19 Quivers and tiltingWe will describe an unexpected connection between representations of quiversand simple Lie algebras. To summarize, the quivers with a finite number ofindecomposable representations correspond to certain semisimple Lie algebras,the indecomposable representations correspond to positive roots, and the irre-ducible representations correspond to simple roots.Definition 241 A quiver is a finite directed graph (possible with multiple edgesand loops). A representation of a quiver (over some fixed field) consists of avector space for each vertex of the graph and a linear map between the corre-sponding vector spaces for each edge.Example 242 Representations of a point are just vector spaces. Representa-tions of a point with a loop are vector spaces with an endomorphism. Overan algebraically closed field the indecomposable representations are classifiedby Jordan blocks. Representations of 2 points joined by a line are just linearmaps of vector spaces. There are 3 indecomposable representations: a mapfrom a 0-dimensional space to a 1-dimensional one, a map from a 1-dimensionalspace to a 0-dimensional one, and a map from a 1-dimensional space onto a1-dimensional one. More generally, stars with n incoming arrows correspond ton maps to a vector space. When n = 2 there are 6 indecomposable representa-tions, and when n = 4 there are 12. When n = 4 there is a qualitative change:there are now infinitely many indecomposables. For example we can take 41-dimensional subspaces of a 2-dimensional space. The first two determine abase (0, 1) and (1, 0), the third is spanned by (1, 1) and determines the ratiobetween the two bases, but nor the 4th space can be spanned by (1, a) for anya, so we get a 1-parameter family of indecomposables. Although there are aninfinite number of indecomposables, it is not hard to classify them explicitly:it is a “tame” problem. For stars with 5 incoming vertices the indecomposablerepresentations are “wild”: there is no neat description of them. We will seethat the cases with a finite number of indecomp osables correspond to Dynkindiagrams of the finite dimensional semisimple Lie algebras with all roots thesame length, and the tame cases correspond to affine Dynkin diagrams.The representations of a quiver are the same as modules over a certain ringassociated with the quiver. This ring has an idempotent for each vertex, withthe idempotents commuting and summing to 1. There is also an element foreach edge, subject to some obvious relations. The algebra is finite dimensionalif the quiver contains no cycles.There are two titling functors we can apply to modules over quivers:• If a vertex a is a source, then we can change all the arrows to point intoa, and change the vector space of a to be Coker(Va7→ ⊕a→bVb).• If a vertex a is a sink, then we can change all the arrows to point out ofa, and change the vector space of a to be Ker(⊕b→aVb7→ Va)The functors take representations of a quiver to representations of a differentquiver, with a source changed to a sink or a sink changed to a source. Theyare almost but not quite inverses of each other. They are inverses provided94Va7→ ⊕a→bVbis injective, or ⊕b→aVb7→ Vais surjective. In particular they areinverses of each other on indecomposable mo dules, except for the special caseof indecomposable modules of total dimension 1.The idea is that we try to classify irreducible modules by repeated applyingtilting functors, trying to make the module vanish. If we succeed then wecan recover the original module from a 1-dimensional module by applying the“almost inverse” tilting functors in the opposite order. We will see that we cando this provided the quiver is one of the diagrams An, Dn, E6, E7, and E8,Take a vector space spanned by the vertices of a quiver, and give it an innerproduct such that the vertices of a quiver have norm 2, and their inner productis minus the number of lines joining them. Then the dimension vector of aquiver can be represented by a point in this space in the obvious way, and theeffect of tilting by a source or sink a is just reflection in the hyperplane a⊥(except on the vector a itself).We want to find a sequence of tiltings so that the dimension vector has anegative coefficient. It is easy enough to find a sequence of reflections of simpleroots that do this: the problem is that we have a constraint that we can onlyuse a reflection of a simple root if it is a source or a sink for the quiver. To dothis we will use Coxeter elements.A Coxeter element of a reflection group is a pro duct of the reflections ofsimple roots in some order.Lemma 243 If a Coxeter element of a reflection group fixes a vector then thevector is orthogonal to all simple roots..Proof If a vector a =Panvnis fixed by a Coxeter element (for simple rootsvi) then the reflection of viis the only one that can change the coefficient of vi,so it must fix a. So a is fixed by all reflections of simple ro ots, and is thereforeorthogonal to all simple roots, so is 0. Corollary 244 If σ is a Coxeter element of a finite reflection group and a is anon-zero vector, then σk(a) has a negative coefficient for some k.Proof Otherwise we could find a non-zero fixed vector a + σ(a) + σ2(a) + · · · +σh−1(a), where h is the order of the Coxeter element. Exercise 245 Show that if a Coxeter diagram of a reflection group is a treethen any two Coxeter elements are conjugate, and in particular have the sameorder (called the Coxeter number).Exercise 246 Find the order of the Coxeter elements of An.We now construct a special Coxeter element σ associated to a given quiver asfollows. First take the reflection of some source, change the source to a sink, andthen mark that vertex as used. Keep repeating this until all vertices have beenused. The result is the original quiver, as each edge has had its direction changedtwice. So we have found a sequence of reflections of sources that preserves thequiver. This means that we can keep on repeating the sequence of reflections ofthe Coxeter element, and every time we will be reflecting in some source.95We can now show that the indecomposable representations of any quiver oftype An, Dn, or Encorrespond to the positive roots of the associated root sys-tem: in fact we can apply tiltings until the dimension vector becomes a simpleroot, when it is trivial to find the unique indecomposable. Take the dimensionvector a of any indecomposable representation. As σk(a) has negative