Foliation by free boundary constant mean curvature leaves

# Foliation by free boundary constant mean curvature leaves

Fabio Montenegro
###### Abstract

Let be a Riemannian manifold of dimension with smooth boundary and . We prove that there exists a smooth foliation around whose leaves are submanifolds of dimension , constant mean curvature and its arrive perpendicular to the boundary of M, provided that is a nondegenerate critical point of the mean curvature function of .

## 1 Introduction

The strategy of the proof of this result was inspired by [2]. In this work, Rugang Ye considered the foliation by geodesic spheres around of small radius and showed that this foliation can be perturbed into a foliation whose leaves are spheres of constant mean curvature, provided that is a nondegenerate critical point of the scalar curvature function of . So we are going to consider a family of foliations whose leaves are submanifolds of with boundary contained in and it’s arriving perpendicular to the boundary of . The idea is then to perturb each leaves to obtain, via implicit function theorem, a foliation whose leaves are hemispheres of constant mean curvature and its arrive perpendicular to the boundary of M, provided that is a nondegenerate critical point of the mean curvature function of .

We refer to [1] for basic terminology in local Riemannian geometry. Let be an (n+1)-dimensional Riemannian manifold with smooth boundary , . We will denote by and the covariant derivatives and by and the full Riemannian curvature tensor of and , respectivily. The trace of second fundamental form of the boundary will be denoted by . We will make use of the index notation for tensors, commas denoting covariant differentiation and we will adopt the summation convention.

###### Definition 1.1

Let denote the inward unit normal vector field along . Let be a submanifold with boundary contained in . The unit conormal of that points outside will be denoted by . is called free boundary when on .

Acknowledgements. Part ot this work was done while the author was visiting the university of Princeton, New Jersey. I would like to thank Fernando C. Marques for his kind invitation and helpful discussions on the subject.

## 2 Fermi Coordinate System

Consider a point and an orthonormal basis for . Let be the open ball in . There are and for which we can define the Fermi coordinate system centered at , , given by

 φ0(x,t)=expMϕ0(x)(tT(x)) (1)

where is the normal coordinate system in centered at and is the inward unit vector normal to the boundary at .

For each we will consider the Fermi coordinate system centered at

 c(τ)=exp∂Mp(τiei),

which we denote by , and it defined by

 φτ(x,t)=expMϕτ(x)(tT(x)) (2)

where

 ϕτ(x)=expc(τ)(xieτi),

are the parallel transport of to along the geodesic in , and is the inward unit vector normal to the boundary at .

We will denote the metric tensor of by , the coefficients of in the coordinates system by , and . The expansion of (up to fourth order) in Fermi coordinates can be found in [1, p.1604].

###### Lemma 2.1

In Fermi coordinates centered at we have , , and

 gijτ(x,t)=δij+2hijt (3)
 +13¯Rikjlxkxl+2hij,k––txk+(Rtitj+3hikhkj)t2+16¯Rikjl,m––xkxlxm+(23Symij(¯Rikmlhmj)+hij,kl––)txkxl+(Rtitj,k––+6Symij(hil,k––hlj))t2xk+(13Rtitj,t+83Symij(Rtitkhkj)+4hikhklhlj)t3+(120¯Rikjl,mp––––+115¯Rikql¯Rjmqp)xkxlxmxp+(13Symij(¯Rilpm,k––hpj)+23Symij(¯Rikplhpj,m––)+13hij,klm–––––)txkxlxm
 +(12Rtitj,kl––+13Symij(¯RikmlRtmtj)+73Symij(¯Rikmlhpj)hmp−43Sympj(¯Rpkmlhmj)hip−43Symip(¯Rikmlhmp)hpj+43¯Rmkplhimhpj+4Symij(him,kl––hmj)−12(himhmj),kl––+4him,k––hmj,l–)t2xkxl+(13Rtitj,tk––+83Symij(Rtitl,k––hlj)+83Symij(Rtitlhlj,k––)+8hjl,k––hlmhim+4hml,k––hljhim)t3xk+(112Rtitj,tt−13Rtitk,tRtktj,t+RtitkRtktj+6Symij(Rtitkhlj)hkl+56Symij(Rtitk,thkj)−83Symlj(Rtktjhlk)hil−83Symil(Rtktlhik)hlj+133Rtktlhikhlj+5hikhklhlmhmj)t4+a1ijlkm(τ,x,t)xixjxkxlxm+a2ijlk(τ,x,t)txixjxkxl+a3ijl(τ,x,t)t2xixjxk+a4ij(τ,x,t)t3xixj+a5i(τ,x,t)t4xi+a6(τ,t)t5

where every coefficient is computed at and , , , , , and are smooth functions. Here underlined indices mean covariant differentiation as tensor on the boundary.

Proof. We write the expansion

 gτ=(gτij)=I+C1+C2+C3+C4+O(r5)

where is the identity matrix and is homogeneous of degree .

For small , we have

 g−1τ=∞∑j=0(−1)j(gτ−I)j

and so

 g−1τ=I−C1+(C21−C2)+(2C1C2−C3−C31)
 +(C41−C4+C22+2C1C3−3C21C2)+O(r5).

The expansion formula (3) now follows immediately by applying the expansion of in [1, p.1604].

## 3 Pertubation by free boundary submanifods

We will work with the following set of functions

 C2,αT(Sn+)={φ∈C2,α(Sn+);∂φ∂et=0in∂Sn+} (4)

where .

For we define

 S+φ={(1+φ(x,t))(x,t);(x,t)∈Sn+}

and

 Sr,τ,φ=φτ(αr(S+φ)) (5)

where is the dilation for and sufficiently small such that

 αr0(B2×[0,2))⊂Brp/2×[0,tp).

There are numbers and such that is an embedded hypersurface in for any and . In addition is a free boundary submanifold of , this is, and its arrive perpendicular to the boundary of M, because . We denote the inward mean curvature function of by .

For we denote the inward mean curvature of the surface at with respect to the metric on , given by , here is the metric tensor in .

For each we have

 H(r,τ,φ)(x,t)=rh(r,τ,φ)(φτ(r(1+φ(x,t))(x,t))). (6)

For and we have

 ds2τ,r(x,t)(v,w)=1r2φ∗r(ds2)(rx,rt)(dαr(x,t)v,dαr(x,t)w)
 =φ∗r(ds2)(rx,tx)(v,w).

But, by the Lemma 2.2 in [1, p.1604],

 ds2τ,r(x,t)(v,w)=⟨v,w⟩Rn+1+O(r)

with when . One readily checks extends smoothly to the euclidean metric when goes to zero. Hence also extends to . Then by a straightforward computation the inward mean curvature function of at with respect to the metric on , can be written as

 H(r,τ,sφ)(¯x,¯t)=1Ψs(Δρ−sΔ¯¯¯¯φ−s22Δ¯¯¯¯φ2) (7)
 −1+s¯¯¯¯φΨ2s[∂Ψs∂t(t−s∂¯¯¯¯φ∂t)+∑i,jgijτ∂Ψs∂xi(xj−s∂¯¯¯¯φ∂xj)]

where , ,

 ¯¯¯¯φ(x,t)=φ⎛⎜ ⎜⎝x√t2+|x|2,t√t2+|x|2⎞⎟ ⎟⎠, (8)
 Ψs=Ψs(r,x,t)= (9)
  ⎷(t−s(1+s¯¯¯¯φ)∂¯¯¯¯φ∂t)2+∑i,jgijτ(rx,rt)(xi−s(1+s¯φ)∂¯¯¯¯φ∂xi)(xj−s(1+s¯¯¯¯φ)∂¯¯¯¯φ∂xj)

and is the standard Laplace operator on relative to the metric .

###### Lemma 3.1

We have

 H(r,τ,0)(x,t)=n+[hτiit−(n+3)hτijtxixj]r+[3n+22hτijhτklt2xixjxkxl (10)
 −(n+4)hτij,k––txixjxk+(−n+42Rtitj−3n+202hτikhτkj−hτijhτkk)t2xixj+13¯Rkikjxixj+2hτji,j–txi+2(hτij)2t2]r2+[∫10(1−η)22Hrrr(ηr,τ,0)dη]r3

where every coefficient is computed at .

###### Corollary 3.2

The following holds true

 H(0,τ,0)=limr→0H(r,τ,0)=n.

Now we consider as a mapping from into and let denote the differential of with respect to . In order to calculate we consider the variation of by smooth maps given by . For each we denote . Note that is an embedded in with . We will denote by a unit vector field normal to and the mean curvature of . We decompose the variational vector field

 ∂s=φ(x,t)(x,t)=∂Ts+vsNs

where is the function on defined by .

By the Proposition 16 in [3, p.14] we have

 Hφ(r,τ,0)φ=(∂sH(r,τ,sφ))∣∣s=0=dH(r,τ,0)(∂T0)−Lr,τv0

and

 ∂sds2τ,r(Ns,et)∣∣s=0=−∂v0∂et+ds2τ,r(N0,∇N0et)v0

where is the Jacobi operator.

In particular

 Hφ(0,τ,0)φ=Lφ:=−(ΔSn++n)φ, (11)

where is the standard Laplace operator in .

###### Lemma 3.3

We have

 Hφr(0,τ,0)φ=2∂¯¯¯¯φ∂xihτijxj[(xn+1)3+(xn+1)2+nxn+1]−2∂2¯¯¯¯φ∂xi∂xjhτijxn+1 (12)
 +∂¯¯¯¯φ∂t(hτii+∂thτijxixj)+(ΔSn+φ+3φ)hτijxixjxn+1−φhτijxixj(xn+1)2

and

 Hφφ(0,τ,0)φφ=2nφ2−(n−2)(∂¯¯¯¯φ∂t)2−(n−2)∑i(∂¯¯¯¯φ∂xi)2 (13)

where was defined in (8).

The Jacob operator

 L:C2,αT(Sn+)→C0,α(Sn+)

has an -dimensional kernel consisting of first order spherical harmonic functions , , which satisfy

 ∂∂etxi∣∣∣Sn+=0in∂Sn+.

In addition we have the -decompositions of spaces and . Let P denote the orthogonal projection from onto , and be the isomorphism sending to , the th coordinate basis. Define , that is,

 ~P(f)=2wn+1(∫Sn+fxi)ei

because

 ∫Sn+xixj=wn+12δij,

where .

###### Lemma 3.4

We have

 ~P(H(r,τ,0))=−2wnr2(n+2)wn+1hτjj,i–ei+O(r3). (14)

Proof. From Lemma 3.1 and the fact that we have

 ~P(H(r,τ,0))=[−(n+4)hτij,k––~P(xn+1xixjxk)+2hτji,j–~P(xn+1xi)]r2+O(r3)

where

 ~P(xn+1xixjxk)=2(δijδkl+δikδjl+δilδjk)wn(n+2)(n+4)wn+1el

and

 ~P(xn+1xi)=wnδijwn+1(n+2)ej.

Hence (14) follows.

###### Lemma 3.5

If is the solution of the Neumann problem

 ⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩−(ΔSn+φ+nφ)=hτiixn+1−(n+3)hτijxixjxn+1∂φ∂en+1=0on∂(Sn+) (15)

then

 ~P(Hφr(0,τ,0)φτ)=0 (16)

and

 ~P(Hφφ(0,τ,0)φτφτ)=0. (17)

for .

Proof. The function is the solution of the problem (15). Now we can use (12), (13) and the fact

 ~P(xi1…xi2kxn+1)=0,

for all interge , to prove (16) and (17).

## 4 Main Theorem

###### Definition 4.1

Consider and let be a neighborhood of on . A smooth codimension 1 foliation of for a neighborhood of is called a free boundary foliation centered at , provided that its leaves are all closed and free boundary.

###### Theorem 4.2

If is a nondegenerate critical point of the mean curvature function of , then there exist and smooth functions and with such that for all . Hence the family is a smooth family of constant mean curvature spheres with having mean curvature . Furthermore is a free boundary foliation centered at p.

Proof. We will use the Taylor’s formula with integral remainder

 H(r,τ,rφ)=n+[Hφ(0,τ,0)φ+Hr(0,τ,0)]r
 +[12Hφφ(0,τ,0)φφ+Hφr(0,τ,0)φ+12Hrr(0,τ,0)]r2+R(r,τ,φ)r3

where

 R(r,τ,φ)=∫10(1−η)Hφrr(ηr,τ,0)φdη+12∫10Hφφr(ηr,τ,0)φφdη
 +12∫10(1−η)2Hφφφ(r,τ,ηφ)φφφdη.

We are interested in solving the equation , but first we are going to treat the equation

 P⊥(H(r,τ,rφ)−n)=0, (18)

where denotes the orthogonal projection from onto .

By (11) and the fact we can write the equation in (18) as follows (after division by )

 Lφ+Hr(0,τ,0)+¯R(r,τ,φ)r=0,

where

 ¯R(r,τ,φ)=12P⊥(Hφφ(0,τ,0)φφ)+P⊥(Hφr(0,τ,0)φ)+P⊥(12Hrr(0,τ,0))
 +P⊥(R(r,τ,φ))r.

Consider the mapping given by

 G(r,τ,φ)=Lφ+Hr(0,τ,0)+¯R(r,τ,φ)r,

where .

For , let be a solution of the equation

 Lφ0+Hr(0,0,0)=0.

One sees that and is a bounded invertible linear transformation. By the implicit function theorem we can solve for a function , for some , with . Furthermore

 φr(0,0)=−Gφ(0,0,φ0)−1Gr(0,0,φ0)
 =−(−Δ−n)−1∂∂r(¯R(r,τ,φ)r)|r=0,φ=φ0=−(−Δ−n)−1¯R(0,0,φ0)

where

 ¯R(0,0,φ0)=12Hφφ(0,0,0)φ0φ0+Hφr(0,0,0)φ0.

Since

 Lφ(r,τ)+Hr(0,τ,0)+O(r)=0,

we have, for ,

 Lφ(0,τ)+Hr(0,τ,0)=0.

Then, by Lemma 3.5,

 ~P(Hφr(0,τ,0)φ(0,τ))=~P(Hφφ(0,τ,0)φ(0,τ)φ(0,τ))=0.

On the other hand,

 φ(r,τ)=φ(0,τ)+r∫10φr(ηr,τ)dη,

so that

 (19)

and