Representations of quivers
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1 Representations of quivers Gwyn Bellamy October 13, 215
2 1 Quivers Let k be a field. Recall that a k-algebra is a k-vector space A with a bilinear map A A A making A into a unital, associative ring. Notice that a (unital) module M for a k-algebra A is automatically a k-vector space. 1.1 What is a quiver? A quiver Q is simply a directed graph. That is, it is a collection Q = {e 1,..., e k } of vertices and a collection Q 1 = { 1,..., l } of arrows between vertices. In order to specify which two vertices an arrow goes between, we define a pair of functions h, t : Q 1 Q such that h() is the vertex at the head of and t() is the tail of i.e. is the arrow beginning at vertex t() and ending at vertex h(). Example 1.1. Here are some examples of quivers. e 2 β e 1 e 2 e 1 e 5 e 3 e 1 γ δ e e 1 e 1 e 2 β n For instance, in the second example, Q = {e 1, e 2, e 3, e 4, e 5 }, Q 1 = {, β, γ, δ} and t(γ) = e 3, h(β) = e 5 etc. Remark 1.2. A quiver is the case, holdall, or tube that an archer uses to hold his/her arrows. 1.2 Representations As representation theorists, we re interested in representations of a quiver. A representation M = {(V i, ϕ ) i Q, Q 1 } is the data of a complex vector space V i at each vertex i of Q and a linear map ϕ : V t() V h() between the vector space at the tail of to the vector space at the head of, for every arrow.
3 Example 1.3. Here are two representations ϕ φ M : C 2 C 3, N : C 4 C (1) φ β ϕ β where ϕ = , ϕ β = π, φ = ( ) (, φ β = ). We can associate to a representation M = {(V i, ϕ )} a dimension vector dim M Z Q by setting (dim M) i = dim V i. Unlike a vector space, the representation M is not uniquely defined by its dimension vector. However, as we shall see, dim M is a very useful invariant of M. Example 1.4. If we consider the two representations M and N in (1), then dim M = (2, 3), dim N = (4, 1). 1.3 Subrepresentations, simple representations etc. Given a representation M = {(V i, ϕ ) i Q, Q 1 } of Q, a subrepresentation N of M is the representation {(U i, ϕ Ut() ) i Q, Q 1 }, where each U i is a subspace of V i, chosen so that ϕ (U t() ) U h() for all Q 1. Example 1.5. Here is an example of a subrepresentation. Let M be the representation of e 1 e 2 given in (1). Let U 1 = C be the subspace of C 2 spanned by v 1 = and U 2 = C 2 the subspace of C 3 spanned by u 1 = 1, u 2 = 1. ( 1 1 ) β ii
4 Then, M restricts to the subrepresentation U 1 U 2, where ϕ ϕ β ϕ = ( 2 9 ), ϕ β = ( π ). If on the other hand, we take the representation N in (1) and U 1 the subspace of C 4 spanned by the vectors 1 u 1 =, u 2 = 1, and U 2 = {} C, then this does not produce a subrepresentation of N since φ (U 1 ) U 2, φ β (U 1 ) U 2, i.e. one cannot just choose arbitrary subspaces to get subrepresentations. Example 1.6. If we consider the example of Q being e 1 then a representation of Q is simply a vector space V 1 together with a linear map ϕ : V 1 V 1. Then a subspace W 1 V 1 is a subrepresentation precisely if ϕ (W 1 ) W 1. We say that the representation M is simple if the only subrepresentations of M are M itself and. Here is a construction of some simple representations. For each i Q, we let E(i) = {(E(i) j, ϕ )} denote the representation where E(i) i = C and E(i) j = for j i. Moreover, ϕ = for all. We leave it to the reader to check that E(i) is simple. Given two representations M = {(V i, ϕ )} and N = {(U i, ψ )} we can construct a new representation M N, the direct sum of M and N, as follows. At each vertex i, we simply take the direct sum of vector spaces V i U i. Then, we define the new map η = (ϕ, ψ ) for each Q 1 i.e. ( ) ϕ η = : V t() U t() V h() U h(). ψ iii
5 Notice that M and N are naturally subrepresentions of the direct sum M N. Of course, to better understand representations of a given quiver, we want to be able to compare them. A homomorphism between representations. Let M = {(C v i, ϕ )} and N = {(C w i, ψ )} be representations of a quiver Q. Then a homomorphism f : M N is a collection of linear maps f i Hom C (C v i, C w i ) for each i Q such that the diagrams C v t() f t() C w t() ϕ ψ C v h() f h() C w h() commute for all Q 1. The space of all homomorphisms from M to N is denoted Hom Q (M, N). 1.4 The path algebra There is a natural algebra we can associate to each quiver Q. Recall that an algebra A is a C- vector space together with a bilinear multiplication map A A A that makes it into a ring with identity. First we define a path in Q to be a tuple of arrows k k 1 1 such that h( i ) = t( i+1 ) for i = 1,..., k 1. The length of the path is k. We think of each vertex e i as being a path of length zero i.e. the path that does nothing. Let CQ be the complex vector space with basis given by all paths in Q. To define a multiplication on CQ, it suffices to do so for paths. Then, { β l β 1 k 1 if h( k ) = t(β 1 ) (β l β 1 ) ( k 1 ) = otherwise. That is, multiplication is simply the concatenation of paths where-ever that makes sense. For the paths e i of length zero, we have { k 1 if t( 1 ) = i ( k 1 ) e i = otherwise. Example 1.7. The path algebra CQ of Q : e 1 e 2 β iv
6 has basis {e 1, e 2,, β} and multiplication is given by e 2 = e 1 =, e 2 β = β e 1 = β, β = β =,... But notice that the path algebra CQ, where Q : e 1 has basis {e 1,, 2, 3,... } and multiplication e 1 n = n e 1 = n, n m = n+m. In particular, CQ is infinite dimensional and commutative. Definition 1.8. An A-module M is a vector space equipped with a bilinear map A M M, (a, m) a m such that 1 m = m for all m M. a (b m) = (a b) m for all a, b A and m M. We will be interested, in particular, in the category of modules for the path algebra CQ. Exercise 1.9. Notice that the definition of a module uses explicitly the identity element 1 of the algebra A. However, we ve made no mention of an identity for the path algebra. Check that 1 := i Q e i is the identity in CQ. Example 1.1. If Q is the quiver e 1 e 2 then we could take M = Cf 1 Cf 2 with e i f j = δ i,j f j and β f 1 = 4f 2, f 2 =, β f 1 = 12f 2,, β f 2 =, or N = C 7, where e 1 v = v and e 2 v = v = β v = for all v C Representations vs. CQ-modules We have seen that we can associated to a quiver its category of representations and also the category of modules over the path algebra. It turns out that these are equivalent notions (formally, v
7 one says that the categories are equivalent). To show that they are equivalent, we explain how to go from a representation to a module for the path algebra and back. Let M = {(V i, ϕ ) e i Q, Q 1 } be a representation of Q. We construct out of M a module F (M) for the path algebra CQ. As a vector space, F (M) := V i, e i Q is just the direct sum of the spaces V i. Since every path in CQ is the composition of a series of arrows, we just have to say how a) the trivial paths e i act on F (M) b) how each of the arrows Q 1 acts on F (M). Firstly, for each i we have a projection map p i : F (M) V i and an inclusion map q i : V i F (M). We define e i = q i p i i.e. e i v = q i p i (v). Next, associated to Q 1 is the linear map ϕ : V t() V h(). We use a similar trick to extend this to a map F (M) F (M) by setting = q h() ϕ p t() : F (M) q h() ϕ p t() V t() Vh() F (M), i.e. v = q h() ϕ p t() (v). Now we want to go in the opposite direction. Given a module N for the path algebra CQ, we want to define a representation G(N) = {(U i, ψ ) e i Q, Q 1 }. Notice first that the identity element 1 in CQ is the sum i Q e i. Also, the rules for multiplication in CQ show that e i e j = δ i,j e i (we say that {e i Q } form a complete set of orthogonal idempotents in CQ). This implies that N = e i N. i Q We set U i = e i N. We just need to define a linear map ψ : e t() N e h() N for each Q 1. For this, notice that Theorem The maps F and G define quasi-inverse equivalences between the category Rep(Q) of all representations of Q and the category CQ-Mod of all CQ-modules. Proof. We have done all the hard work already in defining the maps F and G. It is straightforward to check that G F (M) = M for any representation M and F G(N) = N for any CQ-module N. It is clear how to define F and G on morphisms to make them functors. vi
8 Example Let s take Q to be the quiver e 1 e 2 e β 3 and the representation N : 3 C C C. Then CQ has basis {e 6 1, e 2, e 3,, β} and F (N) is the three-dimensional module with basis f 1, f 2, f 3 such that e i f j = δ i,j, f 1 = 3f 2, β f 3 = 6f 2, and f i = β f j = otherwise. Similarly, if we take the CQ-module M in example 1.1 then G(M) is the representation of Q given by 4 C C. 1.6 Indecomposable representations 7 In order to study the representations of a finite dimensional algebra, the first thing to do is try and break a representation up into a direct sum of simpler representations - we say that M decomposes as a direct sum M 1 M 2. We ve already seen that many representations occur as the direct sum of smaller representations. Repeating, we get smaller and smaller representations. Of course this process must eventually stop. That is, eventually we get to a representation M that cannot be written as the direct sum M 1 M 2 of two subrepresentations. Definition The representation M is said to be indecomposable if it cannot be decomposed into a direct sum of two proper subrepresentations. Notice that we can always decompose M = M, but such a decomposition doesn t help us understand M better, so we ignore these trivial decompositions. Clearly, if we want to understand, or classify all representations of Q, it suffices to classify the indecomposable representations. In general this turns out to be a very difficult task. Theorem 1.14 (Krull-Schmidt). Let A be a finite-dimensional algebra and M a finitely generated A-module. Then, up to permutation of factors, M has a unique decomposition into indecomposables, M = M 1 M k. The key point underlying the Krull-Schmidt Theorem is the fact that, under the hypothesis of the theorem, if M is indecomposable then End A (M) is a local ring i.e. it has a unique maximal right ideal which is also the unique maximal left ideal (and hence two-sided). vii
9 Example Consider A = R[x]/(x 4 1) as a module over itself. Then A R[x] (x 2 + 1) R[x] (x + 1) R[x] (x 1). Example Here is a counter example to the Krull-Schmidt theorem when the ring Λ is not assumed to be Artinian. Let Λ = k[x, y], where k is a field. Let (x, y) denote the ideal in Λ generated by x and y. Take M = (x, y) (x, y 1). Then there is a natural map M Λ, (f, g) f + g. This is surjective. Therefore we get a short exact sequence C M Λ. Since Λ is projective as a Λ-module, this sequence splits i.e. M Λ C. But neither of (x, y) or (x, y 1) are isomorphic to Λ since they are not projective. 1.7 Composition factors Let M be an A-module. A composition series for M is a series M : = M M 1 of submodules, one properly contained inside another. Recall that M is Artinian if every composition series has finite length. If A is finite dimensional and M finitely generated then it is Artinian. The factors of M are the quotients M i /M i 1. Theorem 1.17 (Jordan-Hölder). Let A be a finite dimensional algebra and M a finitely generated A-module. Let M : = M M 1 M k = M N : = N N 1 N l = M be two composition series with simple factors. Then k = l and there exists a bijection σ : {1,..., k} {1,..., k} such that N i /N i 1 M σ(i) /M σ(i 1), i = 1,..., k. Notice that the Jordan-Hölder Theorem implies that M has a well-defined length. viii
10 Example Let k be an infinite field and A = k[x, y]/(x 2, xy, y 2 ). For each [ : β] P 1 k, the ideal (x+βy) in A is one-dimensional and the quotient is a two-dimensional local ring. Therefore, we get a composition series A : (x + βy) (x, y) A with composition factors k, k, k (where k here is thought of as the unique simple module for A). Hence A has an infinite number of composition series, but they all have the same factors. Example If we take Q to be the quiver e 1 e 2 a b, and A = kq/ ab, then k{ba} k{ba, b} k{ba, a, b} k{e 1, ba, a, b} A is a composition series with simples factors S 1, S 1, S 2, S 1, S Other basic examples of finite dimensional algebras a If Q is the quiver e 1 e 2, then kq is three-dimensional, it is isomorphic to ( ) k k, k the algebra of two by two upper triangular matrices. If we take A = k[x, y]/(x 2, y 2 ) then notice that A is isomorphic to kq/ a 2, ab ba, b 2, where Q is the quiver with one vertex and two loops (labeled a and b) at that vertex. Consider A = k[x]/(x n ) as a module over itself. This is an example of a uniserial module - it has a unique composition series with simple factors. ix
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