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Given an algebra A and a scalar λ F , where F is a ground field, a linear operator R: A A is called a RotaBaxter operator∈ (RB-operator, for short) of weight λ if the → identity R(x)R(y)= R(R(x)y + xR(y)+ λxy) (1) holds for all x, y A. An algebra A equipped with a RotaBaxter operator is called ∈ a RotaBaxter algebra (RB-algebra). The notion of a RotaBaxter operator was introduced by G. Baxter [10] in 1960 as for- mal generalization of integration by parts formula and then developed by G.-C. Rota [58] and others [8, 17]. In 1980s, the deep connection between constant solutions of the classical YangBaxter equation from mathematical physics and RB-operators of weight zero on a semisimple finite-dimensional Lie algebra was discovered [12]. In 2000, M. Aguiar stated [3] that solutions of associative YangBaxter equation (AYBE, [73]) and RB-operators of weight zero on any associative algebra are related. In 2000, the connection between RB-algebras and prealgebras was found [3, 24]. Later, this connection was extended and studied for postalgebras, see, e.g., [9, 30, 32]. To the moment, applications of RotaBaxter operators in symmetric polynomials, quantum field renormalization, shuffle algebra, etc. were found [8, 19, 32, 33, 54]. Also, RB-operators have been studied by their own interest. RB-operators were clas- sified on sl2(C) [44, 45, 55, 56], M2(C) [13, 67], sl3(C) [45], the Grassmann algebra Gr2 [13, 39], the 3-dimensional simple Jordan algebra of bilinear form, the Kaplansky superalgebra K3 [13], 3-dimensional solvable Lie algebras [45, 66, 69], low-dimensional Lie superalgebras [42, 50, 60, 61], low-dimensional pre-Lie (super)algebras [1, 47], low- dimensional semigroup algebras [36]. The classification of RB-operators of special kind on polynomials, power series and Witt algebra was found [22, 35, 37, 48, 65, 70]. Author proceeds on the study of RotaBaxter operators initiated in [13, 28]. Let us give a brief outline of the work. In §2 the required preliminaries are stated. In §3 it is proved that Aguiar’s correspondence between the solutions of AYBE and RB-operators of weight zero on the matrix algebra Mn(F ) is bijective (Theorem 3.4, joint result with P. Kolesnikov). Further, in §4, we consider RB-operators of nonzero weight. There is a big difference between RB-operators of nonzero weight and RB-operators of weight zero. There are few known general constructions for the first ones such as splitting and triangular-splitting RB-operators. In contrast, there are a lot of examples for the second ones, and it is not clear which of them are of most interest. Thus, we start exposition and study of RB- operators with the nonzero case. In [13], it was proved that any RB-operator of nonzero weight on an odd-dimensional simple Jordan algebra J of bilinear form is splitting, i.e., it is a projection on a subalgebra along another one provided that J splits into a direct vector-space sum of these two subalgebras. Given an algebra A, an RB-operator R on A of weight λ, and ψ Aut(A), the operator R(ψ) = ψ−1Rψ is an RB-operator of weight λ on A. Thus, it is∈ reasonable to classify RB-operators of any weight on an algebra only up to conjugation with automorphisms.

2 Since the map φ defined as φ(R) = (R + λid) preserves the set of all RB-operators of the given weight λ, we study RB-operators− of nonzero weight up to the action of φ. The combinatorics arisen from the substitution of unit into the RB-identity implies Theorem A (= Theorem 4.8). Let A be a unital power-associative algebra over a field of characteristic zero and A = F 1 N (as vector spaces), where N is a nil-algebra. ⊕ Then each RB-operator R of nonzero weight on A is splitting and up to φ we have R(1) = 0. By Theorem A, we get that all RB-operators of nonzero weight on the Grassmann algebra are splitting (Corollary 4.10, §4.4). The second main result in the case of nonzero weight gives us a powerful tool to study RB-operators of nonzero weight on the matrix algebra. Theorem B (= Theorem 4.17). Given an algebraically closed field F of characteristic zero and an RB-operator R of nonzero weight λ on Mn(F ), there exists a ψ Aut(Mn(F )) such that the matrix R(ψ)(1) is diagonal and up to φ the set of its diagonal∈ elements has a form pλ, ( p + 1)λ, . . . , λ, 0,λ,...,qλ with p, q N. {− − − } ∈ As a corollary, we show that an RB-operator on M3(F ) up to conjugation with an automorphism preserves the subalgebra of diagonal matrices (Corollary 4.20, §4.5). In §5, we study RB-operators of weight zero. In [47], the following question is raised: “Whether we can give a meaningful “classification rules” so that the classification of Rota- Baxter operators can be more “interesting”? In Lemma 5.1 (§5.1), we present, maybe, the first systematic attempt to study possible constructions of RB-operators of weight zero. For example, Lemma (= Lemma 5.1(a)). Given an algebra A, a linear operator R on A is an RB-operator of weight zero, if A = B C (as vector spaces) with Im R having trivial multiplication, R: B C, R: C (0),⊕BR(B), R(B)B C. This construction→ works for any→ variety and any dimension⊆ of algebras. Let us list some examples of RB-operators of weight zero arisen from Lemma directly:

[4, 43] Let A be an associative algebra and e Z(A) be an element such that • ∈ e2 =0. A linear map l : x ex is an RB-operator of weight 0. e → In [30], given a variety Var of algebras and a pre-Var-algebra A, an enveloping • RB-algebra B of weight zero and the variety Var for A was constructed. By the construction, B = A A′ (as vector spaces), where A′ is a copy of A as a vector space, and the RB-operator⊕ R was defined as follows: R(a′)= a, R(a)=0, a A. ∈ In [22], all homogeneous RB-operators on the Witt algebra W = Span L n Z • { n | ∈ } over C with the Lie product [Lm, Ln]=(m n)Lm+n were described. A homoge- neous RB-operator with degree k Z satisfies− the condition R(L ) Span L ∈ m ∈ { m+k} for all m Z. One of the four homogeneous RB-operators on W , ∈ (W2) R(L )= δ L , k =0, m Z, m m+2k,0 m+k 6 ∈ is defined by Lemma.

[13] All RB-operators of weight zero on the Grassmann algebra Gr2 over a field F • with char F =2 are defined by Lemma. 6

3 [4, 13, 67] There are exactly four nonzero RB-operators of weight zero on M (F ) • 2 over an algebraically closed field F up to conjugation with automorphisms M2(F ), transpose and multiplication on a nonzero scalar. One of them,

(M1) R(e21)= e12, R(e11)= R(e12)= R(e22)=0 is defined by Lemma.

As we know, we cover in Lemma 5.1 most of currently known examples of RB- operators of weight zero. In §5.2, we find the general property of RB-operators of weight zero on unital algebras. Theorem C (= Theorem 5.5). Let A be a unital associative (alternative, Jordan) algebraic algebra over a field of characteristic zero. Then there exists N such that RN =0 for every RB-operator R of weight zero on A. Let us compare the obtained results in both zero and nonzero cases. Theorems A and B say that the nature of an RB-operator of nonzero weight on a unital algebra is projective-like, it acts identically on a some subalgebra. To the contrary, Theorem C says that an RB-operator of weight zero on a unital algebra is nilpotent. After Theorem C, we define a new invariant of an algebra A, namely, RotaBaxter index (RB-index) rb(A) as the minimal natural number with such nilpotency property. We prove that rb(M (F )) = 2n 1 over a field F of characteristic zero (§5.4) and n − study RB-index for the Grassmann algebra (§5.3), unital composition algebras and simple Jordan algebras (§5.5). Let us clarify the title of the paper. All three main results (Theorems A, B, and C) come from the combinatorics arisen from multiple substitutions of the unit into the RB-identity. So, these results are not valid when an algebra is not unital. Also, we study typical constructions of RB-operators on unital algebras. For completeness, we add few examples of Lie algebras such that RB-operators on them are arisen from the same constructions. The proofs essentially involve the results on maximal subspaces and subalgebras of special kind in the matrix algebra [2, 51, 68, 71] as well as the structure theory of asso- ciative and nonassociative algebras.

2 Preliminaries

In §2, we give some known examples and state useful properties of RotaBaxter operators. Let F stand for the ground field. By algebra we mean a vector space endowed with a bilinear (but not necessary associative) product. All algebras are considered over F . Trivial RB-operators of weight λ are zero operator and λid. − Statement 2.1 ([32]). Given an RB-operator P of weight λ, (a) the operator P λid is an RB-operator of weight λ, (b) the operator λ−−1P−is an RB-operator of weight 1, provided λ =0. 6

4 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 clear − − that φ2 coincides with the identity map. Statement 2.2 ([13]). Given an algebra A, an RB-operator P on A of weight λ, and ψ Aut(A), the operator P (ψ) = ψ−1P 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. ∈ Statement 2.3 ([32]). Let an algebra A split as a vector space into a direct sum of two subalgebras A1 and A2. An operator P defined as P (a + a )= λa , a A , a A , (2) 1 2 − 2 1 ∈ 1 2 ∈ 2 is RB-operator of weight λ on A. Let us call an RB-operator from Statement 2.3 as splitting RB-operator with subalge- bras A1, A2. Note that the set of all splitting RB-operators on an algebra A is in bijection with all decompositions of A into a direct sum of two subalgebras A1, A2. Remark 2.1. Given an algebra A, let P be a splitting RB-operator of weight λ on A with subalgebras A1, A2. Hence, φ(P ) is an RB-operator of weight λ and φ(P )(a + a )= λa , a A , a A . 1 2 − 1 1 ∈ 1 2 ∈ 2

So φ(P ) is splitting RB-operator with the same subalgebras A1, A2. Example 2.2. (a) [4, 43] Let A be an associative algebra and e A an element such that e2 = λe, λ F . A linear map l : x ex is an RB-operator∈ of weight λ satisfying − ∈ e → l2 + λl = 0. For λ = 0, l is a splitting RB-operator on A with the subalgebras e e 6 e A1 = (1 e)A and A2 = eA, the decomposition A = A1 A2 (as vector spaces) is the Pierce− one. ⊕ (b) [13] In an alternative algebra A with an element e such that e2 = λe, λ F , the − ∈ operator le is an RB-operator on A if e lies in associative or commutative center of A. Example 2.3. In [9], it was proved that every RB-algebra of nonzero weight λ and a variety Var under the operations x y = R(x)y, x y = xR(y), x y = λxy is a post-Var-algebra. In [30], given a post-≻Var-algebra A, its≺ enveloping RB-algebra· B of weight λ and the variety Var was constructed. By the construction, B = A A′ (as ⊕ vector spaces), where A′ is a copy of A as of a vector space, and the RB-operator R was defined as follows: R(a′) = λa, R(a) = λa, a A. Note that R is the splitting − ∈ RB-operator on B with A = Span a + a′ a A and A = A. 1 { | ∈ } 2 Example 2.4. Due to [70], all nontrivial monomial RB-operators of nonzero weight on F [x] (i.e., mapping each monomial to a monomial) up to the action of φ and Aut(F [x]) are splitting with subalgebras F and Id x . h i Example 2.5. By [6], all RB-operators of nonzero weight on the associative algebra A = Span e ,...,e e e = e , e e =0, i =1,...,n,j =2,...,n are splitting. { 1 n | 1 i i j i } 5 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(−). ◦ Statement 2.4 ([13]). Let A be an algebra. (a) If R is an RB-operator of weight λ on A, then R is an RB-operator of weight λ on A(±). (b) If all RB-operators of nonzero weight on A(+) (or A(−)) are splitting, then all RB-operators of nonzero weight on A are splitting.

Lemma 2.5 ([13, 32]). Let A be a unital algebra, P be an RB-operator of weight λ on A. (a) If λ =0 and P (1) F , then P is splitting with one of subalgebras being unital; (b) if λ 6 = 0, then 1 ∈ Im P . Moreover, if A is simple finite-dimensional algebra, dim A> 1, then dim(ker P6∈) 2; ≥ (c) if λ =0 and P (1) F , then P (1) = 0, P 2 =0, and Im P ker P ; (d) if λ =0 and A is a∈ power-associative algebra, then (P (1))⊂n = n!P n(1), n N. ∈

Lemma 2.6. Let an algebra A be equal to a direct sum of two ideals A1, A2, and let R be an RB-operator of weight λ on A. Then PriR is the RB-operator of weight λ on Ai, i =1, 2. Here Pri denotes the projection from A onto Ai. Proof. Straightforward.

Theorem 2.7 ([13]). All RB-operators on a quadratic division algebra are trivial.

For RB-operators of weight zero, the last result may be generalized as follows.

Theorem 2.8. If R is an RB-operator of weight zero on an algebraic power-associative algebra A without zero divisors, then R =0.

Proof. Suppose R(x) =0 for some x A. As A is algebraic, consider the equality 6 ∈ (R(x))m + α (R(x))m−1 + ... + α R(x)+ α =0, α F, m−1 1 0 i ∈ of minimal degree m 1. By Lemma 2.5(b), α0 = 0. Thus, R(x)y = 0 for y = (R(x))m−1 + α (R(x≥))m−2 + ... + α . As R(x) = 0, we have y = 0 and m is not m−1 1 6 minimal.

3 YangBaxter equation

In §3, we consider connections of RotaBaxter operators with different versions of YangBaxter equation. In Theorem 3.4, we state the bijection between the solutions of associative YangBaxter equation and RB-operators of weight zero on the matrix algebra Mn(F ) (joint with P. Kolesnikov).

6 3.1 Classical YangBaxter equation

Let L be a semisimple finite-dimensional Lie algebra over C. For r = ai bi L L, introduce classical YangBaxter equation (CYBE, [12]) as ⊗ ∈ ⊗ P

[r12,r13] + [r12,r23] + [r13,r23]=0, (3) where r = a b 1, r = a 1 b , r = 1 a b 12 i ⊗ i ⊗ 13 i ⊗ ⊗ i 23 ⊗ i ⊗ i are elements fromXU(L)⊗3. X X The switch map τ : L L L L acts in the following way: τ(a b)= b a. The solution r of CYBE is called⊗ → skew-symmetric⊗ if r + τ(r)=0. A linear⊗ map−R⊗: L L → defined by a skew-symmetric solution r of CYBE as

R(x)= a , x b (4) h i i i X is an RB-operator of weight zero on L [12]. Here , denotes the Killing form on L. h· ·i Example 3.1. There exists unique (up to conjugation and scalar multiple) nonzero skew-symmetric solution of CYBE on sl2(C): e h h e [64]. It corresponds to the RB-operator R(e)=0, R(f)=4h, R(h)= 8e. ⊗ − ⊗ − Given a finite-dimensional semisimple Lie algebra L, let us call a linear map P a skew- symmetric one if so is its matrix in the orthonormal basis (with respect to the Killing form).

Statement 3.1 ([32, 56]). Given a finite-dimensional semisimple Lie algebra L, skew- symmetric solutions of CYBE on L are in one-to-one correspondence with skew- symmetric RotaBaxter operators of weight zero on L.

An element r L⊗n, n N, is called L-invariant if [r, y]=0 for all y L. Here L ∈ ∈ n ∈ ⊗n acts on L by the formula [x1 ... xn,y]= x1 ... [xi,y] ... xn, xi,y L. ⊗ ⊗ i=1 ⊗ ⊗ ⊗ ⊗ ∈ P Theorem 3.2 ([26]). Let L be a simple finite-dimensional Lie algebra over C and r = ai bi be a non skew-symmetric solution of CYBE such that r + τ(r) is L-invariant. i ⊗ PThen the linear map R: L L defined by (4) is an RB-operator of nonzero weight. →

Example 3.2 ([26]). Let L = sl2(C) with the Chevalley basis e, f, h. Consider an element 1 r = α(h e e h)+ h h + e f L L, α C. ⊗ − ⊗ 4 ⊗ ⊗ ∈ ⊗ ∈ For any α C, the tensor r is the non skew-symmetric solution of CYBE and r + τ(r) ∈ is L-invariant. Due to Theorem 3.2, we get the RB-operator R on sl2(C) defined by r as follows: R(e)=0, R(h)=2h +8αe, R(f)=4(f αh). The weight of R equals 4. − −

7 3.2 Modified YangBaxter equation In 1983 [59], Semenov-Tyan-Shansky introduced modified YangBaxter equation (MYBE)1 as follows: let L be a Lie algebra, R a linear map on L, then

R(x)R(y) R(R(x)y + xR(y)) = xy. (5) − − It is easy to check that R is a solution of MYBE if and only if R+id is an RB-operator of weight 2. So, there is one-to-one correspondence (up to scalar multiple and action − of φ) between the set of solutions of MYBE and RB-operators of nonzero weight. In [59], the general approach for solving MYBE on a simple finite-dimensional Lie algebra L over C was developed. Applying this method, all solutions of MYBE (i.e., all RB-operators of nonzero weight) on sl2(C) and sl3(C) were found in [45].

3.3 Associative YangBaxter equation Let A be an associative algebra, r = a b A A. The tensor r is a solution of i ⊗ i ∈ ⊗ associative YangBaxter equation (AYBE, [4, 5, 57, 73]) if P r r r r + r r =0, (6) 13 12 − 12 23 23 13 where the definition of r12,r13,r23 is the same as for CYBE. A solution r of AYBE on an algebra A is a solution of CYBE on A(−) provided that r+τ(r) is A-invariant [5]. A tensor u v A A is said to be A-invariant if au v = u va for all a A. In particular, each skew-symmetric⊗ ∈ ⊗ solution of AYBE is a skew-symmetric⊗ ⊗ solution∈ of CYBE on A(−).

Statement 3.3 ([3]). Let r = a b be a solution of AYBE on an associative algebra A. i⊗ i A linear map Pr : A A defined as → P

Pr(x)= aixbi (7) X is an RB-operator of weight zero on A.

Example 3.3 ([4]). Up to conjugation, transpose and scalar multiple, all nonzero solu- tions of AYBE on M (C) are (e + e ) e ; e e ; e e ; e e e e . 2 11 22 ⊗ 12 12 ⊗ 12 22 ⊗ 12 11 ⊗ 12 − 12 ⊗ 11

All RB-operators arisen by Statement 3.3 from the solutions of AYBE on M2(C) are exactly all RB-operators of weight zero on M2(C) (see Theorem 5.18, §5.4). Let us generalize this fact on each matrix algebra Mn(F ), this result is a joint one with P. Kolesnikov.

Theorem 3.4. The map r P is the bijection between the set of the solutions of AYBE → r on Mn(F ) and the set of RB-operators of weight zero on Mn(F ). 1There is at least one another version of YangBaxter equation defined by M. Gerstenhaber [23] which is also called modified YangBaxter equation.

8 ql Proof. Given a linear operator R on Mn(F ), denote R(epq) = tipeil. Then the equa- i,l tion (1) for R could be rewritten in the form P

(tbjtdl tbc tdl tdjtjl )=0. (8) ia jc − ja ij − bc ia j X kl Let r = sij eij ekl be a solution of AYBE. So, i,j,k,l ⊗ P r r = sklsste e e , 13 12 ij pi pj ⊗ kl ⊗ st p X r r = sklsste e e , 12 23 ij lp ij ⊗ kp ⊗ st (9) p X r r = sklslt e e e . 23 13 ij pr pr ⊗ ij ⊗ kt p X

By substituting the summands from (9) into (6) and gathering them on the tensor eij e e , we get the equality ⊗ kl ⊗ st (skl sst skpsst + sspspt)=0. (10) pj ip − ij pl kl ij p X By the interchange of variables, the equations (8) and (10) coincide. Thus, the map χ from the set of the solutions of AYBE on Mn(F ) to the set of RB-operators of weight kl ql zero on Mn(F ) acting as χ(r)= Tr, where r = sij eij ekl and Tr(epq)= sipeil, is i,j,k,l ⊗ i,l the bijection. It remains to note that χ is exactlyP the map r Pr. P → Theorem 3.5 ([63]). Up to conjugation, transpose and scalar multiple all nonzero skew- symmetric solutions of AYBE on M3(C) are (A1) e e e e , 32 ⊗ 31 − 31 ⊗ 32 (A2) e11 e12 e12 e11 + e13 (e12 e21) (e12 e21) e13 +(e11 + e22) e23 e (e + e⊗ ), − ⊗ ⊗ − − − ⊗ ⊗ − 23 ⊗ 11 22 (A3) e22 e23 e23 e22, (A4) e ⊗ (e − e ⊗) (e e ) e +(e + e ) e e (e + e ), 13 ⊗ 12 − 21 − 12 − 21 ⊗ 13 11 22 ⊗ 23 − 23 ⊗ 11 22 (A5) (e + e ) e e (e + e )+ e e e e , 11 22 ⊗ 23 − 23 ⊗ 11 22 21 ⊗ 13 − 13 ⊗ 21 (A6) (e11 + e33) e23 e23 (e11 + e33)+ e11 e13 e13 e11, (A7) e e ⊗e −e +⊗e e e e⊗, − ⊗ 13 ⊗ 21 − 21 ⊗ 13 33 ⊗ 23 − 23 ⊗ 33 (A8) (e + e ) e e (e + e ). 11 33 ⊗ 23 − 23 ⊗ 11 33 We call an RB-operator R on M (F ) a skew-symmetric one if R∗ = R, where R∗ is n − the conjugate operator relative to the trace form.

Corollary 3.6. Up to conjugation, transpose and scalar multiple all nonzero skew-sym- metric RB-operators on M3(C) are (R1) R(e )= e , R(e )= e ; 31 23 32 − 13

9 (R2) R(e11) = e21 e32, R(e12) = e11 + e31, R(e13) = e12 e21, R(e21) = e31, R(e )= e , R(e−)=−e + e ; − − 22 − 32 23 11 22 (R3) R(e )= e , R(e )= e ; 23 22 22 − 32 (R4) R(e13) = e12 e21, R(e12) = R(e21) = e31, R(e23) = e11 + e22, R(e11) = R(e )= e ; − − 22 − 32 (R5) R(e13)= e12, R(e21)= e31, R(e23)= e11 + e22, R(e11)= R(e22)= e32; (R6) R(e )= e , R(e )= −e , R(e )= e + e , R(e )= R(e )= −e ; 33 32 23 − 33 13 11 12 11 21 − 31 (R7) R(e23)= e11 e33, R(e11)= R(e33)= e32; (R8) R(e )= e− , R−(e )= e , R(e )= e , R(e )= e . 13 12 21 − 31 33 32 23 − 33 In 2006, K. Ebrahimi-Fard defined in his Thesis [20, p. 113] the associative Yang Baxter equation of weight λ. Later, this equation was twice rediscovered: in 2010 in [54] and in 2018 in [72]. Given an associative algebra A and a tensor r A A, we say that r is a solution of associative YangBaxter equation of weight λ (wAYBE)∈ ⊗ if r r r r + r r = λr . (11) 13 12 − 12 23 23 13 13 In [20, 72], it was shown that the solutions of wAYBE generate ǫ-unitary bialgebras. Moreover, the following analogues of Statement 3.3 and Theorem 3.4 hold, the last one was stated as the generalization of Theorem 3.4.

Statement 3.7 ([7, 20, 72]). Let r = ai bi be a solution of AYBE of weight λ on an associative algebra A. A linear map P : A⊗ A defined by (7) is an RB-operator of r → weight λ on A. P − Theorem 3.8 ([72]). The map r P is the bijection between the set of the solutions → r of AYBE of weight λ on Mn(F ) and the set of RB-operators of weight λ on Mn(F ). The analogues of classical or associative YangBaxter equations for alternative and Jordan algebras were defined in [25, 73]. The connection between solutions of the Yang Baxter equation and RB-operators on the CayleyDickson algebra C(F ) was found in [13].

4 RotaBaxter operators of nonzero weight

In §4, first, we give a list of algebras whose all RB-operators of nonzero weight are either splitting (odd-dimensional simple Jordan algebras of bilinear form, the simple Jordan Kaplansky superalgebra K3, §4.2) or triangular-splitting (sl2(C), the Witt algebra (for homogeneous RB-operators), §4.1). Second, we prove that all RB-operators of nonzero weight on a unital power- associative algebra A over a field of characteristic zero are splitting provided that A = F 1 N (as vector spaces), where N is a nil-algebra (Theorem 4.8). In Corol- lary 4.10 (§4.4),⊕ we get that all RB-operators of nonzero weight on the Grassmann algebra are splitting. Third, we state the main result about RB-operators of nonzero weight on the ma- trix algebra. Over an algebraically closed field F of characteristic zero, given an RB- operator R of nonzero weight on M (F ), there exists ψ Aut(M (F )) such that R(ψ)(1) n ∈ n 10 is diagonal (Theorem 4.17, §4.5). As a corollary, we show that any RB-operator on M3(F ) up to conjugation with an automorphism preserves the subalgebra of diagonal matrices (Corollary 4.20, §4.5).

4.1 Triangular-splitting RB-operators Given an algebra A and a vector space B, we say that B is a A-module, if the action of A on B is defined, i.e., ab, ba B for all a A, b B. Consider a construction of RB-operators∈ of∈ nonzero∈ weight which generalizes State- ment 2.3.

Statement 4.1 ([31]). Let an algebra A be a direct sum of subspaces A−, A0, A+, more- over, A±, A0 are subalgebras of A, and A± are A0-modules. If R0 is an RB-operator of weight λ on A0, then an operator P defined as

P (a + a + a )= R (a ) λa , a A , a A , (12) − 0 + 0 0 − + ± ∈ ± 0 ∈ 0 is an RB-operator of weight λ on A. Let us call an RB-operator of nonzero weight defined by (12) as triangular-splitting one provided that at least one of A−, A+ is nonzero. If A0 = (0), then P is splitting RB-operator on A. If A0 has trivial product, then every linear map on A0 is suitable as R0. Remark 4.1. Let P be a triangular-splitting RB-operator on an algebra A with subal- gebras A±, A0. Then the operator φ(P ) is the triangular-splitting RB-operator with the same subalgebras,

φ(P )(a + a + a )= λa + φ(R )(a ), a A , a A . − 0 + − − 0 0 ± ∈ ± 0 ∈ 0

Example 4.2. In [22], all homogeneous RB-operators on the Witt algebra W =Span Ln n Z over C with the Lie product [L , L ]=(m n)L were described. A homo-{ | ∈ } m n − m+n geneous RB-operator with degree k Z satisfies the condition R(Lm) Span Lm+k for all m Z. Due to [22], all nonzero∈ homogeneous RB-operators of weigh∈ t 1 on{ W up} to ∈ the action φ and conjugation with automorphisms of W are the following: (WT1) R(Lm)=0, m 1, R(Lm)= Lm, m 2; (WT2) R(L )=0, m ≥ −1, R(L )= −L , m ≤1 −, R(L )= kL for some k C. m ≥ m − m ≤ − 0 0 ∈

The RB-operators (WT1) and (WT2) are triangular-splitting with abelian A0 = L0. Let us consider the simple Lie algebra sl2(C) with the Chevalley basis e, f, h.

Theorem 4.2 ([45, 55]). All nontrivial RB-operators of nonzero weight on sl2(C) up to conjugation with an automorphism are the following: (a) the splitting RB-operator with A1 = Span e + αh , A2 = Span h, f , α =0; (b) the triangular-splitting RB-operator with subalgebra{ } s A =Span{e , A} =Span6 f − { } + { } and A = Span h . 0 { } It is easy to calculate that Example 3.2 is a particular case of (b) (up to conjugation).

11 4.2 The simple Jordan algebra of a bilinear form

Let J = Jn+1(f)= F 1 V be a direct vector-space sum of F and finite-dimensional vector space V , dim V = n>⊕ 1, and f be a nondegenerate symmetric bilinear form on V . Under the product

(α 1+ a)(β 1+ b)=(αβ + f(a, b)) 1+(αb + βa), α,β F, a, b V, (13) · · · ∈ ∈ the space J is a simple Jordan algebra [74]. Theorem 4.3 ([13]). Let J be an odd-dimensional simple Jordan algebra of bilinear form over a field of characteristic not two. Each RB-operator R of nonzero weight on A is splitting and R(1) = 0 up to φ.

Let us choose a basis e1, e2, ..., en of V such that the matrix of the form f in this basis is diagonal with elements d1,d2,...,dn on the main diagonal. As f is nondegenerate, d =0 for each i. i 6 Example 4.3 ([13]). Let J2n(f), n 2, be the simple Jordan algebra of bilinear form f over an algebraically closed field of≥ characteristic not two. Let R be a linear operator 2n−1 on J2n(f) defined by a matrix (rij)i,j=0 in the basis 1, e1, e2,...,en with the following nonzero entries 1 r00 = 3, r01 = d1, r10 = , rjj = 1, j =1,..., 2n 1, − −√d1 − − p di+1 di di rii+1 = , ri+1 i = , i =2,..., 2n 2. di s−di+1 −s−di+1 −

Then R is a non-splitting RB-operator of weight 2 on J2n(f). The analogue of Theorem 4.3 holds for the simple 3-dimensional Jordan superalgebra K3 which is defined over a field of characteristic not two as follows: K3 = A0 A1, A = Span e (even part), A = Span x, y (odd part), ⊕ 0 { } 1 { } x y e e2 = e, ex = xe = , ey = ye = , xy = yx = , x2 = y2 =0. 2 2 − 2 Theorem 4.4 ([13]). All RB-operators of nonzero weight on the simple Jordan Kaplansky superalgebra K3 are splitting.

4.3 Sum of fields

Proposition 4.5 ([6, 15]). Let A = F e1 F e2 ... F en be the direct sum of copies of ⊕ ⊕ ⊕n a field F with eiej = δijei. A linear operator R(ei)= rikek, rik F , is an RB-operator k=1 ∈ of weight 1 on A if and only if the following conditionsP are satisfied: (SF1) rii =0 and rik 0, 1 or rii = 1 and rik 0, 1 for all k = i; (SF2) if r = r =0∈{for i =}k, then r−r =0 for∈{ all l − }i, k ; 6 ik ki 6 il kl 6∈ { } (SF3) if r =0 for i = k, then r =0 and r =0 or r = r for all l i, k . ik 6 6 ki kl il ik 6∈ { } 12 Proof. The RB-identity (1) is equivalent to the equalities

r (1+2r r )=0, r r + r r = r r , i = k, kl kk − kl ik kl ki il il kl 6 from which Proposition 4.5 follows. Example 4.4 ([8, 53]). The following operator is an RB-operator of weight 1 on A:

s n R(e )= e , 1 i

Remark 4.5. From (SF2) and (SF3) it easily follows that rikrki =0 for all i = k. In [6], the statement of Proposition 4.5 was formulated with this equality and (SF1) but6 without (SF2) and (SF3). That is why the formulation in [6] seems to be not complete.

Remark 4.6. The sum of fields in Proposition 4.5 can be infinite. More about RB-operators of nonzero weight on a finite direct sum of copies of a field, including counting of all of them, splitting ones, etc. see in [29].

4.4 Grassmann algebra Lemma 4.6. Let A be a unital power-associative algebra over a field of characteristic zero and let R be an RB-operator of nonzero weight on A. If R(1) is nilpotent, then R(1) = 0 and R is splitting.

Proof. The following formulas hold in all associative RB-algebras of weight λ [32]:

n n!Rn(1) = ( 1)n−kλn−ks(n, k)(R(1))k, (14) − k=1 X n (R(1))n = k!λn−kS(n, k)Rk(1), (15) Xk=1 where s(n, k) and S(n, k) are Stirling numbers of the first and second kind respectively. The proof of the formulas (14), (15) for power-associative algebras is absolutely the same as for associative ones (see, e.g., [32, Thm. 3.1.1]). Suppose that t is maximal nonzero power of R(1). If t = 0, then we are done by Lemma 2.5(a). Suppose that t 1. Then all elements R(1), (R(1))2, ..., (R(1))t are linearly independent. By (14)≥ and the properties of Stirling numbers, elements R(1), R2(1),..., Rt(1) are also linearly independent. The number S(n, k) equals the number of ways of partitioning a set of n elements into k non-empty subsets. It is well-known that

n n n S(n, n)=1, S(n, n 1) = , S(n, n 2) = +3 . (16) − 2 − 3 4       13 Without loss of generality, assume that λ =1. We apply (16) to write down (15) for t +1 and t +2:

t +1 0=(t + 1)!Rt+1(1) + t! Rt(1) 2   t +1 t +1 +(t 1)! +3 Rt−1(1) + ..., (17) − 3 4    

t +2 0=(t + 2)!Rt+2(1)+(t + 1)! Rt+1(1) 2   t +2 t +2 + t! +3 Rt(1) + . . .. (18) 3 4     Act (t + 2)R on (17) and subtract the result from (18) to get the equality

t +2 t +2 t +2 0= t! Rt+1(1)+(t 1)! +6 Rt(1) + ... =0. (19) 2 − 3 4       Let us multiply (17) by (t + 2)/2 and subtract it from (19):

1 t +2 t−1 0= Rt(1) + α Ri(1), α F, −2 3 i i ∈ i=1   X a contradiction with linear independence of the elements R(1),...,Rt(1).

Corollary 4.7. Let A be a unital power-associative algebra over a field of characteristic zero. Any RB-operator R on A of nonzero weight such that Im (R) is a nil-algebra is splitting.

Now, we are ready to state the first of the three main results of the paper. We find sufficient conditions under which a unital algebra has only splitting RB-operators.

Theorem 4.8. Let A be a unital power-associative algebra over a field of characteristic zero and A = F 1 N (as vector spaces), where N is a nil-algebra. Then each RB- ⊕ operator R on A of nonzero weight is splitting and (up to φ) we have R(1) = 0.

Proof. Let R be an RB-operator of nonzero weight λ on A. Suppose there exists x A such that R(x)=1 and let λ = 1. We have ∈ − 1= R(x)R(x)=2R(R(x)) R(x2)=2 R(x2), (20) − − therefore, R(x2)=1. Analogously, R(xk)=1 for all k N. Hence, x = α 1+ a for some α F ∗. From x x2 ker R, we have α(1 α)1∈ + a(1 2α a) · ker R. If α = α(1∈ α)=0, then −α =1∈and ker R contains a +−a2, a + a2 (−a + a2−)2 = ∈a 2a3 a4 1 − − − − and so on. Thus, a ker R and R(x)= αR(1) = 1. By Lemma 2.5(a), we are done. ∈

14 Suppose that α =0. Consider 1 6 x x3 = α(1 α2) 1+ a(1 3α2 3αa a2) ker R − − · − − − ∈ 2 and denote α2 = α(1 α ). By the same reasons as above, we can consider only the case α =0. From − 2 6 (x x3) (1 + α)(x x2)= α a + (1 2α)a2 a3 ker R, − − − 1 − − ∈ we analogously obtain a ker R. ∈ Let us prove that R(A) N. Suppose there exists x A such that R(x)=1+ a for a N. As Im (R) is a subalgebra,⊆ 1+2a + a2 Im (R∈) and hence a + a2 Im (R). ∈ ∈ ∈ It remains to repeat the above arguments to deduce that a Im (R). So, 1 Im (R), a contradiction. ∈ ∈ Applying Lemma 4.6, we obtain that R is splitting.

Corollary 4.9. Let A be a unital power-associative algebra with no idempotents except 0 and 1. Then λid is the only invertible RB-operator of nonzero weight λ on A. − Proof. Suppose that R is invertible RB-operator of weight 1 on A (we rescale by State- − ment 2.1(b). For x A such that R(x)=1, we have by (20) that x is an idempotent in A. Thus, x =1 and∈ R is splitting by Lemma 2.5(a). As R is invertible, R = id. Let Gr (Gr ) denote the Grassmann algebra of a vector space V = Span e ,...,e n ∞ { 1 n} (V = Span e i N ). { i | ∈ +} In [13], it was proved that all RB-operators of nonzero weight on Gr2 are splitting. Now we extend this result as follows.

Corollary 4.10. Given an RB-operator R of nonzero weight on Gr∞ or Grn, we have R is splitting and R(1) = 0 up to φ.

Corollary 4.11. Given the polynomial algebra F [x1,...,xn], let A be its quotient by m1 mn N the ideal I = x1 ,...,xn , mi . Each RB-operator R of nonzero weight on A is splitting and Rh(1) = 0 up toi φ. ∈

4.5 Matrix algebra We need the following bounds on the dimensions of ker(R) and Im (R) for an RB- operator R on Mn(F ). Lemma 4.12. Let F be either an algebraically closed field or a field of characteristic zero and R be an RB-operator of nonzero weight on Mn(F ) which is not splitting. (a) We have n 1 dim(ker R) n2 n. − ≤ ≤ − (b) If dim(ker R)= n 1, then Im (R) contains the identity matrix and it is conjugate to the subalgebra Span e− j > 1 or i = j =1 . { ij | }

15 Proof. In [2], the bound dim A n2 n+1 was proved for a proper maximal subalgebra A ≤ − of Mn(F ) over a field F of characteristic zero. This bound can be proved when F is an algebraically closed field of any characteristic since the required WedderburnMaltsev decomposition over such fields exists [11, p. 143]. (a) As R is not splitting, 1 ker R by Lemma 2.5(a). Hence, dim ker R n2 n. 6∈ ≤ − The bound n 1 dim(ker R) follows from dim(Im R) n2 n +1. (b) Let dim(ker− ≤R)= n 1. Then dim(Im R)= n2 n≤+1 −and we are done by [2]. − − Let Dn denote the subalgebra of all diagonal matrices in Mn(F ) and Ln (Un) the set of all strictly lower (upper) triangular matrices in Mn(F ). Example 4.7 ([13]). Decomposing M (F ) = L D U (as vector spaces), we n n ⊕ n ⊕ n have a triangular-splitting RB-operator defined with A− = Ln, A+ = Un, A0 = Dn or A− = Un, A+ = Ln, A0 = Dn. All RB-operators of nonzero weight on Dn were described in Proposition 4.5.

In some sense, all RB-operators of nonzero weight on M2(F ) when char F = 2 were described in [13]. Let us refine thus result as follows. 6 Theorem 4.13. Let F be an algebraically closed field of characteristic zero. Every non- trivial RB-operator of weight 1 on M2(F ) (up to conjugation with an automorphism of M2(F ), up to φ and transpose) equals to one of the following cases: x x 0 x (M1) R 11 12 = − 12 , x x 0 x  21 22  11  x x x x (M2) R 11 12 = − 11 − 12 , x x 0 0  21 22   x11 x12 α αγ (M3) R = x21 , α,γ F , x21 x22 − 1 γ ∈     x x αx x (M4) R 11 12 = 12 − 12 , α F 0 , x21 x22 x21 (1/α)x21 ∈ \{ }   −  x x x x + αx ( α2/4)x x (M5) R 11 12 = 22 − 11 21 − 21 − 12 , α F . x21 x22 0 0 ∈     Proof. Let R be an RB-operator of weight 1 on M2(F ). In [13], it was stated that either R is defined by Example 4.7 or R is splitting with one subalgebra being unital. It is easy to check all non-splitting RB-operators defined by Example 4.7 are exactly (M1) and (M2). Now, consider the case when R is splitting, i.e., M2(F ) = ker(R) ker(R + id) (as vector spaces). We may assume that dim(ker(R)) dim(ker(R + id)).⊕ ≥ Case I, dim(ker(R)) =3. By Lemma 4.12(b), ker(R) is unital and conjugate to the α β subalgebra of upper-triangular matrices. So, ker(R + id) = F v for v = . Since 1 γ   ker(R + id) is a subalgebra in M2(F ), we have the condition β = αγ and it is (M3). Case II, dim(ker(R))=2. Suppose that ker(R) is unital subalgebra. So, ker(R + id) consists of matrices of rank not greater than 1. By [51], we may assume that ker(R+id)= α β Span e , e . Thus, ker(R) consists of linear combinations of 1 and w = . We { 11 12} 1 0   16 2 have either ker(R) ∼= F F or ker(R) has a one-dimensional radical. Since w = β 1+αw, we have the second variant⊕ if and only if β = α2/4, it is (M5). · − In the case ker(R) = F F , we may suppose that ker(R) consists of diagonal matrices. ∼ ⊕ ′ Denote R(e12)=(δij) and R(e21)=(γij). Since Im (R ) = ker(R), we get δ12 = γ21 = 1 ′ − and δ21 = γ12 =0. Since ker(R )=Im(R) is a subalgebra of M2(F ) not containing unit, the determinants of R(e12) and R(e21) are zero. So, δ11δ22 = γ11γ22 =0. From R(e12)R(e21)= δ22R(e21)+ γ22R(e12), we receive the relations

δ22γ22 =0, δ11γ22 + δ22γ11 = δ11γ11 +1.

Considering R(e21)(e12)= δ11R(e21)+ γ11R(e12), we get δ11γ11 =0, δ11γ22 + δ22γ11 = δ22γ22 +1.

From all these equations, we see that exactly two of δ11, δ22,γ11,γ22 are zero. Up to transpose, we may assume that δ22 = γ11 =0. Finally, δ11γ22 =1 and it is (M4).

Remark 4.8. In general case Mn(F ), not all of RB-operators of weight 1 are splitting or are defined by Example 4.7. For example, a linear map R: M3(F ) M3(F ) defined as follows: R(e ) = e , R(e ) = e , R(e ) = e , R(e )=0 for→ all other matrix 13 − 13 23 − 23 33 22 kl unities ekl, is such an RB-operator of weight 1. However, R is triangular-splitting with subalgebras A = Span e , e , A = Span e , e , A = Span e , e , e , e , e . − { 31 32} + { 13 23} 0 { 11 12 21 22 33} We need some preliminary results before proving the main theorem of the current section. m n Let Fn(m) = j for natural n, m. It is well known that F1(m) = m(m + 1)/2, j=1 2 F2(m)= m(m + 1)(2P m + 1)/6, F3(m)=(F1(m)) . For any n,

1 n n +1 F (m)= ( 1)j B mn+1−j, (21) n n +1 − j j j=0 X   where B0 =1, B1,...,Bn are Bernoulli numbers. Lemma 4.14 ([52, 54]). Let A be a unital power-associative algebra, R be an RB-operator of weight 1 on A, a = R(1). Then R(an)= F (a) for all n N. − n ∈ Proof. For n = 0, we have R(a0) = R(1) = a by the definition. Suppose that n > 0. With the help of (14) and (15), we calculate

n R(an)= k!( 1)n−kS(n, k)Rk+1(1) − Xk=1

17 n k! k+1 = ( 1)n−kS(n, k) s(k +1, t)at (k + 1)! − t=1 Xk=1 X n+1 ( 1)k+1−j = ( 1)n+1−jaj − S(n, k)s(k +1, j) . (22) − k +1 j=1 ! X Xk≥1 Applying the equality (6.99) from [27]

( 1)k+1−j 1 n +1 − S(n, k)s(k +1, j)= B , k +1 n +1 j n+1−j Xk≥1   the formulas (21) and (22) coincide. Let us formulate the useful result about the generalized Vandermonde determinant.

Lemma 4.15 ([46, Thm. 20]). Let t be a positive integer and let A (X) denote the t m t × matrix 10 0 ... 0 X 1 0 ... 0  X2 2X 2 ... 0   X3 3X2 6X... 0    ......    Xt−1 (t 1)Xt−2 (t 1)(t 2)Xt−3 ... (t 1) (t m + 1)Xt−m  − − − − ··· −    where the first column consists of powers of X and then every consequent column is formed by derivation of the previous one with respect to X. Given a partition t = m1 + ... + mk, there holds

k mi−1 det(A (X ) A (X ) ...A (X )) = j! (X X )mimj . (23) m1 1 m2 2 mk k j − i i=1 j=1 ! Y Y 1≤Yi

18 Theorem 4.17. Given an algebraically closed field F of characteristic zero and an RB- operator R of nonzero weight λ on M (F ), there exists a ψ Aut(M (F )) such that n ∈ n the matrix R(ψ)(1) is diagonal and (up to φ) the set of its diagonal elements has a form pλ, ( p + 1)λ, . . . , λ, 0,λ,...,qλ with p, q N. {− − − } ∈ Proof. If R(1) is nilpotent, we are done by Lemma 4.6. Otherwise, consider the algebra A generated by 1 and R(1). If 1 and R(1) are linearly dependent, we are done by Lemma 2.5(a). From (14), (15) it follows that A is closed under the action of R. By Lemma 4.16(a), we may suppose that ker(R) = (0) on A. 6 Up to conjugation, we suppose that R(1) is in the Jordan form J. Let m(x)=(x m1 mk − λ1) ... (x λk) be a minimal polynomial of J, λi = λj for i = j. Express J as a sum of − 6 6 s Jordan blocks J(λ1),...,J(λk) Mn(F ). Introduce matrices ei Mn(F ), i = 1,...,k, 1 2 1 ∈ ∈ s 1 s−1 such that J(λi)= λiei + ei , ei is a diagonal matrix in Mn(F ) and ei =(J(λi) λiei ) s − for all s 2. Note that ei =0 for s > mi. ≥ m Lemma 4.15 implies that e1,...,e i i =1,...,k is a linear basis of A. Moreover, { i i | } one of λi is zero (denote λk =0), otherwise R is invertible on A. 2 m−1 Another linear basis of A is 1, a, a ,...,a , where m = m1 + ... + mk. By k 2 m−2 Lemma 4.14, R(a ) = Fk(a). Hence, R(1), R(a), R(a ),...,R(a ) are linearly inde- pendent. So, dim(ker R)=1 on A. s From Lemma 4.15, all ei , s = 1,...,mi, lie in Im (R) for i = 1,...,k 1. Also, 2 mk − it follows from R(1) Im (R) that ek,...,ek Im (R). Therefore, R(v) has zero projection on e1 for all∈ v A in the decomposition∈ on the basis es, i = 1,...,k, s = k ∈ i 1,...,mi. s For the ideal Ak = Span ek s = 1,...,mk , consider the induced RB-operator Rk { | 1 } 1 from R by Lemma 2.6. We have that Rk(ek) is nilpotent. As ek is a unit in Ak, by 1 ′ ′ s Lemma 4.6 we have Rk(ek)=0. Representing A as A Ak for A = Span ei i = 1,...,k 1; s =1,...,m , we define ⊕ { | − i} ′ 1 R(1) = a + b, R(e )= a1 + b1, R(ek)= a2, (24) k−1 ′ ′ 1 ′ 1 2 where a, a1, a2 A , b, b1 Ak, e = ei . As e +ek =1, we get b1 = b = ek, a1 +a2 = a. ∈ ∈ i=1 1 1 1 Calculating R(ek)R(ek) by (1), weP have a2(a2 +1) = 0, i.e., a2 = ei for some − i∈I ′ ′ I 1,...,k 1 . Introduce λ = λi, if i I, and λ = λi +1, otherwise.P ⊂{ − } i 6∈ i Let x be a nonzero vector from ker(R). Calculating R(x)R(1) by (1), we get xR(1) ker R. As ker R is one-dimensional algebra, x2 = αx for α F . Further, we will consider∈ 2 ∈ ′ three possibilities, Case 1: α =0 and x =0. If α =0, then x is diagonal, i.e., x = x +xk ′ 1 6 1 for x Span ei i = 1,...,k 1 and xk Span ek . Since xR(1) ker R, either 1 ∈ { | 1 − } ∈ { } ∈ ek ker R (Case 2) or ei ker R for some i 1,...,k 1 (Case 3). ∈ ∈ ∈{ −mi} Case 1: x2 =0, then by xR(1) ker R we have x = γe 1 for some i 1,...,k 1 , ∈ i1 1 ∈{ − } mi > 1 and nonzero γ F . Let γ =1 and denote x as x1. As it was noted above, there exists x A such that∈R(x )= x . From 2 ∈ 2 1 0= x2 = R(x )R(x )= R(x x + x x x2)= R(x2), 1 2 2 1 2 2 1 − 2 − 2 we have x2 ker(R). Representing x as x′ + x′′ for x′ Span es s = 1,...,m and 2 ∈ 2 2 2 2 ∈ { i | i} x′′ Span es j = i, s =1,...,m , we have (x′ )2 = εx , ε F , and (x′′)2 =0. 2 ∈ { j | 6 j} 2 1 ∈ 2 19 Also, by

λ x = R(1)x = R(1)R(x )= R(R(1)x + x x )= R(R(1)x ) x , i 1 1 2 2 1 − 2 2 − 1 ′ 2 ′ ′′ ′′ we get R(1)x2 =(λi+1)x2 +δx1, δ F . Thus, x2 +ei1 x2 = δx1 and (λi+1)x2 = R(1)x2. ′ ∈ − ′′ mi2 From these equalities, we get x2 Span x1 and x2 = βei2 for β =0 and i2 such that ∈ 2 { } 6 λi2 = λi1 +1, mi2 > 1. Moreover, x2 =0. On the step t, we find xt such that R(xt)= xt−1. From

0= x2 = R(x )R(x )=2R(x x x2), t−1 t t t t−1 − t R(x + x R(1) x )= R(1)R(x )= R(1)x =(λ + t 2)x + u, t−1 t − t t t−1 i1 − t−1 where u Span x ,...,x , by induction, we get ∈ { 1 t−2} x2 =2x x + ψx , ψ F, (25) t t t−1 1 ∈ R(1)x =(λ + t 1)x + w, w Span x ,...,x . (26) t i1 − t ∈ { 1 t−1} t−1 ∗ s ∗ s Represent xt as xt,j + xt for xt,j Span eij s =1,...,mij , xt Span ep p j=1 ∈ { | } ∈ { | 6∈ P ∗ 2 ∗ ∗ i1,...,it−1 ,s 1 . By (25) and (26), we get (xt ) = 0 and R(1)xt = (λi1 + t 1)xt . { } ∗ ≥ } ∗ mit − Thus, either xt =0 or xt = χeit , χ F 0 , λit = λi1 + t 1 and mit > 1. m m 2 ∈ \{ } ij − ij By (26), we have (j t)xt,j +eij xt,j Span eij , 1 j t 1. So, xt,j Span eij . ∗ 2 − ∈ { } ≤ ≤ − ∈ { } Hence, xt =0 and xt =0. 6 mi We can continue the process endless, as all ei for mi > 1 lie in Im (R). It is a contradiction to the fact that A is a finite-dimensional algebra. 1 1 1 Case 2: ek ker(R). From ekR(1) = ek(a + b) = b ker R, we get that b = 0, ∈ ′ ∈ ′ it means mk = 1. Define R as induced RB-operator R on A (see Lemma 2.6). So, R′(e′)= a and by Lemma 4.15, R′ is invertible on A′. We finish Case 2 by the following result: ′ ′ Lemma 4.18. Given an invertible RB-operator R on A , we have mi = 1 for all i = 1,...,k 1 and λ ,...,λ = 1, 2,...,k 1 . − { 1 k−1} { − } ′ k ′ ′ Proof. Define xk such that (R ) (xk)=1. We have that R (x1)= e . From (20), we get 2 ′ ′ ′ ′ x1 = x1. From a1 = R (x1)R (e ) = a1 + R (x1a1 x1), we have x1a1 = x1. The only − 1 possibility is the following: one of λi equals 1, mi = 1 and x1 = ei . Exchange indices 1,...,k 1 in such way that λ′ = λ =1. − 1 1 For x2, we have

a x = R′(x )R′(1) = R′(x + x a ) R′(x )= R′(x + x a ) x , 1 1 2 1 2 1 − 2 1 2 1 − 1 ′ ′ so, R (x1 + x2a1)= x1(a1 +1)=2x1 =2R (x2), i.e.,

x1 + x2a1 =2x2. (27)

∗ ∗ s Representing x2 as αx1 + x1, x1 Span ei 2 i k 1; s = 1,...,mi , from (27) ∈ { | ≤′ ≤ − ′ } we have α = 1 and either x2 = x1 or one of λi (denote λ2) equals 2, m2 = 1 and

20 1 C∗ ′ ′ x2 = x1 + βe2 for some β . Note that λ2 = λ2, otherwise 2 = λ2 = λ2 +1 or ∈ ′ ′ ′ λ2 = λ1 = 1, a contradiction. If x2 = x1, then R (x2) = x1 = R(x1) = e , so a is diagonal, k =2 and m1 =1. If x2 = x1, we proceed with x3 an so on. In every step t 3 we deal with the equality 6 ≥ a x = R′(x )R′(1) = R′(x + x a ) R′(x )= x + R′(x a ) x . 1 t−1 t t−1 t 1 − t t−2 t 1 − t−1 1 1 s Express xt = yt + zt for yt Span e1,...,et−1 and zt Span ei i t; s =1,...,mi . ∈ { } ′ ∈ {′ | ≥ } Then a1zt = tzt and so either zt = 0 or one of λi (denote λt) equals t, mt = 1 and 1 C∗ xt = y1 + γet for some γ . ∈ p−1 ′ The space A is finite-dimensional, so xp = αjxj for some p. Consider the minimal j=1 such p. So, P p−1 ′ ′ p ′ p−j ′ e =(R ) (xp)= αj(R ) (e ). j=1 X By (14), we get e′ Span a , a2,...,ap−1 . It may happen only if p 1 is a degree ∈ { 1 1 1 } − of the minimal polynomial of a1, the spectrum of a1 coincide with 1, 2,...,p 1 and ′ { − } m1 = ... = mp−1 = 1. We have also proved that λi = λi for all i = 1,...,k 1. Lemma 4.18 is proved. − Case 3: y = e1 ker(R) for some i 1,...,k 1 . From e1 R(1) ker R, we 1 i1 ∈ 1 ∈ { − } i1 ∈ get mi1 =1. Consider y A such that R(y )= y = e1 and y Span es i = i . From 2 ∈ 2 1 i1 2 ∈ { i | 6 1} y = R(y )R(y )=2R(y y ) R(y2), 1 2 2 1 2 − 2 λ y = R(y )R(1) = R(y R(1)) y , i1 1 2 2 − 1 we have y + y2 =0 and (λ + 1)y = y R(1). We conclude that y = e1 for some i . 2 2 i1 2 2 2 − i2 2 Moreover, mi2 =1 and λi2 = λi1 +1. On the step t, we find y Span es i = i such that R(y )= y . From t ∈ { i | 6 1} t t−1 y2 = R(y )R(y )=2R(y y ) R(y2), t−1 t t t−1 t − t y R(1) = R(y )R(1) = R(y + y R(1)) y , t−1 t t−1 t − t−1 2 t by induction arguments, we get that yt +( 1) yt and ytR(1) (λi1 + t 1)yt lie in Span y ,...,y . So, y = ( 1)t+1e1 + w −for w Span y ,...,y− , and− m = 1, { 1 t−1} t − it ∈ { 1 t−1} it λit = λi1 + t 1. − 1 1 1 As A is finite-dimensional and ek is the only element from e1,...,ek which is not in p−1 1 ′ Im (R), we get yp =( 1) ek + u, u Span y1,...,yp−1 A , on a step p. Consider A = Span− e1 ,...,e1 and∈ A {= Span es }i ⊆ i ,...,i ; s =1,...,m . 1 { i1 ip } 2 { i | 6∈ { 1 p} i} We have A = A A . Moreover, A is closed under the action of R and dim(ker R) = 1 ⊕ 2 1 1 on A1. Thus, A2 is closed under R and R is invertible on A2. We are done by Lemma 4.18. Remark 4.9. For every p, q N, we apply Examples 4.4 and 4.7 to construct an RB-operator R on a matrix algebra∈ with diagonal matrix R(1) which has the numbers pλ, ( p + 1)λ, . . . , λ, 0,λ,...,qλ on the diagonal. − − − 21 (ψ) Let us call an RB-operator R on Mn(F ) diagonal, if R (Dn) Dn for some ψ Aut(M (F )). By Theorem 4.13, all RB-operators of nonzero weight⊆ on M (F ) ∈ n 2 are diagonal. Let us apply Theorem 4.17 to prove the following general statement.

Proposition 4.19. Let R be an RB-operator of nonzero weight on Mn(F ) and R(1) = n (aij)i,j=1 be a diagonal matrix with k different values of diagonal elements: aii = λj, if m +1 i m , where m +m +...+m = n and m =0. Define B = B ... B j−1 ≤ ≤ j 1 2 k 0 1 ⊕ ⊕ k (as vector spaces) for B = Span e m +1 i m . Then B is R-invariant j { ii | j−1 ≤ ≤ j} subalgebra of Mn(F ). If k = n, then R is a diagonal RB-operator. Proof. Without loss of generality, assume that R is an RB-operator of weight 1. By (1), − R(1)R(x)= R(R(1)x + R(x) x), R(x)R(1) = R(xR(1) + R(x) x) − − and [R(1), R(x)] = R([R(1), x]). Considering x = eij for eij Bk, we get R(1)R(eij) = R(eij)R(1). From the last equality we have R(e ) B. ∈ ij ∈ If k = n, then B is the subalgebra of diagonal matrices and R is diagonal RB-operator by the definition. Corollary 4.20. Let F be an algebraically closed field of characteristic zero. All RB- operators of nonzero weight on M3(F ) are diagonal.

Proof. Consider an RB-operator R on M3(F ). By Theorem 4.17, we may suppose that R(1) is a diagonal matrix. If R(1) is scalar, then R is splitting with one of subalgebras being unital by Lemma 2.5(a), this case we will consider later. If all diagonal elements of R(1) are pairwise distinct, then R is diagonal by Proposition 4.19. Thus, R(1) has exactly two different values on the diagonal. Applying φ, one of the values is zero. By the proof of Theorem 4.17, we have three cases: 1. R(e11 + e22)= e11 + e22, R(e33)=0, R(1) = 1(e11 + e22)+0e33; 2. R(e11 + e22)=0, R(e33)= e11 e22, R(1) = 1(e11 + e22)+0e33; 3. R(e + e )= e , R(e )=0− ,−R(1) = 0(e +−e ) 1e . 11 22 − 33 33 11 22 − 33 Denote B = Span e 1 i, j 2 , B = Span e , B = B B . By Proposi- 1 { ij | ≤ ≤ } 2 { 33} 1 ⊕ 2 tion 4.19, B is an R-invariant subalgebra of M3(F ). Consider the induced RB-operator R1 from R on B1 ∼= M2(F ). By Theorem 4.13, R1 is diagonal on B1 up to conjugation with an automorphism ψ of B1. Extend the action of ψ on the entire algebra B as follows: ψ ψ(e33) = e33. It is easy to calculate directly that R (D3) D3. Let us show it for the ⊆ ψ ψ case 1, the proofs for the cases 2 and 3 are analogous. Indeed, R1 (e11)= d1, R1 (e22)= d2 ψ ψ for d1,d2 Span e11, e22 . Thus, R (e11) = d1 + αe33, R (e22) = d2 + βe33 for some α, β F .∈ Finally,{Rψ(e )=} ψ−1Rψ(e )= ψ−1R(e )=0. ∈ 33 33 33 Now, let R be a splitting RB-operator of weight 1 with one of subalgebras being unital, i.e., M3(F ) equals a direct vector-space sum of subalgebras A1 = ker(R) and A = ker(R + id) = Im(R), and we may assume that 1 A . By Lemma 4.12(a), we 2 ∈ 2 have 2 dim(A1) 6. Consider all these variants: 1. dim≤ A = 2,≤dim A = 7. By Lemma 4.12(b), up to conjugation D A , we are 1 2 3 ⊂ 2 done.

22 2. dim A1 =3, dim A2 =6. Define B as a linear span of all matrix unities eij except e21 and e31. As the algebra B is a unique up to conjugation maximal subalgebra in M3(F ) [2], we may assume that A2 is a subalgebra in B. Let

M = Span e , e , e , e = M (F ), (28) { 22 23 32 33} ∼ 2 then 3 dim(A M) 4. If A M = M, then e A (as 1 A ) and αe +βe A ≤ 2∩ ≤ 2∩ 11 ∈ 2 ∈ 2 12 13 ∈ 2 for some α, β F which are not zero simultaneously. Hence, (αe12 +βe13)e22 = αe12 A2 and βe A∈. If α =0, then e A as well as e e = e , a contradiction. If β∈=0, 13 ∈ 2 13 ∈ 2 13 32 12 then e12e23 = e13 A2, a contradiction. Thus, dim(A ∈ M)=3. By [2], there exists a matrix S M such that S−1(A 2 ∩ ∈ 2 ∩ M)S = Span e , e , e . Let T = e + S, then B is invariant under the conjugation { 22 23 33} 11 with T . So, we get that A2 = D3 U3 (as vector spaces) and therefore D3 is up to conjugation R-invariant. ⊕ 3. dim A1 = 4, dim A2 = 5. Consider A2 M for M defined by (28). As above, we assume that A is a subalgebra of B. By the∩ dimensional reasons, dim(A M) 2. If 2 2 ∩ ≥ A2 M = M, then D3 A2 and we are done. If dim(A2 M)=3, then we deal with it as in∩ the variant 2 and⊂ get D A . So, dim(A M)=2∩ and as vector spaces 3 ⊂ 2 2 ∩ A = Span e + d i =1, 2, 3; d M (A M). 2 { 1i i | i ∈ } ⊕ 2 ∩

Up to conjugation with the matrix from B, we have either A2 M = Span e22, e33 or A M = Span e + e , e , or A M = Span e , e . In∩ the first case{ D }A 2 ∩ { 22 33 23} 2 ∩ { 22 23} 3 ⊂ 2 and we have done. In the second case, d = 0 as 1 A . Also, e , e A as 1 ∈ 2 12 13 ∈ 2 (e22 +e33)(e12 +d2)= d2 A2 and analogously d3 A2. So, A2 = U3 Span e11, e22 +e33 (as vector spaces). Let us∈ show that the third case∈ either can be reduced⊕ { to the similar} subalgebra or contains D3 up to conjugation. Indeed, we may assume that d ,d ,d Span e , e . As 1 A , we have e +e 1 2 3 ∈ { 32 33} ∈ 2 11 33 ∈ A . Also, (e + d )e = e + d e A . Thus, d = ke for some k F . Since 2 12 2 22 12 2 22 ∈ 2 2 32 ∈ (e12 +d2)e23 = e13 +ke33 A2, we have A2 = Span e11 +e33, e12 +ke32, e13 +ke33, e22, e23 . ∈ { 1 0 1 } For k = 0, we apply the conjugation with the matrix P = 0 1 0 and we get   6 k 0 0 −1 P A2P = D3 Span e12, e21 (as vector spaces), so A2 up to conjugation  contains D3. For k =0, we get⊕ A ={U Span} e +e , e (as vector spaces), the algebra conjugated 2 3 ⊕ { 11 33 22} to the one obtained in the second case. Let us proceed with this variant of A2. As dim A1 = 4 and A1 A2 = (0), we have dim(A1 (D3 U3)) = 1 (as vector spaces). Let t be a nonzero∩ vector from A (D U ) ∩(as vector⊕ spaces). Without 1 ∩ 3 ⊕ 3 loss of generality, we may assume that t = e11 + γe22 + δe33 + d, where δ = 1 and d = xe + ye + ze U . Due to the arguments stated above, t2 = t, δ = 06 and we 12 13 23 ∈ 3 have two possibilities: t = e11 + xe12 + ye13 or t = e11 + ye13 + e22 + ze23. In the first case, 1 x y − − we apply the conjugation with the matrix U = 0 1 0 and we get e U −1A U,   11 1 0 0 1 ∈ −1 dim(D3 U A2U)=2, we are done. In the second case we conjugate with the matrix ∩

23 1 0 y − −1 −1 V = 0 1 z , so, e11 + e22 V A1V and dim(D3 V A2V )=2. Thus, we finish 00− 1  ∈ ∩ this variant.  4. dim A =5, dim A =4. Consider the algebra C = A F 1 (as vector spaces), it 1 2 1 ⊕ is 6-dimensional unital subalgebra in M3(F ). Due to the variant 2, we may assume that C = D3 U3 (as vector spaces). So, D3 is up to conjugation R-invariant. 5. dim⊕ A =6, dim A =3. Consider the algebra C = A F 1 (as vector spaces), it 1 2 1 ⊕ is 7-dimensional unital subalgebra in M3(F ). By Lemma 4.12(b), we are done.

Problem 4.10. Does there exist not diagonal RB-operator on Mn(F )?

Problem 4.11. To classify all diagonal RB-operators on Mn(F ).

4.6 Derivations of nonzero weight Given an algebra A and λ F , a linear operator d: A A is called a derivation of weight λ [34] if d satisfies the identity∈ →

d(xy)= d(x)y + xd(y)+ λd(x)d(y), x,y A. (29) ∈ Let us call zero operator and (1/λ)id (if λ =0) as trivial derivations of weight λ. − 6 Statement 4.21 ([47]). Given an algebra A and invertible derivation d of weight λ on A, the operator d−1 is an RB-operator of weight λ on A.

Corollary 4.22. There are no nontrivial invertible derivations of nonzero weight on (a) [13] Kaplansky superalgebra K3, (b) Grassmann algebra, sl2(C) and any unital simple algebra. Proof. (b) It follows from Corollary 4.10, Theorem 4.2 and Lemma 4.16(b) respectively.

5 RotaBaxter operators of weight zero

In §5, we study RB-operators of weight zero. In Lemma 5.1 (§5.1), we give a list of general constructions of RB-operators of weight zero which cover most of known examples of RB-operators. In §5.2, we state the general property of RB-operators of weight zero on unital alge- bras. Given a unital associative (alternative, Jordan) algebraic algebra A over a field of characteristic zero, there exists N such that RN = 0 for all RB-operators R of weight zero on A (Theorem 5.5, §5.2). So, we define a RotaBaxter index (RB-index) rb(A) on A as the minimal natural number with such nilpotency property. We prove that rb(Mn(F ))=2n 1 over a field F of characteristic zero (Theorem 5.22, §5.4) and study RB-index for the Grassmann− algebra (§5.3), unital composition algebras and simple Jordan algebras (§5.5).

24 5.1 Constructions of RotaBaxter operators Below, we say that an algebra D is abelian if its product is trivial, i.e., zero.

Lemma 5.1. Given an algebra A, described below linear operator R on A is an RB- operator of weight zero, (a) If A = B C (as vector spaces) with abelian Im R, R: B C, R: C (0), BR(B), R(B)B ⊕C. In particular, → → ⊆ (a1) If A is a Z -graded algebra A = A A with abelian Im R and R: A A , 2 0 ⊕ 1 0 → 1 R: A1 (0); (a2)→ If A is a Z -graded algebra A = A A with abelian odd part and R: A A , 2 0 ⊕ 1 0 → 1 R: A1 (0). (b)→ If A = B C (as vector spaces), R(B)C,CR(B) C, R: B C, R: C (0), and R(b)R(b′)= ⊕R(R(b)b′ + bR(b′)) for all b, b′ B. In particular,⊆ → → ∈ (b1) If A is a Z2-graded algebra A = A0 A1, R: A1 A0, R: A0 (0), R(c)R(d)= R(R(c)d + cR(d))=0 for all c,d A ; ⊕ → → ∈ 1 (b2) If A is a Z -graded algebra A = A A with abelian even part, R: A A , 2 0 ⊕ 1 1 → 0 R: A0 (0), and R(c)d + cR(d)=0 for all c,d A1. (c)→ If A = B C (as vector spaces), R: B ∈B, R: C (0), B2 = (0), BC,CB ⊕ → → ⊆ ker R. In particular, (c1) If A is a Z -graded algebra A = A A with abelian even part and R: A A , 2 0 ⊕ 1 0 → 0 R: A (0); 1 → (c2) If A is a Z2-graded algebra A = A0 A1 with abelian odd part, R: A0 (0), R: A A , and A Im R, Im R A ker R⊕. → 1 → 1 0 · · 0 ⊆ (d) If A is a Z -graded algebra A = A A A with A2 = (0), and R: A A , 3 0 ⊕ 1 ⊕ 2 2 1 → 2 R: A0 A2 (0). (e)⊕ If A →= B C D (as vector spaces) with C2 = D2 = (0), CD,DC D, ⊕ ⊕ ⊆ R: B C, R: C D, R: D (0), moreover, R(b)b′ + bR(b′)=0, R(bd) = R(b)d, R(db)=→dR(b) for all→ b, b′ B, d→ D. In particular, Z ∈ ∈ 2 2 (e1) If A is a 3-graded algebra A = A0 A1 A2 with A0 = A1 = (0), R: A2 A0, R: A A , R: A (0), moreover, R⊕(a )b⊕+ a R(b )=0, R(a c ) = R(→a )c , 0 → 1 1 → 2 2 2 2 2 1 2 1 R(c1a2)= c1R(a2) for all a2, b2 A2, c1 A1. Z ∈ ∈ 2 (f) If A is a 3-graded algebra A = A0 A1 A2 with A1A2, A2A1, A2 = (0), R: A0 A , R: A A , R: A (0), moreover,⊕R(c)⊕R(d)= R(R(c)d+cR(d)) for all c,d A→. 1 1 → 2 2 → ∈ 0 (g) If A = B C D (as vector spaces) with CD,DC,D2 D, abelian R(C), R: B C, R: C ⊕ D⊕, R: D (0), moreover, R(x)R(y) = R(R(⊂x)y + xR(y)) for all x B,→y B C →or x B C→, y B. ∈ ∈ ⊕ ∈ ⊕ ∈ (h) [28] If A = B C (as vector spaces), where B is a subalgebra with an RB- ⊕ operator P of weight zero, BC,CB ker P C, and R is defined as follows: R B P , R 0. ⊆ ⊕ | ≡ |C ≡ (i) [47] If A = A1 A2, Ri are RB-operators on Ai, i = 1, 2, and the linear map R: A A is defined by⊕ the formula R(x + x )= R (x )+ R (x ), x A , x A . → 1 2 1 1 2 2 1 ∈ 1 2 ∈ 2 Proof. Straightforward.

25 Let A be an algebra. Note that conjugation with automorphisms of A and scalar multiple does not change the correspondence of an RB-operator of weight zero on A to Lemma 5.1.

2 Example 5.1. By Example 2.2, a linear map le in an associative algebra A with e =0 is an RB-operator of weight zero. If e Z(A), then le corresponds to Lemma 5.1(a1) for A = (1 e)A and A = eA. ∈ 0 − 1 Example 5.2 ([4]). Given an associative algebra A and an element e such that e2 =0, a linear map lere which acts on A as x exe is an RB-operator of weight zero on A. It corresponds to Lemma 5.1(a) for C =→eA + Ae and any B such that A = B C (as ⊕ vector spaces). Example 5.3. In [30], given a variety Var of algebras and a pre-Var-algebra A, an enveloping RB-algebra B of weight zero and the variety Var for A was constructed. By the construction, B = A A′ (as vector spaces), where A′ is a copy of A as a vector space, and the RB-operator⊕ R was defined as follows: R(a′) = a, R(a)=0, a A. By ∈ the definition of the operations on B [30], the operator R corresponds to Lemma 5.1(a2). Example 5.4. In [22], all homogeneous RB-operators on the Witt algebra W over C were described (see Example 4.2 about the definitions). According to [22], all nonzero homogeneous RB-operators of weight zero on W with degree k up to the multiplication on α C are the following: ∈ (W1) R(Lm)=0, m =0, R(L0)= L0; (W2) R(L )= δ 6 L , k =0, m Z; m m+2k,0 m+k 6 ∈ (W3) R(Lm)=(δm+2k,0 +2δm+3k,0)Lm+k, k =0, m Z; k 6 ∈ (W4) R(L )= δ ZL , l doesn’t divide k, m Z. m m+2k m+k,l m+k ∈ Note that the RB-operator (W1) corresponds to Lemma 5.1(c), the RB-operator (W2)  by Lemma 5.1(a), (W3)  by Lemma 5.1(b1) for A1 = Span L−3k, L−2k , A = Span L m 3k, 2k . Finally, the RB-operator (W4) corresponds{ to} 0 { m | 6∈ {− − }} Lemma 5.1(b) for B = Span L m + k lZ and C = Span L m + k lZ . { m | ∈ } { m | 6∈ } Example 5.5. In [47], a classification of RB-operators on all 2- and some 3-dimensional pre-Lie algebras over C was given. In particular, for the simple pre-Lie algebra S2 = Span e , e , e e e = e e = e , e e = e , e e = e , all RB-operators of weight zero { 1 2 3 | 1 2 2 1 3 3 1 1 3 2 − 2} on S2 are of the form R(e3)= αe3, R(e1)= R(e2)=0. Thus, all RB-operators of weight zero on S2 correspond to Lemma 5.1(c1) for A = Span e and A = Span e , e . 0 { 3} 1 { 1 2} Example 5.6. In [47], the RB-operator R on a class of simple pre-Lie algebras In = Span e1,...,en enen = 2en, enej = ej, ejej = en, 1 j n 1 introduced in [16] for { | ≤1 ≤n−1− } n 3 was constructed as follows R(e1) = en + ei, R(ei)=0, 2 i n. ≥ √2 n i=2 ≤ ≤ − It is easy to see that R corresponds to Lemma 5.1(b1) forPA0 = Span e2,...,en and { } A = Span e . 1 { 1} Example 5.7. In [13], it was proved that all RB-operators of weight zero on the simple Kaplansky Jordan superalgebra K3 up to conjugation with Aut(K3) are the following:

26 R(e)= R(x)=0, R(y)= ae + bx for a, b F . Thus, given an RB-operator R of weight zero on K , we have R2 = 0 and R corresponds∈ to Lemma 5.1(b1) for A = Span e, x 3 0 { } and A = Span y . 1 { } In [56], all RB-operators of weight zero on sl2(C) were described as 22 series. In [44] a very simple classification of them was given; up to conjugation with an automorphism of sl2(C) and scalar multiple all nonzero RB-operators on sl2(C) are the following: (L1) R(e)=0, R(f)= te h, R(h)=2e, t C; (L2) R(e)=0, R(f)=2te−+ h, R(h)=2e +∈1 h, t C∗; t ∈ (L3) R(h)= h, R(e)= R(f)=0; (L4) R(f)= h, R(e)= R(h)=0; (L5) R(f)= e, R(e)= R(h)=0. Note that Example 3.1 up to scalar multiple is (L1) for t =0.

Statement 5.2. All RB-operators on the simple Lie algebra sl2(C) correspond to Lemma 5.1. Proof. It is easy to check that (L1) corresponds to Lemma 5.1(e) with A = Span f , 0 { } A1 = Span te h , A2 = Span e . Cases (L2) and (L3) come from Lemma 5.1(c1). RB-operator{ R−of the} type (L4) is{ constructed} by Lemma 5.1(h) for B = Span f, h and { } C = Span e , where R on B corresponds to Lemma 5.1(b2). Finally, (L5) corresponds to Lemma{ 5.1(d).}

5.2 Unital algebras Lemma 5.3. Let A be a unital power-associative algebraic algebra and R be an RB- operator on A of weight zero. Then R(1) is nilpotent. Proof. In the case char F = p> 0, we have (R(1))p = p!Rp(1) = 0. Suppose that char F =0. Denote a = R(1). Assume that a is not nilpotent. As A is algebraic, we can consider a minimal k such that

2 k α1a + α2a + ... + αka =0 (30) and not all of αi are zero. Thus, k> 1 and αk =0. m m 6 N m m+1 am+1 By Lemma 2.5(d), a = m!R (1) for all m . Hence, R(a )= m!R (1) = m+1 . Let us apply the operator (k + 1)R l to the∈ left-hand side of (30): − a (k + 1)a2 (k + 1)a3 (k + 1)ak α + α + ... + α + α ak+1 1 2 2 3 k−1 k k   (α a2 + α a3 + ... + α ak+1) − 1 2 k k +1 k +1 k +1 = α 1 a2 + α 1 a3 + ... + α 1 ak =0. (31) 1 2 − 2 3 − k−1 k −       k If αk−1 = 0, then minimality k implies αi = 0 for all i = 1,...,k 1 and a = 0. If α =0, then the expressions (30) and (31) are proportional by a non−zero scalar. Hence, k−1 6 we obtain α1 =0, α2 =0, ..., αk−1 =0, a contradiction.

27 Lemma 5.4. Let A be a unital power-associative algebra over a field of characteristic zero and let R be an RB-operator of weight zero on A. Suppose that (R(1))m =0. Then (a) R2m =0 if A is associative or alternative, (b) R3m−1 =0 if A is Jordan. Proof. (a) Suppose A is an associative algebra. By Lemma 2.5(d), Rm(1) = 0. By standard computations in RB-algebra (see, e.g., [14]), we write

0= R(x)R(1)Rm−1(1) = R2(x)Rm−1(1) + R(xR(1))Rm−1(1) m−1 m+1 m+1−l l = χR (x)+ βlR (xR (1)) (32) Xl=1 for x A and for some χ, β N . Let t 2m. Multiple substitution of (32) into itself ∈ l ∈ >0 ≥ along with the equality 1 1 1 (k + l)! Rk(1)Rl(1) = (R(1))k (R(1))l = (R(1))k+l = Rk+l(1) (33) k! l! k!l! k!l! leads to

1 m−1 Rt(x)= β Rt−l(xRl(1)) = β Rt−l1−l2 (xRl1+l2 (1)) −χ l l1,l2 Xl=1 2≤l1+Xl2≤m−1 t−l1−...−lm l1+...+lm = ... = βl1,...,lm−1 R (xR (1)) m−1≤l1+...X+lm−1≤m−1 mβ β = β Rt−m+1(xRm−1(1)) = 1 1,...,1 Rt−m(xRm(1)) + ... =0. (34) 1,...,1 − χ For alternative algebras, the proof can be repeated without any changes. Indeed, any two elements in an alternative algebra generate an associative subalgebra. Hence, we can avoid bracketings under the action of R in (32)–(34). b) In the Jordan case, the computations similar to (32)–(34) show that for t m +1 ≥ 1 t−1 Rt(x)= β Rt−l(... ((xyp1 )yp2 ) ...yps ), y = R(1), (35) −χ l,p1,...,ps Xl=1 p1+X...+ps=l where we have some nonassociative words in the alphabet x, y under the action of R in the right-hand side. { } We know that ym = 0. Let us show that any word w(x, y) of the length 2m with a single occurrence of x and 2m 1 occurrences of y equals zero in A. Indeed, by the Shirshov theorem [74, p. 71], the subalgebra− S of A generated by x, y is a special one. It { } means that there exists an associative enveloping algebra E in which S embeds injectively. Calculating the meaning of w in E, we get a linear combination L of associative words of the length 2m with a single occurrence of x and 2m 1 occurrences of y. By the Dirichlet’s principle, any associative word from L has a− subword yl with l m. As ym = 0 in S, we have done. Therefore, to get zero it is enough to execute≥ (35) with t =3m 1. − 28 Theorem 5.5. Let A be a unital associative (alternative, Jordan) algebraic algebra over a field of characteristic zero. Then there exists N such that RN = 0 for every RB- operator R of weight zero on A.

Proof. Each associative, alternative or Jordan algebra is power-associative, thus we may apply Lemma 5.3 for algebras from all of these varieties. We are done by Lemma 5.4. Let A be an algebra (or just a ring). Define RotaBaxter index (RB-index) rb(A) of A as follows

rb(A) = min n N Rn =0 for an RB-operator R of weight zero on A . { ∈ | } If such number is undefined, put rb(A)= . ∞ Lemma 5.6. Let A be a commutative associative (alternative) algebra and e be a nonzero idempotent of A. For an RB-operator R on A of weight zero, e Im (R). 6∈ Proof. Suppose that (0 =)e Im (R), i.e., e = R(x) for some x A. Then R(x) = 6 ∈ ∈ R(x)R(x)=2R(R(x)x)=2R(ex). On the one hand, x 2ex = k ker R. On the other hand, − ∈

R(x)=2R(ex)=2R(e(2ex + k))=4R(ex)+2R(ek) =2R(x)+2R(x)R(k) 2R(xR(k))=2R(x). − Therefore, e = R(x)=0, a contradiction.

Remark 5.8. The analogous statement is not valid for associative and Jordan algebras, see Examples 3.3 and 5.7.

Corollary 5.7. Given a field F of characteristic zero, let A be an algebra over F isomor- phic to a direct (not necessary finite) sum of finite extensions of a field F , in particular A can be a sum of copies of F . Then there are no nonzero RB-operators of weight zero on A.

Proof. Suppose that R is a nontrivial RB-operator of weight zero on A, then Im (R) contains a nonzero element a. Since A has no nilpotent elements, the subalgebra S of Im (R) generated by a is a semisimple finite-dimensional commutative algebra and S contains an idempotent [38, Thm. 1.3.2]. It is a contradiction to Lemma 5.6.

Remark 5.9. Let us explain why there are only trivial RB-operators of weight zero on some semigroup algebras of order 2 and 3 in [36]. We recall some facts about semigroups due to [18]. A set A with a binary operation · is called semigroup if the product is associative. An element a S is called regular if axa = a for some x S. By an· inverse semigroup we mean a∈ semigroup in which ∈ every element is regular and any two idempotents of S commute with each other. The Mashke’s Theorem for semigroups [18, Thm. 5.26] says: the semigroup algebra F [S] of a finite inverse semigroup S is semisimple if and only if the characteristic of F is zero or a prime not dividing the order of any subgroup of S.

29 In [36], there are only trivial RB-operators of weight zero on semigroup algebras of order 2 and 3 exactly for commutative inverse semigroups. In this case, by the Mashke’s Theorem for semigroups, F [S] is a commutative semisimple algebra. By the WedderburnArtin theorem [38, Thm. 2.1.7], F [S] is a direct sum of finite extensions of a field F , and all RB-operators of weight zero on F [S] are trivial by Corollary 5.7. In particular, it implies the affirmative answer on the question from [36]: whether all RB-operators of weight zero are trivial on F [G] over a field F of characteristic zero for any (prime order) cyclic group G. Moreover, all RB-operators of weight zero are trivial on F [G] for every finite abelian group G if char F =0 or char F > G . | | Given an associative algebra A, let us denote its Jacobson radical by Rad(A). In an associative Artinian algebra A, Rad(A) is the largest nilpotent ideal in A [38, Thm. 1.3.1]. The same is true for the radical Rad(A) of an alternative Artinian algebra A [74, p. 250].

Theorem 5.8. Let A be a commutative associative (alternative) finite-dimensional alge- bra over a field of characteristic zero and R be an RB-operator on A. Then (a) Im R Rad(A), (b) rb(A)⊆ 2m 1 if A is unital, where m is the index of nilpotency of Rad(A). ≤ − Proof. (a) Suppose Im R Rad(A), then Im R is not nilpotent algebra. In this case, Im R has an idempotent [38,6⊆ Thm. 1.3.2] (in alternative case, apply [74, p.250] and the standard lifting of an idempotent), a contradiction to Lemma 5.6. So Im R Rad(A). ⊆ (b) By (a), R(1) Rad(A). The remaining proof is analogous to the proof of Lemma 5.4(a) with one∈ change: by (a), we can write down (32) with R(x)(R(1))m−1 instead of R(x)(R(1))m. Thus, it is enough to consider the power t 2m 1. ≥ − It is well-known that in characteristic zero, the solvable radical of a finite-dimensional Lie algebra is preserved by any derivation [41, p. 51] as well as the locally nilpotent radical and the nil-radical of an (not necessary Lie or associative) algebra [62].

Corollary 5.9. Let A be a commutative associative (alternative) finite-dimensional al- gebra over a field of characteristic zero and let R be an RB-operator on A. Then Rad(A) is R-invariant.

Theorem 5.10. Let A be a unital associative algebra equal F 1 N (as vector spaces), where N is nilpotent ideal of index m, char F =0, and let R be an⊕ RB-operator of weight zero on A. Then we have (a) Im R N, (b) rb(A)⊆ 2m 1. ≤ − Proof. (a) For x = α 1+ n Im R, n N, the following is true: (x α 1)m = 0. As Im R is a subalgebra in· A, αm∈ 1 Im R∈. By Lemma 2.5(b), α =0. − · · ∈ (b) Analogous to the proof of Theorem 5.8.

Lemma 5.11. Let A be an algebra and let e be an idempotent of A. Then for an RB- operator R on A of weight zero, e Im (Rk) ker R for k 2. 6∈ ∩ ≥

30 Proof. Suppose that e Im (R2) ker R, i.e., e = R(x) for some x Im R. Then e = R(x) = R(x)R(x) =∈ R(R(x)x∩+ xR(x)) = R(ex + xe)=0, as ker R∈ is an (Im R)- module. It is a contradiction. The proof for k > 2 is analogous.

Corollary 5.12. Let A be an associative (alternative) finite-dimensional algebra and let R be an RB-operator of weight zero on A. Then Im (Rk) ker R is a nilpotent ideal in ∩ Im (Rk) for k 2. ≥ Proof. It follows from Lemma 5.11 and the stated above property of alternative algebras: every finite-dimensional alternative not nilpotent algebra contains an idempotent [74], as Im (Rk) ker R Rad(A) and Rad(A) is nilpotent. ∩ ⊂ 5.3 Grassmann algebra Recall that Gr denotes the Grassmann algebra of the space V = Span e ,...,e n { 1 n} and let A0(n) be its subalgebra generated by V .

Lemma 5.13. Let R be an RB-operator of weight zero on Grn and let char F =0. Then [(n+1)/2]+1 we have (a) R(Grn) A0(n); (b) R(e1 e2 ... en)=0; (c) (R(1)) = 0; (d) rb(Gr ) 2[ n+1 ]+2⊆ . ∧ ∧ ∧ n ≤ 2 Proof. (a) It follows from Theorem 5.10(a). (b) If R(e1 e2 ... en) =0, then there exists x Grn such that R(e1 ... en)x = e ... e . Further,∧ ∧ ∧ 6 ∈ ∧ ∧ 1 ∧ ∧ n R(e ... e )R(x)= R(R(e ... e )x + e ... e R(x)) = R(e ... e ), 1 ∧ ∧ n 1 ∧ ∧ n 1 ∧ ∧ n 1 ∧ ∧ n a contradiction, as R(x) A (n). ∈ 0 (c) The linear basis of A0(n) consists of the vectors eα = eα1 eα2 ... eαs for α = α ,α ,...,α α <α <...<α 1, 2,...,n . The element∧ (R∧(1))k ∧is a sum { 1 2 s | 1 2 s}⊆{ } of summands of the form S = µ eασ(1) ... eασ(k) . By anticommutativity of the σ∈Sk ∧ ∧ generators of the Grassmann algebra,P the necessary condition for S not to be zero is the following: all numbers α , α ,..., α (maybe except only one) are even. Thus, | 1| | 2| | k| (R(1))k 2k−1(V ). Hence, we have (R(1))[(n+1)/2]+1 n+1(V ) = (0). (d) Follows∈ ∧ from (c) and Lemma 5.4(a). ∈ ∧

Example 5.10 ([13]). Let F be a field of characteristic not two. Up to conjugation with an automorphism, all RB-operators of nonzero weight on Gr2 over F with linear basis 1, e , e , e e are the following: R(1), R(e ) Span e , e e , R(e )= R(e e )=0. 1 2 1 ∧ 2 1 ∈ { 2 1 ∧ 2} 2 1 ∧ 2

Corollary 5.14. Given an RB-operator R of weight zero on Gr2 over a field F with 2 char F =2, we have R =0 and R corresponds to Lemma 5.1(a2). Moreover, rb(Gr2)= 2. 6

Statement 5.15. Let char F =2, 3. Given an RB-operator R of nonzero weight on Gr3, we have R3 =0. 6

31 Proof. Applying Lemma 5.13(b) and (1), let us compute the following for x = 1+ y, y A (3), ∈ 0 0= R(x)R(x)R(x)= R(R(x)R(x)x + R(x)xR(x)+ xR(x)R(x)) =3R(R(x)R(x))=3R2(R(x)x + xR(x)) =6R3(x)+3R2(R(x)y + yR(x))=6R3(x).

Using R3(1) = 1/6(R(1))3 =0, we have done.

Example 5.11. Let a linear map R on Gr be such that R(e e ) = e and R equals 3 1 ∧ 2 3 zero on all other basic elements. The operator R is an RB-operator of weight zero on Gr3 corresponds to Lemma 5.1(a1) for A0 = Span 1, e1 e2 and A1 equal to a linear span of all other basic elements. { ∧ }

Example 5.12. Let char F = 2, 3. Define a linear map R on Gr3 as follows: R(1) = e + e e , R(e )= e e 6 e and R equals zero on all other basic elements. Then R 1 2 ∧ 3 1 1 ∧ 2 ∧ 3 is an RB-operator of weight zero, R3 =0 but R2 =0. The operator R is constructed by 6 Lemma 5.1(h) for B = Span 1, e1 + e2 e3, e1 e2 e3 and C = Span e2, e3, e1 e2, e1 e , e e , where R on B corresponds{ ∧ to Lemma∧ ∧ 5.1(f).} { ∧ ∧ 3 2 ∧ 3} Corollary 5.16. Over a field F with char F =2, 3, we have rb(Gr )=3. 6 3 Proof. It follows from Statement 5.15 and Example 5.12.

Problem 5.13. What is the precise value of rb(Grn) for n> 3?

5.4 Matrix algebra

Lemma 5.17. Let R be an RB-operator of weight zero on Mn(F ), char F =0, then (a) [13] Im R consists only of degenerate matrices and dim(Im R) n2 n, ≤ − (b) R2n =0, (c) dim(Im R) (2n 1) dim(Im R ker R). ≤ − ∩ Proof. (b) By Lemma 5.3, R(1) is a nilpotent matrix. Thus, (R(1))n = 0 and by Lemma 5.4(a), R2n =0. (c) It follows from (b).

Theorem 5.18 ([4, 13, 67]). All nonzero RB-operators of weight zero on M2(F ) over an algebraically closed field F up to conjugation with automorphisms M2(F ), transpose and multiplication on a nonzero scalar are the following: (M1) R(e21)= e12, R(e11)= R(e12)= R(e22)=0; (M2) R(e21)= e11, R(e11)= R(e12)= R(e22)=0; (M3) R(e21)= e11, R(e22)= e12, R(e11)= R(e12)=0; (M4) R(e )= e , R(e )= e , R(e )= R(e )=0. 21 − 11 11 12 12 22

Corollary 5.19. All nonzero RB-operators of weight zero on M2(F ) over an algebraically closed field F correspond to Lemma 5.1.

32 Proof. (M1) corresponds to Lemma 5.1(a); (M2), (M3) correspond to Lemma 5.1(b); (M4) corresponds to Lemma 5.1(g) for B = Span e , C = Span e , D = { 21} { 11} Span e , e . { 12 22} C Remark 5.14. If R is an RB-operator on M2(F ), then R is an RB-operator on gl2( ) C C C by Statement 2.4(a). Since gl2( )= sl2( ) Span E , the projection of R on sl2( ) is an RB-operator by Lemma 2.6. Thus, the projection⊕ { } of (M1) gives (L5), the projection of (M2) gives (L4), and the projections of (M3), (M4) coincide with (L1), where t =0.

Example 5.15 ([24]). Let A be a subalgebra of Mn(F ) of matrices with zero n-th row. Denote by A0 the subalgebra of A consisting of matrices with zero n-th column and by A1 the subspace of A of matrices with all zero columns except n-th. A linear map R acting as follows: R: A A , R: A (0) is an RB-operator on A by Lemma 5.1(a1). 0 → 1 1 → Example 5.16. Let R be an RB-operator of weight zero on Mn−1(F ), then a linear operator P of weight zero on M (F ) defined as follows: P (e )= R(e ), 1 i, j n 1, n ij ij ≤ ≤ − P (ein) = P (eni)=0, 1 i n, is an RB-operator of weight zero on Mn(F ) by Lemma 5.1(h). ≤ ≤

In Corollary 3.6 all skew-symmetric RB-operators of weight zero on M3(C) up to conjugation, transpose and scalar multiple were listed.

Statement 5.20. All skew-symmetric RB-operators of weight zero on M3(C) except (R2) correspond to Lemma 5.1. Proof. The RB-operator (R1) corresponds to Lemma 5.1(a) for the space B = Span e , e and the linear span C of all other matrix unities. The RB-operator (R3) { 31 32} corresponds to Lemma 5.1(h), it is the extension of (M4) from Theorem 5.18 by Exam- ple 5.16. Finally, the RB-operators (R4)–(R8) correspond to Lemma 5.1(g) for B = Span e , e , C = Span e , e , e , e , e and D = Span e , e . { 13 23} { 11 12 21 22 33} { 31 32} Modifying and generalizing the RB-operator (R2), we get the following examples. Example 5.17. For A = M (F ), define r A A as n ∈ ⊗ r = (e (e + e + ... + e ) ji ⊗ i,j+1 i+1,j+2 n−j+i−1,n 1≤i≤j≤n−1 X (e + e + ... + e ) e ). (36) − i,j+1 i+1,j+2 n−j+i−1,n ⊗ ji By the definition, r is skew-symmetric. It is easy to check that r is a solution of the asso- ciative YangBaxter equation (6). Up to sign, r coincides with Example 2.3.3 from [5].

Example 5.18. An RB-operator on Mn(F ) obtained from Example 5.17 by State- ment 3.3 is the following one:

e + e + ... + e , i j n 1, i,j+1 i+1,j+2 n−j+i−1,n ≤ ≤ − R(eij)= (e + e + ... + e ), i > j, (37) − i−1,j i−2,j−1 i−j+1,1 0, j = n.   33 Due to the definition, we have R2n−1 =0 and R2n−2 =0, as 6 ( 1)n+1e ( 1)ne ( 1)n−1e ... − n,1 →R − n−1,1 →R − n−2,1 →R e e + e + ... + e e +2e +3e + ... +(n 2)e →R 11 →R 12 23 n−1,n →R 13 24 35 − n−2,n ... e +(n 2)e e 0. →R 1,n−1 − 2,n →R 1,n →R Remark 5.19. Note that the RB-operator from Example 5.18 preserves the well-known n−1 Z-grading on Mn(F ): Mn(F )= Mi, Mi = Span est t s = i . i=−n+1 { | − } P Remark 5.20. The linear operator defined by (37) on the ring of matrices Mn(A) over an algebra A is an RB-operator on M (A). So we have inequality 2n 1 rb(M (A)) n − ≤ n for every nonzero algebra A.

Corollary 5.21. We have 2n 1 rb(Mn(H)) 2n, where H is the algebra of quater- nions. − ≤ ≤

Proof. We consider M Mn(H) as an algebra over R. The lower bound follows from Remark 5.20. Given a nilpotent∈ matrix M M (H), we have M n = 0 [68, 71]. Thus, ∈ n the upper bound follows from Lemma 5.4(a).

Theorem 5.22. Given a field F of characteristic zero, we have rb(M (F ))=2n 1. n − Proof. From Lemma 5.17(b) and Example 5.18, we have 2n 1 rb(M (F )) 2n. Let − ≤ n ≤ us extend the field F to its algebraic closure F¯ and prove the statement over F¯, as the RB-index does not decrease under a field extension. Consider the Jordan normal form J of the matrix R(1). Since R(1) is nilpotent, we n−1 have either J =0 or J = e12 + e23 + ... + en−1n. In the first case, we apply Lemma 5.4 and get R2n−2 =0 as required. It remains to study the second one. Up to the conjugation with an automorphism of Mn(F ) we may assume that R(1) = J. By Lemma 2.5(d),

J k+1 R(J k)= (38) k +1 and so R(enn)=0. Consider the Z-grading on Mn(F ) from Remark 5.19. We deduce Theorem from the following result. n−1 Lemma 5.23. We have R: Mi Mj for all i = n +1,...,n 2. → j=i+1 − − P Proof. We already know that R maps Mn−1 to zero. For i = n +1,...,n 2, we prove the statement by induction on i (decreasing from n 2 to −n +1). − − − Let i = n 2. From − 2 R(1)R(e1n−1)= R (e1n−1)= R(e1n−1)R(1)

34 n for A =(aij)i,j=1 = R(e1n−1), we conclude that

a21 a22 a23 ... a2n 0 a11 a12 ... a1n−1 a31 a32 a33 ... a3n 0 a21 a22 ... a2n−1 ......  = ......  . a a a ... a  0 a a ... a   n1 n2 n3 nn  n−11 n−12 n−1n−1  0 0 000  0 a a ... a     n1 n2 nn−1     n−2 n−1  Thus, R(e1n−1)= a11E + a12J + ... + a1n−1J + a1nJ . Since Im (R) contains only degenerate matrices, a11 =0. Further, 1 1 R(e )R2(1) = R(e )J 2 = (a J 3 + a J 4 + ... + a J n−1), 1n−1 2 1n−1 2 12 13 1n−2

2 R(e1n−1R (1)) + R(R(e1n−1)R(1)) 2 3 n−2 n−1 = R(a12J + a13J + ... + a1n−2J + a1n−1J ) a J 3 a J 4 a J n−1 = 12 + 13 + ... + 1n−2 . 3 4 n 1 − n−1 Comparing coefficients, we have R(e1n−1) = a1nJ Mn−1. Since R(e1n−1 +e2n) = e /n by (38), we have R(e ) M as well. ∈ 1n 2n ∈ n−1 Let us prove the inductive step for i 0. Consider the matrix unity e1i+1, the proof ≥ n for other unities from Mi is the same. Denote A =(akl)k,l=1 = R(e1i+1). By

2 R(1)R(e1i+1)= R (e1i+1), 2 R(e1i+1)R(1) = R (e1i+1)+ R(e1i+2) we get

a21 a22 a23 ... a2n 0 a11 a12 ... a1n−1 a a a ... a 0 a a ... a  31 32 33 3n   21 22 2n−1  ...... = ...... + B (39) a a a ... a  0 a a ... a   n1 n2 n3 nn  n−11 n−12 n−1n−1  0 0 000  0 a a ... a     n1 n2 nn−1  n−1    2 i ′ ′ for B Mj. Thus, R(e1i+1) = a11E + a12J + a13J + ... + a1i+1J + B for B ∈ j=i+2 ∈ n−1 P Mj. Since Im (R) contains only degenerate matrices, a11 =0. Further, j=i+1 P 1 1 R(e )R2(1) = R(e )J 2 = (a J 3 + ... + a J i+2 + B′J 2). (40) 1i+1 2 1i+1 2 12 1i+1 By the inductive hypothesis,

2 ′′ 2 i+1 ′ R(e1i+1R (1)) + R(R(e1i+1)R(1)) = B + R(a12J + ... + a1i+1J + B J) a J 3 a J 4 a J i+2 = 12 + 13 + ... + 1i+1 + B′′′ (41) 3 4 i +2

35 n−1 n−1 ′′ 2 ′′′ ′ ′′ for B = R(e1i+1R (1)) Mj and B = R(B J)+ B Mj. Comparing ∈ j=i+3 ∈ j=i+3 P P n−1 coefficients in (40) and (41), we have a12 = a13 = ... = a1i+1 =0 and R(e1i+1) Mj. ∈ j=i+1 Finally, let us prove the inductive step for n +1 i< 0. Consider the matrixP unity − ≤ n e|i|+11, the proof for other unities from Mi is the same. Denote A =(akl)k,l=1 = R(e|i|+11). From the formulas

2 R(1)R(e|i|+11)= R (e|i|+11)+ R(e|i|1), 2 R(e|i|+11)R(1) = R (e|i|+11)+ R(e|i|+12), we again deduce (39) by the inductive hypothesis. It is easy to see that all coefficients akl corresponding to the matrix unities e|i|+kj, k =1,...,n i , j k, are zero. Thus, n−1 −| | ≤ R(e|i|+11) Mj. We have proved Lemma and the whole Theorem as well. ∈ j=i+1 P For an algebra A equal A1 A2 ... Ak, we have rb(A) max rb(Ai) by ⊕ ⊕ ⊕ ≥ i=1,...,k{ } Lemma 5.1(i). Suppose that A is a semisimple finite-dimensional associative algebra over k a field F of characteristic zero and A = A for A M (F ). By Lemma 5.4(a) and i i ∼= ni i=1 Theorem 5.22, rb(A) 2 max ni = 1 +L max rb(Ai) . ≤ i=1,...,k i=1,...,k{ } Conjecture 5.24. Let A be a semisimple finite-dimensional associative algebra over a field F of characteristic zero. If A = A ... A for simple A , then rb(A) = 1 ⊕ ⊕ k i max rb(Ai) . i=1,...,k{ }

It is not difficult to verify that Conjecture 5.24 holds for the case A = M2(F ) M2(F ) and for the case A = F F ... F (it is Corollary 5.7). ⊕ ⊕ ⊕ ⊕ 5.5 Simple Jordan algebras and composition algebras Theorem 5.25 ([28]). (a) Let J be a (not necessary simple or finite-dimensional) Jordan algebra of a bilinear form over a field of characteristic not two. Then rb(J) 3. ≤ (b) Let F be an algebraically closed field of characteristic not two and Jn+1(f) be the 2, n =2, simple Jordan algebra of a bilinear form f over F . We have rb(Jn+1(f)) = 3, n 3. ( ≥ An algebra A over a field F with char F =2 is called a composition algebra [74] if there 6 exists a nondegenerate quadratic form n on A satisfying n(xy)= n(x)n(y), x, y A. Every unital composition algebra is alternative and quadratic and has a dimension∈ 1, 2, 4 or 8 over F . Moreover, a composition algebra A is either a division algebra or a split algebra, depending on the existence of a nonzero x A such that n(x)=0. A split composition ∈ algebra is either F , or F F , or M (F ), or C(F ), the matrix CayleyDickson algebra. ⊕ 2

36 Let us give the definition of the product on C(F ) = M (F ) vM (F ). We extend the 2 ⊕ 2 product from M2(F ) on C(F ) as follows:

a(vb)= v(¯ab), (vb)a = v(ab), (va)(vb)= ba,¯ a, b M (F ). (42) ∈ 2

a22 a12 Here a¯ for a =(aij) M2(F ) means the matrix − . ∈ a21 a11 −  Theorem 5.26. Given a unital composition algebra C over a field F of characteristic zero, we have

1, C is a division algebra or dim(C) 2, rb(C)= ≤ (43) (3, C is split and dim(C)=4, 8.

Proof. If C is a division composition algebra, then we have rb(C)=1 by Theorem 2.7. If C is a split composition algebra with dim C 2, then C = F or C ∼= F F . By Corollary 5.7, rb(C)=1. ≤ ⊕ If C is a split composition algebra with dim C =4, then C ∼= M2(F ) and we are done by Statement 5.22. Finally, let C be a split composition algebra with dim C = 8, i.e., C = C(F ), the matrix CayleyDickson algebra. By Statement 2.4(a), each RB-operator on C(F ) is an RB-operator on C(F )(+) which is isomorphic to the simple Jordan algebra of a bilinear form [74, p. 57]. By Theorem 5.25(a), rb(C(F )) 3. Extending the RB-operator (M4) ≤ from M2(F ) on C(F ) by Lemma 5.1(h), we have rb(C(F ))=3. Every simple finite-dimensional Jordan algebra is either simple division Jordan algebra or of Hermitian, Albert or Clifford type [49]. If J is a simple finite-dimensional division Jordan algebra, then rb(J)=1 by Theorem 2.8. The case of simple Jordan algebras of Clifford type was considered in [28], see Theorem 5.25. For one of the cases of Hermitian type, we can say the following

Statement 5.27. We have (a) 2n 1 rb(M (F )(+)) 3n 1,where char F =0, − ≤ n ≤ − (b) 2n 1 rb(M (H)(+)) 3n 1. − ≤ n ≤ − Proof. (a) The lower bound follows from Statement 2.4(a) and Example 5.18, the upper bound follows from Lemma 5.4(b). (b) By the theory of matrices over H [68, 71], the proof is analogous. Finally, we study simple Jordan algebras of Albert type. Over an algebraically closed field F , the only simple Jordan algebra of Albert type is H3(C(F )), the space of Hermitian matrices over C(F ) under the product a b = ab + ba. ◦ Theorem 5.28. Let F be an algebraically closed field of characteristic zero, A be the simple Jordan algebra over F of Albert type, then 5 rb(A) 8. ≤ ≤

37 Proof. As every Jordan algebra of Albert type is cubic, i.e., satisfies a cubic equation, so (R(1))3 =0 for any RB-operator R of weight zero on A. By Lemma 5.4(b), rb(A) ≤ 8. Due to the first Tits construction [40, Chap. IX, § 12], we have A = A1 A2 (+) ⊕ (as vector spaces) for A1 ∼= M3(F ) and A1-module A2, Applying Lemma 5.1(h) and Statement 2.4(a), we have the lower bound rb(A) rb(M (F ))=5. ≥ 3

Remark 5.21. By Example 5.7, we have rb(K3)=2 for the Kaplansky superalgebra K3.

Acknowledgements

Author is grateful to P.S. Kolesnikov for the helpful discussions, comments, correc- tions, and the joint Theorem 3.4. Also, author thanks V.N. Zhelyabin, Yu.N. Maltsev, and Yi Zhang for useful comments. The author was supported by the Austrian Science Foundation FWF grant P28079 and by the Program of fundamental scientific researches of the Siberian Branch of Russian Academy of Sciences, I.1.1, project 0314-2016-0001.

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Vsevolod Gubarev University of Vienna Oskar-Morgenstern-Platz 1, 1090 Vienna, Austria Sobolev Institute of Mathematics Acad. Koptyug ave. 4, 630090 Novosibirsk, Russia e-mail: [email protected]

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