Homology, Homotopy and Applications, vol.4(2), 2002, pp.165–190

TREES, FREE RIGHT-SYMMETRIC ALGEBRAS, FREE NOVIKOV ALGEBRAS AND IDENTITIES

ASKAR DZHUMADIL’DAEV and CLAS LÖFWALL

(communicated by Larry Lambe)

Abstract An algebra with the identity x ∘ (y ∘ z − z ∘ y) = (x ∘ y) ∘ z − (x ∘ z) ∘ y is called right-symmetric. A right-symmetric algebra with the identity x ∘ (y ∘ z) = y ∘ (x ∘ z) is called Novikov. We describe bases of free right-symmetric algebras and free Novikov algebras and give realizations of them in terms of trees. The free Lie algebra is obtained as a Lie subalgebra of the free right-symmetric algebra. We use our methods to study identities of Witt algebras.

To Jan–Erik Roos on his sixty–fifth birthday

1. Introduction

An algebra A over a commutative ring R is an R-module with a bilinear operation ∘ : A × A → A. The R-algebra A is called right-symmetric if it satisfies the identity

a ∘ (b ∘ c − c ∘ b) = (a ∘ b) ∘ c − (a ∘ c) ∘ b .

The reason for this name is the following. If (a, b, c) = a ∘ (b ∘ c) − (a ∘ b) ∘ c is the associator, the identity can be rewritten as a right-symmetric condition of associators

(a, b, c) = (a, c, b).

A right-symmetric algebra is called Novikov if it also satisfies the left-commutativity identity

a ∘ (b ∘ c) = b ∘ (a ∘ c).

Right-symmetric algebras first appeared in Cayley’s paper [3]. In this paper the identity QP U = (Q × P)U + (QP)U is mentioned. From this it is easy to obtain a right-symmetric identity for the right-symmetric Witt algebra. Cayley in fact also considered a realization of the right-symmetric Witt algebra as rooted trees. Rightsymmetric algebras were re-examined in [9], [10], [15]. These algebras have many

Partially supported by The Royal Swedish Academy of Sciences Received February 15, 2001, revised February 8, 2002; published on July 12, 2002. 2000 Mathematics Subject Classification: 17D25, 17B01, 17B66. Key words and phrases: rooted trees, Lie-admissable algebras, right-symmetric algebras, Novikov algebras, vector fields algebras, identities, free basis. \begin{matrix}c \\ ◯ \end{matrix} 2002, Askar Dzhumadil’daev and Clas Löfwall. Permission to copy for private use granted.

other names: Vinberg algebras, Koszul algebras, Gerstenhaber algebras and pre-Lie algebras. An algebra is called left-symmetric, if it satisfies the left-symmetric condition for associators

(a, b, c) = (b, a, c).

Any right (respectively left)-symmetric algebra is left (respectively right)-symmetric under the opposite multiplication a ⋆ b = b ∘ a. Novikov algebras firstly appeared in the study of hydrodynamical equations [1] (see also [11], [12], [13]).

Example 1.1. Let U = k[x_{1}, …, x_{n}] and ∂_{i} = ∂/∂_{x_{i}} be the partial derivative. Then W_{n}^{rsym} = {∑ u_{i}∂_{i} : u_{i} ∈ U} under multiplication u∂_{i}∘v∂_{j} = v∂_{j}(u)∂_{i} is rightsymmetric. This algebra is called the right-symmetric Witt algebra (in n variables). When n = 1 this algebra is Novikov. More generally, if A is a commutative algebra and ∂ is a derivation on A, then A is a Novikov algebra under multiplication a ∗ b = ∂(a)b. In fact, we will prove in Section 7 that any Novikov algebra is a quotient of a subalgebra of an algebra of this kind.

In our paper we describe two bases of free right-symmetric algebras. The first uses the concept of rooted trees, and the basis elements are called t-elements. The other basis is obtained by considering a basis for the free (non-associative) algebra modulo the right-symmetric axiom. Here the basis elements are called r-elements. They have the same structure as the t-elements, but the multiplication rule is different. An r-element may be seen as a non-associative monomial and this basis is therefore suitable when the value of algebra homomorphisms are studied. This is indeed the case in section 10, where T -ideals in the free right-symmetric algebra appear in studying identities of the Witt algebra (see below).

On the other hand, the multiplication of t-elements has a nice explicit form, which sometimes makes it easier to use this basis.

The basis for the free right-symmetric algebra in terms of trees was established independently in [4]. Tree algebras also appears in quantum field theory [5] and in numerical analysis [2].

We also give a description of a basis in the free Novikov algebra. We prove that the number of basis elements in degree n of the free Novikov algebra with 1 generator is equal to the number of non-ordered partitions of n − 1.

For the free Novikov algebra on k generators, this number is the coefficient of y^{−1}z^{n} in the series

∏_{j=−1}^{∞} \frac{1}{(1 − y^{j}z)^{k}} .

The identities studied in section 10 are the following, which we call the left and right standard polynomial.

s_{k+1}^{l}(X_{0}, …, X_{k}) = ∑_{σ∈Sym_{k}} sign(σ)X_{σ(1)} ∘ (X_{σ(2)} ∘ ⋯ ∘ (X_{σ(k)} ∘ X_{0}) ⋯)

s_{k}^{r}(X_{1}, …, X_{k}) = ∑_{σ∈Sym_{k}} sign(σ)(⋯ (X_{σ(1)} ∘ X_{σ(2)}) ∘ ⋯ ∘ X_{σ(k−1)}) ∘ X_{σ(k)}

They are obtained from the standard skew-symmetric associative identity

s_{k}(t_{1}, …, t_{k}) = ∑_{σ∈Sym_{k}} sign(σ)t_{σ(1)} ⋯ t_{σ(k)}

in the following way. Let l_{X} and r_{X} be the left and right multiplication operators: l_{X}(Y) = X ∘ Y, r_{X}(Y) = Y ∘ X. Then s_{k+1}^{l} is obtained by substituting l_{X_{i}} for t_{i} in s_{k} and apply the result on X_{0}, while s_{k}^{r} is obtained by substituting r_{X_{i}} for t_{i} in s_{k} and apply the result on a unit e (which is added if necessary).

In [8] some identities of the right-symmetric Witt algebra was found. It was proved that W_{n}^{rsym} satisfies the left standard polynomial identity of degree 2n + 1

s_{2n+1}^{l} = 0.

It is not difficult to prove that W_{n}^{rsym} also satisfies the right standard polynomial identity

s_{n^{2}+2n}^{r} = 0.

A conjecture in [8] states that

s_{n^{2}+2n−1}^{r} = 0

is also a polynomial identity and that this identity is minimal for right polynomial identities. It is easy to see that in any right-symmetric algebra

s_{3}^{l} = 0 ⇒ s_{q}^{r} = 0, q ⩾ 3.

In our paper we prove that, given k > 3, there is a right-symmetric algebra which satisfies the identity s_{k}^{l} = 0 but which does not satisfy the identity s_{q}^{r} = 0 for any q.

In particular, for n > 1, the identity s_{n^{2}+2n}^{r} = 0 does not follow from s_{2n+1}^{l} = 0, so, W_{n}^{rsym} has at least two independent polynomial identities when n > 1.

The key method here is to introduce in the free right-symmetric algebra

• a compatible order and

• a non-archimedian norm The compatible order in the free right-symmetric algebra is very much like the

orders studied in Gröbner theory for associative algebras and Lie algebras. Using this method, we prove that the Lie subalgebra generated by Ω of the free rightsymmetric algebra on Ω under the Lie commutator [X, Y] = X ∘ Y − Y ∘ X is free as a Lie algebra.

The non-archimedian norm allows us to make the free right-symmetric algebra to a topological algebra with a decreasing filtration of balls, such that these balls are closed under the action of algebra endomorphisms.

2. The set of rooted trees

Let Ω be any set. The set of rooted trees with nodes labelled from Ω, denoted T (Ω), is defined in the following way. Define recursively a set T̂(Ω) by the rules given below, where t is just a formal symbol.

• a ∈ Ω ⇒ a ∈ T̂(Ω)

• a ∈ Ω, x_{1}, …, x_{n} ∈ T̂(Ω) ⇒ t(a, x_{1}, …, x_{n}) ∈ T̂(Ω) We do not exclude the possibility that n = 0 in the second clause above, so t(a) ∈ T̂(Ω) if a ∈ Ω. The set T (Ω) is now defined as T̂(Ω)/ ∼, where ∼ is the least equivalence relation that satisfies

• a ∼ t(a)

• t(a, x_{1}, …, x_{n}) ∼ t(a, x_{i_{1}}, …, x_{i_{n}})

if (i_{1}, …, i_{n}) is a permutation of (1, …, n).

• t(a, x_{1}, …, x_{n}) ∼ t(a, y_{1}, …, y_{n})

if x_{i} ∼ y_{i} for i = 1, …, n. An equivalence class in T (Ω) is denoted by any of its representatives in T̂(Ω).

Example 2.1. We have

t(a, b, t(b, a, b)) ∈ T ({a, b}) and t(a, b, t(b, a, b)) = t(a, t(b, b, a), b) .

It is clear that the above definition of T (Ω) coincides with the graph theoretical notion of “rooted labelled trees”, where two such trees are considered to be equal if there is an isomorphism of labelled graphs which sends the root to the root.

Example 2.2. The equality t(a, b, t(b, a, b)) = t(a, t(b, b, a), b) may be illustrated by the following isomorphic rooted trees.

[figure: a; b; b; a; b; ≅; a; b; b; a; b]

Definition 2.3. We make the following definitions for y ∈ T (Ω).

(i) The length of y, denoted |y|, is defined by

|t(a, x_{1}, …, x_{n})| = 1 + |x_{1}| + ⋯ + |x_{n}| for n ⩾ 0.

(ii) The number of branches in y, denoted br(y), is defined by

br(t(a, x_{1}, …, x_{n})) = n for n ⩾ 0.

(iii) The operation • on T (Ω) is defined as follows, where a ∈ Ω and x_{1}, …, x_{n}, y ∈ T (Ω),

t(a, x_{1}, …, x_{n}) • y = t(a, x_{1}, …, x_{n}, y).

The operation • is non-associative and non-commutative. It satisfies however the right-commutative identity (a • b) • c = (a • c) • b (in fact we will prove later that T (Ω) is the free right-commutative magma on Ω, cf. Proposition 6.3).

For a given set X, let Mon(X) denote the free commutative monoid on X; i.e., Mon(X) consists of words (including the empty word) in the alphabet X, where the letters may be written in any order. We have a bijection

T (Ω) → Ω × Mon(T (Ω))  (1)

where the correspondence is t(a, x_{1}, …, x_{n}) ↦ (a, x_{1} ⋯ x_{n}) and a ↦ (a, 1).

Let T (Ω)_{n} denote the subset of T (Ω) consisting of all elements of length n. Suppose Ω is finite, |Ω| = k. Then T (Ω)_{n} is finite and we may form the generating series f_{k}(z) = ∑_{n⩾1} a(n, k)z^{n} where a(n, k) = |T (Ω)_{n}|.

The length of a monomial x_{1} ⋯ x_{j} is defined as |x_{1}| + ⋯ + |x_{j}|. It is well-known that the generating series for Mon(T (Ω)) with respect to length is (|Ω| = k)

exp(f_{k}(z)) =^{def} ∏_{n=1}^{∞} \frac{1}{(1 − z^{n})^{a(n,k)}} .

Hence by (1), f_{k}(z) satisfies the following functional equation, which may be found in [3].

f_{k}(z) = kz exp(f_{k}(z))  (2)

This equation makes it possible to compute the numbers a(n, k) recursively. E.g., if |Ω| = 1 then a(1, 1) = a(2, 1) = 1, a(3, 1) = 2, a(4, 1) = 4, … where the corresponding trees look as follows

[figure]

It is easily seen that a(n, k) is a polynomial in k of degree n for each n. The first polynomials are

a(1, k) = k, a(2, k) = k^{2}

a(3, k) = \frac{3}{2}k^{3} + \frac{1}{2}k^{2}, a(4, k) = \frac{8}{3}k^{4} + k^{3} + \frac{1}{3}k^{2} .

We may write f_{k}(z) as a power series in k:

f_{k}(z) = φ_{1}(z)k + φ_{2}(z)k^{2} + φ_{3}(z)k^{3} + ⋯

We will now give recursive formulas for the functions φ_{1}, φ_{2}, …. We have

exp(f_{k})(z) = ∏_{j=1}^{∞} exp(φ_{j}(z)k^{j}) = ∏_{j=1}^{∞}(exp(φ_{j}(z)))^{k^{j}} .

Hence by (2),

\frac{f_{k}(z)}{kz} = ∏_{j=1}^{∞}(exp(φ_{j}(z)))^{k^{j}} = e^{∑_{j=1}^{∞} log(exp(φ_{j}))k^{j}} .

Definition 2.4. For any series without constant term, g(z) = ∑_{n⩾1} b_{n}z^{n}, let logexp(g) denote the series ∑_{n⩾1} −b_{n} log(1 − z^{n}). Also, let y^{(n)} = y^{n}/n!.

The multinomial theorem may be written

(y_{1} + ⋯ + y_{r})^{(n)} = ∑ ; ∑ n_{i}=n x_{1}^{(n_{1})} · . . . · x_{r}^{(n_{r})}

and we get

\frac{f_{k}(z)}{kz} = ∑_{n=1}^{∞} ∑ ; ∑ jn_{j}=n(ψ_{1}^{(n_{1})}ψ_{2}^{(n_{2})} · . . .)k^{n}

where ψ_{j} = logexp(φ_{j}).

Hence

\frac{1}{z}(φ_{1}(z) + φ_{2}(z)k + φ_{3}(z)k^{2} + ⋯) =

1 + ψ_{1}(z)k + (ψ_{1}^{(2)}(z) + ψ_{2}(z))k^{2} + (ψ_{1}^{(3)}(z) + ψ_{1}(z)ψ_{2}(z) + ψ_{3}(z))k^{3} + ⋯

and we have proved the following theorem

Theorem 2.5. Let Ω be a set with k elements and let f_{k}(z) = φ_{1}(z)k + φ_{2}(z)k^{2} + φ_{3}(z)k^{3} + ⋯ be the generating function for T (Ω). Then

φ_{1}(z) = z
φ_{2}(z) = zlogexp(z) = −z log(1 − z)
φ_{3}(z) = z(\frac{1}{2}(log(1 − z))^{2} + logexp(φ_{2}))
φ_{4}(z) = z(−\frac{1}{6}(log(1 − z))^{3} − log(1 − z)logexp(φ_{2}) + logexp(φ_{3}))
…

and in general, given φ_{1}, …, φ_{n−1}, we have

φ_{n}(z)/z = ∑ ; ∑ jn_{j}=n(ψ_{1}^{(n_{1})}ψ_{2}^{(n_{2})} · . . .)

where ψ_{j} = logexp(φ_{j}).

3. The tree algebra

Let R be a commutative ring. For any set Ω we define the “tree algebra”, T (Ω), as the free R-module on T (Ω). The operation • on T (Ω) is extended to T (Ω) by linearity. A bilinear multiplication ∘ on T (Ω) is defined recursively on basis elements as follows, where a ∈ Ω, x_{1}, …, x_{n}, y ∈ T (Ω).

a ∘ y = a • y = t(a, y)
t(a, x_{1}, …, x_{n}) ∘ y = t(a, x_{1}, …, x_{n}, y) +

∑_{i=1}^{n} t(a, x_{1}, …, x̂_{i}, …, x_{n}) • (x_{i} ∘ y)

In other words, y_{1} ∘ y_{2} is obtained as follows. Add a new branch to the root of y_{2} and “plant” this graph to each node of y_{1} and add the resulting trees. In this way (T (Ω), ∘) becomes a positively graded (non-associative) R-algebra, T (Ω) = ⊕_{n⩾1}T (Ω)_{n}, where T (Ω)_{n} is the free R-module on T (Ω)_{n}. We will use the notation T (Ω) for this algebra (without mentioning the operation ∘).

Proposition 3.1. The operations • and ∘ on T (Ω) satisfy the following identities. (i)

(x • y) • z = (x • z) • y

i.e., (T (Ω), •) is right-commutative. (ii)

(x • y) ∘ z = (x ∘ z) • y + x • (y ∘ z)

i.e., (T (Ω), ∘) is a derivation algebra of (T (Ω), •) (iii)

(x ∘ y) ∘ z = (x ∘ z) ∘ y + x ∘ (y ∘ z) − x ∘ (z ∘ y)

i.e., T (Ω) is right-symmetric.

Proof. The first two identities follow directly from the definitions. We use (i) and (ii) and induction over the length of x to prove (iii). Suppose first that x = a ∈ Ω. Then (a ∘ y) ∘ z − a ∘ (y ∘ z) = (a • y) ∘ z − a • (y ∘ z) and by (ii) this equals (a ∘ z) • y = (a • z) • y. In the same way (a ∘ z) ∘ y − a ∘ (z ∘ y) = (a • y) • z and hence by (i) we get (iii). If x ∉ Ω we may write x = x_{1} • x_{2}, where |x_{1}| < |x| and |x_{2}| < |x|. By (ii)

((x_{1} • x_{2}) ∘ y) ∘ z = ((x_{1} ∘ y) • x_{2}) ∘ z + (x_{1} • (x_{2} ∘ y)) ∘ z
= ((x_{1} ∘ y) ∘ z) • x_{2} + (x_{1} ∘ y) • (x_{2} ∘ z) +
(x_{1} ∘ z) • (x_{2} ∘ y) + x_{1} • ((x_{2} ∘ y) ∘ z).

Hence

(x ∘ y) ∘ z − (x ∘ z) ∘ y = ((x_{1} ∘ y) ∘ z − (x_{1} ∘ z) ∘ y) • x_{2} +
x_{1} • ((x_{2} ∘ y) ∘ z − (x_{2} ∘ z) ∘ y)

and hence by induction and (ii)

(x ∘ y) ∘ z − (x ∘ z) ∘ y = (x_{1} ∘ (y ∘ z − z ∘ y)) • x_{2} +
x_{1} • (x_{2} ∘ (y ∘ z − z ∘ y))
= (x_{1} • x_{2}) ∘ (y ∘ z − z ∘ y) = x ∘ (y ∘ z − z ∘ y).

Now suppose Ω is well-ordered. The order may be extended to a well-order < on T (Ω) in such a way that y_{1} < y_{2} if |y_{1}| < |y_{2}| or if |y_{1}| = |y_{2}| and br(y_{1}) < br(y_{2}). Then the following properties are easily proven.

Proposition 3.2. If a well-order < on T (Ω) satisfies that y_{1} < y_{2} if |y_{1}| = |y_{2}| and br(y_{1}) < br(y_{2}), then the leading term of t(a, x_{1}, …, x_{n}) ∘ y is t(a, x_{1}, …, x_{n}) • y and the leading term of (… (a ∘ x_{1}) ∘ x_{2}) ∘ ⋯ ∘ x_{n}) is t(a, x_{1}, …, x_{n}).

Proof. It follows from the definition that t(a, x_{1}, …, x_{n}) • y is the only term of t(a, x_{1}, …, x_{n}) ∘ y which has n + 1 branches and all terms have the same length (we ignore the coefficients, the leading term has however coefficient 1). In the same way, t(a, x_{1}, …, x_{n}) is the only term with n branches in (… (a ∘ x_{1}) ∘ x_{2}) ∘ ⋯ ∘ x_{n}).

Proposition 3.3. As an R-algebra, T (Ω) is generated by Ω.

Proof. We prove by induction over a well-order on T (Ω) satisfying the assumptions in Proposition 3.2, that any element in T (Ω) is generated by Ω. Suppose this is true for all elements less than t(a, x_{1}, …, x_{n}). Since x_{i} < t(a, x_{1}, …, x_{n}) for all i, it follows by induction that (… (a∘x_{1})∘x_{2})∘⋯∘x_{n}) is generated by Ω. By Proposition 3.2, this expression has t(a, x_{1}, …, x_{n}) as leading term. Since by induction all other terms are generated by Ω, it follows that t(a, x_{1}, …, x_{n}) is generated by Ω.

4. Super-trees

We will shortly describe how signs may be introduced in the notions of tree and tree algebra. Suppose Ω is a super-set; i.e., the elements in Ω are divided into even and odd elements. We will use the notation ϵ(x, y) for the sign introduced when x and y are interchanged in a formula; i.e., ϵ(x, y) = −1 if both x and y are odd and +1 otherwise. The function ϵ is supposed to be bi-additive (with values in the multiplicative group {−1, 1}).

The notion of odd and even elements is naturally extended to trees. A tree is a signed expression ±t(a, x_{1}, …, x_{n}) or 0. The definition of equality for trees is changed to

t(a, x_{1}, …, x_{n}) = ϵ(x_{i}, x_{i+1})t(a, x_{1}, …, x_{i+1}, x_{i}, …, x_{n})

for any i = 1, …, n − 1. Also if y is odd, then for any n ⩾ 0

t(a, y, y, x_{1}, …, x_{n}) = 0.

We let Mon(X), where X is a super-set, denote the monoid of signed monomials and 0, where an odd variable occurs at most once in a monomial. We have a bijection

T (Ω) → Ω × Mon(T (Ω))/(a, 0) = 0.

If Ω is finite, a functional equation for the generating series of all unsigned non-zero trees, T (Ω)^{+}, may be given. The generating series for a graded super-set, which is finite in each degree, is a power series in z and y, where y^{2} = 1. If f(z, y) = ∑ a_{n}z^{n} + y ∑ b_{n}z^{n}, then a_{n} is the number of even elements and b_{n} is the number of odd elements in degree n.

If f(z, y) = ∑_{n⩾1} a_{n}z^{n}+y(∑_{n⩾1} b_{n}z^{n}) is the generating function for a positively graded super-set, we make the following definition.

exp(f(z, y)) =^{def} ∏_{n=1}^{∞} \frac{(1 + yz^{n})^{b_{n}}}{(1 − z^{n})^{a_{n}}}

Suppose Ω consists of k_{0} even elements and k_{1} odd elements and let f_{k_{0},k_{1}}(z, y) be the generating series for T (Ω)^{+}. Then we have

f_{k_{0},k_{1}}(z, y) = (k_{0} + k_{1}y)z exp(f_{k_{0},k_{1}}(z, y)).

The definition of the operation • is the same as before. The tree algebra, T (Ω), is defined as the free R-module on T (Ω)^{+}. The operation • on T (Ω) is extended to T (Ω) by linearity. The definition of ∘ on T (Ω) is changed to

a ∘ y = a • y = t(a, y)
t(a, x_{1}, …, x_{n}) ∘ y = t(a, x_{1}, …, x_{n}, y) +

∑_{i=1}^{n} ϵ(x_{i}, x_{i+1} + … + x_{n})t(a, x_{1}, …, x̂_{i}, …, x_{n}) • (x_{i} ∘ y).

Proposition 4.1. The operations • and ∘ on T (Ω), where Ω is a super-set, satisfy the following identities. (i)

(x • y) • z = ϵ(y, z)(x • z) • y

and

(x • y) • y = 0 if y is odd

i.e., (T (Ω), •) is right-super-commutative. (ii)

(x • y) ∘ z = ϵ(y, z)(x ∘ z) • y + x • (y ∘ z)

(T (Ω), ∘) is a derivation algebra of (T (Ω), •) (iii)

(x ∘ y) ∘ z = ϵ(y, z)(x ∘ z) ∘ y + x ∘ (y ∘ z) − ϵ(y, z)x ∘ (z ∘ y)

i.e., T (Ω) is right-super-symmetric.

5. Compatible orders

In this section we assume that R is an integral domain. In the applications in later sections we will need that the order on T (Ω) is not just a well-order but also satisfies a compatibility condition which is made precise below.

Definition 5.1. Let < be a well-order on T (Ω). We make the following definitions.

• The order < is •-compatible if x < y ⇒ x • z < y • z and z • x < z • y for x, y, z ∈ T (Ω).

• For x ∈ T (Ω)\{0}, lead(x) ∈ T (Ω) is the maximal basis element which occurs in x with non-zero coefficient.

• The order < is ∘-compatible if x < y ⇒ lead(x∘z) < lead(y∘z) and lead(z∘x) < lead(z∘y) for x, y, z ∈ T (Ω).

• The order < is ∘-leading if lead(x ∘ y) = x • y for x, y ∈ T (Ω).

• For x, y ∈ T (Ω) \ {0}, x < y ⇔ lead(x) < lead(y).

Proposition 5.2. Suppose < is a well-order on T (Ω). Then the following holds.

(i) If < is •-compatible and ∘-leading, then < is ∘-compatible.

(ii) If < is ∘-compatible, then

lead(x ∘ y) = lead(lead(x) ∘ lead(y)) for x, y ∈ T (Ω) \ {0}. (iii) If < is •-compatible and ∘-leading, then lead(x ∘ y) = lead(x) • lead(y) for

x, y ∈ T (Ω) \ {0}. (iv) If < is ∘-compatible, then x < y ⇒ x ∘ z < y ∘ z and z ∘ x < z ∘ y for

x, y, z ∈ T (Ω) \ {0}.

Proof. The first statement follows directly from the definitions. To prove (ii), suppose first that x ∈ T (Ω). Let y = r_{1}y_{1}+r_{2}y_{2}+…+r_{n}y_{n}, where r_{i} ∈ R and y_{i} ∈ T (Ω) for i = 1, …, n, r_{1} ≠ 0 and y_{1} > y_{2} > … > y_{n}. Since < is ∘-compatible we have x ∘ y_{1} > x ∘ y_{2} > … > x ∘ y_{n}. Hence lead(x ∘ y) = lead(x ∘ y_{1}) = lead(x ∘ lead(y)) and (ii) is proved when x ∈ T (Ω).

Next suppose x = s_{1}x_{1} + s_{2}x_{2} + … + s_{n}x_{n}, where s_{i} ∈ R and x_{i} ∈ T (Ω) for i = 1, …, n, s_{1} ≠ 0 and x_{1} > x_{2} > … > x_{n}. Then x_{1} ∘ y > x_{2} ∘ y > … > x_{n} ∘ y, since

lead(x_{1} ∘ y) = lead(x_{1} ∘ lead(y)) > lead(x_{2} ∘ lead(y)) = lead(x_{2} ∘ y),

where the equalities follow from above and the inequality follows from the assumption that < is ∘-compatible. Hence lead(x ∘ y) = lead(x_{1} ∘ y) and again from above we get lead(x ∘ y) = lead(x_{1} ∘ lead(y)) = lead(lead(x) ∘ lead(y)).

(iii) follows from (i) and (ii).

To prove (iv), suppose x, y ∈ T (Ω) and x < y. Then by (ii) and since < is ∘-compatible we have

lead(x ∘ z) = lead(lead(x) ∘ lead(z)) < lead(lead(y) ∘ lead(z)) = lead(y ∘ z).

In the same way it is proven that lead(z ∘ x) < lead(z ∘ y).

To find a •-compatible order on T (Ω) we will use a total order on Mon(X) defined for each totally ordered set X, which extends the order on X and is functorial in X. Such an order on Mon(X) is called “natural”. It is called “compatible” if it is compatible with the monoid operation on Mon(X). A natural order on Mon(X) satisfies in particular the following,

a <_{X′} b ⇔ a <_{X} b for all a, b ∈ Mon(X′) whenever X′ ⊂ X as totally ordered sets.

Starting with a well-order on Ω and a compatible natural well-order on Mon(X) (such as the lexicographic order), we may use equation (1) in section 2 to define a total order on all elements in T (Ω) of length n, for n = 1, 2, …. In this way we obtain a •-compatible total order on T (Ω). The order might not be a well-order; e.g., if the number of branches (which corresponds to the degree of a monomial) is taken before the length (which corresponds to an extra weight that the variables have), we have the following infinite sequence in T ({a}).

t(a, a, a) > t(a, t(a, a, a)) > t(a, t(a, t(a, a, a))) > …

On the other hand, if length is considered as the first criterion, then we obviously obtain a well-order on T (Ω). To obtain a ∘-compatible order, we also want the order to be ∘-leading. We give below the definition of two orders on T (Ω), which have all desired properties (proved in Proposition 5.4). The proof is easy for the first one, but the second one will be more useful for us in the applications.

Definition 5.3. Given a well-order on Ω, two orders ≺_{deglex} and ≺_{revlex} on T (Ω) are defined as follows.

(i) We have

x = t(a, x_{1}, …, x_{n}) ≺_{deglex} t(b, y_{1}, …, y_{m}) = y if

• |x| < |y|

• |x| = |y| and n < m

• |x| = |y|, n = m and (x_{n}, …, x_{1}, a) ≺_{deglex} (y_{n}, …, y_{1}, b) in lexicographic sense, where x_{1} ⪯_{deglex} … ⪯_{deglex} x_{n} and y_{1} ⪯_{deglex} … ⪯_{deglex} y_{n}.

(ii) We have

x = t(a, x_{1}, …, x_{n}) ≺_{revlex} t(b, y_{1}, …, y_{m}) = y if

• |x| < |y|

• |x| = |y| and t(a, x_{1}, …, x_{n−1}) ≺_{revlex} t(b, y_{1}, …, y_{m−1}), where x_{i} ⪯_{revlex} x_{n} for i < n and y_{i} ⪯_{revlex} y_{m} for i < m.

• |x| = |y|, t(a, x_{1}, …, x_{n−1}) = t(b, y_{1}, …, y_{m−1}) and x_{n} ≺_{revlex} y_{m}, where x_{i} ⪯_{revlex} x_{n} for i < n and y_{i} ⪯_{revlex} y_{m} for i < m.

Proposition 5.4. Given a well-order on Ω, the orders ≺_{deglex} and ≺_{revlex} defined above are •-compatible, ∘-compatible and ∘-leading well-orders on T (Ω), and hence als

Proof. The first order, ≺_{deglex}, is ∘-leading by Proposition 3.2 and the natural compatible well-order on Mon(X) used in the definition is “weight-degree-lexico-graphic” (so the order should rather be called “weightdeglex”). Hence, by Proposition 5.2, the order is ∘-compatible.

The order on Mon(X) used in the definition of ≺_{revlex} may be described in the following way, which shows that it is a natural compatible well-order. A monomial m is written as a product of monomials m_{1}m_{2} ⋯ m_{n}, where each monomial m_{i} is a product of variables of the same weight (i.e., length) and the weight of the variables in m_{1} is greater than the weight of the variables in m_{2} and so on. Now m = m_{1}m_{2} ⋯ m_{n} > m′_{1}m′_{2} ⋯ m′_{k} = m′ if |m| > |m′|, or |m| = |m′| and |m_{i}| = |m′_{i}| for i = 1, … j − 1 and |m_{j}| < |m′_{j}| for some j, or n = k and |m_{i}| = |m′_{i}| for i = 1, …, n and m > m′ by the reverse lexicographic principle; i.e., the first variable which differs, counted from the right, is greater in m. This proves that the order is •-compatible.

By Proposition 5.2 it is enough to prove that it is also ∘-leading; i.e., given a ∈ Ω and x_{1} ⩽ … ⩽ x_{n}, y ∈ T (Ω), we must prove that

t(a, x_{1}, …, x̂_{i}, …, x_{n}) • (x_{i} ∘ y) < t(a, x_{1}, …, x_{n}, y)

for all i. We do this by induction over |t(a, x_{1}, …, x_{n}, y)|. By the induction hypothesis we get that lead(x_{i} ∘ y) = x_{i} • y. Since < is •-compatible, we get

lead(t(a, x_{1}, …, x̂_{i}, …, x_{n}) • (x_{i} ∘ y)) = t(a, x_{1}, …, x̂_{i}, …, x_{n}) • (x_{i} • y)

and hence the claim to be proven is

t(a, x_{1}, …, x̂_{i}, …, x_{n}) • (x_{i} • y) < t(a, x_{1}, …, x_{n}, y).

Suppose i < n. By induction we get

t(a, x_{1}, …, x̂_{i}, …, x_{n−1}) • (x_{i} • y) < t(a, x_{1}, …, x_{n−1}, y)

and the claim follows by multiplying this to the right with x_{n}, using that < is •-compatible (and that • is right-commutative).

Suppose i = n. We have to prove that

t(a, x_{1}, …, x_{n−1}, x_{n} • y) < t(a, x_{1}, …, x_{n−1}, x_{n}, y).

But this follows from the definition of the order, since |x_{n} • y| > |x_{n}| and |x_{n} • y| > |y|.

6. Free algebras

The free magma on a set Ω, M(Ω), is recursively defined as a set of parenthesized strings in the alphabet Ω in the following way.

• a ∈ Ω ⇒ a ∈ M(Ω)

• x, y ∈ M(Ω) ⇒ (xy) ∈ M(Ω)

Any element in M(Ω) \ Ω may be uniquely written as ((… ((ax_{1})x_{2}) …)x_{n}), where a ∈ Ω and x_{1}, …, x_{n} ∈ M(Ω). This expression will be written as r(a, x_{1}, …, x_{n}), where x_{1}, …, x_{n} are supposed to be of the same form (or belong to Ω) and we say in this case that r(a, x_{1}, …, x_{n}) is written in normal form. To be able to make substitutions, we also allow general expressions r(x_{1}, …, x_{n}) (which is equal to ((… ((x_{1}x_{2})x_{3}) …)x_{n})). The rule for normalizing a general expression is

r(r(x_{1}, …, x_{n}), y_{1}, …, y_{m}) = r(x_{1}, …, x_{n}, y_{1}, …, y_{m}).

The length of an element is defined in the same way as for T (Ω). The multiplication ∗ on M(Ω) (defined by x ∗ y = (xy)) satisfies the rule r(a, x_{1}, …, x_{n}) ∗ y = r(a, x_{1}, …, x_{n}, y) (and a ∗ y = r(a, y) if a ∈ Ω). The free R-module on M(Ω) with multiplication defined as the linear extension of ∗ is called the free (non-associative) algebra on Ω and is denoted M(Ω). Observe that M(Ω) has no unit.

The free right-symmetric R-algebra on Ω, denoted RS(Ω), is defined as the quotient of M(Ω) with the twosided ideal generated by all elements of the form (x ∗ y) ∗ z − (x ∗ z) ∗ y − x ∗ (y ∗ z) + x ∗ (z ∗ y).

The free right-commutative R-algebra on Ω, denoted RC(Ω), is defined as the quotient of M(Ω) by the twosided ideal generated by all elements of the form (x ∗ y) ∗ z − (x ∗ z) ∗ y. In the same way the free right-commutative magma on Ω, denoted RC(Ω), is defined as the quotient of M(Ω) by the congruence relation generated by (x ∗ y) ∗ z ≡ (x ∗ z) ∗ y.

Suppose Ω is well-ordered by < . The order may be extended to all of M(Ω) by taking length first and then for two elements of the same length, (xy) < (x′y′) if x < x′ or x = x′ and y < y′.

Definition 6.1. An element in M(Ω) is called right-symmetric normal if it is of the form a ∈ Ω or r(a, x_{1}, …, x_{n}), where x_{1} ⩽ x_{2} ⩽ … ⩽ x_{n} and x_{1}, …, x_{n} are right-symmetric normal.

Proposition 6.2. The R-algebras RS(Ω) and RC(Ω) are generated as R-modules by elements represented by right-symmetric normal elements in M(Ω). Also any element in RC(Ω) may be written in right-symmetric normal form.

Proof. We prove the statement by induction over the well-order < defined above. Suppose the statement is true for all elements less than r(a, x_{1}, …, x_{n}). Then the statement is true for r(a, x_{1}, …, x_{n−1}) and for x_{n}. We may therefore assume that x_{1}, …, x_{n} are of the right form and x_{1} ⩽ x_{2} ⩽ … ⩽ x_{n−1}. Put x = r(a, x_{1}, …, x_{n−2}), y = x_{n−1} and z = x_{n}. If y > z, we may replace r(a, x_{1}, …, x_{n}) = (x ∗ y) ∗ z by (x ∗ z) ∗ y + x ∗ (y ∗ z) − x ∗ (z ∗ y) in the case of RS(Ω) and by (x ∗ z) ∗ y in the case of RC(Ω) or RC(Ω). The new elements are less than (x ∗ y) ∗ z and we may use the induction hypothesis to get the result.

Theorem 6.3. There exist unique isomorphisms of free R-algebras over Ω extending the identity on Ω,

RS(Ω) ≅ (T (Ω), ∘)
RC(Ω) ≅ (T (Ω), •)

and a unique isomorphism of free magmas over Ω extending the identity on Ω,

RC(Ω) ≅ (T (Ω), •).

Proof. Since obviously T (Ω) is generated by Ω under •, we get a surjective homomorphism y ↦ y^{t} from M(Ω) to T (Ω). The element y^{t} is obtained by replacing all the prefixes r by t in y. Since (T (Ω), •) is right-commutative, the map factors through RC(Ω). To prove injectivity, assume we have two elements x, y ∈ RC(Ω) such that x^{t} = y^{t} in T (Ω). By Proposition 6.2 we may assume that x and y are represented by elements where all branches are in linear order. But then x and y have to be identical, since x^{t} = y^{t}. This proves the last two statements.

By Proposition 3.3 we have a surjective homomorphism φ : M(Ω) → (T (Ω), ∘), and by Proposition 3.1 the map factors through RS(Ω). To prove injectivity, suppose φ(x) = 0. Then there is a finite subset Ω_{0} of Ω such that x ∈ M(Ω_{0}). Hence we may assume that Ω is finite.

Let N be the number of normal right-symmetric elements in M(Ω) of length n. By Proposition 6.2 these elements generate the elements of length n in M(Ω) modulo the right-symmetric axiom. Obviously we also have that N is the number of elements in T (Ω) of length n. Hence we get a surjective R-linear map R^{N} → R^{N} and we may apply the well-known Lemma 6.4 below.

Lemma 6.4. Let R be a commutative ring with unit and f : R^{N} → R^{N} an R-linear surjective map. Then f is an isomorphism.

Proof. Let A be the matrix of f. Since f is surjective, there is a solution to the matrix equation AX = I. Taking determinants gives det(A) · det(X) = 1. Hence det(A) is a unit in R and A is invertible (the standard formula for A^{−1} given in linear algebra is valid).

Remark 6.5. The lemma is true even when R^{N} is replaced by a finitely generated R-module M (consider M as an R[x]-module).

It is possible to tell more about the isomorphisms in the theorem above.

Proposition 6.6. Let f : M(Ω) → (T (Ω), ∘) and g : M(Ω) → (T (Ω), •) be the unique homomorphisms extending the identity on Ω. Given a ∘-compatible, ∘-leading well-order, ≺, on T (Ω) we have lead(f(x)) = g(x) and the compatible order on RS(Ω) which is defined through the isomorphism RS(Ω) ≅ (T (Ω), ∘) (Theorem 6.3) has the same definition on normal right-symmetric elements as ≺ has on T (Ω).

Proof. Let x = r(a, x_{1}, …, x_{n}) ∈ M(Ω). We have that g(x) = x^{t} where x^{t} is obtained from x by replacing all the prefixes r by t. We prove that lead(f(x)) = x^{t} by induction on the length of x. Indeed, by induction and by Proposition 5.2 (ii), we get that lead(f(x)) = lead(t(a, x_{1}^{t}, …, x_{n−1}^{t}) ∘ x_{n}^{t}). Since the order is ∘-leading, this equals t(a, x_{1}^{t}, …, x_{n}^{t}) = x^{t}. The second statement is now a direct consequence.

Remark 6.7. The injectivity of the map φ in the proof of Theorem 6.3 may also be deduced from Proposition 6.6.

7. Free Novikov algebras

A right-symmetric R-algebra is called Novikov if it is also left symmetric; i.e., it satisfies the axiom x(yz) = y(xz). The free Novikov algebra on a set Ω is denoted N (Ω) and is defined as the quotient of M(Ω) by the two-sided ideal generated by all expressions

x(yz) − y(xz) and (xy)z − (xz)y − x(yz) + x(zy) .

We want to find a minimal generating set for N (Ω) as an R-module. A step in this direction will be the “ordered nests”.

A subexpression of x = r(a, x_{1} …, x_{n}) is either x itself or a or a subexpression of some of x_{1}, …, x_{n}.

Definition 7.1. An r-expression x in M(Ω) is called a nest if each subexpression has at most one branch which is not in Ω. This means that either x ∈ Ω or it has the form x = r(a_{0}, a_{1} …, a_{n}, y), where a_{i} ∈ Ω, n ⩾ 0 and y is a nest. Let in the first case d_{1}(x) = 0 and root(x) be undefined and in the second case d_{1}(x) = n and root(x) = a_{0}. A nest x is called ordered (given a total order on Ω) if either x ∈ Ω or, in the second case, y ∈ Ω or y is ordered and either d_{1}(x) < d_{1}(y) or d_{1}(x) = d_{1}(y) and root(x) ⩽ root(y).

In any r-expression the atoms are divided into two groups, the roots and the leaves. A root of x is an element in Ω which occurs in x directly after a left parenthesis. The other atoms are called leaves. An element in Ω has by definition no root and one leaf, so any expression has at least one leaf and the expressions with exactly one leaf are the nests r(a_{1}, r(a_{2}, …, r(a_{n−1}, a_{n})) …), while the expressions with exactly one root are the (ordered) nests r(a_{0}, a_{1}, …, a_{n}). Given a total order on Ω, we now define the notion of a Novikov element.

Definition 7.2. An r-expression x ∈ M(Ω) is called a Novikov element if it is an ordered nest and the sequence of all leaves in x, read from the left, is ordered.

Example 7.3. If a < b < c < d, then r(a, r(b, b, r(b, c, r(a, c, d, d)))) is a Novikov element, while r(a, r(b, b, r(b, c, r(a, b, d, d)))) is not. The sequence of leaves is bccdd in the first case and bcbdd in the second case.

Proposition 7.4. The residue classes in N (Ω) of all Novikov elements constitute a generating set for N (Ω) as an R-module.

Let A be the submodule of M(Ω) generated by the Novikov elements and B be the submodule generated by the ordered nests. Let I be the two-sided ideal in M(Ω) generated by the left-commutative and right-symmetric axioms. Proposition 7.4 may be reformulated as M(Ω) = A + I. We begin the proof by proving the weaker statement that M(Ω) = B + I.

Lemma 7.5. If x ∈ M(Ω) has length N and L leaves, then x ∈ B_{N,⩽L} + I, where B_{N,⩽L} is the submodule of M(Ω) generated by ordered nests of length N and with at most L leaves.

Proof. Suppose x = r(a_{0}, x_{1}, …, x_{n}) is of length N and has L leaves. We prove that x ∈ B_{N,⩽L} + I by induction firstly over N and secondly over n, the number of branches in x. First we use the right-symmetric rule to re-order x_{1}, …, x_{n}. The extra terms which are introduced in this process will have fewer branches and at most L leaves (and length N). Hence, by induction, we may assume that either x_{i} ∈ Ω for all i, and then x is an ordered nest, or x_{n} ∉ Ω. By induction x_{n} ∈ B + I and we may assume that x_{n} = r(b_{0}, b_{1}, …, b_{m}, y) is an ordered nest, such that x has length N and has at most L leaves. Now we use the left-commutative axiom to get that x is equivalent modulo I to r(b_{0}, b_{1}, …, b_{m}, r(a_{0}, x_{1}, …, x_{n−1}, y)). By induction r(a_{0}, x_{1}, …, x_{n−1}, y) ∈ B_{N−m−1,⩽L−m} + I and hence we may assume that x is a nest of length N and with at most L leaves. Using the left-commutative axiom again, we easily get that x ∈ B_{N,⩽L} + I.

Proof. Proof of Proposition 7.4 We will prove that any ordered nest x of length N, with L leaves and with d_{1}(x) = d belongs to A_{N,⩽(L,d)} + I, where A_{N,⩽(L,d)} is the submodule of M(Ω) generated by all Novikov elements x of length N, with < L leaves or with L leaves and with d_{1}(x) ⩽ d. We prove this by induction, firstly over N, secondly over L and thirdly over P = d_{1}(x)+d_{1}(y), where x = r(a_{0}, a_{1}, …, a_{n}, y) and a_{i} ∈ Ω for i = 0, …, n.

First suppose x = r(a_{0}, a_{1}, …, a_{n}) where a_{i} ∈ Ω. We may re-order the elements a_{1}, …, a_{n} using the right-symmetric rule. The extra terms that appear will have < L leaves. Hence by lemma 7.5 they belong to B_{N,<L} + I and hence by induction they belong to A_{N,<(L,d)} + I.

Next suppose x = r(a_{0}, …, a_{d}, r(b_{0}, …, b_{m}, y)) is an ordered nest of length N, with L leaves and d+m = P. By induction r(b_{0}, …, b_{m}, y) ∈ A_{N′,⩽(L′,m)} +I, where N′ +d+1 = N and L′ +d = L. By induction, we do not have to consider terms with < L′ leaves. Hence we may assume that r(b_{0}, …, b_{m}, y)) is a Novikov element with L′ leaves. Also, modulo terms with fewer leaves, we may assume that a_{1} ⩽ … ⩽ a_{d}.

Suppose a_{d} > b_{1}. The proof is finished if we can prove that a_{d} and b_{1} may be interchanged. Indeed, the process may be repeated a finite number of steps, until the set of leaves a_{1}, …, a_{d} is minimal (of course there are only finitely many elements in Ω involved). This is in fact a fourth level of induction.

As before we may move b_{1} to the right of b_{m}, but to pass y an extra term with L leaves will appear if y ∉ Ω, namely

r(a_{0}, …, a_{d}, r(b_{0}, b_{2} …, b_{m}, r(y, b_{1}))) .

By induction r(b_{0}, b_{2} …, b_{m}, r(y, b_{1})) may be replaced by a Novikov element with d_{1} ⩽ m − 1 (modulo terms with fewer leaves). Hence we get an ordered nest with d_{1} ⩽ d and d_{1} + d_{2} < P and hence by induction this term ∈ A_{N,⩽(L,d)} + I. We may now continue with the term

r(a_{0}, …, a_{d}, r(b_{0}, b_{2} …, b_{m}, y, b_{1})) ,

which modulo I is equal to r(b_{0}, b_{2} …, b_{m}, y, r(a_{0}, …, a_{d}, b_{1})). Now interchange a_{d} and b_{1} (which is possible as before) and shift back to

r(a_{0}, …, a_{d−1}, b_{1}, r(b_{0}, b_{2} …, b_{m}, y, a_{d})) .

Finally, by the same argument as above, a_{d} may be moved to the left of y. By induction r(b_{0}, b_{2} …, b_{m}, a_{d}, y) may be replaced by a Novikov element with d_{1} ⩽ m and as was remarked above, we are done.

The Novikov elements may be coded in the following way. Associate to any root in a Novikov element a “weight”, which is n − 1, where n is the number of branches in the subexpression determined by the root. A Novikov element is given by an ordered sequence of roots with weights and an ordered sequence of leaves, such that the total weight of the roots is one less than the number of leaves. Giving the leaves the weight −1, we may hence consider a Novikov element as an ordered sequence of variables a[i], where a ∈ Ω and i = −1, 0, 1, 2, …, such that the total weight of the sequence is −1.

Example 7.6. The Novikov element r(a, r(b, b, r(b, c, r(a, c, d, d)))) is coded by the sequence b[−1]c[−1]c[−1]d[−1]d[−1]a[0]b[1]b[1]a[2].

Hence, it is natural to consider the following R-algebra.

Definition 7.7. N P(Ω)= the (associative) commutative polynomial ring over R with the set of variables equal to {a[i]; a ∈ Ω, i ⩾ −1}. Let ∂ : N P(Ω) → N P(Ω) be the R-derivation defined by ∂(a[i]) = a[i + 1] and let ∘ be the binary operation on N P(Ω) defined by p ∘ q = ∂(p)q.

The elements of degree −1 in N P(Ω) is a free R-module with basis in one-to-one correspondence with the set of Novikov elements. Let N P_{i}(Ω) denote the set of elements in N P(Ω) of weight i − 1. Then p ∘ q ∈ N P_{i+j}(Ω) if p ∈ N P_{i}(Ω) and q ∈ N P_{j}(Ω) and hence (N P(Ω), ∘) is a graded R-algebra and (N P_{0}(Ω), ∘) is a subalgebra. It is a general fact, mentioned in the introduction, that these algebras are Novikov.

Theorem 7.8. The set of residue classes of Novikov elements is a basis for the free Novikov algebra N (Ω) as an R-module. We have that N (Ω) is isomorphic to (N P_{0}(Ω), ∘), where N P_{0}(Ω) consists of the elements of weight −1 in N P(Ω). If Ω has k elements the generating function for the R-basis of N P(Ω) consisting of monomials is

y ∏_{j=−1}^{∞} \frac{1}{(1 − y^{j}z)^{k}} ,

where the coefficient of y^{m}z^{n} is the number of elements in the basis of weight m and length n. The generating function for a basis of N (Ω) is the formal power series in z obtained as the constant term of the series above considered as a series in y.

Proof. Since (N P_{0}(Ω), ∘) is Novikov, we have a homomorphism f : N (Ω) → (N P_{0}(Ω), ∘), which extends the map a ↦ a[−1], a ∈ Ω. Since a[−1] ∈ N P_{0}(Ω), a ∈ Ω, we have im(f) ⊆ N P_{0}(Ω). We claim that equality holds. Suppose p = a_{1}[−1] ⋯ a_{n}[−1]x_{1} ⋯ x_{m} is a monomial in N P_{0}(Ω), where x_{i} has weight ⩾ 0. If n = 1 and m = 0, then p = a[−1] and hence p ∈ im(f). We prove by induction firstly on n and secondly on m, that p ∈ im(f). If x_{1} = a[0], then p = a[−1] ∘ (a_{1}[−1] ⋯ a_{n}[−1]x_{2} ⋯ x_{m}) and hence p ∈ im(f) by induction. If x_{1} = a[k], where k > 0, then n > k and

p = (a_{1}[−1] ⋯ a_{k}[−1]a[k − 1]) ∘ (a_{k+1}[−1] ⋯ a_{n}[−1]x_{2} ⋯ x_{m}) −

∑_{i=1}^{k} a_{1}[−1] ⋯ { \atop a_{i}[−1]} ⋯ a_{n}[−1]a_{i}[0]a[k − 1]x_{2} ⋯ x_{m},

and hence again, p ∈ im(f) by induction.

Now the injectivity follows in the same way as in the proof of Theorem 6.3. We may assume that Ω is finite. Let N be the number of Novikov elements of length n. By Proposition 7.4 their residue classes generate the set of elements of length n in N (Ω). Also N is the number of monomials of length n and weight −1 in N P(Ω). Hence, by the above, we get a surjective R-linear map R^{N} → R^{N} and we may apply Lemma 6.4.

The algebra N (Ω) may also be described in terms of the tree algebra defined in section 3.

Proposition 7.9. Let T (Ω)/ ∼ be defined by the equalities

t(a, x_{1}, …, x_{n}, t(b, y_{1}, …, y_{m}, z)) = t(b, y_{1}, …, y_{m}, t(a, x_{1}, …, x_{n}, z)) .

and let T N (Ω) denote the free R-module on T (Ω)/ ∼ . The operation ∘ on T (Ω) induces an operation on T N (Ω) which makes T N (Ω) isomorphic to N (Ω).

Proof. We just sketch the proof, since it consists of a number of lengthy but rather straight-forward computations. Firstly one proves that the operation ∘ on T (Ω) induces an operation on T N (Ω), secondly that this right-symmetric algebra is also left-commutative. Thirdly one proves that the set of classes of Novikov elements (defined in the same way as for the r-expressions) generate T N (Ω). Then as before we get the isomorphism with N (Ω). Observe that the proof of the last step is easier than the proof of Proposition 7.4, since the branches in t-expressions commute.

8. Lie algebras

The free Lie algebra on a set Ω, denoted L(Ω), is the quotient of M(Ω) by the ideal generated by all expressions

xx and (xy)z − x(yz) − (xz)y.

The first identity implies that xy = −yx in L(Ω). In general a Lie algebra is an R-algebra, which satisfies the two identities above. A simple computation shows that any right-symmetric algebra is a Lie algebra under the product [a, b] = ab−ba. We will consider the Lie subalgebra generated by Ω in RS(Ω) and we will prove that it is free. To do this we will define the set of Hall elements in M(Ω) as follows, given the total order on M(Ω) defined before Definition 6.1 .

Definition 8.1. An r-expression r(a, x_{1}, …, x_{n}) is called a Hall element if x_{1} ⩽ … ⩽ x_{n}, a > x_{1} and r(a, x_{1}, …, x_{i}) > x_{i+1} for i = 1, …, n − 1 and x_{1}, …, x_{n} are Hall elements. Also, an element in Ω is a Hall element.

It is well-known that the set of Hall elements represents a basis for the free Lie algebra but we will not use this. In fact, this result will be a consequence of the theorem below.

Proposition 8.2. The residue classes of the set of Hall elements generate L(Ω) as an R-module.

Proof. Using the axiom xx = 0 in L(Ω), it is enough to consider elements in M(Ω) of the form (xy), where x > y. We will prove by induction over < that any element p ∈ M(Ω) of this form may be written as a linear combination of Hall elements ⩽ p modulo the Lie identities. Suppose p = (uv), where u > v. By induction we may assume that u, v are Hall elements. If u ∈ Ω then p is a Hall element. Suppose u = (xy), where x > y. Then p is a Hall element if also y ⩽ v. If y > v we may replace p by (xv)y + x(yv). Since x > y, we have |xv| > |y| and since v < y we have xv < xy. Hence (xv)y < (xy)v and by induction (xv)y may be replaced by a linear combination of Hall elements < (xy)v.

We have x(yv) < (xy)v and (yv)x < (xy)v and hence, by induction, x(yv) may also be replaced by a linear combination of Hall elements < (xy)v modulo the Lie identities and we are done.

Theorem 8.3. For any set Ω, the Lie subalgebra of RS(Ω) generated by Ω is isomorphic to L(Ω).

Proof. We use Proposition 6.3 and consider RS(Ω) as (T (Ω), ∘). We have a Lie homomorphism φ : L(Ω) → T (Ω) whose image is the Lie subalgebra of T (Ω) generated by Ω under the product [x, y] = x ∘ y − y ∘ x. We will now use the wellorder ≺_{revlex} (written < in this proof) on T (Ω) defined in Definition 5.3. Consider the map x ↦ x^{t} from M(Ω) to T (Ω), where x^{t} is obtained from x by replacing all r by t. Then if x, y are Hall elements in M(Ω), we have x < y ⇔ x^{t} < y^{t}.

We prove the following two statements.

1. lead([x, y]) = lead(x) • lead(y), where x > y
2. lead(φ(x)) = x^{t} if x is a Hall element

Since < is ∘-leading, we have for x, y ∈ T (Ω) that lead([x, y]) = max(x • y, y • x) if x • y ≠ y • x. Suppose

y < x and x = t(a, x_{1}, …, x_{n}), y = t(b, y_{1}, …, y_{m}), where
x_{1} ⩽ … ⩽ x_{n} and y_{1} ⩽ … ⩽ y_{m}.

We have y • x = t(b, y_{1}, …, y_{m}, x) and hence by definition y • x < x • y if |y| < |x|, since |x| > max(|x_{n}|, |y|). If y < x and |y| = |x|, then x • y = t(a, x_{1}, …, x_{n}, y) and again by definition y • x < x • y. Hence we have proved that lead([x, y]) = x • y if x, y ∈ T (Ω) and y < x. By Proposition 5.2, we get

lead([x, y]) = lead(x) • lead(y) if x, y ∈ T (Ω) \ {0} and y < x

and hence the first claim is proved. The second claim is proved by induction on the length of x, where x = r(a, x_{1}, …, x_{n}). By the induction hypothesis we have

lead(φ(r(a, x_{1}, …, x_{n−1}))) = t(a, x_{1}^{t}, …, x_{n−1}^{t})

and lead(φ(x_{n})) = x_{n}^{t}. Since the map x → x^{t} is order-preserving on Hall elements, we have x_{n}^{t} < t(a, x_{1}^{t}, …, x_{n−1}^{t}). Hence from above we get

lead(φ(x)) = lead([φ(r(a, x_{1}, …, x_{n−1})), φ(x_{n})])
= lead(φ(r(a, x_{1}, …, x_{n−1}))) • lead(φ(x_{n}))
= t(a, x_{1}^{t}, …, x_{n−1}^{t}) • x_{n}^{t} = t(a, x_{1}^{t}, …, x_{n}^{t}) = x^{t}.

Now the injectivity of φ easily follows, since the set of all x^{t}, where x is a Hall element, is linearly independent by construction of T (Ω). We may also use Lemma 6.4 and argue as in the proof of Proposition 6.3.

9. A non-archimedian norm

Let Ω be a set. Consider the compatible order ≺_{revlex} on T (Ω). By Proposition 6.6 this transforms to a compatible order on RS(Ω), which we will write as < . Let RS(Ω) be the corresponding basis of RS(Ω); i.e., the set of right-symmetric r-elements x = r(a, x_{1}, …, x_{k}) ∈ M(Ω), such that a ∈ Ω, x_{1} ⩽ x_{2} ⩽ … ⩽ x_{k} and x_{1}, …, x_{k} ∈ RS(Ω). Furthermore, for x ∈ RS(Ω), let |x| denote the length of x.

Definition 9.1. Define maps m^{+}, l^{+} : RS(Ω) → N ∪ {∞} by • m^{+}(0) = l^{+}(0) = ∞

• m^{+}(a) = 0, l^{+}(a) = 1, if a ∈ Ω

• m^{+}(x) = |x_{k}|, l^{+}(x) = |x|, if x = r(a, x_{1}, …, x_{k}) ∈ RS(Ω)

• m^{+}(x) = min_{λ_{x′} ≠0} m^{+}(x′), l^{+}(x) = min_{λ_{x′} ≠0} l^{+}(x′), if x = ∑_{x′∈RS(Ω)} λ_{x′}x′ ∈ RS(Ω)

In terms of trees m^{+}(t) is the number of vertices in the maximal branch of t, if t ∈ T (Ω).

Proposition 9.2. For any x, y ∈ RS(Ω),

m^{+}(x + y) ⩾ min{m^{+}(x), m^{+}(y)}
l^{+}(x + y) ⩾ min{l^{+}(x), l^{+}(y)}
m^{+}(λx) ⩾ m^{+}(x)
l^{+}(λx) ⩾ l^{+}(x)
l^{+}(x) ⩾ m^{+}(x)

Proof. Evident.

Let η_{i} ∈ RS(Ω), i = 1, 2, …. Any element of RS(Ω) can be viewed as a rightsymmetric polynomial on variables a_{1}, a_{2}, …. For f(a_{1}, a_{2}, …) ∈ RS(Ω) denote by ηf ∈ RS(Ω) the polynomial obtained from f by the substitution a_{i} ↦ η_{i}.

Proposition 9.3. Any endomorphism η of RS(Ω) (i.e., a homomorhism of RS(Ω) to itself as an algebra) is uniquely defined by a substitution a ↦ η(a) ∈ RS(Ω), a ∈ Ω.

Proof. Since RS(Ω) is free right-symmetric, any map Ω → RS(Ω) can be prolongated in a unique way to a homomorphism of right-symmetric algebras RS(Ω) → RS(Ω). Inversely, any endomorphsim η : RS(Ω) → RS(Ω) is induced by its restriction Ω → RS(Ω).

Definition 9.4. An ideal in a free algebra (such as M(Ω) or RS(Ω)) is called a T-ideal if it is invariant under all endomorphisms of the algebra.

Example 9.5. The ideal in M(Ω) which defines RS(Ω)) is a T -ideal. More generally, if f_{1}, … f_{k} ∈ M(Ω), then the free algebra on Ω with axioms f_{1} = 0, …, f_{k} = 0 is equal to M(Ω) modulo the T -ideal generated by f_{1}, … f_{k}. If A is any rightsymmetric algebra, then {f ∈ RS(Ω); f = 0 is an identity on A} is a T -ideal in RS(Ω).

It follows from Proposition 9.2 that {x ∈ RS(Ω); m^{+}(x) ⩾ q} and {x ∈ RS(Ω); l^{+}(x) ⩾ q} are ideals in RS(Ω). We will now prove that they are also T -ideals.

Lemma 9.6. For any x, y, x_{1}, … x_{k} ∈ RS(Ω) and a ∈ Ω, we have

(i)  m^{+}(r(a, x_{1}, …, x_{k})) = max_{1⩽i⩽k} |x_{i}|
(ii)  m^{+}(x ∗ y) = max(m^{+}(x), |y|)

For any x, y ∈ RS(Ω) and for any endomorphism η of RS(Ω), we have

(iii)  m^{+}(x ∗ y) ⩾ max(m^{+}(x), l^{+}(y))
(iv)  l^{+}(η(x)) ⩾ l^{+}(x)

Proof. By Proposition 5.4 and 6.6 we have

r(a, x_{1}, …, x_{k}) = r(a, x_{i_{1}}, …, x_{i_{k}}) + ∑ y_{i}

where x_{i_{1}} ⩽ … ⩽ x_{i_{k}}, y_{i} ≺_{revlex} r(a, x_{i_{1}}, …, x_{i_{k}}) and |y_{i}| = |r(a, x_{1}, …, x_{k})|. But, by definition of ≺_{revlex}, we have that m^{+}(y) ⩾ m^{+}(x) if |x| = |y| and y ≺_{revlex} x. Hence,

m^{+}(r(a, x_{1}, …, x_{k})) = m^{+}(r(a, x_{i_{1}}, …, x_{i_{k}})) = |x_{i_{k}}| = max_{1⩽i⩽k} |x_{i}| ,

which proves (i). Now (ii) follows directly, since

r(a, x_{1}, …, x_{k}) ∗ y = r(a, x_{1}, …, x_{k}, y) .

To prove (iii), suppose x = ∑ λ_{i}x_{i} and y = ∑ λ′_{j}y_{j}, where λ_{i}, λ′_{j} ≠ 0 for all i, j. Then

m^{+}(x ∗ y) = min_{i,j}(m^{+}(x_{i} ∗ y_{j})) ⩾ min_{i,j}(max(m^{+}(x_{i}), l^{+}(y_{j})))
⩾ max(min_{i}(m^{+}(x_{i})), min_{j}(l^{+}(y_{j}))) = max(m^{+}(x), l^{+}(y)) .

By Proposition 9.2, it is enough to prove (iv) when x ∈ RS(Ω). Consider x as a polynomial, x = f(a_{0}, a_{1}, …), where a_{i} ∈ Ω. Then η(x) is a linear combination of terms y = f(x_{0}, x_{1}, …), where x_{i} ∈ RS(Ω), and hence |y| ⩾ |x| for all nonzero terms y. Hence

l^{+}(η(x)) ⩾ |x| = l^{+}(x) .

Theorem 9.7. For any x, y ∈ RS(Ω) and for any endomorphism η of RS(Ω), we have

(i)  m^{+}(x ∗ y) ⩾ max(m^{+}(x), m^{+}(y))
(ii)  m^{+}(η(x)) ⩾ m^{+}(x) .

Proof. The first claim follows from Lemma 9.6 (iii) and Proposition 9.2. It is enough to prove the second claim when x ∈ RS(Ω). We prove the inequality by induction over the length of x. If x ∈ Ω, we have m^{+}(η(x)) ⩾ 0 = m^{+}(x). Suppose x, y ∈ RS(Ω) and that the claim is true for x. We have

m^{+}(η(x ∗ y)) = m^{+}(η(x) ∗ η(y)) ⩾ max(m^{+}(η(x)), l^{+}(η(y)))
⩾ max(m^{+}(x), l^{+}(y)) = max(m^{+}(x), |y|) = m^{+}(x ∗ y) ,

where the first inequality follows from Lemma 9.6 (iii), the second inequality follows by induction and Lemma 9.6 (iv) and the last equality follows from Lemma 9.6 (ii).

Corollary 9.8. For any q ∈ N, the set RS_{q}(Ω) = {x ∈ RS(Ω); m^{+}(x) ⩾ q} and the set {x ∈ RS(Ω); l^{+}(x) ⩾ q} are T -ideals in RS(Ω).

Proof. This follows from Proosition 9.2 and Theorem 9.7.

So, the system of T -ideals {RS_{q}(Ω)}, q ⩾ 0, endows RS(Ω) with a filtration,

RS(Ω) = RS_{0}(Ω) ⊃ RS_{1}(Ω) ⊃ RS_{2}(Ω) ⊃ ⋯
RS_{p}(Ω) ∗ RS_{q}(Ω) ⊆ RS_{max(p,q)}(Ω) ,

which defines a topology on RS(Ω), such that ∗ is continuous. As usual this topology is metrizable with a non-archimedean metric, which may be defined in the following way.

Definition 9.9. Let ∥ ∥ : RS(Ω) → Q be the map defined by • ∥0∥ = 0

• ∥x∥ = 1/(m^{+}(x) + 1), if x ≠ 0

The map ∥ ∥ : RS(Ω) → Q has the following properties.

Theorem 9.10. For any x, y ∈ RS(Ω), λ ∈ k and any endomorphism η of RS(Ω), we have

0 ⩽ ∥x∥ ⩽ 1 and ∥x∥ = 0 ⇔ x = 0
∥λx∥ = ∥x∥, if λ ≠ 0
∥x + y∥ ⩽ max(∥x∥, ∥y∥)
∥x ∗ y∥ ⩽ min(∥x∥, ∥y∥)
∥η(x)∥ ⩽ ∥x∥.

Proof. It follows from Theorem 9.7 and Proposition 9.2.

Let B_{q}(Ω) = {x ∈ RS(Ω) : ∥x∥ ⩽ q} be the ball of radius q in RS(Ω). We have

B_{p}(Ω) ∗ B_{q}(Ω) ⊆ B_{min(p,q)}

and from Theorem 9.10 we obtain the following statement.

Corollary 9.11. For any q ⩽ 1, the ball B_{q} is a T -ideal of RS(Ω) and hence, for any T -ideal J generated by elements s_{1}, …, s_{k} ∈ B_{q} we have J ⊆ B_{q}.

10. On the T -ideal generated by right-bracketed polynomials

Let f and g be right-symmetric polynomials in RS(Ω). We say that the identity g = 0 follows from the identity f = 0, and write

f = 0 ⇒ g = 0,

if any right-symmetric algebra A that satisfies the identity f = 0 also satisfies the identity g = 0. The condition f = 0 ⇒ g = 0 is equivalent to the condition that the T -ideal generated by f contains g. In other words, the element g of the free algebra RS(Ω) can be obtained from f(a_{1}, …, a_{k}) using the following operations:

• substitution of a_{i} by any element in the free algebra RS(Ω)

• multiplying f with any element of RS(Ω) from right or left side (or both)

• taking linear combinations of obtained elements.

Details about general facts on polynomial identities, see for example, [14].

Let s_{q+1}^{l} be the left standard polynomial of degree q + 1 in RS(Ω). It is a skewsymmetric polynomial in the last q arguments and defined by

s_{q+1}^{l}(a_{0}, a_{1}, …, a_{q}) = ∑_{σ∈Sym_{q}} sign(σ)r(a_{σ(1)}, r(a_{σ(2)}, …, r(a_{σ(q)}, a_{0}) ⋯).

Let s_{q}^{r} be the right standard polynomial of degree q in RS(Ω). It is a skew-symmetric polynomial and defined by

s_{q}^{r}(a_{1}, …, a_{q}) = ∑_{σ∈Sym_{q}} sign(σ)r(a_{σ(1)}, a_{σ(2)}, …, a_{σ(q)}).

Lemma 10.1. For any q > 0, we have the following identities,

s_{q}^{r}(a_{1}, …, a_{q}) = ∑_{i}(−1)^{i+q}s_{q}^{r}(a_{1}, …, â_{i}, …, a_{q}) ∗ a_{i}
s_{q+2}^{r}(a_{1}, …, a_{q+2}) = ∑_{i<j}(−1)^{i+j+1}s_{q}^{r}(a_{1}, …, â_{i}, …, â_{j}, …, a_{q+2}) ∗ s_{2}^{r}(a_{i}, a_{j})

Proof. The first equality is obvious and the second follows easily from the rightsymmetric identity.

Let L_{q} ∈ RS(Ω) be defined by the following pictures of the corresponding elements in T (Ω).

[figure: L_{q} =; a_{q−2}; a_{2}; a_{q−1}; a_{1}; a_{k+1}; a_{k}; ⋯; ⋯; a_{q}] if q = 2k + 1 is odd, where k ⩾ 0, and

[figure: −L_{q} =; a_{q−1}; a_{q−3}; a_{2}; a_{q−2}; a_{1}; a_{k+1}; a_{k}; ⋯; ⋯; a_{q}]

if q = 2(k + 1) is even, where k ⩾ 0. E.g., L_{6} = −r(a_{6}, a_{5}, r(a_{3}, a_{2}), r(a_{4}, a_{1})). For n ∈ Z, let ⌊n/2⌋ denote the greatest integer no more than n/2. Consider, as in section 9, the compatible order ≺_{revlex} on RS(Ω), which we will write as < .

Lemma 10.2. If a_{1}, …, a_{q} ∈ Ω and a_{1} ⩽ … ⩽ a_{q}, then the leading term of s_{q}^{r}(a_{1}, …, a_{q}) is equal to (−1)^{k}k!L_{q}, where k = ⌊(q − 1)/2⌋.

Proof. We use induction on k. The statement is trivial for k = 0 (i.e., for q = 1 and q = 2).

Suppose q ⩾ 3 and that our statement is true for k − 1. Then, by lemma 10.1,

X = s_{q}^{r}(a_{1}, …, a_{q}) = ∑_{i<j} X_{i,j},

where

X_{i,j} = (−1)^{i+j+1}s_{q}^{r}(a_{1}, …, â_{i}, …, â_{j}, …, a_{q}) ∗ (a_{i} ∗ a_{j} − a_{j} ∗ a_{i}).

To obtain the leading term of X, we should find the maximum of the leading term of X_{i,j} and add these maxima for i < j. If the result is non-zero, this gives the leading term of X. Since the order is compatible and lead(x ∘ y) = x • y, the leading term of X_{i,j} is equal to the •-product of the leading term of

(−1)^{i+j+1}s_{q}^{r}(a_{1}, …, â_{i}, …, â_{j}, …, a_{q})

and −r(a_{j}, a_{i}). Therefore, by the induction hypothesis, the leading term of X_{i,j} as an unlabeled tree (up to a scalar multiple) has the following form

[figure: if q = 2k + 1, and]

[figure: if q = 2(k + 1).]

Note that L_{q} is the maximal element among all normal form elements, such that the underlying unlabeled tree has the form above. Thus, the leading term of X_{i,j} is L_{q} up to a scalar multiple, if and only if i + j = q, in case of q = 2k + 1 and i + j = q − 1, in case of q = 2(k + 1). Also, the scalar multiple is (−1)^{q}(−1)^{k−1}(k − 1)! = (−1)^{k}(k−1)!, in case of q = 2k+1 and (−1)^{q−1}(−1)^{k−1}(k−1)! = (−1)^{k}(k−1)!, in case of q = 2(k + 1). There are exactly k such pairs (i, j), and hence we get that the sum of these terms is (−1)^{k}k!L_{q} (in both cases). Since this is non-zero, it must be the leading term of X, which completes the induction step.

Theorem 10.3. i) The T -ideal of the free right-symmetric algebra RS(Ω) generated by s_{3}^{l} contains s_{q}^{r} for all q ⩾ 3.

ii) The T -ideal of the free right-symmetric algebra RS(Ω) generated by all the polynomials s_{k}^{l} for k > 3 does not contain s_{q}^{r} for any q.

Corollary 10.4. The polynomial s_{2n+1}^{l}, where n > 1, generates a T -ideal which does not contain s_{q}^{r} for any q. Hence s_{2n+1}^{l} = 0 ⇏ s_{q}^{r} = 0 for any q.

Corollary 10.5. The Witt algebra W_{n}^{rsym} has at least two independent polynomial identities when n > 1.

Proof. Proof of Theorem 10.3.

i) Let J be T -ideal generated by s_{3}^{l}. Then by the right-symmetric identity the computation below proves that s_{3}^{r} ∈ J.

s_{3}^{r}(a_{1}, a_{2}, a_{3}) = (a_{1} ∗ a_{2}) ∗ a_{3} − (a_{1} ∗ a_{3}) ∗ a_{2} + (a_{2} ∗ a_{3}) ∗ a_{1}
− (a_{2} ∗ a_{1}) ∗ a_{3} + (a_{3} ∗ a_{1}) ∗ a_{2} − (a_{3} ∗ a_{2}) ∗ a_{1}
= a_{1} ∗ (a_{2} ∗ a_{3}) − a_{1} ∗ (a_{3} ∗ a_{2}) + a_{2} ∗ (a_{3} ∗ a_{1})
−a_{2} ∗ (a_{1} ∗ a_{3}) + a_{3} ∗ (a_{1} ∗ a_{2}) − a_{3} ∗ (a_{2} ∗ a_{1})
= a_{1} ∗ (a_{2} ∗ a_{3}) − a_{2} ∗ (a_{1} ∗ a_{3}) − (a_{1} ∗ (a_{3} ∗ a_{2})
−a_{3} ∗ (a_{1} ∗ a_{2})) + (a_{2} ∗ (a_{3} ∗ a_{1}) − a_{3} ∗ (a_{2} ∗ a_{1})) ∈ J.

By lemma 10.1, we have s_{3}^{r} ∈ J ⇒ s_{q}^{r} ∈ J, for any q > 3.

ii) As an element of RS(Ω), s_{k}^{l}(a_{1}, …, a_{k}) is a sum of r-elements of the form r(b_{1}, r(b_{2}, …, r(b_{k−1}, b_{k}) ⋯)), where b_{i} ∈ Ω. Therefore,

∥s_{k}^{l}(a_{1}, …, a_{k})∥ = 1/k ⩽ 1/4,

if k > 3. By corollary 9.11, the T -ideal generated by all s_{k}^{l} for k > 3 lies in the ball of radius 1/4 :

J ⊆ B_{1/4}.

According to lemma 10.2, the leading term of s_{q}^{r}(a_{1}, …, a_{q}) has a maximal branch with at most 2 vertices. Therefore,

||s_{q}^{r}(a_{1}, …, a_{q})|| ⩾ 1/3.

Hence,

s_{q}^{r}(a_{1}, …, a_{q}) ∉ B_{1/4}

and we get,

s_{q}^{r}(a_{1}, …, a_{q}) ∉ J

for any q.

References

[1] A.A. Balinskii, S.P. Novikov, Poisson bracket of hamiltonian type, Frobenius algebras and Lie algebras, Dokladu AN SSSR, 283(1985), No.5, 1036-1039.

[2] J.C. Butcher, The numerical analysis of ordinary differential equations, Runge-Kutta and general linear methods, John Wiley & Sons, Ltd, Chichester, 1987.

[3] A. Cayley, On the theory of analitical forms called trees, Phil. Mag., 13(1857), p.19-30= Math. Papers, Cambridge, 3(1891), 242-246.

[4] F. Chapoton, M. Livernet, Pre-Lie Algebras and the Rooted Trees Operad, International Mathematics Research Notices, 2001, No.8, 395–408.

[5] A. Connes, D. Kreimer, Hopf algebras, renormalization and noncommutative geometry, Comm. Math.Phys., 93(1999), 836-876.

[6] A. S. Dzhumadil’daev, A remark on the space of invariant differential operators, Moscow Univ. Math. Bull., 37(1982), No.2, p.63-68.

[7] A. S. Dzhumadil’daev, Cohomologies and deformations of right-symmetric algebras, J. Math. Sci., 93(1999), No. 6, 1836-1876. Preprint available math.RT/9807065.

[8] A. S. Dzhumadil’daev, Minimal identities for right-symmetric algebras, J. Algebra, 225(2000), 201-230.

[9] M. Gerstenhaber, The cohomology structure of an associative ring, Ann. Math., 78(1963), No.2, 267-288.

[10] J.L. Koszul, Domaines bornés homogènes et orbites de groupes de transformations affines, Bull. Soc. Math. France 89(1961), No. 4, 515-533.

[11] J.M. Osborn, Novikov algebras, Nova J. Algebra Geom., 1(1992), 1-14.

[12] J.M. Osborn, Simple Novikov algebras with an idempotent, Comm. Algebra, 20(1992), No. 9, 2729-2753.

[13] J.M. Osborn, Infinite-dimensional Novikov algebras of characteristic 0, J.Algebra, 167(1994), 146-167.

[14] C. Procesi, Rings with polynomial identites, Marcel Dekker, 1973.

[15] E.B. Vinberg, Convex homogeneous cones, Transl. Moscow Math. Soc., 12(1963), 340-403.

This article may be accessed via WWW at http://www.rmi.acnet.ge/hha/ or by anonymous ftp at

ftp://ftp.rmi.acnet.ge/pub/hha/volumes/2002/n2a7/v4n2a7.(dvi,ps,pdf)

Askar Dzhumadil’daev askar@math.kz S.Demirel University, Toraygirov 19, Almati, Kazakhstan

Clas Löfwall clas@matematik.su.se

Department of Mathematics, Stockholm University, Sweden