Classification of Lagrangian surfaces of constant curvature in complex projective plane

Journal of Geometry and Physics 53 (2005) 428–460

Classification of Lagrangian surfaces of constant curvature in complex projective plane Bang-Yen Chen∗ Department of Mathematics, Michigan State University, East Lansing, Michigan 48824-1027, USA Received 7 April 2004; received in revised form 22 July 2004; accepted 26 July 2004 Available online 13 September 2004

Abstract From Riemannian geometric point of view, one of the most fundamental problems in the study of Lagrangian submanifolds is the classification of Lagrangian immersions of real space forms into complex space forms. The purpose of this article is thus to classify Lagrangian surfaces of constant curvature in complex projective plane CP 2 . Our main result states that there are 29 families of Lagrangian surfaces of constant curvature in CP 2 . Twenty-two of the 29 families are constructed via Legendre curves. Conversely, Lagrangian surfaces of constant curvature in CP 2 are obtained from the 29 families. As an immediate by-product, many interesting new examples of Lagrangian surfaces of constant curvature in CP 2 are discovered. © 2004 Elsevier B.V. All rights reserved. MSC: 53D12; 53C40; 53C42SC JGS SC: Differential geometry Keywords: Lagrangian surface; Legendre curve; Special Legendre curve; Surface of constant curvature

1. Introduction ˜ is called Lagrangian if the almost complex A submanifold M of a Kaehler manifold M ˜ structure J of M interchanges each tangent space of M with its corresponding normal space. ∗

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Lagrangian submanifolds appear naturally in the context of classical mechanics and mathematical physics. For instance, the systems of partial differential equations of Hamilton– Jacobi type lead to the study of Lagrangian submanifolds and foliations in the cotangent bundle. Furthermore, Lagrangian submanifolds are part of a growing list of mathematically rich special geometries that occur naturally in string theory. For a Lagrangian submanifold M with mean curvature vector H and shape operator A, the dual 1-form of JH is the well-known Maslov form. A Lagrangian submanifold is called Maslovian if it has no minimal points and if its Maslov vector field JH is an eigenvector of AH . In the study of Lagrangian submanifolds, it is important to construct non-trivial new examples. Also, from Riemannian geometric point of view, one of the most fundamental problems in the study of Lagrangian submanifolds is to classify Lagrangian immersions of real space forms into complex space forms. Such submanifolds are either totally geodesic or flat if the immersions were minimal [8,11] (for indefinite case, this was done in a series of articles [9,12,13,15]). For non-minimal Lagrangian immersions, this problem have been studied in [3,4,6,7] among others. In particular, Lagrangian submanifolds of constant curvature c in complex space forms of holomorphic sectional curvature 4c have been determined in [6] by utilizing the notion of twisted products. Moreover, Maslovian Lagrangian immersions of real space forms into complex space forms were classified in [3,4,7]. In particular, Maslovian Lagrangian surfaces of constant curvature in complex projective space were classified in [4]. In this paper, we classify Lagrangian surfaces of constant curvature in the complex projective plane CP 2 without the Maslovian condition. The class of Lagrangian surfaces of constant curvature in CP 2 is much bigger than the class of Maslovian Lagrangian surfaces of constant curvature. In fact, our main result states that there are 29 families of Lagrangian surfaces of constant curvature in CP 2 . Twenty-two of the 29 families are constructed via Legendre curves. Conversely, Lagrangian surfaces of constant curvature in CP 2 are obtained locally from the 29 families. As an immediate by-product, many interesting new examples of Lagrangian surfaces of constant curvature in CP 2 are discovered.

2. Preprimaries ˜ n (4c) denote a complete simply-connected Kaehler n-manifold M ˜ n (4c) with conLet M ˜ n (4c). We stant holomorphic sectional curvature 4c and M a Lagrangian submanifold in M n ˜ ˜ denote the Riemannian connections of M and M (4c) by ∇ and ∇, respectively. The formulas of Gauss and Weingarten are given, respectively, by (cf. [1]) ∇˜ X Y = ∇X Y + h(X, Y ),

(2.1)

∇˜ X ξ = −Aξ X + DX ξ,

(2.2)

for tangent vector fields X, Y and normal vector field ξ, where D is the connection on the normal bundle. The second fundamental form h is related to the shape operator Aξ by h(X, Y ), ξ = Aξ X, Y . The mean curvature vector H of M is defined by H = (1/n) trace h. A point p ∈ M is called minimal if H vanishes at p.

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˜ n (4c) we have (cf. [8]) For Lagrangian submanifolds M in M DX JY = J∇X Y,

(2.3)

h(X, Y ), JZ = h(Y, Z), JX = h(Z, X), JY .

(2.4)

If we denote the Riemann curvature tensor of M by R, then the equations of Gauss and Codazzi are given, respectively, by R(X, Y )Z, W = h(X, W), h(Y, Z) − h(X, Z), h(Y, W) + c(X, WY, Z − X, ZY, W), (∇¯ X h)(Y, Z) = (∇¯ Y h)(X, Z),

(2.5) (2.6)

where X, Y, Z, W are tangent to M and ∇h is defined by (∇h)(X, Y, Z) = DX h(Y, Z) − h(∇X Y, Z) − h(Y, ∇X Z).

(2.7)

We recall the following theorems for later use (cf. [5]). Theorem A. Let (M n ,  . , .) be an n-dimensional simply connected Riemannian manifold. Let σ be a TM-valued symmetric bilinear form on M satisfying (i) σ(X, Y ), Z is totally symmetric, (ii) (∇σ)(X, Y, Z) = ∇X σ(Y, Z) − σ(∇X Y, Z) − σ(Y, ∇X Z) is totally symmetric, (iii) R(X, Y )Z = c(Y, ZX − X, ZY ) + σ(σ(Y, Z), X) − σ(σ(X, Z), Y ), ˜ n (4c) whose second funThen there exists a Lagrangian isometric immersion L : M → M damental form h is given by h = Jσ. ˜ n (4c) be Lagrangian isometric immersions of a RiemanTheorem B. Let L1 , L2 : M → M nian n-manifold with second fundamental forms h1 and h2 , respectively. If h1 (X, Y ), JL1 Z = h2 (X, Y ), JL2 Z

(2.8)

for all vector fields X,Y,Z tangent to M, then there exists a biholomorphic isometry φ of ˜ n (4c) such that L1 = L2 ◦ φ. M

3. Lagrangian and Legendrian submanifolds We recall the following basic relationship between Legendrian submanifolds of S 2n+1 (1) and Lagrangian submanifolds of the complex projective n-space CP n (4) with constant holomorphic curvature 4 (cf. [14]). Let S 2n+1 (r) = {(z1 , . . . , zn+1 ) ∈ Cn+1 : z, z = r2 } be the hypersphere in Cn+1 centered√at the origin with radius r. On Cn+1 we consider the complex structure J induced by i = −1. On S 2n+1 (1) we consider the canonical Sasakian

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structure consisting of φ given by the projection of the complex structure J of Cn+1 on the tangent bundle of S 2n+1 (1) and the structure vector field ξ = Jx with x being the position vector. An isometric immersion f : M → S 2n+1 (1) is called Legendrian, if ξ is normal to f∗ (TM) and φ(f∗ (TM)), f∗ (TM) = 0, where ,  denotes the inner product on Cn+1 . The vectors of S 2n+1 (1) normal to ξ at a point z define the horizontal subspace Hz of the Hopf fibration π : S 2n+1 (1) → CP n (4). Therefore, the condition “ξ is normal to f∗ (TM)” means that f is horizontal; thus it describes an integral manifold of maximal dimension of the contact distribution H. Let ψ : M → CP n (4) be a Lagrangian isometric immersion. Then there exists an isoˆ → M and a Legendrian immersion f : M ˆ → S 2n+1 (1) such metric covering map τ : M that ψ(τ) = π(f ). Hence, every Lagrangian immersion can be lifted locally (or globally if we assume the manifold is simply connected) to a Legendrian immersion of the same Riemannian manifold. ˆ → S 2n+1 (1) is a Legendrian immersion. Then ψ = Conversely, suppose that f : M n π(f ) : M → CP (4) is again an isometric immersion, which is Lagrangian. Under this correspondence, the second fundamental forms hf and hψ of f and ψ satisfy π∗ hf = hψ . Moreover, hf is horizontal with respect to π. We shall denote hf and hψ simply by h. Let L : M → S 2n+1 (1) ⊂ Cn+1 be an isometric immersion. Denote by ∇ˆ and ∇ the Levi–Civita connections of Cn+1 and M, respectively. Let h denote the second fundamental form of M in S 2n+1 (1). Then we have: ∇ˆ X Y = ∇X Y + h(X, Y ) − (X, Y L.

(3.1)

4. Legendre curves A curve z = z(t) is called regular if its speed, v(t) := |z (t)|, is nowhere zero. A regular curve z = z(s) in the hypersphere S 2n−1 (r) ⊂ Cn+1 is called Legendre, if z (t), iz(t) = 0 holds identically. The following lemma follows easily from the definition. Lemma 1. Every horizontal lift of a regular curve in a Lagrangian submanifold of CP n (4) via the Hopf fibration π is a Legendre curve in S 2n+1 (1) ⊂ Cn+1 . It is known that a unit speed Legendre curve z(s) in S 3 (r) ⊂ C2 satisfies: z (s) = iκ(s)z (s) −

z(s) , r2

(4.1)

where κ(s) is the curvature function of z in S 3 (1) (cf. [4]). A unit speed curve z(s) in S 3 (r) ⊂ C2 satisfies z (s) = −z(s)/r 2 if and only if it is a geodesic. A geodesic in S 3 (r) can be Legendre or non-Legendre. For examples, z(s) = (cos s, sin s) is Legendre and z(s) = (eis , 0) is non-Legendre.

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Lemma 2. Let r be a positive number r. Then we have (1) Every Legendre curve z = z(t) : I → S 3 (r) ⊂ C2 satisfies z (t) = iλ(t)z (t) +

v  v2 z (t) − 2 z(t), v r

(4.2)

where v is the speed of z and λ = κv with κ being the curvature of z in S 3 (r). (2) Conversely, if a regular curve z = z(t) in S 3 (r) ⊂ C2 with speed v satisfying differential Eq. (4.2) for some nowhere zero real-valued function λ, then z is a Legendre curve. Let S 2 (1/2) := {(x1 , x2 , x3 ) ∈ R3 : x12 + x22 + x32 = 1/4}. Then the Hopf fibration π : S 3 (1) → CP 1 (4) ≡ S 2 (1/2) is given by
 |z|2 − |w|2 ¯ π(z, w) = zw; , (z, w) ∈ S 3 (1) ⊂ C2 . (4.3) 2 For each Legendre curve γ in S 3 (1) ⊂ C2 , the projection π ◦ γ is a regular curve in Conversely, each regular curve ξ in S 2 (1/2) gives rise to a horizontal lift in S 3 (1) which is unique up to a factor eiθ , θ ∈ R. Each horizontal lift of ξ is a Legendre curve in S 3 (1). Since the Hopf fibration π is a Riemannian submersion, a Legendre curve γ in S 3 is projected to a curve in S 2 (1/2) with the same curvature. In order to explain our methods for constructing Lagrangian surfaces of constant curvature in CP 2 , we recall the notions of special Legendre curves and their associated special Legendre curves which are introduced in [2,4]. Let z = z(t) be a Legendre curve in S 5 (r) ⊂ C3 with speed v. Then z, iz, z , iz are orthogonal vector fields. Differentiating z (t), iz(t) = z (t), z(t) = 0 yields z , iz = 0 and z , z = −1. Thus, there exists a non-zero normal vector field Pz perpendicular to z, iz, z , iz such that S 2 (1/2).

z (t) = iψ(t)z (t) +

v  v2 z (t) − 2 z(t) − a(t)Pz (t) v r

(4.4)

for some real-valued functions ψ and a. The Legendre curve z is called special if Pz in (4.4) is a parallel normal vector field, i.e., Pz (t) = a(t)z (t) for some function a(t). If a special Legendre curve does not lie in any proper linear complex subspace of C3 , then the function a in (4.4) is not identical zero. If z is a special Legendre curve satisfying (4.4), then Pz is also a special Legendre curve on I  = {t ∈ I : a(t) = 0} which is called the associated special Legendre curves of z (see [4]). Special Legendre curves do exist extensively. In fact, it was proved in [2] that, for any given non-zero functions ψ(s), a(s) defined on an open interval I, there exists a unit speed special Legendre curve z : I → S 5 (c) ⊂ C3 satisfying (4.4) with |Pz | = 1.

5. Classification theorem The main result of this paper is following classification theorem.

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Theorem 1. There are 29 families of Lagrangian surfaces of constant curvature in the complex projective plane CP 2 (4): (1) Totally geodesic Lagrangian surfaces of constant curvature one. (2) Flat minimal Lagrangian surface defined by √ √ √ √ √   1 L(s, t) = √ (e−i 2s , 2eis/ 2 cos( 3/2 t), 2eis/ 2 cos( 3/2 t)). 3

(3) Lagrangian surfaces of curvature one defined by π ◦ L with L(s, y) = (cos y, z(s) sin y), where z(s) = (z1 (s), z2 (s)) is an arbitrary unit speed Legendre curve in S 3 (1) ⊂ C2 . (4) Lagrangian surfaces of curvature one defined by π ◦ L with L(s, y) = z(s) cos y + Pz (s) sin y, where z : I → S 5 (1) ⊂ C3 is an arbitrary unit speed special Legendre curve and Pz : I → S 5 (1) ⊂ C3 is the associated special Legendre curve of z. (5) Lagrangian surfaces of positive curvature c2 defined by π ◦ L with L(s, y) = ei(b−c)s z(y) + ei(b+c)s w(y),

b > 0,

c=



1 + b2 ,

√ √ where z : I → S 5 ( b + c/√ 2c) ⊂ √ C3 is an arbitrary special Legendre curve with speed 1/2 and w : I → S 5 ( c − b/ 2c) is the associated special Legendre curve of z with speed 1/2. (6) Lagrangian surfaces of positive curvature a2 defined by π ◦ L, where L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is a unit speed Legendre curve in S 3 (1) ⊂ C2 given by eibs [(a − b)eias − (a + b)e−ias ], 2a  eibs cos(as) z2 = , a = 1 + b2 a for an arbitrary non-zero real number b. (7) The Lagrangian surface of curvature one defined by π ◦ L, where z1 =

L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is the unit speed Legendre curve in S 3 (1) ⊂ C2 given by z1 =

1 − i sin s , √ 2

cos s z2 = √ (sec s + tan s)i . 2

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(8) The flat Lagrangian surface defined by π ◦ L, where L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is the unit speed Legendre curve in S 3 (1) ⊂ C2 given by √ √ 2 b2 − 1 ei b −1s , z2 = , b > 1. z1 = √ 2 b beis/ b −1 (9) The flat Lagrangian surface defined by π ◦ L, where L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is the unit speed Legendre curve in S 3 (1) ⊂ C2 given by 

√ 2 − s2 i√1−s2 −i tan−1 (√1−s2 ) z1 = √ , e 2

√

1−s2

s se  . z2 = √  √ 2 1 + 1 − s2

(10) Lagrangian surfaces of negative curvature −b2 defined by π ◦ L, where L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is a unit speed Legendre curve in S 3 (1) ⊂ C2 given by  −1 z1 = (b − i e−2bs − k2 )ebs+ikb z2 = ebs+ikb

−1

tan−1 (k−1

√

tan−1 (k−1

e−2bs −k2 )−ib−1

√

√

e−2bs −k2 )

,

e−2bs −k2

√ with arbitrary positive number b and k = b2 + 1. (11) Lagrangian surfaces of positive curvature b2 defined by π ◦ L, where L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is a unit speed Legendre curve in S 3 (1) ⊂ C2 given by 

a/b a2 cos2 (bs) − 1 + ia sin(bs) z1 = , √ −1 −1 2 2 a(a2 − 1)a/2b eib tan (sin(bs)/ a cos (bs)−1) cos(bs)

 a/b  a2 − cos2 (bs) a2 cos2 (bs) − 1 + ia sin(bs) z2 = , √ −1 2 2 a(a2 − 1)a/2b ei tan (b sin(bs)/ a cos (bs)−1) with arbitrary positive number b >

i

√ √ 2 and a = b2 − 1 > | sec(bs)|.

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(12) Lagrangian surfaces of positive curvature b2 defined by π ◦ L, where L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is a unit speed Legendre curve in S 3 (1) ⊂ C2 given by  a/b  1 + b2 − cos2 (bs) a2 cos2 (bs) + 1 + ia sin(bs) , z1 = √ √ −1 2 2 ba/b b2 + 1ei tan (b sin(bs)/ a cos (bs)+1)  a/b cos(bs) a2 cos2 (bs) + 1 + ia sin(bs) z2 = , √ √ −1 −1 2 2 ba/b b2 + 1e−ib tanh (sin(bs)/ a cos (bs)+1) √ with arbitrary positive number b > 1 and a = b2 − 1. (13) Lagrangian surfaces of positive curvature b2 < 1 defined by π ◦ L, where L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is a unit speed Legendre curve in S 3 (1) ⊂ C2 given by  √  −i tan−1 b sin(bs)/ 1−a2 cos2 (bs) 2 2 b + sin (bs)e z1 = √   ia/b , b2 + 1 i( 1 − a2 cos2 (bs) + a sin(bs))

z2 =

eib

−1

√

tanh−1 (sin(bs)/

√

b2

1−a2 cos2 (bs))

cos(bs)

−1 −1 −1 + 1 eiab sinh (ab sin(bs))

,

√ with arbitrary positive number b ∈ (0, 1) and a = 1 − b2 . (14) Lagrangian surface of negative curvature −b2 defined by π ◦ L, where L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is a unit speed Legendre curve in S 3 (1) ⊂ C2 given by  i a2 − k2 cosh2 (bs) − b sinh(b)  , z1 = √ √ ib−1 k sin−1 k sinh(bs)/ a2 −k2 2 2 a −b e  √ iab−1 tan−1 a sinh(bs)/ a2 −k2 cosh2 (bs) cosh(bs) e  z2 = √ −1 √ −1 2 ib k sin k sinh(bs)/ a −k2 a2 − b 2 e √ with arbitrary positive number b and a > k := 1 + b2 .

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(15) Lagrangian surfaces of negative curvature −b2 defined by π ◦ L, where L(s, y) = (z1 (s), z2 (s) cos y, z2 (s) sin y) and z = (z1 , z2 ) is a unit speed Legendre curve in S 3 (1) ⊂ C2 given by  k/b  sinh(bs) k cosh(bs) + i a2 − k2 sinh2 (bs) z1 = √ , √ −1 2 −1 2 2 a2 + b2 (1 + a2 + b2 )k/2b eiab tanh (a cosh(bs)/ a −k sinh (bs))  k/b   (i a2 −k2 sinh2 (bs)−b cosh(bs)) k cosh(bs)+i a2 −k2 sinh2 (bs) z2 = √ a2 + b2 (1 + a2 + b2 )k/2b √ with arbitrary positive numbers a, b and k = 1 + b2 . (16) Flat Lagrangian surfaces defined by π ◦ L, with


 ae−is/a , z(v)eias , √ 1 + a2

L(s, v) =

0 = a ∈ R,

√ where z = z(v) is an arbitrary unit speed Legendre curve in S 3 (1/ 1 + a2 ) ⊂ C2 . (17) Flat Lagrangian surface defined by π ◦ L with √

i

L(s, v) = e

b2 −s2

 √ z(v)s1+ib 1 + b2 − s 2 ,√ √ √ −1 b2 −s2 (c + b2 − s2 )ic 1 + b2 ei tan

where b is an arbitrary √ positive number and z = z(v) is an arbitrary unit speed Legendre curve in S 3 (1/ 1 + b2 ) ⊂ C2 . (18) Lagrangian surfaces of positive curvature c2 defined by π ◦ L, with ibs




L=e

  b i sin(cs) − cos(cs), 2z(t) cos(cs) , c = 1 + b2 , c

where b is an arbitrary positive number and z = z(v) is an arbitrary Legendre curve 1 of speed 1/2 in S 3 ( 2c ) ⊂ C2 . (19) Lagrangian surfaces of curvature one defined by π ◦ L, with

 L(s, t) =

 1 + a2 sin2 s −i tan−1 (a sin s) e , z(t)(sec s + tan s)i/a cos s , √ 1 + a2

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where a is an arbitrary positive √ number and z = z(t) is an arbitrary Legendre curve of constant speed a in S 3 (a/ 1 + a2 ) ⊂ C2 . (20) Lagrangian surfaces of positive curvature b2 defined by π ◦ L with √

√ 2 − c2 − cos2 bs( a2 cos2 bs − c2 + ia sin bs)a/b b  , L(s, t) =  √ √ b2 − c2 exp i tan−1 (b sin(bs/ a2 cos2 bs − c2 ))  a   z(t) cos(bs) exp i sin−1 a sin(bs)/ a2 − c2 b   c − tan−1 (c tan(bs)/ a2 − c2 sec2 bs)  ,  b 

where b, c are any positive = z(t) is an arbitrary unit speed √ numbers with b > 1, z√ Legendre curve in S 3 (1/ b2 − c2 ) ⊂ C2 and a = b2 − 1. (21) Lagrangian surfaces of positive curvature b2 defined by π ◦ L with

L(s, t) =

√ b2 + c2 − cos2 bs( a2 cos2 bs + c2 + ia sin bs)a/b , √ b2 + c2 (a2 + c2 )a/2b exp(i tan−1 ((b sin bs)/ a2 cos2 bs + c2 ))  z(t)(cos bs)( a2 cos2 bs + c2 + ia sin bs)a/b
   c tan bs ic × exp tanh−1 √ , b a2 + c2 sec2 bs √

√

where b is an arbitrary positive number√> 1, z = z(t) an arbitrary√ Legendre curve of constant speed (a2 + c2 )−a/2b in S 3 (1/ b2 + c2 ) ⊂ C2 , and a = b2 − 1. (22) Lagrangian surfaces of positive curvature b2 defined by π ◦ L with

 L(s, t) =

 c2 + sin2 s −i tan−1 ((sin s)/c) 2ic tanh−1 (tan(s/2)) e , z(t)e cos s , √ 1 + c2

where z = z(t) is an arbitrary unit speed Legendre curve in S 3 ( √ 1

1+c2

(23) Lagrangian surfaces of positive curvature b2 defined by π ◦ L with
 c sin bs c −1 tanh L(s, t) = z(t) exp i √ b c2 − a2 cos2 bs
 a sin bs a cos bs, − sinh−1 √ b c 2 − a2




) ⊂ C2 .

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 √

 b2 + c2 − cos2 bs exp i √

a2 (2b2 +2c2 −1) 2b2 (a2 −c2 )

tan−1

√

c2 −a2 cos2 bs b sin bs

 2 2 √ b2 +c2 (a sin bs+ c2 −a2 cos2 bs)ia/b exp i 2ba2 (a−2c 2 −c2 )  b sin bs ×tan−1 √

    ,    

c2 −a2 cos2 bs

√ 2 3 2 2 where z = z(t) is an arbitrary unit speed Legendre √ curve in S (1/ b + c ) ⊂ C , b 2 is an arbitrary real number in (0, 1) and a = 1 − b . (24) Lagrangian surfaces of negative curvature −b2 defined by π ◦ L with  √ −ib−1 a sin−1 a(sinh bs)/ c2 −a2 L(s, t) = e

 i c2 − a2 cosh2 bs − b sinh bs × , √ c 2 − b2   √ icb−1 tan−1 c(sinh bs)/ c2 −a2 cosh2 bs z(t)e cosh bs , a=

 1 + b2 ,

√ where z = z(t) is an arbitrary unit speed Legendre curve in S 3 (1/ c2 − b2 ) ⊂ C2 , b an arbitrary positive number, and c an arbitrary number > a. (25) Lagrangian surfaces of curvature one defined by π ◦ L with
 sechbs −1 L(s, t) = √ (1 + 4b2 ) cosh2 bs − 4b2 e−i tan (2b tanh bs) , 2 1 + 4b    2beis/2 cos 1 + 4b2 t/2 , 2beis/2 sin 1 + 4b2 t/2 , where b is an arbitrary non-zero real number. (26) Curvature one Lagrangian surfaces of type (Aρψ , αρψ ) described in Proposition 6.1. K , βK ) described in Propo(27) Lagrangian surfaces with positive curvature K of type (BµΦ µΦ sition 6.2. K , γ K ) described in Propo(28) Lagrangian surfaces of positive curvature K of type (CµΦ µΦ sition 6.3. K , δK ) described in Propo(29) Lagrangian surfaces of positive curvature K of type (DµΦ µΦ sition 6.4. Conversely, up to rigid motions of CP 2 (4), locally (in a neighborhood of each point belonging to an open dense subset), every Lagrangian surface of constant curvature in CP 2 (4) is obtained from one of the 29 families. Proof. Let M be a Lagrangian surface of constant curvature K in CP 2 (4). Denote the tangent bundle of M by TM. If M is minimal in CP 2 (4), then it is totally geodesic or flat (cf. [8,11]).

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So, M is an open portion of a Lagrangian totally geodesic real projective plane RP 2 (1) or a flat minimal surface in CP 2 (4). This gives Cases (1) and (2).
Next, let us assume that M is non-minimal. Then U := {p ∈ M : H(p) = 0} is a nonempty open subset. We shall work on U instead of M. For each point p ∈ U, we define a function γp by γp : Tp1 U → R : v → γp (v) = h(v, v), Jv, where Tp1 U := {v ∈ Tp U : v, v = 1}. Since Tp1 U is a unit circle which is compact, there exists a vector v ∈ Tp1 U such that γp attains an absolute minimum at v. Since p is a non-totally geodesic point, (2.4) implies that γp = 0. So, by applying linearity, we have γp (v) < 0. As γp attains an absolute minimum at v, it follows from (2.4) that h(v, v), Jw = 0 for all w orthogonal to v. Combining this with (2.4) shows that v is an eigenvector of the shape operator AJv . Hence, there exists an orthonormal basis {e1 , e2 } of Tp M with e1 = v which satisfies h(e1 , e1 ) = λJe1 ,

h(e1 , e2 ) = µJe2 ,

h(e2 , e2 ) = µJe1 + ϕJe2 , (λ + µ)2

(5.1)

for some functions λ, µ, ϕ. As H = 0, we have > 0 on U. If ϕ = 0 on U, the the Lagrangian surface is Maslovian. Hence, it follows from Theorem 3 of [4] that the Lagrangian surface, restricted to U, is given by one of the Lagrangian surfaces given in Cases (3)–(15). Next, let us assume that ϕ = 0 on an open subset V ⊂ U. In this case, (5.1) and the equation of Codazzi imply e1 µ = ϕω12 (e1 ) + (λ − 2µ)ω12 (e2 ),

+ ϕ2

e2 λ = (λ − 2µ)ω12 (e1 ),

e2 µ − e1 ϕ = 3µω12 (e1 ) + ϕω12 (e2 ),

(5.2)

where ∇X e1 = ω12 (X)e2 . Also, from (5.1) and the equation of Gauss we have K = λµ − µ2 + 1 = const.

(5.3)

Case (I). ∇e1 e1 = 0 on an open neighborhood V1 of a point in V. In this case, (5.2) and (5.3) imply e1 µ = (λ − 2µ)ω12 (e2 ),

e2 λ = 0,

e2 µ − e1 ϕ = ϕω12 (e2 )

(5.4)

on V1 . By differentiating (5.3) with respect to e2 and by applying (5.4), we obtain (λ − 2µ)e2 µ = 0. Thus, we have either λ = 2µ or e2 µ = 0 at each point of V1 . If λ = 2µ on some connected open subset W ⊂ V1 , then K = µ2 + 1 on W which implies that µ is constant on W. So, e2 µ = 0 also holds on W. Consequently, we have e2 µ = 0 identically on V1 . Therefore, we obtain from (5.4) that e1 µ = (λ − 2µ)ω12 (e2 ),

e2 λ = e2 µ = 0,

e1 ϕ = −ϕω12 (e2 ).

(5.5)

As we have ∇e1 e1 = 0 on V1 , there exists a local coordinate system {s, u} on V1 such that the metric tensor is given by g = ds ⊗ ds + G2 (s, u) du ⊗ du

(5.6)

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for some function G with ∂/∂s = e1 , ∂/∂u = Ge2 . From (5.5) we have λ = λ(s) and µ = µ(s). Also, it follows from (5.6) that ∂ ∂ Gs = (ln G)s , ω12 (e2 ) = . ∂s ∂u G By (5.5), (5.6) and (5.7), we find (ln G)s = −(ln ϕ)s . Thus, (5.6) becomes ∇∂/∂u

g = ds ⊗ ds +

F 2 (u) du ⊗ du, ϕ2

e1 =

∂ , ∂s

e2 =

ϕ ∂ F (u) ∂u

(5.7)

(5.8)

for some positive function F (u). By applying (5.8) and the equation of Gauss, we have ϕϕss − 2ϕs2 = Kϕ2 . After solving this differential equation, we obtain  A(u) sec(bs + B(u)), if K = b2 > 0,     A(u) ϕ= (5.9) , if K = 0,  cs + B(u)    A(u)sech (bs + B(u)), if K = −b2 < 0, for some functions A(u), B(u) and b > 0, where A is nowhere zero on V1 . Let t = t(u) be an antiderivative of F (u)/A(u). Then (5.8) and (5.9) give  ds ⊗ ds + cos2 (bs + θ(t)) dt ⊗ dt, if K = b2 > 0,    if K = 0, g = ds ⊗ ds + (cs + θ(t))2 dt ⊗ dt,    2 ds ⊗ ds + cosh (bs + θ(t)) dt ⊗ dt, if K = −b2 < 0,

(5.10)

some function θ(t). We divide Case (I) into several cases. Case (I.i). λ = 2µ on an open subset U1 ⊂ V1 . In this case, both λ, µ are constant and K = 1 + µ2 ≥ 1 on U1 by (5.3). If λ = µ = 0 on U1 , the Lagrangian surface is Maslovian. So, this reduces to previous case. Hence, we may assume that λ = 2µ = 2b for some positive number b on U1 which gives K = 1 + b2 > 1. From (5.9) and (5.10) we have g = ds ⊗ ds + cos2 (bs + θ(t)) dt ⊗ dt, λ = 2µ = 2b > 0, ϕ = f (t) sec(bs + θ(t)), ∂ ∂ ∂ (5.11) ∇∂/∂s = 0, ∇∂/∂s = −b tan(bs + θ(t)) , ∂s ∂t ∂t b ∂ ∂ ∂ ∇∂/∂t = sin(2bs + 2θ(t)) − θ  (t) tan(bs + θ(t)) , ∂t 2 ∂s ∂t where f is non-zero function. Applying (5.1), (5.11) and the formula of Gauss, we find Lss = 2ibLs − L, Lst = (ib − c tan(bs + θ))Lt , Ltt = (ib cos(bs + θ(t)) + b sin(bs + θ(t))) cos(bs + θ(t))Ls + (if (t) − θ  tan(bs + θ))Lt − cos2 (bs + θ(t))L.

(5.12)

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After solving the first equation of this system, we obtain  L = ei(b−c)s (A(t) + B(t)e2ics ), c = 1 + b2

441

(5.13)

for some C3 -valued functions A(t), B(t). By substituting this into the second equation of (5.12), we discover that B (t) = A (t)e2iθ . Hence, (5.13) becomes Lt = A (t)ei(b−c)s (1 + e2i(bs+θ) ).

(5.14)

If θ is constant on U1 , we have θ = 0 after applying a suitable translation on s. Thus, (5.14) becomes Lt = A (t)ei(b−c)s (1 + e2ics ) which implies that L = A(t)ei(b−c)s (1 + e2ics ) + K(s)

(5.15)

for some C3 -valued function K(s). Substituting (5.15) into the first equation in (5.12) yields K = 2ibK − K. Hence, after solving the last equation, we obtain K(s) = ei(b−c)s (a1 + a2 e2ics ) for some vectors a1 , a2 ∈ C3 . Therefore, we may put L = F (t)(ei(b−c)s + ei(b+c)s ) + c1 ei(b+c)s for some vector function F (t) and vector c1 . Substituting this into the last equation in (5.12) gives 2F  (t) − 2if (t)F  (t) + 2c2 F (t) + c(b + c)c1 = 0. Thus, we get F (t) = z(t) − ((b + c)/2c)c1 , where z = z(t) is a C3 -valued solution of z (t) − if (t)z (t) + c2 z(t) = 0.

(5.16)

Consequently, we obtain 

   b ibs L=e 2z(t) − c1 cos(cs) + ic1 sin(cs) . c

(5.17)

Since |L|2 = 1, (5.17) implies that |c1 | = 1,

|2cz(t) − bc1 |2 = c2 ,

z(t), ic1  = 0.

(5.18)

It follows immediately from (5.17) that Ls =

ieibs {(c1 + 2bcz(t)) cos(cs) + 2ic2 z(t) sin(cs)}, c

Lt = 2 cos(cs)eibs z (t).

(5.19)

Thus, by applying the first equation in (5.11) and (5.19), we also have |z (t)| =

1 , 2

|c1 + 2bcz(t)|2 = c2 ,

|z| =

1 . 2c

(5.20)

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From |c1 | = 1, |z| = 1/2c and the second equation in (5.20) we get z(t), c1 =0. Also, by applying Lemma 2, we know that z = z(t) is a Legendre curve of speed 1/2 in S 3 (1/2c) ⊂ C2 ⊂ C3 . So, if we choose c1 = (1, 0, 0), we obtain from (5.17) that ibs

L=e






b i sin(cs) − cos(cs), 2z1 (t) cos(cs), 2z2 (t) cos(cs) c

Consequently, restricted to U1 , the Lagrangian surface in CP 2 (4) is congruent to the composition π ◦ L, where L is given by Case (18). Next, assume that θ is a non-constant function on an open interval I  0. From (5.14) we find  t  i(b−c)s i(b+c)s L=e A(t) + e A (t)e2iθ dt + K(s), c = b2 + 1 (5.21) 0

for some C -valued function K. Substituting this into the first equation in (5.12) gives K = 2ibK − K. Solving this equation gives K = a1 ei(b−c)s + a2 ei(b+c)s for some vectors a1 , a2 ∈ C3 . Hence, we obtain 3

(5.22) L = ei(b−c)s z(t) + ei(b+c)s w(t),  t  2iθ where z(t) = A(t) + a1 and w(t) = 0 A (t)e dt + a2 . Since |L| = 1, (5.22) implies that 1 = |A(t)|2 + |W(t)|2 + 2ze2ics , w. Hence, by applying z, e2ics w = cos(2cs)z, w + sin(2cs)z, iw, we find z, w = z, iw = 0,

|z(t)|2 + |w(t)|2 = 1.

Also, from (5.22), we have ˜ s = i(b − c)ei(b−c)s z(t) + i(b + c)ei(b+c)s w(t), L Lt = z (t)ei(b−c)s (1 + e2i(cs+θ(t)) ).

(5.23)

Applying these gives |z (t)| = |w (t)| = 1/2, |z(t)|2 = b + c/2c, and |w(t)|2 = c − b/2c. So, after differentiating the last equation, we have z e2iθ , w = 0. Moreover, by applying L, Lt  = Ls , iLt  = 0, we get z, e2i(cs+θ) z  + z , e2ics w = (b − c)z, e2i(cs+θ) z  + (b + c)z , e2ics w = 0. Therefore, we obtain A, e2i(cs+θ) z  = z , e2ics w = 0 which implies that z, iz  = z , w = z , iw = w , iw = 0. √ √ √ √ Thus, z : I → S 5 ( b + c/ 2c) ⊂ C3 and w : I → S 5 ( c − b/ 2c) ⊂ C3 are Legendre curves of constant speed 1/2. Now, by substituting (5.22) into the last equation in (5.13), we find z (t) = i(H(t) − θ  (t))z (t) −

c(c − b) c(b + c) −2iθ w(t). z(t) − e 2 2

(5.24)

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Since √ w (t) = √ e2iθ z (t), w = w(t) is a parallel normal vector field. Consequently, z : 5 I → S (√b + c/√2c) ⊂ C3 is a special Legendre curve of constant speed 1/2 and w : I → S 5 ( c − b/ 2c) ⊂ C3 is an associated special Legendre curve of z with the same constant speed. Therefore, the Lagrangian surface is congruent to the composition π ◦ L, where L is given by Case (5). Case (I.ii). λ = 2µ on an open subset U2 ⊂ V1 . In this case, (5.2), (5.3) and ∇e1 e1 = 0 imply e2 λ = e2 µ = 0. Thus, we obtain from (5.4) that µ (s) ω12 (e2 ) = . (5.25) λ − 2µ If µ = 0 identically on an open subset V of U2 , then (5.3) and (5.25) imply K = 1 and ω12 = 0 on V which is impossible. So, µ = 0 almost everywhere on U2 . Case (I.ii.a). λ = µ = 0 on U2 . From (5.3) and (5.4) we get K = 1,

e1 (ln µ) = −ω12 (e2 ),

e2 λ = e2 µ = 0,

e1 ϕ = −ϕω12 (e2 ).

(5.26)

So, λ and µ depend only on s according to (5.8). Combining (5.7) and the second equation in (5.26) gives G = F (u)/µ(s). Hence, (5.6) reduces to g = ds ⊗ ds +

dt ⊗ dt , µ2 (s)

(5.27)

where t = t(u) is an antiderivative of F (u). Thus, (5.10) yields µ−1 = a cos(s + b) with a = 0. Hence, after making a suitable translation in s, we obtain sec s g = ds ⊗ ds + a2 cos2 s dt ⊗ dt, λ=µ= . (5.28) a Without loss of generality, we may choose a > 0. From (5.28) we find ω12 (e2 ) = − tan s. Thus, we may obtain from the last equation in (5.26) that ϕs = ϕ tan s which gives ϕ = f (t) sec s for some function f. From (5.1), (5.28) and the formula of Gauss, we obtain i sec s Ls − L,  i asec s  Lst = a − tan s Lt ,

Lss =

Ltt = (ia cos s

+ a2

sin s cos s)Ls + iaf (t)Lt

(5.29) − a2

cos2 sL.

Solving the first equation in (5.29) gives  −1 L = z(t)(sec s + tan s)i/a cos s + C(t) 1 + a2 sin2 s e−i tan (a sin s) for some C3 -valued functions z and C. Substituting this into the second equation in (5.29) gives C = 0. So, C is a constant vector, say c1 . Hence, L is given by  −1 L = z(t)(sec s + tan s)i/a cos s + c1 1 + a2 sin2 s e−i tan (a sin s) . (5.30) Substituting this into the last equation of (5.29) yields z (t) = iaf (t)z (t) − (1 + a2 )z(t).

(5.31)

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It follows immediately from (5.30) that Ls =

(a sin s − i){c1 a2 e−i tan

−1 (a

sin s)

 cos s − z(t) 1+a2 sin2 s(sec s+ tan s)i/a }

 a 1 + a2 sin2 s

Lt = z (t)(sec s + tan s)i/a cos s.

,

(5.32) Since |L| = 1, (5.30) and (5.32) imply that |z(t)| = √

a 1 + a2

,

|c1 | = √

1 1 + a2

|z (t)| = a,

,

c1 , z = 0.

 Moreover, from Ls , iLt  = 0, we also have √ c1 , iz(t) 2= z(t), iz (t) = 0. These shows2 3 2 that z = z(t) is a Legendre curve in S (a/ 1 + a ) ⊂ C with constant speed a, where C is a complex hyperplane with c1 as its hyperplane normal. Consequently, the Lagrangian surface, restricted to U2 , is obtained from Case (19).

Case (I.ii.b). λ = µ, µ = 0 on an open subset W1 ⊂ U2 . In this case, (5.3) implies K = 1. Moreover, from (5.3), (5.5) and (5.7), we have λ = µ + K−1 µ , 1 ω2 (e2 ) = e1 (ln |K − µ2 − 1|) = e1 (ln ϕ) = −e1 (ln G) on W1 , where G is defined by (5.6). Hence, we get  ϕG = f (t), G |K − µ2 − 1| = p(t),

(5.33)

(5.34)

for some positive real-valued function p and and non-zero real-valued function f. We divide this case into several cases. Case (I.ii.b.1). K = b2 > µ2 + 1 > 1 on a neighborhood W1,1 of a point p ∈ W1 . Without loss of generality, we may choose b > 1. From (5.10) and (5.34) we get g = ds ⊗ ds + cos2 (bs + θ(t)) dt ⊗ dt, µ2 = b2 − 1 − p2 (t) sec2 (bs + θ(t)),

ϕ = f (t) sec(bs + θ(t)).

(5.35)

It follows from (5.5) that µ = µ(s). Thus, by differentiating the second equation in (5.35), we get (ln p(t)) = ∂(ln cos(bs + θ(t)))/∂t. Hence, p(t) = k(s) cos(bs + θ(t)) for some function k(s). Differentiating the last equation with respect to s gives (ln k(s)) = b tan(bs + θ(t)). Therefore, θ and p are constant. We may assume θ = 0 by applying a suitable translation in s. Hence, we obtain from (5.35) that g = ds ⊗ ds + (cos2 bs) dt ⊗ dt, ∂ ∂ ∂ ∂ b ∂ ∇∂/∂s = 0, ∇∂/∂s = −b tan(bs) , ∇∂/∂t = sin(2bs) , ∂s ∂t ∂t ∂t 2 ∂s  2a2 − c2 sec2 bs λ= √ ϕ = f (t) sec bs, , µ = a2 − c2 sec2 bs, a2 − c2 sec2 bs √ where c = p is a positive number and a = b2 − 1.

(5.36)

(5.37)

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From (5.1), (5.36), (5.37) and the formula of Gauss, we get 2a2 − c2 sec2 bs Lss = i √ Ls − L, a2 − c2 sec2 bs √ Lst = (i a2 − c2 sec2 bs − b tan bs)Lt , √ Ltt = (b sin bs + i a2 cos2 bs − c2 ) cos bsLs + if (t)Lt − cos2 bsL.

(5.38)

After solving the second equation in (5.38) we find  L = B(cos bs) exp i

a sin−1 b




a sin bs √ a2 − c 2

 −

c tan−1 b


√

c tan bs

 +A

a2 − c2 sec2 bs

for some C3 -valued functions A = A(s) and B = B(t). Substituting this into the first equation in (5.38) yields 2a2 − c2 sec2 bs  A (s) = i √ A (s) − A(s). a2 − c2 sec2 bs Solving (5.39) gives



A(s) = c0 cos(bs) exp i

a sin−1 b




a sin bs √ a2 − c 2

(5.39)  −

c tan−1 b


√



c tan bs

a2 − c2 sec2 bs √ √ b2 − c2 − cos2 bs( a2 cos2 bs − c2 + ia sin bs)a/b + c1 √ exp(i tan−1 (b sin bs/ a2 cos2 bs − c2 ))

for some vectors c0 , c1 ∈ C3 . Hence we obtain 

 a c a sin bs c tan bs −1 −1 L = z(t) cos(bs) exp i sin − tan √ √ b b a2 − c 2 a2 − c2 sec2 bs √ √ b2 − c2 − cos2 bs( a2 cos2 bs − c2 + ia sin bs)a/b + c1 , (5.40) √ exp(i tan−1 (b sin bs/ a2 cos2 bs − c2 )) where z(t) = A(t) + c0 . Since |L| = 1, (5.40) implies |z|2 =

1 , b2 − c 2

|c1 |2 =

1 , (b2 − c2 )(a2 − c2 )2a/b

z, c1  = 0.

(5.41)

Also, from (5.36), (5.39) and Ls , iLt  = 0, we find |z (t)| = 1 and z, ic1  = 0. Substituting these into the last equation of (5.38) yields z (t) = if (t)z (t) − (b2 − c2 )z(t). √ Thus, z(t) is a unit speed Legendre curve in S 3 (1/ b2 − c2 ) ⊂ C2 , where C2 is a complex hyperplane with hyperplane normal c1 . Consequently, the Lagrangian surface, restricted to W1,1 , is congruent to π ◦ L with L given by Case (20).

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Case (I.ii.b.2). 0 < K = b2 < µ2 + 1 on a neighborhood W1,2 of a point p ∈ W1 . Without loss of generality, we may assume b > 0. From (5.10) and (5.34) we get g = ds ⊗ ds + cos2 (bs + θ(t)) dt ⊗ dt, µ2 = b2 − 1 + p2 (t) sec2 (bs + θ(t)),

ϕ = f (t) sec(bs + θ(t)).

(5.42)

Since µ = µ(s) and p(t) sec(bs + θ(t)) depend only on s according to the second equation in (5.42), p(t) and θ(t) both are constant as in Case (I.ii.c.1). So, we have θ = 0 after applying a suitable translation in s. Hence, we obtain from (5.42) that g = ds ⊗ ds + cos2 bs dt ⊗ dt, ∂ ∂ ∂ ∇∂/∂s = 0, ∇∂/∂s = −b tan bs , ∂s ∂t ∂t

∇∂/∂t

∂ b ∂ = sin 2bs ∂t 2 ∂s

(5.43)

and 2b2 − 2 + c2 sec2 bs λ= √ , b2 − 1 + c2 sec2 bs ϕ = f (t) sec bs,

µ=



b2 − 1 + c2 sec2 bs, (5.44)

where c = p is a positive number. From (5.1), (5.43) (5.44) and the formula of Gauss, we have 2b2 − 2 + c2 sec2 bs Lss = i √ Ls − L, b2 − 1 + c2 sec2 bs √ Lst = (i b2 − 1 + c2 sec2 bs − b tan bs)Lt ,  Ltt = (b sin bs + i c2 + (b2 − 1) cos2 bs) cos bsLs + ifLt − cos2 bs L.

(5.45)

Case (I.ii.b.2.1). b > 1. In this case, (5.45) becomes 2a2 + c2 sec2 (bs) Lss = i  Ls − L, a2 + c2 sec2 (bs)  Lst = (i a2 + c2 sec2 (bs) − b tan(bs))Lt , √ Ltt = (b sin bs + i c2 + a2 cos2 bs)(cos bs)Ls + ifLt − (cos2 bs)L √ b2 − 1. After solving the first equation in (5.46) we find √ √ b2 + c2 − cos2 bs( a2 cos2 bs + c2 + ia sin bs)a/b L = C(t) √ exp(i tan−1 (b sin bs/ a2 cos2 bs + c2 ))  + z(t)(cos bs)( a2 cos2 bs + c2 + ia sin bs)a/b 
 c tan bs ic tanh−1 √ × exp b a2 + c2 sec2 bs

(5.46)

with a =

(5.47)

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for some C3 -valued functions z and C. Substituting this into the second equation in (5.45) shows that C is a constant vector, say c1 ∈ C3 . Since |L| = 1, (5.47) implies |z|2 = |c1 |2 =

1 (b2

+ c2 )(a2

+ c2 )a/b

,

z, c1  = 0.

(5.48)

Also, from (5.43), (5.47) and Ls , iLt  = 0, we find |z (t)| = (a2 + c2 )−a/b , 2

z, ic1  = 0.

(5.49)

Also, by substituting (5.47) into the last equation of (5.46), we get z (t) = if (t)z (t) − (b2 + c2 )z(t).

(5.50)

√

Therefore, z is a Legendre curve in S 3 (1/ b2 + c2 ) ⊂ C2 with speed 1/(a2 + c2 )a/2b , where C2 is a complex hyperplane with c1 as its normal. Hence the Lagrangian surface, restricted to W1,1 , is congruent to π ◦ L with L given by Case (21). Case (I.ii.b.2.2). b = 1. In this case, (5.44) becomes Lss = ic sec s Ls − L, Lst = (ic sec s − tan s)Lt , Ltt = (sin s + ic)(cos s)Ls + ifLt − (cos2 s)L. After solving the first equation in (5.51) we find  −1 −1 −1 L = z(t)(cos s)e2ic tanh (tan s/2) + C(t) c2 + sin2 s e−i tan (c

(5.51)

sin s)

(5.52)

for some C3 -valued functions z and C. Substituting this into the second equation in (5.45) shows that C is a constant vector, say c1 . Since |L| = 1, (5.52) implies |z|2 = |c1 |2 =

1 , 1 + c2

z, c1  = 0.

(5.53)

From (5.43), (5.52) and Ls , iLt  = 0 we find |z (t)|2 = 1, z, ic1  = 0. Also, by substituting (5.52) into the last equation of (5.51), we get z (t) = if (t)z (t) − (1 + c2 )z(t).

√

(5.54)

Thus, z is a unit speed Legendre curve in S 3 (1/ 1 + c2 ) ⊂ C2 , where C2 is a complex hyperplane with c1 as its hyperplane normal. Consequently, the Lagrangian surface, restricted to W1,1 , is congruent to π ◦ L with L given by Case (22). Case (I.ii.b.2.3). 0 < b < 1. In this case, (5.44) becomes  c2 sec2 bs − 2a2 Ls − L, a = 1 − b2 , Lss = i √ c2 sec2 bs − a2 √ Lst = (i c2 sec2 bs − a2 − b tan bs)Lt , √ Ltt = (b sin bs + i c2 − a2 cos2 bs) cos bsLs + ifLt − cos2 bsL.

(5.55)

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After solving the first and second equations in (5.55) we find √ (cos bs) exp i(b−1 c tanh−1 (c sin bs/ c2 − a2 cos2 bs)) L = z(t) √ exp i(ab−1 sinh−1 (a sin bs/ c2 − a2 )) √ c1 b2 + c2 − cos2 bs exp i{(a2 (2b2 + 2c2 − 1)/2b2 (a2 − c2 )) √ × tan−1 ( c2 − a2 cos2 bs/b sin bs)} + √ (a sin bs + c2 − a2 cos2 bs)ia/b exp i{(a2 − 2c2 /2b2 (a2 − c2 )) √ × tan−1 (b sin bs/ c2 − a2 cos2 bs)} for some C3 -valued function z = z(t) and constant vector c1 . Since |L| = 1, (5.56) implies that 1 |z|2 = |c1 |2 = 2 , z, c1  = 0. (b + c2 )

(5.56)

(5.57)

From (5.43), (5.56) and Ls , iLt  = 0 we find |z (t)| = 1 and z, ic1  = 0. Also, substituting (5.56) into the last equation of (5.55) yields z (t) = if (t)z (t) − (b2 + c2 )z(t). √ Thus, z is a unit speed Legendre curve in S 3 (1/ b2 + c2 ) ⊂ C2 , where C2 is a complex hyperplane with c1 as its hyperplane normal. Consequently, the Lagrangian surface, restricted to W1,1 , is congruent to π ◦ L with L given by Case (23). Case (I.ii.b.3). K = 0 on a neighborhood W1,3 of a point p ∈ W1 . In this case, we obtain from (5.10) and (5.34) that g = ds ⊗ ds + (cs + θ(t))2 dt ⊗ dt, p2 (t) A(t) µ2 = , − 1, ϕ= cs + θ(t) (cs + θ(t))2

(5.58)

If c = 0, we get g = ds ⊗ ds + θ 2 (t) dt ⊗ dt and ϕ = ϕ(t). Thus, we have g = ds ⊗ ds + dv ⊗ dv,

ϕ = ϕ(v),

(5.59)

where v = v(t) is an antiderivative of θ(t). Since λ = λ(s) and µ = µ(s) for Case (I.ii), we know from (5.33) that µ is constant. If µ = 0, then (5.1) and the equation of Gauss imply that K = 1 which is a contradiction. Hence µ is a non-zero constant, say a. Thus, we obtain from (5.3) that λ = a − a−1 . Therefore, (5.1), (5.59) and the formula of Gauss imply that
 1 Lss = i a − Lsv = iaLv , Lvv = iaLs + iϕ(v)Lv − L. Ls − L, a (5.60) Solving the first and the second equations in (5.60) gives L = z(v)eias + c1 e−is/a

(5.61)

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for some C3 -valued function z = z(v) and vector c1 ∈ C3 . Since |L| = |Ls | = |Lv | = 1 and Ls , iLy  = 0, we derive from (5.61) that |z|2 =

1 , 1 + a2

|z |2 = 1,

|c1 |2 =

a2 , 1 + a2

z, c1  = z, ic1  = 0. (5.62)

Also, by substituting (5.61) into the last equation in (5.60), we find z (y) = iϕ(y)z (y) − (1 + a2 )z(y).

(5.63)

Consequently, the surface is congruent to π ◦ L where L is given by Case (16). Next, assume that c = 0. In this case, p(t) and θ(t) are constant due to µ = µ(s) and the second equation in (5.58). So, we have θ = 0 after applying a suitable translation in s. Consequently, we have g = ds ⊗ ds + c2 s2 dt ⊗ dt. So, if we put v = ct and p = b, we find g = ds ⊗ ds + s2 dv ⊗ dv, b2 − 2s2 λ= √ , s b2 − s 2

µ=

(5.64) √

b2 − s 2 , s

ϕ=

f (v) , s

From (5.58), (5.64) and the formula of Gauss, we obtain

 √ b2 − 2s2 1 b2 − s 2 Lss = i √ Ls − L, Lsv = +i Lv , s s s √b2 − s2 Lvv = (is b2 − s2 − s)Ls + if (v)Lv − s2 L. After solving the first and second equations in (5.66) we obtain   √ √ 1+ib 2 − s2 z(v)s 1 + b c 2 2 1 L = ei b −s + √ √ −1 b2 −s2 (c + b2 − s2 )ib ei tan

(5.65)

(5.66)

(5.67)

for some C3 -valued functions z and constant vector c1 ∈ C3 . Since |L| = 1, (5.67) implies |z|2 = |c1 |2 =

1 , 1 + b2

z, c1  = 0.

(5.68)

Also, from (5.43), (5.56) and Ls , iLv  = 0 we find |z (v)| = 1 and z, ic1  = 0. Moreover, by substituting (5.67) into the last equation of (5.66), we get z (v) = if (v)z (v) − (1 + b2 )z(v). Therefore, z is a unit speed Legendre curve in S 3 ( √ 1

1+b2

) ⊂ C2 , where C2 is a complex hy-

perplane with c1 as its hyperplane normal. Consequently, the Lagrangian surface, restricted to W1,1 , is congruent to π ◦ L with L given by Case (17).

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Case (I.ii.b.4). K = −b2 < 0 on a neighborhood W1,4 of a point p ∈ W1 . We may assume b > 0. From (5.10) and (5.34) we get g = ds ⊗ ds + cosh2 (bs + θ(t)) dt ⊗ dt, µ2 = p2 (t) sech 2 (bs + θ(t)) − b2 − 1,

ϕ = f (t) sech (bs + θ(t)),

(5.69)

for some function p(t) and f (t) satisfying p2 > b2 + 1. Since µ = µ(s), the second equation in (5.69) implies that p(t)sech (bs + θ(t)) depends on only on s which is impossible unless both p and θ are constant. Thus, we have θ = 0 after applying a suitable translation in s. If we denote the constant p by c, we obtain c2 ≥ b2 + 1. Moreover, we have g = ds ⊗ ds + cosh2 bs dt ⊗ dt, ϕ = f (t) sech bs, ∂ b ∂ ∂ ∂ ∂ ∇∂/∂s = 0, ∇∂/∂s = b tanh bs , ∇∂/∂t = − sinh 2bs ∂s ∂t ∂t ∂t 2 ∂s   c2 sech 2 bs − 2a2 λ= √ , µ = c2 sech 2 bs − a2 , a = b2 + 1. 2 2 2 c sech bs − a From (5.1), (5.70) and the formula of Gauss, we find  c2 sech 2 bs − 2a2 Lss = i √ Ls − L, a = b2 + 1, √c2 sech 2 bs − a2 Lst = (i c2 sech 2 bs − a2 + b tanh bs)Lt ,  Ltt = (i c2 − a2 cosh2 bs − b sinh bs) cosh bsLs + ifLt − cosh2 bsL. Solving the first and the second equations of this system gives  √ −1 −1 2 2 L = e−iab sin ((a sinh bs)/ c −a ) {c1 (i c2 − a2 cosh2 bs − b sinh bs) √ ib−1 c tan−1 ((c sinh bs)/ c2 −a2 cosh2 bs) cosh bs} + z(t)e

(5.70)

(5.71)

(5.72)

for some vector c1 and vector function z. Since |L| = |Ls | = 1, |Lt | = cosh bs and Ls , iLy  = 0, we obtain from (5.73) that |z|2 = |c1 |2 =

c2

1 , − b2

|z |2 = 1,

z, c1  = z, ic1  = 0.

By substituting (5.72) into the third equation of (5.71) we obtain z (t) = if (t)z (t) − (c2 − b2 )z(t).

√ Thus, z(t) is a unit speed Legendre curve in S 3 (1/ c2 − b2 ) ⊂ C2 . Therefore, the immersion L, restricted W1,4 , is congruent to π ◦ L, where L is given by case (24). Case (II). ∇e1 e1 = 0 on an open subset V2 ⊂ V . In this case, ω12 (e1 ) is never zero on V2 . Since Span {e1 } and Span {e2 } are of rank one, there exists local coordinates {x, y} on V2 such that ∂/∂x, ∂/∂y are parallel to e1 , e2 , respectively. Thus, the metric tensor g takes the form: g = E2 dx ⊗ dx + G2 dy ⊗ dy,

(5.73)

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for some positive functions E, G. We may assume ∂/∂x = Ee1 , ∂/∂y = Ge2 . From (5.73) we have Ey Gx ∂E ∂G ω21 (e1 ) = , ω12 (e2 ) = , Ey = , Gx = . (5.74) EG EG ∂y ∂x If λ = 2µ, (5.3) reduces to K = µ2 + 1 which implies µ is constant. So, the first equation in (5.2) and ω12 (e1 ) = 0 give ϕ = 0 which contradicts to ϕ = 0. Hence, we have λ = 2µ. Also, the second equation in (5.2) and ω12 (e1 ) = 0 give e2 λ = 0. So, λ is a non-trivial function. Case (II.i). µ = 0 on V2 . In this case, we get from (5.3) that K = 1. Moreover, from (5.2) we find ϕω12 (e1 ) = λω21 (e2 ),

e2 λ = λω12 (e1 ),

e1 ϕ = ϕω21 (e2 ),

(5.75)

It follows from (5.74) and (5.75) that λE = η(x) and ϕG = k(y) for some functions η(x) and k(y). Hence, (5.73) becomes g=

η2 (x) k2 (y) dx ⊗ dx + 2 dy ⊗ dy. 2 λ ϕ

(5.76)

If u = u(x) and v = v(y) are antiderivatives of η(x) and k(y) respectively, then (5.1) and (5.76) reduce to g = λ−2 du ⊗ du + ϕ−2 dv ⊗ dv,

 ∂ ∂ ∂ ∂ ∂ h , =J , h , = 0, ∂u ∂u ∂u ∂u ∂v


h

∂ ∂ , ∂v ∂v

(5.77)

 =J

∂ . ∂v

(5.78)

From (5.77) we have ∇∂/∂u

λu ∂ ϕ 2 λv ∂ ∂ =− + 3 , ∂u λ ∂u λ ∂v

∇∂/∂v

∂ λ2 ϕu ∂ ϕv ∂ = 3 − . ∂v ϕ ∂v ϕ ∂u

∇∂/∂u

λv ∂ ϕu ∂ ∂ =− − , ∂v λ ∂u ϕ ∂v (5.79)

By applying (5.77), (5.78), (5.79) and the formula of Gauss, we obtain Luu = (i − (ln λ)u )Lu + (ln ϕ)u Lv − λ12 L, Luv = −(ln λ)v Lu − (ln ϕ)u Lv , Lvv = (ln λ)v Lu + (i − (ln ϕ)v )Lv − ϕ12 L.

(5.80)

By applying (5.75) and (5.77), we find λω12 (e1 ) = ϕλv ,

ϕω21 (e2 ) = λϕu ,

Since K = 1, (5.77) and (5.81) imply that
 
ϕλv λϕu 1 + = . 2 2 λϕ λ ϕ u v

ϕ3 λv = λ3 ϕu .

(5.81)

(5.82)

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If λv = 0, we get ϕu = 0 from (5.81) which contradicts (5.82). Hence, we must have λv =  0. Similarly, we also have ϕu = 0. So, the last equation in (5.81) gives λϕu ϕλv = 2 = f (u, v) (5.83) λ2 ϕ for some non-zero function f. Also, f is non-constant according to (5.82) and (5.83). We divide Case (II.i) into two cases. Case (II.i.a). λ = ϕ = 0 on a neighborhood O1 of a point in W2,1 . In this case, (5.81) reduces to λu = λv . Thus, λ = ϕ is a function of s := u + v. So, (5.82) yields 2λ(s)λ (s) − 2λ2 (s) + 1 = 0. After solving this differential equation and applying a suitable translation in s, we obtain cosh bs λ= √ , 2b

(5.84)

where b is a non-zero real number. Hence, system (5.80) reduces to Luu = (i − b tanh bs)Lu + b tanh bsLv − 2b2 sech2 bsL, Luv = −b tanh bs(Lu + Lv ), Lvv = b tanh bsLu + (i − b tanh bs)Lv − 2b2 sech2 bsL.

(5.85)

If we put t = u − v as well as s = u + v, then (5.85) gives Ls = 21 (Lu + Lv ), Lst = 41 (Luu − Lvv ),

Lt = 21 (Lu − Lv ),

Lss = 41 (Luu + 2Luv + Lvv ),

Ltt = 41 (Luu − 2Luv + Lvv ).

Thus, (5.85) becomes
 i Lss = − b tanh bs Ls − b2 sech2 bsL, 2
 i Ltt = + b tanh bs Ls − b2 sech2 bsL. 2


Lst =

 i − b tanh bs Lt , 2

Solving the first two equations in (5.86) gives

  (1 + 4b2 ) cosh2 bs − 4b2 is/2 L = A(t)e sechbs + c1 −1 ei tan (2b tanh bs)

(5.86)

(5.87)

for some vector function A and vector c1 . Substituting (5.87) into the last equation of 2 )A(t) = 0. After solving this equation we find A(t) = (5.86) √ yields 4A (t) + (1 + 4b√ 2 c2 cos( 1 + 4b t/2) + c3 sin( 1 + 4b2 t/2) for some c2 , c3 ∈ C3 . Hence, we obtain  −i tan−1 (2b tanh bs) (1 + 4b2 ) cosh2 bs − 4b2 sechbs L = c1 e   (5.88) + eis/2 sechbs{c2 cos( 1 + 4b2 t/2) + c3 sin( 1 + 4b2 t/2)} Thus, we obtain Case (25) after choosing suitable initial conditions.

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Case (II.i.b). ϕ = 0 and λ = ϕ on a neighborhood O2 of a point in W2,1 . From (5.75) we know that e2 λ, e1 ϕ, ω12 (e2 ) are nonzero on O2 . By applying (5.77) we get g = ρ2 du ⊗ du + ψ2 dv ⊗ dv,

(5.89)

where ρ = 1/λ and ψ = 1/ϕ. Since we have ϕu , λv = 0 in Case (II.i), we find ρv , ψu = 0. By applying (5.82) and (5.83) we find

 ρv ρv ψu ψu + + ρψ = 0. (5.90) = = −f, ψ ρ ρ u ψ v The first equation in (5.90) implies ρuv = f 2 ρ − fu ψ. If ρ = ρ(v), we obtain fu = f 2 ρ/ψ = −fρρv which implies f (u, v) = H(v)e−uρρv and ψ = −ρv euρρv /H(v) for some function H(v). Thus, we have H 2 (v) = ρv2 e2uρρv by using ψu = −fρ which is impossible. Thus, we must also have ρu = 0. Hence, we obtain ρu , ρv , ψu = 0. √Next, assume that ψv = 0, i.e., ψ = ψ(u). Then the first equation in (5.90) gives ρ = ± 2ψ(u)ψ (u)v + 2h(u) for some function h = h(u). By substituting this into the second equation in (5.90) and by applying the first equation, we get 2ψ (ψψ v + h) + ψ((ψ2 + ψψ )v + h ) + ψ2 ψ + 4ψ(ψψ v + h)2 = 0. Thus, ψ is constant which is a contradiction. Hence, we get ρu , ρv , ψu , ψv = 0. From (5.89) and (5.90), we find ∇∂/∂u

ρu ∂ ψu ∂ ∂ = − , ∂u ρ ∂u ψ ∂v

∇∂/∂v

∂ ρv ∂ ψv ∂ =− + . ∂v ρ ∂u ψ ∂v

∇∂/∂u

ρv ∂ ψu ∂ ∂ = + , ∂v ρ ∂u ψ ∂v

Moreover, from (5.78), we have 

∂ ∂ ∂ ∂ ∂ , =J , h , = 0, h ∂u ∂u ∂u ∂u ∂v

(5.91)
h

∂ ∂ , ∂v ∂v

 =J

∂ . ∂v

By combining (5.78), (5.89), (5.91), (5.92) and the formula of Gauss we obtain
 ρu ψu ρv ψu Luv = Lu + Lu − Lv − ρ2 L, Lv , Luu = i + ρ ψ ρ ψ 
ρv ψv Lvv = − Lu + i + Lv − ψ2 L. ρ ψ

(5.92)

(5.93)

A direct computation shows that the compatibility conditions: Luuv = Luvu and Luvv = Lvvu hold if and only if (5.90) holds true. Thus, according to Proposition 1, the Lagrangian surface is locally given by Case (26). Case (II.ii). µ = 0 and λ = 2µ on a neighborhood V2,3 of a point p ∈ V2 . We divide this case into two cases: λ = µ or λ = µ.

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Case (II.ii.a). λ = µ. Let θ is a solution of λ(1 − 2 cos2 θ) = ϕ sin θ cos θ and put eˆ 1 = cos θe1 + sin θe2 , eˆ 2 = − sin θe1 + cos θe2 , then (5.1) yields ˆ eˆ 1 , h(ˆe1 , eˆ 1 ) = λJ

h(ˆe1 , eˆ 1 ) = 0,

ˆ eˆ 2 , h(ˆe2 , eˆ 2 ) = ϕJ

(5.94)

where λˆ = sin2 θ(2λ cos θ + ϕ sin θ) + λ cos θ, ϕˆ = cos2 θ(ϕ cos θ − 2λ sin θ) − λ sin θ. Hence, this case reduces to Case (I.ii.a) or Case (II.i) according to ∇e1 e1 = 0 or ∇e1 e1 = 0. Case (II.ii.b). λ = µ. The assumption ∇e1 e1 = 0 for Case (II) and the second equation in (5.2) imply e2 λ = 0. Since K = λµ − µ2 + 1 is nonzero, we get µej λ = (2µ − λ)ej µ,

j = 1, 2,

(5.95)

which implies e2 µ = 0 as well. Combining (5.2) with (5.95) gives e1 µ = ϕω12 (e1 ) + (λ − 2µ)ω12 (e2 ), e1 ϕ = 4µω21 (e1 ) + ϕω21 (e2 ), e2 (ln µ) = ω21 (e1 ).

(5.96)

By applying the last equation of (5.96) and Cartan’s structure equation, we find d(µ−1 ω1 ) = 0. Thus, there exists a function u such that du =

ω1 , µ

∂ = µe1 . ∂u

(5.97)

Due to K = λµ − µ2 + 1, the first two equations in (5.96) give 4µe1 µ + ϕe1 ϕ = (4K − 4µ2 − 4 − ϕ2 )ω12 (e2 ).

(5.98)

Case (II.ii.b.1). 4K = 4µ2 + ϕ2 + 4. In this case, we have K > µ2 + 1. So, we may assume  ϕ = 2 K − µ2 − 1. Thus, by K = λµ − µ2 + 1 and (5.96), we have  K − µ2 − 1 2 (5.99) ω1 (e2 ) − 2 K − µ2 − 1e2 (ln µ). e1 µ = µ Let Φ = Φ(u, v) be a solution of (ln Φ)u = 

e2 µ2 K − µ2 − 1

.

(5.100)

Then, by ∂/∂u = µe1 , (5.98), (5.100) and the last equation in (5.96), we obtain   ∂ µ2 e1 µ Φe2 Φ , = {(ln Φ)u + − µω12 (e2 )}e2 = 0. ∂u K − µ2 − 1 K − µ2 − 1 K − µ2 − 1  Hence, there exists a coordinate system {u, v} so that ∂/∂v = Φe2 / K − µ2 − 1. With respect to this coordinate system, we have g = µ2 du ⊗ du + ∂µ2 ∂Φ = = 0, ∂u ∂v

Φ2 dv ⊗ dv K − µ2 − 1

(5.101) (5.102)

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where the second equation is due to (5.100), (5.101) and e2 µ = 0. Also, by applying (5.101) and (5.102), we know that the Gauss curvature K satisfies

   
 −KµΦ µv K − µ 2 − 1 2(K − µ2 − 1)µv + Φµu  . + = Φ (K − µ2 − 1)3/2 K − µ2 − 1 u v

(5.103) From (5.101) we find µu ∂ (K − µ2 − 1)µµv ∂ ∂ = − , ∂u µ ∂u ∂v Φ2
 ∂ µv ∂ µµu Φu ∂ ∇∂/∂u = + , + ∂v µ ∂u Φ ∂v K − µ2 − 1

∇∂/∂u

(5.104)


 ∂ Φ2 µµu + (K − µ2 − 1)ΦΦu ∂ µµv Φv ∂ ∇∂/∂v =− + . + ∂v ∂u Φ ∂v µ2 (K − µ2 − 1)2 K − µ2 − 1

Thus, by (5.1), (5.101), (5.102) and the formula of Gauss, we obtain   µu (K − µ2 − 1)µµv Luu = i(K + µ2 − 1) + Lv − µ2 L, Lu − µ Φ2   µv µu 2µv Luv = Lu + µ iµ + Lv + µ Φ K − µ2 − 1   iΦ Φµu + 2(K − µ2 − 1)µv Lu Lvv = Φ − K − µ2 − 1 µ(K − µ2 − 1)2   µµv Φ2 Φv L + 2iΦ + − + L. v Φ K − µ2 − 1 K − µ2 − 1

(5.105)

A direct computation shows that the compatibility conditions: Luuv = Luvu and Luvv = Lvvu hold if and only if (5.102) and (5.103) hold true. Therefore, the Lagrangian surface is locally given by Case (27). Case (II.ii.b.2). 4K = 4µ2 + ϕ2 + 4. From (5.98) we get ω12 (e2 ) =

4µe1 µ + ϕe1 ϕ . 4(K − µ2 − 1) − ϕ2

(5.106)

Thus, by applying (5.95), (5.96) and (5.106), we find ω21 (e1 ) = e2 (ln µ), G= 

ω12 (e2 ) = e1 (ln G),

1 |4(K − µ2 − 1) − ϕ2 |

.

(5.107)

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which implies [µe1 , Ge2 ] = 0. Thus, there exists a coordinate system {u, v} with ∂/∂u = µe1 , ∂/∂v = Ge2 . With respect to this coordinate system, we have g = µ2 du ⊗ du +

dv ⊗ dv . |4(K − µ2 − 1) − ϕ2 |

(5.108)

If 4(K − µ2 − 1) > ϕ2 , then (5.108) implies ∂ ∂ ∂ = (ln µ)u − {4(K − µ2 − 1) − ϕ2 }µµv , ∂u ∂u ∂v ∂ ∂ ∂ 4µµu + ϕϕu ∇∂/∂u = (ln µ)v + , 2 2 ∂v ∂u 4(K − µ − 1) − ϕ ∂v ∂ 4µµu + ϕϕu ∂ ∂ 4µµv + ϕϕv ∇∂/∂v =− 2 + . ∂v µ (4(K − µ2 − 1) − ϕ2 )2 ∂u 4(K − µ2 − 1) − ϕ2 ∂v

∇∂/∂u

(5.109)

From (5.1), (5.3), (5.108), (5.109) and the formula of Gauss, we have µu }Lu − {4(K − µ2 − 1) − ϕ2 }µµv Lv , µ   µv 4µµu + ϕϕu 2 Luv = Lv Lu + iµ + µ 4(K − µ2 − 1) − ϕ2   i 4µµu + ϕϕu Lvv = Lu − 4(K − µ2 − 1) − ϕ2 µ2 (4(K − µ2 − 1) − ϕ2 )2   iϕ 4µµv + ϕϕv Lv +  + 4(K − µ2 − 1) − ϕ2 4(K − µ2 − 1) − ϕ2 Luu = {iK + iµ2 +

−

4(K

L . − 1) − ϕ2

− µ2

A straightforward long computation shows that the compatibility conditions: (Luu )v = (Luv )u and (Luv )v = (Lvv )u hold if and only if µ and ϕ satisfy µv =

Kϕu + ϕµµu − µ2 ϕu , µ(4(K − µ2 − 1) − ϕ2 )3/2




Gu µ

 u

+

µ v

G

v

= −KµG,

(5.110)

 where G = 1/ 4(K − µ2 − 1) − ϕ2 . From these we conclude that the Lagrangian surface is locally given by Case (28). If 4(K − µ2 − 1) < ϕ2 , (5.108) becomes g = µ2 du ⊗ du +

dv ⊗ dv , ϕ2 − 4(K − µ2 − 1)

(5.111)

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which yields ∂ ∂ µu ∂ = + {4(K − µ2 − 1) − ϕ2 }µµv , ∂u µ ∂u ∂v ∂ ∂ µv ∂ 4µµu + ϕϕu ∇∂/∂u = + , ∂v µ ∂u 4(K − µ2 − 1) − ϕ2 ∂v 4µµu + ϕϕu ∂ ∂ ∂ 4µµv + ϕϕv ∇∂/∂v = 2 + , ∂v µ (4(K − µ2 − 1) − ϕ2 )2 ∂u 4(K − µ2 − 1) − ϕ2 ∂v

∇∂/∂u

(5.112)

From (5.1), (5.3), and (5.112) we have   µu Luu = i(K + µ2 − 1) + Lu + (4(K − µ2 − 1) − ϕ2 )µµv Lv , µ   µv 4µµu + ϕϕu 2 Lv , Luv = Lu + iµ + µ 4(K − µ2 − 1) − ϕ2   i 4µµu + ϕϕu Lu Lvv = + 4(K − µ2 − 1) − ϕ2 µ2 (4(K − µ2 − 1) − ϕ2 )2   iϕ 4µµv + ϕϕv Lv +  + ϕ2 − 4(K − µ2 − 1) 4(K − µ2 − 1) − ϕ2 +

4(K

L . − 1) − ϕ2

− µ2

A straightforward computation shows that the compatibility conditions: (Luu )v = (Luv )u and (Luv )v = (Lvv )u hold if and only if µ and ϕ satisfy
 µ µ2 ϕu − Kϕu − ϕµµu Gu v µv = , + = −KµG, (5.113) µ u G v µ(ϕ2 − 4(K − µ2 − 1))3/2  where G = 1/ ϕ2 − 4(K − µ2 − 1). From these we conclude that the Lagrangian surface is locally given by Case (29). The converse can be verified by very long computations.

6. Some existence results

Proposition 1. Let ρ = ρ(u, v) and ψ = ψ(u, v) be real-valued functions with ρu , ρv , ψu , ψv = 0 defined on a simply-connected open subset U of R2 satisfying

 ρv ρv + + ρψ = 0. (6.1) ρρv = ψψu , ψ u ψ v

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B.-Y. Chen / Journal of Geometry and Physics 53 (2005) 428–460

Then Aρψ := (U, g0 ) with g0 = ρ2 du ⊗ du + ψ2 dv ⊗ dv is of constant curvature one. Moreover, up to rigid motions on CP 2 (4), there exists a unique Lagrangian isometric immersion αρψ : Aρψ → CP 2 (4) whose second fundamental form satisfies 


∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ , =J , h , = 0, h , =J . (6.2) h ∂u ∂u ∂u ∂u ∂v ∂v ∂v ∂v Proof. A direct computation shows that the Riemannian connection of Aρψ satisfies ∇∂/∂u

∂ ρu ∂ ψu ∂ = − , ∂u ρ ∂u ψ ∂v

∇∂/∂v

ρv ∂ ψv ∂ ∂ =− + . ∂v ρ ∂u ψ ∂v

∇∂/∂u

∂ ρv ∂ ψu ∂ = + , ∂v ρ ∂u ψ ∂v

and Aρψ is of curvature one. If we define a symmetric bilinear form σ on Aρψ by


 ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ σ , = , σ , = 0, σ , = , ∂u ∂u ∂u ∂u ∂v ∂v ∂v ∂v

(6.3)

(6.4)

then (6.1), (6.3), (6.4) and the definitions of g0 , ρ, ψ imply that σ(X, Y ), Z and (∇σ)(X, Y, Z) are totally symmetric. Moreover, a direct computation shows that the curvature tensor R and σ satisfy condition (iii) of Theorem A. So, according to Theorems A and B, up to rigid motions there exists a unique Lagrangian immersion αρψ : Aρψ → CP 2 (4) whose second fundamental form is given by (6.2).
Proposition 2. Let µ = µ(u, v) and Φ = Φ(u, v) be real-valued functions defined on a simply-connected open subset U of R2 satisfying ∂Φ ∂µ2 = = 0, ∂u ∂v

  
µv K − µ2 − 1 2(K − µ2 − 1)µv + Φµu −µΦK + = , Φ (K − µ2 − 1)3/2 K − µ2 − 1 u v

(6.5)

K := (U, g ) with where K is a real number ≥ µ2 . Then BµΦ 1

g2 = µ2 du ⊗ du +

Φ2 dv ⊗ dv K − µ2 − 1

is of constant curvature K. Moreover, up to rigid motions on CP 2 (4), there exists a unique K : BK → CP 2 (4) whose second fundamental form Lagrangian isometric immersion βµΦ µΦ satisfies

   ∂ ∂ ∂ ∂ ∂ ∂ 2 h , = K+µ J , h , = µ2 J , ∂u ∂u ∂u ∂u ∂v ∂v (6.6)

 ∂ ∂ ∂ Φ ∂ h , = + 2ΦJ . J ∂v ∂v ∂u ∂v K − µ2 − 1

B.-Y. Chen / Journal of Geometry and Physics 53 (2005) 428–460

459

K satisfies Proof. A direct computation shows that the Riemannian connection of PµΦ

(K − µ2 − 1)µµv ∂ ∂ ∂ = (ln µ)u − , ∂u ∂u ∂v Φ2 
∂ ∂ ∂ µµu ∇∂/∂u = (ln µ)v + , + (ln Φ) u ∂v ∂u ∂v K − µ2 − 1 ∇∂/∂u

Φ2 µµu + (K − µ2 − 1)ΦΦu ∂ ∂ ∇∂/∂v =− ∂v ∂u µ2 (K − µ2 − 1)2
 µµv ∂ + + (ln Φ)v ∂v K − µ2 − 1

(6.7)

K is the positive constant K. and the Gauss curvature of PµΦ K by If we define a symmetric bilinear form σ on PµΦ





 ∂ ∂ ∂ ∂ ∂ ∂ 2 σ , = (K + µ ) , σ , = µ2 , ∂u ∂u ∂u ∂u ∂v ∂v
 
∂ ∂ ∂ ∂ Φ σ , = + 2Φ , 2 ∂v ∂v ∂u ∂v K−µ

(6.8)

then it follows from (6.5), (6.7), (6.8) and the definition of g0 that σ(X, Y ), Z and (∇σ)(X, Y, Z) are totally symmetric. A direct computation shows that the curvature tensor R and σ satisfy condition (iii) of Theorem A. Thus, Theorems A and B imply that, up to K : BK → CP 2 (4) whose rigid motions, there exists a unique Lagrangian immersion βµΦ µΦ second fundamental form is given by (6.6).
Similarly, we have the following. Proposition 3. Let µ = µ(u, v) and ϕ = ϕ(u, v) be real-valued functions defined on a simply-connected open subset U of R2 satisfying
 µ Kϕu + ϕµµu − µ2 ϕu Gu v µv = =  0, + = −KµG, (6.9) µ u G v µ(4(K − µ2 − 1) − ϕ2 )3/2  K := where G = 1/ 4(K − µ2 − 1) − ϕ2 for a real number K > µ2 + 1 + ϕ2 /4. Then Cµϕ (U, g2 ) with g2 = µ2 du ⊗ du + G2 dv ⊗ dv is of constant curvature K. Moreover, up to K : C K → CP 2 (4) rigid motions, there exists a unique Lagrangian isometric immersion γµϕ µϕ whose second fundamental form satisfies 

∂ ∂ ∂ ∂ ∂ ∂ , = (K + µ2 + 1)J , h , = µ2 J , h ∂u ∂u ∂u ∂u ∂v ∂v
 (6.10) ∂ ∂ 1 1 ∂ ∂ h , = + J J . ∂v ∂v 4(K − µ2 − 1) − ϕ2 ∂u 4(K − µ2 − 1) − ϕ2 ∂v

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B.-Y. Chen / Journal of Geometry and Physics 53 (2005) 428–460

Proposition 4. Let µ = µ(u, v) and ϕ = ϕ(u, v) be real-valued functions defined on a simply-connected open subset U of R2 satisfying
 µ µ2 ϕu − Kϕu − ϕµµu Gu v µv = =  0, + = −KµG, (6.11) 2 2 3/2 µ u G v µ(ϕ − 4(K − µ − 1))  K := where G = 1/ ϕ2 − 4(K − µ2 − 1) for a real number K < µ2 + 1 + ϕ2 /4. Then Dµϕ (U, g3 ) with metric g3 = µ2 du ⊗ du + G2 dv ⊗ dv is of constant curvature K. Moreover, K up to rigid motions, there exists a unique Lagrangian isometric immersion δK µϕ : Dµϕ → 2 CP (4) whose second fundamental form satisfies

 ∂ ∂ ∂ ∂ ∂ ∂ 2 h , = (K + µ + 1)J , h , = µ2 J , ∂u ∂u ∂u ∂u ∂v ∂v
 (6.12) ∂ ∂ 1 1 ∂ ∂  h , = 2 + . J J ∂v ∂v ϕ − 4(K − µ2 − 1) ∂u ϕ2 − 4(K − µ2 − 1) ∂v References [1] B.Y. Chen, Geometry of Submanifolds, Marcel Dekker, New York, 1973. [2] B.Y. Chen, Representation of flat Lagrangian H-umbilical submanifolds in complex Euclidean spaces, Tohoku Math. J. 51 (1999) 13–20. [3] B.Y. Chen, Lagrangian surfaces of constant curvature in complex Euclidean plane, Tohoku Math. J. 56 (2004) 289–298. [4] B.Y. Chen, Maslovian Lagrangian surfaces of constant curvature in complex projective or complex hyperbolic planes, to appear. [5] B.Y. Chen, F. Dillen, L. Verstraelen, L. Vrancken, An exotic totally real minimal immersion of S 3 in CP 3 and its characterization, Proc. Roy. Soc. Edinburgh Sec. A 126 (1996) 153–165. [6] B.Y. Chen, F. Dillen, L. Verstraelen, L. Vrancken, Lagrangian isometric immersions of a real-space-form ˜ n (4c), Math. Proc. Cambridge Philos. Soc. 124 (1998) 107–125. M n (c) into a complex-space-form M [7] B.Y. Chen, O.J. Garay, Maslovian Lagrangian isometric immersions of real space forms into complex space forms, Jpn. J. Math. 30 (2004) 2. [8] B.Y. Chen, K. Ogiue, On totally real submanifolds, Trans. Am. Math. Soc. 193 (1974) 257–266. [9] B.Y. Chen, L. Vrancken, Lagrangian minimal isometric immersions of a Lorentzian real space form into a Lorentzian complex space form, Tohoku Math. J. 54 (2002) 121–143. [10] R. Courant, D. Hilbert, Methods of Mathematical Physics, Wiley–Interscience, New York–London, 1962. [11] N. Ejiri, Totally real minimal immersions of n-dimensional totally real space forms into n-dimensional complex space forms, Proc. Am. Math. Soc. 84 (1982) 243–246. [12] M. Kriele, L. Vrancken, Minimal Lagrangian submanifolds of Lorentzian complex space forms with constant sectional curvature, Arch. Math. 72 (1999) 223–232. [13] M. Kriele, L. Vrancken, Lorentzian affine hyperspheres with constant sectional curvature, Trans. Am. Math. Soc. 352 (2000) 1581–1599. [14] H. Reckziegel, Horizontal lifts of isometric immersions into the bundle space of a pseudo-Riemannian submersion, Global Differiantial Geometric and Global Analysis (1984), Lecture Notes Math. 12 (1985) 264–279. [15] L. Vrancken, Minimal Lagrangian submanifolds with constant sectional curvature in indefinite complex space forms, Proc. Am. Math. Soc. 130 (2002) 1459–1466.