Back to Representation Theory Page

On this page:

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

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]

Exact Sequences

In this section, we will review some of the basis of the language of exact sequences.

[definition] [deftitle]Definition %counter% (Exact sequences)[/deftitle]

A sequence of morphisms

is said to be exact if it is exact at each $M_i$ (that is, $\im f_{i-1} = \ker f_{i}$). [/definition]

[definition] [deftitle]Definition %counter% (Short exact sequences)[/deftitle]

A short exact sequence is an exact sequence of the form

[/definition]

These sequences are particularly nice because the exactness of $0 \to L \to M$ tells us that that $\ker f = 0$ so $f$ is injective. The exactness of $M \to N \to 0$ tells us that $\coker g = 0$ so $g$ is surjective. Finally, the exactness at $M$ tells us that $\im f = \ker g$. In some sense, this tells us that $M$ is “built from” $L$ and $N$. As an example, let $M$ be a left $R$-module and $L$ a left submodule of $M$, then

is a short exact sequence. Note that the “built from” is NOT as simple as just saying that $M$ is isomorphic to $L \oplus_R M/L$. A counterexample is

Note that $\ZZ/2\ZZ$ embeds into $\ZZ/4\ZZ$ as $2\ZZ$ contains $4\ZZ$ as a $\ZZ$-module so the Third Isomorphism Theorem implies

It is very clear, however, that $\ZZ/4\ZZ$ is NOT isomorphic to $\ZZ/2\ZZ \oplus_\ZZ \ZZ/2\ZZ$.

[example] [extitle]Example %counter% (Canonical decomposition of a morphism)[/extitle]

Let $f:M \to N$ be a morphism in $\Rep(Q)$. Then the sequence

is exact. Indeed:

• We know that the inclusion morphism $\iota$ is injective because each of the $k$-linear maps it encodes are inclusions of vector spaces (which are obviously injective).
• The sequence is exact at $f$ because $\iota$ embeds $\ker f$ into $f$.
• The sequence is exact at $N$ because the kernel of $\pi$ is literally $\im f$.
• We know that the quotient map $\pi$ is surjective because each of the $k$-linear maps it encodes are quotient maps of vector spaces (which are obviously surjective). [/example]

[example] [extitle]Example %counter% (Quotient SES)[/extitle]

Let $f:M \to N$ be a morphism in $\Rep(Q)$. Then the sequence

is short exact. Clearly exactness at $\ker f$ and $M/\ker f$ holds. To see exactness at $M$, we note let $f = (f_i : M_i \to N_i)$, $\pi = (\pi_i : M_i \to M_i/\ker f_i)$, and $\iota = (\iota_i : \ker f_i \to M)$. Since

and each

is short exact, it follows that

[/example]

[definition] [deftitle]Definition %counter% (Sections and retractions)[/deftitle]

A morphism is called a section if it is a right inverse of some morphism. A morphism is called a retraction if it is a left inverse of some morphism. [/definition]

[definition] [deftitle]Definition %counter% (Split SES)[/deftitle]

A short exact sequence

splits if $f$ is a section. [/definition]

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

Let $Q$ be the quiver $1 \to 2$. Consider the representations

We claim that

and

where $f = (0, 1)$, $g = (1, 0)$, $f’ = (0, 1)$, and $g’ = (1, 0)$ are both SES. Futhermore, the second example splits but the first does not. [/example]

[proof] (First example is SES) Clearly $f$ is injective and $g$ is surjective, and also both are morphisms so it remains to show that the sequence is exact at $M$. Noting that

it follows that

Thus, we have exactness.

(Second example is SES) Clearly $f’$ is injective and $g’$ is surjective, and also both are morphisms so it remains to show that the sequence is exact at $M$. Noting that

it follows that

(First example fails to split) For the first SES to split, there needs to be a left inverse $h:M\to S(2)$ of $f$. There is no such morphism, however, because commutativity of the diagram

implies that $h_2 = 0$ so the only morphism $h:M \to S(2)$ is the zero morphism and $h\circ f = 0$.

(Second example splits) On the other hand, the second example does split because the diagram

commutes for any $k$-linear map $h_2$. In particular, if we set $h_2 = 1$, then $h\circ f$ is the identity mapping on $S(2)$. [/proof]

[theorem] [thmtitle]Theorem %counter% (Splitting Lemma)[/thmtitle]

Let

be a SES in $\Rep(Q)$. Then sequence splits if and only if $g$ is a retraction. [/theorem]

We will skip the proof of this claim. Finally, there is the much more recognizable definition of a split exact sequence:

[theorem] [thmtitle]Theorem %counter% (Splitting Lemma)[/thmtitle]

Let

be a SES in $\Rep(Q)$. If the sequence is split exact, then $\im f$ is a direct summand of $M$. In particular, $M \cong L \oplus N$. [/theorem]

Hom Functors