A local characterization of affine holomorphic immersions with an anti-complex and ∇-parallel shape operator.

Maria Robaszewska

preprint 2001/12

Abstract

We study the complex hypersurfaces f : M^{(n)} → C^{n+1} which together with their transversal bundles have the property that around any point of M there exists a local section of the transversal bundle inducing a ∇-parallel, anti-complex shape operator. We give a class of examples of such hypersurfaces with an arbitrary rank of S from 1 to [\frac{n}{2}] and show that every such hypersurface with tf > 0 and S ≠ 0 is locally of this kind, modulo an affine isomorphism of C^{n+1}.

2000 Mathematics Subject Classification: 53B05, 53B25.

Typeset by AMS-TEX

1. Introduction. Among the connections induced on complex hypersurfaces f : M^{(n)} → C^{n+1} by C^{∞} complex transversal bundles there are two particular kinds of great interest - a holomorphic one and an affine Kähler one. The latter is meant to be a generalisation of the Kähler connection. In terms of the curvature tensor, the holomorphic affine connection is characterized by the condition

R(JX, Y) = JR(X, Y) for all vector fields X, Y,

while for the affine Kähler connection we have, by definition,

R(JX, JY) = R(X, Y) for all X, Y [NS].

Since, if only the affine fundamental form h does not vanish on M, a holomorphic connection is induced by a holomorphic transversal bundle, it is possible to adapt the ideas from the real affine hypersurface geometry for this case. For instance, having a non-degenerate hypersurface one can construct a holomorphic analogue of affine normal vector field [DVV]. Then one can consider the condtion S = λI, λ = const, which describes the affine spheres [DVV].

On the contrary, the non-flat affine Kähler connections induced on hypersurfaces (in particular - the non-flat Kähler ones) can not be treated in this way. Instead of the holomorphic transversal bundles, with the complex shape operator, one has to consider the transversal bundles N having the property that the shape operator corresponding to sections of N is anti-complex. This property of N implies the desired condition for the curvature tensor R of ∇ and is necessary for ∇ to be affine Kähler if tf > 1 at some point of M [O].

Clearly, no section ξ of such a bundle can induce S proportional to the identity, except for the case S = 0. The ∇ - parallelity of S is a weaker condition than S = λI, λ = const. This condition is shown to have some non-trivial realisations even if we require S to be anti-complex. It is worth noting that we need to consider degenerate hypersurfaces, because the non-degeneracy implies S = 0 (Lemma 2).

2. Preliminaries. Let M be an n-dimensional connected complex manifold. We shall consider a holomorphic immersion f : M → C^{n+1} together with a C^{∞} complex transversal bundle N. If ξ : U → C^{n+1} is a local section of N, then the induced connection ∇ on M, the second fundamental form h, the shape operator S and the transversal forms μ and ν are defined by the following Gauss and Weingarten formulas [NS]:

D_{X}f_{∗}Y = f_{∗}∇_{X}Y + h(X, Y)ξ − h(JX, Y)Jξ,
D_{X}ξ = −f_{∗}SX + μ(X)ξ + ν(X)Jξ.

Here D denotes the standard connection on C^{n+1}, and J the complex structure on M and on C^{n+1} as well.

Let m ∈ M. The complex rank of the C-bilinear form h_{m}^{c}(·, ·) = h_{m}(·, ·) − ih_{m}(J·, ·) depends on f only. It is called the type number of f at m and denoted by tf_{m} [O]. We shall assume that it is positive everywhere on M.

Our first requirement on the transversal bundle is that the induced shape operator S is anti-complex, ie SJ = −JS [O]. The fundamental equations satisfied by ∇, h, μ, ν and such an S are the following:

\begin{matrix}R(X, Y)Z = h(Y, Z)SX − h(X, Z)SY + h(JY, Z)SJX− & h(JX, Z)SJY \\ (Gauss), \end{matrix}

\begin{matrix}(∇_{X}h)(Y, Z) + μ(X)h(Y, Z) + ν(X)h(JY, Z) = \\ (∇_{Y} h)(X, Z) + μ(Y)h(X, Z) + ν(Y)h(JX, Z) (Codazzi I), \end{matrix}

\begin{matrix}(∇_{X}S)Y − μ(X)SY + ν(X)SJY = (∇_{Y} S)X− & μ(Y)SX + ν(Y)SJX \\ (Codazzi II), \end{matrix}

h(X, SY) − h(Y, SX) = 2dμ(X, Y) (Ricci I),
h(SX, JY) − h(SY, JX) = 2dν(X, Y) (Ricci II).

Furthermore, we shall assume that for every point m ∈ M there exists a local section ξ : U → C^{n+1} of N with U ∋ m such that ∇S^{ξ} = 0, where S^{ξ} denotes the shape operator induced by ξ. If the points m_{1}, m_{2} ∈ U can be joined by a curve lying in U, and B_{1} is a basis of T_{m_{1}}M, then by parallel displacement we can obtain a basis B_{2} of T_{m_{2}}M with respect to which S_{m_{2}}^{ξ} has the same matrix as S_{m_{1}}^{ξ} with respect to B_{1}. Hence rank S_{m_{1}}^{ξ} = rank S_{m_{2}}^{ξ}, where rank S_{m}^{ξ} := dim_{C} im S_{m}^{ξ}. The assumed connectedness of M and the independence of rank S_{m}^{ξ} of ξ at a fixed m imply that q := rank S is well-defined for the whole M.

Studying the immersions with ∇-parallel shape operator we shall make use of the following remarks:

Remark 1. If ∇S = 0, then for every X, Y, Z

R(X, Y)SZ = S(R(X, Y)Z).

P r o o f. It is an obvious consequence of the commutativity of S and ∇_{W} for any W.

Remark 2. If ∇S = 0 and SJ = −JS, then ker S ⊂ ker μ ∩ ker ν or S = 0.

P r o o f. Let SX = 0. By the second Codazzi equation we have −μ(X)SY + ν(X)SJY = 0 for any Y. If S ≠ 0, then there exists Y such that SY ≠ 0. Since SY and SJY = −JSY are linearly independent over R, it follows that μ(X) = 0 and ν(X) = 0.

Remark 3. If ∇S = 0, SJ = −JS and S ≠ 0, then the section ξ inducing S is anti-holomorphic, ie ν(X) = μ(JX) for any X.

P r o o f. We may assume that SX ≠ 0. The assertion follows easily from the second Codazzi equation, written for Y = JX.

For an anti-holomorphic ξ we can rewrite the first Codazzi equation as

\begin{matrix}(∇_{X}h)(Y, Z) + μ(X)h(Y, Z) + μ(JX)h(JY, Z) \\ = & (∇_{Y} h)(X, Z) + μ(Y)h(X, Z) + μ(JY)h(JX, Z). \end{matrix}

3. Theorem. We can now formulate our main result.

Theorem. Let M be an n-dimensional connected complex manifold, n > 1, f : M → C^{n+1} a holomorphic immersion and ∇ a linear connection induced on M by a transversal bundle N. Let f and M satisfy the following assumptions:

(1) tf > 0 everywhere on M,

(2) for every m ∈ M there exist its neighbourhood U and a local section ξ : U → C^{n+1} of N inducing an anti-complex and ∇-parallel shape operator S,

(3) q := rank S > 0. Under the conditions stated above,

(i) q ≤ \frac{n}{2}; (ii) for every m ∈ M there exist its neighbourhood V, a complex map ϕ̃ : V →

C^{n} ˜=C^{q} × C^{n−2q} × C^{q}, a complex affine isomorphism Ã : C^{n+1} → C^{n+1}, a holomorphic function F̃ such that

\begin{matrix}Ã ∘ f ∘ ϕ̃^{−1} & (x̃^{1}, ..., x̃^{q}, ỹ^{1}, ..., ỹ^{n−2q}, z̃^{1}, ..., z̃^{q}) \\ = & (x̃^{1}, ..., x̃^{q}, ỹ^{1}, ..., ỹ^{n−2q}, z̃^{1}, ..., z̃^{q}, F̃(ỹ^{1}, ..., ỹ^{n−2q}, z̃^{1}, ..., z̃^{q})); \end{matrix}

(iii) if q > 1, then the local section Ã∘ξ : V → C^{n+1} of ÃN iducing the ∇-parallel shape operator in this map is described by the following formula:

Ã ∘ ξ ∘ ϕ̃^{−1}(x̃^{1}, ..., x̃^{q}, ỹ^{1}, ..., ỹ^{n−2q}, z̃^{1}, ..., z̃^{q}) = (\overline{z̃^{1}}, ..., \overline{z̃^{q}}, \underbrace{0, ..., 0}_{n−q times}, 1),

here Ã denotes the linear part of Ã; (iv) if q = 1, then

Ã ∘ ξ ∘ ϕ̃^{−1}(x̃, ỹ^{1}, ..., ỹ^{n−2}, z̃) = (\overline{G̃(z̃)}, \underbrace{0, ..., 0}_{n−1 times}, \overline{e^{M̃(z̃)}})

where G̃ and M̃ are holomorphic functions such that

G̃′(z̃) − M̃′(z̃)G̃(z̃) ≡ 1.

In the real representation, setting x̃^{k} = x^{2k−1} + ix^{2k}, ỹ^{l} = y^{2l−1} + iy^{2l}, z̃^{j} = z^{2j−1}+iz^{2j} for j, k = 1, ..., q; l = 1, ..., n−2q; F̃ = F^{1}+iF^{2}, G̃ = G^{1}+iG^{2}, M̃ = M^{1} + iM^{2}, we have:

\begin{matrix}A ∘ f ∘ ϕ^{−1}(x^{1}, ..., x^{2q}, y^{1}, ..., y^{2n−4q}, z^{1}, ..., z^{2q}) = \\ (x^{1}, ..., x^{2q}, y^{1}, ..., y^{2n−4q}, z^{1}, ..., z^{2q}, F^{1}(y, z), F^{2}(y, z)), \end{matrix}

\begin{matrix}A⃗ ∘ ξ ∘ ϕ^{−1}(x^{1}, ..., x^{2q}, y^{1}, ..., y^{2n−4q}, z^{1}, ..., z^{2q}) = \\ (z^{1}, −z^{2}, ..., z^{2q−1}, −z^{2q}, \underbrace{0, ..., 0}_{2n−2q times}, 1, 0) \end{matrix}

if q > 1

and

A⃗ ∘ ξ ∘ ϕ^{−1}(x^{1}, x^{2}, y^{1}, ..., y^{2n−4}, z^{1}, z^{2}) = (G^{1}(z), −G^{2}(z), \underbrace{0, ..., 0}_{2n−2 times}, e^{M^{1}(z)} cos M^{2}(z), −e^{M^{1}(z)} sin M^{2}(z)) ,

if q = 1.

Remark 4. An easy computation shows that the inverse is also true:

(a)For any holomorphic function F̃ of n − q variables, where q ≤ \frac{n}{2}, the shape operator S induced on the hypersurface

\begin{matrix}f : C^{n} ⊃ U ∋ (x̃^{1}, …, x̃^{q}, ỹ^{1}, …, ỹ^{n−2q}, z̃^{1}, …, z̃^{q}) ↦ \\ ↦ (x̃^{1}, …, x̃^{q}, ỹ^{1}, …, ỹ^{n−2q}, z̃^{1}, …, z̃^{q}, F̃(ỹ, z̃)) ∈ C^{n+1} \end{matrix}

endowed with the transversal field

ξ(x̃^{1}, …, x̃^{q}, ỹ^{1}, …, ỹ^{n−2q}, z̃^{1}, …, z̃^{q}) = (\overline{z̃^{1}}, …, \overline{z̃^{q}}, \underbrace{0, …, 0}_{n-q times}, 1)

is parallel with respect to the induced connection and rank S = q.

(b)For any holomorphic function F̃ of n − 1 variables and for any holomorphic functions of one variable G̃ and M̃ satisfying the equation

G̃′(z̃) − M̃′(z̃)G̃(z̃) ≡ 1.

the shape operator S induced on the hypersurface

f : C^{n} ⊃ U ∋ (x̃, ỹ^{1}, …, ỹ^{n−2}, z̃) ↦ (x̃, ỹ^{1}, …, ỹ^{n−2}, z̃, F̃(ỹ, z̃)) ∈ C^{n+1}

endowed with the transversal field

ξ(x̃, ỹ^{1}, …, ỹ^{n−2}, z̃) = (\overline{G̃(z̃)}, \underbrace{0, …, 0}_{n-1 times}, \overline{e^{M̃(z̃}})

is parallel with respect to the induced connection and rank S = 1.

P r o o f o f t h e t h e o r e m. We begin by proving two lemmas in which we establish some inclusions between im S, ker S and ker h.

Lemma 1. If tf > 0, SJ = −JS, and ∇S = 0, then the following conditions are equivalent:

(1) dμ = 0 and dν = 0,

(2) im S ⊂ ker h,

(3) im S ⊂ ker S.

P r o o f.

(1)⇒(2). Suppose that dμ = 0 and dν = 0. Let m′ ∈ M and X, Y ∈ T_{m′}M. Applying the Ricci equations we have

0 = 2dμ(X, Y) + 2dν(JX, Y)
= h(X, SY) − h(Y, SX) − h(J^{2}X, SY) + h(JY, SJX)
= h(X, SY) − h(Y, SX) + h(X, SY) + h(JY, −JSX)
= h(X, SY) − h(Y, SX) + h(X, SY) + h(Y, SX) = 2h(X, SY),

hence for any X, Y ∈ T_{m′}M we have h(X, SY) = 0.

The Ricci equations make it obvious that (2) implies (1).

(2)⇒(3).

Suppose that im S ⊂ ker h, which yields R(X, Y)SZ = 0 by the Gauss equation. Since tf_{m′} > 0, there exist X_{0}, Y_{0} ∈ T_{m′}M such that h(X_{0}, Y_{0}) ≠ 0. We first show that S^{2}X_{0} = 0. Indeed, making use of Remark 1 we have

0 = S(R(X_{0}, JX_{0})Y_{0}) = 2(h(JX_{0}, Y_{0})S^{2}X_{0} − h(X_{0}, Y_{0})S^{2}JX_{0})

with h(X_{0}, Y_{0}) ≠ 0, which means that S^{2}X_{0} and S^{2}JX_{0} are linearly dependent over R. It is possible only when S^{2}X_{0} = S^{2}JX_{0} = 0.

Now we take an arbitrary Z ∈ T_{m′}M. Then

0 = S(R(X_{0}, Z)Y_{0}) = −h(X_{0}, Y_{0})S^{2}Z − h(JX_{0}, Y_{0})S^{2}JZ,

and by a similar argument S^{2}Z = 0.

(3)⇒(2). If S^{2} = 0, then the right-hand side of the equality R(X, Y)SZ = S(R(X, Y)Z) vanishes for every X, Y, Z. For S = 0 (2) holds, therefore we can assume that S ≠ 0. Let us take X_{0} such that SX_{0} ≠ 0. Since

0 = R(X_{0}, JX_{0})SZ = 2(h(JX_{0}, SZ)SX_{0} − h(X_{0}, SZ)SJX_{0})

and SX_{0}, SJX_{0} are linearly independent over R, we have

h(X_{0}, SZ) = h(JX_{0}, SZ) = 0 for any Z.

Now we can write for any Y, Z

0 = R(X_{0}, Y)SZ = h(Y, SZ)SX_{0} + h(JY, SZ)SJX_{0}.

Hence h(Y, SZ) = 0 for any Y, Z. ■

Lemma 2. Under the assumptions of Lemma 1, the equivalent conditions (1),

(2) and (3) are satisfied.

P r o o f. If ∇S = 0, then rank S is constant on the domain of S. We have to consider three cases.

1^{∘}. rank S = 0. Then, of course , (3) holds. 2^{∘}. rank S = 1. We shall prove that in this case (3) holds, too. Suppose, contrary to our claim, that S^{2} ≠ 0. We shall obtain a contradiction

with the assumption tf > 0.

Let m′ ∈ M. We fix X_{0} ∈ T_{m′}M such that S^{2}X_{0} ≠ 0.

Step 1. ker S ⊂ ker h.

Let Z ∈ ker S. Then

0 = R(X_{0}, JX_{0})SZ = S(R(X_{0}, JX_{0})Z) = 2(h(JX_{0}, Z)S^{2}X_{0} − h(X_{0}, Z)S^{2}JX_{0}),

hence h(X_{0}, Z) = h(JX_{0}, Z) = 0. For any Y we have now

0 = R(X_{0}, Y)SZ = S(R(X_{0}, Y)Z) = h(Y, Z)S^{2}X_{0} + h(JY, Z)S^{2}JX_{0},

which implies h(Y, Z) = 0 for any Z ∈ ker S and for any Y.

Step 2.

(a) T_{m′}M = ker S ⊕ CX_{0} and

(b) T_{m′}M = ker S ⊕ CSX_{0}.

Since dim_{C} ker S = n−1, it is sufficient and easy to check that ker S∩CX_{0} = {0} and ker S ∩ CSX_{0} = {0}.

Step 3. h(X_{0}, SX_{0}) = h(JX_{0}, SJX_{0}) = 0.

If h(X_{0}, X_{0}) = 0 and h(X_{0}, JX_{0}) = 0 then X_{0} ∈ ker h because of Steps 1 and 2(a). The claimed equality holds.

Assume now that h(X_{0}, X_{0}) ≠ 0 or h(X_{0}, JX_{0}) ≠ 0. We have

\begin{aligned}0 & = R(X_{0}, JX_{0})SX_{0} − S(R(X_{0}, JX_{0})X_{0} \\ & = 2S(h(JX_{0}, SX_{0})X_{0} − h(X_{0}, SX_{0})JX_{0} − h(JX_{0}, X_{0})SX_{0} + h(X_{0}, X_{0})SJX_{0}), \end{aligned}

therefore Z_{0} ∈ ker S, where

Z_{0} := h(JX_{0}, SX_{0})X_{0} − h(X_{0}, SX_{0})JX_{0} − h(JX_{0}, X_{0})SX_{0} + h(X_{0}, X_{0})SJX_{0}.

According to Step 1, we have h(Z_{0}, X_{0}) = 0 and h(Z_{0}, JX_{0}) = 0, hence

\begin{matrix}h(JX_{0}, SX_{0})h(X_{0}, X_{0}) − h(X_{0}, SX_{0})h(JX_{0}, X_{0}) \\ − h(JX_{0}, X_{0})h(SX_{0}, X_{0}) + h(X_{0}, X_{0})h(SJX_{0}, X_{0}) = 0 \end{matrix}

and

\begin{matrix}h(JX_{0}, SX_{0})h(X_{0}, JX_{0}) − h(X_{0}, SX_{0})h(JX_{0}, JX_{0}) \\ − h(JX_{0}, X_{0})h(SX_{0}, JX_{0}) + h(X_{0}, X_{0})h(SJX_{0}, JX_{0}) = 0. \end{matrix}

Thus we obtain

h(SX_{0}, X_{0})h(JX_{0}, X_{0}) = 0 and h(SX_{0}, X_{0})h(X_{0}, X_{0}) = 0,

which implies h(SX_{0}, X_{0}) = 0, and consequently, by the anti-complexity of S and the properties of h(· , ·), h(SJX_{0}, JX_{0}) = 0.

Step 4. h(Z, SW) + h(W, SZ) = 0 for any Z, W ∈ T_{m′}M.

By Step 2(a) we have Z = Z_{1} + αX_{0} + βJX_{0}, W = W_{1} + γX_{0} + δJX_{0} with Z_{1}, W_{1} ∈ ker S and α, β, γ, δ ∈ R. An easy computation gives

h(Z, SW) + h(W, SZ)
= h(αX_{0} + βJX_{0}, S(γX_{0} + δJX_{0})) + h(γX_{0} + δJX_{0}, S(αX_{0} + βJX_{0}))
= 2αγh(X_{0}, SX_{0}) + 2βδh(JX_{0}, SJX_{0}) + (αδ + βγ)h(X_{0}, (SJ + JS)X_{0}),

which vanishes by the anti-complexity of S and by Step 3.

Step 5. (∇_{W} h)(X_{0}, SX_{0}) = (∇_{W} h)(JX_{0}, SJX_{0}) = 0 for any W.

We extend X_{0} to the local vector field X_{0} such that S^{2}X_{0} ≠ 0 at any point of the domain of X_{0}. We have

\begin{aligned}(∇_{W} h)(X_{0}, SX_{0}) & = W (h(X_{0}, SX_{0})) − h(∇_{W} X_{0}, SX_{0}) − h(X_{0}, ∇_{W} (SX_{0})) \\ & = W (h(X_{0}, SX_{0})) − (h(∇_{W} X_{0}, SX_{0}) + h(X_{0}, S(∇_{W} X_{0}))) = 0. \end{aligned}

The same is true for JX_{0} in place of X_{0}.

Step 6. ∇h = −2μ ⊗ h.

If Y ∈ ker S_{m′}, then we can extend Y to a local section Y of ker S. For any X, Z ∇_{X}Y ∈ ker S and

(∇_{X}h)(Y, Z)
= X(h(Y, Z)) − h(∇_{X}Y, Z) − h(Y, ∇_{X}Z) = 0
= −2μ(X)h(Y, Z).

Let X ∈ ker S. Then for any Y, Z we have

(∇_{X}h)(Y, Z) =_{Codazzi I}

− μ(X)h(Y, Z) − μ(JX)h(JY, Z)

+ (∇_{Y} h)(X, Z) + μ(Y)h(X, Z) + μ(JY)h(JX, Z)

= 0 = −2μ(X)h(Y, Z),

since ker S ⊂ ker μ ∩ ker ν and ker S ⊂ ker h.

It follows that

(∇_{X}h)(Y, ·) = −2μ(X)h(Y, ·)

if X ∈ ker S or Y ∈ ker S.

From the first Codazzi equation we obtain

\begin{matrix}(∇_{X_{0}}h)(JX_{0}, SX_{0}) + μ(X_{0})h(JX_{0}, SX_{0}) + μ(JX_{0})h(J^{2}X_{0}, SX_{0}) \\ = & (∇_{JX_{0}}h)(X_{0}, SX_{0}) + μ(JX_{0})h(X_{0}, SX_{0}) + μ(J^{2}X_{0})h(JX_{0}, SX_{0}). \end{matrix}

Hence

(∇_{X_{0}}h)(JX_{0}, SX_{0}) = −2μ(X_{0})h(JX_{0}, SX_{0}),

by virtue of Steps 3 and 5.

Similarly,

\begin{matrix}(∇_{X_{0}}h)(JX_{0}, SJX_{0}) + μ(X_{0})h(JX_{0}, SJX_{0}) + μ(JX_{0})h(J^{2}X_{0}, SJX_{0}) \\ = & (∇_{JX_{0}}h)(X_{0}, SJX_{0}) + μ(JX_{0})h(X_{0}, SJX_{0}) + μ(J^{2}X_{0})h(JX_{0}, SJX_{0}), \end{matrix}

which gives

(∇_{JX_{0}}h)(X_{0}, SJX_{0}) = −2μ(JX_{0})h(X_{0}, SJX_{0}).

We have now

(∇_{X_{0}}h)(X_{0}, SX_{0}) = 0 = −2μ(X_{0})h(X_{0}, SX_{0}),
(∇_{X_{0}}h)(X_{0}, SJX_{0}) = −(∇_{X_{0}}h)(JX_{0}, SX_{0})
= 2μ(X_{0})h(JX_{0}, SX_{0}) = −2μ(X_{0})h(X_{0}, SJX_{0})

and

(∇_{X_{0}}h)(X_{0}, Z) = −2μ(X_{0})h(X_{0}, Z)

for Z ∈ ker S. Therefore

(∇_{X_{0}}h)(X_{0}, · ) = −2μ(X_{0})h(X_{0}, · )

by Step 2(b).

In the same manner we can see that

(∇_{X_{0}}h)(JX_{0}, · ) = −2μ(X_{0})h(JX_{0}, · ),

(∇_{JX_{0}}h)(X_{0}, · ) = −2μ(JX_{0})h(X_{0}, · ), and

(∇_{JX_{0}}h)(JX_{0}, · ) = −2μ(JX_{0})h(JX_{0}, · ),

which completes the proof of Step 6.

As a consequence of Step 6 we obtain Step 7. R(X, Y) · h = −4 dμ(X, Y) h for any X, Y. Applying the Ricci equation yields Step 8. R(X, Y) · h = −2 (h(X, SY) − h(Y, SX)) h . In particular, R(X_{0}, JX_{0}) · h = −4 h(X_{0}, SJX_{0}) h. On the other hand, a direct computation gives

(R(X_{0}, JX_{0}) · h)(X_{0}, X_{0})

= −2h(R(X_{0}, JX_{0})X_{0}, X_{0})
= −2h(2(h(JX_{0}, X_{0})SX_{0} − h(X_{0}, X_{0})SJX_{0}), X_{0})
= −4h(JX_{0}, X_{0})h(SX_{0}, X_{0}) + 4h(X_{0}, X_{0})h(SJX_{0}, X_{0})
= 4h(X_{0}, X_{0})h(X_{0}, SJX_{0}),

and

(R(X_{0}, JX_{0}) · h)(X_{0}, JX_{0})
= −h(R(X_{0}, JX_{0})X_{0}, JX_{0}) − h(X_{0}, R(X_{0}, JX_{0})JX_{0})
= −h(2(h(JX_{0}, X_{0})SX_{0} − h(X_{0}, X_{0})SJX_{0}), JX_{0})
− h(X_{0}, 2(h(JX_{0}, JX_{0})SX_{0} − h(X_{0}, JX_{0})SJX_{0}))
= −2h(JX_{0}, X_{0})h(SX_{0}, JX_{0}) + 2h(X_{0}, X_{0})h(SJX_{0}, JX_{0})
− 2h(JX_{0}, JX_{0})h(X_{0}, SX_{0}) + 2h(X_{0}, JX_{0})h(X_{0}, SJX_{0})
= 4h(JX_{0}, X_{0})h(X_{0}, SJX_{0}).

A comparison with Step 8 gives

h(X_{0}, SJX_{0}) = 0 or h(X_{0}, X_{0}) = h(X_{0}, JX_{0}) = 0,

which both together with Steps 1, 2 and 3 lead to a contradiction with the assumption h_{m′} ≠ 0.

3^{∘}. rank S > 1. If SX = 0, then, by Remark 2, μ(X) = ν(X) = 0. Let SX ≠ 0. Then there

exists Y such that SX and SY are linearly independent over C. From the Codazzi II we have

(μ(X) + iν(X))SY − (μ(Y) + iν(Y))SX = 0,

which implies μ(X) = ν(X) = 0. Therefore (1) of the lemma 1. holds. ■ We now go back to the proof of the theorem.

Fix m ∈ M. Let ξ : U → C^{n+1} be defined on a connected neighbourhood U of m and have the property described in the assumption (2) of the theorem.

Lemma 3. There exists a q-dimensional complex subspace W of C^{n+1} such that for every m′ ∈ U f_{∗}im S_{m′} = W.

P r o o f. Let m′ ∈ U. It is sufficient to show that f_{∗}im S_{m} = f_{∗}im S_{m′}. Let γ : [0, 1] → U be a C^{1} curve joining m and m′; γ(0) = m, γ(1) = m′. We choose X_{1}_{m}, ..., X_{q}_{m} ∈ T_{m}M such that SX_{1}_{m}, ..., SX_{q}_{m} form a basis over C of f_{∗}im S_{m}. Let X̃_{1}, ..., X̃_{q} be the vector fields defined along the curve γ, parallel with respect to ∇, X̃_{i}(0) = X_{i}_{m} for i ∈ {1, ..., q}. It is easy to check that the map

[0, 1] ∋ t ↦ f_{∗}S_{γ(t)}X̃_{i}(t) ∈ C^{n+1}

is a constant one. Indeed,

\frac{d}{dt}(t ↦ f_{∗}S_{γ(t)}X̃_{i}(t))

=D_{γ̇(t)}f_{∗}SX̃_{i}
=f_{∗}∇_{γ̇(t)}SX̃_{i} + h(γ̇(t), SX̃_{i}(t))ξ_{γ(t)} − h(Jγ̇(t), SX̃_{i}(t))Jξ_{γ(t)}.

The last two terms vanish because im S ⊂ ker h, and

∇_{γ̇(t)}SX̃_{i} = S∇_{γ̇(t)}X̃_{i} = S0 = 0.

It follows that

span_{C}{f_{∗}S_{m}X̃_{i}(0) : i = 1, ..., q} = span_{C}{f_{∗}S_{m′} X̃_{i}(1) : i = 1, ..., q},

that means, f_{∗}im S_{m} = f_{∗}im S_{m′} =: W. ■

Let Ã_{1} : C^{n+1} → C^{n+1} be a linear isomorphism such that

Ã_{1}W = span_{C}{ẽ_{1}, ..., ẽ_{q}}.

Here and subsequently ẽ_{1}, …, ẽ_{n+1} denotes the standard basis of C^{n+1}, whereas e_{1}, …, e_{2n+2} the standard basis of R^{2n+2}.

Lemma 4. There exists i_{0} ∈ {q + 1, ..., n + 1} such that

ẽ_{i_{0}} ∉ (Ã_{1} ∘ f)_{∗}T_{m}M and the i_{0}-th coordinate of Ã_{1}ξ_{m} does not vanish.

P r o o f. Suppose the assertion of the lemma is false. Then ẽ_{j} ∈ (Ã_{1} ∘ f)_{∗}T_{m}M for every j ∈ {q + 1, ..., n + 1} such that the j-th coordinate of Ã_{1}ξ does not vanish at m. Then obviously Ã_{1}ξ_{m} ∈ (Ã_{1} ∘ f)_{∗}T_{m}M, which contradicts the transversality of ξ. ■

Let Ã_{2}^{0} be the linear isomorphism of C^{n+1}, defined on its basis in the following way:

Ã_{2}^{0}ẽ_{k} := \{ \begin{matrix}ẽ_{k} & if k ∉ \{i_{0}, n + 1\} \\ ẽ_{n+1} & if k = i_{0} \\ ẽ_{i_{0}} & if k = n + 1. \end{matrix}

Let Ã_{2} := Ã_{2}^{0} ∘ Ã_{1}. Now ẽ_{n+1} is transversal to (Ã_{2} ∘ f)_{∗}T_{m}M and Ã_{2}ξ has the non-vanishing n + 1-th coordinate at m. Moreover, the isomorphism Ã_{2}^{0} does not change the subspace (Ã_{1} ∘ f)_{∗}im S = span_{C}{ẽ_{1}, ..., ẽ_{q}}.

We denote by π the projection π : C^{n+1} ∋ (ζ^{1}, ..., ζ^{n+1}) ↦ (ζ^{1}, ..., ζ^{n}) ∈ C^{n}. It is easy to check that

d_{m}(π ∘ Ã_{2} ∘ f) : T_{m}M → C^{n}

is a monomorphism. Indeed, if d_{m}(π ∘ Ã_{2} ∘ f).V = 0, then (Ã_{2} ∘ f)_{∗}V ∈ Cẽ_{n+1}. But (Ã_{2} ∘ f)_{∗}T_{m}M ∩ Cẽ_{n+1} = {0} and (Ã_{2} ∘ f)_{∗} is a monomorphism; therefore V = 0.

We can now take ϕ̃_{1} := π ∘ Ã_{2} ∘ f as a complex map on some neighbourhood U_{1} ⊂ U of m. In this map

Ã_{2} ∘ f ∘ ϕ̃_{1}^{−1}(ζ^{1}, ..., ζ^{n}) = (ζ^{1}, ..., ζ^{n}, φ̃(ζ)),

with a holomorphic function φ̃.

In the real representation, identifying C^{k} with R^{2k}

ι_{k} : R^{2k} ∋ (w^{1}, ..., w^{2k}) ↦ (w^{1} + iw^{2}, ..., w^{2k−1} + iw^{2k}) ∈ C^{k},

we can write

A_{2} ∘ f ∘ ϕ_{1}^{−1}(w^{1}, ..., w^{2n}) = (w^{1}, ..., w^{2n}, φ^{1}(w), φ^{2}(w)).

Here A_{2} := ι_{n+1}^{−1} ∘ Ã_{2} and ϕ_{1} := ι_{n}^{−1} ∘ ϕ̃_{1}.

Lemma 5.

(a) \frac{∂φ^{k}}{∂w^{s}} = 0 for k = 1, 2 and s = 1, ..., 2q.

(b) im S = span_{R}\{\frac{∂}{∂w^{s}} : s = 1, ..., 2q\}.

P r o o f. At any point m′ ∈ U_{1} we have

(A_{2} ∘ f)_{∗}(\frac{∂}{∂w^{s}}) = e_{s} + \frac{∂φ^{1}}{∂w^{s}} e_{2n+1} + \frac{∂φ^{2}}{∂w^{s}} e_{2n+2}.

If s ∈ {1, ..., 2q}, then e_{s} = (A_{2} ∘ f)_{∗}SW_{s} with some W_{s} ∈ T_{m′}M, because

e_{s} ∈ span_{R}{e_{1}, ..., e_{2q}} = (A_{2} ∘ f)_{∗}im S_{m′}.

Therefore we have

(A_{2} ∘ f)_{∗}(\frac{∂}{∂w^{s}} − SW_{s}) = \frac{∂φ^{1}}{∂w^{s}} e_{2n+1} + \frac{∂φ^{2}}{∂w^{s}} e_{2n+2}.

From the transversality of e_{2n+1} and e_{2n+2} to (A_{2} ∘ f)_{∗}T M and from the injectivness of (A_{2} ∘ f)_{∗} it follows that (a) holds, and \frac{∂}{∂w^{s}} − SW_{s} = 0 for s = 1, ..., 2q. According to the last equality,

span_{R}\{\frac{∂}{∂w^{s}} : s = 1, ..., 2q\} ⊂ im S,

which implies (b), because the dimensions are equal. ■

Lemma 6. The transversal field A_{2}ξ does not depend on the variables w^{1}, ..., w^{2q}. P r o o f. We use the Weingarten formula

D_{\frac{∂}{∂w^{s}}} A_{2}ξ = −(A_{2} ∘ f)_{∗}S \frac{∂}{∂w^{s}} + μ(\frac{∂}{∂w^{s}})A_{2}ξ + ν(\frac{∂}{∂w^{s}})JA_{2}ξ.

According to Lemmas 5(b), 2 and Remark 2, for s = 1, ..., 2q

\frac{∂}{∂w^{s}} ∈ im S ⊂ ker S ⊂ ker μ ∩ ker ν,

hence

D_{\frac{∂}{∂w^{s}}} A_{2}ξ = 0.■

We introduce now the functions Ξ^{1}, ..., Ξ^{2n+2} by the following formula

(A_{2}ξ ∘ ϕ_{1}^{−1})(w) = ∑_{k=1}^{n+1}[Ξ^{2k−1}(w)e_{2k−1} − Ξ^{2k}(w)e_{2k}].

Lemma 7.

rank_{R} [\frac{∂Ξ^{k}}{∂w^{j}} (w)]_{k=1,...,2n+2;j=2q+1,...,2n} = 2q

for w ∈ ϕ_{1}(U_{1}).

P r o o f.

rank_{R}[\frac{∂Ξ^{k}}{∂w^{j}} (w)]_{k=1,... ,2n+2;j=2q+1,... ,2n}

= rank_{R}[(−1)^{k−1} \frac{∂Ξ^{k}}{∂w^{j}} (w)]_{k=1,... ,2n+2;j=2q+1,... ,2n}

= dim_{R} span \{D_{\frac{∂}{∂w^{j}}} A_{2}ξ : j = 2q + 1, …, 2n\} = dim_{R} span \{D_{\frac{∂}{∂w^{j}}} A_{2}ξ : j = 1, …, 2n\} = dim_{R} im S_{ϕ_{1}^{−1}(w)}.

The last equality is due to the isomorphism im S and im { X ↦ D_{X}A_{2}ξ}.

A consequence of Lemma 7 is the following

Corollary.

n ≥ 2q.

Lemma 8.

Ξ̃^{k} := Ξ^{2k−1} + iΞ^{2k}

is a holomorphic function for k = 1, …, n + 1.

P r o o f. It is sufficient to show that Ξ^{2k−1} and Ξ^{2k} satisfy the Cauchy-Riemann equations. From Remark 3 it follows that

D_{JX}(A_{2}ξ) = −JD_{X}(A_{2}ξ)

for any X ∈ T_{m′}M. Let X = \frac{∂}{∂w^{2s−1}} , then JX = \frac{∂}{∂w^{2s}} and

D_{JX}(A_{2}ξ) = ∑_{l=1}^{n+1} [\frac{∂Ξ^{2l−1}}{∂w^{2s}} e_{2l−1} − \frac{∂Ξ^{2l}}{∂w^{2s}} e_{2l}] ,

−JD_{X}(A_{2}ξ) = −J ∑_{l=1}^{n+1} [\frac{∂Ξ^{2l−1}}{∂w^{2s−1}} e_{2l−1} − \frac{∂Ξ^{2l}}{∂w^{2s−1}} e_{2l}]

= ∑_{l=1}^{n+1} [−\frac{∂Ξ^{2l−1}}{∂w^{2s−1}} e_{2l} − \frac{∂Ξ^{2l}}{∂w^{2s−1}} e_{2l−1}] .

Therefore

\frac{∂Ξ^{2l−1}}{∂w^{2s}} = −\frac{∂Ξ^{2l}}{∂Ξ^{2s−1}} and \frac{∂Ξ^{2l−1}}{∂w^{2s−1}} = \frac{∂Ξ^{2l}}{∂w^{2s}} .■

Now Ξ̃^{k}, k = 1, …, n + 1 are holomorphic functions of complex variables ζ^{s} = w^{2s−1} + iw^{2s}, s = q + 1, …, n. According to Lemma 7, we have

rank_{C} [\frac{∂Ξ̃^{k}}{∂ζ^{l}} ]_{k=1,... ,n+1; l=q+1,... ,n} = q.

Lemma 9. Let r ≤ N ≤ M. Let U be an open set in K^{N}, where K = R or C. Let F : U → K^{M} be a C^{1} mapping such that for every x ∈ U rank F′(x) = r.

Let x_{0} ∈ U and let i_{1} < i_{2} < ⋯ < i_{r} and j_{1} < j_{2} < ⋯ < j_{r} be chosen so that

det [\frac{∂F^{i_{k}}}{∂x^{j_{l}}} (x_{0})]_{k=1,... ,r; l=1,... ,r} ≠ 0.

Then there exist a neighbourhood U′ ⊂ U of x_{0} and a diffeomorphism Φ : U′ → Φ(U′) ⊂ K^{N} such that

(F ∘ Φ^{−1})^{i_{k}}(y^{1}, …, y^{N}) ≡ y^{j_{k}} for k = 1, …, r

and

\frac{∂(F ∘ Φ^{−1})^{k}}{∂y^{l}} ≡ 0 if l ∉ \{j_{1}, …, j_{r}\}, k ∈ \{1, …, M\}.

P r o o f. We define

Φ̂(x^{1}, …, x^{N})^{k} = \{ \begin{matrix}x^{k} if k ∉ \{j_{1}, …, j_{r}\} \\ F^{i_{s}}(x^{1}, …, x^{N}) if k = j_{s}. \end{matrix}

Since

det [\frac{∂Φ̂^{k}}{∂x^{l}} (x_{0})]_{k=1,... ,N; l=1,... ,N} = det [\frac{∂F^{i_{s}}}{∂x^{j_{s}}} (x_{0})]_{s=1,... ,r} ≠ 0,

there exists a neighbourhood U′ ⊂ U of x_{0} such that Φ := Φ̂|_{U′} is a diffeomorphism.

It remains to prove that F ∘ Φ^{−1} depends only on the variables y^{j_{1}}, …, y^{j_{r}}. Let y ∈ Φ(U′). Then

(F ∘ Φ^{−1})′(y) = F′(Φ^{−1}(y)) ∘ (Φ^{−1})′(y)

and

rank (F ∘ Φ^{−1})′(y) = rank F′(Φ^{−1}(y)) = r,

because (Φ^{−1})′(y) is an isomorphism.

Suppose that for some k_{0} ∈ {1, …, M}, l_{0} ∈ {1, …, N} \ {j_{1}, …, j_{r}}

\frac{∂(F ∘ Φ^{−1})^{k_{0}}}{∂y^{l_{0}}} (y) ≠ 0.

Then k_{0} ∉ {i_{1}, …, i_{r}} and

det [\frac{∂(F ∘ Φ^{−1})^{k}}{∂y^{l}} (y)]_{k∈\{i_{1},... ,i_{r}, k_{0}\}; l∈\{j_{1},... ,j_{r}, l_{0}\}} ≠ 0,

which contradicts the rank assumption. ■

We now restrict our attention to the case q > 1.

Lemma 10. If q > 1, then (a)

rank_{C} [\frac{∂Ξ̃^{k}}{∂ζ^{l}} ]_{k=1,... ,q; l=q+1,... ,n} = q,

(b)

Ξ̃^{k}(·) = b^{k} = const for k = q + 1, …, n + 1,

(c)

b^{n+1} ≠ 0.

P r o o f. Since, for q > 1, μ = 0 and ν = 0 (see proof of Lemma 2,3^{∘}), we have

D_{\frac{∂}{∂w^{l}}} A_{2}ξ ∈ (A_{2} ∘ f)_{∗}im S = span_{R}\{e_{1}, …, e_{2q}\},

therefore

\frac{∂Ξ^{k}}{∂w^{l}} = 0

for k = 2q + 1, …, 2n + 2; l = 2q + 1, …, 2n; which implies

\frac{∂Ξ̃^{k}}{∂ζ^{l}} = 0 for k = q + 1, …, n + 1; l = q + 1, …, n

and

rank_{C} [\frac{∂Ξ̃^{k}}{∂ζ^{l}} ]_{k=1,... ,n+1; l=q+1,... ,n} = rank_{C} [\frac{∂Ξ̃^{k}}{∂ζ^{l}} ]_{k=1,... ,q; l=q+1,... ,n} .

The point (c) is a consequence of Lemma 4 and the definition of Ã_{2}. ■

We may now apply Lemma 9, taking r = q, N = n, M = n+1, U = ϕ̃_{1}(U_{1}), K = C, F = (Ξ̃^{1}, …, Ξ̃^{n+1}), (i_{1}, …, i_{q}) = (1, …, q) and j_{1} < ⋯ < j_{q} from {q + 1, …, n} chosen so that

det ([\frac{∂Ξ̃^{k}}{∂ζ^{j_{s}}} (ϕ̃_{1}(m))]_{k=1,... ,q; s=1,... ,q}) ≠ 0.

In this way we obtain a new map ϕ̃_{2} := Φ∘ϕ̃_{1} on the neighbourhood U_{2} := ϕ̃_{1}^{−1}(U′) of m.

Since

(Ξ̃^{k} ∘ Φ^{−1})(η^{1}, …, η^{n}) = η^{j_{k}}

for k = 1, …, q and

(Ξ̃^{k} ∘ Φ^{−1})(η^{1}, …, η^{n}) = b^{k}

for k = q + 1, …, n + 1, the transversal field is now described by the formula

(Ã_{2}ξ ∘ ϕ̃_{2}^{−1})(η^{1}, …, η^{n}) = ∑_{k=1}^{q} \overline{η^{j_{k}}} ẽ_{k} + ∑_{k=q+1}^{n+1} \overline{b^{k}}ẽ_{k}.

Let Ã_{3}^{0} be the linear isomorphism of C^{n+1} which transforms the basis

(ẽ_{1}, …, ẽ_{n+1})

onto the basis

(ẽ_{1}, …, ẽ_{q}, ẽ_{q+1}, …, ˆ ẽ_{j_{1}}, …, ˆ ẽ_{j_{q}}, …, ẽ_{n}, ẽ_{j_{1}}, …, ẽ_{j_{q}}, ẽ_{n+1}),

and let

ϕ̃_{3}^{0}(η^{1}, …, η^{n}) := (η^{1}, …, η^{q}, η^{q+1}, …, η̂^{j_{1}}, …, η̂^{j_{q}}, …, η^{n}, η^{j_{1}}, …, η^{j_{q}}).

Taking Ã_{3} := Ã_{3}^{0} ∘ Ã_{2}, ϕ̃_{3} := ϕ̃_{3}^{0} ∘ ϕ̃_{2}, we may write:

\begin{matrix}Ã_{3} ∘ f ∘ ϕ̃_{3}^{−1}(s̃^{1}, …, s̃^{q}, t̃^{1}, …, t̃^{n−2q}, ũ^{1}, …, ũ^{q}) \\ = (s̃^{1}, …, s̃^{q}, t̃^{1}, …, t̃^{n−2q}, χ̃^{1}(t̃, ũ), …, χ̃^{q}(t̃, ũ), ρ̃(t̃, ũ)) \end{matrix}

and

Ã_{3}ξ ∘ ϕ̃_{3}^{−1}(s̃^{1}, …, s̃^{q}, t̃^{1}, …, t̃^{n−2q}, ũ^{1}, …, ũ^{q}) = ∑_{k=1}^{q} \overline{ũ^{k}}ẽ_{k} + ∑_{k=q+1}^{n+1} \overline{a^{k}}ẽ_{k}.

Applying now the isomorphism Ã_{4}^{0}, where

Ã_{4}^{0}ẽ_{k} := \{ \begin{matrix}ẽ_{k} & for k = 1, …, n \\ \frac{1}{\overline{a^{n+1}}} (− ∑_{k=q+1}^{n} \overline{a^{k}}ẽ_{k} + ẽ_{n+1}) & for k = n + 1, \end{matrix}

Ã_{4} := Ã_{4}^{0} ∘ Ã_{3}, we obtain

\begin{matrix}Ã_{4} ∘ f ∘ ϕ̃_{3}^{−1}(s̃^{1}, …, s̃^{q}, t̃^{1}, …, t̃^{n−2q}, ũ^{1}, …, ũ^{q}) \\ = (s̃^{1}, …, s̃^{q}, σ̃^{1}(t̃, ũ), …, σ̃^{n−q}(t̃, ũ), Q̃(t̃, ũ)) \end{matrix}

and

Ã_{4}ξ ∘ ϕ̃_{3}^{−1}(s̃^{1}, …, s̃^{q}, t̃^{1}, …, t̃^{n−2q}, ũ^{1}, …, ũ^{q}) = ∑_{k=1}^{q} \overline{ũ^{k}}ẽ_{k} + ẽ_{n+1}.

Here

σ̃^{k}(t̃, ũ) = \{ \begin{matrix}t̃^{k} − \frac{\overline{a^{q+k}}}{\overline{a^{n+1}}} ρ̃(t̃, ũ) & for k = 1, …, n − 2q \\ χ̃^{k−(n−2q)}(t̃, ũ) − \overline{\frac{a^{q+k}}{\overline{a^{n+1}}}} ρ̃(t̃, ũ) & for k = n − 2q + 1, …, n − q; \end{matrix}

and Q̃ = \frac{1}{\overline{a^{n+1}}} ρ̃.

We now turn to the case q = 1.

By virtue of Lemmas 7, 8 and 9 we may define on a connected neighbourhood ˜Ũ_{2} of m a complex map ˜ ϕ̃_{2} in which the coordinates of the transversal field are the functions of one variable η^{i_{0}} only, with i_{0} > 1.

Next we apply the isomorphisms

˜ϕ̃_{3}^{0}(η^{1}, …, η^{n}) := (η^{1}, η^{2}, …, η̂^{i_{0}}, …, η^{n}, η^{i_{0}})

and

˜Ã_{3}^{0}(θ^{1}, …, θ^{n+1}) := (θ^{1}, θ^{2}, …, θ̂^{i_{0}}, …, θ^{n}, θ^{i_{0}}, θ^{n+1}),

to obtain

\begin{matrix}˜Ã_{3} ∘ f ∘ ˜ ϕ̃_{3}^{−1}(s̃, t̃^{1}, …, t̃^{n−2}, ũ) \\ = (s̃, t̃^{1}, …, t̃^{n−2}, ˜ χ̃(t̃, ũ), ˜ ρ̃(t̃, ũ)) \end{matrix}

and

˜Ã_{3}ξ ∘ ˜ ϕ̃_{3}^{−1}(s̃, t̃^{1}, …, t̃^{n−2}, ũ) = ∑_{k=1}^{n+1} \overline{Θ^{k}(ũ)}ẽ_{k}

with Θ^{n+1}(˜ϕ̃_{3}(m)) ≠ 0; ˜ ϕ̃_{3} := ˜ ϕ̃_{3}^{0} ∘ ˜ ϕ̃_{2} and ˜ Ã_{3} := ˜ Ã_{3}^{0} ∘ Ã_{2}.

In the real representation ˜ χ̃ = χ^{1} + iχ^{2}, ˜ ρ̃ = ρ^{1} + iρ^{2}, Θ^{k} = ϑ^{2k−1} + iϑ^{2k}, we have

\begin{matrix}A_{3} ∘ f ∘ ϕ_{3}^{−1}(s^{1}, s^{2}, t^{1}, …, t^{2n−4}, u^{1}, u^{2}) \\ = (s^{1}, s^{2}, t^{1}, …, t^{2n−4}, χ^{1}(t, u), χ^{2}(t, u), ρ^{1}(t, u), ρ^{2}(t, u)) \end{matrix}

and

A_{3}ξ ∘ ϕ_{3}^{−1}(s^{1}, s^{2}, t^{1}, …, t^{2n−4}, u^{1}, u^{2}) = ∑_{k=1}^{n+1} [ϑ^{2k−1}(u)e_{2k−1} − ϑ^{2k}(u)e_{2k}] .

Let π̂ denote the projection

C^{n} ∋ (s̃, t̃^{1}, …, t̃^{n−2}, ũ) ↦ ũ ∈ C.

Lemma 11.There exist c_{2}, …, c_{n+1} ∈ C, c_{n+1} ≠ 0, a neighbourhood ˜ Ũ_{3} of m and a holomorphic function H̃ such that

Θ^{k}(ũ) = c_{k}e^{H̃(ũ)}

for k = 1, …, n + 1; ũ ∈ π̂(˜ϕ̃_{3}( ˜Ũ_{3})).

P r o o f. We fix j ∈ {2, …, n + 1}. Since Θ^{j} is a holomorphic function, and since ˜ Ũ_{2} is assumed to be connected, there are two possibilities - either Θ^{j} ≡ 0 on π̂ ∘ ˜ ϕ̃_{3}( ˜Ũ_{2}) or there exist a neighbourhood W̃_{j} of ũ_{0} := π̂(˜ϕ̃_{3}(m)) such that Θ^{j}(ũ) ≠ 0 for any ũ ∈ W̃_{j} \ {ũ_{0}}.

In the former case we take c_{j} = 0.

Suppose that Θ^{j} ≢ 0. We can find r > 0 such that K(ũ_{0}, r) ⊂ W̃_{j}. Then

W̃′_{j} := K(ũ_{0}, r) \ {ũ : Re ũ = Re ũ_{0}, Im ũ ≤ Im ũ_{0}}

is a simply connected domain and Θ^{j}(ũ) ≠ 0 for any ũ ∈ W̃′_{j}. If this is the case, there exists a holomorphic function λ^{j} on W̃′_{j} such that

e^{λ^{j}} = Θ^{j}|W̃′_{j}

and

(λ^{j})′ = \frac{Θ^{j}′}{Θ^{j}}

on W̃′_{j}.

On the other hand, using the real representation we may write

(Θ^{j})′ = \frac{∂ϑ^{2j−1}}{∂u^{1}} + i\frac{∂ϑ^{2j}}{∂u^{1}} .

From the Weingarten formula

D_{\frac{∂}{∂u^{1}}} A_{3}ξ = −(A_{3} ∘ f)_{∗}S \frac{∂}{∂u^{1}} + μ(\frac{∂}{∂u^{1}} )A_{3}ξ + ν(\frac{∂}{∂u^{1}} )JA_{3}ξ

it follows that for any j ∈ {2, …, n + 1}

\frac{∂ϑ^{2j−1}}{∂u^{1}} = μ(\frac{∂}{∂u^{1}} )ϑ^{2j−1} + ν(\frac{∂}{∂u^{1}} )ϑ^{2j}

and

\frac{∂ϑ^{2j}}{∂u^{1}} = μ(\frac{∂}{∂u^{1}} )ϑ^{2j} − ν(\frac{∂}{∂u^{1}} )ϑ^{2j−1}.

Hence

(Θ^{j})′ = (μ(\frac{∂}{∂u^{1}} ) − iν(\frac{∂}{∂u^{1}} ))(ϑ^{2j−1} + iϑ^{2j}).

We may also assume that ˜ Ũ_{2} is simply connected, and from Lemma 2 we know that the 1-forms μ, ν are closed; therefore there exist functions K and L on ˜ Ũ_{2} such that μ = dK, ν = −dL. The functions K ∘ ϕ_{3}^{−1} and L ∘ ϕ_{3}^{−1} do not depend on the variables s and t, because \frac{∂}{∂s^{i}} , \frac{∂}{∂s^{i}} ∈ ker S ⊂ ker μ ∩ ker ν,

\frac{∂K ∘ ϕ_{3}^{−1}}{∂s^{i}} = dK(\frac{∂}{∂s^{i}}) = μ(\frac{∂}{∂s^{i}}) = 0,

and similarly for L in place of K or t in place of s. It follows that there exist functions H^{1} and H^{2} defined on some open subset of R^{2} such that K∘ϕ_{3}^{−1} = H^{1} ∘π̂ and L ∘ ϕ_{3}^{−1} = H^{2} ∘ π̂. We have now

μ(\frac{∂}{∂u^{s}}) = dK((ϕ_{3}^{−1})_{∗}e_{2n−2+s}) = dH^{1}(π̂_{∗}e_{2n−2+s}) = dH^{1}(e_{s}) = \frac{∂H^{1}}{∂u^{s}}

and

ν(\frac{∂}{∂u^{s}}) = −dL((ϕ_{3}^{−1})_{∗}e_{2n−2+s}) = −dH^{2}(π̂_{∗}e_{2n−2+s}) = −dH^{2}(e_{s}) = −\frac{∂H^{2}}{∂u^{s}} .

According to Remark 3, H^{1} and H^{2} satisfy the Cauchy-Riemann equations, therefore H̃ := H^{1} + iH^{2} is a holomorphic function.

Since

(H̃)′ = \frac{∂H^{1}}{∂u^{1}} + i\frac{∂H^{2}}{∂u^{1}} = μ(\frac{∂}{∂u^{1}} ) − iν(\frac{∂}{∂u^{1}} ),

we may go back to the (Θ^{j})′ and write

(Θ^{j})′ = H̃′Θ^{j}

on some neighbourhood π̂(˜ϕ̃_{3}( ˜Ũ′_{3})) of ũ_{0} (including ũ_{0}). Comparing this with (λ^{j})′ we obtain

H̃′ = (λ^{j})′

on W̃′′_{j} := W̃′_{j} ∩ π̂(˜ϕ̃_{3}( ˜Ũ′_{3})). Hence there exists d_{j} ∈ C such that λ^{j} = H̃ + d_{j} and

Θ^{j} = e^{H̃+d_{j}} = c_{j}eH̃

with a non-zero constant c_{j} := e^{d_{j}}. We may expand this equality from W̃′′_{j} to some

neighbourhood π̂(˜ϕ̃_{3}( ˜Ũ_{3})) of ũ_{0}, because both sides are continuous and well-defined in the neighbourhood of ũ_{0}. Since Θ^{n+1} ≢ 0, we have in particular c_{n+1} ≠ 0. ■

Next we use the following isomorphism of C^{n+1}:

˜Ã_{4}^{0}ẽ_{k} := \{ \begin{matrix}ẽ_{k} & for k = 1, …, n \\ \frac{1}{\overline{c_{n+1}}} (ẽ_{n+1} − ∑_{s=2}^{n} \overline{c_{s}}ẽ_{s}) & for k = n + 1. \end{matrix}

In this way we obtain

\begin{matrix}˜Ã_{4} ∘ f ∘ ˜ ϕ̃_{3}^{−1}(s̃, t̃^{1}, …, t̃^{n−2}, ũ) \\ = (s̃, ˜ σ̃^{1}(t̃, ũ), …, ˜ σ̃^{n−1}(t̃, ũ), ˜ Q̃(t̃, ũ)) \end{matrix}

and

˜Ã_{4}ξ ∘ ˜ ϕ̃_{3}^{−1}(s̃, t̃^{1}, …, t̃^{n−2}, ũ) = \overline{Θ^{1}(ũ)}ẽ_{1} + \overline{e^{H̃(ũ)}}ẽ_{n+1},

with ˜ Ã_{4} := ˜ Ã_{4}^{0} ∘ ˜ Ã_{3}, ˜ σ̃^{k}(t̃, ũ) := t̃^{k} − \overline{(\frac{c_{k+1}}{c_{n+1}} )}˜ρ̃(t̃, ũ) for k = 1, …, n − 2,

˜σ̃^{n−1} := ˜ χ̃(t̃, ũ) − \overline{(\frac{c_{k+1}}{c_{n+1}} )}˜ρ̃(t̃, ũ) and ˜ Q̃(t̃, ũ) := \overline{(\frac{1}{c_{n+1}} )}˜ρ̃(t̃, ũ).

Thus for any q > 0 it is possible to find a map ϕ̂_{3} and an isomorphism Â_{4} of C^{n+1} such that the immersion and the transversal field have the shape

\begin{matrix}Â_{4} ∘ f ∘ ϕ̂_{3}^{−1}(s̃^{1}, …, s̃^{q}, t̃^{1}, …, t̃^{n−2q}, ũ^{1}, …, ũ^{q}) \\ = (s̃^{1}, …, s̃^{q}, σ̂^{1}(t̃, ũ), …, σ̂^{n−q}(t̃, ũ), Q̂(t̃, ũ)) \end{matrix}

and

Â_{4}ξ ∘ ϕ̂_{3}^{−1}(s̃^{1}, …, s̃^{q}, t̃^{1}, …, t̃^{n−2q}, ũ^{1}, …, ũ^{q}) = ∑_{k=1}^{q} \overline{Θ̂^{k}(ũ)}ẽ_{k} + \overline{Θ̂^{n+1}(ũ)}ẽ_{n+1},

here σ̂^{i}, Q̂, Θ̂^{j} are holomorphic functions and Θ̂^{n+1} ≠ 0.

Since Â_{4} ∘ f is an immersion, and Â_{4} ∘ ξ is transversal to (Â_{4} ∘ f)_{∗}T M,

| \begin{matrix}I_{q} & \begin{matrix}0 & ⋯ & 0 \\ ⋮ & ... & ⋮ \\ 0 & ⋯ & 0 \end{matrix} & \begin{matrix}0 & ⋯ & 0 \\ ⋮ & ... & ⋮ \\ 0 & ⋯ & 0 \end{matrix} & \overline{Θ̂^{1}} ; ⋮ ; \overline{Θ̂^{q}} \\ \begin{matrix}0 & ⋯ & 0 \\ ⋮ & ... & ⋮ \\ 0 & ⋯ & 0 \end{matrix} & [\frac{∂σ̂^{k}}{∂t̃^{l}} ]1≤k≤n−q; 1≤l≤n−2q & [\frac{∂σ̂^{k}}{∂ũ^{l}} ]1≤k≤n−q; 1≤l≤q & 0 ⋮ 0 \\ 0 ⋯ 0 & \frac{∂Q̂}{∂t̃^{1}} … \frac{∂Q̂}{∂t̃^{n−2q}} & \frac{∂Q̂}{∂ũ^{1}} … \frac{∂Q̂}{∂ũ^{q}} & \overline{Θ̂^{n+1}} \end{matrix} | ≠ 0.

Therefore

det ([\frac{∂σ̂^{k}}{∂t̃^{l}} ]_{k=1,... ,n−q; l=1,... ,n−2q} [\frac{∂σ̂^{k}}{∂ũ^{l}} ]_{k=1,... ,n−q; l=1,... ,q}) ≠ 0

and there exist i_{1} < ⋯ < i_{n−2q} such that

det ([\frac{∂σ̂^{i_{k}}}{∂t̃^{l}} ]_{k=1,... ,n−2q; l=1,... ,n−2q}) ≠ 0.

By the appropiate isomorphism Â_{5}^{0} we may vary the order of basis vectors in C^{n+1}, putting σ̂^{i_{1}}, …, σ̂^{i_{n−2q}} at the positions q + 1, …, n − q. This permutation does not affect the field Â_{4}ξ, because its coordinates from the q + 1-th to the n-th vanish.

Applying now the local diffeomorphism

\begin{matrix}ϕ̂_{4}^{0}(s̃^{1}, …, s̃^{q}, t̃^{1}, …, t̃^{n−2q}, ũ^{1}, …, ũ^{q}) \\ := (s̃^{1}, …, s̃^{q}, σ̂^{i_{1}}(t̃, ũ), …, σ̂^{i_{n−2q}}(t̃, ũ), ũ^{1}, …, ũ^{q}) \end{matrix}

gives a new map ϕ̂_{4} := ϕ̂_{4}^{0} ∘ ϕ̂_{3} such that Â_{5} ∘ f and Â_{5} ∘ ξ are described by the formulas

\begin{matrix}Â_{5} ∘ f ∘ ϕ̂_{4}^{−1}(x̃^{1}, …, x̃^{q}, ỹ^{1}, …, ỹ^{n−2q}, z̃^{1}, …, z̃^{q}) \\ = (x̃^{1}, …, x̃^{q}, ỹ^{1}, …, ỹ^{n−2q}, ψ̂^{1}(ỹ, z̃), …, ψ̂^{q}(ỹ, z̃), F̂(ỹ, z̃)) \end{matrix}

and

Â_{5} ∘ ξ ∘ ϕ̂_{4}^{−1}(x̃^{1}, …, x̃^{q}, ỹ^{1}, …, ỹ^{n−2q}, z̃^{1}, …, z̃^{q}) = ∑_{k=1}^{q} \overline{Θ̂^{k}(z̃)}ẽ_{k} + \overline{Θ̂^{n+1}(z̃)}ẽ_{n+1}.

Lemma 12. For k = 1, …, q

ψ̂^{k}(ỹ, z̃) = ∑_{s=1}^{q} C^{k}_{s}ỹ^{s} + Π̂^{k}(z̃),

where C^{k}_{s}, s = 1, …, n − 2q are the complex numbers, and Π̂^{k} is a holomorphic function.

P r o o f. We shall now use the real representation of Â_{5} ∘ f and Â_{5} ∘ ξ, setting ψ̂^{l} = ψ̌^{2l−1} + iψ̌^{2l} for l = 1, …, q, F̂ = F̌^{1} + iF̌^{2}, and Θ̂^{s} = Θ̌^{2s−1} + iΘ̌^{2s}:

\begin{matrix}Ǎ_{5} ∘ f ∘ ϕ̌_{4}^{−1}(x^{1}, …, x^{2q}, y^{1}, …, y^{2n−4q}, z^{1}, …, z^{2q}) \\ = (x^{1}, …, x^{2q}, y^{1}, …, y^{2n−4q}, ψ̌^{1}(y, z), …, ψ̌^{2q}(y, z), F̌^{1}(y, z), F̌^{2}(y, z)), \end{matrix}

\begin{matrix}Ǎ_{5} ∘ ξ ∘ ϕ̌_{4}^{−1}(x^{1}, …, x^{2q}, y^{1}, …, y^{2n−4q}, z^{1}, …, z^{2q}) \\ = ∑_{k=1}^{q}[Θ̌^{2k−1}(z)e_{2k−1} − Θ̌^{2k}(z)e_{2k}] + Θ̌^{2n+1}(z)e_{2n+1} − Θ̌^{2n+2}(z)e_{2n+2}. \end{matrix}

At any point m′ of the domain Ǔ of ϕ̌_{4} ker S is spanned by

\frac{∂}{∂x^{1}} , …, \frac{∂}{∂x^{2q}} , \frac{∂}{∂y^{1}} , …, \frac{∂}{∂y^{2n−4q}} .

For any W ∈ T_{m′}M, for any j = 1, …, 2n − 4q

S(∇_{W} \frac{∂}{∂y^{j}}) = ∇_{W} (S \frac{∂}{∂y^{j}}) = 0,

therefore

∇_{\frac{∂}{∂y^{s}}} \frac{∂}{∂y^{j}} = ∑_{k=1}^{2q} α_{sj}^{k} \frac{∂}{∂x^{k}} + ∑_{l=1}^{2n−4q} β_{sj}^{l} \frac{∂}{∂y^{l}}

and

∇_{\frac{∂}{∂z^{s}}} \frac{∂}{∂y^{j}} = ∑_{k=1}^{2q} γ_{sj}^{k} \frac{∂}{∂x^{k}} + ∑_{l=1}^{2n−4q} δ_{sj}^{l} \frac{∂}{∂y^{l}} .

We have

(Ǎ_{5} ∘ f)_{∗}(\frac{∂}{∂x^{k}}) = e_{k},

(Ǎ_{5} ∘ f)_{∗}(\frac{∂}{∂y^{l}}) = e_{2q+l} + ∑_{r=1}^{2q} \frac{∂ψ̌^{r}}{∂y^{l}} e_{2n−2q+r} + \frac{∂F̌^{1}}{∂y^{l}} e_{2n+1} + \frac{∂F̌^{2}}{∂y^{l}} e_{2n+2};

hence

(Ǎ_{5} ∘ f)_{∗}(∇_{\frac{∂}{∂y^{s}}} \frac{∂}{∂y^{j}})

= ∑_{k=1}^{2q} α_{sj}^{k}e_{k} + ∑_{l=1}^{2n−4q} β_{sj}^{l}e_{2q+l} + ∑_{r=1}^{2q}( ∑_{l=1}^{2n−4q} β_{sj}^{l} \frac{∂ψ̌^{r}}{∂y^{l}})e_{2n−2q+r}

+ ( ∑_{l=1}^{2n−4q} β_{sj}^{l} \frac{∂F̌^{1}}{∂y^{l}})e_{2n+1} + ( ∑_{l=1}^{2n−4q} β_{sj}^{l} \frac{∂F̌^{2}}{∂y^{l}})e_{2n+2}

and

(Ǎ_{5} ∘ f)_{∗}(∇_{\frac{∂}{∂z^{s}}} \frac{∂}{∂y^{j}})

= ∑_{k=1}^{2q} γ_{sj}^{k}e_{k} + ∑_{l=1}^{2n−4q} δ_{sj}^{l}e_{2q+l} + ∑_{r=1}^{2q}( ∑_{l=1}^{2n−4q} δ_{sj}^{l} \frac{∂ψ̌^{r}}{∂y^{l}})e_{2n−2q+r}

+ ( ∑_{l=1}^{2n−4q} δ_{sj}^{l} \frac{∂F̌^{1}}{∂y^{l}})e_{2n+1} + ( ∑_{l=1}^{2n−4q} δ_{sj}^{l} \frac{∂F̌^{2}}{∂y^{l}})e_{2n+2}.

On the other hand, using the Gauss formula we may write

(Ǎ_{5} ∘ f)_{∗}(∇_{\frac{∂}{∂y^{s}}} \frac{∂}{∂y^{j}})

= D_{\frac{∂}{∂y^{s}}} (Ǎ_{5} ∘ f)_{∗} \frac{∂}{∂y^{j}} − h(\frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})Ǎ_{5}ξ + h(J \frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})JǍ_{5}ξ

= ∑_{k=1}^{q}[(−h(\frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2k−1}(z) + h(J \frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2k}(z))e_{2k−1}

+ (h(\frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2k}(z) + h(J \frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2k−1}(z))e_{2k}]

+ ∑_{r=1}^{2q} \frac{∂^{2}ψ̌^{r}}{∂y^{s}∂y^{j}} e_{2n−2q+r}

+ (\frac{∂^{2}F̌^{1}}{∂y^{s}∂y^{j}} − h(\frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2n+1}(z) + h(J \frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2n+2}(z))e_{2n+1}

+ (\frac{∂^{2}F̌^{2}}{∂y^{s}∂y^{j}} + h(\frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2n+2}(z) + h(J \frac{∂}{∂y^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2n+1}(z))e_{2n+2}

and similarly

(Ǎ_{5} ∘ f)_{∗}(∇_{\frac{∂}{∂z^{s}}} \frac{∂}{∂y^{j}})

= D_{\frac{∂}{∂z^{s}}} (Ǎ_{5} ∘ f)_{∗} \frac{∂}{∂y^{j}} − h(\frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})Ǎ_{5}ξ + h(J \frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})JǍ_{5}ξ

= ∑_{k=1}^{q}[(−h(\frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2k−1}(z) + h(J \frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2k}(z))e_{2k−1}

+ (h(\frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2k}(z) + h(J \frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2k−1}(z))e_{2k}]

+ ∑_{r=1}^{2q} \frac{∂^{2}ψ̌^{r}}{∂z^{s}∂y^{j}} e_{2n−2q+r}

+ (\frac{∂^{2}F̌^{1}}{∂z^{s}∂y^{j}} − h(\frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2n+1}(z) + h(J \frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2n+2}(z))e_{2n+1}

+ (\frac{∂^{2}F̌^{2}}{∂z^{s}∂y^{j}} + h(\frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2n+2}(z) + h(J \frac{∂}{∂z^{s}} , \frac{∂}{∂y^{j}})Θ̌^{2n+1}(z))e_{2n+2}

Since in the second pair of expressions there are no terms containing the basis vectors e_{t} with t ∈ {2q + 1, …, 2n − 2q}, we conclude that β_{sj}^{l} = 0 and δ_{sj}^{l} = 0 for any l, s, j. Comparing now the coefficients of e_{t′} with t′ ∈ {2n − 2q + 1, …, 2n} we obtain

\frac{∂^{2}ψ̌^{r}}{∂y^{s}∂y^{j}} = 0

and

\frac{∂^{2}ψ̌^{r}}{∂z^{s}∂y^{j}} = 0

for any r, s, j.

It follows that

\frac{∂ψ̌^{r}}{∂y^{j}} = E_{j}^{r} = const

and

ψ̌^{r}(y, z) = ∑_{j=1}^{2n−4q} E_{j}^{r}y^{j} + Π̌^{r}(z).

The Cauchy-Riemann equations for ψ̂^{r} imply that Π̂^{r} := Π̌^{2r−1} + iΠ̌^{2r} is a holomorphic function and E_{2j−1}^{2r−1} = E_{2j}^{2r}, E_{2j}^{2r−1} = −E_{2j−1}^{2r} for r = 1, …, q, j = 1, …, n − 2q. We set C^{k}_{s} := E_{2j−1}^{2k−1} + iE_{2j−1}^{2k} and the lemma follows. ■

We define the isomorphism Â_{6}^{0} of C^{n+1},

Â_{6}^{0}ẽ_{k} := \{ \begin{matrix}ẽ_{k} & if k ∉ \{q + 1, …, n − q\} \\ ẽ_{k} − ∑_{j=1}^{q} C^{j}_{k}ẽ_{n−q+j} & if k ∈ \{q + 1, …, n − q\}. \end{matrix}

Let Â_{6} := Â_{6}^{0} ∘ Â_{5}, then

\begin{matrix}Â_{6} ∘ f ∘ ϕ̂_{4}^{−1}(x̃^{1}, …, x̃^{q}, ỹ^{1}, …, ỹ^{n−2q}, z̃^{1}, …, z̃^{q}) \\ = (x̃^{1}, …, x̃^{q}, ỹ^{1}, …, ỹ^{n−2q}, Π̂^{1}(z̃), …, Π̂^{q}(z̃), F̂(ỹ, z̃)) \end{matrix}

and Â_{6}ξ ∘ ϕ̂_{4}^{−1} is given by the same formula as Â_{5}ξ ∘ ϕ̂_{4}^{−1}.

Lemma 13. If q > 1 then

Π̂^{r}(z̃) = ∑_{s=1}^{q} C_{1}^{r}_{s}z̃^{s} + C_{2}^{r}

with C_{1}^{r}_{s}, C_{2}^{r} ∈ C.

P r o o f. Recall that for q > 1

Θ̂^{k}(z̃^{1}, …, z̃^{q}) = z̃^{k} for k = 1, …, q

and

Θ̂^{n+1}(z̃^{1}, …, z̃^{q}) ≡ 1.

Therefore we have

\begin{matrix}A_{6}ξ ∘ ϕ_{4}^{−1}(x^{1}, …, x^{2q}, y^{1}, …, y^{2n−4q}, z^{1}, …, z^{2q}) \\ = ∑_{k=1}^{q}[z^{2k−1}e_{2k−1} − z^{2k}e_{2k}] + e_{2n+1}, \end{matrix}

D_{\frac{∂}{∂z^{s}}} (A_{6}ξ) = (−1)^{s−1}e_{s} = −(A_{6} ∘ f)_{∗}((−1)^{s} \frac{∂}{∂x^{s}})

which implies

S \frac{∂}{∂z^{s}} = (−1)^{s} \frac{∂}{∂x^{s}} .

Applying the covariant derivative ∇_{W} to the right-hand side of the above formula we obtain zero, therefore

S(∇_{W} \frac{∂}{∂z^{s}}) = ∇_{W} (S \frac{∂}{∂z^{s}}) = 0

and

∇_{W} \frac{∂}{∂z^{s}} ∈ ker S

for any tangent vector W. We can now proceed analogously to the proof of Lemma 12. ■

The matrix (C_{1}^{r}_{s})_{r,s=1,... ,q} is invertible, since Â_{6} ∘ f is an immersion and

Â_{6}ξ ∘ ϕ̂_{4}^{−1}(x̃^{1}, …, x̃^{q}, ỹ^{1}, …, ỹ^{n−2q}, z̃^{1}, …, z̃^{q}) = ∑_{k=1}^{q} \overline{z̃^{k}}ẽ_{k} + ẽ_{n+1}

is a transversal field. Let

(C_{3}^{i}_{j})_{i,j=1,... ,q} = [(C_{1}^{r}_{s})_{r,s=1,... ,q}]^{−1}.

To complete the proof of the theorem in the case q > 1 it remains to apply the affine isomorphism Ã_{7}^{0} of C^{n+1},

\begin{matrix}Ã_{7}^{0}(θ^{1}, …, θ^{n+1}) := \\ (θ^{1}, …, θ^{n−q}, ∑_{j_{1}=1}^{q} C_{3}^{1}_{j_{1}}(θ^{n−q+j_{1}} − C_{2}^{j_{1}}), …, ∑_{j_{q}=1}^{q} C_{3}^{q}_{j_{q}}(θ^{n−q+j_{q}} − C_{2}^{j_{q}}), θ^{n+1}). \end{matrix}

The linear part of Ã_{7}^{0} does not change the transversal field Â_{6}ξ, therefore we have Ã ∘ f and Ã^{→}ξ as claimed, with Ã := Ã_{7}^{0} ∘ Â_{6}, F̃ = F̂, ϕ̃ = ϕ̂_{4}.

For q = 1

Â_{6} ∘ f ∘ ϕ̂_{4}^{−1}(x̃, ỹ^{1}, …, ỹ^{n−2}, z̃) = (x̃, ỹ^{1}, …, ỹ^{n−2}, Π̂(z̃), F̂(ỹ, z̃))

with Π̂′ ≠ 0 in the neighbourhood of ˜ ϕ̃_{4}(m). We define a local diffeomorphism

˜ϕ̃_{5}^{0}(x̃, ỹ^{1}, …, ỹ^{n−2}, z̃) := (x̃, ỹ^{1}, …, ỹ^{n−2}, Π̂(z̃))

and obtain

Â_{6} ∘ f ∘ ˜ ϕ̃_{5}^{−1}(x̃, ỹ^{1}, …, ỹ^{n−2}, z̃) = (x̃, ỹ^{1}, …, ỹ^{n−2}, z̃, ˆ F̂(ỹ, z̃)),

Â_{6}ξ ∘ ˜ ϕ̃_{5}^{−1}(x̃, ỹ^{1}, …, ỹ^{n−2}, z̃) = \overline{˜G̃(z̃)}ẽ_{1} + \overline{e^{˜M̃(z̃)}}ẽ_{n+1}

with ˆ F̂ = F̂ ∘ Π̂^{−1}, ˜ M̃ = H̃ ∘ Π̂^{−1}, ˜ G̃ = Θ^{1} ∘ Π̂^{−1}, ˜ ϕ̃_{5} = ˜ ϕ̃_{5}^{0} ∘ ϕ̂_{4}.

Lemma 14.

˜G̃′ − ˜ M̃′˜G̃ = const ≠ 0.

P r o o f. Let ˜ G̃ = G^{1} + iG^{2}, ˜ M̃ = M^{1} + iM^{2}. Then ˜G̃′ − ˜ M̃′˜G̃

= (\frac{∂G^{1}}{∂z^{1}} + i\frac{∂G^{2}}{∂z^{1}} ) − (\frac{∂M^{1}}{∂z^{1}} + i\frac{∂M^{2}}{∂z^{1}} ) (G^{1} + iG^{2})

= (\frac{∂G^{1}}{∂z^{1}} − \frac{∂M^{1}}{∂z^{1}} G^{1} + \frac{∂M^{2}}{∂z^{1}} G^{2}) + i (\frac{∂G^{2}}{∂z^{1}} − \frac{∂M^{2}}{∂z^{1}} G^{1} − \frac{∂M^{1}}{∂z^{1}} G^{2}) .

It is easily seen that

S \frac{∂}{∂z^{1}} = (−\frac{∂G^{1}}{∂z^{1}} + \frac{∂M^{1}}{∂z^{1}} G^{1} − \frac{∂M^{2}}{∂z^{1}} G^{2}) \frac{∂}{∂x^{1}} +(\frac{∂G^{2}}{∂z^{1}} − \frac{∂M^{2}}{∂z^{1}} G^{1} − \frac{∂M^{1}}{∂z^{1}} G^{2}) \frac{∂}{∂x^{2}} ,

because

D_{\frac{∂}{∂z^{1}}} A_{6}ξ = \frac{∂G^{1}}{∂z^{1}} e_{1} − \frac{∂G^{2}}{∂z^{1}} e_{2} + (\frac{∂M^{1}}{∂z^{1}} e^{M^{1}} cos M^{2} − \frac{∂M^{2}}{∂z^{1}} e^{M^{1}} sin M^{2}) e_{2n+1}

+ (−\frac{∂M^{1}}{∂z^{1}} e^{M^{1}} sin M^{2} − \frac{∂M^{2}}{∂z^{1}} e^{M^{1}} cos M^{2}) e_{2n+2}

= (\frac{∂G^{1}}{∂z^{1}} − \frac{∂M^{1}}{∂z^{1}} G^{1} + \frac{∂M^{2}}{∂z^{1}} G^{2}) e_{1}

+ (−\frac{∂G^{2}}{∂z^{1}} + \frac{∂M^{2}}{∂z^{1}} G^{1} + \frac{∂M^{1}}{∂z^{1}} G^{2}) e_{2} + \frac{∂M^{1}}{∂z^{1}} A_{6}ξ − \frac{∂M^{2}}{∂z^{1}} JA_{6}ξ.

For any vector W

∇_{W} \frac{∂}{∂z^{1}} ∈ span_{R} \{\frac{∂}{∂x^{1}} , \frac{∂}{∂x^{2}} \} ⊂ ker S,

since

D_{W} (A_{6} ∘ f)_{∗} \frac{∂}{∂z^{1}} = W (\frac{∂ ˇF̌^{1}}{∂z^{1}} )e_{2n+1} + W (\frac{∂ ˇF̌^{2}}{∂z^{1}} )e_{2n+2} ∈ span_{R}\{e_{1}, e_{2}, A_{6}ξ, JA_{6}ξ\}.

It follows that ∇_{W} (S \frac{∂}{∂z^{1}} ) = 0 for any W. Therefore

−\frac{∂G^{1}}{∂z^{1}} + \frac{∂M^{1}}{∂z^{1}} G^{1} − \frac{∂M^{2}}{∂z^{1}} G^{2} =: −B_{1} = const

and

\frac{∂G^{2}}{∂z^{1}} − \frac{∂M^{2}}{∂z^{1}} G^{1} − \frac{∂M^{1}}{∂z^{1}} G^{2} =: B_{2} = const.

Moreover, (B_{1})^{2} + (B_{2})^{2} ≠ 0, because S ≠ 0. ■

Let

˜ϕ̃_{6}^{0}(x̃, ỹ^{1}, …, ỹ^{n−2}, z̃) := (x̃, ỹ^{1}, …, ỹ^{n−2}, (B_{1} + iB_{2})z̃),
˜Ã_{7}^{0}(ζ^{1}, …, ζ^{n+1}) := (ζ^{1}, …, ζ^{n−1}, (B_{1} + iB_{2})ζ^{n}, ζ^{n+1}).

Now ˜ Ã_{7} ∘ f ∘ ˜ ϕ̃_{6}^{−1} and ˜ Ã_{7}ξ ∘ ˜ ϕ̃_{6}^{−1} have the same shape as Â_{6} ∘ f ∘ ˜ ϕ̃_{5}^{−1} and Â_{6}ξ ∘ ˜ ϕ̃_{5}^{−1} with ˆ F̂, ˜ G̃, ˜ M̃ replaced by F̃, G̃ M̃;

F̃(z̃) := ˆ F̂(\frac{z̃}{B_{1} + iB_{2}} ), G̃(z̃) := ˜ G̃(\frac{z̃}{B_{1} + iB_{2}} ), M̃(z̃) := ˜ M̃(\frac{z̃}{B_{1} + iB_{2}} ).

It is easy to check that

G̃′ − M̃′G̃ = 1.

This finishes the proof of the theorem. ■

There are many examples of functions G̃ and M̃ satisfying the above equation. In fact, for any holomorphic M̃ (and in this way for any μ, ν, because μ(\frac{∂}{∂z^{k}}) = \frac{∂M^{1}}{∂z^{k}} and ν(\frac{∂}{∂z^{k}}) = −\frac{∂M^{1}}{∂z^{k}} ) there exists G̃ such that G̃′ − M̃′G̃ ≡ 1. For example

G̃(z̃) = z̃, M̃(z̃) = 0

(that is what we obtain also for q > 1, with μ = ν = 0);

G̃(z̃) = 1, M̃(z̃) = −z̃;

G̃(z̃) = ez̃, M̃(z̃) = z̃ + e^{−z̃}.

References.

[DVV] F. D i l l e n, L. V r a n c k e n and L. V e r s t r a e l e n, Complex affine differential geometry, Atti. Acad. Peloritana Pericolanti Cl. Sci. Fis. Mat. Nat., 66 (1988), 231-260.

[NS] K. N o m i z u and T. S a s a k i, Affine differential geometry, Cambridge University Press, 1994.

[O] B. O p o z d a, Equivalence theorems for complex affine hypersurfaces, Result. Math. 27 (1995), 316-327.