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The codimension one homology of a complete manifold with nonnegative Ricci curvature

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The Codimension One Homology of a Complete Manifold with Nonnegative Ricci Curvature Zhongmin Shen and Christina Sormani November 10,2000Abstract In this paper we prove that a complete noncompact manifold with nonnegative Ricci curvature has a trivial codimension one homology unless it is a ?at normal bundle over a compact totally geodesic submanifold.In particular,we prove the conjecture that a complete noncompact manifold with positive Ricci curvature has a trivial codimension one integer homology.We also have a corollary stating when the codimension two integer homology of such a manifold is torsion free.1Introduction In this paper we prove that a complete noncompact manifold with nonnegative Ricci curvature has a trivial codimension one homology unless it is a split or ?at normal bundle over a compact totally geodesic submanifold (Thms 1.1and 1.2).In particular,we prove the conjecture that a complete noncompact manifold with positive Ricci curvature has a trivial codimension one integer homology (Cor 1.1).We also have a corollary stating when the codimension two integer homology of such a manifold is torsion free (Cor 1.2).In 1976,Shing Tung Yau proved that a complete noncompact manifold with positive Ricci curvature has a trivial codimension one real homology [Yau].It was then conjectured that the integer homology might be trivial as well.In 1993,

the ?rst author proved that if the manifold is also proper,then the codimension one integer homology is trivial [Sh1].Recall that a manifold is proper if the manifold’s Busemann function has compact level sets.It is unknown whether all complete noncompact manifolds with positive Ricci curvature are proper.However,the authors have proven that such a manifold is proper when it has Euclidean volume growth [Sh2]or linear volume growth [So1].

Recently Itokawa and Kobayashi have proven that if a complete noncompact manifold with nonnegative Ricci curvature has more than linear volume growth then the codimension one integer homology is trivial [ItKo,Thm 1].Since a manifold with nonnegative Ricci curvature must have at least linear volume

growth[Yau],the three papers[ItKo],[So1]and[Sh1]prove the conjecture re-garding manifolds with positive Ricci curvature.In fact they demonstrate that codimension one integer homology is trivial unless the manifold has linear vol-ume growth and nonnegative Ricci curvature.Note that split or?at normal bundles over a compact manifolds with nonnegative Ricci curvature are exam-ples of such exceptional manifolds.It is not hard to see that these examples can have codimension one integer homology equal to Z.

Itokawa and Kobayashi have also proven that if a manifold with nonnegative Ricci curvature has bounded diameter growth then the integer codimension one homology is either Z,Z2or0[ItKo Thm2].Note that the split or?at normal bundles over compact manifolds have bounded diameter growth,but there also exist examples of manifolds with nonnegative Ricci curvature,linear volume growth and unbounded diameter growth[So1].Itokawa and Kobayashi used minimizing currents to prove their results,and the?rst author used an adaption of Morse theory to prove his result regarding proper manifolds.

We reprove these results and complete the classi?cation of the codimension one integer homology using Poincare Duality,the Cheeger Gromoll Splitting Theorem[ChGl]and properties of noncontractible loops developed by the second author[So2].

Theorem1.1Let M n be an orientable complete noncompact manifold with nonnegative Ricci curvature and G is an abelian group,then one of the following holds:

(i)H n?1(M,G)=0,

(ii)H n?1(M,G)=G,

(iii)H n?1(M,G)=ker(G×2→G).

Case(ii)can only occur when M n is an an isometrically split manifold over a compact totally geodesic orientable submanifold.

Case(iii)can only occur if M n is a one-ended?at normal bundle over a compact totally geodesic unorientable submanifold.

Recall that ker(Z×2→Z)=0and ker(Z k×2→Z k)=0if k is odd and is Z2if k is even.

Theorem1.2Let M n be a complete noncompact unorientable manifold with nonnegative Ricci curvature and G=Z2or Z.Then one of the following holds:

(i)H n?1(M,G)=0,

(ii)H n?1(M,G)=G,

(iii)H n?1(M,G)=ker(G×2→G).

Case(ii)can only occur when M is a one-ended?at normal bundle over a compact totally geodesic orientable submanifold.

Case(iii)can only occur when M is an isometrically split manifold over a compact unorientable submanifold.

Remark:These two cases are opposite to the two cases in Theorem1.1.

For an outline of our proof of Theorems1.1and1.2and the various cases involved,see the following diagram:

Complete Open Manifold With Ric≥0

two or more ends one end

M=N×R

orientable

H n?1(M,G)=G

unorientable

H n?1(M,G)=G0

with loop to

in?nity property

unorientable

H n?1(M,G?)=0

orientable

H n?1(M,G)=0

without loop to

in?nity property

?at normal bundle over

a totally geodesic soul

orientable

H n?1(M,G)=G0

unorientable

H n?1(M,G)=G G0:=ker(G×2→G),G?:=Z2or Z

Corollary1.1Let M n be a complete noncompact manifold with nonnegative Ricci curvature.If there is a point p∈M such that Ricci p(v,v)>0then H n?1(M,Z)is trivial.

Note that we have eliminated the possibility that H n?1(M,Z)=Z2.This was indicated as a possibility in[ItKo,Thm2]if M had a cover of degree one or two isometric toˉM×[0,∞)∪K where K andˉM are compact.Their second theorem also indicated that Z was a possibility only if M had a cover of degree one or two which split isometrically exactly as we have.

The paper is divided into two sections.In the?rst section,we de?ne a topological property called the loops to in?nity property[Defn2.2].We then

prove that orientable manifolds with this property and only one end have trivial codimension one G homology where G is any abelian group[Prop 2.1].This section of the paper is purely topological and uses Poincare Lefschetz Duality and the Universal Coe?cient Theorem(c.f.[Mun]and[Mas]).

In the second section of the paper,we eliminate the topological conditions in Propositions2.1using the properties of manifolds with nonnegative Ricci curvature[Props 3.1-3.3].In particular,we use the Splitting Theorem[ChGl] and a theorem in[So2].

In addition to proving Theorems1.1and1.2,we prove the following corollary in Section2.

Corollary1.2Let M n be an orientable complete noncompact manifold with nonnegative Ricci curvature,then M is either an isometrically split manifold or a?at normal bundle over a compact totally geodesic submanifold or,for any abelian group G,

H n?2(M,Z)?G=0,

in which case H n?2(M,Z)has no elements of?nite order.

The authors would like to thank Dan Christensen for helpful discussions regarding homology theory.

2Topology and H n?1(M,Z)

Before we prove Theorems1.1and1.2we introduce some topological concepts from[So2]concerning noncontractible loops in complete noncompact manifolds. De?nition2.1Given a rayγand a loop C:[0,L]→M based atγ(0),we say that a loopˉC:[0,L]→M is homotopic to C alongγif there exists r>0with

ˉC(0)=ˉC(L)=γ(r)and the loop,constructed by joiningγfrom0to r with ˉC from0to L and then withγfrom r to0,is homotopic to C(see diagram below).

De?nition2.2A complete noncompact manifold M n has the loops to in?nity property if there is a rayγ,such that for any element,g∈π1(M,γ(0)),and any compact set,K?M,there exists a closed loop,ˉC,contained in M\K which is homotopic alongγto a representative loop,C,such that g=[C].

When a ray,γ,is speci?ed,we say M n has the loops to in?nity property along γ.

In [So2]it is proven that a manifold with nonnegative Ricci curvature either has the loops to in?nity property along some ray or the manifold is a ?at normal bundle over a compact totally geodesic soul and has a double cover which splits isometrically.

Recall that a manifold is said to have one end if given any compact set K ?M ,there is only one unbounded connected component of M \K .Note that if a complete manifold has two or more ends then it has a length minimizing geodesic de?ned for all values of R ,called a line .Recall also that by the Cheeger-Gromoll Splitting Theorem [ChGl],if a complete noncompact manifold with nonnegative Ricci curvature contains a line,then it splits isometrically.So a manifold with nonnegative Ricci curvature either splits isometrically or has only one end.

Thus,in some sense,most manifolds with nonnegative Ricci curvature have the loops to in?nity property

and only one end.We now prove the following simpli?ed version of Theorem 1.1relating the loops to in?nity property to the codimension one integer homology of a manifold.

Proposition 2.1Let M n be a complete noncompact manifold with one end and the loops to in?nity property.Then H n ?1(M,Z 2)is trivial.

If M n is also orientable,then H n ?1(M,G )is trivial where G is an abelian group.

Proof:Suppose H n ?1(M,G )is not trivial.Then there is a chain,σ,which has no boundary but which is not a boundary .See the diagram below.The images of the simplices of this chain must be contained in some compact set,K 0=Cl (U 0),containing γ(0).Since M n has only one end,there is a compact set,K ,consisting of K 0and the bounded connected components of M \K 0.Fur-thermore,K can be chosen with smooth boundary.Thus K is an n dimensional manifold with boundary such that ?K is connected and

H n ?1(K,G )=0.(1)

We now proceed to show that(1)cannot hold under the appropriate condi-tions on M.

Letγbe the ray along which M n has the loops to in?nity property.Since γis a ray,it must leave K,so let t0>0be de?ned such thatγ(t0)∈?K.In [So2],it is proven that if M n has the loops to in?nity property alongγand?K is smooth and connected,then

i?:π1(?K,γ(t0))→π1(K,γ(t0))(2) is onto.Here i?is the map induced by the inclusion map i:?K→K.

Now H1(N,Z)is just the abelianization ofπ1(N,p)(c.f.[Mas,Ch.VIII, Thm7.1]).Thus we know

i?:H1(?K,Z)→H1(K,Z)(3) is onto.This is part of a long exact sequence involving the relative homology and the reduced homology,

H1(?K,Z)i?→H1(K,Z)π?→H1(K,?K,Z)??→?H0(?K,Z),(4) (c.f.[Mun,Thm23.3]).So Ker(π?)=Im(i?)=H1(K,Z).Thusπ?=0and ??is one-to-one.Now,?K is connected so the reduced homology,?H0(?K,Z), is trivial.Thus the relative homology,H1(K,?K,Z),is trivial as well.

We now claim that the cohomology H1(K,?K,G)must be trivial as well. To prove this we will apply the Universal Coe?cient Theorem for Cohomology (c.f.[Mun,Thm53.1]),which provides the following short exact sequence:

0→Ext(H0(K,?K;Z),G)→H1(K,?K;G)→Hom(H1(K,?K;Z),G)→0. Since H1(K,?K;Z)is trivial,the homomorphisms from it to G must be trivial as well.Thus

Ext(H0(K,?K;Z),G)=H1(K,?K;G).(5) On the other hand,since K and?K are connected,the long exact relative homology sequence(c.f.[Mun,Thm23.3]),

...→H0(?K,Z)i?→H0(K,Z)π?→H0(K,?K;Z)→0(6) is

...→Z i?→Zπ?→H0(K,?K;Z)→0,(7) where i?is an isomorphism.Thus ker(π?)=Im(i?)=Z andπ?is onto,so H0(K,?K;Z)=0.Thus Ext(H0(K,?K;Z),G)=0and,by(5),H1(K,?K,G) must be trivial as well.

Now by Poincare Lefschetz Duality[Mun,Thm70.7],we know that

H1(K,?K,Z2)→H n?1(K,Z2)(8) is an isomorphism.Since H1(K,?K,Z2)is trivial and,by(1),H n?1(K,Z2)is not trivial,we have a contradiction.Thus H n?1(M,Z2)must be trivial.

If(K,?K)is orientable,then Poincare Lefschetz Duality implies that

H1(K,?K,G)→H n?1(K,G)(9) is an isomorphism.Thus,when M is orientable,we contradict(1),so H n?1(M,G) must be trivial.Q.E.D.

3Ricci Curvature and H n?1(M,Z)

To prove the main theorems,we must deal with complete noncompact manifolds with nonnegative Ricci curvature that either have more than one end,are not orientable or fail to satisfy the loops to in?nity property.We deal with these three cases seperately.Refer to the diagram at the end of the paper. Proposition3.1Let M n be a complete noncompact with nonnegative Ricci curvature and two or more ends.Let G be an abelian group.Then M=N n?1×R and

H n?1(M,G)= G,if M is orientable

ker(G×2→G),if M is not orientable

Proposition3.2Let M n be a one-ended complete noncompact manifold with nonnegative Ricci curvature satisfying the loops to in?nity property.Then,re-gardless of the orientability of M,H n?1(M,Z)=0.

Proposition3.3Let M n be a complete noncompact manifold with nonnegative Ricci curvature,that has one end and doesn’t have a ray with the loops to in?nity property.Let G be an abelian group.Then M is a?at normal bundle over a totally geodesic soul and

H n?1(M,G)= G,if M is unorientable

ker(G×2→G),if M is orientable

Note that Propositions2.1, 3.1and 3.3imply Theorem1.1,and Proposi-tions2.1-3.3imply Theorem1.2.The proof of Corollary1.1will appear at the end of the paper.

Proof of Proposition3.1:By the Cheeger-Gromoll Splitting Theorem,a complete noncompact manifold with nonnegative Ricci curvature and two or more ends must split isometrically,M n=N n?1×R,where N n?1is compact. Thus H n?1(M n,G)=H n?1(N n?1,G),which is

ker(G×2→G)

if N n?1is not orientable and is G if N n?1is orientable(c.f.[Mun Cor65.5]). The proposition then follows from the fact that N n?1is orientable i?M n is orientable.Q.E.D.

Proof of Proposition 3.2:This proposition holds if M is orientable by Propo-sition 2.1.If M is not orientable,then it has a double cover π:?M →M such

that ?M is orientable.We ?rst claim

that ?M has only one end.

Assume on the contrary that ?M has two or more

ends.By the Cheeger-

Gromoll Splitting Theorem,it splits isometrically,?M =N n ?1

×R where N is

compact [ChGl].Since ?M is orientable,so is the

totally geodesic submanifold,

N n ?1.Let g be the deck transform acting on ?M .Then g takes lines to lines

and acts as an isometry on each component,g :R →R and g :N n ?1→N n ?1.Now g :R →R is an isometry so g (r )=±r +r 0.If g (r )=r +r 0,then π:{p 0}×R →π({p 0}×R )would map a line to a line and M itself would split isometrically and have two ends.Thus g (r )=?r +r 0.

Now M has the loops to in?nity property along a ray,γ,which lifts to a ray ?γ.Since N is compact,we can choose ?γ(t )=(q 0,t +t 0).Let C 0be a curve

in ?M joining (q 0,t 0)to g (q 0,t 0)=(gq 0,?t 0+r 0).Then for any compact set K ?M ,including K =π(N n ?1×{r 0/2}),there exists a curve,C 1∈M \K ,based at γ(t 1)which is homotopic along γto π?C 0[Defn 2.2].See the diagram

below.By lifting the homotopy,we lift C 1to a curve,?C 1,which runs from

?γ(t 1)=(q 0,t 1+t 0)to g ?γ(t 1)=(gq 0,?t 1?t 0+r 0)[Defn 2.1].Thus ?C 1passes through N ×{r 0/2},and we have a contradiction.Any unorientable manifold with nonnegative Ricci curvature and the loops to in?nity property,has an orientable double cover with only one

end.

γg ?γγ

r 0

cover?M,and it is easy to see that this gives us a curve,?ˉC∈?M\K which is homotopic along?γto C.

Thus the orientable double cover,?M,satis?es all the conditions of Proposi-tion2.1and H n?1(?M,G)is trivial.We claim that this implies that H n?1(M,Z) is trivial as well.

Given any simplicial mapσ:?k→M,there is a continuous lift ofσ,?σ:?k→?M,which is unique up to deck transform.Thus,there is a unique chain in C k(?M,Z),

f(σ)=?σ+g?σ,(10) where g is the deck transform acting on the double cover.We can extend f to a well de?ned map,f:C k(M,Z)→C k(?M,Z).In fact,f commutes with the boundary operator,?,so f?:H k(M,Z)→H k(?M,Z)is well de?ned.Note also, thatπ??f?([σ])=2[σ].

Thus,given any h∈H n?1(M,Z),f?(h)is in the trivial group,H n?1(?M,Z), so

2h=π?(f?(h))=π?(0)=0.(11) Thus,2H n?1(M,Z)=0and

H n?1(M,Z)

H n?1(M,Z)?Z2=

(c.f.[Mun Cor65.5]).Q.E.D. Proof of Corollary1.2:If M is neither a isometrically split manifold nor a?at normal bundle over a compact totally geodesic submanifold,then by Theorem1.1,we know H n?1(M,G)=0.The Universal Coe?cient Theorem (c.f.[Mun,Thm55.1]),states that there is a short exact sequence,

0→H n?1(M,Z)?G→H n?1(M,G)→H n?2(M,Z)?G→0.(14)

Substituting,H n?1(M,G)=0,we get

H n?2(M,Z)?G=0.(15)

In particular,substituting G=Z k,we have

H n?2(M,Z)

0=H n?2(M,Z)?Z k=

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