Coloured quiver mutation for higher cluster categories

Transcription

1 Advances n Mathematcs 222 (2009) wwwelsevercom/locate/am Coloured quver mutaton for hgher cluster categores Asla Bae Buan a,, Hugh Thomas b a Insttutt for Matematse Fag, Norges Tens-Naturvtensapelge Unverstet, Alfred Getz ve 1, N-7491 Trondhem, Norway b Department of Mathematcs and Statstcs, Unversty of New Brunswc, Fredercton NB, E3B 1J4, Canada Receved 20 September 2008; accepted 12 May 2009 Avalable onlne 6 June 2009 Communcated by Mchel Van den Bergh Abstract We defne mutaton on coloured quvers assocated to tltng obects n hgher cluster categores We show that ths operaton s compatble wth the mutaton operaton on the tltng obects Ths gves a combnatoral approach to tltng n hgher cluster categores and especally an algorthm to determne the Gabrel quvers of tltng obects n such categores 2009 Elsever Inc All rghts reserved Keywords: Cluster algebras; Quver mutaton; Cluster categores; Calab Yau categores; Tltng 0 Introducton A cluster category s a certan 2-Calab Yau orbt category of the derved category of a heredtary abelan category Cluster categores were ntroduced n [5] n order to gve a categorcal model for the combnatorcs of Fomn Zelevnsy cluster algebras [10] They are trangulated [15] and admt (cluster-)tltng obects, whch model the clusters of a correspondng (acyclc) cluster algebra [7] Each cluster n a fxed cluster algebra comes together wth a fnte quver, and n the categorcal model ths quver s n fact the Gabrel quver of the correspondng tltng obect T [6], e the quver Q wth the ndecomposable drect summands of T as vertces and rreducble maps n the addtve closure add T as arrows Both authors were supported by a STORFORSK-grant from the Norwegan Research Councl; the second author was also supported by an NSERC Dscovery Grant * Correspondng author E-mal addresses: aslab@mathntnuno (AB Buan), hthomas@unbca (H Thomas) /$ see front matter 2009 Elsever Inc All rghts reserved do:101016/am

2 972 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) A prncpal ngredent n the constructon of a cluster algebra s quver mutaton It controls the exchange procedure whch gves a rule for producng a new cluster varable and hence a new cluster from a gven cluster Exchange s modelled by cluster categores n the acyclc case [4] n terms of a mutaton rule for tltng obects, e a rule for replacng an ndecomposable drect summand of a tltng obect wth another ndecomposable rgd obect, to get a new tltng obect Quver mutaton descrbes the relaton between the Gabrel quvers of the correspondng tltng obects Analogously to the defnton of the cluster category, for a postve nteger m, t s natural to defne a certan m + 1-Calab Yau orbt category of the derved category of a heredtary abelan category Ths s called the m-cluster category Implctly, m-cluster categores was frst studed n [15], and ther (cluster-)tltng obects have been studed n [1,8,12 14,16 21] Combnatoral descrptons of m-cluster categores n Dynn type A n and D n are gven n [2,3] In cluster categores the mutaton rule for tltng obects s descrbed n terms of certan trangles called exchange trangles By [14] the exstence of exchange trangles generalses to m-cluster categores It was shown n [19,21] that there are exactly m + 1 non-somorphc complements to an almost complete tltng obect, and that they are determned by the m+1 exchange trangles defned n [14] The am of ths paper s to gve a combnatoral descrpton of mutaton n m-cluster categores Apror, one mght expect to be able to do ths by eepng trac of the Gabrel quvers of the tltng obects However, t s easy to see that the Gabrel quvers do not contan enough nformaton We proceed to assocate to a tltng obect a quver each of whose arrows has an assocated colour c {0,,m} The arrows wth colour 0 form the Gabrel quver of the tltng obect We then defne a mutaton operaton on coloured quvers and show that t s compatble wth mutaton of tltng obects A consequence s that the effect of an arbtrary sequence of mutatons on a tltng obect n an m-cluster category can be calculated by a purely combnatoral procedure Our defnton of a coloured quver assocated to a tltng obect maes sense n any m + 1- Calab Yau category, such as for example those studed n [14] We hope that our constructons may shed some lght on mutaton of tltng obects n ths more general settng In Secton 1, we revew some elementary facts about hgher cluster categores In Secton 2, we explan how to defne the coloured quver of a tltng obect, we defne coloured quver mutaton, and we state our man theorem In Sectons 3 and 4, we state some further lemmas about hgher cluster categores, and we prove certan propertes of the coloured quvers of tltng obects We prove our man result n Sectons 5 and 6 In Sectons 7 and 8 we pont out some applcatons In Secton 9 we nterpret our constructon n terms of m-cluster complexes In Secton 10, we gve an alternatve algorthm for computng coloured quver mutaton The descrpton of ths algorthm maes t clear that coloured quver mutaton s a very natural generalsaton of ordnary quver mutaton Secton 11 dscusses the example of m-cluster categores of Dynn type A n,usngthe model developed by Baur and Marsh [3] 1 Hgher cluster categores Let K be an algebracally closed feld, and let Γ be a fnte acyclc quver wth n = n Γ vertces Then the path algebra H = KΓ s a heredtary fnte-dmensonal basc K-algebra Let mod H be the category of fnte-dmensonal left H -modules Let D = D b (H ) be the bounded derved category of H, and let [] be the th shft functor on D Weletτ denote the

3 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Auslander Reten translate, whch s an autoequvalence on D such that we have a bfunctoral somorphsm n D Hom ( A,B[1] ) D Hom(B, τa) (1) where D = Hom K (,K)s the dualty functor In other words ν =[1]τ s a Serre functor Let G = τ 1 [m] Them-cluster category s the orbt category C = C m = D/τ 1 [m] The obects n C are the obects n D, and two obects X, Y are somorphc n C f and only f X G Y n D The maps are gven by Hom Cm (X, Y ) = Z Hom D(X, G Y) By [15], the category C s trangulated and the canoncal functor D C s a trangle functor We denote therefore by [1] the suspenson n C Them-cluster category s also Krull Schmdt and has an AR-translate τ nherted from D, such that the formula (1) stll holds n C It follows that ν =[1]τ s a Serre functor for C and that C s m + 1-Calab Yau, snce ν [m + 1] The ndecomposable obects n D are of the form M[], where M s an ndecomposable H -module and Z We can choose a fundamental doman for the acton of G = τ 1 [m] on D, consstng of the ndecomposable obects M[] wth 0 m 1, together wth the obects M[m] wth M an ndecomposable proectve H -module Then each ndecomposable obect n C s somorphc to exactly one of the ndecomposables n ths fundamental doman We say that M[d] has degree d, denoted δ(m[d]) = d Furthermore, for an arbtrary obect X = X n C m, we let Δ d (X) = X [ d] be the H -module whch s the (shfted) drect sum of all drect summands X of X wth δ(x ) = d In the followng theorem the equvalence between () and () s shown n [19,21] and the equvalence between () and () s shown n [20] Theorem 11 Let T be an obect n C satsfyng Hom C (T, T []) = 0 for = 1,,m Then the followng are equvalent () If Hom C (T, U[]) = 0 for = 1,,mthen U s n add T () If Hom C (U T,U[]) = 0 for = 1,,mthen U s n add T () T has n ndecomposable drect summands, up to somorphsm Here add T denotes the addtve closure of T A (cluster-)tltng obect T n an m-cluster s an obect satsfyng the condtons of the above theorem For a tltng obect T = v =1 T, wth each T ndecomposable, and T an ndecomposable drect summand, we call T = T/T an almost complete tltng obect We let Irr A (X, Y ) denote the K-space of rreducble maps X Y n a Krull Schmdt K-category A The followng crucal result s proved n [21] and [19] Proposton 12 There are, up to somorphsm, m + 1 complements of an almost complete tltng obect Let T be an ndecomposable drect summand of an m-cluster tltng obect T = T T The complements of T are denoted T (c) for c = 0, 1,,m, where T = T (0) By [14], there are m + 1 exchange trangles T (c) f (c) B (c) g (c+1) T (c+1) h (c+1)

4 974 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Here the B (c) are n add(t /T ) and the maps f (c) (resp g (c) ) are mnmal left (resp rght) add(t /T )-approxmatons, and hence not splt mono or splt ep Note that by mnmalty, the maps f (c) and g (c) have no proper zero drect summands 2 Coloured quver mutaton We frst recall the defnton of quver mutaton, formulated n [10] n terms of sew-symmetrc matrces Let Q = (q ) be a quver wth vertces 1,,n and wth no loops or orented twocycles, where q denotes the number of arrows from to Let beavertexnq Then, a new quver μ (Q) = Q = ( q ) s defned by the followng data { q f = or = q = max{0,q q + q q q q } f It s easly verfed that ths defnton s equvalent to the one of Fomn Zelevnsy Now we consder coloured quvers Let m be a postve nteger An m-coloured (mult-)quver Q conssts of vertces 1,,n and coloured arrows (c), where c {0, 1,,m} Letq (c) denote the number of arrows from to of colour (c) We wll consder coloured quvers wth the followng addtonal condtons (I) No loops: q (c) = 0 for all c (II) Monochromatcty: If q (c) 0, then q (c ) (III) Sew-symmetry: q (c) = q (m c) = 0forc c We wll defne an operaton on a coloured quver Q satsfyng the above condtons Let be avertexnq and let μ (Q) = Q be the coloured quver defned by q (c) = q (c+1) q (c 1) max{0,q (c) t c q(t) + (q(c) q (c 1) )q (0) + q(m) (q (c) q(c+1) f = f = )} f In an m-cluster category C, for every tltng obect T = n =1 T, wth the T ndecomposable, we wll defne a correspondng m-coloured quver Q T, as follows Let T,T be two non-somorphc ndecomposable drect summands of the m-cluster tltng obect T and let r (c) of T to have vertces correspondng to ndecomposable drect summands T, and q (c) denote the multplcty of T n B (c) We defne the m-coloured quver Q T = r (c) Note, n partcular, that the (0)-coloured arrows are the arrows from the Gabrel quver for the endomorphsm rng of T By defnton, Q T satsfes condton (I) We show n Secton 3 that (II) s satsfed (ths also follows from [21]), and n Secton 4 that (III) s also satsfed The am of ths paper s to prove the followng theorem, whch s a generalsaton of the man result of [4]

5 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Theorem 21 Let T = n =1 T and T = T/T T (1) be m-tltng obects, where there s an exchange trangle T B (0) T (1) Then Q T = μ (Q T ) In the case m = 1 the coloured quver of a tltng obect T s gven by q (0) = q and q (1) = q where q denotes the number of arrows n the Gabrel quver of T Then coloured mutaton of the coloured quver corresponds to FZ-mutaton of the Gabrel quver Example A 3, m = 2 Let Γ be the quver wth underlyng graph A 3 and wth the followng orentaton The AR-quver of the 2-cluster category of H = KΓ s P 1 I 3 P 3 [1] I 1 [1] P 1 [2] P 2 [2] P 2 I 2 P 2 [1] I 2 [1] P 2 [2] P 3 I 1 P 1 [1] I 3 [1] P 3 [2] The drect sum T = I 1 I 2 P 3 [1] of the encrcled ndecomposable obects gves a tltng obect Its coloured quver s Now consder the exchange trangle I 1 (0) (2) I 2 (0) (2) P 3 [1] I 2 P 3 [1] I 3 [1] and the new tltng obect T = I 1 I 3 [1] P 3 [1] The coloured quver of T s I 1 (1) (1) I 3 [1] (2) (0) P 3 [1] (0) (2) 3 Further bacground on hgher cluster categores In ths secton we summarse some further nown results about m-cluster categores Most of these are from [20] and [21] We nclude some proofs for the convenence of the reader Tltng obects n C = C m gve rse to partal tltng modules n mod H, where a partal tltng module M n mod H, s a module wth Ext 1 H (M, M) = 0

6 976 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Lemma 31 (a) When T s a tltng obect n C m, then each Δ d (T ) s a partal tltng module n mod H (b) The endomorphsm rng of a partal tltng module has no orented cycles n ts ordnary quver Proof (a) s obvous from the defnton See [11, Corollary 42] for (b) In the followng note that degrees of obects, as defned n Secton 1, are always consdered wth a fxed choce of fundamental doman, and sums and dfferences of degrees are always computed modulo m + 1 Lemma 32 (See [20,21]) Assume m>1 (a) End(X) K for any ndecomposable obect X satsfyng Ext 1 (X, X) = 0 (b) We have that δ ( = 1 f δ(t T (c+1) ) ( (c)) (c) ) = m δ T 1 f δ(t (c) )/ {m 1,m} 2 f δ(t (c) ) = m 1 (c) The dstrbuton of degrees of complements s one of the followng there s exactly one complement of each degree, or there s no complement of degree m, two complements n one degree d m, and exactly one complement n all degrees d,m (d) If Hom(T (c),t (c ) ) 0, then c {c,c + 1,c+ 2} (e) For t {1,,m} we have Hom ( T (c),t (c ) [t] ) { K f c = c + t = 0 (mod m + 1) 0 else Proof (a) follows from the fact that Hom H (X, X) = K for ndecomposable H -modules satsfyng Ext 1 H (X, X) = 0, and the defnton of maps n an m-cluster category f (c) (b) follows from the fact that Hom(T (c+1) are not splt mono and (c) follows from (b),t (c) [1]) 0, snce n the exchange trangles, the Consderng the two dfferent possble dstrbutons of complements, we obtan from (c) that f m 3 and c c + 3 and c c 1, then Hom(T (c),t (c ) ) = 0 Consder the case c = c 1 We can assume m>2, snce else the statement s vod Hence we can clearly assume that δ(t (c) ) = δ(t (c 1) ) There s an exchange trangle nduced from an exact sequence n mod H, T (c 1) B (c 1) T (c) T (c 1) [1] It s clear that Hom(T (c 1) [1],T (c 1) ) = 0, snce m > 2 We clam that also Hom(B (c 1),T (c 1) ) = 0 Ths holds snce B (c 1) T (c 1) s a partal tltng obect n H, and so there are no cycles n the endomorphsm rng, by Lemma 31 Hence also Hom(T (c),t (c 1) ) = 0 follows, and ths fnshes the proof for (d)

7 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) For (e) we frst apply Hom(T (c+1),)to the exchange trangle T (c) B (c) T (c+1) and consder the correspondng long-exact sequence, to obtan that Hom ( T (c+1),t (c) [t] ) { K f t = 1 = 0 f t = 0ort {2,,m} Now consder Hom(T (c+u) Hom ( T (c+u),t (c),t (c) [v]) When 0 <v u m, we have that [v] ) Hom ( T (c+u+1) Hom ( T (c 1) When m v>u 0, we have that Hom ( T (c+u),t (c),t (c) [v + 1] ),T (c) [v] ) Hom ( T (c+u 1) [v + m u] ) Hom ( T (c),t (c 1) [1 + u v] ),T (c) [v 1] ) Hom ( T (c),t (c) [v u] ) Combnng these facts, (e) follows Lemma 33 The followng statements are equvalent (a) Hom(T (1),T (1) [1]) = 0 (b) T s not a drect summand of B (m) (c) T s not a drect summand of B (0) Furthermore, Hom(T (c),t (1) [1]) = 0 for c 1 Proof Note that r (0) exact sequence Hom ( T (c),t (0) = r (m) [1] ) Hom ( T (c) = dm Irr add T (T,T ), so (b) and (c) are equvalent Consder the,b (0) [1] ) Hom ( T (c),t (1) comng from applyng Hom(T (c),)to the exchange trangle T (0) B (0) T (1) [1] ) Hom ( T (c),t (0) [2] ) The frst and fourth terms are always zero Usng 32(e) we get that the second term (and hence the thrd) s non-zero f and only f c = 1 and T s a drect summand of B (0) Lemma 34 (See [14,21]) For 0 l m, the composton γ (v,l) = h (v) h (v 1) [1] h (v 2) s non-zero and a bass for Hom(T (v),t (v l) [2] h (v l+1) [l 1] : T (v) [l]) T (v l) [l]

8 978 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Proof For m = 1, see [5] Assume m 2 For the frst clam see [14], whle the second clam then follows from Lemma 32(e) We nclude an ndependent proof of the followng crucal property Proposton 35 (See [21]) B (u) u v and B (v) have no common non-zero drect summands whenever Proof When m = 1, ths s proved n [4] Assume m>1 We consder two cases, u v =1or u v > 1 Consder frst the case u v =1 Wthout loss of generalty we can assume u = 0 and v = 1, ) = 0 Assume that there exsts a (non-zero) ndecomposable T x, whch s a drect summand of B (0) and n B (1) We have that δ(t (1) ) {0, 1} by Lemma 32(b) Assume frst ) = 0 Then the exchange trangle and that δ(t (0) δ(t (1) T (0) B (0) T (1) s nduced from the degree 0 part of the derved category, and hence from an exact sequence n mod H Then the endomorphsm rng of the partal tltng module T x T (1) has a cycle, whch s a contradcton to Lemma 31 Assume now that δ(t (1) ) = 1 Then δ(t (2) ) {0, 1, 2}, where 0 can only occur f m = 2 If δ(t (2) ) {1, 2}, then clearly δ(t x ) = 1, and hence the partal tltng module T (1) T x contans a cycle, whch s a contradcton Assume that δ(t (2) ) = 0 (and hence m = 2) Then δ(t x ) {0, 1} Ifδ(T x ) = 1, we get a contradcton as n the prevous case If δ(t x ) = 0, consder the exchange trangle T (2) B (2) T (0) whch s nduced from an exact sequence n mod H Hence there s a non-zero map T x B (2) obtaned by composng T x T (2) wth the monomorphsm T (2) B (2), and thus there are cycles n the endomorphsm rng of the partal tltng module T x B (2) T (0), a contradcton Ths fnshes the case wth u v =1 Assume now that u v > 1 Then we have m>2 Snce Hom(T (v),t x ) 0 and Hom(T (u),t x ) 0, we have that the degrees of T (u) and T (v) ether dffer by at most 1, or are equal to m 1 and 0 In both cases, by Lemma 32(c), we have v u 2 So wthout loss of generalty we can assume v = u 2 Assume that δ(t (u) ) = 0 Then δ(t (v) ) = m 1 usng Lemma 32(c) and the fact that Hom(T (v),t x ) 0 Then also δ(t (v) ) m But Hom(T x,t (v+1) ) 0, so δ(t x ) m, contradctng the fact that Hom(T (u),t x ) 0 Corollary 36 Q T satsfes condton (II) 4 Symmetry Let T = T T T be a tltng obect In ths secton we show that the coloured quver Q T satsfes condton (III)

9 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Proposton 41 Wth the notaton of the prevous secton, we have r (c) = r (m c) Proof By the remar at the begnnng of the proof of Lemma 33 we only need to consder the case c/ {0,m} It s enough to show that r (c) r (m c) We frst prove Lemma 42 Let α: T (c) α[ c] γ (0,c) [ c] : T (c) T be rreducble n add((t /T ) T (c) ) Then the composton s non-zero [ c] T (m c+1) Proof We have already assumed c 0 Assume s zero Ths means that T (c) α[ c] h (0) [ c] : T (c) [ c] T (0) T must factor through B (m) [ c] T (m) [ c + 1] drect summand of B (c), we have that T s not a drect summand of B (0) = 0, we have that T s not a drect summand of B (m) ), a contradcton So α[ c] h (0) by Proposton 35 Snce Ths means that α s not rre- [ c] : T (c) [ c] T (m) [ c + 1] r (m) = r (0) ducble n add((t /T ) T (c) s non-zero Assume c>1 If the composton α[ c] h (0) h (0) [ c] factors through We clam that Hom(T (c) [ c],b (m 1) f T s not a drect summand of B (m 1) B (m 1) [ c + 1] T (m) [ c + 1] g (0) T Snce T s by assumpton a [ c] h (m) [ c + 1] s zero, then α[ c] [ c + 1]) Hom(T (c) In addton we have that Hom(T (c) c>1, usng Lemma 32(e) Ths s a contradcton, and ths argument can clearly be terated to see that α[ c] γ (0,c) [ c] : T (c) [ c] T (m c+1) s non-zero, usng Lemma 32(e),B (m 1) [1]) = 0 Ths clearly holds,t [1]) = 0 snce We now show that any rreducble map α : T (c) T gves rse to an rreducble map δ : T (m c) T Consder the composton B (c 1) Snce T s a drect summand of B (c) Thus, B (c 1) (c) [ c] g [ c] T (c) s n add T Snce Hom(X, T (m c+1) vanshes Usng the exchange trangle B (c 1) (c) [ c] T (m c+1) by assumpton, t s not a drect summand of B (c 1) [c]) = 0 for any X n add T, the composton (c) [ c] g [ c] T (c) [ c] h [ c] T (c 1) [ c + 1]

10 980 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) we see that α[ c] γ (0,c) T (c 1) [ c] : T (c) [ c] T (m c+1) [ c + 1], e there s a commutatve dagram factors through the map T (c) (c) [ c] h [ c] B (c 1) [ c] g (c) [ c] T (c) [ c] h (c) [ c] T (c 1) [ c + 1] φ 1 T (m c+1) Smlarly, usng the exchange trangle B (c 2) (c 1) (c 1) [ c + 1] g [ c+1] T (c 1) [ c + 1] h [ c+1] T (c 2) [ c + 2] we obtan a map φ 2 : T (c 2) [ c + 2] T (m c+1) Repeatng ths argument c tmes we obtan a map φ c : T T (m c+1) φ c = α[ c] γ (0,c), such that γ (c,c) [ c] T (c) [ c] T (m c+1) h (c) [ c] T (c 1) [ c + 1] φ 2 φ 1 h (c 1) [ c+1] T (c 2) [ c + 2] φ c h (c 2) [ c+2] T We clam that Lemma 43 There s a map β : T T (m c+1) such that β s rreducble n add((t /T ) T (m c+1) ), such that γ (c,c) [ c] β = α[ c] γ (0,c), and

11 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Proof Let be a mnmal left add(t (m c+1) add T (m c+1) ( T (m c+1) ) T T (ψ ψ ) Letφ c be as above, and factor t as T (ψ ψ ) T)-approxmaton, wth T n add T and (T (m c+1) ) n ( T (m c+1) ) T ( ɛ ɛ ) T (m c+1) Snce γ (c,c) factors through T (1) [ 1], we have that γ (c,c) [ c]ψ = 0, so we have γ (c,c) [ c](ψ ɛ + ψ ɛ ) = γ (c,c) [ c]ψ ɛ Hence, we let β = ψ ɛ and snce the drect summands n ɛ are somorphsms, t s clear that β s rreducble to T Then, by Lemma 42 the set {α t γ (0,c) } s also lnearly ndependent For each α t, consder the correspondng map β t, such that γ (c,c) [ c] β t = α t [ c] γ (0,c), and whch we by Lemma 43 can assume s rreducble Assume a non-trval lnear combnaton t β t s zero Then also t (γ (c,c) [ c] β t ) = t α t γ (0,c) = 0 But ths contradcts Lemma 42 snce t α t s rreducble Hence t follows that {β t } s also lnearly ndependent Hence, n the exchange trangle Next, assume {α t } s a bass for the space of rreducble maps from T (c) T (m c) B (m c) So, we have that r (c) T (m c+1) 5 Complements after mutaton, we have that T appears wth multplcty at least r (c) r (m c), and the proof of the proposton s complete n B (m c) In ths secton we show how mutaton n the vertex affects the complements of the almost complete tltng obect T/T Ths s crucal for the proof of our man theorem As before, let T = T T T be an m-tltng obect, and let T = T/T T (1) We need to consder T (e) T (d) T (c) for all possble values of c,d,e However, we have the followng restrcton on the colour of arrows Proposton 51 Assume q (e) > 0,q (0) > 0 and q(c) > 0 Then c {e,e + 1} Proof Consder the exchange trangle T (e) summand of the mddle term B (e) T X T (e+1) Note that T s a drect > 0 Consder also the exchange by the assumpton that q (e)

12 982 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) trangle T (c) T Z T (c+1) Pc an arbtrary non-zero map h : T T, and consder the map ( ) h 0 00 : T X T Z It suffces to show that whenever c/ {e,e + 1}, then h s not rreducble n add T So assume that c/ {e,e + 1} We clam that there s a commutatve dagram T (e) T (c) T X ( h 0) 00 T Z T (e+1) T (c+1) where the rows are the exchange trangles The composton T (e) T h T T (c+1) s zero snce fc e 1, Hom(T (e),t (c+1) ) = 0byusngc/ {e,e + 1} and Lemma 32(e), fc = e 1, there s no non-zero composton T (e) T T T (c+1) = T (e) Hence the leftmost vertcal map exsts, and then the rghtmost map exsts, usng that C s a trangulated category Then, snce Hom(T (e+1) [ 1],T (c) ) = 0 by Lemma 32(e), there s a map T X T (c), such that T (e) T (c) = T (e) T X T (c) Hence there s map T (e+1) T Z such that T X T Z = (T X T (c) T Z) + (T X T (e+1) T Z) By restrcton we get h : T T = ( T T (c) T ) + ( T T (e+1) T ) (2) Under the assumpton c/ {e,e + 1} we have that T (e+1) add((t /T ) T (e+1) ) Hence T (e+1) T = T (e+1) T cannot be rreducble n B (e+1) T, where T s not a d- Also, by Proposton 35 we have that T s not a drect summand of was rreducble n add((t /T ) T (c) ), then there would be an rreducble ), and snce c e + 1, ths does not hold, by Propos- rect summand of B (e+1) B (e+1) map T (c 1) IfT T (c) T n add((t /T ) T (c 1) ton 35 Hence, T T (c) = T B (c 1) Also by Proposton 35 we have that T s not a drect summand of B (c 1) that h : T T s not rreducble n add T T (c), where T s not a drect summand of B (c 1) By (2), ths shows Recall that T = (T /T ) T (1) For, let(t )(u) denote the complements of T /T, where there are exchange trangles ( T ) (u) ( B ) (u) ( T ) (u+1) We frst want to compare the complements (T )(u) of T /T wth the complements T (u) of T/T More precsely, we want to prove the followng proposton

13 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Proposton 52 (a) If q (u) = 0 for u = 0,,m, then (T )(v) T (v) for all v (b) If e m and q (e) > 0, then (T )(v) T (v) for v e + 1 For the proof we need the followng three lemmas Lemma 53 Assume that q (u) = 0 for u = m, 0, 1,,x 1 (a) For u = 0, 1,,x 1, the mnmal left add(t /T )-approxmaton T (u) add(t /T )-approxmaton (b) For u = 0, 1,,x, we have (T )(u) = T (u) B (u) s also an Proof By assumpton T s not a drect summand of any of the B (u) Assume there s a map T (u) T (1) and consder the dagram T (u+1) [ 1] T (u) B (u) T (1) Snce Hom(T (u+1),t (1) [1]) = 0 by Lemma 33, we see that the map T (u) T (1) factors through T (u) B (u) Hence the mnmal left add(t /T )-approxmaton T (u) B (u) s also an add(t /T )-approxmaton, so we have proved (a) Then (b) follows drectly Lemma 54 Assume that e m and there are exchange trangles and T (e) (T ) p X T (e+1) (3) T (T ) q Y T (1) (4) where p = q (e) > 0 and q = q (0) 0, eb(e) = (T ) p X and B (0) = (T ) q Y, where T s not somorphc to any drect summand of Y (a) The composton T (e) (T ) p X (T ) pq Y p X s a left add(t /T )-approxmaton (b) There s a trangle T (e) (T ) pq Y p X ( T ) (e+1) C wth C n add(t /(T T )) and T (e) = (T )(e) (c) There s a trangle T (e+1) (T )(e+1) C (T (1) ) p

14 984 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Proof Consder an arbtrary map f : T (e) U wth U n add(t /T ) We have that Hom(T (e+1),t (1) [1]) = 0, by Lemma 33 Hence, by applyng Hom(,U)to the trangle (3) we get that f factors through T (e) (T ) p X By applyng Hom(,U)to the trangle (4), and usng that Hom(T (1),T (1) [1]) = 0, we get that f factors through T (e) (T ) p X Y p (T ) pq X Ths proves (a) For (b) and (c) we use the exchange trangles (3) and (4) and the octahedral axom to obtan the commutatve dagram of trangles T (e) (T ) p X T (e+1) T (e) (T ) pq Y p X C ( T (1) ) p ( (1)) p T By (a) the map T (e) (T ) pq Y p X s a left add(t /T )-approxmaton, and by Lemma 53 we have that (T )(e) = T (e) Hence C = (T )(e+1) C, where C s n add((t ) pq Y p X) add(t /(T T )), and wth no copes somorphc to T n Y Note that the nduced add(t /T )-approxmaton s n general not mnmal Lemma 55 Assume e m and q (e) > 0 (a) Then there s a trangle ( ) T (e+1) C α B (e+1) ( T (1) ) p (e+2) T where α s a mnmal left add(t /T )-approxmaton, and C s as n Lemma 54 (b) There s an nduced exchange trangle ( ) T (e+1) B (e+1) (T (1) α(c ) ) p T (e+2) where α(c ) C (c) (T )(e+2) T (e+2) Proof Consder the exchange trangle T (e+2) [ 1] T (e+1) B (e+1) and the trangle from Lemma 54(b) T (e+1) ( T ) (e+1) C ( T (1) ) p (5)

15 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Apply the octahedral axom, to obtan the commutatve dagram of trangles T (e+2) [ 1] T (e+1) B (e+1) T (e+2) [ 1] (T )(e+1) C G ( T (1) ) p ( (1)) p T Snce T does not occur as a drect summand of B (e+1) by Proposton 35, we have that Hom(T (1),B (e+1) [1]) = 0 Hence the rghtmost trangle splts, so we have a trangle T (e+2) [ 1] ( T ) (e+1) C B (e+1) ( T (1) ) p (6) [1]) = 0 By Lemma 32(e) we get that Hom(T (e+2),t [1]) = 0, and clearly Hom(T (e+2),t l [1]) = 0, for l We hence get that all maps (T )(e+1) C U, wth U n add T, factor through (T )(e+1) C B (e+1) (T (1) ) p Mnmalty s clear from the trangle (6) Ths proves (a), and (b) follows from the fact that C contans no copes of T, and hence splts off (c) s a drect consequence of (b) By Lemma 33 we have that Hom(T (e+2),t (1) We are now n poston for provng Proposton 52 Proof of Proposton 52 (a) s a drect consequence of Lemma 53 For (b) note that by Lemmas 53 and 55 we have (T )(v) T (v) for v = 0,,e and v = e + 2 For v e + 2 consder the exchange trangles T (v) B (v) T (v+1) Snce Hom(T (v+1),t (1) [1]) = 0 by Lemma 33 and q (v) = 0, t s clear that the map T (v) s a left add T /T -approxmaton Hence (b) follows 6 Proof of the man result B (v) Ths secton contans the proof of the man result, Theorem 21 As before, let T = T T T be an m-tltng obect, and let T = T/T T (1) We wll compare the numbers of (c)-coloured arrows from to, n the coloured quvers of T and T, e we wll compare q (c) and q(c) We need to consder an arbtrary T whose coloured quver locally loos le T (e) T (d) T (c)

16 986 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) for any possble value of c,d,e Our am s to show that the formula q (u) = q (u+1) q (u 1) max{0,q (u) t u q(t) + (q(u) q (u 1) )q (0) + q(m) (q (u) q(u+1) f = f = )} f holds The case where = s drectly from the defnton The case where = follows by condton (II) for Q T For the rest of the proof we assume / {, } We wll dvde the proof nto four cases, where p 0 denotes the number of arrows from to, and q = q (0) I p = 0, II p 0, e m and q = 0, III p 0, e m and q 0, IV p 0 and e = m Note that n the three frst cases, the formula reduces to { q (u) = max 0,q (u) q (t) + ( q (u) t u q (u 1) } ) (0) q and n the frst two cases t further reduces to q (u) = q(u) Case I We frst consder the stuaton where there s no coloured arrow, eq (u) = 0for all u That s, we assume Q T locally loos le ths T T (d) T (c) wth c,d arbtrary It s a drect consequence of Proposton 52 that q (u) shows that the formula holds = q(u) Case II We consder the settng where we assume Q T locally loos le ths for all u whch T (e) T (d) T (c) wth e m and q = 0 We then clam that we have the followng, whch shows that the formula holds Lemma 61 In the above settng q (u) = q(u) for all u

17 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Proof It follows drectly from Proposton 52 that q (u) that q (e) = q(e) = q (u) for u = 0,,e 1 We clam By Lemma 54 we have the (not necessarly mnmal) left add(t /T )-approxmaton T (e) (T ) pq Y p X = Y p X Frst, assume that T does not appear as a drect summand of B (e) = (T ) p X, then the same holds for Y p X, and hence for (B )(e) whch s a drect summand of Y p X Next, assume T appears as a drect summand of B (e), and hence n X Then T s by Proposton 35 not a drect summand of B (e+1), and by Lemma 55 we have that T s also not a drect summand of C Therefore T appears wth the same multplcty n B (e) as n (B )(e),alson ths case We now show that q (u) = q(u) for u>e If q (e) 0, then q(u) = q (u) = 0foru>eand we are fnshed So assume q (e) = 0, e T does not appear as a drect summand of X Consder the map ( ) T (e+1) C B (e+1) ( T (1) ) p (T (1) ) p We have that (B )(e+1) B(e+1) C By assumpton, T s not a drect summand of (T ) pq Y p X = Y p X, and thus not n C From ths t follows that q (e+1) = q (e+1) Snce, by Proposton 52 we have for u = e + 2,,m, that (T )(u) = T (u) and the left add(t /T )-approxmaton concde wth the left add(t /T )-approxmatons of T (u),tnowfollows that q (u) = q(u) for all u Case III We now consder the settng wth p non-zero, q 0 and e m That s, we assume Q T locally loos le ths T (e) T (0) T (c) where c {e,e + 1} by Proposton 51, and where there are z = q (c) 0 arrows from T to T Lemma 62 In the above settng, we have that Q T s gven by q (e) = q (e+1) = q (e) q(0) + q(e) f c = e q (e) q(0) q(e+1) 0 otherwse { (e) q q(0) + q(e+1) 0 otherwse f c = e + 1 and q (e+1) f c = e + 1 and q (e+1) q (e) q(0) >q (e) q(0) (7) (8)

18 988 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) and q (u) = 0 for u/ {e,e + 1} (9) Proof We frst deal wth the case where c = e and z>0 By assumpton X n the trangle (3) has z copes of T,so(T ) pq Y p X has pq + z copes of T Hence to show (7) t s suffcent to show that C n the trangle T (e+1) ( T ) (e+1) C ( T (1) ) p has no copes of T Ths follows drectly from Lemma 55 and the fact that T (by the assumpton that z>0 and Proposton 35) s not a drect summand of B (e+1) In ths case (8) and (9) follow drectly from Proposton 35 Consder the case wth c = e + 1 and 0 z pq We have that X n the trangle (3) does not have T as a drect summand Assume T appears as a drect summand of C wth multplcty z We clam that z = z Assume frst z <z, then on one hand T appears wth multplcty z z > 0n(B )(e+1) On the other hand T appears wth multplcty pq z > 0n(B )(e) Ths contradcts Proposton 35 Hence z = z Therefore (B )(e) has pq z copes of T and (7) and (8) hold If pq z, then (9) follows drectly from the above and Proposton 35 In the case pq = z, we also need to show that T does not appear as a drect summand of (B )(u) for u e + 1 Snce pq 0, we have z 0, and the result follows from Proposton 52 Now assume c = e + 1 and z>pq Assume C = (T ) l C, where T s not a drect summand of C Now snce ( ) T (e+1) T l C ( T (1) ) p T z Y wth T not a drect summand of Y, s a mnmal left add T -approxmaton, we have that l pq < z and T appears wth multplcty z l>0 n the mnmal left add T /T -approxmaton of (T )(e+1), hence T cannot appear as a drect summand of the mnmal left add T /T - approxmaton of (T )(e) Hence l = pq, and we have completed the proof of (7) and (8) n ths case The case (9), e u e,e + 1 follows from Proposton 35 Case IV We now consder the case wth q (m) 0 Assume frst there are no arrows from to Then we can use the symmetry proved n Proposton 41 and reduce to Case I The formula s easly verfed n ths case Assume q (d) 0forsomed Ifd 0, we can agan use the symmetry, ths tme to reduce to Case III It s straghtforward to verfy that the formula holds also n ths case If d = 0 we need to consder the followng stuaton T (m) (0) T (0) (m) T (c) (m c)

19 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Now by Proposton 51 we have that c s n {m, 0} Assume there are z 0 (c)-coloured arrows from T to T The coloured quver of T s of the form T (0) (m) T (1) (m) (0) T (c ) (m c ) and applyng Proposton 51 we have that c {0,m} by Proposton 51 Hence for all u/ {0,m} we have that q (u) = q(u) = 0 Therefore t suffces to show that q(u) = q(u),foru {0,m} Ths s a drect consequence of the followng Lemma 63 Assume we are n the above settng A map T T or T T s rreducble n add T f and only f t s rreducble n add T Proof Assume T T s not rreducble n add T, and that T T = T U T for some U = t U t add T, wth U t the ndecomposable drect summands of U Note that by Lemma 32(a), we can assume that all T U t and all U t T are non-somorphsms If there s some ndex t such that U t T (1),themapU t T factors through some U n add(t /(T T )), snce there are no (1)-coloured arrows or n the coloured quver of T Ths shows that T T s not rreducble n add T Assume T T s not rreducble n add T, and that T T = T V T for some V = t V t add T, wth V t the ndecomposable drect summands of V If there s some ndex t such that V t T,themapT V t factors through B (m), whch s n add(t /(T T T )) add T, snce there are no (0)-coloured arrows or n the coloured quver of T Ths shows that T T s not rreducble n add T By symmetry, the same property holds for maps T T Thus we have proven that the formula holds n all four cases, and ths fnshes the proof of Theorem 21 7 m-cluster-tlted algebras An m-cluster-tlted algebra s an algebra gven as End C (T ) for some tltng obect T n an m-cluster category C = C m Obvously, the subquver of the coloured quver of T gven by the (0)-coloured maps s the Gabrel quver of End C (T ) An applcaton of our man theorem s that the quvers of the m-cluster-tlted algebras can be combnatorally determned va repeated (coloured) mutaton For ths one needs transtvty n the tltng graph of m-tltng obects More precsely, we need the followng, whch s also ponted out n [21] Proposton 71 Any m-tltng obect can be reached from any other m-tltng obect va terated mutaton

20 990 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Proof We setch a proof for the convenence of the reader Let T be a tltng obect n an m-cluster category C of the heredtary algebra H = KQ, and let C 1 be the 1-cluster category of H By [20], there s a tltng obect T of degree 0, e all drect summands of T have degree 0, such that T can be reached from T va mutaton It s suffcent to show that the canoncal tltng obect H can be reached from T va mutaton Snce T s of degree 0, t s nduced from an H -tltng module Especally T s a tltng obect n C 1 Snce T and H are tltng obects n C 1, by [5] there are C 1 -tltng obects T = T 0,T 1,,T r = H, such that T mutates to T +1 (n C 1 ) for = 0,,r 1 Now each T s nduced by a tltng module for some Q where all KQ are derved equvalent to KQ Hence, each T s easly seen to be an m-cluster tltng obect Snce T +1 dffers from T n only one summand the mutatons n C 1 are also mutatons n C Ths concludes the proof A drect consequence of the transtvty s the followng Corollary 72 For an m-cluster category C = C m of the acyclc quver Q, all quvers of m- cluster-tlted algebras are gven by repeated coloured mutaton of Q 8 Combnatoral computaton In ths secton, we dscuss concrete computaton wth tltng obects n an m-cluster tltng category An ndecomposable obect T n mod H s called exceptonal f Ext 1 H (T, T ) = 0 If T s exceptonal, then t s unquely determned by ts mage [T ] n the Grothendec group K 0 (mod H) There s a map from D b (mod H) to K 0 (mod H) whch, for T mod H, taes T [] to ( 1) [T ] An exceptonal ndecomposable n D b (mod H) can be unquely specfed by ts class n K 0 (mod H) together wth ts degree The map from D b (mod H)to K 0 (mod H)does not descend to C However, f we fx our usual choce of fundamental doman n D b (mod H), then we can dentfy the ndecomposable obects n t as above Let us defne the combnatoral data correspondng to a tltng obect T to be Q T together wth ([T ], deg T ) for 1 n Theorem 81 Gven the combnatoral data for a tltng obect T n C, t s possble to determne, by a purely combnatoral procedure, the combnatoral data for the tltng obect whch results from an arbtrary sequence of mutatons appled to T Proof Clearly, t suffces to show that, for any, we can determne the class and degree for T () If we can do that then, by the coloured mutaton procedure, we can determne the coloured quver for (T /T ) T (), and by applyng ths procedure repeatedly, we can calculate the result of an arbtrary sequence of mutatons Snce we are gven Q T, we now B (0), and we can calculate [B 0 ] Now we have the followng lemma: Lemma 82 [T (1) ]=[B (0) ] [T ], and deg(t (1) ) = deg T or deg T +1, whchever s consstent wth the sgn of the class of [T (1) ], unless ths yelds a non-proectve ndecomposable obect n degree m, or an ndecomposable of degree m + 1

21 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Proof The proof s mmedate from the exchange trangle T B (0) T (1) Applyng ths lemma, and supposng that we are not n the case where ts procedure fals, we can determne the class and degree T (1) By the coloured mutaton procedure, we can also determne the coloured quver for μ (T ) We therefore have all the necessary data to apply Lemma 82 agan Repeatedly applyng the lemma, there s some such that we can calculate the class and degree of T () for 1, and the procedure descrbed n the lemma fals to calculate T (+1) We also have the followng lemma: Lemma 83 [T (m) ]=[B (m) ] [T ], and deg T (m) = deg T or deg T 1, whchever s consstent wth the sgn of [T (m) ], unless ths yelds an ndecomposable n degree 1 Applyng ths lemma, startng agan wth T, we can obtan the degree and class for T (m) We can then determne the coloured quver for μ 1 (T ), and we are now n a poston to apply Lemma 83 agan The last complement whch Lemma 83 wll successfully determne s T (+1) It follows that we can determne the degree and class of any complement to T/T 9 The m-cluster complex In ths secton, we dscuss the applcaton of our results to the study of the m-cluster complex, a smplcal complex defned n [9] for a fnte root system Φ We shall begn by statng our results for the m-cluster complex n purely combnatoral language, and then brefly descrbe how they follow from the representaton-theoretc perspectve n the rest of the paper For smplcty, we restrct to the case where Φ s smply laced Number the vertces of the Dynn dagram for Φ from 1 to nthem-coloured almost postve roots, Φ 1 m, consst of m copes of the postve roots, numbered 1 to m, together wth a sngle copy of the negatve smple roots We refer to an element of the th copy of Φ + as havng colour, and we wrte such an element as β () Snce the Dynn dagram for Φ s a tree, t s bpartte; we fx a bpartton {1,,n}= I + I The m-cluster complex, Δ m, s a smplcal complex on the ground set Φ 1 m Its maxmal faces are called m-clusters The defnton of Δ m s combnatoral; we refer the reader to [9] The m-clusters each consst of n elements of Φ 1 m [9, Theorem 29] Every codmenson 1 face of Δ m s contaned n exactly m + 1 maxmal faces [9, Proposton 210] There s a certan combnatorally-defned becton R m : Φ 1 m Φm 1, whch taes faces of Δ m to faces of Δ m [9, Theorem 24] It wll be convenent to consder ordered m-clusters An ordered m-cluster s ust an n-tuple from Φ 1 m, the set of whose elements forms an m-cluster Wrte Σ m for the set of ordered m-clusters For each ordered m-cluster C = (C 1,,C n ), we wll defne a coloured quver Q C We wll also defne an operaton μ : Σ m Σ m, whch taes ordered m-clusters to ordered m-clusters, changng only the th element We wll defne both operatons nductvely The set Π of negatve smple roots forms an m-cluster Its assocated quver s defned by drawng, for each edge n the Dynn dagram, a par of arrows Suppose I + and I Then we draw an arrow from to wth colour 0, and an arrow from to wth colour m

22 992 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Suppose now that we have some ordered m-cluster C, together wth ts quver Q C We wll now proceed to defne μ (C) Wrte q (0) for the number of arrows n Q C of colour 0 from to Defne: β = C + q (0) C Let c be the colour of C We defne μ (C) by replacng C by some other element of Φ m 1, accordng to the followng rules: If C s postve and β s postve, replace C by β (c) If C s postve and β s negatve, replace C by R m ( β (c) ) If C s negatve smple α, defne γ by γ (0) = R m ( α ), and then replace C by β + C γ, wth colour zero Defne the quver for the m-cluster μ (C) by the coloured quver mutaton rule from Secton 2 Snce any m-cluster can be obtaned from Π by a sequence of mutatons, the above suffces to defne μ (C) and Q C for any ordered m-cluster C Proposton 91 The operaton μ defned above taes m-clusters to m-clusters, and the m- clusters μ (C) for 0 m are exactly those contanng all the C for The connecton between the combnatorcs dscussed here and the representaton theory n the rest of the paper s as follows Φ 1 m corresponds to the ndecomposable obects of (a fundamental doman for) C m The cluster tltng obects n C m correspond to the m-clusters The operaton R m corresponds to [1] For further detals on the translaton, the reader s referred to [18,20] The above proposton then follows from the approach taen n Secton 8 10 An alternatve algorthm for coloured mutaton Here we gve an alternatve descrpton of coloured quver mutaton at vertex (1) For each par of arrows (c) (0) wth, the arrow from to of arbtrary colour c, and the arrow from to of colour 0, add a par of arrows: an arrow from to of colour c, and one from to of colour m c (2) If the graph volates condton (II), because for some par of vertces and there are arrows from to whch have two dfferent colours, cancel the same number of arrows of each colour, untl condton (II) s satsfed (3) Add one to the colour of any arrow gong nto and subtract one from the colour of any arrow gong out of Proposton 101 The above algorthm s well-defned and correctly calculates coloured quver mutaton as prevously defned

23 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) Proof Fx a quver Q and a vertex at whch the mutaton s beng carred out To prove that the algorthm s well-defned, we must show that at step 2, there are only two colours of arrows runnng from to for any par of vertces, (Otherwse there would be more than one way to carry out the cancellaton procedure of step 2) Snce n the orgnal quver Q, there was only one colour of arrows from to, n order for ths problem to arse, we must have added two dfferent colours of arrows from to at step 1 Two colours of arrows wll only be added from to f, n Q, there are both (0)-coloured arrows from to and from to, and thus (1) apples twce In ths case, by condton (III), there are (m)-coloured arrows from to and from to It follows that n step 1, we wll add both (0)-coloured and (m)-coloured arrows Applyng Proposton 51, we see that any arrows from to n Q are of colour 0 or m Thus, as desred, after step 1, there are only two colours of arrows n the quver, so step 2 s well-defned We now prove correctness Let Q = μ (Q) Wrte q (c) for the number of c-coloured arrows from to n Q, and smlarly q (c) for Q Wrte ˆQ and ˆq (c) for the result of applyng the above algorthm It s clear that only the fnal step of the algorthm s relevant for ˆq where one of or concdes wth, and therefore that n ths case ˆq (c) = q (c) as desred Suppose now that nether nor concdes wth Suppose further that n Q there are no (0)-coloured arrows from ether or to, and therefore also no m-coloured arrows from to or In ths case, q (c) = q(c) In the algorthm, no arrows wll be added between and n step 1, and therefore no further changes wll be made n step 2 Thus ˆq (c) = q(c) = q(c), as desred Suppose now that there are (0)-coloured arrows from to both and In ths case, q (c) = q (c) In ths case, as dscussed n the proof of well-defnedness, an equal number of (0)-coloured and (m)-coloured arrows wll be ntroduced at step 1 They wll therefore be cancelled at step 2 Thus ˆq (c) = q(c) = q(c) as desred Suppose now that there s a (0)-coloured arrow from to, but not from to Letthe arrows from to, f any, be of colour c At step 1 of the algorthm, we wll add q (c) q(0) arrows of colour c to Q By Proposton 51, the arrows n Q from to are of colour c or c + 1 One verfes that the algorthm yelds the same result as coloured quver mutaton, n the three cases that the arrows from to n Q are of colour c, that they are of colour c + 1 but there are fewer than q (c) q(0), and that they are of colour c + 1 and there are at least as many as q(c) q(0) The fnal case, that there s a (0)-coloured arrow from to but not from to, s smlar to the prevous one 11 Example: type A n In [3], a certan category C BM s constructed, whch s shown to be equvalent to the m-cluster category of Dynn type A n The descrpton of C BM s as follows Tae an nm + 2-gon Π, wth vertces labelled clocwse from 1 to nm + 2 Consder the set X of dagonals γ of Π wth the property that γ dvdes Π nto two polygons each havng a number of sdes congruent to 2 modulo m For each γ X, there s an obect A γ n C BM These obects A γ form the ndecomposables of the addtve category C BM We shall not recall the exact defnton of the morphsms, other than to note that they are generated by the morphsms p : A A whch exst provded that and are both dagonals n X, and that, startng at and movng clocwse around Π, one reaches before

24 994 AB Buan, H Thomas / Advances n Mathematcs 222 (2009) A collecton of dagonals n X s called non-crossng f ts elements ntersect parwse only on the boundary of the polygon An ncluson-maxmal such collecton of dagonals dvdes Π nto m + 2-gons; we therefore refer to such a collecton of dagonals as an m + 2-angulaton If we remove one dagonal γ from an m + 2-angulaton Δ, then the two m + 2-gons on ether sde of γ become a sngle 2m + 2-gon We say that γ s a dameter of ths 2m + 2-gon, snce t connects vertces whch are dametrcally opposte (wth respect to the 2m + 2-gon) If δ s another dameter of ths 2m + 2-gon, then (Δ \ γ) δ s another maxmal non-crossng collecton of dagonals from X (In partcular, δ X) For Δ an m + 2-angulaton, let A Δ = γ Δ A γ Then we have that A Δ s a basc (m-cluster- )tltng obect for C BM, and all basc tltng obects of C BM arse n ths way It follows from the prevous dscusson that f T = A Δ s a basc tltng obect, and γ Δ, then the complements to A Δ\γ wll consst of the obects A δ where δ s a dameter of the 2m+2-gon obtaned by removng γ from the m + 2-angulaton determned by Δ In fact, we can be more precse Defne δ () to be the dameter of the 2m+2-gon obtaned by rotatng the vertces of γ by steps counterclocwse (wthn the 2m + 2-gon) Then A () γ = A δ() Defne Q Δ to be the coloured quver for the tltng obect A Δ, wth vertex set Δ Usngthe setup of [3], t s straghtforward to verfy: Proposton 111 The coloured quver Q Δ of T = A Δ has an arrow from γ to δ f and only f γ and δ both le on some m + 2-gon n the m + 2-angulaton defned by Δ In ths case, the colour of the arrow s the number of edges formng the segment of the boundary of the m + 2-gon whch les between γ and δ, counterclocwse from γ Gven the proposton above, t s straghtforward to verfy drectly that Q Δ satsfes condtons (I), (II), and (III), and that mutaton s ndeed gven by the mutaton of coloured quvers Example A 3, m = 2 We return to the example from Secton 2 The quadrangulaton of a decagon correspondng to the tltng obect T s on the left The quadrangulaton correspondng to T s on the rght Passng from the fgure on the left to the fgure on the rght, the dagonal 27 (whch corresponds to the summand I 2 ) has been rotated one step counterclocwse wthn the hexagon wth vertces 1, 2, 3, 4, 7, 10

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