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X is semi-locally simply connected, if for any point x ∈ M, there is an open neighborhood containing x such that any loop in this neighborhood is contractible in X. In particular, if for all x ∈ M there exists t (depending on x) such that ρ(t, x) is finite, then M must be semi-locally simply connected. Now we state our first result:
Theorem 1.1. Let (Mi,pi) be a sequence of complete Riemannian manifolds con- verging to (X,p) such that for all i, (1) B4(pi) ∩ ∂Mi = ∅ and the closure of B4(pi) is compact, (2) Ric ≥−(n − 1) on B4(pi), V ol(B4(pi)) ≥ v > 0. g Then lim ρ(t, x)=0 holds for all x ∈ B1/800(p). t→0 Theorem 1.1 implies the following.
Corollary 1.2. Let (Mi,pi) be a sequence of complete Riemannian n-manifolds with Ric ≥−(n − 1) and converging to (X,p). If there exist r and v such that for all i and any x ∈ Mi, V ol(Br(x)) ≥ v > 0. Then (X,p) is semi-locally simply connected. g
As mentioned, the universal cover of a Ricci limit space always exists [17, 18]. As a corollary, the universal cover of X in Corollary 1.2 is always simply connected. To prove Theorem 1.1, we establish the existence of a slice for pseudo-group ac- tions on incomplete Ricci limits. Then Theorem 1.1 follows from this slice theorem and Pan-Wei’s result [16].
Theorem 1.3. Let (Mi,pi) be a sequence of connected Riemannian manifolds con- verging to (X,p) such that for all i, B4(pi) ∩ ∂Mi = ∅ and the closure of B4(pi) is ^ compact. Let (B4(pi), p˜i) be the universal cover of B4(pi) and let Γi be the funda- mental group of B4(pi). The pseudo-group
Gi = {γ ∈ Γi|d(γp˜i, p˜1) ≤ 1/100} acts on B1(˜pi). Passing to a subsequence, (B¯1(˜pi), p˜i, Gi) converges to (B¯1(˜p), p,˜ G) in the equivariant Gromov-Hausdorff sense. Then the following holds. For any x˜ ∈ B1/800(˜p), there is a slice S at x˜ for pseudo-group G-action, that is, S satisfies: (1) x˜ ∈ S and S is Gx˜-invariant, where Gx˜ is the isotropy subgroup at x˜; (2) S/Gx˜ is homeomorphic to a neighborhood of x.
Note that we don’t assume B1(˜pi) has a volume lower bound in Theorem 1.3. The idea to find the slice is that we extend G to a Lie group Gˆ, which acts homeomorphically on an extended space. Then we apply Palais’s slice theorem [14] for this Gˆ-action. In this construction, we will show that the extended group and space are locally homeomorphic old ones, thus we can get the slice atx ˜. See Sections 4 and 5 for details. The next result of this paper is about the description of universal covers of Ricci limit spaces. Now we assume that diam(Mi) ≤ D for some D > 0. Let Mi be the f universal cover of Mi with fundamental group Γi. Passing to a subsequence, we obtain the following convergence 3
GH (Mi, p˜i, Γi) −−−−→ (Y, p,˜ G) f π π y GH y (Mi,pi) −−−−→ (X,p), Ennis and Wei showed that there is ǫ = ǫ(X) > 0 such that
ǫ GH ǫ (Mi, p˜i, Γi ) −→ (Y, p,˜ H ) f ǫ ǫ and the universal cover of X is isometric Y/H [8], where Γi is generated by elements
{g ∈ Γi | d(q,gq) ≤ ǫ for some q ∈ Mi}. f We give a description of Hǫ from the G-action on Y .
Theorem 1.4. Let Mi be a sequence of closed Riemannian n-manifolds with
Ric(Mi) ≥−(n − 1), diam(Mi) ≤ D.
Suppose that GH (Mi, p˜i, Γi) −−−−→ (Y, p,˜ G) f πi π y GH y (Mi,pi) −−−−→ (X,p).
Let H be the group generated by G0 and all isotropy subgroups of G, where G0 is the identity component subgroup of G. Then Y/H is the universal cover of X. Consequently, π1(X,p) is isomorphic to G/H.
Note that H is a subgroup of Hǫ. Theorem 1.4 shows that they are indeed the same. This improvement relies on the fact that Isom(Y ) is a Lie group [2, 7]. We organize the paper as follows. In Section 2, we recall some preliminaries, which include Ricci limit spaces and Palais’s slice theorem. In Section 3, we study the limit pseudo-group actions on a closed ball of the Ricci limit space. To prove Theorem 1.3 next, we will use the limit pseudo-group G to construct a Lie group Gˆ in Section 4, then we construct a Gˆ-space and apply Palais’s theorem to find a slice on the new space in Section 5. We prove Theorem 1.4 in Section 6. The authors would like to thank Xiaochun Rong for many helpful discussions.
Contents
1. Introduction 1 2. Preliminaries 4 3. Pseudo-group 6 4. Groupfication of G 9 5. Construction of a Gˆ-space 12 6. Proof of Theorem 1.4 15 References 17 4
2. Preliminaries 2.1. Ricci limit spaces. One important result about Ricci limit spaces that we need in this paper is their isometry groups are always Lie groups. [2] proves the non-collapsing case, then [7] proves the general case. Theorem 2.1. [2, 7] The isometry group of any Ricci limit space is a Lie group. We recall some of elements in the proof of Theorem 2.1 since we need them later in Section 3. Let (X,p) be a Ricci limit space. For any integer k ≤ n, Rk is the set of k points where every tangent cone is isometric to R . We say y ∈ (Rk)ǫ,δ, if for all k k Rk 0
ρH (x) = sup d(x, hx),DH,r(x) = sup ρH (w) h∈H w∈Br(x) be the displacement function of H on Br(x). A standard fact is that there is no k k non-trivial group H ⊂ Isom(R ) such that DH,1(0) ≤ 1/20, where 0 ∈ R . Using certain path connectedness property of (Rk)ǫ,δ, where k is the dimension of X in the sense of [7], the proof of Theorem 2.1 rules out a sequence of nontrivial subgroups Hi with DHi,R(z) → 0 for all R > 0 and all z ∈ X. A similar idea is applied in [15, Lemma 2.1] to prove the following. Proposition 2.2. [15] Let X be a Ricci limit space and let g be an isometry of X. If g fixes every point in B1(x), where x ∈ X, then g fixes every point in X. We will prove a local version of Proposition 2.2 in Section 3, which is essential to study pseudo group actions. The main difference is that we are considering limit space that comes from a sequence of incomplete manifolds. We also recall the main theorem in [16]:
Theorem 2.3. Let (Mi,pi) be a sequence of connected Riemannian n-manifolds (not necessarily complete) converging to (Y,p) such that for all i, (1) B4(pi) ∩ ∂Mi = ∅ and the closure of B4(pi) is compact, (2) Ric ≥−(n − 1) on B4(pi), V ol(B4(pi)) ≥ v > 0. Then lim ρ(t, x)=0 holds for any x ∈ B2(p). t→0 In fact, a stronger result lim ρ(t, x)/t = 1 is proved in [16]. t→0 2.2. Palais’s slice theorem. We review the definition of a slice and Palais’s the- orem [14] in this subsection. Let X be a completely regular topological space and let G be a Lie group. We say that X is a G-space, if G acts on X by homeomorphisms. For any x ∈ X, we define isotropy subgroup Gx = {g ∈ G| gx = x}. If G-action on X is proper, then Gx is compact for all x ∈ X. We say a subset S ⊂ X is Gx-invariant, if GxS = S. For Gx-invariant set S, we can define an equivalence relation on G × S by claiming 5
−1 (g,s) ∼ (gh ,hs) for all g ∈ G, h ∈ Gx,s ∈ S. We denote the corresponding quotient space as
G ×Gx S = G × S/ ∼ . ′ ′ There is a natural left G action on G ×Gx S given by g[g ,s] = [gg ,s], where g ∈ G ′ and [g ,s] ∈ G ×Gx S.
Definition 2.4. We call S ⊂ X a Gx-slice (briefly, a slice) at x if the followings hold: (1) x ∈ S and S is Gx-invariant; (2) GS is an open neighborhood of x; ψ :[g,s] 7→ gs is a G-homeomorphism between G ×Gx S and GS, that is, ψ is a homeomorphism and ψ ◦ g = g ◦ ψ for all g ∈ G.
In particular, if S is a slice at x, then S/Gx is homeomorphic to GS/G. The following slice theorem is from [14]. Theorem 2.5. Let X be a G-space and x ∈ X, where G is a Lie group. Then the following two conditions are equivalent: (1) Gx is compact and there is a slice at x. (2) There is a neighborhood U of x ∈ X such that {g ∈ G|gU ∩ U 6= ∅} has compact closure in G. It is well-known that for an isometric G-action on a locally compact metric space X, (or more generally, proper actions,) the second statement in theorem 2.5 is always fulfilled, thus a slice always exists. We include the proof here for readers’ convenience. Corollary 2.6. Let X be a locally compact metric space. Let G be a Lie group acting isometrically and effectively on X. Then for any x ∈ X, there is a Gx-slice at x. Proof. We claim that G(1) = {g ∈ G| d(gx, x) ≤ 1} is compact. If this claim holds, then we set U = B1/2(x). With this U, {g ∈ G|gU ∩ U 6= ∅} is contained in G(1), thus it has a compact closure in G. Then the result follows from Theorem 2.5. We prove the claim. Let {gj} be any sequence in G(1). We shall find a converging subsequence. Let {xi} be a sequence that is dense in X. For x1, we have
d(x, gj x1) ≤ d(x, gj x)+ d(gj x, gj x1) ≤ 1+ d(x, x1).
So we may assume gj x1 converges to some y1 by passing to a subsequence. For this subsequence, we repeat this process for x2,...,xi, and by a standard diagonal argument, we may assume that for all i, gjxi converges to yi as j → ∞. We define g : X → X by setting g(xi)= yi and extending g to the whole space continuously. By construction, g is an isometry of X and gj converges to g. Moreover, g ∈ G(1) because d(x, gx) = lim d(x, gj x) ≤ 1.
2.3. Loop Liftings. In this subsection, we explain how we apply Theorem 2.3 and the slice theorem to prove semi-local simple connectedness. We consider an easier case that the diameter of Mi is strictly less than 4. In this case, the local universal ^ cover B4(pi) is the same as universal cover Mi of Mi. For consistency, We still use f the notation B4(pi) instead of Mi. Let Gi be the fundamental group of B4(pi). 6
Passing to a subsequence if necessary, we have ^ GH (B4(pi), p˜i, Gi) −−−−→ (X, p,˜ G) e (2.7) π π y GH y (B4(pi),pi) −−−−→ (X = X/G,p), e Since x ∈ B1/800(p), there exists xi ∈ Mi converging to x. We change the base points and consider ^ GH (B4(pi), x˜i, Gi) −−−−→ (X, x,˜ G) e (2.8) π π y GH y (B4(pi), xi) −−−−→ (X = X/G, x). e To prove lim ρ(t, x) = 0, we lift a loop based at x to a loop based atx ˜; see t→0 diagram 2.8. This idea is also used in [16], where the case Gx˜ =x ˜ is considered. Note that the loop is only continuous and may not be rectifiable. For a general non-compact Lie group G-action, we don’t know whether a path-lifting exists. Even though such path-lifting exists, a loop may be lifted to a non-closed path since in general Gx˜ 6=x ˜. We make use of the slice theorem to lift any loop based at x to a loop based atx ˜. Furthermore, a loop in a small ball centered at x must be lifted in a small ball centered atx ˜ since π : S/Gx˜ → GS/G is a homeomorphism. Now we follow this idea to prove Theorem 1.1 when diam(Mi) < 4 for all i:
Proof. Consider (X,˜ x,˜ G) as in the diagram 2.8. Given any ǫ> 0, since limt→0 ρ(t, x˜)= 0, there exists δ′ > 0 such that ρ(t, x˜) <ǫ holds for all t ≤ δ′. G is a Lie group due to [2]. By Corollary 2.6, there is a slice S atx ˜. Since GS is open, GS/G contains a open ball Bδ(x). We show that −1 π (Bδ(x)) ∩ S ⊂ Bδ′ (˜x) −1 when δ is small enough. Otherwise, there exist δj → 0 andx ˜j ∈ π (Bδj (x)) ∩ S ′ butx ˜j ∈/ Bδ (˜x). While π(˜xj ) ∈ Bδj (x) converges to x = π(˜x),x ˜j does not converge tox ˜. This contradicts to that π : S/Gx˜ → GS/G is a homeomorphism. −1 Now we have δ > 0 small such that π (Bδ(x)) ∩ S ⊂ Bδ′ (˜x). Choose any loop γ in Bδ(x). Since slice theorem applies, we can lift γ to loopγ ˜ based at x˜ and contained in S. Actually the image ofγ ˜ is also contained in Bδ′ (˜x) since −1 π (Bδ(x)) ∩ S ⊂ Bδ′ (˜x). Because ρ(˜x,δ′) <ǫ, there exists a homotopy H, of which the image is contained in Bǫ(˜x), fromγ ˜ to the constant loop atx ˜. Then π ◦ H is a homotopy from γ to the constant loop at x and the image of this homotopy is in Bǫ(x). So any loop in Bδ(x) is contractible in Bǫ(x), i.e. ρ(t, x) ≤ ǫ when t<δ. Since ǫ can be arbitrarily small, limt→0 ρ(t, x) = 0.
3. Pseudo-group ^ We no longer assume that Mi has diameter less than 4. B4(pi) may not have a ^ Gromov-Hausdorff limit in general (see [18]); instead, we will use B¯1(˜pi) in B4(pi). ^ Choosep ˜i ∈ B4(pi) with π(˜pi) = pi. We consider the closed ball B¯1(˜pi) with 7
^ intrinsic metric in B4(pi) and
Gi = {g ∈ Γi|d(gpi,pi) ≤ 1/100}.
Passing to a subsequence, (B¯1(˜pi), p˜i) converges to (B¯1(˜p), p˜) by Bishop-Gromov relative volume comparison. We will define the meaning of convergence Gi → G, where the limit G is a pseudo-group, which means gg′ ∈/ G for some g,g′ ∈ G. By the argument of last subsection, to prove Theorem 1.1, it is enough to prove the existence of a slice for pseudo-group actions. Now we start to construct the pseudo-group G. LetΓi be the fundamental group of B4(pi). Define ′ ′ Gi = {g ∈ Γi|d(gp˜i, p˜i) ≤ 1/30}, Gi = {g ∈ Γi|d(gp˜i, p˜i) ≤ 1/100}⊂ Gi. ′ ^ Notice that elements in Gi or Gi preserve the distance on B4(pi) but may not pre- serve intrinsic metric on B¯1(˜pi). However, the intrinsic metric d and restricted met- ^ ¯ ¯ ¯ ric (from B4(pi)) on B1(˜pi) coincide on B1/3(˜pi). So for any z1,z2 ∈ B1/10(˜pi) and ′ ¯ ¯ g ∈ Gi, we have gz1,gz2 ∈ B2/15(˜pi) ⊂ B1/3(˜pi) and thus d(z1,z2) = d(gz1,gz2). Abusing the notation, we use notation d for intrinsic metrics on B¯1(˜pi) and B¯1(˜p). For r ≤ 1, we restrict metric d from B¯1(˜p) (or B¯1(˜pi)) to B¯r(˜p) (or B¯r(˜pi)). ′ ¯ ¯ We see Gi and Gi as sets of distance-preserving maps from B1/10(˜pi) to B1/3(˜pi). ′ ′ After passing to a subsequence, Gi converges to G and Gi converges to G, where ′ ¯ ¯ G and G are closed subsets of distance-preserving maps from B1/10(˜p) to B1/3(˜p). ¯ ¯ ′ Though Theorem 1.3 is stated with B1/800(˜p) and G, we need B1/10(˜p) and G to ¯ provide extra room to study B1/800(˜p) and G. We consider G′ at first. For all g ∈ G′, d(gp,˜ p˜) ≤ 1/30 holds. Notice that any ′ ¯ ′ g ∈ G is only assumed to be defined on B1/10(˜p). G has compact-open topology ¯ ¯ ′ as a set of continuous maps C(B1/10(˜p), B1/3(˜p)). G is compact by the same proof of Corollary 2.6. However, there is no group structure so far since g1g2 may not be ¯ ′ defined on B1/10(˜p) for g1,g2 ∈ G . Before endowing G′ a pseudo-group structure, we prove a lemma as below for later use. ′ Lemma 3.1. For any g ∈ G , if g fix Br(˜p) for some 0 DH,r(x) = sup ρH (w) w∈Br(x) as the displacement function of some subgroup H on Br(x). Proof of Lemma 3.1. Let k be the dimension of B1(˜p) in the sense of [7]. We assume r is the largest t such that g fixes Bt(˜p). Let C be the closure of subset generated by ′ g in G . C fixes Br(˜p) as well and thus C is a group. Let s = (1/10−r)/4. We shall show for all ǫ > 0, there exist x(ǫ) ∈ (Rk)ǫ,τ(ǫ) ∩ Br+3s and s > τ(ǫ) > r(ǫ) > 0, such that (3.2) DC,r(ǫ)(x(ǫ)) = r(ǫ)/20 8 If so, then we can find a sequence ǫl → 0, where l ∈ N, such that −1 k (3.3) (Br(ǫl)(x(ǫl)), (r(ǫl)) d, C) → (B1 (0), d0,H). H is non-trivial with DH,1(0) = 1/20, which is impossible. Fixing ǫ, we verify (3.2). Since r is the largest t such that g fix Bt(˜p), we can find w ∈ Bs+r(˜p) ∩ Rk such that g(w) 6= w. We claim any geodesic from w to a point in Bs(˜p) is contained in B1/10−s(˜p). Otherwise there exists a geodesic from w to x ∈ Bs(˜p) and there is point y on this geodesic with d(˜p,y)=1/10 − s. Then d(x, w)= d(x, y)+d(y, w) > 1/10−2s+2s =1/10 and d(x, w) ≤ d(w, p˜)+d(˜p, x) ≤ 1/10 − 3s + s =1/10 − 2s, which is a contradiction. Choose θ < min{d(w,gw),s} such that w ∈ (Rk)ǫ,2θ. Then we have DC,θ(w) ≥ θ/20. Let z ∈ (Rk)ǫ,2η be a regular point in Bs(˜p) ∩ Br(˜p) for small η. DC,η(z)= 0 ≤ η/20 since C fix Br(˜p). Let λ< min(θ, η). Notice that DC,r(z) is continuous in r and z when Br(z) is away from the bound- ¯ ary of B1/10(p). Suppose that DC,λ(z) > λ/20, by intermediate value theorem, we can find λ ≤ r ≤ η such that DC,r(z)= r/20. Similarly, if DC,λ(w) < λ/20, we can find λ ≤ r ≤ θ,DC,r(w)= r/20. Now we assume DC,λ(z) ≤ λ/20 ≤ DC,λ(w). By [2] for k = n and [7] for general k, we may assume there exists δ and a geodesic c : [0,l] → B1/10−s(˜p) from w to z such that c is contained in (Rk)ǫ,δ. Notice that we just showed that a geodesic from w to z must be contained in B1/10−s(˜p), so c ⊂ B1/10−s(˜p). By intermediate value theorem we can find z(ǫ) ∈ (Rk)ǫ,δ on c such that DC,λ(z(ǫ)) = λ/20. Corollary 3.4. For any δ > 0 and r < 1/10, there exists ǫ so that the follow- ′ ′ ′ ¯ ing holds: if gi,gi ∈ Gi such that d(gix,g˜ ix˜) < ǫ for all x˜ ∈ Br(˜pi), we have ′ ¯ d(giw,giw) <δ for all w ∈ B1/10(˜pi). ′ ′ ′ Proof. Assume there exist δ > 0, r< 1/10, gi,gi ∈ Gi such that d(gix,˜ gix˜) <ǫi → ¯ ′ ¯ 0 for allx ˜ ∈ Br(˜pi), but d(giwi,giwi) ≥ δ for some wi ∈ B1/10(˜pi). For large i, ′ −1 ′ ′ −1 ′ −1 gigi ∈ Gi since d(gigi p,˜ p˜) < ǫi. By passing to a subsequence, gigi → g and ¯ ¯ wi → w ∈ B1/10(˜p). g fix Br(˜p) but d(gw,w) ≥ δ. This contradicts to the lemma above. ′ ′ Now we give G a pseudo-group structure as follows. For any g0,g1,g2 ∈ G , ¯ ¯ ¯ notice that g2(B1/250(˜p)) ⊂ B1/10(˜p); so g1g2(B1/250(˜p)) is always well-defined. We ¯ say g0 = g1g2 if g0(z) = g1g2(z) for all z ∈ B1/250(˜p). By Lemma 3.1, such g0 is unique if it exists. This gives a pseudo-group structure on G′. The identity map ¯ defined on B1/10(˜p), which we write as 1, is the identity element. Remark 3.5. We remind readers that we view elements in G′ as continuous maps ¯ ¯ from B1/10(˜p) to B1/3(˜p). We check the product of the pseudo-group structure on ′ ¯ G by checking their values on B1/250(˜p). ′ −1 ′ ′ We claim that for any g ∈ G , it has an inverse g ∈ G . We can choose gi ∈ Gi −1 −1 ′ such that gi converges to g. Since d(gi p˜i, p˜i) = d(˜pi,gip˜i) ≤ 1/30, gi ∈ Gi. −1 ′ ′ ¯ After passing to a subsequence, we assume gi → g ∈ G . For any z ∈ B1/250(˜p), ′ we can find zi converging to z where zi ∈ B1/240(˜pi). Then gg (z) is well-defined ′ ¯ −1 ′ ′ ¯ since g (z) ∈ B1/10(˜p). zi = gigi zi → gg z, so gg (z) = z for all z ∈ B1/250(˜p). Thus g′ = g−1 in G′. 9 ′ A similar proof shows that for any g1,g2 ∈ G such that d(g1g2(˜p), p˜) < 1/30, we ′ have g1g2 ∈ G . Notice that g1g2(˜p) means g1(g2(˜p)), which is always defined even ′ though g1g2 is not in G . For G, we can define a pseudo-group structure as above. Recall that any g ∈ G satisfies d(gp,˜ p˜) ≤ 1/100, thus G is a subset of G′. It is clear that the followings hold: (1) if g ∈ G, we have g−1 ∈ G; ′ (2) g1g2g3 ∈ G for all g1,g2,g3 ∈ G, since d(g1g2g3p,˜ p˜) < 1/30. We give a simple but useful lemma below: ¯ Lemma 3.6. Let g1,g2 ∈ G. If there exists x ∈ B1/240(˜p) such that g1g2(x) ∈ ¯ B1/240(˜p), then g1g2 ∈ G. ′ ¯ Proof. Since g1,g2 ∈ G, g1g2 ∈ G . Because x ∈ B1/240(˜p) and g1(g2(x)) ∈ ¯ B1/240(˜p), d(g1g2p,˜ p˜) < 1/100, we have g1g2 ∈ G as well. ¯ We define B1/250(˜p)/G as the quotient set, where the equivalent relation is given ¯ ¯ by z ∼ gz for any z ∈ B1/250(˜p) and g ∈ G such that gz ∈ B1/250(˜p). We define ¯ ¯ a natural quotient metric on B1/250(˜p)/G. For any [z], [w] ∈ B1/250(˜p)/G, define d([z], [w]) be the minimum of d(z, w) where z, w are pre-images of [z], [w]. By [9], ¯ ¯ B1/250(˜pi)/Gi, with the quotient metric as above, converges to B1/250(˜p)/G . ¯ ¯ It is plausible that B1/250(˜p)/G is isometric to B1/250(p), but the following obser- vation shows we can only get an isometry between B1/800([˜p]) and B1/800(p) where ¯ [˜p]= π(˜p) ∈ B1/250(˜p)/G. ¯ ¯ Remark 3.7. For any g ∈ Γi with gB1/250(˜pi) ∩ B1/250(˜pi) is non-empty, we have ¯ ¯ g ∈ Gi. As a result, B1/250(˜pi)/Gi is homeomorphic to B1/250(pi). However, the ¯ ¯ quotient metric on B1/250(˜pi)/Gi may not be equal to the metric on B1/250(pi) ¯ for the following reason. For any x1, x2 ∈ B1/250(pi), d(x1, x2) equals min d(˜x1, x˜2) ^ wherex ˜1, x˜2 ∈ B4(pi) and π(˜x1)= x1, π(˜x2)= x2. It’s possible for any pair ofx ˜1, x˜2 ¯ on which we get the minimum, one of them is out of B1/250(˜pi), then the quotient ¯ metric would be larger. But two metric coincide on x1, x2 ∈ B1/800(pi), since we ¯ ¯ ¯ can choosex ˜1 ∈ B1/800(˜pi) and there must be ax ˜2 ∈ B1/400(˜x1) ⊂ B1/250(˜pi) such ¯ that d(x1, x2)= d(˜x1, x˜2). As a result, 1/800-ball at [˜pi]= π(˜pi) ∈ B1/250(˜pi)/Gi is ¯ isometric to B1/800(˜pi). Passing to the limit, B1/800([˜p]) is isometric to B1/800(p). 4. Groupfication of G We need a Lie group action to apply the slice theorem. However, G is only a pseudo-group. To overcome this problem, we will first construct a Lie group Gˆ out of G. This process is related to [9]. Let FG be the free group generated by eg for all g ∈ G. We quotient FG by the −1 normal subgroup generated by all elements of the form eg1 eg2 eg1g2 , where g1,g2 ∈ G with g1g2 ∈ G. The quotient group is denoted as Gˆ. We write elements in Gˆ as gˆ = [eg1 eg2 ...egk ]. Lemma 4.1. The natural map i : G → Gˆ, g 7→ [eg] is injective. Proof. Assuming there exists g ∈ G such that [eg] = 1, we shall show that g is l the identity map on B1/10(˜p). By the definition, there exist gj ,gj1,gj2 ∈ G with 10 gj1gj2 ∈ G such that the following holds in the free group FG: k kj 1 sl −sl j −1 sj j (4.2) eg = Y((Y e l )(egj1 egj2 egj1 gj2 ) ( Y e l )), gj gj j=1 l=1 l=kj l where each sj and sj is 1 or −1. We see the right-hand side as a sequence of words. The equality means we can find a way to keep canceling adjacent terms and get eg eventually. Let φi : G → Gi ֒→ Γi be the equivariant GH approximation. Although φi generally is not a homomorphism onto the image, we claim that k kj 1 l l l sj −1 sj l −sj (4.3) φi(g)= Y((Y φi(gj) )(φi(gj1)φi(gj2)φi(gj1gj2) ) ( Y φi(gj ) )). j=1 l=1 l=kj Notice the right-hand side of (4.3) comes from (4.2) by the following procedure: −1 −1 replace eh by φi(h) and replace eh by φi(h) which is the inverse of φi(h). In particular, if two terms in (4.2) are inverse to each other, the corresponding two terms in (4.3) are inverse to each other. (4.2) means that we can simplify the right hand-side to eg by cancelling adjacent inverse terms. So, we can also simplify the right hand-side of (4.3) to φi(g) by the same procedure as in (4.2). It remains to show that there exists ǫi → 0 such that φi(g) is ǫi-close to the ¯ ¯ identity on B1/250(˜pi), that is, d(φi(g)z,z) ≤ ǫi for all z ∈ B1/250(˜pi). If this is ¯ ¯ true, then φi(g) converges to the identity map on B1/250(˜p) and thus on B1/10(˜p) as well by lemma 3.1; together with φi(g) converging to g, this proves g is the identity. On Γi the association law always holds; we consider −1 hij := φi(gj1)φi(gj2)φi(gj1gj2) first. Since φi are equivariant GH approximations, for any large i and any z ∈ B1/250(˜pi), φi(gj1)φi(gj2)(z) is close to φi(gj1gj2)(z). So hij (z) is close to z. Then ¯ hij is close to identity map 1 on B1/250(˜pi) and hij ∈ Gi. Then hij must be close ¯ to identity map 1 on B1/10(˜pi) by Corollary 3.4. −1 ¯ We claim that h hij h is close to 1 on B1/10(˜pi) for large i and all h ∈ Gi. Choose ¯ any ǫ > 0. We have shown that d(hij x, x) ≤ ǫ for large i and all x ∈ B1/10(˜pi). ′ ¯ Since h ∈ Gi, d(hp˜i, p˜i) ≤ 1/100 always holds. Then hx ∈ B1/10(˜pi) for all ′ ¯ ′ ¯ ′ ′ x ∈ B1/250(˜pi). Let x = hx ∈ B1/10(˜pi), we have d(hij hx , hx ) < ǫ for large i ′ ¯ −1 ¯ and all x ∈ B1/250(˜pi). In particular, h hij h is close to 1 on B1/250(˜pi) when ǫ is −1 ¯ small. Using corollary 3.4, h hij h is close to 1 on B1/10(˜pi) for all large i. l l sj Since φi(hj ) ∈ Gi, by induction we get kj 1 l l l sj −1 sj l −sj (Y φi(gj ) )(φi(gj1)φi(gj2)φi(gj1gj2) ) ( Y φi(gj) ) l=1 l=kj is close to 1 on B1/250(˜pi). Then φi(g) is close to 1 on B1/250(˜pi). Let i → ∞, we have φi(g) converges to the identity map by lemma 3.1. Generally, if [eg] = [eh] for some g,h ∈ G, then it follows from the same method that g = h. Now we construct a topology on Gˆ. Let K = {g′ ∈ G|d(g′p,˜ p˜) < 1/200} 11 and let {Uλ}λ∈Λ be the set of all open sets of G that is contained in K. We use left translations on Gˆ and the map i defined in Lemma 4.1 to define a basis on Gˆ; more precisely, {gˆ · i(Uλ)|gˆ ∈ G,ˆ λ ∈ Λ} generates a topology on Gˆ. ′ Lemma 4.4. For any g ∈ G and λ ∈ Λ, gUλ is a subset of G . Then {gUλ ∩ G|g ∈ G, λ ∈ Λ} generates the compact-open topology of G. Proof. Recall that the compact-open topology of G comes from G being a subset ¯ ¯ of C(B1/10(˜p), B1/3(˜p)). We first show that this is the same topology comes from ¯ ¯ G ⊂ C(B1/250(˜p), B1/3(˜p)). Since for locally compact metric spaces, compact- open topology is same as uniform convergence topology, it is enough prove that ¯ ¯ if gi ∈ G uniformly converges to g ∈ G on B1/250(˜p), then also on B1/10(˜p). By ′ ¯ precompactness, we may assume gi uniformly converges to some g on B1/10(˜p). It follows from Lemma 3.1 that g = g′. From now on, we will use the compact-open ¯ ¯ topology from C(B1/250(˜p), B1/3(˜p)). Next, we show that gUλ ∩ G is open in G for any g ∈ G and λ ∈ Λ. Regarding ¯ this Uλ, we may assume there exist a compact set C ⊂ B1/250(˜p) and an open set ¯ ′ ′ ′ V ⊂ B1/3(˜p) such that O = {g ∈ G |g C ⊂ V } and Uλ = K ∩ O. Recall that ′ ′ ′ d(g p,˜ p˜) < 1/200 for all g ∈ Uλ ⊂ K, we have g C ⊂ B1/100(˜p). So, we can assume V ⊂ B1/100(˜p) by replacing V by V ∩ B1/100(˜p). For any g ∈ G, gV is well-defined. Then ′ ′ ′ gUλ = g(K ∩ O)= {gg |g ∈ K ⊂ G, g C ⊂ V } = {g′′ ∈ G′|d(g′′p,g˜ p˜) < 1/200,g′′C ⊂ gV }, which is open in G′; here we use the fact that gg′ ∈ G′ if g,g′ ∈ G. Then we prove that any open set in G can be generated by {gUλ ∩ G|g ∈ G, λ ∈ Λ}. As above, we can assume that this open set has the form O = {g′ ∈ G′|g′C ⊂ ¯ ¯ V }, where C ⊂ B1/250(˜p) is compact and V ⊂ B1/3(˜p) is open. Let T = {g′ ∈ G′|d(g′p,˜ p˜) < 1/95}, which is an open set containing G. Choose O′ = O ∩ T , then O ∩ G = O′ ∩ G. It ′ suffices to show that O ∩ G can be generated by {gUλ ∩ G|g ∈ G, λ ∈ Λ}. Notice that gO′ ⊂ G′ for all g ∈ G. ′ ′ ′ ′ For any g ∈ O ⊂ T , g C ⊂ B1/40(˜p); therefore, we may assume V ⊂ B1/40(˜p) and g−1V is well-defined for all g−1 ∈ G. Then by the same argument above, g−1O′ = {g′′ ∈ G′|d(g′′p,g˜ −1p˜) < 1/95,g′′C ⊂ g−1V } ′ −1 ′ is open in G . By our choice of {Uλ|λ ∈ Λ}, g O ∩ K is in this family of open sets. It is direct to check ′ −1 ′ O ∩ G = ∪g∈G(g(g O ∩ K) ∩ G). ′ Hence O ∩ G can be generated by {gUλ ∩ G|g ∈ G, λ ∈ Λ}. Lemma 4.5. i : G → i(G) is a homeomorphism. Moreover, the image i(G) con- tains a neighborhood of 1 ∈ Gˆ. 12 Proof. We have proved injection in Lemma 4.1. Also, i : G → i(G) is an open map by the construction of topology on Gˆ. We show the continuity of i. For any −1 giˆ (Uλ), whereg ˆ ∈ gˆ and λ ∈ Λ, we shall show that i (ˆgi(Uλ)) is open in G. −1 −1 Assume i (ˆgi(Uλ)) is non-empty. Let g ∈ i (ˆgi(Uλ)). This means that there exists gλ ∈ Uλ such that [eg]=ˆg[egλ ] ∈ giˆ (Uλ) ∩ i(G). We need to find an open −1 neighborhood of g contained in i (ˆgi(Uλ)). Since Uλ is an open subset of K, −1 U := ggλ Uλ ∩ G is an open neighborhood of g in G by the same proof of Lemma 4.4. In Gˆ, we have − − i(U) ⊂ [eg][e 1 ]i(Uλ)=ˆg[egλ ][e 1 ]i(Uλ)=ˆgi(Uλ). gλ gλ −1 Thus U ⊂ i (ˆgi(Uλ)). Now we show that Gˆ is a Lie group. By Lemma 4.5, it suffices to show G contains no small groups, i.e., a neighborhood of the identity contained in G has no non-trivial group. Lemma 4.6. G contains no small groups. Proof. Suppose that G has a sequence of subgroups Hi such that DHi,1/10(˜p) → 0. Similar to the proof of Lemma 3.1, we shall find DHi(ǫ) ,r(ǫ)(x(ǫ)) = r(ǫ)/20 to derive a contradiction. By Lemma 3.1, there exists a regular point wi ∈ B1/100(˜p) such that h(wi) 6= wi for some h ∈ Hi. By passing to a subsequence, we may assume wi → w and s = (1/10 − d(w, p˜))/3 > 0. Consider U = Bs(w). For any x, y ∈ U, all geodesics from x to y are contained in B1/10−s(˜p). Let z ∈ (Rk)ǫ,2η be a regular point in U, assume η is small enough such that B2η(z) ⊂ U. Then for large i, we have DHi,η(z) ≤ η/20. Since wi is a regular point, there exists θ such that wi ∈ (Rk)ǫ,2θ. We may choose θ < d(h(wi), wi) for one h ∈ Hi with h(wi) 6= wi. Then we have DHi,θ(wi) ≥ θ/20. Let λ< min(θ, η). By the same proof of Lemma 3.1, we can find z such that DHi,λ(z(ǫ)) = λ/20, which will lead to a contradiction. 5. Construction of a Gˆ-space We define ˆ ¯ ˆ ¯ G ×G B1/250(˜p)= G × B1/250(˜p)/ ∼, where the equivalence relation is given by (ˆgi(g), x˜) ∼ (ˆg,gx˜) for all g ∈ G, gˆ ∈ ˆ ¯ ¯ ˆ ¯ G, x˜ ∈ B1/250(˜p) with gx˜ ∈ B1/250(˜p). We endow G ×G B1/250(˜p) with the quotient ˆ ¯ topology. We will always write [ˆg, x˜] as elements in G ×G B1/250(˜p) and (ˆg, x˜) as ˆ ¯ ′ ′ ˆ ˆ ¯ elements in G × B1/250(˜p). Byg ˆ [ˆg, x˜]=[ˆg g,ˆ x˜], G acts on G ×G B1/250(˜p). ˆ ˆ ¯ Lemma 5.1. G acts as homeomorphisms on G ×G B1/250(˜p). Proof. We fix any elementg ˆ′ ∈ Gˆ. The map ′ ˆ ¯ ˆ ¯ ′ gˆ : G ×G B1/250(˜p) → G ×G B1/250(˜p), [ˆg, x˜] 7→ gˆ [ˆg, x˜] is bijective. It suffices to show that this is an open map. Let U be an open subset ˆ ¯ ˆ ¯ ˆ ¯ of G ×G B1/250(˜p). Let π : G × B1/250(˜p) → G ×G B1/250(˜p) be the quotient map. Note that π commutes with Gˆ-action: π(ˆg′(ˆg, x˜)) = [ˆg′g,ˆ x˜]=ˆg′[ˆg, x˜]=ˆg′(π(ˆg, x˜)). 13 Thus π−1(ˆg′U)=ˆg′π−1(U). ′ −1 ˆ ¯ ′ Sinceg ˆ π (U) is open in G × B1/250(˜p), we conclude thatg ˆ U is open as well. The following lemma translates the study of Gˆ to G in certain cases. ′ ˆ ¯ ′ ¯ Lemma 5.2. Suppose [ˆg, x˜] = [1, x˜ ] for given gˆ ∈ G, x˜ ∈ B1/250(˜p), x˜ ∈ B1/250(˜p), ′ then gˆ = [eg] for some g ∈ G with g(˜x)=˜x . ′ ˆ ¯ Proof. By [ˆg, x˜] = [1, x˜ ] and the definition of G ×G B1/250(˜p), there exist k and gj ∈ G for 1 ≤ j ≤ k so that [ˆg, x˜] = [[eg1 eg2 ...egk ], x˜] = [[eg1 eg2 ...egk−1 ],gkx˜] = [[eg1 eg2 ...egk−2 ],gk−1gkx˜] = ... = [1,g1g2...gk−1gkx˜] ′ and g1g2gk−1gkx˜ =x ˜ . Here gk−1gkx˜ means gk−1(gk−1(˜x)). In particular, we have ¯ ¯ ¯ gkx˜ ∈ B1/250(˜p), gk−1gkx˜ ∈ B1/250(˜p),..., and g1g2...gk−1gkx˜ ∈ B1/250(˜p). Since ¯ gk−1gkx˜ ∈ B1/250(˜p), we conclude that gk−1gk ∈ G by Lemma 3.6. Using Lemma ′ 3.6 repeatedly, we see that g = g1g2...gk−1gk ∈ G with g(x)=˜x . ˆ ¯ Now we construct homeomorphisms between opens sets in G ×G B1/250(˜p) and ¯ the ones in B1/250(˜p). Lemma 5.3. For any x˜ ∈ B1/250(˜p), there exists a neighborhood U of [1, x˜] ∈ ˆ ¯ ′ G ×G B1/250(˜p) which is homeomorphic to a neighborhood U of x˜. Proof. Let ǫ =1/250 − d(˜x, p˜) > 0. Recall that we denote K = {g′ ∈ G|d(g′p,˜ p˜) < 1/200}. −1 2 Choose a small neighbourhood U1 of 1 in K with U1 = U1 and (U1) ⊂ K; let U2 ⊂ B1/250(˜p) be a neighborhoodx ˜. We can choose U1 and U2 small enough such ′ ′ that gg z ∈ Bǫ/3(˜x) for all g,g ∈ U1,z ∈ U2. This guarantees that any geodesic between two points in U1(U2)= {gz|g ∈ U1,z ∈ U2} is contained in B1/250(˜p). ˆ ¯ ˆ ¯ Let π : G × B1/250(˜p) → G ×G B1/250(˜p) be the quotient map and let U = π(i(U1) × U2). U is open since −1 −1 ¯ π (U)= [(i(U1)i(g) × (gU2 ∩ B1/250(˜p))) ˆ ¯ is union of some open sets in G × B1/250(˜p), where the union is taken over all g ∈ G ¯ with gU2 ∩ B1/250(˜p) 6= ∅. We define a natural map φ : U¯ = π(i(U¯1) × U¯2) → U¯1(U¯2), [i(g),z] 7→ gz, where U¯i means the completion of Ui. We shall show that φ is a homeomorphism. Then we can restrict φ on U to obtain the desired homeomorphism between U and ′ U = U1(U2). We first prove φ is well defined. Assume [i(g1),z1] = [i(g2),z2] for g1,g2 ∈ U¯1 ¯ and z1,z2 ∈ U2. Since U1(U2) ⊂ B1/250(˜p), we have [1,g1z1] = [1,g2z2]. By lemma 5.2, φ([i(g1),z1]) = g1z2 = g2z2 = φ([i(g2),z2]). 14 Next, we show that φ is bijective. It is clear that φ is surjective. We prove the ¯ ¯ ¯ injection of φ. Since U1(U2) ⊂ B1/250(˜p), if g1z1 = g2z2 ∈ B1/250(˜p) for g1,g2 ∈ U1 and z1,z2 ∈ U¯2, we have [i(g1),z1] = [1,g2z2] = [1,g1z1] = [i(g2),z2]. Hence φ is injective. −1 φ is continuous: Assume gizi → w ∈ U¯1(U¯2), where gi ∈ U¯1 and zi ∈ U¯2. We can write w = gz for some g ∈ U¯1 and z ∈ U¯2. By passing to a subsequence, we ′ ′ ′ ′ may assume gi → g ∈ U¯1 and zi → z ∈ U¯2. Then g z = w = gz. By the injection of φ, we see that [i(g),z] = [i(g′),z′]. Hence −1 ′ ′ −1 φ (gizi) = [i(gi),zi] → [i(g ),z ] = [i(g),z]= φ (w). −1 This shows the continuity of φ on U¯1(U¯2). φ is continuous: Assume [i(gi),zi] → [i(g),z] ∈ π(U¯1 × U¯2), where gi,g ∈ U¯1 ′ and zi,z ∈ U¯2. By passing to a subsequence, we can assume gi → g ∈ U¯1 and ′ ′ ′ zi → z ∈ U¯2. Then [i(gi),zi] → [i(g ),z ] = [i(g),z]. So ′ ′ ′ ′ φ([i(gi),zi]) = gizi → g z = φ([i(g ),z ]) = φ([i(g),z]). This proves the continuity of φ. In conclusion, φ is the desired homeomorphism. We need the following to apply Palais’s theorem. Lemma 5.4. For all [1, x˜] with d(˜x, p˜) < 1/250, let U be a neighborhood of [1, x˜] as described in Lemma 5.3. Then the set {gˆ ∈ Gˆ|gUˆ ∩ U 6= ∅} has compact closure in Gˆ. Proof. We write U as π(i(U1) × U2) as in the proof of Lemma 5.3. We claim that i : G → Gˆ provides a homeomorphism between {gˆ ∈ Gˆ|gUˆ ∩ U 6= ∅} and {g ∈ G|gU2(U1) ∩ U2(U1) 6= ∅} (see Lemma 4.5). The latter one has compact closure in G because G is compact; consequently, the set {gˆ ∈ Gˆ|gUˆ ∩ U 6= ∅} has compact closure in Gˆ as well. We verify the claim. By Lemma 4.5, it suffices to check that i maps one set exactly to the other. IfgU ˆ ∩U 6= ∅, then we can find g1,g2 ∈ U1 andx ˜1, x˜2 ∈ U2 such that [ˆgi(g1), x˜1] = [i(g2), x˜2]. So [ˆg,g1x˜1] = [1,g2x˜2]. By Lemma 5.2,ˆg = [eg]= i(g) for g ∈ G with gg1x˜1 = g2x˜2, that is g satisfies gU1(U2) ∩ U1(U2) 6= ∅. Conversely, if gU1(U2) ∩ U1(U2) 6= ∅, then we have g1,g2 ∈ U1 andx ˜1, x˜2 ∈ U2 such that gg1x˜1 = g2x˜2. Thus i(g)[i(g1), x˜] = [i(g2), x˜], which shows i(g)U ∩ U 6= ∅. ˆ ˆ Corollary 5.5. There is a G[1,x˜]-slice S at [1, x˜]. Proof. This follows directly from Lemma 5.4 and Theorem 2.5. Remark 5.6. We can assume that the slice Sˆ at [1, x˜] in Corollary is contained in U, the neighborhood of [1, x˜] described in Lemma 5.3. To see this, first note that ˆ ˆ by Lemma 5.2 that G[1,x˜] = i(Gx˜), so [1,Br(˜x)] is G[1,x˜]-invariant for any r. Then ˆ ˆ S ∩ [1,Br(˜x)] is still a G[1,x˜]-slice. By choosing r sufficiently small, we may assume Sˆ is contained in U. Remark 5.7. By the proofs of Lemmas 5.2 and 5.4, the homeomorphism φ is equi- variant, that is, compatible with the group actions: for anyg ˆ ∈ Gˆ and z ∈ U, 15 gzˆ ∈ U iffg ˆ = i(g) for some g ∈ G and g(φ(z)) ∈ U ′; moreover g(φ(z)) = φ(i(g)z). As a result, U/Gˆ is homeomorphic to U ′/G. Now we finish the proof of Theorem 1.3. ˆ ˆ ¯ ˆ Proof of Theorem 1.3. We define X = G ×G B1/250(˜p)/G with quotient topology. Letx ˜ ∈ B1/800(˜p). By choosing a smaller neighborhood, we may assume that ′ U = φ(U) in Lemma 5.3 is contained in B1/800(˜p) where φ is the homeomorphism constructed in Lemma 5.3. We will show that S = φ(Sˆ) is a Gx˜-slice atx ˜. Since ˆ ˆ φ([1, x˜]) =x ˜ and S is G[1,x˜]-invariant, we see thatx ˜ ∈ S and S is Gx˜-invariant. It remains to show that S/Gx˜ is homeomorphic to a neighborhood of x. By Remark ˆ ˆ 5.7, S/Gx˜ is homeomorphic to S/G[1,x˜]. The latter one is homeomorphic to a →֒ ˆneighborhood V of [[1, x˜]] ∈ Xˆ by the definition of a slice. Note that U/G Xˆ is homeomorphic onto its image which contains V . U/Gˆ is homeomorphic to ′ ′ ¯ U /G. U /G is homeomorphic to a neighborhood of x in B1/800(p) ⊂ B1/250(˜p)/G (also see Remark 3.7). Combining all these homeomorphisms, we get that S/Gx˜ is homeomorphic to a neighborhood of x. With Theorem 1.3, we finish the proof of Theorem 1.1, which follows the same idea as in Section 2.3. ^ Proof of Theorem 1.1. Let B1(˜pi) be the 1-ball in B4(pi) and let Gi = {g ∈ Γi|d(gp˜i, p˜i) ≤ 1/100}. As seen in Section 3, we can pass to a subsequence and obtain convergence GH GH (B¯1(˜pi), x˜i, Gi) −→ (B¯1(˜p), x,˜ G), (B¯1(pi), xi) −→ (B¯1(p), x), where x ∈ B1/800(p) andx ˜ ∈ B1/800(˜p) with π(˜x) = x. By Theorem 2.3 and the assumption that B1(˜pi) is non-collapsing, for any small ǫ > 0, there is δ > 0 such ′ that ρ(t, x˜) ≤ ǫ for all t ≤ δ ; in other words, any loop in Bδ′ (x) is contractible in Bǫ(x). By Theorem 1.3, there is a slice S ⊂ B1/800(˜p) atx ˜ such that S/Gx˜ is home- omorphic to a neighborhood of x. Using this homeomorphism and the same ar- gument as in Section 2.3, we can find δ > 0 such that Bδ(x) ⊂ B1/800(x) and −1 π (Bδ(x)) ∩ S ⊂ Bδ′ (˜x). For any loop γ in Bδ(x), we can make use of the home- −1 ′ omorphism to lift γ to a loopγ ˜ in π (Bδ(x)) ∩ S ⊂ Bδ(˜x). For thisγ ˜, there is a homotopy H, whose image is in Bǫ(˜x), that contractsγ ˜. Then π ◦ H is a homotopy that contracts γ with imH ⊂ Bǫ(x). This shows that ρ(t, x) ≤ ǫ for all t ≤ δ. Thus limt→0 ρ(t, x) = 0. 6. Proof of Theorem 1.4 We define Gr by the G-action on Y as below: Gr = h{g ∈ G | d(q,gq) ≤ r for some q ∈ Y }i. Note that by definition, Gr is normal in G for any r. We first prove the following gap lemma for Gr. One may compare this with [13] (see Lemma 2.4 and the 3rd paragraph in the proof of Theorem 9.1). Lemma 6.1. There exists a number δ > 0 such that Gr = Gδ for all r ∈ (0,δ]. δ Moreover, for this δ, G is the group generated by G0 and all isotropy subgroups of G. 16 We first prove the single point version. We define G(r, q)= h{g ∈ G | d(q,gq) ≤ r}i. Lemma 6.2. Let q ∈ Y . There exists a number δ = δ(q) > 0 such that G(r, q) = G(δ, q) for all r ∈ (0,δ]. Moreover, for this δ > 0, G(δ, q) is the group generated by G0 and Gq, the isotropy subgroup at q. Proof. By Theorem 2.1, G is a Lie group, thus each connected component of the orbit G · q is both open and closed in G · q. Let Cq be the component containing q. Pick δ > 0 sufficiently small such that Bδ(q) ∩ G · q ⊂ Cq. For any 0 < r ≤ δ, by our choice of δ, we get G(r, q) · q ⊂ Cq. It is also clear that G(r, q) · q is both open and closed in G · q. Therefore, we conclude that G(r, q) · q = Cq. Now we check that G(r, q) = G(δ, q). It is clear that G(r, q) ⊂ G(δ, q). On the other hand, given any g ∈ G(δ, q), then gq ∈ Cq. Since G(r, q) · q = Cq, we get gq = hq for some h ∈ G(r, q). Notice that h−1g fixes q, thus h−1g ∈ G(r, q). Consequently, g = h · (h−1g) ∈ G(r, q). For the second part of the Lemma, we denote Hq = hG0, Gqi. Since G0 is generated by elements in any small neighborhood of the identity element, we have Hq ⊂ G(δ, q). Conversely, for any element g ∈ G(δ, q), we have shown that gq ∈ G(r, q) · q = Cq = G0 · q. Then by the same argument is the former paragraph, g −1 −1 can be written as h · (h g) with h ∈ G0 and h g ∈ Gq. Now we prove Lemma 6.1: Proof of Lemma 6.1. We first prove that Gr stabilizes as r → 0. We argue by ri+1 contradiction. Suppose that there is ri → 0 such that G is a proper subgroup ri ri ri+1 of G for each i. We pick gi ∈ G − G with d(giqi, qi) ≤ ri for some qi ∈ Y . Let H be the group generated by G0 and all isotropy subgroups of G. It is clear r that H is a subgroup of G for any r > 0. Hence each gi is outside H. Let Ci be the connected component of G · qi that contains qi. Since gi cannot be generated by G0 and Gqi , giqi must be outside Ci. Let X be closure of a lift of X in Y and ′ ′ −1 let hi ∈ G such that qi = hiqi is a point in X ∩ (G · qi). Then gi = higihi satisfies ′ ′ ′ d(giqi, qi)= d(higiqi,hiqi)= d(giqi, qi) ≤ ri. r ′ ri ri+1 Recall that G is normal for each r, thus gi ∈ G − G . After passing to a ′ subsequence, qi converges to q ∈ X and gi converges to g ∈ G with gq = q. Let ′ δ(q) > 0 as in Lemma 6.2. By Lemma 6.2, gi ∈ G(δ(q), q)= G(r, q) for all r ≤ δ(q). ′ It follows that gi ∈ H for i sufficiently large. A contradiction. For the second half of the statement, the proof is similar to the one of Lemma 6.2. Let H be the group generated by G0 and all isotropy subgroups of G. It is clear that H is a subgroup of Gδ. Conversely, let g be a generator of Gδ. There is q ∈ Y with gq ⊂ Cq = G0 · q Therefore, g can be generated by elements in G0 and Gq. With Lemma 6.1 and [9, Theorem 3.10], we are ready to prove Theorem 1.4. Proof of Theorem 1.4. We first recall that it was shown in [8] that there is ǫ0 > 0 ǫ ǫ such that for all ǫ ≤ ǫ0, Y/H covers the universal cover of X, where H comes from ǫ GH ǫ (Mi, p,˜ Γi ) −→ (Y, p,H˜ ). f Hence Y/Hǫ is the universal cover of X. 17 Let ǫ0 be the constant above and δ be the constant in Lemma 6.1. Put η = η η min{ǫ0,δ}/2. Since G contains G0, G/G is discrete. Because X = Y/G is com- pact, we can apply [9, Theorem 3.10] to find a sequence of subgroups Ki converging η to G . According to [9], Ki can be constructed as follows. Let (fi, φi, ψi) represent an ǫi pointed equivariant Gromov-Hausdorff approximation (Mi, p˜i, Γi) → (Y, p,˜ G) f for some ǫi → 0. Then we have a map φi : Si(1/ǫi) → S∞(1/ǫi), where Si(r)= {g ∈ Γi | d(˜pi,gp˜i) ≤ r}. Let R be a sufficiently large number compared with 3D, then Ki is generated by η {g ∈ Si(R) | φi(g) ∈ G }. η η We claim that Ki =Γi for all i large. Let gi be a generator of Γi , that is, gi has η+ǫi η displacement less than η at certain point. Consequently, φi(gi) ∈ G = G . The η last equality follows from our choice of η and Lemma 6.1. This shows that Γi ⊂ Ki. η Conversely, suppose that there is a sequence gi ∈ Si(R) with φi(gi) ∈ G . Since η η/4 li η/4 G = G , we can write φi(gi) = Qj=1 gi,j , where gi,j are generators of G . Here we can assume that li is bounded by a constant number depending on R but η not i because G is generated by G0 and isotropy subgroups. Let hi,j ∈ Γi that is η/4 η/2 li the element closest to gi,j ∈ G , then hi,j ∈ Γi for i large. Note that Qj=1 hi,j li η and Qj=1 gi,j share the same limit in G as i → ∞. We conclude that gi ∈ Γi for η all i large. This proves the claim that Γi = Ki. η GH η So far, we have shown that (Mi, p˜i, Γi ) −→ (Y, p,˜ G ). Together with the second part of Lemma 6.1 and the resultf in [8] as explained above, we finish the proof. 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