Index and topology of minimal hypersurfaces in $\mathbb {R}^n$

Abstract

In this paper, we consider immersed two-sided minimal hypersurfaces in $\mathbb {R}^n$ with finite total curvature. We prove that the sum of the Morse index and the nullity of the Jacobi operator is bounded from below by a linear function of the number of ends and the first Betti number of the hypersurface. When $n=4$, we are able to drop the nullity term by a careful study for the rigidity case. Our result is the first effective Morse index bound by purely topological invariants, and is a generalization of Li and Wang (Math Res Lett 9(1):95–104, 2002). Using our index estimates and ideas from the recent work of Chodosh–Ketover–Maximo (Minimal surfaces with bounded index, 2015. arXiv:1509.06724), we prove compactness and finiteness results of minimal hypersurfaces in $\mathbb {R}^4$ with finite index.

Appendix A. Minimal hypersurfaces with finite total curvature

In the appendix we give a brief explanation of proposition 2.4. Namely, we prove that immersed minimal hypersurfaces with finite total curvature in $\mathbb {R}^n$ is regular at infinity. The proof we include here is a generalization of [33] and [1]. We refer the readers to their original papers for more details.

Proof

Let $\Sigma ^{n-1}\subset \mathbb {R}^n$ be a minimal hypersurface with finite total curvature, i.e. $\int |A|^{n-1}< \infty$. We prove that $\Sigma$ has finitely many embedded ends, each of which is a graph over some affine plane of some function u

$$\begin{aligned} |x|^{n-3}|u(x)|+|x|^{n-2}|\nabla u(x)|+|x|^{n-1}|\nabla ^2 u(x)|\le C. \end{aligned}$$

The proof can be divided into several steps as follows.

Step 1:

A $\epsilon$ regularity theorem for |A|.

Consider an annulus $A(R,2R)=B(0,2R)-B(0,R)$. We have the following:

Proposition A.1

There is an $\epsilon _0>0$ such that if $\int _{A(R,2R)\cap \Sigma }|A|^{n-1}<\epsilon _0$ then

$$\begin{aligned} \sup _{x\in \partial B(0,R)\cap \Sigma }|A|^2(x)\le \frac{1}{R^2}\mu \left( \int _{A(R,2R)\cap \Sigma }|A|^{n-1}.\right) \end{aligned}$$

Here $\mu$ is a continuous function satisfying $\mu (\epsilon )\rightarrow 0$ as $\epsilon \rightarrow 0$.

This proposition is a direct consequence of the Simons’ equality and the scaling invariance of $\int |A|^{n-1}$. As a consequence, since the total curvature is finite, we deduce that

$$\begin{aligned} |A|(x)\cdot |x|\rightarrow 0, \quad \hbox { as}\ |x|\rightarrow \infty . \end{aligned}$$

Step 2:

Euclidean volume growth.

Recall that by [33] the Morse index of $\Sigma$ is bounded from above by a constant times the total curvature, and in particular, the index of $\Sigma$ is finite. We then conclude by [19] that $\Sigma$ has finitely many ends. We study the asymptotic behavior of each end.

By the Morse-theoretic argument in [35], we see that for $R>R_0$, the distance function $r={{\mathrm{dist}}}(\cdot , 0)$ is a Morse function with no critical point. Take R sufficiently large such that $\Sigma \cap \partial B_R(0)$ transversely, and each connected component of $\Sigma \cap \partial B_R(0)$ corresponds to an end of $\Sigma$. We first prove that $\Sigma -B_{R_0}(0)$ is a collection of finitely many ends, each diffeomorphic to $S^{n-2}\times [0,\infty )$. In particular, this shows that $\Sigma$ is properly immersed with finite topology. Let us look at one end V.

Let $V_{r_0}=\partial B_{r_0}(0)\cap V$ to be level set of r on V. Since

$$\begin{aligned} (\nabla ^{\Sigma })^2 r(X,Y)=(\nabla ^{\mathbb {R}^n})^2 r(X,Y)-\left\langle A(X,Y),\nabla ^{\mathbb {R}^n}r \right\rangle , \end{aligned}$$

and that $|A|(r)\le \frac{\mu _1(r)}{r}$, where $\mu _1$ is the continuous function obtained in step 1, that converges to zero as r tends to infinity, we conclude that

$$\begin{aligned} |A_{V_r}-\frac{1}{r}I|\le \frac{\mu _1(r)}{r}, \end{aligned}$$

where $A_{V_r}$ is the second fundamental form of $V_r$ inside V. By the Gauss equation we conclude that

$$\begin{aligned} |K_{V_r}(x,P)-\frac{1}{r^2}|\le \frac{2\mu _1(r)}{r} \end{aligned}$$

(A.1)

for every point x on $V_r$ and every tangential two-plane P in $T_xV_r$, where $K_{V_r}$ is the sectional curvature. Since $\mu _1(r)\rightarrow 0$ as $r\rightarrow \infty$, $V_r$ is diffeomorphic to the standard sphere of dimension $n-2$. Combining this with the Morse-theoretic argument, we conclude that each end of $\Sigma$ is diffeomorphic to $S^{n-1}\times [0,1)$. Therefore $\Sigma$ is properly immersed with finite topology.

Let us prove that $\Sigma$ has Euclidean volume growth. By the curvature condition A.1 and the standard volume comparision, we see that for each level surface $V_r$,

$$\begin{aligned} 1-2\mu _1(r)<\frac{{{\mathrm{Vol}}}(V_r)}{\omega _{n-2}r^{n-2}}-1<1+2\mu _1(r). \end{aligned}$$

Hence by the co-area formula we conclude that V has Euclidean volume growth, i.e.

$$\begin{aligned} \lim _{R\rightarrow \infty }\frac{{{\mathrm{Vol}}}(V\cap B_R(0))}{\omega _{n-1}R^{n-1}}=1. \end{aligned}$$

Since $\Sigma$ has only finitely many ends at infinity, $\Sigma$ also has Euclidean volume growth.

Step 3:

Tangent plane at infinity.

Having Euclidean volume growth, consider the varifold limit $\Sigma _\infty$ of rescaled surfaces $\frac{1}{r}\Sigma$. By slight abuse of notation we also use $\Sigma _\infty$ to denote the support of this varifold. For any positive $\delta >0$, $\Sigma _\infty -B_\delta (0)$ is the limit of minimal hypersurfaces whose |A| converges to 0 uniformly. Therefore $\Sigma _\infty$ is totally geodesic outside of 0, i.e. $\Sigma _\infty$ is a union of hyperplanes through 0.

Choose a radius $r_0$ sufficiently large such that for any $r>r_0$ the curvature condition A.1 holds. Consider an end $V\setminus B_{r_0}(0)$ and its rescalings $\{\frac{1}{R}(V\setminus B_{r_0}(0))\}$. Previously we know that as $R\rightarrow \infty$ these rescaled ends converge subsequentially to $P\setminus \{0\}$ for some hyperplane P. By the curvature condition A.1 we know that

$$\begin{aligned} \lim _{R\rightarrow \infty }\frac{{{\mathrm{Vol}}}((V\setminus B_{r_0}(0))\cap B_R(0))}{\omega _{n-1}R^{n-1}}=1. \end{aligned}$$

Since the varifold convergence does not increase the density at infinity, the convergence

$$\begin{aligned} \frac{1}{R}(V\setminus B_{r_0}(0))\rightarrow P\setminus \{0\} \end{aligned}$$

is in multiplicity one.

Next we point out that the plane P does not depend on the choice of the subsequence in the convergence, that is, each end has a unique tangent plane at infinity. To see this, first note that since $\Sigma$ has finite index, the end $V\setminus B_{r_0}(0)$ is stable for sufficiently large radius $r_0$. Also since for each sequence of $R_i\rightarrow \infty$ there is a subsequence of $\{\frac{1}{R_i}(V\setminus B_{r_0}(0))\}$ converging to some hyperplane of multiplicity one, by Lemma 3 of [33] we conclude that this limiting hyperplane is unique.

This means that the each end of the original surface $\Sigma$ has a unique tangent plane at infinity.

Step 4:

Regular at infinity.

Since each end of $\Sigma$ converges to a hyperplane of multiplicity one, we deduce that outside some compact set, each end is a graph of bounded slope over the tangent plane at infinity. According to proposition 3 of [28], this end is regular at infinity. $\square$