Burcu Bektas ¸
email: bektasbu@itu.edu.tr

Marilena Moruz
email: marilena.moruz@gmail.com

AND Joeri Van Der Veken
email: joeri.vanderveken@wis.kuleuven.be

Luc Vrancken
email: luc.vrancken@univ-valenciennes.fr

LAGRANGIAN SUBMANIFOLDS OF THE NEARLY K ÄHLER S 3 × S 3 FROM MINIMAL SURFACES IN S 3

We study non-totally geodesic Lagrangian submanifolds of the nearly Kähler S 3 ×S 3 for which the projection on the first component is nowhere of maximal rank. We show that this property can be expressed in terms of the so called angle functions and that such Lagrangian submanifolds are closely related to minimal surfaces in S 3 . Indeed, starting from an arbitrary minimal surface, we can construct locally a large family of such Lagrangian immersions, including one exceptional example. We also show that locally all such Lagrangian submanifolds can be obtained in this way.

Introduction

The nearly Kähler manifolds are almost Hermitian manifolds with almost complex structure J for which the tensor field ∇J is skew-symmetric, where ∇ is the Levi Civita connection. They have been studied intensively in the 1970's by Gray ([12]). Nagy ([17], [START_REF] Nagy | On nearly-Kähler geometry[END_REF]) made further contribution to the classification of nearly Kähler manifolds and more recently it has been shown by Butruille ([4]) that the only homogeneous 6-dimensional nearly Kähler manifolds are the nearly Kähler 6sphere S 6 , S 3 × S 3 , the projective space CP 3 and the flag manifold SU (3)/U (1) × U (1), where the last three are not endowed with the standard metric. All these spaces are compact 3-symmetric spaces. Note that in 2014 V. Cortés and J. J. Vásquez have discovered the first non homogeneous (but locally homogeneous) nearly Kähler structures in [START_REF] Cortés | Locally homogeneous nearly Kahler manifolds[END_REF], while more recently, the first complete non homogeneous nearly Kähler structures were discovered on S 6 and S 3 × S 3 in [START_REF] Foscolo | New G2 holonomy cones and exotic nearly Kaehler structures on the 6-sphere and the product of a pair of 3-spheres[END_REF]. A natural question for the above mentioned four homogeneous nearly Kähler manifolds is to study their submanifolds. There are two natural types of submanifolds of nearly Kähler (or more generally, almost Hermitian) manifolds, namely almost complex and totally real submanifolds. Almost complex submanifolds are submanifolds whose tangent spaces are invariant under J. For a totally real submanifold, a tangent vector is mapped by the almost complex structure J into a normal vector. In this case, if additionally the dimension of the submanifold is half the dimension of the ambient manifold, then the submanifold is Lagrangian. Note that the Lagrangian submanifolds of nearly Kähler manifolds are especially interesting as they are always minimal and orientable (see [START_REF] Ejiri | Totally real submanifolds in a 6-sphere[END_REF] for S 6 or [START_REF] Schäfer | Decomposition and minimality of Lagrangian submanifolds in nearly Kähler manifolds[END_REF], [START_REF] Gutowski | Deformations of generalized calibrations and compact non-Kähler manifolds with vanishing first Chern class[END_REF] for the general case). Lagrangian submanifolds of S 6 have been studied by many authors (see, amongst others, [START_REF] Dillen | Classification of totally real 3-dimensional submanifolds of S 6 (1) with K ≥ 1/16[END_REF], [START_REF] Dillen | Totally real submanifolds in S 6 (1) satisfying Chen's equality[END_REF], [START_REF] Ejiri | Equivariant minimal immersions of S 2 into S 2m (1)[END_REF], [START_REF] Ejiri | Totally real submanifolds in a 6-sphere[END_REF], [START_REF] Vrancken | Killing vector fields and Lagrangian submanifolds of the nearly Kaehler S 6[END_REF], [START_REF] Vrancken | Special Lagrangian submanifolds of the nearly Kaehler 6-sphere[END_REF], [START_REF] Lotay | Ruled Lagrangian submanifolds of the 6-sphere[END_REF] and [START_REF] Palmer | Calibrations and Lagrangian submanifolds in the six sphere (English summary)[END_REF]), whereas the study of Lagrangian submanifolds of S 3 × S 3 only started recently.

The first examples of those were given in [START_REF] Schäfer | Decomposition and minimality of Lagrangian submanifolds in nearly Kähler manifolds[END_REF] and [START_REF] Moroianu | Generalized Killing spinors and Lagrangian graphs[END_REF]. Moreover, in [START_REF] Zhang | Lagrangian submanifolds in the 6-dimensional nearly Kähler manifolds with parallel second fundamental form[END_REF] and [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF], the authors obtained a classification of the Lagrangian submanifolds, which are either totally geodesic or have constant sectional curvature. An important tool in the study in [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF] and [START_REF] Zhang | Lagrangian submanifolds in the 6-dimensional nearly Kähler manifolds with parallel second fundamental form[END_REF] is the use of an almost product structure P on S 3 × S 3 , which was introduced in [START_REF] Bolton | Almost complex surfaces in the nearly Kähler S 3 × S 3[END_REF]. Its definition is recalled in Section 2. The decomposition of P into a tangential part and a normal part along a Lagrangian submanifold allows us to introduce three principal directions, E 1 , E 2 , E 3 , with corresponding angle functions θ 1 , θ 2 , θ 3 . In this paper we are interested in studying non-totally geodesic Lagrangian submanifolds f : M → S 3 × S 3 : x → f (x) = (p(x), q(x)), for which the first component has nowhere maximal rank. Basic properties of the structure and its Lagrangian submanifolds are given in Section 2. In Section 3, we show that in this case θ 1 = π 3 (Theorem 1) and p(M ) has to be a (branched) minimal surface in S 3 (Theorem 3). Conversely, for a non-totally geodesic minimal surface in S 3 which locally corresponds to a solution of the Sinh-Gordon equation, ∆ω = -8 sinh ω, and for an additional arbitrary solution of the Liouville equation, ∆µ = -e µ , we can construct locally a Lagrangian immersion of S 3 × S 3 . Thus, we obtain a large class of examples of Lagrangian immersions. We also obtain that a similar class of Lagrangian immersions can be associated to a totally geodesic surface in S 3 . This last case contains, in particular, the constant curvature sphere obtained in [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF]. Additionally, for each non-totally geodesic minimal surface, we obtain also one exceptional example. In case of the Clifford torus in S 3 , this additional example is the flat Lagrangian torus in S 3 × S 3 discovered in [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF]. We also show that any non-totally geodesic Lagrangian immersion for which the first component has nowhere maximal rank is obtained by applying one of the three previously mentioned constructions. The main results are summarized in Section 4.

Preliminaries

In this section we recall the homogeneous nearly Kähler structure of S 3 × S 3 and we mention some of the known results from [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF] and [START_REF] Zhang | Lagrangian submanifolds in the 6-dimensional nearly Kähler manifolds with parallel second fundamental form[END_REF]. By the natural identification T (p,q) (S 3 × S 3 ) ∼ = T p S 3 ⊕ T q S 3 , we write a tangent vector at (p, q) as Z(p, q) = (U (p, q), V (p, q)) or simply Z = (U, V ). We regard the 3-sphere as the set of all unit quaternions in H and we use the notations i, j, k to denote the imaginary units of H. In computations it is often useful to write a tangent vector Z(p, q) at (p, q) on S 3 × S 3 as (pα, qβ), with α and β imaginary quaternions. This is possible as for v ∈ T p S 3 we know that v, p = 0 and, in addition, for p ∈ S 3 we can always find ṽ ∈ H such that v = pṽ . Moreover, Re(ṽ) = 0 as 0 = p, v = Re(pv) = Re(ppṽ) = Re(ṽ). We define the vector fields [START_REF] Bolton | Almost complex surfaces in the nearly Kähler S 3 × S 3[END_REF] Ẽ1 (p, q) = (pi, 0), F1 (p, q) = (0, qi), Ẽ2 (p, q) = (pj, 0), F2 (p, q) = (0, qj), Ẽ3 (p, q) = -(pk, 0), F3 (p, q) = -(0, qk), which are mutually orthogonal with respect to the usual Euclidean product metric on S 3 × S 3 . The Lie brackets are [ Ẽi , Ẽj ] = -2ε ijk Ẽk , [ Fi , Fj ] = -2ε ijk Fk and [ Ẽi , Fj ] = 0, where

ε ijk =    1, if (ijk)
is an even permutation of (123), -1, if (ijk) is an odd permutation of (123), 0, otherwise.

The almost complex structure J on the nearly Kähler S 3 × S 3 is defined by

(2) J(U, V ) (p,q) = 1 √ 3 (2pq -1 V -U, -2qp -1 U + V ),
for (U, V ) ∈ T (p,q) (S 3 ×S 3 ). The nearly Kähler metric on S 3 ×S 3 is the Hermitian metric associated to the usual Euclidean product metric on S 3 × S 3 :

g(Z, Z ) = 1 2 ( Z, Z + JZ, JZ ) (3) = 4 3 ( U, U + V, V ) - 2 3 ( p -1 U, q -1 V + p -1 U , q -1 V ),
where Z = (U, V ) and Z = (U , V ). In the first line •, • stands for the usual Euclidean product metric on S 3 × S 3 and in the second line •, • stands for the usual Euclidean metric on S 3 . By definition, the almost complex structure is compatible with the metric g.

From [START_REF] Bolton | Almost complex surfaces in the nearly Kähler S 3 × S 3[END_REF] we have the following lemma.

Lemma 1. The Levi-Civita connection ∇ on S 3 × S 3 with respect to the metric g is given by

∇ Ẽi Ẽj = -ε ijk Ẽk ∇ Ẽi Fj = ε ijk 3 ( Ẽk -Fk ) ∇ Fi Ẽj = ε ijk 3 ( Fk -Ẽk ) ∇ Fi Fj = -ε ijk Fk . Then we have that (4) ( ∇ Ẽi J) Ẽj = -2 3 √ 3 ε ijk ( Ẽk + 2 Fk ), ( ∇ Ẽi J) Fj = -2 3 √ 3 ε ijk ( Ẽk -Fk ), ( ∇ Fi J) Ẽj = -2 3 √ 3 ε ijk ( Ẽk -Fk ), ( ∇ Fi J) Fj = -2 3 √ 3 ε ijk (2 Ẽk + Fk ). Let G := ∇J. Then G is skew-symmetric and satisfies that (5) G(X, JY ) = -JG(X, Y ), g(G(X, Y ), Z) + g(G(X, Z), Y ) = 0,
for any vectors fields X, Y, Z tangent to S 3 × S 3 . Therefore, S 3 × S 3 equipped with g and J, becomes a nearly Kähler manifold.

The almost product structure P introduced in [START_REF] Bolton | Almost complex surfaces in the nearly Kähler S 3 × S 3[END_REF] is defined as ( 6)

P Z = (pq -1 V, qp -1 U ), ∀Z = (U, V ) ∈ T (p,q) (S 3 × S 3 )
plays an important role in the study of the Lagrangian submanifolds of the nearly Kähler S 3 × S 3 . It has the following properties:

P 2 = Id (P is involutive),
P J = -JP (P and J anti-commute), g(P Z, P Z ) = g(Z, Z ) (P is compatible with g), g(P Z, Z ) = g(Z, P Z ) (P is symmetric).

Moreover, the almost product structure P can be expressed in terms of the usual product structure QZ = Q(U, V ) = (-U, V ) and vice versa:

QZ = 1 √ 3 (2P JZ -JZ), P Z = 1 2 (Z - √ 3QJZ).
Next, we recall the relation between the Levi-Civita connections ∇ of g and ∇ E of the Euclidean product metric •, • .

Lemma 2. [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF] The relation between the nearly Kähler connection ∇ and the Euclidean connection

∇ E is ∇ E X Y = ∇X Y + 1 2 (JG(X, P Y ) + JG(Y, P X)).
We recall here a useful formula, already known in [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF]. Let D be the Euclidean connection on R 8 . For vector fields X = (X 1 , X 2 ) and Y = (Y 1 , Y 2 ) on S 3 × S 3 , we may decompose D X Y along the tangent and the normal directions as follows:

(7) D X Y = ∇ E X Y + 1 2 D X Y, (p, q) (p, q) + 1 2 D X Y, (-p, q) (-p, q).
Here, notice the factor 1 2 due to the fact that (p, q) and (-p, q) have length √ 2. Moreover, as Y, (p, q) = 0, ( 7) is equivalent with

D X Y = ∇ E X Y - 1 2 Y, X (p, q) - 1 2 Y, (-X 1 , X 2 ) (-p, q).
In the special case that Y 2 = 0, the previous formula reduces to

(8) D X (Y 1 , 0) = ∇ E X (Y 1 , 0) -X 1 , Y 1 (p, 0).
We find it appropriate here to prove an additional important formula not explicitly mentioned in [START_REF] Bolton | Almost complex surfaces in the nearly Kähler S 3 × S 3[END_REF], that allows us to evaluate G for any tangent vector fields.

Proposition 1. Let X = (pα, qβ), Y = (pγ, qδ) ∈ T (p,q) S 3 × S 3 . Then G(X, Y ) = 2 3 √ 3 (p(β × γ + α × δ + α × γ -2β × δ), q(-α × δ -β × γ + 2α × γ -β × δ)). (9)
Proof. As α is an imaginary unit quaternion, we may write α = α 1 • i + α 2 • j + α 3 • k and similarly for β, γ, δ. Then, using (1), we write for more convenience in computations X = U α + V β , where

U α = α 1 Ẽ1 + α 2 Ẽ2 -α 3 Ẽ3 and V β = β 1 F1 + β 2 F2 -β 3 F3 . Similarly, Y = U γ + V δ .
We now use the relations in (4) and compute

G(U α , V β ) = 2 3 √ 3 (U α×β -V α×β ), G(U α , U β ) = 2 3 √ 3 (U α×β + 2V α×β ).
As P U α = V α , we obtain that

G(V α , V β ) = - 2 3 √ 3 (V α×β + 2U α×β ).
Finally, by linearity we get the relation in [START_REF] Ejiri | Equivariant minimal immersions of S 2 into S 2m (1)[END_REF].

From now on we will restrict ourselves to 3-dimensional Lagrangian submanifolds M of S 3 × S 3 . It is known from [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF] and [START_REF] Zhang | Lagrangian submanifolds in the 6-dimensional nearly Kähler manifolds with parallel second fundamental form[END_REF] that, as the pull-back of T (S 3 × S 3 ) to M splits into T M ⊕ JT M , there are two endomorphisms A, B : T M → T M such that the restriction P | T M of P to the submanifold equals A + JB, that is P X = AX + JBX, for all X ∈ T M . Note that the previous formula, together with the fact that P and J anti-commute, also determines P on the normal space by P JX = -JP X = BX -JAX. In addition, from the properties of J and P it follows that A and B are symmetric operators which commute and satisfy moreover that A 2 + B 2 = Id (see [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF]). Hence A and B can be diagonalised simultaneously at a point p in M and there is an orthonormal basis e 1 , e 2 , e 3 ∈ T p M such that (10) P e i = cos(2θ i )e i + sin(2θ i )Je i .

The functions θ i are called the angle functions of the immersion. Next, for a point p belonging to an open dense subset of M on which the multiplicities of the eigenvalues of A and B are constant (see [START_REF] Szabo | Structure theorems on riemannian spaces satisfying R(X, Y ) • R = 0. I. the local version[END_REF]), we may extend the orthonormal basis e 1 , e

JG(E i , E j ) = 1 √ 3 ε ijk E k .
The following result is known ( [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF]):

Proposition 2. The sum of the angles θ 1 + θ 2 + θ 3 is zero modulo π.

For the Levi-Civita connection ∇ on M we introduce (see [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF]) the functions ω k ij satisfying

∇ Ei E j = 3 k=1 ω k ij E k and ω k ij = -ω j ik .
As usual, we write:

∇X Y = ∇ X Y + h(X, Y ), ∇X JY = -S JY X + ∇ ⊥ X JY
, where h is the second fundamental form on M and S JY is the shape operator in the direction of JY . As for the Lagrangian manifolds of a strict 6-dimensional nearly Kähler manifold we have that G(X, Y ) is normal (see [START_REF] Gutowski | Deformations of generalized calibrations and compact non-Kähler manifolds with vanishing first Chern class[END_REF], [START_REF] Schäfer | Decomposition and minimality of Lagrangian submanifolds in nearly Kähler manifolds[END_REF]), it follows that

∇ ⊥ X JY = J∇ X Y + G(X, Y ), Jh(X, Y ) = -S JY X.
The latter equation implies in particular that the cubic form g(h(X, Y ), JZ) is totally symmetric. We denote by h k ij the components of this cubic form on M : (13)

h k ij = g(h(E i , E j ), JE k ).
3. Results

3.1.

Elementary properties of orientable minimal surfaces in S 3 . We recall some elementary properties of minimal surfaces. Let p : S → S 3 ⊂ R 4 be an oriented minimal surface. We are going to check that the immersion either admits local isothermal coordinates for which the conformal factor satisfies the Sinh-Gordon equation or is totally geodesic. First, we take isothermal coordinates u, v such that ∂u, ∂v is positively oriented, ∂u, ∂u = ∂v, ∂v = 2e ω and ∂u, ∂v = 0 in a neighborhood of a point of S. As it is often more useful to use complex notation we write z = u + Iv and consider ∂z = 1 2 (∂u -I∂v) and ∂ z = 1 2 (∂u + I∂v). Note that we use I here in order to distinguish between the i, j, k introduced in the quaternions. We also extend everything in a linear way in I. This means that ∂z, ∂z = ∂ z, ∂ z = 0 and ∂z, ∂ z = e ω . If we write ∂u = pα and ∂v = pβ, the unit normal is given by N = p α×β 2e ω . It is elementary to check that this is independent of the choice of complex coordinate and that the matrix p ∂u |∂u| ∂v |∂v| N belongs to SO(4). We denote by σ the component of the second fundamental form in the direction of N . Remark that with this choice, the minimality of the surface implies σ(∂z, ∂ z) = 0 and we may determine the components of the connection ∇ on the surface:

(14) ∇ ∂z ∂z = ω z ∂z, ∇ ∂z ∂ z = ∇ ∂ z ∂z = 0 and ∇ ∂ z = ω z ∂ z.
The Codazzi equation of a surface in S 3 states that ∇σ(∂z, ∂ z, ∂z) = ∇σ(∂ z, ∂z, ∂z).

So it follows that ∂ z(σ(∂z, ∂z)) = 0. Hence σ(∂z, ∂z) is a holomorphic function. Then we have two cases: Case 1. If σ(∂z, ∂z) = 0 on an open set, then by conjugation σ(∂ z, ∂ z) = 0 and therefore, using the analyticity of a minimal surface, σ = 0 everywhere. Case 2. If σ(∂z, ∂z) = 0, then there exists a function g(z) such that σ(∂z, ∂z) = g(z). Away from isolated points we can always make a change of coordinates if necessary such that σ(∂z, ∂z) = -1.

Notice that by conjugation we get also σ(∂ z, ∂ z) = -1. Such a change of coordinates is unique up to translations and replacing z by -z.

Next, given the immersions p : S → S 3 (1) i → R 4 , from the Gauss formula we obtain:

p zz = ω z p z -N, p z z = -e ω p, (15)
p z z = ω z p z -N,
where N is the normal on S 3 and N z = e -ω p z , N z = e -ω p z . Therefore

p zz z = (ω z z -e -ω )p z -ω z e ω p, p z zz = -e ω ω z p -e ω p z , which shows that ω satisfies ω z z = -2 sinh ω ⇐⇒ ∆ω = -8 sinh ω (Sinh-Gordon equation). ( 16
)
Notice that by ∆ω we denote the Euclidean Laplacian of ω in R 2 = C. Let P be the lift of the minimal immersion to the immersion of the frame bundle in SO(4), i.e.

P : U S → SO(4) : w → (p w Jw N ),
where U S denotes the unit tangent bundle of S and J denotes the natural complex structure on an orientable surface. In terms of our chosen isothermal coordinate this map can be parametrised by

P(u, v, t) = p(u, v), cos t p u | p u | + sin t p v | p v | , -sin t p u | p u | + cos t p v | p v | , N (u, v) ,
for some real parameter t. Note that we have the frame equations which state that

dP = PΩ t = -PΩ,
where in terms of the coordinates u, v and t the matrix Ω is given by

        0 √ 2e ω 2 (cos(t)du + sin(t)dv) √ 2e ω 2 (cos(t)dv -sin(t)du) 0 - √ 2e ω 2 (cos(t)du+ sin(t)dv) 0 1 2 (ω u dv -ω v du) + dt - √ 2e -ω 2 (cos(t)du- sin(t)dv) - √ 2e ω 2 (cos(t)dv- sin(t)du) -1 2 (ω u dv -ω v du) -dt 0 √ 2e -ω 2 (sin(t)du+ cos(t)dv) 0 √ 2e -ω 2 (cos(t)du -sin(t)dv) - √ 2e -ω 2 (sin(t)du + cos(t)dv) 0         . 3.2.
From the Lagrangian immersion to the minimal surface. Now we will consider Lagrangian submanifolds in the nearly Kähler S 3 × S 3 . We write the Lagrangian submanifold M as

f : M → S 3 × S 3 x → f (x) = (p(x), q(x)),
and we assume that the first component has nowhere maximal rank. We have the following:

Theorem 1. Let f : M → S 3 × S 3 x → f (x) = (p(x), q(x)),
be a Lagrangian immersion such that p : M → S 3 has nowhere maximal rank. Then π 3 is an angle function up to a multiple of π. The converse is also true.

Proof. It is clear that p has nowhere maximal rank if and only if there exists a non zero vector field X such that dp(X) = 0. As usual we identify df (X) with X, so we have that X = df (X) = (dp(X), dq(X)) and QX = (-dp(X), dq(X)). Therefore p has nowhere maximal rank if and only if

X = QX = 1 √ 3 (2P JX -JX) = 1 √ 3 (2BX -2JAX -JX).
Comparing tangent and normal components we see that this is the case if and only if

AX = -1 2 X BX = √ 3 2 X.
So we see that X is an eigenvector of both A and B and that the corresponding angle function is

π 3 (up to a multiple of π).
For the remainder of the paper we will consider Lagrangian immersions for which the map p has nowhere maximal rank. In view of the previous lemma this means that one of the angle functions is constant, namely θ 1 = π 3 . Then using that the angles are only determined up to a multiple of π and given that 2θ 1 + 2θ 2 + 2θ 3 is a multiple of 2π, we may write

2θ 1 = 2π 3 , 2θ 2 = 2Λ + 2π 3 , 2θ 3 = -2Λ + 2π 3 , (17)
for Λ an arbitrary function which takes values in [-π 2 , π 2 ]. If necessary by interchanging E 2 , E 3 with -E 3 , E 2 we may assume that Λ ≥ 0 and therefore Λ takes values only in [0, π 2 ]. Similarly, if necessary interchanging E 1 , E 3 by -E 1 , -E 3 , we may also assume that h 3 13 ≤ 0 (see equation ( 13)). Note however that at the points where Λ is 0 or π 2 modulo π, we have that two of the angle functions coincide. If this is true on an open set, it follows from [START_REF] Zhang | Lagrangian submanifolds in the 6-dimensional nearly Kähler manifolds with parallel second fundamental form[END_REF] that the Lagrangian submanifold is totally geodesic and is congruent either with f :

S 3 → S 3 × S 3 : u → (1, u) or f : S 3 → S 3 × S 3 : u → (uiu -1 , u -1
). So by restricting to an open dense subset of M which we denote by M * , we may actually assume that Λ ∈ (0, π 2 ), in which case the function Λ, as well as the vector fields E 1 , E 2 , E 3 are differentiable. Notice that the case when Λ is constant is treated in [START_REF] Bektaş | Lagrangian submanifolds with constant angle functions in the nearly Kähler S 3 × S 3[END_REF], where such Lagrangian submanifolds are determined to be either totally geodesic or of constant sectional curvature. As we consider here Λ ∈ (0, π 2 ), the only possibility is Λ = π 3 , in which case the Lagrangian submanifold is not totally geodesic, but of constant sectional curvature.

Theorem 2. Let M be a Lagrangian submanifold of constant sectional curvature in the nearly Kähler S 3 ×S 3 . If M is not totally geodesic, then up to an isometry of the nearly Kähler S 3 ×S 3 , M is locally congruent with one of the following immersions:

(1) f :

S 3 → S 3 × S 3 : u → (uiu -1 , uju -1 ), ( 2
) f : R 3 → S 3 ×S 3 : (u, v, w) → (p(u, w), q(u, v))
, where p and q are constant mean curvature tori in S 3 given by p(u, w) = (cos u cos w, cos u sin w, sin u cos w, sin u sin w) ,

q(u, v) = 1 √ 2 (cos v (sin u + cos u) , sin v (sin u + cos u) , cos v (sin u -cos u) , sin v (sin u -cos u)) .
Note that these are precisely the two Lagrangian immersions with constant sectional curvature obtained in [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF]. These two examples will appear as special solutions in respectively Case 2 and Case 3. However we will mainly focus on the case that Λ is not constant. In the following, we will identify a tangent vector X in T x M with its image through df in T (p,q) S 3 × S 3 , that is X ≡ df (X) = (dp(X), dq(X)), and we can write QX ≡ Q(df (X)) = (-dp(X), dq(X)). Therefore, if we see dp(X) projected on the first factor of S 3 × S 3 , that is dp(X) ≡ (dp(X), 0), we can write [START_REF] Nagy | On nearly-Kähler geometry[END_REF] dp(X) = 1 2 (X -QX).

We use relations [START_REF] Ejiri | Totally real submanifolds in a 6-sphere[END_REF] and [START_REF] Nagy | Nearly Kähler geometry and Riemannian foliations[END_REF]

to compute P E 1 = -1 2 E 1 + √ 3 2 JE 1 .
As mentioned before this is equivalent with stating that dp(E 1 ) = 0 and that p has nowhere maximal rank. By straightforward computations we obtain [START_REF] Palmer | Calibrations and Lagrangian submanifolds in the six sphere (English summary)[END_REF] (dp

(E 2 ), 0) = 1 2 -1 √ 3 sin(2Λ + 2π 3 ) E 2 + 1 √ 3 1 2 + cos(2Λ + 2π 3 ) JE 2 , (dp(E 3 ), 0) = 1 2 -1 √ 3 sin(-2Λ + 2π 3 ) E 3 + 1 √ 3 1 2 + cos(-2Λ + 2π 3 ) JE 3 and dp(E 2 ), dp(E 2 ) = sin 2 Λ, dp(E 3 ), dp(E 3 ) = sin 2 Λ, ( 20
)
dp(E 2 ), dp(E 3 ) = 0.
We denote

v 2 := dp(E 2 ) ≡ (dp(E 2 ), 0), v 3 := dp(E 3 ) ≡ (dp(E 3 ), 0), (21) ξ = 1 √ 3 E 1 -JE 1
and we may easily see that Qξ = -ξ, i.e. ξ lies entirely on the first factor of S 3 × S 3 . Moreover, v i , v j = δ ij sin Λ, ξ, v 2 = ξ, v 3 = 0 and ξ, ξ = 1. Therefore, p(M ) is a surface in S 3 and ξ can be seen as a unit normal to the surface. As far as the Lagrangian immersion itself is concerned we also have due to the minimality that

h 1 11 + h 2 12 + h 3 13 = 0, h 2 11 + h 2 22 + h 3 23 = 0, h 3 11 + h 3 22 + h 3 33 = 0. (22)
From [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF] we know that the covariant derivatives of the endomorphisms A and B are

(∇ X A)Y = BS JX Y -Jh(X, BY ) + 1 2 (JG(X, AY ) -AJG(X, Y )), (23) (∇ X B)Y = -AS JX Y + Jh(X, AY ) + 1 2 (JG(X, AY ) -AJG(X, Y )). ( 24
)
We are going to use the definition of ∇A and ∇B in the previous expressions and then evaluate them for different vectors in the basis in order to get information about the functions ω k ij and h k ij . For X = Y = E 1 in ( 23) we obtain that

h 2 12 = -h 3 13 , ω 2 11 = h 2 11 cot Λ, ω 3 11 = -h 3 11 cot Λ. ( 25
)
If we take X = E 1 and Y = E 2 in ( 23) and ( 24), we see that

E 1 (Λ) = h 3 13 , (26)
ω 3 12 = √ 3 6 -h 3 12 cot 2Λ (27)
and, for X = E 2 and Y = E 1 in (23), we obtain 23) and ( 24) and obtain h 3 11 = 0, (31)

h 2 11 = 0, (28) ω 2 21 = -cot Λh 3 13 , (29)
ω 3 21 = - √ 3 6 -h 3 12 cot Λ. (30) Then we choose successively X = E 3 , Y = E 1 , X = E 2 , Y = E 3 and X = E 3 , Y = E 2 in relations (
ω 2 31 = √ 3 6 + cot Λh 3 12 , (32)
ω 3 31 = -cot Λh 3 13 , (33)
ω 3 22 = -cot 2Λh 3 22 , (34)
ω 3 32 = -cot 2Λh 3 23 , (35)
E 2 (Λ) = h 3 23 , (36)
E 3 (Λ) = -h 3 22 . (37)
We can easily see from (25), ( 28) and (31) that ω 2 11 = 0 and ω 3 11 = 0 and, if we consider as well the relations in [START_REF] Vrancken | Killing vector fields and Lagrangian submanifolds of the nearly Kaehler S 6[END_REF], we have that

h 3 33 = -h 3 22 , h 1 11 = 0 and h 2 22 = -h 3 23 .
Later on we will also need to study the Codazzi equations for M . From [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF] we know their general form:

∇h(X, Y, Z) -∇h(Y, X, Z) = 1 3 (g(AY, Z)JBX -g(AX, Z)JBY -g(BY, Z)JAX + g(BX, Z)JAY ). ( 38
)
We are going to use the definition for ∇h in the previous relation and take different values for the vectors X, Y and Z. Thus, we evaluate it successively for E 2 , E 3 , E 3 and we obtain the following relations, respectively:

E 1 , E 2 , E 1 ; E 1 , E 2 , E 2 ; E 1 , E 3 , E 3 ; E 1 , E 3 , E 2 and
E 1 (h 3 13 ) = 1 3 (- √ 3h 3 12 + 6(h 3 13 ) 2 cot Λ -6(h 3 12 ) 2 csc(2Λ) + sin(2Λ)), E 1 (h 3 12 ) = 1 3 h 3 13 ( √ 3 + 9h 3 12 cot Λ + 3h 3 12 tan Λ), (39)
E 2 (h 3 13 ) -E 1 (h 3 23 ) = 1 √ 3 h 3 22 + h 3 12 h 3 22 cot Λ -h 3 13 h 3 23 cot Λ -h 3 12 h 3 22 cot(2Λ), E 1 (h 3 22 ) -E 2 (h 3 12 ) =h 3 13 h 3 22 (2 cot Λ -tan Λ) + 1 6 h 3 23 (2 √ 3 -3h 3 12 cot Λ + 9h 3 12 tan Λ), E 3 (h 3 12 ) -E 1 (h 3 23 ) = 1 √ 3 h 3 22 + (h 3 12 h 3 22 -h 3 13 h 3 23 ) cot Λ -(3h 3 12 h 3 22 + 2h 3 13 h 3 23 ) cot(2Λ), E 3 (h 3 13 ) + E 1 (h 3 22 ) = 1 √ 3 h 3 23 + h 3 13 h 3 22 cot Λ + h 3 12 h 3 23 cot Λ -h 3 12 h 3 23 cot(2Λ), (40) E 2 (h 3 13 ) -E 3 (h 3 12 ) =2(h 3 12 h 3 22 + h 3 13 h 3 23 ) cot(2Λ), E 3 (h 3 22 ) -E 2 (h 3 23 ) = - 1 2 (8(h 3 12 ) 2 + 4(h 3 13 ) 2 + 3((h 3 22 ) 2 + (h 3 23 ) 2 )) cot Λ- 1 3 ( √ 3h 3 12 + sin 4Λ) + 3 2 ((h 3 22 ) 2 + (h 3 23 ) 2 ) tan Λ, E 2 (h 3 22 ) + E 3 (h 3 23 ) = - 1 3 h 3 13 ( √ 3 + 6h 3 12 cot Λ). Theorem 3. Let f : M → S 3 × S 3 x → f (x) = (p(x), q(x)),
be a Lagrangian immersion such that p : M → S 3 has nowhere maximal rank. Assume that M is not totally geodesic. Then p(M ) is a (branched) minimal surface in S 3 . Moreover

P : M * → SO(4) : x → p(x) v 2 sin Λ v 3 sin Λ ξ ,
where v 2 , v 3 and ξ are defined by [START_REF] Szabo | Structure theorems on riemannian spaces satisfying R(X, Y ) • R = 0. I. the local version[END_REF], is a map which is contained into the frame bundle over the minimal surface p.

Proof. Recall that dp(E 1 ) = 0, hence p(M ) is a surface. Denoting the second fundamental form of the surface in the direction of ξ by σ, a straightforward computation yields that (41)

σ(E 2 , E 2 ) = h 3 13 , σ(E 2 , E 3 ) = σ(E 3 , E 2 ) = 1 √ 3 cos Λ sin Λ -h 3 12 , σ(E 3 , E 3 ) = -h 3
13 . As dp(E 2 ) and dp(E 3 ) are orthogonal and have the same length, the above formulas indeed imply that the surface is minimal. Moreover we also see that the surface is totally geodesic if and only if h 3 13 = 0 and h 3 12 =

1 √ 3 cos Λ sin Λ. Note also that if we write (dp(E 2 ), 0) = (pα, 0) and (dp(E 3 ), 0) = (pγ, 0), we have that

G((dp(E 2 ), 0), (dp(E 3 ), 0)) = G((pα, 0), (pγ, 0)) = 2 3 √ 3 (p(α × γ), 2q(α × γ)). Therefore, (p(α × γ), 0) = 3 √ 3 4 (G((dp(E 2 ), 0), (dp(E 3 ), 0)) -Q(G((dp(E 2 ), 0), (dp(E 3 ), 0))))
. A straightforward computation, using [START_REF] Palmer | Calibrations and Lagrangian submanifolds in the six sphere (English summary)[END_REF] and [START_REF] Gray | Nearly Kähler manifolds[END_REF], shows that this gives (p(α × γ), 0) = (sin Λ) 2 ξ.

Therefore ξ corresponds with the normal N on the surface.

The reverse construction.

In the following, we will separate the study of the submanifold into three cases, according to whether the surface is totally geodesic or not and whether the map to the frame bundle is an immersion or not.

3.3.1. Case 1. p(M ) is not a totally geodesic surface and the map P is an immersion.

In that case we can identify M with the frame bundle on the minimal surface induced earlier. Recall that

P : x ∈ M * → p v 2 sin Λ v 3 sin Λ ξ .
Writing again d P = -P Ω, we can express the matrix Ω in terms of

{E 1 , E 2 , E 3 } by                0 sin(Λ)ω 2 sin(Λ)ω 3 0 -sin(Λ)ω 2 0 1 √ 3 + h 3 12 csc(2Λ) ω 1 + h 3 22 csc(2Λ)ω 2 + h 3 23 csc(2Λ)ω 3 h 3 13 csc(Λ)ω 2 + cos Λ √ 3 -h 3 12 csc Λ ω 3 -sin(Λ)ω 3 -( 1 √ 3 + h 3 12 csc(2Λ))ω 1 - h 3 22 csc(2Λ)ω 2 -h 3 23 csc(2Λ)ω 3 0 cos Λ √ 3 -h 3 12 csc(Λ) ω 2 - h 3 13 csc(Λ)ω 3 0 -cos Λ √ 3 + h 3 12 csc Λ ω 3 - h 3 13 csc(Λ)ω 2 -cos Λ √ 3 -h 3 12 csc(Λ) ω 2 - h 3 13 csc(Λ)ω 3 0               
, where ω i (E j ) = δ ij . The above matrix implies that the map P into SO(4) ⊂ R 16 is an immersion if and only if

1 √ 3 + h 3 12 csc(2Λ) = 0.
As it is an immersion, in view of the dimensions, its image is an open part of the frame bundle and we can identify M with an open part of the frame bundle on the minimal surface. Moreover we can write

v 2 sin Λ = cos(t + γ(t, u, v)) p u | p u | + sin(t + γ(t, u, v)) p v | p v | ,
where γ is some function. As P is an immersion, we have that t + γ(t, u, v) depends on t and can be taken as the new variable t on the frame bundle. Doing so, we have that P = P and Ω = Ω (for P, Ω as in subsection 3.1). Comparing both expressions for the matrix Ω we deduce

ω 1 = 1 1 √ 3 + h 3 12 csc 2Λ - √ 2 csc 2Λ sin Λ e ω/2 (h 3 22 cos t -h 3 23 sin t) + 1 2 ω v du- √ 2 csc 2Λ sin Λ e ω/2 (h 3 22 sin t + h 3 23 cos t) - 1 2 ω u dv + dt , ω 2 = 1 sin Λ √ 2e ω/2 (cos(t)du + sin(t)dv), ω 3 = 1 sin Λ √ 2e ω/2 (cos(t)dv -sin(t)du),
as well as

e -ω cos(2t) + h 3 13 1 sin 2 Λ = 0, e -ω sin(2t) + h 3 12 csc Λ -cos Λ √ 3 1 sin Λ = 0, (42)
which implies that

h 3 13 = -e -ω cos(2t) sin 2 Λ, h 3 12 = -e -ω sin(2t) sin Λ + cos Λ √ 3 sin Λ. ( 43
)
We may express E 1 , E 2 , E 3 with respect to the basis {∂t, ∂u, ∂v} as follows. For E i = a i ∂t+b i ∂u+ c i ∂v, we use the previously obtained expressions of ω j in ω j (E i ) = δ ij and by straightforward computations we get

E 1 = 1 √ 3 + h 3 12 csc(2Λ) ∂t, E 2 =(csc(2Λ)h 3 22 + 1 2 √ 2 sin Λe -ω/2 (cos(t)ω v -sin(t)ω u ))∂t+ e -ω/2 cos t sin Λ √ 2 ∂u + e -ω/2 sin t sin Λ √ 2 ∂v, (44)
E 3 = csc(2Λ)h 3 23 - 1 2 √ 2 sin Λ e -ω/2 (cos(t)ω u + sin(t)ω v ) ∂t- e -ω/2 sin t sin Λ √ 2 ∂u + e -ω/2 cos t sin Λ √ 2 ∂v.
In order to be able to proceed with the reverse construction, i.e. in order to be able to construct a Lagrangian immersion starting from the minimal surface we need to express Λ, h 3 22 and h 3 23 in terms of the variables t, u, v. Remark that, as E 1 (Λ) = h 3 13 , we may use (43) and the expression of E 1 in (44) to determine how Λ depends on the variable t. We get (45)

Λ t = - 2 cos(2t) sin 2 Λ √ 3e ω -2 cos t sin t tan Λ .
In order to solve the above differential equation, we use (45) to compute the derivative of the expression where ε 1 = ±1 and, at the same time, the surface is defined on an open set where c 1 -2 cos(4t) ≥ 0. Note that as the above expression contains a square root which would complicate simplifications later on, we will avoid its use as much as possible. For later use, remark that we can write

√ 3e ω tan Λ -sin(2t): ∂t √ 3e ω tan Λ -sin(2t)
(47) 2 √ 3e ω tan Λ -2 sin(2t) 2 = c 1 -2 cos(4t).
As we can rewrite the above equation as

2 √ 3e ω tan Λ -2 sin(2t) 2 + 2 cos(4t) + 2 = c 1 + 2,
we see that c Combining this with the previous expression of c 1 and taking the derivative with respect to u and v, we can compute

Λ u = - sin 2 Λ µ u + e ω cot Λ 3e ω cot Λ(µ u -ω u ) -2 √ 3µ u sin(2t) + ω u 6e 2ω cot Λ -2 √ 3e ω sin(2t) Λ v = - sin 2 Λ µ v + e ω cot Λ 3e ω cot Λ(µ v -ω v ) -2 √ 3µ v sin(2t) + ω v 6e 2ω cot Λ -2 √ 3e ω sin(2t) .
Using this, together with (44), we can solve in (36) and (37), for h 3 22 and h 3 23 . This gives us

h 3 22 = e -3ω/2 sin 2 Λ 6 √ 2 3e ω cos Λ((ω u -µ u ) sin t + (µ v -ω v ) cos t)- √ 3 sin Λ((µ u + ω u ) cos(3t) + (µ v + ω v ) sin(3t)) , h 3 23 = e -3ω/2 sin 2 α 6 √ 2 √ 3 sin Λ((µ u + ω u ) sin(3t) + (-µ v -ω v ) cos(3t))- 3e ω cos Λ(µ u -ω u ) cos t -3e ω cos Λ(µ v -ω v ) sin t) .
In order to determine a differential equation for the function µ we now apply the previously obtained Codazzi equations for M . By (44), it turns out that (39) and the first 5 equations of (40) are trivially satisfied. Recall from ( 16) that ∆ω = -8 sinh ω. The seventh equation of (40) reduces to (48) ∆µ = -4e ω (cos(2Λ) + 2) csc 2 Λ + 8 √ 3 cot Λ sin(2t) + 8 sinh ω.

A straightforward computation, using the definition of µ and (47), shows that this reduces to

(49) ∆µ = -e µ .
Further on, with these new notations, we may see by straightforward computations that the sixth equation of ( 40) is now trivially satisfied.

Reverse construction

We denote by p : S → S 3 ⊂ R 4 a given minimal surface S which is not totally geodesic, on which we take suitable isothermal coordinates as introduced before. Hence we have a solution ω of ∆ω = -8 sinh ω. Additionally, we take a solution of (50) ∆µ = -e µ and we take the open part of the frame bundle such that

(51) 2 √ 3e ω tan Λ -2 sin(2t) 2 = e ω+µ -2 -2 cos(4t)
has a solution for the function Λ on an open domain. We define

h 3 13 = -e ω cos(2t) sin 2 Λ, h 3 12 = (-e -ω sin(2t) sin Λ + cos Λ √ 3 ) sin Λ, h 3 22 = e -3ω/2 sin 2 Λ 6 √ 2 3e ω cos Λ((ω u -µ u ) sin t + (µ v -ω v ) cos t)- √ 3 sin Λ((µ u + ω u ) cos(3t) + (µ v + ω v ) sin(3t)) , h 3 23 = e -3ω/2 sin 2 Λ 6 √ 2 √ 3 sin Λ((µ u + ω u ) sin(3t) + (-µ v -ω v ) cos(3t))- 3e ω cos Λ(µ u -ω u ) cos t -3e ω cos Λ(µ v -ω v ) sin t)
and we define as well a metric on the open part of the frame bundle, by assuming that the vectors

E 1 = 1 2 √ 3 -2e -ω tan Λ sin t cos t ∂t, E 2 = - e -3ω/2 sin Λ 12 √ 2 √ 3 tan Λ((µ u + ω u ) cos(3t) + (µ v + ω v ) sin(3t)) + 3e ω ((µ u + ω u ) sin t+ (-µ v -ω v ) cos t) ∂t + e -ω 2 cos t sin Λ √ 2 ∂u + e -ω 2 sin t sin Λ √ 2 ∂v, (52)
E 3 = e -3ω/2 sin Λ 12 √ 2 √ 3 tan Λ((µ u + ω u ) sin(3t) + (-µ v -ω v ) cos(3t)) -3e ω ((µ u + ω u ) cos t+ (µ v + ω v ) sin t) ∂t - e -ω 2 sin t sin Λ √ 2 ∂u + e -ω 2 cos t sin Λ √ 2 ∂v
form an orthonormal basis.

We now want to determine the Lagrangian immersion

f : S × I → S 3 × S 3 (u, v, t) → f (u, v, t) = (p(u, v, t), q(u, v, t)).
We already know that the first component is the given minimal surface p. We write for both bases

∂ ∂t (q) = qβ 1 , ∂ ∂t (p) = pα 1 , ∂ ∂u (q) = qβ 2 , ∂ ∂u (p) = pα 2 , ∂ ∂v (q) = qβ 3 , ∂ ∂v (p) = pα 3 ,
and

E 1 (q) = q β1 , E 1 (p) = pα 1 , E 2 (q) = q β2 , E 2 (p) = pα 2 , E 3 (q) = q β3 , E 3 (p) = pα 3 .
Note that α 1 = 0 and α 2 and α 3 are determined by the minimal surface. In particular α 2 and α 3 are mutually orthogonal imaginary quaternions with length squared 2e ω . From (52) we then get that

α1 = 0, α2 = e -ω 2 cos t sin Λ √ 2 α 2 + e -ω 2 sin t sin Λ √ 2 α 3 , α3 = - e -ω 2 sin t sin Λ √ 2 α 2 + e -ω 2 cos t sin Λ √ 2 α 3
and from the properties of the minimal surface we obtain

∂α2 ∂u = -∂α3 ∂v = 1 2 ω u α 2 -1 2 ω v α 3 -e ω α 2 × α 3 , ∂α2 ∂v = 1 2 ω v α 2 + 1 2 ω u α 3 + α 2 × α 3 , ∂α3 ∂u = 1 2 ω v α 2 + 1 2 ω u α 3 -α 2 × α 3 .
Using the properties of the vector cross product, this also implies that

∂α2×α3 ∂u = 2α 2 + 2e ω α 3 + ω u α 2 × α 3 , ∂α2×α3 ∂v = -2e ω α 2 -2α 3 + ω v α 2 × α 3 .
Now, in order to find βi , we remark that the vectors E 1 , E 2 and E 3 need to correspond with eigenvectors of the operators A and B with suitable eigenfunctions. We have

E 1 = (0, q β1 ), E 2 = (pα 2 , q β2 ), ( 53
)
E 3 = (pα 3 , q β3 ).
The angle functions are θ 1 = 2π 3 , θ 2 = 2Λ + 2π 3 , θ 3 = -2Λ + 2π 3 and

P E i = cos(2θ i )E i + sin(2θ i )JE i , (54)
for i = 1, 2, 3. At the same time, by the definition of P in [START_REF] Dillen | Classification of totally real 3-dimensional submanifolds of S 6 (1) with K ≥ 1/16[END_REF] and by (53) we have ( 55)

P E 1 = (p β1 , 0), P E 2 = (p β2 , q α2 ), P E 3 = (p β3 , q α3 ).

Next we continue the computations in order to determine β1 . For this, we compute G(E 2 , E 3 ) in two different ways, once using ( 12) and once using [START_REF] Ejiri | Equivariant minimal immersions of S 2 into S 2m (1)[END_REF]. We obtain, respectively

G(E 2 , E 3 ) = - 1 √ 3 JE 1 = - 1 3 (p2 β1 , q β1 ),
and

G(E 2 , E 3 ) = G((pα 2 , q β2 ), (pα 3 , q β3 )) = 2 3 √ 3 (p( β2 × α 3 + α2 × β3 + α2 × α3 -2 β2 × β3 , q(-β2 × α 3 -α2 × β3 + 2 α2 × α3 -β2 × β3 ) = 2 3 √ 3 (p(2 -1 2 (1 -3 cot 2 Λ))α 2 × α3 , q(1 -1 3 (1 -3 cot 2 Λ))α 2 × α3 ) = 1 2 √ 3 (1 + cot 2 Λ)(2pα 2 × α3 , q α2 × α3 ).
Hence, comparing both expressions we get that β1 = -

√ 3 2 csc 2 Λ α2 × α3 = - √ 3 4 e -ω α 2 × α 3 . Moreover, we also obtain that β2 = 1 2 √ 2 (1 - √ 3 cot Λ)e -ω 2 sin Λ(cos tα 2 + sin tα 3 ), β3 = 1 2 √ 2 (1 + √ 3 cot Λ)e -ω 2 sin Λ(-sin tα 2 + cos tα 3 ).
We then take the inverse of (52) and deduce that

β 1 = - √ 3α 2 × α 3 2 √ 3e ω -2 sin(2t) tan(Λ) , β 2 = 1 8 e -ω µ v + ω v - (µ u + ω u ) cos(2t) tan(Λ) √ 3e ω -sin(2t) tan(Λ) α 2 × α 3 -4( √ 3 cot(Λ) cos(2t) + 1)α 2 -4 √ 3 sin(2t) cot Λα 3 , β 3 = 1 8 -e -ω µ u + ω u + (µ v + ω v ) cos(2t) tan(Λ) √ 3e ω -sin(2t) tan(Λ) α 2 × α 3 -4 √ 3 cot(Λ) sin(2t)α 2 + 4(1 + √ 3 cos(2t) cot Λ)α 3 .
By straightforward computations, it now follows that

∂β 1 ∂u - ∂β 2 ∂t -2β 1 × β 2 = 0, ∂β 1 ∂v - ∂β 3 ∂t -2β 1 × β 3 = 0, ∂β 3 ∂u - ∂β 2 ∂v -2β 3 × β 2 = 0,
from which we deduce that the integrability conditions for the immersion q are satisfied.

3.3.2. Case 2. The minimal surface p(M ) is totally geodesic, i.e. σ = 0. As mentioned before this means that h 3 13 = 0, h 3 12 = cos Λ sin Λ √ 3

. The equations following from ( 23) and ( 24), just like in the first case, give (57)

h 2 12 = 0, ω 2 11 = 0, ω 3 21 = -2+cos(2Λ) 2 √ 3 , h 2 11 = 0, ω 3 11 = 0, ω 3 22 = -h 3 22 cot(2Λ), h 3 11 = 0, ω 3 12 = sin 2 Λ √ 3 , ω 2 31 = 2+cos(2Λ) 2 √ 3 , ω 2 21 = 0, ω 3 31 = 0, ω 3 32 = -h 3 23 cot(2Λ) and (58) E 1 (Λ) = 0, E 2 (Λ) = h 3 23 , E 3 (Λ) = -h 3 22 .
In this case, the equations of Codazzi become

(59) E 1 (h 3 23 ) = - √ 3 2 h 3 22 , E 1 (h 3 22 ) = √ 3 2 h 3 23 , E 2 (h 3 22 ) = -E 3 (h 3 23 )
and

(60) -1-(1+12(h 3 22 ) 2 +12(h 3 23 ) 2 ) cos(2Λ)+cos(4Λ)+cos(6Λ)+4(E 2 (h 3 23 )-E 3 (h 3 22 )) sin(2Λ) = 0.
In what follows we are going to introduce new vector fields on M by:

X 1 = 4 √ 3 E 1 , X 2 = - 2h 3 22 csc 2 Λ sec Λ √ 3 E 1 + 2 csc Λ E 2 , (61)
X 3 = - 2h 3 23 csc 2 Λ sec Λ √ 3 E 1 + 2 csc Λ E 3 .
We can easily check that

[X 1 , X 2 ] = 2X 3 , [X 2 , X 3 ] = 2X 1 , (62) [X 3 , X 1 ] = 2X 2 .
Taking a canonical metric on M such that X 1 , X 2 and X 3 have unit length and are mutually orthogonal, it follows from the Koszul formula that all connection components are determined. From (4.1), Proposition 5.2 and its preceeding paragraph in [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF] it follows that we can locally identify M with S 3 and we can consider X 1 , X 2 and X 3 as the standard vector fields on S 3 with

X 1 (x) = xi, X 2 (x) = xj, (63)
X 3 (x) = xk.
Using the above formulas, the component p of the map can now be determined explicitly. First, we write (64) D Xi p = pα i , for i = 1, 2, 3, where D denotes the Euclidean covariant derivative. Of course, by Theorem 1, D X1 p = 0. Then, we may compute by ( 18)

(dp(X 2 ), 0) = ( 2 cos Λ √ 3 + 2 sin Λ)E 2 + (-2 cos Λ + sin Λ √ 3 )JE 2 , (dp(X 3 ), 0) = (- 2 cos Λ √ 3 + 2 sin Λ)E 3 + (2 cos Λ + sin Λ √ 3 )JE 3
Before we can determine the second component q of the Lagrangian immersion, we need to explore the Codazzi equations further. First we look at the system of differential equations for the function Λ in (59) and (60). Notice that by using the relations in (61) we have that (69) 3 22 csc Λ, where the last two equations can be seen as the definition for the functions h 3 23 and h 3 22 . The first one is, of course, a condition for the unknown function of Λ. Three out of the four Codazzi equations then can be seen as integrability conditions for the existence of a solution of this system, whereas the last one reduces to

X 1 (Λ) = 0, X 2 (Λ) = 2h 3 23 csc Λ, X 3 (Λ) = -2h
X 2 (X 2 (Λ)) + X 3 (X 3 (Λ)) = (cot(Λ) -tan(Λ))((X 2 (Λ)) 2 + (X 3 (Λ)) 2 ) + 4(1 + 2 cos(2Λ)) cot(Λ).
Under the change of variable Λ = arctan(e 2β ), this equation simplifies to

(70) X 2 (X 2 (β)) + X 3 (X 3 (β)) = 2(3 -e 4β ) e 4β .
Note also that for Λ = π 3 , we get the solution corresponding to example (1) in Theorem 2, as it follows. From ( 61) and (69) we see that

X 1 = 4 √ 3 E 1 , X 2 = 4 √ 3 E 2 , X 3 = 4 √ 3 E 3 .
This implies that M has constant sectional curvature

√ 3
4 . Hence this corresponds to example (1) in Theorem 2.

Remark 1. Note that there exist at least locally many solutions of the system

X 1 (β) = 0, X 2 (X 2 (β)) + X 3 (X 3 (β)) = 2(3 -e 4β ) e 4β .
This can be seen by choosing special coordinates on the usual S 3 . We take

x 1 = cos v cos(t + u), x 2 = cos v sin(t + u), x 3 = sin v cos(u -t), x 4 = sin v sin(u -t).
As, given (63), at the point x = (x 1 , x 2 , x 3 , x 4 ) the vectors in the basis are

X 1 (x) = (-x 2 , x 1 , x 4 , -x 3 ), X 2 (x) = (-x 3 , -x 4 , x 1 , x 2 ), X 3 (x) = (-x 4 , x 3 , -x 2 , x 1 ), it is straightforward to see that ∂t = X 1 , ∂u = cos(2v)X 1 + sin(2t) sin(2v)X 2 + cos(2t) sin(2v)X 3 , ∂v = cos(2t)X 2 -sin(2t)X 3 ,
and conversely,

X 1 = ∂t, X 2 = sin(2t) sin(2v) ∂u -sin(2t) cos(2v) sin(2v) ∂t + cos(2t)∂v, X 3 = cos(2t) sin(2v) ∂u -cos(2t) cos(2v) sin(2v) ∂t -sin(2t)∂v.
At last, the equations in (70) become ∂ ∂t β = 0 and

(71) csc 2 (2v) ∂ 2 β ∂u 2 + ∂ 2 β ∂v 2 + 2 cot(2v) ∂β ∂v = 2(3e -4β -1).
The above differential equation is an elliptic quasilinear second order PDE. Hence, we can apply the Cauchy-Kowalevskaya theorem (see [START_REF] John | Partial Differential Equations[END_REF]). Therefore, if we start with an analytic regular curve without self intersections and analytic Cauchy data along the curve, we locally have a unique (analytic) solution. Given that we can choose arbitrarily both the curve and the Cauchy data along the curve, locally there exist many solutions for the system in (71).

In the following part we are going to determine the second part of the immersion. We start with an arbitrary solution of

X 1 (β) = 0, X 2 (X 2 (β)) + X 3 (X 3 (β)) = 2(3 -e 4β )
e 4β and we are going to find a system of differential equations determining the immersion q. We define h 3 22 and h 3 23 as in (69) and such that Λ = arctan(e 2β ). First, we can write for each of the bases that we took, {E i } and {X i }, the following:

X 1 (q) = qβ 1 , X 1 (p) = pα 1 , X 2 (q) = qβ 2 , X 2 (p) = pα 2 , X 3 (q) = qβ 3 , X 3 (p) = pα 3 ,
and

E 1 (q) = q β1 , E 1 (p) = pα 1 , E 2 (q) = q β2 , E 2 (p) = pα 2 , E 3 (q) = q β3 , E 3 (p) = pα 3 ,
where α 1 = 0 and α 2 and α 3 are as determined previously. Then, we prove as before that β1 = -

√ 3 2 sin 2 Λ α2 × α3 , β2 = cos(2Λ + 2π 3 ) -1 √ 3 sin(2Λ + 2π 3 ) 1 -2 √ 3 sin(2Λ + 2π 3 ) α2 = 1 2 (1 - √ 3 cot Λ) α2 , (72) β3 = cos(-2Λ + 2π 3 ) -1 √ 3 sin(-2Λ + 2π 3 ) 1 -2 √ 3 sin(-2Λ + 2π 3 ) α3 = 1 2 (1 + √ 3 cot Λ) α3
and we continue the computations in order to find the system of differential equations for the immersion q in terms of the basis {X i }. As we identify df (X 1 ) ≡ X 1 , we have

D X1 f = (X 1 (p), X 1 (q)) = (0, qβ 1 ) ≡ X 1 (61) = 4 √ 3 E 1 = 4 √ 3 (pα 1 , q β1 ).
Therefore, β 1 = 4 √ 3 β1 . We may compute similarly for D X2 f and D X3 f and find

β 2 = 2 csc Λ β2 -2 √ 3 h 3 22 csc 2 Λ sec Λ β1 , β 3 = 2 csc Λ β3 -2 √ 3 h 3 23 csc 2 Λ sec Λ β1 , α2 = -1 csc Λ hxjx -1 h -1 , α3 = -1 csc Λ hxkx -1 h -1 and β1 = - √ 3 2 hxix -1 h -1 .
Using now relations (72) we may express

β 2 = -(1 - √ 3 cot Λ) hxjx -1 h -1 + h 3 22 csc 2 Λ sec Λ hxix -1 h -1 , β 3 = -(1 + √ 3 cot Λ) hxkx -1 h -1 + h 3 23 csc 2 Λ sec Λ hxix -1 h -1 . Finally, as X i (q) = qβ i , we find    X 1 (q) = -2qhxix -1 h -1 , X 2 (q) = q(h 3 22 csc 2 Λ sec Λ hxix -1 h -1 -(1 - √ 3 cot Λ) hxjx -1 h -1 ), X 3 (q) = q(h 3 23 csc 2 Λ sec Λ hxix -1 h -1 -(1 + √ 3 cot Λ) hxkx -1 h -1 ),
which, given (69) and Λ = arctan(e 2β ), is equivalent to

   X 1 (q) = -2qhxix -1 h -1 , X 2 (q) = q -X 3 (β)hxix -1 h -1 -(1 - √ 3e -2β ) hxjx -1 h -1 , X 3 (q) = q X 2 (β) hxix -1 h -1 -(1 + √ 3e -2β ) hxkx -1 h -1 . (73)
By straightforward computations, one may see that X i (X j (q)) -X j (X i (q)) = [X i , X j ](q) hold for i, j = 1, 2, 3. Therefore, the immersion f is completely determined by ( 68) and (73).

3.3.3. Case 3. The minimal surface p(M ) is not totally geodesic, but the map P is not an immersion. As mentioned before this means that (74)

h 3 12 = - sin(2Λ) √ 3 .
Therefore, the equations in subsection 3.2 which follow from ( 23) and ( 24) become ) and

h 2 12 = -h 3 13 , ω 3 12 = 1+2 cos(2Λ) 2 √ 3 , ω 2 21 = ω 3 31 = -h 3 13 cot Λ, h 2 11 = h 3 11 = 0, ω 3 21 = 1+2 cos(2Λ) 2 √ 3 , ω 2 31 = -1+2 cos(2Λ) 2 √ 3 , ω 2
(77) -1 -[1 + 6(h 3 22 ) 2 + 6(h 3 23 ) 2 ] cos 2Λ + cos 4Λ + cos 6Λ + 2[-E 3 (h 3 22 ) + E 2 (h 3
23 )] sin 2Λ = 0, respectively. The Lie brackets of the vector fields

E 1 , E 2 , E 3 give [E 1 , E 2 ] = 0, [E 1 , E 3 ] = 0, [E 2 , E 3 ] = - 1 + 2 cos(2Λ) √ 3 E 1 + h 3 22 cot(2Λ)E 2 + h 3 23 cot(2Λ)E 3 .
Next, we take new vector fields X 1 , X 2 , X 3 of the form

X 1 = E 1 , X 2 = √ 2(h 3 22 -h 3 23 ) 3 3 4 (sin(2Λ)) 3 2 E 1 + √ 2 3 1 4 sin(2Λ) E 2 - √ 2 3 1 4 sin(2Λ) E 3 , (78)
X 3 = √ 2(h 3 22 + h 3 23 ) 3 3 4 (sin(2Λ)) 3 2 E 1 + √ 2 3 1 4 sin(2Λ) E 2 + √ 2 3 1 4 sin(2Λ) E 3 .
We can easily check that [X 1 , X 2 ] = 0, [X 1 , X 3 ] = 0 and [X 2 , X 3 ] = 0, therefore, by the lemma on page 155 in [START_REF] Boothby | An introduction to differentiable manifolds and Riemannian geometry[END_REF], we know that there exist coordinates {t, u, v} on M such that

X 1 = ∂t, X 2 = ∂u, X 3 = ∂v.
Using (75) we obtain:

Λ t = 0, Λ u = h 3 22 + h 3 23 3 1/4 √ cos Λ sin Λ , Λ v = -h 3 22 + h 3 23 3 1/4 √ cos Λ sin Λ .
Furthermore, we express h 3 22 and h 3 23 from the previous relations as

h 3 22 = 1 2 3 1/4 (Λ u -Λ v ) √ cos Λ sin Λ, h 3 23 = 1 2 3 1/4 (Λ u + Λ v ) √ cos Λ sin Λ
and therefore, the expression of (78) becomes

X 1 = E 1 , X 2 = - Λ v csc(2Λ) √ 3 E 1 + 1 3 1 4 √ cos Λ sin Λ E 2 - 1 3 1 4 √ cos Λ sin Λ E 3 , (79)
X 3 = Λ u csc(2Λ) √ 3 E 1 + 1 3 1 4 √ cos Λ sin Λ E 2 + 1 3 1 4 √ cos Λ sin Λ E 3 .
Finally, by straightforward computations, one may see that equation (77) becomes

(80) - √ 3(Λ 2 u + Λ 2 v ) cos(2Λ) + 3 1 4 (E 2 (Λ u ) -E 3 (Λ u ) + E 2 (Λ v ) + E 3 (Λ v )) √ cos Λ sin Λ- -2(sin(2Λ) + sin(4Λ)) = 0.
We compute dp(∂u) and dp(∂v):

dp (∂u) = √ 3 -2 cos 2Λ + π 6 3 3/4 √ 2 sin(2Λ) E 2 + 2 sin 2Λ + π 3 - √ 3 3 3/4 √ 2 sin(2Λ) E 3 + 2 cos 2Λ + 2π 3 + 1 3 3/4 √ 2 sin(2Λ) JE 2 + 2 cos 2Λ + π 3 -1 3 3/4 √ 2 sin(2Λ) JE 3 , dp(∂v) = √ 3 -2 cos 2Λ + π 6 3 3/4 √ 2 sin(2Λ) E 2 - 2 sin 2Λ + π 3 - √ 3 3 3/4 √ 2 sin(2Λ) E 3 + 2 cos 2Λ + 2π 3 + 1 3 3/4 √ 2 sin(2Λ) JE 2 - 2 cos 2Λ + π 3 -1 3 3/4 √ 2 sin(2Λ) JE 3 ,
and we remark that they are mutually orthogonal and that their length is 2 tan Λ √ 3 . So, as u, v are isothermal coordinates on the surface, for which ∂u, ∂u = ∂v, ∂v = 2e ω , we obtain that (81) e ω = tan Λ √ 3 .

On the one hand, for z = x + Iy as in subsection 3.1, we may compute dp(∂z):

dp(∂z) = 1 2 [dp (∂u) -I • dp (∂v)] = 1 2 √ 2 3 3/4 sin(2Λ) (1 -I) √ 3 -2 cos 2Λ + π 6 E 2 - (1 + I) √ 3 -2 sin 2Λ + π 3 E 3 + (1 -I) 2 cos 2Λ + 2π 3 + 1 JE 2 + (1 + I) 2 cos 2Λ + π 3 -1 JE 3 . As √ 3 -2 cos 2Λ + π 6 = 2 sin Λ( √ 3 sin Λ + cos Λ), 2 sin 2Λ + π 3 - √ 3 = 2 sin Λ(cos Λ - √ 3 sin Λ), 2 cos 2Λ + 2π 3 + 1 = 2 sin Λ(sin Λ - √ 3 cos Λ), 2 cos 2Λ + π 3 -1 = -2 sin Λ(sin Λ + √ 3 cos Λ),
we finally have

dp(∂z) = sin Λ √ 2 3 3/4 sin(2Λ) (1 -I) √ 3 sin Λ + cos Λ E 2 + (1 + I) cos Λ - √ 3 sin Λ E 3 + +(1 -I) sin Λ - √ 3 cos Λ JE 2 -(1 + I) sin Λ + √ 3 cos Λ JE 3 .
Moreover, from (81), it follows that ω z = 1 sin(2Λ) (Λ u -iΛ v ). On the other hand, we may compute ∇ E ∂z dp(∂z) using the Euclidean connection ∇ E :

∇ E ∂z dp(∂z) = - 1 √ 3 E 1 + e -iπ 4 sin 2 Λ √ 3 cot Λ + 3 (Λ u -iΛ v ) 3 4 √ 3 sin 3 2 (2Λ) E 2 + e -iπ 4 sin Λ(Λ v + iΛ u ) √ 3 cos Λ -3 sin Λ 3 4 √ 3 sin 3 2 (2Λ) E 3 + JE 1 + e -iπ 4 sin Λ(Λ u -iΛ v ) √ 3 sin Λ -3 cos Λ 3 4 √ 3 sin 3 2 (2Λ) JE 2 - 1 3 + i 3 sin Λ(Λ u -iΛ v ) √ 3 sin Λ + 3 cos Λ √ 2 4 √ 3 sin 3 2 (2Λ) JE 3 .
From the previous computations we see, indeed, that

∇ E ∂z dp(∂z) = -N + ω z dp(∂z),
which corresponds to [START_REF] Lotay | Ruled Lagrangian submanifolds of the 6-sphere[END_REF]. From here, we remark the component in the direction of the normal N = ξ (see subsection 3.2) and we see that the choice of coordinates {t, u, v} following from ( 78) is the right one, as we have indeed σ(∂z, ∂z) = -1, as in subsection 3.1. Using (81) together with the fact that, by taking the inverse in (79), we have

E 1 = ∂t, E 2 = 3 1 4 sin(2Λ) 2 √ 2 Λ v -Λ u √ 3 sin(2Λ) ∂t + ∂u + ∂v , E 3 = - 3 1 4 sin(2Λ) 2 √ 2 Λ v + Λ u √ 3 sin(2Λ) ∂t + ∂u -∂v ,
we may prove that equation ( 80) is equivalent to the Sinh-Gordon equation in [START_REF] Moroianu | Generalized Killing spinors and Lagrangian graphs[END_REF], which characterizes the minimal surface.

Reverse construction

Let S be a minimal surface given by p : S → S 3 ⊂ R 4 , on which we take isothermal coordinates u and v as in subsection 3.1. Hence, we have a solution ω of the Sinh-Gordon equation ∆ω = -8 sinh ω. Next, we define a function Λ ∈ (0, π 2 ) such that e ω = tan Λ √ 3 .

Remark 2. If ω = 0, then Λ = π 3 , which corresponds to example (2) in Theorem 2.

We then define a metric on an open part of the unit frame bundle of the surface by assuming that the vectors (u, v, t) → f (u, v, t) = (p(u, v, t), q(u, v, t)),

E 1 = ∂t, E 2 = √ 3e ω/2 2 √ 1 + 3e 2ω ω v -ω u 2 √ 3 ∂t + ∂u + ∂v , (82)
E 3 = - √ 3e ω/2 2 √ 1 + 3e 2ω
for which we already know that the first component is the given minimal surface p. We write for both bases and E 1 (q) = q β1 , E 1 (p) = pα 1 , E 2 (q) = q β2 , E 2 (p) = pα 2 , E 3 (q) = q β3 , E 3 (p) = pα 3 .

Note that α 1 = 0 and α 2 and α 3 are determined by the minimal surface. In particular α 2 and α 3 are mutually orthogonal imaginary quaternions with length squared 2e ω . From the derivates of p in the latter relations together with (82), we obtain Finally, we take the inverse of the matrix which give {E i } in the basis {∂t, ∂u, ∂v} in (82) and obtain

β 1 = - √ 3e -ω 4 α 2 × α 3 ,
β 2 = e -ω 8 (4e ω α 2 -4α 3 + ω v α 2 × α 3 ),

β 3 = - e -ω 8 (4α 2 -4e ω α 3 + ω u α 2 × α 3 ).
By straightforward computations, it now follows that

∂β 1 ∂u - ∂β 2 ∂t -2β 1 × β 2 = 0, ∂β 1 ∂v - ∂β 3 ∂t -2β 1 × β 3 = 0, ∂β 3 ∂u - ∂β 2 ∂v -2β 3 × β 2 = 0,
from which we deduce that the integrability conditions for the immersion q are satisfied.

Conclusion

The results in Section 3.3 can now be summarized in the following theorems. on V . Then, there exists a Lagrangian immersion f : V → S 3 × S 3 : x → (p(x), q(x)), where q is determined by

∂q ∂t = - √ 3 2 √
3e ω -2 sin(2t) tan Λ q α 2 × α 3 , ∂q ∂u = 1 8 e -ω µ v + ω v -(µ u + ω u ) cos(2t) tan Λ √ 3e ω -sin(2t) tan Λ q α 2 × α 3 -4( √ 3 cot Λ cos(2t) + 1) q α 2 -4 √ 3 sin(2t) cot Λ q α 3 , ∂q ∂v = 1 8 -e -ω µ u + ω u + (µ v + ω v ) cos(2t) tan Λ √ 3e ω -sin(2t) tan Λ q α 2 × α 3 -4 √ 3 cot Λ sin(2t) q α 2 + 4(1 + √ 3 cos(2t) cot Λ) q α 3 , where α 2 = pp u and α 3 = pp v .

Theorem 5. Let X 1 , X 2 , X 3 be the standard vector fields on S 3 . Let β be a solution of the differential equations and q is determined by

X 1 (β) = 0,
X 1 (q) = -2qhxix -1 h -1 , X 2 (q) = q -X 3 (β)hxix -1 h -1 -(1 - √ 3e -2β ) hxjx -1 h -1 , X 3 (q) = q X 2 (β) hxix -1 h -1 -(1 + √ 3e -2β ) hxkx -1 h -1 .
Note that in the previous theorem the image of p is a totally geodesic surface in S 3 . Theorem 6. Let ω be a solution of the Sinh-Gordon equation ∆ω = -8 sinh ω on an open connected domain of U in C and let p : U → S 3 be the associated minimal surface with complex coordinate z such that σ(∂z, ∂z) = -1. Then, there exist a Lagrangian immersion f : U × I → S 3 × S 3 : x → (p(x), q(x)), where q is determined by ∂q ∂t = -√ 3e -ω 4 q α 2 × α 3 , ∂q ∂u = e -ω 8 (4e ω qα 2 -4qα 3 + ω v q α 2 × α 3 ), ∂q ∂v = -e -ω 8 (4qα 2 -4e ω qα 3 + ω u q α 2 × α 3 ).

where α 2 = pp u and α 3 = pp v .

Theorem 7. Let f : M → S 3 × S 3 : x → (p(x), q(x)) be a Lagrangian immersion such that p has nowhere maximal rank. Then every point x of an open dense subset of M has a neighborhood U such that f | U is obtained as described in Theorem 4, 5 or 6.

2 = 4 ,ε 1 c 1 -

 241 where c 1 does not depend on t. 2 cos(4t) + 2 sin(2t) ,

1 ≥ - 2 } and 2 √

 122 and equality can hold if t ∈ {± π 4 , ± 5π 4 3e ω tan Λ ± 2 = 0. So on an open dense subset we can write c 1 = e ω+µ -2.

E 1 (

 1 Λ) = h3 13 , E 2 (Λ) = h3 23 , E 3 (Λ) = -h3 22 . Moreover, the equations of Codazzi in (39) yield h 3 13 = 0 and, therefore, ω 2 21 = ω 3 31 = 0. The first two equations in (40) imply that E 1 (h 3 23 ) = 0 and E 1 (h 3 22 ) = 0, while the next three ones vanish identically. The last two equations in (40) become (76) E 2 (h 3 22 ) = -E 3 (h 3 23

ω v + ω u 2 √ 3

 3 ∂t + ∂u -∂v form an orthonormal basis. Next, we want to determine the Lagrangian immersionf : S × I → S 3 × S 3

∂

  ∂t (q) = qβ 1 , ∂ ∂t (p) = pα 1 , ∂ ∂u (q) = qβ 2 , ∂ ∂u (p) = pα 2 , ∂ ∂v (q) = qβ 3 , ∂ ∂v (p) = pα 3

(α 2 -

 2 α 3 ).We then follow the same steps as in Case 1 and obtainβ1 = -√ 3e -ω 4 α 2 × α 3 , β2 = √ 3(e ω/2 -e -ω/2 ) 4 √ 1 + 3e 2ω (α 2 + α 3 ), (84) β3 = -√ 3(e ω/2 -e -ω/2 ) 4 √ 1 + 3e 2ω (α 2 -α 3 ).

Theorem 4 .

 4 Let ω and µ be solutions of, respectively, the Sinh-Gordon equation ∆ω = -8 sinh ω and the Liouville equation ∆µ = -e µ on an open simply connected domain U ⊆ C and let p : U → S 3 be the associated minimal surface with complex coordinate z such that σ(∂z, ∂z) = -1.Let V = {(z, t) | z ∈ U, t ∈ R, e ω+µ -2 -2 cos(4t) > 0} and let Λ be a solution of 2 √3e ω tan Λ -2 sin(2t) = e ω+µ -2 -2 cos(4t)

X 2 (

 2 X 2 (β)) + X 3 (X 3 (β)) = 2(3 -e 4β )e 4β , on a connected, simply connected open subset U of S 3 . Then there exist a Lagrangian immersion f : U → S 3 × S 3 : x → (p(x), q(x)), where p(x) = xix -1

This work was partially supported by the project 3E160361 (Lagrangian and calibrated submanifolds) of the KU Leuven research fund and part of it was carried out while the first author visited KU Leuven, supported by The Scientific and Technological Research Council of Turkey (TUBITAK), under grant 1059B141500244.

Now we use the definition of J to write out JE i :

(p(2 β2 -α2 ), q(-2α 2 + β2 )), (56)

(p(2 β3 -α3 ), q(-2α 3 + β3 )).

Then, by using (56), ( 53) and the values of θ i in [START_REF] Nagy | Nearly Kähler geometry and Riemannian foliations[END_REF], we rewrite equation (54) and, by comparing it to (55), we obtain

and we see that (65)

Next, we want to determine a system of differential equations satisfied by α 2 and α 3 . For this, we consider S 3 × S 3 ∈ R 4 × R 4 . On the one hand, we use (64) together with D X (dp(Y ), 0) = (D X dp(Y ), 0). On the other hand, we use [START_REF] Dioos | Lagrangian submanifolds in the nearly Kahler S 3 × S 3[END_REF] and, therefore, we obtain (67)

We choose a unit quaternion h such that at the point p(x) = 1 we have

Using (63), we can check that α

are the unique solutions for the system of differential equations in (67):

This in its turn implies that (68)

is the unique solution of X i (p) = pα i with initial conditions p(1) = 1. Indeed we have