A gap theorem of self-shrinkers

arXiv:1212.6028v1 [math.DG] 25 Dec 2012

A GAP THEOREM OF SELF-SHRINKERS QING-MING CHENG AND GUOXIN WEI

Abstract. In this paper, we study complete self-shrinkers in Euclidean space and prove that an n-dimensional complete self-shrinker with√ polynomial volume growth √ in Euclidean space Rn+1 is isometric to either Rn , S n ( n), or Rn−m × S m ( m), 1 ≤ m ≤ n − 1, if the squared norm S of the second fundamental form is constant and satisfies S < 10 7 .

1. Introduction Let X : M → Rn+1 be a smooth n-dimensional immersed hypersurface in the (n + 1)-dimensional Euclidean space Rn+1 . The immersed hypersurface M is called a self-shrinker if it satisfies the quasilinear elliptic system: H = −X N ,

where H denotes the mean curvature vector of M, X N denotes the orthogonal projection of X onto the normal bundle of M. It it known that self-shrinkers play an important role in the study of the mean curvature flow because they describe all possible blow up at a given singularity of a mean curvature flow. For n = 1, Abresch and Langer [1] classified all smooth closed self-shrinker curves in R2 and showed that the round circle is the only embedded self-shrinkers. For n ≥ 2, Huisken [9] studied compact self-shrinkers. He proved that if M is an nn+1 dimensional compact √ self-shrinker with non-negative mean curvature H in R , n then X(M) = S ( n). We should notice that the condition of non-negative mean curvature is essential. In fact, let ∆ and ∇ denote the Laplacian and the gradient operator on the self-shrinker, respectively and h·, ·i denotes the standard inner product of Rn+1 . Because ∆H − hX, ∇Hi + SH − H = 0, we obtain H > 0 from the maximum principle if the mean curvature is non-negative. Furthermore, Angenent [2] has constructed compact embedded self-shrinker torus S 1 × S n−1 in Rn+1 . Huisken [10] and Colding and Minicozzi [5] have studied complete and noncompact self-shrinkers in Rn+1 . They have proved that if M is an n-dimensional complete embedded self-shrinker in Rn+1 with H ≥ 0 and with polynomial volume 2001 Mathematics Subject Classification: 53C44, 53C42. Key words and phrases: the second fundamental form, elliptic operator, self-shrinkers. The first author was partially supported by JSPS Grant-in-Aid for Scientific Research (B): No. 24340013. The second author was partly supported by grant No. 11001087 of NSFC. 1

2

QING-MING CHENG AND GUOXIN WEI

√ growth, then M is√isometric to either the hyperplane Rn , the round sphere S n ( n), or a cylinder S m ( m) × Rn−m , 1 ≤ m ≤ n − 1. Without the condition H ≥ 0, Le and Sesum [11] proved that if M is an ndimensional complete embedded self-shrinker with polynomial volume growth and S < 1 in Euclidean space Rn+1 , then S = 0 and M is isometric to the hyperplane Rn , where S denotes the squared norm of the second fundamental form. Furthermore, Cao and Li [3] have studied the general case. They have proved that if M is an n-dimensional complete self-shrinker with polynomial volume growth and S ≤ 1 in Euclidean space Rn+1 , then M is √ isometric to either the hyperplane Rn , the round √ n m sphere S ( n), or a cylinder S ( m) × Rn−m , 1 ≤ m ≤ n − 1. Recently, Ding and Xin [6] have studied the second gap on the squared norm of the second fundamental form and they have proved that if M is an n-dimensional complete self-shrinker with polynomial volume growth in Euclidean space Rn+1 , there exists a positive number δ = 0.022 such that if 1 ≤ S ≤ 1 + 0.022, then S = 1. Motivated by the above results of Le and Sesum, Cao and Li, Ding and Xin, we consider the second gap for the squared norm of the second fundamental form and prove the following classification theorem for self-shrinkers: Theorem 1.1. Let M be an n-dimensional complete self-shrinker with polynomial volume growth in Rn+1 . If the squared norm S of the second fundamental form is constant and satisfies 3 S ≤1+ , 7 then M is isometric to one of the following: (1) the hyperplane Rn , √ (2) a cylinder Rn−m × S m ( √ m), for 1 ≤ m ≤ n − 1, n (3) the round sphere S ( n). 2. Preliminaries In this section, we give some notations and formulas. Let X : M → Rn+1 be an n-dimensional self-shrinker in Rn+1 . Let {e1 , · · · , en , en+1 } be a local orthonormal basis along M with dual coframe {ω1 , · · · , ωn , ωn+1}, such that {e1 , · · · , en } is a local orthonormal basis of M and en+1 is normal to M. Then we have ωn+1 = 0, ωn+1i =

n X

hij ωj , hij = hji ,

j=1

where Pn hij denotes the component of the second fundamental Pnform of M. H = h e is the mean curvature vector field, H = |H| = jj n+1 j=1 j=1 hjj is the mean P curvature and II = i,j hij ωi ⊗ ωj en+1 is the second fundamental form of M. The Gauss equations and Codazzi equations are given by (2.1)

Rijkl = hik hjl − hil hjk ,

(2.2)

hijk = hikj ,

A GAP THEOREM OF SELF-SHRINKERS

3

where Rijkl is the component of curvature tensor, the covariant derivative of hij is defined by n n n X X X hijk ωk = dhij + hkj ωki + hik ωkj . k=1

k=1

k=1

Let

Fi = ∇i F, Fij = ∇j ∇i F, hijk = ∇k hij , and hijkl = ∇l ∇k hij ,

where ∇j is the covariant differentiation operator, we have (2.3)

hijkl − hijlk =

n X

him Rmjkl +

n X

hmj Rmikl .

m=1

m=1

The following elliptic operator L is introduced by Colding and Minicozzi in [5]: (2.4)

Lf = ∆f − < X, ∇f >,

where ∆ and ∇ denote the Laplacian and the gradient operator on the self-shrinker, respectively and < ·, · > denotes the standard inner product of Rn+1 . By a direct calculation, we have (2.5) Lhij = (1 − S)hij , LH = H(1 − S), LXi = −Xi , L|X|2 = 2(n − |X|2), X 1 LS = h2ijk + S(1 − S). 2 i,j,k

(2.6)

If S is constant, then we obtain from (2.4) and (2.6) X h2ijk = S(S − 1), (2.7) i,j,k

hence one has either (2.8)

S = 0, or S = 1, or S > 1.

We can choose a local field of orthonormal frames on M n such that, at the point that we consider, λi , if i = j, hij = 0, if i 6= j. then

S=

X

h2ij =

i,j

X

λ2i ,

i

where λi is called the principal curvature of M. From (2.1) and (2.3), we get (2.9)

hijij − hjiji = (λi − λj )λi λj .

By a direct calculation, we obtain X h2ijkl = S(S − 1)(S − 2) + 3(A − 2B), (2.10) i,j,k,l

where A =

2 2 i,j,k λi hijk , B =

P

P

i,j,k

λi λj h2ijk .

4


We define two functions f3 and f4 as follows: f3 =

X

hij hjk hki =

n X

λ3j ,

f4 =

j=1

i,j,k

X

hij hjk hkl hli =

n X

λ4j ,

j=1

i,j,k,l

then we have Lemma 2.1. Let M be an n-dimensional complete self-shrinker without boundary and with polynomial volume growth in Rn+1 . Then X λi h2ijk , (2.11) Lf3 = 3(1 − S)f3 + 6 i,j,k

(2.12)

Lf4 = 4(1 − S)f4 + 4(2A + B).

Proof. By the definition of f3 , f4 , Lf3 and Lf4 , we have the following calculations: X hijm hjk hki , f3m = 3 i,j,k

f3mm = 3

X

hjk hki hijmm + 3

X

hijm hjkm hki + 3

∆f3 =

X

f3mm = 3

m

X

hjk hki ∆hij + 6

X

λi h2ijm ,

i,j,m

i,j,k

< X, ∇f3 >= 3

hijm hjk hkim ,

i,j,k

i,j,k

i,j,k

X

X i,j,k

hjk hki < X, ∇hij >,

Lf3 = ∆f3 − < X, ∇f3 > X X =3 hjk hki Lhij + 6 λi h2ijm i,j,m

i,j,k

= 3(1 − S)f3 + 6 and f4m = 4

X

X

λi h2ijk ,

i,j,k

hijm hjk hkl hli ,

i,j,k,l

f4mm = 4

X

hijmm hjk hkl hli + 4

i,j,k,l

+4

X

X m

f4mm = 4

X

hijm hjkm hkl hli

i,j,k,l

hijm hjk hklm hli + 4

i,j,k,l

∆f4 =

X

hijm hjk hkl hlim ,

i,j,k,l

hjk hkl hli ∆hij + 4

X

i,j,m

i,j,k,l

< X, ∇f4 >= 4

X

X

i,j,k,l

λ2i h2ijm + 4

X

λi λj h2ijk + 4

i,j,k

hjk hkl hli < X, ∇hij >,

X

i,j,m

λ2j h2ijm ,


5

Lf4 = ∆f4 − < X, ∇f4 > X X X hjk hkl hli Lhij + 8 λ2i h2ijm + 4 λi λj h2ijm =4 i,j,m

i,j,k,l

i,j,m

= 4(1 − S)f4 + 4(2A + B).

✷ 3. Some estimates In this section, we will give some estimates which are needed to prove our theorem. From now on, we denote S − 1 = tS, where t is a positive constant if we assume that S is constant and S > 1, then X (1 − t)S = 1, h2ijk = tS 2 . i,j,k

By a direct calculation, one obtains X X 3X 3X h2ijkl ≥ h2iiii + (hijij + hjiji )2 + (hijij − hjiji )2 4 i6=j 4 i6=j i i,j,k,l X 3X 2 = h + (hijij + hjiji )2 iiii (3.1) 4 i6=j

i

X 3 X 4 3 2 + S λi − ( λi ) , 2 i i P P we next have to estimate S i λ4i − ( i λ3i )2 since we want to give the estimate of P 2 i,j,k,l hijkl . Define !2 X X 2 1 1 λ3i f3 . = f4 − f≡ λ4i − S S i i Firstly, we have

Lemma 3.1. There is one point x ∈ M such that the following identity holds at the point.   !2 X X c tS 2  λ3i − λ4i  S i i (3.2) ! !2 X X X X =c 2λi h2ijk λ3i − (2A + B)S + 3c λ2i hiij , i,j,k

where c is a real number.

i

j

i

6


Proof. Define a function 1 X 4 1 X 3 F = S λi − c λi 4 6 i i

we have from Lemma 2.1,

!2

2 1 1 = Sf4 − c f3 , 4 6

2 1 1 Sf4 − c f3 4 6 X = S(1 − S)f4 − c (1 − S)f32 + 2 λi h2ijk f3

LF = L (3.3)

i,j,k

XX 2 2 +3 ( λi hiij ) +(2A + B)S. j

i

On the other hand, we have from Stokes formula, Z |X|2 LF e− 2 dv = 0, M

hence there is a point x ∈ M such that X XX 2 2 2 2 S(1 − S)f4 − c (1 − S)f3 + 2 λi hijk f3 + 3 ( λi hiij ) +(2A + B)S = 0 i,j,k

j

i

at the point because of the continuity of the function.

✷ Secondly, we have Lemma 3.2. f32 (λ1 − λ2 )2 (λ1 λ2 )2 , ≥ 2 2 S λ1 + λ2 where λ1 = max{λi }, λ2 = min{λi }. f = f4 −

i

i

Proof. Since Sf4 − f32 = then

1X 2 (λi S − f3 λi )2 , S i

2 1 2 2 1 2 λ1 S − f3 λ1 + λ2 S − f3 λ2 S S 2 2 f (λ + λ22 ) = Sλ41 + Sλ42 + 3 1 − 2(λ31 + λ32 )f3 S 2 λ + λ22 (λ31 + λ32 )2 S 2 ≥ S(λ41 + λ42 ) − 1 S (λ21 + λ22 )2 S λ2 λ2 (λ1 − λ2 )2 . = 2 λ1 + λ22 1 2

Sf4 − f32 ≥

✷ Thirdly, one has


7

Lemma 3.3. (3.4) P 2 h where α = P i iii2 i,j,k hijk

1 A − B ≤ (λ1 − λ2 )2 tS 2 (1 − α), 3 P 2 h = i 2iii . tS

Proof. By means of symmetry, we have X (λ2i − λi λj )h2ijk A−B = i,j,k

1X 2 (λ + λ2j + λ2k − λi λj − λj λk − λk λi )h2ijk 3 i,j,k i 1X 3(λi − λj )2 h2iij = 3 i,j 1 X (λ2 + λ2j + λ2k − λi λj − λj λk − λk λi )h2ijk . + 3 i6=j6=k6=i i

=

Without loss of generality, we can assume that λi ≤ λj ≤ λk and consider z = λ2i + λ2j + λ2k − λi λj − λj λk − λk λi

as a function of λj , which takes its maximum at one of the boundary points λi or λk . On the other hand, zλj =λi = zλj =λk = (λi − λk )2 ≤ (λ1 − λ2 )2 . Hence we get # " X 1 X A−B ≤ (λ1 − λ2 )2 h2ijk 3(λi − λj )2 h2iij + 3 i,j i6=j6=k6=i ! X X 1 ≤ (λ1 − λ2 )2 h2ijk − h2iii . 3 i i,j,k Combining (2.7) and the definition of α, we get 0 ≤ α < 1 and (3.4). ✷ P P 2 P From Lemma 3.1, one knows that the estimates of k ( i λi hiik )2 and ( i,j,k h2ijk λi )2 are needed. Lemma 3.4. (3.5)

X X k

where α =

h2iii . tS 2

P

i

i

λ2i hiik

!2

≤

1 + 2α 2 tS f, 3

8


Proof. Since S =

P h2ij is constant, we have i λi hiik = 0, then !2 " #2 X X X X λ2i hiik = (λ2i − aλi )hiik

P

ij

i

k

i

k

≤

X

(λ2i

i

− aλi )2

X

h2iik ,

i,k

for any constant a. Let a = S1 f3 = S1 i λ3i , we have !2  !2  X X X X 1 X 3 X 2 λ2i hiik ≤ λ4i − (3.6) λi hiik = f h2iik . S i i i k i,k i,k P

Since

X

h2ijk =

(3.7)

X

h2iik

i,k

h2iii + 3

i

i,j,k

then

X

X

h2iij +

i6=j

X

h2ijk ,

i6=j6=k6=i

X 1 X 1 1 X 2 hijk + 2 h2iii = (1 + 2α) h2ijk = (1 + 2α)tS 2 , ≤ 3 i,j,k 3 3 i i,j,k

combining (3.6) and (3.7), we get (3.5).

✷ Lemma 3.5. (3.8)

X

λi h2ijk

i,j,k

!2



4X

1 ≤  (A + 2B) − 3 3

k

1 S + 2λ2k

Proof. A straightforward computation gives !2 X λi h2ijk

X i

!2  λ2i hiik  tS 2 .

i,j,k

)2 1X = [(λi + λj + λk )hijk − (ai hjk + aj hki + ak hij )] hijk 3 i,j,k X 1X [(λi + λj + λk )hijk − (ai hjk + aj hki + ak hij )]2 h2ijk ≤ 9 i,j,k i,j,k # " X X 1 (S + 2λ2k )a2k tS 2 , ak λ2i hiik + 3 3(A + 2B) − 12 = 9 k i,k (

for any constant ak ∈ R. Let

ak = 2 then (3.8) follows.

λ2i hiik , S + 2λ2k

P

i

✷


9

4. Proof of Theorem 1.1 In this section, we will prove the Theorem 1.1. The proof has three parts. In the first part of proof, we will show that S > 1 + 51 = 1.2 if S > 1. In the second part, 1 > 1.24688 if S > 65 = 1.2. In the third part, we will we will prove that S > 0.802 1 . show that S > 1 + 73 if S > 0.802 Proof of Theorem 1.1. Part I: Claim: S > 1 +

1 5

=

6 5

if S > 1.

Letting c = 2 and applying Lemma 3.1, we get X 1 0 = (S − 1)[ Sf4 − f32 ] + (2 λi h2ijk )f3 2 i,j,k −

XX S (2A + B) + 3 ( λ2i hiij )2 2 j i

X 1 1 (2 λi h2ijk )2 ≤ (S − 1)[ Sf4 − f32 ] + 2 2(S − 1) i,j,k

XX S−1 2 S f3 − (2A + B) + 3 ( λ2i hiij )2 2 2 j i S−1 8 XX 2 2 2 2 ( λi hiik ) ≤ (Sf4 − f3 ) + S (A + 2B) − 2 3 9S k i XX S − (2A + B) + 3 ( λ2i hiij )2 2 j i +

(4.1)

S−1 S 19 (Sf4 − f32 ) − [2A − 5B] + (1 + 2α)tS 2 f 2 6 27 65 38 S = − (2A − 5B) + [ + α]tS 2 f, 6 54 2

≤

at the point x, then it follows that

(4.2)

−

65 65(2A − 5B) tSf ≤ − . 9 65 + 76α

On the other hand, we have

(4.3)

3 Sf ≤ S(S − 1)(S − 2) + 3(A − 2B). 2

10


Combining (4.2) and (4.3), we obtain 130 3 [1 − t]Sf 2 27 65(2A − 5B) 65 + 76α 65(3A − 3B) 76α(A + 2B) = S(S − 1)(S − 2) + 4(A − B) − − 65 + 76α 65 + 76α 195 ≤ S(S − 1)(S − 2) + [4 − ](A − 2B) (Since A + 2B ≥ 0) 65 + 76α 195 1−α ≤ S(S − 1)(S − 2) + [4 − ] (λ1 − λ2 )2 tS 2 . 65 + 76α 3 Letting y = 65 + 76α, we get 195 1 195 (4 − )(1 − α) = (4 − )(141 − y) 65 + 76α 76 y √ 1 195 × 141 1 = (564 + 195 − − 4y) ≤ (759 − 2 4 × 195 × 141) ≡ 3γ1 , 76 y 76 where γ1 = 0.4198 · · · < 0.42. Since we assume t ≤ 16 , that is, 1 ≤ S ≤ 1 + 51 = 65 , then ≤ S(S − 1)(S − 2) + 3(A − 2B) −

3γ1 130 (λ1 − λ2 )2 3 2 2 (4.4) S(S − 1)(S − 2) + (λ1 λ2 )2 S. (λ1 − λ2 ) tS ≥ (1 − t) 2 2 3 2 27 λ1 + λ2 We next consider two cases: Case 1: λ1 (x)λ2 (x) ≥ 0. We see from (4.4) that that is,

S(S − 1)(S − 2) ≥ −γ1 (λ1 − λ2 )2 tS 2 ≥ −Sγ1 tS 2 , S − 2 ≥ −γ1 S,

then S≥

2 6 2 ≥ > 1.4 > 1.2 = . 1 + γ1 1 + 0.42 5

Case 2: λ1 (x)λ2 (x) < 0. From (4.4), we obtain

that is,

(S − 1)(S − 2) + γ1 StS ≥ (S − 1)(S − 2) + γ1 (λ21 + λ22 )tS 3 130 ≥ 2γ1λ1 λ2 tS + (1 − t)(λ1 λ2 )2 2 27 3 130 1 ≥ 2γ1λ1 λ2 tS + (1 − )(λ1 λ2 )2 2 27 6 4γ12 t2 S 2 , ≥− 3 130 4[ 2 × (1 − 27×6 )] S≥

6 16 + 27γ12 > 1.286 > 1.2 = . 2 8 + 8γ1 + 27γ1 5


Hence we have proved S >1+ Part II: Claim: S > Letting c =

9 5

1 0.802

6 1 = . 5 5

> 1.24688 if S > 56 .

and applying Lemma 3.1, we have

X 5 λi h2ijk )f3 0 = (S − 1)[ Sf4 − f32 ] + (2 9 i,j,k

XX 5S (2A + B) + 3 ( λ2i hiij )2 − 9 j i

X 5 9 ≤ (S − 1)[ Sf4 − f32 ] + λi h2ijk )2 (2 9 16(S − 1) i,j,k

XX 4(S − 1) 2 5S f3 − (2A + B) + 3 ( λ2i hiij )2 9 9 j i X XX 5tS 5 9 ≤ λi h2ijk )2 − S(2A + B) + 3 (Sf ) + (2 ( λ2i hiij )2 9 16tS i,j,k 9 j i X 1 5 5 (2 λi h2ijk )2 − S(2A + B) ≤ tS 2 f + 9 9 × 16tS i,j,k 9 XX 16 X X 2 5 4 2 ( λ2i hiij )2 ( λi hiik ) +3 (A + 2B)S − + 9 3 9 k j i i X 5 1 20 ≤ tS 2 f + (2 λi h2ijk )2 + (A + 2B)S 9 144tS 27 +

(4.5)

i,j,k

163 5 (1 + 2α)tS 2 f − (2A + B)S + 9 81 × 3 X 5 2 1 5(2A − 5B)S = tS f + (2 λi h2ijk )2 − 9 144tS i,j,k 27 +

163 (1 + 2α)tS 2 f, 81 × 3

at the point x, that is, (4.6)

0 ≤ 3tSf +

X 3 163 27 (2 λi h2ijk )2 − (2A − 5B) + × (1 + 2α)tSf, 2 80tS 81 × 3 5 i,j,k

then (4.7)

−

X 3 298 + 326α tSf ≤ (2 λi h2ijk )2 − (2A − 5B). 45 80tS 2 i,j,k

11

12


Since (2

(4.8)

X 1X (hijij + hjiji )2 ] λi h2ijk )2 ≤ 4S 2 [ h2iiii + 4 i6=j i i,j,k X 3X ≤ 4S 2 [ h2iiii + (hijij + hjiji )2 ] 4 i i6=j X 3 h2ijkl − Sf ] ≤ 4S 2 [ 2 i,j,k,l

X

3 = 4S 2 [S(S − 1)(S − 2) + 3(A − 2B) − Sf ], 2

then one obtains P (2 i,j,k λi h2ijk )2 3 Sf + ≤ S(S − 1)(S − 2) + 3(A − 2B). (4.9) 2 4S 2 1 , then we will get a contradiction. We now assume 61 < t ≤ 0.198, that is, S ≤ 0.802 From (4.7), we have X 298 9 298 (4.10) − Sf ≤ (2 λi h2ijk )2 − (2A − 5B). 2 225 40S 298 + 326α i,j,k Noting A + 2B ≥ 0, we see from (4.9) and (4.10) that 79 3 × 298 (A − B) Sf ≤ S(S − 1)(S − 2) + 4 − 450 298 + 326α (4.11) (λ1 − λ2 )2 2 3 × 298 tS (1 − α). ≤ S(S − 1)(S − 2) + 4 − 298 + 326α 3

On the other hand,

(4.12)

1 3 × 298 4− (1 − α) 3 298 + 326α 3 × 298 1 4− (624 − z) = 3 × 326 Z 1 3 × 298 × 624 = (2496 + 894 − 4z − ) 3 × 326 Z √ 1 ≤ (2496 + 894 − 2 4 × 3 × 298 × 624) 3 × 326 ≡ γ2 = 0.41146 · · · < 0.4115,

where Z = 298 + 326α. From (4.11), we have

0 ≤ S(S − 1)(S − 2) + γ2 (λ1 − λ2 )2 S 2 t − (4.13)

≤ S(S − 1)(S − 2) + γ2 (λ1 − λ2 )2 S 2 t −

79 (λ1 − λ2 )2 (λ1 λ2 )2 , S 450 λ21 + λ22

79 Sf 450


13

then it follows that (4.14)

79 2 2 (λ1 λ2 ) . S(S − 1)(S − 2) ≥ (λ1 − λ2 ) −γ2 tS + 450 2

We next consider two cases: Case 1: λ1 (x)λ2 (x) > 0. From (4.14), we have S(S − 1)(S − 2) ≥ (λ1 − λ2 )2 (−γ2 tS 2 ) ≥ −Sγ2 tS 2 that is, S − 2 ≥ −γ2 S, then S≥

2 2 1 ≥ > . 1 + γ2 1 + 0.42 0.802

Case 2: λ1 (x)λ2 (x) ≤ 0. From (4.13), we obtain S(S − 1)(S − 2) + γ2 StS 2 ≥ S(S − 1)(S − 2) + γ2 (λ21 + λ22 )S 2 t

79 (λ1 − λ2 )2 S (λ1 λ2 )2 + γ2 (2λ1 λ2 )tS 2 450 λ21 + λ22 79 ≥ S(λ1 λ2 )2 + 2γ2 λ1 λ2 tS 2 450 450 2 2 3 (2γ2 tS 2 )2 γ tS , =− ≥− 79 79 2 4 × 450 S

≥

that is, 2 + 450 γ2 1 79 2 S≥ . 450 2 > 1.247456 > 1.2469 > 0.802 1 + γ2 + 79 γ2 It is a contradiction, hence we have proved S>

Part III: Claim: S >

10 7

if S >

1 . 0.802

1 . 0.802

Before we prove the above Claim, we will prove the following Lemma. Lemma 4.1. Let M be an n-dimensional complete self-shrinker without boundary and with polynomial volume growth in Rn+1 . If the squared norm S of the second fundamental form is constant, then for any constant δ > 0, c0 ≥ 0 and c1 satisfying (4.15)

(β + t)c0 δ = (δ − 1 + δc0 )2 ,

and β ≥ 0, there exists a point p0 ∈ M such that, at p0 ,

14


tS 2 (S − 2)

(4.16)

β 2 ≥ (2 − δt + c1 δ)Sf − (5 − 2δ + c1 δ + )A + (6 + δ + 2c1 δ − β)B 3 3 XX r 1 2β 2 ( λ2i hiik )2 . − − 3(1 + c0 )δ + 4 3t t S k i

Proof. From [6], we have Z

(4.17)

M

(A − 2B − Sf )e−

|X|2 2

dv = 0,

then for any constant c1 , we have Z Z 2 |X|2 − |X| c1 S 2 f e− 2 dv (4.18) c1 S(A − 2B)e 2 dv = M

M

since S is constant. From (3.3), we have Z |X|2 (1 − S)(cf32 − Sf4 )e− 2 dv ZM (4.19) X XX |X|2 = [(2A + B)S − 2cf3 λi h2ijk − 3c ( λ2i hiij )2 ]e− 2 dv, M

j

i,j,k

i

then Z

(4.20)

|X|2

(c1 S 2 f − tS 2 f4 + ctSf32 )e− 2 dv M Z X XX = 2cf3 λi h2ijk − (2A + B)S + 3c ( λ2i hiij )2 M

j

i,j,k

i

|X|2 + c1 S(A − 2B) e− 2 dv,

thus we have that there exists a point p0 ∈ M such that, at p0 , (4.21)

c1 S 2 f − tS 2 f4 + ctSf32 X XX = 2cf3 λi h2ijk − (2A + B)S + 3c ( λ2i hiij )2 + c1 S(A − 2B), j

i,j,k

i

then,

(4.22)

c1 S 2 f − tS(Sf4 − f32 ) X = (1 − c)tSf32 + 2cf3 λi h2ijk − (2A + B)S i,j,k

XX + 3c ( λ2i hiij )2 + c1 S(A − 2B). j

i


Putting c = 1 + c0 with c0 ≥ 0, we get

(c1 S 2 − tS 2 )f = −c0 tSf32 + 2(c0 + 1)f3

(4.23)

X i,j,k

λi h2ijk − (2A + B)S

XX + 3(1 + c0 ) ( λ2i hiij )2 + c1 S(A − 2B). j

i

For any positive constant δ > 0, we have from (4.22), δtSf = c1 δSf + c0 δtf32 − 2(1 + c0 ) (4.24)

Putting

f3 X δ λi h2ijk S i,j,k

+ (2δ − c1 δ)A + (δ + 2c1 δ)B 1XX 2 λi hiij )2 . ( − 3(1 + c0 )δ S j i

1 uijkl = (hijkl + hjkli + hklij + hlijk ), 4 by a direct computation, we have X X 3X (hiijj − hjjii )2 u2ijkl + h2ijkl ≥ 4 i,j i,j,k,l i,j,k,l X 3X 2 = u + (λi − λj )2 λ2i λ2j ijkl (4.26) 4 i,j i,j,k,l X 3 = u2ijkl + (Sf4 − f32 ). 2

(4.25)

i,j,k,l

From (2.10) and (4.26), we obtain (4.27)

S(S − 1)(S − 2) + 3(A − 2B) ≥

X

i,j,k,l

3 u2ijkl + Sf. 2

By a direct calculation, we can get X 1 4 X 2 X 2 u2ijkl ≥ Sf + 2 λ ( λ hjji )2 − 2A 2 tS i i j j i,j,k,l (4.28) 1 X 2f3 X λi h2ijk + 2 ( + λi h2ijk )2 . S i,j,k S i,j,k Combining (4.27) and (4.28), we have

(4.29)

S(S − 1)(S − 2) + 3(A − 2B) 4 X 2 X 2 λ( λ hjji )2 ≥ 2Sf − 2A + 2 tS i i j j 2f3 X 1 X + λi h2ijk + 2 ( λi h2ijk )2 . S i,j,k S i,j,k

15

16


From (4.24), one has δtSf +

1 X 2f3 X 4 X 2 X 2 2 2 λ h + λ ( λ h ) + ( λi h2ijk )2 i ijk jji tS 2 i i j j S i,j,k S 2 i,j,k

= c1 δSf + c0 δtf32 − +

2[(1 + c0 )δ − 1]f3 X λi h2ijk S i,j,k

1 X ( λi h2ijk )2 + (2δ − c1 δ)A + (δ + 2c1 δ)B S2 i,j,k

(4.30)

X 4λ2j 1 X 2 − 3(1 + c0 )δ ( λi hiij )2 tS S i j [(1 + c0 )δ − 1]2 1 X ≥ c1 δSf + t − ( λi h2ijk )2 2 c0 δ tS +

i,j,k

+ (2δ − c1 δ)A + (δ + 2c1 δ)B X 1 X 4λ2j ( − 3(1 + c0 )δ)( λ2i hiij )2 . + S j tS i

Taking δ and c, such that, (β + t)c0 δ = [(c0 + 1)δ − 1]2 ,

(4.31)

with β ≥ 0, we have from Lemma 3.5 4 X 2 X 2 1 X 2f3 X δtSf + 2 λi h2ijk + 2 ( λi ( λj hjji )2 + λi h2ijk )2 tS i S S j i,j,k i,j,k X X 1 4 1 ( λ2 hiik )2 ≥ c1 δSf − β (A + 2B) − 3 3 k S + 2λ2k i i

(4.32)

+ (2δ − c1 δ)A + (δ + 2c1 δ)B X 1 X 4λ2j + − 3(1 + c0 )δ ( λ2i hiij )2 S j tS i

2β β )B = c1 δSf + (2δ − c1 δ − )A + (δ + 2c1 δ − 3 3 X 4 1 4λ2k 3(1 + c0 )δ X 2 + ( λi hiik )2 β − + 2 2 3 S + 2λ tS S k i k

β 2β ≥ c1 δSf + (2δ − c1 δ − )A + (δ + 2c1 δ − )B 3 3 XX r 2β 2 1 ( λ2i hiik )2 . − − 3(1 + c0 )δ + 4 3t t S k i From (4.29), we have


17

tS 2 (S − 2) + 3(A − 2B)

(4.33)

2β β )B ≥ 2Sf − 2A − δtSf + c1 δSf + (2δ − c1 δ − )A + (δ + 2c1 δ − 3 3 r XX 2β 2 1 + 4 ( λ2i hiik )2 , − − 3(1 + c0 )δ 3t t S k i

that is, tS 2 (S − 2) ≥ (2 − δt + c1 δ)Sf − (5 − 2δ + c1 δ + (4.34)

β )A 3

2β )B + (6 + δ + 2c1 δ − 3 r XX 2β 2 1 + 4 ( λ2i hiik )2 . − − 3(1 + c0 )δ 3t t S k i ✷

Taking 6 + δ + 2c1 δ − 2β = 5 − 2δ + c1 δ + β3 , we have from (4.15) that β = 3 6 , c0 = 17 and applying c1 δ + 3δ + 1, (β + t)c0 δ = ((c0 + 1)δ − 1)2 . Taking δ = 17 5 54 2 5 Lemma 4.1, we obtain β = 5 − t, c1 = − 17 − 17 t,

(4.35)

Putting g1 (t) = 4

tS 2 (S − 2) 7 8 22t )Sf − ( − ≥( − 5 r 5 5 1 2 54 + 4 ( − 1) − S 3 5t q

2 54 ( 3 5t

− 1) − 2t −

69 , 5

4t )(A − B) 3 2 69 X X 2 ( λi hiik )2 . − t 5 k i

we can obtain that

g1 (t) < 0, when t > 0.1978. Since (4.36)

√ 2 36 6 (g1 (t)) = 2 − q t > 0.14, then we have g1 (t) ≤ g1 (0.1978) < 0 when 0.1978 < t ≤

3 . 10

18


From Lemma 3.3 and Lemma 3.4, we have tS 2 (S − 2) 7 8 22t )Sf − ( − ≥( − 5 r 5 5 2 54 1 + 4 ( − 1) − S 3 5t

(4.37)

4t )(A − B) 3 2 69 X X 2 ( λi hiik )2 − t 5 k i

8 22t 7 4t 1 − α ≥( − )Sf − ( − ) (λ1 − λ2 )2 tS 2 5 r5 5 3 3 2 54 2 69 1 + 2α + 4 ( − 1) − − tSf 3 5t t 5 3 4t 7 = −( − )(1 − α)(λ1 − λ2 )2 tS 2 15 9 r 8 4 2 54 2 + + ( − 1) − − 9 5t 3 3 5t 3t r 4 46 8 2 54 α tSf. ( − 1) − − + 3 3 5t 3t 5

3 3 , the result is obvious true. If t ≤ 10 , we will obtain a contradiction. In If t > 10 3 this case, we have 0.198 ≤ t ≤ 10 . Putting r 8 4 2 54 2 (4.38) a(t) = + ( − 1) − − 9, 5t 3 3 5t 3t

8 b(t) = − 3

(4.39)

r

2 54 4 46 ( − 1) + + , 3 5t 3t 5

then we have (4.40)

tS 2 (S − 2) 7 4t ≥ −( − )(1 − α)(λ1 − λ2 )2 tS 2 + [a(t) − b(t)α]tSf. 15 9 √

′ 14 √12 Since a (t) = − 15t 2 −

5

(4.41)

6 −1+ 54 t2 5t

< 0, we have

52 4 3 a(t) ≥ a( ) = − + 10 9 3 √

′ Since b (t) = − 3t42 + √24

5

6 −1+ 54 t2 5t

r

70 ≈ 0.662834 > 0. 3

> 0 if t > 0.14, we have that b(t) is an increasing

3 ], then function of t ∈ [0.198, 10

(4.42)

3 614 8 b(t) ≤ b( ) = − 10 45 3

r

70 ≈ 0.763221 > 0. 3


19

Therefore we get tS 2 (S − 2) (4.43) 7 4t 3 3 ≥ −( − )(1 − α)(λ1 − λ2 )2 tS 2 + [a( ) − b( )α]tSf. 15 9 10 10 We next consider two cases: 3 3 ) − b( 10 )α ≤ 0. Case 1: a( 10 a(

3

)

In this case b( 103 ) ≤ α ≤ 1. Since λ1 , λ2 are the maximum and minimum of the 10 principal curvatures at any point of M, we obtain, for any j, λj + λ1 ≥ λ2 + λ1 ,

(λ1 − λj )(λ1 + λj ) ≥ (λ1 − λj )(λ1 + λ2 ).

So we get

λ2j − (λ1 + λ2 )λj ≤ −λ1 λ2 ,

and

f4 − (λ1 + λ2 )f3 ≤ −λ1 λ2 S,

then

Sf = Sf4 − f32 ≤ −f32 + (λ1 + λ2 )Sf3 − λ1 λ2 S 2 ,

(4.44) (4.45)

Sf ≤

(λ1 − λ2 )2 2 S . 4

From (4.43) and (4.45), we have (4.46)

tS 2 (S − 2)

7 4t 3 3 (λ1 − λ2 )2 2 − )(1 − α)(λ1 − λ2 )2 tS 2 + [a( ) − b( )α] tS . 15 9 10 10 4 3 3 ) − b( 10 )α ≤ 0, using −2λ1 λ2 ≤ λ21 + λ22 ≤ S, we see from (4.46) Since a( 10 7 4t 1 3 S − 2 ≥ −2( − ) + a( ) 15 9 2 10 (4.47) 7 4t 1 3 + 2( − ) − b( ) α S. 15 9 2 10 Since 7 4t 1 3 2( − ) − b( ) 15 9 2 10 (4.48) 4 3 1 2 1 7 ≥ 2( − × ) − × 0.77 = − × 0.77 > 0, 15 9 10 2 3 2 we have from (4.47) 3 a( 10 ) 7 4t (4.49) S − 2 ≥ −2( − ) 1 − 3 S. 15 9 b( 10 ) ≥ −(

On the other hand,

3 a( 10 ) ≈ 0.86847 > 0.86, 3 b( 10 )

20


then from (4.49), we see 7 − 15 7 ≥ −2( − 15 > 0.1061S,

S − 2 ≥ −2( (4.50)

4t )(1 − 0.86)S 9 4 × 0.198) × 0.14S 9

hence (4.51)

S>

10 2 > 1.8 > . 1 + 0.1061 7

This is impossible. 3 3 ) − b( 10 )α > 0. Case 2: a( 10 In this case

3 ) a( 10 3 b( 10 )

> α ≥ 0. From Lemma 3.2 and (4.43), we obtain

tS 2 (S − 2)

(4.52)

4t 3 3 7 2 2 ≥ −( − )(1 − α)(λ1 − λ2 ) tS + a( ) − b( )α tSf 15 9 10 10 4t 7 ≥ −( − )(1 − α)(λ1 − λ2 )2 tS 2 9 15 3 3 (λ1 − λ2 )2 + a( ) − b( )α (λ1 λ2 )2 tS. 10 10 λ21 + λ22

Putting y = − λ1Sλ2 , we have − 12 ≤ y = − λ1Sλ2 ≤ (4.52) that

2

2

1 λ1 +λ2 2 S

≤ 12 , then we infer from

S−2

(4.53)

4t 3 3 7 ≥ −( − )(1 − α)(1 + 2y)S + a( ) − b( )α (1 + 2y)(−y)2S 15 9 10 10 7 4t 3 = −( − )(1 + 2y) + a( )(1 + 2y)y 2 15 9 10 7 4t 3 2 + ( − )(1 + 2y) − b( )(1 + 2y)y α S. 15 9 10

Defining two functions ρ(y) and ̺(y) by (4.54)

ρ(y) = −(

(4.55)

̺(y) = (

4t 3 7 − )(1 + 2y) + a( )(1 + 2y)y 2, 15 9 10

7 4t 3 − )(1 + 2y) − b( )(1 + 2y)y 2. 15 9 10


21

Since n > 2, we have 1 + 2y > 0, then 4t 3 7 ̺(y) = ( − )(1 + 2y) − b( )(1 + 2y)y 2 15 9 10 7 4t 3 2 = (1 + 2y) − − b( )y 15 9 10 (4.56) 7 4 3 1 > (1 + 2y) − × − 0.7633 × ] 15 9 10 4 = (1 + 2y) × 0.142508 > 0. 4t 3 7 − )(1 + 2y) + a( )(1 + 2y)y 2 15 9 10 4t 3 2 7 = (1 + 2y)[− + + a( )y ] 15 9 10 7 4 3 1 < (1 + 2y)[− + × + 0.663 × ] 15 9 10 4 = (1 + 2y) × (−0.1676) < 0.

ρ(y) = −( (4.57)

By a direct calculation, we obtain 7 4t 3 ′ ρ (y) = −2( − ) + a( )[2y + 6y 2] 15 9 10 4t 3 3 7 = −2[ − + a( )y + 3a( )y 2] 15 9 10 10 (4.58) 7 4 3 1 3 < −2[ − × − a( ) 15 9 10 12 10 1 1 × 0.663] < 0, < −2[ − 3 12 it follows that 1 7 4t 1 3 (4.59) ρ(y) ≥ ρ( ) = −2( − ) + a( ). 2 15 9 2 10 From the above arguments, we have S − 2 ≥ (ρ(y) + ̺(y)α)S 7 4t 1 3 ≥ (−2( − ) + a( ))S 15 9 2 10 (4.60) 14 8 1 > − S + (S − 1) + × 0.66 × S 15 9 2 8 8 2 = − S + 0.33S − > 0.28S − , 45 9 9 then 10 10 9 (4.61) S> > 1.54 > . 1 − 0.28 7 It is a contradiction. 3 , that is, S > 10 if S > 1. This completes the proof of Hence, we have t > 10 7 Theorem 1.1. ✷

22


References [1] U. Abresch and J. Langer, The normalized curve shortening flow and homothetic solutions, J. Diff. Geom., 23 (1986), 175-196. [2] S. Angenent, Shrinking doughnuts, In nonlinear diffusion equations and their equilibrium states, Birkha¨ user, Boston-Basel-Berlin, 7, 21-38, 1992. [3] H. -D. Cao and H. Li, A gap theorem for self-shrinkers of the mean curvature flow in arbitrary codimension, to appear in Calc. Var. Partial Differential Equations, arXiv:1101.0516. [4] Q. -M. Cheng and H. C. Yang, Chern’s conjecture on minimal hypersurfaces, Math. Z., 227 (1998), 377-390. [5] T. H. Colding and W. P. Minicozzi II, Generic mean curvature flow I; Generic singularities, Ann. of Math., 175 (2012), 755-833. [6] Q. Ding and Y. L. Xin, The rigidity theorems of self shrinkers, to appear in Trans. Amer. Math. Soc., arXiv:1105.4962. [7] Q. Ding and Y. L. Xin, Volume growth, eigenvalue and compactness for self-shrinkers, to appear in Asia J. Math., arXiv:1101.1411. [8] G. Huisken, Flow by mean curvature of convex surfaces into spheres, J. Diff. Geom., 22 (1984), 237-266. [9] G. Huisken, Asymptotic behavior for singularities of the mean curvature flow, J. Diff. Geom., 31 (1990), 285-299. [10] G. Huisken, Local and global behaviour of hypersurfaces moving by mean curvature, Differential geometry: partial differential equations on manifolds, 175-191, Proc. Sympos. Pure Math., 54 (1993), Amer. Math. Soc.. [11] Nam Q. Le and N. Sesum, Blow-up rate of the mean curvature during the mean curvature flow and a gap theorem for self-shrinkers, Comm. Anal. Geom., 19 (2011), 1-27. [12] H. Li and Y. Wei, Classification and rigidity of self-shrinkers in the mean curvature flow, to appear in J. Math. Soc. Japan, arXiv:1201.4623. Qing-Ming Cheng, Department of Applied Mathematics, Faculty of Sciences , Fukuoka University, 814-0180, Fukuoka, Japan, [email protected] Guoxin Wei, School of Mathematical Sciences, South China Normal University, 510631, Guangzhou, China, [email protected]