Study of some holomorphic curves in C3 and their projection into the complex projectve space CP2
Fathi Haggui and Abdessami Jalled
Abstract.
We study holomorphic curves f:C⟶C3 avoiding four complex hyperplanes and a real subspace of real dimension four or five in C3. We show that the projection of f into the complex projective space CP2 is not necessarily constant.
keywords
Complex projective space, Holomorphic curves, Kobayashi hyperbolicity
1. Introduction
The classical Picard Theorem [1] (see also [2]) states that every holomorphic map from the complex Euclidean space C to CP1 that avoids three points, is constant. This Theorem has been extended to higher dimension by M.Green [3] who provided with examples of complex Kobayashi hyperbolic manifolds. We note that if H1,...,Hm are complex hyperplanes in CPn, then they are said to be in general position if m⩾n+1 and any (n+1) of these hyperplanes are linearly independent. Let us recall the Green Theorem:
Theorem (Green, [3]). Let C be a union of 2n+1 complex hyperplanes in general position in CPn. Then, any holomorphic curve f:C→CPn∖C is constant.
In particular, for n=2, any holomorphic curve f:C→CP2∖C is constant, where C is a union of five complex lines in general position in CPn.
As a direct consequence of the Green Theorem, the canonical projection into the complex projective space CP2 of any holomorphic map f:C→C3 which avoids five complex hyperplanes in C3 is constant, since its image avoids the projections of the five complex hyperplanes, which are complex projective lines in general position in CP2 (see Lemma 2.1). Our main goal is to study the projection into CP2 of a holomorphic curve f:C→C3 which avoids four complex hyperplanes in general position in C3 and a real subspace H of real dimension four or five and check if the projection remains constant.
Throughout the paper we identify R6, endowed with its standard complex structure Jst, to C3.
Definition 1.1**.**
Let n⩾3 and let H=(H1,...,Hn) be a family of real subspaces of R6 such that codimRHj=2 for j=1,...,n. Then H is said to be in general position if for every 3-tuple (i,j,k) of distinct integers i,j,k∈{1,...,n},
[TABLE]
Here, if H is a real subspace in R6, then H⊥ denotes the orthogonal complement of H with respect to the Euclidean metric.
We first study the case of four real dimensional subspaces in C3. We have the following
Theorem 1**.**
(i)* Let H1,...,Hn be n complex hyperplanes in C3 in general position (n⩾5), then there exists a non constant holomorphic curve f:C→C3 which avoid these n hyperplanes and π(f) is constant.
(ii) Let H1,H2,H3,H4 be four complex hyperplanes in C3. Then there exists a real subspace H of R6, of real dimension four, such that (H,Hj,Hk) are in general position for all j=k, j,k∈{1,...,4}, and there exists f:C→C3 holomorphic, such that*
[TABLE]
and π(f) is non constant.
Remark 1.1**.**
Here π denotes the canonical projection from C3∖{0} into CP2 and π(f):=π∘f. Notice that π(f) is well-defined in Theorem 1 (ii) since f(C)⊂C3∖{0}.
In case (i), according to the Green Theorem and to Lemma 2.1 (see below), π(f) is constant.
We study then the case of a subspace in C3 of real dimension five. We have the following:
Theorem 2**.**
*Let H1,H2,H3,H4 be four complex hyperplanes in C3 and let H be a real subspace of R6 of real dimension five. Let H~ be a complex hyperplane of C3 such that H~⊂H. Then:
(1) If (H~,Hj,Hk) are in general position for all j=k, j,k∈{1,...,4}, then every holomorphic map f:C→C3 such that f(C)⋂(i=1⋃4Hi⋃H)=∅ is constant.
(2) If there exist Hj,Hk, j=k, j,k∈{1,...,4}, such that (H~,Hj,Hk) are not in general position, then there exists f:C→C3, holomorphic, such that \displaystyle f(\mathbb{C})\bigcap\big{(}\bigcup_{i=1}^{4}H_{i}\bigcup H\big{)}=\emptyset and π(f) is non constant.*
Remark 1.2**.**
(a)* The existence and uniqueness of H~⊂H is explained in the proof of Theorem 2.
(b) The condition ”(H~,Hj,Hk) are not in general position” is equivalent to the condition ”dimRSpanR(H~⊥,Hj⊥,Hk⊥)=4”.
(c) The fact of considering four complex hyperplanes is an optimal condition (see the end of section two for more details).*
The paper is organized as follows. In the first section, we give some results and properties. In section two, we prove Theorem 1. Finally, in section three, we prove Theorem 2.
2. Preliminaries and properties
In 1972, Fujimoto [4] (see also M.Green[3] and [5]) showed a statement that characterizes the image of a holomorphic map f:C→CPn omitting (n+p) hyperplanes in general position. He proved the following
Theorem (Fujimoto [4], Serge Lang [5] pp 196).
Let f:C→CPn be holomorphic. Assume that the image of f lies in the complement of n+p hyperplanes in general position, then this image is contained in a complex projective subspace of complex dimension ⩽[n/p].
The version of the Green Theorem stated in the introduction is a particular case of the previous Theorem, with p=n+1.
In the remaining of the paper we will need the following properties satisfied by the canonical projection in CP2 of a holomorphic curve f:C→C3. For H a real subspace of R6, we denote by H⋆ the set H∖{0}. Then, we have the following Lemma
Lemma 2.1**.**
Let π:C3∖{0}→CP2 be the canonical projection. Then:
- (1)
If H is a complex hyperplane in C3, then π(H⋆) is a complex projective line in CP2.
2. (2)
If f:C→C3 is holomorphic and H is a complex hyperplane in C3, then
[TABLE]
3. (3)
If H1,H2,H3 are complex hyperplanes in general position in C3, then π(H1⋆),π(H2⋆),π(H3⋆) are in general position in CP2.
Notation: if Z∈CP2, we denote [z1:z2:z3] its homogeneous coordinates, where
(z1,z2,z3)∈C3.
Proof.
Point (1). We may assume that H={(z1,z2,z3)∈C3/a1z1+a2z2+a3z3=0}, with a1,a2,a3∈C, a3=0. Then
[TABLE]
[TABLE]
We notice that [0:1:−a3a2] corresponds to [∞1:1:−a3∞a1+a2∞]. Hence π(H⋆) is a projective complex line in CP2.
Point (2). We first notice that π(f) is well defined since, by assumption f(C)∩H=∅, which implies that f(C)⊂C3∖{0}. Assume now, to get a contradiction, that π(f)(C)∩π(H⋆)=∅. Then there are two possibilities.
Case (α). There exists z∈C and there exists λ∈C such that
[TABLE]
Then, there exists cz∈C∗ such that \displaystyle f(z)=\big{(}c_{z},\lambda c_{z},-\frac{a_{1}+a_{2}\lambda}{a_{3}}c_{z}\big{)}. In particular a1f1(z)+a2f2(z)+a3f3(z)=0, where f=(f1,f2,f3). Hence, f(z)∈H. This is a contradiction.
Case (β). There exists z∈C such that
[TABLE]
Then, there exists cz∈C∗ such that \displaystyle f(z)=\big{(}0,c_{z},-\frac{a_{2}}{a_{3}}c_{z}\big{)} and a1f1(z)+a2f2(z)+a3f3(z)=0. We obtain again that f(z)∈H: this is a contradiction.
Point (3). Since H1,H2,H3 are complex hyperplanes in C3, then there is a linear change of coordinates such that the hyperplanes are defined by equations
[TABLE]
Now by projection into CP2, we get
[TABLE]
Hence π(H1⋆)∩π(H2⋆)∩π(H3⋆)=∅, meaning that π(H1⋆),π(H2⋆),π(H3⋆) are in general position since there is no triple point.
∎
3. Proof of Theorem 1
To prove theorem 1, we need the following Lemma which characterize the image of a holomorphic map f:C→CPn avoiding 2n complex hyperplanes in general position. This precises the result of H.Fujimoto [4], [5] pp 196.
Definition 3.1**.**
Let H1,...,Hm, m⩾2n, be hyperplanes of CPn. We call diagonal, a line passing through the two points i∈I⋂Hi and j∈J⋂Hj, where card(I)=card(J)=n and I∩J=∅.
Lemma 3.1**.**
Let H1,...,H2n be (2n) projective hyperplanes in general position in CPn. Then there are 21C2nn diagonals Δ1,....,Δ21C2nn such that for every holomorphic curve f:C⟶CPn∖i=1⋃2nHi, there exists kf∈{1,...,21C2nn} such that f(C)⊂Δkf.
Proof.
The proof is inspired by the Fujimoto Theorem, [5] pp 196.
Let f:C⟶CPn be holomorphic, such that f(C)⋂(i=1⋃2nHi)=∅.
Let L1,…, L2n be linear forms defining the hyperplanes H1,...,H2n, namely Hk=Lk−1({0}) for k=1,...,2n. If f=[f1:...:fn+1], we denote
[TABLE]
Let I={1,⋯,2n} be the set of indices and ∼ be the equivalence relation defined by i∼j if hi/hj is constant. We take a partition of the set of indices according to ∼. First, we know that the complement of a given class S has at most n elements (see [5] pp 197). Hence S has at least n elements and there are at most two classes.
The case of one class is not possible. In fact, There exists α2,...,α2n∈C such that
[TABLE]
Hence f(C)⊂(k=2⋂n+1Hk)⋂H1=k=1⋂n+1Hk=∅, which is impossible. Hence there are exactly two classes S1 and S2.
We know that each of the two classes S1,S2 contains n elements. Then there exists a permutation σ:{1,...,2n}→{1,...,2n} such that
[TABLE]
Hence There exists α2,...,αn,βn+1,...,β2n−1∈C such that h1,...,h2n satisfy the systems:
[TABLE]
Hence
[TABLE]
Then f(C)⊂Δσ, where Δσ is the unique diagonal (line) passing through the two points k=1⋂nHσ(k) and k=n+1⋂2nHσ(k).
Now the two points, and consequently Δσ, are completely determined by S1={σ(1),...,σ(n)} since S2 is automatically fixed once S1 is chosen. Hence Δσ is completely determined by a choice of a partition of {1,...,2n} into two subsets, each of them containing n elements. There are exactly 21C2nn such partitions. This proves the Lemma.
∎
We may prove now Theorem 1.
We denote by z=(z1,z2,z3) the coordinates in C3, where zj=xj+iyj, j=1,2,3. Hence (x1,y1,x2,y2,x3,y3) denote the coordinates in R6.
Point (i). Consider first the case n=5. By a linear change of coordinates, we take the hyperplanes H1,H2,H3,H4 and H5 in standard form defined by the following equations
[TABLE]
By hypothesis \displaystyle f(\mathbb{C})\cap\big{(}\bigcup_{i=1}^{5}H_{i}\big{)}=\emptyset. Then there exists h1,h2,h3:C→C, holomorphic, such that
[TABLE]
Moreover, since π(f)(C) omits π(Hi) for i=1,...,5 (see Lemma 2.1) and π∘f is constant by Green (see [3]), there exists (ω1,ω2,ω3)=(0,0,0) such that for all z∈C,
[TABLE]
Therefore
[TABLE]
which implies that
[TABLE]
Hence f=(eh1,c2eh1,c3eh1), with 1+c2+c3=0, and f is not constant.
Essentially the same type of argument works in general. Let H1,...,Hn, n⩾5, be n hyperplanes defined by:
[TABLE]
By hypothesis \displaystyle f(\mathbb{C})\bigcap\big{(}\bigcup_{i=1}^{n}H_{i}\big{)}=\emptyset, then in particular \displaystyle f(\mathbb{C})\cap\big{(}\bigcup_{i=1}^{5}H_{i}\big{)}=\emptyset and f=(eh,c2eh,c3eh) is not constant, where h is holomorphic from C to C.
Hence, in order that f avoids H1,...,Hn, it is sufficient to choose c2,c3∈C such that for every k=1,...,n
[TABLE]
We point out that what preceeds proves more generally that given a countable set of complex hyperplanes in C3 passing through the origin, there exists f:C→C3 not constant and avoiding each hyperplane. This proves Point (i).
Point (ii). Let H1,H2,H3 and H4 be four complex hyperplanes in general position in C3. We know that there is a linear change of coordinate such that H1,H2,H3 and H4 are defined in standard form by :
[TABLE]
Then
[TABLE]
We pose now
[TABLE]
Then \displaystyle H^{\perp}=Span_{\mathbb{R}}\big{[}(1,0,1,0,0,0);(1,0,0,0,1,0)\big{]}, which of course satisfies the condition SpanR(H⊥,Hj⊥,Hk⊥)=R6 for all j=k, j,k∈{1,...,4}.
Since f(C)⋂(i=1⋃4Hi)=∅, then there exists holomorphic functions fi:C→C, i=1,2,3 such that
[TABLE]
Then, by Lemma 2.1 (2), g:=π(f) satisfies g(C)⊂CP2∖j=1⋃4π(Hj⋆). Hence g has the following form
[TABLE]
where g2=f2−f1 and g3=f3−f1. According to Lemma 3.1 there exists 21C42=3 diagonals Δ12,34,Δ13,24,Δ14,23 such that g=π(f(C)) is contained in one of these diagonals, where Δij,kl is the diagonal line passing through \displaystyle\big{(}\pi(H^{\star}_{i})\bigcap\pi(H^{\star}_{j})\big{)} and \displaystyle\big{(}\pi(H^{\star}_{k})\bigcap\pi(H^{\star}_{l})\big{)}.
We recall that
[TABLE]
Hence Δ12,34,Δ13,24,Δ14,23 are given by
[TABLE]
Suppose that g(C) is contained in Δ12,34, the cases g(C)⊂Δ13,24 or g(C)⊂Δ14,23 being similar. Then
eg2+1=0⇒eg2=−1⇒g=[1:−1:eg3], where g3=f3−f1. Hence
[TABLE]
On another hand
f(C)∩H=∅⇔∀z∈C, \displaystyle\left\{\begin{array}[]{cllll}Re(e^{f_{1}(z)})&\neq&0\\
or\\
Re(e^{f_{1}(z)}-e^{f_{3}(z)})&\neq&0.\\
\end{array}\right.
We pose f3=2f1, then f=(ef1,−ef1,e2f1) avoids H. In fact
[TABLE]
Now if Re(ef1(z))=0 for some z∈C, then Im(ef1(z))=0 and consequently Re(ef3(z))=0. Hence f(C)∩H=∅.
Finally, π(f)=[1:−1:ef1] is not constant and \displaystyle f(\mathbb{C})\bigcap\big{(}\bigcup_{j=1}^{4}H_{j}\bigcup H\big{)}=\emptyset. This concludes the proof of Theorem 1.
4. Proof of Theorem 2
Let H be a real subspace of C3 such that dimRH=5, then H contains a unique complex hyperplane H~ of C3. Indeed, there exists (a1,b1,a2,b2,a3,b3)∈R6∖{0} such that
[TABLE]
Hence H~:={z∈C3/ j=1∑3(aj−ibj)zj=0} is a complex hyperplane in C3, contained in H.
Point (1). Assume that (H~,Hj,Hk) are in general position for some j=k, j,k∈{1,...,4}. Since H~⊂H, where H~ is a complex hyperplane of C3, and
[TABLE]
then it follows from Theorem 1 (i) that there is (c1,c2)∈(C∗)2 which satisfies 1+c2+c3=0 and there exists h:C→C holomorphic such that
[TABLE]
On another hand H:={(x1,y1,...,x3,y3)∈R6/ j=1∑3(aixi+biyi)=0}. By hypothesis f(C)∩H=∅ then for every z∈C we have,
[TABLE]
Thus, for every z∈C
[TABLE]
We denote
[TABLE]
then
[TABLE]
However {(x,y)∈R2 / ax+by=0} is either a real line or R2, depending on the values of a and b. Then by the little Picard Theorem eh is constant because it avoids an infinite number of points. Hence h is constant and f is then constant. We point out that the projection of f into CP2 is also constant.
Point (2). Suppose there exists j=k, j,k∈{1,...,4}, such that
dimRSpanR(H~⊥,Hj⊥,Hk⊥)=4. Then:
[TABLE]
In fact for all i=1,...,4, dimRHi⊥=2, then dimRSpanR(Hi⊥,Hj⊥)=4.
Suppose H~⊥⊂SpanR(H1⊥,H2⊥) then there exists α1,α2∈C such that
[TABLE]
Since f(C)⋂(i=1⋃4Hi)=∅, then by 3
[TABLE]
We take f1=c, c∈C∖{0}, such that Re(α1ec−α2ec)=0 and f3 not constant. Then
[TABLE]
avoids i=1⋃4Hi⋃H, and π(f) is not constant.
This concludes the proof of Theorem 2.
∎
By the end of the paper, we show the optimality of considering four complex hyperplanes. Let H1,H2,H3 be three complex hyperplanes in C3, then there exists H a real hyperplane in R6 and a complex hyperplane H~ contained in H, \displaystyle\big{(}H_{1},H_{2},H_{3},\tilde{H}\big{)} are in general position, and there exists f:C→C3, holomorphic, such that \displaystyle f(\mathbb{C})\cap\big{(}\bigcup_{j=1}^{3}H_{j}\bigcup H\big{)}=\emptyset and π∘f is not constant. In fact:
We pose H={x1+x2+x3=0} and H~={z1+z2+z3=0}, which is clearly contained in H. Since \displaystyle f(\mathbb{C})\bigcap\big{(}\bigcup_{j=1}^{3}H_{j}\bigcup H\big{)}=\emptyset, then \displaystyle f(\mathbb{C})\bigcap\big{(}\bigcup_{j=1}^{3}H_{j}\bigcup\tilde{H}\big{)}=\emptyset and f=(ef1,ef2,ef3). Hence
[TABLE]
where g2=f2−f1 and g3=f3−f1. By lemma 3.1, g:=π(f) is contained in one of diagonals Δ12,34,Δ13,24,Δ14,23 (see 2). Suppose π(f)(C)⊂Δ13,24, then
[TABLE]
Hence f=(1,eg2,−eg2) avoids \displaystyle\big{(}\bigcup_{j=1}^{3}H_{j}\bigcup H\big{)} and π(f) is not constant.