Isotropic automorphisms of non-quadratic elliptic CR-manifolds in ℂ^{4} ^{∗}^{†}

Vladimir V. Ežov ^{‡} University of Adelaide Department of Pure Mathematics Adelaide, SA 5005, South Australia vezhov@maths.adelaide.edu.au

Gerd Schmalz ^{‡‡} Mathematisches Institut der Universität Bonn Beringstraße 1, D-53115 Bonn schmalz@math.uni-bonn.de

July 5, 1999

Abstract

We use an analogue (formerly constructed by the authors) of the Chern-Moser normal form for 2-codimensional real-analytic non-quadratic CR-manifolds in ℂ^{4} with elliptic Levi form to describe local isotropic holomorphic automorphisms. We prove that apart from very special explicit family of exceptional manifolds any isotropic automorphism is uniquely determined by the restriction of its Jacobian at the origin to the complex tangent space.

∗Mathematics Subject Classification: primary 32C16; secondary 32C05, 32H02 †Keywords and Phrases: CR-manifolds, invariants, classification ‡Research of the first author was supported by ARC and SFB 256.

‡‡Research of the second author was supported by SFB 256 of Deutsche Forschungsgemeinschaft.

1 Introduction

Let M be an embedded real-analytic CR-manifold in ℂ^{N}. We are interested in a description of the isotropic automorphisms of M with respect to some point p ∈ M, i.e., the germs of biholomorphic mappings at p that preserve p and M. According to a result by Belošapka these germs are uniquely determined by their 2-jets if M is Levi non-degenerate at p [B2]. In the case when M is a Levi-nondegenerate hypersurface, not equivalent to a quadric, Belošapka and Loboda [B1, L1] proved that any such automorphism is uniquely determined by the pseudo-unitary part of its first order derivatives at p (see [E2] for a simple proof). According to a result by Ežov [E1], for a given isotropic automorphism Φ there exist coordinates such that the equation of M has Chern-Moser normal form and Φ is linear. Kružilin and Loboda [KL] proved that, if M, in addition, is strictly pseudoconvex, then there exist normal coordinates where all isotropic automorphisms are linear, simultaneously. The main tool in the study of these problems was the Chern-Moser normal form. Analogues of the normal form for CR-manifolds of higher codimension were constructed in the following situations:

1. M is rigid at p, i.e., the osculating quadric to M at p has only the trivial automorphisms (see [ES3]). If the CR-codimension k and the CR-dimension n of M satisfy 3 ≤ k ≤ n^{2} − 3 rigidity is a property of general position. It follows easily that any isotropic automorphism is automatically linear if the equation of M has normal form (see [ES2]).

2. M ⊂ ℂ^{4} has codimension 2. This case is exceptional because there exist only three types of (non-degenerate) Levi forms up to linear coordinate change. These three types are called hyperbolic, elliptic and parabolic. This is similar to the situation of hypersurfaces where one has finitely many Levi forms that are characterised by their signature. In difference to the above described rigid CR-manifolds, and again in resemblance with hypersurfaces, the osculating quadrics have large automorphism groups.

M is of hyperbolic type at p if, in suitable coordinates z_{1}, z_{2}, w_{1}, w_{2}, the equation of M takes the form

v_{1} = |z_{1}|^{2} + O(3)
v_{2} = |z_{2}|^{2} + O(3),

where w_{j} = u_{j} + iv_{j} and p = 0. The normal form was constructed by Loboda [L2]. Schmalz proved in [S1] that there exist normal coordinates where the isotropic automorphisms are simultaneously linear if M is not semi-quadratic, i.e., of the form

v_{1} = |z_{1}|^{2}
v_{2} = |z_{2}|^{2} + N^{2}(z_{2}, z̄_{2}, u_{2}).

In the latter case any isotropic automorphism splits into an automorphism of the sphere v_{1} = |z_{1}|^{2} and an automorphism of the strictly pseudoconvex hypersurface v_{2} = |z_{2}|^{2} + N^{2}(z_{2}, z̄_{2}, u_{2}) in ℂ^{2}.

The hyperbolicity is a property that is stable with respect to small perturbations. In particular, the set of hyperbolic points at M is open.

3. The only other stable situation in the said sense is the elliptic situation. A point p ∈ M ⊂ ℂ^{4} is of elliptic type if, in suitable coordinates z_{1}, z_{2}, w_{1}, w_{2}, the equation of M takes the form

v_{1} = Re z_{1}z̄_{2} + O(3)
v_{2} = Im z_{1}z̄_{2} + O(3).

The automorphisms of elliptic CR-manifolds will be studied in this paper. We will make use of the normal form constructed by the authors in [ES1]. Thus, M will be given by one complex equation

V = z_{1}z̄_{2} + N(z, z̄, η, η̄),

where z_{1}, z_{2}, ω_{1}, ω_{2} are coordinates in ℂ^{4} and

V = \frac{1}{2i}(ω_{1} − ω̄_{2}), η = \frac{1}{2}(ω_{1} + ω̄_{2}),

The expansion of the complex valued function N into polynomials N_{kl} of degree k with respect to z and l with respect to z̄ and with coefficients that are analytic functions of η equals

N = ∑_{k>1}(N_{k1} + N_{k1}) + N_{22} + N_{32} + N_{23} + N_{33} + ∑_{k,l} N_{kl},

where the polynomials N_{kl} meet the conditions

\frac{∂N_{k1}}{∂z̄_{2}} = 0  \frac{∂N_{1k}}{∂z_{1}} = 0, für k ≥ 2 (1)
\frac{∂^{3}N_{21}}{∂z_{1}∂z̄_{1}∂z_{2}} = 0  \frac{∂^{3}N_{12}}{∂z̄_{2}∂z̄_{1}∂z_{2}} = 0
\frac{∂^{4}N_{22}}{∂z_{1}∂z_{2}∂z̄_{1}∂z̄_{2}} = 0  \frac{∂^{4}N_{22}}{(∂z_{1})^{2}(∂z̄_{2})^{2}} |_{η̄=0} = 0
\frac{∂^{5}N_{32}}{(∂z_{1})^{3}(∂z̄_{2})^{2}} |_{η̄=0} = 0  \frac{∂^{5}N_{23}}{(∂z_{1})^{2}(∂z̄_{2})^{3}} |_{η̄=0} = 0
\frac{∂^{6}N_{33}}{(∂z_{1})^{3}(∂z̄_{2})^{3}} |_{η̄=0} = 0.

By N we will denote the space of all polynomials that satisfy these conditions. The space R spanned by all monomials that do not satisfy these conditions is then a direct complement of N. Thus, one of the properties of this decomposition N ⊕R is that any monomial either belongs to N or to R. Another property, that we will frequently employ, is that the N -space is stable under multiplication by any function in (z, z̄, η, η̄). It is worth mentioning that the latter is not true for the R-space. In this paper we attempt to find out when any isotropic automorphism ϕ = (f, g) : z^{∗} = f(z, w), w^{∗} = g(z, w) of an elliptic CR-manifold M is uniquely determined by the first order derivatives of f with respect to z at 0. In other words, if we denote by Aut_{0,id} M the subgroup of germs of isotropic automorphisms ϕ = (f, g) satisfying

\frac{∂f}{∂z} |\begin{matrix}z=0 \\ w=0 \end{matrix} = id,

then the latter statement reduces to: For which M Aut_{0,id} M = {id}?

Though the technical details are very different from the hyperbolic case, considered in [S1], the rough scheme of our approach is similar, namely

1. One shows that any automorphism ϕ ∈ Aut_{0,id} M can be embedded into some one-parametric subgroup ϕ^{t}. We have

Im g^{t}(z, ω) − f_{1}^{t}(z, ω)\overline{f_{2}^{t}(z, ω)} − N(f^{t}, f̄^{t}, Re g^{t})|_{Im ω=z_{1}z̄_{2}+N} ≡ 0, (2)

where Im ω = \frac{1}{2i}(ω_{1} − ω̄_{2}), Im g^{t} = \frac{1}{2i}(g_{1}^{t} − ḡ_{2}^{t}), Re g^{t} = \frac{1}{2}(g_{1}^{t} + ḡ_{2}^{t}).

It follows that the derivative of the left hand side with respect to t for t = 0 must vanish. Denote this function by P(z, z̄, η, η̄). It can be expressed by

P(z, z̄, η) = 2 Re χ(z_{1}z̄_{2} − V + N(z, z̄, η, η̄))|_{Im ω=z_{1}z̄_{2}+N} , (3)

where χ is a vector field that is called infinitesimal automorphism. It is now sufficient to prove that there are no non-vanishing vector fields such that

2 Re χ(V − z_{1}z̄_{2} − N(z, z̄, η, η̄))|_{Im ω=z_{1}z̄_{2}+N} ≡ 0.

Since this first step is a literal repetition of the arguments from the hyperbolic case [S1] the details will be omitted.

2. At the second step we will study the action of vector fields on N. Similar to the hyperbolic case, we will introduce the notions of weight and defect. If the vector field χ does not vanish its lower weighted components coincide with those from infinitesimal automorphisms of the quadric. To study the action of χ on N we introduce the socalled weight-defect diagrams (see Section 3 below). This construction is designed to show that if N_{γ} is the non-vanishing term of lowest weight of N and χ ≢ 0 then P_{γ_{0}+1} does not vanish.

2 Preliminaries

The elliptic quadric Q is given by the equation V = z_{1}z̄_{2}. The group Aut_{0,id} Q consists of the mappings

z_{j}^{∗} = \frac{z_{j} + a_{j}ω_{j}}{1 − δ_{j}}
ω_{j}^{∗} = \frac{ω_{j}}{1 − δ_{j}} ,

where δ_{j} = 2iz_{j}â_{j} − (r_{j} + ia_{j}â_{j})ω_{j}; a_{j} ∈ ℂ, r_{1} = r̄_{2} = r ∈ ℂ, j = 1, 2. By â = { \atop (a_{1}, a_{2})} we denote the vector (ā_{2}, ā_{1}) ([ES1]). The corresponding infinitesimal automorphisms are

∑_{j=1,2}(a_{j}ω_{j} + 2iâ_{j}z_{j}^{2} + r_{j}z_{j}ω_{j})\frac{∂}{∂z_{j}} + (2iâ_{j}z_{j}ω_{j} + r_{j}ω_{j}^{2})\frac{∂}{∂ω_{j}} .

Definition 1 By the following convention we assign a weight γ to each monomial in a formal power series or vector field:

γ(z_{j}) = 1,  γ(ω_{j}) = γ(η) = γ(η̄) = 2
γ (\frac{∂}{∂z_{j}} ) = −1,  γ (\frac{∂}{∂ω_{j}} ) = γ (\frac{∂}{∂η} ) = γ (\frac{∂}{∂η̄}) = −2,

and the weight of a product is the sum of the weights of its factors.

Let M be an elliptic CR-manifold given in normal form. Then any automorphism ϕ ∈ Aut_{0,id} M can be considered as a renormalisation, i.e., a change to (the same) normal coordinates. Therefore, ϕ has the form

z_{j}^{∗} = \frac{z_{j} + a_{j}ω_{j} + f̃_{j}}{1 − δ_{j}}
ω_{j}^{∗} = \frac{ω_{j} + g̃_{j}}{1 − δ_{j}} .

Let N_{γ_{0}+2} be the non-vanishing component of N with minimal weight γ_{0}+2 > 2 then f̃_{j} contains only components of weight ≥ γ_{0} + 2 and g̃_{j} contains only components of weight ≥ γ_{0} + 3.

Analogously, an infinitesimal automorphism χ equals to an infinitesimal automorphism of the quadric up to components of weight ≤ γ_{0}. The Lie algebra of infinitesimal automorphisms, corresponding to Aut_{0,id} M will be denoted by 𝔞𝔲𝔱 _{0,id}.

Definition 2 For any monomial Π in z_{1}, z_{2}, z̄_{1}, z̄_{2}, η, η̄ the 1-defect d_{1}(Π) is defined as the difference of the degrees with respect to z_{1} and with respect to z̄_{2}. By ¯1-defect d¯1(Π) we will denote the difference of the degrees with respect to z̄_{1} and with respect to z_{2}. The total defect d(Π) is the sum of d_{1} and d¯1.

We split the component

2 Re χ_{1} = 2 Re ∑_{j=1,2}(a_{j}ω_{j} + 2iâ_{j}z_{j}^{2})\frac{∂}{∂z_{j}} + (2iâ_{j}z_{j}ω_{j})\frac{∂}{∂ω_{j}}

of weight 1 in 2 Re χ into four parts

2 Re χ_{1} = χ_{1}^{1} + χ¯1 1 + χ_{1}^{2} + χ¯2 1,

with

χ_{1}^{1} = iā_{2} (2z_{1}^{2} \frac{∂}{∂z_{1}} + 2(η + iV)z_{1} \frac{∂}{∂ω_{1}} − i(η − iV)\frac{∂}{∂z̄_{2}} )  (4)
χ¯1 1 = \overline{χ_{1}^{1}} = −ia_{2} (2z̄_{1}^{2} \frac{∂}{∂z̄_{1}} − 2(η̄ − iV̄)z̄_{1} \frac{∂}{∂ω̄_{1}} + i(η̄ + iV̄)\frac{∂}{∂z_{2}} )
χ_{1}^{2} = iā_{1} (2z_{2}^{2} \frac{∂}{∂z_{2}} + 2(η̄ + iV̄)z_{2} \frac{∂}{∂ω_{2}} − i(η̄ − iV̄)\frac{∂}{∂z̄_{1}} )
χ¯2 1 = \overline{χ_{1}^{2}} = −ia_{1} (2z̄_{2}^{2} \frac{∂}{∂z̄_{2}} + 2(η − iV)z̄_{2} \frac{∂}{∂ω̄_{2}} + i(η + iV)\frac{∂}{∂z_{1}} ) .

Each of these fields increases by 1 the defect indicated by the upper index when acting on a monomial. Set ν = z_{1}z̄_{2}. Then any monomial Π(z_{1}, z_{2}, z̄_{1}, z̄_{2}, η, η̄) can be uniquely written (up to a constant multiplier) as

Π = z_{1}^{d_{1}} z̄_{1}^{d¯1}ν^{α}η^{β}ν̄ᾱη̄β̄.

Notice that the defects d_{1}, d¯1 might be negative. We consider ν as a new independent variable, i.e. we performed a non-linear coordinate change. Denote by F_{(d_{1},d¯1)} the linear space of polynomials z_{1}^{d_{1}} z̄_{1}^{d¯1}p(η, η̄, ν, ν̄). We introduce an isomorphism

π_{(d_{1},d¯1)} :  F_{(d_{1},d¯1)} → F_{(0,0)}
z_{1}^{d_{1}} z̄_{1}^{d¯1}p(η, η̄, ν, ν̄) ↦ p(η, η̄, ν, ν̄).

Thus π_{(d_{1},d¯1)} acts by multiplying with z_{1}^{−d_{1}}z̄_{1}^{−d¯1}. We will write π instead of π_{(d_{1},d¯1)} if it is clear or not important what are the defects. The defects measure the “disbalance” of a polynomial. Thus the linear mapping π can also be considered as a projection of the space of all polynomials onto the space of balanced polynomials. A monomial Π can be recovered from its balanced part if one memorises the defects of Π. For polynomials from F_{(0,0)} it is convenient to pass to coordinates

s = η + iν,
t = η − iν.

Then

Π = (\frac{s − t}{2i})^{α} (\frac{s + t}{2})^{β} (\frac{s̄ − t̄}{−2i})ᾱ (\frac{s̄ + t̄}{2})β̄ .

We will need the push-forwards of certain vector fields (acting as linear mappings from F_{(d_{1},d¯1)} to F_{(d′_{1},d′ ; ¯1)} ) with respect to π.

Lemma 1

π_{∗} (\frac{∂}{∂z_{1}} ) = (d_{1} id +\frac{s − t}{2} (\frac{∂}{∂s} − \frac{∂}{∂t}))
π_{∗} (\frac{∂}{∂z̄_{2}} ) = i (\frac{∂}{∂s} − \frac{∂}{∂t})
π_{∗} (\frac{∂}{∂ω_{1}} ) =\frac{1}{2} (\frac{∂}{∂s} + \frac{∂}{∂t} − i\frac{∂}{∂V} )
π_{∗} (\frac{∂}{∂ω̄_{2}} ) =\frac{1}{2} (\frac{∂}{∂s} + \frac{∂}{∂t} + i\frac{∂}{∂V} ) .

Proof. We only verify the first identity, the other verifications are similar. Let Π be a polynomial of defect (d_{1}, d¯1) and Ψ(s, t, s̄, t̄) = π(Π). Then

π_{∗} (\frac{∂}{∂z_{1}} ) Ψ = π_{(d_{1}−1,d¯1)} ∘ (\frac{∂}{∂z_{1}} ) ∘ π_{(d_{1},d¯1)}^{−1}Ψ =

=π_{(d_{1}−1,d¯1)} ( \frac{∂}{∂z_{1}} z_{1}^{d_{1}}z̄_{1}^{d¯1}Ψ(s, t, s̄, t̄)|\begin{matrix}s=η+iz_{1}z̄_{2} \\ t=η−iz_{1}z̄_{2} \end{matrix} ) =

=π_{(d_{1}−1,d¯1)} (d_{1}z_{1}^{d_{1}−1}z̄_{1}^{d¯1}Ψ(s, t, s̄, t̄) + z_{1}^{d_{1}}z̄_{1}^{d¯1}iz̄_{2} (\frac{∂}{∂s} − \frac{∂}{∂t}) Ψ(s, t, s̄, t̄)|\begin{matrix}s=η+iz_{1}z̄_{2} \\ t=η−iz_{1}z̄_{2} \end{matrix})
= (d_{1} id +\frac{s − t}{2} (\frac{∂}{∂s} − \frac{∂}{∂t})) Ψ.
□

In particular we obtain

π_{∗}(χ_{1}^{1}) = iā_{2} (2d_{1} id +2s\frac{∂}{∂s} + [2iV − (s − t)]\frac{∂}{∂t})  (5)

and

π_{∗}(χ_{1}^{1})|_{V =ν} =2iā_{2} (d_{1} id +s\frac{∂}{∂s}) .  (6)

The introduced projection π is adapted to the study of action of the vector fields that increase the 1- and ¯1-defects χ_{1}^{1} and χ¯1 1 (this corresponds to the parameter a_{2} in χ_{1}, cp. (4)). For the study of the action of the vector fields χ_{1}^{2} and χ¯2 1 we would use similar mappings

π′_{(d_{1},d¯1)} :  F_{(d_{1},d¯1)} → F_{(0,0)}
z_{2}^{−d¯1} z̄_{2}^{−d_{1}}p(η, η̄, ν, ν̄) ↦ p(η, η̄, ν, ν̄).

Analogously as above, one obtains

π′_{∗}(χ¯2 1)|_{V =ν} = 2 (t\frac{∂}{∂t} − d_{1} id) .

3 Weight-defect diagrams

Let M be a non-quadratic elliptic manifold with first non-quadratic term N_{γ_{0}+2}. Then the difference of any infinitesimal automorphism χ and the corresponding infinitesimal automorphism of the quadric begins, at earliest, with the component of weight γ_{0} + 1. Consider the infinitesimal automorphism

χ = χ_{1} + χ_{2} + χ_{γ_{0}+1} + ⋯

Suppose, the parameter a in χ_{1} was different from 0. Then, by a linear coordinate change one can achieve that a_{2} = i. Let Δ be the maximal and δ be the minimal total defect that occur in the lowest non-quadratic term N_{γ_{0}+2}. Furthermore, let δ_{1} and δ¯1 be the maximal 1-defect resp. ¯1-defect in N_{γ_{0}+2,Δ} and −δ′_{1} resp. −δ′ ; ¯1 the minimal 1-defect resp. ¯1-defect in N_{γ_{0}+2,δ}. We will analyse when the hypothesis a ≠ 0 leads to a contradiction to identity (3) in its component of weight γ_{0} + 3

P_{γ_{0}+3} ≡ 0.

It is convenient to write this in the form of a diagram:

[figure: P̃_{γ_{0}+3,δ_{1}+1,Δ−δ_{1}}; N_{γ_{0}+2,δ_{1},Δ−δ_{1}}; χ_{1}^{1}; χ¯1 1; P̃_{γ_{0}+3,δ_{1},Δ−δ_{1}+1}; N_{γ_{0}+2,δ_{1}−1,Δ−δ_{1}+1}; χ_{1}^{1}; …; N_{γ_{0}+2,Δ−δ¯1+1,δ¯1−1}; χ¯1 1; P̃_{γ_{0}+3,Δ−δ¯1+1,δ¯1}; N_{γ_{0}+2,Δ−δ¯1,δ¯1}; χ_{1}^{1}; χ¯1 1; P̃_{γ_{0}+3,Δ−δ¯1,δ¯1+1}; χ_{1}^{2}; χ¯2 1; χ_{1}^{2}; P̃_{γ_{0}+3,δ_{1},Δ−δ_{1}−1}; P̃_{γ_{0}+3,δ_{1},Δ−δ_{1}}; P̃_{γ_{0}+3,Δ−δ¯1,δ¯1−1}; P̃_{γ_{0}+3,Δ−δ¯1−1,δ¯1}; χ¯2 1; χ_{1}^{2}; χ¯2 1; χ_{1}^{2}; χ¯2 1; χ_{1}^{2}; P̃_{γ_{0}+3,δ_{1},Δ−δ_{1}−1}; P̃_{γ_{0}+3,δ_{1},Δ−δ_{1}}; P̃_{γ_{0}+3,Δ−δ¯1,δ¯1−1}; P̃_{γ_{0}+3,Δ−δ¯1−1,δ¯1}; χ_{1}^{1}; χ¯1 1; χ_{1}^{1}; χ¯1 1; χ_{1}^{1}; χ¯1 1; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; ·; N_{γ_{0}+2,δ+δ′ ; ¯1,−δ′ ; ¯1}; N_{γ_{0}+2,δ+δ′ ; ¯1−1,−δ′ ; ¯1+1}; …; N_{γ_{0}+2,−δ′_{1}+1,δ+δ′_{1}−1}; N_{γ_{0}+2,−δ′_{1},δ+δ′_{1}}; χ_{1}^{2}; χ¯2 1; χ_{1}^{2}; P̃_{γ_{0}+3,δ+δ′ ; ¯1,−δ′ ; ¯1−1}; P̃_{γ_{0}+3,δ+δ′ ; ¯1−1,−δ′ ; ¯1}; P̃_{γ_{0}+3,−δ′_{1},δ+δ′_{1}−1}; P̃_{γ_{0}+3,−δ′_{1}−1,δ+δ′_{1}}; χ¯2 1; χ_{1}^{2}; χ¯2 1]

Here,

P̃_{γ_{0}+3,α+1,d−α} = N_{γ_{0}+2,α,d−α} \frac{∂}{∂V} χ_{1}^{1}(ν − V)|_{V =ν} + χ_{γ_{0}+1,α+1,d−α}(ν − V)|_{V =ν} .

By χ_{γ_{0}+1,α+1,d−α} we mean the component of 2 Re χ_{γ_{0}+1} that produces by applying to ν − V a term of defect (α + 1, d − α). Since ν − V has defect (0, 0), the part χ_{γ_{0}+1,α+1,d−α} is itself the vector field of the weight γ_{0} + 1 and defect (α + 1, d − α). This diagram illustrates that P_{γ_{0}+3,α+1,d−α} is a sum of P̃_{γ_{0}+3,α+1,d−α} and of the result of applying the vector fields indicated at the arrows to the term at the base point of the arrow pointing to P̃_{γ_{0}+3,α+1,d−α}. The dots mean that this diagram should be continued for all total defects δ ≤ d ≤ Δ. For δ < d < Δ there are (at most) four arrows pointing to P̃_{γ_{0}+3,α+1,d−α} corresponding to the results of applying χ_{1}^{1}, χ_{1}^{2}, χ¯1 1, χ¯2 1. Of course, the length and the place of the lines in the diagram depends on d since maximal and minimal 1- and ¯1-defects are different for different total defects d. Also we should mention that the introduced diagram consists either of terms with odd total defect or of terms with even total defect. Later (see Lemma 4) we will show that the assumption a_{2} ≠ 0, that produces non-vanishing ↖ and ↗ arrows in the diagram ultimates a_{1} = 0, that makes the ↘ and ↙ arrows vanish. Thus, the weight-defect diagram always splits into a number of single-line diagrams with only ↖ and ↗ (or, ↘ and ↙) arrows remaining. We will work with a version of the weight-defect diagram that is adapted to the coordinates s, t. Denote πN_{γ_{0}+2,α,d−α} by 𝔑_{γ_{0}+2,α,d−α} and π_{∗}χ_{γ_{0}+1,α+1,d−α} by X_{γ_{0}+1,α+1,d−α}. Since

πN_{γ_{0}+2,α,d−α} \frac{∂}{∂V} χ_{1}^{1}(ν − V)|_{V =ν} = πN_{γ_{0}+2,α,d−α} \frac{∂}{∂V} 2z_{1}(1 − V)|_{V =ν}
= − 2𝔑_{γ_{0}+2,α,d−α}

we will include this term into the result of the operator at the arrow applied to 𝔑_{γ_{0}+2,α,d−α}. Then at the tips of the arrows there remain only the terms 𝔛_{γ_{0}+3,α+1,d−α} = X_{γ_{0}+1,α+1,d−α}(ν − V)|_{ν=V}.

[figure: 𝔛_{γ_{0}+3,δ_{1}+1,Δ−δ_{1}}; 𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}}; 2s \frac{∂}{∂s} +; 2(δ_{1}−1)id; 2s̄\frac{∂}{∂s̄} +; 2(Δ−δ_{1})id; 𝔛_{γ_{0}+3,δ_{1}+1,Δ−δ_{1}+1}; 𝔑_{γ_{0}+2,δ_{1}−1,Δ−δ_{1}+1}; 2s \frac{∂}{∂s} +; 2(δ_{1}−2)id; …; 𝔑_{γ_{0}+2,Δ−δ¯1+1,δ¯1−1}; 2s̄\frac{∂}{∂s̄} +; 2(δ¯1−1)id; 𝔛_{γ_{0}+3,Δ−δ¯1+1,δ¯1}; 𝔑_{γ_{0}+2,Δ−δ¯1,δ¯1}; 2s \frac{∂}{∂s} +; 2(Δ_{1}−δ¯1−1)id; 2s̄\frac{∂}{∂s̄}; +2δ¯1id; 𝔛_{γ_{0}+3,Δ−δ¯1,δ¯1}]

We interpret the upper left corner of the diagram as

2 (s\frac{∂}{∂s} + (δ_{1} − 1) id) 𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}} + X_{γ_{0}+1,δ_{1}+1,Δ−δ_{1}}(ν − V)|_{V =ν} = 0 (7)

and the upper right corner as

2 (s̄\frac{∂}{∂s̄} + δ¯1 id) 𝔑_{γ_{0}+2,Δ−δ¯1,δ¯1} + X_{γ_{0}+1,Δ−δ¯1,δ¯1+1}(ν − V)|_{V =ν} = 0. (8)

For certain defects the operator in front of 𝔑_{γ_{0}+2,Δ−δ¯1,δ¯1} is non-degenerate. This implies immediately that, for 𝔑_{γ_{0}+2,Δ−δ¯1,δ¯1} ≠ 0 The second summand in the equations above must be different from zero. It is now a crucial observation that

2 Re χ_{γ_{0}−1}|_{V =ν}(ν − V)

always contains a component that belongs to the space R, namely for

χ = ∑_{j=1,2} f_{j} \frac{∂}{∂z_{j}} + g_{j} \frac{∂}{∂ω_{j}}

2 Re χ|_{V =ν}(ν − V) contains, for example,

f_{1}(z, η)z̄_{2}, \overline{f_{2}(z, η)}z_{1}, \frac{1}{2i}g_{1}(z, η), \frac{i}{2}\overline{g_{2}(z, η)}.

Moreover, the field χ can be completely recovered from the R-part of

2 Re χ|_{V =ν}(ν − V).

Our strategy is to analyse when the first summand in the equations can contain non-normal terms. This will either lead to a contradiction with the assumption a_{2} ≠ 0, or else, will determine the terms N_{γ_{0}+2} and χ_{γ_{0}+1}.

Lemma 2 If M has an infinitesimal automorphism χ with a_{2} ≠ 0 then for the maximal 1-defect δ_{1} and the maximal ¯1-defect δ¯1 that occur in N_{γ_{0}+2,Δ} one has

−2 ≤δ_{1} ≤ 1  (9)
−1 ≤δ¯1 ≤ 0  (10)

Proof. Suppose δ_{1} ≥ 2. Then the operator

s\frac{∂}{∂s} + (δ_{1} − 1) id

is invertible. It follows from (7) and from N_{γ_{0}+2,δ_{1},Δ−δ_{1}} ≠ 0 that X_{γ_{0}+1} ≠ 0 and

χ_{1}^{1}(N_{γ_{0}+2,δ_{1},Δ−δ_{1}}) ∉ N .

According to (4), χ_{1}^{1} consists of 3 summands. The first two of them preserve N, and only the third one

−i(η − iV)\frac{∂}{∂z̄_{2}}

can produce non-normal terms from normal ones. This happens exactly when

\frac{∂}{∂z̄_{2}} N_{γ_{0}+2,δ_{1},Δ−δ_{1}} ∉ N .

Since δ_{1} ≥ 2 > 0, N_{γ_{0}+2,δ_{1},Δ−δ_{1}} is divisible by z_{1}^{δ_{1}}. The part N_{R} in N_{γ_{0}+2,δ_{1},Δ−δ_{1}} that contributes to R must be divisible by z̄_{2}^{2}, hence by ν^{2} and not by z̄_{1}. Otherwise, the terms N_{R} and \frac{∂}{∂z̄_{2}} N_{R} would be simultaneously normal or non-normal. On the other hand it cannot be divisible by a higher power of ν because then \frac{∂}{∂z̄_{2}} N_{R} would be normal. Thus, the only possibility (cp. Table 2 in Appendix) is

N_{R} = z_{1}^{δ_{1}}z_{2}^{δ_{1}−Δ}ν^{2}P (η, η̄),

where P (η, η̄) is a homogeneous polynomial of degree (γ_{0} − 2δ_{1} + Δ + 2)/2. Consequently,

−iη \frac{∂}{∂z̄_{2}} N_{R} = −iz_{1}^{δ_{1}+1}νηP (η, η̄).

Since this cancels with the non-normal part of 2 Re χ_{γ_{0}+1}(ν − V), we obtain

2 Re χ_{γ_{0}−1}(ν − V)|_{V =ν} =iz_{1}^{δ_{1}+1}z_{2}^{δ_{1}−Δ}ν(η + iν)P (η + iν̄, η̄ + iν̄),

and

χ_{γ_{0}+1} =iz_{1}^{δ_{1}+2}z_{2}^{δ_{1}−Δ}ω_{1}P (ω_{1}, ω_{2})\frac{∂}{∂z_{1}} .

This means that X_{1}|_{V =ν}(ν − V) contains a monomial

#(s̄ − t̄)^{δ_{1}−Δ}(s − t)s^{n−1}t̄^{k+Δ−δ_{1}}

of a certain bidegree in (s, t) and (s̄, t̄) variables which we set equal to (n, k). According to (7), in 𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}} is then a summand

# (s\frac{∂}{∂s} + (δ_{1} − 1) id)^{−1} [(s̄ − t̄)^{δ_{1}−Δ}(s − t)s^{n+1}t̄^{k}] =

# (\frac{s^{n}}{n + δ_{1} − 1} − \frac{ts^{n−1}}{n + δ_{1} − 2}) (s̄ − t̄)^{δ_{1}−Δ}t̄^{k+Δ−δ_{1}},

i.e., N_{γ_{0}+2,δ_{1},Δ−δ_{1}} contains a monomial

\frac{c}{2(n + δ_{1} − 1)(n + δ_{1} − 2)}z_{1}^{δ_{1}}z_{2}^{δ_{1}−Δ}η^{n}η̄^{k+Δ−δ_{1}},

that is harmonic and does not belong to N. The contradiction proves the statement of the lemma for δ_{1} ≥ 2. Suppose now that δ¯1 ≥ 1. Instead of equation (7) from the left upper corner of the weight-defect diagram we use now equation (8) from the right upper corner. Similar reasoning as above will lead to a contradiction to P_{γ_{0}+3,Δ−δ¯1,δ¯1+1} ≠ 0. For δ¯1 ≥ 1 the operator

s̄\frac{∂}{∂s̄} + δ¯1 id

is invertible and consequently

2 (s̄\frac{∂}{∂s̄} + δ¯1 id) 𝔑_{γ_{0}+2,Δ−δ¯1} ≠ 0.

The latter expression must cancel with 2 Re X̄_{γ_{0}+1}(ν − V)|_{V =ν} that has nonnormal components. Therefore, χ_{1}^{1}N_{γ_{0}+2,Δ−δ¯1,δ¯1} also contains non-normal components that can arise only from

iη̄\frac{∂}{∂z_{2}} N_{γ_{0}+2,Δ−δ¯1,δ¯1}.

According to Table 3 in the Appendix, there are the following possibilities: i. if β = δ¯1 − Δ ≥ 0 then it is a component

Q = z̄_{1}^{δ¯1}z̄_{2}^{β}ν̄(P_{1}(η, η̄) + νP_{2}(η, η̄))

in N_{γ_{0}+2,Δ−δ¯1,δ¯1}. ii. if β = −1 then it is a component

Q = z_{1}z̄_{1}^{δ¯1} ν̄P_{2}(η, η̄).

Consider the first option. Then

iη̄\frac{∂}{∂z_{2}} Q = iz̄_{1}^{δ¯1+1}z̄_{2}^{β}η̄(P_{1}(η, η̄) + νP_{2}(η, η̄))

and

χ_{γ_{0}+1,Δ−δ¯1,δ¯1} = 2z̄_{1}^{δ¯1+1}z̄_{2}^{β}ω̄_{1}P_{1}(ω̄_{2}, ω̄_{1})\frac{∂}{∂ω̄_{2}} − iz̄_{1}^{δ¯1+1}z̄_{2}^{β+1}ω̄_{1}P_{2}(ω̄_{2}, ω̄_{1})\frac{∂}{∂z̄_{2}}

i.e.,

X_{γ_{0}+1,Δ−δ¯1,δ¯1}(ν − V)|_{V =ν} = 2 (\frac{s − t}{2i})^{β} s̄P_{1}(t, s̄) − i (\frac{s − t}{2i})^{β+1} s̄P_{2}(t, s̄).

Then 𝔑_{γ_{0}+2,Δ−δ¯1,δ¯1}) contains a monomial

\frac{1}{2} (s̄\frac{∂}{∂s̄} + δ¯1 id)^{−1} c_{1}(s − t)^{β}s̄^{k}t^{n−β} = \frac{c_{1}}{2(k + δ¯1)}(s − t)^{β}s̄^{k}t^{n−β}

or,

\frac{1}{2} (s̄\frac{∂}{∂s̄} + δ¯1 id)^{−1} c_{2}(s − t)^{β+1}s̄^{k}t^{n−β−1} = \frac{c_{2}}{2(k + δ¯1)}(s − t)^{β+1}s̄^{k}t^{n−1}.

It follows that there is a non-normal component

\frac{c_{1}}{2(k + δ¯1)}z̄_{1}^{δ¯1}z̄_{2}^{β}η^{n}η̄^{k}, resp., \frac{c_{2}}{2(k + δ¯1)}z̄_{1}^{δ¯1}z̄_{2}^{β+1}η^{n−1}η̄^{k}

in N_{γ_{0}+2,Δ−δ¯1,δ¯1}. Analogously, in case ii. we would obtain a non-normal component

\frac{c_{2}}{2(k + δ¯1)}z̄_{1}^{δ¯1}νη^{n}η̄^{k}

in N_{γ_{0}+2,1,δ¯1}. The contradiction proves the assertion of the lemma for δ¯1 ≥ −1. Suppose now δ_{1} ≤ −3. Then 𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}} is divisible by ν^{−δ_{1}}. So χ_{1}^{1} does not produce non-normal terms and therefore

χ_{γ_{0}+1,δ_{1}+1,Δ−δ_{1}} = 0.

It follows that 𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}} belongs to the kernel of s\frac{∂}{∂s} + (δ_{1} − 1) id and, therefore, has the form

𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}} = s^{−δ_{1}+1}Π(t, s̄, t̄).

But this polynomial is never divisible by s − t = 2iν, because it does not vanish for s = t. The contradiction proves the lemma for this case. Notice that this argument remains true for δ_{1} ≤ −1 if 𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}} does not contain monomials that produce non-normal components. Analogously we deal with situation δ¯1 ≤ −2. Then 𝔑_{γ_{0}+2,Δ−δ¯1,δ¯1} is divisible by ν̄^{−δ¯1}. Consequently, χ¯1 1 does not produce non-normal which implies

χ_{γ_{0}+1,Δ−δ¯1,δ¯1+1} = 0.

Then, 𝔑_{γ_{0}+2,Δ−δ¯1,δ¯1} belongs to the kernel of s̄\frac{∂}{∂s̄} + δ¯1, i.e.,

𝔑_{γ_{0}+2,Δ−δ¯1,δ¯1} = s̄^{−δ¯1}Π(s, t, t̄),

that is, though, never divisible by s̄− t̄= −2iν. This completes the proof of the lemma. □

We obtain an analogous result concerning the minimal 1 and ¯1 defects that occur in the polynomial of minimal total defect by a simple dualisation: The (anti-holomorphic) coordinate change (z_{1}, z_{2}, ω_{1}, ω_{2}) ↦ (z̄_{2}, z̄_{1}, ω̄_{2}, ω̄_{1}) transforms the normalised equation V = ν + N into a normalised equation with respect to the new coordinates (and the new complex structure). Denote by δ¯2 the maximal 1-defect and by δ_{2} the the maximal ¯1-defect that occur in the component of maximal total defect in the lowest non-quadratic term. Then −δ¯2 is the minimal 1-defect and −δ_{2} the minimal ¯1-defect in the component of minimal total defect in the lowest non-quadratic term of the initial manifold. Reformulating Lemma 2 for δ_{2} and δ¯2 we obtain

Lemma 3 If M has an infinitesimal automorphism with a_{1} ≠ 0 then

−2 ≤δ¯2 ≤ 1  (11)
−1 ≤δ_{2} ≤ 0  (12)

Remark. In the latter proofs we have only used the fact that there is only one arrow, ↖, at the left corner of the upper line (upper left corner) of the weight-defect diagram pointing at P_{γ_{0}+3,δ_{1}+1,Δ−δ_{1}}, and, also, that there is only one arrow, ↗, at the right corner of the upper line (upper right corner) of the weight-defect diagram pointing at P_{γ_{0}+3,Δ−δ¯1,δ_{¯1+1}}. The same situation happens, of course, at the most upper line that provides the most left entry N_{γ_{0}+2,d} (left upper corner) and, analogously, for the right upper corner and, respectively in Lemma 3) , for the right lower and left lower corners of the diagram. Thus, inequalities (9) - (12) applied to those corners are also valid.

4 Results

Theorem 1 For any real-analytic, non-quadratic, elliptic CR-manifold M the group Aut_{0,id} M ≅ id or ℂ.

In dependence on whether Aut_{0,id} M ≅ id or ℂ we will distinguish two types of manifolds.

Definition 3 M is called non-exceptional if Aut_{0,id} M ≅ id. If Aut_{0,id} M ≅ ℂ the manifold M is called exceptional.

In particular, if inequalities (9) or (10) and (11) or (12) are not satisfied then M is non-exceptional. Thus, the property to be non-exceptional is a property of general position. In fact we do not know whether exceptional manifolds exist. For the time being we can only spot the candidates for exceptional manifolds and determine their first non-quadratic terms, which, as we shall see, determine the entire defining equation of the manifold. It turns out that the first non-quadratic term N_{γ_{0}+2} = ∑_{d=δ}^{Δ} N_{γ_{0}+2,d} of the defining equation in normal form of an exceptional M consists of solutions of the equation

χ_{γ_{0}+1,d+1}(ν − V)|_{V =ν} + χ_{1}N_{γ_{0}+2,d}|_{V =ν} + χ_{1}(ν − V)|_{V =N_{γ_{0}+2,d}} = 0 (13)

that can be interpreted as a single line of the weight-defect diagram corresponding to the total defect d with ↖ and ↗ arrows only. We consider (13) as a system of equations for

χ_{γ_{0}+1,d+1} = ∑_{j=0}^{δ_{1}+δ¯1−d+1} χ_{γ_{0}+1,δ_{1}−j+1,d−δ_{1}+j}

and

N_{γ_{0}+2,d} = ∑_{j=0}^{δ_{1}+δ¯1−d} N_{γ_{0}+2,δ_{1}−j,d−δ_{1}+j}.

It is more convenient to consider system (13) in coordinates (s, t, s̄, t̄) in F_{(0,0)}. Let 𝔑_{γ_{0}+2,δ_{1}−j,d−δ_{1}+j}^{n,k} be the component of degree n in s, t and degree k in s̄, t̄ of 𝔑_{γ_{0}+2,δ_{1}−j,d−δ_{1}+j}. Analogously, let X_{γ_{0}+1,δ_{1}−j+1,d−δ_{1}+j}^{n−1,k} be the component of bidegree (n − 1, k) of X_{γ_{0}+1,δ_{1}−j+1,d−δ_{1}+j} (the degrees of \frac{∂}{∂s} and \frac{∂}{∂t} are set equal to −1).

Theorem 2 The expressions below form a basis of solutions of (13)

1. For d = −2, and bidegree (n, 0), n ≥ 5

X_{γ_{0}+1,−1,0}^{n−1,0} = − (n − 3)(n − 4)[(η + iV)^{n−1} (d_{1} id −\frac{s − t}{2} (\frac{∂}{∂s} − \frac{∂}{∂t})) +

+ \frac{1}{2}(n − 1)(η − iV)^{n−2}(s − t)^{2} (\frac{∂}{∂s} − \frac{∂}{∂t}) −

− \frac{1}{2}(η − iV)^{n−1}(s − t) (\frac{∂}{∂s} + \frac{∂}{∂t} + i\frac{∂}{∂V} ) ]

here and below in this theorem a vector field X is considered as a differential operator applied to the polynomials obtained by homomorphism π_{(d−1,d−2)} from a polynomial of defect (d_{1}, d − 2)).

𝔑_{γ_{0}+2,−2,0}^{n,0} = − \frac{i}{12}(n − 4)(n − 3)(n − 2)t^{n} + \frac{i}{8}(n − 4)(n − 3)(2n − 3)st^{n−1} −

− \frac{i}{4}(n − 4)(n − 3)(n − 1)s^{2}t^{n−2} +

+ \frac{i}{24}(n − 2)(n − 1)(2n − 9)s^{3}t^{n−3} +

+ \frac{i}{4}(n − 3)s^{n−1}t − \frac{i}{4}(n − 4)s^{n}  (14)

2. For d = −1, and bidegree (n, 0), n ≥ 4

X_{γ_{0}+1,0,0}^{n−1,0} =4(n − 3)[n(η + iV)^{n−1} (d_{1} id +\frac{s − t}{2} (\frac{∂}{∂s} − \frac{∂}{∂t})) +

+ n(η − iV)^{n−1} \frac{s − t}{2} (\frac{∂}{∂s} − \frac{∂}{∂t}) +

+ (η + iV)^{n} (\frac{∂}{∂s} + \frac{∂}{∂t} − i\frac{∂}{∂V} ) +

+ (η − iV)^{n} (\frac{∂}{∂s} + \frac{∂}{∂t} + i\frac{∂}{∂V} ) ]

𝔑_{γ_{0}+2,−1,0}^{n,0} =\frac{1}{2}(n − 3)(n − 2)t^{n} − (n − 3)nst^{n−1} +
+ \frac{1}{2}(n − 1)ns^{2}t^{n−2} − ns^{n−1}t + (n − 3)s^{n}  (15)

3. For d = −m and bidegree (n, k), n ≥ m + 3, k > m ≥ 0

X_{γ_{0}+2,−m+1}^{n−1,k} =X_{γ_{0}+2,1,−m}^{n−1,k} =
=\frac{(n − 1)!}{(n − m − 3)!}(η + iV)^{n−1}(η̄ + iV̄)^{k−m} ×

× (t̄− s̄)^{m} (−2δ_{1} id +(s − t) (\frac{∂}{∂s} − \frac{∂}{∂t}))

𝔑_{γ_{0}+2,0,−m}^{n,k} =

= \frac{(n − 3)!}{2(n − m − 3)!} [−(n − 2)s^{n} + (n − 1)s^{n−1}t − st^{n−1}] (t̄− s̄)^{m}t̄^{k−m}

= (s − t)^{2}P_{n−2}^{(0)}(s, t)(t̄− s̄)^{m}t̄^{k−m}

𝔑_{γ_{0}+2,−j,−m+j}^{n,k} = (s − t)^{j+2}P_{n−j−2}^{(j)}(s, t)(t̄− s̄)^{m−j}t̄^{k−m+j},  (16)

where

\frac{1}{2} (s\frac{∂}{∂s} − (j + 1) id) (s − t)^{j+2}P_{n−j−2}^{(j)}(s, t) + (s − t)^{j+1}P_{n−j−1}^{(j−1)}(s, t) = 0.

4. For d = −m + 1 and bidegree (n, k), n ≥ m + 2, k > m ≥ 0

X_{γ_{0}+2,−m+2}^{n−1,k} = X_{γ_{0}+2,2,−m}^{n−1,k} =

= \frac{n!}{(n − m − 2)!}(η + iV)^{n−1}(t̄− s̄)^{m} (−2δ_{1} id +(s − t) (\frac{∂}{∂s} − \frac{∂}{∂t}))

𝔑_{γ_{0}+2,1,−m}^{n,k} =

= \frac{(n − 2)!}{2(n − m − 2)!} [−(n − 1)s^{n} + ns^{n−1}t − t^{n}] (t̄− s̄)^{m}t̄^{k−m}

= (s − t)^{2}P_{n−2}^{(0)}(s, t)(t̄− s̄)^{m}t̄^{k−m}

𝔑_{γ_{0}+2,−j+1,−m+j}^{n,k} = (s − t)^{j+2}P_{n−j−2}^{(j)}(s, t)(t̄− s̄)^{m−j}t̄^{k−m+j},  (17)

where

\frac{1}{2} (s\frac{∂}{∂s} − j id) (s − t)^{j+2}P_{n−j−2}^{(j)}(s, t) + (s − t)^{j+1}P_{n−j−1}^{(j−1)}(s, t) = 0.

Theorem 3 For exceptional M each polynomial 𝔑_{γ_{0}+2,d_{1},d_{2}} is a linear combination of the elements of the basis listed in Theorem 2.

In fact, only polynomials (16) and (17) can occur in a common linear combination. Because of of the defect-bidegree partition, if (14) or (15) occurs in a linear combination then the whole combination is a multiple of such particular term. One can extend a weight-defect diagram for all weights γ ≥ γ_{0}. The system of equations analogous to (13) for arbitrary γ ≥ γ_{0} takes the form

χ_{γ+1,d+1}(ν − V)|_{V =ν} + χ_{1}N_{γ+2,d}|_{V =ν} + χ_{1}(ν − V)|_{V =N_{γ+2,d}} = P (N_{<γ+2}, χ_{<γ+1}). (18)

We consider (18) as a system of non-homogeneous equations with the right hand side P. Having a basis of solutions of the homogeneous system determined in Theorem 2, we conclude with the following Theorem 4 For exceptional M each polynomial 𝔑_{γ+2,d_{1},d_{2}} is a sum of a linear combination of the elements of the basis listed in Theorem 2 (with γ_{0} substituted by γ) and the unique solution Ñ_{γ+2,d} of (18) with the conditions that Ñ_{γ+2,d} does not contain monomials of the form #z̄_{2}^{2}ν^{2}η^{n−4}, #z̄_{2}ν^{3}η^{n−4}, #ν^{2}η^{n−2}z_{2}^{m}η̄^{k−m}, #z_{1}ν^{2}η^{n−2}z_{2}^{m}η̄^{k−m}.

The uniqueness of the particular solution in the above theorem means that there could be none (and, most likely, to happen so), in case there is a solution, we generally determine how to choose the particular one.

5 The case δ_{1} = −2

In this section we consider the case when N_{γ_{0}+2,Δ} has maximal 1-defect δ_{1} = −2. It follows from Lemma 2 that we may restrict ourselves to the case d¯1 = Δ + 2 ≤ δ¯1 ≤ 0. Then N_{γ_{0}+2,−2,Δ+2} consists of terms

z̄_{2}^{2}z_{2}^{−Δ−2}P (ν, ν̄, η, η̄).

According to Table 2 in the Appendix, the only monomials Ψ in this expression such that \frac{∂}{∂z̄_{2}} Ψ belongs to R are

Ψ_{1} = z̄_{2}^{2}z_{2}^{m}η^{n−2}η̄^{k−m}, m ≥ 1  (19)
Ψ_{2} = z̄_{2}^{2}ν̄η^{n−2}η̄^{k−1}  (20)
Ψ_{3} = z̄_{2}^{2}ν^{2}η^{n−4}.  (21)

If N_{γ_{0}+2,−2,Δ+2} does not contain monomials of the type (19), (20) or (21) then N_{γ_{0}+2,−2,Δ+2} belongs to the kernel of (s\frac{∂}{∂s} − 3 id), and we obtain a contradiction by the same argument as in Lemma 2. Suppose now that N_{γ_{0}+2,−2,Δ+2} contains a monomial

Ψ_{1} = #z̄_{2}^{2}z_{2}^{m}η^{n−2}η̄^{k−m}.

Then Δ = −m − 2 and therefore there are no monomials of type Ψ_{2}, Ψ_{3} in N_{γ_{0}+2,−2,−m}. Let us remark that we set n for the degree with respect to the holomorphic variables s, t and k for the degree with respect to the antiholomorphic variables s̄, t̄ in π(Ψ_{1}). These degrees are preserved by the vector fields that we apply below. Further on we will always refer to the pair (n, k) as to the bidegree of a polynomial. The homogeneous part of bidegree (n, k) of a polynomial Ψ will be denoted by Ψ^{n,k}. We consider the part of bidegree (n, k) of the upper line in the weight-defect diagram starting from the left corner. It consists of the terms 𝔑_{γ_{0}+2,−2,−m}^{n,k}, …, 𝔑_{γ_{0}+2,−2−m,0}^{n,k}. According to Tables 2 and 3 in the Appendix, for 1 ≤ j ≤ m−1, both

\frac{∂}{∂z̄_{2}} N_{γ_{0}+2,−2−j,−m+j}^{n,k}, and \frac{∂}{∂z_{2}} N_{γ_{0}+2,−3−j,−m+j+1}^{n,k}

always belong to N. Therefore, 𝔛_{γ_{0}+3,−2,−m+1}^{n,k}, …, 𝔛_{γ_{0}+3,−m,−1}^{n,k} vanish and the field 𝔛_{γ_{0}+3,−1,−m}^{n,k}, is determined by the R-contribution of 𝔑_{γ_{0}+2,−2,−m}^{n,k} and 𝔛_{γ_{0}+3,−1−m,0}^{n,k} and 𝔛_{γ_{0}+3,−2−m,1}^{n,k} are determined by the R-contributions of 𝔑_{γ_{0}+2,−2−m,0}^{n,k}. Thus, we will be able to calculate all contributions to P_{γ_{′}+∋,−∞−⇕,′}^{\,∥} in two different ways, starting from the left, and, also, from the right corner. This will lead to a contradiction. The R-term produced by iη \frac{∂}{∂z̄_{2}} equals

#z̄_{2}z_{2}^{m}η^{n−1}η̄^{k−m}.

Then

χ_{γ_{0}+1,−1,−m}^{n−1,k} = #z_{2}^{m}ω_{1}^{n−1}ω_{2}^{k−m} \frac{∂}{∂z_{1}}
X_{γ_{0}+1,−1,−m}(ν − V)|_{V =ν} = #(s − t)s^{n−1}(s̄ − t̄)^{m}t̄^{k−m}.

¿From the equation

2 (s\frac{∂}{∂s} − 3 id) 𝔑_{γ_{0}+2,−2,−m}^{n,k} = −X_{γ_{0}+1,−1,−m}^{n−1,k}(ν − V)|_{V =ν}

we obtain that n ≠ 4. Only is this case there exist a solution that is uniquely determined by the condition that 𝔑_{γ_{0}+2,−2,−m}^{n,k} must be divisible by s − t (because of the negative 1-defect). It follows

𝔑_{γ_{0}+2,−2,−m}^{n,k} = #[(n − 4)s^{n} − (n − 3)s^{n−1}t + s^{3}t^{n−3}](t̄− s̄)^{m}t̄^{k−m}.

Notice that this term is divisible by (s−t)^{2} but not by (s−t)^{3} (corresponding to the 1-defect −2). For convenience in calculations we set the coefficient

# = \frac{(n − 5)!}{(n − m − 5)!}

Since the terms 𝔛_{γ_{0}+3,−2,−m+1}^{n,k}, …, 𝔛_{γ_{0}+3,−m,−1}^{n,k} vanish, we obtain

𝔑_{γ_{0}+2,−2−j,−m+j}^{n,k} =

− (s\frac{∂}{∂s} − (3 + j) id)^{−1} (s̄\frac{∂}{∂s̄} − (m − j) id) 𝔑_{γ_{0}+2,−3−j,−m+j+1}^{n,k} +

+ λs^{j+3}t^{n−j−3}P (s̄, t̄),

where the inverse operator is defined for polynomials that do not contain monomials of degree 3 + j in s. The number λ will be uniquely determined by the condition

𝔑_{γ_{0}+2,−2−j,−m+j}|_{s=t} = 0

(divisibility by ν). It follows by induction that, for 1 ≤ j ≤ m − 1,

𝔑_{γ_{0}+2,−2−j,−m+j}^{n,k} = \frac{m!(n − 5 − j)!}{2(m − j)!(n − m − 5)!}[(n − 4 − j)s^{n} − (n − 3)s^{n−1}t +

+ λ_{j+3}^{(j)}s^{j+3}t^{n−j−3} + ⋯ + λ_{3}^{(j)}s^{3}t^{n−3}](t̄− s̄)^{m−j}t̄^{k−m+j}

and, moreover, that 𝔑_{γ_{0}+2,−2−j,−m+j}^{n,k} is divisible exactly by (s − t)^{j+2}. For j = m − 1 we write the polynomial in brackets in the form

(s − t)^{m+1} (c_{0}η^{n−m−1} + c_{1}(s − t)η^{n−m−2} + (s − t)^{2}P_{n−m−2}(s, t)) ,

with c_{0} ≠ 0. Then the R-component of

iη̄\frac{∂}{∂z_{2}} N_{γ_{0}+2,−1−m,−1}^{n,k}

equals

ic_{0}z̄_{2}^{m+1}η^{n−m−1}η̄^{k} + ic_{1}z̄_{2}^{m+1}νη^{n−m−2}η̄^{k}.

This implies

χ_{γ_{0}+1,−m−1,0}^{n−1,k} = 2c_{0}z̄_{2}^{m+1}ω̄_{2}^{n−m−1}ω̄_{1}^{k} \frac{∂}{∂ω̄_{2}} − ic_{1}z̄_{2}^{m+2}ω̄_{2}^{n−m−2}ω̄_{1}^{k} \frac{∂}{∂z̄_{2}}

and, therefore,

X_{γ_{0}+1,−m−1,0}^{n−1,k}(ν − V)|_{V =ν} = c′_{0}(s − t)^{m+1}t^{n−m−1}s̄^{k} + c′_{1}(s − t)^{m+2}t^{n−m−2}s̄^{k}.

On the other hand, one can start to determine N_{γ_{0}+2,Δ}^{n,k} from the right upper corner in the weight-defect diagram. We observe that the only terms in N_{γ_{0}+2,−m−2,0}^{n,k} that contribute into R are

\begin{matrix}z̄_{2}^{m+2}ν̄Π(η, η̄) & ⟶^{η̄\frac{∂}{∂z_{2}}} & z̄_{2}^{m+2}z̄_{1}η̄Π(η, η̄) \\ z̄_{2}^{m+2}νν̄Π(η, η̄) & ⟶^{η̄\frac{∂}{∂z_{2}}} & z̄_{2}^{m+2}νz̄_{1}η̄Π(η, η̄). \end{matrix}

This leads to

X_{γ_{0}+1,−m−2,1}^{n−1,k}(ν − V)|_{V =ν} = b′_{0}(s − t)^{m+2}t^{n−m−2}s̄^{k} + b′_{1}(s − t)^{m+3}t^{n−m−3}s̄^{k}

and

𝔑_{γ_{0}+2,−m−2,0}^{n,k} = b_{0}(s − t)^{m+2}t^{n−m−2}s̄^{k} + b_{1}(s − t)^{m+3}t^{n−m−3}s̄^{k} + #(s, t)t̄^{k}.

Finally, we consider the second equation from the right upper corner in the diagram

2 (s̄\frac{∂}{∂s̄} − id) 𝔑_{γ_{0}+2,−m−1,−1}^{n,k} + 2 (s\frac{∂}{∂s} − (m + 3) id) 𝔑_{γ_{0}+2,−m−2,0} +

+ 𝔛_{γ_{0}+3,−m−1,0}^{n,k} = 0.

It suffices to take into account the part of degree k in s̄.

2b_{0}(s − t)^{m+1}[−s + (m + 3)t]t^{n−m−2}s̄^{k} + 2b_{1}(m + 3)(s − t)^{m+2}t^{n−m−2}s̄^{k} +

+ c_{0}(s − t)^{m+1}t^{n−m−1}s̄^{k} + c_{1}(s − t)^{m+2}t^{n−m−2}s̄^{k} = 0.

Since c_{0} ≠ 0, the term at the left hand side cannot vanish. The contradiction completes the prove in the case N_{γ_{0},−2,−m} contained a polynomial Ψ_{1}.

We consider now the case when N_{γ_{0}+2,−2,Δ+2} contains monomials of type

Ψ_{2} = z̄_{2}z_{2}z̄_{1}η^{n−2}η̄^{k−1}.

We have Δ = −2 and, therefore, N_{γ_{0}+2,Δ} = N_{γ_{0}+2,−2,0}. The part of 𝔑_{γ_{0}+2,−2,0} with bidegree (n, k) (n ≥ 2, k ≥ 1) does not depend on occurrences of Ψ_{3} because the bidegree of πΨ_{3} is (n, 0) with n ≥ 2. We will determine 𝔑_{γ_{0}+2,−2,0}^{n,k} in two ways and compare the results.

The R-component of χ_{1}^{1}N_{γ_{0}+2,}^{n,k} equals

cz̄_{2}z_{2}z̄_{1}η^{n−1}η̄^{k−1}.

The above expression can be compensated only by the following components of χ_{γ_{0}+1,−1,0}^{n−1,k}(ν − V)|_{V =ν}:

c^{(1)}ω_{1}^{n−1}ω_{2}^{k} \frac{∂}{∂z_{1}} (ν − V)|_{V =ν}
c^{(2)}z̄_{2}ω̄_{1}^{k}ω̄_{2}^{n−1} \frac{∂}{∂ω̄_{2}} (ν − V)|_{V =ν}

These terms also contain the non-normal expressions

c^{(1)}z̄_{2}η^{n−1}η̄^{k}, resp. \frac{−c^{(2)}}{2i} z̄_{2}η^{n−1}η̄^{k}

that must cancel each other. Therefore, c^{(2)} = 2ic^{(1)} = 2kc and

χ_{γ_{0}+1,−1,0}^{n−1,k} = c (ω_{1}^{n−1}ω_{2}^{k} \frac{∂}{∂z_{1}} + 2iz̄_{2}ω̄_{1}^{k}ω̄_{2}^{n−1} \frac{∂}{∂ω̄_{2}} ) .

One computes

πχ_{γ_{0}+1,−1,0}^{n−1,k}(ν − V)|_{V =ν} = c(s − t)(s^{n−1}t̄^{k} + t^{n−1}s̄^{k}).

It follows

2 (s\frac{∂}{∂s} − 3 id) 𝔑_{γ_{0}+2,−2,0}^{n,k} − c(s − t)(s^{n−1}t̄^{k} + t^{n−1}s̄^{k}).  (22)

This equation determines 𝔑_{γ_{0}+2,−2,0}^{n,k} up to summands of degree 3 in s. It has no solution for n = 3, 4, thus, in this case we obtain a contradiction. Since N_{γ_{0}+2,−2,0}^{n,k} must be divisible by ν̄ (thus, vanish for s̄ = t̄), one has to add to the solution uniquely determined terms of degree 3 in t. This leads to the explicit expression

𝔑_{γ_{0}+2,−2,0}^{n,k} = c′[ − ((n − 4)s^{n} + (n − 3)ts^{n−1} − s^{3}t^{n−3})t̄^{k} +

+ \frac{(n − 3)(n − 4)}{6}(−s^{3}t^{n−3} + 3st^{n−1} − 2t^{n})s̄^{k}]. (23)

We construct now a contradiction by an analogous study of the R-component of χ¯1 1N_{γ_{0}+2,−2,0}^{n,k} that has the form

iη̄\frac{∂}{∂z_{2}} N_{γ_{0}+2,−2,0}^{n,k} = #z̄_{2}^{2}η^{n−2}z̄_{1}η̄^{k}.

This determines

χ_{γ_{0}+1,−2,1}^{n−1,k} = −#z̄_{2}^{3}ω̄_{2}^{n−2}z̄_{1}ω̄_{1}^{k−1} \frac{∂}{∂z_{2}} .

and,

X_{γ_{0}+1,−2,1}^{n−1,k}(ν − V)|_{V =ν} = −#(s − t)^{3}t^{n−3}s̄^{k}.

For 𝔑_{γ_{0}+2,−2,0}^{n,k} we have the equation

2s̄\frac{∂}{∂s̄}𝔑_{γ_{0}+2,−2,0}^{n,k} = #(s − t)^{3}t^{n−1}s̄^{k}.

The solution is unique up to some term that does not contain s̄ and that is determined by the condition that 𝔑_{γ_{0}+2,−2,0}^{n,k} is divisible by (s − t)^{2} . The explicit expression is

𝔑_{γ_{0}+2,−2,0}^{n,k} = #(s − t)^{2}t^{n−2}(t̄^{k} − s̄^{k}).

This term does not contain summands of degree 3 with respect to s whereas the above obtained expression (23) does. The desired contradiction proves that in this situation also a_{2} = 0.

It remains to study the case when N_{γ_{0}+2,−2,Δ+2} contains terms of type (21) and does not contain terms of type (19) or (20). Then N_{γ_{0}+2,−2,Δ+2} has ¯1defect 0. Since χ_{γ_{0}+1,−1,0} and therefore N_{γ_{0}+2,−2,0} are determined by the R contribution of

η \frac{∂}{∂z̄_{2}} Ψ_{3}

the bidegree of 𝔑_{γ_{0}+2,−2,0} equals (n, 0) (the bidegree of πΨ_{3}). It follows that N_{γ_{0}+2,−2,0} does not depend on z_{2}, z̄_{1}, η̄. Thus N_{γ_{0}+2,−2,Δ+2} = N_{γ_{0}+2,−2}(z_{1}, z̄_{2}, η). The R-component of χ_{1}^{1}(N_{γ_{0}+2,−2}^{n,0})|_{V =ν} equals

κz̄_{2}ν^{2}η^{n−3}.  (24)

By means of a linear coordinate change

z′ = (\begin{matrix}\frac{1}{c} & 0 \\ 0 & c̄ \end{matrix}) z, ω′ = ω

one can control the coefficient κ in (24). We obtain

χ_{γ_{0}+1,−1,Δ+2}^{n−1,0} = μ [ω_{1}^{n−1} \frac{∂}{∂z_{1}} + 2i(n − 1)z̄_{2}^{2}ω̄_{2}^{n−2} \frac{∂}{∂z̄_{2}} − 2iz̄_{2}ω̄_{2}^{n−1} \frac{∂}{∂ω̄_{2}} ] .

Here, we used that there are no non-normal terms of the type z̄_{2}η^{n−1}, or z̄_{2}νη^{n−2}. The coefficient κ evaluates to

κ = 2(n − 1)(n − 2)μ.

The expression X_{γ_{0}+1,−1,Δ+2}^{n−1,0}(ν − V)|_{V =ν} equals

\frac{μ}{2i}(s − t)(s^{n−1} − (n − 1)(s − t)t^{n−2} − t^{n−1}).

It follows

𝔑_{γ_{0}+2,−2}^{n,0} = (s\frac{∂}{∂s} − 3 id)^{−1} \frac{μi}{4}(s − t)[s^{n−1} − (n − 1)(s − t)t^{n−2} − t^{n−1}].

Without loss of generality we set μ = −(n − 3)(n − 4).

𝔑_{γ_{0}+2,−2}^{n,0} = −\frac{i}{12}(n − 4)(n − 3)(n − 2)t^{n} + \frac{i}{8}(n − 4)(n − 3)(2n − 3)st^{n−1} −

− \frac{i}{4}(n − 4)(n − 3)(n − 1)s^{2}t^{n−2} + \frac{i}{4}(n − 3)s^{n−1}t − \frac{i}{4}(n − 4)s^{n} + λs^{3}t^{n−3},

where λ is determined by the fact that 𝔑_{γ_{0}+2,−2}^{n,0}|_{s=t} = 0 (divisibility by ν). Thus,

λ = \frac{i}{24}(n − 2)(n − 1)(2n − 9).

It follows that 𝔑_{γ_{0}+2,−2}^{n,0} = 0 for n = 4 and 𝔑_{γ_{0}+2,−2}^{n,0} ≠ 0 for n ≥ 5. This construction leads to the solution (14) in Theorem 2.

6 The case δ_{1} = −1

In this section we study the case δ_{1} = −1. The only monomials Ψ in N_{γ_{0}+2,−1,δ+1} for that η \frac{∂}{∂z̄_{2}} Ψ does not belong to the space N are

Ψ_{1} = z̄_{2}νz_{2}^{m}η^{n−2}η̄^{k−m}, m ≥ 1, n ≥ 2, k ≥ m  (25)
Ψ_{2} = z̄_{2}νη^{n−2}ν̄η̄^{k−1},  n ≥ 2, k ≥ 1  (26)
Ψ_{3} = z̄_{2}ν^{3}η^{n−4},  n ≥ 4.  (27)

If N_{γ_{0}+2,−1,δ+1} does not contain such monomials then again, by the argument from Lemma 2, we obtain a contradiction to P ≡ 0. Suppose now that N_{γ_{0}+2,−1,δ+1} contains a monomial of type

Ψ_{1} = #z̄_{2}νz_{2}^{m}η^{n−2}η̄^{k−m}.

Then Δ = −m − 1 ≤ −2. It follows that there are no monomials of type Ψ_{2}, Ψ_{3} in N_{γ_{0}+2,−1,−m}. As in the previous section, we have to consider components of bidegree (n, k) of the upper line in the weight-defect diagram. It consists of the terms 𝔑_{γ_{0}+2,−1,−m}^{n,k}, …, 𝔑_{γ_{0}+2,−1−m,0}^{n,k}. The polynomial 𝔑_{γ_{0}+2,−1−m,0}^{n,k} can be determined as in the previous section with m substituted by m − 1:

𝔑_{γ_{0}+2,−m−1,0}^{n,k} = b_{0}(s − t)^{m+1}t^{n−m−1}s̄^{k} + b_{1}(s − t)^{m+2}t^{n−m−2}s̄^{k} + P (s, t)t̄^{k}.

The contribution into R of

η \frac{∂}{∂z̄_{2}} N_{γ_{0}+2,−m−1,0}^{n,k}

equals zero. Therefore,

2 (s\frac{∂}{∂s} − (m + 2) id) 𝔑_{γ_{0}+2,−m−1,0}^{n,k} = 2b_{0}(s−t)^{m}[−s+(m+2)t]t^{n−m−1}s̄^{k} +

+ 2b_{1}(m + 2)(s − t)^{m+1}t^{n−m−1}s̄^{k} + #(s, t)t̄^{k}

is the complete contribution of N_{γ_{0}+2,−m−1,0}^{n,k} into P_{γ_{0}+3,−m,0}^{n,k}. It must cancel with 𝔛_{γ_{0}+3,−m,0}^{n,k} and

2 (s̄\frac{∂}{∂s̄} − id) 𝔑_{γ_{0}+2,−m,−1}^{n,k}.

To determine 𝔑_{γ_{0}+2,−m,−1}^{n,k} we start from the left corner of the diagram. As in the previous section we find

𝔑_{γ_{0}+2,−1,−m}^{n,k} =#(s^{n} − s^{2}t^{n−2})(t̄− s̄)^{m}t̄^{k−m}
𝔑_{γ_{0}+2,−2,−m+1}^{n,k} =#(s^{n} − (n − 2)s^{3}t^{n−3} + (n − 3)s^{2}t^{n−2})(t̄− s̄)^{m−1}t̄^{k−m+1} +

+ #′((n − 4)s^{n} − (n − 3)s^{n−1}t + s^{3}t^{n−3})(t̄− s̄)^{m−1}t̄^{k−m+1}.

We observe that both summands in 𝔑_{γ_{0}+2,−2,−m+1}^{n,k} are divisible by (s − t)^{2} and 𝔑_{γ_{0}+2,−2,−m+1}^{n,k} is divisible by (s − t)^{3} for

#(n − 2) + #′(n − 4) = 0.  (28)

By induction we obtain

𝔑_{γ_{0}+2,−m,−1}^{n,k} = (s − t)^{m}[c_{0}η^{n−m} + c_{1}(s − t)η^{n−m−1} + (s − t)^{2}#(s, t)](t̄− s̄)t̄^{k−1},

where c_{0} = 0 exactly under the condition (28). Thus, the contribution of 𝔑_{γ_{0}+2,−m,−1}^{n,k} into P_{γ_{0}+3,−m,0}^{n,k} equals to

2 (s̄\frac{∂}{∂s̄} − id) 𝔑_{γ_{0}+2,−m,−1}^{n,k} = (s−t)^{m}[c_{0}η^{n−m}+c_{1}(s−t)η^{n−m−1}+(s−t)^{2}#′(s, t)]t̄^{k}.

As in the previous section

X_{γ_{0}+1,−m,0}^{n−1,k}(ν − V)|_{V =ν} = c′_{0}(s − t)^{m}t^{n−m}s̄^{k} + c′_{1}(s − t)^{m+1}t^{n−m−1}s̄^{k}.

Adding up all listed contributions to P_{γ_{0}+3,−m,0} we obtain the identity:

c′_{0}(s−t)^{m}t^{n−m}s̄^{k}+c′_{1}(s−t)^{m+1}t^{n−m−1}s̄^{k}+2b_{0}(s−t)^{m}[−s+(m+2)t]t^{n−m−1}s̄^{k}+

+ 2b_{1}(m + 2)(s − t)^{m+1}t^{n−m−1}s̄^{k} + P (s, t)t̄^{k} +

+ (s − t)^{m}[c_{0}η^{n−m} + c_{1}(s − t)η^{n−m−1} + (s − t)^{2}#(s, t)]t̄^{k} ≡ 0. (29)

Comparing the coefficients at (s−t)^{m}t^{n−m}s̄^{k} we conclude that c′_{0} = 0, b_{0} = 0. This implies that

P (s, t) = −2b_{1}(s − t)^{m+1}t^{n−m−1} + (s − t)^{m+2}#(s, t).

Identity (29) takes the form

c′_{1}(s−t)^{m+1}t^{n−m−1}s̄^{k}+2b_{1}(m+2)(s−t)^{m+1}t^{n−m−1}s̄^{k}−2b_{1}(s−t)^{m+1}t^{n−m−1}t̄^{k}+

+ (s − t)^{m+2}#(s, t)t̄^{k} + (s − t)^{m+1}[c_{1}η^{n−m−1} + (s − t)^{2}#(s, t)]t̄^{k} ≡ 0.

We see that the coefficient at (s − t)^{m+1}t̄^{k} does not vanish unless c_{1} = c′_{1} = b_{1} = 0 because η^{n−m−1} cannot cancel with t̄^{k}. Contradiction. Let’s now consider the the situation when N_{γ_{0}+2,Δ} contains a monomial

Ψ_{2} = z̄_{2}νη^{n−2}ν̄η̄^{k−1}.

Then Δ = −1 and, therefore, N_{γ_{0}+2,Δ} = N_{γ_{0},−1,0}. The R-component of χ_{1}^{1}N_{γ_{0}+2,−1}^{n,k} equals #z̄_{1}z̄_{2}νη^{n−2}η̄^{k}, and, hence,

χ_{γ_{0}+1,−1,1}^{n−1,k} = (#z̄_{1}z̄_{2}^{2}ω̄_{1}^{k}ω̄_{2}^{n−2} \frac{∂}{∂z̄_{2}} ) .

It follows that

X_{γ_{0}+1,0,0}^{n−1,k}(ν − V)|_{V =ν} = #(s − t)^{2}t^{n−2}s̄^{k}.

Therefore, the equation

2 (s̄\frac{∂}{∂s̄}) 𝔑_{γ_{0}+2,−1}^{n,k}) + X_{γ_{0}+1,0,0}^{n−1,k}(ν − V)|_{V =ν} = 0

implies

𝔑_{γ_{0}+2,−1}^{n,k} = #(s − t)^{2}t^{n−2}(s̄^{k} − t̄^{k}).

(we have added −#(s − t)^{2}t^{n−2}t̄^{k} from the kernel, in order to satisfy the normal form condition N_{γ_{0}+2,−1}^{n,k}|_{s̄=t̄} = 0). On the other hand χ¯1 1Ψ_{2} contains the R-component cνν̄η^{n−1}η̄^{k−1} which implies

χ_{γ_{0}+1,0,0}^{n−1,k} = c′z_{1}ω_{1}^{n−1}ω_{2}^{k} \frac{∂}{∂z_{1}} − c′z̄_{2}ω̄_{1}^{k}ω̄_{2}^{n−1} \frac{∂}{∂z̄_{2}} .

The contradiction proves that P_{γ_{0}+3} ≠ 0 if N_{γ_{0}+2,−1} contains a monomial of the form (26). Suppose now that N_{γ_{0}+2,−1} contains a monomial Ψ_{3}. Since Ψ_{3} does not depend on the variables z_{2}, z̄_{1}, η̄ one can repeat the argument from the previous section to show that the entire term N_{γ_{0}+2,−1} does not depend on them. The R-component of χ_{1}^{1}N_{γ_{0}+2,−1} equals to cν^{3}η^{n−3}. It follows that

χ_{γ_{0}+1}^{n−1,0} = μ (nω_{1}^{n−1}z_{1} \frac{∂}{∂z_{1}} + nω̄_{2}^{n−1}z̄_{2} \frac{∂}{∂z̄_{2}} + 2ω_{1}^{n} \frac{∂}{∂ω_{1}} + 2ω̄_{2}^{n} \frac{∂}{∂ω̄_{2}} ) ,

with μ = \frac{3c}{n(n−1)(n−2)}. By an appropriate coordinate change, as in the previous section, we can achieve that X_{γ_{0}+1}^{n−1,0}(ν − V)|_{V =ν} takes the form

2(n − 3) [(n − 2)s^{n} − ns^{n−1}t + nst^{n−1} − (n − 2)t^{n}] .

Then

𝔑_{γ_{0},−1}^{n,0} = (s\frac{∂}{∂s} − 2 id)^{−1} (n − 3)[(n − 2)s^{n} − ns^{n−1}t + nst^{n−1} − (n − 2)t^{n}]
=(n − 3)s^{n} − ns^{n−1}t − (n − 3)nst^{n−1} + \frac{(n − 3)(n − 2)}{2}t^{n} + λs^{2}t^{n−2},

where λ = \frac{n(n−1)}{2} is determined by the condition that 𝔑_{γ_{0}+2,−1} is divisible by (s − t). For n = 3 the polynomial 𝔑_{γ_{0}+2,−1}^{n,0} = 0 whereas, for n ≥ 4, 𝔑_{γ_{0},−1}^{n,0} ≠ 0. This represents the solution (15) of Theorem (2).

7 The case δ_{1} = 0

Suppose δ_{1} = 0. According to Table 2, χ_{1}^{1}N_{γ_{0},0,Δ} has a non-trivial Rcomponent if and only if N_{γ_{0},0,Δ} contains a monomial

Ψ = #z_{2}^{m}ν^{2}η^{n−2}η̄^{k−m}, m = −Δ ≥ 0, n ≥ 2, k ≥ m.

Any solution (𝔑_{γ_{0}+2,Δ}^{n,k}, X_{γ_{0}+1,Δ+1}^{n−1,k}) of the upper line in the weight-defect diagram of bidegree (n, k)

X_{γ_{0}+1,Δ+1}^{n−1,k}(ν − V)|_{V =ν} + (X_{1}^{1} + X¯1 1 − 2 id)𝔑_{γ_{0}+1,Δ}^{n,k} = 0  (30)

is the sum of a particular solution and of a solution of the system (30) that satisfies the extra condition

(s\frac{∂}{∂s} − id) 𝔑_{γ_{0}+1,0,Δ}^{n,k} = 0  (31)

which is equivalent to the absence of Ψ in N_{γ_{0}+1,0,Δ}^{n,k}. We start with the construction of a particular solution. The field X_{γ_{0}+1,1,−m} is uniquely determined by the R-contribution

Ψ ⟶^{η̄\frac{∂}{∂z_{2}}} #z_{1}z_{2}^{m}νη^{n−1}η̄^{k−m}

and equals

X_{γ_{0}+1,1,−m} = #z_{1}^{2}z_{2}^{m}ω_{1}^{n−1}ω_{2}^{k−m} \frac{∂}{∂z_{1}} ,

thus,

X_{γ_{0}+1,1,−m}^{n−1,k}|_{V =ν} = #(s − t)s^{n−1}(s̄ − t̄)^{m}t̄^{k−m}.

Now we determine 𝔑_{γ_{0}+2,0,−m}^{n,k} up to a summand from the kernel of s\frac{∂}{∂s} − id by resolving the equation

2 (s\frac{∂}{∂s} − id) 𝔑_{γ_{0}+2,0,−m}^{n,k} + c(s^{n} − ts^{n−1})(s̄ − t̄)^{m}t̄^{k−m} = 0.

For c = \frac{(n−1)!}{(n−m−3)!} we obtain the particular solution

𝔑_{γ_{0}+2,0,−m}^{n,k} = \frac{(n − 3)!}{2(n − m − 3)!}[−(n − 2)s^{n} + (n − 1)s^{n−1}t − st^{n−1}](s̄ − t̄)^{m}t̄^{k−m}.

Then

2 (s̄\frac{∂}{∂s̄} − m id) 𝔑_{γ_{0},0,−m}^{n,k} =

= \frac{m(n − 3)!}{(n − m − 3)!}[−(n − 2)s^{n} + (n − 1)s^{n−1}t − st^{n−1}](s̄ − t̄)^{m−1}t̄^{k−m+1}.

Since χ¯1 1N_{γ_{0}+2,0,−m}^{n,k} belongs to N there is no contribution of the field to the terms of defect (−1, −m + 1) and we have

𝔑_{γ_{0}+2,−1,−m+1}^{n,k} = − (s\frac{∂}{∂s} − id)^{−1} (s̄\frac{∂}{∂s̄} − m id) 𝔑_{γ_{0}+2,0,−m}.

¿From this we obtain

𝔑_{γ_{0},−1,−m+1}^{n,k} = \frac{m(n − 4)!}{2(n − m − 3)!}[(n − 3)s^{n} − (n − 1)s^{n−1}t + (n − 1)s^{2}t^{n−2} −

(n − 3)st^{n−1}](s̄ − t̄)^{m−1}t̄^{k−m+1}.

Consecutively, this procedure leads to the expressions for all terms

𝔑_{γ_{0}+2,−j,−m+j}^{n,k} = (μ_{n}^{(j)}s^{n} + μ_{n−1}^{(j)}s^{n−1}t + ∑_{α=1}^{j+1} μ_{α}^{(j)}s^{α}t^{n−α}) (s̄ − t̄)^{m−j}t̄^{k−m+j}
= (s − t)^{j+2}P_{n−j−2}(s, t)(s̄ − t̄)^{m−j}t̄^{k−m+j},  (32)

in particular

𝔑_{γ_{0}+2,−m,0}^{n,k} = (s − t)^{m+2}P_{n−m−2}(s, t)t̄^{k}.

Since s̄\frac{∂}{∂s̄}𝔑_{γ_{0}+2,−m,0}^{n,k} = 0, hence the sequence terminates at this term. Hence, we have obtained a particular solution of (30), that leads to (16) of Theorem 2. We consider now the situation when (31) is satisfied, i.e., 𝔑_{γ_{0}+2,0,−m}^{n,k} is linear in s. Then n − 1 is the degree in t. Since st^{n−1} contains η^{n−1} the term N_{γ_{0}+2,0,−m}^{n,k} must be divisible by z_{2}^{m}ν̄ if m > 0 in order to have normal form. For m = 0 it must be divisible by ν̄^{2}. Thus, we obtain

𝔑_{γ_{0}+2,0,−m}^{n,k} = st^{n−1}(s̄ − t̄)^{m+1}Q_{k−m−1}(s̄, t̄), m > 0  (33)
𝔑_{γ_{0}+2,0,0}^{n,k} = st^{n−1}(s̄ − t̄)^{2}Q_{k−2}(s̄, t̄),  m = 0.  (34)

Consider, at first, the case m = 0. Since δ¯1 ≤ 0, the term N_{γ_{0}+2,0}^{n,k} consists of just one component N_{γ_{0}+2,0,0}^{n,k}. According to (34), η̄\frac{∂}{∂z_{2}} N_{γ_{0}+2,0}^{n,k} is divisible by z̄_{1}^{2}z_{2} and therefore χ¯1 1N_{γ_{0}+2,0}^{n,k} is normal. This implies that χ_{γ_{0}+1,0,1}^{n−1,k} = 0. But, since

s̄\frac{∂}{∂s̄}(s̄ − t̄)^{2}Q_{k−2}(s̄, t̄) = (s̄ − t̄)Q_{k−1}(s̄, t̄) ≠ 0

we have P_{γ_{0}+3} ≢ 0. Now, let m = 1. Because of δ¯1 ≤ 0

N_{γ_{0}+2,−1}^{n,k} = N_{γ_{0}+2,0,−1} + N_{γ_{0}+2,−1,0}.

According to the conditions of the normal form we obtain that the expression η̄\frac{∂}{∂z_{2}} N_{γ_{0}+2,−1,0}^{n,k} is also normal. In order to achieve P_{γ_{0}+3} ≡ 0 the term χ¯1 1N_{γ_{0},−1,0} must vanish. Therefore, N_{γ_{0}+2,−1,0}^{n,k} is divisible by z̄_{2}ν^{2}. This implies

𝔑_{γ_{0}+2,−1,0}^{n,k} = (s − t)^{3}P_{n−3}(s, t)t̄^{k}.  (35)

Then η \frac{∂}{∂z̄_{2}} N_{γ_{0}+2,−1,0}^{n,k} and χ_{1}^{1}N_{γ_{0}+2,−1,0}^{n,k} are also normal. Therefore

2 (s\frac{∂}{∂s} − 2 id) (s − t)^{3}P_{n−3}(s, t)t̄^{k} = (s − t)^{2}P_{n−2}(s, t)t̄^{k}.  (36)

On the other hand,

2 (s̄\frac{∂}{∂s̄} − id) st^{n−1}(s̄ − t̄)^{2}Q_{k−2}(s̄, t̄) = st^{n−1}(s̄ − t̄)Q_{k−1}(s̄, t̄)  (37)

is not divisible by (s − t)^{2} nor by t̄^{k}. This means that the expressions (36) and (37) cannot cancel each other. Therefore, χ_{γ_{0}+1,0,0}^{n−1,k} must be different from zero which implies that N_{γ_{0},0,−1} contains

Ψ = #η^{n}z_{2}ν̄η̄^{k−2},

so that

−iη̄\frac{∂}{∂z_{2}} Ψ = cη^{n}ν̄η̄^{k−1}

does not belong to N. This defines χ_{γ_{0}+1} uniquely, up to a constant factor

χ_{γ_{0}+1}^{n−1,k} = c′ [nω_{1}^{n−1}ω_{2}^{k}z_{1} \frac{∂}{∂z_{1}} + nω̄_{2}^{n−1}ω̄_{1}^{k}z̄_{2} \frac{∂}{∂z̄_{2}} + 2ω_{1}^{n}ω_{2}^{k} \frac{∂}{∂ω_{1}} + 2ω̄_{2}^{n}ω̄_{1}^{k} \frac{∂}{∂ω̄_{2}} ]

with c = 2kc′. Then

X_{γ_{0}+1,0}^{n−1,k}(ν − V)|_{V =ν} =

= \frac{−ic′}{2} [(n − 1)s^{n}t̄^{k} − ns^{n−1}tt̄^{k} + nst^{n−1}s̄^{k} − (n − 1)t^{n}s̄^{k}]

= \frac{−ic′}{2} [s^{n−1}t̄^{k} ((n − 1)s − nt) + t^{n−1}s̄^{k} (ns − (n − 1)t)] .

We see that

P_{γ_{0}+3,0,0}^{n,k} = (s − t)^{2}P_{n−2}(s, t)t̄^{k} + st^{n−1}(s̄ − t̄)Q_{k−1}(s̄, t̄) +

+ \frac{−ic′}{2} [s^{n−1}t̄^{k} ((n − 1)s − nt) + t^{n−1}s̄^{k} (ns − (n − 1)t)] ≢ 0.

Contradiction. Let m ≥ 2. Then N_{γ_{0}+2,0,−m}^{n,k} is divisible by z_{2}^{m}ν̄, so η̄\frac{∂}{∂z_{2}} N_{γ_{0}+2,0,−m}^{n,k} is divisible by z_{2}^{m−1}ν̄ ∈ N. It follows that χ¯1 1N_{γ_{0}+2,0,−m}^{n,k} ∈ N. According to Table 2, the ↖ arrows of the upper line of the weight-defect diagram may contribute into R only if they are applied to N_{γ_{0}+2,−1,−m+1}^{n,k} and N_{γ_{0}+2,−2,−m+2}^{n,k}. The corresponding solutions 𝔑_{γ_{0}+2,−j,−m+j}^{n,k} until j = m − 1 can be extracted from the already considered cases δ_{1} = −1 (with m substituted by m − 1) and δ_{1} = −2 (with m substituted by m − 2). In particular,

𝔑_{γ_{0}+2,−m+1,−1}^{n,k} = (s − t)^{m−1}(c_{0}η^{n−m+1} + c_{1}(s − t)η^{n−m} +

+ (s − t)^{2}P_{n−m−1}(s, t))(t̄− s̄)t̄^{k−1} + 𝔑̃_{γ_{0}+2,−m+1,−1}^{n,k}, where 𝔑̃_{γ_{0}+2,−m+1,−1}^{n,k} is a solution of the upper line of the diagram when the ↖ arrows do not contribute into R. Starting with

𝔑̃_{γ_{0}+2,0,−m}^{n,k} = 𝔑_{γ_{0}+2,0,−m}^{n,k} = st^{n−1}(s̄ − t̄)^{m+1}Q_{k−m−1}(s̄, t̄)

and inductively solving the equation

(s\frac{∂}{∂s} − (j + 2) id) 𝔑̃_{γ_{0}+2,−j−1,−m+j+1}^{n,k}+(s̄\frac{∂}{∂s̄} − (m − j) id) 𝔑̃_{γ_{0}+2,−j,−m+j}^{n,k} = 0,

we find that

𝔑̃_{γ_{0}+2,−j−1,−m+j+1}^{n,k} = (s − t)^{j+1}st^{n−j−2}(s̄ − t̄)^{m−j}Q_{k−m+j}(s̄, t̄)

and therefore

𝔑̃_{γ_{0}+2,−m+1,−1}^{n,k} = (s − t)^{m−1}st^{n−m}(s̄ − t̄)^{2}Q_{k−2}(s̄, t̄)

Hence,

2 (s̄\frac{∂}{∂s̄} − id) 𝔑_{γ_{0}+2,−m+1,−1}^{n,k} = (s − t)^{m−1}(c_{0}η^{n−m+1} + c_{1}(s − t)η^{n−m} +

+ (s − t)^{2}P_{n−m−1}(s, t))t̄^{k} + (s − t)^{m−1}st^{n−m}(s̄ − t̄)Q_{k−1}(s̄, t̄), (38)

with Q_{k−1}(s̄, t̄) ≠ 0. As we have previously computed, the contribution of 𝔑_{γ_{0}+2,−m,0}^{n,k} and X_{γ_{0}+1,−m+1,0}^{n−1,k} into P_{γ_{0}+3,−m+1,0}^{n,k} equals to

c′_{0}(s − t)^{m−1}t^{n−m+1}s̄^{k} + c′_{1}(s − t)^{m}t^{n−m}s̄^{k} +

+2b_{0}(s−t)^{m−1}[−s+(m+1)t]t^{n−m}s̄^{k} +2b_{1}(m+1)(s−t)^{m}t^{n−m}s̄^{k} +P (s, t)t̄^{k}.

We see that this expression cannot cancel with (38) because nothing can cancel with (s̄ − t̄)Q_{k−1}(s̄, t̄) in the last summand of (38). Contradiction.

8 The case δ_{1} = 1

As in the previous section we distinguish two cases in dependence whether the ↖-arrow in the left upper corner of the diagram contributes into R or not, i.e.,

η \frac{∂}{∂z̄_{2}} N_{γ_{0}+2,1,−m}^{n,k} ∈ N  (39)

holds or not. First, suppose (39) is not valid. According to Table 2, this implies that N_{γ_{0}+2,1,−m}^{n,k} must contain a monomial of the form

Ψ = #z_{1}ν^{2}η^{n−2}z_{2}^{m}η̄^{k−m},

where m = −Δ + 1 ≥ 0. The R-component of

χ_{1}^{1}N_{γ_{0}+2,1,−m}^{n,k}

is then

#z_{1}^{2}z_{2}^{m}νη^{n−1}η̄^{k−m}.

It follows

χ_{γ_{0}+1,2,−1}^{n−1,k} = #z_{1}^{3}z_{2}^{m}ω_{1}^{n−1}ω_{2}^{k−m} \frac{∂}{∂z_{1}} ,

and furthermore,

X_{γ_{0}+1,2,−1}^{n−1,k}|_{V =ν}) = #(s − t)s^{n−1}(s̄ − t̄)^{m}t̄^{k−m}.

Resolving the equation

2s\frac{∂}{∂s}𝔑_{γ_{0}+2,0,−m}^{n,k} + #(s^{n} − ts^{n−1})(s̄ − t̄)^{m}t̄^{k−m} = 0

with # = n(n − 1) we find

𝔑_{γ_{0}+2,1,−m}^{n,k} = \frac{1}{2}[ns^{n−1}t − (n − 1)s^{n} − t^{n}](s̄ − t̄)^{m}t̄^{k−m}.

¿From the equations

(s\frac{∂}{∂s} − j + 1 id) 𝔑_{γ_{0}+2,−j,−m+j+1}^{n,k})+ (s̄\frac{∂}{∂s̄} − (m − j) id) 𝔑_{γ_{0}+2,1−j,−m+j}^{n,k}) = 0

one determines by induction the expressions

𝔑_{γ_{0}+2,−j,−m+j+1}^{n,k} =

(λ_{n}^{(j+1)}s^{n} + λ_{n−1}^{(j+1)}s^{n−1}t + ∑_{α=0}^{j+1} λ_{α}^{(j+1)}s^{α}t^{n−α}) (s̄ − t̄)^{m−j−1}t̄^{k−m+j+1}

that must be divisible exactly by (s − t)^{j+2}. These expressions are uniquely defined up to a multiple of the solution (32) (with m substituted by m − 1). Eventually, we obtain for j = m

𝔑_{γ_{0}+2,1−m,0}^{n,k} = (s − t)^{m+1}P_{n−m−1}(s, t)t̄^{k}  (40)

This belongs to the kernel of s̄\frac{∂}{∂s̄} and, hence, we have found a particular solution of the upper line of the weight-defect diagram that corresponds to non-trivial χ_{γ_{0}+1,2,−m}^{n−1,k} and leads to (17) of the Theorem 2. Suppose now that (39) is satisfied, i.e.

s\frac{∂}{∂s}𝔑_{γ_{0}+2,1,−m}^{n,k} = 0.

It follows from (39) and the condition of the normal form that

𝔑_{γ_{0}+2,1,−m}^{n,k} =#t^{n}(s̄ − t̄)^{m+1}Q_{k−m−1}(s̄, t̄) for m ≥ 1

and

𝔑_{γ_{0}+2,1,0}^{n,k} =#t^{n}(s̄ − t̄)^{2}Q_{k−2}(s̄, t̄)  for m = 0.

Consider the case m = 0. Then N_{γ_{0}+2,1,0}^{n,k} takes the form indicated above and χ¯1 1N_{γ_{0}+2,1,0}^{n,k} is divisible by z_{2}ν̄ and, therefore belongs to N. This implies that χ_{γ_{0}+1,1,1}^{n−1,k} vanishes. But

s̄\frac{∂}{∂s̄}𝔑_{γ_{0}+2,1,0}^{n,k} = #t^{n}(s̄ − t̄)Q_{k−1}(s̄, t̄) ≠ 0

implies then P_{γ_{0}+3} ≠ 0. Now, let m = 1. Then, because of δ¯1 ≤ 0,

N_{γ_{0}+2,0}^{n,k} = N_{γ_{0}+2,1,−1}^{n,k} + N_{γ_{0}+2,0,0}^{n,k},

where

𝔑_{γ_{0}+2,1,−1}^{n,k} = t^{n}(s̄ − t̄)^{2}Q_{k−2}(s̄, t̄).

Therefore

(s̄\frac{∂}{∂s̄} − 1 id) 𝔑_{γ_{0}+2,1,−1}^{n,k} = t^{n}(s̄ − t̄)Q_{k−1}(s̄, t̄).  (41)

For N_{γ_{0}+2,0,0}^{n,k}, Table 3 indicates that η̄\frac{∂}{∂z_{2}} N_{γ_{0}+2,0,0}^{n,k} ∈ N. Therefore,

s̄\frac{∂}{∂s̄}𝔑_{γ_{0}+2,0,0}^{n,k} = 0.

Thus, 𝔑_{γ_{0}+2,0,0}^{n,k} must be divisible by ν^{2}. By subtracting a solution of type (32) with m = 0 we can achieve that it is divisible even by ν^{3}. Therefore,

𝔑_{γ_{0}+2,0,0}^{n,k} =(s − t)^{3}P_{n−3}(s, t)t̄^{k}
(s\frac{∂}{∂s} − 1 id) 𝔑_{γ_{0}+2,0,0}^{n,k} =(s − t)^{2}P_{n−2}(s, t)t̄^{k}.  (42)

Since the expressions (41) and (41) cannot cancel each other this implies that χ_{γ_{0}+1,1,0}^{n−1,k} ≠ 0. As N_{γ_{0}+2,0,0}^{n,k} is divisible by ν^{3} the expression η \frac{∂}{∂z̄_{2}} N_{γ_{0}+2,0,0}^{n,k} belongs to N. Thus, the R-component can arise only from η̄\frac{∂}{∂z_{2}} N_{γ_{0}+2,1,−1}^{n,k}. According to Table 2 this can only happen if N_{γ_{0}+2,1,−1}^{n,k} contains a monomial

Ψ = #z_{1}z_{2}ν^{2}η^{n−2}.

But then k = 0 that contradicts to the assumption that 𝔑_{γ_{0}+2,1,−1}^{n,k} is divisible by s̄t̄^{2}, i.e., k ≥ 2. We study now the case m ≥ 2. Then N_{γ_{0}+2,1,−m}^{n,k} is divisible by z_{1}z_{2}^{m}ν̄, so η̄\frac{∂}{∂z_{2}} N_{γ_{0}+2,0,−m}^{n,k} is divisible by z_{1}z_{2}^{m−1}ν̄ ∈ N. It follows that χ¯1 1N_{γ_{0}+2,1,−m}^{n,k} ∈ N. After subtracting, if needed, a multiple of the solution (32) we may assume that the ↖ arrows of the upper line of the weight-defect diagram may contribute into R only if they are applied to N_{γ_{0}+2,−1,−m+2}^{n,k} and N_{γ_{0}+2,−2,−m+3}^{n,k}. The corresponding solutions 𝔑_{γ_{0}+2,−j+1,−m+j}^{n,k} until j = m−1 can be extracted from the already considered cases δ_{1} = −1 (with m substituted by m − 2) and δ_{1} = −2 (with m substituted by m − 3). In particular,

𝔑_{γ_{0}+2,−m+2,−1}^{n,k} = (s − t)^{m−2}(c_{0}η^{n−m+2} + c_{1}(s − t)η^{n−m+1} +

+ (s − t)^{2}P_{n−m}(s, t))(t̄− s̄)t̄^{k−1} + 𝔑̃_{γ_{0}+2,−m+2,−1}^{n,k},

where 𝔑̃_{γ_{0}+2,−m+2,−1}^{n,k} is a solution of the upper line of the diagram when the ↖ arrows do not contribute into R. Starting with

𝔑̃_{γ_{0}+2,1,−m}^{n,k} = 𝔑_{γ_{0}+2,1,−m}^{n,k} = t^{n}(s̄ − t̄)^{m+1}Q_{k−m−1}(s̄, t̄)

and inductively solving the equation

(s\frac{∂}{∂s} − (j + 1) id) 𝔑̃_{γ_{0}+2,−j,−m+j+1}^{n,k}+(s̄\frac{∂}{∂s̄} − (m − j) id) 𝔑̃_{γ_{0}+2,−j+1,−m+j}^{n,k} = 0,

we find that

𝔑̃_{γ_{0}+2,−j,−m+j+1}^{n,k} = (s − t)^{j+1}t^{n−j−1}(s̄ − t̄)^{m−j}Q_{k−m+j}(s̄, t̄)

and therefore

𝔑̃_{γ_{0}+2,−m+2,−1}^{n,k} = (s − t)^{m−2}t^{n−m+2}(s̄ − t̄)^{2}Q_{k−2}(s̄, t̄)

Hence,

2 (s̄\frac{∂}{∂s̄} − id) 𝔑_{γ_{0}+2,−m+2,−1}^{n,k} = (s − t)^{m−2}(c_{0}η^{n−m+2} + c_{1}(s − t)η^{n−m+1} +

+ (s − t)^{2}P_{n−m}(s, t))t̄^{k} + (s − t)^{m−2}t^{n−m−2}(s̄ − t̄)Q_{k−1}(s̄, t̄), (43)

with Q_{k−1}(s̄, t̄) ≠ 0. As we have previously computed, the contribution of 𝔑_{γ_{0}+2,−m+1,0}^{n,k} and X_{γ_{0}+1,−m+2,0}^{n−1,k} into P_{γ_{0}+3,−m+2,0}^{n,k} equals to

c′_{0}(s − t)^{m−2}t^{n−m+2}s̄^{k} + c′_{1}(s − t)^{m−1}t^{n−m+1}s̄^{k} +

+ 2b_{0}(s − t)^{m−2}[−s + mt]t^{n−m+1}s̄^{k} + 2b_{1}m(s − t)^{m−1}t^{n−m+1}s̄^{k} + P (s, t)t̄^{k}.

We see that this expression cannot cancel with (43) because nothing can cancel with (s̄ − t̄)Q_{k−1}(s̄, t̄) in the last summand of (43). Contradiction.

9 Description of Aut_{0,id}

In this section we prove that the dimension of Aut_{0,id} M equals to zero or two and in case it is equal to two then Aut_{0,id} M ≅ ℂ.

Lemma 4 If for some χ ∈ 𝔞𝔲𝔱 _{0,id}M, a_{2} ≠ 0 then a_{1} = 0.

Proof. If in one of the lines of the weight-defect diagram d_{1} attains values ≤ −2 then δ¯2 ≥ 2 and this would contradict to inequality (11) and would imply a_{1} = 0. Therefore, since δ¯2 ≤ 1, the solution N_{γ_{0}+2,Δ} of the upper line of the weight-defect diagram does not contain a multiple of (14).

Analogously, using inequality (12) we conclude that N_{γ_{0}+2,Δ} does not contain neither a multiple of (32) for m ≥ 1, nor a multiple of (40) for m ≥ 1. Thus, 𝔑_{γ_{0}+2,Δ}^{n,0} can be equal

𝔑_{γ_{0}+2,Δ}^{n,0} =𝔑_{γ_{0}+2,−1,0}^{n,0} = (n − 3)s^{n} − ns^{n−1}t − (n − 3)nst^{n−1} +

+ \frac{(n − 3)(n − 2)}{2}t^{n} + \frac{n(n − 1)}{2}s^{2}t^{n−2},

𝔑_{γ_{0}+2,Δ}^{n,k} =𝔑_{γ_{0}+2,0,0}^{n,k} = \frac{1}{2}[(n − 1)s^{n−1}t − (n − 2)s^{n} − st^{n−1}]t̄^{k}
𝔑_{γ_{0}+2,Δ}^{n,k} =𝔑_{γ_{0}+2,1,0}^{n,k} = \frac{1}{2}[ns^{n−1}t − (n − 1)s^{n} − t^{n}]t̄^{k}

Consider the first option. We see that η \frac{∂}{∂z_{1}} N_{γ_{0}+2,−1,0}^{n,0} belongs to N, and also χ¯2 1N_{γ_{0}+2,−1,0}^{n,0} ≠ 0. From the weight-defect diagram it follows that in order to cancel χ¯2 1N_{γ_{0}+2,−1,0}^{n,0} the term N_{γ_{0}+2,−3,0}^{n,0} ≠ 0 (The term N_{γ_{0}+2,−2,1}^{n,0} does not contribute into P_{γ_{0}+3,−2,0}^{n,0} because of the bidegree (n, 0). Thus, δ¯2 ≥ 3 that contradicts to (11). For the second and the third options 0 ≠ χ_{1}^{2}N_{γ_{0}+2,ϵ,0}^{n,k} ∈ N (ϵ = 0, 1). According to the weight-defect diagram, this implies that at least one of the polynomials N_{γ_{0}+2,ϵ,−2}^{n,k} or N_{γ_{0}+2,ϵ+1,−3}^{n,k} does not vanish. This means that δ_{2} ≤ −2. Contradiction with (12). □

Corollary 1 The weight-defect diagram always separates into one-line diagrams.

Lemma 5 For any χ ∈ 𝔞𝔲𝔱 _{0,id}, the parameter r = 0.

Proof. If a = (a_{1}, a_{2}) = 0 then Φ = exp χ is by Theorem 3 in [ES1] is a birational transformation

Φ :  z′_{1} = \frac{z_{1}}{1 − rω_{1}}  ω′_{1} = \frac{ω_{1}}{1 − rω_{1}}
z′_{2} = \frac{z_{2}}{1 − rω_{2}}  ω′_{2} = \frac{ω_{2}}{1 − rω_{2}} .

Considering N_{γ_{0}+2} and N_{γ_{0}+4} we immediately conclude that, for r ≠ 0, the transformation Φ changes the equation of M, so Φ ∉ Aut M. For, say, a_{2} ≠ 0, according to the previous lemma, we have a_{1} = 0. We now consider the next level of the weight-defect diagram, that corresponds to the weight γ_{0} +4. We observe that the lowest weighted component in χ that depends on r is χ_{2} with

2 Re χ_{2} = 2 Re ∑_{j=1,2} r_{j}z_{j}ω_{j} \frac{∂}{∂z_{j}} + r_{j}ω_{j}^{2} \frac{∂}{∂ω_{j}} ,

where r_{1} = r = r̄_{2}. The operator 2 Re χ_{2} preserves the space N and is, obviously, nondegenerate for r ≠ 0. We distinguish two components of 2 Re χ_{2}, namely:

χ_{2}^{1} = rω_{1}z_{1} \frac{∂}{∂z_{1}} + rω̄_{2}z̄_{2} \frac{∂}{∂z̄_{2}} + rω_{1}^{2} \frac{∂}{∂ω_{1}}

and

χ¯1 2 = r̄ω̄_{1}z̄_{1} \frac{∂}{∂z̄_{1}} + r̄ω_{2}z_{2} \frac{∂}{∂z_{2}} + r̄ω_{2}^{2} \frac{∂}{∂ω_{2}} ,

that, respectively, increase by 1 the degrees in (s, t) and (s̄, t̄) variables. Starting with admissible 𝔑_{γ_{0}+2,Δ}^{n,k}, provided by Theorem 2, we are going to show that it will be impossible to determine 𝔑_{γ_{0}+3,Δ−1}^{n+1,k} via the identity P_{γ_{0}+4,Δ−1}^{n+1,k} ≡ 0. Without loss of generality we may assume that 𝔑_{γ_{0}+3,δ_{1}+j,Δ−δ_{1}−j−1}^{n+1,k} = 0 for j ≥ 0 because, otherwise we could have subtracted an appropriate solution of the homogeneous equation analogous to the upper line of the weight-defect diagram that we considered above. We see that the operator π_{∗}(χ_{2}^{1}) increases the degree in s by one. Therefore, applied to 𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}}^{n,k} it produces #s^{4}t^{n−3} for 𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}}^{n,k} of the type (14), #s^{3}t^{n−2} for the type (15), #s^{2}t^{n−1} for (16) and #st^{n} for (17) in P_{γ_{0}+4,δ_{1},Δ−δ_{1}}^{n+1,k}. Each of these terms belongs to the kernel of (s\frac{∂}{∂s} − (δ_{1} − 1) id), the operator representing the ↖-arrow of the weightdefect diagram, that is applied to 𝔑_{γ_{0}+3,δ_{1}−1,Δ−δ_{1}}^{n+1,k} and points at P_{γ_{0}+4,δ_{1},Δ−δ_{1}}^{n+1,k}. Thus we cannot solve the equation P_{γ_{0}+4,δ_{1},Δ−δ_{1}}^{n+1,k} ≡ 0 because

(s\frac{∂}{∂s} − (δ_{1} − 1) id) 𝔑_{γ_{0}+3,δ_{1}−1,Δ−δ_{1}}^{n+1,k}

does not contain terms from the kernel of (s\frac{∂}{∂s} − (δ_{1} − 1) id) but on the other hand

π_{∗}(χ_{2}^{1})𝔑_{γ_{0}+2,δ_{1},Δ−δ_{1}}^{n,k}

does. Contradiction. Therefore, r = 0. □ This lemma demonstrates that an isotropic automorphism of a non-quadratic elliptic manifold can be uniquely reconstructed by its Jacobian matrix at the origin.

Corollary 2 Aut_{0,id} M ≅ {id} or ℂ.

Proof of Theorem 1 If 𝔞𝔲𝔱 _{0,id} ∋ χ ≠ 0 then one of the parameters a_{1} or a_{2} in χ does not vanish. Then the other one vanishes, and so does r. Then Aut_{0,id} ≅ ℂ. We conclude with a few questions that remain open. Can any isotropic automorphisms of a non-exceptional manifold be linearised in normal coordinates? Can the whole group Aut_{0} M be equivariantly linearised? Which “sprouts” from the list in Theorem 2 give rise to exceptional manifolds? Are there exceptional manifolds that are torsion-free in the sense of [SS] (In [EIS] also the notion semi-flat was used)?

10 Appendix

Below we display the list of all monomials belonging to R. By Π we denote an arbitrary polynomial in the indicated variables. The index is the degree of the polynomial.

Bidegree in (z, z̄) Monomial in R
(1, 1)  all but ν
(2, 1)  z_{1}z_{2}z̄_{1}Π(η, η̄), Π_{2}(z)z̄_{2}Π(η, η̄)
(1, 2)  z_{2}z̄_{1}z̄_{2}Π(η, η̄), z_{1}Π_{2}(z̄)Π(η, η̄)
(2, 2)  ν^{2}Π(η), νν̄Π(η, η̄)
(3, 2)  z_{1}ν^{2}Π(η)
(2, 3) z̄_{2}ν^{2}Π(η)
(3, 3)  ν^{3}Π(η)
(l, 1)  Π_{l}(z)z̄_{2}Π(η, η̄)
l ≥ 3
(1, l)  z_{1}Π_{l}(z̄)Π(η, η̄)
l ≥ 3
(l, 0)  Π_{l}(z)Π(η, η̄)
(0, l)  Π_{l}(z̄)Π(η, η̄)

Table 1.

By direct inspection one obtains a list of normal monomials that produce non-normal monomials after applying \frac{∂}{∂z̄_{2}} resp. \frac{∂}{∂z_{2}} . Just consider when the pre-image of a monomial from R does not belong to R. We summarise the results in the following tables.

Monomial Ψ  \frac{∂}{∂z̄_{2}} Ψ  Defects
Π_{k}(z)z̄_{2}^{2}Π(η, η̄)  Π_{k}(z)z̄_{2}Π(η, η̄) d_{1} ≥ −2, d¯1 ≤ 0
Π_{k} ≠ z_{1}  d_{1} = −2 + deg_{z_{1}} Π_{k}

(Π_{k}, Π) ≠ (z_{1}^{2}, Π(η)) (Π_{k}, Π) ≠ (z_{1}^{3}, Π(η))

ν̄z̄_{2}^{2}Π(η, η̄)  ν̄z̄_{2}Π(η, η̄)  d_{1} = −2, d¯1 = 0
νν̄z̄_{2}Π(η, η̄)  νν̄Π(η, η̄)  d_{1} = −1, d¯1 = 0
z̄_{2}^{2}ν^{2}Π(η)  z̄_{2}ν^{2}Π(η)  d_{1} = −2, d¯1 = 0
z̄_{2}ν^{3}Π(η)  ν^{3}Π(η)  d_{1} = −1, d¯1 = 0

Table 2.

Monomial Ψ  \frac{∂}{∂z_{2}} Ψ  Defects
z_{2}νΠ(η, η̄)  νΠ(η, η̄)  d_{1} = 0, d¯1 = −1
z_{2}ν̄Π(η, η̄)  ν̄Π(η, η̄)  d_{1} = 0, d¯1 = −1
z_{1}z_{2}ν̄Π(η, η̄)  z_{1}ν̄Π(η, η̄)  d_{1} = 1, d¯1 = −1
z_{2}z̄_{2}ν̄Π(η, η̄)  z̄_{2}ν̄Π(η, η̄)  d_{1} = −1, d¯1 = −1
z_{1}z_{2}Π_{k}(z̄)Π(η, η̄) z_{1}Π_{k}(z̄)Π(η, η̄) d_{1} ≤ 1, d¯1 ≥ −1

Π_{k} ≠ z̄_{1}z̄_{2}, k ≥ 2

z_{2}ν^{2}Π(η)  ν^{2}Π(η)  d_{1} = 0, d¯1 = −1
z_{2}νν̄Π(η, η̄)  νν̄Π(η, η̄)  d_{1} = 0, d¯1 = −1
z_{1}z_{2}ν^{2}Π(η)  z_{1}ν^{2}Π(η)  d_{1} = 1, d¯1 = −1
z_{2}z̄_{2}ν^{2}Π(η)  z̄_{2}ν^{2}Π(η)  d_{1} = −1, d¯1 = −1
z_{2}ν^{3}Π(η)  ν^{3}Π(η)  d_{1} = 0, d¯1 = −1
z_{2}Π_{k}(z̄)Π(η, η̄) Π_{k}(z̄)Π(η, η̄) d_{1} ≤ 0, d¯1 ≥ −1

Π_{k} ≠ z̄_{1}z̄_{2}, k ≥ 2 Table 3.

Acknowledgement. We would like to thank I. Lieb for interesting discussion and useful comments.

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