Extrinsic Geometry of a Riemannian Manifold and Ricci Solitons

Abstract

The object of this paper is to find a vector field $\xi$ and a constant $\lambda$ on an n-dimensional compact Riemannian manifold $\left(M^n,g\right)$ such that we obtain the Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$. In order to achieve this objective, we choose an isometric embedding provided in the work of Kuiper and Nash in the Euclidean space $\left({\mathbb{R}}^m,\overline{g}\right)$ and choose $\xi$ as the tangential component of a constant unit vector on ${\mathbb{R}}^m$ and call it a Kuiper–Nash vector. If $\tau$ is the scalar curvature of the compact Riemannian manifold $\left(M^n,g\right)$ with a Kuiper–Nash vector $\xi$, we show that if the integral of the function $\xi \left(\tau \right)$ has a suitable lower bound containing a constant $\lambda$, then $\left(M^n,g,\xi ,\lambda \right)$ is a Ricci soliton; we call this a Kuiper–Nash Ricci soliton. We find a necessary and sufficient condition involving the scalar curvature $\tau$ under which a compact Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ is a trivial soliton. Finally, we find a characterization of an n-dimensional compact trivial Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ using an upper bound on the integral of ${\left(div\xi \right)}^2$ containing the scalar curvature $\tau$.

1. Prologue

Given an n-dimensional Riemannian manifold $(M^n,g)$, in general there are two methods for addressing the geometry of $(M^n,g)$. The first is the intrinsic geometry of $(M^n,g)$, and the other is the extrinsic geometry of $(M^n,g)$. In the study of the intrinsic geometry of $(M^n,g)$, among others, some tools are distance functions, geodesics, and Jacobi fields on $(M^n,g)$, and these basic tools yield global results on the geometry of $(M^n,g)$ such as the Theorem of Hadamard, the Hopf–Rinow theorem, the Bonnet–Myers theorem, and the Morse Index theorem (cf. [1,2,3,4]).

An important aspect of the intrinsic geometry of $(M^n,g)$ deals with the existence of certain vector fields on $(M^n,g)$, such as Killing vector fields, conformal vector fields, potential fields of a Ricci soliton, and almost Ricci solitons and these vector fields influence the geometry as well as the topology of $(M^n,g)$. These are not only rich due to their elegance but also are influential in physics and general relativity (cf. [1,4,5,6,7,8]). Also, an equally important component of the intrinsic geometry deals with certain partial differential equations such as the Fischer–Marsden equation and Hamilton’s Ricci flow. This component is highly influential, as exhibited by its use in resolving the century-old Poincare conjecture [9,10].

Recall that classical differential geometry originated with the study of curves and surfaces in the Euclidean space ${\mathbb{R}}^3$, which took a very magnanimous shape after the contributions of Kuiper and Nash (cf. [4,7,11,12]), which showed that an n-dimensional Riemannian manifold $(M^n,g)$ can be isometrically embedded in a Euclidean space ${\mathbb{R}}^m$ for a sufficiently large $m>n$. This paved the way for studying extrinsic geometry of $(M^n,g)$ under the title submanifold geometry (cf. [8,13,14,15,16,17,18,19,20,21,22,23,24]).

The extrinsic geometry under submanifold geometry is vast and it encompasses several global results such as the Chern–Lashof theorem using the notion of total absolute curvature (cf. [21,22,25,26]) to local results on submanifold geometry (cf. [8,13,14,15,16,17,18,19,23,24]). The vastness of this subject is also due to the fact that it includes the minimal submanifolds of a unit sphere, which has several open problems (cf. [27,28]). An important aspect of extrinsic geometry is the question of analyzing the obstructions to an embedding of the Riemannian manifold $(M^n,g)$ into a Euclidean space ${\mathbb{R}}^m$, and there are interesting results such as that of Tompkins [29], who proved that a flat closed $(M^n,g)$ cannot be isometrically embedded in the Euclidean space ${\mathbb{R}}^{2n}$. This finding was further followed by several important contributions that can be found in (cf. [30,31]).

One of the most important structures on a Riemannian manifold $(M^n,g)$ is the Ricci soliton; it is given by a smooth vector field $\xi$ called the potential field and a constant $\lambda$ satisfying:

$${\displaystyle \frac{1}{2}}{\pounds }_{\xi }g+Ric=\lambda g,$$

(1)

where ${\pounds }_{\xi }$ is the Lie derivative with respect to $\xi$, and $Ric$ is the Ricci tensor on M (cf. [5,6,9]). A Ricci soliton is denoted by $(M,g,\xi ,\lambda )$. A Ricci soliton $(M,g,\xi ,\lambda )$ is a trivial Ricci soliton if the potential field $\xi$ is a Killing field, that is, ${\pounds }_{\xi }g=0$, and we observe that a trivial Ricci soliton $(M,g,\xi ,\lambda )$ is an Einstein manifold. A Ricci soliton $(M,g,\xi ,\lambda )$ is a stable solution of the following heat equation known as Hamilton’s Ricci flow:

$${\displaystyle \frac{\partial g}{\partial t}}=-2Ric\left(g\left(t\right)\right),$$

(2)

where the evolving metric $g\left(t\right)=\rho \left(t\right){\varphi }_t^{\ast }\left(g\right)$ satisfies $g\left(0\right)=g$, the scaling function $\rho$ satisfies $\rho \left(0\right)=1$, and the one-parameter diffeomorphism group $\left\{{\phi }_t\right\}$ induces the potential field $\xi .$

It is a well-known fact about the heat equation that it distributes the temperature potential evenly; using this as a clue, Hamilton used Ricci flow to obtain an even distribution of curvature on the Riemannian manifold through a Ricci soliton. It is for this reason that the geometry of Ricci solitons has been subject of immense interest (cf. [5,6,9]). An important aspect of Ricci soliton structure on a Riemannian manifold $\left(M^n,g\right)$ is that we see a union of geometry and global analysis through it. Through this union, Perelman conceived of the idea of settling the famous century-old Poincare conjecture (cf. [9]). We also observe that a Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ becomes an Einstein manifold if the potential field $\xi$ is a Killing vector field; therefore, a Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ can be considered to be a generalization of an Einstein manifold. The potential field $\xi$ and the constant $\lambda$ of a Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ is induced through the Ricci flow as a stable solution. A natural question follows: Could there be another way to obtain the vector field $\xi$ and the constant $\lambda$ on a compact Riemannian manifold $\left(M^n,g\right)$ such that it becomes a Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$? This article is devoted to answering this question.

Given an n-dimensional smooth Riemannian manifold $(M^n,g)$, to measure the appetite of $(M^n,g)$ to acquire the structure of a Ricci soliton $(M^n,g,\xi ,\lambda )$, we require that the potential field $\xi$ and the constant $\lambda$ satisfy Equation (1). In this article, we wish to approach this question through Kuiper–Nash isometric embedding $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right)$, where $\left({\mathbb{R}}^m,\overline{g}\right)$ is the Euclidean space for sufficiently large $m>n$, and $\overline{g}$ is the Euclidean metric. We use this embedding to bring a smooth vector field $\xi$ on $(M^n,g)$ that will assume the role of the potential field for the prospective Ricci soliton $(M^n,g,\xi ,\lambda )$. There are several ways to achieve this vector field. However, our choice is to pick up a constant unit vector field $\overrightarrow{u}$ on the Euclidean space $\left({\mathbb{R}}^m,\overline{g}\right)$, which gives us the following decomposition of $\overrightarrow{u}$

$$\overrightarrow{u}=\xi +\Gamma ,$$

where $\xi$ is tangential to $M^n$ and $\Gamma$ is normal to $M^n$. This vector field $\xi$ is called the Kuiper–Nash vector on $(M^n,g)$ and $\Gamma$ is the Kuiper–Nash normal on $(M^n,g)$. Note that the choice of the pair $\xi$, $\Gamma$ is not unique, and it changes with the choice of the constant unit vector $\overrightarrow{u}$ on the Euclidean space $\left({\mathbb{R}}^m,\overline{g}\right)$.

In this article, we explore the condition under which on an n-dimensional Riemannian manifold $(M^n,g)$ and the Kuiper–Nash vector $\xi$ together with a constant $\lambda$ makes $(M^n,g,\xi ,\lambda )$ a Ricci soliton. Indeed, we prove that an n-dimensional compact Riemannian manifold $(M^n,g)$ with a scalar curvature $\tau$ and a Kuiper–Nash vector $\xi$—if the integral of the function $\xi \left(\tau \right)$ has a suitable lower bound containing the constant $\lambda$—is necessarily a Ricci soliton $(M^n,g,\xi ,\lambda )$ (cf. Theorem 1). We call this Ricci soliton $(M^n,g,\xi ,\lambda )$ a Kuiper–Nash Ricci soliton and study its properties (cf. Proposition 1). Recall that a trivial Ricci soliton $(M^n,g,\xi ,\lambda )$ is an Einstein manifold, and it is for this reason that Ricci solitons are considered to be a generalization of an Einstein manifold. Moreover, finding conditions under which a Ricci soliton $(M^n,g,\xi ,\lambda )$ is trivial is an important challenge in the geometry of Ricci solitons. In this article, we find two results that give conditions under which a Kuiper–Nash Ricci soliton $(M^n,g,\xi ,\lambda )$ is trivial (cf. Theorems 2 and 3).

Example: Consider the Euclidean spaces $\left(R^n,g\right)$ and $\left(R^{n+1},\overline{g}\right)$, where $g,\overline{g}$ are Euclidean metrics. Then, we have the isometric embedding $\Psi :\left(R^n,g\right)\to \left(R^{n+1},\overline{g}\right)$, given by

$$\Psi \left(x\right)=\left(x.0\right),x\in R^n.$$

It is a totally geodesic embedding with unit normal $N={\displaystyle \frac{\partial }{\partial x^{n+1}}}$, where $x^1,\dots ,x^{n+1}$ are Euclidean coordinates on $R^{n+1}$. For a constant unit vector $\overrightarrow{u}$, we express $\overrightarrow{u}$ as

$$\overrightarrow{u}=\xi +\Gamma ,$$

where $\xi \in \mathfrak{X}\left(R^n\right)$ and $\Gamma =\overline{g}\left(\overrightarrow{u},N\right)N=\sigma N$. Differentiating the above equation with respect to $U\in \mathfrak{X}\left(R^n\right)$ and equating tangential and normal components, we confirm

$${\nabla }_U\xi =0,U\left(\sigma \right)=0,$$

where ∇ is the Riemannian connection on the Euclidean space $\left(R^n,g\right)$. As a result, we obtain

$${\pounds }_{\xi }g=0$$

and consequently, on the Euclidean space, we have

$${\displaystyle \frac{1}{2}}{\pounds }_{\xi }g+Ric=\lambda g,$$

where $\lambda =0$, as the Euclidean space $\left(R^n,g\right)$, is Ricci flat. Thus, $\left(R^n,g,\xi ,\lambda \right)$ is a Kuiper–Nash Ricci soliton with a Kuiper–Nash vector $\xi$.

2. Preliminaries

On an n-dimensional Riemannian manifold $\left(M^n,g\right)$, let ∇ be the Riemannian connection. Then, the curvature tensor of $\left(M^n,g\right)$ is given by

$$R(U,V)W=\left[{\nabla }_U,{\nabla }_V\right]W-{\nabla }_{[U,V]}W,U,V,W\in \mathfrak{U}\left(M^n\right),$$

(3)

where $\mathfrak{U}\left(M^n\right)$ is the space of smooth vector fields on $M^n$. Contracting the curvature tensor field gives a Ricci tensor $Ric$ of $\left(M^n,g\right)$ and is a symmetric tensor

$$Ric\left(U,V\right)=\sum _{j=1}^{n}g\left(R\left(e_j,U\right)V,e_j\right),U,V\in \mathfrak{U}\left(M^n\right),$$

(4)

where $\left\{e_1,\dots ,e_n\right\}$ is a local orthonormal frame on $\left(M^n,g\right)$. The Ricci operator Q of $\left(M^n,g\right)$ is a symmetry operator $Q:\mathfrak{U}\left(M^n\right)\to \mathfrak{U}\left(M^n\right)$ defined by

$$Ric\left(U,V\right)=g\left(QU,V\right),U,V\in \mathfrak{U}\left(M^n\right),$$

(5)

and the scalar curvature $\tau$ of $\left(M^n,g\right)$ is given by

$$\tau =\sum _{l=1}^{n}Ric\left(e_l,e_l\right).$$

(6)

The following formula is well known (cf. [13], trace in Equation (3.1) p. 58)

$${\displaystyle \frac{1}{2}}\nabla \tau =\sum _{l=1}^n\left({\nabla }_{e_l}Q\right)\left(e_l\right),$$

(7)

where $\nabla \tau$ is the gradient of the scalar curvature $\tau$ and

$$\left({\nabla }_{U}Q\right)\left(V\right)={\nabla }_{U}QV-Q\left({\nabla }_{U}V\right).$$

For an n-dimensional Riemannian manifold $(M^n,g)$, we have the Kuiper–Nash isometric embedding $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right)$, where $\left({\mathbb{R}}^m,\overline{g}\right)$ is the Euclidean space for sufficiently large $m>n$ and $\overline{g}$ is the Euclidean metric. We denote the Euclidean connection on $\left({\mathbb{R}}^m,\overline{g}\right)$ with $\overline{\nabla }$, denote the normal bundle of this isometric $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right)$ with $\nu$, and denote the space of smooth sections of the normal bundle $\nu$ with $\Pi \left(\nu \right)$. Then, we have the following fundamental equations for the isometric embedding $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right),$

$${\overline{\nabla }}_{U}Y={\nabla }_{U}V+h(U,V),U,V\in \mathfrak{U}\left(M^n\right),$$

(8)

$${\overline{\nabla }}_{U}N=-S_{N}U+{\nabla }_U^{\perp }N,N\in \Pi \left(\nu \right),$$

(9)

where h is the second fundamental form, $S_N$ is the shape operator with respect to the normal vector field N, and they are related by

$$\overline{g}\left(h\left(U,V\right),N\right)=g\left(S_{N}U,V\right),U,V\in \mathfrak{U}\left(M^n\right).$$

(10)

Also, we have the following fundamental equations of the isometric embedding $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right)$, the curvature tensor of $(M^n,g)$ has the expression

$$R\left(U,V\right)W=S_{h(V,W)}U-S_{h(U,W)}V,U,V\in \mathfrak{U}\left(M^n\right),$$

(11)

and the Ricci tensor of $(M^n,g)$ has the expression

$$Ric\left(U,V\right)=n\overline{g}\left(H,h\left(U,V\right)\right)-\sum _{j=1}^n\overline{g}\left(h\left(e_j,U\right),h\left(e_j,V\right)\right),U,V\in \mathfrak{U}\left(M^n\right),$$

(12)

where H is the mean curvature vector field defined by

$$H={\displaystyle \frac{1}{n}}\sum _{j=1}^{n}h\left(e_j,e_j\right),$$

(13)

for a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$.

Given an n-dimensional smooth Riemannian manifold $(M^n,g)$, we have the Kuiper–Nash isometric embedding $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right)$ (cf. [12,23]). Fixing a constant unit vector $\overrightarrow{u}$, we are interested in the tangential and normal parts of $\overrightarrow{u}$ as described in the following:

Definition 1.

Given the Kuiper–Nash isometric embedding $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right)$ and a fixed constant unit vector $\overrightarrow{u}$ on the Euclidean space $\left({\mathbb{R}}^m,\overline{g}\right)$ expressed as $\overrightarrow{u}=\xi +\Gamma$, the tangential vector field ξ is called a Kuiper–Nash vector and the normal component Γ is called the Kuiper–Nash normal.

Definition 2.

If ξ is the Kuiper–Nash vector and Γ is the Kuiper–Nash normal on a Riemannian manifold $(M^n,g)$ with respect to the Kuiper–Nash isometric embedding $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right)$ and constant unit vector $\overrightarrow{u}$ on the Euclidean space $\left({\mathbb{R}}^m,\overline{g}\right)$, then the function $\phi :M^n\to R$ defined by $\phi =\overline{g}\left(H,\Gamma \right)$ is called a Kuiper–Nash function on $(M^n,g)$, and the operator $K:\mathfrak{U}\left(M^n\right)\to \mathfrak{U}\left(M^n\right)$ defined by

$$K=S_{\Gamma }$$

is called the Kuiper–Nash operator of the Riemannian manifold $(M^n,g)$.

Differentiating the expression $\overrightarrow{u}=\xi +\Gamma$ with respect to $U\in \mathfrak{U}\left(M^n\right)$ and using Equations (8) and (9) while equating like parts, we arrive at

$${\nabla }_U\xi =KU,{\nabla }_U^{\perp }\Gamma =-h\left(\xi ,U\right),U\in \mathfrak{U}\left(M^n\right).$$

(14)

Lemma 1.

On an n-dimensional Riemannian manifold $(M^n,g)$ with a Kuiper–Nash vector ξ, a Kuiper–Nash function φ, and a Kuiper–Nash operator K, the following equations hold:

(i) $TrK=n\phi ,$

(ii) $Q\left(\xi \right)={\displaystyle \sum _{j=1}^n}\left({\nabla }_{e_J}K\right)\left(e_j\right)-n\nabla \phi$,

• where $\left\{e_1,\dots ,e_n\right\}$ is a local orthonormal frame, $TrK$ is the trace of K and $\nabla \phi$ is the gradient of the Kuiper–Nash function φ.

Proof. 

Note that using the definition $K=S_{\Gamma }$, we find

$$\begin{array}{ccc}\hfill Tr.K& =& \sum _{j=1}^{n}g\left(K{e}_j,e_j\right)=\sum _{j=1}^{n}g\left(S_{\Gamma }{e}_j,e_j\right)=\sum _{j=1}^n\overline{g}\left(h\left(e_j,e_j\right),\Gamma \right)\hfill \\ & =& n\overline{g}\left(H,\Gamma \right)=n\phi ,\hfill \end{array}$$

which proves (i). Now, differentiating Equation (14), we obtain

$${\nabla }_U{\nabla }_V\xi ={\nabla }_{U}KV=\left({\nabla }_{U}K\right)\left(V\right)+K\left({\nabla }_{U}V\right),U,V\in \mathfrak{U}\left(M^n\right).$$

Consequently, using Equation (3), we conclude

$$R\left(U,V\right)\xi =\left({\nabla }_{U}K\right)\left(V\right)-\left({\nabla }_{V}K\right)\left(U\right)$$

and using a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$ and the symmetry of the Kuiper–Nash operator K in the above equation, we arrive at

$$Ric\left(V,\xi \right)=\sum _{j=1}^{n}g\left(V,\left({\nabla }_{e_j}K\right)\left(e_j\right)\right)-\sum _{j=1}^{n}g\left(\left({\nabla }_{V}K\right)\left(e_j\right),e_j\right).$$

(15)

Note that

$$\begin{array}{cc}\hfill \sum _{j=1}^{n}g\left(\left({\nabla }_{V}K\right)\left(e_j\right),e_j\right)& =\sum _{j=1}^{n}g\left({\nabla }_{V}K{e}_j-K\left({\nabla }_V{e}_j\right),e_j\right)\hfill \\ \hfill & =\sum _{j=1}^{n}g\left({\nabla }_{V}K{e}_j,e_j\right)-\sum _{j=1}^{n}g\left(K\left({\nabla }_V{e}_j\right),e_j\right)\hfill \\ \hfill & =\sum _{j=1}^{n}Vg\left(K{e}_j,e_j\right)-2\sum _{j=1}^{n}g\left({\nabla }_V{e}_j,K{e}_j\right)\hfill \\ \hfill & =nV\left(\phi \right)-2\sum _{j=1}^{n}g\left({\nabla }_V{e}_j,K{e}_j\right).\hfill \end{array}$$

Using the fact that the Kuiper–Nash operator K is symmetric and ${\nabla }_V{e}_j={\displaystyle \sum _k}{\omega }_j^k\left(V\right)e_k$, where ${\omega }_j^k$ are skew-symmetric connection forms, we conclude

$$\sum _{j=1}^{n}g\left({\nabla }_V{e}_j,K{e}_j\right)=\sum _{j,k=1}^n{\omega }_j^k\left(V\right)g\left(K{e}_j,e_k\right)=0,$$

and inserting the above equation into the previous equation gives

$$\sum _{j=1}^{n}g\left(\left({\nabla }_{V}K\right)\left(e_j\right),e_j\right)=nV\left(\phi \right).$$

Combining the above equation with Equation (15) yields

$$Ric\left(V,\xi \right)=\sum _{j=1}^{n}g\left(V,\left({\nabla }_{e_j}K\right)\left(e_j\right)\right)-nV\left(\phi \right),$$

which proves (ii). □

Recall that the action of the Laplace operator on a smooth function $f:M^n\to R$ on a Riemannian manifold $\left(M^n,g\right)$ is given by

$$\Delta f=div\left(\nabla f\right).$$

(16)

where $\nabla f$ is the gradient of f and the divergence is defined by

$$divU=\sum _{l=1}^{n}g\left({\nabla }_{e_l}U,e_l\right).$$

On a compact Riemannian manifold $\left(M^n,g\right)$, the Stokes’s Theorem implies

$$\underset{M^n}{\int }\left(divU\right)d{V}_g=0,$$

(17)

where $d{V}_g$ is the volume element of $\left(M^n,g\right)$.

Lemma 2.

On an n-dimensional compact Riemannian manifold $(M^n,g)$ with a Kuiper–Nash vector ξ and a Kuiper–Nash function φ, the following hold:

(i) $\underset{M^n}{\int }\phi d{V}_g=0,$

(ii) $\underset{M^n}{\int }\xi \left(\phi \right)d{V}_g=-n\underset{M^n}{\int }{\phi }^{2}d{V}_g$.

Proof. 

Using Equation (14) and (i) in Lemma 1, we obtain $div\xi =n\phi$, which leads to (i) upon integration. Now, using $div\xi =n\phi$ and $div\left(\phi \xi \right)=\xi \left(\phi \right)+\phi div\xi =\xi \left(\phi \right)+n{\phi }^2$, which yields (ii) upon integration. □

Note that if F is a $(1,1)$ tensor field on an n-dimensional Riemannian manifold $\left(M^n,g\right)$, then we define

$${∥F∥}^2=\sum _{j=1}^{n}g\left(F{e}_j,F{e}_j\right).$$

(18)

Also, for a (0,2)-type symmetric tensor $\Omega$, we have

$${\left|\Omega \right|}^2=\sum _{j,k=1}^n\Omega {\left(e_j,e_k\right)}^2.$$

(19)

3. Transforming a Riemannian Manifold $\mathbf{(}{\mathit{M}}^{\mathit{n}}\mathbf{,}\mathit{g}\mathbf{)}$ to a Ricci Soliton

Let $\left(M^n,g\right)$ be an n-dimensional Riemannian manifold $\left(M^n,g\right)$ with a Kuiper–Nash vector field $\xi$, a Kuiper–Nash function $\phi$, and a Kuiper Nash operator K. In this section, we wish to find conditions under which a compact $\left(M^n,g\right)$ becomes a Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ and study the properties of this inherited Ricci soliton structure on $\left(M^n,g\right)$.

Now, we prove the main result of this section:

Theorem 1.

If an n-dimensional compact and connected Riemannian manifold $(M^n,g)$ with a scalar curvature τ, a Ricci operator Q, a Kuiper–Nash vector ξ, a Kuiper–Nash function φ, and a Kuiper–Nash operator K satisfies

$$\underset{M^n}{\int }\xi \left(\tau \right)d{V}_g\ge \underset{M^n}{\int }\left({∥K∥}^2+\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)+{\displaystyle \frac{1}{n}}{\left(n\lambda -\tau \right)}^2\right)d{V}_g$$

for a constant λ, then $\left(M^n,g,\xi ,\lambda \right)$ is a Ricci soliton.

Proof. 

Using symmetry of the Kuiper–Nash operator K and Equation (14), we compute

$${\displaystyle \frac{1}{2}}\left({\pounds }_{\xi }g\right)\left(U,V\right)=g\left(KU,V\right),U,V\in \mathfrak{U}\left(M^n\right).$$

(20)

Thus, for a constant $\lambda$, we have

$${\displaystyle \frac{1}{2}}\left({\pounds }_{\xi }g\right)\left(U,V\right)+Ric\left(U,V\right)-\lambda g\left(U,V\right)=g\left(KU,V\right)+g\left(QU,V\right)-\lambda g\left(U,V\right)$$

and treating the above equation with a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$, using Equation (19) and (i) in Lemma 1, we have

$$\begin{array}{cc}\hfill {\left|{\displaystyle \frac{1}{2}}{\pounds }_{\xi }g+Ric-\lambda g\right|}^2& ={∥K∥}^2+{∥Q∥}^2+n{\lambda }^2+2{\displaystyle \sum _j}g\left(K{e}_j,Q{e}_j\right)\hfill \\ \hfill & -2n\lambda \phi -2\lambda \tau ,\hfill \end{array}$$

(21)

where we have used

$$\sum _{j,k=1}^{n}Ric{\left(e_j,e_k\right)}^2=\sum _{j,k=1}^{n}g{\left(Q{e}_j,e_k\right)}^2=\sum _{j=1}^{n}g\left(Q{e}_j,Q{e}_j\right)={∥Q∥}^2.$$

Next, we need to compute the divergence of the vector $Q\xi$, for which we use a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$ and Equations (7) and (14) and arrive at the following

$$\begin{array}{ccc}\hfill divQ\left(\xi \right)& =& \sum _{j=1}^{n}g\left({\nabla }_{e_j}Q\xi ,e_j\right)=\sum _{j=1}^{n}g\left(\left({\nabla }_{e_j}Q\right)\left(\xi \right)+Q\left({\nabla }_{e_j}\xi \right),e_j\right)\hfill \\ & =& {\displaystyle \frac{1}{2}}\xi \left(\tau \right)+\sum _{j=1}^{n}g\left(K{e}_j,Q{e}_j\right),\hfill \end{array}$$

where we used the symmetry of Q and Equation (7) to obtain

$$\sum _{j=1}^{n}g\left(\left({\nabla }_{e_j}Q\right)\left(\xi \right),e_j\right)=\sum _{j=1}^{n}g\left(\xi ,\left({\nabla }_{e_j}Q\right)\left(e_j\right)\right)={\displaystyle \frac{1}{2}}g\left(\xi ,\nabla \tau \right)={\displaystyle \frac{1}{2}}\xi \left(\tau \right).$$

Inserting the above equation into Equation (21), we confirm that

$$\begin{array}{ccc}\hfill {\left|{\displaystyle \frac{1}{2}}{\pounds }_{\xi }g+Ric-\lambda g\right|}^2& =& {∥K∥}^2+{∥Q∥}^2+n{\lambda }^2+2divQ\left(\xi \right)-\xi \left(\tau \right)\hfill \\ & & -2n\lambda \phi -2\lambda \tau .\hfill \end{array}$$

Integrating the above equation while using Lemma 2, we conclude

$$\begin{array}{cc}\hfill \underset{M^n}{\int }{\left|{\displaystyle \frac{1}{2}}{\pounds }_{\xi }g+Ric-\lambda g\right|}^{2}d{V}_g& =\underset{M^n}{\int }\left({∥K∥}^2+\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2+{\displaystyle \frac{1}{n}}{\left(n\lambda -\tau \right)}^2\right)\right)d{V}_g\hfill \\ \hfill & -\underset{M^n}{\int }\xi \left(\tau \right)d{V}_g.\hfill \end{array}$$

(22)

Now, using the condition in the statement, we obtain

$$\underset{M^n}{\int }{\left|{\displaystyle \frac{1}{2}}{\pounds }_{\xi }g+Ric-\lambda g\right|}^{2}d{V}_g\le 0,$$

which proves

$${\left|{\displaystyle \frac{1}{2}}{\pounds }_{\xi }g+Ric-\lambda g\right|}^2=0,$$

that is,

$${\displaystyle \frac{1}{2}}{\pounds }_{\xi }g+Ric=\lambda g,$$

(23)

making $\left(M^n,g,\xi ,\lambda \right)$ a Ricci soliton. □

Definition 3.

The Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ of Theorem 1 is called the Kuiper–Nash Ricci soliton.

Proposition 1.

An n-dimensional compact Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ satisfies

$${\displaystyle \frac{1}{2}}\underset{M^n}{\int }\xi \left(\tau \right)d{V}_g=\underset{M^n}{\int }\left(\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)+{\displaystyle \frac{1}{n}}{\left(n\lambda -\tau \right)}^2\right)d{V}_g.$$

Proof. 

If $\left(M^n,g,\xi ,\lambda \right)$ is an n-dimensional compact Kuiper–Nash Ricci soliton, then Equation (22) takes the form

$$\underset{M^n}{\int }\xi \left(\tau \right)d{V}_g=\underset{M^n}{\int }\left({∥K∥}^2+\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)+{\displaystyle \frac{1}{n}}{\left(n\lambda -\tau \right)}^2\right)d{V}_g.$$

(24)

Combining Equations (20) and (23), we obtain

$$K=\lambda I-Q,$$

(25)

which confirms

$${∥K∥}^2=n{\lambda }^2-2\lambda \tau +{∥Q∥}^2,$$

that is,

$${∥K∥}^2=\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)+{\displaystyle \frac{1}{n}}{\left(\lambda n-\tau \right)}^2.$$

Inserting the above equation in Equation (24), we obtain

$$\underset{M^n}{\int }\xi \left(\tau \right)d{V}_g=2\underset{M^n}{\int }\left(\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)+{\displaystyle \frac{1}{n}}{\left(n\lambda -\tau \right)}^2\right)d{V}_g,$$

which proves the result. □

In the rest of this section, we find the conditions under which the Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ is a trivial Ricci soliton. First, we prove the following:

Theorem 2.

The scalar curvature τ of an n-dimensional compact Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ satisfies

$$\underset{M^n}{\int }{\tau }^{2}d{V}_g\le n\lambda \underset{M^n}{\int }\tau d{V}_g,$$

if and only if $\left(M^n,g,\xi ,\lambda \right)$ is a trivial Ricci soliton.

Proof. 

Suppose that the scalar curvature $\tau$ of an n-dimensional compact Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ satisfies

$$\underset{M^n}{\int }{\tau }^{2}d{V}_g\le n\lambda \underset{M^n}{\int }\tau d{V}_g.$$

(26)

Using Equations (14) and (25), we have

$${\nabla }_U\xi =\lambda U-QU.$$

(27)

We use Equations (7) and (27) and a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$ to compute the divergence of $Q\left(\xi \right)$ and find

$$\begin{array}{ccc}\hfill divQ\left(\xi \right)& =& \sum _{j=1}^{n}g\left({\nabla }_{e_j}Q\xi ,e_j\right)=\sum _{j=1}^{n}g\left(\left({\nabla }_{e_j}Q\right)\left(\xi \right)+Q\left({\nabla }_{e_j}\xi \right),e_j\right)\hfill \\ & =& {\displaystyle \frac{1}{2}}\xi \left(\tau \right)+\sum _{j=1}^{n}g\left(\lambda e_j-Q{e}_j,Q{e}_j\right)={\displaystyle \frac{1}{2}}\xi \left(\tau \right)+\lambda \tau -{∥Q∥}^2,\hfill \end{array}$$

that is,

$$divQ\left(\xi \right)={\displaystyle \frac{1}{2}}\xi \left(\tau \right)-\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)+\lambda \tau -{\displaystyle \frac{1}{n}}{\tau }^2.$$

Integrating above equation, we arrive at

$${\displaystyle \frac{1}{2}}\underset{M^n}{\int }\xi \left(\tau \right)d{V}_g=\underset{M^n}{\int }\left(\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)-{\displaystyle \frac{1}{n}}\left(n\lambda \tau -{\tau }^2\right)\right)d{V}_g$$

Combining it with Proposition 1, we obtain

$$\underset{M^n}{\int }{\displaystyle \frac{1}{n}}{\left(n\lambda -\tau \right)}^{2}d{V}_g=-{\displaystyle \frac{1}{n}}\underset{M^n}{\int }\left(n\lambda \tau -{\tau }^2\right)d{V}_g.$$

Using the condition in the statement, we conclude

$$\underset{M^n}{\int }{\displaystyle \frac{1}{n}}{\left(n\lambda -\tau \right)}^{2}d{V}_g=0,$$

that is,

$$\tau =n\lambda .$$

(28)

It shows that the scalar curvature $\tau$ is a constant and using Equation (28) in Proposition 1, we conclude

$$\underset{M^n}{\int }\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)d{V}_g=0.$$

(29)

Using the Cauchy–Schwartz inequality

$${∥Q∥}^2\ge {\displaystyle \frac{1}{n}}{\tau }^2$$

in Equation (29), we confirm

$${∥Q∥}^2={\displaystyle \frac{1}{n}}{\tau }^2$$

It being an equality in a Cauchy–Schwartz inequality, it holds if and only if

$$Q={\displaystyle \frac{\tau }{n}}I.$$

(30)

Combining Equations (28) and (30) with Equation (27), we confirm that

$${\pounds }_{\xi }g=0,$$

that is, $\left(M^n,g,\xi ,\lambda \right)$ is trivial. The converse is trivial, for if $\left(M^n,g,\xi ,\lambda \right)$ is a trivial Ricci soliton, then $Ric=\lambda g$, which implies that $\tau =n\lambda$ is a constant. Hence, the condition in the statement holds. □

Note that in the statement of Theorem 2, we used the inequality

$$\underset{M^n}{\int }{\tau }^{2}d{V}_g\le n\lambda \underset{M^n}{\int }{\tau }^{2}d{V}_g,$$

which ultimately turns out to be an equality. Therefore, it should be noted that a strict inequality cannot occur. Since our goal is to reach a trivial Ricci soliton, which requires $\tau =n\lambda$ with constant $\tau$, that is, ${\tau }^2=n\lambda \tau$, the considered inequality is justified.

Finally, we prove the following characterization of a compact trivial Ricci soliton using a Kuiper–Nash Ricci soliton.

Theorem 3.

An n-dimensional compact Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$, $n>2$, with a scalar curvature τ satisfies

$$\underset{M^n}{\int }{\left(div\xi \right)}^{2}d{V}_g\le {\displaystyle \frac{1}{n}}\underset{M^n}{\int }\left(\tau -n\lambda \right)\left(2\tau -n\lambda \right)d{V}_g,$$

if and only if $\left(M^n,g,\xi ,\lambda \right)$ is a trivial Ricci soliton.

Proof. 

Suppose $\left(M^n,g,\xi ,\lambda \right)$ is an n-dimensional compact Kuiper–Nash Ricci soliton with a scalar curvature $\tau$ that satisfies the inequality

$$\underset{M^n}{\int }{\left(div\xi \right)}^{2}d{V}_g\le {\displaystyle \frac{1}{n}}\underset{M^n}{\int }\left(\tau -n\lambda \right)\left(2\tau -n\lambda \right)d{V}_g.$$

(31)

Then, using Equations (20) and (23), we have

$$KU=\lambda U-QU,U\in \mathfrak{U}\left(M^n\right),$$

(32)

which in view of Equation (14) yields

$${\nabla }_U\xi =\lambda U-QU,U\in \mathfrak{U}\left(M^n\right).$$

(33)

Using a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$ with the above equation, we compute

$${∥\nabla \xi ∥}^2=\sum _{j=1}^{n}g\left({\nabla }_{e_j}\xi ,{\nabla }_{e_j}\xi \right)=n{\lambda }^2-2\lambda \tau +{∥Q∥}^2.$$

(34)

Also, using Equation (23), we compute

$${\displaystyle \frac{1}{4}}{\left|{\pounds }_{\xi }\right|}^2=\left(n{\lambda }^2-2\lambda \tau +{∥Q∥}^2\right).$$

(35)

Note that the trace of Equation (32) gives $n\phi =n\lambda -\tau$, that is

$$n\nabla \phi =-\nabla \tau .$$

(36)

Also, the Equation (32) implies

$$\left({\nabla }_{U}K\right)\left(U\right)=-\left({\nabla }_{U}Q\right)\left(U\right),$$

which in view of Equation (7) gives

$$\sum _{j=1}^n\left({\nabla }_{e_j}K\right)\left(e_j\right)=-{\displaystyle \frac{1}{2}}\nabla \tau .$$

(37)

Using Equations (36) and (37) in (ii) of Lemma 1, we obtain

$$Q\left(\xi \right)={\displaystyle \frac{1}{2}}\nabla \tau ,$$

that is,

$$Ric\left(\xi ,\xi \right)={\displaystyle \frac{1}{2}}\xi \left(\tau \right).$$

(38)

Now, we use Equation (32) in computing the divergence of the vector field $Q\xi$ and arrive at

$$\begin{array}{ccc}\hfill divQ\xi & =& \sum _{j=1}^{n}g\left({\nabla }_{e_j}Q\xi ,e_j\right)=\sum _{j=1}^{n}g\left(\left({\nabla }_{e_j}Q\right)\left(\xi \right)+Q\left({\nabla }_{e_j}\xi \right),e_j\right)\hfill \\ & =& \sum _{j=1}^{n}g\left(\xi ,\left({\nabla }_{e_j}Q\right)\left(e_j\right)\right)+\sum _{j=1}^{n}g\left(\lambda e_j-Q{e}_j,Q{e}_j\right).\hfill \end{array}$$

Using Equation (7) in the above equation, we conclude

$$divQ\xi ={\displaystyle \frac{1}{2}}\xi \left(\tau \right)+\lambda \tau -{∥Q∥}^2.$$

Integrating the above equation while using Equations (18) and (38), we conclude

$$\underset{M^n}{\int }\left(Ric\left(\xi ,\xi \right)+\lambda \tau -{∥Q∥}^2\right)d{V}_g=0.$$

(39)

Next, we recall the following integral formula (cf. [32])

$$\underset{M^n}{\int }\left(Ric\left(\xi ,\xi \right)+{\displaystyle \frac{1}{2}}{\left|{\pounds }_{\xi }\right|}^2-{∥\nabla \xi ∥}^2-{\left(div\xi \right)}^2\right)d{V}_g=0.$$

Inserting Equations (34) and (35) into the above equation yields

$$\underset{M^n}{\int }\left(Ric\left(\xi ,\xi \right)+\left(n{\lambda }^2-2\lambda \tau +{∥Q∥}^2\right)-{\left(div\xi \right)}^2\right)d{V}_g=0.$$

Substituting Equation (39) into the above equation gives

$$\underset{M^n}{\int }\left(2{∥Q∥}^2+n{\lambda }^2-3\lambda \tau -{\left(div\xi \right)}^2\right)d{V}_g=0,$$

that is,

$$2\underset{M^n}{\int }\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)d{V}_g=\underset{M^n}{\int }\left({\left(div\xi \right)}^2-{\displaystyle \frac{1}{n}}\left(2{\tau }^2-3n\lambda \tau +n^2{\lambda }^2\right)\right)d{V}_g.$$

The above equation is

$$2\underset{M^n}{\int }\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)d{V}_g=\underset{M^n}{\int }{\left(div\xi \right)}^{2}d{V}_g-{\displaystyle \frac{1}{n}}\underset{M^n}{\int }\left(\tau -n\lambda \right)\left(2\tau -n\lambda \right)d{V}_g,$$

which in view of the inequality (31), confirms

$$\underset{M^n}{\int }\left({∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2\right)d{V}_g\le 0.$$

However, due to the Chauchy–Schwartz inequality ${∥Q∥}^2\ge {\displaystyle \frac{1}{n}}{\tau }^2$, the above inequality gives

$${∥Q∥}^2-{\displaystyle \frac{1}{n}}{\tau }^2=0$$

and finally, we obtain

$$Q={\displaystyle \frac{\tau }{n}}I,$$

(40)

that is,

$$\left({\nabla }_{U}Q\right)\left(U\right)={\displaystyle \frac{1}{n}}U\left(\tau \right)U.$$

Summing the above equation over a local orthonormal frame $\left\{e_1,\dots ,e_n\right\}$, we obtain

$$\sum _{j=1}^n\left({\nabla }_{e_j}Q\right)\left(e_j\right)={\displaystyle \frac{1}{n}}\nabla \tau .$$

Combining the above equation with Equation (7),

$${\displaystyle \frac{1}{2}}\nabla \tau ={\displaystyle \frac{1}{n}}\nabla \tau$$

and as $n>2$, we obtain $\nabla \tau =0$, that is, $\tau$ is a constant. Now, using Equation (33), we have $div\xi =n\lambda -\tau$, which on integration and the fact that $\tau$ is a constant gives $\tau =n\lambda$. Thus, Equation (40) yields

$$Ric=\lambda g$$

and in turn it proves

$${\pounds }_{\xi }g=0,$$

that is, $\left(M^n,g,\xi ,\lambda \right)$ is a trivial Ricci soliton. The converse is trivial, as if $\left(M^n,g,\xi ,\lambda \right)$ is a trivial Ricci soliton, then $\tau =n\lambda$; therefore, $div\xi =0$, and consequently, the given condition in the statement holds. □

4. Epilogue

In this article, we have analyzed an n-dimensional Riemannian manifold $\left(M^n,g\right)$ to acquire the structure of a Ricci soliton by lending it a smooth vector field $\xi$ using the work of Kuiper and Nash on the isometric embedding $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right)$ into the Euclidean space $\left({\mathbb{R}}^m,\overline{g}\right)$ for a sufficiently large $m>n$ choosing a constant unit vector field $\overrightarrow{u}$ that has a tangential component $\xi$. We called the Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ a Kuiper–Nash Ricci soliton. The scope for this type of study is quite modest, as there could be several questions, some of which could be the following:

(1)

Can we extend this type of study to other solitons, for instance, the quest for an n-dimensional Riemannian manifold $\left(M^n,g\right)$ becoming a Ricci–Bourguignon soliton (cf. [9]) through the Kuiper–Nash isometric embedding $\Psi :(M^n,g)\to \left({\mathbb{R}}^m,\overline{g}\right)$, $m>n$?

(2)

We have seen that each constant unit vector field $\overrightarrow{u}$ under the condition of Theorem 1 on an n-dimensional compact Riemannian manifold $\left(M^n,g\right)$ obtains a Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$. Note that we change the constant unit vector field on the Euclidean space $\left({\mathbb{R}}^m,\overline{g}\right)$ such that if $\overrightarrow{v}$ we will obtain yet another Kuiper–Nash Ricci soliton $\left(M^n,g,\zeta ,\mu \right)$. Then, assuming $\overline{g}\left(\overrightarrow{u},\overrightarrow{v}\right)=0$, one would be interested in analyzing the relation between the corresponding potential fields $\xi$ and $\zeta$, respectively. In particular, finding conditions under which these potential fields $\xi$ and $\zeta$ are orthogonal will be quite interesting, which will enable us to introduce the notion of orthogonal Kuiper–Nash Ricci solitons. If this proposed question has an affirmative answer and since there are m number of choices of choosing constant unit vector fields on the Euclidean space $\left({\mathbb{R}}^m,\overline{g}\right)$ that are mutually orthogonal, we could obtain an m number of orthogonal Kuiper–Nash Ricci solitons.

(3)

Using Theorem 1 on an n-dimensional compact Riemannian manifold $\left(M^n,g\right)$, we have that the Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ is a submanifold of the Euclidean space $\left({\mathbb{R}}^m,\overline{g}\right)$. Although in this article we obtained interesting results on the Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$, we did not explore the full power of the submanifold geometry of the Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$. For instance, it will be interesting to see the influence of the condition where the mean curvature vector H of the Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ is parallel in the normal bundle, which will in turn have an impact on the Kuiper–Nash function $\phi$ and will give the gradient

$$\nabla \phi =-S_H\xi .$$

This may lead to the challenge of classifying the compact Kuiper–Nash Ricci soliton $\left(M^n,g,\xi ,\lambda \right)$ with a parallel mean curvature vector H and a harmonic Kuiper–Nash function $\phi$.