Báo cáo toán học: "An Extension of Uniqueness Theorems for Meromorphic Mappings"

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Trong bài báo này, chúng tôi cung cấp cho một số kết quả về số lượng của ánh xạ meromorphic của Cm vào CP n trong điều kiện là những hình ảnh nghịch đảo của hyperplanes trong CP n. Đồng thời, chúng tôi đưa ra một câu trả lời cho một câu hỏi mở được đặt ra bởi H. Fujimoto vào năm 1998.

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Vietnam Journal of Mathematics 34:1 (2006) 71–94 9LHWQD P -RXUQDO RI 0$ 7+ (0$ 7, &6 9$67 An Extension of Uniqueness Theorems for Meromorphic Mappings Gerd Dethloﬀ1 and Tran Van Tan2 1 Universit´ de Bretagne Occidentale UFR Sciences et Techniques e D´partement de Math´matiques 6, avenue Le Gorgeu, e e BP 452 29275 Brest Cedex, France 2 Dept. of Math., Hanoi University of Education, 136 Xuan Thuy Road Cau Giay, Hanoi, Vietnam Received February 22, 2005 Revised June 20, 2005 Abstract. In this paper, we give some results on the number of meromorphic map- pings of Cm into CP n under a condition on the inverse images of hyperplanes in CP n . At the same time, we give an answer for an open question posed by H. Fujimoto in 1998. 1. Introduction 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 tran- formation of f if they have the same inverse images, counted with multiplicities, for four distinct values. In 1975, Fujimoto [2] generalized Nevanlinna’s result to the case of mero- morphic mappings of Cm into CP n . This problem continued to be studied by Smiley [9], Ji [5] and others. Let f be a meromorphic mapping of Cm into CP n and H be a hyperplane in CP n such that imf H. Denote by v(f,H ) the map of Cm into N0 such that v(f,H ) (a) (a ∈ Cm ) 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 0 if v(f,H ) (a) > k, k) v(f,H ) (a) = v(f,H ) (a) if v(f,H ) (a) k. Let f be a linearly nondegenerate meromorphic mapping of Cm into CP n and {Hj }q=1 be q hyperplanes in general position with j k) k) m − 2 for all 1 (a) dim z : v(f,Hi ) (z ) > 0 and v(f,Hj ) (z ) > 0 i 0 . j =1 In [5], Ji proved the following Theorem J. [5] If q = 3n + 1 and k = +∞, then for three mappings f1 , f2 , f3 ∈ Fk {Hj }q=1 , f, 1 , the mapping f1 × f2 × f3 : Cm −→ CP n × CP n × CP n is j algebraically degenerate, namely, {(f1 (z ), f2 (z ), f3 (z )), z ∈ Cm } is contained in 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 dis- tinct values. However in 1988, Steinmetz [10] gave examples which showed that Cartan’s declaration is false. On the other hand, in 1998, Fujimoto [4] showed that Cartan’s declaration is true if we assume that meromorphic functions on C share four distinct values counted with multiplicities truncated by 2. He gave the following theorem Theorem F. [4] If q = 3n + 1 and k = +∞ then Fk {Hj }q=1 , f, 2 contains at j most two mappings. 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 concerning truncating multiplicities. 2 Denote by Ψ the Segre embedding of CP n × CP n into CP n +2n which is deﬁned by sending the ordered pair ((w0 , ..., wn ), (v0 , ..., vn )) to (..., wi vj , ...) (in lexicographic order). Let h : Cm −→ CP n × CP n be a meromorphic mapping. Let (h0 : ... : hn2 +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 , ..., hn2 +2n are linearly dependent over C. Our main results are stated as follows: Theorem 1. There are at most two distinct mappings in Fk {Hj }q=1 , f, p in j each of the following cases:
An Extension of Uniqueness Theorems 73 3, q = 3n + 1, p = 2 and 23n k +∞ i) 1 n (6n − 1)n k +∞ ii) 4 n 6, q = 3n, p = 2 and n−3 (6n − 4)n iii) n ≥ 7, q = 3n − 1, p = 1 and k +∞ n−6 5(n + 1) +∞, where Theorem 2. Assume that q = , (65n + 171)n k 2 [x] := max{d ∈ N : d x} for a positive constant x. Then one of the following assertions holds : i) #Fk {Hj }q=1 , f, 1 2. j ii) For any f1 , f2 ∈ Fk {Hj }q=1 , f, 1 , the mapping f1 × f2 : Cm −→ CP n × CP n j is linearly degenerate (with the algebraic structure in CP n × CP n given by the Segre embedding). We ﬁnally remark that we obtained similar uniqueness theorems with moving targets in [11], but only with a bigger number of targets and with much bigger truncations. 2. Preliminaries We set z := (|z1 |2 + · · · + |zm |2 )1/2 for z = (z1 , . . . , zm ) ∈ Cm , B (r) := z : √ −1 (∂ − ∂ ), υ := (ddc z 2)m−1 and c z < r , S (r) := z : z = r , d := 4π σ := dc log z 2 ∧ (ddc log z 2 )m−1 . Let F be a nonzero holomorphic function on Cm . For an m-tuple α := (α1 , . . . , αm ) of nonnegative integers, set |α| := α1 + · · · + αm and Dα F := ∂ |α| F α . We deﬁne the map vF : C → N0 by vF (z ) := max p : m α1 ∂z1 . . . ∂zmm Dα F (z ) = 0 for all α with |α| < p . Let k be a positive integer or +∞. k) Deﬁne the map vF of Cm into N0 by 0 if vF (z ) > k, k) vF (z ) := vF (z ) if vF (z ) k . k) Let ϕ be a nonzero meromorphic function on Cm . We deﬁne the map vϕ as follows. For each z ∈ Cm , choose nonzero holomorphic functions F and G on a neighborhood U of z such that ϕ = G on U and dim F −1 (0) ∩ G−1 (0) m − 2. F k) k) Then put vϕ (z ) := vF (z ). Set k) vϕ) := z : vϕ (z ) > 0 . k Deﬁne r nk) (t) k) dt, (1 < r < +∞) N (r, vϕ ) := t2m−1 1 where
74 Gerd Dethloﬀ and Tran Van Tan nk) (t) := vϕ) υ for m ≥ 2, k k) vϕ ∩B (t) and nk) (t) := vϕ) (z ) for m = 1. k |z | t Set N (r, vϕ ) := N +∞) (r, vϕ ). For l a positive integer or +∞, set r k) nl (t) k) dt, (1 < r < +∞) Nl (r, vϕ ) := t2m−1 1 k) k) min vϕ) , l υ for m ≥ 2 and nl (t) := k where nl (t) := |z | t k) ∩B (t) vϕ k) k) +∞) min vϕ (z ), l for m = 1. Set N (r, vϕ ) := N1 (r, vϕ ) and N (r, vϕ ) := k) N1 (r, vϕ ). For a closed subset A of a purely (m − 1)-dimensional analytic subset of Cm , we deﬁne r n(t) dt, (1 < r < +∞), N (r, A) := t2m−1 1 ⎧ where for m ≥ 2, ⎨ υ A∩B (t) n(t) := ⎩ #(A ∩ B (t)) for m = 1. Let f : C → CP be a meromorphic mapping. For arbitrarily ﬁxed homo- m n geneous coordinates (w0 : · · · : wn ) on CP n , we take a reduced representation f = (f0 : · · · : fn ), which means that each fi is a holomorphic function on Cm and f (z ) = (f0 (z ) : · · · : fn (z )) outside the analytic set {f0 = · · · = fn = 0} of codimension ≥ 2. Set f := (|f0 |2 + · · · + |fn |2 )1/2 . The characteristic function of f is deﬁned by log f σ − Tf (r) := log f σ, r > 1. S (r ) S (1) For a nonzero meromorphic function ϕ on C m , the characteristic function Tϕ (r) of ϕ is deﬁned by considering ϕ as a meromorphic mapping of Cm into CP 1 . Let H = {a0 w0 +· · ·+an wn = 0} be a hyperplane in CP n such that imf H . Set (f, H ) := a0 f0 + · · · + an fn . We deﬁne k) k) k) Nf (r, H ) := N k) (r, v(f,H ) ) and Nl,f (r, H ) := Nl (r, v(f,H ) ). k) k) +∞) Sometimes we write N f (r, H ) for N1,f (r, H ), Nl,f (r, H ) for Nl,f (r, H ) and +∞) Nf (r, H ) for N+∞,f (r, H ).
An Extension of Uniqueness Theorems 75 1/2 |a0 |2 + · · · + |an |2 f Set ψf (H ) := . We deﬁne the proximity function (f, H ) by log | ψf (H ) | σ − log | ψf (H ) | σ. mf (r, H ) := S (r ) S (1) For a nonzero meromorphic function ϕ, the proximity function is deﬁned by log+ | ϕ | σ. m(r, ϕ) := S (r ) We note that m(r, ϕ) = mϕ (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 a hyperplane in CP n such that im f H . Then Nf (r, H ) + mf (r, H ) = Tf (r). For a nonzero meromorphic function ϕ we have N (r, v ϕ ) + m(r, ϕ) = Tϕ (r) + O(1). 1 Second Main Theorem. Let f : Cm → CP n be a linearly nondegenerate meromorphic mapping and H1 , ..., Hq be hyperplanes in general position in CP n . Then q (q − n − 1)Tf (r) Nn,f (r, Hj ) + o(Tf (r)) j =1 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 ∂ ∂zi ϕ = o(Tϕ (r)) as r → ∞, r ∈ E, m r, / ϕ 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 1 1 1 1 1 1 Φl (F, G, H ) := F · G · H · . F G H 1 1 1 ∂ ∂ ∂ ∂zl F ∂zl G ∂zl H By [4] (Proposition 3.4) we have the following
76 Gerd Dethloﬀ and Tran Van Tan Theorem 2.2. Let F, G, H be nonzero meromorphic functions on Cm . Assume 111 that Φl (F, G, H ) ≡ 0 and Φl ≡ 0 for all l, 1 l m. Then one of ,, FGH the following assertions holds i) F = G or G = H or H = F . FGH ii) ,, are all constant. GH F 3. Proof of the Theorems First of all, we need the following lemmas: Lemma 1. Let f1 , ..., fd be linearly nondegenerate meromorphic mappings of Cm into CP n and {Hj }q=1 be hyperplanes in CP n .Then there exists a dense j subset C ⊂ Cn+1 {0} such that for any c = (c0 , ..., cn ) ∈ C , the hyperplane Hc deﬁned by c0 ω0 + · · · + cn ωn = 0 satisﬁes dim (fi−1 (Hj ) ∩ fi−1 (Hc )) m − 2 for all i ∈ {1, . . . , d} and j ∈ {1, ..., q }. Proof. We refer to [5], Lemma 5.1. Let f1 , f2 , f3 ∈ Fk {Hj }q=1 , f, 1 , for q ≥ n + 1. Set j T (r) := Tf1 (r) + Tf2 (r) + Tf3 (r). (fi , Hj ) j For each c ∈ C , set Fic := for i ∈ {1, 2, 3} and j ∈ {1, . . . , q }. (fi , Hc ) Lemma 2. Assume that there exist j0 ∈ {1, ..., q }, c ∈ C , l ∈ {1, ..., m} and a closed subset A of a purely (m − 1) -dimensional analytic subset of Cm satisfying j0 j0 j0 1) Φl := Φl F1c , F2c , F3c ≡ 0, and c k) k) k) 2) min v(f1 ,Hj ) , p = min v(f2 ,Hj ) , p = min v(f3 ,Hj ) , p on Cm \ A, where 0 0 0 p is a positive integer. Then q k) k) N (r, vΦl ) + (p − 1)N (r, A) 2 N fi (r, Hj )+ Np−1,fi (r, Hj0 ) c j =1,j =j0 k+2 T (r) + (p + 2)N (r, A) + o(T (r)) k+1 for all i ∈ {1, 2, 3}. Proof. Without loss of generality, we may assume that l = 1. For an arbitrary k) k) point a ∈ Cm \ A satisfying v(f1 ,Hj ) (a) > 0, we have v(fi ,Hj ) (a) > 0 for all 0 0 3 fi−1 (Hc ). i ∈ {1, 2, 3}. We choose a such that a ∈ / We distinguish two cases, i=1 which lead to equations (1) and (2).
An Extension of Uniqueness Theorems 77 Case 1. If v(f1 ,Hj0 ) (a) ≥ p, then v(fi ,Hj0 ) (a) ≥ p, i ∈ {1, 2, 3}. This means that j a is a zero point of Fic0 with multiplicity ≥ p for i ∈ {1, 2, 3}. We have ∂ 1 ∂ 1 Φ1 = F1c F3c j0 j0 j0 j0 − F1c F2c c j0 j0 ∂z1 F3c ∂z1 F2c j0 j0 ∂ 1 j0 j0 ∂ 1 − F2c F3c + F2c F1c j0 j0 ∂z1 F1c ∂z1 F3c ∂ 1 ∂ 1 j0 j0 j0 j0 − F3c F1c + F3c F2c j0 . j0 ∂z1 F2c ∂z1 F1c j0 ∂ j0 −F1c ∂z1 F3c ∂ 1 j0 j0 On the other hand F1c F3c = , so a is a zero point of j0 j0 ∂z1 F3c F3c ∂ 1 j0 j0 with multiplicity ≥ p − 1. By applying the same argument F1c F3c j0 ∂z1 F3c also to all other combinations of indices, we see that a is a zero point of Φ1 with multiplicity ≥ p − 1 . (1) c Case 2. If v(f1 ,Hj0 ) (a) p, then p0 := v(f1 ,Hj0 ) (a) = v(f2 ,Hj0 ) (a) = v(f3 ,Hj0 ) (a) p. There exists a neighborhood U of a such that v(f1 ,Hj0 ) p on U. In- deed, there exists otherwise a sequence {as }∞ ⊂ Cm , with lim as = a and s=1 s→∞ v(f1 ,Hj0 ) (as ) ≥ p +1 for all s. By the deﬁnition, we have Dβ (f1 , Hj0 )(as ) = 0 for all |β | < p + 1. So Dβ (f1 , Hj0 )(a) = lim Dβ (f1 , Hj0 )(as ) = 0 for all |β | < p + 1. s→∞ Thus v(f1 ,Hj0 ) (a) ≥ p + 1. This is a contradiction. Hence v(f1 ,Hj0 ) p on U. We can choose U such that U ∩ A = ∅ , v(fi ,Hj0 ) p on U and (fi , Hc ) has no zero point on U for all i ∈ {1, 2, 3}. Then vF j0 = vF j0 = vF j0 p on U. So 1c 2c 3c j0 j0 j0 U ∩ {F1c = 0} = U ∩ {F2c = 0} = U ∩ {F3c = 0}. Choose a such that a is j0 regular point of U ∩{F1c = 0}. By shrinking U we may assume that there exists j a holomorphic function h on U such that dh has no zero point and Fic0 = hp0 ui on U, where ui (i = 1, 2, 3) are nowhere vanishing holomorphic functions on U (note that vF j0 (a) = vF j0 (a) = vF j0 (a) = p0 ). We have 1c 2c 3c − u1 ∂z1 u3 − u3 ∂z1 u1 hp0 ∂ ∂ ∂ ∂ hp0 u3 ∂z1 u2 u2 ∂z1 u3 Φ1 = u 1 + u2 c u2 u3 u3 u1 − u1 ∂z1 u2 hp0 ∂ ∂ u2 ∂z1 u1 + u3 . u1 u2 So, we have a is a zero point of Φ1 with mulitplicity ≥ p0 . (2) c By (1), (2) and our choice of a, there exists an analytic set M ⊂ Cm with codimension ≥ 2 such that vΦ1 ≥ min{v(f1 ,Hj0 ) , p − 1} on c k) z : v(f1 ,Hj ) (z ) > 0 \ (M ∪ A). (3) 0 For each j ∈ {1, . . . , q } \ {j0 }, let a (depending on j ) be an arbitrary point k) k) in Cm such that v(f1 ,Hj ) (a) > 0 (if there exist any). Then v(fi ,Hj ) (a) > 0
78 Gerd Dethloﬀ and Tran Van Tan for all i ∈ {1, 2, 3}, since f1 , f2 , f3 ∈ Fk {Hj }q=1 , f, 1 . We can choose a ∈ / j fi−1 (Hc ) ∪ fi−1 (Hj0 ) , i = 1, 2, 3. Then there exists a neighborhood U of a such that v(fi ,Hj ) k on U and (fi , Hj0 ), (fi , Hc ) ( i = 1, 2, 3 ) have no zero point on U . We have B := f1 1 (Hj ) ∩ U = f2 1 (Hj ) ∩ U = f3 1 (Hj ) ∩ U and − − − 1 1 1 j0 = j0 = j0 on B. Choose a such that a is a regular point of B . By F1c F2c F3c shrinking U, we may assume that there exists a holomorphic function h on U 1 1 such that dh has no zero point and U ∩ {h = 0} = B . Then j0 − j0 = hϕ2 F2c F1c 1 1 and j0 − j0 = hϕ3 on U where ϕ2 , ϕ3 are holomorphic functions on U . F3c F1c Hence, we get 1 0 0 1 hϕ2 hϕ3 Φ1 j0 j0 j0 j0 = F1c F2c F3c F1c c 1 ∂ ∂ ∂ ∂ ∂ ϕ2 ∂z1 h + h ∂z1 ϕ2 ϕ3 ∂z1 h + h ∂z1 ϕ3 j0 ∂z1 F1c ϕ2 ϕ3 = F1c F2c F3c h2 j0 j0 j0 . ∂ ∂ ∂z1 ϕ2 ∂z1 ϕ3 Therefore, a is a zero point of Φ1 with multiplicity ≥ 2. Thus, for each j ∈ c {1, . . . , q } \ {j0 }, there exists an analytic set N ⊂ Cm with codimension ≥ 2 such that vΦ1 ≥ 2 on c k) z : v(f1 ,Hj ) (z ) > 0 \ N. (4) By (3) and (4), we have q k) k) N (r, vΦ1 ) + (p − 1)N (r, A). 2 N f1 (r, Hj ) + Np−1,f1 (r, Hj0 ) c j =1,j =j0 Similarly, we have q k) k) N (r, vΦ1 )+(p−1)N (r, A), i = 1, 2, 3. 2 N fi (r, Hj )+Np−1,fi (r, Hj0 ) c j =1,j =j0 (5) Let a be an arbitrary zero point of some Fic0 , a ∈ fi−1 (Hc ), say i = 1. We have j / ∂ 1 j0 ∂ 1 Φ1 = F2c − F3c F1c j0 j0 j0 j0 j0 j0 + F3c − F1c F2c ∂z c j0 ∂z1 F1c 1 F2c j0 ∂ 1 j0 j0 + F1c − F2c F3c j0 . (6) ∂z1 F3c So we have
An Extension of Uniqueness Theorems 79 1 + max{v (a), i = 2, 3} v (a) 1+ v (a) + v (a). 1 1 1 1 Φ1 j j j F0 F0 F0 c 2c 3c ic k) Furthermore, if 0 < vF j0 (a) k and, hence, v(f1 ,Hj ) (a) > 0 and a ∈ A, then / 0 1c by (3) we may assume that v 11 (a) = 0 (outside an analytic set of codimension Φc ≥ 2). (7) j Let a be an arbitrary pole of all Fic0 , i = 1, 2, 3. By (6) we have 3 max{v (a), i = 1, 2, 3} + 1 < v (a) v (a) (8) 1 1 1 Φ1 j j F0 F0 c i=1 ic ic It follows from (6) that a pole of Φ1 is a zero or a pole of some Fic0 . Thus, by j c (6), (7) and (8), we have 3 3 k) N r, vF j0 − N N r, v N r, v + r, vF j0 + 3N (r, A) 1 1 Φ1 j F0 ic ic c i=1 i=1 ic 3 3 1 N r, v + N r, vF j0 + 3N (r, A) 1 k+1 j F0 ic i=1 i=1 ic 3 3 1 N r, v + TF j0 (r) + 3N (r, A) 1 k+1 j F0 ic i=1 i=1 ic 3 1 N r, v + T (r) + 3N (r, A) + O(1). (9) 1 k+1 j F0 i=1 ic We have ∂ 1 ∂ 1 Φ1 = F1c F3c j0 j0 j0 − F2c c j0 j0 ∂z1 F3c ∂z1 F2c j0 ∂ 1 j0 ∂ 1 j0 − F3c + F2c F1c j0 j0 ∂z1 F1c ∂z1 F3c ∂ 1 ∂ 1 j0 j0 j0 − F1c + F3c F2c j0 j0 ∂z1 F2c ∂z1 F1c 3 3 1 so m(r, Φ1 ) j j∂ m(r, Fic0 ) + 2 m r, Fic0 ∂z1 + 0(1). By Theorem 2.1, j0 c Fic i=1 i=1 we have ∂ 1 j m r, Fic0 = o TF j0 (r) . j ∂z1 Fic0 ic Thus, we get 3 m(r, Φ1 ) j m(r, Fic0 ) + o(T (r)), (10) c i=1
80 Gerd Dethloﬀ and Tran Van Tan (note that TF j0 (r) Tfi (r) + O(1)). ic By (9), (10) and by First Main Theorem, we have + m(r, Φ1 ) + O(1) N (r, vΦ1 ) TΦ1 (r) + O(1) = N r, v 1 c c c Φ1c 3 1 j + m(r, Fic0 ) + N r, v T (r) + 3N (r, A) + o(T (r)) 1 k+1 j F0 i=1 ic 3 1 TF j0 (r) + T (r) + 3N (r, A) + o(T (r)) k+1 ic i=1 3 1 Tfi (r) + T (r) + 3N (r, A) + o(T (r)) k+1 i=1 k+2 = T (r) + 3N (r, A) + o(T (r)). (11) k+1 By (5) and (11) we get Lemma 2. 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 {Hj }q=1 (q ≥ n + 2) be hyperplanes in CP n in general position. j Take a positive integer k with q−qn 1 k +∞. Then n− q k k) Tf (r) Nn,f (r, Hj )+o(T f (r)) (q − n − 1)(k + 1) − qn j =1 q nk k) N f (r, H j ) + o(T f (r)) (q − n − 1)(k + 1) − qn j =1 for all r > 1 except a set E of ﬁnite Lebesgue measure. Proof. By First and Second Main Theorems, we have q (q − n − 1)Tf (r) Nn,f (r, Hj ) + o Tf (r) j =1 q q k n k) Nn,f (r, Hj ) + Nf (r, Hj ) + o Tf (r) k+1 k+1 j =1 j =1 q k qn k) ≤ Tf (r) + o Tf (r) , r ∈ E, Nn,f (r, Hj ) + / k+1 k+1 j =1 which impies that q qn k k) q−n−1− Tf (r) Nn,f (r, Hj ) + o Tf (r) . k+1 k+1 j =1
An Extension of Uniqueness Theorems 81 Thus, we have q k k) Tf (r) Nn,f (r, Hj )+o(T f (r)) (q − n − 1)(k + 1) − qn j =1 q nk k) N f (r, H j ) + o(T f (r)) (q − n − 1)(k + 1) − qn j =1 Proof of Theorem 1. Assume that there exist three distinct mappings f1 , f2 , f3 ∈ Fk ({Hj }q=1 , f, p). Denote by Q the set which contains all indices j ∈ {1, . . . , q } j j j j satisfying Φl F1c , F2c , F3c ≡ 0 for some c ∈ C and some l ∈ {1, ..., m}. We now prove that #({1, ..., q }\Q) ≥ 3n − 1. (12) For the proof of (12) we distinguish three cases: 3, q = 3n + 1, p = 2, k ≥ 23n. Suppose that (12) does not Case 1. 1 n hold, then #Q ≥ 3. For each j0 ∈ Q, by Lemma 2 (with A = ∅, p = 2) we have q k+2 k) k) 2 N fi (r, Hj ) + N fi (r, Hj0 ) T (r) + o(T (r)), i = 1, 2, 3. (13) k+1 j =1,j =i0 By (13) and Lemma 3 we have q qn nk k) q−n−1− Tfi (r) N fi (r, Hj ) + o Tfi (r) k+1 k+1 j =1 nk (k + 2) nk k) T (r) + N fi (r, Hj0 ) + o T (r) , i = 1, 2, 3. 2 2(k + 1) 2(k + 1) Thus, we obtain 3 qn 3nk (k + 2) nk k) q −n−1− T (r) T (r) + N fi (r, Hj0 ) + o T (r) , 2(k + 1)2 k+1 2(k + 1) i=1 which implies 2(q − n − 1)(k + 1)2 − 2qn(k + 1) − 3nk (k + 2) T (r) 3 k) k) nk (k + 1) N fi (r, Hj0 ) + o(T (r)) = 3nk (k + 1)N fi (r, Hj0 ) + o(T (r)). i=1 Hence, we have k) 2(q − n − 1)(k + 1)2 − 2qn(k + 1) − 3nk (k + 2) N fi (r, Hj0 ) ≥ lim inf r →∞ r ∈E T (r) 3nk (k + 1) 2 k − 6nk − 6n + 2 , i ∈ {1, 2, 3}. = (14) 3k (k + 1)
82 Gerd Dethloﬀ and Tran Van Tan Set , i ∈ {1, 2, 3}. Ai := r > 1 : Tfi (r) = min Tf1 (r), Tf2 (r), Tf3 (r) Then A1 ∪ A2 ∪ A3 = (1, +∞). Without loss of generality, we may assume that the Lebesgue measure of A1 is inﬁnite. By (14) we have k) k 2 − 6nk − 6n + 2 N f1 (r, Hj0 ) ≥ , j0 ∈ Q. lim inf Tf1 (r) k (k + 1) r →∞ r ∈A1 \E Take three distinct indices j1 , j2 , j3 ∈ Q (note that #Q ≥ 3). Then we have k) k) k) 3(k 2 − 6nk − 6n + 2) N f1 (r, Hj1 ) + N f1 (r, Hj2 ) + N f1 (r, Hj3 ) ≥ lim inf , Tf1 (r) k (k + 1) r →∞ r ∈A1 \E which implies that q k) N f1 (r, Hj ) 3(k 2 − 6nk − 6n + 2) j =1 ≥ lim inf . (15) Tf1 (r) k (k + 1) r →∞ r ∈A1 \E (f1 , H1 ) (f2 , H1 ) Since f1 ≡ f2 there exists c ∈ C such that ≡ . Indeed, (f1 , Hc ) (f2 , Hc ) (f1 , H1 ) (f2 , H1 ) ≡ otherwise by Lemma 1 we have that for all hyperplanes (f1 , H ) (f2 , H ) (f1 , H1 ) (f2 , H1 ) H in CP n . In particular ≡ for all j = 2, ..., n + 1. We (f1 , Hj ) (f2 , Hj ) choose homogeneous coordinates (ω0 : · · · : ωn ) on CP n with Hj = {ωj = 0} (1 j n + 1) and take reduced representations: f1 = (f11 : · · · : f1n+1 ), f2 = (f21 : · · · : f2n+1 ). Then ⎧ ⎨ f1j = f2j f1 f11 = · · · = n+1 , so f1 ≡ f2 . ⇒ f1 f21 ⎩1 f21 f2n+1 (j = 2, . . . , n + 1) This is a contradiction. Since dim(fi−1 (H1 ) ∩ fi−1 (Hc )) m − 2 we have log (|(fi , H1 )|2 + |(fi , Hc )|2 ) 2 σ + O(1) 1 T (fi ,H1 ) (r) = (fi ,Hc ) S (r ) log fi σ + O(1) = Tfi (r) + O(1), i = 1, 2, 3. S (r ) q k) Since f1 = f2 on z : v(f1 ,Hj ) (z ) > 0 and j =1
An Extension of Uniqueness Theorems 83 k) k) m − 2 for all i = j, dim z : v(f1 ,Hi ) (z ) > 0 and v(f1 ,Hj ) (z ) > 0 we have q k) N f1 (r, Hj ) N r, v (f1 ,H1 ) − (f2 ,H1 ) T (f1 ,H1 ) − (f2 ,H1 ) (r) + 0(1) (f1 ,Hc ) (f2 ,Hc ) (f1 ,Hc ) (f2 ,Hc ) j =1 T (f1 ,H1 ) (r) + T (f2 ,H1 ) (r) + 0(1) Tf1 (r) + Tf2 (r) + 0(1), (f1 ,Hc ) (f2 ,Hc ) which implies Tf1 (r) + Tf2 (r) ≥ 1. lim inf q r →∞ k) N f1 (r, Hj ) j =1 On the other hand, by Lemma 3, we have q qn nk k) q−n−1− Tfi (r) N fi (r, Hj ) + o Tfi (r) k+1 k+1 j =1 q nk k) = N f1 (r, Hj ) + o Tfi (r) , k+1 j =1 which implies Tfi (r) nk lim sup , i = 1, 2, 3. (q − n − 1)(k + 1) − qn q k) r →∞ r ∈E N f1 (r, Hj ) j =1 Hence, we obtain Tf1 (r) Tf1 (r) + Tf2 (r) Tf2 (r) − lim sup = lim sup q q q k) k) k) r →∞ r ∈A1 \E r →∞ r ∈A1 \E N f1 (r, Hj ) N f1 (r, Hj ) N f1 (r, Hj ) j =1 j =1 j =1 Tf1 (r) + Tf2 (r) Tf2 (r) ≥ − lim inf lim sup q q k) k) r →∞ r ∈A1 \E r →∞ r ∈A1 \E N f1 (r, Hj ) N f1 (r, Hj ) j =1 j =1 nk ≥1− . (q − n − 1)(k + 1) − qn Consequently, we get q k) N f1 (r, Hj ) (q − n − 1)(k + 1) − qn j =1 lim inf (q − n − 1)(k + 1) − qn − nk Tf1 (r) r →∞ r ∈A1 \E 2k + 1 − 3n = . (16) k + 1 − 3n
84 Gerd Dethloﬀ and Tran Van Tan By (15) and (16) we have 3(k 2 − 6nk − 6n + 2) 2k + 1 − 3n . k + 1 − 3n k (k + 1) This contradicts k ≥ 23n. Thus, we get (12) in this case. Case 2. 4 n 6, q = 3n, p = 2, k ≥ (6n−1)n . n−3 Suppose that (12) does not hold, then there exists j0 ∈ Q. By Lemma 2 (with A = ∅, p = 2) we have 3n k+2 k) k) 2 N fi (r, Hj ) + N fi (r, Hj0 ) T (r) + o(T (r)), i = 1, 2, 3. k+1 j =1,j =j0 On the other hand, by Lemma 3 we have 3n (2n − 2)(k + 1) − (3n − 1)n k) N fi (r, Hj ) + o(Tfi (r)) ≥ Tfi (r), nk j =1,j =j0 and 3n (2n − 1)(k + 1) − 3n2 k) N fi (r, Hj ) + o(Tfi (r)) ≥ Tfi (r), nk j =1 which implies that 3n (4n − 3)(k + 1) − (6n − 1)n k) k) N fi (r, Hj )+N fi (r, Hj0 )+o(Tfi (r)) ≥ 2 Tfi (r). nk j =1,j =j0 Hence, we have (4n − 3)(k + 1) − (6n − 1)n k+2 Tfi (r) T (r) + o(T (r)), i = 1, 2, 3. nk k+1 Consequently, we get (4n − 3)(k + 1) − (6n − 1)n 3(k + 2) T (r) T (r) + o(T (r)), nk k+1 which implies that 3nk (k + 2) (4n − 3)(k + 1) − (6n − 1)n T (r) T (r) + o(T (r)) k+1 3n(k + 1)T (r) + o(T (r). (6n−1)n Hence, we obtain k + 1 n−3 . This is a contradiction. Thus, we get (12) in this case. (6n − 4)n Case 3. n ≥ 7, q = 3n − 1, p = 1, k ≥ . n−6
An Extension of Uniqueness Theorems 85 Suppose that (12) does not hold, then there exists j0 ∈ Q. By Lemma 2 (with A = ∅, p = 1) we have 3n−1 k+2 k) 2 N fi (r, Hj ) T (r) + o(T (r)), i = 1, 2, 3, k+1 j =1,j =j0 k) (note that N0,fi (r, Hj0 ) = 0). On the other hand, by Lemma 3, we have 3n−1 (2n − 3)(k + 1) − (3n − 2)n k) N fi (r, Hj ) + o(Tfi (r)) ≥ 2 2 Tfi (r). nk j =1,j =j0 Hence, we get 2[(2n − 3)(k + 1) − (3n − 2)n)] k+2 Tfi (r) T (r) + o(T (r)), nk k+1 which implies nk (k + 2) ((4n − 6)(k + 1) − (6n − 4)n)Tfi (r) T (r) + o(T (r)), i = 1, 2, 3. k+1 Hence, we have 3nk (k + 2) ((4n − 6)(k + 1) − (6n − 4)n)T (r) T (r) + o(T (r)) k+1 3n(k + 1)T (r) + o(T (r)). Thus, we obtain (4n − 6)(k + 1) − (6n − 4)n 3n(k + 1) implying (6n − 4)n k+1 , n−6 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 / j j j Φl F1c , F2c , F3c ≡ 0 for all c ∈ C , l ∈ {1, ..., m}, j ∈ {1, . . . , 3n − 1}. On the other hand, C is dense in Cn+1 . Hence, Φl F1c , F2c , F3c ≡ 0 for all j j j c ∈ Cn+1 \ {0}, l ∈ {1, ..., m}, j ∈ {1, . . . , 3n − 1}. In particular (for Hc = Hi ) we have (f1 , Hj ) (f2 , Hj ) (f3 , Hj ) ≡0 Φl , , (f1 , Hi ) (f2 , Hi ) (f3 , Hi ) 3 n − 1 , l ∈ { 1 , . . . , m} . for all 1 i=j (17)
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 aj := Hj (j = 1, 2, 3, 4) are distinct points in CP 1 . We have that (f1 , a1 ) (f2 , a1 ) (f3 , a1 ) g1 := , g2 := , g3 := (f1 , a2 ) (f2 , a2 ) (f3 , a2 ) 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 ∈ {1, ∞, 0}, α = β ). / (18) We have v(f1 ,a3 ) ≥ k + 1 on {z : (f1 , a3 )(z ) = 0}. Indeed, otherwise there exists k) z0 such that 0 < v(f1 ,a3 ) (z0 ) k . Then v(fi ,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 3 a∗ := (aj1 : −aj0 ) for every point aj = (aj0 : aj1 ) ∈ CP 1 . So g1 (z0 ) = g2 (z0 ) = j (a∗ , a1 ) 3 = 0, ∞ (note that a3 = a1 , a3 = a2 ). So, by (18) we have α = 1. This (a∗ , a2 ) 3 is a contradiction. Thus v(f1 ,a3 ) ≥ k + 1 on {z : (f1 , a3 )(z ) = 0}. Similarly, v(fi ,aj ) ≥ k + 1 on {z : (fi , aj )(z ) = 0} for i ∈ {1, 2, 3}, j ∈ {3, 4}. (a∗ , a1 ) α (a∗ , a1 ) (a∗ , a1 ) Set b1 = α 3 3 , b3 = 3 , b2 = . Then we have (a∗ , a2 ) β (a∗ , a2 ) (a∗ , a2 ) 3 3 3 vg2 −b3 = v (f2 ,a3 )(a∗ ,a2 ) ≥ k + 1 on {z : (g2 − b3 )(z ) = 0}, 1 (f2 ,a2 )(a∗ ,a2 ) 3 vg2 −b1 = vg1 − α b1 = v (f1 ,a3 )(a∗ ,a2 ) ≥ k + 1 on {z : (g2 − b1 )(z ) = 0}, and 1 1 (f1 ,a2 )(a∗ ,a2 ) 3 vg2 −b2 = vg3 − β b2 = v (f3 ,a3 )(a∗ ,a2 ) ≥ k + 1 on {z : (g2 − b2 )(z ) = 0}. 1 α (f3 ,a2 )(a∗ ,a2 ) 3 Since the points b1 , b2 , b3 are distinct, by First and Second Main Theorems, we have 3 Tg2 (r) N r, vg2 −bj + o Tg2 (r) j =1 3 1 N r, vg2 −bj + o Tg2 (r) k+1 j =1 3 Tg (r) + o Tg2 (r) . k+1 2 This contradicts k ≥ 23. Case 2. If n ≥ 2, for each 1 i = j 3n − 1, by (17) and Theorem 2.2, there exists a constant αij such that (f2 , Hj ) (f1 , Hj ) (f3 , Hj ) (f1 , Hj ) = αij or = αij (f2 , Hi ) (f1 , Hi ) (f3 , Hi ) (f1 , Hi )
An Extension of Uniqueness Theorems 87 or (f3 , Hj ) (f2 , Hj ) = αij . (19) (f3 , Hi ) (f2 , Hi ) 3n − 1. Indeed, if there We now prove that αij = 1 for all 1 i=j (f2 , Hj0 ) exists αi0 j0 = 1, we may assume without loss of generality that = (f2 , Hi0 ) q (f1 , Hj0 ) k) αi0 j0 . On the other hand f1 = f2 on Ω := z : v(f1 ,Hj ) (z ) > 0 . (f1 , Hi0 ) j =1 Hence, (f1 , Hj0 ) = (f2 , Hj0 ) = 0 on Ω \ f1 1 (Hi0 ). So we have − q k) k) N r, v (f1 ,Hj0 ) + N r, v(f1 ,Hi0 ) − N N f1 (r, Hj ) r, v(f1 ,Hi0 ) . (f1 ,Hi ) j =1,j =i0 0 Thus, by First and Second Main Theorems, we have q (q − n − 2)Tf1 (r) Nn,f1 (r, Hj ) + o(Tf1 (r)) j =1,j =i0 q n N1,f1 (r, Hj ) + o(Tf1 (r)) j =1,j =i0 q q nk n k) N f1 (r, Hj ) + Nf1 (r, Hj ) + o(Tf1 (r)) k+1 k+1 j =1,j =i0 j =1,j =i0 nk nk k) N r, v(f1 ,Hi0 ) − N r, v(f1 ,Hi0 ) N r, v (f1 ,Hj0 ) + k+1 k+1 (f1 ,Hi ) 0 (q − 1)n + Tf1 (r) + o(Tf1 (r)) k+1 (q − 1)n nk nk T (f1 ,H 0 ) (r) + Nf (r, Hi0 ) + Tf1 (r) + o Tf1 (r) k + 1 (f1 ,Hj0 ) (k + 1)2 1 k+1 i (q − 1)n nk nk + + Tf1 (r) + o Tf1 (r) . k + 1 (k + 1)2 k+1 (q − 1)n nk nk qn Thus, we get (q − n − 2) + + n+ . This (k + 1)2 k+1 k+1 k contradicts any of the following cases: i) 2 n 3, q = 3n + 1 and k ≥ 23n, ii) 4 n 6, q = 3n and k ≥ (6n−1)n ,n−3 (6n−4)n iii) n ≥ 7, q = 3n − 1 and k ≥ n−6 . Thus αij = 1 for all 1 i = j 3n − 1. By (19), for i = 3n − 1, j ∈ {1, . . . , 3n − 2}, we may asssume without loss of generality that (f1 , Hj ) (f2 , Hj ) = , j = 1, . . . , n. (20) f1 , H3n−1 f2 , H3n−1
88 Gerd Dethloﬀ and Tran Van Tan 3, denote by Lsv the set of all j ∈ {1, ..., 3n − 2} such that For 1 s
An Extension of Uniqueness Theorems 89 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). 5n+4 Case 2. If n is even, then q = . 2 We now prove that #Q 1. (22) 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 q k+2 k) 2 N fi (r, Hj ) T (r) + o(T (r)), i = 1, 2, 3, k+1 j =1,j =j0 which implies that, for i = 1, 2, 3 q 1 k) k+2 k) N fi (r, Hj ) − 2 N (r, Hj ) T (r) + o(T (r)) n n,fi k+1 j =1,j =j0 q 2 k) − Nn,fi (r, Hj ), i = 1, 2, 3. n j =1,j =i0 Hence, we get 3 q 1 k) 3(k + 2) k) N fi (r, Hj )− 2 N (r, Hj ) T (r) + o(T (r)) n n,fi k+1 j =1,j =j0 i=1 3 q 2 k) − Nn,fi (r, Hj ), (23) n j =1,j =j0 i=1 5n+4 By Lemma 3 (with q = 2 ), we have q 3n(k + 1) − (5n + 2)n k) Nn,fi (r, Hj ) + o(Tfi (r)) ≥ 2 Tfi (r), i = 1, 2, 3. k j =1,j =j0 Hence, we have 3 q 3n(k + 1) − (5n + 2)n 2 k) Nn,fi (r, Hj ) + o(T (r)) ≥ T (r). (24) n nk j =1,j =j0 i=1 By (23) and (24) we have
90 Gerd Dethloﬀ and Tran Van Tan 3 q (5n + 2)n(k + 1) − 3n 1 k) k) N fi (r, Hj ) − 2 N (r, Hj ) T (r) + o(T (r)) n n,fi nk (k + 1) j =1,j =j0 i=1 5n + 2 T (r) + o(T (r)). k On the other hand, we obtain 1 k) k) N fi (r, Hj ) − N (r, Hj ) ≥ 0 for all i ∈ {1, 2, 3}, j ∈ {1, ..., q }. n n,fi Hence, we get 3 1 k) k) N fi (r, Hj ) − N (r, Hj ) n n,fi i=1 5n + 2 T (r) + o(T (r)), j ∈ {1, ..., q } \ {j0 }. k In particular, we get 3 1 k) 5n + 2 k) N fi (r, Hj1 ) − N (r, Hj1 ) T (r) + o(T (r)). (25) n n,fi k i=1 Set Ai := {z ∈ Cm : v(fi ,Hj1 ) (z ) = 1} for i = 1, 2, 3. For each i ∈ {1, 2, 3}, we have Ai \ Ai ⊆ sing fi−1 (Hj1 ). Indeed, otherwise there exists a ∈ Ai \ Ai ∩ reg fi−1 (Hj1 ). Then p0 := v(fi ,Hj1 ) (a) ≥ 2. Since a is a regular point of fi−1 (Hj1 ) we can choose nonzero holomorphic functions h and u on a neigh- borhood U of a such that dh and u have no zeroes and (fi , Hj1 ) ≡ hp0 u on U . Since a ∈ Ai there exists b ∈ Ai ∩ U . Then, we get 1 = v(fi ,Hj1 ) (b) = vhp0 u (b) = p0 ≥ 2. This is a contradiction. Thus, we see that Ai \ Ai ⊆ sing fi−1 (Hj1 ). 3 sing fi−1 (Hj1 ). This means that Set B := A1 ∪ A2 ∪ A3 . Then B \ B ⊆ i=1 B \ B is included in an analytic set of codimension ≥ 2. So we have 3 k) k) (n − 1)N (r, B ) nN fi (r, Hj1 ) − Nn,fi (r, Hj1 ) . i=1 By (25) we have (5n + 2)n N (r, B ) T (r) + o(T (r)), (n − 1)k where we note that n ≥ 2, since n is even. It is clear that k) k) k) on Cm \B (⊆ Cm min v(f1 ,Hj ) , 2 = min v(f2 ,Hj ) , 2 = min v(f3 ,Hj ) , 2 B ). 1 1 1