Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras

Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras

We define and investigate Lie algebras associated with quadratic forms. We also present their connections with Lie algebras and Ringel-Hall algebras associated with representation directed algebras.

Gespeichert in:
Bibliographische Detailangaben
Veröffentlicht in:Algebra and Discrete Mathematics
Datum:2008
1. Verfasser: Kosakowska, J.
Format: Artikel
Sprache:English
Veröffentlicht: Інститут прикладної математики і механіки НАН України 2008
Online Zugang:https://nasplib.isofts.kiev.ua/handle/123456789/153372
Tags: Tag hinzufügen
Keine Tags, Fügen Sie den ersten Tag hinzu!
Назва журналу:Digital Library of Periodicals of National Academy of Sciences of Ukraine
Zitieren:Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras / J. Kosakowska // Algebra and Discrete Mathematics. — 2008. — Vol. 7, № 4. — С. 49–79. — Бібліогр.: 15 назв. — англ.

Institution

Digital Library of Periodicals of National Academy of Sciences of Ukraine
id nasplib_isofts_kiev_ua-123456789-153372
record_format dspace
spelling Kosakowska, J.
2019-06-14T03:39:30Z
2019-06-14T03:39:30Z
2008
Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras / J. Kosakowska // Algebra and Discrete Mathematics. — 2008. — Vol. 7, № 4. — С. 49–79. — Бібліогр.: 15 назв. — англ.
1726-3255
2000 Mathematics Subject Classification: 17B20.
https://nasplib.isofts.kiev.ua/handle/123456789/153372
We define and investigate Lie algebras associated with quadratic forms. We also present their connections with Lie algebras and Ringel-Hall algebras associated with representation directed algebras.
Partially supported by Research Grant No. N N201 269 135 of Pol ish Ministry of Science and High Education
en
Інститут прикладної математики і механіки НАН України
Algebra and Discrete Mathematics
Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras
Article
published earlier
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection DSpace DC
title Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras
spellingShingle Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras
Kosakowska, J.
title_short Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras
title_full Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras
title_fullStr Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras
title_full_unstemmed Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras
title_sort lie algebras associated with quadratic forms and their applications to ringel-hall algebras
author Kosakowska, J.
author_facet Kosakowska, J.
publishDate 2008
language English
container_title Algebra and Discrete Mathematics
publisher Інститут прикладної математики і механіки НАН України
format Article
description We define and investigate Lie algebras associated with quadratic forms. We also present their connections with Lie algebras and Ringel-Hall algebras associated with representation directed algebras.
issn 1726-3255
url https://nasplib.isofts.kiev.ua/handle/123456789/153372
citation_txt Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras / J. Kosakowska // Algebra and Discrete Mathematics. — 2008. — Vol. 7, № 4. — С. 49–79. — Бібліогр.: 15 назв. — англ.
work_keys_str_mv AT kosakowskaj liealgebrasassociatedwithquadraticformsandtheirapplicationstoringelhallalgebras
first_indexed 2025-11-25T07:17:18Z
last_indexed 2025-11-25T07:17:18Z
_version_ 1850510104032444416
fulltext Algebra and Discrete Mathematics RESEARCH ARTICLE Number 4. (2008). pp. 49 – 79 c© Journal “Algebra and Discrete Mathematics” Lie algebras associated with quadratic forms and their applications to Ringel-Hall algebras Justyna Kosakowska Communicated by V. V. Kirichenko Abstract. We define and investigate Lie algebras associated with quadratic forms. We also present their connections with Lie algebras and Ringel-Hall algebras associated with representation directed algebras. 1. Introduction Let q be a unit integral quadratic form q(x) = q(x(1), . . . , x(n)) = n∑ i=1 x(i)2 + ∑ i,j aijx(i)x(j), where aij ∈ {−1, 0, 1}. In [4], with q complex Lie algebras G(q), G̃(q) are associated, where G̃(q) is the extension of G(q) by the C-dual of the radical of q and C is the complex number field. The following facts were proved in [4]. • If q is positive definite and connected, then G(q) = G̃(q) is a finite dimensional simple Lie algebra. • If q is connected and non-negative of corank one or two, then G̃(q) is isomorphic to an affine Kac-Moody algebra (if the corank of q equals one) or to elliptic (if the corank of q equals two and q is not of Dynkin type An). Partially supported by Research Grant No. N N201 269 135 of Polish Ministry of Science and High Education 2000 Mathematics Subject Classification: 17B20. Key words and phrases: quadratic form, Lie algebra, Ringel-Hall algebra. 50 Lie algebras associated with quadratic forms In [4], the Lie algebra G(q) was defined by generators and relations. Unfortunately, the set of relations defining G(q) is infinite. In [5], the authors give a finite and small set of relations sufficient to define G(q) for positive definite forms q. In this paper, for any integral quadratic form q (2.1), we define by generators and relations, a Lie algebra L(q, r). For a positive definite form q, we describe a minimal set of relations defining L(q, r). Moreover we show that, for any representation directed C-algebra A with Tits form qA, there are isomorphisms of Lie algebras L(qA, r) ∼= L(A) ∼= K(A), where L(A) is the Lie algebra associated with A in [11] by Ch. Riedtmann and K(A) is the Lie algebra associated with A in [13] by C. M. Ringel. The isomorphism L(A) ∼= K(A) is proved in [7]. Results of the present paper allow us to define Lie algebras L(A) and K(A) in a combinatorial way. Similar results are presented in [9] and [10] for Tits forms of posets of finite prinjective type. The paper is organised as follows. In Section 2 we give basic definitions and facts concerning weakly positive and positive definite quadratic forms and their roots. In Sections 3, 4 we give a definition and prove basic properties of the Lie algebra L(q, r). Moreover (for some class of quadratic form q) we show that the Lie algebra L(q, r) is a Lie subalgebra of G(q). In Section 5 we prove the existence of isomorphisms of Lie algebras L(qA, r) ∼= L(A) ∼= K(A), for any representation directed C-algebra A. Moreover we present appli- cations of these results to Ringel-Hall algebras of representation directed algebras. In Section 6 we give a minimal set of relations sufficient to define the Lie algebra L(q, r), where q is a positive definite quadratic form. More precisely we prove the following theorem. Theorem 1.1. Let q be a positive definite quadratic form (2.1). There is an isomorphism of Lie algebras L(q, r) ∼= L(q)/(j), where L(q) is the free complex Lie algebra generated by the set {v1, . . . , vn} and (j) is the ideal of L(q) generated by the set j, which is consisted of the following elements J. Kosakowska 51 • [vi, vj ] for all i, j ∈ {1, . . . , n} such that i < j and aij 6= −1, • [vi, [vi, vj ]] for all i, j ∈ {1, . . . , n} such that i < j and aij = −1, • [vj , [vi, vj ]] for all i, j ∈ {1, . . . , n} such that i < j and aij = −1, • [vi1 , . . . , vim ] for all positive chordless cycles (i1, . . . , im) (see Sec- tion 6 for definitions). Finally, in Section 7, we present some examples and remarks. 2. Preliminaries on weakly positive and positive definite quadratic forms Let e1, . . . , en be the standard basis of the free abelian group Zn. Let q : Zn → Z be a connected unit integral quadratic form defined by q(x) = q(x(1), . . . , x(n)) = n∑ i=1 x(i)2 + ∑ i,j aijx(i)x(j), (2.1) where aij ∈ {−1, 0, 1}. Let B(q) be the bigraph associated with q (i.e. the set of vertices of B(q) is {1, . . . , n}; for i 6= j there exists a solid edge i j if and only if aij = −1 and a broken edge i ___ j if and only if aij = 1). An integral quadratic form q is said to be weakly positive, if q(x) > 0 for any 0 6= x ∈ Nn, where N is the set of non-negative integers. A vector x ∈ Zn is called a root of q, if q(x) = 1; if in addition x(i) ≥ 0, for any i = 1, . . . , n, then we call x a positive root. Denote by Rq = {x ∈ Zn ; q(x) = 1}, R+ q = {x ∈ Nn ; q(x) = 1} (2.2) the set of all roots and all positive roots of q, respectively. We associate with q the symmetric Z-bilinear form 〈−,−〉q : Zn × Zn → Z, (2.3) where 〈x, y〉q = q(x+y)−q(x)−q(y), for all x, y ∈ Z. It is straightforward to check that 〈ei, x〉q = 2 · x(i) + ∑ i6=j aijx(j), (2.4) for any i = 1, . . . , n. Let us recall the following useful facts concerning the Z-linear form 〈ei,−〉q (see [12]). 52 Lie algebras associated with quadratic forms Lemma 2.5. Let q : Zn → Z be an integral quadratic form (2.1) and let i ∈ {1, . . . , n}. (a) 〈ei, ei〉q = 2. (b) Let x be a root of q and 0 6= d ∈ Z. The vector x − dei is a root of q if and only if d = 〈ei, x〉q. (b′) Let x, y be roots of q. The vector x + y is a root of q if and only if 〈x, y〉q = −1. Assume that q is positive definite. (c) If x is a root of q such that x 6= ei, then −1 ≤ 〈ei, x〉q ≤ 1. (d) The set Rq of all roots of q is finite. Assume that q is weakly positive. (e) If x is a root of q, then −1 ≤ 〈ei, x〉q. (f) The set R+ q of all positive roots of q is finite. Lemma 2.6. Let q : Zn → Z be a weakly positive quadratic form and let z 6= 0 be a positive root of q. Then z is a Weyl root, i.e. there exists a chain x(1), . . . , x(m) of positive roots of q such that (a) x(1) = z, x(i) = x(i−1) − eji for i = 1, . . . , m and for some ji ∈ {1, . . . , n}, (b) x(m) = ej, for some j ∈ {1, . . . , n}. 3. Lie algebras associated with quadratic forms In a complex Lie algebra L, we use the following multibracket notation [y1, y2, . . . , yn] = [y1, [y2, [. . . [yn−1, yn]]]], (3.1) for all y1, . . . , yn ∈ L. Let L be a complex Lie algebra generated by a set {v1, . . . , vn}. Elements of the form [vi1 , . . . , vim ] we call standard multibrackets. All Lie algebras considered in this paper are assumed to be complex finitely generated Lie algebras and all quadratic forms are assumed to be integral quadratic forms (2.1). Let Sn be the symmetric group of n-elements. J. Kosakowska 53 Lemma 3.2. (a) Let L be a Lie algebra and let y1, . . . , yn, x ∈ L. There exists a subset S ⊆ Sn and for any σ ∈ S there exists εσ ∈ {0, 1}, such that [[y1, . . . , yn], x] = ∑ σ∈S (−1)εσ [yσ(1), . . . , yσ(n), x]. (b) Let L be a Lie algebra generated by a set {y1, . . . , yn}. Any ele- ment y ∈ L is a linear combination of standard multibrackets [yi1 , . . . , yim ], where ij ∈ {1, . . . , n}. Proof. (a) Apply recursively the Jacobi identity, see also [1, Lemma 1.1]. The statement (b) follows from (a). With a quadratic form q (2.1), we associate the complex free Lie algebra L(q) = LieC〈v1, . . . , vn〉 (3.3) generated by the set {v1, . . . , vn}. Note that the Lie algebra L(q) has a Nn-gradation, if we define the degree of vi to be ei, for any i = 1, . . . , n. Let a ⊆ L(q) be a subset consisting of some standard multibrackets and let (a) be the homogeneous ideal of the Lie algebra L(q) generated by the set a. Let L(q, a) = L(q)/(a) (3.4) be the quotient Lie algebra with induced Nn-grading. Denote by π : L(q) → L(q, a) the natural epimorphisms. Let v = [vi1 , . . . , vim ] ∈ L(q). For the sake of simplicity, we will denote by v = [vi1 , . . . , vim ] the element π(v). • We call an element v = [vi1 , . . . , vim ] ∈ L(q) a root, if m ≥ 1 and 〈eik , eik+1 + . . . + eim〉v = −1, for all k = 1, . . . , m − 1. • Let v = [vi1 , . . . , vim ] ∈ L(q), we set ℓ(v) = m and we call this number the length of the element v. • We set ev = ei1 + . . . + eim ∈ Nn. For e ∈ Nn, denote by L(q, a)e the homogeneous space spanned by all standard multibrackets v = [vi1 , . . . , vim ] ∈ L(q, a), of degree ev = e. Moreover, for any integer m, let L(q, a)m = ⊕ e∈Nn e(1)+...+e(n)≤m L(q, a)e and (a)m = (a) ∩ L(q)m. (3.5) The following lemma shows connections between roots in the complex Lie algebra L(q) and positive roots of the quadratic form q. 54 Lie algebras associated with quadratic forms Lemma 3.6. Let u, w, v = [vi1 , . . . , vim ] ∈ L(q) be roots. (a) At least one of the elements [vi1 , vi2 ], [vi2 , vi1 , vi3 , . . . , vim ] is not a root. (b) If 〈eu, ev + ew〉q = −1, then 〈eu, ev〉q ≥ 0 or 〈eu, ew〉q ≥ 0. (c) The vector ev = ei1 + . . . + eim is a positive root of the quadratic form q. Let v = [vi1 , . . . , vim ] ∈ L(q) and let q be weakly positive. (d) If ev = ei1 + . . . + eim is a positive root of q, then there exists a permutation σ ∈ Sm such that the element vσ = [viσ(1) , . . . , viσ(m) ] is a root in L(q). Proof. (a) Let v = [vi1 , . . . , vim ] ∈ L(q) be a root and assume that [vi1 , vi2 ] is a root. It follows that 〈ei1 , ei2〉q = −1 and 〈ei1 , ei3 + . . . + eim〉q = 〈ei1 , ei2 + . . . + eim〉q − 〈ei1 , ei2〉q = −1 − (−1) = 0. Therefore [vi2 , vi1 , vi3 , . . . , vim ] is not a root. The statement (b) is obvious. (c) Obviously eim is a root of q. Let 2 ≤ k ≤ m and assume that eik +. . .+eim is a root of q. We have 〈eik−1 , eik +. . .+eim〉q = −1, because v is a root in L(q). By Lemma 2.5(b), the vector eik−1 + eik + . . . + eim is a root of q. Inductively we finish the proof of the statement (c). (d) Let v be a positive root of q. From Lemma 2.5(b) it follows that v + ei is a positive root of q if and only if 〈ei, v〉q = −1. Therefore the statement (d) follows easily from Lemma 2.6 and Lemma 2.5(b), because q is weakly positive. Definition 3.7. (a) Let r be the set of all standard multibrackets [vi1 , . . . , vim ] ∈ L(q), such that [vi1 , . . . , vim ] is not a root and [vi2 , . . . , vim ] is a root, where i1, . . . , im ∈ {1, . . . , n} and m ∈ N. (b) Let L(q, r) = LieC〈v1, . . . , vn〉/(r) be a complex Lie algebra generated by the set {v1, . . . , vn} modulo the ideal (r) generated by the set r. We consider L(q, r) as a Lie algebra with a Nn- gradation, where we define the degree of vi to be ei, for any i = 1, . . . , n. Lemma 3.8. Let a ⊆ L(q) be a subset consisting of some standard multibrackets. Let v = [vi1 , . . . , vim ], y ∈ L(q, a). Assume that for any z ∈ L(q, a), such that ℓ(z) ≤ ℓ(v) + ℓ(y), the following condition is satis- fied: J. Kosakowska 55 if 〈ei, ez〉q 6= −1, then [vi, z] = 0 in L(q, a). (3.9) Then [v, y] = 0 in L(q, a) or there exists σ ∈ Sm and ε ∈ {0, 1} such that [v, y] = (−1)ε[viσ(1) , . . . , viσ(m) , y] in L(q, a). Proof. Let v = [vi1 , . . . , vim ], y ∈ L(q, a). We precede with induction on m = ℓ(v). For m = 1, the lemma is obvious. Let m > 1. We apply the Jacobi identity and get [v, y] = −[y, v] = [vi1 , [[vi2 , . . . , vim ], y]] − [[vi2 , . . . , vim ], [vi1 , y]]. Note that 〈ei1 , ei2 + . . . + eim + ey〉q = 〈ei1 , ei2 + . . . + eim〉q + 〈ei1 , ey〉q. Therefore 〈ei1 , ei2 + . . .+ eim〉q 6= −1 or 〈ei1 , ey〉q 6= −1 or 〈ei1 , ei2 + . . .+ eim + ey〉q 6= −1. If 〈ei1 , ei2 + . . . + eim〉q 6= −1, then, by (3.9), v = [vi1 , . . . , vim ] = 0 and [v, y] = 0. We are done. If 〈ei1 , ey〉q 6= −1, then, by (3.9), we have [vi1 , y] = 0 and [v, y] = [vi1 , [[vi2 , . . . , vim ], y]]. We finish by induction on m applied to [[vi2 , . . . , vim ], y]]. In the case 〈ei1 , ei2 +. . .+eim +ey〉q 6= −1, we have [vi1 , [[vi2 , . . . , vim ], y]] = 0. Finally [v, y] = (−1)ε[[vi2 , . . . , vim ], [vi1 , y]] and we finish by induction on m. 4. Grading of L(q, r) and positive roots of q Lemma 4.1. Let q be a weakly positive quadratic form (2.1) and let a ⊆ L(q) be a subset consisting of some standard multibrackets. Let m be a positive integer. Assume that the following conditions are satisfied. (a) If v = [vi1 , . . . vis ] is not a root in L(q) and ℓ(v) = s ≤ m, then L(q, a)ev = 0. (b) If v = [vi1 , . . . vis ] is a root in L(q) and ℓ(v) = s < m, then dimC L(q, a)ev ≤ 1. Then dimC L(q, a)ev ≤ 1, if v = [vi1 , . . . vis ] is a root in L(q) and ℓ(v) = s = m. 56 Lie algebras associated with quadratic forms Proof. Let v = [vi1 , . . . , vim ] be a root of L(q). If v = 0 in L(q, a), then we are done. Assume that 0 6= v ∈ L(q, a). We have to prove that dimC L(q, a)ev ≤ 1. If m = 1, the lemma is obvious. Let m > 1, and let 0 6= w ∈ L(q, a)ev be a standard multibracket. It follows that there exists a permutation σ ∈ Sm such that w = vσ = [viσ(1) , . . . , viσ(m) ] 6= 0. Since vσ 6= 0 in L(q, a), ℓ(vσ) = m, then the assumption (a) of our lemma yields that vσ is a root. It is enough to prove that there exists a ∈ C such that vσ = av. Since v, vσ ∈ L(q, a)ev , there exists k = 1, . . . , m such that ik = iσ(1). Note that we may assume, without loss of the generality, that k < m, because [vim−1 , vim ] = −[vim , vim−1 ] and we may replace v by −v. If k = 1, then v = [viσ(2) , . . . , viσ(m) ], [vi2 , . . . , vim ] ∈ L(q, a)ev , ℓ(v) < m, and the condition (b) yields that dimC L(q, a)ev ≤ 1. Then there exists a ∈ C such that [viσ(2) , . . . , viσ(m) ] = a[vi2 , . . . , vim ]. Therefore vσ = av and we are done. Let k > 1. Consider the following set Y = {vτ ; for all τ ∈ Sm such that there exists c ∈ C such that vτ = cv}. Note that for all vτ ∈ Y there exists l such that iτ(l) = iσ(1). We choose an element vτ ∈ Y such that l is minimal with this property. Without loss of the generality, we may assume that τ = id, vτ = v and k = l. Let v = [vi1 , [vi2 , [. . . , vik−1 , [vik , y] . . .]]] = [vi1 , vi2 , . . . , vik−1 , vik , y], where we set y = [vik+1 , . . . , vim ]. By the choice of k it follows that ij 6= iσ(1) for all j < k. Our assumptions yield: 〈eik , ey〉q = −1, because v is a root and 〈eik , ey + ei1 + . . . + eik−1 〉q = 〈eiσ(1) , eiσ(2) + . . . + eiσ(m) 〉q = −1, because vσ is a root. By the bilinearity of 〈−,−〉q, it follows that 〈eik , ei1+ . . . + eik−1 〉q = 0. We prove that v = [[vi1 , . . . , vik ], y] in L(q, a). Applying the Jacobi identity, we get [vi1 , . . . , vik−1 , [vik , y]] = = −[vi1 , . . . , vik−2 , [vik , [y, vik−1 ]]] − [vi1 , . . . , vik−2 , [y, [vik−1 , vik ]]]. Note that 〈eik−1 , ey〉q = 〈eik−1 , ey + eik〉q − 〈eik−1 , eik〉q = −1 − 〈eik−1 , eik〉q. J. Kosakowska 57 If 〈eik−1 , ey〉q = −1, then 〈eik−1 , eik〉q = 0. Therefore the condition (a) yields [vik−1 , vik ] = 0 and v = [vi1 , . . . , vik−2 , [vik , [vik−1 , y]]] which is the contradiction with the choice of k. Therefore v = [vi1 , . . . , vik−2 , [[vik−1 , vik ] y]]. Inductively, applying the Jacobi identity to [vi1 , . . . , vis , [[vis+1 , . . . , vik ], y]] we get [vi1 , . . . , vis , [[vis+1 , . . . , vik ], y]] = = −[vi1 , . . . , vis−1 , [[vis+1 , . . . , vik ], [y, vis ]]]− − [vi1 , . . . , vis−1 , [y, [vis , [vis+1 , . . . , vik ]]]]. Consider 〈eis , ey〉q = 〈eis , ey + eis+1 + . . . + eik〉q − 〈eis , eis+1 + . . . + eik〉q = = −1 − 〈eis , eis+1 + . . . + eik〉q. If 〈eis , ey〉q = −1, then 〈eis , eis+1 + . . . + eik〉q = 0. The condition (a) yields [vis , [vis+1 , . . . , vik ]] = 0 and v = [vi1 , . . . , vis−1 , [[vis+1 , . . . , vik ], [vis , y]]]. Applying Lemma 3.8, we get the contradiction with the choice of k. Therefore v = [vi1 , . . . , vis−1 , [[vis , [vis+1 , . . . , vik ]], y]]. Inductively we get v = [[vi1 , . . . , vik ], y]. Since v 6= 0, k < m, then (by the assumption (a) of the lemma) the element [vi1 , . . . , vik ] is a root. Therefore, by Lemma 3.6 (c), we have q(ei1 + . . . + eik) = 1. Now consider 1 = q(ei1 + . . .+eik) = q(ei1 + . . .+eik−1 )+q(eik)+〈eik , ei1 + . . .+eik−1 〉q. Since we proved above that 〈eik , ei1 + . . . + eik−1 〉q = 0, we have q(ei1 + . . . + eik−1 ) = 1 − q(eik) = 1 − 1 = 0. We get a contradiction, because q is weakly positive. This shows that k = 1 and vσ ∈ Y. This finishes the proof of lemma. 58 Lie algebras associated with quadratic forms Let L(q, r) be the Lie algebra introduced in Definition 3.7 Proposition 4.2. Let q be a weakly positive quadratic form (2.1). The following conditions hold. (a) If e = ei1 + . . . + eim is not a root of q, then L(q, r)e = 0. (b) If e = ei1 + . . . + eim is a root of q, then dimC L(q, r)e ≤ 1. Proof. (a) Assume that e = ei1 + . . . + eim is not a root of q. By Lemma 3.6, v = [vi1 , . . . , vim ] is not a root in L(q, r). Let k be maximal with the property that [vik , . . . , vim ] is a root. Since v = [vi1 , . . . , vim ] is not a root, then k > 1. Therefore [vik−1 , vik , . . . , vim ] is not a root and [vik−1 , vik , . . . , vin ] ∈ r. Finally [vik−1 , vik , . . . , vim ] = 0 and v = 0 in L(q, r). The statement (b) follows easily by induction on the length ℓ(v) of v, if we apply Lemma 4.1. As a consequence we get the following corollary. Corollary 4.3. Let q be a weakly positive quadratic form and let R+ q be the set of all positive roots of q. Then L(q, r) is a nilpotent Lie algebra and L(q, r) = ⊕ e∈R+ q L(q, r)e and dimC L(q, r) ≤ |R+ q |. Let q : Zn → Z quadratic form (2.1). With q we associate Cartan matrix C = (cij) ∈ Mn(Z) defined by cij = q(ei + ej) − q(ei) − q(ej). Following [4], to q we attach a Zn-graded complex Lie algebra G(q) with generators xi, x−i, hi, i = 1, . . . , n, which are homogeneous of degree ei,−ei, 0, respectively, and subject to the following relations: 1. [hi, hj ] = 0, for all i, j = 1, . . . , n, 2. [hi, xεj ] = εcijxεj , for all i, j = 1, . . . , n and ε ∈ {−1, 1}, 3. [xεi, x−εi] = εhi, for all i = 1, . . . , n and ε ∈ {−1, 1}, 4. [xε1i1 , . . . , xεnin ] = 0, if q(ε1ei1 + . . . + εnein) > 1 for εj ∈ {−1, 1}. Denote by G+(q) a Lie subalgebra of G(q) generated by the elements x1, . . . , xn. Proposition 4.4. If q is weakly positive and positive semi-definite, then L(q, r) ≃ G+(q). J. Kosakowska 59 Proof. By [4, Proposition 2.2] and Corollary 4.3, we have dimC G+(q) ≥ |R+ q | ≥ dimC L(q, r). On the other hand, it is easy to see that all relations r are satisfied in G+(q). Therefore we may define a homomorphism of Lie algebras Ψ : L(q, r) → G+(q) by Ψ(ui) = xi for all i = 1, . . . , n. Since G+(q) is generated by the elements x1, . . . , xn, the homomorphism Ψ is surjective. Therefore Ψ is an isomorphism, because dimC G+(q) ≥ dimC L(q, r). 5. Connections with Ringel-Hall algebras We present applications of Lie algebras L(q, r) to Lie algebras and Ringel- Hall algebras associated with representation directed algebras. We get a description of these Ringel-Hall algebras by generators and relations. For the basic concepts of representation theory the reader is referred to [2], [3] and for the basic concepts of Ringel-Hall algebras to [13], [14]. Let Q = (Q0, Q1) be a finite quiver without oriented cycles. Let CQ be the complex path algebra of Q. Assume that I is an admissible ideal of CQ such that A = CQ/I is a representation directed algebra. By mod (A) we denote the category of all right finite dimensional A-modules and by ind(A) we denote the set of all representatives of isomorphism classes of indecomposable A-modules. For any A-module M denote by dimM ∈ NQ0 the dimension vector of M (i.e. (dimM)(i) equals the number of composition factors of M which are isomorphic to the simple A-module Si corresponding to the vertex i ∈ Q0). Let qA : ZQ0 → Z be the Tits form of A (see [6]). By [6, Theorem 3.3], qA is weakly positive and there is a bijection (given by dim) between the set ind(A) and the set R+ qA . Let K(A) be the corresponding complex Lie algebra defined in [13]. Recall that, for a representation directed algebra A, the C-Lie algebra K(A) is the free C-linear space with basis {uX ; X ∈ ind(A)}. If X, Y are non-isomorphic indecomposable A-modules such that Ext1A(X, Y ) = 0, then the Lie bracket in K(A) is defined by [uY , uX ] =    ϕZ Y X(1) · uZ if there is an indecomposable A − module Z and a short exact sequence 0 → X → Z → Y → 0, 0 otherwise, , 60 Lie algebras associated with quadratic forms where ϕZ Y X are Hall polynomials (see [13]). In [7] it is proved that the Lie algebra K(A) is isomorphic to L(A), where L(A) is the Lie algebra associated with A in [11]. Let H(A) be the universal enveloping algebra of the Lie algebra K(A). Recall that H(A) = H1(A), where Hq(A) is the generic Ringel-Hall algebra associated in [13] with the algebra A. In fact, in [13], generic Ringel-Hall algebras were associated with directed Auslander-Reiten quivers. However, it is possible to associate generic Ringel-Hall algebras with representation directed C-algebras. The reader is referred to [7] for details. Proposition 5.1. Let A be a representation directed C-algebra. (a) The Lie algebra K(A) is generated by the set {ui ; i ∈ Q0}, where ui = uSi and Si is a simple A-module corresponding to the vertex i ∈ Q0. (b) In the Lie algebra K(A) the relations from the set r hold, if we interchange ui’s by vi’s. Proof. The statement (a) is proved in [14, Proposition 6]. Let [vi1 , . . . , vin ] be an element from the set r. It follows that [vi2 , . . . , vin ] is a root and the element [vi1 , vi2 , . . . , vin ] is not a root. By Lemma 3.6, the vector m = ei2 + . . . + eim is a positive root of the Tits form qA of A. If [vi2 , . . . , vin ] = 0 in K(A), then we are done. Otherwise [vi2 , . . . , vin ] = a · uM for some 0 6= a ∈ C and the unique indecomposable A-module M ∈ ind(A) with dimM = m = ei2 + . . . + ein , because A is represen- tation directed C-algebra. Since qA is weakly positive, then, by Lemma 3.6, the vector ei1 + m is not a root of qA. Therefore there exists no indecomposable A-module with dimension vector ei1 + m. Then, by [13, Theorem 2], [vi1 , . . . , vin ] = 0 in K(A) and we are done. Corollary 5.2. If A is a representation directed C-algebra, then there is an isomorphism of C-algebras F : L(qA, r) → K(A) ∼= L(A) given by F (vi) = ui, in particular dimC L(qA, r) = |R+ q |. If, in addition, qA is positive semi-definite, then L(qA, r) ∼= G+(qA). Proof. By Proposition 5.1, F is a well-defined homomorphism of graded Lie algebras. Since the Lie algebra L(qA, r) is generated by the set {vi ; i ∈ Q0} and the Lie algebra K(A) is generated by the set {ui ; i ∈ Q0}, the homomorphism F is an epimorphism. By Corollary 4.3, F is a monomorphism, because dimC K(A) = |R+ qA |. Finally F is an isomor- phism of Lie algebras. The final assertion follows from Proposition 4.4. J. Kosakowska 61 6. A minimal set of relations defining L(q, r) for a positive definite form q The set r usually is not a minimal set generating the ideal (r) of the Lie algebra L(q). In this section we describe a minimal set of elements defining the ideal (r) of L(q) for a positive definite form q (2.1). In this section all quadratic forms are assumed to be positive definite. Remark 6.1. The following easily verified facts are essentially used in this section. 1. Let i, j ∈ {1, . . . , n} be such that 〈ei, ej〉q 6= −1, then [. . . , [vi, [vj , . . .]]] = [. . . , [vj , [vi, . . .]]] in L(q, r). Indeed, apply the Jacobi identity and note that in this case [vi, vj ] ∈ (r). 2. If a ∈ L(q) is a standard multibracket such that ea is not a root of q, then a ∈ (r). Indeed, apply Lemma 3.2 (b) and Proposition 4.2 (a). 3. Let a, b ∈ L(q) be standard multibrackets such that 〈ea, eb〉q ≥ 0, then [a, b] ∈ (r). Indeed, q(ea + eb) = q(ea) + q(eb) + 〈ea, eb〉q ≥ 1 + 1 = 2, then ea + eb is not a root of q. Therefore [a, b] ∈ (r). 4. Let a, b ∈ L(q) be standard multibrackets such that 〈ea, eb〉q ≥ 0, then [. . . , [a, [b, . . .]]] = [. . . , [b, [a, . . .]]] in L(q, r). Indeed, apply the Jacobi identity and the fact that in this case [a, b] ∈ (r). 5. If a ∈ L(q) and 〈ei, ea〉q ≤ −2, for some i = 1, . . . , n, then a ∈ (r). Indeed, by Lemma 2.5, ea is not a root of q and therefore a ∈ (r). 6.1. The first step of reduction Let r1 ⊆ r ⊆ L(q) be the set consisting of the following elements: • [vi, vj ] for all i, j ∈ {1, . . . , n} such that i < j and 〈ei, ej〉q 6= −1, • [vi, [vi, vj ]] for all i, j ∈ {1, . . . , n} such that i < j and 〈ei, ej〉q = −1, 62 Lie algebras associated with quadratic forms • [vj , [vi, vj ]] for all i, j ∈ {1, . . . , n} such that i < j and 〈ei, ej〉q = −1. Let r0 ⊆ r ⊆ L(q) be the set consisting of all elements [vi1 , . . . , vim ] of r such that 〈ei1 , ei2 + . . . + eim〉q = 0. Define p ⊆ r to be p = r1 ∪ r0. (6.2) Proposition 6.3. If q is a positive definite quadratic form (2.1), then the ideals (p) and (r) of the Lie algebra L(q) are equal. Proof. The inclusion (p) ⊆ (r) is obvious. It is enough to prove that (r) ⊆ (p). Let v = [vi2 , . . . , vim ] be a root in L(q) and let i1 ∈ {1, . . . , n} be such that [vi1 , vi2 , . . . , vim ] ∈ r and 〈ei1 , ei2 + . . . + eim〉q 6= 0. Since [vi1 , vi2 , . . . , vim ] is not a root, we have 〈ei1 , ei2 + . . . + eim〉q ≥ 1. We claim that [vi1 , v] ∈ p. We precede with induction on ℓ(v) = m − 1. For ℓ(v) < 3 our state- ment easily follows by a case by case inspection on all possible cases. Let ℓ(v) ≥ 3, then m ≥ 4, v = [vi2 , [vi3 , v]], 〈ei2 , ei3 + ev〉q = −1, 〈ei3 , ev〉q = −1 and ℓ(v) ≥ 1. Note that 〈ei1 , ei2〉q = aij ∈ {−1, 0, 1}, if i1 6= i2. Therefore it is enough to consider the following three cases. 1) If 〈ei1 , ei2〉q ∈ {0, 1}, then 〈ei1 , ei3 + . . . + eim〉q ≥ 0. Therefore [vi1 , vi2 ], [vi1 , [vi3 . . . , vim ]] ∈ r and by the induction hypothesis we have [vi1 , [vi3 . . . , vim ]], [vi1 , vi2 ] ∈ (p). Finally [vi1 , . . . , vim ] = −[vi2 , [[vi3 . . . , vim ], vi1 ]] − [[vi3 , . . . , vim ], [vi1 , vi2 ]] ∈ (p). 2) Let i1 = i2. Applying the Jacobi identity to [vi1 , v] we get [vi1 , [vi1 , [vi3 , v]]] = −[vi1 , [vi3 , [v, vi1 ]]] − [vi1 , [v, [vi1 , vi3 ]]] = = [vi3 , [[v, vi1 ], vi1 ]] + [[v, vi1 ], [vi1 , vi3 ]] + [v, [[vi1 , vi3 ], vi1 ]] + [[v, vi1 ], [vi1 , vi3 ]]. Note that, we have 〈ei1 , ei1 + ev〉q = 2 + 〈ei1 , ev〉q ≥ 2 + (−1) = 1, and therefore by the induction hypothesis [vi1 , [vi1 , v]] ∈ (p). More- over, [vi1 , [vi1 , vi3 ]] ∈ (r1) ⊆ (p). Finally [vi1 , v] = 2[[vi1 , vi3 ][vi1 , v]]. If [vi1 , vi3 ] ∈ (p), then [vi1 , v] ∈ (p). Assume that [vi1 , vi3 ] 6∈ (p). In this case 〈ei1 , ei3〉q = −1 and 〈ei1 , ev〉q = 〈ei1 , ei2 +ei3 +ev〉q−〈ei1 , ei2〉q−〈ei1 , ei3〉q ≥ 1−2−(−1) = 0. Then [vi1 , v] = 2[[vi1 , vi3 ][vi1 , v]] ∈ (r0) ⊆ (p) and we are done. 3) Let 〈i1, i2〉q = −1. By Lemma 2.5 (c), 〈ei1 , ei3 + . . . + eim〉 ∈ {−1, 0, 1}, because q is positive definite. On the other hand 1 ≤ 〈ei1 , ei2 + . . . + eim〉q = −1 + 〈ei1 , ei3 + . . . + eim〉q ≤ 0. This contradiction shows that the case 3) does not hold. This finishes the proof. J. Kosakowska 63 Corollary 6.4. If q is a positive definite quadratic form (2.1), then L(q, r) ∼= L(q, p). 6.2. The second step of reduction Let i1, . . . , im ∈ {1, . . . , n}. Following [5], we call the sequence (i1, . . . , im) a chordless cycle of the form q (2.1), if the following conditions are satisfied: • the elements i1, . . . , im are pairwise different, • aij = 〈eij , eik〉q 6= 0 if and only if |k − j| = 1 mod m. Chordless cycles are playing an important role in [5], where Lie alge- bras associated with positive definite quadratic forms are investigated. A chordless cycle (i1, . . . , im) is called positive, if 〈ei1 , eim〉q = 1 and 〈eij , eik〉q = −1 for all j, k such that {j, k} 6= {1, m} and |j − k| = 1 mod m. Remark 6.5. We note that if (i1, . . . , im) is a chordless cycle, then (i1, . . . , im) is a simple cycle in the bigraph B(q). Moreover, if the chord- less cycle (i1, . . . , im) is positive, then the cycle (i1, . . . , im) in B(q) has exactly one broken edge i1 ___ im . Let r2 ⊆ L(q) be the set consisting of all elements [vi1 , . . . , vim ] such that (i1, . . . , im) is a positive chordless cycle. Lemma 6.6. r2 ⊆ p. Proof. Let v = [vi1 , . . . , vim ] ∈ r2. From the definition of a positive chordless cycle, it follows easily that [vik , . . . , vim ] is a root for any k > 1. Moreover 〈ei1 , ei2 + . . . + eim〉q = 0, and therefore v ∈ p Set j = r1 ∪ r2. (6.7) For all elements x, y ∈ L(q) we write x ≡ y if x − y ∈ (j). Obviously ≡ is an equivalence relation. Before we prove that the ideals (p) and (j) of L(q) are equal, we need to prove two technical lemmata. 64 Lie algebras associated with quadratic forms Lemma 6.8. Let q be a positive definite quadratic form (2.1). Let m ≥ 3 be an integer, let v = [vi2 , . . . , vim ] be a root and let (j)m−1 = (p)m−1. Let i1 ∈ {1, . . . , n} be such that [vi1 , vi2 , . . . , vim ] ∈ p, 〈ei1 , ei2〉q = −1 and 〈ei1 , ei2 + . . . + eim〉q = 0. Then [vi1 , v] ∈ (j) or there exists ε ∈ {0, 1} such that [vi1 , v] ≡ (−1)ε[vi1 , [a, x]], where (a) a = [vik , . . . , vi2 ], x = [vik+1 , . . . , vim ], for some 2 ≤ k < m, (b) 〈ei1 , eij 〉q = 0, for all j = 3, . . . , k, and (c) 〈ei1 , eik+1 〉q = 1. Proof. Let m ≥ 3 and let v = [vi2 , . . . , vim ] be a root. Let i1 ∈ {1, . . . , n} be such that [vi1 , vi2 , . . . , vim ] ∈ p, 〈ei1 , ei2+. . .+eim〉q = 0 and 〈ei1 , ei2〉q = −1. Note that [vi1 , v] ≡ [vi1 , [a, x]], where a = [vil , . . . , vi2 ], x = [vil+1 , . . . , vim ] and 〈ei1 , eij 〉q = 0, for all j = 3, . . . , l (i.e. the conditions (a), (b) are sat- isfied). Indeed, it is enough to set l = 2, a = xi2 and x = [vi3 , . . . , vim ]. Fix a and x such that [vi1 , v] ≡ [vi1 , [a, x]], where a = [vil , . . . , vi2 ], x = [vil+1 , . . . , vim ] and 〈ei1 , eij 〉q = 0, for all j = 3, . . . , l. As we noted above there exists at least one such a presentation of [vi1 , v]. Consider the element il+1. By Lemma 2.5, 〈ei1 , eil+1 〉q ∈ {−1, 0, 1, 2}. • If 〈ei1 , eil+1 〉q = 1, then we set k = l and note that [vi1 , v] has the required form, i.e. [vi1 , v] ≡ [vi1 , [a, x]], where a = [vik , . . . , vi2 ], x = [vik+1 , . . . vim ], 〈ei1 , eij 〉q = 0, for all j = 3, . . . , k, and 〈ei1 , eik+1 〉q = 1. • If 〈ei1 , eil+1 〉q = −1, then 〈ei1 , eil+2 + . . . + eim〉q = 〈ei1 , ev〉q − 〈ei1 , ea〉q − 〈ei1 , eil+1 〉q = 0 − (−1) − (−1) = 2. Therefore, by Lemma 2.5, m = l + 2 and i1 = il+2. Note that [vi1 , [vi1 , vil+1 ]] ∈ (r1) ⊆ (j), 〈ei1 , ei1 + ea〉q = 1, then [vi1 , [vi1 , a]] ∈ (p)(m−1) = (j)(m−1) ⊆ (j), 〈ei1 , eil+1 + ea〉q = −2, therefore [vil+1 , a] ∈ (p)(m−1) = (j)(m−1) ⊆ (j). Then we have [vi1 , v] ≡ [vi1 , [a, [vil+1 , vi1 ]]] = −[a, [[vil+1 , vi1 ], vi1 ]] − [[vil+1 , vi1 ], [vi1 , a]] ≡ [[vi1 , a], [vil+1 , vi1 ]] = −[vil+1 , [vi1 , [vi1 , a]]] − [vi1 , [[vi1 , a], vil+1 ]] ≡ [vi1 , [vil+1 , [vi1 , a]]] = −[vi1 , [vi1 , [a, vil+1 ]]] − [vi1 , [a, [vil+1 , vi1 ]]] ≡ −[vi1 , [a, [vil+1 , vi1 ]]] ≡ −[vi1 , v]. J. Kosakowska 65 Therefore 2 · [vi1 , v] ∈ (j) and [vi1 , v] ∈ (j). • If 〈ei1 , eil+1 〉q = 2, then i1 = il+1 and [vi1 , v] ≡ [vi1 , [a, [vi1 , y]]], where y = [vil+2 , . . . , vim ]. By the bilinearity of 〈−,−〉q and as- sumptions, we have 〈ei1 , ey〉q = −1. Moreover 〈ei1 , ei1 + ey〉q = 1, then [vi1 , [vi1 , y]] ∈ (p)(m−1) = (j)(m−1) ⊆ (j), 〈ei1 , ei1 + ea〉q = 1, then [vi1 , [vi1 , a]] ∈ (p)(m−1) = (j)(m−1) ⊆ (j), 〈ei1 , ey + ea〉q = −2, therefore [vil+1 , a] ∈ (p)(m−1) = (j)(m−1) ⊆ (j). Similarly as above we can prove that [vi1 , v] ∈ (j). • Let 〈ei1 , eil+1 〉q = 0 and y = [vil+2 , . . . , vim ]. Consider [vi1 , [a, [vil+1 , y]]] = = −[vi1 , [vil+1 , [y, a]]] − [vi1 , [y, [a, vil+1 ]]] = = [vil+1 , [[y, a], vi1 ]] + [[y, a], [vi1 , vil+1 ]] − [vi1 , [[vil+1 , a], y]]. Note that [vi1 , vil+1 ] ∈ (r1) ⊆ (j), 〈ei1 , ea + ey〉q = 0 and therefore [[y, a], vi1 ] ∈ (p)m−1 = (j)m−1 ⊆ (j). Then [vi1 , v] ≡ [vi1 , [a, [vil+1 , y]]] ≡ −[vi1 , [[vil+1 , a], y]]. We may set a := [vil+1 , a], x := y and continue this procedure inductively. Note that there exists k such that 〈ei1 , eik+1 〉q ≥ 1, because 〈ei1 , ei2 + . . . + eim〉q = 0 and 〈ei1 , ei2〉q = −1. Therefore continuing this procedure inductively, we prove that [vi1 , v] ∈ (j) or [vi1 , v] ≡ (−1)ε[vi1 , [a, x]], where a = [vik , . . . , vi2 ], x = [vik+1 , . . . vim ], 〈ei1 , eij 〉q = 0, for all j = 3, . . . , k, and 〈ei1 , eik+1 〉q = 1. Lemma 6.9. Assume that q is a positive definite quadratic form (2.1). Let m ≥ 3 be an integer, v = [vi2 , . . . , vim ] be a root and (j)m−1 = (p)m−1. Let i1 ∈ {1, . . . , n} be such that [vi1 , vi2 , . . . , vim ] ∈ p, 〈ei1 , ei2〉q = −1 and 〈ei1 , ei2 +. . .+eim〉q = 0. Moreover assume that [vi1 , v] ≡ (−1)ε[vi1 , [a, x]] and the conditions (a)-(c) of Lemma 6.8 are satisfied. Then [vi1 , v] ∈ (j) or [vi1 , v] ≡ [vi1 , [a, x]] ≡ [vi1 , [a, [b, y]]], where (i) a = [vik , . . . , vi2 ], b = [vis , vis−1 . . . , vik+1 ], y = [vis+1 , . . . , vim ], (ii) 〈ei1 , eij 〉q = 0, for all j = 3, . . . , k and j = k + 2, . . . , s, (iii) 〈ei1 , ei2〉q = −1, 〈ei1 , eik+1 〉q = 1, (iv) 〈eb, ea〉q = 0, (v) if s < m then 〈eis+1 , ea〉q = −1 and 〈eis+1 , eb〉q = −1. 66 Lie algebras associated with quadratic forms Proof. Note that [vi1 , v] ≡ [vi1 , [a, x]] ≡ [vi1 , [a, [b, y]]], where b = vik+1 , y = [vik+2 , . . . , vim ] and the conditions (i), (ii), (iii) are satisfied, if we put s = k + 1. We may assume that the condition (iv) is also satisfied. Indeed, it is enough to show that 〈eik+1 , ea〉q = 0. • If 〈eik+1 , ea〉q = −1, then 〈eik+1 , ea + ey〉q = −2. Therefore [a, y] ∈ (r)m−1 = (p)m−1 = (j)m−1 ⊆ (j). Then [vi1 , [a, [vik+1 , y]]] = −[vi1 , [vik+1 , [y, a]]] − [vi1 , [y, [a, vik+1 ]]] ≡ [vi1 , [[a, vik+1 ], y]] = −[[a, vik+1 ], [y, vi1 ]] − [y, [vi1 , [a, vik+1 ]]]. Since 〈ei1 , ey〉q = 0 = 〈ei1 , eik+1 + ea〉q, then [y, vi1 ] ∈ (p)m−1 = (j)m−1 and [vi1 , [a, vik+1 ]] ∈ (p)m−1 = (j)m−1.Therefore we have [vi1 , [a, [vik+1 , y]]] ∈ (j). • If 〈eik+1 , ea〉q = 2, then a = vik+1 . It is a contradiction, because 〈ei1 , eik+1 〉q = 1 6= −1 = 〈ei1 , ea〉q. • If 〈eik+1 , ea〉q = 1, then 〈eik+1 , ea+ey〉q = 0. Therefore [vik+1 , [y, a]], [a, vik+1 ] ∈ (p)m−1 = (j)m−1 and [vi1 , [a, [vik+1 , y]]] = −[vi1 , [vik+1 , [y, a]]] − [vi1 , [y, [a, vik+1 ]]] ∈ (j). Finally, we can assume that 〈eik+1 , ea〉q = 〈eb, ea〉q = 0 and the condi- tion (iv) is satisfied. Therefore [vi1 , [a, [b, y]]] ≡ [vi1 , [b, [a, y]]], because 〈eb, ea〉q = 0. If k +1 = m, then we are done. Assume that k +1 < m and consider the element ik+2. 1. Let 〈eik+2 , eb〉q = −1. (a) If 〈eik+2 , ea〉q = −1, then we put s = k + 1 and we are done. (b) If 〈eik+2 , ea〉q ≥ 0, then [vik+2 , a] ∈ (j)m−1. If m = k + 2, then [vi1 , v] ≡ −[vi1 , [b, [vik+2 , a]]] − [vi1 , [vik+2 , [a, b]]] ∈ (j) and we are done. Assume that m > k + 2. We can assume that 〈eik+1 , ea〉q = 0. Indeed, since [vik+2 , a] ∈ (j)m−1, we have [vi1 , v] ≡ [vi1 , [b, [a, [vik+2 , vik+3 , . . . , vim ]]]] ≡ [vi1 , [b, [vik+2 , [a, [vik+3 , . . . , vim ]]]]]. J. Kosakowska 67 If 〈eik+2 , ea〉q ≥ 1, then 〈eik+2 , ea +eik+3 + . . .+eim〉q ≥ 1−1 = 0. It follows that [vik+2 , [a, [vik+3 , . . . , vim ]]] ∈ (p)m−1 = (j)m−1. Therefore we can assume that 〈eik+2 , ea〉q = 0. Moreover [vi1 , v] ≡ [vi1 , [b, [vik+2 , [a, [vik+3 , . . . , vim ]]]]] = −[vi1 , [vik+1 , [[a, [vik+3 , . . . , vim ]], b]]] −[vi1 , [[a, [vik+3 , . . . , vim ]], [b, vik+2 ]]] ≡ −[vi1 , [[vik+2 , b], [a, [vik+3 , . . . , vim ]]]], because 〈eik+2 , eb〉q = −1 and 〈eik+2 , ea + eb + eik+3 + . . . + eim〉q = −2. We can assume that 〈ei1 , eik+2 〉q = 0. Indeed, if 〈ei1 , eik+2 〉q = −1, then 0 = 〈ei1 , ev〉q = 〈ei1 , ea + eb + eik+2 〉q + 〈ei1 , eik+3 + . . . + eim 〉q = −1 + 〈ei1 , eik+3 + . . . + eim 〉q. It follows that 〈ei1 , eik+3 + . . . + vim〉q = 1, 〈ei1 , ea + eik+3 + . . .+vim〉q = 0 and [vi1 , [a, [vik+3 , . . . , vim ]]] ∈ (p)m−1 = (j)m−1. Therefore [vi1 , v] ≡ −[vi1 , [[vik+2 , b], [a, [vik+3 , . . . , vim ]]]] ≡ −[[vik+2 , b], [vi1 , [a, [vik+3 , . . . , vim ]]]] ∈ (j), because 〈ei1 , eik+2 + eb〉q = 0. If 〈ei1 , eik+2 〉q ≥ 1, then 0 = 〈ei1 , ev〉q = 〈ei1 , ea + eb + eik+2 〉q + 〈ei1 , eik+3 + . . . + eim 〉q ≥ 1 + 〈ei1 , eik+3 + . . . + eim 〉q. It follows that 〈ei1 , eik+3 + . . . + vim〉q ≤ −1, 〈ei1 , ea + eik+3 + . . . + vim〉q ≤ −2 and [a, [vik+3 , . . . , vim ]] ∈ (p)m−1 = (j)m−1. Therefore, [vi1 , v] ∈ (j) and we are done. Finally, 〈eik+2 , ea〉q = 0, 〈ei1 , eik+2 〉q = 0, 〈eik+2 + eb, ea〉q = 0 and [vi1 , v] ≡ −[vi1 , [[vik+2 , b], [a, [vik+3 , . . . , vim ]]]] ≡ −[vi1 , [a, [[vik+2 , b], [vik+3 , . . . , vim ]]]]. We set a = a, b = [vik+2 , b], y = [vik+3 , . . . , vim ] and con- tinue this procedure inductively using [vi1 , [a, [b, y]]] instead of [vi1 , [a, [b, y]]]. 2. Let 〈eik+2 , eb〉q 6= −1, then [vik+2 , b] ∈ (j)m−1. If m = k + 2, then [vi1 , v] ≡ [vi1 , [a, [b, vik+2 ]]] ∈ (j) 68 Lie algebras associated with quadratic forms and we are done. Assume that m > k + 2. Since [vik+2 , b] ∈ (j)m−1, we have [vi1 , [a, [b, [vik+2 , [vik+3 , . . . , vim ]]]]] ≡ [vi1 , [a, [vik+2 , [b, [vik+3 , . . . , vim ]]]]]. For the sake of simplicity we present partial results in tables. In the first column of the following table we consider all possible values of 〈eik+2 , eb〉q. In the second column we give the corresponding value of 〈eik+2 , eb +eik+3 + . . .+eim〉q. The third column contains the sign "+", if we can deduce that X = [vik+2 , [b, [vik+3 , . . . , vim ]]] ∈ (j)m−1, and the sign "−", otherwise. 〈eik+2 , eb〉q 〈eik+2 , eb + eik+3 + . . . + eim 〉q X ∈ (j)m−1 0 -1 − 1 0 + 2 1 + Therefore we may assume that 〈eik+2 , eb〉q = 0, because otherwise [vi1 , v] ≡ [vi1 , [a, [vik+2 , [b, [vik+3 , . . . , vim ]]]]] ∈ (j). (a) Assume that 〈eik+2 , ea〉q 6= −1. Then [vik+2 , a] ∈ (j)m−1 and [vi1 , v] ≡ [vi1 , [a, [vik+2 , [b, [vik+3 , . . . , vim ]]]]] ≡ [vi1 , [vik+2 , [a, [b, [vik+3 , . . . , vim ]]]]]. In the first column of the following table we consider all pos- sible values of 〈eik+2 , ea〉q. In the second column we give the corresponding value of x = 〈eik+2 , ea + eb + eik+3 + . . . + eim〉q. The third column contains the sign "+", if we can deduce that X = [vik+2 , [a, [b, [vik+3 , . . . , vim ]]]] ∈ (j)m−1, and the sign "−", otherwise. 〈eik+2 , ea〉q x X ∈ (j)m−1 0 -1 − 1 0 + 2 1 + Therefore we may assume that 〈eik+2 , ea〉q = 0, because oth- erwise [vi1 , v] ∈ (j). Moreover, [vi1 , [vik+2 , [a, [b, [vik+3 , . . . , vim ]]]]] ≡ ≡ [vi1 , [vik+2 , [b, [a, [vik+3 , . . . , vim ]]]]], J. Kosakowska 69 because 〈ea, eb〉q = 0. In the first column of the following table we consider all pos- sible values of 〈ei1 , eik+2 〉q. In the second column we present a consequences of the information contained in the first col- umn. Finally, in the second table we present conclusions of the results presented in the first table. 〈ei1 , eik+2 〉q consequences -1 〈ei1 , eb + eik+3 + . . . + eim〉q = 2 0 〈ei1 , ea + eb + eik+3 + . . . + eim〉q = 0 1 〈ei1 , ea + eik+3 + . . . + eim〉q = −2 2 〈ei1 , ea + eb + eik+3 + . . . + eim〉q = −2 〈ei1 , eik+2 〉q conclusions -1 [b, [vik+3 , . . . , vim ]] ∈ (j)m−1 0 [vi1 , [a, [b, [vik+3 , . . . , vim ]]]] ∈ (j)m−1 1 [a, [vik+3 , . . . , vim ]] ∈ (j)m−1 2 [a, [b, [vik+3 , . . . , vim ]]] ∈ (j)m−1 All these cases imply that [vi1 , v] ∈ (j). (b) Assume that 〈eik+2 , ea〉q = −1. In this case 〈eik+2 , ea + eb + eik+3 + . . . + eim〉q = −2. It follows that [a, [b, [vik+3 , . . . , vim ]]] ∈ (j)m−1 and [vi1 , v] ≡ [vi1 , [a, [vik+2 , [b, [vik+3 , . . . , vim ]]]]] = [vi1 , [vik+2 , [a, [b, [vik+3 , . . . , vim ]]]]] −[vi1 , [[b, [vik+3 , . . . , vim ]], [a, vik+2 ]]] ≡ −[vi1 , [[vik+2 , a], [b, [vik+3 , . . . , vim ]]]]. Consider 〈ei1 , eik+2 〉q and − [vi1 , [[vik+2 , a], [b, [vik+3 , . . . , vim ]]]] = = [[vik+2 , a], [[b, [vik+3 , . . . , vim ]], vi1 ]]+ + [[b, [vik+3 , . . . , vim ]], [vi1 , [vik+2 , a]]]. 70 Lie algebras associated with quadratic forms We present again partial results in tables. 〈ei1 , eik+2 〉q consequences -1 〈ei1 , eb + eik+3 + . . . + eim〉q = 2 1 〈ei1 , eb + eik+3 + . . . + eim〉q = 0 and 〈ei1 , eik+2 + ea〉q = 0 2 i1 = ik+2 〈ei1 , eik+2 〉q conclusions -1 [b, [vik+3 , . . . , vim ]] ∈ (j)m−1 1 [[b, [vik+3 , . . . , vim ]], vi1 ] ∈ (j)m−1 and [vi1 , [vik+2 , a]] ∈ (j)m−1 2 contradiction, because 〈ei1 , eb〉q = 1 6= 0 = 〈eik+2 , eb〉q Therefore, we can assume that 〈ei1 , eik+2 〉q = 0, because oth- erwise [vi1 , v] ∈ (j). Moreover [vi1 , v] ≡ −[vi1 , [[vik+2 , a], [b, [vik+3 , . . . , vim ]]]] ≡ −[vi1 , [b, [[vik+2 , a], [vik+3 , . . . , vim ]]]], because 〈eik+2 + ea, eb〉q = 0. Therefore [vi1 , v] ≡ −[vi1 , [a, [b, y]]], where a = [vik+2 , a], b = b, y = [vik+3 , . . . , vim ] and the condi- tions (i)-(iv) are satisfied. Continuing this procedure inductively we show that [vi1 , v] ∈ (j) or [vi1 , v] ≡ [vi1 , [a, x]] ≡ [vi1 , [a, [b, y]]] and the conditions (i)-(v) are satis- fied. Proposition 6.10. Let q be a positive definite quadratic form. The ideals (j) and (p) of the Lie algebra L(q) are equal. Proof. The inclusion (j) ⊆ (p) is obvious. It is enough to prove that (p) ⊆ (j). Let v = [vi2 , . . . , vim ] be a root in L(q) and let i1 ∈ {1, . . . , n} be such that [vi1 , vi2 , . . . , vim ] ∈ p and 〈ei1 , ei2 + . . .+eim〉q = 0. We claim that [vi1 , v] ∈ j. We prove our claim by induction on ℓ(v) = m − 1. For ℓ(v) < 3 our statement easily follows by a case by case inspection on all possible cases. Let ℓ(v) ≥ 3, then m ≥ 4, v = [vi2 , [vi3 , v]], 〈ei2 , ei3 + ev〉q = −1, 〈ei3 , ev〉q = −1 and ℓ(v) ≥ 1. By Lemma 2.5, it is enough to consider the J. Kosakowska 71 following three cases. 1) If 〈ei1 , ei2〉q = 0, then by the bilinearity of 〈−,−〉q, we have 〈ei1 , ei3 + . . . + eim〉q = 0. Moreover [vi1 , v] = −[vi2 , [[vi3 , . . . , vim ], vi1 ]] − [[vi3 , . . . , vim ], [vi1 , vi2 ]]. By definitions, [vi1 , vi2 ] ∈ j and [vi1 , [vi3 , . . . , vim ]] ∈ r. Then, by Propo- sition 6.3, [vi1 , [vi3 , . . . , vim ]] ∈ (p) and by the induction hypothesis we have [vi1 , [vi3 , . . . , vim ]] ∈ (j). Finally, [vi1 , v] ∈ (j). 2) If i1 = i2, then 〈ei1 , ei2〉q = 2 and 〈ei1 , ei3 + . . . + eim〉q = −2. This is a contradiction with Lemma 2.5 and therefore the case 2) does not hold. 3) Let 〈ei1 , ei2〉q ∈ {1,−1}. In this case we apply the Jacobi identity and develop Lemmata 6.8, 6.9 to find an element w ∈ L(q) such that [vi1 , v] − w ∈ (j) (i.e. [vi1 , v] ≡ w). Finally we show that w ∈ (j), which implies that [vi1 , v] ∈ (j). 3.1) Let 〈ei1 , ei2〉q = 1. We reduce this case to the case 3.2) presented below. Since 〈ei1 , ev〉q = 0 and 〈ei1 , ei2〉q = 1, then there exists k ∈ {3, . . . , m} such that 〈ei1 , eik〉q = −1. Choose k minimal with this prop- erty. We may assume that k < m, because [vim−1 , vim ] = −[vim , vim−1 ] and we can work with −v instead of v. Note that for all s = 3, . . . , k− 1, we have 〈ei1 , eis〉q = 0. Indeed, if there exists s = 3, . . . , k − 1 such that 〈ei1 , eis〉q 6= 0, then by the choice of k, we have 〈ei1 , eis〉q ≥ 1. Then 〈ei1 , eik + . . .+ eim〉q ≤ 〈ei1 , ev〉q −〈ei1 , ei2〉q −〈ei1 , eis〉q = −2 and we get a contradiction, because [vik , . . . , vim ] is a root. Now applying the Jacobi identity we get [vi2 , . . . , vik−1 , [vik , y]] = = [vi2 , . . . , vik , [vik−1 , y]] + [vi2 , . . . , vik−2 , [[vik−1 , vik ], y]]. By Lemma 3.6 (a), the element [vik , [vik−1 , y]] or the element [vik−1 , vik ] is not a root, then [vik−1 , vik ] ∈ (p) or [vik , [vik−1 , y]] ∈ (p). By the in- duction hypothesis [vik , [vik−1 , y]] ∈ (j) or [vik−1 , vik ] ∈ (j). Therefore v ≡ [vi2 , . . . , vik−2 , [x, z]], where x = vik and z = [vik−1 , y] or x = [vik−1 , vik ], z = y. In both cases 〈ei1 , ex〉q = −1. Continuing this procedure (i.e. x plays a role of vik and z plays a role of y), we get [vi1 , v] ≡ [vi1 , [vi2 , [x, z]]], where 〈ei1 , ex〉q = −1. Applying the Jacobi identity, we get [vi2 , [x, z]] = [x, [vi2 , z]] + [[vi2 , x], z]. 72 Lie algebras associated with quadratic forms By Lemma 3.6 (b), [vi2 , z] or [vi2 , x] is not a root, then [x, [vi2 , z]] ∈ (p) or [vi2 , x] ∈ (p). By the induction hypothesis [x, [vi2 , z]] ∈ (j) or [vi2 , x] ∈ (j). Therefore [vi1 , v] ≡ [vi1 , [x, [vi2 , z]]] or [vi1 , v] ≡ [vi1 , [[vi2 , x], z]]. If [vi1 , v] ≡ [vi1 , [x, [vi2 , z]]], then applying Lemma 3.8 we get a reduction to the case 1) or to the case 3.2) below. If [vi1 , v] ≡ [vi1 , [[vi2 , x], z]], then [vi1 , v] ≡ [vi1 , [[vi2 , x], z]] = −[[vi2 , x], [z, vi1 ]] − [z, [vi1 , [vi2 , x]]]. Note that ℓ(z) ≥ 1, because we choose k with the property k < m. Then 〈ei1 , ez〉q = 〈ei1 , ev〉q − 〈ei1 , ex〉q − 〈ei1 , ei2〉q = 0 − (−1) − 1 = 0 and 〈ei1 , ei2 + ex〉q = 1 − 1 = 0, and therefore by the induction hypothesis [vi1 , v] = −[[vi2 , x], [z, vi1 ]] − [z, [vi1 , [vi2 , x]]] ∈ (j). 3.2) Let 〈ei1 , ei2〉q = −1. Applying Lemma 6.8 and 6.9 we get [vi1 , v] ∈ (j) or [vi1 , v] ≡ [vi1 , [a, x]] ≡ [vi1 , [a, [b, y]]], where (i) a = [vik , . . . , vi2 ], b = [vis , vis−1 . . . , vik+1 ], y = [vis+1 , . . . vim ], (ii) 〈ei1 , eij 〉q = 0, for all j = 3, . . . , k and j = k + 2, . . . , s, (iii) 〈ei1 , ei2〉q = −1, 〈ei1 , eik+1 〉q = 1, (iv) 〈eb, ea〉q = 0, (v) if s < m then 〈eis+1 , ea〉q = −1 and 〈eis+1 , eb〉q = −1. Consider the following cases. (a) If s = m, then [vi1 , v] ≡ [vi1 , [a, b]] and q(ea + eb) = q(ea) + q(eb) + 〈ea, eb〉q = 1 + 1 + 0 = 2. Therefore ea+eb is not a root of q, by the induction hypothesis [a, b] ∈ (j) and [vi1 , v] ≡ [vi1 , [a, b]] ∈ (j). We may assume that s < m and consider 〈ei1 , eis+1〉q. Partial results J. Kosakowska 73 are presented in the following tables. 〈ei1 , eis+1〉q consequence -1 q(ei1 + (eis+1 + ea)) = q(ei1) + q(eis+1 + ea) + 〈ei1 , eis+1 + ea〉q = 1 + 1 − 1 − 1 = 0 1 q(−ei1 + (eis+1 + eb)) = q(ei1) + q(eis+1 + eb) − 〈ei1 , eis+1 + eb〉q = 1 + 1 − 1 − 1 = 0 2 i1 = is+1 〈ei1 , eis+1〉q conclusions -1 contradiction, because q is positive definite 1 contradiction, because q is positive definite 2 contradiction, because 〈eis+1 , eb〉q = −1 6= 1 = 〈ei1 , eb〉q Therefore, we may assume that 〈ei1 , eis+1〉q = 0. (b) Let m ≥ s + 2. Then [vi1 , v] ≡ [vi1 , [a, [b, [vis+1 , z]]]], where ℓ(z) ≥ 1. Moreover 〈eis+1 , eb + ez〉q = −1 − 1 = −2, then [z, b] is not a root and by the induction hypothesis [z, b] ∈ (j). Therefore [vi1 , [a, [b, [vis+1 , z]]]] = −[vi1 , [a, [vis+1 , [z, b]]]] − [vi1 , [a, [z, [b, vis+1 ]]]] ≡ [vi1 , [a, [[b, vis+1 ], z]]]. Applying the Jacobi identity, we get [vi1 , [a, [[b, vis+1 ], z]]] = −[vi1 , [[b, vis+1 ], [z, a]]] − [vi1 , [z, [a, [b, vis+1 ]]]]. Note that 〈eis+1 , ea + ez〉q = −1 − 1 = −2, then [z, a] is not a root and by the induction hypothesis [z, a] ∈ (j). Therefore [vi1 , [a, [[b, vis+1 ], z]]] ≡ [vi1 , [[a, [b, vis+1 ]], z]] = −[[a, [b, vis+1 ]], [z, vi1 ]] − [z, [vi1 , [a, [b, vis+1 ]]]]. Note that 〈ei1 , ea+eb+eis+1〉q = −1+1+0 = 0 and 〈ei1 , ez〉q = 〈ei1 , ev〉q− 〈ei1 , ea+eb+eis+1〉q = 0. By the induction hypothesis [vi1 , [a, [b, vis+1 ]]] ∈ (j) and [z, vi1 ] ∈ (j). Therefore [vi1 , v] ∈ (j). (c) Let m = s+1. We recall that 〈ei1 , ei2〉q = −1, 〈ei1 , eis+1〉q = 1 and 〈ei1 , eij 〉q = 0 for all j = 3, . . . , s. Applying 1) and the Jacobi identity, it is straightforward to prove the following conditions: 74 Lie algebras associated with quadratic forms (i) [vi1 , v] ≡ (−1)ε[vi1 , vi2 , . . . , vik , vis+1 , vis , . . . , vik+1 ], (ii) [vi1 , v] ≡ (−1)ε[vi1 , [[vij , . . . , vi2 ], vij+1 , . . . , vik+1 ]], for all j = 2, . . . , k, s + 1, s . . . , k + 2. Without loss of generality, we can assume that in both cases ε = 0. It follows from (i) that there exists a numbering of elements {i2, . . . , is+1}, such that [vi1 , v] ≡ [vi1 , vi2 , . . . , vis+1 ], where, [vi2 , . . . , vis+1 ] is a root, 〈ei1 , ei2〉q = −1, 〈ei1 , eis+1〉q = 1 and 〈ei1 , eij 〉q = 0 for j = 3, . . . , s. Moreover, it follows from (ii) that the elements [vij , . . . , vi2 ] for all j = 2, . . . , k, s + 1, s . . . , k + 2, are roots, because otherwise [vij , . . . , vi2 ] ∈ j and [vi1 , v] ∈ j. We claim that 〈eij , eij+1〉q = −1 for all j = 2, . . . , s. Assume, for the contrary, that there exists j = 2, . . . , s, such that 〈eij , eij+1〉 6= −1. If j = s, then [vi2 , . . . , vis+1 ] is not a root. If j = 2, then [vi2 , vi3 ] ∈ j. Applying the Jacobi identity we get [vi1 , [vi2 , . . . , vis+1 ]] ≡ [vi1 , [vi3 , [vi2 , [vi4 , . . . , vis+1 ]]]] and, by the case 1), [vi1 , v] ∈ j. Therefore we can assume that j = 3, . . . , s− 1. Since [vi2 , . . . , vis+1 ] is a root, 〈eij , eij+1 + . . . + eis+1〉q = −1 and 〈eij , eij+2 + . . . + eis+1〉q ≥ −1. By the bilinearity of 〈−,−〉q, Lemma 2.5(c) and our assumptions, we have 〈eij , eij+2 + . . . + eis+1〉 = −1 and 〈eij , eij+1〉 = 0. By assumptions and (ii), the elements [vij , vij−1 , . . . , vi2 ] and [vij+1 , vij+2 , . . . , vis+1 ] are roots. Moreover 〈eij+1 , ei2 + . . .+ eij−1〉q = 〈eij+1 , ei2 + . . . + eij−1 + eij 〉q = −1. Set y = [vij−1 , . . . , vi2 ] and x = [vij+2 , . . . , vis+1 ], then 1 = q(v) = q(ey + eij + eij+1 + ex) = q(ey + eij ) + q(eij+1 + ex) + 〈ey + eij , eij+1 + ex〉q = 2 + 〈ey, eij+1〉q + 〈ey, ex〉q + 〈eij , eij+1〉q + 〈eij , ex〉q = 2 + (−1) + 〈ey, ex〉q + 0 + (−1) = 〈ea, ex〉q. It follows q(−ex + (ey + ei1)) = q(ex) + q(ey + ei1) − 〈ex, ey + ei1〉q = 1 + 1 − 〈ex, ey〉q − 〈ex, ei1〉q = 2 − 1 − 1 = 0, because 〈ex, ei1〉q = 〈eis+1 , ei1〉q = 1. This is a contradiction, because q is positive definite. Finally, we proved that 〈eij , eij+1〉q = −1 for all j = 2, . . . , s. If 〈eij , eil〉q ≤ 0 for all 2 ≤ j < l ≤ s + 1, then (i1, i2, . . . , is+1) is a positive chordless cycle and therefore [vi1 , vi2 , . . . , vis+1 ] ∈ j. Indeed, if J. Kosakowska 75 (i1, i2, . . . , is+1) is not a positive chordless cycle, then there exists 2 ≤ j < l ≤ s + 1 such that l 6= j, j + 1 and 〈eij , eil〉q = −1. Therefore q(ei2 + . . . + eis+1) ≤ s − (s − 1) + 〈eij , eil〉q = 0 and q is not positive definite. Assume that 〈eij , eil〉q > 0 for some 2 ≤ j < l ≤ s + 1. Choose j, l such that 2 ≤ j < l − 1 ≤ s + 1 and l − j is minimal with the property 〈eij , eil〉q 6= 0. If 〈eij , eil〉q = −1, then q(eij + . . . + eil) = 0. If 〈eij , eil〉q = 2, then q(eij + . . . + eil) = −1. In both cases q is not positive definite. Therefore 〈eij , eil〉q = 1. Note that in this case (ij , ij+1, . . . , il) is a positive chordless cycle and [vil , vil−1 , . . . , vij+1 ] is a root. If l = s + 1, then v ≡ [vi2 , . . . , vij , . . . , vil ] ∈ j, by the definition. Therefore we can assume that l < s + 1. If j = 2, then [vi1 , v] ≡ [vij , [[vil , vil−1 , . . . , vij+1 ], [vil+1 , . . . , vis+1 ]] ≡ [[vil , vil−1 , . . . , vij+1 ], [vij [vil+1 , . . . , vis+1 ]], because [vij , [vil , vil−1 , . . . , vij+1 ]] ∈ j. It follows by 1), that [vi1 , v] ∈ j, because 〈ei1 , eij+1 + . . . + eil〉q = 0. Therefore we can assume that 2 < j < l < s + 1 and [vi1 , v] ≡ [vi1 , [x, [vij , [b, y]]]], where x = [vij−1 , . . . , vi2 ], b = [vil , vil−1 , . . . , vij+1 ] and y = [vil+1 , . . . , vis+1 ]. Since ex, eij + eb + ey and ex + eij + eb + ey are roots of q, by Lemma 2.5(b′) we have 〈ex, eij + eb + ey〉q = −1. Consider q(−ey + (ei1 + ex)) = q(ey) + q(ei1 + ex) − 〈ei1 + ex, ey〉q = 1 + 1 − 〈ei1 , ey〉q − 〈ex, ey〉q = 1 − 〈ex, ey〉q, because 〈ei1 , ey〉q = 〈ei1 , eis+1〉q = 1 and 〈ei1 , ex〉q = 〈ei1 , ei2〉q = −1. On the other hand [vi1 , v] ≡ [vi1 , [x, [vij , [b, y]]]] = −[vi1 , [vij , [[b, y], x]]] − [vi1 , [[b, y], [x, vij ]]]. Therefore 〈eij , ex〉q = −1, because otherwise by the induction hypothesis we have [x, vij ] ∈ j, and by 1), [vi1 , v] ≡ [vi1 , [x, [vij , [b, y]]]] = −[vi1 , [vij , [[b, y], x]]] ∈ j. Similarly we have 〈ex, eb〉q = −1, because otherwise [vi1 , v] ≡ [vi1 , [x, [vij , [b, y]]]] ≡ [vi1 , [x, [b, [vij , y]]]] ≡ [vi1 , [b, [[y, vij ], x]]] ∈ j, 76 Lie algebras associated with quadratic forms by the case 1). Finally 〈ex, ey〉q = −1 − 〈ex, eij + eb〉q = −1 − 〈ex, eij 〉q − 〈ex, eb〉q = 1 and q(−ey + (ex + ei1)) = 1 − 〈ex, ey〉q = 0. This is a contradiction, because q is positive definite. This finishes the proof. 7. Examples and final remarks In this section we present some examples and remarks that illustrate basic results of this paper. Theorem 7.1. If A = CQ/I is a representation directed C-algebra, such that its Tits form qA is positive definite, then the map Φ : L(qA, j) → K(A) (7.2) given by vi 7→ ui is an isomorphism of Lie algebras. Moreover L(qA, j) ∼= G+(qA). Proof. By Corollary 5.2, the map Φ : L(qA, r) → K(A), given by Φ(vi) = ui, is an isomorphism of Lie algebras. By Propositions 6.3 and 6.10, we have L(qA, j) = L(qA, r), because qA is positive definite. The isomorphism L(qA, j) ∼= G+(qA) follows from Proposition 4.4. Remark 7.3. Let A be a representation directed C-algebra and let qA be its Tits form. It is well-known (see [6]) that qA is weakly positive. It follows that the set R+ qA of positive roots of qA is finite. Therefore dimC K(A) = |R+ qA | is finite. In this case, the subset r of L(qA) is finite, even if qA is not positive definite. Moreover, we are able to describe an algorithm that constructs the set r. Indeed, it is enough to develop Definition 3.7 and construct all Weyl roots of q (see [8, Remark 4.15]). If qA is positive definite, then L(qA, j) = L(qA, r). The set j is a min- imal set generating the ideal (r) and j is smaller than r (see Example 7.4). If qA is not positive definite, then (j) ( (r) in general (see Example 7.5). Example 7.4. Let L be the following poset 4 L : 2 @@������� 3 ^^======= 1 @@������� ^^======= J. Kosakowska 77 and let KL be the incidence algebra of the poset L (see [15]). It is easy to see that KL is representation directed, qKL is positive definite and B(qKL) has the form 4 2 ������� 3 >>>>>>> 1 ������� >>>>>>> � � � � � � � Then j = {[u2, u3], [u1, u4], [u1, [u1, u2]], [u1, [u1, u3]], [u2, [u1, u2]], [u3, [u1, u3]], [u2, [u2, u4]], [u3, [u3, u4]], [u4, [u2, u4]], [u4, [u3, u4]], [u1[u2, u4]], [u1, [u3, u4]]} and L(qKL, j) ∼= K(KL). Note that r = {[u2, u3], [u1, u4], [u1, [u1, u2]], [u1, [u1, u3]], [u2, [u2, u1]], [u3, [u3, u1]], [u2, [u2, u4]], [u3, [u3, u4]], [u4, [u4, u2]], [u4, [u4, u3]], [u1[u2, u4]], [u1, [u3, u4]], [u3, u2], [u4, u1], [u1, [u2, u1]], [u1, [u3, u1]], [u2, [u1, u2]], [u3, [u1, u3]], [u2, [u4, u2]], [u3, [u4, u3]], [u4, [u2, u4]], [u4, [u3, u4]], [u1[u4, u2]], [u1, [u4, u3]], [u1, u2, u1, u3], [u2, u2, u1, u3], [u3, u2, u1, u3], [u1, u3, u1, u2], [u2, u3, u1, u2], [u3, u3, u1, u2], [u1, u2, u3, u1], [u2, u2, u3, u1], [u3, u2, u3, u1], [u1, u3, u2, u1], [u2, u3, u2, u1], [u3, u3, u2, u1], [u4, u2, u4, u3], [u2, u2, u4, u3], [u3, u2, u4, u3], [u4, u3, u4, u2], [u2, u3, u4, u2], [u3, u3, u4, u2], [u4, u2, u3, u4], [u2, u2, u3, u4], [u3, u2, u3, u4], [u4, u3, u2, u4], [u2, u3, u2, u4], [u3, u3, u2, u4], [u1, u4, u2, u3, u1], [u2, u4, u2, u3, u1], [u3, u4, u2, u3, u1], [u4, u4, u2, u3, u1], [u1, u4, u3, u2, u1], [u2, u4, u3, u2, u1], [u3, u4, u3, u2, u1], [u4, u4, u3, u2, u1], [u1, u1, u2, u3, u4], [u2, u1, u2, u3, u4], [u3, u1, u2, u3, u4], [u4, u1, u2, u3, u4], [u1, u1, u3, u2, u4], [u2, u1, u3, u2, u4], [u3, u1, u3, u2, u4], [u4, u1, u3, u2, u4], [u1, u4, u2, u1, u3], [u2, u4, u2, u1, u3], [u3, u4, u2, u1, u3], [u4, u4, u2, u1, u3], [u1, u4, u3, u1, u2], [u2, u4, u3, u1, u2], [u3, u4, u3, u1, u2], [u4, u4, u3, u1, u2], [u1, u1, u2, u4, u3], [u2, u1, u2, u4, u3], [u3, u1, u2, u4, u3], [u4, u1, u2, u4, u3], [u1, u1, u3, u4, u2], [u2, u1, u3, u4, u2], [u3, u1, u3, u4, u2], [u4, u1, u3, u4, u2]} Example 7.5. Consider the following graph 4 Q : 2 a @@�������� 3 c ^^======== 1 d @@�������� b ^^======== 78 Lie algebras associated with quadratic forms Let A = CQ/I, where I = (ab, cd). The form qA is not positive definite and B(qA) has the following form 4 B(qA) : 2 �������� 3. ======== 1 �������� <<<<<<<< � � � � � � � � � � � � � � Note that [u4, [u3, [u2, u1]]] ∈ (r), but [u4, [u3, [u2, u1]]] 6∈ (j). On the other hand, the algebra A is representation directed and qA is weakly positive. By Corollary 5.2, K(A) ∼= L(qA, r). References [1] R. K. Amayo and I. Stewart, Infinite dimensional Lie algebras, Noordhoff Inter- national Publishing, 1974. [2] I. Assem, D. Simson and A. Skowroński, "Elements of the Representation Theory of Associative Algebras", Vol. I: Techniques of Representation Theory, London Mathematical Society Student Texts, 65. Cambridge University Press, Cambridge, 2006. [3] M. Auslander, I. Reiten and S. Smalø, "Representation theory of Artin algebras", Cambridge Studies in Advanced Mathematics 36, Cambridge University Press, 1995. [4] M. Barot, D. Kussin and H. Lenzing, The Lie algebra associated to a unit form, J. Algebra 296 (2007), 1-17. [5] M. Barot and D. Rivera, Generalized Serre relations for Lie algebras associated to positive unit forms, J. Pure Appl. Algebra 211 (2007), 360-373. [6] K. Bongartz, Algebras and quadratic forms, J. London Math. Soc. 28 (1983), 461-469. [7] S. Kasjan and J. Kosakowska, On Lie algebras associated with representation di- rected algebras, arXiv:0902.3348v1 [math.RT]. [8] J. Kosakowska, A classification of two-peak sincere posets of finite prinjective type and their sincere prinjective representations, Coll. Math. 87 (2001), 27-77. [9] J. Kosakowska, Prinjective Ringel-Hall algebras for posets of finite prinjective type, Acta Math. Sinica, in press. [10] J. Kosakowska, A specialization of prinjective Ringel-Hall algebra and the associ- ated Lie algebra, Acta Math. Sinica, English Series, 24(2008), 1687-1702. [11] Ch. Riedtmann, Lie algebras generated by indecomposables, J. Algebra 170 (1994), 526-546. [12] C. M. Ringel, Tame Algebras and Integral Quadratic Forms, Lecture Notes in Mathematics, Vol. 1099 (Springer-Verlag, Berlin, Heidelberg, New York, Tokyo 1984). J. Kosakowska 79 [13] C. M. Ringel, Hall algebras, Banach Center Publications, Vol. 26, Warsaw 1990, 433-447. [14] C. M. Ringel, From representations of quivers via Hall and Loewy algebras to quantum groups, Proceedings Novosibirsk Conference 1989. Contemporary Math- ematics 131.2 (1992), 381-401. [15] D. Simson, "Linear Representations of Partially Ordered Sets and Vector Space Categories", Algebra, Logic and Applications, Vol. 4, Gordon & Breach Science Publishers, 1992. Contact information J. Kosakowska Faculty of Mathematics and Computer Sci- ence, Nicolaus Copernicus University, ul. Chopina 12/18, 87-100 Toruń, Poland E-Mail: justus@mat.uni.torun.pl URL: www.mat.uni.torun.pl/~justus Received by the editors: 15.08.2008 and in final form 27.09.2008.

Ähnliche Einträge