aa r X i v : . [ m a t h . DG ] S e p Riemannian submersions from compact four manifolds
Xiaoyang Chen ∗ Abstract
We show that under certain conditions, a nontrivial Riemannian submersion from posi-tively curved four manifolds does not exist. This gives a partial answer to a conjecture dueto Fred Wilhelm. We also prove a rigidity theorem for Riemannian submersions with totallygeodesic fibers from compact four-dimensional Einstein manifolds.
A smooth map π : ( M, g ) → ( N, h ) is a Riemannian submersion if π ∗ is surjective andsatisfies the following property: g p ( v, w ) = h π ( p ) ( π ∗ v, π ∗ w )for any v, w that are tangent vectors in T M p and perpendicular to the kernel of π ∗ .A fundamental problem in Riemannian geometry is to study the interaction between curva-ture and topology. A lot of important work has been done in this direction. In this paper westudy a similar problem for Riemannian submersions: Problem:
Explore the structure of π under additional curvature assumptions of ( M, g ) . When (
M, g ) has constant sectional curvature, we have the following classification results([8], [21], [22]).
Theorem 1.1.
Let π : ( M m , g ) → ( N, h ) be a nontrivial Riemannian submersion (i.e.
M, g ) has positive sectional curvature. ∗ The author is supported in part by NSF DMS-1209387.
1. (
M, g ) is an Einstein manifold.When (
M, g ) has positive sectional curvature, we have the following important conjecturedue to Fred Wilhelm (although never published anywhere).
Conjecture 1
Let π : ( M, g ) → ( N, h ) be a nontrivial Riemannian submersion, where ( M, g ) is a compact Riemannian manifold with positive sectional curvature. Then dim ( F ) < dim ( N ) ,where F is the fiber of π . By Frankel’s theorem [7], it is not hard to see that Conjecture 1 is true if at least two fibersof π are totally geodesic. In fact, since any two fibers do not intersect with each other, Frankel’stheorem implies that 2 dim ( F ) < dim ( M ). Hence dim ( F ) < dim ( N ). If all fibers of π aretotally geodesic, we have the following much stronger result: Proposition 1.2.
Let π : ( M, g ) → ( N, h ) be a nontrivial Riemannian submersion such that allfibers of π are totally geodesic, where ( M, g ) is a compact Riemannian manifold with positivesectional curvature. Then dim ( F ) < ρ ( dim ( N )) + 1 , where F is any fiber of π and ρ ( n ) is themaximal number of linearly independent vector fields on S n − . Notice that we always have ρ ( dim ( N )) + 1 ≤ dim ( N ) − dim ( N ) and equality holdsif and only dim ( N ) = 2 , dim ( M ) = 4, Conjecture 1 is equivalent to the following conjecture. Conjecture 2
There is no nontrivial Riemannian submersion from any compact four manifold ( M , g ) with positive sectional curvature. In fact, suppose there exists such a Riemannian submersion π : ( M , g ) → ( N, h ). Then Con-jecture 1 would imply dim ( N ) = 3. Hence the Euler number of M is zero. On the other hand,since ( M , g ) has positive sectional curvature, H ( M , R ) = 0 by Bochner’s vanishing theorem([15], page 208). By Poincar´ e duality, the Euler number of M is positive. Contradiction.Let π : ( M, g ) → ( N, h ) be a Riemannian submersion. We say that a function f defined on M is basic if f is constant along each fiber. A vector field X on M is basic if it is horizontal andis π -related to a vector field on N . In other words, X is the horizontal lift of some vector field on N . Let H be the mean curvature vector field of the fibers and A be the O’Neill tensor of π . Wedenote by k A k the norm of A , i.e., k A k = P i,j k A X i X j k , where { X i } is a local orthonormalbasis of the horizontal distribution of π . The next theorem gives a partial answer to Conjecture2. Theorem 1.3.
There is no nontrivial Riemannian submersion from any compact four manifoldwith positive sectional curvature such that either k A k or H is basic. We emphasize that in Conjecture 1 the assumption that (
M, g ) has positive sectional curva-ture can not be replaced by (
M, g ) has positive sectional curvature almost everywhere, namely,(
M, g ) has nonnegative sectional curvature everywhere and has positive sectional curvature onan open and dense subset of M . Indeed, Let g be the metric on S × S constructed by B.Wilking which has positive sectional curvature almost everywhere [23]. Then by a theorem2f K. Tapp [18], g can be extended to a nonnegatively curved metric ˜ g on S × R such that( S × S , g ) becomes the distance sphere of radius 1 about the soul. By Proposition 5.1, weget a Riemannian submersion π : ( S × S , g ) → ( S , h ), where h is the induced metric on thesoul S from ˜ g . This example shows that in Conjecture 1 the assumption that ( M, g ) has pos-itive sectional curvature can not be replaced by (
M, g ) has positive sectional curvature almost everywhere.Riemannian submersions are also important in the study of compact Einstein manifolds, forexample, see [3]. Our next theorem gives a complete classification of Riemannian submersionswith totally geodesic fibers from compact four-dimensional Einstein manifolds.
Theorem 1.4.
Suppose π : ( M , g ) → ( N, h ) is a Riemannian submersion, where ( M , g ) isa compact four-dimensional Einstein manifold. If all fibers of π are totaly geodesic and havedimension , then locally π is the projection of a metric product B ( c ) × B ( c ) onto one of thefactors, where B ( c ) is a two-dimensional compact manifold with constant curvature c . If the dimension of the fibers of π is 1 or 3 (all fibers are not necessarily totally geodesic),then the Euler number of M is zero. By a theorem of Berger [2, 13], ( M , g ) must be flat. Henceby a theorem of Walschap [21], locally π is the projection of a metric product B ( c ) × B ( c )onto one of the factors. Acknowledgment
This paper is a part of my Ph.D thesis at University of Notre Dame [5]. Theauthor would like to express gratitude to his advisor, Professor Karsten Grove, for many helpfuldiscussions. He also thanks Professor Anton Petrunin for discussing the proof of Theorem 3.1.The author benefits a lot from his ”Exercises in orthodox geometry” [16].
In this section we recall some definitions and facts on Riemannian submersions which willbe used in this paper. We refer to [14] for more details.Let π : ( M, g ) → ( N, h ) be a Riemannian submersion. Then π induces an orthogonal splitting T M = H ⊕ V , where V is tangent to the fibers and H is the orthogonal complement of V . Wewrite Z = Z h + Z v for the corresponding decomposition of Z ∈ T M . The O’Neill tensor A isgiven by A X Y = ( ∇ X Y ) v = 12 ([ X, Y ]) v , where X, Y ∈ H and are π -related to some vector field on N , respectively.Fix X ∈ H , define A ∗ X by A ∗ X : V → H V
7→ − ( ∇ X V ) h . Then A ∗ X is the dual of A X .Define the mean curvature vector field H of π by H = X i ( ∇ V i V i ) h , { V i } ki =1 is any orthonormal basis of V and k = dim V .Define the mean curvature form ω of π by ω ( Z ) = g ( H, Z ) , where Z ∈ T M . It is clear that i V ω = ω ( V ) = 0 for any V ∈ V .We say that a function f defined on M is basic if f is constant along each fiber. A vectorfield X on M is basic if it is horizontal and is π -related to a vector field on N . In other words, X is the horizontal lift of some vector field on N . A differential form α on M is called to bebasic if and only i V α = 0 and L V α = 0 for any V ∈ V , where L V α is the Lie derivative of α .The set of basic forms of M , denoted by Ω b ( M ), constitutes a subcomplex d : Ω rb ( M ) → Ω r +1 b ( M )of the De Rham complex Ω( M ). The basic cohomology of M , denoted by H ∗ b ( M ), is defined tobe the cohomology of (Ω b ( M ) , d ). Proposition 2.1.
The inclusion map i : Ω b ( M ) → Ω( M ) induces an injective map H b ( M ) → H DR ( M ) . Proof.
See pages 33 −
34, Proposition 4 . We first give a proof of Proposition 1.2.
Proof.
Fix p ∈ M and choose X p to be any point in the unit sphere of H p . Extend X p to be aunit basic vector field X . Since all fibers of π are totally geodesic, by O’Neill’s formula ([14]), K ( X, V ) = k A ∗ X V k for any unit vertical vector field V . Since K ( X, V ) > A ∗ X V = 0 for any V = 0. Let v , v , · · · v k be any orthonormal basis of V p , where k = dim ( F p ) and F p is the fiber passing through p . Then A ∗ X ( v ) , A ∗ X ( v ) , · · · A ∗ X ( v k ) are linearlyindependent and are perpendicular to X p . Since X p could be any point in the unit sphere of H p , then we get k linearly independent vector fields on the unit sphere of H p . By the definitionof ρ ( dimN ), we see that dim ( F p ) = k ≤ ρ ( dim ( N )) < ρ ( dim ( N )) + 1. Remark 1.
It would be very interesting to know whether one can replace dim ( F ) < dim ( N ) by dim ( F ) < ρ ( dim ( N )) + 1 in Conjecture . It would be the Riemannian analogue of Toponogov’sConjecture (page in [17]) and would imply that dim ( N ) must be even (In fact, if dim ( N ) is odd, then ρ ( dim ( N )) = 0 . Hence dim ( F ) < ρ ( dim ( N )) + 1 implies dim ( F ) = 0 and hence π is trivial, contradiction). In particular, there would be no Riemannian submersion with one-dimensional fibers from even-dimensional manifolds with positive sectional curvature. Let ( M m , g ) be an m -dimensional compact manifold with positive sectional curvature, m ≥ N , h ) be a 2-dimensional compact Riemannian manifold. Now we are going to prove thefollowing theorem which implies Theorem 1.3. 4 heorem 3.1. There is no Riemannian submersion π : ( M m , g ) → ( N , h ) such that1. the Euler numbers of the fibers are nonzero and2. either k A k or H is basic. Remark 2.
If Conjecture is true, then there would be no Riemannian submersion π : ( M m , g ) → ( N , h ) , where ( M m , g ) has positive sectional curvature and m ≥ . Before we prove Theorem 3.1, we firstly show how to derive Theorem 1.3. The proof is bycontradiction. Suppose there exists a nontrivial Riemannian submersion π : ( M , g ) → ( N, h )such that either k A k or H is basic, where ( M , g ) is a compact four manifold with positivesectional curvature. Since ( M , g ) has positive sectional curvature, H ( M , R ) = 0 by Bochner’svanishing theorem ([15], page 208). By Poincar´ e duality, χ ( M ) = 2 + b ( M ) is positive. By atheorem of Hermann [12], π is a locally trivial fibration. Then χ ( M ) = χ ( N ) χ ( F ), where F isany fiber of π . It follows that dim ( N ) = 2 and χ ( F ) is nonzero (hence all fibers have nonzeroEuler numbers), which is a contradiction by Theorem 3.1.The proof of Theorem 3.1 is again by contradiction. Suppose π : ( M m , g ) → ( N , h ) be aRiemannian submersion satisfying the conditions in Theorem 3.1. By passing to its orienteddouble cover, we can assume that N is oriented. The idea of the proof of Theorem 3.1 is toconstruct a nowhere vanishing vector field (or line field) on some fiber of π , which will implythe Euler numbers of the fibers are zero. Contradiction.Since ( M, g ) has positive sectional curvature, by a theorem of Walschap [21], k A k can notbe identical to zero on M . Hence there exists p ∈ M such that k A k ( p ) = 0.If k A k is basic, then k A k 6 = 0 at any point on F p , where F p is the fiber at p . Let X, Y be anyorthonormal oriented basic vector fields in some open neighborhood of F p . Then k A X Y k = k A k = 0 at any point on F p . Define a map s by s : F p → T F p x A X Y k A X Y k ( x ) . Let
Z, W be another orthonormal oriented basic vector fields. Then Z = aX + bY and W = cX + dY , ad − bc >
0. Then A Z W = ( ad − bc ) A X Y. Hence s does not depend on the choice of X, Y . Then s is a nowhere vanishing vector field on F p . Thus the Euler number of F p is zero. Contradiction.If H is basic, the construction of such nowhere vanishing vector field (or line field) is muchmore complicated. Under the assumption that H is basic, we firstly construct a metric ˆ g on M m such that π : ( M m , ˆ g ) → ( N , h ) is still a Riemannian submersion and all fibers are minimalsubmanifolds with respect to ˆ g . Of course, in general ˆ g can not have positive sectional curvatureeverywhere. However, the crucial point is that there exists some fiber F such that ˆ g has positivesectional curvature at all points on F . Pick any fiber F which is close enough to F . Thenusing the classical variational argument, we construct a continuous codimension one distributionon F . Thus the Euler number of F is zero. Contradiction.5ow we are going to explain the proof of Theorem 3.1 in details. We firstly need the followinglemmas: Lemma 3.2.
Suppose ω is the mean curvature form of a Riemannian submersion from compactRiemannian manifolds. If ω is a basic form, then it is a closed form.Proof. See page 82 in [20] for a proof.
Lemma 3.3.
Suppose π : ( M m , g ) → ( N, h ) is a Riemannian submersion such that H is basic,where ( M m , g ) is a compact Riemannian manifold with positive sectional curvature. Then thereexists a metric ˆ g on M m such that π : ( M m , ˆ g ) → ( N, h ) is still a Riemannian submersion andall fibers are minimal submanifolds with respect to ˆ g . Furthermore, there exists some fiber F such that ˆ g has positive sectional curvature at all points on F .Proof. The idea is to use partial conformal change of metrics along the fibers, see also page 82in [20]. Let ω be the mean curvature form of π . Since H is basic, ω is a basic form. Then ω isclosed by Lemma 3.2. So [ ω ] defines a cohomological class in H b ( M m ). Because ( M m , g ) haspositive sectional curvature, H DR ( M m ) = 0 by Bochner’s vanishing theorem ([15], page 208).By Proposition 2.1, we see that H b ( M m ) = 0. Then there exists a basic function f globallydefined on M m such that ω = df . Define ˆ f = f − max p ∈ M m f ( p ). Then max p ∈ M m ˆ f ( p ) = 0 and ω = d ˆ f . Let λ = e ˆ f and define ˆ g = ( λ k g v ) ⊕ g h , where k = dim ( M m ) − dim ( N ), g v /g h are the vertical / horizontal components of g , respectively.Since the horizontal components of g remains unchanged, π : ( M m , ˆ g ) → ( N, h ) is still aRiemannian submersion. Now the mean curvature form ˆ ω associated to ˆ g is computed to beˆ ω = ω − dlogλ = 0 . Hence all fibers of π are minimal submanifolds with respect to ˆ g .Let φ ( p ) = λ k ( p ) , p ∈ M m . Then ˆ g = ( φg v ) ⊕ g h . Note for any p ∈ M m , 0 < φ ( p ) ≤
1. Moreover, we have max p ∈ M m φ ( p ) = 1. Let p ∈ M m suchthat φ ( p ) = 1 and F be the fiber of π passing through p . Since f is a basic function on M m , φ is also basic. Then φ ≡ F , which will play a crucial role in our argument below. Ofcourse, in general ˆ g can not have positive sectional curvature everywhere. However, by Lemma3.4 below, we see that ˆ g still has positive sectional curvature at all points on F . (The readershould compare it to the following fact: Let ˆ h = e f h be a conformal change of h , where h is aRiemannian metric on M with positive sectional curvature. Then ˆ h still has positive sectionalcurvature at those points where f attains its maximum value.)Indeed, by Lemma 3.4 below, for any basic vector fields X, Y and vertical vector fields
V, W ,we have ˆ K ( X + V, Y + W ) k ( X + V ) ∧ ( Y + W ) k = ˆ R ( X + V, Y + W, Y + W, X + V )= R ( X + V, Y + W, Y + W, X + V ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )6 Q ( ∇ φ, φ, X, Y, V, W ) + [ − g ( W, W ) g ( ∇ V ∇ φ, X )+ g ( V, W ) g ( ∇ W ∇ φ, X ) + g ( V, W ) g ( ∇ V ∇ φ, Y ) − g ( V, V ) g ( ∇ W ∇ φ, Y )] + 12 [ − Hess ( φ )( X, X ) g ( W, W )+2
Hess ( φ )( X, Y ) g ( V, W ) − Hess ( φ )( Y, Y ) g ( V, V )] , where ˆ K ( X + V, Y + W ) is the sectional curvature of the plane spanned by X + V, Y + W withrespect to ˆ g and k ( X + V ) ∧ ( Y + W ) k = ˆ g ( X + V, X + V )ˆ g ( Y + W, Y + W ) − [ˆ g ( X + V, Y + W )] . Moreover, ∇ is the Levi-Civita connection and Hess ( φ ) is the Hessian of φ with respect to g .Also ˆ R/R are the Riemannian curvature tensors with respect to ˆ g/g , respectively. Furthermore, P ( ∇ φ, φ, X, Y, V, W ), Q ( ∇ φ, φ, X, Y, V, W ) are two functions depending on ∇ φ, φ, X, Y, V, W and Q (0 , φ, X, Y, V, W ) ≡ φ ≡ max p ∈ M m φ ( p ) on F , we see that ∇ φ ≡ F . Hence Q ( ∇ φ, φ, X, Y, V, W ) ≡ Q (0 , φ, X, Y, V, W ) ≡ ∇ V ∇ φ ≡ , ∇ W ∇ φ ≡ F . Then at any point on F , we haveˆ R ( X + V, Y + W, Y + W, X + V ) = R ( X + V, Y + W, Y + W, X + V )+ 12 [ − Hess ( φ )( X, X ) g ( W, W ) + 2
Hess ( φ )( X, Y ) g ( V, W ) − Hess ( φ )( Y, Y ) g ( V, V )] . On the other hand, let A = (cid:18) Hess ( φ )( X, X ) Hess ( φ )( X, Y ) Hess ( φ )( X, Y ) Hess ( φ )( Y, Y ) , (cid:19) , B = (cid:18) g ( W, W ) − g ( V, W ) − g ( V, W ) g ( V, V ) (cid:19) . Then ˆ R ( X + V, Y + W, Y + W, X + V ) = R ( X + V, Y + W, Y + W, X + V ) + 12 tr ( − AB ) . Since φ attains its maximum at any point on F , we see that − A is nonnegative definite on F .It is easy to check that B is also nonnegative definite. Hence tr ( − AB ) ≥ − AB isnot nonnegative definite if AB = BA ). Since g has positive sectional curvature everywhere on M m by assumption, then at any point on F , we see thatˆ R ( X + V, Y + W, Y + W, X + V ) ≥ R ( X + V, Y + W, Y + W, X + V ) > . Hence ˆ g still has positive sectional curvature at all points on F . Lemma 3.4.
Let π : ( M m , g ) → ( N, h ) be a Riemannian submersion and g = g v ⊕ g h , where g v /g h are the vertical / horizontal components of g , respectively. Suppose φ is a positive basicfunction defined on M m . Let ˆ g = ( φ g v ) ⊕ g h . Suppose ˆ ∇ / ∇ are the Levi-Civita connectionsand ˆ R/R are the Riemannian curvature tensors with respect to ˆ g/g , respectively. Moreover, let Hess ( φ ) be the Hessian of φ with respect to g . Then for any horizontal vector fields X, Y ( X, Y are not necessarily basic vector fields) and vertical vector fields
V, W , we have ˆ ∇ X Y = ∇ X Y. ∇ V W = ∇ V W − g ( V, W )2 ∇ φ + ( φ − ∇ V W ) h . ˆ ∇ V X = ∇ V X + g ( X, ∇ φ )2 φ V + 1 − φ n X i =1 g ([ X, ε i ] , V ) ε i . ˆ ∇ X V = ∇ X V + g ( X, ∇ φ )2 φ V + 1 − φ n X i =1 g ([ X, ε i ] , V ) ε i , where { ε i } ni =1 is any orthonormal basis of the horizontal distribution with respect to g and n = dim ( N ) .Moreover, if X, Y are basic vector fields and
V, W are vertical vector fields, then ˆ R ( X + V, Y + W, Y + W, X + V ) = R ( X + V, Y + W, Y + W, X + V )+( φ − P ( ∇ φ, φ, X, Y, V, W ) + Q ( ∇ φ, φ, X, Y, V, W )+[ − g ( W, W ) g ( ∇ V ∇ φ, X ) + g ( V, W ) g ( ∇ W ∇ φ, X )+ g ( V, W ) g ( ∇ V ∇ φ, Y ) − g ( V, V ) g ( ∇ W ∇ φ, Y )]+ 12 [ − Hess ( φ )( X, X ) g ( W, W ) + 2
Hess ( φ )( X, Y ) g ( V, W ) − Hess ( φ )( Y, Y ) g ( V, V )] , where P ( ∇ φ, φ, X, Y, V, W ) , Q ( ∇ φ, φ, X, Y, V, W ) are two functions which depend on ∇ φ, φ, X, Y, V, W and Q (0 , φ, X, Y, V, W ) ≡ .Proof. The proof is based on a lengthy computation and the following
Koszul ′ sf ormula :2ˆ g ( ˆ ∇ X Y, Z ) = X (ˆ g ( Y, Z )) + Y (ˆ g ( Z, X )) − Z (ˆ g ( X, Y ))+ˆ g ([ X, Y ] , Z ) − ˆ g ([ Y, Z ] , X ) − ˆ g ([ X, Z ] , Y ) . We just prove the fourth-fifth equalities in Lemma 3.4, others are left to the readers. In thecomputation below, we will use the following trick very often: If we encounter with anythinglike φX , we will rewrite φX = X + ( φ − X . By rewriting it in this way, we can compare newcurvature terms with odd terms. We will also use the fact that φ is a basic function very often.Now let X, Y be horizontal vector fields (not necessarily basic) and { ε i } ni =1 be any orthonor-mal basis of the horizontal distribution with respect to g . By Koszul ′ sf ormula , we see that2ˆ g ( ˆ ∇ X V, ε i ) = X (ˆ g ( V, ε i )) + V (ˆ g ( ε i , X )) − ε i (ˆ g ( X, V ))+ˆ g ([ X, V ] , ε i ) − ˆ g ([ V, ε i ] , X ) − ˆ g ([ X, ε i ] , V )= V (ˆ g ( ε i , X )) + ˆ g ([ X, V ] , ε i ) − ˆ g ([ V, ε i ] , X ) − ˆ g ([ X, ε i ] , V ) . Since ˆ g h = g h and ˆ g v = φg v , we get2 g ( ˆ ∇ X V, ε i ) = V g ( ε i , X )) + g ([ X, V ] , ε i ) − g ([ V, ε i ] , X ) − φg ([ X, ε i ] , V )= V g ( ε i , X )) + g ([ X, V ] , ε i ) − g ([ V, ε i ] , X ) − g ([ X, ε i ] , V ) + (1 − φ ) g ([ X, ε i ] , V ) . Koszul ′ sf ormula again, we see that2 g ( ∇ X V, ε i ) = V g ( ε i , X )) + g ([ X, V ] , ε i ) − g ([ V, ε i ] , X ) − g ([ X, ε i ] , V ) . Then 2 g ( ˆ ∇ X V, ε i ) = 2 g ( ∇ X V, ε i ) + (1 − φ ) g ([ X, ε i ] , V ) . Hence ( ˆ ∇ X V ) h = ( ∇ X V ) h + 1 − φ n X i =1 g ([ X, ε i ] , V ) ε i . Note that − φ P ni =1 g ([ X, ε i ] , V ) ε i does not depend on the choice of { ε i } ni =1 . By the similarargument above, we see that ( ˆ ∇ X V ) v = ( ∇ X V ) v + g ( X, ∇ φ )2 φ V. Hence ˆ ∇ X V = ∇ X V + g ( X, ∇ φ )2 φ V + 1 − φ n X i =1 g ([ X, ε i ] , V ) ε i . The similar argument will also establish the first-third equalities in Lemma 3.4. We just mentionthat in the proof of these equalities, the fact that φ is a basic function and hence V φ = 0 willbe used very often.Now we are going to prove the fifth equality in Lemma 3.4. In the following we alwaysassume that
X, Y are basic vector fields. First of all, we haveˆ R ( X + V, Y + W, Y + W, X + V ) = ˆ R ( X, Y, Y, X ) + ˆ R ( V, W, W, V )+ ˆ R ( X, W, W, X ) + ˆ R ( Y, V, V, Y ) + 2 ˆ R ( X, Y, Y, V ) + 2 ˆ R ( Y, X, X, W )+2 ˆ R ( X, Y, W, V ) + 2 ˆ R ( X, W, Y, V ) + 2 ˆ R ( V, W, W, X ) + 2 ˆ R ( W, V, V, Y ) . Since ˆ g h = g h , ( M m , ˆ g ) → ( N, h ) is still a Riemannian submersion. Then by O’Neill’s formula[14], we have I = ˆ R ( X, Y, Y, X ) = R N ( X, Y, Y, X ) −
34 ˆ g ([ X, Y ] v , [ X, Y ] v )= R N ( X, Y, Y, X ) − g ([ X, Y ] v , [ X, Y ] v ) + 34 (1 − φ ) g ([ X, Y ] v , [ X, Y ] v )= R ( X, Y, Y, X ) + 34 (1 − φ ) g ([ X, Y ] v , [ X, Y ] v ) , where R N is the Riemannian curvature tensor of ( N, h ). On the other hand, by the first-fourthequalities in Lemma 3.4, I = ˆ R ( V, W, W, V ) = ˆ g ( ˆ ∇ V ˆ ∇ W W − ˆ ∇ W ˆ ∇ V W − ˆ ∇ [ V,W ] W, V )= φg ( ˆ ∇ V [ ∇ W W − g ( W, W ) ∇ φ + ( φ − ∇ W W ) h ] , V ) − φg ( ˆ ∇ W [ ∇ V W − g ( V, W ) ∇ φ + ( φ − ∇ V W ) h ] , V )9 φg ( ∇ [ V,W ] W − g ([ V, W ] , W ) ∇ φ + ( φ − ∇ [ V,W ] W ) h , V )= φg ( ˆ ∇ V ( ∇ W W ) − ˆ ∇ W ( ∇ V W ) − ∇ [ V,W ] W, V ) − φ [ g ( W, W ) g ( ˆ ∇ V ∇ φ, V ) − g ( V, W ) g ( ˆ ∇ W ∇ φ, V )]+( φ −
1) ˜ P ( ∇ φ, φ, X, Y, V, W ) + ˜ Q ( ∇ φ, φ, X, Y, V, W ) . = φg ( ˆ ∇ V ( ∇ W W ) v + ˆ ∇ V ( ∇ W W ) h , V ) − φg ( ˆ ∇ W ( ∇ V W ) v + ˆ ∇ W ( ∇ V W ) h , V ) − φg ( ∇ [ V,W ] W, V ) − φ [ g ( W, W ) g ( ˆ ∇ V ∇ φ, V ) − g ( V, W ) g ( ˆ ∇ W ∇ φ, V )]+( φ −
1) ˜ P ( ∇ φ, φ, X, Y, V, W ) + ˜ Q ( ∇ φ, φ, X, Y, V, W ) . Since φ is a basic function, g ( ∇ φ, V ) = V φ = 0. Hence ∇ φ is a horizontal vector field. Then bythe first-fourth equalities in Lemma 3.4, we see that g ( ˆ ∇ W ∇ φ, V ) = − g ( ∇ W V, ∇ φ ) + g ( ∇ φ, ∇ φ )2 φ g ( W, V ) , and I = ˆ R ( V, W, W, V ) = φR ( V, W, W, V )+( φ −
1) ˇ P ( ∇ φ, φ, X, Y, V, W ) + ˇ Q ( ∇ φ, φ, X, Y, V, W )= R ( V, W, W, V ) + ( φ − R ( V, W, W, V )+( φ −
1) ˇ P ( ∇ φ, φ, X, Y, V, W ) + ˇ Q ( ∇ φ, φ, X, Y, V, W )= R ( V, W, W, V ) + ( φ − P ( ∇ φ, φ, X, Y, V, W ) + Q ( ∇ φ, φ, X, Y, V, W ) , where P ( ∇ φ, φ, X, Y, V, W ) , Q ( ∇ φ, φ, X, Y, V, W ) are two functions depending on ∇ φ, φ, X, Y, V, W and Q (0 , φ, X, Y, V, W ) ≡ X is a basic vector field, [ X, W ] is vertical. Hence by the first-fourth equalities inLemma 3.4, I = ˆ R ( X, W, W, X ) = ˆ g ( ˆ ∇ X ˆ ∇ W W − ˆ ∇ W ˆ ∇ X W − ˆ ∇ [ X,W ] W, X )= g ( ˆ ∇ X [ ∇ W W − g ( W, W ) ∇ φ + ( φ − ∇ W W ) h ] , X ) − g ( ˆ ∇ W [ ∇ X W + g ( X, ∇ φ )2 φ W + 1 − φ n X i =1 g ([ X, ε i ] , W ) ε i ] , X ) − g ( ∇ [ X,W ] W − g ([ X, W ] , W ) ∇ φ + ( φ − ∇ [ X,W ] W ) h , X )= R ( X, W, W, X ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )+ Q ( ∇ φ, φ, X, Y, V, W ) − Hess ( φ )( X, X ) g ( W, W ) , where P ( ∇ φ, φ, X, Y, V, W ) , Q ( ∇ φ, φ, X, Y, V, W ) are two functions depending on ∇ φ, φ, X, Y, V, W and Q (0 , φ, X, Y, V, W ) ≡
0. 10y the similar argument, we see that I = ˆ R ( Y, V, V, Y ) = R ( Y, V, V, Y ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )+ Q ( ∇ φ, φ, X, Y, V, W ) − Hess ( φ )( Y, Y ) g ( V, V ) .I = ˆ R ( X, Y, Y, V ) = R ( X, Y, Y, V ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )+ Q ( ∇ φ, φ, X, Y, V, W ) .I = ˆ R ( Y, X, X, W ) = R ( Y, X, X, W ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )+ Q ( ∇ φ, φ, X, Y, V, W ) .I = ˆ R ( X, Y, W, V ) = R ( X, Y, W, V ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )+ Q ( ∇ φ, φ, X, Y, V, W ) .I = ˆ R ( X, W, Y, V ) = R ( X, W, Y, V ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )+ Q ( ∇ φ, φ, X, Y, V, W ) + 12 Hess ( φ )( X, Y ) g ( V, W ) .I = ˆ R ( V, W, W, X ) = R ( V, W, W, X ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )+ Q ( ∇ φ, φ, X, Y, V, W ) + 12 g ( V, W ) g ( ∇ W ∇ φ, X ) − g ( W, W ) g ( ∇ V ∇ φ, X ) .I = ˆ R ( W, V, V, Y ) = R ( W, V, V, Y ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )+ Q ( ∇ φ, φ, X, Y, V, W ) + 12 g ( V, W ) g ( ∇ V ∇ φ, Y ) − g ( V, V ) g ( ∇ W ∇ φ, Y ) , where P i ( ∇ φ, φ, X, Y, V, W ) , Q i ( ∇ φ, φ, X, Y, V, W ) are two functions depending on ∇ φ, φ, X, Y, V, W and Q i (0 , φ, X, Y, V, W ) ≡ i = 3 , , · · · . Henceˆ R ( X + V, Y + W, Y + W, X + V ) = I + I + I + I + 2 X i =4 I i = R ( X + V, Y + W, Y + W, X + V ) + ( φ − P ( ∇ φ, φ, X, Y, V, W )+ Q ( ∇ φ, φ, X, Y, V, W ) + [ − g ( W, W ) g ( ∇ V ∇ φ, X ) + g ( V, W ) g ( ∇ W ∇ φ, X )+ g ( V, W ) g ( ∇ V ∇ φ, Y ) − g ( V, V ) g ( ∇ W ∇ φ, Y )]+ 12 [ − Hess ( φ )( X, X ) g ( W, W ) + 2
Hess ( φ )( X, Y ) g ( V, W ) − Hess ( φ )( Y, Y ) g ( V, V )] , where P ( ∇ φ, φ, X, Y, V, W ) , Q ( ∇ φ, φ, X, Y, V, W ) are two functions which depend on ∇ φ, φ, X, Y, V, W and Q (0 , φ, X, Y, V, W ) ≡ Proof of Theorem 3.1: roof. We prove it by contradiction. We already proved it if k A k is basic. Hence it suffices toshow it if H is basic. We prove it by contradiction. Let π : ( M m , g ) → ( N , h ) be a Riemanniansubmersion such that H is basic and the fibers have nonzero Euler numbers, where ( M m , g ) haspositive sectional curvature and m ≥
4. By Lemma 3.3, there exists a metric ˆ g on M m suchthat π : ( M m , ˆ g ) → ( N , h ) is still a Riemannian submersion and all fibers of π are minimalsubmanifolds with respect to ˆ g . Furthermore, there exists some fiber F such that ˆ g has positivesectional curvature at all points in F . Let r be a fixed positive number such that the normalexponential map of F is a diffeomorphism when restricted to the tubular neighborhood of F with radius r . By continuity of sectional curvature, there exists ǫ , 0 < ǫ < r such that ˆ g haspositive sectional curvature at the ǫ neighborhood of F . Choose another fiber F such that0 < ˆ d ( F , F ) < ǫ , where ˆ d ( F , F ) is the distance between F and F with respect to ˆ g . Since π : ( M m , ˆ g ) → ( N , h ) is a Riemannian submersion, F and F are equidistant. On the otherhand, since 0 < ˆ d ( F , F ) < ǫ , then for any point q ∈ F , there is a unique point p ∈ F suchthat ˆ d ( p, q ) = ˆ d ( F , F ). Let L = ˆ d ( p, q ) and γ : [0 , L ] → M m , γ (0) = p, γ ( L ) = q be the unique minimal geodesic with unit speed realizing the distance between p and q . Let V ⊆ T q ( M m ) bethe subspace of vectors v = X ( L ) where X is a parallel field along γ such that X (0) ∈ T p ( F ).Then dim ( V ∩ T q ( F )) = dim ( V ) + dim ( T q ( F )) − dim ( V + T q ( F )) ≥ ( m −
2) + ( m − − ( m −
1) = m − . We claim that dim ( V ∩ T q ( F )) = m −
3. If not, then dim ( V ∩ T q ( F )) = m − . Let X i , i =1 , · · · m − , be orthonormal parallel fields along γ such that X i (0) ∈ T p ( F ) , X i ( L ) ∈ T q ( F ).For each i , choose a variation f i ( s, t ) of γ such that f i ( s, ∈ F , f i ( s, L ) ∈ F for small s and ∂f i (0 ,t ) ∂s = X i ( t ). By construction, ˙ X i ( t ) = ˆ ∇ ˙ γ X i ( t ) = 0 for all t , where ˆ ∇ is the Levi-Civitaconnection with respect to ˆ g . By the second variation formula, for i = 1 , · · · m − , we have12 d E i ( s ) ds | s =0 = Z L (ˆ g ( ˙ X i , ˙ X i ) − ˆ R ( X i , ˙ γ, ˙ γ, X i )) dt +ˆ g ( ˆ B ( X i , X i ) , ˙ γ )( L ) − ˆ g ( ˆ B ( X i , X i ) , ˙ γ )(0)= − Z L ˆ R ( X i , ˙ γ, ˙ γ, X i ) dt + ˆ g ( ˆ B ( X i , X i ) , ˙ γ )( L ) − ˆ g ( ˆ B ( X i , X i ) , ˙ γ )(0) , where E i ( s ) = R L ˆ g ( ∂f i ( s,t ) ∂t , ∂f i ( s,t ) ∂t ) dt , ˆ R is the curvature tensor of ˆ g and ˆ B j is the secondfundamental form of F j with respect to ˆ g , j = 0 , F and F are minimal submanifolds in ( M m , ˆ g ), we have m − X i =1 ˆ B j ( X i , X i ) = 0 , j = 0 , . Then 12 m − X i =1 d E i ( s ) ds | s =0 = − m − X i =1 Z L ˆ R ( X i , ˙ γ, ˙ γ, X i ) dt. g has positive sectional curvature at the ǫ neighborhood of F and 0 < ˆ d ( F , F ) < ǫ , wesee that ˆ R ( X i , ˙ γ, ˙ γ, X i ) < . Hence 12 m − X i =1 d E i ( s ) ds | s =0 < . Then there exists some i such that d E i ( s ) ds | s =0 <
0, which contradicts that γ is a minimalgeodesic realizing the distance between F and F . So dim ( V ∩ T q ( F )) = m −
3. Since dim ( T q ( F )) = m −
2, then V ∩ T q ( F ) is a codimension one subspace of T q ( F ). Since q isarbitrary on F , by doing the same construction as above for any q , then we get a continuouscodimension one distribution on F . Thus the Euler number of F is zero. Contradiction. In this section we prove Theorem 1.4. Suppose π : ( M , g ) → ( N , h ) is a Riemanniansubmersion with totally geodesic fibers, where ( M , g ) is a compact four-dimensional Einsteinmanifold. We are going to show that the A tensor of π vanishes and then locally π is theprojection of a metric product onto one of the factors. We firstly need the following lemmas: Lemma 4.1.
Let π be a Riemannian submersion with totally geodesic fibers from compactRiemannian manifolds, then all fibers are isometric to each other.Proof. See [12].
Lemma 4.2.
Suppose π : ( M , g ) → ( N , h ) is a Riemannian submersion with totally geodesicfibers, where ( M , g ) is a compact four-dimensional Einstein manifold. Let c , c be the sectionalcurvature of ( F , g | F ) and ( N , h ) , respectively, where g | F is the restriction of g to the fibers F . Let Ric ( g ) = λg for some λ . Then ( i ) 2 c + k A k = 2 λ ;( ii ) 2 c ◦ π − k A k = 2 λ ;( iii ) k A k = ( c ◦ π − c ) , where k A k = k A ∗ X U k + k A ∗ X V k + k A ∗ Y U k + k A ∗ Y V k . Here X, Y /U, V is an orthonormalbasis of H / V , respectively.Proof. See page 250, Corollary 9 .
62 in [3]. For completeness, we give a proof here.Let
U, V / X, Y are orthonormal basis of V / H , respectively. Then by O’Neill’s formula([14]) , we have λ = Ric ( U, U ) = c + k A ∗ X U k + k A ∗ Y U k ; λ = Ric ( V, V ) = c + k A ∗ X V k + k A ∗ Y V k ; λ = Ric ( X, X ) = c ◦ π − k A X Y k + k A ∗ X U k + k A ∗ X V k ; λ = Ric ( Y, Y ) = c ◦ π − k A X Y k + k A ∗ Y U k + k A ∗ Y V k .
13n the other hand, by direct calculation, we see that 2 k A X Y k = k A k . Hence2 c + k A k = 2 λ ;2 c ◦ π − k A k = 2 λ ; k A k = 23 ( c ◦ π − c ) . By Lemmas 4.1 and 4.2, we see that c , k A k are constants on M and c is a constant on N .Fix p ∈ M . Locally we can always choose basic vector fields X, Y such that
X, Y isan orthonormal basis of the horizontal distribution. At point p , since the image of A ∗ X isperpendicular to X and dim V = dim H = 2, A ∗ X must have nontrivial kernel. Then there existssome v ∈ V such that k v k = 1 and A ∗ X ( v ) = 0. Extend v to be a local unit vertical vector field V and choose U such that U, V is a local orthonormal basis of V . Lemma 4.3. A ∗ X V ( p ) = 0; A ∗ Y V ( p ) = 0 . Proof.
We already see A ∗ X V ( p ) = A ∗ X,p ( v ) = 0. On the other hand, at point p , we have A ∗ Y V = g ( A ∗ Y V, X ) X = − g ( ∇ Y V, X ) X = g ( V, ∇ Y X ) X = g ( V, A Y X ) X = − g ( V, A X Y ) X = − g ( V, ∇ X Y ) X = g ( ∇ X V, Y ) X = − g ( A ∗ X V, Y ) X = 0 . Since all fibers of π are totally geodesic, by O’Neill’s formula ([14]), we see that K ( X, U ) = k A ∗ X U k . Because ( M , g ) is Einstein, at point p , we have λ = Ric ( U, U ) = c + k A ∗ X U k + k A ∗ Y U k ; λ = Ric ( V, V ) = c + k A ∗ X V k + k A ∗ Y V k ;Combined with Lemma 4.3, we see that λ = c and k A ∗ X U k ( p ) = 0, k A ∗ Y U k ( p ) = 0. Then k A k ( p ) = 0. Hence k A k ≡ M and c = c . Let c = c = c . Then locally π isthe projection of a metric product B ( c ) × B ( c ) onto one of the factors, where B ( c ) is atwo-dimensional compact manifold with constant curvature c .14 Conjecture 1 and the Weak Hopf Conjecture
In this section we point out several interesting corollaries of Conjecture 1.Suppose (
E, g ) is a complete, open Riemannian manifold with nonnegative sectional cur-vature. By a well known theorem of Cheeger and Gromoll [4], E contains a compact totallygeodesic submanifold Σ, called the soul, such that E is diffeomorphic to the normal bundle ofΣ. Let Σ r be the distance sphere to Σ of radius r . Then for small r >
0, the induced metricon Σ r has nonnegative sectional curvature by a theorem of Guijarro and Walschap [10]. In [9],Gromoll and Tapp proposed the following conjecture: Weak Hopf Conjecture
Let k ≥ . Then for any complete metric with nonnegative sectionalcurvature on S n × R k , the induced metric on the boundary of a small metric tube about the soulcan not have positive sectional curvature. The case n = 2 , k = 3 is of particular interest since the metric tube of the soul is diffeomorphicto S × S .Recall that a map between metric spaces σ : X → Y is a submetry if for all x ∈ X and r ∈ [0 , r ( x )] we have that f ( B ( x, r )) = B ( f ( x ) , r ), where B ( p, r ) denotes the open metric ballcentered at p of radius x and r ( x ) is some positive continuous function. If both X and Y are Riemannian manifolds, then σ is a Riemannian submersion of class C , by a theorem ofBerestovskii and Guijarro [1]. Proposition 5.1.
Suppose Σ is a soul of ( E, g ) , where ( E, g ) is a complete, open Riemannianmanifold with nonnegative sectional curvature. If the induced metric on Σ r has positive sectionalcurvature at some point for some r > , then there is a Riemannian submersion from Σ r to Σ with fibers S l − , where l = dim ( E ) − dim (Σ) . Proof.
In fact, by a theorem of Guijarro and Walschap in [11], if Σ r has positive sectionalcurvature at some point, the normal holonomy group of Σ acts transitively on Σ r . By Corollary5 in [24], we get a submetry π : ( E, g ) → Σ × [0 , + ∞ ) with fibers S l − , where Σ × [0 , + ∞ ) isendowed with the product metric. Then π : ( π − (Σ × (0 , + ∞ )) , g ) → Σ × (0 , + ∞ ) is also asubmetry. By a theorem of Berestovskii and Guijarro in [1], π is a C , Riemannian submersion.Then Σ r = π − (Σ × { r } ) and π : Σ r → Σ is also a C , Riemannian submersion with fibers S l − ,where Σ r is endowed with the induced metric from ( E, g ). Proposition 5.2.
When k > n , Conjecture implies Weak Hopf Conjecture.Proof. Suppose for some complete metric g on S n × R k with nonnegative sectional curvature,the induced metric on Σ r has positive sectional curvature for some r >
0, where Σ is a soul.Since S n × R k is diffeomorphic to the normal bundle of Σ, we see that Σ is a homotopy sphereand dim (Σ) = n . By Proposition 5.1, we get a Riemannian submersion from Σ r to Σ with fibers S k − , where Σ r is endowed with the induced metric from g and hence has positive sectionalcurvature. Since k > n , we see k − ≥ n , which is impossible if Conjecture 1 is true for C , Riemannian submersions.
Remark 3.
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