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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2021; 61(1): 191-203

Published online March 31, 2021

Copyright © Kyungpook Mathematical Journal.

### Second Order Parallel Tensor on Almost Kenmotsu Manifolds

Venkatesha Venkatesha*, Devaraja Mallesha Naik, Aysel-Turgut Vanli

Department of Mathematics, Kuvempu University, Shivamogga 577-451, India
e-mail : vensmath@gmail.com

Department of Mathematics, Kuvempu University, Shivamogga 577-451, India Department of Mathematics, CHRIST (Deemed to be University), Bengaluru 560029, Karnataka, India
e-mail : devaraja.mallesha@christuniversity.in

Department of Mathematics, Gazi University, Ankara, Turkey
e-mail : avanli@gazi.edu.tr

Received: November 14, 2019; Revised: July 18, 2020; Accepted: July 21, 2020

### Abstract

Let M be an almost Kenmotsu manifold of dimension 2n+1 having non-vanishing ξ-sectional curvature such that $trl>−2n−2$. We prove that any second order parallel tensor on M is a constant multiple of the associated metric tensor and obtained some consequences of this. Vector fields keeping curvature tensor invariant are characterized on M.

Keywords: almost Kenmotsu manifold, second order parallel tensor, nullity distribution, homothetic vector field.

### 1. Introduction

In 1923, Eisenhart [13] proved that if a positive definite Riemannian manifold (M, g) admits a second order parallel symmetric covariant tensor other than the constant multiple of metric tensor, then it is reducible. In 1926, Levy [18] proved that a second order parallel symmetric non-singular tensor in a space of constant curvature is a constant multiple of the metric tensor. Using Ricci identities, Sharma in [24] gives a global approach to the Eisenhart problem, and generalized the Levy's theorem. This problem was studied by the same author in contact geometry [25, 26, 27] on different manifolds, for example for K-contact manifolds in [26]. Since then many geometers have investigated the Eisenhart problem on various contact manifolds: nearly Sasakian[30], P-Sasakian [7, 29], f-Kenmotsu manifold [4], N(κ)-quasi Einstein manifold [6], 3-dimensional normal paracontact geometry [1], contact manifolds having non-vanishing ξ-sectional curvature [14], $(κ,μ)$-contact metric manifold [19], almost Kenmotsu manifolds [34] and 3-dimensional non-cosymplectic normal almost contact pseudo-metric manifold of non-vanishing ξ-sectional curvature [32].

In contact geometry, Kenmotsu manifolds, introduced by Kenmotsu in [17], are one of the important classes of manifolds. Such manifolds were observed to be normal. Let $(M,φ,ξ,η,g)$ be an almost Kenmotsu structure (see Section 2) on a (2n+1)-dimensional differentiable manifold. The purpose of this paper is to study second order parallel tensors on M under certain conditions. Throughout the paper, we suppose that the almost Kenmotsu manifold is of dimension 2n+1. Denoting the Ricci tensor by S, the tensor $12£ξφ$ by h, where $£$ denotes the Lie differentiation, and the operator $R(⋅,ξ)ξ$ by $l$, we prove the following.

### Theorem 1.1.

Let α be a second order symmetric parallel tensor, and A be the (1,1)-tensor metrically equivalent to α on an almost Kenmotsu manifold M. The following hold

• (i) $Aξ=α(ξ,ξ)ξ$, if M has non-vanishing ξ-sectional curvature;

• (ii) $trA=α(ξ,ξ)−tr(A(h2−2φh−φ(∇ξh)))+α(ξ,ξ)S(ξ,ξ)$;

• (iii) $tr(Al)=α(ξ,ξ)S(ξ,ξ),$

where tr denotes the trace.

Since the Ricci tensor S is a second order tensor, we have:

### Corollary 1.1.

If an almost Kenmotsu manifold M is Ricci symmetric, and if M has non-vanishing ξ-sectional curvature, then

• (i) $Qξ=−(2n+trh2)ξ$;

• (ii) the scalar curvature $r=‖Qξ‖2−2n−tr(h2)−tr(Q(h2−2φh−φ(∇ξh)));$

• (iii) $tr(Ql)=‖Qξ‖2,$

where Q is the Ricci operator determined by S(X,Y)=g(QX,Y).

### Theorem 1.2.

Let M be an almost Kenmotsu manifold having non-vanishing ξ-sectional curvature such that $trl>−2n−2$. The second order parallel tensor on M is a constant multiple of the associated metric tensor.

### Corollary 1.2.

If M is an almost Kenmotsu manifold having non-vanishing ξ-sectional curvature such that $trl>−2n−2$ is Ricci symmetric, then it is Einsteinian.

Since $trl=S(ξ,ξ)=−2n$ and the ξ-sectional curvature $K(ξ,X)=−1$ for Kenmotsu manifold, we have the following.

### Corollary 1.3.

Any Ricci symmetric Kenmotsu manifold is Einsteinian.

### Corollary 1.4.

An affine Killing vector field on M which has non-vanishing ξ-sectional curvature such that $trl>−2n−2$ is homothetic.

Blair, Koufogiorgos and Papantoniou [3] introduced $(κ,μ)$-nullity distributions on a contact metric manifold generalizing the notion of $κ$-nullity distributions by defining

$Np(κ,μ)={Z∈TpM:R(U,V)Z=κ[g(V,Z)U−g(U,Z)V] +μ[g(V,Z)hU−g(U,Z)hV]},$

for any p ∈ M, where $κ,μ∈ℝ$.

### Corollary 1.5.

A second order parallel tensor on an almost Kenmotsu manifold with $ξ∈N(κ,μ)$ is a constant multiple of the associated metric tensor.

The above corollary has been proved by Wang and Liu in [34]. Recently, Dileo-Pastore [12] introduced $(κ,μ)′$-nullity distribution on an almost Kenmotsu manifold which is defined by

$Np(κ,μ)′={Z∈TpM:R(U,V)Z=κ[g(V,Z)U−g(U,Z)V] +μ[g(V,Z)h′U−g(U,Z)h′V]},$

for any p∈ M, where $h′=h∘φ$ and $κ,μ∈ℝ$. Here we recall the following results due to Dileo-Pastore [11, 12].

### Lemma 1.1.

([12, Proposition 4.1]) Let M be an almost Kenmotsu manifold such that $h′≠0$ with $ξ∈N(κ,μ)′$. Then $κ<−1$, $μ=−2$ and $spect(h′)={0,λ,−λ}$ with 0 as simple eigenvalue and $λ=−1−k.$

### Lemma 1.2.

([11, Theorem 6]) Let M be a locally symmetric almost Kenmotsu manifoldsuch that $h′≠0$ and $R(X,Y)ξ=0$ for any $X,Y∈D$, where $D=ker(η).$ Then, M is locally isometric to $ℍn+1(−4)×ℝn$.

### Lemma 1.3.

([12, Corollary 4.2]) Let M be an almost Kenmotsu manifold such that $h′≠0$ with $ξ∈N(κ,μ)′$. Then M is locally symmetric if and only if $spect(h′)={0,1,−1}$ that is if and only if k=-2.

### Lemma 1.4.

([12, Proposition 4.3]) Let M be an almost Kenmotsu manifold such that $h′≠0$ with $ξ∈N(κ,μ)′$. Then the ξ-sectional curvature satisfies

$K(ξ,X)=κ−2λ, if X∈[λ]′κ+2λ, if X∈[−λ]′,$

where $[λ]′$ denotes the eigenspace of h' related to the eigenvalue $λ$.

We use these to prove:

### Theorem 1.3.

Let M be an almost Kenmotsu manifold with $ξ∈N(κ,μ)′$ and $h′≠0$. If M admits a second order parallel tensor, then either the second order parallel tensor is a constant multiple of the associated metric tensor or M is locally isometric to $ℍn+1(−4)×ℝn$.

In [34], Wang and Liu proved the above theorem in another way. Now waving the hypothesis non-vanishing ξ-sectional curvature in Theorem 1.2 by $∇ξh=0$, we prove:

### Theorem 1.4.

Let M be an almost Kenmotsu manifold with $∇ξh=0$ such that $trl>−2n−2$. Then the second order parallel tensor on M is a constant multiple of the associated metric tensor.

In [23] Naik et al. proved that every vector field which leaves the curvature tensor invariant are Killing in a $(κ,μ)′$-almost Kenmotsu manifold with $h′≠0$ and $κ≠−2$. Here, as an application of Theorem 1.4, we prove the following.

### Theorem 1.5.

Let M be an almost Kenmotsu manifold with $∇ξh=0$ such that $trl>−2n−2$. Then every vector field keeping curvature tensor invariant are homothetic.

The m-Bakry-Emery Ricci tensor is a natural extension of the Ricci tensor to smooth metric measure spaces and is given by

$Sfm=S+Hessf−1mdf⊗df,$

where f is a smooth function on M and m is an integer such that $0. If $Sfm$ is a constant multiple of the metric g, then the Riemannian manifold (M,g) is called m-quasi-Einstein manifold (see [5, 15] and the references therein). Now applying Theorem 1.2, Theorem 1.3 and Theorem 1.4, we deduce the following statement.

### Theorem 1.6.

Let M be an almost Kenmotsu manifold either having non-vanishing ξ-sectional curvature such that $trl>−2n−2$ or $ξ∈N(κ,μ)′$ and $h′≠0$ or $∇ξh=0$ such that $trl>−2n−2$. Then the m-Bakry-Emery Ricci tensor $Sfm=S+Hessf−1mdf⊗df$ is parallel if and only if M is m-quasi-Einstein manifold.

A Ricci soliton on a Riemannian manifold (M,g) is defined by

$£Vg+2S+2ρg=0,$

where V is a smooth vector field and ρ is a constant. In the context of contact geometry and paracontact geometry, Ricci solitons are studied in [8, 9, 10, 20, 21, 22, 28, 31, 33]. In the similar vein as Theorem 1.6. we state the following.

### Theorem 1.7.

Let M be an almost Kenmotsu manifold either having non-vanishing ξ-sectional curvature such that $trl>−2n−2$ or $ξ∈N(κ,μ)′$ and $h′≠0$ or $∇ξh=0$ such that $trl>−2n−2$. Then the the second order tensor $£Vg+2S$ is parallel if and only if M admits a Ricci soliton.

### 2. Preliminaries

An almost contact metric structure on a (2n+1)-Riemannian manifold M is a quadruple $(φ,ξ,η,g)$, where ⊎ is an endomorphism, ξ a global vector field, η a 1-form and g a Riemannian metric, such that

$φ2=−I+η⊗ξ, η(ξ)=1,$
$g(φX,φY)=g(X,Y)−η(X)η(Y).$

We easily obtain from (2.1) that $φξ=0$ and $η∘φ=0$ (see [2]).

A manifold M with $(φ,ξ,η,g)$ structure is said to be an almost contact metric manifold. We define the fundamental 2-form on $(M,φ,ξ,η,g)$ by $Φ(X,Y)=g(X,φY)$.

If the 1-form $η$ is closed and $dΦ=2η∧Φ$, then M is said to be an almost Kenmotsu manifold [16]. It is well known that a normal almost Kenmotsu manifold is a Kenmotsu manifold.

The two tensor fields $l:=R(⋅,ξ)ξ$ and $h:12£ξφ$ are known to be symmetric and satisfy

$lξ=0 hξ=0, trh=0, hφ+φh=0.$

Further, one has the following formulas:

$∇ξφ=0,$

$(∇ξh)X=−φX−2hX−φh2X−φR(X,ξ)ξ,$

$trl=S(ξ,ξ)=−2n−trh2.$

### 3. Proof of Theorems

Now we prove the results stated in Section 1.

Proof of Theorem 1.1.

Suppose that α is a symmetric (0, 2)-tensor and A be the (1, 1)-tensor metrically equivalent to α, that is, $g(AX,Y)=α(X,Y)$. Note that $∇α=0$ implies $∇A=0$, and so

$R(X,Y)AZ=AR(X,Y)Z.$

Therefore

Thus we get

$g(R(Z,AW)X,Y)=g(R(X,Y)Z,AW) =−g(R(X,Y)W,AZ)=−g(R(W,AZ)X,Y).$

Setting $X=Z=ξ$ in the above equation, we get

$g(R(ξ,AW)ξ,Y)+g(R(W,Aξ)ξ,Y)=0.$

For $W=ξ$ and $Y=Aξ$, equation (3.1) becomes

$g(R(ξ,Aξ)ξ,Aξ)=0.$

Since the ξ-sectional curvature $K(ξ,X)$ is non-vanishing, equation (3.2) implies that $Aξ=fξ$, for some scalar function f on M. As $g(ξ,ξ)=1$, we have

$f=g(Aξ,ξ)=α(ξ,ξ).$

This proves (i). Applying $φ$ to (2.6) shows that

$R(ξ,X)ξ=h2X−φ2X−2φhX−φ(∇ξh)X.$

Using (3.3) in 3.1, we get

$g(h2AW+AW−η(AW)ξ−2φhAW−φ(∇ξh)AW,Y)+g(R(W,Aξ)ξ,Y)=0.$

If ${ei}$ is a local orthonormal basis, then putting $W=Y=ei$ in the above equation and summing it over i, leads to

$tr(h2A)+trA−α(ξ,ξ)−2tr(φhA)−tr(φ(∇ξh)A)+α(ξ,ξ)S(ξ,ξ)=0.$

This gives (ii). Now plugging X by AX in 3.3 and then contracting with respect to X gives

$tr(Al)=tr(φ(∇ξh)A)−trA+α(ξ,ξ)+2tr(φhA)−tr(h2A).$

Using this in (ii) yields (iii). This finishes the proof.

Proof of Corollary 1.1. As the Ricci tensor S is parallel, it follows from Theorem 1.1 that

$Qξ=S(ξ,ξ)ξ.$

Now (i) follows from (2.8). Note that from (3.4), we have

$S(ξ,ξ)S(ξ,ξ)=g(Qξ,Qξ)=‖Qξ‖2.$

Hence (ii) and (iii) follows directly from (ii) and (iii) of Theorem 1.1.

Proof of Theorem 1.2. Let ${ei,φei,ξ}i=1n$ be a local orthonormal basis such that $hei=λiei$. Then $hφei=−λiφei$. Hence as $hξ=0$, we have

$trh2=2∑ i=1nλi2.$

But by hypothesis, $trl=−2n−trh2>−2n−2,$ and so $trh2<2$. Therefore $∑ i=1nλi2<1$ which means $λi2<1$ for each i. This fact will be used in the rest of analysis for this theorem.

Let α be a (0,2)-tensor such that $∇α=0$, and A be the dual (1, 1)-type tensor which is metrically equivalent to α, that is, α(X,Y)=g(AX,Y). We will analyse the symmetric and anti-symmetric cases of second order tensor separately.

First, suppose that α is symmetric. Then from item (i) of Theorem 1.1, we get

$Aξ=α(ξ,ξ)ξ.$

To show $α(ξ,ξ)$ is constant, we differentiate it along X to obtain

$X(α(ξ,ξ))=2α(∇Xξ,ξ)=2g(−φ2X−φhX,Aξ) =2α(ξ,ξ)g(−φ2X−φhX,ξ)=0.$

Now differentiating (3.5) along X yields

$A(−φ2X−φhX)=α(ξ,ξ)(−φ2X−φhX).$

Replacing X by $φX$, it follows that

$A(φX−hX)=α(ξ,ξ)(φX−hX).$

Now putting $X=ei$ and $X=φei$ in (3.6) respectively gives

$A(φei)−λiA(ei)=α(ξ,ξ){φei−λiei},$

and

$λiA(φei)−A(ei)=α(ξ,ξ){λiφei−ei}.$

Multiplying $λi$ to (3.7) and then subtracting it with (3.8) shows

$(λi2−1)A(ei)=α(ξ,ξ)(λi2−1)ei.$

Similarly, we can find

$(λi2−1)A(φei)=α(ξ,ξ)(λi2−1)φei.$

Since $λi2<1$, we have $A(ei)=α(ξ,ξ)ei$ and $A(φei)=α(ξ,ξ)φei$ for every $i=1,…,n$. But $Aξ=α(ξ,ξ)ξ$, and so

$AX=α(ξ,ξ)X,$

for any X ∈ TM, that is, α is a constant multiple of g. Now suppose that α is skew-symmetric. Note that $∇α=0$ implies

$g(R(X,Y)AZ,W)=g(AR(X,Y)Z,W).$

The skew-symmetry of α then gives

$g(R(X,Y)AZ,W)+g(R(X,Y)Z,AW)=0.$

For $X=A2ξ$, $Y=Z=ξ$ and $W=Aξ$, we find

$g(R(A2ξ,ξ)ξ,A2ξ)=0.$

From (3.10), the hypothesis $K(ξ,X)$ is non-vanishing imply that $A2ξ=fξ$. Clearly,

$f=−g(Aξ,Aξ)=−‖Aξ‖2.$

One can easily verify the constancy of f by differentiating (3.11) along any vector field and getting the derivative 0. Now differentiating $A2ξ=−‖Aξ‖2ξ$ along $φX$ yields

$A2(φX−hX)=−‖Aξ‖2(φX−hX).$

Next, we take $X=ei$ and $φei$ successively in the above equation and argue as before in order to obtain $A2ei=−‖Aξ‖2ei$ and $A2(φei)=−‖Aξ‖2φei$, for every $i=1,2,…,n$. So, $A2X=−‖Aξ‖2X$ for any X orthogonal to ξ. Hence, we obtain

$A2=−‖Aξ‖2I.$

If $‖Aξ‖≠0$, then $J=‖Aξ‖−1A$ defines a Kaehlerian structure on M leading to a contradiction that M is odd dimensional. Thus $‖Aξ‖=0$, and so $Aξ=0$. Differentiating it along $φX$ gives

$A(φX−hX)=0.$

Now, as before we put $X=ei$ and $φei$ successively in the above equation to conclude that AX=0, for any $X⊥ξ$. Since $Aξ=0$, we obtain AX=0 for any X∈ TM. This completes the proof.

Proof of Corollary 1.4. A vector field V such that $£V∇=0$ is called an affine Killing vector field. Note that $£V∇=0$ is equivalent to $∇(£Vg)=0.$ Now the result follows from Theorem 1.2.

Proof of Corollary 1.5. Let M be an almost Kenmotsu manifold with $ξ∈N(κ,μ)′$, that is

$R(U,V)ξ=κ{η(V)U−η(U)V}+μ{η(V)hU−η(U)hV},$

for all $U,V∈TM$. It then follows from Dileo-Pastore [12] that k=-1 and h=0. Hence (3.12) gives $trl=−2n$, $K(ξ,X)=−1$ and so the conclusion follows from Theorem 1.2.

Proof of Theorem 1.3. Observe that, if X is an eigenvector of h with eigenvalue $λ$, and thus $hφX=−λφX,$then $X+φX$ is an eigenvector of h' with eigenvalue $−λ$, while $X−φX$ is eigenvector with eigenvalue $λ$. Thus, it follows that h and h' admit the same eigenvalues.

First suppose that $κ≠−2$. Note that, if X is such that $X⊥ξ$ and $hX=λX$, then from Lemma we see $λ$ is different from +1 and -1. Thus Lemma implies that K(ξ,X) is non-vanishing. If the second order parallel tensor α is symmetric, then part (i) of Theorem~1.1 shows that

$Aξ=α(ξ,ξ)ξ.$

That $α(ξ,ξ)$ is constant, can be verified by differentiating it and getting the derivative equal to 0. Now differentiating (3.13) along X and $φX$, where $X⊥ξ$ and $hX=λX$, gives

$(λ2−1)A(X)=α(ξ,ξ)(λ2−1)X,$

and

$(λ2−1)A(φX)=α(ξ,ξ)(λ2−1)φX.$

Since $λ$ is different from +1 and -1, we obtain

$AX=α(ξ,ξ)X,$

for any X ∈ TM, that is, α is a constant multiple of g. Now suppose that α is skew-symmetric. Then as in the proof of Theorem 1.2, we obtain $A2ξ=−‖Aξ‖2$, where $‖Aξ‖2$ is constant. Differentiating it along X and $φX$ successively, where $X⊥ξ$ and $hX=λX$, gives

$(λ2−1)A2(X)=−‖Aξ‖2(λ2−1)X,$

and

$(λ2−1)A2(φX)=−‖Aξ‖2(λ2−1)φX.$

Thus, we obtain

$A2=−‖Aξ‖2I.$

Proceeding the similar manner as in the Theorem 1.2 one gets AX=0 for any X∈ TM.

Now suppose that $κ=−2$ and $ξ∈N(κ,μ)′$, that is,

$R(U,V)ξ=κ{η(V)U−η(U)V}+μ{η(V)h′U−η(U)h′V}.$

Then Lemma shows M is locally symmetric. Now it follows from Lemma that M is locally isometric to $ℍn+1(−4)×ℝn$, since $R(X,Y)ξ=0$ for any $X,Y∈D$. This completes the proof.

Proof of Theorem 1.4. Let ${ei,φei,ξ}i=1n$ be a local orthonormal basis as considered in Theorem 1.2. If the second order parallel tensor α is symmetric, then

$g(R(X,Y)Z,AW)+g(R(X,Y)W,AZ)=0.$

Putting $Y=Z=W=ξ$ in above, we get

$g(R(X,ξ)ξ,Aξ)=0.$

Note that 2.6 takes the form

$R(X,ξ)ξ=φ2X+2φhX−h2X.$

Using (3.15) in (3.14), and putting $X=ei$ and $φei$ successively in the resulting equation gives

$(1+λi2)g(ei,Aξ)−2λig(φei,Aξ)=0,$

and

$(1+λi2)g(φei,Aξ)−2λig(ei,Aξ)=0,$

from which we obtain

${(1+λi2)2−4λi2}g(ei,Aξ)=0,$

and

${(1+λi2)2−4λi2}g(φei,Aξ)=0.$

Since $λi$ is different from +1 and -1, we get $g(X,Aξ)=0$ for any $X⊥ξ$. Hence $Aξ=α(ξ,ξ)ξ$. Now arguing in the similar manner as in Theorem 1.2, one can conclude that

$AX=α(ξ,ξ)X$

for any X∈ TM, that is, α is a constant multiple of g. If α is skew-symmetric, then we have equation (3.9). Putting $Y=Z=ξ$ and $W=Aξ$ in (3.9), we find

$g(R(X,ξ)ξ,A2ξ)=0.$

Then using (3.15) in above, and putting X=ei and $φei$ successively we get $g(X,Aξ)=0$ for any $X⊥ξ$. Hence $Aξ=α(ξ,ξ)ξ$, and similar to the proof of Theorem 1.2, we obtain AX=0 for any X ∈ TM. This finishes the proof.

Proof of Theorem 1.5. Let ${ei,φei,ξ}i=1n$ be a local orthonormal basis such that $hei=λiei$. The hypothesis $trl>−2n−2$ shows that $∑ i=1nλi2<1$ which means $λi2<1$, that is, $λi≠(+1,−1)$ for each i.

Now the condition $£VR=0$ implies

$(£Vg)(R(X,Y)Z,W)+(£Vg)(R(X,Y)W,Z)=0.$

Let G be a (1,1)-tensor field defined by $g(GX,Y)=(£Vg)(X,Y)$. Then

$g(R(X,Y)Z,GW)+g(R(X,Y)W,GZ)=0.$

Taking $Y=Z=W=ξ$ in (3.16), we get

$g(R(X,ξ)ξ,Gξ)=0.$

Using (3.15) in above, and putting $X=ei$ and $φei$ successively in the resulting equation gives

$(1+λi2)g(ei,Gξ)−2λig(φei,Gξ)=0,$

and

$(1+λi2)g(φei,Gξ)−2λig(ei,Gξ)=0,$

from which we obtain

${(1+λi2)2−4λi2}g(ei,Gξ)=0,$

and

${(1+λi2)2−4λi2}g(φei,Gξ)=0.$

Since $λi$ is different from +1 and -1, we get $g(X,Gξ)=0$ for any $X⊥ξ$. Hence $Gξ=g(Gξ,ξ)ξ$. Now putting $Y=Z=ξ$ in 3.16, using 3.15, and taking X=ei and $φei$ successively, we obtain

${(1+λi2)2−4λi2}(g(Gξ,ξ)g(ei,W)−g(Gei,W)=0,$

and

${(1+λi2)2−4λi2}(g(Gξ,ξ)g(φei,W)−g(Gφei,W)=0.$

So that, we have $GX=g(Gξ,ξ)X$ for any $X⊥ξ$, and hence

$G=g(Gξ,ξ)I,$

that is,

$£Vg=2fg$

for some function f. Thus V is a conformal vector field, and we have

Since $£VR=0$ implies $£VS=0$, the above equation gives $∇df=0$, and hence

$∇(df⊗df)=0.$

Since $df⊗df$ is a (0,2)-tensor, it follows from Theorem 1.2 that $df⊗df=cg$, for some constant c. Thus

$(Yf)gradf=cY,$

which for $Y=gradf$ gives . Now, if ${ei}$ is an orthonormal basis then putting $Y=ei$ in (3.17), taking inner product with ei and summing over i yields

$‖gradf‖2=c(2n+1).$

Consequently, we obtain f is constant and hence V is homothetic.

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