Pierre Bonneau

Emmanuel Mazzilli

ALMOST HOLOMORPHIC CURVES IN REAL ANALYTIC HYPERSURFACES

Using the theory of exterior differential systems, we study the existence of germ of pseudo-holomorphic disk in a real analytic hypersurface locally defined in a complex manifold equipped with J a real analytic almost complex structure. The integrable case in C n with J the multiplication by i has been intensively studied by several authors [DF], [DA1] and [DA2] for example. The non integrable case is drastically different essentially due to the following fact : in generic case, there is no J-invariant objects of dimension bigger than one. This simple observation leads to the non existence of some equivalents of Segree varieties or ideals of holomorphic functions which play a fundamental role in the complex case. Nevertheless in the almost complex case, we adopt the exterior differential system point of view of E.Cartan developed and clarified in [BCGGG].

introduction

In this paper we study the existence of germ of pseudo-holomorphic disk in a real analytic hypersurface locally defined in a complex manifold equipped with J a real analytic almost complex structure. In the integrable case, the first motivation to study such existence in the boundary of a bounded real analytic domain of C n was the existence of subelliptic estimates for the ∂neumann problem in the domain (see [DF]). More precisely, the non-existence of germ of disk in the boundary is a sufficient condition for the subelliptic estimates for ∂-neumann problem in the domain. Nevertheless, in the non integrable context, we think that such problem is a nice geometric problem linked with the existence of peudo-holomorphic foliation on real analytic hypersurface embedded in almost complex manifold equipped with J a real analytic almost complex structure. The main difference with the integrable case is the non existence of J-invariant manifold of dimension bigger than one for generic structure J and so, it is not possible to localize the germ of disk in the intersection of complex hypersurface defined locally and not in sense of germs. This approach is the key trick in the complex case.

Namely in this paper, we adopt the exterior differential system for pfaffian system point of view which is, to our knowledge, original and of interest for this problem. Roughly speaking, for the pfaffian system -which is the case writing the PDF system associated to the existence of pseudo-holomorphic disk in an hypersurface -, we have two fundamental objects for the existence of solutions of the PDF system: the torsion and the dimension of the "tableaux". The nullity of the torsion gives a necessary conditions to obtain a solution of the system and to obtain sufficient conditions, the dimension of the tableau and the dimension of its prolongation play a fundamental role. In our context, we investigate carefully the two intrinsic notions and we prove that the system is in involution in the sense of Cartan ( [BCGGG]).

The paper is organized as follows: in the section 2, we recall all the materials for almost holomorphic curves and for elementary almost complex geometry used in the paper.

In the section 3, we compute carefully the torsion which is" like an intrinsic way to see the Levi form" and the dimensions of the tableaux. We give necessary conditions for which the PDF system associated to our problem is in involution at the first step (see corollary 3.5 and 3.6). In fact, we study more general system than the previous one and we compute the prolongation of the tableau for which this tableau is in involution (see theorem 3.1 for precise statements). At the end of the section 3, we study some examples in the complex case to illustrate the abstract results of the beginning of the section (see example 3.7) and we investigate the successive torsion fo the prolongation of the more general system until that its tableau is in involution.

In the section 4, we give directly a necessary and sufficient condition to obtain an ordinary integral element of the exterior differential system and so, by Cartan-Kähler theorem, the existence of germ of disk in the hypersurface (see [BCGGG] pp. 81-86 and theorem 4.1, section 4).

In the section 5, we explain how to construct the PDF system on finite union of manifold "quasiequivalent" to the system of existence of germ of disk, for which the tableau is in involution on each manifold and free torsion (see theorem 5.5 for precise statements).

Finally in the section 6, we investigate the case of the existence of germs of harmonic disks in a real analytic hypersurface locally defined.

Almost holomorphic curves

We first provide R 2 and R 2n with almost complex structures. On an open set D of R 2 , we define an almost complex structure J 0 , that is to say, for all x = (x 1 , x 2 ) ∈ D, we have an isomorphism J 0 (x) = J 0 of (the tangent space to) R 2 , which depends R-analytically to x and verifies J 2 0 = -Id (Id is the identity). We note A 0 (x) = A 0 the marix of J 0 (x) = J 0 , so we have A 2 0 = -I 2 , (I 2 is the unity matrix in dimension 2), that is to say

A 0 (x) = a(x) b(x) -1+a(x) 2 b(x) -a(x)
with b(x) = 0.

(1)

In the same way, we define an almost complex structure J(y) = J on an open set D of R 2n . J is R-analytically dependind in y ∈ D and J 2 = -Id. So, the matrix A associated to J verifies A 2 = -I 2n (I 2n is the unity matrix in dimension 2n). Then, a function f : D -→ D is sayed almost holomorphic, for the almost complex structures (J 0 , J) if for all x ∈ D, df (x)

• J 0 (x) = J(f (x)) • df (x).
This gives, if we note ∂f ∂x (x) ∈ M 2n,2 (R) the matrix ∂f ∂x (x) = ∂f j (x) ∂x i j=1,...,,2n ; i=1,2

, ∂f ∂x (x)A 0 (x) = A(f (x)) ∂f ∂x (x), (2)
and, therefore, the system ∀j = 1, ..., 2n, 1+a 2 b ∂f j ∂x 2 = (aa j,j )

∂f j ∂x 1 -i =j a j,i ∂f i ∂x 1 b ∂f j ∂x 1 = (a + a j,j ) ∂f j ∂x 2 + i =j a j,i ∂f i ∂x 2 , (3)
or

∂f j ∂x 1 = a+a j,j b ∂f j ∂x 2 + i =j a j,i b ∂f i ∂x 2 ∂f j ∂x 2 = b(a-a j,j ) 1+a 2 ∂f j ∂x 1 -i =j ba j,i 1+a 2 ∂f i ∂x 1 . (4)
The first equation can be written, with obvious notations,

∂f ∂x 1 = aI + A b ∂f ∂x 2 , (5)
and the second one

∂f ∂x 2 = b 1 + a 2 (aI -A) ∂f ∂x 1 . (6)
But, b 1+a 2 (aI -A) aI+A b = I. Therefore, the two equations are the sames. Finally, f is almost

holomorphic if ∂f ∂x 2 = A ∂f ∂x 1 with A = b(aI-A) 1+a 2
In [BCGGG],

∂f j ∂x i is noted p j i and, therefore, the condition of almost holomorphicity is p 2 = Ap 1 . One of the aims of this paper is to study the possibility to have an almost holomorphic curve in a R-analytic hypersurface. Let be H = {y ∈ D : ρ(y) = 0} an hypersurface in D with ρ R-analytic. Is it possible to have f : D -→ D almost complex such that ρ(D) ⊂ H ? In other words, have we a solution f : D -→ D for the system of PDE ρ(f (x)) = 0 p 2 = Ap 1 ?

But, in a first time, we study a more general problem. In a second time, we shall return to the almost complex case, and, also, to the complex case.

Necessary conditions

We are now looking for necessary conditions to have a curve f : D -→ D such that ρ(D) ⊂ H = {ρ = 0} with ρ R-analytic and verifying p 2 = Ap 1 where A = (α j,i (y)) i,j=1,...,2n ∈ M 2n (R) is any matrix of order 2n whose the coefficients α j,i are R-analytic functions of y ∈ D. Therefore, we have a solution for the system of PDE

ρ(f (x)) = 0 p 2 = Ap 1 (8)
which, in terms of [BCGGG] (see page 131, example 5.4), is expressed by the Pfaffian differential system      ρ(f (x)) = 0 p 2 = Ap 1 θ j = df jp j 1 dx 1p j 2 dx 2 = 0, ∀j = 1, ..., 2n,

differential system which lives on the space M 1 of variables (x 1 , x 2 , f 1 , ..., f 2n , p j i ), j = 1, ..., 2n, i = 1, 2.

On account of the second line of 9, we have p j 2 = i α j,i p i 1 , so we have to restrict M 1 by deleting the p j 2 . And this differential system can be written ρ(f (x)) = 0 θ j = df jp j 1 dx 1i α j,i p i 1 dx 2 = 0, ∀j = 1, ..., 2n, (10) on the space M 2 of variables (x 1 , x 2 , f 1 , ..., f 2n , p 1 1 , ..., p 2n 1 ).

By deriving the first line of 10, we obtain, if we note ρ j = ∂ρ ∂y j (f (x)), the system

ρ 1 p 1 1 + ρ 2 p 2 1 = -2n j=3 ρ j p j 1 2n j ′ =1 ρ j ′ α j ′ ,1 p 1 1 + 2n j ′ =1 ρ j ′ α j ′ ,2 p 2 1 = -2n j=3 2n j ′ =1 ρ j ′ α j ′ ,j p j 1 (11)
and, when the determinant of this system D = 2n j ′ =1 ρ j ′ (ρ 1 α j ′ ,2ρ 2 α j ′ ,1 ) is = 0, we have

p 1 1 = -1 D 2n j=3 p j 1 2n j ′ =1 ρ j ′ (ρ j α j ′ ,2 -ρ 2 α j ′ ,j ) := 2n j=3 γ 1 j p j 1 p 2 1 = -1 D 2n j=3 p j 1 2n
j ′ =1 ρ j ′ (ρ 1 α j ′ ,jρ j α j ′ ,1 ) := 2n j=3 γ 2 j p j 1 .

(

) 12
Using this formulas, the structure equations of the last line of 10 become

θ 1 = df 1 - 2n j=3 γ 1 j p j 1 dx 1 - 2n j ′ =1 α 1,j ′ p j ′ 1 dx 2 = df 1 - 2n j=3 γ 1 j p j 1 dx 1 -(α 1,1 p 1 1 + α 1,2 p 2 1 + 2n j=3 α 1,j ′ p j 1 )dx 2 = df 1 - 2n j=3 γ 1 j p j 1 dx 1 - 2n j=3 (α 1,1 γ 1 j + α 1,2 γ 2 j + α 1,j )p j 1 dx 2 , (13)
θ 2 = df 2 - 2n j=3 γ 2 j p j 1 dx 1 - 2n j ′ =1 α 2,j ′ p j ′ 1 dx 2 = df 2 - 2n j=3 γ 2 j p j 1 dx 1 -(α 2,1 p 1 1 + α 2,2 p 2 1 + 2n j=3 α 2,j ′ p j 1 )dx 2 = df 2 - 2n j=3 γ 2 j p j 1 dx 1 - 2n j=3 (α 2,1 γ 1 j + α 2,2 γ 2 j + α 2,j )p j 1 dx 2 , (14)
and, ∀i = 3, ..., 2n,

θ i = df i -p i 1 dx 1 - 2n j ′ =1 α i,j ′ p j ′ 1 dx 2 = df i -p i 1 dx 1 -(α i,1 2n j=3 γ 1 j p j 1 + α i,2 2n j=3 γ 2 j p j 1 + 2n j=3 α i,j p j 1 )dx 2 = df i -p i 1 dx 1 - 2n j=3 (α i,1 γ 1 j + α i,2 γ 2 j + α i,j )p j 1 dx 2 . ( 15
)
We define β i,j = α i,1 γ 1 j + α i,2 γ 2 j + α i,j , and, finally, the system 10 becomes

     θ 1 = df 1 -2n j=3 γ 1 j p j 1 dx 1 -2n j=3 β 1,j p j 1 dx 2 θ 2 = df 2 -2n j=3 γ 2 j p j 1 dx 1 -2n j=3 β 2,j p j 1 dx 2 θ i = df i -p i 1 dx 1 -2n j=3 β i,j p j 1 dx 2 ∀i = 3, ..., 2n, (16)
on the space M of variables (x 1 , x 2 , f 1 , ..., f 2n , p 3 1 , ..., p 2n 1 ). ( 17)

Until now, we calculated with the variables x 1 , x 2 . From now on, we calculate with the variables 17. We remark the functions α i,j , γ i j , β i,j , ρ are functions of f = (f 1 , ..., f 2n ) only. So, ρ i = ∂ρ ∂y i (f ) shall be note, sometimes, ρ i = ∂ρ ∂f i . And also, from now on, d is the exterior derivative relative to the variables 17.

In T * M, we take the basis θ 1 , ..., θ 2n , dx 1 , dx 2 , dp 3 1 , ..., dp 2n 1 .

We also consider the dual basis noted

∂ ∂θ 1 , ..., ∂ ∂θ 2n , ∂ ∂x 1 , ∂ ∂x 2 , ∂ ∂p 3 1 , ..., ∂ ∂p 2n 1 of T M.
Let I ⊂ T ⋆ M be the sub-bundle generated by (θ i , i = 1, ..., 2n), J ⊂ T ⋆ M be the sub-bundle generated by (dx 1 , dx 2 , θ i , i = 1, ..., 2n), and let {I} ⊂ Ω ⋆ (M ) be the algebraic ideal generated by the C ∞ -sections of I.

Conformably to [BCGGG], p. 130, for k = 1, ..., 2n, we write

dθ k = 2n j=3 2 i=1 A k j,i dp j 1 ∧ dx i + c k 1,2 dx 1 ∧ dx 2 modulo {I}. ( 18
)
For an equality modulo {I}, we shall write ≈ .

From 16, we obtain

     df 1 ≈ 2n j=3 γ 1 j p j 1 dx 1 + 2n j=3 β 1,j p j 1 dx 2 df 2 ≈ 2n j=3 γ 2 j p j 1 dx 1 + 2n j=3 β 2,j p j 1 dx 2 df i ≈ p i 1 dx 1 + 2n j=3 β i,j p j 1 dx 2 ∀i = 3, ..., 2n, (19)
So, using 19, we have to calculate

                                                                         dθ 1 = -2n j=3 2n i=1 ∂ ∂f i γ 1 j p j 1 df i ∧ dx 1 -2n j=3 2n i=1 ∂ ∂f i β 1,j p j 1 df i ∧ dx 2 -2n j=3 γ 1 j dp j 1 ∧ dx 1 -2n j=3 β 1,j dp j 1 ∧ dx 2 ≈ 2n j=3 ∂γ 1 j ∂f 1 p j 1 2n j ′ =3 β 1,j ′ p j ′ 1 dx 1 ∧ dx 2 + 2n j=3 ∂γ 1 j ∂f 2 p j 1 2n j ′ =3 β 2,j ′ p j ′ 1 dx 1 ∧ dx 2 + 2n j,j ′ =3 ∂γ 1 j ∂f j ′ p j 1 β j,j ′ p j ′ 1 dx 1 ∧ dx 2 -2n j=3 γ 1 j dp j 1 ∧ dx 1 -2n j=3 ∂β 1,j ∂f 1 p j 1 2n j ′ =3 γ 1 j ′ p j ′ 1 dx 1 ∧ dx 2 -2n j=3 ∂β 1,j ∂f 2 p j 1 2n j ′ =3 γ 2 j ′ p j ′ 1 dx 1 ∧ dx 2 + 2n j,j ′ =3 ∂β 1,j ∂f j ′ p j 1 p j ′ 1 dx 1 ∧ dx 2 -2n j=3 β 1,j dp j 1 ∧ dx 2 ≈ dx 1 ∧ dx 2 2n j,j ′ =3 p j 1 p j ′ 1 ∂γ 1 j ∂f 1 β 1,j ′ + ∂γ 1 j ∂f 2 β 2,j ′ + ∂γ 1 j ∂f j ′ β j,j ′ - ∂β 1,j ∂f 1 γ 1 j ′ - ∂β 1,j ∂f 2 γ 2 j ′ - ∂β 1,j ∂f 1j ′ -2n j=3 γ 1 j dp j 1 ∧ dx 1 -2n j=3 β 1,j dp j 1 ∧ dx 2 dθ 2 = -2n j=3 2n i=1 ∂γ 2 j ∂f i p j 1 df i ∧ dx 1 -2n j=3 2n i=1 ∂β 2,j ∂f i p j 1 df i ∧ dx 2 -2n j=3 γ 1 j dp j 1 ∧ dx 1 -2n j=3 β 2,j dp j 1 ∧ dx 2 ≈ dx 1 ∧ dx 2 2n j,j ′ =3 p j 1 p j ′ 1 ∂γ 2 j ∂f 1 β 1,j ′ + ∂γ 2 j ∂f 2 β 2,j ′ + ∂γ 2 j ∂f j ′ β j,j ′ - ∂β 2,j ∂f 1 γ 1 j ′ - ∂β 2,j ∂f 2 γ 2 j ′ - ∂β 2,j ∂f 1j ′ -2n j=3 γ 2 j dp j 1 ∧ dx 1 -2n j=3 β 2,j dp j 1 ∧ dx 2 ∀i = 3, ..., 2n, dθ i = -dp i 1 ∧ dx 1 -2n j=3 2n j ′ =1 ∂β i,j ∂f j ′ p j 1 df j ′ ∧ dx 2 -2n j=3 β i,j dp j 1 ∧ dx 2 ≈ -2n j,j ′ =3 p j 1 p j ′ 1 ∂β i,j ∂f 1 γ 1 j ′ + ∂β i,j ∂f 2 γ 2 j ′ + ∂β i,j ∂f j ′ dx 1 ∧ dx 2 -dp i 1 ∧ dx 1 -2n j=3 β i,j dp j 1 ∧ dx 2 .
(20) With the notation 18 adapted to our situation, we have

c 1 1,2 = 2n j,j ′ =3 p j 1 p j ′ 1 ∂γ 1 j ∂f 1 β 1,j ′ + ∂γ 1 j ∂f 2 β 2,j ′ + ∂γ 1 j ∂f j ′ β j,j ′ - ∂β 1,j ∂f 1 γ 1 j ′ - ∂β 1,j ∂f 2 γ 2 j ′ - ∂β 1,j ∂f 1j ′ c 2 1,2 = 2n j,j ′ =3 p j 1 p j ′ 1 ∂γ 2 j ∂f 1 β 1,j ′ + ∂γ 2 j ∂f 2 β 2,j ′ + ∂γ 2 j ∂f j ′ β j,j ′ - ∂β 2,j ∂f 1 γ 1 j ′ - ∂β 2,j ∂f 2 γ 2 j ′ - ∂β 2,j ∂f 1j ′ c i 1,2 = - 2n j,j ′ =3 p j 1 p j ′ 1 ∂β i,j ∂f 1 γ 1 j ′ + ∂β i,j ∂f 2 γ 2 j ′ + ∂β i,j ∂f j ′ A 1 (j,1),1 = -γ 1 j , A 2 (j,1),1 = -γ 2 j , A i (j,1),2 = -δ j i , A 1 (j,1),2 = -(α 1,1 γ 1 j + α 1,2 γ 2 j + α 1,j ), A 2 (j,1),2 = -(α 2,1 γ 1 j + α 2,2 γ 2 j + α 2,j ), A i (j,1),2 = -(α i,1 γ 1 j + α i,2 γ 2 j + α i,j ), (21)
which allow to define (see [BCGGG] p. 133)

π : J ⊥ = Span ∂ ∂p 3 1 , ..., ∂ ∂p 2n 1 -→ I * ⊗ (J/I) * = Span ∂ ∂θ 1 , ..., ∂ ∂θ 2n ⊗ Span(x 1 , x 2 )
and we have, for

v = 2n i=3 v i ∂ ∂p i 1 ∈ J ⊥ , (22)
π(v) = - 2n j=3 γ 1 j v j ∂ ∂θ 1 ⊗ x 1 - 2n j=3 β 1,j v j ∂ ∂θ 1 ⊗ x 2 - 2n j=3 γ 2 j v j ∂ ∂θ 2 ⊗ x 1 - 2n j=3 β 2,j v j ∂ ∂θ 2 ⊗ x 2 - 2n i=3 v i ∂ ∂θ i ⊗ x 1 - 2n i=3 2n j=3 β i,j v j ∂ ∂θ i ⊗ x 2 = - 2n j=3 v j γ 1 j ∂ ∂θ 1 ⊗ x 1 + β 1,j ∂ ∂θ 1 ⊗ x 2 + γ 2 j ∂ ∂θ 2 ⊗ x 1 + β 2,j ∂ ∂θ 2 ⊗ x 2 + ∂ ∂θ j ⊗ x 1 + 2n i=3 β i,j ∂ ∂θ i ⊗ x 2 := 2n j=3 v j U j . (23)
The image of π is the tableau A = A (0) associated to the system 16, so (U 3 , ..., U 2n ) is a basis of A (the independence of the vectors U j is obvious). So dim(A) = 2n -2.

We also have to remark that A 1 = P ∈ A : ∂P ∂x 1 = 0 = {0} (see [BCGGG] p. 119 for the definition of A 1 ) and therefore dim(A 1 ) = 0. We now need the prolongations of A (see [BCGGG] p. 117). The first prolongation is

A (1) = P = P 1,1 ⊗ x 2 1 + P 1,2 ⊗ x 1 x 2 + P 2,2 ⊗ x 2 2 : ∂P ∂x 1 and ∂P ∂x 2 ∈ A
(sometimes, we shall remove the sign ⊗). To obtain P ∈ A (1) , we want

∂P ∂x 1 = 2P 1,1 x 1 + P 1,2 x 2 = 2n j=3 v j U j ∂P ∂x 2 = P 1,2 x 1 + 2P 2,2 x 2 = 2n j=3 v ′j U j (24) that is to say, if we explicit U j ,          P 1,1 = 1 2 j v j [γ 1 j ∂ ∂θ 1 + γ 2 j ∂ ∂θ 2 + ∂ ∂θ j ] P 1,2 = 2n j=3 v j [β 1,j ∂ ∂θ 1 + β 2,j ∂ ∂θ 2 + 2n i=3 β i,j ∂ ∂θ i ] = 2n j=3 v ′j [γ 1 j ∂ ∂θ 1 + γ 2 j ∂ ∂θ 2 + ∂ ∂θ j ] P 2,2 = 1 2 2n j=3 v ′j [β 1,j ) ∂ ∂θ 1 + β 2,j ∂ ∂θ 2 + 2n i=3 β i,j ∂ ∂θ i ] (25)
The equality of the second and third lines gives

     v ′j = Σ 2n i=3 v i β j,i ∀j = 3, ..., 2n. 2n j=3 v ′j γ 2 j = Σ 2n i=3 v i β 2,i 2n
j=3 v ′j γ 1 j = Σ 2n i=3 v i 1, i. (26)
We can translate this matricially. We shall note α = α j,i j,i=3,...,2n

∈ M 2n-2 (R). α is the matrix obtained from A if we remove the two first lines and columns.

We also define the one column matrix α .,i = α j,i j=3,...,2n

∈ M 2n-2,1 (R) and the one line matrix

γ j = γ j i i=3,...,2n ∈ M 1,2n-2 (R). Let β = α .,1 γ 1 + α .,2 γ 2 + α = (β j,i ) i,j=3,...,2n ∈ M 2n-2 (R) be the matrix defined by β j,i = α j,1 γ 1 i + α j,2 γ 2 i + α j,i , β 2 = (β 2,i ) i=3,...,2n ∈ M 1,2n-2 (R), the matrix defined by β 2,i = α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i . and β 1 = (β 1,i ) i=3,...,2n ∈ M 1,2n-2 (R), the matrix defined by β 1,i = α 1,1 γ 1 i + α 1,2 γ 2 i + α 1,i . Then, 26 can be written      v ′ = βv γ 2 v ′ = β 2 v γ 1 v ′ = β 1 v, (27)
which implies (γ 2 β -β 2 )v = 0 and (γ 1 β -β 1 )v = 0.
We also consider the linear applications

L : R 2n-2 -→ R 2n-2 , l 1 : R 2n-2 -→ R, l 2 : R 2n-2 -→ R, l 1 : R 2n-2 -→ R, l 2 : R 2n-2 -→ R respectively associated to the matrix β, β 2 , γ 2 , β 1 , γ 1 , and 26 can be written      v ′ = L(v) l 2 (v ′ ) = l 1 (v) l 2 (v ′ ) = l 1 (v), (28)
and, therefore, we have (l 2 • Ll 1 )v = 0 and ( l 2 • Ll 1 )v = 0, that is to say, v ∈ Ker(l 2 • Ll 1 ) ∩ Ker( l 2 • Ll 1 ) := S, and S is isomorphic to A.

It is easy to verify A

(1)

1 = {0}. Of course, we have dim(A (1) ) = dim(S) ≤ 2n -2 = dim(A) + dim(A 1 )
. Following [BCGGG], p. 120, A is involutive if this inequality becomes an equality, that is to say

A involutive ⇔ dim(A (1) ) = dim(A) + dim(A 1 ) ⇔ dim(A (1) ) = dim(S) = 2n -2 ⇔ S = R 2n-2 ⇔ Ker(l 2 • L -l 1 ) = Ker( l 2 • L -l 1 ) = R 2n-2 ⇔ l 2 • L = l 1 and l 2 • L = l 1 ⇔ γ 2 β = β 2 and ⇔ γ 1 β = β 1 ⇔ 2n j=3 γ 2 j β j,i = β 2,i and 2n j=3 γ 1 j β j,i = β 1,i ∀i = 3, ..., 2n (29)
Inductively, we define (see [BCGGG], p. 117) the q-prolongation A (q) of the tableau A (q ≥ 1)

A (q) = P = q+1 l=0 P [q+1-l,l] x q+1-l 1 x l 2 : ∀ | J |= q, ∂P ∂x J ∈ A . ( 30
)
To obtain P ∈ A (q) , we shall write ∂P ∂x J in the basis (U 3 , ..., U 2n ) of A. Let J = [q -k, k]; 0 ≤ k ≤ q be a multi-index with length | J |= q. Then,

∂P ∂x J = ∂P ∂x q-k 1 ∂x k 2 = k!(q -k + 1)!P [q-k+1,k] x 1 + (q -k)!(k + 1)!P [q-k,k+1] x 2 = 2n j=3 v k j U j , (31)
so, P ∈ A (q) if and only if ∀k = 0, ..., q, k!(q

-k + 1)!P [q-k+1,k] = 2n j=3 v k j [γ 1 j ∂ ∂θ 1 + γ 2 j ∂ ∂θ 2 + ∂ ∂θ j ] (q -k)!(k + 1)!P [q-k,k+1] = 2n j=3 v k+1 j 2n i=1 β i,j ∂ ∂θ i , (32)
and, therefore,

           P [q+1,0] = 1 (q+1)! 2n j=3 v 0 j [γ 1 j ∂ ∂θ 1 + γ 2 j ∂ ∂θ 2 + ∂ ∂θ j ] P [q-k,k+1] = 1 (k+1)!(q-k)! 2n j=3 v k+1 j [γ 1 j ∂ ∂θ 1 + γ 2 j ∂ ∂θ 2 + ∂ ∂θ j ] = 1 (k+1)!(q-k)! 2n j=3 v k j 2n i=1 β i,j ∂ ∂θ i ∀k = 0, ..., q -1 P [0,q+1] = 1 (q+1)! 2n j=3 v q j 2n i=3 (β i,j ∂ ∂θ i (33) So, we have ∀k = 0, ..., q -1,      v k+1 = βv k γ 2 v k+1 = β 2 v k γ 1 v k+1 = β 1 v k , (34)
and

     v k+1 = β (k+1) v 0 γ 2 β (k+1) v 0 = β 2 β k v 0 γ 1 β (k+1) v 0 = β 1 β k v 0 , ( 35
) namely      v k+1 = L k+1 v 0 (l 2 • L -l 1 ) • L k v 0 = 0 ( l 2 • L -l 1 ) • L k v 0 = 0. ( 36
)
Thus, we have

L k v 0 ∈ Ker(l 2 • L -l 1 ) ∩ Ker( l 2 • L -l 1 ) = S or v 0 ∈ L -k (S) and so dim(A (q) ) = dim(∩ q-1 k=0 L -k (S)
). We verify easily A (q) 1 = {0}. For each A (q-1) we have two possibilities : either ∩ q-2 k=0 L -k (S) ⊂ L -(q-1) (S), then dim(∩ q-1 k=0 L -k (S)) = dim(∩ q-2 k=0 L -k (S)), so dim(A (q) ) = dim(A (q-1) ) + dim(A (q-1) 1

) (this later term is zero) and A (q-1) is involutive, or ∩ q-2 k=0 L -k (S) L -(q-1) (S), then dim(∩ q-1 k=0 L -k (S)) < dim(∩ q-2 k=0 L -k (S)), and A (q-1) is not involutive. We note D 2 = γ 2 ββ 2 = (D 2,3 , ..., D 2,2n ) the one line matrix of l 2 • Ll 1 , defined by

D 2,i = 2n j=3 γ 2 j β j,i -β 2,i = 2n j=3 (α j,1 γ 1 i + α j,2 γ 2 i + α j,i )γ 2 j -(α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i ), (37)
and D 1 = γ 1 ββ 1 = (D 1,3 , ..., D 1,2n ) the one line matrix of l 2 • Ll 1 , defined by

D 1,i = 2n j=3 γ 1 j β j,i -β 1,i = 2n j=3 (α j,1 γ 1 i + α j,2 γ 2 i + α j,i )γ 1 j -(α 1,1 γ 1 i + α 1,2 γ 2 i + α 1,i ). ( 38
)
We want to calculate D 2 more precisely. We define µ i = 2n j=1 ρ j α j,i , and µ

(2)

i = 2n j=1 ρ j α (2) j,i
, where α

(2) j,i is the generic term of the matrix

A 2 = α (2) j,i i,j=1,...,2n . From 12, D = ρ 1 µ 2 -ρ 2 µ 1 , γ 1 i = -1 D (ρ i µ 2 -ρ 2 µ i ) and γ 2 i = -1 D (ρ 1 µ i -ρ i µ 1 ), (39)
so we have ∀i = 3, ..., 2n,

2n j=3 γ 2 j α j,i = -1 D 2n j=3 α j,i ρ 1 2n i ′ =1 ρ i ′ α i ′ ,j -ρ j 2n i ′ =1 ρ i ′ α i ′ ,1 = -1 D ρ 1 2n i ′ =1 ρ i ′ 2n j=3 α i ′ ,j α j,i - 2n j=3 ρ j α j,i 2n i ′ =1 ρ i ′ α i ′ ,1 = -1 D ρ 1 2n i ′ =1 ρ i ′ (α (2) i ′ ,i -α i ′ ,1 α 1,i -α i ′ ,2 α 2,i ) -µ 1 (µ i -ρ 1 α 1,i -ρ 2 α 2,i ) = -1 D ρ 1 µ (2) i -ρ 1 µ 2 α 2,i -µ 1 µ i + ρ 2 µ 1 α 2,i . (40)
Therefore,

D 2,i = γ 1 i -1 D ρ 1 µ (2) 1 -ρ 1 µ 2 α 2,1 -µ 2 1 + ρ 2 µ 1 α 2,1 + γ 2 i -1 D ρ 1 µ (2) 2 -ρ 1 µ 2 α 2,2 -µ 1 µ 2 + ρ 2 µ 1 α 2,2 - 1 D ρ 1 µ (2) i -ρ 1 µ 2 α 2,i -µ i 1 + ρ 2 µ 1 α 2,i -α 2,1 γ 1 i -α 2,2 γ 2 i -α 2,i = -ρ 1 D µ (2) 1 γ 1 i + µ (2) 2 γ 2 i + µ (2) i + ρ 1 µ 2 D α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i + µ 1 D µ 1 γ 1 i + µ 2 γ 2 i + µ i ρ 2 µ 1 D α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i -α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i = α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i ρ 1 µ 2 -ρ 2 µ 1 D -1 + 1 D ρ 1 µ (2) 1 γ 1 i + µ (2) 2 γ 2 i + µ (2) i + µ 1 µ 1 γ 1 i + µ 2 γ 2 i + µ i (41)
From 39, in the previous line, the first term is zero, so, we only have

D 2,i = 1 D ρ 1 µ (2) 1 -1 D (ρ i µ 2 -ρ 2 µ i) + µ (2) 2 -1 D (ρ 1 µ i -ρ i µ 1) + µ (2) i + µ 1 µ 1 -1 D (ρ i µ 2 -ρ 2 µ i) + µ 2 -1 D (ρ 1 µ i -ρ i µ 1) + µ i = -1 D 2 µ (2) 1 ρ 1 ρ i µ 2 -µ (2) 1 ρ 1 ρ 2 µ i + µ (2) 2 ρ 2 1 µ i -µ (2) 2 ρ 1 ρ i µ 1 -Dρ 1 µ (2) i -µ 1 µ i -ρ 2 µ 1 + ρ 1 µ 2 -D . (42)
Always from 39, the last term between { } is zero. So,

D 2,i = -ρ 1 D 2 µ (2) 1 ρ i µ 2 -ρ 2 µ i + µ (2) 2 ρ 1 µ i -ρ i µ 1 + µ (2) i ρ 2 µ 1 -ρ 1 µ 2 = -ρ 1 D 2 µ 1 ρ 2 µ (2) i -ρ i µ (2) 2 + µ 2 ρ i µ (2) 1 -ρ 1 µ (2) 2 + µ i ρ 1 µ (2) 2 -ρ 2 µ (2) 1 = -ρ 1 D 2 ρ 1 µ i µ (2) 2 -µ 2 µ (2) i + ρ 2 µ 1 µ (2) i -µ i µ (2) 1 + ρ i µ 2 µ (2) 1 -µ 1 µ (2) 2 . ( 43
)
Analogous calculations allow to express D 1 more precisely also. We obtain ∀i = 3, ..., n,

D 1,i = -ρ 2 D 2 ρ 1 µ i µ (2) 2 -µ 2 µ (2) i + ρ 2 µ 1 µ (2) i -µ i µ (2) 1 + ρ i µ 2 µ (2) 1 -µ 1 µ (2) 2 . ( 44
)
Thus, if we note the one line matrix ρ • = (ρ 3 , ..., ρ 2n ), µ • = (µ 3 , ..., µ 2n ), µ

(2)

• = (µ (2) 3 , ..., µ (2)
2n ), we have

D 1 = -ρ 2 D 2 ρ 1 µ (2) 2 µ • -µ 2 µ (2) • + ρ 2 µ 1 µ (2) • -µ (2) 1 µ • + µ 2 µ (2) 1 -µ 1 µ (2) 2 ρ • := ρ 2 D 0 D 2 = -ρ 1 D 2 ρ 1 µ (2) 2 µ • -µ 2 µ (2) • + ρ 2 µ 1 µ (2)
•µ

(2)

1 µ • + µ 2 µ (2) 1 -µ 1 µ (2) 2 ρ • := ρ 1 D 0 . (45) So, ρ 1 D 1 = ρ 2 D 2 and D 1 = D 2 = 0 if and only if ρ 1 µ i µ (2) 2 -µ 2 µ (2) i + ρ 2 µ 1 µ (2) i -µ i µ (2) 1 + ρ i µ 2 µ (2) 1 -µ 1 µ (2) 2 = -D 2 D 0,i = 0 ∀i = 3, ..., 2n. ( 46
)
Remark 3.1. Of course, D 1 and D 2 depend on A, so we note them D 1 (A) and D 2 (A).Using the previous expression of D 1 (A) and D 2 (A)), we prove easily, if I is the unit (2n, 2n) matrix, A an other (2n, 2n) matrix and α, β functions of f, then

D k α(f )I + β(f )A = β 3 (f )D k (A), ∀k = 1, 2. ( 47
)
Moreover, if A = (a i,j ) i,j=1,...,2n = λ(f )I, then µ A,i = 2n j=1 ρ j a j,i = λρ i and µ

(2)

A,i = 2n j=1 ρ j a (2) j,i = λ 2 ρ i , so D k (A) = 0. Therefore, if A 2 = λ(f )I, then A = α(f )I + β(f )A verifies D k (A) = 0 (48) (because µ (2) A,i = λ 2 ρ i implies D k (A) = 0).
We now return after 36. As explain there, A (q-1) is involutive if and only if

∩ q-2 k=0 L -k (S) ⊂ L -(q-1) (S) (49) that is (l 2 • L -l 1 ) • L k (v) = 0 ∀k = 0, ..., q -2 ( l 2 • L -l 1 ) • L k (v) = 0 (50) implies (l 2 • L -l 1 ) • L (q-1) (v) = 0 ( l 2 • L -l 1 ) • L (q-1) (v) = 0, ( 51
) namely                                      (l 2 • L -l 1 )(v) = 0 (l 2 • L 2 -l 1 • L)(v) = 0 ....
• L (q-1) -l 1 • L (q-2) )(v) = 0 ( l 2 • L -l 1 )(v) = 0 ( l 2 • L 2 -l 1 • L)(v) = 0 ....
( l 2 • L (q-1) -l 1 • L (q-2) )(v) = 0 (52) implies (l 2 • L -l 1 ) • L (q-1) (v) = 0 ( l 2 • L -l 1 ) • L (q-1) (v) = 0, (53)
This means that the two equations 53 are linear combinations of the 52 ones, that is to say

(l 2 • L -l 1 ) • L (q-1) = q-2 k=0 α k (l 2 • L -l 1 ) • L k + β k ( l 2 • L -l 1 ) • L k ( l 2 • L -l 1 ) • L (q-1) = q-2 k=0 α ′ k (l 2 • L -l 1 ) • L k + β ′ k ( l 2 • L -l 1 ) • L k (54)
or, matricially,

D 2 β q-1 = α 0 D 2 + α 1 D 2 β + ... + α q-2 D 2 β q-2 + β 0 D 1 + β 1 D 1 β + ... + β q-2 D 1 β q-2 D 1 β q-1 = α ′ 0 D 2 + α ′ 1 D 2 β + ... + α ′ q-2 D 2 β q-2 + β ′ 0 D 1 + β ′ 1 D 1 β + ... + β ′ q-2 D 1 β q-2 . ( 55
)
Finally, we obtained the following theorem and corollary:

Theorem 3.2. A (q-1) is involutive if

D 0 β q-1 ∈ Span(D 0 β q-2 , D 0 β q-3 , ..., , D 0 β, D 0 ) (56) Corollary 3.3. A is involutive if (see 29) D 0 = 0. ( 57
)
We can remark that, if A (q 0 ) is involutive, then A (q 0 +k ) is also involutive for k 0. In fact, if

D 0 β q-1 ∈ Span(D 0 , D 0 β, ..., D 0 β q-2 ), D 0 β q-1 = q-2 i=0 α i D 0 β i , then D 0 β q = q-2 i=0 α i D 0 β i+1 = α q-2 D 0 β q-1 + q-3 i=0 α i D 0 β i+1 ∈ Span(D 0 , D 0 β, ..., D 0 β q-2
). Thus, we have the

Corollary 3.4. A (q) is involutive if q Rank(D 0 , D 0 β, ..., D 0 β 2n-3 ).
This is the Cartan-Kuranishi theorem in the present case. Now, we want to calculate the torsion of the system 16 (see [BCGGG], p. 138). With the notations of [BCGGG], we define (see 21)

c = c 1 1,2 ∂ ∂θ 1 + c 2 1,2 ∂ ∂θ 2 + 2n i=3 c i 1,2 ∂ ∂θ i ⊗ dx 1 ∧ dx 2 (58)
which is a section of the bundle I * ⊗ 2 (J/I) and has to be quotiented by Image(π) where

π : J ⊥ ⊗ J/I = Span ∂ ∂p 3 1 , ..., ∂ ∂p 2n 1 ⊗ Span(dx 1 , dx 2 ) -→ I * ∧ 2 (J/I) = Span ∂ ∂θ 1 , ..., ∂ ∂θ 2n ⊗ Span(dx 1 ∧ dx 2 ) (59) is defined by v = 2n i=3 2 k=1 v i k ∂ ∂p i 1 ⊗ dx k -→ π(v) = 2n j=2 2n i=3 A j (i,1),1 v i 2 -A j (i,1),2 v i 1 ∂ ∂θ j ⊗ dx 1 ∧ dx 2 (60)
and therefore, by 21,

π(v) = 2n i=3 γ 1 i v i 2 -β 1,i v i 1 ∂ ∂θ 1 + 2n i=3 γ 2 i v i 2 -β 2,i v i 1 ∂ ∂θ 2 + 2n j=3 v j 2 - 2n i=3 β j,i v i 1 ∂ ∂θ j ⊗ dx 1 ∧ dx 2 .
(61)

Thus, the torsion vanishes if there exists v such that π

(v) = c, i.e.      2n i=3 γ 1 i v i 2 -β 1,i v i 1 = c 1 1,2 2n i=3 γ 2 i v i 2 -β 2,i v i 1 = c 2 1,2 v j 2 -2n i=3 β j,i v i 1 = c j 1,2 ∀j = 3, ..., 2n. (62)
From the last lines, we obtain, ∀j = 3, ..., 2n,

v j 2 = c j 1,2 + 2n i=3 β j,i v i 1 , (63)
and we carry over the first lines

2n i=3 γ 1 i [c i 1,2 + 2n i ′ =3 β i,i ′ v i ′ 1 ] -β 1,i v i 1 = c 1 1,2 2n i=3 γ 2 i [c i 1,2 + 2n i ′ =3 β i,i ′ v i ′ 1 ] -β 2,i v i 1 = c 2 1,2 , (64)
or 2n i ′ =3 v i ′ 1 2n i=3 β i,i ′ γ 1 i -β 1,i ′ = c 1 1,2 -2n i=3 γ 1 i c i 1,2 2n i ′ =3 v i ′ 1 2n i=3 β i,i ′ γ 2 i -β 2,i ′ = c 2 1,2 -2n i=3 γ 2 i c i 1,2 , (65)
that is, matricialy,

D 1 v 1 = c 1 1,2 -2n i=3 γ 1 i c i 1,2 D 2 v 1 = c 2 1,2 -2n i=3 γ 2 i c i 1,2 , (66)
that is to say

ρ 2 D 0 v 1 = c 1 1,2 -2n i=3 γ 1 i c i 1,2 ρ 1 D 0 v 1 = c 2 1,2 -2n i=3 γ 2 i c i 1,2 , (67)
which implies

ρ 1 c 1 1,2 - 2n i=3 γ 1 i c i 1,2 = ρ 2 c 2 1,2 - 2n i=3 γ 2 i c i 1,2 (68)
To summarize, we have the theorem:

Theorem 3.5. We have two possibilities, the torsion can be absorbed if D 0 = 0 and if 68 is satisfied, or, when

D 0 = 0 if c 1 1,2 = 2n i=3 γ 1 i c i 1,2 c 2 1,2 = 2n i=3 γ 2 i c i 1,2 .
(69)

In the almost complex case, the previous theorem gives immediately

Corollary 3.6. The condition D 0 = 0 is satisfied in the almost complex case. So, in almost complex analysis, A (0) is involutive, and the torsion may be absorbed if 69 is verified.

In fact, in the almost complex case, as explain in the paragraph 2, we have

A = A = b(aI-A)
1+a 2 , and 48 gives the corollary.

In the complex case, we can specify a little more Corollary 3.7. In complex analysis, we have D 0 = 0 and the torsion may be absorbed if

c 1 1,2 = c 2 1,2 = 0.
In the complex case, we have f :

D -→ D holomorphic i.e. ∀i = 1, ..., n, p 2i-1 2 = -p 2i 1 and p 2i 2 = p 2i-1 1 with ρ(f (x)) = 0. Therefore, we have, for i = 1, ..., n,      α 2i-1,2i = -1 α 2i,2i-1 = 1 α j,j ′ = 0 in the other cases. (70)
Deriving ρ(f (x)) = 0 relatively to x 1 and x 2 gives

ρ 1 p 1 1 + ρ 2 p 2 1 = -n j=2 (ρ 2j-1 p 2j-1 1 + ρ 2j p 2j 1 ) ρ 2 p 1 1 -ρ 1 p 2 1 = -n j=2 (ρ 2j p 2j-1 1 -ρ 2j-1 p 2j 1 ), (71)
system whose the determinant

D = -(ρ 2 1 + ρ 2 2 ) is supposed non zero, from which    p 1 1 = Σ n j=2 p 2j-1 1 ρ 2 ρ 2j +ρ 1 ρ 2j-1 D + p 2j 1 ρ 1 ρ 2j -ρ 2 ρ 2j-1 D p 2 1 = Σ n j=2 p 2j-1 1 ρ 2 ρ 2j-1 -ρ 1 ρ 2j D + p 2j 1 ρ 1 ρ 2j-1 +ρ 2 ρ 2j D . (72) So, we have          γ 1 2j-1 = γ 2 2j = ρ 1 ρ 2j-1 +ρ 2 ρ 2j D := D 2,j D γ 1 2j = -γ 2 2j-1 = ρ 1 ρ 2j -ρ 2 ρ 2j-1 D := D 1,j D β 1,j = -γ 2 j and β 2,j = γ 1 j β 2i-1,j = -δ j 2i and β 2i,j = δ j 2i-1 . (73)
From 16 and 72, we obtain

               θ 1 = df 1 -2n j=3 γ 1 j p j 1 dx 1 + 2n j=3 γ 2 j p j 1 dx 2 θ 2 = df 2 -2n j=3 γ 2 j p j 1 dx 1 -2n j=3 γ 1 j p j 1 dx 2 ∀j = 2, ..., n θ 2j-1 = df 2j-1 -p 2j-1 1 dx 1 + p 2j 1 dx 2 θ 2j = df 2j -p 2j 1 dx 1 -p 2j-1 1 dx 2 . (74)
Consequently,

∀j = 2, ..., n, dθ 2j-1 = -dp 2j-1 1 ∧dx 1 +dp 2j 1 ∧dx 2 and dθ 2j = -dp 2j 1 ∧dx 1 -dp 2j-1 1 ∧dx 2 , ( 75
)
and so, c 2j 1,2 = c 2j-1 1,2 = 0 ∀j = 2, ..., n, and, therefore the torsion may be absorbed if

c 1 1,2 = c 2 1,2 = 0.
After this remark, we, now, continue to study the complex case. From the two first lines of 74, we have

                                                           dθ 1 = -2n j=3 γ 1 j dp j 1 ∧ dx 1 + 2n j=3 γ 2 j dp j 1 ∧ dx 2 -2n j=3 2n i=1 p j 1 ∂γ 1 j ∂f i df i ∧ dx 1 -p j 1 ∂γ 2 j ∂f i df i ∧ dx 2 ≈ -2n j=3 γ 1 j dp j 1 ∧ dx 1 + 2n j=3 γ 2 j dp j 1 ∧ dx 2 -dx 1 ∧ dx 2 2n j=3 p j 1 ∂γ 1 j ∂f 1 2n j ′ =3 γ 2 j ′ p j ′ 1 - ∂γ 1 j ∂f 2 2n j ′ =3 γ 1 j ′ p j ′ 1 + n i=2 ∂γ 1 j ∂f 2i-1 p 2i 1 - ∂γ 1 j ∂f 2i p 3i-1 1 - ∂γ 2 j ∂f 1 2n j ′ =3 γ 1 j ′ p j ′ 1 - ∂γ 2 j ∂f 2 2n j ′ =3 γ 2 j ′ p j ′ 1 -2n j ′ =3 ∂γ 2 j ∂f j ′ p j ′ 1 dθ 2 = -2n j=3 γ 2 j dp j 1 ∧ dx 1 -2n j=3 γ 1 j dp j 1 ∧ dx 2 -2n j=3 2n i=1 p j 1 ∂γ 2 j ∂f i df i ∧ dx 1 + p j 1 ∂γ 1 j ∂f i df i ∧ dx 2 ≈ -2n j=3 γ 2 j dp j 1 ∧ dx 1 -2n j=3 γ 1 j dp j 1 ∧ dx 2 -dx 1 ∧ dx 2 2n j=3 p j 1 ∂γ 2 j ∂f 1 2n j ′ =3 γ 2 j ′ p j ′ 1 - ∂γ 2 j ∂f 2 2n j ′ =3 γ 1 j ′ p j ′ 1 + n i=2 ∂γ 2 j ∂f 2i-1 p 2i 1 - ∂γ 2 j ∂f 2i p 3i-1 1 + ∂γ 1 j ∂f 1 2n j ′ =3 γ 1 j ′ p j ′ 1 + ∂γ 1 j ∂f 2 2n j ′ =3 γ 2 j ′ p j ′ 1 + 2n j ′ =3 ∂γ 1 j ∂f j ′ p j ′ 1 (76)
and, by a long but easy calculation, we obtain

c 1 1,2 = n j,j ′ =2 [p 2j-1 1 p 2j ′ -1 1 + p 2j 1 p 2j ′ 1 ][P 2 j ′ (γ 2 2j ) -P 1 j ′ (γ 1 2j )] + [p 2j 1 p 2j ′ -1 1 -p 2j-1 1 p 2j ′ 1 ][P 2 j ′ (γ 1 2j ) + P 1 j ′ (γ 2 2j )] (77)
and

c 2 1,2 = n j,j ′ =2 -[p 2j-1 1 p 2j ′ -1 1 + p 2j 1 p 2j ′ 1 ][P 2 j ′ (γ 1 2j ) + P 1 j ′ (γ 2 2j )] + [p 2j 1 p 2j ′ -1 1 -p 2j-1 1 p 2j ′ 1 ][P 2 j ′ (γ 2 2j ) -P 1 j ′ (γ 1 2j )] (78)
where, for all k = 2, ..., n, we have introduced the differential polynomials

P 1 k = γ 2 2k ∂ ∂y 1 -γ 1 2k ∂ ∂y 2 + ∂ ∂y 2k-1 P 2 k = γ 1 2k ∂ ∂y 1 + γ 2 2k ∂ ∂y 2 + ∂ ∂y 2k . (79)
Thus, if we note, for j, k = 2, ..., n,

B j,k = P 2 k (γ 1 2j ) + P 1 k (γ 2 2j) B j,k = P 2 k (γ 2 2j ) -P 1 k (γ 1 2j ), (80)
then

c 1 1,2 = 2n[ j,j ′ =2 [p 2j-1 1 p 2j ′ -1 1 + p 2j 1 p 2j ′ 1 ]B j,j ′ + [p 2j 1 p 2j ′ -1 1 -p 2j-1 1 p 2j ′ 1 ]B j,j ′ c 2 1,2 = 2n[ j,j ′ =2 -[p 2j-1 1 p 2j ′ -1 1 + p 2j 1 p 2j ′ 1 ]B j,j ′ + [p 2j 1 p 2j ′ -1 1 -p 2j-1 1 p 2j ′ 1 ]B j,j ′ . (81)
We have, here, two quadratic polynomials in p j 1 which have to be 0 in order that the torsion vanishes. If one of the two polynomials is non degenerated positive or non degenerated negative, then the only solution is p j 1 = 0. Therefore, the only holomorphic discs in H are points. We have no proper holomorphic disc in H.

To have really an holomorphic disc in H, we need, for our quadratic polynomials, non zero solutions.

Example 3.8. The complex case in dimension 6.

We have, with the previous notations,

c 1 1,2 = B 2,2 (p 3 1 ) 2 + (p 4 1 ) 2 + B 3,3 (p 5 1 ) 2 + (p 6 1 ) 2 + p 3 1 p 5 1 + p 4 1 p 6 1 B 2,3 + B 3,2 + p 4 1 p 5 1 -p 3 1 p 6 1 B 2,3 -B 3,2 = B 2,2 p 3 1 + B 2,3 + B 3,2 2B 2,2 p 5 1 - B 2,3 -B 3,2 2B 2,2 p 6 1 2 + p 4 1 + B 2,3 + B 3,2 2B 2,2 p 6 1 + B 2,3 -B 3,2 2B 2,2 p 5 1 2 + 4B 2,2 B 3,3 -(B 2,3 -B 3,2 ) 2 -(B 2,3 + B 3,2 ) 2 4B 2,2 (p 5 1 ) 2 + (p 6 1 ) 2 c 2 1,2 = -B 2,2 (p 3 1 ) 2 + (p 4 1 ) 2 -B 3,3 (p 5 1 ) 2 + (p 6 1 ) 2 -p 3 1 p 5 1 + p 4 1 p 6 1 B 2,3 + B 3,2 + p 4 1 p 5 1 -p 3 1 p 6 1 B 2,3 -B 3,2 = -B 2,2 p 3 1 + B 2,3 + B 3,2 2B 2,2 p 5 1 + B 2,3 -B 3,2 2B 2,2 p 6 1 2 + p 4 1 + B 2,3 + B 3,2 2B 2,2 p 6 1 - B 2,3 -B 3,2 2B 2,2 p 5 1 2 + -4B 2,2 B 3,3 + (B 2,3 + B 3,2 ) 2 + (B 2,3 -B 3,2 ) 2 4B 2,2 (p 5 1 ) 2 + (p 6 1 ) 2 . (82)
Now, the existence of a non trivial (non constant) solution for 8 implies that the torsion c 1 1,2 = c 2 1,2 = 0 with some p i j = 0. So, the quadratic forms c 2 1,2 andc 2 1,2 , quadratic forms in p i j , are not definite positive (or negative). Therefore

4B 2,2 B 3,3 -(B 2,3 + B 3,2 ) 2 -(B 2,3 -B 3,2 ) 2 ≤ 0 4B 2,2 B 3,3 -(B 2,3 + B 3,2 ) 2 -(B 2,3 -B 3,2 ) 2 ≤ 0. ( 83
)
An other method would be to look for an orthogonal basis for the quadratic form c 2 1,2 . This, of course, gives the same result.

Example 3.9. The pseudo-ellipsoïds

To end with the complex case, we give an example : suppose now we have for ρ the polynomial

ρ(y) = α 1 y 2k 1 1 + α 2 y 2k 2 2 + α 3 y 2k 3 3 + α 4 y 2k 4 4 + α 5 y 2k 5 5 + α 6 y 2k 6 6 , (84)
with α 1 , ..., α 6 ∈ R, and k 1 , ..., k 6 ∈ N.

We note

v i = ρ i = 2α i k i y 2k i -1 i , i = 1, ..., 6
, and also

w i = ∂v i ∂y i = ∂ 2 ρ ∂ 2 y i = 2k i (2k i -1)α i y 2k i -2 i ,
and we want to explicit the coefficients B j,k and B j,k of 83.

We have, with the previous definitions applied to this example,

D = -(v 2 1 + v 2 2 ), γ 1 2j-1 = γ 2 2j = ρ 1 ρ 2j-1 +ρ 2 ρ 2j D = D 2,j D , γ 1 2j = -γ 2 2j-1 = ρ 1 ρ 2j -ρ 2 ρ 2j-1 D = D 1,j D and B 2,3 = P 2 3 (γ 1 4 ) + P 1 3 (γ 2 4 ) = 1 D 2 γ 1 6 ∂D 1,2 ∂y 1 D -D 1,2 ∂D ∂y 1 + γ 2 6 ∂D 1,2 ∂y 2 D -D 1,2 ∂D ∂y 2 + ∂D 1,2 ∂y 6 D -D 1,2 ∂D ∂y 6 + γ 2 6 ∂D 2,2 ∂y 1 D -D 2,2 ∂D ∂y 1 -γ 1 6 ∂D 2,2 ∂y 2 D -D 2,2 ∂D ∂y 2 + ∂D 2,2 ∂y 5 D -D 1,2 ∂D ∂y 5 = 1 D 3 {(v 1 v 6 -v 2 v 5 )[-v 4 w 1 (v 2 1 + v 2 2 ) + (v 1 v 4 -v 2 v 3 )2v 1 w 1 ] + (v 1 v 5 + v 2 v 6 )[v 3 w 1 (v 2 2 + v 2 2 ) + (v 1 v 4 -v 2 v 3 )2v 2 w 2 ] + (v 1 v 5 + v 2 v 6 )[-v 3 w 1 (v 2 1 + v 2 1 ) + (v 1 v 3 + v 2 v 4 )2v 1 w 1 ] -(v 1 v 6 -v 2 v 5 )[-v 4 w 2 (v 2 1 + v 2 1 ) + (v 1 v 3 + v 2 v 4 )2v 2 w 2 ]} = 1 D 3 {w 1 [(v 2 1 + v 2 2 ) -v 4 (v 1 v 6 -v 2 v 5 ) -v 3 (v 1 v 5 + v 2 v 6 ) + 2v 1 (v 1 v 6 -v 2 v 5 )(v 1 v 4 -v 2 v 3 ) + (v 1 v 5 + v 2 v 6 )(v 1 v 3 + v 2 v 4 ) ] + w 2 [(v 2 1 + v 2 2 ) v 3 (v 1 v 5 + v 2 v 6 ) + v 4 (v 1 v 6 -v 2 v 5 ) + 2v 2 (v 1 v 5 + v 2 v 6 )(v 1 v 4 -v 2 v 3 ) + (v 1 v 6 -v 2 v 5 )(v 1 v 3 + v 2 v 4 ) ] = 1 D 2 [v 2 (v 3 v 6 -v 4 v 5 ) -v 1 (v 3 v 5 + v 4 v 6 )][w 1 + w 2 ]. ( 85
)
Analogous calculations give

B 3,2 = 1 D 2 [-v 1 (v 3 v 5 + v 4 v 6 ) -v 2 (v 3 v 6 -v 4 v 5 )][w 1 + w 2 ] B 2,3 = 1 D 2 [-v 1 (v 3 v 6 -v 4 v 5 ) -v 2 (v 3 v 5 + v 4 v 6 )][w 1 + w 2 ] B 3,2 = 1 D 2 [v 1 (v 3 v 6 -v 4 v 5 ) -v 2 (v 3 v 5 + v 4 v 6 )][w 1 + w 2 ] B 2,2 = -v 1 D 2 [(v 2 3 + v 2 4 )(w 1 + w 2 ) + (v 2 1 + v 2 2 )(w 3 + w 4 )] B 3,3 = -v 1 D 2 [(v 2 5 + v 2 6 )(w 1 + w 2 ) + (v 2 1 + v 2 2 )(w 5 + w 6 )] B 2,2 = -v 2 D 2 [(v 2 3 + v 2 4 )(w 1 + w 2 ) + (v 2 1 + v 2 2 )(w 3 + w 4 )] B 3,3 = -v 2 D 2 [(v 2 5 + v 2 6 )(w 1 + w 2 ) + (v 2 1 + v 2 2 )(w 5 + w 6 )] (86)
So, the first equation of 83 becomes

4v 2 1 (v 2 1 + v 2 2 ) D 4 [(v 2 3 + v 2 4 )(w 1 + w 2 )(w 5 + w 6) + (v 2 5 + v 2 6 )(w 1 + w 2 )(w 3 + w 4) + (v 2 1 + v 2 2 )(w 3 + w 4 )(w 5 + w 6) ] ≤ 0, (87)
and the second line

4v 2 2 (v 2 1 + v 2 2 ) D 4 [(v 2 3 + v 2 4 )(w 1 + w 2 )(w 5 + w 6) + (v 2 5 + v 2 6 )(w 1 + w 2 )(w 3 + w 4) + (v 2 1 + v 2 2 )(w 3 + w 4 )(w 5 + w 6) ] ≤ 0, (88)
and, in fact, we have only one condition

(v 2 1 + v 2 2 )(w 3 + w 4 )(w 5 + w 6 ) + (v 2 3 + v 2 4 )(w 1 + w 2 )(w 5 + w 6 ) + (v 2 5 + v 2 6 )(w 1 + w 2 )(w 3 + w 4 ) 0, (89)
or, more explicitly,

[α 2 1 k 2 1 y 2k 1 -2 1 + α 2 2 k 2 2 y 2k 2 -2 ][α 3 k 3 (2k 3 -1)y 2k 3 -2 3 + α 4 k 4 (2k 4 -1)y 2k 4 -2 4 ][α 5 k 5 (2k 5 -1)y 2k 5 -2 5 + α 6 k 6 (2k 6 -1)y 2k 6 -2 6 ] + [α 2 3 k 2 3 y 2k 3 -2 3 + α 2 4 k 2 4 y 2k 4 -2 4 ][α 1 k 1 (2k 1 -1)y 2k 1 -2 1 + α 2 k 2 (2k 2 -1)y 2k 2 -2 2 ][α 5 k 5 (2k 5 -1)y 2k 5 -2 5 + α 6 k 6 (2k 6 -1)y 2k 6 -2 6 ] + [α 2 5 k 2 5 y 2k 5 -2 5 + α 2 6 k 2 6 y 2k 6 -2 ][α 1 k 1 (2k 1 -1)y 2k 1 -2 1 + α 2 k 2 (2k 4 -1)y 2k 4 -2 4 ][α 3 k 3 (2k 3 -1)y 2k 3 -2 3 + α 4 k 4 (2k 4 -1)y 2k 4 -2 4 ] ≤ 0. ( 90
)
This inequality express 83. But, in 83, B j,k and B j,k are functions of (x 1 , x 2 ) defined on D ⊂ R 2 .

In particular,

y i = y i (f (x)) = f i (x). First case If (y 1 , y 2 ) = (f 1 (x), f 2 (x)) and (f 3 (x), f 4 (x)
) and (f 5 (x), f 6 (x)) are not identically zero on D, (we remark, by holomorphy, if, for example, f 3 (x) no zero on D, f 4 (x) has the same property) then the condition 90 implies that w 1 + w 2 , w 3 + w 4 , w 5 + w 6 are not all strictly positive or strictly negative. And, therefore, α 1 , α 2 , α 3 , α 4 , α 5 , and α 6 are not all strictly positive or strictly negative. So, the hypersurface ρ = 0 is not compact, as the theorem of Diederich and Fornaess says. Nevertheless, the necessary condition 89 is strictly stronger than the Diederich-Fornaess condition (that is to say {ρ = 0} no compact).

Second case If (f 1 (x), f 2 (x)) = 0 on D, or (f 3 (x), f 4 (x)) = 0 on D, or (f 5 (x), f 6 (x)) = 0 on D, for example (f 5 (x), f 6 (x)) = 0 on D, then f is, in fact, an holomorphic function from (an open set in) R 2 to R 4 .
We can resume the preceding calculations in this case, or, more simply, consider the quadratic forms 81 when p 5 1 = p 6 = 0. Then, the quadratic forms 81 are only

-B 2,2 [(p 3 1 ) 2 + (p 4 1 ) 2 ] and B 2,2 [(p 3 1 ) 2 + (p 4 1 ) 2 ] ( 91
)
which have to be neither positive definite nor negative definite. Therefore,

B 2,2 = B 2,2 = 0, (92)
which gives, from 86,

v 1 [(v 2 3 + v 2 4 )(w 1 + w 2 ) + (v 2 1 + v 2 2 )(w 3 + w 4 )] = 0 v 2 [(v 2 3 + v 2 4 )(w 1 + w 2 ) + (v 2 1 + v 2 2 )(w 5 + w 6 )] = 0. ( 93
)
If v 1 = 0 or v 2 = 0, for example v 1 = 0, then α 1 = 0, (so H = {ρ = 0} is not compact) or

y 1 = f 1 (x) = 0. If y 1 = f 1 (x) = 0, then y 2 = f 2 (x) = C a constant, so (f 3 , f 4 ) is an holomorphic function from D ⊂ R 2 to H ∩ {y 1 = 0} ∩ {y 2 = C} which is of dimension 1 : impossible (except if (f 3 , f 4 ) constant).
We have, of course, similar results if f 3 (x) = 0 or f 5 (x) = 0. Therefore, we suppose v 1 = 0 and the first line in 93 becomes

(v 2 3 + v 2 4 )(w 1 + w 2 ) + (v 2 1 + v 2 2 )(w 3 + w 4 ) = 0, (94) that is to say [2α 1 k 1 y 2k 1 -2 1 (2k 1 -1) + 2α 2 k 2 y 2k 2 -2 2 (2k 2 -1)](v 2 3 + v 2 4 ) + [2α 3 k 3 y 2k 3 -2 3 (2k 3 -1) + 2α 4 k 4 y 2k 4 -2 4 (2k 4 -1)](v 2 1 + v 2 2 ) = 0 (95) whereas v 2 1 + v 2 2 > 0 and v 2 3 + v 2 4 > 0. So, 2α 1 k 1 y 2k 1 -2 1 (2k 1 -1) + 2α 2 k 2 y 2k 2 -2 2 (2k 2 -1) v 2 1 + v 2 2 = - 2α 3 k 3 y 2k 3 -2 3 (2k 3 -1) + 2α 4 k 4 y 2k 4 -2 4 (2k 4 -1) v 2 3 + v 2 4 , (96)
so, α 1 , α 2 , α 3 , α 4 , are not all strictly positive or all strictly negative. Consequently, H = {ρ = 0} is not compact, as Diederich and Fornaess say.

3.1. The successive torsion. In this paragraph, we attempt to compute the successive torsions when A (0) is not in involution. Recall that we have the following alternative : D 0 = 0 or D 0 = 0 (see 66).

First case : D 0 = 0. Then, A (0) is involutive, and, if there is a solution to 8, the torsion at the order 1 vanishes and therefore

c 1 1,2 -2n i=3 γ 1 i c i 1,2 = 0 c 2 1,2 -2n i=3 γ 2 i c i 1,2 = 0. (97)
A fundamental result ensures that all the successive torsions are zero for all the orders higher than 1 if we have 97.

Second case : D 0 = 0. Then A (0) is not involutive, and if q 0 = Rank(D 0 , D 0 β, ..., D 0 β 2n-3 ), then A (q) is non involutive if q < q 0 , and A (q) is involutive if q ≥ q 0 . So (see [BCGGG], p. 333), the torsion of the q-prolongation of 8 is vanishing if q > q 0 . As necessary conditions to have a solution of 8, we first have 68,

ρ 1 c 1 1,2 - 2n i=3 γ 1 i c i 1,2 = ρ 2 c 2 1,2 - 2n i=3 γ 2 i c i 1,2 (98)
and, also, we write the nullity of the torsion of the q-prolongation of 8, with q ≤ q 0 . Of course, this is not easy to explicit in the general case, but we can precise this. We are first looking for the torsion of the first prolongation.

From

p 2 = Ap 1 , we deduce      p 1,2 = ∂A ∂x 1 p 1 + Ap 1,1 p 2,2 = ∂A ∂x 2 p 1 + Ap 1,2 = ∂A ∂x 2 p 1 + A ∂A ∂x 1 p 1 + A 2 p 1,1 . (99)
Deriving, in x 1 and x 2 , for the second time, the first line of 10, that is to say, deriving 11, we obtain

     2n i,j=1 ρ i,j p 1 1 p j 1 + 2n i=1 ρ i p i 1,1 = 0 2n i,j=1 ρ i,j p 1 1 p j 2 + 2n i=1 ρ i p i 1,2 = 0 2n i,j=1 ρ i,j p 1 2 p j 2 + 2n i=1 ρ i p i 2,2 = 0, (100)
and, using 99, expressing all the derivatives of f as functions of the derivatives in x 1 , we have

           2n i=1 ρ i p i 1,1 + 2n i,j=1 ρ i,j p i 1 p j 1 = 0 2n i=1 ρ i 2n k=1 ∂α i,k ∂x 1 p k 1 + α i,k p k 1,1 + 2n i,j=1 ρ i,j p i 1 α j,k p k 1 = 0 2n i,j,k,l=1 ρ i,j α i,k p k 1 α j,l p l 1 + 2n i=1 ρ i 2n j=1 ∂α i,j ∂x 2 p i 1 + 2n j,k=1 α i,j ∂α j,k ∂x 1 p k 1 + 2n j,k=1 α i,j α j,k p k 1,1 = 0. ( 101
)
Using 12, we obtain the system

     p 1 1,1 ρ 1 + p 2 1,1 ρ 2 + p 3 1,1 ρ 3 = h 1 (p 3 1 , ..., p 2n 1 , p 4 1,1 , ..., p 2n 1,1 ) p 1 1,1 2n
i=1 ρ i α i,1 + p 2 1,1 2n
i=1 ρ i α i,2 + p 3 1,1 2n
i=1 ρ i α i,3 = h 2 (p 3 1 , ..., p 2n 1 , p 4 1,1 , ..., p 2n 1,1 ) p 1 1,1 2n i,j=1 ρ i α i,j α j,1 + p 2 1,1 2n i,j=1 ρ i α i,j α j,2 + p 3 1,1 2n i,j=1 ρ i α i,j α j,3 = h 3 (p 3 1 , ...,
     p 1 1,1 = 2n k=4 γ 1,1,1 k,1,1 p k 1,1 + 2n k=3 γ 1,1,1 k,1 p k 1 + 2n i,k=4 γ 1,1,1 i1,k1 p i 1 p k 1 p 2 1,1 = 2n k=4 γ 2,1,1 k,1,1 p k 1,1 + 2n k=3 γ 2,1,1 k,1 p k 1 + 2n i,k=4 γ 2,1,1 i1,k1 p i 1 p k 1 p 3 1,1 = 2n k=4 γ 3,1,1 k,1,1 p k 1,1 + 2n k=3 γ 3,1,1 k,1 p k 1 + 2n i,k=4 γ 3,1,1 i1,k1 p i 1 p k 1 . (103)
For the initial system 16, we had the structure equations

     θ 1 = df 1 -2n j=3 γ 1 j p j 1 dx 1 -2n j=3 β 1,j p j 1 dx 2 θ 2 = df 2 -2n j=3 γ 2 j p j 1 dx 1 -2n j=3 β 2,j p j 1 dx 2 θ i = df i -p i 1 dx 1 -2n j=3 β i,j p j 1 dx 2 ∀i = 3, ..., 2n, (104)
on the space of variables M = x 1 , x 2 , f 1 , ..., f 2n , p 3 1 , ..., p 2n 1 . For the first prolongation, we now have the new structure equations

     θ 3,1 = dp 3 1 - 2n k=4 γ 3,1,1 k,1,1 p k 1,1 + 2n k=3 γ 3,1,1 k,1 p k 1 + 2n j,k=3 p j 1 p k 1 γ 3,1,1 j1,k1 dx 1 - 2n k=1 ∂α 3,k ∂x 1 p k 1 + α 3,k p k 1,1 dx 2 θ i,1 = dp i 1 -p i 1,1 dx 1 -2n k=1 ∂α i,k ∂x 1 p k 1 + α i,k p k 1,1 dx 2 ∀i = 4, ..., 2n, (105)
on the space of variables

M 1 = x 1 , x 2 , f 1 , ..., f 2n , p 3 1 , ..., p 2n 1 , p 4 1,1 , p 5 1,1 , ..., p 2n 1,1 .
In fact, in these structure equations, we should have to express p 1 1 , p 2 1 , p 1 1,1 , p 2 1,1 , p 3 1,1 in function of the variables in M 1 , as explain before. From 105, we deduce

           dθ 3,1 = 2n k=4 A 3,1 (k,1,1),1 dp k 1,1 ∧ dx 1 + 2n k=4 A 3,1 (k,1,1),2 dp k 1,1 ∧ dx 2 + 2n j=3 A 3,1 (j,1),1 dp j 1 ∧ dx 1 + 2n j=3 A 3,1 (j,1),2 dp j 1 ∧ dx 2 + c 3,1 1,2 dx 1 ∧ dx 2 ∀i = 4, ..., 2n, dθ i,1 = -dp i 1,1 ∧ dx 1 + 2n k=4 A i,1 (k,1,1),2 dp k 1,1 ∧ dx 2 + 2n j=3 A i,1 (j,1),2 dp j 1 ∧ dx 2 + c i,1 1,2 dx 1 ∧ dx 2 , ( 106
)
where we do not explicit the coefficients A • • which are defined in accordance with the notations of [BCGGG], p. 130. The outstanding point is

A i,1 (k,1,1),1 = -δ i k (107)
so (see [BCGGG], p. 138), in the calculation of

π : Span ∂ ∂p 3 1 , ..., ∂ ∂p 2n 1 , ∂ ∂p 4 1,1 , ..., ∂ ∂p 2n 1,1 ⊗ Span(x 1 , x 2 ) -→ Span ∂ ∂θ 1 , ..., ∂ ∂θ 2n , ∂ ∂θ 3,1 , ..., ∂ ∂θ 2n,1 ⊗ Span(dx 1 ∧ dx 2 ), (108) we have, if v = 2 i=1 2n j=3 v j,1 i ∂ ∂p j 1 + 2n k=4 v k,1,1 i ∂ ∂p k 1,1 ⊗ x i , π(v) = 2n i=1 2n j=3 A i (j,1),1 v j,1 2 -A i (j,1),2 v j,1 1 ∂ ∂θ i + 2n j=3 A 3,1 (j,1),1 v j,1 2 -A 3,1 (j,1),2 v j,1 1 + 2n k=4 A 3,1 (k,1,1),1 v k,1,1 2 -A 3,1 (k,1,1),2 v k,1,1 1 ∂ ∂θ 3,1 + 2n i=4 2n j=3 A i,1 (j,1),1 v j,1 2 -A i,1 (j,1),2 v j,1 1 + v i,1,1 2 - 2n k=4 A i,1 (k,1,1),2 v k,1,1 1 ∂ ∂θ i,1 . (109)
The torsion vanishes if there exists v such that

π(v) = c = 2n i=1 c i 1,2 ∂ ∂θ i + 2n j=3 c j,1 1,2 ∂ ∂θ j,1 , (110)
which gives

     2n j=3 A i (j,1),1 v j,1 2 -A i (j,1),2 v j,1 1 = c i 1,2 ∀i = 1, ..., 2n 2n j=3 A 3,1 (j,1),1 v j,1 2 -A 3,1 (j,1),2 v j,1 1 + 2n k=4 A 3,1 (k,1,1),1 v k,1,1 2 -A 3,1 (k,1,1),2 v k,1,1 1 = c 3,1 1,2 2n j=3 A i,1 (j,1),1 v j,1 2 -A i,1 (j,1),2 v j,1 1 + v i,1,1 2 -2n k=4 A i,1 (k,1,1),2 v k,1,1 1 = c i,1 1,2 ∀i = 4, ..., 2n. (111)
In this system, the 2n first lines are exactly 62 and give the torsion of the initial system 16. The 2n -3 last lines give ∀i = 4, ..., 2n,

v i,1,1 2 = c i,1 1,2 + 2n k=4 A i,1 (k,1,1),2 v k,1,1 1 + 2n j=3 A i,1 (j,1),1 v j,1 2 -A i,1 (j,1),2 v j,1 1 (112)
and, transferring to the line 2n + 1, we obtain

2n k ′ =4 v k ′ ,1,1 1 2n k=4 A 3,1 (k,1,1),1 A k,1 (k ′ ,1,1),2 -A 3,1 (k ′ ,1,1),2 = c 3,1 1,2 - 2n k=4 c k,1 1,2 A 3,1 (k,1,1),1 - 2n j=3 v j,1 2 A 3,1 (j,1),1 + 2n k=4 A 3,1 (k,1,1),1 A k,1 (j,1),1 -v j,1 1 A 3,1 (j,1),2 + 2n k=4 A 3,1 (k,1,1),1 A k,1 (j,1),2 . (113)
We obtain only one condition, and, if there exist k ′ = 4, ..., 2n such that 2n k=4 A 3,1 (k,1,1),1 A k,1 (k ′ ,1,1),2 -A 3,1 (k ′ ,1,1),2 = 0, then the torsion may be absorbed. This result is true for all the prolongations. We note [k 1 , k 2 ] the multi-index containing k 1 times the number 1, and k 2 times the number 2. If we have the prolongation of order k 0 , of the initial system, we have the space of variables

M k 0 = {x 1 , x 2 , f 1 , ..., f 2n , p 3 1 , ..., p 2n 1 , p 4 1,1 , ..., p 2n 1,1 , p 5 [3,0] , ..., p 2n [3,0] , ..., p k 0 +3 [k 0 +1],0] , ..., p 2n [k 0 +1,0] }
and all the p i J with J of length | J |≤ k 0 + 1 are expressed as functions of the variables p in M k 0 . Now, we are looking for the prolongation of order k 0 + 1 of the initial system 16, and we carry out the same calculation as, before, for the first prolongation. We have derived ρ • f (x) = 0 k 0 + 1 times and obtained ∀j = 0, ..., k 0 + 1

2n i=1 ρ i p i [j,k 0 +1-j] = g j (p 1 , p 1,1 , ..., p [k 0 ,0] ). ( 114
)
Deriving this one time more in x 1 and x 2 , we obtain a system of k 0 + 3 equations

2n i=1 ρ i p i [k 0 +2,0] = g k 0 +2 (p 1 , ..., p [k 0 +1,0] ) 2n i=1 ρ i p i [j,k 0 +2-j] = g j (p 1 , ..., p [k 0 +1,0] ) ∀j = 0, ..., k 0 + 1, (115)
and, if h j ( p) refers to a function of the variables p belonging to M k 0 , this can be written

           p 1 [k 0 +2,0] ρ 1 + p 2 [k 0 +2,0] ρ 2 + ... + p k 0 +3 [k 0 +2,0] ρ k 0 +3 = h 0 ( p) p 1 [k 0 +2,0] µ 1 1 + p 2 [k 0 +2,0] µ 1 2 + ... + p k 0 +3 [k 0 +2,0] µ 1 k 0 +3 = h 1 ( p) ................... p 1 [k 0 +2,0] µ k 0 +2 1 + p 2 [k 0 +2,0] µ k 0 +2 2 + ... + p k 0 +3 [k 0 +2,0] µ k 0 +3 k 0 +3 = h k 0 +2 ( p) (116)
where µ i ′ j ′ are coefficients depending on ρ i and α i,j , we do not explicit. When the determinant of this system of k 0 + 3 equations is not zero, we can solve and express p 1

[k 0 +2,0] , ..., p k 0 +3 [k 0 +2,0]
as functions of the variables p in M k 0 . For the prolongation of order k 0 + 1 of the initial system 16, we have a new space of variables M k 0 +1 which contains all the variables in M k 0 plus the variables ,0] , and we have new structure equations relative to

p k 0 +4 [k 0 +2,0] , p k 0 +5 [k 0 +2,0] , ..., p 2n [k 0 +2
θ k 0 +2,[k 0 +1,0] , ...θ 2n,[k 0 +1,0] , θ k 0 +3,[k 0 +1,0] = dp k 0 +3 [k 0 +1,0] -p k 0 +3 [k 0 +2,0] dx 1 -p k 0 +3 [k 0 +1,1] dx 2 θ k 0 +j,[k 0 +1,0] = dp k 0 +j [k 0 +1,0] -p k 0 +j [k 0 +2,0] dx 1 -p k 0 +j [k 0 +1,1] dx 2 if j = 4, ..., 2n -k 0 . (117)
In fact, ,1] has to be expressed as a function of the variables p in M k 0 +1 , and ,0] is given by the Cramer system before. But the important point is that, if j = 4, ..., 2nk 0 ,

p k 0 +j [k 0 +1
p k 0 +3 [k 0 +2
dθ k 0 +j,[k 0 +1,0] = -dp k 0 +j [k 0 +2,0] ∧ dx 1 + ... (118)
and, therefore,

A k 0 +j,[k 0 +1,0] (k 0 +i,[k 0 +2,0]
),1 = -δ i j . This is the same result as 107 for the first prolongation. The calculations to obtain π are similar to the case of the first prolongation.

Then, we write that the torsion vanishes if there exist v such that ,0] (119) which gives

π(v) = c = 2n i=1 c i 1,2 ∂ ∂θ i + 2n j=3 c j,1 1,2 ∂ ∂θ j,1 + ... + 2n k=k 0 +3 c k,[k 0 +1,0] 1,2 ∂ ∂θ k,[k 0 +1
c k 0 +3,[k 0 +1,0] 1,2 = .... c k 0 +j,[k 0 +1,0] 1,2 = v k 0 +j,[k 0 +2,0] 2 -2n k=k 0 +4 A k 0 +j,[k 0 +1,0] k,[(k 0 +2,0]),2 v k 0 +2,0] 1 ∀j = 4, ..., 2n -k 0 . ( 120
)
From this, if j = 4, ..., 2nk 0 ,, then

v k 0 +j,[k 0 +2,0] 2 = c k 0 +j,[k 0 +1,0] 1,2 + 2n k=k 0 +4 A k 0 +j,[k 0 +1,0] k,[(k 0 +2,0]),2 v k 0 +2,0] 1 (121)
and, transferring these values to the first line of 120, we obtain only one condition. This condition is satisfied if one of the coefficients of v k,[k 0 +2,0] 1 is different of zero, and, then, the torsion may be absorbed. This situation is exactly similar with the case of the first prolongation. Remember, in all the cases, that the torsion vanishes for a prolongation of order q if q > q 0 = Rank(D, Dβ, ..., Dβ 2n-3 ) because A (q 0 ) is then involutive.

Sufficient conditions

A system of PDE is said involutive if there exists an ordinary integral element (see [BCGGG], p. 107). This notion is very important because the Cartan-Kähler theorem, in its useful version (see [BCGGG], p. 86) says that the existence of an ordinary integral element in a point implies the existence of an integral manifold tangent, at this point, to the ordinary integral element. So, a sufficient condition to have a solution of 8 is to have an ordinary integral element. The exterior differential system I is generated as a differential ideal by the sections of the subbundle I = Span(θ 1 , ..., θ 2n ) ⊂ T ⋆ M. T * M = Span(dx 1 , dx 2 , θ 1 , ..., θ 2n , dp 3 1 , ..., dp 2n 1 ) and we cosider also the dual basis of

T M T M = Span( ∂ ∂x 1 , ∂ ∂x 2 , ∂ ∂θ 1 , ..., ∂ ∂θ 2n , ∂ ∂p 3 1 , ..., ∂ ∂p 2n 1
). We are looking for an almost holomorphic curve verifying 8, so we search for a ordinary integral element E ⊂ T m M with m ∈ M and dim(E) = 2. So, let ( e 1 , e 2 ) be a basis of E, with

e 1 = 2 i=1 a 1 i ∂ ∂x i + 2n i ′ =1 b 1 i ′ ∂ ∂θ i ′ + 2n i ′′ =3 c 1 i ′′ ∂ ∂p i ′′ 1 e 2 = 2 i=1 a 2 i ∂ ∂x i + 2n i ′ =1 b 2 i ′ ∂ ∂θ i ′ + 2n i ′′ =3 c 2 i ′′ ∂ ∂p i ′′ 1 . (122)
First E has to be an integral element (see [BCGGG], p. 65), that is ∀ϕ ∈ I, ϕ/E = 0 i.e.

∀i ′ = 2, ..., 2n, θ i ′ ( e 1 ) = θ i ′ ( e 2 ) = 0, or b 1 i ′ = b 2 i ′ = 0. Thus, e 1 = 2 i=1 a 1 i ∂ ∂x i + 2n i ′′ =3 c 1 i ′′ ∂ ∂p i ′′ 1 e 2 = 2 i=1 a 2 i ∂ ∂x i + 2n i ′′ =3 c 2 i ′′ ∂ ∂p i ′′ 1 . (123)
But, E has to be an ordinary integral element (see [BCGGG], p. 73). m is an ordinary zero of I ∩ Ω 0 (M ) = ∅, and we need an integral flag 0 ⊂ E 1 ⊂ E ⊂ T m M with E 1 of dimension 1 and Kähler regular (see [BCGGG] p. 68).

Let α e 1 + β e 2 = 2 i=1 (αa

1 i + βa 2 i ) ∂ ∂x i + 2n i ′′ =3 (αc 1 i ′′ + βc 2 i ′′ ) ∂ ∂p i ′′ 1 a basis of E 1 . We are looking for a 1-form Ω 1 = γ 1 dx 1 + γ 2 dx 2 + γ 3 θ 2 + ... + γ 2n+1 θ 2n + γ 2n+2 dp 3 1 + ... + γ 4n-1 dp 2n 1 such that Ω 1 /E 1 = 0 and E 1 is an ordinary zero of F Ω 1 (I) (see [BCGGG] p. 64). First, Ω 1 /E 1 = 2 i=1 γ i (αa 1 i + βa 2 i ) + 2n i ′′ =3 γ 2n-1+i ′′ (αc 1 i ′′ + βc 2 i ′′ ) = 0 gives α 2 i=1 γ i a 1 i + 2n i ′′ =3 γ 2n-1+i ′′ c 1 i ′′ + β 2 i=1 γ i a 2 i + 2n i ′′ =3 γ 2n-1+i ′′ c 2 i ′′ = 0. ( 124
)
We now have to define

F Ω 1 (I). If D = Span a 1 ∂ ∂x 1 + a 2 ∂ ∂x 2 + a 3 ∂ ∂θ 2 + ... + a 2n+1 ∂ ∂θ 2n + a 2n+2 ∂ ∂p 3 1 + ... + a 4n-1 ∂ ∂p 2n 1 is a straight line in T M, such that Ω 1 /E 1 = 0, we write, for ϕ = 2n i ′ =2 α ′ i ′ θ i ′ ∈ I ∩ Ω 1 (M ), ϕ/D = ϕ Ω 1 (D)Ω 1 .
This makes sense as explained in [BCGGG] p. 68. This last equality gives α ′ 2 a 3 + ... + α ′ 2n a 2n+1 = ϕ Ω 1 (D)(γ 1 a 1 + ... + γ 4n-1 a 4n-1 ), i.e.

ϕ Ω 1 (D) = α ′ 2 a 3 + ... + α ′ 2n a 2n+1 γ 1 a 1 + ... + γ 4n-1 a 4n-1 , (125)
and F Ω 1 (I) is the space of these functions ϕ Ω 1 with γ 1 a 1 + ... + γ 4n-1 a 4n-1 = 0. Evidently, ϕ Ω 1 (E 1 ) = 0, but, is E 1 an ordinary zero for F Ω 1 (I) ?

Let D ′ a straight line neighbouring E 1 = Span 2 i=1 (αa 1 i + βa 2 i ) ∂ ∂x i + 2n i ′′ =3 (αc 1 i ′′ + βc 2 i ′′ ) ∂ ∂p i ′′ 1 , D ′ = Span 2 i=1 (αa 1 i + βa 2 i + ε i ) ∂ ∂x i + 2n i ′ =1 ε ′ i ′ ∂ ∂θ i ′ + 2n i ′′ =3 (αc 1 i ′′ + βc 2 i ′′ + ε ′′ i ′′ ) ∂ ∂p i ′′ 1 (126)
with ε, ε ′ ε ′′ small enough. We want to obtain a finite number q of functions ϕ 1 Ω 1 , ..., ϕ q Ω 1 ∈ F Ω 1 (I) such that dϕ 1 Ω 1 , ..., dϕ q Ω 1 are independent and

∀ϕ ∈ I, ϕ(D ′ ) = 0 ⇔ ϕ 1 Ω 1 (D ′ ) = ..., = ϕ q Ω 1 (D ′ ) = 0 . ( 127
)
We choose q = 2n -1, and, for k = 1, ..., 2n -1, we define

ϕ k = c k (γ 1 a 1 + ... + γ 4n-1 a 4n-1 )θ k+1 with 0 = c k ∈ Ω 0 (M ). (128)
Then,

ϕ 1 Ω 1 (D) = c 1 a 3 , ϕ 2 Ω 1 (D) = c 2 a 4 , ...... ϕ 2n-1 Ω 1 (D) = c 2n-1 a 2n+1 , (129)
and, as said before, we want

dϕ 1 Ω 1 ∧ ... ∧ dϕ 2n-1 Ω 1
= 0 at the point E 1 , and, consequently, in a a neighbourhood,

ϕ 1 Ω 1 (D ′ ) = ϕ 2 Ω 1 (D ′ ) = ... = ϕ 2n-1 Ω 1 (D ′ ) = 0 ⇒ ∀ϕ ∈ I, ϕ Ω 1 (D ′ ) = 0 . ( 130
)
The second condition is obviously satisfied. In fact, ∀ϕ ∈ I, ϕ = α 2 θ 2 + ...

+ α 2n θ 2n , so ϕ Ω 1 (D ′ ) = 2n j=2 α j θ j Ω 1 (D ′ ) = 2n j=2 α j c j-1 (γ 1 a 1 +...+γ 4n-1 a 4n-1 ) ϕ j-1 Ω 1 (D ′ ) = 0.
For the first condition, according to [BCGGG], p. 68, we note

G 1 (T M, Ω 1 ) the open set of the straight lines D in T M such that Ω 1 /D = 0. Then ϕ k Ω 1 : G 1 (T M, Ω 1 ) -→ R is defined before : ϕ k Ω 1 (D) = c k a k+2 , so dϕ k Ω 1 = c k da k+2 , and
dϕ 1 Ω 1 ∧ ... ∧ dϕ 2n-1 Ω 1 = c 1 ...c 2n-1 da 3 ∧ ... ∧ da 2n+1 = 0. ( 131
)
So, E 1 is an ordinary zero of F Ω 1 (I). Then, E 1 is said a Kählerordinary point. We have also to prove that r is locally constant in a neighbourhood of E 1 in V 0 1 (I) (see [BCGGG] p. 67 and 68), where V 0 1 (I) denotes the subspace of Kählerordinary points. We have to precise

. Let u = 2 i=1 u i ∂ ∂x i + 2n i ′′ =3 u i ′′ ∂ ∂p i ′′ 1 be a basis of E 1 , and let D ′ be a straight line neighbouring E 1 , D ′ = Span(u ε ) = Span 2 i=1 (u i + ε i ) ∂ ∂x i + 2n i ′ =1 ε ′ i ′ ∂ ∂θ i ′ + 2n i ′′ =3 (u i ′′ + ε ′′ i ′′ ) ∂ ∂p i ′′ 1 .
If we take ε = 0, then

D ′ = E 1 .
Following [BCGGG], p. 67, we define

H(D ′ ) = {v ∈ T m M : ϕ(u ε , v) = 0 ∀ϕ ∈ I ∩ Ω 2 (M )
} the polar space of D ′ , and r(D ′ ) = dim(H(D ′ )) -2. We take

ϕ = ω 1 ∧ θ 1 + ... + ω 2n ∧ θ 2n + α 1 dθ 1 + ... + α 2n dθ 2n ∈ I ∩ Ω 2 (M ), (132)
with

ω i = d i 1 dx 1 + ... + d i 4n-1 dp 2n 1 ∈ Ω 1 (M )
and α i ∈ Ω 0 (M ), and

v = v 1 ∂ ∂x 1 + v 2 ∂ ∂x 2 + v 3 ∂ ∂θ 1 + ... + v 2n+2 ∂ ∂θ 2n + v 2n+3 ∂ ∂p 3 1 + ... + v 4n ∂ ∂p 2n 1 . ( 133
)
For simplifying notations, the basis (dx 1 , dx 2 , θ 1 , ..., θ 2n , dp 3 1 , ..., dp 2n 1 ) we have seen before, is now noted (µ 1 , µ 2 , µ 3 , ..., µ 2n+2 , µ 2n+3 , ..., µ 4n ). So, we have

ϕ = 2n k=1 4n-1 i=1 d k i µ i ∧ θ k + 2n k=1 α k dθ k , (134)
and, therefore,

ϕ(u ε , v) = 2n k=1 4n-1 i=1 d k i [µ i (u ε )θ k (v) -µ i (v)θ k (u ε )] + 2n k=1 α k dθ k (u ε , v). (135)
But, this is 0 for all ϕ ∈ I, i.e. ∀d k i and ∀α k . So ∀k = 1, ..., 2n, ∀i = 1, ..., 4n -1, we have

µ i (u ε )θ k (v) -µ i (v)θ k (u ε ) = 0 dθ k (uε, v) = 0. ( 136
)
or, more explicitly, ∀k = 1, ..., 2n,

               (αa 1 i + βa 2 i + ε i )v k+1 -v i ε ′ k = 0 ∀i = 1, 2 ε i ′ v k+1 -v i ′ +1 ε ′ k = 0 ∀i ′ = 1, ..., 2n (αc 1 i ′′ + βc 2 i ′′ + ε ′′ i ′′ )v k+1 -v 2n-1+i ′′ ε ′ k = 0 ∀i ′′ = 3, ..., 2n dθ 2 (u ε , v) = 0 dθ j (u ε , v) = 0 ∀j = 3, ..., 2n.
(137) α e 1 + β e 2 is a basis of E 1 , so there exists i or i ′′ such that αa 1 i + βa 2 i = 0 or αc 1 i ′′ + βc 2 i ′′ = 0. For example, without particularizing, we suppose αa 1 1 +βa 2 1 = 0 (but, if αa 1 2 +βa 2 2 = 0 or αa 1 3 +βa 2 3 = 0 or ... or αa 1 2n + βa 2 2n = 0 or αc 1 1 + βc 2 1 = 0 or ... or αc 1 2n + βc 2 2n = 0, then an analogous demonstration can be obtain). And we suppose ε small enough to have αa 1 1 + βa 2 1 + ε 1 = 0. Then, from the three first lines of 137, ∀k = 1, ..., 2n, ∀i ′ = 1, ..., 2n, and ∀i ′′ = 3, ..., 2n,

v k+1 = v 1 ε ′ k αa 1 1 + βa 2 1 + ε 1 = v 2 ε ′ k αa 1 2 + βa 2 2 + ε 2 = v i ′ +1 ε ′ k ε i ′ = v 2n-1+i ′′ ε ′ k αc 1 i ′′ + βc 2 i ′′ + ε ′′ i ′′ . ( 138
)
(In the three last terms, if the denominator is 0, the numerator is also 0.) Now, we study two cases.

First case.

There exists k 0 ∈ {1, ..., 2n} such that ε ′ k 0 = 0. Then ∀i ′ = 1, ..., 2n, and ∀i ′′ = 3, ..., 2n,

v 1 αa 1 1 + βa 2 1 + ε 1 = v 2 αa 1 2 + βa 2 2 + ε 2 = v i ′ +1 ε i ′ = v 2n-1+i ′′ αc 1 i ′′ + βc 2 i ′′ + ε ′′ i ′′ , ( 139
)
and therefore,

         v 2 = v 1 (αa 1 2 +βa 2 2 +ε 2 ) αa 1 1 +βa 2 1 +ε 1 v i ′ +1 = v 1 ε ′ i ′ αa 1 1 +βa 2 1 +ε 1 ∀i ′ = 1, ..., 2n v 2n-1+i ′′ = v 1 (αc 1 i ′′ +βc 2 i ′′ +ε ′′ i ′′ ) αa 1 1 +βa 2 1 +ε 1 ∀i ′′ = 3, ..., 2n. (140)
Using 20, we can calculate

dθ 2 (u ε , v) = 2n i=3 ∂γ 2 i ∂x 2 - ∂(α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i ∂x 1 p i 1 (αa 1 1 + βa 2 1 + ε 1 )v 2 -(αa 1 2 + βa 2 2 + ε 2 )v 1 - 2n i=3 γ 2 i (αc 1 i + βc 2 i + ε ′′ i )v 1 -(αa 1 1 + βa 2 1 + ε 1 )v 2n-1+i - 2n i=3 (α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i ) (αc 1 i + βc 2 i + ε ′′ i )v 2 -(αa 1 2 + βa 2 2 + ε 2 )v 2n-1+i (141)
and, ∀j = 3, ..., 2n,

dθ j (u ε , v) = -(αc 1 j + βc 2 j + ε ′′ j )v 1 -(αa 1 1 + βa 2 1 + ε 1 )v 2n-1+j - 2n i=3 ∂(α j,1 γ 1 i + α j,2 γ 2 i + α j,i ) ∂x 1 p i 1 (αa 1 1 + βa 2 1 + ε 1 )v 2 -(αa 1 2 + βa 2 2 + ε 2 )v 1 - 2n i=3 (α j,1 γ 1 i + α j,2 γ 2 i + α j,i ) (αc 1 j + βc 2 j + ε ′′ j )v 2 -(αa 1 2 + βa 2 2 + ε 2 )v 2n-1+i . (142)
But, from 140, we obtain

     (αa 1 1 + βa 2 1 + ε 1 )v 2 -v 1 (αa 1 2 + βa 2 2 + ε 2 ) = 0 (αa 1 1 + βa 2 1 + ε 1 )v 2n-1+i ′′ -v 1 (αc 1 i ′′ + βc 2 i ′′ + ε ′′ i ′′ ) = 0 (αc 1 i ′′ + βc 2 i ′′ + ε ′′ i ′′ )v 2 -v 2n-1+i ′′ (αa 1 2 + βa 2 2 + ε 2 ) = 0. ( 143
)
These quantities appear in 141 and 142 in the expressions between [ ], so 141 and 142 are satisfied and v = (v 1 , ..., v 4n-1 ) given by 140 are the solutions of 137. In other words, in this case, dim H(D ′ ) = 1.

We recall that we want r(D ′ ) locally constant, i.e. we want dim H(D ′ ) locally constant, and therefore we want dim H(D ′ ) = 1 if ε is small enough.

Second case. Now, ∀k = 1, ..., 2n, ε ′ k = 0. We note ε = (ε 1 , ε 2 , 0, ..., 0, ε ′′ 3 , ..., ε ′′ 2n ). Then, 137 can be written

     v 3 = ... = v 2n+1 = 0 dθ 2 (u ε , v) = 0 dθ j (u ε , v) = 0 ∀j = 3, ..., 2n.
(144) and we want dim H(D ′ ) = 1. The two last equation are calculated in 141 and 142, and, if we note

     X 2 = (αa 1 1 + βa 2 1 + ε 1 )v 2 -(αa 1 2 + βa 2 2 + ε 2 )v 1 X i = (αc 1 i + βc 2 1 + ε ′′ i )v 1 -(αa 1 1 + βa 2 1 + ε 1 )v 2n-1+i ∀i = 3, ..., 2n X 2n-2+i = (αc 1 i + βc 2 1 + ε ′′ i )v 2 -(αa 1 2 + βa 2 2 + ε 2 )v 2n-1+i ∀i = 3, ..., 2n, (145)
the two last lines of 144 become

           2n i=3 ∂γ 2 i ∂x 2 - ∂(α 2,1 γ 1 i +α 2,2 γ 2 i +α 2,i ) ∂x 1 p i 1 X 2 -2n i=3 γ 1 i X i -2n i=3 (α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i )X 2n-2+i = 0 ∀j = 3, ..., 2n, 2n i=3 ∂(α j,1 γ 1 i +α j,2 γ 2 i +α j,i ) ∂x 1 p i 1 X 2 + X j + 2n i=3 (α j,1 γ 1 i + α j,2 γ 2 i + α j,i )X 2n-2+i = 0. (146)
But, the variables X 2 , ..., X 4n-2 are not independent. In fact, ∀i = 3, ..., 2n, (αc

1 i + βc 2 i + ε ′′ i )X 2 + (αa 1 2 + βa 2 2 + ε 2 )X i -(αa 1 1 + βa 2 1 + ε 1 )X 2n-2+i = 0. ( 147
)
We now define linear applications

f ε = f : R 2n -→ R 4n-3
defined by f (v 1 , v 2 , v 2n+2 , ..., v 4n-1 ) = (X 2 , X 3 , ..., X 2n , X 2n+1 , ..., X 4n-2 ) where the X i are defined by 145, and

g ε = g : R 4n-3 -→ R 2n-1 derfined by g(X 2 , X 3 , ..., X 4n-2 ) = (Y 1 , Y 2 , ..., Y 2n-1 )
where Y 1 , Y 2 , ..., Y 2n-1 are successively defined by the first members of the equations 146. Then, 144 can be written

∀v = (v 1 , v 2 , v 2n+2 , ..., v 4n-1 ) ∈ R 4n , g • f (v) = 0,
and we want dim Ker(g

• f ) = 1.
First, we remark that X 2 , ..., X 4n-2 are linked by 147, so dim Im(f ) 2n -1. But, it is easy to extract, from the matrix of f, a (2n -1, 2n -1) determinant equal to A 2n-1 1 = 0. Therefore, dim Im(f ) = 2n -1, and 147 are the equation which define Im(f ) as a subspace of R 4n-3 . So, dim(Ker(f )) = 1. We now have two cases.

• If Im(f ) ∩ Ker(g) = {0}, then Ker(g • f ) = Ker(f ), and we have dim Ker(g • f ) = 1. • If Im(f )∩Ker(g) = {0}, then Ker(g•f ) is strictly greater than Ker(f ), and dim Ker(g•f ) > 1. Finally, E 1 is Kähler -regular if Ker(g) ∩ Im(f ) = {0} when αa 1 1 + βa 2 1 = 0.
But, as explained before, analogous demonstrations can be obtained if αa 1 i + βa 2 i = 0 for some i, or αc 1 i ′′ + βc 2 i ′′ = 0 for some i ′′ . To precise this, we give, succinctly, the case αc 1 3 + βc 2 3 = 0. If there exists k 0 ∈ {2, ..., 2n} such that ε ′ k 0 = 0, then, from the three first lines of 137,

         v i = v 2n+2 (αa 1 i +βa 2 i +ε i ) αc 1 3 +βc 2 3 +ε ′′ 3 ∀i = 1, 2 v i ′ +1 = v 2n+2 ε ′ i ′ αc 1 3 +βc 2 3 +ε ′′ 3 ∀i ′ = 2, ..., 2n v 2n-1+i ′′ = v 2n+2 (αc 1 i ′′ +βc 2 i ′′ +ε ′′ i ′′ ) αc 1 3 +βc 2 3 +ε ′′ 3 ∀i ′′ = 4, ..., 2n. (148)
From this, we deduce

     X 2 = (αa 1 1 + βa 2 1 + ε 1 )v 2 -(αa 1 2 + βa 2 2 + ε 2 )v 1 = 0 X i = (αc 1 i + βc 2 1 + ε ′′ i )v 1 -(αa 1 1 + βa 2 1 + ε 1 )v 2n-1+i = 0 ∀i = 3, ..., 2n X 2n-2+i = (αc 1 i + βc 2 1 + ε ′′ i )v 2 -(αa 1 2 + βa 2 2 + ε 2 )v 2n-1+i = 0 ∀i = 3, ..., 2n, (149)
So, 148 gives the solutions of 137, and dim H(D ′ ) = 1.

If, now, ∀k = 2, ..., 2n, ε ′ k = 0, 137 becomes 144. We then define f and g as previously and obtain the same conclusion : if αc 1 3 +βc 2 3 = 0, then E 1 is Kähler-regular if Ker(g)∩Im(f ) = {0}. We obtained the theorem below Theorem 4.1. E 1 is Kählerregular if Ker(g) ∩ Im(f ) = {0}, so, then, E is an ordinary integral element and we have an almost holomorphic curve.

Using the Cartan-Kähler theorem (see [BCGGG] pp. 81-86), we have directly the existence of germ of a disk in the hypersurface associated to the ordinary integral element.

For ending, we can precise that Ker(g) ∩ Im(f ) = {0} if and only if the system

                 2n i=3 ∂γ 2 i ∂x 2 - ∂(α 2,1 γ 1 i +α 2,2 γ 2 i +α 2,i ) ∂x 1 p i 1 X 2 -2n i=3 γ 1 i X i -2n i=3 (α 2,1 γ 1 i + α 2,2 γ 2 i + α 2,i )X 2n-2+i = 0 ∀j = 3, ..., 2n, 2n i=3 ∂(α j,1 γ 1 i +α j,2 γ 2 i +α j,i ) ∂x 1 p i 1 X 2 + X j + 2n i=3 (α j,1 γ 1 i + α j,2 γ 2 i + α j,i )X 2n-2+i = 0 (αc 1 j + βc 2 j + ε ′′ j )X 2 + (αa 1 2 + βa 2 2 + ε 2 )X j -(αa 1 1 + βa 2 1 + ε 1 )X 2n-2+j = 0 (150)
is a Cramer system with unknowns X 2 , X 3 , ..., X 4n-2 , whose the determinant is = 0.

A test for finding curves in a real analytic hypersurface

All the results in this paragraph can be certainly adapted in a more general setting but we chose to write its in the nearly complex setting for the convenience of the reader. Furthermore, to our knowledge, the procedure is new.

In the section 3 we give a necessary condition to find a germ of curve in a real analytic hypersurface. Clearly this condition is far to be sufficient but we can obtain a more precise condition. With the same notations as in section 3,

f = (f 1 , • • • , f 2n ) and p i 1 = (p 1 1 , • • • , p 2n
1 ) the necessary conditions can be viewed (in a less intrinsic way) as the following : the torsion is roughly speaking the obstruction to find a 2-jet to complete a 1-jet satisfying the equations (with the Einstein notation) :

     p i 2 = J(p i 1 ) ∂ i ρ(f )p i 1 = 0 ∂ i ρ(f )J(p i 1 ) = 0, ( 151
)
in other words, if we derive the previous equations with respect to x 1 and x 2 :

               -Dρ(f )(p i 11 ) = D 2 ρ(f )(p i 1 , p i 1 ) -Dρ(f )(J(p i 11 )) = Dρ(DJ(p i 1 )(p i 1 )) + D 2 ρ(f )(p i 1 , J(p i 1 )) -Dρ(f )(J(p i 11 )) = Dρ(DJ(p i 1 )(p i 1 )) + D 2 ρ(J(p i 1 ), p i 1 ) Dρ(f )(p i 11 ) = Dρ(DJ(Jp i 1 )(p i 1 )) + D 2 ρ(J(p i 1 ), J(p i 1 )) . (152)
It is obvious that the last system has solutions if and only if

D 2 ρ(f )(p i 1 , p i 1 ) + Dρ(DJ(Jp i 1 )(p i 1 )) + D 2 ρ(J(p i 1 ), J(p i 1 )) = 0. ( 153
)
The last quantity is called the Levi form which coincides to the usual Levi form in the integrable case. Unfortunately, the nullity of the Levi form does not give precise informations for the existence of germ of curve in {ρ = 0} := H. Now the equations 151 and 153 define a real analytic set perhaps with singularities. Nevertheless, we can stratify this set and the strates are real analytic manifolds. Each strate can be defined by equations on (f, p i 1 ) ∈ T J H which define a PDE system. We then derive these equations with respect to x 1 and x 2 and thus compute the torsion of the above system. More precisely, suppose that the strate is defined by the system of equations g j (f, p i 1 ) = 0 (154) with 1 ≤ j ≤ k. To compute the torsion, we differentiate with respect to x 1 and x 2 and we want that there exists p i 1 1 such that

∂ f l g j (f, p i 1 )p l 1 = -∂ p l 1 g j (f, p i 1 )P l 11 ∂ f l g j (f, p i 1 )(J(p i 1 ) l + ∂ p l 1 g j (f, p i 1 )(DJ(p i 1 )(p i 1 )) l = -∂ p l 1 g j (f, p i 1 )J(P i 11 ) l (155)
Remark that for (f, p i 1 ) fixed in the strate, the above equations are endomorphims in p i 11 . Therefore the previous linear system (for (f, p i 1 ) fixed on the strate) has solutions if and only if the terms at left hand the equality are in the image of it. So these terms have to satisfy a certain number of linear equations (with (f, p i 1 ) fixed) with coefficients depending analytically of (f, p i 1 ). The linear equations depend of the rank of the system 155 and then of the nullity of some minors determinant of the matrix defining 155 and the non nullity of others.

Remark 5.1. We can note that if the previous endomorphism is surjective for (f, p i 1 ) fixed on each strate, the PDE system defined by 155 is free torsion on each strate of the stratification of the variety defined by the equations 151 and 153. In the following we will see that it is always possible to obtain a PDE system with free torsion equivalent to 151 in finite steps.

Then we have a partition of the variety defined by 154 obtained by the nullity of some minors determinant and the non nullity of others. The elements of this partition are sub-analytic sets (even semi-analytic sets) and so we can stratify again each element of the partition by smooth real analytic variety. On each smooth strate, we consider the PDE system consisting of linear equations defining the image of 155 on the strate. Reproducing the previous arguments, we compute the torsion of this system of PDE (on each strate of a stratification). The process will stop in finite time when the ad-hoc endomorphism is surjective on each strate.

In fact, we have proved the following lemma:

Lemma 5.2. We have a (finite) collection of PDE systems (S t ) defined by h t l (f, ∂ 1 (f j )) = 0 and ∂ 2 f = J(∂ 1 f ) such that X t := {h t l (f, p i 1 ) = 0} is a submanifold of the manifold defined by the equations 151 which can be identified to T J H and the closure of each X t is a semi-analytic set. Furthermore the systems (S t ) are free torsion and equivalent to the system associated 151 on X t .

Remark 5.3. If we have a solution of the system associated to 151 around a point p ∈ H then, in any neighborhood of p there exists p 0 such that there exits t 0 with S t 0 has a solution passing through p 0 . For all t, the varieties X t contain (f, 0) for all f in H but in general nothing else. Moreover, suppose that for all t the tableaux associated of S t are in involution and the dimension of the tableau associated to S t is locally constant around (f, p i 1 ), therefore, for all (f, p i 1 ) in X t , there exists a solution of S t passing through (f, p i 1 ) and then a solution of 151. If for all t the tableaux associated to S t are in involution and the dimension of the tableau associated to S t are locally constant around each point of X t , since S t are free torsion, the systems S t are in involution. Consider the set M of points f in H such that there exists a non trivial disk passing through f . Suppose that this set is non empty and pick up a point f 0 in the closure of M . Therefore we can choose f n in M satisfying: there exists a regular disk passing through f n contained in H and f n → f 0 . Using the last remark, we can construct t n with the following properties: t n → f 0 , there exists a regular pseudo-holomorphic disk passing through t n with derivative v n of unit norm 1, there exists t 0 such that (t n , v n ) ∈ X t 0 . Extracting a subsequence, we can assume (t n , v n ) tends to (f 0 , v 0 ). If the dimension of the tableau associated to S t 0 is locally constant on X t 0 and if X t 0 is closed then there exists a holomorphic curve passing through (f 0 , v 0 ). It is not the case in general (see the next section).

Unfortunately the "tableaux" of the systems (S t ) are not in involution but we have an estimate on the dimension of its.

Lemma 5.4. The dimension de A St on any point of X t is less than 2n -2.

The estimates is obvious : the dimension of the tableau of the system associated to 151 is, on any point of T J H, 2n-2 (see section 3). The structure forms of S t are the same than the structure forms of 151, furthermore X t are submanifolds of 151 and so J ⊥ St ⊂ J ⊥ 151 , the two previous facts guarantee the estimate.

Theorem 5.5. We have a (finite) collection of PDE systems (λ t ) defined by

h t l (f, ∂ 1 (f j ), ∂ 2 1 (f j ), • • • , , ∂ k 1 (f j )) = 0 such that Y t := {h t l (f, ∂ 1 (f j ), ∂ 2 1 (f j ), • • • , , ∂ k 1 (f j )) = 0}
is a submanifold of the manifold defined by the derivatives of the equations 151 and the closure of each Y t is a semi-analytic set. Furthermore, the systems (λ t ) are free torsion with tableaux in involution and equivalent to the system associated 151 on Y t .

Applying the lemma 5.2 and omit the index t, we have a system S 1 with free torsion but perhaps without the tableau in involution. Thanks to the lemma 5.4, the tableau of the (2n -2) prolongation of S 1 are in involution but torsion can appear by prolongation procedure (the tableau of S 1 was not in involution). Using the procedure to obtain the lemma 5.2, we get a system S 2 := (S 2n-2 1 , T 1 ) with free torsion but perhaps with the tableaux not in involution. The lemma 5.4 guarantees that the dimension of S 2 is less than 2n -2, so by induction we have a family of system S k such that the following estimates is satisfied for all k:

dimA S 2n-2 k ≤ dimA S k ≤ dimA S 2n-2 k-1 ≤ dimA S 1 k-1 ≤ dimA S k-1 .
But all these integers are less than 2n -2 and then, there exists k 0 such that dimA S 1 k = dimA S k .

Some examples. Consider the hypersurface model

H = {ρ(z) = Re(z 3 ) + |f (z 1 , z 2 )| 2 = 0}
in R 6 , equipped with the standard complex structure defined by i, and f the holomorphic function defined by f (z 1 , z 2 ) = z 2 1z 3 2 . The previous construction gives at the first step, the following real analytic subset of T C H:

     ρ(z) = 0 w 3 = 0 i ∂ i f (z 1 , z 2 )w i = 0. (156)
There are two strates in the Withney's stratification:

         ρ(z) = 0 w 3 = 0 i ∂ i f (z 1 , z 2 )w i = 0 |∂ z,w i ∂ i f (z 1 , z 2 )w i | 2 > 0 (157) and          ρ(z) = 0 w 3 = 0 i ∂ i f (z 1 , z 2 )w i = 0 ∂ z,w i ∂ i f (z 1 , z 2 )w i = 0. (158)
The system associated to 158 can be written:

                   Re(z 3 ) = 0 w 3 = 0 z 1 = 0 z 2 = 0 w 1 = 0 w 2 = 0, (159)
and it is clear that the system associated to this manifold is free torsion and in involution so there exists holomorphic curves contained in 158 passing through (z, 0) but there is no non-trivial holomorphic curves passing through (Im(z 3 ), 0, 0) contained in 158. The manifold 157 is more interesting. We want to calculate the torsion of this system and therefore we differentiate with respect ∂ ∂t where t is the variable in the unit disk: (160) this system in (w

                   ρ(z) = 0 w 3 = 0 w (1) 3 = 0 2z 1 w 1 + 3z 2 2 w 2 = 0 2w 2 1 + 2z 1 w (1) 1 + 6z 2 w 2 2 + 3z 2 2 w (1) 2 = 0 |z 1 | 2 + |z 2 | 2 + |w 1 | 2 > 0,
(1) 1 , w (1) 2 , w (1)
3 ) has always solutions if z ′ := (z 1 , z 2 ) = (0, 0) and therefore the PDF system associated to 157 is free torsion. Furthermore, the tableau associated to the linear system in (w

(1) 1 , w (1) 2 , w (1)
3 ) (when (z, w) are fixed and satisfied 157) is in involution, dimA = dimA 1 = 1 if z ′ = 0 and dimA = dimA 1 = 2 if not. Then for all (z, w) with z ′ = 0 such that 157 is satisfied, there is an holomorphic curve passing through (z, w) at the time zero. Also, there exists a non trivial holomorphic curve contained in H at the point (z, 0) with z ′ = 0 and z 3 = ia with a real number a. But at this point the dimension of A is not locally constant so we cannot apply the previous theory. Nevertheless, if we have a curve it stays on z 3 equal constant (thanks to the equations 160). Therefore the curve stays in the set where |z 2 1z 3 2 | is constant, equal to Re(z 3 ) = 0. Now, it is obvious that the curve is t → (t 3 , t 2 , ia).

Consider the hypersurface H

= {ρ(z) = 2Re(z 3 ) + |z 1 | 2 -|z 2 | 2 = 0} in R 6 .
In this case, it is obvious that the first step of the previous construction gives :

     ρ(z) = 0 w 3 + w 1 z1 -w 2 z2 = 0 |w 1 | 2 -|w 2 | 2 = 0; (161)
we have two strates in the Withney's stratification:

         ρ(z) = 0 w 3 + w 1 z1 -w 2 z2 = 0 |w 1 | 2 -|w 2 | 2 = 0 |w 1 | 2 + |w 2 | 2 > 0 (162)
As before, by deriving the first line of 165, we obtain, if we note ρ j = ∂ρ ∂y j (f (x)), the system

n j=1 ρ j ∂f j ∂x 1 = 0 n j=1 ρ j ∂f j ∂x 2 = 0, (166)
or

∂f 1 ∂x 1 = -1 ρ 1 n j=2 ρ j ∂f j ∂x 1 ∂f 1 ∂x 2 = -1 ρ 1 n j=2 ρ j ∂f j ∂x 2 .
(167) and, deriving one time more, we obtain

∂ 2 f 1 ∂x 2 1 = 1 ρ 2 1 n l=1 ρ l,1 ∂f l ∂x 1 n k=2 ρ k ∂f k ∂x 1 - 1 ρ 1 n k=2 ∂f k ∂x 1 n l=1 ρ k,l ∂f l ∂x 1 - 1 ρ 1 n k=2 ρ k ∂ 2 f k ∂x 2 1 = -1 ρ 3 1 n k=2 ρ k ∂f k ∂x 1 ρ 1,1 n j=2 ρ j ∂f j ∂x 1 + 1 ρ 2 1 n k,l=2 ρ k ρ l,1 ∂f k ∂x 1 ∂f l ∂x 1 + 1 ρ 2 1 n k=2 ρ k,1 ∂f k ∂x 1 n l=2 ρ l ∂f l ∂x 1 - 1 ρ 1 n k=2 ∂f k ∂x 1 n l=2 ρ k,l ∂f l ∂x 1 - 1 ρ 1 n k=2 ρ k ∂ 2 f k ∂x 2 1 = -1 ρ 1 n k=2 ρ k ∂ 2 f k ∂x 2 1 + 1 ρ 2 1 n k,l=2 E k,l ∂f k ∂x 1 ∂f l ∂x 1 (168)
where E k,l = -

ρ 1,1 ρ 1 ρ k ρ l + ρ k ρ 1,l + ρ 1,k ρ l -ρ 1 ρ k,l
, and, in the same way,

∂ 2 f 1 ∂x 1 ∂x 2 = -1 ρ 1 n k=2 ρ k ∂ 2 f k ∂x 1 ∂x 2 + 1 ρ 2 1 n k,l=2 E k,l ∂f k ∂x 1 ∂f l ∂x 2 , (169)
and also

∂ 2 f 1 ∂x 2 2 = -1 ρ 1 n k=2 ρ k ∂ 2 f k ∂x 2 2 + 1 ρ 2 1 n k,l=2 E k,l ∂f k ∂x 2 ∂f l ∂x 2 . ( 170
)
From the second equation of 165, we have the necessary condition

∂ 2 f 1 ∂x 2 1 + ∂ 2 f 1 ∂x 2 2 = 0, f or all x ∈ D, (171)
that is to say

n k,l=2 E k,l ∂f k ∂x 1 ∂f l ∂x 1 + ∂f k ∂x 2 ∂f l ∂x 2 = 0 f or all x ∈ D. ( 172
)
Remark 6.1. We remark that if k = 1, ..., n, then E k,1 = E 1,k = 0. So, using 166, this equality may written

n k,l=2 E k,l ∂f k ∂x 1 ∂f l ∂x 1 + ∂f k ∂x 2 ∂f l ∂x 2 = n k,l=1 E k,l ∂f k ∂x 1 ∂f l ∂x 1 + ∂f k ∂x 2 ∂f l ∂x 2 = n k,l=1 - ρ 1,1 ρ 1 ρ k ρ l + +ρ k ρ 1,l + ρ 1,k ρ l -ρ 1 ρ k,l ∂f k ∂x 1 ∂f l ∂x 1 + ∂f k ∂x 2 ∂f l ∂x 2 = -ρ 1,1 ρ 1 n k=1 ρ k ∂f k ∂x 1 2 + n k=1 ρ k ∂f k ∂x 2 2 + 2 n k=1 ρ k ∂f k ∂x 1 n l=1 ρ 1,l ∂f l ∂x 1 + n k=1 ρ k ∂f k ∂x 2 n l=1 ρ 1,l ∂f l ∂x 2 + n k,l=1 ρ k,l ∂f k ∂x 1 ∂f l ∂x 1 + ∂f k ∂x 2 ∂f l ∂x 2 = n k,l=1 ρ k,l ∂f k ∂x 1 ∂f l ∂x 1 + ∂f k ∂x 2 ∂f l ∂x 2 = 0. ( 173
)
This equality may be obtain directly by deriving the first line of 166 in x 1 , the second line in x 2 and adding.

When 172 can be satisfied with some ∂f j ∂x i = 0 ? If there exists j such that E j,j = 0 then, of course there exists ∂f j ∂x i = 0 such that 172 is verified. So, we suppose all the E j,j = 0. In particular, E 2,2 = 0, and 172 can be written

E 2,2 ∂f 2 ∂x 1 2 + 2 ∂f 2 ∂x 1 n l=3 E 2,l ∂f l ∂x 1 + n k,l=3 E k,l ∂f k ∂x 1 ∂f l ∂x 1 + n k,l=2 E k,l ∂f k ∂x 2 ∂f l ∂x 2 = 0, ∀x ∈ D, (174)
which implies

δ ′ = n l=3 E 2,l ∂f l ∂x 1 2 -E 2,2 n k,l=3 E k,l ∂f k ∂x 1 ∂f l ∂x 1 + n k,l=2 E k,l ∂f k ∂x 2 ∂f l ∂x 2 ≥ 0 ∀x ∈ D, (175)
or

n k,l=3 (E 2,k E 2,l -E 2,2 E k,l ) ∂f k ∂x 1 ∂f l ∂x 1 n k,l=2 E 2,2 E k,l ∂f k ∂x 2 ∂f l ∂x 2 ∀x ∈ D. ( 176
)
For a bilinear symmetric form B on R n associated to a symmetric matrix A, we said that B > 0 (respectively B < 0) or A > 0 (resp. A < 0) if ∀u ∈ R n , u = 0, we have

B(u, u) > 0 (resp. B(u, u) < 0)
that is to say t uAu > 0 (resp. t uAu < 0).

We define the matrix associated to the two bilinear forms in 176, therefore E ∈ M n-1 (R) with entries E k,l , k, l = 2, ..., n, and

E ∈ M n-2 (R) with entries E k,l = E 2,k E 2,l -E 2,2 E k,l , k, l = 3, ..., n.
With evident notations, 176 can be written where u ∈ R n-2 and v ∈ R n-1 .

If ∃v = 0 : t vE 2,2 Ev ≤ 0, 177 can be satisfied (taking u = 0), and also if ∃v = 0 : t u Eu 0 (taking v = 0). The only case where 177, and therefore 176, cannot be satisfied is when

∃x ∈ D : E 2,2 E > 0 and E < 0. (178) But, E 2,2 E > 0 implies E < 0. In fact, if we write v = (v 2 , u 3 , ..., u n ) = (v 2 , u), then E 2,2 E > 0, says ∀u ∈ R n-2 , ∀v 2 ∈ R, t vE 2,2 Ev = E 2,2 n k,l=2 E k,l v k v l = E 2,2 [E 2 2,2 v 2 2 + 2 n k=3 E 2,k u 2 v k + n k,l=3 E k,l v k v l ] = E 2 2,2 v 2 2 + 2v 2 E 2,2 n k=3 E 2,k u k + E 2,2 n k,l=3 E k,l u k u l > 0. (179) So, ∀u ∈ R n-2 δ ′ = E 2,2 n k=3 E 2,k u k 2 -E 3 2,2 n k,l=3 E k,l u k u l < 0, (180)
or

∀u ∈ R n-2 n k,l=3 E 2,k E 2,l -E 2,2 E k,l u k u l < 0, (181)
that is to say E < 0. So, 176 cannot be satisfied if ∃x ∈ D : E 2,2 E > 0.

Of course, this implies E > 0 or E < 0. But, if E > 0, then all the E j,j > 0, so E 2,2 > 0, and E 2,2 E > 0.

In the same way, if E < 0, then E j,j < 0, and E 2,2 < 0, so E 2,2 E > 0. Therefore, 176 cannot be satisfied if ∃x ∈ D : E > 0 or < 0, that is to say if E is non degenerated positive or negative on D.

We now suppose the condition 176 is satisfied (otherwise, 165 has only trivial (constant) solutions), that is to say, the bilinear symmetric form E associated to the matrix E verifies

E has neither (n -1, 0) nor (0, n -1) signature. (182)
So, we shall use the method of [BCGGG] to study 165 and note p i j for ∂f i ∂x j , p i k,l for ∂ 2 f i ∂x k ∂x l and so on. From 174, which is a trinomial with unknown p 2 1 = ∂f 2 ∂x 1 , we obtain

p 2 1 = -n l=3 E 2,l p l 1 + ε √ δ ′ E 2,2 = - n l=3 E 2,l E 2,2 p l 1 + ε δ 1 (183)
with ε = +1 or -1, and

δ 1 = n k=3 E 2,l E 2,2 p l 1 2 -n k,l=3 E k,l E2,2 p k 1 p l 1 -n k,l=2 E k,l E2,2 p k 2 p l 2 .
We define

S 1 = n k,l=3 E k,l E 2,2 p k 1 p l 1 , S 2 = n k,l=2 E k,l E 2,2 p k 2 p l 2 , S 2,1 = n k=3 E 2,k E 2,2 p k 1 , S 2,2 = n l=2 E 2,l E 2,2 p l 2 .
So, for example,

∂S 2 ∂x i = n k,l=2 ∂ ∂x i E k,l E 2,2 p k 2 p l 2 + E k,l E 2,2 p k 2,i p l 3 + p k 2 p l 2,i = n k,l=2 ∂ ∂x i E k,l E 2,2 p k 2 p l 2 + 2 E k,l
E 2,2 p k 2,i p l 2 , and we have similar formulas for S 2 , S 2,1 , S 2,2 .

From 183 which is

p 2 1 = -S 2,1 + ε S 2 2,1 -S 1 -S 2 , (184)
by deriving, we obtain

   p 2 1,1 = - ∂S 2,1 ∂x 1 + ε 2 √ δ 1 2S 2,1 ∂S 2,1 ∂x 1 -∂S 1 ∂x 1 -∂S 2 ∂x 1 p 2 1,2 = - ∂S 2,1 ∂x 2 + ε 2 √ δ 1 2S 2,1 ∂S 2,1 ∂x 2 -∂S 1 ∂x 2 -∂S 2 ∂x 2 .
(185)

This system, where p 2 1,1 and p 2 1,2 appear in the two members of the first equation and also of the second equation (because p 2 2,2 = -p 2 1,1 ), can be written (with neither p 2 1,1 nor p 2 1,2 in the second member),

   p 2 1,1 + p 2 1,2 εS 2,2 √ δ 1 = -A ∂S 2,1 ∂x 1 -ε 2 √ δ 1 ∂S 1 ∂x 1 + S 2,2,1 + 2S 1,2,2 -p 2 1,1 εS 2,2 √ δ 1 + p 2 1,2 = -A ∂S 2,1 ∂x 2 -ε 2 √ δ 1 ∂S 1 ∂x 2 + S 2,2,2 -2S 1,1,2 , (186)
where

A = 1 - εS 2,1 √ δ 1 , S 1,2,2 = n k=3 n l=2 E k,l E 2,2 ∂ 2 f k ∂x 1 ∂x 2 ∂f l ∂x 2 S 1,1,2 = n k=3 n l=2 E k,l E 2,2 ∂ 2 f k ∂x 2 1 ∂f l ∂x 2 , S 2,2,1 = n k,l=2 ∂ ∂x 1 E k,l E 2,2 ∂f k ∂x 2 ∂f l ∂x 2 , and S 2,2,2 = n k,l=2 ∂ ∂x 2 E k,l E 2,2 ∂f k ∂x 2 ∂f l ∂x 2 , From this Cramer system, whose the determinant is D = 1 + S 2 2,2 δ 1 , we deduce ∂ 2 f 2 ∂x 2 1 = -1 D A ∂S 2,1 ∂x 1 + ε 2 √ δ 1 ∂S 1 ∂x 1 + S 2,2,1 + 2S 1,2,2 -2AS 2,2 ∂S 2,1 ∂x 2 - S 2,2 2δ 1 ∂S 1 ∂x 2 + S 2,2,2 -2S 1,1,2 (187)
and

∂ 2 f 2 ∂x 1 ∂x 2 = -1 D A ∂S 2,1 ∂x 2 + ε 2 √ δ 1 ∂S 1 ∂x 2 + S 2,2,2 -2S 1,1,2 + 2AS 2,2 ∂S 2,1 ∂x 1 + S 2,2 2δ 1 ∂S 1 ∂x 1 + S 2,2,1 + 2S 1,2,2 . (188)
Now, on the space M of the variables

x 1 , x 2 , f 1 , ..., f n , p 3 1 , ..., p n 1 , p 2 2 , ..., p n 2 , p 3 1,1 , ..., p n 1,1 , p 3 1,2 , ..., p n 1,2 , (189)
to translate the system 165 (with tne condition 182), we only have the equations of structure

                   θ 1 = df 1 -{p 1 1 }dx 1 -{p 1 2 }dx 2 = 0 θ 2 = df 2 -{p 2 1 }dx 1 -p 2 2 dx 2 = 0 θ i = df i -p i 1 dx 1 -p i 2 dx 2 = 0 f or i = 3, ..., n θ 2,2 = dp 2 2 -{p 2 1,2 }dx 1 + {p 2 1,1 }dx 2 = 0 θ i,1 = dp i 1 -p i 1,1 dx 1 -p i 1,2 dx 2 = 0 f or i = 3, ..., n θ i,2 = dp i 2 -p i 1,2 dx 1 + p i 1,1 dx 2 = 0 f or i = 3, ..., n, (190)
where we write {p i j } (or { p i j,k }) to emphasize the fact that, in this case, {p i j } (or {p i j,k }) is not a variable, but the expression p i j = ∂f i ∂x j , or p i j,k = ∂ 2 f i ∂x j ∂x k , expression explicited before in 167, 168, 169, 170, 183, 187 and 188. The equations of structure 190 are θ a = 0 with a ∈ {1, ..., n, (3, 1), ..., (n, 1), (2, 2), ..., (n, 2)} = A. If M is the space of the variables 189, we have (see [BCGGG], p. 129) to choose for T ⋆ M the basis

(θ 1 , ..., θ n , θ 2,2 θ 3,1 , ..., θ n,1 , θ 3,2 , ..., θ n,2 , dx 1 , dx 2 , dp 3 1,1 , ..., dp n 1,1 , dp 3 1,2 , ..., dp n 1,2 ) = (θ a (a ∈ A), dx 1 , dx 2 , dp 3 1,1 , ..., dp n 1,1 , dp 3 1,2 , ..., dp n 1,2 ) (191)
and, for the dual space T M, the dual basis

∂ ∂θ a (a ∈ A), ∂ ∂x 1 , ∂ ∂x 2 , ∂ ∂p 3 1,1 , ..., ∂ ∂p n 1,1 , ∂ ∂p 3 1,2 , ..., ∂ ∂p n 1,2 . ( 192
)
Let I ⊂ T ⋆ M be the sub-bundle generated by θ a , a ∈ A and let {I} ⊂ Ω ⋆ (M ) be the algebraic ideal generated by the C ∞ -sections of I.

Until now, we calculated with the variables x 1 , x 2 . From now on, we calculate with the variables 189. We remark the functions ρ, ρ i , ρ i,j , E k,l are functions of f = (f 1 , ..., f n ) only. So, ρ i = ∂ρ ∂y i (f ) shall be note, sometimes,

ρ i = ∂ρ ∂f i . The functions S 1 , S 2 , S 2,1 , S 2,2 , S 1,2,2 , {p 2 1 }, {p 2 1,1 }, {p 2
1,2 } are functions of the variables f 1 , ..., f n , p k j , p k i,j with p k i and p k i,j in the list 189, and not of x = (x 1 , x 2 ). Also, from now on, d is the exterior derivative relative to the variables 189. So, (193) Conformably to [BCGGG], p. 130, for a ∈ A, we write

       d = n i=1 ∂ ∂f i df i + ∂ ∂x 1 dx 1 + ∂ ∂x 2 dx 2 + n j=3 ∂ ∂p j 1 dp j 1 + n k=2 ∂ ∂p k 2 dp k 2 + n j=3 ∂ ∂p j 1,1 dp j 1,1 + n j=3 ∂ ∂p j 1,2 dp j 1,2 = d f + d x + d p .
dθ a = n j=3 2 i=1 A a (j,1,1),i dp j 1,1 ∧ dx i + A a (j,1,2),i dp j 1,2 ∧ dx i + c a 1,2 dx 1 ∧ dx 2 modulo {I}. ( 194
)
For an equality modulo {I}, we shall write ≈ . From 190, we have

df 1 ≈ - ρ 2 ρ 1 {p 2 1 } + 1 ρ 1 n k=3 ρ k p k 1 dx 1 - 1 ρ 1 n k=2 ρ k p k 2 dx 2 df 2 ≈ {p 2 1 }dx 1 + p 2 2 dx 2 df i ≈ p i 1 dx 1 + p i 2 dx 2 f or i = 3, ..., n dp 2 2 ≈ {p 2 1,2 }dx 1 -{p 2 1,1 }dx 2 dp i 1 ≈ p i 1,1 dx 1 + p i 1,2 dx 2 f or i = 3, ..., n dp i 2 ≈ p i 1,2 dx 1 -p i 1,1 dx 2 f or i = 3, ..., n. (195)
From 190,

                   dθ 1 = -d{p 1 1 } ∧ dx 1 -d{p 1 2 } ∧ dx 2 dθ 2 = -d{p 2 1 } ∧ dx 1 -dp 2 2 ∧ dx 2 dθ i = -dp i 1 ∧ dx 1 -dp i 2 ∧ dx 2 f or i = 3, ..., n dθ 2,2 = -d{p 2 1,2 } ∧ dx 1 + d{p 2 1,1 } ∧ dx 2 dθ i,1 = -dp i 1,1 ∧ dx 1 -dp i 1,2 ∧ dx 2 f or i = 3, ..., n dθ i,2 = -dp i 1,2 ∧ dx 1 + dp i 1,1 ∧ dx 2 f or i = 3, ..., n, (196)
Now, we shall calculate the tableau bundle. So, we need the coefficients A a (j,1,.),i , and, therefore, the components of dθ a containing dp j 1,1 ∧ dx i or dp j 1,2 ∧ dx i . From 195 and 194, the terms in dθ a containing df i or dp k l have no contribution to A a (j,1,.),i , but only to c a 1,2 . We obtain , for i = 3, ..., n,

dθ i = -dp i 1 ∧ dx 1 -dp i 2 ∧ dx 2 ≈ -(p i 1,1 dx 1 + p i 1,2 dx 2 ) ∧ dx 1 -(p i 1,2 dx 1 -p i 1,1 dx 2 ) ∧ dx 2 = 0 dθ i,1 = -dp i 1,1 ∧ dx 1 -dp i 1,2 ∧ dx 2 = A i,1 (i,1,1),1 dp i 1,1 ∧ dx 1 + A i,1 (i,1,2),2 dp i 1,2 ∧ dx 2 dθ i,2 = -dp i 1,2 ∧ dx 1 + dp i 1,1 ∧ dx 2 = A i,2 (i,1,2),1 dp i 1,2 ∧ dx 1 + A i,2 (i,1,1),2 dp i 1,1 ∧ dx 2 , (197)
so, for i = 3, ..., n, we obtain A i .,. = 0, and

A i,1 (i,1,1),1 = A i,1 (i,1,2),2 = A i,2 (i,1,2),1 = -1, A i,2 (i,1,1),2 = 1.
As explain before, we also have

dθ 1 ≈ c 1 1,2 dx 1 ∧ dx 2 dθ 2 ≈ c 2 1,2 dx 1 ∧ dx 2 dθ 2,2 ≈ n j=3 2 i=1 A 2,2 (j,1,1),i dp j 1,1 ∧ dx i + A 2,2 (j,1,2),i dp j 1,2 ∧ dx i + c 2,2 1,2 dx 1 ∧ dx 2 , (198)
so A 1 .,. = A 2 .,. = 0 and we have to calculate A 2,2 (j,1,.),i . For this, we look for the coefficients of dp j 1,k ∧ dx i in the expression of dp 2 1,l . For example, from 194, to obtain the coefficients A 2,2 (j,1,k),2 , we have to extract in the expression of d p p 2 1,1 the terms containing dp j 1,k .

Using these remarks, and noting d p•• f the part of df containing dp i 1,1 or dp i 1,2 , we obtain from 187 and 188,

d p•• p 2 1,1 = -1 D A n k=3 E 2,k E 2,2 dp k 1,1 + ε 2 √ δ 1 n k,l=3 E k,l E 2,2 p l 1 dp k 1,1 + 2 n k=3 n l=2 E k,l E 2,2 p l 2 dp k 1,2 -AS 2,2 n k=3 E 2,k E 2,2 dp k 1,2 - S 2,2 2δ 1 n k,l=3 E k,l E 2,2 p l 1 dp k 1,2 -2 n k=3 n l=2 E k,l E 2,2 p l 2 dp k 1,1 d p•• p 2 1,2 = -1 D A n k=3 E 2,k E 2,2 dp k 1,2 + ε 2 √ δ 1 n k,l=3 E k,l E 2,2 p l 1 dp k 1,2 -2 n k=3 n l=2 E k,l E 2,2 p l 2 dp k 1,1 + AS 2,2 n k=3 E 2,k E 2,2 dp k 1,1 + S 2,2 2δ 1 n k,l=3 E k,l E 2,2 p l 1 dp k 1,1 + 2 n k=3 n l=2 E k,l E 2,2 p l 2 dp k 1,2 , (199)
and, from 194 and 196, we have

A 2,2 (k,1,1),1 = 1 D ε 2 √ δ 1 AS 2,2 E 2,k E 2,2 -2 n l=2 E k,l E 2,2 p l 2 + S 2,2 δ 1 n l=3 E k,l E 2,2 p l 1 A 2,2 (k,1,2),1 = 1 D A E 2,k E 2,2 + ε 2 √ δ 1 n l=3 E k,l E 2,2 p l 1 + S 2,2 δ 1 n l=2 E k,l E 2,2 p l 2 A 2,2 (k,1,1),2 = -1 D A E 2,k E 2,2 + ε 2 √ δ 1 n l=3 E k,l E 2,2 p l 1 + S 2,2 δ 1 n l=2 E k,l E 2,2 p l 2 A 2,2 (k,1,2),2 = -1 D -ε 2 √ δ 1 AS 2,2 E 2,k E 2,2 -2 n l=2 E k,l E 2,2 p l 2 - S 2,2 δ 1 n l=3 E k,l E 2,2 p l 1 (200) So, we remark A 2,2 (k,1,1),1 = A 2,2 (k,1,2),2 A 2,2 (k,1,1),2 = -A 2,2 (k,1,2),1 . (201)
Now, we are able to write the tableau of the system 165 (with condition 182). We have to calculate (see [BCGGG], p. 133) the application π

Span ∂ ∂p 3 1,1 , ..., ∂ ∂p n 1,1 , ∂ ∂p 3 1,2 , ..., ∂ ∂p n 1,2 → Span ∂ ∂θ a : a ∈ A ⊗ Span X 1 , X 2 (202)
defined by : if

v = n k=3 v k,1 ∂ ∂p k 1,1 + v k,2 ∂ ∂p k 1,2 , then π(v) = a∈A 2 i=1 n k=3 A a (k,1,1),i v k,1 + A a (k,1,2),i v k,2 ∂ ∂θ a ⊗ X i = n k=3 -v k,1 ∂ ∂θ k,1 ⊗ X 1 -v k,2 ∂ ∂θ k,1 ⊗ X 2 + v k,1 ∂ ∂θ k,2 ⊗ X 2 -v k,2 ∂ ∂θ k,2 ⊗ X 1 + n k=3 A 2,2 (k,1,1),1 v k,1 + A 2,2 (k,1,2),1 v k,2 ∂ ∂θ 2,2 ⊗ X 1 + A 2,2 (k,1,1),2 v k,1 + A 2,2 (k,1,2),2 v k,2 ∂ ∂θ 2,2 ⊗ X 2 = n k=3 v k,1 - ∂ ∂θ k,1 ⊗ X 1 + ∂ ∂θ k,2 ⊗ X 2 + A 2,2 (k,1,1),1 ∂ ∂θ 2,2 ⊗ X 1 + A 2,2 (k,1,1),2 ∂ ∂θ 2,2 ⊗ X 2 + v k,2 - ∂ ∂θ k,1 ⊗ X 2 - ∂ ∂θ k,2 ⊗ X 1 + A 2,2 (k,1,2),1 ∂ ∂θ 2,2 ⊗ X 1 + A 2,2 (k,1,2),2 ∂ ∂θ 2,2 ⊗ X 2 = n k=3 v k,1 U k,1 + v k,2 U k,2 . (203)
The tableau A (0) is the image of π, therefore A (0) = Span U k,1 , U k,2 and its dimension is 2n -4.

It is easy to see that the tableau A (0) 1 = P ∈ A (0) : ∂P ∂X 1 = 0 is {0} (see [BCGGG], p. 119) . The tableau A (0) is in involution if dim(A (1) ) = dim(A (0) ) + dim(A (0) 1 ) = dim(A (0) ) ( [BCGGG], p. 120), where A (1) is the first prolongation of A (0) . So, we need to know A (1) = P : ∂P ∂X 1 and ∂P ∂X 2 ∈ A (0) .

Let P be P = P 1,1 X 2 1 + P 1,2 X 1 X 2 + P 2,2 X 2 2 . For P ∈ A (1) , we want

∂P ∂X 1 = 2P 1,1 X 1 + P 1,2 X 2 = n j=3 v 1 j U j,1 + v 2 j U j,2 ∂P ∂X 2 = P 1,2 X 1 + 2P 2,2 X 2 = n j=3 v 1 j U j,1 + v 2 j U j,2 , (204)
that is to say, (205) and, therefore, from the equality of the second and third lines, for all j = 3, ..., n,

               2P 1,1 = n j=3 v 1 j -∂ ∂θ j,1 + A 2,2 (j,
     v 2 j = -v 1 j v 1 j = v 2 j n j=3 v 1 j A 2,2
(j,1,1),2 + v 2 j A 2,2 (j,1,2),2v 1 j A 2,2 (j,1,1),1v 2 j A 2,2 (j,1,2),1 = 0.

(206)

Carrying the two first lines in the third one, we obtain n j=3 v 1 j A 2,2 (j,1,1),2 + A 2,2 (j,1,2),1 + v 2 j A 2,2 (j,1,2),2 -A 2,2 (j,1,1),1 = 0, (207) which is verified, from 201. So, the equations of A (2) are

v 2 j = -v 1 j v 1 j = v 2 j , (208)
and, therefore, dimA (1) = 2n -4 = dimA (0) . More precisely, we have proved the following:

Theorem 6.2. The first tableau, A (0) , associated to the partial differential system 165 is in involution.

Now, we want to calculate the torsion of the system. Remember 182 is already a condition of torsion. From 197, we know ∀i = 3, ..., n, c i 1,2 = c i,1 1,2 = c i,2 1,2 = 0. So, we have to obtain c 1 1,2 , c 2 1,2 , c 2,2 1,2 . We have

{p 2 1 } = -S 2,1 + ε S 2 2,1 -S 1 -S 2 = - n k=3 E 2,k E 2,2 p k 1 + ε n k=3 E 2,k E 2,2 p k 1 2 - n k,l=3 E k,l E 2,2 p k 1 p l 1 - n k,l=2 E k,l E 2,2 p k 2 p l 2 (209) so d{p 2 1 } = - n k=3 n i=1 ∂ ∂f i E 2,k E 2,2 p k 1 df i + E 2,k E 2,2 dp k 1 + ε 2 √ δ 1 2S 2,1 n k=3 n i=1 ∂ ∂f i E 2,k E 2,2 p k 1 df i + E 2,k E 2,2 dp k 1 - n k,l=3 n i=1 ∂ ∂f i E k,l E 2,2 p k 1 p l 1 df i + 2 E k;l E 2,2 dp k 1 p l 1 - n k,l=2 n i=1 ∂ ∂f i E k,l E 2,2 p k 2 p 2 1 df i + 2 E k;l E 2,2 dp k 2 p l 2 (210)
and also

d{p 1 1 } = d - ρ 2 ρ 1 {p 2 1 } - n j=3 ρ j ρ 1 p j 1 = - n i=1 ∂ ∂f i ρ 2 ρ 1 {p 2 1 }df i - ρ 2 ρ 1 d{p 2 1 } - n j=3 n i=1 ∂ ∂f i ρ j ρ 1 p j 1 - n j=3 ρ j ρ 1 dp j 1 . (211)
Therefore,

d x {p 2 1 } ≈ ( ρ 2 ρ 1 ){p 2 1 } + n k=3 ρ k ρ 1 p k 1 dx 1 - n k=2 ρ k ρ 1 p k 2 dx 2 n k=3 ∂ ∂f 1 E 2,k E 2,2 p k 1 + ε 2 √ δ 1 2S 2,1 n k=3 ∂ ∂f 1 E 2,k E 2,2 p k 1 - n k,l=3 ∂ ∂f 1 E k,l E 2,2 p k 1 p l 1 - n k,l=2 ∂ ∂f 1 E k,l E 2,2 p k 2 p l 2 -{p 2 1 }dx 1 + p 2 2 dx 2 n k=3 ∂ ∂f 2 E 2,k E 2,2 p k 1 + ε 2 √ δ 1 2S 2,1 n k=3 ∂ ∂f 2 E 2,k E 2,2 p k 1 - n k,l=3 ∂ ∂f 2 E k,l E 2,2 p k 1 p l 1 - n k,l=2 ∂ ∂f 2 E k,l E 2,2 p k 2 p l 2 - n i=3 p i 1 dx 1 + p i 2 dx 2 n k=3 ∂ ∂f i E 2,k E 2,2 p k 1 + ε 2 √ δ 1 2S 2,1 n k=3 ∂ ∂f i E 2,k E 2,2 p k 1 - n k,l=3 ∂ ∂f i E k,l E 2,2 p k 1 p l 1 - n k,l=2 ∂ ∂f i E k,l E 2,2 p k 2 p l 2 - n k=3 p k 1,1 dx 1 + p k 1,2 dx 2 E 2,k E 2,2 + ε √ δ 1 2S 2,1 E 2,k E 2,2 -2 n l=3 E k,l E 2,2 p l 1 -{p 2 1,2 }dx 1 -{p 2 1,1 }dx 2 n l=2 E 2,l E 2,2 p l 2 - n k=3 {p k 1,2 }dx 1 -{p k 1,1 }dx 2 n l=2 E k,l E 2,2 p l 2 .
(212) and

d x {p 1 1 } ≈ - ρ 2 ρ 1 d x {p 2 1 } - ρ 2 ρ 1 {p 2 1 } + n k=2 ρ k ρ 1 p k 1 dx 1 + n k=2 ρ k ρ 1 p k 2 dx 2 ∂ ∂f 1 ρ 2 ρ 1 {p 2 1 } + n j=3 ∂ ∂f 1 ρ j ρ 1 p j 1 -{p 2 1 }dx 1 + p 2 2 dx 2 ∂ ∂f 2 ρ 2 ρ 1 {p 2 1 } + n j=3 ∂ ∂f 2 ρ j ρ 1 p j 1 - n i=3 p i 1 dx 1 + p i 2 dx 2 ∂ ∂f i ρ 2 ρ 1 {p 2 1 } + n j=3 ∂ ∂f i ρ j ρ 1 p j 1 - n j=3 ρ j ρ 1 p j 1,1 dx 1 + p j 1,2 dx 2 . ( 213
)
From 196,

c 1 1,2 dx 1 ∧ dx 2 = -d x {p 1 1 } ∧ dx 1 -d x {p 1 2 } ∧ dx 2 (214)
and so

c 1 1,2 = -- ρ 2 ρ 1 n k=2 ρ k ρ 1 p k 2 n k=3 ∂ ∂f 1 E 2,k E 2,2 p k 1 + ε 2 √ δ 1 2S 2,1 n k=3 ∂ ∂f 1 E 2,k E 2,2 p k 1 - n k,l=3 ∂ ∂f 1 E k,l E 2,2 p k 1 p l 1 - n k,l=2 ∂ ∂f 1 E k,l E 2,2 p k 2 p l 2 + p 2 2 n k=3 ∂ ∂f 2 E 2,k E 2,2 p k 1 + ε 2 √ δ 1 2S 2,1 n k=3 ∂ ∂f 2 E 2,k E 2,2 p k 1 - n k,l=3 ∂ ∂f 2 E k,l E 2,2 p k 1 p l 1 - n k,l=2 ∂ ∂f 2 E k,l E 2,2 p k 2 p l 2 + n i=3 p i 2 n k=3 ∂ ∂f i E 2,k E 2,2 p k 1 + ε 2 √ δ 1 2S 2,1 n k=3 ∂ ∂f i E 2,k E 2,2 p k 1 - n k,l=3 ∂ ∂f i E k,l E 2,2 p k 1 p l 1 - n k,l=2 ∂ ∂f i E k,l E 2,2 p k 2 p l 2 + n k=3 p 2 1,2 E 2,k E 2,2 + ε √ δ 1 2S 2,1 E 2,k E 2,2 -2 n l=3 E k,l E 2,2 p l 1 -{p 2 1,1 } n l=2 E 2,l E 2,2 p l 2 - n k=3 p k 1,1 n l=2 E k,l E 2,2 p l 2 + n k=2 ρ k ρ 1 p k 2 ∂ ∂f 1 ρ 2 ρ 1 {p 2 1 } + n j=3 ∂ ∂f 1 ρ j ρ 1 p j 1 + p 2 2 ∂ ∂f 2 ρ 2 ρ 1 {p 2 1 } + n j=3 ∂ ∂f 2 ρ j ρ 1 p j 1 + n i=3 p i 2 ∂ ∂f i ρ 2 ρ 1 {p 2 1 } + n j=3 ∂ ∂f i ρ j ρ 1 p j 1 + n j=3 ρ j ρ 1 p j 1,2 + n j=3 ∂ ∂f 1 ρ j ρ 1 p j 2 ρ 2 ρ 1 {p 2 1 } + n k=3 ρ k ρ 1 p k 1 + n j=2 ∂ ∂f 2 ρ j ρ 1 p j 2 {p 2 1 } + n i=3 n j=2 ∂ ∂f i ρ j ρ 1 p j 2 p i 1 + ρ 2 ρ 1 {p 2 1,2 } + n j=3 ρ j ρ 1 p j 1,2 . (215)
And, in the same way,

c 2 1,2 dx 1 ∧ dx 2 = -d x {p 2 1 } ∧ dx 1 , ( 216
) so c 2 1,2 = -1 ρ 1 n k=2 ρ k p k 2 n k=3 ∂ ∂f 1 E 2,k E 2,2 p k 1 + ε 2 √ δ 1 2S 2,1 n k=3 ∂ ∂f 1 E 2,k E 2,2 p k 1 - n k,l=3 ∂ ∂f 1 E k,l E 2,2 p k 1 p l 1 - n k,l=2 ∂ ∂f 1 E k,l E 2,2 p k 2 p l 2 -p 2 2 n k=3 ∂ ∂f 2 E 2,k E 2,2 p k 1 + ε 2 √ δ 1 2S 2,1 n k=3 ∂ ∂f 2 E 2,k E 2,2 p k 1 - n k,l=3 ∂ ∂f 2 E k,l E 2,2 p k 1 p l 1 - n k,l=2 ∂ ∂f 2 E k,l E 2,2 p k 2 p l 2 - n i=3 p i 2 n k=3 ∂ ∂f i E 2,k E 2,2 p k 1 + ε 2 √ δ 1 2S 2,1 n k=3 ∂ ∂f i E 2,k E 2,2 p k 1 - n k,l=3 ∂ ∂f i E k,l E 2,2 p k 1 p l 1 - n k,l=2 ∂ ∂f i E k,l E 2,2 p k 2 p l 2 - n k=3 p k 1,2 E 2,k E 2,2 + ε √ δ 1 2S 2,1 E 2,k E 2,2 -2 n l=3 E k,l E 2,2 p l 1 + {p 2 1,1 } n l=2 E 2,

  ........................... ................................

  So, when the determinant of this system of three equations is non zero, we can solve and obtain

	p 2n 1 , p 4 1,1 , ..., p 2n 1,1 ),
	(102)
	where h 1 , h 2 , h 3 are functions in the variables p 3 1 , p 4 1 , ..., p 2n 1 , p 4 1,1 , p 5 1,1 , ..., p 2n 1,1 which are linear in p 4 1,1 , p 5 1,1 , ..., p 2n 1,1 .

and w 1 = w 2 = 0. Obviously, the second strate does not contain any non trivial holomorphic curves; the first case is more interesting, the first prolongation of the system is done by:

(163)

The last system has always solutions in (w 1 1 , w 1 2 , w 1 3 ) then 162 is free torsion; but, since the dimension of the tableau associated to 162 is 3 and the dimension of the tableau associated to 163 is 2 (the equation |w

2 w2 = 0) the tableau associated to 162 is not in involution. Consider the second prolongation of the system:

w(1) 2 = 0 is contained in the three equations w

(1) 1 w1 -w

(1)

2 | 2 = 0, and the equation |w

2 w2 = 0, the system defined by 163 is free torsion. Furthermore, for the same reasons, the dimension of the tableau associated to 164 is 2 and so the system associated to 163 is involutive and we have holomorphic disks passing through all (z, w) such that 162 is satisfied.

The harmonic case

Now, if D is an open set in R 2 , and D an open set in R n , we are looking for necessary or sufficient conditions to have a curve f : D -→ D such that f (D) ⊂ H = {ρ = 0} with ρ Ranalytic and all the components of f harmonic. Therefore, we want a solution for the system of PDE ρ(f (x)) = 0

(the space 2 (J/I) in [BCGGG] is, here, of dimension 1) such that, if

then

The torsion c = a∈A c a 1,2 ∂ ∂θ a may be absorbed if there exists v such that π

(i,1,1),2 = 1. So, the torsion may be absorbed if for all i = 3, ..., n,

Carrying the third and the fourth lines in the last one, we obtain 0 for the left member (by 201). To summarize, we have proved the theorem below: Theorem 6.3. The conditions of torsion of the partial differential system 165 are c 1 1,2 = c 2 1,2 = c 2,2 1,2 = 0 E has neither (n -1, 0) nor (0, n -1) signature.

(225)