Pollution: A Silent Killer

STRUCTURE THEORY FOR ONE CLASS OF LOCALLY FINITE LIE ALGEBRAS. arXiv:math/0310208v1 [math.RA] 14 Oct 2003 1 L.A. Simonian Abstract In this paper I consider locally finite Lie algebras of characteristic zero sat- isfying the condition that for every finite number of elements x1 , x2 , · · · , xk of such an algebra L there is finite-dimensional subalgebra A which contains these elements and L(adA)n ⊂ A for some integer n. For such algebras I prove several structure theorems that can be regarded as generalizations of the classical structure theorems of the finite-dimensional Lie algebras theory. ——————— INTRODUCTION. The subject of this article is similar to that of Refs.[1, 2, 3]. I consider locally finite Lie algebras of characteristic zero. A Lie al- gebra is called locally finite if every its finite subset is contained in a finite- dimensional subalgebra. We will study representations R = (M, L) of such algebras. A representation R = (M, L) is a homomorphism of a Lie algebra L into the algebra of linear transformations of a linear space M. We will assume that R = (M, L) satisfies the following condition: (1) for every x ∈ L the linear transformation xR has the Fitting null component of a finite co-dimension. Here xR is the linear transformation that corresponds to x ∈ L in the representation R. For such linear transformation the following is true: LEMMA 1. Let A be a linear transformation of a linear space M of an infinite dimension, M be the Fitting null component of A, that is, the set of all x ∈ M with xAm = 0 for some integer m. If M/M0 has finite dimension, then M is a direct sum M0 ⊕ M1 , where M1 is a finite-dimensional space invariant under A, and the transformation induced by A in M1 is an automorphism. Lemma 1 allows us to determine the trace trA of A as a trace of the linear transformation, which is induced by A in the finite-dimensional space 1 Send correspondence to the author at [email protected] or 1060 Ocean Avenue F8, Brooklyn, NY 11226, USA 1 M/M0 . Besides, given a representation R = (M, L) satisfying condition (1), this Lemma allows us to construct a decomposition M = Mρ ⊕ Mσ ⊕ . . . ⊕ Mτ ⊕ M0 into weight spaces relative to a nilpotent subalgebra H of L. This decompo- sition has the same properties as in the finite-dimensional case and is used in the proof of THEOREM 3. Let L be a locally finite Lie algebra of characteristic zero. Suppose L has a representation R = (M, L) in a vector space M satisfying the condition (1) and the associative span A of L in the representation R = (M, L) is locally finite. If the kernel of the representation R is locally solvable and tr(xR )2 = 0 for every x ∈ L′ , then L is locally solvable. Theorem 3 can be considered as a generalization of the Cartan’s criterion for solvability. Subsequent results are obtained for Lie algebras and their representations that satisfy the following conditions: (2) representations R = (M, L): for every finite-dimensional subagebra A of L there is such an integer n that MAn has finite dimension. (3) Lie algebra L: for every finite set x1 , x2 , · · · , xk ∈ L there is a finite-dimensional subagebra A which contains these elements and for which L(adA)n ⊂ A for some integer n. COROLLARY. An algebra L satisfying the condition (3) is locally solv- able if and only if tr(adx)2 = 0 for every x ∈ L′ . Since the intersection of a finite number of finite-codimensional subspaces has finite codimension, the trace can be defined simultaneously for any finite number of transformations for algebras that satisfy condition (2). All prop- erties of the usual finite-dimensional trace are true in this case. Therefore, for the representation R = (M, L) satisfying the condition (2) we can define a trace form f (a, b) = traR bR , a, b ∈ L. In particular, if L satisfies the condition (3), we obtain the form K(a, b) = tr(ada)(adb) It is natural to name this form the Killing form of L. Lie algebra L is said to be semi-simple if its locally solvable ideal equals to 0, that is, if L has no non-zero locally solvable ideals. 2  THEOREM 4. Let L be a locally finite semi-simple Lie algebra of characteristic 0, and R = (M, L) be an arbitrary faithful representation satisfying the condition (2). Then the trace form f (a, b) is non-degenerate. If L satisfies the condition (3) and the Killing form K(a, b) is non-degenerate, then L is semi-simple. This theorem can be regarded as a generalization of the Cartan’s criterion for semi-simplicity. Finally, the following theorem generalizes the structure theorem. THEOREM 6 (Structure Theorem). Let L be a semi-simple Lie algebra that satisfies condition (3). Then L is a subdirect sum of a set of finite- dimensional simple algebras. 1. WEIGHT SPACES. PROOF OF LEMMA 1. Let N be a finite- dimensional subspace of M such that M = N ⊕ M0 and let e1 , e2 , . . . , en be n a basis of the N. We have ei A = aij ej + mi , where mi ∈ M0 . For any mi P j=1 there is an integer number ni such that mi Ani = 0. Let P denote the linear subspace spanned by mi , mi A,. . . , mi An1 −1 , i = 1, 2, . . . , n. It is clear that P is invariant under A and has a finite dimension. From the equality ei A = n aij ej + mi it follows that NA ⊂ P + N. Hence K = P + N is invariant P j=1 under A. We have M = K + M0 . Since K is finite-dimensional, it can be represented as K = K1 ⊕ K0 , where K1 and K0 are, respectively, the Fitting one and the Fitting null components of K relative to the transformation induced by A in K. Then M = K1 + K0 + M0 . Since K0 ⊂ M0 , then M = K1 + M0 . We shall show now that this sum is direct. Since K is finite-dimensional there is t such that yAt = 0 for any y ∈ K ∩ M0 . On the other hand, there exists s such that K1 = KAs = KAs+1 = · · ·. Let r = max(s, t). Then K1 = KAr and yAr = 0 for any y ∈ K ∩ M0 . Now let x ∈ K1 ∩ M0 . Then from the equality K1 = KAr it follows that x = yAr for some y ∈ K. On the other hand, since x ∈ K1 ∩ M0 , it holds xAr = 0. But then 0 = xAr = yA2r . Hence y ∈ K ∩ M0 and, consequently, yAr = 0. But then x = yAr is equal to zero and K1 ∩ M0 = 0. From the construction of K1 it follows, that A is a linear transformation acting in K1 as an automorphism. Therefore, the only possibility is to put M1 = K1 . This lemma may be considered as a generalization of the well-known Fitting’s lemma. Let M0 and M1 are the Fitting null and the Fitting one 3 components of M relative to A, respectively. As in [4], a linear transformation A will be called algebraic, if every vector x ∈ M is contained in a finite-dimensional subspace that is invariant under A. LEMMA 2. If the Fitting null component M0 of M relative to A has finite codimension, then A is an algebraic linear transformation. PROOF. By Lemma 1, M = M1 ⊕ M0 , where M1 has finite dimension and A acts in M1 as an automorphism. Let x ∈ M. We shall show that the dimension of the smallest subspace that contains x and is invariant relative to A is finite. Let x = y +z, where y ∈ M1 and z ∈ M0 . Since M1 is invariant under A, yA ∈ M1 . On the other hand, zAm = 0 for some integer m. Let N be the subspace, which is generated by M1 , z, zA, . . . , zAm−1 . It is clear that x ∈ N, N is invariant under A, and N has finite dimension. The lemma is proved. Let the characteristic roots of A be in the base field Φ and let M1 = Mα ⊕ Mβ ⊕ · · · ⊕ Mγ be the decomposition of M1 into the weight spaces relative to A. Then M = M0 ⊕ ( Mα ). Also we have that all Mα with L α6=0 α 6= 0 are of finite dimension. It is worth recalling that by definition x ∈ Mα , if and only if x(A − αE)m = 0 for some integer m. We will require some known results which I outline here for completeness. Let A be an associative algebra, and a ∈ A. Let us consider the inner derivation Da : x → x′ = [x, a] in A. If we denote x(k) = (x(k−1) )′ , x(0) = x, then the following formulas hold: ! ! k k−1 ′ k k−2 ′′ xak = ak x + a x + a x + · · · + x(k) 1 2 ! ! k ′ k−1 k ′′ k−2 k a x=a x− k xa + x a + · · · + (−1)k x(k) 1 2 xφ(a) = φ(a)x + φ1 (a)x′ + φ2 (a)x′′ + · · · + x(r) , where φ(λ) is a polynomial of degree r and φk (λ) = φ(k) (λ)/k!. LEMMA 3 [6]. Let A, B be linear transformations in a vector space M satisfying B(adA)u = 0 for some integer u. Let µ(λ) be a polynomial and let MµA = {x|xµ(A)m = 0 for some integer m}. Then MµA is invariant under B. PROOF. Let x ∈ MµA and suppose that xµ(A)m = 0. Putting φ(λ) = µ(λ)mu , we obtain Bφ(A) = φ(A)B + φ1 (A)B ′ + · · · + φu−1 (A)B (u−1) . Since 4 φ0 (λ) = φ(λ), φ1(λ), . . . , φu−1 (λ) are divisible by µ(λ)m , xφj (A) = 0, 0 ≤ j ≤ u − 1. Therefore xBφ(A) = 0 and xB ∈ MµA . COROLLARY. If B(adA)u = 0 then the weight spaces Mα are invariant under B. Let R = (M, L) be a representation of a Lie algebra L in a vector space M of infinite dimension, satisfying the condition (1) and let characteristic roots of every AR , A ∈ L, lie in the base field. We shall also assume that for every finite-dimensional subalgebra H of L there is such an integer m that L(adH)m = 0. THEOREM 1. If H is an finite-dimensional subalgebra of L, then M can be decomposed as ( Mα ) ⊕ M0H where Mα , α 6= 0, are finite-dimensional L α6=0 weight spaces relative to L with the weights α, and M0H is a weight space relative to H with the weight α = 0. The dimension of M0H is infinite. PROOF. First lets show that for every x ∈ M the smallest subspace N that is invariant under H and contains x has finite dimension. Indeed (see also [4]), if B1 , B2 , . . . , Br is a basis of H, then N is the linear span of the set of all elements of the form xBim1 1 Bim2 2 . . . Bimk k , i1 ≤ i2 ≤ . . . ≤ ik . Since all Bi are algebraic, there is only a finite set of linearly independent elements of a given form and all of them may be found among the elements xBim1 1 Bim2 2 . . . Bimk k for which mj ≤ sj , where sj are integers and j = 1, 2, . . . , k. Suppose now that every element A of a finite subset F ⊂ H is locally nilpotent, that is, for A the following condition is satisfied: for any x ∈ M there exists m such that xAm = 0. We shall show that this condition holds for every element of the subalgebra {F } generated by the set F . Indeed, since H is nilpotent, F is contained in Jacobson radical of representation (N, H) [5]. Hence {F } is contained in it. But this means that {F } consists of nilpotent relative to N transformations and hence the given condition holds for the elements of {F }. Since H is nilpotent, there exists a chain of ideals 0 ⊂ H1 ⊂ H2 ⊂ . . . Hn−1 ⊂ Hn = H such that dim Hi+1 /Hi = 1. Take an arbitrary element A ∈ H1 and let M = MαA be a decomposition of M into the weight L αA R spaces relative to A . From Lemma 3 it follows that all MαA are invariant under L. Besides there is just a finite number of the subspaces MαA , and dim MαA < ∞ if αA 6= 0. Therefore M1A = MαA can be decomposed L αA 6=0 into a direct sum of a finite number of weight spaces relative to L (see [6, p. 5 43]). We recall that a map α : A → α(A) of L into the base field Φ is called the weight of M relative to L if there exists a nonzero element x ∈ M such that x(AR − a(A))m = 0 for all A ∈ L. Here m is an integer which depends on x and A. The set of elements (zero included) satisfying this condition forms the subspace that is called the weight subspace. It should be recalled that M0A also is invariant under L. Let B be an element of H2 \ H1 and M0A = MαB be a decomposition of M0A into the weight spaces relative L αB ′ to B R . The subspace M1B = MαB has a finite dimension, is invariant L αB 6=0 under L and can be decomposed into a direct sum of a finite number of weight spaces relative to L. If we add this decomposition to decomposition ′ of M1A , we obtain that M = ( Mα ) ⊕ M0B ′ . AR and B R act in M0B as L α6=0 locally nilpotent transformations. Therefore, the subalgebra {A, B} consists ′ of the locally nilpotent in M0B transformations [7]. Continuing in this way we obtain - by virtue of finite dimensionality of H - the statement of the theorem. If L is nilpotent we may combine Lie’s theorem with Theorem 1 to obtain the following THEOREM 2. If L is finite-dimensional, then M is a direct sum of weight spaces Mα , and the matrices in the weight space Mα , α 6= 0 can be taken simultaneously in the form α(A) 0 ... 0    ∗ α(A) . . . 0  Aα =    ... ... ... ...    ∗ ∗ . . . α(A) This theorem is proved in exactly the same way as in Ref.[6]. In a similar fashion we obtain COROLLARY. The weights α : A → α(A) are linear functions on L which vanish on L′ . 2. CARTAN’S CRITERION. Let L be a finite-dimensional Lie algebra, H be a nilpotent subalgebra of L, R = (M, L) be a representation of L in a vector space M satisfying the condition (1). PROPOSITION 1. Let M = Mρ ⊕ Mσ · · · Mτ ⊕ M0 6  L = Lα ⊕ Lβ ⊕ · · · ⊕ Lγ ⊕ L0 be the decompositions of M and L into weight spaces relative to H. (The existence of the first decomposition was proved in the previous section). Then Mρ Lα ⊂ Mρ+α if ρ+ α is the weight of M relative to H; otherwise Mρ Lα = 0. PROOF. For every x ∈ M and A, B ∈ L we have the equality xA(B − ρI − αI) = x(B − ρI)A + x(A(adB − αI)). If x(B − ρI)m = 0 and A(adB − αI)n = 0 then by repeating this equality we obtain xA(B−ρI−αI)m+n+1 = 0. Here ρ = ρ(B), α = α(B), and I is an identity operator of M. It is also true that [Lα , Lβ ] ⊂ Lα+β if α + β is a root of L and [Lα , Lβ ] = 0 otherwise (see [6, p. 64]). Suppose now that H is a Cartan subalgebra. Then H = L0 , the root module corresponding to the root 0. Also, we have L′ = [L, L] = [Lα , Lβ ], P α,β where the sum is taken over all roots α, β, and L′ ∩ H = [Lα , L−α ], where P α the summation is taken over all α such that −α is also a root (see [6, p. 67]). Let A be a linear transformation of M with the Fitting null component M0A of a finite codimension. Then, as noted in Introduction, trA can be defined as the trace of the linear transformation induced by A in the quotient space M/M0A . It is easy to see that trA is equal to the trace of the linear transformation induced by A in the quotient space of M by any invariant under A subspace of a finite codimension contained in M0A . LEMMA 4. Let Φ be algebraically closed of characteristic 0. Under the assumptions of this section let H be a Cartan subalgebra of L and let α be a root such that −α is also a root. Let eα ∈ Lα , e−α ∈ L−α , hα = [eα , e−α ]. Then r(hα ) is a rational multiple of α(hα ) for every weight ρ of H in M. PROOF. Let M0α = M0 + Miα , i = 0, ±1, ±2, . . .. Lets turn to the P i quotient space M = M/M0α . If M0α is invariant under x ∈ L, then the operator induced by x in M is denoted as xR . Consider functions of the form ρ(h) + iα(h), i = 0, ±1, ±2, . . ., which are weights, and form the subspace N = M ρ+iα where M ρ+iα = Mρ+iα + M0α /M0α and the sum is taken over P i the corresponding weight spaces of the representation R = (M, L). N is invariant relative to H and, by Proposition 1, it is also invariant relative to the linear transformations eR R α and e−α . Thus, if trN denotes the trace of an induced mapping in N, then trN hR R R α = trN [eα , e−α ] = 0. On the other hand, the restriction of hR α α α to M σ = Mσ +M0 /M0 has the single characteristic root 7 σ(hα ). Hence 0 = trN hR α = nρ+iα (ρ + iα)(hα ) where nρ+iα = dim Mρ+iα . P i Thus we have ( nρ+iα )ρ(hα ) + ( inρ+iα )α(hα ) = 0. Since nρ+iα is a P P P i i i positive integer, this shows that ρ(hα ) is a rational multiple of α(hα ). PROOF OF THEOREM 3. Assume first that the base field Φ is alge- braically closed. It suffices to prove that C ′ ⊂ C for every finite-dimensional subalgebra C of the algebra L. Hence we shall have that C ⊃ C ′ ⊃ C ′′ ⊃ C (k) = 0. We therefore suppose that there exists a finite-dimensional subal- gebra C such that C ′ = C. Let H be a Cartan subalgebra of C and let M = Mρ ⊕ Mσ ⊕ · · · ⊕ Mτ ⊕ M0 C = Cα ⊕ Cβ ⊕ · · · ⊕ Cγ ⊕ C0 . be the decomposition of M and C into weight spaces relative to H. Then the formula H ∩ C ′ = [Cα , C−α ] implies that H = [Cα , C−α ] summed on P P α α α such that −α is also a root. Choose such α, let eα ∈ Cα , e−α ∈ C−α , and consider the element hα = [eα , e−α ]. The formula H = [Cα , C−α ] P α implies that every element of H is a sum of terms of the form [eα , e−α ]. Let us turn to the quotient space M = M/M0 and denote by hR α the operator R induced by hα in M . The restriction of hα to M ρ = Mρ + M0 /M0 has 2 the single characteristic root ρ(hα ). Hence the restriction of (hR α ) has the single characteristic root ρ(hα )2 . Let nρ be the dimension of Mρ . Then we 2 have tr(hRα )) = nρ (ρ(hα ))2 . On the other hand, tr(hR R α ) = tr(hα ). Thus P ρ6=0 nρ (ρ(hα ))2 = 0 since tr(hR 2 α ) = 0. By the Lemma 4, ρ(hα ) = rρ α(hα ), P ρ6=0 where rρ is rational. Hence α(hα )2 ( nρ rρ2 ) = 0. Since nρ are positive P ρ6=0 integers, this implies that α(hα ) = 0 and ρ(hα ) = 0. Since ρ are linear functions and every h ∈ H is a sum of elements of the form hα , hβ , . . ., we see that ρ(h) = 0. Thus 0 is the only weight for M, that is, we have M = M0 . If α is a root, then the condition ( 0 if ρ + α is not a weight of M Mρ Cα = ⊂ Mρ+α if ρ + α is a weight implies that MCα = 0 for every α 6= 0. Hence Cα ⊕Cβ ⊕· · ·⊕Cγ , α, β, . . . , γ 6= 0 is contained in the kernel K of representation (M, C). Hence C/K is a 8 homomorphic image of H. It follows that the C/K is nilpotent. According to our assumptions the kernel K is solvable, and it follows that C is solvable which contradicts C ′ = C. If the base field is not algebraically closed, then let Ω be its algebraic closure. Then (MΩ , LΩ ) = RΩ is the representation of LΩ in MΩ and KΩ is the kernel of this representation if K is the kernel of R = (M, L). Since K is locally solvable, KΩ is locally solvable. Next we note that tr(xR )2 = 0 and trxR y R = try RxR imply that trxR y R = (1/2)tr(xR y R + y R xR ) = (1/2)(tr(xR + y R )2 − tr(xR )2 − tr(y R )2 ) = 0. Hence if xi ∈ L and ωi ∈ Ω, 2 then tr( ωi xR i ) = ωi ωj trxR R i xj = 0. To prove that the condition (1) P P i i holds we use the fact that associative span A of L in the representation m R = (M, L) is locally finite. We need to show that ( ωi xi )R has the Fit- P i=1 ting null component of finite codimension for every x1 , x2 , . . . , xm ∈ L and ω1 , ω2 , . . . , ωm ∈ Ω. Since A is locally finite, the subalgebra A in A gen- erated by x1 , x2 , . . . , xm has a finite dimension. Therefore there exists an integer u such that y(xR )u = 0. Here x is an arbitrary linear combination of x1 , x2 , . . . , xm with coefficients from the base field Φ, and y is an arbitrary element from the Fitting null component M0X of the linear transformation xR . Take (u + 1)m elements of the form ki1 x1 + mi2 x2 + ni3 x3 + · · · + sim xm , where i1 , i2 , i3 , . . . , im receive their values 1, 2, 3, . . . , u+1 independently, and k1 , k2 , . . . , ku+1, m1 , m2 , . . . , mu+1 , n1 , n2 , . . . , nu+1 , . . . , s1 , s2 , . . . , su+1 are pair- wise different nonzero integers. The intersection N of the Fitting null com- ponents of these elements has a finite codimension. Let us prove that for any y ∈ N the equality y(ω1x1 + ω2 x2 + · · · + ωm xm )u = 0 takes place for every ω1 , ω2 , . . . , ωm ∈ Ω. We have (kj x1 + mi2 x2 + ni3 x3 + · · · + sim xm )u = 0, j = 1, 2, . . . , u + 1 for any mi2 , ni3 , . . . , sim which are taken from the set of integers shown above. It follows immediately that yP0 + kj yP1 + kj2 yP2 + · · · + kju yPu = 0, j = 1, 2, . . . , u + 1 where Pi = Pi (x1 , mi2 x2 , ni3 x3 , . . . , sim xm ) is the homogeneous component of i-th degree relative to x1 of (kj x1 + mi2 x2 + ni3 x3 + · · · + sim xm )u . Since the 9 determinant of this system is Vandermonde’s determinant, it follows that yP0 = yP1 = yP2 = · · · = yPu = 0. Next for Pi , i = 0, 1, 2, . . . , u, we have yPi (x1 , ml x2 , ni3 x3 , . . . , sim xm ) = 0, l = 1, 2, . . . , u − i + 1. It immediatelly follows that yPi0 + ml yPi1 + m2l yPi2 + . . . + mu−i l yPi(u−i) = 0, l = 1, 2, . . . , u − i + 1 where Pij (x1 , x2 , ni3 x3 , . . . , sim xm ) is a component of Pi which is homogeneous of degree i relative to x1 and of degree j relative to x2 . Since the determinant of this system is Vandermonde’s determinant, it follows that yPi0 = yPi1 = · · · = yPi(u−i) = 0. Continuing in this way we obtain that yPi1 i2 ...im = 0, i1 , i2 , . . . , im = 1, 2, . . . , u, i1 + i2 + · · · + im = u , where Pi1 i2 ...im = Pi1 i2 ...im (x1 , x2 , . . . xm ) are homogeneous polynomials of degree i1 relative to x1 , of degree i2 relative to x2 and so on, and finally, of degree im relative to xm , which arise in comput- ing of the power (x1 + x2 + · · · + xm )u , and which are its multihomogeneous components. On the other hand, y(ω1x1 + ω2 x2 + · · · + ωm )u = ω1i1 ω2i2 . . . ωm im X yPi1 i2 ...im (x1 , x2 , . . . , xm ). Hence y(ω1x1 + ω2 x2 + · · · + ωm xm )u = 0. Thus we have proved that the conditions of the theorem hold in RΩ = (MΩ , LΩ ). The first part of the proof, therefore, implies that LΩ is locally solvable. LEMMA 5. Let R = (M, L) and A is a finite-dimensional subalgebra of L such that MAn has finite dimension for some integer n. Then a subspase of all x ∈ L for which xAn = 0 has finite codimension. PROOF. Let z1 , z2 , · · · , zk is a basis of A. Then any product of n elements of the basis transfers M into finite-dimensional subspase of M. Thus the kernel of this product has finite codimension. Since there is only a finite number of different product of n elements of the basis, the intersection of all kernels of such products has a finite codimension as well. The lemma is proved. From Lemma 5 it follows that the associative algebra generated by L in R = (M, L) is locally finite. Therefore we can apply the results of [5] about the Jacobson radical of Lie algebra. Thus we obtain the following COROLLARY. Let L be a locally finite Lie algebra over a field of char- acteristic 0. Suppose L satisfies the condition (3). Then L is locally solvable if and only if tr(ada)2 = 0 for every a ∈ L′ . 10  PROOF. The sufficiency of the condition is a consequence of Theorem 3, since the kernel of the adjoint representation is the centre. Conversely, assume L is locally solvable. Then from [5] it follows that ada, a ∈ L′ , belongs to radical JadL∗ (adL). Hence ada is a nilpotent linear transformation and tr(ada)2 = 0. 3. A TRACE FORM. Let R = (M, L) be a representation of a Lie alge- bra L in a vector space M which satisfies condition (2). Then we can define a trace form f (a, b) = traR bR , a, b ∈ L. The function f (a, b) is evidently a symmetric bilinear form on M with values in the base field Φ. In particular, if L satisfies (3), we obtain the Killing form K(a, b) = tr(ada)(adb). If f (a, b) is the trace form defined by the representation R = (M, L), then f ([a, c], b) + f ([a, [b, c]) = tr([a, c]R bR + aR [b, c]R ) = tr([aR , cR ]bR + aR [bR , cR ]) = tr([aR bR , cR ]) = 0. As noted in the Introduction, we can calculate the trace simultaneously for every finite number of elements of L, since these elements can be consid- ered as linear transformations acting on a common quotient space of M of a finite dimension. Therefore, the last chain of equalities is correct. A bilinear form f (a, b) on L that satisfies the condition f ([a, c], b) + f (a, [b, c]) = 0 is called an invariant form on L. Hence the trace form is invari- ant. We note next that if f (a, b) is any symmetric invariant form on L, then the radical L⊥ of the form - that is, the set of elements z such that f (a, z) = 0 for all a ∈ L - is an ideal. This is clear since f (a, [z, b]) = −f ([a, b], z) = 0. PROOF OF THEOREM 4. Let f (a, b) be a trace form of R = (M, L). Then L⊥ is an ideal of L and f (a, a) = tr(aR )2 = 0 for every a ∈ L⊥ . Hence L⊥ is locally solvable by Theorem 3. Since L is semi-simple, L⊥ = 0, and f (a, b) is non-degenerate. Next suppose that the Killing form is non- degenerate. If R(L)′ 6= 0 then by the Theorem 7 from [5] R(L)′ ⊂ J(L), Jacobson radical of L. This implies that for every a ∈ R(L)′ , it is true that ada ∈ J(adL∗ ). Hence for every b ∈ L we have ada · adb ∈ J(adL∗ ). Since by Levitzki theorem J(adL∗ ) is locally nilpotent ideal, then ada · adb is a nilpotent linear transformation and therefore tr(ada · adb) = 0. This contradicts our assumption that the trace form tr(ada·adb) is non-degenerate. Therefore, it is required of R(L) that R(L)′ = 0. Let a ∈ R(L), b ∈ L and N is a subspase of L such that L/N is finite- dimensional and ada and adb act in N as nil transformations. Let us denote 11 L = L/N and R(L) = R(L) + N/N and choose a basis for L such that the first vectors form a basis for R(L). The matrices of linear transformations induced by ada and adb in L, respectively, are of the forms ! ! 0 0 ∗ 0 and . ∗ 0 ∗ ∗ This implies that tr(ada)(adb) = 0. Hence R(L) ⊂ L⊥ , and the Killing form is degenerate. Using Killing form the following characterization of the locally solvable radical in the characteristic 0 case (cf.[6, p. 73]) can be obtained. THEOREM 5. If a locally finite Lie algebra L over a field of character- istic 0 satisfies the condition (3), then the locally solvable radical R(L) of L is the orthogonal complement L′⊥ of L′ relative to the Killing form K(a, b). PROOF. The algebra B = L′⊥ is an ideal. Further, if b ∈ B ′ , then tr(adb)2 = K(b, b) = 0. The kernel of the representation a → ada, a ∈ B, is abelian. Hence B is locally solvable, by Corollary of theorem 2. Hence B ⊂ R(L). Next, let x ∈ R(L), a, b ∈ L. Then K(x, [a, b]) = K([x, a], b). By [5] the element [x, a] belongs to the Jacobson radical J(L) of L. Consequently, ad[x, a] belongs to J(adL∗ ) and ad[x, a] · adb is nilpotent for every b. Hence K([x, a], b) = trad[x, a] · adb = 0. Thus K(x, [a, b]) = 0 and x ∈ L′⊥ . Thus R(L) ⊂ L′⊥ and so R(L) = L′⊥ . Let Ω be an extension of the base field Φ of L. Then the Killing form fΩ of LΩ is obtained from the Killing form f of L by the extension. Therefore, fΩ is non-degenerate if and only if f is non-degenerate [8]. Consequently, LΩ is semi-simple if and only if L is semi-simple (see also [9]). 4. STRUCTURE OF SEMI-SIMPLE ALGEBRAS. We continue the consideration of locally finite Lie algebras L which satisfy the condition (3). LEMMA 6 (cf. [6, p. 29]). Let A be a finite-dimensional subalgera of L ∞ such that L(adA)n ⊂ A. Then Aω = Ak is an ideal of L. T k=1 PROOF. If Aω = 0 then the assertion of Lemma is trivial. Let Aω 6= 0. We have [L, An ] ⊂ L(adA)n . A is finite-dimensional. Therefore, Aω = Am for some integer m. Thus Aω = An+m−1 . Now we have [L, Aω ] = [L, An+m−1 ] ⊂ L(adA)n+m−1 ⊂ A(adA)m−1 = Am = Aω , which completes the proof. 12  Let f (a, b) be any symmetric invariant bilinear form on L and let A be a subspace of L. Denote by A⊥ the subspace of L that consists of all elements b ∈ L such that f (a, b) = 0 for all a ∈ A. LEMMA 7. If A is a finite-dimensional subspace of L such that A∩A⊥ = 0, then L = A ⊕ A⊥ . PROOF. Since A ∩ A⊥ = 0, A is a non-degenerate subspace. Let us show that L = A + A⊥ . Take any basis a1 , a2 , . . . , am in A and let c be any element of L. Find a decomposition c = a + a⊥ , where a ∈ A and a⊥ ∈ A⊥ . We will look for a in the form a = x1 a1 + x2 a2 + · · · + xm am . Then c will look like: c = x1 a1 + x2 a2 + · · · + xm am + a⊥ . From f (ai , a⊥ ) = 0 it follows m that f (ai , c) = xk f (ai , ak ), i = 1, 2, . . . , m. This system of equations P k=1 has exactly one solution, since its determinant is the Gram’s determinant of the system a1 , a2 , . . . , am . Since A is a non-degenerate subspace, this determinant is non-zero. The vector a = x1 a1 + x2 a2 + · · · + xm am , where xk were just found, satisfies the conditions f (ai , c−a) = 0. Indeed, f (ai , c−a) = m f (ai , c) − xm f (ai , ak ) = 0. From the equalities f (ai , c − a) = 0, it follows P k=1 that c − a ∈ A⊥ . To complete the proof it remains to put a⊥ = c − a. PROOF OF THEOREM 6. Let A be a finite-dimensional subalgebra for which L(adA)n ⊂ A. By Lemma 6, subalgebra Aω is an ideal of L. By Theorem 4, the Killing form K(a, b) is non-degenerate. Since K(a, b) is invariant, (Aω )⊥ is an ideal of L. Indeed, for every a ∈ Aω , every b ∈ (Aω )⊥ and every c ∈ L we have K(a, [b, c]) = −K([a, c], b) = 0. Let us prove that Aω ∩ (Aω )⊥ = 0. If, on the contrary, b1 , b2 ∈ Aω ∩ (Aω )⊥ and a is any element of L, then K([b1 , b2 ], a) = −K(b1 , [a, b2 ]) = 0. Since K(a, b) is non-degenerate, [b1 , b2 ] = 0. Hence Aω ∩ (Aω )⊥ is an abelian ideal of L. Since L is semi-simple, Aω ∩ (Aω )⊥ = 0. By Lemma 7 this implies that L = Aω ⊕ (Aω )⊥ . Since Aω and (Aω )⊥ are direct summands, every ideal of Aω or (Aω )⊥ is an ideal of L. Therefore Aω and (Aω )⊥ are semi-simple subalgebras. Moreover, since Aω is finite-dimensional, Aω is a direct sum of ideals which are simple. Let us denote by Π the set of all finite-dimensional simple ideals of L. This set is not empty; otherwise we have that Aω = 0 for every x1 , x2 , · · · , xk ∈ L. The previous implies that L is a locally nilpotent algebra. But by condition of the Theorem, L is semi-simple. Since M ω = M for every finite-dimensional simple ideal M, L = M ⊕M ⊥ . Now denote by N the intersection of all M ⊥ where M ∈ Π. Let us prove that N = 0. Suppose 13 not, and let x1 , x2 , · · · , xk be a non-zero element of N. Then by condition of the Theorem, there exists a finite-dimensional subalgebra A, such that x1 , x2 , · · · , xk ∈ A and L(adA)n ⊂ A. Let B = A ∩ N. Then L(adB)n ⊂ A and L(adB)n ⊂ N, since N is an ideal of L. Consequently, L(adB)n ⊂ B and B ω is ideal of L. If B ω 6= 0, then B ω is a semi-simple ideal contained in N. Next, B ω is a direct sum of simple ideals contained in N and, consequently, these ideals do not belong to Π. But this contradicts the definition of Π. Hence B ω = 0 for every finite subset x1 , x2 , · · · , xk ∈ N and N is a locally nilpotent ideal of L. But L is a semi-simple Lie algebra. Hence N = 0. Then by Remac’s theorem L is a subdirect sum of simple finite-dimensional ideals M ≈ L/M ⊥ . The proof of the theorem is complete. It is well known that every derivation of a semi-simple finite-dimensional Lie algebra is inner. What about the infinite-dimensional case? Let L = Lα , where Lα are finite-dimensional simple ideals of L. In this case Stew- L α art [10] proved the theorem that we give here for completeness. THEOREM 7. Let L be a semi-simple Lie algebra and let L is a direct sum, L = Lα , where Lα are finite-dimensional simple ideals of L. Then L α 1) every derivation of L is an element of the complete direct sum L̃ of Lα and the algebra of all derivations D(L) of L is isomorphic to L̃. 2) a derivation D is inner if and only if Lα D = 0 for every Lα with the exception of a finite number of it 3) L has outer derivations 4) every derivation of L is locally finite. It follows from Theorem 7 that any semi-simple Lie algebra satisfying condition (3) is a subalgebra of the algebra of all derivations of such an algebra which is a direct sum of its simple finite dimensional ideals. References [1] L. A. Simonian, Cartan’s criteria for infinite-dimensional Lie algebras, Izvestija vuzov. Matematika, No 4, 1992, 104, (Russian) [2] L. A. Simonian, The theorems of Weyl, Levi and Malcev - Harish - Chandra for locally finite Lie algebras, Izvestija vuzov. Matematika, No 6, 1993, 39 - 45, (Russian) 14  [3] L. A. Simonian, On one class of locally finite Lie algebras, Proc.Latvian Acad.of Scien.,vol. 50, 1996,No 1,35-36 [4] B. I. Plotkin, On algebraic sets of elements in groups and Lie algebras, Uspehi Mat.Nauk, 13, No 6, 1958, 133 - 138, (Russian) [5] L. A. Simonian, On the Jacobson radical of Lie algebras, Latv. Mat. Ezhegodnik, 34 (1993), 230 - 234, (Russian) [6] N. Jacobson, Lie algebras, Intersc. Publishers, New York ·London, 1962 [7] L. A. Simonian, Some questions of the theory of Lie algebras represen- tations, Latv. Univ. Zinatn. Raksti, 58, 1964, 5 - 20, (Russian) [8] N. Bourbaki, Algebra. Modules. Rings. Forms, Moscow, Nauka, 1966, (Russian) [9] N. Bourbaki, Lie groups and algebras. Lie algebras. Free Lie algebras and Lie groups, Moscow, Mir, 1976, (Russian) [10] I. Stewart, Structure theorems for a class of locally finite Lie algebras, Proc. London Math. Soc., 1972, v 24, 79 -100 15