# Conformal Metrics to a Product or Doubly Warped Product onS2×S2 and the Hopf Conjecture

Authors: Thierno Seck; Athoumane Niang
Journal: Journal of Mathematics
 (2022)

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## Abstract

Hopf’s well-known conjecture states that there exists no metric of positive sectional curvature in the product manifoldS2×S2. In this paper, we show that the Hopf conjecture is true for conformal metrics to the product metric or doubly warped products on S2×S2.


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## Body

## 1. Introduction

A classical topic in Riemannian geometry is to study manifolds with positive sectional curvature. The most rigid classical curvature concept (consequently containing most information), which then gives rise to the other ones, is sectional curvature (see [1]). Let M,gM be a Riemannian manifold and let TpM denote the tangent vector space of M at p∈M. The sectional curvature is the most natural generalization to higher dimensions of the Gaussian curvature of a surface, given that it controls the behavior of geodesics (see [2]). We will say that M,gM has positive sectional curvature if for every point p∈M, the sectional curvature KΠ of every 2-plane Π⊂TpM is positive. An example of such manifolds are the n-dimensional spheres of Sn with the metric induced by ℝn+1. For 4-dimensional Riemannian manifolds, very few topological obstructions to positive sectional curvature are known, and many conjectures about this subject remain open, as, for example, the Hopf conjecture on S2×S2, which is one of the oldest conjectures in global Riemannian geometry (see [3]).Conjecture (H. Hopf):S2×S2 does not admit a Riemannian metric of positive sectional curvature [4]By Synge’s lemma (see [5]), one knows that on ℝℙ2×ℝℙ2 there is no Riemannian metric of positive sectional curvature. The only known examples of positively curved compact connected 4-manifolds are S4, ℝℙ4, and ℂℙ2 (see [4]).There have been various attempts to prove or disprove Hopf’s conjecture; one was to start with the standard product metric (which is nonnegatively curved) and try to deform it to a positively curved metric (see [1, 4, 6–8] for more details). Although one can make the curvature of mixed planes positive, there appear new planes of zero or even negative curvature; hence, this method seems not to answer the question. From work of Bourguignon [7], it is known that in the neighborhood of the product metric of S2×S2, there is no metric of positive curvature. In [9], Weinstein shows that a metric of positive sectional curvature on S2×S2 cannot be induced by an immersion in ℝ6. Now, it follows from a result of Hopf that an embedding of such a Riemannian structure in ℝ5 is not possible as then the manifold has to be a sphere.Similarly, using the metricg0, the Riemannian metric is induced by ℝ6=ℝ3×ℝ3 on M=S2×S2; the authors of [10] gave a positive answer to the conjecture in the class of conformal metrics g=expfg0 where f is a smooth function on S2×S2.In [3, 11], Oscar Perdomo suggested an idea that may be useful for trying to prove Hopf’s conjecture. This idea relies on the main two theorems proven in [3]. The first one is a small variation of a theorem by Nash that states that any Riemannian manifold can be isometrically embedded in some ℝn. The second one states that if M=S1×S2, in particular k1+k2=n (Si⊂ℝki be two compact Riemannian surfaces with the metric induced by the euclidean spaces), then for any smooth function F:M⟶ℝ, the manifold x,Fx∈ℝn+1:x∈M with the metric induced by ℝn+1 does not have positive sectional curvature. So, he makes the following remark:Remark 1.
If for any smooth mapF:S2×S2⟶ℝkthe manifoldx,Fx∈ℝ6+k:x∈S2×S2with the metric induced byℝ6+kdoes not have positive sectional curvature, then Hopf’s conjecture would be true.The casek=1 is solved in [3].More generally, one might ask whether any nontrivial product manifold admits a metric with positive curvature. It is easy to see that ifM and N admit nonnegative curvature metrics, then M×N with the product metric has nonnegative curvature. This generalized conjecture is open. Thus, in [12], Amann and Kennard have proven that no product manifold N×N can carry a metric of positive sectional curvature admitting a certain degree of torus symmetry. Also, Hsiang and Kleiner proved that (see [13, 14]) S2×S2 does not admit a Riemannian metric with positive sectional curvature and an isometric circle action.Singly warped products were first introduced by Bishop and O’Neil in their attempt to construct a class of Riemannian manifolds with negative curvature [15]. Warped products have many applications, in general relativity [16], in the studies related to solutions of Einstein’s equations (see [17] and references therein).The study of relativity theory demands a wider class of manifold, and the idea of doubly twisted product as a generalization of warped product was introduced and studied by many authors (see [18] and references therein).LetM1,gM1 and M2,gM2 be two Riemannian manifolds of dimensions m1 and m2, respectively, and let π1:M1×M2⟶M1 and π2:M1×M2⟶M2 be the canonical projections. Also, let f1:M1×M2⟶ℝ+ and f2:M1×M2⟶ℝ+ be smooth functions. Then, the doubly twisted product ([18, 19]) of M1,gM1 and M2,gM2 with twisting functions f1 and f2 is defined to be the product manifold M=M1×M2 with metric tensor gM=f12gM1⊕f22gM2 given by(1)gM=f22π1∗gM1+f12π2∗gM2.We denote this Riemannian manifoldM,gM by M1×f1,f2M2. In particular, if f2=1, then M1×f1,f2M2 is called the twisted product of M1,gM1 and M2,gM2 with twisting function f1. Moreover, if f1 only depends on the points of M1, then M1×f1M2 is called the warped product of M1,gM1 and M2,gM2 with warping function f1. Also, as a generalization of the warped product of the two Riemannian manifolds M1,gM1 and M2,gM2, M1×f1,f2M2 is called the doubly warped product of Riemannian manifolds M1,gM1 and M2,gM1 with the warping functions f1 and f2 if f1 and f2 and only depend on the points of M1 and M2, respectively (see [18]).The Hopf conjecture is true for the singly warped product manifold (see [20]). In this paper, we study the Hopf conjecture in the class of doubly warped products and conformal metrics to the product metrics on S2×S2. Let S2×f1,f2S2 be a doubly twisted product manifolds with twisting functions f1 and f2. Next, we are going to prove the following results.Theorem 1.
A doubly warped product manifoldS2×f1,f2S2does not have positive sectional curvature.Theorem 2.
A conformal metricgS2×S2to a product metricgS2⊕gS2onS2×S2does not have positive sectional curvature.The paper is organized as follows: After preliminaries, Section2 contains the basic notions of doubly twisted products, and the proofs of the two theorems above will be given in Section 3.


## 2. Preliminaries

LetM1,gM1 and M2,gM2 be two Riemannian manifolds with Levi-Civita connections ∇M1 and ∇M2, respectively. Denote by ∇ the Levi-Civita connection and grad the gradient of the doubly twisted products M1×f1,f2M2 of M1,gM1 and M2,gM2 with twisting functions f1 and f2. Let σ1=logf1 and σ2=logf2. For a Z vector field of M1, the lift of Z to M1×f1,f2M2 is the vector field Z˜ whose value at each p,q is the lift of Zp to p,q. Thus, the lift of Z is the unique vector field on M1×f1,f2M2, that is, π1-related to Z and π2-related to the zero vector field on M2 (see [19]). We denote the set of lifts of vector fields on M1 and M2 to M1×M2 by LM1 and LM2, respectively. The following two propositions are results from [18] (Proposition 1 and Proposition 2).Proposition 1 (see [18]).
IfX,Y∈LM1andV∈LM2, then(2)∇XY=∇XM1Y+Xσ2Y+Yσ2X−gX,Ygradσ2,(3)∇XV=Vσ2X+Xσ1V.In a doubly twisted productM1×f1,f2M2, we can interchange the roles of the base manifold M1,gM1 and the fiber manifold M2,gM2. In the above proposition, we have the expression of the Levi-Civita connection for vector fields X,Y∈LM1 in (2). Similarly, by making corresponding changes, one can easily see the expression of Levi-Civita connection for vector fields V,W∈LM2 as in (2). Obviously, the analogue expressions hold for vector fields in LM2 by making corresponding changes [18].We will denote byRM1, RM2, and R the curvature tensors of M1,gM1, M2,gM2, and M1×f1,f2M2 , respectively.ForX,Y∈LM1, define hM1σ1X,Y=XYσ1−∇XM1Yσ1. First note that if X,Y∈LM1 and V,∈LM2, we have XVσ1=VXσ2 and by [18] the Hessian form hσ1 of σ1 on M1×f1,f2M2 satisfies(4)hσ1X,V=XVσ1−Xσ1Vσ2−Xσ1Vσ1,hσ1X,V=hBσ1X,Y−Xσ2Yσ1−Xσ1Yσ2+gX,Yggradσ1,gradσ2.Then, we have the following relations.Proposition 2 (see [18]).
LetX,Y,Z∈LM1andV∈LM2. Then,(5)RX,YZ=RM1X,YZ−gY,ZHσ2X+Xσ2gradσ2+gX,ZHσ2Y+Yσ2gradσ2+hM1σ2X,Z−Xσ2Zσ2Y−hM1σ2Y,Z−Yσ2Zσ2X,RX,YV=hσ2X,VY−hσ2Y,VX+Vσ2Xσ2Y−Vσ2Yσ2X,RX,VY=hM1σ1X,Y+Xσ1Yσ1−Xσ2Yσ1−Xσ1Yσ2V+Yσ1Vσ2−VYσ2X+Vσ2gradσ2+Hσ2VgMX,Y,where Hσ2 is the Hessian tensor of σ2 on M1×f1,f2M2, i.e., hσ2X,Y=gMHσ2X,Y.By using Propositions1 and 2, we get the following lemma for later use.Lemma 1.
LetM1×f1,f2M2be a doubly twisted product manifold with metric tensorgMdenoted by⋅,⋅. ForX∈LM1,V∈LM2such thatXandVare orthonormal on some neighborhoodUof a point for the product metricgB⊕gF. Then, at any pointpofU, the sectional curvature of the mixed 2-plane spanned byXpandVpis given by(6)KX,Vp=1f12f22XXσ1−∇XM1Xσ1+Xσ12−2Xσ1Xσ2V2+1f12f22VVσ2−∇VM2Vσ2+Vσ22−2Vσ1Vσ2X2+1f12f22gradσ1,gradσ2V2X2atp.Proof.
By definition,(7)KX,Vp=RX,VX,VX,XV,V−X,V2.
By using Proposition2, we have that(8)RX,VX,V=hM1σ1X,X+Xσ12−2Xσ1Xσ2V2+Vσ2gradσ2,V+Hσ2V,VX2,=XXσ1−∇XM1Xσ1+Xσ12−2Xσ1Xσ2V2+Vσ2gradσ2,V+Hσ2V,VX2.
Since(9)Vσ2=gradσ2,Vand by using (4), we have(10)H2σV,V=hσ2V,V=VVσ2−∇VM2Vσ2−2Vσ2Vσ1+gradσ1,gradσ2V2.
Then, we have(11)RX,VX,V=XXσ1−∇XM1Xσ1+Xσ12−2Xσ1Xσ2V2+VVσ2−∇VM2Vσ2+Vσ22−2Vσ1Vσ2X2+gradσ1,gradσ2V2X2.
Moreover, we have(12)X,X=f22gM1X,X=f22,V,V=f12gM2V,V=f12,X,V=0.
Thus, one obtains the result.


## 3. The Main Results

In this section, we give the proof of Theorem1 and Theorem 2.


### 3.1. Proof of Theorem1

Proof.
We will assume that the sectional curvature of the manifoldM1×f1,f2M2 is everywhere positive. We consider the case where M1 and M2 are the standard unit sphere S2 and the doubly warped product manifolds S2×f1,f2S2 with the warping functions f1 and f2. We keep the hypotheses and notations of Lemma 1. Since S2×f1,f2S2 is a doubly warped product manifolds, then(13)Xσ2=Vσ1=0,gradσ1,gradσ2=gradσ2σ1=0.
By using the above relations with (6) and the fact that the mixed sectional curvature KX,V is positive, we have(14)XXσ1−∇XM1Xσ1+Xσ12V2+VVσ2−∇VM2Vσ2+Vσ22X2>0.
Sinceσ1 (respectively, σ2) depends only of the points of the left sphere M1 of M (respectively, depends only of the points of the right sphere M2 of M), we can define a smooth function σ˜1 in M1 (respectively, a smooth function σ˜2 in M2) such that σ1=σ˜1∘π1 and σ2=σ˜2∘π2. So, we have the following remarks:(1)
If a pointp∈M1 is a maximum for σ˜1, then for any y∈M2, p,y is a maximum of σ1(2)
If a pointq∈M2 is a maximum for σ˜2, then for any z∈M1, z,q is a maximum of σ2(3)
Consequently, ifp is a maximum for σ˜1 and q a maximum for σ2˜, then p,q is a maximum for σ1 and σ2
Then, at the critical pointx=p,q of σ1 and σ2, we have(15)Xσ12=Vσ22=∇XM1Xσ1=∇VM2Vσ2=0atx=p,q.
Since the critical pointx=p,q of σ1 and σ2 is a maximum, then we have that(16)XXσ1≤0atx=p,q,(17)VVσ2≤0atx=p,q.
Using (15)–(17) in (14), we get a contradiction.


### 3.2. Proof of Theorem2

Proof.
We takeM1=S2 and M2=S2. If f1=f2=f, M1×f1,f2M2 is a doubly twisted product with the twisting function f and σ=logf. Then, the tensor metric gM=f12π1∗gM1+f12π2∗gM2=f2π∗1gM1+π2∗gM2=f2gM1⊕gM2 is a conformal metric to the product metric on M1×M2.
Now, forX∈LM1,V∈LM2 such that X and V are orthonormal with respect to the product metric gM1⊕gM2 on some neighborhood U of a point, then at any point p of U the sectional curvature of the mixed 2-plane spanned by X and V is given by(18)KX,Vp=1f4XXσ−∇XM1Xσ−Xσ2V2+1f4VVσ−∇VM2Vσ−Vσ2X2+1f4gradσ2V2X2atp.
Finally, SinceS2×f,fS2 is a compact manifold, the smooth function σ=:S2×S2⟶ℝ has a maximum q=q1,q2. At the critical point q of σ, we have Xσ2=gradσ2=∇XM1Xσ=∇VM2Vσ=0. And the sectional curvature of the mixed 2-plane spanned by Xq and Vq is given by(19)KX,V=1f4XXσV2+VVσX2.
Since the critical pointq of σ is a maximum then, we have that(20)hσX,X=HσX,X=XXσ≤0atq,hσV,V=HσV,V=VVσ≤0atq,where Hσ denotes the Hessian of σ with respect to the metric f2gM1⊕gM2. In this case, the sectional curvature KX,V at q is not positive [21, 22].


## 3.1. Proof of Theorem1

Proof.
We will assume that the sectional curvature of the manifoldM1×f1,f2M2 is everywhere positive. We consider the case where M1 and M2 are the standard unit sphere S2 and the doubly warped product manifolds S2×f1,f2S2 with the warping functions f1 and f2. We keep the hypotheses and notations of Lemma 1. Since S2×f1,f2S2 is a doubly warped product manifolds, then(13)Xσ2=Vσ1=0,gradσ1,gradσ2=gradσ2σ1=0.
By using the above relations with (6) and the fact that the mixed sectional curvature KX,V is positive, we have(14)XXσ1−∇XM1Xσ1+Xσ12V2+VVσ2−∇VM2Vσ2+Vσ22X2>0.
Sinceσ1 (respectively, σ2) depends only of the points of the left sphere M1 of M (respectively, depends only of the points of the right sphere M2 of M), we can define a smooth function σ˜1 in M1 (respectively, a smooth function σ˜2 in M2) such that σ1=σ˜1∘π1 and σ2=σ˜2∘π2. So, we have the following remarks:(1)
If a pointp∈M1 is a maximum for σ˜1, then for any y∈M2, p,y is a maximum of σ1(2)
If a pointq∈M2 is a maximum for σ˜2, then for any z∈M1, z,q is a maximum of σ2(3)
Consequently, ifp is a maximum for σ˜1 and q a maximum for σ2˜, then p,q is a maximum for σ1 and σ2
Then, at the critical pointx=p,q of σ1 and σ2, we have(15)Xσ12=Vσ22=∇XM1Xσ1=∇VM2Vσ2=0atx=p,q.
Since the critical pointx=p,q of σ1 and σ2 is a maximum, then we have that(16)XXσ1≤0atx=p,q,(17)VVσ2≤0atx=p,q.
Using (15)–(17) in (14), we get a contradiction.


## 3.2. Proof of Theorem2

Proof.
We takeM1=S2 and M2=S2. If f1=f2=f, M1×f1,f2M2 is a doubly twisted product with the twisting function f and σ=logf. Then, the tensor metric gM=f12π1∗gM1+f12π2∗gM2=f2π∗1gM1+π2∗gM2=f2gM1⊕gM2 is a conformal metric to the product metric on M1×M2.
Now, forX∈LM1,V∈LM2 such that X and V are orthonormal with respect to the product metric gM1⊕gM2 on some neighborhood U of a point, then at any point p of U the sectional curvature of the mixed 2-plane spanned by X and V is given by(18)KX,Vp=1f4XXσ−∇XM1Xσ−Xσ2V2+1f4VVσ−∇VM2Vσ−Vσ2X2+1f4gradσ2V2X2atp.
Finally, SinceS2×f,fS2 is a compact manifold, the smooth function σ=:S2×S2⟶ℝ has a maximum q=q1,q2. At the critical point q of σ, we have Xσ2=gradσ2=∇XM1Xσ=∇VM2Vσ=0. And the sectional curvature of the mixed 2-plane spanned by Xq and Vq is given by(19)KX,V=1f4XXσV2+VVσX2.
Since the critical pointq of σ is a maximum then, we have that(20)hσX,X=HσX,X=XXσ≤0atq,hσV,V=HσV,V=VVσ≤0atq,where Hσ denotes the Hessian of σ with respect to the metric f2gM1⊕gM2. In this case, the sectional curvature KX,V at q is not positive [21, 22].



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