1 introduction↩︎

This paper is concerned with study on the possible singularities of the mean curvature flow, which is one of the most important problems in the research on the mean curvature flow. By making use of Huisken’s monotonicity formula, we know that a solution to the flow is asymptotically self-similar near a given type I singularity. Thus, it is modeled by self-shrinking solutions of the flow. An \(n\)-dimensional hypersurface \(X: M\rightarrow \mathbb{R}^{n+1}\) in the \((n+1)\)-dimensional Euclidean space \(\mathbb{R}^{n+1}\) is called a self-shrinker if it satisfies \[H+ \langle X, N\rangle=0,\] where \(N\) and \(H\) denote the unit normal vector and mean curvature of this hypersurface. Since self-shrinkers describe all possible blow-ups at a given singularity, self-shrinkers play an important role in the study on singularities of the mean curvature flow.
Abresch and Langer [1] classified closed self-shrinking curves in \(\mathbb{R}^2\) and showed that the round circle is the only embedded self-shrinker. Huisken [2] proved an \(n\)-dimensional compact self-shrinker in \(\mathbb{R}^{n+1}\) with mean curvature \(H\geq 0\) is isometric to the sphere \(S^n(\sqrt{n})\). Furthermore, Drugan [3] constructed an immersed, non-embedded self-shrinker of genus \(0\). Hence, we know that self-shrinkers do not share common features with the Hopf theorem: a topological sphere with constant mean curvature in \(\mathbb{R}^3\) is the round sphere. But according to the theorem of Brendle [4]: if \(X:M^2 \to \mathbb{R}^{3}\) is a compact embedded self-shrinker in \(\mathbb{R}^3\) with genus \(0\), then \(X:M^2 \to \mathbb{R}^{3}\) is the round sphere. Thus self-shrinkers share common features with Alexandrov theorem on the embedded sphere with constant mean curvature in \(\mathbb{R}^3\). On the other hand, Huisken [5], Colding and Minicozzi [6] gave a complete classification for \(n\)-dimensional complete embedded self-shrinkers in \(\mathbb{R}^{n+1}\) with mean curvature \(H\geq 0\) and with polynomial volume growth.
Ding and Xin [7] and Cheng and Zhou [8] have proven that an \(n\)-dimensional complete self-shrinker has polynomial volume growth if and only if it is proper. Furthermore, it is also known that there exist complete self-shrinkers without polynomial volume growth in \(\mathbb{R}^{n+1}\) in Halldorsson [9].
Since many formulas on self-shrinkers are very similar to formulas on minimal hypersurfaces in the unit sphere in the some sense, one hopes that self-shrinkers share some common properties of minimal hypersurfaces in the unit sphere. It is well-known, that for minimal hypersurfaces in the unit sphere, the following Chern problems are very important: Chern problems. For \(n\)-dimensional compact minimal hypersurfaces in \(S^{n+1}(1)\) with constant squared norm \(S\) of the second fundamental form, is the following true?

1. \(S\leq c(n)\), where \(c(n)\) is a constant depending only on dimension \(n\),

2. the values of \(S\) of the squared norm of the second fundamental form are discrete,

3. the values of \(S\) should determine the hypersurfaces up to a rigid motion in the ambient sphere \(S^{n+1}(1)\).

For self-shrinkers, the following conjecture is well-known: Conjecture. An \(n\)-dimensional complete self-shrinker \(X: M\rightarrow \mathbb{R}^{n+1}\) with constant squared norm of the second fundamental form is isometric to one of

1. \(S^{n}(\sqrt{n})\),

2. \(\mathbb{R}^{n}\),

3. \(S^k (\sqrt k)\times \mathbb{R}^{n-k}\), \(1\leq k\leq n-1\).

Cheng and Ogata [10] confirmed this conjecture for \(n=2\). Namely, they have proven the following:

We will resolve the above conjecture, affirmatively, if the scalar curvature \(R\) satisfies \(R\geq -\dfrac{7}{5}\).

For minimal hypersurfaces in the unit sphere, we know that the scalar curvature is constant if and only if the squared norm of the second fundamental form is constant, thanks to the Gauss equation. But self-shrinkers do not share this property. In [15], Guo proved that compact self-shrinkers with constant scalar curvature in \(\mathbb{R}^{n+1}\) are isometric to the sphere \(S^{n}(\sqrt n)\). Since an \(n\)-dimensional compact self-shrinker in \(\mathbb{R}^{n+1}\) must have a convex point, the constant scalar curvature must be positive at this point. By making use of Stokes formula, Guo [15] proved the scalar curvature \(R= (n-1)\). Thus, Chern type problems on compact self-shrinkers were resolved by Guo. On the other hand, study on \(n\)-dimensional complete non-compact self-shrinkers in \(\mathbb{R}^{n+1}\) is more important. Luo, Sun and Yin [16] have proven that an \(n\)-dimensional complete self-shrinker in \(\mathbb{R}^{n+1}\) with polynomial volume growth and positive constant scalar curvature is isometric to one of

1. \(S^{k}(\sqrt k)\times \mathbb{R}^{n-k}, \;1\leq k\leq n-1\),

2. \(S^{n}(\sqrt{n})\).

In fact, according to the Gauss equation, if the scalar curvature \(R\) is positive, we have \(H\neq 0\) because of \(H^2-S=R\). Hence, according to the results and proof due to Colding and Minicozzi [6], we can remove the condition that scalar curvature is constant, thanks to the Gauss equation.

Furthermore, we will prove the following:

By making use of the generalized maximum principle due to Cheng and Peng [17], we obtain the following:

Furthermore, we get a pinching theorem on the mean curvature of complete self-shrinkers with non-negative scalar curvature.

2 Preliminaries↩︎

Let \(X: M\rightarrow\mathbb{R}^{n+1}\) be an \(n\)-dimensional connected hypersurface in the \((n+1)\)-dimensional Euclidean space \(\mathbb{R}^{n+1}\). We choose a local orthonormal frame field \(\{e_A\}_{A=1}^{n+1}\) in \(\mathbb{R}^{n+1}\) with dual coframe field \(\{\omega_A\}_{A=1}^{n+1}\), such that, restricted to \(M\), \(e_1,\cdots, e_n\) are tangent to \(M^n\). Then we have \[dX=\sum_i\limits \omega_i e_i, \; \; de_i=\sum_j\limits \omega_{ij}e_j+\omega_{i n+1}e_{n+1}, \; de_{n+1}=\sum_i\omega_{n+1 i}e_i,\] where \(\omega_{ij}\) is the Levi-Civita connection of \(X: M\rightarrow\mathbb{R}^{n+1}\). Because of \(\omega_{n+1}=0\) along \(M\), one has \[\label{2461-2} \omega_{in+1}=\sum_j h_{ij}\omega_j,\quad h_{ij}=h_{ji}.\tag{1}\] \[H= \sum_i\limits h_{ii}, \; \;\vec{h}=\sum_{i,j}h_{ij}\omega_i\otimes\omega_je_{n+1}\] are called the mean curvature and the second fundamental form, respectively. Setting \(S=\sum_{i,j}\limits (h_{ij})^2\), components \(R_{ijkl}\) of the curvature tensor, components \(R_{ij}\) of the Ricci curvature tensor and the scalar curvature \(R\) are given by \[\label{eq:2462} R_{ijkl}=h_{ik}h_{jl}-h_{il}h_{jk}, \quad R_{ij}=Hh_{ij}-\sum_kh_{ik}h_{kj}, \quad R=H^2-S.\tag{2}\]

Defining the covariant derivative of \(h_{ij}\) by \[\sum_{k}h_{ijk}\omega_k=dh_{ij}+\sum_kh_{ik}\omega_{kj} +\sum_k h_{kj}\omega_{ki},\] we obtain the Codazzi equations \[\label{eq:2463} h_{ijk}=h_{ikj}.\tag{3}\] Defining \[\sum_lh_{ijkl}\omega_l=dh_{ijk}+\sum_lh_{ljk}\omega_{li} +\sum_lh_{ilk}\omega_{lj}+\sum_l h_{ijl}\omega_{lk},\] we have the following Ricci identities: \[\label{eq:2464} h_{ijkl}-h_{ijlk}=\sum_m h_{mj}R_{mikl}+\sum_m h_{im}R_{mjkl}.\tag{4}\] For a smooth function \(f\), the \(\mathcal{L}\)-operator is defined by \[\mathcal{L}f=\Delta f-\langle X,\nabla f\rangle,\] where \(\Delta\) and \(\nabla\) denote the Laplacian and the gradient operator, respectively. By a direct calculation, we can derive the following formulas, which can also be found in [6], [13].

The following generalized maximum principle for \(\mathcal{L}\)-operator due to Cheng and Peng [17] will play an important role. Generalized maximum principle for \(\mathcal{L}\)-operator. Let \(X : M^{n}\to \mathbb{R}^{n+p}\) be a complete self-shrinker with Ricci curvature bounded from below. Let \(f\) be any \(C^{2}\)-function bounded from above on this self-shrinker. Then, there exists a sequence of points \(\{p_{m}\}\subset M^{n}\), such that \[\lim_{m\rightarrow\infty} f(p_{m})=\sup f,\quad \lim_{m\rightarrow\infty} |\nabla f|(p_{m})=0,\quad \limsup_{m\rightarrow\infty}\mathcal{L}f(p_{m})\leq 0.\]

3 Proof of theorem 1.2↩︎

In order to prove our theorem 1.2, we need to remove the condition of polynomial volume growth in the theorem 1.1 of Cheng and Wei [13] as follows.

We will use the same notation as in [13]. The following point-wise estimates can be found in Cheng and Wei [13]. Define \(\lambda_1=\max\limits_i\{\lambda_i\}\) and \(\lambda_2=\min\limits_i\{\lambda_i\}\) at each point.

Since \(S\) is constant, we have \[\label{eq:3462} \begin{align} &\sum_{i,j,k}h_{ijk}^2= S(S-1),\;\;\mathcal{L} h_{ij}=(n-S)h_{ij},\\ &\sum_{i,j,k,l}h_{ijkl}^2=S(S-1)(S-2) +3(A-2B), \end{align}\tag{5}\] Defining \(u_{ijkl}\) by \[u_{ijkl}:=\frac{1}{4}(h_{ijkl}+h_{jkli}+h_{klij}+h_{lijk} ),\] a direct computation yields \[\label{eq:3463} \begin{align} \sum_{i,j,k,l}h_{ijkl}^2 = \sum_{i,j,k,l}u_{ijkl}^2 +\frac{3}{2}\bigl( Sf_4-f_3^2\bigl). \end{align}\tag{6}\] If \(S> 1\), defining \(S-1=tS\), we know \(0<t\leq \dfrac27\). Thus, we obtain the following:

Proof of theorem 3.1. Since \(S\) is constant, we know that Ricci curvature is bounded from the Gauss equation and principal curvatures \(\lambda_i\), \(h_{ij}\), \(h_{ijk}\) and \(h_{ijkl}\) for any \(i, j, k, l\) are bounded according to the formula (3.2). Thus, it is obvious that the function \(F=\dfrac14Sf_4-\dfrac 16 c f_3^2\) for a given constant \(c\) is bounded and \[\begin{align} \mathcal{L}F&= S(1-S)f_4+S(2A+B)\\ &-c\bigl((1-S)f_3^2+2f_3\sum_{i,j,k}\lambda_ih_{ijk}^2 +3\sum_{j}(\sum_i\lambda_i^2h_{iij})^2\bigl). \end{align}\] By applying the generalized maximum principle for \(\mathcal{L}\)-operator to the function \(F\), there exists a sequence \(\{p_m\} \subset M\) such that \[\lim_{m\to\infty}F(p_m)=\sup F, \;\;\lim_{m\to\infty}|\nabla F|(p_m)=0,\] \[\lim\sup_{m\to\infty} \mathcal{L}F(p_m)\leq 0.\] Since \(\lambda_i\), \(h_{ijk}\) and \(h_{ijkl}\) are bounded, we can assume that \(\{\lambda_i(p_m)\}\), \(\{h_{ijk}(p_m)\}\) and \(\{h_{ijkl}(p_m)\}\) converge if necessary taking a subsequence of \(\{p_m\}\). For simple, we still use \(\lambda_i\), \(h_{ijk}\) and \(h_{ijkl}\) to denote limits of \(\{\lambda_i(p_m)\}\), \(\{h_{ijk}(p_m)\}\) and \(\{h_{ijkl}(p_m)\}\), respectively and all computations are processed under limits. Hence, we have \[\label{eq:3469} -(S-1)\bigl(Sf_4-cf_3^2\bigl)\leq 2cf_3\sum_{i,j,k}\lambda_ih_{ijk}^2- S(2A+B)+3c\sum_{j}(\sum_i\lambda_i^2h_{iij})^2.\tag{7}\] If \(S\leq 1\), then our assertion is true. Next we assume \(1<S\leq \dfrac75\).
Taking \(c=2\) in (7 ), we obtain \[0\leq (S-1)\bigl(\dfrac12Sf_4-f_3^2\bigl)+2\sum_{i,j,k}\lambda_ih_{ijk}^2 f_3-\dfrac S2(2A+B)+3\sum_{j}(\sum_ih_{iij})^2.\] Therefore, we can make use of the same arguments (word by word) as in the part I of proof of the theorem 1.1 of Cheng and Wei [13] (page 4903) in place of their lemma 3.1 by the above inequality to prove \(S>\dfrac65\) if \(S>1\).
From now, we prove \(S>\dfrac75\) if \(S>\dfrac65\). We define a function \(\phi\) by \[\phi=-\dfrac{13}{3}tSf-\frac{2f_3}{S} \sum_{i,j,k}\lambda_ih_{ijk}^2 -\frac{1}{S^2}\bigl(\sum_{i,j,k}\lambda_ih_{ijk}^2 \bigl)^2 -\dfrac{4}{tS^2}\sum_{j}\lambda_j^2(\sum_i\lambda_i^2h_{iij})^2.\\\] Letting \(c=1+c_1\) in (7 ) with \(c_1=\dfrac{22+3t-\sqrt{(22+3t)^2-400}}{26}\), we have \[\label{eq:34611} (\dfrac{10}{13}+c_1)^2=\dfrac{3(14+t)}{13}c_1.\tag{8}\] From (7 ), we obtain \[\begin{align} \phi&\leq -\dfrac{13}{3}(2A+B)-\dfrac{13c_1t}{3}f_3^2+(\dfrac{10}3+\dfrac{13}3c_1)\frac{2f_3}{S} \sum_{i,j,k}\lambda_ih_{ijk}^2 \\ &+\dfrac{13(1+c_1)}S\sum_{j}(\sum_i\lambda_i^2h_{iij})^2-\frac{1}{S^2}\bigl(\sum_{i,j,k}\lambda_ih_{ijk}^2 \bigl)^2 -\dfrac{4}{tS^2}\sum_{j}\lambda_j^2(\sum_i\lambda_i^2h_{iij})^2\\ &\leq -\dfrac{13}{3}(2A+B)+\bigl\{\dfrac{13}{3tc_1}(\dfrac{10}{13}+c_1)^2-1\bigl\} \dfrac{1}{S^2}\bigl(\sum_{i,j,k}\lambda_ih_{ijk}^2\bigl)^2 \\ &+\dfrac{13(1+c_1)}S\sum_{j}(\sum_i\lambda_i^2h_{iij})^2 -\dfrac{4}{tS^2}\sum_{j}\lambda_j^2(\sum_i\lambda_i^2h_{iij})^2.\\ \end{align}\] According to the lemma 3.1, we get \[\begin{align} \phi&\leq -\dfrac{13}{3}(2A+B)+\dfrac{13(1+c_1)}S\sum_{j}(\sum_i\lambda_i^2h_{iij})^2 -\dfrac{4}{tS^2}\sum_{j}\lambda_j^2(\sum_i\lambda_i^2h_{iij})^2\\ &+\bigl\{\dfrac{13}{3c_1}(\dfrac{10}{13}+c_1)^2-t\bigl\} \biggl[\frac{1}{3}(A+2B) -\frac{4}{3}\sum_j\frac{1}{S+2\lambda_j^2}\big( \sum_i \lambda_i^2h_{iij}\bigl)^2\biggl] \\ &=-4A+5B+\sum_{j}\bigl(\dfrac{13(1+c_1)}S -\dfrac{4\lambda_j^2}{tS^2}-\dfrac {56}{3(S+2\lambda_j^2)}\bigl)\big( \sum_i \lambda_i^2h_{iij}\bigl)^2,\\ \end{align}\] where in the last equality, we used (8 ). By making use of \(a^2+b^2\geq 2ab\), we have \[\label{eq:34612} \begin{align} \phi&\leq -4A+5B+\dfrac{\eta(t)}S\sum_{j}\big( \sum_i \lambda_i^2h_{iij}\bigl)^2.\\ \end{align}\tag{9}\] where \(\eta(t)\) is defined by \[\begin{align} \eta(t)&= 13(1+c_1)+\dfrac{2}{t}-8\sqrt{\dfrac{7}{3t}}\\ &=\bigl(24+\dfrac{3t-\sqrt{(22+3t)^2-400}}{2}+\dfrac{2}{t} -\dfrac{8\sqrt {21}}{3}\dfrac1{\sqrt{t}}\bigl). \end{align}\] Since \(\dfrac{d\eta(t)}{dt}=0\) only has one real root \(t_1=0.109566\cdots\), then \(\eta(t)\) is an increasing function if \(\dfrac16\leq t \leq \dfrac 27\). Thus, we have \(1.1628<\eta(1/6)\leq \eta(t)\leq \eta(2/7)\simeq 3.033833811\). From the lemma 3.1 and (9 ), we obtain \[\label{eq:34613} \begin{align} \phi&\leq -4A+5B+t\eta(t) \frac{1+2\alpha}{3}Sf.\\ \end{align}\tag{10}\] Hence, we have \[\begin{align} &-(\dfrac{13+\eta(t)}{3}+ \dfrac{2\eta(t)\alpha}3)tSf-\frac{2f_3}{S} \sum_{i,j,k}\lambda_ih_{ijk}^2 -\frac{1}{S^2}\bigl(\sum_{i,j,k}\lambda_ih_{ijk}^2 \bigl)^2 \\&-\dfrac{4}{tS^2}\sum_{j}\lambda_j^2(\sum_i\lambda_i^2h_{iij})^2\leq -4A+5B.\\ \end{align}\] Letting \(a=\dfrac{13+\eta(t)}{3}\) and \(b=\dfrac{2\eta(t)}3\), we have \[\label{eq:34614} \begin{align} &-(a+ b\alpha)tSf-\frac{2f_3}{S} \sum_{i,j,k}\lambda_ih_{ijk}^2 -\frac{1}{S^2}\bigl(\sum_{i,j,k}\lambda_ih_{ijk}^2 \bigl)^2 \\&-\dfrac{4}{tS^2}\sum_{j}\lambda_j^2(\sum_i\lambda_i^2h_{iij})^2\leq -4A+5B.\\ \end{align}\tag{11}\] If \[-\frac{2f_3}{S} \sum_{i,j,k}\lambda_ih_{ijk}^2 -\frac{1}{S^2}\bigl(\sum_{i,j,k}\lambda_ih_{ijk}^2 \bigl)^2 -\dfrac{4}{tS^2}\sum_{j}\lambda_j^2(\sum_i\lambda_i^2h_{iij})^2> 0,\] \[\label{eq:34615} \begin{align} &-atSf\leq \dfrac{a}{a+ b\alpha}(-4A+5B),\\ \end{align}\tag{12}\] because of \(1.1628<\eta(1/6)\). From (6 ) and the above inequality, we obtain \[\begin{align} &(\dfrac32-at)Sf\leq \dfrac{a}{a+ b\alpha}(-4A+5B)+tS^2(S-2)+3(A-2B)\\ &=-\dfrac{a}{a+ b\alpha}(\dfrac{13}{3}(A-B)-\dfrac13(A+2B))\\ &+4(A-B)-(A+2B)+tS^2(S-2),\\ \end{align}\] that is, \[\begin{align} &(\dfrac32-at)Sf \leq (4-\dfrac{a}{a+ b\alpha}\dfrac{13}{3})(A-B)+tS^2(S-2).\\ \end{align}\] The lemma 3.1 and the above inequality yield \[\label{hskwomjl} \begin{align} &(\dfrac3{2t}-a)\dfrac{(\lambda_1-\lambda_2)^2}{\lambda_1^2+\lambda_2^2}(\lambda_1\lambda_2)^2 \leq (4-\dfrac{a}{a+ b\alpha}\dfrac{13}{3})\dfrac {1-\alpha}3(\lambda_1-\lambda_2)^2S+S(S-2).\\ \end{align}\tag{13}\] Setting \(w=\dfrac{b}{a}\), we have \[\begin{align} &\bigl(12-\dfrac{13}{1+w\alpha}\bigl)(1-\alpha)\\ &=\bigl(12-\dfrac{13}{1+ w\alpha}\bigl)\bigl(1+w-(1+ w\alpha)\bigl)\dfrac{1}{w}\\ &=\dfrac1w\bigl(12(1+w)+13-12(1+w\alpha)-\dfrac{13(1+w)}{1+ w\alpha}\bigl)\\ &\leq \dfrac1w\bigl(12(1+w)+13-4\sqrt{39(1+w)}\bigl)=\dfrac1w(2\sqrt{3(1+w)}-\sqrt{13})^2. \end{align}\] Hence, we get from (10 ) \[\label{eq:34619} \begin{align} &(\dfrac3{2t}-a)\dfrac{(\lambda_1-\lambda_2)^2}{\lambda_1^2+\lambda_2^2}(\lambda_1\lambda_2)^2\\ &\leq\dfrac1{9w}(2\sqrt{3(1+w})-\sqrt{13})^2(\lambda_1-\lambda_2)^2S+S(S-2). \end{align}\tag{14}\] From \(\eta(2/7)\simeq3.033833811\), we have \[a=\dfrac{13+\eta(2/7)}{3}\simeq5.34461127, \;\;b=\dfrac{2\eta(2/7)}3\simeq2.02255587, \;\; w\simeq0.378429.\] \[\begin{align} &-0.095\dfrac{(\lambda_1-\lambda_2)^2}{\lambda_1^2+\lambda_2^2}(\lambda_1\lambda_2)^2 \leq0.06255(\lambda_1-\lambda_2)^2S+S(S-2). \end{align}\] If \(\lambda_1\lambda_2\geq 0\), we have from (3.15) \[\begin{align} &-0.095\dfrac14S \leq0.06255S+(S-2). \end{align}\] Hence, \[S> 1.84>\dfrac75.\] If \(\lambda_1\lambda_2< 0\), we have from (3.15) \[\begin{align} &-0.095\dfrac12S \leq0.1251S+(S-2). \end{align}\] We infer \[S> 1.7>\dfrac75.\] If \[-\frac{2f_3}{S} \sum_{i,j,k}\lambda_ih_{ijk}^2 -\frac{1}{S^2}\bigl(\sum_{i,j,k}\lambda_ih_{ijk}^2 \bigl)^2 -\dfrac{4}{tS^2}\sum_{j}\lambda_j^2(\sum_i\lambda_i^2h_{iij})^2\leq 0,\] since \(1.1628<\eta(1/6)\), we infer \[\label{qvesuazi} \begin{align} &-atSf-\frac{2f_3}{S} \sum_{i,j,k}\lambda_ih_{ijk}^2 -\frac{1}{S^2}\bigl(\sum_{i,j,k}\lambda_ih_{ijk}^2 \bigl)^2 \\&-\dfrac{4}{tS^2}\sum_{j}\lambda_j^2(\sum_i\lambda_i^2h_{iij})^2 \leq \dfrac{a}{a+ b\alpha}(-4A+5B).\\ \end{align}\tag{15}\] In view of the proposition 3.1 and the definition of \(\phi\), we conclude \[\begin{align} &(2-at)Sf\\ &\leq \dfrac{a}{a+ b\alpha}(-4A+5B)+S(S-1)(S-2) +5A-6B\\ &=\dfrac13\bigl(16-\dfrac{13a}{a+ b\alpha}\bigl)(A-B)+tS^2(S-2) -\dfrac13(1-\dfrac{a}{a+ b\alpha})(A+2B)\\ &\leq\dfrac13\bigl(16-\dfrac{13a}{a+ b\alpha}\bigl)(A-B) +tS^2(S-2). \end{align}\] In the second equality and the last inequality, we used \[-4A+5B=-\dfrac{13}3(A-B)+\dfrac13(A+2B), \quad 5A-6B=\dfrac{16}{3}(A-B)-\dfrac{1}3(A+2B)\] and \(A+2B\geq 0\), respectively. Thus, we have \[\begin{align} (2-at)Sf &\leq\dfrac13\bigl(16-\dfrac{13a}{a+ b\alpha}\bigl)(A-B) +tS^2(S-2). \end{align}\] From the lemma 3.1, we obtain \[\label{eq:34617} \begin{align} &(2-at)Sf\\ &\leq\dfrac13\bigl(16-\dfrac{13a}{a+ b\alpha}\bigl)\dfrac {1-\alpha}3(\lambda_1-\lambda_2)^2tS^2 +tS^2(S-2). \end{align}\tag{16}\] From \(w=\dfrac{b}{a}\), we have \[\label{eq:34618} \begin{align} &\bigl(16-\dfrac{13}{1+w\alpha}\bigl)(1-\alpha)\\ &=\dfrac1w\bigl(16(1+w)+13-16(1+w\alpha)-\dfrac{13(1+w)}{1+ w\alpha}\bigl)\\ &\leq \dfrac1w\bigl(16(1+w)+13-8\sqrt{13(1+w)}\bigl)=\dfrac1w(4\sqrt{1+w}-\sqrt{13})^2. \end{align}\tag{17}\] In view of \[f=f_4-\frac{f_3^2}{S}\geq \dfrac{(\lambda_1-\lambda_2)^2}{\lambda_1^2+\lambda_2^2}(\lambda_1\lambda_2)^2,\] in the lemma 3.1, (16 ) and (17 ), we have \[\label{zrqobwvk} \begin{align} &(\dfrac2t-a)\dfrac{(\lambda_1-\lambda_2)^2}{\lambda_1^2+\lambda_2^2}(\lambda_1\lambda_2)^2\\ &\leq\dfrac1{9w}(4\sqrt{1+w}-\sqrt{13})^2(\lambda_1-\lambda_2)^2S+S(S-2). \end{align}\tag{18}\] From \(\eta(2/7)\simeq3.033833811\), we have \[a=\dfrac{13+\eta(2/7)}{3}\simeq5.34461127, \;\;b=\dfrac{2\eta(2/7)}3\simeq2.02255587, \;\; w\simeq0.378429.\] If \(\lambda_1\lambda_2\leq 0\), we get from (14 ) \[\begin{align} &(\dfrac2t-a)(\lambda_1\lambda_2)^2 +\dfrac2{9w}(4\sqrt{1+w}-\sqrt{13})^2S\lambda_1\lambda_2\leq \dfrac1{9w}(4\sqrt{1+w}-\sqrt{13})^2S^2+S(S-2). \end{align}\] Thus, we have \[\begin{align} &-\dfrac{\bigl\{\dfrac1{9w}(4\sqrt{1+w}-\sqrt{13})^2\bigl\}^2}{(\dfrac2t-a)}S \leq \dfrac1{9w}(4\sqrt{1+w}-\sqrt{13})^2S+(S-2). \end{align}\] Therefore, we conclude \[\begin{align} &-0.07377S \leq 0.349294S+(S-2), \end{align}\] that is, \(S\geq 1.40541873>1+\dfrac25\).
If \(\lambda_1\lambda_2>0\), we get \[\begin{align} &0\leq \dfrac1{9w}(4\sqrt{1+w}-\sqrt{13})^2S^2+S(S-2). \end{align}\] Hence, we have \[\begin{align} &0\leq 0.349294S+(S-2).\\ \end{align}\] Hence \(S\geq 1.4822566>1+\dfrac25\). We complete the proof of the theorem 3.1.

Proof of theorem 1.2. According to the above theorem 3.1, we can assume \(S>\dfrac75\). Next, we will prove that \(S>\dfrac75\) does not happen. If \(S>\dfrac75\), since \(S\) is constant, for this fixed self-shrinker, \(S>\dfrac75+\epsilon\) for a very small \(\epsilon>0\) depending on this self-shrinker. According to the Gauss equation (2 ), we infer that the Ricci curvature is bounded from below and \(H^2=S+R>\epsilon\). Thus, we can assume \(H>\sqrt{\epsilon}\) on \(M\). From (?? ), we have \[\label{mvobjigc} \begin{align} \dfrac12\mathcal{L}\dfrac {S} {H^2} &=\dfrac{1}{H^4}\sum_{i, j, k}|h_{ij}\nabla_kH-h_{ijk}H|^2-\dfrac1H\langle\nabla H,\nabla \dfrac {S}{H^2}\rangle. \end{align}\tag{19}\] Applying the generalized maximum principle for \(\mathcal{L}\)-operator to the function \(\dfrac{S}{H^2}\), we know there exists a sequence \(\{p_m\} \subset M^n\) such that

1. \(\lim_{m\to\infty}\dfrac{S}{H^2}(p_m)=\sup \dfrac{S}{H^2}\),

2. \(\lim_{m\to\infty}|\nabla \dfrac{S}{H^2}|(p_m)=0\),

3. \(\lim\sup_{m\to\infty} \mathcal{L}\dfrac{S}{H^2}(p_m)\leq 0\).

Since \(S\) is constant, we obtain \[|\nabla \dfrac{S}{H^2}|^2=\dfrac{4S^2}{H^6}|\nabla H|^2.\] We derive \[\label{ntqhjvsz} \lim_{m\to\infty}|\nabla H(p_m)=0.\tag{20}\] Hence, we get, from (16 ) and (17 ), \(h_{ijk}=0\) for any \(i, \;j, \;k\). In view to (?? ) and \(S=\)constant, we have \(\lim_{m\to\infty}{S}(p_m)=1\) or \(\lim_{m\to\infty}{S}(p_m)=0\). This is impossible because of \(S>\dfrac{7}{5}\). We complete our proof of the theorem 1.2.

4 Bounds of scalar curvature↩︎

In this section, we will discuss upper bounds of scalar curvature and prove the theorem 1.3. Proof of theorem 1.3. Since the scalar curvature is a positive constant, we know \[0<R=H^2-S\leq H^2-\frac{H^2}{n}=\frac{n-1}{n}H^2,\] that is, \[H^2\geq \frac{n}{n-1}R>0.\] We can assume \(H>0\). From the Gauss equation (2 ), we obtain \[1>\dfrac{S}{H^2}=1-\dfrac{R}{H^2}\geq \dfrac{1}{n}.\] Since the scalar curvature is constant, we have \[\nabla_iH^2=\nabla_iS\] for any \(i\), that is, \[H\nabla_i H=\sum_{j}\lambda_jh_{jji},\] where \(\lambda_j\)’s denote the principal curvatures. Therefore, we obtain from the Schwarz inequality \[H^2|\nabla H|^2\leq S\sum_{i,j}h_{iij}^2.\] From \(\dfrac{S}{H^2}<1\), we have \[|\nabla H|^2\leq \sum_{i,j}h_{iij}^2.\] In view of \[\label{eq:4463} 0=\dfrac12\mathcal{L} R=\dfrac{1}{2}\mathcal{L}(H^2-S) =|\nabla H|^2-\sum_{i,j,k}h_{ijk}^{2}+(1-S)R,\tag{21}\] we have \(S\leq 1\). Hence \(H^2\leq nS\leq n\). Thus, we conclude \[\dfrac{n}{n-1}R\leq H^2\leq n,\] which yields \[0<R\leq n-1.\]

5 Proofs of theorems 14 and 1.5↩︎

In this section, we will prove theorems 1.4 and 1.5. Proof of theorem 1.4. If the scalar curvature \(R=0\), the result has been proven by Luo, Sun and Yin [16]. Hence, we only consider the case that the scalar curvature is a positive constant. According to the theorem 1.3 and its proof, we know \[0<R\leq n-1, \quad H^2\geq \frac{n}{n-1}R>0.\] We can assume \(H>0\). Hence, we obtain \[\label{eq:5461} \nabla_i\dfrac{S}{H^2}=-\dfrac{R}{H^4}\nabla_iH^2, \quad |\nabla\dfrac{S}{H^2}|^2=\dfrac{4R^2}{H^6}|\nabla H|^2.\tag{22}\] From the formula (?? ), we have \[\label{eq:5462} \begin{align} \dfrac12\mathcal{L}\dfrac {S} {H^2} &=\dfrac{1}{H^4}\sum_{i, j, k}|h_{ij}\nabla_kH-h_{ijk}H|^2-\dfrac1H\langle\nabla H,\nabla \dfrac {S}{H^2}\rangle. \end{align}\tag{23}\] In view of the theorem 1.3, we know \(S\leq 1\). Hence, from the Gauss equation (2 ), we know that the Ricci curvature is bounded from below. Applying the generalized maximum principle due to Cheng and Peng [17] to the function \(\dfrac{S}{H^2}\), we know that there exists a sequence \(\{p_k\} \subset M^n\) such that \[\lim_{k\to\infty}\dfrac{S}{H^2}(p_k)=\sup \dfrac{S}{H^2}, \; \;\lim_{k\to\infty}|\nabla \dfrac{S}{H^2}|(p_k)=0, \; \; \lim\sup_{k\to\infty} \mathcal{L}\dfrac{S}{H^2}(p_k)\leq 0.\] In view of (22 ), we obtain \[\lim_{k\to\infty}\dfrac1H\langle\nabla H,\nabla \dfrac {S}{H^2}\rangle(p_k)=\lim_{k\to\infty}\dfrac{2R}{H^4}|\nabla H|^2(p_k)=0.\] From (23 ), we derive \[\lim_{m\to\infty}\dfrac{1}{H^4}\sum_{i, j, k}|h_{ij}\nabla_kH-h_{ijk}H|^2(p_m)= 0.\] Hence, we know for any \(i, j, k\), \[\lim_{m\to\infty}|\nabla H|(p_m)=0, \quad \lim_{m\to\infty}h_{ijk}(p_m)=0.\] Since the scalar curvature is constant, from the Gauss equation (2 ) and (6 ), we have \[\label{eq:} 0=\dfrac12\mathcal{L} R=\dfrac{1}{2}\mathcal{L}(H^2-S) =|\nabla H|^2-\sum_{i,j,k}h_{ijk}^{2}+(1-S)R.\tag{24}\] Thus, we get \[\lim_{k\to\infty}S(p_k)=1=\sup S\] from \(S\leq 1\). Because of \[\sup \dfrac{S}{H^2}=\lim_{k\to\infty}\dfrac{S}{H^2}(p_k)=\dfrac{1}{R+1},\] we have \[\dfrac{S}{H^2}\leq \dfrac{1}{R+1} \;{\rm and}\;S\leq \dfrac{H^2}{R+1}.\] If \(S(p)=1\) for some \(p\in M\), letting \(u=1-S\geq 0\), from (?? ), we have \[\dfrac{1}{2}\mathcal{L}u =-\sum_{i,j,k}h_{ijk}^{2}-uS\leq 0.\] According to the strong maximum principle, we know \(S\equiv 1\). Hence, \(X : M^{n}\to \mathbb{R}^{n+1}\) is isometric to \(S^{n}(\sqrt{n})\) or \(S^k (\sqrt k)\times \mathbb{R}^{n-k}\), \(1\leq k\leq n-1\).
Otherwise, we have \(S<1\) and \(\sup S=1\). From \(H^2\geq \frac{n}{n-1}R>0\) and \(H^2\leq nS<n\), we obtain \(0<R<n-1\). The Gauss equation \(H^2-S=R\) implies \[1>S\geq \dfrac{R}{n-1}\;\;{\rm and} \;\;\sup S+R =\sup H^2\leq n\] since \(R\) is constant. For any \(j\), from \[\begin{align} H^2-R&=S=\sum_{i=1}^n\lambda_i^2\geq \dfrac{1}{n-1}(\lambda_j-H)^2+\lambda_j^2, \end{align}\] we obtain \[\begin{align} 0\geq n\lambda_j^2-2H\lambda_j-(n-2)H^2+(n-1)R. \end{align}\] Thus, we derive \[\label{eq:5467} \begin{align} &\dfrac{1}n\bigl(H-\sqrt{(n-1)^2H^2-n(n-1)R}\bigl)\\ &\leq\lambda_j\leq \dfrac{1}n\bigl(H+\sqrt{(n-1)^2H^2-n(n-1)R}\bigl). \end{align}\tag{25}\] Hence, we have \[\begin{align} &-\dfrac{1}n\bigl(\dfrac{n-2}2H+\sqrt{(n-1)^2H^2-n(n-1)R}\bigl)\\ &\leq\lambda_j-\dfrac H2\leq \dfrac{1}n\bigl(-\dfrac{n-2}2H+\sqrt{(n-1)^2H^2-n(n-1)R}\bigl). \end{align}\] According to the Gauss equation (2 ), we get \[\begin{align} &R_{jj}=H\lambda_j-\lambda_j^2=\dfrac{H^2}4-(\lambda_j-\dfrac{H}2)^2\\ &\geq \dfrac{H^2}4-\dfrac{1}{n^2}\bigl(\dfrac{n-2}2H+\sqrt{(n-1)^2H^2-n(n-1)R}\bigl)^2\\ &=\dfrac{1}{n^2}\bigl((n-1)H+\sqrt{(n-1)^2H^2-n(n-1)R}\bigl)\bigl((H-\sqrt{(n-1)^2H^2-n(n-1)R}\bigl)\\ &=\dfrac{1}{n^2}\dfrac{(n-1)H+\sqrt{(n-1)^2H^2-n(n-1)R}}{(H+\sqrt{(n-1)^2H^2-n(n-1)R}}\bigl(n(n-1)R-n(n-2)H^2\bigl)\\ \end{align}\] If \(R>(n-2)\), we have \[n(n-1)R-n(n-2)H^2=nR-n(n-2)S>n(R-(n-2))>0.\] We know \[R_{jj}>\dfrac{R-(n-2)}{n}>0.\] In view of the Myers theorem, \(X : M^{n}\to \mathbb{R}^{n+1}\) is compact. It is a contradiction. Therefore, \(R\leq n-2\). From the Gauss equation (2 ), we conclude \[H^2= S+R< n-1.\] If \(R=n-2\) and there exists a point \(p\) such that some principal curvature \(\lambda_j\) at \(p\) is non-positive, then, at \(p\), \[H-\sqrt{(n-1)^2H^2-n(n-1)R}\leq 0.\] Hence, \[H^2\geq \dfrac{n-1}{n-2}R=n-1.\] It is impossible because of \(H^2<n-1\). Thus, every principal curvature is positive on \(M\). From the theorem of Stoker [18]–[20], we know that \(X : M^{n}\to \mathbb{R}^{n+1}\) is the boundary of a convex body in \(\mathbb{R}^{n+1}\) and diffeomorphic to \(\mathbf{R}^n\). Hence, \(X : M^{n}\to \mathbb{R}^{n+1}\) is proper. According to the theorem of Colding and Minicozzi [6], it is also impossible because of \(S<1\). Hence, \(R<n-2\) and \(H^2<R+1<n-1\).
From \[\nabla_i|X|^2=2\langle X,e_i\rangle,\] we have \[\label{eq:5468} |\nabla |X|^2|^2=4\sum_i\langle X,e_i\rangle^2.\tag{26}\] Applying the generalized maximum principle for \(\mathcal{L}\)-operator to \(-|X|^2\), we know there exists a sequence \(\{p_k\} \subset M^n\) such that \[\lim_{k\to\infty}|X|^2(p_k)=\inf |X|^2, \;\;\lim_{k\to\infty}|\nabla |X|^2|(p_k)=0, \;\; \lim\inf_{k\to\infty} \mathcal{L}|X|^2(p_k)\geq 0.\] In view of (?? ) and (26 ), we derive \[\lim_{k\to\infty}|\nabla |X|^2|^2(p_k)=4\lim_{k\to\infty}\sum_i\langle X,e_i\rangle^2(p_k)=0\] and \[\inf |X|^2= \lim_{k\to\infty}|X|^2(p_k)=\lim_{k\to\infty}H^2(p_k)\geq \frac{n}{n-1}R,\] that is, \[\inf |X|^2\geq \frac{n}{n-1}R.\] Furthermore, if \(\sup |X|^2<\infty\), by applying the generalized maximum principle for \(\mathcal{L}\)-operator to \(|X|^2\), we obtain \[\lim_{k\to\infty}|X|^2(p_k)=\sup|X|^2, \;\;\lim_{k\to\infty}|\nabla |X|^2|(p_k)=0,\] \[\lim\sup_{k\to\infty} \dfrac12\mathcal{L}|X|^2(p_k)=n-\sup |X|^2\leq 0.\] According to (?? ) and (26 ), we have \[\sup |X|^2= \lim_{k\to\infty}|X|^2(p_k)=\lim_{k\to\infty}H^2(p_k)\geq n.\] \(H^2<n-1\) yields a contradiction. Hence, we conclude \(\sup |X|^2=\infty\). We finish our proof of the theorem 1.4.

Proof of theorem 1.5. If \(R=0\), then \(H\equiv 0\) and \(S\equiv 0\). Hence, \(X : M^{n}\to \mathbb{R}^{n+1}\) is totally geodesic. Thus, \(X : M^{n}\to \mathbb{R}^{n+1}\) is isometric to \(\mathbb{R}^n\). Next, we consider \(R>0\). According to the theorem 1.3, we have that \(X : M^{n}\to \mathbb{R}^{n+1}\) is isometric to \(S^{n}(\sqrt{n})\) in this case \(H^2=\dfrac{n}{n-1}R\) or \(S^{n-1}(\sqrt {n-1})\times \mathbb{R}\) in this case \(H^2=\dfrac{n-1}{n-2}R\) or \(X : M^{n}\to \mathbb{R}^{n+1}\) satisfies \(R< n-2\), \(\dfrac{n}{n-1}R\leq H^2\), \(S<1\) and \(\sup S=1\). From (25 ), for any \(j\), we have \[\begin{align} &\dfrac{1}n\bigl(H-\sqrt{(n-1)^2H^2-n(n-1)R}\bigl)\leq\lambda_j\leq \dfrac{1}n\bigl(H+\sqrt{(n-1)^2H^2-n(n-1)R}\bigl). \end{align}\] If there exists a point \(p\in M\) such that for some \(j\), \(\lambda_j\leq 0\) at \(p\), we have \[H-\sqrt{(n-1)^2H^2-n(n-1)R}\leq0.\] Hence, we get \[(n-1)R\leq (n-2)H^2.\] Since \(H^2\leq \dfrac{n-1}{n-2}R\) holds, we know that \(H^2\) attains its maximum at \(p\), \[H^2(p)=\sup H^2=\dfrac{n-1}{n-2}R.\] From \(S<1\) on \(M\) and \(\sup S=1\), we know that \(S\) does not get its maximum on \(M\). Therefore, from the Gauss equation (2 ) \(H^2=S+R\), we conclude that \(H^2\) can not attain its maximum on \(M\). This is a contradiction. Hence, we conclude that every principal curvature is positive on \(M\). \(X : M^{n}\to \mathbb{R}^{n+1}\) is locally, strictly convex. From the theorem of Stoker [18]–[20], we know that \(X : M^{n}\to \mathbb{R}^{n+1}\) is the boundary of a convex body in \(\mathbb{R}^{n+1}\) and diffeomorphic to \(\mathbf{R}^n\). Hence, \(X : M^{n}\to \mathbb{R}^{n+1}\) is proper. According to the theorem of Colding and Minicozzi [6], it is also impossible because of \(S<1\). Thus, \(X : M^{n}\to \mathbb{R}^{n+1}\) is isometric to \(S^{n}(\sqrt{n})\) or \(S^{n-1}(\sqrt {n-1})\times \mathbb{R}\).

Acknowledgement. This work was partly supported by MEXT Promotion of Distinctive Joint Research Center Program JPMXP0723833165 and Osaka Metropolitan University Strategic Research Promotion Project (Development of International Research Hubs).

References↩︎