Characterizations of Euclidean spheres

1.   Introduction

Geometry of hypersurfaces of a Riemannian manifold is one of the important branches of differential geometry. In that, one of important questions is characterizing spheres among compact hypersurfaces of a Euclidean space ^{[1,3,4,5,6,7,8]}. On a Riemannian manifold $(M, g)$, the Ricci operator $S$ is defined using Ricci tensor $Ric$, namely $Ric(X, Y) = g(SX, Y)$, $X\in \mathfrak{X}(M)$, where $\mathfrak{X}(M)$ is the Lie algebra of smooth vector fields on $M$. Similarly, the rough Laplace operator on the Riemannian manifold $(M, g)$, $\Delta :$ $\mathfrak{X} (M)\rightarrow \mathfrak{X}(M)$ is defined by

where $\nabla$ is the Riemannian connection and $\left\{ e_{1}, ..., e_{m}\right\}$ is a local orthonormal frame on $M$, $m = \dim M$. Rough Laplace operator is used in finding characterizations of spheres as well as of Euclidean spaces ^[11,12]. Recall that de-Rham Laplace operator $\square :$ $\mathfrak{X}(M)\rightarrow \mathfrak{X}(M)$ on a Riemannian manifold $(M, g)$ is defined by (^[10], p83)

and is used to characterize a Killing vector field on a compact Riemannian manifold. Recall that a vector field $\mathbf{u}$ on a Riemannian manifold $(M, g)$ is said to be a geodesic vector field ^[9] if

Let $M$ be an orientable immersed hypersurface of the Euclidean space and $\psi :M\rightarrow \mathbf{E}^{m+1}$ be the immersion. We denote the unit normal to the hypersurface $M$ by $\xi$ and support function of the hypersurface by $\sigma$ defined by $\sigma = \langle \psi, \xi \rangle$, where $\langle, \rangle$ is the Euclidean metric on $\mathbf{E}^{m+1}$. Then treating $\psi$ as position vector field of the hypersurface $M$, we have $\psi = \psi ^{T}+\sigma \xi$.

Consider the sphere $\mathbf{S}^{m}(c)$ of constant curvature $c$ as hypersurface of the Euclidean space $\mathbf{E}^{m+1}$ with unit normal $\xi$ and shape operator $A = -\sqrt{c}I$. Now, consider the embedding $\psi : \mathbf{S}^{m}(c)\rightarrow \mathbf{E}^{m+1}$. Then it follows that the tangential component $\psi ^{T}$ for the sphere $\mathbf{S}^{m}(c)$ satisfies $\square \psi ^{T} = 0$ as well as $\psi ^{T}$ is a geodesic vector field. These raise two questions: (i) Under what condition on a compact hypersurface $M$ of a Euclidean space $\mathbf{E}^{m+1}$ with immersion $\psi :M\rightarrow \mathbf{E}^{m+1}$ such that $\psi ^{T}$ is a geodesic vector field, $M$ is isometric to a sphere? (ii) Under what conditions on a compact hypersurface $M$ of a Euclidean space $\mathbf{E}^{m+1}$ with immersion $\psi :M\rightarrow \mathbf{E}^{m+1}$ such that $\psi ^{T}$ satisfying $\square \psi ^{T} = 0$, $M$ is isometric to a sphere? In this paper, we answer these questions, for first question by showing that under the condition $Ric\left(\psi ^{T}, \psi ^{T}\right) \geq \frac{m-1}{m}\left(\text{div }\psi ^{T}\right) ^{2}$ the hypersurface is isometric to a sphere, where as for the second question, it requires the condition $\left\vert \sigma \alpha \right\vert \leq 1$, where $\alpha$ is the mean curvature (Theorem 3.1 and Theorem 3.2).

2.   Preliminaries

Let $M$ be an orientable immersed hypersurface of of the Euclidean space $\mathbf{E}^{m+1}$ with immersion $\psi :M\rightarrow \mathbf{E}^{m+1}$ with unit normal $\xi$ and shape operator $A$. Then we have the following Gauss-Weingarten formulae

where $D$, $\nabla$ are Riemannian connections on $\mathbf{E}^{m+1}$, $M$ respectively, $g$ is the induced metric on $M$ and $\mathfrak{X}(M)$ is the Lie algebra of smooth vector fields on $M$. The curvature tensor field $R$ and the Ricci curvature $Ric$ of the hypersurface are given by

and

where, $\alpha = \frac{1}{m}Tr.A$ is the mean curvature of the hypersurface ^[2]. Using Eq (2.3), we see that the Ricci operator $S$ of the hypersurface $M$ is given by

Also, as the Euclidean space $\mathbf{E}^{m+1}$ is space of constant curvature, the Codazzi equation for the hypersurface $M$ is

where $\left(\nabla A\right) (X, Y) = \nabla _{X}AY-A\left(\nabla _{X}Y\right)$. Using Eq (2.5) and symmetry of the shape operator $A$, the gradient $\text{grad }\alpha$ of the mean curvature $\alpha$ is given by

where $\left\{ e_{1}, ...e_{n}\right\}$ is a local orthonormal frame on $M$.

Let $\psi ^{T}$ be the tangential component of the immersion $\psi :M\rightarrow \mathbf{E}^{m+1}$ and $\sigma = \langle \psi, \xi \rangle$ be the support function of the hypersurface $M$. Then, we have $\psi = \psi ^{T}+\sigma \xi$ and using Eq (2.1), we get

Using above equation, we have

Thus, for a compact hypersurface $M$ of the Euclidean space $\mathbf{E}^{m}$, on integrating the above equation, we have the following Minkowski's formula

On a compact Riemannian manifold $(M, g)$, the Laplace operator $\Delta$ acting on a smooth function $h:M\rightarrow \mathbf{R}$ is defined by $\Delta h = \text{div }\left(\text{grad }h\right)$ and the Hessian operator $H_{h}$ for the smooth function $h$ is a symmetric operator defined by

On a compact Riemannian manifold $(M, g)$, we have the following formula known as Bochner's formula

On the other hand, given a Riemannian manifold $\left (M, g\right)$ and a vector field $X \in \mathfrak{X} (M)\text{, }$, we let $\theta _{X}$ denote the dual one form of $X$ (that is, defined by $\theta _{X} \left (Y\right) = g \left (X, Y\right))$ and $A_{X}$ be the $\left (1, 1\right)$-tensor (viewed as an endomorphism) defined by

Write as usual

for all $Y, Z \in \mathfrak{X} (M)\text{.}$

Let $B$ and $\phi$ be the symmetric and anti-symmetric parts of $A_{X}\text{.}$ In other words, we have

Now, formula $\nabla _{X}\psi ^{T} = X +\sigma A X$ in Eq (2.7) is nothing but $A_{\psi ^{T}} = I +\sigma A\text{.}$ It follows that $B = I +\sigma A$ and $\phi = 0\text{, }$ and this implies that $\psi ^{T}$ is gradient.

3.   Characterizations of spheres

Let $M$ be an orientable compact immersed hypersurface of the Euclidean space $\mathbf{E}^{m+1}$ with immersion $\psi :M\rightarrow \mathbf{E}^{m+1}$ and unit normal $\xi$, shape operator $A$. In this section, we answer the questions raised in the introduction and find two new characterizations of the Euclidean spheres.

Theorem 1. Let $\psi :M \rightarrow \mathbf{E}^{m +1}$ be an immersion of a compact simply connected hypersurface with $\psi ^{T}$ a non-trivial geodesic vector field, $m \geq 2$. Then the Ricci curvature satisfies

if and only if, the mean curvature $\alpha$ is a constant, $\psi ^{T}$ is a non-homothetic conformal vector field, and $M$ is\ isometric to the sphere $\mathbf{S}^{m} \left (\alpha ^{2}\right)$.

Proof. Suppose $\psi ^{T}$ is a geodesic vector field and the Ricci curvature of the hypersurface $M$ satisfies

Then, using Eqs (1.3) and (2.7), we have $\sigma A\psi ^{T} = -\psi ^{T}$. Taking covariant derivative with respect to $X\in \mathfrak{X}(M)$ in this equation and using Eq (2.7), we get

that is

Now, for a local orthonormal frame $\left\{ e_{1}, ..., e_{m}\right\}$ on $M$, choosing $X = e_{i}$ in above equation and taking the inner product with $e_{i}$ in above equation and summing the resulting equation, we conclude

where we used symmetry of the shape operator $A$ and Eq (2.6). Now, using Eq (2.7) in above equation, we have

Note that $\text{div }\left(\alpha \left(\sigma \psi ^{T}\right) \right) = \sigma \psi ^{T}\left(\alpha \right) +\alpha \text{div }\left(\sigma \psi ^{T}\right)$ and using Eqs (2.7), (2.8), we get

Inserting the above equation in Eq (3.2) and using Eq (2.3), we conclude

Integrating the above equation while using Minkowski's formula (2.9), we have

that is,

Now, we use $\text{div }\psi ^{T} = m(1+\sigma \alpha)$ and Eq (2.9), to arrive at

and inserting the above equation in Eq (3.4), we have

Using inequality (3.1) in Eq (3.5), we get

and above inequality in view of the Schwart's inequality $\left\Vert A\right\Vert ^{2}\geq m\alpha ^{2}$ implies

If $\sigma = 0$, then by Minkowski's formula (2.9), we get a contradiction. Thus, we have $\left\Vert A\right\Vert ^{2} = m\alpha ^{2}$ and this equality holds, if and only if, $A = \alpha I$. In other words, $M$ is shown to be totally umbilical. Moreover, using $A = \alpha I$, we have

and we get

Using Eq (2.6), we get $(m-1)\text{grad }\alpha = 0$ and with restriction on dimension $m$, we conclude $\alpha$ is a constant. Moreover, this constant $\alpha \neq 0$ due to the fact that the Euclidean space does not have a compact minimal hypersurface. Inserting $A = \alpha I$ in Eq (2.2), we see that the simply connected hypersurface $M$ has constant curvature $\alpha ^{2}$ and $M$ being compact, it is complete. Hence, $M$ is complete simply connected hypersurface of constant positive curvature $\alpha ^{2}$ and is therefore isometric to the sphere $\mathbf{S}^{m}(\alpha ^{2})$. Since $B = \left (1 +\sigma \alpha \right)I,$ we get $L_{\psi^{T} }g = 2\left(1 +\sigma \alpha \right)g$. In other words, $\psi ^{T}$ is a conformal vector field which is non-homothetic, given that the function $\left.1 +\sigma \alpha \right.$ is not constant as $\psi ^{T}$ is supposed to be non-trivial. The converse is trivial as $\psi ^{T}$ for the natural embedding $\psi :$ $\mathbf{S}^{m}(\alpha ^{2})\rightarrow \mathbf{E}^{m+1}$ has $\psi ^{T} = 0$, which satisfies the hypothesis of the Theorem.

Theorem 2. Let $\psi :M\rightarrow \mathbf{E}^{m+1}$ be an immersion of a compact simply connected hypersurface with $\square \psi ^{T} = 0$, $m\geq 2$. Then the mean curvature $\alpha$ and support function $\sigma$ satisfies

if and only if, the mean curvature $\alpha$ is a constant, $\psi ^{T}$ is a parallel vector field (i.e., the covariant derivative of $\psi ^{T}$ vanishes), and $M$ is isometric to the sphere $\mathbf{S}^{m}\left(\alpha ^{2}\right)$.

Proof. Let $M$ be a compact simply connected hypersurface of the Euclidean space $\mathbf{E}^{m+1}$ with $\square \psi ^{T} = 0$, and

We use Eqs (2.6) and (2.7) in computing $\Delta \psi ^{T}$, where $\Delta$ is rough Laplace operator, and obtain

Using Eq (2.7), we get

Also, in view of Eq (2.4), we have

Now, using Eqs (1.2), (3.7) and (3.8), we conclude

and taking the inner product in above equation with $\psi ^{T}$ and using $\square \psi ^{T} = 0$, we arrive at

Using Eq (3.3), we conclude

and in view of Eq (2.3), we get

Integrating the above equation, while using Eq (2.9), we get

Define a smooth function $h$ on the hypersurface $M$, by $h = \frac{1}{2} \left\Vert \psi \right\Vert ^{2}$, which has gradient $\text{grad }h = \psi ^{T}$ and $\Delta h = m(1+\sigma \alpha)$. Also using Eq (2.7), the Hessian $H_{h}$ is given by

Thus, using Bochner's formula (2.10), we have

Inserting $\Delta h = m(1+\sigma \alpha)$ in the last term of the right hand side of above equation, we have

and using Eq (2.9), we get

Combining Eqs (3.9) and (3.11), we arrive at

Using Schwarz' inequality $\left\Vert H_{h}\right\Vert ^{2}\geq \frac{1}{m} \left(\Delta h\right) ^{2}$ and inequality (3.6) with $m\geq 2$, in above equation, we conclude the equality $\left\Vert H_{h}\right\Vert ^{2} = \frac{1 }{m}\left(\Delta h\right) ^{2}$ and this equality holds, if and only if

Now, using Eq (3.10) and $\Delta h = m(1+\sigma \alpha)$ in above equation, we get

If $\sigma = 0$, we get a contradiction by Eq (2.9). Thus, we get $A = \alpha I$ and following the proof of Theorem 3.1, we get $M$ is isometric to the sphere $\mathbf{S}^{n}(\alpha ^{2})$. We also deduce that $L_{\psi^{T} }g = 2\left(1 +\sigma \alpha \right)g$, with both $\sigma$ and $\alpha$ constants. It follows that $1 +\sigma \alpha = 0$, since otherwise $\psi ^{T}$ becomes a homothetic vector field and consequently $M$ is isometric to the Euclidean space, a contradiction. Thus, $\nabla X = A_{\psi ^{T}} = 0,$ that is $\psi ^{T}$ is parallel. The converse is trivial as on $\mathbf{S}^{m}(\alpha ^{2})$ as hypersurface of the Euclidean space $\mathbf{ E}^{m+1}$, we have $\psi ^{T} = 0$ and $\sigma = -\frac{1}{\alpha }$, and $\left\vert \sigma \alpha \right\vert = 1$.

Acknowledgments

This work was supported by NSTIP strategic technologies program number (13-MAT874-02) in the Kingdom of Saudi Arabia.

Conflict of interest

The authors declare that there is no conflict of interest.