Obstructions to nonnegative scalar and mean curvatures

# Topological obstructions to nonnegative scalar curvature and mean convex boundary

Ezequiel Barbosa Universidade Federal de Minas Gerais (UFMG), Caixa Postal 702, 30123-970, Belo Horizonte, MG, Brazil  and  Franciele Conrado Instituto de Ciências Exatas-Universidade Federal de Minas Gerais
30161-970-Belo Horizonte-MG-BR
###### Abstract.

We study topological obstructions to the existence of a Riemannian metric on manifolds with boundary such that the scalar curvature is nonnegative and the boundary is mean convex. In particular, we construct many examples of compact manifolds with boundary which admit no Riemannian metric with nonnegative scalar curvature and mean convex boundary.

Partially supported by CNPq

## 1. Introduction

A central problem in modern differential geometry concerns the connection between curvatures and topology of a manifold. Consider, firstly, the case of surfaces. Let be a compact two-dimensional Riemannian manifold with nonempty boundary . The Gauss-Bonnet theorem states that

 ∫MKda+∫∂Mkgds=2πχ(M),

where denotes the Gaussian curvature, is the geodesic curvature of the boundary, is the Euler characteristic, is the element of area and is the element of length. Note that the topological invariant gives a topological obstruction to the existence of a certain type of Riemannian metrics on the surface . For instance, a compact surface with negative (nonpositive) Euler characteristic does not admit a Riemannian metric with nonnegative (positive) Gaussian curvature and nonnegative geodesic curvature.

In higher dimensions, the relationship between curvatures and the topology of a manifold is much more complicated. The scalar curvature of the manifold and the mean curvature of its boundary are commonly studied. A classical theorem due to Gromov [Grom2], assuming no control on the boundary, states that every compact manifold with non-empty boundary admits a Riemannian metric of positive sectional curvature. However, there are topological obstructions if one further imposes convexity restrictions on the boundary. For instance, it is a result of Gromoll [Grom1] that a compact Riemannian manifold of positive sectional curvature with non-empty convex boundary is diffeomorphic to the standard disc. However, positive sectional curvature is more restrictive than the escalar curvature. In dimension 3, a 3-dimensional manifold with positive Ricci curvature and convex boundary (positive definite second fundamental form) is diffeomorphic to a 3-ball. For the scalar curvature and mean curvature of the boundary, its not clear which topological obstructions can appear. In particular, the existence of compact, orientable and essential surface properly embedded in which is not a disk or a cylinder is a topological obstruction for existence of a metric with nonnegative scalar curvature and mean convex boundary in . For instance, define , where is a compact, connected and orientable surface with boundary which is not a disk or a cylinder. Note that is essential in . Hence, there is no metric on with nonnegative scalar curvature and convex boundary. However, if has an essential cylinder, can exists a such metric on . Again, for instance, consider the manifold with the product metric , where is the standard metric on and is the flat metric on . Such Riemannian manifold is flat with totally geodesic boundary. In this notes, we show that the existence of a compact, orientable and essential surface properly embedded in which is not a disk, is a topological obstruction for the existence of a metric with nonnegative scalar curvature and strictly mean convex boundary or with positive scalar curvature and mean convex boundary in . Furthermore, we show that the existence of an essential cylinder properly embedded in a compact and orientable -dimensional Riemannian manifold with nonnegative scalar curvature and mean convex boundary forces the metric to be flat with totally geodesic boundary. To be more precise, we prove the following result:

###### Theorem 1.1.

Let be a connected, compact and orientable 3-dimensional manifold with non-empty boundary . Assume that the connected components of are spheres or incompressible tori, but at least one of the components is a torus. Then there is no Riemannian metric on with nonnegative scalar curvature and strictly mean convex boundary or positive scalar curvature and mean convex boundary. In particular, if there exists a Riemannian metric on with nonnegative scalar curvature and mean convex boundary then should be flat with totally geodesic boundary.

As a direct consequence of the theorem above, we obtain that the 3-dimensional manifolds and have no metric with nonnegative scalar curvature and mean convex boundary, where is a torus minus an open disk, and is a closed, connected and orientable 3-dimensional manifold. Moreover, the 3-dimensional manifolds , , , , have no metric with nonnegative scalar curvature and strictly mean convex boundary or with positive scalar curvature and mean convex boundary. Also, let be a closed 3-dimensional manifold. Then the manifold has no metric with nonnegative scalar curvature and strictly mean convex boundary. If it has a metric with nonnegative scalar curvature and mean convex boundary, it should be flat with totally geodesic boundary.Thus, from this last claim, we can glue two copies of along the boundary and build a flat closed 3-dimensional manifold which is a connected sum of a 3-dimensional torus and a closed 3-dimensional manifold.

For high dimension , we want to prove that, if is a closed -dimensional manifold, then does not admit a metric of nonnegative scalar curvature and mean convex boundary and does not admit a metric of positive scalar curvature and mean convex boundary.

Finally, we study free-boundary -slicing in order to extend the results above for any dimension .

## 2. Preliminaries and Technical Results

Let , , be a compact orientable Riemannian manifold with non-empty boundary . Assume that and , where denote the mean curvature of with respect to the outward unit normal vector. For each Riemannian metric on consider and satisfying:

 ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩−Δ~gΦ~g+cnR~gΦ~g=λ(~g)Φ~g∂Φ~g∂ν~g=−2cnH~gΦ~g∫MΦ~gdv~g=1

where denote the outward unit normal vector of the boundary in and . Note that, as we are considering, we can assume that .

Moreover, note that

 λ(~g) = −∫MΔ~gΦ~gdv~g+cn∫MR~gΦ~gdv~g = −∫∂M∂Φ~g∂ν~gdσ~g+cn∫MR~gΦgdv~g.

Therefore,

 λ(~g)=2cn∫∂MΦ~gH~gdσ~g+cn∫MR~gΦ~gdv~g.
###### Lemma 2.1.

Let , , be a compact orientable Riemannian manifold with non-empty boundary such that and . If then

 Dλg(h)=−cn∫∂M⟨h,Bg⟩dσg−cn∫M⟨h,Ricg⟩dvg,

for every symmetric tensor .

###### Proof.

Firstly, note that implies that and . Let be a symmetric tensor. Consider for each a smooth family of Riemannian metrics on in a such way that e . Denote by

 λ(t):=λ(g(t)), R(t):=Rg(t)and H(t):=Hg(t).

As and , we obtain that

 Dλg(h)=λ′(0)=2cn∫∂MH′(0)dσg+cn∫MR′(0)dvg.

We divide the proof in some steps.

1. We have that

 R′(t)=−⟨h(t),Ricg(t)⟩+divg(t)(divg(t)(h(t))−d(trg(t) h(t))), h(t):=g′(t).

Hence,

 Dλg(h) = 2cn∫∂MH′(0)dσg−cn∫M⟨h,Ricg⟩dvg + cn∫Mdivg(divg(h)−d(trg h))dvg = 2cn∫∂MH′(0)dσg−cn∫M⟨h,Ricg⟩dvg + cn∫∂M⟨(divg(h))#−(d(trg h))#,ν⟩dσg = cn∫∂M(2H′(0)+X)dσg−cn∫M⟨h,Ricg⟩dvg,

where and .

Einstein convention and notation:

• Without a summation symbol, lower and upper index indicate a summation from to .

• denote the Riemannian connection of , ;

• denote the second fundamental form of in

Consider a local chart on such that is a local chart on and .

1. Computation of in .

We have that

 d(trg h)=n∑k=1∂k(n∑i,j=1gijhij)dxk and divg(h)=n∑k=1(divg(h))kdxk.

It follows that

 (d(trg h))#=n∑i,j,k,l=1glk∂k(gijhij)∂l,

and

 (divg(h))#=n∑k,l=1glk(divg(h))k∂l=n∑i,j,k,l=1glkgij(∇ih)jk∂l.

Thus,

 (divg(h))#−(d(trg h))#=n∑l=1⎧⎨⎩n∑i,j,k,=1(glkgij(∇ih)jk−glk∂k(gijhij))⎫⎬⎭∂l.

In , we get that and , for every . Hence,

 X = n∑i,j=1(gij(∇ih)jn−ν(gijhij)) = gij(∇ih)jn+ν(hnn)−ν(gijhij)−ν(hnn) = gij(∇ih)jn−ν(gij)hij−gijν(hij) = gij(∇ih)jn+gikgjlν(gkl)hij−gijν(hij) = gij(∇ih)jn+2gikgjl(Bg)kl(h)ij−gijν(hij) = gij(∇ih)jn+2⟨h,Bg⟩−gijν(hij)
1. Computation of :

We have . Hence­,

 H′(t) = ddt(gijt)(Bt)ij+gijtddt(Bt)ij = −giktgjlt(ht)kl(Bt)ij+tr (ddtBt) = −⟨h(t),B(t)⟩+tr (ddtBt).

Lets focus our attention on . We have that

 (Bt)ij=−gt(νgt,∇ti∂j).
 ⇒ddt∣∣∣t=0(Bt)ij=−h(ν,∇i∂j)−⟨ddt∣∣∣t=0(νgt),∇i∂j⟩−⟨ν,ddt∣∣∣t=0(∇ti∂j)⟩,

where

###### Claim 1.

For every , we obtain that

 2⟨ddt∣∣∣t=0(∇tXY),Z⟩=(∇Xh)(Y,Z)+(∇Yh)(X,Z)−(∇Zh)(X,Y).

It follows from the claim 1 that

 ddt∣∣∣t=0(Bt)ij = −h(ν,∇i∂j)−⟨ddt∣∣∣t=0(νgt),∇i∂j⟩−12(∇ih)jn −12(∇jh)in+12(∇νh)ij.
###### Claim 2.

In ,

 ddt∣∣∣t=0(νgt)=−gklhnk∂l−12hnnν.
###### Proof.

In , we have and , for all and . Thus,

 0=ddt∣∣∣t=0(gt)nk=hnk+⟨ddt∣∣∣t=0(νgt),∂k⟩,

and

 0=ddt∣∣∣t=0(gt)nn=hnn+2⟨ddt∣∣∣t=0(νgt),ν⟩.

Moreover,­ for all , we have that

 ⟨ddt∣∣∣t=0(νgt),ν⟩=−12hnn,

and

Denote by

 ddt∣∣∣t=0(νgt)=n∑l=1al∂l.

Note that

Also,

 −hnk=⟨ddt∣∣∣t=0(νgt),∂k⟩=n−1∑i=1aigki, ∀k=1,⋯,n−1.
 ⇒al=−glkhnk, ∀l=1,⋯,n−1.

Hence,

 ddt∣∣∣t=0(νgt)=n−1∑l=1al∂l+anν=−glkhnk∂l−12hnnν.

It follows from the claim 2 that

 ⟨ddt∣∣∣t=0(νgt),∇i∂j⟩ = −glkhnk⟨∇i∂j,∂l⟩−12hnn⟨∇i∂j,ν⟩ = −glkhnkΓmijgml+12hnn(Bg)ij = −hnkΓkij+12hnn(Bg)ij

However,

 −h(∇i∂j,ν)=−hnkΓkij−hnnΓnij=(Bg)ijhnn−hnkΓkij,

since

 Γnij=12n∑k=1gnk{∂igjk+∂jgik−∂kgij}=−12ν(gij)=−(Bg)ij.

It implies that

 ⟨ddt∣∣∣t=0(νgt),∇i∂j⟩ = −h(∇i∂j,ν)−(Bg)ijhnn+12hnn(Bg)ij = −h(∇i∂j,ν)−12hnn(Bg)ij.

Hence,

 ddt∣∣∣t=0(Bt)ij=−12(∇ih)jn−12(∇jh)in+12(∇νh)ij+12hnn(Bg)ij.

Consequently,

 tr (ddt∣∣∣t=0(Bt))=−gij(∇ih)jn+12gij(∇νh)ij+12hnnHg.

As , we obtain that

 2H′(0) = −2⟨h,Bg⟩+2tr (ddt∣∣∣t=0(Bt)) = −2⟨h,Bg⟩−2gij(∇ih)jn+gij(∇νh)ij = −2⟨h,Bg⟩−2gij(∇ih)jn+gijν(hij)−2gijh(∇iν,∂j).
###### Claim 3.

In ,

 gijh(∇iν,∂j)=⟨h,Bg⟩.
###### Proof.

Denote by

 ∇iν=n∑k=1Γkin∂k.

Note that, in , we have

 Γnin=0 e Γkin=gmk(Bg)im, ∀k=1,⋯,n−1.
 ⇒∇iν=gmk(Bg)im∂k

Then, in , we obtain that

 gijh(∇iν,∂j)=gijgmk(Bg)imhkj=⟨h,Bg⟩.

It follows from the claim 3 that

 2H′(0)=−4⟨h,Bg⟩−2gij(∇ih)jn+gijν(hij).

Therefore,

 (2.1) 2H′(0)+X|∂M=−2⟨h,Bg⟩−gij(∇ih)jn.
###### Claim 4.

In , we have

 gij(∇ih)jn=−⟨h,Bg⟩+div∂Mg(ω),

for some

###### Proof.

It follows from the claim 3 that, in ,

 gij(∇ih)jn = gij∂i(hjn)−gijh(∇i∂j,ν)−gijh(∇iν,∂j) = gij∂i(hjn)−gijh(∇i∂j,ν)−⟨h,Bg⟩.

For , in , we can write

 ∇i∂j=(Bg)ijν+¯¯¯¯¯∇i∂j,

where is the Riemannian connection of

Hence, since , we obtain that

 gijh(∇i∂j,ν)=hnnHg+gijh(¯¯¯¯¯∇i∂j,ν)=gijh(¯¯¯¯¯∇i∂j,ν).

This implies that, in ,

 gij(∇ih)jn=gij∂i(hjn)−gijh(¯¯¯¯¯∇i∂j,ν)−⟨h,Bg⟩.

Define as

 ω:=h(.,ν)|∂M.

Note that

 div∂Mg(ω) = gij(¯¯¯¯¯∇iω)j=gij∂i(ωj)−gijω(¯¯¯¯¯∇i∂j) = gij∂i(hjn)−gijh(¯¯¯¯¯∇i∂j,ν).

Then, in ,

 gij(∇ih)jn=−⟨h,Bg⟩+div∂Mg(ω).

It follows from (2.1) and 4 that

 2H′(0)+X|∂M=−⟨h,Bg⟩−div∂Mg(ω).

Hence,

 Dλg(h)=−cn∫∂M⟨h,Bg⟩dσg−cn∫M⟨h,Ricg⟩dvg−cn∫∂Mdiv∂Mg(ω)dσg.

We conclude, since is a closed manifold, that

 Dλg(h)=−cn∫∂M⟨h,Bg⟩dσg−cn∫M⟨h,Ricg⟩dvg.

###### Proposition 2.2.

Let , , be a compact orientable Riemannian manifold with boundary such that , and . The metric is a critical point of the functional if and only if is Ricci flat with totally geodesic boundary.

###### Remark 2.3.

Let , , be a compact orientable Riemannian manifold with boundary. Define the following pair of operators acting in :

 ⎧⎪⎨⎪⎩Lg=−Δgφ+cnRgφ in MTg=∂φ∂ν+2cnH∂Mgφ on ∂M

where denotes the outward unit normal vector of the boundary in and . Consider the first eigenvalue of with boundary condition :

 (2.2) {Lg(φ)=λ1(M,g)φ in MTg(φ)=0 on ∂M

We have that,

 λ1(M,g)=inf0≢φ∈H1(M)∫M(|∇gφ|2+cnRgφ2)dvg+2cn∫∂MH∂Mgφ2dσg∫Mφ2dvg.

We can choose a positive function solution of (2.2). The conformal metric is such that

 {Rh=λ1(M,g)φ−4n−2 in MH∂Mh≡0 on ∂M

In particular, this implies that if then and .

###### Proposition 2.4.

Let , , be a compact orientable Riemannian manifold with boundary such that and . Then either admits a metric with positive scalar curvature and minimal boundary or is Ricci flat with totally geodesic boundary.

###### Proof.

We can assume that . Hence,­ we obtain . If , then there exists a metric on with positive scalar curvature and minimal boundary (See remark 2.3).

Then, assume that . If , we have that is a critical point of the functional . Consequently, and . If , there exists a symmetric tensor such that . Consider a family of metrics on , , Since , we obtain that there exists such that the function is an increase function. Since , we get that for all . Therefore, for each there is a metric on such that and (See remark 2.3). ∎

###### Proposition 2.5.

Let , , be a compact orientable Riemannian manifold with non-empty boundary such that and . Then every free-boundary stable minimal hypersurface in has a metric with positive scalar curvature and minimal boundary.

###### Proof.

Consider a compact orientable free-boundary stable minimal in . It follows from the second variation formula for the volume that

 ∫Σ|∇φ|2dvg≥∫Σφ2(Ricg(N,N)+|BΣg|2)dvg+∫∂Σφ2B∂Mg(N,N)dσg

for every , where denote a unit vector field on in . As , it follows from the Gauss equation that

 Ricg(N,N)+|BΣg|2=12(Rg−RΣg+|BΣg|2)>−12RΣg.

Hence

 ∫Σ|∇φ|2dvg>−12∫Σφ2RΣgdvg+∫∂Σφ2B∂Mg(N,N)dσg

for every . Since , and teis a free-boundary hypersurface in , we obtain

 B∂Mg(N,N)=H∂Mg−H∂Σg≥−H∂Σg.

Thus,

 ∫Σ|∇φ|2dvg>−12∫Σφ2RΣgdvg−∫∂Σφ2H∂Σgdσg

for every . Consequently,

 ∫Σ|∇φ|2dvg+cn∫Σφ2RΣgdvg+2cn∫∂Σφ2H∂Σgdσg>(1−2cn)∫Σ|∇φ|2dvg

for every