Decomposition Methods for Representations of Quivers appearing in Topological Data Analysis

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1 Decomposition Methods for Representations of Quivers appearing in Topological Data Analysis Erik Lindell SA114X Degree Project in Engineering Physics, First Level Supervisor: Wojtek Chacholski Project Collaborators: Daniel Nyman & Axel Siberov Department of Mathematics School of Engineering Sciences Royal Institute of Technology (KTH) Stockholm, Sweden 1

2 Abstract This paper describes methods to compute the decomposition type of representations of some quivers that appear in topological data analysis. Specifically, these are the A n-quivers, and the quivers CL(f) and CL(f, f). A background is given to explain the appearance of these quivers in the data analysis setting. The theoretical background covers the basic category theoretical concepts used to define decomposition of representations, as well as the necessary tools used to find indecomposable representations. A brief review of how these tools can be applied on A n-quivers is given. The main results in the paper are the matrices presented at the ends of sections 4 and 5, which give linear relations between vectors representing the decomposition types of representations of these quivers and vectors of numerical invariants of the representations. These invariants are simply the dimensions of some subspaces of the vector spaces in the representation, and are therefore easily calculated. Finally, more general ladders and grids are considered, and it it proven that the only grids of finite representation type are the ladders with at most four rungs. 2

3 Contents 1 Introduction 4 2 Theoretical background Quivers in topological data analysis Category theory Decomposition of quiver representations Review of A n -quivers Indecomposable representations of the A n -quiver Computing the bar code of a representation The commuting square Indecomposable representations of CL(f) Computing the decomposition The commuting figure eight 29 6 Larger commuting grids Ladders General grids Discussion 35 8 Acknowledgements 36 References 36 3

4 1 Introduction Topological data analysis is the field of mathematics where the topological properties of data sets are analysed. When comparing different data sets to each other in this setting, objects known as quivers naturally appear as underlying structures. A quiver is a finite directed graph, such as the example illustrated in figure 1. Figure 1: An example of a quiver Quivers are often used to parametrize vector spaces and linear maps. A vector space representation of a quiver assigns a (finite-dimensional) vector space, over some given field k, to each vertex of the quiver and a linear map to each arrow. Just as it is possible to represent a quiver using vector spaces and linear maps, it can be represented in the analogous manner using other structures, such as sets and functions, or groups and group homomorphisms. The purpose of this paper is to explore the vector space representations of some specific quivers which appear in topological data analysis and in particular how to decompose them. The main results are presented in sections 4 and 5, and consist of numerical invariants which completely characterize the structure of the representations of these quivers, together with linear equations which connect these invariants to the decomposition type of the representations. To understand what this means, and to understand the relevance of these results, some theoretical background is needed. 2 Theoretical background 2.1 Quivers in topological data analysis The connection between quivers and data analysis is not obvious. Initially, we need to specify what we mean mathematically by a data set. In general, we can define a data set as a finite metric space. This means that a data set is a finite set of data points with a well defined distance function (i.e. a metric) between the points. If, for example, the age of a certain population of people is measured, the distance is simply given by the age difference. A natural problem which arises given a set of data is finding ways to compare it to other data. If, for example, similar measurements have been made in two different locations, or at different times, it might be relevant to see how similar these two data sets are. With our previous example in mind, an example could be to compare the age of students at two 4

5 different universities. One way of doing such a comparison is using statistical methods, such as averages, variances, and so on. In topological data analysis, however, what we are interested in are the topological properties of the set, which are given by the metric space structure that we have given to the set. How can this structure be compared? Initially, we can make the following definition: Definition 1. Given two metric spaces (M, ρ) and (N, σ), an isometry between them is a map f : M N which preserves distance. Specifically, for all x, y M we have ρ(x, y) = σ(f(x), f(y)) An isometry thus preserves the structure of a metric space. If there exists a bijective isometry between two metric spaces their structure as metric spaces is identical and they are said to be isometric. A way of measuring the similarity of two data sets is thus to determine how close they are to being isometric. This can be done using something called Gromov-Hausdorff distance. To define Gromov-Hausdorff distance we need to start by defining the Hausdorff distance between two subsets of a metric space. Definition 2. Let X and Y be two non-empty subsets of a metric space (M, d). The Hausdorff distance between X and Y is defined as d H (X, Y ) = max{sup inf x X y Y d(x, y), sup y Y inf x X d(x, y)} To find this distance one thus finds the point in X which lies the furthest from any point in Y, then does the same thing in Y. The Hausdorff distance is then the maximum of these two distances. We can now define the Gromov-Hausdorff distance between compact metric spaces. Definition 3. Let X and Y be compact metric spaces. The Gromov-Hausdorff distance between X and Y is defined as inf d H (f(x), g(y )) where the infimum is taken over all metric spaces M and all isometric embeddings f : X M and g : Y M. Gromov-Hausdorff distance thus gives us a tool to compare the topological structure of data sets to each other. However, this approach turns out to have one flaw which renders it useless in practice: it has a staggering computational complexity. To compute it, we have to go through all embeddings of X into M to see which are isometric. If f is an embedding X M, we can create a new embedding by permuting the image of f. There are n! permutations of n elements, so the computations complexity of Gromov-Hausdorff distance will thus be at least n!. Hence the Gromov-Hausdorff distance is practically incalculable for even quite small sets of data. Instead of giving up the geometric approach to compare data sets, because of this problem, we can search for ways of simplifying the data sets to render them practically comparable, while still preserving some information about the geometric structure of the data. A first attempt at simplification is to use the data to generate something called a dendrogram, which is a diagram that represents how the data is clustered. To do this, we start by taking 5

6 some small distance ɛ, and introducing the following equivalence relation on our space (M, ρ): Definition 4. Two points x, y M are equivalent if there exists a sequence of points {x i } n i=0 M, such that x 0 = x, x n = y and ρ(x i, x i+1 ) ɛ for all 0 i n 1. This is easily verified to be an equivalence relation. This captures the way the data points are clustered in the way that different equivalence classes are separated by a minimal distance between points which is larger than ɛ. Using this procedure repeatedly, with increasing distances, which can for example be taken as ɛ, 2ɛ, 3ɛ and so on, we get a sequence of sets (partitions of M, where the equivalence classes are the elements) M 0, M 1, M 2,... where M 0 = M. For n large enough, all points of M will be equivalent, since it is a finite set. This means that the sequence will terminate. Thus we are only concerned with the first n sets of the sequence. For each i we can define a function f i : M i M i+1 by mapping each equivalence class of M i to the equivalence class it belongs to in M i+1. We thus get a sequence of finite sets and functions M 0 M 1 M n. As can be seen in figure 2, the plot has a treelike structure, which is why it is called a dendrogram. 0 ɛ 2ɛ 3ɛ 4ɛ 5ɛ Figure 2: A basic example of a dendrogram Recalling the definition of a set representation of a quiver given at the beginning of the introduction, we can realize that a dendrogram is simply a set representation of the so called A n -quiver with n vertices, since it assigns to each vertex i a set M i and to each arrow i i + 1 a function f i. We thus see how a set representation of this quiver can be generated from a set of data. Since ɛ is some finite distance, it is clear that two different data sets can generate identical dendrograms. Some information was thus lost when generating the dendrogram, so it is indeed a simplification of the data set from which it was generated. Can the dendrograms of two data sets be compared in a more practical way than the data sets themselves could be? To answer this question, we first need to specify some notion 6

7 of distance between dendrograms. A dendrogram gives us a new metric on the data set, where we define distance as follows: if x and x are data points, then the distance between them is the smallest integer n such that x and x belong to the same equivalence class under nɛ. This can be verified to be a metric. A dendrogram is actually a special kind of metric space, called an ultrametric space, which has several very special properties. In fact, it is proven by Carlsson and Mémoli [2, Theorem 9] that there is a one-to-one correspondance between finite ultrametric spaces and dendrograms. For more information on dendrograms and ultrametric spaces, see their paper. Dendrograms are thus metric spaces, so the natural way to measure distance between dendrograms is therefore to use Gromov-Hausdorff distance, which will once again have a computational complexity of at least n!. Comparison of dendrograms in this way is thus a useless method in practice, which means that further simplification is needed. As previously noted, a dendrogram is simply a set representation of an A n -quiver. A possible further simplification is therefore to instead look at the vector space representations of the same quiver. There is a natural way of generating a vector space representation of the A n -quiver from the dendrogram generated by the data set. For the set M i, which we assume has m elements, let us label the elements by x i. This set generates the vector space V i = m kx i, i=1 where the x i form a basis. This is called the free vector space over M i. Since the map f i goes from the basis of V i to the basis of V i+1, this defines a linear map from V i V i+1. In this way we can assign to each vertex of the A n -quiver the vector space V i and to each arrow the linear map defined by f i and thus get a vector space representation. We shall see later on that this is actually a further simplification from a dendrogram, and that this simplification is sufficient to make comparison possible. Before going there, we are going to introduce one more type of quiver that is relevant in data analysis. We have now seen how representations of A n -quivers can be used as a way of simplifying sets of data with one measured variable. Another natural situation in data analysis is to make measurements of several variables that might be correlated and to compare different such data sets. What happens when two variables are measured? This corresponds to having a data set with two different metrics, say ρ and σ. Analogously to the procedure with one metric, we can now introduce two equivalence relations on M. Combining these two equivalence relations gives us a new equivalence relation, where points are equivalent if they are equivalent under both original equivalence relations. Taking some small distance ɛ for the first metric and some small distance δ for the second metric, we can thus take steps in both m and n and get one equivalence relation for every pair (mɛ, nδ), where m, n are positive integers. By plotting the equivalence classes against two axes, we get a two dimensional dendrogram, with an underlying quiver which is a directed grid in two 7

8 dimensions, as in the following figure:.. Since we already know that one dimensional dendrograms cannot be effectively compared, the same must be true in two dimensions, so we can immediately go to the vector space representation, which is generated in the same way as in the previous case. Here it is important to note that the dendrogram must be commutative, so the same follows for the vector space representation. We may go even further, and do the same thing with n metrics, where n is a positive integer. This would would yield a vector space representation of an n-dimensional grid. The vector space representations of such grids thus have a natural connection to correlation between n variables. In this paper, however, we will only concern ourselves with A n -quivers (1-dimensional grids) and 2-dimensional grids. Since the main subject in this paper is 2-dimensional grids, the word grid will from now on be used to refer to ones of the 2-dimensional variety, unless otherwise is explicitly stated. The main focus of this project was to investigate the commutative vector space representations of these types of two-dimensional grids. In particular, the focus was on the two following quivers, which we will call the commutative square and the commutative figure eight. We shall denote them by CL(f) (commutative ladder forward) and CL(f, f), respectively: These are the simplest cases of so called ladders, i.e. grids where only one step is taken in one of the directions. Much about these two has already been proven by Escolar and Hiraoka [3]. The methods there are beyond the scope of this project and the problems here will be treated using much more elementary methods. From now on we shall only be concerned with vector space representations of quivers, so for brevity the term representation shall be used to mean vector space representation, unless otherwise is explicitly stated.. 8

9 2.2 Category theory We have now seen how two types of quivers, A n -quivers and two dimensional grids, appear in data analysis and why their representations are of interest to us. The basic problem is comparing data sets to each other, so what we need to answer is how we can compare representations of a quiver to each other. One strategy for doing this is to decompose representations into smaller parts, and then comparing the decomposition. What is a natural way of decomposing a representation? Furthermore, to be able to compare representations to each other in the first place, we must start by specifying a notion of structural similarity between representations. How could that be done? Both of these question can be given meaningful answers using the language of category theory, in which these notions are formalized in a much more general sense, which we can then apply to our more specific problems. We shall start by introducing the basic concepts of category theory and seeing how they can be related to quivers and representations. The fundamental definition in category theory is that of a category. Categories. To define a category, which we will denote by C, we start with a collection of objects O, and a collection of arrows A (or morphisms), together with two functions dom : A O and cod : A O, which assign to each arrow their domain (where they start) and their codomain (where they go), respectively. If f A satisfies dom f = a and cod f = b, for a, b O, we usually write f : a b, or a f b. If dom f = cod f we say that f is an endomorphism. For C to be a category, we need two more functions. First we need an identity function id : O A, which assigns to each object c its identity arrow id c, which satisfies dom(id c ) = c = cod (id c ) This means that we have an arrow which allows us stay at an object. Note that the identity is an endomorphism, so every object has at least one endomorphism. As we shall see, however, an identity arrow must satisfy more than just being an endomorphism. Apart from the identity function, we also need a function which we call composition. This function takes any pair of arrows (g, f), which satisfies dom g = cod f and assigns to it a new arrow g f, for which dom(g f) = dom(f), cod(g f) = cod(g) What this means is that going from an object to another via two arrows defines an arrow in its own right. Pictorially, this can be illustrated by the commutative diagram: a f g f b c g Having defined composition, we can now complete the definition of a category. An identity arrow must satisfy the unit law: for any object c and arrows f, g such that dom f = c = cod g we have f id c = f, id c g = g 9

10 Composition is associative: Given objects a, b, c, d and arrows a f b g c h d, we have the equality (h g) f = h (g f) This completes the definition of a category. With the definition of a category complete, we can define what it means for two objects in a category to be isomorphic. An isomorphism is an arrow f : a b which has an inverse arrow, which means that there exists an arrow f : b a such that f f = id a and f f = id b. In this case, the objects a and b are said to be isomorphic. This notion will allow us to classify representations of quivers in a natural way, i.e. into isomorphism classes. Quivers as categories. There are several ways in which a quiver can be viewed as a category. The most simple way is to simply take the vertices as objects and the arrows as arrows. This simple construction can be applied on an A n -quiver to turn it into a category, by simply adding identity arrows and imposing that composition of arrows defines a new arrow. Considering, for example, grids, it is easily realized that things cannot generally be done this simply, since we are assuming that the grid is commutative. In general we must therefore introduce relations on a quiver to turn it into a category. A relation is simply some equation equating different arrows (which may be composite) or objects to each other. The category is now constructed just as in the basic case, but taking the imposed relations into account. For the commutative grid, the set of relations is given by equating all arrows with the same codomain and domain to each other. Given a field k, another example of a category is the vector spaces over k, in which the arrows are linear maps. We denote this category by Vect k. The collection of all sets, with functions as arrows, is also a category. To relate these categories to representations of quivers, we will need the notion of a functor. Functors. A functor can be viewed as a morphism between different categories. Let C and D be two categories. A functor F between C and D is a pair of functions, between the objects of C and D and between the arrows of C and D, respectively. We call the two functions the object function and the arrow function. The object function assigns to each object c in C an object F c in D and the arrow function assigns to each arrow f : c c in C an arrow F f : F c F c in D. For this pair to be a functor, it must preserve identity and composition, which means that for any object c in C and any composable pair of arrows f, g in C, we have F (id c ) = id F c, F (g f) = F g F f Given a quiver Q, we can define a functor to Vect k by assigning to each vertex a vector space and to each arrow a linear map. The identity arrows are naturally sent to the identity maps of each vector space. But this is precisely how we earlier defined a representation of Q. This enables us to make the following definition. Definition 5. A representation of a quiver Q is a functor from Q to Vect k. In the corresponding way, we can define a set representation of a quiver Q to be functor from Q to the category of sets. We shall see that the representations of Q (functors from Q to Vect k ), can be viewed as the objects in a category, which we will call the category of functors from Q to Vect k. This is the type of category we will primarily be interested in, since it is the isomorphisms in such a category that will allow us to classify representations 10

11 of a given quiver. Before we get there, we must however identify what the arrows would be in this type of category. The notion needed is that of a natural transformation, or a morphism of functors. Natural transformations. Let C, D be two categories. Given functors F, G : C D, a natural transformation T : F G is a function that to each object c of C assigns an arrow T c : F c Gc such that given an arrow f : c c in C, the diagram F c F f T c T c Gc F c Gc is commutative. If, for every c in C, the arrow T c is an isomorphism in D, we say that T is a natural equivalence. Taking as objects the functors from C to D and as arrows the natural transformations between the functors, it is easily verified that this has the structure of a category. In this category, the isomorphisms are simply the natural equivalences. We shall denote the category of functors from C to D by Fun(C, D). Natural transformations in Fun(Q, Vect k ). This might seem quite involved, so let us take a step back, and see what the definition of a natural transformation means in Fun(Q, Vect k ), i.e. the category of representations of Q. Given two representations (functors) V and W, a natural transformation from V to W is a family of linear maps u i : V i W i such that given a linear map f α : V s(α) V t(α) in V, we get a commuting diagram Gf V s(α) u s(α) W s(α) f α V t(α) u t(α) g α W t(α) If all the u i are invertible linear maps, this is a natural equivalence, and the representations are isomorphic. What this means in practice is the the u i are a change of basis in each vector space, which transforms V to W. In that way the representations behave in the same way, but are only viewed in different coordinates. Thus it is natural to classify them as the same representation. In the category of set representations of a quiver, the morphisms can be defined in the same way, exchanging linear map for function and invertible linear map for bijection. With the notion of isomorphism between representation defined, we can explain why the vector space representation of the underlying quiver for a data set is a further simplification from the dendrogram from which it is generated. The reason is simply that there are more isomorphisms in the category of vector space representations of a quiver than in the category of set representations of the same quiver. Two dendrograms which are not isomorphic can therefore generate isomorphic vector space representations. We shall illustrate this with an example. 11

12 Example. Let us look at the two dendrograms 0 ɛ 2ɛ 0 ɛ 2ɛ It is clear that we can find no set of bijections between the sets of each node such that the resulting diagram is commutative. The vector space representations generated by these two dendrograms are k 4 f 1 k 2 f 2 k k 4 g 1 k 2 g 2 k where the maps are given by the matrices [ ] f 1 =, f = g 2 = [ 1 1 ], g 1 = [ ] We define linear maps F 1 : k 4 k 4, F 2 : k 2 k 2 and F 3 : k k by F 1 = , F 2 = I 2, F 3 = I 1, where we denote the identity matrix k n k n by I n. It is clear that F 2 g 2 = f 2 F 3 and it is easily verified that F 1 g 1 = f 1 F 2. Thus the diagram is commutative, which means that the representations are isomorphic. With a well-defined notion of what it means for two representations to be isomorphic, we have at least some way of comparing two representations to each other, i.e. by determining if they are isomorphic. However, if two representations are not isomorphic, we still want to have some measure of proximity between them. If none of the representations W or U of the quiver Q are isomorphic to the representation V of the same quiver, we want to determine which is the most similar to V. To do this, let us therefore move on to the problem of decomposing a representation to smaller components. For this we need the notion of a coproduct in a category. Coproducts. The coproduct is the category theoretical way of summing objects in a category to a larger object. Let C be a category and a, b be objects in C. A third object c is a coproduct of a and b if there exist arrows, which we call inclusion arrows, i : a c, j : b c that satisfy the following property: if d is a fourth object of C, for any arrows f : a d and g : b d, there is a unique arrow h : c d such that the diagram 12

13 a i f c h d b j g is commutative. Then we write c = a b. If a coproduct a b exists, it must be unique, up to isomorphism. To prove this, suppose that c, d are both coproducts of a and b. Using the properties of c and d, we get the commutative diagram a i i i h h c d c b j j j where h and h are unique. By uniqueness the arrow h h must be unique. Furthermore, the identity arrow c c makes the diagram commute, so we must have h h = Id c. In the same manner we can show that h h = Id d, so it follows that c and d are isomorphic. If we take the category Vect k as a first example, the coproduct is simply the direct sum of vector spaces. We can note that the category contains one object, namely the zero vector space, such that V 0 = V for all vector spaces V over k. The zero vector space is actually a special case of a type of object in a category called a zero object, which has this property. Since zero objects give us a way of trivially decomposing an object in a category, it will be useful to define them stringently. Furthermore, zero objects give us one way of defining a special type of morphisms called zero morphisms, which will play a crucial role in some of our later proofs. Definition 6. An object a of a category C is called initial if, for every object b Ob C, there exists a unique morphism a b. Similarly, a is called terminal, if for every object b Ob C, there exists a unique morphism b a. A zero object is simply an object which is both initial and terminal. If a and b are zero objects in a category, it follows by the uniqueness criteria and the following diagram id a a f g that g f = id a and f g = id b. Thus any two zero objects are isomorphic. Furthermore, suppose that C has a zero object, which we shall denote by 0, and let a be an object such that a 0 exists. Looking at the diagram b id b 13

14 a i id a i h a 0 a a 0 i 0 j j j it is clear by uniqueness that h i = id a and i h = id a 0, so a and a 0 are isomorphic. Definition 7. In a category with a zero object we can define a morphism f : a b to be a zero morphism if it can be written as a composite f = h g, where g h a 0 b As seen previously, the coproduct in the category Vect k is simply the direct sum of vector spaces, while the zero object is the zero vector space. Naturally, the zero morphisms are simply zero maps. Returning to our main category of interest, i.e. the category of representations of a quiver Q, this gives a strong hint to what the coproducts in this category are. For simplicity, we shall use the term sum of representations, instead of coproduct. We make the following definition: Definition 8. If V and W are representations of a quiver Q, their sum, which we shall denote by V W, is defined by taking the direct sum of vector spaces at each vertex,, i.e. (V W ) i = V i W i, and defining the new maps component wise from the maps of V and W. The example of vector spaces also makes it clear what the zero objects and zero morphisms are in the category of representations Q. The zero object is the representation where we assign the zero vector space to each vertex. We will refer to this representation as the zero representation of Q. Similarly, the zero morphisms are simply the morphisms where every linear map is a zero map. Since the category of representations of Q will be our primary category of interest, we will reserve the term morphisms to refer to natural transformations between representations. To distinguish these from the morphisms in Vect k, we will simply denote these as maps. We now have a well defined concept of a sum of representations. However, our goal was to decompose a representation into smaller parts. The point here is that the sum of representations gives meaning to the question if a representation can be decomposed, i.e. if it is isomorphic to a sum of other representations. We say that a representation is indecomposable if it cannot be written as a sum of other, nonzero representations. If a representation V is isomorphic to a sum of other representations, we call these representations summands of V. 14

15 2.3 Decomposition of quiver representations The concepts introduced in the previous section gives us the adequate tools to compare quiver representations up to isomorphism class. We start by introducing some useful terminology. We define the total dimension of a representation as the sum of the dimensions of its vector spaces. It follows that every representation has a finite total dimension, since it consists of a finite number of finite vector spaces. It is clear that the total dimension of a sum of representations is the sum of their total dimensions. Note, in particular, that this implies that every representation can be decomposed into a finite number of indecomposables. This decomposition must be unique up to isomorphism. Definition 9. If a quiver has only a finite number of indecomposable representations, we say that it is of finite representation type. The quivers of this type are of primary interest, since it is possible to actually find all indecomposable representations for such a quiver explicitly. If a quiver only has a finite number of indecomposables, every representation can be written as a sum of these indecomposable representations. Working with a quiver of finite representation type, this makes it possible to develop methods to determine the decomposition of a representation effectively. The decomposition can be described using the number of how many copies of each indecomposable it contains. Quivers of finite representation type thus have a natural finite set of numerical invariants which determine their isomorphism class, i.e. these numbers. In most cases, there is no simple way of directly seeing how a representation can be decomposed. The goal in this paper is therefore to develop methods for finding the decomposition for representations of our particular quivers of interest. The strategy will be to find other invariants than the number of indecomposables, which can be more easily determined, and then find ways of computing the decomposition from those instead. As it will turn out, it will be beneficial to represent decomposition types using vectors. If a quiver is of finite representation type, the number of each indecomposable in the decomposition of a particular representation can naturally be represented using a vector in N m, where m is the number of indecomposables. Making this identification between representations and vectors also gives us a way to define distance between representations, by for example using regular Euclidean distance, or some other suitable metric, on these vectors. This property of quivers of finite representation type thus enables us to actually compare representations in the way that we wanted. It is clear that quivers of finite representation type have properties that make them very practical to work with. A basic and important question, in general and in this setting in particular, is thus whether a given quiver is of finite representation type. For quivers without relations, this question is answered by a fundamental theorem of representation theory, known as Gabriel s theorem: 15

16 Gabriel s theorem. A quiver without relations is of finite representation type if and only if each connected component of its underlying undirected graph is one of the following types (known as simply-laced Dynkin diagrams): A n : (n vertices, n 1) D n : (n vertices, n 1) E 6 : E 7 : E 8 : Proof. See [1, Theorem ]. This theorem immediately tells us that A n -quivers are of finite representation type, but a priori nothing about grids. Fortunately, it is proven by Escolar and Hiraoka [3] that CL(f) and CL(f, f) are both of finite representation type. Pursuing the strategy of decomposition to compare representations of these quivers is thus a reasonable approach. This leads us on to the next problem: how can we go about finding summands of a representation? To enable us to this, we shall prove the following theorem: Theorem. Suppose V and W are a representations of the same quiver Q. Suppose there exist morphisms f : W V and g : W A such that Then W is a summand of V. f id W W V W The identity on W is defined in the obvious way, i.e. by taking the identity map from every space to itself (the zero map if the space is zero). To prove the theorem, we first make the following definition. Definition 10. Let f = {f i } be a morphism V W, which are representations of a quiver Q. We then define the kernel of f as the representation of Q which assigns to each vertex g 16

17 i of Q the kernel of f i. The maps of this representation are simply the restrictions of the maps in V to ker f i. Since f is a morphism, commutativity ensures that this is well defined, since the map from V i V j must map the kernel of f i into the kernel of f j. Proof of the theorem. Let U be a third representation of Q. Let h be a morphism W U. We then have the following diagram: W f id W h ker g i g h V W U 0 where i is the inclusion map ker g V and 0 denotes the morphism where every map is the zero map. It is clear that h g is the unique morphism making this diagram commute, which means that V must be the sum of W and ker g and thus that W is a summand of V. This theorem provides us with a powerful tool for finding summands of representations and thereby to decompose them. Since our goal is to decompose representations into indecomposables, we need some tool for determining if a given representation is indecomposable. Fortunately, a classical result of representation theory gives us a solution to this problem. To understand the theorem, we first make the following definition: Definition 11. An endomorphism f is said to be nilpotent if its composition with itself a sufficient number of times becomes a zero morphism, i.e. if 0 f f f = 0, where the composition is done some finite number of times. This definition makes sense in all categories with a zero object, so we will use it both concerning morphisms between representations and linear maps between individual vector spaces. Which is intended will hopefully be clear from context. We may now state the theorem. Theorem. A representation V of a quiver Q is indecomposable if and only if every endomorphism V V is either an isomorphism or nilpotent. Proof. See [1, Lemma ]. With these two theorems, we have the adequate arsenal for our purposes. To illustrate how to use the concepts and tools which have been introduced here in practice, we shall start by applying them on A n -quivers, before moving on to our main objects of interest. 17

18 3 Review of A n -quivers The A n -quivers are the most basic quivers that we are interested in. How to classify and decompose them is already known, but since this nicely illustrates the concepts introduced in the previous chapter and the methods that will be used for CL(f) and CL(f, f), a brief review will be given here. The spaces and maps of a representation V of an A n -quiver will be labeled as follows: f 1 f 2 f n 1 V 1 V 2 V n Furthermore, we shall denote the composite map from V i V j by f ij. 3.1 Indecomposable representations of the A n -quiver The first step towards decomposing a given representation of an A n quiver is finding which representations are indecomposable. We will start by defining a special type of representation, which we shall call a k-bar. Definition 12. A k-bar is a representation of an A n -quiver of the form 0 0 k k 0 0 where k denotes the one dimensional vector space over k and all nonzero maps are assumed to be the identity on k. In a k-bar all nonzero vector spaces are thus one dimensional and they lie in a consecutive sequence, connected by identity maps. Proposition 1. All k-bars are indecomposable. Proof. Suppose that f = {f i } is an endomorphism of a k-bar. Every map f i is either 0 or invertible, since all nonzero linear maps from k k are given by multiplication with some scalar. It is clear that if V i = 0 then f i, so we need only concern ourselves with the nonzero spaces. We can thus WLOG assume that all V i = k. If all f i are nonzero we are done, because then the endomorphism is an isomorphism. If all are zero we are also done, because this implies that f is nilpotent. Assume, therefore, that some, but not all, f i = 0. We can thus find some V j such that f j 0. If j i we can start at V j and go to V i by the path f i id k = 0 or by the path id k f j 0. This is a contradiction. In the similar fashion we can get a contradiction if j i. Thus is follows that either all f i are zero, or none are, which proves that the k-bar is indecomposable. We have thus found at least some indecomposables of the A n -quiver. Now we want to show that these are the only indecomposables. From now on, we shall denote the k-bar with its first nonzero space at the i:th vertex and last nonzero space at the j:th vertex by [i, j]. Proposition 2. If a representation V of the A n -quiver is indecomposable, it is a k- bar. Proof. Let f ij be the nonzero map of V such that j i is maximal. In other words, f ij is the longest nonzero map of V. This implies that in every space V k, such that i k j, 18

19 there exists a nonzero element v k such that f k (v k ) = v k+1 and thus that f ij (v i ) = v j. We can now define a morphism g from the k-bar [i, j] to V by { v k if i k j g k (1) = 0 otherwise which clearly commutes and is thus a morphism. In a similar fashion we will construct a morphism h : V [i, j]. As in the previous case, we define all maps to be zero if k < i or k > j. Next, we expand {v j } to a basis for V j and define a map h j : V j k by mapping v j to 1 and the other basis vectors to 0. For i k < j we now define h k : V k k by h j f kj. This definition ensures commutativity and furthermore, it is clear that h k (v k ) = 1. Thus it follows that the composition h g is the identity on [i, j], which shows that [i, j] is a summand of V. Since V was assumed to be indecomposable, it follows that V must actually be isomorphic to [i, j]. Thus the k-bars are the only indecomposable representations of the A n -quiver. Now we know precisely which representations of an A n -quiver are indecomposable. Every other representation must thus be a sum of k-bars. The number of a certain k-bar in a given representation is an invariant for that isomorphism class. The number of possible k-bars for a given A n -quiver is finite, so this gives us a finite number of invariants which completely determines the isomorphism class of a given representation. The precise number of k-bars is n n(n + 1) i =, 2 i=1 since we get 1 bar of length n, two of length n 1 and so on, down to n bars of length 1. If we number the k-bars, we can represent the decomposition with a vector in N n(n+1)/2, as previously described. In this case we call this vector the bar code of the representation. 3.2 Computing the bar code of a representation How can we determine the bar code of a given representation? As stated previously, our strategy will be to find invariants which are easy to calculate and then compute the decomposition from these. These invariants will be the dimensions of certain subspaces of the spaces in a representation. We can note that the number of maps in a representation of a given A n -quiver is the same as the number of k-bars of length 2, counting all possible composite maps as well as the f i :s themselves. If we look at the kernels of all these maps, together with the spaces themselves, we have as many vector spaces as we have k-bars. Since the dimension of a direct sum of vector spaces is the same the sum of the dimensions of the summands, we know that the dimension of a space V i is the same as the total number of k-bars in V which contains the i:th vertex. Summing over the indices of these bars, we thus get a linear equation n j = dim V i j where n j is the number of bar number j. Similarly, for the maps it is clear that ker(f i, g i ) = ker f i ker g i, 19

20 where f i is the i:th map of one representation, and g i is the i:th map of another. Thus the dimensions of the kernels depend linearly on the dimensions of the corresponding kernels in the constituent k-bars. This means that the dimension of each kernel in a representation must be the same as the total number of k-bars with a nonzero space at the vertex of the map s codomain and a zero at the domain vertex, so we can create a linear equation for the dimension of every kernel just as for the dimensions of the spaces themselves. In this way we get as many equations as we have k-bars. These equations are easily verified to be independent, which they should be, since the dimension of no kernel or space can be determined from the others. Thus this system of equations has a unique solution, which is the bar code of V. Since the dimensions of all kernels and spaces are easily determined, this gives us an efficient method for calculating bar codes. We shall illustrate the method with a basic example; the A 2 -quiver. Example. The k-bars of the A 2 -quiver are k k 2. k k Every representation of A 2 can be written as a sum of these three. We denote by n j the number of bar j (1 j 3) in the decomposition of a representation V of A 2. Then we can see that the dimension of V 1 is precisely n 1 + n 2, the dimension of V 2 is n 1 + n 3 and the dimension of ker f 1 is n 2. Thus we get the matrix equation n 1 n 2 n 3 dim V 0 = dim V 1 dim ker f 1 The matrix is invertible, so from this equation the decomposition of V can be computed directly. With the knowledge of how to calculate bar codes, the question of how to compare them remains. As mentioned earlier, representing the decomposition as a vector makes it possible to define a natural distance between representation. This gives an idea of how similar two representations are to each other. However, for bar coding to be a truly powerful method of data analysis, the ability to do statistics on bar codes is required. What would, for example, be a meaningful way of taking averages of bar codes? Simply averaging a set of bar codes as vectors does not necessarily return an integer vector, and having a fractional number of an indecomposable is meaningless. The statistics of bar codes is a current area of research. The process of finding the indecomposable representations for the A n -quiver, and of computing the decomposition type of a given representation, was fairly simple. Furthermore, it clearly illustrated how to use the theorems from the theoretical background and thus gives us some hint about how one might approach the same problems for our other quivers. Let us move on to the simplest example of these: the commuting square. 20

21 4 The commuting square CL(f) is our first step from grids in one dimension, i.e. A n -quivers, to grids in two dimensions. The vector spaces and linear maps of a representation of CL(f) will be labeled as follows: V 0 f 01 V 1 f 02 f 13 V 2 f 23 V 3 Figure 3: A general representation of CL(f) Furthermore, we shall denote the composite map f 13 f 01 = f 23 f 02 by f Indecomposable representations of CL(f) The indecomposable representations of this quiver are already known, and can for example be found in [3, Figure 13]. However, we shall try find these indecomposables in an independent and more elementary way. Furthermore, we shall develop an efficient method for explicit computation of the decomposition of a given representation. Just as for the A n -quivers, we shall start with a representation and assume it is indecomposable and try to use this to determine its structure completely. Here, however, we will have to consider some different cases. First we can note that if two nearby spaces of the square are zero, what remains is in practice the A 2 -quiver, attached to two 0-dimensional spaces. These spaces add no conditions on the decomposibilty of the representation, so in these cases the representation must decompose as the A 2 -quiver. This gives us 8 different indecomposable representations, which we will number as follows: k 0 0 k k 0 0 k k k k 0 0 k k 0 0 k k k Figure 4: The first 8 indecomposables of CL(f) Here all unlabeled maps k k are once again the identity. That these are indecomposable follows from the proof for A n -quivers. When searching for more indecomposables, we can therefore assume that no nearby spaces are zero if our representation is assumed to be 21

22 indecomposable. We can also clearly see that if V is indecomposable, then both V 0 and V 3 cannot be zero, because in that case the representation can be decomposed in an obvious way. We want to prove the following proposition: Proposition. The indecomposables of CL(f) are those in figure 4, together with the three representations k k 0 k k k k 0 k k k k Figure 5: The last three indecomposables of CL(f) Proof. In the previous indecomposables, the composite map f 03 is zero. Are there any more indecomposables where this is true? Let us start by showing that if f 03 is zero and V is indecomposable, then either V 0 or V 3 is zero. Since f 03 = 0, it follows that im f 01 ker f 13 and im f 02 ker f 23. It always holds that V 1 = im f13 ker f 13 and V 2 = im f23 ker f 23, so it follows that V must be isomorphic to the sum V 0 ker f 13 0 im f 13 ker f 23 0 im f 23 V 3 If V is indecomposable it must thus either hold that V 0 = 0 or V 1 = 0. If V 0 is zero, it must also hold that ker f 13 = ker f 23 = 0, i.e. that these two maps are injective. Let us look at these cases separately. We start with the case where V 3 = 0. In general, we have the isomorphisms V 1 = im f01 V 1 / im f 01 and V 2 = im f02 V 2 / im f 02. In this case it thus follows that V is isomorphic to the sum V 0 im f 01 0 V 1 / im f 01 im f 02 0 V 2 / im f 02 0 Since V is assumed to be indecomposable it must thus hold that im f 01 = V 1 and im f 02 = V 2, so f 01 and f 02 must both be surjective. Next we want to show that both f 01 and f 02 are injective, i.e. that they are isomorphisms. To show this let as assume that it is not true. First we assume that the intersection 22

23 ker f 01 ker f 02 is nonzero. In that case, it is clear that V is isomorphic to the sum ker f 01 ker f 02 0 V 0 /(ker f 01 ker f 02 ) V V 2 0 Since we can assume that neither V 1 nor V 2 is zero, the second summand cannot be zero. Thus it follows that ker f 01 ker f 02 = 0. Next we want to show that both kernels must actually be zero. To derive a contradiction, let us therefore assume that there is some nonzero vector v 0 ker f 01 \ ker f 02. This means that f 02 (v 0 ) is nonzero. We can then take some linear map g 2 : V 2 k which satisfies g 2 (f 02 (v 0 )) 0. In addition, we define a linear map g 0 : V 0 k by g 0 = g 2 f 02. Lastly, we define linear maps h 0 : k V 0 and h 2 : k v 2 by h 0 (1) = v 0 and h 2 (1) = f 02 (v 0 ). This gives us the following commutative diagram: k 0 h 0 V 0 V 1 k 0 k 0 g 0 h 2 V 2 0 g 2 k 0 It clearly holds that g h is the identity on indecomposable 6. Thus V is this indecomposable, which is a contradiction. It thus follows that ker f 02 must be zero, and by symmetry that this also holds for ker f 01. Thus both maps are indeed injective, and thereby isomorphisms. Since V 0 is nonzero, we can find some surjective linear map g 0 : V 0 k. Since f 01 and f 02 are both isomorphisms, we can define surjective linear maps g 1 : V 1 k and g 2 : V 2 k simply by g 1 = g 0 f01 1 and g 2 = g 0 f02 1. Next we let v 0 be some vector in V 0 such that g 0 (v 0 ) = 1. We then define a linear map h 0 : k V 0 by 1 v 0. Finally, we define linear maps h 1 : k V 1 and h 2 : k V 2 by h 1 (1) = f 01 h 0 (1) and h 2 (1) = f 02 h 0 (1). This 23

24 gives us the commutative diagram k k h 0 h 1 V 0 V 1 g 0 g 1 k 0 k k h 2 V 2 0 g 2 k 0 for which it clearly holds that g h is the identity on the representation k k k 0 i.e. the representation we label as number 9. Thus number 9 is a summand of V, which means that 9 and V are isomorphic. It is easily realized that 9 is indecomposable. If it were not, any decomposition would necessarily need to decompose one the sides of the square. Every side of the square is, however, an indecomposable representation of the A 2 -quiver. Thus 9 must be indecomposable. Since V was assumed to be indecomposable it thus follows that V and 9 are isomorphic. This means that we have found every possible indecomposable where V 0 = 0. This completes the first case. Let us move on to the case where V 0 = 0. In that case we have the indecomposable representation 0 V 1 V 2 V 3 which does not belong to our previous list of 9 indecomposables. We want to show that f 13 and f 23 are both bijections. Suppose that f 13 is not bijective. Then im f 13 V 3. We can thus find some nonzero linear map p : V 3 V 3 / im f 13. Since the map is nonzero we must then also be possible to find a linear map p : V 3 / im f 13 k such that p p is a surjection V 3 k. We denote this composition by g 3. Since we go through V 3 / im f 13 it follows that g 3 f 13 = 0. It follows that the following diagram is commutative: V 1 0 f 13 g 3 V 3 k 24

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