Margulis Lemma

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14 VITALI KAPOVITCH AND BURKHARD WILKING holds for all small r. Hence it suffices to prove that if it holds for r ≤ 1/10, then it holds for 10r. By assumption ∫ − dt fi 10r (p, q)dµ(p)dµ(q) ≤ 10rε i. B 10r(x)×B 10r(x) Furthermore, as long as C 1 ε i ≤ 1/2 our induction assumption implies that there is a subset S ⊂ B r (x) with vol(S) ≥ 1 2 vol(B r(x)) and f i (S) ⊂ B 2r (f i (x)). By Bishop–Gromov ∫ − dt fi 10r (p, q)dµ(p)dµ(q) ≤ C 2(n)rε i S×B 10r(x) with some universal constant C 2 (n). Therefore there is a subset B 10r (x) ′′ ⊂ B 10r (x) with vol(B 10r (x)) ′′ ≥ (1 − C 2 (n)ε i /2) vol(B 10r (x)) and ∫ − dt fi 10r (p, q)dµ(p) ≤ 2r for all q ∈ B 10r(x) ′′ . S Using that f i (S) ⊂ B 2r (f i (x)) this clearly implies that f i (B 10r (x) ′′ ) ⊂ B 20r (f i (x)). Thus the sublemma is valid if we put C 1 (n) = C 2 (n)/2 provided we know in addition that C 1 (n)ε i ≤ 1/2. We can remove the upper bound on ε i by just putting C 1 (n) = C 2 (n). □ Of course, a similar inequality holds for f −1 i . Proof of Lemma 3.2. Clearly, we can finish the proof of the lemma by establishing Claim. Given δ ∈ (0, 1/10) there exists i 0 such that for all x i , y i ∈ B ri (p 1 i )′ with d(x i , y i ) < 1/2 we have |d(f i (x i ), f i (x i )) − d(x i , y i )| ≤ 10δ for all i ≥ i 0 (δ). Consider subsets B δ (x i ) ′′ and B δ (y i ) ′′ as in the sublemma. Then vol(B δ (x i ) ′′ ) ≥ 1 2 vol(B δ(x i )) for large i and combining with Bishop–Gromov gives vol(B 1 (x i )) 2 ≤ C 2 (n, δ) vol(B δ (x i ) ′′ ) vol(B δ (y i ) ′′ ). Thus, ∫ ∫ − dt fi 1 (p, q) ≤ C 2− dt fi 1 (p, q) ≤ C 2ε i . B δ (x i) ′′ ×B δ (y i) ′′ B 1(x i) 2 Choose i 0 so large that C 2 ε i ≤ δ for i ≥ i 0 . Then for such i we can find x ′ i ∈ B δ(x i ) ′′ and y i ′ ∈ B δ(y i ) ′′ with dt fi 1 (x′ i , y′ i ) ≤ δ. Combining with d(f i(x ′ i ), f i(x i )) ≤ 2δ and d(f i (y i ′), f i(y i )) ≤ 2δ we deduce dt fi 1 (x i, y i ) ≤ 7δ as claimed. □ Lemma 3.4. Consider (M i , p 1 i ), (N i, p 2 i ), (P i, p 3 i ) like above and two sequences of diffeomorphisms f i : M i → N i and g i : N i → P i with the zooming in property. Then f i ◦ g i also has the zooming in property. Proof. By Sublemma 3.3 there is an i 0 such that for all q i ∈ B ri (p) and r ≤ 1/2 there is a subset B r (q i ) ′′ ⊂ B r (q i ) ′ with vol(B r (q i ) ′′ ) ≥ (1−C 1 ε i ) and f i (B r (q i ) ′′ ) ⊂ B 2r (q i ). Thus, ∫ ∫ − dt fi◦gi r (a, b) ≤ C 2 ε i r + e 2nCεi − dt gi r (f i (a), f i (b)) B r(q) 2 (B r(q) ′′ ) ∫ 2 ≤ C 2 ε i r + C 3 − r (x, y) ≤ C 4 ε i , B 2r(f i(q)) 2 dt gi
STRUCTURE OF FUNDAMENTAL GROUPS 15 where we used that the differential of f i at q ∈ B r (q i ) ′ has a bilipschitz constant e Cεi and that the ratio of vol(B 2r (f i (q))) and vol(B r (q)) is for large i bounded by a universal constant. We are left with the case of r ∈ [1/2, 1]. By Lemma 3.2, f i and g i converge in the weakly measured sense to isometries. Clearly this implies that f i ◦ g i also converges in the weakly measured sense to an isometry. And this implies that the distortion function dt fi◦gi 1 has averaged L 1 norm ≤ δ i on every unit ball in B ρi (p i ) for some ρ i → ∞ and δ i → 0. □ The next lemma explains the notion zooming in property. Lemma 3.5. Let (M i , p 1 i ), (N i, p 2 i ) and f i be as above. Then there is a ρ i → ∞, δ i → 0 and Ti 1 ⊂ B ρi (p 1 i ) such that the following holds • vol(B 1 (q) ∩ Ti 1) ≥ (1 − δ i) vol(B 1 (q)) for all q ∈ B ρi/2(p 1 i ). • For any sequence of real numbers λ i → ∞ and any sequence q i ∈ Ti 1 f i : (λ i M i , q i ) → (λ i N i , f i (q i )) has the zooming in property. We say that f i is good on all scales at q i . Proof. Let G j i = B 2r i (p j i )′ and B j i = B 2r i (p i ) \ B 2ri (p j i )′ . After adjusting r i → ∞ and ε i → 0 we may assume vol(B j i ) ≤ ε i vol(B 1 (q)) for all q ∈ B 2ri (p j i ). Let χ ij be the characteristic function of B j i . By the weak 1-1 inequality there exists a universal C such that the set H j i := { x ∈ B ri/2(p j i ) | Mx(χ ij) ≥ √ } ε i satisfies vol(H j i ) ≤ C√ ε i vol(B 1 (q)) for all q ∈ B 2ri (p j i ). We put T i 1 := ( ) ( ) B ri/2(p 1 i )\H1 i ∩ f −1 i Bri/2(p 2 i )\H2 i and Ti 2 := f i (Ti 1). Using Lemma 3.2 we can find ρ i → ∞ and δ i → 0 such that vol ( B ρi (p j i ) \ T j ) i ≤ δi vol(B 1 (q)) for all q ∈ B ri (p j i ), j = 1, 2. By definition of T j i vol(B r (q) ∩ G j i ) vol(B r (q)) ≥ 1 − √ ε i for all q ∈ T j i and all r ≤ 1, j = 1, 2. Let dt λif r denote the distortion on scale r of f i : λ i M i → λ i N i . Clearly dt λif r (p, q) = λ i dt fi r/λ i (p, q). Thus for all λ i → ∞ and all q i ∈ T j i the zooming in property. the map f i : (λ i M i , q i ) → (λ i N i , f i (q i )) has □ Proposition 3.6 (First main example). Let α > 1. Consider a sequence of manifolds (M i , p i ) with a fixed lower Ricci curvature bound and a sequence of time dependent vector fields Xi t (piecewise constant in time) with compact support. Let c i : [0, 1] → B ri (p i ) be an integral curve of Xi t with c i(0) = p i
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Margulis Lemma

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