# Decompositions of Infinite-Dimensional $A_{\infty,\infty}$ Quiver
Representations

Nathaniel Gallup and Stephen Sawin

###### Abstract

Using linear algebraic methods we show that every (possibly infinite-
dimensional) representation of a quiver with underlying graph
$A_{\infty,\infty}$ is Krull-Schmidt, as long as the arrows in the quiver
eventually point outward.

##  1 Introduction

Most early techniques in the study of representations of quivers (e.g.
Auslander-Reiten theory, the Euler form, etc.) assume that the quiver has
finitely many vertices and that the vector spaces of these representations are
finite-dimensional. In particular, Gabriel’s Theorem [Gab72] characterizes the
quivers of finite-type as exactly those with underlying graph the ADE Dynkin
diagrams. A nice survey of Gabriel’s Theorem can be found in [Bri08], and we
adopt most of Brion’s notation in this paper. More recently there have been
several results which weaken these finiteness hypotheses.

In [BLP11], Bautista, Liu, and Paquette observed that any locally finite-
dimensional representation (meaning each vector space is finite-dimensional)
of various infinite quivers including an eventually outward
$A_{\infty,\infty}$ quiver is Krull-Schmidt (a direct sum of indecomposables)
and identified the indecomposables in tht case as being in bijection with the
connected full subquivers.

In [Rin16], Ringel was able to remove the finite-dimensional hypothesis from
one direction of Gabriel’s Theorem. There he proved that infinite-dimensional
representations of (finite) ADE Dynkin quivers are Krull-Schmidt, and showed
that these indecomposables are the same as those given in Gabriel’s Theorem
which correspond to the positive roots of the corresponding root system.

In [EER09], Enochs, Estrada, and Rozas studied injective quiver
representations over a general (not necessarily commutative) ring $R$. They
proved that the indecomposable injective representations of certain infinite
quivers (which include eventually outward $A_{\infty,\infty}$ quivers) called
_transfinite tree quivers_ are in bijection with a subset of the connected
full subquivers. Note that as they are working over a general ring, there is
no locally finite-dimensional hypothesis. Here the term “injective” refers to
injective objects in the category of representations and in fact (see Prop.
2.1 in [EER09]) the linear maps in these representations are necessarily
surjective.

In this paper, we simultaneously weaken the two finiteness conditions (those
requiring finite quivers and locally finite-dimensional representations) as
well as the injective representation requirement in the type-A case over a
field. More specifically we show that every (possibly infinite-dimensional)
representation of an eventually outward quiver with underlying graph
$A_{\infty,\infty}$ is Krull-Schmidt. This eventually outward condition is in
fact necessary: there exist representations of a not eventually outward
$A_{\infty,\infty}$ quiver which are not Krull-Schmidt, an example of which is
given at the end of the paper in Section 8. We furthermore give a description
of the indecomposables in this case, which are in bijection with the connected
full subquivers which we can think of as “extended simple roots” which are
limits of the classical simple roots. These match those from [BLP11] and also
those from [EER09] if we restrict to injective representations. Our proof
furthermore gives an algorithm for decomposing a given quiver representation
into indecomposables.

In Sections 2 we review the necessary quiver representation theory background.
The remainder of the paper can be divided into two parts.

In the first part we discuss representations of a general quiver $\Omega$
which is an eventually outward finitely branched tree, i.e. a union of
finitely many journeys, the arrows of which eventually point away from its
start. More specifically, in Section 3 we define a certain partial order on
the set $\mathcal{C}$ of connected full subquivers of $\Omega$ by taking the
poset closure of two “moves” which we call _reduction_ and _enhancement_.
Using the eventually outward condition we prove that this is in fact a well-
founded poset.

In Section 4 we then define a poset filtration of any representation $V$ of
$\Omega$, i.e. an order-preserving function $F$ from $\mathcal{C}$ to the
poset of subrepresentations of $V$. We will show that the successive quotients
$F^{\alpha}/\sum_{\beta<\alpha}F^{\beta}$ of this poset filtration are
isotypic, meaning a direct sum of indecomposables all of which are isomorphic.

In Section 5 we prove that for any connected full subquiver
$\alpha\in\mathcal{C}$, the successive quotient
$F^{\alpha}/\sum_{\beta<\alpha}F^{\beta}$ is supported on $\alpha$ and under
mild conditions the morphism $F^{\alpha}\to
F^{\alpha}/\sum_{\beta<\alpha}F^{\beta}$ lifts, i.e. that the
subrepresentation $\sum_{\beta<\alpha}F^{\beta}$ is complemented, say by
$W^{\alpha}$, inside of $F^{\alpha}$. Thus we obtain an _almost gradation_ $W$
of $F$, i.e. a function $W:\mathcal{C}\to\text{Sub}(V)$ such that for all
$\alpha\in\mathcal{C}$ we have
$F^{\alpha}=W^{\alpha}\oplus\sum_{\beta<\alpha}F^{\beta}$. We then prove that
each $W_{\alpha}$ is an isotypic subrepresentation. Finally, using the fact
that $\mathcal{C}$ is a well-founded poset, we use induction to prove that our
poset filtration _spans_ $V$. This means that
$V=\sum_{\alpha\in\mathcal{C}}W^{\alpha}$.

In the second part of the paper, we specialize to the case where $\Omega$ is
an eventually outward quiver of type $A_{\infty,\infty}$, and prove that any
representation of such a quiver is Krull-Schmidt. We furthermore classify the
indecomposables. Specifically, in Section 6 we show that if $\Omega$ is
eventually outward and of type $A_{\infty,\infty}$, then the partial order on
the set $\mathcal{C}$ of connected subquivers of $\Omega$ is actually the
product of two total orders on the set $\Omega_{1}$ of arrows of $\Omega$. It
follows that, if $V$ is any representation of $\Omega$, then the poset
filtration $F:\mathcal{C}\to\text{Sub}(V)$ is the intersection of two linear
filtrations $L,R:\Omega_{1}\to\text{Sub}(V)$. We then prove the general
theorem that any almost gradation of a poset filtration which is the
intersection of two linear filtrations is independent. All together this
proves our main result that $V=\bigoplus_{\alpha\in\mathcal{C}}W_{\alpha}$ and
thus every (possibly infinite dimensional) representation of an eventually
outward type $A_{\infty,\infty}$ quiver $\Omega$ is Krull-Schmidt (a direct
sum of indecomposables).

In Section 7 we describe the isomorphism classes of the indecomposable
subrepresentations of an eventually outward type $A_{\infty,\infty}$ quiver.
As for any finite quiver of type $A_{n}$, they are in bijection with the
connected subquivers, and are _thin_ , meaning one-dimensional at every
vertex. Note that this means as long as the quiver is eventually outward, the
indecomposables are independent of the directions of the arrows. This agrees
with the indecomposables found in [BLP11] for a locally finite-dimensional
representation of an eventually outward type $A_{\infty,\infty}$ quiver, and
with those found in [EER09] for a (possibly infinite-dimensional) injective
representation of an eventually outward type $A_{\infty,\infty}$ quiver. Note
the dimensions of these are $1$ at each vertex contained in the subquiver and
$0$ otherwise. The finite subquivers correspond to what are normally called
positive roots, which are finite roots of Tits length $1$, or equivalently the
image of simple roots under finitely many Weyl reflections. The infinite
quivers are naturally limits of these in the appropriate topology.

Finally in Section 8 we give an example of a representation of an
$A_{\infty,\infty}$ quiver which is not Krull-Schmidt, i.e. is not a direct
sum of indecomposable subrepresentations. This quiver is not eventually
outward, as is required by the results of Section 6.

##  2 Background

In this section we review the standard definitions in the theory of quivers
and their representations and fix the notation that will be used throughout
the paper.

###  2.1 Quivers

A _quiver_ , denoted by $\Omega$, consists of a set $\Omega_{0}$ of _vertices_
, a set $\Omega_{1}$ of _arrows_ , a _source function_
$s:\Omega_{1}\to\Omega_{0}$, and a _target function_
$t:\Omega_{1}\to\Omega_{0}$. We say that the quiver has an _underlying graph_
if every pair of vertices in $\Omega_{0}$ are the source and target of at most
one arrow in $\Omega_{1}$ (in some order), and no vertex is the source and
target of the same arrow. Then the underlying graph has a vertex for each
vertex of $\Omega$ and an undirected edge for each arrow of $\Omega$. The
quiver is a tree if the underlying graph is a tree.

A vertex $x\in\Omega_{0}$ is a _source_ if it is not the target of any arrow,
and a _sink_ if it is not the source of any arrow.

###  2.2 Subquivers

A _subquiver_ of a quiver $\Omega$ is a quiver $\Omega^{\prime}$ such that
$\Omega^{\prime}_{0}\subseteq\Omega_{0}$,
$\Omega^{\prime}_{1}\subseteq\Omega_{1}$,
$s^{\prime}=s|_{\Omega^{\prime}_{1}}$, and
$t^{\prime}=t|_{\Omega^{\prime}_{1}}$. A _full_ subquiver is one that contains
every arrow whose source and target are contained in it.

From now on, we will assume that the underlying graphs of all quivers in this
paper are trees. Notice that means the removal of any arrow $e$ of a quiver
divides the quiver into two connected subquivers, the one including $t(e)$
which we call the subquiver _in front_ of $e$ and the one containing $s(e)$
which we call the subquiver _behind_ $e$. In general, any vertex, arrow, or
subquiver contained in the former will be said to be _in front of_ $e$ and $e$
will be said to _point towards it_ , while any in the latter will be said to
be _behind_ $e$, and $e$ will be said to _point away from it_.

###  2.3 Walks and Paths

A (finite or infinite) _walk_ $w$ in a quiver with an underlying graph is a
map from a connected (finite or infinite) interval of the integers to
$\Omega_{0}$ so that if $i$ and $i+1$ are in the interval, there is an arrow
whose source and target are $w_{i}$ and $w_{i+1}$. We then say $w$ _contains_
that arrow $e$. We say $e$ is _oriented consistently_ if $s(e)=w_{i}$ and
$t(e)=w_{i+1}$, and _oriented inconsistently_ if $s(e)=w_{i+1}$ and
$t(e)=w_{i}$. Adding an integer to every element of the interval results in a
new map we will consider the same walk. If the interval has a minimal $i$ we
define $s(w)=w_{i}$ and call the walk a _journey_ , if the interval has a
maximal $j$ we define $t(e)=w_{j}$. If $w$ has a target and $w^{\prime}$ has a
source and these two vertices are the same we can define the _concatenation_
walk $w^{\prime}w$ in the obvious way. An _injective walk_ is one where the
map is injective.

###  2.4 Types of Quivers and Walks

Since $\Omega$ is a tree then every two vertices are connected by a unique
minimal walk. We say that $\Omega$ is a _finitely-branched_ tree if it is a
union of finitely many injective paths, or equivalently for any vertex $x$ it
is the union of finitely many injective journeys starting at $x$. Finally, if
every injective journey contains at most finitely many arrows oriented
inconsistently, we say that $\Omega$ is _eventually outward_.

These two types of quivers can be helpfully redefined in terms of a strict
partial order $\prec$ on the set $\Omega_{1}$ of arrows of $\Omega$ defined by
$f\prec e$ if and only if $f$ is behind $e$ and $e$ is in front of $f$.

###### Lemma 2.1

$\Omega$ is eventually outward if and only if its $\prec$ is well-founded (no
infinite descending chains). If $\Omega$ is a tree it is finitely branched if
and only if its $\prec$ has no infinite antichains (no two elements are
related).

Proof: If $\Omega$ contains an injective journey with infinitely many arrows
inconsistently oriented, the set of inconsistently oriented arrows in that
journey is an infinite descending chain. If $\cdots\prec e_{3}\prec e_{2}\prec
e_{1}$ is an infinite downward chain, there is a unique injective journey
$p_{i}$ from the target of each $e_{i}$ to the source of each $e_{i+1}$, and
these are all distinct because if $j>i$ then $p_{j}$ is behind $e_{i+1}$ and
$p_{i}$ is in front. Concatenating $\cdots p_{2}e_{2}^{-1}p_{1}e_{1}^{-1}$
gives a journey with each $e_{i}$ oriented inconsistently.

If $\Omega$ is finitely branched, let $p_{1},\ldots,p_{n}$ be a set of
injective paths covering it. If $E$ is an antichain, there can be at most two
elements of $E$ contained in any given path, or else one would be in front of
the other. Thus $E$ is finite. On the other hand if $\Omega$ is not finitely
branched, then for some $x$ there is an infinite sequence $p_{1},p_{2},\ldots$
of injective journeys from x such that each journey $p_{i}$ contains a first
arrow $e_{i}$ not in any of the previous journeys, and after that all the
arrows are not in any other journeys. Of those infinitely many arrows, either
an infinite set of them point away from $x$ or an infinite set point towards
$x$. Each of these sets separately forms an antichain, because the path
connecting $e_{i}$ to $e_{j}$ will follow the reverse of $p_{i}$ to the final
point of shared intersection with $p_{j}$, then out $p_{j}$. Thus there is an
infinite antichain. $\diamond$

###### Lemma 2.2

If $\alpha$ is a connected full nonempty subquiver of an eventually outward
finitely branching tree, then there are only finitely many arrows that point
towards $\alpha$.

Proof: Being finitely branching, the quiver is a union of finitely many
journeys from an $x\in\alpha$. Since $\alpha$ is connected all arrows that
point towards $\alpha$ point towards $x$, therefore the result follows from
eventually outward. $\diamond$

###  2.5 Quiver Representations

A _representation_ $(V,f)$ of a quiver $Q$ over a field $\mathbb{F}$ consists
of an $\mathbb{F}$-vector space $V_{i}$ at every vertex $i\in Q_{0}$ and a
linear map $f_{e}:V_{s(e)}\to V_{t(e)}$ for every arrow $e\in Q_{1}$. If
$(V,f)$ and $(W,g)$ are representations of $Q$, a _morphism of
representations_ $T:V\to W$ consists of a linear map $T_{i}:V_{i}\to W_{i}$
for all vertices $i\in Q_{0}$ making the appropriate diagram commute.

Direct sums of representations of $Q$ are defined in the obvious way. A quiver
representation is called _indecomposable_ if it is nonzero and is not a direct
sum of two proper, non-zero subrepresentations, and is called _Krull-Schmidt_
if it is a direct sum of indecomposable subrepresentations.

###  2.6 The Transport of a Subspace

Let $\Omega$ be a quiver, $V$ a representation of $\Omega$, $w$ a finite
injective walk in $\Omega$, and $W\subset V_{s(w)}$ a subspace. We define the
_transport_ of $W$ via $w$ to $t(w)$ recursively as follows:

  * •

If $w$ contains a single arrow $e$ oriented consistently then the transport of
$W$ is $f_{e}[W]\subseteq V_{t(w)}$.

  * •

If $w$ contains a single arrow $e$ oriented inconsistently then the transport
of $W$ is $f_{e}^{-1}[W]\subseteq V_{t(w)}$.

  * •

If $w$ is the concatenation of path $w_{1}$ and $w_{2}$ then the transport of
$w$ is the transport via $w_{2}$ of the transport via $w_{1}$ of $W$.

Notice that if $W^{\prime}\subseteq W\subseteq V_{s(w)}$ then the transport of
$W^{\prime}$ is contained in the transport of $W$.

###### Lemma 2.3

Let $\Omega$ be a quiver which is a tree, $V$ a representation of $\Omega$,
$i\in\Omega_{0}$ a vertex, and $W_{i}\subseteq V_{i}$ a subspace. For each
vertex $j\in\Omega_{0}$, define $W_{j}$ to be the transport of $W_{i}$ to $j$
along the unique injective walk starting at $i$ and ending at $j$. Then $W$ is
the maximal subrepresentation of $V$ whose value at $i$ is $W_{i}$ and such
that for each arrow $e$ pointing away from $i$, $f_{e}$ restricted to $W$ is
onto. It is the minimal subrepresentation $W$ of $V$ whose value at $i$ is
$W_{i}$ and such that for each arrow $e$ pointing towards $i$, $f_{e}$
projected onto $V/W$ is one-to-one.

Proof: Elementary.

$\diamond$

##  3 The Partial Order of Connected Components

Let $\Omega$ be an eventually outward finitely branched tree quiver. Let
$\mathcal{C}$ be the set of connected full subquivers of $\Omega$. If
$\alpha\in\mathcal{C}$, the _complement_ of $\alpha$ in $\Omega$, denoted by
$\alpha^{\prime}$, is the full subquiver of $\Omega$ containing the vertices
which are not in $\alpha$. Then $\alpha^{\prime}$ is a disjoint union of
finitely many components, with each component connected to $\alpha$ by a
single arrow, which we call a _boundary arrow_ of $\alpha$. If $e$ connects
one of the components of $\alpha^{\prime}$ to $\alpha$, we denote this
component by $\alpha^{\prime}_{e}$. Let $\text{ba}(\alpha)$ denote the set of
boundary arrows of $\alpha$. Thus we have the following:

$\alpha^{\prime}=\bigsqcup_{e\in\text{ba}(\alpha)}\alpha^{\prime}_{e}$

Furthermore, we denote the subquiver of $\Omega$ which includes
$\alpha^{\prime}_{e}$ and $e$ and both the source and target of $e$ (one of
which is already in $\alpha^{\prime}_{e}$) by $\alpha^{\prime\prime}_{e}$.

If $e$ is any arrow not contained in $\alpha$ then $\alpha$ is contained
entirely in one of the two components of $\Omega$ created by the removal of
$e$, and therefore $e$ either points towards it or away from it. Let the set
of boundary arrows of $\alpha$ which point towards $\alpha$ be
$\text{iba}(\alpha)$ and the set pointing away be $\text{oba}(\alpha)$. if it
points from $\alpha$ to $\alpha^{\prime}_{e}$. More generally, if $e$ is an
arrow not in $\alpha$, then all of $\alpha$ is contained in one of the two
subquivers into which the removal of $e$ separates $\Omega$, and therefore
$\alpha$ is either in front of $e$ or behind it.

Define a relation on the set $\mathcal{C}$ of connected subquivers of $\Omega$
by letting $\alpha>\beta$ if one can get from $\alpha$ to $\beta$ by a
sequence of the following moves:

  1. 1.

We say that $\beta$ is a _reduction_ of $\alpha$ if there exists an arrow $e$
in $\alpha$ such that $\beta$ is the portion of $\alpha$ behind $e$ (in terms
of its orientation in $\Omega$). Thus $e$ is a boundary arrow of $\beta$
pointing away.

  2. 2.

We say that $\beta$ is an _enhancement_ of $\alpha$ if there exists an arrow
$e$ in $\beta$ such that $\alpha$ is the portion of $\beta$ in front of $e$.
Thus $e$ is a boundary arrow of $\alpha$ pointing towards it.

###### Proposition 3.1

There is no infinite sequence $\alpha_{1},\alpha_{2},...$ such that each
$\alpha_{n+1}$ is a nontrivial reduction or enhancement of $\alpha_{n}$.

Proof: Suppose for contradiction that such a sequence exists. For each
$n\in\mathbb{N}$, let $S_{n}$ be the set of arrows $e$ not in $\alpha_{n}$
which point towards $\alpha_{n}$ and such that the journey from $e$ to
$\alpha$ goes through a boundary arrow of $\alpha_{n}$ that also points
towards $\alpha_{n}$. By Lem. 2.2, $S_{n}$ is finite. Notice first that if
$\alpha_{n+1}$ is a reduction or enhancement of $\alpha$ and $e\in S_{n+1}$,
then $e\in S_{n}$, because in the first case the same journey works, and in
the second the same journey works perhaps extended to go through the arrow of
enhancement. So $S_{n+1}\subseteq S_{n}$, and in the case of an enhancement
$S_{n+1}\subset S_{n}$ because if we enhanced on boundary arrow $e$ of
$\alpha_{n}$ then $e\in S_{n}-S_{n+1}$. Therefore there can be only finitely
many enhancements, and there exists some $N\in\mathbb{N}$ such that for all
$n\geq N$, $\alpha_{n+1}<\alpha_{n}$ is a reduction. For each $n\geq N$ let
$e_{n-N}$ be the arrow in $\alpha_{n}$ along which the reduction
$\alpha_{n+1}<\alpha_{n}$ occurs. Then we obtain a sequence of arrows
$e_{1},e_{2},\ldots$ such that for all $m>n$, $e_{m}$ is behind $e_{n}$.

Since $\Omega$ is finitely branched, it is the union of finitely many
injective journeys, and each $e_{n}$ must be on one of these journeys, so
infinitely many $e_{n_{1}},e_{n_{2}},\ldots$ lie on a single journey. If any
$e_{n_{i}}$ is oriented conistently with the journey, then only finitely many
other $e_{n_{j}}$ can be behind it, which contradicts the fact that it was an
infinite sequence. But if all the $e_{n_{i}}$ are oriented inconsistently,
this contradicts the fact that $\Omega$ was evenetually outward.

$\diamond$

###### Corollary 3.1

The relation $>$ on $\mathcal{C}$ defined above is a well-founded strict
partial order.

Proof: Transitivity of the relation is trivially satisfied. To show asymmetry,
if $\alpha<\beta<\alpha$ then we obtain an infinite decreasing sequence of
moves, contradicting Prop. 3.1. Irreflexivity follows from asymmetry. Now that
we know $<$ is a strict partial order, the aforementioned proposition also
implies that it is well-founded as it shows there are no infinite strictly
decreasing chains. $\diamond$

##  4 Poset Filtrations of Subrepresentations

Let $\Omega$ be a quiver which is a tree, and $V$ a representation of
$\Omega$. If $e$ is any arrow of $\Omega$, define two subrepresentations of
$V$ as follows. Define the subrepresentation $V^{e,+}\subset V$ by

$V^{e,+}_{i}=\begin{cases}V_{i}&\text{ if }i\text{ is behind }e\\\
\text{transport of }V_{t(e)}&\text{ else}.\end{cases}$

Define $V^{e,-}\subset V$ by

$V^{e,-}_{i}=\begin{cases}\\{0\\}_{i}\subset V_{i}&\text{ if }i\text{ is in
front of }e\\\ \text{transport of }\\{0\\}_{t(e)}&\text{ else}.\end{cases}$

It is easy to check that each of these is indeed a subrepresentation.

Define a _virtual arrow_ $E$ to be a path, i.e. an oriented injective journey
with all arrows oriented away from the source, which is either infinite or
ends in a leaf (thus it could consist of a single leaf vertex). If
$\alpha\in\mathcal{C}$ is a full connected subquiver and $E$ lies entirely
within $\alpha$ we say $E$ is a _virtual boundary arrow_ of $\alpha$, and
denote by $\text{vba}(\alpha)$ the set of all virtual boundary arrows of
$\alpha$. If $E$ is a virtual arrow, define a subrepresentation
$V^{E,+}\subset V$ by

$V^{E,+}_{i}=\begin{cases}V_{i}&\text{ if }i\in E\\\ \text{transport of
}V_{j}&\text{ else}\end{cases}$

where $j$ is the closest vertex in $E$ to $i$, in the sense that the unique
injective walk from $j$ to $i$ does not contain any other vertices of $E$.

###### Remark 4.1

Notice that this is a subrepresentation, and also that if $E^{\prime}\subset
E$ is also a virtual arrow, then $V^{E,+}=V^{E^{\prime},+}$, so this does not
depend on where $E$ starts.

If $\alpha\in\mathcal{C}$ is a connected subquiver and
$e\in\text{ba}(\alpha)$, define $V^{e,\alpha}$ to be $V^{e,+}$ if
$e\in\text{iba}(\alpha)$ and $V^{e,-}$ if $e\in\text{oba}(\alpha)$, and if
$E\in\text{vba}(\alpha)$ define $V^{E,\alpha}=V^{E,+}$. Finally, let
$\text{b}(\alpha)=\text{ba}(\alpha)\cup\text{vba}(\alpha)$.

Define a function $F:\mathcal{C}\to\text{Sub}(V)$, denoted $\alpha\mapsto
F^{\alpha}$, by

$F^{\alpha}=\bigcap_{d\in\text{b}(\alpha)}V^{d,\alpha}.$ (1)

In general a _poset filtration_ of an $R$ module $M$ consists of a partially
ordered set $(P,\leq)$, and a function $F:P\to\text{Sub}(M)$ which is order-
preserving, meaning $p\leq q$ implies $F^{p}\subseteq F^{q}$.

###### Proposition 4.1

$F$ is a poset filtration of $V$, i.e. if $\beta\leq\alpha$ are connected
subquivers then $F^{\beta}\subseteq F^{\alpha}$.

Proof:

It suffices to check this if $\beta$ is obtained from $\alpha$ by a single
reduction or enhancement. First consider a reduction, with $e$ an arrow
contained in $\alpha$ and $\beta$ the portion of $\alpha$ behind $e$. Then
$\beta$ and $\alpha$ share all the same boundary arrows except that $\beta$
has one outward pointing boundary arrow $e$ which $\alpha$ does not share and
$\alpha$ has some number of boundary arrows which $\beta$ does not share, all
of which are in front of $e$. Furthermore, every virtual boundary arrow of
$\beta$ is also a virtual boundary arrow of $\alpha$ but $\alpha$ has some
number of virtual boundary arrows which are not contained in $\beta$, and
hence by Remark 4.1 we can assume are entirely contained in
$\beta_{e}^{\prime}$.

Therefore for any vertex $i$ in $\Omega$ we have the following:

$F^{\alpha}_{i}=\left[\bigcap_{d\in\text{b}(\alpha)\cap\text{b}(\beta)}V^{d,\alpha}_{i}\right]\cap\left[\bigcap_{d\in\text{b}(\alpha)\smallsetminus\text{b}(\beta)}V^{d,\alpha}_{i}\right]$
(2)
$F^{\beta}_{i}=\left[\bigcap_{d\in\text{b}(\alpha)\cap\text{b}(\beta)}V^{d,\beta}_{i}\right]\cap
V^{e,\beta}_{i}.$ (3)

First notice that because $\beta\subseteq\alpha$, for any
$d\in\text{ba}(\alpha)\cap\text{ba}(\beta)$, $d$ is outward pointing with
respect to $\alpha$ if and only if it is outward pointing with respect to
$\beta$, and hence $V^{d,\alpha}_{i}=V^{d,\beta}_{i}$. Hence the first terms
of the intersections in Equations 2 and 3 are equal. Now we consider two
cases.

Case 1: If $i$ is in front of $e$, then $V_{i}^{e,\beta}=\\{0\\}$ hence
$F_{i}^{\beta}=\\{0\\}$ so trivially we have $F_{i}^{\beta}\subseteq
F_{i}^{\alpha}$.

Case 2: If $i$ is behind $e$, then $V_{i}^{e,\beta}$ is the transport of
$\\{0\\}_{t(e)}$ to $i$. But notice that for
$d\in\text{b}(\alpha)\smallsetminus\text{b}(\beta)$, $V^{d,\alpha}_{i}$ is the
transport of $V^{d,\alpha}_{t(e)}$ to $i$. Since $\\{0\\}_{t(e)}\subseteq
V^{d,\alpha}_{t(e)}$ and transport of subspaces preserves inclusion, it
follows that $V_{i}^{e,\beta}\subseteq V^{d,\alpha}_{i}$ for all
$d\in\text{b}(\alpha)\smallsetminus\text{b}(\beta)$, and therefore the second
term Equation 3 is contained in the second term of Equation 2.

Now consider an enhancement from $\alpha$ to $\beta$ along an inward boundary
arrow $e$ of $\alpha$ contained in $\beta$. Then $\alpha$ and $\beta$ share
all the same boundary/ virtual boundary arrows except that $\alpha$ has one
arrow $e$ pointing towards it which $\beta$ does not share and $\beta$ has
boundary/virtual boundary arrows all behind $e$ which $\alpha$ does not share.
Therefore for a vertex $i$ in $\Omega$ we have the following:

$F^{\beta}_{i}=\left[\bigcap_{d\in\text{b}(\alpha)\cap\text{b}(\beta)}V^{d,\beta}_{i}\right]\cap\left[\bigcap_{d\in\text{b}(\beta)\smallsetminus\text{b}(\alpha)}V^{d,\beta}_{i}\right]$
(4)
$F^{\alpha}_{i}=\left[\bigcap_{d\in\text{b}(\alpha)\cap\text{b}(\beta)}V^{d,\alpha}_{i}\right]\cap
V^{e,\alpha}_{i}$ (5)

As before, the first terms of the intersections in Equations 4 and 5 are equal
and the third terms in Equations 4 and 5 are equal. Now we consider two cases.

Case 1: If $i$ is behind $e$, then $V_{i}^{e,\alpha}=V_{i}$, hence the second
term of Equation 4 is trivially contained in the second term of Equation 5,
yielding $F_{i}^{\beta}\subseteq F_{i}^{\alpha}$.

Case 2: If $i$ is in front of $e$, then $V_{i}^{e,\beta}$ is the transport of
$V_{t(e)}$ to $i$. But notice that for
$d\in\text{b}(\beta)\smallsetminus\text{b}(\alpha)$, $V^{d,\beta}_{i}$ is the
transport of $V^{d,\beta}_{t(e)}$ to $i$. Since $V^{d,\beta}_{t(e)}\subseteq
V_{t(e)}$ and transport of subspaces preserves inclusion, it follows that
$V_{i}^{e,\beta}\subseteq V^{d,\alpha}_{i}$ for all
$d\in\text{b}(\alpha)\smallsetminus\text{b}(\beta)$, and therefore the second
term in Equation 4 is contained in the second term of Equation 5. So
$F_{i}^{\beta}\subseteq F_{i}^{\alpha}$.

$\diamond$

###### Lemma 4.1

If $e$ is an arrow in $\alpha$ with exactly one literal or virtual boundary
arrow of $\alpha$ behind it, then $f_{e}$ is onto when restricted to
$F^{\alpha}$.

Proof: Let $b$ be the boundary behind $e$, and let $X=V^{b,\alpha}_{s(e)}$.
For each literal or virtual boundary $b_{i}$ of $\alpha$ other than $b$, let
$Y_{i}=V^{b_{i},\alpha}_{t(e)}$. Then we have

$F^{\alpha}_{s(e)}=X\cap\bigcap_{i}f_{e}^{-1}[Y_{i}]\qquad
F^{\alpha}_{t(e)}=f[X]\cap\bigcap_{i}Y_{i}.$

If $y\in f[X]\cap\bigcap_{i}Y_{i}$ then $y=f(x)$ where $x\in X$, and since
$y\in\bigcap_{i}Y_{i}$ necessarily $x\in\bigcap_{i}f_{e}^{-1}[Y_{i}].$ So
$f_{e}$ is onto on $F^{\alpha}$. $\diamond$

##  5 Quotient and Lift

For each connected subgraph $\alpha\in\mathcal{C}$ define

$W^{\alpha}=F^{\alpha}/\sum_{\beta<\alpha}F^{\beta}.$ (6)

###### Proposition 5.1

For each arrow $e$ in $\alpha$, if there is exactly one virtual boundary arrow
of $\Omega$ behind $e$, then $f_{e}$ is bijective on $W^{\alpha}$. For each
vertex $i$ not in $\alpha$, $W^{\alpha}_{i}=\\{0\\}$.

Proof: Suppose $e$ is in $\alpha$ and there is exactly one virtual boundary
arrow of $\Omega$ behind it. Then either that virtual boundary is a virtual
boundary of $\alpha$ as well, or there is a boundary arrow of $\alpha$ behind
$e$, so by Prop. 4.1, $f_{e}$ is onto when restricted to $F^{\alpha}$, and
similarly when restricted to $\sum_{\beta<\alpha}F^{\beta}$. Thus it suffices
to show that the kernel of $f_{e}$ in $F^{\alpha}$ is contained in
$\sum_{\beta<\alpha}F^{\beta}$. Let $\beta$ be the reduction of $\alpha$ by
the arrow $e$, This is nonempty, and $F^{\beta}-{s(e)}=f_{e}^{-1}[0]\cap
F^{\alpha}-{s(e)}$. This implies the first sentence.

Suppose $i\in\alpha^{\prime}_{e}$ for some boundary $e$. If $\alpha$ is behind
$e$ then $F^{\alpha}_{i}=V^{e,-}_{i}=0$ and the second sentence follows. If
$\alpha$ is in front of $e$ then $F^{\alpha}_{i}$ is $F^{\alpha}_{t(e)}$
transported to $i$. We will enhance $\alpha$ as follows. Take $\beta$ to
contain all the vertices of $\alpha$, and since there is a unique injective
walk from $i$ to $t(e)$, add all the vertices in that path to $\beta$. For any
vertex you add to $\beta$, if it is the source of an arrow, add the target to
$\beta$. The resulting $\beta$ is an enhancement of $\alpha$ by $e$, and has
the following properties: it contains $i$, all the literal boundary arrows of
$\beta$ which are not in $\alpha$ are inward, and the injective journey from
$i$ to any of those boundary arrows, or to virtual boundary in $\beta$ but not
$\alpha$, agrees in orientation with all its component arrows. The
intersection of all the representations associated to literal and virtual
boundary that $\beta$ shares with $\alpha$ at the vertex $i$ will be
$F^{\alpha}_{i}$. The representation associated to a new boundary $e^{\prime}$
or a new virtual boundary $E$ will be $V^{e^{\prime},+}$ or $V^{E,+}$, and
because there is a consistently oriented injective journey from $i$ to this
boundary the value of that representation at $i$ is $V_{i}$. Thus
$F^{\beta}_{i}=F^{\alpha}_{i}$, and thus
$\sum_{\beta<\alpha}F^{\beta}_{i}=F^{\alpha}_{i}$, and the quotient is
trivial. $\diamond$

In general we define an _almost gradation_ of a poset filtration $(F,P)$ of an
$R$ module $M$ to be a function $C:P\to\text{Sub}(M)$ satisfying the condition
that for all $p\in P$, $F^{p}=F^{<p}\oplus C^{p}$, where we define
$F^{<p}=\sum_{q<p}F^{q}$.

###### Proposition 5.2

For each connected subquiver $\alpha\in\mathcal{C}$, if every edge $e$ in
$\alpha$ has exactly one virtual boundary of $\Omega$ behind it, the map
$F^{\alpha}\to W^{\alpha}$ lifts, i.e. there exists an almost gradation of the
poset filtration $F$.

Proof: We need to lift each $W^{\alpha}_{i}$ to $F^{\alpha}_{i}$ consistently
with the maps $f_{e}$. Since all $F^{\alpha}_{i}$ outside of $\alpha$ are $0$,
this is trivial except for vertices and arrows in $\alpha$. So it suffices to
choose lifts for each $i$ in $\alpha$, in such a way that for each arrow $e$
in $\alpha$ between $i$ and $j$ the following diagram commutes (using Lm. 4.1
and Prop. 5.1)

${0}$${X^{\alpha}_{j}}$${F^{\alpha}_{j}}$${W^{\alpha}_{j}}$${0}$${0}$${X^{\alpha}_{i}}$${F^{\alpha}_{i}}$${W^{\alpha}_{i}}$${0}$$\scriptstyle{f_{e}}$$\scriptstyle{f_{e}}$
(7)

where $X^{\alpha}=\sum_{\beta<\alpha}F^{\beta}$. In the category of vector
spaces we can lift one particular $W^{\alpha}_{i}$ to $F^{\alpha}_{i}$. We
will extend this lift recursively to each other vertex across each arrow as
follows. If the lift $l_{i}:W_{i}^{\alpha}\to F_{i}^{\alpha}$ is chosen in the
above diagram, then on $W_{j}^{\alpha}$ define the lift $f_{e}\circ l_{i}\circ
f_{e}^{-1}$ (the inverse exists by Prop. 5.1). If on the other hand the the
lift $l_{j}:W_{j}^{\alpha}\to F_{j}^{\alpha}$ is chosen in the above diagram,
the lift in $i$ is $f_{e}^{-1}\circ l_{j}\circ f_{e}$, where $f_{e}^{-1}$ is
any lift of $f_{e}$. $\diamond$

In general we say that an almost gradation $C:P\to\text{Sub}(M)$ of a poset
filtration $F:P\to\text{Sub}(M)$ of an $R$ module $M$ _spans_ if
$M=\bigoplus_{p\in P}C^{p}$.

###### Proposition 5.3

The sum of the subrepresentations $F^{\alpha}$ spans $V$.

Proof: Let $i$ be a vertex. Define $\alpha$ recursively by including $i$ in
$\alpha$, and including the target of every arrow whose source is in $\alpha$.
$\alpha$ has the property that every literal boundary arrow necessarily has
$\alpha$ behind it and for every vertex $j$, there is a path from $i$ to $j$.
Then $F^{\alpha}_{i}$ is the intersection of the transport of $V_{j}$ for
various $j$ in $\alpha$, which is then $V_{i}$. Thus $F^{\alpha}_{i}=V_{i}$.
$\diamond$

###### Proposition 5.4

For each connected subquiver $\alpha\in\mathcal{C}$, if every edge $e$ in
$\alpha$ has exactly one virtual boundary of $\Omega$ behind it, then
$W_{\alpha}$ is an isotypic $\Omega$-representation.

Proof: Recall that $W$ is an isotypic $\Omega$-representation if there is an
indecomposable $\Omega$-representation $X$, a vector space $Y$, and for each
vertex $i$ an isomorphism $W_{i}\cong X_{i}\otimes Y$ so that each arrow
$f^{W}_{i}$ is taken under the isomorphism to $f^{X}_{i}\otimes 1$. If
$\alpha$ satisfies the assumptions of the proposition then by Prop. 5.1 every
$f_{e}$ is an isomorphism if $e$ is in $\alpha$ and $0$ otherwise. Let
$X^{\alpha}$ be the $\Omega$-representation which assigns the one-dimensional
vector space $F$ to each vertex in $\alpha$ and the vector space $0$ to each
vertex not in $\alpha$, the identity to each arrow in $\alpha$ and the trivial
map to each arrow not in $\alpha$. $X^{\alpha}$ is clearly indecomposable.
Choose one vertex $i$ in $\alpha$ and let $Y$ be the vector space
$W^{\alpha}_{i}$. Notice that for any other vertex $j$ of $\alpha$ there is a
unique isomorphism $f$ between $W^{\alpha}_{i}$ and $W^{\alpha}_{j}$
consisting of a product of $f_{e}$’s and their inverses. Define the
isomorphism $X^{\alpha}_{j}\otimes Y=W^{\alpha}_{i}\to W^{\alpha}_{j}$ to be
that isomorphism. Check that this writes $W^{\alpha}$ as a representation with
multiplicity. $\diamond$

##  6 Complete Decomposition of $A_{\infty,\infty}$.

Let $A_{\infty,\infty}$ be the graph with a vertex for each integer $i$ and an
edge between each two adjacent integers, and let $\Omega$ be an eventually
outward quiver with underlying graph $A_{\infty,\infty}$.

${\ldots}$${x_{-2}}$${x_{-1}}$${x_{0}}$${x_{1}}$${x_{2}}$${\ldots}$$\scriptstyle{a_{-3}}$$\scriptstyle{a_{-2}}$$\scriptstyle{a_{-1}}$$\scriptstyle{a_{0}}$$\scriptstyle{a_{1}}$$\scriptstyle{a_{2}}$
Figure 1: The Graph $A_{\infty,\infty}$

Notice that every connected subquiver $\alpha$ of $\Omega$ has exactly two
literal/virtual boundary arrows, a left boundary which is either the literal
left boundary arrow or the left virtual arrow of $\Omega$, and a right
boundary which is either the literal right boundary arrow or the right virtual
arrow of $\Omega$.

We will define two total orderings, one $<_{L}$ on the set of all arrows of
$A_{\infty,\infty}$ together with a symbol $-\infty$ (representing the left
virtual boundary of $\Omega$), and the other $<_{R}$ on the set of all arrows
in $A_{\infty,\infty}$ together with the symbol $\infty$ representing the
right virtual arrow. If $e$ and $e^{\prime}$ are two arrows oriented the same
way, then $e<_{L}e^{\prime}$ and $e<_{R}e^{\prime}$ if $e$ is behind
$e^{\prime}$. If they are oriented oppositely then
$e<_{L}-\infty<_{L}e^{\prime}$ and $e^{\prime}<_{R}\infty<_{R}e$ if $e$ is
pointing to the left and $e^{\prime}$ to the right. Note that the restriction
of either $<_{R}$ or $<_{L}$ to the set of arrows $\Omega_{1}$ is a total
order extension of the partial order $\prec$ from 2.4, in the sense that for
any pair of edges $e,e^{\prime}$ if $e\prec e^{\prime}$ then we have that
$e<_{R}e^{\prime}$ and $e<_{L}e^{\prime}$.

Recall that if $(P,\leq_{P})$ and $(Q,\leq_{Q})$ are two partially ordered
sets, their _product_ $(P,\leq_{P})\times(Q,\leq_{Q})$ is the partially
ordered set $(P\times Q,\leq_{P\times Q})$ where $(p,q)\leq_{P\times
Q}(p^{\prime},q^{\prime})$ if and only if $p\leq_{P}p^{\prime}$ and
$q\leq_{Q}q^{\prime}$.

###### Lemma 6.1

There is an order-embedding $\iota$ (i.e. order -preserving, injective, and
order-reflecting) of the poset $\mathcal{C}$ of connected subquivers of
$\Omega$ with the reduction/enhancement partial order into the poset
$(\Omega_{1}\cup\\{-\infty\\},<_{L})\times(\Omega_{1}\cup\\{\infty\\},<_{R})$
given by sending a connected subquiver $\alpha$ to the ordered pair of
consisting of its left and right literal or virtual boundary arrows
respectively.

Proof: The map is clearly injective since a connected subgraph of
$A_{\infty,\infty}$ is uniquely determined by its boundary arrows. To prove it
is order-preserving, it suffices to show that if $\alpha,\beta\in\mathcal{C}$
are connected components and $\alpha>\beta$ with $\beta$ a reduction or
enhancement of $\alpha$ then the left/right boundary arrow of $\alpha$ is
smaller than the left/right boundary arrow of $\beta$.

If $\beta$ is a reduction of $\alpha$ along an arrow $e$ in $\alpha$ then
$\alpha$ and $\beta$ agree behind $e$, so each has a boundary arrow behind $e$
and these arrows are equal. The remaining boundary arrow $a$ of $\alpha$ is in
front of $e$ which is the remaining boundary arrow of $\beta$. Regardless of
the direction of $e$, we obtain that $a$ is strictly larger than $e$ in the
$<_{R}$ or $<_{L}$ order, so the desired result follows. The case when $\beta$
is an enhancement of $\alpha$ is similar.

To see that it is order-reflecting, assume $\iota(\alpha)>\iota(\beta)$ and
show $\alpha>\beta$. If $\iota(\alpha)=(i,j)$ and $\iota(\beta)=(k,\ell)$,
then $k\leq i$ and $\ell\leq j$. Let $\gamma$ be the unique connected subset
such that $\iota(\gamma)=(k,j)$. Then by the definition of $<_{L}$ $\gamma$ is
an enhancement of $\alpha$ if $k$ is to the left of $i$ and an reduction if
$k$ is to the right, in either case $\gamma<\alpha$ (or equal if $k=i$.
Similarly we see by the definition of $<_{R}$ that $\beta<\gamma$.

$\diamond$

Associate to any arrow $e$ a subrepresentation of $V$ called $L^{e}$ which is
$V^{e,+}$ if $e$ points to the right and $V^{e,-}$ if $e$ points to the left,
and a subrepresentation of $V$ called $R^{e}$ which is $V^{e,+}$ if $e$ points
to the left and $V^{e,-}$ if $e$ points to the right. Associate to $-\infty$ a
subrepresentation $L^{-\infty}=V^{E,+}$ where $E$ is the left virtual arrow of
$A_{\infty,\infty}$ and associate to $\infty$ a subrepresentation
$R^{\infty}=V^{E,+}$ where $E$ is the right virtual arrow of
$A_{\infty,\infty}$.

###### Lemma 6.2

$L$ is a poset filtration of $V$ with respect to the $<_{L}$ order and $R$ is
a poset filtration of $V$ with respect to the $<_{R}$ order. For each
connected subquiver $\alpha$ the subrepresentation $V^{\alpha}$ is the
intersection of $L^{e}$ and $R^{e^{\prime}}$, where $e$ and $e^{\prime}$ are
respectively the left and right virtual or literal boundary arrows of
$\alpha$.

Proof: We check the first sentence only for the $L$, the argument for $R$ is
the same. If $e$ is a literal arrow oriented to the right then
$L^{e}=V^{e,+}$, and thus $L^{e}_{i}$ is $V_{i}$ to its left and a transport
of $V_{t(e)}$ to its right. $L^{-\infty}_{i}$ is the transport of $V_{j}$ for
$j$ sufficiently far to the left of it. If $e$ is oriented to the left,
$L^{e}=V^{e,-}$, hence $L^{e}_{i}$ is $0$ to its left and a transport of
$\\{0\\}_{t(e)}$ to its right. Thus if $e$ is oriented to the right,
$L^{e}_{i}$ is bigger than any $L^{e^{\prime}}_{i}$ to its left (behind it),
including the left virtual arrow, and if $e$ is oriented to the left
$L^{e}_{i}$ is smaller than any arrow to its left (in front of it), and
smaller than $L^{-\infty}$.

That $V^{\alpha}$ is the intersection of $L$ and $R$ is just a restatement of
the definition of $V^{\alpha}$ in the case when there are always two boundary
arrows. $\diamond$

Let $M$ be an $R$ module. We say that a poset filtration $(F,P)$ of $M$ is
_distributive_ if for all finite subsets $Q\subseteq P$ and all $p\in Q$ which
is maximal in $Q$ we have $F_{p}\cap\sum_{q\in
Q\smallsetminus\\{p\\}}F_{q}\subseteq F_{<p}$.

###### Proposition 6.1

Let $R$ be a ring and $M$ an $R$ module. If $(F,P)$ is a distributive poset
filtration of $M$ then every almost gradation of $F$ is independent.

Proof: Let $C:P\to\text{Sub}(M)$ be an almost gradation of $F$, and suppose
that $Q\subseteq P$ is a finite subset and for each $q\in Q$ we have $c_{q}\in
C_{q}$ such that $\sum_{q\in Q}c_{q}=0$. Since $Q$ is finite, there exists a
maximal element $p\in Q$ of $Q$ (i.e. if $p\leq q$ for some $q\in Q$ then
$p=q$). Then we write

$c_{p}=-\sum_{q\in Q\smallsetminus\\{p\\}}c_{q}.$

Note that the LHS is contained in $F_{p}$, and the RHS is contained in
$\sum_{q\in Q-\\{p\\}}F_{q}$, so both sides are contained in
$F_{p}\cap\sum_{q\in Q\smallsetminus\\{p\\}}F_{q}\subset F_{<p}$ by
distributivity. Thus $c_{p}\in F_{<p}$ and since $c_{p}\in C_{p}$ and
$V_{p}=C_{p}\oplus F_{<p}$, it follows $c_{p}=0$, and then by induction on the
size of $Q$, all $c_{q}=0$ as desired. $\diamond$

###### Proposition 6.2

If $(E,I)$, and $(F,J)$ are linear filtrations of $M$ then for each $(i,j)\in
I\times J$, we have that

$[E\cap F]_{<(i,j)}=E_{i}\cap F_{<j}+E_{<i}\cap F_{j}.$ (8)

and therefore their intersection $(E\cap F,I\times J)$ is distributive.

Proof: By definition, we have that $[E\cap
F]_{<(i,j)}=\sum_{(k,\ell)<(i,j)}E_{k}\cap F_{\ell}$. We now show mutual
inclusion of the desired equality.

For the containment ($\subseteq$), suppose that $(k,\ell)<(i,j)$, and thus
either $k<i$ and $\ell\leq j$, or $k\leq i$ and $\ell<j$. Without loss of
generality, suppose that $k<i$ and $\ell\leq j$. Then we have that
$E_{k}\subseteq E_{<i}$, and $F_{\ell}\subseteq F_{j}$. Therefore $E_{k}\cap
F_{\ell}\subseteq E_{<i}\cap F_{j}$.

To show the containment ($\supseteq$), we prove only that $E_{i}\cap
F_{<j}\subseteq[E\cap F]_{<(i,j)}$, the other case being similar. Consider
$v\in E_{i}\cap F_{<j}\subset\sum_{\ell<j}F_{\ell}$. in particular we have
that $v\in F_{<j}=\sum_{\ell<j}F_{\ell}$. Since $(J,\leq)$ is totally ordered
and the sum is manifestly nonempty,
$\sum_{\ell<j}F_{\ell}=\bigcup_{\ell<j}F_{\ell}$. Therefore there exists some
$\ell^{\prime}<j$ such that $v\in F_{\ell^{\prime}}$. Hence $v\in E_{i}\cap
F_{\ell^{\prime}}\subseteq[E\cap F]_{<(i,j)}$.

Let $Q\subseteq I\times J$ and let $p=(i,j)\in Q$ be maximal. If $x\in[E\cap
F]_{p}\cap\sum_{Q\smallsetminus\\{p\\}}[E\cap F]_{q}$, then we can write
$x=\sum_{(\ell,m)\in Q\smallsetminus\\{p\\}}x_{\ell,m}$ where $x_{\ell,m}\in
E_{\ell}\cap F_{m}$. Because $p$ is maximal either $\ell<i$ or $m<j$. Since
each of the latter lie in $F_{<j}$ we have

$x-\sum_{\begin{subarray}{c}(\ell,m)\in Q\\\
\ell<i\end{subarray}}x_{\ell,m}\in F_{<j}.$

But of course ever term in the left hand side is in $E_{i}$ so

$x-\sum_{\begin{subarray}{c}(\ell,m)\in Q\\\
\ell<i\end{subarray}}x_{\ell,m}\in E_{i}\cap F_{<j}.$

On the other hand each $x_{\ell,m}\in E_{<i}$, so

$x\in E_{<i}+E_{i}\cap F_{<j}.$

and noting that $x\in F_{j}$ and $E_{i}\cap F_{<j}\subset F_{j}$ gives

$x\in E_{<i}\cap F_{j}+E_{i}\cap F_{<j}\subset[E\cap F]_{<(i,j)}$

by Eq. (8) $\diamond$

A _morphism of poset filtrations_ $\varphi:(E,P)\to(F,Q)$ of an $R$ module $V$
is a order-preserving map $\varphi:P\to Q$ such that $F\circ\varphi=E$.

###### Lemma 6.3

Suppose that $\varphi:(E,P)\to(F,Q)$ is a morphism of poset filtrations which
is an order-embedding, that $(F,Q)$ is distributive, and that and for all
$p\in P$ we have that $F_{<\varphi(p)}\subseteq E_{<p}$. Then $(E,P)$ is
distributive as well.

Proof: Let $S\subseteq P$ be a finite subset and $p\in S$ be maximal. Then
$\varphi(S)\subseteq Q$ is finite and $\varphi(p)\in\varphi(S)$ is maximal
because $\varphi$ is order-reflecting. Then

$E_{p}\cap\sum_{s\in
S\smallsetminus\\{p\\}}E_{s}\overset{(1)}{=}F_{\varphi(p)}\cap\sum_{s\in
S\smallsetminus\\{p\\}}F_{\varphi(s)}\overset{(2)}{=}F_{\varphi(p)}\cap\sum_{q\in\varphi(S)\smallsetminus\\{\varphi(p)\\}}F_{q}\overset{(3)}{\subseteq}F_{<\varphi(p)}\overset{(4)}{\subseteq}E_{<p}$

where (1) follows because $\varphi$ is a poset filtration morphism, (2)
follows because order-embeddings are injective, (3) follows because $(F,Q)$ is
distributive, and (4) follows by hypothesis. $\diamond$

###### Lemma 6.4

The order-embedding
$\iota:\mathcal{C}\to(\Omega_{1}\cup\\{-\infty\\},<_{L})\times(\Omega_{1}\cup\\{\infty\\},<_{R})$
defined in Lem. 6.1 satisfies the condition that for all
$\alpha\in\mathcal{C}$, $[L\cap R]_{<\iota(\alpha)}\subseteq F_{<\alpha}$.

Proof: Suppose that $\iota(\alpha)=(e_{L},e_{R})$, and by Equation 6.2,
$[L\cap R]_{<\iota(\alpha)}=L_{e_{L}}\cap R_{<e_{R}}+L_{<e_{L}}\cap
R_{e_{R}}$. It suffices to show that $L_{e_{L}}\cap R_{<e_{R}}$ and
$L_{<e_{L}}\cap R_{e_{R}}$ are contained in $F_{<\alpha}$, and we only show
the former, the latter being similar.

Note that $R_{<e_{R}}=\sum_{e<_{R}e_{R}}R_{e}=\bigcup_{e<_{R}e_{R}}R_{e}$, the
last equality holding because $\Omega_{1}\cup\\{\infty\\}$ is linearly ordered
and some $R_{e}$ is nonempty. Therefore $L_{e_{L}}\cap
R_{<e_{R}}=L_{e_{L}}\cap\bigcup_{e<_{R}e_{R}}R_{e}=\bigcup_{e<_{R}e_{R}}L_{e_{L}}\cap
R_{e}$. Define $\beta=\iota^{-1}(e_{L},e)$ since $\iota$ is injective and note
$\beta<\alpha$ since $\iota(\beta)<\iota(\alpha)$ and $\iota$ is order-
reflecting. So $F_{\beta}\subset F_{<\alpha}$ and hence $L_{e_{L}}\cap
R_{<e_{R}}\subset F_{<\alpha}$. $\diamond$

###### Corollary 6.1

Let $V$ be an $A_{\infty,\infty}$-representation. Every almost gradation of
the poset filtration $(F,C)$ is independent.

Proof: By Lem. 6.1 and Lem. 6.2 $\iota:(F,C)\to(L\cap
R,\Omega_{1}\cup\\{-\infty\\}\times\Omega_{1}\cup\\{\infty\\})$ is an order-
embedding morphism. By Prop. 6.2 $(L\cap
R,\Omega_{1}\cup\\{-\infty\\}\times\Omega_{1}\cup\\{\infty\\})$ is
distributive. By Lem. 6.4, $\iota$ satisfies the necessary conditions to apply
Lem. 6.3, which implies that $(F,C)$ is also distributive. Hence Prop. 6.1
gives the desired result. $\diamond$

###### Theorem 1

If $V$ is an $A_{\infty,\infty}$-representation then $V$ can be written as a
direct sum of representations, each of which is isomorphic to $W^{\alpha}$ for
some connected subquiver $\alpha$, and each $W^{\alpha}$ is an isotypic.

Proof: By Prop. 4.1, $(F,C)$ is a poset filtration of subrepresentations of
$V$. By Prop. 5.2 $W^{\alpha}$ is an almost gradation of $(F,C)$, by Cor. 6.1
$W^{\alpha}$ is independent, and by Prop: 5.3, $W^{\alpha}$ spans $V$. By
Prop. 5.4 each $W^{\alpha}$ is an isotypic. $\diamond$

##  7 Description of the Indecomposables

###### Proposition 7.1

For each full connected subquiver $\alpha\in\mathcal{C}$ the isomorphism class
of representations $X^{\alpha}$ with $X^{\alpha}_{i}$ being zero dimensional
if $i$ is not in $\alpha$ and $1$-dimensional if $i$ is in $\alpha$ and with
$f_{e}$ a bijection for each arrow in $\alpha$ is indecomposable, and thus
every representation of an eventually outward $A_{\infty,\infty}$ quiver is a
direct sum of indecomposables $X^{\alpha}$.

Proof: Each $X^{\alpha}$ is clearly indecomposable, because if it were a sum
of two subrepresentations then each is either one or zero dimensional at each
vertex, and thus each is the sum of the $X^{\alpha}_{i}$ for some subset of
the vertices and the other is the sum for the complementary set of vertices.
If each is a nonempty set, since $\alpha$ is connected there must be a pair of
vertices connected by an arrow from different subrepresentations. Since the
morphism associated to that edge is a bijection, this is impossible.

By Props. 5.1 and 5.4, each $W^{\alpha}$ is a sum of copies of $X^{\alpha}$,
and by Thm. 1 any representation is a sum of $X^{\alpha}$. $\diamond$

Note that the indecomposables in Prop. 7.1 are exactly the indecomposable
representations found by Bautista et. al. [BLP11] in their Prop. 5.9, although
they deal only with locally finite representations.

There is a useful interpretation of this with the Euler form. Let us define
the _root space_ of the quiver to be the space of functions from $\Omega_{0}$
to the real numbers which are zero on all but finitely many vertices. If $n$
and $m$ are two such functions define the Euler form to be

$\left\langle
n,m\right\rangle_{\Omega}=\sum_{i\in\Omega_{0}}n_{i}m_{i}-\frac{1}{2}\sum_{e\in\Omega_{1}}\left(n_{s(e)}m_{t(e)}+n_{t(e)}m_{s(e)}\right)$
(9)

where of course $n_{i}$ is the value of $n$ at $i$.

Observe that the Euler form is positive definitive because the corresponding
Tits quadratic form satisfies

$\left\langle
n,n\right\rangle_{\Omega}=\frac{1}{2}\sum_{i\in\Omega_{0}}(n_{i}+n_{i+1})^{2}>0$

unless every $n_{i}=-n_{i+1}$, which contradicts the finiteness.

Define the _weight space_ of the quiver to be the space of functions from
$\Omega_{0}$ to the real numbers, with no restriction on the values. Observe
that if $n$ is in the weight space and $m$ is in the root space then Eq. (9)
still makes sense and thus the weight space can be thought of as functionals
on the root space. Define a uniform topology on the weight space with an
entourage for each finite subset $S$ of the vertices, the open neighborhood of
each $n$ being the weights that assign the same value as $n$ to every vertex
in $S$. Equivalently, Cauchy sequences are those that for each such $S$ are
eventually constant on each vertex of $S$. It is easy to check that the weight
space is the completion of the root space in this topology and that each root
gives through the Euler form a continuous map from the weight space to the
real numbers (the reals are given the discrete topology).

Observe that every representation of a quiver which is locally finite-
dimensional (that is the vector space at each vertex is finite-dimensional)
determines a natural number-valued weight called its dimension vector that
assigns to each vertex the dimension of the associated vector space.

###### Proposition 7.2

For the quiver $A_{\infty,\infty}$ a weight valued in the natural numbers is
the dimension of an indecomposable representation if and only if it is in the
closure of the set of all $\mathbb{N}$-valued roots of length $1$.

Proof: By Prop. 7.1 the representations $X^{\alpha}$ are the only
indecomposables. One checks that if $\alpha$ consists of finitely many
vertices its dimension is of length $1$. It is also clear that if $\alpha$ is
infinite it is in the closure of the set of all finite connected subquivers
$\alpha^{\prime}$ contained in it.

Suppose first $n$ is a root valued in the natural numbers, whose support
$\alpha$ (vertices $i$ with $n_{i}>0$) is connected. Then if $i$ and $j$ are
its left and right endpoints

$\left\langle
n,n\right\rangle_{\Omega}=\frac{1}{2}(n_{i}^{2}+n_{j}^{2})+\sum_{e\in\alpha}\frac{1}{2}\left(n_{s(e)}-n_{t(e)}\right)^{2}.$

The first term is at least $1$ and only $1$ if $n+i=n_{j}=1$. The second term
is at least $0$ and only $0$ if all $n_{k}$ values in between are equal.
Therefore $\left\langle n,n\right\rangle_{\Omega}=1$ if $n$ is all $1$s on
$\alpha$ and $\left\langle n,n\right\rangle_{\Omega}\geq 2$ otherwise. If $n$
has disconnected support $\left\langle n,n\right\rangle_{\Omega}$ is the sum
of the values of each connected component, and therefore at least $2$. So if
$n$ is a root it is length $1$ only when it corresponds to $\alpha$.

If $n$ is in the closure of the set of all $\mathbb{N}$-valued roots of length
$1$, it cannot assign a number bigger than $1$ to any vertex, because an open
set around it based on a finite set of vertices containing that vertex would
include no $\mathbb{N}$-valued roots of length $1$. So it is the indicator
function of a set of vertices. If that set were not connected it has a finite
gap, and a finite set of vertices containing that gap and the points on either
side would give an open set that includes no $\mathbb{N}$-valued roots of
length $1$. So every weight in the closure is of the desired form.

The dimensions of course determine the indecomposable representation, because
if a representation has the same dimensions as $X^{\alpha}$ but is not
isomorphic, then for one edge $e\in\alpha$ we must have $f_{e}=0$. In that
case the representation decomposes as the sum of the subrepresentations
supported respectively behind and in front of $e$.

$\diamond$

##  8 A Representation which is not Krull-Schmidt

In this section we give an example of a representation of a type
$A_{\infty,\infty}$ quiver which is not Krull-Schmidt. It follows from Thm. 1
that this quiver is not eventually outward.

${\ldots}$${x_{-2}}$${x_{-1}}$${x_{0}}$${x_{1}}$${x_{2}}$${\ldots}$$\scriptstyle{a_{-3}}$$\scriptstyle{a_{-2}}$$\scriptstyle{a_{-1}}$$\scriptstyle{a_{0}}$$\scriptstyle{a_{1}}$$\scriptstyle{a_{2}}$
Figure 2: The Quiver $\Omega$

Let $\Omega$ be the $A_{\infty,\infty}$ quiver in Figure 2 with all arrows
pointing towards the vertex $x_{0}$, and note that this quiver is not
eventually outward. Let $V$ be the representation of $\Omega$ with $V_{n}=0$
for all $n<0$ and for $n\geq 0$, $V_{n}$ is the vector space of all sequences
$(b_{0},b_{1},\ldots)$ with entries in $\mathbb{F}$ and $b_{m}=0$ for all
$m<n$. Let $f_{n}:V_{n+1}\to V_{n}$ be the inclusion map for all $n\geq 0$.

Suppose that $V$ is Krull-Schmidt, i.e. that $V=\bigoplus_{i\in I}V^{i}$ where
for each $i\in I$, $V^{i}$ is an indecomposable representation of $\Omega$.

###### Lemma 8.1

For all $k\geq 0$ there exists $i_{k}\in I$ and nonzero $x\in V^{i_{k}}_{k}$
such that $x\in V_{k}\smallsetminus f_{k}(V_{k+1})$.

Proof: Consider the standard basis vector $e_{k+1}=(0,\ldots,0,1,0,\ldots)\in
V_{k}\smallsetminus f_{k}(V_{k+1})$. Since $V=\bigoplus_{i\in I}V^{i}$, we can
write $e_{k+1}=v_{1}+\ldots+v_{n}$ where $v_{m}\in V^{i_{m}}_{k}$ and
$i_{m}\in I$. It is not possible that $v_{m}\in f_{k}(V_{k+1})$ for all $1\leq
m\leq n$ because then $e_{k+1}\in f_{k}(V_{k+1})$, which is false. Hence there
must be some $v_{m}\in V_{k}\smallsetminus f_{k}(V_{k+1})$, and taking
$x=v_{m}$ and $i_{k}=i_{m}$ yields the desired result. $\diamond$

###### Lemma 8.2

If there exists $i\in I$ and nonzero $x\in V^{i}_{k}$ such that $x\in
V_{k}\smallsetminus f_{k}(V_{k+1})$ then $V^{i}_{m}=0$ if $m<0$ or $m>k$ and
$V^{i}_{m}=\operatorname{Span}(x)$ for all $0\leq m\leq k$.

Proof: We prove this by induction on $k$. Suppose $k\geq 0$ and assume the
result holds for all $0\leq n<k$.

First prove that given $n<k$, $f_{n}:V^{i}_{n+1}\to V^{i}_{n}$ is surjective.
Given $v\in V^{i}_{n}$, if $v\in V_{n}\smallsetminus f_{n}(V_{n+1})$, then by
induction $V^{i}_{m}$ is $0$ for $m>n$ and in particular $V^{i}_{k}=0$
contradicting that $x\neq 0$. Therefore it must be that $v\in f_{n}(V_{n+1})$,
and we call its preimage $v^{\prime}$. Then by our direct sum decomposition of
$V$ we can write $v^{\prime}=\sum_{t=1}^{k}v_{i_{t}}$ where $v_{i_{t}}\in
V^{i_{t}}_{n+1}$. But then
$v=f_{n}(v^{\prime})=\sum_{t=1}^{k}f_{n}(v_{i_{t}})$ and $f_{n}(v_{i_{t}})\in
V^{i_{t}}_{n}$. Since $v\in V^{i}_{n}$ it must be that there is some $s$ such
that $i_{s}=i$ and $v=f_{n}(v_{i_{s}})$ but $f_{n}(v_{i_{t}})=0$ for all
$t\neq s$. Since $f_{n}$ is injective, it follows that
$v^{\prime}=v_{i_{s}}\in V^{i}_{n+1}$ and $v_{i_{t}}=0$ for all $t\neq s$,
hence $v\in f_{n}(V^{i}_{n+1})$ as desired.

Given $x\in V^{i}_{k}$ such that $x\in V_{k}\smallsetminus f_{k}(V_{k+1})$ let
$C$ and $D$ be the subrepresentations of $\Omega$ given respectively below.

$\displaystyle\ldots\to 0\to$ $\displaystyle
f\left(\operatorname{Span}(x)\right)\overset{f_{0}}{\longleftarrow}\ldots\overset{f_{k-1}}{\longleftarrow}\operatorname{Span}(x)\overset{f_{k}}{\longleftarrow}0\ldots$
$\displaystyle\ldots\to 0\to$ $\displaystyle
ff_{k}(V_{k+1})\overset{f_{0}}{\longleftarrow}\ldots\overset{f_{k-1}}{\longleftarrow}f_{k}(V_{k+1})\overset{f_{k}}{\longleftarrow}V_{k+1}\overset{f_{k+1}}{\longleftarrow}\ldots$

Here $f=f_{0}\circ\ldots\circ f_{k-1}$. Note that $C$ is a subrepresentation
of $V^{i}$ and thus $C\oplus[D\cap V^{i}]\subset V_{i}$ (since $x\notin
f_{k}(V_{k+1})$, it follows that $C_{k}$ and $D_{k}\cap V^{i}_{k}$ are
independent). In fact $V^{i}=C\oplus[D\cap V^{i}]$. For $n>k$ it is obvious
that $V^{i}_{n}=C_{n}\oplus[D\cap V^{i}]_{n}$. If $v\in V^{i}_{k}\subset
V_{K}$ then $v=\alpha x+w$ where $\alpha\in\mathbb{F}$ and $w\in
f_{k+1}\left(V_{k+1}\right)$, but in fact $w=v-\alpha x\in V^{i}$ and thus
$w\in V^{i}_{k}\cap f_{k+1}\left(V_{k+1}\right)$, and so $v\in C\oplus(D\cap
V^{i})$. Thus $V^{i}_{k}=C_{k}\oplus(D_{k}\cap V^{i}_{k})$. Since $f_{n}$ is
injective and surjective on $V^{i}$ for $0\leq n<k$ the same is true for every
$n\geq 0$, and both side are $0$ for $n<0$, so $V^{i}=C\oplus(D\cap V^{i})$.

Since $V^{i}$ is indecomposable, this implies that $D\cap V^{i}$ is empty and
$V^{i}=C$. $\diamond$

###### Corollary 8.1

All $V^{i}$ are of the form

$\ldots\to
0\to\operatorname{Span}(x)\overset{f_{0}}{\longleftarrow}\ldots\overset{f_{k-1}}{\longleftarrow}\operatorname{Span}(x)\overset{f_{k}}{\longleftarrow}0\ldots$

for some $k\geq 0$.

Proof: Given $i\in I$, since $V^{i}$ is indecomposable, it is nonzero.
Therefore there exists a nonzero vector $x$ in $V^{i}_{\ell}$ for some
$\ell,\geq 0$. If $x\in V_{\ell}\smallsetminus f_{\ell}(V_{\ell+1})$ then by
Lem. 8.2 we are done. If $x=f_{\ell}(y)$ for $y\in V_{\ell+1}$ write $y=\sum
y_{t}$ for $y_{t}\in V^{i_{t}}$ and thus $f_{\ell}(y_{t})\in V^{i_{t}}$ and by
independence of the direct summands one $i_{t}$ must equal $i$ and $y\in
V^{i}$. Repeating this process we must come to a $z\in V^{i}_{k}\smallsetminus
f_{k}\left(V_{k}\right)$, since every nonzero sequence has a first nonzero
entry. $\diamond$

###### Lemma 8.3

For each $k\geq 0$ there is at most one $i\in I$ such that
$V^{i}_{k}\cap[V_{k}\smallsetminus f_{k}(V_{k+1})]$ is nonempty.

Proof: Suppose that $i,j\in I$ are such that there exists $x\in V^{i}_{k}$,
$y\in V^{j}_{k}$ and $x,y\in V_{k}\smallsetminus f_{k}(V_{k+1})$. By Lem. 8.2,
we have that $V^{i}$ and $V^{j}$ are equal, respectively, to the following
subrepresentations.

$\displaystyle\ldots\to 0\to$
$\displaystyle\operatorname{Span}(x)\overset{f_{0}}{\longleftarrow}\ldots\overset{f_{k-1}}{\longleftarrow}\operatorname{Span}(x)\overset{f_{k}}{\longleftarrow}0\ldots$
$\displaystyle\ldots\to 0\to$
$\displaystyle\operatorname{Span}(y)\overset{f_{0}}{\longleftarrow}\ldots\overset{f_{k-1}}{\longleftarrow}\operatorname{Span}(y)\overset{f_{k}}{\longleftarrow}0\ldots$

Now by hypothesis, $V_{k+1}=\bigoplus_{\ell\in I}V^{\ell}_{k+1}$, and then
because $f_{k}$ is injective, we have $f_{k}(V_{k+1})=\bigoplus_{\ell\in
I}f_{k}(V^{\ell}_{k+1})$. By Cor. 8.1, either $V^{\ell}_{k+1}=0$ or $f_{k}$ is
an isomorphism when restricted to $V^{\ell}_{k+1}$, hence we can write
$f_{k}(V_{k+1})=\bigoplus_{\ell\in L}V^{\ell}_{k+1}$ with $L=\\{\ell\in
I:V^{\ell}_{k+1}\neq 0\\}$.

Again by hypothesis, $V_{k}=\bigoplus_{\ell\in I}V^{\ell}_{k}$ and hence the
family $(V^{i}_{k},V^{j}_{k},V^{\ell}_{k}:\ell\in L)$ of $V_{k}$ is
independent. But this contradicts the fact that $V_{k}/V_{k+1}$ is
$1$-dimensional.

$\diamond$

###### Corollary 8.2

$V$ is not Krull-Schmidt.

Proof: If $V$ were a direct sum of indecomposable subrepresentations, say
$V=\bigoplus_{i\in I}V^{i}$, then by Cor. 8.1, each of the indecomposables is
of the form described in said corollary, which means for all $i\in I$ there
exists $k\geq 0$ such that $V^{i}_{k}\cap[V_{k}\smallsetminus f_{k}(V_{k+1})]$
is nonempty. But by Lem. 8.3 there is at most one such $i$ for each $k$, hence
$I$ must be countable. However $V^{i}_{0}$ is one-dimensional for each $i$ and
$V_{0}$ has uncountable dimension, yielding a contradiction. $\diamond$

## References

  * [BLP11] Raymundo Bautista, Shiping Liu, and Charles Paquette.  Representation theory of an infinite quiver.  arXiv preprint arXiv:1109.3176, 2011.
  * [Bri08] Michel Brion.  Representations of quivers.  https://www-fourier.ujf-grenoble.fr/~mbrion/notes_quivers_rev.pdf, 2008\.
  * [EER09] E Enochs, Sergio Estrada, and JR García Rozas.  Injective representations of infinite quivers. applications.  Canadian Journal of Mathematics, 61(2):315–335, 2009.
  * [Gab72] Peter Gabriel.  Unzerlegbare darstellungen i.  Manuscripta mathematica, 6(1):71–103, 1972.
  * [Rin16] Claus Michael Ringel.  Representation theory of Dynkin quivers. Three contributions.  Frontiers of Mathematics in China, 11(4):765–814, 2016.