Remarks on Hypersurfaces in a Unit Sphere

Abstract

In 1973, D. A. Hoffman proved that a complete immersed surface in $\mathbb {S}^3$ with constant mean curvature $H\ne 0$ and Gauss curvature which does not change sign must be a sphere or a flat torus. For the n-dimensional case with $n\ge 3$, we prove that Hoffman’s theorem cannot be extended when we consider constant higher-order mean curvature $H_r$, $2\le r<n$, and Gauss–Kronecker curvature K, instead of the mean curvature H and Gauss curvature, respectively. More precisely, we show the existence of non-isoparametric hypersurfaces $\Sigma$ in $\mathbb {S}^{n+1}$, $n\ge 3$, with constant higher-order mean curvature and whose Gauss–Kronecker curvature K does not change sign. Besides, we also derive estimates for the infimum and the supremum of the principal curvatures of a hypersurface with constant higher-order mean curvature.

About Theorem 1

In this appendix, we obtain Theorem 1 from [11, Theorem 4.1]. Let $\Sigma ^n$ and ${\overline{M}}^{n+k}$ be Riemannian manifolds of dimension n and $n + k$, respectively. Throughout the Appendix we use $\Sigma ^n \hookrightarrow {\overline{M}}^{n+k}$ to denote an isometric immersion of $\Sigma$ in ${\overline{M}}.$

Remember that the Gauss formula of $\Sigma ^n \hookrightarrow {\overline{M}}^{n+k}$ is

$$\begin{aligned} {\overline{\nabla }}_X Y=\nabla _X Y+B(X,Y), \end{aligned}$$

which is valid for all $X,Y \in {\mathfrak {X}}(\Sigma )$. Here, $\nabla , {\overline{\nabla }}$ are the Riemannian connections of $\Sigma ^n$ and ${\overline{M}}^{n+k}$, respectively, and B is the second fundamental form of $\Sigma$. The mean curvature vector field $\textbf{H}\in {\mathfrak {X}}^\perp (\Sigma )$ of $\Sigma ^n$ in ${\overline{M}}$ is given by

$$\begin{aligned} \textbf{H}=\frac{1}{n}\sum _{i=1}^n B(e_i,e_i), \end{aligned}$$

where $\{e_1,\dots ,e_n\}$ is a local orthonormal frame on $\Sigma$.

Definition 14

An isometric immersion $\Sigma ^n \hookrightarrow {\overline{M}}^{n+k}$ is said to have parallel mean curvature if $\textbf{H}$ is parallel in the normal bundle, i.e., ${\overline{\nabla }}^\perp \textbf{H} =0$.

Remark 15

The following properties follow directly from the definition:

• 1. If $\textbf{H}$ is parallel, then $\vert \textbf{H} \vert$ is constant.

• 2. If $\textbf{H}\ne 0$, $\textbf{H}$ is parallel if and only if $\vert \textbf{H}\vert$ is constant and $\textbf{H}/\vert \textbf{H}\vert$ is parallel.

• 3. If the codimension $k=1$, $\textbf{H}$ is parallel if and only if $\vert \textbf{H} \vert$ is constant.

Now we consider a surface $\Sigma ^2$ isometrically immersed in ${\overline{M}}^4(c)$, a 4-manifold with constant sectional cuavature c.

Theorem 16

(Theorem 4.1 in [11]). A complete immersed surface $\Sigma ^2 \hookrightarrow {\overline{M}}^4(c)$, $c\ge 0$, with parallel mean curvature and Gauss curvature K which does not change sign must be minimal ($\textbf{H}\equiv 0$), a sphere of radius $(\vert \textbf{H} \vert ^2 + c)^{-\frac{1}{2}}$ or a product of circles ${\mathbb {S}}^1(s) \times {\mathbb {S}}^1(t)$, $0<s \le \infty$, $0< t < \infty$, with the standard product immersion.

Remark 17

In particular, consider the immersion $\Sigma ^2 \hookrightarrow {\mathbb {S}}^4$ with parallel mean curvature $\textbf{H} \ne 0$. If $\Sigma$ is not totally umbilical and its Gaussian curvature K does not change sign, then $K=0$ and $\Sigma$ must be a product of circles ${\mathbb {S}}^1(s) \times {\mathbb {S}}^1(t)$, $s^2+t^2=1$.

We first look at a surface $\Sigma ^2$ isometrically immersed in${\mathbb {S}}^3$. Then, we use the canonical embedding of ${\mathbb {S}}^3$ in ${\mathbb {S}}^4$ to obtain the immersion of $\Sigma$ in ${\mathbb {S}}^4$. It is well known that ${\mathbb {S}}^3$ is totally geodesic submanifold of ${\mathbb {S}}^4$, that is, their Riemannian connections coincide. Let $A =\{e_1, e_2, e_3, e_4\}$ be an adapted frame to $\Sigma$ (in ${\mathbb {S}}^4$) such that

$$\begin{aligned} e_1, e_2 \in T\Sigma , \qquad \quad e_3 \in T{\mathbb {S}}^3\subset T{\mathbb {S}}^4 \qquad \text {and} \qquad e_4 \in T^\perp {\mathbb {S}}^3 \subset T{\mathbb {S}}^4. \end{aligned}$$

Then, the mean curvature vector field of $\Sigma \hookrightarrow {\mathbb {S}}^4$ is given by

$$\begin{aligned} \textbf{H}= \frac{1}{2} (\lambda _{11}^3+\lambda _{22}^3) e_3 + \frac{1}{2} (\lambda _{11}^4+\lambda _{22}^4) e_4 \end{aligned}$$

(29)

where $\lambda _{ij}^\alpha = \langle B(e_i,e_j), e_\alpha \rangle$, $\alpha =3,4$, and B is the second fundamental form of the immersion $\Sigma \hookrightarrow {\mathbb {S}}^4$.

Now suppose that the immersion $\Sigma \hookrightarrow {\mathbb {S}}^3$ has constant nonzero mean curvature. In view of Theorem 16 (see also Remark 17), it is sufficient to show that $\textbf{H}$ is parallel (see also Section 3 in [15] for more general cases). Let

$$\begin{aligned} \mu _{ij}^{3} = \langle B_\Sigma ^{{\mathbb {S}}^3} (e_i,e_j), e_3 \rangle \end{aligned}$$

where $B_\Sigma ^{{\mathbb {S}}^3}$ is the second fundamental form of $\Sigma \hookrightarrow {\mathbb {S}}^3$. One can check directly that

for $X,Y \in T\Sigma$. It follows that

$$\begin{aligned} \lambda _{ij}^{3} = \langle B(e_i,e_j), e_3 \rangle = \langle B_\Sigma ^{{\mathbb {S}}^3} (e_i,e_j), e_3 \rangle = \mu _{ij}^{3}. \end{aligned}$$

On the other hand, since $B_\Sigma ^{{\mathbb {S}}^3} \in T {\mathbb {S}}^3$ and $e_4 \in T^\perp {\mathbb {S}}^3$ we get that

$$\begin{aligned} \lambda _{ij}^{4} = \langle B(e_i,e_j), e_4 \rangle = \langle B_\Sigma ^{{\mathbb {S}}^3} (e_i,e_j), e_4 \rangle =0. \end{aligned}$$

Thus, (29) reduces to

$$\begin{aligned} \textbf{H}= \frac{1}{2} \left( \lambda _{11}^3+\lambda _{22}^3\right) e_3 =H e_3, \qquad \text {where }H=\frac{1}{2} \left( \lambda _{11}^3+\lambda _{22}^3\right) \text { is constant.} \end{aligned}$$

From this we notice that $\vert \textbf{H}\vert = \vert H\vert$ is also constant. Furthermore, since $\textbf{H} \in T {\mathbb {S}}^4$ and ${\mathbb {S}}^3$ is a totally geodesic submanifold of ${\mathbb {S}}^4$,

$$\begin{aligned} {\overline{\nabla }}_{X}^{\perp }\, \textbf{H} = \langle {\overline{\nabla }}_X \, \textbf{H}, e_3\rangle e_3 = H \langle {\overline{\nabla }}_X \, e_3, e_3\rangle e_3 =0. \end{aligned}$$

Thus, the immersion $\Sigma ^2 \hookrightarrow {\mathbb {S}}^{4}$ has parallel mean curvature $\textbf{H}\ne 0$.