Margulis Lemma

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10 VITALI KAPOVITCH AND BURKHARD WILKING Lemma 2.2. Suppose a sequence of complete inner metric spaces (Y i , p i ) converges to a locally compact metric space (Y ∞ , p ∞ ). Suppose G i acts isometrically on Y i and that the action of G i converges to an action of a closed subgroup G ∞ ⊂ Iso(Y ∞ ). Let G i (ε) denote the subgroup generated by those elements that displace p i by at most ε, i ∈ N ∪ {∞}. Suppose G ∞ (ε) = G ∞ ( a+b 2 Then there is some sequence ε i (a + ε i , b − ε i ). ) for all ε ∈ (a, b) and 0 ≤ a < b. ) for all ε ∈ → 0 such that G i (ε) = G i ( a+b 2 Proof. Suppose on the contrary we can find g i ∈ G i (ε 2 ) \ G i (ε 1 ) for fixed ε 1 < ε 2 ∈ (a, b). Without loss of generality d(p i , g i p i ) ≤ ε 2 . Because of g i ∉ G i (ε 1 ) it follows that for any finite sequence of points p i = x 1 , . . . , x h = g i p i ∈ G i ⋆ p i there is one j with d(x j , x j+1 ) ≥ ε 1 . Clearly this property carries over to the limit and implies that g ∞ ∈ G ∞ (ε 2 ) is not contained in G((ε 1 + a)/2) – a contradiction. □ Lemma 2.3. Suppose (Mi n, q i) converges to (R k × K, q ∞ ) where Ric Mi ≥ −1/i and K is compact. Assume the action of π 1 (M i ) on the universal cover ( ˜M i , ˜q i ) converges to a limit action of a group G on some limit space (Y, ˜q ∞ ). Then G(r) = G(r ′ ) for all r, r ′ > 2 diam(K). Proof. Since Y/G is isometric to R k ×K, it follows that there is a submetry σ : Y → R k . It is immediate from the splitting theorem that this submetry has to be linear, that is, for any geodesic c in Y the curve σ◦c is affine linear. Hence we get a splitting Y = R k × Z such that G acts trivially on R k and on Z with compact quotient K. We may think of ˜q ∞ as a point in Z. For g ∈ G consider a mid point x ∈ Z of ˜q ∞ and g˜q ∞ . Because Z/G = K we can find g 2 ∈ G with d(g 2˜q ∞ , x) ≤ diam(K). Clearly d(˜q ∞ , g 2˜q ∞ ) ≤ 1 2 d(˜q ∞, g˜q ∞ ) + diam(K) d(˜q ∞ , g −1 2 g˜q ∞) = d(g 2˜q ∞ , g˜q ∞ ) ≤ 1 2 d(˜q ∞, g˜q ∞ ) + diam(K). This proves G(r) ⊂ G ( r/2 + diam(K) ) and the lemma follows. Lemma 2.4 (Gap Lemma). Suppose we have a sequence of manifolds (M i , p i ) with a lower Ricci curvature bound converging to some limit space (X, p ∞ ) and suppose that the limit point p ∞ is regular. Then there is a sequence ε i → 0 and a number δ > 0 such that the following holds. If γ 1 , . . . , γ li is a short basis of π 1 (M i , p i ) then either |γ j | ≥ δ or |γ j | < ε i . Moreover, if the action of π 1 (M i ) on the universal cover ( ˜M i , ˜p i ) converges to an action of the limit group G on (Y, ˜p ∞ ), then the orbit G ⋆ ˜p ∞ is locally path connected. Proof. That the orbit is locally path connected is a consequence of the following Claim. There is a δ > 0 such that for all points x ∈ G ⋆ ˜p ∞ with d(˜p ∞ , p) < δ we can find y ∈ G ⋆ ˜p ∞ with max{d(˜p ∞ , y), d(y, x)} ≤ 0.51 · d(˜p ∞ , x). To prove the claim we argue by contradiction and assume that for some δ h > 0 converging to 0 we can find g h ∈ G with d(g h ˜p ∞ , ˜p ∞ ) = δ h such that for all a ∈ G with d(a˜p ∞ , ˜p ∞ ) ≤ 0.51δ h we have d(a˜p ∞ , g˜p ∞ ) > 0.51δ h . After passing to a subsequence we can assume that ( 1 δ h Y, ˜p ∞ ) converges to a tangent cone (C p∞ Y, o) and that the action of G on 1 δ h Y converges to an action of □
STRUCTURE OF FUNDAMENTAL GROUPS 11 some group K on C p∞ Y . Let g ∞ ∈ K be the limit of g h . Clearly d(o, g ∞ o) = 1 and for all a ∈ K with d(o, ao) ≤ 0.51 we have d(go, ao) ≥ 0.51. In particular the orbit K ⋆ o is not convex. Because X = Y/G the quotient C˜p∞ Y/K is isometric to lim h→∞ ( 1 ε h X, p ∞ ) which by assumption is isometric to some Euclidean space R k . As in the proof of Lemma 2.3 it follows that C˜p∞ Y is isometric to R k × Z and the orbits of the action of K are given by v × Z where v ∈ R k . In particular, the K orbits are convex – a contradiction. Clearly, the claim implies that G ⋆ ˜p ∞ ∩ B r (˜p ∞ ) is path connected for r < δ. In fact, it is now easy to construct a Hölder continuous path from ˜p ∞ to any point in G ⋆ ˜p ∞ ∩ B r (˜p ∞ ). Of course, the claim also implies G(ε) = G(δ) for all ε ∈ (0, δ]. Therefore the first part of Lemma 2.4 follows from Lemma 2.2. □ Theorem 3 will follow from the following slightly more general result. Theorem 2.5. Given n and R there is a constant C such that the following holds. Suppose (M, g) is a manifold, p ∈ M, B 2R (p) is compact and Ric > −(n − 1) on B 2R (p). Furthermore, we assume that π 1 (M, g) is generated by loops of length ≤ R. Then π 1 (M, p) can be generated by C loops of length ≤ R. Moreover, there is a point q ∈ B R/2 (p) such that any Gromov short generator system of π 1 (M, q) has at most C elements. Proof. It is clear that it suffices to show that for a point q ∈ B R/2 (p) the number of elements in a Gromov short generator system of π 1 (M, q) is bounded above by some a priori constant. We consider a Gromov short generator system γ 1 , . . . , γ k of π 1 (M, q). With |γ i | ≤ |γ i+1 | for some q ∈ B R/2 (p). Clearly |γ i | ≤ 2R and it easily follows from Bishop–Gromov relative volume comparison that there are effective a priori bounds (depending only on n, R and r) for the number of short generators with length ≥ r. We will argue by contradiction. It will be convenient to modify the assumption on π 1 being boundedly generated. We call (M i , g i ) a contradicting sequence if the following holds • B 3 (p i ) is compact and Ric > −(n − 1) on B 3 (p i ). • For all q i ∈ B 1 (p i ) the number of short generators of π 1 (M i , q i ) of length ≤ 4 is larger than 2 i . Clearly it suffices to rule out the existence of a contradicting sequence. We may assume that (B 3 (p i ), p i ) converges to some limit space (X, p ∞ ). We put dim(X) = max { k | there is a regular p ∈ B 1/4 (p ∞ ) with C x X ∼ = R k} . We argue by reverse induction on dim(X). We we start our induction at dim(X) ≥ n + 1. It is well known that this can not happen so there is nothing to prove. The induction step is subdivided in two substeps. Step 1. For any contradicting sequence (M i , p i ) converging to (X, p ∞ ) there is a new contradicting sequence converging to (R dim(X) , 0). Choose q i ∈ B 1/4 (p i ) converging to some point q ∞ ∈ B 1/4 (p ∞ ) with C q∞ X =
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