symplectic mean curvature flow in cp2

Calc. Var. (2013) 48:111–129DOI 10.1007/s00526-012-0544-x Calculus of Variations

Symplectic mean curvature flow in CP2

Xiaoli Han · Jiayu Li · Liuqing Yang

Received: 15 July 2011 / Accepted: 31 May 2012 / Published online: 11 July 2012© Springer-Verlag 2012

Abstract Let � be an immersed symplectic surface in CP2 with constant holomorphicsectional curvature k > 0. Suppose � evolves along the mean curvature flow in CP2. Inthis paper, we show that the symplectic mean curvature flow exists for long time and con-verges to a holomorphic curve if the initial surface satisfies |A|2 ≤ λ|H |2 + 2λ−1

λk and

cos α ≥√

7λ−33λ

( 12 < λ ≤ 2

3

)or |A|2 ≤ 2

3 |H |2 + 45 k cos α and cos α ≥ 1 − ε, for some ε.

Mathematics Subject Classification 53C44 · 53C21

1 Introduction

Let M be a Kähler surface, ω be the Kähler form on M and J be a complex structurecompatible with ω. The Riemannian metric g on M is defined by

g(U, V ) = ω(U, J V ).

Communicated by J. Jost.

X. HanDepartment of Mathematical Sciences, Tsinghua University, Beijing 100084, People’s Republic of Chinae-mail: [email protected]

J. Li (B)School of Mathematical Sciences, University of Science and Technology of China,Hefei 230026, People’s Republic of Chinae-mail: [email protected]

J. Li · L. YangAcademy of Mathematics and Systems Sciences, Chinese Academy of Sciences,Beijing 100190, People’s Republic of Chinae-mail: [email protected]

L. Yange-mail: [email protected]

123

112 X. Han et al.

For a compact oriented real surface � which is smoothly immersed in M, the kähler angleα of � in M was defined by [6]

ω|� = cos αdμ� (1.1)

where dμ� is the area element of the induced metric on �. We say that � is a symplecticsurface if cos α > 0 and � is a holomorphic curve if cos α ≡ 1. The problem is whether onecan deform a symplectic surface to a holomorphic curve in a Kähler surface. One way is to usemean curvature flows (cf. [2,9]), the other way is to use variational method [11]. Chen–Tian[4] and Chen–Li [2] have proved that, in a Kähler–Einstein surface, if the initial surface issymplectic, then along the mean curvature flow, at every time t the surface �t is symplectic,which we call a symplectic mean curvature flow. They also showed that, there is no typeI singularity along a symplectic mean curvature flow. The symplectic mean curvature flowexists globally and converges at infinity in graphic cases (cf. [5]). Han–Li [9] proved that, ina Kähler–Einstein surface with positive scalar curvature, if the initial surface is sufficientlyclose to a holomorphic curve, then the symplectic mean curvature flow exists globally andconverges to a holomorphic curve at infinity. The second type singularities were also studiedby Chen–Li [3], Han–Li [10], Neves [15], Neves–Tian [16], etc.

Even though one thinks the mean curvature flows may produce minimal surfaces, thereare rather few results on the global existence and convergence to a minimal surface at infinity.

In this paper, we find the condition that |A|2 ≤ λ|H |2 + 2λ−1λ

k and cos α ≥√7λ−3

3λ

( 12 ≤ λ ≤ 2

3

)or |A|2 ≤ 2

3 |H |2 + 45 k cos α and cos α ≥ 1 − ε, for small ε > 0

is preserved by the mean curvature flow, and consequently, we show that the symplecticmean curvature flow exists for long time and converges to a holomorphic curve at infinityif the initial surface satisfies one of the conditions. As we know that it is the first long timeexistence and convergence result without graphic structure or small initial data conditions.Because k > 0, one can see that the surface close to CP1 satisfies the pinching condition.The main point is to prove the pinching estimate in our theorem, which was inspired byAndrews–Baker [1] and Huisken [13].

We believe that, the symplectic mean curvature flow exists globally and converges to aholomorphic curve at infinity in a Kähler-Einstein surface with positive scalar curvature.

Throughout this paper we will adopt the following ranges of indices:

A, B, . . . = 1, . . . , 4,

α, β, γ, . . . = 3, 4,

i, j, k, . . . = 1, 2.

2 Preliminaries

Suppose that � is a submanifold in a Riemannian manifold M, we choose an orthonormalbasis {ei } for T � and {eα} for N�. Recall the evolution equation for the second fundamentalform hα

i j and |A|2 along the mean curvature flow (see [2,17,19])

Lemma 2.1 For a mean curvature flow F : � ×[0, t0) → M, the second fundamental formhα

i j satisfies the following equation

∂

∂thα

i j = hαi j + (∇∂k K

)αi jk + (∇∂ j K

)αkik

−2Kli jkhαlk + 2Kαβ jkhβ

ik + 2Kαβikhβjk

123

Symplectic mean curvature flow in CP2 113

−Klkikhαl j − Klk jkhα

il + Kαkβkhβi j

−Hβ(

hβikhα

jk + hβjkhα

ik

)

+hαimhβ

mkhβk j − 2hβ

imhαmkhβ

k j + hβikhβ

kmhαmj

+hαkmhβ

mkhβi j + hβ

i j < eβ,∇H eα >, (2.1)

where K ABC D is the curvature tensor of M and ∇ is the covariant derivative of M. Therefore

∂

∂t|A|2 = |A|2 − 2|∇ A|2 +

[(∇∂k K)αi jk + (∇∂ j K

)αkik

]hα

i j

−4Kli jkhαlkhα

i j + 8Kαβ jkhβikhα

i j − 4Klkikhαl j h

αi j + 2Kαkβkhβ

i j hαi j

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i, j

hαi j h

βi j

⎞⎠

2

. (2.2)

Corollary 2.2 Along the mean curvature flow, the length of the mean curvature vectorsatisfies

∂

∂t|H |2 = |H |2 − 2|∇ H |2 + 2Kαkβk Hα Hβ + 2

∑

i, j

(∑α

Hαhαi j

)2

. (2.3)

Suppose that M is a compact Kähler surface. Let � be a smooth surface in M. The Kählerangle of � in M is defined by (1.1). Recall the evolution equation of cos α ([2]),

Lemma 2.3 Along the symplectic mean curvature flow, cos α satisfies(

∂

∂t−

)cos α = |∇ J�t |2 cos α + Ric (Je1, e2) sin2 α. (2.4)

where |∇ J�t |2 = |h31k − h4

2k |2 + |h32k + h4

1k |2, {e1, e2, e3, e4} is any orthonormal basis forT M such that {e1, e2} is the basis for T � and {e3, e4} is the basis for N�.

It is easy to see that |∇ J�t |2 is independent of the choice of the frame and only dependon the orientation of the frame. It is proved in [2,11] that

|∇ J�t |2 ≥ 1

2|H |2 (2.5)

and

|∇ cos α|2 ≤ sin2 α|∇ J�t |2. (2.6)

Now suppose M is a Kähler surface with constant holomorphic sectional curvature k, thenfrom Theorem 2.1 and Theorem 2.3 in [20], we have

Lemma 2.4 M has a curvature tensor of the form

Kkjih = −k

4

[(gkh g ji − g jh gki

) + (Jkh J ji − J jh Jki

) − 2Jk j Jih]. (2.7)

Thus M is symmetric. Furthermore, M is Einstein

K ji = 3

2kgi j . (2.8)

123

114 X. Han et al.

3 Pinching estimate

In this section we want to show the pinching estimate is preserved by the symplectic meancurvature flow. Before proving this key pinching estimate, we deduce the local expressionof the complex structure of the Kähler surface. Let M be a Kähler surface with the Kählermetric g and � be a real surface in M. Suppose ω is the associated Kähler form and J is thecomplex structure compatible with g and ω, i.e.,

ω(X, Y ) = g(J X, Y ) = 〈J X, Y 〉for any X, Y ∈ T M. Fix p ∈ M. We choose the local frame of M around p{e1, e2, e3, e4}such that {e1, e2} is the frame of the tangent bundle T � and {e3, e4} is the frame of the normalbundle N�. Suppose

Je1 = xe2 + ye3 + ze4.

Then using 〈J X, Y 〉 = −〈X, JY 〉 and

−e1 = x Je2 + y Je3 + z Je4,

we have⎧⎨⎩

x〈Je2, e4〉 + y〈Je3, e4〉 = 0x〈Je2, e3〉 − z〈Je3, e4〉 = 0.

−y〈Je2, e3〉 − z〈Je2, e4〉 = 0

Suppose y = 0. Set 〈Je2, e4〉 = A. Then we have

〈Je2, e3〉 = − z

yA, 〈Je3, e4〉 = − x

yA.

Thus

J =

⎛⎜⎜⎝

0 x y z−x 0 − z

y A A−y z

y A 0 − xy A

−z −A xy A 0

⎞⎟⎟⎠ .

As J is isometric, we have{

x2 + y2 + z2 = 1x2 + ( z

y )2 A2 + A2 = 1 ,

we can obtain that A2 = y2, i.e., A = ±y. Thus we see that

J =

⎛⎜⎜⎝

0 x y z−x 0 −z y−y z 0 −x−z −y x 0

⎞⎟⎟⎠ , (3.1)

or

J =

⎛⎜⎜⎝

0 x y z−x 0 z −y−y −z 0 x−z y −x 0

⎞⎟⎟⎠ . (3.2)

123


If y = 0, then by the same argument we can see that J also has the form (3.1) or (3.2).By the definition of the Kähler angle, we know that

x = cos α = ω (e1, e2) = 〈Je1, e2〉.If we assume the Kähler form is anti-self-dual, then J has the form (3.2).

We begin by estimating the gradient terms:

Lemma 3.1 For any η > 0 we have the inequality

|∇ A|2 ≥(

3

4− η

)|∇ H |2 −

(1

4η−1 − 1

)|w|2, (3.3)

where wαi = ∑

l Kαlil , |wα|2 = ∑i |wα

i |2 and |w|2 = ∑α |wα|2.

Proof Similar as [8] and [13] we decompose the tensor ∇ A into

∇i hαjk = Eα

i jk + Fαi jk,

where

Eαi jk = 1

4

(∇i Hα · g jk + ∇ j Hα · gi k + ∇k Hα · gi j)

−1

2wα

i g jk + 1

2

(wα

j gik + wαk gi j

).

It is easy to get that, 〈Eαi jk, Fα

i jk〉 = 0. Furthermore,

|Eα|2 = 3

4|∇ H |2 + |wα|2 + 〈wα

i ,∇i Hα〉

≥(

3

4− η

)|∇ Hα|2 −

(1

4η−1 − 1

)|wα|2.

We finish the proof of the Lemma. ��Theorem 3.2 Suppose � is a symplectic surface in CP2 with constant holomorphic sec-

tional curvature k > 0. Assume that |A|2 ≤ λ|H |2 + 2λ−1λ

k and cos α ≥√

7λ−33λ

holds

on the initial surface for any 12 < λ ≤ 2

3 , then it remains true along the symplectic meancurvature flow.

Proof From (2.4) and (2.8), we know that(

∂

∂t−

)cos α = |∇ J�t |2 cos α + 3k

2cos α sin2 α.

Thus at any time t, cos α ≥√

7λ−33λ

if it holds on the initial surface.

Since CP2 is symmetric, by (2.2) we know that

∂

∂t|A|2 = |A|2 − 2|∇ A|2

−4Kli jkhαlkhα

i j + 8Kαβ jkhβikhα

i j − 4Klkikhαl j h

αi j + 2Kαkβkhβ

i j hαi j

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i, j

hαi j h

βi j

⎞⎠

2

.

123

116 X. Han et al.

Now in our case the first four terms reduce to

−4Kli jkhαlkhα

i j = −4K1212(hα

12

)2 − 4K1221hα11hα

22

−4K2112hα11hα

22 − 4K2121(hα

12

)2

= −4K1212

(2

(hα

12

)2 − 2hα11hα

22

)

= −4K1212(|A|2 − |H |2) ,

and

8Kαβ jkhβikhα

i j = 8K3412h3i1h4

i2 + 8K3421h3i2h4

i1

+8K4312h4i1h3

i2 + 8K4321h4i2h3

i1

= 16K1234(h3

1i h42i − h3

i2h41i

)

= 8K1234(|A|2 − |∇ J�t |2

),

and

−4Klkikhαl j h

αi j = −4K1212

(hα

1 j

)2 − 4K2121

(hα

2 j

)2

= −4K1212|A|2,and

2Kαkβkhβi j h

αi j = 2K3k3k

(h3

i j

)2 + 2K4k4k

(h4

i j

)2 + 4K3k4kh3i j h

4i j

= 2K33

(h3

i j

)2 − 2K3434

(h3

i j

)2 + 2K44

(h4

i j

)2

−2K3434

(h4

i j

)2 + 4K34h3i j h

4i j

= 3k|A|2 − 2K3434|A|2,where we have used the equality (2.8). Therefore,

∂

∂t|A|2 = |A|2 − 2|∇ A|2

+8 (K1234 − K1212) |A|2 + 3k|A|2 − 2K3434|A|2+4K1212|H |2 − 8K1234|∇ J�t |2

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i, j

hαi j h

βi j

⎞⎠

2

. (3.4)

Similarly, the evolution equation of |H |2 becomes

∂

∂t|H |2 = |H |2 − 2|∇ H |2 + 3k|H |2 − 2K3434|H |2

+2∑

i, j

(∑α

Hαhαi j

)2

. (3.5)

123


Using (3.2) and (2.7) we get that,

K1212 = K3434 = k

4

(3 cos2 α + 1

) ;

K1234 = −k

4

(z2 + y2 − 2x2) = k

4

(3 cos2 α − 1

). (3.6)

Putting (3.6) into (3.4), we get that

∂

∂t|A|2 = |A|2 − 2|∇ A|2 − k|A|2 − k

2(3 cos2 α + 1)|A|2

+k(3 cos2 α + 1

) |H |2 − 2k(3 cos2 α − 1

) |∇ J�t |2

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i j

hαi j h

βi j

⎞⎠

2

.

Using the inequality (2.5) and cos α ≥√

7λ−33λ

≥√

33 , we obtain that

∂

∂t|A|2 ≤ |A|2 − 2|∇ A|2 − k|A|2 − k

2(3 cos2 α + 1)|A|2 + 2k|H |2

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i, j

hαi j h

βi j

⎞⎠

2

. (3.7)

Similarly,

∂

∂t|H |2 = |H |2 − 2|∇ H |2 + 3k|H |2 − k

2

(3 cos2 α + 1

) |H |2

+2∑

i, j

(∑α

Hαhαi j

)2

. (3.8)

Set Q = |A|2 − λ|H |2 − bk. Therefore,

∂

∂tQ ≤ Q − 2

(|∇ A|2 − λ|∇ H |2) − k|A|2

−k

2

(3 cos2 α + 1

) (|A|2 − λ|H |2) + (2 − 3λ)k|H |2

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+2∑α,β

⎛⎝∑

i, j

hαi j h

βi j

⎞⎠

2

−2λ∑

i, j

(∑α

Hαhαi j

)2

≤ Q − 2(|∇ A|2 − λ|∇ H |2) − k

2

(3 cos2 α + 1

)Q

−bk2

2

(3 cos2 α + 1

) − k|A|2 + (2 − 3λ) k|H |2

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i, j

hαi j h

βi j

⎞⎠

2

−2λ∑

i, j

(∑α

Hαhαi j

)2

. (3.9)

123

118 X. Han et al.

First we estimate the gradient terms in (3.9). In (3.3) we choose η = 34 − λ, then

|∇ A|2 ≥ λ|∇ H |2 − 4λ − 2

3 − 4λ|w|2,

where

|w|2 = K 23212 + K 2

3121 + K 24121 + K 2

4212.

Using (3.2) and (2.7) again, we obtain that

|w|2 = 9k2

8

(x2z2 + x2 y2) = 9k2

8x2 (

1 − x2) = 9k2

8cos2 α sin2 α.

Thus

|∇ A|2 ≥ λ|∇ H |2 − 9(2λ − 1)

4(3 − 4λ)k2 cos2 α sin2 α. (3.10)

In order to estimate the other terms in (3.9) we do the same way as in [1]. Set

R1 = ∑α,β,i, j

(∑k

(hα

ikhβjk − hα

jkhβik

))2, R2 = ∑

α,β

(∑i, j hα

i j hβi j

)2, R3 = ∑

i, j

(∑α Hα

hαi j

)2. At the point |H | = 0, we choose {e3, e4} for N� such that e3 = H/|H | and choose

{e1, e2} for T � such that h3i j = λiδi j . Set hα

i j = hαi j + 1

2 Hαgi j , then h4i j = h4

i j , h3i j =

h3i j − 1

2 |H |gi j . Since(

h3i j

)is diagonal, we see

(h3

i j

)is also diagonal. Set

(h3

i j

)= λiδi j .

Denote the norm of(

hαi j

),(

hαi j

)by |hα|, |hα| respectively. R1, R2, R3 reduce to

R2 =∑α,β

⎛⎝∑

i j

hαi j h

βi j

⎞⎠

2

= |h3|4 + |h4|4 + |h3|2|H |2 + 1

4|H |4 + 2

⎛⎝∑

i j

h3i j h

4i j

⎞⎠

2

;

R1 =∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

= 2∑

i, j

(∑

k

(h3

ik h4jk − h3

jk h4ik

))2

;

R3 =∑

i j

(∑α

Hαhαi j

)2

= |h3|2|H |2 + 1

2|H |4.

Using the fact that(

h3i j

)is diagonal, then we have

⎛⎝∑

i j

h3i j h

4i j

⎞⎠

2

=(∑

i

λi h4i i

)2

≤(∑

i

λi2) (∑

i

(h4

i i

)2)

= |h3|2∑

i

(h4

i i

)2,

∑

i, j

(∑

k

(h3

ik h4jk − h3

jk h4ik

))2

=∑

i = j

(λi − λ j

)2(

h4i j

)2

=∑

i = j

(λi − λ j

)2 (h4

i j

)2

123


≤∑

i = j

2(λ2

i + λ2j

)2 (h4

i j

)2

≤ 2|h3|2∑

i = j

(h4

i j

)2

= 2|h3|2(

|h4|2 −∑

i

(h4

i i

)2)

,

so⎛⎝∑

i j

h3i j h

4i j

⎞⎠

2

+∑

i, j

(∑

k

(h3

ik h4jk − h3

jk h4ik

))2

≤ 2|h3|2|h4|2.

Therefore,

2R1 + 2R2 − 2λR3 ≤ 2|h3|4 + 2|h4|4 + (2 − 2λ)|h3|2|H |2

−2λ − 1

2|H |4 + 8|h3|2|h4|2. (3.11)

Putting (3.10), (3.11) into (3.9), we obtain that

∂

∂tQ ≤ Q + 9 (2λ − 1)

2 (3 − 4λ)k2 cos2 α sin2 α − k

2

(3 cos2 α + 1

)Q

−bk2

2

(3 cos2 α + 1

) − k|A|2 + (2 − 3λ) k|H |2

+2|h3|4 + 2|h4|4 + (2 − 2λ) |h3|2|H |2 − 2λ − 1

2|H |4 + 8|h3|2|h4|2.

Sine |H |2 = 22λ−1

(|h3|2 + |h4|2 − Q − bk

), putting it into the above inequality we obtain

that

∂

∂tQ ≤ Q − 2

2λ − 1Q2

+[

4λ

2λ − 1|h3|2 + 4

2λ − 1|h4|2 − 4

2λ − 1bk + 3k − k

2

(3 cos2 α + 1

)]Q

−bk2

2

(3 cos2 α + 1

) + 9 (2λ − 1)

2 (3 − 4λ)k2 cos2 α sin2 α + 3bk2 − 2

2λ − 1b2k2

−4 − 4λ

2λ − 1|h4|4 + 12λ − 8

2λ − 1|h3|2|h4|2 +

(−4k + 4λbk

2λ − 1

)|h3|2

+(

−4k + 4bk

2λ − 1

)|h4|2

= Q − 2

2λ − 1Q2

+[

4λ

2λ − 1|h3|2 + 4

2λ − 1|h4|2 − 4

2λ − 1bk + 3k − k

2

(3 cos2 α + 1

)]Q

−4 − 4λ

2λ − 1

(|h4|2 − 1

2bk

)2

+ 12λ − 8

2λ − 1|h3|2|h4|2

123

120 X. Han et al.

+(

−4k + 4λbk

2λ − 1

) (|h3|2 + |h4|2

)+ 5

2bk2 − λ + 1

2λ − 1b2k2

+9 (2λ − 1)

2 (3 − 4λ)k2 cos2 α sin2 α − 3

2bk2 cos2 α.

If 12 < λ ≤ 2

3 , we want to choose b such that −4k + 4λbk2λ−1 ≤ 0 and 5

2 b − λ+12λ−1 b2 ≤ 0, that is

5(2λ − 1)

2(λ + 1)≤ b ≤ 2λ − 1

λ.

We also need 9(2λ−1)2(3−4λ)

sin2 α − 32 b ≤ 0, that is,

sin2 α ≤ 3 − 4λ

3(2λ − 1)b.

We choose

b = 2λ − 1

λ,

and assume

sin2 α ≤ 3 − 4λ

3(2λ − 1)× 2λ − 1

λ= 3 − 4λ

3λ,

that is

cos2 α ≥ 7λ − 3

3λ.

At the point |H | = 0, we use the following inequality (see [14,7]),

2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i, j

hαi j h

βi j

⎞⎠

2

≤ 3|A|4. (3.12)

Thus, using (3.10) and (3.9) we have

∂

∂tQ ≤ Q + 9 (2λ − 1)

2 (3 − 4λ)k2 cos2 α sin2 α − k|A|2

−k

2

(3 cos2 α + 1

) |A|2 + 3|A|4.

Since |H | = 0, we have |A|2 = Q + bk. Thus,

∂

∂tQ ≤ Q + 9(2λ − 1)

2(3 − 4λ)k2 cos2 α sin2 α

−3k

2

(cos2 α + 1

)(Q + bk) + 3(Q + bk)2.

≤ Q + 9 (2λ − 1)

2 (3 − 4λ)k2 cos2 α sin2 α

+[

3(|A|2 + bk

) − 3k

2

(cos2 α + 1

)]Q

−3bk2

2

(cos2 α + 1

) + 3b2k2

123


≤ Q +[

3(|A|2 + bk) − 3k

2

(cos2 α + 1

)]Q

+3bk2(

b − 1

2

)+ k2 cos2 α

(9 (4λ − 2)

4 (3 − 4λ)sin2 α − 3b

2

). (3.13)

Thus we need choose b ≤ 12 and assume that sin2 α ≤ 3−4λ

3(2λ−1)b.

Therefore, we choose b = 2λ−1λ

and suppose that cos α ≥√

7λ−33λ

, then we have

∂

∂tQ ≤ Q + C Q.

Applying the maximum principle for parabolic equation, we know that

Q ≤ 0

along the flow, if it is true on initial surface for 12 < λ ≤ 2

3 . ��

4 Long time existence and convergence

In this section we prove the long time existence of the symplectic mean curvature flow underthe assumption of Theorem 3.2 by constructing a function which depend on cos α, whichwas used in [12,18].

Theorem 4.1 Under the assumption of Theorem 3.2, the symplectic mean curvature flowexists for long time.

Proof Suppose f is a positive increasing function which will be determined later. Now wecompute the evolution equation of |H |2 f

( 1cos α

).

(∂

∂t−

)(|H |2 f

(1

cos α

))=

(∂

∂t−

)|H |2 f

(1

cos α

)

+|H |2(

∂

∂t−

) (f

(1

cos α

))

−2∇|H |2 · ∇ f

(1

cos α

)

It follows that (∂

∂t−

)cos α = |∇ J�t |2 cos α + 3

2k sin2 α cos α

≥ |∇ J�t |2 cos α.

By (3.8) and cos α ≥ δ > 0, we have(

∂

∂t−

)|H |2 ≤ −2|∇ H |2 +

(5 − 3δ2

2

)|H |2 + 2|H |2|A|2

≤ −2|∇ H |2 +(

5 − 3δ2

2

)|H |2 + 2|H |2

(λ|H |2 + 2λ − 1

λk

)

= −2|∇ H |2 +(

5 − 3δ2

2+ 4λ − 2

λ

)k|H |2 + 2λ|H |4.

123

122 X. Han et al.

Putting the above inequality into the evolution equation of |H |2 f ( 1cos α

), we get that(

∂

∂t−

) (|H |2 f

(1

cos α

))

≤ f

(−2|∇ H |2 +

(5 − 3δ2

2+ 4λ − 2

λ

)k|H |2 + 2λ|H |4

)

−|H |2(

f ′ |∇ J�t |2cos α

+ 2 f ′ |∇ cos α|2cos3 α

+ f ′′ |∇ cos α|2cos4 α

)

−2∇ (

f |H |2) − |H |2∇ f

f· ∇ f

(1

cos α

)

=(

5 − 3δ2

2+ 4λ − 2

λ

)k f |H |2 + |H |2 f

(−2

|∇ H |2|H |2 + 2λ|H |2 − f ′

f

|∇ J�t |2cos α

)

+|H |2(

− f ′′ + 2

(f ′)2

f− 2 f ′ cos α

)|∇ cos α|2

cos4 α

−2|H |2 ∇ (f |H |2)

f |H |2 · ∇ f

(1

cos α

). (4.1)

Set φ = f |H |2. At the point where φ = 0, it is easy to see that

∇φ = f ∇|H |2 + |H |2∇ f = f ∇|H |2 − |H |2 f ′ ∇ cos α

cos2 α,

i.e.,

∇ cos α

cos2 α= f

f ′

(∇|H |2|H |2 − ∇φ

φ

). (4.2)

Plugging (4.2) into (4.1), we obtain(

∂

∂t−

)φ ≤

(5 − 3δ2

2+ 4λ − 2

λ

)kφ + φ

(−2

|∇ H |2|H |2 + 2λ|H |2 − f ′

f

|∇ J�t |2cos α

)

+ φ f

( f ′)2

(− f ′′ + 2

(f ′)2

f− 2 f ′ cos α

)

×( |∇|H |2|2

|H |4 − 2∇|H |2|H |2 · ∇φ

φ+ |∇φ|2

φ2

)

+2|H |2 f ′ ∇φ

φ· ∇ cos α

cos2 α

≤(

5 − 3δ2

2+ 4λ − 2

λ

)kφ + φ

(− f ′

2 f

|H |2cos α

+ 2λ|H |2)

+φ

(−2

|∇ H |2|H |2 − 4

f f ′′

( f ′)2

|∇|H ||2|H |2 + 8

|∇|H ||2|H |2 − 8

f

f ′ cos α|∇|H |||H |2

)

+2|H |2 f ′ ∇φ

φ· ∇ cos α

cos2 α

+φ

(− f f ′′

( f ′)2 − 2f

f ′ cos α + 2

) ( |∇φ|2φ2 − 2

∇|H |2|H |2 · ∇φ

φ

)

123


=(

5 − 3δ2

2+ 4λ − 2

λ

)kφ + φ

(− f ′

2 f

1

cos α+ 2λ

)|H |2

+φ

(−4

f f ′′

( f ′)2 − 8f

f ′ cos α + 6

) |∇|H ||2|H |2

+φ

(− f f ′′

( f ′)2 − 2f

f ′ cos α + 2

) ( |∇φ|2φ2 − 2

∇|H |2|H |2 · ∇φ

φ

)

+2|H |2 f ′ ∇φ

φ· ∇ cos α

cos2 α. (4.3)

Set ff ′ = g, we choose g such that for x ∈ [

1, 1δ

]{

x/g ≥ 4λ,

−4g′ + 8g/x − 2 = 0.

Let g(x) = xc(x), then c(x) need satisfy{

0 < c(x) ≤ 14λ

,

−2xc′ = 1 − 2c.

We choose c(x) = 12 − ax by solving the last equation, where a will be defined later. It

reduces to solve the inequality

0 <1

2− ax ≤ 1

4λ, x ∈

[1,

1

δ

],

i.e.,(

1

2− 1

4λ

)1

x≤ a <

1

2x, x ∈

[1,

1

δ

].

Thus if δ > 1 − 12λ

, we can choose a = 12 − 1

4λ(note that

√7λ−3

3λ> 1 − 1

2λ, for λ ∈ ( 1

2 , 23

]).

Then

g = x

(1

2− ax

)= x

2−

(1

2− 1

4λ

)x2,

and

f (x) = (1 − 2a)2x2

(1 − 2ax)2 = x2

(2λ − (2λ − 1) x)2 , x ∈[

1,1

δ

].

It is evident that for x ∈ [1, 1

δ

],

1 ≤ f (x) ≤ 1

(2λδ − (2λ − 1))2 .

By (4.3), we have(

∂

∂t−

)φ ≤

(5 − 3δ2

2+ 4λ − 2

λ

)kφ

+φ

(− f f ′′

( f ′)2 − 2f

f ′ cos α + 2

)( |∇φ|2φ2 − 2

∇|H |2|H |2 · ∇φ

φ

)

+2|H |2 f ′ ∇φ

φ· ∇ cos α

cos2 α. (4.4)

123

124 X. Han et al.

This implies that

|H |2 ≤ |H |2 f

(1

cos α

)≤ e

(5−3δ2

2 + 4λ−2λ

)kt |H |2(0) f

(1

cos α

)(0),

we have

|H |2 ≤ C0eC1t ,

where C0 depends only on max�0 |H |2 and λ. Pinching inequality implies |A|2 ≤ C2eC1t +2λ−1

λk. We finish the proof of the theorem. ��

Theorem 4.2 Under the assumption of Theorem 3.2, the symplectic mean curvature flowconverges to a holomorphic curve.

Proof We can rewrite the evolution equation of cos α

(∂

∂t−

)cos α = |∇ J�t |2 cos α + 3k

2cos α sin2 α

as(

∂

∂t−

)sin2 (α/2) = −1

2|∇ J�t |2 cos α − 3k sin2 (α/2) cos2 (α/2) cos α (4.5)

≤ −c sin2 (α/2) , (4.6)

where c > 0 depends only on k and the lower bound of cos α. Applying the maximumprinciple, we get that sin2(α/2) ≤ e−ct . By Theorem 4.1 we know that the symplectic meancurvature flow exists for long time. Thus for any ε > 0, there exists T such that as t > T,

we have

cos α ≥ 1 − ε,

sin α ≤ 2ε,

|∇ cos α|2 ≤ 2ε|∇ J�t |2 ≤ 4ε|A|2. (4.7)

Therefore,(

∂

∂t−

)cos α ≥ 1

2|H |2 cos α + 3

2k sin2 α cos α

≥(

1

2λ|A|2 − 2λ − 1

2λ2 k

)cos α + 3

2k sin2 α cos α

≥ 1

2λ(1 − ε) |A|2 − 2λ − 1

2λ2 k. (4.8)

From (3.4) we see that(

∂

∂t−

)|A|2 ≤ −2|∇ A|2 + C1|A|4 + C2|A|2,

where C1, C2 are constants that depend on the bounds of the curvature tensor of CP2.

123


Let p > 1 be a constant to be fixed later. For simplicity, we set u = cos α. Now we

consider the function |A|2epu .

(∂

∂t−

) |A|2epu

= 2∇( |A|2

epu

)· ∇epu

epu+ 1

e2pu[epu

(∂

∂t−

)|A|2 − |A|2

(∂

∂t−

)epu]

≤ 2p∇( |A|2

epu

)· ∇u

+ 1

e2pu

[epu (

C1|A|4+C2|A|2)− p|A|2epu[

1

2λ(1−ε) |A|2− 2λ−1

2λ2 k − p|∇u|2]]

.

Using (4.7) we obtain that,(

∂

∂t−

) |A|2epu

≤ 2p∇( |A|2

epu

)· ∇u

+ 1

epu

[(C1 − 1

2λp (1 − ε) + 4p2ε

)|A|4 + C3|A|2

].

Set p2 = 1/ε, then

C1 − 1

2λp(1 − ε) + 4p2ε = C1 − 1

2λε− 1

2 + 1

2λε

12 + 4.

As t is sufficiently large, i.e. ε is sufficiently close to 0, we have(

C1 − 1

2λε− 1

2 + 1

2λε

12 + 4

)≤ −1.

So,(

∂

∂t−

) |A|2epu

≤ 2p∇( |A|2

epu

)· ∇u − |A|4

epu+ C3

|A|2epu

≤ 2p∇( |A|2

epu

)· ∇u − |A|4

e2pu+ C3

|A|2epu

Applying the maximum principle for parabolic equations, we conclude that |A|2epu is uni-

formly bounded, thus |A|2 is also uniformly bounded. Thus F(·, t) converges to F∞ in C2

as t → ∞. Since sin2(α/2) ≤ e−ct , we have cos α ≡ 1 at infinity. Thus the limiting surfaceF∞ is a holomorphic curve. ��

5 Another pinching estimate

In this section, we will derive another pinching condition.

Theorem 5.1 Suppose � is a symplectic surface in CP2 with constant holomorphic sec-tional curvature k > 0. If there exists a small ε > 0 such that |A|2 ≤ 2

3 |H |2 + 45 k cos α and

cos α ≥ 1 − ε holds on the initial surface, then it remains true along the symplectic meancurvature flow, where ε ≤ 14

265 is an absolute constant. Furthermore, the symplectic meancurvature flow exists for long time and converges to a holomorphic curve at infinity.

123

126 X. Han et al.

Proof From (3.7), (3.8), we see that,

∂

∂t|A|2 ≤ |A|2 − 2|∇ A|2 − k|A|2 − k

2

(3 cos2 α + 1

) |A|2 + 2k|H |2

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i j

hαi j h

βi j

⎞⎠

2

,

and

∂

∂t|H |2 = |H |2 − 2|∇ H |2 + 3k|H |2 − k

2

(3 cos2 α + 1

) |H |2

+2∑

i j

(∑α

Hαhαi j

)2

.

Set Q = |A|2 − 23 |H |2 − bk cos α. Then we have,

∂

∂tQ ≤ Q − 2

(|∇ A|2 − 2

3|∇ H |2

)− k|A|2

−k

2

(3 cos2 α + 1

) (|A|2 − 2

3|H |2

)

−bk

(|∇ J�t |2 cos α + 3

2k cos α sin2 α

)

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i j

hαi j h

βi j

⎞⎠

2

−4

3

∑

i j

(∑α

Hαhαi j

)2

≤ Q − 2

(|∇ A|2 − 2

3|∇ H |2

)− k

2

(3 cos2 α + 1

)Q

−2bk2 cos α − k|A|2 − bk

2cos α|H |2

+2∑

α,β,i, j

(∑

k

(hα

ikhβjk − hα

jkhβik

))2

+ 2∑α,β

⎛⎝∑

i j

hαi j h

βi j

⎞⎠

2

−4

3

∑

i j

(∑α

Hαhαi j

)2

.

By an argument similar to the one used in the proof of Theorem 3.2, at the point H = 0 wecan get that

∂

∂tQ ≤ Q + 9k2

2cos2 α sin2 α − k

2

(3 cos2 α + 1

)Q

−2bk2 cos α − k|A|2 − bk

2cos α|H |2

+2|h3|4 + 2|h4|4 + 2

3|h3|2|H |2 − 1

6|H |4 + 8|h3|2|h4|2.

123


Sine |H |2 = 6(|h3|2 + |h4|2 − Q − bk cos α

), putting it into the above inequality we

obtain that

∂

∂tQ ≤ Q − 6Q2

+[

8|h3|2 + 12|h4|2 − 9bk cos α + 3k − k

2

(3 cos2 α + 1

)]Q

+9k2

2cos2 α sin2 α + bk2 cos α − 3b2k2 cos2 α

−4|h4|2 + k (5b cos α − 4) |h3|2 + k(9b cos α − 4)|h4|2≤ Q − 6Q2

+[

8|h3|2 + 12|h4|2 − 9bk cos α + 3k − k

2

(3 cos2 α + 1

)]Q

+k (5b cos α − 4) |h3|2 −(

2|h4|2 − k

4(9b cos α − 4)

)2

+ k2

16(9b cos α − 4)2

+9k2

2cos2 α sin2 α + bk2 cos α − 3b2k2 cos2 α

≤ Q − 6Q2

+[

8|h3|2 + 12|h4|2 − 9bk cos α + 3k − k

2

(3 cos2 α + 1

)]Q

+k (5b cos α − 4) |h3|2 −(

2|h4|2 − k

4(9b cos α − 4)

)2

+ k2

16

(33b2 cos2 α − 56b cos α + 16 + 72 cos2 α sin2 α

)(5.1)

Now we need choose b and the lower bound of cos α such that 5b cos α − 4 ≤ 0 and

33b2 cos2 α − 56b cos α + 16 + 72 cos2 α sin2 α ≤ 0.

First we choose b = 45 , then we need

33 × 16

25cos2 α − 56 × 4

5cos α + 16 + 72 cos2 α sin2 α ≤ 0. (5.2)

Assume that cos α ≥ δ. If

33 × 16

25cos2 α − 56 × 4

5cos α + 16 + 72

(1 − cos2 α

) ≤ 0 (5.3)

holds, then (5.2) holds. Solving (5.3), we get that

δ ≥ 251

265. (5.4)

At the point H = 0, using (3.12), we have(

∂

∂t−

)Q ≤ 9

2k2 sin2 α cos2 α − k

2

(3 cos2 α + 1

)Q

−2bk2 cos α − k|A|2 + 3|A|4.

123

128 X. Han et al.

Putting |A|2 = Q + bk cos α into the above inequality, we get that,(

∂

∂t−

)Q ≤

[3

(|A|2 − bk cos α) + 6bk cos α − k − k

2

(3 cos2 α + 1

)]Q

+9

2k2 sin2 α cos2 α − 3bk2 cos α + 3b2k2 cos2 α.

Choose b = 45 and assume that cos α ≥ δ, then we need

9

2

(1 − cos α2) − 12

5cos α + 48

25cos2 α ≤ 0.

Solving it we get that

δ ≥ 121

129. (5.5)

Compare (5.4) and (5.5), we choose δ ≥ 251265 and b = 4

5 .

The global existence and convergence of the symplectic mean curvature flow can beproved in a similar manner as the one used in the proof of Theorem 4.1 and Theorem 4.2. ��Remark 5.2 Comparing these two pinching estimates, it is clear that the pinching condition|A|2 ≤ 2

3 |H |2 + 45 k cos α is better than the condition that |A|2 ≤ 2

3 |H |2 + 12 k. On the other

hand, the condition cos α ≥√

306 is better than cos α ≥ 251

265 .

Acknowledgments The research was supported by the National Natural Science Foundation of China,No. 10901088, No. 11071236, No. 11131007. The authors thank Jun Sun and Chao Wu for their valuablesuggestions. The authors also thank the referee for his helpful comments.

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