Structure of Degenerate Block Algebras - Springer Link

Algebra Colloquium 10:1 (2003) 53–62

Algebra Colloquium c AMSS CAS 2003

Structure of Degenerate Block Algebras∗ Linsheng Zhu Department of Mathematics, Nanjing University, Nanjing 210093, China Department of Mathematics, Changshu College Changshu, Jiangsu 215500, China E-mail: [email protected]

Daoji Meng Department of Mathematics, Nankai University, Tianjin 300071, China Received 24 April 2001 Revised 22 August 2001 Communicated by Nanqing Ding Abstract. Given a non-trivial torsion-free abelian group (A, +, 0), a field F of characteristic 0, and a non-degenerate bi-additive skew-symmetric map φ : A×A → F , we define a Lie algebra L = L(A, φ) over F with basis {ex | x ∈ A\{0}} and Lie product [ex , ey ] = φ(x, y)ex+y . We show that L is endowed uniquely with a non-degenerate symmetric invariant bilinear form and the derivation algebra Der L of L is a complete Lie algebra. We describe the double extension D(L, T ) of L by T , where T is spanned by the locally finite derivations of L, and determine the second cohomology group H 2 (D(L, T ), F ) using anti-derivations related to the form on D(L, T ). Finally, we compute the second Leibniz cohomology groups HL2 (L, F ) and HL2 (D(L, T ), F ). 2000 Mathematics Subject Classification: 17B05, 17B30 Keywords: quadratic Lie algebra, double extension, derivation, Leibniz cohomology

1 Introduction Let F be a field of characteristic 0, A a torsion-free abelian group, L the vector space over F with basis {ex | x ∈ A}, and φ : A × A → F a skewsymmetric bi-additive function. Then L becomes a Lie algebra under the bracket [ex , ey ] = φ(x, y)ex+y .

∗

This work was supported by the NNSF of China (19971044), the Doctoral Programme Foundation of Institution of Higher Education (97005511), and the Foundation of Jiangsu Educational Committee.

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For convenience, we suppose Kφ = {x ∈ A | φ(x, A) = 0} = {0}. Then L(A, P φ) = F e0 ⊕ L(A, φ), where F e0 is the center of L and L(A, φ) = x∈A\{0} F ex is the derived algebra of L. We denote L(A, φ) simply by L. In [6], Dokovic and Zhao proved that the Lie algebra L(A, φ) is simple. They also determined the derivations of L(A, φ), described all isomorphisms between two such algebras, and computed H 2 (L, F ). The Lie algebra L is the degenerate case of the generalized Block algebra studied in [5]. For the more general case, refer to [19]. In Sec. 2 of this paper, we prove that L is quadratic and dim B(L) = 1. Suppose rank(A) < ∞. Let T be the unique torus of Der L. Then we have the root space decomposition of the double extension D(L, T ) of L by T . In Sec. 3, we describe the derivation algebras of L and D(L, T ), and prove that Der L is an infinite-dimensional complete Lie algebra. In Sec. 4, we compute the second cohomology groups H 2 (L, F ), H 2 (D(L, T ),F ) and the second Leibniz cohomology groups HL2 (L,F ), HL2 (D(L,T ),F ) (cf. [8, 9, 14]). 2 Quadratic Lie Algebras A Lie algebra g endowed with a non-degenerate symmetric invariant bilinear form B is called a quadratic Lie algebra (cf. [2, 24]). Denote by B(g) the vector space spanned by all non-degenerate symmetric invariant bilinear forms on g (see [25] for detail). We denote Dera g = {D ∈ Der g | B(Dx, y) = −B(x, Dy)}, the anti-derivation algebra of g (cf. [8]). Now we give the definition of double extension (see, e.g., [15]). Let (g, B) be a quadratic Lie algebra and h a Lie algebra. Assume ψ : h → Dera (g) is a homomorphism of Lie algebras. Define the 2-cocycle φ : g × g → h∗ by φ(x, y)(z) = B(ψ(z)(x), y) ˙ φ h∗ be the central extension of g by h∗ for any x, y ∈ g and z ∈ h. Let g + relative to φ. The homomorphism ψ can be extended to a homomorphism (still de˙ h∗ ) by noted by ψ) from h to Der (g + ψ(z)(x + f ) = ψ(z)(x) + ad∗ z(f ) for x ∈ g, z ∈ h, and f ∈ h∗ , where ad∗ is the coadjoint representation of h, i.e., ad∗ z(f ) = f ◦ ad z. Let G = h⊕g⊕h∗ be a direct sum of vector spaces. Define a Lie product [(y1 , x1 , f1 ), (y2 , x2 , f2 )] to be [y1 , y2 ]h , [x1 , x2 ]g +ψ(y1 )(x2 )−ψ(y2 )(x1 ), ad∗ y1 (f2 )−ad∗ y2 (f1 )+φ(x1 , x2 ) and a bilinear form on G by ¯ (y1 , x1 , f1 ), (y2 , x2 , f2 ) = B(x1 , x2 ) + f1 (y2 ) + f2 (y1 ) B

Structure of Degenerate Block Algebras

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¯ h + h∗ ) = for any xi ∈ g, yi ∈ h, and fi ∈ h∗ (i = 1, 2). Clearly, B(g, ∗ ∗ ¯ h) = B(h ¯ , h ) = 0 and B| ¯ (h⊕h∗ )×(h⊕h∗ ) is non-degenerate. Then one B(h, ¯ is a quadratic Lie algebra. We call it the double can easily check that (G, B) extension of g by h, denoted by D(g, h). Theorem 2.1. L is a quadratic Lie algebra and dim B(L) = 1. Proof. We establish this assertion in several steps. Step 1. L is quadratic. Define a bilinear form B(·, ·) on L by setting B(ex , ey ) = δx,−y

(1)

for all x, y ∈ A\{0}. It follows from (1) that B([ex , ey ], ez ) = φ(x, y)δx+y,−z and B(ex , [ey , ez ]) = φ(y, z)δx,−y−z for all x, y, z ∈ A\{0} with x + y 6= 0 and y + z 6= 0. Clearly, x + y = −z if and only if x = −y − z. Since φ(y, ±y) = 0 and φ is skew-symmetric bi-additive, we have φ(x, y)δx+y,−z = φ(y, z)δx,−y−z for x, y ∈ A. Hence, B is invariant on L. The non-degeneracy and symmetry are obvious. Step 2. If B 0 is a non-degenerate symmetric invariant bilinear form on L, then B 0 (ex , e−x ) 6= 0 and B 0 (ex , ey ) = 0 for any y 6= x. Assume φ(x, y) 6= 0. Since B 0 is invariant, we have B 0 ([ex , e−x+y ], ex ) = B 0 (ex , [e−x+y , ex ]), i.e., φ(x, −x + y)B 0 (ey , ex ) = φ(−x + y, x)B 0 (ex , ey ). Since φ is skewsymmetric and φ(x, y) 6= 0, B 0 (ey , ex ) = −B 0 (ex , ey ). Since B 0 is symmetric, we obtain B 0 (ex , ey ) = 0. Now assume x + y 6= 0 but φ(x, y) = 0. Since φ is non-degenerate, we can choose z ∈ A such that φ(z, x)φ(z, y)φ(z, x + y) 6= 0. From the above discussion, we see that 0 = φ(x, z)B 0 (ex+z , ey−z ) = B 0 ([ex , ez ], ey−z ) = B 0 (ex , [ez , ey−z ]) = φ(z, y)B 0 (ex , ey ), and so B 0 (ex , ey ) = 0. Since B 0 is non-degenerate and L is generated by {ex | x ∈ A\{0}}, we have B 0 (ex , e−x ) 6= 0 for any x ∈ A\{0}. Step 3. If B 0 is a non-degenerate symmetric invariant bilinear form on L, then B 0 = cB for some scalar c ∈ F . Denote B 0 (ex , e−x ) = ax for x ∈ A\{0}. Then we have ax = a−x 6= 0. Consider the invariant symmetric bilinear form B 00 = B 0 − ax0 B for some x0 6= 0. Then ex0 is in the radical of B 00 . Since the radical of B 00 is an ideal of L and L is simple, L is in the radical of B 00 , i.e., B 00 = 0. Our assertion follows. 2 From now on, we assume rank(A) = n < ∞. Suppose {xi | i = 1, 2, . . . , n} is a minimal set which generates the group A. Define µi ∈ Hom(A, F ) by setting µi (xj ) = δij for i, j = 1, 2, . . . , n. Moreover, define the derivations ti (i = 1, 2, . . . , n) by ti (ex ) = µi (x)ex for any x ∈ A\{0}, and denote

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Pn T = i=1 F ti . Let K = span{c1 , c2 , . . . , cn } be the vector space with basis ˜ = L ⊕ K by {c1 , c2 , . . . , cn }. Define the Lie product on L [ex , ey ] = φ(x, y)ex+y if x + y 6= 0, Pn Pn [ex , e−x ] = i=1 ai ci for any x = i=1 ai xi ∈ A\{0}, [c, u] = 0 for any c ∈ K and u ∈ L. ˜ by setting ti (K) = 0. We extend the derivation ti of L to a derivation of L ˆ =T ⊕L ˜ be a direct sum of vector spaces and define the Lie product Let L ˆ by setting on L [ti , x] = ti · x (2) ˜ for x ∈ L. Theorem 2.2. With above notations, we have ˜ is the universal central extension of L, (i) L ˆ coincides with the double extension D(L, T ) of L by T , and h = (ii) L ˜ ˙ K is a Cartan subalgebra of L, T+ P ˙ (iii) D(L, T )P = h + x∈∆ Gx , where ∆ = A\{0}, and Pn Gx = {u ∈ D(L, T ) | n [t, u] = i=1 ai µi (t)u ∀ t ∈ h} = F ex for x = i=1 ai xi . Proof. (i) follows from [6], while (ii) and (iii) are obvious.

2

ˆ is an extended affine Lie Remark. From [1], we can easily check that L algebra with only isotropic roots. We conjecture that any (Z × Z)-graded infinite-dimensional extended affine Lie algebra with only isotropic roots must be of this form or its q-deformation. 3 Derivations of Der L and D(L, T ) A Lie algebra g is called complete if its center is zero and all the derivations of g are inner. This notion was first given by Jacobson [11]. Some developments in the area of complete Lie algebras were obtained in [16, 17, 23, 24]. Theorem 3.1. Assume rank(A) = n < ∞. Then the derivation algebra Der L of L is a complete Lie algebra. Pn Proof. Denote g = Der L. Recall that T = i=1 F ti is the set of degree ˙ ad L, where ad L is the inner derivations of L. By [6], we know that g = T + derivation algebra. Since L is simple, L ' ad L. Clearly, g is isomorphic to ˙ L, where the Lie product on T + ˙ L is as in (2). Then we can suppose T+ ˙ L. g=T+ Let D ∈ Der g. Since L = [g, g], we see that DL = [D(g), g] ⊂ L and so D|L ∈ Der L. There exists x ∈ g such that (D − ad x)(L) = 0. Let D0 = D−ad x. Applying D0 to [T, g] = g, we see that D0 (T ) = 0. Therefore, D0 = 0, i.e., D = ad x. So g is complete. 2

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Let γ ∈ Hom(T, K). We extend γ to a linear transformation γ¯ on D(L, T ) by setting γ¯ (L+K) = 0. One can easily check that γ¯ ∈ Der D(L, T ). Denote a = span {¯ γ | γ ∈ Hom(T, K)}. ˙ ad D(L, T ). Theorem 3.2. Der D(L, T ) = a + ˙ Proof. By Theorem 2.2, we P have D(L, T ) = h + n A\{0} and Gx = F ex for x = i=1 ai xi . From [18] or [22], we have

P

x∈∆

Gx , where ∆ =

Der G = D0 + ad G, where D0 = {D ∈ Der G | D(Gx ) ⊂ Gx } and D0 ∩ ad G = ad h. For any D ∈ D0 , one may easily check that D(h) ⊂ K. Since dim Gx = 1 for any x ∈ A, we may assume Dex = λx ex . Applying D to [ex , ey ] = φ(x, y)ex+y gives λx+y = λx + λy for any x, y ∈ A if φ(x, y) 6= 0. If φ(x, y) = 0 and x + y 6= 0, since rad φ = 0, we can choose z ∈ A such that φ(x, z) 6= 0, φ(y, z) 6= 0, and φ(x + y, z) 6= 0. It follows that φ(y + z, x) 6= 0. Applying D to [ey , ez ] = φ(y, z)ey+z , [ex+y , ez ] = φ(x + y, z)ex+y+z , and [ey+z , ex ] = φ(y + z, x)ex+y+z , we have λx + λy + λz = λx λy+z = λx+y+z = λx+y + λz . Thus, λx+y = λx + λy for all x, y ∈ A. In particular, λx = −λ−x for any x ∈ A. Applying D to ci = [exi , e−xi ] (i = 1, 2, . . . , n) gives D(K) = 0. ˙ K) = 0. Thus, D1 ∈ a. Define D1 ∈ End G so that D1 |h = D|h and D1 (L + Let D2 = D − D1 and t = λx1 t1 + λx2 t2 + · · · + λxn tn . It is clear that D2 = ad t. Hence, Der G = a + ad G. It is easy to prove that a ∩ ad G = {0}, which concludes the assertion. 2 Remark. One of the most efficient tools in the investigation of derivations of finite-dimensional Lie algebras is the root space decomposition with respect to a Cartan subalgebra (cf. [11]). The fact that the method bears fruit even for infinite-dimensional algebras was first observed by Berman [4], and the method has been developed by many authors (cf. [3, 5, 6, 21, 22]). 4 Second Cohomology Group and Second Leibniz Cohomology Group In [6], Dokovic and Zhao determined the second cohomology group H 2 (L, F ). In this section, we obtain a general result on the second cohomology of a quadratic Lie algebra with a commutative diagonalizable subalgebra (we call such a subalgebra a torus) by using the idea of Kac [12]. As a consequence of the result, we obtain the second cohomology group H 2 (D(L, T ), F ). Us˜ is the universal central ing the methods in [6, 9, 14], we prove that L extension of the Lie algebra L in the category of Leibniz algebras, i.e., HL2 (L, F ) = H 2 (L, F ), but HL2 (D(L, T ), F ) 6= H 2 (D(L, T ), F ). We begin with the following two facts. Lemma 4.1. [12] Let (g, B) be a quadratic Lie algebra and φ an F -valued 2-cocycle on g. Assume there exists an orthonormal basis {e1 , e2 , . . . , en , . . . }

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of g such that for each i, φ(ei , ej ) = 0 for all but a finite number of j. Then there is an anti-derivation d on g such that φ(x, y) = B(dx, y) for any x, y ∈ g. P Lemma 4.2. [12] Let g be a Lie algebra, h ⊂ g a torus, and g = α gα the root space decomposition with respect to h. Then every F -valued 2-cocycle φ on g is equivalent to a cocycle φ0 (i.e., φ(x, y) − φ0 = f ([x, y]) for all x, y ∈ g and some f ∈ g∗ ) such that φ0 (gα , gβ ) = 0 for α + β 6= 0. Let (g, B) be a quadratic Lie algebra with a torus h which equals its own centralizer. Then g has a root space decomposition with respect to h X ˙ g = g0 + gα , (3) α∈∆

where h = g0 and Bgα +g−α is non-degenerate. If dim gα < ∞ for all α ∈ ∆ ∪ {0}, we call (3) a regular decomposition of g (cf. [18]). Recall that for any quadratic Lie algebra (g, B) and an anti-derivation d, φ(x, y) := B(dx, y) is a 2-cocycle on g. If any F -valued 2-cocycle on g is of this form, then g is called good. Proposition 4.3. If a quadratic Lie algebra (g, B) has a regular decomposition, then g is good. Proof. It follows from Lemmas 4.1 and 4.2.

2

Corollary 4.4. Quadratic Lie algebras with a regular triangular decomposition (e.g., Kac–Moody algebras and extended affine Lie algebras) and finite-dimensional quadratic Lie algebras are all good. Proposition 4.5. (i) H 2 (Der L, F ) = 0. (ii) dim H 2 (D(L, T ), F ) = n(n − 1)/2, where n = rank(A). Proof. (i) By Theorem 3.1, the Lie algebra Der L is complete. From [16], we know that for any complete Lie algebra, its central extension is split, i.e., its second cohomology group is trivial. Now the assertion follows. (ii) By Theorem 2.2 and Proposition 4.3, D(L, F ) is good. One may easily check that for d1 , d2 ∈ Dera g, φ1 (x, y) = B(d1 x, y) is equivalent to φ2 (x, y) = B(d2 x, y) if and only if d1 − d2 ∈ ad g is an inner derivation. By ˙ ad D(L, T ). Assume {t1 , t2 , . . . , tn } and Theorem 3.2, Der D(L, T ) = a + {c1 , c2 , . . . , cn } are bases of T and K, respectively, and B(ti , cj ) = δij for i, j = 1, 2, . . . , n. Define P dij ∈ a (1 ≤ i < j ≤ n) such that dij (ti ) = cj , ˙ K) = 0. Denote by a1 the vector space dij (tj ) = −ci , and dij ( l6=i,j F tl + spanned by all dij . Then one may easily check that each dij is an anti˙ ad D(L, T ). Moreover, if we derivation on D(L, T ) and Dera D(L, T ) = a1 + denote φij (x, y) = B(dij x, y), then φij (ti , tj ) = 1, X X ˙ L+ ˙ ˙ L+ ˙ φij (K + F tk , K + F tl ) = 0, k6=i,j

l6=i,j


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and {φij } is a basis of H 2 (D(L, T ), F ).

2

Remark. Let g(A) be a Kac–Moody algebra associated to the generalized Cartan matrix A. Then g is good by Corollary 4.4. By Theorem 2.1 in [8], one may easily check that the dimension of the second cohomology group H 2 (g, C) of g is l(l − 1)/2 using the method in Proposition 4.5, where l is the corank of A (cf. [20]). A Leibniz algebra g over F is a vector space equipped with a bilinear map, [·, ·] : g × g → g satisfying the Leibniz identity [x, [y, z]] = [[x, y], z] − [[x, z], y] for all x, y, z ∈ g. Clearly, for any Leibniz algebra g, there is an associated Lie algebra gLie obtained by factoring by the relation [x, x] = 0 for all x ∈ g. On the other hand, any Lie algebra is a Leibniz algebra. The Leibniz cohomology is a non-commutative analog of the Lie algebra cohomology developed by Loday (refer to [9, 10, 14] for detail). Let g be a ˆ is Leibniz algebra over F . A central extension of g is a pair (ˆ g, π), where g ˆ → g is a surjective homomorphism whose kernel a Leibniz algebra and π : g ˆ. The pair (ˆ lies in the center of g g, π) is called a universal central extension of g if for any central extension (˜ g, τ ) of g, there is a unique homomorphism ˆ → g ˜ such that τ ◦ ψ = π. A Leibniz algebra g is called perfect if ψ : g [g, g] = g. It is well known that the universal central extension of a Leibniz algebra g exists if and only if g is perfect. ˜ and L be as in Sec. 2. Then L ˜ is the universal Theorem 4.6. Let L central extension of the Lie algebra L in the category of Leibniz algebras. Proof. It is sufficient to show that H 2 (L, F ) → HL2 (L, F ) is an isomorphism. Since it is already injective, we only need to prove the surjectivity. We know that {ex | x ∈ A\{0}} forms a basis for L and [ex , ey ] = φ(x, y)ex+y . Let (ex , ey ) 7→ ψ(ex , ey ) for x, y 6= 0 be a Leibniz 2-cocycle of L. Then ψ(x, [y, z]) = ψ([x, y], z) − ψ([x, z], y) (4) for all x, y, z ∈ L. For sake of simplicity, we set λ(x, y) = ψ(ex , ey ) for x, y 6= 0. By setting x = eu , y = ev , and z = ew in (4), we have φ(v, w)λ(u, v + w) = φ(u, v)λ(u + v, w) − φλ(u + w, v). Claim 1. If φ(x, y) 6= 0, then λ(x, y) = −λ(y, x). By setting u = v = x and w = y − x in (5), we have φ(x, y − x)λ(x, y) = −φ(x, y − x)λ(y, x). It follows from φ(x, y) 6= 0 that λ(x, y) = −λ(y, x). Claim 2. If φ(x, y) = 0 and y 6= −x, then λ(x, y) = −λ(y, x).

(5)

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For µ ∈ A, define λµ (x) = λ(x, µ − x) for x 6= 0, µ. By the above claim, λµ (x) = −λµ (µ − x) when φ(x, µ) 6= 0. Claim 2 is equivalent to that if µ 6= 0 and φ(x, µ) = 0, then λµ (x) = −λµ (µ − x). By setting x = u, y = v, and w = µ − x − y in (5), we obtain [φ(y, µ) + φ(x, y)]λ(x, µ − x) = φ(x, y)λ(x + y, µ − x − y) − [φ(x, µ) − φ(x, y)]λ(µ − y, y), i.e., φ(x, y)[λµ (x)−λµ (x+y)−λµ (µ−y)]+φ(y, µ)λµ (x)+φ(x, µ)λµ (µ−y) = 0. (6) Setting y = 2x in (6), we have 2φ(x, µ)λµ (x) + φ(x, µ)λµ (µ − 2x) = 0. Assume φ(x, µ) 6= 0. By Claim 1, we have λµ (µ − 2x) = −λµ (2x). It follows that if φ(x, µ) 6= 0, then λµ (2x) = 2λµ (x). (7) If φ(µ, x), φ( µ, y), and φ(µ, x + y) are all non-zero, then by (6) and Claim 1, we have φ(x, y)[λµ (x + y)−λµ (x)−λµ (y)] − φ(y, µ)λµ (x) + φ(x, µ)λµ (y) = 0. (8) By replacing x and y in (8) with 2x and 2y, respectively, and using (7), we have φ(x, y)[λµ (x + y) − λµ (x) − λµ (y)] = 0 (9) and φ(y, µ)λµ (x) = φ(x, µ)λµ (y).

(10)

If φ(x, µ), φ(y, µ) 6= 0 and φ(x + y, µ) = 0, then φ(µ, x + 2y) 6= 0, and so (10) is valid if we replace y with 2y. Using (7), we conclude that (10) is valid even if φ(x + y, µ) = 0. Then by (8), we see that (9) holds if φ(x, µ), φ(y, µ) 6= 0 and φ(x + y, µ) = 0. It follows from (10) that the ratio aµ =

λµ (x) φ(µ, x)

is independent of x, provided that φ(µ, x) 6= 0. In other words, there exists a constant aµ ∈ F such that λµ (x) = aµ φ(µ, x)

(11)

whenever φ(µ, x) 6= 0. Let x 6= 0, µ and φ(µ, x) = 0. Choose y ∈ A such that φ(x, y) and φ(µ, y) are both non-zero. Then by (9), λµ (x) = λµ [(x − y) + y] = λµ (x − y) + λµ (y) = aµ (φ(µ, x − y) + φ(µ, y)) = 0, λµ (µ − x) = λµ (µ − x − y + y) = λ(µ − x − y) + λµ (y) = aµ (φ(µ, µ − x − y) + φ(µ, y)) = aµ φ(µ, µ − x) = 0.


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So λµ (x) = −λµ (µ − x) holds if x 6= 0, µ and φ(x, µ) = 0. Claim 3. If µ = 0, then λ0 (z) = −λ0 (−z). If µ = 0 and φ(x, y) 6= 0, then by (6), we have λ0 (x) − λ0 (x + y) − λ0 (−y) = 0.

(12)

For any z ∈ A − {0}, we may choose x ∈ A\{0} such that φ(x, z) 6= 0. It follows from (12) that λ0 (x) = λ0 (x + z − z) = λ0 (x + z) − λ0 (z) = λ0 (x) − λ0 (−z) − λ0 (z). (13) Hence, λ0 (z) = −λ0 (−z) holds for any 0 6= z ∈ A. By the above, we see that ψ(ex , ey ) = −ψ(ey , ex ) holds for any x, y ∈ A\{0} and so ψ is also a Lie cocycle. This completes the proof of the theorem. 2 Theorem 4.7. dim HL2 (D(L, T ), F ) = n2 , where n = rank A < ∞. ˜ F ) = 0. For any ψ ∈ HL2 (D(L, T ), F ), Proof. By Theorem 4.6, HL2 (L, 2 ˜ ˜ L) ˜ = 0. Let we know that ψ|L× ∈ HL ( L, F ). So we may assume ψ(L, ˜ L ˜ {x1 , x2 , . . . , xn } be a basis of A and assume ti , ci are as in Sec. 2. Note that for any z ∈ A\{0}, there exists x ∈ A\{0} such that φ(z, x) 6= 0. By setting x = t, y = ez , and z = ex in (3), we have φ(z, x)ψ(t, ez ) = ψ(t, [ez , ex ]) = ψ([t, ez ], ex ) − ψ([t, ex ], ez ) = 0, i.e., ψ(t, ez ) = 0 for any t ∈ T and z ∈ A\{0}. Since [exi , e−xi ] = ci for i = 1, 2, . . . , n, ψ(t, ci ) = ψ(t, [exi , e−xi ]) = ψ([t, exi ], e−xi ) − ψ([t, e−xi ], exi ) = 0. Hence, ψ(T, K) = 0. Clearly, ψ(T, T ) is a bilinear function. On the other hand, since T + K is abelian, for any bilinear function ψ on T , we may extend it to a bilinear function on D(L, T ) (still denoted by ψ) ˜ = 0. One may check that ψ is a Lebniz 2-cocycle. by setting ψ(D(L, T ), L) Now the assertion follows. 2 Acknowledgements. This paper was finished during the first author’s visit to the Morningside Center of Mathematics, Chinese Academy of Sciences. He would like to thank the hospitality of this center and thank Prof. K.M. Zhao for giving the main ideas of this paper and simplifying the proof of Theorem 3.1.

References [1] B.N. Allison, S. Azam, S. Berman, Y. Gao, A. Pianzola, Extended Affine Lie Algebras and Their Root Systems, Mem. Amer. Math. Soc., Vol. 603, 1997. [2] S. Benayadi, A new characterization of semisimple Lie algebras, Proc. Amer. Math. Soc. 125 (1997) 685–688. [3] G. Benkart, R.V. Moody, Derivations, central extensions, and affine Lie algebras, Algebras, Groups Geom. 3 (1986) 456–492. [4] S. Berman, On derivations of Lie algebras, Canad. J. Math. 28 (1976) 174– 183.

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