On the Genus of the Graph of Tilting Modules

Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 45 (004), No., 45-47. On the Genus of the Graph of Tilting Modules Dedicated to Idun Reiten on the occasion of her 60th birthday Luise Unger Michael Ungruhe FernUniversität Hagen, Fachbereich Mathematik D 58084 Hagen, Germany MSC 000: 05C0, 6G0, 6G70 Let Λ be a finite dimensional, connected, associative algebra with unit over a field k. Let n be the number of isomorphism classes of simple Λ-modules. By mod Λ we denote the category of finite dimensional left Λ-modules. A module Λ T mod Λ is called a tilting module if (i) the projective dimension pd Λ T of Λ T is finite, and (ii) Ext i Λ(T, T ) = 0 for all i > 0, and (iii) there is an exact sequence 0 Λ Λ Λ T Λ T d 0 with Λ T i add Λ T for all i d. Here add Λ T denotes the category of direct sums of direct summands of Λ T. Tilting modules play an important role in many branches of mathematics such as representation theory of Artin algebras or the theory of algebraic groups. m Let T i be the decomposition of Λ T into indecomposable direct summands. We call i= ΛT basic if Λ T i Λ T j for all i j. A basic tilting module has n indecomposable direct summands. A direct summand Λ M of a basic tilting module Λ T is called an almost complete tilting module if Λ M has n indecomposable direct summands. Let T (Λ) be the set of all non isomorphic basic tilting modules over Λ. We associate with T (Λ) a quiver K(Λ) as follows: The vertices of K(Λ) are the tilting modules in T (Λ), and there is an arrow Λ T Λ T if Λ T and Λ T have a common direct summand which is an 038-48/93 $.50 c 004 Heldermann Verlag

46 L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules almost complete tilting module and if Ext Λ(T, T ) 0. We call K(Λ) the quiver of tilting modules over Λ. With K(Λ) we denote the underlying graph of K(Λ). It has been recently shown [7] that K(Λ) is the Hasse diagram of a partial order of tilting modules which was basically introduced in [0]. From this it follows, that K(Λ) has no oriented cycles. If K(Λ) is finite, then it is connected. Examples show that K(Λ) may be rather complicated. One measure for the complicatedness of a graph G is its genus γ(g). This is the minimal genus of an orientable surface on which G can be embedded. The aim of these notes is to show that there are finite quivers of tilting modules of arbitrary genus. To be precise, we prove: Theorem. For all integers r 0 there is a representation finite, connected algebra such that γ(k( )) = r. The proof of the theorem is constructive. For each r N we give an explicit example of an algebra and embed K( ) in an orientable surface of genus r. This gives an upper bound for γ(k( )). Then we use general results from graph theory to show that the bound is sharp. This will be done in Section 3. In Section we recall some basic facts about tilting modules and embeddings of graphs. In Section we introduce the algebras and derive some properties of K( ). For unexplained terminology and results from representation theory we refer to [], and from graph theory to [8]. Acknowledgement. Part of this work was done while the first author was visiting the Centre of Advanced Studies in Oslo. She wants to thank the staff of the Centre for the warm hospitality and the possibility to enjoy the stimulating atmosphere at the Centre.. Preliminaries.. The construction of K(Λ) Let Λ M be a direct summand of a tilting module. A basic Λ-module Λ X is called a complement to Λ M if Λ M Λ X is a tilting module and if add M add X = 0. It was proved in [5] that every direct summand of a tilting module has a distinguished complement Λ X which is characterized by the fact that there is no epimorphism Λ E Λ X with Λ E add Λ M. The module Λ X is unique up to isomorphism, and it is called the source complement to Λ M. There is the dual concept of a source complement. A complement Λ Y to Λ M is called a sink complement to a direct summand Λ M of a tilting module, if there is no monomorphism ΛY Λ E with Λ E add Λ M. In contrast to source complements, sink complements do not always exist. If Λ M has a sink complement then it is unique up to isomorphism [6]. The source and the sink complement to an almost complete tilting module Λ M coincide if and only if Λ M is not faithful [4]. The following result is basically contained in [4], compare [6]. Proposition. Let Λ M be a faithful almost complete tilting module. Let Λ X be a complement to Λ M which is not the sink complement to Λ M. Then () there is a complement Λ Y to Λ M which is not isomorphic to Λ X, () there is an exact sequence η : 0 Λ X Λ E Λ Y 0 with Λ E add Λ M,

L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules 47 (3) Ext i Λ(X, Y ) = 0 for all i > 0, and Ext i Λ(Y, X) = 0 for all i >, (4) the module Λ Y is uniquely determined by the property (). We call η the sequence connecting the complements Λ X and Λ Y to Λ M. This result allows an alternative definition of the quiver K(Λ) which is more useful for calculations. The vertices are the elements from T (Λ) as above. There is an arrow Λ T Λ T in K(Λ) if Λ T = Λ M Λ X and Λ T = Λ M Λ Y where Λ X and Λ Y are indecomposable, and if there is an exact sequence 0 Λ X Λ E Λ Y 0 with Λ E add Λ M. If K(Λ) is finite, then it is connected. Then the definition of K(Λ) yields an algorithm to construct K(Λ). We write the tilting module Λ Λ as a direct sum of indecomposable modules n ΛΛ = ΛΛ i. Then Λ Λ i is the source complement to Λ Λ[i] = ΛΛ j. If Λ Λ i is not the i= j i sink complement to Λ Λ[i] we construct the exact sequence 0 Λ Λ i Λ E i Λ Y i 0 with ΛE i add Λ Λ[i] connecting the complements Λ Λ i and Λ Y i to Λ Λ[i]. In this way we construct all neighbors of Λ Λ. We now proceed analogously with the neighbors of Λ Λ and all vertices we constructed. Since K(Λ) is finite and connected and has no oriented cycles this algorithm stops when we constructed all basic tilting modules over Λ... Embeddings of graphs Let G be a connected, finite graph with p vertices and q edges. We think of G as embedded on a surface S. Then G forms a polyhedron of genus γ(g). From the Euler polyhedron formula Beinecke and Harary [3] deduce the following lower bound for γ(g) which we shall use in Section 3. Proposition. If G is connected and has no triangles, then γ(g) q (p ). 4 In general this bound is not sharp. As an example we consider the following graph G which will become important in Section 3. This graph has 8 vertices and 9 edges, hence the formula yields γ(g) 3 4.

48 L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules But G is not even planar, namely it contains the subgraph which is homeomorphic to 3 3 This graph is isomorphic to the complete bigraph K 3,3 : 3 3. Kuratowski s theorem [9] implies γ(g). Conversely, we draw G differently and shade some of its faces: 5 7 6 8 9 5 6 8 0 7 3 4 3 4 We push a cylinder through the lower cube, close it under the upper square, adjust the vertices and edges accordingly and obtain an embedding of G on a torus. To be precise, the

L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules 49 following figure shows an embedding of G on a torus: 3 5 7 4 7 5 6 8 0 9 3 8 6 4 The parallel dotted lines have to be identified. Hence γ(g) =.. The algebras ΛPSfrag r and replacements properties of.. The algebras Let Λ be the path algebra of the quiver : K( ) a α γ b c β d δ bound by the relation αβ = γδ. For all r > let be the path algebra of the quiver r : a α γ b c β d δ 3 r bound by the relations αβ = γδ and rad = 0, i.e. the composition of two consecutive arrows in r \ {a} is zero. The Auslander-Reiten quivers Γ Λr of are as follows: ΓΛ P b S c I b P a X P d Y S d P c S b I c

40 L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules and for r > ΓΛr P b S c I b P S P a X P d Y S d I d P c S b I c P S S 3 S r S r P 3 P r Here S x denotes the simple module corresponding to the vertex x, the module P x is the projective cover of S x and I x denotes the injective hull of I x. Moreover, X is the radical of P d = I a and Y = I a / soc I a, where soc I a is the socle of I a. For all i r we identify an indecomposable Λ i -module Λi M with the corresponding Λ j -module Λj M, j i, whose support is Λ i. With this identification Γ Λi is a full, convex subquiver of Γ Λj for all i < j r. We have gl dimλ i = i + for all i r, where gl dimλ denotes the global dimension of an algebra Λ. The simple module S d is the unique indecomposable module of projective dimension, the modules I d, S, S are the unique indecomposable modules of projective dimension 3, and for all 3 j r the module S j is the unique indecomposable module of projective dimension j +. These observations show: Remark. Let j r. A non projective indecomposable module Λj X lies in mod Λ j \ mod Λ j if and only if pd Λj X = j +... Properties of the quiver K( ) The following technical lemmas roughly describe the structure of the quiver K( ). Let r be an integer, r, and let i < j r. We decompose the projective module Λ j Λ j into Λj Λ j = Λj Λ i Λj P ij. Hence Λj P ij is the maximal direct summand of Λj Λ j with add Λj P ij add Λj Λ i = 0. Lemma. Let i < j r, and let Λi T and Λi T be tilting modules over Λ i. Then (a) Λj T Λj M is a tilting module over Λ j if and only if Λj M = Λj P ij. (b) Λj T Λj P ij Λj T Λj P ij is an arrow in K(Λ j ) if and only if Λi T Λi T is an arrow in K(Λ i ). Proof. (a) Since Λj P ij is projective, Ext k Λ j (P ij, T ) = 0 for all k > 0. Since no indecomposable direct summand of Λj T is a successor of an indecomposable direct summand of Λj P ij in the Auslander-Reiten quiver of Λ j, it follows that Ext k Λ j (T, P ij ) = 0 for all k > 0. Hence Λ j T Λj P ij is a tilting module. The module Λj P ij is the source and the sink complement to Λ j T, hence the unique complement. (b) There is an arrow Λj T Λj P ij Λj T Λj P ij if and only if Ext Λ j (T P ij, T P ij ) 0 and if Λj T Λj P ij and Λj T Λj P ij have a common direct summand which is an almost

L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules 4 complete tilting module. Equivalently, Ext Λ i (T, T ) 0 and Λi T and Λi T have a common direct summand which is an almost complete tilting module, hence if and only if there is an arrow Λi T Λi T in K(Λ i ). In particular, we may identify K(Λ i ) with the full convex subquiver of K( ), i < r, with vertices Λr T Λr P ir where Λi T are the tilting modules over Λ i. With this identification, the building blocks of K( ) are the subquivers K(Λ i ) \ K(Λ i ) of K( ) with i r. To simplify the notation we denote by K(Λ ) \ K(Λ 0 ) the subquiver K(Λ ) of K( ). The next lemma gives an algebraic description of the vertices in K(Λ i ) \ K(Λ i ) of K( ) with < i r. Lemma. For all < i r, the subquiver K(Λ i ) \ K(Λ i ) of K( ) has as vertices all tilting modules of projective dimension i +. Proof. With the previous lemma, Λr T K(Λ i ) if and only if Λr T = Λr T Λr P ir with Λi T a tilting module over Λ i. Using the lemma again, Λi T K(Λ i ) if and only if there is an indecomposable, non projective direct summand Λr X of Λr T with Λr X mod Λ i \ mod Λ i. With the remark in., this holds if and only if pd Λi X = i +. Next we study arrows in K( ) between vertices in different building blocks of K( ). Lemma 3. Let i < j r. Let Λr T K(Λ i ) \ K(Λ i ) and Λr T K(Λ j ) \ K(Λ j ) be tilting modules over. (a) There are no arrows Λr T Λr T in K( ). (b) If there is an arrow Λr T Λr T in K( ) then pd Λr T = i + and pd Λr T = i +. In particular, j = i +. Proof. (a) Assume there is an arrow Λr T Λr T in K( ) with Λr T K(Λ i ) \ K(Λ i ) and T K(Λ j ) \ K(Λ j ) and i < j. Then Λr T = Λr T Λr P ir and Λr T = Λr T Λr P jr, where Λ i T and Λj T are tilting modules over Λ i respectively Λ j. Note that Λr P jr is a direct summand of Λr P ir. Then 0 Ext (T, T ) = Ext (T, T ) = Ext Λ j (T, T ). Since Λr T K(Λ j )\ K(Λ j ) there is an indecomposable direct summand Λj X mod Λ j \Λ j of Λj T with Ext Λ j (T, X) 0. This is a contradiction since Λj X is not a predecessor of an indecomposable direct summand of Λj T in Γ Λj. (b) Let Λr T Λr T in K( ) be an arrow in K( ) with Λr T K(Λ i ) \ K(Λ i ) and T K(Λ j ) \ K(Λ j ). Let η : 0 Λr X Λr E Λr Y 0 be the corresponding sequence connecting the complements Λr X and Λr Y, where Λr T = Λr X Λr M and Λr T = Λr Y Λr M. Since Λr T K(Λ j ) \ K(Λ j ) it follows that Λr Y mod Λ j \ mod Λ j, hence pd Λr Y = j +. Let Λr Z mod with Ext j+ (Y, Z) 0. We apply Hom Λr (, Z) to η and obtain pd Λr X = j. Since Λr T K(Λ i ) \ K(Λ i ) and i j we get j = i +, the assertion. As a consequence we obtain

4 L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules Lemma 4. Let r >. (a) There is an arrow Λr T Λr T in K( ) with Λr T K(Λ ) and Λr T K(Λ ) \ K(Λ ) if and only if Λr T = Λr S d Λr M and Λr T = Λr I d Λr M. (b) Let 3 i r. There is an arrow Λr T Λr T in K( ) with Λr T K(Λ i ) and T K(Λ i ) \ K(Λ i ) if and only if Λr T = Λr S i Λr M and Λr T = Λr S i Λr M. Proof. (a) Let Λr T Λr T be an arrow in K( ) with Λr T K(Λ ) and Λr T K(Λ )\ K(Λ ). Then pd Λr T = 3 with Lemma and pd Λr T =. Then Λr S d is a direct summand of T. Moreover, Lemma shows that Λr P Λr P is a direct summand of Λr T. Hence the sequence connecting the complements is the Auslander-Reiten sequence, which implies that Λr T = Λr S d Λr M and Λr T = Λr I d Λr M. Conversely, if Λr T = Λr S d Λr M and T = Λr I d Λr M, the Auslander-Reiten sequence starting in Λr S d lies in add( Λr T Λr T ). Hence we obtain an arrow Λr T Λr T in K( ) with Λr T K(Λ ) and Λr T K(Λ )\ K(Λ ). (b) Let 3 i r and let Λr T Λr T be an arrow in K( ) with Λr T K(Λ i ) and T K(Λ i ) \ K(Λ i ). Then pd Λr T = i + and pd Λr T = i. Since < i it follows that S i is a direct summand of of Λr T and Λr S i is a direct summand of of Λr T. Conversely, if T = Λr S i Λr M and Λr T = Λr S i Λr M then the Auslander-Reiten sequence starting in S i lies in add( Λr T Λr T ). This yields an arrow Λr T Λr T in K( ) with Λr T K(Λ i ) and Λr T K(Λ i ) \ K(Λ i ). To summarize our observations in this section we obtain the following structure of K( ): K(Λ ) K(Λ ). K(Λ )\ K(Λ ). K(Λ 3)\ K(Λ )..... K(Λ r)\ K( ) K(Λ 3) There are arrows from vertices in K(Λ i ) \ K(Λ i ) to vertices in K(Λ j ) \ K(Λ j ) if and only if j = i +. 3. The proof of the theorem 3.. An embedding of K( ) We use induction on r to embed K( ) on a surface of genus r.

L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules 43 Let r =. Direct calculation shows that K(Λ ) equals X X 3 X 5 X 7 X4 X7 X5 X 6 X 8 X0 X X 9 X X3 X 8 X 6 X X 4 where the parallel dotted lines have to be identified. We saw in. that the underlying graph K(Λ )) of K(Λ ) has genus, hence can be embedded on a torus T. The vertices of K(Λ ) are the tilting modules X = P a P b P c P d, X = P a P c I c P d, X3 = P a P b I b P d, X4 = P a I b I c P d, X5 = P b P c X P d, X6 = P c S c X P d, X7 = P b S b X P d, X8 = S b S c X P d, X9 = P c I c S c P d, X0 = S b S c Y P d, X = S c I c Y P d, X = S b I b Y P d, X3 = I b I c Y P d, X4 = P b I b S b P d, X5 = P b P c S d P d, X6 = P c I c S d P d, X7 = P b I b S d P d, X8 = I b I c S d P d. Let r =. The quiver K(Λ ) is the full convex subquiver of K(Λ ) with vertices Λ Xi = Λ Xi Λ P, where Λ P = Λ P Λ P. The quiver K(Λ )\ K(Λ ) is the full convex subquiver of K(Λ ) with vertices the tilting modules of projective dimension 3. Direct calculations show that K(Λ ) \ PSfrag K(Λ ) replacements is Y Y Y3 Y4 Y Y5 Y6 Y7 Y8 Y 5 Y9 Y0 Y Y Y9 Y3 Y4 Y5 Y6 Y3 Y Y Y 3 Y 4 Y 5

44 L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules where we identify along the parallel horizontal, respectively vertical lines. It follows that K(Λ ) \ K(Λ ) can be embedded on a torus T. The vertices of K(Λ ) \ K(Λ ) are the tilting modules Y = P d I d I c I b P S, Y = P d I d I c P c P S, Y3 = P d I d I c P c P P, Y4 = P d I d I b I c P P, Y5 = P d I d I c I b S S, Y6 = P d I d I c P c S S, Y7 = P d I d I c P c S P, Y8 = P d I d I c I b S P, Y9 = P d I d P b I b S S, Y0 = P d I d P b P c S S, Y = P d I d P b P c S P, Y = P d I d P b I b S P, Y3 = P d I d P b I b P S, Y4 = P d I d P b P c P S, Y5 = P d I d P b P c P P, Y6 = P d I d P b I b P P. X The subquivers PSfrag 7 X5 replacements Q : and Q : Y 6 Y5 X8 X6 Y4 Y3 bound squares on T respectively T. In K(Λ ) they are joint as follows: Y 3 Y 6 Y3 Y 5 α α α 4 X 5 X 7 α 3 X 8 X 6 We cut out the interiors of Q on T and Q on T and insert a cylinder connecting T and T. We obtain a surface of genus on which K(Λ ) can be embedded: T Y6 Y3 X7 X 8 Y5 Y3 X 5 X 6 T

L. Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules 45 Let r >. We abbreviate PSfrag the replacements injective -module by Λr P d Λr I d by Λr I and the projectiveinjective -module r P i by Λr P r. Direct calculations show that K( ) \ K( ) is i=3 Z 5 Z 6 Z 7 Z 8 Z Z Z 3 Z 4 with Z = I P r S r I c P c S, Z = I P r S r P b I c S, Z 3 = I P r S r P b P c P, Z 4 = I P r S r I c P c P, Z 5 = I P r S r I c I b S, Z 6 = I P r S r P b I b P Z 7 = I P r S r P b I b P, Z 8 = I P r S r I c I b P. We assume by induction that K( ) is embedded on a surface S r of genus r such that a) Z 5 Z Q : and Z 4 Z 8 Q : b) or Z 6 Z Z 3 Z 7 Z 6 Z 7 Q 3 : and Z 5 Z 8 Q 4 : Z Z 3 Z Z 4 bound squares on S r. Here Z i, i 4, denotes the -module which we obtain when we replace the direct summand S r of Z i by S r and the direct summand P r by P,r. For r =, let P,r = 0. Note that this assumption is satisfied for r =. We embedded K(Λ ) on a surface S of genus and the subquivers Z 5 = Y5 Z = Y6 Z 6 = Y 9 Z = Y 0 and Z 4 = Y7 Z 8 = Y8 Z 3 = Y Z 7 = Y bound squares on S. The quiver K( ) is the full convex subquiver of K( ) with vertices the tilting modules over of the form Λr T Λr P r, where Λr T is a tilting module over. Note that there is an arrow Λr Z i = Λr Z i Λr P r Λr Z i in K( ).

46 PSfragL. replacements Unger, M. Ungruhe: On the Genus of the Graph of Tilting Modules Let us assume first that we are in the situation (a). We cut out the interiors of the squares Q and Q and insert a handle Z 5 6 Z 7 8 Z Z Z 3 On this handle we embed K( ) \ K( ) and the arrows joining i and Z i: 6 5 Z 5 Z 6 Z 7 Z Z Z 4 Z Z 3 Z Z 3 This yields an embedding of K( ) on a surface S r of genus r and the squares Z 8 7 4 8 4 Z 6 Z 7 and Z 5 Z 8 Z Z 3 Z Z 4 bound squares on S r. We proceed analogously in case (b), and it follows that γ(k( )) r. 3.. A lower bound for γ(k( )) If r =, then γ(k(λ )) = as it was shown in.. Hence we may assume that r >. Consider K( ) \ K(Λ ). We embedded this quiver on a surface of genus r, hence γ(k( ) \ K(Λ )) r. The graph K(Λ ) \ K(Λ ) has 6 vertices and 3 edges. For all i r, the graphs K(Λ i ) \ K(Λ i ) have 8 vertices and edges. Moreover, there are 8 edges joining vertices in K(Λ i ) \ K(Λ i ) with vertices in K(Λ i+ ) \ K(Λ i ), < i < r. Hence K( )\K(Λ ) has p = 6+8(r ) vertices and q = 3+0(r ) edges. Since K( )\K(Λ ) has no triangles we may use the formula in. which gives γ(k( )\K(Λ ) q (p ) = 4 r, hence γ(k( ) \ K(Λ )) = r. We saw above that there are 4 arrows α, α, α 3, α 4 joining vertices in K(Λ ) with vertices in K(Λ ) \ K(Λ ). Let K( ) be the subquiver of K( ) which we obtain by deleting three of these arrows, say α, α 3, α 4. Then γ(k( )) γ(k( ) ). The blocks of K( ), i.e. the maximal connected subgraphs of K( ) which are connected, non trivial and have no α cutpoints are K(Λ ), and K(Λr ) \ K(Λ ). Since the genus of a graph is the sum of the genera of its blocks [], we obtain that γ(k( )) γ(k( ) ) = γ(k(λ )) + γ( ) + γ(k( ) \ K(Λ )) = + 0 + r. Hence γ(k( )) = r. To finish the proof of the theorem we have to show that there is an algebra Λ 0 with γ(k(λ 0 )) = 0. If Λ 0 is the ground field, then K(Λ 0 ) consists of a single vertex, hence it has genus 0.

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