Examples of self-iterating Lie algebras

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1 Journal of Algebra ) Examples of self-iterating Lie algebras V.M. Petrogradsky Faculty of Mathematics, Ulyanovsk State University, Lev Tolstoy 42, Ulyanovsk, Russia Received 28 June 2005 Communicated by Efim Zelmanov Abstract We construct examples of self-iterating Lie algebras similar to the Grigorchuk group. In case of characteristic 2, we construct a two-generated restricted Lie algebra of polynomial growth that has a nil-p-mapping Published by Elsevier Inc. 1. Fibonacci Lie algebra Let K be the ground field. Denote I ={0, 1, 2,...}. Consider the formal power series ring R = K[[T I ]] = K[[t i i = 0, 1, 2,...]]. Denote by τ : R R the shift given by τt i ) = t i+1 for i I.Let i = t i, i I, denote the partial derivatives of this ring. Denote by v t the action of v Der R onto t R. We define the following two derivations of R: v 1 = 1 + t t t t t ) )))), v 2 = 2 + t t t t ) ))). Remark that we can write these derivations recursively: v 1 = 1 + t 0 τv 1 ), v 2 = τv 1 ). The research was partially supported by Grant RFBR addresses: petrogradsky@rambler.ru, petrogradsky@hotbox.ru /$ see front matter 2005 Published by Elsevier Inc. doi: /j.jalgebra

2 882 V.M. Petrogradsky / Journal of Algebra ) Let L = algv 1,v 2 ) be the Lie subalgebra of Der R generated by v 1 and v 2. Similarly, define v i = τ i 1 v 1 ) = i + t i 1 i+1 + t i i+2 + t i+1 i+3 + ) )), i = 1, 2,... 1) We also can write Lemma 1. The following commutation relations hold: 1) [v i,v i+1 ]=v i+2 for i = 1, 2,...; 2) [v i,v j ]=t i 1 t i t j 3 v j+1 for all i<j; 3) [v i,v i+2 ]=t i 1 v i+3 for i = 1, 2,... Proof. We have v i = i + t i 1 v i+1, i = 1, 2,... 2) [v 1,v 2 ]= [ 1 + t 0 τv 1 ), τv 1 ) ] = [ 1,τv 1 ) ] = [ 1, 2 + t 1 τ 2 v 1 ) ] = τ 2 v 1 ) = v 3. Consider the general case. Let i<j, then [v i,v j ]= [ i + t i 1 i+1 + t i +tj 3 j 1 + t j 2 v j ) )) ],v j = [ i + t i 1 i+1 + t i +t j 3 j 1 )... ) ],v j = [ i + t i 1 i+1 + t i +t j 3 j 1 )... ) ], j + t j 1 v j+1 = [ i + t i 1 i+1 + t i +t j 3 j 1 )... ) ],t j 1 v j+1 = ti 1 t i t j 3 v j+1. The third claim is a partial case of the second. We consider the first relation as a partial case as well. Lemma 2. Consider the Lie subalgebra L = algv 1,v 2 ) Der R. Then L is Z Z-graded by means of the weight function wt v n = wt t n = λ n, n= 1, 2,..., λ= Proof. Let us introduce a grading on L such that v i are homogeneous. Suppose that we have a weight function wt v i = a i R, where i = 1, 2... Since it is natural to have homogeneous summands in 2), we assume that a i = wt v i = wt i = wt t i 1 + wt v i+1 = a i 1 + a i+1. Hence, we get the Fibonacci relation a i+1 = a i + a i 1. So, we set a i = λ i, i = 0, 1, 2,...

3 V.M. Petrogradsky / Journal of Algebra ) Remark that the weights wt v 1 = λ and wt v 2 = λ 2 are linearly independent over Z. Hence, L = a,b 0 L a,b, where L a,b is spanned by products that contain a factors v 1 and b factors v 2. We recall the notion of growth. Let A be an associative or Lie) algebra generated by a finite set X. Denote by A X,n) the subspace of A spanned by all monomials in X of length not exceeding n.ifa is a restricted Lie algebra, then we define [7] A X,n) = [x 1,...,x s ] pk x i X, sp k n K. In either situation, one considers the growth function defined by γ A n) = γ A X, n) = dim K A X,n). The growth function clearly depends on the choice of the generating set X. Furthermore, it is easy to see that exponential growth is the highest possible growth for Lie and associative algebras. The growth function γ A n) is compared with the polynomial functions n k, k R +, by computing the upper and lower Gelfand Kirillov dimensions [6], namely ln γ A n) GKdim A = lim n ln n, GKdim A = lim n ln γ A n) ln n. This setting assumes that all elements of X have the same weight equal to 1. In our situation, we have X ={v 1,v 2 }. We shall use a little bit different growth function. Namely, we set γ L n) = dim K y y L, wty n, n N. The standard arguments [6] prove that we can use this function to compute the Gelfand Kirillov dimensions. The growth of L will be studied somewhere else. We remark that L is a self-similar Lie algebra. Namely, consider subalgebras L i = algv i,v i+1 ), i = 1, 2,... Then, clearly, L i = L1 = L for all i = 2, 3,... On the other hand, we have the embedding L 1 K K[t 0 ] L 2, L 2 = L, where the semidirect product is defined via the action 1 v 2 = v 3 and 1 v j = 0forj 3. These properties resemble those of the Grigorchuk group, Gupta Sidki group, etc., [1 4]. 2. Fibonacci restricted Lie algebra, char K = 2 The goal of the paper is to consider a particular case. Now we suppose that char K = 2. Consider the truncated polynomial ring R = K[t i i = 0, 1, 2,...]/ t 2 i i = 0, 1, 2,... ).

4 884 V.M. Petrogradsky / Journal of Algebra ) Let v i Der R, i = 1, 2,..., be as above. Basic facts on restricted Lie algebras can be found in [5]. We denote by L = alg p v 1,v 2 ) Der R the restricted subalgebra generated by v 1,v 2, it will also be referred to as the Fibonacci restricted Lie algebra. Let H = algv 1,v 2 ) be the Lie subalgebra generated by brackets only. Remark that H is contained in the span of monomials H H = v 1,v 2,v 3,t α 0 0 tα 1 1 tα n 4 n 4 v n n 4, α i {0, 1} K. 3) Indeed, we apply Lemma 1 to check that the product of two monomials in the write hand side is expressed via these monomials. Let n<m, then [ α t 0 0 tα n 4 n 4 v n,t β 0 0 tβ m 4 m 4 v ] m 3 α m = t 0 0 tα n 4 n 4 tβ 0 0 tβ m 4 m 4 t i )v m+1 + t α 0 0 tα n 4 n 4 β j 0 m 4 i=n 1 i=0, i j t β i i ) v n t j )v m, where we use that v m = m + t m 1 m+1 + ) acts on t i s trivially because i n 4 <m. The action v n t j can be nontrivial only in the case n j m 4, in which case v n t j = n + t n 1 n+1 + +t j 2 j + ) )) t j = t n 1 t j 2. In all cases we obtain monomials of type 3). Hence, H Der R is a Lie subalgebra. Let A be an associative algebra over the field K and char K = 2, then a + b) 2 = a 2 + b 2 +[a,b], a,b A. 4) We get v 2 1 = 1 + t 0 v 2 ) 2 = t 0 [ 1,v 2 ]=t 0 [ 1, 2 + t 1 v 3 ]=t 0 v 3. We apply τ and obtain v 2 i = t i 1v i+2, i = 1, 2,... 5) Let H be the restricted subalgebra generated by H. It is sufficient to add pth powers of the basis of H [5], moreover only powers 5) are nonzero. These are linearly independent with 3) and we obtain L H = H t i 3 v i i = 3, 4,... K. 6) Lemma 3. Let char K = 2 and let L be the Fibonacci restricted Lie algebra. Then: 1) L has a polynomial growth. ln 2 2) GKdim L ln )/2)

5 V.M. Petrogradsky / Journal of Algebra ) Proof. Fix a number m. Consider a homogeneous element g H = algv 1,v 2 ) of weight not exceeding m. By remark above, g is a sum of monomials of type w = t α 0 0 tα 1 1 tα n 4 n 4 v n. Then n 4 m wtg) = wtw) = wtv n ) + i=0 n 4 α i wt t i = λ n i=0 n 4 α i λ i λ n >λ n λn 4 1 1/λ = λn 4 C, C = λ /λ We obtain λ n 4 C<m. Hence, n<n 0 = 4 + lnm/c)/ ln λ. The number of monomials 3) of weight not exceeding m is bounded by n n n=4 n x 3 dx 3 + 2n 0 2 ln ) m ln 2/ ln λ C 0 m ln 2/ ln λ. ln 2 C i= lnm/C)/ ln λ ln 2 To evaluate the growth of the whole of the restricted Lie algebra L we need also to take into account the squares 6), i.e. we count the number of elements t n 3 v n of weight not exceeding m. Wehavewtt n 3 + wt v n = λ n λ n 3 = λ n 3 λ 3 1) m. The number of such elements is bounded by C 1 ln m and we are done. Lemma 4. Let char K = 2 and let L be the Fibonacci restricted Lie algebra. Then L has a nil-p-mapping. Proof. Consider v L.Lets be the maximal number such that v s appear in the decomposition of v. From 6) and 3) we have s 1 v = g i t 0,...,t i 3 )v i + ht 0,...,t s 3 )v s, 7) i=1 where g i = g i t 0,...,t i 3 ) and h = ht 0,...,t s 3 ) are polynomials from R. We assume that h has zero constant term. Otherwise we take the number s + 1 and consider the decomposition v = +hv s+1, where h = 0.) We apply the p-mapping to 7) and use 4). Consider v i s with the highest value of i that might appear. The commutators yield at most s 1 [g i v i,g s 1 v s 1 ]=g i g s 1 t i 1 t s 4 v s + f j v j, f j R, 1 i s 2, [g i v i,hv s ]=hg i t i 1 t s 3 v s+1 + j=1 λ i s f j v j, f j R, 1 i s 1. j=1

6 886 V.M. Petrogradsky / Journal of Algebra ) In the second case h is a polynomial without constant term, thus the term with v s+1 goes to h of a relation similar to 7). Consider the squares. We have hv s ) 2 = h 2 v 2 s = 0 and the squares arising from the sum yield at most g s 1 v s 1 ) 2 = g 2 s 1 t s 2v s+1, this term also belongs to h. Thus, we obtain the same presentation as 7): We iterate the process v 2 = s g i t 0,...,t i 3 )v i + ht 0,...,t s 2 )v s+1. i=1 s+m 1 v 2m = i=1 g i t 0,...,t i 3 )v i + ht 0,...,t s+m 3 )v s+m. 8) The weight of any homogeneous monomial of v is at least λ. Hence, weights of monomials of v 2m are at least λ2 m. Since polynomials only reduce the weight, weights of monomials in 8) are at most wtv s+m ) = λ s+m.ifλ2 m >λ s+m, then v 2m = 0. Therefore, it is sufficient to take m>s 1) ln λ ln2/λ). Remark that ln λ ln2/λ) 2.27.) Remark that the nil-index of the p-mapping is unbounded. It is sufficient to consider the powers v 1 + v 2 + +v s ) 2m. References [1] R.I. Grigorchuk, On the Burnside problem for periodic groups, Funktsional. Anal. i Prilozhen. 14 1) 1980) [2] R.I. Grigorchuk, Just infinite branch groups, in: New Horizons in Pro-p Groups, in: Progr. Math., vol. 184, Birkhäuser, Boston, MA, 2000, pp [3] R.I. Grigorchuk, Branch groups, Mat. Zametki 67 6) 2000) in Russian); translation in: Math. Notes ) 2000) [4] N. Gupta, S. Sidki, On the Burnside problem for periodic groups, Math. Z ) 1983) [5] N. Jacobson, Lie Algebras, Interscience, New York, [6] G.R. Krause, T.H. Lenagan, Growth of Algebras and Gelfand Kirillov Dimension, Amer. Math. Soc., Providence, RI, [7] D.S. Passman, V.M. Petrogradsky, Polycyclic restricted Lie algebras, Comm. Algebra 29 9) 2001)