Complex Contact Manifolds

with

Legendrian Vector Fields

ルジャンドリアンなベクトル場を持つ複素接触多様体

HIROKI KODAMA 児玉大樹

論文内容の要旨

論文題目 Complex Contact Manifolds with

Legendrian Vector Fields

ルジャンドリアンなベクトル場を持つ複素接触多様体

氏名児玉大樹

実 3 次元多様体において, 余次元 1 の葉層構造は完全可積分な接平面場と見なすことができ, 一方で接触構造は至る所で積分不可能な接平面場と見なすことができる. この二つは研究の手法において, 様々な類似点が見られる. 特に Yasha Eliashberg 氏と William Thurston 氏はコンフォリエーションという概念を導入することにより, 葉層構造と接触構造を統一的に扱うことに成功し, 向きづけられた閉 3 次元多様体上の余次元 1 の C^{2} 級葉層構造は, S^{2} × S^{1} 上の S^{2} × {pt} を葉とする積葉層構造を除き, 常に正および負の接触構造で C^{0}-近似出来ることを示した. 彼らは更に, 近似よりも強い条件として, 線型変形とよばれる概念を考えた. これは平面場の族 {ξ_{t}} で, t = 0 のとき葉層構造, t > 0 のとき正の接触構造, t < 0 のとき負の接触構造を定めるようなものである.

一方, 三松佳彦氏は彼らに先立って, アノソフ葉層の線型変形を研究していた. 良く知られているように, アノソフ流の安定及び不安定葉層は, 互いを定める 1 形式によって正と負の接触構造に線型変形される. このような接触構造の組は双接触構造と呼ばれる. しかしながら, このような双接触構造は必ずしもアノソフ流から付随して得られるとは限らない. 三松佳彦氏はアノソフ流の概念を拡張して得られる射影的アノソフ流を定義した. 射影的アノソフ流には常に双接触構造が付随し, 逆に双接触構造があれば必ずその双方にルジャンドリアンな射影的アノソフ流が存在する.

実 3 次元多様体の場合と異なり, 複素 3 次元多様体においては, アノソフ性は必ずしも正則接触構造の存在を保証しない. Étienne Ghys 氏が定義した正則アノソフ流や, それを一般化した正則射影的アノソフ流のなかに, それをルジャンドル流として持つような正則接触構造を許容しない物が多数あることを, 筆者は修士論文において示した.

複素 3 次元多様体における接触構造と葉層構造の変形を考えよう. 正則平面場の正則な族 {ξ_{t}}(t ∈ U ⊂ ℂ) であり, 有限個の t を除けば {ξ_{t}} は正則接触構造を定めるようなものを考える. 特に, ξ_{0} は接触構造であるとする. このとき, 多様体の各点 p の近傍において, ξ_{0} にルジャンドリアンなベクトル場 X̃ を次のように定める; X̃ の定める ℂ 作用 ϕ^{s} による ξ_{0} の push-forward を ξ^{s} とするとき, \frac{∂}{∂t}(ξ_{t}|_{p})|_{t=0} と\frac{∂}{∂s}(ξ^{s}|_{p})|_{s=0} が一致する. X̃ の取り方には自由度があるが, 点 p におけるベクトル X̃_{p} はこれによって一意に定まる. これらの各点 p でベクトル X̃_{p} を対応させるベクトル場を X で表す. このとき X は多様体上の正則なベクトル場で ξ_{0} にルジャンドリアンであり, この対応により, X は正則接触構造 ξ_{0} の正則な変形の 1-jet を表すことが分かる. 本論文では, \frac{∂}{∂t}(ξ_{t}|_{p})|_{t=0} が至る所消えない場合, つまりルジャンドリアンなベクトル場 X が至るところ 0 にならない場合を扱う.

さて, 実 3 次元上の C^{r} という条件と異なり, 複素 3 次元上の正則的という条件は非常に厳しいため, そのような構造を許容する多様体は限られると考えられる. 本論文において, 筆者はまず, 次のような定理を示した.

定理 (M, ξ) は複素 3 次元接触多様体で, X は ξ にルジャンドリアンな非特異ベクトル場であるとする. このとき, 次のいずれかが成立する.

(i) (parabolic)

M の座標近傍近傍系 U = {U = {(x, y, z)}} が存在して, 各座標近傍 U 上で, ξ と X が ξ = ker(dy − xdz) および X = ∂/∂x によって与えられる. (ii) (loxodromic)

M の座標近傍近傍系 U = {U = {(x, y, z)}} が存在して, 各座標近傍 U 上で, ξ と X が ξ = ker(dy − exp(ax)dz) および X = ∂/∂x によって与えられる.

ルジャンドリアンなベクトル場が loxodromic であるならば, そのベクトル場は正則射影的アノソフ流を定め, 逆に, 正則射影的アノソフ流が接触構造を許容するならば, それに対応するベクトル場は接触構造に対して loxodromic である.

ルジャンドリアンなベクトル場を持つ接触構造の最も典型的な例は非可換な 3 次元リー群の等質空間上の接触構造である. 多様体上の非可換な二つの非特異なベクトル場 X, Z が [X, Z] ⊄ CX ⊕ ℂZ を満たせば, ℂX ⊕ ℂZ は接触構造を定め, aX + bZ はその接触構造にルジャンドリアンなベクトル場である. このとき Y = [X, Z] とすれば Y もまた非特異なベクトル場となるので, 元の多様体は等質空間であることが分かる. この場合, 接触構造を定める正則な 1 形式を ω(X) = ω(Z) = 0, ω(Y) = 1 で定めることが出来る.

本論文では逆に, ルジャンドリアンなベクトル場を許容する接触多様体で, 接触構造を定める正則な 1 形式が存在しないもの, つまり, 3 次元リー群の等質空間ではない例を構成した.

次に, 接触構造を定める 1 形式 ω が存在する場合は, ルジャンドリアンなベクトル場 X が存在すれば元の多様体は等質空間であるか, という問題が考えられる. 接触形式 ω のレーブベクトル場を Y としたときに [X, Y] について次のような条件で場合分けが出来る.

命題次の 3 つの条件のうちいずれかが成立する;

(i) [X, Y] が ℂX ⊕ ℂY と横断的. (ii) 0 でない定数 c が存在して [X, Y] = cX. (iii) [X, Y] = 0.

(i) のときには自動的に, (ii) のときにはÉtienne Ghys 氏と Alberto Verjovsky 氏の結果を使うことにより, 問題は肯定的に解決される. 一方, (iii) のときには問題は未解決であるが, この論文では次のように局所的, および大域的構造を得た.

定理 M の座標近傍近傍系 U = {U = {(x, y, z)}} が存在して, 各座標近傍 U 上で, ω と X が ω = ker dy −xdz および X = ∂/∂x によって与えられる. M の普遍被覆は ℂ 上の ℂ^{2}-バンドルであり, 各ファイバー ℂ^{2} の接空間は ℂX ⊕ℂY に一致する.

また, 特に M の普遍被覆が ℂ^{3} で, ω = ker dy − xdz および X = ∂/∂x のときを考察した. まず, (ℂ, ω, X) の自己同型群は

Aut(ℂ^{3}, ω, X) = {(f, C) | f ∈ O(ℂ), C ∈ ℂ},
(f, C): (x, y, z) ↦ (x + f′(z), y + f(z), z + C).

で与えられることを示した. この条件の下での M の分類は Γ\ℂ^{3} が 3 次元閉多様体になるような Aut(ℂ^{3}, ω, X) の部分群 Γ の分類と対応する. そのような部分群 Γ に対して, 次の定理を示した.

定理 Γ が Aut(ℂ^{3}, ω, X) の部分群で Γ\ℂ^{3} が 3 次元閉多様体になるようなとき, うまく元 σ をとれば, 任意の (f, C) ∈ σ^{−1}Γσ について f は r(z) exp(αz) (r(z) は 4 次以下の多項式, α は定数) の形の式の, 高々16 個の線形結合と定数の和で表される.

Complex Contact Manifolds

with

Legendrian Vector Fields

Hiroki Kodama

Contents

1 Introduction 6

2 General case 11

3 If there is a global contact form 19 3.1 Case (i); [X, Y ] = Z . . . . . . . . . . . . . . . . . . . . . . . 20 3.2 Case (ii); [X, Y ] = cX . . . . . . . . . . . . . . . . . . . . . 21 3.3 Case (iii); [X, Y ] = 0 . . . . . . . . . . . . . . . . . . . . . . 21

3.3.1 proof of the proposition 22 . . . . . . . . . . . . . . . . 22 3.3.2 proof of the proposition 24 . . . . . . . . . . . . . . . . 24

3.4 some remarks on the case M̃ ≈ ℂ^{3} . . . . . . . . . . . . . . . 29

Keywords: complex contact manifold, Legendrian vector field 2000 Mathematics Subject Classification: Primary 37F75; Secondary 37D20, 37F15, 53C15. This work is supported by Research Fellowship of the Japan Society for the Promotion of Science for Young Scientists.

1 Introduction

A complex contact manifold is a closed complex manifold M of dimension 2n + 1 with an open atlas U = {U} such that on each open set U there is a holomorphic 1-form ω satisfying ω ∧(dω)^{n} ≠ 0 and on the intersection U ∩U′ there exists a holomorphic function f: U ∩ U′ → ℂ^{∗} such that ω = fω′. In particular, ξ = ⋃_{U∈U}(ker ω) is a well-defined holomorphic sub-bundle of T^{1,0}M. We denote this complex contact manifold by (M, ξ).

A nonsingular Legendrian vector field of (M, ξ) is a holomorphic section of ξ ∖ {0}, where {0} is the zero-section of ξ. In other words, X is a nonvanishing holomorphic vector field satisfying ω(X) = 0 on each open set U.

Let us investigate deformations of contact structures. Suppose that {ξ_{t}} (t ∈ U ⊂ ℂ) is a holomorphic family of contact structures. On a neighborhood U of each point p ∈ M, there is a holomorphic vector field X̃ on U such that X is Legendrian to ξ_{0} and \frac{∂}{∂s}(ξ^{s}|_{p})|_{s=0} coincides with \frac{∂}{∂t}(ξ_{t}|_{p})|_{t=0}, where ξ^{s} is a push-forward of ξ_{0} by the action ϕ^{s} which is generated by the vector field X̃. The choice of X̃ is not unique, but the vector X̃_{p} on the point p is uniquely determined. Let X be the vector field such that on each point p it determines the vector X̃_{p}. The vector X is holomorphic and Legendrian to ξ_{0}. The condition X_{p} ≠ 0 corresponds to \frac{∂}{∂t}(ξ_{t}|_{p})|_{t=0} ≠ 0. Thus the Legendrian vector fields correspond to the 1-jet of the holomorphic non-degenerate deformations of the contact structures.

The purpose of this paper is to investigate complex contact manifolds of dimension three with Legendrian vector fields. In the section 2, We prove the following theorem.

Theorem 1 Suppose (M, ξ) is a connected complex contact manifold of

complex dimension three and X is a Legendrian vector field of (M, ξ). Then

(M, ξ, X) satisfies one of the following conditions;

(i) There is an open atlas U = {U = {(x, y, z)}} such that on each coor-

dinated open set U, ξ = ker(dy − xdz) and X = ∂/∂x.

(ii) There is an open atlas U = {U = {(x, y, z)}} such that on each coor-

dinated open set U, ξ = ker(dy − exp(ax)dz) and X = ∂/∂x, where

a ∈ ℂ^{∗} is a constant, which does not depend on the choice of U.

We say that the Legendrian vector field X is parabolic (resp. loxodromic) if (M, ξ, X) satisfies the condition (i) (resp. condition (ii)).

In the section 3, we consider a special case, so that the contact plane field ξ is defined by a global holomorphic 1-form ω on M. The Reeb vector field of ω is a vector field Y on M satisfying ω(Y) = 1 and Y ⌋dω = 0. The Reeb vector field Y is a uniquely defined holomorphic vector field.

Proposition 2 The Legendrian vector field X and the Reeb vector field Y satisfy one of the following conditions;

(i) [X, Y] = Z and TM = ℂX ⊕ ℂY ⊕ ℂZ.

(ii) [X, Y] = cX, where c is a nonzero complex number.

(iii) [X, Y] = 0.

We introduce some examples so that M is a homogeneous space of a Lie group.

Example 3 Let SL(2, ℂ) be the Lie group of the 2 × 2 complex matrices with determinants 1. The space of the left invariant vector fields of SL(2, ℂ) forms a Lie algebra sl(2, ℂ) of the 2 × 2 complex traceless matrices. Let us set

T = (\begin{matrix}1/2 & 0 \\ 0 & −1/2 \end{matrix}) , U = (\begin{matrix}0 & 0 \\ 1 & 0 \end{matrix}) , S = (\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) .

They form a basis of sl(2, ℂ) and satisfy

[T, U] = −U, [T, S] = S, [U, S] = −2T.

Remind that each left invariant vector field induces a holomorphic ℂ-action by the right translation by

(\begin{matrix}e^{t/2} & 0 \\ 0 & e^{−t/2} \end{matrix}) , (\begin{matrix}1 & 0 \\ u & 1 \end{matrix}) , (\begin{matrix}1 & s \\ 0 & 1 \end{matrix}) .

Let Γ be a cocompact discrete subgroup of SL(2, ℂ) and M = Γ\ SL(2, C). The vector fields T, U and S are also well-defined in M.

Example 3-1 (geodesic) Let a and b be non-zero complex numbers. Define a 1-form ω and a vector field X by ω(T) = 0, ω(U) = ω(S) = b and X = aT. The Reeb vector field Y of ω equals to (1/2b)(U + S). The bracket [X, Y] = (a/2b)(−U + S) is transverse to ℂX ⊕ ℂY. This is case (i).

Example 3-2 (horocycle) Let a and b be non-zero complex numbers. Define a 1-form ω and a vector field X by ω(T) = b, ω(U) = ω(S) = 0 and X = aU. The Reeb vector field Y of ω equals to (1/b)T. The bracket is [X, Y] = (a/b)U = (1/b)X. This is case (ii).

Example 4 Let Solv = {(x, y, z) | x, y, z ∈ ℂ} be the Lie group defined by

(a, b, c) · (x, y, z) = (x + a, e^{−a}y + b, e^{a}z + c).

The space of the left invariant vector fields of the Lie group Solv forms a Lie algebra solv, spanned by

H = \frac{∂}{∂x}, U = e^{−x} \frac{∂}{∂y} , S = e^{x} \frac{∂}{∂z}

which satisfy

[H, U] = −U, [H, S] = S, [U, S] = 0.

These left invariant vector fields H, U and S induce holomorphic ℂactions by the right translations by

(h, 0, 0), (0, u, 0) and (0, 0, s).

Let Γ be a cocompact subgroup of Solv and M = Γ\Solv. The vector fields H, U and S are also well-defined in M.

Example 4-1 Let a ∈ ℂ be a non-zero complex number. Define a 1-form ω and a vector field X by ω(H) = 0, ω(U) = ω(S) = 1 and X = aH. The Reeb vector field Y of ω equals to (1/2)(U + S). The bracket [X, Y] = (a/2)(−U + S) is transverse to ℂX ⊕ ℂY. This is case (i).

Example 4-2 Let a ∈ ℂ be a non-zero complex number. Define a vector field X and 1-form ω by X = (a/2)(U +S) and ω(H) = 0, ω(U) = ω(S) = 1. The Reeb vector field Y of ω equals to (1/2)(U + S). Thus [X, Y] = 0. This is case (iii).

Example 5 Let Nil be the Lie group of the 3 × 3 complex upper triangular matrices with all eigenvalues 1, that is,

Nil = \{( \begin{matrix}1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{matrix} ) | x, y, z ∈ ℂ\}.

The space of the left invariant vector fields of the Lie group Nil forms a Lie algebra nil, spanned by

P = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) , Q = ( \begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) , R = ( \begin{matrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} ) .

They satisfy

[P, Q] = 0, [P, R] = Q, [Q, R] = 0.

These left invariant vector fields P, Q and R induce holomorphic ℂ-actions by the right translations by

( \begin{matrix}1 & p & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) , ( \begin{matrix}1 & 0 & q \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) , ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & r \\ 0 & 0 & 1 \end{matrix} ) .

Let Γ be a cocompact discrete subgroup of Nil and M = Γ\Nil. The vector fields P, Q and R are also well-defined in M. Define a 1-form ω and a vector field X by ω(P) = ω(R) = 0, ω(Q) = 1 and X = P. The Reeb vector field Y of ω equals to Q. Thus, [X, Y] = 0. This is case (iii).

We give the classification for each cases. The case (i) is the easiest. The space A of holomorphic vector fields on M is spanned by the vector fields X, Y and Z; and it forms a 3-dimensional Lie algebra. 3-dimensional Lie algebras are classified. The case (ii) is not easy, but this case has been already classified by Étienne Ghys and Alberto Verjovsky [GV]. These two cases are classified into example 3-1, example 3-2 and example 4-1.

For the case (iii), we got the following result.

Proposition 6 Suppose that X is a Legendrian vector field of a complex contact manifold (M, ω), Y is the Reeb vector field of ω and [X, Y] = 0. Then there is an open atlas U = {U = {(x, y, z)}} such that on each U, ω = dy − xdz, X = ∂/∂x and Y = ∂/∂y. The universal cover M̃ of M is a ℂ^{2}-bundle over ℂ and the plane field ℂX ⊕ ℂY is tangent to the fibers ℂ^{2}.

Pulling these three cases together, we have the following theorem.

Theorem 7 Suppose X is a Legendrian flow of a complex contact manifold

(M, ω), then either it is given in the examples 3-1, 3-2 or 4-1; or it has the

structure stated in the proposition 6.

We also investigate the case that [X, Y] = 0 and the bundle ℂ^{2} → M̃ → ℂ is trivial.

Proposition 8 The group Aut(ℂ^{3}, ω, X) of automorphisms on ℂ^{3} that pre-

serve ω = dy − xdz and X = ∂/∂x is given by

Aut(ℂ^{3}, ω, X) = {(f, C) | f ∈ O(ℂ), C ∈ ℂ},
(f, C): (x, y, z) ↦ (x + f′(z), y + f(z), z + C).

Theorem 9 Suppose Γ is a subgroup of Aut(ℂ^{3}, ω, X) such that Γ\ℂ^{3} is

a closed 3-dimensional complex manifold. Then there exist an element σ ∈

Aut(ℂ^{3}, ω, X), an integer N ≤ 16 and complex numbers α_{1}, …, α_{N} such

that for any (f, C) in the conjugate σ^{−1}Γσ of the subgroup, the function f

is written as

f(z) = ∑_{k=1}^{N} r_{k}(z) exp(α_{k}z) + b_{0},

where r_{k} is a polynomial of degree 4 or less and b_{0} is a constant.

We remark that complex contact structures are widely investigated. For example, Yun-Gang Ye classified the projective threefold that admits a contact structure by using some algebraic topological methods [Ye]. Holomorphic ℂ-actions on complex manifolds with some structure is also well investigated. Frédéric Bosio investigate higher rank actions, for example ℂ^{2}-actions on 3-dimensional manifolds [Bo]. These works motivated the author to investigate Legendrian flows on complex contact manifolds.

The author thanks Professor Takashi Tsuboi for his comments and encouragement in this work. The author thanks Professor Shigenori Matsumoto, who inspired the author of the proof of theorem 1 and gave many examples. Part of this work has been done during the stay at ENS-Lyon. The author thanks for its hospitality and express his gratitude especially to Professor Étienne Ghys, who gave the author of many suggestions during the stay.

2 General case

Suppose that (M, ξ) is a connected 3-dimensional complex contact manifold and X is a Legendrian vector field of (M, ξ). Let ϕ: ℂ × M → M; (t, p) ↦ ϕ^{t}(p) denote the holomorphic ℂ-action generated by the vector field X.

Fix a point p ∈ M. The tangent space T_{p}M at the point p is a 3dimensional complex vector space. Let P_{p}X be a set of complex 2-dimensional subspaces ν_{p} of T_{p}M that satisfy X_{p} ∈ ν_{p}. Clearly P_{p}X ≈ ℂP^{1}, so the bundle PX = ⋃_{p∈M} P_{p}X is a ℂP^{1}-bundle over M. The bundle PX naturally bears a holomorphic structure and the contact structure ξ is a holomorphic section of the holomorphic bundle PX.

The holomorphic ℂ-action ϕ on the base space M induces a holomorphic ℂ-action Φ on the bundle PX in the following way. For ν_{p} = span_{ℂ}{X_{p}, v_{p}} ∈ P_{p}X ⊂ PX, we define Φ^{t}(ν_{p}) ∈ P_{ϕ^{t}(p)}X by span_{ℂ}{dϕ^{t}(X_{p}), dϕ^{t}(v_{p})}. This ℂ-action on PX is well-defined and holomorphic.

Lemma 10 The ℂP^{1}-bundle PX over M is trivial.

Proof. Let ξ_{t} be a pull-back of the contact structure ξ by a holomorphic

diffeomorphism ϕ^{−t}: M → M. In other words, ξ_{t} = Φ^{t}(ξ). For any point

p ∈ M, we define a holomorphic map A_{p}: ℂ → P_{p}X by t ↦ ξ_{t}|_{p}. Since the

plane field ξ is contact, the derivative \frac{d}{dt}A_{p} does not vanish. We can take a

sufficiently small number ε, so that at any point p ∈ M, three 2-planes ξ|_{p},

ξ_{ε}|_{p} and ξ_{2ε}|_{p} are transverse to each other.

Fix three distinct points b_{0}, b_{1} and b_{2} ∈ ℂP^{1}. The biholomorphic map

I_{p}: P_{p}X → ℂP^{1} satisfying

ξ|_{p} ↦ b_{0}, ξ_{ε}|_{p} ↦ b_{1}, ξ_{2ε}|_{p} ↦ b_{2}

is uniquely defined. We define a bundle isomorphism I: PX → M × ℂP^{1} by

I(ν_{p}) = (p, I_{p}(ν_{p})) for any 2-plane ν_{p} ∈ P_{p}X.

Lemma 11 The ℂ-action Φ on the ℂP^{1}-bundle PX splits. In other words,

there exist a holomorphic ℂ-action ψ on ℂP^{1} such that I ∘ dϕ^{τ} ∘ I^{−1}(p, x) =

(ϕ^{τ}(p), ψ^{τ}(x)).

Proof. Put Φ^{t}(p, x) = (ϕ^{t}(p), Ψ^{t}(p, x)). Since the map Ψ^{t}(p, ·) (t ∈ ℂ) is a holomorphic ℂ-action on ℂP^{1}, it is enough to show that Ψ^{t}(p, x) does not depend on the choice of p ∈ M.

Fix a point x ∈ ℂP^{1}. Let us consider a map f^{t}: M → ℂP^{1}, defined by f^{t}(p) = Ψ^{t}(p, x). Obviously f^{0}(p) = x, thus f^{0} : M → ℂP^{1} is a fixed-point map. Since M is compact, f^{t} is a fixed point map for any t ∈ ℂ because of the maximum principle. This means that Ψ^{t}(p, x) does not depend on the choice of p ∈ M.

Since ψ is a holomorphic ℂ-action on ℂP^{1}, it is represented by an element of sl(2, ℂ) = Lie(Aut(ℂP^{1})). An element of sl(2, ℂ) is conjugate to either

( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ), ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) or ( \begin{matrix}a/2 & 0 \\ 0 & −a/2 \end{matrix} ) (a ≠ 0).

If the element is ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ), it means that the action ψ is trivial and it contradicts the proof of the lemma 10.

We say the Legendrian vector field is parabolic (resp. loxodromic) if the element is ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) (resp. ( \begin{matrix}a/2 & 0 \\ 0 & −a/2 \end{matrix} ) (a ≠ 0)). For each case, the following lemmas give the local structures.

Lemma 12 If the action ψ is represented by the conjugate class of ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ), there is an open atlas U = {U = {(x, y, z)}} such that on each coordinate open set U, ξ = ker(dy − xdz) and X = ∂/∂x.

Proof. For each t ∈ ℂ, the map ψ^{t}: ℂP^{1} → ℂP^{1} is conjugate to

exp(t ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} )) = ( \begin{matrix}1 & t \\ 0 & 1 \end{matrix} ) ,

so it has one fixed point b_{0} ∈ ℂP^{1}. Define a holomorphic section ν ∈ PX ≈ M × ℂP^{1} by ν = M × {b_{0}}. Since ν is invariant under the ℂ-action of Φ, it means that ν is holomorphic plane field, invariant under the action of ϕ and X ⊂ ν. So it defines a holomorphic foliation and locally denoted by ν = ker dz.

On a small open set U, we choose a coordinate (x, y, z) so that ν = ker dz and X = ∂/∂x. Let f(x, y, z) be a holomorphic map U → ℂ so that ξ = ker ω where ω = dy + f(x, y, z)dz.

By definition,

Φ^{t}(ξ) = ker(dy + f(x − t, y, z)dz)

where f(x − t, y, z) is defined. Fix a point p = (x_{0}, y_{0}, z_{0}). The action of ψ^{t} on P_{p}X ≈ ℂP^{1} is denoted by

\begin{cases}[0 : 1] ↦ [0 : 1], \\ [1 : f(x_{0}, y_{0}, z_{0})] ↦ [1 : f(x_{0} − t, y_{0}, z_{0})]. \end{cases}

Since this action is conjugate to ( \begin{matrix}1 & t \\ 0 & 1 \end{matrix} ), we have

f(x_{0} − t, y, z) = a(y, z)t + b(y, z)

where a and b are holomorphic functions defined on π(U) ⊂ ℂ^{2}. In this notation, π: ℂ^{3} → ℂ^{2} is the projection (x, y, z) ↦ (y, z).

We can assume x_{0} = 0 to obtain ω = dy+(−a(y, z)x+b(y, z))dz on U. Let ω̃ = ω/a(y, z), so that ξ is denoted also by ker ω̃. Let us take a holomorphic function g: ℂ^{2} → ℂ; (y, z) ↦ g(y, z) that satisfies g_{y}(y, z) = 1/a(y, z).

We set a new coordinate (x̃, ỹ, z) on U by

\begin{cases}x̃ = x − \frac{b(y,z)}{a(y,z)} + g_{z}(y, z), \\ ỹ = g(y, z). \end{cases}

Since dy = a(y, z)dỹ − a(y, z)g_{z}(y, z)dz, it is easy to check

ω̃ = \frac{dy}{a(y, z)} − (x − \frac{b(y, z)}{a(y, z)}) dz

= dỹ − (x − \frac{b(y, z)}{a(y, z)} + g_{z}(y, z)) dz

= dỹ − x̃dz.

We also have X = ∂/∂x̃.

Lemma 13 If the action ψ is represented by the conjugate class of ( \begin{matrix}a/2 & 0 \\ 0 & −a/2 \end{matrix} ), there is an open atlas U = {U = {(x, y, z)}} such that on each coordinate open set U, ξ = ker(dy − exp(ax)dz) and X = ∂/∂x.

Proof. For each t ∈ ℂ, the map ψ^{t}: ℂP^{1} → ℂP^{1}, which is conjugate to

exp (t (\begin{matrix}a/2 & 0 \\ 0 & −a/2 \end{matrix})) = (\begin{matrix}e^{at/2} & 0 \\ 0 & e^{−at/2} \end{matrix}) ,

has two fixed point b_{0}, b_{1} ∈ ℂP^{1}. Consider holomorphic sections ν_{0} = M × {b_{0}} and ν_{1} = M × {b_{1}} ⊂ PX.

On a small open set U, we choose coordinate (x, y, z) so that ν_{0} = ker dy, ν_{1} = ker dz and X = ∂/∂x. Let f be a holomorphic map U → ℂ^{∗} so that ξ = kerω where ω = dy + f(x, y, z)dz. Then we have

Φ^{t}(ξ) = ker(dy + f(x − t, y, z)dz)

where f(x − t, y, z) is defined. Fix a point p = (x_{0}, y_{0}, z_{0}). The action of ψ^{t} on P_{p}X ≈ ℂP^{1} is denoted by

\begin{cases}[0 : 1] ↦ [0 : 1], \\ [1 : 0] ↦ [1 : 0], \\ [1 : f(x_{0}, y_{0}, z_{0})] ↦ [1 : f(x_{0} − t, y_{0}, z_{0})]. \end{cases}

Since this action is conjugate to ( \begin{matrix}exp(at/2) & 0 \\ 0 & exp(−at/2) \end{matrix} ), we have

f(x_{0} − t, y, z) = b(y, z) e^{±at}

where b: π(U) → ℂ^{∗} is a holomorphic function. Because of the holomorphicity, the double sign is uniquely determined (say it is minus). We can assume x_{0} = 0, so ω = dy +b(y, z) exp(ax)dz on U. We set a new coordinate (x̃, y, z) on U by

x̃ = x − \frac{log(−b(y, z))}{a}.

Then ω = dy − exp(ax̃)dz and X = ∂/∂x̃.

The lemma 12 and the lemma 13 show the theorem 1.

In the next section, we will investigate the case that the contact structure is defined by a global holomorphic 1-form ω. In advance of that, it is important to introduce some examples of contact structures that admit Legendrian vector fields, but are not defined by global holomorphic 1-forms.

Example 14 Suppose a matrix A = ( \begin{matrix}λ_{1} & 0 \\ 0 & λ_{2} \end{matrix} ) ∈ GL(2, ℂ) preserves a lattice Λ ⊂ ℂ^{2} and |λ_{1}|, |λ_{2}| ≠ 1. Let Ā be the corresponding diffeomorphism of a complex torus N = ℂ^{2}/Λ. Choose n ∈ ℤ and η ∈ ℂ so that exp(nη) = λ_{2}/λ_{1} and exp(mη) ≠ λ_{1} for any m ∈ ℤ.

An equivalence relation in ℂ × N is defined by

(x, [y, z]) ∼ (x − η, Ā([y, z])) ∼ (x − 2πi, [y, z])

and we define a manifold M by M = ℂ × N/∼. The vector field X = ∂/∂x is well-defined on M. The universal cover M̃ of M is isomorphic to ℂ^{3}. Let ω̃ = dy − exp(nx)dz be a holomorphic contact structure on M̃. The contact structure ξ = ker ω̃ is well-defined on M, thus X is a Legendrian flow of the contact manifold (M, ξ). It is easy to check that Legendrian flow X is loxodromic.

Proposition 15 In this example, there is no global holomorphic 1-form ω such that ξ = ker ω.

Proof. Let Y be an arbitrary holomorphic vector field on M and Ỹ = f(x, [y, z])dx + g(x, [y, z])dy + h(x, [y, z])dz be the pull-back of Y on ℂ × N. Since N is compact, f, g and h do not depend on [y, z]. We denote f(x, [y, z]) by f(x) and so on.

Because of the definition of M, it follows that

f(x) = f(x − η) = f(x − 2πi)

g(x) = \frac{1}{λ_{1}} g(x − η) = g(x − 2πi)

h(x) = \frac{1}{λ_{2}} h(x − η) = h(x − 2πi).

The solution of the difference equation f(x) = f(x − 2πi) is given by

f(x) = ∑_{l∈ℤ}a_{l}e^{lx}.

Remark: {a_{l}} must satisfy the Cauchy-Hadamard condition

lim sup_{l→∞} \sqrt[l]{|a_{l}|} = lim sup_{l→∞} \sqrt[l]{|a_{−l}|} = 0.

In the rest of this paper, we omit this condition.

Substituting this into f(x) = f(x − η), we obtain a_{l} = 0 for l ≠ 0. Thus f(x) = a_{0}. Similarly,

g(x) = ∑_{l∈ℤ}b_{l}e^{lx}.

Substituting this into g(x) = (1/λ_{1})g(x − η), we obtain

∑_{l∈ℤ}b_{l}e^{lx} = ∑_{l∈ℤ}b_{l} \frac{1}{λ_{1}e^{lη}} e^{lx}.

Since λ_{1}e^{lη} ≠ 1 for any l ∈ ℤ, b_{l} = 0 for any l.

Thus we have

f = a_{0}, g = 0, h = 0.

So, Y = a_{0}X.

Assume that there exists a holomorphic contact 1-form ω so that ξ = ker ω. Let Y be the Reeb vector field of ω, then Y = a_{0}X and ω(Y) = 0. This is contradiction.

Example 16 Let a^{+}, a^{−}, λ and η be non-zero complex numbers and n be a positive integer satisfying that Re η > 0, |λ| > 1 and a^{−}/a^{+} = exp(nη).

Let

M̃ = {(x, y, z) ∈ ℂ^{3} | (y, z) ≠ (0, 0)} = ℂ × (ℂ^{2} ∖ {(0, 0)})

and define a manifold M by M̃/ ∼, where ∼ is an equivalence relation defined by

(x, y, z) ∼ (x, λy, λz) ∼ (x − η, a^{+}y, a^{−}z) ∼ (x − 2πi, y, z).

The vector field X = ∂/∂x and the contact structure ξ = ker(dy − exp(nx)dz) on M̃ are well-defined on M, thus X is a Legendrian vector field of (M, ξ).

To prove that there is no holomorphic 1-form ω on M such that ξ = ker ω, we show the following lemma.

Lemma 17 There is no holomorphic function

f: ℂ^{2} ∖ {(0, 0)} → ℂ ∖ {0}

such that f(λy, λz) = λf(y, z) for any (y, z) ∈ ℂ^{2} ∖ {(0, 0)}, where λ is a complex number satisfying |λ| > 1.

Proof. Let S = {(y, z) | y ≠ 0, z = 0} ⊂ ℂ^{2}∖{(0, 0)} and g: S → ℂ∖{0} be the restriction of f. Let γ^{r}: [0, 1] → S be a path defined by γ^{r}(t) = exp(2πit). Since the path γ is null-homotopic in ℂ^{2} ∖ {(0, 0)}, the image g(γ^{r}) is nullhomotopic in ℂ ∖ {0}.

Fix a path γ^{l}: [0, 1] → S such that γ^{l}(0) = 1 and γ^{l}(1) = λ. The image g(γ^{l}) is a path in ℂ ∖ {0} so that g(γ^{l})(1) = λg(γ^{l})(0). Let us consider the universal covering map exp: ℂ → ℂ ∖ \{0\}. Let { \atop g(γ^{r})} and { \atop g(γ^{l})} be some lift of g(γ^{r}) and g(γ^{l}). Since the path g(γ^{r}) is null-homotopic in ℂ ∖ {0}, the initial and end points of { \atop g(γ^{r})} are the same.

Let η = { \atop g(γ^{l})}(1) − { \atop g(γ^{l})}(0). Obviously, exp η = λ. Let T = S/ ∼; y ∼ λy and A = ℂ/ ∼; x ∼ x + η.

π_{1}(T) ≅ ℤ ⊕ ℤ is generated by the homotopy classes of γ^{r} and γ^{l}. Both of these paths define closed paths on A, thus the holomorphic map g: S → ℂ∖{0} induces the holomorphic map g̃: T → A. Since T is compact and A is non-compact, the map g̃ is constant because of the maximum principal. Thus the function g is also constant, but it contradicts the assumption g(λy) = λg(y).

Proposition 18 In the example 16, there is no holomorphic 1-form ω on M such that ξ = ker ω.

Proof. Suppose that there is a holomorphic 1-form ω on M such that ξ = ker ω. Let ω̃ = f(x, y, z)(dy − exp(nx)dz) be the pull-back of ω on ℂ × (ℂ^{2} ∖ {(0, 0)}. By the definition of the manifold M,

f(x, y, z) = λf(x, λy, λz).

Let g(y, z) = 1/f(0, y, z). It follows that λg(y, z) = g(λy, λz). This contradicts the previous lemma.

3 If there is a global contact form

In this section, we suppose that the contact structure ξ is defined by a global holomorphic 1-form ω on M. We denote this contact manifold by (M, ω).

The Reeb vector field Y of ω is the holomorphic vector field satisfying ω(Y) = 1 and Y ⌋dω = 0. The second condition is equivalent to L_{Y} ω = 0, which means that the contact form ω is invariant under the ℂ-action generated by Y.

Here we state the proposition 2 again.

Proposition 2 The Legendrian vector field X and the Reeb vector field Y satisfy one of the following conditions;

(i) [X, Y] = Z and TM = ℂX ⊕ ℂY ⊕ ℂZ.

(ii) [X, Y] = cX, where c is a nonzero complex number.

(iii) [X, Y] = 0.

Before giving the proof, here is a very important notice.

Remark 19 Since M is compact and ℂ is open, any holomorphic map f: M → ℂ is a constant map because of the maximum principal.

This trivial fact will be used in the rest of paper repeatedly, and this is the reason why our theory does not works on another (for example, real analytic) category.

proof of the proposition 2. First of all, we point out that ω([X, Y]) = Xω(Y)− Y ω(X) − dω(X, Y) = 0. Let α = X⌋dω. It follows that α(X) = α(Y) = 0 and α ≠ 0 anywhere, so the plane field ker α coincides with ℂX ⊕ ℂY. Let k satisfy α ∧ dα = k ω ∧ dω. Since holomorphic functions on the closed manifold M are constant, so is k. If k ≠ 0, the bracket [X, Y] is not included in ℂX ⊕ ℂY anywhere. This is case (i). If k = 0, [X, Y] ⊂ ℂX ⊕ ℂY. Since [X, Y] also belongs to ker ω, we obtain [X, Y] = cX. Again c is constant, thus (ii) or (iii) holds.

Before giving the classification for each case, we show that any holomorphic ℂ-action is volume preserving.

Lemma 20 For any holomorphic action f: ℂ × M → M, the non-vanishing

(3, 0)-form ω ∧ dω is invariant under the action of f.

Proof. Let ω_{t} = (f^{−t})^{∗}ω and ω_{t} ∧ dω_{t} = g(t)ω ∧ dω. Since the (3, 0)-form

ω ∧ dω is non-vanishing, it defines a volume form

dvol = \sqrt{−1} ω ∧ dω ∧ \overline{ω ∧ dω}.

Clearly, (f^{−t})^{∗}(dvol) = |g(t)|^{2} dvol. Integrating both volume forms on M, we

obtain

∫_{M} dvol = |g(t)|^{2} ∫_{M} dvol,

so |g(t)| = 1. Since g: ℂ → {|z| = 1} ⊂ ℂ is holomorphic and g(0) = 1, it is

constant equal to 1. It follows that ω_{t} ∧ dω_{t} = ω ∧ dω for any t.

3.1 Case (i); [X, Y ] = Z

The space of the holomorphic vector fields

A = {aX + bY + cZ | (a, b, c) ∈ ℂ^{3}}

is a 3-dimensional Lie algebra.

Let [X, Y] = Z, [Y, Z] = aX + bY + cZ and [Z, X] = dX + eY + fZ. Since ker ω = ℂX ⊕ ℂZ is a contact plane field, [X, Z] ∉ ℂX ⊕ ℂZ, so e ≠ 0. Since the ℂ-action generated by Y preserves the contact plane field ker ω = ℂX ⊕ ℂZ, [Z, Y] ∈ ℂX ⊕ ℂZ, that is b = 0. Since X (resp. Y) preserves a volume form, f = 0 (resp. c = 0)

By the Jacobi identity,

[[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y]
= [Z, Z] + [aX, X] + [dX + eY, Y]
= dZ = 0.

So, d = 0. Thus [X, Y] = Z, [Y, Z] = aX and [Z, X] = eY where e ≠ 0. If a = 0, the A is isomorphic to the Lie algebra solv and the universal

cover M̃ of M is isomorphic to the Lie group Solv. This is given in the example 4-1.

If a ≠ 0, the Lie algebra A is isomorphic to the Lie algebra sl(2, ℂ) and the universal cover M̃ is isomorphic to SL(2, ℂ). This is given in the example 3-1.

3.2 Case (ii); [X, Y ] = cX

In this subsection we investigate the case [X, Y] = cX. Normalizing the contact form ω by ω̃ = cω, the Reeb vector field Ỹ of ω̃ equals to (1/c)Y and [X, Ỹ] = X.

Suppose that [X, Y] = X. Let g^{α} be a holomorphic ℂ-action generated by Y. It follows that (g^{α})_{∗}X = exp(α)X and (g^{α})_{∗}Y = Y. Étienne Ghys and Alberto Verjovsky showed the following proposition by using Anosov property of the action g^{α}.

Proposition 21 ([GV]) There uniquely exists a holomorphic line field E^{s}

such that it is transverse to the plane field ℂX ⊕ ℂY and invariant under

the action of (g^{α})_{∗}.

We can define a complex vector field Z so that it satisfies E^{s} = ℂZ and ω ∧ dω(X, Y, Z) = 1. Clearly, the vector field Z is well-defined and holomorphic.

The bracket [Y, Z] is tangent to ℂZ, because the line field ℂZ is invariant under the action of (g^{α})_{∗}. Since (g^{α})_{∗} preserves the volume and [Y, X] = −X, the bracket [Y, Z] equals to Z. By the Jacobi identity,

[[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y]
= [X, Z] + [Z, X] + [[Z, X], Y]
= [[Z, X], Y] = 0.

Thus [X, Z] = aY.

Suppose that ker ω = ℂX ⊕ℂ(bY +Z). Since ω is invariant by the action of g^{α}, [Y, bY + Z] ∈ ℂX ⊕ ℂ(bY + Z). [Y, bY + Z] = Z, so b = 0 and ker ω = ℂX ⊕ ℂZ. We conclude a ≠ 0 because ℂX ⊕ ℂZ defines contact plane field. Thus the space of holomorphic vector fields

A = {aX + bY + cZ | (a, b, c) ∈ ℂ^{3}}

is isomorphic to sl(2, ℂ). This is given in the example 3-2.

3.3 Case (iii); [X, Y ] = 0

First of all, we will complete the case so that M is a homogeneous space of a 3-dimensional Lie group. Suppose that Z is a holomorphic vector field on M such that ℂX ⊕ ℂY ⊕ ℂZ = TM. Replacing Z by k(Z − (ω(Z)Y)), we can assume that ω(Z) = 0 and ω∧dω(X, Y, Z) = 1 without loss of generality. Let A be the space of holomorphic vector fields on M, which is spanned by X, Y and Z. Suppose that [Y, Z] = aX + bY + cZ and [Z, X] = dX + eY + fZ. 1 = ω ∧ dω(X, Y, Z) = dω(Z, X) = −ω([Z, X]) = −e, thus e = −1. Since the action generated by Y preserves the contact plane field ℂX ⊕ ℂZ, b = 0. Since the actions generated by X, Y and Z preserve the volume, it follows that f = 0, c = 0 and d = 0. Thus we obtain [Y, Z] = aX and [Z, X] = −Y.

If a = 0, A ≅ nil and the universal cover M̃ of M is isomorphic to the Lie group Nil. This is given in the example 5.

If a ≠ 0, A ≅ solv and the universal cover M̃ is isomorphic to Solv. This is given in the example 4-2.

Thus we have shown that if M is a homogeneous space, it is either one of the examples 3-1, 3-2, 4-1, 4-2 or 5.

Let us consider the general case. We begin by investigating the local structure.

Proposition 22 There is an open atlas U = {U = {(x, y, z)}} such that on

each open set U, ω = dy − xdz, X = ∂/∂x and Y = ∂/∂y.

3.3.1 proof of the proposition 22

Let us define a holomorphic 1-form ω_{t} by ω_{t} = (ϕ^{−t})^{∗}ω for t ∈ ℂ. The Reeb vector field of ω_{t} equals to the vector field ϕ^{t}_{∗}Y = Y.

Lemma 23 Any Legendrian vector field X of (M, ω) is parabolic.

Proof. Assume that a Legendrian vector field X is loxodromic. Take a suffi-

ciently small number ε so that 1-forms ω and ω_{ε} are linearly independent. We

can write η_{0} = ker(a_{0}ω+b_{0}ω_{ε}) and η_{1} = ker(a_{1}ω+b_{1}ω_{ε}) where a_{0}b_{1}−b_{0}a_{1} ≠ 0.

Thus we can take two holomorphic 1-forms α, β and holomorphic functions

f, g such that η_{0} = ker α, η_{1} = ker β, ω_{t} = f(t)α+g(t)β and f(0) = g(0) = 1.

Since the Legendrian vector field X is loxodromic,

f(t)/g(t) = exp(at).

ω_{t} ∧ dω_{t} = f(t)g(t)(α ∧ dβ + β ∧ dα) = f(t)g(t)ω ∧ dω, so

f(t)g(t) = 1 for any t.

Thus f(t) = exp(at/2) and g(t) = exp(−at/2). Since Y is the Reeb vector

field of ω_{t},

f(t)α(Y) + g(t)β(Y) = 1.

Since α(Y) and β(Y) are constants, it is contradiction.

That is why the Legendrian vector field X is parabolic. Although we have already known that X is parabolic, here we repeat a similar argument. Let us take a holomorphic 1-form α and holomorphic functions f, g so that η = ker α, ω_{t} = f(t)α + g(t)ω, f(0) = 0 and g(0) = 1.

Since the Legendrian vector field X is parabolic, f(t)/g(t) = bt for some non-zero constant b. By replacing α by bα, we can assume that

f(t)/g(t) = t.  (1)

ω_{t} ∧ dω_{t} = f(t)g(t)(dα ∧ ω + α ∧ dω) + g(t)^{2}ω ∧ dω. Take a constant k such that dα ∧ ω + α ∧ dω = k ω ∧ dω, we have

k f(t)g(t) + g(t)^{2} = 1.  (2)

Since Y is the Reeb vector field of ω_{t},

f(t)α(Y) + g(t) = 1.  (3)

The solution of this system of equations (1), (2) and (3) is given by g(t) = 1, f(t) = t, α(Y) = 0 and k = 0.

Let us start the calculation again. Let U = {(x, y, z)} be a coordinate neighborhood such that X = ∂/∂x, Y = ∂/∂y, α = f(x, y, z)dz and ω = dy + g(x, y, z)dz. Substituting

α = f(x, y, z)dz
ω = dy + g(x, y, z)dz
ω_{t} = dy + g(x − t, y, z)dz

into ω_{t} = ω + tα, we obtain

dy + g(x − t, y, z)dz = dy + (g(x, y, z) + tf(x, y, z))dz.

Thus f(x, y, z) = f(0, y, z) and g(x, y, z) = g(0, y, z) − xf(0, y, z), so we have

dα = f_{y}(0, y, z)dy ∧ dz

and

dω = −f(0, y, z)dx ∧ dz + (g_{y}(0, y, z) − xf_{y}(0, y, z))dy ∧ dz.

Since Y ⌋dω = 0, g_{y}(0, y, z) − xf_{y}(0, y, z) = 0 everywhere on U. Thus g_{y}(0, y, z) = f_{y}(0, y, z) = 0. Let f(z) = f(0, 0, z) and g(z) = g(0, 0, z). It follows that α = f(z)dz and ω = dy + (g(z) − xf(z))dz.

Let us take a new coordinate U = {(x̃, y, z̃)} in the following manner. First, define z̃ so that dz̃ = f(z)dz, we have α = dz̃ and ω = dy+(g̃(z̃)−x)dz̃. Let x̃ = x − g̃(z̃), we got ω = dy − x̃dz̃ and X = ∂/∂x̃ with respect to the new coordinate (x̃, y, z̃). The proposition 22 is proved.

Thus we got the local structure. We are going to investigate the global structure of (M, ω, X).

Proposition 24 The universal cover M̃ of M is a ℂ^{2}-bundle over ℂ and the

plane field ℂX ⊕ ℂY is tangent to the fibers ℂ^{2}.

3.3.2 proof of the proposition 24

Fix a point p ∈ M. Let Ω be the set of paths γ: [0, 1] → M such that γ(0) = p. Obviously,

M̃ = Ω/homotopy .

Define a map π: M̃ → ℂ by

π([γ]) = ∫_{γ} α .

It is enough to show that this map is well-defined, surjective and each fiber is isomorphic to ℂ^{2}.

Take a finite open atlas U = {U_{1}, …, U_{n}} so that on each open set U_{i}, there is a coordinate (x, y, z_{i}) satisfying X = ∂/∂x and ω = dy − xdz_{i}.

On every coordinate open set U_{i} = {(x, y, z_{i})},

∫_{C} α = ∫_{C} dz_{i} = 0

for any closed curve C. It means that

∫_{γ} α = 0

for any null-homotopic loop γ. So the map π is well-defined.

On each open set U_{i}, define a holomorphic vector field Z_{i} by Z_{i} = ∂/∂z_{i}. Let {ϕ_{i}: U_{i} → ℝ} be a partition of unity with respect to the open covering {U_{1}, …, U_{n}}. Define a complex vector field Z by Z = ∑_{i} ϕ_{i}Z_{i}.

On each non-empty intersection U_{i} ∩ U_{j}, take holomorphic functions f_{ij} and g_{ij} so that Z_{j} = f_{ij}∂/∂x + g_{ij}∂/∂y + ∂/∂z_{i}. Obviously, f_{ii} = g_{ii} = 0. Since the vector field Z_{j} is commutative with X = ∂/∂x and Y = ∂/∂y, f_{ij} and g_{ij} depend only on z_{i}. The vector field Z is denoted by

Z = ∑_{j=1}^{n} ϕ_{j}(x, y, z_{i})f_{ij}(z_{i})\frac{∂}{∂x} + ∑_{j=1}^{n} ϕ_{j}(x, y, z_{i})g_{ij}(z_{i})\frac{∂}{∂y} + \frac{∂}{∂z_{i}}

on the open set U_{i}.

Let us consider the underlying real 6-dimensional C^{∞}-manifold, which we also denote by M, and denote the coordinates and functions by

x = x^{1} + \sqrt{−1} x^{2}, y = y^{1} + \sqrt{−1} y^{2}, z_{i} = z_{i}^{1} + \sqrt{−1} z_{i}^{2},
f_{ij} = f_{ij}^{1} + \sqrt{−1} f_{ij}^{2}, g_{ij} = g_{ij}^{1} + \sqrt{−1} g_{ij}^{2}.

Let J: TM → TM be an almost complex structure defined by complex structure of M. Six real vector fields X, JX, Y, JY, Z and JZ form a frame of T M. On the open set U_{i}, they are denoted by

X = \frac{∂}{∂x^{1}} , JX = \frac{∂}{∂x^{2}} , Y = \frac{∂}{∂y^{1}} , JY = \frac{∂}{∂y^{2}} ,

Z = ∑_{j=1}^{n} ϕ_{j}(x, y, z_{i})f_{ij}^{1}(z_{i})\frac{∂}{∂x^{1}} + ∑_{j=1}^{n} ϕ_{j}(x, y, z_{i})f_{ij}^{2}(z_{i})\frac{∂}{∂x^{2}}

+ ∑_{j=1}^{n} ϕ_{j}(x, y, z_{i})g_{ij}^{1}(z_{i})\frac{∂}{∂y^{1}} + ∑_{j=1}^{n} ϕ_{j}(x, y, z_{i})g_{ij}^{2}(z_{i})\frac{∂}{∂y^{2}}

+ \frac{∂}{∂z_{i}^{1}} ,

JZ = − ∑_{k=1}^{n} ϕ_{k}(x, y, z_{i})f_{ik}^{2}(z_{i})\frac{∂}{∂x^{1}} + ∑_{k=1}^{n} ϕ_{k}(x, y, z_{i})f_{ik}^{1}(z_{i})\frac{∂}{∂x^{2}}

− ∑_{k=1}^{n} ϕ_{k}(x, y, z_{i})g_{ik}^{2}(z_{i})\frac{∂}{∂y^{1}} + ∑_{k=1}^{n} ϕ_{k}(x, y, z_{i})g_{ik}^{1}(z_{i})\frac{∂}{∂y^{2}}

+ \frac{∂}{∂z_{i}^{2}} .

Consider a real C^{∞} splitting TM = E^{1} ⊕ E^{2} ⊕ E^{3}, where E^{1} = ℝZ, E^{2} = ℝJZ and E^{3} = ℝX ⊕ ℝJX ⊕ ℝY ⊕ ℝJY.

For any two paths γ and γ′ on M such that γ(1) = γ′(0), we denote their concatenation by γ · γ′;

γ · γ′(t) = \begin{cases}γ(2t) & (0 ≤ t ≤ 1/2) \\ γ′(2t − 1) & (1/2 ≤ t ≤ 1). \end{cases}

Remark 25 For any path γ, there exist a path

γ′ = γ_{1}^{1} · γ_{1}^{2} · γ_{1}^{3} · γ_{2}^{1} · γ_{2}^{2} · · · · · γ_{m}^{2} · γ_{m}^{3}

so that γ′ ≃ γ, γ′ is C^{0}-close to γ and each γ_{j}^{i} is Legendrian to E^{i}.

Let ∥ · ∥ be the hermitian metric on the original complex manifold M so that ∥X∥ = ∥Y ∥ = ∥Z∥ = 1 and X, Y, Z are perpendicular to each other. We regard ∥ · ∥ as a Riemann metric on the underlying manifold M.

Lemma 26 Suppose that a path γ = γ^{3} ·γ^{1} is given so that γ^{1} is Legendrian to E^{1} and γ^{3} is Legendrian to E^{3}. Then there exist paths γ̃^{1} and γ̃^{3} so that γ ≃ γ̃^{1} · γ̃^{3}, where γ̃^{1} is Legendrian to E^{1} and γ̃^{3} is Legendrian to E^{3}. Moreover, there exists a constant C that does not depend on γ so that

∥γ̃^{1}∥ ≦ ∥γ^{1}∥
∥γ̃^{3}∥ ≦ ∥γ^{3}∥ exp(C∥γ^{1}∥)

where ∥γ∥ denotes the length of path γ.

Proof. It is enough to calculate [X, Z]. On each open set U_{i},

\begin{aligned}[X, Z] & = ∑_{j=1}^{n} \frac{∂ϕ_{j}}{∂x^{1}} (x, y, z_{i})f_{ij}^{1}(z_{i})\frac{∂}{∂x^{1}} + ∑_{j=1}^{n} \frac{∂ϕ_{j}}{∂x^{1}} (x, y, z_{i})f_{ij}^{2}(z_{i})\frac{∂}{∂x^{2}} \\ & + ∑_{j=1}^{n} \frac{∂ϕ_{j}}{∂x^{1}} (x, y, z_{i})g_{ij}^{1}(z_{i})\frac{∂}{∂y^{1}} + ∑_{j=1}^{n} \frac{∂ϕ_{j}}{∂x^{1}} (x, y, z_{i})g_{ij}^{2}(z_{i})\frac{∂}{∂y^{2}} . \end{aligned}

Since f_{ij}^{1} is defined on U_{i} ∩ U_{j} and supp(ϕ_{j}|_{U_{i}}) ⊂ U_{i} ∩ U_{j}, the coefficients are well-defined C^{∞}-functions on U_{i}. Thus [X, Z] is denoted by linear combination of X, JX, Y and JY with coefficient C^{∞}-functions on each open set U_{i}, so it is on M.

[X, Z] = ψ^{1}X + ψ^{2}JX + ψ^{3}Y + ψ^{4}JY,

where ψ^{1}, …, ψ^{4} : M → ℝ are C^{∞}-functions. Since M is compact, ψ^{1}, …, ψ^{4} are bounded.

Similarly, the same lemma holds for the pair E^{2} and E^{3}, where we replace C by larger number if needed.

Lemma 27 Suppose that a path γ = γ^{2} ·γ^{1} is given so that γ^{1} is Legendrian to E^{1} and γ^{2} is Legendrian to E^{2}. Then there exist paths γ̃^{1}, γ̃^{2} and γ̃^{3} so that γ ≃ γ̃^{1} · γ̃^{2} · γ̃^{3}, where γ̃^{i} is Legendrian to E^{i}. Moreover, there exists a constant D that does not depend on γ so that

∥γ̃^{1}∥ ≦ ∥γ^{1}∥
∥γ̃^{2}∥ ≦ ∥γ^{2}∥
∥γ̃^{3}∥ ≦ D∥γ^{1}∥∥γ^{2}∥ exp(C(∥γ^{1}∥ + ∥γ^{2}∥)).

Proof. On each open set U_{i},

[Z, JZ] = (( − ∑_{j=1}^{n} ∑_{k=1}^{n} ϕ_{j}f_{ij}^{1} \frac{∂ϕ_{k}}{∂x^{1}} f_{ik}^{2} − ∑_{j=1}^{n} ∑_{k=1}^{n} ϕ_{j}f_{ij}^{2} \frac{∂ϕ_{k}}{∂x^{2}} f_{ik}^{2} − ∑_{j=1}^{n} ∑_{k=1}^{n} ϕ_{j}g_{ij}^{1} \frac{∂ϕ_{k}}{∂y^{1}} f_{ik}^{2}

− ∑_{j=1}^{n} ∑_{k=1}^{n} ϕ_{j}g_{ij}^{2} \frac{∂ϕ_{k}}{∂y^{2}} f_{ik}^{2} − ∑_{k=1}^{n} \frac{∂ϕ_{k}}{∂z^{1}} f_{ik}^{2} − ∑_{k=1}^{n} ϕ_{k} \frac{∂f_{ik}^{2}}{∂z^{1}} )

+ ( ∑_{j=1}^{n} ∑_{k=1}^{n} ϕ_{k}f_{ik}^{2} \frac{∂ϕ_{j}}{∂x^{1}} f_{ij}^{1} − ∑_{j=1}^{n} ∑_{k=1}^{n} ϕ_{k}f_{ik}^{1} \frac{∂ϕ_{j}}{∂x^{2}} f_{ij}^{1} + ∑_{j=1}^{n} ∑_{k=1}^{n} ϕ_{k}g_{ik}^{2} \frac{∂ϕ_{j}}{∂y^{1}} f_{ij}^{1}

− ∑_{j=1}^{n} ∑_{k=1}^{n} ϕ_{k}g_{ik}^{1} \frac{∂ϕ_{j}}{∂y^{2}} f_{ij}^{1} − ∑_{j=1}^{n} \frac{∂ϕ_{j}}{∂z^{2}} f_{ik}^{1} − ∑_{j=1}^{n} ϕ_{j} \frac{∂f_{ik}^{1}}{∂z^{2}} ))\frac{∂}{∂x^{1}}

+ ( ⋯)\frac{∂}{∂x^{2}} + ( ⋯)\frac{∂}{∂y^{1}} + ( ⋯)\frac{∂}{∂y^{2}} .

Since C^{∞}-functions f_{ij}^{1} (resp. f_{ik}^{2}) is defined on U_{i} ∩ U_{j} (resp. U_{i} ∩ U_{k}) and supp(ϕ_{j} \frac{∂ϕ_{k}}{∂x^{1}} |_{U_{i}}) ⊂ U_{i} ∩ U_{j} ∩ U_{k},

∑_{j=1}^{n} ∑_{k=1}^{n} ϕ_{j}f_{ij}^{1} \frac{∂ϕ_{k}}{∂x^{1}} f_{ik}^{2}

is well-defined C^{∞}-functions on U_{i}. The latter part

− ∑_{k=1}^{n} \frac{∂ϕ_{k}}{∂z^{1}} f_{ik}^{2} − ∑_{k=1}^{n} ϕ_{k} \frac{∂f_{ik}^{2}}{∂z^{1}}

is also well-defined C^{∞}-functions on U_{i} for the similar reason. Thus [Z, JZ] is denoted by linear combination of X, JX, Y and JY with coefficient C^{∞}- functions on each open set U_{i}, so it is on M.

[Z, JZ] = ψ^{1}X + ψ^{2}JX + ψ^{3}Y + ψ^{4}JY,

where ψ^{1}, …, ψ^{4} : M → ℝ are C^{∞}-functions. Since M is compact, functions ψ^{1}, …, ψ^{4} are bounded.

Getting these two lemmas together, we conclude the following proposition.

Proposition 28 Let

γ = γ_{1}^{1} · γ_{1}^{2} · γ_{1}^{3} · γ_{2}^{1} · γ_{2}^{2} · · · · · γ_{m}^{2} · γ_{m}^{3}

be a path so that each γ_{j}^{i} is Legendrian to E^{i}. Then there exist paths γ̃^{1}, γ̃^{2} and γ̃^{3} so that γ ≃ γ̃^{1} · γ̃^{2} · γ̃^{3} and γ̃^{i} is Legendrian to E^{i}. Moreover,

∥γ̃^{1}∥ ≦ l_{1}
∥γ̃^{2}∥ ≦ l_{2}
∥γ̃^{3}∥ ≦ (Dl_{1}l_{2} + l_{3}) exp(C(l_{1} + l_{2})))

where

\begin{aligned}l_{1} & = ∥γ_{1}^{1}∥ + ∥γ_{2}^{1}∥ + ⋯ + ∥γ_{m}^{1}∥ \\ l_{2} & = ∥γ_{1}^{2}∥ + ∥γ_{2}^{2}∥ + ⋯ + ∥γ_{m}^{2}∥ \\ l_{3} & = ∥γ_{1}^{3}∥ + ∥γ_{2}^{3}∥ + ⋯ + ∥γ_{m}^{3}∥. \end{aligned}

Because of the remark 25 and the proposition 28, for any path γ there exist three paths γ^{1}, γ^{2}, γ^{3} so that γ ≃ γ^{1} · γ^{2} · γ^{3} and γ^{i} is Legendrian to E^{i}. Obviously,

∫_{γ^{1}} α = Re (∫_{γ} α) , ∫_{γ^{2}} α = \sqrt{−1} Im (∫_{γ} α) , ∫_{γ^{3}} α = 0.

Suppose that there exist different three paths γ̃^{1}, γ̃^{2}, γ̃^{3} so that γ ≃ γ̃^{1} · γ̃^{2} · γ̃^{3} and γ̃^{i} is Legendrian to E^{i}. Two paths γ^{1} and γ̃^{1} are homotopic to each other because

∫_{γ^{1}} α = ∫_{γ^{1}} α,

E^{1} is real 1-dimensional and transverse to ker α. Similarly γ^{2} ≃ γ̃^{2}, thus γ^{3} ≃ γ̃^{3} and the initial point p′ of the path γ^{3} depends only on the complex number ∫_{γ} α.

Thus for any c ∈ ℂ, π^{−1}(c) is isomorphic to the space

{γ^{3} | γ^{3}(0) = p′ and γ^{3} is Legendrian to E^{3}}/homoptopy,

which is clearly ℂ^{2}. It is obvious that π: M̃ → ℂ is surjective, so the proposition 24 is proved.

3.4 some remarks on the case M̃ ≈ ℂ^{3}

Here we give some examples. Since we know that the universal cover M̃ is ℂ^{2}-bundle over ℂ, the most natural example is ℂ^{3}.

Suppose that M̃ ≈ ℂ^{3}, X = ∂/∂x and ω = dy − xdz. First of all, we investigate the structure of the group Aut(ℂ^{3}, ω, X) of automorphisms on M̃ ≈ ℂ^{3} that preserve ω and X. Let us take an arbitrary element γ ∈ Aut(ℂ^{3}, ω, X) such that

γ: (x, y, z) ↦ (γ^{1}(x, y, z), γ^{2}(x, y, z), γ^{3}(x, y, z)).

Since γ preserves the Legendrian vector field X = ∂/∂x and the Reeb vector field Y = ∂/∂y, it follows that

γ^{1}_{x} = 1, γ^{2}_{x} = 0, γ^{3}_{x} = 0,
γ^{1}_{y} = 0, γ^{2}_{y} = 1, γ^{3}_{y} = 0.

Since γ preserves α = dz, it follows that γ^{3}_{z} = 1. Thus we can write γ(x, y, z) = (x+f(z), y+g(z), z+C), where f and g: ℂ → ℂ are holomorphic functions and C ∈ ℂ is a constant. So we obtain that

\begin{aligned}γ^{∗}ω & = d(y + g(z)) − (x + f(z))d(z + C) \\ & = dy − (x + f(z) − g′(z))dz. \end{aligned}

Since γ preserves ω, it follows that f(z) = g′(z). Thus,

Aut(ℂ^{3}, ω, X) = {(f, C) | f ∈ O(ℂ), C ∈ ℂ},
(f, C): (x, y, z) ↦ (x + f′(z), y + f(z), z + C).

Thus, the proposition 8 is proved. Let us introduce some calculation in the group Aut(ℂ^{3}, ω, X). product The product is defined by the composition, namely

(g(z), D) ∘ (f(z), C) = (f(z) + g(z + C), C + D).

unit element The unit element of Aut(ℂ^{3}, ω, X) is obviously (0, 0). inverse The inverse element is given by

(f(z), C)^{−1} = (−f(z − C), −C).

conjugate

(g(z), D)^{−1} ∘ (f(z), C) ∘ (g(z), D) = (f(z + D) + g(z) − g(z + C), C).

commutator

\begin{matrix}(f(z), C)^{−1} ∘ (g(z), D)^{−1} ∘ (f(z), C) ∘ (g(z), D) \\ = (f(z + D) − f(z) + g(z) − g(z + C), 0). \end{matrix}

Here we state the theorem 9 again.

Theorem 9 Suppose Γ is a subgroup of Aut(ℂ^{3}, ω, X) such that Γ\ℂ^{3} is a closed 3-dimensional complex manifold. Then there exist a element σ ∈ Aut(ℂ^{3}, ω, X), an integer N ≤ 16 and complex numbers α_{1}, …, α_{N} such that for any (f, C) in the conjugate σ^{−1}Γσ of the subgroup, the function f is written as

f(z) = ∑_{k=1}^{N} r_{k}(z) exp(α_{k}z) + b_{0},

where r_{k} is a polynomial of degree 4 or less and b_{0} is a constant.

We are going to prove the theorem step by step. Let Γ be such a subgroup of Aut(ℂ^{3}, ω, X). Obviously, there exist elements (f_{1}, η_{1}), (f_{2}, η_{2}) ∈ Aut(ℂ^{3}, ω, X) such that η_{1}ℤ ⊕ η_{2}ℤ is a lattice on ℂ.

Lemma 29 For any (f, η) ∈ Aut(ℂ^{3}, ω, X) such that η ≠ 0, there exits a holomorphic function g: ℂ → ℂ so that

(−g, 0) ∘ (f, η) ∘ (g, 0) = (0, η).

Proof. The left hand side of the equation is (f(z) + g(z) − g(z + η), η), so it is enough to find a solution of the difference equation

g(z + η) − g(z) = −f(z).

It is classically known that this difference equation has solutions for any f (for example, see a paper of Guichard [Gu]).

Without loss of generality, we can assume (0, η_{1}), (f_{2}, η_{2}) ∈ Aut(ℂ^{3}, ω, X) such that η_{1}ℤ ⊕ η_{2}ℤ is a lattice on ℂ. The commutator

(0, η_{1})^{−1} ∘ (f_{2}, η_{2})^{−1} ∘ (0, η_{1}) ∘ (f_{2}, η_{2})

is equal to

(f_{2}(z) − f_{2}(z + η_{1}), 0)

.

Lemma 30 If (0, η_{1}) and (f_{2}, η_{2}) are commutative, there exists a holomorphic function g: ℂ → ℂ so that

\begin{aligned}(−g, 0) ∘ (0, η_{1}) ∘ (g, 0) & = (0, η_{1}) \\ (−g, 0) ∘ (f_{2}, η_{2}) ∘ (g, 0) & = (a, η_{2}), \end{aligned}

where a is a constant.

Proof. Since both (f_{2}, η_{2}) and (g, 0) are commutative to (0, η_{1}), it follows that f_{2}(z + η_{1} = f_{2}(z) and g(z + η_{1}) = g(z), thus

\begin{aligned}f_{2}(z) & = ∑_{n∈ℤ} a_{n} exp (\frac{2nπiz}{η_{1}} ) , \\ g(z) & = ∑_{n∈ℤ} b_{n} exp (\frac{2nπiz}{η_{1}} ) . \end{aligned}

(−g, 0) ∘ (f_{2}, η_{2}) ∘ (g, 0) is equal to (f_{2}(z) − (g(z + η_{2}) − g(z)), η_{2}). The first component is equal to

∑_{n∈ℤ} (a_{n} − b_{n} (exp (2nπi\frac{η_{2}}{η_{1}} ) − 1)) exp (\frac{2nπiz}{η_{1}} ) .

Since η_{1} and η_{2} span a lattice,

exp (2nπi\frac{η_{2}}{η_{1}} ) − 1

is zero if and only if n = 0. For any n ≠ 0 we can take b_{n} so that

a_{n} − b_{n} (exp (2nπi\frac{η_{2}}{η_{1}} ) − 1)

is equal to zero. Thus we obtain

f_{2}(z) − (g(z + η_{2}) − g(z)) = a_{0}.

We define the difference operator Δ_{1} and Δ_{2} by

Δ_{1}f(z) = f(z + η_{1}), Δ_{2}f(z) = f(z + η_{2}).

Lemma 31 Suppose that (0, η_{1}) and (f_{2}, η_{2}) ∈ Γ. If (f, 0) ∈ Γ, the function f is denoted by

f(z) = ∑_{k=1}^{N} p_{k}(z) exp(α_{k}z),

where p_{k} is a polynomial of degree 3 or less, N is an integer less than or equal to 16 and α_{k} ∈ ℂ.

Proof. If f = 0, the lemma is trivial. Suppose f ≠ 0. The commutator

\begin{aligned}(f, 0)^{−1} ∘ (0, η_{1})^{−1} ∘ (f, 0) ∘ (0, η_{1}) & = (f(z + η_{1}) − f(z), 0) \\ & = (Δ_{1}f, 0) \end{aligned}

is in Γ. Similarly, (Δ_{1}^{2}f, 0), (Δ_{1}^{3}f, 0) and (Δ_{1}^{4}f, 0) are in Γ. Since Γ\ℂ^{3} is a well-defined manifold; f, Δ_{1}f, Δ_{1}^{2}f, Δ_{1}^{3}f and Δ_{1}^{4}f have a nontrivial linear relation with all coefficients in ℤ;

(n_{4}Δ_{1}^{4} + n_{3}Δ_{1}^{3} + n_{2}Δ_{1}^{2} + n_{1}Δ_{1} + n_{0})f = 0,
(n_{4}, n_{3}, n_{2}, n_{1}, n_{0}) ∈ ℤ^{5} ∖ {(0, 0, 0, 0, 0)}.

Let P (x) = n_{d}x^{d} + ⋯ + n_{0} ∈ ℤ[x] be a non-zero polynomial satisfying P (Δ_{1})f = 0 such that it minimize the degree d. Clearly, d ≤ 4. Since (Δ_{1} − 1)g = 0 is equivalent to g = 0, the polynomial P (x) is not factored by x − 1. Let

P (x) = n_{d}(x − k_{1})^{e_{1}} ⋯ (x − k_{m})^{e_{m}}

be the factorization of P (x). Let us solve the difference equation

(Δ_{1} − k_{1})^{e_{1}} ⋯ (Δ_{1} − k_{m})^{e_{m}}f = 0.

For any k ≠ −1, the general solutions of the difference equation

(Δ_{1} − k)g = 0 ⟺ g(z + η_{1}) = (k + 1)f(z)

are given by

g(z) = ∑_{l∈ℤ}a_{l} exp (\frac{2lπi + log(k + 1)}{η_{1}} z) ,

where a_{l} ∈ ℂ.

A particular solution of the difference equation

(Δ_{1} − k)g = exp(αz)

is given by

g(z) = \begin{cases}\frac{1}{exp(αη_{1}) − (k + 1)} exp(αz) & if exp(αη_{1}) ≠ k + 1, \\ \frac{1}{η_{1}} z exp(αz) & if exp(αη_{1}) = k + 1. \end{cases}

Similarly, a particular solution of the difference equation

(Δ_{1} − k)g = z^{n} exp(αz)

is given by

g(z) = p(z) exp(αz)

where p(z) is a polynomial of degree n if exp(bη_{1}) ≠ k + 1, or of degree n + 1 if exp(bη_{1}) = k + 1.

Thus, the solution of the difference equation

(Δ_{1} − k_{1})^{e_{1}} ⋯ (Δ_{1} − k_{m})^{e_{m}}f = 0

is given by

f(z) = ∑_{j=1}^{m} ∑_{l∈ℤ}p_{jl}(z) exp (\frac{2lπi + log(k_{j} + 1)}{η_{1}} z) ,  (4)

where p_{jl}(z) is a polynomial of degree e_{j} − 1 or less.

By the way, the commutator

(f, 0)^{−1} ∘ (f_{2}, η_{2})^{−1} ∘ (f, 0) ∘ (f_{2}, η_{2})

is equal to

(f(z + η_{2}) − f(z), 0) = (Δ_{2}f, 0).

So, we can execute the similar argument.

Let P̃(x) = n′_{d′}x^{d′} + ⋯ + n′_{0} ∈ ℤ[x] be a non-zero polynomial satisfying P̃(Δ_{2})f = 0 such that it minimize the degree d′. Let

P̃(x) = n′_{d′}(x − k′_{1})^{e′_{1}} ⋯ (x − k′_{m′})^{e′_{m′}}

be the factorization of P̃(x). The solution of the difference equation

(Δ_{2} − k′_{1})^{e′_{1}} ⋯ (Δ_{2} − k_{m′})^{e′_{m′}}f = 0

is given by

f(z) = ∑_{j′=1}^{m′} ∑_{l′∈ℤ}p′_{j′l′}(z) exp (\frac{2l′πi + log(k′_{j′} + 1)}{η_{2}} z) ,  (5)

where p′_{j′l′}(z) is a polynomial of degree e′_{j′} − 1 or less.

Let us compare (4) and (5). Since η_{1}/η_{2} ∉ ℝ, the equation

α = \frac{2lπi + log(k_{j} + 1)}{η_{1}} = \frac{2l′πi + log(k′_{j′} + 1)}{η_{2}},

where l, l′ ∈ ℤ, 1 ≦ j ≦ m and 1 ≦ j′ ≦ m′ has at most mm′ solutions. Remind that mm′ ≤ 16, thus f is written by the sum of finite (at most 16) terms. Here we denote the possible value of α by α_{1}, α_{2}, …, α_{N} (N ≤ 16). We finally obtain

f(z) = ∑_{k=1}^{N} p_{k}(z) exp(α_{k}z).

Lemma 32 Suppose that (0, η_{1}) and (f_{2}, η_{2}) ∈ Γ. Then there exists a holomorphic map g: ℂ → ℂ so that

\begin{aligned}(−g, 0) ∘ (0, η_{1}) ∘ (g, 0) & = (0, η_{1}), \\ (−g, 0) ∘ (f_{2}, η_{2}) ∘ (g, 0) & = (f̃_{2}, η_{2}), \end{aligned}

where

f̃_{2}(z) = ∑_{k=1}^{N} p̃_{k}(z) exp(α_{k}z) + a_{0},

where p̃_{k} is a polynomial of degree 4 or less and a_{0} ∈ ℂ is a constant.

Proof. Since (0, η_{1}) and (f_{2}, η_{2}) ∈ Γ, the commutator

\begin{aligned}(f_{2}, η_{2})^{−1} ∘ (0, η_{1})^{−1} ∘ (f_{2}, η_{2}) ∘ (0, η_{1}) & = (f_{2}(z + η_{1}) − f_{2}(z), 0) \\ & = (Δ_{1}f_{2}, 0) \end{aligned}

also belongs to Γ. Thus we can write

Δ_{1}f_{2}(z) = ∑_{k=1}^{N} p_{k}(z) exp(α_{k}z),

where N ≤ 16 and deg(p_{k}) ≤ 3.

By solving this difference equation, we obtain

f_{2}(z) = ∑_{k=1}^{N} q_{k}(z) exp(α_{k}z) + ∑_{l∈ℤ}a_{l} exp (\frac{2lπi}{η_{1}} z) ,

where q_{k} is a polynomial of degree 4 or less, α_{k} ∈ ℂ and a_{l} ∈ ℂ.

As in the lemma 30, let

g(z) = ∑_{n∈ℤ} b_{l} exp (\frac{2lπiz}{η_{1}} ) ,

where {b_{l}} are numbers so that

a_{l} − b_{l} (exp (2lπi\frac{η_{2}}{η_{1}} ) − 1) = 0

for any l ≠ 0. The calculation given in the lemma 30 shows that

\begin{aligned}(−g, 0) ∘ (0, η_{1}) ∘ (g, 0) & = (0, η_{1}), \\ (−g, 0) ∘ (f_{2}, η_{2}) ∘ (g, 0) & = (f̃, η_{2}), \end{aligned}

where

f̃_{2}(z) = ∑_{k=1}^{N} q_{k}(z) exp(α_{k}z) + a_{0}.

Thus we can assume, without loss of generality, that (0, η_{1}), (f_{2}, η_{2}) ∈ Γ such that η_{1}ℤ ⊕ η_{2}ℤ is a lattice on ℂ and

f_{2}(z) = ∑_{k=1}^{N} q_{k}(z) exp(α_{k}z) + a_{0},

where q_{k} is a polynomial of degree 4 or less and a_{0} is a constant. Lemma 33 Under the assumption above, any element (f, C) ∈ Γ has a following representation;

f(z) = ∑_{k=1}^{N} r_{k}(z) exp(α_{k}z) + b_{0},

where r_{k} is a polynomial of degree 4 or less and c is a constant. Proof. Since the commutator

(f, C)^{−1} ∘ (0, η_{1})^{−1} ∘ (f, C) ∘ (0, η_{1}) = (Δ_{1}f, 0) belongs to Γ,

f(z) = ∑_{k=1}^{N} r_{k}(z) exp(α_{k}z) + ∑_{l∈ℤ}b_{l} exp (\frac{2lπi}{η_{1}} z) ,

where r_{k} is a polynomial of degree 4 or less and b_{l} ∈ ℂ. We can assume b_{l} = 0 if \frac{2lπi}{η_{1}} = α_{k} for some k.

By the way, the commutator

(f, C)^{−1} ∘ (f_{2}, η_{2})^{−1} ∘ (f, C) ∘ (f_{2}, η_{2}) = (f_{2}(z) − f_{2}(z + C) + Δ_{2}f(z), 0)

also belongs to Γ.

f_{2}(z) − f_{2}(z + C) + Δ_{2}f(z)

= ∑_{k=1}^{N}((1 − exp(α_{k}C))q_{k}(z) + (exp(α_{k}η_{2}) − 1)q̃_{k}(z)) exp(α_{k}z)

+ ∑_{l∈ℤ}(1 − exp (\frac{2lπiη_{2}}{η_{1}} )) b_{l} exp (\frac{2lπi}{η_{1}} z) .

Because of the lemma 31,

(1 − exp (\frac{2lπiη_{2}}{η_{1}} )) b_{l} = 0

for any l. Since η_{1}/η_{2} ∉ ℝ, b_{l} = 0 for any l ≠ 0.

This lemma complete the proof of the theorem 9.

Example 34 Let Γ be the subgroup of Aut(ℂ^{3}, ω, X) defined by

\{(az + b, c) | a, b, c ∈ ℤ[\sqrt{−1}]\}.

Then Γ\(ℂ^{3}, ω, X) is a complex manifold with Legendrian vector field. In fact, this subgroup Γ preserves the vector field Z = x ∂/∂y + ∂/∂z, which is transverse to ℂX ⊕ ℂY. So this is an example of homogeneous spaces. Since [X, Y] = [Y, Z] = 0 and [X, Z] = Y, this is given in the example 5.

Example 35 Let n_{1} and n_{2} be integers such that |n_{1}| > 2 > |n_{2}|. Let α_{j} and β_{j} be solutions of the quadratic equation t^{2} − n_{j}t + 1 = 0 (j = 1, 2). Let η_{1} and η_{2} be complex numbers such that exp(η_{1}) = α_{1}, exp(η_{2}) = α_{2} and η_{1}ℤ ⊕ η_{2}ℤ is a lattice of ℂ.

For example,

n_{1} = 4, α_{1} = 2 + \sqrt{3},  β_{1} = 2 − \sqrt{3},  η_{1} = log (2 + \sqrt{3});
n_{2} = −1, α_{2} = exp (\frac{2πi}{3}) , β_{2} = exp (\frac{−2πi}{3}) , η_{2} = \frac{2πi}{3}.

Define four functions f_{0}, …, f_{3} by

\begin{aligned}f_{0}(z) & =  e^{z} +  e^{−z}, \\ f_{1}(z) & = α_{1}e^{z} + β_{1}e^{−z}, \\ f_{2}(z) & = α_{2}e^{z} + β_{2}e^{−z}, \\ f_{3}(z) & = α_{1}α_{2}e^{z} + β_{1}β_{2}e^{−z}. \end{aligned}

Let Γ be the subgroup of Aut(ℂ^{3}, ω, X) defined by

{(a_{0}f_{0} + a_{1}f_{1} + a_{2}f_{2} + a_{3}f_{3}, b_{1}η_{1} + b_{2}η_{2}) | a_{j}, b_{k} ∈ ℤ}.

This Γ actually forms a subgroup of Aut(ℂ^{3}, ω, X) because

f_{2}(z + η_{1}) = α_{1}α_{2}e^{z} + β_{1}β_{2}e^{−z}
= f_{3}(z),
f_{2}(z + η_{2}) = α_{2}^{2}e^{z} + β_{2}^{2}e^{−z}
= (n_{2}α_{2} − 1)e^{z} + (n_{2}β_{2} − 1)e^{−z}
= n_{2}f_{2}(z) − f_{0}(z)

and so on.

The action of this subgroup Γ preserves the vector field

Z = y \frac{∂}{∂x} + x\frac{∂}{∂y} + \frac{∂}{∂z} .

[X, Y] = 0, [X, Z] = Y, [Y, Z] = X. So this is given in the example 4-2.

References

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[Gh] E. Ghys, Holomorphic Anosov systems, Invent. Math., 119 (1995), No. 3, 584-614.

[Gu] M. C. Guichard, L’équation aux Différences Fines (French), Ann. de l’Éc. Normale, 3^{e} Série, Tome IV, 1887, 361-380.

[GV] E. Ghys and A. Verjovsky, Locally free holomorphic actions of the complex affine group, Proceedings of Geometric Study of Foliations, held in Tokyo, November 1993, World Scientific, Singapore, 1994, 201-217.

[Ye] Y. G. Ye, A note on complex projective threefolds admitting holomorphic contact structures, Invent. Math., 115 (1994), No. 2, 311-314.

email address: kodama@ms.u-tokyo.ac.jp

address: Graduate School of Mathematical Sciences, University of Tokyo, 3-8-1 Komaba, Meguro-ku, Tokyo 153-9814, Japan.