HALL ALGEBRAS ASSOCIATED TO TRIANGULATED CATEGORIES, II: ALMOST ASSOCIATIVITY 0 1 FANXU 0 2 Dedicated to Professor YingBo Zhang. n a Abstract. Byusingtheapproachin[8]toHallalgebrasarisinginhomologi- J callyfinitetriangulatedcategories, wefindan‘almost’associativemultiplica- 5 tionstructureforindecomposableobjectsina2-periodictriangulatedcategory. 2 As an application, we give a new proof of the theorem of Peng and Xiao in [5]whichprovidesawayofrealizingsymmetrizableKac-Moodyalgebras and ] ellipticLiealgebrasvia2-periodictriangulatedcategories. T R . h Introduction t a m Let U be the universal enveloping algebra of a simple Lie algebra of type A,D or E over the field of rational numbers Q. There are many interesting results in- [ volving the categorification and the geometrization of U. The work of Gabriel [1] 3 strongly suggested the possibility of the categorification. He showed that there v exists a bijection between isomorphism classes of all indecomposable modules over 8 a hereditary algebra of Dynkin type and the positive roots of the corresponding 8 5 semisimple Lie algebra. In [6], Ringel explicitly realized the positive part of U 5 through the Hall algebra approach. A different but somewhat parallel realization . was given by Lusztig. He showed that the negative part of U can be geometrically 0 1 realized by using constructible functions on affine spaces of representations of a 7 preprojective algebra in [3]. One may naturally consider to recover the whole Lie 0 algebras and the whole (quantized) enveloping algebras [6]. : v Nakajima[4]showedthatanarbitrarilylargefinite-dimensionalquotientofU can i berealizedintermsofthehomologyofatriplevariety. Adifferentconstructionwas X givenby Lusztig in [3] in terms of constructible functions on the triple variety. On r a the other hand, Peng and Xiao [5] defined a Lie bracket between two isomorphism classes of indecomposable objects in a k-additive triangulated category with the translation functor T satisfying T2 = 1 for a finite field k with the cardinality q. It induces a Lie algebra over Z/(q−1), while it is still unknown which associative multiplication induces the Lie bracket over Z/(q −1). However, it seems to be hopeless to realize the whole enveloping algebra by the constructions of Nakajima [4] and Lusztig [3]. Recently, To¨en gave a multiplication formula which defines an associative alge- bra (called the derived Hall algebra) corresponding to a dg category [7]. In [8], Date:October 28,2007. Lastmodified: October 09,2009. 2000 Mathematics Subject Classification. Primary16G20, 17B67;Secondary17B35,18E30. Key words and phrases. 2-periodictriangulatedcategory, envelopingalgebra,Liealgebra. TheresearchwassupportedinpartbythePh.D.ProgramsFoundationofMinistryofEducation ofChina(No. 200800030058). 1 2 FANXU we extended to prove that To¨en’s formula can be applied to define an associative algebra for any triangulated category with some homological finiteness conditions. Unfortunately, a 2-periodic triangulated category does not satisfy these homolog- ical finiteness conditions in general. Hence, To¨en’s formula can not supply the realization of quantum groups. However, the approach in [8] strongly suggests the possibility to construct an associative multiplication over Z/(q−1). Inspiredbythe methoddiscussedin[8],inthis paper,weprovethatthereexists an ‘almost’ associative multiplication over Z[1]/(q − 1) for isomorphism classes q of indecomposable objects in a 2-periodic triangulated category (Corollary 1.8). The associativity of the multiplication heavily depends on the choice of structure constants (Hall numbers) for defining the multiplication. The key techniques in this paper are to substitute derived Hall numbers in [7] or [8] for Hall numbers in [5] and introduce new variants associated to indecomposable objects. As a direct application, we obtain a new proof of the theorem of Peng and Xiao in [5], i.e., Theorem 2.4 in Section 3. 1. The ‘almost’ associativity Given a finite field k with q elements, let C be a k-additive triangulated cate- 2 gory with the translation functor T = [1] satisfying (1) the homomorphism space Hom (X,Y) for any two objects X andY in C is a finite dimensional k-space,(2) C2 the endomorphism ring EndX for any indecomposable object X is finite dimen- sional local k-algebra and (3) T2 ∼= 1. Then the category C is called a 2-periodic 2 triangulated category. For any M ∈C , we set dimM to be the canonical image of 2 M intheGrothendieckgroupofC . Throughoutthispaper,weassumeC isproper, 2 2 i.e.,foranynonzeroindecomposableobjectX inC ,dimX 6=0.ByindC wedenote 2 2 the set of representatives of isomorphism classes of all indecomposable objects in C . ForanyindecomposableobjectX ∈C , wesetd(X)=dim (EndX/radEndX). 2 2 k Recall that (see [5, Lemma 8.1]) |AutX|=|radEndX|(qd(X)−1). For any X,Y and Z in C , we will use fg to denote the composition of morphisms 2 f :X →Y andg :Y →Z,and|A|todenotethecardinalityofafinitesetA.Given X,Y;L∈C , put 2 W(X,Y;L)={(f,g,h)∈hom(X,L)×hom(L,Y)×(Y,X[1])| f g h X −→L−→Y −→X[1] is a triangle}. ThereisanaturalactionofAutX×AutY onW(X,Y;L).Theorbitspaceisdenoted by V(X,Y;L). The orbit of (f,g,h)∈W(X,Y;L) is denoted by (f,g,h)∧. Then (f,g,h)∧ ={(af,gc−1,ch(a[1])−1)|(a,c)∈AutX ×AutY}. We also write FL = |V(X,Y;L)|. Throughout this section, we fix a triple pair XY (X,Y,Z) such that X,Y,Z are nonzero indecomposable objects in C and none of 2 the following conditions holds (1) X ∼=Z ∼=Y[1], (2) X ∼=Y ∼=Z[1], (3) Y ∼=Z ∼=X[1]. HALL ALGEBRAS ASSOCIATED TO TRIANGULATED CATEGORIES, II: ALMOST ASSOCIATIVITY3 Lemma 1.1. Let X,Y,Z and M 6= 0 be in C with X,Y,Z indecomposable and 2 M ⊕Z ≇X ⊕Y. Then we have FM⊕Z 1 XY ∈Z[ ]. |AutZ| q where Z[1] is the polynomial ring for 1 with coefficients in Z. q q Proof. We define the action of AutZ on V(X,Y;M ⊕Z) as follows. For any α = g ( f f , 1 ,h)∧ ∈V(X,Y;M ⊕Z) and d∈AutZ, define 1 2 (cid:18) g2 (cid:19) (cid:0) (cid:1) g g d.( f f , 1 ,h)∧ =( f f d−1 , 1 ,h)∧. 1 2 (cid:18) g2 (cid:19) 1 2 (cid:18) dg2 (cid:19) (cid:0) (cid:1) (cid:0) (cid:1) The orbit space is denoted by V(X,Y;M ⊕Z). Let g g G ={d∈AutZ |( f f d−1 , 1 ,h)∧ =( f f , 1 ,h)∧}. α 1 2 (cid:18) dg2 (cid:19) 1 2 (cid:18) g2 (cid:19) (cid:0) (cid:1) (cid:0) (cid:1) Then by definition, G is equal to α g g b−1 {d∈AutZ |( f f d−1 , 1 ,h)=( af af , 1 ,bh(a[1])−1) 1 2 (cid:18) dg2 (cid:19) 1 2 (cid:18) g2b−1 (cid:19) (cid:0) (cid:1) (cid:0) (cid:1) for some (a,b)∈AutX×AutY}, or equivalently, equal to {d∈AutZ |dg =g b,g b=g ,af =f and f d=af 2 2 1 1 1 1 2 2 for some (a,b)∈AutX×AutY}. Given(b,d)∈AutY×AutZ suchthatdg =g bandg b=g ,wehavethefollowing 2 2 1 1 diagram with the middle square being commutative g  1  f f g (cid:16) 1 2 (cid:17)  2  h X // M ⊕Z //Y // X[1] OO OO 1 0   b 0 d   X // M ⊕Z //Y // X[1] f f g h (cid:16) 1 2 (cid:17)  1  g  2  Bytheaxiomsoftriangulatedcategories,thereexistsa∈AutX suchthataf =f 1 1 and f d=af . Hence, 2 2 G ={d∈AutZ |dg =g b and g b=g for some b∈AutY}. α 2 2 1 1 The map g naturally induces a triangle 1 M g1 //Y h1 // C(g1) //M[1] where C(g ) is the cone of the map g . Let b′ =1−b. Then we have 1 1 {b′ |b∈AutY |g b=g }={b′ ∈EndY |b′ ∈h Hom(C(g ),Y),1−b′ ∈AutY}. 1 1 1 1 4 FANXU We claim that for any t∈Hom(C(g ),Y), both th andh t are nilpotent. Assume 1 1 1 that h t is not nilpotent, then it is invertible since Y is indecomposable. This 1 implies g =0. Consider the following diagram 1 M //M // 0 //M[1] u v  0  (cid:15)(cid:15) (cid:16) f1 f2 (cid:17) (cid:15)(cid:15)  g2  (cid:15)(cid:15) h X // M ⊕Z //Y // X[1] where v = 1 0 . The middle square is commutative, then there exists u : M →X such(cid:0) that th(cid:1)e diagram is commutative. This implies uf1 =1M. However, X is indecomposable and then u is an isomorphism. The morphism g is also an 2 isomorphism by the octahedral axiom as showed in the following diagram. Z Z  0  g2 (cid:16) f1 f2 (cid:17) (cid:15)(cid:15)  g2  (cid:15)(cid:15) h X //M ⊕Z //Y //X[1] X f1 // M(cid:15)(cid:15) //C(g(cid:15)(cid:15) 2) // X[1] where C(g ) is the cone of g . It contradicts to the assumption M ⊕Z ∼= X ⊕Y. 2 2 Hence, g 6= 0. In the same way, we have g 6= 0. This implies that for any 1 2 t ∈ Hom(C(g ),Y), both th and h t are nilpotent. We have 1 −h t ∈ AutY. 1 1 1 1 Therefore, G is isomorphic to α G′ ={d′ ∈EndZ |1−d′ ∈AutZ,d′g =g b′ for some b′ ∈h Hom(C(g ),Y)}. α 2 2 1 1 Let d′ ∈ EndZ satisfy d′g = g b′ for some b′ ∈ h hom(C(g ),Y)}. Since b′ is 2 2 1 1 nilpotent, we assume (b′)k = 0 for some k ∈ N, then (d′)kg = 0. However, Z is 2 indecomposableandg 6=0. Wededucethatd′ isnilpotentandthen1−d′ ∈AutZ. 2 Hence, G′ is a vector space. Finally, we obtain α FM⊕Z 1 1 XY = ∈Z[ ]. |AutZ| |G | q X α α∈V(X,Y;M⊕Z) (cid:3) Note that the conclusion of the lemma may not hold if M ⊕ Z ∼= X ⊕ Y in Lemma 1.1, then . Indeed, if X ≇ Y, FX⊕Y = |Hom(X,Y)| = qn where n = XY dimkHom(X,Y). If X ∼=Y, FXXY⊕Y =|EndX|+|radEndX|. Denoteby(X,Y) thesubsetofHom (X,Y)consistingofthemorphismswhose Z C2 conesareisomorphictoZ.Thefollowingproposition([8,Proposition2.5])alsoholds for 2-periodic triangulated categories. Proposition 1.2. For any Z,L,M ∈C , we have 2 (1) Any α=(l,m,n)∧ ∈V(Z,L;M) has the representative of the form: 0 n 0    11  l 0 m 0 n  2  (cid:16) 2 (cid:17)  22  Z //M // L //Z[1] HALL ALGEBRAS ASSOCIATED TO TRIANGULATED CATEGORIES, II: ALMOST ASSOCIATIVITY5 where Z = Z (α) ⊕ Z (α), L = L (α) ⊕ L (α), n is an isomorphism 1 2 1 2 11 between L (α) and Z (α)[1] and n ∈radHom(L (α),Z (α)[1]). 1 1 22 2 2 (2) |(M,L) | |EndL (α)| Z[1] 1 = |AutL| |n(α)Hom(Z[1],L)||AutL (α)| α∈VX(Z,L;M) 1 and |(Z,M) | |EndZ (α)| L 1 = |AutZ| |Hom(Z[1],L)n(α)||AutZ (α)| α∈VX(Z,L;M) 1 n 0 where n(α)= 11 . (cid:18) 0 n22 (cid:19) For α∈(l,m,n)∧, define s(α)=dim nHom(Z[1],L), t(α)=dim Hom(Z[1],L)n. k k For any objects U,V and W in C , define 2 (1) gW := |(U,W)V|, if U ≇W ⊕V[1], UV |AutU| (2) g¯W := |(W,V)U[1]|, if V ≇W ⊕U[1], UV |AutV| (3) gW :=|Hom(W,V[1])|· |(W⊕V[1],W)V| = 1 , W⊕V[1],V |Aut(W⊕V[1])| |AutV| (4) g¯W :=|Hom(U[1],W)|· |(W,W⊕U[1])U[1]| = 1 . U,W⊕U[1] |Aut(W⊕U[1])| |AutU| Different from[5], we will consider the image of numbers in Z[1]/(q−1) instead q of Z/(q−1) where Z[1] is the polynomial ring for 1 with coefficient in Z. q q We have the following corollary of Proposition 1.2. Proposition1.3. LetX,Y,Z,L,L′andM beinC withX,Y,Z andM 6=0being 2 indecomposable. Then we have the following properties. (1) If L≇M ⊕Z[1] and L6=0, then the number gL gM belongs to Z[1]. XY ZL q (2) IfL∼=M⊕Z[1]andL≇X⊕Y,thenthenumbergL gM belongstoZ[1]. XY ZL q (3) If L′ ≇M ⊕Y[1] and L′ 6=0, then the number gL′ gM belongs to Z[1]. ZX L′Y q (4) If L′ ∼=M ⊕Y[1] and L′ ≇X ⊕Z, then gZLX′ gLM′Y belongs to Z[q1]. (5) If X ≇ Y and X ≇ Y[1], then the numbers gX⊕YgY −g0 gY XY X[1],X⊕Y X[1],X 0,Y and gX⊕YgX −gX⊕Y[1]gX belong to Z[1]. XY Y[1],X⊕Y Y[1],X X⊕Y[1],Y q (6) IfZ ≇X andZ ≇X[1],thenthenumbergZ⊕XgZ −g0 gZ and ZX Z⊕X,X[1] X,X[1] Z,0 gZ⊕XgX −gX⊕Z[1]gX belong to Z[1]. ZX Z⊕X,Z[1] X,Z[1] Z,X⊕Z[1] q (7) gL gM −gL g¯M ∈Z[1]. In Z[1]/(q−1), gL gM −gL g¯M =0. XY ZL XY ZL q q XY ZL XY ZL (8) gL′ gM −g¯L′ gM ∈Z[1]. In ∈Z[1]/(q−1), gL′ gM −g¯L′ gM =0. ZX L′Y ZX L′Y q q ZX L′Y ZX L′Y Proof. If L≇M ⊕Z[1], then by Proposition 1.2, we have L (α)=0 and 1 1 1 gM = ∈Z[ ]. ZL |n(α)Hom(Z[1],L)| q α∈VX(Z,L;M) 6 FANXU In the same way, gL ∈ Z[1]. This proves (1). Next, we prove (2). In this case, XY q gM = 1 . As in the proof of Lemma 1.1, gM⊕Z[1] is equal to ZL |AutZ| XY 1 1 |AutZ| = . |h(α)Hom(X[1],Y)| |h(α)Hom(X[1],Y)| |G | α∈V(X,XY;M⊕Z[1]) α∈V(X,XY;M⊕Z[1]) α This proves (2). The proofs of (3) and (4) are similar. As for (5), the number gX⊕YgY −g0 gY is equal to XY X[1],X⊕Y X[1],X 0,Y |Hom(X,Y)|−1 qdimkHom(X,Y)−1 = . |AutX| |radEndX|(qd(X)−1) Since dimkHom(X,Y) = l (Hom(X,Y)) ∈ Z, the number belongs to Z[1]. The d(X) EndX q proofs of the rest part of (5) and (6) are similar. We prove (7). If L = 0, then gL gM −gL g¯M = 1 − 1 = 0. If L ∼= M ⊕Z[1], then gM = g¯M = XY ZL XY ZL |AutX| |AutX| ZL ZL 1 . Hence, gL gM −gL g¯M = 0. If L ≇ M ⊕ Z[1] and L 6= 0, then by |AutZ| XY ZL XY ZL Proposition1.2,wehavegL ∈Z[1]andgM −g¯M = ( 1 − 1 )∈ XY q ZL ZL α∈V(Z,L;M) qs(α) qt(α) Z[1]. This proves (7). The proof of (8) is similar. P (cid:3) q Corollary 1.4. Let X,Y,Z and M be in C with X,Y,Z and M 6= 0 being inde- 2 composable. Then we have gXLYgZML− gZLX′ gLM′Y ∈Z[1q] [L]X,L∈C2 [L′]X,L′∈C2 and in Z[1]/(q−1), q gXLYgZML− gZLX′ gLM′Y = gXLYg¯ZML− g¯ZL′XgLM′Y. [L]X,L∈C2 [L′]X,L′∈C2 [L]X,L∈C2 [L′]X,L′∈C2 Proof. As for the firststatement of the corollary,by Proposition1.3 (1) and (2), it is sufficient to prove that the number 1 gX⊕YgM +δ g0 gM −gZ⊕XgM −δ g0 gM ∈Z[ ]. XY Z,X⊕Y X,Y[1] XY Z,0 ZX Z⊕X,Y X,Z[1] ZX 0,Y q If X ⊕Y ≇M ⊕Z[1] and Z⊕X ≇M ⊕Y[1], by Proposition1.3 (1) and (3), this number belongs to Z[1]. Hence, we only need to check the following cases: q (1) M ∼=X,Y ∼=Z[1],X ≇Y,X ≇Y[1], (2) M ∼=Y,X ∼=Z[1],Y ≇X,Y ≇X[1], (3) M ∼=Z,X ∼=Y[1],Z ≇X,Z ≇X[1]. We note that the cases (1),(2) and (3) are symmetric to each other. Proposition 1.3 (5) and (6) correspond to the cases (1),(2) and (3), respectively. This proves the first statement of the corollary. The second statement of the corollary can be deduced by the first statement and Proposition 1.3 (7) (8). (cid:3) Here we recall some notations in [8]. Let X,Y,Z,L,L′ and M be in C . Then 2 define Hom(M ⊕X,L)Y,Z[1] :={ m ∈Hom(M ⊕X,L)| L′[1] (cid:18) f (cid:19) m Cone(f)≃Y,Cone(m)≃Z[1] and Cone ≃L′[1]} (cid:18) f (cid:19) HALL ALGEBRAS ASSOCIATED TO TRIANGULATED CATEGORIES, II: ALMOST ASSOCIATIVITY7 and Hom(L′,M ⊕X)Y,Z[1] :={(f′,−m′)∈Hom(L′,M ⊕X)| L Cone(f′)≃Y,Cone(m′)≃Z[1] and Cone(f′,−m′)≃L}. The orbit space of Hom(M ⊕X,L)Y,Z[1] under the action of AutL is just the orbit L′[1] space of Hom(L′,M⊕X)Y,Z[1] under the actionof AutL′ (see [8]). We denoted by L V(L′,L;M⊕X) theorbitspace. Thefollowingdiagramillustratestherelation Y,Z[1] amongV(X,Y;L),V(Z,L;M),V(Z,X;L′),V(L′,Y;M)andV(L′,L;M⊕X) . Y,Z[1] (1.1) Z Z l′ l L(cid:15)(cid:15)′ f′ // M(cid:15)(cid:15) g′ //Y h′ // L′[1] m′ m m′[1] (cid:15)(cid:15) f (cid:15)(cid:15) g h (cid:15)(cid:15) X //L // Y //X[1] n′ n (cid:15)(cid:15) (cid:15)(cid:15) Z[1] Z[1] Lemma 1.5. Let α∈V(L′,L;M ⊕X) has the representative of the form as Y,Z[1] follows: 0 0 0 m θ 0      1  f′ −m′ 0 f 0 θ      2  L′ ⊕L′ //M ⊕X //L ⊕L // L′[1]⊕L′[1] 1 2 1 2 1 2 such that θ is a nonzero isomorphism where L=L ⊕L and L′ =L′ ⊕L′. Then 1 1 2 1 2 we have M ∼=X, Z ∼=Y[1], L∼=X⊕Y, L′ ∼=X⊕Y[1]. Proof. By definition, the triangle α induces the triangles g  1  0 f g (cid:16) (cid:17)  2  X //L1⊕L2 // Y //X[1] and 0   f′   L′ ⊕L′ //M // Y //L′[1]⊕L′[1]. 1 2 1 2 Consider the following diagram g  1  0 f g (cid:16) (cid:17)  2  X //L1⊕L2 //Y //X[1]. v (cid:15)(cid:15) (cid:15)(cid:15) 1 (cid:15)(cid:15) 0 // L //L //0 1 1 The left square is commutative. Then there exists a map v such that g v = 1. 2 Since Y is indecomposable, we deduce Y ∼= L and X ∼= L . In the same way, 1 2 8 FANXU we deduce L′ ∼= Y[1],L′ ∼= M. If we consider the other two induced triangles, we 1 2 obtain L ∼=Z[1],L ∼=M and L′ ∼=Z,L′ ∼=X. (cid:3) 1 2 1 2 We define the action of AutX on V(L′,L;M ⊕X) by Y,Z[1] m m a.((f′,−m′), ,θ))∧ =((f′,−m′a−1), ,θ))∧ (cid:18) f (cid:19) (cid:18) af (cid:19) m foranya∈AutX and((f′,−m′), ,θ))∧ ∈V(L′,L′;M⊕X) .Theorbit (cid:18) f (cid:19) Y,Z[1] space is denoted by V¯(L′,L′;M ⊕X) . Y,Z[1] Lemma1.6. Foranyα∈V(L′,L;M⊕X) ,thestablesubgroupG ofαunder Y,Z[1] α the action of AutX is isomorphic to a vector space if L6=0 and L′ 6=0. Proof. By definition, G ={a∈AutX |af =fb for some b∈AutL such that mb=m}. α (1) If f = 0, then Y ∼=L⊕X[1]. However, Y is indecomposable, so L=0. Hence, Y ∼=X[1] and Z ∼=M. This imply L′ ∼=Z⊕X. In the following, we assume f 6=0. (2) If L ≇ M ⊕Z[1], then mb = b implies b = 1−b′ with b′ ∈ nhom(Z[1],L) nilpotent. In this case, G is isomorphic to α G′ ={a′ ∈EndX |1−a′ ∈AutX,a′f =fb′ for some b′ ∈nhom(Z[1],L)}. α Sinceb′ isnilpotent,a′ isnilpotent. AndsinceX isindecomposable,1−a′ ∈AutX. We claim that G′ is a vector space. Indeed, 0∈G′ . For any a′,a′ ∈G′ , we have α α 1 2 α a′f =fb′ for i=1,2 and some b′,b′ ∈nhom(Z[1],L). So (a′ +a′)f =f(b′ +b′). i i 1 2 1 2 1 2 It is clear that b′ +b′ ∈nhom(Z[1],L) and 1−(a′ +a′)∈AutX since a′ +a′ is 1 2 1 2 1 2 nilpotent. This shows a′ +a′ ∈G′ . 1 2 α (3) If L∼=M ⊕Z[1], α has the representative of the following form: 1 0   f′ −m′ f f L′(cid:16) (cid:17)//M ⊕X  1 2 //M ⊕Z[1] θ //L′[1] . The stable subgroup is 1 0 G ={a∈AutX |af =fb for some b= ∈AutL} α (cid:18) b21 b22 (cid:19) where f =(f ,f ). Equivalently, we have 1 2 af =f +f b and af =f b . 1 1 2 21 2 2 22 Set a′ =a−1 and b′ =b −1. If a′ ∈AutX, then 22 22 1 0 a′ f f = 0 a′f . (cid:0) 1 2 (cid:1)(cid:18) −(b′22)−1b21 1 (cid:19) (cid:0) 2 (cid:1) Since Y is indecomposable, this shows X ∼= Z[1] and Y ∼= M. In this case, L′ = 0. (cid:3) For any X ∈ C , we denote by u by the isomorphism class of X. Define the 2 X multiplication by u u := gL u X Y YX L X[L] and we set u u (L)=gL . X Y XY HALL ALGEBRAS ASSOCIATED TO TRIANGULATED CATEGORIES, II: ALMOST ASSOCIATIVITY9 Theorem 1.7. Let X,Y,Z and M 6= 0 be indecomposable objects in C . Then in 2 Z[1]/(q−1) we have q (1) If X ∼=Y[1],X ≇Z, X ≇Z[1] and M ∼=Z, then dim Hom(Z,X) k [(u u )u −u (u u )](M)=− . Y X Z Y X Z d(X) (2) If X ∼=Z[1],X ≇Y,X ≇Y[1] and M ∼=Y, then dim Hom(X,Y) k [(u u )u −u (u u )](M)= . Y X Z Y X Z d(X) (3) Otherwise, [(u u )u −u (u u )](M)=0. Y X Z Y X Z Proof. Itisequivalenttocomputerthenumber gL gM − gL′ gM . [L],L∈C2 XY ZL [L′],L′∈C2 ZX L′Y By Corollary 1.4, we have P P gXLYgZML− gZLX′ gLM′Y [L]X,L∈C2 [L′]X,L′∈C2 = gXLYg¯ZML− g¯ZLX′ gLM′Y [L]X,L∈C2 [L′]X,L′∈C2 |(X,L) | |(M,L) | Y Z[1] = · |AutX| |AutL| X [L],L≇M⊕Z[1]∈C2 |(L′,X) | |(L′,M) | Z[1] Y − · |AutX| |AutL′| [L′],L′≇XM⊕Y[1]∈C2 |(X,M ⊕Z[1]) | |(M,M ⊕Z[1]) | Y Z[1] +|Hom(Z[1],M)|· · |AutX| |Aut(M ⊕Z[1])| |(M ⊕Y[1],X) | |(M ⊕Y[1],M) | Z[1] Y −|Hom(M,Y[1])|· · |AutX| |Aut(M ⊕Y[1])| 1 |Hom(M ⊕X,L)YL,′Z[1[]1]| = ·δ · |AutX| L,M⊕Z[1] |AutL| [L],[L′X];L,L′∈C2 1 |Hom(L′,M ⊕X)Y,Z[1]| − ·δ′ · L |AutX| L′,M⊕Y[1] |AutL′| [L],[L′X];L,L′∈C2 whereδL,M⊕Z[1] =|Hom(Z[1],M)|forL∼=M⊕Z[1]and1otherwise,δL′,M⊕Y[1] = |Hom(M,Y[1])|forL′ ∼=M⊕Y[1]and1otherwise. ByProposition1.2andLemma 1.5, unless L∼=X ⊕Y and L′ ∼=X ⊕Y[1], the above sum is equal to 1 1 1 ( ( − ))· qs(α) qt(α) |AutX| [L],[L′X];L,L′∈C2 α∈V(L′,LX;M⊕X)Y,Z[1] where s(α)=dim θHom(L′[1],L)−δ dim Hom(Z[1],M) and k L,M⊕Z[1] C t(α)=dimkHom(L′[1],L)θ−δL′,M⊕Y[1]dimCHom(M,Y[1]). Here, θ is a morphism from L to L′[1] in a triangle belonging to α. Consider the action of AutX on V(L′,L;M ⊕X) , we always have s(α) = s(a.α) and Y,Z[1] 10 FANXU t(α)=t(a.α) for any a∈AutX and α∈V(L′,L;M ⊕X) . Hence, the sum is Y,Z[1] again equal to 1 1 |AutX| 1 ( ( − )· )· qs(α) qt(α) |G | |AutX| [L],[L′X];L,L′∈C2 α¯∈V¯(L′,LX;M⊕X)Y,Z[1] α 1 1 1 = ( ( − )· ) qs(α) qt(α) |G | [L],[L′X];L,L′∈C2 α¯∈V¯(L′,LX;M⊕X)Y,Z[1] α whereGα isthestablesubgroupforα∈V(L′,L;M⊕X)Y,Z[1].IfL=0,L′ ∼=X⊕Z or L∼=X⊕Y,L′ =0, then it is easy to know |Hom(M ⊕X,L)YL,′Z[1[]1]| = |Hom(L′,M ⊕X)YL,Z[1]| =1. |AutL| |AutL′| Hence, if X ∼= Y[1],X ≇ Z, X ≇ Z[1] and M ∼= Z (i.e., L = 0 and L′ ∼= X ⊕Z), then 1 1 [(u u )u −u (u u )](M)= − ·|Hom(Z,X)|. Y X Z Y X Z |AutX| |AutX| By [5, Lemma 8.1], we have 1 1 dim Hom(Z,X) k − ·|Hom(Z,X)|=− . |AutX| |AutX| d(X) If X ∼=Z[1],X ≇Y,X ≇Y[1] and M ∼=Y (i.e., L∼=X ⊕Y and L′ =0), then 1 1 [(u u )u −u (u u )](M)= ·|Hom(X,Y)|− . Y X Z Y X Z |AutX| |AutX| Otherwise, G is isomorphic to a vector space by Lemma 1.6. Therefore, the sum α vanishes in Z[1]/(q−1). Now we assume that L∼=X ⊕Y and L′ ∼=X ⊕Y[1]. We q note that Lemma 1.5 shows that M ∼= X and Z ∼= Y[1] in this case. By Lemma 1.2, we have |Hom(M ⊕X,L)YL,′Z[1[]1]| |V(L′,L;M ⊕X)Y,Z[1]| = |AutL| |AutX|·|Hom(Y,X)| and |Hom(L′,M ⊕X)Y,Z[1]| |V(L′,L;M ⊕X)Y,Z[1]| L = |AutL′| |AutX|·|Hom(X[1],Y)| for L∼=X⊕Y, L′ ∼=X⊕Y[1],M ∼=X and Z ∼=Y[1]. In this case, we deduce that |V(L′,L;M ⊕X) | |V(L′,L;M ⊕X) | Y,Z[1] Y,Z[1] |Hom(Z[1],M)|· −|Hom(M,Y[1])|· |AutX|·|Hom(Y,X)| |AutX|·|Hom(X[1],Y)| vanishes. This completes the proof of the theorem. (cid:3) We note that the theorem is a refinement of Proposition 7.5 in [5] which is a crucial point to prove the Jacobi identity in [5]. For any X ∈ indC , we introduce a new variant θ associated to X. Define 2 X a new multiplication between indecomposable objects by the following rules. For X,Y ∈C , 2 (1) if X ≇Y[1], u ·u := gL u , X Y [L] YX L (2) u ·u := gL P+θ , X X[1] [L] X[1]X X[1] P