Journal of Lie Theory Volume 12 (2002) 503–514 \begin{matrix}c \\ ◯ \end{matrix} 2002 Heldermann Verlag

On a Family of Operators and their Lie Algebras

Jan A. Sanders and Jing Ping Wang^{∗}

Communicated by P. Olver

Abstract. An infinite family of differential operators is constructed. Each of these operators defines a Lie bracket and the operator is a homomorphism from the new Lie algebra to the standard Lie algebra. An interesting feature of these operators is that they factorize into first order operators with integer coefficients. This generalizes recent results of Zhiber and Sokolov

1. Introduction

In a recent paper Zhiber and Sokolov [6] study integrable hyperbolic equations of Liouville type. They found a family of special operators with the property that they define a new Lie bracket and are homomorphisms from the Lie algebra with the newly induced bracket to the original Lie algebra. These operators have the form

£_{··}^{1} = D
£_{··}^{2} = D(D + u)
£_{··}^{3} = D(D + u)(D + u)
£_{0·}^{4} = D(D + u)(D + u)(D + u)
£_{1·}^{4} = D(D + u)(D + u)(D + 2u)
£_{00}^{5} = D(D + u)(D + u)(D + u)(D + u)
£_{01}^{5} = D(D + u)(D + u)(D + 2u)(D + 3u)
£_{00}^{6} = D(D + u)(D + u)(D + u)(D + u)(D + u)
£_{01}^{6} = D(D + u)(D + u)(D + u)(D + u)(D + 2u)
£_{11}^{6} = D(D + u)(D + u)(D + 2u)(D + 3u)(D + 3u)
£_{10}^{6} = D(D + u)(D + u)(D + 2u)(D + 3u)(D + 4u),

where D is the total derivative operator with respect to the independent variable x and u is the dependent variable. They are the polynomial homogeneous operators

ISSN 0949–5932 / $2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag with D and u of equal weight of order < 7 with the property that for all P and Q, one has

∗ J. P. Wang gratefully acknowledges the support from Netherlands Organization for Scientific Research (NWO) for this research.

[£^{n}P, £^{n}Q] ∈ im £^{n},  (1)

where P, Q are functions of x and the derivatives of u. Here the bracket is given by

[P, Q] = Q′P − P′Q,

where P′ is the Fréchet derivative of P. According to [6] these operators form a complete list all homogeneous operators of the weigth (1, 1) and of order < 7 satisfying the homomorphism condition. In particular, contrary to our somewhat restricted approach, no a priori assumption is made on the factorization of these operators. The Lie algebra of such P and Q with this bracket is denoted by 𝔤. These operators are claimed to lead to a not necessarily anti-symmetric generalization of Hamiltonian operators. Of course, the occurrence of all these regular sequences of integers in a problem that starts from a rather innocent looking condition (1) on the operator is rather startling and the authors of [6] suspect that there is some deep mathematics behind this. In this paper we show how this sequence continues by explicitly producing the family of operators and the corresponding Lie brackets. We show that for n > 4 there exist operators £_{αβ}^{n} where α = 0, 1 and β = 0 for n odd, and β = 0, 1 for n even, see Figure 1. We do this by studying multiplicative deformations, that is, when we have an operator that works, we multiply it from the right with a first order operator and derive the properties of this first order operator assuming the product is in the family. Thus our study does not answer the question of Zhiber and Sokolov whether these operators are unique. But we do prove that they are unique under multiplicative deformations. No deep mathematics is needed to get to these results, which of course does not imply that there is no deep mathematics behind it. Although the proofs in this paper are fairly straightforward, the reader should appreciate the amount of work that went into guessing the right form of the operators and Lie brackets. This was done using Maple V [1] and Form 3.0 [5] programs.

2. Main results

We consider the Lie algebra of evolutionary vectorfields based on a the independent variable x and the dependent variable u. Let D be the total derivative operator with respect to x. We write u_{i} = D^{i}u. Given any two expressions P and Q depending on x and a finite number of the u_{i}, i = 0, ⋯, their bracket is defined to be [P, Q] = Q′P − P′Q, where P′ is the Fréchet derivative of P, cf. [3] for the theoretical foundations. We study a kind of special local differential operators ℜ, which produce a Liesubalgebra, that is, for any P and Q, there is a 𝔹_{ℜ}(P, Q) such that

[ℜP, ℜQ] = ℜ𝔹_{ℜ}(P, Q)  (2)

and so the domain of ℜ is a subalgebra. If we view 𝔹_{ℜ} as a 2-form on the domain of ℜ, then this form is antisymmetric, and obeys the Jacobi identity modulo the

9 8 7 6 5 4 321

Figure 1: The operators to order 9

kernel of ℜ, so it defines a (possibly new) Lie bracket on a suitable domain. Before we give the results, we define the following notation, which will use throughout the paper.

Definition 2.1. Let Γ_{i} = D + α_{i}u, with α_{i} ∈ ℂ. Define 𝕃_{k}^{l} = ∏_{i=k}^{l} Γ_{i}, Δ_{n}^{0} = α_{n}, Δ_{n}^{1} = Δ_{n}^{0} − Δ_{n−1}^{0} and Δ_{n}^{2} = Δ_{n}^{1} − Δ_{n−1}^{1}.

Theorem 2.2. For every even n there exist 4 and for every odd n there exist 2 operators of the type ℜ = 𝕃_{1}^{n}, as long as the order n is larger than 4, such that for every P, Q there is a 𝔹_{n}(P, Q) satisfying [𝕃_{1}^{n}P, 𝕃_{1}^{n}Q] = 𝕃_{1}^{n}𝔹_{n}(P, Q). The sequences of α_{i}, i = 1, ⋯, n are

\begin{matrix}£_{00}^{n+2} : 0, 1, 1, 1, 1, ⋯, 1, 1, & n ∈ ℤ, \\ £_{01}^{n+2} : 0, 1, 1, 1, 1, ⋯, 1, 2, & n ∈ 2ℤ, \\ £_{11}^{n+2} : 0, 1, 1, 2, 3, ⋯, n − 1, n − 1, & n ∈ 2ℤ, \\ £_{10}^{n+2} : 0, 1, 1, 2, 3, ⋯, n − 1, n, & n ∈ ℤ. \end{matrix}

Here the first subindex is Δ_{4}^{1}, the second one is |Δ_{n+2}^{2}|. Moreover, the bracket 𝔹_{n}(P, Q) = ℤ_{n}(P, Q) + D_{Q}[𝕃_{1}^{n}P] − D_{P} [𝕃_{1}^{n}Q], where, for any odd n,

ℤ_{n}(P, Q) = Δ_{n}^{0}(𝕃_{2}^{n}P · Q − 𝕃_{2}^{n}Q · P) − Δ_{n}^{1}(𝕃_{3}^{n}P · 𝕃_{n}^{n}Q − 𝕃_{3}^{n}Q · 𝕃_{n}^{n}P), (3)

and for any even n = 2m,

ℤ_{n}(P, Q) = Δ_{n}^{0}(𝕃_{2}^{n}P · Q − 𝕃_{2}^{n}Q · P) − Δ_{n}^{1}(𝕃_{3}^{n}P · 𝕃_{n}^{n}Q − 𝕃_{3}^{n}Q · 𝕃_{n}^{n}P)

+Δ_{n}^{2} ∑_{i=2}^{m−1} (−1)^{i}(𝕃_{i+2}^{n}P · 𝕃_{n−i+1}^{n}Q − 𝕃_{i+2}^{n}Q · 𝕃_{n−i+1}^{n}P).  (4)

Proof. The existence of the operators and Lie brackets is proved in section 3, combining the results in the Lemmas. The uniqueness under multiplication, that is to say, proving that given ℜ there exist one (if the order of ℜ is even) or two (if the order of ℜ is odd) uniquely determined operators Γ such that ℜΓ again satisfies the conditions, is the subject of section 4, as well as proving the nonexistence of any such Γ in case the sequence of ℜ ends with 1, 1, 2 or n − 4, n − 3, n − 3.

Corollary 2.3. The generating function for the number of operators of given order is

\frac{1 + t^{4}}{1 − t} + \frac{2t^{6}}{1 − t^{2}} .

3. Existence of the operators

Lemma 3.1. ℜ is an operator satisfying (2) if and only if for any P, Q ∈ 𝔤, there is a ℤ_{ℜ} ∈ C^{2}(𝔤, 𝔤) such that

ℜ′[ℜP]Q − ℜ′[ℜQ]P = ℜℤ_{ℜ}(P, Q).

Proof Let us compute the following bracket

[ℜP, ℜQ] = D_{ℜQ}[ℜP] − D_{ℜP} [ℜQ]
= ℜ(D_{Q}[ℜP] − D_{P} [ℜQ]) + ℜ′[ℜP]Q − ℜ′[ℜQ]P.

So, if and only if the lemma is valid, there exists 𝔹_{ℜ} ∈ C^{2}(𝔤, 𝔤) satisfying (2) with 𝔹_{ℜ}(P, Q) = ℤ_{ℜ}(P, Q) + D_{Q}[ℜP] − D_{P} [ℜQ].

Lemma 3.2. Assume ℜ is an operator satisfying (2). For any operator Γ, ℜΓ is also such an operator if and only if

O(P, Q) = ℤ_{ℜ}(ΓP, ΓQ) + Γ′[ℜΓP]Q − Γ′[ℜΓQ]P  (5)

is in the image of the operator Γ for all P, Q ∈ 𝔤.

Proof. According to Lemma 3.1, we need to compute

(ℜΓ)′[ℜΓP]Q − (ℜΓ)′[ℜΓQ]P
= ℜ′[ℜΓP]ΓQ − ℜ′[ℜΓQ]ΓP + ℜ(Γ′[ℜΓP]Q − Γ′[ℜΓQ]P)
= ℜ (ℤ_{ℜ}(ΓP, ΓQ) + Γ′[ℜΓP]Q − Γ′[ℜΓQ]P)

Therefore, we only need to check whether the expression O(P, Q) is in the image of the operator Γ.

Lemma 3.3. Let α_{1} = 0 and α_{2} = α_{3} = 1. Suppose n ≥ 3. Define ℜ = 𝕃_{1}^{n} and

ℤ_{n}(P, Q) = Δ_{n}^{0}(𝕃_{2}^{n}P · Q − 𝕃_{2}^{n}Q · P) − Δ_{n}^{1}(𝕃_{3}^{n}P · Γ_{n}Q − 𝕃_{3}^{n}Q · Γ_{n}P),

Then, if Δ_{n+1}^{1} = Δ_{n}^{1} = α, with α = 0 or 1 (and therefore Δ_{n+1}^{2} = 0),

ℤ_{n}(Γ_{n+1}P, Γ_{n+1}Q) + Γ′_{n+1}[𝕃_{1}^{n}Γ_{n+1}P]Q − Γ′_{n+1}[𝕃_{1}^{n}Γ_{n+1}Q]P = Γ_{n+1}ℤ_{n+1}(P, Q).

Proof. We directly compute

Γ_{n+1}ℤ_{n+1}(P, Q)−Γ′_{n+1}[𝕃_{1}^{n+1}P]Q+Γ′_{n+1}[𝕃_{1}^{n+1}Q]P −ℤ_{n}(Γ_{n+1}P, Γ_{n+1}Q)

= Γ_{n+1}(Δ_{n+1}^{0}(𝕃_{2}^{n+1}P · Q − 𝕃_{2}^{n+1}Q · P))
−Γ_{n+1}(Δ_{n+1}^{1}(𝕃_{3}^{n+1}P · Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n+1}P))
−Γ′_{n+1}[𝕃_{1}^{n+1}P]Q + Γ′_{n+1}[𝕃_{1}^{n+1}Q]P
−Δ_{n}^{0}(𝕃_{2}^{n+1}P · Γ_{n+1}Q − 𝕃_{2}^{n+1}Q · Γ_{n+1}P)
+Δ_{n}^{1}(𝕃_{3}^{n+1}P · Γ_{n}Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n}Γ_{n+1}P)
= Δ_{n+1}^{1}(𝕃_{2}^{n+1}P · Γ_{n+1}Q − 𝕃_{2}^{n+1}Q · Γ_{n+1}P)
−Δ_{n+1}^{1}(𝕃_{2}^{n+1}P · Γ_{n+1}Q − 𝕃_{2}^{n+1}Q · Γ_{n+1}P)
−Δ_{n+1}^{1}(𝕃_{3}^{n+1}P · Γ_{n}Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n}Γ_{n+1}P)
+Δ_{n}^{1}(𝕃_{3}^{n+1}P · Γ_{n}Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n}Γ_{n+1}P)
= −Δ_{n+1}^{2}(𝕃_{3}^{n+1}P · Γ_{n}Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n}Γ_{n+1}P)
= 0,

where we used the fact that either Δ_{n+1}^{0} = Δ_{n}^{0} + 1 or Δ_{n+1}^{0} = 1 = α_{2}. This shows that we have an inductive relation for such a family of 𝕃_{1}^{n} and ℤ_{n} for α = 0, 1. We have not yet shown how the induction starts, but for this we can refer to [6], as listed in section 1.

Suppose now all the Γ_{i} satisfy the relations given in Lemma 3.3 up till Γ_{n}, but Γ_{n+1} does not, i.e., Δ_{n+1}^{2} ≠ 0. We say that Γ_{n+1} splits over 𝕃_{1}^{n} if

Γ′_{n+1} = Γ′_{i+1} + Γ′_{n−i+1}, ∀i = 2, ⋯, n − 2.

Lemma 3.4. If Γ_{n+1} splits over 𝕃_{1}^{n} then

Γ_{n+1}(P · Q) = Γ_{i+1}P · Q + P · Γ_{n−i+1}Q.

Proof. This is obvious, since α_{n+1}u = α_{i+1}u + α_{n−i+1}u.

Lemma 3.5. Suppose n ≥ 5 and Γ_{n+1} splits over 𝕃_{1}^{n}. If 𝕃_{1}^{n} = £_{00}^{n} this implies Γ′_{n+1} = Γ′_{n} + 1 = 2 and if 𝕃_{1}^{n} = £_{10}^{n} then Γ′_{n+1} = Γ′_{n} = n − 2. Let, for n = 2m + 1 odd and m ≥ 2,

ℤ_{n+1}(P, Q) = Δ_{n+1}^{0}(𝕃_{2}^{n+1}P · Q − 𝕃_{2}^{n+1}Q · P)

−Δ_{n+1}^{1}(𝕃_{3}^{n+1}P · Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n+1}P)

+Δ_{n+1}^{2} ∑_{i=2}^{m} (−1)^{i}(𝕃_{i+2}^{n+1}P · 𝕃_{n−i+2}^{n+1}Q − 𝕃_{i+2}^{n+1}Q · 𝕃_{n−i+2}^{n+1}P)

Then

ℤ_{n}(Γ_{n+1}P, Γ_{n+1}Q) + Γ′_{n+1}[𝕃_{1}^{n+1}P]Q − Γ′_{n+1}[𝕃_{1}^{n+1}Q]P = Γ_{n+1}ℤ_{n+1}(P, Q).

Proof. We compute

Γ_{n+1}ℤ_{n+1}(P, Q) − Γ′_{n+1}[𝕃_{1}^{n+1}P]Q + Γ′_{n+1}[𝕃_{1}^{n+1}Q]P − ℤ_{n}(Γ_{n+1}P, Γ_{n+1}Q) =

= Γ_{n+1}(Δ_{n+1}^{0}(𝕃_{2}^{n+1}P · Q − 𝕃_{2}^{n+1}Q · P))

−Γ_{n+1}(Δ_{n+1}^{1}(𝕃_{3}^{n+1}P · Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n+1}P))

+Γ_{n+1}(Δ_{n+1}^{2} ∑_{i=2}^{m} (−1)^{i}(𝕃_{i+2}^{n+1}P · 𝕃_{n−i+2}^{n+1}Q − 𝕃_{i+2}^{n+1}Q · 𝕃_{n−i+2}^{n+1}P))

−Γ′_{n+1}[𝕃_{1}^{n+1}P]Q + Γ′_{n+1}[𝕃_{1}^{n+1}Q]P

−Δ_{n}^{0}(𝕃_{2}^{n}Γ_{n+1}P · Γ_{n+1}Q − 𝕃_{2}^{n}Γ_{n+1}Q · Γ_{n+1}P) +Δ_{n}^{1}(𝕃_{3}^{n}Γ_{n+1}P · Γ_{n}Γ_{n+1}Q − 𝕃_{3}^{n}Γ_{n+1}Q · Γ_{n}Γ_{n+1}P)

= +Δ_{n}^{1}(𝕃_{3}^{n+1}P · Γ_{n}Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n}Γ_{n+1}P)

−Δ_{n+1}^{1}(𝕃_{3}^{n+1}P · Γ_{n}Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n}Γ_{n+1}P)

+Δ_{n+1}^{2} ∑_{i=2}^{m} (−1)^{i}(𝕃_{i+1}^{n+1}P · 𝕃_{n−i+2}^{n+1}Q − 𝕃_{i+1}^{n+1}Q · 𝕃_{n−i+2}^{n+1}P)

+Δ_{n+1}^{2} ∑_{i=2}^{m} (−1)^{i}(𝕃_{i+2}^{n+1}P · 𝕃_{n−i+1}^{n+1}Q − 𝕃_{i+2}^{n+1}Q · 𝕃_{n−i+1}^{n+1}P)

= −Δ_{n+1}^{2}(𝕃_{3}^{n+1}P · Γ_{n}Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n}Γ_{n+1}P)

+Δ_{n+1}^{2} ∑_{i=2}^{m} (−1)^{i}(𝕃_{i+1}^{n+1}P · 𝕃_{n−i+2}^{n+1}Q − 𝕃_{i+1}^{n+1}Q · 𝕃_{n−i+2}^{n+1}P)

−Δ_{n+1}^{2} ∑_{i=3}^{m+1} (−1)^{i}(𝕃_{i+1}^{n+1}P · 𝕃_{n−i+2}^{n+1}Q − 𝕃_{i+1}^{n+1}Q · 𝕃_{n−i+2}^{n+1}P)

= −Δ_{n+1}^{2}(𝕃_{3}^{n+1}P · Γ_{n}Γ_{n+1}Q − 𝕃_{3}^{n+1}Q · Γ_{n}Γ_{n+1}P)

+Δ_{n+1}^{2}(𝕃_{3}^{n+1}P · 𝕃_{n}^{n+1}Q − 𝕃_{3}^{n+1}Q · 𝕃_{n}^{n+1}P)

= 0,

where we used the fact that either Γ′_{n+1} = 1 + Γ′_{n} or Δ_{n+1}^{1} = 0.

4. Uniqueness of the operators under multiplicative deformation

In this section we show, using the symbolic method [2, 4], which choice we have if we want to go from ℜ = 𝕃_{1}^{n}, with n ≥ 4, to ℜΓ, with Γ = Γ_{n+1} = D + α_{n+1}u. According to Lemma 3.2 we need to check whether the expression (5) is in the image of the operator Γ. We do this term by term. First we pick out the terms in O(P, Q) without u and its derivatives, denoted by H_{0}^{n+1} and integrate it. The image part is denoted by Z_{0} and the rest is put as zero to obtain the condition on α_{n+1}. On the next step we pick out the term in O(P, Q) − ΓZ_{0} linear in u or its derivatives, denoted by H_{1}^{n+1} and treat it the same way as before. We continue this procedure until we either get the obstruction, meaning no such operator exists or the operator ℜΓ. To prove the uniqueness of the operators, we start to show there is at most one solution for each case from even order to odd order (cf. Lemma 4.1) and two solutions for each case from odd order to even order (cf. Lemma 4.2). Then we prove that there are no such operators starting from £_{α1}^{n}, where α = 0, 1 and n ∈ 2ℤ by a long computation (cf. after Lemma 4.3), see Figure 1. We carry out the above procedure by the symbolic method by mapping u_{i} ↦ z^{i}, P_{i} ↦ x^{i} and Q_{i} ↦ y^{i}. Notice that 𝕃_{k}^{n} = D^{l−k+1} + ∑_{j=k}^{n} α_{j}D^{j−k}uD^{n−j} + O(u^{2}). The symbolic expression of u-linear terms in 𝕃_{k}^{n}P is

ψ_{k}^{n}(x, z) = ∑_{j=k}^{n} α_{j}(x + z)^{j−k}x^{n−j}.  (6)

From formula (3) and (4), we obtain, for any n, H_{0}^{n+1}(x, y) = α_{n+1}(x^{n+1} − y^{n+1}) + α_{n}(x^{n}y − xy^{n}) − Δ_{n}^{1}(x^{n−1}y^{2} − x^{2}y^{n−1})

Lemma 4.1. The function H_{0}^{2m+1}(x, y) has a factor x + y if and only if

Δ_{2m+1}^{2} = 0.

Proof. It is easy to see that

H_{0}^{2m+1}(x, −x) = 2(α_{2m+1} − α_{2m} − Δ_{2m}^{1})x^{2m+1} = 2Δ_{2m+1}^{2}x^{2m+1}.

This has to be zero.

Notice that the function H_{0}^{2m}(x, y) has a factor x + y. We go to the next step. Let us write out the symbolic expression of the linear term in u.

H_{1}^{2m}(x, y, z) = ℘_{2m}(x, y, z) − ℘_{2m}(y, x, z) − α_{2m} \frac{H_{0}^{2m}(x,y)}{x+y},

where (cf. formula (6))

℘_{2m}(x, y, z) = α_{2m}ψ_{1}^{2m}(x, z) + α_{2m}α_{2m−1}x^{2m−1} + α_{2m−1}ψ_{2}^{2m}(x, z)y

−Δ_{2m−1}^{1} (x^{2m−2}(α_{2m}(y + z) + α_{2m−1}y) + ψ_{3}^{2m}(x, z)y^{2}) .

Lemma 4.2. The function H_{1}^{2m}(x, y, z) has a factor x + y + z if and only if

1. α_{2m} = 1 or 2 when α_{1} = 0 and α_{j} = 1 for j = 2, ⋯, 2m − 1.

2. α_{2m} = 2m − 2 or 2m − 3 when α_{1} = 0, α_{2} = 1 and α_{j} = j − 2 for j = 3, ⋯, 2m − 1.

Proof. Let us compute H_{1}^{2m}(x, y, −x − y) in both cases. For the first case

ψ_{k}^{2m}(x, −x − y) = ∑_{j=k}^{2m} α_{j}(−y)^{j−k}x^{2m−j}

= ∑_{j=k}^{2m−1}(−y)^{j−k}x^{2m−j} + α_{2m}(−y)^{2m−k} − δ_{k}^{1}x^{2m−1}

= \frac{x^{2m+1−k} − x(−y)^{2m−k}}{x + y} + α_{2m}(−y)^{2m−k} − δ_{k}^{1}x^{2m−1}.

510 Sanders and Wang Therefore

H_{1}^{2m}(x, y, −x − y) =

= (α_{2m} − 1)\frac{xy^{2m−1} − yx^{2m−1}}{x + y} − (α_{2m}^{2} − α_{2m})y^{2m−1} + α_{2m}x^{2m−1}

−(α_{2m} − 1)\frac{yx^{2m−1} − xy^{2m−1}}{x + y} + (α_{2m}^{2} − α_{2m})x^{2m−1} − α_{2m}y^{2m−1}

−α_{2m} \frac{α_{2m}(x^{2m} − y^{2m}) + (x^{2m−1}y − xy^{2m−1})}{x + y}

= (α_{2m} − 1)(α_{2m} − 2)\frac{xy(x^{2m−2} − y^{2m−2})}{x + y}.

This proves the first case of the lemma. For the second case

ψ_{k}^{2m}(x, −x − y) = ∑_{j=k}^{2m} α_{j}(−y)^{j−k}x^{2m−j}

= ∑_{j=k}^{2m−1}(j − 2)(−y)^{j−k}x^{2m−j} + α_{2m}(−y)^{2m−k}
+δ_{k}^{1}(x^{2m−1} − yx^{2m−2}) + δ_{k}^{2}x^{2m−2}
= \frac{x^{2m+1−k} (kx − 2x + ky − 3y) − x(−y)^{2m−k} (2mx − 2x + 2my − 3y)}{(x + y)^{2}}
+α_{2m}(−y)^{2m−k} + δ_{k}^{1}(x^{2m−1} − yx^{2m−2}) + δ_{k}^{2}x^{2m−2}.

Therefore H_{1}^{2m}(x, y, −x − y) =

= (α_{2m} − 2m + 2)\frac{y^{2}x^{2m−1} + xy^{2m−1} (2mx − 2x + 2my − 3y)}{(x + y)^{2}}

−(α_{2m} − 2m + 2)\frac{x^{2}y^{2m−1} + yx^{2m−1} (2my − 2y + 2mx − 3x)}{(x + y)^{2}}

+α_{2m}^{2}(x^{2m−1} − y^{2m−1}) + A_{2m}(xy^{2m−2} − yx^{2m−2})

−α_{2m} \frac{α_{2m}(x^{2m} − y^{2m}) + (2m − 3)(x^{2m−1}y − xy^{2m−1}) − (x^{2m−2}y^{2} − x^{2}y^{2m−2})}{x + y}

= (α_{2m} − 2m + 2)(2m − 3)\frac{xy^{2m−1} − yx^{2m−1}}{x + y}

+α_{2m} \frac{(α_{2m} − 2m + 2)(x^{2m−1}y − xy^{2m−1})}{x + y}

= ((α_{2m} − 2m + 2)(α_{2m} − 2m + 3)\frac{x^{2m−1}y − xy^{2m−1}}{x + y} .

This completes the proof of the Lemma. Now we prove that for the first case, starting from 𝕃_{1}^{2m} with α_{2m} = 2 and for the second case, with α_{2m} = 2m − 3, we can not find such 𝕃_{1}^{2m+1}. To do so, we need

Sanders and Wang 511

to write out H_{0}^{2m−1}(x, y) using formula (4).

H_{0}^{2m+1}(x, y) =

= α_{2m+1}(x^{2m+1} − y^{2m+1}) + α_{2m}(x^{2m}y − xy^{2m})

−Δ_{2m}^{1}(x^{2m−1}y^{2} − x^{2}y^{2m−1}) + Δ_{2m}^{2} ∑_{i=2}^{m−1}(−1)^{i}(x^{2m−i}y^{i+1} − x^{i+1}y^{2m−i}).

Lemma 4.3. The function H_{0}^{2m+1}(x, y) has a factor x + y if and only if

Δ_{2m+1}^{2} = (m − 2)Δ_{2m}^{2}.

Proof. It is easy to see that

H_{0}^{2m+1}(x, −x) = 2(Δ_{2m+1}^{2} − (m − 2)Δ_{2m}^{2})x^{2m+1}.

This has to be zero. The proof of the lemma follows. Now let write out the the possible sequences of α_{j} (m > 1):

1. α_{1} = 0, α_{j} = 1 for j = 2, ⋯, 2m − 1, α_{2m} = 2 and α_{2m+1} = m + 1.

2. α_{1} = 0, α_{2} = 1 and α_{j} = j − 2 for j = 3, ⋯, 2m − 1 α_{2m} = 2m − 3 and

α_{2m+1} = m − 1. We show that for such α_{j} ’s one has H_{1}^{2m+1}(x, y, −x−y) ≠ 0, that is, such operators do not exist.

H_{1}^{2m+1}(x, y, z) =
= ℘_{2m+1}(x, y, z) − ℘_{2m+1}(y, x, z) − α_{2m+1} \frac{H_{0}^{2m+1}(x, y)}{x + y},

where (cf. formula (4))

℘_{2m+1}(x, y, z) =

= α_{2m+1}ψ_{1}^{2m+1}(x, z) + α_{2m}α_{2m+1}x^{2m} + α_{2m}ψ_{2}^{2m+1}(x, z)y

−Δ_{2m}^{1} (x^{2m−1}ψ_{2m}^{2m+1}(y, z) + ψ_{3}^{2m+1}(x, z)y^{2})

+Δ_{2m}^{2} ∑_{i=2}^{m−1}(−1)^{i}(x^{2m−i}ψ_{2m−i+1}^{2m+1}(y, z) + ψ_{i+2}^{2m+1}(x, z)y^{i+1}).

For the first case,

ψ_{k}^{2m+1}(x, −x − y) = ∑_{j=k}^{2m+1} α_{j}(−y)^{j−k}x^{2m+1−j}

= ∑_{j=k}^{2m}(−y)^{j−k}x^{2m+1−j} + (m + 1)(−y)^{2m+1−k} − δ_{k}^{1}x^{2m} + (−y)^{2m−k}x
= \frac{x^{2m+2−k} − x(−y)^{2m+1−k}}{x + y} + (m + 1)(−y)^{2m+1−k} − δ_{k}^{1}x^{2m} + (−y)^{2m−k}x.

512 Sanders and Wang

So ℘_{2m+1}(x, y, −x − y) =

= (m + 1)\frac{x^{2m+1} − xy^{2m}}{x + y} + (m + 1)^{2}y^{2m} − (m − 2)y^{2m−1}x

+2(m + 1)x^{2m} + 2\frac{x^{2m}y + xy^{2m}}{x + y} − 3(m + 1)y^{2m}

−2x^{2m−1}y − \frac{x^{2m−1}y^{2} − xy^{2m}}{x + y}

+ ∑_{i=2}^{m−1}(\frac{y^{i+1}(−x)^{2m−i} − yx^{2m}}{x + y} + (m + 1)x^{2m} − x^{2m−1}y)

− ∑_{i=2}^{m−1}(\frac{x^{2m−i}(−y)^{i+1} − xy^{2m}}{x + y} + (m + 1)y^{2m} − y^{2m−1}x)

= \frac{(m + 1)^{2}x^{2m+1} + (m^{2} − m + 4)x^{2m}y − (m + 1)x^{2m−1}y^{2}}{x + y}

+2\frac{y^{3}x^{2m−1} + y^{m+1}(−x)^{m+1}}{(x + y)^{2}} .

Finally, H_{1}^{2m+1}(x, y, −x − y) =

= \frac{(m + 1)^{2}x^{2m+1} + (m^{2} − m + 4)x^{2m}y − (m + 1)x^{2m−1}y^{2}}{x + y}

+2\frac{y^{3}x^{2m−1} − x^{3}y^{2m−1}}{(x + y)^{2}}

−\frac{(m + 1)^{2}y^{2m+1} + (m^{2} − m + 4)y^{2m}x − (m + 1)y^{2m−1}x^{2}}{x + y}

−(m + 1)\frac{(m + 1)(x^{2m+1} − y^{2m+1}) + 2(x^{2m}y − xy^{2m}) − (x^{2m−1}y^{2} − x^{2}y^{2m−1})}{x + y}

−(m + 1)\frac{x^{3}y^{3}(x^{2m−4} − y^{2m−4})}{(x + y)^{2}}

= (m − 1) ((m − 2)\frac{(x^{2m}y − xy^{2m})}{x + y} − \frac{x^{3}y^{3}(x^{2m−4} − y^{2m−4})}{(x + y)^{2}}) .  (7)

For the second case,

ψ_{k}^{2m+1}(x, −x − y) = ∑_{j=k}^{2m+1} α_{j}(−y)^{j−k}x^{2m+1−j}

= ∑_{j=k}^{2m}(j − 2)(−y)^{j−k}x^{2m+1−j} + α_{2m+1}(−y)^{2m+1−k}
−(−y)^{2m−k}x + δ_{k}^{1}(x^{2m} − yx^{2m−1}) + δ_{k}^{2}x^{2m−1}
= \frac{x^{2m+2−k} (kx − 2x + ky − 3y) − x(−y)^{2m+1−k} (2mx − x + 2my − 2y)}{(x + y)^{2}}
+(m − 1)(−y)^{2m+1−k} − (−y)^{2m−k}x + δ_{k}^{1}(x^{2m} − yx^{2m−1}) + δ_{k}^{2}x^{2m−1}.

So ℘_{2m+1}(x, y, −x − y) =

= (m − 1)\frac{x^{2m+1} (−x − 2y) − xy^{2m} (2mx − x + 2my − 2y)}{(x + y)^{2}}

+(m − 1) ((m − 1)y^{2m} + y^{2m−1}x + x^{2m} − yx^{2m−1})

+(2m − 3)y \frac{−x^{2m}y + xy^{2m−1} (2mx − x + 2my − 2y)}{(x + y)^{2}}

+(2m − 3) (yx^{2m−1} − (m − 1)y^{2m} − y^{2m−1}x + (m − 1)x^{2m})

− ∑_{i=2}^{m−1} \frac{(−x)^{2m−i}y^{i+1} ((2m − i − 1)y + (2m − i − 2)x)}{(x + y)^{2}}

+ ∑_{i=2}^{m−1}(\frac{yx^{2m} (2my − y + 2mx − 2x)}{(x + y)^{2}} − (m − 1)x^{2m} − x^{2m−1}y)

+ ∑_{i=2}^{m−1} \frac{x^{2m−i}(−y)^{i+1} (ix + iy − y)}{(x + y)^{2}}

− ∑_{i=2}^{m−1}(\frac{xy^{2m} (2mx − x + 2my − 2y)}{(x + y)^{2}} − (m − 1)y^{2m} − y^{2m−1}x)

= \frac{(m − 1)^{2}x^{2m+2} + (m − 1)(2m − 3)x^{2m+1}y + (m − 1)(3m − 5)x^{2m}y^{2}}{(x + y)^{2}}

+(2 − 2m)\frac{y^{3}x^{2m−1} + (−x)^{m+1}y^{m+1}}{(x + y)^{2}} .

Finally,

H_{1}^{2m+1}(x, y, −x − y) =

= \frac{(m − 1)^{2}x^{2m+2} + (m − 1)(2m − 3)x^{2m+1}y + (m − 1)(3m − 5)x^{2m}y^{2}}{(x + y)^{2}}

+(2 − 2m)\frac{y^{3}x^{2m−1} − x^{3}y^{2m−1}}{(x + y)^{2}}

−\frac{(m − 1)^{2}y^{2m+2} + (m − 1)(2m − 3)y^{2m+1}x + (m − 1)(3m − 5)y^{2m}x^{2}}{(x + y)^{2}}

−(m − 1)\frac{(m − 1)(x^{2m+1} − y^{2m+1}) + (2m − 3)(x^{2m}y − xy^{2m})}{x + y}

+(m − 1)\frac{x^{3}y^{3}(x^{2m−4} − y^{2m−4})}{(x + y)^{2}}

= (m − 1)\frac{(m − 2)(x^{2m+1}y − y^{2m+1}x) + (m − 2)(x^{2m}y^{2} − y^{2m}x^{2})}{(x + y)^{2}}

−(m − 1)\frac{x^{3}y^{3}(x^{2m−4} − y^{2m−4})}{(x + y)^{2}}

= (m − 1) ((m − 2)\frac{(x^{2m}y − xy^{2m})}{x + y} − \frac{x^{3}y^{3}(x^{2m−4} − y^{2m−4})}{(x + y)^{2}}) .  (8)

5. Open problems

We have shown in this paper the existence of a certain infinite family of operators. Notice that the £_{0α}^{n}–family, α = 0, 1, behaves exactly as the £_{1α}^{n}–family, and the obstruction expressions (7) and (8) are in both cases equal. However, we have no explanation for this symmetry. Another question is whether the Lie brackets that are defined are mutually compatible. Finally it remains to be shown that these operators, characterized by the property (1) are truly unique, not only up to multiplicative deformations. This would imply that they always factorize into first order operators, as is the case for order < 7, c.f. [6] Acknowledgement. The authors thank Professor V. V. Sokolov for drawing their attention to this problem.

References

[1] Char, B. W., K. O. Geddes, G. H. Gonnet, B. L. Leong, M. B. Monagan, and S. M. Watt, “Maple V Language Reference Manual,” Springer–Verlag, Berlin, 1991.

[2] Gel’fand, I. M., and L. A. Dikiĭ, Asymptotic properties of the resolvent of Sturm-Liouville equations, and the algebra of Korteweg-de Vries equations, Uspehi Mat. Nauk, 30(5(185)) (1975), 67–100. English translation: Russian Math. Surveys, 30 no. 5 (1975), 77–113.

[3] Olver, P. J., “Applications of Lie Groups to Differential Equations,” Graduate Texts in Mathematics 107 Springer-Verlag, New York, second edition, 1993.

[4] Sanders, J. A., and J. P. Wang, On the integrability of homogeneous scalar evolution equations, J. Differential Equations 147(2) (1998), 410–434.

[5] Vermaseren, J. A. M., “New Features of FORM,” Technical report, Nikhef, Amsterdam, 2000; math-ph/0010025.

[6] Zhiber, A. V., and V. V. Sokolov, Exactly integrable hyperbolic equations of Liouville type, Uspihi mat. Nauk 56(1) (2001), 63–106. English translation: Russian Math. Surveys 56 (2001).

Jan Sanders and Jing Ping Wang Vrije Universiteit Faculty of Sciences Division of Mathematics and Computer Science De Boelelaan 1081a 1081 HV Amsterdam The Netherlands

Received June 14, 2001 and in final form July 30, 2001