Back to Representation Theory Page

On this page:

• Introduction
• Quiver Representations
• Indecomposable Representations
• $\Rep(Q)$ is an Abelian $k$-Category

Introduction

On this page, our goal will be to define the objects and morphisms of the category of representations of a quiver $Q$.

Quiver Representations

[definition] [deftitle]Definition %counter% (Quivers)[/deftitle]

A quiver is a directed multigraph $Q = (Q_0, Q_1, s, t)$ where:

• $Q_0$ is the set of vertices.
• $Q_1$ is the set of arrows.
• $s:Q_1 \to Q_0$ is the source map that sends an arrow to its source vertex.
• $h:Q_1 \to Q_0$ is the target map that sends an arrow to its target vertex.

A quiver is finite given $Q_0$ and $Q_1$ are finite sets. Unless otherwise stated, my quivers are always finite. [/definition]

[example] [extitle]Remark %counter%[/extitle]

Another common notation in the quivers community is $(Q_0, Q_1, h, t)$ where $h$ is the “head” of the arrow and $t$ is the “tail” of the arrow. [/example]

[example] [extitle]Remark %counter%[/extitle]

In my experience, a lot of combinatorialists/graph theorists call arrows arcs. In the areas I am personally in (representation theory and cluster algebras), everyone seems to use “arrows” instead, so I do as well. [/example]

In what follows, we will assume that $k$ is an algebraically closed field. It is common to assume $k = \CC$.

[definition] [deftitle]Definition %counter% (Quiver representations)[/deftitle]

A (quiver) representation $M = (M_i, \phi_\alpha)_{i\in Q_0, \alpha\in Q_1}$ of a quiver $Q$ is a collection consisting of:

• A $k$-vector space $M_i$ for every $i \in Q_0$.
• A $k$-linear map $\phi_a:M_{s(a)} \to M_{t(a)}$ for every $\alpha \in Q_1$.

We say that $M$ is finite-dimensional if $\dim_k M_i < \infty$ for every $M_i$. We define the dimension vector

Finally, an element of a representation $M$ is a tuple $(m_i)_{i\in Q_0}$ with $m_i \in M_i$. [/definition]

On my pages, unless I say otherwise, all representations are assumed to be finite-dimensional.

[example] [extitle]Example %counter%[/extitle]

Let $Q$ be the quiver $1 \rightarrow 2 \leftarrow 3$. Then

is a representation of $Q$ with dimension vector $(2, 1, 1)$. [/example]

[definition] [deftitle]Definition %counter% (Morphism of quiver representations)[/deftitle]

Let $M = (M_i, \phi_\alpha)$ and $M’ = (M_i’, \phi_\alpha’)$ be representations of the same quiver $Q$. A morphism $f:M \to M’$ of representations $M$ and $M’$ is a collection $(f_i:M_i \to M_i’)_{i\in Q_0}$ of $k$-linear maps such that for every $i \overset{\alpha}{\to} j$, the diagram

commutes. If each $f_i$ is a $k$-linear isomorphism, then we say that $f$ is an isomorphism. The class of all representations isomorphic to a given representation $M$ is called the isoclass of $M$. [/definition]

[example] [extitle]Example %counter%[/extitle]

The zero representation is the representation consisting only of the zero vector space. Furthermore, there is an obvious identity morphism in $\Hom(0, 0)$. This says that the zero representation is the zero object of $\Rep(Q)$. [/example]

[example] [extitle]Example %counter% (Inclusion morphism)[/extitle]

There is a natural notion of inclusion for quiver representations. Say that $M = (M_i, \phi_\alpha)$ and $N = (N_i, \psi_\alpha)$ are representations of $Q$ for which $M_i$ is a subspace of $N_i$ for every $i \in Q_0$ and $\phi_\alpha = \psi_\alpha|_{M_{s(\alpha)}}$ for every $\alpha \in Q_1$. Then there is an inclusion morphism $M \injto N$ given by sending $M_i \mapsto N_i$ via inclusion for every $i\in Q_0$. Obviously, the corresponding diagram commutes, so the inclusion morphism is actually a morphism. [/example]

[example] [extitle]Example %counter% (Projection morphism)[/extitle]

There is a natural notion of projection morphisms for quiver representations defined in the obvious way. [/example]

[definition] [deftitle]Definition %counter% (Category $\Rep(Q)$)[/deftitle]

Fix a quiver $Q$ along with an algebraically closed field $k$. Then $\Rep(Q)$ is the category of finite-dimensional representations of $Q$ where the objects are the finite-dimensional representations of $Q$ and the morphisms are as above. [/definition]

Indecomposable Representations

[definition] [deftitle]Definition %counter% (Direct sums)[/deftitle]

Let $M = (M_i, \phi_a)$ and $M’ = (M_i’, \phi_a’)$ be representations of $Q$. Then the direct sum of $M$ and $M’$ is defined as

[/definition]

It is immediate that the direct sum is a representation of $Q$ so we omit the proof of this fact. One further note: we define arbitrary finite direct sums inductively:

[example] [extitle]Example %counter%[/extitle]

Let $Q$ be the quiver $1 \rightarrow 2 \leftarrow 3$. Given the representations

Then the direct sum $M \oplus M’$ is the representation

We claim that representation is isomorphic to

We begin by constructing the diagram

and each map (going downward) is a $k$-linear isomorphism so it follows that $(I_3, I_3, \mqty[0 & 1])$ is an isomorphism of representations as desired. [/example]

Just like any other branch of algebra, our interest is to classify the objects we are working with up to isomorphism. In representation theory of quivers, this amounts to classifying the representations and the morphism up to isomorphism.

[definition] [deftitle]Definition %counter% (Indecomposable represenations)[/deftitle]

A representation $M$ of a quiver $Q$ is called indecomposable if $M\neq 0$ and $M$ is not the direct sum of two nonzero representations. [/definition]

[theorem] [thmtitle]Theorem %counter% (Krull-Schmidt Theorem)[/thmtitle]

Let $M$ be a representation of the quiver $Q$. Then $M \cong \bigoplus_{i=1}^t M_i$ where each $M_i\in \Rep(Q)$ is indecomposable. Furthermore, the decomposition is unique up to reordering of the summands. [/theorem]

$\Rep(Q)$ is an Abelian $k$-Category

Our goal in this section will be to show that $\Rep(Q)$ is an abelian $k$-category. Recall the requirements of an abelian category:

[definition] [deftitle]Definition %counter% (Abelian category)[/deftitle]

We say that a category $\mathcal{C}$ is abelian provided that

• $\Hom(M, N)$ is an abelian group for every objects $A$ and $B$ of $\mathcal{C}$.
• Morphisms distribute over the operation of $\Hom(M, N)$.
• $\mathcal{C}$ has a zero object.
• $\mathcal{C}$ has finite products and coproducts.
• Every morphism has a kernel and cokernel.
• Every monomorphism is a kernel and every epimorphism is a cokernel.

If, in addition, $\Hom(M, N)$ is an $R$-module for all objects $M$ and $N$, and the composition of morphisms is $R$-bilinear, then $\mathcal{C}$ is an abelian $R$-category. [/definition]

It is fairly immediate that $\Hom(M, N)$ (in the category $\Rep(Q)$) is a $k$-vector space and the zero object, of course, is just the zero representation. The direct sum satisfies the finite products and coproducts requirement as well. Thus, we at the very least have an additive $k$-category. Let us now show the remaining requirements.

[theorem] [thmtitle]Theorem %counter% (Kernels in $\Rep(Q)$)[/thmtitle]

Let $M \xrightarrow{g} N$ be a morphism of representations of $Q$. Then there is morphism $L \xrightarrow{f} M$ (called the kernel of $g$) such that

• $gf = 0$.
• Given any morphism $X \xrightarrow{v} M$ such that $gv = 0$, $v$ factors uniquely through $f$.

[/theorem]

[proof] Let $M = (M_i, \phi_\alpha)$. We define $L = (L_i, \psi_\alpha)$ by setting each $L_i \coloneqq \ker g_i$ and each $\psi_\alpha \coloneqq \phi_\alpha|_{L_{s(\alpha)}}$. Furthermore, let $f:L \injto M$ be the inclusion morphism. Note that $\psi_\alpha$ is actually well-defined as the commutativity of the diagram of $g$ implies the image of $L_{s(\alpha)}$ under $\psi_\alpha$ is a subset of $L_{t(\alpha)}$. The inclusion of $L_i$ into $M_i$ makes it clear that $gf = 0$ as well.

To show the uniqueness claim, we define $u$ by setting it equal to $v$ which is fine as $f$ is the inclusion morphism. From there, the commutativity of the diagram forces the uniqueness of $u$. [/proof]

[definition] [deftitle]Definition %counter% (Kernel of quiver representation)[/deftitle]

Let $(M_i, \phi_\alpha) \xrightarrow{f} (N_i, \psi_\alpha)$ be a morphism of representations of $Q$. Then the kernel of $f = (f_i)$ is the representation

[/definition]

[theorem] [thmtitle]Theorem %counter% (Cokernels in $\Rep(Q)$)[/thmtitle]

Let $L \xrightarrow{f} M$ be a morphism of representations of $Q$. Then there is a morphism $M \xrightarrow{g} N$ (called the cokernel of $f$) such that:

• $gf = 0$.
• Given any morphism $M \xrightarrow{v} X$ such that $vf = 0$, $v$ factors uniquely through $g$.

[/theorem]

[proof] Denote $M$ by $(M_i, \phi_\alpha)$. We define $N = (N_i, \psi_\alpha)$ by setting $N_i \coloneqq \coker f_i$ and

to be the induced map on the cokernels. That is,

where the bar denotes the represented coset. Since the proof of this map being well-defined is exactly the same for doing this for groups/modules/rings/etc., we will omit the proof.

It is immediate that $gf = 0$ as we quotient out each $M_i$ by the image of $f_i$ and each $g_i$ is the canonical projection map onto the quotient. Therefore, $gf = 0$. The commutativity of the diagram above also ensures that $u = v$. [/proof]

[definition] [deftitle]Definition %counter% (Cokernel of quiver representation)[/deftitle]

Let $(M_i, \phi_\alpha) \xrightarrow{f} (N_i, \psi_\alpha)$ be a morphism of representations of $Q$. Then the cokernel of $f = (f_i)$ is the representation

where each $\psi_\alpha$ is the composition of $\phi_\alpha$ and the canonical projection. [/definition]

So now we have kernels and cokernels. Before we start making statements about the normality (the final condition in abelian categories) of monomorphisms and epimorphisms, we characterize them in terms of mono/epimorphicity of the underlying $k$-linear maps.

Recall that $f:M \to N$ is an epimorphism provided that

for every $g_1, g_2:N \to P$. Similarly, we call that $f:M \to N$ is a monomorphism provided that

for every $g_1, g_2:L \to M$.

[definition] [deftitle]Definition %counter% (Injective and surjective)[/deftitle]

Let $f:M \to N$ be a morphism of quiver representations. We say that $f$ is injective provided that every $f_i$ is. Similarly, we say that $f$ is surjective provided that every $f_i$ is. [/definition]

[theorem] [thmtitle]Theorem %counter% (Epi = surjection and mono = injection in $\Rep(Q)$)[/thmtitle]

Let $f:M \to N$ be a morphism of quiver representations. Then

1. $f$ is a monomorphism if and only if $f$ is injective.
2. $f$ is an epimorphism if and only if $f$ is surjective. [/theorem]

[proof] (First claim) Suppose that $f\circ g_1 = f\circ g_2$ where $g_1, g_2: L \to M$ are morphisms quiver represenations. Then

So then, $f_i$ is a monomorphism in the category of $k$-vector spaces which we know to be an equivalent condition to $f_i$ being injective.

(Second claim) Same argument as above. [/proof]

Now we just need a couple more definitions:

[definition] [deftitle]Definition %counter% (Subrepresentations and quotients)[/deftitle]

A representation $M$ is called a subrepresentation of the represenation N if there is an injective morphism $M \injto N$. Furthermore, we define the quotient representation M/L to be the cokernel of the injection $M \injto N$. [/definition]

With this terminology in place, the following theorem is immediate:

[theorem] [thmtitle]Theorem %counter% (First Isomorphism Theorem for $\Rep(Q)$)[/thmtitle]

If $f:M \to N$ is a morphism of representations, then there is a unique isomorphism

where $\im f$ is the representation $\im f \coloneqq (f(M_i), \psi_\alpha)$ where

for every arrow $i \xrightarrow{a} j$ in $Q$. [/theorem]

[proof] There is an obvious choice of isomorphism induced by the one we see from abstract algebra. [/proof]

[theorem] [thmtitle]Theorem %counter% (Normality in $\Rep(Q)$)[/thmtitle]

All monomorphims and epimorphisms in $\Rep(Q)$ are normal. [/theorem]

[proof] Consider the diagram

where $f$ has the property that $\pi\circ f = 0$. Since $\pi$ is just the canonical projection, this implies that $X$ can be embedded into $L$. So we define the map $g:X \to L$ by the embedding of $X$ into $L$. Thus, we see that $L \injto M$ is a kernel. A very similar argument shows that surjections are cokernels and we will omit the proof. [/proof]

Thus, we finally arrive at the result we were wanted to come to!

[theorem] [thmtitle]Theorem %counter% ($\Rep(Q)$ is an abelian $k$-category)[/thmtitle]

$\Rep(Q)$ is an abelian $k$-category. [/theorem]