Local estimate of fundamental groups

TL;DR

This paper provides a quantitative proof that the fundamental group of a complete nonnegatively Ricci curved manifold can be generated by a number of generators bounded by a function of the dimension, using advanced geometric analysis tools.

Contribution

It offers a new quantitative bound on the number of generators needed for the fundamental group, improving understanding of its structure in nonnegative Ricci curvature manifolds.

Findings

Bound C(n) ≤ n^{n^{20n}} for generators of the fundamental group.

Utilizes quantitative Cheeger-Colding's almost splitting theory.

Employs the squeeze lemma for covering groups.

Abstract

For any complete $n$-dim Riemannian manifold $M^n$ with nonnegative Ricci curvature, Kapovitch and Wilking proved that any finitely generated subgroup of the fundamental group $\pi_1(M^n)$ can be generated by $C(n)$ generators. Inspired by their work, we give a quantitative proof of the above theorem and show that $C(n)\leq n^{n^{20n}} $. Our main tools are quantitative Cheeger-Colding's almost splitting theory, and the squeeze lemma for covering groups between two Riemannian manifolds with nonnegative Ricci curvature.

Equations

$$\displaystyle f(x_{0})=y_{0}\ ,\quad\quad\quad\quad\mathbf{Y}\subset\mathbf{U}_{\epsilon}\big{(}f(\mathbf{X})\big{)},$$

$$\displaystyle\Big{|}d_{\mathbf{Y}}\big{(}f(x_{1}),f(x_{2})\big{)}-d_{\mathbf{X}}(x_{1},x_{2})\Big{|}<\epsilon\ ,\quad\quad\quad\quad\forall x_{1},x_{2}\in\mathbf{X},$$

$$\displaystyle\sup_{B_{s}(p)\atop i=1,\cdots,k+1}|\nabla\mathbf{b}_{i}|\leq 1+\epsilon\quad\quad and\quad\quad\sup_{t\leq s}\fint_{B_{t}(p)}\sum_{i,j=1}^{k+1}\big{|}\langle\nabla\mathbf{b}_{i},\nabla\mathbf{b}_{j}\rangle-\delta_{ij}\big{|}\leq\epsilon.$$

$$\underset{B_R(p)}{\sup }|\nabla f|\le \frac{60n}{R}\underset{B_{2R}(p)}{\sup }|f|.$$

$$f_{+}(x)=\max \{f(x),0\}.$$

$$\displaystyle\frac{1}{V\big{(}B_{R}(p)\big{)}}\int_{B_{R}(p)}|\Delta b^{+}|\leq\frac{3n}{R},$$

$$d(q,x)\ge d(q,p)-d(p,x)>2R-R=R.$$

$$\Delta b^{+}=\Delta d(q,\cdot )\le \frac{n-1}{d(q,\cdot )}<\frac{n-1}{R},on{B}_R(p)$$

$$\displaystyle\sup_{x\in B_{R}(p)}\big{(}\Delta b^{+}\big{)}_{+}(x)\leq\frac{n-1}{R}.$$

$$\displaystyle\frac{V\big{(}\partial B_{R}(p)\big{)}}{V\big{(}B_{R}(p)\big{)}}\leq\frac{n}{R}.$$

$${\int }_{B_R(p)}|\Delta b^{+}|$$

$$\displaystyle\leq 2V\big{(}B_{R}(p)\big{)}\max_{B_{R}(p)}\big{(}\Delta b^{+}\big{)}_{+}+\Big{|}\int_{B_{R}(p)}\Delta b^{+}\Big{|}$$

$$\displaystyle\leq\frac{2(n-1)}{R}V\big{(}B_{R}(p)\big{)}+\Big{|}\int_{\partial B_{R}(p)}\frac{\partial b^{+}}{\partial\vec{n}}\Big{|}$$

$$\displaystyle\leq\frac{2(n-1)}{R}V\big{(}B_{R}(p)\big{)}+V\big{(}\partial B_{R}(p)\big{)}$$

$$\displaystyle\leq\frac{3n}{R}V\big{(}B_{R}(p)\big{)}.$$

$$\left\{\begin{array}[]{rl}\Delta\mathbf{b}&=0\quad\quad\quad\quad\quad on\ B_{4R}(x)\\ \mathbf{b}&=b^{+}\quad\quad\quad\quad\quad on\ \partial B_{4R}(x),\end{array}\right.$$

$${\text{fint}}_{B_{4R}(x)}|\nabla (\mathbf{b}-b^{+}){|}^2$$

$${\text{fint}}_{B_{2R}(x)}|{\nabla }^2\mathbf{b}{|}^2$$

$$\mathbf{b}(z_1)=\underset{z\in B_{4R}(x)}{\sup }\mathbf{b}(z),\mathbf{b}(z_2)=\underset{z\in B_{4R}(x)}{\min }\mathbf{b}(z).$$

$$\mathbf{b}(y)-b^{+}(y)$$

$$\le d(z_1,y)\le d(z_1,x)+d(x,y)\le 8R.$$

$$\underset{B_{4R}(x)}{\sup }|\mathbf{b}-b^{+}|\le 8R.$$

$$\displaystyle\int_{B_{4R}(x)}\big{|}\nabla(\mathbf{b}-b^{+})\big{|}^{2}$$

$$\le \underset{B_{4R}(x)}{\sup }|\mathbf{b}-b^{+}|{\int }_{B_{4R}(x)}|\Delta b^{+}|\le 8R{\int }_{B_{4R}(x)}|\Delta b^{+}|.$$

$$\displaystyle\frac{1}{2}\Delta\big{(}|\nabla\mathbf{b}|^{2}\big{)}=|\nabla^{2}\mathbf{b}|^{2}+\langle\nabla\Delta\mathbf{b},\nabla\mathbf{b}\rangle+Rc(\nabla\mathbf{b},\nabla\mathbf{b})\geq|\nabla^{2}\mathbf{b}|^{2}.$$

$$\phi(s)=\left\{\begin{array}[]{rl}&1\quad\quad\quad\quad\quad 0\leq s\leq 2R\\ &0\quad\quad\quad\quad\quad s\geq 3R,\end{array}\right.$$

$$-\frac{2}{R}\le {\varphi }^{\prime }\le 0,|{\varphi }^{\prime \prime }|\le \frac{16}{R^2}.$$

$$\displaystyle\sup_{B_{3R}(x)}|\nabla\mathbf{b}|\leq\frac{60n}{R}\sup_{B_{4R}(x)}|\mathbf{b}|\leq\frac{60n}{R}\sup_{\partial B_{4R}(x)}|b^{+}|\leq 60n\cdot\Big{[}4+\frac{d_{p}(x)}{R}\Big{]}.$$

$${\int }_{B_{2R}(x)}|{\nabla }^2\mathbf{b}{|}^2$$

$$\displaystyle=\frac{1}{2}\int_{B_{3R}(x)}|\nabla\mathbf{b}|^{2}\cdot\Delta\big{(}\phi\circ d_{x}\big{)}\leq 10^{6}n^{2}\Big{[}1+\frac{d_{p}(x)}{R}\Big{]}^{2}\int_{B_{3R}(x)}\Big{|}\Delta\big{(}\phi\circ d_{x}\big{)}\Big{|}.$$

$$\displaystyle\int_{B_{3R}(x)}\big{|}\Delta(\phi\circ d_{x})\big{|}$$

$$\displaystyle\leq 2V\big{(}B_{3R}(x)\big{)}\max_{B_{3R}(x)-B_{2R}(x)}\big{(}\Delta(-\phi\circ d_{x})\big{)}_{+}+\Big{|}\int_{B_{3R}(x)}\Delta(\phi\circ d_{x})\Big{|}$$

$$\displaystyle\leq 2V\big{(}B_{3R}(x)\big{)}\max_{B_{3R}(x)-B_{2R}(x)}\big{(}-\phi^{\prime}\Delta d_{x}-\phi^{\prime\prime}\big{)}_{+}+\int_{\partial B_{3R}(x)}\Big{|}\frac{\partial(\phi\circ d_{x})}{\partial\vec{n}}\Big{|}$$

$$\displaystyle\leq 2V\big{(}B_{3R}(x)\big{)}\max_{B_{3R}(x)-B_{2R}(x)}\Big{[}\frac{2}{R}\cdot\frac{n-1}{d_{x}}+|\phi^{\prime\prime}|\Big{]}+\sup_{\partial B_{3R}(x)}|\phi^{\prime}|\cdot V\big{(}\partial B_{3R}(x)\big{)}$$

$$\displaystyle\leq 2V\big{(}B_{3R}(x)\big{)}\cdot\frac{20n}{R^{2}}+\frac{2n}{R^{2}}V\big{(}B_{3R}(x)\big{)}$$

$$\displaystyle\leq\frac{50n}{R^{2}}V\big{(}B_{2R}(x)\big{)}.$$

$$\left\{\begin{array}[]{rl}V\big{(}\cup_{i\in\mathbf{J}}B_{\delta R}(x_{i})\big{)}&\geq\big{[}(1-2\delta)^{n}-\frac{2\cdot 10^{n}}{D}\big{]}V\big{(}B_{R}(\tilde{p})\big{)}\\ \sup\limits_{i\in\mathbf{J}}\fint_{B_{2\delta R}(x_{i})}|f|&\leq DK\quad\quad and\quad\quad\sup\limits_{i\in\mathbf{J}}\fint_{B_{2\delta R}(x_{i})}|h|\leq D\hat{K}.\end{array}\right.$$

Full text

The author was partially supported by NSFC-11771230

\titlecontents

section[0em]  \titlerule*[1pc].\contentspage \titlecontentssubsection[1.5em]  \titlerule*[1pc].\contentspage

1. 1 Almost orthonormal local Busemann functions
2. 2 Existence of almost orthonormal linear (A.O.L.) harmonic functions
3. 3 Existence of almost linear function and Hessian estimate
4. 4 Segment inequality and measure of ‘good’ points
5. 5 Quantitative almost splitting theorem
6. 6 Squeeze Lemma and Dimension induction on harmonic functions

Introduction

It is well-known that any compact Riemannian manifold $(M^{n},g)$ has finitely generated fundamental group. For non-compact complete Riemannian manifolds, the conclusion is not always right. For example, the surface with infinite genus has infinitely generated fundamental group. A natural question is:

Question 0.1**.**

For which complete Riemannian manifold $(M^{n},g)$, the fundamental group $\pi_{1}(M^{n})$ is finitely generated?

Note the above example has no non-negative sectional curvature metric, to obtain the finitely generatedness of fundamental group, we may consider adding some curvature assumption on complete Riemannian manifolds.

For the group $\mathbf{G}$, we define that $\mathfrak{ng}(\mathbf{G})$ is the minimal number of generators needed. In $1911$ Bieberbach proved: For complete flat Riemannian manifold $M^{n}$, the fundamental group $\pi_{1}(M^{n})$ is finitely generated and \mathfrak{ng}\big{(}\pi_{1}(M^{n})\big{)}\leq C(n).

Bieberbach reduced the study of fundamental groups of flat manifolds to the study of the discrete subgroup of the isometry group of $\mathbb{R}^{n}$, which was later developed into a more general theory about the discrete subgroups of Lie groups (see [Rag]).

Later for hyperbolic manifolds, the following theorem was obtained (see [BGS]): Any finite volume hyperbolic manifold has finitely generated fundamental group.

Remark 0.2**.**

The above theorem is a corollary of the following general result in Lie group theory: Any lattice $\Gamma$ of a Lie group $\mathbf{G}$ is finitely generated, where lattice means that $\Gamma$ is a discrete subgroup of $\mathbf{G}$ and $\mathbf{G}/\Gamma$ has finite volume.

The study of the above non-positive curvature case, has more algebraic flavor, which has strong contrast to the following non-negative curvature case.

In $1972$, Cheeger and Gromoll [Soul] studied non-negative sectional curvature Riemannian manifolds. Among other things, they obtained the following result: If $M^{n}$ has sectional curvature $sec\geq 0$, then $\pi_{1}(M^{n})$ is finitely generated.

In fact, in [Soul] it was showed that $M^{n}$ is homotopic to a compact totally geodesic submanifold of $M^{n}$, through the deformation by the gradient flow of Busemann function. The above theorem follows as a corollary of this more general structure result.

In $1978$, Gromov [Gromov-almost-flat] used Toponogov Comparison Theorem to study the fundamental group directly, and proved: For $M^{n}$ with sectional curvature $sec\geq 0$, \mathfrak{ng}\big{(}\pi_{1}(M^{n})\big{)}\leq C(n)=\frac{\mathrm{Vol}\big{(}\mathbb{S}^{n-1}\big{)}}{\mathrm{Vol}\big{(}\mathbb{D}^{n-1}(\frac{\pi}{6})\big{)}}, where $\mathbb{S}^{n-1}$ is the unit sphere in $\mathbb{R}^{n}$, $\mathbb{D}^{n-1}(r)$ is the geodesic ball with radius $r$ in $\mathbb{S}^{n-1}$.

For compact Riemannian manifolds with $Rc\geq 0$, under the additional assumption of conjugate radius, Guofang Wei [Wei] gave a uniform estimate on the generators of the fundamental group similar as in Gromov’s result.

Back in $1968$, Milnor [Milnor] proved that for complete Riemannian manifold $M^{n}$ with $Rc\geq 0$, any finitely generated subgroup of $\pi_{1}(M^{n})$, is polynomial growth of order $\leq n$. Furthermore, he posed the following conjecture:

Conjecture 0.3** (Milnor).**

For complete Riemannian manifold $(M^{n},g)$ with $Rc\geq 0$, $\pi_{1}(M^{n})$ is finitely generated.

In the $1980s$, Peter Li [Li] used the heat kernel and Anderson [Anderson] used the property of covering maps, to prove Milnor conjecture for Euclidean (maximal) volume growth case respectively (moreover, they proved the fundamental group is finite in fact).

In $2000$, Sormani [Sormani] used the excess estimate of Abresch and Gromoll [AG] on the universal cover of manifolds, successfully proved Milnor conjecture for linear (minimal) volume growth case.

Also in $2000$, B. Wilking [Wilking] used Milnor’s Theorem and the theory of discrete subgroup in Lie group to prove: If for any complete Riemannian manifold $(M^{n},g)$ with $Rc(g)\geq 0$ and $\pi_{1}(M^{n})$ is an abelian group, we have $\pi_{1}(M^{n})$ is finitely generated; then for any complete Riemannian manifold $(N^{n},\tilde{g})$ with $Rc(\tilde{g})\geq 0$, $\pi_{1}(N^{n})$ is finitely generated.

Note the proof of the above Wilking’s Theorem do not need to use Bochner formula in Riemannian geometry, only relies on the Bishop-Gromov Volume Comparison Theorem.

Recently, in $2011$, V. Kapovitch and B. Wilking [KW] proved the following local estimate of fundamental groups among other things :

Theorem 0.4** (Kapovitch and Wilking).**

For complete Riemannian manifold $(M^{n},g)$ with $Rc\geq 0$, any finitely generated subgroup $\Gamma$ of $\pi_{1}(M^{n})$ satisfies $\mathfrak{ng}(\Gamma)\leq C(n)$.

Their proof was inspired by Fukaya and Yamaguchi’s work [FY], started by contradiction, used the equivariant pointed Gromov-Hausdorff convergence and reduced the problem to the study of Ricci limit space with group actions, then do the induction on the dimension of Ricci limit space. The main technical tools are Cheeger-Colding’s theory of Ricci limit space and almost rigidity results. However this proof can not give the explicit estimate of the above $C(n)$.

One main purpose of this paper is to give an explicit uniform estimate of $\mathfrak{ng}(\Gamma)$ for finitely generated subgroup $\Gamma\subseteq\pi_{1}(M^{n})$. More precisely, we prove the following theorem, which can be thought as the quantitative version of Theorem 0.4.

Theorem 0.5**.**

For complete Riemannian manifold $(M^{n},g)$ with $Rc\geq 0$, any finitely generated subgroup $\Gamma$ of $\pi_{1}(M^{n})$ satisfies $\mathfrak{ng}(\Gamma)\leq n^{n^{20n}}$.

Remark 0.6**.**

The above upper bound $n^{n^{20n}}$ is not sharp, and our method can not provide the sharp bound either. Also, comparing the concise proof of Theorem 0.4 in [KW], our proof is sort of lengthy.

One advantage of our proof is self-contained. Basically, our argument only use Bishop-Gromov Volume Comparison Theorem and the Bochner formula for complete Riemannian manifolds with $Rc\geq 0$. We try to reveal the relation between discrete isometry group actions and $Rc\geq 0$ in an intrinsic way. Even the Compactness Theorem of Gromov-Hausdorff convergence is not used, let alone the theory of Ricci limit spaces, we only need the concept of $\epsilon$-Gromov-Hausdorff approximation.

There are three key ingredients of our proof. The first one tells us how to transfer from Gromov-Hausdorff approximation (geometry assumption) to almost orthonormal linear harmonic functions (analysis result). The second one is doing reverse argument, i.e. transferring analysis to geometry. And the third one studies the generating set of the covering groups through analysis and geometry. We will describe those three key ingredients in the rest of this section, and conclude the section with a sketchy description of our proof.

Firstly, we hope to reveal more explicit relation between group actions and the assumption $Rc\geq 0$, through the concrete analysis on distance function, in the similar spirit of Gromov’s proof about estimate of fundamental group’s generators. This hope starts from the following classical global result of Cheeger and Gromoll [Cheeger-Gromoll] obtained in $1971$:

Theorem 0.7** (Cheeger and Gromoll).**

If $(M^{n},g)$ is a complete $n$-dim Riemannian manifold with $Rc\geq 0$, and $M^{n}$ contains a line, then $M^{n}$ is isometric to $N^{n-1}\times\mathbb{R}$, where $N^{n-1}$ is a complete $(n-1)$-dim Riemannian manifold with $Rc\geq 0$.

The proof of Theorem 0.7 used the harmonic function $\mathbf{b}$ globally defined on $M^{n}$, which is sort of the limit of the distance function of $M^{n}$ (more precisely, the Busemann function). In fact, the above manifold $N^{n-1}$ is a level set of $\mathbf{b}$ and the splitting lines $\mathbb{R}$ are the gradient flow lines of $\mathbf{b}$. One crucial technical point is to get the gradient estimate of the harmonic function $\mathbf{b}$, which is $|\nabla\mathbf{b}|\equiv 1$.

In $1975$, Cheng and Yau [CY] established the well-known local gradient estimate of harmonic functions on complete Riemannian manifolds with $Rc\geq 0$, which enables us to obtain the gradient estimate of harmonic functions from the $C^{0}$-bound of harmonic functions locally. On such manifolds, in $1990$, Abresch and Gromoll [AG] obtained the excess estimate, which gives the local estimate of the sum of two local Busemann functions with respect to a segment (such sum is [math] for two Busemann functions with reverse directions in the proof of Theorem 0.7).

Then in the $1990$s, based on the gradient estimate and the excess estimate, Colding initiated the study of local properties of distance function, through harmonic function locally defined on manifolds with Ricci lower bound, in a series paper [Colding-shape], [Colding-large], [Colding-volume]; while solving several important problems in the theory of Gromov-Hausdorff convergence, which was established by Gromov in the $1980$s (for more details see [Gromov-book]). More precisely, among other things, Colding constructed the locally defined harmonic function based on the local Busemann function with respect to a segment, and proved such harmonic functions are ‘almost linear’ harmonic functions with bounded gradient, where the ‘almost linear’ is in average integral sense.

The first ingredient of our proof is the above existence of almost orthonormal linear harmonic functions, which was originally established by Colding in [Colding-shape], [Colding-large], [Colding-volume]. For our purpose, we need the explicit quantitative version, so we give all the details of the proof here. Although some calculation is sort of tedious, these explicit estimates possibly give an intrinsic expression of the transfer from geometry (Gromov-Hausdorff approximation) to analysis (existence of almost orthonormal linear harmonic functions).

In $1996$, Cheeger and Colding [CC-Ann] established the almost splitting theorem among other things, which transfers from analysis (existence of almost orthonormal linear harmonic functions) to geometry (Gromov-Hausdorff approximation). Very roughly, they proved that if there exist $k$ almost orthonormal linear harmonic functions on a geodesic ball $B_{r}(p)\subset M^{n}$ with $Rc\geq 0$, then a smaller concentric geodesic ball is close to a ball of $\mathbb{R}^{k}\times\mathbf{X}_{k}$ in the sense of Gromov-Hausdorff distance.

This almost splitting theorem is the second ingredient of our proof. For the same reason as the above, we need the quantitative version. During the proof of the almost splitting theorem, one crucial thing is the existence of a suitable Gromov-Hausdorff approximation under the suitable Hessian integral bound assumption. Because the original proof of this existence result in [CC-Ann] is concise, one of our contribution is a different detailed proof by modifying some argument in [CN], for more details see Section 5 of this paper.

It is a natural question whether the topology of two metric spaces are the same when they are very close in Gromov-Hausdorff distance sense. Generally, the answer is no, although there is an intrinsic Reifenberg type theorem when one of the spaces is $n$-dim Euclidean space and the other space is $n$-dim Riemannian manifolds (for details, see [CC1, Appendix]).

However, for a family of Riemannian manifolds converging to a metric space in Gromov-Hausdorff distance sense, using the equivariant Gromov-Hausdorff convergence theory developed by Fukaya and Yamaguchi in the $1980$s (see [Fukaya-Japan], [FY]), Kapovitch and Wilking [KW] obtained the results, which relate the fundamental groups of those converging Riemannian manifolds, to the limit group, which acts on the limit space of the universal covers of those Riemannian manifolds.

The study of the fundamental groups can be put into a more general context, i.e. the covering group of a covering map between two Riemannian manifolds with $Rc\geq 0$. The third ingredient of our proof is to characterize the change of the covering groups by the Gromov-Hausdorff distance between two metric spaces in quantitative form, where one metric space is a geodesic ball in Riemannian manifolds with $Rc\geq 0$ and the other one is a ball in a product metric space $\mathbb{R}^{k}\times\mathbf{X}_{k}$. Our squeeze lemma provides a bridge linking analysis with group actions and geometry. Very roughly, if there are $k$ almost orthonormal linear harmonic functions on a geodesic ball $B_{r}(p)\subset M^{n}$ with $Rc\geq 0$ (analysis), then from the almost splitting theorem in the second ingredient, we know that $B_{cr}(p)$ is close to $B_{cr}(0,\hat{p})\subset\mathbb{R}^{k}\times\mathbf{X}_{k}$ in Gromov-Hausdorff sense (geometry). Then the group actions on $B_{cr}(p)$ are almost generated by the group actions on $B_{cr}(\hat{p})\subset\mathbf{X}_{k}$, for more details see Lemma 6.4.

Now we describe our proof in a rough way. We start with a geodesic ball $B_{r}(p)\subset M^{n}$ with $Rc\geq 0$, firstly we use the first ingredient tool to find one almost linear harmonic function, then apply Squeeze Lemma to shrink the group action to a group action on a ball $B_{cr}(0,\hat{p})\subset\mathbf{X}_{n-1}$. Now we apply the first ingredient tool again to find two almost orthonormal linear harmonic functions on a smaller geodesic ball $B_{r_{1}}(p_{1})\subset M^{n}$, and the group action on $B_{r}(p)$ can be ‘controlled’ or generated by the group action on $B_{r_{1}}(p_{1})$. We repeat the above procedure by induction on the dimension of almost orthonormal linear harmonic functions. When the dimension is $n$, the group action is shown to be trivial, and we get our conclusion.

Part I. G-H approximation yields A.O.L. harmonic functions

In Part I of this paper, we will prove the following version result about analytic characterization of Gromov-Hausdorff approximation, which was implied in the argument of Cheeger and Colding in a series of papers, [Colding-shape], [Colding-large], [Colding-volume], [CC-Ann] and [Cheeger-GAFA].

Let us recall the definition of the pointed $\epsilon$-Gromov-Hausdorff approximation.

Definition 0.8**.**

Let $(\mathbf{X},d_{\mathbf{X}},x_{0})$ and $(\mathbf{Y},d_{\mathbf{Y}},y_{0})$ be two pointed metric spaces. For $\epsilon>0$, a map f:\big{(}\mathbf{X},x_{0}\big{)}\rightarrow(\mathbf{Y},y_{0}) is called a pointed $\epsilon$-Gromov-Hausdorff approximation if

[TABLE]

where \mathbf{U}_{\epsilon}\big{(}f(\mathbf{X})\big{)}\vcentcolon=\Big{\{}z\in\mathbf{Y}:d\big{(}z,f(\mathbf{X})\big{)}\leq\epsilon\Big{\}}. For simplicity, we also call that $f$ is an $\epsilon$-Gromov-Hausdorff approximation when the base points are fixed and clear.

In the rest of the paper, unless otherwise mentioned, we use $\mathbf{X},\mathbf{Y},\mathbf{X}_{k}$ to denote metric spaces. Also, we are only interested in geometry and analysis on $n$-dim manifolds with $n\geq 3$, so we will always assume the dimension of any manifolds $\mathbf{n\geq 3}$ in the rest of this paper.

For our application, we need the quantitative estimate, which relate the existence of the Gromov-Hausdorff approximation to the existence of almost orthonormal linear harmonic functions. Although many results in Part I are well-known to some experts in this field, we made the contribution to establish the suitable statement and the quantitative estimates. Also we elaborate the concise argument of Cheeger-Colding to present the proof of some known results in all the details for self-contained reason, and also hope to provide a backup reference for future study, besides the original works of Cheeger-Colding.

Theorem 0.9**.**

For $B_{10r}(q)\subset(M^{n},g)$ with $Rc(g)\geq 0$, any $0<\epsilon<1,\delta=n^{-3400n^{3}}\epsilon^{110n}$ and integer $0\leq k\leq n$, assume there is an $(\delta r)$-Gromov-Hausdorff approximation, $f:B_{10r}(q)\rightarrow B_{10r}(0,\hat{q})\subset\mathbb{R}^{k}\times\mathbf{X}_{k}$, and \mathrm{diam}\big{(}B_{r}(\hat{q})\big{)}\geq\frac{1}{4}r where $B_{r}(\hat{q})\subset\mathbf{X}_{k}$. Then there are harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k+1} defined on $B_{s}(p)\subset B_{10r}(q)$, where $s=n^{-320n^{3}}\epsilon^{10n}r$, such that

[TABLE]

1. Almost orthonormal local Busemann functions

Cheng-Yau’s gradient estimates was originally proved in [CY], the following form include an explicit form of constant, which is needed in the later proof. The proof is the same as in [CY], so we omit it.

Theorem 1.1** (Cheng-Yau’s gradient estimates).**

Assume $Rc(M^{n})\geq 0,p\in M^{n}$, $B_{2R}(p)\subset M^{n}$, $f:B_{2R}(p)\rightarrow\mathbb{R}$ is a harmonic function and f\in C\big{(}\overline{B_{2R}(p)}\big{)}, then

[TABLE]

If $\Delta f=c_{0}\geq 0$ for f\in C\big{(}\overline{B_{2R}(p)}\big{)}, then \sup\limits_{B_{R}(p)}|\nabla f|\leq\frac{200n(R+1)}{R}\Big{[}\sup_{B_{2R}(p)}|f|+c_{0}\Big{]}.

∎

Set $b^{+}(\cdot)=d(q,\cdot)-d(q,p):M^{n}\rightarrow\mathbb{R}$, the function $b^{+}$ is called the local Busemann function with respect to the couple points $[p,q]$. And we define the positive part of a function $f$ as

[TABLE]

Lemma 1.2**.**

Given $R>0$, $Rc(M^{n})\geq 0$ and $p,q\in M$ with $d(p,q)>2R$, then

[TABLE]

where $b^{+}(\cdot)=d(q,\cdot)-d(q,p):M^{n}\rightarrow\mathbb{R}$.

Proof: Note for any $x\in B_{R}(p)$,

[TABLE]

By Laplace Comparison Theorem,

[TABLE]

hence the positive part of $\Delta b^{+}$ satisfies

[TABLE]

By Bishop-Gromov Comparison Theorem,

[TABLE]

Now from (1.1), (1.2), we have

[TABLE]

∎

Lemma 1.3**.**

Suppose $Rc(M^{n})\geq 0$, $x\in M$, and the function $\mathbf{b}$ satisfies

[TABLE]

where $b^{+}(\cdot)=d(q,\cdot)-d(q,p):M^{n}\rightarrow\mathbb{R}$. Then

[TABLE]

Proof: From Maximum principle, one have $z_{1},z_{2}\in\partial B_{4R}(x)$ such that

[TABLE]

Then for any $y\in B_{4R}(x)$,

[TABLE]

Similarly, we have $\mathbf{b}(y)-b^{+}(y)\geq-d(z_{2},y)\geq-8R$. Hence

[TABLE]

From integration by parts, we get

[TABLE]

From $\Delta\mathbf{b}=0$ and Bochner’s formula,

[TABLE]

Let $\phi\in C^{\infty}(R_{+})$ be a nonnegative cut-off function such that

[TABLE]

and

[TABLE]

We use the notation $d_{p}(x)=d(p,x)$, similarly $d_{x}(\cdot)=d(x,\cdot)$. From Theorem 1.1 and Maximum principle,

[TABLE]

Then from (1.6) and (1.8),

[TABLE]

On the other hand, from (1.7) and Laplace Comparison Theorem,

[TABLE]

The inequality (1.4) follows from (1.9) and (1.10).

∎

Lemma 1.4**.**

Let $0<\delta<\frac{1}{2}$, $R>0$, $D>1$ be given, suppose $Rc(M^{n})\geq 0$, $\tilde{p}\in M$, and f,h\in L^{1}\big{(}B_{R}(\tilde{p})\big{)} with $\fint_{B_{R}(\tilde{p})}|f|\leq K$ , $\fint_{B_{R}(\tilde{p})}|h|\leq\hat{K}$. Then there exists finite many disjoint balls $\{B_{\frac{1}{2}\delta R}(x_{j})\}_{j\in\mathbf{J}}$ such that we have the following:

1. (I)

For any $j\in\mathbf{J}$, $B_{\frac{1}{2}\delta R}(x_{j})\subset B_{(1-\frac{3}{2}\delta)R}(\tilde{p})$ and $B_{2\delta R}(x_{j})\subset B_{R}(\tilde{p})$. 2. (II)

[TABLE]

Proof: Choose a maximal set of disjoint balls $\{B_{\frac{1}{2}\delta R}(x_{i})\}_{i\in\mathbf{S}}$ contained in $B_{(1-\frac{3}{2}\delta)R}(\tilde{p})$ and $B_{2\delta R}(x_{i})\subset B_{R}(\tilde{p})$. Assume $\tilde{q}\in\cap_{i=1}^{\lambda}B_{2\delta R}(x_{i})$, where $\{1,\cdots,\lambda\}\subset\mathbf{S}$, and note $B_{\frac{1}{2}\delta R}(x_{i})\subset B_{\frac{5}{2}\delta R}(\tilde{q})$,

[TABLE]

Hence $\lambda\leq 10^{n}$, and for all $\tilde{q}\in B_{R}(\tilde{p})$, there exists at most $10^{n}$-many $i$ with $\tilde{q}\in B_{2\delta R}(x_{i})$.

Note $B_{(1-2\delta)R}(\tilde{p})\subset\cup_{i\in\mathbf{S}}B_{\delta R}(x_{i})$ (otherwise, if $y\in B_{(1-2\delta)R}(\tilde{p})-\cup_{i\in\mathbf{S}}B_{\delta R}(x_{i})$, then $B_{\frac{1}{2}\delta R}(y)\cap B_{\frac{1}{2}\delta R}(x_{i})=\emptyset$ for any $i\in\mathbf{S}$ and $B_{\frac{1}{2}\delta R}(y)\subset B_{(1-\frac{3}{2}\delta)R}(\tilde{p})$, contradicting the choice of \big{\{}B_{\frac{1}{2}\delta R}(x_{i})\big{\}}_{i\in\mathbf{S}}).

Set

[TABLE]

Let $\mathbf{J}=\mathbf{S}-\mathbf{I}$, then

[TABLE]

Now we have

[TABLE]

From (1.11) and (1.12), we have

[TABLE]

Then \big{\{}B_{\frac{1}{2}\delta R}(x_{j})\big{\}}_{j\in\mathbf{J}} is our choice satisfying all properties required.

∎

Lemma 1.5**.**

Suppose $Rc(M^{n})\geq 0$, and $p,q\in M$ with $d(p,q)>2r$. For all $\epsilon>0$, there exists finitely many balls $B_{\delta r}(x_{i})\subset B_{r}(p),\forall i\in\mathbf{A}$, where $\mathbf{A}$ is a finite set, and harmonic functions $\mathbf{b}_{i}$ with $\delta=2^{-150n}\epsilon^{18}$ such that

[TABLE]

where $b^{+}(\cdot)=d(q,\cdot)-d(q,p):M^{n}\rightarrow\mathbb{R}$.

Proof: By scaling the metric $g$ to the new metric $r^{-2}g$, we only need to prove the conclusion for $r=1$.

From Lemma 1.2, we have $\fint_{B_{1}(p)}|\Delta b^{+}|\leq 3n$. Combining this with Lemma 1.4, for $D>1$, and $\frac{1}{2}>\delta_{1}>0$ (to be determined later), we can choose finitely many balls \Big{\{}B_{\delta_{1}}(y_{j})\Big{\}}_{j\in\mathbf{J}} and $B_{2\delta_{1}}(y_{j})\subset B_{1}(p)$ such that

[TABLE]

For each $j\in\mathbf{J}$, let $\mathbf{b}_{j}$ satisfy

[TABLE]

Apply Lemma 1.3 to $B_{2\delta_{1}}(y_{j})$, we have

[TABLE]

For any $j\in\mathbf{J}$, we can now apply Lemma 1.4 to the functions $|\nabla(\mathbf{b}_{j}-b^{+})|^{2}$ and $|\nabla^{2}\mathbf{b}_{j}|^{2}$ on $B_{\delta_{1}}(y_{j})$, $j\in\mathbf{J}$. We get balls \Big{\{}B_{\delta_{2}\delta_{1}}(x_{i}^{j})\Big{\}}_{i,j}, where $B_{2\delta_{2}\delta_{1}}(x_{i}^{j})\subset B_{\delta_{1}}(y_{j})$ and let $\delta=\delta_{1}\delta_{2}$, such that

[TABLE]

Furthermore, we set $\Omega_{j}=B_{\delta_{1}}(y_{j})-\bigcup_{i}B_{\delta}(x_{i}^{j})$, then

[TABLE]

And we get

[TABLE]

If we choose $\delta_{2}=\delta_{1}^{2}$, $D=\delta_{1}^{-\frac{1}{3}}\cdot 10^{n}$. Using $(1-2\delta_{1})^{n}\geq 1-2n\delta_{1}$, then for suitable $C(n)$, when $\delta_{1}=2^{-50n}\epsilon^{6}$, $\delta=\delta_{2}\delta_{1}=2^{-150n}\epsilon^{18}$, we get

[TABLE]

From (1.15), (1.16) and (1.17), we have

[TABLE]

Using Cauchy-Schwartz inequality, the conclusion follows.

∎

Let $SM^{n}$ be the unit tangent bundle of $M^{n}$, if $\pi:SM^{n}\rightarrow M^{n}$ is the projection map, for any $\Omega\subset SM^{n}$, the Liouville measure of $\Omega$, denoted by $V(\Omega)$, is defined by V(\Omega)=\mu\big{(}\pi(\Omega)\big{)}\cdot V(\mathbb{S}^{n-1}), where $V(\mathbb{S}^{n-1})$ is the volume of the conical Euclidean $(n-1)$-sphere, and $\mu$ is the volume measure of $(M^{n},g)$ determined by the metric $g$.

Definition 1.6**.**

For $\nu\in S_{x}M^{n}$, let $\gamma_{\nu}(\cdot)$ be the geodesic starting from $x$ with $\gamma_{\nu}^{\prime}(0)=\nu$, the geodesic flow $\mathfrak{g}^{t}(x,\nu):[0,\infty)\times SM^{n}\rightarrow SM^{n}$ is defined by

[TABLE]

Theorem 1.7** (Liouville’s Theorem).**

For any region $D\subset SM^{n}$ we have $V(\mathfrak{g}^{t}D)=V(D)$, where $\mathfrak{g}^{t}:SM^{n}\rightarrow SM^{n}$ is the geodesic flow on $M^{n}$, and the measure on $D,\mathfrak{g}^{t}D$ is the Liouville measure.

Proof: [Arnold].

∎

Lemma 1.8**.**

Let $x\in M^{n}$, $l,r>0$, suppose f\in C^{\infty}\big{(}B_{r+l}(x)\big{)}, and $g$ is a Lipschitz function on $M^{n}$, then for any $0\leq t\leq l$,

[TABLE]

Proof: Let $h(\tau)=f\circ\gamma_{v}(\tau)$, then for any $0\leq t\leq l$

[TABLE]

Then

[TABLE]

For any $0\leq t\leq l$, applying Theorem 1.7 and (1.18),

[TABLE]

∎

Now we prove Colding’s integral Toponogov theorem in quantitative form.

Theorem 1.9**.**

Suppose $Rc(M^{n})\geq 0$, $p,q\in M$ with $d(p,q)>2r$. For all $1>\epsilon>0$, there exists $\delta=2^{-240n}\epsilon^{18}$ such that for all $0\leq t\leq\delta$,

[TABLE]

where $b^{+}(x)=d(x,q)-d(p,q)$.

Proof: From Lemma 1.5, for $\epsilon_{1}>0$ (to be determined later), we can find finitely many balls $B_{\delta r}(x_{i})\subset B_{r}(p)$ with $\delta=2^{-150n}\epsilon_{1}^{18}$ such that

[TABLE]

Let $h(\nu)=(b^{+}\circ\gamma_{v})^{\prime}(tr)-\frac{(b^{+}\circ\gamma_{v})(\delta r)-(b^{+}\circ\gamma_{v})(0)}{\delta r}$, apply Lemma 1.8, we get

[TABLE]

Note $|\nabla b^{+}|\leq 1$, then we have

[TABLE]

Let $\epsilon_{1}=2^{-5n}\epsilon$, then we have the conclusion for $\delta=2^{-240n}\epsilon^{18}$.

∎

Proposition 1.10**.**

Given $0<\epsilon<1,0\leq k\leq n,B_{10r}(q)\subset(M^{n},g)$ with $Rc(g)\geq 0$, any \delta\leq n^{-1250n}\cdot\epsilon^{100n}\big{(}\frac{r_{1}}{r}\big{)} and $\frac{r_{1}}{r}\leq n^{-110n}\epsilon^{10n}$. Assume there is an $(\delta r)$-Gromov-Hausdorff approximation $\Phi:B_{10r}(0,\hat{q})\rightarrow B_{10r}(q)$, where $B_{10r}(0,\hat{q})\subset\mathbb{R}^{k}\times\mathbf{X}_{k}$, and there are $\hat{q}_{0},\hat{q}_{1}^{+}\in B_{3r}(\hat{q})\subset\mathbf{X}_{k}$ satisfying d(\hat{q}_{0},\hat{q}_{1}^{+})=r_{0}\in\big{[}\frac{1}{16}r,2r\big{]}. Then we have

[TABLE]

where \big{\{}\mathbf{e}_{i}\big{\}}_{i=1}^{k} is the standard basis for $\mathbb{R}^{k},q_{1}=\Phi(0,\hat{q}_{0})$ and

[TABLE]

Proof: Step (1). Firstly we have

[TABLE]

which implies $(r_{0}\mathbf{e}_{i},\hat{q}_{0})\in B_{10r}(0,\hat{q})$, and $p_{i}^{+}$ is well-defined. Also we can easily see $B_{r_{1}}(q_{1})\subset B_{10r}(q)$.

In this step we always assume that $x,y\in B_{r_{1}}(q_{1})$. Because $\Phi$ is an $(\delta r)$-Gromov-Hausdorff approximation, there exists $(\tilde{x},\hat{x}),(\tilde{y},\hat{y})\in B_{10r}(0,\hat{q})$ such that

[TABLE]

Let d_{0}=d\big{(}(\tilde{x},\hat{x}),(\tilde{y},\hat{y})\big{)}, then

[TABLE]

Assume $r_{1}=\zeta r$, we have

[TABLE]

which implies

[TABLE]

Similarly we have

[TABLE]

We define $\tilde{x}=(\tilde{x}_{1},\cdots,\tilde{x}_{k}),\tilde{y}=(\tilde{y}_{1},\cdots,\tilde{y}_{k})$ and

[TABLE]

From (1.21), (1.22) and (1.23),

[TABLE]

in the last inequality above we used the assumption $r_{0}\geq\frac{1}{16}r$. Similarly, we have

[TABLE]

Then we get

[TABLE]

For $i\leq k$, we have

[TABLE]

Let $\Psi=(b_{1}^{+},\cdots,b_{k}^{+},b_{k+1}^{+}):M^{n}\rightarrow\mathbb{R}^{k+1}$, from (1.21), (1.25), (1.26) and (1.24),

[TABLE]

If we assume $d(x,y)=lr_{1}$, where $0<l<1$ is to be determined later, then

[TABLE]

For $0<\epsilon_{1}<1$ (to be determined later), if the following holds:

[TABLE]

from (1.27) and (1.28), we obtain

[TABLE]

From (1.29), for all $\nu\in SB_{r_{1}}(q_{1})$ except a zero-measure set, we have

[TABLE]

where $\gamma_{\nu}$ is the geodesic satisfying $\gamma_{\nu}^{\prime}(0)=\nu$.

Step (2). From Theorem 1.9, let

[TABLE]

then we have

[TABLE]

where $f(\nu)\vcentcolon=\frac{(b_{i}^{+}\circ\gamma_{\nu})(lr_{1})-(b_{i}^{+}\circ\gamma_{\nu})(0)}{lr_{1}}$ for $\nu\in SB_{r_{1}}(q_{1})$.

For fixed $i$, and some $\theta\in(0,\frac{\pi}{2})$ (to be determined later), set

[TABLE]

then for any $\nu\in C_{\theta}$,

[TABLE]

From (1.32),

[TABLE]

Now from (1.32), (1.34) and (1.30), for $j\neq i$, we have

[TABLE]

Note (1.33), then

[TABLE]

Now note $\langle\nabla b_{i}^{+},\nabla b_{j}^{+}\rangle$ is constant on $T_{x}M^{n}$ for any fixed $x\in M^{n}$, we have

[TABLE]

Let \epsilon_{1}\leq\Big{(}\frac{V(C_{\theta})}{4V\big{(}SB_{r_{1}}(q_{1})\big{)}}\Big{)}^{2}\cdot\theta and $\theta=\frac{\epsilon^{2}}{16}\cdot\frac{1}{(n+1)^{4}}$, then \fint_{B_{r_{1}}(q_{1})}\big{|}\langle\nabla b_{i}^{+},\nabla b_{j}^{+}\rangle\big{|}\leq 4\sqrt{\theta}=\frac{\epsilon}{(n+1)^{2}}. So we let $\epsilon_{1}=2n^{-50n}\epsilon^{4n+2}$, the above conclusion follows. Plug into (1.31) and (1.28), the corresponding $\zeta,\delta$ can be determined.

∎

2. Existence of almost orthonormal linear (A.O.L.) harmonic functions

Definition 2.1**.**

For $q^{+},q^{-},p\in\mathbf{X}$, where $\mathbf{X}$ is a metric space, we say that $[q^{+},q^{-},p]$ is an AG-triple on $\mathbf{X}$ with the excess $s$ and the scale $t$ if

[TABLE]

where $\mathbf{E}(\cdot)=d(\cdot,q^{+})+d(\cdot,q^{-})-d(q^{+},q^{-})$.

For $p\in M^{n},r_{2}\geq r_{1}\geq 0$, we define $A_{r_{1},r_{2}}(p)$ as the following:

[TABLE]

We have the following Abresch-Gromoll lemma.

Lemma 2.2**.**

On complete Riemannian manifold $(M^{n},g)$ with $Rc\geq 0$, assume that $[q^{+},q^{-},p]$ is an AG-triple with the excess $\leq\frac{1}{n}\frac{r^{2}}{R}$ and the scale $\geq R$, furthermore assume $R\geq 2^{2n}r$, then \sup_{B_{r}(p)}\mathbf{E}\leq 2^{6}\cdot\big{(}\frac{r}{R}\big{)}^{\frac{1}{n-1}}r, where $\mathbf{E}(\cdot)=d(\cdot,q^{+})+d(\cdot,q^{-})-d(q^{+},q^{-})$.

Proof: Define the function \varphi(\rho)=\int_{\rho}^{2}\int_{t}^{2}\big{(}\frac{s}{t}\big{)}^{n-1}dsdt, and $\epsilon$ solves

[TABLE]

From R\geq 2^{2n}r>\big{[}3+4(n-1)\varphi(1)\big{]}r, it is easy to see that $\epsilon$ exists and is unique.

By $R\geq 2^{2n}r$ and (2.1), we have

[TABLE]

Note for $\rho\leq 1$,

[TABLE]

From (2.2) and (2.3), we obtain 2^{n}\big{(}\frac{\epsilon}{4}\big{)}^{2-n}\geq\frac{\epsilon}{8n}\frac{R}{r}, which implies \epsilon\leq 2^{6}\cdot\big{(}\frac{r}{R}\big{)}^{\frac{1}{n-1}}.

To prove the conclusion, we only need to prove that $\sup_{B_{r}(p)}\mathbf{E}\leq\epsilon r$. By contradiction, if $\sup_{B_{r}(p)}\mathbf{E}>\epsilon r$, then there exists $x_{0}\in B_{r}(p)$, such that

[TABLE]

From $|\nabla\mathbf{E}|\leq 2$ and (2.1), we have

[TABLE]

which implies $p\in A_{\frac{\epsilon}{4}r,2r}(x_{0})$.

Define the function $h:A_{\frac{\epsilon}{4}r,2r}(x_{0})\rightarrow[0,\infty)$ by \epsilon\leq 2^{6}\cdot\big{(}\frac{r}{R}\big{)}^{\frac{1}{n-1}}, and it is easy to check $(\mathbf{E}-h)\geq 0$ on $\partial B_{2r}(x_{0})$.

For any $x\in\partial B_{\frac{\epsilon}{4}r}(x_{0})$, using $|\nabla\mathbf{E}|\leq 2$, (2.4) and (2.1), we have

[TABLE]

hence we obtain $\min_{x\in\partial A_{\frac{\epsilon}{4}r,2r}(x_{0})}(\mathbf{E}-h)(x)\geq 0$.

On the other hand, let $d=d(x,x_{0})$, then $\varphi^{\prime\prime}(\frac{d}{r})+\frac{n-1}{d}r\cdot\varphi^{\prime}(\frac{d}{r})=1$. From $Rc\geq 0$ and Laplace Comparison Theorem $\Delta d\leq\frac{n-1}{d}$, note $\varphi^{\prime}\leq 0$, for any $x\in A_{\frac{\epsilon}{4}r,2r}(x_{0})$, we have the following inequality in weak sense:

[TABLE]

Then we get $\min_{x\in A_{\frac{\epsilon}{4}r,2r}(x_{0})}\Delta(\mathbf{E}-h)(x)\leq 0$.

From Weak Maximum Principle for weak superharmonic function (c.f. [GT, Theorem $8.1$]), note $p\in A_{\frac{\epsilon}{4}r,2r}(x_{0})$, we have $(\mathbf{E}-h)(p)\geq\min_{\partial A_{\frac{\epsilon}{4}r,2r}(x_{0})}(\mathbf{E}-h)\geq 0$. And note $d(p,x_{0})<r$, we obtain

[TABLE]

which is contradicting the assumption on the excess, the conclusion is proved.

∎

The following lemma provides the existence of good cut-off function on manifolds with $Rc\geq 0$, which will be used later.

Lemma 2.3**.**

If $Rc(M^{n})\geq 0$ and $p\in M^{n}$, for any $\tau\in(0,1)$, there is a nonnegative smooth function $\phi:M^{n}\rightarrow[0,1]$

[TABLE]

satisfying $\sup\limits_{x\in B_{r}(p)}|\Delta\phi(x)|\leq\frac{10^{15}n^{10}}{\tau^{2n}(1-\tau)^{8}}r^{-2}$.

Proof: Scaling $(M^{n},g)$ to $(M^{n},r^{-2}g)$, then we only to construct $\phi(x)$ for $r=1$. We define $h_{1}(\rho)=\int_{\rho}^{1}\int_{t}^{1}(\frac{s}{t})^{n-1}dsdt$ and

[TABLE]

It is easy to see

[TABLE]

Using Laplace Comparison Theorem, it is straightforward to get $\Delta W_{1}\geq\frac{1}{h_{1}(\tau)}$. From the theory of elliptic equations of second order (c.f. [GT]), we can define the function $W:\overline{B_{1}(p)}-B_{\tau}(p)\rightarrow\mathbb{R}$ satisfying

[TABLE]

Apply the Maximum Principle to $W-W_{1}$ on $\overline{B_{1}(p)}-B_{\tau}(p)$, we get

[TABLE]

which implies that $W\geq 0$.

For any $x_{0}\in\partial B_{\frac{1}{2}(1+\tau)}(p)$, we define W_{2}(x)=\frac{h_{2}\big{(}d(x_{0},x)\big{)}}{h_{1}(\tau)}. Note B_{\frac{1}{2}(1-\tau)}(x_{0})\subset\Big{(}\overline{B_{1}(p)}-B_{\tau}(p)\Big{)}, from Laplace Comparison Theorem again,

[TABLE]

Then Maximum Principle yields

[TABLE]

From the definition of $W$ and Maximum Principle, we know

[TABLE]

On $\partial B_{\frac{1}{2}(1-\tau)}(x_{0})$, we have $W_{2}(x)=\frac{h_{2}(\frac{1}{2}(1-\tau))}{h_{1}(\tau)}$. Also note $W_{2}(x_{0})=0$, then we get

[TABLE]

By $\Delta W(x)\geq 0$, apply Maximum Principle to $W$ on $\overline{B_{1}(p)}-B_{\frac{1}{2}(1+\tau)}(p)$, we have

[TABLE]

We define $\delta_{0}=\frac{(1-\tau)^{2}\tau}{16n^{2}}$, then we can get

[TABLE]

We can find a smooth function $f:[0,1]\rightarrow[0,1]$ as the following:

[TABLE]

which satisfies

[TABLE]

Note when $\tau<d(x,p)\leq\tau+\delta_{0}$, from (2.6), we have $W(x)\geq W_{1}(x)\geq\frac{h_{1}(\tau+\delta_{0})}{h_{1}(\tau)}$, then f\big{(}W(x)\big{)}=1. And when $\frac{1}{2}(1+\tau)\leq d(x,p)<1$, from (2.7), we get $W(x)\leq 1-\frac{h_{2}(\frac{1}{2}(1-\tau))}{h_{1}(\tau)}$, hence f\big{(}W(x)\big{)}=0.

Now we can define smooth function $\phi$ as the following,

[TABLE]

From $0\leq W\leq 1$ and Theorem 1.1, for any $x$ satisfying $\tau+\delta_{0}\leq d(x,p)\leq\frac{1}{2}(1+\tau)$,

[TABLE]

Then from (2.8) and (2.5), on $M^{n}$,

[TABLE]

∎

Lemma 2.4**.**

On complete Riemannian manifold $M^{n}$ with $Rc\geq 0$, for harmonic function $\mathbf{b}$ defined on $B_{r}(p)$ and any $\tau\in(0,1)$, we have

[TABLE]

Proof: From Bochner formula and $\Delta\mathbf{b}=0$, we have

[TABLE]

From Lemma 2.3, one can choose a nonnegative cut-off function $\phi$ such that

[TABLE]

satisfying $\sup_{x\in B_{r}(p)}|\Delta\phi(x)|\leq\frac{10^{15}n^{10}}{\tau^{2n}(1-\tau)^{8}}r^{-2}$. Now we have

[TABLE]

From Cauchy-Schwartz inequality and the above inequality, we have

[TABLE]

∎

Theorem 2.5** (Li-Yau).**

Let $(M^{n},g)$ be a complete Riemannian manifold with $Rc\geq 0$, then the heat kernel $H(x,y,t)$ satisfies

[TABLE]

Proof: By choosing suitable $\epsilon$ in [LY, Corollary $3.1$], the conclusion follows.

∎

Now we have the following existence result of almost linear harmonic function $\mathbf{b}$ with respect to the local Busemann function $b^{+}$.

Lemma 2.6**.**

On complete Riemannian manifold $M^{n}$ with $Rc\geq 0$, assume that $[q^{+},q^{-},p]$ is an AG-triple with the excess $\leq\frac{4}{n}\frac{r^{2}}{R}$ and the scale $\geq R$, also assume $R\geq 2^{2n+1}r$. Then there exists harmonic function $\mathbf{b}$ defined on $B_{2r}(p)$ such that

[TABLE]

where $b^{+}(x)=d(x,q^{+})-d(p,q^{+})$.

Remark 2.7**.**

The bound of $|\nabla\mathbf{b}|$ was only a uniform bound $C(n)$ in [CC-Ann], it was observed in [ChN] that $|\nabla\mathbf{b}|$ can have the improved bound $1+\epsilon$. The argument to get |\nabla\mathbf{b}|\leq 1+2^{51n^{2}}\big{(}\frac{r}{R}\big{)}^{\frac{1}{4(n-1)}}, comes from $(3.42)-(3.45)$ of [ChN], which was suggested to us by A. Naber.

Proof: Step (1). We define the function $\mathbf{b}$ by

[TABLE]

Define \psi(\rho)=\int_{\rho}^{4}\int_{t}^{4}\big{(}\frac{s}{t}\big{)}^{n-1}dsdt, choose $q\in B_{6r}(p)-\overline{B_{4r}(p)}$, and define the function $\tilde{h}(x):A_{2r,8r}(q)\rightarrow[0,\infty)$ by \tilde{h}(x)=\frac{2(n-1)r}{\frac{R}{2r}-7}\psi\Big{(}\frac{d(x,q)}{2r}\Big{)}. From Laplace Comparison Theorem, it is straightforward to get

[TABLE]

Note $B_{2r}(p)\subset A_{2r,8r}(q)$, hence $\min_{x\in B_{2r}(p)}\Delta(\mathbf{b}-b^{+}+\tilde{h})(x)\geq 0$.

From Maximum principle, for any $x\in B_{2r}(p)$, we get

[TABLE]

Let $b^{-}(x)=d(x,q^{-})-d(p,q^{-})$, then note $\mathbf{E}(x)=b^{+}(x)+b^{-}(x)+\mathbf{E}(p)$, then from $R\geq 2^{2n+1}r$ and Lemma 2.2, \sup_{B_{2r}(p)}\mathbf{E}(x)\leq 2^{8}\big{(}\frac{r}{R}\big{)}^{\frac{1}{n-1}}r, and we have

[TABLE]

From Laplace Comparison Theorem again,

[TABLE]

By Maximum Principle, note $\mathbf{E}(x)\geq 0$ and the assumption on the excess, we get

[TABLE]

From (2.10) and $\tilde{h}\geq 0$, we obtain

[TABLE]

By (2.9) and (2.11), it yields

[TABLE]

which implies

[TABLE]

From Theorem 1.1 and (2.13), note $\frac{R}{2r}\geq 1$, we get

[TABLE]

Note $d(p,q^{+})\geq R>4r$, then from Lemma 1.2, $\fint_{B_{2r}(p)}|\Delta b^{+}|\leq\frac{3n}{2r}$. Now do integration by parts, from (2.12), we get

[TABLE]

From (2.15) and the Bishop-Gromov Comparison Theorem, we have

[TABLE]

Step (2). Also (2.15) implies the following

[TABLE]

From (2.14), (2.17) and apply Lemma 2.4 for $\tau=\frac{14}{15}$ there, we have

[TABLE]

From the Bochner formula and $\Delta\mathbf{b}=0$, using $Rc\geq 0$, we have $\Delta|\nabla\mathbf{b}|^{2}\geq 2|\nabla^{2}\mathbf{b}|^{2}$, combining

[TABLE]

we get

[TABLE]

Let $\phi\in C^{\infty}(M^{n})$ be a nonnegative cut-off function such that

[TABLE]

and

[TABLE]

Now for any $y\in B_{r}(p)$, from (2.19), (2.20) and (2.21), we have

[TABLE]

From Theorem 2.5, for $x\in A_{\frac{3r}{2},\frac{7}{4}r}(p),y\in B_{r}(p)$, we have

[TABLE]

From (2.22), (2.23), (2.16) and (2.18), we get

[TABLE]

Take the integration of (2.24), from (2.17) and (2.23), we get

[TABLE]

∎

On Riemannian manifolds, if there is a segment $\gamma_{p,q}$ between two points $p,q$, we can choose the middle point of the segment $\gamma_{p,q}$, denoted as $z$. Then $[p,q,z]$ is an AG-triple with the excess [math] and the scale $\frac{1}{2}d(p,q)$.

For a metric space $\mathbf{X}_{k}$, generally we can not find the middle point as in Riemannian manifolds. However, if there exists a suitable Gromov-Hausdorff approximation from $\mathbb{R}^{k}\times\mathbf{X}_{k}$ to manifold $M^{n}$ locally, the following lemma provides the existence of almost middle point and AG-triple in metric space $\mathbf{X}_{k}$.

Lemma 2.8**.**

For $0\leq k\leq n,0<\delta<\frac{1}{3}$, $B_{10r}(0,\hat{q})\subset\mathbb{R}^{k}\times\mathbf{X}_{k}$ and $B_{10r}(q)\subset(M^{n},g)$, if there is an $(\delta r)$-Gromov-Hausdorff approximation

[TABLE]

then for $\hat{q}_{1}^{+},\hat{q}_{1}^{-}\in B_{r}(\hat{q})\subset\mathbf{X}_{k}$, there exists $\hat{q}_{0}\in B_{3r}(\hat{q})$ such that

[TABLE]

And if $\sqrt{\delta}r\leq\frac{d(\hat{q}_{1}^{+},\hat{q}_{1}^{-})}{100}$, then \big{[}p_{k+1}^{+},p_{k+1}^{-},q_{1}\big{]} is an AG-triple with the excess $\leq 16\sqrt{\delta}r$ and the scale $\geq\frac{1}{4}d(\hat{q}_{1}^{+},\hat{q}_{1}^{-})$, where

[TABLE]

Proof: From the assumption that $\Phi$ is an $(\delta r)$-Gromov-Hausdorff approximation,

[TABLE]

It is easy to see that $p_{k+1}^{+},p_{k+1}^{-}\in B_{r+\delta r}(q)$. In fact the segment $\gamma_{p_{k+1}^{+},p_{k+1}^{-}}\subset B_{3r}(q)$, otherwise, note $d(\hat{q}_{1}^{+},\hat{q}_{1}^{-})\leq 2r$, we will get

[TABLE]

which is contradicting (2.26).

Then we can choose the middle point of $\gamma_{p_{k+1}^{+},p_{k+1}^{-}}$, denoted as $z_{0}\in B_{3r}(q)$, then there exists $(s_{0},\hat{q}_{0})\in B_{10r}(0,\hat{q})$ such that

[TABLE]

Now using $\frac{1}{2}d(p_{k+1}^{+},p_{k+1}^{-})=d(p_{k+1}^{+},z_{0})$ and (2.27), we have

[TABLE]

Similarly, by using $\frac{1}{2}d(p_{k+1}^{+},p_{k+1}^{-})=d(p_{k+1}^{-},z_{0})$, we have

[TABLE]

From (2.26), (2.28) and (2.29),

[TABLE]

Take the sum of the above two inequalities, we have

[TABLE]

Simplify the above inequality, using $d(\hat{q}_{1}^{+},\hat{q}_{1}^{-})\leq 2r$, we get

[TABLE]

From (2.28), (2.29) and (2.30), we have

[TABLE]

From (2.26), (2.31) and (2.32), we get (2.25) and the following

[TABLE]

Similarly we have $d(q_{1},p_{k+1}^{-})\geq\frac{1}{4}d(\hat{q}_{1}^{+},\hat{q}_{1}^{-})$. Hence the scale is $\geq\frac{1}{4}d(\hat{q}_{1}^{+},\hat{q}_{1}^{-})$.

Finally from (2.25), we get

[TABLE]

which implies $\hat{q}_{0}\in B_{3r}(\hat{q})$.

∎

Proposition 2.9**.**

For $B_{10r}(q)\subset(M^{n},g)$ with $Rc(g)\geq 0$ and any $\epsilon\in(0,1)$, any $\frac{r_{1}}{r}\leq n^{-300n^{3}}\epsilon^{10n}$ and \delta\leq n^{-2000n^{3}}\epsilon^{70n}\big{(}\frac{r_{1}}{r}\big{)}^{4}. If there is an $(\delta r)$-Gromov-Hausdorff approximation for $0\leq k\leq n$ and $B_{10r}(0,\hat{q})\subset\mathbb{R}^{k}\times\mathbf{X}_{k}$,

[TABLE]

where \mathrm{diam}\big{(}B_{r}(\hat{q})\big{)}=r_{0}\geq\frac{1}{4}r. Then there are harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k+1} defined on some geodesic ball $B_{r_{1}}(q_{1})\subset B_{10r}(q)$, such that

[TABLE]

Proof: We firstly assume $\delta\leq(400)^{-2}$, then from assumption \mathrm{diam}\big{(}B_{r}(\hat{q})\big{)}=r_{0} and Lemma 2.8, there are $\hat{q}_{1}^{+},\hat{q}_{1}^{-}\in B_{r}(\hat{q})$ and $\hat{q}_{0}\in B_{3r}(\hat{q})$ such that

[TABLE]

Then we have

[TABLE]

Let \big{\{}\mathbf{e}_{i}\big{\}}_{i=1}^{k} be the standard basis for $\mathbb{R}^{k}$, put $q_{1}=\Phi(0,\hat{q}_{0})$ and

[TABLE]

From (2.33), (2.34), let $r_{1}=\zeta\cdot r$, if

[TABLE]

then we can apply Proposition 1.10 to get

[TABLE]

Define $p_{k+1}^{-}=\Phi(0,\hat{q}_{1}^{-}),p_{i}^{-}=\Phi(-\tilde{r}_{0}\cdot\mathbf{e}_{i},\hat{q}_{0})$, then from Lemma 2.8, \big{[}p_{k+1}^{+},p_{k+1}^{-},q_{1}\big{]} is an AG-triple with the excess $\leq 16\sqrt{\delta}r$ and the scale $\geq\frac{1}{4}r_{0}$. Because $\Phi$ is an $(\delta r)$-Gromov-Hausdorff approximation, it is easy to show that $[p_{i}^{+},p_{i}^{-},q_{1}]$ is AG-triple with the excess $\leq 16\sqrt{\delta}r$ and the scale $\geq\frac{1}{4}r_{0}$ for $i=1,\cdots,k$.

If we assume

[TABLE]

then $16\sqrt{\delta}r\leq\frac{4}{n}\frac{r_{1}^{2}}{\frac{1}{4}r_{0}}$. Also from $\frac{1}{4}r_{0}\geq\frac{1}{16}r$ and (2.35), we have $\frac{1}{4}r_{0}\geq 2^{2n+1}r_{1}$.

Now we can apply Lemma 2.6 to obtain harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k+1} satisfying

[TABLE]

From (2.39) and (2.40), we get

[TABLE]

From (2.37) and the above inequality, we have

[TABLE]

To get the conclusion, it is easy to check that $\epsilon_{1}=2^{-25n^{2}}\epsilon$ is enough for the need.

From (2.35), (2.36) and (2.38), we get the conclusion.

∎

Lemma 2.10**.**

Suppose $(M^{n},g)$ has $Rc\geq 0$, for $z\in M^{n}$, let $f:B_{R}(z)\rightarrow\mathbb{R}$ be a nonnegative function and f\in L^{1}\big{(}B_{R}(z)\big{)}, then there exists $p\in B_{\frac{R}{2}}(z)$ such that

[TABLE]

Proof: Let $c=15^{n}\fint_{B_{R}(z)}f$, define

[TABLE]

and $\mathcal{I}(f)(x)=\sup_{x\in B\subset B_{R}(z)}\fint_{B}f$, where the supremum is taken over all geodesic balls $B$ containing $x$ and $B\subset B_{R}(z)$.

Let T_{c}=\Big{\{}x\in B_{R}(z):\mathcal{I}(f)(x)>c\Big{\}}, then $\mathcal{S}_{c}\subset T_{c}$. Choose any $S\subset\subset T_{c}$, for any $x\in S$, there exists geodesic ball $B_{x}\subset B_{R}(z)$ such that $x\in B_{x}$ and

[TABLE]

By compactness of $S$, we can select a finite collection of such balls $\{B_{\alpha}\}_{\alpha\in\mathtt{F}}$ that cover $S$. Set R_{0}=\sup\big{\{}r_{\alpha}|\ \alpha\in\mathtt{F}\big{\}}, and \mathrm{F}_{j}=\big{\{}\alpha\in\mathtt{F}|\ \frac{R_{0}}{2^{j}}<r_{\alpha}\leq\frac{R_{0}}{2^{j-1}}\big{\}}, $j=1,2,\cdots$.

We define $\mathrm{G}_{j}\subset\mathrm{F}_{j}$ as follows:

1. (1)

Let $\mathrm{G}_{1}$ be any maximal collection of $\alpha$ in $\mathrm{F}_{1}$, such that $\{B_{r_{\alpha}}(x_{\alpha})\}_{\alpha\in\mathrm{G}_{1}}$ are disjoint. 2. (2)

Assume $\mathrm{G}_{1},\cdots,\mathrm{G}_{k-1}$ have been selected, choose $\mathrm{G}_{k}$ to be any maximal subcollection of

[TABLE]

which also satisfies $B_{r_{\alpha}}(x_{\alpha})\cap B_{r_{\beta}}(x_{\beta})=\emptyset$ for any $\alpha,\beta\in G_{k},\alpha\neq\beta$.

Now we define $\mathtt{G}=\bigcup_{k=1}^{\infty}\mathrm{G}_{k}\subset\mathtt{F}$. For any $\alpha\in\mathtt{F}$, there is $j\in\mathbb{N}$, such that $\alpha\in\mathrm{F}_{j}$. If $\alpha\in\mathrm{G}_{j}$, we get $B_{r_{\alpha}}(x_{\alpha})\subset\bigcup_{\alpha^{\prime}\in\mathtt{G}}B_{5r_{\alpha^{\prime}}}(x_{\alpha^{\prime}})$.

Otherwise $\alpha\notin\mathrm{G}_{j}$, by the definition of $\mathrm{G}_{j}$, there is $\alpha^{\prime}\in\bigcup_{i=1}^{j}\mathrm{G}_{i}$ such that $B_{r_{\alpha}}(x_{\alpha})\cap B_{r_{\alpha^{\prime}}}(x_{\alpha^{\prime}})\neq\emptyset$. Note $r_{\alpha}\leq\frac{R_{0}}{2^{j-1}}<2r_{\alpha^{\prime}}$, hence $B_{r_{\alpha}}(x_{\alpha})\subset B_{r_{\alpha^{\prime}}+2r_{\alpha}}(x_{\alpha^{\prime}})\subset\bigcup_{\beta\in\mathtt{G}}B_{5r_{\beta}}(x_{\beta})$.

Then we find $\mathtt{G}\subset\mathtt{F}$, such that for any $\alpha^{\prime}\neq\beta^{\prime}\in\mathtt{G}$, $B_{r_{\alpha^{\prime}}}(x_{\alpha^{\prime}})\cap B_{r_{\beta^{\prime}}}(x_{\beta^{\prime}})=\emptyset$ and

[TABLE]

From Bishop-Gromov Comparison Theorem and (2.42), we get

[TABLE]

If one takes the supremum over all such $S\subset\subset T_{c}$, we have

[TABLE]

If (2.41) does not hold for any point in $B_{\frac{R}{2}}(z)$, then we have $B_{\frac{R}{2}}(z)\subset\mathcal{S}_{c}$, from Bishop-Gromov Volume Comparison Theorem, \frac{\mu(\mathcal{S}_{c})}{\mu\big{(}B_{R}(z)\big{)}}\geq\frac{\mu\big{(}B_{\frac{R}{2}}(z)\big{)}}{\mu\big{(}B_{R}(z)\big{)}}\geq 2^{-n}, it is the contradiction, the conclusion follows.

∎

The following lemma is well known so we omit its proof here.

Lemma 2.11**.**

Let $(\mathbf{X},d_{\mathbf{X}},x_{0})$ and $(\mathbf{Y},d_{\mathbf{Y}},y_{0})$ be two pointed metric spaces, if there is a pointed $\epsilon$-Gromov-Hausdorff approximation f:\big{(}\mathbf{X},x_{0}\big{)}\rightarrow(\mathbf{Y},y_{0}), then there exists a pointed $(3\epsilon)$-Gromov-Hausdorff approximation h:\big{(}\mathbf{Y},y_{0}\big{)}\rightarrow(\mathbf{X},x_{0}).

∎

Theorem 2.12**.**

For $B_{10r}(q)\subset(M^{n},g)$ with $Rc(g)\geq 0$, any $0<\epsilon<1,\delta=n^{-3400n^{3}}\epsilon^{110n}$ and integer $0\leq k\leq n$, assume there is an $(\delta r)$-Gromov-Hausdorff approximation ,

[TABLE]

and \mathrm{diam}\big{(}B_{r}(\hat{q})\big{)}=r_{0}\geq\frac{1}{4}r. Then there are harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k+1} defined on some geodesic ball $B_{s}(p)\subset B_{10r}(q)$, where $s=n^{-320n^{3}}\epsilon^{10n}r$, such that

[TABLE]

Proof: From Lemma 2.11, there is an $(3\delta r)$-Gromov-Hausdorff approximation,

[TABLE]

Let $r_{1}=n^{-300n^{3}}\epsilon_{1}^{10n}r$, assume $\delta\leq\frac{1}{3}n^{-3200n^{3}}\epsilon_{1}^{110n}$, where $\epsilon_{1}>0$ is to be determined later. Apply Proposition 2.9, we obtain harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k+1} defined on $B_{r_{1}}(q_{1})\subset B_{10r}(q)$, such that

[TABLE]

Apply Lemma 2.10, we get $B_{s}(p)\subset B_{r_{1}}(q_{1})$, where $s=\frac{r_{1}}{2}=\frac{1}{2}n^{-300n^{3}}\epsilon_{1}^{10n}r$, such that

[TABLE]

Choose $\epsilon_{1}=15^{-n}\epsilon$, let $\delta=n^{-3400n^{3}}\epsilon^{110n}$ and $s=n^{-320n^{3}}\epsilon^{10n}r$, the conclusion follows.

∎

Part II. A.O.L. harmonic functions produce G-H approximation

In Part II of this paper, we will prove the following quantitative version of almost splitting theorem, which was implied in the argument of Cheeger and Colding in a series of papers, [Colding-shape], [Colding-large], [Colding-volume], [CC-Ann] and [Cheeger-GAFA].

For our application, we need the quantitative estimate, which relate the Gromov-Hausdorff distance to the average integral bound of almost orthonormal linear harmonic functions. Although we believe that many results in Part II are well-known to some experts in this field, but we can not find the reference providing those quantitative estimates exactly. So we elaborate the concise argument of Cheeger-Colding to present the proof in all the details for self-contained reason.

Theorem 2.13**.**

For $\epsilon>0$ and $1\leq k\leq n$, there is $\delta=n^{-700n^{4}}\epsilon^{18n^{4}}$ such that for complete Riemannian manifold $(M^{n},g)$ with $Rc\geq 0$, if there exist harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k}, defined on $B_{r}(p)$, satisfying $\mathbf{b}_{i}(p)=0$ and

[TABLE]

then we can find a metric space $\mathbf{X}_{k}$ and an $(\epsilon s)$-Gromov-Hausdorff approximation $f_{k}=(\mathbf{b}_{1},\cdots,\mathbf{b}_{k},\mathcal{P}_{k}):B_{s}(p)\rightarrow B_{s}(0,\hat{p})\subset\mathbb{R}^{k}\times\mathbf{X}_{k}$, where $s=\frac{1}{1280}r$.

Remark 2.14**.**

Colding and Minicozzi [CM] gave a characterization of Gromov-Hausdorff distance through the integral estimate of Hessian of harmonic functions among other things.

3. Existence of almost linear function and Hessian estimate

When there is a harmonic function $\mathbf{b}$ defined locally on manifold $M^{n}$, with bounded gradient and the average integral of \big{|}|\nabla\mathbf{b}|-1\big{|} is small enough, we will show the existence of an almost linear function, which is a generalization of linear function in $\mathbb{R}^{n}$. The proof of Proposition 3.4 has close relationship with the argument in [Cheeger-GAFA].

Definition 3.1**.**

For $\mathbf{X}\subset M^{n},t\in\mathbb{R}$, the function $\rho(x)=d(x,\mathbf{X})+t:M^{n}\rightarrow\mathbb{R}$ is called almost linear function, which is the generalization of function $f(x)=d(x,\mathbb{R}^{n-1})+t$ defined on $\mathbb{R}^{n}$.

Definition 3.2**.**

For Lipschitz function $f$ defined on metric space $\mathbf{M}$, the pointwise Lipschitz constant function $\mathscr{L}(f)$ is defined by

[TABLE]

and the Lipschtiz constant $\mathbf{L}(f)$ is defined by \mathbf{L}(f)=\sup_{z\in\mathbf{M}}\big{\{}\mathscr{L}(f)(z)\big{\}}.

Remark 3.3**.**

From the classical Rademacher theorem, the Lipschitz function is almost differentiable on manifolds, for the general argument on metric measure spaces see [Cheeger-GAFA]. Hence when the pointwise Lipschitz constant function $\mathscr{L}(f)(z)$ appears as the integrand function in an integral, we can replace it by $|\nabla f|(z)$, and we will use this fact freely in the following argument.

Proposition 3.4**.**

For any $\epsilon>0$, there is $\delta=2^{-100n^{2}}\epsilon^{2n^{2}}$ such that for any complete Riemannian manifold $(M^{n},g)$ with $Rc\geq 0$, if there exists one harmonic function $\mathbf{b}$ defined on $B_{r}(p)$ satisfying $\sup_{B_{r}(p)}|\nabla\mathbf{b}|\leq 2$ and \fint_{B_{r}(p)}\big{|}|\nabla\mathbf{b}|-1\big{|}\leq\delta. Then one can find $t_{0},t_{1}\in\mathbb{R}$ and two functions $\rho,\tilde{\rho}$ defined on $B_{\frac{r}{10}}(p)$, such that

[TABLE]

Proof: Step (1). For $1>\theta>0$ (to be determined later), we define

[TABLE]

then from assumption, we get

[TABLE]

We define $\mathbf{b}_{*}$ as the following:

[TABLE]

It is easy to get \sup_{x\in B_{r}(p)}\big{|}\mathscr{L}(\mathbf{b}_{*})(x)\big{|}\leq 1+\frac{\delta}{\theta}.

Put $h(x)=\frac{\theta}{\delta+\theta}\mathbf{b}_{*}(x)+\frac{\delta}{\delta+\theta}\mathbf{b}_{*}(p)$, then

[TABLE]

Note $h(p)=\mathbf{b}_{*}(p)$ and \sup\limits_{B_{r}(p)}\big{|}\mathscr{L}(h-\mathbf{b}_{*})\big{|}\leq\frac{\delta}{\theta}, we get $\sup_{B_{r}(p)}|h-\mathbf{b}_{*}|\leq\frac{\delta}{\theta}r$.

For any $x\in\overline{B_{\frac{r}{5}}(p)}$, from (3.2), there exists $x^{\prime}\in B_{\frac{7}{5}\theta^{\frac{1}{n}}r}(x)$ such that $x^{\prime}\in\mathbf{A}$. Note $\mathbf{b}(x^{\prime})-\mathbf{b}_{*}(x^{\prime})=0$ and

[TABLE]

Then for any $x\in\overline{B_{\frac{r}{5}}(p)}$, we have

[TABLE]

From above we have

[TABLE]

From (3.2), note $h=\frac{\theta}{\delta+\theta}\mathbf{b}+\frac{\delta}{\delta+\theta}\mathbf{b}_{*}(p)$ on $\mathbf{A}$, we have

[TABLE]

Now from (3.3) and (3.4), we can choose $\theta=2^{-12n^{2}}\epsilon_{1}^{n}$, where $\epsilon_{1}>0$ is to be determined later, and

[TABLE]

Then we have

[TABLE]

Step (2). Now for $x\in\overline{B_{\frac{1}{5}r}(p)}$, we consider

[TABLE]

For $x\in B_{\frac{r}{10}}(p)$, assume

[TABLE]

where $x^{*}\in\partial B_{\frac{r}{5}}(p)$. Let $\gamma_{x,x^{*}}(s)$ be the minimizing geodesic from $x$ to $x^{*}$, parametrized by arc-length. From $\mathbf{L}(\rho)\leq 1$, we get

[TABLE]

which implies

[TABLE]

But $d(x,x^{*})\geq\frac{r}{10}$, we have

[TABLE]

Let $t_{0}=\rho(p)-\frac{r}{160}$, note $0\leq\rho(x)-t_{0}\leq\frac{r}{10}$ for $x\in B_{\frac{r}{160}}(p)$, then from (3.7),

[TABLE]

Now we have

[TABLE]

By the above, for $x\in B_{\frac{r}{160}}(p)$, we get

[TABLE]

Similarly, let $t_{1}=\tilde{\rho}(p)+\frac{r}{160}$, for $x\in B_{\frac{r}{160}}(p)$, we get

[TABLE]

For any $x\in B_{\frac{r}{320}}(p)$, using $\mathbf{L}(\rho)\leq 1$, we have

[TABLE]

Step (3). Let $\epsilon_{2}=2^{11}\epsilon_{1}^{\frac{1}{n}}$, we will prove that

[TABLE]

By contradiction, if there exists $w\in B_{\frac{1}{10}r}(p)$ such that

[TABLE]

where $\theta_{1}=2^{10}\epsilon_{1}^{\frac{1}{n}}\leq\frac{1}{20}$. Let K=\frac{1}{2}\big{[}\tilde{\rho}(w)+\rho(w)\big{]}, since \big{|}\mathbf{L}(\rho)\big{|}\leq 1,\big{|}\mathbf{L}(\tilde{\rho})\big{|}\leq 1 and $\tilde{\rho}\leq\rho$, also note $B_{\theta r}(w)\subset B_{\frac{1}{5}r}(p)$, we have

[TABLE]

Now we define

[TABLE]

then \big{|}\mathbf{L}(\check{h})\big{|}\leq 1, also note $\check{h}=\tilde{\rho}=\rho=h$ on $\partial B_{\frac{1}{5}r}(p)$, from (3.6), we get

[TABLE]

We define the Lipschitz function $\phi(x):\overline{B_{\frac{1}{5}r}(p)}\rightarrow\mathbb{R}$ satisfying

[TABLE]

also

[TABLE]

Now define $\tilde{\mathbf{b}}=\check{h}+(1-\phi)(\mathbf{b}-\check{h})$, we have

[TABLE]

Let $\chi_{E}$ denote the characteristic function of a set $E$, then from above and the Bishop-Gromov Comparison Theorem we get

[TABLE]

From the fact that $\tilde{\mathbf{b}}|_{\partial B_{\frac{r}{5}}(p)}=\mathbf{b}|_{\partial B_{\frac{r}{5}}(p)}$ and $\mathbf{b}$ is harmonic, note the harmonic function has the smallest energy, we have

[TABLE]

which implies

[TABLE]

In the last inequality above, we used (3.5).

On the other hand, note $\mathscr{L}(\check{h})=0$ on $B_{\theta_{1}r}(w)$, we have

[TABLE]

which implies $\theta_{1}<2^{10}\epsilon_{1}^{\frac{1}{n}}$, it is the contradiction with the choice of $\theta_{1}$.

Step (4). Note $\tilde{\rho}(x)\leq h(x)\leq\rho(x)$ and (3.9), then we get

[TABLE]

From (3.6) and (3.10) we have

[TABLE]

From (3.5) and (3.6),

[TABLE]

then

[TABLE]

If we choose $\epsilon_{1}\leq 2^{-23n}\epsilon^{n}$ and also $2^{10}\epsilon_{1}^{\frac{1}{n}}\leq\frac{1}{20}$, then we obtain the conclusion. From (3.5), we only need to choose $\delta\leq 2^{-100n^{2}}\epsilon^{2n^{2}}$.

∎

4. Segment inequality and measure of ‘good’ points

In the proofs of this section, when the context is clear, for simplicity, we use $B_{r}$ instead of $B_{r}(p)$, similar for $B_{4r}$ etc.

Lemma 4.1** (Segment Inequality).**

Assume $(M^{n},g)$ is a complete Riemannian manifold with $Rc\geq 0$, then for any nonnegative function $f$ defined on $B_{2r}(p)\subset M^{n}$,

[TABLE]

Proof: In the proof, we assume $y_{1},y_{2}\in B_{r}$, and set

[TABLE]

Then $E=E_{1}+E_{2}$. Along any geodesic $\gamma$ starting from $y_{1}$, write the volume element of $M^{n}$ in geodesic polar coordinate as $ds\wedge\mathcal{A}(s)$. Then $\mathcal{A}(s)=J(\theta,s)d\theta$, from Bishop-Gromov Comparison Theorem,

[TABLE]

For $y\in B_{r},v\in S_{y}M^{n}$, we define

[TABLE]

Then we have

[TABLE]

where \big{|}I(y,v)\big{|} denotes the measure of $I(y,v)$.

Assume $\gamma_{y_{1}}^{v_{1}}$ is the geodesic starting from $y_{1}$ with $(\gamma_{y_{1}}^{v_{1}})^{\prime}(0)=v_{1}$, then for any $y_{1}\in B_{r}$, $s\in I(y_{1},v_{1})$,

[TABLE]

where $\mathcal{T}(y_{1},v_{1})=\max\limits_{t\in I(y_{1},v_{1})}t$.

Thus from (4.1), for any $y_{1}\in B_{r},v_{1}\in S_{y_{1}}M^{n}$,

[TABLE]

Integrating (4.2) with respect to $v_{1}$ over the unit tangent space $S_{y_{1}}M^{n}$, and note \Big{(}\bigcup\limits_{y_{1}\in B_{r}(p)\atop y_{2}\in B_{r}(p)}\gamma_{y_{1},y_{2}}\Big{)}\subset B_{2r}, where $\gamma_{y_{1},y_{2}}$ is the minimal geodesic connecting $y_{1}$ with $y_{2}$ in $M^{n}$, we have

[TABLE]

And we integrate (4.3) with respect to $y_{1}$ over $B_{r}$,

[TABLE]

Similarly, we get

[TABLE]

Take the sum of (4.4) and (4.5), the conclusion follows.

∎

For $x\in B_{2r}(p)\subset M^{n}$ and a closed subset $\mathbf{X}\subseteq M^{n}$, we define

[TABLE]

where $t_{0}\in\mathbb{R}$ is some constant, we define $\mathfrak{P}(x)\in\mathbf{X}$ by d\big{(}x,\mathbf{X}\big{)}=d\big{(}x,\mathfrak{P}(x)\big{)} (if there are two points $y_{1},y_{2}$ satisfying d\big{(}x,\mathbf{X}\big{)}=d(x,y_{1})=d(x,y_{2}), then define $\mathfrak{P}(x)=y_{1}$ or $y_{2}$ freely). We assume

[TABLE]

For $0<\eta<\frac{1}{2}$, we have \big{|}\frac{\hat{\rho}(y)-\hat{\rho}(x)}{\hat{\rho}(x)-\eta r}\big{|}\leq 4. And we also define

[TABLE]

For $x,y\in B_{r}(p)$, define

[TABLE]

Definition 4.2**.**

For $0<\eta<\frac{1}{2}$, we define

[TABLE]

For non-negative function $f$ defined on $B_{r}(p)$, we define

[TABLE]

Lemma 4.3**.**

Assume (4.6), then for any non-negative function $f$ satisfying $\fint_{B_{4r}(p)}f\leq\delta$, we have \frac{V(\mathfrak{Q}_{\eta,f}^{r,\rho})}{V\big{(}B_{r}(p)\big{)}}\geq 1-3^{n}\eta^{-n}\delta.

Proof: Firstly we have

[TABLE]

Assume $\theta_{s}(x)$ is the gradient flow of $\rho(\cdot)$ starting from $\mathfrak{P}(x)$ at time $s+\eta r$, using Co-Area formula and Bishop-Gromov volume comparison theorem,

[TABLE]

From the above, we obtain \frac{V(\mathfrak{Q}_{\eta,f}^{r,\rho})}{V\big{(}B_{r}\big{)}}\geq 1-3^{n}\eta^{-n}\delta.

∎

Lemma 4.4**.**

Assume (4.6), $\mathbf{b}$ is harmonic function on $B_{16r}(p)$ satisfying $\sup_{B_{16r}(p)}|\nabla\mathbf{b}|\leq 2$ and $\fint_{B_{16r}(p)}|\nabla\mathbf{b}-\nabla\rho|\leq\delta$. Then we have

[TABLE]

Proof: From the assumption and Bishop-Gromov Comparison Theorem,

[TABLE]

From the above inequality, apply Lemma 4.3 to $|\nabla\mathbf{b}-\nabla\rho|$, we get the first inequality of the conclusion. To prove the $3$rd inequality of the conclusion, we note

[TABLE]

On the other hand, note $x\in T_{\eta,\mathbf{b}}^{r,\rho}$, we have

[TABLE]

Hence we obtain \frac{V\big{(}T_{\eta,\mathbf{b}}^{r,\rho}(x)\big{)}}{V\big{(}B_{r}\big{)}}\geq 1-\sqrt{\eta}. Finally we prove the $2$nd inequality. From assumption and Lemma 2.4, we get

[TABLE]

Note we have

[TABLE]

On the other hand, from Lemma 4.1, we have

[TABLE]

We used (4.7) in the last inequality above.

From above, we get

[TABLE]

∎

For $x,y\in B_{r}(p)$, for each $\mathbf{b}_{i}$ from Theorem 2.13 and the corresponding $\rho_{i},\mathfrak{P}_{i}$, we define

[TABLE]

Definition 4.5**.**

For $0<\eta<\frac{1}{2}c_{1}$, we define

[TABLE]

Lemma 4.6**.**

Assume $0<r\leq\hat{\rho}_{j}(x)\leq 3r$ for any $x\in B_{r}(p)$, if

[TABLE]

then we have

[TABLE]

Proof: Apply Lemma 4.3 to \big{|}\nabla(\mathbf{b}_{i}-\rho_{i})\big{|} and \Big{|}\big{\langle}\nabla\mathbf{b}_{k},\nabla\mathbf{b}_{l}\big{\rangle}\Big{|} respectively, we get our conclusion.

∎

5. Quantitative almost splitting theorem

The main results of this section were sort of implied in Cheeger-Colding’s work (see [CC-Ann] and [CC1]), however we will not follow their argument there. Instead, we adapt the argument of Colding-Naber in [CN] to prove the main result of this section. Although our argument has close relationship with [CC-Ann] and [CC1], the main difference is that the angle between two segments is not involved into our argument, and the first variation formula is applied to the case of both end points are moving.

Lemma 5.1**.**

For any $0<\eta<\frac{1}{2}$, assume (4.6), $\mathbf{b}$ is harmonic function on $B_{16r}(p)$ satisfying $\sup_{B_{r}(p)}|\nabla\mathbf{b}|\leq 2$ and $\sup_{B_{r}(p)}|\mathbf{b}-\rho|\leq\eta r$. Then for any $x\in T_{\eta,\mathbf{b}}^{r,\rho}\cap Q_{\eta,\mathbf{b}}^{r,\rho}$, $y\in T_{\eta,\mathbf{b}}^{r,\rho}(x)\cap Q_{\eta,\mathbf{b}}^{r,\rho}$, we have \Big{|}d(x,y)-d\big{(}\mathfrak{G}_{\rho}(x),\mathfrak{G}_{\rho}(y)\big{)}\Big{|}\leq 5000\eta^{\frac{1}{8}}\cdot r.

Proof: Using the first variation formula for arc length, and note

[TABLE]

[TABLE]

Step (1). If

[TABLE]

From (5.1) and $x,y\in Q_{\eta,\mathbf{b}}^{r,\rho}$, we have

[TABLE]

Hence

[TABLE]

Note d(x,y)\geq\big{|}\hat{\rho}(y)-\hat{\rho}(x)\big{|}, by d\big{(}\mathfrak{G}_{\rho}(x),\mathfrak{G}_{\rho}(y)\big{)}=\sqrt{d\big{(}\mathfrak{P}(x),\mathfrak{P}(y)\big{)}^{2}+\big{|}\hat{\rho}(y)-\hat{\rho}(x)\big{|}^{2}}, (5.2) and (5.3), we have

[TABLE]

Step (2). In the rest of the proof, we assume that

[TABLE]

From (5.1) and $x,y\in Q_{\eta,\mathbf{b}}^{r,\rho}$, for $t\in[0,\hat{\rho}(x)-\eta r]$, we get

[TABLE]

Now we estimate \langle\nabla\mathbf{b},\tau_{s}^{\prime}\rangle\big{(}\tau_{s}(l_{s})\big{)}. For any $0\leq t_{1}\leq l_{s}$, we have

[TABLE]

Take the integral of the above inequality from [math] to $l_{s}$ with respect to $t_{1}$, we get

[TABLE]

In the second inequality above we used the assumption $\sup_{B_{r}(p)}|\mathbf{b}-\rho|\leq\eta r$. Then

[TABLE]

Similarly, we can also have

[TABLE]

Hence we have

[TABLE]

Step (3). We will show the uniform lower bound of $l_{t}$ when t\in\big{[}0,\hat{\rho}(x)-\eta r\big{]}. There are two cases to be discussed.

1. (3.A)

If \big{|}\hat{\rho}(y)-\hat{\rho}(x)\big{|}\leq\frac{1}{4}l_{0}. From (5.5) and $|\nabla\mathbf{b}|\leq 2$,

[TABLE]

From (5.4), we get

[TABLE] 2. (3.B)

If \big{|}\hat{\rho}(y)-\hat{\rho}(x)\big{|}>\frac{1}{4}l_{0}. Let \Big{(}\frac{\hat{\rho}(y)-\hat{\rho}(x)}{\hat{\rho}(x)-\eta r}\Big{)}^{2}=\alpha_{1}, from (5.6), (5.7) and (5.5), we can get

[TABLE]

Note for any $0\leq s\leq\frac{2\eta r}{\sqrt{\alpha_{1}}}$, using $\sqrt{\alpha_{1}}\geq 2^{-2}\cdot\frac{l_{0}}{\hat{\rho}(x)-\eta r}\geq\frac{l_{0}}{12r}$ and (5.4), we have

[TABLE]

From (5.9) and (5.10), we get

[TABLE]

Then by (5.4) again,

[TABLE]

From above two cases, we always have

[TABLE]

Step (4). From (5.8), (5.11) and $\eta\geq\delta$, we obtain

[TABLE]

From (5.5) and (5.12), we get

[TABLE]

Define $\mathfrak{L}(t)=\int_{0}^{t}\frac{s}{l_{s}}ds$, then $l_{t}=\frac{t}{\mathfrak{L}^{\prime}(t)}$ and we have

[TABLE]

On the other hand, let $f(s)=\sqrt{\alpha_{1}s^{2}+l_{0}^{2}}$, and define

[TABLE]

Then it is easy to get

[TABLE]

From (5.13) and (5.14), we obtain

[TABLE]

Take the integral of the above inequality, also note $\mathfrak{L}(0)=F(0)=0$ and $0\leq t\leq\hat{\rho}(x)-\eta r\leq 3r$, we have

[TABLE]

Simplify the above inequality, note $\mathfrak{L}(t)\geq 0$ and $F(t)\geq 0$, using $l_{0}\geq\eta^{\frac{1}{8}}r$,

[TABLE]

From (5.13), (5.14) and (5.15), we have

[TABLE]

Let $t=\hat{\rho}(x)-\eta r$ in (5.16), note \Big{|}l_{0}-d\big{(}\mathfrak{P}(x),\mathfrak{P}(y)\big{)}\Big{|}\leq 2\eta r and $l_{\hat{\rho}(x)-\eta r}=d(x,y)$,

[TABLE]

∎

Corollary 5.2**.**

For any $\eta\in(0,1)$, assume (4.6) and

[TABLE]

then there exists $\delta_{1}=n^{21}\eta^{\frac{1}{2n}}$ such that \sup\limits_{x,y\in T_{\eta,\mathbf{b}}^{r,\rho}\cap Q_{\eta,\mathbf{b}}^{r,\rho}\cap B_{(1-\delta_{1})r}(p)}\Big{|}d(x,y)-d\big{(}\mathfrak{G}_{\rho}(x),\mathfrak{G}_{\rho}(y)\big{)}\Big{|}\leq n^{22}\eta^{\frac{1}{3n}}r.

Proof: From Lemma 4.4, for any $x,y\in T_{\eta,\mathbf{b}}^{r,\rho}\cap Q_{\eta,\mathbf{b}}^{r,\rho}\cap B_{(1-\delta_{1})r}(p)$, we have

[TABLE]

We claim that $T_{\eta,\mathbf{b}}^{r,\rho}\cap Q_{\eta,\mathbf{b}}^{r,\rho}\cap T_{\eta,\mathbf{b}}^{r,\rho}(x)\cap T_{\eta,\mathbf{b}}^{r,\rho}(y)\cap B_{\delta_{1}r}(x)\neq\emptyset$, otherwise

[TABLE]

which implies

[TABLE]

On the other hand, from Bishop-Gromov Comparison Theorem and $B_{r}(p)\subset B_{2r}(x)$, using the definition of $\delta_{1}$, we have

[TABLE]

which is the contradiction. Hence we can find

[TABLE]

Now apply Lemma 5.1 to $y,z$ and $x,z$ respectively, then we have

[TABLE]

∎

Proof: [of Theorem 2.13] Step (1). We firstly deal with the case $k=1$. From Proposition 3.4, there exists $\delta\leq 2^{-100n^{2}}\epsilon_{1}^{2n^{2}}$, where $\epsilon_{1}>0$ is a constant to be determined later, such that if (2.43) holds for $\delta$, we can find two functions $\rho_{1},\tilde{\rho}_{1}$ satisfying the following

[TABLE]

From (5.19), (2.43) and (5.20), for $\eta\in(0,1)$ to be determined later, if we assume

[TABLE]

apply Corollary 5.2, there exist $\delta_{1}=n^{21}\eta^{\frac{1}{2n}}$ such that

[TABLE]

Now choose a maximal collection of disjoint balls with radius $\frac{\delta_{1}r}{320}$ centered at $B_{\frac{(1-\delta_{1})r}{320}}(p)$, denoted as \Big{\{}B_{\frac{\delta_{1}r}{320}}(x_{i})\Big{\}}_{i=1}^{m}, then

[TABLE]

From Lemma 4.4, (5.19), (5.22) and (5.20), we have

[TABLE]

On the other hand, from Bishop-Gromov Comparison Theorem and (5.24),

[TABLE]

From (5.25) and (5.26), we get V\big{(}B_{\frac{\delta_{1}r}{320}}(x_{i})\big{)}>V\big{(}B_{\frac{r}{320}}(p)\backslash(T_{\eta,\mathbf{b}_{1}}^{\frac{r}{320},\rho_{1}}\cap Q_{\eta,\mathbf{b}_{1}}^{\frac{r}{320},\rho_{1}})\big{)}. So there exists $y_{i}$ such that $y_{i}\in B_{\frac{\delta_{1}r}{320}}(x_{i})\cap T_{\eta,\mathbf{b}_{1}}^{\frac{r}{320},\rho_{1}}\cap Q_{\eta,\mathbf{b}_{1}}^{\frac{r}{320},\rho_{1}}$, combining (5.24) yields

[TABLE]

Now we define $\mathbf{Y}=\{y_{1},\cdots,y_{m}\}\cap B_{\frac{r}{1280}-n^{22}\eta^{\frac{1}{3n}}r}(p)$ and $\mathbf{X}_{1}=\bigcup\limits_{y_{i}\in\mathbf{Y}}\mathfrak{P}_{1}(y_{i})\subset\mathbf{X}_{\mathbf{b}_{1}}$, which is a metric space with distance function $d$ inherited from the metric $g$ of the Riemannian manifold $M^{n}$.

Step (2). For any $x\in B_{\frac{r}{1280}}(p)$, we define

[TABLE]

where \hat{p}=\mathfrak{P}_{1}\big{(}\mathfrak{n}_{1}(p)\big{)} and

[TABLE]

If $\mathcal{D}_{x}$ contains more than one elements, we choose $\mathfrak{n}_{1}(x)$ from it freely.

If $x=y_{i}$ for some $i$, then $\mathfrak{n}_{1}(x)=x$. If $x\neq y_{i}$ for any $i$, then there is $\tilde{x}$ such that

[TABLE]

From (5.27), there exists $y_{i}$ such that $\tilde{x}\in B_{\frac{\delta_{1}r}{80}}(y_{i})$, then $d(y_{i},x)<\frac{\delta_{1}r}{40}$ and

[TABLE]

which implies $y_{i}\in\mathbf{Y}$. From the definition of $\mathfrak{n}_{1}(x)$,

[TABLE]

From the definition of $\mathfrak{g}_{1}$ and (5.23), we have

[TABLE]

which implies that \mathfrak{g}_{1}\big{(}B_{\frac{r}{1280}}(p)\big{)}\subset B_{\frac{r}{1280}}(0,\hat{p}).

Now for any $x_{1},x_{2}\in B_{\frac{r}{1280}}(p)$, from (5.23) and (5.29), we have

[TABLE]

Now to show that $\mathfrak{g}_{1}:B_{\frac{1}{1280}r}(p)\rightarrow B_{\frac{1}{1280}r}(0,\hat{p})$ is an pointed $(\frac{\epsilon r}{3})$-Gromov-Hausdorff approximation, we only need to show that

[TABLE]

Step (3). For any $(t,\hat{x})\in B_{\frac{r}{1280}}(0,\hat{p})\subset\mathbb{R}\times\mathbf{X}_{1}$, there is $y_{i}\in\mathbf{Y}$ such that $\hat{x}=\mathfrak{P}_{1}(y_{i})$.

If t+\hat{\rho}_{1}\big{(}\mathfrak{n}_{1}(p)\big{)}>\hat{\rho}_{1}(y_{i}), then

[TABLE]

For $t_{1}=\tilde{\rho}_{1}(p)+\frac{r}{20}$, note

[TABLE]

There is $\tilde{y}\in\tilde{\rho}_{1}^{-1}(t_{1})$ such that d\big{(}y_{i},\tilde{\rho}_{1}^{-1}(t_{1})\big{)}=d(y_{i},\tilde{y}). Define

[TABLE]

then

[TABLE]

And from (5.18), we have

[TABLE]

Using (5.34) and (5.21), we get

[TABLE]

Note

[TABLE]

From (5.27), there is $y_{i_{1}}$ such that $d(z,y_{i_{1}})<\frac{\delta_{1}}{80}r$, using (5.35) and (5.33),

[TABLE]

Using (5.29), (5.23), (5.35) and (5.36), we have

[TABLE]

Hence we can find $y$ such that

[TABLE]

From (5.27), there is $y_{i_{2}}$ such that $y\in B_{\frac{\delta_{1}r}{80}}(y_{i_{2}})$, then $y_{i_{2}}\in B_{\frac{r}{1280}}(p)$ and

[TABLE]

Now using (5.35), similar as (5.36), we get

[TABLE]

If t+\hat{\rho}_{1}\big{(}\mathfrak{n}_{1}(p)\big{)}\leq\hat{\rho}_{1}(y_{i}), we get \mathfrak{G}_{\rho_{1}}\Big{(}\gamma_{\mathfrak{P}_{1}(y_{i}),y_{i}}\big{(}t+\hat{\rho}_{1}\circ\mathfrak{n}_{1}(p)\big{)}\Big{)}=(t+\hat{\rho}_{1}\big{(}\mathfrak{n}_{1}(p)\big{)},\hat{x}), let z_{0}=\gamma_{\mathfrak{P}_{1}(y_{i}),y_{i}}(t+\hat{\rho}_{1}\big{(}\mathfrak{n}_{1}(p)\big{)}), then

[TABLE]

Now using (5.29) and (5.39), we have

[TABLE]

Then using (5.40), similar as (5.37) we get $d(z_{0},p)<\frac{r}{1280}+n^{23}\eta^{\frac{1}{3n}}r$. The rest argument is similar to get (5.38), we can find $y_{j_{2}}\in B_{\frac{r}{1280}}(p)$ such that

[TABLE]

From above, choose $\eta=n^{-100n}\epsilon^{3n}$, $\epsilon_{1}=n^{-301n^{2}}\epsilon^{9n^{2}}<2^{-n}\eta^{3n}$, and finally $\delta=n^{-700n^{4}}\epsilon^{18n^{4}}<2^{-100n^{2}}\epsilon_{1}^{2n^{2}}$, we get the pointed $(\frac{\epsilon r}{3})$-Gromov-Hausdorff approximation $\mathfrak{g}_{1}:B_{\frac{r}{1280}}(p)\rightarrow B_{\frac{r}{1280}}(0,\hat{p})$.

Step (4). Note $\mathbf{b}_{1}(p)=0$, from (5.20), we have \big{|}\rho_{1}(p)\big{|}<\epsilon_{1}r, then

[TABLE]

Define $\mathcal{P}_{1}=\mathfrak{P}_{1}\circ\mathfrak{n}_{1}$, for $f_{1}=(\mathbf{b}_{1},\mathcal{P}_{1})$, then from (5.42) and (5.20),

[TABLE]

Now from (5.30) and (5.43),

[TABLE]

From (5.38), (5.41) and (5.43), we have

[TABLE]

From (5.44) and (5.45), we obtain that $f_{1}$ is an $(\frac{\epsilon}{3}r)$-Gromov-Hausdorff approximation.

Step (5). We will only prove the case $k=2$, the rest cases have similar argument. From the case $k=1$, for $\mathbf{b}_{i},i=1,2$, there is corresponding pointed \big{(}\frac{\epsilon r}{3}\big{)}-Gromov-Hausdorff approximation $f_{\mathbf{b}_{i}}:B_{\frac{r}{1280}}(p)\rightarrow B_{\frac{r}{1280}}(0,\hat{p}_{i})\subset\mathbb{R}\times\mathbf{X}_{\mathbf{b}_{i}}$. We assume $f_{\mathbf{b}_{i}}=(\mathbf{b}_{i},\mathcal{P}_{\mathbf{b}_{i}})$. We will prove the map f_{2}(x)=\Big{(}\mathbf{b}_{1}(x),\mathbf{b}_{2}\big{(}x\big{)},\mathcal{P}_{\mathbf{b}_{2}}\circ\mathcal{P}_{\mathbf{b}_{1}}(x)\Big{)}:B_{\frac{r}{1280}}(p)\rightarrow B_{\frac{r}{1280}}(0,\hat{p}_{i})\subset\mathbb{R}^{2}\times\mathbf{X}_{2} is an pointed $(\frac{2\epsilon r}{3})$-Gromov-Hausdorff approximation.

From the first variation formula of arc length, for $x\in P_{\eta,\mathbf{b}_{1},\mathbf{b}_{2}}^{r,\rho_{1}}\cap Q_{\eta,\mathbf{b}_{1}}^{r,\rho_{1}}\cap Q_{\eta,\mathbf{b}_{2}}^{r,\rho_{2},\rho_{1}}$, let $\check{x}=\mathfrak{n}_{1}(x)$, then we have

[TABLE]

Let \mathfrak{w}(x)=\Big{(}\mathbf{b}_{1}(x),\mathbf{b}_{2}\big{(}\mathcal{P}_{\mathbf{b}_{1}}(x)\big{)},\mathcal{P}_{\mathbf{b}_{2}}\big{(}\mathcal{P}_{\mathbf{b}_{1}}(x)\big{)}\Big{)} for any $x,y\in B_{\frac{r}{1280}}(p)$,

[TABLE]

Now we have

[TABLE]

The rest argument is similar as in the case $k=1$.

∎

Part III. Covering groups of Riemannian manifolds with $Rc\geq 0$

6. Squeeze Lemma and Dimension induction on harmonic functions

In this section, unless otherwise mentioned, we assume $\varphi:\tilde{M}^{n}\rightarrow M^{n}$ is the covering map with covering group $\Gamma$ such that $M^{n}=\tilde{M}^{n}/\Gamma$, where $(\tilde{M}^{n},\tilde{g})$ and $(M^{n},g)$ are two complete Riemannian manifolds and the metric $g$ is the quotient metric of $\tilde{g}$ with respect to group action of $\Gamma$.

We include the definition of quotient metric here for convenience.

Definition 6.1**.**

Consider a subgroup $G\subset\mathrm{Isom}(\mathbf{X})$, where $\mathbf{X}$ is a metric space, for every $\bar{x},\bar{y}\in\mathbf{X}/G$, set the quotient metric $\bar{d}$ on $\mathbf{X}/G$ by:

[TABLE]

For any function $f$ defined on a domain $\mathbf{D}\subset M^{n}$, we define

[TABLE]

For $\tilde{p},\tilde{q}\in\tilde{M}^{n}$ and any $s>0$, we define

[TABLE]

Similarly we can define $\Gamma_{\tilde{q}}(s)$. When the point is fixed and clear in the context, we use $\Gamma(s)$ instead of $\Gamma_{\tilde{p}}(s)$ for simplicity.

The following Lemma is motivated by the use of the canonical fundamental domain $\mathcal{F}$ in [Anderson], and is needed for the proof of the squeeze lemma.

Lemma 6.2**.**

For any function $f\geq 0$ defined on $B_{r}(p)\subset M^{n}$, $\tilde{p}$ is one lift of $p$ and $B_{r}(\tilde{p})$ is the geodesic ball centered at $\tilde{p}$ with radius $r$ in $\tilde{M}^{n}$. Then

[TABLE]

Proof: Let \tilde{\Omega}=\varphi^{-1}\big{(}B_{r}(p)\big{)}. For any $\tilde{y}\in B_{r}(\tilde{p})$,

[TABLE]

which implies \varphi\big{(}B_{r}(\tilde{p})\big{)}\subset B_{r}(p), then

[TABLE]

We choose a measurable section $s:B_{r}(p)\rightarrow B_{r}(\tilde{p})$, such that $\varphi(s(x))=x$ for any $x\in B_{r}(p)$. Let $T=s(B_{r}(p))$, then we have $\fint_{B_{r}(p)}f=\fint_{T}\mathcal{L}(f)$. Let $S$ be the union of $g(T)$ over all $g\in\Gamma$ such that $g(T)\cap B_{r}(\tilde{p})\neq\emptyset$. Then from $\mathrm{diam}(T)\leq 2r$, we obtain $S\subseteq B_{3r}(\tilde{p})$. And we also have $\fint_{B_{r}(p)}f=\fint_{S}\mathcal{L}(f)$ and $B_{r}(\tilde{p})\subseteq S$.

From the Bishop-Gromov volume comparison theorem, we know $V(B_{3r}(\tilde{p}))\leq 3^{n}V(B_{r}(\tilde{p}))$. Note $B_{r}(\tilde{p})\subseteq S\subseteq B_{3r}(\tilde{p})$, we have $V(S)\leq 3^{n}V(B_{r}(\tilde{p}))$, hence

[TABLE]

∎

Before we state and prove our squeeze lemma, we would like to include the following well-known result and its proof here, because the squeeze lemma can be looked at the Gromov-Hausdorff perturbation version of the following Lemma.

Lemma 6.3**.**

For a locally compact, pointed length space $(\mathcal{Z},q)$, assume $G$ is a closed subgroup of the isometry group $\mathrm{Isom}(\mathcal{Z})$ and $\mathcal{Z}/G=\mathbf{K}$, where $\mathbf{K}$ is a compact metric space with $\mathrm{diam}(\mathbf{K})=r_{0}$. Then for any $\epsilon>0$, we have

[TABLE]

Proof: For any $g_{1}\in G$, because $\mathcal{Z}$ is a length space, there is a segment $\gamma_{q,g_{1}q}$ from $q$ to $g_{1}q$. Then one can find a middle point $x\in\mathcal{Z}$ of the segment $\gamma_{q,g_{1}q}$ such that $d(x,q)=d(x,g_{1}q)=\frac{1}{2}d(q,g_{1}q)$. Since $\mathcal{Z}/G=\mathbf{K}$ and $G$ is a closed subgroup, one can find $g_{2}\in G$ such that $d(g_{2}q,x)\leq r_{0}$. Then we have

[TABLE]

Assume $r=d(q,g_{1}q)$, then we have

[TABLE]

which implies g_{1}\in\langle G\big{(}\frac{r}{2}+r_{0}\big{)}\rangle. Using that $G$ is a closed subgroup, by induction on $r$ we get g_{1}\in\big{\langle}G(2r_{0})\big{\rangle}, and the conclusion follows.

∎

Lemma 6.4** (Squeeze Lemma).**

For any $\epsilon>0,\delta=n^{-900n^{4}}\epsilon^{40n^{4}}$, integer $0\leq k\leq n$, if there exist harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k} defined on $B_{r}(p)$, satisfying $\mathbf{b}_{i}(p)=0$ and

[TABLE]

then there is a family of $(\epsilon s)$-Gromov-Hausdorff approximation for $s\in(0,10r_{c}]$,

[TABLE]

where $r_{c}=\frac{r}{12800}$. And let \mathrm{diam}\big{(}B_{r_{c}}(\hat{p}_{10r_{c}})\big{)}=r_{0}, we have

[TABLE]

Proof: In the proof, we can assume that $r_{0}\leq\frac{1}{2}r_{c}$, otherwise the conclusion follows directly. Let $\tilde{\mathbf{b}}_{i}=\mathcal{L}(\mathbf{b}_{i}),i=1,\cdots,k$, from (6.2) and Lemma 6.2, we have $\mathbf{b}_{i}(p)=\tilde{\mathbf{b}}_{i}(\tilde{p})=0$ and

[TABLE]

For $0<\epsilon_{1}<1$ to be determined later, from Theorem 2.13, if we assume

[TABLE]

there are two family of $(\epsilon_{1}s)$-Gromov-Hausdorff approximation for $s\in(0,10r_{c}]$,

[TABLE]

From Lemma 2.11, there is an $(30\epsilon_{1}r_{c})$-Gromov-Hausdorff approximation

[TABLE]

where $B_{10r_{c}}(0,\check{p}_{10r_{c}})\subset\mathbb{R}^{k}\times\tilde{\mathbf{X}}_{k,10r_{c}}$.

For any $\gamma\in\Gamma(r_{c})$, we have

[TABLE]

Also note $\check{p}_{10r_{c}}=\tilde{\mathcal{P}}_{10r_{c}}(\tilde{p})$, then we have $\tilde{\mathcal{P}}_{10r_{c}}(\tilde{p}),\tilde{\mathcal{P}}_{10r_{c}}(\gamma\tilde{p})\in B_{2r_{c}}(\check{p}_{10r_{c}})$.

Apply the argument of Lemma 2.8 to $\tilde{\mathcal{P}}_{10r_{c}}(\tilde{p})$ and $\tilde{\mathcal{P}}_{10r_{c}}(\gamma\tilde{p})$, we can obtain $\tilde{z}\in B_{4r_{c}}(\check{p}_{10r_{c}})$ such that

[TABLE]

Then note $\tilde{\mathbf{b}}_{i}(\tilde{p})=\tilde{\mathbf{b}}_{i}(\gamma\tilde{p})=0$, we have

[TABLE]

Similarly, we have

[TABLE]

From (6.7), we get that \big{(}0,\tilde{z}\big{)}\in B_{\frac{1}{2}r_{c}+70\sqrt{\epsilon_{1}}r_{c}}(0,\check{p}_{10r_{c}}). Because $\tilde{f}_{10r_{c}}$ is an $(10\epsilon_{1}r_{c})$-Gromov-Hausdorff approximation, there is $z_{1}\in B_{10r_{c}}(\tilde{p})$ such that

[TABLE]

Now we have

[TABLE]

Assume $z_{0}=\varphi(z_{1})$, we have $d(z_{0},p)\leq d(z_{1},\tilde{p})<\frac{1}{2}r_{c}+90\sqrt{\epsilon_{1}}r_{c}$. Then

[TABLE]

which implies $\mathcal{P}_{10r_{c}}(z_{0})\in B_{r_{c}}(\hat{p}_{10r_{c}})$. From the assumption \mathrm{diam}\big{(}B_{r_{c}}(\hat{p}_{10r_{c}})\big{)}=r_{0},

[TABLE]

Let $\tilde{\mathbf{b}}=(\tilde{\mathbf{b}}_{1},\cdots,\tilde{\mathbf{b}}_{k})$ and $\mathbf{b}=(\mathbf{b}_{1},\cdots,\mathbf{b}_{k})$, from (6.11) and (6.9), we obtain

[TABLE]

There exists $\gamma_{2}\in\Gamma$ such that $d(z_{1},\gamma_{2}\tilde{p})=d(z_{0},p)<r_{c}$, then from (6.10),

[TABLE]

We have $\gamma_{2}\tilde{p}\in B_{2r_{c}}(\tilde{p})$, from (6.9) and (6.12), we obtain

[TABLE]

Now from (6.6) and (6.13), we obtain

[TABLE]

Similarly from (6.8) and (6.13), we can get

[TABLE]

Let $\epsilon_{1}=\frac{\epsilon^{2}}{90000}$, from (6.14) and (6.15), we get \gamma\in\Big{\langle}\Gamma\big{(}\frac{1}{2}d(\tilde{p},\gamma\tilde{p})+r_{0}+\frac{120}{300}\epsilon r_{c}\big{)}\Big{\rangle}.

From (6.3), we can choose $\delta=n^{-900n^{4}}\epsilon^{40n^{4}}$. Now by induction, we have \gamma\in\big{\langle}\Gamma(\epsilon r_{c}+2r_{0})\big{\rangle}, which implies the conclusion.

∎

Lemma 6.5**.**

If $\epsilon<\frac{1}{10n^{2}}$, there do not exist $\{\nu_{i}\}_{i=1}^{n+1}\subset\mathbb{R}^{n}$ such that

[TABLE]

Proof: To prove the conclusion, we only need to show that $\nu_{1},\cdots,\nu_{n+1}$ are linearly independent. By contradiction, otherwise, without loss of generality, one can assume $\nu_{n+1}=\sum_{i=1}^{n}a_{i}\nu_{i}$ and $|a_{1}|=\max_{i=1,\cdots,n}|a_{i}|$, where $a_{i}\in\mathbb{R}$. Then from the assumption,

[TABLE]

which implies

[TABLE]

On the other hand, from (6.16) and $\epsilon<\frac{1}{10n^{2}}$, we can get

[TABLE]

which is the contradiction to the assumption.

∎

Proposition 6.6**.**

For complete Riemannian manifold $(M^{n},g)$ with $Rc\geq 0$, if there are harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{n}, defined on $B_{r}(p)$, satisfying $\mathbf{b}_{i}(p)=0$ and

[TABLE]

then $\Gamma_{\tilde{p}}(\frac{r}{12800})=\{e\}$.

Proof: Let $\delta=n^{-80000n^{7}}$ in the rest proof. From (6.17) and Lemma 6.2, let $\tilde{\mathbf{b}}_{i}=\mathcal{L}(\mathbf{b}_{i})$, then $\tilde{\mathbf{b}}_{i}(\tilde{p})=\mathbf{b}_{i}(p)=0$ and

[TABLE]

From (6.17), (6.18) and Theorem 2.13, if $\delta_{1}>0$ is to be determined later such that

[TABLE]

then there are two family of $(\delta_{1}\cdot s)$-Gromov-Hausdorff approximation for any $0<s\leq 10r_{c}$,

[TABLE]

where $r_{c}=\frac{1}{12800}r$.

For any $0<t\leq r_{c}$, let \mathrm{diam}\big{(}B_{t}(\hat{p}_{10t})\big{)}=t_{1}. If $t_{1}\geq\frac{1}{4}t$, note

[TABLE]

is an $(10\delta_{1}t)$-Gromov-Hausdorff approximation.

For $\epsilon>0$ to be determined later, assume

[TABLE]

Apply Theorem 2.12, there exist harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{n+1} defined on $B_{l_{1}}(q)\subset B_{10t}(p)$ such that

[TABLE]

where $l_{1}=n^{-320n^{3}}\epsilon^{10n}t$.

Let $l\rightarrow 0$ in the above, we get $\{\nu_{i}\}_{i=1}^{n+1}$, where $\nu_{i}\in T_{q}M^{n}$ such that

[TABLE]

Choose suitable $\epsilon=\frac{1}{20n^{2}}$, then (6.19) holds because of (6.20) and the definition of $\delta$.

Note $\epsilon<\frac{1}{10n^{2}}$, from Lemma 6.5, it is impossible. Hence we have

[TABLE]

Note the definition of $\delta$ implies

[TABLE]

For any $0<s\leq r_{c}$, from (6.22), (6.17), (6.21) and Lemma 6.4, we have

[TABLE]

From (6.23), by induction on $s$, for any positive integer $m$, we have

[TABLE]

Note the group action is discrete, hence for big enough $m$, we have \Big{\langle}\Gamma_{\tilde{p}}\Big{(}\big{(}\frac{3}{4}\big{)}^{m}r_{c}\Big{)}\Big{\rangle}=\{e\}, which implies \big{\langle}\Gamma_{\tilde{p}}(r_{c})\big{\rangle}=\{e\}.

∎

We define \mathfrak{ng}\big{\langle}\Gamma(r)\big{\rangle} as the minimal number of generators in $\Gamma(r)$ needed to generated \big{\langle}\Gamma(r)\big{\rangle}.

Proposition 6.7**.**

For $\tilde{p},\tilde{q}\in(\tilde{M}^{n},\tilde{g})$ with $Rc(\tilde{g})\geq 0$, assume $d(\tilde{p},\tilde{q})\leq\delta<\frac{s}{2}$, then

[TABLE]

Proof: Note $\Gamma_{\tilde{p}}(\lambda_{2})$ is a finite set, then there are only finite number of subsets $\mathcal{B}_{1},\cdots,\mathcal{B}_{m}\subseteq\Gamma_{\tilde{p}}(\lambda_{2})$ satisfying

[TABLE]

Let $\mathcal{A}_{\lambda_{1},\lambda_{2}}$ be one of the above subsets, such that there does not exist $\mathcal{B}_{i}$ with $\mathcal{A}_{\lambda_{1},\lambda_{2}}\subsetneqq\mathcal{B}_{i}$. Then $\mathcal{A}_{\lambda_{1},\lambda_{2}}\subset\Gamma_{\tilde{p}}(\lambda_{2})$ satisfy:

[TABLE]

And for any $\gamma\in\Gamma_{\tilde{p}}(\lambda_{2})$, there is $h\in\mathcal{A}_{\lambda_{1},\lambda_{2}}$ such that $h^{-1}\gamma\in\Gamma_{\tilde{p}}(\lambda_{1})$.

For any $h\in\mathcal{A}_{\lambda_{1},\lambda_{2}}$, we get $h\tilde{p}\in B_{\lambda_{2}}(\tilde{p})$. From (6.24),

[TABLE]

Now let $\#(\mathcal{A}_{\lambda_{1},\lambda_{2}})$ denote the number of the elements in $\mathcal{A}_{\lambda_{1},\lambda_{2}}$, we have

[TABLE]

In the last inequality the Bishop-Gromov Comparison Theorem is used. Then

[TABLE]

It is easy to see $\Gamma_{\tilde{p}}(\lambda_{2})\subset\mathcal{A}_{\lambda_{1},\lambda_{2}}\cdot\Gamma_{\tilde{p}}(\lambda_{1})$, from (6.25), we have

[TABLE]

For $\gamma\in\Gamma_{\tilde{p}}(s)$, we have

[TABLE]

which implies $\Gamma_{\tilde{p}}(s)\subset\Gamma_{\tilde{q}}(s+2\delta)$. Then

[TABLE]

∎

Proposition 6.8**.**

For any $\epsilon\in(0,1)$ and $0\leq k<n$, there exists $\delta=(n^{-30}\epsilon)^{5000n^{7}}$, such that if there exist harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k}, satisfying

[TABLE]

Then there are harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k+1} defined on $B_{r_{1}}(q_{1})\subset B_{r}(q)$, such that

[TABLE]

Proof: Set $r_{c}=\frac{1}{12800}r$, from Proposition 6.7,

[TABLE]

Define \hat{r}_{1}=\min\Big{\{}s\geq 0|\ \Gamma_{\tilde{q}}(r_{c})\subset\big{\langle}\Gamma_{\tilde{q}}(s)\big{\rangle}\Big{\}}, then $\hat{r}_{1}\leq r_{c}$ and

[TABLE]

For $\delta_{1}>0$ to be determined later, we choose $\delta\leq n^{-900n^{4}}\delta_{1}^{40n^{4}}$. From (6.26) and apply Lemma 6.4 on $B_{10\hat{r}_{1}}(q)$, we have a family of \big{(}\delta_{1}\cdot s\big{)}-Gromov-Hausdorff approximation for any $0<s\leq 10\hat{r}_{1}$,

[TABLE]

And assume \mathrm{diam}\big{(}B_{\hat{r}_{1}}(\hat{q}_{10\hat{r}_{1}})\big{)}=r_{0}, we have

[TABLE]

From (6.29) and (6.31), we have \Gamma_{\tilde{q}}(r_{c})\subset\big{\langle}\Gamma_{\tilde{q}}(\delta_{1}\hat{r}_{1}+2r_{0})\big{\rangle}. Then by the definition of $\hat{r}_{1}$, we get $\hat{r}_{1}\leq\delta_{1}\hat{r}_{1}+2r_{0}$. If we assume $\delta_{1}\leq\frac{1}{32}$, it yields $r_{0}\geq\frac{1}{2}\hat{r}_{1}-\frac{1}{2}\delta_{1}\hat{r}_{1}\geq\frac{1}{4}\hat{r}_{1}$. Note \mathrm{diam}\big{(}B_{\hat{r}_{1}}(\hat{q}_{10\hat{r}_{1}})\big{)}=r_{0}, then combining (6.30), choose $\delta_{1}=n^{-3400n^{3}}\epsilon^{110n}$. Apply Theorem 2.12, we can find harmonic functions \big{\{}\mathbf{b}_{i}\big{\}}_{i=1}^{k+1} defined on $B_{r_{1}}(q_{1})\subset B_{10\hat{r}_{1}}(q)$, such that

[TABLE]

where $r_{1}=n^{-320n^{3}}\epsilon^{10n}\hat{r}_{1}>0$.

From Proposition 6.7 and $q_{1}\in B_{10\hat{r}_{1}}(q)$, we have

[TABLE]

From (6.28), (6.29) and (6.32), we have

[TABLE]

∎

Theorem 6.9**.**

Assume $\varphi:\tilde{M}^{n}\rightarrow M^{n}$ is the covering map with covering group $\Gamma$ such that $M^{n}=\tilde{M}^{n}/\Gamma$, where $(\tilde{M}^{n},\tilde{g})$ and $(M^{n},g)$ are two complete Riemannian manifolds and the metric $g$ is the quotient metric of $\tilde{g}$ with respect to group action of $\Gamma$, furthermore $Rc\geq 0$, then \mathfrak{ng}\big{\langle}\Gamma_{\tilde{p}}(1)\big{\rangle}\leq n^{n^{20n}} for any $\tilde{p}\in\tilde{M}^{n}$.

Proof: Apply Proposition 6.8, firstly when $k=0$, we get

[TABLE]

where $\epsilon_{1}=(n^{-30}\epsilon_{2})^{\tau},\tau\vcentcolon=5000n^{7}$ and $\epsilon_{2}$ is to be determined later.

By induction, apply Proposition 6.8, for $0\leq k\leq(n-1)$, we get

[TABLE]

Let $\epsilon_{j}=(n^{-30}\epsilon_{j+1})^{\tau},j=1,\cdots,n-1$.

We let $\epsilon_{n}=n^{-200\tau}$, then from (6.33) and (6.34) for $k=(n-1)$, apply Proposition 6.6, we have $\Gamma_{\tilde{q}_{n}}(\frac{r_{n}}{12800})=\{e\}$.

From the induction formula for $\epsilon_{j}$, we have

[TABLE]

Then from (6.33), (6.35) and Proposition 6.7 yields

[TABLE]

∎

Theorem 6.10**.**

Suppose $(M^{n},g)$ is a complete Riemannian manifold with $Rc\geq 0$, then for any finitely generated subgroup $\Gamma$ of $\pi_{1}(M^{n})$, we have $\mathfrak{ng}(\Gamma)\leq n^{n^{20n}}$.

Proof: From [Munkres, Theorem $82.1$], there exists a covering map $\varphi:N^{n}\rightarrow M^{n}$ such that \varphi_{*}\big{(}\pi_{1}(N^{n},\hat{p})\big{)}=\Gamma, where $\varphi(\hat{p})=p$. Now from [Munkres, Theorem $54.6(a)$], we get $\pi_{1}(N^{n},\hat{p})\simeq\Gamma$. Hence there exists a complete Riemannian manifold $(N^{n},\tilde{g})$ with $Rc(N^{n})\geq 0$ and $\pi_{1}(N^{n})=\Gamma$.

Assume $\Gamma=\langle\gamma_{1},\cdots,\gamma_{k}\rangle$ and $\sup_{i=1,\cdots,k}d(\gamma_{i}\hat{p},\hat{p})\leq C_{1}$. Then by scaling the metric $\tilde{g}$ to $\hat{g}=C_{1}^{-1}\tilde{g}$, we get that \Gamma=\big{\langle}\Gamma_{\hat{p}}(1)\big{\rangle}, where $\hat{p}\in(N^{n},\hat{g})$. From Theorem 6.9, we have \mathfrak{ng}(\Gamma)\leq\mathfrak{ng}\big{\langle}\Gamma_{\hat{p}}(1)\big{\rangle}\leq n^{n^{20n}}.

∎

Acknowledgments

The author thank Aaron Naber for helpful suggestion, which greatly improves the quantitative estimates in the earlier version of this paper. We are grateful to Jiaping Wang for continuous encouragement, William P. Minicozzi II and Christina Sormani for comments. We are indebted to Vitali Kapovitch for several email reply about [KW], which help us to understand the results there better.