Tilting Modules over the Path Algebra of Type A, Polytopes, and Catalan Numbers
aa r X i v : . [ m a t h . R T ] M a y Contemporary Mathematics
Tilting Modules over the Path Algebra of Type A ,Polytopes, and Catalan Numbers Lutz Hille
Dedicated to Helmut Strade on the occasion of his 70th birthday
Abstract.
It is well known that the number of tilting modules over a pathalgebra of type A n coincides with the Catalan number C n . Moreover, thenumber of support tilting modules of type A n is the Catalan number C n +1 .We show that the convex hull C ( A n ) of all roots of a root system of type A n isa polytope with integral volume ( n + 1) C n +1 = (cid:0) nn (cid:1) . Moreover, we associateto the set of tilting modules and to the set of support tilting modules certainpolytopes and show that their volumes coincide with the number of thosemodules, respectively. Finally, we show that these polytopes can be definedjust using the root system and relate their volumes so that we can derive theabove results in a new way.
1. Introduction
We consider a quiver Q of type A with n vertices, more details can be found inSection 2. An indecomposable representation of Q can be identified with an interval[ i, j ] representing the support of the dimension vector (0 , ..., , , . . . , , , ..., Q two series of polytopes, the C – and the P –series, which both comein three versions. The first series of polytopes C + ( Q ) ⊂ C clus ( Q ) ⊂ C ( Q ) consistsjust of the convex hulls of certain roots in the root system of type A . Thus thesepolytopes are defined independently of the orientation of Q .All modules in this note should be understood as modules over the path algebraof type A . A tilting module is a particular module satisfying certain genericityproperties. We always identify modules over the path algebra with representationsof the corresponding quiver.The second series of polytopes P + ( Q ) ⊂ P clus ( Q ) ⊂ P ( Q ) consists of the unionof certain simplices σ T associated to tilting modules T or some generalizationsof these. Note that the first series of polytopes only depends on the underlying Mathematics Subject Classification.
Primary 16G20, 16G99, 05A19, 05A10, 05A99;Secondary 52B20, 52B99, 17B99.
Key words and phrases.
Quiver, path algebra, Dynkin diagram, root system, convex hull, tilt-ing module, support tilting module, 2-support tilting module, polytope, volume, Catalan number,tilting sequence, rooted tree, Dyck path. c (cid:13) graph of Q , whereas the second one depends on a chosen orientation of the quiver.However, we will show that the polytopes in the second series also only depend onthe underlying graph of Q . Thus we will write C ( Q ) or C ( A n ) interchangeably,but we have to write P ( Q ) until we haven proven that the latter definition isindependent of the orientation of the quiver.The principal goal of this paper is to compare both types of polytopes. In fact,we show that they coincide (see Theorem 1.2). Moreover, we obtain the numberof certain versions of tilting modules as the volume of the corresponding polytope,where we use a certain normalization of the euclidean volume. A second aim ofthis paper is to further simplify the counting by passing to tilting sequences. Thisapproach is explained in Section 5. Here we consider an additional order on theindecomposable direct summands, and then the counting gets even easier, namely,we just obtain a bijection between the tilting sequences and a certain symmetricgroup (see Theorem 5.1).In fact, in this paper we do not use any representation theory other than somegeometric interpretations of representation-theoretic notions. For any dimensionvector d there is exactly one rigid (or generic) module that is dense in the corre-sponding representation space (see Section 2 or also [ ] for more details), and wewill work with this rigid module.Note that we use the integral volume in this paper (see also Section 6 for moreexplanations), that is, the volume vol ∆ of any simplex ∆ generated by an integralbasis is 1. Thus our volume is just n ! times the usual euclidean volume. Theadvantage of this definition is that the volume of any lattice polytope is an integer.To be precise we define certain positive numbers (not depending on the orien-tation by Theorem 1.2) t + ( Q ) = vol P + ( Q ) , t clus ( Q ) = vol P clus ( Q ) , t ( Q ) = vol P ( Q ) . In order to get an interpretation of these numbers, we need to consider severalversions of tilting modules. Note that a tilting module T = ⊕ ni =1 T ( i ) is a direct sumof n pairwise non-isomorphic indecomposable modules T ( i ) satisfying Ext ( T, T ) =0. For an indecomposable representation [ i, j ] the support is just the interval [ i, j ],for a direct sum of such modules the support is just the union of the supportsof the indecomposable direct summands. Thus, the support of a module (or arepresentation) is the support of its dimension vector (see Section 2 for details).A support (or cluster) tilting module is a module T that is a tilting module ifrestricted to its support. A 2–support tilting module is a module T together with adecomposition T = T + ⊕ T − such that both T + and T − are support tilting modulesand the supports are a disjoint union of the vertices of the quiver Q of type A n .Thus, any 2-support tilting module T defines a subset I of the set Q of verticesof Q such that I is the support of T + and Q \ I is the support of T − . We define T + ( Q ), T clus ( Q ), and T ( Q ), respectively, as the set (of isomorphism classes) oftilting modules, the set (of isomorphism classes) of support tilting modules, andthe set (of isomorphism classes) of 2-support tilting modules. By using the results in[ ] we obtain the following interpretation of the volumes of the polytopes associatedto tilting modules and their generalizations. Theorem . t + ( Q ) = ♯ T + ( Q ) , t clus ( Q ) = ♯ T clus ( Q ) , t ( Q ) = ♯ T ( Q ) . ILTING MODULES, POLYTOPES, AND CATALAN NUMBERS 3
Note that the classification of tilting modules T + ( Q ) over quivers Q of type A n is well known. A description using trees for the directed orientation can be foundin [ ]. This leads to a recursion formula for t + ( Q ). In particular, the recursionformula in Theorem 1.3 is the same as the recursion formula for the number of3–regular trees with n + 1 leaves and one root (see Section 6).The key point for the correspondence in type A is Theorem 1.2 that does notgeneralize to the other Dynkin quivers apart from type C and is certainly wrong foreuclidean and wild quivers. Let us briefly comment on the non-simply laced types.There is the notion of a path algebra, where we use two fields, one being a fieldextension of degree two (respectively, three for Dynkin type G ) of the other one.This is certainly more technical and will be explained in our forthcoming paper[ ] in detail. We also note that most of the combinatorics related to the Catalannumbers is already known for type A , but in all other cases it is unknown.The next theorem shows that for type A the polytopes in the P –series areindependent of the chosen orientation. Thus we can simply write P ( A n ) instead of P ( Q ). Theorem . C ( Q ) = P ( Q ) , C clus ( Q ) = P clus ( Q ) , C + ( Q ) = P + ( Q ) . By using decompositions of the polytopes we get several recursion formulasrelating the three different polytopes. Note that t ( A ) = t clus ( A ) = 2 t + ( A ) = 2.Consequently, already these formulas determine the numbers t ( A n ), t clus ( A n ), and t + ( A n ) uniquely, by induction. However, we can also compute t ( A n ) = (cid:0) nn (cid:1) =(2 n )! /n ! n ! directly, which gives even more ways to determine these numbers.The next result is a standard decomposition that holds for any quiver Q . InSection 2 we will present two examples illustrating these formulas. Theorem . t ( A n ) = X I ⊆ Q t + ( A n | I ) t + ( A n | Q \ I ) ,t clus ( A n ) = X i ∈ Q t + ( A n | Q \{ i } ) = n − X i =0 t + ( A i ) t + ( A n − − i ) . For the C –series of polytopes we can prove the following formulas directly bydetermining the facets and their volumes, where the polytope C ( A n ) is the simplestone to consider. Theorem . vol C ( A n ) = (cid:18) nn (cid:19) = ( n + 1) C n , vol C clus ( A n ) = (cid:18) n + 2 n + 1 (cid:19) / ( n + 2) = C n +1 , vol C + ( A n ) = (cid:18) nn (cid:19) / ( n + 1) = C n . In type A we have one dimension vector (1 , , . . . ,
1) corresponding to the in-terval [1 , n ]. This provides us with yet another recursion formula which is easy toprove for the directed orientation.
LUTZ HILLE
Theorem . t + ( A n ) = n X i =1 t + ( A n | Q \{ i } ) . Since there are several ways to prove these theorems, we give a short outlineof their proofs. One way to prove the results uses induction over the facets andanother way uses the fact that a quiver of type A has precisely one sincere root.However, by using a purely combinatorial approach, we obtain a different proof bysimply using counting arguments as explained in Stanley’s book on enumerativecombinatorics [ ].We also like to mention that the same idea works for arbitrary Dynkin quivers.However, in this case the polytope P ( Q ) is not convex, and thus does not coincidewith C ( Q ), which is the convex hull of all the roots of the root system. This makesthe above formulas more complicated, and for the details we refer the reader toour forthcoming paper [ ]. We would also like to point out that the number oftilting modules for arbitrary Dynkin quivers has recently been computed in [ ] bycompletely different methods. Moreover, the polytope P ( Q ) (and its variations) canbe defined even for arbitrary infinite quivers Q . This way we would get a simplicialcomplex that is a triangulation of a certain quadric in the real Grothendieck group K ( Q ) R of the category of representations of Q .The outline of the paper is as follows. After the introduction we start withsome basic representation theory in Section 2. Here we only recall a few facts thatare well known. Details and further references can be found in [ ]. In Section 3we prove the first three theorems and in Section 4 we proceed with the last twotheorems. In Section 5 we modify the problem slightly. Instead of tilting moduleswe will consider tilting (or full strongly exceptional) sequences. Finally, in the lastsection we will give some combinatorial interpretation of the results using Stanley’sbook [ ]. Acknowledgments.
This work started during a stay of the author in Bielefeld atthe SFB 701 ’Spectral Structures and Topological Methods in Mathematics’. Hewould like to thank Henning Krause for the invitation and the stimulating workingconditions. Moreover, this work was supported by SPP 1388 ’Representation The-ory’. The author is also indebted to Friedrich Knop for several hints concerning thecombinatorics of the Catalan numbers and to Claus Michael Ringel for discussingfurther combinatorial aspects of the Catalan numbers. Finally, he is grateful toKarin Baur, J¨org Feldvoss, and the referee for many helpful comments on variousdrafts of this paper.
2. Representations of A n In this section we always consider a quiver Q of type A n , that is, a Dynkindiagram of type A n , where we choose an orientation of any edge between the vertices i and i + 1. For every oriented edge α ∈ Q , called arrow, we define its startingpoint to be s ( α ) and its terminal point to be t ( α ). A representation of Q consistsof n finite dimensional vector spaces V i over a fixed field k , where i = 1 , . . . , n ,with linear maps V ( α ) : V s ( α ) −→ V t ( α ) . Note that either s ( α ) + 1 = t ( α ) or s ( α ) − t ( α ). The dimension vector dim V of the representation V = { V i } isdefined as dim V = (dim V , . . . , dim V n ). A sincere root is a dimension vector withdim V i = 0 for all vertices i . The set of all representations of a quiver with the usual ILTING MODULES, POLYTOPES, AND CATALAN NUMBERS 5 homomorphisms forms an abelian category of global dimension one that has enoughprojective and also enough injective representations. Note that any representation V has a short projective resolution 0 −→ P −→ P −→ V −→ ( V, W )is defined as the cokernel of the induced map Hom ( P , W ) −→ Hom ( P , W ). Thekernel of this map is the set of all homomorphisms Hom ( V, W ). A representation V is simple if it has no proper subrepresentations, it is indecomposable if it has nonon-trivial decomposition into a direct sum of two representations, and it is rigid ifExt ( V, V ) = 0. The latter condition has also a natural interpretation in the space R ( Q ; d ) of all representations of dimension vector d consisting just of all possiblelinear maps R ( Q ; d ) = M α ∈ Q Hom ( k d ( s ( α )) , k d ( t ( α )) ) . The group G ( d ) = Q GL( d i ) acts on R ( Q ; d ) via base change, and V is an elementof the dense orbit over an algebraic closure precisely when Ext ( V, V ) = 0. Since R ( Q ; d ) is irreducible (as an affine space), there is at most one rigid representation M ( d ) of dimension vector d . Conversely, when Q is a Dynkin quiver, in particular,when Q is of type A , there are only finitely many orbits. Consequently, for anydimension vector d up to isomorphism there is precisely one representation thathas dimension vector d and is rigid. Using this fact, we can define an equivalencerelation on the possible dimension vectors as follows. We say d and d ′ are equiv-alent provided the indecomposable direct summands of M ( d ) and M ( d ′ ) coincide(up to positive multiplicity). Any module M ( d ) can have at most n pairwise non-isomorphic indecomposable direct summands. Let us assume M ( d ) has just n − M and M , meaning that M ( d ) ⊕ M is rigid, M ( d ) ⊕ M is rigid, and neither M nor M is already a direct summand of M ( d ). Then it isknown that (up to renumbering) there is an exact sequence M −→ M −→ M ,with M consisting of direct summands of M ( d ). Such a sequence is called exchangesequence . In type A , M has at most two indecomposable summands. These se-quences play a crucial role for the recursive construction of all tilting modules. Infact, for any equivalence class of dimension vectors d there exists a unique d ′ in thisclass so that M ( d ′ ) has no multiple indecomposable direct summands. Such a mod-ule is called basic . The maximal ones among those modules are the tilting modulesthat contribute to the volume of the polytope P ( Q ). All the other ones correspondto certain faces. We are dealing with the maximal ones, all others contribute withvolume 0.The principal aim of this paper is to determine the number of maximal equiv-alence classes M ( d ) by using polytopes together with their basic representations.The situation becomes quite elementary if Q is a quiver of type A n with its directedorientation. The details can already be found in [ ]. If Q is of type A n with anotherorientation, the situation is slightly different, however the approach in [ ] can bemodified as follows. Instead of diagrams with all connections on the top, we use forarrows from i + 1 to i connections at the bottom of the diagram. This even givesa constructive way to compute M ( d ) for any d and any orientation of the quiver.Finally, we define the cones σ associated to tilting modules, support tiltingmodules, and 2-support tilting modules, respectively. We start with any module M and its decomposition into indecomposable direct summands M = ⊕ M ( i ) a ( i ) ,where a ( i ) is the multiplicity of the indecomposable direct summand M ( i ) in M . LUTZ HILLE
For such an M we define σ M = conv { dim M ( i ) , | i ∈ I } to be the convex hullof zero and the dimension vectors of the indecomposable direct summands. Notethat the multiplicities a ( i ) > T form an integral basis of Z n , and thus σ T is a simplex which has volume 1by our definition of the volume. If T is a support tilting module, we add thenegative standard basis of the complement of the support of T , thus we define σ T = conv { dim T ( i ) , − e j , | i ∈ I, j ∈ J } , where T = ⊕ T ( i ) a ( i ) and J is thecomplement of the support of T . Finally, for a 2-support tilting module T wedecompose both T + and T − as T + = ⊕ i ∈ I T ( i ) a ( i ) , T − = ⊕ j ∈ J T ( j ) a ( j ) and define σ T = conv { dim T ( i ) , − dim T ( j ) , | i ∈ I, j ∈ J } .We illustrate the construction of the simplices and the two series of polytopesby two examples, namely, quivers of types A and A . Example 1.
Let Q be a quiver of type A . Then there is just one orientation upto a permutation of the two vertices. The dimension vectors of the indecomposablerepresentations are (1 , , , + = { (1 , , (1 , , (0 , } , Φ clus = { (1 , , (1 , , (0 , , ( − , , (0 , − } , andΦ = { (1 , , (1 , , (0 , , ( − , , ( − , − , (0 , − } . The convex hull C + ( Q ) of Φ + has volume 2, the convex hull C clus ( Q ) of Φ clus is apentagon of volume 5, and the convex hull C ( Q ) of Φ is a hexagon of volume 6. Thefollowing pairs of roots, together with zero, form a simplex in P + ( Q ): (1 , , (1 , , , (0 , P clus ( Q ) we have three additional simplices definedby the pairs (0 , , ( − , − , , (0 , − , − , (1 , P ( Q ) by two simplices defined by ( − , , ( − , −
1) and( − , − , (0 , − Example 2.
The case A is more complicated, since we have two, essentiallydifferent, orientations in the quiver Q . We first describe the parts independent ofthe orientation, a difference only occurs for the polytopes of the P –series. For theroots we obtainΦ + = { (1 , , , (0 , , , (0 , , , (1 , , , (0 , , , (1 , , } , Φ clus = Φ + ∪ { ( − , , , (0 , − , , (0 , , − } , Φ = Φ + ∪ − Φ + . The corresponding convex hulls have volume 5 for C + ( Q ), volume 14 for C clus ( Q ),and volume 20 for C ( Q ). This can easily be seen from the decomposition of the P –series. We have to chose an orientation for this and describe the triples defininga simplex for the directed orientation first: { (1 , , , (1 , , , (1 , , } , { (0 , , , (1 , , , (1 , , } , { (1 , , , (1 , , , (0 , , } , { (0 , , , (0 , , , (1 , , } , { (1 , , , (0 , , , (0 , , } . For the other orientation we get two simplices replaced by two others, however, theunion of both pairs is the same. For this we replace the second and the fourth oneby { (1 , , , (0 , , , (0 , , } and { (1 , , , (1 , , , (0 , , } . For the polytope P clus ( Q ) we need to add for any pair of roots, obtained by deleting(1 , , { (1 , , , (1 , , , (0 , , − } for the first simplex in P + ( Q ), and ILTING MODULES, POLYTOPES, AND CATALAN NUMBERS 7 we get (1 , , , (0 , − , , (0 , , −
1) for the simple root (1 , , − , , , (0 , − , , (0 , , −
1) and obtain for the volume 5 + 5 + 3 + 1 = 14. In asimilar way we get for the volume of P ( Q ) the sum 5 + 5 + 5 + 5 = 20. Alternatively,using the computation of the volume with counting the facets and their volume for C ( Q ), the volume of P ( Q ) is 2 × C ( Q )has 6 squares and 8 triangles as facets.
3. Proofs of Theorems 1.1–1.3
We first prove Theorem 1.1: It is well known that for any tilting module T = ⊕ T ( i ) the dimension vectors dim T ( i ) form a Z –basis of Z n . Thus vol σ T = 1. Thuswe have proven the following lemma. Lemma . For any tilting module, any support tilting module, and any –support tilting module T we have vol σ T = 1 . It is therefore sufficient to show that vol ( σ T ∩ σ T ′ ) = 0 for any two differenttilting modules T and T ′ . This follows from the definition of the volume in [ ].Crucial for our computation is now Theorem 1.2, which is not true for arbitraryDynkin quivers. Note that any non-trivial exchange sequence0 −→ T ( i ) −→ ⊕ T ( j ) −→ T ( i ) ′ −→ T ( i )+dim T ( i ) ′ = P dim T ( j ), we see that P + ( Q ) is convex, strictly convex at the common facet forone middle term, and flat for two middle terms. Consequently, P + ( Q ) is convexprecisely when there are at most two middle terms for any exchange relation.Note that such a relation corresponds to two simplices σ T and σ T ′ with acommon facet. If P + ( Q ) is convex, then so are P ( Q ) clus and P ( Q ). Since P ( Q )is convex and has the roots as its vertices, it must coincide with C ( Q ). The sameargument works for P + ( Q ) and P clus ( Q ).The proof of Theorem 1.3 follows from the decomposition of the polytope P clus ( Q ) with respect to the possible quadrants. Any subset I of the vertices Q of Q defines the quadrant consisting of non-negative entries, whenever the index isnot in I , and non-positive otherwise. The volume of P clus ( A n ) intersected with thisquadrant has the same volume as P + ( A n ) intersected with the corresponding face.Thus its volume coincides with the volume of P + ( A n | Q \ I ). A similar argumentholds for P ( A n ). Here we need to determine again the volume in the quadrantdefined by I . The volume in this case coincides with the product of the volume of P + ( A n | Q \ I ) and the volume of P + ( A n | I ).Theorems 1.4 and 1.5 can also be proven by using the combinatorial interpre-tation in Section 6. Alternatively, one could have used induction over the facets.Since these proofs need some detailed computations, we defer them to the nextsection.
4. Proofs of Theorems 1.4 and 1.5
We first recall the recursion formula t + ( A n ) = P n − i =0 t + ( A i ) t + ( A n − − i ). Fromthis we get one of the standard recursions of the Catalan numbers C n , since t + ( A ) = t + ( A ) = 1. This is the same recursion as for the number of trees in B n which will be considered in the last section. Further details can be found inStanley’s book [ ]. We start with the analogous formula for the polytope P + ( A n )for the directed orientation. LUTZ HILLE
Lemma . vol P + ( A n ) = n P i =1 vol P + ( A n | Q \{ i } ) . This simply uses the fact that there exists precisely one sincere root for A n .Thus, the volume of P + ( A n ) is just the sum of the volumes of the facets of P + ( A n )not containing 0. This is obvious for the quiver of type A n with its directed ori-entation, since every tilting module contains the projective injective (having thesincere root as dimension vector) as a direct summand. Since P ( A n ) = C ( A n )(independent of the orientation of the arrows) we get the same formula for thevolume. Taking away the projective injective direct summand, we obtain a partialtilting module with support at A n | Q \{ i } for precisely one vertex i of the quiver A n . Such a partial tilting module corresponds to a facet (defined by d i = 0). Thus,each tilting module corresponds to precisely one facet in P + ( A n ) not containing 0.Moreover, the volume of σ T and the volume of each of its faces (in particular, ofthe facet from which we delete the sincere root) is always 1. Hence we obtain theformula in Lemma 4.1, and the last formula in Theorem 1.4 follows directly.In a next step we compute the volumes of all the polytopes P ∗ ( A n ) = C ∗ ( A n ).This can be done in several ways. We give a combinatorial approach using certainpaths from (0 ,
0) to (0 , n ) later. Firstly, we use the volume of the facets of C ( A n ). Lemma . vol C ( A n ) = n P i =1 (cid:0) n − i − (cid:1)(cid:0) n +1 i (cid:1) = (cid:0) nn (cid:1) . We start by explaining the first equality. Each hyperplane d i = 1 containsprecisely one facet F i , and each facet is in the orbit under the symmetric group S n +1 (that is, the Weyl goup of the root system A n ) of precisely one such facet.Thus we need to compute the orbit of F i and the volume of the facet F i . Thevolume is just (cid:16) n − i − (cid:17) and the orbit has exactly (cid:0) n +1 i (cid:1) elements. This shows thefirst equality. The second one is a simple recursion using binomial coefficients: (cid:18) nn (cid:19) = (cid:18) n − n − (cid:19) + (cid:18) n − n (cid:19) = (cid:18) n − n − (cid:19) + 2 (cid:18) n − n − (cid:19) + (cid:18) n − n (cid:19) = . . . This proves the first formula in Theorem 1.4. The formula in Theorem 1.5 is justthe Catalan recursion, as well as the second formula in Theorem 1.4. This finishesthe proofs of Theorems 1.4 and 1.5.In the next section we will need another formula which will be used to determinethe number of tilting (or full strongly exceptional) sequences and to relate them toour combinatorial description. Note that the right hand side is n + 1 times n !. Lemma . ( n + 1)! = 0! n ! + 1!( n − n + 2!( n − n ( n − + . . .
5. Tilting sequences
If we replace a tilting module by an ordered tuple of modules compatible withnon-vanishing homomorphisms, we even get an easier formula. We define T + ( Q )to be the set of tilting sequences, i.e., ( T (1) , ..., T ( n )) satisfying two conditions: T = ⊕ T ( i ) has no self extensions and Hom ( T ( j ) , T ( i )) = 0 for all j > i . Note thata tilting sequence is also called a full strongly exceptional sequence of modules. Ina similar way we define support tilting sequences and 2–support tilting sequences.Then we get the following formulas for the corresponding numbers, as we will seebelow. ILTING MODULES, POLYTOPES, AND CATALAN NUMBERS 9
Theorem . t ( A n ) = ( n + 1)! = 0! n ! + n n − n ( n − n − . . . ,t + ( A n ) = n ! . The result follows from the interpretation of the map considered in [ ]. Definethe set B n as the set of all 3–regular trees with n + 1 leaves and one root. There isa natural map S n −→ B n from the symmetric group to the set of all those trees.The number of elements in the preimage of this map is just the number of tiltingsequences (interpreted as a tree with a compatible order on the inner vertices)defining the same tilting module (interpreted as a tree). Thus we have a bijectionbetween tilting sequences and elements of the symmetric group.In order to define the corresponding polytope, we extend the positive rootsto Φ + consisting of the positive roots for A n together with all sums of orthog-onal roots . In case Q is of type A we just add the vector (1 , , P + ( A n ) as the convexhull of zero and Φ. In a similar way, we define P ( A n ) as the convex hull ofΦ + and − Φ + , and P clus ( A n ) as the convex hull of Φ + and the negative simpleroots. To complete the picture, we also need to define a simplex σ T for ev-ery tilting sequence T . This can be done as follows. Whenever we have a se-quence ( T ( i (1)) , T ( i (2)) , . . . T ( i ( r ))) of indecomposable direct summands of T with i (1) < i (2) < i (3) < . . . < i ( r ) and all components being incomparable (no ho-momorphisms and no extensions between different members), then we consider thevertices dim T ( i (1)) , dim T ( i (1)) + dim T ( i (2)) , . . . , dim T ( i (1)) + . . . + dim T ( i ( r )).In this way we get different simplices for different tilting sequences, and the unionof all simplices σ of the tilting sequences for a given tilting module T is just thesimplex σ of T . Thus, the number of tilting sequences is just the volume of P + ( A n ),the number of support tilting sequences is the volume of P clus ( A n ), and the numberof 2–support tilting sequences is the volume of P ( A n ). The first and the last ofthese numbers have been computed in Theorem 5.1.For the corresponding polytope P + ( A n ) we can form P ( A n ) just as the joinof P + ( A n ) with − P + ( A n ). This defines a polytope P ( A n ), and its volume is thenumber of 2–support tilting sequences. The counting in the above theorem thencomputes the volume of P ( A n ) from the volume of P + ( A n ) using the formula withthe volume of the facets. So we have to compute the facets (in particular, in allquadrants except the positive and the negative) and their volumes.The facets in the positive quadrant correspond to elements of the symmetricgroup, and each facet contributes with volume 1. A facet in a quadrant correspond-ing to a subset I of the vertices of the quiver corresponds to a facet for A n | I and anopposite facet for A n | Q \ I . Thus Lemma 4.3 just computes the volume of P ( A n )from P + ( A ) and Theorem 5.1 is proven. Summarizing this we have the followingresult. Theorem . P ( A n ) = C ( A n ) have as volume the number of –support tiltingsequences ( n + 1)! . Moreover, the number of tilting sequences for A n is n ! , that is,the volume of P + ( A n ) = C + ( A n ) .
6. Some further comments
In the final section we present some further explanation for our use of the notionof the volume. This method is inspired by toric geometry and lattice polytopes.Moreover, we give another interpretation of the computation of the volume usingStanley’s exercise of a combinatorial interpretation of the Catalan numbers (seeExercise 6.19 in [ ]). Very surprisingly, it does not only give an interpretation ofthe volume of P + ( A n ) and P clus ( A n ) (where the Catalan numbers occur naturally),but also for the volume of P ( A n ), if we modify Dyck paths so that they correspondto the 2–support tilting modules. Note that any cube of the form [0 , n has volumeone in the euclidean metric and can be decomposed into n ! many simplices, all ofvolume 1. This is just a recursive computation. For n = 1,2 the claim is obvious.Then proceed by induction, and observe that the cube has n facets containing 0.(In fact, we could use any vertex instead of 0). By induction, the formula holdsfor the facets, and consequently, for the convex hull of the facet and (1 , . . . ,
1) (apyramid over the facet). Now one checks that the n pyramids over the n facetsdecompose the cube into n pyramids of volume ( n − n !. We consider the set B n of rooted3–regular trees with one root and n + 1 leaves (examples can be found in [ ]). If weconsider the quiver A n with its directed orientation, then we can identify T + ( Q )with B n (see [ ]). Thus, we can compute the number of tilting modules using astandard recursion formula for the number of trees. Take such a tree and take theunique vertex connected to the root. Decompose the tree S into the two connectedcomponents S + and S − obtained from deleting this vertex. Then we get C n = ♯B n = n − X i =0 ♯B i ♯B n − − i , ♯B = ♯B = 1 . This is one of the standard recursion formulas for the Catalan numbers C n . Dyck paths can be used for a combinatorial description ofthe volume of the polytopes. A
Dyck path is a path from (0 ,
0) to (0 , n ) using onlysteps (1 , −
1) or (1 ,
1) so that the path never goes below the x –axes (meaning thatthe first coordinate of a point is non-negative). We denote the set of Dyck paths by D + n . For given n the number of Dyck paths coincides with the number of possiblebracketings of an expression with n inputs. Moreover, this can be identified with theelements in B n and with the vertices of the associahedron. For the combinatoricswe refer to the famous exercise in Stanley’s book [ ], where we use only 5 interpre-tations of the 66 (in fact, even more can be found on Stanley’s homepage). If weconsider arbitrary paths from (0 ,
0) to (0 , n ) with steps (1 ,
1) or (1 , −
1) (withoutthe condition to be above the x –axes) we obtain a set D n that has (cid:0) nn (cid:1) elements.An interpretation of 2-support tilting sequences is obtained as follows. Wheneverthe path stays above the x –axes, we identify the corresponding Dyck path with itstree, and thus with a direct summand T + of T . Whenever the path stays below the x –axes, we identify the corresponding path with T − . This bijection identifies pathsin D n with 2–support tilting modules for A n . Consequently, we have computed thevolume of P ( A n ) as the number of elements in D n which is (cid:0) nn (cid:1) . ILTING MODULES, POLYTOPES, AND CATALAN NUMBERS 11
References [1] Karin Baur and Lutz Hille,
On the complement of the dense orbit for a quiver of type A ,Comm. Algebra (2014), 2871–2889.[2] Lutz Hille, On the volume of a tilting module , Abh. Math. Sem. Univ. Hamburg (2006),261–277.[3] Lutz Hille, Root systems, polytopes, and the number of tilting modules for Dynkin quivers ,in preparation.[4] Mustafa A. A. Obaid, S. Khalid Nauman, Wafaa M. Fakieh, and Claus Michael Ringel,
Thenumbers of support-tilting modules for a Dynkin algebra , Preprint, arXiv:1403.5827.[5] Richard P. Stanley,
Enumerative combinatorics. Vol. 2 , Cambridge Studies in AdvancedMathematics, vol. 62, Cambridge University Press, Cambridge, 1999.
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