## versão impressa ISSN 0001-3765versão On-line ISSN 1678-2690

### An. Acad. Bras. Ciênc. vol.88 no.4 Rio de Janeiro out./dez. 2016

#### http://dx.doi.org/10.1590/0001-3765201620150085

Mathematical Sciences

On reduced L2 cohomology of hypersurfaces in spheres with finite total curvature

1School of Mathematics and Physics, Jiangsu University of Technology, Changzhou, Jiangsu, 213001, China

Abstract

In this paper, we prove that the dimension of the second space of reduced L2 cohomology of M is finite if is a complete noncompact hypersurface in a sphere 𝕊n+1and has finite total curvature (n≥3).

Key words: total curvature; reduced L2 cohomology; hypersurface in sphere; L2 harmonic 2-form

Introduction

For a complete manifold Mn, the p-th space of reduced L2-cohomology is defined, for 0pn in Carron (2007). It is interesting and importantto discuss the finiteness of the dimension of these spaces.Carron (1999) proved that if Mn (n3) is a complete noncompact submanifold of n+pwith finite total curvature and finite mean curvature (i. e., the Ln-norm of the mean curvature vector is finite),then each p-th space of reduced L2-cohomology on M has finite dimension, for 0pn.The reduced L2 cohomology is related with the L2 harmonic forms (Carron 2007). In fact, several mathematicians studied the space of L2harmonic p-forms for p=1,2.If Mn (n3) is a complete minimal hypersurface in n+1 with finite index, Li and Wang (2002) proved that the dimension of the space of the L2 harmonic 1-forms M is finite and M has finitely many ends.More generally, Zhu (2013) showed that:suppose that Nn+1 (n3) is a complete simply connected manifold with non-positive sectional curvature andMn is a complete minimal hypersurface in N with finite index. If the bi-Ricci curvature satisfies

b-Ric¯(X,Y)+1n|A|20,

for all orthonormal tangent vectors X,Y in TpN for pM, then the dimension of the space of the L2 harmonic 1-forms M is finite. Furthermore, following the idea of Cheng and Zhou (2009); Zhu (2013) gave a result on finitely many ends of complete manifolds with a weighted Poincaréinequality by use of the space of L2 harmonic functions.Cavalcante et al. (2014) discussed a complete noncompact submanifold Mn (n3) isometrically immersed in a Hadamard manifoldNn+p with sectional curvature satisfying -k2KN0 for some constant k and showed that if the total curvature is finiteand the first eigenvalue of the Laplacian operator of M is bounded from below by a suitable constant, then the dimension of the space of the L2 harmonic 1-forms on M is finite.Fu and Xu (2010) studied a complete submanifold Mn in a sphere 𝕊n+p with finite total curvature and bounded mean curvature and proved that the dimension of the space of the L2 harmonic 1-forms on M is finite. Zhu and Sw. (2014) proved Fu-Xu’s result without the restriction on the mean curvature vector and therefore obtained that the first space of reduced L2-cohomology on M has finite dimension.Zhu (2011) studied the existence of the symplectic structure and L2 harmonic 2-forms on complete noncompact manifolds by use of a special version of Bochner formula.

Motivated by above results, we discuss a complete noncompact hypersurface Mn in a sphere 𝕊n+1 with finite total curvature in this paper.We obtain the following finiteness results on the space of all L2 harmonic 2-forms and the second space of reduced L2cohomology:

Theorem 1. Let Mn (n3) be an n-dimensional complete noncompact oriented manifold isometrically immersed in an (n+1)-dimensional sphere𝕊n+1. If the total curvature is finite, then the space of all L2 harmonic 2-forms has finite dimension.

Corollary 2. Let Mn (n3) be an n-dimensional complete noncompact oriented manifold isometrically immersed in𝕊n+1. If the total curvature is finite, then the dimension of the second space of reduced L2cohomology of M is finite.

Remark 3. Under the same condition of Corollary 2, we conjecture thatthe p-th space of reduced L2 cohomology of M has finite dimension for 3pn-3.

PRELIMINARIES

In this section, we recall some relevant definitions and results. Suppose that Mn is an n-dimensional complete Riemannian manifold.The Hodge operator*:p(M)n-p(M) is defined by

*ei1eip=sgnσ(i1,i2,,in)eip+1ein,

where σ(i1,i2,,in) denotes a permutation ofthe set (i1,i2,,in) and sgnσ is the sign ofσ. The operator d*:p(M)p-1(M)is given by

d*ω=(-1)(nk+k+1)*d*ω.

The Laplacian operator is defined by

ω=-dd*ω-d*dω.

A p-form ω is called L2 harmonic if ω=0and

Mω*ω<+.

We denote by Hp(L2(M))the space of all L2 harmonic p-forms on M.Let

Z2p(M)={αL2(p(T*M)):dα=0}

and

Dp(d)={αL2(p(T*M)):dαL2(p+1(T*M))}.

We define the p-th space of reduced L2 cohomology by

H2p(M)=Z2p(M)Dp-1(d)¯.

Suppose that x:Mn𝕊n+1 is an isometric immersion of an n-dimensional manifold M in an (n+1)-dimensionalsphere.Let A denote the second fundamental form and H the mean curvature of the immersion x.Let

Φ(X,Y)=A(X,Y)-HX,Y,

for all vector fields X and Y, where , is the induced metric of M.We say the immersion x has finite total curvature if

ΦLn(M)<+.

We state several results which will be used to prove Theorem 1.

Proposition 4. (Carron 2007)Let (M,g) is a complete Riemannian manifold, then the space of L2 harmonic p-forms Hp(L2(M)) is isomorphic tothe p-th space of reduced L2 cohomology H2p(M).

Lemma 5. (Li 1993) If (Mn,g) is a Riemannianmanifold and ω=aIωIp(M), then

|ω|2=2ω,ω+2|ω|2+2E(ω),ω,

where E(ω)=Rkβiβjαiαai1kβipeipejαei1.

Proposition 6. (Hoffman and Spruck 1974, Zhu and Fang 2014)Let Mn be a complete noncompact oriented manifold isometrically immersed in a sphere𝕊n+1. Then

(M|f|2nn-2)n-2nC0(M|f|2+n2M(H2+1)f2)

for each fC01(M), where C0 depends only on n and H is the mean curvature of M in 𝕊n+1.

AN INEQUALITY FOR L2 HARMONIC 2-FORMS

In this section, we show an inequality for L2 harmonic 2-forms on hypersurfaces in a sphere 𝕊n+1,which plays an important role in the proof of main results.

Proposition 7. Let Mn (n3) be an n-dimensional complete noncompact hypersurface isometrically immersed in an (n+1)-dimensional sphere 𝕊n+1. If ωH2(L2(M)), then

hh|h|2+2h2-|Φ|2h2+32H2h2,

for n=3 and

hh1n-2|h|2+2(n-2)h2-n-22|Φ|2h2+nH2h2,

for n4, where h=|ω|.

Proof. Suppose that ωH2(L2(M)). Then we have

|ω|2=2||ω||2+2|ω||ω|. (1)

By Lemma 5, we get that:

|ω|2
=2ω,ω+2|ω|2+2E(ω),ω
=2|ω|2+2E(ω),ω. (2)

Combining (1) with (AN INEQUALITY FOR L2 HARMONIC 2-FORMS), we obtain that

|ω||ω|=|ω|2-||ω||2+E(ω),ω. (3)

There exists the Kato inequality for L2 harmonic 2-forms as follows (Cibotaru and Zhu 2012, Wang 2002):

n-1n-2||ω||2|ω|2. (4)

By (3) and (4), we get that

|ω||ω|1n-2||ω||2+E(ω),ω. (5)

Now, we give the estimate of the term E(ω),ω. Let ω1=bi1i2ei2ei12(M) andω2=ci1i2ei2ei12(M), wherebi1i2=-bi2i1 and ci1i2=-ci2i1. By Lemma 5, we obtain that

E(ω1)
=Rk1i1j1i1bk1i2ei2ej1+Rk2i2j2i2bi1k2ej2ei1
+Rk2i2j1i1bi1k2ei2ej1+Rk1i1j2i2bk1i2ej2ei1
=Rick1j1bk1i2ei2ej1+Rick2j2bi1k2ej2ei1
+Rk2i2j1i1bi1k2ei2ej1+Rk1i1j2i2bk1i2ej2ei1.

So, we get that

E(ω1),ω2=
Rick1j1bk1i2cj1i2+Rick2j2bi1k2ci1j2
+Rk2i2j1i1bi1k2cj1i2+Rk1i1j2i2bk1i2ci1j2,

which implies that

E(ω),ω=
Rick1j1ak1i2aj1i2+Rick2j2ai1k2ai1j2
+Rk2i2j1i1ai1k2aj1i2+Rk1i1j2i2ak1i2ai1j2. (6)

By Gauss equation, we have that

Rijkl=(δikδjl-δilδjk)+hikhjl-hilhjk.

A direct computation shows that

Rick1j1=(n-1)δk1j1+nHhk1j1-hk1ihij1; (7)
Rick2j2=(n-1)δk2j2+nHhk2j2-hk2ihij2; (8)
Rk2i2j1i1=(δk2j1δi2i1-δk2i1δi2j1)+hk2j1hi2i1-hk2i1hi2j1 (9)

and

Rk1i1j2i2=(δk1j2δi1i2-δk1i2δi1j2)+hk1j2hi1i2-hk1i2hi1j2. (10)

Since the curvature operator E is linear andzero order, and hence tensorial, it is sufficient to compute E(ω),ω at a point p.We can choose an orthonormal frame {ei} such thathij=λiδij at p. Obviously,

nH=λ1++λn.

By (6)-(10), we have

E(ω),ω=
(n-1)(aj1i2)2+nHλk1(ak1i2)2-λk12(ak1i2)2
+
(n-1)(ai1j2)2+nHλk2(ai1k2)2-λk22(ai1k2)2
+
ai1j1aj1i1-λk2λi2(ak2i2)2
+
aj2i2ai2j2-λj2λi2(aj2i2)2
=
2ij((n-2)+(λ1++λn)λi-λi2-λiλj)(aij)2.

Note that

|A|2=|Φ|2+nH2.

For n=3, we have that

E(ω),ω=2ij(1+(λ1+λ2+λ3)λi-λi2-λiλj)(aij)2
=ij(2+(λ1+λ2+λ3)(λi+λj)-(λi2+λj2)-2λiλj)(aij)2
=ij(2+12(3H)2-12k=1,ki,j3λk2-12(λi+λj)2)(aij)2
ij(2+12(3H)2-12k=1,ki,j3λk2-(λi2+λj2))(aij)2
ij(2+92H2-|A|2)(aij)2
=(2+32H2-|Φ|2)|ω|2.

For n4, we obtain that

E(ω),ω=2ij((n-2)+(λ1++λn)λi-λi2-λiλj)(aij)2
=ij(2(n-2)+(λ1++λn)(λi+λj)-(λi2+λj2)-2λiλj)(aij)2
=ij(2(n-2)+(λ1++λi^++λj^++λn)(λi+λj))(aij)2
=ij(2(n-2)+12(nH)2-12(k=1,ki,jnλk)2-12(λi+λj)2)(aij)2
ij(2(n-2)+12(nH)2-n-22(k=1,ki,jnλk2)-(λi2+λj2))(aij)2
ij(2(n-2)+12(nH)2-n-22|A|2)(aij)2
=(2(n-2)+12(nH)2-n-22|A|2)|ω|2
=(2(n-2)+nH2-n-22|Φ|2)|ω|2.

By (5), we have that:

hh|h|2+2h2-|Φ|2h2+32H2h2,

for n=3 and

hh1n-2|h|2+2(n-2)h2-n-22|Φ|2h2+nH2h2,

for n4.∎

Remark 8. If ω is 1-form , then the term E(ω,ω) is equal to Ric(ω,ω).The corresponding estimate for this term was given by Leung (1992).

PROOF OF MAIN RESULTS

In this section, we prove Theorem 1 and Corollary 2.

If η is a compactly supported piecewise smooth function on M, then

div(η2hh)
=η2hh+(η2h),h
=η2hh+η2|h|2+2ηhη,h.

Integrating by parts on M, we obtain that

Mη2hh+Mη2|h|2+2Mηhη,h=0. (11)

Case I: n=3. By Proposition 7 and (11), we obtain that

-
2Mηhη,h-2Mη2|h|2-2Mη2h2
+M|Φ|2η2h2-32MH2h2η20. (12)

Note that

-2Mηhη,ha1Mη2|h|2+1a1Mh2|η|2, (13)

for any positive real number a1.Now we give an estimate of the term M|Φ|2η2h2 as follows:set ϕ1(η)=(Suppη|Φ|3)13. Then there exists

M|Φ|2η2h2(Suppη(|Φ|2)32)23(M(η2h2)3)13
=ϕ1(η)2(M(ηh)6)13
C0ϕ1(η)2(M|(ηh)|2+9M(H2+1)(ηh)2)
C0ϕ1(η)2((1+1b1)Mh2|η|2+(1+b1)Mη2|h|2+9M(H2+1)(ηh)2), (14)

for any positive real number b1, where the second inequality holds because of Proposition 6.By (12)-(14), we obtain that

𝒜1Mη2|h|2+1MH2η2h2+𝒞1Mη2h2𝒟1Mh2|η|2, (15)

where

𝒜1:
=(2-C0ϕ1(η)2)-(a1+b1C0ϕ1(η)2),
1:
=32-9C0ϕ1(η)2,
𝒞1:
=2-9C0ϕ1(η)2

and

𝒟1:=1a1+C0ϕ1(η)2(1+1b1).

Since the total curvature ΦL3(M) is finite, we can choose a fixedr0 such that

ΦL3(M-Br0)<δ1=112C0.

Set

𝒜~1:
=(2-C0δ12)-(a1+b1C0δ12),
~1:
=32-9C0δ12,
𝒞~1:
=2-9C0δ12

and

𝒟~1:=1a1+C0δ12(1+1b1).

Thus,

𝒜~1Mη2|h|2+~1MH2η2h2+𝒞~1Mη2h2𝒟~1Mh2|η|2, (16)

for any ηC0(M-Br0).By Proposition 6, we have

1C0(M(ηh)6)13M|(ηh)|2+9M(H2+1)(ηh)2
(1+1c1)Mh2|η|2+(1+c1)Mη2|h|2+9M(H2+1)(ηh)2, (17)

for any positive real number c1.By (16) and (17), we have

1C0(M(ηh)6)13
(1+1c1)Mh2|η|2+(1+c1)Mη2|h|2+9M(H2+1)(ηh)2
(1+1c1+(1+c1)𝒟~1𝒜~1)Mh2|η|2+(9-(1+c1)~1𝒜~1)MH2η2h2
+(9-(1+c1)𝒞~1𝒜~1)Mη2h2. (18)

Choose a sufficient large c1 such that

9-(1+c1)~1𝒜~1<0

and

9-(1+c1)𝒞~1𝒜~1<0.

Then (18) implies that

(M(ηh)6)13A~Mh2|η|2, (19)

for any ηC0(M-Br0). where A~ is a positive constant.

Case II: n4. By Proposition 7 and (11), we obtain that

-
2Mηhη,h-n-1n-2Mη2|h|2-2(n-2)Mη2h2
+n-22M|Φ|2η2h2-nMH2h2η20. (20)

Note that

-2Mηhη,ha2Mη2|h|2+1a2Mh2|η|2, (21)

for any positive real number a2. Weset ϕ2(η)=(Suppη|Φ|n)1n and obtain that

M|Φ|2η2h2(Suppη(|Φ|2)n2)2n(M(η2h2)nn-2)n-2n
=ϕ2(η)2(M(ηh)2nn-2)n-2n
C0ϕ2(η)2(M|(ηh)|2+n2M(H2+1)(ηh)2)
C0ϕ2(η)2(M(1+1b2)h2|η|2+(1+b2)η2|h|2+n2M(H2+1)(ηh)2), (22)

for any positive real number b2, where the second inequality holds because of Proposition 6.By (20)-(22), there exists

𝒜2Mη2|h|2+2MH2η2h2+𝒞2Mη2h2𝒟2Mh2|η|2, (23)

where

𝒜2:
=(n-1n-2-n-22C0ϕ2(η)2)-(a2+n-22b2C0ϕ2(η)2),
2:
=n-n2(n-2)2C0ϕ2(η)2,
𝒞2:
=2(n-2)-n2(n-2)2C0ϕ2(η)2

and

𝒟2:=1a2+n-22(1+1b2)C0ϕ2(η)2.

Since the total curvature ΦLn(M) is finite, we can choose a fixedr0 such that

ΦLn(M-Br0)<δ2=1n(n-2)C0.
𝒜~2:
=(n-1n-2-n-22C0δ22)-(a2+n-22b2C0δ22),
~2:
=n-n2(n-2)2C0δ22,
𝒞~2:
=2(n-2)-n2(n-2)2C0δ22

and

𝒟~2:=1a2+n-22(1+1b2)C0δ22.

Obviously, 𝒜~2, ~2, 𝒞~2 and 𝒟~2 are positive.Thus,

𝒜~2Mη2|h|2+~2MH2η2h2+𝒞~2Mη2h2𝒟~2Mh2|η|2, (24)

for any ηC0(M-Br0).Combining with Proposition 6, we get that

1C0
(M|ηh|2nn-2)n-2nM|(ηh)|2+n2M(H2+1)(ηh)2
(1+c2)Mη2|h|2+(1+1c2)Mh2|η|2+n2M(H2+1)η2h2, (25)

for any positive real number c2.By (24) and (25), we have

1C0(M|ηh|2nn-2)n-2n
(1+1c2+(1+c2)𝒟~2𝒜~2)Mh2|η|2+(n2-(1+c2)~2𝒜~2)MH2η2h2
+(n2-(1+c2)𝒞~2𝒜~2)Mη2h2. (26)

We choose a sufficient large c2 such that

n2-(1+c2)~2𝒜~2<0

and

n2-(1+c2)𝒞~2𝒜~2<0.

Then (26) implies that

(M(ηh)2nn-2)n-2nA~Mh2|η|2, (27)

for any ηC0(M-Br0), where A~ is a positive constant depending only on n.

By Case I and Case II, we have that

(M(ηh)2nn-2)n-2nA~Mh2|η|2, (28)

for any ηC0(M-Br0), where A~ is a positive constant depending only on n (n3).

Next, the proof follows standard techniques (after inequality (33) in Cavalcante et al. (2014 and uses a Moser iteration argument (lemma 11 in Li (1980)).We include a concise proof here for the sake of completeness.Choose r>r0+1 and ηC0(M-Br0) such that

{η=0 on Br0(M-B2r),η=1 on Br-Br0+1,|η|<c~ on Br0+1-Br0,|η|c~r-1on B2r-Br,

for some positive constant c~. Then (28) becomes that

(Br-Br0+1h2nn-2)n-2nA~Br0+1-Br0h2+A~r2B2r-Brh2.

Letting r and noting that hL2(M), we obtain that

(M-Br0+1h2nn-2)n-2nA~Br0+1-Br0h2. (29)

By Hölder inequality

Br0+2-Br0+1h2(Br0+2-Br0+1h2nn-2)n-2n(Br0+2-Br0+11n2)2n,

we get that

Br0+2h2(1+A~Vol(Br0+2)2n)Br0+1h2. (30)

Set

Ψ={|2-|Φ|2+32H2|,for n=3,|2(n-2)-n-22|Φ|2+nH2|,for n4.

Fix xM and take τC01(B1(x)). Proposition 7 implies that

hhα|h|2-Ψh2,

where

α={12,for n=3,1n-2,for n4.

Then, for p>2, there exists

Mτ2hp-1hαMτ2hp-2|h|2-Mτ2Ψhp.

That is,

-2B1(x)τhp-1τ,h
(α+(p-1))B1(x)τ2hp-2|h|2
-B1(x)τ2Ψhp. (31)

Note that

-2τhp-1τ,h
=-2hp2τ,τhp2-1h
1αhp|τ|2+ατ2hp-2|h|2.

Combining with (31), we obtain that

(p-1)B1(x)τ2hp-2|h|2B1(x)Ψτ2hp+1αB1(x)|τ|2hp. (32)

Combining Cauchy-Schwarz inequality with (32), we obtain that

B1(x)|(τhp2)|2B1(x)𝒜Ψτ2hp+|τ|2hp, (33)

where 𝒜=1p-1(p24+p2) and =(1+p2)+1α(p-1)(p24+p2).Choose f=τhp2 in Proposition 6. Combining with (33), we obtain that

(B1(x)(τhp2)2nn-2)n-22p𝒞B1(x)(τ2+|τ|2)hp, (34)

where 𝒞 depends on n and supB1(x)Ψ.Set pk=2nk(n-2)k and ρk=12+12k+1 for k=0,1,2,.Take a function τkC0(Bρk(x)) satisfying:

{0τk1,τk=1 on Bρk+1(x),|τk|2k+3.

Choosing p=pk and τ=τk in (34), we obtain that

(Bρk+1(x)hpk+1)1pk+1(𝒞pk4k+4)1pk(Bρk(x)hpk)1pk. (35)

By recurrence, we have

hLpk+1(B12(x))i=0kpi1pi4ipi(𝒞44)1pihL2(B1(x))𝒟hL2(B1(x)), (36)

where 𝒟 is a positive constant depending only on n, Vol(Br0+2) and supBr0+2Ψ. Letting k,we get

hL(B12(x))𝒟hL2(B1(x)). (37)

Now, choose yB¯r0+1 such that supBr0+1h2=h(y)2. Note that B1(y)Br0+2.(37) implies that

supBr0+1h2𝒟hL2(B1(y))2𝒟hL2(Br0+2)2. (38)

By (30), we have

supBr0+1h2hL2(Br0+1)2, (39)

where depends only on n, Vol(Br0+2) and supBr0+2Ψ.In order to show the finiteness of the dimension of H2(L2(M)), it suffices to provethat the dimension of any finite dimensional subspaces of H2(L2(M)) is bounded above by a fixed constant. Combining (39)with Lemma 11 in Li (1980), we show that dimH2(L2(M))<+.By Proposition 4, we obtain that the dimension of the second space of reduced L2cohomology of M is finite.

Remark 9. For the case of n=3, Theorem 1 can also be obtained by a different method. In fact, Yau (1976) proved that if ωH2(L2(M)), thenω is closed and coclosed. By use of the Hodge-* operator, we obtain the dimensions of H2(L2(M)) and H1(L2(M)) are equal.By Theorem 1.1 in Zhu and Fang (2014), we obtain the desired result.

ACKNOWLEDGMENTS

The author would like to thank professor Detang Zhou for useful suggestions.The work was partially supported by NSFC Grants 11471145, 11371309 and Qing Lan Project.

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Received: February 6, 2015; Accepted: June 17, 2016

* AMS Classification: 53C20, 53C40. E-mail: zhupeng2004@126.com