Margulis Lemma

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22 VITALI KAPOVITCH AND BURKHARD WILKING We will call elements of G 4 (p i ) ”good points” in B 4 (p i ). We put G r (q) = B r (q) ∩ G 4 (p i ) for q ∈ B 2 (p i ) and r ≤ 2. It is immediate from the weak 1-1 inequality (Lemma 1.4) that vol(G 1(p i)) vol(B 1(p i)) → 1. By the Product Lemma (2.1), for any q i ∈ G 4 (p i ) and any h ≥ 1 we have that hB 1/h (q i ) is ˜ε i close to a ball in a product R k × Y where ˜ε i → 0 can be chosen independently of h and q i . Note that the space Y may depend on h and q i . For any point q i ∈ G 1 (p i ) let ρ i (q i ) be the largest number ρ < 1 such that the map b i : B ρ (q i ) → R k has distance distortion exactly equal to ρ · 10 −n2 . For any choice q i ∈ G 1 (p i ) we know that ρ i (q i ) → 0 and by the Product Lemma (2.1) ( 1 ρ M ) i(q i) i, q i subconverges to R k × K where K is compact with diam(K) = 10 −n2 . Again note, that a priori, the constants ρ i (q i ) and the space K depend on the choice of q i ∈ B 1 (p i ) ′ . We will show that, in fact, both are essentially independent of the choice of good points in B 1 (p i ). For each i we define the number ρ i as the supremum of ρ i (q i ) with q i ∈ G 1 (p i ) and put λ i = 1 ρ i . The statement of the theorem will follow easily from the following sublemma. Sublemma 5.2. There exist constants ¯C 1 , ¯C 2 independent of i such that the following holds. For any good point p ∈ G 1 (p i ) and any r ≤ 2 there exists B r (p) ′ ⊂ B r (p) such that (11) vol B r (p) ′ ≥ (1 − ¯C 1 rε i ) vol B r (p) and such that for any q ∈ B r (p) ′ there exists p ′ ∈ B Lρi (p) (with L = 9 n ) and a time dependent (piecewise constant in time) divergence free vector field X t (dependence on q is suppressed) and its integral curve c q (t): [0, 1] → B 2 (p i ) such that c(0) = p ′ , c(1) = q and 1 vol(B r(p)) ∫B r(p) ′ ∫ 1 0 ( Mx(‖∇·X t ‖ 3/2 ) ) 2/3 (cq (t)) dt dµ(q) ≤ ¯C 2 rε i . Proof. Despite the fact that L can be chosen as L = 9 n we treat it for now as a large constant L ≥ 10 which needs to be determined. Throughout the proof we will denote by C j various constants depending on n. Sometimes these constants will also depend on L in which case that dependence will always be explicitly stated. Although we consider a fixed i for the major part of the proof, we take the liberty of assuming that i is large. This way we can ensure that for any r ≥ ρ i 1 and any good point p ∈ G 1 (p i ) the ball Lr B 2Lr(p) is by the Product Lemma in the measured Gromov–Hausdorff sense arbitrarily close to a ball in a product R k × K where by our choice of ρ i , K has diameter ≤ 10 −n2 . This implies, for example, that we can assume that ( 1 (12) r1 ) k vol(B 2 r 2 ≤ r1 (q 1)) vol(B r2 (q ≤ 2( ) r 1 k 2)) r 2 for all q1 , q 2 ∈ B Lr (p), r 1 , r 2 ∈ [r/20, 2Lr]. The sublemma is trivially true for r ≤ Lρ i with X ≡ 0. By induction it suffices to show that if the sublemma holds for some r ∈ (ρ i , 2 L ] then it holds for Lr.
STRUCTURE OF FUNDAMENTAL GROUPS 23 Let q 1 , . . . q l be a maximal r/2-separated net in B Lr (p). Note that l⋃ B Lr (p) ⊂ B r/2 (q m ). m=1 By a standard volume argument we can deduce from (12) that (13) ∑ l m=1 vol B r/2(q m ) ≤ 3 k+1 ≤ 3 n . vol B Lr (p) Fix q m and consider the following vector field k∑ X(x) = (b i α(p) − b i α(q m ))∇b i α(x). α=1 Since b i α are Lipschitz with a universal Lipschitz constant by (1), X satisfies (14) |X(x)| ≤ C(n) · Lr. Also, by construction, X satisfies (15) Mx ‖∇·X‖ 2 (p) ≤ C 4 ε 2 i L 2 r 2 . Note that we get an extra L 2 r 2 factor as compared to (10) because |b i (p)−b i (q m )| ≤ C(n)Lr. By applying (6) we get ( Mx [Mx(‖∇·X‖ 3/2 )] 4/3) (p) ≤ C(n) Mx(‖∇·X‖ 2 )(p) ≤ (C 5 Lε i r) 2 . In particular, for R 1 = 2C(n)Lr ∫ − [Mx(‖∇·X‖ 3/2 )] 4/3 (p) ≤ (C 5 Lε i r) 2 B R1 (p) provided R 1 ≤ 1. However, in the case of R 1 ∈ [1, 4C(n)] the same follows more directly from our initial assumptions combined with Lemma 1.4 for all large i. By Cauchy inequality the last estimate gives ∫ ( (16) − Mx(‖∇·X‖ 3/2 ) ) 2/3 (p) ≤ C5 Lε i r. B R1 (p) Consider the measure preserving flow φ of X on [0, 1]. Because of (14) the flow lines φ t (q) with q ∈ B r (p) stay in the ball B R1 (p) for t ∈ [0, 1]. We choose a universal constant C 6 (L) with C 5 L · vol(B 2C(n)Lr(p)) vol(B r/10 (q m) ≤ C 6 (L) Combining that the flow is measure preserving, inequality (16) and our choice of C 6 we see that (17) ∫ 1 vol(B r (q 10 m)) B r(q m) ∫ 1 0 ( Mx(‖∇·X‖ 3/2 ) ) 2/3 (φt (x)) dt dµ(x) ≤ C 6 (L)rε i . By the Product Lemma (2.1) applied to the rescaled balls 1 Lr B Lr(p) we know that 1 Lr B Lr(p) is measured Gromov–Hausdorff close to a unit ball in R k ×K 3 where diam K 3 ≤ 10 −n2 . Moreover, φ 1 is measured close to a translation by b i (p)−b i (q m ) in R k , see proof of Proposition 3.8. Since we only need to argue for large i, we may
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Margulis Lemma

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