^{0}.$

Note that the condition of irreducibility in Definition 6 in the categorical
language (see (8)) amounts to saying that for $u\not=v\in F^{0}$, $\phi({\rm
St}(u))\not=\phi({\rm St}(v))$.

######  Example 7.

Consider the representation graph $H$ below of the graph $E$ with one vertex
and three loops (2). One can check that the conditions of Definition 6 are not
satisfied and therefore $H$ is not an irreducible representation for $E$. On
the other hand the representation graphs (3) and (4), respectively, are
irreducible.

|
---|---
|  |  |
$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H:\,\,\,\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}e}}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}e}}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}e}}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}e}}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}e}}$$\scriptstyle{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}e}}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$
(9)

###  2.3. The category of the representation graphs of a given graph

Let $E$ be a weighted graph. Denote by $\operatorname{\mathbf{RG}}(E)$ the
category of nonempty, connected representation graphs for $E$. A morphism
$\alpha:(F,\phi)\to(G,\psi)$ in $\operatorname{\mathbf{RG}}(E)$ is a
homomorphism $\alpha:F\to G$ of directed graphs such that $\psi\alpha=\phi$.
One checks easily that a morphism $\alpha:(F,\phi)\to(G,\psi)$ is an
isomorphism if and only if $\alpha^{0}$ and $\alpha^{1}$ are bijective. Lemma
4 states that if $T\rightarrow E$ is a covering of the weighted graph $E$,
then we have a natural functor
$\operatorname{\mathbf{RG}}(T)\rightarrow\operatorname{\mathbf{RG}}(E)$.

In this section we show that the category $\operatorname{\mathbf{RG}}(E)$ is a
disjoint union of certain subcategories, each of which contains a unique
universal representation and a unique irreducible representation of $E$. In §3
we will show that the universal representations of $E$ give us indecomposable
$L_{K}(E)$-modules, whereas irreducible representations of $E$ give rise to
simple $L_{K}(E)$-modules.

We need to establish several lemmas which allows us to define equivalence
relations on representations.

######  Lemma 8.

Let $(F,\phi)$ and $(G,\psi)$ be objects in $\operatorname{\mathbf{RG}}(E)$.
Let $u\in F^{0}$ and $v\in G^{0}$. If
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))\subseteq\psi({}_{v}\\!\operatorname{Path}(G_{d}))$,
then
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{v}\\!\operatorname{Path}(G_{d}))$.

###### Proof.

Suppose that
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))\subseteq\psi({}_{v}\\!\operatorname{Path}(G_{d}))$.
It follows that $\phi^{0}(u)=\psi^{0}(v)$ since
$u\in{}_{u}\\!\operatorname{Path}(F_{d})$. We have to show that
$\psi({}_{v}\\!\operatorname{Path}(G_{d}))\subseteq\phi({}_{u}\\!\operatorname{Path}(F_{d}))$.
Let $p\in{}_{v}\\!\operatorname{Path}(G_{d})$. If $p=v$, then
$\psi(p)=\psi^{0}(v)=\phi^{0}(u)\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$.
Suppose now that $p$ is nontrivial. Then $p=y_{1}\dots y_{n}$ for some
$y_{1},\dots,y_{n}\in G_{d}^{1}$ where $n\geq 1$. We proceed by induction on
$n$.

Case $n=1$: Suppose that $p=g$ for some $g\in G^{1}$. Then $\psi^{1}(g)=e_{i}$
for some $e\in s^{-1}(\psi^{0}(v))$ and $1\leq i\leq w(e)$. Since $(F,\phi)$
satisfies Condition (1) in Definition 2, there is precisely one $f\in
s_{F}^{-1}(u)$ such that $\operatorname{tag}(\phi^{1}(f))=i$. Hence
$\phi^{1}(f)=e^{\prime}_{i}$ for some $e^{\prime}\in s^{-1}(\phi^{0}(u))$.
Assume that $e\neq e^{\prime}$. Since
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))\subseteq\psi({}_{v}\\!\operatorname{Path}(G_{d}))$,
there is a $g^{\prime}\in s_{G}^{-1}(v)$ such that
$\psi^{1}(g^{\prime})=\phi^{1}(f)=e^{\prime}_{i}$. But this is absurd since
$(G,\psi)$ satisfies Condition (1) in Definition 2. Thus
$\psi(g)=e_{i}=\phi(f)\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$.

Suppose now that $p=g^{*}$ for some $g\in G^{1}$. Then $\psi^{1}(g)=e_{i}$ for
some $e\in r^{-1}(\psi^{0}(v))$ and $1\leq i\leq w(e)$. Since $(F,\phi)$
satisfies Condition (2) in Definition 2, there is precisely one $f\in
r_{F}^{-1}(u)$ such that $\operatorname{st}(\phi^{1}(f))=e$. Hence
$\phi^{1}(f)=e_{j}$ for some $1\leq j\leq w(e)$. Assume that $i\neq j$. Since
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))\subseteq\psi({}_{v}\\!\operatorname{Path}(G_{d}))$,
there is a $g^{\prime}\in r_{G}^{-1}(v)$ such that
$\psi^{1}(g^{\prime})=\phi^{1}(f)=e_{j}$. But this is absurd since $(G,\psi)$
satisfies Condition (2) in Definition 2. Thus
$\psi(g^{*})=e^{*}_{i}=\phi(f^{*})\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$.

Case $n\rightarrow n+1$: Suppose $p=y_{1}\dots y_{n}y_{n+1}$. By the induction
assumption we know that $\psi(y_{1}\dots
y_{n})\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$. Hence $\psi(y_{1}\dots
y_{n})=\phi(x_{1}\dots x_{n})$ for some path $x_{1}\dots
x_{n}\in{}_{u}\\!\operatorname{Path}(F_{d})$. Set
$u^{\prime}:=r_{F_{d}}(x_{n})$ and $v^{\prime}:=r_{G_{d}}(y_{n})$. Clearly
$\phi^{0}(u^{\prime})=\psi^{0}(v^{\prime})$ since $\phi$ and $\psi$ are graph
homomorphisms.
Suppose that $y_{n+1}=g$ for some $g\in G^{1}$. Then $\psi^{1}(g)=e_{i}$ for
some $e\in s^{-1}(\psi^{0}(v^{\prime}))$ and $1\leq i\leq w(e)$. Since
$(F,\phi)$ satisfies Condition (1) in Definition 2, there is precisely one
$f\in s_{F}^{-1}(u^{\prime})$ such that $\operatorname{tag}(\phi^{1}(f))=i$.
Hence $\phi^{1}(f)=e^{\prime}_{i}$ for some $e^{\prime}\in
s^{-1}(\phi^{0}(u^{\prime}))$. Assume that $e\neq e^{\prime}$. Since
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))\subseteq\psi({}_{v}\\!\operatorname{Path}(G_{d}))$,
there is a $g^{\prime}\in s_{G}^{-1}(v^{\prime})$ such that $\phi(x_{1}\dots
x_{n}f)=\psi(y_{1}\dots y_{n}g^{\prime})$, which implies
$\psi^{1}(g^{\prime})=\phi^{1}(f)=e^{\prime}_{i}$. But this is absurd since
$(G,\psi)$ satisfies Condition (1) in Definition 2. Thus
$\psi(g)=e_{i}=\phi(f)$ and hence $\psi(y_{1}\dots y_{n}g)=\phi(x_{1}\dots
x_{n}f)\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$.

Suppose now that $y_{n+1}=g^{*}$ for some $g\in G^{1}$. Then
$\psi^{1}(g)=e_{i}$ for some $e\in r^{-1}(\psi^{0}(v^{\prime}))$ and $1\leq
i\leq w(e)$. Since $(F,\phi)$ satisfies Condition (1) in Definition 2, there
is precisely one $f\in r_{F}^{-1}(u^{\prime})$ such that
$\operatorname{st}(\phi^{1}(f))=e$. Hence $\phi^{1}(f)=e_{j}$ for some $1\leq
j\leq w(e)$. Assume that $i\neq j$. Since
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))\subseteq\psi({}_{v}\\!\operatorname{Path}(G_{d}))$,
there is a $g^{\prime}\in r_{G}^{-1}(v^{\prime})$ such that $\phi(x_{1}\dots
x_{n}f^{*})=\psi(y_{1}\dots y_{n}(g^{\prime})^{*})$, which implies
$\psi^{1}(g^{\prime})=\phi^{1}(f)=e_{j}$. But this is absurd since $(G,\psi)$
satisfies Condition (2) in Definition 2. Thus
$\psi(g^{*})=e^{*}_{i}=\phi(f^{*})$ and hence $\psi(y_{1}\dots
y_{n}g^{*})=\phi(x_{1}\dots
x_{n}f^{*})\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$. ∎

######  Proposition 9.

Let $\alpha:(F,\phi)\to(G,\psi)$ be a morphism in
$\operatorname{\mathbf{RG}}(E)$. If $u\in F^{0}$, then
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{\alpha^{0}(u)}\\!\operatorname{Path}(G_{d}))$.

###### Proof.

Let $u\in F^{0}$ and $p\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$. Then
there is a $q\in{}_{u}\\!\operatorname{Path}(F_{d})$ such that $\phi(q)=p$.
Clearly $\alpha(q)\in{}_{\alpha^{0}(u)}\\!\operatorname{Path}(G_{d})$. Hence
$p=\phi(q)=\psi(\alpha(q))\in\psi({}_{\alpha^{0}(u)}\\!\operatorname{Path}(G_{d}))$.
We have shown that
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))\subseteq\psi({}_{\alpha^{0}(u)}\\!\operatorname{Path}(G_{d}))$.
It follows from Lemma 8 that
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{\alpha^{0}(u)}\\!\operatorname{Path}(G_{d}))$.
∎

######  Proposition 10.

Let $\alpha:(F,\phi)\to(G,\psi)$ be a morphism in
$\operatorname{\mathbf{RG}}(E)$. Then $\alpha$ is a covering map.

###### Proof.

We first show that $\alpha^{0}$ is surjective. Choose a $u\in F^{0}$. If $v\in
G^{0}$, then there is a path
$q\in_{\alpha^{0}(u)}\\!\operatorname{Path}_{v}(G_{d})$ since $G$ is
connected. By Proposition 9 we have
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{\alpha^{0}(u)}\\!\operatorname{Path}(G_{d}))$.
Hence there is a $p\in{}_{u}\\!\operatorname{Path}(F_{d})$ such that
$\phi(p)=\psi(q)$. Since $\phi=\psi\alpha$, we obtain
$\psi(\alpha(p))=\psi(q)$. It follows from Lemma 3 that $\alpha(p)=q$. Thus
$\alpha^{0}(r(p))=r(\alpha(p))=r(q)=v$.

We next show that $\alpha^{1}$ is surjective. Suppose $g\in G^{1}$ with
$v:=s(g)$, $w:=s(\psi(g))$ and $i:=\operatorname{tag}(\psi(g))$. Since
$\alpha^{0}$ is surjective, there is a $u\in F^{0}$ such that
$\alpha^{0}(u)=v$. Since $\phi(u)=\psi(\alpha(u))=w$, by the definition of a
representation graph, there is a unique edge $f\in F^{1}$ with $s(f)=u$ such
that $\operatorname{tag}(\phi(f))=i$. Since $s(\alpha(f))=s(g)$ and
$\operatorname{tag}(\psi(g))=\operatorname{tag}(\phi(f))=\operatorname{tag}(\psi(\alpha(f)))=i$,
it follows that $\alpha(f)=g$. This shows that $\alpha^{1}$ is surjective.

Next we show that that for any $u\in F^{0}$, the map
$\alpha^{1}:s^{-1}(u)\rightarrow s^{-1}(\alpha^{0}(u))$ is injective. Let
$f_{1},f_{2}\in F^{1}$ are distinct, with $s(f_{1})=s(f_{2})=u$. By the
definition of the representation,
$\operatorname{tag}(\phi(f_{1}))\not=\operatorname{tag}(\phi(f_{2}))$ and thus
$\operatorname{tag}(\psi(\alpha(f_{1})))\not=\operatorname{tag}(\psi(\alpha(f_{1})))$.
Hence $\alpha(f_{1})\not=\alpha(f_{2})$.

A similar argument shows that for any $u\in F^{0}$, the map
$\alpha^{1}:r^{-1}(u)\rightarrow r^{-1}(\alpha^{0}(u))$ is bijective. ∎

###  2.4. Quotients of representation graphs

For any object $(F,\phi)$ in $\operatorname{\mathbf{RG}}(E)$ we define an
equivalence relation $\sim$ on $F^{0}$ by $u\sim v$ if
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\phi({}_{v}\\!\operatorname{Path}(F_{d}))$.
Recall that if $\sim$ and $\approx$ are equivalence relations on a set $X$,
then one writes $\approx~{}\leq~{}\sim$ and calls $\approx$ finer than $\sim$
(and $\sim$ coarser than $\approx$) if $x\approx y$ implies that $x\sim y$,
for any $x,y\in X$.

######  Definition 11.

Let $(F,\phi)\in\operatorname{\mathbf{RG}}(E)$ be a representation graph of
$E$. An equivalence relation $\approx$ on $F^{0}$ is called admissible if the
following hold:

  1. (1)

$\approx~{}\leq~{}\sim$.

  2. (2)

If $u\approx v$, $p\in{}_{u}\\!\operatorname{Path}_{x}(F_{d})$,
$q\in{}_{v}\\!\operatorname{Path}_{y}(F_{d})$ and $\phi(p)=\phi(q)$, then
$x\approx y$.

The admissible equivalence relations on $F^{0}$ (with partial order $\leq$)
form a bounded lattice whose maximal element is $\sim$ and whose minimal
element is the equality relation $=$.

Let $(F,\phi)\in\operatorname{\mathbf{RG}}(E)$ be a representation graph of
$E$ and $\approx$ an admissible equivalence relation on $F^{0}$. If $f,g\in
F^{1}$ we write $f\approx g$ if $s(f)\approx s(g)$ and $\phi(f)=\phi(g)$. This
defines an equivalence relation on $F^{1}$. Define a representation graph
$(F_{\approx},\phi_{\approx})$ for $E$ by

$\displaystyle F_{\approx}^{0}$ $\displaystyle=F^{0}/\approx,$ $\displaystyle
F_{\approx}^{1}$ $\displaystyle=F^{1}/\approx,$ $\displaystyle s([f])$
$\displaystyle=[s(f)],$ $\displaystyle r([f])$ $\displaystyle=[r(f)],$
$\displaystyle\phi_{\approx}^{0}([v])$ $\displaystyle=\phi^{0}(v),$
$\displaystyle\phi_{\approx}^{1}([f])$ $\displaystyle=\phi^{1}(f).$

We call $(F_{\approx},\phi_{\approx})$ a quotient of $(F,\phi)$. It is easy to
check that $(F_{\approx},\phi_{\approx})$ is a representation graph of $E$.

######  Theorem 12.

Let $(F,\phi)$ and $(G,\psi)$ be objects in $\operatorname{\mathbf{RG}}(E)$.
Then there is a morphism $\alpha:(F,\phi)\to(G,\psi)$ if and only if
$(G,\psi)$ is isomorphic to a quotient of $(F,\phi)$.

###### Proof.

$(\Rightarrow)$ Suppose there is a morphism $\alpha:(F,\phi)\to(G,\psi)$. If
$u,v\in F^{0}$, we write $u\approx v$ if $\alpha^{0}(u)=\alpha^{0}(v)$.
Clearly $\approx$ defines an equivalence relation on $F^{0}$. Below we check
that $\approx$ is admissible.

  1. (i)

Suppose $u\approx v$. Then
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{\alpha^{0}(u)}\\!\operatorname{Path}(G_{d}))=\psi({}_{\alpha^{0}(v)}\\!\operatorname{Path}(G_{d}))=\phi({}_{v}\\!\operatorname{Path}(F_{d}))$
by Proposition 9. Hence $u\sim v$.

  2. (ii)

Suppose $u\approx v$, $p\in{}_{u}\\!\operatorname{Path}_{x}(F_{d})$,
$q\in{}_{v}\\!\operatorname{Path}_{y}(F_{d})$ and $\phi(p)=\phi(q)$. Clearly
$\alpha(p)\in{}_{\alpha^{0}(u)}\\!\operatorname{Path}_{\alpha^{0}(x)}$
$(G_{d})$ and
$\alpha(q)\in{}_{\alpha^{0}(v)}\\!\operatorname{Path}_{\alpha^{0}(y)}(G_{d})$.
Moreover, $\psi(\alpha(p))=\phi(p)=\phi(q)=\psi(\alpha(q))$. Since
$\alpha^{0}(u)=\alpha^{0}(v)$, it follows from Lemma 3 that
$\alpha(p)=\alpha(q)$. Hence
$\alpha^{0}(x)=r(\alpha(p))=r(\alpha(q))=\alpha^{0}(y)$ and therefore
$x\approx y$.

Note that by Lemma 3 we have $f\approx g$ if and only if
$\alpha^{1}(f)=\alpha^{1}(g)$, for any $f,g\in F^{1}$. Define a graph
homomorphism $\beta:F_{\approx}\to G$ by $\beta^{0}([v])=\alpha^{0}(v)$ and
$\beta^{1}([f])=\alpha^{1}(f)$. Clearly $\psi\beta=\phi_{\approx}$ and
therefore $\beta:(F_{\approx},\phi_{\approx})\to(G,\psi)$ is a morphism. In
view of Proposition 10, $\beta^{0}$ and $\beta^{1}$ are bijective and hence
$\beta$ is an isomorphism.

$(\Leftarrow)$ Suppose now that $(G,\psi)\cong(F_{\approx},\phi_{\approx})$
for some admissible equivalence relation $\approx$ on $F^{0}$. In order to
show that there is a morphism $\alpha:(F,\phi)\to(G,\psi)$ it suffices to show
that there is a morphism $\beta:(F,\phi)\to(F_{\approx},\phi_{\approx})$. But
this is obvious (define $\beta^{0}(v)=[v]$ and $\beta^{1}(f)=[f]$). ∎

###  2.5. The subcategories $\operatorname{\mathbf{RG}}(E)_{C}$

Let $(F,\phi)$ and $(G,\psi)$ be objects in $\operatorname{\mathbf{RG}}(E)$.
We write $(F,\phi)\leftrightharpoons(G,\psi)$ if there is a $u\in F^{0}$ and a
$v\in G^{0}$ such that
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{v}\\!\operatorname{Path}(G_{d}))$.
One checks easily that $\leftrightharpoons$ defines an equivalence relation on
$\operatorname{Ob}(\operatorname{\mathbf{RG}}(E))$. If $C$ is an
$\leftrightharpoons$-equivalence class, then we denote by
$\operatorname{\mathbf{RG}}(E)_{C}$ the full subcategory of
$\operatorname{\mathbf{RG}}(E)$ such that
$\operatorname{Ob}(\operatorname{\mathbf{RG}}(E)_{C})=C$. If
$\alpha:(F,\phi)\to(G,\psi)$ is a morphism in $\operatorname{\mathbf{RG}}(E)$,
then $(F,\phi)\leftrightharpoons(G,\psi)$ by Proposition 9. Thus
$\operatorname{\mathbf{RG}}(E)$ is the disjoint union of the subcategories
$\operatorname{\mathbf{RG}}(E)_{C}$, where $C$ ranges over all
$\leftrightharpoons$-equivalence classes.

Fix an $\leftrightharpoons$-equivalence class $C$, a representation graph
$(F,\phi)\in C$ and a vertex $u\in F^{0}$. We denote by
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}$ the set of all
paths in $\phi({}_{u}\\!\operatorname{Path}(F_{d}))$ that are reduced (see
§2.1). Define a representation graph $(T,\xi)=(T_{C},\xi_{C})$ for $E$ by

$\displaystyle T^{0}$ $\displaystyle=\\{v_{p}\mid
p\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}\\},$
$\displaystyle T^{1}$ $\displaystyle=\\{e_{p}\mid
p\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}},p\neq\phi(u)\\},$
$\displaystyle s(e_{x_{1}\dots x_{n}})$
$\displaystyle=\begin{cases}v_{x_{1}\dots x_{n-1}},&\text{ if
}x_{n}\in\hat{E}^{1},\\\ v_{x_{1}\dots x_{n}},&\text{ if
}x_{n}\in(\hat{E}^{1})^{*},\end{cases}$ $\displaystyle r(e_{x_{1}\dots
x_{n}})$ $\displaystyle=\begin{cases}v_{x_{1}\dots x_{n}},&\text{ if
}x_{n}\in\hat{E}^{1},\\\ v_{x_{1}\dots x_{n-1}},&\text{ if
}x_{n}\in(\hat{E}^{1})^{*},\end{cases}$ $\displaystyle\xi^{0}(v_{x_{1}\dots
x_{n}})$ $\displaystyle=\begin{cases}\phi(u),&\text{ if }n=1\text{ and
}x_{1}=\phi(u),\\\ r_{\hat{E}}(x_{n}),&\text{ if }x_{n}\in\hat{E}^{1},\\\
s_{\hat{E}}(x_{n}^{*}),&\text{ if }x_{n}\in(\hat{E}^{1})^{*},\end{cases}$
$\displaystyle\xi^{1}(e_{x_{1}\dots x_{n}})$
$\displaystyle=\begin{cases}x_{n},&\text{ if }x_{n}\in\hat{E}^{1},\\\
x_{n}^{*},&\text{ if }x_{n}\in(\hat{E}^{1})^{*}.\end{cases}$

Here we use the convention that if $x_{1}\dots
x_{n}\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}$, where
$n=1$, then $x_{1}\dots x_{n-1}=\phi(u)$. Clearly $T$ is nonempty and
connected.

######  Example 13.

Suppose $E$ is a weighted graph with one vertex and two loops of weight $2$:

$E:\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
37.25858pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{}{{}{{}{{}{{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}}}}}}}}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{}{{}{{}}}\ignorespaces\ignorespaces{\hbox{\kern-37.25858pt\raise
0.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{\mathtt{e_{1},e_{2}}}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}{}{{}}}}}}}{\hbox{\kern-5.60223pt\raise
4.70912pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{}{{}{{}{{}{{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}}}}}}}}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{}{{}{{}}}\ignorespaces\ignorespaces{\hbox{\kern
19.0919pt\raise 0.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{\mathtt{f_{1},f_{2}}}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}{}{{}}}}}}}{\hbox{\kern
5.60223pt\raise 4.70912pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise 0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}{\hbox{\kern
29.603pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}\ignorespaces}}}}\ignorespaces$

and $(F,\phi)$ is a representation graph for $E$, where $F$ is given by

$F:\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
28.68182pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{u\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{}{{}{{}{{}{{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}}}}}}}}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{}{{}{{}}}\ignorespaces\ignorespaces{\hbox{\kern-28.68182pt\raise
0.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern 0.0pt\raise-0.8264pt\hbox{$\scriptstyle{g}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}}{}{{}{{}}{}{{}{{}}{}{{}{{}{{}}}}}}}}{\hbox{\kern-5.85959pt\raise
4.87825pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{}{{}{{}{{}{{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}}}}}}}}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{}{{}{{}}}\ignorespaces\ignorespaces{\hbox{\kern
19.0919pt\raise 0.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern 0.0pt\raise-2.43056pt\hbox{$\scriptstyle{h}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}}{}{{}{{}}{}{{}{{}}{}{{}{{}{{}}}}}}}}{\hbox{\kern
5.85959pt\raise 4.87825pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise 0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}{\hbox{\kern
29.86229pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}\ignorespaces}}}}\ignorespaces$

and $\phi:F\to\hat{E}$ is defined by $\phi^{0}(u)=v$, $\phi^{1}(g)=e_{1}$,
$\phi^{1}(h)=f_{2}$. Then
${}_{u}\\!\operatorname{Path}(F_{d})=\operatorname{Path}(F_{d})$ and
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}$ consists of
all nonbacktracking paths in $\operatorname{Path}(\hat{E}_{d})$ whose letters
come from the alphabet $\\{v,e_{1},e_{1}^{*},f_{2},f_{2}^{*}\\}$. Let $C$ be
the $\leftrightharpoons$-equivalence class of $(F,\phi)$. Then $T_{C}$ is the
graph

$\textstyle{v_{v}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{e_{1}}}$$\scriptstyle{e_{f_{2}}}$$\textstyle{v_{e_{1}^{*}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{e_{1}^{*}}}$$\scriptstyle{e_{e_{1}^{*}f_{2}}}$$\textstyle{v_{f_{2}^{*}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{f_{2}^{*}}}$$\scriptstyle{e_{f_{2}^{*}e_{1}}}$$\textstyle{v_{e_{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{e_{1}e_{1}}}$$\scriptstyle{e_{e_{1}f_{2}}}$$\textstyle{v_{f_{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{f_{2}e_{1}}}$$\scriptstyle{e_{f_{2}f_{2}}}$$\textstyle{v_{e_{1}^{*}e_{1}^{*}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{e_{1}^{*}e_{1}^{*}}}$$\textstyle{v_{e_{1}^{*}f_{2}^{*}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{e_{1}^{*}f_{2}^{*}}}$$\textstyle{v_{e_{1}^{*}f_{2}}}$$\textstyle{v_{f_{2}^{*}e_{1}^{*}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{f_{2}^{*}e_{1}^{*}}}$$\textstyle{v_{f_{2}^{*}f_{2}^{*}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{f_{2}^{*}f_{2}^{*}}}$$\textstyle{v_{f_{2}^{*}e_{1}}}$$\textstyle{v_{e_{1}f_{2}^{*}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{e_{1}f_{2}^{*}}}$$\textstyle{v_{e_{1}e_{1}}}$$\textstyle{v_{e_{1}f_{2}}}$$\textstyle{v_{f_{2}e_{1}^{*}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{f_{2}e_{1}^{*}}}$$\textstyle{v_{f_{2}e_{1}}}$$\textstyle{v_{f_{2}f_{2}}.}$$\textstyle{\vdots}$$\textstyle{\vdots}$$\textstyle{\vdots}$$\textstyle{\vdots}$

######  Proposition 14.

If $(G,\psi)\in C$, then there is a morphism
$\alpha:(T_{C},\xi_{C})\to(G,\psi)$.

###### Proof.

Since $(G,\psi)\leftrightharpoons(F,\phi)$, there is a $v\in G^{0}$ such that
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{v}\\!\operatorname{Path}(G_{d}))$.
Define a homomorphism $\alpha:T_{C}\to G$ as follows. Let $x_{1}\dots
x_{n}\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}$. Since
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{v}\\!\operatorname{Path}(G_{d}))$,
there is a (uniquely determined) path $y_{1}\dots
y_{n}\in{}_{v}\\!\operatorname{Path}(G_{d})$ such that $\psi(y_{1}\dots
y_{n})=x_{1}\dots x_{n}$. Define $\alpha^{0}(v_{x_{1}\dots x_{n}})=r(y_{n})$,
$\alpha^{1}(e_{x_{1}\dots x_{n}})=y_{n}$ if $y_{n}\in G^{1}$ and respectively
$\alpha^{1}(e_{x_{1}\dots x_{n}})=y_{n}^{*}$ if $y_{n}\in(G^{1})^{*}$. We
leave it to the reader to check that $\alpha$ is a graph homomorphism and that
$\psi\alpha=\xi_{C}$. ∎

We are in a position to show that in each equivalence class $C$ of the
representation graphs of the weighted graph $E$, there is only one irreducible
representation graph up to isomorphism.

######  Corollary 15.

Up to isomorphism the representation graphs in $C$ are precisely the quotients
of $(T_{C},\xi_{C})$, and consequently

$(S_{C},\zeta_{C}):=((T_{C})_{\sim},(\xi_{C})_{\sim})$

is the unique irreducible representation graph in $C$.

###### Proof.

The first statement follows from Theorem 12 and Proposition 14. The second
statement now follows since a quotient
$((T_{C})_{\approx},(\xi_{C})_{\approx})$ satisfies the equivalent conditions
in Definition 6 if and only if $\approx$ equals $\sim$. ∎

Recall that an object $X$ in a category $C$ is called repelling (resp.
attracting) if for any object $Y$ in $C$ there is a morphism $X\rightarrow Y$
(resp. $Y\rightarrow X$). By Proposition 14 $(T_{C},\xi_{C})$ is a repelling
object in $C$. On the other hand, if $(G,\psi)$ is an object in $C$, then
clearly $(S_{C},\zeta_{C})$ is isomorphic to a quotient of $(G,\psi)$. It
follows from Theorem 12 that $(S_{C},\zeta_{C})$ is an attracting object in
$C$.

######  Example 16.

Suppose $E$ is a weighted graph with one vertex and two loops of weight $2$:

$E:\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
37.25858pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{}{{}{{}{{}{{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}}}}}}}}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{}{{}{{}}}\ignorespaces\ignorespaces{\hbox{\kern-37.25858pt\raise
0.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{\mathtt{e_{1},e_{2}}}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}{}{{}}}}}}}{\hbox{\kern-5.60223pt\raise
4.70912pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{}{{}{{}{{}{{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}}}}}}}}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{}{{}{{}}}\ignorespaces\ignorespaces{\hbox{\kern
19.0919pt\raise 0.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{\mathtt{f_{1},f_{2}}}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}}{}{{}{{}}{}{{}}{}{{}}{}{{}{{}{{}}{}{{}}}}}}}{\hbox{\kern
5.60223pt\raise 4.70912pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise 0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}{\hbox{\kern
29.603pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

Consider the representation graphs $(F_{1},\phi_{1}),\dots,(F_{7},\phi_{7})$
for $E$ given below.

$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(F_{1},\phi_{1}):}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$.$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$
---
$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(F_{2},\phi_{2}):}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$.$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$

$\textstyle{(F_{3},\phi_{3}):}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$. | $\textstyle{(F_{4},\phi_{4}):}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$.
---|---
$\textstyle{(F_{5},\phi_{5}):}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{e_{1}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{e_{1}}$. | $\textstyle{(F_{6},\phi_{6}):}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$.
$\textstyle{(F_{7},\phi_{7}):}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$. |

All the representation graphs $(F_{i},\phi_{i})~{}(1\leq i\leq 7)$ lie in the
same $\leftrightharpoons$-equivalence class $C$. One checks easily that
$(F_{1},\phi_{1})\cong(T_{C},\xi_{C})$ and
$(F_{7},\xi_{7})\cong(S_{C},\zeta_{C})$ (cf. Example 13). Moreover, we have

$\textstyle{(F_{1},\xi_{1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(F_{2},\xi_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(F_{3},\xi_{3})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(F_{4},\xi_{4})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(F_{5},\xi_{5})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(F_{6},\xi_{6})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(F_{7},\xi_{7})}$

where an arrow $(F_{i},\phi_{i})\longrightarrow(F_{j},\phi_{j})$ means that
$(F_{j},\phi_{j})$ is a quotient of $(F_{i},\phi_{i})$.

##  3\. Weighted Leavitt path algebras and their representations

Weighted Leavitt path algebras were introduced in [17], as a generalisation of
Leavitt path algebras. They are a model for the algebras $L_{K}(n,m)$ which
could not be obtained via the classical theory of Leavitt path algebras. The
structure of weighted Leavitt path algebras was further investigated in [18,
27, 26, 29]. Further generalisations of these algebras were also carried out
in [28, 24].

We recall the notion of weighted Leavitt path algebras.

######  Definition 17.

Let $(E,w)$ be a weighted graph. The $K$-algebra $L_{K}(E,w)$ presented by the
generating set $\\{v,e_{i},e_{i}^{*}\mid v\in E^{0},e\in E^{1},1\leq i\leq
w(e)\\}$ and the relations

  1. (1)

$uv=\delta_{uv}u\quad(u,v\in E^{0})$,

  2. (2)

$s(e)e_{i}=e_{i}=e_{i}r(e),~{}r(e)e_{i}^{*}=e_{i}^{*}=e_{i}^{*}s(e)\quad(e\in
E^{1},1\leq i\leq w(e))$,

  3. (3)

$\sum\limits_{1\leq i\leq w(v)}e_{i}^{*}f_{i}=\delta_{ef}r(e)\quad(v\in
E^{0},e,f\in s^{-1}(v))$ and

  4. (4)

$\sum\limits_{e\in s^{-1}(v)}e_{i}e_{j}^{*}=\delta_{ij}v\quad(v\in E^{0},1\leq
i,j\leq w(v))$

is called the weighted Leavitt path algebra of $(E,w)$. In relations (3) and
(4) we set $e_{i}$ and $e_{i}^{*}$ zero whenever $i>w(e)$.

Throughout the paper, we continue to denote a weighted graph $(E,w)$ by $E$
and the weighted Leavitt path algebra $L_{K}(E,w)$ by $L_{K}(E)$. We also
interchangeably use the terminology the Leavitt path algebra of a weighted
graph instead of a weighted Leavitt path algebra. If all the edges of the
weighted graph $E$ have weight one, then $L_{K}(E,w)$ coincides with the
classical Leavitt path algebra $L_{K}(E)$ (see Example 19).

Weighted Leavitt path algebras are involutary graded rings with unit if
$E^{0}$ is finite and local units otherwise. In fact, the weighted Leavitt
path algebra $L_{K}(E)$ is a $\mathbb{Z}^{n}$-graded ring, where
$n=\max\\{w(e)\mid e\in E\\}$. The grading is defined as follows: Consider the
free ring $F_{E}$ generated by $\\{v\mid v\in E^{0}\\}$, $\\{e_{1},\dots
e_{w(e)}\mid e\in E^{1}\\}$ and $\\{e_{1}^{*},\dots e_{w(e)}^{*}\mid e\in
E^{1}\\}$, with the coefficients in $K$, set for $v\in E^{0}$, $\deg(v)=0$,
for $e\in E^{1}$, $1\leq i\leq w(e)$, $\deg(e_{i})=(0,\dots,0,1,0,\dots)$ and
$\deg(e_{i}^{*})=(0,\dots,0,-1,0,\dots)\in\mathbb{Z}^{n}$, where $1$ and $-1$
are in the $i$-th component, respectively. This defines a
$\mathbb{Z}^{n}$-grading on the free ring $F_{E}$. (Note that $n$ could be
infinite). Since all the relations in Definition 17 involve homogeneous
elements, so the quotient of $F_{E}$ by the homogeneous ideal generated by
these relations, i.e., $L_{K}(E)$ is also a $\mathbb{Z}^{n}$-graded ring.

######  Example 18.

The Leavitt path algebra of a weighted graph consisting of one vertex and
$n+k$ loops of weight $n$ is isomorphic to the Leavitt algebra $L_{K}(n,n+k)$.
To show this, let $E^{1}=\\{y_{1},\dots,y_{n+k}\\}$ with $w(y_{i})=n$, $1\leq
i\leq n+k$. Denote the $n$ edges corresponding to the (structure) edge
$y_{i}\in E^{1}$ by $\\{y_{1i},\dots,y_{ni}\\}$. We visualise this data as
follows:

$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{y_{11},\dots,y_{n1}}$$\scriptstyle{y_{12},\dots,y_{n2}}$$\scriptstyle{y_{13},\dots,y_{n3}}$$\scriptstyle{y_{1,n+k},\dots,y_{n,n+k}}$

Set $x_{sr}=y_{rs}^{*}$ for $1\leq r\leq n$ and $1\leq s\leq n+k$ and arrange
the $y$’s and $x$’s in the matrices

$Y=\left(\begin{matrix}y_{11}&y_{12}&\dots&y_{1,n+k}\\\
y_{21}&y_{22}&\dots&y_{2,n+k}\\\ \vdots&\vdots&\ddots&\vdots\\\
y_{n1}&y_{n2}&\dots&y_{n,n+k}\end{matrix}\right),\qquad
X=\left(\begin{matrix}x_{11\phantom{n+{},}}&x_{12\phantom{n+{},}}&\dots&x_{1n\phantom{n+{},}}\\\
x_{21\phantom{n+{},}}&x_{22\phantom{n+{},}}&\dots&x_{2n\phantom{n+{},}}\\\
\vdots&\vdots&\ddots&\vdots\\\
x_{n+k,1}&x_{n+k,2}&\dots&x_{n+k,n}\end{matrix}\right)$

Then Condition (3) of Definition 17 precisely says that $Y\cdot X=I_{n,n}$ and
Condition (4) is equivalent to $X\cdot Y=I_{n+k,n+k}$ which are the generators
of $L_{K}(n,n+k)$.

######  Example 19.

Let $(E,w)$ be a weighted graph where $w:E\rightarrow\mathbb{N}$ is the
constant map $w(e)=1$, for all $e\in E$. Then $L_{K}(E,w)$ coincides with the
usual Leavitt path algebra $L_{K}(E)$.

######  Example 20.

Consider a weighted graph with one vertex and one edge $e$ of weight $n$ and
an unweighted graph $F$ with one vertex and $n$ edges
$\\{e_{1},\dots,e_{n}\\}$. Then the map

$\displaystyle L_{K}(E,w)$ $\displaystyle\rightarrow L_{K}(F),$ $\displaystyle
e_{i}$ $\displaystyle\mapsto e_{i}^{*}$

induces an isomorphism, i.e.,

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Note that this isomorphism is not graded as $L(E,w)$ is
$\mathbb{Z}^{n}$-graded, whereas $L(F)$ is just $\mathbb{Z}$-graded.

###  3.1. Representations of weighted Leavitt path algebras

Let $E$ be a weighted graph and $L_{K}(E)$ the weighted Leavitt path algebra
associated to $E$. In this section, for a representation graph $(F,\phi)$ of
$E$, we construct a $L_{K}(E)$-module $V_{F}$. This gives rise to a functor
from the category of representations of the graph $E$ (see §2.3) to the
category of (right) $L_{K}(E)$-modules:

$\displaystyle V:\operatorname{\mathbf{RG}}(E)$
$\displaystyle\longrightarrow\operatorname{Mod}L_{K}(E),$
$\displaystyle(F,\phi)$ $\displaystyle\longmapsto V_{F}.$

We will then investigate the functor $V$. In Theorem 28, we show that $V_{F}$
is a simple $L_{K}(E)$-module if and only if $F$ is an irreducible
representation. In Theorem 43 we further show that a universal representation
$T$ of $E$ gives an indecomposable $L_{K}(E)$-module $V_{T}$.

For a representation graph $(F,\phi)$ of the weighted graph $E$, let $V_{F}$
be the $K$-vector space with basis $F^{0}$. We show that there is a natural
action of the weighted Leavitt path algebra $L_{K}(E)$ on the $K$-vector space
$V_{F}$, resulting $V_{F}$ to be a $L_{K}(E)$-module. For any $u\in E^{0}$,
$e\in E^{1}$ and $1\leq i\leq w(e)$, define endomorphisms
$\sigma_{u},\sigma_{e_{i}},\sigma_{e_{i}^{*}}\in\operatorname{End}_{K}(V_{F})$
by

$\displaystyle\sigma_{u}(v)$
$\displaystyle=\begin{cases}v,\quad\quad\quad&\text{if }\phi^{0}(v)=u\\\
0,&\text{else}\end{cases},$ $\displaystyle\sigma_{e_{i}}(v)$
$\displaystyle=\begin{cases}r_{F}(f),\quad&\text{if }\exists f\in
s_{F}^{-1}(v):~{}\phi^{1}(f)=e_{i}\\\ 0,&\text{else}\end{cases},$
$\displaystyle\sigma_{e_{i}^{*}}(v)$
$\displaystyle=\begin{cases}s_{F}(f),\quad&\text{if }\exists f\in
r_{F}^{-1}(v):~{}\phi^{1}(f)=e_{i}\\\ 0,&\text{else}\end{cases},$

where $v\in F^{0}$. It follows from the universal property of $L_{K}(E)$ that
there is an algebra homomorphism

$\pi:L_{K}(E)\longrightarrow\operatorname{End}_{K}(V_{F})^{{\rm op}},$ (10)

such that $\pi(u)=\sigma_{u}$, $\pi(e_{i})=\sigma_{e_{i}}$ and
$\pi(e_{i}^{*})=\sigma_{e_{i}^{*}}$. We refer to this representation, the
representation of $L_{K}(E)$ defined by $(F,\phi)$. Clearly $V_{F}$ becomes a
right $L_{K}(E)$-module by defining $x\cdot a:=\pi(a)(x)$, for any $a\in
L_{K}(E)$ and $x\in V_{F}$.

The following lemma describes the action of monomial elements of the weighted
Leavitt path algebra $L_{K}(E)$ on the $K$-vector space $V_{F}$. Note that by
Lemma 3, for any $p\in\operatorname{Path}(\hat{E}_{d})$ and $u\in F^{0}$ there
is at most one $v\in F^{0}$ such that
$p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d}))$.

######  Lemma 21.

Let $E$ be a weighted graph and $(F,\phi)$ a representation of $E$ with the
induced map
$\phi:\operatorname{Path}(F_{d})\rightarrow\operatorname{Path}(\hat{E}_{d})$.
If $p\in\operatorname{Path}(\hat{E}_{d})$ and $u\in F^{0}$, then

$u\cdot p=\begin{cases}v,\quad&\text{if
}p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d})),\text{ for some }v\in
F^{0},\\\ 0,\quad&\text{otherwise.}\end{cases}$

###### Proof.

Let $p\in\operatorname{Path}(\hat{E}_{d})$ and $u\in F^{0}$. Suppose that
$p=u^{\prime}$ for some $u^{\prime}\in E^{0}$. If
$p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d}))$, for some $v\in F^{0}$,
then clearly $u=v$ and $u^{\prime}=\phi^{0}(v)$. Hence $u\cdot p=u=v$ as
desired. If there is no $v\in F^{0}$ such that
$p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d}))$, then clearly
$u^{\prime}\neq\phi^{0}(v)$. Hence $u\cdot p=0$ as desired.
Suppose now that $p$ is nontrivial, say $p=x_{1}\dots x_{n}$ where $n\geq 1$
and $x_{1},\dots,x_{n}\in\hat{E}_{d}^{1}$. We proceed by induction on $n$.

Case $n=1$: Suppose $p=e_{i}$ for some $e\in E^{1}$ and $1\leq i\leq w(e)$. If
$p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d}))$ for some $v\in F^{0}$,
then clearly $p=\phi^{1}(f)$ for some $f\in F^{1}$ such that $s_{F}(f)=u$ and
$r_{F}(f)=v$. Hence $u\cdot p=r_{F}(f)=v$ as desired. If there is no $v\in
F^{0}$ such that $p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d}))$, then
clearly there is no $f\in s_{F}^{-1}(v)$ such that $\phi^{1}(f)=e_{i}$. Hence
$u\cdot p=0$ as desired.

Suppose $p=e_{i}^{*}$ for some $e\in E^{1}$ and $1\leq i\leq w(e)$. If
$p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d}))$ for some $v\in F^{0}$,
then clearly $e_{i}=\phi^{1}(f)$ for some $f\in F^{1}$ such that $s_{F}(f)=v$
and $r_{F}(f)=u$. Hence $u\cdot p=s_{F}(f)=v$ as desired. If there is no $v\in
F^{0}$ such that $p\in\phi^{-1}({}_{u}\\!\operatorname{Path}_{v}(F_{d}))$,
then clearly there is no $f\in r_{F}^{-1}(u)$ such that $\phi^{1}(f)=e_{i}$.
Hence $u\cdot p=0$ as desired.

Case $n\to n+1$: Suppose $p=x_{1}\dots x_{n}x_{n+1}$. If
$p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d}))$ for some $v\in F^{0}$,
then clearly $x_{1}\dots
x_{n}\in\phi({}_{u}\\!\operatorname{Path}_{u^{\prime}}(F_{d}))$ for some
$u^{\prime}\in F^{0}$ and moreover
$x_{n+1}\in\phi({}_{u^{\prime}}\\!\operatorname{Path}_{v}(F_{d}))$. By the
induction assumption we have $v\cdot x_{1}\dots x_{n}=u^{\prime}$. It follows
from the case $n=1$ that $u\cdot p=u^{\prime}\cdot x_{n+1}=v$.
Suppose now that there is no $v\in F^{0}$ such that
$p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d}))$. If $x_{1}\dots
x_{n}\in\phi^{-1}({}_{u}\\!\operatorname{Path}_{u^{\prime}}(F_{d}))$ for some
$u^{\prime}\in F^{0}$, then there is no $v\in F^{0}$ such that
$x_{n+1}\in\phi({}_{u^{\prime}}\\!\operatorname{Path}_{v}(F_{d}))$. Hence, by
the induction assumption, $u\cdot p=u^{\prime}\cdot x_{n+1}=0$ as desired. If
there is no $u^{\prime}\in F^{0}$ such that $x_{1}\dots
x_{n}\in\phi({}_{u}\\!\operatorname{Path}_{u^{\prime}}(F_{d}))$, then, by the
induction assumption, $u\cdot p=(u\cdot x_{1}\dots x_{n})\cdot x_{n+1}=0\cdot
x_{n+1}=0$ as desired. ∎

######  Corollary 22.

If $a=\sum_{p\in\operatorname{Path}(\hat{E}_{d})}k_{p}p\in L_{K}(E)$ and $u\in
F^{0}$, then

$u\cdot a=\sum_{v\in
F^{0}}\Big{(}\sum_{p\in\phi({}_{u}\\!\operatorname{Path}_{v}(F_{d}))}k_{p}\Big{)}v.$

Let $T\rightarrow E$ be a covering of the weighted graph $E$ and $F$ a
representation for $T$. By Lemma 4, the $L_{K}(T)$-module $V_{F}$ is also a
$L_{K}(E)$-module. In fact, for a particular kind of coverings, namely
universal coverings, we can say more.

######  Corollary 23.

Let $E$ be a weighted graph, $T$ the universal cover of $E$ and
$F\hookrightarrow\hat{T}$ a connected representation graph of $T$. Then any
$L_{K}(T)$-submodule $A\subseteq V_{F}$ is also an $L_{K}(E)$-module.

###### Proof.

Since by Lemma 4, $F$ is a representation for $E$, $V_{F}$ is a an
$L_{K}(E)$-module as well as an $L_{K}(T)$-module. Let $A\subseteq V_{F}$ be a
$L_{K}(T)$-submodule and $a=\sum_{i}k_{i}v_{i}\in A$. It is enough to show
that for an edge $e\in L_{K}(E)$, we have $ae,ae^{*}\in A$. We have
$ae=\sum_{i}k_{i}v_{i}e=\sum_{i}k_{i}r(f_{i})$, for those $i$ such that
$\sigma(f_{i})=e$. Since $F\hookrightarrow T$, we then have
$af_{i}=k_{i}r(f_{i})\in A$ as $A$ is a $L_{K}(T)$-module. Thus
$ae=\sum_{i}k_{i}r(f_{i})\in A$. The argument for $ae^{*}$ is similar. ∎

Recall that a weighted Leavitt path algebra is a $\mathbb{Z}^{n}$-graded ring,
where $n$ is the maximum weight in the graph (see §3). In particular, Leavitt
path algebras (associated to graphs of weight one) are $\mathbb{Z}$-graded.
This grading has played a crucial role in determining the algebraic structure
of these algebras. We next characterise the representation graphs whose
associated $L_{K}(E)$-modules are graded. To do this we will use the following
statement whose proof is easy and we will leave it to the reader.

######  Lemma 24.

Let $A$ be a $\Gamma$-graded ring and $M$ a right $A$-module, where $\Gamma$
is a totally ordered abelian group. If there is a homogeneous element $a\in A$
with $\deg(a)\not=0$ and $0\not=m\in M$ such that $ma=m$, then $M$ cannot be
$\Gamma$-graded.

Let $F$ be a connected representation graph for the weighted graph $E$. Since
$F$ is an immersion into $\hat{E}$, we have a monomorphism
$\pi(F)\rightarrow\pi(\hat{E})$. Combining this with the length map (7), we
obtain a homomorphism $|\,|:\pi(F)\rightarrow\mathbb{Z}^{n}$. This allows us
to give a criterion on when the $L_{K}(E)$-module $V_{F}$ is a graded module.

######  Theorem 25.

Let $E$ be a weighted graph and $F$ a connected representation graph of $E$.
Then the $L(E)$-module $V_{F}$ is graded if and only if for (reduced) paths
$p,q$ of $F$ with $s(p)=s(q)$ and $r(p)=r(q)$, we have $|p|=|q|$.

###### Proof.

Suppose for any paths $p,q$ of $F$, with $s(p)=s(q)$ and $r(p)=r(q)$ we have
$|p|=|q|$. Choose a base vertex $v$ in $F$ and assigne $\deg(v)=0$. Define a
map $\deg:F^{0}\rightarrow\mathbb{Z}^{n}$, by $\deg(w)=|q|$, where $q$ is a
(reduced) path with $s(q)=v$ and $r(q)=w$. Since $F$ is connected and the
length of paths connecting $v$ to $w$ are the same, the map is well-defined.
For $\alpha\in\mathbb{Z}^{n}$, define

$(V_{F})_{\alpha}:=\Big{\\{}\sum k_{i}v_{i}\,|\deg(v_{i})=\alpha\Big{\\}}.$

Clearly $V_{F}=\bigoplus_{\alpha\in\mathbb{Z}^{n}}(V_{F})_{\alpha}$. Next we
check that $(V_{F})_{\alpha}L_{K}(E)_{\beta}\subseteq(V_{F})_{\alpha+\beta}$,
where $\alpha,\beta\in\mathbb{Z}^{n}$. It is enough to show that
$wp\in(V_{F})_{\alpha+\beta}$, where $w\in(V_{F})_{\alpha}\cap F^{0}$ and $p$
is a monomial in $L_{K}(E)_{\beta}$. Either $wp=0$ or by the definition of
representation, one can lift $p$ to a unique path $q$ in $F$ such that
$|q|=|p|$ and by Lemma 21, $wq=z$, where $z=r(q)$. Since there is a path $t$
with $s(t)=v$, $r(t)=w$ and $|t|=\alpha$, we have $s(tq)=v$, $r(tq)=z$. Since
$|tq|=|t|+|q|$ we have $z\in(V_{F})_{\alpha+\beta}$.

Conversely, suppose $V_{F}$ is a graded $L_{K}(E)$-module. If there are paths
$p,q$ in $F$ with $s(p)=s(q)$ and $r(p)=r(q)$ but $|p|\not=|q|$, then
$s(p)\not=pq^{*}\in\pi_{1}(F,s(p))$. Let $t$ be the image of $pq^{*}$ in
$\hat{E}$. Then by the definition of representation, $vt=v$. Since
$|pq^{*}|\not=0$, it follows from Lemma 24 that $V_{F}$ is not graded, which
is a contradiction. ∎

######  Example 26.

Consider the weighted graph
$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathtt{e_{1},e_{2}}}$$\scriptstyle{\mathtt{f_{1},f_{2}}}$
from Example 16. Theorem 25 will then give the $L_{K}(2,2)$-modules
$V_{F_{1}}$ and $V_{F_{2}}$ are $\mathbb{Z}^{2}$-graded whereas
$V_{F_{3}},V_{F_{4}},V_{F_{5}},V_{F_{6}}$ and the simple module $V_{F_{7}}$
are not graded.

Consider the graph $E=\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{}{{}{{}{{}{{}{{}}{}{{}{{}}{}{{}{{}{{}}{}{{}}{}{{}}{}{{}}}}}}}}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{}{{}{{}}}\ignorespaces\ignorespaces{\hbox{\kern
13.5pt\raise 18.00694pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{\mathtt{e}}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{}{{}{{}}{}{{}}{}{{}}{}{{}}{}{{}{{}{{}{{}}{}{{}{{}}{}{{}{{}}{}{{}}}}}}}}{\hbox{\kern
5.49542pt\raise 0.33377pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{}{{}{{}{{}{{}{{}}{}{{}{{}}}}}}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{}{{}{{}}}\ignorespaces\ignorespaces{\hbox{\kern
19.0919pt\raise 0.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.43056pt\hbox{$\scriptstyle{\mathtt{f}}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{}{{}{{}}{}{{}}{}{{}}{}{{}}{}{{}{{}}{}{{}{{}{{}{{}}{}{{}}{}{{}}}}}}}{\hbox{\kern
3.2935pt\raise-2.99599pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{}{{}{{}{{}{{}{{}{{}}{}{{}}{}{{}}{}{{}{{}}{}{{}{{}}{}{{}}}}}}}}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{}{{}{{}}}\ignorespaces\ignorespaces{\hbox{\kern
13.5pt\raise-18.68748pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{\mathtt{g}}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{}{{}{{}}{}{{}}{}{{}}{}{{}}{}{{}}{}{{}{{}{{}{{}{{}}{}{{}}{}{{}}}}}}}{\hbox{\kern
0.09149pt\raise-3.00475pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise 0.0pt\hbox{{}{}{{}}{{}{}{}{}}{}}}}\ignorespaces{}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}\ignorespaces}}}}\ignorespaces$
from the Introduction and its representation graphs $F$ and $G$ (see (3) and
(4)). Theorem 25 will then give the $L_{K}(E)$-module $V_{G}$ is
$\mathbb{Z}$-graded whereas $V_{F}$ is not a graded module.

######  Corollary 27.

If $T$ is a universal representation of $E$ then $V_{T}$ is graded.

###### Proof.

If $T$ is a universal cover, then $\pi(T)=1$. This implies the condition of
Theorem 25. ∎

###  3.2. Irreducible representations of weighted Leavitt path algebras

In this section we characterise those representation graphs that induce simple
modules on the level of Leavitt path algebras. We continue to denote by $E$ a
weighted graph, $(F,\phi)$ a representation graph for $E$ and $V_{F}$ the
$L_{K}(E)$-module defined by $(F,\phi)$. In the following theorem we show that
a representation graph $F$ is irreducible (see Definition 6) if and only if
$V_{F}$ is a simple $L_{K}(E)$-module.

######  Theorem 28.

Let $E$ be a weighted graph, $(F,\phi)$ a representation graph for $E$ and
$V_{F}$ the $L_{K}(E)$-module defined by $F$. Then the following are
equivalent.

  1. (i)

The $L_{K}(E)$-module $V_{F}$ is simple.

  2. (ii)

The representation graph $F$ is connected and

$\text{for any }x\in V_{F}\setminus\\{0\\},\text{ there is }a\in
L_{K}(E),\text{ such that }x\cdot a\in F^{0}.$

  3. (iii)

The representation graph $F$ is connected and

$\text{for any }x\in V_{F}\setminus\\{0\\},\text{ there is a }k\in K\text{ and
a }p\in\operatorname{Path}(\hat{E}_{d}),\text{ such that }x\cdot kp\in F^{0}.$

  4. (iv)

The representation graph $F$ is an irreducible representation.

###### Proof.

(i) $\Longrightarrow$ (iv). If $C$ is a connected component of $F$, then the
subspace of $V_{F}$ generated by $C^{0}$ is clearly invariant under the action
of $L:=L_{K}(E)$. Hence $F$ must be connected. Now assume that there are
$u\neq v\in F^{0}$ such that
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\phi({}_{v}\\!\operatorname{Path}(F_{d}))$.
Consider the submodule $(u-v)L\subseteq V_{F}$. Since $V_{F}$ is simple by
assumption, we have $(u-v)L=V_{F}$. Hence there is an $a\in L$ such that
$(u-v)a=v$. Clearly there is an $n\geq 1$, $k_{1},\dots,k_{n}\in K^{\times}$
and pairwise distinct $p_{1},\dots,p_{n}\in\operatorname{Path}(\hat{E}_{d})$
such that $a=\sum_{s=1}^{n}k_{s}p_{s}$. We may assume that $(u-v)p_{s}\neq 0$,
for any $1\leq s\leq n$. It follows from Lemma 21 that
$p_{s}\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\phi({}_{v}\\!\operatorname{Path}(F_{d}))$,
for any $s$ and moreover, that $(u-v)p_{s}=u_{s}-v_{s}$ for some distinct
$u_{s},v_{s}\in F^{0}$. Hence

$v=(u-v)a=(u-v)(\sum_{s=1}^{n}k_{s}p_{s})=\sum_{s=1}^{n}k_{s}(u_{s}-v_{s})$

which contradicts Lemma 63.

(iv) $\Longrightarrow$ (iii). Let $x\in V_{F}\setminus\\{0\\}$. Then there is
an $n\geq 1$, pairwise disjoint $v_{1},\dots,v_{n}\in F^{0}$ and
$k_{1},\dots,k_{n}\in K^{\times}$ such that $x=\sum_{s=1}^{n}k_{s}v_{s}$. If
$n=1$, then $x\cdot k_{1}^{-1}\phi^{0}(v_{1})=v_{1}$. Suppose now that $n>1$.
By assumption, we can choose a
$p_{1}\in\phi({}_{v_{1}}\\!\operatorname{Path}(F_{d}))$ such that
$p_{1}\not\in\phi({}_{v_{2}}\\!\operatorname{Path}(F_{d}))$. Clearly $x\cdot
p_{1}\neq 0$ is a linear combination of at most $n-1$ vertices from $F^{0}$.
Proceeding this way, we obtain paths $p_{1},\dots,p_{m}$ such that $x\cdot
p_{1}\dots p_{m}=kv$ for some $k\in K^{\times}$ and $v\in F^{0}$. Hence
$x\cdot k^{-1}p_{1}\dots p_{m}=v$.

(iii) $\Longrightarrow$ (ii). Trivial.

(ii) $\Longrightarrow$ (i). Let $U\subseteq V_{F}$ be a nonzero
$L_{K}(E)$-submodule and $x\in U\setminus\\{0\\}$. By assumption, there is an
$a\in L_{K}(E)$ and a $v\in F^{0}$ such that $v=x\cdot a\in U$. Let now
$v^{\prime}$ be an arbitrary vertex in $F^{0}$. Since by assumption $F$ is
connected, there is a $p\in{}_{v}\\!\operatorname{Path}_{v^{\prime}}(F_{d})$.
It follows from Lemma 21 that $v^{\prime}=v\cdot\phi(p)\in U$. Hence $U$
contains $F^{0}$ and thus $U=V_{F}$. ∎

###  3.3. The fullness of the functor $V$

Let $\alpha:(F,\phi)\rightarrow(G,\psi)$ be a morphism in
$\operatorname{\mathbf{RG}}(E)$. Then it induces a surjective
$L_{K}(E)$-module homomorphism $V_{\alpha}:V_{F}\rightarrow V_{G}$ such that
$V_{\alpha}(u)=\alpha^{0}(u)$, for any $u\in F^{0}$.

The example below shows that $V$ is not full, namely, there can be
$L_{K}(E)$-module homomorphisms $V_{F}\rightarrow V_{G}$ that are not induced
by a morphism $(F,\phi)\rightarrow(G,\psi)$.

######  Example 29.

Suppose $E$ is the weighted graph

$E:\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.86229pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{u\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{{}{}}\ignorespaces\ignorespaces{\hbox{\kern
8.64932pt\raise 20.78754pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{\mathtt{e_{1},e_{2}}}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}}{\hbox{\kern
30.6642pt\raise 7.30396pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}}{}}}}\ignorespaces{}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{}{}{{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}}{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{{}}{{}{}{}}{}}}}\ignorespaces{}\ignorespaces{}{}{}{{}{}}\ignorespaces\ignorespaces{\hbox{\kern
9.62154pt\raise-21.71114pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{\mathtt{f_{1},f_{2}}}$}}}\kern
3.0pt}}}}}}\ignorespaces{}{}{}{{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}}{\hbox{\kern
33.67302pt\raise-2.9942pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{{}{}{}{{}}{{}{}{}\lx@xy@spline@}{}}}}\ignorespaces{}\ignorespaces\hbox{\hbox{\kern
0.0pt\raise 0.0pt\hbox{{}{}{{}}{{}{}{}}{}}}}\ignorespaces{}{\hbox{\kern
29.86229pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{v}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

Consider the representation graphs $(F,\phi)$ and $(G,\psi)$ for $E$ given
below.
$\textstyle{u_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v_{1}}$$\textstyle{(F,\phi):}$$\textstyle{u_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v_{2}.}$
$\textstyle{(G,\psi):}$$\textstyle{u\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{1}}$$\scriptstyle{f_{2}}$$\textstyle{v.}$

Since $(G,\psi)$ is a quotient of $(F,\phi)$, by Theorem 12, there is a
morphism $\alpha:(F,\phi)\rightarrow(G,\psi)$, which induces a homomorphism
$V_{\alpha}:V_{F}\rightarrow V_{G}$. Although $(F,\phi)$ is not a quotient of
$(G,\psi)$, there is an $L_{K}(E)$-module homomorphism
$\sigma:V_{G}\rightarrow V_{F}$ in the opposite direction such that
$\sigma(u)=u_{1}+u_{2}$ and $\sigma(v)=v_{1}+v_{2}$. Note that $V_{\alpha}$
and $\sigma$ are not inverse to each other.

######  Question 30.

Can it happen that $V_{F}\cong V_{G}$ as $L_{K}(E)$-modules although
$(F,\phi)\not\cong(G,\psi)$ in $\operatorname{\mathbf{RG}}(E)$?

The authors do not know the answer to Question 30. But we will show that if
$V_{F}\cong V_{G}$, then $(F,\phi)$ and $(G,\psi)$ must be equivalent.

######  Lemma 31.

Let $(F,\phi)$ and $(G,\psi)$ be objects in $\operatorname{\mathbf{RG}}(E)$
and let $\sigma:V_{F}\rightarrow V_{G}$ be a $L_{K}(E)$-module homomorphism.
Let $u\in F^{0}$ and $\sigma(u)=\sum_{s=1}^{n}k_{s}v_{s}$, where $n\geq 1$,
$k_{1},\dots,k_{n}\in K^{\times}$ and $v_{1},\dots,v_{n}$ are pairwise
distinct vertices from $G^{0}$. Then
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{v_{s}}\\!\operatorname{Path}(G_{d}))$,
for any $1\leq s\leq n$.

###### Proof.

Let $p\in\operatorname{Path}(\hat{E}_{d})$ such that
$p\not\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$. Then

$0=\sigma(0)=\sigma(u\cdot p)=\sigma(u)\cdot p=\sum_{s=1}^{n}k_{s}v_{s}\cdot
p=\sum_{s=1}^{n}k_{s}(v_{s}\cdot p)$

by Lemma 21. One more application of Lemma 21 gives that $v_{s}\cdot p=0$, for
any $1\leq s\leq n$, whence
$p\not\in\psi({}_{v_{s}}\\!\operatorname{Path}(G_{d}))$ for any $1\leq s\leq
n$. Hence we have shown that
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))\supseteq\psi({}_{v_{s}}\\!\operatorname{Path}(G_{d}))$,
for any $1\leq s\leq n$. It follows from Lemma 8 that
$\phi({}_{u}\\!\operatorname{Path}(F_{d}))=\psi({}_{v_{s}}\\!\operatorname{Path}(G_{d}))$,
for any $1\leq s\leq n$. ∎

######  Proposition 32.

Let $(F,\phi)$ and $(G,\psi)$ be objects in $\operatorname{\mathbf{RG}}(E)$.
If there is a nonzero $L_{K}(E)$-module homomorphism $\sigma:V_{F}\rightarrow
V_{G}$, then $(F,\phi)\leftrightharpoons(G,\psi)$.

###### Proof.

The proposition follows from Lemma 31 and the definition of the equivalence
$\leftrightharpoons$. ∎

######  Proposition 33.

Let $(F,\phi)$ and $(G,\psi)$ be irreducible representation graphs for the
weighted graph $E$. Then $V_{F}\cong V_{G}$ as $L_{K}(E)$-modules if and only
if $(F,\phi)\cong(G,\psi)$.

###### Proof.

Clearly isomorphic objects in $\operatorname{\mathbf{RG}}(E)$ yield isomorphic
$L_{K}(E)$-modules. Hence we only have to show that $V_{F}\cong V_{G}$ implies
$(F,\phi)\cong(G,\psi)$. Suppose that $V_{F}\cong V_{G}$. Then
$(F,\phi)\leftrightharpoons(G,\psi)$ by Proposition 32, i.e., $(F,\phi)$ and
$(G,\psi)$ are in the same $\leftrightharpoons$-equivalence class $C$. Since
they are irreducible, it follows from Corollary 15 that
$(F,\phi)\cong(S_{C},\zeta_{C})\cong(G,\psi)$. ∎

###  3.4. Indecomposability of $L_{K}(E)$-modules $V_{F}$

Recall that for a ring $R$, an $R$-module is called _indecomposable_ if it is
non-zero and cannot be written as a direct sum of two non-zero submodules. It
is easy to see that an $R$-module $M$ is indecomposable if and only if the
only idempotent elements of the endomorphism ring $\operatorname{End}_{R}(M)$
are $0$ and $1$.

We will show in Theorem 43 that for any universal representation $T$ of the
graph $E$, the $L_{K}(E)$-module $V_{T}$ is indecomposable. However, Example
34 below shows that in general the indecomposibility of $L_{K}(E)$-module
$V_{F}$, for a representation $(F,\phi)$, depends on the ground field $K$.

######  Example 34.

Let $E$ be the graph

$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathtt{e}}$$\scriptstyle{\mathtt{f}}$

and the graph $F$ below the representation graph for $E$:

---
|  |  |
$\textstyle{v_{7}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e}$$\textstyle{v_{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e}$$\textstyle{v_{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{v_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e}$$\textstyle{v_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e}$$\textstyle{v_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{v_{8}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{v_{5}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$

If $\epsilon\in\operatorname{End}_{L}(V_{F})$, then by Lemma 31 there are
$k,l\in K$ such that $\epsilon(v_{1})=kv_{1}+lv_{2}=v_{1}\cdot(kv+le)$.
Conversely, if $k,l\in K$, then there is a uniquely determined endomorphism
$\epsilon\in\operatorname{End}_{L}(V_{F})$ such that
$\epsilon(v_{1})=kv_{1}+lv_{2}$. This yields a bijection between
$\operatorname{End}_{L}(V_{F})$ and $K\times K$.

Let now $\epsilon\in\operatorname{End}_{L}(V_{F})$ be the endomorphism
corresponding to a pair $(k,l)\in K\times K$. Since $v_{1}$ generates the
$L_{K}(E)$-module $V_{F}$, the endomorphism $\epsilon$ is idempotent if and
only if $\epsilon(v_{1})=\epsilon^{2}(v_{1})$. Clearly

$\epsilon^{2}(v_{1})=v_{1}\cdot(kv+le)^{2}=v_{1}\cdot(k^{2}v+2kle+l^{2}e^{2})=(k^{2}+l^{2})v_{1}+2klv_{2}.$

Thus $\epsilon$ is idempotent if and only if

$k=k^{2}+l^{2}\text{ and }l=2kl.$ (11)

If $2=0$ in $K$, then the only solutions for (11) are $(k,l)=(0,0)$ and
$(k,l)=(1,0)$. The corresponding endomorphisms are $\epsilon=0$ and
$\epsilon=\operatorname{id}_{V_{F}}$. Thus $V_{F}$ is an indecomposable
$L_{K}(E)$-module in this case. If $2\neq 0$, then there are two more
solutions for (11), namely $(k,l)=(1/2,1/2)$ and $(k,l)=(1/2,-1/2)$. Thus
$V_{F}$ is not indecomposable in this case.

Fix an $\leftrightharpoons$-equivalence class $C$ and define
$(T,\xi)=(T_{C},\xi_{C})$ and $(S,\zeta)=(S_{C},\zeta_{C})$ as in Subsection
2.5. Since the $L_{K}(E)$-module $V_{S}$ is simple, it is indecomposable. We
will show that the $L_{K}(E)$-module $V_{T}$ is indecomposable too.

Let $p,p^{\prime}\in\operatorname{Path}(\hat{E}_{d})$ be non-backtracking
paths such that $r(p)=s(p^{\prime})=v$. We define a non-backtracking path
$p\ast p^{\prime}\in\operatorname{Path}(\hat{E}_{d})$ as follows. If
$p,p^{\prime}\in\operatorname{Path}(\hat{E}_{d})\setminus\\{v\\}$ let $p\ast
p^{\prime}$ be the element of $\operatorname{Path}(\hat{E}_{d})$ one gets by
removing all subwords of the form $e_{i}e_{i}^{*}$ and $e_{i}^{*}e_{i}$ from
the juxtaposition $pp^{\prime}$ (if $p^{\prime}=p^{*}$, then $p\ast
p^{\prime}:=v)$. Moreover define $v\ast p:=p$ and $p\ast v:=p$.

Define the group

$G:=\big{\\{}p\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}\mid
v_{p}\sim v_{\phi(u)}\big{\\}}.$ (12)

In Proposition 39, we will show that $(G,\ast)$ is a free group. We need
several lemmas.

######  Lemma 35.

If $p\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}$, then
$v_{p}=v_{\phi(u)}\cdot p$.

###### Proof.

Clearly $v_{p}=v_{\phi(u)}\cdot p$ if $p=\phi(u)$. Suppose now that
$p=x_{1}\dots x_{n}$ for some
$x_{1},\dots,x_{n}\in\hat{E}^{1}\cup(\hat{E}^{1})^{*}$. For $1\leq i\leq n$
define

$\hat{e}_{x_{1}\dots x_{i}}:=\begin{cases}e_{x_{1}\dots x_{i}},&\text{ if
}x_{i}\in\hat{E}^{1},\\\ e^{*}_{x_{1}\dots x_{i}},&\text{ if
}x_{i}\in(\hat{E}^{1})^{*}.\end{cases}.$

One checks easily that
$q:=\hat{e}_{x_{1}}\hat{e}_{x_{1}x_{2}}\dots\hat{e}_{x_{1}\dots
x_{n}}\in{}_{v_{\phi(u)}\\!\operatorname{Path}_{v_{p}}}(T_{d})$ and
$\xi(q)=p$. It follows from Lemma 21 that $v_{\phi(u)}\cdot p=v_{p}$. ∎

######  Lemma 36.

Let $p,p^{\prime}\in G$. Then $p\ast p^{\prime}$ is defined and $p\ast
p^{\prime}\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}$.

###### Proof.

Let $r,r^{\prime}\in{}_{u}\\!\operatorname{Path}(F_{d})$ such that $\phi(r)=p$
and $\phi(r^{\prime})=p^{\prime}$. By the proof of Lemma 35 there are
(uniquely determined) paths
$q\in{}_{v_{\phi(u)}\\!\operatorname{Path}_{v_{p}}}(T_{d})$ and
$q^{\prime}\in{}_{v_{\phi(u)}\\!\operatorname{Path}_{v_{p^{\prime}}}}(T_{d})$
such that $\xi(q)=p$ and $\xi(q^{\prime})=p^{\prime}$. Let
$\alpha:(T,C)\to(F,\phi)$ be the morphism defined in the proof of Proposition
14. One checks easily that $\alpha(q)=r$ and $\alpha(q^{\prime})=r^{\prime}$.
Since $v_{p}\sim v_{\phi(u)}$, there is a
$q^{\prime\prime}\in{}_{v_{p}}\\!\operatorname{Path}(T_{d})$ such that
$\xi(q^{\prime\prime})=p^{\prime}=\xi(q^{\prime})$. Set
$r^{\prime\prime}:=\alpha(q^{\prime\prime})$. Clearly
$s(r^{\prime\prime})=s(\alpha(q^{\prime\prime}))=\alpha(s(q^{\prime\prime}))=\alpha(v_{p^{\prime}})=\alpha(r(q))=r(\alpha(q))=r(r)$.
Hence $rr^{\prime\prime}\in{}_{u}\\!\operatorname{Path}(F_{d})$. It follows
that
$pp^{\prime}=\phi(rr^{\prime\prime})\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$.
In particular $r(p)=s(p^{\prime})=\phi(u)$ and hence $p\ast p^{\prime}$ is
defined.
Write $pp^{\prime}=x_{1}\dots x_{n}$ and $rr^{\prime\prime}=y_{1}\dots y_{n}$.
Suppose that $x_{j}x_{j+1}=e_{i}e_{i}^{*}$ for some $1\leq j\leq n-1$, $e\in
E^{1}$ and $1\leq i\leq w(e)$. Since $\phi(y_{j})=x_{j}$ and
$\phi(y_{j+1})=x_{j+1}$, there are $f,g\in F^{1}$ such that $y_{j}=f$ and
$y_{j}=g^{*}$. Moreover $r(f)=r(g)$ and $\phi(f)=e_{i}=\phi(g)$. It follows
from Definition 2(2) that $f=g$. Hence $y_{1}\dots y_{j-1}y_{j+1}\dots
y_{n}\in{}_{u}\\!\operatorname{Path}(F_{d})$ and $\phi(y_{1}\dots
y_{j-1}y_{j+1}\dots y_{n})=x_{1}\dots x_{j-1}x_{j+1}\dots x_{n}$. Similarly,
if $x_{j}x_{j+1}=e_{i}^{*}e_{i}$ for some $1\leq j\leq n-1$, $e\in E^{1}$ and
$1\leq i\leq w(e)$, then $y_{1}\dots y_{j-1}y_{j+1}\dots
y_{n}\in{}_{u}\\!\operatorname{Path}(F_{d})$ and $\phi(y_{1}\dots
y_{j-1}y_{j+1}\dots y_{n})=x_{1}\dots x_{j-1}x_{j+1}\dots x_{n}$ (follows from
Definition 2(1)). Thus $p\ast
p^{\prime}\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}$. ∎

######  Lemma 37.

Let $p,p^{\prime}\in G$. Then $p\ast p^{\prime}\in G$.

###### Proof.

By the previous lemma $p\ast p^{\prime}$ is defined and $p\ast
p^{\prime}\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))_{\operatorname{nb}}$.
It remains to show that $v_{p\ast p^{\prime}}\sim v_{\phi(u)}$. By the proof
of Lemma 35 there are paths
$q\in{}_{v_{\phi(u)}}\\!\operatorname{Path}_{v_{p}}(T_{d})$ and
$q^{\prime}\in{}_{v_{\phi(u)}}\\!\operatorname{Path}_{v_{p^{\prime}}}(T_{d})$
such that $\xi(q)=p$ and $\xi(q^{\prime})=p^{\prime}$. Since $v_{p}\sim
v_{\phi(u)}$, there is a path
$q^{\prime\prime}\in{}_{v_{p}}\\!\operatorname{Path}(T_{d})$ such that
$\xi(q^{\prime\prime})=p^{\prime}=\xi(q^{\prime})$. By Lemmas 35 and 21 we
have

$v_{p\ast p^{\prime}}=v_{\phi(u)}\cdot(p\ast
p^{\prime})=v_{\phi(u)}\cdot(pp^{\prime})=r(qq^{\prime\prime})=r(q^{\prime\prime})$

and hence $q^{\prime\prime}\in{}_{v_{p}}\\!\operatorname{Path}_{v_{p\ast
p^{\prime}}}(T_{d})$.

We will show that $v_{p\ast p^{\prime}}\sim v_{p^{\prime}}$ which implies
$v_{p\ast p^{\prime}}\sim v_{\phi(u)}$. Let
$p^{\prime\prime}\in\xi({}_{v_{p^{\prime}}}\\!\operatorname{Path}(T_{d}))$.
Then there is an $r\in{}_{v_{p^{\prime}}}\\!\operatorname{Path}(T_{d})$ such
that $\xi(r)=p^{\prime\prime}$. Since
$q^{\prime}r\in{}_{v_{\phi(u)}}\\!\operatorname{Path}(T_{d})$, we have
$p^{\prime}p^{\prime\prime}=\xi(q^{\prime}r)\in\xi({}_{v_{\phi(u)}}\\!\operatorname{Path}(T_{d}))$.
Since $v_{\phi(u)}\sim v_{p}$, there is an
$s\in{}_{v_{p}}\\!\operatorname{Path}(T_{d})$ such that
$\xi(s)=p^{\prime}p^{\prime\prime}$. Write $s=s_{1}s_{2}$ such that
$\xi(s_{1})=p^{\prime}$ and $\xi(s_{2})=p^{\prime\prime}$. Clearly
$s(s_{1})=v_{p}$. It follows from Lemma 3 that $s_{1}=q^{\prime\prime}$. Since
$q^{\prime\prime}\in{}_{v_{p}}\\!\operatorname{Path}_{v_{p\ast
p^{\prime}}}(T_{d})$, we obtain $s_{2}\in{}_{v_{p\ast
p^{\prime}}}\\!\operatorname{Path}(T_{d})$. Hence
$p^{\prime\prime}=\xi(s_{2})\in\xi({}_{v_{p\ast
p^{\prime}}}\\!\operatorname{Path}(T_{d}))$. We have shown that
$\xi({}_{v_{p^{\prime}}}\\!\operatorname{Path}(T_{d}))\subseteq\xi({}_{v_{p\ast
p^{\prime}}}\\!\operatorname{Path}(T_{d}))$. It follows from Lemma 8 that
$\xi({}_{v_{p^{\prime}}}\\!\operatorname{Path}(T_{d}))=\xi({}_{v_{p\ast
p^{\prime}}}\\!\operatorname{Path}(T_{d}))$ and thus $v_{p\ast p^{\prime}}\sim
v_{p^{\prime}}$. ∎

######  Lemma 38.

Let $p\in G$. Then $p^{*}\in G$.

###### Proof.

By the proof of Lemma 35 there is a path
$q\in{}_{v_{\phi(u)}}\\!\operatorname{Path}_{v_{p}}(T_{d})$ such that
$\xi(q)=p$. Clearly
$q^{*}\in{}_{v_{p}}\\!\operatorname{Path}_{v_{\phi(u)}}(T_{d})$ and
$\xi(q^{*})=p^{*}$. Since $v_{p}\sim v_{\phi(u)}$, there is a path
$q^{\prime}\in{}_{v_{\phi(u)}}\\!\operatorname{Path}(T_{d})$ such that
$\xi(q^{\prime})=p^{*}$. Let $\alpha:(T,\xi)\to(F,\phi)$ be the morphism
defined in the proof of Proposition 14. Then
$\alpha(q^{\prime})\in{}_{u}\\!\operatorname{Path}(F_{d})$ and
$\phi(\alpha(q^{\prime}))=\xi(q^{\prime})=p^{*}$. Hence
$p^{*}\in\phi({}_{u}\\!\operatorname{Path}(F_{d}))$. Moreover, $p^{*}$ is not
backtracking since $p$ is not backtracking.

It remains to show that $v_{p^{*}}\sim v_{\phi(u)}$. Let
$p^{\prime}\in\xi({}_{v_{p^{*}}}\\!\operatorname{Path}(T_{d}))$. Then there is
an $r\in{}_{v_{p^{*}}}\\!\operatorname{Path}(T_{d})$ such that
$\xi(r)=p^{\prime}$. By the proof of Lemma 35 there is a path
$q^{\prime\prime}\in{}_{v_{\phi(u)}}\\!\operatorname{Path}_{v_{p^{*}}}(T_{d})$
such that $\xi(q^{\prime\prime})=p^{*}$. Clearly
$q^{\prime\prime}r\in{}_{v_{\phi(u)}}\\!\operatorname{Path}(T_{d})$ and hence
$p^{*}p^{\prime}=\xi(q^{\prime\prime}r)\in\xi({}_{v_{\phi(u)}}\\!\operatorname{Path}(T_{d}))$.
Since $v_{\phi(u)}\sim v_{p}$, there is an
$s\in{}_{v_{p}}\\!\operatorname{Path}(T_{d})$ such that
$\xi(s)=p^{*}p^{\prime}$. Write $s=s_{1}s_{2}$ such that $\xi(s_{1})=p^{*}$
and $\xi(s_{2})=p^{\prime}$. Clearly
$s_{1}\in{}_{v_{p}}\\!\operatorname{Path}(T_{d})$. It follows from Lemma 3
that $s_{1}=q^{*}$. Since
$q^{*}\in{}_{v_{p}}\\!\operatorname{Path}_{v_{\phi(u)}}(T_{d})$, we obtain
$s_{2}\in{}_{v_{\phi(u)}}\\!\operatorname{Path}(T_{d})$. Hence
$p^{\prime}=\xi(s_{2})\in\xi({}_{v_{\phi(u)}}\\!\operatorname{Path}(T_{d}))$.
We have shown that
$\xi({}_{v_{p^{*}}}\\!\operatorname{Path}(T_{d}))\subseteq\xi({}_{v_{\phi(u)}}\\!\operatorname{Path}(T_{d}))$.
It follows from Lemma 8 that
$\xi({}_{v_{p^{*}}}\\!\operatorname{Path}(T_{d}))=\xi({}_{v_{\phi(u)}}\\!\operatorname{Path}(T_{d}))$
and thus $v_{p^{*}}\sim v_{\phi(u)}$. ∎

######  Proposition 39.

The group $(G,\ast)$ defined in (12) is a free group.

###### Proof.

By Lemma 37, $\ast$ defines a binary operation on $G$. Clearly this operation
is associative. Moreover, $\phi(u)\ast p=p\ast\phi(u)$ for any $p\in G$. If
$p\in G$, then $p^{*}\in G$ by Lemma 38. Clearly $pp^{*}=p^{*}p=\phi(u)$.
Hence $(G,\ast)$ is a group.

Let $F$ be the free group on $X:=\\{e_{i}\mid e\in E^{1},1\leq i\leq w(e)\\}$.
We identify $F$ with the set of all reduced words over the alphabet $X\cup
X^{-1}$ (a word over $X\cup X^{-1}$ is reduced if it does contain any subwords
of the form $xx^{-1}$ or $x^{-1}x$ where $x\in X$). The product of two reduced
words $w$ and $w^{\prime}$ is the reduced word one gets by removing all
subwords of the form $xx^{-1}$ or $x^{-1}x$ from the juxtaposition
$ww^{\prime}$.

Define a map $\theta:G\to F$ as follows. If $p=x_{1}\dots x_{n}\in
G\setminus\\{\phi(u)\\}$, let $\theta(p)$ be the word one gets by replacing
any letter $e_{i}^{*}$ by $e_{i}^{-1}$. Clearly $\theta(p)$ is a reduced word
since $p$ is not backtracking. Moreover, we define $\theta(\phi(u))$ as the
identity element of $F$, namely the empty word. Clearly $\theta$ is a group
homomorphism. Moreover, $\theta$ is injective and hence $G$ is isomorphic to a
subgroup of a free group. It follows from the Nielsen-Schreier theorem that
$G$ is a free group. ∎

Let $W\subseteq V_{T}$ be the linear span of the $\sim$-equivalence class of
$v_{\phi(u)}$, i.e. $W=\langle v_{p}\mid p\in G\rangle$. Let $A$ be the
subalgebra of $L_{K}(E)$ generated by the image of the group $G$ in
$L_{K}(E)$.

######  Lemma 40.

The $K$-vector space $W$ is a right $A$-module where the action of $A$ on $W$
is induced by the action of $L_{K}(E)$ on $W$.

###### Proof.

It suffices to show that if $p\in G$ and $v_{p^{\prime}}\sim v_{\phi(u)}$,
then $v_{p^{\prime}}\cdot p=v_{p^{\prime\prime}}$ for some
$v_{p^{\prime\prime}}\sim v_{\phi(u)}$. Since $v_{p^{\prime}}\sim
v_{\phi(u)}$, we have $p^{\prime}\in G$. It follows from Lemmas 35 and 36 that

$v_{p^{\prime}}\cdot p=(v_{\phi(u)}\cdot p^{\prime})\cdot p=v_{\phi(u)}\cdot
p^{\prime}p=v_{\phi(u)}\cdot p^{\prime}\ast p=v_{p\ast p^{\prime}}.$

By Lemma 37 we have $p\ast p^{\prime}\in G$. Thus $v_{p\ast p^{\prime}}\sim
v_{\phi(u)}$. ∎

Since $W$ is a right $A$-module, there is a $K$-algebra homomorphism
$\delta:A\to\operatorname{End}_{K}(W)^{{\rm op}}$ defined by
$\delta(a)(w)=w\cdot a$. Define $\bar{A}:=A/\ker(\delta)$. Let
$~{}\bar{}:A\to\bar{A}$ be the canonical $K$-algebra homomorphism and
$\bar{\delta}:\bar{A}\to\operatorname{End}_{K}(W)^{{\rm op}}$ the $K$-algebra
homomorphism induced by $\delta$. Then the following diagram commutes.

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
13.33096pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
15.58653pt\raise 5.43056pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.43056pt\hbox{$\scriptstyle{\delta}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 30.75pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx<EMAIL_ADDRESS>0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.83888pt\hbox{$\scriptstyle{\bar{}~{}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 0.0pt\raise-29.76389pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
30.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{\operatorname{End}_{K}(W)^{{\rm
op}}}$}}}}}}}{\hbox{\kern-5.5pt\raise-38.44165pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\bar{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
26.94774pt\raise-25.05971pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.83888pt\hbox{$\scriptstyle{\bar{\delta}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 49.31393pt\raise-5.5pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces{\hbox{\lx@xy@drawline@}}{\hbox{\kern
54.54782pt\raise-38.44165pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

$W$ becomes a right $\bar{A}$-module by defining
$w\cdot\bar{a}=\bar{\delta}(\bar{a})(w)=\delta(a)(w)=w\cdot a$.

######  Proposition 41.

The algebra $\bar{A}$ is isomorphic to the group algebra $K[G]$.

###### Proof.

Let $a\in A$. Then $a=\sum_{i=1}^{n}k_{i}p_{i,1}\dots p_{i,m_{i}}$ for some
$n,m_{1},\dots,m_{n}\in\mathbb{N}$, $k_{1},\dots,k_{n}\in K$ and $p_{i,j}\in
G~{}(1\leq i\leq n,1\leq j\leq m_{i})$. Clearly $p_{i,1}\dots
p_{i,m_{i}}-p_{i,1}\ast\dots\ast p_{i,m_{i}}\in\ker(\delta)$ for any $1\leq