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In this paper, we give some results on the number of meromorphic pings ofCm In 1926, Nevanlinna showed that for two nonconstant meromorphic functions f and g on the complex plane C, if t

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An Extension of Uniqueness Theorems

for Meromorphic Mappings

1Universit´ e de Bretagne Occidentale UFR Sciences et Techniques

D´ epartement de Math´ ematiques 6, avenue Le Gorgeu,

BP 452 29275 Brest Cedex, France

2Dept of Math., Hanoi University of Education, 136 Xuan Thuy Road

Cau Giay, Hanoi, Vietnam

Received February 22, 2005Revised June 20, 2005

Abstract. In this paper, we give some results on the number of meromorphic pings ofCm

In 1926, Nevanlinna showed that for two nonconstant meromorphic functions

f and g on the complex plane C, if they have the same inverse images for five

distinct values, then f = g, and that g is a special type of a linear fractional formation of f if they have the same inverse images, counted with multiplicities,

tran-for four distinct values

In 1975, Fujimoto [2] generalized Nevanlinna’s result to the case of morphic mappings of Cm into CP n This problem continued to be studied by

mero-Smiley [9], Ji [5] and others.

in CP n such that imf  H Denote by v (f,H)the map ofCminto N0 such that

v (f,H) (a) (a ∈ C m ) is the intersection multiplicity of the image of f and H at

f (a) Let k be a positive interger or +∞ We set

72 Gerd Dethloff and Tran Van Tan

j=1, f, p ) the set of all linearly

nondegenerate meromorphic mappings g ofCmintoCP n such that:

In [5], Ji proved the following

Theorem J [5] If q = 3n + 1 and k = + ∞, then for three mappings f1, f2, f3∈

a proper algebraic subset of CP n × CP n × CP n

In 1929, Cartan declared that there are at most two meromorphic functions

on C which have the same inverse images (ignoring multiplicities) for four tinct values However in 1988, Steinmetz [10] gave examples which showed thatCartan’s declaration is false On the other hand, in 1998, Fujimoto [4] showedthat Cartan’s declaration is true if we assume that meromorphic functions onCshare four distinct values counted with multiplicities truncated by 2 He gavethe following theorem

dis-Theorem F [4] If q = 3n + 1 and k = + ∞ then F k

He also proposed an open problem asking if the number q = 3n+1 in Theorem

F can be replaced by a smaller one Inspired by this question, in this paper we

will generalize the above results to the case where the number q = 3n + 1 is

in fact replaced by a smaller one We also obtain an improvement concerningtruncating multiplicities

Denote by Ψ the Segre embedding of CP n × CP n into CP n2+2n which is

defined by sending the ordered pair ((w0, , w n ), (v0, , v n )) to ( , w i v j , ) (in

lexicographic order)

h n2+2n) be a representation of Ψ◦ h We say that h is linearly degenerate

(with the algebraic structure in CP n × CP n given by the Segre embedding) if

h0, , h n2+2n are linearly dependent overC

Our main results are stated as follows:

Theorem 1 There are at most two distinct mappings in F k

(α1, , α m) of nonnegative integers, set |α| := α1+· · · + α m and D α F :=

D α F (z) = 0 for all α with |α| < p Let k be a positive integer or + ∞.

Define the map v F k) ofCm intoN0by

v k F)(z) :=



v F (z) if v F (z)  k.

We define the map v ϕ k) as

follows For each z ∈ C m

, choose nonzero holomorphic functions F and G on a neighborhood U of z such that ϕ = F G on U and dim

F −1(0)∩G −1(0)

 m−2 Then put v ϕ k)(z) := v F k)(z) Set

74 Gerd Dethloff and Tran Van Tan

for m = 1 Set N (r, v ϕ ) := N1+∞) (r, v ϕ ) and N k)(r, v ϕ) :=

N1k)(r, v ϕ ) For a closed subset A of a purely (m −1)-dimensional analytic subset

homo-geneous coordinates (w0 : · · · : w n) onCP n, we take a reduced representation

f = (f0 :· · · : f n ), which means that each f i is a holomorphic function onCm

and f (z) = (f0(z) : · · · : f n (z)) outside the analytic set {f0=· · · = f n= 0} of

For a nonzero meromorphic function ϕ on C m , the characteristic function T ϕ (r)

of ϕ is defined by considering ϕ as a meromorphic mapping ofCm intoCP1.

Let H = {a0w0+· · ·+a n w n = 0} be a hyperplane in CP n such that imf  H Set (f, H) := a0f0+· · · + a n f n We define

N f k)(r, H) := N k)(r, v (f,H) ) and N l,f k)(r, H) := N l k)(r, v (f,H) ).

Sometimes we write N k f)(r, H) for N 1,f k)(r, H), N l,f (r, H) for N l,f +∞) (r, H) and

N f (r, H) for N +∞,f +∞) (r, H).

We note that m(r, ϕ) = m ϕ (r, + ∞) + O(1) ([4], p 135).

We state First and Second Main Theorems of Value Distribution Theory

First Main Theorem Let f :Cm → CP n

be a meromorphic mapping and H

meromorphic mapping and H1, , H q be hyperplanes in general position in CP n Then

except for a set E ⊂ (1, +∞) of finite Lebesgue measure.

The following so-called logarithmic derivative lemma plays an essential role

in Nevanlinna theory

Theorem 2.1 ([5], Lemma 3.1) Let ϕ be a non-constant meromorphic function

on Cm Then for any i, 1  i  m, we have



= o(T ϕ (r)) as r → ∞, r /∈ E, where E ⊂ (1, +∞) of finite Lebesgue measure.

Let F, G and H be nonzero meromorphic functions on Cm For each l, 1

l  m, we define the Cartan auxiliary function by

76 Gerd Dethloff and Tran Van Tan

Theorem 2.2 Let F, G, H be nonzero meromorphic functions onCm Assume that Φ l (F, G, H) ≡ 0 and Φ l1

F are all constant.

3 Proof of the Theorems

First of all, we need the following lemmas:

Lemma 1 Let f1, , f d be linearly nondegenerate meromorphic mappings of

i (H c)) m − 2 for all i ∈ {1, , d} and j ∈ {1, , q}.

Proof Without loss of generality, we may assume that l = 1 For an arbitrary

point a ∈ C m \ A satisfying v k)

(f1,H j0)(a) > 0, we have v

k)

(f i ,H j0)(a) > 0 for all

i ∈ {1, 2, 3} We choose a such that a /∈ 3

i=1f

−1

i (H c) We distinguish two cases,which lead to equations (1) and (2)

Case 1 If v (f1,H j0)(a) ≥ p, then v (f i ,H j0)(a) ≥ p, i ∈ {1, 2, 3} This means that

a is a zero point of F ic j0 with multiplicity≥ p for i ∈ {1, 2, 3} We have

also to all other combinations of indices, we see that

a is a zero point of Φ1c with multiplicity ≥ p − 1 (1)

Case 2 If v (f1,H j0)(a)  p, then p0:= v (f1,H j0)(a) = v (f2,H j0)(a) = v (f3,H j0)(a)

p There exists a neighborhood U of a such that v (f1,H j0)  p on U

In-deed, there exists otherwise a sequence {a s } ∞

s=1 ⊂ C m, with lim

s →∞ a s = a and

v (f1,H j0)(a s)≥ p+1 for all s By the definition, we have D β

(f1, H j0)(a s) = 0 forall|β| < p+1 So D β (f1, H j0)(a) = lim

s →∞ D

β (f1, H j0)(a s) = 0 for all|β| < p+1.

Thus v (f1,H j0)(a) ≥ p + 1 This is a contradiction Hence v (f1,H j0) p on U.

We can choose U such that U ∩ A = ∅ , v (f i ,H j0) p on U and (f i , H c) has

no zero point on U for all i ∈ {1, 2, 3} Then v F j0

1c = 0} By shrinking U we may assume that there exists

a holomorphic function h on U such that dh has no zero point and F j0

a is a zero point of Φ1c with mulitplicity ≥ p0. (2)

By (1), (2) and our choice of a, there exists an analytic set M ⊂ C mwithcodimension≥ 2 such that vΦ 1 ≥ min{v (f1,H j0), p − 1} on

For each j ∈ {1, , q} \ {j0}, let a (depending on j) be an arbitrary point

in Cm such that v (f k),H )(a) > 0 (if there exist any) Then v (f k),H )(a) > 0

78 Gerd Dethloff and Tran Van Tan

for all i ∈ {1, 2, 3}, since f1, f2, f3 ∈ F k

v 1

Φ1c (a)  1 + max{v 1

F j0 ic

by (3) we may assume that v 1

Φ1c (a) = 0 (outside an analytic set of codimension

(a), i = 1, 2, 3 } + 1 <3

i=1

v 1

F j0 ic

+

80 Gerd Dethloff and Tran Van Tan

The following lemma is a version of Second Main Theorem without taking

account of multiplicities of order > k in the counting functions.

Lemma 3 Let f be a linearly nondegenerate meromorphic mapping of Cm into CP n and {H j } q

j=1(q ≥ n + 2) be hyperplanes in CP n in general position Take a positive integer k with q −n−1 qn  k  +∞ Then

for all r > 1 except a set E of finite Lebesgue measure.

Proof By First and Second Main Theorems, we have

Case 1 1  n  3, q = 3n + 1, p = 2, k ≥ 23n Suppose that (12) does not hold, then #Q ≥ 3 For each j0∈ Q, by Lemma 2 (with A = ∅, p = 2) we have

82 Gerd Dethloff and Tran Van Tan

Then A1∪ A2∪ A3= (1, + ∞) Without loss of generality, we may assume that

the Lebesgue measure of A1is infinite By (14) we have

choose homogeneous coordinates (ω0 : · · · : ω n) on CP n with H j ={ω j = 0}

(1 j  n + 1) and take reduced representations:

84 Gerd Dethloff and Tran Van Tan

By (15) and (16) we have

3(k2− 6nk − 6n + 2) k(k + 1) 2k + 1 − 3n

Suppose that (12) does not hold, then there exists j0 ∈ Q By Lemma 2

which is a contradiction Thus, we get (12) in this case

So, in any case we have #({1, , q} \ Q) ≥ 3n − 1 Without loss of

generality, we may assume that 1, , 3n − 1 /∈ Q Then we have

86 Gerd Dethloff and Tran Van Tan

In the following we distinguish the cases n = 1 and n ≥ 2.

Case 1 If n = 1, then a j := H j (j = 1, 2, 3, 4) are distinct points in CP1 We

are distinct nonconstant meromorphic functions By (17) and by Theorem 2.2,

there exist constants α, β such that

g2= αg1, g3= βg1, (α, β / (18)

We have v (f1,a3 )≥ k + 1 on {z : (f1, a3)(z) = 0 } Indeed, otherwise there exists

z0such that 0 < v (f1,a3 )(z0  k Then v k)

(f i ,a3 )(z0) > 0, for all i ∈ {1, 2, 3} We

have (f1, a3)(z0) = (f2, a3)(z0) = 0 so f1(z0) = f2(z0) = a ∗ , where we denote

a ∗ j := (a j1 :−a j0) for every point a j = (a j0 : a j1)∈ CP1 So g

1(z0) = g2(z0) =

(a ∗ , a1

is a contradiction Thus v (f1,a3 ) ≥ k + 1 on {z : (f1, a3)(z) = 0 } Similarly,

Thus α ij = 1 for all 1

By (19), for i = 3n − 1, j ∈ {1, , 3n − 2}, we may asssume without loss

88 Gerd Dethloff and Tran Van Tan

For 1  s < v  3, denote by L sv the set of all j ∈ {1, , 3n − 2} such that

This is a contradiction Thus, for any case we have that f1, f2, f3 can not be

Proof of Theorem 2 Assume that #F k({H j } q

j=1, f, 1) ≥ 3 Take arbitrarily

three distinct mappings f1, f2, f3 ∈ F k({H j } q

j=1, f, 1) We have to prove that

f s × f v:Cm −→ CP n × CP n is linearly degenerate for all 1 s < v  3 Denote by Q the set which contains all indices j ∈ {1, , q} satisfying

Φl

F 1c j , F 2c j , F 3c j

even

Case 1 If n is odd, then q = 5(n+1)2 .

Indeed, otherwise there exists j0∈ Q Then by Lemma 2 (with A = ∅, p =

Hence, we obtain

(2q − 2n − 4)(k + 1) − 2(q − 1)n  3n(k + 1)

implying

k + 1  (5n + 3)n.

This is a contradiction Thus, we get (21)

Case 2 If n is even, then q = 5n+42

Indeed, suppose that this assertion does not hold, then there exist two

distinct indices j0, j1∈ Q By Lemma 2 (with A = ∅, p = 1) we have

90 Gerd Dethloff and Tran Van Tan

i (H j1) Then p0 := v (f i ,H j1)(a) ≥ 2 Since a is a regular point

of f i −1 (H j1) we can choose nonzero holomorphic functions h and u on a borhood U of a such that dh and u have no zeroes and (f i , H j1)≡ h p0u on U

neigh-Since a ∈ A i there exists b ∈ A i ∩U Then, we get 1 = v (f i ,H j1)(b) = v h p0 u (b) =

i (H j1) This means that

B \ B is included in an analytic set of codimension ≥ 2 So we have

By Lemma 2 (with A = B , p = 2) we have

By (21) and (22) we have #({1, , q}\Q) ≥ q−1 Without loss of generality

we may assume that 1, , q − 1 /∈ Q For any j ∈ {1, , q − 1} we have

92 Gerd Dethloff and Tran Van Tan

If there exists some L sv = ∅, we may assume without loss of generality

that L13 =∅ Then L12∪ L23 ={1, , q − 2} Since q = 5(n + 1)

get that f s × f v : Cm −→ CP n × CP n is linearly degenerate We thus have

Acknowledgements The second author would like to thank Professor Do Duc Thai

for valuable discussions, the Universit´e de Bretagne Occidentale for its hospitality andsupport, and the PICS-CNRS For MathVietnam for its support

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