Remarks on the Warped Product Structure from the Hessian of a Function

Abstract

The warped product structure of a gradient Yamabe soliton and a Ricci soliton with a concircular potential field is proved in another way.

1. Introduction

Let M be an $(n+1)$-dimensional complete and simply connected Riemannian manifold with metric g. Given a gradient vector field $\nabla f$ for a smooth function $f:M⟶\mathbb{R}$, if a one-parameter family of diffeomorphisms generated by the integral curves of $\nabla f$ on M satisfies

$$\mathrm{Hess}f=(R-\rho )g$$

(1)

for the scalar curvature R and some constant $\rho$, then M is called a gradient steady, expanding and shrinking Yamabe soliton for $\rho =0$, $\rho <0$ and $\rho >0$, respectively. In [1], the warped product structure of a gradient Yamabe soliton is shown. But is well-known that the equation

$$\mathrm{Hess}f=\mu g$$

(2)

for smooth functions $f\prime \mu :M⟶\mathbb{R}$ determines the warped product structure $M=(-\infty ,\infty ){\times }_{\left(f^{\prime }\right)}F$ where $F=\exp ({\left(\nabla f\right)}^{\perp })$ and $f^{\prime }$ is the warping function [2]. Thus this fact can be applied to a gradient Yamabe soliton [1] and a Ricci soliton with a concircular potential field [3].

In another way by using the Jacobi differential equation, we show that the Equation (2) determines the warped product structure. Note that each fiber of a warped product is called totally umblic if the shape operator is a multiple of the identity at each point. A gradient Yamabe soliton with $\mu =R-\rho$ is a warped product with the totally umblic fibers $p\times F$ by the following shape operator $S_N$ (4).

Let $H_s=\{x\in M|f\left(x\right)=s\}$ be a regular hypersurface of M with a unit normal vector field $N=-\frac{\nabla f}{\left|\nabla f\right|}$ for some $s\in \mathbb{R}$. For orthogonal distributions $\nabla f$ and ${\left(\nabla f\right)}^{\perp }$ on $TM$, we see that

$$g(D_{\nabla f}\nabla f,V)=\mathrm{Hess}f(\nabla f,V)=\mu g(\nabla f,V)=0$$

for all $V\in {\left(\nabla f\right)}^{\perp }$. Put $-{\gamma }^{\prime }=N=-\frac{\nabla f}{\left|\nabla f\right|}$. Since $g(D_{\nabla f}\nabla f,V)=0$ for all $V\in {\left(\nabla f\right)}^{\perp }$, we get

$$D_{\nabla f}\nabla f=\frac{g(D_{\nabla f}\nabla f,\nabla f)}{{\left|\nabla f\right|}^2}\nabla f.$$

So it follows from

$$\nabla f(\frac{1}{\left|\nabla f\right|})\nabla f=-\frac{g(D_{\nabla f}\nabla f,\nabla f)}{\sqrt{g(\nabla f,\nabla f)}{\left|\nabla f\right|}^2}\nabla f$$

that

$$\frac{1}{\left|\nabla f\right|}{D}_{\nabla f}(\frac{1}{\left|\nabla f\right|}\nabla f)=0,\mathrm{that}\mathrm{is},D_{N}N=D_{{\gamma }^{\prime }}{\gamma }^{\prime }=0.$$

(3)

So we can consider each level hypersurface $H_t$ of $H_s$ along a unit-speed geodesic $\gamma \left(t\right)$ orthogonal to $H_s$ with $N_{\gamma \left(s\right)}=-{\gamma }^{\prime }\left(s\right)$ and $\gamma \left(s\right)\in H_s$ by (3). The shape operator of each level hypersurface $H_t$ of $H_s$ along a geodesic $\gamma$ is given by

$$\begin{array}{c}\hfill \begin{array}{ccc}g(S_{N}V,W)& =& -g(D_{V}N,W)\\ & =& \frac{1}{\left|\nabla f\right|}g(D_V\nabla f,W)\\ & =& \frac{1}{\left|\nabla f\right|}\mathrm{Hess}f(V,W)\\ & =& \frac{\mu }{\left|\nabla f\right|}g(V,W)\end{array}\end{array}$$

(4)

for all $V,W\in {\left(\nabla f\right)}^{\perp }$. The trace of the shape operator is the mean curvature

$$\theta =n\frac{\mu }{\left|\nabla f\right|}.$$

(5)

Since $Vg(\nabla f,\nabla f)=2g(D_V\nabla f,\nabla f)=2\mathrm{Hess}f(\nabla f,V)=2\mu g(\nabla f,V)=0$ for all $V\in {\left(\nabla f\right)}^{\perp }$, $\left|\nabla f\right|$ is constant on each level hypersurface.

Differentiation of the Equation (2) ${\nabla }_i{\nabla }_{j}f=\mu g_{ij}$ with the Einstein convention gives

$${\nabla }_k{\nabla }_i{\nabla }_{j}f={\nabla }_k\mu g_{ij}$$

which implies

$${\nabla }_i{\nabla }_k{\nabla }_{j}f+R_{kil}^j{\nabla }^{l}f={\nabla }_k\mu g_{ij}.$$

Take the trace in k and j, then we have

$${\nabla }_i\Delta f+R_{il}{\nabla }^{l}f={\nabla }_i\mu .$$

The trace of (2) $\mathrm{\Delta }f=(n+1)\mu$ gives

$$(n+1){\nabla }_i\mu +R_{il}{\nabla }^{l}f={\nabla }_i\mu .$$

So we obtain

$$\begin{array}{c}\hfill n{D}_X\mu =-\mathrm{Ric}(\nabla f,X)\end{array}$$

(6)

for all $X\in TM$ as in [4]. Here we show that the mean curvature $\theta =n\frac{\mu }{\left|\nabla f\right|}$ (5) and the Equation (6) determine the warped product structure. Since

$$\mu =\mathrm{Hess}f(N,N)=-g(D_N\left(\right|\nabla f\left|N\right),N)=-N\left(\right|\nabla f\left|\right),$$

(7)

We get

$$\begin{array}{c}\hfill \begin{array}{ccc}\left|\nabla f\right|\mathrm{Ric}(N,N)\stackrel{\left(6\right)}{=}n{D}_N\mu & \stackrel{\left(5\right)}{=}& N\left(\right|\nabla f\left|\right)\theta +\left|\nabla f\right|D_N\theta \\ & \stackrel{\left(7\right)}{=}& -\mu \theta +\left|\nabla f\right|D_N\theta \\ & \stackrel{\left(5\right)}{=}& -\left|\nabla f\right|\frac{{\theta }^2}{n}-\left|\nabla f\right|D_{{\gamma }^{\prime }}\theta \\ & =& \left|\nabla f\right|(-\frac{{\theta }^2}{n}-{\theta }^{\prime }).\end{array}\end{array}$$

(8)

Hence we obtain the Raychaudhuri Equation (12) along a geodesic $\gamma \left(t\right)$ with the zero shear tensor $\sigma \left(t\right)$

$$\begin{array}{c}\hfill {\theta }^{\prime }\left(t\right)+\frac{{\theta }^2\left(t\right)}{n}+\mathrm{Ric}({\gamma }^{\prime }\left(t\right),{\gamma }^{\prime }\left(t\right))=0.\end{array}$$

(9)

Thus we can show Theorem 1 by the Jacobi differential equation.

Theorem 1.

Let M be a complete and simply connected Riemannian manifold. If $Hessf=\mu g$ for smooth functions $f,\mu :M⟶\mathbb{R}$, then M is a warped product $M=(-\infty ,\infty ){\times }_{\left(f^{\prime }\right)}F$ where $F=\exp ({\left(\nabla f\right)}^{\perp })$ and $f^{\prime }$ is the warping function.

For a gradient Yamabe soliton, we get

$$\begin{array}{c}\hfill \mu =R-\rho \stackrel{\left(1\right)}{=}\mathrm{Hess}f(N,N)=\mathrm{Hess}f({\gamma }^{\prime },{\gamma }^{\prime })=g(D_{{\gamma }^{\prime }}\nabla f,{\gamma }^{\prime })={\gamma }^{\prime }g(\nabla f,{\gamma }^{\prime })=f^{″}.\end{array}$$

(10)

Thus a gradient Yamabe soliton with a regular hypersurface $H_s=\{x\in M|f\left(x\right)=s\}$ for some $s\in \mathbb{R}$ is a warped product with the totally umblic fibers whose scalar curvature are constant by (10) (cf. [1]). Differentiation of the above equation

$$\left|\nabla f\right|\mathrm{Ric}(N,N)\stackrel{\left(8\right)}{=}n{D}_N(R-\rho )=-n{D}_{{\gamma }^{\prime }\left(t\right)}(R-\rho )=-n{\left(f^{″}\left(t\right)\right)}^{\prime }=-n{g}^{″}\left(t\right)$$

for $f^{\prime }\left(t\right)=g\left(t\right)$ shows that if $\mathrm{Ric}(N,N)=0$ or $D_{{\gamma }^{\prime }\left(t\right)}R=0$, then we get $g^{″}\left(t\right)=0$. Thus under the assumption $\mathrm{Ric}(N,N)=0$ or $D_{{\gamma }^{\prime }\left(t\right)}R=0$, if a singular point is allowed, then $g\left(t\right)=at+b$ for some $a,b\in \mathbb{R}$. Otherwise M is a product manifold.

A vector field v on M is said to be concircular if it satisfies

$$D_{X}v=\mu X$$

for all $X\in TM$ and a non-trivial function $\mu$ on M. The warped product structure of a Ricci soliton with a concircular potential field is shown in [3]. It is also pointed out that the gradient $\nabla f$ of f is a concircular vector field if and only if $\mathrm{Hess}f=\mu g$. Then a gradient Ricci soliton equations $\mathrm{Ric}+\mathrm{Hess}f=\lambda g$ with $\mathrm{Hess}f=\mu g$ becomes

$$\mathrm{Ric}=(\lambda -\mu )g.$$

So M is Einstein. Thus $\lambda -\mu$ is constant under $n\ge 3$. In Theorem 5.1 in [3], M is turned out to be Ricci-flat. So we have $\lambda =\mu$. Therefore we have

$$0=\left|\nabla f\right|\mathrm{Ric}(N,N)\stackrel{\left(8\right)}{=}n{D}_N\mu =-n{g}^{″}.$$

If a singular point is allowed, then $g\left(t\right)=at+b$ for some $a,b\in \mathbb{R}$. A Gaussian gradient Ricci soliton with $f\left(x\right)=\frac{\lambda }{2}{\left|x\right|}^2$ on ${\mathbb{R}}^n$ is an example for it.

We show the warped product structure with the base B whose dimension is $\dim B=m>1$.

Theorem 2.

Let M be a complete and simply connected Riemannian manifold with $dimM=m+n$. Assume that $TB$ and $TF$ are orthogonal distributions on $TM$ and

$$\mathrm{Hess}{f}_{\nu }=\mu g\nu =1,\dots ,m$$

such that $N_{\nu }=-\frac{\nabla f_{\nu }}{|\nabla f_{\nu }|}$ are unit normal vectors of $\exp \left(TF\right)$ for smooth functions $f_{\nu },\mu :M⟶\mathbb{R}$ and $dimB=m>1$. If $|\nabla f_1|=|\nabla f_2|=\cdots =|\nabla f_m|$ and $f_1^{″}=f_2^{″}=\cdots =f_m^{″}$, then M is a warped product $M=B{\times }_{\left(f^{\prime }\right)}F$ with the warping function $f^{\prime }$.

2. The Raychaudhuri and Jacobi Equation

Let M be an $(n+1)$-dimensional Riemannian manifold and $H_s=\{x\in M|f\left(x\right)=s\}$ be a regular hypersurface of M with a unit normal vector field $N=-\frac{\nabla f}{\left|\nabla f\right|}$ for some $s\in \mathbb{R}$. Consider each level hypersurface $H_t$ of $H_s$ along a unit-speed geodesic $\gamma \left(t\right)$ orthogonal to $H_s$ with $N_{\gamma \left(s\right)}=-{\gamma }^{\prime }\left(s\right)$ and $\gamma \left(s\right)\in H_s$. The Riemannian curvature tensor of M along $\gamma \left(t\right)$, the shape operator of $H_s$ is denoted by $R(·,{\gamma }^{\prime }\left(t\right)){\gamma }^{\prime }\left(t\right)$, $S_N$, respectively. If a smooth $(1,1)$ tensor field $A:{\left({\gamma }^{\prime }\left(t\right)\right)}^{\perp }\to {\left({\gamma }^{\prime }\left(t\right)\right)}^{\perp }$ satisfies

$$\begin{array}{c}\hfill A^{″}\left(t\right)+R(A,{\gamma }^{\prime }\left(t\right)){\gamma }^{\prime }\left(t\right)=0,\ker A\left(t\right)\cap \ker A^{\prime }\left(t\right)=\left\{0\right\}\end{array}$$

with initial conditions $A\left(s\right)=\mathrm{Id}$ and $A^{\prime }\left(s\right)=S_N$ for the identity endomorphism $\mathrm{Id}$ of ${\left({\gamma }^{\prime }\left(t\right)\right)}^{\perp }$, then A is said to be an H-Jacobi tensor along $\gamma \left(t\right)$.

Put $B=A^{\prime }{A}^{-1}$ and $R(A,{\gamma }^{\prime }){\gamma }^{\prime }=R_{{\gamma }^{\prime }}A$. Then it follows from [5] that

$$B^{\prime }=A^{″}{A}^{-1}-A^{\prime }{A}^{-1}{A}^{\prime }{A}^{-1}=-R_{{\gamma }^{\prime }}-B\circ B$$

(11)

and the shape operator of each level hypersurface $H_t$ is

$$A^{\prime }{A}^{-1}\left(t\right)=S_{-{\gamma }^{\prime }\left(t\right)}=S_t.$$

Thus we get the mean curvature $\theta \left(t\right)=\mathrm{tr}{S}_t$ of $H_t$. For the adjoint ∗, the vorticity $\frac{1}{2}(B-B^{\ast })$ becomes zero, since a variation tensor field A is a Lagrange tensor (Proposition 1 in [5]). The trace of (11) gives the Raychaudhuri equation

$${\theta }^{\prime }+\frac{{\theta }^2}{n}+\mathrm{Ric}({\gamma }^{\prime },{\gamma }^{\prime })+\mathrm{tr}{\sigma }^2=0$$

(12)

for the shear tensor $\sigma =B-\frac{\theta }{n}\mathrm{Id}$ and the Ricc tensor $\mathrm{Ric}({\gamma }^{\prime },{\gamma }^{\prime })$. The expansion $\theta =\mathrm{tr}\left(B\right)$ satisfies $\theta =\frac{{\left(\det \left(A\right)\right)}^{\prime }}{\det \left(A\right)}$ as in [5]. Differentiation of $x={(\det A)}^{\frac{1}{n}}$ gives

$$x^{\prime }=\frac{1}{n}x\theta \mathrm{and}{x}^{″}=\frac{1}{n}({\theta }^{\prime }+\frac{{\theta }^2}{n})x.$$

(13)

By (12) and (13), we have the Jacobi differential equation

$$x^{″}+\frac{1}{n}(\mathrm{Ric}({\gamma }^{\prime },{\gamma }^{\prime })+\mathrm{tr}{\sigma }^2)x=0.$$

(14)

3. Proofs of Theorems

Proof of Theorem 1.

We have the Raychaudhuri equation with the zero shear tensor $\sigma \left(t\right)$ by (9) along a geodesic $\gamma \left(t\right)$. The shear tensor $\sigma \left(t\right)$ along a geodesic $\gamma \left(t\right)$ is zero if and only if the shape operator of the hypersurface $H_s$ is given by $S_{-{\gamma }^{\prime }\left(s\right)}=c\mathrm{Id}$ for some real constant c together with the isotropic curvature tensor $R_{{\gamma }^{\prime }\left(t\right)}=h\left(t\right)\mathrm{Id}$ [6]. Now we find the isotropic curvature tensor, that is, $h\left(t\right)$. By (10) and $f^{\prime }\left(t\right)=g({\gamma }^{\prime },\nabla f)=\left|\nabla f\right|$, the shape operator of each level hypersurface $H_t$ of $H_s$ along a geodesic $\gamma$ is

$$A^{\prime }{A}^{-1}\left(t\right)=S_{-{\gamma }^{\prime }\left(t\right)}\stackrel{\left(4\right)}{=}\frac{f^{″}\left(t\right)}{f^{\prime }\left(t\right)}\mathrm{Id}$$

with the initial conditions $A\left(s\right)=\mathrm{Id}$ and $A^{\prime }{A}^{-1}\left(s\right)=S_{-{\gamma }^{\prime }\left(s\right)}=\frac{f^{″}\left(s\right)}{f^{\prime }\left(s\right)}\mathrm{Id}$. The uniqueness of the first order differential equation shows that a Jacobi tensor is

$$A\left(t\right)=f^{\prime }\left(t\right)\mathrm{Id}.$$

The curvature tensor along a geodesic $\gamma \left(t\right)$ is $R_{{\gamma }^{\prime }\left(t\right)}=-\frac{g^{″}\left(t\right)}{g\left(t\right)}=-A^{″}{A}^{-1}\left(t\right)$ for $g\left(t\right)=f^{\prime }\left(t\right)$. The Jacobi Equation (14) is

$$g^{″}\left(t\right)+\frac{1}{n}(\mathrm{Ric}({\gamma }^{\prime }\left(t\right),{\gamma }^{\prime }\left(t\right))g\left(t\right)=0,$$

since the shear tensor $\sigma \left(t\right)$ is zero. The line element determines uniquely a Jacobi equation with the suitable initial conditions and vice versa. So we have

$$d{s}^2=d{t}^2+{\left(f^{\prime }\left(t\right)\right)}^{2}d{g}_0^2,$$

where $d{g}_0^2$ is a metric of a totally umblic hypersurface as fiber. □

Proof of Theorem 2.

Let M be a complete and simply connected Riemannian manifold with metric g and $\dim M=m+n$. Let $TB$ and ${\left(TB\right)}^{\perp }=TF$ be orthogonal distributions on $TM$ with $\dim TB=m$ and $\dim TF=n$. Assume that for smooth functions $f_{\nu },\mu :M⟶\mathbb{R}$

$$\mathrm{Hess}{f}_{\nu }=\mu g\nu =1,...,m$$

(15)

such that $N_{\nu }=-\frac{\nabla f_{\nu }}{|\nabla f_{\nu }|}$ are unit normal vectors of $\exp \left(TF\right)$. Then we have

$$g(D_X\nabla f_{\nu },V)=\mathrm{Hess}{f}_{\nu }(X,V)=\mu g(X,V)=0$$

for all $X\in TB$, $V\in TF$ and all $\nu$. Hence we get a totally geodesic integrable distribution $TB$. Put $-{\gamma }_{\nu }^{\prime }=N_{\nu }=-\frac{\nabla f_{\nu }}{|\nabla f_{\nu }|}$ for all $\nu$. Since $g(D_{\nabla f_{\nu }}\nabla f_{\nu },Y)=\mu g(\nabla f_{\nu },Y)=0$ for all $Y\in {\left(\nabla f_{\nu }\right)}^{\perp }$, we get

$$D_{\nabla f_{\nu }}\nabla f_{\nu }=\frac{g(D_{\nabla f_{\nu }}\nabla f_{\nu },\nabla f_{\nu })}{|\nabla f_{\nu }{|}^2}\nabla f_{\nu }.$$

So it follows from

$$\nabla f_{\nu }(\frac{1}{|\nabla f_{\nu }|})\nabla f_{\nu }=-\frac{g(D_{\nabla f_{\nu }}\nabla f_{\nu },\nabla f_{\nu })}{\sqrt{g(\nabla f_{\nu },\nabla f_{\nu })}{\left|\nabla f_{\nu }\right|}^2}\nabla f_{\nu }$$

that

$$\frac{1}{|\nabla f_{\nu }|}{D}_{\nabla f_{\nu }}\left(\frac{1}{|\nabla f_{\nu }|}\nabla f_{\nu }\right)=0.$$

Hence we see $D_{{\gamma }_{\nu }^{\prime }}{\gamma }_{\nu }^{\prime }=0$. So let ${\gamma }_{\nu }\left(t\right)$ be a geodesic with $-{\gamma }_{\nu }^{\prime }\left(t\right)=N_{\nu }=-\frac{\nabla f_{\nu }}{|\nabla f_{\nu }|}$ for $\nu =1,\dots ,m$.

The second fundamental form of $\exp \left(TF\right)$ with respect to $N_{\nu }$ is given by

$$\begin{array}{c}\hfill \begin{array}{ccc}g(\alpha (V,W),N_{\nu })=g(D_{V}W,N_{\nu })& =& -g(D_V{N}_{\nu },W)\\ & =& g(S_{N_{\nu }}V,W)\\ & =& \frac{1}{|\nabla f_{\nu }|}g(D_V\nabla f_{\nu },W)\\ & =& \frac{1}{|\nabla f_{\nu }|}\mathrm{Hess}{f}_{\nu }(V,W)\\ & =& \frac{\mu }{|\nabla f_{\nu }|}g(V,W)\end{array}\end{array}$$

(16)

for all $V,W\in TF$. For each $f_{\nu }$ whose gradient vector is $\nabla f_{\nu }$, we see

$$\mu =\mathrm{Hess}{f}_{\nu }({\gamma }_{\nu }^{\prime },{\gamma }_{\nu }^{\prime })=f_{\nu }^{″}.$$

(17)

The mean curvature is given by the trace of (16)

$${\theta }^F=n\frac{\mu }{|\nabla f_{\nu }|}=n\frac{f_{\nu }^{″}}{|\nabla f_{\nu }|}.$$

(18)

Since

$$Vg(\nabla f_{\nu },\nabla f_{\nu })=2g(D_V\nabla f_{\nu },\nabla f_{\nu })=2\mathrm{Hess}{f}_{\nu }(\nabla f_{\nu },V)=2\mu g(\nabla f_{\nu },V)=0$$

for all $V\in TF$, we see that $|\nabla f_{\nu }|$ is constant on each $\exp \left(TF\right)$.

If $|\nabla f_1|=|\nabla f_2|=\cdots =|\nabla f_m|(:=\nabla f)$ and $f_1^{″}=f_2^{″}=\cdots =f_m^{″}(:=f^{″})$, then the shape operator $S_{N_{\nu }}=\frac{f^{″}}{\left|\nabla f\right|}\mathrm{Id}$ is totally umblic (16) for all $\nu$ on $\exp \left(T_{{\gamma }_{\nu }\left(t\right)}F\right)$.

From the Equation (15) ${\nabla }_i{\nabla }_j{f}_{\nu }=\mu g_{ij}$, it follows that

$${\nabla }_k{\nabla }_i{\nabla }_j{f}_{\nu }={\nabla }_k\mu g_{ij}$$

and

$${\nabla }_i{\nabla }_k{\nabla }_j{f}_{\nu }+R_{kil}^j{\nabla }^l{f}_{\nu }={\nabla }_k\mu g_{ij}.$$

Take the trace in k and j on $\tilde{F}:=\mathrm{Span}\{TF,N_{\nu }\}$, then we have

$${\nabla }_i{\mathrm{\Delta }}^{\tilde{F}}{f}_{\nu }+R_{il}^{\tilde{F}}{\nabla }^l{f}_{\nu }={\nabla }_i\mu .$$

The trace ${\mathrm{\Delta }}^{\tilde{F}}{f}_{\nu }=(n+1)\mu$ of (15) on $\tilde{F}$ gives

$$(n+1){\nabla }_i\mu +R_{il}^{\tilde{F}}{\nabla }^l{f}_{\nu }={\nabla }_i\mu .$$

So we obtain

$$\begin{array}{c}\hfill n{D}_V\mu =-{\mathrm{Ric}}^{\tilde{F}}(\nabla f_{\nu },V)\end{array}$$

(19)

for all $V\in \tilde{F}$. Since

$$\mu =\mathrm{Hess}{f}_{\nu }(N_{\nu },N_{\nu })=-g(D_{N_{\nu }}\left(\right|\nabla f_{\nu }|N_{\nu }),N_{\nu })=-N_{\nu }\left(\right|\nabla f_{\nu }\left|\right),$$

(20)

We get

$$\begin{array}{c}\hfill \begin{array}{ccc}|\nabla f_{\nu }|{\mathrm{Ric}}^{\tilde{F}}(N_{\nu },N_{\nu })\stackrel{\left(19\right)}{=}n{D}_{N_{\nu }}\mu & \stackrel{\left(18\right)}{=}& N_{\nu }\left(\right|\nabla f_{\nu }\left|\right){\theta }^F+\left|\nabla f_{\nu }\right|D_{N_{\nu }}{\theta }^F\\ & \stackrel{\left(20\right)}{=}& -\mu {\theta }^F+\left|\nabla f_{\nu }\right|D_{N_{\nu }}{\theta }^F\\ & \stackrel{\left(18\right)}{=}& -|\nabla f_{\nu }|\frac{{\left({\theta }^F\right)}^2}{n}-\left|\nabla f_{\nu }\right|D_{{\gamma }^{\prime }}{\theta }^F\\ & =& |\nabla f_{\nu }|(-\frac{{\left({\theta }^F\right)}^2}{n}-{\left({\theta }^F\right)}^{\prime }).\end{array}\end{array}$$

(21)

Hence we obtain

$$\begin{array}{c}\hfill {\left({\theta }^F\right)}^{\prime }\left(t\right)+\frac{{\left({\theta }^F\right)}^2\left(t\right)}{n}+{\mathrm{Ric}}^{\tilde{F}}(N_{\nu }\left(t\right),N_{\nu }\left(t\right))=0.\end{array}$$

(22)

The shear tensor $\sigma \left(t\right)$ along a geodesic ${\gamma }_{\nu }\left(t\right)$ is zero for all $\nu$. Then we get a warped product $I{\times }_{f^{\prime }\left(t\right)}F$ for $t\in I=(-\infty ,\infty )$ by the same arguments in Theorem 1 for each ${\gamma }_{\nu }\left(t\right)$. The tangent space of the base $\exp \left(TB\right)$ at all ${\gamma }_{\nu }\left(t\right)$ is $\mathrm{Span}{\left\{{\gamma }_{\nu }^{\prime }\left(t\right)\right\}}_{i=1}^m$. So we get Theorem 2.  □

4. Remark

Here we show that the change of the sign of $N=\frac{\nabla f}{\left|\nabla f\right|}$ gives the same Raychaudhuri equation. For a smooth function $f:M\to \mathbb{R}$ and some $s\in \mathbb{R}$, let $H_s=\{x\in M|f\left(x\right)=s\}$ be a regular hypersurface with a unit normal vector field $N=\frac{\nabla f}{\left|\nabla f\right|}$ on $H_s$. We have

$$\mu =\mathrm{Hess}f(N,N)=g(D_N\left(\right|\nabla f\left|N\right),N)=N\left(\right|\nabla f\left|\right).$$

(23)

The second fundamental form of $H_s$ is given by

$$\alpha (V,W)=-g(D_{V}N,W)=g(S_{N}V,W)=-\frac{1}{\left|\nabla f\right|}g(D_V\nabla f,W)$$

$$\begin{gathered} =-\frac{1}{\left|\nabla f\right|}\mathrm{Hess}f(V,W)=-\frac{\mu }{\left|\nabla f\right|}g(V,W) \\ . \end{gathered}$$

The trace of the second fundamental form is the mean curvature

$$\theta =-n\frac{\mu }{\left|\nabla f\right|}.$$

(24)

Since

$$Vg(\nabla f,\nabla f)=2g(D_V\nabla f,\nabla f)=2\mathrm{Hess}f(\nabla f,V)=2\mu g(\nabla f,V)=0$$

for all $V\in {\left(\nabla f\right)}^{\perp }$, we see that $\left|\nabla f\right|$ is constant on $H_s$. Differentiation of (24) gives

$$\begin{array}{c}\hfill \begin{array}{ccc}-\left|\nabla f\right|\mathrm{Ric}(N,N)\stackrel{\left(6\right)}{=}n{D}_N\mu & =& -\left(N\right(\left|\nabla f\right|)\theta +\left|\nabla f\right|{\theta }^{\prime })\\ & =& -\mu \theta +\left|\nabla f\right|{\theta }^{\prime })\\ & =& -(-|\nabla f|\frac{{\theta }^2}{n}+|\nabla f|{\theta }^{\prime })\\ & =& -\left|\nabla f\right|(-\frac{{\theta }^2}{n}+{\theta }^{\prime }).\end{array}\end{array}$$

(25)

Hence we obtain

$${(-\theta )}^{\prime }\left(t\right)+\frac{{(-\theta )}^2\left(t\right)}{n}+\mathrm{Ric}({\gamma }^{\prime }\left(t\right),{\gamma }^{\prime }\left(t\right))=0.$$