A NOTE ON SIMPLE-MINDED SYSTEMS OVER SELF-INJECTIVE NAKAYAMA ALGEBRAS

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1 A NOTE ON SIMPLE-MINDED SYSTEMS OVER SELF-INJECTIVE NAKAYAMA ALGEBRAS JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG Abstract Let A = A l n be a self-njectve Nakayama algebra wth n smples and Loewy length l +, and let sms(a) be the set of smple-mnded systems n the stable module category A-mod Accordng to Redtmann, the objects n sms(a) correspond bjectvely to the τ n -stable non-crossng parttons of order l, where the τ n -stable non-crossng parttons are defned usng coverng theory of stable translaton quvers Recently, Chan also classfed the objects n sms(a) n terms of two-term tltng complexes of the homotopy category K b (A-proj) We gve a new constructon of objects n sms(a) n terms of non-crossng parttons of order e = gcd(n, l) Introducton In her famous work on classfcaton of representaton-fnte self-njectve algebras over an algebracally closed feld k, Redtmann defned the noton of (combnatoral) confguraton for any stable translaton quver Defnton ([9, Defnton 3]) Let be a stable translaton quver, and let k( ) be ts mesh category A confguraton C s a set of vertces of whch satsfes the followng condtons: ß () For any e, f C, Hom k( ) (e, f) 0 (e f), = k (e = f) () For any e 0, there exsts some f C such that Hom k( ) (e, f) 0 It was shown n [9, Proposton 3] that f π : Γ s a coverng of stable translaton quvers, then C s a confguraton of Γ f and only f π (C) s a confguraton of Accordng to [8], for an (ndecomposable, basc, fnte dmensonal) representaton-fnte self-njectve algebra A over an algebracally closed feld k, ts stable Auslander-Reten quver s Γ A (or stable AR-quver for short) has the form ZQ/Π, where Q s a Dynkn quver of type A n, D n, E 6, E 7 or E 8, and ZQ s the correspondng stable translaton quver whch s the unversal cover of s Γ A, Π s an admssble group of ZQ Ths mples the followng statement: C s a confguraton of s Γ A f and only f π (C) s a Π-stable confguraton of the unversal cover ZQ In partcular, f A = A l n s the self-njectve Nakayama algebra gven by the cycle quver Q α n α n wth the n possble relatons α l+ = 0, then there s the followng unversal cover π : ZA l s Γ A = ZAl / τ n, where τ s the translaton of the stable translaton quver ZA l and concdes wth the Auslander-Reten translaton n s Γ A under π For example, f l = 3, the correspondng stable translaton quver ZA 3 has the Mathematcs Subject Classfcaton(00): 6G0, 8Gxx Keywords: Non-crossng partton; Self-njectve Nakayama algebra; Sms of long-type; Sms of short-type Date: verson of September 8, 08 The authors are supported by NSFC (No33006, No5739) We would lke to thank Aaron Chan for comments and suggestons on a prelmnary verson of ths paper The revson of ths paper was done when the second author was vstng Unversty of Stuttgart The second author would lke to thank Prof Steffen Koeng for hs hosptalty and the Alexander von Humboldt Foundaton for support α

2 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG followng fgure (where the dashed arrows ndcate the translaton τ): (, 3) (0, 3) (, 3) (, ) (0, ) (, ) (, ) (0, ) (, ) Redtmann classfed confguratons of ZA l n terms of Brauer relatons Defnton ([9, Defnton 6]) A Brauer relaton of order n s an quvalence relaton on the set n mπ = {e n m Z} of complex numbers such that the convex hulls of dstnct equvalence classes are dsjont For example, the followng fgure gves a Brauer relaton of order 6: {{, 6, 4}, {, 3}, {5}} Proposton 3 (cf [9, Proposton 6]) Let A = A l n be the self-njectve Nakayama algebra defned as before Then we have the followng bjectons () {confguratons of ZA l } {Brauer relatons of order l}; () {confguratons of s Γ A } {τ n -stable Brauer relatons of order l} Notce that the number of Brauer relatons of order n equals the Catalan number n+( n n ) and that a Brauer relaton of order n s equvalent to a non-crossng partton on the set n = {,,, n}; the latter noton wll be defned n Secton 4 On the other hand, the authors n [4] proved that for a representaton-fnte self-njectve algebra A over an algebracally closed feld k, there s a bjecton: {Confguratons of s Γ A } {smple-mnded systems of A-mod} The noton of smple-mnded system (sms for short) was ntroduced by Koeng and Lu (see [6]) n the stable module category A-mod of any artn algebra A Later, Dugas [5] defned ths noton n any Hom-fnte Krull-Schmdt trangulated k-category The two defntons are compatble wth each other n the stable module category A-mod of any self-njectve algebra A (The precse defnton s gven n Remark 4) It was shown n [6] that when A s representaton-fnte self-njectve, the sms s n A-mod can be defned as follows Defnton 4 (cf [6, Theorem 56]) Let A be a representaton-fnte self-njectve algebra over an algebracally closed feld k A famly of objects S of A-mod s a sms f and only f the followng two condtons are satsfed: ß () For any S, T S, Hom A (S, T ) 0 (S T ), = k (S = T ) () For any ndecomposable non-projectve A-module X, there exsts S S such that Hom A (X, S) 0 It follows that, f A = A l n s the self-njectve Nakayama algebra, then the sms s of A-mod correspond precsely the confguratons of s Γ A, and both are classfed by τ n -stable Brauer relatons of order l Recently, Chan [3] gves a new classfcaton of sms s over self-njectve Nakayama algebras n terms of two-term tltng complexes In order to state hs result, we frst recall some notatons from [3] For an algebra A, we denote by tlt(a) the set of basc two-term tltng complexes concentrated n non-postve degrees, up to homotopy

3 A NOTE ON SIMPLE-MINDED SYSTEMS 3 equvalence and shfts n the bounded homotopy category K b (A-proj) of complexes of fntely generated projectve modules, by sms(a) the set of sms s, up to somorphsm n the stable module category A-mod Moreover, we denote by T (n) the set of trangulatons of a punctured regular convex n-gon (punctured n-dsc) Theorem 5 ([3]) Let A = A l n be the self-njectve Nakayama algebra defned as before, and let e = gcd(n, l) Then we have the followng () sms(a) = sms (A) sms + (A) () tlt(a) = tlt (A) tlt + (A) (3) If l n, then there are bjectons: (4) If l n, then there are surjectons: sms ± (A) T (e) tlt ± (A) tlt ± (A) sms ± (A) Both Redtmann s and Chan s classfcatons are mplct and t s not easy to wrte down the sms s explctly from these classfcatons Let l, n, e be as above Redtmann used the non-crossng parttons of order l, n order to bjectvely map non-crossng parttons to sms s she needed to put τ n -stable condton usng coverng theory of stable translaton quvers Chan set up a bjecton between two-term tltng complexes and sms s, usng mutaton theores on K b (A-proj) and on A-mod and ther relatons whch are developed n [5] and [4] The man result of the present paper s Theorem 56, whch gves an explct classfcaton of sms s over any self-njectve Nakayama algebra More precsely, for each par (p, k) where p s a non-crossng partton of order e and k s an nteger between and e, we assocate two types of sms s: sms s of long type and of short type, and Theorem 56 tells us how to dstngush two sms s of the same type There are two man ngredents n the proof of Theorem 56 One s Theorem 3, whch gves a suffcent and necessary condton for an orthogonal system n stable category A-mod to be a sms The other s Proposton 44, whch descrbes the orthogonalty condton n A-mod n terms of the arcs We remark that Dugas torson par theory n stable module category plays a key role n the proof of Theorem 3 Ths paper s organzed as follows In Secton, we state some useful facts on module categores and stable categores (n partcular, for self-njectve Nakayama algebras), ncludng the torson par theory on stable categores of self-njectve algebras developed by Dugas, and some coverng theory In Secton 3, we apply the torson par theory to gve a suffcent and necessary condton for an orthogonal system n stable category of self-njectve Nakayama algebra to be a sms In Secton 4, we frst descrbe the orthogonalty condton n stable category of symmetrc Nakayama algebra n terms of arcs and then show how sms s of symmetrc Nakayama algebra are closely related to (classcal) non-crossng parttons In Secton 5, we construct explctly two types of sms s over self-njectve Nakayama algebras: sms s of long type and sms s of short type In last secton, we study the behavors of our constructon under (co)syzygy functors and gve some applcatons of our results Prelmnares Throughout ths paper all algebras wll be fnte dmensonal algebras wth over an algebracally closed feld k For an k-algebra A, we denote by A-mod the category of fnte dmensonal left A-modules and by A-mod the stable category of A-mod Recall that A-mod has the same objects as A-mod but the morphsm space between two objects M and N s a quotent space Hom A (M, N) := Hom A (M, N)/P(M, N), where P(M, N) s the subspace of Hom A (M, N) consstng of those homomorphsms from M to N whch factor through a projectve A-module For a A-module M, we denote by soc(m) and rad(m) the socle and the radcal of M, respectvely We defne rad 0 (M) := M, rad k+ (M) := rad(rad k (M)) (k N) and top(m) := M/rad(M) Basc propertes of self-njectve Nakayama algebra A l n We denote by A l n the self-njectve Nakayama algebra wth n smples and Loewy length l + More precsely, A l n = kq/i s gven by the

4 4 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG followng quver Q n wth admssble deal I = rad l+ (kq) Let A = A l n be a self-njectve Nakayama algebra defned as above For any nteger Z, we denote by ī the postve nteger n {,, n} wth = ī mod n As usual, we denote by D, ν = DHom A (, A), Ω, and τ = DT r the k-dual functor, the Nakayama functor, the syzygy functor and the Auslander-Reten translate of A, respectvely Let S, S,, S n be the smple A-modules correspondng to the vertces,,, n of the quver Q For any ndecomposable A-module M, the Loewy length of M s denoted by l(m) and t means the number of composton factors n any composton seres of M Notce that any ndecomposable A-module M s unseral and completely determned up to somorphsm by top(m), soc(m) and l(m); f top(m) s somorphc to S, soc(m) s somorphc to S j, and the number of composton factors of M whch are somorphc to S s t +, then we denote M by Mj,t The followng two results are well known Theorem ([, Secton V35]) Let A and M be as above There exsts an ndecomposable projectve A-module P and an nteger t such that M = P/rad t (P ) In partcular, A s of fnte representaton type Theorem ([, Secton V4]) Let A and M = P/rad t (P ) be as above The sequence ä Ä q 0 rad(p )/rad t+ (P ) (rad(p )/rad t (P )) (P/rad t+ (P )) ( j, p) P/rad t (P ) 0 (where q and p are the canoncal epmorphsms and and j are the ncluson homomorphsms) s an almost splt sequence In partcular, l(m) = l(τ(m)), where τ s the Auslander-Reten translate of A The Nakayama functor ν of A s mportant n the present paper and we gve a descrpton for t n the followng two lemmas Lemma 3 Let A l n = kq/i be a self-njectve Nakayama algebra If M s an ndecomposable nonprojectve A l n-module, then ν(m) = τ l (M) Proof As before, we denote A l n by A We frst prove ths result for smple modules For smple module S, we have a short exact sequence: 0 K P π S 0, where π s the projectve cover of S Applyng Hom A (, A) to the short exact sequence we get an njectve homomorphsm Hom A (S, A) Hom A (P, A) = e A, where e s the prmtve dempotent correspondng to the vertex n Q Then Hom A (S, A) = soc(e A) and ν(s ) = D(Hom A (S, A)) = D(soc(e A)) = top(ae l ) = τ l (S ) Snce any ndecomposable A-module s unseral and ν s an (exact) equvalence functor, we know from the pcture of AR-quver of A that ν(m) = τ l (M) for any ndecomposable nonprojectve A-module M Lemma 4 For any ndecomposable non-projectve A l n-module M, we denote by O ν (M) the ν-orbt of M Then the numbers of objects n O ν (M) s n/e and O ν (M) = {M, τ e (M),, τ n+e (M)}, where e s the greatest common dvsor of n and l Proof By Lemma 3, we have O ν (M) = {M, τ l (M),, τ (k )l (M)}, where k s the mnmum postve nteger such that n dvdes kl Snce n/e and l/e are coprme, we have k = n/e Thus, the numbers of objects n O ν (M) s n/e and O ν (M) = {M, τ e (M),, τ n+e (M)} In the rest of ths subsecton, we prove several elementary results on homomorphsm spaces n the stable category of a self-njectve Nakayama algebra For f Hom A (M, N), we wll denote ts mage n Hom A (M, N) by f Lemma 5 Let A = A l n be a self-njectve Nakayama algebra, and let M, N be two ndecomposable non-projectve A-modules Suppose that there exsts a non-zero morphsm f Hom A (M, N) satsfyng Imf = rad t (N) Then f = 0 f and only f l(m) + t l + In partcular, f we denote by (respectvely j) the number of composton factors of N/rad t (N) (respectvely M) whch are somorphc to top(n), then f = 0 mples + j [l/n] +, where [l/n] s the maxmum nteger of no more than l/n

5 A NOTE ON SIMPLE-MINDED SYSTEMS 5 Proof = Snce f = 0, we have the followng commutatve dagram n A-mod: M f N g P N where π s the projectve cover of N Then Img = rad t (P N ) and l(m) l(rad t (P N )) = l(p N ) t = l+ t, that s, l(m) + t l + = Suppose that l(m) + t l + and let π : P N N be the projectve cover of N Then we can defne a morphsm g from M to P N satsfyng Img = rad t (P N ) and f = πg, that s, f factors through a projectve module Remark 6 Notce that + j [l/n] + s not a necessary and suffcent condton for f = 0 n general However, f A l n s a symmetrc Nakayama algebra (that s, there exsts an nteger d such that l = dn), then the condton + j d + s a necessary and suffcent condton for f = 0 Lemma 7 Let M and N be two ndecomposable non-projectve A l n-modules Let f Hom A l n (M, N) be a non-zero homomorphsm such that Imf = rad t (N), where t s an nteger such that there s no epmorphsms from M to rad s (N) for s < t Then f = 0 f and only f Hom A l n (M, N) = 0 Proof = When Hom A l n (M, N) = 0, t s clear that f = 0 = If f = 0, by Lemma 5, then l(m) + t l + For any morphsm g n Hom A l n (M, N), snce there s no epmorphsms from M to rad s (N)(s < t), we have Img = rad s (N), where s t Therefore l(m) + s l +, and agan by Lemma 5, g = 0 Ths shows Hom A l n (M, N) = 0 Lemma 8 Let A l n be a self-njectve Nakayama algebra, For any two ndecomposable non-projectve A l n-modules M and N (M N), f Hom A l n (M, N) = 0 and Hom A l n (N, M) = 0, then top(m) top(n) and soc(m) soc(n) Proof Ths s clear We now descrbe when the ν-orbt O ν (M) of an ndecomposable non-projectve A l n-module M forms a set of parwse orthogonal stable brcks (a termnology due to ZPogorza ly) Defnton 9 Let A be a self-njectve algebra A set S of objects n the stable category A-mod s called a set of parwse orthogonal stable brcks f for any objects M, N n S, the followng condton s satsfed: Hom A (M, N) 0, M N, = k, M = N In partcular, for an ndecomposable A-module M, f Hom A (M, M) = k, then M s called a stable brck Proposton 0 Let A = A l n be a self-njectve Nakayama algebra and M be an ndecomposable nonprojectve A-module Then the ν-orbt O ν (M) of M s a set of parwse orthogonal stable brcks f and only f l(m) e or l + e l(m) l, where e s the greatest common dvsor of n and l Proof = when l(m) e, snce any two composton factors of M are not somorphc and top(m) s not a composton factor of the objects except M n O ν (M), t s clear that O ν (M) s a set of parwse orthogonal stable brcks when l + e l(m) l, for any object N n O ν (M), consder the morphsms f from N to τ e (N) satsfyng Imf = rad e (τ e (N)) and g from N to N satsfyng Img = rad n (N) Notce that by Lemma 4, l(n) = l(m) So by Lemma 5, f = 0 and g = 0 Furthermore, by Lemma 7, Hom A (N, N) = k and Hom A (N, τ e (N)) = 0 (τ e (N) N) There s a smlar proof between N and τ ke (N) (τ ke (N) N, k N) Therefore O ν (M) s a set of parwse orthogonal stable brcks because of the arbtrarness of the module N = Consder the morphsm f from M to τ e (M) satsfyng Imf = rad e (τ e (M)) If f = 0, then l(τ e (M)) = l(m) e If f 0, snce Hom A (M, τ e (M)) = 0, by Lemma 5, we have l + e l(m) l π,

6 6 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG For any symmetrc Nakayama algebra, the Nakayama functor s somorphc to the dentty functor and therefore we have the followng corollary Corollary Let A dn n non-projectve A dn () Hom A dn n = kq/i be a symmetrc Nakayama algebra and M = Mj,t be an ndecomposable n -module Then the followng are equvalent: (M, M) = k (that s, M s a stable brck); () l(m) n or (d )n + l(m) dn; (3) t = 0 or t = d The torson par theory n stable module category We brefly recall the torson par theory on a Hom-fnte Krull-Schmdt trangulated k-category n the sense of Dugas Let T be a Hom-fnte Krull- Schmdt trangulated k-category wth suspenson denoted by [] For any famles S, S of objects n T, we defne a famly of objects S S := {X T There s a dstngushed trgngle S X S S [], S S, S S } It s easy to show that (S S ) S 3 = S (S S 3 ) for S, S, S 3 T For a famly S of objects n T, we denote (S) 0 = {0}, and for any postve nteger n, we nductvely defne (S) n = (S) n (S {0}) Then (S) n (S) m = (S) n+m for any non-negatve ntegers m and n (cf [5, Lemma 3]) Smlarly, one can defne n (S), and we have (S) n = n (S) We say that S s extenson-closed, f S S S We denote the extenson closure of a famly S of objects n T as F(S) := n 0(S) n, whch s the smallest extenson closed full subcategory of T contanng S Notce that we dentfy S wth the correspondng full subcategory of T Lemma ([5, Lemma 7]) If S T s a set of parwse orthogonal stable brcks, then (S) n s closed under drect summands for each postve nteger n In partcular, F(S) s closed under drect summands For any famly S of objects n T, we set S = {Y T T (X, Y ) = 0, X S}, S = {Y T T (Y, X) = 0, X S} We know that both S and S are extenson closed subcategores of T as well as closed under drect summands We shall denote S S by S Defnton 3 ([5, Defnton 3]) A par (X, Y) of full, addtve subcategores of T, whch are closed under drect summands, forms a torson par f the followng condtons hold: () T (X, Y) = 0; () T = X Y, that s, for each T T, there exsts a dstngushed trangle where X X, Y Y X f T g Y X[], The above dstngushed trangle n () s called a (X, Y)-trangle of T It s easy to show that for any (X, Y)-trangle of T, f s a rght X -approxmaton and g s a left Y-approxmaton It s true that for a (X, Y)-trangle, f s a mnmal rght X -approxmaton f and only f g s a mnmal left Y-approxmaton (cf [5, Lemma 3]) Furthermore, we can choose a rght mnmal verson of f and ths resultng trangle s unque up to somorphsm, we call t the mnmal (X, Y)-trangle In the present paper, we wll apply the above torson par theory n a specal case where T s the stable category A-mod of a self-njectve algebra A In ths case, the suspenson functor s the cosyzygy functor Ω (also denoted by [] f there s no confuson) and dstngushed trangles n A-mod are nduced by short exact sequences n A-mod Notce also that A-mod has Serre functor νω = ν[ ], that s, for all M, N A-mod, we have the followng k-lnear somorphsms: Hom A (M, N) = DHom A (N, νωm) Remark 4 Let A be a self-njectve algebra and S be a set of parwse orthogonal stable brcks n A-mod Then by the orgnal defnton n [6] or [5], S s a sms f and only f F(S) = A-mod

7 A NOTE ON SIMPLE-MINDED SYSTEMS 7 Now we take a set S of parwse orthogonal stable brcks n A-mod wth ν(s) = S and assume that both ( S, F(S)) and (F(S), S ) are torson pars n A-mod We defne operators a : T S, b, c : T F(S) and d : T S va the mnmal trangles ax X bx and cx X dx correspondng to these two torson pars respectvely Notce that n general these operators are not functors, see [5, Secton 3] for more nformaton about them Lemma 5 ([5, Lemma 43]) Assume that S s a set of parwse orthogonal stable brcks n A-mod wth ν(s) = S Then ν(f(s)) = F(S) Furthermore, ν(ax) = a(νx) and ν(bx) = b(νx) for all X A-mod, and smlarly for c and d Lemma 6 ([5, Lemma 46]) Let S be as n Lemma 5 For any mnmal ( S, F(S))-trangle ay Y by and any X S, we have the followng () The map Hom A (g, X) : Hom A (by, X) Hom A (Y, X) s an somorphsm; () The map Hom A (X, f) : Hom A (X, ay ) Hom A (X, Y )s a monomorphsm; (3) If Y X, then ay X 3 Some coverng theory In ths subsecton, we use coverng theory to prove two useful lemmas n the stable category of a self-njectve Nakayama algebra The coverng of translaton quvers was ntroduced by Redtmann ([8, 9]), and t was extended to coverng functors of k-categores by Bongartz and Gabrel ([]) There s a bref ntroducton n [4] on some coverng theory we are makng use of Let A = A l n be a self-njectve Nakayama algebra, and let s Γ A be the stable AR-quver of A Recall that sγ A has the form ZA l / τ n, where ZA l s the stable translaton quver assocated to the Dynkn quver of type A l (as mentoned n Introducton) and τ s ts translaton Accordng to [9], we have A-mod = k( s Γ A ), where k( s Γ A ) s the mesh category of the stable AR-quver s Γ A As a result, there s a coverng functor F : k(za l ) A-mod, where k(za l ) s the mesh category of ZA l In partcular, there are the followng bjectons (for e, f, h ZA l ): Hom k(zal )(e, h) = Hom k(sγ A )(F e, F f), Hom k(zal )(e, h) = Hom A (F e, F f), F h=f f F e=f h Hom k(zal )(e, f) = Hom k(sγ A )(F h, F f), F h=f f F e=f h Hom k(zal )(e, f) = Hom A (F h, F f) As n [9], t s often convenent to wrte a vertex of ZA l as ts coordnate (p, q), where p, q Z, q l We recall the followng lemma Lemma 7 ([9, Lemma 6]) For any vertces (p, q) and (r, s) n ZA l, we have dm k (Hom k(zal )((p, q), (r, s))) In partcular, dm k (Hom k(zal )((p, q), (r, s))) = f and only f p r < p + q r + s p + l Gven a category C and a functor F : C k-mod, we set the support Supp(F ) := {X C X s ndecomposable and F (X) 0} The above lemma means that, for a gven vertex a = (r, s) n ZA l, Supp(Hom k(zal )(, a)) les n the followng parallelogram (ncludng boundary and nner): (r + s l, l) (r + s l, l s + ) (r, s) (r, ) Smlarly, for any ndecomposable non-projectve A-module M, the support Supp(Hom A (, M)) s the set of ndecomposable non-projectve A-modules N satsfyng Hom A (N, M) 0 Now combnng Lemma 7 and the coverng theory, t s easy to see that the support Supp(Hom A l n (, M)) les n a parallelogram n the stable AR-quver s Γ A of A That s, we have the followng lemma

8 8 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG Lemma 8 Let A be a self-njectve Nakayama algebra and M be an ndecomposable non-projectve A- module Then the support Supp(Hom A (, M)) les n the followng parallelogram wth vertces V, N, M, W (ncludng boundary and nner) n s Γ A : V W M N The above lemma can be proved drectly, but we prefer to the short proof usng coverng theory The followng lemma s also clear from coverng theory Lemma 9 Let A = A l n and B = A l e, where e s the greatest common dvsor of n and l There s a coverng of stable translaton quvers π : s Γ A s Γ B = s Γ A / ν (where ν s the Nakayama automorphsm on s Γ A ), whch nduces a coverng functor F : A-mod B-mod Consequently, S s a sms of B-mod f and only f F (S) s a sms of A-mod 3 A necessary and suffcent condton for an orthogonal system to be a sms In ths secton we gve a necessary and suffcent condton for an orthogonal system n the stable category of a self-njectve Nakayama algebra to be a sms Ths result s the bass of our constructon of sms s over a self-njectve Nakayama algebra n Secton 5 We know from [4] that for a representaton-fnte self-njectve algebra A, f S s a sms (see Defnton 4) n A-mod, then the followng condtons hold: () S s a set of parwse orthogonal stable brcks n A-mod, () The number of objects n S s equal to the number of non-somorphc non-projectve smple A-modules, (3) S s Nakayama-stable, that s, ν(s) = S, where ν s the Nakayama functor of A It s natural to ask f condtons (), (), (3) can be suffcent for S to be a sms The followng theorem gves a postve answer for self-njectve Nakayama algebras Theorem 3 Let A be a self-njectve Nakayama algebra If a famly S of objects n A-mod satsfes the above condtons (), (), (3), then S s a sms Remark 3 We cannot delete any condton from (), (), (3) A counterexample for deletng (3) comes from the Nakayama algebra A = A 4 4 It s easy to check that S = {, 3, 3, 4 } satsfes (), () but not (3), 4 and S s not a sms The proof of Theorem 3 s based on the followng proposton, whch generalzes the result [5, Theorem 33] of Dugas n our specal case Proposton 33 Let A be a self-njectve Nakayama algebra If S s a set of parwse orthogonal stable brcks n A-mod, then both ( S, F(S)) and (F(S), S ) are torson pars n A-mod Proof of Theorem 3 Accordng to Remark 4, we only need to prove that F(S) = A-mod Snce (F(S), S ) s a torson par, t s enough to show that S = {0} Snce ( S, F(S)) s also a torson par, by Lemma 6, f 0 Y S, then ay S However, the condtons () and () of S show that S = {0} Therefore Y = by F(S) Snce Y S, we have Y F(S) It follows that Hom A (Y, Y ) = 0 and Y = 0 n A-mod, ths contradcts the assumpton on Y The rest of ths secton s devoted to the proof of Proposton 33 Let A = A l n be a self-njectve Nakayama algebra In ths secton, we wll use a dfferent notaton to denote ndecomposable A-module: we denote by M j (j) the ndecomposable A-module whch has socle S, Loewy length j + and top S j

9 A NOTE ON SIMPLE-MINDED SYSTEMS 9 for all {,,, n}, j {0,,, l} Ths notaton makes t easer to see short exact sequences Notce that M j (j) s the same as the ndecomposable A-module M j,k wth the notaton n Secton, where k s the maxmum nteger wth the property kn j Lemma 34 Let M j (j) be an ndecomposable A-module, where {,, n}, j {0,,, l} () For any k {0,,, j } and an ncluson homomorphsm ι : M k followng dstngushed trangle n A-mod: (k) M j (j), there exsts the M k (k) ι M j (j) M j (j k ) k M k (k)[] M k (k) M j (j) 0 M j (j k ) k () Gven a canoncal epmorphsm π : M j j (l k) M +l j k (j), where l k j, there exsts the followng dstngushed trangle n A-mod: M j π (l k) +l j k M j (j) M j k (j + k) M j (l k)[] +l j k M j (l k) +l j k M j k (l) +l j k M j k (j + k) M j (j) (3) Denote by Supp (Hom A (, M j nether submodules of M j (j))) the collecton of objects n Supp(Hom A (, M j (j))) whch are For any Mp p q (q) Supp (Hom A (, M j (j))), nor factor modules of M j let f be the composton of canoncal epmorphsm π : M p q p (q) M q+k (q k) and ncluson homomorphsm ι : M q+k (q k) M j (j) Then there exsts the followng dstngushed trangle n A-mod: M p q p (q) f M j (j) Å ã π ι M j p q p l (j q + k ) M (l k) M p q (q)[], where j {,, l}, p = + k, k {,,, l j}, π s the canoncal epmorphsm and ι s the ncluson homomorphsm p

10 0 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG Mp p q (q) Mp p l (l) M p l (l k) M j (j) M j (j q + k ) p q 0 (4) For any k {,, j }, h {0,, k }, an ncluson homomorphsm ι : M k a canoncal epmorphsm π : M k A-mod: M k (k) ι M j (k) M k k+h (j) M k k+h (h) (k) M j (j) and (h), there exsts the followng dstngushed trangle n M j (h + j k) k+h M k (k)[] M k (k) M j (j) M j (h + j k) k+h M k k+h (h) Remark 35 The above graphs show the short exact sequences nducng the correspondng dstngushed trangles 0 appearng n the above graphs represents the zero module Proof of Lemma 34 We only prove (3), the proofs of (), () and (4) are smlar Snce any short exact sequence n A-mod nduces a dstngushed trangle n A-mod, t suffces to show that the followng sequence n A-mod s a short exact sequence: ( ) 0 Mp p q (q) where ι : M p q p and π : M p l p Ä f ι ä M j (q) Mp p l (l) M p l (j) Mp p l (l) Ä ä π 0 ι π M j p q (l) s the njectve envelope of M p q p p l (j q + k ) M (l k) 0, (q), π : M j (j) Mp q(j q+k ) j (l k) are the canoncal epmorphsms Å Å ã f π 0 Snce ι s njectve, s njectve; snce both π ιã and π are surjectve, s surjectve ι π Å ã Å ã Å π 0 f 0 Moreover, t s straghtforward to check that = and ths shows that ( ) s a complex π ι 0ã ι We also know that dm k (Mp p q (q)) + dm k (M j p l (j q + k ) M p q (l k)) = dm k (M j (j) Mp p l (l)) = l + j + Therefore, ( ) s a short exact sequence

11 A NOTE ON SIMPLE-MINDED SYSTEMS Proof of Proposton 33 We only prove that (F(S), S ) s a torson par, the case for ( S, F(S)) can be proved by a dual argument (usng the dual functor D = Hom k (, k)) We know that both F(S) and S are closed under drect summands, and clearly Hom A (F(S), S ) = 0 It remans to check the second condton of torson pars: for any M A-mod, there exsts the followng dstngushed trangle: X f M g Y X[] where X F(S), Y S Wthout loss of generalty we can assume that M s ndecomposable We wll prove our statement by nducton on the Loewy length m = l(m) of M If m =, we show that any smple module S ( n) belongs to F(S) S Case If S F(S) (resp S S ), then S d S 0 S [] (resp 0 S d S 0 ) s the desred dstngushed trangle Case If S / F(S) and S / S, snce S / S, there exsts some X F(S) such that Hom A (X, S ) 0 Let T = {Y F(S) Hom A (Y, S ) 0}, then T s non-empty Take X T wth l(x) l(y ) for all Y T For the canoncal epmorphsm π : X S, by Lemma 34, we have the followng dstngushed trangle: X π S Y X[] Clam: Y S Snce X π S Y X[] s a dstngushed trangle, we have a short exact sequence n A-mod: 0 X S I(X) Y 0, where I(X) s the njectve envelope of X By Lemma 8, Supp(Hom A (, Y ) les n the parallelogram Y ZUS (ncludng the boundary): I(X) V X U Z Y X R If Y / S, by Lemma 34, there exsts X F(S) n Supp(Hom A (, Y )) such that we have a short exact sequence n A-mod as follows: S 0 X V R X 0, where top(r) = top(x), Hom A (R, S ) 0 and l(r) < l(x) Snce F(S) s closed under both extensons and drect summands, R F(S), ths contradcts wth the choce of X Therefore Y S and X π S Y X[] s the desred trangle Now we assume that m > and M F(S) S for l(m) m, and we consder the case l(m) = m Case If M F(S)(resp M S ), then M d M 0 M[] (resp 0 M d M 0 ) s the desred trangle Case If M / F(S) and M / S, let T = {Y F(S) Y Supp(Hom A (, M))}, snce M / S, we have that T s non-empty Take W T wth l(w ) l(y ) for any object Y T Then W s an ndecomposable module and there are three subcases () If there exsts an ncluson homomorphsm W M, by Lemma 34, we have a dstngushed trangle W M W W [], where l(w ) < l(m) By nducton, there exsts the followng dstngushed

12 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG trangle C W D C[], where C F(S), D S Then by octahedral axom, we have the followng commutatve dagram n A-mod: M W W [] M[] d M D K[] d M[] C[] d C[] W [] W [] W [] Rotatng the second row we get the desred dstngushed trangle K M D K[], where K F(S) and D S () If there exsts a canoncal epmorphsm W π M, by Lemma 34, we have the followng dstngushed trangle: W π M N W [] Clam: N S Snce W π M N W [] s a dstngushed trangle, there s a short exact sequence n A-mod as follows: 0 W M I(W ) N 0, where I(W ) s the njectve envelope of W Consder the followng pcture: I(W ) U L S N X W V G T M K R where Supp(Hom A (, M)) les n the parallelogram MUV R (ncludng boundary) and Supp(Hom A (, N)) les n the parallelogramn LKR (ncludng boundary) Assume that N / S Then there s an object X n F(S) such that X les n the parallelogram NLKR Accordng to the choce of W, X does not le n the parallelogram MGKR So f X belongs to the parallelogram NLGM, then we have the followng short exact sequence n A-mod: 0 W S T X 0 Snce F(S) s closed under extenson and drect summands, we have T F(S), Hom A (T, M) 0 and l(t ) < l(w ) Ths contradcts to the choce of W Therefore, N S, W π M N W [] s the desred dstngushed trangle (3) If there s nether an ncluson homomorphsm nor a canoncal epmorphsm from W to M, then by Lemma 34, we have the followng trangle: W M M M W [], ( )

13 A NOTE ON SIMPLE-MINDED SYSTEMS 3 where M M s the canoncal epmorphsm Snce l(m ) < l(m), by nducton, we have the followng dstngushed trangle: N M N N [], ( ) where N F(S), N S Next we show that M also les n S By Lemma 34, there s a short exact sequence n A-mod: Consder the followng pcture: 0 W M I(W ) M M 0 I(W ) Y M M W X M R V S Snce the parallelogram V XRS (ncludng boundary) les n Supp(Hom A (, M)), accordng to the choce of W, there s no module of F(S) lyng n parallelogram V XRS (ncludng boundary) Suppose that M / S, usng the same method as n (), we can easly show that there s a module of F(S) lyng n the parallelogram M Y XV (ncludng boundary), whch gves a smlar contradcton as n () Therefore, M S Now, we show that there exsts a desred dstngushed trangle for M Frst, rotatng ( ) and ( ) and usng octahedral axom we have the followng commutatve dagram n A-mod: M M M W [] M[] d M N M U d M[] N [] d N [] (M M )[] (M M )[] W [] By rotatng the thrd column we have the followng dstngushed trangle: W U[ ] N W [] Snce both W and N belong to F(S), we have U[ ] F(S) Rotatng the second row we get the followng dstngushed trangle: U[ ] M N M U ( ) Snce both N and M belong to S, ( ) s the desred dstngushed trangle 4 The arcs and non-crossng parttons In ths secton, we show how the orthogonalty condtons n A dn n -mod closely related to (classcal) noncrossng parttons of the set {,, n} Under the notatons n Secton, we frst ntroduce the arc for any ndecomposable A l n-module

14 4 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG Defnton 4 Let A l n = kq/i be a self-njectve Nakayama algebra For any ndecomposable A l n-module M = M j,t, the arc of M s defned to be the arc Ùj from the vertex to the vertex j n (undrected graph) Q wth the drecton of arrows n Q In partcular, f = j, then the arc of M s the vertex n Q Example 4 Let A 5 4 = kq/i be a self-njectve Nakayama algebra gven by the followng quver Q 4 3 and admssble deal I = rad 6 (kq) Consder the ndecomposable non-projectve A 5 4-modules M = M 3,0 and N = N 4,0, ther arcs are shown together as follows: 4 We now use arcs to descrbe orthogonal relatons between stable brcks over A dn n 3 Lemma 43 Let M = M k,l ( k, k ) and N = N j k j,l j be two stable brcks over A dn n If ther arcs ntersect as follows (ths means that j ı k, k ĵk j, k j Ùj and k j ): k j j then Hom A dn n (N, M) 0 k Proof If l = 0, then l(m) n and there exsts a unque nteger t satsfyng top(rad t (M)) = S j Therefore, there s a morphsm f from N to M satsfyng Imf = rad t (M) and the number of composton factors of M/rad t (M) whch are somorphc to S s By Corollary, there are two cases for l j : When l j = 0, we read from the pcture that S s not a composton factor of N, and l(n)+t n < l+ By Lemma 5, f 0 and therefore Hom A dn n (N, M) 0; When l j = d, the number of composton factors of N whch are somorphc to S s d, and l(n) + t dn < l + By Lemma 5, f 0 and therefore Hom A dn n (N, M) 0 If l = d, then there exst a mnmum nteger t satsfyng top(rad t (M)) = S j and a maxmum nteger t satsfyng top(rad t (M)) = S j Agan we consder two cases for l j : When l j = 0, we also read from the pcture that S s not a composton factor of N and there s a morphsm f from N to M satsfyng Imf = rad t (M), and l(n) + t dn < l + By Lemma 5, f 0 and therefore Hom A dn n (N, M) 0; When l j = d, there s a morphsm f from N to M satsfyng Imf = rad t (M), and l(n)+t nd < l+ By Lemma 5, f 0 and therefore Hom A dn n (N, M) 0 Proposton 44 Let A dn n = kq/i (d ) be a symmetrc Nakayama algebra and M = Mk,l, N = N j k j,l j be two ndecomposable non-projectve A dn n -modules Then {M, N} s a set of parwse orthogonal stable brcks f and only f t satsfes the followng condtons: (a) j, k k j ; (b) l = 0 or l = d, l j = 0 or l j = d ; (c) Ther arcs belong to one of the four cases:,

15 A NOTE ON SIMPLE-MINDED SYSTEMS 5 () () k j k k j j l + l j > 0 j k j j l + l j d k (3) (4) j l j l j l l j, k kj k j k where the two arcs n () are dsjont and the two arcs n other cases are ntersected Proof = (a) and (b) follow from Lemma 8 and Corollary The four pctures about arcs n ()-(4) can follow from Lemma 43, we just need to verfy the condtons for l and l j n four cases Case If l = l j = 0, then there s a unque nteger t satsfyng top(rad t (M)) = S j such that the number of composton factors of M/rad t (M) whch are somorphc to S s and a morphsm f : N M satsfyng Imf = rad t (M) Snce the number of composton factors of N whch are somorphc to S s and d, by Remark 6, we have that f 0 Ths contradcton shows that l + l j > 0 Case If l = d, l j = d, then the number of composton factors of N whch are somorphc to S s d and the number of composton factors of M whch are somorphc to S j s d There exsts a mnmum nteger t satsfyng top(rad t (M)) = S j such that the number of composton factors of M/rad t (M) whch are somorphc to S s There s a morphsm f : N M satsfyng Imf = rad t (M) By Remark 6, f 0 Ths contradcton shows that l + l j d Case 3 If l = 0, l j = d, then the number of composton factors of N whch are somorphc to S s d and the number of composton factors of M whch are somorphc to S j s There exsts a unque nteger t satsfyng top(rad t (M)) = S j such that the number of composton factors of M/rad t (M) whch are somorphc to S s, and there s a morphsm f : N M satsfyng Img = rad t (M) By Remark 6, f 0 Ths contradcton shows that l j l Case 4 If l = d, l j = 0, then we can show smlarly as Case 3 that l l j = By Corollary, we can assume that M = Mk,l and N = N j k j,l j are two stable brcks n A dn n -mod under the followng condtons: j, k k j and l = 0 or l = d, l j = 0 or l j = d We now prove that Hom A dn n (M, N) = 0 and Hom A dn n (N, M) = 0 by checkng the four cases In each case, we consder three subcases accordng to the values of l and l j Case () When l = 0, l j = d, the number of composton factors of N whch are somorphc to S s d and the number of composton factors of M whch are somorphc to S j s There s a maxmum nteger t satsfyng top(rad t (N)) = S such that the number of composton factors of N/rad t (N) whch are somorphc to S j s d There exsts a morphsm f : M N satsfyng Imf = rad t (N) By Remark 6, f = 0 and by Lemma 7, Hom A dn n (M, N) = 0 Moreover, there s a unque nteger t satsfyng top(rad t (M)) = S j and a morphsm g : N M satsfyng Img = rad t (M) By Remark 6, g = 0 and by Lemma 7, Hom A dn n (N, M) = 0 () When l = d, l j = 0, the number of composton factors of N whch are somorphc to S s and the number of composton factors of M whch are somorphc to S j s d There s a smlar descrpton as () for ths case Then Hom A dn n (N, M) = 0 and Hom A dn n (M, N) = 0 () When l = d, l j = d, the number of composton factors of N whch are somorphc to S s d and the number of composton factors of M whch are somorphc to S j s d There s a mnmum nteger t satsfyng top(rad t (N)) = S such that the number of composton factors of N/rad t (N) whch are somorphc to S j s There exsts a morphsm f : M N satsfyng Imf = rad t (N) By Remark 6, f = 0 and by Lemma 7, Hom A dn n (M, N) = 0 Smlarly, there s a mnmum nteger t satsfyng

16 6 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG top(rad t (M)) = S j such that the number of composton factors of M/rad t (M) whch are somorphc to S s There s a morphsm g : N M satsfyng Img = rad t (M) By Remark 6, g = 0 and by Lemma 7, Hom A dn n (N, M) = 0 Case () When l = 0, l j = 0, S j s not a composton factor of M and S s not a composton factor of N Then Hom A dn n (M, N) = 0, Hom A dn n (N, M) = 0 and therefore Hom A dn n (M, N) = 0, Hom A dn n (N, M) = 0 () When l = 0, l j = d, S j s not a composton factor of M and the number of composton factors of N whch are somorphc to S s d Then Hom A dn n (N, M) = 0 and there s a maxmum nteger t satsfyng top(rad t (N)) = S, however, l(rad t (N)) > l(m), we have Hom A dn n (M, N) = 0 Therefore, Hom A dn n (M, N) = 0, Hom A dn n (N, M) = 0 () When l = d, l j = 0, S s not a composton factor of N and the number of composton factors of M whch are somorphc to S j s d It follows smlarly as above that Hom A dn n (M, N) = 0, Hom A dn n (N, M) = 0 Case 3 () When l = 0, l j = 0, S s not a composton factor of N Then Hom A dn n (M, N) = 0 and therefore Hom A dn n (M, N) = 0 There s a unque nteger t satsfyng top(rad t (M)) = S j, however, l(rad t (M)) > l(n), we have Hom A dn n (N, M) = 0 and therefore Hom A dn n (N, M) = 0 () When l = d, l j = 0, we have that S s not a composton factor of N Then Hom A dn and therefore Hom A dn n l(rad t (M)) > l(n) Then Hom A dn n n (M, N) = 0 (M, N) = 0 There s a maxmum nteger t satsfyng top(rad t (M)) = S j, however, (N, M) = 0 and Hom A dn n (N, M) = 0 () When l = d, l j = d, the number of composton factors of N whch are somorphc to S s d and the number of composton factors of M whch are somorphc to S j s d There exsts a mnmum nteger t satsfyng top(rad t (N)) = S such that the number of composton factors of N/rad t (N) whch are somorphc to S j s, and there s a morphsm f : M N satsfyng Imf = rad t (N) By Remark 6, f = 0 and by Lemma 7, Hom A dn n (M, N) = 0 There exsts an nteger t satsfyng top(rad t (M)) = S j such that the number of composton factors of M/rad t (M) whch are somorphc to S s, and there s a morphsm g : N M satsfyng Img = rad t (M) By Remark 6, g = 0 and by Lemma 7, Hom A dn n (N, M) = 0 Case 4 Ths s smlar to Case 3 Summarzng the above dscusson we get that {M, N} s a set of parwse orthogonal stable brcks Remark 45 Notce that n Proposton 44, we assume d In fact, f d =, then d = 0 and we should remove the condtons for l and l j n (c) 4 Non-crossng partton (Classcal) non-crossng parttons of the set n := {,, n} was ntroduced by Kreweas n [7], see [0] for a further ntroducton Although non-crossng partton of n s equvalent to Brauer relaton of order n, ths former language s more convenent for our purpose In ths subsecton, we descrbe the relatonshp between sms s of A dn n and non-crossng parttons of n Defnton 46 A partton of the set n s a map p from n to ts power set wth the followng propertes: () p() for all n; () p() = p(j) or p() p(j) = for all, j n We call p() a block of p A non-crossng partton of the set {,, n} s a partton p that no two blocks cross each other, that s, f a and b belong to one block and x and y belong to another, we cannot have a < x < b < y We show how a sms S of A dn n the socle seres of S gve the complete set {S,, S n } of smple A dn there s a subset p() = {, k, k,, ks top(m j ) = S k j partton p of the set n, soc(m j ) = S k j+ relates to a non-crossng partton Frst notce that both the top and n -modules For each n, } of n such that there exsts M j (dependng on ) n S wth for 0 j s, where k 0 = k s =, k = k In ths way, we get a Remark 47 For the above partton p, we can defne a permutaton σ p on n, where σ p () = k for any n n Moreover k j = σj p() for j s By Proposton 44, we have that the partton p satsfes the followng ant-clockwse property Corollary 48 Let S be a sms of A dn n p() = {, k, k,, ks },where s 3 Then k t t s = kq/i and p be the partton obtaned as above Suppose that s a vertex on the arc ĭkt n the quver Q for each

17 A NOTE ON SIMPLE-MINDED SYSTEMS 7 Proof Frstly, we consder the objects M 0 and M n S, where {M 0, M } s a set of parwse orthogonal stable brcks and top(m 0 ) = S, soc(m 0 ) = S k, top(m ) = S k, soc(m ) = S k By Proposton 44, we have that ther arcs must belong to the frst case Then k s a vertex on the arc ı k from the vertex to the vertex k Smlarly, when s 4, we have that k t s a vertex on the arc ĭkt for each 3 t s We are ready to prove that the above partton p correspondng to S s actually a non-crossng partton Corollary 49 Let S be a sms of A dn n partton of n and p be the partton correspondng to S Then p s a non-crossng Proof Use the above notatons, we have the block p() = {, k, k,, ks M j n S satsfyng top(m j ) = S k j and soc(m j ) = S k j+ } such that there exsts object = k s = for each 0 j s, where k 0, k = k Gven two blocks p() = {, k, k,, ks } and p(j) = {j, k j, kj,, ksj j 48, wthout loss of generalty we can assume that the vertex j les on the arc k ı, that s, j } By Corollary k k s We clam that k j, kj,, ksj j are also vertces on the arc k ı Otherwse, wthout loss of generalty we can assume that the vertex k j s a vertex on the arc k ı Moreover, there exst objects M 0 n S satsfyng top(m 0 ) = S, soc(m 0 ) = S k and M j0 n S satsfyng top(m j0 ) = S j, soc(m j0 ) = S kj By Lemma 43, Hom A dn n (M 0, M j0 ) 0 Ths s a contradcton! Therefore, p s a non-crossng partton In the next two results, we use non-crossng parttons to descrbe some propertes of sms s Frst we fx some notatons: S s a sms of A dn n, p s the correspondng non-crossng partton; For n, the block p() = {, k, k,, ks } s determned as follows, there exsts object M j n S satsfyng top(m j ) = S k j and soc(m j ) = S k j+ for each 0 j s, where k 0 = ks =, k = k Lemma 40 Let S be a sms of A dn n (d ) and p be the correspondng non-crossng partton For n, let p() = {, k, k,, ks } be defned as above If we denote M j by M kj for each k j+,l j 0 j s, then there s at most one l j satsfyng l j = 0 for 0 j s Proof If s =, then p() = {} and therefore we have our desred result If s, wthout loss of generalty we can assume l 0 = 0 When s 3, we use Corollary 48 Notce that k = when s = Then, whenever s 3 or s =, we have that the arcs of M 0 and M j are as follows for any j s : k j+ k Snce M 0 and M j are parwse orthogonal stable brcks, by Proposton 44, we must have l j = d for j s Lemma 4 Let S be a sms of A dn n (d ) and p be the correspondng non-crossng partton For n, let p() = {, k, k,, ks } be defned as above Denote M j by M kj for each 0 j s k j+,l j Suppose that there exsts some satsfyng l j = d for all 0 j s Then, for any other block p(s) dfferent from p() there s only one t such that l st = 0 k j

18 8 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG Proof Wthout loss of generalty we can assume the vertces n p() and n p(s) as follows: k s s k Consder the modules Mk s s,l s0 and Mk,l 0 Snce {Mk s s,l s0, Mk,l 0 } s a set of parwse orthogonal stable brcks, by Proposton 44, we must have l s0 = 0 Moreover, by Lemma 40, l s0 s the only one 5 An explct constructon of sms s over self-njectve Nakayama algebras In ths secton, we frst construct sms s of symmetrc Nakayama algebra A dn n theory to deal wth the general case and then use coverng We denote by P the set of non-crossng parttons of n = {,, n} and by ī the postve nteger n n wth = ī mod n for any Z For p P, let p() = {, k, k,, ks } (wth the orderng as n Corollary 48, when s 3) be the block for n and let î be the set {, +,, k } Under these notatons, we ntroduce the followng defnton Defnton 5 Let A dn n be a symmetrc Nakayama algebra and P be the set of non-crossng parttons of the set n For any non-crossng partton p n P and any k n, we defne two types of sets of ndecomposable A dn n -modules as follows 0, î p(k) =, () L p,k = {Mk,l =,, n}, where l = d, otherwse 0, î p(k) =, () S p,k = {Mk,l =,, n}, where l = 0, = k, d, otherwse Remark 5 From the above defnton, f d, we have the followng facts about L p,k and S p,k Let p P and k n be gven The modules Mk,l wth p(k) n L p,k satsfy l = d and for each block p(t) dfferent from p(k), there exsts a unque module Mk,l n L p,k satsfyng p(t) and l = 0 Moreover, for each block p(t), there exsts a unque module Mk,l n S p,k satsfyng p(t) and l = 0 We now prove the followng theorem whch states that both L p,k and S p,k are sms s of A dn n of A dn n are of these forms and all sms s Theorem 53 Let A dn n be a symmetrc Nakayama algebra and P be the set of non-crossng parttons of the set n Then we have the followng (a) For any p P and any k n, L p,k and S p,k are sms s (b) All sms s of A dn n are of these forms (c) If d, then for p, p P and k, k n, we have the followng results: () L p,k S p,k () L p,k = L p,k f and only f p = p and p(k) = p(k ) (3) S p,k = S p,k f and only f the followng three condtons hold: () p = p ; () k = k or k k = ; p(k) p(k () p k k P, where p k k () ), p(k) p(k ), = p(), otherwse Proof (a) We only prove that L p,k s a sms, snce the proof for S p,k s smlar By Theorem 3, t s enough to show that any two objects M k,l and M k,l ( ) n L p,k (p P, k n) form a set of parwse orthogonal stable brcks When d =, l = l = 0, snce p s a non-crossng partton, there are four cases about the arcs of M k,l and M k,l correspondng to the four dagrams of Proposton 44 It follows from Remark 45 that {M k,l, M k,l } s a set of parwse orthogonal stable brcks

19 A NOTE ON SIMPLE-MINDED SYSTEMS 9 When d, by the defnton of L p,k, we consder four cases ()-(4) In each case t s straghtforward to check by Proposton 44 that {M k,l, M k,l } s a set of parwse orthogonal stable brcks We now lst all the cases as follows: () l = l = 0, that s, p(k) =, p(k) = There are three subcases about the arcs of M k,l and M k,l : k k k k k k () l = 0, l = d, that s, p(k) =, p(k) There are three subcases about the arcs of and M k,l : M k,l k k k k k k (3) l = d, l = 0, that s, p(k), p(k) = Ths s smlar to Case () (4) l = l = d, that s, p(k), p(k) There are three subcases about the arcs of and M k,l : M k,l k k k k k k (b) By Corollary 49, any sms S of A dn n determnes a non-crossng partton p n P For n, we denote by p() the block whch belongs to Then we can assume that p() = {, k, k,, ks } such for each 0 j s, that there exsts object M j n S satsfyng top(m j ) = S k j and soc(m j ) = S k j+ where k 0 = ks =, k = k Notce that by our notaton M j = M kj for 0 j s k j+,l (cf Lemma j 40 and Lemma 4) If there s a block p() satsfyng l j = d for each 0 j s, then from the proof of Lemma 4, 0, k t we know for each block p(s) that l st = s p() =, Therefore S = L p, d, otherwse If there s no block p() satsfyng l j = d for each 0 j s, suppose that p( ),, p( k ) are all blocks, by Lemma 40, wthout loss of generalty, we assume l t0 = 0 for any block p( t ) We have l tj = d for j dfferent from 0 Then there exsts some n {,, k } satsfyng î t p() = for any t dfferent from It follows easly that S must have the form S p, (c) () Ths follows easly from Remark 5 More precsely, f d, for each block p (t), there exsts a unque object Mk,l n S p,k wth p (t) and l = 0, however, all the modules Mk,l n L p,k whch corresponds to the block p(k) satsfy l = d for any n ths block Therefore L p,k S p,k for any p, p, k, k () If p = p, p(k) = p(k ), then by defntons of L p,k and L p,k, we have L p,k = L p,k If L p,k = L p,k, then p = p Otherwse, there exst modules Mk,l n L p,k and Mk n L,l p,k wth k k, and therefore M k,l Mk ths contradcts the fact that L, p,k = L,l p,k Assume now that

20 0 JING GUO, YUMING LIU, YU YE AND ZHEN ZHANG L p,k = L p,k We have that l = d for the modules Mk,l n L p,k, where s n p(k) or s n p(k ) By Lemma 4, there s only one such block for L p,k Therefore p(k) = p(k ) (3) If S p,k = S p,k, then p = p Otherwse, there exst modules Mk,l S p,k and Mk,l S p,k wth k k, and therefore M k,l Mk ths contradcts the fact that S, p,k = S,l p,k Assume now that S p,k = S p,k and k k By defnton of S p,k, S p,k = S p,k f and only f the followng condtons hold: k p(k ) = ; k p(k) = ; î p(k) = f and only f î p(k ) = for k, k The frst two condtons are equvalent to k k = Moreover, the last condton mples that p k k s also a non-crossng partton Conversely, f p k k s a non-crossng partton, then clearly the last condton holds Remark 54 () For an equvalent formulaton of S p,k = S p,k, see Remark 65 () For k, k n, f d =, then L p,k = L p,k = S p,k = S p,k Example 55 We descrbe the sms s of the symmetrc Nakayama algebr A 6 usng the set P of non-crossng parttons of Snce P = {p, p }, where p = {{}, {}}, p = {{, }}, we can drectly wrte down all sms s of A 6 from the defntons of L p,k and S p,k : L p, =,, S p, =, L p, =,, S p, =, L p, = L p, =,, S p, = S p, = {, } For general self-njectve Nakayama algebra A l n, by Lemma 9, we know that there s a coverng functor F : A l n-mod A l e-mod, where e s the greatest common dvsor of n and l Consequently, S s a sms of A l e-mod f and only f F (S) s a sms of A l n-mod For the symmetrc Nakayama algebra A l e, we have defned two types of sms s L p,k and S p,k, where p P, k e, and P s the set of non-crossng parttons of e = {,, e} Usng the above coverng functor we defne two classes of objects n A l n-mod as follows: L p,k := F (L p,k ), S p,k := F (S p,k ) Notce that the coverng functor F s nduced from a coverng of stable Auslander-Reten quvers π : sγ A l n s Γ A l e = s Γ A l n / ν (where ν s the Nakayama automorphsm on s Γ A l n ), therefore t s very easy to construct L p,k and S p,k from L p,k and S p,k n practce The followng theorem s clear from coverng theory Theorem 56 Let A l n be a self-njectve Nakayama algebra and P be the set of non-crossng parttons of e, where e s the greatest common dvsor of n and l Then we have the followng (a) For any p P and any k e, L p,k and S p,k are sms s (b) All sms s of A l n are of these forms (c) If l/e, then for p, p P and k, k e, we have the followng results: () L p,k S p,k () L p,k = L p,k f and only f p = p and p(k) = p(k )