Margulis Lemma

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30 VITALI KAPOVITCH AND BURKHARD WILKING Thus, the normalizer of Υ i has finite index ≤ [G : G ′ ] in Γ i for all large i and Step 4 is established. Step 5. The desired contradiction arises if, after passing to a subsequence, the indices [Γ i : Γ iε ] ∈ N ∪ {∞} converge to ∞. By Step 4 there is a subgroup Υ i ⊂ Γ iε of index ≤ H which is normalized by a subgroup of Γ i of index ≤ H. Similarly to the beginning of the proof of Step 3, one can reduce the situation to the case of Υ i ⊳ Γ i . Moreover, Υ i ⊳ Γ i is uniformly open and hence the action of Γ i /Υ i on M i /Υ i is uniformly discrete and converges to a properly discontinuous action of the virtually abelian group G/Υ ∞ on the space R k × (R l × ˜K)/Υ ∞ . It is now easy to see that Γ iε /Υ i contains an abelian subgroup of controlled finite index for large i. After replacing Γ i once more by a subgroup of controlled finite index we may assume that Γ i /Υ i is abelian. This also shows that without loss of generality Υ i = Γ iε . Recall that by Step 2 we have assumed that f j i converges to the identity in the weakly measured sense for every j = 1, . . . , k. Therefore, the same is true for the commutator [f j i , γ i] ∈ Γ i if γ i ∈ Γ i has bounded displacement. This in turn implies that [f j i , γ i] ∈ Γ iε for all large i and j = 1, . . . , k. By Theorem 2.5, we can find for some large σ elements d i 1, . . . , d i σ ∈ Γ i with bounded displacement generating Γ i . We also assume that the first τ elements generate Γ iε for some τ < σ. Let ˆΓ i := 〈d i 1, . . . , d i σ−1〉. Note that ˆΓ i ⊳ Γ i since ˆΓ i contains Γ iε = Υ i ⊳ Γ i and Γ i /Γ iε is abelian. If for some subsequence and some integer H the group ˆΓ i has index ≤ H in Γ i then we may replace Γ i by ˆΓ i . Thus, without loss of generality, the order of the cyclic group Γ i /ˆΓ i tends to infinity. The element f k+1 i := d i σ ∈ Γ i represents a generator of this factor group which has bounded displacement. We have seen above that f j i normalizes ˆΓ i and [f k+1 i , f j i ] ∈ Γ iε ⊂ ˆΓ i , j = 1, . . . , k. Clearly f k+1 i , being an isometry with bounded displacement, has the zooming in property. We replace M i by B i := ˜M i /ˆΓ i . Notice that (B i , ˆp i ) converges to (R k × (R l × ˜K)/Ĝ, ˆp ∞) where Ĝ is the limit group of ˆΓi . Since G/Ĝ is a noncompact group acting cocompactly on (R l × ˜K)/Ĝ, we see that (Rl × ˜K)/Ĝ splits as Rp × K ′ with K ′ compact and p > 0. If p > 1, we put f j i = id for j = k + 2, . . . , k + p. Thus the induction assumption applies: There is a positive integer C such that the following holds for all large i: We can find a subgroup N i of π 1 (B i ) of index ≤ C and a nilpotent chain of normal subgroups {e} = N 0 ⊳ . . . ⊳ N n−k−1 = N i such that the quotients are cyclic. Moreover, the groups are invariant under the automorphism induced by conjugation of (f j i )C! and the induced automorphism of N h+1 /N h is the identity, j = 1, . . . , k+1. We put d = C!, and consider the group ¯N i ⊂ π 1 (M i ) generated by (f k+1 i ) d and N i . Clearly, the index of ¯N i in π 1 (M i ) is bounded by C · d. Moreover, the chain {e} = N 0 ⊳ . . . ⊳ N n−k−1 ⊳ N n−k := ¯N i is normalized by the elements (f j i )d (j = 1, . . . , k) and the action on the cyclic quotients is trivial. In other words, the sequence fulfills the conclusion of our theorem – a contradiction. □
STRUCTURE OF FUNDAMENTAL GROUPS 31 Remark 6.2. If one was only interested in proving that the fundamental group contains a polycylic subgroup of controlled index, the proof would simplify considerably. First of all, one would not need any diffeomorphisms f j i to make the induction work. In the proof of Step 1 the use of the rescaling theorem could be replaced with the more elementary Product Lemma 2.1. Step 2 would become unnecessary. The core of the remaining arguments would be the same although they would simplify somewhat. 7. Margulis Lemma Proof of Theorem 1. Let B 1 (p) be a metric ball in complete n-manifold with Ric > −(n − 1) on B 1 (p), Ñ the universal cover of B 1 (p) and let ˜p ∈ Ñ be a lift of p. Step 1. There are universal positive constants ε 1 (n) and C 1 (n) such that the group 〈 {g Γ := ∈ π1 (B 1 (p), p) ∣ d(q, gq) ≤ ε1 (n) for all q ∈ B 1/2 (˜p) }〉 has a subgroup of index ≤ C 1 (n) which has a nilpotent basis of length at most n. Put N = Ñ/Γ and let ˆp denote the image of ˜p. By the definition of Γ, for each point q ∈ B 1/2 (ˆp), the fundamental group π 1 (N, q) is generated by loops of length ≤ ε 1 (n). Moreover, B 3/4 (ˆp) is compact. Assume, on the contrary, that the statement is false. Then we can find a sequence of pointed n-dimensional manifolds (N i , p i ) satisfying • Ric Ni ≥ −(n − 1), • B 3/4 (p i ) is compact, • for each point q ∈ B 1/2 (p i ) the fundamental group π 1 (N i , q) is generated by loops of length ≤ 2 −i , and • π 1 (N i , p i ) does not contain a subgroup of index ≤ 2 i which has a nilpotent basis of length ≤ n. After passing to a subsequence we can assume that (N i , p i ) converges to (X, p ∞ ). We choose q i ∈ B 1/4 (p i ) such that q i converges to a regular point q ∈ X. Choose λ i → ∞ very slowly such that π 1 (λ i N i , q i ) is still generated by loops of length ≤ 1 and such that (λ i N i , q i ) converges to C q X ∼ = R k for some k ≥ 0. Notice that the rescaled sequence M i = λ i N i satisfies Ric Mi ≥ − n−1 → 0 λ 2 i and B ri (q i ) is compact with r i = λi 2 → ∞. Thus the existence of this sequence contradicts the Induction Theorem 6.1 with fi 1 = · · · = f i k = id. We can now finish the proof of the theorem by establishing. Step 2. Consider ε 1 (n) > 0 and Γ from Step 1. Then there are ε 2 (n), C 2 (n) > 0 such that the group 〈 {g H := ∈ π1 (B 1 (p), p) | d(˜p, g˜p) ≤ ε 2 (n) }〉 satisfies: Γ ∩ H has index at most C 2 (n) in H. We will provide (in principle) effective bounds on ε 2 and C 2 depending on n and the (ineffective) bound ε 1 (n). Put Γ ′ := 〈 {g ∈ π1 (B 1 (p), p) | d(q, gq) ≤ ε 1 (n) for all q ∈ B 3/4 (˜p) }〉 ⊂ Γ.
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Margulis Lemma

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