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

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


Introduction
Let M be an (n + 1)-dimensional complete and simply connected Riemannian manifold with metric g.Given a gradient vector field ∇ f for a smooth function f : M −→ R, if a one-parameter family of diffeomorphisms generated by the integral curves of ∇ f on M satisfies Hess f = (R − ρ)g (1) for the scalar curvature R and some constant ρ, then M is called a gradient steady, expanding and shrinking Yamabe soliton for ρ = 0, ρ < 0 and ρ > 0, respectively.In [1], the warped product structure of a gradient Yamabe soliton is shown.But is well-known that the equation for smooth functions f , µ : M −→ R determines the warped product structure M = (−∞, ∞) × ( f ) F where F = exp((∇ f ) ⊥ ) and f 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 µ = R − ρ is a warped product with the totally umblic fibers p × F by the following shape operator S N (4).
Let H s = {x ∈ M| f (x) = s} be a regular hypersurface of M with a unit normal vector field So we can consider each level hypersurface H t of H s along a unit-speed geodesic γ(t) orthogonal to H s with N γ(s) = −γ (s) and γ(s) ∈ H s by (3).The shape operator of each level hypersurface H t of H s along a geodesic γ is given by for all V, W ∈ (∇ f ) ⊥ .The trace of the shape operator is the mean curvature Differentiation of the Equation (2) ∇ i ∇ j f = µ g ij with the Einstein convention gives Take the trace in k and j, then we have The trace of (2) ∆ f = (n + 1)µ gives for all X ∈ TM as in [4].Here we show that the mean curvature θ = n µ |∇ f | (5) and the Equation (6) determine the warped product structure.Since We get = nD N µ Hence we obtain the Raychaudhuri Equation ( 12) along a geodesic γ(t) with the zero shear tensor σ(t) Thus we can show Theorem 1 by the Jacobi differential equation.

Theorem 1.
Let M be a complete and simply connected Riemannian manifold.If Hess f = µ g for smooth functions f , µ : ) and f is the warping function.
For a gradient Yamabe soliton, we get Thus a gradient Yamabe soliton with a regular hypersurface H s = {x ∈ M| f (x) = s} for some s ∈ R is a warped product with the totally umblic fibers whose scalar curvature are constant by (10) (cf.[1]).Differentiation of the above equation shows that if Ric(N, N) = 0 or D γ (t) R = 0, then we get g (t) = 0. Thus under the assumption Ric(N, N) = 0 or D γ (t) R = 0, if a singular point is allowed, then g(t) = at + b for some a, b ∈ R. Otherwise M is a product manifold.
A vector field v on M is said to be concircular if it satisfies for all X ∈ TM and a non-trivial function µ 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 ∇ f of f is a concircular vector field if and only if Hess f = µ g.Then a gradient Ricci soliton equations Ric + Hess f = λ g with Hess f = µ g becomes Ric = (λ − µ)g.
So M is Einstein.Thus λ − µ is constant under n ≥ 3.In Theorem 5.1 in [3], M is turned out to be Ricci-flat.So we have λ = µ.Therefore we have If a singular point is allowed, then g(t) = at + b for some a, b ∈ R. A Gaussian gradient Ricci soliton with f (x) = λ 2 |x| 2 on R n is an example for it.We show the warped product structure with the base B whose dimension is dimB = 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 F with the warping function f .

The Raychaudhuri and Jacobi Equation
Let M be an (n + 1)-dimensional Riemannian manifold and H s = {x ∈ M| f (x) = s} be a regular hypersurface of M with a unit normal vector field N = − ∇ f |∇ f | for some s ∈ R. Consider each level hypersurface H t of H s along a unit-speed geodesic γ(t) orthogonal to H s with N γ(s) = −γ (s) and γ(s) ∈ H s .The Riemannian curvature tensor of M along γ(t) , the shape operator of H s is denoted by with initial conditions A(s) = Id and A (s) = S N for the identity endomorphism Id of (γ (t)) ⊥ , then A is said to be an H-Jacobi tensor along γ(t).
Put B = A A −1 and R(A, γ )γ = R γ A. Then it follows from [5] that and the shape operator of each level hypersurface H t is Thus we get the mean curvature θ(t) = tr S t of H t .For the adjoint * , the vorticity 1 2 (B − B * ) becomes zero, since a variation tensor field A is a Lagrange tensor (Proposition 1 in [5]).The trace of (11) gives the Raychaudhuri equation for the shear tensor σ = B − θ n Id and the Ricc tensor Ric(γ , γ ).The expansion θ = tr(B) satisfies θ = (det(A)) det(A) as in [5].Differentiation of x = (detA) By ( 12) and (13), we have the Jacobi differential equation

Proofs of Theorems
Proof of Theorem 1.We have the Raychaudhuri equation with the zero shear tensor σ(t) by ( 9) along a geodesic γ(t).The shear tensor σ(t) along a geodesic γ(t) is zero if and only if the shape operator of the hypersurface H s is given by S −γ (s) = c Id for some real constant c together with the isotropic curvature tensor R γ (t) = h(t) Id [6].Now we find the isotropic curvature tensor, that is, h(t).By (10) and f (t) = g(γ , ∇ f ) = |∇ f |, the shape operator of each level hypersurface H t of H s along a geodesic γ is The uniqueness of the first order differential equation shows that a Jacobi tensor is since the shear tensor σ(t) is zero.The line element determines uniquely a Jacobi equation with the suitable initial conditions and vice versa.So we have where dg 2 0 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 dimM = m + n.Let TB and (TB) ⊥ = TF be orthogonal distributions on TM with dimTB = m and dimTF = n.Assume that for smooth functions such that N ν = − ∇ f ν |∇ f ν | are unit normal vectors of exp(TF).Then we have for all X ∈ TB, V ∈ TF and all ν.Hence we get a totally geodesic integrable distribution TB.
Hence we see D γ ν γ ν = 0.So let γ ν (t) be a geodesic with −γ The second fundamental form of exp(TF) with respect to N ν is given by for all V, W ∈ TF.For each f ν whose gradient vector is ∇ f ν , we see The mean curvature is given by the trace of ( 16) Take the trace in k and j on F := Span{TF, N ν }, then we have The trace ∆ F f ν = (n + 1)µ of (15) on F gives So we obtain for all V ∈ F. Since we get Hence we obtain The shear tensor σ(t) along a geodesic γ ν (t) is zero for all ν.Then we get a warped product I × f (t) F for t ∈ I = (−∞, ∞) by the same arguments in Theorem 1 for each γ ν (t).The tangent space of the base exp(TB) at all γ ν (t) is Span{γ ν (t)} m i=1 .So we get Theorem 2.

Remark
Here we show that the change of the sign of N = ∇ f |∇ f | gives the same Raychaudhuri equation.For a smooth function f : M → R and some s ∈ R, let H s = {x ∈ M| f (x) = s} be a regular hypersurface with a unit normal vector field N = ∇ f |∇ f | on H s .We have The second fundamental form of H s is given by The trace of the second fundamental form is the mean curvature = Hence we obtain (−θ) (t) + (−θ) 2 (t) n + Ric(γ (t), γ (t)) = 0.
we see that |∇ f | is constant on H s .Differentiation of (24) gives −|∇ f |Ric(N, N)