A Note on Incompressible Vector Fields

Abstract

In this paper, we use incompressible vector fields for characterizing Killing vector fields. We show that on a compact Riemannian manifold, a nontrivial incompressible vector field has a certain lower bound on the integral of the Ricci curvature in the direction of the incompressible vector field if, and only if, the vector field $\xi$ is Killing. We also show that a nontrivial incompressible vector field $\xi$ on a compact Riemannian manifold is a Jacobi-type vector field if, and only if, $\xi$ is Killing. Finally, we show that a nontrivial incompressible vector field $\xi$ on a connected Riemannian manifold has a certain lower bound on the Ricci curvature in the direction of $\xi$, and if $\xi$ is also a geodesic vector field, it necessarily implies that $\xi$ is Killing.

1. Introduction

It is well know in hydromechanics that there are many fluids which are difficult to compress (in other words their volumes do not change under a pressure) and their density is essentially constant. This fact is expressed as the following equation of continuity:

$$\mathrm{div}\left(V\right)=0,$$

with V being flow velocity. The above equation also says that the velocity of the fluid flow is incompressible. Furthermore, in electromagnetism, the magnetic field F is divergence free and is expressed as

$$\mathrm{div}\left(F\right)=0.$$

Incompressible vector fields are very important in magnetohydrodynamics and they are often used in modern technology, especially in electronic engineering and electrodynamics (cf. [1,2,3,4,5,6]).

On a Riemannian manifold $(N,h)$, a smooth vector field $\mathbf{t}$ is said to be an incompressible vector field if

$$\mathrm{div}\left(\mathbf{t}\right)=0,$$

(1)

where

$$\mathrm{div}\left(\mathbf{t}\right)={\Sigma }_{i=1}^{n}h\left({\nabla }_{u_i}\mathbf{t},u_i\right),$$

for a local frame $\{u,\dots ,u_i\}$ on $(N,h)$, $n=\dim N$, and ∇ is the Riemannian connection on N with respect to the Riemannian metric. It follows that all parallel vector fields on $(N,h)$ satisfy Equation (1) and are, therefore, incompressible vector fields. If a vector field does not have a source or a sink, this property is equivalent to the fact that the vector field is incompressible.

It is worth noting that Killing fields on $(N,h)$ (cf. [7,8,9]) play a very important role in geometry as well as in physics. A vector field $\xi$ on $(N,h)$ is said to be a Killing if

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

(2)

${\pounds }_{\xi }$ being the Lie derivative in the direction of $\xi$. Note that for a Killing vector field $\xi$, Equation (2) is equivalent to

$$h\left({\nabla }_{E_1}\xi ,E_2\right)+h\left({\nabla }_{E_2}\xi ,E_1\right)=0,E_1,E_2\in \mathfrak{X}\left(N\right)$$

(3)

where $\mathfrak{X}\left(N\right)$ is the Lie algebra of vector fields on N. Taking a local frame $\{u_1,\dots ,u_n\}$ on N and using $E_1=E_2=u_i$ in Equation (3) and taking the sum, we obtain

$$\mathrm{div}\left(\xi \right)=0,$$

Thus, we observe that a Killing vector field is an incompressible vector field. However, the converse is not true. There are incompressible vector fields which are not Killing vector fields. For instance, consider a vector field $\mathbf{t}=u^2\frac{\partial }{\partial u^1}+u^1\frac{\partial }{\partial u^2}$ defined on the Euclidean space $({\mathbf{R}}^2,h)$, where $h=⟨⟩$ is the Euclidean metric. Then, it is easy to see that $\mathrm{div}\left(\mathbf{t}\right)=0$ and ${\pounds }_{\mathbf{t}}h\ne 0$, that is, $\mathbf{t}$ is an incompressible vector field that is not Killing.

Apart from incompressible vector fields, in fluid mechanics there is yet another class of important vector fields, namely, irrotational vector fields. In three-dimensional Euclidean space, a vector field is said to be irrotational if its curl is zero everywhere and if, in addition, the vector field is smooth, it will be conservative (as Euclidean space is simply connected), that is, it will be equal to the gradient of a smooth function. On a Riemannian manifold $(N,h)$, a smooth vector field t is said to be conservative if it is a gradient of a smooth function. Moreover, by Helmholtz’s theorem, any vector field on $(N,h)$ can be written as the sum of a conservative vector field and an incompressible vector field.

Note that on a compact Riemannian manifold, owing to the presence of a Killing vector field, its geometry as well as topology is influenced. A compact Riemannian manifold possessing a Killing vector field cannot have negative Ricci curvature as well, as its fundamental group has a cyclic subgroup of constant index [7,8,9]. Furthermore, it is interesting to note that a Killing vector field on a Riemannian manifold of even dimension and positive curvature should vanish at some point.

Note that incompressible vector fields are important objects in fluid mechanics and that Killing vector fields are important in shaping the geometry and topology of the space on which they are defined. Moreover, a Killing vector field is an incompressible vector field, while the converse is false. Therefore, a natural question arises: Under what condition is an incompressible vector field $\mathbf{t}$ on a Riemannian manifold $(N,h)$ a Killing vector field? In this paper, we consider this question and prove three results in Section 3. The first two results give characterizations of Killing vector fields using incompressible vector fields on a compact Riemannian manifold and the third result elucidates the conditions under which an incompressible vector field on a connected Riemannian manifold is Killing.

2. Preliminaries

Let $(N,h)$ be an n-dimensional Riemannian manifold and $\mathbf{t}$ be an incompressible vector field on $(N,h)$. Then, we have

$$\mathrm{div}\left(\mathbf{t}\right)={\Sigma }_{i=1}^{n}h\left({\nabla }_{u_i}\mathbf{t},u_i\right)=0,$$

where $\{u_1,\dots ,u_n\}$ is a local frame on N and ∇ is the Riemannian connection on $(N,g)$. We denote by $\mathfrak{X}\left(N\right)$ the Lie algebra of the smooth vector fields on N. We define a symmetric operator $B:\mathfrak{X}\left(N\right)\mapsto \mathfrak{X}\left(N\right)$ by

$$\frac{1}{2}\left({\pounds }_{\mathbf{t}}h\right)(E_1,E_2)=h\left(B{E}_1,E_2\right)E_1,E_2\in \mathfrak{X}\left(N\right),$$

(4)

where ${\pounds }_{\mathbf{t}}$ is the Lie derivative with respect to the incompressible vector field $\mathbf{t}$. Let $\alpha$ be 1- form on N, defined by

$$\alpha \left(E\right)=h(E,\mathbf{t}),E\in \mathfrak{X}\left(N\right).$$

Then, we define a skew-symmetric operator $\phi :\mathfrak{X}\left(N\right)\mapsto \mathfrak{X}\left(N\right),$ by

$$\frac{1}{2}d\alpha \left(E_1,E_2\right)=h\left(\phi E_1,E_2\right)E_1,E_2\in \mathfrak{X}\left(N\right),$$

(5)

where $d\alpha$ is the exterior derivative of $\alpha$. Using Equations (4) and (5) together with Koszul’s formula (cf. [10]), we obtain

$${\nabla }_E\mathbf{t}=B\left(E\right)+\phi \left(E\right),E\in \mathfrak{X}\left(N\right).$$

(6)

Since $\phi$ is skew-symmetric, we have

$$\mathrm{tr}·\phi ={\Sigma }_{i=1}^{n}h\left(\phi u_i,u_i\right)=0,$$

for a local frame $\{u_1,\dots ,u_n\}$ on N. If we define a smooth function $b:N\mapsto \mathbf{R}$ by

$$b=\mathrm{tr}B={\Sigma }_{i=1}^{n}h\left(B{u}_i,u_i\right),$$

then as $\mathbf{t}$ is an incompressible vector field we obtain

$$\mathrm{div}\left(\mathbf{t}\right)=b=0.$$

(7)

Using Equation (6), we have

$$\begin{array}{ccc}\hfill {\nabla }_{E_1}{\nabla }_{E_2}\xi & =& {\nabla }_{E_1}\left(B{E}_2+\phi E_2\right)={\nabla }_{E_1}B{E}_2+{\nabla }_{E_1}\phi E_2\hfill \\ & =& \left(\nabla B\right)\left(E_1,E_2\right)+B\left({\nabla }_{E_1},E_2\right)+\left(\nabla \phi \right)\left(E_1,E_2\right)+\phi \left({\nabla }_{E_1},E_2\right).\hfill \end{array}$$

where $\left(\nabla B\right)\left(E_1,E_2\right)={\nabla }_{E_1}B{E}_2-B\left({\nabla }_{E_1}{E}_2\right)$. Thus, we have the following for the expression for the curvature tensor $\mathbf{R}\left(E_1,E_2\right)\xi$ of $\left(N,H\right)$ (cf. [10]).

$$\begin{array}{cc}\hfill R\left(E_1,E_2\right)\xi & ={\nabla }_{E_1}{\nabla }_{E_2}\xi -{\nabla }_{E_2}{\nabla }_{E_1}\xi -{\nabla }_{[E_1,E_2]}\hfill \\ \hfill & =\left(\nabla B\right)\left(E_1,E_2\right)-\left(\nabla B\right)\left(E_1,E_2\right)+\left(\nabla \phi \right)\left(E_1,E_2\right)-\left(\nabla \phi \right)\left(E_1,E_2\right).\hfill \end{array}$$

(8)

Note that the Ricci tensor $\mathrm{Ric}$ of $\left(N,h\right)$ is given by (cf. [10])

$$\mathrm{Ric}\left(E_1,E_2\right)={\Sigma }_{i=1}^n\left(\mathbf{R}(u_i,E_1)E_2,u_i\right),$$

for a local frame $\{u_1,\dots ,u_n\}$ on N, $n=\dim N$.

In addition, note that as $b=0$, we have

$$\begin{array}{cc}\hfill 0& =E{\Sigma }_{i=1}^{n}h\left(B{u}_i,u_i\right)={\Sigma }_{i=1}^n\left[h({\nabla }_{E}B{u}_i,u_i)+h(B{u}_i,{\nabla }_E{u}_i)\right]\hfill \\ \hfill & ={\Sigma }_{i=1}^n\left[h(\nabla B)(E,u_i)+B({\nabla }_E{u}_i,u_i)+h(B{\nabla }_E{u}_i,u_i)\right]\hfill \\ \hfill & ={\Sigma }_{i=1}^{n}h\left((\nabla B)(E,u_i),u_i\right)+2{\Sigma }_{i=1}^{n}h\left(B{\nabla }_E{u}_i,u_i\right).\hfill \end{array}$$

(9)

On using a normal coordinate local frame, we have

$${\Sigma }_{i=1}^{n}h\left(B{\nabla }_E{u}_i,u_i\right)=0,$$

this can also be achieved using symmetry of B and skew-symmetry of connection forms ${\omega }_i^j$ defined by ${\nabla }_E{u}_i={\Sigma }_{j=1}^n{\omega }_i^j\left(E\right)u_j$. Thus, from Equation (9), we conclude that

$${\Sigma }_{i=1}^{n}h\left((\nabla B)(E,u_i),u_i\right)=0.$$

(10)

Similarly, as $\mathrm{tr}\phi =0$, we have

$$\begin{array}{ccc}\hfill 0& =& E{\Sigma }_{i=1}^{n}h\left(\phi u_i,u_i\right)={\Sigma }_{i=1}^n\left[h({\nabla }_E\phi u_i,u_i)+h(\phi u_i,{\nabla }_E{u}_i)\right]\hfill \\ & =& {\Sigma }_{i=1}^n\left[h((\nabla \phi )(E,u_i)+\phi \left({\nabla }_E{u}_i\right),u_i)+h(\phi u_i,{\nabla }_E{u}_i)\right].\hfill \end{array}$$

Using the skew-symmetry of $\phi$, we have

$${\Sigma }_{i=1}^n\left[h((\nabla \phi )(E,u_i),u_i)-h({\nabla }_E{u}_i,\phi u_i)+h(\phi u_i,{\nabla }_E{u}_i)\right]=0,$$

that is,

$${\Sigma }_{i=1}^{n}h((\nabla \phi )(E,u_i),u_i)=0.$$

(11)

Using Equation (8), we obtain

$$\begin{array}{ccc}\hfill \mathrm{Ric}(E,\xi )& =& {\Sigma }_{i=1}^{n}h\left(\mathbf{R}(u_i,E)\xi ,u_i\right)\hfill \\ & =& {\Sigma }_{i=1}^{n}h\left((\nabla B)(u_i,E)-(\nabla B)(E,u_i)+(\nabla \phi )(u_i,E)-(\nabla \phi )(E,u_i),u_i\right).\hfill \end{array}$$

Using Equations (10) and (11), we conclude

$$\mathrm{Ric}\left(E,\xi \right)={\Sigma }_{i=1}^{n}h\left((\nabla B)(u_i,E)+(\nabla \phi )(u_i,E),u_i\right).$$

Note that B is symmetric and $\phi$ is skew-symmetric, we have

$$h\left((\nabla B)(E_1,E_2),E_3\right)=h\left(E_2,(\nabla B)(E_1,E_3)\right)$$

and

$$h\left((\nabla \phi )(E_1,E_2),E_3\right)=-h\left(E_2,(\nabla \phi )(E_1,E_3)\right).$$

Thus, we have

$$\mathrm{Ric}(E,\xi )={\Sigma }_{i=1}^n\left[h(E,(\nabla B)(u_i,u_i))-h(E,(\nabla \phi )(u_i,u_i))\right].$$

(12)

3. Characterization of Killing Vector Fields

In this section, we find conditions under which an incompressible vector field on a compact Riemannian manifold $\left(N,h\right)$ becomes a Killing vector field. Note that if $\xi$ is a parallel vector field then it is both incompressible as well as a Killing vector field. Therefore, a non-parallel incompressible vector field is called a nontrivial incompressible vector field. For the symmetric operator B and skew-symmetric operator $\phi$, we define

$${\parallel B\parallel }^2={\Sigma }_{i=1}^{n}h\left(B{u}_i,B{u}_i\right),{\parallel \phi \parallel }^2={\Sigma }_{i=1}^{n}h(\phi u_i,\phi u_i),$$

where $\{u_1,\cdots ,u_n\}$ is a local frame on $\left(N,h\right)$, $n=\dim N$.

Now, we are ready to prove the following

Theorem 1. 

A nontrivial incompressible vector field ξ on a compact Riemannian manifold $\left(N,h\right)$ is a Killing vector field if and only if

$${\int }_M\mathrm{Ric}(\xi ,\xi )\ge {\int }_M{\parallel \phi \parallel }^2.$$

Proof. 

Let $\xi$ be a nontrivial incompressible vector field on an n-dimensional Riemannian manifold $\left(N,h\right)$. Therefore, using Equation (12), we obtain

$$\mathrm{Ric}(\xi ,\xi )={\Sigma }_{i=1}^{n}h\left(\xi ,(\nabla B)(u_i,u_i)\right)-{\Sigma }_{i=1}^{n}h\left(h(\xi ,(\nabla \phi )(u_i,u_i))\right),$$

(13)

where $\{u_1,\cdots ,u_n\}$ is a local frame on $\left(N,h\right)$. Note that

$$\begin{array}{ccc}\hfill \mathrm{div}B\xi & =& {\Sigma }_{i=1}^{n}h\left({\nabla }_{u_i}B\xi ,u_i\right)\hfill \\ & =& {\Sigma }_{i=1}^{n}h\left((\nabla B)(u_i,\xi )+B\left({\nabla }_{u_i}\xi \right),u_i\right)\hfill \\ & =& {\Sigma }_{i=1}^{n}h\left((\nabla B)(u_i,\xi ),u_i\right)+{\Sigma }_{i=1}^{n}h\left({\nabla }_{u_1}\xi ,B{u}_i\right).\hfill \end{array}$$

Using the symmetry of B and Equation (6), we obtain

$$\begin{array}{ccc}\hfill \mathrm{div}B\xi & =& {\Sigma }_{i=1}^{n}h\left(\xi ,(\nabla B)(u_i,u_i)\right)+{\Sigma }_{i=1}^{n}h\left(B{u}_i+\phi u_i,B{u}_i\right)\hfill \\ & =& {\Sigma }_{i=1}^{n}h\left(\xi ,(\nabla B)(u_i,u_i)\right)+{\parallel B\parallel }^2+{\Sigma }_{i=1}^{n}h\left(\phi u_i,B{u}_i\right).\hfill \end{array}$$

Note that B is a symmetric operator and, therefore, diagonalized by a local frame $\{u_1,\dots ,u_n\}$ as

$$B\left(u_i\right)={\lambda }_i{u}_i$$

and as such, we have

$${\Sigma }_{i=1}^{n}h\left(\phi u_i,B{u}_i\right)={\Sigma }_{i=1}^{n}h\left(\phi u_i,{\lambda }_i{u}_i\right)={\Sigma }_{i=1}^n{\lambda }_{i}h\left(\phi u_i,u_i\right)=0,$$

where we have used the skew-symmetry of $\phi$ that guarantees $h\left(\phi u_i,u_i\right)=0$. Hence, we have

$$\mathrm{div}B\xi ={\Sigma }_{i=1}^{n}h\left(\xi ,(\nabla B)(u_i,u_i)\right)+{\parallel B\parallel }^2$$

(14)

Furthermore, we have

$$\begin{array}{ccc}\hfill \mathrm{div}\phi \xi & =& {\Sigma }_{i=1}^{n}h\left({\nabla }_{u_i}\xi \phi ,u_i\right)={\Sigma }_{i=1}^{n}h\left((\nabla \phi )(u_i,\phi )+\phi \left({\nabla }_{u_i}\xi \right),u_i\right)\hfill \\ & =& {\Sigma }_{i=1}^{n}h\left((\nabla \phi )(u_i,\xi ),u_i\right)-{\Sigma }_{i=1}^{n}h\left(\left({\nabla }_{u_i}\xi \right),\phi u_i\right)\hfill \end{array}$$

where we used the skew-symmetry of $\phi$. Again using the skew-symmetry of $\phi$, and Equation (6), we obtain

$$\mathrm{div}\phi \xi =-{\Sigma }_{i=1}^{n}h\left(\xi ,(\nabla \phi )(u_i,u_i)\right)-{\parallel \phi \parallel }^2.$$

(15)

Now, we use Equations (14) and (15) in Equation (13) to confirm

$$\mathrm{Ric}(\xi ,\xi )=\mathrm{div}B\xi -{\parallel B\parallel }^2+\mathrm{div}\phi \xi +{\parallel \phi \parallel }^2,$$

that is,

$${\parallel B\parallel }^2=\mathrm{div}B\xi +\mathrm{div}\phi \xi +{\parallel \phi \parallel }^2-\mathrm{Ric}(\xi ,\xi ).$$

Integrate the above equation, from which, keeping in mind Stokes’s Theorem, we conclude

$${\int }_M{\parallel B\parallel }^2={\int }_M\left({\parallel \phi \parallel }^2-\mathrm{Ric}(\xi ,\xi )\right).$$

The statement of the theorem implies

$${\int }_M{\parallel B\parallel }^2\le 0,$$

that is, $B=0$. Hence, we have

$${\nabla }_E\xi =\phi E$$

and this gives $\left({\pounds }_{\xi }h\right)\left(E_1,E_2\right)=g\left(\phi E_1,E_2\right)+g\left(E_1,\phi E_2\right)=0,$ where the skew-symmetry of $\phi$ is used. Hence, $\xi$ is a Killing vector field.

Conversely, assume $\xi$ is a Killing vector field on $\left(N,h\right)$, then $\xi$ is incompressible and $B=0$. Then, Equation (12) gives

$$\mathbf{R}(\xi ,\xi )=-{\Sigma }_{i=1}^{n}h\left(\xi ,(\nabla \phi )(u_i,u_i)\right)$$

and using Equation (15) in the above equation, we obtain

$$\mathrm{Ric}(\xi ,\xi )=\mathrm{div}\phi \xi +{\parallel \phi \parallel }^2.$$

Integrating the above equation while using Stokes’s Theorem, we conclude

$${\int }_M\mathrm{Ric}(\xi ,\xi )={\int }_M{\parallel \phi \parallel }^2.$$

Hence, the condition holds and this completes the proof. □

We observe that in the above theorem we used a geometric constraint, namely, the integral of the $\mathrm{Ric}(\xi ,\xi )$ has a lower bound containing ${\parallel \phi \parallel }^2$.

In our next result, we are going to use an analytic constraint, namely, in addition to $\xi$ being an incompressible vector field, it is also a Jacobi-type vector field, to obtain a characterization of a Killing vector field on a compact Riemannian manifold $\left(N,h\right)$.

Recall that a vector field u on a Riemannian manifold $\left(N,h\right)$ is said to be a Jacobi-type vector field if it satisfies (cf. [11])

$${\nabla }_{E_1}{\nabla }_{E_2}u-{\nabla }_{{\nabla }_{E_1{E}_2}}u+\mathbf{R}(u,E_1)E_2=0,E_1,E_2\in \mathfrak{X}\left(N\right).$$

Moreover, a Killing vector field is a Jacobi-type vector field and the converse is false (cf. [11]). In addition, it is known that a Jacobi-type vector field u on $\left(N,h\right)$ satisfies (cf. [11]).

$$Q\left(u\right)+▵u=0,$$

(16)

where Q is the Ricci operator defined by

$$\mathrm{Ric}\left(E_1,E_2\right)=\left(Q{E}_1,E_2\right),E_1,E_2\in \mathfrak{X}\left(N\right),$$

and $▵:\mathfrak{X}\left(N\right)\mapsto \mathfrak{X}\left(N\right)$ is the Laplace operator defined by

$$▵u={\Sigma }_{i=1}^n\left({\nabla }_{u_i}{\nabla }_{u_i}-{\nabla }_{{\nabla }_{u_i}{u}_i}u\right),$$

for a local frame $\{u_1,\dots ,u_n\}$ on N, $n=\dim N$. Next, we prove the following:

Theorem 2. 

Let ξ be a nontrivial incompressible vector field on an n-dimensional compact Riemannian manifold $\left(N,h\right)$. Then, ξ is a Jacobi-type vector field if and only if ξ is a Killing vector field.

Proof. 

Suppose $\xi$ is an incompressible vector field and is also a Jacobi-type vector field on $\left(N,h\right)$. Then, we have (cf. [11]).

$$Q\left(\xi \right)+▵\xi =0$$

(17)

Using Equation (12), we have

$$Q\left(\xi \right)={\Sigma }_{i=1}^n\left((\nabla B)(u_i,u_i)-(\nabla \phi )(u_i,u_i)\right),$$

(18)

where $\{u_1,\dots ,u_n\}$ is a local frame on N. Now, using Equation (6), we have

$${\nabla }_E{\nabla }_E\xi ={\nabla }_E(BE+\phi E)={\nabla }_{E}BE+{\nabla }_E\phi E,$$

and

$${\nabla }_{{\nabla }_{E}E}\xi =B\left({\nabla }_{E}E\right)+\phi \left({\nabla }_{E}E\right).$$

Thus, we have

$${\nabla }_E{\nabla }_E\xi -{\nabla }_{{\nabla }_{E}E}\phi =(\nabla B)(E,E)+(\nabla \phi )(E,E).$$

Usinga local frame $\{u_1,\dots ,u_n\}$ in the above equation, we obtain

$${\Sigma }_{i=1}^n\left({\nabla }_{u_i}{\nabla }_{u_i}\xi -{\nabla }_{{\nabla }_{u_i}{u}_i}\xi \right)={\Sigma }_{i=1}^n\left((\nabla B)(u_i,u_i)+(\nabla \phi )(u_i,u_i)\right),$$

that is,

$$▵\xi ={\Sigma }_{i=1}^n\left((\nabla B)(u_i,u_i)+(\nabla \phi )(u_i,u_i)\right).$$

(19)

Inserting Equations (18) and (19) into Equation (17), we conclude

$${\Sigma }_{i=1}^n(\nabla B)(u_i,u_i)=0.$$

(20)

Finally, Equation (14) in view of Equation (20) implies

$$\mathrm{div}B\xi ={\parallel B\parallel }^2.$$

(21)

Integration the above equation, we obtain $B=0$ and, therefore, $\xi$ is a Killing vector field.

Conversely, if $\xi$ is Killing, it is incompressible and also a Jacobi-type vector field (cf. [11]). □

On a Riemannian manifold, there is one more important type of vector field, namely, a geodesic vector field (cf. [3]). A vector field u on a Riemannian manifold $\left(N,h\right)$ is called a geodesic vector field if

$${\nabla }_{u}u=0.$$

(22)

Now, we have the following result:

Theorem 3. 

Let ξ be an incompressible vector field on an n-dimensional connected Riemannian manifold $\left(N,h\right)$. If ξ is also a geodesic vector field and the Ricci curvature $\mathrm{Ric}(\xi ,\xi )$ satisfies

$$\mathrm{Ric}(\xi ,\xi )\ge {\parallel \phi \parallel }^2$$

then ξ is a Killing vector field.

Proof. 

Suppose $\xi$ is an incompressible vector field on an n-dimensional connected Riemannian manifold $\left(N,h\right)$ that is also a geodesic vector field. Then, by using Equations (6) and (22), we obtain

$$B\xi +\phi \xi =0.$$

(23)

Using Equations (14) and (15) in Equation (23), we conclude

$${\Sigma }_{i=1}^{n}h\left(\xi ,(\nabla B)(u_i,u_i)\right)+{\parallel B\parallel }^2-{\Sigma }_{i=1}^{n}h\left(\xi ,(\nabla \phi )(u_i,u_i)\right)-{\parallel \phi \parallel }^2=0.$$

(24)

Using Equation (13) in Equation (24), we conclude

$$\mathrm{Ric}(\xi ,\xi )-{\parallel \phi \parallel }^2+{\parallel B\parallel }^2=0,$$

that is,

$${\parallel B\parallel }^2={\parallel \phi \parallel }^2-\mathrm{Ric}(\xi ,\xi ).$$

Using the condition in the statement, we have

$${\parallel B\parallel }^2\le 0,$$

proving that $B=0$, that is, $\xi$ is a Killing vector field. □

4. Conclusions

In Theorems 1 and 2 we found characterizations of a Killing vector field using an incompressible vector field on a compact Riemannian manifold. However, in Theorem 3 we found only the necessary conditions for an incompressible vector field $\xi$ on a connected Riemannian manifold $\left(N,h\right)$ to be a Killing vector field. We have shown that if in addition $\xi$ is a geodesic vector field and the geometric constraint $Ric\left(\xi ,\xi \right)\ge {∥\phi ∥}^2$ holds, then $\xi$ is Killing. The converse of this result is not true, for example, the vector field $\xi =J\Psi$ on the Euclidean space $\left({\mathbf{R}}^{2n},⟨,⟩\right)$ is a Killing vector field, where J is the complex structure and $\Psi$ is the position vector field on ${\mathbf{R}}^{2n}$. For the Euclidean connection ∇ on the Euclidean space $\left({\mathbf{R}}^{2n},⟨,⟩\right)$, we have

$${\nabla }_E\xi =JE,E\in \mathfrak{X}\left({\mathbf{R}}^{2n}\right).$$

Consequently, for a local frame $\left\{u_1,\dots ,u_{2n}\right\}$ on ${\mathbf{R}}^{2n}$, we have

$$\mathrm{div}\left(\xi \right)={\Sigma }_{i=1}^{2n}⟨{\nabla }_{u_i}\xi ,u_i⟩={\Sigma }_{i=1}^{2n}⟨J{u}_i,u_i⟩=0,$$

that is, $\xi$ is incompressible. However,

$${\nabla }_{\xi }\xi =J\xi \ne 0,$$

that is, $\xi$ is not a geodesic vector field.

Hence, it will be an interesting question to use an incompressible vector field on a connected noncompact Riemannian manifold in order to find a characterization of a Killing vector field.

Recall that there is a typical relationship between singularity theory and submanifold theory (cf. [7,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29]). Note that when a straight line moves along a curve, it sweeps out a surface known as a ruled surface in Euclidean 3-space [15]. Ruled surfaces are an important research subject in classical differential geometry and commonly applied in a few areas, such as CAD, moving geometry, and line geometry. For more details of these relationships, we refer to an interesting paper [17]. We point out that it will be interesting to look at the geometry of ruled submanifolds possessing incompressible vector fields, and we wish to take up this issue in our future studies.