Agostino Martinelli
email: agostino.martinelli@ieee.org

Nonlinear Unknown Input Observability: Analytical expression of the observable codistribution in the case of a single unknown input

This paper investigates the unknown input observability problem in the nonlinear case. Specifically, the systems here analyzed are characterized by dynamics that are nonlinear in the state and linear in the inputs and characterized by a single unknown input and multiple known inputs. Additionally, it is assumed that the unknown input is a differentiable function of time (up to a given order). The goal of the paper is not to design new observers but to provide a simple analytic condition in order to check the weak local observability of the state. In other words, the goal is to extend the well known observability rank condition to these systems. Specifically, the paper provides a simple algorithm to directly obtain the entire observable codistribution. As in the standard case of only known inputs, the observable codistribution is obtained by recursively computing the Lie derivatives along the vector fields that characterize the dynamics. However, in correspondence of the unknown input, the corresponding vector field must be suitably rescaled. Additionally, the Lie derivatives must be computed also along a new set of vector fields that are obtained by recursively performing suitable Lie bracketing of the vector fields that define the dynamics. In practice, the entire observable codistribution is obtained by a very simple recursive algorithm. However, the analytic derivations required to prove that this codistribution fully characterizes the weak local observability of the state are complex and, for the sake of brevity, are provided in a separate technical report. The proposed analytic approach is illustrated by checking the weak local observability of several nonlinear systems driven by known and unknown inputs.

INTRODUCTION

The problem of state observability for systems driven by unknown inputs (UI) is a fundamental problem in control theory. This problem was introduced and firstly investigated in the seventies [START_REF] Basile | On the observability of linear, time invariant systems with unknown inputs[END_REF][START_REF] Bhattacharyya | Observer design for linear systems with unknown inputs[END_REF][START_REF] Guidorzi | On Wonham stabilizability condition in the synthesis of observers for unknowninput systems[END_REF][START_REF] Wang | Observing the states of systems with unmeasurable disturbance[END_REF]. A huge effort has then been devoted to design observers for both linear and nonlinear systems in presence of UI, e.g., [START_REF] Barbot | An algebraic framework for the design of nonlinear observers with unknown inputs[END_REF][START_REF] Barbot | An observation algorithm for nonlinear systems with unknown inputs[END_REF][START_REF] Chen | Unknown input observer design for a class of nonlinear systems: an LMI approach[END_REF][START_REF] Darouach | Fullorder observers for linear systems with unknown inputs[END_REF][START_REF] Floquet | A sliding mode approach of unknown input observers for linear systems[END_REF][START_REF] Floquet | On sliding mode observers for systems with unknown inputs[END_REF][START_REF] Guan | A novel approach to the design of unknown input observers[END_REF][START_REF] Ha | Trinh State and input simultaneous estimation for a class of nonlinear systems[END_REF][START_REF] Hammouri | Unknown input observer for state affine systems: A necessary and sufficient condition[END_REF][START_REF] Hou | Design of observers for linear systems with unknown inputs[END_REF][START_REF] Koenig | Design of a class of reduced order unknown inputs nonlinear observer for fault diagnosis[END_REF][START_REF] Koening | Jacquet Unknown input observers for switched nonlinear discrete time descriptor system[END_REF][START_REF] Yang | Observer for linear systems with unknown inputs[END_REF].

The goal of this paper is not to design new observers for systems driven by UI but to provide a simple analytic condition in order to check the weak local observability of the state. The obtained results hold for systems whose dynamics are nonlinear in the state and linear in both the known and the unknown inputs. Additionally, we restrict the analysis to the case of a single unknown input (or disturbance) and we assume that this unknown input is a differentiable function of time (up to a given order).

In [START_REF] Hermann | Nonlinear Controllability and Observability[END_REF] the observability properties of a nonlinear system are derived starting from the definition of indistinguishable states. In section 2 we introduce a new definition of indistinguishable states for the case UI. Then, in section 3 we introduce the approach to check if the state is weakly locally observable at a given point. For the brevity sake, we only provide the method. All the analytical derivations are provided separately in a technical report [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF].

As in the standard case of only known inputs, the observable codistribution is obtained by recursively computing the Lie derivatives along the vector fields that characterize the dynamics. However, in correspondence of the unknown input (denoted with w), the corresponding vector field (denoted with g) must be suitably rescaled. In particular, it must be divided by the first order Lie derivative of the output along g (the result is the vector field g L 1 g that appears in the second line of the algorithm in definition 2). Additionally, the Lie derivatives must be computed also along a new set of vector fields that are obtained by recursively performing suitable Lie bracketing of the vector fields that define the dynamics (the vectors φ i m in definition 2). In practice, the entire observable codistribution is obtained by a very simple recursive algorithm. However, the analytic derivations required to prove that this codistribution fully characterizes the weak local observability of the state are complex and, for the sake of brevity, are provided in a separate technical report [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF]. Finally, the recursive algorithm converges in a finite number of steps and the criterion to establish that the convergence has been reached is provided.

The proposed analytic approach is illustrated in section 4 by checking the weak local observability of several nonlinear systems driven by a single unknown input.

The considered system

In the sequel we will refer to a nonlinear control system with m u known inputs (u ≡ [u 1 , • • • , u mu ] T ) and a single unknown input or disturbance (w). The state is the vector x ∈ M , with M an open set of R n . We assume that the dynamics are nonlinear with respect to the state and linear with respect to the inputs (both known and unknown). Finally, for the sake of simplicity, we will refer to the case of a single output y (the extension to multiple outputs is straightforward). Our system is characterized by the following equations:

(2.1)

     ẋ = mu i=1 f i (x)u i + g(x)w y = h(x)
where f i (x), i = 1, • • • , m u , and g(x) are vector fields in M and the function h(x) is a scalar function defined on the open set M . For the sake of simplicity, we will assume that all these functions are analytic functions in M .

Let us consider the time interval I ≡ [0, T ]. Note that, since the equations in (2.1) do not depend explicitly on time, this can be considered as a general time interval of length T . In the sequel, we will assume that the solution of (2.1) exists in I and we will denote by x(t; x 0 ; u; w) the state at a given time t ∈ I, when x(0) = x 0 and the known input and the disturbance are u(t) and w(t), respectively, ∀t ∈ I. We introduce the following definition: Definition 1 (Indistinguishable states) Two states x a and x b are indistinguishable if, for any u(t) (the known input vector function), there exist w a (t) and w b (t) (i.e., two unknown inputs in general, but not necessarily, different from each other) such that h(x(t; x a ; u; w a )) = h(x(t; x b ; u; w b )) ∀t ∈ I. This definition states that, if x a and x b are indistinguishable, then, for any known input, by looking at the output during the time interval I, we cannot conclude if the initial state was x a and the disturbance w a or if the initial state was x b and the disturbance w b . We remark that, contrary to the definition of indistinguishable states in the case without disturbances, the new definition does not establish an equivalence relation. Indeed, we can have x a and x b indistinguishable, x b and x c indistinguishable but x a and x c are not indistinguishable. As in the case of known inputs, given x 0 , the indistinguishable set I x0 is the set of all the states x such that x and x 0 are indistinguishable. Starting from this definition, we can use exactly the same definitions of observability and weak local observability adopted in the case without disturbances.

3 The observable codistribution Ω This section introduces the analytical method to check the weak local observability of the state x that satisfies (2.1) at a given x 0 ∈ M . This is obtained by computing a codistribution (the observable codistribution). In the sequel, we will denote by L 1 g the first order Lie derivative of the function h(x) along the vector field g(x), i.e., L 1

g ≡ L g h The analytical computation of the observable codistribution is based on the assumption that L 1 g = 0 on a given neighbourhood of x 0 . In the appendix we show that, when this assumption does not hold, it is possible to introduce new local coordinates and the observability properties can be investigated starting from a new output that satisfies the aforementioned assumption. We will denote with the symbol d x the gradient with respect to the state x. Additionally, for a given codistribution Λ and a given vector field θ, we will denote by L θ Λ the codistribution whose covectors are the Lie derivatives along θ of the covectors in Λ (we are obviously assuming that the dimension of these covectors coincides with the dimension of θ). Finally, given two vector spaces V 1 and V 2 , we denote with V 1 + V 2 their sum, i.e., the span of all the generators of both V 1 and V 2 . We define the Ω codistribution recursively, as follows Definition 2 (Ω codistribution) This codistribution is defined recursively by the following algorithm:

1. Ω 0 = d x h; 2. Ω m = Ω m-1 + mu i=1 L fi Ω m-1 + L g L 1 g Ω m-1 + mu i=1 L φ i m-1 d x h
where the vectors

φ i m ∈ R n (i = 1, • • • , m u )
are defined by the following algorithm:

1. φ i 0 = f i ; 2. φ i m = [φ i m-1 , g] L 1 g
where the parenthesis [•, •] denote the Lie brackets of vector fields.

In [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF] we prove that the first algorithm in definition 2 converges. The proof is carried out by first considering the case of a single known input (i.e., m u = 1) and is given by theorem 2 in section 7. Then, its validity is extended to the case of multiple inputs (m u > 1) in section 8. In [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF] it is also provided the criterion to check that the convergence has been reached. This criterion needs first of all to compute the following function:

ρ = L 2 g h (L 1 g ) 2
In [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF] we prove that it exists m such that d x ρ ∈ Ω m (and therefore d x ρ ∈ Ω m ∀m ≥ m ). Additionally, we prove that the convergence of the algorithm has been reached when Ω m+1 = Ω m , and m ≥ m . From the derivations in [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF] it is possible to see that the required number of steps is at most 2n + 2.

In [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF] it is also shown that the computed codistribution is the entire observable codistribution. Also in this case, the proof is given by first considering the case of a single known input (see theorem 1 in section 6) and then, its validity is extended to the case of multiple inputs in section 8. Note that this proof is based on the assumption that the unknown input (w) is a differentiable function of time, up to a given order (the order depends on the specific case).

We conclude this section by outlining the steps to investigate the weak local observability at a given point x 0 of a nonlinear system driven by a single disturbance and several known inputs. In other words, to investigate the weak local observability of a system defined by a state that satisfies the dynamics in (2.1). The validity of the following procedure is a consequence of the theoretical results obtained in [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF]:

1. For the chosen x 0 , compute L 1 g (= L 1 g h) and ρ = L 2 g h (L 1 g ) 2 .
In the case when L 1 g = 0, introduce new local coordinates, as explained in the appendix and re-define the output. 5. If the gradient of a given state component (

x j , j = 1, • • • , n) belongs to Ω * (namely if d x x j ∈ Ω * )
on a given neighbourhood of x 0 , then x j is weakly locally observable at x 0 . If this holds for all the state components, the state x is weakly locally observable at x 0 . Finally, if the dimension of Ω * is smaller than n on a given neighbourhood of x 0 , then the state is not weakly locally observable at x 0 .

Applications

We apply the method described in section 3 in order to investigate the observability properties of several nonlinear systems characterized by the equations given in (2.1).

We consider a vehicle that moves on a 2Denvironment. The configuration of the vehicle in a global reference frame, can be characterized through the vector [x v , y v , θ] T where x v and y v are the cartesian vehicle coordinates, and θ is the vehicle orientation. We assume that the dynamics of this vector satisfy the unicycle differential equations:

(4.2)    ẋv = v cos θ ẏv = v sin θ θ = ω
where v and ω are the linear and the rotational vehicle speed, respectively, and they are the system inputs. We consider the following three cases of output (see also figure 1 for an illustration):

1. the distance from the origin (e.g., a landmark is at the origin and its distance is measured by a range sensor);

2. the bearing of the origin in the local frame (e.g., a landmark is at the origin and its bearing angle is measured by an on-board camera);

3. the bearing of the vehicle in the global frame (e.g., a camera is placed at the origin).

We can analytically express the output in terms of the state. We remark that the expressions become very simple if we adopt polar coordinates:

r ≡ x 2 v + y 2 v , φ = atan yv xv .
We have, for the three cases, y = r, y = β = π -(θ -φ) and y = φ, respectively. For each of these three cases, we consider the following two cases: v is known, ω is unknown; v is unknown, ω is known. The dynamics in these new coordinates become:

(4.3)     ṙ = v cos(θ -φ) φ = v r sin(θ -φ) θ = ω
f =   0 0 1   g =   cos(θ -φ) sin(θ-φ) r 0  
We follow the five steps mentioned at the end of section 3. We have

L 1 g = cos(θ -φ) and ρ ≡ L 2 g (L 1 g ) 2 = tan 2 (θ-φ)
r . Additionally:

d x ρ = tan(θ -φ) r - tan(θ -φ) r , - 2 cos 2 (θ -φ) , 2 cos 2 (θ -φ)
We also have Ω 0 = span{[1, 0, 0]}. Hence, d x ρ / ∈ Ω 0 . Additionally, Ω 1 = Ω 0 . We need to compute Ω 2 and, in order to do this, we need to compute φ 1 . We obtain:

φ 1 =   -tan(θ -φ) 1 r 0   and Ω 2 = span [1, 0, 0], 0, 1 cos 2 (θ-φ) , - 1 cos 2 (θ-φ) . It is imme- diate to check that d x ρ ∈ Ω 2 , meaning that m = 2.
Additionally, by a direct computation, it is possible to check that Ω 3 = Ω 2 meaning that m * = 2 and Ω * = Ω 2 , whose dimension is 2. We conclude that the dimension of the observable codistribution is equal to 2 and the state is not weakly locally observable.

4.2 y = r, u = v, w = ω In this case we have

f =   cos(θ -φ) sin(θ-φ) r 0   g =   0 0 1  
We follow the five steps mentioned at the end of section 3. We easily obtain L 1 g = 0. Hence, we have to introduce new local coordinates, as explained in the appendix. We obtain L 1 f h = cos(θ -φ) and we obtain that the relative degree of the associated system in (5.6) is r = 2. Let us denote the new coordinates by x 1 , x 2 , x 3 . In accordance with (5.7) and (5.8) we should set x 1 = r and x 2 = cos(θ -φ). On the other hand, to simplify the computation, we set x 2 = θ -φ. Finally, we set x 3 = θ. We compute the new vector fields that characterize the dynamics in the new coordinates. We have:

(4.4) f ≡    cos(x 2 ) - sin(x 2 ) x 1 0    g ≡   0 1 1  
Additionally, we set h = cos(x 2 ) and Ω 1 = span{[1, 0, 0], [0, -sin(x 2 ), 0]}. In the new coordinates we obtain: L 1 g = -sin(x 2 ) and ρ = -

cos(x 2 ) sin 2 (x 2 ) . It is im- mediate to check that d x ρ ∈ Ω 1 , meaning that m = 1.
Additionally, by a direct computation, it is possible to check that Ω 2 = Ω 1 meaning that m * = 1 and Ω * = Ω 1 , whose dimension is 2. We conclude that the dimension of the observable codistribution is equal to 2 and the state is not weakly locally observable.

y

= θ -φ, u = ω, w = v In this case we have f =   0 0 1   g =   cos(θ -φ) sin(θ-φ) r 0  
We follow the five steps mentioned at the end of section 3. We have L 1 g = -sin(θ-φ) r and ρ = 2 cot(θ -φ). Additionally:

d x ρ = 2 sin 2 (θ -φ) [0, 1, -1]
We also have Ω 0 = span{[0, -1, 1]}. Hence, d x ρ ∈ Ω 0 , meaning that m = 0. Additionally, by a direct computation, it is possible to check that Ω 1 = Ω 0 meaning that m * = 0 and Ω * = Ω 0 , whose dimension is 1. We conclude that the dimension of the observable codistribution is equal to 1 and the state is not weakly locally observable.

4.4 y = θ -φ, u = v, w = ω In this case we have

f =   cos(θ -φ) sin(θ-φ) r 0   g =   0 0 1  
We follow the five steps mentioned at the end of section 3. We have L 1 g = 1 and ρ = 0. Hence, d x ρ = [0, 0, 0] and we do not need to check if

d x ρ ∈ Ω m .
In other words, we can set m = 0. By a direct computation we obtain:

Ω 0 = span{[0, -1, 1]}, Ω 1 = span [0, -1, 1], -sin(θ-φ) r 2
, -cos(θ-φ) r , cos(θ-φ) r . Additionally, we obtain Ω 2 = Ω 1 , meaning that m * = 1 and Ω * = Ω 1 , whose dimension is 2. We conclude that the dimension of the observable codistribution is equal to 2 and the state is not weakly locally observable.

4.5 y = φ, u = ω, w = v In this case we have

f =   0 0 1   g =   cos(θ -φ) sin(θ-φ) r 0  
We follow the five steps mentioned at the end of section 3. We have L 1 g = sin(θ-φ) r and ρ = -2 cot(θ -φ). Additionally:

d x ρ = 2 sin 2 (θ -φ) [0, -1, 1]
We also have Ω 0 = span{[0, 1, 0]}. Hence, d x ρ / ∈ Ω 0 . Additionally, Ω 1 = Ω 0 . We need to compute Ω 2 and, in order to do this, we need to compute φ 1 . We obtain:

φ 1 =   -r cot(θ -φ) 0   and Ω 2 = span [0, 1, 0], 1 sin 2 (θ-φ) [0, 1, -1] . It is immediate to check that d x ρ ∈ Ω 2 ,
meaning that m = 2. Additionally, by a direct computation, it is possible to check that Ω 3 = Ω 2 meaning that m * = 2 and Ω * = Ω 2 , whose dimension is 2. We conclude that the dimension of the observable codistribution is equal to 2 and the state is not weakly locally observable.

4.6 y = φ, u = v, w = ω In this case we have f =   cos(θ -φ) sin(θ-φ) r 0   g =   0 0 1  
We follow the five steps mentioned at the end of section 3. We easily obtain L 1 g = 0. Hence, we have to introduce new local coordinates, as explained in the appendix. We obtain L 1 f h = sin(θ-φ) r and we obtain that the relative degree of the associated system in (5.6) is r = 2. Let us denote the new coordinates by x 1 , x 2 , x 3 . In accordance with (5.7) and (5.8) we set x 1 = φ and x 2 = sin(θ-φ) r . Finally, we set x 3 = cos(θ-φ) r . We compute the new vector fields that characterize the dynamics in the new coordinates. We obtain:

(4.5) f ≡   x 2 -2x 2 x 3 x 2 2 -x 2 3   g ≡   0 x 3 -x 2   Additionally, we set h = x 2 and Ω 1 = span{[1, 0, 0], [0, 1, 0]}.
In the new coordinates we obtain: L 1 g = x 3 and ρ = -

x 2 x 2 3
. Since ρ depends on x 3 , d x ρ / ∈ Ω 1 . Since the dimension of Ω 1 is already 2 and since we know that it exists a given integer m such that d x ρ ∈ Ω m , we conclude that the dimension of Ω m (and consequently the dimension of Ω * ) is larger than 2. Hence, the entire state is weakly locally observable.

We conclude this section by remarking that, the results obtained for the six cases previously investigated, agree with what we expected. By using the observability rank condition in [START_REF] Hermann | Nonlinear Controllability and Observability[END_REF], we easily obtain that, when both the inputs are known, the dimension of the observable codistribution is 2 for the first two observations (y = r and y = θ -φ) and 3 for the last one (y = φ). In particular, for the first two observations, all the initial states rotated around the vertical axis are indistinguishable. When one of the inputs misses, this unobservable degree of freedom obviously remains. Hence, for the first four cases previously investigated, it is non surprising that the dimension of the observable codistribution does not exceed 2. Additionally, for y = θ-φ and y = φ, when the first input (v) is unknown, we lose a further degree of freedom, which corresponds to the absolute scale.

CONCLUSIONS

In this paper we investigated the problem of nonlinear observability when one of the system inputs is unknown. The goal was not to design new observers but to provide simple analytic conditions in order to check the weak local observability of the state. An unknown input was also called disturbance. We introduced a simple analytic condition to check the weak local observability of the state at a given point x 0 . This condition is based on the computation of a codistribution (the observable codistribution).

As in the standard case of only known inputs, the observable codistribution is obtained by recursively computing the Lie derivatives along the vector fields that characterize the dynamics. However, in correspondence of the unknown input (denoted with w), the corresponding vector field (denoted with g) must be suitably rescaled. In particular, it must be divided by the first order Lie derivative of the output along g at x 0 (the result is the vector field g L 1 g that appears in the second line of the algorithm in definition 2; note that when L 1 g = 0 at x 0 , the coordinates must be changed, as explained in the appendix). Additionally, the Lie derivatives must be computed also along a new set of vector fields that are obtained by recursively performing suitable Lie bracketing of the vector fields that define the dynamics (the vectors φ i m in definition 2). In practice, the entire observable codistribution was obtained by a very simple recursive algorithm. However, the analytic derivations required to prove that this codistribution fully characterizes the weak local observability of the state are complex and, for the sake of brevity, they were provided in a separate technical report [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF]. Finally, in [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF] we prove that the recursive algorithm converges in a finite number of steps and the criterion to establish that the convergence has been reached was also provided. Also this proof is based on several tricky analytical steps (the details are provided in [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF]). To this regard, we really wish to emphasize that almost all the properties derived in [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF] have been proved by induction. Obtaining these properties, and not simply their proofs, has required a huge effort in terms of analytical computation and number of trials.

The proposed analytic approach has been illustrated by checking the weak local observability of several nonlinear systems driven by a single unknown input and a known input.

We are extending the analytic results presented in this paper to the case of multiple unknown inputs. So far, we have introduced a tool, the extended observability rank condition, that only provides sufficient conditions for the weak local observability in the case of multiple disturbances (see definition 2 and proposition 4 in [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF]).

In particular, we have successfully adopted this tool to investigate the observability properties of the visualinertial structure from motion problem in the case of missing inputs [START_REF] Martinelli | Visual-inertial structure from motion: observability vs minimum number of sensors[END_REF][START_REF] Martinelli | Observability Properties and Deterministic Algorithms in Visual-Inertial Structure from Motion[END_REF][START_REF] Martinelli | Minimalistic sensor design in visualinertial structure from motion[END_REF].

APPENDIX

Let us suppose that L 1 g = 0 on a given neighbourhood of x 0 . We introduce the following system associated with the system in (2.1):

(5.6) ẋ = f (x) + g(x)u y = h(x)

where f (x) = f i (x) is one of the vector field f i (x) in (2.1). This is a system without disturbances and with a single known input u. Let us denote by r the relative degree of this system at x 0 . Since L 1 g = 0 on a given neighbourhood of x 0 , we have r > 1. Additionally, we can introduce the following new local coordinates (see proposition 4.1.3 in [START_REF] Isidori | Nonlinear Control Systems[END_REF]):

(5.7)

x = Q(x) =   Q 1 (x) • • • Q n (x)  
such that the first new r coordinates are:

(5.8)

Q 1 (x) = h(x), Q 2 (x) = L 1 f h(x), • • • , Q r (x) = L r-1 f h(x)
Now let us derive the equations of the original system (i.e., the one in (2.1)) in these new coordinates. We have:

(5.9)

     ẋ = mu i=1 fi (x )u i + g(x )w y = x 1
where fi (i = 1, • • • , m) and g can be analytically computed starting from the expression of

Q 1 , • • • , Q n .
In particular, g has the following structure:

(5.10)

g ≡           0 0 • • • 0 gr (x ) • • • gn (x )          
It is possible to check that the first r components of x are weakly locally observable. Indeed, the first r components of x are the output and its Lie derivatives along f up to the (r -1)-order. Additionally, the first (r -1) Lie derivatives are constant on the indistinguishable sets because all the Lie derivatives up to the (r -1)-order that includes at least one direction along g vanish automatically (see propositions 1 and 3 in [START_REF] Martinelli | Non linear Unknown Input Observability: Extension of the observability rank condition and the case of a single unknown input[END_REF]). In order to investigate the observability properties of the remaining components, we consider the new output h(x ) = x r and we set L 1 g = gr = L g L r-1 f h = 0.

2 .

 2 Build the codistribution Ω m (at x 0 ) by using the algorithm provided in definition 2, starting from m = 0 and, for each m, check if d x ρ ∈ Ω m . 3. Denote by m the smallest m such that d x ρ ∈ Ω m . 4. For each m ≥ m check if Ω m+1 = Ω m and denote by Ω * = Ω m * where m * is the smallest integer such that m * ≥ m and Ω m * +1 = Ω m * (note that m * ≤ 2n + 2).

Figure 1 :

 1 Figure 1: The vehicle state in cartesian and polar coordinates together with the three considered outputs.

This work was supported by the French National Research Agency ANR through the project VIMAD † A.


Tuan Dinh

Huynh

EXAMPLES OF HYPERBOLIC HYPERSURFACES OF LOW DEGREE IN PROJECTIVE SPACES

Keywords: Kobayashi conjecture, hyperbolicity, Brody Lemma, Nevanlinna Theory

We construct families of hyperbolic hypersurfaces of degree 2n in the projective space P n (C) for 3 ≤ n ≤ 6.

Introduction and the main result

The Kobayashi conjecture states that a generic hypersurface X d ⊂ P n (C) of degree d ≥ 2n -1 is hyperbolic. It is proved by Demailly and El Goul [START_REF] Demailly | Hyperbolicity of generic surfaces of high degree in projective 3-space[END_REF] for n = 3 and a very generic surface of degree at least 21. In [START_REF] Pȃun | Vector fields on the total space of hypersurfaces in the projective space and hyperbolicity[END_REF], Pȃun improved the degree to 18. In P 4 (C), Rousseau [START_REF] Rousseau | Weak analytic hyperbolicity of generic hypersurfaces of high degree in P 4[END_REF] was able to show that a generic three-fold of degree at least 593 contains no Zariski-dense entire curve, a result from which hyperbolicity follows, after removing divisorial components [START_REF] Diverio | A remark on the codimension of the Green-Griffiths locus of generic projective hypersurfaces of high degree[END_REF]. In P n (C), for any n and for d ≥ 2 (n-1) 5 , Diverio, Merker and Rousseau [START_REF] Diverio | Effective algebraic degeneracy[END_REF] established algebraic degeneracy of entire curves in X d . Also, Siu [START_REF] Siu | Hyperbolicity of generic high-degree hypersurfaces in complex projective space[END_REF] proposed a positive answer for arbitrary n and degree d = d(n) 1 very large. Most recently, Demailly [START_REF] Demailly | Proof of the Kobayashi conjecture on the hyperbolicity of very general hypersurfaces[END_REF] has announced a strategy that is expected to attain Kobayashi's conjecture for very generic hypersurfaces of degree d ≥ 2n.

Concurrently, many authors tried to find examples of hyperbolic hypersurfaces of degree as low as possible. The first example of a compact Kobayashi hyperbolic manifold of dimension 2 is a hypersurface in P 3 (C) constructed by Brody and Green [START_REF] Brody | A family of smooth hyperbolic hypersurfaces in P 3[END_REF]. Also, the first examples in all higher dimensions n -1 ≥ 3 were discovered by Masuda and Noguchi [START_REF] Masuda | A construction of hyperbolic hypersurface of P n (C)[END_REF], with degree large. So far, the best degree asymptotic is the square of dimension, given by Siu and Yeung [START_REF] Siu | Defects for ample divisors of abelian varieties, Schwarz lemma, and hyperbolic hypersurfaces of low degrees[END_REF] with d = 16(n -1) 2 and by Shiffman and Zaidenberg [START_REF] Shiffman | Hyperbolic hypersurfaces in P n of Fermat-Waring type[END_REF] with d = 4(n -1) 2 . In P 3 (C) many examples of low degree were given (see the reference of [START_REF] Zaidenberg | Hyperbolic surfaces in P 3 : examples[END_REF]). The lowest degree found up to date is 6, given by Duval [START_REF] Duval | Une sextique hyperbolique dans P 3 (C)[END_REF]. There are not so many examples of low degree hyperbolic hypersurfaces in P 4 (C). We mention here an example of a hypersurface of degree 16 constructed by Fujimoto [START_REF] Fujimoto | A family of hyperbolic hypersurfaces in the complex projective space[END_REF]. Various examples in P 5 (C) and P 6 (C) only appear in the cases of arbitrary dimension mentioned above.

Before going to introduce the main result, we need some notations and conventions. A family of hyperplanes {H i } 1≤i≤q with q ≥ n+1 in P n (C) is said to be in general position if any n+1 hyperplanes in this family have empty intersection. A hypersurface S in P n (C) is said to be in general position with respect to {H i } 1≤i≤q if it avoids all intersection points of n hyperplanes in {H i } 1≤i≤q , namely if: S ∩ ∩ i∈I H i = ∅, ∀I ⊂ {1, . . . , q}, |I| = n. Now assume that {H i } 1≤i≤q is a family of hyperplanes of P n (C) (n ≥ 2) in general position. Let {H i } i∈I be a subfamily of n + 2 hyperplanes. Take a partition I = J ∪ K such that |J|, |K| ≥ 2. Then there exists a unique hyperplane H JK containing ∩ j∈J H j and ∩ k∈K H k . We call H JK a diagonal hyperplane of {H i } i∈I . The family {H i } 1≤i≤q is said to be generic if, for all disjoint subsets I, J, J 1 , . . . , J k of {1, . . . , q} such that |I|, |J i | ≥ 2 and |I| + |J i | = n + 2, 1 ≤ i ≤ k, for every subset {i 1 , . . . , i l } of I, the intersection between the |J| hyperplanes H j , j ∈ J, the k diagonal hyperplanes H IJ 1 , . . . , H IJ k , and the l hyperplanes H i 1 , . . . , H i l is a linear subspace of codimension min{k + l, |I|} + |J|, with the convention that when min{k + l, |I|} + |J| > n, this intersection is empty. Such a generic condition naturally appears in our constructions, and it has the virtue of being preserved when passing to smaller-dimensional subspaces Our aim in this article is to prove that, for 3 ≤ n ≤ 6, a small deformation of a union of generic 2n hyperplanes in P n (C) is hyperbolic.

Main Theorem. Let n be an integer number in {3, 4, 5, 6}. Let {H i } 1≤i≤2n be a family of 2n generic hyperplanes in P n (C), where H i = {h i = 0}. Then there exists a hypersurface S = {s = 0} of degree 2n in general position with respect to {H i } 1≤i≤2n such that the hypersurface

Σ = s + Π 2n i=1 h i = 0
is hyperbolic for sufficiently small complex = 0.

Our proof is based on the technique of Duval [START_REF] Duval | Around Brody lemma[END_REF] in the case n = 3. By the deformation method of Zaidenberg and Shiffman [START_REF] Shiffman | New examples of Kobayashi hyperbolic surfaces in CP 3[END_REF], the problem reduces to finding a hypersurface S such that all complements of the form ∩ i∈I H i \ ∪ j ∈I H j \ S are hyperbolic. This situation is very close to Theorem 2.5. To create such S, we proceed by deformation in order to allow points of intersection of S with more and more linear subspaces coming from the family {H i } 1≤i≤2n .

Nevanlinna theory and some applications

We recall some facts from Nevanlinna theory in the projective space P n (C). Let E = µ ν a ν be a divisor on C and let k ∈ N ∪ {∞}. Summing the k-truncated degrees of the divisor on disks by

n [k] (t, E) := |aν |<t min {k, µ ν } (t > 0),
the truncated counting function at level k of E is defined by

N [k] (r, E) := r 1 n [k] (t, E) t d t (r > 1).
When k = ∞, we write n(t, E), N (r, E) instead of n [∞] (t, E), N [∞] (r, E). We denote the zero divisor of a nonzero meromorphic function ϕ by (ϕ) 0 . Let f : C → P n (C) be an entire curve having a reduced

representation f = [f 0 : • • • : f n ] in the homogeneous coordinates [z 0 : • • • : z n ] of P n (C). Let D = {Q = 0} be a hypersurface in P n (C) defined by a homogeneous polynomial Q ∈ C[z 0 , . . . , z n ] of degree d ≥ 1. If f (C) ⊂ D,
we define the truncated counting function of f with respect to D as

N [k] f (r, D) := N [k] r, (Q • f ) 0 .
The proximity function of f for the divisor D is defined as

m f (r, D) := 2π 0 log f (re iθ ) d Q Q(f )(re iθ ) d θ 2π ,
where Q is the maximum absolute value of the coefficients of Q and

f (z) = max {|f 0 (z)|, . . . , |f n (z)|}. Since Q(f ) ≤ Q • f d , one has m f (r, D) ≥ 0.
Finally, the Cartan order function of f is defined by

T f (r) := 1 2π 2π 0 log f (re iθ ) d θ.
It is known that [START_REF] Eremenko | Brody curves omitting hyperplanes[END_REF] if f is a Brody curve, then its order

ρ f := lim sup r→+∞ T f (r) log r
is bounded from above by 2. Furthermore, Eremenko [START_REF] Eremenko | Brody curves omitting hyperplanes[END_REF] showed the following.

Theorem 2.1. If f : C → P n (C) is a Brody curve omitting n hyperplanes in general position, then it is of order 1.

Consequently, we have the following theorem.

Theorem 2.2. If f : C → P n (C) is a Brody curve avoiding the first n coordinate hyperplanes {z i = 0} n-1 i=0
, then it has a reduced representation of the form

[1 : e λ 1 z+µ 1 : • • • : e λ n-1 z+µ n-1 : g],
where g is an entire function and λ i , µ i are constants. If f also avoids the remaining coordinate hyperplane {z n = 0}, then g is of the form e λn z+µn .

The core of Nevanlinna theory consists of two main theorems.

First Main Theorem. Let f : C → P n (C) be a holomorphic curve and let D be a hypersurface of degree d in P n (C) such that f (C) ⊂ D. Then for every r > 1, the following holds

m f (r, D) + N f (r, D) = d T f (r) + O(1), hence N f (r, D) ≤ d T f (r) + O(1). (2.1)
A holomorphic curve f : C → P n (C) is said to be linearly nondegenerate if its image is not contained in any hyperplane. For non-negatively valued functions ϕ(r), ψ(r), we write ϕ(r) ≤ ψ(r) if this inequality holds outside a Borel subset E of (0, +∞) of finite Lebesgue measure. Next is the Second Main Theorem of Cartan [START_REF] Cartan | Sur les zéros des combinaisons linéaires de p fonctions holomorphes données[END_REF].

Second Main Theorem. Let f : C → P n (C) and let {H i } 1≤i≤q be a family of hyperplanes in general position in P n (C). Then the following estimate holds:

(q -n -1) T f (r) ≤ q i=1 N [n] f (r, H i ) + S f (r),
where S f (r) is a small term compared with T f (r)

S f (r) = o(T f (r)) .
The next three theorems can be deduced from the Second Main Theorem.

Theorem 2.3. Let {H i } 1≤i≤n+2 be a family of hyperplanes in general position in

P n (C) with n ≥ 2. If f : C → P n (C) \ ∪ n+2
i=1 H i is an entire curve, then its image lies in one of the diagonal hyperplanes of {H i } 1≤i≤n+2 .

The following strengthened theorem is due to Dufresnoy [START_REF] Dufresnoy | Théorie nouvelle des familles complexes normales. Applications à l'étude des fonctions algébroïdes[END_REF].

Theorem 2.4. If a holomorphic map f : C → P n (C) has its image in the complement of n + p hyperplanes H 1 , . . . , H n+p in general position, then this image is contained in a linear subspace of dimension n p .

As a consequence, we have the classical generalization of Picard's Theorem (case n = 1), due to Fujimoto [START_REF] Fujimoto | On holomorphic maps into a taut complex space[END_REF] (see also [START_REF] Lee Green | Holomorphic maps into complex projective space omitting hyperplanes[END_REF]).

Theorem 2.5. The complement of a collection of 2n + 1 hyperplanes in general position in P n (C) is hyperbolic.

For hyperplanes that are not in general position, we have the following result (see [START_REF] Kobayashi | Hyperbolic complex spaces, volume 318 of Grundlehren der Mathematischen Wissenschaften[END_REF], Theorem 3.10.15). Theorem 2.6. Let {H i } 1≤i≤q be a family of q ≥ 3 hyperplanes that are not in general position in

P n (C). If f : C → P n (C) \ ∪ q
i=1 H i is an entire curve, then its image lies in some hyperplane.

Starting lemmas

Let us introduce some notations before going to other applications. Let {H i } 1≤i≤q be a family of generic hyperplanes of P n (C), where H i = {h i = 0}. For some integer 0 ≤ k ≤ n -1 and some subset

I k = {i 1 , . . . , i n-k } of the index set {1, . . . , q} having cardinality n -k, the linear subspace P k,I k = ∩ i∈I k H i P k (C) is called a subspace of dimension k. For a holomorphic mapping f : C → P n (C), we define n f (t, P k,I k ) := |z|<t,f (z)∈P k,I k min i∈I k ord z (h i • f ) (t > 0),
where we take the sum only for z in the preimage of P k,I k , and

N f (r, P k,I k ) := r 1 n f (t, P k,I k ) t d t (r > 1).
(3.1)

We denote by P * k,I k the complement P k,I k \ ∪ i ∈I k H i which will be called a star-subspace of dimension k. We can also define n f (t, P * k,I k ) and N f (r, P * k,I k ). Assume now q = 2n + 1 + m with m ≥ 0. Consider complements of the form

P n (C) \ (∪ 2n+1+m i=1 H i \ A m,n ), (3.2)
where A m,n is a set of at most m elements of the form P * k,I k (0 ≤ k ≤ n -2). We note that if m = 0, these complements are hyperbolic by Theorem 2.5.

In P 2 (C), a union of lines ∪ q i=1 H i is in general position if any three lines have empty intersection, and it is generic if in addition any three intersection points between three distinct pairs of lines are not collinear. Lemma 3.1. In P 2 (C), if m ≤ 3, all complements of the form (3.2) are hyperbolic.

Proof. Without loss of generality, we can assume that A m,2 is a set of m distinct points belonging to

∪ 1≤i 1 <i 2 ≤5+m H i 1 ∩ H i 2 .
When m = 1, an entire curve f :

C → P 2 (C) \ (∪ 6 i=1 H i \ A 1,2
H i 3 H i 4 H i 2 L L H i 1 A H i 2 = H i 3 B H i 1 A B H i 4 Let f : C → P 2 (C) \ (∪ 7 i=1 H i \ A 2,2 ) be an entire curve. If z ∈ f -1 (A), we have ord z (h i 1 • f ) ≥ 1, ord z (h i 2 • f ) ≥ 1.
This implies

min {ord z (h i 1 • f ), 2} + min {ord z (h i 2 • f ), 2} ≤ 3 min 1≤j≤2 ord z (h i j • f ). (3.3)
Hence, by summing this inequality

|z|<t,f (z)=A min {ord z (h i 1 • f ), 2} + |z|<t,f (z)=A min {ord z (h i 2 • f ), 2} ≤ 3 |z|<t,f (z)=A min 1≤j≤2 ord z (h i j • f ). (3.4)
Similarly for h i 3 , h i 4 and z ∈ f -1 (B), we have

|z|<t,f (z)=B min {ord z (h i 3 • f ), 2} + |z|<t,f (z)=B min {ord z (h i 4 • f ), 2} ≤ 3 |z|<t,f (z)=B min 3≤j≤4 ord z (h i j • f ).
(3.5) By taking the sum of both sides of these inequalities and by integrating, we obtain

i∈I N [2] f (r, H i ) ≤ 3 N f (r, A) + N f (r, B) . (3.6)
Now, let L = { = 0} be the line passing through A and B. Since = α 1 h i 1 +α 2 h i 2 = α 3 h i 3 +α 4 h i 4 for some α 1 , . . . , α 4 ∈ C, the following inequalities hold

min 1≤j≤2 ord z (h i j • f ) ≤ ord z ( • f ) (z ∈ f -1 (A)), min 3≤j≤4 ord z (h i j • f ) ≤ ord z ( • f ) (z ∈ f -1 (B)).
(3.7) Since f -1 (A) and f -1 (B) are two disjoint subsets of f -1 (L), by taking the sum of both sides of these inequalities on discs and by integrating, we obtain

N f (r, A) + N f (r, B) ≤ N f (r, L). (3.8)
If f would be linearly nondegenerate, then starting from Cartan Second Main Theorem, and using (3.6), (3.8), we would get

4 T f (r) ≤ 7 i=1 N [2] f (r, H i ) + S f (r) = i∈I N [2] f (r, H i ) + S f (r) ≤ 3 N f (r, A) + N f (r, B) + S f (r) ≤ 3 N f (r, L) + S f (r) [Use (2.1)] ≤ 3 T f (r) + S f (r), (3.9)
which is absurd. Thus, any entire curve f :

C → P 2 (C) \ (∪ 7 i=1 H i \ A 2,2
) must be contained in some line L. Furthermore, the number of points of intersection between L and ∪ 7 i=1 H i \ {A, B} is at least 3 by the generic condition. By Picard's Theorem, this contradicts the assumption that f is nonconstant.

When m = 3, A 3,2 is a set consisting of three points A, B, C, where

A = H i 1 ∩ H i 2 , B = H i 3 ∩ H i 4 , C = H i 5 ∩ H i 6 .
In this case, the index set J = {i 1 , i 2 , i 3 , i 4 , i 5 , i 6 } may contain 4, 5 or 6 elements.

H i 6 H i 2 C B H i 1 = H i 3 H i 4 = H i 5 C H i 1 = H i 3 = H i 5 A B H i 4 H i 2 H i 6 A H i 6 A C H i 5 H i 5 A H i 1 = H i 3 B C H i 2 H i 4 H i 6 H i 2 H i 1 B H i 3 H i 4
Suppose to the contrary that there is an entire curve f :

C → P 2 (C) \ (∪ 8 i=1 H i \ A 3,2
). Similarly as above, cf. (3.4), (3.5), (3.6), we can show in all the four illustrated cases

i∈J N [2] f (r, H i ) ≤ 3 N f (r, A) + N f (r, B) + N f (r, C) .
Next, let C = {c = 0} be the degenerate cubic consisting of the three lines AB = { AB = 0}, BC = { BC = 0}, and CA = { CA = 0}. Similarly as in (3.7), we have

2 min 1≤j≤2 ord z (h i j • f ) ≤ ord z ( AB • f ) + ord z ( CA • f ) = ord z (c • f ) (z ∈ f -1 (A)).
We also have two other inequalities for h i 3 , h i 4 , z ∈ f -1 (B) and for h i 5 , h i 6 , z ∈ f -1 (C). Summing these inequalities and integrating, we get

2 N f (r, A) + N f (r, B) + N f (r, C) ≤ N f (r, C).
If the curve f is linearly nondegenerate, then by proceeding as we did in (3.9), we also get a contradiction.

5 T f (r) ≤ 8 i=1 N [2] f (r, H i ) + S f (r) ≤ 3 N f (r, A) + N f (r, B) + N f (r, C) + S f (r) ≤ 3 2 N f (r, C) + S f (r) ≤ 9 2 T f (r) + S f (r).
Thus the curve f must be contained in some line. By analyzing the position of this line with respect to {H i } 1≤i≤8 \ {A, B, C} and by using Picard's theorem, we conclude as above.

In P 3 (C), the generic condition for the family of planes {H i } 1≤i≤q excludes the following cases.

(1) There are three disjoint subsets I, J, K of {1, . . . , q} with |I| = 3, |J| = 2, |K| = 3 such that the diagonal (hyper)plane H IJ contains the point ∩ k∈K H k .

(2) There are four disjoint subsets

I, J, K 1 , K 2 of {1, . . . , q} with |I| = 3, |J| = 2, |K 1 | = |K 2 | = 2 such that the three points (∩ k 1 ∈K 1 H k 1 ) ∩ H IJ , (∩ k 2 ∈K 2 H k 2 ) ∩ H IJ and ∩ i∈I H i are collinear. ∩ i∈I H i ∩ j∈J H j ∩ k∈K H k H IJ H IJ ∩ k 1 ∈K 1 H k 1 ∩ i∈I H i ∩ k 2 ∈K 2 H k 2 ∩ j∈J H j Lemma 3.2. In P 3 (C), if m ≤ 2, all complements of the form (3.2) are hyperbolic.
Proof. Without loss of generality, we can assume that A m,3 is a set of m elements belonging to:

∪ 1≤i 1 <i 2 ≤7+m (H i 1 ∩ H i 2 ) * ∪ 1≤i 1 <i 2 <i 3 ≤7+m H i 1 ∩ H i 2 ∩ H i 3 .
Suppose to the contrary that there exists a Brody curve f :

C → P 3 (C) \ (∪ 7+m i=1 H i \ A m,3
). When m = 1, the curve f must avoid at least five planes. By Theorem 2.4, its image is contained in some line L. By the generic condition, the number of intersection points between L and ∪ 8 i=1 H i \ A 1,3 is at least 3. By Picard's theorem, f must be constant, which is a contradiction.

Next, we consider the case

m = 2. If A 2,3 = {l * 1 , l * 2 }
where l 1 , l 2 are lines, then the curve f avoids five planes, say {H i } 1≤i≤5 . By Theorems 2.4 and 2.3, its image lands in some line L, which is contained in a diagonal plane P of the family {H i } 1≤i≤5 . We may assume that the plane P passes through the point H 1 ∩ H 2 ∩ H 3 and contains the line H 4 ∩ H 5 . If the line L does not pass through the point

H 1 ∩ H 2 ∩ H 3 , then it intersects {H i } 1≤i≤3 in three distinct points, hence f is constant by Picard's theorem. Thus L must pass through the point H 1 ∩ H 2 ∩ H 3 .
In the plane P, the curve f can pass through the points l 1 ∩ P and l 2 ∩ P. But by the generic condition, cf. (2) above, the three points H 1 ∩ H 2 ∩ H 3 , l 1 ∩ P, l 2 ∩ P are not collinear. Hence, f (C) is contained in a complement of at least three points in the line L, which is impossible by Picard's theorem.

Two substantial cases remain:

(a) A 2,3 = {A, l * }, where A is a point and l is a line;

(b) the set A 2,3 consists of two points.

We treat case (a). If both A and l are contained in some common plane H i , then f avoids five planes. By Theorem 2.3, its image must be contained in some diagonal plane, which does not contain the point A by the generic condition. Hence f must avoid seven planes in general position, which is absurd by Theorem 2.5. Thus, we can assume that

A = H 1 ∩ H 2 ∩ H 3 and l * = (H 4 ∩ H 5 ) \ ∪ i =4,5 H i . Hence f avoids four planes H i (6 ≤ i ≤ 9).
First, we show that f is linearly nondegenerate. Suppose to the contrary that f (C) is contained in some plane P. If A ∈ P, then f also avoids H 1 , H 2 , H 3 , which is impossible by Theorem 2.5. Hence the plane P must pass through the point A. If f (C) is contained in some line L ⊂ P, then L must also pass through A, for the same reason. Note that the number of intersection points between L and {H i } 6≤i≤9 is at least 2, and it equals 2 only if either L passes through some point

H i 1 ∩ H i 2 ∩ H i 3 (6 ≤ i 1 < i 2 < i 3 ≤ 9) or L intersects two lines H i 1 ∩ H i 2 , H i 3 ∩ H i 4 ({i 1 , i 2 , i 3 , i 4 } = {6, 7, 8, 9}).
If L has empty intersection with H 4 ∩ H 5 , then f avoids at least four points in the line L, hence it is constant. If L intersects H 4 ∩ H 5 , then by considering the diagonal plane passing through A and containing H 4 ∩ H 5 , the two cases where |L ∩ {H i } 6≤i≤9 | = 2 are excluded by the generic condition. Thus f always avoids three distinct points in L, hence it is constant.

Consequently, we can assume that f does not land in any line in the plane P. There are two possible positions of P:

(a1) it is a diagonal plane containing A and some line in

∪ 6≤i 1 <i 2 ≤9 H i 1 ∩ H i 2 ;
(a2) it does not contain any line in

∪ 6≤i 1 <i 2 ≤9 H i 1 ∩ H i 2 .
In case (a1), assume that P contains the line H 6 ∩ H 7 . Among {H i ∩ P} 1≤i≤9 , two lines H 6 ∩ P, H 7 ∩ P coincide, and dropping the line H 1 ∩ P, by the generic condition, it remains seven lines {H i ∩ P} i =1,7 in general position in P.

H 6 ∩ H 7 H 4 ∩ H 5 A B P
Letting B be the intersection point of the line l = H 4 ∩ H 5 with the plane P, the curve f lands in P \ (∪ 1≤i≤9 H i ∩ P) \ {A, B})). As in (3.9), f (C) is contained in some line, which is a contradiction.

Next, consider case (a2).

A H 6 ∩ H 7 ∩ H 8 P H 4 ∩ H 5 H 8 ∩ H 9 A P H 6 ∩ H 7 If P contains some point in ∪ 6≤i 1 <i 2 <i 3 ≤9 H i 1 ∩ H i 2 ∩ H i 3 ,
N f (r, A) + N f (r, l * ) ≤ N f (r, D). (3.11)
From the elementary inequality

min {ord z (h 4 • f ), 3} + min {ord z (h 5 • f ), 3} ≤ 4 min 4≤i≤5 ord z (h i • f ) (z ∈ f -1 (l * )),
by taking the sum on disks and then by integrating, we get

N [3] f (r, H 4 ) + N [3] f (r, H 5 ) ≤ 4 N f (r, l * ). (3.12)
Next, we try to bound

N [3] f (r, H i ) (1 ≤ i ≤ 3) from above in terms of N f (r, A). Since f is of the form (3.10), for any z 1 , z 2 ∈ f -1 (A), we have f (k) (z 1 ) = f (k) (z 2 ) (k ∈ N), hence ord z 1 (h i • f ) = ord z 2 (h i • f ) (1 ≤ i ≤ 3).
(3.13) Thus, it suffices to consider the two cases:

(a3) ord z (h i • f ) ≤ 2 for all 1 ≤ i ≤ 3 and for all z ∈ f -1 (A); (a4) ord z (h i • f ) ≥ 3 for some i with 1 ≤ i ≤ 3 and for all z ∈ f -1 (A).
In case (a3), the elementary inequality

3 i=1 min {ord z (h i • f ), 3} ≤ 5 min 1≤i≤3 ord z (h i • f ) (z ∈ f -1 (A)), yields N [3] f (r, H 1 ) + N [3] f (r, H 2 ) + N [3] f (r, H 3 ) ≤ 5 N f (r, A). (3.14)
Since f is linearly nondegenerate, we can proceed similarly as in (3.9)

5 T f (r) ≤ 9 i=1 N [3] f (r, H i ) + S f (r) ≤ 5 N f (r, A) + 4 N f (r, l * ) + S f (r) = 5 N f (r, A) + N f (r, l * ) -N f (r, l * ) + S f (r) ≤ 5 N f (r, D) -N f (r, l * ) + S f (r) ≤ 5 T f (r) -N f (r, l * ) + S f (r). (3.15)
This implies

N f (r, l * ) = S f (r)
and hence, by (3.12), we have

N [3] f (r, H 4 ) + N [3] f (r, H 5 ) = S f (r).
Therefore, the first inequality of (3.15) can be rewritten as

5 T f (r) ≤ 3 i=1 N [3] f (r, H i ) + S f (r).
By the First Main Theorem, the right-hand side of the above inequality is bounded from above by 3 T f (r) + S f (r). Thus we get 5

T f (r) ≤ 3 T f (r) + S f (r),
which is absurd. Next, we consider case (a4). Assume that ord z (h

1 • f ) ≥ 3 for all z ∈ f -1 (A). Since f is of the form (3.10), we claim that ord z (h i • f ) ≤ 2 (z∈ f -1 (A), 2 ≤ i ≤ 3).
(3.16) Indeed, if ord z (h i • f ) ≥ 3 for some z ∈ f -1 (A) and for some 2 ≤ i ≤ 3, say i = 2, then (e λ 1 z+µ 1 , e λ 2 z+µ 2 , e λ 3 z+µ 3 ) is a solution of a system of six linear equations of the form

                     0 = a 10 + a 11 u + a 12 v + a 13 w, 0 = a 11 λ 1 u + a 12 λ 2 v + a 13 λ 3 w, 0 = a 11 λ 2 1 u + a 12 λ 2 2 v + a 13 λ 2 3 w, 0 = a 20 + a 21 u + a 22 v + a 23 w, 0 = a 21 λ 1 u + a 22 λ 2 v + a 23 λ 3 w, 0 = a 21 λ 2 1 u + a 22 λ 2 2 v + a 23 λ 2 3 w,
where u, v, w are unknowns, and where a ij (0

≤ i ≤ 3) are the coefficients of h i (1 ≤ i ≤ 2)
(h 2 • f ), 3} + min {ord z (h 3 • f ), 3} ≤ 3 min 1≤i≤3 ord z (h i • f ) (z∈ f -1 (A)).
By taking the sum on disks and by integrating, we get

N [3] f (r, H 2 ) + N [3] f (r, H 3 ) ≤ 3 N f (r, A). (3.17)
We may therefore proceed similarly as in (3.15)

5 T f (r) ≤ 9 i=1 N [3] f (r, H i ) + S f (r) ≤ N [3] f (r, H 1 ) + 3 N f (r, A) + 4 N f (r, l * ) + S f (r) ≤ N f (r, H 1 ) + 4 N f (r, A) + N f (r, l * ) -N f (r, A) + S f (r) ≤ T f (r) + 4 N f (r, D) -N f (r, A) + S f (r) ≤ 5 T f (r) -N f (r, A) + S f (r). (3.18) This implies N f (r, A) = S f (r).
By (3.17), we have

N [3] f (r, H 2 ) + N [3]
f (r, H 3 ) = S f (r). Hence we can rewrite the first inequality of (3.18) and use First Main Theorem to get a contradiction

5 T f (r) ≤ N [3] f (r, H 1 ) + N [3] f (r, H 4 ) + N [3] f (r, H 5 ) + S f (r) ≤ 3 T f (r) + S f (r).
Let us consider case (b). Assume now A 2,3 = {A, B}, where A, B are two points contained in

∪ 1≤i 1 <i 2 <i 3 ≤9 H i 1 ∩ H i 2 ∩ H i 3 .
There are three possibilities for the positions of A and B:

(b1) both A and B are contained in some line H i ∩ H j ;

(b2) both A and B are contained in some plane H i but they are not contained in any line H i ∩ H j ;

(b3) there is no plane H i containing both points A and B.

In case (b1), the curve f avoids a family of five planes and, therefore, its image is contained in some diagonal plane of this family, which contains neither A nor B by the generic condition. Hence f avoids all planes H i , which is absurd by Theorem 2.5.

Next, we consider case (b2).

Assume that A = H 1 ∩ H 2 ∩ H 3 and B = H 1 ∩ H 4 ∩ H 5 , hence f avoids the 4 planes H i (6 ≤ i ≤ 9
). Similarly as in case (a), the generic condition allows us to assume that f is linearly nondegenerate.

Since f avoids four planes, it is of the form (3.10) in some affine coordinates on P 3 (C). Since f has no singular point, we have

min i∈{1,2,3} ord z (h i • f ) = 1 (z∈ f -1 (A)), min i∈{1,4,5} ord z (h i • f ) = 1 (z∈ f -1 (B)). (3.19)
Hence by using these two equalities together with (3.16),

i∈{1,2,3} min {ord z (h i • f ), 3} ≤ 6 = 6 min i∈{1,2,3} ord z (h i • f ), (z ∈ f -1 (A)), i∈{1,4,5} min {ord z (h i • f ), 3} ≤ 6 = 6 min i∈{1,4,5} ord z (h i • f ), (z ∈ f -1 (B)).
Thus, by taking the sum on disks of both sides of these inequalities and by integrating,

5 i=1 N [3] f (r, H i ) ≤ 6 N f (r, A) + N f (r, B) .
Next, using again that f is of the form (3.10), one can find two planes

P 1 = {p 1 = 0}, P 2 = {p 2 = 0} containing the line AB such that ord z (p 1 • f ) ≥ 2 (z ∈ f -1 (A)), ord z (p 2 • f ) ≥ 2 (z ∈ f -1 (B)).
Let Q = {q = p 1 p 2 = 0} be the degenerate quadric P 1 ∪ P 2 . We have

3 = 3 min i∈{1,2,3} ord z (h i • f ) ≤ ord z (p 1 • f ) + ord z (p 2 • f ) = ord z (q • f ) (z ∈ f -1 (A)), 3 = 3 min i∈{1,4,5} ord z (h i • f ) ≤ ord z (p 1 • f ) + ord z (p 2 • f ) = ord z (q • f ) (z ∈ f -1 (B)),
which implies, by integrating, that

3 N f (r, A) + N f (r, B) ≤ N f (r, T ).
We proceed similarly as above to get a contradiction

5 T f (r) ≤ 9 i=1 N [3] f (r, H i ) + S f (r) ≤ 6 N f (r, A) + N f (r, B) + S f (r) ≤ 2 N f (r, T ) + S f (r) ≤ 4 T f (r) + S f (r).

Now, we consider case (b3). Assume that

A = H 1 ∩ H 2 ∩ H 3 , B = H 4 ∩ H 5 ∩ H 6 ,
when f avoids the three planes H 7 , H 8 , H 9 . If f (C) is contained in some plane P, then it is not hard to see that P must pass through both A and B. Furthermore, by using Theorem 2.6, one can show that P does not pass through the point C = H 7 ∩ H 8 ∩ H 9 . One can then always find 7 lines in general position in P among {H i ∩ P} 1≤i≤9 . Hence one can use similar arguments as in Lemma 3.1, case m = 2, to get a contradiction. Thus, we can suppose that f is linearly nondegenerate.

Assume that the omitted planes H 7 , H 8 , H 9 are given in the homogeneous coordinates [z 0 : z 1 : z 2 : z 3 ] by the equations {z 0 = 0}, {z 1 = 0}, {z 2 = 0}. Since {H i } 1≤i≤9 is a family of planes in general position, the planes H i (1 ≤ i ≤ 6) are given by

h i = 3 j=0 a ij z j = 0, with a i3 = 0 (1 ≤ i ≤ 6). Set l i 1 ,i 2 = H i 1 ∩ H i 2 (1 ≤ i 1 < i 2 ≤ 3), l j 1 ,j 2 = H j 1 ∩ H j 2 (4 ≤ j 1 < j 2 ≤ 6).
For 1 ≤ i < j ≤ 3 or 4 ≤ i < j ≤ 6, let R i,j = {r i,j = 0} be the plane containing the lines AB, l i,j and let T i,j = {t i,j = a j3 h i -a i3 h j = 0} be the plane passing through the point C = [0 : 0 : 0 : 1] and containing the line l i,j . We note that all r i,j , t i,j are linear combinations of h i and h j with nonzero coefficients.

Since f avoids three planes, by Theorem 2.2 it has a reduced representation of the form

[1 : e λ 1 z+µ 1 : e λ 2 z+µ 2 : g], (3.20)
where λ 1 , λ 2 , µ 1 , µ 2 are constants with λ 1 = λ 2 , λ 1 , λ 2 = 0 and where g is an entire function. Since f has no singular point, we have

min 1≤i≤3 ord z (h i • f ) = 1 (z∈ f -1 (A)), min 4≤j≤6 ord z (h j • f ) = 1 (z∈ f -1 (B)).
(3.21)

Since f is of the form (3.20), we claim that

min {ord z (h i 1 • f ), ord z (h i 2 • f )} ≤ 2 (z∈ f -1 (A) 1 ≤ i 1 < i 2 ≤ 3), min {ord z (h j 1 • f ), ord z (h j 2 • f )} ≤ 2 (z∈ f -1 (B), 4 ≤ j 1 < j 2 ≤ 6). (3.22)
Indeed, if one of these inequalities does not hold, say min {ord z (h

1 • f ), ord z (h 2 • f )} ≥ 3 for some z ∈ f -1 (A)
, then z is a solution of the following system of equations

     0 = (t 1,2 • f )(z), 0 = (t 1,2 • f ) (z), 0 = (t 1,2 • f ) (z).
Equivalently, (e λ 1 z+µ 1 , e λ 2 z+µ 2 ) is a solution of a system of three linear equations of the form

     0 = (
min {ord z (h i • f ), 3} ≤ 6 (z ∈ f -1 (A)), 6 j=4 min{ord z (h j • f ), 3} ≤ 6 (z ∈ f -1 (B)). (3.23)
Now we prove the following equality

Claim 3.1. T f (r) = N f (r, A) + N f (r, B) + S f (r). (3.24)
Proof. Since f is of the form (3.20) and since t i,j does not contain the term x 3 , we have

ord z 1 (t i 1 ,i 2 • f ) = ord z 2 (t i 1 ,i 2 • f ) (z 1 ,z 2 ∈ f -1 (A), 1 ≤ i 1 <i 2 ≤ 3), ord z 1 (t j 1 ,j 2 • f ) = ord z 2 (t j 1 ,j 2 • f ) (z 1 ,z 2 ∈ f -1 (B), 4 ≤ j 1 <j 2 ≤ 6).
(3.25) Thus, it suffices to consider the four cases depending on f and t i,j :

(b3.1) ord z (t i 1 ,i 2 • f ) = 1 for all 1 ≤ i 1 < i 2 ≤ 3, for all z ∈ f -1 (A) and ord z (t j 1 ,j 2 • f ) = 1 for all 4 ≤ j 1 < j 2 ≤ 6, for all z ∈ f -1 (B); (b3.2) ord z (t i 1 ,i 2 • f ) ≥ 2 for some 1 ≤ i 1 < i 2 ≤
3, for all z ∈ f -1 (A) and ord z (t j 1 ,j 2 • f ) = 1 for all 4 ≤ j 1 < j 2 ≤ 6, for all z ∈ f -1 (B);

(b3.3) ord z (t i 1 ,i 2 • f ) = 1 for all 1 ≤ i 1 < i 2 ≤ 3, for all z ∈ f -1 (A) and ord z (t j 1 ,j 2 • f ) ≥ 2 for some 4 ≤ j 1 < j 2 ≤ 6, for all z ∈ f -1 (B); (b3.4) ord z (t i 1 ,i 2 • f ) ≥ 2 for some 1 ≤ i 1 < i 2 ≤ 3, for all z ∈ f -1 (A) and ord z (t j 1 ,j 2 • f ) ≥ 2 for some 4 ≤ j 1 < j 2 ≤ 6, for all z ∈ f -1 (B).
Consider case (b3.1). Since t i,j is a linear combination of h i and h j with nonzero coefficients, we have

min {ord z (h i 1 • f ), ord z (h i 2 • f )} = 1 (z∈ f -1 (A) 1 ≤ i 1 < i 2 ≤ 3), min {ord z (h j 1 • f ), ord z (h j 2 • f )} = 1 (z∈ f -1 (B), 4 ≤ j 1 < j 2 ≤ 6).
Using these equalities together with (3.21), we get

3 i=1 min {ord z (h i • f ), 3} ≤ 5 = 5 min 1≤i≤3 ord z (h i • f ) (z∈ f -1 (A)), 6 i=4 min {ord z (h i • f ), 3} ≤ 5 = 5 min 4≤i≤6 ord z (h i • f ) (z∈ f -1 (B)).
(3.26)

By taking the sum on disks and by integrating these two inequalities, we obtain

N [3] f (r, H 1 ) + N [3] f (r, H 2 ) + N [3] f (r, H 3 ) ≤ 5 N f (r, A), N [3] f (r, H 4 ) + N [3] f (r, H 5 ) + N [3] f (r, H 6 ) ≤ 5 N f (r, B).
Letting B be a plane passing through A and B, we proceed similarly as before

5 T f (r) ≤ 9 i=1 N [3] f (r, H i ) + S f (r) ≤ 5 N f (r, A) + 5 N f (r, B) + S f (r) ≤ 5 N f (r, B) + S f (r) ≤ 5 T f (r) + S f (r).
(3.27)

Here, S f (r) = o(T f (r)) is negligible, hence all inequalities are equalities modulo S f (r). This gives (3.24), as wanted.

Next, we consider case (b3.2). Let us set

E t,A = {z ∈ C : |z| < t, f (z) = A}, E 1 t,A,i = {z ∈ C : |z| < t, f (z) = A, ord z (h i • f ) = 1} (1 ≤ i ≤ 3), E ≥2 t,A,i = {z ∈ C : |z| < t, f (z) = A, ord z (h i • f ) ≥ 2} (1 ≤ i ≤ 3), E t,B = {z ∈ C : |z| < t, f (z) = B}, E 1 t,B,i = {z ∈ C : |z| < t, f (z) = B, ord z (h i • f ) = 1} (4 ≤ i ≤ 6), E ≥2 t,B,i = {z ∈ C : |z| < t, f (z) = B, ord z (h i • f ) ≥ 2} (4 ≤ i ≤ 6).
Assume that ord z (t 1,2 • f ) ≥ 2 for all z ∈ f -1 (A). Since t 1,2 , r 1,2 are linear combinations of h 1 and h 2 with nonzero coefficients, we have

E ≥2 t,A,1 = E ≥2 t,A,2 , ord z (r 1,2 • f ) ≥ 2 (z ∈ E ≥2 t,A,1
).

(3.28)

For the same reason (3.30) Furthermore, it follows from (3.28) that

E 1 t,A,1 = E 1 t,A,2 , which yields 3 i=1 min {ord z (h i • f ), 3} ≤ 5 (z ∈ E 1 t,A,1
ord z (r • f ) ≥ 5 (z ∈ E ≥2 t,A,1 ).
Using this inequality together with (3.23) and (3.21), we get

3 i=1 min {ord z (h i • f ), 3} ≤ 6 = 6 min 1≤i≤3 ord z (h i • f ) ≤ 6 5 ord z (r • f ) (z ∈ E ≥2 t,A,1 ).
Combining (3.29), (3.21) and (3.30), we receive

3 i=1 min {ord z (h i • f ), 3} ≤ 5 = 5 min 1≤i≤3 ord z (h i • f ) ≤ 5 4 ord z (r • f ) (z ∈ E 1 t,A,1 ).
Since ord z (t j 1 ,j 2 • f ) = 1 for all 4 ≤ j 1 < j 2 ≤ 6, for all z ∈ f -1 (B), by similar arguments as in (3.26) and by using (3.30), we also have

6 i=4 min {ord z (h i • f ), 3} ≤ 5 = 5 min 4≤i≤6 ord z (h i • f ) 5 4 ord z (r • f ) (z∈ E t,B ).
By taking the sum on disks and by integrating these three inequalities, we obtain

3 i=1 r 1 z∈E ≥2 t,A,1 min {ord z (h i • f ), 3} t d t ≤ 6 r 1 z∈E ≥2 t,A,1 min 1≤i≤3 ord z (h i • f ) t d t ≤ 6 5 r 1 z∈E ≥2 t,A,1 ord z (r • f ) t d t, (3.31) 3 i=1 r 1 z∈E 1 t,A,1 min {ord z (h i • f ), 3} t d t ≤ 5 r 1 z∈E 1 t,A,1 min 1≤i≤3 ord z (h i • f ) t d t ≤ 5 4 r 1 z∈E 1 t,A,1 ord z (r • f ) t d t, (3.32) 6 i=4 r 1 z∈E t,B min {ord z (h i • f ), 3} t d t ≤ 5 N f (r, B) ≤ 5 4 r 1 z∈E t,B ord z (r • f ) t d t. (3.33)
We then proceed similarly as before:

5 T f (r) ≤ 9 i=1 N [3] f (r, H i ) + S f (r) ≤ 3 i=1 N [3] f (r, H i ) + 6 i=4 N [3] f (r, H i ) + S f (r) = 3 i=1 r 1 z∈E t,A min {ord z (h i • f ), 3} t d t + 6 i=4 r 1 z∈E t,B min {ord z (h i • f ), 3} t d t + S f (r) = 3 i=1 r 1 z∈E ≥2 t,A,1 min {ord z (h i • f ), 3} t d t + r 1 z∈E 1 t,A,1 min {ord z (h i • f ), 3} t d t + 6 i=4 r 1 z∈E t,B min {ord z (h i • f ), 3} t d t + S f (r) ≤ 6 5 r 1 z∈E ≥2 t,A,1 ord z (r • f ) t d t + 5 4 r 1 z∈E 1 t,A,1 ord z (r • f ) t d t + 5 4 r 1 z∈E t,B ord z (r • f ) t d t + S f (r) = 5 4 r 1 z∈E t,A ord z (r • f ) t d t + r 1 z∈E t,B ord z (r • f ) t d t + 6 5 - 5 4 r 1 z∈E ≥2 t,A,1 ord z (r • f ) t d t + S f (r) ≤ 5 4 N f (r, R) - 1 20 r 1 z∈E ≥2 t,A,1 ord z (r • f ) t d t + S f (r) ≤ 5 T f (r) - 1 20 r 1 z∈E ≥2 t,A,1 ord z (r • f ) t d t + S f (r). (3.34) This implies r 1 z∈E ≥2 t,A,1 ord z (r • f ) t d t = S f (r) (3.35)
and whence all inequalities in (3.34) become equalities modulo S f (r), which gives

6 i=1 N [3] f (r, H i ) = 5 T f (r) + S f (r), (3.36) 3 i=1 N [3] f (r, H i ) = 5 4 r 1 z∈E t,A ord z (r • f ) t d t + S f (r), (3.37) 6 i=4 N [3] f (r, H i ) = 5 4 r 1 z∈E t,B ord z (r • f ) t d t + S f (r). ( 3
min {ord z (h i • f ), 3} t d t = S f (r) r 1 z∈E ≥2 t,A,1 min 1≤i≤3 ord z (h i • f ) t d t = S f (r).
Hence

3 i=1 N [3] f (r, H i ) = 3 i=1 r 1 z∈E 1 t,A,1 min {ord z (h i • f ), 3} t d t + S f (r), (3.40)
N f (r, A) = r 1 z∈E 1 t,A,1 min 1≤i≤3 ord z (h i • f ) t d t + S f (r). (3.41)
Combining (3.32), (3.40), (3.41), we get

3 i=1 N [3] f (r, H i ) = 5 N f (r, A) + S f (r). (3.42)
The equality (3.24) follows from (3.36), (3.39), (3.42).

Case (b3.3) can be treated by similar arguments as for case (b3.2).

Next, we consider case (b3.4). Assume that

ord z (t 1,2 • f ) ≥ 2 (z ∈ f -1 (A)), ord z (t 4,5 • f ) ≥ 2 (z ∈ f -1 (B)).
By similar argument as in (3.28), we have

E ≥2 t,A,1 = E ≥2 t,A,2 , E ≥2 t,B,4 = E ≥2 t,B,5 , E 1 t,A,1 = E 1 t,A,2 , E 1 t,B,4 = E 1 t,B,5 , which implies ord z (r 1,2 • f ) ≥ 2 (z ∈ E ≥2 t,A ,1 ), (3.43)
ord z (r 4,5 • f ) ≥ 2 (z ∈ E ≥2 t,B ,4 ), (3.44)
3 i=1 min {ord z (h i • f ), 3} ≤ 5 (z ∈ E 1 t,A,1 ), 6 i=4 min {ord z (h i • f ), 3} ≤ 5 (z ∈ E 1 t,B ,4 ).
Letting S = {s = r 12 r 4,5 = 0} be the degenerate quadric R 1,2 ∪ R 4,5 , we see that

ord z (s • f ) = ord z (r 1,2 • f ) + ord z (r 4,5 • f ) ≥ 2 (z ∈ E t,A ∪ E t,B ).
Furthermore, by using (3.43) and (3.44), we have

ord z (s • f ) = ord z (r 1,2 • f ) + ord z (r 4,5 • f ) ≥ 3 (z ∈ E ≥2 t,A,1 ∪ E ≥2 t,B,4 ).
Similarly as in the previous case, by using these inequalities together with (3.21) and (3.23), we receive

3 i=1 min {ord z (h i • f ), 3} ≤ 6 = 6 min 1≤i≤3 ord z (h i • f ) ≤ 6 3 ord z (s • f ) (z ∈ E ≥2 t,A,1 ), 3 i=1 min {ord z (h i • f ), 3} ≤ 5 = 5 min 1≤i≤3 ord z (h i • f ) ≤ 5 2 ord z (s • f ) (z ∈ E 1 t,A,1 ), 6 i=4 min {ord z (h i • f ), 3} ≤ 6 = 6 min 4≤i≤6 ord z (h i • f ) ≤ 6 3 ord z (s • f ) (z ∈ E ≥2 t,B,4 ), 6 i=4 min {ord z (h i • f ), 3} ≤ 5 = 5 min 4≤i≤6 ord z (h i • f ) ≤ 5 2 ord z (s • f ) (z ∈ E 1 t,B,4 ).
By taking the sum on disks and by integrating these four inequalities, we obtain

3 i=1 r 1 z∈E ≥2 t,A,1 min {ord z (h i • f ), 3} t d t ≤ 6 r 1 z∈E ≥2 t,A,1 min 1≤i≤3 ord z (h i • f ) t d t ≤ 2 r 1 z∈E ≥2 t,A,1 ord z (s • f ) t d t, 3 i=1 r 1 z∈E 1 t,A,1 min {ord z (h i • f ), 3} t d t ≤ 5 r 1 z∈E 1 t,A,1 min 1≤i≤3 ord z (h i • f ) t d t ≤ 5 2 r 1 z∈E 1 t,A,1 ord z (s • f ) t d t, 6 i=4 r 1 z∈E ≥2 t,B,4 min {ord z (h i • f ), 3} t d t ≤ 6 r 1 z∈E ≥2 t,B,4 min 4≤i≤6 ord z (h i • f ) t d t ≤ 2 r 1 z∈E ≥2 t,B,4 ord z (s • f ) t d t, 6 i=4 r 1 z∈E 1 t,B,4 min {ord z (h i • f ), 3} t d t ≤ 5 r 1 z∈E 1 t,B,4 min 4≤i≤6 ord z (h i • f ) t d t ≤ 5 2 r 1 z∈E 1 t,B,4 ord z (s • f ) t d t.
Now, we proceed similarly as above

5 T f (r) ≤ 9 i=1 N [3] f (r, H i ) + S f (r) = 3 i=1 r 1 z∈E t,A min {ord z (h i • f ), 3} t d t + 6 i=4 r 1 z∈E t,B min {ord z (h i • f ), 3} t d t + S f (r) = 3 i=1 r 1 z∈E 1 t,A,1 min {ord z (h i • f ), 3} t d t + r 1 z∈E ≥2 t,A,1 min {ord z (h i • f ), 3} t d t + 6 i=4 r 1 z∈E 1 t,B,1 min {ord z (h i • f ), 3} t d t + r 1 z∈E ≥2 t,B,4 min {ord z (h i • f ), 3} t d t + S f (r) ≤ 5 2 r 1 z∈E t,A ord z (s • f ) t d t + r 1 z∈E t,B ord z (s • f ) t d t + 2 - 5 2 r 1 z∈E ≥2 t,A,1 ord z (s • f ) t d t + r 1 z∈E ≥2 t,B,4 ord z (s • f ) t d t + S f (r) ≤ 5 2 N f (r, S) - 1 2 r 1 z∈E ≥2 t,A,1 ord z (s • f ) t d t + r 1 z∈E ≥2 t,B,4 ord z (s • f ) t d t + S f (r) ≤ 5 T f (r) - 1 2 r 1 z∈E ≥2 t,A,1 ord z (s • f ) t d t + r 1 z∈E ≥2 t,B,4 ord z (s • f ) t d t + S f (r).
This implies

r 1 z∈E ≥2 t,A,1 ord z (s • f ) t d t = S f (r), r 1 z∈E ≥2 t,B,4 ord z (s • f ) t d t = S f (r), 6 i=1 N [3] f (r, H i ) = 5 T f (r) + S f (r), 3 i=1 N [3] f (r, H i ) = 5 2 r 1 z∈E t,A ord z (r • f ) t d t + S f (r), 6 i=4 N [3] f (r, H i ) = 5 2 r 1 z∈E t,B ord z (r • f ) t d t + S f (r).
By proceeding similarly as in (3.42), we receive

3 i=1 N [3] f (r, H i ) = 5 N f (r, A) + S f (r), 6 i=4 N [3] f (r, H i ) = 5 N f (r, B) + S f (r).
Hence, the equality (3.24) also holds in this case. Claim 3.1 is thus proved.

Next, since f is of the form (3.20), one can find a plane K = {k = 0} passing through A and C such that ord z (k

• f ) ≥ 2 (z ∈ f -1 (A)).
Let B i = {b i = 0} be the plane containing the two AB, H i ∩ K (1 ≤ i ≤ 3). Since b i is a linear combination of h i and k with nonzero coefficients, we have

ord z (b i • f ) ≥ 2 (z ∈ E ≥2 t,A,i ),
which yields

3 i=1 ord z (b i • f ) ≥ 4 (z ∈ ∪ 3 i=1 E ≥2 t,A,i ).
z (h i • f ) = 1 ≤ 1 4 3 i=1 ord z (b i • f ) = 1 4 ord z (c • f ) (z ∈ ∪ 3 i=1 E ≥2 t,A,i ), min 1≤i≤3 ord z (h i • f ) = 1 ≤ 1 3 3 i=1 ord z (b i • f ) = 1 3 ord z (c • f ) (z ∈ E t,A \ ∪ 3 i=1 E ≥2 t,A,i ), min 4≤i≤6 ord z (h i • f ) = 1 ≤ 1 3 3 i=1 ord z (b i • f ) = 1 3 ord z (c • f ) (z ∈ E t,B ).
By taking the sum on disks and by integrating these inequalities,

r 1 z∈∪ 3 i=1 E ≥2 t,A,i min 1≤i≤3 ord z (h i • f ) t d t ≤ 1 4 r 1 z∈∪ 3 i=1 E ≥2 t,A,i ord z (c • f ) t d t, r 1 z∈E t,A \∪ 3 i=1 E ≥2 t,A,i min 1≤i≤3 ord z (h i • f ) t d t ≤ 1 3 r 1 z∈E t,A \∪ 3 i=1 E ≥2 t,A,i ord z (c • f ) t d t, N f (r, B) = r 1 z∈E t,B min 4≤i≤6 ord z (h i • f ) t d t ≤ 1 3 r 1 z∈E t,B ord z (c • f ) t d t.
By using these inequalities together with (3.24), we receive

5 T f (r) = 5 N f (r, A) + 5 N f (r, B) + S f (r) = 5 r 1 z∈∪ 3 i=1 E ≥2 t,A,i min 1≤i≤3 ord z (h i • f ) t d t + r 1 z∈E t,A \∪ 3 i=1 E ≥2 t,A,i min 1≤i≤3 ord z (h i • f ) t d t + 5 r 1 z∈E t,B min 4≤i≤6 ord z (h i • f ) t d t + S f (r) ≤ 5 3 r 1 z∈E t,A ord z (c • f ) t d t + r 1 z∈E t,B ord z (c • f ) t d t + 5 4 - 5 3 r 1 z∈∪ 3 i=1 E ≥2 t,A,i ord z (c • f ) t d t + S f (r) ≤ 5 3 N f (r, C) - 5 12 r 1 z∈∪ 3 i=1 E ≥2 t,A,i ord z (c • f ) t d t + S f (r) ≤ 5 T f (r) - 5 12 r 1 z∈∪ 3 i=1 E ≥2 t,A,i ord z (c • f ) t d t + S f (r).
This implies

r 1 z∈∪ 3 i=1 E ≥2 t,A,i ord z (c • f ) t d t = S f (r). (3.46)
By using (3.23) and (3.45), we get

3 i=1 min {ord z (h i • f ), 3} ≤ 6 ≤ 3 2 ord z (c • f ) (z ∈∪ 3 i=1 E ≥2 t,A,i ),
which yields

r 1 z∈∪ 3 i=1 E ≥2 t,A,i min 1≤i≤3 ord z (h i • f ) t d t ≤ 3 2 r 1 z∈∪ 3 i=1 E ≥2 t,A,i ord z (c • f ) t d t [Use (3.46)]
= S f (r).

(3.47)

Moreover, we also have

3 i=1 min {ord z (h i • f ), 3} = 3 = 3 min 1≤i≤3 ord z (h i • f ) (z ∈ E t,A \ ∪ 3 i=1 E ≥2 t,A,i ),
which implies, by integrating, that

3 i=1 r 1 z∈E t,A \∪ 3 i=1 E ≥2 t,A,i min {ord z (h i • f ), 3} t d t ≤ 3 r 1 z∈E t,A \∪ 3 i=1 E ≥2 t,A,i min 1≤i≤3 ord z (h i • f ) t d t.
(3.48) By combining (3.47) and (3.48), we get

3 i=1 N [3] f (r, H i ) = 3 i=1 r 1 z∈E t,A min {ord z (h i • f ), 3} t d t = 3 i=1 r 1 z∈E t,A \∪ 3 i=1 E ≥2 t,A,i min {ord z (h i • f ), 3} t d t + r 1 z∈∪ 3 i=1 E ≥2 t,A,i min {ord z (h i • f ), 3} t d t + S f (r) ≤ 3 r 1 z∈E t,A \∪ 3 i=1 E ≥2 t,A,i min 1≤i≤3 ord z (h i • f ) t d t + S f (r) ≤ 3 N f (r, A) + S f (r).
This is a contradiction, because the complement of n + 1 (n ≥ 3) points in a line is hyperbolic by Picard's theorem. Now, let I be the largest subset of Q such that the curve f (C) lands in ∩ i∈I H i . We have |I| ≤ n-2. By stability of intersections, f (C) ∩ H j is contained in S for all l ∈ Q \ I. Therefore the curve f (C) lands in

∩ i∈I H i \ ∪ j∈Q\I H j \ S . (4.1)
So, the problem reduces to finding a hypersurface S of degree 2n such that all complements of the form (4.1) are hyperbolic, where I is a subset of Q of cardinality at most n -2. For example when n = 3 ([10]), we need to find a sextic curve S such that all complements of the form H i \ ∪ j =i H j \ S are hyperbolic. In this case, we have the complement of five lines in the hyperplane H i on which all points of intersection with S are deleted.

We will construct such S by deformation, step by step. For 2 ≤ l ≤ n -1, let ∆ l be a finite collection of subspaces of dimension n -l, in the sense of section 3. Let D l ∈ ∆ l be another subspace of dimension n -l, defined as D l = ∩ i∈I D l H i . For a hypersurface S = {s = 0} in general position with respect to the family {H i } 1≤i≤2n and = 0, we set

S = s + Π i ∈I D l h n i i = 0 ,
where n i ≥ 1 are chosen (freely) so that i ∈I D l n i = 2n. It is not hard to see that the hypersurface S is also in general position with respect to the family {H i } 1≤i≤2n . We denote by ∆ l the family of all subspaces of dimension n -l (2 ≤ l ≤ n) with the convention ∆ n = ∅.

Lemma 4.1. Assume that all complements of the form

∩ i∈I H i \ ∪ j∈J H j \ (((∆ l ∪ ∆ l+1 ) ∩ S) ∪ A m,n-|I| ) ( 4
H i ∩ H j } j∈J in ∩ i∈I H i ∼ = P n-|I| (C).
Then all complements of the form

∩ i∈I H i \ ∪ j∈J H j \ (((∆ l ∪ D l ∪ ∆ l+1 ) ∩ S ) ∪ A m,n-|I| ) (4.3)
are also hyperbolic for sufficiently small = 0.

Proof. By the definition of S , we see that

S ∩ ∩ m∈M H m = S ∩ ∩ m∈M H m when M ∩(Q\I D l ) = ∅, hence (∆ l ∪ D l ∪ ∆ l+1 ) ∩ S = ((∆ l ∪ ∆ l+1 ) ∩ S) ∪ (D l ∩ S ).
When |I| ≥ l, using this, we observe that two complements (4.2), (4.3) coincide.

Assume therefore |I| ≤ l -1. Suppose by contradiction that there exists a sequence of entire curves

(f k (C)) k , k → 0 contained in the complement ∩ i∈I H i \ ∪ j∈J H j \ (((∆ l ∪ D l ∪ ∆ l+1 ) ∩ S k ) ∪ A m,n-|I| ) .
By the Brody Lemma, we may assume that (f k ) converges to an entire curve f (C) ⊂ ∩ i∈I H i . Our aim is to prove that the curve f (C) lands in some complement of the form (4.2). Let ∩ k∈K H k be the smallest subspace containing f (C). It is clear that K ⊃ I. Take an index j in J \ K. By stability of intersections, one has

f (C) ∩ H j ⊂ lim f k (C) ∩ H j ⊂ ((∆ l ∪ ∆ l+1 ) ∩ S) ∪ A m,n-|I| ∪ lim(D l ∩ S k ). ( 4.4)
If the index j does not belong to

I D l , then H j ∩ D l ∩ S k ⊂ ∆ l+1 ∩ S. It follows from (4.4) that f (C) ∩ H j ⊂ ((∆ l ∪ ∆ l+1 ) ∩ S) ∪ A m,n-|I| . (4.5)
If the index j belongs to

I D l , noting that lim(D l ∩ S k ) is contained in D l ∩ (∪ i ∈I D l H i ), again from (4.4), one has f (C) ∩ H j ⊂ ((∆ l ∪ ∆ l+1 ) ∩ S) ∪ A m,n-|I| ∪ (D l ∩ (∪ i ∈I D l H i )). ( 4.6)
Assume first that K = I. We claim that (4.5) also holds when the index j ∈ J \ I belonging to I D l . Indeed, for the supplementary part in (4.6), we have

f (C) ∩ H j ∩ D l ∪ i ∈I D l H i ⊂ ∪ i ∈I D l (f (C) ∩ H j ∩ H i ),
so that (4.5) applies here to all i ∈ I D l . Hence, the curve f (C) lands inside

∩ i∈I H i \ ∪ j∈J H j \ (((∆ l ∪ ∆ l+1 ) ∩ S) ∪ A m,n-|I| ) , contradicting the hypothesis. Assume now that I is a proper subset of K. Let us set A m,n-|I|,K = {X ∩ (∩ k∈K H k )|X ∈ A m,n-|I| }.
This set consists of star-subspaces of ∩ k∈K H k ∼ = P n-|K| (C). Let B m,K be the subset of A m,n-|I|,K containing all star-subspaces of dimension n -|K| -1 (i.e., of codimension 1 in ∩ k∈K H k ), and let C m,K be the remaining part. A star-subspace in B m,K is of the form (∩ k∈K H k ∩ H j ) * for some index j ∈ J \ K. Then let R denote the set of such indices j, so that

|R| = |B m,K |.
We consider two cases separately, depending on the dimension of the subspace

Y = ∩ k∈K H k ∩ D l .
Case 1. Y is a subspace of dimension n -|K| -1. In this case, Y is of the form (∩ k∈K H k ) ∩ H y for some index y in I D l . It follows from (4.4), (4.5), (4.6) that the curve f (C) lands inside the set

∩ k∈K H k \ ∪ j∈(J\K)\(R∪{y}) H j \ (((∆ l ∪ ∆ l+1 ) ∩ S) ∪ C m,K ) .
Now we need to show that this set is of the form (4.2). First, we verify the corresponding required inequality between cardinalities

|(J \ K) \ (R ∪ {y})| ≥ |J \ K| -|B m,K | -1 ≥ |J| -|J ∩ K| -|J| + 2|I| -2n -1 -|C m,K | -1 = 2 (n -|K|) + |C m,K | + 2|K \ I| -|J ∩ K| ≥ 2 (n -|K|) + 1 + |C m,K |,
where the last inequality holds because I and J are two disjoint sets and I is a proper subset of K. Secondly, we verify that the set K is of cardinality at most n -2. Indeed, if |K| = n -1, then since S is in general position with respect to {H i } 1≤i≤2n , we see that

∩ k∈K H k \ ∪ j∈(J\K)\(R∪{y}) H j \(((∆ l ∪∆ l+1 )∩S)∪C m,K ) = ∩ k∈K H k \ ∪ j∈(J\K)\(R∪{y}) H j \C m,K .
Owing to the inequality |(J \ K) \ (R ∪ {y})| ≥ 3 + |C m,K |, the curve f lands in a complement of at least three points in a line. By Picard's theorem, f is constant, which is a contradiction.

Case 2. Y is a subspace of dimension at most n -|K| -2. In this case, the curve f (C) lands inside

∩ k∈K H k \ ∪ j∈(J\K)\R H j \ (((∆ l ∪ ∆ l+1 ) ∩ S) ∪ C m,K ∪ Y * ) .
This set is of the form (4. The lemma is thus proved.

End of proof of the Main Theorem. We now come back to the proof of the Main Theorem. Keep the notation as in Lemma 4.1. We claim that {∩ i∈I H i ∩ H j } j∈J is also a family of generic hyperplanes in the projective space ∩ i∈I H i ∼ = P n-|I| (C). Indeed, let I, J , J 1 , . • when n = 3, we start with the hyperbolicity of all complements H i \ ∪ j =i H j , which follows from Theorem 2.5 in P 2 (C);

• when n = 4, we start with the hyperbolicity of all complements

H i \ ∪ j =i H j , ∩ i∈I H i \ ∪ j∈J H j \ A 1,2 (|I| = 2 , 5+|A 1,2 | ≤ |J| ≤ 6),
which follows from Theorem 2.5 in P 3 (C) and Lemma 3.1 for m = 1;

• when n = 5, we start with the hyperbolicity of all complements Indeed, the degree of freedom for the choice of a curve of degree d is

H i \ ∪ j =i H j , ∩ i∈I H i \ ∪ j∈J H j \ A 1
(d + 1)(d + 2) 2 -1.
We want C m to pass through all points in A m,2 with multiplicity at least k. The number of equations (with the coefficients of C m as unknowns) for this is not greater than m k(k + 1) 2 .

Thus, for the existence of C m , it is necessary that A 3

E 1 j H i A i 2 A i 1 A i 3 A i 4 A 1 A 2 A 4 H 1 j
If there exists a collinear subset {A i 1 , A i 2 , A i 3 } of A 4,2 , then by the generic condition, it must be contained in some line H i . Let A i 4 be the remaining point of the set A 4,2 and let C 4 be the degenerate quintic consisting of the three lines A i j A i 4 (1 ≤ j ≤ 3) and of the line H i with multiplicity 2. Since C 4 passes through all points in A 4,2 with multiplicity at least 3, the inequality (5.2) is satisfied. By using (5.1), we get a contradiction. Now we assume that any subset of A 4,2 containing three points is not collinear. Let E i j = {e i j = 0} (1 ≤ i ≤ 4, 1 ≤ j ≤ 2) be the eight conics passing through all points of A 4,2 , tangent to the line H i j at the point A i (1 ≤ i ≤ 4, 1 ≤ j ≤ 2). Let E = {e = 0} be the degenerate curve of degree 16 consisting of all these E i j . We claim that f does not land in E. Otherwise, it lands in some conic E i j . Since the number of intersection points between E i j and ∪ 9 i=1 H i \ A 4,2 is > 3 and since any complement of three distinct points in an irreducible curve is hyperbolic, f must be constant, which is a contradiction.

f

  (r, H i ) = 5 N f (r, B) + S f (r). (3.39) Owing to (3.35), the two inequalities (3.31) become

( 3 .

 3 45) Let C = {c = b 1 b 2 b 3 = 0} be the degenerate cubic ∪ 1≤i≤3 B i . It follows from (3.21) and (3.45) that min 1≤i≤3 ord

  2) since |{j ∈ (J \ K) \ R}| ≥ 2 (n -|K|) + 1 + |C m,K ∪ Y * |,which also implies |K| ≤ n -2 by similar argument as in Case 1.

, 3 ( 2 (|I| = 3 ,N [ 2 ]NClaim 5 . 1 .

 323251 |I| = 2 , 7+|A 1,3 | ≤ |J| ≤ 8), ∩ i∈I H i \ ∪ j∈J H j \ A 2,5+|A 2,2 | ≤ |J| ≤7),which follows from Theorem 2.5 in P 4 (C), Lemma 3.2 for m = 1, and Lemma 3.1 for m = 2;We may then proceed similarly as in (3.9)(m + 2) T f (r) ≤ 5+m i=1 f (r, H i ) + S f (r) f (r, A i ) + S f (r) ≤ 3 k N f (r, C m ) + S f (r) ≤ 3d k T f (r) + S f (r).(5.1)When m ≥ 5, the following claim yields a concluding contradiction. If m ≥ 5, we can find some curve C m which does not contain f (C) such that

( 3 )

 3 We try to find two natural numbers k, d satisfying (5.2) and(5.3). This can be done by choosing d = (m + 2)M and k = 3M + 1 for large enough M . Using the remaining freedom in the choice of C m , we can choose it not containing f (C), which proves the claim.Next, we consider the remaining case where m = 4.

  A 2,2 is a set consisting of two points A, B, whereA = H i 1 ∩ H i 2 , B = H i 3 ∩ H i 4 .We denote by I the index set {i 1 , i 2 , i 3 , i 4 }, which has three elements if both A and B belong to a single line H i and which has four elements otherwise.

		), if it exists, must avoid at least four
	lines.	000 000 111 111 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 1111111 0 1 1111111 1111111 1111111 1111111 1111111 000 000 000 000 111 111 A 1,2 00 11 111 111 0 0 1 1 1111111 1111111 f (C)	0 0 0 1 1 1 0 1
	By Theorem 2.3, its image lies in a diagonal line, which does not contain the intersection point of the
	two remaining lines by the generic condition. Hence, f must be contained in the complement of four
	points in a line. By Picard's theorem, f is constant, which is contradiction.
	When m = 2,

  Assume that the omitted planes H 6 , H 7 , H 8 , H 9 are given in the homogeneous coordinates [z 0 : z 1 : z 2 : z 3 ] by equations {z i = 0} (0 ≤ i ≤ 3). By Theorems 2.2, f has a reduced representation of the form [1 : e λ 1 z+µ 1 : e λ 2 z+µ 2 : e λ 3 z+µ 3 ],

say H 6 ∩ H 7 ∩ H 8 , then the curve f avoids three lines H i ∩ P (6 ≤ i ≤ 8), which are not in general position. By Theorem 2.6, f (C) must be contained in some line, which is a contradiction. Therefore, P does not contain any point in

∪ 6≤i 1 <i 2 <i 3 ≤9 H i 1 ∩ H i 2 ∩ H i 3 .

But then the curve f avoids a collection of four lines {H i ∩ P} 6≤i≤9 , which are in general position. By Theorem 2.3, its image must land in some diagonal line of this family, which is a contradiction.

Still in case (a), we can therefore assume that f is linearly nondegenerate. (3.10) where λ i , µ i are constants with λ i = 0 (1 ≤ i ≤ 3 and λ i = λ j (i = j). Let D be the diagonal plane passing through the point A = H 1 ∩ H 2 ∩ H 3 and containing the line l = H 4 ∩ H 5 . By similar arguments as in Lemma 3.1, cf. (3.7), (3.8), we can show that

  in the homogeneous coordinate [z 0 : z 1 : z 2 : z 3 ]. Since λ i are nonzero distinct constants, this forces the two linear forms h 1 , h 2 to be linearly dependent, which is a contradiction. It follows from (3.16) that min {ord z

  = {r = r 1,2 r 4,5 r 5,6 r 4,6 = 0} be the degenerate quartic R 1,2 ∪ R 4,5 ∪ R 5,6 ∪ R 4,6 whose four components pass through A and B, we have ord z (r • f ) ≥ 4 (z∈E t,A ∪E t,B ).

).

(3.29) Letting R

  .2) are hyperbolic where I, J are two disjoint subsets of {1, . . . , 2n} such that |I| ≤ n-2, |J|+2|I| ≥ 2n+1 and m ≤ |J| + 2|I| -(2n + 1). Here, A m,n-|I| is a set of at most m star-subspaces coming from the family of hyperplanes {∩ i∈I

  . . , J k be disjoint subsets of J such that |I|, |J i | ≥ 2, |I| + |J i | = (n -|I|) + 2, 1 ≤ i ≤ k and let {i 1 , . . . , i l } be a subset of I. Let us set I = I ∪ I; then the intersection between the |J | hyperplanes H j , j ∈ J , the k diagonal hyperplanes H IJ 1 , . . . , H IJ k , and the |I| + l hyperplanes H i (i ∈ I), H i 1 , . . . , H i l is a linear subspace of codimension min{k +|I|+l, |I|}+|J |, with the convention that when min{k +|I|+l, |I|}+|J | > n, this intersection is empty. Since min{k + |I| + l, |I|} + |J | = min{k + l, |I|} + |I| + |J | we deduce that in the projective space ∩ i∈I H i , the intersection between the |J | hyperplanes H j , j ∈ J , the k diagonal hyperplanes H IJ 1 , . . . , H IJ k , and the l hyperplanes H i 1 , . . . , H i l is a linear subspace of codimension min{k + l, |I|} + |J |, with the convention that when min{k + l, |I|} + |J | > n -|I|, this intersection is empty. Starting point of the process by deformation: We start with the hyperbolicity of all complements of the forms ∩ i∈I H i \ ∪ j∈J H j \ A m,n-|I| , where I, J, A m,n-|I| are as in Lemma 4.1. More precisely,

Acknowledgments

his support, his very careful readings and many inspiring discussions on the subject. I am specially thankful to my thesis co-advisor Joël Merker for his encouragements, his prompt helps in L A T E X and his comments that greatly improved the manuscript. I would like to thank Junjiro Noguchi for his bibliographical help. Finally, I acknowledge support from Hue University -College of Education.

By symmetry, we also have

Hence we can rewrite (3.27) to get a contradiction:

In P 4 (C), by the generic condition for the family of hyperplanes {H i } 1≤i≤q , when q ≥ 10, we see that, for all three disjoint subsets I, J, K of {1, . . . , q} with |I| ≥ 2, |J| ≥ 2, |I| + |J| = 6, |K| = 4, the diagonal hyperplane H IJ does not contain the point ∩ k∈K H k . Lemma 3.3. In P 4 (C), all complements of the form (3.2) are hyperbolic if m = 1.

Proof. We can assume that A 1,4 is a set consisting of one element in

Suppose to the contrary that there is an entire curve f :

is not a set of a point, then f avoids at least seven hyperplanes. By Theorem 2.4, its image is contained in a line L and we can continue to analyze the position of L with respect to ∪ 10 i=1 H i \ A 1,4 to get a contradiction. Consider the remaining case where A 1,4 consists of a point, say ∩ 4 i=1 H i . By Theorem 2.3, the curve f lands in some diagonal hyperplane of the family {H i } 5≤i≤10 , which does not contain the point ∩ 4 i=1 H i by the generic condition. Hence, f must avoid all H i (1 ≤ i ≤ 10), which is impossible by Theorem 2.5.

Stability of intersections

We will also invoke the following known complex analysis fact.

Stability of intersections. Let X be a complex manifold and let H ⊂ X be an analytic hypersurface. Suppose that a sequence (f n ) of entire curves in X converges toward an entire curve f . If

Proof of the Main Theorem

We keep the notation of the previous section. Let S be a hypersurface of degree 2n, which is in general position with respect to the family {H i } 1≤i≤2n . We would like to determine what conditions S should satisfy for Σ to be hyperbolic. Suppose that Σ k is not hyperbolic for a sequence ( k ) converging to 0. Then we can find entire curves f k : C → Σ k . By the Brody lemma, after reparameterization and extraction, we may assume that the sequence (f k ) converges to an entire curve f : C → ∪ 2n i=1 H i . The curve f (C) lands inside some hyperplane H i . Moreover, it cannot land inside any subspace of dimension 1 (a line). Indeed, if f (C) ⊂ ∩ i∈I H i for some subset I of the index set Q = {1, . . . , 2n} having cardinality n -1, then for all j ∈ Q \ I, by stability of intersections, one has

Thus f (C) and H j have empty intersection by the general position. Hence the curve f (C) lands in

• when n = 6, we start with the hyperbolicity of all complements

which follows from Theorem 2.5 in P 5 (C), Lemma 3.3 for m = 1, Lemma 3.2 for m = 2, and Lemma 3.1 for m = 3.

Details of the process by deformation:

In the first step, we apply inductively Lemma 4.1 for l = n -1 and get at the end a hypersurface S 1 such that all complements of the forms

are hyperbolic. Considering this as the starting point of the second step, we apply inductively Lemma 4.1 for l = n -2. Continuing this process, we get at the end of the (n -2)th step a hypersurface S = S n-2 satisfying the required properties.

Some discussion

Actually, our method works for a family of at least 2n generic hyperplanes in P n (C). We hope that the Main Theorem is true for all n ≥ 3. As we saw above, the problem reduces to proving the following conjecture.

Conjecture. All complements of the form (3.2) are hyperbolic.

We already know it to be true for n = 2, since Lemma 3.1 holds generally, without restriction on m.

Lemma 5.1. In P 2 (C), all complements of the form (3.2) are hyperbolic Proof. Assume now m ≥ 4 and A m,2 = {A 1 , . . . , A m }, where

We denote by I the index set {i j : 1 ≤ i ≤ m, 1 ≤ j ≤ 2}. Suppose to the contrary that there exists an entire curve f :

. By the generic condition, we can assume that f is linearly nondegenerate. By similar arguments as in Lemma 3.1 (cf. (3.6)), we have

Let C m = {c m = 0} be an algebraic curve in P 2 (C) of degree d passing through all points in A m,2 with multiplicity at least k which does not contain the curve f (C). Starting from the inequality

and proceeding as in (3.8), we get

Letting z be a point in f -1 (A i ), we have

By the construction of E i j , if ord z (h i j • f ) ≥ 2 for some 1 ≤ j ≤ 2, then we also have ord z (e i j • f ) ≥ 2. Furthermore, if ord z (h i j • f ) ≥ 2 for all 1 ≤ j ≤ 2, then ord z (e i j • f ) ≥ 2 for all 1 ≤ i ≤ 4, 1 ≤ i ≤ 2. Thus, the following inequality holds:

This implies

We proceed similarly as before to derive a contradiction 6 T f (r) ≤

T f (r) + S f (r).

Lemma 5.1 is thus proved.