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

Vladimir V. Ežov ^{‡} University of South Australia School of Mathematics Mawson Lakes, SA 5095, South Australia vladimir.ejov@unisa.edu.au

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

January 30, 2001

Abstract

For non-exceptional elliptic CR-manifolds in ℂ^{4} (introduced in [ES4]) we prove that all isotropic automorphisms can be simultaneously linearised in normal coordinates. This result is a higher codimensional analogue of theorems by Webster, Kružilin-Loboda for nonumbilic and strictly pseudoconvex hypersurfaces ([W1, KL]). Together with the result by Schmalz [S1] in the hyperbolic case it establishes the linearisation property for real-analytic CR-manifolds of codimension 2 in ℂ^{4} with stable Levi form.

∗Mathematics Subject Classification: primary 32C16; secondary 32C05, 32H02 †Keywords and Phrases: CR-manifolds, invariants, classification ‡Research of the first author was supported by ARC, MPI Bonn, 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, elliptic, non-quadratic CR-manifold in ℂ^{4}, i.e., in normal coordinates (z^{1}, z^{2}, ω^{1}, ω^{2}) it can be written as a complex equation

V − z^{1}z̄^{2} = N(z^{1}, z̄^{1}, z^{2}, z̄^{2}, η, η̄),

where V = \frac{ω^{1}−ω̄^{2}}{2 i} , η = \frac{ω^{1}+ω̄^{2}}{2} and N ≢ 0 uniquely splits into the sum

N = N^{CM}(z^{1}, z̄^{2}, η) + N^{tor}(z^{1}, z̄^{1}, z^{2}, z̄^{2}, η, η̄).

Here, the restriction of the Chern-Moser component N^{CM} to z^{1} = z^{2} and η ∈ ℝ is a power series in (z^{1}, z̄^{1}, η) whose real and imaginary part are in classical Chern-Moser normal form (in ℂ^{2}) and in the torsion component

N^{tor} = ∑\begin{matrix}k,ℓ≥1 \\ k+ℓ≥3 \end{matrix} N_{kℓ}^{tor}

the homogeneous polynomials N_{kℓ}^{tor} of degree k in z and ℓ in z̄ meet the conditions

N_{k,0}^{tor} = 0, k ≥ 0,  N_{1,1}^{tor} = 0
\frac{∂N_{k1}^{tor}}{∂z̄^{2}} = 0, k ≥ 0,  \frac{∂^{2}N_{21}^{tor}}{∂z^{1}∂z^{2}} = 0
\frac{∂^{4}N_{22}^{tor}}{∂z^{1}∂z^{2}∂z̄^{1}∂z̄^{2}} = 0

(see [ES1], and further, [EIS]). We note that it was proved in [CS] that partially integrable abstract elliptic CR-manifolds of CR dimension and codimension 2 without torsion are automatically real-analytic, embeddable and N^{tor} vanishes identically. The object of our main interest is the group Aut_{0} M of germs of biholomorphic isotropic (at 0) automorphisms of M. In [ES4] we were considering the subgroup Aut_{0,id} M of nil-isotropic automorphisms ψ = (f, g) : z^{∗} = f(z, ω), ω^{∗} = g(z, ω) with

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

and proved that Aut_{0,id} M = {id} or ℂ. The case Aut_{0,id} M = ℂ is extremely restrictive and occurs only for exceptional manifolds parametrised by at most one complex and two integer parameters. The existence and explicit description of such exceptional manifolds are yet to be established. Note that for all hypersurfaces or 2-codimensional hyperbolic manifolds in ℂ^{4} the group Aut_{0,id} is trivial which is a necessary condition for the linearisability of the entire group Aut_{0}. For strictly pseudoconvex or non-umbilic hypersurfaces and hyperbolic manifolds it was proved that this condition is also sufficient ([KL], [W1], [S1]). However, the construction in [E2] shows that there exist hypersurfaces with indefinite Levi-form such that Aut_{0} is not linearisable (even in projective sense). In this paper we prove the linearisability of Aut_{0} for non-exceptional elliptic manifolds. Let us introduce some useful notations. Recall that for any ψ ∈ Aut_{0} M the linear map

L_{ψ} : (z, ω) ↦ (Cz, ρω)

with

C = \frac{∂f}{∂z} |_{0} , ρ = \frac{∂g}{∂ω} |_{0}

is an automorphism of the quadric Q : V − z^{1}z̄^{2} = 0. By Aut_{0}^{0} M we denote the subgroup of Aut_{0} M that consists of automorphisms ψ such that L_{ψ} belongs to the connected component of the identity in Aut_{0} Q. We note that for ψ ∈ Aut_{0}^{0} M the matrices C and ρ are diagonal. Let ψ = (f, g) ∈ Aut_{0}^{0} M. Define a = (a^{1}, a^{2}) ∈ ℂ^{2} and r ∈ ℂ via

\frac{∂f}{∂ω} |_{0} = C (\begin{matrix}a^{1} & 0 \\ 0 & a^{2} \end{matrix})  (1)

\frac{1}{2} ( \frac{∂^{2}g^{1}}{(∂ω^{1})^{2}} |_{0} + \overline{\frac{∂^{2}g^{2}}{(∂ω^{2})^{2}} |_{0}}) = ρ_{11}r.

Since M is given in normal form we can consider ψ as a normalisation with parameters C, ρ, a, r (see [ES1]). Therefore, ψ equals up to terms of order 3 in z, ω to a fractional linear automorphisms of the elliptic CR-quadric V − z^{1}z̄^{2} = 0:

z̃^{1} = \frac{z^{1} + a^{1}ω^{1}}{1 − 2 i ā^{2}z^{1} − (r + i a^{1}ā^{2})ω^{1}}  (2)
z̃^{2} = \frac{z^{2} + a^{2}ω^{2}}{1 − 2 i ā^{1}z^{2} − (r̄ + i a^{2}ā^{1})ω^{2}}
ω̃^{1} = \frac{ω^{1}}{1 − 2 i ā^{2}z^{1} − (r + i a^{1}ā^{2})ω^{1}}
z̃^{2} = \frac{ω^{2}}{1 − 2 i ā^{1}z^{2} − (r̄ + i a^{2}ā^{1})ω^{2}} .

If we plug ψ into the equation of M we obtain an identity which is satisfied for all points of M. To analyse this identity we introduce the notion of weight γ and defects:

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. 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 δ(Π) is the sum of d_{1} and d¯1. According to these definitions we expand

N(z, z̄, η, η̄) = ∑_{γ=γ_{0}}^{∞} N_{γ}(z, z̄, η, η̄),

where N_{γ} is the component weight γ in N and N_{γ_{0}} ≠ 0. We can further split N_{γ} into components N_{γ,d_{1},d¯1} of 1-defect d_{1} and ¯1-defect d¯1. We are interested in certain properties of 3 consecutive components N_{γ_{0}}, N_{γ_{0}+1}, N_{γ_{0}+2} that arise from the components of weights γ_{0}, γ_{0} + 1, γ_{0} + 2 in the mentioned above identity. For N_{γ_{0}} the identity implies

ρ_{11}N_{γ_{0}}(z, z̄, η, η̄) = N_{γ_{0}}(Cz, C̄z̄, ρ_{11}η, ρ_{22}η̄).  (3)

Suppose that a_{1} = 0 and let N_{γ_{0},δ} be the component of maximal total defect in N_{γ_{0}}. Making use of the notation of the vector field

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}} )

we derive from the component of weight-defect (γ_{0} + 1, δ + 1) in the identity

ρ_{11}N_{γ_{0}+1,δ+1}(z, z̄, η, η̄) = N_{γ_{0}+1,δ+1}(Cz, C̄z̄, ρ_{11}η, ρ_{22}η̄) + ρ_{11}C_{11}χ_{1}^{1}N_{γ_{0},δ}(z, z̄, η, η̄). (4)

And, finally, if a = 0 then the (γ_{0} + 2, δ)-component of the identity provides

ρ_{11}N_{γ_{0}+2,δ}(z, z̄, η, η̄) = N_{γ_{0}+2,δ}(Cz, C̄z̄, ρ_{11}η, ρ_{22}η̄) + 2ρ_{11}^{2}(Re χ_{2})N_{γ_{0},δ} (5)

where χ_{2} is the vector field

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

with r_{1} = r = r̄_{2}. Now we are ready to prove the individual linearisation of isotropic automorphisms.

2 Linearisation of a single automorphism

Let M be a non-exceptional manifold. We may assume that M is given in normal form. Then any automorphism

ψ : z^{∗} = f(z, ω) ω^{∗} = g(z, ω)

from Aut_{0}^{0} M can be considered as a normalisation, and, therefore, it is uniquely determined by the parameters λ ∈ ℝ_{>0}, μ ∈ ℝ_{>0}, −π < θ ≤ π, −\frac{π}{2} < ϕ ≤ \frac{π}{2} defined by

C = \frac{∂f}{∂z} |_{0} = λ e^{i ϕ} (\begin{matrix}μ e^{i θ} & 0 \\ 0 & \frac{1}{μ} e^{− i θ} \end{matrix})

and by the parameters a, r defined in (1). Though, all these parameters are free for the quadric, this is no longer the case for automorphisms of nonquadratic manifolds.

Lemma 1 Parameters (a, r, λ) are uniquely determined by (μ, ϕ, θ).

Proof. Suppose that two automorphisms in Aut_{0}^{0} M have the same parameters λ, μ, ϕ, θ. Then according to the Main Theorem of [ES4] automorphism ψ_{0} = ψ_{1} ∘ ψ_{2}^{−1} ∈ Aut_{0,id} is the identical map. Thus, ψ_{1} = ψ_{2} and the corresponding parameters a and r coincide. To show that λ is also determined by (μ, ϕ, θ) we consider the equation of M in normal form

V = z^{1}z̄^{2} + N(z, z̄, η, η̄)  (6)

By T_{C} we denote the action of C on N (or homogeneous polynomials from the expansion of N) induced by the linear mapping

z^{∗} = Cz, ω^{∗} = ρω,

where ρ = λ^{2} e^{2 i θ} id, i.e. the operator

T_{C}N(z, z̄, η, η̄) = \frac{e^{−2 i θ}}{λ^{2}} N(Cz, \overline{Cz}, λ^{2} e^{2 i θ} η, λ^{2} e^{−2 i θ} η̄).

Let N_{γ_{0}} be the first non-vanishing weighted component of N. Since ψ preserves the equation (6) then

T_{C}N_{γ_{0}} = N_{γ_{0}}.

Any monomial

Π = c(z^{1})^{α}(z̄^{1})ᾱ(z^{2})^{β}(z̄^{2})β̄η^{δ}η̄δ̄

in N_{γ_{0}} is an eigenvector of T_{C} with the eigenvalue

e^{i(α+β−ᾱ−β̄)ϕ} e^{i(α+β̄−ᾱ−β+2δ−2δ̄−2)θ} λ^{γ_{0}−2}μ^{α+ᾱ−β−β̄}.

Hence, λ is determined by λ = μ^{\frac{α+ᾱ−β−β̄}{2−γ_{0}}}. □

Remark 1 If α + β̄ + 2δ = ᾱ + β + 2δ̄ + 2 then, in general, it is impossible to determine θ via ϕ. If α + β = ᾱ + β̄ then ϕ cannot be determined by θ.

Corollary 1 The dimension of Aut_{0} M does not exceed 3.

Since parameters a, r depend on the other parameters in C it will be our first aim to choose normal coordinates so that in these coordinates a = 0 and r = 0 (this is the simplest possible dependence).

Lemma 2 For each ψ ∈ Aut_{0}^{0} M there exist normal coordinates such that a = 0.

Proof. According to Theorem 3 in [ES1] it suffices to find coordinates such that the chain Γ_{0} = {z = 0, v = 0} is invariant with respect to ψ. Therefore, we will look for an invariant chain Γ and pass to such normal coordinates where Γ will be mapped to Γ_{0}. Since Γ is uniquely defined by T_{0}Γ, ψ-invariance of Γ means that

dψ|_{0}T_{0}Γ = T_{0}Γ.  (7)

Applying Jacobian

dψ|_{0} = (\begin{matrix}λμ e^{i(ϕ+θ)} & 0 & λμ e^{i(ϕ+θ)} a^{1} & 0 \\ 0 & \frac{λ}{μ} e^{i(ϕ−θ)} & 0 & \frac{λ}{μ} e^{i(ϕ−θ)} a^{2} \\ 0 & 0 & λ^{2} e^{2 i θ} & 0 \\ 0 & 0 & 0 & λ^{2} e^{−2 i θ} \end{matrix})

to the basis τ_{1} = (b^{1}, 0, 1, 0)^{t}, τ_{2}(0, b^{2}, 0, 1)^{t} of T_{0}Γ we obtain

dψ|_{0} (\begin{matrix}b^{1} & 0 \\ 0 & b^{2} \\ 1 & 0 \\ 0 & 1 \end{matrix}) = (\begin{matrix}λμ e^{i(ϕ+θ)}(b^{1} + a^{1}) & 0 \\ 0 & \frac{λ}{μ} e^{i(ϕ−θ)}(b^{2} + a^{2}) \\ λ^{2} e^{2 i θ} & 0 \\ 0 & λ^{2} e^{−2 i θ} \end{matrix})

Now, equation (7) implies that the columns in the matrix at the right hand side are linear combinations of the two basis vectors. This can only happen if the first one is τ_{1} scaled by λ^{2} e^{2 i θ} and the second one is τ_{2} scaled by λ^{2} e^{−2 i θ} and therefore

a^{1} = (\frac{λ}{μ} e^{i(θ−ϕ)} −1) b^{1}
a^{2} = (λμ e^{− i(θ+ϕ)} −1) b^{2}.

Using the splitting C = ΛU with Λ = λ (\begin{matrix}e^{i θ} & 0 \\ 0 & e^{− i θ} \end{matrix}), U = e^{i ϕ} (\begin{matrix}μ & 0 \\ 0 & \frac{1}{μ} \end{matrix})

this can be written as

(ΛU^{−1} − id)b = a.  (8)

This system is always soluble for b if ΛU^{−1} does not have eigenvalue 1. Then, solving (8) for b we find the unique ψ-invariant chain tangent to the span of τ_{1}, τ_{2}. Now, suppose that 1 is an eigenvalue of ΛU^{−1}. Then either λ = \frac{1}{μ} and θ = −ϕ, or λ = μ and θ = ϕ. Without loss of generality we consider the first possibility. System (8) will have a (non-unique) solution if and only if a^{2} = 0. In this case we find a complex 1-parameter family of ψ-invariant chains. We are going to show by contradiction that, in fact, a^{2} = 0 if ψ is an automorphism of a non-exceptional manifold. Then (3) and (4) takes the form

T_{C}N_{γ_{0}} =N_{γ_{0}}  (9)
(T_{C} − id)N_{γ_{0}+1,δ+1} =#χ_{1}^{1}N_{γ_{0},δ}.  (10)

Here and everywhere below # substitutes a non-vanishing constant.

¿From the form of χ_{1}^{1} we read that χ_{1}^{1}N_{γ_{0}} ≠ 0 belongs to the kernel of T_{C} −id since N_{γ_{0}} belongs to that kernel by (9). It follows that the right hand side of (10) equals #χ_{1}^{1}N_{γ_{0},δ} and lies in the kernel of T_{C} − id whereas the left hand side of (10) lies in the range of T_{C} − id. As T_{C} − id is diagonalisable (the monomials form an eigenbasis) its range is spanned by the eigenvectors with eigenvalues different from 0 and therefore intersects with the kernel by {0}. Thus, χ_{1}^{1}N_{γ_{0},δ} = 0. On the other hand, in [ES4] it was proved that for a non-exceptional manifold χ_{1}^{1}N_{γ_{0}} ≠ 0 if a^{2} ≠ 0. Contradiction. □

Having chosen coordinates where Γ_{0} is an ψ-invariant chain, we can now change the coordinates corresponding to a reparametrisation of this chain in order to make r = 0:

Lemma 3 Let ψ ∈ Aut_{0}^{0} M and a = 0 then in some normal coordinates r = 0.

Proof. Suppose that λ^{2} e^{2 i θ} ≠ 1. Consider the transformation

z^{∗} = (id −R)^{−1}z  (11)
ω^{∗} = (id −R)^{−1}ω

where

R = (\begin{matrix}r′ω^{1} & 0 \\ 0 & r̄′ω^{2} \end{matrix}) , z = (\begin{matrix}z^{1} \\ z^{2} \end{matrix}) , ω = (\begin{matrix}ω^{1} \\ ω^{2} \end{matrix}) .

One computes that in ∗-coordinates the new parameter

r^{∗} = r − (λ^{2} e^{2 i θ} −1)r′.

Taking

r′ = \frac{r}{1 − λ^{2} e^{2 i θ}}

we obtain r^{∗} = 0. Suppose now that

λ^{2} e^{2 i θ} = 1,  (12)

i.e., λ = 1, and e^{2 i θ} = 1. We show that then r = 0 automatically. Identity (5) can be written as

T_{C}N_{γ_{0}} = N_{γ_{0}}
(T_{C} − id)N_{γ_{0}+2} = # Re χ_{2}N_{γ_{0}}.

It is a consequence of (12) that the right hand side of the last equation belongs to the kernel of T_{C} − id though the left hand side belongs to the range of T_{C} − id. By the same argument as in the proof of the previous lemma this implies Re χ_{2}N_{γ_{0}} = 0 which is impossible if r ≠ 0. □

Combining Lemmata 2 and 3 we easily prove that any ψ ∈ Aut_{0}^{0} M can be linearised.

Theorem 1 For each individual automorphism ψ ∈ Aut_{0}^{0} M there exist normal coordinates such that ψ takes the form of a linear transformation

z^{∗} = ΛUz
ω^{∗} = Λ^{2}ω

Proof. According to Lemmata 2 and 3 we can choose normal coordinates such that the parameters a and r of ψ considered as a normalisation of M vanish. By Theorem 3 in [ES1] the vanishing of a means that ψ is a mapping of the form

z^{∗} = ΛU(id −R)^{−1}z
ω^{∗} = Λ^{2}(id −R)^{−1}ω.

Condition r = 0 implies that R = 0 and therefore ψ has the required form. □

3 Simultaneous linearisation

The goal of this section is to prove

Theorem 2 For every non-exceptional manifold M there exist normal coordinates (z, ω) such that all the elements of Aut_{0} M take the form

z^{∗} = ΛUz  (13)
ω^{∗} = Λ^{2}ω

The proof of this Theorem requires several preliminary results.

Lemma 4 The group Aut_{0}^{0} M is Abelian.

Proof. This is an immediate consequence of the fact that an automorphism from Aut_{0}^{0} M is uniquely determined by the parameter C which is the restriction of the differential in 0 to the complex tangent space and that the matrices C are diagonal. □

Lemma 5 Every ψ ∈ Aut_{0}^{0} M splits into the composition of two automorphisms ψ_{1} with parameter

C_{1} = (\begin{matrix}e^{i(ϕ+θ)} & 0 \\ 0 & e^{i(ϕ−θ)} \end{matrix}) ,  (14)

referred to as a rotation, and ψ_{2} with parameter

C_{2} = (\begin{matrix}λμ & 0 \\ 0 & \frac{λ}{μ} \end{matrix})

referred to as a scaling.

Proof. Let ψ ∈ Aut_{0}^{0} M and (λ, μ, ϕ, θ, a, r) be the corresponding set of parameters. According to the Theorem on individual linearisation from the previous section we can start with a normal coordinate system such that a = 0, r = 0 and ψ is a linear map (13). Analysing the action of the linear map ψ on the equation of M we observe that ψ splits into the required composition of two automorphisms

ψ_{1} : z^{∗} = (\begin{matrix}e^{i(ϕ+θ)} & 0 \\ 0 & e^{i(ϕ−θ)} \end{matrix}) z, ω^{∗} = ω  (15)

and

ψ_{2} : z^{∗} = (\begin{matrix}λμ & 0 \\ 0 & \frac{λ}{μ} \end{matrix}) z, ω^{∗} = λ^{2}ω.  (16)
□

It follows now that the group Aut_{0}^{0} M decomposes into the direct product of the subgroup S of all scalings and the subgroup K of all rotations.

Lemma 6 The subgroup S is either trivial or one-parametric.

Proof. Let id ≠ ψ ∈ S. We choose normal coordinates such that ψ is linear of the form (16). Set ℓ := ln λ and m := ln μ. Consider the one-parametric group

ψ^{t} : z^{∗} = \binom{e^{(ℓ+m)t} 0}{0 e^{(ℓ−m)t} 0} z, ω^{∗} = e^{2ℓt} ω.

ψ^{t} acts on a monomial Π of the equation of M by multiplication with e^{ht} where h is an integers that depends only on Π. Since ψ^{1} = ψ is an automorphism this integer must vanish. Thus, all ψ^{t} are automorphisms. Since the dimension of S does not exceed 1 it follows that S is either trivial or one-parametric. □

We prove that all automorphisms from K can be simultaneously linearised.

Lemma 7 There exist normal coordinates such that all automorphisms ψ ∈ K take the form (15).

Proof. At first we prove that all automorphisms in K have a common invariant chain. If K contains ψ with parameter C such that neither e^{i(ϕ+θ)} = 1 nor e^{i(ϕ−θ)} = 1 then there is a unique ψ-invariant chain Γ. Since K is Abelian Γ is then also invariant with respect to the other automorphisms from K. If such ψ does not exists then either for all elements in K the parameter ϕ equals θ or for all elements ϕ equals −θ, i.e. the group K is either one-parametric or finite cyclic. In both cases it suffices to find an invariant chain for a generator. Without loss of generality we may now assume that the standard chain Γ_{0} is invariant for all ψ ∈ K. For all automorphisms with θ = 0 the parameter r vanishes automatically, according to Lemma 3, regardless to parametrisation of Γ_{0}. These automorphisms form, obviously, a subgroup of K. The factor group of K by this subgroup is either one-parametric or finite cyclic. In both cases we can find coordinates such that the parameter r vanishes for an an automorphism from a generating coset. Then r = 0 for all elements of K. □

Corollary 2 K is a compact Lie group.

We consider now the situation when Aut_{0}^{0} M contains a scaling σ since, otherwise, by the lemma above we know that Aut_{0}^{0} M = K can be linearised. Let Γ_{K} be the family of all K-invariant chains.

Lemma 8 The family Γ_{K} contains a σ-invariant chain Γ_{σ}.

Proof. Assume K is not trivial (otherwise there is nothing to prove). We choose normal coordinates such that K is linearised. Family Γ_{K} may consist of single chain Γ_{0} (this is the case if for some ψ ∈ K ϕ ≠ θ and ϕ ≠ −θ ) or else

T_{0}Γ = span \{(\begin{matrix}0 \\ 0 \\ 1 \\ 0 \end{matrix}) , (\begin{matrix}0 \\ b \\ 0 \\ 1 \end{matrix})\} , b ∈ ℂ,  (17)

(this is the case if ϕ = θ for every ψ ∈ K), or

T_{0}Γ = span \{(\begin{matrix}b \\ 0 \\ 1 \\ 0 \end{matrix}) , (\begin{matrix}0 \\ 0 \\ 0 \\ 1 \end{matrix})\} , b ∈ ℂ,

(this is the case if ϕ = −θ for every ψ ∈ K). If Γ_{0} is the only K-invariant chain we are done because σ commutes with K and, therefore, preserves Γ_{0}. Since the two remaining cases are similar we consider only the case ϕ = θ for all ψ ∈ K. The Jacobian dσ|_{0} has the form

dψ|_{0} = (\begin{matrix}λμ & 0 & λμa^{1} & 0 \\ 0 & \frac{λ}{μ} & 0 & \frac{λ}{μ}a^{2} \\ 0 & 0 & λ^{2} & 0 \\ 0 & 0 & 0 & λ^{2} \end{matrix}) .

Since σ commutes with K it maps any chain from Γ_{K} into a chain from Γ_{K}. From the first component of

dψ|_{0} (\begin{matrix}0 \\ 0 \\ 1 \\ 0 \end{matrix}) = (\begin{matrix}0 \\ 0 \\ λ^{2} \\ 0 \end{matrix}) ,

one reads that a^{1} = 0. If a^{2} = 0 then Γ_{σ} = Γ_{0} and we are done. If a^{2} ≠ 0 then, as it had been shown in the proof of Lemma 2, λ ≠ μ^{−1}, and Γ_{σ} is the chain tangent to (17) for b = \frac{a^{2}}{λμ−1}. □

Let now Γ_{0} be a common invariant chain for Aut_{0}^{0} M.

Lemma 9 There exist normal coordinates such that Aut_{0}^{0} M is linearised.

Proof. If the parameter λ of the scaling σ is equal to 1 then r = 0 automatically. So, it remains to choose coordinates such that Γ_{0} is a common invariant chain and K is linearised. Let λ≠ 1. Choose normal coordinates such that Γ_{0} remains a common invariant chain and σ is linearised. It turns out that K is then also linearised automatically. Indeed, if for ψ ∈ K the parameter r ≠ 0 then N_{γ_{0}} and N_{γ_{0}+2} satisfy the conditions:

T_{C_{σ}}N_{γ_{0}} = N_{γ_{0}}
T_{C_{ψ}}N_{γ_{0}} = N_{γ_{0}}
T_{C_{σ}}N_{γ_{0}+2} = N_{γ_{0}+2}
T_{C_{ψ}}N_{γ_{0}+2} − N_{γ_{0}+2} = #T_{C_{ψ}} Re χ_{2}N_{γ_{0}}

Since 2 Re χ_{2}N_{γ_{0}} ≠ 0 the right hand side of the last equation contains monomials Π such that T_{C_{σ}}Π = λ^{2}Π, though, the left hand side contains only monomials form the kernel of T_{C_{σ}} −id, by the third equation. Contradiction. □

Lemma 10 Aut_{0} M has a common invariant chain.

Proof. Aut_{0} consists at most of two cosets with respect to Aut_{0}^{0} M. Therefore, a chain that is invariant for Aut_{0}^{0} M and one element of ψ ∈ Aut_{0} M \ Aut_{0}^{0} M will be invariant for the whole group. We start with normal coordinates where Aut_{0}^{0} M is linearised. The Jacobian of ψ has the form

dψ|_{0} = (\begin{matrix}0 & λμ e^{i(ϕ+θ)} & 0 & λμ e^{i(ϕ+θ)} a^{1} \\ \frac{λ}{μ} e^{i(ϕ−θ)} & 0 & \frac{λ}{μ} e^{i(ϕ−θ)} a^{2} & 0 \\ 0 & 0 & 0 & λ^{2} e^{2 i θ} \\ 0 & 0 & λ^{2} e^{−2 i θ} & 0 \end{matrix})  (18)

Notice that ψ^{2} already belongs to Aut_{0}^{0} and therefore it is linear. So, the right upper block of the square of (18) must vanish. This gives the equations

a_{1} + \frac{λ}{μ} e^{i(θ−ϕ)}a_{2} = 0  (19)
λμ e^{− i(θ+F)}a_{1}+  a_{2} = 0.

If λ ≠ 1 or ϕ ≠ 0 this immediately implies that a_{1} = a_{2} = 0, i.e., the standard chain is invariant for ψ. Assume now λ = 1 and ϕ = 0. Then

dψ|_{0} = (\begin{matrix}0 & μ e^{i θ} & 0 & μ e^{i θ} a^{1} \\ \frac{1}{μ} e^{− i θ} & 0 & e^{−2 i θ}a^{1} & 0 \\ 0 & 0 & 0 & e^{2 i θ} \\ 0 & 0 & e^{−2 i θ} & 0 \end{matrix})

. Since Aut_{0}^{0} M is a normal subgroup of Aut_{0} M then dψ|_{0}T_{0}Γ_{0} is invariant for Aut_{0} M. It follows that either a^{1} = 0 (and therefore Γ_{0} is invariant for ψ, as well), or Aut_{0}^{0} M = {id} (because we would obtain more than a oneparametric family of invariant chains). In this case we find an invariant chain for ψ by solving the equation (8). □

It follows that there exist normal coordinates such that Aut_{0}^{0} M is linearised and any ψ ∈ Aut_{0} M \ Aut_{0}^{0} M has the form

z^{∗} = JΛU(id −R)^{−1}z  (20)
ω^{∗} = JΛ^{2}(id −R)^{−1}ω

where

J = (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) , R = (\begin{matrix}rω^{1} & 0 \\ 0 & r̄ω^{2} \end{matrix}) .

It suffices to find a transformation of the form (11) such that Aut_{0}^{0} M remains linearised and one (and therefore every) element from Aut_{0} M \ Aut_{0}^{0} M becomes linear.

Lemma 11 If for some ψ ∈ Aut_{0} M the parameter λ is different from 1 then Aut_{0} M is automatically linearised if Aut_{0}^{0} M is linearised.

Proof. Without loss of generality we assume that λ ≠ 1 for some ψ ∈ Aut_{0} M \ Aut_{0}^{0} M. Taking the square of ψ in the form (20) we obtain a fractional linear mapping with parameter

r′ = r + λ^{2} e^{−2 i θ} r̄  (21)

On the other hand we know that ψ^{2} is already linear, thus r′ = 0. But this is only possible if r was already 0. □

Lemma 12 If e^{2 i θ} is different from 1 for some ψ ∈ Aut_{0}^{0} M then Aut_{0} M is automatically linearised if Aut_{0}^{0} M is linearised.

Proof. Let ψ_{1} ∈ Aut_{0} M \ Aut_{0}^{0} M have parameters θ_{1} and r. Take ψ_{2} ∈ Aut_{0}^{0} M with e^{2 i θ_{2}} ≠ 1. The condition r′ = 0 in (21) implies that 2(θ_{1} − arg r) = π, or r = 0 . Considering (21) for the composition ψ_{1} ∘ ψ_{2} leads to 2(θ_{1} − θ_{2} − arg r) = π, or r = 0. By the assumption of the lemma it follows that r = 0. □

Theorem 2 is now a consequence of the lemma below.

Lemma 13 If λ = 1 and e^{2 i θ} = 1 for all ψ ∈ Aut_{0}^{0} M then Aut_{0} M is can be linearised.

Proof. According to the proof of Lemma 3, any coordinate change of the form (11) preserves the linear form of Aut_{0}^{0}. Therefore it suffices to find a transformation (11) that linearises one ψ ∈ Aut_{0} M \ Aut_{0}^{0} M. This changes the parameter r of ψ into r + r′ + e^{−2 i θ} r̄′. If r ≠ 0 then (21) ensures that there exists r′ such that r + r′ + e^{−2 i θ} r̄′ = 0 □

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