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Gluing Harmonic Maps

Shaozong Wang
Abstract.

In this paper, we consider harmonic maps from closed, two-dimensional Riemannian manifolds into a closed, Riemannian target manifold of dimension two or higher. We develop a gluing theory for such harmonic maps. In addtion, we develop the properties of this gluing map and apply them to the phenomenon of energy bubbling.

1. Introduction

There is extensive work on gluing in various contexts. Taubes [26] [25] [24] discussed gluing for anti-self-dual (ASD) connections and Yang-Mills connections. Gluing for ASD connections has also been explored by Donaldson [6], Mrowka [19], Feehan and Leness [8]. Gluing for Seiberg-Witten monopoles is discussed by Frøyshov [9] and Parker [20], and for Non-Abelian monopoles by Feehan and Leness [7]. Brendle [3] discussed gluing for yang-Mills connections. Brendle and Kapouleas [4] studied the gluing method for Eguchi-Hanson metrics. Additionally, gluing in the context of pseudoholomorphic curves is covered in Fukaya [10], Abouzaid [1], McDuff and Salamon [15], as well as McDuff and Wehrheim [17] [18]. Malchiodi, Rupflin, and Sharp [14] and Rupflin [22] considered gluing for almost-harmonic maps.

In this paper we consider two harmonic maps, f1:Σ1Nf_{1}:\Sigma_{1}\rightarrow N and f2:Σ2Nf_{2}:\Sigma_{2}\rightarrow N, where Σ1\Sigma_{1} and Σ2\Sigma_{2} are Riemannian manifolds of dimension 2 and NN is a Riemannian manifold. We consider under what conditions can we glue these two maps and, when the gluing map exists, what properties does the gluing map possess. In particular, we show surjectivity of the gluing map, and thus in the bubbling sequence will lie in the image of the gluing map.

1.1. Gluing of Manifolds and Maps

Loosely speaking, we are given two compact Riemannian manifolds of dimension 2, Σ1\Sigma_{1} and Σ2\Sigma_{2}, a Riemannian manifold NN, and two harmonic maps f1:Σ1Nf_{1}:\Sigma_{1}\rightarrow N and f2:Σ2Nf_{2}:\Sigma_{2}\rightarrow N. Suppose f1(x1)=f2(x2)f_{1}(x_{1})=f_{2}(x_{2}), then we connect Σ1\Sigma_{1} and Σ2\Sigma_{2} by punching holes at x1x_{1} and x2x_{2}, then gluing them by a neck. Denote this by Σ1#δ,RΣ2\Sigma_{1}\#_{\delta,R}\Sigma_{2}, where δ\delta and RR are parameters of neck. We glue the two maps using cutoff functions.

Here are the details: Consider gluing f0f^{0} and ff^{\infty} whose domains are both S2S^{2}. We denote

y:=f0(0)=f()y:=f^{0}(0)=f^{\infty}(\infty)

Since f0f^{0} and ff^{\infty} are harmonic maps, we know they are smooth. Since S2S^{2} is compact, we know there exists c>0c>0 such that df0Lc\|df^{0}\|_{L^{\infty}}\leq c and dfLc\|df^{\infty}\|_{L^{\infty}}\leq c (Here the norms are in the sense of the round metric).

Let ϵ\epsilon be less than the injective radius of NN. By the upper bound of the differential of the harmonic maps, we can compute that when cotϵc<|z|<tanϵc\cot\frac{\epsilon}{c}<|z|<\tan\frac{\epsilon}{c}, we have d(f0(z),y)<ϵd(f^{0}(z),y)<\epsilon and d(f(z),y)<ϵd(f^{\infty}(z),y)<\epsilon. Thus there exists ζ0(z),ζ(z)TyN\zeta^{0}(z),\zeta^{\infty}(z)\in T_{y}N such that f0(z)=expy(ζ0(z))f^{0}(z)=\exp_{y}(\zeta^{0}(z)) and f(z)=expy(ζ(z))f^{\infty}(z)=\exp_{y}(\zeta^{\infty}(z)).

Consider some fixed nondecreasing smooth function ρ\rho satisfying:

ρ(z)={0,|z|11,|z|2\rho(z)=\left\{\begin{aligned} &0,|z|\leq 1\\ &1,|z|\geq 2\\ \end{aligned}\right.

We define our pre-gluing:

(1) fR(z):=fδ,R(z)={f0(z),|z|2δRexpy(ρ(δRz)ζ0(z)+ρ(δRz)ζ(R2z)),δ2R|z|2δRf(R2z),|z|δ2Rf^{R}(z):=f^{\delta,R}(z)=\left\{\begin{aligned} &f^{0}(z),&|z|\geq\frac{2}{\delta R}\\ &\exp_{y}\left(\rho(\delta Rz)\zeta^{0}(z)+\rho\left(\frac{\delta}{Rz}\right)\zeta^{\infty}(R^{2}z)\right),&\frac{\delta}{2R}\leq|z|\leq\frac{2}{\delta R}\\ &f^{\infty}(R^{2}z),&|z|\leq\frac{\delta}{2R}\end{aligned}\right.

We will need to specify which weighted norm we are using. For the reason mentioned in section 10.3 in [16], we use the same weight as in chapter 10 in [16]. Namely,

θR(z)={R2+R2|z|2|z|1R1+|z|2|z|1R\theta^{R}(z)=\left\{\begin{aligned} &R^{-2}+R^{2}|z|^{2}&|z|\leq\frac{1}{R}\\ &1+|z|^{2}\ &|z|\geq\frac{1}{R}\\ \end{aligned}\right.

and we let the metric be

(2) g=(θR)2(ds2+dt2)g=(\theta^{R})^{-2}(ds^{2}+dt^{2})

Note that similar setup can be done for general closed Riemannian manifolds of dimension 2.

1.2. Existence of the Gluing Map

Given Riemannian manifolds Σ\Sigma and NN, where Σ\Sigma is closed and of dimension 2. Consider a smooth map f:ΣNf:\Sigma\rightarrow N. For ξf1TN\xi\in f^{-1}TN, we can consider the perturbation of ff by ξ\xi under the exponential map, which we write as expf(ξ)\exp_{f}(\xi). We define Wf2,pW^{2,p}_{f} to be the space consisting all ξ\xi that is W2,pW^{2,p} in each coordinate chart, with the usual Sobolev space structure. We can define LpL^{p} spaces similarly.

We define the harmonic map operator \mathcal{F} from the space Wf2,pW^{2,p}_{f} to LfpL^{p}_{f} as

f(ξ):=Φf(ξ)1((g(expf(ξ))k+gαβ(ΓN)ijk(expf(ξ))(expf(ξ))αi(expf(ξ))βj)yk)\mathcal{F}_{f}(\xi):=\Phi_{f}(\xi)^{-1}\left((\triangle_{g}(\exp_{f}(\xi))^{k}+g^{\alpha\beta}(\Gamma^{N})^{k}_{ij}(\exp_{f}(\xi))(\exp_{f}(\xi))^{i}_{\alpha}(\exp_{f}(\xi))^{j}_{\beta})\frac{\partial}{\partial y^{k}}\right)\\

where gg denotes the metric on Σ\Sigma, and Φ\Phi is the parallel transport from expf(ξ)\exp_{f}(\xi) to ff along the geodesic expf(tξ)\exp_{f}(t\xi). See Section 2.1 for details. We will deal with a second order nonlinear system of PDEs.

First, we define the space we will be working with, which is a space of pairs of harmonic maps satisfying certain conditions. These conditions will allow us to carry out a construction similar to that in chapter 10 of [16].

Definition 1.1 (Space of Harmonic Map Pairs).

Fix domain manifolds (closed, of dimension 2, with Riemannian metric) Σ1\Sigma_{1} and Σ2\Sigma_{2} and target Riemannian manifold NN of dimension no less than 2. Fix a constant 1<p<21<p<2. Fix points x1Σ1x_{1}\in\Sigma_{1} and x2Σ2x_{2}\in\Sigma_{2}. Let (c,p)\mathcal{M}(c,p) denote all pairs of harmonic maps (f1,f2)(f_{1},f_{2}) such that

  1. (1)

    f1(x1)=f2(x2)f_{1}(x_{1})=f_{2}(x_{2})

  2. (2)

    dfiLc\|df_{i}\|_{L^{\infty}}\leq c and d2fiLc\|d^{2}f_{i}\|_{L^{\infty}}\leq c for i=1,2i=1,2.

  3. (3)

    for the harmonic map operator \mathcal{F}, Df1:=df1(0)D_{f_{1}}:=d\mathcal{F}_{f_{1}}(0) and Df2:=df2D_{f_{2}}:=d\mathcal{F}_{f_{2}} are both surjective.

  4. (4)

    Furthermore, let Wf1,22,p:={(ξ0,ξ)Wf12,p×Wf22,pξ0(0)=ξ()}W^{2,p}_{f_{1,2}}:=\{(\xi^{0},\xi^{\infty})\in W^{2,p}_{f_{1}}\times W^{2,p}_{f_{2}}\mid\xi^{0}(0)=\xi^{\infty}(\infty)\}. Df1×Df2:Wf1,22,pLf1p×Lf2pD_{f_{1}}\times D_{f_{2}}:W^{2,p}_{f_{1,2}}\rightarrow L^{p}_{f_{1}}\times L^{p}_{f_{2}} is surjective.

In this paper, we will need to specify the parameters of the neck in connected sum. To avoid repeating too much, we define the following symbol.

Definition 1.2 (Set of Parameter Pairs).

For any 0<δ0<10<\delta_{0}<1, we define 𝒜(δ0)\mathcal{A}(\delta_{0}) to be the set of all pairs of (δ,R)(\delta,R) such that 0<δ<δ00<\delta<\delta_{0}, δR>1/δ0\delta R>1/\delta_{0}.

We will show that for pairs of harmonic maps in definition 1.1, we can glue the pair and then perturb it to get another harmonic map. This is stated as the following theoerm, which is proved in Section 6:

Theorem 1.1 (Existence of Gluing Map).

Given any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p). There exists δ0=δ0(c,p,Σ1,Σ2,x1,x2,N,𝒰)>0\delta_{0}=\delta_{0}(c,p,\Sigma_{1},\Sigma_{2},x_{1},x_{2},N,\mathcal{U})>0, such that for each pair of (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), there exists a gluing map ıδ,R:𝒰W2,p(Σ1#δ,RΣ2,N)\imath_{\delta,R}:\mathcal{U}\rightarrow W^{2,p}(\Sigma_{1}\#_{\delta,R}\Sigma_{2},N) such that each element in ıδ,R(𝒰)\imath_{\delta,R}(\mathcal{U}) is a smooth harmonic map, where Σ1#δ,RΣ2\Sigma_{1}\#_{\delta,R}\Sigma_{2} is the glued manifold as defined in 2. Furthermore, for any σ>0\sigma>0, we can choose δ0=δ0(c,p,Σ1,Σ2,x1,x2,N,𝒰)\delta_{0}=\delta_{0}(c,p,\Sigma_{1},\Sigma_{2},x_{1},x_{2},N,\mathcal{U}) such that, for any (f1,f2)𝒰(f_{1},f_{2})\in\mathcal{U}, there exists ξRW2,p(Σ1#δ,RΣ2)\xi^{R}\in W^{2,p}(\Sigma_{1}\#_{\delta,R}\Sigma_{2}) satisfying

ıδ,R((f1,f2))=expfRξR,ξR2,p,R<σ\imath_{\delta,R}((f_{1},f_{2}))=\exp_{f^{R}}\xi^{R},\quad\|\xi^{R}\|_{2,p,R}<\sigma

where fRf^{R} denotes the pregluing of f1,f2f_{1},f_{2} defined in (1).

1.3. Surjectivity of the Gluing Map and Bubbling

Consider bubbling described in [21]. By derivative estimates and facts about the bubbling phenomenon, we will eventually show the following using the result in Section 8:

Theorem 1.2 (Bubble Lies in Image of Gluing Map).

Let Σ\Sigma and NN be closed Riemannian manifolds and the dimension of Σ\Sigma be 22, let {fν}:ΣN\{f_{\nu}\}:\Sigma\rightarrow N be a sequence of harmonic maps such that each bubble point only contains a single bubble. Denote the bubble points by {xi}\{x_{i}\} and the corresponding bubble map by hih_{i}. Denote the limit modulo bubble points by ff. Then there exists 1<p^=p^(N)<21<\hat{p}=\hat{p}(N)<2 such that, suppose for each ii, (f,hi)(c,p)(f,h_{i})\in\mathcal{M}(c,p) for some c>0c>0 and pp^p\leq\hat{p}. Then there exists δ0=δ0(c,Σ,N,f,{hi},{xi},p)\delta_{0}=\delta_{0}(c,\Sigma,N,f,\{h_{i}\},\{x_{i}\},p) such that the following holds:

Let ı\imath denote the gluing map of W2,p(Σ,f1TN)W^{2,p}(\Sigma,f^{-1}TN) and the W2,pW^{2,p} maps of the bubbles with (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}). Then there exists a subsequence of {fν}\{f_{\nu}\} that lies in the image of the gluing map ı\imath.

1.4. Outline

In Section 2, we introduce the setup of the problem. We define the harmonic map operator and compute the linearization. Then we explore the structure of the space of harmonic maps, which behaves locally like a smooth manifold. In Section 3, we introduce the implicit function theorem needed in the gluing construction. In Section 4, we explain how to choose the appropriate metric and the cutoff functions to glue the domain manifolds. In Section 5, we construct an approximate right inverse, and then get the exact right inverse using Newton iteration. In Section 6, we apply the implicit function theorem to find a way to perturb the preglued map into a harmonic map. In Section 7, we cite a way to ’suspend’ the harmonic map into a conformal harmonic map, which will be used in the proof of surjectivity. In Section 8, by derivative estimates and properties of conformal harmonic maps, we show that the gluing map constructed in Section 6 is locally surjective under certain conditions, and finally we will be able to apply this to the situation of bubbling to show that, under certain conditions, the bubble lies in the image of the gluing map. The appendix contains technical details including proof of Lemma 2.1, norm and derivative estimates, as well as related facts in elliptic PDE theory and functional analysis.

The framework for Sections 4, 5, and the first part of 8 is similar to chapter 10 of [16]. However, we are dealing with a second order nonlinear system of PDEs on Riemannian manifolds while in [16] they have a first order PDE on symplectic manifolds.

1.5. Acknowledgement

The author expresses deep gratitude to Professor Paul Feehan for his invaluable guidance and unwavering support, and to Professor Dan Ketover for his insightful comments and suggestions. Special thanks are also extended to Professor Jason Lotay, Zilu Ma, Gregory Parker, Junsheng Zhang, Xiao Ma, Liuwei Gong, and Jiakai Li for their helpful discussions. This work is also based in part on research supported by the National Science Foundation under Grant No. 1440140, while the author was in residence at the Simons Laufer Mathematical Sciences Institute in Berkeley, California, during Fall 2022 as an associate of the program Analytic and Geometric Aspects of Gauge Theory.

2. Basic Setup

2.1. The operator and its Linearization

Let’s follow chapter 10 in [16] and consider gluing harmonic maps on S2S^{2} (The Riemann sphere) or any Riemann manifolds of dimension 2. For simplicity, we consider the case when Σ1=S2,Σ2=S2\Sigma_{1}=S^{2},\Sigma_{2}=S^{2}. However, we will see that the gluing can be used for general Riemann manifolds of dimension 2.

First, let’s consider the general definition of harmonic maps. Suppose MM, NN are Riemannian manifolds and ff is a smooth map from the domain manifold MM to the target manifold NN. We say ff is a harmonic map if it satisfies the harmonic map equation. There are multiple ways of writing the harmonic map equation. One would be

(3) P(f):=gf+A(f)(df,df)P(f):=\triangle_{g}f+A(f)(df,df)

where we consider an isometric embedding NNN\subset\mathbb{R}^{N} and AA is the second fundamental from of the embedding.

However, the above will depend on the ambient Euclidean space. In particular, since we are using the implicit function theorem later, the above will make it difficult to utilize the surjectivity onto the tangent space. An alternate form would be

trg(df)=0tr_{g}(\nabla df)=0

In coordinates, this is

gαβ(fαβk(ΓM)αβγfγk+(ΓN)ijk(f)fαifβj)yk=0g^{\alpha\beta}\left(f^{k}_{\alpha\beta}-(\Gamma^{M})^{\gamma}_{\alpha\beta}f^{k}_{\gamma}+(\Gamma^{N})^{k}_{ij}(f)f^{i}_{\alpha}f^{j}_{\beta}\right)\frac{\partial}{\partial y^{k}}=0

where we are denoting fxα\frac{\partial f}{\partial x^{\alpha}} by fαf_{\alpha}. We know that the above is a well defined vector, independent of choice of coordinates on MM and NN. We can also verify this by directly computing the transformation law in different coordinates.

In this paper we use another form obtained from considering coordinate charts in MM and computing the Euler-Lagrange equation of the energy. See section 1.1 from [13] for details. The equation is

gfk+gαβ(ΓN)ijk(f)fαifβj=0\triangle_{g}f^{k}+g^{\alpha\beta}(\Gamma^{N})^{k}_{ij}(f)f^{i}_{\alpha}f^{j}_{\beta}=0

or, equivalently,

gαβ2fkxαxβ+gαβxαfkxβ+gαβ12detgdetgxαfkxβ+gαβ(ΓN)ijk(f)fαifβj=0g^{\alpha\beta}\frac{\partial^{2}f^{k}}{\partial x^{\alpha}\partial x^{\beta}}+\frac{\partial g^{\alpha\beta}}{\partial x^{\alpha}}\frac{\partial f^{k}}{\partial x^{\beta}}+g^{\alpha\beta}\frac{1}{2\det g}\frac{\partial\det g}{\partial x^{\alpha}}\frac{\partial f^{k}}{\partial x^{\beta}}+g^{\alpha\beta}(\Gamma^{N})^{k}_{ij}(f)f^{i}_{\alpha}f^{j}_{\beta}=0

We define the operator PP as

P(f):=(gfk+gαβ(ΓN)ijk(f)fαifβj)ykP(f):=(\triangle_{g}f^{k}+g^{\alpha\beta}(\Gamma^{N})^{k}_{ij}(f)f^{i}_{\alpha}f^{j}_{\beta})\frac{\partial}{\partial y^{k}}

It can be verified directly by coordinate change that the above is a tensor.

Next, we consider the linearization of the above differential operator. The idea is that, for a perturbation of ff and xMx\in M, the operator will spit out a vector that belongs to a different fiber in the tangent bundle. Thus we have to pull back the vector before comparing them.

Here are the details: Given ξW2,p(Σ,f1TN)\xi\in W^{2,p}(\Sigma,f^{-1}TN), we know that for 1<p<21<p<2, W2,pW^{2,p} is embedded into C0,γC^{0,\gamma} where γ=22p\gamma=2-\frac{2}{p}. We define the following:

(4) Φf(ξ):f1TN(expf(ξ))1TN\displaystyle\Phi_{f}(\xi):f^{-1}TN\rightarrow(\exp_{f}(\xi))^{-1}TN
(5) f(ξ):=Φf(ξ)1((g(expf(ξ))k+gαβ(ΓN)ijk(expf(ξ))(expf(ξ))αi(expf(ξ))βj)yk)\displaystyle\mathcal{F}_{f}(\xi):=\Phi_{f}(\xi)^{-1}\left((\triangle_{g}(\exp_{f}(\xi))^{k}+g^{\alpha\beta}(\Gamma^{N})^{k}_{ij}(\exp_{f}(\xi))(\exp_{f}(\xi))^{i}_{\alpha}(\exp_{f}(\xi))^{j}_{\beta})\frac{\partial}{\partial y^{k}}\right)
(6) Df:=df(0)\displaystyle D_{f}:=d\mathcal{F}_{f}(0)

We would like to compute DfD_{f}. Consider the path:

W2,p(Σ,N)\displaystyle\mathbb{R}\rightarrow W^{2,p}(\Sigma,N)
λexpf(λξ)\displaystyle\lambda\mapsto\exp_{f}(\lambda\xi)
Φf(λξ)f(λξ)=(g(expf(ξ))k+gαβ(ΓN)ijk(expf(ξ))(expf(ξ))αi(expf(ξ))βj)yk\displaystyle\Phi_{f}(\lambda\xi)\mathcal{F}_{f}(\lambda\xi)=(\triangle_{g}(\exp_{f}(\xi))^{k}+g^{\alpha\beta}(\Gamma^{N})^{k}_{ij}(\exp_{f}(\xi))(\exp_{f}(\xi))^{i}_{\alpha}(\exp_{f}(\xi))^{j}_{\beta})\frac{\partial}{\partial y^{k}}
Dfξ\displaystyle D_{f}\xi =ddλf(λξ)λ=0\displaystyle=\frac{d}{d\lambda}\mathcal{F}_{f}(\lambda\xi)\mid_{\lambda=0}
=λP(expf(λξ))\displaystyle=\nabla_{\lambda}P(\exp_{f}(\lambda\xi))

We use normal coordinates for f(x)Nf(x)\in N to get an explicit expression for DfD_{f}:

Consider {yi}\{\frac{\partial}{\partial y^{i}}\} a basis of TNTN in normal coordinates. Using ff we can view it as a basis of f1TNf^{-1}TN. P(f)Γ(f1TN)P(f)\in\Gamma(f^{-1}TN) is locally given by

(gfi+gαβΓjli(f(x))flxαfjxβ)yi(\triangle_{g}f^{i}+g^{\alpha\beta}\Gamma_{jl}^{i}(f(x))\frac{\partial f^{l}}{\partial x^{\alpha}}\frac{\partial f^{j}}{\partial x^{\beta}})\frac{\partial}{\partial y^{i}}

where Γ\Gamma is the Christoffel symbol on NN.

In normal coordinates, expf(ξ)\exp_{f}(\xi) is just f+ξf+\xi. We have

P(expf(λξ))=(g(fi+λξi)+gαβΓjli(f(x)+λξ)(fj+λξj)xα(fl+λξl)xβ)yiP(\exp_{f}(\lambda\xi))=(\triangle_{g}(f^{i}+\lambda\xi^{i})+g^{\alpha\beta}\Gamma_{jl}^{i}(f(x)+\lambda\xi)\frac{\partial(f^{j}+\lambda\xi^{j})}{\partial x^{\alpha}}\frac{\partial(f^{l}+\lambda\xi^{l})}{\partial x^{\beta}})\frac{\partial}{\partial y^{i}}

We denote (g(fi+λξi)+gαβΓjli(f(x)+λξ)(fj+λξj)xα(fl+λξl)xβ)(\triangle_{g}(f^{i}+\lambda\xi^{i})+g^{\alpha\beta}\Gamma_{jl}^{i}(f(x)+\lambda\xi)\frac{\partial(f^{j}+\lambda\xi^{j})}{\partial x^{\alpha}}\frac{\partial(f^{l}+\lambda\xi^{l})}{\partial x^{\beta}}) by ϕi\phi_{i} and yi\frac{\partial}{\partial y^{i}} by eie_{i}. Since we are using normal coordinates, we have

ξ(ϕiei)\displaystyle\nabla_{\xi}(\phi_{i}e_{i}) =(ξϕi)ei\displaystyle=(\nabla_{\xi}\phi_{i})\cdot e_{i}
=(gξi+gαβsΓjliξsfjxαflxβ)ei\displaystyle=\left(\triangle_{g}\xi^{i}+g^{\alpha\beta}\partial_{s}\Gamma_{jl}^{i}\xi^{s}\frac{\partial f^{j}}{\partial x^{\alpha}}\frac{\partial f^{l}}{\partial x^{\beta}}\right)e_{i}

Thus we get

(7) Dfξ=(gξi+gαβsΓjli(f(x))ξsfjxαflxβ)eiD_{f}\xi=\left(\triangle_{g}\xi^{i}+g^{\alpha\beta}\partial_{s}\Gamma_{jl}^{i}(f(x))\xi^{s}\frac{\partial f^{j}}{\partial x^{\alpha}}\frac{\partial f^{l}}{\partial x^{\beta}}\right)e_{i}

Note that in the above we are only able to compute the exact form of the operators when considering the normal coordinates at exactly that point. For certain estimates, we will need to consider the operator’s coordinate form in a single coordinate chart. When we change the coordinates, it will be different for every point in the neighborhood, so we will need some uniform bound on the change of coordinates.

Lemma 2.1 (Uniform Boundedness of Coordinate Change).

Given a Riemannian manifold NN, for any p0Np_{0}\in N, consider the normal coordinates centered at p0p_{0}, which we denote by yy, there exists a neighborhood UU of p0p_{0} and a corresponding family of normal coordinates {yp}pU\{y_{p}\}_{p\in U} such that each coordinate ypy_{p} is a normal coordinate at pUp\in U. Furthermore, any function of the form |𝛂|yyp𝛂\frac{\partial^{|\boldsymbol{\alpha}|}y}{\partial y_{p}^{\boldsymbol{\alpha}}} and |𝛂|ypy𝛂\frac{\partial^{|\boldsymbol{\alpha}|}y_{p}}{\partial y^{\boldsymbol{\alpha}}} has bounded C0C^{0} norm on UU only depending on NN and p0p_{0}.

The proof is in the appendix.

Using (7) and lemma 2.1, we get (note that above we are considering a different coordinate system at every point, but below we locally use the same coordinate chart):

(8) Dfξ=(gξk+F1k(f,fx,2fx2,ξ)+F2k(f,fx,ξx))ek\displaystyle D_{f}\xi=\left(\triangle_{g}\xi^{k}+F_{1}^{k}\left(f,\frac{\partial f}{\partial x},\frac{\partial^{2}f}{\partial x^{2}},\xi\right)+F_{2}^{k}\left(f,\frac{\partial f}{\partial x},\frac{\partial\xi}{\partial x}\right)\right)e_{k}

where

F1k(f,fx,2fx2,ξ)=\displaystyle F_{1}^{k}\left(f,\frac{\partial f}{\partial x},\frac{\partial^{2}f}{\partial x^{2}},\xi\right)= gαβ(3y~iysylymflxαfmxβ+2y~iylys2flxαxβ)yky~iξs+\displaystyle g^{\alpha\beta}\left(\frac{\partial^{3}\tilde{y}^{i}}{\partial y^{s}\partial y^{l}\partial y^{m}}\frac{\partial f^{l}}{\partial x^{\alpha}}\frac{\partial f^{m}}{\partial x^{\beta}}+\frac{\partial^{2}\tilde{y}^{i}}{\partial y^{l}\partial y^{s}}\frac{\partial^{2}f^{l}}{\partial x^{\alpha}\partial x^{\beta}}\right)\frac{\partial y^{k}}{\partial\tilde{y}^{i}}\xi^{s}+
gαβxα2y~iylysfsxβyky~iξl+gαβdetgxα12detg2y~iylysfsxβyky~iξl+\displaystyle\frac{\partial g^{\alpha\beta}}{\partial x^{\alpha}}\frac{\partial^{2}\tilde{y}^{i}}{\partial y^{l}\partial y^{s}}\frac{\partial f^{s}}{\partial x^{\beta}}\frac{\partial y^{k}}{\partial\tilde{y}^{i}}\xi^{l}+g^{\alpha\beta}\frac{\partial\det g}{\partial x^{\alpha}}\frac{1}{2\det g}\frac{\partial^{2}\tilde{y}^{i}}{\partial y^{l}\partial y^{s}}\frac{\partial f^{s}}{\partial x^{\beta}}\frac{\partial y^{k}}{\partial\tilde{y}^{i}}\xi^{l}+
gαβξryr(y~iyηyθy~jyϕy~lΓθϕη+2yηy~jy~ly~iyη)y~jymfmxαy~lysfsxβ\displaystyle g^{\alpha\beta}\xi^{r}\frac{\partial}{\partial y^{r}}\left(\frac{\partial\tilde{y}^{i}}{\partial y^{\eta}}\frac{\partial y^{\theta}}{\partial\tilde{y}^{j}}\frac{\partial y^{\phi}}{\partial\tilde{y}^{l}}\Gamma^{\eta}_{\theta\phi}+\frac{\partial^{2}y^{\eta}}{\partial\tilde{y}^{j}\partial\tilde{y}^{l}}\frac{\partial\tilde{y}^{i}}{\partial y^{\eta}}\right)\frac{\partial\tilde{y}^{j}}{\partial y^{m}}\frac{\partial f^{m}}{\partial x^{\alpha}}\frac{\partial\tilde{y}^{l}}{\partial y^{s}}\frac{\partial f^{s}}{\partial x^{\beta}}

and

F2k(f,fx,ξx)=gαβflxαξsxβ2y~iylysyky~iF_{2}^{k}\left(f,\frac{\partial f}{\partial x},\frac{\partial\xi}{\partial x}\right)=g^{\alpha\beta}\frac{\partial f^{l}}{\partial x^{\alpha}}\frac{\partial\xi^{s}}{\partial x^{\beta}}\frac{\partial^{2}\tilde{y}^{i}}{\partial y^{l}\partial y^{s}}\frac{\partial y^{k}}{\partial\tilde{y}^{i}}

Note that by lemma 2.1, the coordinate change functions are uniformly bounded.

2.2. Structure of the Moduli Space

This part follows from section 4.2.4 of [5] or section A.4 in [16]. Suppose Df0:W2,p(S2,(f0)1TN)Lp(S2,(f0)1TN)D_{f^{0}}:W^{2,p}(S^{2},(f^{0})^{-1}TN)\rightarrow L^{p}(S^{2},(f^{0})^{-1}TN) is Fredholm. It follows that the kernel and image of Df0D_{f^{0}} are closed and admit topological complements. For simplicity, we write U:=W2,p(S2,(f0)1TN)U:=W^{2,p}(S^{2},(f^{0})^{-1}TN) and V:=Lp(S2,(f0)1TN)V:=L^{p}(S^{2},(f^{0})^{-1}TN). So we can write U=U0FU=U_{0}\oplus F, V=V0GV=V_{0}\oplus G, where FF and GG are finite-dimensional and Df0D_{f^{0}} is a linear isomorphism from U0U_{0} to V0V_{0}.

Let NN be a connected open neighburhood of 0 in UU. By definition we know \mathcal{F} is Fredholm. If Df0D_{f^{0}} is surjective, then by the implicit function theorem of Banach spaces we know there is a diffeomorphism ψ\psi from one neighborhood of 0 in UU to another, such that ψ=Df0\mathcal{F}\circ\psi=D_{f^{0}}. Furthermore, we know that

Now consider the general case when Df0D_{f^{0}} is not necessarily surjective. Consider projection of VV onto V0V_{0}, we will have the derivative be surjective. Then we obtain

Proposition 2.1.

The Fredholm map Df0D_{f^{0}} from a neighborhood of 0 is locally right equivalent to a map of the form

~:U0×FV0×G,F~(ξ,η)=(Df0(ξ),α(ξ,η))\tilde{\mathcal{F}}:U_{0}\times F\rightarrow V_{0}\times G,\ \tilde{F}(\xi,\eta)=(D_{f^{0}}(\xi),\alpha(\xi,\eta))

where Df0D_{f^{0}} is a linear isomorphism from U0U_{0} to V0V_{0}, FF and GG are finite-dimensional, and the derivative of α\alpha vanishes at 0.

As an immediate corollary we obtain a finite-dimensional model for a neighborhood of 0 in the zero set Z()Z(\mathcal{F}). Note that elements in Z()Z(\mathcal{F}) are smooth by elliptic regularity (use partition of unity to reduce the case on manifolds to that on Euclidean space).

Explicitly, there exists a CC^{\infty} diffeomorphism gg from some open set WW containing 0W2,p(S2,(f0)1TN)0\in W^{2,p}(S^{2},(f^{0})^{-1}TN), such that

g(0)=0,dg(0)=idg(0)=0,\quad dg(0)=id

and

1(0)g(W)=g(WkerD)\mathcal{F}^{-1}(0)\cap g(W)=g(W\cap\ker D)

From the above we can get a coordinate chart for g(W)g(W) by the isomorphism between kerD\ker D and m\mathbb{R}^{m}, where mm is the dimension of the kernel. Thus we can view the vectors in the kernel as tangent vectors of the moduli space. Furthermore, a smooth path ftf_{t} in the open neighborhood can be represented as g(γ1(t)ξ1++γm(t)ξm)g(\gamma_{1}(t)\xi_{1}+\cdots+\gamma_{m}(t)\xi_{m}). Similarly, for a sufficiently small open neighborhood, we can find a smooth frame.

In the setting for theorem 1.1, since the linearizations are assumed to be surjective, together with proposition C.2 in the appendix, we know that the above can be applied.

Consider (c,p)\mathcal{M}(c,p), note that for linear operators between Banach spaces, surjective operators form open sets in the sense of operator norm. Thus locally we can still find charts as above.

3. The implicit function theorem

We first state a version of implicit function theorem for general Banach spaces:

Proposition 3.1 (Implicit Function Theorem for General Banach Spaces).

(See proposition A.3.4 in [16]) Let XX and YY be Banach spaces, UXU\subset X be an open set, and f:UYf:U\rightarrow Y be a continuously differentiable map. Let x0Ux_{0}\in U be such that D:=df(x0):XYD:=df(x_{0}):X\rightarrow Y is surjective and has a (bounded linear) right inverse Q:YXQ:Y\rightarrow X. Choose positive constants δ\delta and cc such that Qc\|Q\|\leq c, Bδ(x0;X)UB_{\delta}(x_{0};X)\subset U, and

(9) xx0<δdf(x)D12c.\|x-x_{0}\|<\delta\ \Rightarrow\|df(x)-D\|\leq\frac{1}{2c}.

Suppose that x1Xx_{1}\in X satisfies

(10) f(x1)<δ4c,x1x0<δ8.\|f(x_{1})\|<\frac{\delta}{4c},\ \|x_{1}-x_{0}\|<\frac{\delta}{8}.

Then there exists a unique xXx\in X such that

(11) f(x)=0,xx1imQ,xx0<δ.f(x)=0,\ x-x_{1}\in\text{im}Q,\ \|x-x_{0}\|<\delta.

Moreover, xx12cf(x1)\|x-x_{1}\|\leq 2c\|f(x_{1})\|.

Then we can get a version of implicit function theorem that can be used in our setting:

Theorem 3.1 (Implicit Function Theorem).

Let Σ\Sigma be a Riemann surface and NN a Riemannian manifold. Let ff be a map from Σ\Sigma to NN. f\mathcal{F}_{f} defined in (5) is a continuously differentiable map from W2,p(Σ,f1TN)W^{2,p}(\Sigma,f^{-1}TN) to Lp(Σ,f1TN)L^{p}(\Sigma,f^{-1}TN). Let ξ0W2,p(Σ,f1TN)\xi_{0}\in W^{2,p}(\Sigma,f^{-1}TN) be such that D:=df(ξ0):W2,p(Σ,f1TN)Lp(Σ,f1TN)D:=d\mathcal{F}_{f}(\xi_{0}):W^{2,p}(\Sigma,f^{-1}TN)\rightarrow L^{p}(\Sigma,f^{-1}TN) is surjective and has a (bounded linear) right inverse QQ. Choose positive constants δ\delta and cc such that Qc\|Q\|\leq c, and

(12) ξξ0<δdf(ξ)D12c.\|\xi-\xi_{0}\|<\delta\ \Rightarrow\ \|d\mathcal{F}_{f}(\xi)-D\|\leq\frac{1}{2c}.

Suppose that ξ1W2,p(Σ,f1TN)\xi_{1}\in W^{2,p}(\Sigma,f^{-1}TN) satisfies

(13) f(ξ1)<δ4c,ξ1ξ0<δ8.\|\mathcal{F}_{f}(\xi_{1})\|<\frac{\delta}{4c},\ \|\xi_{1}-\xi_{0}\|<\frac{\delta}{8}.

Then there exists a unique ξW2,p(Σ,f1TN)\xi\in W^{2,p}(\Sigma,f^{-1}TN) such that

(14) f(ξ)=0,ξξ1imQ,ξξ0<δ.\mathcal{F}_{f}(\xi)=0,\ \xi-\xi_{1}\in\text{im}Q,\ \|\xi-\xi_{0}\|<\delta.

Moreover, ξξ12cf(ξ1)\|\xi-\xi_{1}\|\leq 2c\|\mathcal{F}_{f}(\xi_{1})\|.

4. Pregluing

From now on we consider gluing two harmonic maps from S2S^{2} to some closed manifold NN of dimension no less than 2. However, we will see that the same procedure can be applied in the general case.

More specifically, consider (c,p)\mathcal{M}(c,p) defined in definition 1.1. Let f1,f2,x1,x2f_{1},f_{2},x_{1},x_{2} be f0,f,0,f^{0},f^{\infty},0,\infty respectively (here we are using the projective plane as the coordinate chart for S2S^{2}). Furthermore, suppose ϵ\epsilon is less than the injective radius of NN. Consider the setup in Section 1.1.

Let r:=δRr:=\delta R. Note that this definition is only for the brevity of notations. In the process of estimates, rr will denote the radius for polar coordinates. We will need the W2,pW^{2,p}-small perturbations f0,r,f,rf^{0,r},f^{\infty,r} of f0,ff^{0},f^{\infty}:

(15) f0,r(z):={fR(z)|z|1rf0(0)|z|1rf,r(z):={fR(zR2)|z|rf()|z|rf^{0,r}(z):=\left\{\begin{aligned} &f^{R}(z)\ &|z|\geq\frac{1}{r}\\ &f^{0}(0)\ &|z|\leq\frac{1}{r}\end{aligned}\right.\quad f^{\infty,r}(z):=\left\{\begin{aligned} &f^{R}\left(\frac{z}{R^{2}}\right)\ &|z|\leq r\\ &f^{\infty}(\infty)\ &|z|\geq r\end{aligned}\right.

Note that f0,r(z)=fR(z)f^{0,r}(z)=f^{R}(z) for all |z|δ/R|z|\geq\delta/R and f,r(z)=fR(z/R2)f^{\infty,r}(z)=f^{R}(z/R^{2}) for all |z|R/δ|z|\leq R/\delta.

We set up some simplified notation. We define

Wf2,p:=W2,p(S2,f1TN),Lfp:=Lp(S2,f1TN)W^{2,p}_{f}:=W^{2,p}(S^{2},f^{-1}TN),\ L^{p}_{f}:=L^{p}(S^{2},f^{-1}TN)

for f:S2Nf:S^{2}\rightarrow N. Given f0,f:S2Nf^{0},f^{\infty}:S^{2}\rightarrow N such that f0(0)=f()f^{0}(0)=f^{\infty}(\infty), denote

Wf0,2,p:={(ξ0,ξ)Wf02,p×Wf2,pξ0(0)=ξ()}W^{2,p}_{f^{0,\infty}}:=\left\{(\xi^{0},\xi^{\infty})\in W^{2,p}_{f^{0}}\times W^{2,p}_{f^{\infty}}\mid\xi^{0}(0)=\xi^{\infty}(\infty)\right\}

We define D0,:Wf0,2,pLf0p×LfpD_{0,\infty}:W^{2,p}_{f^{0,\infty}}\rightarrow L^{p}_{f^{0}}\times L^{p}_{f^{\infty}}

D0,(ξ0,ξ):=(Df0ξ0,Dfξ)D_{0,\infty}(\xi^{0},\xi^{\infty}):=(D_{f^{0}}\xi^{0},D_{f^{\infty}}\xi^{\infty})

Let 𝒲f0,Wf0,2,p\mathcal{W}_{f^{0,\infty}}\subset W^{2,p}_{f^{0,\infty}} be the L2L^{2} orthogonal complement of the kernel of D0,D_{0,\infty}, and define

Q0,:=Qf0,f:=(D0,𝒲f0,)1Q_{0,\infty}:=Q_{f^{0},f^{\infty}}:=(D_{0,\infty}\mid_{\mathcal{W}_{f^{0,\infty}}})^{-1}

Similarly, we have Q0,,r:=Qf0,r,f,rQ_{0,\infty,r}:=Q_{f^{0,r},f^{\infty,r}}.Since the operators D0,,rD_{0,\infty,r} are small perturbations of D0,D_{0,\infty} (See proposition B.2), we know these right inverses still exist.

We will need to show that DfD_{f} for any harmonic map ff is a Fredholm operator. This will be shown in section C.

and from that get an estimate: There are positive constants δ0,c0\delta_{0},c_{0} only depending on such that

(16) Q0,,rηW2,pc0ηLp,\|Q_{0,\infty,r}\eta\|_{W^{2,p}}\leq c_{0}\|\eta\|_{L^{p}},

for all (f0,f)(c)(f^{0},f^{\infty})\in\mathcal{M}(c), (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), and (η0,η)Lf0,rp×Lf,rp(\eta^{0},\eta^{\infty})\in L^{p}_{f^{0,r}}\times L^{p}_{f^{\infty,r}}. This will be shown in lemma C.3.

5. Approximate right inverse

The idea of our proof is that, first find an approximate right inverse, then we use this to find the real right inverse and apply the implicit function theorem. In this section, we will define the approximate right inverse and prove certain estimates that will be useful later.

In the definition below, note that for any pair of positive numbers δ1,δ2\delta_{1},\delta_{2} such that δ1<δ2\delta_{1}<\delta_{2}, we have 𝒜(δ1)𝒜(δ2)\mathcal{A}(\delta_{1})\subset\mathcal{A}(\delta_{2}). Thus we can always shrink δ0\delta_{0} if necessary.

Definition 5.1.

Let 𝒰\mathcal{U} be any precompact subset of (c,p)\mathcal{M}(c,p). Let δ0\delta_{0} be the same as in C.3. For any (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U} and any (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), we define

TfR:Lp(S2,(fR)1TN)W2,p(S2,(fR)1TN)T_{f^{R}}:L^{p}(S^{2},(f^{R})^{-1}TN)\rightarrow W^{2,p}(S^{2},(f^{R})^{-1}TN)

along the preglued map fR:S2Nf^{R}:S^{2}\rightarrow N defined by (1) as follows:

Given ηLfRp\eta\in L_{f^{R}}^{p} we first define the pair

(η0,η)Lf0,rp×Lf,rp(\eta^{0},\eta^{\infty})\in L^{p}_{f^{0,r}}\times L^{p}_{f^{\infty,r}}

by cutting off η\eta along the circle |z|=1/R|z|=1/R:

(17) η0(z):={η(z), if |z|1/R,0, if |z|1/R,η(z):={η(z/R2), if |z|R,0, if |z|R.\eta^{0}(z):=\left\{\begin{aligned} &\eta(z),&\text{ if }|z|\geq 1/R,\\ &0,&\text{ if }|z|\leq 1/R,\end{aligned}\right.\ \eta^{\infty}(z):=\left\{\begin{aligned} &\eta(z/R^{2}),&\text{ if }|z|\leq R,\\ &0,&\text{ if }|z|\geq R.\end{aligned}\right.

Second, define

(ξ0,ξ):=Q0,,r(η0,η)(\xi^{0},\xi^{\infty}):=Q_{0,\infty,r}(\eta^{0},\eta^{\infty})

and note that the vector fields ξ0,ξ\xi^{0},\xi^{\infty} have the same central value ξ0\xi_{0}:

ξ0(0)=ξ()=:ξ0Tf0(0)N\xi^{0}(0)=\xi^{\infty}(\infty)=:\xi_{0}\in T_{f^{0}(0)}N

Third, let 1βδ,R:1-\beta_{\delta,R}:\mathbb{C}\rightarrow\mathbb{R} denote a cutoff function defined as follows: βδ,R(z)=0\beta_{\delta,R}(z)=0 for |z|δ/R|z|\leq\delta/R, βδ,R(z)=1\beta_{\delta,R}(z)=1 for |z|1/R|z|\geq 1/R, and

βδ,R(z):=κ(log(R|z|/δ)log(1/δ)),δR|z|1R\beta_{\delta,R}(z):=\kappa\left(\frac{\log(R|z|/\delta)}{\log(1/\delta)}\right),\ \frac{\delta}{R}\leq|z|\leq\frac{1}{R}

where κ\kappa: [0,1]\mathbb{R}\rightarrow[0,1] is a CC^{\infty} cut-off function such that κ(t)=1\kappa(t)=1 if t1t\geq 1 and κ(t)=0\kappa(t)=0 if t0t\leq 0.

Fourth, define TfRη:=ξRT_{f^{R}}\eta:=\xi^{R} by

(18) ξR(z):={ξ0(z), if |z|1δR,ξ0(z)+βδ,R(1R2z)(ξ(R2z)ξ0), if 1R|z|1δR,ξ0(z)+ξ(R2z)ξ0, if |z|=1R,ξ(R2z)+βδ,R(z)(ξ0(z)ξ0), if δR|z|1R,ξ(R2z) if |z|δR.\xi^{R}(z):=\left\{\begin{aligned} &\xi^{0}(z),\ &\text{ if }|z|\geq\frac{1}{\delta R},\\ &\xi^{0}(z)+\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)(\xi^{\infty}(R^{2}z)-\xi_{0}),\ &\text{ if }\frac{1}{R}\leq|z|\leq\frac{1}{\delta R},\\ &\xi^{0}(z)+\xi^{\infty}(R^{2}z)-\xi_{0},\ &\text{ if }|z|=\frac{1}{R},\\ &\xi^{\infty}(R^{2}z)+\beta_{\delta,R}(z)(\xi^{0}(z)-\xi_{0}),\ &\text{ if }\frac{\delta}{R}\leq|z|\leq\frac{1}{R},\\ &\xi^{\infty}(R^{2}z)\ &\text{ if }|z|\leq\frac{\delta}{R}.\end{aligned}\right.

Let us consider why η\eta^{\infty} should be defined as such. We would like to have ξ=Q,rη\xi^{\infty}=Q_{\infty,r}\eta^{\infty}. Hence D,rξ=ηD_{\infty,r}\xi^{\infty}=\eta^{\infty}. Also, we want DfRξ(R2z)=ηD_{f^{R}}\xi^{\infty}(R^{2}z)=\eta. Note that f,r(z)=fR(zR2)f^{\infty,r}(z)=f^{R}(\frac{z}{R^{2}}) holds for |z|R/δ|z|\leq R/\delta.

For |z|R/δ|z|\leq R/\delta, we use coordinates (s,t)(s,t) for the domain, and pick normal coordinates at the point f,r(z)f^{\infty,r}(z). Due to the special structure of the coordinates in the domain, after calculation, we find that the linearization is simpler than expected because certain terms in the Laplace operator cancel out. We have

Df,rξ=\displaystyle D_{f^{\infty,r}}\xi^{\infty}= (g(ξ)i+gαβkΓjli(f,r(z))(ξ)k(f,r)jxα(f,r)lxβ)ei\displaystyle\left(\triangle_{g}(\xi^{\infty})^{i}+g^{\alpha\beta}\partial_{k}\Gamma^{i}_{jl}(f^{\infty,r}(z))(\xi^{\infty})^{k}\frac{\partial(f^{\infty,r})^{j}}{\partial x^{\alpha}}\frac{\partial(f^{\infty,r})^{l}}{\partial x^{\beta}}\right)e_{i}
=\displaystyle= (1+|z|2)2((2s2+2t2)(ξ)i+\displaystyle(1+|z|^{2})^{2}\left(\left(\frac{\partial^{2}}{\partial s^{2}}+\frac{\partial^{2}}{\partial t^{2}}\right)(\xi^{\infty})^{i}+\right.
kΓjli(f,r(z))(ξ)k((f,r)js(f,r)ls+(f,r)jt(f,r)lt))ei\displaystyle\left.\partial_{k}\Gamma^{i}_{jl}(f^{\infty,r}(z))(\xi^{\infty})^{k}\left(\frac{\partial(f^{\infty,r})^{j}}{\partial s}\frac{\partial(f^{\infty,r})^{l}}{\partial s}+\frac{\partial(f^{\infty,r})^{j}}{\partial t}\frac{\partial(f^{\infty,r})^{l}}{\partial t}\right)\right)e_{i}

where eie_{i} is yi\frac{\partial}{\partial y^{i}} in normal coordinates at f,r(z)f^{\infty,r}(z).

Now consider |w|1/(δR)|w|\leq 1/(\delta R). We have

DfRξ(R2)=\displaystyle D_{f^{R}}\xi(R^{2}\cdot)= (R2+R2|w|2)2R4((2(ξ)is2(R2w)+2(ξ)it2(R2w))+\displaystyle(R^{-2}+R^{2}|w|^{2})^{2}R^{4}\left(\left(\frac{\partial^{2}(\xi^{\infty})^{i}}{\partial s^{2}}(R^{2}w)+\frac{\partial^{2}(\xi^{\infty})^{i}}{\partial t^{2}}(R^{2}w)\right)+\right.
kΓjli(f,r(R2w))(ξ(R2w))k((f,r)js(R2w)(f,r)ls(R2w)+\displaystyle\partial_{k}\Gamma^{i}_{jl}(f^{\infty,r}(R^{2}w))(\xi^{\infty}(R^{2}w))^{k}\left(\frac{\partial(f^{\infty,r})^{j}}{\partial s}(R^{2}w)\frac{\partial(f^{\infty,r})^{l}}{\partial s}(R^{2}w)+\right.
(f,r)jt(R2w)(f,r)lt(R2w)))ei\displaystyle\left.\left.\frac{\partial(f^{\infty,r})^{j}}{\partial t}(R^{2}w)\frac{\partial(f^{\infty,r})^{l}}{\partial t}(R^{2}w)\right)\right)e_{i}

Since we want DfRξ(R2)=ηD_{f^{R}}\xi(R^{2}\cdot)=\eta, for |z|R|z|\leq R, let w=z/R2w=z/R^{2}, we have η(z)=η(z/R2)\eta^{\infty}(z)=\eta(z/R^{2}).

In the rest of this section, we will get estimates that will be useful in our construction of the right inverse. Let us denote the (0,p,R)(0,p,R)-norm (resp. (2,p,R)(2,p,R)-norm) on a certain region Ω\Omega by 0,p,R,Ω\|\cdot\|_{0,p,R,\Omega} (resp. 2,p,R,Ω\|\cdot\|_{2,p,R,\Omega}).

Lemma 5.1.

For any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p), we can choose δ0\delta_{0} small enough only depending on c,p,N,𝒰c,p,N,\mathcal{U}, such that for any (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U} and any (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), the approximate right inverse TfRT_{f^{R}} defined in 5.1 satisfies:

DfRTfRηη0,p,R,Ω114η0,p,R,Ω1\|D_{f^{R}}T_{f^{R}}\eta-\eta\|_{0,p,R,\Omega_{1}}\leq\frac{1}{4}\|\eta\|_{0,p,R,\Omega_{1}}

for every ηLp(S2,(fR)1TN)\eta\in L^{p}(S^{2},(f^{R})^{-1}TN), where Ω1={δ/R|z|1/R}\Omega_{1}=\{\delta/R\leq|z|\leq 1/R\}.

Proof.

In this region,

f0,r(z)=f,r(R2z)=fR(z)=yf^{0,r}(z)=f^{\infty,r}(R^{2}z)=f^{R}(z)=y

Therefore over this annulus the vector field ξR\xi^{R} takes values in the fixed vector space TyNT_{y}N. By (7), we see that the operators are all equal to the usual Laplace-Beltrami operator. In particular, they vanish on the constant function ξ0\xi_{0}. Furthermore, the definition of ξ\xi^{\infty} implies that DfRξ(R2)=ηD_{f^{R}}\xi^{\infty}(R^{2}\cdot)=\eta in the region |z|1/R|z|\leq 1/R.

Let’s consider the polar coordinates. Given the weighted metric, we have

g=(R2+R2r2)22r2+(R2+R2r2)2rr+(R2+R2r2)2r22θ2\triangle_{g}=(R^{-2}+R^{2}r^{2})^{2}\frac{\partial^{2}}{\partial r^{2}}+\frac{(R^{-2}+R^{2}r^{2})^{2}}{r}\frac{\partial}{\partial r}+\frac{(R^{-2}+R^{2}r^{2})^{2}}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}

We also recall that, by our construction of the cutoff function, we have

|rβ(z)|C|z|log(1/δ) and |2r2β(z)|C|z|2log(1/δ)\left|\frac{\partial}{\partial r}\beta(z)\right|\leq\frac{C}{|z|\log(1/\delta)}\text{\ and\ }\left|\frac{\partial^{2}}{\partial r^{2}}\beta(z)\right|\leq\frac{C}{|z|^{2}\log(1/\delta)}

where the constants are universal.

Hence, when δ/R|z|1/R\delta/R\leq|z|\leq 1/R, note that Df0,rξ0=η0=0D_{f^{0,r}}\xi^{0}=\eta^{0}=0, we find

DfRξRη\displaystyle D_{f^{R}}\xi^{R}-\eta =Df0,r(βδ,R(ξ0ξ0))\displaystyle=D_{f^{0,r}}(\beta_{\delta,R}(\xi^{0}-\xi_{0}))
=g(βδ,R)(ξ0ξ0)+(R2+R2r2)2βδ,Rrξ0r\displaystyle=\triangle_{g}(\beta_{\delta,R})(\xi^{0}-\xi_{0})+(R^{-2}+R^{2}r^{2})^{2}\frac{\partial\beta_{\delta,R}}{\partial r}\frac{\partial\xi^{0}}{\partial r}

We would like to estimate the LpL^{p} norm of DfRξRηD_{f^{R}}\xi^{R}-\eta under the weighted norm. For the second part, we have:

(R2+R2r2)2βδ,Rrξ0r0,p,R\displaystyle\left\|(R^{-2}+R^{2}r^{2})^{2}\frac{\partial\beta_{\delta,R}}{\partial r}\frac{\partial\xi^{0}}{\partial r}\right\|_{0,p,R}
=\displaystyle= (δ/R|z|1/R(R2+R2|z|2)2|(R2+R2r2)2βδ,Rrξ0r|p)1/p\displaystyle\left(\int_{\delta/R\leq|z|\leq 1/R}(R^{-2}+R^{2}|z|^{2})^{-2}\left|(R^{-2}+R^{2}r^{2})^{2}\frac{\partial\beta_{\delta,R}}{\partial r}\frac{\partial\xi^{0}}{\partial r}\right|^{p}\right)^{1/p}
=\displaystyle= (δ/R|z|1/R(R2+R2|z|2)2p2|βδ,Rrξ0r|p)1/p\displaystyle\left(\int_{\delta/R\leq|z|\leq 1/R}(R^{-2}+R^{2}|z|^{2})^{2p-2}\left|\frac{\partial\beta_{\delta,R}}{\partial r}\frac{\partial\xi^{0}}{\partial r}\right|^{p}\right)^{1/p}
\displaystyle\leq (1+δ04)22pβδ,Rrξ0rLp(δ/R|z|1/R)\displaystyle(1+\delta_{0}^{4})^{2-\frac{2}{p}}\left\|\frac{\partial\beta_{\delta,R}}{\partial r}\frac{\partial\xi^{0}}{\partial r}\right\|_{L^{p}(\delta/R\leq|z|\leq 1/R)}

We can consider using the Sobolev embedding on the manifold, which, in our case, is the 2 dimensional sphere. First, we use the Hölder’s inequality to get:

βδ,Rrξ0rpβδ,RrL2(δR|z|1R)ξ0rLq(δR|z|1R)\displaystyle\left\|\frac{\partial\beta_{\delta,R}}{\partial r}\frac{\partial\xi^{0}}{\partial r}\right\|_{p}\leq\left\|\frac{\partial\beta_{\delta,R}}{\partial r}\right\|_{L^{2}(\frac{\delta}{R}\leq|z|\leq\frac{1}{R})}\left\|\frac{\partial\xi^{0}}{\partial r}\right\|_{L^{q}(\frac{\delta}{R}\leq|z|\leq\frac{1}{R})}

where q=2p2p>2q=\frac{2p}{2-p}>2.

For the LqL^{q} norm part, we want to use the Sobolev embedding on the sphere. From chapter 2 in [2] we know that for compact manifold the Sobolev embedding holds. Note that here we can treat ξ0\xi^{0} in the same way as real-valued functions on the sphere. |ξ0r|C|ξ0|\left|\frac{\partial\xi^{0}}{\partial r}\right|\leq C\left|\nabla\xi^{0}\right|, where CC is univeral. Thus we have

ξ0rLq(δR|z|1R)Cξ0W2,p(S2)C(c,p,N,𝒰)η0LpC(c,p,N,𝒰)η0,p,R\displaystyle\left\|\frac{\partial\xi^{0}}{\partial r}\right\|_{L^{q}(\frac{\delta}{R}\leq|z|\leq\frac{1}{R})}\leq C\|\xi^{0}\|_{W^{2,p}(S^{2})}\leq C(c,p,N,\mathcal{U})\|\eta^{0}\|_{L^{p}}\leq C(c,p,N,\mathcal{U})\|\eta\|_{0,p,R}

if we apply lemma C.3. Here we are considering S2S^{2} with the round metric.

We also know that

βδ,RrL2(δR|z|1R)Clog(1/δ)\left\|\frac{\partial\beta_{\delta,R}}{\partial r}\right\|_{L^{2}(\frac{\delta}{R}\leq|z|\leq\frac{1}{R})}\leq\frac{C}{\sqrt{\log(1/\delta)}}

For the first part (g(βδ,R)(ξ0ξ0)\triangle_{g}(\beta_{\delta,R})(\xi^{0}-\xi_{0})), note that we have control over the C0,γC^{0,\gamma} norm of ξ0\xi^{0} by the W2,pW^{2,p} norm. We consider the Sobolev embedding of C0,μC^{0,\mu} into W2,pW^{2,p} where 1<p<21<p<2 and μ=22p\mu=2-\frac{2}{p}. For every ball B2B\subset\mathbb{R}^{2} and every ξW2,p(B)\xi\in W^{2,p}(B), we have

(19) z0,z1B|ξ(z1)ξ(z0)|C(p)ξW2,p(B)|z1z0|22p\displaystyle z_{0},z_{1}\in B\ \Rightarrow\ |\xi(z_{1})-\xi(z_{0})|\leq C(p)\|\xi\|_{W^{2,p}(B)}|z_{1}-z_{0}|^{2-\frac{2}{p}}

Note that the constant comes from the extension. Thus for balls of different radius, the constant remains the same.

Thus we have the estimate:

g(βδ,R)(ξ0ξ0)0,p,R\displaystyle\|\triangle_{g}(\beta_{\delta,R})(\xi^{0}-\xi_{0})\|_{0,p,R}
=\displaystyle= (R2+R2r2)22βδ,Rr2(ξ0ξ0)+(R2+R2r2)2rβδ,Rr(ξ0ξ0)0,p,R\displaystyle\left\|(R^{-2}+R^{2}r^{2})^{2}\frac{\partial^{2}\beta_{\delta,R}}{\partial r^{2}}(\xi^{0}-\xi_{0})+\frac{(R^{-2}+R^{2}r^{2})^{2}}{r}\frac{\partial\beta_{\delta,R}}{\partial r}(\xi^{0}-\xi_{0})\right\|_{0,p,R}
\displaystyle\leq (R2+R2r2)2Cr2log(1/δ)(ξ0ξ0)0,p,R\displaystyle\left\|(R^{-2}+R^{2}r^{2})^{2}\frac{C}{r^{2}\log(1/\delta)}(\xi^{0}-\xi_{0})\right\|_{0,p,R}
=\displaystyle= (δR|z|1R(R2+R2r2)2|(R2+R2r2)2Cr2log(1/δ)(ξ0ξ0)|p)1p\displaystyle\left(\int_{\frac{\delta}{R}\leq|z|\leq\frac{1}{R}}(R^{-2}+R^{2}r^{2})^{-2}\left|(R^{-2}+R^{2}r^{2})^{2}\frac{C}{r^{2}\log(1/\delta)}(\xi^{0}-\xi_{0})\right|^{p}\right)^{\frac{1}{p}}
\displaystyle\leq (δR|z|1R(1+δ04)2p2Cr2p(log(1/δ))p|ξ0ξ0|p)1p\displaystyle\left(\int_{\frac{\delta}{R}\leq|z|\leq\frac{1}{R}}\frac{(1+\delta_{0}^{4})^{2p-2}C}{r^{2p}(\log(1/\delta))^{p}}|\xi^{0}-\xi_{0}|^{p}\right)^{\frac{1}{p}}
\displaystyle\leq (1+δ04)22/pC(p)log(1/δ)11pξ0W2,p(B1/R)\displaystyle\frac{(1+\delta_{0}^{4})^{2-2/p}C(p)}{\log(1/\delta)^{1-\frac{1}{p}}}\|\xi^{0}\|_{W^{2,p}(B_{{1}/{R}})}

Furthermore, note that,

ξ0W2,p(B1/R)Cξ0W2,p(S2)C(c,p,N,𝒰)η0LpC(c,p,N,𝒰)η0,p,R\displaystyle\|\xi^{0}\|_{W^{2,p}(B_{1/R})}\leq C\|\xi^{0}\|_{W^{2,p}(S^{2})}\leq C(c,p,N,\mathcal{U})\|\eta^{0}\|_{L^{p}}\leq C(c,p,N,\mathcal{U})\|\eta\|_{0,p,R}

Thus by choosing small enough δ0\delta_{0} only depending on c,p,N,𝒰c,p,N,\mathcal{U}, we have the desired result. ∎

Lemma 5.2.

For any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p), we can choose δ0\delta_{0} small enough only depending on c,p,N,𝒰c,p,N,\mathcal{U}, such that for any (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U} and any (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), the approximate right inverse TfRT_{f^{R}} defined in 5.1 satisfies:

DfRTfRηη0,p,R,Ω214η0,p,R,Ω2\|D_{f^{R}}T_{f^{R}}\eta-\eta\|_{0,p,R,\Omega_{2}}\leq\frac{1}{4}\|\eta\|_{0,p,R,\Omega_{2}}

for every ηLp(S2,(fR)1TN)\eta\in L^{p}(S^{2},(f^{R})^{-1}TN), where Ω2={1/R|z|1/δR}\Omega_{2}=\{1/R\leq|z|\leq 1/\delta R\}.

Proof.

We still use the polar coordinates. Since the weight is different, here we have

g=(1+r2)22r2+(1+r2)2rr+(1+r2)2r22θ2\displaystyle\triangle_{g}=(1+r^{2})^{2}\frac{\partial^{2}}{\partial r^{2}}+\frac{(1+r^{2})^{2}}{r}\frac{\partial}{\partial r}+\frac{(1+r^{2})^{2}}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}

and

ξR(z)=ξ0(z)+βδ,R(1R2z)(ξ(R2z)ξ0)\xi^{R}(z)=\xi^{0}(z)+\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)(\xi^{\infty}(R^{2}z)-\xi_{0})

We know that DfRξ0=Df0,rξ0=η0=ηD_{f^{R}}\xi^{0}=D_{f^{0,r}}\xi^{0}=\eta^{0}=\eta. We also know that DfRξ(R2z)=Df,r(R2)ξ(R2z)D_{f^{R}}\xi^{\infty}(R^{2}z)=D_{f^{\infty,r}(R^{2}\cdot)}\xi^{\infty}(R^{2}z). Again, we use (s,t)(s,t) coordinates for 1/R|z|1/δR1/R\leq|z|\leq 1/\delta R. We use (s~=R2s,t~=R2t)(\tilde{s}=R^{2}s,\ \tilde{t}=R^{2}t) coordinates for RR2|z|R/δR\leq R^{2}|z|\leq R/\delta. We know that in the annulus 1/R|z|1/δR1/R\leq|z|\leq 1/\delta R, fR(z)=f,r(R2z)=f0(0)=f()=yf^{R}(z)=f^{\infty,r}(R^{2}z)=f^{0}(0)=f^{\infty}(\infty)=y. We choose the normal coordinates at yy and we can write ξ(R2z)\xi^{\infty}(R^{2}z) as (ξ(R2z))iei(\xi^{\infty}(R^{2}z))^{i}e_{i}, where eie_{i} is yi\frac{\partial}{\partial y^{i}} for the normal coordinates at yy. Note that in the annulus 1/R|z|1/δR1/R\leq|z|\leq 1/\delta R, DfR=Df,r(R2)=gD_{f^{R}}=D_{f^{\infty,r}(R^{2}\cdot)}=\triangle_{g}. We have

g=\displaystyle\triangle_{g}= 1detgxα(gαβdetgxβ)\displaystyle\frac{1}{\sqrt{\det g}}\frac{\partial}{\partial x^{\alpha}}\left(g^{\alpha\beta}\sqrt{\det g}\frac{\partial}{\partial x^{\beta}}\right)
=\displaystyle= (1+|z|2)2(2s2+2t2)\displaystyle(1+|z|^{2})^{2}\left(\frac{\partial^{2}}{\partial s^{2}}+\frac{\partial^{2}}{\partial t^{2}}\right)

Thus

g(ξ(R2))(z)=\displaystyle\triangle_{g}(\xi^{\infty}(R^{2}\cdot))(z)= (1+|z|2)2R4(2(ξ)is~2(R2z)+2(ξ)it~2(R2z))ei\displaystyle(1+|z|^{2})^{2}R^{4}\left(\frac{\partial^{2}(\xi^{\infty})^{i}}{\partial\tilde{s}^{2}}(R^{2}z)+\frac{\partial^{2}(\xi^{\infty})^{i}}{\partial\tilde{t}^{2}}(R^{2}z)\right)e_{i}

Note that for R|w|R/δR\leq|w|\leq R/\delta, we have f,r(w)=yf^{\infty,r}(w)=y, thus D,r=gD_{\infty,r}=\triangle_{g}. Furthermore, we have Df,rξ(w)=η(w)=0D_{f^{\infty,r}}\xi^{\infty}(w)=\eta^{\infty}(w)=0. Using the same coordinates as above (s~,t~\tilde{s},\tilde{t} for the domain, and normal coordinates at yNy\in N), we have

gξ(w)=(1+|w|2)2(2(ξ)is~2(w)+2(ξ)it~2(w))ei=0\displaystyle\triangle_{g}\xi^{\infty}(w)=(1+|w|^{2})^{2}\left(\frac{\partial^{2}(\xi^{\infty})^{i}}{\partial\tilde{s}^{2}}(w)+\frac{\partial^{2}(\xi^{\infty})^{i}}{\partial\tilde{t}^{2}}(w)\right)e_{i}=0

Thus we know that g(ξ(R2))(z)=0\triangle_{g}(\xi^{\infty}(R^{2}\cdot))(z)=0.

Now we have

gξRη=(g(βδ,R(1R2z)))(ξ(R2z)ξ0)+(1+|z|2)2r(βδ,R(1R2z))r(ξ(R2z))\displaystyle\triangle_{g}\xi^{R}-\eta=\left(\triangle_{g}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\right)(\xi^{\infty}(R^{2}z)-\xi_{0})+(1+|z|^{2})^{2}\frac{\partial}{\partial r}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\frac{\partial}{\partial r}\left(\xi^{\infty}(R^{2}z)\right)

We want to estimate the weighted LpL^{p} norm of the above formula.

We can compute that

|r(βδ,R(1R2z))|\displaystyle\left|\frac{\partial}{\partial r}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\right|\leq Crlog(1/δ)\displaystyle\frac{C}{r\log(1/\delta)}
|2r2(βδ,R(1R2z))|\displaystyle\left|\frac{\partial^{2}}{\partial r^{2}}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\right|\leq Cr2log(1/δ)\displaystyle\frac{C}{r^{2}\log(1/\delta)}

where CC is a universal constant. Thus,

|g(βδ,R(1R2z))|C(1+r2)2r2log(1/δ)\displaystyle\left|\triangle_{g}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\right|\leq\frac{C(1+r^{2})^{2}}{r^{2}\log(1/\delta)}

Let’s first estimate

(1+|z|2)2r(βδ,R(1R2z))r(ξ(R2z))0,p,R\displaystyle\left\|(1+|z|^{2})^{2}\frac{\partial}{\partial r}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\frac{\partial}{\partial r}\left(\xi^{\infty}(R^{2}z)\right)\right\|_{0,p,R}
=\displaystyle= (02π𝑑θ1/R1/(δR)r𝑑r(1+r2)2p2|r(βδ,R(1R2z))r(ξ(R2z))|p)1p\displaystyle\left(\int_{0}^{2\pi}d\theta\int_{1/R}^{1/(\delta R)}rdr\ (1+r^{2})^{2p-2}\left|\frac{\partial}{\partial r}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\frac{\partial}{\partial r}\left(\xi^{\infty}(R^{2}z)\right)\right|^{p}\right)^{\frac{1}{p}}
\displaystyle\leq (1+(1δR)2)22p(02π𝑑θ1/R1/(δR)r𝑑r|r(βδ,R(1R2z))|p|r(ξ(R2z))|p)1p\displaystyle\left(1+\left(\frac{1}{\delta R}\right)^{2}\right)^{2-\frac{2}{p}}\left(\int_{0}^{2\pi}d\theta\int_{1/R}^{1/(\delta R)}rdr\ \left|\frac{\partial}{\partial r}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\right|^{p}\left|\frac{\partial}{\partial r}\left(\xi^{\infty}(R^{2}z)\right)\right|^{p}\right)^{\frac{1}{p}}
\displaystyle\leq (1+(1δR)2)22p(A(|r(βδ,R(1R2z))|p)2p)12(A(|r(ξ(R2z))|p)22p)2p2p\displaystyle\left(1+\left(\frac{1}{\delta R}\right)^{2}\right)^{2-\frac{2}{p}}\left(\int_{A}\ \left(\left|\frac{\partial}{\partial r}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\right|^{p}\right)^{\frac{2}{p}}\right)^{\frac{1}{2}}\left(\int_{A}\left(\left|\frac{\partial}{\partial r}\left(\xi^{\infty}(R^{2}z)\right)\right|^{p}\right)^{\frac{2}{2-p}}\right)^{\frac{2-p}{2p}}

where AA denotes the annulus 1/R|z|1/(δR)1/R\leq|z|\leq 1/(\delta R). The last inequality comes from Hölder’s inequality.

For the part of the cutoff function, we have

(A(|r(βδ,R(1R2z))|p)2p)12\displaystyle\left(\int_{A}\ \left(\left|\frac{\partial}{\partial r}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\right|^{p}\right)^{\frac{2}{p}}\right)^{\frac{1}{2}}\leq (02π𝑑θ1/R1/(δR)r𝑑rCr2(log(1/δ))2)12\displaystyle\left(\int_{0}^{2\pi}d\theta\int_{1/R}^{1/(\delta R)}rdr\ \frac{C}{r^{2}(\log(1/\delta))^{2}}\right)^{\frac{1}{2}}
\displaystyle\leq C2πlog(1/δ)\displaystyle C\sqrt{\frac{2\pi}{\log(1/\delta)}}

where CC is a universal constant.

Before we look at the part involving ξ\xi^{\infty}, here are some thoughts about measuring the norm of the derivative of ξ\xi^{\infty}. Let us note that with the coordinate chart given by stereographic projection, there is a weight for the derivative of ξ\xi^{\infty}, since in the coordinate chart, at the points away from the origin, say w2\{0}w\in\mathbb{R}^{2}\backslash\{0\}, the vector is dilated with a ratio of (1+|w|2)1(1+|w|^{2})^{-1}. We have

r(ξ(R2z))=R2ξr(R2z)(1+|R2z|2)ξr(R2z)\displaystyle\frac{\partial}{\partial r}(\xi^{\infty}(R^{2}z))=R^{2}\frac{\partial\xi^{\infty}}{\partial r}(R^{2}z)\leq(1+|R^{2}z|^{2})\frac{\partial\xi^{\infty}}{\partial r}(R^{2}z)

Recall that using the coordinates from the stereographic projection of the Riemann sphere S2S^{2}, the W2,pW^{2,p} norm should be defined as

ξW2,p(S2,f1TN)=\displaystyle\|\xi\|_{W^{2,p}(S^{2},f^{-1}TN)}= (S2|ξ|p+|ξ|p+|2ξ|p)1p\displaystyle\left(\int_{S^{2}}|\xi|^{p}+|\nabla\xi|^{p}+|\nabla^{2}\xi|^{p}\right)^{\frac{1}{p}}
=\displaystyle= (2(1+r2)2|ξ|p+(1+r2)p2(|sξ|p+|tξ|p)+\displaystyle\left(\int_{\mathbb{R}^{2}}(1+r^{2})^{-2}|\xi|^{p}+(1+r^{2})^{p-2}(|\nabla_{s}\xi|^{p}+|\nabla_{t}\xi|^{p})+\right.
(1+r2)2p2(|2ξs2|p+|2ξst|p+|2ξt2|p))1p\displaystyle\left.(1+r^{2})^{2p-2}\left(\left|\frac{\partial^{2}\xi}{\partial s^{2}}\right|^{p}+\left|\frac{\partial^{2}\xi}{\partial s\partial t}\right|^{p}+\left|\frac{\partial^{2}\xi}{\partial t^{2}}\right|^{p}\right)\right)^{\frac{1}{p}}

Note that here we cannot directly consider the range of ξ\xi to be N\mathbb{R}^{N}. By the definition of the W2,pW^{2,p} norm for maps mapping to vector bundles, we need to consider the local coordinates of that vector bundle. However, for δR|z|1δR\frac{\delta}{R}\leq|z|\leq\frac{1}{\delta R}, since f(z)yf(z)\equiv y, these two definitions are equivalent.

Now let’s come back to the part involving ξ\xi^{\infty}. Denote 2p2p\frac{2p}{2-p} by pp^{*}. Since 1<p<21<p<2, we know p>2p^{*}>2. Thus it is easy to see that, in the annulus 1R|z|1δR\frac{1}{R}\leq|z|\leq\frac{1}{\delta R},

R2pR4(1+R2)p2R4(1+R4|z|2)p2\displaystyle R^{2p^{*}}\leq R^{4}(1+R^{2})^{p^{*}-2}\leq R^{4}(1+R^{4}|z|^{2})^{p^{*}-2}

We know that

(A(|r(ξ(R2z))|p)22p)2p2p=\displaystyle\left(\int_{A}\left(\left|\frac{\partial}{\partial r}\left(\xi^{\infty}(R^{2}z)\right)\right|^{p}\right)^{\frac{2}{2-p}}\right)^{\frac{2-p}{2p}}= (A|R2ξr(R2z)|p)1p\displaystyle\left(\int_{A}\left|R^{2}\frac{\partial\xi^{\infty}}{\partial r}(R^{2}z)\right|^{p^{*}}\right)^{\frac{1}{p^{*}}}
\displaystyle\leq (AR4(1+R4|z|2)p2|ξr(R2z)|p)1p\displaystyle\left(\int_{A}R^{4}(1+R^{4}|z|^{2})^{p^{*}-2}\left|\frac{\partial\xi^{\infty}}{\partial r}(R^{2}z)\right|^{p^{*}}\right)^{\frac{1}{p^{*}}}
=\displaystyle= (AR4(1+|R2z|2)2|(1+|R2z|2)ξr(R2z)|p)1p\displaystyle\left(\int_{A}R^{4}(1+|R^{2}z|^{2})^{-2}\left|(1+|R^{2}z|^{2})\frac{\partial\xi^{\infty}}{\partial r}(R^{2}z)\right|^{p^{*}}\right)^{\frac{1}{p^{*}}}
\displaystyle\leq (R|w|R/δ(1+|w|2)2|(1+|w|2)ξr(w)|p)1p\displaystyle\left(\int_{R\leq|w|\leq R/\delta}(1+|w|^{2})^{-2}\left|(1+|w|^{2})\frac{\partial\xi^{\infty}}{\partial r}(w)\right|^{p^{*}}\right)^{\frac{1}{p^{*}}}
\displaystyle\leq CξLp(S2)\displaystyle C\|\nabla\xi^{\infty}\|_{L^{p^{*}}(S^{2})}

where CC is universal.

Then we use the Sobolev embedding theorem for closed Riemann manifolds, as well as C.3:

ξLpCξW2,p(S2)C(c,p,N,𝒰)ηLp(S2)\displaystyle\|\nabla\xi^{\infty}\|_{L^{p^{*}}}\leq C\|\xi^{\infty}\|_{W^{2,p}(S^{2})}\leq C(c,p,N,\mathcal{U})\|\eta^{\infty}\|_{L^{p}(S^{2})}

From the definition of η\eta^{\infty}, we have

ηLp(S2)=\displaystyle\|\eta^{\infty}\|_{L^{p}(S^{2})}= (BR(0)|η(zR2)|p(1+|z|2)2)1p\displaystyle\left(\int_{B_{R}(0)}\left|\eta\left(\frac{z}{R^{2}}\right)\right|^{p}(1+|z|^{2})^{-2}\right)^{\frac{1}{p}}
=\displaystyle= (R4B1/R(0)|η(w)|p(1+R4|w|2)2)1p\displaystyle\left(R^{4}\int_{B_{{1}/{R}}(0)}\left|\eta(w)\right|^{p}(1+R^{4}|w|^{2})^{-2}\right)^{\frac{1}{p}}
=\displaystyle= (B1/R(0)|η(w)|p(R2+R2|w|2)2)1p\displaystyle\left(\int_{B_{{1}/{R}}(0)}\left|\eta(w)\right|^{p}(R^{-2}+R^{2}|w|^{2})^{-2}\right)^{\frac{1}{p}}
\displaystyle\leq η0,p,R\displaystyle\|\eta\|_{0,p,R}

Thus we have

(1+|z|2)2r(βδ,R(1R2z))r(ξ(R2z))0,p,R\displaystyle\left\|(1+|z|^{2})^{2}\frac{\partial}{\partial r}\left(\beta_{\delta,R}(\frac{1}{R^{2}z})\right)\frac{\partial}{\partial r}\left(\xi^{\infty}(R^{2}z)\right)\right\|_{0,p,R}
\displaystyle\leq C(c,p,N,𝒰)(1+(1δR)2)22p2πlog(1/δ)η0,p,R\displaystyle C(c,p,N,\mathcal{U})\left(1+\left(\frac{1}{\delta R}\right)^{2}\right)^{2-\frac{2}{p}}\sqrt{\frac{2\pi}{\log(1/\delta)}}\|\eta\|_{0,p,R}

On the other hand,

(g(βδ,R(1R2z)))(ξ(R2z)ξ0)0,p,R\displaystyle\left\|\left(\triangle_{g}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\right)(\xi^{\infty}(R^{2}z)-\xi_{0})\right\|_{0,p,R}
\displaystyle\leq C(1+r2)2r2log(1/δ)(ξ(R2z)ξ0)0,p,R\displaystyle\left\|\frac{C(1+r^{2})^{2}}{r^{2}\log(1/\delta)}(\xi^{\infty}(R^{2}z)-\xi_{0})\right\|_{0,p,R}
=\displaystyle= (A(1+r2)2|C(1+r2)2r2log(1/δ)(ξ(R2z)ξ0)|p)1p\displaystyle\left(\int_{A}(1+r^{2})^{-2}\left|\frac{C(1+r^{2})^{2}}{r^{2}\log(1/\delta)}(\xi^{\infty}(R^{2}z)-\xi_{0})\right|^{p}\right)^{\frac{1}{p}}
\displaystyle\leq Clog(1/δ)(1+(1δR)2)22p(A1r2p|ξ(R2z)ξ0|p)1p\displaystyle\frac{C}{\log(1/\delta)}\left(1+\left(\frac{1}{\delta R}\right)^{2}\right)^{2-\frac{2}{p}}\left(\int_{A}\frac{1}{r^{2p}}|\xi^{\infty}(R^{2}z)-\xi_{0}|^{p}\right)^{\frac{1}{p}}

We can consider changing coordinates w=1R2zw=\frac{1}{R^{2}z}. Since 1R|z|1δR\frac{1}{R}\leq|z|\leq\frac{1}{\delta R}, we have δR1R2z1R\frac{\delta}{R}\leq\frac{1}{R^{2}z}\leq\frac{1}{R}. Also, in the new coordinate system, ξ0=ξ(0)\xi_{0}=\xi^{\infty}(0). We have

(20) |ξ(w)ξ(0)|CξW2,p(S2)|w|22p\displaystyle|\xi^{\infty}(w)-\xi^{\infty}(0)|\leq C\|\xi^{\infty}\|_{W^{2,p}(S^{2})}|w|^{2-\frac{2}{p}}

where CC is a universal constant.

Thus we have

(A1r2p|ξ(R2z)ξ0|p)1p\displaystyle\left(\int_{A}\frac{1}{r^{2p}}|\xi^{\infty}(R^{2}z)-\xi_{0}|^{p}\right)^{\frac{1}{p}}
\displaystyle\leq CξW2,p(S2)(1R1δR1r2p|1R2r|2p2r𝑑r)1p\displaystyle C\|\xi^{\infty}\|_{W^{2,p}(S^{2})}\left(\int_{\frac{1}{R}}^{\frac{1}{\delta R}}\frac{1}{r^{2p}}\left|\frac{1}{R^{2}r}\right|^{2p-2}rdr\right)^{\frac{1}{p}}
=\displaystyle= CξW2,p(S2)(1R1δR(rR)44p1r𝑑r)1p\displaystyle C\|\xi^{\infty}\|_{W^{2,p}(S^{2})}\left(\int_{\frac{1}{R}}^{\frac{1}{\delta R}}(rR)^{4-4p}\frac{1}{r}dr\right)^{\frac{1}{p}}
\displaystyle\leq CξW2,p(S2)(1R1δR1r𝑑r)1p\displaystyle C\|\xi^{\infty}\|_{W^{2,p}(S^{2})}\left(\int_{\frac{1}{R}}^{\frac{1}{\delta R}}\frac{1}{r}dr\right)^{\frac{1}{p}}
\displaystyle\leq CξW2,p(S2)log(1/δ)1p\displaystyle C\|\xi^{\infty}\|_{W^{2,p}(S^{2})}\log(1/\delta)^{\frac{1}{p}}

Similar as in the proof of 5.1, we know

ξW2,p(S2)C(c,p,N,𝒰)η0,p,R\displaystyle\|\xi^{\infty}\|_{W^{2,p}(S^{2})}\leq C(c,p,N,\mathcal{U})\|\eta\|_{0,p,R}

Thus we get

(g(βδ,R(1R2z)))(ξ(R2z)ξ0)0,p,R\displaystyle\left\|\left(\triangle_{g}(\beta_{\delta,R}(\frac{1}{R^{2}z}))\right)(\xi^{\infty}(R^{2}z)-\xi_{0})\right\|_{0,p,R}
\displaystyle\leq C(c,p,N,𝒰)log(1/δ)11p(1+(1δR)2)22pη0,p,R\displaystyle\frac{C(c,p,N,\mathcal{U})}{\log(1/\delta)^{1-\frac{1}{p}}}\left(1+\left(\frac{1}{\delta R}\right)^{2}\right)^{2-\frac{2}{p}}\|\eta\|_{0,p,R}

Hence we can choose δ0\delta_{0} small enough only depending on c,p,N,𝒰c,p,N,\mathcal{U} to get the desired inequality. ∎

Proposition 5.1.

For any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p), we can choose δ0>0\delta_{0}>0 and c0>0c_{0}>0 only depending on c,p,N,𝒰c,p,N,\mathcal{U}, such that for any (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U} and any (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), the approximate right inverse TfRT_{f^{R}} defined in 5.1 satisfies:

(21) DfRTfRηη0,p,R12η0,p,R,TfRη2,p,Rc02η0,p,R\|D_{f^{R}}T_{f^{R}}\eta-\eta\|_{0,p,R}\leq\frac{1}{2}\|\eta\|_{0,p,R},\quad\|T_{f^{R}}\eta\|_{2,p,R}\leq\frac{c_{0}}{2}\|\eta\|_{0,p,R}

for every ηLp(S2,(fR)1TN)\eta\in L^{p}(S^{2},(f^{R})^{-1}TN).

Proof.

We have TfRη=ξRT_{f^{R}}\eta=\xi^{R} for each ηLfRp\eta\in L^{p}_{f^{R}}, and must prove that

(22) DfRξRη0,p,R12η0,p,R.\|D_{f^{R}}\xi^{R}-\eta\|_{0,p,R}\leq\frac{1}{2}\|\eta\|_{0,p,R}.

Since Df0,rξ0=η0D_{f^{0,r}}\xi^{0}=\eta^{0} and Df,rξ=ηD_{f^{\infty,r}}\xi^{\infty}=\eta^{\infty}, the term on the left hand side vanishes for |z|1/δR|z|\geq 1/\delta R and for |z|δ/R|z|\leq\delta/R. For δ/R|z|1/δR\delta/R\leq|z|\leq 1/\delta R, we can apply the previous two lemmas. The first equality is proved.

Before we prove the second inequality, we take a closer look of how the W2,pW^{2,p} norm is defined on the weighted sphere.

We still consider the stereographic projection of the weighted S2S^{2} (which is the connected sum of the original two S2S^{2}s). We consider ξW2,p(S2,(fR)1TN)\xi\in W^{2,p}(S^{2},(f^{R})^{-1}TN).

For |z|1R|z|\geq\frac{1}{R}, we have

fR(z)={f0(z),|z|2δRexpy(ρ(δRz)ζ0(z)),1R|z|2δRf^{R}(z)=\left\{\begin{aligned} &f^{0}(z),\ &&|z|\geq\frac{2}{\delta R}\\ &\exp_{y}(\rho(\delta Rz)\zeta^{0}(z)),\ &&\frac{1}{R}\leq|z|\leq\frac{2}{\delta R}\end{aligned}\right.

Since 𝒰\mathcal{U} is precompact, we know that the norm of dexpyd\exp_{y} and dexpy1d\exp_{y}^{-1} are uniformly bounded. Thus, there exists c1,c2c_{1},c_{2} only depending on c,p,N,𝒰c,p,N,\mathcal{U} such that

c1dexpyc2,c1dexpy1c2c_{1}\leq\|d\exp_{y}\|\leq c_{2},\quad c_{1}\leq\|d\exp_{y}^{-1}\|\leq c_{2}

Furthermore, since f0(z)=expyζ0(z)f^{0}(z)=\exp_{y}\zeta^{0}(z), we have ζ0(0)=0\zeta^{0}(0)=0, |ζ0(z)|Csupdf0|z|C(c)|z||\zeta^{0}(z)|\leq C\sup\|df^{0}\||z|\leq C(c)|z|, and |ζ0(z)|C(c,p,N,𝒰)supdf0|\nabla\zeta^{0}(z)|\leq C(c,p,N,\mathcal{U})\sup\|df^{0}\|. Thus for 1R|z|2δR\frac{1}{R}\leq|z|\leq\frac{2}{\delta R} we have

supdfRC(c,p,N,𝒰)\displaystyle\sup\|df^{R}\|\leq C(c,p,N,\mathcal{U})

We know that the image of |z|1R|z|\geq\frac{1}{R} under fRf^{R} is the same as the image of S2S^{2} under f0,rf^{0,r}, which is the same as the image of S2S^{2} under f0f^{0}. Since S2S^{2} is compact and 𝒰\mathcal{U} is precompact, we know there is a uniform injective radius only depending on c,p,N,𝒰c,p,N,\mathcal{U}.

We can choose coordinate charts as follows: We pick a set of points on S2S^{2} and a radius σ<π/2\sigma<\pi/2 such that the geodesic balls of radius σ\sigma and centered at those points yield a covering of S2S^{2}. Furthermore, since dfR\|df^{R}\| is bounded, we can choose σ\sigma small enough that for any of these geodesic balls, say Bσ(x1)B_{\sigma}(x_{1}), let the radius of the image be less than the injective radius of f0(S2)f^{0}(S^{2}), then considering the normal coordinates at fR(x1)f^{R}(x_{1}), we have a coordinate chart for the image of Bσ(x1)B_{\sigma}(x_{1}). We know that this coordinate chart will only depend on c,p,N,𝒰c,p,N,\mathcal{U}.

Now on each of these coordinate charts, we can consider the vectors in coordinates, and thus talk about the derivatives in coordinates. That is how we define the W2,pW^{2,p} space for |z|1R|z|\geq\frac{1}{R}. Since we choose σ<π/2\sigma<\pi/2, we know that in all these coordinate charts, if we consider the Riemann metric matrix gNg_{N} of NN, we have 0<C1gNC0<C^{-1}\leq\|g_{N}\|\leq C where CC is a universal constant. Thus we know that this norm is equivalent to the 2,p,R2,p,R norm, which is formed by using the stereographic projection coordinate chart with the weight.

Next, we consider |z|1R|z|\leq\frac{1}{R}:

fR(z)={f(R2z),|z|δ2Rexpy(ρ(δRz)ζ(R2z)),δ2R|z|1Rf^{R}(z)=\left\{\begin{aligned} &f^{\infty}\left(R^{2}z\right),\ &&|z|\leq\frac{\delta}{2R}\\ &\exp_{y}\left(\rho\left(\frac{\delta}{Rz}\right)\zeta^{\infty}(R^{2}z)\right),\ &&\frac{\delta}{2R}\leq|z|\leq\frac{1}{R}\end{aligned}\right.

By definition,

f,r(z)={f(),|z|δRexpy(ρ(δRz)ζ(z)),δR2|z|δRf(z),|z|δR2f^{\infty,r}(z)=\left\{\begin{aligned} &f^{\infty}(\infty),\ &&|z|\geq\delta R\\ &\exp_{y}\left(\rho\left(\frac{\delta R}{z}\right)\zeta^{\infty}(z)\right),\ &&\frac{\delta R}{2}\leq|z|\leq\delta R\\ &f^{\infty}(z),\ &&|z|\leq\frac{\delta R}{2}\end{aligned}\right.

For f,r(z)f^{\infty,r}(z), we can consider the coordinate change w=1zw=\frac{1}{z} and do the same as above for f0,rf^{0,r}. We want to show that this norm is equivalent to the 2,p,R2,p,R norm for |z|1R|z|\leq\frac{1}{R}. Namely, for ξW2,p\xi^{\infty}\in W^{2,p} defined on |z|R|z|\leq R (in S2S^{2}, not 2\mathbb{R}^{2}), we want to show that ξ(R2z)2,p,R\|\xi^{\infty}(R^{2}z)\|_{2,p,R} is equivalent to ξW2,p\|\xi^{\infty}\|_{W^{2,p}}.

From the construction of the W2,pW^{2,p} norm (details written in the case |z|1R|z|\geq\frac{1}{R}), we know the W2,pW^{2,p} norm is equivalent to

ξW2,p(|z|R)=\displaystyle\|\xi\|_{W^{2,p}(|z|\leq R)}= (|z|R(1+r2)2|ξ|p+(1+r2)p2(|sξ|p+|tξ|p)+\displaystyle\left(\int_{|z|\leq R}(1+r^{2})^{-2}|\xi|^{p}+(1+r^{2})^{p-2}(|\nabla_{s}\xi|^{p}+|\nabla_{t}\xi|^{p})+\right.
(1+r2)2p2(|2ξs2|p+|2ξst|p+|2ξt2|p))1p\displaystyle\left.(1+r^{2})^{2p-2}\left(\left|\frac{\partial^{2}\xi}{\partial s^{2}}\right|^{p}+\left|\frac{\partial^{2}\xi}{\partial s\partial t}\right|^{p}+\left|\frac{\partial^{2}\xi}{\partial t^{2}}\right|^{p}\right)\right)^{\frac{1}{p}}

We can use change of variables in the integration to directly verify that this is equivalent to the 2,p,R2,p,R norm. For example, consider ξ\nabla\xi and the change of variables w=R2zw=R^{2}z we have

|z|1R(R2+|z|2R2)2+p|R2ξ(R2z)|p\displaystyle\int_{|z|\leq\frac{1}{R}}(R^{-2}+|z|^{2}R^{2})^{-2+p}|R^{2}\nabla\xi^{\infty}(R^{2}z)|^{p}
=\displaystyle= R4|w|R(R2+|w|2R2)2+pR2p|ξ(w)|p\displaystyle R^{-4}\int_{|w|\leq R}(R^{-2}+|w|^{2}R^{-2})^{-2+p}R^{2p}|\nabla\xi^{\infty}(w)|^{p}
=\displaystyle= |w|R(1+|w|2)2+p|ξ(w)|p\displaystyle\int_{|w|\leq R}(1+|w|^{2})^{-2+p}|\nabla\xi^{\infty}(w)|^{p}

Now we have shown that W2,p(S2)W^{2,p}(S^{2}) norm is equivalent to 2,p,R2,p,R norm, where S2S^{2} is the weighted sphere. In particular, Lp(S2)L^{p}(S^{2}) norm is equivalent to 0,p,R0,p,R norm.

Now let’s come back to showing

TfRη2,p,Rc02η0,p,R\displaystyle\|T_{f^{R}}\eta\|_{2,p,R}\leq\frac{c_{0}}{2}\|\eta\|_{0,p,R}

From (18) and what we proved above, we know that we are only left to consider βδ,R(1R2z)(ξ(R2z)ξ0)\beta_{\delta,R}(\frac{1}{R^{2}z})(\xi^{\infty}(R^{2}z)-\xi_{0}) for 1R|z|1δR\frac{1}{R}\leq|z|\leq\frac{1}{\delta R} and βδ,R(z)(ξ0(z)ξ0)\beta_{\delta,R}(z)(\xi^{0}(z)-\xi_{0}) for δR|z|1R\frac{\delta}{R}\leq|z|\leq\frac{1}{R}.

For δR|z|1R\frac{\delta}{R}\leq|z|\leq\frac{1}{R}, when we consider the 2,p,R2,p,R norm, for parts where there is no derivative on β\beta, the second order derivative part can be controlled directly. The other parts can be estiamted in the same way as when there’s derivative on β\beta. Thus we only need to consider the parts (note that we will have to power by 1p\frac{1}{p}, which is not written out in the formula)

δR|z|1R\displaystyle\int_{\frac{\delta}{R}\leq|z|\leq\frac{1}{R}} (R2+R2|z|2)p2|βδ,R(z)(ξ0(z)ξ0)|p\displaystyle(R^{-2}+R^{2}|z|^{2})^{p-2}|\nabla\beta_{\delta,R}(z)(\xi^{0}(z)-\xi_{0})|^{p}
+(R2+R2|z|2)2p2(|2βδ,R(z)(ξ0(z)ξ0)|p+|βδ,R(z)ξ0(z)|p)\displaystyle+(R^{-2}+R^{2}|z|^{2})^{2p-2}(|\nabla^{2}\beta_{\delta,R}(z)(\xi^{0}(z)-\xi_{0})|^{p}+|\nabla\beta_{\delta,R}(z)\nabla\xi^{0}(z)|^{p})

For the first two parts, we use estimate for derivatives of β\beta as well as (19). For the last part, the estimate is similar to the estimate immediately above (19).

For 1R|z|1δR\frac{1}{R}\leq|z|\leq\frac{1}{\delta R}, for parts where there is no derivative of β\beta, we can just estimate

1R|z|1δR\displaystyle\int_{\frac{1}{R}\leq|z|\leq\frac{1}{\delta R}} (1+|z|2)2|ξ(R2z)ξ0|p+(1+|z|2)p2R2p|ξ(R2z)|p+\displaystyle(1+|z|^{2})^{-2}|\xi^{\infty}(R^{2}z)-\xi_{0}|^{p}+(1+|z|^{2})^{p-2}R^{2p}|\nabla\xi^{\infty}(R^{2}z)|^{p}+
(1+|z|2)2p2R4p|2ξ(R2z)|p\displaystyle(1+|z|^{2})^{2p-2}R^{4p}|\nabla^{2}\xi^{\infty}(R^{2}z)|^{p}

Similar as before, the second derivative part can be controlled directly. The other parts can be controlled in the same way as when there’s derivative on β\beta.

Now we are only left with

1R|z|1δR\displaystyle\int_{\frac{1}{R}\leq|z|\leq\frac{1}{\delta R}} (1+|z|2)p2|βδ,R(1R2z)1R2|z|2(ξ(R2z)ξ0)|p\displaystyle(1+|z|^{2})^{p-2}\left|\nabla\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\frac{1}{R^{2}|z|^{2}}(\xi^{\infty}(R^{2}z)-\xi_{0})\right|^{p}
+(1+|z|2)2p2|βδ,R(1R2z)2R2|z|3(ξ(R2z)ξ0)|p\displaystyle+(1+|z|^{2})^{2p-2}\left|\nabla\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\frac{2}{R^{2}|z|^{3}}(\xi^{\infty}(R^{2}z)-\xi_{0})\right|^{p}
+(1+|z|2)2p2(|2βδ,R(1R2z)1R4|z|4(ξ(z)ξ0)|p+|βδ,R(1R2z)1|z|2ξ(R2z)|p)\displaystyle+(1+|z|^{2})^{2p-2}\left(\left|\nabla^{2}\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\frac{1}{R^{4}|z|^{4}}(\xi^{\infty}(z)-\xi_{0})\right|^{p}+\left|\nabla\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\frac{1}{|z|^{2}}\nabla\xi^{\infty}(R^{2}z)\right|^{p}\right)

The estimate for the first three part is the same: we use (20) and change of coordinates. The estimate for the last part is the same as estimating

(1+|z|2)2r(βδ,R(1R2z))r(ξ(R2z))0,p,R\displaystyle\left\|(1+|z|^{2})^{2}\frac{\partial}{\partial r}\left(\beta_{\delta,R}\left(\frac{1}{R^{2}z}\right)\right)\frac{\partial}{\partial r}\left(\xi^{\infty}(R^{2}z)\right)\right\|_{0,p,R}
\displaystyle\leq C(c,p,N,𝒰)(1+(1δR)2)22p2πlog(1/δ)η0,p,R\displaystyle C(c,p,N,\mathcal{U})\left(1+\left(\frac{1}{\delta R}\right)^{2}\right)^{2-\frac{2}{p}}\sqrt{\frac{2\pi}{\log(1/\delta)}}\|\eta\|_{0,p,R}

The proof is complete. ∎

From the above proposition 5.1 we can construct the ’real’ right inverse:

Definition 5.2.

For any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p), we can choose δ0>0\delta_{0}>0 only depending on c,p,N,𝒰c,p,N,\mathcal{U}, such that for any (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U} and any (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), we define

QfR:=TfR(DfRTfR)1=k=0TfR(𝟙DfRTfR)kQ_{f^{R}}:=T_{f^{R}}(D_{f^{R}}T_{f^{R}})^{-1}=\sum_{k=0}^{\infty}T_{f^{R}}(\mathbbm{1}-D_{f^{R}}T_{f^{R}})^{k}

and we have

DfRQfR=\displaystyle D_{f^{R}}Q_{f^{R}}= 𝟙\displaystyle\mathbbm{1}
QfRη2,p,R\displaystyle\|Q_{f^{R}}\eta\|_{2,p,R}\leq c0η0,p,R\displaystyle c_{0}\|\eta\|_{0,p,R}

where c0c_{0} only depends on c,p,N,𝒰c,p,N,\mathcal{U}.

6. Construction of the gluing map

Let us further tailor the implicit function theorem for our setting:

Theorem 6.1.

For any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p) and δ0\delta_{0} as in definition 5.1, consider (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}). Let Σ\Sigma denote S2S^{2} with the weighted metric defined in the pregluing. Let f=(f0,f)𝒰f=(f^{0},f^{\infty})\in\mathcal{U}, consider Banach spaces X=W2,p(Σ,(fR)1TN)X=W^{2,p}(\Sigma,\left(f^{R}\right)^{-1}TN) and Y=Lp(Σ,(fR)1TN)Y=L^{p}(\Sigma,\left(f^{R}\right)^{-1}TN). Let UU be an open subset of XX, fR:UY\mathcal{F}_{f^{R}}:U\rightarrow Y be a continuously differentiable map. Suppose we have the following:

  1. (1)

    Consider 0U0\in U, DfR:=dfR(0)D_{f^{R}}:=d\mathcal{F}_{f^{R}}(0) is surjective and has a linear right inverse QQ such that Qc~\|Q\|\leq\tilde{c} for some constant c~\tilde{c}.

  2. (2)

    There exists a positive constant σ\sigma such that Bσ(0,X)UB_{\sigma}(0,X)\subset U, and

    dfR(ξ)DfR12c~\|d\mathcal{F}_{f^{R}}(\xi)-D_{f^{R}}\|\leq\frac{1}{2\tilde{c}}

    for all ξ<σ\|\xi\|<\sigma.

  3. (3)

    There exists some ξ1X\xi_{1}\in X that satisfies

    fR(ξ1)<σ4c~,ξ1<σ8\|\mathcal{F}_{f^{R}}(\xi_{1})\|<\frac{\sigma}{4\tilde{c}},\|\xi_{1}\|<\frac{\sigma}{8}

Then there exists a unique ξ\xi such that

fR(ξ)=0,ξξ1imQ,ξ<σ\mathcal{F}_{f^{R}}(\xi)=0,\ \xi-\xi_{1}\in imQ,\ \|\xi\|<\sigma

Moreover, ξξ12c~fR(ξ1)\|\xi-\xi_{1}\|\leq 2\tilde{c}\|\mathcal{F}_{f^{R}}(\xi_{1})\|

We can choose ξ1\xi_{1} to be 0, and in this section we will show that the conditions in the above theorem are satisfied.

First, let us estimate the norm of fR(0)\mathcal{F}_{f^{R}}(0) so that we know what σ/4c~\sigma/4\tilde{c} should be in the third condition of the theorem.

From (1) we know that fR(0)=0\mathcal{F}_{f^{R}}(0)=0 for |z|2δR|z|\geq\frac{2}{\delta R} and |z|δ2R|z|\leq\frac{\delta}{2R}.

Recall that for normal coordinates, we have the Taylor expansion of the metric:

gij(y1,,yn)=δij13ykylRiklj+O(|y|3)g_{ij}(y^{1},\cdots,y^{n})=\delta_{ij}-\frac{1}{3}y^{k}y^{l}R_{iklj}+O(|y|^{3})

and \mathcal{F} can be written as:

fR(0)k=\displaystyle\mathcal{F}_{f^{R}}(0)^{k}= g(fR)k+gαβ(ΓN)ijk(fR)(fR)ixα(fR)jxβ\displaystyle\triangle_{g}(f^{R})^{k}+g^{\alpha\beta}(\Gamma^{N})^{k}_{ij}(f^{R})\frac{\partial(f^{R})^{i}}{\partial x^{\alpha}}\frac{\partial(f^{R})^{j}}{\partial x^{\beta}}
=\displaystyle= gαβ2(fR)kxαxβ+gαβxα(fR)kxβ+gαβ12detgdetgxα(fR)kxβ+gαβ(ΓN)ijk(fR)(fR)ixα(fR)jxβ\displaystyle g^{\alpha\beta}\frac{\partial^{2}(f^{R})^{k}}{\partial x^{\alpha}\partial x^{\beta}}+\frac{\partial g^{\alpha\beta}}{\partial x^{\alpha}}\frac{\partial(f^{R})^{k}}{\partial x^{\beta}}+g^{\alpha\beta}\frac{1}{2\det g}\frac{\partial\det g}{\partial x^{\alpha}}\frac{\partial(f^{R})^{k}}{\partial x^{\beta}}+g^{\alpha\beta}(\Gamma^{N})_{ij}^{k}(f^{R})\frac{\partial(f^{R})^{i}}{\partial x^{\alpha}}\frac{\partial(f^{R})^{j}}{\partial x^{\beta}}

For δ2R|z|2δR\frac{\delta}{2R}\leq|z|\leq\frac{2}{\delta R}, we can choose normal coordinates at yNy\in N and have that all fR(z)f^{R}(z) lie in the image of the normal coordinate chart by making δ0=δ0(c,p,N,𝒰)\delta_{0}=\delta_{0}(c,p,N,\mathcal{U}) small enough.

For δR|z|1δR\frac{\delta}{R}\leq|z|\leq\frac{1}{\delta R}, we know that fRf^{R} is constant, so we have

fR(0)=P(fR)=0\displaystyle\mathcal{F}_{f^{R}}(0)=P(f^{R})=0

For 1δR|z|2δR\frac{1}{\delta R}\leq|z|\leq\frac{2}{\delta R}, we have:

fR(z)=\displaystyle f^{R}(z)= expy(ρ(δRz)ζ0(z))\displaystyle\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))
fRs=\displaystyle\frac{\partial f^{R}}{\partial s}= dexpy(ρ(δRz)ζ0(z))[δRρs(δRz)ζ0(z)+ρ(δRz)ζ0(z)s]\displaystyle d\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))\cdot\left[\delta R\frac{\partial\rho}{\partial s}(\delta Rz)\zeta^{0}(z)+\rho(\delta Rz)\frac{\partial\zeta^{0}(z)}{\partial s}\right]
2fRs2=\displaystyle\frac{\partial^{2}f^{R}}{\partial s^{2}}= dexpy(ρ(δRz)ζ0(z))[δ2R22ρs2(δRz)ζ0(z)+2δRρs(δRz)ζ0(z)s+ρ(δRz)2ζ0(z)s2]+\displaystyle d\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))\cdot\left[\delta^{2}R^{2}\frac{\partial^{2}\rho}{\partial s^{2}}(\delta Rz)\zeta^{0}(z)+2\delta R\frac{\partial\rho}{\partial s}(\delta Rz)\frac{\partial\zeta^{0}(z)}{\partial s}+\rho(\delta Rz)\frac{\partial^{2}\zeta^{0}(z)}{\partial s^{2}}\right]+
d2expy(ρ(δRz)ζ0(z))[δRρs(δRz)ζ0(z)+ρ(δRz)ζ0(z)s]2\displaystyle d^{2}\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))\left[\delta R\frac{\partial\rho}{\partial s}(\delta Rz)\zeta^{0}(z)+\rho(\delta Rz)\frac{\partial\zeta^{0}(z)}{\partial s}\right]^{2}
fR(0)k=\displaystyle\mathcal{F}_{f^{R}}(0)^{k}= (1+|z|2)2(2(fR)ks2+2(fR)kt2)+(1+|z|2)2s(fR)ks+(1+|z|2)2t(fR)kt+\displaystyle(1+|z|^{2})^{2}\left(\frac{\partial^{2}(f^{R})^{k}}{\partial s^{2}}+\frac{\partial^{2}(f^{R})^{k}}{\partial t^{2}}\right)+\frac{\partial(1+|z|^{2})^{2}}{\partial s}\frac{\partial(f^{R})^{k}}{\partial s}+\frac{\partial(1+|z|^{2})^{2}}{\partial t}\frac{\partial(f^{R})^{k}}{\partial t}+
(1+|z|2)62((1+|z|2)4s(fR)ks+(1+|z|2)4t(fR)kt)+\displaystyle\frac{(1+|z|^{2})^{6}}{2}\left(\frac{\partial(1+|z|^{2})^{-4}}{\partial s}\frac{\partial(f^{R})^{k}}{\partial s}+\frac{\partial(1+|z|^{2})^{-4}}{\partial t}\frac{\partial(f^{R})^{k}}{\partial t}\right)+
(1+|z|2)2(ΓN)ijk(fR)((fR)is(fR)js+(fR)it(fR)jt)\displaystyle(1+|z|^{2})^{2}(\Gamma^{N})_{ij}^{k}(f^{R})\left(\frac{\partial(f^{R})^{i}}{\partial s}\frac{\partial(f^{R})^{j}}{\partial s}+\frac{\partial(f^{R})^{i}}{\partial t}\frac{\partial(f^{R})^{j}}{\partial t}\right)
fR(0)=\displaystyle\|\mathcal{F}_{f^{R}}(0)\|= fR(0)0,p,R=(1δR|z|2δR(1+|z|2)2|fR(0)|p𝑑s𝑑t)1p\displaystyle\|\mathcal{F}_{f^{R}}(0)\|_{0,p,R}=\left(\int_{\frac{1}{\delta R}\leq|z|\leq\frac{2}{\delta R}}(1+|z|^{2})^{-2}|\mathcal{F}_{f^{R}}(0)|^{p}dsdt\right)^{\frac{1}{p}}

We have δR>1δ0\delta R>\frac{1}{\delta_{0}}. For δ0=δ0(c,p,N,𝒰)\delta_{0}=\delta_{0}(c,p,N,\mathcal{U}) small enough, the exponential map is a smooth isometry for |z|2δR|z|\leq\frac{2}{\delta R}. We have

0<c1(c,p,N,𝒰,δ0)dexpy(ζ0(z))c2(c,p,N,𝒰,δ0),|z|2δR\displaystyle 0<c_{1}(c,p,N,\mathcal{U},\delta_{0})\leq\|d\exp_{y}(\zeta^{0}(z))\|\leq c_{2}(c,p,N,\mathcal{U},\delta_{0}),\ \forall|z|\leq\frac{2}{\delta R}
d2expy(ζ0(z))c3(c,p,N,𝒰,δ0),|z|2δR\displaystyle\|d^{2}\exp_{y}(\zeta^{0}(z))\|\leq c_{3}(c,p,N,\mathcal{U},\delta_{0}),\ \forall|z|\leq\frac{2}{\delta R}
0<c1(c,p,N,𝒰,δ0)dexpy(ρ(δRz)ζ0(z))c2(c,p,N,𝒰,δ0),|z|2δR\displaystyle 0<c_{1}(c,p,N,\mathcal{U},\delta_{0})\leq\|d\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))\|\leq c_{2}(c,p,N,\mathcal{U},\delta_{0}),\ \forall|z|\leq\frac{2}{\delta R}
d2expy(ρ(δRz)ζ0(z))c3(c,p,N,𝒰,δ0),|z|2δR\displaystyle\|d^{2}\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))\|\leq c_{3}(c,p,N,\mathcal{U},\delta_{0}),\ \forall|z|\leq\frac{2}{\delta R}

Note that we may further assume c1c_{1} is decreasing with respect to δ0\delta_{0} while c2c_{2} and c3c_{3} are increasing.

Since f0(z)=expy(ζ0(z))f^{0}(z)=\exp_{y}(\zeta^{0}(z)), and the exponential map is a smooth isometry for |z|2δR|z|\leq\frac{2}{\delta R}, we have ζ0(z)=expy1(f0(z))\zeta^{0}(z)=\exp_{y}^{-1}(f^{0}(z)), and

dexpy1(f0(z))1/c1(c,p,N,𝒰,δ0),|z|2δR\displaystyle\|d\exp_{y}^{-1}(f^{0}(z))\|\leq 1/c_{1}(c,p,N,\mathcal{U},\delta_{0}),\ \forall|z|\leq\frac{2}{\delta R}
d2expy1(f0(z))c4(c,p,N,𝒰,δ0),|z|2δR\displaystyle\|d^{2}\exp_{y}^{-1}(f^{0}(z))\|\leq c_{4}(c,p,N,\mathcal{U},\delta_{0}),\ \forall|z|\leq\frac{2}{\delta R}
|ζ0(z)|cc1(c,p,N,𝒰,δ0)|z|,|z|2δR\displaystyle\left|\zeta^{0}(z)\right|\leq\frac{c}{c_{1}(c,p,N,\mathcal{U},\delta_{0})}|z|,\ \forall|z|\leq\frac{2}{\delta R}
ζ0(z)s=expy1(f0(z))s=dexpy1(f0(z))f0(z)s\displaystyle\frac{\partial\zeta^{0}(z)}{\partial s}=\frac{\partial\exp_{y}^{-1}(f^{0}(z))}{\partial s}=d\exp_{y}^{-1}(f^{0}(z))\frac{\partial f^{0}(z)}{\partial s}
2ζ0(z)s2=s(dexpy1(f0(z))f0(z)s)=d2expy1(f0(z))(f0(z)s)2+dexpy1(f0(z))2f0(z)s2\displaystyle\frac{\partial^{2}\zeta^{0}(z)}{\partial s^{2}}=\frac{\partial}{\partial s}\left(d\exp_{y}^{-1}(f^{0}(z))\frac{\partial f^{0}(z)}{\partial s}\right)=d^{2}\exp^{-1}_{y}(f^{0}(z))\left(\frac{\partial f^{0}(z)}{\partial s}\right)^{2}+d\exp_{y}^{-1}(f^{0}(z))\frac{\partial^{2}f^{0}(z)}{\partial s^{2}}

Similar to the above paragraph, we may assume c4c_{4} is increasing with respect to δ0\delta_{0}.

We now start to divide fR(0)\mathcal{F}_{f^{R}}(0) into parts and estimate:

(Note that here ρ\rho is a function defined on 2\mathbb{R}^{2} instead of \mathbb{R}, which is different from the ρ\rho we started with. Actually it is ρ\rho (that we started with) composed with absolute value function)

|2fRs2|=\displaystyle\left|\frac{\partial^{2}f^{R}}{\partial s^{2}}\right|= |dexpy(ρ(δRz)ζ0(z))[δ2R22ρs2(δRz)ζ0(z)+2δRρs(δRz)ζ0(z)s+ρ(δRz)2ζ0(z)s2]+\displaystyle\left|d\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))\cdot\left[\delta^{2}R^{2}\frac{\partial^{2}\rho}{\partial s^{2}}(\delta Rz)\zeta^{0}(z)+2\delta R\frac{\partial\rho}{\partial s}(\delta Rz)\frac{\partial\zeta^{0}(z)}{\partial s}+\rho(\delta Rz)\frac{\partial^{2}\zeta^{0}(z)}{\partial s^{2}}\right]+\right.
d2expy(ρ(δRz)ζ0(z))[δRρs(δRz)ζ0(z)+ρ(δRz)ζ0(z)s]2|\displaystyle\left.d^{2}\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))\left[\delta R\frac{\partial\rho}{\partial s}(\delta Rz)\zeta^{0}(z)+\rho(\delta Rz)\frac{\partial\zeta^{0}(z)}{\partial s}\right]^{2}\right|
\displaystyle\leq C(c,p,N,𝒰,δ0)[δR+1]\displaystyle C(c,p,N,\mathcal{U},\delta_{0})\left[\delta R+1\right]

where CC is increasing with respect to δ0\delta_{0}.

Things are the same for derivatives with respect to tt instead of ss.

|(fR)s|=\displaystyle\left|\frac{\partial(f^{R})}{\partial s}\right|= dexpy(ρ(δRz)ζ0(z))[δRρs(δRz)ζ0(z)+ρ(δRz)ζ0(z)s]\displaystyle d\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))\cdot\left[\delta R\frac{\partial\rho}{\partial s}(\delta Rz)\zeta^{0}(z)+\rho(\delta Rz)\frac{\partial\zeta^{0}(z)}{\partial s}\right]
\displaystyle\leq C(c,p,N,𝒰,δ0)\displaystyle C(c,p,N,\mathcal{U},\delta_{0})

where CC is increasing with respect to δ0\delta_{0}.

We can estimate from the formula of fR(0)\mathcal{F}_{f^{R}}(0) that

|fR(0)|\displaystyle|\mathcal{F}_{f^{R}}(0)|\leq C(c,p,N,𝒰,δ0)(δR+1)\displaystyle C(c,p,N,\mathcal{U},\delta_{0})(\delta R+1)
fR(0)0,p,R|{1δR|z|2δR}=\displaystyle\|\mathcal{F}_{f^{R}}(0)\|_{0,p,R}\bigg{\rvert}_{\{\frac{1}{\delta R}\leq|z|\leq\frac{2}{\delta R}\}}= (1δR|z|2δR(1+|z|2)2|fR(0)|p)1p\displaystyle\left(\int_{\frac{1}{\delta R}\leq|z|\leq\frac{2}{\delta R}}(1+|z|^{2})^{-2}|\mathcal{F}_{f^{R}}(0)|^{p}\right)^{\frac{1}{p}}
\displaystyle\leq C(c,p,N,𝒰,δ0)(δR+1)(δR)2p\displaystyle C(c,p,N,\mathcal{U},\delta_{0})\frac{(\delta R+1)}{(\delta R)^{\frac{2}{p}}}

where C(c,p,N,𝒰,δ0)C(c,p,N,\mathcal{U},\delta_{0}) here is increasing with respect to δ0\delta_{0}.

For δ2R|z|δR\frac{\delta}{2R}\leq|z|\leq\frac{\delta}{R}, we have:

fR(z)=\displaystyle f^{R}(z)= expy(ρ(δRz)ζ(R2z))\displaystyle\exp_{y}\left(\rho\left(\frac{\delta}{Rz}\right)\zeta^{\infty}(R^{2}z)\right)
|fRs|\displaystyle\left|\frac{\partial f^{R}}{\partial s}\right|\leq |dexpy(ρ(δRz)ζ(R2z))|[|δRz2||ρs(δRz)||ζ(R2z)|+ρ(δRz)R2|ξs(R2z)|]\displaystyle\left|d\exp_{y}\left(\rho\left(\frac{\delta}{Rz}\right)\zeta^{\infty}(R^{2}z)\right)\right|\left[\left|\frac{\delta}{Rz^{2}}\right|\left|\frac{\partial\rho}{\partial s}\left(\frac{\delta}{Rz}\right)\right|\left|\zeta^{\infty}(R^{2}z)\right|+\rho\left(\frac{\delta}{Rz}\right)R^{2}\left|\frac{\partial\xi^{\infty}}{\partial s}(R^{2}z)\right|\right]
|2fRs2|\displaystyle\left|\frac{\partial^{2}f^{R}}{\partial s^{2}}\right|\leq |dexpy(ρ(δRz)ζ(R2z))|[|2δRz3||ρs(δRz)||ζ(R2z)|+\displaystyle\left|d\exp_{y}\left(\rho\left(\frac{\delta}{Rz}\right)\zeta^{\infty}(R^{2}z)\right)\right|\left[\left|\frac{2\delta}{Rz^{3}}\right|\left|\frac{\partial\rho}{\partial s}\left(\frac{\delta}{Rz}\right)\right|\left|\zeta^{\infty}(R^{2}z)\right|+\right.
|δ2R2z4||2ρs2(δRz)||ζ(R2z)|+|δRz2||ρs(δRz)|R2|ζs(R2z)|+ρ(δRz)R4|2ξs2(R2z)|]+\displaystyle\left.\left|\frac{\delta^{2}}{R^{2}z^{4}}\right|\left|\frac{\partial^{2}\rho}{\partial s^{2}}\left(\frac{\delta}{Rz}\right)\right|\left|\zeta^{\infty}(R^{2}z)\right|+\left|\frac{\delta}{Rz^{2}}\right|\left|\frac{\partial\rho}{\partial s}\left(\frac{\delta}{Rz}\right)\right|R^{2}\left|\frac{\partial\zeta^{\infty}}{\partial s}(R^{2}z)\right|+\rho\left(\frac{\delta}{Rz}\right)R^{4}\left|\frac{\partial^{2}\xi^{\infty}}{\partial s^{2}}(R^{2}z)\right|\right]+
|d2expy(ρ(δRz)ζ(R2z))|[|δRz2||ρs(δRz)||ζ(R2z)|+ρ(δRz)R2|ξs(R2z)|]2\displaystyle\left|d^{2}\exp_{y}\left(\rho\left(\frac{\delta}{Rz}\right)\zeta^{\infty}(R^{2}z)\right)\right|\left[\left|\frac{\delta}{Rz^{2}}\right|\left|\frac{\partial\rho}{\partial s}\left(\frac{\delta}{Rz}\right)\right|\left|\zeta^{\infty}(R^{2}z)\right|+\rho\left(\frac{\delta}{Rz}\right)R^{2}\left|\frac{\partial\xi^{\infty}}{\partial s}(R^{2}z)\right|\right]^{2}
fR(0)k=\displaystyle\mathcal{F}_{f^{R}}(0)^{k}= (R2+R2|z|2)2(2(fR)ks2+2(fR)kt2)+\displaystyle(R^{-2}+R^{2}|z|^{2})^{2}\left(\frac{\partial^{2}(f^{R})^{k}}{\partial s^{2}}+\frac{\partial^{2}(f^{R})^{k}}{\partial t^{2}}\right)+
(R2+R2|z|2)2s(fR)ks+(R2+R2|z|2)2t(fR)kt+\displaystyle\frac{\partial(R^{-2}+R^{2}|z|^{2})^{2}}{\partial s}\frac{\partial(f^{R})^{k}}{\partial s}+\frac{\partial(R^{-2}+R^{2}|z|^{2})^{2}}{\partial t}\frac{\partial(f^{R})^{k}}{\partial t}+
(R2+R2|z|2)212(R2+R2|z|2)4((R2+R2|z|2)4s(fR)ks+(R2+R2|z|2)4t(fR)kt)+\displaystyle(R^{-2}+R^{2}|z|^{2})^{2}\frac{1}{2(R^{-2}+R^{2}|z|^{2})^{-4}}\left(\frac{\partial(R^{-2}+R^{2}|z|^{2})^{-4}}{\partial s}\frac{\partial(f^{R})^{k}}{\partial s}+\frac{\partial(R^{-2}+R^{2}|z|^{2})^{-4}}{\partial t}\frac{\partial(f^{R})^{k}}{\partial t}\right)+
(R2+R2|z|2)2(ΓN)ijk(fR)((fR)is(fR)is+(fR)it(fR)it)\displaystyle(R^{-2}+R^{2}|z|^{2})^{2}(\Gamma^{N})_{ij}^{k}(f^{R})\left(\frac{\partial(f^{R})^{i}}{\partial s}\frac{\partial(f^{R})^{i}}{\partial s}+\frac{\partial(f^{R})^{i}}{\partial t}\frac{\partial(f^{R})^{i}}{\partial t}\right)

Similar as we did for 1δR|z|2δR\frac{1}{\delta R}\leq|z|\leq\frac{2}{\delta R}, for δ0=δ0(c,p,N,𝒰)\delta_{0}=\delta_{0}(c,p,N,\mathcal{U}) small enough, we have:

0<c1(c,p,N,𝒰,δ0)dexpy(ζ(R2z))c2(c,p,N,𝒰,δ0),δ2R|z|δR,\displaystyle 0<c_{1}(c,p,N,\mathcal{U},\delta_{0})\leq\|d\exp_{y}(\zeta^{\infty}(R^{2}z))\|\leq c_{2}(c,p,N,\mathcal{U},\delta_{0}),\ \forall\frac{\delta}{2R}\leq|z|\leq\frac{\delta}{R},
d2expy(ζ(R2z))c3(c,p,N,𝒰,δ0),δ2R|z|δR,\displaystyle\|d^{2}\exp_{y}(\zeta^{\infty}(R^{2}z))\|\leq c_{3}(c,p,N,\mathcal{U},\delta_{0}),\ \forall\frac{\delta}{2R}\leq|z|\leq\frac{\delta}{R},
0<c1(c,p,N,𝒰,δ0)dexpy(ρ(δRz)ζ(R2z))c2(c,p,N,𝒰,δ0),δ2R|z|δR,\displaystyle 0<c_{1}(c,p,N,\mathcal{U},\delta_{0})\leq\left\|d\exp_{y}\left(\rho\left(\frac{\delta}{Rz}\right)\zeta^{\infty}(R^{2}z)\right)\right\|\leq c_{2}(c,p,N,\mathcal{U},\delta_{0}),\ \forall\frac{\delta}{2R}\leq|z|\leq\frac{\delta}{R},
d2expy(ρ(δRz)ζ(R2z))c3(c,p,N,𝒰,δ0),δ2R|z|δR,\displaystyle\|d^{2}\exp_{y}\left(\rho\left(\frac{\delta}{Rz}\right)\zeta^{\infty}(R^{2}z)\right)\|\leq c_{3}(c,p,N,\mathcal{U},\delta_{0}),\ \forall\frac{\delta}{2R}\leq|z|\leq\frac{\delta}{R},

where c1c_{1} is decreasing with respect to δ0\delta_{0} and c2,c3c_{2},c_{3} are increasing.

Since f(R2z)=expy(ζ(R2z))f^{\infty}(R^{2}z)=\exp_{y}(\zeta^{\infty}(R^{2}z)) for δ2R|z|δR\frac{\delta}{2R}\leq|z|\leq\frac{\delta}{R}, similar as the case when 1δR|z|2δR\frac{1}{\delta R}\leq|z|\leq\frac{2}{\delta R}, we have:

ζ(R2z)=\displaystyle\zeta^{\infty}(R^{2}z)= expy1(f(R2z))\displaystyle\exp_{y}^{-1}(f^{\infty}(R^{2}z))
dexpy1(f(R2z)\displaystyle\|d\exp_{y}^{-1}(f^{\infty}(R^{2}z)\|\leq 1/c1(c,p,N,𝒰,δ0),|z|δ2R,\displaystyle 1/c_{1}(c,p,N,\mathcal{U},\delta_{0}),\ \forall|z|\geq\frac{\delta}{2R},
d2expy1(f(R2z))\displaystyle\|d^{2}exp_{y}^{-1}(f^{\infty}(R^{2}z))\|\leq c4(c,p,N,𝒰,δ0),|z|δ2R,\displaystyle c_{4}(c,p,N,\mathcal{U},\delta_{0}),\ \forall|z|\geq\frac{\delta}{2R},
|ζ(R2z)|\displaystyle|\zeta^{\infty}(R^{2}z)|\leq cc1(c,p,N,𝒰,δ0)1R2|z|,|z|δ2R,\displaystyle\frac{c}{c_{1}(c,p,N,\mathcal{U},\delta_{0})}\frac{1}{R^{2}|z|},\ \forall|z|\geq\frac{\delta}{2R},
|ζ(R2z)s|=\displaystyle\left|\frac{\partial\zeta^{\infty}(R^{2}z)}{\partial s}\right|= |R2ζs(R2z)|=|R2dexpy1(f(R2z))fs(R2z)|\displaystyle\left|R^{2}\frac{\partial\zeta^{\infty}}{\partial s}(R^{2}z)\right|=\left|R^{2}d\exp_{y}^{-1}(f^{\infty}(R^{2}z))\frac{\partial f^{\infty}}{\partial s}(R^{2}z)\right|
\displaystyle\leq cc1(c,p,N,𝒰,δ0)R21+δ2R24,|z|δ2R\displaystyle\frac{c}{c_{1}(c,p,N,\mathcal{U},\delta_{0})}\frac{R^{2}}{1+\frac{\delta^{2}R^{2}}{4}},\ \forall|z|\geq\frac{\delta}{2R}
|2ζ(R2z)s2|=\displaystyle\left|\frac{\partial^{2}\zeta^{\infty}(R^{2}z)}{\partial s^{2}}\right|= |R42ζs2(R2z)|\displaystyle\left|R^{4}\frac{\partial^{2}\zeta^{\infty}}{\partial s^{2}}(R^{2}z)\right|
=\displaystyle= |R4d2expy1(f(R2z))(fs(R2z))2+R4dexpy1(f(R2z))2fs2(R2z)|\displaystyle\left|R^{4}d^{2}\exp^{-1}_{y}(f^{\infty}(R^{2}z))\left(\frac{\partial f^{\infty}}{\partial s}(R^{2}z)\right)^{2}+R^{4}d\exp^{-1}_{y}(f^{\infty}(R^{2}z))\frac{\partial^{2}f^{\infty}}{\partial s^{2}}(R^{2}z)\right|
\displaystyle\leq c4(c,p,N,𝒰,δ0)(c2+c)R4(1+δ2R24)2,|z|δ2R\displaystyle c_{4}(c,p,N,\mathcal{U},\delta_{0})(c^{2}+c)\frac{R^{4}}{\left(1+\frac{\delta^{2}R^{2}}{4}\right)^{2}},\ \forall|z|\geq\frac{\delta}{2R}

We now divide fR(0)\mathcal{F}_{f^{R}}(0) into parts and estimate, eventually we will get the same estimate as in the case 1δR|z|2δR\frac{1}{\delta R}\leq|z|\leq\frac{2}{\delta R}.

Now we can take a look at what we should choose to be σ\sigma and cc in the condition of the implicit function theorem. First, from the previous section, we have

QfR2,p,Rc0(c,p,N,𝒰)\displaystyle\|Q_{f^{R}}\|_{2,p,R}\leq c_{0}(c,p,N,\mathcal{U})

So we should choose c~=c0(c,p,N,𝒰)\tilde{c}=c_{0}(c,p,N,\mathcal{U}).

Since we want

fR(0)0,p,R<σ4c~\displaystyle\|\mathcal{F}_{f^{R}}(0)\|_{0,p,R}<\frac{\sigma}{4\tilde{c}}

while we have proved

fR(0)0,p,RC(c,p,N,𝒰,δ0)(δR+1)(δR)2p<C(c,p,N,𝒰,δ0)(1+δ0)(δR)12p\displaystyle\|\mathcal{F}_{f^{R}}(0)\|_{0,p,R}\leq C(c,p,N,\mathcal{U},\delta_{0})\frac{(\delta R+1)}{(\delta R)^{\frac{2}{p}}}<C(c,p,N,\mathcal{U},\delta_{0})(1+\delta_{0})(\delta R)^{1-\frac{2}{p}}

where CC is increasing with respect to δ0\delta_{0}.

We first choose c0c_{0} and δ0\delta_{0} from proposition 5.1. We know that the results of the proposition still hold if we make δ0\delta_{0} smaller. Furthermore, we have seen that C(c,p,N,𝒰,δ0)C(c,p,N,\mathcal{U},\delta_{0}) is increasing with respect to δ0\delta_{0}.

For σ\sigma, we first require Bσ(0,X)UB_{\sigma}(0,X)\subset U. This upper bound for σ\sigma only depends on c,p,N,𝒰c,p,N,\mathcal{U}.

Then we can consider the (0,p,R)(0,p,R)-norm of fR(0)\mathcal{F}_{f^{R}}(0) being smaller than σ/c~\sigma/\tilde{c}. By making δ0\delta_{0} small enough only depending on c,p,N,𝒰c,p,N,\mathcal{U}, we can make σ\sigma as small as we like. Now we want to have

ξ<σdfR(ξ)DfR12c~\displaystyle\|\xi\|<\sigma\Rightarrow\|d\mathcal{F}_{f^{R}}(\xi)-D_{f^{R}}\|\leq\frac{1}{2\tilde{c}}

Consider dfR(ξ)DfR\|d\mathcal{F}_{f^{R}}(\xi)-D_{f^{R}}\|. Let’s recall how we computed the linearization. In coordinates, we can write the harmonic map equation as

P(f)=(gfi+gαβΓjli(f(x))flxαfjxβ)yiP(f)=\left(\triangle_{g}f^{i}+g^{\alpha\beta}\Gamma^{i}_{jl}(f(x))\frac{\partial f^{l}}{\partial x^{\alpha}}\frac{\partial f^{j}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}

(f)\mathcal{F}(f) = Φ(f)1P(f)\Phi(f)^{-1}P(f). We choose normal coordinates at yy,

P(expf(ξ))=(g(fi+ξi)+gαβΓjli(f+ξ)(f+ξ)lxα(f+ξ)jxβ)yi\displaystyle P(\exp_{f}(\xi))=\left(\triangle_{g}(f^{i}+\xi^{i})+g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi)\frac{\partial(f+\xi)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{j}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}
P(expf(ξ+λζ))=(g(fi+ξi+λζi)+gαβΓjli(f+ξ+λζ)(f+ξ+λζ)lxα(f+ξ+λζ)jxβ)yi\displaystyle P(\exp_{f}(\xi+\lambda\zeta))=\left(\triangle_{g}(f^{i}+\xi^{i}+\lambda\zeta^{i})+g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi+\lambda\zeta)\frac{\partial(f+\xi+\lambda\zeta)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi+\lambda\zeta)^{j}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}

To compute the linearization, note that the coordinate vectors are parallel along radial geodesics under normal coordinates

df(ξ)(ζ)=\displaystyle d\mathcal{F}_{f}(\xi)(\zeta)= ddλf(ξ+λζ)λ=0\displaystyle\frac{d}{d\lambda}\mathcal{F}_{f}(\xi+\lambda\zeta)\mid_{\lambda=0}
=\displaystyle= limλ0Φf(ξ+λζ)1P(expf(ξ+λζ))Φf(ξ)1P(expf(ξ))λ\displaystyle\lim_{\lambda\rightarrow 0}\frac{\Phi_{f}(\xi+\lambda\zeta)^{-1}P(\exp_{f}(\xi+\lambda\zeta))-\Phi_{f}(\xi)^{-1}P(\exp_{f}(\xi))}{\lambda}
=\displaystyle= limλ01λ(g(λζi)+gαβΓjli(f+ξ+λζ)(f+ξ+λζ)lxα(f+ξ+λζ)jxβ\displaystyle\lim_{\lambda\rightarrow 0}\frac{1}{\lambda}\left(\triangle_{g}(\lambda\zeta^{i})+g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi+\lambda\zeta)\frac{\partial(f+\xi+\lambda\zeta)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi+\lambda\zeta)^{j}}{\partial x^{\beta}}\right.
gαβΓjli(f+ξ)(f+ξ)lxα(f+ξ)jxβ)yi\displaystyle\left.-g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi)\frac{\partial(f+\xi)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{j}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}
=\displaystyle= (gζi+gαβsΓjli(f+ξ)ζs(f+ξ)lxα(f+ξ)jxβ+2gαβΓjli(f+ξ)ζlxα(f+ξ)jxβ)yi\displaystyle\left(\triangle_{g}\zeta^{i}+g^{\alpha\beta}\partial_{s}\Gamma^{i}_{jl}(f+\xi)\zeta^{s}\frac{\partial(f+\xi)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{j}}{\partial x^{\beta}}+2g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi)\frac{\partial\zeta^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{j}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}

Also recall that

Dfζ=(gζi+gαβsΓjli(f)ζsflxαfjxβ)yi\displaystyle D_{f}\zeta=\left(\triangle_{g}\zeta^{i}+g^{\alpha\beta}\partial_{s}\Gamma^{i}_{jl}(f)\zeta^{s}\frac{\partial f^{l}}{\partial x^{\alpha}}\frac{\partial f^{j}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}

Let’s estimate df(ξ)(ζ)Dfζ0,p,R\|d\mathcal{F}_{f}(\xi)(\zeta)-D_{f}\zeta\|_{0,p,R}:

df(ξ)(ζ)Dfζ=\displaystyle d\mathcal{F}_{f}(\xi)(\zeta)-D_{f}\zeta= (gαβζs(sΓjli(f+ξ)(f+ξ)lxα(f+ξ)ixβsΓjli(f)flxαfjxβ)\displaystyle\left(g^{\alpha\beta}\zeta^{s}\left(\partial_{s}\Gamma^{i}_{jl}(f+\xi)\frac{\partial(f+\xi)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{i}}{\partial x^{\beta}}-\partial_{s}\Gamma^{i}_{jl}(f)\frac{\partial f^{l}}{\partial x^{\alpha}}\frac{\partial f^{j}}{\partial x^{\beta}}\right)\right.
+2gαβΓjli(f+ξ)ζlxα(f+ξ)jxβ)yi\displaystyle\left.+2g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi)\frac{\partial\zeta^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{j}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}

Note that at every point, we are considering the normal coordinates at that point, so we will need to consider a coordinate change to get the real norm. By lemma 2.1, we can directly use the above formula to estimate df(ξ)(ζ)Dfζ\|d\mathcal{F}_{f}(\xi)(\zeta)-D_{f}\zeta\|:

df(ξ)(ζ)Dfζ=\displaystyle d\mathcal{F}_{f}(\xi)(\zeta)-D_{f}\zeta= (gαβζs(sΓjli(f+ξ)(f+ξ)lxα(f+ξ)ixβsΓjli(f)flxαfjxβ)\displaystyle\left(g^{\alpha\beta}\zeta^{s}\left(\partial_{s}\Gamma^{i}_{jl}(f+\xi)\frac{\partial(f+\xi)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{i}}{\partial x^{\beta}}-\partial_{s}\Gamma^{i}_{jl}(f)\frac{\partial f^{l}}{\partial x^{\alpha}}\frac{\partial f^{j}}{\partial x^{\beta}}\right)\right.
+2gαβΓjli(f+ξ)ζlxα(f+ξ)jxβ)yi\displaystyle\left.+2g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi)\frac{\partial\zeta^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{j}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}
gαβζs(sΓjli(f+ξ)(f+ξ)lxα(f+ξ)ixβsΓjli(f)flxαfjxβ)\displaystyle g^{\alpha\beta}\zeta^{s}\left(\partial_{s}\Gamma^{i}_{jl}(f+\xi)\frac{\partial(f+\xi)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{i}}{\partial x^{\beta}}-\partial_{s}\Gamma^{i}_{jl}(f)\frac{\partial f^{l}}{\partial x^{\alpha}}\frac{\partial f^{j}}{\partial x^{\beta}}\right)
=\displaystyle= gαβζs(sΓjli(f+ξ)(f+ξ)lxα(f+ξ)ixβsΓjli(f)(f+ξ)lxα(f+ξ)ixβ\displaystyle g^{\alpha\beta}\zeta^{s}\left(\partial_{s}\Gamma^{i}_{jl}(f+\xi)\frac{\partial(f+\xi)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{i}}{\partial x^{\beta}}-\partial_{s}\Gamma^{i}_{jl}(f)\frac{\partial(f+\xi)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{i}}{\partial x^{\beta}}\right.
+sΓjli(f)(f+ξ)lxα(f+ξ)ixβsΓjli(f)flxαfjxβ)\displaystyle+\left.\partial_{s}\Gamma^{i}_{jl}(f)\frac{\partial(f+\xi)^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{i}}{\partial x^{\beta}}-\partial_{s}\Gamma^{i}_{jl}(f)\frac{\partial f^{l}}{\partial x^{\alpha}}\frac{\partial f^{j}}{\partial x^{\beta}}\right)

note that we are using coordinate chart centered at f(x)f(x), so we have:

2gαβΓjli(f+ξ)ζlxα(f+ξ)jxβ\displaystyle 2g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi)\frac{\partial\zeta^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{j}}{\partial x^{\beta}}
=\displaystyle= 2gαβΓjli(f+ξ)ζlxα(f+ξ)jxβ2gαβΓjli(f)ζlxα(f+ξ)jxβ\displaystyle 2g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi)\frac{\partial\zeta^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{j}}{\partial x^{\beta}}-2g^{\alpha\beta}\Gamma^{i}_{jl}(f)\frac{\partial\zeta^{l}}{\partial x^{\alpha}}\frac{\partial(f+\xi)^{j}}{\partial x^{\beta}}

For |z|2δR|z|\geq\frac{2}{\delta R}, we have fR=f0f^{R}=f^{0}, so everything does not depend on δ\delta or RR. So for δ0\delta_{0} sufficiently small only depending on c,p,N,𝒰c,p,N,\mathcal{U}, we can assume df(ξ)Df18c~\|d\mathcal{F}_{f}(\xi)-D_{f}\|\leq\frac{1}{8\tilde{c}}.

For 1δR|z|2δR\frac{1}{\delta R}\leq|z|\leq\frac{2}{\delta R}, note that the C0C^{0} norm is controlled by the 2,p,R2,p,R norm of ζ\zeta, where the constant only depends on c,p,N,𝒰c,p,N,\mathcal{U}. Thus we can also estimate that for δ0\delta_{0} sufficiently small, we have df(ξ)Df18c~\|d\mathcal{F}_{f}(\xi)-D_{f}\|\leq\frac{1}{8\tilde{c}}.

For δR|z|1δR\frac{\delta}{R}\leq|z|\leq\frac{1}{\delta R}, we have fR(z)yf^{R}(z)\equiv y, and we also have df(ξ)Df18c~\|d\mathcal{F}_{f}(\xi)-D_{f}\|\leq\frac{1}{8\tilde{c}} for δ0\delta_{0} sufficiently small. The rest is similar.

Thus we can apply the implicit function theorem. In particular, we know there exists a unique ξ\xi such that expfR(ξ)\exp_{f^{R}}(\xi) is a harmonic map.

Although we are considering S2S^{2}s, everything can be done the same way for general closed Riemannian manifolds of dimension 2. Combining the above, we arrive at the following theorem:

Theorem 1.1 (Existence of Gluing Map).

Given any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p). There exists δ0=δ0(c,p,Σ1,Σ2,x1,x2,N,𝒰)>0\delta_{0}=\delta_{0}(c,p,\Sigma_{1},\Sigma_{2},x_{1},x_{2},N,\mathcal{U})>0, such that for each pair of (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), there exists a gluing map ıδ,R:𝒰W2,p(Σ1#δ,RΣ2,N)\imath_{\delta,R}:\mathcal{U}\rightarrow W^{2,p}(\Sigma_{1}\#_{\delta,R}\Sigma_{2},N) such that each element in ıδ,R(𝒰)\imath_{\delta,R}(\mathcal{U}) satisfies the harmonic map equation, where Σ1#δ,RΣ2\Sigma_{1}\#_{\delta,R}\Sigma_{2} is the glued manifold as defined in 2. Furthermore, for any σ>0\sigma>0, we can choose δ0=δ0(c,p,Σ1,Σ2,x1,x2,N,𝒰)\delta_{0}=\delta_{0}(c,p,\Sigma_{1},\Sigma_{2},x_{1},x_{2},N,\mathcal{U}) such that, for any (f1,f2)𝒰(f_{1},f_{2})\in\mathcal{U}, there exists ξRW2,p(Σ1#δ,RΣ2)\xi^{R}\in W^{2,p}(\Sigma_{1}\#_{\delta,R}\Sigma_{2}) satisfying

ıδ,R((f1,f2))=expfRξR,ξR2,p,R<σ\imath_{\delta,R}((f_{1},f_{2}))=\exp_{f^{R}}\xi^{R},\quad\|\xi^{R}\|_{2,p,R}<\sigma

where fRf^{R} denotes the pregluing of f1,f2f_{1},f_{2} defined in (1).

7. A trick for estimate

From [21] we know there is a trick to ’suspend’ a harmonic map to get a conformal harmonic map. Indeed, consider a harmonic map f:DNf:D\rightarrow N from a disk DD to a closed Riemannian manifold NN and the Hopf differential

|ψf|2dz2=[|fx|2|fy|22ifx,fy]dz2|\psi_{f}|^{2}dz^{2}=[|f_{x}|^{2}-|f_{y}|^{2}-2i\langle f_{x},f_{y}\rangle]dz^{2}

Note that ψ\psi is holomorphic, and vanishes \iff ff is conformal. The suspension FF of ff is defined by finding the unique solution to

¯ξ=0,ξ=14ψf,ξD=0\overline{\partial}\xi=0,\quad\partial\xi=-\frac{1}{4}\psi_{f},\quad\xi\mid_{\partial D}=0

and let

F=(f,z¯+ξ):DN×/2F=(f,\bar{z}+\xi):D\rightarrow N\times\mathbb{C}/\mathbb{Z}^{2}

Then FF is harmonic and conformal. In [21], it’s \mathbb{C} instead of /2\mathbb{C}/\mathbb{Z}^{2}. This change is due to our desire to make target manifold of FF compact, so that we can apply estimates from [23]. We will also use estimates given in [21]:

For any subdomain ADA\subset D,

(23) Ae(F)Ae(f)+Area(A)[1+c12E2(f)]\int_{A}e(F)\leq\int_{A}e(f)+Area(A)[1+c_{1}^{2}E^{2}(f)]

For the circle SrDS_{r}\subset D,

(24) length(F(Sr))length(f(Sr))+2r(1+c12E(f))length(F(S_{r}))\leq length(f(S_{r}))+2r(1+c_{1}^{2}E(f))
Lemma 7.1 (Isopermetric Inequality).

(See lemma 2.4 in [21]) Let (M,g)(M,g) be a closed Riemannian manifold and let Σ\Sigma be a minimal surface in (M×2/2,g+dx2+dy2)(M\times\mathbb{R}^{2}/\mathbb{Z}^{2},\ g+dx^{2}+dy^{2}). Then there are constants cM,ϵM>0c_{M},\epsilon_{M}>0 such that if Area(Σ)<ϵMArea(\Sigma)<\epsilon_{M}, then Area(Σ)cMlength2(Σ)Area(\Sigma)\leq c_{M}length^{2}(\partial\Sigma).

The only difference is that here we are considering 2/2\mathbb{R}^{2}/\mathbb{Z}^{2} instead of 2\mathbb{R}^{2}, but that doesn’t matter. The lemma essentially follows from the isoperimetric inequality. The next two lemmas are analogous to lemma 4.7.3 and 4.7.5 in [16]. One major difference between our case and the case of symplectic geometry is that, in the latter, there is the existence of Darboux coordinate charts, which gives better constant for the isoperimetric inequality. We will see that this will affect our choice of parameters later.

Proposition 7.1 (Propositions of Harmonic Maps).

(See proposition 1.1 in [21]) There are positive constants C1C_{1} and ϵ0\epsilon_{0}, depending only on (Σ,h)(\Sigma,h) and (M,g)(M,g) such that

  1. (1)

    (Sup Estimate) If f:ΣMf:\Sigma\rightarrow M is harmonic and D(2r)D(2r) is a geodesic disk of radius 2r2r with energy E(2r)=12D(2r)|df|2ϵ0E(2r)=\frac{1}{2}\int_{D(2r)}|df|^{2}\leq\epsilon_{0}, then

    supD(r)|df|2C1r2E(2r).\underset{D(r)}{\sup}{|df|^{2}}\leq C_{1}r^{-2}E(2r).
  2. (2)

    (Uniform Convergence) If {fn}\{f_{n}\} is a sequence of harmonic maps from a disk D(2r)D(2r) with E(2r)<ϵ0E(2r)<\epsilon_{0} for all nn, then there is a subsequence that converges in C1C^{1}.

  3. (3)

    (Energy Gap) Any non-trivial harmonic map f:S2Mf:S^{2}\rightarrow M has energy E(f)ϵ0E(f)\geq\epsilon_{0}.

  4. (4)

    (Removable Singularities) Any smooth finite-energy harmonic map from a punctured disk D{0}D-\{0\} to MM extends to a smooth harmonic map on DD.

Lemma 7.2 (Compactness of Harmonic Maps).

(See lemma 1.2 in [21]) Let {hn}\{h_{n}\} be a sequence of metrics on Σ\Sigma converging in C2C^{2} to hh, and {fn}\{f_{n}\} a sequence of hnh_{n}-harmonic maps ΣM\Sigma\rightarrow M with E(fn)E0E(f_{n})\leq E_{0}. Then there is a subsequence of {fn}\{f_{n}\}, a finite set of ”bubble points” {x1,,xk}Σ\{x_{1},\cdots,x_{k}\}\in\Sigma, and an hh-harmonic map f:ΣMf_{\infty}:\Sigma\rightarrow M such that

  1. (1)

    fnff_{n}\rightarrow f_{\infty} in C1C^{1} uniformly on compact sets in Σ{x1,,xk}\Sigma-\{x_{1},\cdots,x_{k}\},

  2. (2)

    the energy densities e(fn)=12|dfn|2dvhne(f_{n})=\frac{1}{2}|df_{n}|^{2}dv_{h_{n}} converge as measures to e(f)e(f_{\infty}) plus a sum of point measures with mass miϵ0m_{i}\geq\epsilon_{0}:

    e(fn)e(f)+i=1kmiδ(xi).e(f_{n})\rightarrow e(f_{\infty})+\sum_{i=1}^{k}m_{i}\delta(x_{i}).

Now we are ready to prove the following lemmas. These are analogues to lemma 4.7.3 and 4.7.5 in [16].

Lemma 7.3.

Let (Σ,g)(\Sigma,g) be a compact Riemannian manifold of dimension 2. Then there exist constants 0<μ<1,δ>0,c>00<\mu<1,\delta>0,c>0 such that the following holds:

If 0<r<R0<r<R and F:A(r,R)NF:A(r,R)\rightarrow N is a conformal harmonic map satisfying

E(F):=E(F;A(r,R))<δE(F):=E(F;A(r,R))<\delta

then

  1. (1)

    E(F;A(eTr,eTR))4μe2μTE(F)E(F;A(e^{T}r,e^{-T}R))\leq\frac{4^{\mu}}{e^{2\mu T}}E(F)

  2. (2)

    supz,zA(eTr,eTR)d(F(z),F(z))ceμTE(F)\underset{z,z^{\prime}\in A(e^{T}r,e^{-T}R)}{\sup}d(F(z),F(z^{\prime}))\leq\frac{c}{e^{\mu T}}\sqrt{E(F)}

for log2TlogR/r\log 2\leq T\leq\log\sqrt{R/r}.

Proof.

For every conformal harmonic map FF on (Σ,g)(\Sigma,g), we have

E(F;Br)<δE(F)length(F(Br))24πμE(F;B_{r})<\delta\quad\Rightarrow\quad E(F)\leq\frac{length(F(\partial B_{r}))^{2}}{4\pi\mu}

As long as cN1/(4πμ)c_{N}\leq 1/(4\pi\mu) where cNc_{N} is as in the previous lemma, and δ\delta is small enough as in lemma 7.1. In other words, we can choose μ\mu such that 0<μ<10<\mu<1 and μ1/(4πcN)\mu\leq 1/(4\pi c_{N}).

Without loss of generality, we consider F:A(1/R,R)NF:A(1/R,R)\rightarrow N, where FF is a conformal harmonic map and R>1R>1. Let E(F)<δE(F)<\delta. Define v:[logR,logR]×/2πNv:[-\log R,\log R]\times\mathbb{R}/2\pi\mathbb{Z}\rightarrow N such that

v(s,t):=F(es+it),logR<s<logR,t/2πv(s,t):=F(e^{s+it}),\quad-\log R<s<\log R,\quad t\in\mathbb{R}/2\pi\mathbb{Z}

This is a conformal map composite with FF, so it is still a conformal harmonic map.

For s[logR,logR]s\in[-\log R,\log R] define γs:/2πN\gamma_{s}:\mathbb{R}/2\pi\mathbb{Z}\rightarrow N by

γs(t):=F(es+it)=v(s,t)\gamma_{s}(t):=F(e^{s+it})=v(s,t)

Let r=log2r=\log 2, we have

|γs˙(t)|2=|sv(s,t)|2C(log2)2E(F),|s|log(R/2)|\dot{\gamma_{s}}(t)|^{2}=|\partial_{s}v(s,t)|^{2}\leq\frac{C}{(\log 2)^{2}}E(F),\quad|s|\leq\log(R/2)

Thus

length(γs)2\displaystyle length(\gamma_{s})^{2} =(02π|γ˙s(t)|𝑑t)2\displaystyle=\left(\int_{0}^{2\pi}|\dot{\gamma}_{s}(t)|dt\right)^{2}
2π02π|γ˙s(t)|2𝑑t\displaystyle\leq 2\pi\int_{0}^{2\pi}|\dot{\gamma}_{s}(t)|^{2}dt
4π2C(log2)2E(F)\displaystyle\leq\frac{4\pi^{2}C}{(\log 2)^{2}}E(F)

Now consider the function

ϵ(T):=E(F;A(eT/R,eTR)),log2TlogR\epsilon(T):=E(F;A(e^{T}/R,e^{-T}R)),\quad\log 2\leq T\leq\log R

Abbreviate ρ:=logR\rho:=\log R. Then

ϵ(T)=\displaystyle\epsilon(T)= E(v;[ρ+T,ρT]×/2π)\displaystyle E(v;[-\rho+T,\rho-T]\times\mathbb{R}/2\pi\mathbb{Z})
=\displaystyle= ρ+TρT02π|sv(s,t)|2𝑑t𝑑s\displaystyle\int_{-\rho+T}^{\rho-T}\int_{0}^{2\pi}|\partial_{s}v(s,t)|^{2}dtds

On the other hand,

ϵ(T)=\displaystyle\epsilon(T)= E(F;B(eT/R))E(F;B(eTR))\displaystyle E(F;B(e^{T}/R))-E(F;B(e^{-T}R))
\displaystyle\leq 14πμ(length(γρ+T)2+length(γρT)2)\displaystyle\frac{1}{4\pi\mu}(length(\gamma_{-\rho+T})^{2}+length(\gamma_{\rho-T})^{2})
\displaystyle\leq 12μ02π|sv(ρ+T,t)|2𝑑t+12μ02π|sv(ρT,t)|2𝑑t\displaystyle\frac{1}{2\mu}\int_{0}^{2\pi}|\partial_{s}v(-\rho+T,t)|^{2}dt+\frac{1}{2\mu}\int_{0}^{2\pi}|\partial_{s}v(\rho-T,t)|^{2}dt
=\displaystyle= 12μϵ˙(T)\displaystyle-\frac{1}{2\mu}\dot{\epsilon}(T)

Thus

ϵ˙(T)2μϵ(T)<0\dot{\epsilon}(T)\leq-2\mu\epsilon(T)<0
ϵ(T)e2μ(Tlog2)ϵ(log2)4μe2μTE(F)\epsilon(T)\leq e^{-2\mu(T-\log 2)}\epsilon(\log 2)\leq 4^{\mu}e^{-2\mu T}E(F)

for log2TlogR\log 2\leq T\leq\log R.

To prove the seond part, we will need the following estimate:

|sv(s,t)|2C4μe2μ(log2)2e2μ(ρ|s|)E(v),|s|ρlog2|\partial_{s}v(s,t)|^{2}\leq\frac{C4^{\mu}e^{2\mu}}{(\log 2)^{2}}e^{-2\mu(\rho-|s|)}E(v),\quad|s|\leq\rho-\log 2

First assume that

ρlog21|s|ρlog2\rho-\log 2-1\leq|s|\leq\rho-\log 2

Then we have

|sv(s,t)|2C(log2)2E(v)C4μe2μ(log2)2e2μ(ρ|s|)E(v)|\partial_{s}v(s,t)|^{2}\leq\frac{C}{(\log 2)^{2}}E(v)\leq\frac{C4^{\mu}e^{2\mu}}{(\log 2)^{2}}e^{-2\mu(\rho-|s|)}E(v)

Next assume

|s|ρlog21|s|\leq\rho-\log 2-1

and denote T:=ρ|s|T:=\rho-|s|. Then T1log2T-1\geq\log 2. Hence

|sv(s,t)|2\displaystyle|\partial_{s}v(s,t)|^{2}\leq CE(v;[ρ+T1,ρT+1]×/2π)\displaystyle CE(v;[-\rho+T-1,\rho-T+1]\times\mathbb{R}/2\pi\mathbb{Z})
=\displaystyle= Cϵ(T1)\displaystyle C\epsilon(T-1)
\displaystyle\leq C4μe2μe2μTE(v)\displaystyle C4^{\mu}e^{2\mu}e^{-2\mu T}E(v)

With the above estimate, we can prove the second part of the lemma. For s0s\geq 0 the above estimate implies

0sρlog2|sv(s,t)|2elog2Ceμ(ρs)E(v)0\leq s\leq\rho-\log 2\quad\Rightarrow|\partial_{s}v(s,t)|\leq\frac{2e}{\log 2}\sqrt{C}e^{-\mu(\rho-s)}\sqrt{E(v)}

Fix a point (s0,t0)2(s_{0},t_{0})\in\mathbb{R}^{2} such that 0s0ρT0\leq s_{0}\leq\rho-T and 0t02π0\leq t_{0}\leq 2\pi, we have

d(v(0,0),v(s0,t0))\displaystyle d(v(0,0),v(s_{0},t_{0}))\leq 0s0|sv(s,0)|𝑑s+0t0|tv(s0,t)|𝑑t\displaystyle\int_{0}^{s_{0}}|\partial_{s}v(s,0)|ds+\int_{0}^{t_{0}}|\partial_{t}v(s_{0},t)|dt
\displaystyle\leq 2elog2C(0s0eμ(ρs)𝑑s+0t0eμ(ρs0)𝑑t)E(v)\displaystyle\frac{2e}{\log 2}\sqrt{C}\left(\int_{0}^{s_{0}}e^{-\mu(\rho-s)}ds+\int_{0}^{t_{0}}e^{-\mu(\rho-s_{0})}dt\right)\sqrt{E(v)}
\displaystyle\leq 2elog2C(Teμs𝑑s+2πeμT)E(v)\displaystyle\frac{2e}{\log 2}\sqrt{C}\left(\int_{-\infty}^{-T}e^{\mu s}ds+2\pi e^{-\mu T}\right)\sqrt{E(v)}
=\displaystyle= 2elog2C(1μ+2π)E(v)\displaystyle\frac{2e}{\log 2}\sqrt{C}\left(\frac{1}{\mu}+2\pi\right)\sqrt{E(v)}

We have a similar estimate for s0s\leq 0. Thus

d(v(s0,t0),v(s1,t1))4elog2C(1μ+2π)eμTE(v)d(v(s_{0},t_{0}),v(s_{1},t_{1}))\leq\frac{4e}{\log 2}\sqrt{C}\left(\frac{1}{\mu}+2\pi\right)e^{-\mu T}\sqrt{E(v)}

for log2Tρ\log 2\leq T\leq\rho and s0,s1[ρ+T,ρT]s_{0},s_{1}\in[-\rho+T,\rho-T] and t0,t1[0,2π]t_{0},t_{1}\in[0,2\pi]. Thus the proof is complete. ∎

Lemma 7.4.

Let r>0r>0, δν>0\delta^{\nu}>0, ν+\nu\in\mathbb{Z}_{+}, and δν0\delta^{\nu}\rightarrow 0 as ν\nu\rightarrow\infty. Let Fν:A(δνr,r)NF_{\nu}:A(\frac{\delta^{\nu}}{r},r)\rightarrow N be a sequence of conformal harmonic maps. Suppose

  1. (1)

    FνF:BrNF_{\nu}\rightarrow F:B_{r}\rightarrow N, where FF is a conformal harmonic map, and the convergence is uniform on compact subsets of Br\{0}B_{r}\backslash\{0\}

  2. (2)

    Fν(δν)H:S2\intB1/rNF_{\nu}(\delta^{\nu}\cdot)\rightarrow H:S^{2}\backslash intB_{1/r}\rightarrow N, where HH is a conformal harmonic map, and the convergence is uniform on compact subsets of \intB1/r\mathbb{C}\backslash intB_{1/r}

  3. (3)

    limρ0limνE(Fν,A(δν/ρ,ρ))=0\underset{\rho\rightarrow 0}{\lim}\underset{\nu\rightarrow\infty}{\lim}E(F_{\nu},A(\delta^{\nu}/\rho,\rho))=0

Then,

  1. (1)

    F(0)=H()F(0)=H(\infty)

  2. (2)

    limρ0limνsupδν/ρ|z|ρd(Fν(z),F(0))=0\underset{\rho\rightarrow 0}{\lim}\underset{\nu\rightarrow\infty}{\lim}\underset{\delta^{\nu}/\rho\leq|z|\leq\rho}{\sup}d(F_{\nu}(z),F(0))=0

Proof.

Abbreviate

Eν(ρ):=E(Fν,A(δν/ρ,ρ)),E(ρ):=limνEν(ρ)E_{\nu}(\rho):=E(F_{\nu},A(\delta^{\nu}/\rho,\rho)),\quad E(\rho):=\underset{\nu\rightarrow\infty}{\lim}E_{\nu}(\rho)

By choosing ρ\rho small enough (and ν\nu large enough), we can let the energy be arbitrarily small and thus apply the previous lemma.

For ρ>0\rho>0 such that 2ρr2\rho\leq r, we know A(δν/2ρ,2ρ)A(δν/r,r)A(\delta^{\nu}/2\rho,2\rho)\subset A(\delta^{\nu}/r,r). Let T=log2T=\log 2 in the previous lemma, there exists δ>0\delta>0 and cc such that

Eν(2ρ)<δsupz,zA(δν/ρ,ρ)d(Fν(z),Fν(z))<cEν(2ρ)E_{\nu}(2\rho)<\delta\quad\Rightarrow\quad\underset{z,z^{\prime}\in A(\delta^{\nu}/\rho,\rho)}{\sup}d(F_{\nu}(z),F_{\nu}(z^{\prime}))<c\sqrt{E_{\nu}(2\rho)}

Let ν\nu\rightarrow\infty,

E(2ρ)<δd(F(ρ),H(1/ρ))=limνd(Fν(ρ),Fν(δν/ρ))<cE(2ρ)E(2\rho)<\delta\quad\Rightarrow\quad d(F(\rho),H(1/\rho))=\underset{\nu\rightarrow\infty}{\lim}{d(F_{\nu}(\rho),F_{\nu}(\delta^{\nu}/\rho))}<c\sqrt{E(2\rho)}

Take the limit ρ0\rho\rightarrow 0, we get F(0)=H()F(0)=H(\infty).

To prove the second assertion, fix an arbitrary ϵ>0\epsilon>0. We only need to prove that, when ρ\rho is small enough, there exists an integer ν0>0\nu_{0}>0 (ν0\nu_{0} can depend on ρ\rho) such that,

νν0,δνρ|z|ρd(Fν(z),F(0))<ϵ\nu\geq\nu_{0},\quad\frac{\delta^{\nu}}{\rho}\leq|z|\leq\rho\quad\Rightarrow\quad d(F_{\nu}(z),F(0))<\epsilon

First choose δ\delta and cc as in the first part of this proof. Then choose ρ\rho small enough such that 0<2ρr0<2\rho\leq r and

E(2ρ)<δ,cE(2ρ)<ϵ/3,d(F(ρ),F(0))<ϵ/3E(2\rho)<\delta,\quad c\sqrt{E(2\rho)}<\epsilon/3,\quad d(F(\rho),F(0))<\epsilon/3

Then choose ν0\nu_{0} large enough such that, for νν0\nu\geq\nu_{0}, we have

Eν(2ρ)<δ,cEν(2ρ)<ϵ/3,d(Fν(ρ),F(ρ))<ϵ/3E_{\nu}(2\rho)<\delta,\quad c\sqrt{E_{\nu}(2\rho)}<\epsilon/3,\quad d(F_{\nu}(\rho),F(\rho))<\epsilon/3

Now suppose νν0\nu\geq\nu_{0} and δν/ρ|z|ρ\delta^{\nu}/\rho\leq|z|\leq\rho, then

d(Fν(z),F(0))\displaystyle d(F_{\nu}(z),F(0))\leq d(Fν(z),Fν(ρ))+d(Fν(ρ),F(ρ))+d(F(ρ),F(0))\displaystyle d(F_{\nu}(z),F_{\nu}(\rho))+d(F_{\nu}(\rho),F(\rho))+d(F(\rho),F(0))
\displaystyle\leq cEν(2ρ)+d(Fν(ρ),F(ρ))+d(F(ρ),F(0))\displaystyle c\sqrt{E_{\nu}(2\rho)}+d(F_{\nu}(\rho),F(\rho))+d(F(\rho),F(0))
<\displaystyle< ϵ\displaystyle\epsilon

The proof is complete. ∎

8. Surjectivity of the gluing map

The theorems in this section are analogues to section 10.7 in [16]. From now on, we always assume our target manifold NN to be closed.

Theorem 8.1.

Fix two constants c>1c>1 and 1<p<21<p<2. Consider any precompact open subset 𝒰(c1,p)\mathcal{U}\subset\mathcal{M}(c-1,p). Then, depending only on c,p,N,𝒰c,p,N,\mathcal{U}, we can choose positive constants δ0\delta_{0}, c0c_{0}, and ϵ\epsilon such that the following holds:

Given (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U}, (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), and vv is a harmonic map whose domain is the glued sphere, satisfying ξvR2,p,Rϵ\|\xi_{v}^{R}\|_{2,p,R}\leq\epsilon, where v=expfR(ξvR)v=\exp_{f^{R}}(\xi_{v}^{R}). Then vv belongs to the image of the gluing map.

Proof.

Fix a pair (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}) and an element vv as in the statement of the theorem (We can let ϵ\epsilon be smaller later if necessary). Let us denote by

𝒰1𝒰\mathcal{U}_{1}\subset\mathcal{U}

the set of all pairs f:=(f0,f)𝒰f:=(f^{0},f^{\infty})\in\mathcal{U} that satisfy

(25) expfR(ξR)=v,ξR2,p,R<ϵ1\exp_{f^{R}}(\xi^{R})=v,\quad\|\xi^{R}\|_{2,p,R}<\epsilon_{1}

for some vector field ξR=ξR(f)\xi^{R}=\xi^{R}(f). (Note that by lemma D.1 we know ξRLC(c,p,N,𝒰)ξR2,p,RC(c,p,N,𝒰)ϵ1\|\xi^{R}\|_{L^{\infty}}\leq C(c,p,N,\mathcal{U})\|\xi^{R}\|_{2,p,R}\leq C(c,p,N,\mathcal{U})\epsilon_{1}, and hence by choosing ϵ1\epsilon_{1} small enough only depending on c,p,N,𝒰c,p,N,\mathcal{U}, there is at most one such vector field ξR\xi^{R}.) Since (δ,R)(\delta,R) is fixed, we know 𝒰1\mathcal{U}_{1} is an open set.

We must find a constant ϵ2>0\epsilon_{2}>0 depending only on c,p,N,𝒰c,p,N,\mathcal{U} with the following significance: if there is an element f𝒰1f\in\mathcal{U}_{1} such that ξR(f)2,p,R<ϵ2\|\xi^{R}(f)\|_{2,p,R}<\epsilon_{2} then there is an element f¯:=(f¯0,f¯)𝒰1\bar{f}:=(\bar{f}^{0},\bar{f}^{\infty})\in\mathcal{U}_{1} such that the corresponding vector field ξ¯R:=ξR(f¯)\bar{\xi}^{R}:=\xi^{R}(\bar{f}) belongs to the image of the operator Qf¯RQ_{\bar{f}^{R}}. To prove this, note that (c,p)\mathcal{M}(c,p) is locally a manifold of dimension mm, where mm is some finite number. From lemma D.3 we know that mm is also the dimension of the kernel of DfRD_{f^{R}} for ff such that (f0,f)(c,p)(f^{0},f^{\infty})\in\mathcal{M}(c,p) and δ0\delta_{0} small enough only depending on c,p,N,𝒰c,p,N,\mathcal{U}.

Next we choose a smooth framing

fΞi(f)kerDf=kerDf0,f,i=1,,m,f\mapsto\Xi_{i}(f)\in\ker D_{f}=\ker D_{f^{0},f^{\infty}},\quad i=1,\cdots,m,

over f𝒰1f\in\mathcal{U}_{1}. We can make 𝒰1\mathcal{U}_{1} a small enough open neighborhood (how small only depends on c,p,N,𝒰c,p,N,\mathcal{U}) of ff and choose framing using the coordinate chart. Define

ΞiR(f):=(idQfRDfR)dgR(f)Ξi(f),i=1,,m\Xi^{R}_{i}(f):=(id-Q_{f^{R}}D_{f^{R}})dg^{R}(f)\Xi_{i}(f),\quad i=1,\cdots,m

where gRg^{R} is the pregluing (gR(f)=fRg^{R}(f)=f^{R}). We show that this is a framing of kerDfR\ker D_{f^{R}} for f𝒰1f\in\mathcal{U}_{1} for δ0\delta_{0} sufficiently small only depending on c,p,N,𝒰c,p,N,\mathcal{U}. In fact, by lemma D.3 we know that, in this case, the linear map

(idQfRDfR)dgR(f):kerDfkerDfR(id-Q_{f^{R}}D_{f^{R}})dg^{R}(f):\ker D_{f}\rightarrow\ker D_{f^{R}}

is an isomorphism from the kernel of DfD_{f} to the kernel of DfRD_{f^{R}} for f𝒰1f\in\mathcal{U}_{1} and fR=gR(f)f^{R}=g^{R}(f), and that there is a uniform bound on the inverse of this isomorphism only depending on c,p,N,𝒰c,p,N,\mathcal{U}. Hence there are positive constants c1c_{1} and δ1\delta_{1} only depending on c,p,N,𝒰c,p,N,\mathcal{U} such that the following holds whenever (δ,R)𝒜(δ1)(\delta,R)\in\mathcal{A}(\delta_{1}): If f𝒰1f\in\mathcal{U}_{1} and r=(r1,,rm)mr=(r_{1},\cdots,r_{m})\in\mathbb{R}^{m}, then

|r|c1(c,p,N,𝒰)i=1mriΞiR(f)2,p,R|r|\leq c_{1}(c,p,N,\mathcal{U})\left\|\sum_{i=1}^{m}r_{i}\Xi_{i}^{R}(f)\right\|_{2,p,R}

Hence f(ΞiR(f))i=1mf\mapsto(\Xi_{i}^{R}(f))_{i=1}^{m} is a framing. Increasing c1c_{1} if necessary, we also have that every smooth path tftt\mapsto f_{t} in 𝒰1\mathcal{U}_{1} satisfies the inequality

tΞiR(ft)2,p,Rc1(c,p,N,𝒰)tftL2,i=1,,m\|\nabla_{t}\Xi_{i}^{R}(f_{t})\|_{2,p,R}\leq c_{1}(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{L^{2}},\quad i=1,\cdots,m

where \nabla denotes the Levi-Civita connection on NN. To see this, assume first that ξt=(ξt0,ξt)Tft𝒰\xi_{t}=(\xi^{0}_{t},\xi^{\infty}_{t})\in T_{f_{t}}\mathcal{U} is any smooth vector field along ftf_{t}. Then by lemma D.4 the (2,p,R)(2,p,R)-norm of t(dgR(ft)ξt)\nabla_{t}(dg^{R}(f_{t})\xi_{t}) can be estimated by a constant times tξtW2,p+ξtW2,ptftW2,p\|\nabla_{t}\xi_{t}\|_{W^{2,p}}+\|\xi_{t}\|_{W^{2,p}}\|\partial_{t}f_{t}\|_{W^{2,p}}. Now apply this inequality to ξt:=Ξi(ft)\xi_{t}:=\Xi_{i}(f_{t}) and use lemma D.3 ((ξ0,ξ)L2c0(c,p,N,𝒰)(idQfRDfR)dgR(f)(ξ0,ξ)2,p,R\|(\xi^{0},\xi^{\infty})\|_{L^{2}}\leq c_{0}(c,p,N,\mathcal{U})\|(id-Q_{f^{R}}D_{f^{R}})dg^{R}(f)(\xi^{0},\xi^{\infty})\|_{2,p,R}), the result of lemma D.2, as well as the proof of lemma D.2 (t(Dtξt)DttξtLpC(c,p,N,𝒰)tftW2,ξtW2,p\|\nabla_{t}(D_{t}\xi_{t})-D_{t}\nabla_{t}\xi_{t}\|_{L^{p}}\leq C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\xi_{t}\|_{W^{2,p}}).

Define the smooth map hR=(h1R,,hmR):𝒰1mh^{R}=(h_{1}^{R},\cdots,h_{m}^{R}):\mathcal{U}_{1}\rightarrow\mathbb{R}^{m} by

i=1mhiR(f)ΞiR(f):=ξRQfRDfRξR\sum_{i=1}^{m}h^{R}_{i}(f)\Xi^{R}_{i}(f):=\xi^{R}-Q_{f^{R}}D_{f^{R}}\xi^{R}

where fR:=gR(f)f^{R}:=g^{R}(f) and ξR=ξR(f)\xi^{R}=\xi^{R}(f) is the unique vector field along fRf^{R} that satisfies (25). Thus hR(f)mh^{R}(f)\in\mathbb{R}^{m} is the coordinate vector of the projection of ξR(f)\xi^{R}(f) onto the kernel of DfRD_{f^{R}} along the image of QfRQ_{f^{R}}. We would like to find f¯\bar{f} where hR(f¯)h^{R}(\bar{f}) vanishes. We have

|hR(f)|c2(c,p,N,𝒰)ξR(f)2,p,R|h^{R}(f)|\leq c_{2}(c,p,N,\mathcal{U})\|\xi^{R}(f)\|_{2,p,R}

Hence our assumption implies that there is f𝒰1f\in\mathcal{U}_{1} such that |hR(f)|<c2ϵ2|h^{R}(f)|<c_{2}\epsilon_{2}, for some small ϵ2\epsilon_{2} we can choose. To find an actual zero of hh we aim to apply the implicit function theorem. So we will need to estimate the inverse of dhRdh^{R}.

Let us denote by

H(f):=(H1(f),,Hm(f)):kerDfmH(f):=(H_{1}(f),\cdots,H_{m}(f)):\ker D_{f}\rightarrow\mathbb{R}^{m}

the isomorphism given by the framing Ξ1(f),,Ξm(f)\Xi_{1}(f),\cdots,\Xi_{m}(f), that is

i=1m(Hi(f)ξ)Ξi(f)=ξ\sum_{i=1}^{m}(H_{i}(f)\xi)\Xi_{i}(f)=\xi

Then, by definition of ΞiR(f)\Xi_{i}^{R}(f), we have

i=1m(Hi(f)ξ)ΞiR(f)=(idQfRDfR)dgR(f)ξ\sum_{i=1}^{m}(H_{i}(f)\xi)\Xi^{R}_{i}(f)=(id-Q_{f^{R}}D_{f^{R}})dg^{R}(f)\xi

for ξkerDf\xi\in\ker D_{f}. We shall prove that there is a constant C=C(c,p,N,𝒰)>0C=C(c,p,N,\mathcal{U})>0 such that

dhR(f)+H(f)C(c,p,N,𝒰)ϵ1\|dh^{R}(f)+H(f)\|\leq C(c,p,N,\mathcal{U})\epsilon_{1}

for every f:=(f0,f)𝒰1f:=(f^{0},f^{\infty})\in\mathcal{U}_{1}. Note that the above is the operator norm. Since H(f)H(f) is an isomorphism, we can apply proposition 3.1 for dhRdh^{R} after proving the above inequality.

Along a smooth path tft:=(ft0,ft)𝒰1t\mapsto f_{t}:=(f^{0}_{t},f^{\infty}_{t})\in\mathcal{U}_{1}. Denote ftR:=gR(ft)f^{R}_{t}:=g^{R}(f_{t}) and let ξtR\xi^{R}_{t} be the unique smooth family of vector fields along ftRf_{t}^{R} satisfying

expftR(ξtR)=v,ξtR2,p,Rϵ1\exp_{f^{R}_{t}}(\xi^{R}_{t})=v,\quad\|\xi^{R}_{t}\|_{2,p,R}\leq\epsilon_{1}

Then we have

t(ξtRQDξtR)=i=1mhiR(ft)tΞiR(ft)+i=1m(thiR(ft))ΞiR(ft)\nabla_{t}(\xi^{R}_{t}-QD\xi^{R}_{t})=\sum_{i=1}^{m}h^{R}_{i}(f_{t})\nabla_{t}\Xi^{R}_{i}(f_{t})+\sum_{i=1}^{m}(\partial_{t}h^{R}_{i}(f_{t}))\Xi^{R}_{i}(f_{t})

where Q:=QftRQ:=Q_{f^{R}_{t}} and D:=DftRD:=D_{f^{R}_{t}}. Differentiate expftR(ξtR)=v\exp_{f^{R}_{t}}(\xi^{R}_{t})=v to obtain

E1tftR+E2tξtR=0E_{1}\partial_{t}f^{R}_{t}+E_{2}\nabla_{t}\xi^{R}_{t}=0

Thus

tξtR=E21E1tftR=E21E1dgR(ft)tft\nabla_{t}\xi^{R}_{t}=-E_{2}^{-1}E_{1}\partial_{t}f^{R}_{t}=-E_{2}^{-1}E_{1}dg^{R}(f_{t})\partial_{t}f_{t}

Combining the above, we have

i=1m(thiR(ft)+Hi(ft)tft)ΞiR(ft)\displaystyle\sum_{i=1}^{m}(\partial_{t}h_{i}^{R}(f_{t})+H_{i}(f_{t})\partial_{t}f_{t})\Xi^{R}_{i}(f_{t})
=\displaystyle= (idQD)(idE21E1)dgR(ft)tft+QDtξtRt(QDξtR)i=1mhiR(ft)tΞiR(ft)\displaystyle(id-QD)(id-E_{2}^{-1}E_{1})dg^{R}(f_{t})\partial_{t}f_{t}+QD\nabla_{t}\xi^{R}_{t}-\nabla_{t}(QD\xi^{R}_{t})-\sum_{i=1}^{m}h^{R}_{i}(f_{t})\nabla_{t}\Xi_{i}^{R}(f_{t})

Since |hR(ft)|c2(c,p,N,𝒰)ξR2,p,Rc2(c,p,N,𝒰)ϵ1|h^{R}(f_{t})|\leq c_{2}(c,p,N,\mathcal{U})\|\xi^{R}\|_{2,p,R}\leq c_{2}(c,p,N,\mathcal{U})\epsilon_{1}, we have

i=1mhiR(ft)tΞiR(ft)2,p,RmC(c,p,N,𝒰)ϵ1tftL2\|\sum_{i=1}^{m}h_{i}^{R}(f_{t})\nabla_{t}\Xi_{i}^{R}(f_{t})\|_{2,p,R}\leq\sqrt{m}C(c,p,N,\mathcal{U})\epsilon_{1}\|\partial_{t}f_{t}\|_{L^{2}}

On the other hand, we have the pointwise estimate

|idE2(ftR,ξtR)1E1(ftR,ξtR)|C(c,p,N,𝒰)|ξtR|\left|id-E_{2}(f^{R}_{t},\xi^{R}_{t})^{-1}E_{1}(f^{R}_{t},\xi^{R}_{t})\right|\leq C(c,p,N,\mathcal{U})\left|\xi^{R}_{t}\right|

Similarly, we have an inequality for the first and second derivatives of idE21E1id-E_{2}^{-1}E_{1} in terms of the first and second derivatives of ξtR\xi^{R}_{t} for a fixed tt. Hence we have

(idE21E1)dgR(ft)tft2,p,RC(c,p,N,𝒰)ϵ1tftL2\|(id-E_{2}^{-1}E_{1})dg^{R}(f_{t})\partial_{t}f_{t}\|_{2,p,R}\leq C(c,p,N,\mathcal{U})\epsilon_{1}\|\partial_{t}f_{t}\|_{L^{2}}

For the remaining term, we claim that

t(QDξtR)QDtξtR2,p,RC(c,p,N,𝒰)ξtR2,p,RtftW2,\left\|\nabla_{t}(QD\xi^{R}_{t})-QD\nabla_{t}\xi^{R}_{t}\right\|_{2,p,R}\leq C(c,p,N,\mathcal{U})\|\xi^{R}_{t}\|_{2,p,R}\|\partial_{t}f_{t}\|_{W^{2,\infty}}

We proved a similar result for TT instead of QQ in the appendix. Here we can also use D.2 in that proof. Combining these we get an estimate for the commutator [DT,t][DT,\nabla_{t}] and hence for [(DT)1,t][(DT)^{-1},\nabla_{t}]. Note that Q=T(DT)1Q=T(DT)^{-1}, thus we get an estimate for the commutator of the operator QD=T(DT)1DQD=T(DT)^{-1}D with t\nabla_{t}.

Thus for ξtR2,p,Rϵ1\|\xi^{R}_{t}\|_{2,p,R}\leq\epsilon_{1} we obtain

t(QDξtR)QDtξtR2,p,Rcϵ1tftL2\left\|\nabla_{t}(QD\xi^{R}_{t})-QD\nabla_{t}\xi^{R}_{t}\right\|_{2,p,R}\leq c\epsilon_{1}\|\partial_{t}f_{t}\|_{L^{2}}

Hence we have proved

i=1m(thiR(ft)+Hi(ft)tft)ΞiR(ft)cϵ1tftL2\sum_{i=1}^{m}(\partial_{t}h_{i}^{R}(f_{t})+H_{i}(f_{t})\partial_{t}f_{t})\Xi^{R}_{i}(f_{t})\leq c\epsilon_{1}\|\partial_{t}f_{t}\|_{L^{2}}

Now we apply the implicit function theorem. ∎

With the previous result, now we can consider when there is a single bubble for a sequence of harmonic maps on S2S^{2}.

Theorem 8.2.

There exists p0=p0(N)>1p_{0}=p_{0}(N)>1 such that if we fix 1<p<p01<p<p_{0} and c>1c>1, consider any precompact 𝒰(c1,p)\mathcal{U}\subset\mathcal{M}(c-1,p), then there exists an ϵ>0\epsilon>0 and δ0>0\delta_{0}>0 only depending on c,p,N,𝒰c,p,N,\mathcal{U} such that the following holds:

Given (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U}, (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), and if vv is a harmonic map whose domain is the glued sphere, satisfying

E(v)c,E(v;D(0,2/δR)D(0,δ/2R))<ϵ,\displaystyle E(v)\leq c,\quad E(v;D(0,2/\delta R)-D(0,\delta/2R))<\epsilon,
sup|z|1/ϵd(v(z),f0(z))<ϵ,sup|z|1/ϵd(v(z/R2),f(z))<ϵ\displaystyle\sup_{|z|\geq 1/\epsilon}d(v(z),f^{0}(z))<\epsilon,\quad\sup_{|z|\leq 1/\epsilon}d(v(z/R^{2}),f^{\infty}(z))<\epsilon

where v=expf(ξv)v=\exp_{f}(\xi_{v}). Then vv belongs to the image of the gluing map.

Proof.

From the previous theorem, we only need to prove that there exists δ3\delta_{3} and ϵ3\epsilon_{3} such that for every (δ,R)𝒜(δ3)(\delta,R)\in\mathcal{A}(\delta_{3}), with the given conditions, we have ξR2,p,Rϵ2\|\xi^{R}\|_{2,p,R}\leq\epsilon_{2} where v=expfR(ξR)v=\exp_{f^{R}}(\xi^{R}).

Suppose, by contradiction, that the assertion is wrong for some constant ϵ2>0\epsilon_{2}>0. Then there are sequences

ϵν0,δν0,Rν,(f0ν,fν)𝒰,hν\epsilon_{\nu}\rightarrow 0,\quad\delta_{\nu}\rightarrow 0,\quad R_{\nu}\rightarrow\infty,\quad(f^{0}_{\nu},f^{\infty}_{\nu})\in\mathcal{U},\quad h_{\nu}

such that δνRν\delta_{\nu}R_{\nu} diverges to infinity, hνh_{\nu} is a harmonic map on the glued sphere with parameters δν,Rν\delta_{\nu},R_{\nu}, and

E(hν)c,E(hν;D(0,2/δνRν)D(0,δν/2Rν))<ϵν,\displaystyle E(h_{\nu})\leq c,\quad E(h_{\nu};D(0,2/\delta_{\nu}R_{\nu})-D(0,\delta_{\nu}/2R_{\nu}))<\epsilon_{\nu},
sup|z|1/ϵνd(hν(z),f0ν(z))<ϵν,sup|z|1/ϵνd(hν(z/Rν2),fν(z))<ϵν\displaystyle\sup_{|z|\geq 1/\epsilon_{\nu}}d(h_{\nu}(z),f^{0}_{\nu}(z))<\epsilon_{\nu},\quad\sup_{|z|\leq 1/\epsilon_{\nu}}d(h_{\nu}(z/R_{\nu}^{2}),f^{\infty}_{\nu}(z))<\epsilon_{\nu}

and, for every ν\nu,

inf{ξ2,p,RνξΓ(S2,(fνRν)1TN),hν=expfRνν(ξ)}ϵ2\inf\left\{\|\xi\|_{2,p,R_{\nu}}\mid\xi\in\Gamma(S^{2},(f_{\nu}^{R_{\nu}})^{-1}TN),h_{\nu}=\exp_{f^{R_{\nu}}_{\nu}}(\xi)\right\}\geq\epsilon_{2}

By passing to a subsequence, we can assume (f0ν,fν)(f^{0}_{\nu},f^{\infty}_{\nu}) converges in C1C^{1} to a pair (f0,f)(f^{0},f^{\infty}). Since the LL^{\infty} norm of first order derivatives are bounded, no bubbles form. Furthermore, away from the neck, hνh_{\nu} converges uniformly on compact subsets to h0h^{0} on the complement of a finite set of bubble points.

By assumption, h0(z)=f0(z)h^{0}(z)=f^{0}(z), |z|1|z|\geq 1. By unique continuation of harmonic maps, this implies that h0=f0h^{0}=f^{0} on S2S^{2}. Similarly, hν(/Rν2)h_{\nu}(\cdot/R_{\nu}^{2}) converges to ff^{\infty} on the complement of a finite set of bubble points.

First, we would like to prove that hνh_{\nu} converges to f0f^{0} uniformly on compact subsets of S2\{0}S^{2}\backslash\{0\}, and hν(/Rν2)h_{\nu}(\cdot/R_{\nu}^{2}) converges to ff^{\infty}, uniformly on compact subsets of S2\{}S^{2}\backslash\{\infty\}. Let’s take f0f^{0} for example. To avoid writing too many superscripts, we write f0f^{0} as ff below. Since f0f^{0} and hνh_{\nu} are harmonic maps, they satisfy the harmonic map equations:

(gfi+gαβΓijl(f)fjxαflxβ)yi=0\left(\triangle_{g}f^{i}+g^{\alpha\beta}\Gamma^{i}_{jl}(f)\frac{\partial f^{j}}{\partial x^{\alpha}}\frac{\partial f^{l}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}=0
(g(expfξ)i+gαβΓjli(expfξ)(expfξ)jxα(expfξ)lxβ)yi=0\left(\triangle_{g}(\exp_{f}\xi)^{i}+g^{\alpha\beta}\Gamma_{jl}^{i}(\exp_{f}\xi)\frac{\partial(\exp_{f}\xi)^{j}}{\partial x^{\alpha}}\frac{\partial(\exp_{f}\xi)^{l}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}=0

where

gfi=gαβ2fixαxβ+gαβxαfixβ+gαβ12detgdetgxαfixβ\triangle_{g}f^{i}=g^{\alpha\beta}\frac{\partial^{2}f^{i}}{\partial x^{\alpha}x^{\beta}}+\frac{\partial g^{\alpha\beta}}{\partial x^{\alpha}}\frac{\partial f^{i}}{\partial x^{\beta}}+g^{\alpha\beta}\frac{1}{2\det g}\frac{\partial\det g}{\partial x^{\alpha}}\frac{\partial f^{i}}{\partial x^{\beta}}

First pull back the vector on the left hand side of the second equation and then subtracting the two equations, we will get a quasilinear system of second order elliptic equations for ξ\xi. Here we cite the result in [12], section 8.4:

Theorem 8.3.

Let u(x)C2(Ω)u(x)\in C^{2}(\Omega) and that u(x)u(x) satisfies the system:

aij(x,u)ulxixj+bi(x,u,ux)ulxi+bl(x,u,ux)=0,l=1,,N.a_{ij}(x,u)u^{l}_{x_{i}x_{j}}+b_{i}(x,u,u_{x})u^{l}_{x_{i}}+b^{l}(x,u,u_{x})=0,\quad l=1,\cdots,N.

Suppose the following inequalities hold for xΩ¯,|u|Mx\in\overline{\Omega},|u|\leq M, and the arbitrary pp:

(26) μ1(M)i=1nξ2iaij(x,u)ξiξjμ2(M)i=1nξ2i,\displaystyle\mu_{1}(M)\sum_{i=1}^{n}\xi^{2}_{i}\leq a_{ij}(x,u)\xi_{i}\xi_{j}\leq\mu_{2}(M)\sum_{i=1}^{n}\xi^{2}_{i},
(27) |bi(x,u,p)|μ2(M)(1+|p|)\displaystyle|b_{i}(x,u,p)|\leq\mu_{2}(M)(1+|p|)
(28) |b(x,u,p)|[ϵ(M)+P(p,M)](1+|p|2)\displaystyle|b(x,u,p)|\leq[\epsilon(M)+P(p,M)](1+|p|^{2})
(29) |aij(x,u)xk;aijul|μ2(M)\displaystyle\left|\frac{\partial a_{ij}(x,u)}{\partial x_{k}};\frac{\partial a_{ij}}{\partial u^{l}}\right|\leq\mu_{2}(M)

where ϵ(M)\epsilon(M) is a sufficiently small quantity determined only by n,N,M,μ1,μ2n,N,M,\mu_{1},\mu_{2}, and P(p,M)0P(p,M)\rightarrow 0 as |p||p|\rightarrow\infty. Also suppose the above constant ϵ(M)\epsilon(M) satisfies

(2M+10N)ϵ(M)<μ1(2M+10N)\epsilon(M)<\mu_{1}

Then, for arbitrary ΩΩ\Omega^{\prime}\subset\Omega, the quantity maxΩ|u|\underset{\Omega^{\prime}}{\max}|\nabla u| is bounded in terms of n,N,M,μ1(M),μ2(M),ϵ(M),P(p,M),n,N,M,\mu_{1}(M),\mu_{2}(M),\epsilon(M),P(p,M), distance from Ω\Omega^{\prime} to Ω\partial\Omega, and mes Ω\Omega (measure of Ω\Omega).

Now let’s show that the elliptic system for ξ\xi actually satisfies the conditions in the theorem. We construct the system as follows: first, we write out the equations for ξ\xi at each point using the normal coordinates at that point. Then, we consider disks of small enough radius and do coordinate change to get the equations under the same coordinates for the small disk.

Consider normal coordinates at y=f(x)Ny=f(x)\in N, the ξ\xi satisfies:

(gξi+gαβ(Γijl(f+ξ)Γijl(f))fjxαflxβ+\displaystyle\left(\triangle_{g}\xi^{i}+g^{\alpha\beta}(\Gamma^{i}_{jl}(f+\xi)-\Gamma^{i}_{jl}(f))\frac{\partial f^{j}}{\partial x^{\alpha}}\frac{\partial f^{l}}{\partial x^{\beta}}+\right.
gαβΓjli(f+ξ)(fjxαξlxβ+flxβξjxα)+gαβΓijl(f+ξ)(ξjxαξlxβ))yi=0\displaystyle\left.g^{\alpha\beta}\Gamma_{jl}^{i}(f+\xi)\left(\frac{\partial f^{j}}{\partial x^{\alpha}}\frac{\partial\xi^{l}}{\partial x^{\beta}}+\frac{\partial f^{l}}{\partial x^{\beta}}\frac{\partial\xi^{j}}{\partial x^{\alpha}}\right)+g^{\alpha\beta}\Gamma^{i}_{jl}(f+\xi)\left(\frac{\partial\xi^{j}}{\partial x^{\alpha}}\frac{\partial\xi^{l}}{\partial x^{\beta}}\right)\right)\frac{\partial}{\partial y^{i}}=0

where

gξi=gαβ2ξixαxβ+gαβxβξixβ+gαβ12detgdetgxαξixβ\displaystyle\triangle_{g}\xi^{i}=g^{\alpha\beta}\frac{\partial^{2}\xi^{i}}{\partial x^{\alpha}x^{\beta}}+\frac{\partial g^{\alpha\beta}}{\partial x^{\beta}}\frac{\partial\xi^{i}}{\partial x^{\beta}}+g^{\alpha\beta}\frac{1}{2\det g}\frac{\partial\det g}{\partial x^{\alpha}}\frac{\partial\xi^{i}}{\partial x^{\beta}}

Then, consider the coordinate change. Suppose this point yy is in the disk centered at y0y_{0}, denote the normal coordinates at y0y_{0} as y~\tilde{y}. We can use our construction in lemma 2.1. We have ξi=ξ~lyiy~l\xi^{i}=\tilde{\xi}^{l}\frac{\partial y^{i}}{\partial\tilde{y}^{l}}.

For the Laplacian part,

(gξi)yi\displaystyle(\triangle_{g}\xi^{i})\frac{\partial}{\partial y^{i}} =(gαβ2ξixαxβ+gαβxαξixβ+gαβ12detgdetgxαξixβ)yi\displaystyle=\left(g^{\alpha\beta}\frac{\partial^{2}\xi^{i}}{\partial x^{\alpha}x^{\beta}}+\frac{\partial g^{\alpha\beta}}{\partial x^{\alpha}}\frac{\partial\xi^{i}}{\partial x^{\beta}}+g^{\alpha\beta}\frac{1}{2\det g}\frac{\partial detg}{\partial x^{\alpha}}\frac{\partial\xi^{i}}{\partial x^{\beta}}\right)\frac{\partial}{\partial y^{i}}
=(gαβ2(ξ~lyiy~l)xαxβ+gαβxα(ξ~lyiy~l)xβ+gαβ12detgdetgxα(ξ~lyiy~l)xβ)y~myiy~m\displaystyle=\left(g^{\alpha\beta}\frac{\partial^{2}\left(\tilde{\xi}^{l}\frac{\partial y^{i}}{\partial\tilde{y}^{l}}\right)}{\partial x^{\alpha}\partial x^{\beta}}+\frac{\partial g^{\alpha\beta}}{\partial x^{\alpha}}\frac{\partial\left(\tilde{\xi}^{l}\frac{\partial y^{i}}{\partial\tilde{y}^{l}}\right)}{\partial x^{\beta}}+g^{\alpha\beta}\frac{1}{2\det g}\frac{\partial detg}{\partial x^{\alpha}}\frac{\partial\left(\tilde{\xi}^{l}\frac{\partial y^{i}}{\partial\tilde{y}^{l}}\right)}{\partial x^{\beta}}\right)\frac{\partial\tilde{y}^{m}}{\partial y^{i}}\frac{\partial}{\partial\tilde{y}^{m}}

In the above, for the parts where ξ\xi does not take derivatives, we can make the C0C^{0} norm of ξ\xi arbitrarily small by taking large enough ν\nu. Thus (28)(\ref{quasi3}) is satisfied. For the parts where the change of coordinates yy~\frac{\partial y}{\partial\tilde{y}} does not take any derivatives, note that

yiy~ly~myi=δml\frac{\partial y^{i}}{\partial\tilde{y}^{l}}\frac{\partial\tilde{y}^{m}}{\partial y^{i}}=\delta_{m}^{l}

Thus the part with second order derivative remains the same after the coordinate change, and the rest belongs to bi(x,u,ux)uxilb_{i}(x,u,u_{x})u_{x_{i}}^{l} in the above theorem, which has the required bound.

The only part that is slightly less obvious is the part:

gαβξ~lxα(xβ(yiy~l))y~myiy~mg^{\alpha\beta}\frac{\partial\tilde{\xi}^{l}}{\partial x^{\alpha}}\left(\frac{\partial}{\partial x^{\beta}}\left(\frac{\partial y^{i}}{\partial\tilde{y}^{l}}\right)\right)\frac{\partial\tilde{y}^{m}}{\partial y^{i}}\frac{\partial}{\partial\tilde{y}^{m}}

Note that for imi\neq m, we can make the disk small enough such that y~myi\frac{\partial\tilde{y}^{m}}{\partial y^{i}} is arbitrarily small. On the other hand,

xβ(yiy~l)y~myi+yiy~lxβ(y~myi)=xβδml=0\frac{\partial}{\partial x^{\beta}}\left(\frac{\partial y^{i}}{\partial\tilde{y}^{l}}\right)\frac{\partial\tilde{y}^{m}}{\partial y^{i}}+\frac{\partial y^{i}}{\partial\tilde{y}^{l}}\frac{\partial}{\partial x^{\beta}}\left(\frac{\partial\tilde{y}^{m}}{\partial y^{i}}\right)=\frac{\partial}{\partial x^{\beta}}\delta^{m}_{l}=0

Thus we also have control when ili\neq l. When i=l=mi=l=m, the term belongs to bib_{i} part in the theorem. Other terms belong to blb^{l} part. Furthermore, all conditions in the theorem are met.

For the parts involving Christoffle symbols, it is probably the easiest if we replace the Γ(f+ξ)\Gamma(f+\xi) by Γ(f+ξ)Γ(f)\Gamma(f+\xi)-\Gamma(f). We can do this since originally this is the normal coordinates. Then do the coordinate change. We can make the norm of these terms arbitrarily small since C0C^{0} norm of ξ\xi can be arbitrarily small. Thus the condition in the theorem is still satisfied.

From the theorem, we get a bound for the maximum of |ξ||\nabla\xi|. Thus there can be no bubble anywhere other than the neck. Furthermore, by the main estimate 3.2 in [23], we get a uniform bound for W2,pW^{2,p} norm of ξ\xi away from the neck for any 1<p<1<p<\infty.

Next, we consider the neck. Fix r>0r>0, Consider hνh_{\nu} restricted to Dr(0)D_{r}(0). We know this is a harmonic map, thus we can apply the trick of making it into a conformal harmonic map, which we denote by HνH_{\nu} (See section 7).

We would like to apply lemma 7.4, so we have to verify that the conditions are met. First, we show that {Hν}\{H_{\nu}\} converges to some conformal harmonic map HH, uniformly on compact subsets of Br\{0}B_{r}\backslash\{0\}. Let’s revisit how the conformal harmonic maps are obtained. We are given hνh_{\nu} and they converge to hh on every compact subset. Then we consider the weighted sphere which represents two spheres glued together. The sphere for hνh_{\nu} and hh is represented in the weighted sphere by D1/Rν(0)D_{1/R_{\nu}}(0). Then we define the ζ\zeta by setting

¯ζ=0,ζ=14ψhν,ζD=0\bar{\partial}\zeta=0,\quad\partial\zeta=-\frac{1}{4}\psi_{h_{\nu}},\quad\zeta\mid_{\partial D}=0

on the disk of a fixed radius rr and centered at 0, where

ψhν=|xhν|2|yhν|22ixhν,yhν\psi_{h_{\nu}}=|\partial_{x}h_{\nu}|^{2}-|\partial_{y}h_{\nu}|^{2}-2i\langle\partial_{x}h_{\nu},\partial_{y}h_{\nu}\rangle

Note that here we are simply considering the canonical norm on the disk. The process is similar for hh. We would like to show that ζν\zeta_{\nu} converge to ζ\zeta on compact subsets. We may not be able to do this directly because we have hνh_{\nu} converge on the (standard) sphere while ζ\zeta is defined on the disk. A direct esitmate will encounter multiplication by RνR_{\nu} to some exponent. One possible way around this is to use compactness for the conformal harmonic map sequence and use energy estimate by the energy of the original harmonic maps to show that there exists a converging subsequence.

Here is the argument: Since the energy of the conformal harmonic maps are controlled by the energy of the original harmonic maps, we know the energy is bounded. By Uhlenbeck compactness, there exists a subsequence, which we still denote by HνH_{\nu}, that converges in C1C^{1} to some limiting harmonic map HH on every compact subset of Dr(0)\{0}D_{r}(0)\backslash\{0\}. Since the energy is uniformly bounded, we know that 0 is a removable singularity, so HH is a harmonic map on Dr(0)D_{r}(0). We have to show that there is no bubble point. We know that the maps restricted to NN, which are the hνh_{\nu}’s, have no bubble point on Dr(0)\{0}D_{r}(0)\backslash\{0\}. By the energy estimate in [21], we know

Ae(Hν)Ae(hν)+Area(A)[1+c1E2(hν)]\int_{A}e(H_{\nu})\leq\int_{A}e(h_{\nu})+Area(A)[1+c_{1}E^{2}(h_{\nu})]

for any compact ADr(0)\{0}A\subset D_{r}(0)\backslash\{0\}.

We also need to consider Hν(z/Rν2)H_{\nu}(z/R_{\nu}^{2}) on S2\intB1/rS^{2}\backslash intB_{1/r}. Consider any compact subset of \B1/r\mathbb{C}\backslash B_{1/r}, The corresponding Area(A)Area(A) shrinks as ν\nu\rightarrow\infty. The energy is conformal invariant. Thus, since hνh_{\nu}’s have no bubble point, we know HνH_{\nu}’s have no bubble point. Denote the limit as H~\tilde{H}. We have H~\tilde{H} defined on \B1/r\mathbb{C}\backslash B_{1/r}. By the property of removable singularities, we know H~\tilde{H} is a harmonic map on S2\B1/rS^{2}\backslash B_{1/r}.

Furthermore, by looking at ψH=|xH|2|yH|22ixH,yH\psi_{H}=|\partial_{x}H|^{2}-|\partial_{y}H|^{2}-2i\langle\partial_{x}H,\partial_{y}H\rangle, since the convergence is in the sense of C1C^{1}, we know the limits HH and H~\tilde{H} will also be conformal.

To apply lemma 7.4, we still need to show that limρ0limνE(Hν,A(1/(Rν2ρ),ρ))=0\underset{\rho\rightarrow 0}{\lim}\underset{\nu\rightarrow\infty}{\lim}E(H_{\nu},A(1/(R_{\nu}^{2}\rho),\rho))=0. First, given a small positive number ϵ\epsilon that only depends on the target manifold, we will need to show that the energy will be below the ϵ\epsilon for small enough ρ\rho and large enough ν\nu. We can get this from the conditions.

Then we know from the lemma limρ0limνsup1/(Rν2ρ)|z|ρd(Hν(z),H(0))=0\underset{\rho\rightarrow 0}{\lim}\underset{\nu\rightarrow\infty}{\lim}\underset{1/(R_{\nu}^{2}\rho)\leq|z|\leq\rho}{\sup}d(H_{\nu}(z),H(0))=0. We know from the construction of the pregluing that, for the pregluing HRνH^{R_{\nu}}, we have

limνsupzS2d(Hν(z),HRν(z))=0\underset{\nu\rightarrow\infty}{\lim}\underset{z\in S^{2}}{\sup}d(H_{\nu}(z),H^{R_{\nu}(z)})=0

Thus for large enough ν\nu, there exists a unique small vector field ξν\xi_{\nu} along HRνH^{R_{\nu}} such that

Hν=expHRν(ξν)H_{\nu}=\exp_{H^{R_{\nu}}}(\xi_{\nu})

We claim that

limνξν1,p,Rν=0\underset{\nu\rightarrow\infty}{\lim}\|\xi_{\nu}\|_{1,p,R_{\nu}}=0

We can focus on the ’neck’ part, which is 1/rRν2|z|r1/rR_{\nu}^{2}\leq|z|\leq r. Since f(z)=ezf(z)=e^{z} is conformal, we know Wν(z):=Hν(Rν1ez)W_{\nu}(z):=H_{\nu}(R_{\nu}^{-1}e^{z}) is harmonic and conformal. The neck part is transformed to |s|logr+logRν|s|\leq\log r+\log R_{\nu}, where z=s+itz=s+it. This sequence of harmonic maps converge to x:=H(0)x:=H(0) in all derivatives, uniformly in all compact subsets.

We know from lemma 7.4 that there exists 0<μ0<10<\mu_{0}<1 such that for any μμ0\mu\leq\mu_{0},

|sWν(s,t)|Ceμ(logRν|s|),|s|logr+logRνlog2|\partial_{s}W_{\nu}(s,t)|\leq Ce^{-\mu(\log R_{\nu}-|s|)},\quad|s|\leq\log r+\log R_{\nu}-\log 2

for ν\nu sufficiently large. Since

2|sWν(s,t)|=esRν1|dHν(Rν1es+it)|,\sqrt{2}|\partial_{s}W_{\nu}(s,t)|=e^{s}R_{\nu}^{-1}|dH_{\nu}(R_{\nu}^{-1}e^{s+it})|,

thus, if we first focus on |z|1/Rν|z|\geq 1/R_{\nu}, we have

1Rν|z|r2|dHν(z)|C|z|1μ\frac{1}{R_{\nu}}\leq|z|\leq\frac{r}{2}\quad\Rightarrow\quad|dH_{\nu}(z)|\leq\frac{C}{|z|^{1-\mu}}

For any pp such that μ0>12/p\mu_{0}>1-2/p, we have pμp+2>0p\mu-p+2>0. Thus

dHνpLp(A(1/Rν,ρ))C1/Rν|z|ρ|z|pμpCρpμp+2\|dH_{\nu}\|^{p}_{L^{p}\left(A(1/R_{\nu},\rho)\right)}\leq C\int_{1/R_{\nu}\leq|z|\leq\rho}|z|^{p\mu-p}\leq C\rho^{p\mu-p+2}

This shows that

limρ0limνdHνLp(A(1/Rν,ρ))=0\underset{\rho\rightarrow 0}{\lim}\underset{\nu\rightarrow\infty}{\lim}\|dH_{\nu}\|_{L^{p}\left(A(1/R_{\nu},\rho)\right)}=0

We know from the construction of the pregluing that

limρ0limνdHRνLp(A(1/Rν,ρ))=0\underset{\rho\rightarrow 0}{\lim}\underset{\nu\rightarrow\infty}{\lim}\|dH^{R_{\nu}}\|_{L^{p}\left(A(1/R_{\nu},\rho)\right)}=0

Thus we have the estimate for ξν\xi_{\nu} when |z|1/Rν|z|\geq 1/R_{\nu}. The estimate for when |z|1/Rν|z|\leq 1/R_{\nu} is similar. Note that here pp can be slightly larger than 2.

Next, we try to show

limνξν2,p,Rν=0\underset{\nu\rightarrow\infty}{\lim}\|\xi_{\nu}\|_{2,p,R_{\nu}}=0

First, we have to show

limρ0limνd2HνLp(A(1/Rν,ρ))=0\underset{\rho\rightarrow 0}{\lim}\underset{\nu\rightarrow\infty}{\lim}\|d^{2}H_{\nu}\|_{L^{p}\left(A(1/R_{\nu},\rho)\right)}=0

Same as before, consider Wν(w):=Hν(Rν1ew)W_{\nu}(w):=H_{\nu}(R_{\nu}^{-1}e^{w}). We have Wν(logRν+logz)=Hν(z)W_{\nu}(\log R_{\nu}+\log z)=H_{\nu}(z). Note that although logarithm is multivalued, it doesn’t matter in our case since we only need to consider the local values of the derivatives. To write out the calculations explicitly, we adopt another expression:

Wν(s,t):=Wν(s+it)=Wν(w)\displaystyle W_{\nu}(s,t):=W_{\nu}(s+it)=W_{\nu}(w)
Hν(x,y):=Hν(x+iy)=Hν(x,y)\displaystyle H_{\nu}(x,y):=H_{\nu}(x+iy)=H_{\nu}(x,y)

Let l(z):=logz+logRν=log|z|+logRν+iargz=:(l1(z),l2(z))l(z):=\log z+\log R_{\nu}=\log|z|+\log R_{\nu}+i\arg z=:(l_{1}(z),l_{2}(z)). By direct calculation

xH=sW(l(z))l1x+tW(l(z))l2x\partial_{x}H=\partial_{s}W(l(z))\frac{\partial l_{1}}{\partial x}+\partial_{t}W(l(z))\frac{\partial l_{2}}{\partial x}
x2H=\displaystyle\partial_{x}^{2}H= s2W(l(z))(l1x)2+2tsW(l(z))l1xl2x+t2W(l(z))(l2x)2+\displaystyle\partial_{s}^{2}W(l(z))\left(\frac{\partial l_{1}}{\partial x}\right)^{2}+2\partial_{t}\partial_{s}W(l(z))\frac{\partial l_{1}}{\partial x}\frac{\partial l_{2}}{\partial x}+\partial_{t}^{2}W(l(z))\left(\frac{\partial l_{2}}{\partial x}\right)^{2}+
sW(l(z))2l1x2+tW(l(z))2l2x2\displaystyle\partial_{s}W(l(z))\frac{\partial^{2}l_{1}}{\partial x^{2}}+\partial_{t}W(l(z))\frac{\partial^{2}l_{2}}{\partial x^{2}}

To estimate the second order derivatives of l1l_{1} and l2l_{2}, it is easier to look at z\partial_{z} and z¯\partial_{\bar{z}}, since derivatives with respect to xx and yy can be obtained by linear combination (both first and second order derivatives). From this, we know that

|l(z)|1|z|,|2l(z)|1|z|2\left|\partial l(z)\right|\leq\frac{1}{|z|},\quad\left|\partial^{2}l(z)\right|\leq\frac{1}{|z|^{2}}

We would like to show that

limρ0limν0A(1/Rν,ρ)|x2H(z)|pdxdy=0\underset{\rho\rightarrow 0}{\lim}\underset{\nu\rightarrow 0}{\lim}\int_{A(1/R_{\nu},\rho)}|\partial_{x}^{2}H(z)|^{p}dxdy=0

First, consider the term

sW(l(z))2l1x2\partial_{s}W(l(z))\frac{\partial^{2}l_{1}}{\partial x^{2}}

From previous estimates, we have

|sWν(s,t)|Ceμ(logRνs)=C|z|μ|\partial_{s}W_{\nu}(s,t)|\leq Ce^{-\mu(\log R_{\nu}-s)}=C|z|^{\mu}

Furthermore,

|2l1x2|1|z|2\left|\frac{\partial^{2}l_{1}}{\partial x^{2}}\right|\leq\frac{1}{|z|^{2}}

Thus

A(1/Rν,ρ)|sW(l(z))2l1x2|pdxdyρ(μ2)p+2\int_{A(1/R_{\nu},\rho)}\left|\partial_{s}W(l(z))\frac{\partial^{2}l_{1}}{\partial x^{2}}\right|^{p}dxdy\leq\rho^{(\mu-2)p+2}

We will get the desired result if we let p<22μp<\frac{2}{2-\mu}.

Next, consider the term

s2W(l(z))(l1x)2\partial_{s}^{2}W(l(z))\left(\frac{\partial l_{1}}{\partial x}\right)^{2}

We have

|l1x|21|z|2\left|\frac{\partial l_{1}}{\partial x}\right|^{2}\leq\frac{1}{|z|^{2}}

We divide the annulus into parts as follows: We are considering A(1/Rν,r)A(1/R_{\nu},r), here we can change rr to CrCr for any fixed constant CC and it won’t affect the argument, since we only care about the limit as r0r\rightarrow 0. We choose NN depending on ν\nu and rr, such that r/42N/Rν<r/2r/4\leq 2^{N}/R_{\nu}<r/2. By Hölder’s inequality, we have

A(2kRν,2k+1Rν)|s2W(l(z))(l1x)2|pdxdy\displaystyle\int_{A\left(\frac{2^{k}}{R_{\nu}},\frac{2^{k+1}}{R_{\nu}}\right)}\left|\partial_{s}^{2}W(l(z))\left(\frac{\partial l_{1}}{\partial x}\right)^{2}\right|^{p}dxdy\leq A(2kRν,2k+1Rν)|s2W(l(z))|p|z|2pdxdy\displaystyle\int_{A\left(\frac{2^{k}}{R_{\nu}},\frac{2^{k+1}}{R_{\nu}}\right)}\left|\partial_{s}^{2}W(l(z))\right|^{p}|z|^{-2p}dxdy
=\displaystyle= A(2kRν,2k+1Rν)|s2W(l(z))|p|z|t|z|t2pdxdy\displaystyle\int_{A\left(\frac{2^{k}}{R_{\nu}},\frac{2^{k+1}}{R_{\nu}}\right)}\left|\partial_{s}^{2}W(l(z))\right|^{p}|z|^{-t}|z|^{t-2p}dxdy
\displaystyle\leq (Ak|s2W(l(z))|pr|z|tr)1/r(Ak|z|(t2p)q)1/q\displaystyle\left(\int_{A_{k}}\left|\partial_{s}^{2}W(l(z))\right|^{pr}|z|^{-tr}\right)^{1/r}\left(\int_{A_{k}}|z|^{(t-2p)q}\right)^{1/q}

where we abbreviate A(2kRν,2k+1Rν)A\left(\frac{2^{k}}{R_{\nu}},\frac{2^{k+1}}{R_{\nu}}\right) as AkA_{k}. For the inequality to hold, we will have to choose tt carefully. We must satisfy the following:

tr=1\displaystyle tr=1
1r+1q=1\displaystyle\frac{1}{r}+\frac{1}{q}=1

For the part (Ak|s2W(l(z))|pr|z|tr)1/r\left(\int_{A_{k}}\left|\partial_{s}^{2}W(l(z))\right|^{pr}|z|^{-tr}\right)^{1/r}, we have

(Ak|s2W(l(z))|pr|z|tr)1/r=\displaystyle\left(\int_{A_{k}}\left|\partial_{s}^{2}W(l(z))\right|^{pr}|z|^{-tr}\right)^{1/r}= ({klog2s(k+1)log2}|s2W(s,t)|pr)1/r\displaystyle\left(\int_{\{k\log 2\leq s\leq(k+1)\log 2\}}|\partial_{s}^{2}W(s,t)|^{pr}\right)^{1/r}
\displaystyle\leq (C(pr,N)({klog2s(k+1)log2}|sW(s,t)|2)1/2)p\displaystyle\left(C(pr,N)\left(\int_{\{k\log 2\leq s\leq(k+1)\log 2\}}|\partial_{s}W(s,t)|^{2}\right)^{1/2}\right)^{p}
=\displaystyle= (C(pr,N)(Ak|xH|2)1/2)p\displaystyle\left(C(pr,N)\left(\int_{A_{k}}|\partial_{x}H|^{2}\right)^{1/2}\right)^{p}

Here we have used the main estimate 3.2 in [23] (This only allows us to estimate a real subset, but this is not essential. We can consider a stripe wider on both sides by log2\log 2.), as well as the conformal invariance of energy.

Again, using Hölder’s inequality,

Ak|xH|2(Ak|xH|γ)2/γ(Area(Ak))12γ\int_{A_{k}}|\partial_{x}H|^{2}\leq\left(\int_{A_{k}}|\partial_{x}H|^{\gamma}\right)^{2/\gamma}(Area(A_{k}))^{1-\frac{2}{\gamma}}

From previous part of the proof,

Ak|xH|γC(2k+1Rν)γμγ+2\int_{A_{k}}|\partial_{x}H|^{\gamma}\leq C\left(\frac{2^{k+1}}{R_{\nu}}\right)^{\gamma\mu-\gamma+2}

if we choose γ>2\gamma>2 small enough such that γμγ+2>0\gamma\mu-\gamma+2>0.

Furthermore, we know

(Area(Ak))12γ(2k+1Rν)2(12γ)(Area(A_{k}))^{1-\frac{2}{\gamma}}\leq\left(\frac{2^{k+1}}{R_{\nu}}\right)^{2(1-\frac{2}{\gamma})}

On the other hand,

Ak|z|(t2p)qC(2k+1Rν)(t2p)q+2\int_{A_{k}}|z|^{(t-2p)q}\leq C\left(\frac{2^{k+1}}{R_{\nu}}\right)^{(t-2p)q+2}

as long as (t2p)q+11(t-2p)q+1\neq-1. Note that this holds even when (t2p)q+1<1(t-2p)q+1<-1 due to the way we cut the rings. The constant only depends on t,p,qt,p,q.

Combining the above, we have

Ak|s2W(l(z))(l1x)2|pdxdy(2k+1Rν)(γμγ+2)pγ+2(12γ)p2+t2p+2q\int_{A_{k}}\left|\partial_{s}^{2}W(l(z))\left(\frac{\partial l_{1}}{\partial x}\right)^{2}\right|^{p}dxdy\leq\left(\frac{2^{k+1}}{R_{\nu}}\right)^{(\gamma\mu-\gamma+2)\frac{p}{\gamma}+2(1-\frac{2}{\gamma})\frac{p}{2}+t-2p+\frac{2}{q}}

We consider

(γμγ+2)pγ+2(12γ)p2+t2p+2q\displaystyle(\gamma\mu-\gamma+2)\frac{p}{\gamma}+2(1-\frac{2}{\gamma})\frac{p}{2}+t-2p+\frac{2}{q}
=\displaystyle= (μ1)p+2pγ+p2pγ+t2p+2q\displaystyle(\mu-1)p+\frac{2p}{\gamma}+p-\frac{2p}{\gamma}+t-2p+\frac{2}{q}
=\displaystyle= (μ2)p+t+2q\displaystyle(\mu-2)p+t+\frac{2}{q}
=\displaystyle= (μ2)p+1+1q\displaystyle(\mu-2)p+1+\frac{1}{q}

From previous argument, we need

1<p<22μ1<p<\frac{2}{2-\mu}

Let

ϵμ:=22μ1>0\epsilon_{\mu}:=\frac{2}{2-\mu}-1>0

We can write p=1+ϵμap=1+\frac{\epsilon_{\mu}}{a}, where a>1a>1. We will need to show that there exists q>1q>1 such that

(μ2)p+1+1q>0(\mu-2)p+1+\frac{1}{q}>0

That is

1+1q(2ϵμ+1)(1+ϵμa)>01+\frac{1}{q}-\left(\frac{2}{\epsilon_{\mu}+1}\right)\left(1+\frac{\epsilon_{\mu}}{a}\right)>0

We only need to show

(2ϵμ+1)(1+ϵμa)<2\left(\frac{2}{\epsilon_{\mu}+1}\right)\left(1+\frac{\epsilon_{\mu}}{a}\right)<2

This is obviously true.

Thus we have

Ak|s2W(l(z))(l1x)2|pdxdyC(2k+1Rν)τ\int_{A_{k}}\left|\partial_{s}^{2}W(l(z))\left(\frac{\partial l_{1}}{\partial x}\right)^{2}\right|^{p}dxdy\leq C\left(\frac{2^{k+1}}{R_{\nu}}\right)^{\tau}

where τ>0\tau>0. Since

k=0N1(2kRν)τ<(2NRν)τCrτ\sum_{k=0}^{N-1}\left(\frac{2^{k}}{R_{\nu}}\right)^{\tau}<\left(\frac{2^{N}}{R_{\nu}}\right)^{\tau}\leq Cr^{\tau}

We have shown

limρ0limνd2HνLp(A(1/Rν,ρ))=0\underset{\rho\rightarrow 0}{\lim}\underset{\nu\rightarrow\infty}{\lim}\|d^{2}H_{\nu}\|_{L^{p}\left(A(1/R_{\nu},\rho)\right)}=0

We still have to consider the pregluing. This estimate has already been done in the section ’Construction of the gluing map’.

For the part A(1/(Rν2ρ),1/Rν)A(1/(R_{\nu}^{2}\rho),1/R_{\nu}), we can use symmetry.

Thus the proof is finished. ∎

9. In the case of bubbling

We know that bubbling can occur, for example, from [21] and [23]. Let’s now consider the case when a bubble occurs and try to apply the gluing theorem. For the bubbling, we can use the same construction as in [21]. We can take the sequence that ”bubbles off” and glue the limit. Denote the bubble in the limit as ff^{\infty} and the rest of the limit as f0f^{0}. We can glue f0f^{0} and ff^{\infty}. Note that in the construction, we have control of the energy on the neck. We can follow the same procedure as in [21] to show that the conditions of lemma 7.4 is met. Then we can use a proof that is similar to the proof of sujectivity to show that we can choose a subsequence of the bubbling and that sequence, except for finitly many maps, will lie in the image of the gluing map. Thus we arrive at theorem 1.2.

Appendix A Uniform Boundedness of Coordinate Change

Consider a smooth Riemannian manifold NN and a point p0Np_{0}\in N. By the uniformly normal neighborhood lemma, there exists a neighborhood UU containing p0p_{0} and δ>0\delta>0 such that

  1. (1)

    For all p1,p2Up_{1},p_{2}\in U, there exists a unique geodesic γ\gamma of length less than δ\delta joining p1p_{1} to p2p_{2}. Moreover, γ\gamma is minimizing.

  2. (2)

    For any pUp\in U, Uexpp(Bp(δ))U\subset\exp_{p}(B_{p}(\delta)), and expp\exp_{p} is a diffeomorphism on Bp(δ)B_{p}(\delta).

Consider such a uniformly normal neighborhood UU of p0p_{0} and any point p1Up_{1}\in U, there exists a unique smooth geodesic γ\gamma such that γ(0)=p0\gamma(0)=p_{0} and γ(1)=p1\gamma(1)=p_{1}.

Consider a fixed normal coordinate chart centered at p0p_{0}. Denote the corresponding coordinates of p1p_{1} by y1y_{1}. Note that y1y_{1} is equal to the coordinates of γ˙(0)\dot{\gamma}(0). Consider a vector VTp0NV\in T_{p_{0}}N and let V(t)V(t) for t[0,1]t\in[0,1] be the parallel transport along γ\gamma. We have the following system of ODEs:

γ¨k(t)+Γkij(γ(t))γ˙i(t)γ˙j(t)=0\displaystyle\ddot{\gamma}^{k}(t)+\Gamma^{k}_{ij}(\gamma(t))\dot{\gamma}^{i}(t)\dot{\gamma}^{j}(t)=0
V˙k(t)+Γkij(γ(t))Vi(t)Vj(t)=0\displaystyle\dot{V}^{k}(t)+\Gamma^{k}_{ij}(\gamma(t))V^{i}(t)V^{j}(t)=0

Consider X(t):=γ˙(t)X(t):=\dot{\gamma}(t), we can transform the above system into a system of first order ODEs. Thus we know V(1)V(1) will be a smooth function of V(0)V(0), γ(0)\gamma(0), and γ˙(0)\dot{\gamma}(0). Here, γ(0)p0\gamma(0)\equiv p_{0}. Denote the parallel transport by

(30) PT:Tp0N×Tp0NTpNPT:T_{p_{0}}N\times T_{p_{0}}N\rightarrow T_{p}N

In coordinates (normal coordinates at p0p_{0} we fixed above) we have:

PT:\displaystyle PT: n×n\displaystyle\mathbb{R}^{n}\times\mathbb{R}^{n} n\displaystyle\rightarrow\mathbb{R}^{n}
(V(0),γ˙(0))\displaystyle(V(0),\dot{\gamma}(0)) V(1)\displaystyle\mapsto V(1)

Equivalently, we can write PTPT as

PT:\displaystyle PT: n\displaystyle\mathbb{R}^{n} n×n\displaystyle\rightarrow\mathbb{R}^{n}\times\mathbb{R}^{n}
γ˙(0)\displaystyle\dot{\gamma}(0) matrix representing the parallel transport\displaystyle\mapsto\text{matrix representing the parallel transport}

Note that for the matrix, the kk-th column is simply PT(ek,γ˙)PT(e_{k},\dot{\gamma}) written in coordinates, so the matrix is a smooth function of γ˙(0)\dot{\gamma}(0). Also note that PT(0)PT(0) is the identity matrix. We can write PT(w)=id+M(w)PT(w)=id+M(w) where MM is some matrix whose norm goes to 0 uniformly as |w||w| goes to 0. Similarly, we can control PT1PT^{-1}.

Now we are ready to prove the following lemma:

Lemma 2.1 (Uniform Boundedness of Coordinate Change).

Given a Riemannian manifold NN, for any p0Np_{0}\in N, consider the normal coordinates centered at p0p_{0}, which we denote by yy, there exists a neighborhood UU of p0p_{0} and a corresponding family of normal coordinates {yp}pU\{y_{p}\}_{p\in U} such that each coordinate ypy_{p} is a normal coordinate at pUp\in U. Furthermore, any function of the form |𝛂|yyp𝛂\frac{\partial^{|\boldsymbol{\alpha}|}y}{\partial y_{p}^{\boldsymbol{\alpha}}} and |𝛂|ypy𝛂\frac{\partial^{|\boldsymbol{\alpha}|}y_{p}}{\partial y^{\boldsymbol{\alpha}}} has bounded C0C^{0} norm on UU only depending on NN and p0p_{0}.

Proof.

Consider the uniformly normal neighborhood UU of p0p_{0} as above. Pick an orthonormal frame at p0p_{0}. For any point pUp\in U, consider the parallel transport along the unique minimizing geodesic from p0p_{0} to pp, then we get an orthonormal frame for all points in UU. Then we consider the normal coordinates determined by these frames. Now we have constructed the family of coordinates. Next we prove the properties.

Consider a point pUp\in U and two normal coordinates: One centered at p0p_{0}, which we denote by yy, and the other centered at p~U\tilde{p}\in U, which we denote by y~\tilde{y}. We know that from p0p_{0} to pp there exists a unique minimizing geodesic γ0\gamma_{0}, and the vector γ˙0(0)\dot{\gamma}_{0}(0) written in coordinates at p0p_{0} is exactly the coordinate y(p)y(p). Similarly, we can find γ~\tilde{\gamma} from p~\tilde{p} to pp and γ~˙(0)\dot{\tilde{\gamma}}(0) written in coordinates at p~\tilde{p} is the coordinate y~(p)\tilde{y}(p).

PTPT defined previously gives us relation between the frame at p0p_{0} and the frame at p~\tilde{p}. We can make UU smaller such that the closure of UU is a subset of a uniformly normal neighborhood. Thus we know PT(w)PT(w) and PT1(w)PT^{-1}(w) are smooth functions of ww, and the CkC^{k} norms of PTPT and PT1PT^{-1} are bounded on UU for any nonnegative integer kk.

Suppose γ~˙(0)=aiy~i\dot{\tilde{\gamma}}(0)=a^{i}\frac{\partial}{\partial\tilde{y}^{i}}. Note that

(y~1y~n)T=PT(y1yn)T\left(\frac{\partial}{\partial\tilde{y}^{1}}\cdots\frac{\partial}{\partial\tilde{y}^{n}}\right)^{T}=PT\left(\frac{\partial}{\partial y^{1}}\cdots\frac{\partial}{\partial y^{n}}\right)^{T}

Thus γ~˙(0)=aiPTijyj\dot{\tilde{\gamma}}(0)=a^{i}PT_{ij}\frac{\partial}{\partial y^{j}}. We know that (a1,,an)(a^{1},\cdots,a^{n}) is the coordinates at p~\tilde{p} of pp.

Consider γ~\tilde{\gamma} in coordinates at p0p_{0}. We know γ~(0)=y(p~)\tilde{\gamma}(0)=y(\tilde{p}) and γ~(1)=y(p)\tilde{\gamma}(1)=y(p). γ~˙(0)=y~(p)PT\dot{\tilde{\gamma}}(0)=\tilde{y}(p)PT. Also, γ~(1)=y(p)\tilde{\gamma}(1)=y(p), as the solution of the geodesic equations, is a smooth function of the initial values γ~(0)\tilde{\gamma}(0) and γ~˙(0)\dot{\tilde{\gamma}}(0). Furthermore, denote the geodesic from p0p_{0} to p~\tilde{p} by γ\gamma, we have γ~˙(0)=y~(p)PT\dot{\tilde{\gamma}}(0)=\tilde{y}(p)PT, where each entry in PTPT is a smooth function of γ˙(0)=y(p~)\dot{\gamma}(0)=y(\tilde{p}). Thus γ~(1)\tilde{\gamma}(1) is a smooth function of γ~(0)\tilde{\gamma}(0) and (a1,,an)(a^{1},\cdots,a^{n}). That is, y(p)y(p) is a smooth function of y(p~)y(\tilde{p}) and y~(p)\tilde{y}(p). Shrinking UU if necessary, we know yy~(p)\frac{\partial y}{\partial\tilde{y}}(p) and y~y(p)\frac{\partial\tilde{y}}{\partial y}(p) have bounded C0C^{0} norm. We also know the higher order derivatives of yy with respect to y~\tilde{y} have bounded norms. The derivatives of y~\tilde{y} with respect to yy can be obtained iteratively in terms of lower order derivatives and the derivatives of yy with respect to y~\tilde{y}. The proof is complete. ∎

Appendix B Perturbations in the Pregluing

First, let’s identify the Sobolev spaces regarding the perturbed map with the space regarding the original map. Namely, we identify W2,p(M,(f0)1TN)W^{2,p}(M,(f^{0})^{-1}TN) (resp. Lp(M,(f0)1TN)L^{p}(M,(f^{0})^{-1}TN)) with W2,p(M,(f0,r)1TN)W^{2,p}(M,(f^{0,r})^{-1}TN) (resp. Lp(M,(f0,r)1TN)L^{p}(M,(f^{0,r})^{-1}TN)). For simplicity, we only write out the case for f0f^{0}. The case for ff^{\infty} is completely the same.

Proposition B.1 (Equivalence of Sobolev Spaces under Perturbations).

For any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p) and (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U}, there exists δ0=δ0(c,p,N,𝒰)\delta_{0}=\delta_{0}(c,p,N,\mathcal{U}) such that for any (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), we have:

(\romannum1) Lp(M,(f0)1TN)Lp(M,(f0,r)1TN)L^{p}(M,(f^{0})^{-1}TN)\cong L^{p}(M,(f^{0,r})^{-1}TN).

(\romannum2) W2,p(M,(f0)1TN)W2,p(M,(f0,r)1TN)W^{2,p}(M,(f^{0})^{-1}TN)\cong W^{2,p}(M,(f^{0,r})^{-1}TN).

where the constants in the equivalence relation only depend on c,p,N,𝒰c,p,N,\mathcal{U}.

Proof.

Recall y=f0(0)=f()y=f^{0}(0)=f^{\infty}(\infty). For |z|2δR|z|\geq\frac{2}{\delta R}, we have f0=f0,rf^{0}=f^{0,r}. For |z|2δR|z|\leq\frac{2}{\delta R}, f0(z)=expyζ0(z)f^{0}(z)=\exp_{y}\zeta^{0}(z) whereas f0,r(z)=expy(ρ(δRz)ζ0(z))f^{0,r}(z)=\exp_{y}(\rho(\delta Rz)\zeta^{0}(z)). For any ξW2,p(M,(f0)1TN)\xi\in W^{2,p}(M,(f^{0})^{-1}TN), for |z|2δR|z|\leq\frac{2}{\delta R}, we can use parallel transport from expyζ0(z)\exp_{y}\zeta^{0}(z) to expy(ρ(δRz)ζ0(z))\exp_{y}(\rho(\delta Rz)\zeta^{0}(z)) to get ξ~\tilde{\xi} (For |z|2δR|z|\geq\frac{2}{\delta R} we can just take ξ~=ξ\tilde{\xi}=\xi). In other words:

ξ~(z):=Pγξ(z)\tilde{\xi}(z):=P_{\gamma}\xi(z)

where PγP_{\gamma} is parallel transport along γ\gamma, γ:[0,1]N\gamma:[0,1]\rightarrow N is a geodesic from γ(0)=f0(z)=expyζ0(z)\gamma(0)=f^{0}(z)=\exp_{y}\zeta^{0}(z) to γ(1)=f0,r(z)=expy(ρ(δRz)ζ0(z))\gamma(1)=f^{0,r}(z)=\exp_{y}(\rho(\delta Rz)\zeta^{0}(z)). Since df0\|df^{0}\| is bounded by cc, for small enough δ0\delta_{0}, there always exists such a unique geodesic γ\gamma, whose image is a subset of expy(tζ0(z))\exp_{y}(t\zeta^{0}(z)).

Now we have defined the map, we have to show that the map will induce the isomorphisms. First we have to show that this map maps LpL^{p} (resp. W2,pW^{2,p}) to LpL^{p} (resp. W2,pW^{2,p}) functions, then we have to show that after identifying the two spaces, the two norms are equivalent. We know the map is identity if we only consider |z|2δR|z|\geq\frac{2}{\delta R}. Recall that for the norms, we consider a finite cover of MM such that on each set of the cover there is a coordinate chart for MM and the image of that set has a coordinate chart on NN. We can choose a cover such that there is a coordinate chart in the cover that contains 0S20\in S^{2}. For small enough δ0\delta_{0}, this chart will contain all |z|4δR|z|\leq\frac{4}{\delta R}. We can choose the cover such that all other charts in this cover will only contain |z|2δR|z|\geq\frac{2}{\delta R}. Then we only have to consider the chart containing 0. We also make the δ0\delta_{0} small enough so that, for this chart, we can choose the coordinates on NN to be the normal coordinates at yy.

Now we can only consider the chart centered at yy. First, we identify the LpL^{p} and W2,pW^{2,p} spaces (both for f0f^{0} and f0,rf^{0,r}) with Lp(M,TyN)L^{p}(M,T_{y}N) and W2,p(M,TyN)W^{2,p}(M,T_{y}N). We construct the map by parallel transport along the geodesic ending at yy. Recall the parallel transport map PTPT defined in 30. Since we can write PT(w)=id+M(w)PT(w)=id+M(w) where the norm of MM goes to 0 as w0w\rightarrow 0, we have

Lp(M,(f0)1TN)Lp(M,TyN)Lp(M,(f0,r)1TN)L^{p}(M,(f^{0})^{-1}TN)\cong L^{p}(M,T_{y}N)\cong L^{p}(M,(f^{0,r})^{-1}TN)

for δ0\delta_{0} small enough.

Next, we consider the case for W2,pW^{2,p}. We can consider a fixed cutoff function on this chart such that the function is equal to 11 for all |z|2δR|z|\leq\frac{2}{\delta R} for δ0\delta_{0} small enough. The section multiplied by this cutoff function together with the parts on other charts will be an equivalent norm. This way, we can use approximation by compactly supported smooth sections for W2,pW^{2,p}. For a smooth section with compact support ϕ\phi,

xαPT(ϕ)=xαϕ+xα(Mϕ)=PT(ϕxα)+Mxαϕ\displaystyle\frac{\partial}{\partial x^{\alpha}}PT(\phi)=\frac{\partial}{\partial x^{\alpha}}\phi+\frac{\partial}{\partial x^{\alpha}}(M\phi)=PT\left(\frac{\partial\phi}{\partial x^{\alpha}}\right)+\frac{\partial M}{\partial x^{\alpha}}\phi

Note that Mxα=Myifixα\frac{\partial M}{\partial x^{\alpha}}=\frac{\partial M}{\partial y^{i}}\frac{\partial f^{i}}{\partial x^{\alpha}}, where ff can be f0f^{0} or f0,rf^{0,r}. we want to show that the norm of Mxα\frac{\partial M}{\partial x^{\alpha}} is bounded by some constant that does not depend on δ\delta or RR. This is obviously true for f0f^{0}. For f0,r(z)=expy(ρ(δRz)ζ0(z))f^{0,r}(z)=\exp_{y}(\rho(\delta Rz)\zeta^{0}(z)), we want to show the norm of f0,rxα\frac{\partial f^{0,r}}{\partial x^{\alpha}} is bounded:

|Mxα|\displaystyle\left|\frac{\partial M}{\partial x^{\alpha}}\right| =|Myifixα|\displaystyle=\left|\frac{\partial M}{\partial y^{i}}\frac{\partial f^{i}}{\partial x^{\alpha}}\right|
=|Myiexpyi(ρ(δRz)ζ0(z))xα|\displaystyle=\left|\frac{\partial M}{\partial y^{i}}\frac{\partial\exp_{y}^{i}(\rho(\delta Rz)\zeta^{0}(z))}{\partial x^{\alpha}}\right|
|Myi||dexpy(ρ(δRz)ζ0(z))||δRρxα(δRz)ζ0(z)+ρ(δRz)ζ0(z)xα|\displaystyle\leq\left|\frac{\partial M}{\partial y^{i}}\right|\left|d\exp_{y}(\rho(\delta Rz)\zeta^{0}(z))\right|\left|\delta R\frac{\partial\rho}{\partial x^{\alpha}}(\delta Rz)\zeta^{0}(z)+\rho(\delta Rz)\frac{\partial\zeta^{0}(z)}{\partial x^{\alpha}}\right|

By definition of ζ0\zeta^{0}, we have

f0(z)=expy(ζ0(z))\displaystyle f^{0}(z)=\exp_{y}(\zeta^{0}(z))

Since the C0C^{0} norm of ζ0(z)\zeta^{0}(z) is bounded by some constant times 1/δR1/\delta R, we know we can control the C0C^{0} norm of Mxα\frac{\partial M}{\partial x^{\alpha}}.

For the second order derivatives, we have

2xαxβPT(ϕ)=PT(2xαxβϕ)+(2xαxβM)ϕ+xαMxβϕ+xβMxαϕ\displaystyle\frac{\partial^{2}}{\partial x^{\alpha}\partial x^{\beta}}PT(\phi)=PT\left(\frac{\partial^{2}}{\partial x^{\alpha}\partial x^{\beta}}\phi\right)+\left(\frac{\partial^{2}}{\partial x^{\alpha}\partial x^{\beta}}M\right)\phi+\frac{\partial}{\partial x^{\alpha}}M\frac{\partial}{\partial x^{\beta}}\phi+\frac{\partial}{\partial x^{\beta}}M\frac{\partial}{\partial x^{\alpha}}\phi

We only need to show that 2MxαxβϕLpCϕW2,p\|\frac{\partial^{2}M}{\partial x^{\alpha}\partial x^{\beta}}\phi\|_{L^{p}}\leq C\|\phi\|_{W^{2,p}}, where CC only depends on δ0\delta_{0}. Again, this is obviously true for f0f^{0}, so we only need to consider f0,rf^{0,r}. We write out 2Mxαxβ\frac{\partial^{2}M}{\partial x^{\alpha}\partial x^{\beta}}:

|2Mxαxβ|=|2Myiyjfixαfjxβ+Myi2fixαxβ|\displaystyle\left|\frac{\partial^{2}M}{\partial x^{\alpha}\partial x^{\beta}}\right|=\left|\frac{\partial^{2}M}{\partial y^{i}\partial y^{j}}\frac{\partial f^{i}}{\partial x^{\alpha}}\frac{\partial f^{j}}{\partial x^{\beta}}+\frac{\partial M}{\partial y^{i}}\frac{\partial^{2}f^{i}}{\partial x^{\alpha}\partial x^{\beta}}\right|

The only term that matters is 2fixαxβ\frac{\partial^{2}f^{i}}{\partial x^{\alpha}\partial x^{\beta}}. We can get by direct computation:

|2expyi(ρ(δRz)ζ0(z))xαxβ|\displaystyle\left|\frac{\partial^{2}\exp_{y}^{i}(\rho(\delta Rz)\zeta^{0}(z))}{\partial x^{\alpha}\partial x^{\beta}}\right|\leq C(δ0)(δR+1)\displaystyle C(\delta_{0})(\delta R+1)

Then we have

2MxαxβϕLpC(δ0)ϕC0C(δ0)ϕW2,p\displaystyle\left\|\frac{\partial^{2}M}{\partial x^{\alpha}\partial x^{\beta}}\phi\right\|_{L^{p}}\leq C(\delta_{0})\|\phi\|_{C^{0}}\leq C(\delta_{0})\|\phi\|_{W^{2,p}}

by the Sobolev embedding.

So far we have shown that ξ~W2,pξW2,p\|\tilde{\xi}\|_{W^{2,p}}\leq\|\xi\|_{W^{2,p}}. The proof for ff^{\infty} is the same. For the other direction, consider P1=(id+M)1P^{-1}=(id+M)^{-1}. For δ0\delta_{0} small enough so that the entries of MM are sufficiently small, (id+M)1(id+M)^{-1} exists and is just a function of the entries of MM. Thus we have finished the proof. ∎

Proposition B.2.

For any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p) and any ϵ>0\epsilon>0, there exists δ0=δ0(c,p,N,𝒰,ϵ)>0\delta_{0}=\delta_{0}(c,p,N,\mathcal{U},\epsilon)>0 such that for any (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}) and any (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U}, under the identification in proposition B.1,

D0,rD0<ϵ\|D_{0,r}-D_{0}\|<\epsilon
Proof.

Still, we only have to consider the chart centered at yy and |z|2δR|z|\leq\frac{2}{\delta R}. As in the proof of the proposition B.1, let ξ~\tilde{\xi} and ξ\xi be sections of W2,p(M,(f0)1TN)W^{2,p}(M,(f^{0})^{-1}TN) and W2,p(M,(f0)1TN)W^{2,p}(M,(f^{0})^{-1}TN) respectively such that they are the same after the identification. We want to prove that for δ0\delta_{0} small enough, D0,rξ~D0ξLp<ϵξW2,p\|D_{0,r}\tilde{\xi}-D_{0}\xi\|_{L^{p}}<{\epsilon}\|\xi\|_{W^{2,p}} for any such pair of sections. We have

D0,rξ~D0ξLpD0,rξ~D0,rξLp+D0,rξD0ξLp\|D_{0,r}\tilde{\xi}-D_{0}\xi\|_{L^{p}}\leq\|D_{0,r}\tilde{\xi}-D_{0,r}\xi\|_{L^{p}}+\|D_{0,r}\xi-D_{0}\xi\|_{L^{p}}

Consider the right hand side. For the first part, by (7) and lemma 2.1, we only need to control ξ~ξW2,p\|\tilde{\xi}-\xi\|_{W^{2,p}}. In the proof of proposition B.1, we have ξ~i=ξk(δik+Mik)\tilde{\xi}^{i}=\xi^{k}(\delta_{i}^{k}+M_{ik}), where the norm of MM converge to 0 as δ0\delta_{0} converge to 0. So the second order derivative part of ξ~ξ\tilde{\xi}-\xi can be controlled. For the first order derivative part, we can use Sobolev embedding to control the LpL^{p^{*}} norm (of first order derivative), where p=2p/(2p)>2pp^{*}=2p/(2-p)>2p. Then consider its multiplication with the characteristic function of the neck, and use Holder’s inequality. Since the measure of the neck converge to 0, we can control the norm. The second part can be controlled similarly. ∎

Appendix C Apriori Estimates for the Differential Operators

We try to get an LpL^{p} estimate for the operators D0,DD_{0},D_{\infty}.

Proposition C.1.

Let ff be a smooth map from MM to NN. Then DfD_{f} defined in (6) is locally a second order strongly elliptic system.

Proof.

Say we use the normal coordinates centered at f(x0)f(x_{0}) on NN. From (7) we get operator in coordinates, then consider the coordinate change and apply lemma 2.1, we get the result. ∎

Lemma C.1 (LpL^{p} estimate).

Consider any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p). For every (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U} and ξ=(ξ0,ξ)W2,pf0,\xi=(\xi^{0},\xi^{\infty})\in W^{2,p}_{f^{0,\infty}}, we have

ξW2,pc0(D0,ξLp+ξLp)\displaystyle\|\xi\|_{W^{2,p}}\leq c_{0}(\|D_{0,\infty}\xi\|_{L^{p}}+\|\xi\|_{L^{p}})

where c0c_{0} only depends on c,p,N,𝒰c,p,N,\mathcal{U}.

Proof.

Essentially we only need to show that ξ0W2,pc0(Df0ξ0Lp+ξ0Lp)\|\xi^{0}\|_{W^{2,p}}\leq c_{0}(\|D_{f^{0}}\xi^{0}\|_{L^{p}}+\|\xi^{0}\|_{L^{p}}).

From (8) we know D0:W2,p(S2,(f0)1TN)Lp(S2,(f0)1TN)D_{0}:W^{2,p}(S^{2},{(f^{0})}^{-1}TN)\rightarrow L^{p}(S^{2},{(f^{0})}^{-1}TN) is a linear elliptic system, where each second order derivative term only contains a single component. We can apply the proof for LpL^{p} estimates in [11] (See section 9.5) to prove the same for our case of elliptic system. ∎

Lemma C.2.

Let XX and YY be Banach spaces and L:XYL:X\rightarrow Y be a bounded linear operator. Suppose XX is compactly embedded into YY and

uXC(X,Y,L)(LuY+uY)\|u\|_{X}\leq C(X,Y,L)(\|Lu\|_{Y}+\|u\|_{Y})

for all uXu\in X. Then LL is semi-Fredholm. That is, the kernel of LL is finite dimensional and the range of LL is closed.

Proof.

First, if we choose any sequence in the closed unit ball of kerL\ker L, by compact embedding of XX in YY, we can choose a converging subsequence in YY. Furthermore, in kerL\ker L, the norm in XX is bounded by norm in YY, thus we get a converging subsequence in XX. Thus the closed unit ball of kerL\ker L is compact, which means that kerL\ker L is finite dimensional.

Next, we prove that the range of LL is closed. Consider any sequence {un}n=1\{u_{n}\}_{n=1}^{\infty} such that Lun=fnfLu_{n}=f_{n}\rightarrow f. We would like to show there exists uu such that Lu=fLu=f. Since kerL\ker L is finite dimensional, there exists X1XX_{1}\subset X such that X=X1kerLX=X_{1}\oplus\ker L. Furthermore, we know that X1X_{1} is closed. Without loss of generality, we can assume that unX1u_{n}\in X_{1} for all nn.

First, we assume that the sequence unX\|u_{n}\|_{X} is bounded. Since XX is compactly embedded in YY, we can choose a subsequence, still denoted {un}\{u_{n}\}, that converges in YY. Then

ukulXC(fkflY+ukulY)\|u_{k}-u_{l}\|_{X}\leq C(\|f_{k}-f_{l}\|_{Y}+\|u_{k}-u_{l}\|_{Y})

for all positive integers k,lk,l. Thus {un}\{u_{n}\} converges in XX, thus we can find uXu\in X as the limit and we will have Lu=fLu=f. In fact, we know that uX1u\in X_{1}.

we show that unX\|u_{n}\|_{X} is bounded. If not, then by choosing a subsequence, we can assume without loss of generality that unX\|u_{n}\|_{X}\rightarrow\infty. Thus

limnLununX=limnfnunX=0\underset{n\rightarrow\infty}{\lim}L\frac{u_{n}}{\|u_{n}\|_{X}}=\underset{n\rightarrow\infty}{\lim}\frac{f_{n}}{\|u_{n}\|_{X}}=0

Consider the sequence {un/unX}\{u_{n}/\|u_{n}\|_{X}\} and apply the above argument and get uu such that uX1u\in X_{1} and uX=1\|u\|_{X}=1. However, we have Lu=0Lu=0, a contradiction. ∎

Proposition C.2.

Let MM be a closed Riemannian manifold of dimension 2 and NN another Riemannian manifold. Let ff be a smooth map from MM to NN. Then DfD_{f} defined in 6 is semi-Fredholm.

Proof.

We know that W2,p(M,f1TN)W^{2,p}(M,f^{-1}TN) is compactly embedded into Lp(M,f1TN)L^{p}(M,f^{-1}TN). Furthermore, same as in C.1, we have the estimate

ξW2,pC(M,N,f,p)(DfξLp+ξLp)\|\xi\|_{W^{2,p}}\leq C(M,N,f,p)(\|D_{f}\xi\|_{L^{p}}+\|\xi\|_{L^{p}})

for any ξW2,p(M,f1TN)\xi\in W^{2,p}(M,f^{-1}TN). Thus we can apply the above lemma to DfD_{f} and the proof is complete. ∎

The following is an analog of lemma 10.6.1 in [16]:

Lemma C.3.

Consider any precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p). Then there are positive constants δ0\delta_{0} and c0c_{0} only depending on c,p,N,𝒰c,p,N,\mathcal{U} such that, for all (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U} and (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), the following holds for r:=δRr:=\delta R:

(\romannum1) For every ξ=(ξ0,ξ)W2,pf0,,r\xi=(\xi^{0},\xi^{\infty})\in W^{2,p}_{f^{0,\infty,r}}, we have

ξW2,pc0(D0,,rξLp+ξLp).\|\xi\|_{W^{2,p}}\leq c_{0}(\|D_{0,\infty,r}\xi\|_{L^{p}}+\|\xi\|_{L^{p}}).

(\romannum2) For every ξ=(ξ0,ξ)W2,pf0,,r\xi=(\xi^{0},\xi^{\infty})\in W^{2,p}_{f^{0,\infty,r}}, we have

D0,,rξ=0ξW2,pc0ξL2.D_{0,\infty,r}\xi=0\ \Rightarrow\ \|\xi\|_{W^{2,p}}\leq c_{0}\|\xi\|_{L^{2}}.

(\romannum3) For every η=(η0,η)Lpf0,r×Lpf,r\eta=(\eta^{0},\eta^{\infty})\in L^{p}_{f^{0,r}}\times L^{p}_{f^{\infty,r}} we have

Q0,,rηW2,pc0ηLp\|Q_{0,\infty,r}\eta\|_{W^{2,p}}\leq c_{0}\|\eta\|_{L^{p}}
Proof.

Note that the part for f0f^{0} and the part for ff^{\infty} are symmetric, so we only need to consider one, and the other can be proved in the same way. Let’s consider f0f^{0}. For any ξ0W2,pf0,r\xi^{0}\in W^{2,p}_{f^{0,r}}, note that f0,rf^{0,r} only differ from f0f^{0} in |z|2δR|z|\leq\frac{2}{\delta R}. In |z|2δR|z|\leq\frac{2}{\delta R}, we know that f0f^{0} and f0,rf^{0,r} are both close to yy, so we can do a parallel transport of ξ0\xi^{0} to get ξ0~W2,pf0\tilde{\xi^{0}}\in W^{2,p}_{f^{0}}. From lemma C.1 we know that

ξ0~W2,pc0(D0ξ0~Lp+ξ0~Lp)\displaystyle\|\tilde{\xi^{0}}\|_{W^{2,p}}\leq c_{0}(\|D_{0}\tilde{\xi^{0}}\|_{L^{p}}+\|\tilde{\xi^{0}}\|_{L^{p}})

From the previous section we know that the first estimate is proved.

To prove the remaining estimates we consider the following abstract functional analytic setting. Suppose we have a surjective Fredholm operator

D:𝒲D:\mathcal{W}\rightarrow\mathcal{L}

of index dd between two Banach spaces. (Think of the case 𝒲=W2,p\mathcal{W}=W^{2,p}, =Lp\mathcal{L}=L^{p}, and D=D0,D=D_{0,\infty}) We assume 𝒲\mathcal{W} is equipped with an inner product ,\langle\cdot,\cdot\rangle and denote the corresponding norm by

ξL2:=ξ,ξ.\|\xi\|_{L^{2}}:=\sqrt{\langle\xi,\xi\rangle}.

(Think of the L2L^{2} inner product or W1,2W^{1,2}) We assume further that there are positive constants cc and cDc_{D} such that

(31) ξL2cξ𝒲\|\xi\|_{L^{2}}\leq c\|\xi\|_{\mathcal{W}}

for every ξ𝒲\xi\in\mathcal{W} (This is true because the L2L^{2} or W1,2W^{1,2} norm is controlled by the W2,pW^{2,p} norm) and

(32) Dξ=0ξ𝒲cDξL2.D\xi=0\ \Rightarrow\ \|\xi\|_{\mathcal{W}}\leq c_{D}\|\xi\|_{L^{2}}.

This holds in our setting because the kernel of DD is finite dimensional. Denote by Q:𝒲Q:\mathcal{L}\rightarrow\mathcal{W} the right inverse of DD whose image is the orthogonal complement of the kernel with respect to the above inner product. (Note that here W2,pW^{2,p} is just a subspace of the Hilbert space L2L^{2}. We can consider kerD\ker D, which is finite dimensional, and its orthogonal space in W2,pW^{2,p}, which we denote by (kerD)(\ker D)^{\perp}. It is easy to show that (kerD)(\ker D)^{\perp} is closed (in W2,pW^{2,p}), kerD(kerD)={0}\ker D\cap(\ker D)^{\perp}=\{0\}, and kerD+(kerD)=W2,p\ker D+(\ker D)^{\perp}=W^{2,p}. Thus we have W2,p=kerD(kerD)W^{2,p}=\ker D\oplus(\ker D)^{\perp}. Then we can construct the QQ the same way as for Hilbert spaces)

Now we prove the norm of QQ is bounded: Since we already have W2,p=kerD(kerD)W^{2,p}=\ker D\oplus(\ker D)^{\perp}, we can consider DD on (kerD)(\ker D)^{\perp} and we know the inverse will be bounded by the open mapping theorem.

Now suppose that D:𝒲D^{\prime}:\mathcal{W}\rightarrow\mathcal{L} is another bounded linear operator (D0,,rD_{0,\infty,r} for example) such that

ϵ:=DDQ<1.\epsilon:=\|D^{\prime}-D\|\|Q\|<1.

Since DQ=idDQ=id we have DQid<1\|D^{\prime}Q-id\|<1 and so DD^{\prime} is surjective with right inverse Q(DQ)1Q(D^{\prime}Q)^{-1}. However, we wish to understand the right inverse Q:𝒲Q^{\prime}:\mathcal{L}\rightarrow\mathcal{W} whose image is the orthogonal complement of the kernel of DD^{\prime}.

As a first step we observe that, if Dζ=0D^{\prime}\zeta=0, then QDζ𝒲ϵζ𝒲\|QD\zeta\|_{\mathcal{W}}\leq\epsilon\|\zeta\|_{\mathcal{W}} and ζQDζ𝒲(1ϵ)ζ𝒲\|\zeta-QD\zeta\|_{\mathcal{W}}\geq(1-\epsilon)\|\zeta\|_{\mathcal{W}}. Hence

(33) Dζ=0QDζ𝒲ϵ1ϵζQDζ𝒲.D^{\prime}\zeta=0\ \Rightarrow\ \|QD\zeta\|_{\mathcal{W}}\leq\frac{\epsilon}{1-\epsilon}\|\zeta-QD\zeta\|_{\mathcal{W}}.

Since ζQDζkerD\zeta-QD\zeta\in\ker D, we find ζQDζ𝒲cD|ζQDζ|cD|ζ|\|\zeta-QD\zeta\|_{\mathcal{W}}\leq c_{D}|\zeta-QD\zeta|\leq c_{D}|\zeta|. The last inequality holds because ζQDζ\zeta-QD\zeta is the orthogonal projection of ζ\zeta onto the kernel of DD (Note that the image of QQ is (kerD)(\ker D)^{\perp}). But we saw above that (1ϵ)ζ𝒲ζQDζ𝒲\|(1-\epsilon)\zeta\|_{\mathcal{W}}\leq\|\zeta-QD\zeta\|_{\mathcal{W}}. Hence

Dζ=0ζ𝒲cD1ϵζL2.D^{\prime}\zeta=0\ \Rightarrow\ \|\zeta\|_{\mathcal{W}}\leq\frac{c_{D}}{1-\epsilon}\|\zeta\|_{L^{2}}.

This proves (\romannum2).

Now assume that ξ𝒲\xi\in\mathcal{W} is orthogonal to the kernel of DD^{\prime}. Let e1,,ede_{1},\cdots,e_{d} be an orthonormal basis of kerD\ker D and consider the basis e1,,ede_{1}^{\prime},\cdots,e_{d}^{\prime} of kerD\ker D^{\prime} defined by

eiQDei=ei,i=1,,d.e_{i}^{\prime}-QDe_{i}^{\prime}=e_{i},\ i=1,\cdots,d.

(The map kerDkerD:ζζQDζ\ker D^{\prime}\rightarrow\ker D:\zeta\mapsto\zeta-QD\zeta is an isomorphism between the two kernels. We can see that since in the above argument, we have, if Dζ=0D^{\prime}\zeta=0, then QDζ𝒲ϵζ𝒲\|QD\zeta\|_{\mathcal{W}}\leq\epsilon\|\zeta\|_{\mathcal{W}} and ζQDζ𝒲(1ϵ)ζ𝒲\|\zeta-QD\zeta\|_{\mathcal{W}}\geq(1-\epsilon)\|\zeta\|_{\mathcal{W}}. Note that the index of Fredholm operators stays the same if we have DD<ϵ\|D^{\prime}-D\|<\epsilon for fixed DD and small enough ϵ\epsilon. Since the index is the same for DD^{\prime} and DD, we know the map is surjective.)

Since ξQDξkerD\xi-QD\xi\in\ker D we have

ξQDξ=i=1dξ,eiei.\xi-QD\xi=\sum_{i=1}^{d}\langle\xi,e_{i}\rangle e_{i}.

Moreover, ξ,ei=ξ,eiei=ξ,QDei\langle\xi,e_{i}\rangle=\langle\xi,e_{i}-e_{i}^{\prime}\rangle=\langle\xi,-QDe_{i}^{\prime}\rangle and hence, by (33),

|ξ,ei|cQDei𝒲ξ𝒲cϵ1ϵei𝒲ξ𝒲ccDϵ1ϵξ𝒲.|\langle\xi,e_{i}\rangle|\leq c\|QDe_{i}^{\prime}\|_{\mathcal{W}}\|\xi\|_{\mathcal{W}}\leq\frac{c\epsilon}{1-\epsilon}\|e_{i}\|_{\mathcal{W}}\|\xi\|_{\mathcal{W}}\leq\frac{cc_{D}\epsilon}{1-\epsilon}\|\xi\|_{\mathcal{W}}.

Combining this with the previous identity we find

ξQDξ𝒲cDξQDξL2=cDi=1dξ,ei2dccD2ϵ1ϵξ𝒲.\|\xi-QD\xi\|_{\mathcal{W}}\leq c_{D}\|\xi-QD\xi\|_{L^{2}}=c_{D}\sqrt{\sum_{i=1}^{d}\langle\xi,e_{i}\rangle^{2}}\leq\frac{\sqrt{d}cc_{D}^{2}\epsilon}{1-\epsilon}\|\xi\|_{\mathcal{W}}.

Hence

ξ𝒲\displaystyle\|\xi\|_{\mathcal{W}} QDξ𝒲+ξQDξ𝒲\displaystyle\leq\|QD\xi\|_{\mathcal{W}}+\|\xi-QD\xi\|_{\mathcal{W}}
QDξ+QDDξ𝒲+ξQDξ𝒲\displaystyle\leq\|Q\|\|D^{\prime}\xi\|_{\mathcal{L}}+\|Q\|\|D^{\prime}-D\|\|\xi\|_{\mathcal{W}}+\|\xi-QD\xi\|_{\mathcal{W}}
QDξ+ϵ(1+dccD21ϵ)ξ𝒲.\displaystyle\leq\|Q\|\|D^{\prime}\xi\|_{\mathcal{L}}+\epsilon\left(1+\frac{\sqrt{d}cc_{D}^{2}}{1-\epsilon}\right)\|\xi\|_{\mathcal{W}}.

If ϵ1/2\epsilon\leq 1/2 and ϵ(1+2dccD2)1/2\epsilon(1+2\sqrt{d}cc_{D}^{2})\leq 1/2 we deduce that ξ𝒲2QDξ\|\xi\|_{\mathcal{W}}\leq 2\|Q\|\|D^{\prime}\xi\|_{\mathcal{L}} for every ξ𝒲\xi\in\mathcal{W} that is orthogonal to the kernel of DD^{\prime}.

Now recall that DD=ϵ/Q\|D^{\prime}-D\|=\epsilon/\|Q\| and that DD^{\prime} is surjective, so that DξD^{\prime}\xi runs over all elements in \mathcal{L}. It follows that there is a constant δ>0\delta>0 such that

DD<δQ2Q.\|D^{\prime}-D\|<\delta\ \Rightarrow\ \|Q^{\prime}\|\leq 2\|Q\|.

How small δ\delta must be chosen depends only on the operator norm of QQ and the constants d,c,cDd,c,c_{D}. This finishes the proof. ∎

Appendix D Derivative Estimates

First, we consider Sobolev embeddings on the weighted sphere. We show that the constant in the inequality does not depend on δ\delta or RR, which will be needed in certain proofs. The following lemma is an analog of lemma 10.3.1 in [16].

Lemma D.1.

For any precompact 𝒰(c,p)\mathcal{U}\subset\mathcal{M}(c,p), there exists δ0=δ0(c,p,N,𝒰)\delta_{0}=\delta_{0}(c,p,N,\mathcal{U}) such that for any (f0,f)𝒰(f^{0},f^{\infty})\in\mathcal{U}, and any (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), we have

ξLC(c,p,N,𝒰)ξ2,p,R\|\xi\|_{L^{\infty}}\leq C(c,p,N,\mathcal{U})\|\xi\|_{2,p,R}

for any ξW2,p(S2,(fR)1TN)\xi\in W^{2,p}(S^{2},(f^{R})^{-1}TN), where S2S^{2} is the weighted sphere defined in (2) and fRf^{R} is the pregluing defined in (1).

Proof.

We let δ0\delta_{0} be small enough as in section 4.

For |z|1/R|z|\geq 1/R, the metric is the Fubini-Study metric on S2\B1/RS^{2}\backslash B_{1/R}. Each point z0S2\B1/Rz_{0}\in S^{2}\backslash B_{1/R} is contained in a disc DD of radius π/4\pi/4. Then we get the result from the usual Sobolev embedding.

For |z|1/R|z|\leq 1/R, we can consider the coordinate change w=1/R2zw=1/R^{2}z and the result follows in the same way as above. ∎

The following two lemmas are analogs of lemma 10.6.2 and lemma 10.6.3 in [16]. In this subsection, we always assume that the target manifold NN is closed.

Lemma D.2.

Fix two constants c>0c>0 and 1<p<21<p<2. Consider any precompact 𝒰(c1,p)\mathcal{U}\subset\mathcal{M}(c-1,p). Then, depending only on c,p,N,𝒰c,p,N,\mathcal{U}, we can choose positive constants δ0\delta_{0} and c0c_{0} as well as an open set 𝒰1(c,p)\mathcal{U}_{1}\subset\mathcal{M}(c,p) containing the closure of 𝒰\mathcal{U} such that the following holds:

Fix a pair (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), let (t1,t2)𝒰1:t(f0t,ft)(t_{1},t_{2})\rightarrow\mathcal{U}_{1}:t\mapsto(f^{0}_{t},f^{\infty}_{t}) be a smooth path and tfRtt\mapsto f^{R}_{t} be the corresponding path of preglued maps. Let tηRtt\mapsto\eta^{R}_{t} be a smooth family of sections of the pull-back bundle along fRtf^{R}_{t}, that is, ηRtΓ(S2,(fRt)1TN)\eta^{R}_{t}\in\Gamma(S^{2},(f^{R}_{t})^{-1}TN) for all t(t1,t2)t\in(t_{1},t_{2}). Then

t(TfRtηRt)TfRttηRt2,p,R\displaystyle\|\nabla_{t}(T_{f^{R}_{t}}\eta^{R}_{t})-T_{f^{R}_{t}}\nabla_{t}\eta^{R}_{t}\|_{2,p,R}\leq c0tfRt2,,RηtR0,p,R\displaystyle c_{0}\|\partial_{t}f^{R}_{t}\|_{2,\infty,R}\|\eta_{t}^{R}\|_{0,p,R}
t(TfRtηRt)TfRttηRt2,p,R\displaystyle\|\nabla_{t}(T_{f^{R}_{t}}\eta^{R}_{t})-T_{f^{R}_{t}}\nabla_{t}\eta^{R}_{t}\|_{2,p,R}\leq c0(tf0t,tft)W2,ηtR0,p,R\displaystyle c_{0}\|(\partial_{t}f^{0}_{t},\partial_{t}f^{\infty}_{t})\|_{W^{2,\infty}}\|\eta_{t}^{R}\|_{0,p,R}

for t(t1,t2)t\in(t_{1},t_{2}).

Proof.

First, we simplify by not considering the pregluing map. Suppose we have a path ft:S2Nf_{t}:S^{2}\rightarrow N. Thus, let

(t1,t2)×S2N:(t,z)ft(z)(t_{1},t_{2})\times S^{2}\rightarrow N:(t,z)\mapsto f_{t}(z)

be a smooth map, denote by DftD_{f_{t}} the corresponding family of linearized harmonic map operators, and let ηtLp(S2,(ft)1TN)\eta_{t}\in L^{p}(S^{2},(f_{t})^{-1}TN) be a smooth family of sections of pull-back bundles. Denote by Qt:=QftQ_{t}:=Q_{f_{t}} the right inverse whose image is the L2L^{2}-orthogonal complement of the kernel of DtD_{t}. Define ξt\xi_{t} by ξt:=Qtηt\xi_{t}:=Q_{t}\eta_{t}, ηt=Dtξt\eta_{t}=D_{t}\xi_{t}.

We first attempt to get the estimate

(34) tξtQttηtW2,pC(c,p,N,𝒰)tftW2,ηtLp\|\nabla_{t}\xi_{t}-Q_{t}\nabla_{t}\eta_{t}\|_{W^{2,p}}\leq C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\eta_{t}\|_{L^{p}}

We will need the following:

(35) Dtζ=0ζW2,pc1(c,p,N,𝒰)ζL2D_{t}\zeta=0\ \Rightarrow\ \|\zeta\|_{W^{2,p}}\leq c_{1}(c,p,N,\mathcal{U})\|\zeta\|_{L^{2}}

To prove the above inequality, we would like to make sure that the constant does not depend on the choice of the path. In fact, we can revisit lemma C.1, and get ζW2,pC(c,p,N,𝒰)ζLp\|\zeta\|_{W^{2,p}}\leq C(c,p,N,\mathcal{U})\|\zeta\|_{L^{p}}. Since 1<p<21<p<2, we have ζLpC(c,p,N,𝒰)ζL2\|\zeta\|_{L^{p}}\leq C(c,p,N,\mathcal{U})\|\zeta\|_{L^{2}}.

QtηW2,pc2(c,p,N,𝒰)ηLp\|Q_{t}\eta\|_{W^{2,p}}\leq c_{2}(c,p,N,\mathcal{U})\|\eta\|_{L^{p}}

From the proof of boundedness of QQ^{\prime} in lemma C.3), we know there is a uniform bound since 𝒰\mathcal{U} is precompact.

ζL2c3(c,p,N,𝒰)ζW2,p\|\zeta\|_{L^{2}}\leq c_{3}(c,p,N,\mathcal{U})\|\zeta\|_{W^{2,p}}
t(Dtξt)DttξtLpc4(c,p,N,𝒰)tftW2,ξtW2,p\|\nabla_{t}(D_{t}\xi_{t})-D_{t}\nabla_{t}\xi_{t}\|_{L^{p}}\leq c_{4}(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\xi_{t}\|_{W^{2,p}}

To prove the above inequality, recall (8). We also have

tξt=(tξkt+Γkij(ft(x))ξittfjt)yk\nabla_{t}\xi_{t}=\left(\partial_{t}\xi^{k}_{t}+\Gamma^{k}_{ij}(f_{t}(x))\xi^{i}_{t}\partial_{t}f^{j}_{t}\right)\frac{\partial}{\partial y^{k}}

We can calculate

|tDtξDttξ|C(c,p,N,𝒰)(|ξ|(|tft|+|xtft|+|x2tft|)+|ξx|(|tft|+|xtft|))|\nabla_{t}D_{t}\xi-D_{t}\nabla_{t}\xi|\leq C(c,p,N,\mathcal{U})\left(|\xi|(|\partial_{t}f_{t}|+|\partial_{x}\partial_{t}f_{t}|+|\partial_{x}^{2}\partial_{t}f_{t}|)+\left|\frac{\partial\xi}{\partial x}\right|(|\partial_{t}f_{t}|+|\partial_{x}\partial_{t}f_{t}|)\right)

Thus

tDtξDttξLpC(c,p,N,𝒰)tftW2,ξW1,p\|\nabla_{t}D_{t}\xi-D_{t}\nabla_{t}\xi\|_{L^{p}}\leq C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\xi\|_{W^{1,p}}

From the above inequalities,

QtDttξtQttηtW2,p\displaystyle\|Q_{t}D_{t}\nabla_{t}\xi_{t}-Q_{t}\nabla_{t}\eta_{t}\|_{W^{2,p}}\leq C(c,p,N,𝒰)Dttξtt(Dtξt)Lp\displaystyle C(c,p,N,\mathcal{U})\|D_{t}\nabla_{t}\xi_{t}-\nabla_{t}(D_{t}\xi_{t})\|_{L^{p}}
\displaystyle\leq C(c,p,N,𝒰)tftW2,ξtW1,p\displaystyle C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\xi_{t}\|_{W^{1,p}}
\displaystyle\leq C(c,p,N,𝒰)tftW2,ηtLp\displaystyle C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\eta_{t}\|_{L^{p}}

Now suppose that tζtt\mapsto\zeta_{t} is a path in the kernel of DtD_{t} (the existence is guaranteed by ODE theory) such that ζtL2=1\|\zeta_{t}\|_{L^{2}}=1. Then, since Dtζt=0D_{t}\zeta_{t}=0,

QtDttζtLp\displaystyle\|Q_{t}D_{t}\nabla_{t}\zeta_{t}\|_{L^{p}}\leq C(c,p,N,𝒰)Dttζtt(Dtζt)Lp\displaystyle C(c,p,N,\mathcal{U})\|D_{t}\nabla_{t}\zeta_{t}-\nabla_{t}(D_{t}\zeta_{t})\|_{L^{p}}
\displaystyle\leq C(c,p,N,𝒰)tftW2,ζtW1,p\displaystyle C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\zeta_{t}\|_{W^{1,p}}
\displaystyle\leq C(c,p,N,𝒰)tftW2,\displaystyle C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}

Moreover, 0=tξt,ζt=tξt,ζt+ξt,tζt0=\partial_{t}\langle\xi_{t},\zeta_{t}\rangle=\langle\nabla_{t}\xi_{t},\zeta_{t}\rangle+\langle\xi_{t},\nabla_{t}\zeta_{t}\rangle, hence

|tξt,ζt|=|tζt,ξt|=|ξt,QtDttζt|C(c,p,N,𝒰)tftW2,ξtW1,p\displaystyle|\langle\nabla_{t}\xi_{t},\zeta_{t}\rangle|=|\langle\nabla_{t}\zeta_{t},\xi_{t}\rangle|=|\langle\xi_{t},Q_{t}D_{t}\nabla_{t}\zeta_{t}\rangle|\leq C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\xi_{t}\|_{W^{1,p}}

Now choose an orthonormal (in L2L^{2} norm) frame ζ1t,,ζmt\zeta_{1t},\cdots,\zeta_{mt} of kerDt\ker D_{t} (First get linear independence by ODE theory and then use Gram-Schmidt process) where mm denotes the Fredholm index of DD. Then

tξtQtDttξtW2,p\displaystyle\|\nabla_{t}\xi_{t}-Q_{t}D_{t}\nabla_{t}\xi_{t}\|_{W^{2,p}}\leq C(c,p,N,𝒰)tξtQtDttξtL2\displaystyle C(c,p,N,\mathcal{U})\|\nabla_{t}\xi_{t}-Q_{t}D_{t}\nabla_{t}\xi_{t}\|_{L^{2}}
=\displaystyle= C(c,p,N,𝒰)i=1mtξt,ζit2\displaystyle C(c,p,N,\mathcal{U})\sqrt{\sum_{i=1}^{m}\langle\nabla_{t}\xi_{t},\zeta_{it}\rangle^{2}}
\displaystyle\leq C(c,p,N,𝒰)tftW2,ξtW1,p\displaystyle C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\xi_{t}\|_{W^{1,p}}
\displaystyle\leq C(c,p,N,𝒰)tftW2,ηtLp\displaystyle C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\eta_{t}\|_{L^{p}}

Next, instead of considering ftf_{t}, we must consider the pair (f0,rt,f,rt)(f^{0,r}_{t},f^{\infty,r}_{t}) and we are interested in the operators D,TD,T on the preglued maps fRtf^{R}_{t}. Let tηtRt\mapsto\eta_{t}^{R} be a family of LpL^{p} sections of the pull-back bundle, denote by

ηt=(η0,rt,η,rt)\eta_{t}=(\eta^{0,r}_{t},\eta^{\infty,r}_{t})

the section of pull-back bundle along ft0,rf_{t}^{0,r} and ft,rf_{t}^{\infty,r} corresponding to ηtR\eta_{t}^{R} as in (17). Abbreviate Dt,r:=Dft0,r,ft,rD_{t,r}:=D_{f_{t}^{0,r},f_{t}^{\infty,r}}, Qt,r:=Qft0,r,ft,rQ_{t,r}:=Q_{f_{t}^{0,r},f_{t}^{\infty,r}}, and let ξt:=(ξ0,rt,ξ,rt):=Qt,r(ηt0,r,ηt,r)\xi_{t}:=(\xi^{0,r}_{t},\xi^{\infty,r}_{t}):=Q_{t,r}(\eta_{t}^{0,r},\eta_{t}^{\infty,r}), ξRt:=TfRtηRt\xi^{R}_{t}:=T_{f^{R}_{t}}\eta^{R}_{t}.

We verify that in this case we still have the above inequalities:

(36) Dt,rζ=0ζW2,pc1(c,p,N,𝒰)ζL2D_{t,r}\zeta=0\ \Rightarrow\ \|\zeta\|_{W^{2,p}}\leq c_{1}(c,p,N,\mathcal{U})\|\zeta\|_{L^{2}}

Under proposition B.1, we know

ζW2,p\displaystyle\|\zeta\|_{W^{2,p}}\leq C(c,p,N,𝒰)(DtζLp+ζLp)\displaystyle C(c,p,N,\mathcal{U})(\|D_{t}\zeta\|_{L^{p}}+\|\zeta\|_{L^{p}})
\displaystyle\leq C(c,p,N,𝒰)(Dt,rζLp+(Dt,rDt)ζLp+ζLp)\displaystyle C(c,p,N,\mathcal{U})(\|D_{t,r}\zeta\|_{L^{p}}+\|(D_{t,r}-D_{t})\zeta\|_{L^{p}}+\|\zeta\|_{L^{p}})
\displaystyle\leq C(c,p,N,𝒰)ζLp\displaystyle C(c,p,N,\mathcal{U})\|\zeta\|_{L^{p}}

and we have ζLpC(c,p,N,𝒰)ζL2\|\zeta\|_{L^{p}}\leq C(c,p,N,\mathcal{U})\|\zeta\|_{L^{2}} (Use Hölder’s inequality and note that the integration of characteristic function of the domain, which is the area, is bounded uniformly).

Qt,rηW2,pc2(c,p,N,𝒰)ηLp\|Q_{t,r}\eta\|_{W^{2,p}}\leq c_{2}(c,p,N,\mathcal{U})\|\eta\|_{L^{p}}

The uniform bound can be found as in the proof of lemma C.3.

ζL2c3(c,p,N,𝒰)ζW2,p\|\zeta\|_{L^{2}}\leq c_{3}(c,p,N,\mathcal{U})\|\zeta\|_{W^{2,p}}

For the L2L^{2} norm, it is easy to show the equivalence under perturbations as in proposition B.1. For W2,pW^{2,p}, the equivalence is already proved in that proposition. Thus we get the bound independent of δ\delta or RR.

t(Dt,rξt)Dt,rtξtLpc4(c,p,N,𝒰)tftW2,ξtW2,p\|\nabla_{t}(D_{t,r}\xi_{t})-D_{t,r}\nabla_{t}\xi_{t}\|_{L^{p}}\leq c_{4}(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{W^{2,\infty}}\|\xi_{t}\|_{W^{2,p}}

The proof is the same as before.

Furthermore, we have:

ηtLp2ηRt0,p,R,tft0,rW2,+tf,rtW2,2tfRt2,,R\|\eta_{t}\|_{L^{p}}\leq 2\|\eta^{R}_{t}\|_{0,p,R},\quad\|\partial_{t}f_{t}^{0,r}\|_{W^{2,\infty}}+\|\partial_{t}f^{\infty,r}_{t}\|_{W^{2,\infty}}\leq 2\|\partial_{t}f^{R}_{t}\|_{2,\infty,R}

Hence, by what we have proved,

tξtQt,rtηtW2,pC(c,p,N,𝒰)tftR2,,RηtR0,p,R\|\nabla_{t}\xi_{t}-Q_{t,r}\nabla_{t}\eta_{t}\|_{W^{2,p}}\leq C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}^{R}\|_{2,\infty,R}\|\eta_{t}^{R}\|_{0,p,R}

The pregluing map W2,pf0,,rW2,pfR:ξ=(ξ0,r,ξ,r)ξRW^{2,p}_{f^{0,\infty,r}}\rightarrow W^{2,p}_{f^{R}}:\xi=(\xi^{0,r},\xi^{\infty,r})\mapsto\xi^{R} defined as in (18) is a bounded linear operator (boundedness can be shown using the estimates in the proof of proposition 5.1, and the bound only depends on c,p,N,𝒰c,p,N,\mathcal{U}.) that commutes with the covariant derivative t\nabla_{t} along every path tft:=(f0t,ft)t\mapsto f_{t}:=(f^{0}_{t},f^{\infty}_{t}) (because the cutoff function β\beta depends only on the points zz in the domain and so is independent of tt). Hence

tξRtTftRtηRt2,p,RC(c,p,N,𝒰)tfRt2,,RηRt0,p,R\|\nabla_{t}\xi^{R}_{t}-T_{f_{t}^{R}}\nabla_{t}\eta^{R}_{t}\|_{2,p,R}\leq C(c,p,N,\mathcal{U})\|\partial_{t}f^{R}_{t}\|_{2,\infty,R}\|\eta^{R}_{t}\|_{0,p,R}

This proves the first inequality of the lemma. For the second inequality, we go back to (D). The critical part is 1/δR|z|2/δR1/\delta R\leq|z|\leq 2/\delta R and δ/2R|z|δ/R\delta/2R\leq|z|\leq\delta/R.

For A:=1/δR|z|2/δRA:=1/\delta R\leq|z|\leq 2/\delta R, we have to estimate

tfRt2..R(AχA(|xξ|p+|ξ|p))1/p\displaystyle\|\partial_{t}f^{R}_{t}\|_{2.\infty.R}\left(\int_{A}\chi_{A}(|\partial_{x}\xi|^{p}+|\xi|^{p})\right)^{1/p}\leq (tf0t,tft)W2,(δR)|A|1/2ξW1,p\displaystyle\|(\partial_{t}f^{0}_{t},\partial_{t}f^{\infty}_{t})\|_{W^{2,\infty}}(\delta R)|A|^{1/2}\|\xi\|_{W^{1,p^{*}}}
\displaystyle\leq C(c,p,N,𝒰)(tf0t,tft)W2,ξW2,p\displaystyle C(c,p,N,\mathcal{U})\|(\partial_{t}f^{0}_{t},\partial_{t}f^{\infty}_{t})\|_{W^{2,\infty}}\|\xi\|_{W^{2,p}}

where p=2p/(2p)p^{*}=2p/(2-p).

The rest is same as before. ∎

Denote the pregluing of two harmonic maps by gRg^{R}.

Lemma D.3.

Fix any c>0c>0 and any 1<p<21<p<2, for every precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p), we can choose positive constants δ0,c0\delta_{0},c_{0} only depending on c,p,N,𝒰c,p,N,\mathcal{U} such that the following holds:

For any f:=(f0,f)𝒰f:=(f^{0},f^{\infty})\in\mathcal{U}, let gRg^{R} denote the pregluing (gR(f)=fR)(g^{R}(f)=f^{R}), and (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), then

(37) (ξ0,ξ)L2c0(idQfRDfR)dgR(f)(ξ0,ξ)2,p,R\|(\xi^{0},\xi^{\infty})\|_{L^{2}}\leq c_{0}\|(id-Q_{f^{R}}D_{f^{R}})dg^{R}(f)(\xi^{0},\xi^{\infty})\|_{2,p,R}

for every (ξ0,ξ)kerDf(\xi^{0},\xi^{\infty})\in\ker D_{f}. Thus the operator (idQfRDfR)dgR(f)(id-Q_{f^{R}}D_{f^{R}})dg^{R}(f) is an isomorphism from the kernel of Df:=Df0,fD_{f}:=D_{f^{0},f^{\infty}} to the kernel of DfRD_{f^{R}}.

Proof.

Similar as in (36), we have

(ξ0,ξ)W2,pC(c,p,N,𝒰)(ξ0,ξ)L2\|(\xi^{0},\xi^{\infty})\|_{W^{2,p}}\leq C(c,p,N,\mathcal{U})\|(\xi^{0},\xi^{\infty})\|_{L^{2}}

We shall prove the estimate

DfRdgR(f)ξ0,p,R\displaystyle\|D_{f^{R}}dg^{R}(f)\xi\|_{0,p,R}\leq C(c,p,N,𝒰)(δR)12pξL2\displaystyle C(c,p,N,\mathcal{U})(\delta R)^{1-\frac{2}{p}}\|\xi\|_{L^{2}}

for f=(f0,f)f=(f^{0},f^{\infty}) and ξ=(ξ0,ξ)kerDf\xi=(\xi^{0},\xi^{\infty})\in\ker D_{f}.

We have

ζ0L2+ζL2C(c,p,N,𝒰)dgR(f)ζ2,p,R\|\zeta^{0}\|_{L^{2}}+\|\zeta^{\infty}\|_{L^{2}}\leq C(c,p,N,\mathcal{U})\|dg^{R}(f)\zeta\|_{2,p,R}

for ζ=(ζ0,ζ)kerDf\zeta=(\zeta^{0},\zeta^{\infty})\in\ker D_{f}. To prove this, first note that the kernel is finite dimensional and thus the L2L^{2} norm of ζ0\zeta^{0} and ζ\zeta^{\infty} on an open set of the ball control the L2L^{2} norm on all of S2S^{2} (Note that they cannot be identically zero on any open set). Then we must find a uniform bound for all f𝒰f\in\mathcal{U}. We can show that the bound is continuous on 𝒰\mathcal{U} by considering ζit\zeta_{it} of kerDft\ker D_{f_{t}} for any smooth path ftf_{t} on 𝒰\mathcal{U} as in the proof of D.2.

Furthermore, we have QfRη2,p,Rc0(c,p,N,𝒰)η0,p,R\|Q_{f^{R}}\eta\|_{2,p,R}\leq c_{0}(c,p,N,\mathcal{U})\|\eta\|_{0,p,R} from the proof of proposition 5.1. Combining the above inequalities, we get

(idQfRDfR)dgR(f)ξ2,p,R\displaystyle\|(id-Q_{f^{R}}D_{f^{R}})dg^{R}(f)\xi\|_{2,p,R}\geq dgR(f)ξ2,p,Rc0(c,p,N,𝒰)DfRdgR(f)(ξ)0,p,R\displaystyle\|dg^{R}(f)\xi\|_{2,p,R}-c_{0}(c,p,N,\mathcal{U})\|D_{f^{R}}dg^{R}(f)(\xi)\|_{0,p,R}
\displaystyle\geq C(c,p,N,𝒰)(1(δR)12p)ξL2\displaystyle C(c,p,N,\mathcal{U})(1-(\delta R)^{1-\frac{2}{p}})\|\xi\|_{L^{2}}

Now we prove the estimate

DfRdgR(f)ξ0,p,R\displaystyle\|D_{f^{R}}dg^{R}(f)\xi\|_{0,p,R}\leq C(c,p,N,𝒰)(δR)12pξL2\displaystyle C(c,p,N,\mathcal{U})(\delta R)^{1-\frac{2}{p}}\|\xi\|_{L^{2}}

Abbreviate ξR:=dgR(f)ξ\xi^{R}:=dg^{R}(f)\xi. Note that by construction, DfRξRD_{f^{R}}\xi^{R} vanishes outside 1/δR|z|2/δR1/\delta R\leq|z|\leq 2/\delta R and δ/2R|z|δ/R\delta/2R\leq|z|\leq\delta/R.

To estimate for 1/δR|z|2/δR1/\delta R\leq|z|\leq 2/\delta R, recall

f0t(z):=expyt(ζ0t(z)),yt:=f0t(0),f^{0}_{t}(z):=\exp_{y_{t}}(\zeta^{0}_{t}(z)),\quad y_{t}:=f^{0}_{t}(0),

for 1/δR|z|2/δR1/\delta R\leq|z|\leq 2/\delta R. Then fRt:=gR(f0t,ft)f^{R}_{t}:=g^{R}(f^{0}_{t},f^{\infty}_{t}) is given by

fRt(z)=expyt(ρ(δRz)ζ0t(z))f^{R}_{t}(z)=\exp_{y_{t}}(\rho(\delta Rz)\zeta^{0}_{t}(z))

Differentiating with respect to tt, we get

tfRt=\displaystyle\partial_{t}f^{R}_{t}= E1(yt,ρ(δRz)ζ0t(z))(tyt)+ρ(δRz)E2(yt,ρ(δRz)ζt0(z))(tζt0(z)),\displaystyle E_{1}(y_{t},\rho(\delta Rz)\zeta^{0}_{t}(z))(\partial_{t}y_{t})+\rho(\delta Rz)E_{2}(y_{t},\rho(\delta Rz)\zeta_{t}^{0}(z))(\nabla_{t}\zeta_{t}^{0}(z)),
tf0t=\displaystyle\partial_{t}f^{0}_{t}= E1(yt,ζ0t(z))(tf0t(0))+E2(yt,ζ0t(z))(tζ0t(z))\displaystyle E_{1}(y_{t},\zeta^{0}_{t}(z))(\partial_{t}f^{0}_{t}(0))+E_{2}(y_{t},\zeta^{0}_{t}(z))(\nabla_{t}\zeta^{0}_{t}(z))

Solve the second equation for tζ0t(z)\nabla_{t}\zeta^{0}_{t}(z) and insert into the first equation, then set t=0t=0, we get

(38) ξR(z)=E1(y,ρ(δRz)ζ0(z))ξ0(0)+E2(y,ρ(δRz)ζ0(z))ξ0^(z)\xi^{R}(z)=E_{1}(y,\rho(\delta Rz)\zeta^{0}(z))\xi^{0}(0)+E_{2}(y,\rho(\delta Rz)\zeta^{0}(z))\hat{\xi^{0}}(z)

where y:=f0(0)y:=f^{0}(0), f0(z)=expx(ζ0(z))f^{0}(z)=\exp_{x}(\zeta^{0}(z)) and

ξ0^(z):=ρ(δRz)E2(y,ζ0(z))1(ξ0(z)E1(y,ζ0(z))ξ0(0))\hat{\xi^{0}}(z):=\rho(\delta Rz)E_{2}(y,\zeta^{0}(z))^{-1}(\xi^{0}(z)-E_{1}(y,\zeta^{0}(z))\xi^{0}(0))

for 1/δR|z|2/δR1/\delta R\leq|z|\leq 2/\delta R. This shows that the C2C^{2} norm of ξR\xi^{R} in 1/δR|z|2/δR1/\delta R\leq|z|\leq 2/\delta R is bounded by C(c,p,N,𝒰)(δR)C(c,p,N,\mathcal{U})(\delta R) times the C2C^{2} norm of ξ0\xi^{0}. The area of the annuli (in the weighted metric) are bounded by C(c,p,N,𝒰)(δR)2C(c,p,N,\mathcal{U})(\delta R)^{-2}. Thus

DfRξR0,p,RC(c,p,N,𝒰)(δR)12p(ξ0,ξ)C2C(c,p,N,𝒰)(δR)12p(ξ0,ξ)L2\|D_{f^{R}}\xi^{R}\|_{0,p,R}\leq C(c,p,N,\mathcal{U})(\delta R)^{1-\frac{2}{p}}\|(\xi^{0},\xi^{\infty})\|_{C^{2}}\leq C(c,p,N,\mathcal{U})(\delta R)^{1-\frac{2}{p}}\|(\xi^{0},\xi^{\infty})\|_{L^{2}}

For δ/2R|z|δ/R\delta/2R\leq|z|\leq\delta/R, the same can be proved by direct computation, or we can consider the coordinate change w=1/R2zw=1/R^{2}z and use symmetry.

To show that the operator is an isomorphism as stated in the lemma, we also show

dgR(f)(ξ0,ξ)2,p,RC(c,p,N,𝒰)(ξ0,ξ)W2,p\|dg^{R}(f)(\xi^{0},\xi^{\infty})\|_{2,p,R}\leq C(c,p,N,\mathcal{U})\|(\xi^{0},\xi^{\infty})\|_{W^{2,p}}

Again, we only have to consider 1/δR|z|2/δR1/\delta R\leq|z|\leq 2/\delta R and δ/2R|z|δ/R\delta/2R\leq|z|\leq\delta/R, and we can proceed in the same way as above.

Then, together with previous estimates regarding QfRQ_{f^{R}} and DfRD_{f^{R}}, we know

(idQfRDfR)dgR(f)(ξ0,ξ)2,p,RC(c,p,N,𝒰)(ξ0,ξ)W2,p\|(id-Q_{f^{R}}D_{f^{R}})dg^{R}(f)(\xi^{0},\xi^{\infty})\|_{2,p,R}\leq C(c,p,N,\mathcal{U})\|(\xi^{0},\xi^{\infty})\|_{W^{2,p}}

Lemma D.4.

Fix any c>0c>0 and any 1<p<21<p<2, for every precompact subset 𝒰\mathcal{U} of (c,p)\mathcal{M}(c,p), we can choose positive constants δ0,c0\delta_{0},c_{0} only depending on c,p,N,𝒰c,p,N,\mathcal{U} such that the following holds:

For any f:=(f0,f)𝒰f:=(f^{0},f^{\infty})\in\mathcal{U}, let gRg^{R} denote the pregluing (gR(f)=fR)(g^{R}(f)=f^{R}), and (δ,R)𝒜(δ0)(\delta,R)\in\mathcal{A}(\delta_{0}), then

t(dgR(ft)ξt)2,p,RC(c,p,N,𝒰)(tξtW2,p+ξtW2,ptfW2,p)\|\nabla_{t}(dg^{R}(f_{t})\xi_{t})\|_{2,p,R}\leq C(c,p,N,\mathcal{U})(\|\nabla_{t}\xi_{t}\|_{W^{2,p}}+\|\xi_{t}\|_{W^{2,p}}\|\partial_{t}f\|_{W^{2,p}})

for every (ξ0,ξ)kerDf(\xi^{0},\xi^{\infty})\in\ker D_{f}.

Proof.

dgR(ξ)dg^{R}(\xi) can be calculated as in the proof of lemma D.3. Note that for δ/R|z|1/δR\delta/R\leq|z|\leq 1/\delta R, gR(f(z))y:=f0(0)=f()g^{R}(f(z))\equiv y:=f^{0}(0)=f^{\infty}(\infty), thus dgR=0dg^{R}=0. For |z|1/δR|z|\geq 1/\delta R, gR(f(z))=f0(z)g^{R}(f(z))=f^{0}(z), thus dgRdg^{R} is also the identity map. For |z|δ/2R|z|\leq\delta/2R, gR(f(z))=f(R2z)g^{R}(f(z))=f^{\infty}(R^{2}z), thus dgR(f(R2z))ξ(R2z)=ξ(R2z)dg^{R}(f^{\infty}(R^{2}z))\xi^{\infty}(R^{2}z)=\xi^{\infty}(R^{2}z). For these parts, the (2,p,R)(2,p,R)-norm can be estimated by tξtW2,p\|\nabla_{t}\xi_{t}\|_{W^{2,p}}.

We are now left with δ/2R|z|δ/R\delta/2R\leq|z|\leq\delta/R and 1/δR|z|2/δR1/\delta R\leq|z|\leq 2/\delta R.

For 1/δR|z|2/δR1/\delta R\leq|z|\leq 2/\delta R, with the same calculations in the proof of lemma 10.5,

ξtR(z)=E1(yt,ρ(δRz)ζt0(z))ξt0(0)+E2(yt,ρ(δRz)ζt0(z))ξt0^(z)\xi_{t}^{R}(z)=E_{1}(y_{t},\rho(\delta Rz)\zeta_{t}^{0}(z))\xi_{t}^{0}(0)+E_{2}(y_{t},\rho(\delta Rz)\zeta_{t}^{0}(z))\hat{\xi_{t}^{0}}(z)

where ξtR(z):=dgR(ft(z))ξt(z)\xi_{t}^{R}(z):=dg^{R}(f_{t}(z))\xi_{t}(z), yt:=ft0(0)y_{t}:=f_{t}^{0}(0), ft0(z)=expyt(ζt0(z))f_{t}^{0}(z)=\exp_{y_{t}}(\zeta_{t}^{0}(z)), and

ξt0^(z):=ρ(δRz)E2(yt,ζt0(z))1(ξt0(z)E1(yt,ζt0(z))ξt0(0))\hat{\xi_{t}^{0}}(z):=\rho(\delta Rz)E_{2}(y_{t},\zeta_{t}^{0}(z))^{-1}(\xi_{t}^{0}(z)-E_{1}(y_{t},\zeta_{t}^{0}(z))\xi_{t}^{0}(0))
tξtR(z)=(t(ξtR)k+Γijk(ft(z))(ξtR)i(ξtR)j)yk\nabla_{t}\xi_{t}^{R}(z)=\left(\partial_{t}(\xi_{t}^{R})^{k}+\Gamma_{ij}^{k}(f_{t}(z))(\xi_{t}^{R})^{i}(\xi_{t}^{R})^{j}\right)\frac{\partial}{\partial y^{k}}

We would like to control

(1/δR|z|2/δR(1+|z|2)2|tξtR|p+(1+|z|2)p2|tξtR|p+(1+|z|2)2p2|2tξtR|p)1/p\left(\int_{1/\delta R\leq|z|\leq 2/\delta R}(1+|z|^{2})^{-2}|\nabla_{t}\xi_{t}^{R}|^{p}+(1+|z|^{2})^{p-2}|\partial\nabla_{t}\xi_{t}^{R}|^{p}+(1+|z|^{2})^{2p-2}|\partial^{2}\nabla_{t}\xi_{t}^{R}|^{p}\right)^{1/p}

where \partial denote all possible combinations of s\partial_{s} and t\partial_{t} where z=s+itz=s+it.

Now consider tξRt=tdgR(ft)ξt=(tdgR(ft))ξt+dgR(ft)tξt\nabla_{t}\xi^{R}_{t}=\nabla_{t}dg^{R}(f_{t})\xi_{t}=(\partial_{t}dg^{R}(f_{t}))\xi_{t}+dg^{R}(f_{t})\nabla_{t}\xi_{t}. From the proof of lemma D.3, we know

dgR(ft)tξt2,p,RC(c,p,N,𝒰)tξtW2,p\|dg^{R}(f_{t})\nabla_{t}\xi_{t}\|_{2,p,R}\leq C(c,p,N,\mathcal{U})\|\nabla_{t}\xi_{t}\|_{W^{2,p}}

Furthermore, from the formula for ξR\xi^{R}, we know

(tdgR(ft))ξt2,p,RC(c,p,N,𝒰)tftC0ξtW2,p\|(\partial_{t}dg^{R}(f_{t}))\xi_{t}\|_{2,p,R}\leq C(c,p,N,\mathcal{U})\|\partial_{t}f_{t}\|_{C^{0}}\|\xi_{t}\|_{W^{2,p}}

Thus the estimate holds.

For δ/2R|z|δ/R\delta/2R\leq|z|\leq\delta/R, we can change the coordinates z1/R2zz\mapsto 1/R^{2}z and then the proof is the same (Note that we can calculate that the two coordinates are equivalent in the sense of Riemannian metric, and the 2,p,R2,p,R norm for δ/2R|z|δ/R\delta/2R\leq|z|\leq\delta/R is equivalent to the W2,pW^{2,p} norm of ξ\xi^{\infty} on S2S^{2}). ∎

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