Some generic hypersurfaces in a Euclidean space

1.   Introduction

The geometry of hypersurfaces lies at the foundation of differential geometry, it started with the theory of curves and surfaces in the Euclidean $3$-space $R^{3}$ ^[11]. Given an orientable immersed hypersurface $N$ in the Euclidean space $R^{m+1}$ with immersion $\varphi :N\rightarrow R^{m+1}$, we have the unit normal $\zeta$, the shape operator $T$, the support $\rho = \langle \varphi, \zeta \rangle$ a smooth function defined on the hypersurface $N$ and the mean curvature $\alpha$, given by $m\alpha = trT$ being trace of the shape operator $T$ ^[11]. If the hypersurface $N$ of the Euclidean space $R^{m+1}$ is compact, then we have the following well-known Minkowski's formula:

As an outcome of Minkowski's formula, we conclude that there are no compact minimal hypersurfaces (hypersurfaces with mean curvature $\alpha = 0$) in the Euclidean space $R^{m+1}$.

Among compact hypersurfaces of Euclidean spaces, important are the Euclidean spheres $S^{m}(c)$ of constant curvature $c$, with the imbedding $\varphi :S^{m}(c)\rightarrow R^{m+1}$, $\varphi \left(x\right) = x$, shape operator $T = -\sqrt{c}I$, and unit normal $\zeta = \sqrt{c}\varphi$. Taking $\mathbf{a}$ as a nonzero constant vector field on $R^{m+1}$, we can express it as $\mathbf{a} = \mathbf{u}+f\zeta$, where $f = \langle \mathbf{a}, \zeta \rangle$ and $\mathbf{u}$ is the tangential projection of $\mathbf{a}$ on the sphere $S^{m}(c)$. Letting $g$ be the induced metric and $\nabla$ the Riemannian connection on the sphere $S^{m}(c)$ and differentiating the equation $\mathbf{a} = \mathbf{u}+f\zeta$ with respect to the vector field $E$ on $S^{m}(c)$, we have

where $\nabla f$ is the gradient of $f$.

On an odd dimensional sphere $S^{2m-1}(c)$ with imbedding $\varphi :S^{2m-1}(c)\rightarrow R^{2m}$ with unit normal $\zeta = \sqrt{c}\varphi$, shape operator $T = -\sqrt{c}I$, apart from the above vector field $\mathbf{u}$, there is a unit vector field $\mathbf{v}$ defined on $S^{2m-1}(c)$ by

where $J$ is the complex structure on the Euclidean space $R^{2m}$. Differentiating the above equation using the Euclidean connection $D$ with respect to a vector field $E$ on $S^{2m-1}(c)$, one confirms

that is,

where $\left(JE\right) ^{T}$ is the tangential projection of $JE$ on $S^{2m-1}(c)$.

Given an immersed hypersurface $N$ of the Euclidean space $R^{m+1}$, the natural tools for studying the geometry of $N$ are the shape operator $T$, the mean curvature $\alpha$, the curvature tensor $R$, the Ricci tensor $Ric$, the Ricci operator $S$, and the scalar curvature $\tau$ of $N$. In ^[8], it is shown that a compact hypersurface $M$ of the Euclidean space $R^{m+1}$ satisfies the inequality

if and only if $\alpha$ is a constant and $N$ is isometric to the $n$ -sphere $S^{m}\left(\alpha ^{2}\right)$. Also, in ^[9], the position vector field $\varphi$ of a compactly immersed hypersurface $N$ in the Euclidean space $R^{m+1}$ with immersion $\varphi :N\rightarrow R^{m+1}$ and unit normal $\zeta$ was used to define a vector field $\mathbf{u}$ on the hypersurface $N$ as the tangential projection of the position vector field $\varphi$ that leads to the integral formula

where $\rho = \langle \varphi, \zeta \rangle$ is the support of $N$. In ^[7,9], the above integral was used, which led to many important geometric implications on the compact hypersurface $N$ of the Euclidean space $R^{m+1}$. Moreover, in ^[8], it is shown that a compact hypersurface $N$ of positive Ricci curvature in the Euclidean space $R^{m+1}$ with scalar curvature $\tau \leq \lambda _{1}(m-1)$ is necessarily isometric to the sphere $S^{m}(c)$, where $\lambda _{1}$ is the first nonzero eigenvalue of the Laplace operator $\Delta$ of $N$ with respect to the induced metric.

Recently, there has been a trend toward studying the geometry of the hypersurfaces in $R^{m+1}$, as the graphs of the smooth functions $h:R^{m+1}\rightarrow R$ are called the translation hypersurfaces. The focus, in translation hypersurface $N$ of the Euclidean space $R^{m+1}$, is on the property function $\ h:R^{m+1}\rightarrow R$, whose graph is $N$. In ^[18], translation hypersurfaces of $R^{m+1}$ are studied, whose Gauss-Kronecker curvature depends on either its first $p$ variables or on the rest $q$ variables, where $m = p+q$, and conditions on a translation hypersurface to have Gauss-Kronecker zero curvature are found. If a translation hypersurface $N$ is defined as the graph of the function $h:R^{m+1}\rightarrow R$ with $h$ satisfying certain additional conditions, then it is called a separable hypersurface. Separable hypersurfaces in the Euclidean space $R^{m+1}$ have an interesting geometry, as studied in ^[6,12,13,19]. A complete classification of separable hypersurfaces with zero Gauss-Kronecker curvature in the Euclidean space $R^{m+1}$ is obtained in ^[6].

In this paper, we are interested in studying the impact of the existence of a concircular vector field as well as a Killing vector field on the immersed hypersurface $N$ of the Euclidean space $R^{m+1}$. A vector field ${\omega }$ on a Riemannian manifold $(N, g)$ is a concircular vector field if

where $\sigma$ is a function on $\ N$ and $\Psi (N)$ is the space of smooth vector fields on $N$. We shall use the abbreviation CLVF for a concircular vector field. It is known that a CLVF ${\omega }$ on a Riemannian manifold $(N, g)$ influences the geometry of $(N, g)$ ^[4,5]. Moreover, a CLVF ${ \omega }$ has a role in general relativity ^[3].To understand the role of CLVF in relativity, recall that $m$ -dimensional generalized Robertson-Walker space-time, $m > 3$, is the warped product $I\times _{h^{2}}M$, with Lorentz metric $g = -dt^{2}+h^{2}g^{\ast }$, where $I$ is an interval $h:I\rightarrow R$ is a positive smooth function and $\left(M, g^{\ast }\right)$ is a Riemannian manifold with $\dim M = (m-1)$. In ^[3], Chen has proved a very significant result involving a CLVF, namely: A Lorentzian manifold admits a nontrivial timelike CLVF if and only if it is a generalized Robertson-Walker space-time. Note that Eq (1.2) shows that the vector field $\mathbf{u}$ is a CLVF on the sphere $S^{m}(c)$ with potential function $\sigma = -\sqrt{c}f$ and naturally the shape operator $T$ of the sphere $S^{m}(c)$ as a hypersurface of the Euclidean space $R^{m+1}$ satisfies $T\left(\mathbf{u}\right) = \alpha \mathbf{u}$, where $\alpha = -\sqrt{c}$ is the mean curvature of $S^{m}(c)$. This naturally raises a question: Is a compact and connected hypersurface $N$ with shape operator $T$ and mean curvature $\alpha$ of the Euclidean space $R^{m+1}$ admitting a nonzero CLVF $\mathbf{u}$ satisfying $T\left(\mathbf{u}\right) = \alpha \mathbf{u}$, $\mathbf{u}\left(\alpha \right) = 0$, necessarily isometric to $S^{m}(c)$? In Section 3, we show that this question has an affirmative answer, and indeed, we show that the converse is also true.

Similarly, a vector field ${\omega }$ on an $m$-dimensional Riemannian manifold $(N, g)$ is said to be a Killing vector field if

and we shall use the abbreviation KGVF for a Killing vector field. Note that the presence of a KGVF${\omega }$ on $(N, g)$ influences its geometry as well as topology ^[2,14,17,21]. Note that the unit vector field $\mathbf{v}$ on the sphere $S^{2m-1}(c)$ satisfies Eq (1.4), which leads to

that is, $\mathbf{v}$ is a unit KGVFon the sphere $S^{2m-1}(c)$. We see that $\sigma = g\left(T\mathbf{v}, \mathbf{v}\right) = -\sqrt{c}$ is a constant, and the following holds:

This raises the next question: Does a compact and connected hypersurface $N$ with shape operator $T$, mean curvature $\alpha$, induced metric $g$, admitting a unit KGVF $\mathbf{v}$, of a Euclidean space $R^{m+1}$ with nonzero function $\sigma = g\left(T\mathbf{v}, \mathbf{v}\right)$ satisfying Eq (1.7) necessarily imply $m$ is odd, $\alpha$ a constant, and $M$ isometric to $S^{2m-1}(c)$? In Section 4 of this paper, we answer this question and find a characterization of the sphere $S^{2m-1}(c)$.

Finally, in the last section, we consider an immersed compact and connected hypersurface $N$ in the Euclidean space $R^{m+1}$ with immersion $\varphi :N\rightarrow R^{m+1}$, unit normal $\zeta$, and shape operator $T$. Then, we express the position vector field $\varphi$ as $\varphi = \mathbf{u} +f\zeta$, where $f = \langle \varphi, \zeta \rangle$ is the support function of the hypersurface. In the last section, we shall prove that for a compact and connected hypersurface $N$ with nonzero support function and if the following condition holds

then the mean curvature $\alpha$ is a constant and $N$ is the sphere $S^{m}\left(\alpha ^{2}\right)$.

2.   Preliminaries

Let $N$ be an orientable hypersurface of the Euclidean space $R^{m+1}$ with unit normal $\zeta$, shape operator $T$. We denote the Euclidean metric by $\langle, \rangle$ and by $g$ the induced metric on $N$, and by $\nabla$ and $D$, the Riemannian connection with respect to $g$ and the Euclidean connection, respectively. Then, we have ^[11]

where $\Psi \left(N\right)$ is the space of smooth vector fields on $N$. The curvature tensor field of the hypersurface $N$ is given by

and the Ricci tensor of $N$ has the expression

where $\alpha$ is the mean curvature of the hypersurface $N$, given by $m\alpha = trT$, the trace of the shape operator $T$. For a local orthonormal frame $\left\{ w_{k}\right\} _{1}^{m}$ on the hypersurface, the scalar curvature $\tau$ of the hypersurface $N$ is given by

and combining the above equation with (2.3), gives

where

The Codazzi equation of the hypersurface $N$ is given by

where $\left(\nabla _{E}T\right) F = \nabla _{E}TF-T\left(\nabla _{E}F\right)$. Note that, as the shape operator $T$ is symmetric, we have for $E\in \Psi (N)$ and a local frame $\left\{ w_{k}\right\} _{1}^{m}$,

and using the facts that

where $\left(\lambda _{k}^{j}\right)$ is a symmetric matrix and $\omega _{k}^{i}$ are connection forms, which are skew symmetric, that is, $\omega _{k}^{i}+\omega _{i}^{k} = 0$; in Eq (2.6), we conclude

Therefore, the gradient of $\alpha$ has the expression

Let ${\omega }$ be a CLVF on an $m$-dimensional Riemannian manifold $(N, g)$. Then, we have

where $\sigma$ is the potential function of the CLVF${ \omega }$. A CLVF${\omega }$ on $(N, g)$ is said to be nontrivial if it is not parallel. We have the following expression for the curvature tensor field of $(N, g)$ involving the CLVF${ \omega }$

Taking the trace in the above equation, we see that the Ricci tensor of $\left(N, g\right)$ is given by

The Ricci operator $S$ of the Riemannian manifold $\left(N, g\right)$ is given by

and thus, using Eq (2.9), we see that the Ricci operator $S$ operating on the CLVF${\omega }$ is given by

where $\nabla \sigma$ is the gradient of $\sigma$.

Now, consider a KGVF $\mathbf{v}$ on an $m$-dimensional Riemannian manifold $(N, g)$ that satisfies ^[11]

Note that the flow of a KGVF on a Riemannian manifold consists of isometries, and therefore, its presence influences both the topology and geometry of the manifold on which they live. For instance, if $\mathbf{v}$ is a KGVFon a Riemannian manifold $(N, g)$, then the scalar curvature $\tau$ of $(N, g)$ is constant along the integral curves of $\mathbf{v}$. It is known that, if a positively curved Riemannian manifold $\left(N, g\right)$ admits a nontrivial KGVF, then its fundamental group contains a cyclic subgroup of constant index depending on $\dim N$ ^[17]. Also, the presence of a nontrivial KGVFinfluences the dimension of the Riemannian manifold on which they live. For instance, on the even-dimensional unit sphere $S^{2m}$ there does not exist a unit KGVF, where as on $S^{2m+1}$ a unit KGVF exists ^[2,11]. Moreover, the presence of a nontrivial KGVFon a compact Riemannian manifold $(N, g)$ does not allow it to have a non-positive Ricci curvature ^[11].

There is a skew-symmetric operator $\phi$ associated with the KGVF $\mathbf{v}$ on $(N, g)$ that satisfies

and that the covariant derivative of the operator $\phi$ is given by

It is clear from Eq (2.12) that $\mathbf{v}$, being a unit KGVF on $(N, g)$, satisfies

Note that the flow of a KGVF $\mathbf{v}$ on an $m$-dimensional Riemannian manifold $(N, g)$ consists of isometries of $(N, g)$. Now suppose that $N$ is an orientable hypersurface of the Euclidean space $R^{m+1}$ with shape operator $T$, mean curvature $\alpha$, and induced metric. Suppose that there is a unit KGVF $\mathbf{v}$ on the hypersurface $N$. We say that the shape operator $T$ of the hypersurface is invariant under the unit KGVF $\mathbf{v}$ if

where $\left\{ \psi _{t}\right\}$ is the flow of the unit KGVF $\mathbf{v}$.

Lemma 1. Let $\mathbf{v}$ be a unit KGVF on the hypersurface $N$ of the Euclidean space $R^{m+1}$ such that the shape operator $T$ is invariant under $\mathbf{v}$. Then the shape operator satisfies

Proof. Since $T$ is invariant under $\mathbf{v}$, Eq (2.15) implies

which gives

that is, in view of Eq (2.12), we have

Combining the above equation with Eq (2.5), we get the result. □

3.   Hypersurfaces with a concircular vector field

In this section, we are interested in studying the impact of a nonzero CRVF $\mathbf{\omega }$ with potential $\sigma$ on a compact hypersurface $N$ of the Euclidean space $R^{m+1}$. We would like to recall that given a smooth curve $\beta :I\rightarrow N$ on the hypersurface $N$ with mean curvature $\alpha$, we get a smooth function $f:I\rightarrow R$ defined by $f = \alpha \circ \beta$ and if $f$ is a constant function, we say the mean curvature $\alpha$ is a constant along the curve $\beta$ on the hypersurface. Naturally, if the mean curvature $\alpha$ is a constant, then it will be constant along each curve on the hypersurface. However, mean curvature $\alpha$ being constant along some curves on hypersurface $N$ does not imply that $\alpha$ is a constant on $N$. In the following result, we shall assume that the mean curvature $\alpha$ is a constant along the integral curves of the CRVF $\mathbf{\omega }$, which is a weaker condition than asking if the mean curvature $\alpha$ is a constant. Indeed, we prove the following:

Theorem 1. A compact and connected hypersurface $N$ of the Euclidean space $R^{m+1}$, $m > 1$, admits a nonzero nontrivial CRVF ${\omega }$ such that the mean curvature $\alpha$ is constant along the integral curves of ${\omega }$ and the shape operator $T$ satisfies $T\left({\omega }\right) = \alpha {\omega }$, if and only if $\alpha$ is a constant and $N$ is isometric to $S^{m}\left(\alpha ^{2}\right)$.

Proof. Suppose that the compact and connected hypersurface $N$ of $R^{m+1}$, $m > 1$, admits a nonzero nontrivial CRVF ${\omega }$ with potential $\sigma$, such that the mean curvature $\alpha$ is constant along the integral curves of ${\omega }$ and the shape operator $T$ satisfies

Then we have

Using Eqs (2.8) and (3.1), we get

that is,

Now, using a local frame $\left\{ w_{k}\right\} _{1}^{m}$ on the hypersurface $N$, we have

and employing Eq (3.3) in the above equation leads to

Moreover, Eqs (3.1) and (3.2) give

Next, using Eq (3.1), we compute

which, on using Eq (3.2), on some simplifications, gives

Thus, Eqs (3.4)–(3.6), yield

Also, we have

Substituting this last equation in Eq (3.7), we arrive at

and as ${\omega }$ is a nonzero vector field on the connected hypersurface $N$, we conclude that $\alpha$ is a constant. Now, using Eq (3.1) in the expression of the Ricci operator $S$ of the hypersurface $N$, we get

Combining this equation with Eq (2.10), we have

Differentiating the above equation with respect to a vector field $E$ on $N$, and using Eq (2.8), we get

The mean curvature $\alpha$ is a constant; it has to be a nonzero constant as $N$ is a compact hypersurface by virtue of the fact that there are no compact minimal hypersurfaces in the Euclidean space $R^{m+1}$, which is guaranteed by Minkowski's formula (1.1). Now, it remains to show that the potential $\sigma$ cannot be a constant. To achieve it, we see that Eq (2.8) implies ${div}{ \omega } = m\sigma$, which, on integration, yields

and if $\sigma$ were a constant, it should give $\sigma = 0$, which would make $\omega$ a trivial CRVF, which is a contradiction. Hence, $\sigma$ is a non-constant function. Hence, Eq (3.8) is Obata's differential equation ^[15,16], which confirms that $N$ is isometric to $S^{n}\left(\alpha ^{2}\right)$.

Conversely, suppose $N$ is isometric to $S^{n}\left(c\right)$. Then, by Eq (1.2), there is a CRVF$\mathbf{u}$ on $S^{m}(c)$ with potential $\sigma = -\sqrt{c}f$. We claim that $\mathbf{u}$ is a nonzero and nontrivial CRVF$\mathbf{\ }$on $S^{m}(c)$. If $\mathbf{u} = 0$, then by Eq (1.2), it will follow that $f = 0$, and consequently, the constant vector $\mathbf{a} = 0$, which is contrary to our assumption that $\mathbf{a}$ is a nonzero constant vector field on the Euclidean space $R^{m+1}$. Similarly, if $\mathbf{u}$ is parallel, then by Eq (1.2), we have $f = 0$, and the second equation in Eq (1.2) will imply $\mathbf{u} = 0$, which is a contradiction. Hence, $\mathbf{u}$ is a nonzero and nontrivial CRVF$\mathbf{\ }$ on $S^{m}(c)$, which satisfies $T\left(\mathbf{u}\right) = \alpha \mathbf{u}$ and $\mathbf{u}\left(\alpha \right) = 0$. This completes the proof. □

4.   Hypersurfaces with a Killing vector field

In this section, we are interested in studying hypersurfaces of the Euclidean space $R^{m+1}$, which admit a unit KGVF. Let $N$ be an orientable hypersurface of the Euclidean space $R^{m+1}$ with shape operator $T$, mean curvature $\alpha$, and $\mathbf{v}$ be a unit KGVFon $N$ with respect to which the shape operator $T$ is invariant. We prove the following:

Theorem 2. A compact and connected hypersurface $N$ of the Euclidean space $R^{m+1}$, $m > 1$, with mean curvature $\alpha$ and shape operator $T$, admits a unit KGVF $\mathbf{v}$ such that the shape operator $T$ is invariant under $\mathbf{v}$ and the function $\sigma = g\left(T\mathbf{v}, \mathbf{v}\right)$ is nonzero and satisfies

if and only if $m$ is odd, $m = (2n-1)$, $\alpha$ is a constant, and $N$ is isometric to $S^{2n-1}\left(\alpha ^{2}\right)$.

Proof. Suppose $N$ is a compact and connected hypersurface of the Euclidean space $R^{m+1}$, $m > 1$, that admits a unit KGVF $\mathbf{v}$ such that the shape operator $T$ is invariant under $\mathbf{v}$ and the function $\sigma = g\left(T\mathbf{v}, \mathbf{v}\right)$ is nonzero and satisfies the condition

Define a vector field $\mathbf{u} = T\mathbf{v}-\sigma \mathbf{v}$; it follows that $g\left(\mathbf{u}, \mathbf{v}\right) = 0$, that is, the vector field $\mathbf{u}$ is orthogonal to the unit KGVF $\mathbf{v}$. Now, using Eq (2.12) and Lemma 1, we compute

that is,

Taking the inner product in the above equation with the vector field $\mathbf{v}$ and using $g\left(\mathbf{u}, \mathbf{v}\right) = 0$ and Eqs (2.12) and (2.14), we get

that is,

Differentiating the above equation with respect to $E\in \Psi \left(N\right)$ and using Eqs (4.2), (2.13), and (2.14), we get

Note that by Eqs (2.13) and (2.14), we have

and using it in Eq (4.4), we conclude

Now, using $\mathbf{u} = T\mathbf{v}-\sigma \mathbf{v}$ to plug the first and last terms in the right-hand side of the above equation, we confirm

which, using Eq (2.2), yields

Taking the trace in the above equation and using $\Delta \sigma = {div} \left(\nabla \sigma \right)$, we conclude

that is,

Using Eq (2.3), we have

and inserting it in Eq (4.6), gives

Integrating the above equation, yields

which is rearranged as

Using the inequality (4.1) in the above equation, it confirms

However, by Schwartz's inequality, we have $\left\Vert T\right\Vert ^{2}\geq m\alpha ^{2}$ and therefore, the integrand on the left-hand side of the above inequality is non-negative. Hence, we have

with the function $\sigma$ nonzero on connected $N$, which implies $\left(\left\Vert T\right\Vert ^{2}-m\alpha ^{2}\right) = 0$. The equality $\left\Vert T\right\Vert ^{2} = m\alpha ^{2}$ in Schwartz's inequality holds if and only if

which gives

Taking a local frame $\left\{ w_{k}\right\} _{1}^{m}$ on the hypersurface $N$, in the above equation, we have

which, in view of Eq (2.7), implies

and as $m > 1$, it confirms that $\alpha$ is a constant. Then, by Eqs (2.2) and (4.7), we have

Note that $\alpha \neq 0$, because compact minimal hypersurfaces in Euclidean space do not exist. Hence, $\alpha ^{2} > 0$, and $N$ is isometric to $S^{m}\left(\alpha ^{2}\right)$. Note that a Killing vector field on an even-dimensional compact Riemannian manifold of positive sectional curvature must vanish at some point ^[11]. Therefore, as $\mathbf{v}$ is a unit vector field, it never vanishes, and it announces that $m$ cannot be even. Hence, $m = 2n-1$, that is, $N$ is isometric to $S^{2n-1}\left(\alpha ^{2}\right)$.

Conversely, suppose $N$ is isometric to $S^{2n-1}\left(\alpha ^{2}\right)$. Then by Eqs (1.3) and (1.4), there is a unit vector field $\mathbf{v} = J\zeta$ on $S^{2n-1}\left(\alpha ^{2}\right)$ that satisfies

where $J$ is the complex structure of the ambient Euclidean space $R^{2n}$, and $\zeta$ is the unit normal, and $\left(JE\right) ^{T}$ is the tangential projection of the vector field $JE$ to $S^{2n-1}\left(\alpha ^{2}\right)$. Taking the inner product in Eq (4.8) by the vector field $F$ on the sphere $S^{2n-1}\left(\alpha ^{2}\right)$, we have

and we conclude

by virtue of the skew symmetry of the complex structure, that is, the Euclidean metric is a Hermitian metric. Hence, $\mathbf{v}$ is a unit KGVF on $S^{2n-1}\left(\alpha ^{2}\right)$. Note that, in this case the shape operator is $T = \alpha I$, and the function $\sigma = g\left(T\mathbf{v}, \mathbf{v}\right) = \alpha$ is a nonzero constant. Moreover, with $m = 2n-1$

and

as $\nabla \sigma = 0$. Hence, by Eqs (4.9) and (4.10), we get

and this finishes the proof. □

5.   Hypersurfaces with a generic bound on Ricci curvature

Let $N$ be an immersed hypersurface in the Euclidean space $R^{m+1}$ with unit normal $\zeta$, shape operator $T$, and mean curvature $\alpha$. Let $\varphi :N\rightarrow R^{m+1}$ be the immersion and $\rho = \langle \varphi, \zeta \rangle$ be the support of $N$. The position vector field $\varphi$ is expressed as

and we call $\mathbf{u}$ the basic vector field of the hypersurface $N$. Differentiating Eq (5.1), using Eq (2.1), and equating similar components, we get

The first equation in Eq (5.2), gives

In this section, we prove the following result:

Theorem 3. A compact and connected immersed hypersurface $N$ of the Euclidean space $R^{m+1}$, $m > 1$, with nonzero support $\rho$ and basic vector field $\mathbf{u}$ satisfies$_{{}}$

if and only if, the mean curvature $\alpha$ is a constant and $N$ is isometric to $S^{m}(\alpha ^{2})$.

Proof. Suppose that the immersed hypersurface $N$ of the Euclidean space $R^{m+1}$, $m > 1$, has nonzero support $\rho$ and the basic vector field $\mathbf{u}$ satisfy

Using Eq (5.2), we have

and using a local frame $\left\{ w_{k}\right\} _{1}^{m}$ on the hypersurface $N$ with the above equation, we get

Using Eq (5.3) in the above equation, we have

Note that on using Eq (5.2), we have

which gives

Integrating the last equation, while using Minkowski's formula, we have

Next, we recall the following integral formula ^[20]

which holds for any vector field on the compact Riemannian manifold (N, g).

Using the above integral formula with the integral of Eq (5.5), we get

Now, using Eq (1.1) in Eq (5.6), we have

Employing (5.1), in the above equation, we conclude

Inserting this equation in Eq (5.7), we find that

Finally, observe that

that is,

and utilizing the above equation in Eq (5.8), we obtain

Using inequality (5.4) in the above equation, we get

which gives $\rho ^{2}\left\Vert T-\alpha I\right\Vert ^{2} = 0$. However, the support $\rho \neq 0$ on connected $N$ implies

and as in the proof of Theorem 2, we realize that $\alpha$ is a constant, and by Eq (2.2), the curvature tensor of $N$ is given by

with constant $\alpha \neq 0$ as there are no compact minimal hypersurfaces in the Euclidean space. Hence, $N$ is isometric to $S^{m}\left(\alpha ^{2}\right)$.

Conversely, suppose $N$ is isometric to $S^{m}\left(\alpha ^{2}\right)$. Then, the embedding $\varphi :S^{m}\left(\alpha ^{2}\right) \rightarrow R^{m+1}$ has shape operator $T = \alpha I$, unit normal $\zeta = -\alpha \varphi$ and support $\rho = -\frac{1}{\alpha }\neq 0$. Moreover, the basic vector field $\mathbf{u} = 0$. Hence, the condition (5.4) vacuously holds as an equality. □

6.   Conclusions

In Sections 3 and 4, we have employed a CLVF and a KGVF on a compact hypersurface $N$, respectively, of the Euclidean space $R^{m+1}$ to find a characterization of spheres $S^{m}\left(c\right)$ and $S^{2n-1}(c)$, respectively. This further increases the scope of the study of hypersurfaces in the Euclidean space $R^{m+1}$; for instance, one would be interested in analyzing the impact of the presence of a geodesic vector field $\xi$ on an orientable hypersurface $N$ of the Euclidean space $R^{m+1}$ ^[10]. A vector field $\xi$ on a Riemannian manifold $\left(N, g\right)$ is said to be a geodesic vector field, if its integral curves are geodesics of $\left(N, g\right)$. A unit Killing vector field on $\left(N, g\right)$ is a geodesic vector field, and the converse is not true. To support this fact that a geodesic vector field need not be a KGVF, we need to introduce a $3$-dimensional trans-Sasakian manifold $\left(N, g, \phi, \zeta, \eta, f, h\right)$, where $(N, g)$ is a $3$-dimensional Riemannian manifold, $\phi$ is a $(1, 1)$ tensor field, $\zeta$ is a unit vector field (called Reeb vector field), $\eta$ is $1$-form dual to $\zeta$, and $f$, $h$ are smooth functions on $M$ satisfying ^[1]

and

$E, F\in \Psi (N)$. A trans-Sasakian manifold $\left(N, g, \phi, \zeta, \eta, f, h\right)$ is said to be proper, if neither of the functions $f$ nor $h$ are zero. It is easy to see that $\nabla _{\zeta }\zeta = 0$, that is, $\zeta$ is a geodesic vector field. However, on a proper trans-Sasakian manifold $\left(N, g, \phi, \zeta, \eta, f, h\right)$

that is, $\zeta$ is not a Killing vector field. Hence, on a proper trans-Sasakian manifold $\left(N, g, \phi, \zeta, \eta, f, h\right)$, the Reeb vector field $\zeta$ is a geodesic vector field that is not a KGVF. Thus, a geodesic vector field being a nontrivial generalization of a Killing vector field makes it a potential case for studying the impact of the presence of a geodesic vector field on the geometry of an orientable hypersurface of the Euclidean space $R^{m+1}$.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors would like to extend their sincere appreciation to Supporting Project number (RSPD2024R860) King Saud University, Riyadh, Saudi Arabia.

Conflict of interest

The authors declare that they have no conflicts of interest.