c Allerton Press, Inc., 2008. ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2008, Vol. 52, No. 10, pp. 1–12.  c Yu.B. Ermolayev, 2008, published in Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 2008, No. 10, pp. 3–16. Original Russian Text 

A Realization of Cartan Extensions Yu. B. Ermolayev1 1

Kazan State University, ul. Kremlyovskaya 18, Kazan, 420008 Russia Received November 02, 2004; in final form, revised March 24, 2008

Abstract—We revise the notion of a Cartan extension and consider in detail simple examples of known Cartan extensions. DOI: 10.3103/S1066369X08100010 Key words: graded Lie algebra, Cartan extension, transitivity, creeping mapping, homogeneous element.

1. Basic definitions. A grading of a Lie algebra L = transitive, if the subalgebra L− =

−1 

∞ 

Lk over an arbitrary field K is called

k=−∞

Lk is generated (as a Lie algebra) by the homogeneous

k=−∞

component L−1 and [L−1 , a] = 0 for every nonzero element a ∈ L+ =

∞ 

Lk . An algebra L will be

k=0

called transitive if the grading under consideration is clear from the context. In particular, a Lie algebra −1  Lk is transitive if it is generated by L−1 . L− = k=−∞

Let L be a Lie algebra with transitive grading. A maximal Lie algebra R =

∞ 

Rk with transitive

k=−∞

grading such that Rk = Lk for all k < 0 and Rk ⊇ Lk for all k  0 is called a Cartan extension of L. Obviously, if L and L are two transitive Lie algebras whose negative parts coincide (i.e., L− = L − ), then their Cartan extensions also coincide. In particular, this takes place for Cartan extensions of L and its negative part, i.e., the subalgebra L− . A Cartan extension can be defined inductively. By virtue of the above remark, we let L = L− . −1  Li be a transitive Lie algebra over a field K. Consider the graded Lie algebra R = Let L = ∞ 

i=−d

Ri over the same field K defined by the following conditions:

i=−d

1) Ri = Li for i = −d, . . . , −1 (i.e., one can identify R− = L− ); 2) R0 is the set of all homogeneous derivations h of L (i.e., such derivations that Li h ⊆ Li , i = −d, . . . , −1, and [x, y]h = [xh, y] + [x, yh] ∀ x, y ∈ L); 3) for i > 0, Ri is the set of all differential operators f mapping the algebra L− into R =

i−1 

Rj and

j=i−d

such that Lj f ⊆ Rj+i for j = −d, . . . , −1. ∞  Li is also defined inductively: For f ∈ Ri , g ∈ Rj , and The Lie multiplication on R = i=−d

x ∈ L = R− , the element [f, g] ∈ Ri+j is defined by x[f, g] = [xf, g] + [f, xg] (since a homogeneous element is uniquely defined by its action on L). 1

2

ERMOLAYEV

The Lie algebra R defined above is a Cartan extension of L. In fact, R is transitive by construction, it is maximal since every homogeneous element f  (more precisely, ad f  ) of degree k  0 of any transitive extension of L is a differential operator satisfying condition 2) or 3). −1  Li Let L be a free Lie algebra with generators ξ1 , . . . , ξr over an arbitrary field K, and let L = i=−∞

be its standard grading, but with negative powers, in which L−1 = ξ1 , . . . , ξr . Consider the Kantor algebra ([1]; the universal Cartan extension [2]), i.e., the Cartan extension R of this algebra. Let T1 = L∗−1 = x1 , . . . , xr  — be the conjugate space with conjugate basis (ξi xj = δij , i, j = 1, . . . , r), and let T be the maximal ideal of the tensor algebra generated by T1 . We have R = L ⊕ F,1) where ∞  Fk , where Fk = xi1 · · · xik+1 ξj | it , j = 1, . . . , r. F = T ⊗ L−1 . On F, we define the grading F = The action of F on L−1 is given by the formula

k=0

ξxi1 · · · xim μ = ξ(xi1 )xi2 · · · xim μ

(1.1)

and extends to the whole L (as a derivation). On L, T , F, and R, we will also use the grading by the composition σ for which σ(ξi ) = (0, . . . , −1, . . . , 0), σ(xi ) = (0, . . . , 1, . . . , 0) (±1 at i th place). We will denote by L(i1 ,...,ir ) , T (i1 ,...,ir ) , F (i1 ,...,ir ) , and R(i1 ,...,ir ) the subspaces homogeneous with respect to the composition. Obviously, the degree of an element v ∈ R(i1 ,...,ir ) homogeneous with respect to the composition is  R(i1 ,...,ir ) . In particular, for a monomial f = xi1 · · · xin ξj , we deg v = i1 + · · · + ir , and Rn = i1 +···+ir =n

have σ(f ) = σ(xi1 ) + · · · + σ(xi1 ) + σ(ξj ) and deg f = n − 1. 2. The subspace F . Let L =

∞ 

Lk be a transitive Lie algebra over an arbitrary field K, and L

k=−d

be the free Lie algebra generated by L−1 over the same field with negative grading. Consider the universal Cartan extension R = R(r, K) = L ⊕ F, where r = dim L−1 . Since L is transitive, we have a pair of homomorphisms (ϕ− , ϕ+ ), where ϕ− : L → L− is the epimorphism of Lie algebras (by the first transitivity condition) the restriction of which to L−1 is the linear isomorphism L−1 ∼ = L−1 and ϕ+ : L+ → F is the monomorphism of Lie algebras (by the second transitivity condition) which assigns to each homogeneous element f ∈ L+ the differential operator on L extending the action of ad f on L−1 ∼ = L−1 to the whole L. Let I = ker ϕ− , an ideal in L.  be the subspace of all elements of R which preserve the subspace I. Then Lemma 2.1. Let R  is a Cartan extension of L. R = R/I  is a subalgebra containing the subspace I, and I is an ideal in R.  By the Proof. Obviously, R − − + ∼ ∼ construction, L/I = L = R . In addition R can be identified with the subalgebra in F consisting of all elements preserving I, i.e., R ∼ = L− ⊕ R+ . Finally, if f is a homogeneous element of the Cartan extension of positive degree, then it preserves I (since f can be regarded as a differential operator on L/I, where I is the zero) and, therefore, belongs to R+ . Let I (−d−1) be the subspace of elements of I whose degree  −d−1 and F ={f ∈F | I (−d−1) f ∈ I} (i.e., the subspace in F consisting of all elements preserving I (−d−1) in I, in particular, all generators of I). F induces on F gradings with respect to degree and composition (the corresponding homogeneous subspaces will be denoted by Fk and F (i1 ,...,ir ) ).  We consider elements of Tk as linear mappings L−1 → K T letting ξj xi1 · · · xik = ξj (xi1 )xi2 · · · xik and extending their actions to the whole algebra L as differential operators. 1)

We use different designations for the negative and the positive parts of the algebra letting R− = L and R+ = F since they play different roles in the Cartan extension. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52

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Lemma 2.2. Let W be the subspace in Td+1 consisting of all elements which turn I (−d−1) into zero. Then Fn = W · Fn−d−1 (=W ⊗ Fn−d−1 ) for n  d. Proof. For b ∈ I (−d−1) and f ∈ Fn when n  d, the inclusion bf ∈ I means that bf = 0 (since if an element bf= 0, then deg(bf )  −1 and, therefore, this element cannot lie in I). Let us represent f in the form f = gi fi , where deg gi = d + 1, deg fi = n − d − 1, and fi are all different (and, therefore, linearly i

independent) monomials. Therefore, for a homogeneous b ∈ I (−d−1) , bf = 0 if and only if bgi fi = 0 for each i (bf = hi fi , where deg hi = d + 1 + deg b). In turn, by (1.1), bgi fi = 0 if and only if bgi = 0, i.e., i

gi ∈ W . Hence Fn ⊆ W · Fn−d−1 . The reverse inclusion is obvious by virtue of (1.1). The subspace W in Td+1 is homogeneous with respect the composition, i.e., W = ⊕W (i1 ,...,ir ) , where ik  0, i1 + · · · + ir = d + 1, and each summand lies in W . Lemma 2.3. The subspace F0 preserves I (i.e., R0 = F0 ). Proof. Let η1 , . . . , ηk be generators of I such that deg ηi  −d − 1 for all i = 1, . . . , k. Every element γ of I is a linear combination over K of elements in the form [ηi , a], where a is an element of L. Therefore, it suffices to show that elements of F0 preserve such elements of I. For h ∈ F0 , we have [[ηi , α], h] = [[ηi , h], α] + [ηi , [α, h]] ∈ I since [α, h] ∈ L and [ηi , h] ∈ I by the definition of F0 . Proposition 2.1. Let L =

−1 

Lk be a transitive Lie algebra over an arbitrary field K. In order

k=−d

for f ∈ Fk with k > 0 to lie in Rk , it is necessary and sufficient that the following two conditions hold: f lies in Fk and ξf lies in Rk−1 for any ξ ∈ L−1 . Proof. Since, by Lemma 2.3, R0 = F0 , it follows that the statement of the proposition gives an inductive (on degree) construction of R+ , which in essence is another form of the inductive definition of R given above. In fact, an element f ∈ Fk can be regarded as a homogeneous differential operator which maps −1 k−1  Lk into ⊕ Rj if and only if f ∈ Fk , because, for −d  i  −1, we have Li ∼ L= = Li /Ii , and k=−d

j=k−d

f ∈ Fk acts on Ii , by definition, as zero (and by virtue of the fact that Fk is the set of all homogeneous differential operators of L of degree k). Therefore, condition 3) of the inductive definition holds. By the same reason and by Lemma 2.3, condition 2) holds. Corollary 2.1. Each homogeneous element f ∈ Rm of composition (m1 , . . . , mr ) (for m1 + · · · + mr = m) appears as f = x1 g1 + · · · + xr gr ,

(2.1)

where gi ∈ Rm−1 is a homogeneous element of composition (m1 , . . . , mi − 1, . . . , mr ) for all i = 1, . . . , r. Proof. By the definition of F, each element f ∈ Fm of composition (m1 , . . . , mr ) appears in the form (2.1), where gi ∈ Fm−1 with composition (m1 , . . . , mi − 1, . . . , mr ). Since in this case [ξi , f ] = gi , it follows, by Proposition 2.1, that, for f ∈ R, we have gi ∈ Ri−1 . 3. Creeping mapping on F. Introduce mappings πi for i = 0, . . . , d. Let Vi be a space over K such (i ,...,i ) that Vi = ⊕Vi 1 r , where the summation runs over all compositions of the space Fi (i1 + · · · + ir = i), (i ,...,i )

(i ,...,i )

(i ,...,i )

dim Vi 1 r = dim Fi 1 r − dim Fi 1 r for i = 0, . . . , d − 1, and Vd = V = ⊕V (d1 ,...,dr ) , where the summation runs over all compositions of the space Td+1 (d1 + · · · + dr = d + 1) and dim V (d1 ,...,dr ) = (d ,...,dr ) − dim W (d1 ,...,dr ) . dim Td+11 RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 10 2008

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ERMOLAYEV

Note that if (i1 , . . . , ir ) is the composition of a nonzero element f ∈ F (i1 ,...,ir ) , then all indices ik  −1. In addition, only one index can be equal to −1, and if f ∈ T (d1 ,...,dr ) , f = 0, then all dk  0, k = 1, . . . , r. (i1 ,...,ir )

For i = 0, . . . , d − 1, define πi : Fi → Vi such that πi : Fi

(i1 ,...,ir )

→ Vi

(i ,...,i )

and (ker πi )(i1 ,...,ir ) =

(d ,...,d )

(d ,...,dr )

r 1 → Vd+1 Fi 1 r ; and a (creeping) mapping π = πd : Td+1 → Vd+1 such that π : Td+11 (ker π)(d1 ,...,dr ) = W (d1 ,...,dr ) . We will also use the term creeping for the sequence of mappings P = {P1 , P2 , . . . } such that

and

Pk (xi1 · · · xin+1 ξj ) = xi1 · · · xik−1 · π(xik · · · xik+d ) · xik+d+1 · · · xin+1 ξj for 1  k  n − d + 1, Pk (xi1 · · · xin+1 ξj ) = xi1 · · · xik−1 · πn−k+1 (xik · · · xin+1 ξj ) if n − d + 1 < k  n + 1, Pk (xi1 · · · xim xin+1 ξj ) = 0 if n + 1 < k. Thus, Pk : Fn → Tk−1 ⊗ V ⊗ Fn−k−d for k = 1, . . . , n − d + 1 and Pk : Fn → Tk−1 ⊗ Vk for k = n − d + 2, . . . , n + 1. We also note that Pk (xi1 xi2 · · · xin+1 ξj ) = xi1 Pk−1 (xi2 · · · xin+1 ξj ) for d  n and 2  k  n + 1.

(3.1)

Theorem 3.1. Let R = ker P (i.e., the set of all f ∈ F such that Pk (f ) = 0 for all k = 1, 2, . . . ). Then L ⊕ R is a Cartan extension of L. Proof. Using an induction on k, we will prove that Rk = (ker P )k (the k th homogeneous component of ker P ). Since, by Lemma 2.3, R0 = F0 , it follows that R0 = (ker P )0 by the definition of π0 and P . Let Rj = (ker P )j for j < k. Each element f ∈ Fk of composition (k1 , . . . , kr ) is of the form f = x1 g1 + · · · + xr gr , where (k1 ,...,ki −1,...,kr ) , i = 1, . . . , r. The equality P1 (f ) = 0 is equivalent to the fact that f ∈ Fk (by gi ∈ Fk−1 Lemma 2.1), and the equalities Pt (f ) = 0, t = 2, 3, . . . , mean that gj ∈ (ker P )k−1 = Rk−1 for all j = 1, . . . , r (by the induction assumption). In fact, since Pt (xj gj ) = xj Pt−1 (gj ), j = 1, . . . , r, are  Pt (xi gi ) = 0 implies Pt (xi gi ) = 0, it follows that Pt−1 (gi ) = 0 linearly independent and, therefore, i

for t = 2, 3, 4, . . . , i.e., all gi ∈ Rk−1 by the induction assumption. Hence, by Proposition 2.1, f ∈ Rk . Remark. The main difference of the determination of a Cartan extension given by this theorem from the determination from Proposition 2.1 is only in the fact that here elements of each successive component Rk (passing under the action of L−1 into Rk−1 ) are chosen not from the entire space Fk but only from Fk . 4. Examples. We will not consider properties of a creeping sequence P but give several examples of its use. Our aim is to illustrate the above developed approach to construction of Cartan extensions. By way of examples, we consider well-known algebras. At the same time we prove some of statements of paper [2] for algebras L(r; q) (mainly for r = 2) which were presented without proofs. L(r; q) is the free Lie algebra with r generators truncated by q, i.e., as before, L is the free Lie algebra generated by the space L−1 = ξ1 , . . . , ξr  over an arbitrary field K with standard but negative grading, and I is the ideal in L generated by the homogeneous component L−q−1 . By L(r; q) we mean the algebra L(r; q) ∼ = L/I. Note that, for all such algebras, we have R0 = F0 = F0 = xi ξj , i, j = 1, . . . , r and R1 = F1 (this follows immediately from the definition of the subspaces F0 , F0 , R0 and F1 , R1 respectively). We start with the following simplest case. 5. The algebra L(r; 1). In the case q = 1, the ideal I is generated by the component L−2 , i.e., by the elements [ξi , ξj ], 1  i < j  r. We have [ξs , ξt ]xi xj = (−δsi δtj + δti δsj ). Hence W = x2i , xi xj + xj xi ; 1  i < j  r. The sequence of mappings P is defined by the linear mappings π0 = 0 (since F0 = F0 ) and π : T2 → V2 , where V2 is the set of homogeneous polynomials of degree 2 in commuting variables v1 , . . . , vr over the same field K which is given by the equalities π(x2i ) = 0 for i = 1, . . . , r and π(xi xj ) = vi vj , π(xj xi ) = −vi vj for i < j; i, j = 1, . . . , r. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52

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One can easily see that the kernel P is linearly generated by the elements in the form f (x1 , . . . , xr )ξj , where f (x1 , . . . , xr ) is a symmetric element in T . This allows us to establish an isomorphism between R and the Witt algebra (Wr is an infinite-dimensional algebra of Cartan type) with the use of the correspondence f (x1 , . . . , xr )ξj → f (t1 , . . . , tr )∂j , where f is the corresponding symmetric polynomial in commuting variables t1 , . . . , tr . 6. The algebra L(2; 2). In the case r = q = 2, the ideal I is generated by the elements [ξ1 , ξ2 , ξ1 ], [ξ1 , ξ2 , ξ2 ]. The initial components of F are: (0,0)

F0 = F0 = g(−1,1) = x2 ξ1 , g1 (0,1)

F1 = g(−1,2) = x22 ξ1 ; g1 (1,0)

g1

(0,0)

= x1 ξ1 , g2

= x2 ξ2 , g(1,−1) = x1 ξ2 ; (0,1)

= 2x1 x2 ξ1 + x2 x1 ξ1 , g2 (1,0)

= x21 ξ1 + x2 x1 ξ2 , g2

= x1 x2 ξ1 + x22 ξ2 ;

= x1 x2 ξ2 + 2x2 x1 ξ2 ; g(2,−1) = x21 ξ2 .

Elements of R homogeneous by composition will be written with indication of the composition as a superscript. The space W is linearly generated by the elements x32 (0, 3); 2x1 x22 + x2 x1 x2 , x1 x22 − x22 x1 (1, 2); 2x21 x2 + x1 x2 x1 , x21 x2 − x2 x21 (2, 1); x31 , (3, 0). The mappings πi are defined as follows. Since F0 = F0 , we have π0 = 0 (on F0 ), and π1 is defined on F1 by equalities π1 (x22 ξ1 ) = 0; π1 (x1 x2 ξ1 ) = v1 v2 ∂1 , π1 (x2 x1 ξ1 ) = −2v1 v2 ∂1 , π1 (x22 ξ2 ) = −v1 v2 ∂1 ; π1 (x21 ξ1 ) = v1 v2 ∂2 , π1 (x1 x2 ξ2 ) = −v1 v2 ∂2 , π1 (x2 x1 ξ2 ) = −2v1 v2 ∂2 ; π1 (x21 ξ2 ) = 0. Finally, π is given on T3 by the formulas π(x32 ) = 0 (0, 3); π(x1 x22 ) = v1 v22 , π(x2 x1 x2 ) = −2v1 v22 , π(x22 x1 ) = v1 v22 (1, 2); π(x21 x2 ) = v12 v2 , π(x1 x2 x1 ) = −2v12 v2 , π(x2 x21 ) = v12 v2 (2, 1); π(x31 ) = 0 (3, 0). One can easily see that the kernel of π1 is F1 and the kernel of π is W . Proposition 6.1. Let R =

∞ 

Rn be a Cartan extension of L(2; 2). Then

n=−2

1) Rn = R(−1,n+1) ⊕ R(0,n) ⊕ R(1,n−1) ⊕ · · · ⊕ R(n−1,1) ⊕ R(n,0) ⊕ R(n+1,−1) , where R(k,m) is the space of elements of composition (k, m), 2) in this case, dim R(k,m) = min{k, m} + 2 and the generators are determined by induction (on n = k + m), 3) for k < m, the space R(k,m) is generated by the elements in the form (k,m)

gi

=

k+1 

(k−1,m)

Ak,m ij x1 gj

(k,m−1)

+ x2 gi

,

i = 1, . . . , k + 2, where Ak,m ij ∈ K,

j=1

4) the space R(k,k) is generated by the elements in the form (k,k)

gi

=

k+1  (k−1,k) (k,k−1) k k (Akij x1 gj + Bij x2 gj ), where Akij , Bij ∈ K, j=1

5) for k > m, the space R(k,m) is generated by the elements in the form (k,m)

gi

(k−1,m)

= x1 gi

+

m+1 

(k,m−1)

k,m Bij x2 gj

,

k,m i = 1, . . . , m + 2, where Bij ∈ K.

j=1

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For Cartan extensions of algebras L(r; q), one can naturally define the automorphisms induced by permutations of the variables. In particular, for r = 2, on R, we have the automorphism ρ extending the correspondence: ρ(ξ1 ) = ξ2 , ρ(ξ2 ) = ξ1 , ρ(x1 ) = x2 , ρ(x2 ) = x1 . It is clear that ρ(R(k,m) ) = R(m,k) . If k < m and {gik,m ; i = 1, . . . , k + 2} is a basis of Rk,m, then m,k . This assumption a basis in Rm,k , {gim,k ; i = 1, . . . , k + 2}, will be chosen so that ρ(gik,m ) = gk+3−i k,k with concerning the basis will also take place for m = k, i.e., it will be assumed that ρ(gik,k ) = gk+3−i

the exception of the case when k + 3 − i = i (k is odd) and when ρ(gik,k ) = −gik,k (only for i = 12 (k + 3)). The automorphism ρ will be called a symmetry. By virtue of the above assumptions, for j = 1, . . . , k + 1, we have m,k m,k k,m = Ak,m Bij k+3−i,k+2−j , Aij = Bk+3−i,k+2−j for k < m and i = 1, . . . , k + 2; k k = Akk+3−i,k+2−j , Aki,j = Bk+3−i,k+2−j for i = 1, . . . , k + 1 but i = k + 3 − i, Bi,j k k = −Aki,k+2−j , Aki,j = −Bi,k+2−j for i = k + 3 − i. Bi,j

The statements of Proposition 6.1 follow from Propositions 6.2 and 6.3 stated below. Now we only note that dim R(k,m)  dim R(k−1,m) + dim R(k,m−1) (since R(k,m) = x1 R(k−1,m) + x2 R(k,m−1) ). However a direct computation of the initial components shows that dim R(k,m) for k < m does not depend on m. By virtue of the symmetry, this reasoning allows us to formulate first the statements of Proposition 6.1 as a conjecture. (k,m)

Proposition 6.2. The elements gi following equalities hold:

defined in Proposition 6.1 belong to R(k,m) if and only if the

k,m−1 + Ak,m−2 = 0, Ak,m ij − 2Aij ij k − 2Akij + Bij =0 Ak,k+1 ij

j = 1, . . . , k + 1; 1  k  m − 2,

(m = k + 1),

k+1  k−1,m−2 (−Ak,m + Ak,m−1 Ak−1,m−1 ) = 0, ij Ajs ij js

(A) (A )

j = 1, . . . , k + 1;

s = 1, . . . , k; 1  k  m − 1,

(B)

j=1 k+1  k−1 k k−1 (−Akij Bjt + Bij Ajt ) = 0, t = 1, . . . , k.

(B  )

j=1 (k,m)

If this condition is fulfilled, the elements gi

, i = 1, . . . , k + 2, form a basis in R(k,m) .

Proof. Let the first statement of the proposition hold for elements of degree less than n, and let, for k+1  k,m (k,m) (k−1,m) (k,m−1) = Aij x1 gj + εx2 gi , where ε = 0 or 1. Then, iterating k + m = n and k < m, gi j=1

this formula three times, we obtain (k,m)

gi

=

k  k+1  k−1 

(k−3,m)

k−1,m k−2,m 3 Ak,m Ast x1 gt ij Ajs

j=1 s=1 t=1

 k  k+1  k,m k−1,m 2 k,m k−1,m−1 k,m−1 k−1,m−1 2 + (Aij Ajs x1 x2 + Aij Ajs x1 x2 x1 + εAij Ajs x2 x1 ) gs(k−2,m−1) s=1

j=1

+

k+1  (k−1,m−2) (k,m−3) k,m−1 2 (Ak,m x2 x1 x2 + εAk,m−2 x22 x1 )gj + εx32 gi . ij x1 x2 + εAij ij j=1

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Let us find P1 (gi (k,m)

P1 (gi

)=

7

). We have

 k  k+1  k−1,m k,m k−1,m−1 k,m−1 k−1,m−1 (Ak,m A − 2A A + εA A ) v12 v2 gs(k−2,m−1) ij js ij js ij js s=1

j=1

+

k+1 

(k−1,m−2)

k,m−1 (Ak,m + εAk,m−2 )v1 v22 gj ij − 2εAij ij

.

j=1 (k,m)

The equality P1 (gi k+1 

) = 0, for each i = 1, . . . , k + 2, is equivalent to the two systems

k−1,m k−1,m−1 (Ak,m − 2Ak,m + εAk,m−1 Ak−1,m−1 ) = 0, ij Ajs ij Ajs ij js

s = 1, . . . , k;

(Bε )

j=1 k,m−1 + εAk,m−2 = 0, Ak,m ij − 2εAij ij

(Aε )

j = 1, . . . , k + 1.

For ε = 1, (Aε ) coincides with (A) and (Bε ) can be replaced by (B) (with the use of (A)): k+1  k−1,m−2 (−Ak,m + Ak,m−1 Ak−1,m−1 ) = 0, s = 1, . . . , k, which proves the first statement of of the ij Ajs ij js j=1

proposition for k < m − 2. In fact, each element f (k,m) ∈ F (k,m) is uniquely represented in the form f (k,m) = x1 f (k−1,m) + x2 f (k,m−1) . If, in this case, f (k−1,m) , f (k,m−1) ∈ R, then in order for f (k,m) to belong to R, it suffices that the equality P1 (f (k,m) ) = 0 hold (since Pt is a linear function and Pt (xi g) = xi Pt−1 (g)). (k,m)

This, in particular, implies that dim R(k,m)  dim R(k,m−1) (the elements gi

, i=1, . . . , k + 2, are

(k,m−1) , i = 1, . . . , k + 2, is a basis). Therefore, dim R(k,m) can be greater linearly independent since gi (k,m−1) (k,m) only if R contains nonzero elements from x1 R(k−1,m) . However, equalities (A ), than dim R (k,m) , which proves the for ε = 0, give Ak,m ij = 0, j = 1, . . . , k + 1, i.e., there are no such elements in R

second statement of the proposition for k < m − 2.

The case m = k + 2 is no different from that just considered since if, in the last iteration (when (k,m−3) is an arbitrary element of R(k,m−3) , then all the elements of R(k,k) are used), we assume that gi same it will occur in the expression of the initial element with the factor x32 on the left and will vanish in the process of finding of P1 . The cases m = k + 1 and m = k are proved in the same manner. Proposition 6.3. 1) The space R(−1,m) is generated by the element g(−1,m) = x2 g(−1,m−1) = xm 2 ξ1 . 2) The space R(k,m) , for 0  k < m, is generated by the elements (k,m)

(k−1,m)

+ (m − k + 1)x1 gi

(k−1,m)

− (m − k + 1)x1 gi0 +1

= (m − k)x1 gi−1

gi

(k,m)

gi0 +1 = (m − k)x1 gi0 (k,m)

gi

(k−1,m)

= −(m − k)x1 gi−1

(k−1,m)

+ x2 gi

(k−1,m)

+ x2 gi0 +1

(k−1,m)

− (m − k + 1)x1 gi

(k,m−1)

when 1  i  i0 ;

(k,m−1)

when i0 < k + 1;

(k,m−1)

+ x2 gi

when i0 + 1 < i < k + 2;

and, for m = k, by the elements (k,k)

gi

(k−1,k)

= x1 gi

(k,k−1)

− x2 gi−1

(k−1,m)

where i0 = [ k+3 2 ] and g0

, 1  i  i0 ; (−1,m)

= 0, g1

(k,k)

gi

(k−1,k)

= −x1 gi+1 (−1,m)

= g(−1,m) , g2

(k−1,m)

= 0, gk+2

3) The space R(k,m) , for k > m, is determined by the symmetry ρ. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 10 2008

(k,k−1)

+ x2 gi

= 0.

, i0 < i  k + 2,

8

ERMOLAYEV

To prove these statements, it suffices to verify that the coefficients of the indicated formulas satisfy relations (A), (A ), (B), (B  ).  7. The algebra L(2; 3) = L/I. Here L is again the free Lie algebra generated by L−1 = ξ1 , ξ2  over an arbitrary field K, and I is the ideal in L generated by [ξ1 , ξ2 , ξ1 , ξ1 ], [ξ1 , ξ2 , ξ1 , ξ2 ], [ξ1 , ξ2 , ξ2 , ξ2 ]. The initial components of the space F are as follows: (0,0)

F0 = g(−1,1) = x2 ξ1 , g1

(0,0)

= x1 ξ1 , g2

= x2 ξ2 , g(1,−1) = x1 ξ2 ,

F1 = F (0,1) ⊕ F (1,0) = g(0,1) = 3x1 x2 ξ1 − x2 x1 ξ1 + 2x22 ξ2 , g(1,0) = 2x21 ξ1 − x1 x2 ξ2 + 3x2 x1 ξ2 , F2 = F (−1,3) ⊕ F (0,2) ⊕ F (1,1) ⊕ F (2,0) ⊕ F (3,−1) , where F (−1,3) = x32 ξ1 ,

F (0,2) = 3x1 x22 ξ1 + x2 x1 x2 ξ1 , x2 x1 x2 ξ1 + x22 x1 ξ1 , x22 x1 ξ1 − 3x32 ξ2 ,

F (1,1) = 2x21 x2 ξ1 + x1 x2 x1 ξ1 , x2 x21 ξ1 , x1 x2 x1 ξ1 − x2 x1 x2 ξ2 , x1 x22 ξ2 , x2 x1 x2 ξ2 + 2x22 x1 ξ2 , F (2,0) = 3x31 ξ1 − x21 x2 ξ2 , x21 x2 ξ2 + x1 x2 x1 ξ2 , x1 x2 x1 ξ2 + 3x2 x21 ξ2 ,

F (3,−1) = x31 ξ2 .

Finally, W = W (0,4) ⊕ W (1,3) ⊕ W (2,2) ⊕ W (3,1) ⊕ W (4,0) , where W (0,4) = x42 ,

W (1,3) = 3x1 x32 + x2 x1 x22 , x2 x1 x22 + x22 x1 x2 , x22 x1 x2 + 3x32 x1 ,

W (2,2) = 2x21 x22 + (x1 x2 )2 , x1 x22 x1 , (x1 x2 )2 + (x2 x1 )2 , x2 x21 x2 , (x2 x1 )2 + 2x22 x21 , W (3,1) = 3x31 x2 + x21 x2 x1 , x21 x2 x1 + x1 x2 x21 , x1 x2 x21 + 3x2 x31 ,

W (4,0) = x41 .

The sequence P is defined here by the mappings π0 = 0, π1 : F1 → V1 = v22 ∂1 (−1, 2); v1 v2 ∂1 , (0, 1); v12 ∂1 , v1 v2 ∂2 (1, 0); v12 ∂2 (2, −1)K (after the elements, we indicate their compositions) and is given by the formulas

v22 ∂2

π1 (x22 ξ1 ) = v22 ∂1 , π1 (x1 x2 ξ1 ) = v1 v2 ∂1 , π1 (x2 x1 ξ1 ) = 3v1 v2 ∂1 + 2v22 ∂2 , π1 (x22 ξ2 ) = v22 ∂2 , π1 (x21 ξ1 ) = v12 ∂1 , π1 (x1 x2 ξ2 ) = 2v12 ∂1 + 3v1 v2 ∂2 , π1 (x2 x1 ξ2 ) = v1 v2 ∂2 , π1 (x21 ξ2 ) = v12 ∂2 . The mapping π2 : F2 → V2 = v12 , v1 v2 , v22 K is defined by the formulas π2 (x32 ξ1 ) = 0, π2 (x1 x22 ξ1 ) = v22 , π2 (x2 x1 x2 ξ1 ) = −3v22 , π2 (x22 x1 ξ1 ) = 3v22 , π2 (x32 ξ2 ) = v22 , π2 (x21 x2 ξ1 ) = v1 v2 , π2 (x1 x2 x1 ξ1 ) = −2v1 v2 , π2 (x1 x22 ξ2 ) = 0, π2 (x2 x21 ξ1 ) = 0, π2 (x2 x1 x2 ξ2 ) = −2v1 v2 , π2 (x22 x1 ξ2 ) = v1 v2 , π2 (x31 ξ1 ) = v12 , π2 (x21 x2 ξ2 ) = 3v12 , π2 (x1 x2 x1 ξ2 ) = −3v12 , π2 (x2 x21 ξ2 ) = v12 , π2 (x31 ξ2 ) = 0. Finally, π : T4 → V = v14 , v13 v2 , v12 v22 , v1 v23 , v24 K , and it is given by the formulas π(x42 ) = 0, π(x1 x32 ) = v1 v23 , π(x2 x1 x22 ) = −3v1 v23 , π(x22 x1 x2 ) = 3v1 v23 , π(x32 x1 ) = −v1 v23 , π(x21 x22 ) = v12 v22 , π((x1 x2 )2 ) = −2v12 v22 , π(x1 x22 x1 ) = 0, π(x2 x21 x2 ) = 0, π((x2 x1 )2 ) = 2v12 v22 , π(x22 x21 ) = −v12 v22 , π(x31 x2 ) = v13 v2 , π(x21 x2 x1 ) = −3v13 v2 , π(x1 x2 x21 ) = 3v13 v2 , π(x2 x31 ) = −v13 v2 , π(x41 ) = 0. Passing to constructing of the Cartan extension L(2; 3), note that R0 = F0 , R1 = F1 , and the following lemma holds. Lemma 7.1. For m  2, we have R(−1,m) = 0 for a field K of any characteristic p and R(0,m) = 0 for p = 2, 5 (by virtue of the symmetry, for k  2, we have R(k,−1) = 0 for any p and R(k,0) = 0 for p = 2, 5). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52

No. 10 2008

A REALIZATION OF CARTAN EXTENSIONS

9

m Proof. Every element f ∈ F (−1,m) is proportional to xm 2 ξ1 . However, if m  2, then Pm−1 (λx2 ξ1 ) = λxm−2 π1 (x22 ξ1 ) = λxm−2 v22 ∂1 = 0 (λ ∈ K) only if λ = 0, i.e., R(−1,m) = 0. Then, by virtue of what 2 2 has been proved for m  2, we have R(0,m) ⊆ x2 R(0,m−1) , whence, using an induction on m, we g(0,1) and Pm−1 (f ) = λxm−2 π2 (3x2 x1 x2 ξ1 − x22 x1 ξ1 + 2x32 ξ2 ) = −10λxm−2 v22 , i.e., obtain f = λxm−1 2 2 2 R(0,m) = 0 if p = 2, 5.

Lemma 7.2. The following equalities hold: R(0,2) = 0 if p = 2, 5 and R(0,2) = x2 g(0,1)  if p = 2, 5; R(1,1) = g(1,1) = x1 g(0,1) − x2 g(1,0)  if p = 5 and (1,1)

R(1,1) = g1

(1,1)

= x1 g(0,1) , g2

= x2 g(1,0)  if p = 5;

R(2,0) = 0 if p = 2, 5 and R(2,0) = x1 g(1,0)  if p = 2, 5. Proof. By virtue of Lemma 7.1, it remains to find R(1,1) . Let f ∈ R(1,1) . Then f = Ax1 g(0,1) + Bx2 g(1,0) = A(3x21 x2 ξ1 − x1 x2 x1 ξ1 + 2x1 x22 ξ2 ) + B(2x2 x21 ξ1 − x2 x1 x2 ξ2 + 3x22 x1 ξ2 ) and P1 (f ) = A(3π2 (x21 x2 ξ1 ) − π2 (x1 x2 x1 ξ1 ) + 2π2 (x1 x22 ξ2 )) + B(2π2 (x2 x21 ξ1 ) − π2 (x2 x1 x2 ξ2 ) + 3π2 (x22 x1 ξ2 )) = 5(A + B)v1 v2 . Hence it follows that P1 (f ) = 0 if p = 5 and f ∈ R(1,1) for any A, B ∈ K. If p = 5, then f ∈ R(1,1) holds only if A + B = 0. Lemma 7.3. For p > 5, we have R3 = R(1,2) ⊕ R(2,1) , where R(1,2) = g(1,2) = x2 g(1,1)  and R(2,1) = g(2,1) = x1 g(1,1) . Proof. Let f ∈ R(1,2) . Then f = x2 g(1,1) = x2 (x1 (3x1 x2 ξ1 − x2 x1 ξ1 + 2x22 ξ2 ) − x2 (2x21 ξ1 − x1 x2 ξ2 + 3x2 x1 ξ2 )) (since R(0,2) = 0 and R(1,1) , for p = 2, 5, is one-dimensional). We have P1 (f ) = 3π(x2 x21 x2 )ξ1 −π((x2 x1 )2 )ξ1 +2π(x2 x1 x22 )ξ2 −2π(x22 x21 )ξ1 + π(x22 x1 x2 )ξ2 − 3π(x32 x1 )ξ2 = −2v12 v22 ξ1 + 2v12 v22 ξ1 = 0, i.e., f is proportional to x2 g(1,1) . By virtue of the symmetry, we have the second equality. Proposition 7.1. The Cartan extension of the Lie algebra L(2, 3) for p > 5 terminates after the component of degree 3 and is isomorphic to the classical Lie algebra G2 . Proof. We have R4 = R(1,3) ⊕ R(2,2) ⊕ R(3,1) . If f ∈ R(1,3) , then f = x2 g(1,2) = x2 (3x2 x21 x2 ξ1 − and P1 (g(1,3) ) = 3π(x22 x21 )x2 ξ1 − (x2 x1 )2 ξ1 + 2x2 x1 x22 ξ2 − 2x22 x21 ξ1 + x22 x1 x2 ξ2 − 3x32 x1 ξ2 ) 2 2 3 3 for π(x2 x1 x2 )x1 ξ1 + 2π(x2 x1 x2 )x2 ξ2 − 2π(x2 x1 )x1 ξ1 + π(x2 x1 )x2 ξ2 − 3π(x42 )x1 ξ2 = −3v12 v22 = 0 p = 3. Hence, for p = 2, 3, 5, we have R(1,3) = 0. By virtue of the symmetry, we also have R(3,1) = 0 for p = 2, 3, 5. If f ∈ R(2,2) , then, for some A, B ∈ K, we have f = Ax1 g(1,2) + Bx2 g(2,1) = A(3x2 x21 x2 ξ1 − (x2 x1 )2 ξ1 + 2x2 x1 x22 ξ2 − 2x22 x21 ξ1 + x22 x1 x2 ξ2 − 3x32 x1 ξ2 ) + Bx2 (3x31 x2 ξ1 − x21 x2 x1 ξ1 + 2x21 x22 ξ2 − 2x1 x2 x21 ξ1 + (x1 x2 )2 ξ2 − 3x1 x22 x1 ξ2 ) and P1 (f ) = A(3π(x1 x2 x21 )x2 ξ1 − π((x1 x2 )2 )x1 ξ1 + 2π(x1 x2 )2 )x2 ξ2 − 2π(x1 x22 x1 )x1 ξ1 + π(x1 x22 x1 )x2 ξ2 − 3π(x1 x32 )x1 ξ2 + B(3π(x2 x31 )x2 ξ1 − π(x2 x21 x2 )x1 ξ1 + 2π(x2 x21 x2 )x2 ξ2 − 2π((x2 x1 )2 )x1 ξ1 + π((x2 x1 )2 )x2 ξ2 RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 10 2008

10

ERMOLAYEV

− 3π(x2 x1 x22 )x1 ξ2 ) = (2A − 4B)v12 v22 x1 ξ1 + (−4A + 2B)v12 v22 x2 ξ2 . Consequently, P1 (f ) = 0 is equivalent to the system 2A − 4B = 0, −4A + 2B = 0. Since the determinant of this system equals −12, it follows that, for p = 2, 3, the system has only zero solution, i.e., R(2,2) = 0. 8. The case L(2; 3) for p = 5. Proposition 8.1. The Cartan extension of the algebra L(2, 3) is given by the components R3m = 2m+1 2m+1 2m+2   (k,3m−k+1)  (k,3m−k+2) R(k,3m−k) , R3m+1 = R , R3m+2 = R , where k=m−1

k=m

R(m−1,2m+1) = R(k,3m−k) = R(2m+1,m−1) = R(k,3m−k+1) = R(m,2m+2) = R(k,3m−k+2) = R(2m+2,m) =

k=m (m−1,2m+1) g1 , (k,3m−k) (k,3m−k) g1 , g2 , k = m, . . . , 2m, (2m+1,m−1) g2 , (k,3m−k+1) g1 , k = m, . . . , 2m + 1, (m,2m+2) g1 , (k,3m−k+2) (k,3m−k+2) g1 , g2 , k = m + 1, . . . , 2m (2m+2,m) g2 .

(8.1)

+ 1,

In addition we have (k,3m−k)

= x1 g1

(k,3m−k)

= x1 (2g1

g1 g2

(k,3m−k+1)

g1

(k,3m−k+2) g1 (k,3m−k+2) g2

(k−1,3m−k)

(k,3m−k−1)

+ x2 (−g1

(k−1,3m−k)

(k−1,3m−k+1)

= x1 g1 = =

(k−1,3m−k)

− g2

(k,3m−k−1)

) + x2 g2

(k,3m−k)

+ x2 g2

(k,3m−k−1)

+ 2g2

),

, (8.2)

,

(k,3m−k+1) x2 g1 , (k−1,3m−k+2) −x1 g1 .

All the other components are zero. Proof. We proceed by induction on the degree. The components of small degrees are found immediately. Assume that, for degrees < n, all the statements of Proposition 8.1 and Eqs. (8.1) and (8.2) hold. 1) Let n = 3m and g(k,3m−k) ∈ R(k,3m−k) . Since, for degrees < n, relations (8.1) are assumed to be true, for some A, B, C, D ∈ K, we have (k−1,3m−k)

g(k,3m−k) = x1 (Ag1

(k−1,3m−k)

+ Bg2

(k,3m−k−1)

) + x2 (Cg1

(k,3m−k−1)

+ Dg2

).

Let us transform the right-hand side of this equality using (8.2) three times (k−3,3m−k−1)

g(k,3m−k) = A((x1 x2 x21 g1

(k−2,3m−k−2)

+ (x1 x22 x1 (2g1

(k−2,3m−k−2)

+ (x1 x2 )2 (−g1

(k−2,3m−k−2)

− g2

(k−2,3m−k−2)

+ 2g2

(k−1,3m−k−3)

) + x1 x32 g2

))

))

(k−4,3m−k) (k−3,3m−k−1) (k−3,3m−k−1) + x31 x2 (−g1 + 2g2 )) − B((x41 g1 (k−3,3m−k−1) (k−3,3m−k−1) (k−2,3m−k−2) − g2 ) + x21 x22 g2 )) + (x21 x2 x1 (2g1 (k−2,3m−k−2) (k−1,3m−k−3) (k−1,3m−k−3) + x22 x1 x2 (−g1 + 2g2 )) + C((x22 x21 g1 (k−1,3m−k−3) (k−1,3m−k−3) 3 4 (k,3m−k−4) − g2 ) + x2 g2 )) + (x2 x1 (2g1 (k−2,3m−k−2) (k−2,3m−k−2) 3 (k−3,3m−k−1) 2 + x2 x1 x2 (−g1 + 2g2 )) − D((x2 x1 g1 (k−2,3m−k−2) (k−2,3m−k−2) (k−1,3m−k−3) − g2 ) + x2 x1 x22 g2 )). + ((x2 x1 )2 (2g1

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52

No. 10 2008

A REALIZATION OF CARTAN EXTENSIONS

11

The formula obtained allows us to find (k−3,3m−k−1)

P1 (g(k,3m−k) ) = (3A + 7B + D)v13 v2 g1 (k−1,3m−k−3)

− 5Cv1 v23 g1

(k−2,3m−k−2)

+ (2A − C − 4D)v12 v22 g1

(k−3,3m−k−1)

− 5Bv13 v2 g2

(k−2,3m−k−2)

+ (−4A − B + 2D)v12 v22 g2

(k−1,3m−k−3)

+ (A + 7C + 3D)v1 v23 g2

.

Hence P1 (g(k,3m−k) ) = 0 is equivalent to the system 2A − C + 2D = 0, 2A − B + 2C = 0, which has two independent solutions A = 1, B = 0, C = −1, D = 2 and A = 2, B = −1, C = 0, D = 1 verifying (k−1,3m−k) (k−1,3m−k) + Bg2 )+ formulas (8.1) and (8.2) for n = 3m. In addition, g(k,3m−k) = x1 (Ag1 (k,3m−k−1)

(k,3m−k−1)

+ Cg2 ) = 0 for k = −1, 0, . . . , m − 2; 2m + 2, . . . , 3m + 1 since, by the x2 (Cg1 induction assumption, all the corresponding elements in the right-hand side are zero. What is more, (m−1,2m) and g(2m+1,m−1) = for k = m − 1 and k = 2m + 1, we have, respectively, g(m−1,2m+1) = Cx2 g1 (2m,m−1)

Bx1 g2

R(2m+1,m−1)

.

Therefore,

=

(2m+1,m−1) g2

the =

subspaces

(2m,m−1) x1 g2 

(m−1,2m+1)

R(m−1,2m+1) = g1

(m−1,2m)

= −x2 g1



and

are one-dimensional.

In the cases n = 3m + 1 and n = 3m + 2, the proofs are similar. By this reason, we omit the corresponding calculations. For the given algebra R, we give the structure formulas, which are determined by the general multiplication formulas for R ([2], (8)). Of two symmetric formulas, we write down only one. As before, by the symmetry we understand the automorphism ρ of R which extends the cor(k,3m−k) )= respondence ξ1 ↔ ξ2 , x1 ↔ x2 . Basis elements of R are chosen in such a way that ρ(g1 (k,3m−k)

(k,3m−k+1)

(−1)m+1 g2

(3m−k+1,k)

(k,3m−k+2)

(3m−k+2,k)

, ρ(g1 ) = (−1)m+1 g1 , ρ(g1 ) = (−1)m g2 

k+l−m−n+1  2(m+n)−k−l (k+l,3(m+n)−k−l) (k,3m−k) (l,3n−l) , g1 ] = k+l−m−n+1 , − g1 [g1 k−m k−m−1 2m−k (k,3m−k)

[g1

=

,

(l,3n−l)

, g2

] k+l−m−n+1 2(m+n)−k−l

k−m+1

2m−k−1

(k+l,3(m+n)−k−l)

g1

−

k+l−m−n 2(m+n)−k−l+1

k−m+1

2m−k

(k+l,3(m+n)−k−l)

g2



 2m+2n−k−l+1 2m+2n−k−l+1  (k+l,3(m+n)−k−l+1) (k,3m−k) (l,3n−l+1) , g1 ] = k+l−m−n , 2 − g1 [g2 k−m 2m−k 2m−k+1

,



2m+2n−k−l+1 (k+l,3(m+n)−k−l+2) ] = − k+l−m−n+1 , g1 k−m+1 2m−k



(k,3m−k) (l,3n−l+2) 2m+2n−k−l+1 (k+l,3m+3n−k−l+2) , g2 ] = k+l−m−n g1 [g1 k−m+1 2m−k−1  

k+l−m−n−1 2m+2n−k−l+2 (k+l,3m+3n−k−l+2) − k+l−m−n , + g2 k−m+1 k−m 2m−k

 2m+2n−k−l+1 2m+2n−k−l+1  (k+l,3m+3n−k−l+2) (k,3m−k+1) (l,3n−l+1) , g1 ] = −2 k+l−m−n − g1 [g1 k−m 2m−k 2m−k+1  

k+l−m−n−1 2m+2n−k−l+2 (k+l,3m+3n−k−l+2) +2 k+l−m−n−1 , − g2 k−m−1 k−m 2m−k+1 (k,3m−k)

[g1

(l,3n−l+2)

, g1

(k,3m−k+1)

(l,3n−l+2)

]=−

k+l−m−n 2m+2n−k−l+2

(k+l,3m+3n−k−l+3)

, g1

2m+2n−k−l+3 (k+l,3m+3n−k−l+4) (k,3m−k+2) (l,3n−l+2) , g2 ] = − k+l−m−n−1 , g1 [g1 k−m 2m−k+1 [g1

(k,3m−k+2)

[g1

, g1

(l,3n−l+2)

, g1

k−m

2m−k+1

] = 0.

The formulas can also be verified by induction using the equality ξ[u, v] = [ξu, v] + [u, ξv] (ξ ∈ R1 , u, v ∈ R+ ). This algebra is isomorphic to the special Melikyan algebra [3, 4]. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 10 2008

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ERMOLAYEV

9. The case L(2; 3) for p = 3. ∞ 

Proposition 9.1. If L = L(2; 3) and p = 3, then R(L) =

n=−3 (0,0)

g1

(0,0)

= x1 ξ1 , g2

(1,−1)

= x2 ξ2 , g1

(−1,1)

Rn , where R0 = g1

= x2 ξ1 ,

= x1 ξ2 , and, for n  1, we have

R2k = g(k,k)  (k  1),

R2k+1 = g(k+1,k) , g(k,k+1)  (k  0);

the elements g(k,k) , g(k+1,k) , g(k,k+1) are defined by the equalities g(k,k) = −x1 g(k−1,k) + x2 g(k,k−1) ,

g(k,k+1) = x2 g(k,k) ,

g(k+1,k) = x1 g(k,k) .

In particular, g(0,1) = x2 (−x1 ξ1 + 2x2 ξ2 ), g(1,0) = x1 (2x1 ξ1 − x2 ξ2 ), and R(i,2k−i) = 0 for i = −1, . . . , k − 1; k + 1, . . . , 2k + 1; R(i,2k+1−i) = 0 for i = −1, . . . , k − 1; k + 2, . . . , 2k + 2. The structure formulas are as follows: (i+j+1)! (i+1)!(j+1)! (j

− i)g(i+j,i+j) ;

i+j+1,i+j+1 ; [gi+1,i , gj+1,j ] = 0, [gi,i+1 , gj,j+1 ] = 0, [gi+1,i , gj,j+1 ] = − i+j+2 i+1 g



(i+j+1,i+j) , [g (i,i) , g (j,j+1) ] = i+j+2 g (i+j,i+j+1) . [g(i,i) , g(j+1,j) ] = i+j+2 i+1 g i+1 [g(i,i) , g(j,j) ] =

10. The case L(2; 3) for p = 2. ∞ 

Proposition 10.1. If L = L(2; 3) and p = 2, then R(L) =

Rn , where

n=−3

R3k−1 = g(i,3k−i−1) , i = k − 1, . . . , 2k; (i,3k−i)

R3k = g1

(i,3k−i)

, i = k, . . . , 2k + 1; g2

R3k+1 = g(i,3k−i+1) , i = k, . . . , 2k + 1,

, i = k − 1, . . . , 2k;

(10.1)

k = 0, 1, 2, . . . ; j = 1, . . . , i + 2,

and the elements g(i,n−i) are determined inductively by the formulas g(i,3k−i−1) = x1 g(i−1,3k−i−1) + x2 g(i,3k−i−2) , i = k − 1, . . . , 2k; (i,3k−i)

= x1 g(i−1,3k−i) , i = k, . . . , 2k + 1;

(i,3k−i)

= x2 g(i,3k−i−1) , i = k − 1, . . . , 2k;

g1 g2

(i−1,3k−i+1)

g(i,3k−i+1) = x1 g2

In addition we assume that Rn =

n+1 

(i,3k−i)

+ x2 g1

, i = k, . . . , 2k + 1.

R(i,n−i) , where the components are generated by the

i=−1

corresponding basis elements gi,n−i which are assumed to be equal to zero outside the limits indicated in (10.1). A direct verification shows that the algebra L(2; 4) has F1 = 0, i.e., the Cartan extension terminates after the zero component. REFERENCES 1. I. L. Kantor, “Graded Lie Algebras,” Trudy semin. po vektorn. i tenzorn. analizu (Moscow University, 1970), No. XV, 227–266. 2. Yu. B. Ermolayev, “Universal Cartan Extension,” Izv. Vyssh. Uchebn. Zaved. Mat., No. 11, 22–32 (1997) [Russian Mathematics (Iz. VUZ) 41 (11), 20–30 (1997)]. 3. I. M. Kuznetsov, “The Melikyan Algebras as Lie Algebras of the Type G2 ,” Communications in Algebra 19 (4), 1281–1312 (1991). 4. G. M. Melikyan, “On Simple Lie Algebras of Characteristic 5,” Usp. Mat. Nauk 35 (1), 203–204 (1980).

Translated by V. V. Shurygin RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52

No. 10 2008