Fundamental tone of minimal hypersurfaces with finite index in hyperbolic space

Abstract

Let M be a complete minimal hypersurface in hyperbolic space $\mathbb {H}^{n+1}(-1)$ with constant sectional curvature −1. We prove that if M has a finite index and finite $L^{2}$ norm of the second fundamental form, then the fundamental tone $\lambda_{1} (M)$ is bounded above by $n^{2}$.

1 Introduction

McKean [1] proved that the fundamental tone of an n-dimensional complete simply connected Riemannian manifold M with sectional curvature bounded above by $-\kappa^{2}<0$ is bigger than or equal to $\frac{(n-1)^{2}\kappa^{2}}{4}$, where κ is a real number. Moreover, his result is sharp since the equality is attained by the hyperbolic space $\mathbb{H}^{n}(-\kappa^{2})$ with constant sectional curvature $-\kappa^{2}$. We recall that the fundamental tone $\lambda_{1}(M)$ is defined by

$$\begin{aligned} \lambda_{1} (M) = \inf \biggl\{ \frac{\int_{M} \vert \nabla f\vert ^{2}}{\int_{M} f^{2}}: 0\neq f \in W^{1,2}_{0} (M) \biggr\} . \end{aligned}$$

Interestingly, Cheung and Leung [2] obtained the same lower bound for the fundamental tone of complete submanifold in $\mathbb {H}^{m}(-\kappa^{2})$ with bounded mean curvature as follows (see also [3, 4]).

Theorem

[2]

Let M be an n-dimensional complete noncompact submanifold in $\mathbb{H}^{m}(-\kappa^{2})$ with the mean curvature vector H. If $\vert H\vert \leq\alpha< n-1$, then

$$\begin{aligned} \lambda_{1}(M) \geq\frac{(n-1-\alpha)^{2} \kappa^{2}}{4}. \end{aligned}$$

There have been extensive investigations to obtain an upper bound for the fundamental tone of complete minimal submanifolds in hyperbolic space. Castillon [5] proved that the spectrum of the Laplacian on a complete minimal hypersurface with finite $L^{n}$ norm of the second fundamental form in $\mathbb{H}^{n+1}$, denoted by $\operatorname {Spec}(\Delta)$, is given by $\operatorname {Spec}(\Delta) = [\frac{(n-1)^{2}}{4}, +\infty )$. Candel [6] was able to prove that the fundamental tone of complete simply connected stable minimal surfaces in $\mathbb{H}^{3} (-1)$ is at most $\frac{4}{3}$. In [7], the author proved that if M is a complete stable minimal hypersurface in $\mathbb {H}^{n+1}(-1)$ with finite $L^{2}$ norm of the second fundamental form, then $\frac{(n-1)^{2}}{4} \leq\lambda_{1} (M) \leq n^{2}$. Later, Bérard et al. [8] improved the upper bound for complete stable minimal surfaces in $\mathbb{H}^{3}(-1)$. Indeed, they proved that the fundamental tone of complete stable minimal surfaces in $\mathbb{H}^{3} (-1)$ is at most $\frac{4}{7}$. Fu and Tao [9] showed that if M is an n-dimensional complete submanifold in $\mathbb{H}^{m}(-1)$ with parallel mean curvature vector H and with finite $L^{p}$ norm of the traceless second fundamental form for $p \geq n$, then $\lambda_{1} (M)$ is less than or equal to $\frac {(n-1)^{2}(1-\vert H\vert ^{2})}{4}$. Recently, Gimeno [10] proved that if $M^{2}$ is a complete minimal surface in $\mathbb{H}^{m} (-1)$ with finite $L^{2}$ norm of the second fundamental form, then $\lambda_{1} (M)=\frac{1}{4}$.

The aim of this paper is to obtain an upper bound for the fundamental tone of complete minimal hypersurfaces in $\mathbb{H}^{n+1}(-1)$ with finite index and finite $L^{2}$ norm of the second fundamental form. More precisely, we prove the following.

Theorem 1.1

Let M be a complete orientable minimal hypersurface in $\mathbb {H}^{n+1}(-1)$ with $\int_{M} \vert A\vert ^{2} < \infty$. Suppose M has finite index. Then we have

$$\begin{aligned} \frac{(n-1)^{2}}{4} \leq \lambda_{1}(M) \leq n^{2}. \end{aligned}$$

It is obvious that a complete stable minimal hypersurface in $\mathbb{H}^{n+1}(-1)$ has index 0. Hence our theorem can be regarded as an extension of the results in [6–8]. When $n=2$, we remark that the finite index condition can be omitted, since the finiteness of the $L^{2}$ norm of the second fundamental form implies that M has finite index, which was proved by Bérard et al. [11]. However, in this case, our theorem is weaker than Theorem 4.1 in [5] or Theorem A in [10].

2 Proof of Theorem 1.1

In this section, we prove our main theorem.

Proof of Theorem 1.1

The lower bound of $\lambda_{1} (M)$ is given by $\frac{(n-1)^{2}}{4}$, which was done by Cheung and Leung [2] as mentioned in the Introduction. Thus it suffices to prove that the upper bound of $\lambda_{1}(M)$ is $n^{2}$.

Since M has a finite index, there exists a compact subset $K \subset M$ such that $M \setminus K$ is stable (see [12] for example), i.e., for any compactly supported Lipschitz function f on $M\setminus K$,

$$\begin{aligned} \int_{M \setminus K} \vert \nabla f\vert ^{2} - \bigl( \vert A\vert ^{2}-n\bigr)f^{2}\,dv \geq0, \end{aligned}$$

(1)

where $\vert A\vert ^{2}$ denotes the squared length of the second fundamental form on M and dv denotes the volume form for the induced metric on M. Note that, for some geodesic ball $B(R_{0}) \subset M$ centered at $p\in M$ of radius $R_{0}$ containing the compact set K, the region $M \setminus B(R_{0})$ is still stable. Thus, without loss of generality, we may assume that $K=B(R_{0})$.

Choose a geodesic ball $B(R)\subset M$ centered at $p\in M$ of radius $R>R_{0}$ and take a cut-off function $0\leq\phi\leq1$ on M satisfying

$$ \phi= \textstyle\begin{cases} 0& \text{on }B(R_{0}),\\ 1& \text{on }B(2R+R_{0})\setminus B(R+R_{0}),\\ 0& \text{on }M \setminus B(3R+R_{0}), \end{cases} $$

and $\vert \nabla\phi \vert \leq\frac{1}{R}$ on M. By the definition of the fundamental tone and the domain monotonicity of the eigenvalue, we see that

$$\begin{aligned} \lambda_{1} (M) \leq\lambda_{1} \bigl(M\setminus B(R_{0})\bigr) \leq\frac{\int _{M\setminus B(R_{0})} \vert \nabla f\vert ^{2}}{\int_{M\setminus B(R_{0})} f^{2}} \end{aligned}$$

for any $f \in W^{1,2}_{0} (M \setminus B(R_{0}))$. Substituting f with $\vert A\vert \phi$ gives

$$\begin{aligned} &\lambda_{1}(M) \int_{M\setminus B(R_{0})} \vert A\vert ^{2} \phi^{2} \\ &\quad \leq \int_{M\setminus B(R_{0})} \bigl\vert \nabla\bigl(\vert A\vert \phi\bigr) \bigr\vert ^{2} \\ &\quad = \int_{M\setminus B(R_{0})} \phi^{2} \bigl\vert \nabla \vert A \vert \bigr\vert ^{2} + \int_{M\setminus B(R_{0})} \vert A\vert ^{2}\vert \nabla\phi \vert ^{2} +2 \int_{M\setminus B(R_{0})} \vert A\vert \phi \bigl\langle \nabla \vert A \vert , \nabla\phi\bigr\rangle . \end{aligned}$$

Using the Schwarz inequality and the geometric-arithmetic mean inequality, we get

$$\begin{aligned} 2 \int_{M\setminus B(R_{0})} \vert A\vert \phi \bigl\langle \nabla \vert A \vert , \nabla\phi \bigr\rangle \leq\varepsilon \int_{M\setminus B(R_{0})} \vert A\vert ^{2}\vert \nabla\phi \vert ^{2} + \frac{1}{\varepsilon} \int_{M\setminus B(R_{0})} \phi^{2}\bigl\vert \nabla \vert A\vert \bigr\vert ^{2} \end{aligned}$$

for any $\varepsilon>0$. Therefore

$$\begin{aligned} \lambda_{1}(M) \int_{M\setminus B(R_{0})} \vert A\vert ^{2} \phi^{2} \leq& (1+\varepsilon) \int_{M\setminus B(R_{0})} \vert A\vert ^{2}\vert \nabla\phi \vert ^{2} \\ &{}+ \biggl(1+\frac{1}{\varepsilon} \biggr) \int_{M\setminus B(R_{0})} \phi^{2}\bigl\vert \nabla \vert A\vert \bigr\vert ^{2}. \end{aligned}$$

(2)

On the other hand, a Simons-type inequality [13, 14] for minimal hypersurfaces in $\mathbb{H}^{n+1}$ asserts that

$$\begin{aligned} \vert A\vert \Delta \vert A\vert + \vert A\vert ^{4} + n \vert A\vert ^{2} = \vert \nabla A\vert ^{2} - \bigl\vert \nabla \vert A\vert \bigr\vert ^{2}. \end{aligned}$$

Applying the Kato inequality [15],

$$\begin{aligned} \vert \nabla A\vert ^{2} - \bigl\vert \nabla \vert A\vert \bigr\vert ^{2} \geq\frac{2}{n} \bigl\vert \nabla \vert A \vert \bigr\vert ^{2}, \end{aligned}$$

we have

$$\begin{aligned} \vert A\vert \Delta \vert A\vert + \vert A\vert ^{4} + n \vert A\vert ^{2} \geq\frac{2}{n} \bigl\vert \nabla \vert A\vert \bigr\vert ^{2}. \end{aligned}$$

Multiplying both sides by the function $\phi^{2}$ and integrating over $B(3R+R_{0})\setminus B(R_{0})$, we get

$$\begin{aligned} \frac{2}{n} \int_{M\setminus B(R_{0})} \phi^{2} \bigl\vert \nabla \vert A \vert \bigr\vert ^{2} \leq& \int_{M\setminus B(R_{0})} \phi^{2} \vert A\vert ^{4} + n \int_{M\setminus B(R_{0})} \phi^{2} \vert A\vert ^{2} \\ &{}- \int_{M\setminus B(R_{0})} \phi^{2} \bigl\vert \nabla \vert A \vert \bigr\vert ^{2} -2 \int_{M\setminus B(R_{0})} \vert A\vert \phi\bigl\langle \nabla \vert A \vert , \nabla\phi\bigr\rangle , \end{aligned}$$

(3)

where we used the divergence theorem.

Replacing f with $\phi \vert A\vert$ in the stability inequality (1) on $M \setminus B(R_{0})$ gives

$$\begin{aligned} \int_{M\setminus B(R_{0})} \bigl\vert \nabla\bigl(\phi \vert A\vert \bigr) \bigr\vert ^{2} \geq \int_{M\setminus B(R_{0})} \bigl(\vert A\vert ^{2} -n\bigr)\vert A\vert ^{2}\phi^{2}, \end{aligned}$$

which implies

$$\begin{aligned} &\int_{M\setminus B(R_{0})} \vert A\vert ^{2}\vert \nabla\phi \vert ^{2} + \int_{M\setminus B(R_{0})} \phi^{2}\bigl\vert \nabla \vert A\vert \bigr\vert ^{2} + 2 \int_{M\setminus B(R_{0})} \vert A\vert \phi\bigl\langle \nabla \vert A \vert , \nabla\phi\bigr\rangle \\ &\quad \geq \int_{M\setminus B(R_{0})} \vert A\vert ^{4} \phi^{2} - n \int_{M\setminus B(R_{0})} \vert A\vert ^{2} \phi^{2} . \end{aligned}$$

(4)

Combining (3) with (4), we obtain

$$\begin{aligned} \frac{2}{n} \int_{M\setminus B(R_{0})} \phi^{2}\bigl\vert \nabla \vert A\vert \bigr\vert ^{2} \leq \int _{M\setminus B(R_{0})} \vert A\vert ^{2}\vert \nabla\phi \vert ^{2} + 2n \int_{M\setminus B(R_{0})} \vert A\vert ^{2}\phi^{2}. \end{aligned}$$

(5)

Hence, using (2) and (5), we have

$$\begin{aligned} 2 \biggl\{ \frac{1}{n} - \frac{n(1+\frac{1}{\varepsilon})}{\lambda_{1}(M)} \biggr\} \int _{M\setminus B(R_{0})} \phi^{2}\bigl\vert \nabla \vert A \vert \bigr\vert ^{2} \leq \biggl\{ 1 + \frac {2n(1+\varepsilon)}{\lambda_{1}(M)} \biggr\} \int_{M\setminus B(R_{0})} \vert A\vert ^{2}\vert \nabla\phi \vert ^{2}. \end{aligned}$$

(6)

We now suppose that $\lambda_{1} (M) >n^{2}$. For a sufficiently large $\varepsilon>0$, letting $R\rightarrow\infty$ in (6) shows that $\vert \nabla \vert A\vert \vert \equiv0$ on $M\setminus B(R_{0})$, which implies that $\vert A\vert$ is constant on $M\setminus B(R_{0})$. Since the volume of any complete minimal hypersurface in hyperbolic space is infinite and $L^{2}$ norm of $\vert A\vert$ is finite by our assumption, we see that $\vert A\vert \equiv0$ outside the compact subset $B(R_{0})$. It follows from the maximum principle for minimal hypersurfaces in $\mathbb{H}^{n+1}$ that M must be totally geodesic. However, due to McKean [1], the fundamental tone of totally geodesic hyperplanes in $\mathbb{H}^{n+1}$ is equal to $\frac{(n-1)^{2}}{4}$, which gives a contradiction. Therefore we get the conclusion. □

Remark 2.1

The proof of Theorem 1.1 relies on the inequality (6), which is called a Caccioppoli-type inequality. In [16], Ilias et al. intensively studied a Caccioppoli-type inequality on constant mean curvature hypersurfaces in Riemannian manifolds.