Lectures on Representations of Quivers by William Crawley-Boevey ...

A ’quiver’ is a directed graph, and a representation is defined by a vectorspace for each vertex and a linear map for each arrow. The theory ofrepresentations of quivers touches linear algebra, invariant theory, finitedimensional algebras, free ideal rings, Kac-Moody Lie algebras, and manyother fields.These are the notes for a course of eight lectures given in Oxford inspring 1992. My aim was the classification of the representations for the~ ~ ~ ~ ~Euclidean diagrams A , D , E , E , E . It seemed ambitious for eightn n 6 7 8lectures, but turned out to be easier than I expected.The Dynkin case is analysed using an argument of J.Tits, P.Gabriel andC.M.Ringel, which involves actions of algebraic groups, a study of rootsystems, and some clever homological algebra. The Euclidean case is treatedusing the same tools, and in addition the Auslander-Reiten translations-t,t , and the notion of a ’regular uniserial module’. I have avoided theuse of reflection functors, Auslander-Reiten sequences, and case-by-caseanalyses.The prerequisites for this course are quite modest, consisting of the basic1notions about rings and modules; a little homological algebra, up to Extnand long exact sequences; the Zariski topology on A ; and maybe some ideasfrom category theory.In the last section I have listed some topics which are the object ofcurrent research. I hope these lectures are a useful preparation forreading the papers listed there.William Crawley-Boevey,Mathematical Institute, Oxford University24-29 St. Giles, Oxford OX1 3LB, England2

§1. Path algebrasOnce and for all, we fix an algebraically closed field k.DEFINITIONS.(1) A quiver Q = (Q ,Q ,s,t:Q ----------LQ ) is given by0 1 1 0a set Q of vertices, which for us will be {1,2,...,n}, and0a set Q of arrows, which for us will be finite.1An arrow r starts at the vertex s(r) and terminates at t(r). We sometimesrindicate this as s(r) --------------------L t(r).(2) A non-trivial path in Q is a sequence r ...r (m>1) of arrows which1 msatisfies t(r )=s(r ) for 1

EXAMPLES.(1) If Q consists of one vertex and one loop, then kQ = k[T]. If Q has onevertex and r loops, then kQ is the free associative algebra on r letters.(2) If there is at most one path between any two points, then kQ can beidentified with the subalgebra{C e M (k) | C =0 if no path from j to i}n ijof M (k). If Q is 1----------L2----------L...----------Ln this is the lower triangular matrices.nIDEMPOTENTS. Set A=kQ.2(1) The e are orthogonal idempotents, ie e e = 0 (i$j), e = e .i i j i in(2) A has an identity given by 1 = S e .i=1 i(3) The spaces Ae , e A, and e Ae have as bases the paths starting at ii j j iand/or terminating at j.n(4) A = s Ae , so each Ae is a projective left A-module.i=1 i i(5) If X is a left A-module, then Hom (Ae ,X) = e X.A i i(6) If 0$feAe and 0$gee A then fg$0.iiPROOF. Look at the longest paths x,y involved in f,g. In the product fg thecoefficient of xy cannot be zero.(7) The e are primitive idempotents, ie Ae is a indecomposable module.ii2PROOF. If End (Ae )=e Ae contains idempotent f, then f =f=fe , soA i i i if(e -f)=0. Now use (6).i(8) If e eAe A then i=j.i jPROOF. Ae A has as basis the paths passing through the vertex j.j(9) The e are inequivalent, ie Ae @Ae for i$j.i i jPROOF. Thanks to (5), inverse isomorphisms give elements fee Ae , gee Aei j j iwith fg=e and gf=e . This contradicts (8).ij4

PROPERTIES OF PATH ALGEBRAS.These are exercises, but some are rather testing.(1) A is finite dimensional 5 Q has no oriented cycles.(2) A is prime (ie IJ$0 for two-sided ideals I,J$0) 5 Ai,j E path i to j.(3) A is left (right) noetherian 5 if there is an oriented cycle through i,then only one arrow starts (terminates) at i.(4) rad A has basis {paths i to j | there is no path from j to i}.(5) The centre of A is k*k*...*k[T]*k[T]*..., with one factor for eachconnected component C of Q, and that factor is k[T] 5 C is an orientedcycle.REPRESENTATIONS.We define a category Rep(Q) of representations of Q as follows.A representation X of Q is given by a vector space X for each ieQ and ai 0linear map X :X ----------LX for each reQ .r s(r) t(r) 1A morphism q:X----------LX’ is given by linear maps q :X ----------LX’ for each ieQi i i 0satisfying X’q = q X for each reQ .r s(r) t(r) r 1The composition of q with f:X’----------LX" is given by (fqq) = f qq .i i iEXAMPLE. Let S(i) be the representation with( k (j=i)S(i) = { S(i) =0 (all reQ ).j r 19 0 (else)EXERCISE. It is very easy to compute with representations. For example letQ be the quiver •J----------•----------L•, and let X and Y be the representations1 1 1kJ----------k----------Lk kJ----------k----------L0.Show that Hom(X,Y) is one-dimensional, and that Hom(Y,X)=0.5

LEMMA. The category Rep(Q) is equivalent to kQ-Mod.PROOF. We only give the construction. IfX is a kQ-module, define arepresentation X withX = eXi iX (x) = rx = e rx e X for xeX .r t(r) t(r) s(r)If X is a representation, define a moduleX viae pn i iX=s X . Let X ----------LX----------LX be the canonical maps.i=1 i i ir ...r x = e X ...X p (x)1 m t(r ) r r s(r )1 1 m me x = e p (x),ii iIt is straightforward, but tedious, to check that these are inverses andthat morphisms behave, etc. We can now use the same letter for a module andthe corresponding representation, ignoring the distinction.EXAMPLE. Under this correspondence, the representations S(i) are simplemodules. Moreover, if Q has no oriented cycles, it is easy to see that theS(i) are the only simple modules.DEFINITIONS.(1) The dimension vector of a finite dimensional kQ-module X is the vectorndim X e N , with(dim X) = dim X = dim e X = dim Hom(Ae ,X).i i i inThus dim X = S (dim X) .i=1 inn(2) The Euler form is = S a b - S a b for a,b e Z .i=1 i i reQ1 s(r) t(r)nThis is a bilinear form on Z .n(3) The Tits form is q(a) = . This is a quadratic form on Z .(4) The Symmetric bilinear form is (a,b) = + .6

THE STANDARD RESOLUTION.Let A=kQ. If X is a left A-module, there is an exact sequencenfg0 ----------L s Ae t e X ----------L s Ae t e X ----------L X ----------L 0t(r) k s(r)i k ireQ1 i=1where g(atx) = ax for aeAe , xee X, andi if(atx) = artx - atrx for aeAe and xee Xt(r) s(r)in s(r) t(r) component.PROOF. Clearly gqf=0 and g is onto. If x is an element of the middle termof the sequence, we can write it uniquely in the formnx = S S atx (x ee X almost all zero)a a s(a)i=1 paths a with s(a)=iand define degree(x) = length of the longest path a with x $0.aIf a is a non-trivial path with s(a)=i, then we can express it as a producta=a’r with r an arrow with s(r)=i, and a’ another path. Viewing a’tx as anaelement in the r’th component of left hand term, we havef(a’tx ) = atx - a’trx .a a aWe claim that x + Im(f) always contains an element of degree 0. Namely, ifx has degree d>0, thennx - f( S S a’tx ) ai=1 paths a with s(a)=i and length dhas degree < d, so the claim follows by induction.Im(f)=Ker(g): If xeKer(g), let x’ex+Im(f) have degree zero. Thus0 = g(x) = g(x’) = g(S e tx’ ) = S x’i i e ei inNow this belongs to s X , so each term in the final sum must be zero.i=1 iThus x’=0, and the assertion follows.Ker(f)=0: we can write an element xeKer(f) in the formx = S S atx (x ee X almost all 0).reQ paths a with s(a)=t(r) r,a r,a s(r)1Let a be a path of maximal length such that x $0 (some r). Nowr,af(x) = S S artx - S S atrxr a r,a r a r,aso the coefficient of ar in f(x) is x . A contradiction.r,a7

CONSEQUENCES.i(1) If X is a left A-module, then proj.dim X < 1, ie Ext (X,Y)=0 AY,i>2.PROOF. f and g are A-module maps and Ae tV is isomorphic to the direct sumiof dim V copies of Ae , so is a projective left A-module. Thus the standardiresolution is a projective resolution for X.(2) A is hereditary, ie if XcP with P projective, then X is projective.1 2PROOF. Ext (X,Y) = Ext (P/X,Y) = 0 AY.1(3) If X,Y are f.d., then dim Hom(X,Y) - dim Ext (X,Y) = .PROOF. Apply Hom(-,Y) to the standard resolution:10-----LHom(X,Y)-----LHom(s Ae t e X,Y)-----LHom(s Ae t e X,Y)-----LExt (X,Y)-----L0.i i k i r t(r) k s(r)Now dim Hom(Ae te X,Y) = (dim e X)(dim Hom(Ae ,Y)) = (dim X) (dim Y) .i j j i j i1(4) If X is f.d., then dim End(X) - dim Ext (X,X) = q(dim X).PROOF. Put X=Y in (3).REMARK.Let i be a vertex in Q and suppose that either no arrows start at i, or noarrows terminate at i. Let Q’ be the quiver obtained by reversing thedirection of every arrow connected to i. We say that Q’ is obtained from Qbe reflecting at the vertex i. The two categories Rep(Q) and Rep(Q’) areclosely related, by means of so-called reflection functors. SeeI.N.Bernstein, I.M.Gelfand and V.A.Ponomarev, Coxeter functors andGabriel’s Theorem, Uspekhi Mat. Nauk. 28 (1973), 19-33, English TranslationRuss. Math. Surveys, 28 (1973), 17-32.8

§2. BricksIn this section we consider finite dimensional left A-modules with A anhereditary k-algebra. In particular the results hold when A is a pathalgebra. We recall the Happel-Ringel Lemma and another lemma due to Ringel.INDECOMPOSABLE MODULES.Recall Fitting’s Lemma, that X is indecomposable 5 End(X) is a local ring,ie End(X) = k1 +rad End(X), since the field k is algebraically closed.XAny module can be written as a direct sum of indecomposable modules, and bythe Krull-Schmidt Theorem the isomorphism types of the summands and theirmultiplicities are uniquely determined.We say that X is a brick if End(X)=k. Thus a brick is indecomposable.1LEMMA 1. Suppose X,Y are indecomposable. If Ext (Y,X)=0 then any non-zeromap q:X----------LY is mono or epi.PROOF. We have exact sequencesx:0----------LIm(q)----------LY----------LCok(q)----------L0 and h:0----------LKer(q)----------LX----------LIm(q)----------L0.1From Ext (Cok(q),h) we get1 f 1... ----------L Ext (Cok(q),X) ----------L Ext (Cok(q),Im(q)) ----------L 0.so x = f(z) for some z. Thus there is commutative diagramaz:0 ----------L X ----------L Z ----------L Cok(q) ----------L 0b1

If Im(q)$0 then X or Y is summand of Im(q) by Krull-Schmidt. But if q isnot mono or epi, then dim Im(q) < dim X, dim Y, a contradiction.SPECIAL CASE. If X is indecomposable with no self-extensions (ie1Ext (X,X)=0), then X is a brick.LEMMA 2. If X is indecomposable, not a brick, then X has a submodule and aquotient which are bricks with self-extensions.PROOF. It suffices to prove that if X is indecomposable and not a brickthen there is a proper submodule UCX which is indecomposable and withself-extensions, for if U is not a brick one can iterate, and a dualargument deals with the case of a quotient.Pick qeEnd(X) with I=Im(q) of minimal dimension $ 0. We have IcKer(q), for2X is indecomposable and not a brick so q is nilpotent. Now q =0 byrminimality. Let Ker(q) = s K with K indecomposable, and pick j suchi=1 i ithat the composition a : I(-----LKer(q)----------)K is non-zero. Now a is mono, forjathe map X----------)I----------LK (-----LX has image Im(a)$0 so a mono by minimality.j1We have Ext (I,K )$0, for otherwise the pushoutjr0 ----------L s K ----------L X ----------L I ----------L 0i=1 i11 <

§3. The variety of representationsIn this section Q is a quiver and A=kQ. We define the variety ofnrepresentations of Q of dimension vector aeN , and describe some elementaryproperties. We use elementary dimension arguments from algebraic geometry.The properties we need are listed below.ALGEBRAIC GEOMETRY.rA is affine r-space with the Zariski topology. We consider locally closedr -----subsets U in A , ie subsets U which are open in their closure U.A non-empty locally closed subset U is irreducible if any non-empty subsetrof U which is open in U, is dense in U. The space A is irreducible.The dimension of a non-empty locally closed subset U issup{n | E Z CZ C...CZ irreducible subsets closed in U}.0 1 n----- rWe have dim U = dim U; if W=UuV then dim W = max{dim U,dim V}; the space Ahas dimension r.rIf an algebraic group G acts on A , then the orbitsO are locally closed;------O\O is a union of orbits of dimension strictly smaller than dimO; and ifxeO then dimO=dim G-dim Stab (x).GnDEFINITIONS. Let Q be a quiver and aeN . We defines(r) t(r)Rep(a) = p Hom (k ,k ).reQ k1rThis is isomorphic to A where r = S a a .reQ1 t(r) s(r)aiAn element xeRep(a) gives a representation R(x) of Q with R(x) = k fori1

THE ACTION.GL(a) acts on Rep(a) by conjugation. Explicitly-1(gx) = g x gr t(r) r s(r)for geGL(a) and xeRep(a).If x,yeRep(a), then the set of A-module isomorphisms R(x)-----LR(y) can beidentified with {geGL(a)|gx=y}. It follows that(1) Stab (x) = Aut (R(x)).GL(a)A(2) There is a 1-1 correspondence between isoclasses of representations Xwith dimension vector a and orbits, given byO = {xeRep(a) | R(x)=X}. ToXsee this we only need to realize that every representation of dimensionvector a is isomorphic to some R(x), which follows on choosing a basis.REMARKS.(1) Invariant Theory is about polynomial and rational maps f:Rep(a)----------Lkwhich are constant on GL(a)-orbits. For example, if a = r ...r is an1 moriented cycle, we have a polynomial invariantf (x) = Trace(x x ...x ),a r r r1 2 mand more generally if c (T) is the characteristic polynomial of q, we haveqif (x) = Coefficient of T in c (T).ai x x ...xr r r1 2 m(2) If char k=0, then any polynomial invariant can be expressed as apolynomial in the f . This has been proved by Sibirski and Procesi in caseaQ has only one vertex, and in general can be found in L.Le Bruyn &C.Procesi, Semisimple representations of quivers, Trans. Amer. Math. Soc.317 (1990), 585-598.(3) If char k>0 and Q has only one vertex, any polynomial invariant can beexpressed as a polynomial in the f . This is recent work of S.Donkin.aiPresumably the restriction on Q is unnecessary.12

1LEMMA 1. dim Rep(a) - dimO = dim End (X) - q(a) = dim Ext (X,X).XAPROOF. Say X=R(x). We havedimO = dim GL(a) - dim Stab(x) = dim GL(a) - dim Aut (X)XAsNow GL(a) is non-empty and open in A , so dense, so dim GL(a) = s.Similarly Aut (X) is non-empty and open in End (X), so dense, soAAdim Aut(x) = dim End(X). The assertion follows.CONSEQUENCES.(1) If a$0 and q(a)

&u 0r*| |0 v7 r8which corresponds to UsV.Finally Hom(x,U) gives an exact sequencesof 10----------LHom(V,U)----------LHom(X,U)----------LHom(U,U)----------LExt (V,U),dim Hom(V,U) - dim Hom(X,U) + dim Hom(U,U) - dim Im(f) = 0,but f(1 )=x$0, so dim Hom(X,U) $ dim Hom(UsV,U), and hence X @ UsV.UCONSEQUENCES.(1) IfO is an orbit in Rep(a) of maximal dimension, and X=UsV, thenX1Ext (V,U)=0.A------PROOF. If there is non-split extension 0----------LU----------LE----------LV----------L0 thenO cO \O , soX E EdimO < dimO .X E(2) IfO is closed then X is semisimple.XREMARKS.(1) Suppose Q has no oriented cycles. Let zeRep(a) be the element with allmatrices z =0. We can easily show that z is in the closure of every orbit,rand it follows that there are no non-constant polynomial invariants.Moreover, an orbitO is closed 5 X is semisimple, for the only semisimpleXmodule of dimension a is R(z), and {z} is clearly a closed orbit.(2) If Q is allowed to have oriented cycles, x,x’eRep(a) and R(x) and R(x’)are non-isomorphic semisimple modules, then there is a polynomial invariantf (of the form f ) with f(x)$f(y). In case Q has only one vertex andaichar k=0 this is proved in §12.6 of M.Artin, On Azumaya algebras and finitedimensional representations of rings, J.Algebra 11 (1969), 532-563, but itseems to be true in general. It follows thatO is closed 5 X isXsemisimple.14

§4. Dynkin and Euclidean diagramsIn this section we give the classification of graphs into Dynkin,Euclidean, and ’wild’ graphs, and in the first two cases we study thecorresponding root system.DEFINITIONS.Let G be finite graph with vertices {1,..,n}. We allows loops and multipleedges, so that G is given by any set of natural numbersn = n = the number of edges between i and j.ij jin 2Let q(a) = S a - S n a ai=1 i i0 for all 0$aeZ .nWe say q is positive semi-definite if q(a)>0 for all aeZ .nThe radical of q is rad(q) = {aeZ | (a,-) = 0}.nnWe have a partial ordering on Z given by a0 is a non-zero radical vector, then b isnsincere and q is positive semi-definite. For aeZ we haveq(a)=0 5 aeQb 5 aerad(q).15

PROOF. By assumption 0 = (e ,b) = (2-2n )b - S n b .i ii i j$i ij jIf b =0 then S n b = 0, and since each term in >0 we have b =0i j$i ij j jwhenever there is an edge i----------j. Since G is connected it follows that b=0, acontradiction. Thus b is sincere. Nowb b a a 2i j & iSj*n -------------------- ---------- - ----------i

PROOF.(2) By inspection the given vector d is radical, eg if there are no loopsor multiple edges, we need to check that2d = S d .i neighbours j of i jNow q is positive semi-definite by the lemma. Finally, since some d =1,inrad(q) = Qd n Z = Zd.~(1) Embed the Dynkin diagram in the corresponding Euclidean diagram G, and~note that the quadratic form for G is strictly positive on non-zero,non-sincere vectors.(3) It is not hard to show that G has a Euclidean subgraph G’, say withradical vector d. If all vertices of G are in G’ take a=d. If i is a vertexnot in G’, connected to G’ by an edge, take a=2d+e iEXTENDING VERTICES.If G is Euclidean, a vertex e is called an extending vertex if d =1. Notee(1) There always is an extending vertex.(2) The graph obtained by deleting e is the corresponding Dynkin diagram.NOW SUPPOSE that G is Dynkin or Euclidean, so q is positive semi-definite.ROOTS.nWe define D = {aeZ | a$0, q(a)

PROPERTIES.(1) Each e is a root.i(2) If aeDu{0}, so are -a and a+b with berad(q).PROOF. q(b+a) = q(b)+q(a)+(b,a) = q(a).( o (Dynkin)(3) {imaginary roots} = {9 {rd | 0$reZ} (Euclidean)PROOF. Use the lemma.(4) Every root a is positive or negative.+ - + -PROOF. Let a = a -a where a ,a >0 are non-zero and have disjoint support.+ -Clearly we have (a ,a ) q(a) = q(a ) + q(a ) - (a ,a ) > q(a ) + q(a ).+ -Thus one of a ,ais an imaginary root, and hence is sincere. This meansthat the other is zero, a contradiction.(5) If G is Euclidean then (Du{0})/Zd is finite.PROOF. Let e be an extending vertex. If a is a root with a =0, then d-a anded+a are roots which are positive at the vertex e, and hence are positiveroots. Thusn{aeDu{0}|a =0} c {aeZ |-d

§5. Finite representation typeIn this section we combine almost everything that we have done so far inorder to prove Gabriel’s Theorem. The proof given here is due to J.Tits,P.Gabriel, and the key step to C.M.Ringel, Four papers on problems inlinear algebra, in I.M.Gelfand, ’Representation Theory’, London Math. Soc.Lec. Note Series 69 (1982).THEOREM 1. Suppose Q is a quiver with underlying graph G Dynkin. Theassignment X9-----Ldim X induces a bijection between the isoclasses ofindecomposable modules and the positive roots of q.PROOF.If X is indecomposable, then X is a brick, for otherwise by §2 Lemma 2there is YcX a brick with self-extensions, and then10 < q(dim Y) = dim End(Y) - dim Ext (Y,Y) < 0.If X is indecomposable then it has no self-extensions and dim X is a1positive root, for 0 < q(dim X) = 1 - dim Ext (X,X).If X,X’ are two indecomposables with the same dimension vector, then X=X’by §3 Lemma 1.If a is a positive root, then there is an indecomposable X with dim X = a.To see this, pick an orbitO of maximal dimension in Rep(a). If XX1 1decomposes, X=UsV then Ext (U,V)=Ext (V,U)=0 by §3 Lemma 2. Thus1 = q(a) = q(dim U) + q(dim V) + + = q(dim U) + q(dim V) + dim Hom(U,V) + dim Hom(V,U) > 2,a contradiction.THEOREM 2. If Q is a connected quiver with graph G, then there are onlyfinitely many indecomposable representations 5 G is Dynkin.19

PROOF. If G is Dynkin then the indecomposables correspond to the positiveroots, and there are only a finite number of roots.Conversely, suppose there are only a finite number of indecomposables. Anymodule is a direct sum of indecomposables, so it follows that there arenonly finitely many isoclasses of modules of dimension a for all aeN . Thusthere are only finitely many orbits in Rep(a). By §3 Lemma 1 we have q(a)>0nfor 0$aeN . Now the classification of graphs shows that G is Dynkin.20

§6. More homological algebraFROM NOW ON we suppose that Q is a quiver without oriented cycles, so thepath algebra A=kQ is finite dimensional. We still consider f.d. A-modules.We study the properties of projective, injective, non-projective, andnon-injective modules. We give a little bit of Auslander-Reiten theory.DUALITIES.(1) If X is a left or right A-module, then DX = Hom (X,k), Hom(X,A) andk1Ext (X,A) are all A-modules on the other side.(2) D is duality between left and right A-modules.PROOF. Hom(X,Y)=Hom(DY,DX) and DDX=X.(3) D gives a duality between injective left modules and projective rightmodules.1 1PROOF. Ext (DX,DY) = Ext (Y,X). This is zero for all Y if and only if DX isprojective, if and only if X is injective.(4) Hom(-,A) gives a duality between projective left modules and projectiveright modules.n n nPROOF. If P is a summand of A then Hom(P,A) is a summand of Hom(A ,A)=A ,so is projective. Now the map P----------LHom(Hom(P,A),A) is an iso for all P,since it is for P=A.(5) The Nakayama functor n(-) = DHom(-,A) gives an equivalence fromprojective left modules to injective left modules. The inverse functor is-n (-) = Hom(D(-),A) = Hom(DA,-).(6) Hom(X,nP) = DHom(P,X) for X,P left A-modules, P projective.PROOF. The compositionHom(P,A)t X = Hom(P,A)t Hom(A,X) ----------L Hom(P,X)AAis an isomorphism, since it is for P=A. ThusDHom(P,X) = Hom (Hom(P,A)t X,k) = Hom(X,Hom (Hom(P,A),k)) = Hom(X,nP).k A k21

DEFINITION. The Auslander-Reiten translate of a left A-module X is1 - 1 1tX = DExt (X,A). We also define t X = Ext (DX,A) = Ext (DA,X).If 0----------LL----------LM----------LN----------L0 is an exact sequence then since A is hereditary thereare long exact sequences0----------LtL ----------LtM ----------LtN ----------LnL ----------LnM ----------LnN ----------L0- - - - - -0----------Ln L----------Ln M----------Ln N----------Lt L----------Lt M----------Lt N----------L0.1 -LEMMA 1. Hom(Y,tX) = DExt (X,Y) = Hom(t Y,X).-(Thus t is left adjoint to t)PROOF. Let 0----------LP----------LQ----------LX----------L0 be a projective resolution. The sequencetQ----------LtX----------LnP----------LnQis exact, and tQ=0, so we have a commutative diagram with exact rows0----------LHom(Y,tX) ----------LHom(Y,nP)----------LHom(Y,nQ)@@10----------LDExt (X,Y)----------LDHom(P,Y)----------LDHom(Q,Y)1and hence Hom(Y,tX) = DExt (X,Y). The other isomorphism is dual.LEMMA 2. Let X be indecomposable.-(1) If X is non-projective then Hom(X,P)=0 for P projective, and t tX=X.-(2) If X is non-injective then Hom(I,X)=0 for I injective, and tt X=X.PROOF OF (1). If q:X----------LP is non-zero, then Im(q) is projective since A ishereditary. Now X----------)Im(q) is epi, so Im(q) is summand of X. But X isindecomposable so X=Im(q), a contradiction.Let 0----------LP----------LQ----------LX----------L0 be a projective resolution. Now0----------LtX----------LnP----------LnQ----------LnXis exact, and nX=0 since Hom(X,A)=0. Thus we have a commutative diagram- - - -n nP----------Ln nQ----------Lt tX----------Lt nP@ @P ----------L Q ----------L X ----------L 0- -with exact rows. Since nP is injective, t nP=0, and hence t tX = X.22

-LEMMA 3. t and t give inverse bijections--------------------Ltnon-projective indecomposablesJ---------------------tnon-injective indecomposablesPROOF. Let X be a non-projective indecomposable, and write tX as a directrsum of indecomposables, say tX = s Y . Each Y is non-injective, sincei=1 i iotherwise Hom(Y ,tX)=0 by Lemma 1. By part (2) of Lemma 2 it follows thati- - r -each t (Y )$0. By part (1) of Lemma 2 we have X = t tX = s t (Y ), andi i=1 isince X is indecomposable we must have r=1. Thus tX is a non-injective-indecomposable. Dually for t .REMARKS.-(1) For any f.d. algebra there are more complicated constructions t,tgiving the bijection above, which involve D and a transpose operator Tr. In-general, however, t and t are not functors, Lemma 1 needs to be modified,and Lemma 2 is nonsense.1(2) If X is indecomposable and non-projective, then Ext (X,tX) = DEnd(X),and this space contains a special element, the map feHom (End(X),k) withkf(1 )=1 and f(rad End(X))=1. The corresponding short exact sequenceX0----------LtX----------LE----------LX----------L0 is an Auslander-Reiten sequence, which has very specialproperties.Auslander-Reiten sequences exist for any f.d. algebra, and (under the name’almost split sequences’ and together with the transpose) have been definedand studied by M.Auslander & I.Reiten, Representation theory of artinalgebras III,IV,V,VI, Comm. in Algebra, 3(1975) 239-294, 5(1977) 443-518,5(1977) 519-554, 6(1978) 257-300.(3) The translate t can also be defined as a product of reflectionfunctors, see the remark in §1 and the paper by Bernstein, Gelfand andPonomarev. The equivalence of the two definition was proved by S.Benner andM.C.R.Butler, The equivalence of certain functors occuring in therepresentation theory of artin algebras and species, J. London Math. Soc.,14 (1976), 183-187.23

INDECOMPOSABLE PROJECTIVES AND INJECTIVES.(1) The modules P(i) = Ae are a complete set of non-isomorphiciindecomposable projective left A-modules.nPROOF. The e are inequivalent primitive idempotents and A = s Ae . Nowi i=1 iuse Krull-Schmidt.(2) The modules I(i) = n(P(i)) = D(e A) are a complete set ofinon-isomorphic indecomposable injective left A-modules.PROOF. Use Hom(-,A) and D.(3) = a = for any a.iPROOF. If X has dimension a, then1 = dim Hom(P(i),X) - dim Ext (P(i),X) = dim e X = a . = dim Hom(X,I(i)) = dim Hom(P(i),X) = a . inn(4) The vectors dim P(i) are a basis of Z . The dim I(i) are a basis of Z .PROOF. The module S(i) with dimension vector e has a projective resolutioni0----------LP ----------LP ----------LS(i)----------L0 and an injective resolution 0----------LS(i)----------LI ----------LI ----------L0.1 0 0 1iiCOXETER TRANSFORMATION.n n(1) There is an automorphism c:Z ----------LZ with dim nP = - c(dim P) for Pprojective.PROOF. Define c via c(dim P(i)) = - dim I(i).(2) If X is indecomposable and non-projective then dim tX = c(dim X).PROOF. Let 0----------LP----------LQ----------LX----------L0 be a projective resolution. We have an exactsequence 0----------LtX----------LnP----------LnQ----------L0 and sodim tX = dim nP - dim nQ = -c(dim P - dim Q) = -c(dim X).(3) = - = .PROOF. = = -.(4) ca=a 5 aerad(q).PROOF. =-=(b,a).REMARK. When t is written as a product of reflections, one sees that theCoxeter transformation is a Coxeter element in the sense of Coxeter groups.24

§7. Euclidean case. Preprojectives and preinjectivesFROM NOW ON we set A=kQ where Q is a quiver without oriented cycles andwith underlying graph G Euclidean. We denote by d the minimal positiveimaginary root for G. In this section we describe the three classes ofpreprojective, regular and preinjective modules.DEFINITIONS. If X is indecomposable, theni -m(1) X is preprojective 5 t X=0 for i>>0 5 X=t P(j) some m>0, j.-i m(2) X is preinjective 5 t X=0 for i>>0 5 X=t I(j) some m>0, j.i(3) X is regular 5 t X$0 for all ieZ.We say a decomposable module X is preprojective, preinjective or regular ifeach indecomposable summand is.The defect of a module X is = -.NLEMMA 1. There is N>0 such that c dim X = dim X for regular X.PROOF. Recall that ca=a if and only if a is radical, and that q(ca)=q(a).Thus c induces a permutation of the finite set Du{0}/Zd. Thus there is someNNN>0 with c the identity on Du{0}/Zd. Since e eD it follows that c is theinidentity on Z /Zd.NiNLet c dim X - dim X = rd. An induction shows that c dim X = dim X + irdfor all ieZ. If r>0, so X must bepreprojective. If r>0 this is not positive for i

LEMMA 2. If X is indecomposable, then X is preprojective, regular orpreinjective according as the defect of X is -ve, zero or +ve.PROOF. If X is preprojective then defect < 0, since-m -m m====d >0.iSimilarly preinjectives have defect > 0. If X is regular with dimensionN N-1vector a, then c a=a. Let b=a+..+c a. Clearly cb=b, so that b=rd. NowN-1 i0 = = S = N,i=0so =0, ie X has defect zero.LEMMA 3. Let X,Y be indecomposable.1(1) If Y is preprojective and X is not, then Hom(X,Y)=0 and Ext (Y,X)=0.1(2) If Y is preinjective and X is not, then Hom(Y,X)=0 and Ext (X,Y)=0.-i iPROOF. (1) As X is not preprojective, X=t t X for i>0. Thus-i i i iHom(X,Y) = Hom(t t X,Y) = Hom(t X,t Y) = 0 for i>>0.1 -Also Ext (Y,X) = DHom(t X,Y) = 0. (2) is dual.REMARK. We draw a pictureu-------------------------------------------------------------------------------- u--------------------------------------------------o ---------------------------------------------------------------------------o1 preprojectives 1 regulars 1 preinjectives 1m-------------------------------------------------------------------------------- m--------------------------------------------------. ---------------------------------------------------------------------------.by drawing a dot for each indecomposable module. We draw the projectives at- -2the extreme left, then the modules t P(j), then the t P(j), etc. We draw2the injectives at the extreme right, then the modules tI(j), then t I(j),etc. Finally we draw all the regular indecomposables in the middle.The lemma above, and §6 Lemma 2, say that non-zero maps tend to go from theleft to the right in the picture.26

LEMMA 4. If a is a positive real root, and either $0 or ad. This isimpossible for either X has dimension a

§8. Euclidean case. Regular modulesIn this section we study the category of regular modules. We show that itsbehaviour is completely determined by certain ’regular simple’ modules.PROPERTIES OF REGULAR MODULES.(1) If q:X----------LY with X,Y regular, then Im(q) is regular.PROOF. Im(q)cY, so it has no preinjective summand. Also X----------)Im(q), so ithas no preprojective summand.(2) In the situation above Ker(q) and Coker(q) are also regular.PROOF. 0----------LKer(q)----------LX----------LIm(q)----------L0 is exact, so Ker(q) has defect zero. NowKer(q)cX, so Ker(q)=preprojectivessregulars. If there were anypreprojective summand, then the defect would have to be negative. Similarlyfor Coker(q).(3) If 0----------LX----------LY----------LZ----------L0 is exact and X,Z are regular, then so is Y.PROOF. The long exact sequence shows Hom(Z,Preproj) = 0 = Hom(Preinj,Z).(4) The regular modules form an extension-closed abelian subcategory of thecategory of all modules.-(5) t and t are inverse equivalences on this category.DEFINITION.A module X is regular simple if it is regular, and has no proper non-zeroregular submodule. Equivalently if defect(X)=0, and defect(Y)

PROPERTIES. Let X be regular simple, dim X = a.(1) X is a brick, so a is a root.i(2) t X is regular simple for all ieZ.(3) tX=X 5 a is an imaginary root.PROOF. If tX=X then ca=a so a is radical. Conversely, if q(a)=0, then1Hom(X,tX)=DExt (X,X)$0, so X=tX since X and tX regular simple.N(4) t X=X.NPROOF. We may assume a is a real root. Now ==1, soNNHom(X,t X)$0, so X=t X.DEFINITION.X is regular uniserial if there are regular submodules0 = X C X C ... C X = X0 1 rand these are the ONLY regular submodules of X. We say X has regularcomposition factors X ,X /X ,..,X /X (which are clearly regular1 2 1 r r-1simples), regular length r, regular socle X and regular top X/X .1 r-1LEMMA 1. If X is regular uniserial, S is regular simple, andfx:0----------LS----------LE----------LX----------L0 is non-split, then E is regular uniserial.PROOF. It suffices to prove that if UcE is regular and U is not containedin S, then ScU. Thus f(U)$0, so Tcf(U) where T is the regular socle of X,-1 -1and so f (T) = S + Unf (T).-Since t S is regular simple the inclusion T(-----LX gives an isomorphism- -Hom(t S,T)----------LHom(t S,X). Thus it gives an isomorphism1 - - 1Ext (X,S) = DHom(t S,X) = DHom(t S,T) = Ext (T,S),so the pullback sequence-10 ----------L S ----------L f (T) ----------L T ----------L 0@ 1

LEMMA 2. For each regular simple T and r>1 there is a unique regularuniserial module with regular top T and regular length r. Its regularr-1composition factors are (from the top) T, tT, ..., t T.PROOF. Induction on r. Suppose X is regular uniserial of regular length rr-1with regular top T and regular socle t T. Let S be regular simple. Now1 - - r-1r( k (S=t T)Ext (X,S) = Hom(t S,X) = Hom(t S,t T) = {9 0 (else)rso there is a non-split sequence x:0----------LY----------LE----------LX----------L0 if and only if S=t T,and in this case, since the space of extensions is 1-dimensional, any1non-zero xeExt (X,S) gives rise to the same module E. It is regularuniserial by the previous lemma.THEOREM. Every indecomposable regular module X is regular uniserial.PROOF. Induction on dim X. Let ScX be a regular simple submodule of X. Byrinduction X/S = s Y is a direct sum of regular uniserials. Nowi=1 ir1 1Ext (X/S,S) = s Ext (Y ,S), 0----------LS----------LX----------LX/S----------L0 J-----L (x )iii=1Since X is indecomposable, all x $0. Nowi-( k (if Y has regular socle t S)1 iExt (Y ,S) = {i9 0 (else)-so all Y have regular socle t S.iIf r=1 then X is regular uniserial, so suppose r>2, for contradiction. Wemay assume that dim Y

DEFINITION.Given a t-orbit of regular simples, the corresponding tube consists of theindecomposable regular modules whose regular composition factors belong tothis orbit.PROPERTIES.(1) Every regular indecomposable belongs to a unique tube.(2) Every indecomposable in a tube has the same period p under t.PROOF. If X is regular uniserial with regular top T and regular length r,iithen t X is regular uniserial with regular top t T and regular length r. Ifiit T=T we must have t X=X.i(3) If the regular simples in a tube of period p are S =t S, then theimodules in the tube can be displayed as below. The symbol obtained bystacking various S ’s is the corresponding regular uniserial. We indicateithe inclusion of the maximal proper regular submodule Y of X by Y(-----LX, andthe map of X onto the quotient Z of X by its regular socle as X----------)Z. Thetranslation t acts as a shift to the left, and the two vertical dottedlines must be identified.31

§9. Euclidean case. Regular simples and rootsIn this section we show that the tubes are indexed by the projective line,and that the dimension vectors of indecomposable representations areprecisely the positive roots for G.CONSTRUCTION.Let e be an extending vertex, P=P(e), p=dim P. Clearly =1=.By §7 Lemma 4 there is a unique indecomposable L of dimension d+p.P and L are preprojective, are bricks, and have no self-extensions.Hom(L,P)=0 for if q:L----------LP then Im q is a summand of L, a contradiction.1Ext (L,P)=0 since ==-=0.dim Hom(P,L)=2 since = 2.LEMMA 1. If 0$qeHom(P,L) then q is mono, Coker q is a regularindecomposable of dimension d, and reg.top(Coker q) $0.ePROOF. Suppose q is not mono. Now Ker q and Im q are preprojective (sincethey embed in P and L), and so they have defect

LEMMA 2. If X is regular, X $0 then Hom(Coker q,X)$0 for some 0$qeHom(P,L).e1PROOF. Ext (L,X)=0, sodim Hom(L,X) = = = dim Hom(P,X) $ 0.Let a,b be a basis of Hom(P,L). These give maps a,b:Hom(L,X)----------LHom(P,X).-1If a is an iso, let l be an eigenvalue of a b and set q=b-la.If a is non-iso, set q=a.-----Either way, there is 0$feHom(L,X) with fqq=0. Thus f:Coker q----------LX.LEMMA 3. If X is regular simple of period p, thenp-1dim X + dim tX + ... + dim t X = d.PROOF. Let dim X=a.If a $0 there is a map Coker q----------LX which must be onto.eIf a =0 then d-a is a root, and (d-a) =1, so d-a is a positive root.eeEither way a

THEOREM 1. The assignment q9-----LCoker q gives a bijection PHom(P,L)----------LW, sothe set of tubes is indexed by the projective line.PROOF. If U is indecomposable regular of dimension d and reg.top(U) $0,ethen there is a map Coker q----------LU for some q. This map must be epi, since anyproper regular submodule of U is zero at e. Thus the map is an isomorphism.If 0$q,q’eHom(P,L) and Coker q=Coker q’, then1Hom(L,P)----------LHom(L,L)----------LHom(L,Coker q)----------LExt (L,P)=0so the composition L----------LCoker q’ = Coker q lifts to map g:L----------LL. Thus oneobtains a commutative diagramq’0----------LP----------LL----------LCoker q’----------L0f1 1g<

REMARKS.~(1) For the Kronecker quiver A 1•---------------L•,---------------Lthe regular simples all have period one. They are---------------Ll1k k l:meP .---------------Lm~(2) For the 4-subspace quiver, D4with the following orientationu----------LuL•JoJ----------o1 1 1 1• • • •the real regular simples have period 2, and have dimension vectors1 1 1 1 1 11100 0011 1001 0110 1010 0101The regular simples of dimension vector d are2u----------LuLkJoJ----------o1 1 1 1k k k k&1*with maps ,708&0*,718&1*,718&1*7l8where lek, l$0,1.(3) One can find lists of regular simples in the tables in the back ofV.Dlab & C.M.Ringel, Indecomposable representations of graphs and algebras,Mem. Amer. Math. Soc., 173 (1976). For the different graphs G the tubeswith period $ 1 have period as follows~Amp,q if p>0 arrows go clockwise and q>0 go anticlockwise.~D m-2,2,2m~E 3,3,26~E 4,3,27~E 5,3,28One always has Stubesmore analysis.(period-1) = n-2, which can be proved with a little35

§10. Further topicsIn this section I want to list some of the topics which have attractedinterest in the past, and which are areas of present research. The lists ofpapers are only meant to be pointers: you should consult the references inthe listed papers for more information.(1) Kac’s Theorem: for any quiver the dimension vectors of theindecomposables are the positive roots of the graph.V.Kac, Infinite root systems, representations of graphs and invarianttheory I,II, Invent. Math 56 (1980), 57-92, J. Algebra 77 (1982),141-162.V.Kac, Root systems, representations of quivers and invariant theory, inSpringer Lec. Notes 996 (1983), 74-108.H.Kraft & Ch.Riedtmann, Geometry of representations of quivers, inRepresentations of algebras (ed. P.Webb) London Math. Soc. Lec. NoteSeries 116 (1986), 109-145.(2) Invariant theory and geometry for the action of the group GL(a) on thevariety Rep(a).C.Procesi, The invariant theory of n*n matrices, Adv. Math. 19(1976),306-381.C.M.Ringel, The rational invariants of the tame quivers, Invent. Math.,58(1980), 217-239.L.Le Bruyn & C.Procesi, Semisimple representations of quivers, Trans. Amer.Math. Soc. 317 (1990), 585-598.Ch.Riedtmann & A.Schofield, On open orbits and their complements, J.Algebra130 (1990), 388-411.A.Schofield, Semi-invariants of quivers, J. London Math. Soc. 43 (1991),385-395.A.Schofield, Generic representations of quivers, preprint.(3) Construction of the Lie algebra and quantum group of type G from therepresentations of a quiver with graph G.C.M.Ringel, Hall polynomials for the representation-finite hereditaryalgebras, Adv. Math. 84 (1990), 137-178.C.M.Ringel, Hall algebras and quantum groups, Invent. Math. 101 (1990),583-592.G.Lusztig, Quivers, perverse sheaves, and quantized enveloping algebras, J.Amer. Math. Soc. 4 (1991), 365-421.A.Schofield, Quivers and Kac-Moody Lie algebras, preprint.36

(4) Auslander-Reiten theory for wild quivers. In particular, the behaviouriof the functions dim Hom(X,t Y) for fixed X,Y.C.M.Ringel, Finite dimensional hereditary algebras of wild representationtype, Math.Z., 161 (1978), 235-255.V.Dlab and C.M.Ringel, Eigenvalues of Coxeter transformations and theGelfand-Kirillov dimension of the preprojective algebras, Proc. Amer.Math. Soc. 83 (1981), 228-232.D.Baer, Wild hereditary artin algebras and linear methods, ManuscriptaMath. 55 (1986), 68-82.O.Kerner, Tilting wild algebras, J.London Math.Soc, 39(1989), 29-47.J.A.de la Peña & M.Takane, Spectral properties of Coxeter transformationsand applications, Arch. Math. 55 (1990), 120-134.O.Kerner & F.Lukas, Regular modules over wild hereditary algebras,preprint.(5) Tame algebras of global dimension 2, but with properties analogous tothose of path algebras: the tame concealed and tubular algebras.C.M.Ringel, Tame algebras and integral quadratic forms, Springer Lec. Notes1099 (1984).C.M.Ringel, Representation theory of finite-dimensional algebras, inRepresentations of algebras (ed. P.Webb) London Math. Soc. Lec. NoteSeries 116 (1986), 7-79.I.Assem & A.Skowronski, Algebras with cycle finite derived categories,Math. Ann., 280 (1988), 441-463.(6) Interpretation of the representation theory of quivers asnon-commutative algebraic geometry.H.Lenzing, Curve singularities arising from the representation theory oftame hereditary artin algebras, in Springer Lec.Notes 1177 (1986),199-231.W.Geigle & H.Lenzing, A class of weighted projective curves arising inrepresentation theory of finite dimensional algebras, in Springer Lec.Notes 1273 (1987), 265-297.(7) Tame hereditary algebras when the field is not algebraically closed,and the more ring-theoretic aspects of hereditary algebras.V.Dlab & C.M.Ringel, Indecomposable representations of graphs and algebras,Mem. Amer. Math. Soc., 173 (1976).C.M.Ringel, Representations of K-species and bimodules, J. Algebra, 41(1976), 269-302.V.Dlab & C.M.Ringel, Real subspaces of a vector space over the quaternions,Can. J. Math. 30 (1978), 1228-1242.A.Schofield, Universal localization for hereditary rings and quivers, inSpringer Lec. Notes 1197 (1986).D.Baer, W.Geigle & H.Lenzing, The preprojective algebra of a tamehereditary Artin algebra, Comm. Algebra 15 (1987), 425-457.W.Crawley-Boevey, Regular modules for tame hereditary algebras, Proc.London Math. Soc., 62 (1991), 490-508.37

(8) Infinite dimensional representations, and contrast with the theory ofabelian groups.C.M.Ringel, Infinite dimensional representations of finite dimensionalhereditary algebras, Symposia Math., 23 (1979), 321-412.F.Okoh, Indecomposable pure-injective modules over hereditary artinalgebras of tame type, Commun. in Algebra 8 (1980), 1939-1941.F.Okoh, Separable modules over finite-dimensional algebras, J. Algebra, 116(1988), 400-414.~A.Dean & F.Zorzitto, Infinite dimensional representations of D4, GlasgowMath. J. 32 (1990), 25-33.F.Lukas, A class of infinite-rank modules over tame hereditary algebras,preprint.D.Happel & L.Unger, A family of infinite-dimensional non-selfextendingbricks for wild hereditary algebras, preprint38