2. Finite dimensional algebras In this chapter, A will always denote an algebra (usually finite dimensional) over an algebraically closed field k and

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1 2. Finite dimensional algebras In this chapter, A will always denote an algebra (usually finite dimensional) over an algebraically closed field k and all modules will be finitely generated. Thus, for a finite dimensional irreducible A-module M, Schur s lemma says that End A (M) = k. Many of the results can also be formulated when the field k is not algebraically closed but things become more technical because one has to keep track of the division algebras that arise in Schur s lemma, see e.g. chapter 4 of Benson. My main reference for much of the material in this chapter is the beautiful lecture notes by William Crawley- Boevey which are available from his web page, supplemented by extra tidbits from Benson s book, plus the classic paper of Bernstein-Gelfand-Pomonarev Projective indecomposable modules. Assume that A is a finite dimensional algebra. A projective indecomposable module (PIM for short) means a projective indecomposable module in A -mod. By the Krull-Schmidt theorem, such modules are isomorphic to the indecomposable summands of the regular module A itself. In this section, we want to understand the structure of the PIMs in a little more detail. We will work for a while in the entirely equivalent language of idempotents. Recall an idempotent e A means a non-zero element such that e 2 = e. Two idempotents e 1, e 2 are orthogonal if e 1 e 2 = e 2 e 1 = 0. An idempotent is primitive if it cannot be written as a sum of two orthogonal idempotents. If e is an idempotent, then so is (1 e), and e(1 e) = 0. It follows from this that A = Ae A(1 e). Thus whenever you have an idempotent in A it defines a summand of A. Conversely if you have a direct sum decomposition A = M N, then the element e End A (A) op = A defined to be the projection of A onto M along the direct sum is an idempotent in A. Under this correspondence between idempotents and summands of A A, primitive idempotents correspond to indecomposable summands, i.e. to PIMs. Decompositions 1 A = e 1 + +e r of the identity as a sum of primitive orthogonal idempotents correspond to decompositions A = Ae 1 Ae r of A A as a direct sum of PIMs. Two idempotents e, e A are conjugate if there exists an invertible element u A such that ueu 1 = e. Lemma 2.1. Idempotents e, e A are conjugate if and only if the A- modules Ae and Ae are isomorphic. Proof. ( ). Suppose ueu 1 = e. Then, Ae = Aeu = Aue = Ae as left A-modules. ( ). Suppose that Ae = Ae. Let µ : Ae Ae be an isomorphism. Since Hom A (Ae, Ae ) = eae we get that µ corresponds to an element m eae. Similarly, µ 1 corresponds to an element m e Ae, and mm = e, m m = e. 1

2 2 By the Krull-Schmidt theorem we also have that A(1 e) = A(1 e ). Let ν : A(1 e) A(1 e ) be an isomorphism. Like before, we get elements n (1 e)a(1 e ) and n (1 e )A(1 e) such that nn = (1 e), n n = (1 e ). Now consider m + n A. Then, (m + n)(m + n ) = mm + nn + mn + nm = e + (1 e) = 1, and similarly (m + n )(m + n) = 1, so m + n is invertible. Also (m + n)e = m = e(m + n) so (m + n)e (m + n) 1 = e and e, e are conjugate. The crucial result about idempotents is the following: Theorem 2.2 (Lifting idempotents). Let N be a nilpotent ideal in A. Let e A/N be an idempotent. Then, there exists an idempotent f A lifting e, i.e. such that f = e. Moreover, if e, e A/N are conjugate idempotents f, f A are idempotents lifting e, e respectively, then f, f are conjugate too. Proof. Take e A/N. We e 1 = e and inductively define an idempotent e i A/N i such that ē i = e i 1. Since N is a nilpotent ideal, N n = 0 for some n and taking f = e n proves the first part of the theorem. Given e i 1, define e i as follows. First, let a A/N i be any pre-image of e i 1. Then a 2 a is a pre-image of 0, so a 2 a N i 1, so (a 2 a) 2 = 0 in A/N i. Let e i = 3a 2 2a 3. This still has image e i 1 in A/N i 1, and e 2 i e i = (3a 2 2a 3 )(3a 2 2a 3 1) = (3 2a)(1 + 2a)(a 2 a) 2 = 0/ So e i is an idempotent. For the uniqueness statement, take conjugate idempotents e, e A/N. Say µe = e µ for µ A with µ invertible in A/N. Suppose f, f A are idempotents lifting e, e respectively. Let ν = f µf + (1 f )µ(1 f). Then, νf = f µf = f ν. To complete the proof, we need to show that ν is invertible. Observe µ ν = f µ + µf 2f µf = (f µ µf)(1 2f) N. Since µ is invertible, there exists x A such that 1 µx N. µ ν N too, we deduce that 1 νx = n for some n N. Then, νx(1 + n + n ) = (1 n)(1 + n + n ) = 1 so that ν has a right inverse. done. Since Similarly, ν has a left inverse and we are Corollary 2.3. Let 1 = e e r be a decomposition of 1 as a sum of orthogonal idempotents in A/N. Then there exist orthogonal idempotents 1 = f f r A such that f i = e i for each i. Proof. Proceed by induction on r, the case r = 1 being obvious. For r > 1, let e = e e r 1 and f A be an idempotent lifting e. By induction applied to the quotient e(a/n)e of the ring faf, there are orthogonal

3 idempotents f = f f r 1 faf lifting e = e e r 1. Set f r = 1 f 1 f r 1. Now what is the point? Well, since A is a finite dimensional algebra over an algebraically closed field, J(A) is a nilpotent ideal and Wedderburn s theorem shows that A/J(A) = M n1 (k) M nr (k) where n 1,..., n r are the dimensions of the r inequivalent irreducible A- modules L 1,..., L r. Let e 1,1,..., e 1,n1,..., e r,1,..., e r,nr be the diagonal matrix units, giving us a decomposition 1 = (e 1,1 + + e 1,n1 ) + + (e r,1 + + e r,nr ) of the identity in A/J(A) as a sum of orthogonal primitive idempotents. Note (A/J(A))e i,j = Li for every j. Apply the corollary to lift to a decomposition 1 = (f 1,1 + + f 1,n1 ) + + (f r,1 + + f r,nr ) in A. Since the e i,j are primitive, the f i,j obviously are too. As e i,1,..., e i,ni are all conjugate in A/J(A), so are all of f i,1,..., f i,ni. So the modules Af i,j are isomorphic for every j = 1,..., n i are all isomorphic PIMs to P i say. Hence A = P n 1 1 Pr nr as a direct sum of PIMs. Finally, P i / rad P i = L i, hence the PIM P i has a unique maximal submodule. We have now proved: Theorem 2.4. The map P P/ rad P induces a 1 1 correspondence between the isomorphism classes of PIMs and the isomorphism classes of irreducible modules. Moreover, letting L 1,..., L r be a complete set of inequivalent irreducibles and P 1,..., P r be the corresponding PIMs, we have that AA = P dim L 1 1 P dim Lr r. If you like, this decomposition generalizes Wedderburn s theorem to the non-semisimple case. Here is an important consequence. Corollary 2.5. Let P be a PIM and L = P/ rad P be the corresponding simple module. For any f.d. A-module M, the composition multiplicity [M : L] = dim Hom A (P, M). Proof. Proceed by induction on the length of a composition series of M. The base case is when M is irreducible, which follows by Schur s lemma (in its strong form when the vector space is finite dimensional and the field is algebraically closed). For the induction step, pick a proper submodule K of M. Then there is a short exact sequence 0 K M Q 0. 3

4 4 Apply the exact functor Hom A (P,?) to get a short exact sequence of vector spaces 0 Hom A (P, K) Hom A (P, M) Hom A (P, Q) 0. We deduce by induction and the Jordan-Holder theorem that dim Hom A (P, M) = [K : L] + [Q : L] = [M : L]. We are done. Recalling that A = End A (A) op, the next theorem generalizes the results of the section a little bit. Theorem 2.6. Let M be an f.d. A-module and let B = End A (M) op. Let M = M n 1 1 Mr nr be a decomposition of M into inequivalent indecomposables M 1,..., M r. Let e i B be an idempotent projecting M onto one of the summands isomorphic to M i, and let P i = Be i and L i = P i / rad P i. Then, P 1,..., P r is a complete set of inequivalent PIMs for B, L 1,..., L r are the corresponding irreducible B-modules, and n i = dim L i. Proof. Just observe that decompositions of M as a direct sum of indecomposable A-modules are in 1 1 correspondence to decompositions of 1 B as a sum of primitive orthogonal idempotents, then apply the results above. The next exercise is really important: the remainder of the chapter will be devoted to generalizing this example! Exercise 5. Let A = T n (k) be the algebra of all upper triangular n n matrices over a field k. What is J(A)? How many inequivalent irreducible A-modules are there? What are their dimensions? What do their projective covers look like? One more definition in this section: a finite dimensional algebra A is called a basic algebra if all its irreducible modules are one dimensional. For example, the above exercise shows that T n (k) is a basic algebra. Suppose for a moment that A is any finite dimensional algebra, and let P 1,..., P n be a complete set of inequivalent PIMs. Then, P = P 1 P n is a projective generator for A. By the Morita theorem, A is Morita equivalent to the algebra End A (P ). By Theorem 2.6, the latter algebra IS a basic algebra. In other words, if you are studying representations of finite dimensional algebras, you may as well restrict yourself straight away to studying just the basic ones.

5 2.2. Quivers and path algebras. A quiver Q = (Q 0, Q 1, s, t : Q 1 Q 0 ) is a set Q 0 of vertices, which for us will be {1, 2,..., n} (in particular, finite); a set Q 1 of arrows which for us will be finite. An arrow ρ starts at the vertex s(ρ) and terminates at the vertex t(ρ). A non-trivial path in Q is a sequence ρ 1... ρ m (m 1) of arrows satisfying t(ρ i+1 ) = s(ρ i ). Pictorially: ρ 1 ρ 2 ρ m We also have the trivial paths e i for each vertex i, i.e. the path starting at vertex i and going nowhere. For a path x, we write s(x) for the vertex where it starts, t(x) for the vertex where it terminates. The path algebra kq is the k-algebra with basis the paths in Q and the product of two paths x, y is defined by { the concatenation if t(y) = s(x), xy = 0 otherwise. Note that kq is obviously an associative algebra. The trivial paths e 1,..., e n are mutually orthogonal idempotents, and the 1 is the element e e n. The elements e 1,..., e n together with the paths of length one defined by each of the finitely many arrows in Q generate Q as an algebra, so kq is a finitely generated algebra. If we had allowed infinitely many vertices in our definition, then kq would not have an identity. If we had allowed infinitely many edges then kq would not be finitely generated. Example 2.7. Let Q be the quiver 1 ρ 2 σ 3 Then kq has basis {e 1, e 2, e 3, ρ, σ, σρ}. Some products: ρσ = ρρ = e 1 σρ = σρe 3 = 0; e 3 σρ = σρ. Exercise 6. Suppose Q is a quiver with at most one path between any two points. Then, kq is isomorphic to the subalgebra of M n (k) consisting of all matrices with ij-entry equal to zero if there is no path from j to i. For example, 1 2 n is the lower triangular matrices. Example 2.8. Take Q to be the quiver with one vertex and one loopy arrow. Then, kq = k[t ], the polynomial algebra in one variable. If Q has one vertex and r loops, then kq is the free associative algebra on r generators. I don t know very much about the latter algebra when there is more that one generator, but apparently its module category is a completely wild thing. For instance, I read somewhere that there exist simple modules of every possible dimension. 5

6 6 Let A = kq. Here are some increasingly difficult remarks which I am not going to attempt to prove. (1) Spaces like Ae i, e j A, e j Ae i,... are easy to describe. For instance, Ae i has a basis of all paths starting at i. (2) A is a finite dimensional algebra if and only if Q has no oriented cycles, which is if and only if A is Artinian. (3) A is noetherian if and only if for each vertex i, there is at most one oriented cycle passing through i exactly once. (4) J(A) has a basis consisting of all paths from i to j for all pairs of vertices i, j such that there is no path from j to i. (It is easy at least to see that this is a two-sided nilopotent ideal, hence that it is contained in J(A); I didn t see yet how to prove that it was equal to J(A).) I do want to prove some basic facts about the idempotents e 1,..., e n. Note that they are orthogonal idempotents summing to 1, so A = Ae 1 Ae n is a decomposion of A into projectives. In fact: Lemma 2.9. The e i are primitive idempotents, i.e. each Ae i is a PIM. Moreover, for i j, Ae i = Aej, hence e i, e j are not conjugate to each other. Proof. Suppose Ae i is decomposable. Then End A (Ae i ) = e i Ae i contains an idempotent f e i. Then f 2 = f = fe i so f(e i f) = 0. Since f Ae i, f is some linear combination of paths starting at i. Let x be a path of maximal length appearing with non-zero coefficient in f. Similarly, (e i f) e i A, so it is some linear combination of paths ending at i. Let y be a path of maximal length appearing with non-zero coefficient in (e i f). Now think about f(e i f). It is some linear combination of paths starting and ending at i. Moreover it must contain xy with non-zero coefficient, since no other paths arising in the product can cancel with that one by choice of x, y. This shows f(e i f) 0, a contradiction. Now suppose that Ae i = Aej. Since Hom A (Ae i, Ae j ) = e i Ae j, inverse isomorphisms give us elements f e i Ae j and g e j Ae i such that fg = e i, gf = e j. But gf is a linear combination of paths starting and ending at j and going through i. The trivial path e j is not such a path, since j i. So there is no way gf could equal e j. We are usually going to be concerned with the case that A is a finite dimensional algebra (i.e. Q has no oriented cycles). Then we ve decomposed A = Ae 1 Ae n A into a direct sum of inequivalent PIMs, and by Krull-Schmidt these are all the PIMs. Thus, A has exactly n simple modules, all of which are one dimensional, namely the modules L 1,..., L n where L i = Ae i / rad Ae i. In particular, A is a basic algebra, as defined at the end of the previous section.

7 On the other hand if A is not a finite dimensional algebra, it is not Artinian, so Krull-Schmidt may not hold, so we can t really say too much from this analysis. For instance A might be a free algebra on more than one generator, which has infinitely many simple modules Representations of quivers. Let Q = (Q 0, Q 1, s, t) be a quiver. A representation V of Q is an assignment of a vector space V i to each vertex and a linear map V ρ : V s(ρ) V t(ρ) to each edge. (There is no assumption about anything commuting). We ll always just consider finite dimensional representations of Q, that is, representations for which each V i is a finite dimensional vector space. The dimension vector of a representation V of Q is the vector dimv = (dim V i ) i Q0. Given two representations V, V of Q, a morphism θ : V V is given by linear maps θ i : V i V i for each i Q 0 such that the obvious diagrams commute. You can obviously compose morphisms, so we have a nice category: the category of all finite dimensional representations of Q. There are notions of subrepresentations, quotient representations, direct sums of representations,.... I think they re all pretty obvious, so you should formulate them for yourselves. So the category of (finite dimensional) representations of Q is an abelian category. Example Fix i Q 0. Let L i be the representation defined by putting k on the ith vertex, and 0 on all other vertices, with all maps being zero. Then, L i is an irreducible representation obviously there are no subrepresentations other than 0 and itself. Example Let Q be the quiver Let V and V be the representations and k 1 k 1 k k 1 k 0 respectively. Then, Hom(V, V ) is one dimensional, Hom(V, V ) is zero. Theorem The category of finite dimensional representations of Q is equivalent to the category kq -mod of finite dimensional modules over the path algebra. Proof. It is traditional just to explain how to turn representations of Q into kq-modules and vice versa, and then to omit all the tedious checks. Spose (V i ) n i=1 is a representation of Q. We associate a kq-module V as follows. As a vector space, n V = V i. i=1 7

8 8 The action of e i kq on V is as the projection of V onto the summand V i. The action of ρ kq for ρ Q 1 is as the linear map V ρ : V s(ρ) V t(ρ) on V s(ρ) and as zero on all other V i s. Conversely, given a kq-module V, let V i = e i V and let V ρ be the linear map defined by the action of ρ kq on V s(ρ) noting that ρe i V e j V since ρ = e j ρe i. Now assume for a moment that kq is finite dimensional. In the previous section, we showed that the e i s are mutually orthogonal, non-conjugate primitive idempotents summing to 1, so that P 1,..., P n defined from P i = (kq)e i is a complete set of PIMs. In this section we ve constructed irreducible modules L i for each i = 1,..., n: namely one dimensional on the ith vertex and 0 everywhere else. Now multiplication obviously defines a non-zero map P i L i. Hence L i is the unique irreducible quotient of P i. Thus, in the case that Q has no oriented cycles, L 1,..., L n is a complete set of inequivalent irreducible representations. Example Here I will discuss how basic problems in linear algebra translate into the language of quivers: (1) (Equivalence of matrices) Consider the quiver with two vertices and one edge. Giving a finite dimensional representation means giving a linear map f : V 1 V 2 between two finite dimensional vector spaces. An isomorphism between f : V 1 V 2 and g : W 1 W 2 means vector space isomorphisms θ : V 1 W 1 and φ : V 2 W 2 such that φ f = g θ. In other words, V 1 and W 1 have the same dimension, V 2 and W 2 have the same dimension, and the rectangular matrices representing f and g in some bases are equivalent matrices [f] = [φ] 1 [g][θ]. So you are reduced to studying equivalence classes of matrices which by linear algebra are completely determined by their size and their rank. In particular you see from this that there are three indecomposable representations up to isomorphism (not counting the zero module as an indecompsable), namely k 0, k 1 k and 0 k. Note there are just finitely many indecomposables this is finite. (2) (Similarity of matrices) Consider the quiver with one vertex and one loop. Giving a finite dimensional representation means simply giving an endomorphism of a finite dimensional vector space. Two representations f : V V and g : W W are isomorphic if there is a vector space isomorphism θ : V W such that θ f = g θ. In other words, they are isomorphic if and only if they have the same dimension and the square matrices representing f and g in some bases are similar matrices [f] = [θ] 1 [g][θ]. So you see you are reduced to studying Jordan normal forms to solve the problem.

9 In particular, the indecomposable finite dimensional representations correspond to the Jordan blocks J n (λ). This time there are infinitely many indecomposables but we can classify them this is tame. (Of course you know that the path algebra is the polynomial algebra in one variable k[x], so finite dimensional representations of this quiver are just finitely generated torsion k[x]-modules which we completely understand by anyway). (3) Consider the Kronecker quiver with two vertices and two arrows both from 1 to 2. A representation means a pair of linear maps f, g : V 1 V 2. It is a difficult but solvable (hence important!) problem in linear algebra to classify equivalence classes of pairs of rectangular matrices. We ll discuss this one later on and manage to classify them: it is again tame though non-trivial. (4) Consider the quiver with one vertex and two arrows. A representation of this means a pair of endomorphisms f, g : V V of a vector space. It is a wild problem to classify similarity classes of pairs of square matrices, just as it is a wild problem to study representations of k x, y. (5) Here s one we can solve. Consider the quiver Suppose that V is an indecomposable representation. Either V is or k k or else the maps V 1 V 2 and V 2 V 3 are injective. So to classify the remaining indecomposable representations of this quiver is exactly the problem of classifying indecomposable pairs of subspaces of a vector space V. But you just pick a basis for the intersection, then extend it to the first one then to their sum... You deduce that the indecomposable representations are the two above together with 0 k 0, k 1 k 0, 0 k 1 k, k 1 k 1 k. Thus there are 6 indecomposables up to isomorphism with dimension vectors (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1) and (1, 1, 1). It is not a coincidence that the picture here is just like the picture for lower triangular 3 3 matrices, i.e. the quiver Even though these algebras are not Morita equivalent there seem to be lots of similarities between their representations. We will explain this later on when we talk about reflection functors. 9

10 10 It is perhaps worth pointing out how to translate things into matrices in general. We will use this observation more fundamentally later on. Suppose V is a representation of a quiver Q with dimension vector α. Picking a basis for each V i we can identify V i = k α i and the map V ρ with a α t(ρ) α s(ρ) - matrix x ρ. Given another representation V of the same dimension, we can pick bases again to identify V i with kα i and the maps then are matrices x ρ. Now saying that V = V means that there are linear maps g i GL αi (k) for each i such that x ρ = g t(ρ) x ρ g 1 s(ρ) for each ρ Q 1. This should help in the following exercises. Note these exercises and the above examples contain all the main examples that we ll use to understand everything else we are doing... Exercise 7. (Some indecomposable representations of the Kronecker quiver) Recalling Example 3 above, let Q be the Kronecker quiver. Fix λ k and n 1. Take the representation V = V (λ, n) defined by letting V 1 = V 2 = k n and the linear maps on the two arrows V 1 V 2 being the maps I n : k n k n (the identity matrix) and J n (λ) : k n k n (the Jordan block of size n with eigenvalue λ). Show that V (λ, n) s are (infinitely many) inequivalent indecomposable representations of Q. Up to isomorphism, there is exactly one other indecomposable representation in which the vector spaces V 1 and V 2 have dimension n. What is it? Exercise 8. (three subspace problem) Find 12 non-isomorphic indecomposable representations (not counting the zero representation as an indecomposable) of the quiver with one vertex in the middle and three vertices around the edge, with three arrows all pointing inwards. The general theory developed in a while will show these are all the indecomposables, so you don t need to prove that extra thing right now. Exercise 9. (four subspace problem) Find infinitely many non-isomorphic indecomposable representations of the quiver with one vertex in the middle and four vertices around the edge, with four arrows all pointing inwards. To end the section here are a few more definitions to get familiar with these will play the crucial role later on. (1) We ll work with vectors in Z n. For instance the dimension vector dimv is such a vector. Let ɛ i = diml i, so ɛ 1,..., ɛ n is the standard basis for Z n. (2) Define a bilinear form on Z n by n α, β = α i β i α s(ρ) β t(ρ). i=1 ρ Q 1 I ll call this the asymmetric bilinear form to remind you its not symmetric.

11 (3) Let q(α) = α, α. This is a quadratic form on Z n which I ll call the quadratic form! (4) Let (α, β) be the symmetric bilinear form associated to the quadratic form q. Thus, (α, β) = α, β + β, α. Note: (.,.) doesn t depend on orientation. Exercise 10. Take the two quivers discussed in Example 2.11(5) above (both had underlying graph 1 2 3). For each, write down the matrix of the asymmetric bilinear form.,. and of the symmetric bilinear form (.,.) with respect to the basis ɛ 1, ɛ 2, ɛ 3 for Z 3. This is the Cartan matrix of type A 3! Note that for i j, ɛ i, ɛ j is This will be important shortly. (the number of paths from i to j) The standard resolution. I m now going to start using some basic homological algebra. Here is a rapid review hopefully enough for those of you who haven t taken a course in homological algebra to follow what is going on. Let A be a k-algebra and M, N be A-modules. I want to define the k-vector spaces Ext i A (M, N). The usual way to do this is to take a projective resolution P 1 P 0 M 0 of M (thus it is an exact sequence and all the P i s are projectives). There are many ways to do this, but it doesn t matter which you choose for our purposes since the Comparison Theorem shows that any two projective resolutions are chain homotopy equivalent. Now apply the contravariant functor Hom A (?, N) to P to get a complex 0 Hom A (P 0, N) Hom A (P 1, N)... This complex is not necessarily exact. So we take cohomology: let Ext i A(M, N) be the kernel (cocycles) over the image (coboundaries) of the differential in the ith slot. Since Hom A (?, N) is left exact, we do at least have that Ext 0 A (M, N) = Hom A(M, N). The fact that the complex is unique up to chain homotopy equivalence means that its cohomology is well-defined independent of the choice. There is another way entirely to define Ext i A (M, N). I think it will be enough for us to understand this for Ext 1 A (M, N). This can be identified with the set of all short exact sequences 0 N E M 0 under a natural equivalence relation. The identification goes as follows. Start with a projective resolution of M as above. Using the comparison theorem, we define homomorphisms P 0 E, P 1 N and P 2 0 so the 11

12 12 diagrams commute. Note the map P 1 N is a cocycle since the left hand rectangle commutes. Its class in Ext 1 A (M, N) is then the element of ext corresponding to the s.e.s... The only other thing you need to know is the long exact sequence. If 0 K M L 0 is a short exact sequence, then applying the left exact functor Hom A (?, N) gives the long exact sequence 0 Hom A (L, N) Hom A (M, N) Hom A (K, N) Ext 1 A(L, N)... There is a similar long exact sequence for Hom A (N,?). Now lets look at what happens if A is the path algebra of a quiver. Theorem Let A = kq and V be a kq-module. There is a projective resolution 0 n Ae t(ρ) k e s(ρ) V f Ae i k e i V g V 0 ρ Q 1 where g is the natural multiplication map, and f(a v) = aρ v a ρv for a v Ae t(ρ) e s(ρ) V. Proof. The terms in the sequence are obviously projective. Also it is obvious that g f = 0 and that g is onto. So we just need to show that ker f = 0 and that ker g im f. (1) ker f = 0. Let ξ ker f. We can write ξ = a v ρ,a ρ Q 1 where the second sum is over all paths a with s(a) = t(ρ) and the elements v ρ,a e s(ρ) V are almost all 0. Then, f(ξ) = aρ v ρ,a a ρv ρ,a. ρ a Let a be a path of maximal length such that v ρ,a 0 for some ρ. Then f(ξ) involves aρ v ρ,a, and nothing can cancel this, so f(ξ) 0. (2) ker g im f. This is a little more tricky. First some observations. Given ξ i Ae i k e i V, we can represent it as n ξ = a x a i=1 where the second sum is over all paths a starting at i and almost all the x a e s(a) V are zero. Define deg(ξ) to be the length of the longest path a with x a 0. a a i=1

13 Now if a is a non-trivial path with s(a) = i, we can express it as a product a = a ρ with ρ an arrow starting at i and a some other path. Then f(a x a ) = a x a a ρx a, viewing a x a as an element of the ρth component. Now we claim that ξ + im f always contains an element of degree 0. If ξ has degree d > 0 then ( n ) ξ = f a x a i=1 where the sum is over all paths a starting at i of length d has degree < d. Now the claim follows by induction. We are ready to prove that ker g im f. Take ξ ker g. Let ξ ξ + im f be of degree 0. Then, 0 = g(ξ) = g(ξ ) = g( e i x e i ) = x e i. i This belongs to n i=1 V i so each term on RHS is zero, so ξ = 0. Hence ξ im f. We are done. Corollary Let V, W be A-modules. i 2. a 13 Then Ext i A (V, W ) = 0 for all Proof. We ve just constructed a projective resolution with 0 in degrees 2 and higher! Corollary Submodules of projectives are projective. Proof. Let V be a submodule of a projective P. Take the s.e.s. 0 V P Q 0 and apply Hom A (?, W ). The long exact sequence shows that Ext 1 A (V, W ) = 0. Since this is for all W, V is projective. Corollary If V, W are finite dimensional representations of Q, then In particular, dim Hom(V, W ) dim Ext 1 (V, W ) = dimv, dimw. dim End(V ) dim Ext 1 (V, V ) = q(dimv ). Discussion. An algebra is hereditary if every submodule of a projective module is projective. We have shown in the second corollary above that path algebras of quivers are hereditary algebras. We also know: A is finite dimensional if and only if Q has no oriented cycles, in which case A is a basic algebra. Suppose for a moment that A is any finite dimensional algebra. The Ext quiver of A is the quiver with vertices 1,..., n corresponding to the isomorphism classes of irreducible A-modules L 1,..., L n, and with dim Ext 1 A (L i, L j ) arrows from i to j.

14 14 In case A is a path algebra of Q, the last corollary above shows that dim Ext 1 A(L i, L j ) = ɛ i, ɛ j which we noticed before was the number of arrows in Q from i to j. Thus in the case that A is a path algebra of a quiver with no oriented cycles, the Ext quiver allows us to recover the original quiver in an invariant way! This seems like a good moment to state Gabriel s theorem, though I am not going to write out a proof. (See Benson Propositions and 4.2.5). Theorem 2.18 (Gabriel, circa 1970). Suppose that A is a finite dimensional basic algebra over k. Let Q be its Ext quiver. Then there is a surjective map π : kq A with kernel contained in the ideal of all paths of length at least two. If A is hereditary, this map is an isomorphism. Thus you see that of all finite dimensional basic algebras, the hereditary ones are precisely the ones you get from quivers; but for any finite dimensional algebra you get a first approximation to its representation theory by looking at the representations Ext quiver. This is particularly effective if the Ext quiver happens to have no oriented cycles, since then ker π is contained in the square of the Jacobson radical of kq so you can transfer quite a lot of the information A little algebraic geometry. To proceed, we need a little bit of algebraic geometry. Everything in this section will be familiar to those of you who took my course last year on algebraic groups, but I will try to be careful about stating all the things we are using so that those of you who didn t can follow the main idea. First remember that affine N-space is the algebraic variety A N. The points of A N are just the N-tuples from the field k. The coordinate ring k[a N ] of functions on A N is the polynomial ring k[x 1,..., x n ], where x i is the ith coordinate function. The space A N is endowed with the Zariski topology defined by declaring that the closed sets are the sets V (I) of common zeros of the ideals in k[x 1,..., x n ] (equivalently, the common zeros of a finite collection of polynomials generating the function I). Conversely, given a closed subset X of A N, the set of all functions vanishing on X is a radical ideal of the coordinate ring. The Nullstellensatz says that I(V (I)) = I. This implies that the maps V and I give a 1 1 inclusion reversing correspondence between the closed sets in A N and the radical ideals in k[x 1,..., x n ]. An affine variety is a closed set X in A N. Its coordinate ring k[x] is the ring k[x 1,..., x n ]/I(X), i.e. the functions obtained by restricting polynomial functions down to X. The Nullstellensatz extends to this situation to give a 1 1 inclusion reversing correspondence between the closed sets in X and the radical ideals in k[x]. In particular, the points in X correspond to the maximal ideals in k[x], so you can completely recover X from its coordinate ring. In this way you get a functor Spec from the category of affine algebras (reduced, finitely generated commutative algebras) to the category

15 of affine varieties which is a contravariant equivalence of categories. Products exist in the category of affine varieties: X Y is the affine variety with coordinate ring k[x] k[y ]. A topological space is irreducible if any non-empty open subset is dense. The dimension of an irreducible topological space is sup{n there exist irreducible closed subsets Z 0 Z 1 Z n }. For an affine variety X, it is irreducible if and only if k[x] is an integral domain, in which case its dimension dim X is equal to the Krull dimension of the ring k[x] (defined in terms of chains of prime ideals). For example the dimension of the irreducible variety A N is N. If X is a closed subvariety of A N and Y is a closed subvariety of A M, a morphism f : X Y means a function with f(x 1,..., x N ) = (f 1 (x),..., f M (x)) where each f i is a polynomial function, i.e. a function belonging to k[x]. Equivalenly, a function f : X Y is a morphism if for each θ k[y ] the function f θ : X k defined by f (x) = θ(f(x)) belongs to k[x], in which case f : k[y ] k[x] is an algebra homomorphism. This latter definition makes it clear that morphisms don t depend on the particular choice of the embedding of X, Y into affine spaces. This will be good enough for our purposes: we won t ever think about morphisms of non-affine varieties. A quasi-affine variety is an open subset U of an affine variety X, equivalently, a locally closed (a.k.a. open in its closure) subset of some A N. It has the subspace topology induced by the Zariski topology on X. The dimension of U in this topology is the same as the dimension U if its closure in X. Special case: let X be an affine variety and 0 f k[x]. Then the set D(f) of all non-zeros of f is called a principal open subset of X. I want to convince you that D(f) can be given the structure of an affine variety in a natural way. The trick is to consider homeomorphism D(f) X A 1, x (x, 1/f(x)). The image of D(f) is the set of all (x, c) such that f(x)c = 1. This is a closed subset of X A 1. So we have identified D(f) with an affine variety. Its coordinate ring k[d(f)] = k[x, x]/(fx 1), so the image of f in k[d(f)] is invertible: f x = 1. You get that k[d(f)] = k[x]f, the coordinate ring of X localized at the function f. Return to the general case that U is an arbitrary open subset of an affine variety X. The principal open subsets form a base for the Zariski topology, so U is a (finite) union of D(f) s, hence a finite union of affine varieties. In this way we have made sense of the idea that a quasi-affine variety is a topological space that looks locally like an affine variety. This leads to the general notion of variety: something obtained by gluing affine varieties along open sets. But we won t need such generality. An algebraic group is an affine variety G endowed with a multiplication µ : G G G and an inverse i : G G which are morphisms of affine 15

16 16 varieties. The main example is the group GL n (k) of all n n invertible matrices over k. This is the principal open subset of M n (k) = A n2 defined by the non-vanishing of determinant det, so it is an irreducible affine variety of dimension n 2. To see that µ, i are morphisms, just note that they are polynomial functions in the n 2 coordinate functions and det 1. Any algebraic group is isomorphic to a closed subgroup of some GL n (k). An action of an algebraic group G on an affine variety X means a morphism ρ : G X X of affine varieties that is a group action in the usual sense. In that case a fundamental theorem shows: (1) the orbits of G on X are irreducible and locally closed; (2) if O is an orbit, its boundary O O is a union of orbits of strictly smaller dimension than O; (3) the stabilizer G x of a point x O is a closed subgroup of G, and dim O = dim G dim G x (note G x need not be connected but it has finitely many connected components, each of which is irreducible of the same dimension, so dim G x should be understood as meaning the dimension of any one of these components). Now we are ready to apply these facts The variety of representations. Throughout the section, Q is a quiver and A = kq. We also fix a dimension vector α N n. We will discuss the algebraic geometry arising from the representations of Q of dimension α. Note Hom k (k s, k t ) is just the space of all t s matrices. So we can identify it with the affine variety A ts. Thus it is an irreducible affine variety of dimension ts. Let Rep(α) = Hom k (k αs(ρ), k αt(ρ) ). ρ Q 1 This is an irreducible variety of dimension ρ Q 1 α s(ρ) α t(ρ). Given a point x Rep(α), we define a representation R(x) of Q of dimension α by declaring that R(x) i = k α(i) and that R(x) ρ is the linear map with matrix x ρ. Let GL(α) = n i=1 GL α i (k). This is an algebraic group of dimension n i=1 α2 i. We make GL(α) act on the variety Rep(α) by conjugation. Thus, if x = (x ρ ) ρ Q1 is an element of the product (so each x ρ is an α t(ρ) α s(ρ) - matrix) and g = (g i ) n i=1 (so each g i is an invertible α i α i -matrix) we define gx by (gx) ρ = g t(ρ) x ρ g 1 s(ρ). Lemma There is a 1 1 correspondence V O V between the set of isomorphism classes of representations of Q of dimension α and the set of GL(α)-orbits on Rep(α). The correspondence maps a representation V of dimension α to the set O V = {x Rep(α) R(x) = V }. Moreover, the space Aut A (R(x)) of all A-module automorphisms of R(x) is isomorphic to the stabilizer of x in GL(α).

17 17 Proof. If V is a representation of dimension α, picking a basis in each V i allows us to identify each V i with the vector space k α i and each V ρ with a map from k α s(ρ) to k α t(ρ). The action of GL(α) on Rep(α) corresponds to changing to a different choice of basis. So: representations of dimension α are in 1 1 correspondence with orbits. Now lets look at the geometry of the orbits and try to relate that to representation theory. Recall: orbits are locally closed and irreducible, and the boundary of an orbit is a union of orbits of strictly smaller dimension. The next lemma explains how the first basic geometric notion dimension relates to the representation theory. Lemma Suppose V is a representation of dimension α. Then, dim Rep(α) dim O V = dim End A (V ) q(α) = dim Ext 1 (V, V ). Proof. Suppose V R(x). Then, dim GL(α) dim O V = dim G x. Now, the previous lemma shows that G x = AutA (V ). It is a principal open subset of End A (V ), so has the same dimension. Hence, dim Rep(α) dim O V = ρ The last equality here is α s(ρ) α t(ρ) i α 2 i + dim End A (V ) = dim End A (V ) q(α) = dim Ext 1 (V, V ). Right away we can get somewhere with this: Corollary Suppose α 0 and q(α) 0. Then, for any representation V of dimension α, dim O V < dim Rep(α). Hence, there are infinitely many orbits in Rep(α), i.e. infinitely many non-isomorphic representations of dimension α. For instance, take the quiver in Exercise 9 above. Labelling the outside vertices 1, 2, 3, 4 and the inside vertex 0, we have that q(1, 1, 1, 1, 2) = 8 8 = 0. So we see right away that there are infinitely many non-isomorphic representations of this quiver of this dimension. In particular this cannot have finitely many indecomposables i.e. it is not of finite representation type. Corollary Let V be a representation of dimension α. Then, O V is open in Rep(α) if and only if Ext 1 A (V, V ) = 0, i.e. V has no self-extensions. Proof. A proper closed subvariety of an irreducible variety has strictly smaller dimension. Since orbits are open in their closure and Rep(α) is irreducible, O V is open if and only if O V = Rep(α) which is if and only if dim O V = dim Rep(α) which is if and only if dim Ext 1 A (V, V ) = 0. Corollary There is at most one representation of Q of dimension α (up to isomorphism) with no self-extensions.

18 18 Proof. Any two non-empty opens inside an irreducible variety intersect, but orbits are disjoint. Let us continue in this vein: relating geometry to representation theory. Lemma If 0 U V W 0 is a non-split short exact sequence, then O U W is contained in the boundary of O V. Proof. View each U i as subspaces of V i, and pick basis for V i extending basis of U i. Then each V ρ looks like a matrix [ uρ x ρ 0 w ρ ] for matrices u ρ, x ρ and w ρ of appropriate size. For λ k, let g λ GL(α) be the element with ith component [ ] λ 0 0 I Then, [ ] uρ λx (g λ V ) ρ = ρ. 0 w ρ These are contained in O V for each 0 λ. Evidently, since some x ρ is nonzero, the closure of the set of all such tuples matrices for all 0 λ contains the tuple in which λ = 0. Hence, O U W O V. Corollary If V is an orbit in Rep(α) of maximal dimension and V = U W, then Ext 1 (W, U) = 0. Proof. Suppose 0 U E W 0 is a non-split extension. Then by the lemma, O V is contained in the boundary of O E. But that means that dim O V < dim O E contradicting the maximality of dim O V. Corollary If O V is closed then V is completely reducible. Proof. Let U be a submodule, W = V/U. We need to show the extension is split. If not, then O U W lies in the boundary of O V. But O V is closed so its boundary is empty. Contradiction. Comments. Suppose that Q has no oriented cycles. Let Z Rep(α) be the unique element with all matrices being zero. This is obviously in its own orbit, hence closed. Iterating the lemma above shows that Z is contained in the closure of any other orbit, hence {Z} is the unique closed orbit. Since Z is semisimple, we see in this case that V is completely reducible if and only if O V is closed. Let us pause to think what is going on in some examples. Suppose Q is the quiver with one vertex and one arrow. The orbits in Rep(d) are just the conjugacy classes of d d matrices, parametrized by Jordan forms. The Lemma shows that J d (λ) contains J d1 (λ) J d2 (λ) (d 1 + d 2 = d) in its closure. In particular, iterating, it contains J 1 (λ) d in its closure. Thus

19 the only conceivably closed conjugacy classes are the ones corresponding to diagonalizable matrices: to see that these are indeed closed orbits, look at Exercise 1 below. Now we compute the dimension of some of the orbits. Suppose λ 1,..., λ s are distinct eigenvalues and d d s = d. Consider the orbit J d1 (λ 1 ) J ds (λ s ). The endomorphism ring is of dimension d (it is the direct sum k[x]/(x d 1 ) k[x]/(x ds )). Since q(d) = 0 we deduce by the Lemma that the dimension of the orbit is d 2 d. These are the orbits of maximal dimension everything else has a larger endomorphism ring. For instance if you take s = 2, then Corollary 2.25 implies that there are no extensions between Jordan blocks of different eigenvalues. Of course we know this already, but it illustrates the point. Exercise 1. Let M be the variety of all n n matrices and let GL n (k) act on M by conjugation. Prove that if m M is a semisimple (a.k.a. diagonalizable) matrix, then the conjugacy class of m is closed. Exercise 2. Suppose Q is the quiver with two vertices and one arrow from 1 to 2. The orbits in Rep(α 1, α 2 ) are the equivalence classes of α 2 α 1 matrices, thus parametrized by their rank 0 r min(α 1, α 2 ). Compute the dimensions of each of these orbits. Describe the partial order on the orbits defined by O r O s if O r O s. The closures of these orbits are called rank varieties. These and their analogs for other type A quivers are an important source of tractable examples in algebraic geometry. Exercise 3. Suppose Q is the quiver considered in Exercise 8 of the previous problem set (one vertex in the middle, 3 round the edge, arrows pointing inwards). You should already calculated the dimensions of the twelve different indecomposables. Let α = (1, 1, 1, 2) (where the 2 is on the inside vertex). Question: how many orbits does GL(α) have on Rep(α) in this case? Try to describe the partial order on the orbits given by containment of closures in this case. Note Corollary 2.21 indicates that if we are looking for quivers with finitely many indecomposables, we should study which quivers can have q(α) > 0 for every dimension vector α. This is the purpose of the next section Dynkin and Euclidean diagrams. Recall that q doesn t depend on orientation. If we forget the orientation on a quiver, we get a graph Γ with vertices 1,..., n and n i,j = n j,i edges between i and j. Given such a graph and α Z n, let q(α) = n i=1 α2 i i j n i,jα i α j. Also let (.,.) be the symmetric bilinear form defined by (ɛ i, ɛ j ) = { ni,j i j, 2 2n i,i i = j. Thus q and (.,.) are what they were before if Γ is the graph underlying a quiver. We say q is positive definite if q(α) > 0 for all 0 α Z n, and positive semi-definite if q(α) 0 for all α. The radical of q is {α Z n (α, β) = 19

20 20 0 for all β Z n }. Call α Z n strict if no α i is zero. Finally, let be the partial ordering on Z n defined by α β if α i β i for each i = 1,..., n. Say α is positive if it is 0, negative if it is 0. Let us record a technical lemma. Lemma Suppose Γ is connected and β > 0 is a vector in the radical of q. Then, β is strict and the form q is positive semi-definite. For α Z n, we have that q(α) = 0 if and only if α Qβ, which is if and only if α lies in the radical of q. Proof. Note 0 = (ɛ i, β) = (2 2n i,i )β i j i n i,j β j. If β i = 0 then j i n i,jβ j = 0 and since each term is 0 we get that β j = 0 whenever there is an edge i j. Since Γ is connected, it follows that β = 0, a contradiction. Thus, β is strict. Now, i<j n i,j β i β j 2 ( αi β i α j β j ) 2 = i<j = i j = i β j n i,j αi 2 n i,j ( α i α j ) + β i n i,j αj 2 2β i 2β i<j i<j j β j n i,j αi 2 + n i,j α i α j 2β i i<j 1 (2 2n i,i )β i αi 2 + n i,j α i α j = q(α). 2β i i<j Hence q is positive semi-definite. If q(α) = 0 then α i /β i = α j /β j whenever there is an edge i j. Since Γ is connected it follows that α Qβ. But that implies that α is in the radical of q, since β is. Finally, if α is in the radical of q then q(α) = 0. This completes the proof. Now we can prove the main classification theorem. Theorem Suppose that Γ is connected. (1) If Γ is a Dynkin diagram (which I ll draw on the board) then q is positive definite. (2) If Γ is a Euclidean diagram, then q is positive semi-definite and the radical of q is Zδ, where δ is the vector indicated by the numbers on the graph. Note that δ is strict and > 0. (3) Otherwise, there is a vector α > 0 with q(α) < 0 and (α, ɛ i ) 0 for all i. Proof. Suppose that Γ is a Euclidean diagram. First check that the vector δ = δ i ɛ i belongs to the radical. This amounts to checking that (δ, ɛ i ) = j i δ jn i,j + 2δ i 2δ i n i,i = 0 for each i. For example for Ẽ8, one has to observe that the sum of the δ j s for each neighbor j of i is equal to 2δ i. Now the lemma implies that q is positive semi-definite and that the radical of q is Qδ Z n. Since one of the δ i s is one, that is Zδ. This completes the proof of (2).

21 Now suppose that Γ is a Dynkin diagram. Add one more vertex to get a Euclidean diagram Γ. Note that q Γ is the restriction to Z n of the quadratic form q Γ. By the lemma, if q(α) = 0 for a non-zero α Z m then α Zδ, hence α is strict. So q is strictly positive on all the α s in Z n coming from Γ, so q Γ is positive definite. This proves (1). Finally suppose that Γ is neither Euclidean nor Dynkin. Then Γ has a subgraph Γ that is Euclidean, say with radical vector δ. If all the vertices of Γ are in Γ, take α = δ. Else, let i be a vertex not in Γ but connected to Γ by an edge, and take α = 2δ + ɛ i. Now check that q(α) < 0 and (α, ɛ i ) 0 for all i. From now on we will focus on Γ either Dynkin or Euclidean (it is possible to extend most of these definitions to the general case, but that will not be worth it for us!). Let s associate a few more invariants. First the set of roots is = {0 α Z n q(α) 1}. A root is called real if q(α) = 1 and imaginary if q(α) = 0. Note each ɛ i is a root. These are known as the simple roots. Obviously in the Dynkin case, there are no imaginary roots, while in the Euclidean case the imaginary roots are the integer multiples of δ, by the lemma. Here are some further properties of the set of roots: Lemma (i) If α and β belongs to the radical of q, then α and α + β are roots. (ii) Every root is > 0 or < 0. (iii) If Γ is Euclidean, then ( {0})/Zδ is a finite subset of Z m /Zδ (quotient abelian group). (iv) If Γ is Dynkin, then is finite. Proof. (i) q(β ± α) = q(β) + q(α) ± (β, α) = q(α) 1. (ii) Write α = α + α where α +, α 0 and they have disjoint support. Then obviously, (α +, α ) 0 so 1 q(α) = q(α + ) + q(α ) (α +, α ) q(α + ) + q(α ). Since q is positive semi-definite, one of q(α + ) or q(α ) must therefore be zero, i.e. one of them is an imaginary root. But imaginary roots are strict, so the other one must be zero. (iii) Pick i with δ i = 1. If α is a root with α i = 0 then δ α and δ + α are roots by (i). Since their ith component is 1, they are positive roots, so δ α δ. Therefore {α α i = 0} {α Z n δ α δ so it is finite. Now take β. Then, β β i δ belongs to this finite set. (iv) Embed Γ into Euclidean diagram Γ by adding a vertex i with δ i = 1. Then roots α of Γ are roots of Γ with α i = 0. So done by (iii). 21

22 22 Example Let Γ be the graph The roots are (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1) plus the negatives of these. Let Γ be the graph with one vertex and one edge. The roots are Zδ where δ = (1). Let Γ be the Kronecker graph with two vertices and two edges. The positive roots are (a + 1, a), (a, a + 1) for a 0 and (a, a) for a > Gabriel s theorem. We now want to classify the quivers with finite representation type, i.e. with finitely many finitely generated indecomposables up to isomorphism. By the Krull Schmidt theorem that happens if and only if there are finitely many orbits in Rep(α) for each dimension vector α. We know right away by Corollary 2.21 that if Q is such a quiver, then q is positive definite, hence the underlying graph of Q is a Dynkin diagram. So our main task will be to analyse the Dynkin quivers and prove that they do all indeed have finitely many indecomposables. The main result: Theorem 2.31 (Gabriel, circa 1970). A quiver Q has finite representation type if and only if Q is Dynkin. In that case, the map V dimv defines a 1 1 correspondence between the isomorphism classes of indecomposable representations and the positive roots in (which is finite). For example the Dynkin diagram D 4 has exactly 12 positive roots, so the 12 indecomposable representations found in Exercise are all the indecomposables. To prove the theorem, we need one more technical lemma. Lemma Suppose that V is indecomposable and dim End(V ) > 1. There is a proper indecomposable submodule U V with dim Ext 1 (U, U) > 0. Proof. Note that since V is indecomposable, Fitting s lemma shows that E = End(V ) is a local ring, i.e. E/J(E) is a simple E-module. Since k is algebraically closed, Schur s lemma shows that E/J(E) = End E (E/J(E)) is one dimensional. Hence our assumption that dim E > 1 means exactly that J(E) 0. Pick θ J(E) (i.e. a nilpotent endomorphism of V ) such that im θ is of minimal dimension this guarantees in particular that θ 2 = 0. Let I = im θ and ker θ = K 1 K r with K i s indecomposable. Note I K 1 K r. Choose j so that the composite α of the inclusion I K 1 K r and the projection K 1 K r K j is non-zero. We note that α : I K j is injective: the composite X θ I α K j X has image im α hence dimension dim I by the minimality assumption. Now we claim that Ext 1 (I, K j ) 0. Once we have established that, the lemma follows for applying Hom(, K j ) to 0 I K j Q 0 and using