A Classification Theorem for Complete PMC Surfaces with Non ...

TAIWANESE JOURNAL OF MATHEMATICS Vol. 20, No. 1, pp. 205–226, February 2016 DOI: 10.11650/tjm.20.2016.5766 This paper is available online at http://journal.tms.org.tw

A Classification Theorem for Complete PMC Surfaces with Non-negative Gaussian Curvature in M n (c) × R Zhong Hua Hou* and Wang Hua Qiu Abstract. Let M n (c) be an n-dimensional space form with constant sectional curvature c. Alencar-do Carmo-Tribuzy [5] classified all parallel mean curvature (abbrev. PMC) surfaces with non-negative Gaussian curvature K in M n (c) × R with c < 0. Later on, Fetcu-Rosenberg [28] generalized their results for c 6= 0. However, the classification to PMC surfaces in M n (c) × R with K ≡ 0 is still open. In this paper, we give a complete classification to the PMC surfaces in M n (c) × R with K ≡ 0 whose tangent plane spans the constant angle with factor R.

1. Introduction Let M n (c) be an n-dimensional space form with constant sectional curvature c 6= 0. In the past two decades, the submanifolds theory in product manifold M n (c) × R were widely studied. There have been lots of interesting and significant results (cf. [6,10–12,14,18,23– 27, 29, 30] etc). For instance, Dillen etc in [19, 22] characterized surfaces with a canonical principal direction and in [20,21] completely classified constant angle surfaces in M 2 (c)×R. Abresch and Rosenberg [1] introduced a quadratic form Q(X, Y ) = 2Hh(X, Y ) − c hX, ∂t i hY, ∂t i , on a surface Σ2 with constant mean curvature (abbrev. CMC) immersed M 2 (c)×R, where X, Y are tangent vectors on Σ2 and ∂t is the unit tangent vector to R. Denote by Q(2,0) the (2, 0)-part of Q and it is proved to be holomorphic. Then they completely classified CMC surfaces with vanishing Q(2,0) as four classes in M 2 (c) × R. The scholars call these kinds of surfaces Abresch-Rosenberg surfaces. Alencar, do Carmo and Tribuzy [4] extended the quadratic form Q to immersed surface 2 Σ with parallel mean curvature vector (abbrev. PMC) in M n (c) × R, which is defined by D → −E (1.1) Q(X, Y ) = 2 h(X, Y ), H − c hX, ∂t i hY, ∂t i . Received January 8, 2015, accepted July 27, 2015. Communicated by Sun-Yung Alice Chang. 2010 Mathematics Subject Classification. Primary: 53B25, 53C42. Key words and phrases. Non-negative Gaussian curvature, Parallel mean curvature, Product spaces. This work was supported in part by NSFC (No. 61473059). *Corresponding author.

205

206

Zhong Hua Hou and Wang Hua Qiu

And they concluded that Q(2,0) vanishes for surfaces of genus zero, if |dH| ≤ g Q(2,0) , where g is a continuous, non-negative real function and |dH| is the norm of the differential dH of the mean curvature H of Σ2 in M 2 (c)×R. We call the surfaces with Q(2,0) vanishing Abresch-Rosenberg type surfaces. Batista [9] introduced a (1, 1)-tensor S on a CMC surface in M 2 (c) × R, which is given by c S = 2HA − c hT, ·i T + |T |2 I − 2H 2 I, 2 where A is the shape operator, I the identity transform and T the tangent part to the surface of the unit vertical vector field ∂t of R. He proved that a complete surface in M 2 (c) × R with S = 0 is an Abresch-Rosenberg surface. Fetcu and Rosebberg [28] defined a more general (1, 1)-tensor S on immersed PMC surface Σ2 in M n (c) × R, say X c (1.2) S=2 H α Aα − c hT, ·i T + |T |2 I − 2H 2 I. 2 α They showed that |S| = 0 if and only if Q(2,0) = 0 (cf. Lemma 3.2 below). Therefore the surface with |S| = 0 is an Abresch-Rosenberg type surface. Alencar, do Carmo and Tribuzy [5] obtained a well-known Hopf theorem and classified the complete PMC surfaces immersed in M n (c) × R. Theorem 1.1. (cf. [5]) Let x : Σ2 → M n (c) × R with c 6= 0 be a complete PMC surface. Then one of the following holds: (1) Σ2 is a minimal surface of a totally umbilical hypersurface of M n (c); (2) Σ2 is a CMC surface in a 3-dimensional totally umbilical or totally geodesic submanifold of M n (c); (3) Σ2 lies in M 4 (c) × R. In [5], the following classification theorem is proved for c < 0, and is generalized by Fetcu and Rosebberg [28] for c 6= 0. Theorem 1.2. (cf. [5,28]) Let x : Σ2 → M n (c) × R with c 6= 0 be a complete non-minimal PMC surface with Gaussian curvature K ≥ 0. Then one of the following holds: (1) K ≡ 0; (2) Σ2 is a minimal surface of a totally umbilical hypersurface of M n (c); (3) Σ2 is a CMC surface in a totally umbilical 3-dimensional submanifold of M n (c); (4) Σ2 lies in M 4 (c) × R ⊂ R6 and there exists a plane P such that the level lines of the height function p 7→ hx(p), ∂t i are curves lying in planes parallel to P .

A Classification Theorem for Complete PMC Surfaces in M n (c) × R

207

Then the authors in [5] proposed an interesting and open problem: How to characterize those surfaces with K ≡ 0 in the above theorem. It seems to be difficult even for the case of complete surfaces immersed in M 2 (c) × R. In the present paper, we intend to solve this problem under the additional condition that |T | is constant. Precisely, we proceed to prove the following Main Theorem 1.3. Let Σ2 be a complete non-minimal PMC surface with non-negative Gaussian curvature in M n (c) × R, c 6= 0. Then one of the following holds: (1) |S| ≡ 0, the surface is an Abresch-Rosenberg type surface; (2) |S| is a nonzero constant and K ≡ 0. In addition, if |T | = sin θ with constant θ ∈ [0, π/2], then (i) θ = 0, Σ2 lies in M n (c); or (ii) θ = π/2, Σ2 = γ × R, where γ is a curve of M n (c); or (iii) θ ∈ (0, π/2) and Σ2 lies in M 4 (c) × R ⊂ R6 with c > 0. Up to an isometry of M 4 (c) × R, Σ2 is parameterized by cos θ sin(av) cos(av) 2H cos θ cos(bu), sin(bu), , , , u sin θ , (1.3) x(u, v) = b b a a ab √ √ where b = c + c cos2 θ, a = b2 + 4H 2 and H is the mean curvature of Σ2 .

2. Preliminaries n+1

Let Σ2 be a surface in an (n + 1)-dimensional Riemannian manifold M . Choose a local n+1 2 orthonormal frame field {e1 , e2 , . . . , en+1 } in T M along Σ so that {e1 , e2 } are tangent 2 2 to Σ and the others are normal to Σ . Denote the dual frame by ω 1 , ω 2 , . . . , ω n+1 . n+1 ). We use the following Let ∇ (resp. ∇) be the Riemannian connection of Σ2 (resp. M convention on index ranges in the whole paper: 1 ≤ i, j, k, l ≤ 2;

3 ≤ α, β, γ ≤ n + 1;

1 ≤ A, B, C ≤ n + 1.

A Let ωB be the connection form. The second fundamental form h and the mean curvature → − vector H are defined by X X X → − (2.1) h= ω i ⊗ ωiα ⊗ eα = hαij ω i ⊗ ω j ⊗ eα , H = H α eα , α,i

α

α,i,j

where ! (2.2)

ωiα =

X j

hαij ω j ,

hαij = hαji ,

Hα =

X i

hαii

/2.

208


The mean curvature H is defined to be " #1/2 → X − H = H = (H α )2 .

(2.3)

α

The Gauss-Ricci equations are expressed as (2.4)

i

i Rjkl = Rjkl +

X

(hαik hαjl − hαil hαjk ),

α

α Rβkl = Rβkl +

X (hαik hβil − hαil hβik ).

α

i

We define the covariant differential of h by ∇hαij =

(2.5)

X

hαijk ω k = dhαij −

X

k

hαik ωjk −

X

k

hαkj ωik +

X

hγij ωγα .

γ

k

Then the Codazzi equation is α

hαijk − hαikj = Rikj .

(2.6) n+1

Let M = M n (c)×R with c 6= 0. Naturally, the ambient space M n (c)×R is endowed with the metric ds2M n (c)×R = ds2M n (c) + dt2 . The induced metric on Σ2 is denoted by h , i. Let ∂t = T + N , where T is the tangent part and N the normal part of ∂t . Then (2.7)

T =

X

T i ei ,

T i = hei , ∂t i ;

N=

X

N α eα ,

N α = heα , ∂t i .

α

i n+1

For any X ∈ Γ(T M ), we denote X|M n (c) = X − hX, ∂t i ∂t , which is the projection of n X onto the factor M (c). Then ei |M n (c) = ei − T i ∂t ,

eα |M n (c) = eα − N α ∂t .

It follows that A

RBCD = c

eA |M n (c) , eC |M n (c) eB |M n (c) , eD |M n (c)

− eA |M n (c) , eD |M n (c) eB |M n (c) , eC |M n (c) ,

from which we obtain n o i (2.8) Rjkl = c δik δjl − δil δjk + δil T j T k + δjk T i T l − δik T j T l − δjl T i T k , (2.9)

α

Rikj = cN α (T j δik − T k δij ),

α

Rβkl = 0.


209

3. Some lemmas In this section, we introduce several lemmas needed for the proof of the Main Theorem. The fundamental formulae of submanifolds are given by (3.1)

∇ei ej = ∇ei ej + h(ei , ej ),

(3.2)

∇ei eα = −Aα (ei ) + ∇⊥ ei eα ,

where Aα is the shape operator with respect to eα , and (3.3)

β ∇⊥ ei eα = ωα (ei )eβ ,

∇ei ej = ωjk (ei )ek ,

Aα (ei ) = hαij ej ,

h(ei , ej ) = hαij eα .

Lemma 3.1. (cf. [14, 33]) Let Σ2 be a surface in M n (c) × R. Then we have ∇T i =

(3.4)

X

N α hαij ω j ,

α,j ⊥

α

∇ N =−

(3.5)

X

T i hαij ω j .

i,j

Let Σ2 be a PMC surface in M n (c) × R. According to (1.2), the coefficients {Sij } of S are given by (3.6)

Sij = 2

X

H α hαij − cT i T j +

α

c |T |2 δij − 2H 2 δij , 2

for any i, j. The covariant derivatives {Sijk } of {Sij } are defined by X

(3.7)

Sijk ω k = dSij −

k

X

Sik ωjk −

k

X

Skj ωik .

k

→ − It is known that ∇⊥ H = ∇⊥ H α eα , where α k ∇⊥ H α = dH α + H β ωβα = H,k ω .

(3.8)

→ − α = 0, for all k and α. Taking the covariant derivatives of both Then ∇⊥ H = 0 implies H,k sides of equation (3.6) and using (3.4), we get c H α hαijk − cT i ∇ek T j − cT j ∇ek T i + δij ∇ek (|T |2 ) 2 α X X X X N α hαjk − cT j N α hαik + cδij N α T l hαlk . =2 H α hαijk − cT i

Sijk = 2 (3.9)

X

α

α

α

α,l

Lemma 3.2. (cf. [9, 28]) Let Σ2 be a complete PMC surface immersed M n (c) × R. Then |S| = 0 if and only if Q(2,0) = 0.

210


Using the results in [39], the authors in [28] obtained the following Simons type equation for |S|2 : Theorem 3.3. (cf. [28]) Let Σ2 be a PMC surface immersed M n (c) × R. Then 1 ∆(|S|2 ) = |∇S|2 + 2K |S|2 , 2

(3.10) where |∇S|2 =

P

i,j,k

2 and K is the Gaussian curvature of Σ2 . Sijk

At the end of this section, we prove the following lemma: Lemma 3.4. Let Σ2 be a complete PMC surface in M n (c) × R. Let {e1 , e2 } be the local orthonormal tangent frame field on Σ2 such that {Sij } is diagonalized. If |∇S| = 0, then |S| = 0 or |S| is nonzero constant and (ωji ) = 0. Proof. From (3.6), it is easy to see that S is symmetric and trace-free. Choose the orthonormal tangent frame filed {e1 , e2 } on Σ2 such that Sij = µi δij . Then µ1 = −µ2 = µ. So |S|2 = 2µ2 . |∇S| = 0 implies Sijk = 0, for any i, j, k. Setting i = j = 1 and i = 1, j = 2 in (3.7), we obtain 0=

X

0=

X

S11k ω k = dµ −

X

k

k

S1k ω1k −

X

k k

S12k ω = −

X k

S1k ω2k

Sk1 ω1k = dµ,

k

−

X

Sk2 ω1k = 2µω12 ,

k

which imply µ = 0 or µ is nonzero constant and ω12 = 0. The proof of Lemma 3.4 is completed.

4. A classification theorem In this section, we firstly prove a classification theorem. Then, we solve the problem proposed in [5] under the condition that |T | is constant. Theorem 4.1. Let x : Σ2 → M n (c) × R be a complete PMC surface with Gaussian curvature K ≥ 0. Then |S| = 0 or |S| is nonzero constant and K = 0. Proof. According to (3.10) and the hypothesis, it follows that ∆ |S|2 ≥ 0. By a result of Huber [34], a complete surface with non-negative Gaussian curvature is a parabolic space. Therefore, |S|2 is harmonic. One can immediately get |∇S| = 0 by (3.10) again. Note that dω21 = Kω 1 ∧ ω 2 . Following Lemma 3.4, we complete the proof.


211

It is known that the surface with |S| = 0 is an Abresch-Rosenberg type surface. In the sequel, we proceed to consider the rest case. Denote |T | = sin θ with θ ∈ [0, π/2]. It is obvious that Σ2 lies in M n (c) when θ = 0 and Σ2 = γ × R in case θ = π/2, where γ is a curve in M n (c). So we need only to treat the problem for θ ∈ (0, π/2). Lemma 4.2. Let x : Σ2 → M n (c) × R with c 6= 0 be a complete non-minimal PMC surface with non-negative Gaussian curvature so that |S| is nonzero constant. If θ ∈ (0, π/2) is constant, then Σ2 lies in M 4 (c) × R and the mean curvature vector is orthogonal to ∂t . Proof. According to Theorem 1.1 and the fact that θ is a non-zero constant, it follows that the surface Σ2 lies in M 4 (c) × R. By Lemma 3.4, |S| is nonzero constant and ωji = 0,

(4.1)

under the chosen tangent frame {eo 1 , e2 }. We choose a local orthonormal normal frame n→ − → − field {e3 , e4 , e5 } so that span H , N = span {e3 , e4 }. Then H and N can be decomposed as → − H = H 3 e3 + H 4 e4 , N = N 3 e3 + N 4 e4 . Using (3.4), we get ∇ej |T |2 = 2

(4.2)

X

N α T i hαij = 0,

α,i

for j ∈ {1, 2}. From (4.2), we have a linear system of linear equations on T 1 , T 2 , ! X

N α hα1j

! 1

T +

X

α

N α hα2j

T 2 = 0,

α

for j ∈ {1, 2}, which has non-zero solutions. Hence we obtain ! X

N α hα11

α

!2

! X

N α hα22

−

α

X

N α hα12

= 0,

α

which is equivalent to (4.3)

(N 3 )2 det(A3 ) + (N 4 )2 det(A4 ) + N 3 N 4 (h311 h422 + h411 h322 − 2h312 h412 ) = 0.

Following (4.2) again, we get another system of linear equations on N 3 , N 4 , ! X i

T i h3ik

  X N3 +  T j h4jk  N 4 = 0, j

212


for j ∈ {1, 2}, which has also non-zero solutions. And we also get X (4.4) T i T j (h3i1 h4j2 − h4i1 h3j2 ) = 0. i,j

→ − Suppose that ∇⊥ H = 0. From the first equality in (3.8) we have, for every α, dH α + H β ωβα = 0.

(4.5)

Taking the exterior derivative on both-sides of (4.5) and using the structure equation, we obtain X X α H β (dωβα − ωβγ ∧ ωγα ) = H β Rβij ωi ∧ ωj = 0. β,i

β,i,j

It follows from (2.4), (2.9) and the third equality in (3.3) that   h i X α  → , Aα = A− → Aα − Aα A− → = (4.6) A− H β Rβij H H H β

= 0,

2×2

for any α. Setting α = 3, 4 respectively, one has H 3 (A3 A4 − A4 A3 ) = H 4 (A3 A4 − A4 A3 ) = 0. Since (H 3 )2 + (H 4 )2 > 0, we obtain (4.7)

A3 A4 = A4 A3 .

According (4.7), one can obtain h312 (h411 − h422 ) = h412 (h311 − h322 ),

(4.8) which is equivalent with

h311 h412 − h411 h312 = h322 h412 − h422 h312 .

(4.9)

Define four normal vectors in the normal space span {e3 , e4 } as follows: A=

4 X α=3

hα11 eα ,

B=

4 X α=3

hα22 eα ,

C=

4 X

hα12 eα ,

α=3

D =A−B =

4 X

(hα11 − hα22 )eα .

α=3

→ − → −E In order to prove H ⊥∂t , it suffices to prove N, H = 0. Now, we divide our proof in two cases. Case 1. D is nonzero. It is clear that C is parallel to D by (4.8). So we have C = λD for some function λ on Σ2 , which implies D

(4.10)

hα12 = λ(hα11 − hα22 ),


213

for α ∈ {3, 4}. Substituting (4.9) and (4.10) into (4.4), we have 0 = (T 1 )2 − (T 2 )2 (h311 h412 − h411 h312 ) + T 1 T 2 (h311 h422 − h411 h322 ) (4.11) = λ (T 1 )2 − (T 2 )2 (h411 h322 − h311 h422 ) + T 1 T 2 (h311 h422 − h411 h322 ) = λ (T 1 )2 − (T 2 )2 − T 1 T 2 (h411 h322 − h311 h422 ), which implies λ (T 1 )2 − (T 2 )2 = T 1 T 2 or h411 h322 = h311 h422 . We discuss it in two subcases. Subcase 1. λ (T 1 )2 − (T 2 )2 = T 1 T 2 . If T 1 T 2 6= 0, then (T 1 )2 6= (T 2 )2 and λ 6= 0. From (3.6) and (4.10), we have X X H α λ(hα11 − hα22 ) − cT 1 T 2 S12 = 2 H α hα12 − cT 1 T 2 = 2 α

α

! = 4λ

X

H α hα11 − H 2

− cT 1 T 2 = 0,

α

from which we obtain µ = S11 = 2

X

α 1 2 cT T

H α hα11 − c(T 1 )2 +

c |T |2 − 2H 2 2

c 2 2 (T ) − (T 1 )2 − 2H 2 4λ 2 2 2 c 1 2 = T T + λ (T ) − (T 1 )2 = 0, 2λ

(4.12)

=2

+ H2

+

that contradicts to the assumption |S| 6= 0. Therefore, T 1 T 2 = 0. Without loss of generality, we assume T 2 = 0 and T 1 = sin θ. It follows from (4.1) that X ∇T i = dT i + T k ωki = T 1 ω1i = 0, k

for any i. Using (3.4), we obtain ∇ej T i =

(4.13)

X

N α hαij = 0,

α

D → −E P for any i, j. Setting i = j and taking summation, we have N, H = α N α H α = 0. Subcase 2. h411 h322 = h311 h422 . In this case, A is parallel to B. Owing to the non-minimal property, A and B do not vanish simultaneously. Without loss of generality, we assume that hα11 = νhα22 for α ∈ {3, 4}, where ν 6= −1 is a function on Σ2 since H 6= 0. Using (4.10), we obtain 2ν 2 2λ(ν − 1) α H α , hα22 = H α , hα12 = H , ν+1 ν+1 ν+1 which is equivalent with Aα = H α A0 , where   ν λ(ν − 1) 2  . (4.15) A0 = ν + 1 λ(ν − 1) 1 (4.14)

hα11 =

214


By (4.14) and (4.15), we have det(Aα ) = (H α )2 det(A0 ),

(4.16)

det(A0 ) =

4 ν − λ2 (ν − 1)2 . 2 (ν + 1)

Substituting (4.14) and (4.16) into (4.3), we get D → − E2 0 = (N 3 H 3 + N 4 H 4 )2 det(A0 ) = N, H det(A0 ).

(4.17)

D → −E It follows that N, H = 0 or det(A0 ) = 0. In latter case, we have ν = λ2 (ν − 1)2 . By applying (3.6) and (4.14), we obtain X α α 1 2 2 ν−1 (4.18) S12 = 2 H h12 − cT T = 4λH − cT 1 T 2 = 0. ν + 1 α Moreover, we have µ = S11 = 2H

2

ν−1 ν+1

− c(T 1 )2 +

c |T |2 2

which implies (4.19)

c c(T ) = |T |2 + 2H 2 2 1 2

ν−1 ν+1

− µ,

c c(T ) = |T |2 − 2H 2 2 2 2

ν−1 ν+1

+ µ.

Using (4.18) and (4.19), we obtain via a straightforward calculation 2 16νH 4 16λ2 (ν − 1)2 H 4 c2 2(ν − 1)H 2 4 2 1 2 2 2 = = c (T ) (T ) = |T | − µ − , (ν + 1)2 (ν + 1)2 4 ν+1 where we used ν = λ2 (ν − 1)2 in the first equality. Simplifying the above equation, we find (4.20)

ν−1 c2 |T |4 − 16H 4 − 4µ2 = , ν+1 16µH 2

which is constant. Therefore, both T 1 and T 2 are constant by (4.19) and By D (4.20). → −E similar argument as in Subcase 1, we obtain (4.13), from which we obtain N, H = 0. Case 2. D is a zero vector. → − In this case we have A = B = H . Applying (4.4), we have (4.21)

0 = (T 1 )2 (H 3 h412 − H 4 h312 ) + (T 2 )2 (h321 H 4 − h421 H 3 ) + T 1 T 2 (H 3 H 4 − H 4 H 3 ) = (T 1 )2 − (T 2 )2 (H 3 h412 − H 4 h312 ),

from which we get (T 1 )2 = (T 2 )2 or H 3 h412 = HD4 h312 . When (T 1 )2 = (T 2 )2 , we can see E → − that T 1 and T 2 are constant, which implies that N, H = 0.


215

→ − → − Suppose that H 3 h412 = H 4 h312 . Then C is parallel to H . Set C = ρ H . Then hα12 = ρH α for α ∈ {3, 4}. It follows that det(A3 ) = (H 3 )2 (1 − ρ2 ) and det(A4 ) = (H 4 )2 (1 − ρ2 ). According to (4.3), we find

(4.22)

0 = (N 3 H 3 )2 (1 − ρ2 ) + (N 4 H 4 )2 (1 − ρ2 ) + 2N 3 N 4 H 3 H 4 (1 − ρ2 ) D → − E2 = (1 − ρ2 ) N, H ,

D → −E from which we have N, H = 0 or ρ2 = 1. In case ρ2 = 1, we have from (3.6) that 0 = S12 = 2

X

H α hα12 − cT 1 T 2 = 2ρH 2 − cT 1 T 2 ,

α

which implies (4.23)

(cT 1 T 2 )2 = (2ρH 2 )2 = 4H 4 .

According (4.23) and the fact (T 1 )2 + (T 2 )2 = sin2 θ, it follows that T 1 and T 2 are D that → −E constant, which means that N, H = 0. → − Summarizing the above cases, we claim that H is orthogonal to N . Our statement is proved. Then we present an example that satisfies all the conditions described in Lemma 4.2. Example 4.3. Define the map x : R2 → R6 as follows: x(u, v) = (x1 , . . . , x6 ) (4.24) cos θ cos θ sin(av) cos(av) 2H = cos(bu), sin(bu), , , , u sin θ , b b a a ab √ √ where b = c + c cos2 θ, a = b2 + 4H 2 and H, c > 0, θ are constant. Let Σ2 = x(R2 ). Then Σ2 is a surface described as in Lemma 4.2. P Firstly, 5i=1 (xi )2 = 1/c, thus Σ2 lies in M 4 (c) × R. Let ∂t = (0, . . . , 0, 1) and (4.25)

∂u = xu = (− cos θ sin(bu), cos θ cos(bu), 0, 0, 0, sin θ) ,

(4.26)

∂v = xv = (0, 0, cos(av), − sin(av), 0, 0) .

Then we have (4.27)

(4.28)

|xu |2 = |xu |2 = 1,

hxu , xv i = 0,

ωji = 0,

T = h∂t , xu i xu + h∂t , xv i xv = sin θxu ,

K = 0,

|T |2 = sin2 θ.

216


Let us consider M 4 (c) × R as a hypersurface in R6 . Its normal vector e6 = (x1 , . . . , x5 , 0). √ Denote its unit normal vector by e6 = ce6 . Let D be Euclidean connection and h be the second fundamental form of M 4 (c) × R in R6 . Then (4.29)

h(X, Y ) = −c X|M 4 (c) , Y |M 4 (c) e6 = c(hX, ∂t i hY, ∂t i − hX, Y i e6 ,

for any X, Y ∈ X (M 4 (c) × R). From (3.1), (4.29) and the third fact in (4.27), we have

h(X, Y ) = DX Y − ∇X Y − h(X, Y ) = DX Y + c X|M 4 (c) , Y |M 4 (c) e6 ,

(4.30)

for any tangent vectors X, Y ∈ X (Σ2 ). Using (4.25), (4.26) and (4.30), we obtain h(∂u , ∂u ) = xuu + c cos2 θ · e6

(4.31)

= (−b cos θ cos(bu), −b cos θ sin(bu), 0, 0, 0, 0) cos θ cos θ sin(av) cos(av) 2H 2 + c cos θ cos(bu), sin(bu), , , ,0 b b a a ab cos(bu) sin(bu) cos θ sin(av) cos θ cos(av) 2H cos θ = c cos θ − ,− , , , ,0 , b b a a ab

h(∂u , ∂v ) = xuv + c ∂u |M 4 (c) , ∂v |M 4 (c) e6 = 0, h(∂v , ∂v ) = xvv + ce6 = (0, 0, −a sin(av), −a cos(av), 0, 0) cos θ cos θ sin(av) cos(av) 2H +c cos(bu), sin(bu), , , ,0 b b a a ab 2 c cos θ c cos θ c−a c − a2 2cH = cos(bu), sin(bu), sin(av), cos(av), ,0 . b b a a ab

Thus (4.32)

→ − 1 H H = [h(∂u , ∂u ) + h(∂v , ∂v )] = (0, 0, −2H sin(av), −2H cos(av), b, 0) , 2 a

and (4.33)

D→ E D→ E − − H , h(∂u , ∂u ) = H , h(∂u , ∂v ) = 0,

D→ E − H , h(∂v , ∂v ) = 2H 2 .

From (3.6) and (4.33), we get c hS∂u , ∂u i = − sin2 θ − 2H 2 = − hS∂v , ∂v i , 2

hS∂u , ∂v i = 0,

which imply (4.34)

|S|2 = 2

c 2

sin2 θ + 2H 2

2

,

which is a nonzero constant. From (4.25), (4.26) and (4.32), we have D→ E D→ E D→ E − − − H , ∂t = H , ∂u = H , ∂v = 0.


Using (4.29), we get

217

→ − → − h(∂u , H ) = h(∂v , H ) = 0,

from which, together with (3.1), (3.2) and (4.33), we obtain → − → − → − → − → ∂u − h(∂u , H ) = ( H )u + 0 + 0 = 0, ∇⊥ ∂u H = D ∂u H + A− H (4.35) → − → − → − → ∂v − h(∂v , H ) = −2H 2 ∂v + 2H 2 ∂v + 0 = 0, ∇⊥ ∂v H = D∂v H + A− H → − which imply ∇⊥ H = 0. Therefore, Σ2 defined by (4.24) satisfies all conditions described in Lemma 4.2. In the sequel, we proceed to characterize the surfaces in Lemma 4.2 and prove that the kind of surfaces in Example 4.3 is the only class, up to an isometry of M 4 (c) × R. Lemma 4.4. Under the same conditions as in Lemma 4.2, Σ2 lies in M 4 (c) × R with n o → − → − c > 0 and H ⊥N . With respect to the normal frame field e3 = H /H, e4 = N/ |N | , e5 , A ) are represented as the shape operators Aα and the connection form matrix (ωB     √ c cos θ 0 0 0 ,  , A4 = 0, A5 =  (4.36) A3 =  √ 0 − c cos θ 0 2H

(4.37)



0

0

0

0

    A (ωB ) =    

0

0

2Hω 2

0

0

−2Hω 2

0

0

0 √

0

− c cos θ ω

1

√

0

c cos θ ω

2

0

0 √

c sin θ ω 1

√



c cos θ ω 1  √ − c cos θ ω 2  

 .  √ − c sin θ ω 1   0

0

→ − Proof. Owing to ∇⊥ H = 0, we have 0 = dH α +

X

H β ωβα = Hω3α ,

β

which implies ω3α = ω4α = 0, for α ∈ {3, 4}. Thus by (3.5), we have X X T i hαij ω j = −∇⊥ N α = −dN α − N β ωβα = − |N | ω4α = 0, i,j

β

which is equivalent with (4.38)

T 1 hα11 + T 2 hα12 = 0

and

− T 2 hα11 + T 1 hα12 = −2H α T 2 ,

for α ∈ {3, 4}. By Cramer’s Rule, we obtain   2 )2 1T 2 (T −T 2H  , (4.39) A3 = |T |2 −T 1 T 2 (T 1 )2

218


and A4 = 0 which is the desired form. From (3.6) and (4.39), we get (4.40) (4.41)

µ = S11

4H 2 c 1 2 2 2 2 1 2 c+ , = − c(T ) + |T | − 2H = (T ) − (T ) 2 2 |T |2 4H 2 3 1 2 1 2 0 = S12 = 2Hh12 − cT T = −T T c + , |T |2 2Hh311

1 2

which imply T 1 T 2 = 0. Without loss of generality, we assume T 2 = 0 and T 1 = |T |. Then, A3 becomes the desired form. From (4.6), we get A3 A5 = A5 A3 , which implies h512 = 0. It follows that X (4.42) 0 = K = c cos2 θ + det(Aα ) = c cos2 θ + det(A5 ) = c cos2 θ − (h511 )2 , α

√ which implies that c > 0. We can set h511 = c cos θ (one can change the direction of e5 if necessarily). Consequently, A5 has the desired form. At last, by (3.5), we obtain X cos θ ω45 = dN 5 + N α ωα5 = ∇⊥ N 5 (4.43)

α

=−

X

T i h5ij ω j

√ = −T 1 h511 ω 1 = − c sin θ cos θ ω 1 ,

i,j

√ which implies ω45 = − c sin θ ω 1 . To summarize what we have proven, and applying P α j ωiα = j hij ω , we can see that the connection form matrix is given by (4.37). This completes the proof of Lemma 4.4. In Lemma 4.4, we obtain the connection form of the surface and, following the existence and uniqueness theorem of submanifolds, the surface is unique (up to an isometry of M 4 (c) × R) for a fixed θ, and given by Example 4.3. Next, we will give a way of looking for the surface of Example 4.3, which is similar to that in [20, 21]. Theorem 4.5. Let x : Σ2 → M n (c) × R with c 6= 0 be a complete non-minimal PMC surface with non-negative Gaussian curvature and |S| 6= 0. If θ ∈ (0, π/2) is constant, then Σ2 lies in M 4 (c) × R with c > 0. Up to an isometry of M 4 (c) × R, the immersion is given by (4.24). Proof. By Lemma 4.4, Σ2 lies in M 4 (c) × R with c > 0. Let x(u, v) = (x1 , . . . , x6 ) be the P position vector satisfying 5i=1 (xi )2 = 1/c. We choose the same frame {e1 , . . . , e5 } as in Lemma 4.4. Thus, one can take coordinates (u, v) on Σ2 with ∂u = e1 , ∂v = e2 . Then we have → − T = sin θ ∂u , H = He3 , N = cos θ e4 . Regard M 4 (c) × R as a hypersurface in R6 . The normal vector of M 4 (c) × R in R6 is √ e6 = (x1 , . . . , x5 , 0). Let e6 = ce6 and denote eα = (ξ1α , . . . , ξ6α ),


219

for α ∈ {3, 4, 5}. Then {eα }6α=3 forms a normal frame of Σ2 in R6 . We denote by D (resp. D⊥ ) the Euclidean connection (resp. the normal connection). Then, for any X ∈ X (Σ2 ), we have (4.44) 5 X ⊥ DX e6 = X|M 4 (c) = X − hX, ∂t i ∂t , DX e6 = hDX e6 , eα i eα = − cos θ hX, T i e4 , α=3

from which one gets (4.45)

⊥ Ae6 X = −DX e6 + DX e6 = −X M 4 (c) − cos θ hX, T i e4 = −X + hX, T i T.

By applying (4.45), we obtain − cos2 θ ∂u = −∂u + h∂u , T i T = Ae6 (∂u ) = −∂u M 4 (c) − cos θ sin θ e4 , −∂v = −∂v + h∂v , T i T = Ae6 (∂v ) = −∂v M 4 (c) . It follows that e4 =

1 −∂u M 4 (c) + cos2 θ ∂u , sin θ cos θ

∂v = ∂v M 4 (c) ,

from which we have, by rewriting them into components of e4 , (4.46)

(xj )u = −ξj4 cot θ,

(4.47)

(x6 )u = ξ64 tan θ = tan θ he4 , ∂t i = sin θ,

1 ≤ j ≤ 5, (x6 )v = 0.

From (4.47), we can take x6 = u sin θ. Using (4.29), we have (4.48)

h(X, eα ) = c hX, T i heα , N i e6 = cδ4α cos θ hX, T i e6 ,

for any α ∈ {3, 4, 5} and X ∈ X (Σ2 ). By the fundamental formulae of submanifolds, we get DX eα = −Aα X + ∇⊥ X eα + h(X, eα ) (4.49)

= −Aα X +

5 X

ωαβ (X)eβ + cδ4α cos θ hX, T i e6 ,

β=3

from which we obtain D∂u e3 = −A3 (∂u ) = 0, (4.50)

D∂v e3 = −A3 (∂v ) = −2H∂v ,

D∂u e4 = ω45 (∂u ) e5 + c cos θ h∂u , T i e6 √ = − c sin θ e5 + c sin θ cos θ e6 , D∂v e4 = 0, √ √ D∂u e5 = −A5 (∂u ) + ω54 (∂u ) e4 = − c cos θ ∂u + c sin θ e4 , √ D∂v e5 = −A5 (∂v ) + ω54 (∂v ) e4 = c cos θ ∂v .

220


By rewriting (4.50) into components of eα ’s, we have 3 (ξA )u = 0,

(4.51)

3 (ξA )v = −2H(xA )v ,

(4.53)

√ (ξj4 )u = − c sin θ ξj5 + c sin θ cos θ xj , √ (ξ64 )u = − c sin θ ξ65 ,

(4.54)

4 (ξA )v = 0,

(4.52)

√ √ 5 4 (ξA )u = − c cos θ (xA )u + c sin θ ξA , √ 5 (ξA )v = c cos θ (xA )v ,

(4.55) (4.56)

where 1 ≤ A ≤ 6 and 1 ≤ j ≤ 5. Substituting (4.46) into (4.55), we have √ √ √ c 4 4 4 5 ξ . (4.57) (ξj )u = c cos θ cot θ ξj + c sin θ ξj = sin θ j Taking the derivative of (4.52) with respect to u and using (4.46) and (4.57), we obtain √ √ c 4 4 (4.58) (ξj )uu = − c sin θ ξ + c sin θ cos θ (− cot θ ξj4 ) = −c(1 + cos2 θ)ξj4 . sin θ j By solving (4.58) and using (4.54), we get ξj4 = Cj1 sin(bu) + Cj2 cos(bu) + Cj6 ,

(4.59) where b = have

√

c + c cos2 θ and Cj1 , Cj2 , Cj6 are constants. Combing (4.46) and (4.59), we (xj )u = − cot θ Cj1 sin(bu) + Cj2 cos(bu) + Cj6 ,

from which, we obtain xj =

(4.60)

cot θ 1 Cj cos(bu) − Cj2 sin(bu) − cot θ Cj6 u + ψj (v), b

where ψj (v) is a function with respect to v, for all 1 ≤ j ≤ 5. From (4.29), we have h(∂v , ∂v ) = −ce6 . From D∂v ∂v = ∇∂v ∂v +

5 X

hAα (∂v ), ∂v i eα + h(∂v , ∂v ) = 2He3 −

α=3

it follows that (4.61)

(xj )vv = 2Hξj3 −

√

c cos θ ξj5 − cxj .

Substituting (4.60) into the second equality of (4.51), we obtain (ξj3 )v = −2H(xj )v = −2Hψj0 (v),

√

c cos θ e5 − ce6 ,


221

from which, combining with the first equality of (4.51), we get ξj3 = −2Hψj (v) + Cj3 ,

(4.62)

where Cj3 is a constant. According to (4.52) and (4.59), we have 1 c sin θ 1 =√ c sin θ

ξj5 = √ (4.63)

c sin θ cos θ xj − (ξj4 )u

c sin θ cos θ xj − bCj1 cos(bu) + bCj2 sin(bu) .

Substituting (4.60), (4.62), (4.63) into (4.61), we obtain ψj00 (v) = −4H 2 ψj (v) + 2HCj3 − cot θ c sin θ cos θ xj − bCj1 cos(bu) + bCj2 sin(bu) − cxj = −4H 2 ψj (v) + 2HCj3 + b cot θ Cj1 cos(bu) − Cj2 sin(bu) − b2 xj = −(4H 2 + b2 )ψj (v) + 2HCj3 + b2 cot θ Cj6 u, which is equivalent with ψj00 (v) + a2 ψj (v) = 2HCj3 + b2 cot θ Cj6 u,

(4.64) where a = that

√

4H 2 + b2 . Since u, v are two independent parameters, we have from (4.64)

(4.65)

Cj6 = 0,

(4.66)

ψj00 (v) + a2 ψj (v) = 2HCj3 ,

for any 1 ≤ j ≤ 5. From (4.66), we have (4.67)

ψj (v) = Cj4 sin(av) + Cj5 cos(av) +

2H 3 C , a2 j

where Cj4 , Cj5 are constants. Putting (4.65), (4.67) into (4.60), we get xj = ϕj (u) + ψj (v),

(4.68) for any 1 ≤ j ≤ 5, where (4.69)

ϕj (u) =

cot θ 1 Cj cos(bu) − Cj2 sin(bu) . b

Next, let us characterize Cji for 1 ≤ i, j ≤ 5. Denote C j = (C1j , . . . , C5j ), and suppose that cot θ cos(bu)C 1 − sin(bu)C 2 , ϕ(u) = b (4.70) 2H ψ(v) = sin(av)C 4 + cos(av)C 5 + 2 C 3 . a

222


Then from (4.59), (4.62), (4.63) and (4.68), we have e1 = xu = (ϕ0 (u), sin θ), (4.71)

e2 = xv = (ψ 0 (v), 0),

e3 = (−2Hψ(v) + C 3 , 0), e4 = (− tan θ ϕ0 (u), cos θ), √ √ e5 = ( c(cos θψ(v) − cos−1 θϕ(u)), 0), e6 = c(ϕ(u) + ψ(v), 0).

From he1 , e1 i = 1, we get

1 1 2 2 2 1 2 2 1 2 C + C + C − C cos(2bu) + C 1 , C 2 sin(2bu) = sin2 θ, 2 2 which implies (4.72)

1 2 2 2 C = C = sin2 θ,

1 2 C , C = 0.

Similarly, by he2 , e2 i = 1, we have

1 4 2 5 2 1 4 2 5 2 1 C + C C − C + cos(2av) + C 4 , C 5 sin(2av) = 2 , 2 2 a which implies 4 2 5 2 C = C = 1 , a2

(4.73)

C 4 , C 5 = 0.

By he1 , e6 i = 0, we obtain

1 4

1 5 2H 1 3 0 = C , C sin(av) + C , C cos(av) + 2 C , C sin(bu) a

2 4

2 5 2H 2 3 cos(bu). + C , C sin(av) + C , C cos(av) + 2 C , C a It follows that

(4.74)

C 1, C α = C 2, C α = 0

for any α ∈ {3, 4, 5}. In the same way, we have from he2 , e6 i = 0 and he3 , e6 i = 0 that (4.75)

C 3 , C 4 = C 3 , C 5 = 0,

3 2 a2 C = . b2

5 Using (4.72) to (4.75), we conclude that C i i=1 is orthogonal with each other and 1 2 C = C 2 2 = sin2 θ, C 3 2 = a2 /b2 , C 4 2 = C 5 2 = 1/a2 . Through an orthogonal transformation, the surface is given by (4.24) and it is unique for a given θ, up to an isometry of M 4 (c) × R. Remark 4.6. For any fixed θ ∈ (0, π/2) and given real number H, we construct a class of surfaces with constant mean curvature H, as described in Lemma 4.2. Therefore, we get a kind of surface which is not Abresch-Rosenberg type surfaces.


223

Proof of Main Theorem. Following Theorem 4.1, we obtain case (1) and case (2). In the latter case, θ = 0 implies that Σ2 lies in M n (c), and θ = π/2 means that Σ2 = γ × R where γ is a curve of M n (c). When θ ∈ (0, π/2), we have from Theorem 4.5 that Σ2 lies in M 4 (c) × R with c > 0 and is parameterized by (4.24), up to an isometry of M 4 (c) × R. This completes the proof of Main Theorem.

Acknowledgments The authors would like to express their hearty gratitude to referee for the valuable suggestions and constructive comments.

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