aa r X i v : . [ m a t h . R A ] N ov Rota—Baxter operators on a sum of fields
V. Gubarev
Abstract
We count the number of all Rota—Baxter operators on a finite direct sum A = F ⊕ F ⊕ . . . ⊕ F of fields and count all of them up to conjugation with anautomorphism. We also study Rota—Baxter operators on A corresponding to adecomposition of A into a direct vector space sum of two subalgebras. We showthat every algebra structure induced on A by a Rota—Baxter of nonzero weight isisomorphic to A . Keywords : Rota—Baxter operator, (un)labeled rooted tree, 2-coloring, subtreeacyclic digraph, transitive digraph.
Given an algebra A and a scalar λ ∈ F , where F is a ground field, a linear operator R : A → A is called a Rota—Baxter operator (RB-operator, for short) on A of weight λ ifthe following identity R ( x ) R ( y ) = R ( R ( x ) y + xR ( y ) + λxy ) (1)holds for any x, y ∈ A . The algebra A is called Rota—Baxter algebra (RB-algebra).G. Baxter in 1960 introduced the notion of Rota—Baxter operator [3] as natural gen-eralization of by parts integration formula. In 1960–1970s such operators were studied byG.-C.Rota [19], P. Cartier [10], J. Miller [17], F. Atkinson [2] and others.In 1980s, the deep connection between constant solutions of the classical Yang—Baxterequation from mathematical physics and RB-operators on a semisimple finite-dimensionalLie algebra was discovered by A. Belavin and V. Drinfel’d [4] and M. Semenov-Tyan-Shanskii [20].About different connections of Rota—Baxter operators with symmetric polynomials,quantum field renormalization, Loday algebras, shuffle algebra see in the monograph [14]written by L. Guo in 2012.In the paper, we study Rota—Baxter operators on a finite direct sum A = F ⊕ F ⊕ . . . ⊕ F of n copies of a field F . We continue investigations fulfilled by S. de Bragan¸ca in1975 [6] and by H. An and C. Bai in 2008 [1]. Since all RB-operators on A of weight zeroare trivial [12], i.e., equal to 0, we study only RB-operators on A of nonzero weight λ .In §2, we formulate some preliminaries about RB-operators, including splitting RB-operators which are projections on a subalgebra A parallel to another one A providedthe direct vector space sum decomposition A = A ˙+ A .In §3, we show that RB-operators on A of nonzero weight λ are in bijection with2-colored transitive subtree acyclic digraphs (subtree acyclic digraphs were defined byF. Harary et al. in 1992 [15]) or equivalently with labeled rooted trees on n + 1 vertices1ith 2-colored non-root vertices. For the last, we apply the result of R. Castelo andA. Siebes [11]. Thus, the number of all RB-operators on A of nonzero weight λ equals n ( n + 1) n − . With the help of the bijection, we show that splitting RB-operators on A of nonzero weight λ are in one-to-one correspondence with labeled rooted trees on n + 1 vertices with properly 2-colored non-root vertices. We also study the number of all RB-operators and all splitting RB-operators on A up to conjugation with an automorphismof A .In 2012, D. Burde et al. initiated to study so called post-Lie algebra structures [7]. Oneof the questions arisen in the area [7, 8, 9] is the following one: starting with a semisimpleLie algebra endowed RB-operator of weight 1 what kind of Lie algebras we will get underthe new Lie bracket [ R ( x ) , y ] + [ x, R ( y )] + [ x, y ] ? Such problems could be stated not onlyfor Lie algebras but also for associative or commutative ones. In §4, we show that everyalgebra structure induced on a finite direct sum A of fields by a Rota—Baxter operatorof nonzero weight is isomorphic to A itself. Trivial RB-operators of weight λ are zero operator and − λ id. Statement 1 [14]. Given an RB-operator R of weight λ ,a) the operator − R − λ id is an RB-operator of weight λ ,b) the operator λ − R is an RB-operator of weight 1, provided λ = 0 .Given an algebra A , let us define a map φ on the set of all RB-operators on A as φ ( R ) = − R − λ ( R ) id. It is clear that φ coincides with the identity map. Statement 2 [5]. Given an algebra A , an RB-operator R on A of weight λ , and ψ ∈ Aut( A ) , the operator R ( ψ ) = ψ − Rψ is an RB-operator on A of weight λ . Statement 3 [14]. Let an algebra A to split as a vector space into the direct sum oftwo subalgebras A and A . An operator R defined as R ( a + a ) = − λa , a ∈ A , a ∈ A , (2)is RB-operator on A of weight λ .Let us call an RB-operator from Statement 3 as splitting RB-operator with subalgebras A , A . Note that the set of all splitting RB-operators on an algebra A is in bijection withall decompositions A into a direct sum of two subalgebras A , A . Remark 1 . Given an algebra A , let R be a splitting RB-operator on A of weight λ with subalgebras A , A . Hence, φ ( R ) is an RB-operator of weight λ and φ ( R )( a + a ) = − λa , a ∈ A , a ∈ A . So φ ( R ) is splitting RB-operator with the same subalgebras A , A . Lemma 1 [5]. Let A be a unital algebra, R be an RB-operator on A of nonzeroweight λ . If R (1) ∈ F , then R is splitting.We call an RB-operator R satisfying the conditions of Lemma 1 as inner-splitting one.2 emma 2 [12]. Let A = A ⊕ A be an algebra, R be an RB-operator on A ofweight λ . Then the induced linear map P : A → A defined by the formula P ( x + x ) =Pr A ( R ( x )) , x ∈ A , x ∈ A , is an RB-operator on A of weight λ . Statement 4 [1, 6, 12]. Let A = F e ⊕ F e ⊕ . . . ⊕ F e n be a direct sum of copies ofa field F . A linear operator R ( e i ) = n P k =1 r ik e k , r ik ∈ F , is an RB-operator on A of weight 1if and only if the following conditions are satisfied:(SF1) r ii = 0 and r ik ∈ { , } or r ii = − and r ik ∈ { , − } for all k = i ;(SF2) if r ik = r ki = 0 for i = k , then r il r kl = 0 for all l
6∈ { i, k } ;(SF3) if r ik = 0 for i = k , then r ki = 0 and r kl = 0 or r il = r ik for all l
6∈ { i, k } . Example [2, 17]. The following operator is an RB-operator on A of weight 1: R ( e i ) = s X l = i +1 e l , ≤ i < s, R ( e s ) = 0 , R ( e i ) = − n X l = i e l , s + 1 ≤ i ≤ n. Remark 2 . It follows from (SF3) that r ik r ki = 0 for all i = k . In [1], the statementof Statement 4 was formulated with this equality and (SF1) but without (SF2) and thegeneral version of (SF3). That’s why the formulation in [1] seems to be not complete. Remark 3 . The sum of fields in Statement 4 can be infinite.In advance, we will identify an RB-operator on A with its matrix.Let us calculate the number of different RB-operators of nonzero weight λ on A = F e ⊕ F e ⊕ . . . ⊕ F e n . By Statement 1a, we may assume that λ = 1 . For n = 1 , we haveonly two RB-operators { , − id } . For n = 2 we have 12 cases [1]: (cid:18) (cid:19) , (cid:18) − − (cid:19) , (cid:18) (cid:19) , (cid:18) − − − (cid:19) , (cid:18) − (cid:19) , (cid:18) − (cid:19) , (cid:18) − − (cid:19) , (cid:18) − (cid:19) , (cid:18) − −
10 0 (cid:19) , (cid:18) − (cid:19) , (cid:18) (cid:19) , (cid:18) − − − (cid:19) . Here we identify an RB-operator with its matrix R ∈ M ( F ) by the rule R ( e i ) = n P k =1 r ik e k .For n = 3 , we have ·
16 = 128 variants [1]: a b
00 0 c , a b c + 1 2 c + 1 c , a b c + 1 0 c , a b
00 2 c + 1 c , a b + 1 b
00 0 c , a b + 1 b c + 1 2 c + 1 c , a a + 1 00 b
00 0 c , a a + 1 00 b c + 1 2 c + 1 c , a a + 1 2 a + 10 b
00 0 c , a a + 1 2 a + 10 b
00 2 c + 1 c , a b + 1 b b + 10 0 c , a b b + 10 0 c , a b + 1 b b + 12 c + 1 0 c , a a + 1 2 a + 10 b b + 10 0 c , a a + 10 b
00 0 c , a a + 12 b + 1 b b + 10 0 c for a = r , b = r , c = r ∈ { , − } .For n = 4 , computer can help to state that there are exactly 2000 RB-operators ofweight 1 on A . Thus, we get the first four terms from the sequence A097629 [18]. Theorem 1 . Let A = F e ⊕ F e ⊕ . . . ⊕ F e n be a direct sum of copies of a field F .The number of different RB-operators on A of nonzero weight λ equals n ( n + 1) n − . Proof . Let R be an RB-operator on A of weight λ . We may assume that λ = 1 . Wefollow the previous notations. We have n variants to choose the values of the elements r ii , i = 1 , . . . , n . The choice of any of them, say r ii , influences only on the possible signsof all elements r ik , k = i . So, we may put r ii = 0 for all i and fix the factor n for theanswer.Now, we want to construct a directed graph G on n vertices by any matrix R =( r ij ) ni,j =1 with chosen r ii = 0 . We consider the matrix R as the adjacency matrix ofa directed graph G . Let us interpretate conditions (SF2) and (SF3) in terms of digraphs.Firstly, we rewrite (SF3) as two conditions:(SF3a) if r ik = 0 for i = k , then r ki = 0 ;(SF3b) if r ik = 0 for i = k , then r kl = 0 or r il = r ik for all l
6∈ { i, k } .The condition (SF3a) says that if we have an edge between two vertices i = k , thenthe direction of such edge is well-defined, so, it is a correctness of getting a digraph by thematrix R . In graph theory, the condition (SF3b) is called transitivity , i.e., if have edges ( i, k ) ∈ E and ( k, l ) ∈ E , then we have an edge ( i, l ) ∈ E .Secondly, we read the condition (SF2) in terms of digraphs in such way: there are noin G induced subgraphs isomorphic to H with V ( H ) = { i, k, l } and E ( H ) = { ( i, l ) , ( k, l ) } (see Pict. 1). In [11] the subgraph H was called immorality , thus, a digraph withoutimmoralities is called moral digraph [16]. i l k Picture 1 . The forbidden induced subgraph H on three vertices { i, k, l } due to (SF2)We may reformulate our problem of counting the number N of different RB-operatorson A of nonzero weight λ in such way: What is the number of all transitive moraltransitive digraphs on n vertices? In terms of [11], the last is the same as the number of4ll moral TDAGs on n vertices, here TDAG is the abbreviation for Transitive DirectedAcyclic Graph (we are interested on transitive digraphs which are surely acyclic). In thegraph-theoretic context, moral DAGs are known as subtree acyclic digraphs [15]. Thus, N/ n = { moral TDAGs on n vertices } = { transitive subtree acyclic digraphs on n vertices } . (3)In [11], the authors constructed a bijection between the set of moral TDAGs on n ver-tices and the set of labeled rooted trees on n + 1 vertices as follows (see Pict. 2). Definethe function f ( i ) for a vertex i by induction. For a source i (i.e., such a vertex i thatthere are no edges ( j, i ) in a digraph), we put f ( i ) = 0 . For a not-source vertex j , we mayfind the unique source i such that there exists a directed path p from i to j . So, we define f ( j ) as the length of p . Now, we construct a labeled rooted tree T = ( U, F ) by a moralTDAG G = G ( V, E ) : U = V ∪ { } , F = { (0 , i ) | f ( i ) = 0 } ∪ { ( i, j ) | ( i, j ) ∈ E, f ( i ) = f ( j ) − } . Picture 2 . The corresponding graph G and tree T to the RB-operator R ( e ) = e + e + e , R ( e ) = − e − e − e , R ( e ) = − e , R ( e ) = 0 , R ( e ) = − e .Applying the above constructed correspondence, the number of moral TDAGs on n vertices equals ( n + 1) n − by the Cayley theorem, and so N = 2 n ( n + 1) n − . Theoremis proved.Below we will apply the easy fact that Aut( A ) ∼ = S n . It could be derived, e.g., fromthe Molin—Wedderburn—Artin theory, in particular from the uniqueness up to a rear-rangement of summands of decomposition of a semisimple finite-dimensional associativealgebra into a finite direct sum of simple ones. Corollary 1 [6]. Let A = F e ⊕ F e ⊕ . . . ⊕ F e n be a direct sum of copies of a field F and R be an RB-operator on A of nonzero weight 1. There exists an automorphism ψ of A such that the matrix of the operator R ( ψ ) in the basis e , . . . , e n is an upper-triangularmatrix with entries r ij ∈ { , ± } and r ii ∈ { , − } .5 roof . As we did in the proof of Theorem 1, we define by R a labeled rooted tree T .Define t = max { f ( i ) | i ∈ V ( T ) } and k j = { i | f ( i ) = j } . We may reorder indexes , , . . . , n by action of a permutation from S n ∼ = Aut( A ) in a way such that f (1) = . . . = f ( k ) = 0 ,f ( k + 1) = . . . = f ( k + k ) = 1 ,. . .f ( n − k t + 1) = . . . = f ( n ) = t. Due to the definition of T , we get the upper-triangular matrix. The restrictions on thevalues of elements immediately follow from Statement 4. Corollary 2 . There is a bijection between the set of RB-operators of nonzero weight λ on F e ⊕ F e ⊕ . . . ⊕ F e n anda) the set of 2-colored subtree acyclic digraphs on n vertices;b) the set of labeled rooted trees on n + 1 vertices with 2-colored non-root vertices.Now, we want to compute the number r n of RB-operators of nonzero weight λ on A = F e ⊕ . . . ⊕ F e n which lie in different orbits under the action of the automorphismgroup Aut( A ) ∼ = S n . The group Aut( A ) acts on the set of RB-operators of weight λ inthe way described in Statement 2, ψ : R → R ( ψ ) = ψ − Rψ .In a light of Corollary 2b, we may interpretate the number r n as the number of unlabeled rooted trees on n + 1 vertices with 2-colored non-root vertices. It is exactlythe sequence A000151 [18], the first eight values are 2, 7, 26, 107, 458, 2058, 9498, 44947etc. Let us fix that in advance we will use two colors: white and black, white colorcorresponds to the case r ii = 0 and black color corresponds to r ii = − λ . Considering therooted tree T with n + 1 vertices, we may assume that the root is colored in the thirdcolor, say grey.Note that the map φ acts on a labeled (or unlabeled) rooted tree T on n + 1 verticeswith 2-colored non-root vertices as follows. The φ interchanges a color in every non-rootvertex.Let us describe splitting RB-operators of nonzero weight λ on A . Theorem 2 . An RB-operator R of nonzero weight λ on A = F e ⊕ . . . ⊕ F e n issplitting if and only if the corresponding (labeled) rooted tree T = T ( R ) on n + 1 verticesis properly colored. Proof . Wuthout loss of generality, we put λ = 1 . For simplicity, let us consider thegraph T ′ = T \ { root } , which is a forest in general case.Let us prove the statement by induction on n . For n = 1 , we have either R = 0 (theonly non-root vertex is white) or R = − λ id (the only non-root vertex is black), bothRB-operators are splitting with subalgebras F and (0) .Suppose that we have proved Theorem 2 for all natural numbers less than n . Leta graph T ′ with n vertices be disconnected, denote by T , . . . , T k the connected componentsof T ′ . So, A = A ⊕ . . . A k for A s = Span { e j | j ∈ V ( T s ) } . Define R s as the inducedRB-operator R | A s (see Lemma 2). By the definition, R is splitting if and only if A = er( R ) ˙+ ker( R + id ) or equivalently A s = ker( R s ) ˙+ ker( R s + id ) , s = 1 , . . . , k . By theinduction hypothesis, we have such decomposition for every s if and only if the coloringof T s is proper.Now consider the case when T ′ is connected. We may assume that e corresponds tothe vertex 1, the only source in G , and { , . . . , k } is the set of all vertices of G with thevalue of f ( x ) equal to 1. We also define T s for s = 2 , . . . , k as the connected componentof T ′ \ { } which contains the vertex s . Note that R induces the RB-operator of weight λ on the subalgebra A s = Span { e j | j ∈ V ( T s ) } for all s by Lemma 2.The condition of R to be splitting is equivalent to the conditionrank ( R ) + rank ( R + id ) = n. (4)Analysing the e -coordinate, we have n = rank ( R ) + rank ( R + id ) ≥ rank ( R ′ ) + rank ( R ′ + id ) for R ′ , the induced RB-operator on the subalgebra Span { e j | j ≥ } . Thus, rank ( R ′ ) + rank ( R ′ + id ) = n − , i.e. R ′ is splitting or equivalently R | A s is spplitting for every s = 2 , . . . , k . By the induction hypothesis, the graph T ′ \ { } is properly 2-colored. Itremains to prove that the vertices , . . . , k are colored in the same color and the vertex 1is colored in another one.Up to the action of φ , which preserves the splitting structure of an RB-operator (seeRemark 1), we may assume that the vertex 1 is colored in white. Since we know thatrank ( R + id ) = rank ( R ′ + id ) + 1 , we have to state the equality rank ( R ) = rank ( R ′ ) . So,the condition (4) is fulfilled if and only if the first row (0 , , , . . . , of the matrix R islinearly expressed via other rows. By the definition of the matrix R , the vertices , . . . , k have to be colored in black. Theorem is proved. Corollary 3 . An RB-operator R of nonzero weight λ on A = F e ⊕ . . . ⊕ F e n isinner-splitting if and only if in T = T ( R ) all vertices with even value of f are colored inone color and all vertices with odd value of f are colored in another color. Proof . Up to φ , we may assume that R (1) = 0 . Thus, any vertex with the value of f ( x ) equal to 0 has to be colored in white. By Theorem 2, T ′ = T \ { root } is properly2-colored, so, all vertices with the value of f ( x ) equal to 1 are colored in black, all verticeswith the value of f ( x ) equal to 2 are colored in white and so on.Now, we collect all our knowledges about all RB-operators (in Table 1) and allnonisomorphic RB-operators (in Table 2) of nonzero weight on a sum of fields A = F e ⊕ F e ⊕ . . . ⊕ F e n .We have noticed that the first values of number of splitting RB-operators coin-cides with the sequence A007830 [18] (in labeled case) and coincides with the sequenceA000106 [18] (in unlabeled case). Actually it should be proven for all n . Remark 4 . Counting rooted trees on n + 1 vertices with properly 2-colored non-rootvertices is not the same as counting properly 2-colored forests on n vertices.7 able 1 . Number of RB-operators of nonzero weight on a sum of n fieldsClass of Description formula and firstRB-operators OEIS [18] 5 valuesall labeled rooted trees on n + 1 vertices n ( n + 1) n − , , ,with 2-colored non-root vertices A097629 , splitting labeled rooted trees on n + 1 vertices n + 2) n − ?! , , ,with properly 2-colored non-root vertices A007830 ?! , inner-splitting labeled rooted trees on n + 1 vertices n + 1) n − , , ,(twice) · A000272 , non-splitting labeled rooted trees on n + 1 vertices with — , , ,improperly 2-colored non-root vertices , Table 2 . Number of RB-operators of nonzero weight on a sum of n fields(up to conjugation with an automorphism)Class of Description OEIS [18] first 5 valuesRB-operatorsall rooted trees on n + 1 vertices A000151 , , , , with 2-colored non-root verticessplitting rooted trees on n + 1 vertices with A000106 ?! , , , , properly 2-colored non-root verticesinner-splitting rooted trees on n + 1 vertices (twice) · A000081 , , , , non-splitting rooted trees on n + 1 vertices with — , , , , improperly 2-colored non-root verticesLet us write down all non-splitting pairwise nonisomorphic RB-operators for n = 2 , . Statement 5 . Up to φ , we have the following non-splitting pairwise nonisomorphicRB-operatorsa) for n = 2 : R ( e ) = e , R ( e ) = 0 ;b) for n = 3 :(RB1) R ( e ) = e + e , R ( e ) = e , R ( e ) = 0 ,(RB2) R ( e ) = e + e , R ( e ) = e , R ( e ) = − e ,(RB3) R ( e ) = e + e , R ( e ) = − e − e , R ( e ) = − e ,(RB4) R ( e ) = e + e , R ( e ) = R ( e ) = 0 ,(RB5) R ( e ) = e + e , R ( e ) = − e , R ( e ) = 0 ,(RB6) R ( e ) = e , R ( e ) = R ( e ) = 0 ,(RB7) R ( e ) = e , R ( e ) = 0 , R ( e ) = − e . Proof . a) Non-splitting case appears only when the graph T ′ is non-empty andimproperly 2-colored. Up to φ , we may assume that two vertices are colored in white.8) Cases (RB1)–(RB3) correspond to improperly 2-colorings of the graph T ′ with V ( T ′ ) = { , , } and E ( T ′ ) = { (1 , , (2 , } . Cases (RB4), (RB5) correspond to improp-erly 2-colorings of the graph T ′ with E ( T ′ ) = { (1 , , (1 , } . Finally, cases (RB6), (RB7)correspond to improperly 2-colorings of the graph T ′ with E ( T ′ ) = { (1 , } . Statement 6 . Up to φ , we have the following splitting but not inner-splitting pairwisenonisomorphic RB-operators:a) for n = 2 : R ( e ) = − e , R ( e ) = 0 ;b) for n = 3 :(RB1 ′ ) R ( e ) = e , R ( e ) = 0 , R ( e ) = − e ,(RB2 ′ ) R ( e ) = − e , R ( e ) = R ( e ) = 0 . Let C be an associative algebra and R be an RB-operator on C of weight λ . Then thespace C under the product x ◦ R y = R ( x ) y + xR ( y ) + λxy (5)is an associative algebra [14, 13]. Let us denote the obtained algebra as C R . It is easy tosee that C φ ( R ) ∼ = C R .Let us denote by Ab n the n -dimensional algebra with zero (trivial) product. Theorem 3 . Given an algebra A = F e ⊕ . . . ⊕ F e n and an RB-operator R of weight λ on A , we have A R ∼ = ( Ab n , λ = 0 ,A, λ = 0 . Proof . If λ = 0 , then R = 0 [12] and x ◦ R y = 0 . For λ = 0 , we may assume that λ = 1 , since rescalling of the product does not exchange the algebraic structure.Let us prove the statement by induction on n . For n = 1 , we have either R = 0 or R = − id. Due to (5) we get either x ◦ y = xy or x ◦ y = − xy , in both cases A R ∼ = A .Suppose that we have proved Theorem 3 for all numbers less n . Let a graph T ′ = T ′ ( R ) with n vertices be disconnected, denote by T , . . . , T k the connected components of T ′ .As earlier, we define A = A ⊕ . . . A k for A s = Span { e j | j ∈ V ( T s ) } and define R s asthe induced RB-operator R | A s . By the induction hypothesis, A Rs ∼ = A s for every s and so A = A ⊕ . . . ⊕ A k ∼ = A R ⊕ . . . ⊕ A Rk = A R .Now consider the case when T ′ is connected. We may assume that e corresponds tothe vertex 1, the only source in G . Note the space I = Span { e j | j ≥ } is an ideal in A R which is isomorphic to F e ⊕ . . . ⊕ F e n by the induction hypothesis. Up to φ , we mayassume that the vertex 1 in T ′ is colored in white and , . . . , t is a list of all neighboursof 1 in T ′ . Let us consider the one-dimensional space I in A R generated by the vector a = e − c (2) e − . . . − c ( t ) e t , where c ( i ) = ( , i is colored in white , − , i is colored in black .
9n terms of the matrix entries, c ( i ) = 1 + 2 r ii . We may assume that c (2) = c (3) = . . . = c ( s ) = 1 and c ( s + 1) = . . . = c ( t ) = − for some s ∈ { , . . . , t } .By (5) we compute the product of a with e k for k > t : a ◦ e k = ( e + e + . . . + e s − e s +1 − . . . − e t ) ◦ e k = R ( e + e + . . . + e s − e s +1 − . . . − e t ) e k . Since k is connected with only one vertex from , . . . , t (due to (SF2)), say j , we have a ◦ e k = R ( e − c ( j ) e j ) e k = e k − c ( j )(1 + 2 r jj ) e k = (1 − ( c ( j )) ) e k = 0 . Analogously we can check that a ◦ e k = 0 for all k > . Thus, I is an ideal in A R .Now, we calculate a ◦ a = e ◦ ( e + e + . . . + e s − e s +1 − . . . − e t )= R ( e )( e + e + . . . + e s − e s +1 − . . . − e t ) + e = ( e + . . . + e s + e s +1 + . . . + e t )( e + e + . . . + e s − e s +1 − . . . − e t ) + e = e + e + . . . + e s − e s +1 − . . . − e t = a and so I is isomorphic to F .Summarising, we have A R = I ⊕ I ∼ = ( F e ⊕ . . . ⊕ F e n ) ⊕ F ∼ = A . Theorem is proved. Acknowledgements
The main part of the paper was done while working in Sobolev Institute of Mathe-matics in 2017. The research is supported by RSF (project N 14-21-00065).
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