aa r X i v : . [ m a t h . R A ] J un Spectrum of Rota—Baxter operators
V. Gubarev
Abstract
We prove that the spectrum of every Rota—Baxter operator of weight λ ona unital algebraic (not necessarily associative) algebra over a field of characteristiczero is a subset of { , − λ } . For a finite-dimensional unital algebra the same state-ment is shown to hold without a restriction on the characteristic of the ground field.Based on these results, we define the Rota—Baxter λ -index rb λ ( A ) of an algebra A as the infimum of the degrees of minimal polynomials of all Rota—Baxter operatorsof weight λ on A . We calculate the Rota—Baxter λ -index for the matrix algebra M n ( F ) , char F = 0 : it is shown that rb λ ( M n ( F )) = 2 n − . Keywords : Rota—Baxter operator, matrix algebra, simple Jordan algebra ofa bilinear form, matrix Cayley—Dickson algebra, Rota—Baxter module.
Given an algebra A over a ground field F and a scalar λ ∈ F , a linear operator R : A → A is called a Rota—Baxter operator (RB-operator, for short) of weight λ on A if the following identity holds for all x, y ∈ AR ( x ) R ( y ) = R ( R ( x ) y + xR ( y ) + λxy ) . (1)An algebra A endowed with an RB-operator is called Rota—Baxter algebra (RB-algebra).Rota—Baxter operators appeared in the work of G. Baxter [6] in 1960 in the studyof fluctuation theory as a formal generalization of integration by parts formula. Further,they were studied by G.-C. Rota [31] and others [4,10,26]. The deep connections betweensolutions of different versions of the classical Yang—Baxter equation in mathematicalphysics and RB-operators were discovered in [1, 7, 11]. The subject of double algebras isalso tightly connected to Rota—Baxter operators, see [14, 27]. Since 2000, the connectionbetween RB-algebras and so called Loday algebras (prealgebras and postalgebras) hasbeen studied [1, 5, 12, 17].Rota—Baxter operators have a broad area of application in symmetric polynomials,shuffle algebra, etc. [4, 19, 20].There is a plenty of papers devoted to a partial or complete classification of Rota—Baxter operators on a given small-dimensional algebra: sl ( F ) [22, 28, 29], sl ( F ) [23], M ( F ) [2, 8, 34], M ( F ) [13, 33], the Grassmann algebra Gr [8], and others, see thesurvey [16] and the references therein.The classifications obtained in these works are often bulky and hard to observe, sothere is a need to investigate the general properties of Rota—Baxter operators at least onfinite-dimensional (and close to them) algebras. In an attempt of such a study, the authorproved in [16] the following result. Given a unital algebraic associative (alternative,Jordan) algebra A over a field of characteristic zero, every RB-operator of weight zeroon A is nilpotent. Moreover, if A is algebraic of a bounded degree, then there exists N such that R N = 0 for every RB-operator R of weight zero on A . (By an algebraic algebra1n the nonassociative case we mean such an algebra that every finite subset generatesa finite-dimensional subalgebra.)The statement above leads to a new invariant of an algebra A : the Rota—Baxter index(RB-index) rb ( A ) defined as the lower bound of nilpotency indices of all RB-operatorsof weight zero on A . This invariant was calculated for different finite-dimensional alge-bras [15, 16], including the matrix algebra M n ( F ) .In the present work, we generalize and extend this result in different directions. First,we consider nonzero weight case of Rota—Baxter operators (when λ = 0 ). Further, weremove the restriction on a variety of algebras, we do note assume that A is associa-tive, alternative, or Jordan. Finally, we state the result in the case of a field of anycharacteristic.Before stating the main result of the work, we need a new definition. Given analgebra A , denote the set of all RB-operators of weight λ on A by RB λ ( A ) . Define theRota—Baxter λ -index (RB ( λ ) -index) of A asrb λ ( A ) = min { n ∈ N | for all R ∈ RB λ ( A ) exists k : R k ( R + λ id ) n − k = 0 } . If such a number does not exist, then put rb λ ( A ) = ∞ .If λ = 0 , then rb ( A ) coincides with the earlier defined Rota—Baxter index rb ( A ) [16].In §3 (Theorem 1, over a field of characteristic zero) and in §4 (Theorem 2, over a fieldof positive characteristic), we prove Main Theorem . Let A be a unital finite-dimensional algebra over a field F . Let usfix λ ∈ F .a) Given an RB-operator R of weight λ on A , there exist k, l ≥ such that R k ( R + λ id ) l = 0 . In particular, Spec ( R ) ⊂ { , − λ } .b) There exists a natural n such that for every RB-operator R of weight λ on A onecan find ≤ k ≤ n such that R k ( R + λ id ) n − k (1) = 0 holds. Moreover, rb λ ( A ) ≤ n .When char F = 0 , the part a) of the Theorem holds even if A is algebraic (see §3).In §5 (Theorem 3), we state another important property of Rota—Baxter operators.Given a unital algebraic algebra A and an RB-operator R of nonzero weight on A , theminimal polynomial of the element R (1) has no multiple roots. This result allows us toextend and reprove some results from [16] including the diagonalizibilty of R (1) for an RB-operator R of nonzero weight on the matrix algebra M n ( F ) over a field of characteristiczero.In §6, we compute the Rota—Baxter λ -index for different algebras: a sum of fields,the matrix algebra M n ( F ) , its subalgebra t n ( F ) of not strictly upper-triangular matrices,the simple Jordan algebra J n ( f ) of a nondegenerate symmetric bilinear form f , and thematrix (split) Cayley—Dickson algebra C ( F ) .Let us collect all results obtained earlier (see [8,15,16]) and in the current work aboutRB ( λ ) -index of particular algebras in the following table (see the next page). By Gr n wedenote the Grassmann algebra of a vector space V = Span { e , . . . , e n } . The sign “(?)”denotes the hypothesis, see Conjectures 1 and 2 in §6.2.In §7, we apply the Main Theorem for Rota—Baxter modules (§7.1) and parameterizedRota—Baxter operators (§7.2).Finally, in §7.3, we generalize the construction which is already known for differentvarieties of algebras [9,19,24]. Given an algebra A of a variety Var and an RB-operator R λ -index of different algebrasAlgebra Field rb rb λ , λ = 0 F n any 1 nM n ( F ) char ( F ) = 0 2 n − n − t n ( F ) char ( F ) = 0 n (?) n − n char ( F ) = 0 ? 2 J n ( f ) ¯ F = F , char ( F ) =2 , ( , n = 33 , n > ( , n , | nC ( F ) ¯ F = F , char ( F ) =2 , A = k L i =1 M n i ( F ) char ( F ) = 0 2 max i =1 ,...,k { n i }− (?) k P i =1 n i + max i =1 ,...,k { n i } − (?)on A , one get a countable sequence of algebras A i belonging to the same variety Var .Moreover, all of A i considered with R are Rota—Baxter algebras.We also apply the Main Theorem to state the structure of the limiting algebra A ∞ (see §7.3). Trivial RB-operators of weight λ are zero operator and − λ id. Lemma 1 [19]. Given an RB-operator P of weight λ ,(a) the operator − P − λ id is an RB-operator of weight λ ,(b) the operator λ − P 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 ( P ) = − P − λ ( P ) id, where λ ( P ) denotes the weight of an RB-operator P . It is clearthat φ coincides with the identity map. Lemma 2 [8]. Given an algebra A , an RB-operator P on A of weight λ , and ψ ∈ Aut( A ) , the operator P ( ψ ) = ψ − P ψ is an RB-operator of weight λ on A .The same result is true when ψ is an antiautomorphism of A , i.e., a bijection from A to A satisfying ψ ( xy ) = ψ ( y ) ψ ( x ) for all x, y ∈ A ; e.g., transpose on the matrix algebra. Lemma 3 [19]. Let an algebra A split as a vector space into a direct sum of twosubalgebras A and A . An operator P defined as P ( a + a ) = − λa , a ∈ A , a ∈ A , (2)is RB-operator of weight λ on A .Let us call an RB-operator from Lemma 3 as splitting RB-operator with subalgebras A , A . Note that the set of all splitting RB-operators on an algebra A is in bijectionwith all decompositions of A into a direct sum of two subalgebras A , A . Lemma 4 [8]. Let A be a unital algebra and let R be an RB-operator of nonzeroweight λ on A . If R (1) ∈ F , then R is splitting with one of subalgebras containing 1.The following construction of RB-operators of nonzero weight generalizes Lemma 3.3 emma 5 [18]. Let an algebra A be a direct sum of subspaces A − , A , A + , moreover, A ± , A are subalgebras of A , and A ± are A -modules. If R is an RB-operator of weight λ on A , then an operator P defined as P ( a − + a + a + ) = R ( a ) − λa + , a ± ∈ A ± , a ∈ A , (3)is an RB-operator of weight λ on A .Let us call an RB-operator of nonzero weight defined by (3) as triangular-splittingone provided that at least one of A − , A + is nonzero. Lemma 6 . Let A be a unital algebra and R be an RB-operator of weight λ on A .a) Then the subalgebra S generated by the set { R k (1) | k ∈ N } is commutative andassociative.b) [19, §3.2.2] for all k, l ∈ N the following formula holds, R k (1) R l (1) = min { k,l } X j =0 λ j (cid:18) kj (cid:19)(cid:18) k + l − jk (cid:19) R k + l − j (1) . (4) Proof . a) We prove that the associator h ( k, l, m ) = ( R k (1) , R l (1) , R m (1)) = ( R k (1) R l (1)) R m (1) − R k (1)( R l (1) R m (1)) is zero for all natural k, l, m by induction on r = k + l + m . For r = 0 , it is true. Supposethat the statement holds true for all p < r . Note that h ( k, l, m ) = 0 if at least one of thenumbers k, l, m is zero. So, we may assume that k, l, m ≥ .On the one hand, we have ( R k (1) R l (1)) R m (1)= R ( R k − (1) R l (1) + R k (1) R l − (1) + λR k − (1) R l − (1)) R m (1)= R [( R k − (1) R l (1)) R m (1) + ( R k (1) R l − (1)) R m (1) + λ ( R k − (1) R l − (1)) R m (1)+ R ( R k − (1) R l (1) + R k (1) R l − (1) + λR k − (1) R l − (1)) R m − (1)+ λ ( R k − (1) R l (1)) R m − (1) + λ ( R k (1) R l − (1)) R m − (1) + λ ( R k − (1) R l − (1)) R m − (1)] . On the other hand, R k (1)( R l (1) R m (1))= R k (1) R ( R l − (1) R m (1) + R l (1) R m − (1) + λR l − (1) R m − (1))= R [ R k (1)( R l − (1) R m (1)) + R k (1)( R l (1) R m − (1)) + λR k (1)( R l − (1) R m − (1))+ R k − (1) R ( R l − (1) R m (1) + R l (1) R m − (1) + λR l − (1) R m − (1))+ λR k − (1)( R l − (1) R m (1)) + λR k − (1)( R l (1) R m − (1)) + λ R k − (1)( R l − (1) R m − (1))] . Applying the induction hypothesis and (1), we compute h ( k, l, m ) = R (∆) , where ∆ = R k − (1) R l (1) R m (1) + R k (1) R l − (1) R m (1) + λR k − (1) R l − (1) R m (1)+ R k (1) R l (1) R m − (1) + λR k − (1) R l (1) R m − (1)+ λR k (1) R l − (1) R m − (1) + λ R k − (1) R l − (1) R m − (1) − R k (1) R l − (1) R m (1) − R k (1) R l (1) R m − (1) − λR k (1) R l − (1) R m − (1) R k − (1) R l (1) R m (1) − λR k − (1) R l − (1) R m (1) − λR k − (1) R l (1) R m − (1) − λ R k − (1) R l − (1) R m − (1) = 0 . Finally, we prove that R k (1) R l (1) = R l (1) R k (1) by induction on r = k + l . For r = 0 , it is trivial. Suppose commutativity holds for all p < r . If k = 0 or l = 0 , then R k (1) R l (1) = R l (1) R k (1) . Assume that k, l ≥ . We have R k (1) R l (1) − R l (1) R k (1)= R ( R k − (1) R l (1) + R k (1) R l − (1) + λR k − (1) R l − (1) − R l − (1) R k (1) − R l (1) R k − (1) − λR l − (1) R k − (1)) = 0 by the induction hypothesis. (cid:3) Let A be a unital associative algebra and R be an RB-operator of weight λ on A . Itis known [19] that the following formula n ! R n (1) = n X k =1 ( − n − k λ n − k s ( n, k )( R (1)) k (5)holds. Here s ( n, k ) is a Stirling number of the first kind. By Lemma 6, we may removethe restriction on A to be associative. Lemma 7 . Let A be a unital algebraic algebra over a field of characteristic zero andlet R be an RB-operator of a nonzero weight λ on A . Then there exist r, s such that ( R + λ id ) r R s (1) = 0 . Proof . Denote a = R (1) . We put λ = − for simplicity. Since A is algebraic,there exists a minimal polynomial m for a of the degree n . The formula (5) allows us topresent m in the form m ( a ) = R n (1) + α n − R n − (1) + . . . + α R (1) + α = 0 , α i ∈ F. (6)Applying the formula R k (1) R (1) = ( k + 1) R k +1 (1) + λkR k (1) (7)holding by Lemma 6b, we calculate m ( a ) R (1) = ( n + 1) R n +1 (1) + n ( α n − − R n (1) + ( n − α n − − α n − ) R n − (1)+ . . . + α R (1) . (8)Subtracting (8) from ( n + 1) R ( m ( a )) , we obtain α n − + n ) R n (1) + (2 α n − + ( n − α n − ) R n − (1) + . . . + ( nα + α ) R (1) . (9)We have two cases. Case 1 : all coefficients of (9) are zero. Then it is easy to derive the formulas α n − = − n, α n − = n ( n − , . . . , α n − i = ( − i (cid:18) ni (cid:19) , . . . , α = ( − n . So, we get the equality ( R − id ) n (1) = 0 , and we are done.5 ase 2 : not all coefficients of (9) are zero. By minimality of m , it means that m and (9) are proportional. Find a minimal k such that α = . . . = α k − = 0 and α k = 0 . Thus, coefficients by R i (1) in km ( a ) and (9) are equal. Again, we write downthe equalities α k +1 = − ( n − k ) α k , α k +2 = ( n − k )( n − k − α k , . . . ,α n − i = ( − n − k − i (cid:18) n − ki (cid:19) α k , . . . , α n − = ( − n − k − ( n − k ) α k . Thus, we get ( − n − k α k m ( a ) = ( R − id ) n − k R k (1) = 0 . (cid:3) Lemma 8 . Let A be an algebra and R be an RB-operator of weight λ on A .a) If λ = 0 , then ker( R k ) is an Im ( R k ) -module for each k ≥ .b) [9] If λ = 0 , then ker( R k ) is an ideal in Im ( R + λ id ) k . Proof . We prove Lemma 8 by induction on k . The following equality R ( x ( R ( y ) + λy )) = R ( x ) R ( y ) − R ( R ( x ) y ) (10)holds by (1) for all x, y ∈ A . If x ∈ ker( R ) , then x Im ( R + λ id ) ∈ ker( R ) . Analogously, Im ( R + λ id ) x ∈ ker( R ) .Suppose we have proved Lemma 8 for all numbers less or equal to k . Let x ∈ ker( R k +1 ) , y ∈ A and z = R ( y ) . Then by (10) we get R k +1 ( x ( R + λ id ) k +1 ( y )) = R k [ R ( x )( R + λ id ) k ( z ) − R ( R ( x )( R + λ id ) k ( y ))] . By the induction hypothesis, x Im ( R + λ id ) k +1 ∈ ker( R k +1 ) . Analogously, we obtain Im ( R + λ id ) k +1 x ∈ ker( R k +1 ) . (cid:3) Given an algebra A , denote the set of all RB-operators of weight λ on A by RB λ ( A ) .Define Rota—Baxter λ -index (RB( λ )-index) of A asrb λ ( A ) = min { n ∈ N | for all R ∈ RB λ ( A ) exists k : R k ( R + λ id ) n − k = 0 } . If such a number does not exist, then put rb λ ( A ) = ∞ .When λ = 0 , rb ( A ) = min { n ∈ N | R n = 0 for all R ∈ RB ( A ) } which coincides with the earlier defined Rota—Baxter index of A [16] (it was denotedthere as rb ( A ) ).Now, we prove the Main Theorem over a field of characteristic zero, generalizing [16]. Theorem 1 . Let A be a unital algebraic algebra over a field F of characteristic zero.Let us fix λ ∈ F .a) Given an RB-operator R of weight λ on A , there exist k, l ≥ such that R k ( R + λ id ) l = 0 . In particular, Spec ( R ) ⊂ { , − λ } .b) Suppose that A is algebraic of the restricted degree N , then there exists a natural n ≤ N such that for every RB-operator R of weight λ on A one can find ≤ k ≤ n suchthat R k ( R + λ id ) n − k (1) = 0 holds. Moreover, rb λ ( A ) ≤ n .6 roof . Let λ = 0 . In [16], it was proved that R n (1) = 0 for some n ∈ N under thecondition that A is power-associative. By Lemma 6, we may avoid this condition. When A is algebraic of the restricted degree N , then n ≤ N . Let x ∈ Im ( R n ) . By Lemma 8a, · x = x ∈ ker( R n ) . Thus, R n ( A ) = (0) , we are done.Let λ = 0 . We may assume that λ = − . Since A is algebraic, we may consider theminimal (unital) polynomial f ( x ) of the element R (1) = a of the degree n . By Lemma 7,there exists k such that R k ( R − id ) n − k (1) = 0 . When A is algebraic of the restricteddegree N , then n ≤ N .We want to prove that ∈ ker( R k ) ⊕ ker( R − id ) n − k . First, it is easy to check that ker( R k ) ∩ ker( R − id ) n − k = (0) . Second, let e γ , γ ∈ Γ , and f δ , δ ∈ ∆ , be a linear basis of ker( R k ) and of ker( R − id ) n − k respectively. We may complete the set { e γ | γ ∈ Γ } ∪ { f δ | δ ∈ ∆ } to a basis of A by adding vectors { g π | π ∈ Π } . So, we represent 1 as e + 1 f + 1 g meaning that e ∈ Span { e γ } , f ∈ Span { f δ } , and g ∈ Span { g π } . Let us show that g = 0 and so, ∈ ker( R k ) ⊕ ker( R − id ) n − k . Indeed, R k (1) = R k (1 f ) + R k (1 g ) , and R k (1 f ) ∈ ker( R − id ) n − k , since ker( R − id ) n − k is R -invariant. As ( R − id ) n − k ( R k (1)) = 0 ,we conclude that R k (1 g ) ∈ ker( R − id ) n − k . The map R k is invertible on ker( R − id ) n − k ,so, g ∈ ker( R − id ) n − k and g = 0 .Finally, let x ∈ Im R n − k ( R − id ) k . By Lemma 8b, (ker( R k ) ⊕ ker( R − id ) n − k ) · Im R n − k ( R − id ) k ⊂ ker( R k ) ⊕ ker( R − id ) n − k . So, · x = x ∈ ker( R k ) ⊕ ker( R − id ) n − k . As a consequence, R n ( R − id ) n ( A ) = (0) . (cid:3) Remark 1 . Let us show that all conditions of Theorem 1 are necessary. • If A is not unital, then we may consider an algebra A with zero product. Thenevery linear map on A is an RB-operator of a weight λ . So, we may find a linearmap satisfying Spec ( R )
6⊂ { , − λ } . • If A is not algebraic, then A = F [ x ] with the RB-operator R ( x n ) = x n +1 n +1 is a coun-terexample to the conclusion of Theorem 1 when λ = 0 . Regardless of weight λ ,we may consider A equal to the free commutative unital RB-algebra generated bya non-empty set X = { } (see [20]), it is also a counterexample to the conclusionof Theorem 1. • If char ( F ) = p > , then let A be the free commutative unital RB-algebra ofweight 0 generated by unit. Due to [20], A = L i ≥ F e i with the product e i e j = (cid:0)(cid:0) i + ji (cid:1) mod p (cid:1) e i + j and the RB-operator R acting as follows, R ( e i ) = e i +1 . Notethat A is algebraic. Thus, the conclusion of Theorem 1 does not hold. Lemma 9 . Let A be a finite-dimensional unital algebra over a field F of positivecharacteristic p . Then given an RB-operator R of weight λ on A , there exist k, l ≥ such that R k ( R + λ id ) l = 0 . 7 roof . Let λ = 0 . Since A is finite-dimensional, there exists a minimal l such that R l (1) + α l − R l − (1) + . . . + α R (1) = 0 , α i ∈ F. (11)It is well-known that unit does not lie in the image of R , thus we take a zero free coefficientin (11). Multiplying (11) by R (1) and applying the formula (7) for λ = 0 , we get ( l + 1) R l +1 (1) + lα l − R l (1) + . . . + 2 α R (1) = 0 . (12)Acting on (11) by R , we have R l +1 (1) + α l − R l (1) + . . . + α R (1) = 0 . (13)From (12) and (13), we conclude α l − R l (1) + 2 α l − R l − (1) + . . . + ( l − α R (1) = 0 . Since l was chosen as a minimal number satisfying (11), we have two cases. Case 1 . The vectors (1 , α l − , . . . , α , α ) and ( α l − , α l − . . . , ( l − α , are pro-portional (with some nonzero coefficient). It means that α = α = . . . = α l − = 0 and R l (1) = 0 , as required. Case 2 . The vector ( α l − , α l − . . . , ( l − α , is zero. If l ≤ p , then we derive α = α = . . . = α l − = 0 and R l (1) = 0 . If l > p , then only coefficients α l − p , α l − p , . . . may be nonzero. Anyway, we can find an equality R k (1) + β R k (1) + . . . + β s R k s (1) = 0 , k > k > . . . > k s , p | k , . . . , p | k s (14)with a minimal number of nonzero coefficients (here s ). If s = 1 , we are done.Suppose that s > . Consider ∆ = k − k s = p t q , where ( p, q ) = 1 and t ≥ . Actingby the operator R δ on (14) for a number δ ∈ p N , we may get k = p N − p t , k s = p N − p t − ∆ for some sufficiently big N .Now, we mutiply (14) by R p t (1) . Applying Lemma 6b for λ = 0 R a (1) R b (1) = (cid:18) a + ba (cid:19) R a + b (1) , we have R k (1) R p t (1) = 0 , R k s (1) R p t (1) = (cid:18) p t ( p N − t − q ) p t (cid:19) R p N − ∆ (1) , where the coefficient (cid:0) p t ( p N − t − q ) p t (cid:1) is nonzero by the Lucas’ Theorem. So, we have foundan expression like (14) with less number of nonzero coefficients. It is a contradiction, so s = 1 , and we are done.Let λ be nonzero, so we fix λ = − . Since A is finite-dimensional, there exists anexpression m of the form R n (1) + α n − R n − (1) + . . . + α R (1) + α = 0 , α i ∈ F, of the minimal degree n . 8nalogously to the proof of Lemma 7, we derive the formula q = ( α n − + n ) R n (1) + (2 α n − + ( n − α n − ) R n − (1) + . . . + ( nα + α ) R (1) = 0 . If all α i are zero, R n (1) = 0 and we are done by Lemma 8.Otherwise, find a minimal s such that α s = 0 . So, q = sm , and we have the relations sα s +1 = ( n − s ) α s + ( s + 1) α s +1 ,sα s +2 = ( n − s − α s +1 + ( s + 2) α s +2 ,. . . ,sα n − = 2 α n − + ( n − α n − ,s = α n − + n. (15)Define n − s = tp + k for ≤ k < p and t ≥ . Due to the system (15), we have m = α s (cid:18) R s (1) − (cid:18) k (cid:19) R s +1 (1) + (cid:18) k (cid:19) R s +2 (1) . . . + ( − ) k (cid:18) kk (cid:19) R s + k (1) (cid:19) + α s + p (cid:18) R s + p (1) − . . . + ( − ) k (cid:18) kk (cid:19) R s + p + k (1) (cid:19) + . . . + ( − k (cid:18) R n − k (1) − . . . + ( − k (cid:18) kk (cid:19) R n (1) (cid:19) = R s ( R − id ) k ( α s + α s + p R p + . . . + ( − k R tp )(1) = 0 . Acting on the last expression by the operator R δ ( R − id ) ε with suitable δ, ε , we get theequality of the form R pk ( R − id ) p ( R pk + β R pk + . . . + β e R pk e )(1) = 0 , (16)where k > k > . . . > k e , k > , and β i = 0 , i = 2 , . . . , e .Since ( R − id ) p = R p − id modulo p , the equality (16) can be rewritten in the form R pm (1) + γ R p ( m − m ) (1) + . . . + γ f R p ( m − m f ) (1) = 0 , (17)where < m < m < . . . < m f < m .We may assume, that m = p c − for some c ≥ . Let d be any natural not less than c .Acting by the operator R p d − p c on (17), we get the equation R p d (1) + γ R p d − pm (1) + . . . + γ f R p d − pm f (1) = 0 . (18)By Lemma 6b, we have for λ = − R ap (1) R bp (1) = min { a,b } X c =0 ( − cp (cid:18) apcp (cid:19)(cid:18) p ( a + b − c ) ap (cid:19) R p ( a + b − c ) (1) (19)over a field of characteristic p . 9ith the help of the formula (19) and the Lucas’ Theorem, we calculate R p d (1) + γ R p d − pm (1) + . . . + γ f R p d − pm f (1)) R p d (1)= E ( d ) := 2 R p d (1) + ( − p d R p d (1) + f X j =2 γ j (cid:18) p d − pm j p d (cid:19) R p d − pm j (1) . (20)By the pigeonhole principle, we may find such d > d ≥ c that (cid:0) p d − pm j p d (cid:1) = (cid:0) p d − pm j p d (cid:1) for all j = 2 , . . . , f . Thus, R p d − p d ( E ( d )) − E ( d ) = ( − p R p d ( R p d − p d − id )(1) = 0 . Fix ∆ = p d − − p d − . Then we have R p k (1) − R p k − ∆ p (1) = 0 (21)for all sufficiently big k . Multiplying (21) by R p k (1) , we obtain R p k (1) + ( − p R p k (1) − (cid:18) p k − ∆ pp k (cid:19) R p k − ∆ p (1) = 0 . (22)Acting by (cid:0) p k − ∆ pp k (cid:1) R p k on (21) and subtracting the result from (22), we get (cid:18) − (cid:18) p k − ∆ pp k (cid:19)(cid:19) R p k (1) + ( − p R p k (1) = 0 . If the number in the brackets is zero, we are done. Otherwise, we find an expression ofthe form R p k (1) − lR p k (1) = 0 , l ∈ Z ∗ p . (23)If char ( F ) = p = 2 , then R p k (1) − R p k (1) = R p k ( R − id ) p k (1) = 0 , and have proved the statement.Suppose that p > . It is easy to show that consecutive applications of (23) implythe equality R p k +1 (1) − l p − R p k (1) = R p k +1 (1) − R p k (1) . (24)Finally, with the help of (24) we compute R p k +1 (1) − R p k (1)) = 2 R p k +1 (1) − R p k +1 (1) − R p k +1 + p k (1) + 2 R p k (1) − R p k (1)= 2( R p k (1) − R p k (1)) = 2 R p k ( R − id ) p k (1) , so, R p k ( R − id ) p k (1) = 0 , and we are done. (cid:3) heorem 2 . Let A be a unital finite-dimensional algebra over a field F of positivecharacteristic. Let us fix λ ∈ F .a) Given an RB-operator R of weight λ on A , there exist k, l ≥ such that R k ( R + λ id ) l = 0 . In particular, Spec ( R ) ⊂ { , − λ } .b) There exists a natural n such that for every RB-operator R of weight λ on A onecan find ≤ k ≤ n such that R k ( R + λ id ) n − k (1) = 0 holds. Moreover, rb λ ( A ) ≤ n . Proof . a) It is Lemma 9.b) We may find such n by Lemma 9 and by the condition that A is finite-dimensional.For the bound rb λ ( A ) ≤ n , we use Lemma 8 and repeat the end of the proof of Theo-rem 1. (cid:3) Define polynomials H r,s ( x ) = ( x − r )( x − r + 1) . . . x ( x + 1) . . . ( x + s − (25)for natural r, s ≥ . Lemma 10 . Let A be a unital algebra and let R be an RB-operator of weight − on A . Denote a = R (1) . Then for all r, s ≥ , ( r + s + 1) R ( H r,s ( a )) = H r,s +1 ( a ) . (26) Proof . Let us prove (26) by induction on r . For r = 0 , we want to state the formula ( n + 1)! R n ( a ) = H ,n +1 ( a ) (27)by induction on n . For n = 0 , we have a = a , it is true. Suppose that we have proved that n ! R n − ( a ) = H ,n ( a ) or, equivalently, a n = n ! R n − ( a ) + f ( a ) , where f ( x ) = x n − H ,n ( x ) .By (7), we calculate a n +1 = a n · a = ( n ! R n (1) + f ( a )) R (1)= ( n + 1)! R n +1 (1) − n · n ! R n (1) + af ( a )= ( n + 1)! R n +1 (1) − nH ,n ( a ) + a ( a n − H ,n ( a ))= ( n + 1)! R n +1 (1) + a n +1 − ( a + n ) H ,n ( a ) . So, ( n + 1)! R n +1 (1) = H ,n +1 ( a ) as required. Thus, we get (26) for r = 0 .Suppose we have proved (26) for all numbers less than r . Represent H r,s ( x ) = H r − ,s +1 ( x ) − ( r + s ) H r − ,s ( x ) . Applying the induction hypothesis for the last equality, we get ( r + s + 1) R ( H r,s ( a )) = ( r + s + 1) R ( H r − ,s +1 ( a )) − ( r + s + 1)( r + s ) R ( H r − ,s ( a ))= H r − ,s +2 ( a ) − ( r + s + 1) H r − ,s +1 ( a ) = H r,s +1 ( a ) , as required. (cid:3) emma 11 . Let A be a unital algebra and let R be an RB-operator of weight − on A . Denote a = R (1) . Then for all r, s ≥ , ( r + s )!( R − id ) r R s (1) = H r,s ( a ) . (28) Proof . We prove the statement by induction on r . For r = 0 , it follows from (27).Suppose we have proved (28) for all numbers less than r . Applying the induction hy-pothesis and (26), we get the following equalities ( r + s )!( R − id ) r R s (1) = ( R − id )( r + s ) H r − ,s ( a )= H r − ,s +1 ( a ) − ( r + s ) H r − ,s ( a ) = H r − ,s ( a )( a + s − ( r + s )) = H r,s ( a ) . So, Lemma is proved. (cid:3)
Theorem 3 . Let A be a unital algebra over a field F and let R be an RB-operatorof weight − on A . Denote a = R (1) .a) If char ( F ) = 0 and A is algebraic, then there exist r, s such that H r,s ( a ) = 0 .b) If char ( F ) = p > , then H ,p ( a ) = 0 .In particular, we get that the minimal polynomial of a under the conditions has nomultiple roots. Proof . a) It follows from Lemma 7 and Lemma 11.b) We are done by (28). (cid:3)
Now, we get easily the following result which requires in [16] several pages.
Corollary 1 [16, Thm. 4.17]. Given a field F of characteristic 0, and an RB-operatorof weight − on M n ( F ) , the matrix R (1) is diagonalizable and the set of diagonal elementsin an appropriate basis has the form {− s, − s + 1 , . . . , , , . . . , r − , r } for some naturalnumbers r, s such that r + s + 1 ≤ n .We call an algebra A as a nil-algebra if for every x ∈ A there exists k such that x k = 0 for each bracketing of the nonassociative monomial x k . We generalize and simplify theproof of another statement from [16]. Corollary 2 [16, Thm. 4.8]. Let A be a unital algebra over a field F and A = F · ⊕ N (direct vector-space sum), where N is a nil-algebra. Then each Rota—Baxter operator R of nonzero weight on A is splitting and (up to φ ) we have R (1) = 0 . Proof . Denote a = R (1) . If a ∈ N , then a k = 0 for some k . By Theorem 3, we get a = 0 . Then by Lemma 4, R is splitting.If a N , then a = α · n , where α ∈ F and n ∈ N . So, there exists k such that ( a − α · k = 0 . By Theorem 3, a − α · or a ∈ F . By Lemma 4, R is splitting and φ ( R )(1) = 0 . (cid:3) By Corollary 2, we have rb λ (Gr n ) = 2 for the Grassmann algebra Gr n [16]. Corollary 3 . Let A be a unital algebra over a field F of characteristic zero and let R be an RB-operator of weight λ on A . Denote by S a subalgebra generated by the set { R k (1) | k ∈ N } .a) If dim( S ) = ∞ , then S is a free commutative RB-algebra of weight λ generatedby 1.b) If dim( S ) = n < ∞ and λ = 0 , then S is isomorphic to the quotient of F [ x ] by theideal generated by x n . Moreover, R acts on S as follows, R ( x i ) = x i +1 i +1 for i = 0 , . . . , n − .12) If dim( S ) < ∞ , λ = − , F is algebraically closed, and H r,s ( x ) is a minimalpolynomial of R (1) , then S ∼ = F r + s . Proof . a) It follows directly.b) It follows from Theorem 1 and formula (5).c) Note that S is isomoprhic to the quotient of F [ x ] by the ideal generated by thepolynomial H r,s of degree n = r + s which has no multiple roots. So, S is semisimple.Since F is algebraically closed, S ∼ = F r + s . (cid:3) The notion of capacity plays a crucial role in the structural theory of Jordan alge-bras [25]. Let us generalize it for all unital algebras (of every variety). Given a unitalalgebra A , define a capacity of A as the maximal number of pairwise orthogonal idem-potents in A . If capacity of A equals n , then there exist e , . . . , e n ∈ A such that e i = e i and e i e j = 0 if i = j . Corollary 4 . Let A be a unital algebraic algebra over an algebraically closed field F of characteristic zero. If capacity of A equals n < ∞ , then rb λ ( A ) ≤ n for λ = 0 . Proof . Let R be an RB-operator of nonzero weight λ on A . By Theorem 1a, thereexist k, l such that R k ( R + λ id ) l = 0 on A . In particular, it means that the subalgebra S in A generated by the set { R j (1) | j ∈ N } is finite-dimensional. By Corollary 3c, S ∼ = F m for some m . Clearly, m ≤ n by the definition of capacity. So, minimal polynomial of P (1) for every RB-operator P of weight λ on A has degree not greater than n . By theproof of Theorem 1b, rb λ ( A ) ≤ n . (cid:3) Let F n = F e ⊕ F e ⊕ . . . ⊕ F e n denote a direct sum of n copies of a field F . Here e i = e i . Proposition 1 . rb λ ( F n ) = ( , λ = 0 ,n, λ = 0 . Proof . For λ = 0 , it follows from Corollary 5.7 [16].Let λ = 0 and let R be an RB-operator of weight λ on F n . By the dimensionalreasons and the Main Theorem, R k ( R + λ id ) n − k = 0 for some ≤ k ≤ n . On the otherhand, define R on F n as follows: R ( e i ) = e i +1 + . . . + e n . It is the RB-operator of weight 1on F n . Moreover, R n = 0 and R n − = 0 . (cid:3) Let us generalize the RB-operator (M1) on M ( F ) from [16, Thm. 4.13] and theRB-operator 1-I on M ( F ) from [13, Thm. 3]. Example 1 . A linear map R defined on M n ( F ) as follows, R ( e ij ) = P k ≥ e i + k,j + k , i ≥ j, − P k ≥ e i − k,j − k , i < j,
13s an RB-operator of weight 1 on M n ( F ) . Moreover, R n ( R + id ) n − = 0 but we have ( R ( R + id )) n − = 0 and R n ( R + id ) n − = 0 .The following theorem was stated in [16] in the case of λ = 0 . Theorem 4 . Let F be a field of characteristic zero. Then rb λ ( M n ( F )) = 2 n − forall λ ∈ F . Proof . For λ = 0 , it was proved in [16].Let λ = 0 , we may assume that λ = 1 . By Example 1, rb λ ( M n ( F )) ≥ n − .Let R be an RB-operator of weight − on M n ( F ) . By the Main Theorem, Spec ( R ) ⊂{ , } . So, we extend the action of R to the algebra M n ( ¯ F ) . We want to show that either R n ( R − id ) n − = 0 or R n − ( R − id ) n = 0 . Thus, it is enough to state this for R over ¯ F .Define a = R (1) . If the degree of the minimal polynomial m t ( a ) of a is less than n ,then by Theorem 1, ( R ( R − id )) n − = 0 , and we are done. Suppose that deg( m t ( a )) = n and m t ( a ) coincides with the characteristic polynomial of a . By Corollary 1, a is conjugateto a diagonal matrix D = diag {− r, − r + 1 , . . . , s − , s − } with r + s = n . By Lemma 1from [13], all subspaces V q = Span { e ij | i − j = q } are R -invariant. Since all V q for q = 0 have dimension less or equal to n − , we have ( R ( R − id )) n − = 0 on each of such V q .Further, V which is the subalgebra of diagonal matrices from M n ( ¯ F ) has the dimensionequal to n . So, either R n = 0 or ( R − id ) n or ( R ( R − id )) n − = 0 on V . In any case,rb λ ( M n ( ¯ F )) = 2 n − as well as of M n ( F ) . (cid:3) Recall that t n ( F ) denotes the subalgebra of not strictly upper-triangular matricesfrom M n ( F ) . Corollary 5 . Let F be a field of characteristic zero and let λ ∈ F be nonzero. Thenrb λ ( t n ( F )) = 2 n − . Proof . For simplicity, let λ = 1 . The restriction of the RB-operator R from Ex-ample 1 on t n ( F ) implies rb λ ( t n ( F )) ≥ n − . We may assume that F is algebraicallyclosed, since rb λ ( t n ( F )) ≤ rb λ ( t n ( ¯ F )) . If the degree of the minimal polynomial m x ( a ) of a = R (1) is less than n , we are done by Theorem 1. Suppose that deg( m x ( a )) = n .Find ψ ∈ Aut( M n ( F )) such that R ( ψ ) (1) equals to the D = diag {− r, − r + 1 , . . . , s − , s − } with r + s = n . Now, the RB-operator P = R ( ψ ) acts on the subalgebra ψ ( t n ( F )) .Define V q = Span { e ij | i − j = q } . Applying the proof of Lemma 1 from [13], we concludethat a subspace V q ∩ ψ ( t n ( F )) is P -invariant for every q . It remains to repeat the proofof Theorem 4. (cid:3) Conjecture 1 . Let F be a field of characteristic zero. Then rb ( t n ( F )) = n .Let us extend Conjecture 5.24 from [16] on the case of nonzero weight. Conjecture 2 . Let A = k L i =1 M n i ( F ) be a semisimple finite-dimensional associativealgebra over a field F of characteristic zero. Thena) rb ( A ) = 2 max i =1 ,...,k { n i } − .b) rb λ ( A ) = k P i =1 n i + max i =1 ,...,k { n i } − for nonzero λ .The following example shows that there exists an RB-operator reaching the boundfrom Conjecture 2b.Let D n denote the subalgebra of all diagonal matrices in M n ( F ) and L n ( U n ) the setof all strictly lower (upper) triangular matrices in M n ( F ) .14 xample 2 . We define a triangular-splitting RB-operator R of nonzero weight λ onthe algebra A = k L i =1 M n i ( F ) with the subalgebras A = D , A − = U , A + = L , where D = k M i =1 D n i , L = k M i =1 L n i , U = k M i =1 U n i , and R is defined on D ∼ = F N for N = k P i =1 n i in the same way as the RB-operator fromExample 1. So, x N ( x + λ ) n − = 0 is the minimal polynomial for R . Here n = max i =1 ,...,k { n i } . Let J = J n +1 ( f ) = F ⊕ V be a direct vector-space sum of F and finite-dimensionalvector space V , dim V = n > , and f be a nondegenerate symmetric bilinear form on V .Under the product ( α · a )( β · b ) = ( αβ + f ( a, b )) · αb + βa ) , α, β ∈ F, a, b ∈ V, (29)the space J is a simple Jordan algebra [35]. We assume that char F = 2 , .Let us choose a basis e , e , . . . , e n of V such that the matrix of the form f in thisbasis is diagonal with elements d , d , . . . , d n on the main diagonal. As f is nondegenerate, d i = 0 for each i . Example 3 [8]. Let J n ( f ) , n ≥ , be the simple Jordan algebra of bilinear form f over an algebraically closed field F with char ( F ) = 2 , . Let R be a linear operatoron J n ( f ) defined by a matrix ( r ij ) n − i,j =0 in the basis , e , e , . . . , e n with the followingnonzero entries r = − , r = p d , r = − √ d , r jj = − , j = 1 , . . . , n − ,r i i +1 = d i +1 d i s − d i d i +1 , r i +1 i = − s − d i d i +1 , i = 2 , . . . , n − . Then R is an RB-operator of weight on J n ( f ) . Moreover, R ( R + 2 id ) = 0 but R ( R + 2 id ) = 0 and ( R + 2 id ) = 0 . Proposition 2 . Let λ ∈ F be nonzero and let F be an algebraically closed field withchar F = 2 , . Then rb λ ( J n ( F )) = ( , n is odd , , n is even . Proof . When n is odd, the statement follows from [8, Thm. 4.1].Suppose that n is even. By Example 3, we have rb λ ( J n ( F )) ≥ . Since F is al-gebraically closed, we may assume that all d i equal to 1. Let R be an RB-operatorof weight λ on J n ( F ) . Assume that R is defined by a matrix ( r ij ) ni,j =0 in the basis15 , e , e , . . . , e n . Due to the proof of Theorem 4.1 from [8], we have the following relations r ii = − λ , r kl = − r lk , r i = zr i , n X p =1 ¯ r p = λ , n X p =1 ¯ r pi = ¯ r i , n X p =1 ¯ r pk ¯ r pl = ¯ r k ¯ r l , n X p =1 ¯ r pi ¯ r p = − λz r i , where i, k, l > and k = l , and we have three subcases:(I) z = 1 , ¯ r = − λ/ ,(II) z = − , ¯ r = λ/ ,(III) z = − , ¯ r = − λ/ .It is easy to check that an RB-operator R from (I) is splitting. Further, φ maps RB-operators from (II) to (III) and vice versa. Finally, an RB-operator from (III) satisfiesthe equation R ( R + λ id ) = 0 . (cid:3) Cayley—Dickson algebras are known to be only simple nonassociative alternativefinite-dimensional algebras, they are either division algebras or so called split ones. Thefirst ones have only trivial RB-operators [8]. Let us study the rb ( λ ) -index of the matrix(split) Cayley—Dickson algebra C ( F ) , which is simple alternative 8-dimensional algebra.Recall the definition of the product on C ( F ) = M ( F ) ⊕ vM ( F ) . We extend theproduct from M ( F ) on C ( F ) as follows: a ( vb ) = v (¯ ab ) , ( vb ) a = v ( ab ) , ( va )( vb ) = b ¯ a, a, b ∈ M ( F ) . (30)Here ¯ denotes the involution on M ( F ) which maps a matrix a = ( a ij ) ∈ M ( F ) to thematrix (cid:18) a − a − a a (cid:19) .Given an algebra A with the product · , define the operations ◦ , [ , ] on the space A : a ◦ b = a · b + b · a, [ a, b ] = a · b − b · a. We denote the space A endowed with ◦ as A (+) and the space A endowed with [ , ] as A ( − ) . Lemma 12 . Let A be an algebra. Then(a) [8] If R is an RB-operator of weight λ on A , then R is an RB-operator of weight λ on A ( ± ) .(b) rb λ ( A ) ≤ rb λ ( A ± ) . Proof . (b) It follows from the definition of rb ( λ ) -index of an algebra and (a). (cid:3) Proposition 3 . Let F be an algebraically closed field with char ( F ) = 2 , . Thenrb λ ( C ( F )) = 3 for all λ ∈ F . Proof . When λ = 0 , it was proved in [16, Thm. 5.26].Let λ = 0 . It is known [35, p. 57] that C ( F ) (+) is isomorphic to the simple Jordanalgebra of a bilinear form. By Proposition 2 and Lemma 12b, rb λ ( C ( F )) ≤ .Consider a triangular-splitting RB-operator R of weight 1 on C ( F ) with subalgebras A = Span { e , e , ve , ve } , A + = Span { e , ve } , and A − = Span { e , ve } . Fur-ther, R is defined on A again as a triangular-splitting RB-operator with subalgebras16 = Span { e , e } , B + = Span { ve } , and B − = Span { ve } . Finally, R is defined on B ∼ = F ⊕ F as follows: R ( e ) = e , R ( e ) = 0 . Thus, R ( R + id ) = 0 on the entire C ( F ) but R = 0 and R ( R + id ) = 0 . So, rb λ ( C ( F )) = 3 . (cid:3) Let A be an algebra and let R be an RB-operator R of weight λ on A . An A -bimo-dule M with a linear map p : M → M is called a Rota—Baxter bimodule over ( A, R ) andsimply an ( A, R ) -bimodule [21], if the following equalities hold for all x ∈ A and for all m ∈ M , R ( x ) p ( m ) = p ( R ( x ) m + xp ( m ) + λxm ) ,p ( m ) R ( x ) = p ( mR ( x ) + p ( m ) x + λmx ) . (31)We call an ( A, R ) -bimodule ( M, p ) a unital one, if M is a unital A -bimodule.The relations (31) are equivalent to the condition that R + p is an RB-operator ofweight λ on A ⋉ M . Proposition 4 . Let A be a unital algebraic algebra over a field F of characteristiczero and let R be an RB-operator of weight λ on A . Also, ( M, p ) is a unital ( A, R ) -bimo-dule. Thena) If λ = 0 , then p is nilpotent on M .b) If λ = 0 , then there exist s, t such that p s ( p + λ id ) t = 0 on M . So, M = M ⊕ M ,where M = ker( p s ) and M = ker( p + λ id ) t . For k, l such that R k ( R + λ id ) l = 0 define A = ker( R k ) and A = ker( R + λ id ) l . We get that ( M i , p ) is a (not necessary unital) ( A i , R ) -bimodule for i = 0 , .c) rb λ ( A ⋉ M ) ≤ rb λ ( A ) for all λ ∈ F . Proof . a) By Theorem 1a, R + p is nilpotent on A ⋉ M , so p is nilpotent on M .b) By Theorem 1a, there exist n, m such that ( R + p ) n ( R + p + λ id ) m = 0 on A . So, p n ( p + λ id ) m = 0 on M , and M = M ⊕ M for M = ker( p s ) and M = ker(( p + λ id ) t ) .Also, ker( R + p ) n and ker( R + p + λ id ) m are subalgebras of A ⋉ M . It implies that ( M , p ) is an ( A , R ) -bimodule and ( M , p ) is an ( A , R ) -bimodule.c) Suppose that rb λ ( A ) < ∞ . Then there exists a minimal n such that for everyRB-operator R of weight λ on A one can find ≤ k ≤ n satisfying the equality R k ( R + λ id ) n − k (1) = 0 . On the one hand, n ≤ rb λ ( A ) by the definition of the RB ( λ ) -index.On the other hand, since ( R + p ) k ( R + λ id ) n − k (1) = R k ( R + λ id ) n − k (1) = 0 , we getrb λ ( A ) ≤ n by Theorem 1b. (cid:3) Remark 2 . The same result holds if A is a unital finite-dimensional algebra over anyfield (not necessary of characterstic zero). The study of the parameterized associative Yang—Baxter equation was initiated byA. Polishchuk [30] in 2002. Further, the definition of parameterized associative Yang—Baxter equation of weight λ appeared [3, 11]. Given an associative algebra A , a linear17ap r : F → A ⊗ A is called a solution of the parameterized associative Yang—Baxterequation of weight λ if the equation r ( z ) r ( z ) − r ( z ) r ( z ) + r ( z ) r ( z ) + λr ( z ) = 0 (32)holds for all z , z , z ∈ F . Here z ij = z i − z j and given a decomposition r ( z ) = P i a i ⊗ b i , r = X a i ⊗ b i ⊗ , r = X a i ⊗ ⊗ b i , r = X ⊗ a i ⊗ b i . It is known [11] that a solution r of (32) gives rise to a parameterized Rota—Baxteroperator R of weight λ on A . More precisely, a linear map R : F ⊗ A → A is calledparameterized Rota—Baxter operator R of weight λ on A if R satisfies the relation R b ( x ) R c ( y ) = R c ( R b − c ( x ) y ) + R b ( xR c − b ( y )) + λR b ( xy ) , (33)where b, c ∈ F and x, y ∈ A . By R b ( x ) we denote the element R ( b, x ) ∈ A .So, a parameterized RB-operator R on A is a family of linear operators R b : A → A , b ∈ F . Note that R is a usual RB-operator of weight λ on A .Let us apply the results of Theorem 1 to get some information about operators R b . Proposition 5 . Let A be a unital algebraic associative algebra over a field of char-acteristic zero. Let R be a parameterized RB-operator of weight λ on A . Thena) If λ = 0 , then there exists n such that ( R b (1)) n = 0 for all b ∈ F ,b) If λ = − , then there exist n, m such that H n,m ( R b (1)) = 0 for all b ∈ F . Proof . a) By Theorem 1a, there exists n such that R n (1) = 0 . By (33), we get ( R b (1)) n +1 = R b ( R n (1)) = 0 .b) Denote a = R (1) and c = R b (1) . Note that by (33), R b ( x ) R b ( y ) = R b ( R ( x ) y + xR ( y ) − xy ) . So, the right-hand side has the form R b ( Z ) , where Z is a linear combination of theproducts involving only the action of R . The same holds for the product of everynumber of elements R b ( x i ) . This fact allows us analogously to the formula ( R (1)) n = a n = R n X k =1 k ! S ( n, k )( − n − k R k − (1) ! , from [19, §3.3.2] to state the fomula ( R b (1)) n = c n = R b n X k =1 k ! S ( n, k )( − n − k R k − (1) ! . Here S ( n, k ) denotes a Stirling number of the second kind.So, if we have an equility g ( a ) = R ( q ( a )) for some polynomials g, q ∈ F [ x ] , then wemay derive the equality g ( c ) = R b ( q ( a )) . In particular, applying Lemma 10, we get H r,s +1 ( a ) r + s + 1 = R ( H r,s ( a )) , H r,s +1 ( c ) r + s + 1 = R b ( H r,s ( a )) .
18y Theorem 1a and Lemma 11, there exist r, s such that H r,s ( a ) = 0 . Thus, H r,s +1 ( c ) = 0 ,we are done. (cid:3) Suppose that R is a parameterized RB-operator of weight λ on M n ( C ) . So, Proposi-tion 5 says that if λ = 0 , then matrices R b (1) are nilpotent for all b ∈ F . If λ = 0 , thenmatrices R b (1) are diagonalizable for all b ∈ F . Let us generalize the following construction already known for different varieties ofalgebras, see [9, 19, 24].Let A be an algebra of a variety Var and let R be an RB-operator of weight λ on A .Denote by V the vector space A . We define by induction the algebra A i , i ≥ , on V asfollows x ◦ y = xy,x ◦ i +1 y = R ( x ) ◦ i y + x ◦ i R ( y ) + λx ◦ i y. (34)It is known [5, 17] that R is an RB-operator of weight λ on all A i and all algebras A i belong to the same variety Var . By the definition, R and − ( R + λ id ) are homomorphismsfrom A n to A n − , and we get compositions of such homomorphisms, . . . R −−−−→ − R − λ id A i +1 R −−−−→ − R − λ id A i R −−−−→ − R − λ id A i − R −−−−→ − R − λ id · · · R −−−−→ − R − λ id A . Let us call A i as the i th derived Rota—Baxter algebra of the RB-algebra A . If thereexists a minimal natural number m such that A n ∼ = A m for all n > m , then we define thelimiting derived algebra A ∞ as A m . Otherwise, A ∞ is not defined.The following result generalizes Corollary 4.10 from [24] in the case of weight zero. Proposition 6 . Let A be a unital algebraic algebra over a field F of characteristiczero. Let R be an RB-operator of weight λ on A . Then A ∞ is well-defined.a) If λ = 0 , then A ∞ has trivial product.b) If λ = 0 , then A ∞ ∼ = ker( R n ) ⊕ ker( R + λ id ) n for sufficiently big n . Proof . a) By Theorem 1, we know that R n = 0 for some natural n . By (34), we get x ◦ k y = k X i =0 (cid:18) ki (cid:19) R i ( x ) R k − i ( y ) . (35)Thus, x ◦ i y = 0 for all i ≥ n − .b) By Theorem 1, there exists n such that R n ( R + λ id ) n = 0 . So, ker( R n ) and ker( R + λ id ) n are ideals in A t for all t ≥ n as kernels of the homomorphisms. So, A t = ker( R n ) ⊕ ker( R + λ id ) n for all t ≥ n . It remains to note that R is invertible on ker( R + λ id ) n as well as R + λ id on ker( R n ) . Thus, h ker( R n ) , ◦ s i ∼ = h ker( R n ) , ◦ n i , h ker( R + λ id ) n , ◦ s i ∼ = h ker( R + λ id ) n , ◦ n i for all s ≥ . It implies that A t ∼ = A n for all t ≥ n . (cid:3) Remark 3 . The same result holds if A is a unital finite-dimensional algebra over anyfield (not necessary of characterstic zero). Question . Let A = M n ( F ) . What is a minimal m such that A m has trivial productfor every RB-operator R of weight zero on A ?19 cknowledgements Author is grateful to the participants of the scientific seminar in Altai State Pedagog-ical University (Barnaul) for the helpful discussion. Author is thankful to P. Kolesnikovfor the helpful remarks.Author is supported by the Program of fundamental scientific researches of theSiberian Branch of Russian Academy of Sciences, I.1.1, project 0314-2019-0001.
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