Algebraical Dependence and Uniqueness Problem for Meromorphic Mappings with Few Moving Targets

Abstract

The purpose of this article is twofold. The first is to prove algebraical dependence of meromorphic mappings from $\mathbb {C}^m$ into $P^n(\mathbb {C})$ sharing few moving hyperplanes. The second is to show uniqueness theorem for meromorphic mappings from the viewpoint of dependence. These results improve and extend some earlier works.

1 Introduction and Main Results

In 1926, Nevanlinna [6] showed that for two non-constant meromorphic functions f and g on the complex plane $\mathbb {C}$, if they have the same inverse images for five distinct values, then $f=g$, and if they have the same inverse images, counted with multiplicities, for four distinct values, then g is a special type of a linear fractional transformation of f. These results are usually called the five-value theorem and four-value theorem.

In 1929, as an improvement of the above mentioned Nevanlinna’s results, Cartan [1] declared that there are at most two meromorphic functions on $\mathbb {C}$ which have the same inverse images regardless of multiplicities for four distinct values. However, Steinmetz [14] gave examples which showed that Cartan’s declaration is false. In relation to this, Ji [5] obtained algebraic dependence of meromorphic mappings by the use of Cartan’s original idea in 1988. Later, Stoll [15] generalized the results of Ji to parabolic covering spaces. Ru [12] studied the case of holomorphic curves for moving targets. In 2010, Thoan and Duc [7] proved some results on algebraic dependence of meromorphic mappings. Recently, Thoan et al. [8] improved their work and proved that

Theorem A

(Thoan-Duc-Quang [8]) Let $f_1,\ldots , f_\lambda : \mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be non-constant meromorphic mappings. Let $\{a_j\}_{j=1}^q$ be meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ in general position such that $T_{a_j}(r)=o(\max _{1\le i\le \lambda }\{T_{f_i}(r)\})(1\le j\le q)$ and $(f_i,a_j)\not \equiv 0$ for each $1\le i\le \lambda ,\ 1\le j\le q$. Assume that the following conditions are satisfied.

• (a) $\min \{\nu _{(f_1,a_j)},1\}=\cdots =\min \{\nu _{(f_\lambda ,a_j)},1\}$ for each $1\le j\le q$.

• (b) $\dim \{z:(f,a_i)(z)=(f,a_j)(z)=0\}\le m-2\ (1\le i<j\le q)$.

• (c) There exists an integer number l, $2\le l\le \lambda$, such that for any increasing sequence $1\le j_1<\cdots <j_l\le \lambda$, $f_{j_1}(z)\wedge \cdots \wedge f_{j_l}(z)=0$ on $\cup _{j=1}^q\{z: (f_1,a_j)(z)=0\}$.

If $q>\frac{n(2n+1)\lambda -(n-1)(\lambda -1)}{\lambda -l+1}$, then $f_1,\ldots ,f_\lambda$ are algebraically dependent over $\mathbb {C}$, i.e., $f_1\wedge \cdots \wedge f_\lambda \equiv 0$ on $\mathbb {C}$.

Let $f:\mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be a meromorphic mapping and d be a positive integer or $+\infty$. Let $\{a_j\}_{j=1}^q$ be “small” (with respect to f) meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ in general position such that

$$\begin{aligned} \dim \{z:(f,a_i)(z)=(f,a_j)(z)=0\}\le m-2\ (1\le i<j\le q). \end{aligned}$$

Consider the set $\mathscr {G}(f,\{a_j\}_{j=1}^q,d)$ of all meromorphic mappings $g: \mathbb {C}^m\rightarrow P^n(\mathbb {C})$ satisfying the conditions:

• (i) $\min \{\nu _{(f,a_j)},d\}=\min \{\nu _{(g,a_j)},d\}\ (1\le j\le q)$,

• (ii) $f(z)=g(z)$ on $\cup _{j=1}^q\{z: (f,a_j)(z)=0\}$.

For brevity, we will denote by $\sharp S$ the cardinality of set S. In 2013, Quang [10] proved the following theorem about algebraic dependence of three maps.

Theorem B

(Quang [10]) Assume $f_1,f_2,f_3\in \mathscr {G}(f,\{a_j\}_{j=1}^q,1)$ and $n\ge 2$.

• (a) If $q\ge 3n^2+3/2$, then $f_1\wedge f_2\wedge f_3\equiv 0$.

• (b) If f is linearly non-degenerate over $\mathscr {R}$ and $q\ge (3n^2+3n+3)/2$, then $f_1\wedge f_2\wedge f_3\equiv 0$.

The first purpose of this article is to give an improvement of Theorem A. Namely, applying the new second main theorems given by Quang [11], we will show that

Theorem 1

Let $f_1,\ldots , f_\lambda : \mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be non-constant meromorphic mappings. Let $\{a_j\}_{j=1}^q$ be meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ in general position such that $T_{a_j}(r)=o(\max _{1\le i\le \lambda }\{T_{f_i}(r)\})(1\le j\le q)$ and $(f_i,a_j)\not \equiv 0$ for each $1\le i\le \lambda ,\ 1\le j\le q$. Assume that the following conditions are satisfied:

• (a) $\min \{\nu _{(f_1,a_j)},1\}=\cdots =\min \{\nu _{(f_\lambda ,a_j)},1\}$ for each $1\le j\le q$.

• (b) $\dim \{z:(f,a_i)(z)=(f,a_j)(z)=0\}\le m-2\ (1\le i<j\le q)$.

• (c) There exists an integer number l, $2\le l\le \lambda$, such that for any increasing sequence $1\le j_1<\cdots <j_l\le \lambda$, $f_{j_1}(z)\wedge \cdots \wedge f_{j_l}(z)=0$ on $\cup _{j=1}^q\{z: (f_1,a_j)(z)=0\}$.

If $q>\frac{3n(n+1)\lambda -2(n-1)(\lambda -1)}{2(\lambda -l+1)}$, then $f_1,\ldots ,f_\lambda$ are algebraically dependent over $\mathscr {R}$, i.e. $f_1\wedge \cdots \wedge f_\lambda \equiv 0$ on $\mathscr {R}$.

For the case $\lambda =3$ and $l=2$ in Theorem 1, we have the following corollary, which is better than Theorem B(a).

Corollary 1

Assume $f_1,f_2,f_3\in \mathscr {G}(f,\{a_j\}_{j=1}^q,1)$ and $n\ge 2$. If $q\ge \frac{9n^2+5n+4}{4}$, then $f_1\wedge f_2\wedge f_3\equiv 0$.

On the other hand, the uniqueness problem with truncated multiplicities for meromorphic mappings from $\mathbb {C}^m$ into $P^n(\mathbb {C})$ sharing a finite set of fixed (or moving) targets in $P^n(\mathbb {C})$ has been studied very intensively by many authors in the last few decades, and they related to many problems in Nevanlinna theory and hyperbolic complex analysis (see ref. [2, 4, 13]). In [3], Chen-Li-Yan studied the uniqueness problem without the assumption of the linearly non-degeneracy for meromorphic mappings.

Theorem C

(Chen-Li-Yan [3]) If $q=4n^2+2n, n\ge 2$, then $\sharp \mathscr {G}(f,\{a_j\}_{j=1}^q,1)=1$.

Later, Thoan et al. [8] showed the result still valid if $q=4n^2+2$.

Set $\text {rank}_\mathscr {R}(f):=\text {rank}\{f_0,\ldots ,f_n\}\ \text {over}\ \mathscr {R}$. It is easy to see that the definition of $\text {rank}_\mathscr {R}(f)$ does not depend on the choice of the reduced representation of f. In 2013, Quang and An [9] got the following uniqueness theorem with fewer moving hyperplanes.

Theorem D

(Quang-An [9]) If $q>4nk+2, n\ge 2$, where $k+1=\text {rank}_\mathscr {R}(f)$, then $\sharp \mathscr {G}(f,\{a_j\}_{j=1}^q,1)=1$.

Quang [11] established some new second main theorem for meromorphic mappings intersecting moving hyperplanes. For their application, Quang showed the following result.

Theorem E

(Quang [11])

• (a) If $q>\frac{9n^2+9n+4}{4},n\ge 2$, then $\sharp \mathscr {G}(f,\{a_j\}_{j=1}^q,1)\le 2$.

• (b) If $q>3n^2+n+2$ and $n\ge 2$, then $\sharp \mathscr {G}(f,\{a_j\}_{j=1}^q,1)=1$.

The following question arises naturally: are there uniqueness theorem with few moving hyperplanes? Based on the propagation of dependence in Theorem 1, the second purpose of this paper is to give an answer to the above question. Namely, we will prove the following result.

Theorem 2

Let $f:\mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be a meromorphic mapping and let $\{a_j\}_{j=1}^q$ be “small” (with respect to f) meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ in general position such that

$$\begin{aligned} \dim \{z:(f,a_i)(z)=(f,a_j)(z)=0\}\le m-2\ (1\le i<j\le q). \end{aligned}$$

Set $k+1=\text {rank}_\mathscr {R}(f)$, then the following assertions hold:

• (i) If $k\ge \frac{n+1}{2}$ and $q >2(2n+1)k-2k^2+2\ (n\ge 2)$, then $\sharp \mathscr {G}(f,\{a_j\}_{j=1}^q,1)=1$.

• (ii) If $k<\frac{n+1}{2}$ and $q> \frac{2k^2+2(n-1)k+(n+3)+\sqrt{(2k^2+2(n-1)k+n-1)^2+8(3n+1)k}}{2}(n\ge 2)$, then $\sharp \mathscr {G}(f,\{a_j\}_{j=1}^q,1)=1$.

Put $Q=\frac{2k^2+2(n-1)k+n+3+\sqrt{(2k^2+2(n-1)k+n-1)^2+8(3n+1)k}}{2}$. For $1\le k<\frac{n+1}{2}$, we have

$$\begin{aligned}&2(2n+1)k-2k^2+2- Q\\ =&\frac{8k(k-\frac{n+1}{2})(2k^2-(4n+3)k+2n)}{-6k^2+6(n+1)k-(n-1)+\sqrt{(2k^2+2(n-1)k+n-1)^2+8(3n+1)k}}>0. \end{aligned}$$

By Theorem 2, we get the following corollary.

Corollary 2

Let $f:\mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be a meromorphic mapping and let $\{a_j\}_{j=1}^q$ be “small” (with respect to f) meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ in general position such that

$$\begin{aligned} \dim \{z:(f,a_i)(z)=(f,a_j)(z)=0\}\le m-2\ (1\le i<j\le q). \end{aligned}$$

If $q >2(2n+1)k-2k^2+2, n\ge 2$, where $k+1=\text {rank}_\mathscr {R}(f)$, then $\sharp \mathscr {G}(f,\{a_j\}_{j=1}^q,1)=1$.

Since $1\le k\le n$, Corollary 2 implies that

Corollary 3

If $q>2n^2+2n+2$ and $n\ge 2$, then $\sharp \mathscr {G}(f,\{a_j\}_{j=1}^q,1)=1$.

We would like to notice that Theorem 2 is an improvement of not only the above mentioned theorems, but also of many uniqueness theorem of meromorphic mappings for moving targets without counting multiplicity.

2 Basic Notions and Preliminaries in Nevanlinna Theory

We set $\Vert z\Vert =(|z_1|^2+\cdots +|z_m|^2)^\frac{1}{2}$ for $z=(z_1,\ \cdots ,\ z_m)\in \mathbb {C}^m$, $B(r):=\{z:\Vert z\Vert \le r\}$, $S(r):=\{z:\Vert z\Vert = r\}, d^c:=\frac{\sqrt{-1}}{4\pi }(\overline{\partial }-\partial ), \upsilon :=(dd^c\Vert z\Vert ^2)^{m-1}$ and $\sigma := d^c\log \Vert z\Vert ^2\wedge (dd^c\log \Vert z\Vert ^2)^{m-1}$.

Let F(z) be a nonzero holomorphic function on a domain ${\varOmega }$ in $\mathbb {C}^m$. For $\alpha =(\alpha _1,\cdots , \alpha _m)$ with $\alpha _i \in \mathbb {Z}_+$ $(1\le i \le m)$, set $|\alpha |=\sum _{i=1}^{m}\alpha _i$ and $D^\alpha F=\frac{\partial ^{|\alpha |}F}{\partial ^(\alpha _1)z_1\cdots \partial ^(\alpha _m)z_m}$, where $\mathbb {Z}_+$ denote the set of nonnegative integers. We define the map $\nu _F: {\varOmega }\rightarrow \mathbb {Z}_+$ by $\nu _F(a):=\max \{m: D^\alpha F=0\ for\ all\ \alpha \ with\ |\alpha |<m\}(a\in {\varOmega })$.

We mean by a divisor on a domain ${\varOmega }$ in $\mathbb {C}^m$ a map $\nu :{\varOmega }\rightarrow \mathbb {Z}$ such that, for each $a \in {\varOmega }$, there are nonzero holomorphic functions F and G on a connected neighborhood $U \subset {\varOmega }$ of a such that $\nu (z)=\nu _F(z)-\nu _G(z)$ for each $z \in U$ outside an analytic set of dimension $\le m-2$. Two divisors are regarded as the same if they are identical outside an analytic set of dimension $\le m-2$. For a divisor $\nu$ on ${\varOmega }$, we set $|\nu |:=\overline{\{z:\nu (z)\ne 0\}}$, which is a purely $(m-1)$-dimensional analytic subset of ${\varOmega }$ or empty. For a divisor $\nu$ on $\mathbb {C}^m$ and a positive integer M or $M= +\infty$, we define the truncated divisor $\nu ^{[M]}$ by $\nu ^{[M]}(z):=\min \{M,\nu (z)\}$.

Let $\varphi$ be a nonzero meromorphic function on ${\varOmega }$ in $\mathbb {C}^m$. We define the divisor $\nu _{\varphi }$ as follows: for each $a\in \mathbb {C}^m$, choose nonzero holomorphic functions F and G on a neighborhood U of a such that $\varphi =\frac{F}{G}$ on U and $\dim (F^{-1}(0)\cap G^{-1}(0))\le m-2$. Then, put $\nu _\varphi (a):=\nu _F(a)$, $\nu _{\varphi }(z):=\nu _F(z)$ and $\nu _\varphi ^\infty :=\nu _G$, which are independent of choices of F and G and so globally well-defined on ${\varOmega }$.

Define the counting function of $\nu$ by

$$\begin{aligned} N(r,\nu ):=\int _1^r\frac{n(t)}{t^{2m-1}}dt,\ \ (1<r<+\infty ) \end{aligned}$$

where

$$\begin{aligned} n(t):= \begin{array}{l@{\quad }l} \int _{|\nu |\cap B(r)}\nu (z)\upsilon &{} \text{ if } \quad m\ge 2,\\ \sum _{|z|\le t}\nu (z) &{} \text{ if } \quad m=1. \end{array} \end{aligned}$$

Similarly, we define $n^{[M]}(t)$ and $N(r,\nu ^{[M]})$ and denote by $N^{[M]}(r,\nu )$, respectively.

Let $\varphi :\mathbb {C}^m\rightarrow \mathbb {C}$ be a meromorphic function. Define

$$\begin{aligned} N_\varphi (r):=N(r,\nu _\varphi ), N_\varphi ^{[M]}(r):= N^{[M]}(r,\nu _\varphi ). \end{aligned}$$

For brevity, we will omit the character $^M$ if $M=+\infty$.

Let $f:\mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be a meromorphic mapping. We can choose holomorphic functions $f_0,\ \cdots ,\ f_n$ on $\mathbb {C}^m$ such that $I_f:=\{z:f_0(z)=\cdots =f_n(z)\}$ is of dimension at most $m-2$, and $f(z)=(f_0:\ \cdots \ :f_n)$ outside $I_f$. Usually, $f(z)=(f_0:\ \cdots \ :f_n)$ is called a reduced representation of f.

Set $\Vert f\Vert :=(|f_0|^2+\cdots +|f_n|^2)^\frac{1}{2}$, the characteristic function of f is defined by

$$\begin{aligned} T_f(r):=\int _{S(r)}\log \Vert f\Vert \sigma -\int _{S(1)}\log \Vert f\Vert \sigma ,\ \ r>1. \end{aligned}$$

Note that $T_f(r)$ is independent on the choice of the reduced representation.

Let a be a meromorphic mapping of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ with reduced representation $a=(a_0:\cdots :a_n)$. We define

$$\begin{aligned} m_{(f,a)}(r):=\int _{S(r)}\log \frac{\Vert f\Vert \Vert a\Vert }{|(f,a)|}\sigma -\int _{S(1)}\log \frac{\Vert f\Vert \Vert a\Vert }{|(f,a)|}\sigma , \end{aligned}$$

where $\Vert a\Vert :=(|a_0|^2+\cdots +|a_n|^2)^\frac{1}{2}$ and $r>1$. Throughout this paper, we always assume that the homogeneous coordinates of $P^n(\mathbb {C})$ are chosen so that for a given meromorphic mapping $a=(a_0:\cdots :a_n)$ of $\mathbb {C}^m$ into $P^n(\mathbb {C})$, $a_0\not \equiv 0$. Then, we set $\tilde{a}=(\frac{a_0}{a_0}:\cdots :\frac{a_n}{a_0})$.

Let $a_1,\ldots ,a_q\ (q\ge n+1)$ be q meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ with reduced representations $a_j=(a_{j0}:\cdots :a_{jn})(1\le j\le q)$. We say that $a_1,\ldots ,a_q$ located in general position if $\det (a_{j_k})\not \equiv 0$ for any $1\le j_0<j_1<\cdots <j_n\le q$. Denote by $\mathscr {M}$ the field of meromorphic functions on $\mathbb {C}^m$ and denote by $\mathscr {R}$ the smallest subfield of $\mathscr {M}$ which contains $\mathbb {C}$ and all $\frac{a_{j_k}}{a_{j_l}}$ with $a_{j_l}\not \equiv 0$.

Let f and a be meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ with reduced representations $f=(f_0:\cdots :f_n)$ and $a=(a_0:\cdots :a_n)$. We say that f is linearly non-degenerate over $\mathscr {R}$ if $f_0,\ldots ,f_n$ are linearly independent over $\mathscr {R}$. Put $(f,a):=\sum _{i=0}^na_if_i$, then we say that a is “small” with respect to f if $T_a(r)=o(T_f(r))$ as $r\rightarrow +\infty$. Denote by $\nu _{(f,a)}$ the map of $\mathbb {C}^m$ into $\mathbb {Z}$ whose value $\nu _{(f,a)}(z)$ is the intersection of the images of f and a at f(z).

If $f,a:\mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be meromorphic mappings such that $(f,a)\not \equiv 0$, then the First Main Theorem for moving targets in value distribution theory states that

$$\begin{aligned} T_f(r)+T_a(r)=m_{(f,a)}(r)+N_{(f,a)}(r)+O(1). \end{aligned}$$

For a nonzero meromorphic function $\varphi$ on $\mathbb {C}^m$, the proximity function $m(r,\varphi )$ is defined by

$$\begin{aligned} m(r,\varphi )=\int _{S(r)}\log ^+|\varphi |\sigma , \end{aligned}$$

where $\log ^+|x|=\max \{\log x,0\}$ for $x\ge 0$. The Nevanlinna’s characteristic function is defined by

$$\begin{aligned} T(r,\varphi ):=N(r,\nu _\varphi ^\infty )+m(r,\varphi ). \end{aligned}$$

We regard $\varphi$ as a meromorphic mapping of $\mathbb {C}^m$ into $P^1(\mathbb {C})$, then

$$\begin{aligned} T_\varphi (r)=T(r,\varphi )+O(1). \end{aligned}$$

As usual, by the notation $``\Vert P,''$ we mean the assertion P holds for all $r\in [0,+\infty )$ excluding a Borel subset E of the interval $[0,+\infty )$ with $\int _E dr<+\infty$.

The First Main Theorem for general position (see [15], p. 326) Assume that $1\le \lambda \le n+1$, $f_i:\mathbb {C}^m\rightarrow P^n(\mathbb {C})(1\le i\le \lambda )$ are $\lambda$ meromorphic mappings located in general position. Then

$$\begin{aligned} N(r,\mu _{f_1\wedge \cdots \wedge f_\lambda })+m(r, f_1\wedge \cdots \wedge f_\lambda )\le \sum _{1\le i\le \lambda } T_{f_i}(r) +O(1). \end{aligned}$$

Let V be a complex vector space of dimension $n\ge 1$. The vectors $\{v_1,\ldots ,v_k\}$ are said to be in general position if for each selection of integers $1\le i_1<\cdots <i_p\le k$ with $p\le n$, $v_{i_1}\wedge \cdots \wedge v_{i_p}\ne 0$. The vectors $\{v_1,\ldots ,v_k\}$ are said to be in special position if they are not in general position. Take $1\le p\le k$, then $\{v_1,\ldots ,v_k\}$ are said to be in p-special position if for each selection of integers $1\le i_1<\cdots < i_p\le k$ the vectors $v_{i_1},\ldots ,v_{i_p}$ are in special position.

The Second Main Theorem for general position (see [15], Theorem 2.1, P.326) Let M be a connected complex manifold of dimension m. Let A be a pure $(m-1)$-dimensional analytic subset of M. Let V be a complex vector space of dimension $n+ 1 > 1$. Let p and k be integers with $1\le p\le k\le n+1$. Let $f_j : M \rightarrow P(V)$, $1\le j\le k$ be meromorphic mappings. Assume that $f_1,\ldots ,f_k$ are in general position. Also assume that $f_1,\ldots ,f_k$ are in p-special position on A. Then we have

$$\begin{aligned} \mu _{f_1\wedge \cdots \wedge f_k}\ge (k-p+1)\nu _A. \end{aligned}$$

The following is the “second main theorem type” for meromorphic mappings intersecting moving targets with truncated counting function according to S.D.Quang [11].

Second Main Theorem for moving targets (Quang [11]) Let $f:\mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be a meromorphic mapping and $\{a_i\}_{i=1}^q\ (q\ge 2n-k+2$ ) be meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ located in general position such that $(f,a_i)\not \equiv 0 (1\le i\le q)$, where $k+1=\text {rank}_{\mathscr {R}}(f)$. Then the following assertions hold:

• (a) $\Vert \frac{q}{2n-k+2}T_f(r)\le \sum _{i=1}^qN_{(f,a_i)}^{[k]}(r) +o(T_f(r))+O(\max _{1\le i\le q}T_{a_i}(r))$,

• (b) $\Vert \frac{q-n+2k-1}{n+k+1}T_f(r)\le \sum _{i=1}^qN_{(f,a_i)}^{[k]}(r) +o(T_f(r))+O(\max _{1\le i\le q}T_{a_i}(r))$.

Remark:

• (1) If $k\ge \frac{n+1}{2}$ then the above theorem (a) is stronger than (b), otherwise, if $k<\frac{n+1}{2}$ then (b)is stronger than (a).

• (2) If $k\ge 1$, we have the following estimates:

$$\begin{aligned} \min _{1\le k\le n}\{\max \{\frac{q}{2n-k+2},\frac{q-n+2k-1}{n+k+1}\}\}\ge {\left\{ \begin{array}{ll} \frac{2q}{3(n+1)} \quad \text {if}\ q\ge 3n+3, \\ \frac{q-n+1}{n+2} \quad \text {if}\ q<3n+3. \end{array}\right. } \end{aligned}$$

Therefore, S.D.Quang [11] obtained the following corollary.

Corollary 2.1

Let $f:\mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be a meromorphic mapping and $\{a_i\}_{i=1}^q\ (q\ge 2n+1)$ be meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ located in general position such that $(f,a_i)\not \equiv 0 (1\le i\le q)$.

• (a) Then we have

  $$\begin{aligned} \Vert \frac{2q-n+1}{3(n+1)}T_f(r)\le \sum _{i=1}^qN_{(f,a_i)}^{[n]}(r) +o(T_f(r))+O(\max _{1\le i\le q}T_{a_i}(r)). \end{aligned}$$

• (b) If $q\ge 3n+3$, then

  $$\begin{aligned} \Vert \frac{2q}{3(n+1)}T_f(r)\le \sum _{i=1}^qN_{(f,a_i)}^{[n]}(r) +o(T_f(r))+O(\max _{1\le i\le q}T_{a_i}(r)). \end{aligned}$$

• (c) If $q < 3n+3$, then

  $$\begin{aligned} \Vert \frac{q-n+1}{n+2}T_f(r)\le \sum _{i=1}^qN_{(f,a_i)}^{[n]}(r) +o(T_f(r))+O(\max _{1\le i\le q}T_{a_i}(r)). \end{aligned}$$

3 Proof of Theorem 1

Proof

It suffices to prove Theorem 1 in the case of $\lambda \le n+1$. Assume that $f_1\wedge \cdots \wedge f_\lambda \not \equiv 0$. Considering $\lambda -1$ arbitrary moving targets $a_{i_1},\ldots ,a_{i_{\lambda -1}}$, then there exists $a_{i_0}$ with $i_0\not \in \{i_1,\ldots ,i_{\lambda -1}\}$ such that the matrix $\begin{pmatrix} (f_1,a_{i_0}) &{} \cdots &{} (f_\lambda ,a_{i_0})\\ (f_1,a_{i_1}) &{} \cdots &{} (f_\lambda ,a_{i_1}) \\ \vdots &{} \vdots &{} \vdots \\ (f_1,a_{i_{\lambda -1}}) &{} \cdots &{} (f_\lambda ,a_{i_{\lambda -1}}) \\ \end{pmatrix}$ is nondegenerate.

Indeed, suppose on contrary, the matrix $\begin{pmatrix} (f_1,a_1) &{} \cdots &{} (f_\lambda ,a_1) \\ (f_1,a_2) &{} \cdots &{} (f_\lambda ,a_2) \\ \vdots &{} \vdots &{} \vdots \\ (f_1,a_{n+1}) &{} \cdots &{} (f_\lambda ,a_{n+1}) \\ \end{pmatrix}$ is of rank $\le \lambda -1(< n+1)$. Let $a_j=(a_{j0}:\cdots :a_{jn})(1\le j\le n+1)$ be the reduced representations of $a_1,\ldots ,a_{n+1}$ and $f_j=(f_{j0}:\cdots :f_{jn})(1\le j\le \lambda )$ be the reduced representations of $f_1,\ldots ,f_\lambda$. Then

$$\begin{aligned} \begin{pmatrix} (f_1,a_1) &{} \cdots &{} (f_\lambda ,a_1) \\ (f_1,a_2) &{} \cdots &{} (f_\lambda ,a_2) \\ \vdots &{} \vdots &{} \vdots \\ (f_1,a_{n+1}) &{} \cdots &{} (f_\lambda ,a_{n+1}) \\ \end{pmatrix}=\begin{pmatrix} a_{10} &{} \cdots &{} a_{1n} \\ a_{20} &{} \cdots &{} a_{2n} \\ \vdots &{} \vdots &{} \vdots \\ a_{n+10} &{} \cdots &{} a_{n+1n} \\ \end{pmatrix}\cdot \begin{pmatrix} f_{10} &{} \cdots &{} f_{\lambda 0} \\ f_{11} &{} \cdots &{} f_{\lambda 1} \\ \vdots &{} \vdots &{} \vdots \\ f_{1n} &{} \cdots &{} f_{\lambda n} \\ \end{pmatrix}. \end{aligned}$$

Since $a_1,\cdots ,a_{n+1}$ are in general position, we get $f_1\wedge \cdots \wedge f_\lambda \equiv 0$. This is a contradiction.

We denote by $\mu _{\widetilde{f_1}\wedge \cdots \wedge \widetilde{f_\lambda }}$ the divisor associated with $\widetilde{f_1}\wedge \cdots \wedge \widetilde{f_\lambda }$, $N(r,\mu _{\widetilde{f_1}\wedge \cdots \wedge \widetilde{f_\lambda }})$ the counting function associated with the divisor $\mu _{\widetilde{f_1}\wedge \cdots \wedge \widetilde{f_\lambda }}$, where $\tilde{f}_t:=((f_t,a_{i_0}):\cdots :(f_t,a_{i_{\lambda -1}}))\ (1\le t\le \lambda )$. We will prove the following claim.

Claim 3.1

For every $z\in \mathbb {C}^m$ outside an analytic set of codimension $\ge 2$, we have

$$\begin{aligned} \sum _{j=0}^{\lambda -1}\left( \lambda \mathop {\min }_{1\le t\le \lambda }\{\nu _{(f_t,a_{i_j})}(z)\} + (\lambda -l)\mathop {\sum }_{t=1}^\lambda \min \{\nu _{(f_t,a_{i_j})}(z),1\}\right)&\\ +\mathop {\sum }_{j\in \{1,\ldots ,q\}\setminus \{i_0,i_1,\ldots ,i_{\lambda -1}\}}\mathop {\sum }_{t=1}^\lambda (\lambda -l+1)\min \{\nu _{(f_t,a_j)}(z),1\}&\le \lambda \mu _{\tilde{f}_1\wedge \cdots \wedge \tilde{f}_\lambda }(z). \end{aligned}$$

In fact, the Claim is the Claim 3.2 in [8], we give the proof here for completeness of our proof. Let $I=\{i_0,i_1,\ldots , i_{\lambda -1}\}$, $\overline{I}=\{1,2,\ldots ,q\}\setminus I$, $\mathscr {A}:=\cup _{i\in I}(f_1,a_i)^{-1}(0)$, $\overline{\mathscr {A}}:=\cup _{i\in \bar{I}}(f_1,a_i)^{-1}(0)$ and $A=\bigcup _{1\le i<j\le q}((f_1,a_i)^{-1}(0)\cap (f_1,a_j)^{-1}(0))$.

Case 1 Let $z_0\in \mathscr {A}\setminus (A\cup \cup _{i=1}^\lambda I(f_i)\cup (a_{i_0}\wedge \cdots \wedge a_{i_{\lambda -1}})^{-1}(0))$ be a regular point of $\mathscr {A}$. Without loss of generality, we assume that $z_0$ is a zero of $(f_1,a_{i_0})$ where $i_0\in I$. Let S be an irreducible analysis subset of $\mathscr {A}$ containing $z_0$ and U be an open neighborhood of $z_0$ in $\mathbb {C}^m$, such that $U\cap (\mathscr {A}\setminus S)=\emptyset$. Choosing a holomorphic function h on neighborhood $U^\prime \subset U$ of $z_0$ such that $\nu _h(z)=\min _{1\le t\le \lambda }\{\nu _{(f_t,a_{i_0})}(z)\}$ if $z\in S$ and $\nu _h(z)= 0$ if $z\not \in S$. Then $(f_t,a_{i_0})=hu_t\ (1\le t\le \lambda )$, where $u_t$ is holomorphic function.

Since the matrix $\begin{pmatrix} (f_1,a_{i_1}) &{} \cdots &{} (f_\lambda ,a_{i_1}) \\ \vdots &{} \vdots &{} \vdots \\ (f_1,a_{i_{\lambda -1}}) &{} \cdots &{} (f_\lambda ,a_{i_{\lambda -1}}) \\ \end{pmatrix}$ is of rank $\le \lambda -1$. Therefore, there exist not all zero holomorphic function $b_1,\ldots ,b_\lambda$ such that

$$\begin{aligned} \sum _{t=1}^\lambda b_t(f_t,a_{i_j})=0\ (1\le j\le \lambda -1). \end{aligned}$$

It implies that

$$\begin{aligned} \sum _{t=1}^\lambda b_t\tilde{f_t}=\left( \sum _{t=1}^\lambda b_t(f_t,a_{i_0}),0,\ldots ,0\right) . \end{aligned}$$

Without loss of generality, we may assume that the set of the common zeros of $\{b_t\}_{t=1}^\lambda$ is an analysis subset of codimension $\ge 2$. Then there exists an index, say $\lambda$, such that $S\not \subset b_\lambda ^{-1}(0)$.

Thus, for each $z\in (U^\prime \cap S)\setminus b_\lambda ^{-1}(0)$, we have

$$\begin{aligned} \tilde{ f_1}(z)\wedge \cdots \wedge \tilde{ f_\lambda }(z) =&\tilde{ f_1}(z)\wedge \cdots \wedge \widetilde{f_{\lambda -1}}(z)\wedge (\tilde{ f_\lambda }(z)+\sum _{t=1}^{\lambda -1}\frac{b_t}{b_\lambda }\tilde{f_t}(z))\\ =&\tilde{ f_1}(z)\wedge \cdots \wedge \widetilde{f_{\lambda -1}}(z)\wedge (V(z)h(z))\\ =&h(z)\tilde{ f_1}(z)\wedge \cdots \wedge \widetilde{f_{\lambda -1}}(z)\wedge V(z), \end{aligned}$$

where $V(z)=(u_\lambda +\sum _{t=1}^{\lambda -1} \frac{b_t}{b\lambda }u_t,0,\ldots ,0)$. On the other hand,

$$\begin{aligned} \begin{pmatrix} (f_{j_1},a_{i_0}) &{} \cdots &{} (f_{j_l},a_{i_0}) \\ (f_{j_1},a_{i_1}) &{} \cdots &{} (f_{j_l},a_{i_1}) \\ \vdots &{} \vdots &{} \vdots \\ (f_{j_1},a_{i_{\lambda -1}}) &{} \cdots &{} (f_{j_1},a_{i_{\lambda -1}}) \\ \end{pmatrix}=\begin{pmatrix} a_{i_00} &{} \cdots &{} a_{i_0n} \\ a_{i_10} &{} \cdots &{} a_{i_1n} \\ \vdots &{} \vdots &{} \vdots \\ a_{i_{\lambda -1}0} &{} \cdots &{} a_{i_{\lambda -1}n} \\ \end{pmatrix}\cdot \begin{pmatrix} f_{{j_1}0} &{} \cdots &{} f_{{j_l}0} \\ f_{{j_1}1} &{} \cdots &{} f_{{j_l}1} \\ \vdots &{} \vdots &{} \vdots \\ f_{{j_1}n} &{} \cdots &{} f_{{j_l} n} \\ \end{pmatrix}. \end{aligned}$$

Since $a_1,\cdots ,a_q$ are in general position, $\text {rank}\{a_{i_0},\cdots ,a_{i_{\lambda -1}}\}=\lambda$. By the assumption, for any increasing sequence $1\le j_1<\cdots <j_l\le \lambda -1$, $f_{j_1}(z)\wedge \cdots \wedge f_{j_l}(z)=0$ on S. Thus,

$$\begin{aligned} \text {rank}\{\widetilde{f_{j_1}},\cdots ,\widetilde{f_{j_l}}\}&\le \min \{\text {rank}\{f_{j_1},\cdots ,f_{j_l}\},\text {rank}\{a_{i_0},\cdots ,a_{i_{\lambda -1}}\}\}\\&\textsc {}=\text {rank}\{f_{j_1},\cdots ,f_{j_l}\}. \end{aligned}$$

Hence $\tilde{f_{j_1}}(z)\wedge \cdots \wedge \tilde{f_{j_l}}(z)=0$ on S. It yields that the family $\{\tilde{f_1},\ldots ,\tilde{f_l},V(z)\}$ is in $(l+1)-$ special position on S. By the Second Main Theorem for general position, we have

$$\begin{aligned} \mu _{\tilde{f_1}\wedge \cdots \wedge \widetilde{f_{\lambda -1}}\wedge V}(z)\ge \lambda -l,\ z\in S. \end{aligned}$$

Hence, $\mu _{\tilde{f_1}\wedge \cdots \wedge \tilde{f_\lambda }}(z)\ge \nu _h(z)+\lambda -l,\ z\in (U^\prime \cap S)\setminus b_\lambda ^{-1}(0)$. In particular,

$$\begin{aligned} \mu _{\tilde{f_1}\wedge \cdots \wedge \tilde{f_\lambda }}(z_0)\ge \min _{1\le t\le \lambda }\{\nu _{(f_t,a_{i_0})}(z_0)\}+\lambda -l. \end{aligned}$$

Therefore

$$\begin{aligned}&\sum _{i\in I}(\lambda \mathop {\min }_{1\le t\le \lambda }\{\nu _{(f_t,a_i)}(z_0)\} + (\lambda -l)\mathop {\sum }_{t=1}^\lambda \min \{\nu _{(f_t,a_i)}(z_0),1\})\\&\qquad + \mathop {\sum }_{i\in \overline{I}} \mathop {\sum }_{t=1}^\lambda (\lambda -l+1)\min \{\nu _{(f_t,a_i)}(z_0),1\}\\&\quad =\lambda (\min _{1\le t\le \lambda }\{\nu _{(f_t,a_{i_0})}(z_0)\}+\lambda -l) \le \lambda \mu _{\tilde{f}_1\wedge \cdots \wedge \tilde{f}_\lambda }(z_0). \end{aligned}$$

Case 2 Let $z_0\in \widetilde{\mathscr {A}}\setminus (A\cup \cup _{i=1}^\lambda I(f_i)\cup (a_{i_0}\wedge \cdots \wedge a_{i_{\lambda -1}})^{-1}(0))$ be a regular point of $\widetilde{\mathscr {A}}$. Then $z_0$ is a zero of one holomorphic mapping $(f_1,a_t)$ where $t\in \bar{I}$. According to the assumption and the proof in Case 1, the family $\tilde{f}_1,\ldots ,\tilde{f}_\lambda$ are in l-special position on an irreducible analytic subset of codimension 1 of $\widetilde{\mathscr {A}}$ which contains $z_0$. By the Second Main Theorem for general position, we have

$$\begin{aligned} \mu _{\tilde{f_1}\wedge \cdots \wedge \widetilde{f_\lambda }}(z_0)\ge \lambda -l+1. \end{aligned}$$

Hence

$$\begin{aligned}&\sum _{i\in I}(\lambda \mathop {\min }_{1\le t\le \lambda }\{\nu _{(f_t,a_i)}(z_0)\} + (\lambda -l)\mathop {\sum }_{t=1}^\lambda \min \{\nu _{(f_t,a_i)}(z_0),1\})\\&\quad \quad + \mathop {\sum }_{i\in \overline{I}} \mathop {\sum }_{t=1}^\lambda (\lambda -l+1)\min \{\nu _{(f_t,a_i)}(z_0),1\}\\&\quad =\lambda (\lambda -l+1) \le \lambda \mu _{\tilde{f}_1\wedge \cdots \wedge \tilde{f}_\lambda }(z_0). \end{aligned}$$

From the above two cases, Claim 3.1 is proved.

For nonnegative integers $c_1,c_2,\ldots ,c_\lambda$, it is easy to check that

$$\begin{aligned} \mathop {\min }_{1\le t\le \lambda }c_t\ge \mathop {\sum }_{t=1}^\lambda \min \{c_t, n\}-(\lambda -1)n. \end{aligned}$$

Since $\min \{\nu _{(f_1,a_{i_0})},1\}=\cdots =\min \{\nu _{(f_\lambda ,a_{i_0})},1\}$ and $\nu _{(f_t,a_i)}(z_0)\ge \min \{\nu _{(f_t,a_{i_0})}(z_0),1\}$. Claim 3.1 implies that

$$\begin{aligned}&\sum _{j=1}^{\lambda -1}\sum _{t=1}^\lambda \left( \lambda \min \{\nu _{(f_t,a_{i_j})}(z),n\} - [(\lambda -1)n- (\lambda -l)]\min \{\nu _{(f_t,a_{i_j})}(z),1\}\right) \\&\qquad +\mathop {\sum }_{j\in \{1,\ldots ,q\}\setminus \{i_1,\ldots ,i_{\lambda -1}\}}\sum _{t=1}^\lambda (\lambda -l+1)\min \{\nu _{(f_t,a_j)}(z),1\}\le \lambda \mu _{\tilde{f}_1\wedge \cdots \wedge \tilde{f}_\lambda }(z). \end{aligned}$$

Thus,

$$\begin{aligned}&\sum _{j=1}^{\lambda -1}\sum _{t=1}^\lambda \left( \lambda N_{(f_t,a_{i_j})}^{[n]}(r) - [(\lambda -1)n- (\lambda -l)]N_{(f_t,a_{i_j})}^{[1]}(r)\right) \\&\qquad +\mathop {\sum }_{j\in \{1,\ldots ,q\}\setminus \{i_1,\ldots ,i_{\lambda -1}\}}\sum _{t=1}^\lambda (\lambda -l+1)N_{(f_t,a_j)}^{[1]}(r)\\&\quad \le \lambda N_{\tilde{f}_1\wedge \cdots \wedge \tilde{f}_\lambda }(r). \end{aligned}$$

From $T_{\tilde{f}_t}(r)\le T_{f_t}(r)+o(\mathop {\max }_{1\le i\le \lambda }T_{f_i}(r))$ $(1\le t\le \lambda )$ and the First Main Theorem for general position, we get

$$\begin{aligned}&\sum _{j=1}^{\lambda -1}\sum _{t=1}^\lambda \left( \lambda N_{(f_t,a_{i_j})}^{[n]}(r) - [(\lambda -1)n- (\lambda -l)]N_{(f_t,a_{i_j})}^{[1]}(r)\right) \\&+\qquad \mathop {\sum }_{j\in \{1,\ldots ,q\}\setminus \{i_1,\ldots ,i_{\lambda -1}\}} \sum _{t=1}^\lambda (\lambda -l+1)N_{(f_t,a_j)}^{[1]}(r)\le \lambda \mathop {\sum }_{t=1}^\lambda T_{f_t}(r)+ o( \mathop {\max }_{1\le t\le \lambda }T_{f_t}(r)). \end{aligned}$$

Summing both sides of the above inequality over all sequences $1\le i_1<\cdots <i_{\lambda -1}\le q$, we have

$$\begin{aligned}&\sum _{i=1}^q\sum _{t=1}^\lambda \left( \lambda (\lambda -1) N_{(f_t,a_i)}^{[n]}(r) +((\lambda -l+1)q-(\lambda -1)((\lambda -1)n+1))N_{(f_t,a_i)}^{[1]}(r)\right) \\&\quad \le q\lambda \mathop {\sum }_{t=1}^\lambda T_{f_t}(r)+ o( \mathop {\max }_{1\le t\le \lambda }T_{f_t}(r)). \end{aligned}$$

Since $N_{(f_t,a_i)}^{[1]}(r)\ge \frac{1}{n}N_{(f_t,a_i)}^{[n]}(r)$, then

$$\begin{aligned} \frac{(\lambda -l+1)q+(\lambda -1)(n-1)}{n}\sum _{i=1}^q\sum _{t=1}^\lambda N_{(f_t,a_i)}^{[n]}(r)\le q\lambda \mathop {\sum }_{t=1}^\lambda T_{f_t}(r)+ o( \mathop {\max }_{1\le t\le \lambda }T_{f_t}(r)). \end{aligned}$$

For $2\le l\le \lambda \le n$,

$$\begin{aligned} q>\frac{3n(n+1)\lambda -2(n-1)(\lambda -1)}{2(\lambda -l+1)}\ge \frac{3n^2+4n+5}{2}\ge 3n+3 \end{aligned}$$

Applying the Second Main Theorem for moving targets (Corollary 2.1(b)), we set

$$\begin{aligned}&\Vert \frac{(\lambda -l+1)q+(\lambda -1)(n-1))}{n}\cdot \frac{2q}{3(n+1)}\sum _{t=1}^\lambda T_{f_t}(r)\\&\quad \le q\lambda \mathop {\sum }_{t=1}^\lambda T_{f_t}(r)+ o( \mathop {\max }_{1\le t\le \lambda }T_{f_t}(r)). \end{aligned}$$

Letting $r\rightarrow +\infty$, we have

$$\begin{aligned} \frac{(\lambda -l+1)q+(\lambda -1)(n-1))}{n}\cdot \frac{2q}{3(n+1)}\le q\lambda . \end{aligned}$$

Therefore

$$\begin{aligned} q\le \frac{3n(n+1)\lambda -2(n-1)(\lambda -1)}{2(\lambda -l+1)}. \end{aligned}$$

This is a contradiction. Thus, $f_1\wedge \cdots \wedge f_\lambda \equiv 0$. Theorem 1 is proved completely. $\square$

4 Proof of Theorem 2

Let $f:\mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be a meromorphic mapping and $\{a_1,\ldots , a_q\}$ be meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$. Take a reduced representation $f=(f_1:\cdots :f_q)$ of f, we set

$$\begin{aligned} V(f):=\{(c_1,\ldots ,c_n)\in \mathscr {R}^{n+1}: \sum _{i=0}^nc_if_i=0\}. \end{aligned}$$

It is easy to see that V(f) does nor depend on the choice of the reduced representation of f and $\text {rank}_{\mathscr {R}}(f)=n+1-\dim _{\mathscr {R}}V(f)$.

In order to prove Theorem 2, we need the following lemma.

Lemma 4.1

Let $f:\mathbb {C}^m\rightarrow P^n(\mathbb {C})$ be a meromorphic mapping and d be a positive integer or $+\infty$. Let $\{a_j\}_{j=1}^q$ be “small” (with respect to f) meromorphic mappings of $\mathbb {C}^m$ into $P^n(\mathbb {C})$ in general position such that

$$\begin{aligned} \dim \{z:(f,a_i)(z)=(f,a_j)=0\}\le m-2\ (1\le i<j\le q). \end{aligned}$$

Assume that one of the following conditions satisfies:

• (i) $k\ge \frac{n+1}{2}$ and $q> 2(n+1)k-k^2$, or

• (ii) $k< \frac{n+1}{2}$ and $q> k^2+(n-1)k+(n+1)$.

Then for every $g\in \mathscr {G}(f,\{a_j\}_{j=1}^q,1)$, the following hold:

• (a) $V(f)=V(g)$ and $\text {rank}_{\mathscr {R}}(f)=\text {rank}_{\mathscr {R}}(g)$.

• (b) $\Vert T_{f}(r)=O(T_g(r))$ and $\Vert T_{g}(r)=O(T_f(r))$.

Proof

Without loss of generality, we may assume that $\text {rank}_{\mathscr {R}}(f)\ge \text {rank}_{\mathscr {R}}(g)$. Taking reduced representations $f=(f_{0}:\cdots :f_{n})$ and $g=(g_{0}:\cdots :g_{n})$, respectively.

We will show that $V(g)\subseteq V(f)$.

Indeed, suppose that there exists an element $c=(c_0,\ldots ,c_n)\in \mathscr {R}^{n+1}$ such that $\sum _{i=1}^nc_ig_i=0$, but $\sum _{i=1}^nc_if_i\ne 0$. Since $f(z)=g(z)$ on $\cup _{j=1}^q\{z: (f,a_j)(z)=0\}$, we have $\cup _{i=1}^q\{z: (f,a_j)(z)=0\}\subset \{z: (f,c)(z)=0\}$. This implies that

$$\begin{aligned} \sum _{i=1}^qN_{(f, a_i)}^{[1]}(r)\le N_{(f, c)}^{[1]}(r)+o(T_f(r)). \end{aligned}$$

By the Second Main Theorem for moving hyperplanes, we have

• If $k\ge \frac{n+1}{2}$, then

  $$\begin{aligned} \Vert \frac{q}{2n-k+2}T_f(r)\le&\sum _{i=1}^qN_{(f,a_i)}^{[k]}(r) +o(T_f(r)) \\ \le&\,k\sum _{i=1}^qN_{(f,a_i)}^{[1]}(r) +o(T_f(r))\\ \le&kN_{(f, c)}^{[1]}(r)+o(T_f(r))\le kT_f(r)+o(T_f(r)). \end{aligned}$$

  Let $r\rightarrow +\infty$, we get

  $$\begin{aligned} q\le 2(n+1)k-k^2. \end{aligned}$$

  This is in contradiction with (i).

• If $k< \frac{n+1}{2}$, then

  $$\begin{aligned} \Vert \frac{q-n+2k-1}{n+1+k}T_f(r)\le&\sum _{i=1}^qN_{(f,a_i)}^{[k]}(r) +o(T_f(r)) \\ \le&k\sum _{i=1}^qN_{(f,a_i)}^{[1]}(r) +o(T_f(r))\\ \le&kN_{(f, c)}^{[1]}(r)+o(T_f(r))\le kT_f(r)+o(T_f(r)). \end{aligned}$$

  Let $r\rightarrow +\infty$, we get

  $$\begin{aligned} q\le k^2+(n-1)k+(n+1). \end{aligned}$$

  This is in contradiction with (ii).

Hence $V(g)\subseteq V(f)$. It implies that

$$\begin{aligned} \text {rank}_\mathscr {R}(f)=n+1-\dim _\mathscr {R}V(f)\le n+1-\dim _\mathscr {R}V(g)=\text {rank}_\mathscr {R}(g). \end{aligned}$$

Therefore, $\text {rank}_{\mathscr {R}}(f)=\text {rank}_{\mathscr {R}}(g)$, which yields $V(f)=V(g)$.

By the Second Main Theorem for moving hyperplanes, $q\ge 2n-k+2$, we have

$$\begin{aligned} \Vert \frac{q}{2n+2-k}T_f(r)\le&\sum _{i=1}^qN_{(f,a_i)}^{[k]}(r) +o(T_f(r)) \\ \le&k\sum _{i=1}^qN_{(g,a_i)}^{[1]}(r) +o(T_f(r))\le kq(T_g(r))+o(T_f(r)). \end{aligned}$$

Thus, $\Vert T_f(r)=O(T_g(r))$. Since $\text {rank}_{\mathscr {R}}(f)=\text {rank}_{\mathscr {R}}(g)$, similarly we get $\Vert T_g(r)=O(T_f(r))$. The Lemma is proved. $\square$

Proof of Theorem 2

Suppose that $g\in \mathscr {G}(f,\{a_j\}_{j=1}^q,1)$ and $f\ne g$. By changing indices if necessary, we may assume that

$$\begin{aligned} \underbrace{\frac{(f,a_1)}{(g,a_1)}\equiv \frac{(f,a_2)}{(g,a_2)}\equiv \cdots \equiv \frac{(f,a_{k_1})}{(g,a_{k_1})}}_{group\ 1}&\not \equiv \underbrace{\frac{(f,a_{k_1+1})}{(g,a_{k_1+1})}\equiv \cdots \equiv \frac{(f,a_{k_2})}{(g,a_{k_2})}}_{group\ 2} \\ \not \equiv \cdots&\not \equiv \underbrace{\frac{(f,a_{k_{s-1}+1})}{(g,a_{k_{s-1}+1})}\equiv \cdots \equiv \frac{(f,a_{k_s})}{(g,a_{k_s})}}_{group\ s}, \end{aligned}$$

where $k_s=q$. For each $1\le i\le q$, we set

$$\begin{aligned} \sigma (i)= \left\{ \begin{array}{ll} i+k &{} \quad \hbox {if} \quad i+k\le q; \\ i+k-q, &{} \quad \hbox {if} \quad i+k> q. \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} P_i:=(f,a_i)(g,a_{\sigma (i)})-(g,a_i)(f,a_{\sigma (i)}). \end{aligned}$$

We claim that the number of elements in each group is at most k.

Indeed, suppose this claim does not hold. Without loss of generality, we may assume that $k_1\ge k+1$. Since rank$_{\mathscr {R}}(f)=\dim (\{(f,\widetilde{a}_i)\}_{i=1}^{k+1})_{\mathscr {R}}$, for every $0\le j\le n$, there exist meromorphic mappings $c_{ji}\in {\mathscr {R}}$ such that

$$\begin{aligned} f_j=\sum _{i=1}^{k+1}c_{ji}(f,\widetilde{a}_i). \end{aligned}$$

By Lemma 4.1, we have

$$\begin{aligned} V(f)=V(g). \end{aligned}$$

It yields that

$$\begin{aligned} g_j=\sum _{i=1}^{k+1}c_{ji}(g,\widetilde{a}_i)=\frac{(f,a_1)}{(g,a_1)}f_j \end{aligned}$$

for every $0\le j\le n$. Hence $f=g$. This is a contradiction. Thus, the number of elements of each group is at most k.

Hence $\frac{(f,a_i)}{(g,a_i)}$ and $\frac{(f,a_{\sigma (i)})}{(g,a_{\sigma (i)})}$ belong to distinct groups. This implies that $P_i\not \equiv 0(1\le i\le q)$.

By Claim 3.1 for $\lambda =l=2$ and $i_0=\sigma (i)$, we have

$$\begin{aligned} \sum _{j=i,\sigma (i)}2\min \{\nu _{(f,a_j)}(z),\nu _{(g,a_j)}(z)\}+2\sum _{j=1.j\ne i,\sigma (i)}^q\min \{\nu _{(f,a_j)}(z),1\}\le 2\mu _{\tilde{f}\wedge \tilde{g}}(z). \end{aligned}$$

for every $z\in \mathbb {C}^m$ outside an analytic set of codimension $\ge 2$. Because

$$\begin{aligned}&\min \left\{ \nu _{(f,a_j)}(z),\nu _{(g,a_j)}(z)\right\} \\ \ge&\min \left\{ \nu _{(f,a_j)}(z),k\}+\min \{\nu _{(g,a_j)}(z),k\right\} -k\min \{\nu _{(f,a_j)}(z),1\}. \end{aligned}$$

Thus,

$$\begin{aligned}&2\sum _{j=i,\sigma (i)}\left( \min \{\nu _{(f,a_j)}(z),k\}+\min \{\nu _{(g,a_j)}(z),k\} -k\min \{\nu _{(f,a_j)}(z),1\}\right) \\&\qquad +2\sum _{j=1.j\ne i,\sigma (i)}^q\min \{\nu _{(f,a_j)}(z),1\}\le 2\mu _{\tilde{f}\wedge \tilde{g}}(z). \end{aligned}$$

for every $z\in \mathbb {C}^m$ outside an analytic set of codimension $\ge 2$. It yields that

$$\begin{aligned}&\sum _{j=i,\sigma (i)}\sum _{u=f,g}(2N_{(u,a_j)}^{[k]}(r)-kN_{(u,a_j)}^{[1]}(r))\\&\qquad +\sum _{j=1.j\ne i,\sigma (i)}^q\sum _{u=f,g}N_{(u,a_j)}^{[1]}(r)\le 2T(r)+o(T(r)), \end{aligned}$$

where $T(r):=T_f(r)+T_g(r)$. By summing up over all i, we have

$$\begin{aligned} \sum _{j=1}^q\sum _{u=f,g}\left( 4N_{(u,a_j)}^{[k]}(r)+(q-2k-2)N_{(u,a_j)}^{[1]}(r)\right) \le 2qT(r)+o(T(r)). \end{aligned}$$

Since $N_{(u,a_j)}^{[1]}(r)\ge \frac{1}{k}N_{(u,a_j)}^{[k]}(r)$, we obtain

$$\begin{aligned} \frac{q+2k-2}{k}\sum _{j=1}^q\sum _{u=f,g}N_{(u,a_j)}^{[k]}(r)\le 2qT(r)+o(T(r)). \end{aligned}$$

By the Second Main Theorem for moving targets, we get

• If $k\ge \frac{n+1}{2}$,

  $$\begin{aligned} \frac{q+2k-2}{k}\cdot \frac{q}{2n-k+2}T(r)\le 2qT(r)+o(T(r)). \end{aligned}$$

  Letting $r\rightarrow +\infty$, we have

  $$\begin{aligned} q\le -2k^2+2(2n+1)k+2. \end{aligned}$$

  It is a contradiction.

• If $k<\frac{n+1}{2}$,

  $$\begin{aligned} \frac{q+2k-2}{k}\cdot \frac{q-n+2k-1}{n+k+1}T(r)\le 2qT(r)+o(T(r)). \end{aligned}$$

  Letting $r\rightarrow +\infty$, we have

  $$\begin{aligned} q^2-[2k^2+2(n-1)k+n+3]q-2(n+1-2k)(k-1)\le 0. \end{aligned}$$

  Thus,

  $$\begin{aligned} q\le \frac{2k^2+2(n-1)k+(n+3)+\sqrt{(2k^2+2(n-1)k+n-1)^2+8(3n+1)k}}{2}. \end{aligned}$$

  This is a contradiction.

Therefore, we have $f=g$. This completes the proof of Theorem 2. $\square$