Indecomposable Quiver Representations

Indecomposable Quiver Representations Summer Project 2015 Laura Vetter September 2, 2016

Introduction The aim of my summer project was to gain some familiarity with the representation theory of finite-dimensional algebras, developing my knowledge from linear algebra. In this report, I will discuss some of the theory I have studied and outline some important results I came across. In this first chapter, I will introduce the basic notions of the representation theory of quivers along with some examples. In the next, I will concentrate on the classification problem in representation theory. I will show how to find the indecomposable representations of quivers of type A n and look at some quivers of infinite representation type. In the last chapter, I will show that quiver representations are in some sense the same as modules over the path algebra. As part of my project, I also looked at a more sophisticated tool for describing representations of finite-dimensional algebras and morphisms between them, the so-called Auslander-Reiten quiver, but I decided not to include it in this report. Representation theory studies representations of sets with certain algebraic structures, such as groups or algebras. These representations associate linear maps between vector spaces to each member of the set. In this way, they provide information about these often complicated algebraic structures through the relatively well-understood tools from linear algebra. In some sense, doing representation theory is very similar to doing linear algebra. But instead of finding canonical forms for individual linear maps, representation theory often involves considering several linear maps simultaneously and finding canonical forms for such families of maps. I specifically concentrated on trying to understand representations of quivers, which, as we shall see later, is in an sense the same as trying to understand representations of finite-dimensional algebras over an algebraically closed field. Quivers and their representations We will start with the definition of a quiver and that of a quiver representation. The definitions in this chapter are all based on Schiffler, (2014, Chapter 1). 1

Definition 1. A quiver Q is a set of vertices connected by arrows (or directed graph). More formally, it is a 4-tuple (Q 0, Q 1, s, t), where Q 0 is a set of vertices Q 1 is a set of arrows s : Q 1 Q 0 is a map sending arrows to their starting points, and t : Q 1 Q 0 is a map sending arrows to their terminal points. Naturally, each arrow α Q 1 runs from its starting point s(α) to its terminal point t(α). The quiver defined by Q 0 = {1, 2, 3}, Q 1 = {α, β, γ}, s(α) = 1, s(β) = 2, s(γ) = 2 and t(α) = 2, t(β) = 1, t(γ) = 2 looks like this: γ 1 α 2. β Definition 2. A representation M = (M i, φ α ) i Q0,α Q 1 of a quiver Q associates to each vertex a vector space M i over a field k and to each arrow a k-linear map φ α : M s(α) M t(α). A representation of the quiver given above is: ( ) 1 0 k 2 k. 0 1 1 From now on I will assume that the field k is algebraically closed and that the vector spaces associated to each vertex are finite dimensional. 2

Morphisms In linear algebra, there exist linear maps between vector spaces. In the representation theory of quivers, similar structure-preserving maps exist between two representations. They are called morphisms of representations. Definition 3. If M = (M i, φ α ) i Q0,α Q 1 and M = (M i, φ α) i Q0,α Q 1 are representations of Q, then a morphism between them is a family of linear maps f = (f i ) : M i M i such that f j φ α (m) = φ α f i (m) for all m M i and all α : i j Q 1. This is the same as saying that the following diagram commutes: M i φ α M j f i f j φ α M i M j. We have an isomorphism of representations if and only if each f i has an inverse which can be denoted by f 1 i as it must be unique. Let us look at the following two representations of the quiver 1 2 and a morphism between them: A g B f 1 f 2 h C D. If f is an isomorphism, f 1, f 2 must have inverses and we must have f 2 g = h f 1. This implies that A must be isomorphic to C and B to D. Moreover, we need that g = f2 1 h f 1. This is the same as saying that these representations are isomorphic if one representation can be obtained from the other by changing the bases of the vector spaces in the representation. In fact, this is true in general. Definition 4. A representation K is a subrepresentation of a representation M if we can find a morphism f : K M such that each of the f i is injective. Definition 5. If Q is a quiver and M, M are representations of Q with M = (M i, φ α ) i Q0,α Q 1 and M = (M i, φ α) i Q0,α Q 1, then their direct sum is given by: M M = ( [ ]) M i M i, φα 0 0 φ. α i Q 0,α Q 1 3

If we let Q be the quiver 1 2 3 and consider representations: 1 0 0 1 M = k k 2 k 1 0 1 0 M = k k 2 k Then M = k 1 k 0 0 0 0 k 1 k and M = k 1 k 1 k 0 0 k 0 0. Definition 6. A quiver representation M is said to be indecomposable if M is not isomorphic to the direct sum of two nonzero representations M, M. The aim of representation theory is to classify all representations of a given quiver Q and all morphisms between these representations up to isomorphism. There is an important theorem about representation decomposition which makes this classification process a lot easier. It implies that once we have decomposed a representation of a quiver into indecomposables, we don t have to look for any other decompositions since every decomposition into indecomposables is essentially unique. Therefore, it is important to understand indecomposable representations up to isomorphism as these are the building blocks for all the other representations. Theorem 1 (Krull-Schmidt Theorem). Let Q be a quiver and let M be a representation of Q. Then M = M 1 M 2... M s where the M i are indecomposable representations of Q. Moreover, these indecomposables are unique up to order. Consider again the above representations M and M. Using the Krull-Schmidt Theorem, we can see that they are not isomorphic, because their decompositions are different. 4

Decomposing representations Quivers of type A n In the last chapter, we saw that one of the main aims of representation theory is to classify indecomposable representations. With only tools from linear algebra at one s disposal, this is often a difficult or even impossible task. In these cases, the more refined tools from Auslander-Reiten theory prove useful. In this chapter, however, we will present some examples of relatively simple quivers and explicitly decompose their representations using only theory from linear algebra. In the first section, we will look at an interesting subclass of quivers of finite representation type, namely those of Dynkin type A. We will decompose some simple quiver representations of this type and then do this for the general case, using the method discussed in C. M. Ringel, (2013). For a discussion of quivers of other Dynkin types, see C. Ringel, (2016). In the second section, we will consider two quivers of infinite representation type. Definition 7. A quiver Q is of type A n if and only if the graph obtained from Q by forgetting the direction of the arrows (its underlying graph ), is of the following form: 1 2 3 n 1 n. Example 1 The simplest quiver one can think of consists of one single vertex. Clearly, this is a quiver of type A 1. Its representations will consist of one single vector space V without any maps from or onto it. We know from linear algebra that each vector space admits a basis, which implies either that it can be decomposed into indecomposable one-dimensional vector spaces or that it is already an indecomposable one-dimensional vector space. Therefore, we can conclude that this quiver has only one indecomposable representation (up to isomorphism), namely a one-dimensional vector space, k. 5

Example 2 Now let us look at the A 2 quiver, a representation of which looks like this: φ V W. From linear algebra, we know that kerφ is a subspace of V so we can find a basis B V of kerφ and extend it to a basis B V of V. Now φ(b V ) forms a basis B W of the subspace imφ of W. Finally, we can extend B W to a basis of W. This yields the following decomposition: 0 kerφ 0 V 0 imφ 0 W. where V is a complement of kerφ in V and W is a complement of imφ in W. Both φ and the zero map are linear so they respect the bases found for kerφ, V, imφ, W. In other words, they send each basis element on the left of the arrow to a basis element on the right (or to 0 in the case of the zero map). Therefore, above direct summands can be further decomposed into multiples of k 1 k, k 1 k and 0 0 k respectively. Since we started with an arbitrary representation, we can conclude that any indecomposable representation of this quiver must be isomorphic to one of these three representations. In the next three examples I ll consider the indecomposable representations of quivers of type A 3. There are three different quivers of this type since there are three ways in which the arrows may be oriented. In a similar way to example (2), we will start with an arbitrary representation of a quiver and proceed by splitting off indecomposable direct summands until we reach some representation that cannot be split up further (that is, an indecomposable representation). Since we are working with finite-dimensional representations, this process is finite. Therefore, we can then conclude that any indecomposable representation of the given quiver must be one of the indecomposable representations found in this process. In the following examples, will mean an injective linear map and a surjective linear map. The general method to find indecomposables of a given representation involves splitting off direct summands of that representation that are compatible with the linear maps assigned to the edges of the quiver. After splitting off a direct summand, we will be left with a new representation and corresponding new vector spaces on the vertices of the given quiver. We will denote each new vector space on a given vertex by [name of the previous vector space on that vertex] (e.g. V becomes V ). However, the mappings between these vector spaces will not be renamed. Instead, we will assume that they are the original maps restricted to the new vector spaces. 6

Example 3 First of all, consider the quiver with both arrows pointing inwards and an arbitrary representation: φ ψ 1 2 3 U V W. for kerφ and kerψ to find the following direct sum- We can choose bases B U and B W mands: 0 0 0 0 kerφ 0 0 0 0 kerψ. These direct summands are multiples of the indecomposables k 0 0 0 0 and 0 0 0 0 k. We are left with the following representation: U φ ψ V W. Since we have split off the kernel of both φ and ψ, this direct summand contains two injective maps and is of the form U V W. This representation is isomorphic to a representation in which U and W are subspaces of V and the maps φ and ψ are the inclusion maps from U and W into V. The indecomposable direct summands of these representations are isomorphic. Therefore, we can proceed by thinking of U and W as subspaces of V. Doing so, we can see that U W is a subspace of V so we can find a basis B for it and find the following decomposition: U V W U W id U W id U W. The second representation is a direct sum of indecomposable k 1 k 1 k. Now we can assume that U W = 0, we extend our basis B in V by a linearly independent set B to get a complete basis for U and by a linearly independent set B to get a complete basis for W. The disjoint union B B is then a basis for the subspace U W. This leads to the following decomposition: U U W W 0 V 0. Applying the natural projections from U W onto U and W yields a decomposition 7

U U W W = (U U 0 0 W W ) These may be written as direct sums of indecoposables k 1 k 0 0 and 0 0 k 1 k. Finally, we extend B B B by a linearly independent set B of V to obtain a complete basis of V. This corresponds to splitting up the last direct summand 0 V 0 into indecomposables of the form 0 0 k 0 0. Thus, there are six indecomposable representations of this quiver: k 0 0 0 0 0 0 0 0 k k 1 k 1 k k 1 k 0 0 0 0 k 1 k 0 0 k 0 0. Example 4 Let us consider the quiver with both arrows pointing outwards and an arbitrary representation of it: φ ψ 1 2 3 U V W. This quiver is obtained by reversing the arrows in the quiver in the last example. Its representations are dual to the representations of the quiver from example 3, which suggests that we should proceed in a very similar way. The first step is splitting off cokerφ and cokerψ to get multiples of the indecomposables k 0 0 0 0 and 0 0 0 0 k and two surjective maps: U V W. kerφ kerψ is a subspace of V so we can find a basis B for it to find: U V W 0 kerφ kerψ 0. The second direct summand is a multiple of 0 0 k 0 0. Since we can assume that kerφ kerψ = 0 now, we can find a basis B for kerφ kerψ, and extend it by B to a basis of V : U kerφ kerψ W φ(v ) V ψ(v ). The second direct summand is a multiple of k 1 k 1 k. The first direct summand can be further decomposed into: 8

φ(kerψ) kerψ 0 0 kerψ ψ(kerφ). These are multiples of the indecomposable representations k 1 k 0 0 and 0 0 k 1 k. Again, we have found six indecomposables: k 0 0 0 0 0 0 0 0 k k 1 k 1 k k 1 k 0 0 0 0 k 1 k 0 0 k 0 0. Example 5 Next, I ll consider the quiver with the following orientation and an arbitrary representation of it: Splitting off kerφ and cokerψ we get: φ ψ 1 2 3 U V W U V W U 0 0 0 0 0 W The last two direct summands are multiples of k 0 0 0 0 and 0 0 0 0 k respectively. Now, we deal with a representation such that φ is injective and ψ is surjective. Again, up to isomorphism we may assume that φ is the inclusion of a subspace U into V. Moreover, kerψ is a subspace of V. Similar to the first case, we find a basis of V which is compatible with its two subspaces U and kerψ. U kerψ is a subspace of V so we can construct a basis B for it and find the following decomposition: U kerψ U 0 kerψ 0 U V W where the first direct summand corresponds to the indecomposable k 1 k 0 0. Now we can assume that U kerψ = 0, we extend our basis B in V by a linearly independent set B to get a complete basis for U and by a linearly independent set B to get a complete basis for kerψ. The disjoint union B B is then a basis for the subspace U kerψ. Also, we find a basis B W for ψ(u ) in W. This leads to the following decomposition: 9

U U ψ(u ) 0 V W The representation of the left decomposes into direct sums of k 1 k 1 k. Completing a basis for W by extending B W by a linearly independent set B W and completing a basis for V by extending B B B by B gives us: 0 V imv 0 V 0 splitting up these direct summands gives indecomposables 0 0 k 1 k and 0 0 k 0 0. Thus, there are six indecomposable representations of this quiver. k 0 0 0 0 0 0 0 0 k k 1 k 1 k k 1 k 0 0 0 0 k 1 k 0 0 k 0 0. We will now try to find out what the indecomposable representations of A n look like. But first, we need to introduce some definitions: Definition 8. A thin representation of a quiver of type A is a representation which associates to each vertex a vector space of dimension at most one, that is, either 0 or k. Moreover, the indecomposable representations we found all have connected support quivers. Definition 9. The support quiver Q(M) of a quiver Q is given by the vertices x with M x 0 and the arrows α with M α 0. For example, the support quiver defined relative to the representation k 1 2. 1 k 0 0 is It turns out to be the case that all indecomposable representations of quivers of type A n are thin representations of this form. The other implication holds as well: every thin representation with a connected support quiver is an indecomposable representation. Proposition 1. Let M be a representation of a quiver of type A n. Then M is an indecomposable representation if and only if M is a thin representation whose support quiver is connected. Let us first show the left implication: if M is a thin representation of a quiver Q of type A n whose support quiver is connected, then M is an indecomposable representation. 10

Proof. Suppose that M is not indecomposable. This implies we can find at least two direct summands of this representation such that M = M M. If the support quiver of M has only one vertex, then M must assign a one-dimensional vector space to that vertex (since it is thin) and the zero space to all the other vertices. It is clear that in that case we cannot find a decomposition for M because a onedimensional vector space is indecomposable. Suppose then that the support quiver of M has more than one vertex. Since M is a thin representation, and we must have that M x = M x M x, we can find vertices x, y in the support quiver of M, connected by an arrow α in Q such that M x = 0, M y = k, M x = k and M y = 0. Then both M, and M must assign the zero map to α. Taking the direct sum of M and M shows that M must also assign the zero map to α. But the support quiver of M is connected and so M cannot assign the zero map to α. Thus, M must be indecomposable. This proves that the indecomposable representations listed in the five examples above are really indecomposables (this was more or less assumed above). In order to prove the right implication, which says that if M is an indecomposable representation of a quiver of type A n then M is a thin representation whose support quiver is connected, we need to introduce two more definitions and a lemma (that I will leave to the reader to prove): Definition 10. A subquiver Q of a quiver Q is given by subsets (Q 0, Q 1) of (Q 0, Q 1 ) such that s(α), t(α) Q 0 for each α Q 1. Definition 11. A vertex x is a peak for a representation M if all maps M α towards M x are injective and all maps M β away from M x are surjective. So in the following representation, 3 is a peak: M = V 1 V 2 V 3 V 4 V 5. Lemma 1. An indecomposable representation M of Q of type A n with M x = k has x as a peak. Moreover, a representation of Q has x as a peak if and only if all its direct summands have x as a peak. Now we are in a position to prove the right implication of proposition 1. The proof follows C. M. Ringel, (2013). Proof. Since we have already seen that the indecomposable representations of A 1, A 2, A 3 are thin representations with connected support quivers, we can let n > 3 and argue by induction. Let us assume that the indecomposables of quivers of type A 4, A 5,..., 11

A n 1 are thin representations with connected support and look at the representations of quivers of type A n. We are dealing with a representation of the form: M = V 1 V 2 V 3 V n 1 V n. In order to apply the inductive hypothesis and the results we already found above, we need to restrict the original quiver to subquivers of type A n 1 or smaller. Let Q be the full subquiver with vertices 1, 2. The restricted representation is of the form: M = V 1 V 2. We have already seen that M can we written as a direct sum of indecomposable representations, which are thin and have connected support quivers. Let A be the direct sum of those indecomposables that have 2 as a peak then, by lemma 1, A must have 2 as a peak. Let B be the direct sum of the indecomposables with M 2 = 0. Then it follows that M = A B. Let Q be the subquiver with vertices 2,..., n 1, n and representation of the form: M = V 2 V n 1 V n. This time by the inductive hypothesis, we can write M as a direct sum of indecomposables that are thin and have connected support quivers and M = A B, with A has 2 as a peak and B 2 = 0. To complete the proof, we need to bring the subquivers and their representations back together in order to say something about our original quiver of type A n. We will define three subrepresentations of M: Let A be the subrepresentation of M defined in the following way: its restriction to Q is A and its restriction to Q is A. Let B be the subrepresentation of M such that its restriction to Q is B and B x = 0 for all x 2. Let C be the subrepresentation of M such that its restriction to Q is B and C x = 0 for all x 2. These are indeed subrepresentations and we can write M = A B C. Since we constructed B, B, A, A in such a way that B 2 = 0 = B 2, we have that M 2 = A 2 = A 2, and so the support of B is contained in {1}, the support of C is contained in {3, 4,..., n} and A is possibly supported everywhere and has 2 as a peak. It follows by construction that the direct sum of A, B and C does in fact equal M. 12

Now we repeat this argument on our direct summand A, but this time for vertex n 1 rather than vertex 2. We can see that A = D E F, with D is supported in {1, 2,..., n 2} and E has both 2 and n 1 as peaks (since all direct summands of A must already have 2 as a peak by our lemma) and F is supported in {n}. But since A has 2 as a peak, maps towards 2 are injective and maps away from 2 are surjective. This means that A cannot be of the form: 0 0 0 k an thus F must be zero. Going back to our original representation M, we can see that M = B C D E. As the support of B, C and D is contained in proper subquivers of Q, we can apply the inductive hypothesis on these direct summands. Since E has both 2 and n 1 as peaks, it is of the following form: V 1 V 2 V 3 V n 1 V n. Identifying V 2,..., V n 1 using these isomorphisms yields a representation of a quiver of type A 3. We know that a quiver of type A 3 has indecomposable representations that are thin and have connected support quivers. Therefore, E may be written as a direct sum of thin connected representations. Now M = B C D E and all these direct summands are sums of indecomposables that are thin and have connected support quivers. So we can conclude that the indecomposable direct summands of M must also be thin representations with connected support quivers. Therefore, all indecomposable representations of A n are of this form. Infinite representation types From what I have shown in the previous section, we can conclude that quivers of type A n have a finite number of indecomposable representations up to isomorphism. There also exist quivers for which this is not true. These quivers are said to be of infinite representation type. Here, I will consider two examples of quivers that are of infinite representation type. Example 6 Consider the quiver consisting of a single vertex with one arrow to itself: the loop quiver. A representation of this quiver looks as follows: 13

φ V The Jordan Normal Form Theorem in linear algebra states that we can choose a basis for V such that the endomorphism φ : V V can be represented by a matrix consisting of a direct-sum composition of Jordan blocks. This theorem also says that this representation is unique (op to the ordering of the Jordan blocks). Thus, the indecomposable representations of this quiver are given by the distinct Jordan blocks together with the corresponding vector space. As there exist k = different Jordan blocks for each possible dimension of V (assuming that k is algebraically closed), we can see that there are also infinitely many isomorphism classes of indecomposable representations of this quiver. Example 7 Our last example is the so-called Kronecker quiver. It is a quiver with two vertices and two arrows between these vertices with the same orientation: 1 2. It is not easy to classify all indecomposable representations of this quiver - although this can be done using only knowledge from linear algebra. However, we only need to look at a special subclass of indecomposable representations of the Kronecker quiver to see that it is of infinite representation type. Consider the following representations of the Kronecker quiver: V λ = k 1 k for nonzero λ k. λ Each V λ is indecomposable. To see this, suppose V λ = V λ V λ. Suppose without loss of generality that V λ,1 = k. Then V λ,1 = 0. Moreover, V λ,2 = 0 and V λ,2 = k, because otherwise either V λ or V λ would be the zero representation and they would not form a decomposition of V λ. But then V λ and V λ must assign the zero map to both arrows and so their direct sum would contain two zero maps. But V λ has two nonzero maps, a contradiction. Therefore, V λ cannot be decomposable. Furthermore, V λ = Vµ for nonzero λ, µ k only if λ = µ. To see this, suppose V λ = Vµ. Then there must be bijections f 1, f 2 : k k such that the following diagram commutes: 14

k 1 k λ f 1 f 2 1 k k. µ This means we have 1 f 1 = f 2 1 so f 1 = f 2. But then µ f 1 = f 2 λ = f 1 λ and thus λ = f 1 1 µ f 1 = µ. We can conclude that the V λ are indecomposable and pairwise nonisomorphic, which means that they live in different isomorphism classes. But then the number of isomorphism classes of indecomposable representations of the Kronecker quiver must be at least {V λ : λ k} = k =. 15

Quiver representations are modules over the path algebra In this chapter, we will see that a quiver representation of a quiver Q is the same as a module over a special kind of finite-dimensional k-algebra A, called the path algebra of Q. It also turns out that the category of finitely generated modules over any finitedimensional k-algebra is equivalent to that of finitely generated modules over a quotient of a path algebra (I will not discuss this last claim here, though). This explains why studying quiver representations is so useful: quiver representations can (to some extent) visualise representations over any finite-dimensional k-algebra. Basic notions First of all, we need to understand what is meant by a module over a path algebra. The definitions in this section follow (Etingof, 2011). I will begin with the definition of an algebra: Definition 12. A k-algebra is a vector space A over k together with a bilinear map A A A such that for all a, b, c A, we have (ab)c = a(bc) and we have a unit 1 A such that 1a = a1 = a The path algebra of a quiver Q is a special kind of algebra that has as its basis the set of paths in Q and multiplication defined by the concatenation of paths. Before I can introduce the path algebra, I need to explain what a path is. Informally, a path from vertex i to vertex j in Q is a way to get from i to j by following the direction of the arrows in between i and j. Below is a formal definition. Definition 13. A path p in Q is a sequence of arrows α 1, α 2,..., α n Q such that s(α i ) = t(α i+1 ). Moreover, for every vertex i Q 0 we define a constant (or trivial) path e i with s(e i ) = i = t(e i ). 16

Let Q be the quiver: β α 1 2 3. γ Then (β, α) is a path as well as (β, γ), but (α, γ) is not a path. Definition 14. The path algebra kq of a quiver Q is the k-algebra with basis the set {p : p is a path in Q} and multiplication defined as follows: { p p pp if s(p) = t(p ) = 0 otherwise Lemma 2. The unit of the path algebra is defined as the sum of all constant paths, that is: 1 kq = i Q 0 e i. Proof. For all p kq we have: Therefore, we have that algebra. p e i = i Q 0 pe i = p = { p if s(p) = i 0 otherwise i Q 0 e i p and hence i Q 0 e i is the unit in the path Let Q be the quiver 1 2, that is, A 2. Then the basis of kq is {e 1, e 2, α} and we can define an isomorphism f : kq U(2, k), where U is the algebra of upper triangular 2 2 matrices, defined by: ( ) b c (ae 1 + be 2 + cα) 0 a for all a, b, c k. Since algebras are vector spaces, they are abelian groups under addition. Moreover, it follows from the definition that they are also monoids under multiplication. It follows that an algebra is a special kind of ring and so we can define modules over them. In the rest of this chapter, we will always be concerned with left modules. The following lemma proves useful when considering modules over a k-algebra A. Lemma 3. If A is a k-algebra and M a left A-module, then M is also a k-vector space. 17

Proof. Since A is a k-vector space, for all λ k, m M we can find an a A such that am = λ1 A m = λm. As M satisfies the module axioms with respect to a, it satisfies them with respect to λ. Thus, M is also a k-module, that is, a vector space. This is called the underlying vector space of M. Let us now consider an example of a module over over a path-algebra to get an idea of what these actually look like. α β Let Q be the quiver 1 2 3. Then we can construct a module M(α) whose underlying vector space has basis {e i, α} and which is acted on by the path algebra kq in the following way: λe 1 if p = e 1 κα if p = e 2 p(λe 1 + κα) = λpe 1 + κpα = λα if p = α 0 otherwise for all paths p kq and all κ, λ k. It seems that this module is similar to the representation M = k k 0 of Q. In the next section, we will see that there is in fact a correspondence between modules over kq and representations of Q. Quiver representations and kq-modules In this section, we will show that each quiver representation of a quiver Q is the same as some module over the path algebra kq. We only need to make precise what is meant by the same here. In order to do this, we will have to introduce some basic category theory. The definitions are based on Leinster, (2014a, Chapter 1). Definition 15. A category A consists of a collection of objects ob(a ) a collection of morphisms between objects A and B for all A, B ob(a ), denoted as Hom A (A, B) an identity element 1 A Hom A (A, A) for each A oba a binary operation on morphisms (called composition): Hom A (B, C) Hom A (A, B) Hom A (A, C) for each A, B, C oba that is associative and satisfies 1 A f = f = f 1 B for all f Hom A (A, B). 18

In order to compare two categories, we need a notion of a map between them. A functor is a map between two categories. Definition 16. A functor F: A B consists of a map from objects in A to objects in B F : ob(a ) ob(b); A F (A) a map from morphisms in A to morphisms in B for all A, A ob(a). F : Hom A (A, A ) Hom B (F (A), F (A )); f F (f) such that F (g f) = F (g) F (f) for all morphisms g, f that can be composed in A and such that F (1 A ) = 1 F (A) for all A A. We have just defined maps between objects within a category and maps between categories. We can go one level up and also define maps between functors. Such maps are called natural transformations. Definition 17. Suppose A and B are categories and A B natural transformation α : F G is a family of functions F G are functors. A (α A : F (A) G(A)) A A such that for all A, B A and all f A (A, B), the following diagram commutes: F (A) F (f) F (B) α A G(A) G(f) α B G(B). Remark 1. For any two categories A and B there is a category whose objects are the functors between A and B and whose maps are the natural transformations between these functors. This category is called the functor category from A to B. In this section, we are interested in two categories. The first one is the category modkq consisting of modules over the path algebra kq and module homomorphisms between these. The second category we will be considering is rep(q,k): the category of representations of the quiver Q together with the morphisms between these representations. Category theorists often view these categories as functor categories - see Leinster, (2014b). In this section, however, we will stick with the intuitive representation theoretical conceptions of these categories that we have built up in the previous sections. 19

At the beginning of this section we said that a given quiver representation of a quiver Q is the same as some module over the path algebra kq. They are not exactly isomorphic, however. The appropriate notion of sameness here is slightly different: it is that the two categories rep(q,k) and modkq are equivalent. Definition 18. Let A and B be categories. Then A and B are equivalent if and only if there exist functors A B F G and natural isomorphisms η : 1 A G F and ɛ : 1 B F G, where 1 A and 1 B are identity functors on A and B respectively. Remark 2. It can be deduced from the above definition that a natural transformation A F G α B for all A A. is a natural isomorphism if and only if α A : F (A) G(A) is an isomorphism Theorem 2. There exists an equivalence of categories rep(q,k) modkq. Proof. Given the definitions above, in order to show that our categories modkq and rep(q,k) are equivalent, we need to construct functors F : modkq rep(q,k) and G : rep(q,k) modkq and show that there are natural isomorphisms η : F G 1 modkq and ɛ : G F 1 rep(q,k). Let us first construct F. Let F act on objects in the following way. F sends a module M modkq to a representation (M i, φ α ) i Q0,α Q 1 in rep(q,k) such that: for each i Q 0, M i is the image of the idempotent map e i, which coincides with the fixed points of this map. So we let M i = e i M. Then each M i is indeed a vector space because it is the image of the linear map e i under the k-vector space M. for each α Q 1, φ α is the map defined by for all m M s(α). φ α (m) = αm This is indeed a map M s(α) M t(α), because αm = e t(α) αm e t(α) M = M t(α). Moreover, each φ α is a linear map: if m, m M s(α) then φ α (m + m ) = α (m + m ) = αm + αm, 20

because M is a kq-module. For the same reason, we have: for any m M s(α) λ k. φ α (λm) = α(λm) = (αλ)m = λ(αm) Let F act on morphisms as follows: if f : M M is a morphism in modkq, we set F (f) = (f i ) i Q0 rep(q,k) with f i = f (e i ). Then each f i is really a map M i M i since if m M i, then f i (m) = (f (e i ))(m) = f(e i m) = e i f(m) e i M = M i. To check that (f i ) i Q0 is really a morphism of quiver representations, let α be an arrow from i to j and m M i. Then f j φ α (m) = f (e j )(αm) = f(αm) = αf(e i m) = φ α (f (e i ))(m) = φ α f i (m). Finally, we need to check that F is indeed a functor. Firstly, we have for all f : M M and g : M M and all m M i, F (gf)(m) = (gf) i (m) = gf (e i )(m) = g (e i ) f (e i m) (since f(e i m) M i) = g i f i (m) = F (g)f (f)(m). Secondly, for all m M i and so F (1 M ) = 1 F (M). F (1 M ) i (m) = 1 M (e i )(m) = 1 M (m) = m Let us now construct G. Let G act on objects in the following way. It sends a representation R = (M i, φ α ) i Q0,α Q 1 in repq to a module M modkq such that its underlying vector space is i Q 0 M i, multiplication by a path p is defined by { φ αk... φ αh (m) if s(α h ) = h = i pm = 0 otherwise for any nontrivial p = (h α h,..., α k k) and any m M i, { m if i = j for all trivial paths e i and all m M j, e i m = 0 otherwise. 21

We define the action of a path in the path algebra on an arbitrary m M by extending the definition above linearly. In order to see that M is really a kq-module, we need to check the module axioms (we will check only two of them below - the other two are satisfied in a very similar way): The multiplicative identity 1 kq of the path algebra is a unit with respect to the module. We can write any m M in the form i Q 0 m i, where m i M i. Thus, we have for all m M: 1 kq m = ( j Q 0 e j )( i Q 0 m i ) = k Q 0 e k m k = m. Moreover, let m, m M h and p = (h α h,..., α k k), then p(m+m ) = φ αk... φ αh (m+m ) = φ αk... φ αh (m)+φ αk... φ αh (m ) = pm+pm since all maps φ αh,..., φ αk are linear by definition. Let G act on morphisms as follows: if f = (f i ) i Q0 : (M i, φ α ) (M i, φ α) is a morphism in repq, then G(f) is the module homomorphism f i : M i M i. i Q 0 i Q 0 i Q 0 We need to check that given two representations R = (M i, φ α ) i Q0,α Q 1 and S = (M i, φ α) i Q0,α Q 1 and a morphism of representations f : R S, G(f) is really a module homomorphism: For all m, m M = i Q 0 M i, we have G(f)(m + m ) = ( f i )(m + m ) = (f i (m i + m i)) = (fi (m i ) + f i (m i)) = f i (m i ) + f i (m i) = G(f)(m) + G(f)(m ), where m i is the ith component of m and we have used the linearity of the morphism of representations f. Moreover, we have for all paths p kq and m M, G(f)(pm) = ( f i )(pm) = f i (pm i ) = pf i (m i ) = p f i (m i ) = pg(f)(m) using the fact that f is a morphism of representations. Finally, we check that G is indeed a functor. Firstly, let f : R S and g : S T be morphisms of representations (with R and S defined as above) and m M = i Q 0 M i. Then G(gf)(m) = ( (gf) i )(m) = ( (g i f i ))(m) = ( g i f i )(m) = G(f)G(g)(m) 22

and G(1 R )(m) = ( 1 Mi )(m) = 1 Mi (m i ) = m i = m and so G(1 R ) = 1 G(R). The next step is to show that these functors give rise to an equivalence. That is, we need to find natural isomorphisms η : F G 1 rep(q,k) and ɛ : G F 1 modkq. We will only construct η here, the other natural isomorphism is found in a very similar way. In order to do so, we need to define η R for each representation R rep(q,k). By our construction of F and G, and given a representation R = (M i, φ α ) i Q0, this means that we would like to construct a morphism of representations η R sending F G(R) = (e i j Q 0 M j, φ α ) R. Let us define (η R ) j = π j, where π j : e j Mi M j is the projection onto the jth component. Then the diagrams φ α e j i Q 0 M i e k i Q 0 M i (η R ) j (η R ) k M i φ α M j commute for all α : j k since we have for all m e j Mi : φ α (η R ) j (m) = φ α π j (m) = φ α (m) = π k φ α (m) = (η R ) k φ α (m), where by m in the second line we mean m M j. Moreover, (η R ) j is an isomorphism for all j Q 0. Let ι j : M j e j Mi be the obvious inclusion function. Then ι j (η R ) j (m) = m for all m e j Mi and (η R ) j ι j (m) = m for all m M j. We can conclude that each η R is an isomorphism in rep(q,k). Finally, we need to check that η as defined above is indeed a natural transformation by verifying that the diagram (F G(f)) j (F G(R)) j (F G(S)) j (η R ) j (η S ) j M j f j M j 23

commutes for all j Q 0 and all f : R S rep(q,k) (where R and S are as defined before). Let m (F G(R)) j. Then (η S ) j (F G(f)) j (m) = π j f i (e j m) = f j (m) = f j π j (m) = f j (η R ) j (m). 24

Bibliography Etingof, P. et al. (2011). Introduction to Representation Theory. url: http://arxiv. org/abs/0901.0827v5. Leinster, Tom (2014a). Basic category theory. Vol. 143. Cambridge studies in advanced mathematics. url: http : / / www. loc. gov / catdir / enhancements / fy1512 / 2014451716-b.html. (2014b). Categories vs. Algebras. [Accessed 20/01/2016]. url: https://golem.ph. utexas.edu/category/2014/05/categories_vs_algebras.html. Ringel, C. M. (2013). The representations of quivers of type A n. A fast approach. url: https://www.math.uni-bielefeld.de/~ringel/. Ringel, C.M. (2016). The Representation Theory of Dynkin Quivers. Three Contributions. url: https://www.math.uni-bielefeld.de/~ringel/. Schiffler, Ralf (2014). Quiver representations. Canadian Mathematical Society. Springer International Publishing Switzerland. url: http://www.loc.gov/catdir/enhancements/ fy1411/2014945363-b.html. 25