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Margulis Lemma

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24 VITALI KAPOVITCH AND BURKHARD WILKING<br />

assume that by volume 3/4 of the points in B r/10 (q m ) are mapped by φ 1 to points<br />

in B r/9 (p).<br />

We choose such a point q ∈ B r/10 (q m ). In view of (17) we may assume in<br />

addition that<br />

∫ 1<br />

By <strong>Lemma</strong> 3.7 this implies<br />

0<br />

Mx(‖∇·X‖ 3/2 ) 2/3 (φ t (q)) dt ≤ 4 3 C 6(L)rε i .<br />

∫<br />

(18) − dt r (1)(x, y) dµ(x) dµ(y) ≤ C 7 (L)r · ε i .<br />

B r(q)×B r(q)<br />

Note that by definition, dt r (1)(x, y) ≥ min{r, |d(φ 1 (x), φ 1 (y)−d(x, y)|}. Combining<br />

this inequality with the knowledge that 3/4 of the points in B r/10 (q m ) end up in<br />

B 1/9r (p), implies that we can find a subset B r/2 (q m ) ′ ⊂ B r/2 (q m ) with<br />

(19)<br />

(20)<br />

vol(B r/2 (q m ) ′ ) ≥ (1 − C 8 (L)ε i r) vol(B r/2 (q m )) and<br />

φ 1 (B r/2 (q m ) ′ ) ⊂ B r (p).<br />

Set B r/2 (q m ) ′′ := B r/2 (q m ) ′ ∩ φ −1<br />

1 (B r(p) ′ ). Then we get<br />

φ 1 (B r/2 (q m ) ′′ ) ⊂ B r (p) ′<br />

and<br />

vol(B r/2 (q m ) ′′ ) ≥ vol(B r/2 (q m ) ′ ) − (vol B r (p) − vol B r (p) ′ )<br />

(21)<br />

by (19) and (11)<br />

≥ (1 − C 8 (L)rε i ) vol B r/2 (q m ) − ¯C 1 rε i vol B r (p)<br />

by (12)<br />

≥ (1 − C 8 (L)rε i ) vol B r/2 (q m ) − 2 k+1 ¯C1 rε i vol B r/2 (q m )<br />

≥ (1 − (C 8 (L) + 2 n ¯C1 )rε i ) vol B r/2 (q m ),<br />

where we used that φ 1 is volume-preserving in the first inequality.<br />

Using the induction assumption, we can prolong each integral curve from any<br />

q ∈ B r/2 (q m ) ′′ to a point x ∈ B r (p) ′ by extending it by a previously constructed<br />

integral curve from x to p ′ ∈ B Lρi (p) of a vector field Xold t (which depends on p′ ).<br />

We set Xnew t = 2Xold 2t for 0 ≤ t ≤ 1/2 and Xt new = −2X for 1/2 ≤ t ≤ 1. Let c q (t)<br />

be the integral curve of Xnew t with c q (1) = q and c q (0) = p ′ .<br />

By the induction assumption and (17), we get<br />

(22)<br />

∫B r/2 (q m) ′′ ∫ 1<br />

0<br />

(Mx ‖∇·X t new‖ 3/2 ) 2/3 (c q (t)) dt dµ(q) ≤<br />

≤ C 6 (L)(rε i ) vol B r/2 (q m ) + ¯C 2 (rε i ) vol B r (p)<br />

(12)<br />

≤ C 6 (L)(rε i ) vol B r/2 (q m ) + 2 n ¯C2 (rε i ) vol B r/2 (q m )<br />

= ( C 9 (L) + 2 n ¯C2<br />

)<br />

(rεi ) vol B r/2 (q m ).<br />

Recall that the balls B r/2 (q m ) cover B Lr (p). We put<br />

B Lr (p) ′ := B Lr (p) ∩ ⋃ m<br />

B r/2 (q m ) ′′ .

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