Margulis Lemma
Margulis Lemma
Margulis Lemma
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STRUCTURE OF FUNDAMENTAL GROUPS 19<br />
Covering <strong>Lemma</strong> (1.6) we have<br />
∫ 1<br />
0<br />
Mx(‖∇· ˜Xt i ‖ α ) 1/α (˜c i (t)) dt ≤ C<br />
∫ 1<br />
0<br />
Mx(‖∇·X t i ‖ α ) 1/α (c i (t)) dt<br />
with some universal constant C. Thus ˜φ i1 : (λ i ˜Mi , ˜p i ) → (λ i ˜Mi , φ i1 (˜p i )) has the<br />
zooming in property as well. Any other lift ˜f i of f i is obtained by composing ˜φ i1<br />
with a deck transformation and thus the result carries over to any lift of f i .<br />
□<br />
Proposition 3.8 (Second main example). Let (M i , g i ) be a sequence of manifolds<br />
with Ric > −1/i on B i (p i ) and B i (p i ) compact. Suppose that (M i , p i ) converges to<br />
(R k × Y, p ∞ ).<br />
Then for each v ∈ R k there is a sequence of diffeomorphisms f i : [M i , p i ] →<br />
[M i , p i ] with the zooming in property which converges in the weakly measured sense<br />
to an isometry f ∞ of R k × Y that acts trivially on Y and by w ↦→ w + v on R k .<br />
Moreover, f i is isotopic to the identity and there is a lift ˜f i : [ ˜M i , ˜p i ] → [ ˜M i , ˜p i ] of<br />
f i to the universal cover which has the zooming in property as well.<br />
Proof. Using the splitting R k = Rv ⊕(v) ⊥ and replacing Y by Y ×(v) ⊥ we see that<br />
it suffices to prove the statement for k = 1.<br />
By the work of Cheeger and Colding [CC96] we can find sequences ρ i → ∞ and<br />
ε i → 0 and harmonic functions b i : B 4ρi (p i ) → R such that<br />
|∇b i | ≤ L(n) for all i and<br />
∫<br />
(∣<br />
− ∣|∇bi | − 1 ∣ + ‖Hessbi ‖ ) 2<br />
≤ ε i for any R ∈ [1/4, 4ρ i ].<br />
B 4R (p i)<br />
Let X i be a vector field with compact support with X i = ∇b i on B 3ρi (p i ), and<br />
let φ it denote the flow of X i . Clearly for any t we can find r i → ∞ such that<br />
φ it|Bri (p i) is measure preserving.<br />
Put ψ i := ∣ ∣|∇b i | − 1 ∣ + ‖Hess(b i )‖. We deduce from <strong>Lemma</strong> 1.4 that<br />
∫<br />
− Mx(ψ i ) 2 ≤ C(n, R)ε i<br />
and Cauchy Schwarz gives<br />
∫<br />
−<br />
B 2R (p i)<br />
B R (p i)<br />
Mx(ψ i ) ≤ √ C(n, R)ε i .<br />
Suppose now that t 0 ≤ R<br />
4L(n) . Then φ t(q) ∈ B 3R (p i ) for all q ∈ B<br />
4<br />
R/2 (p i )<br />
and all t ∈ [−t 0 , t 0 ]. Combining that φ t is measure preserving and vol(B R (p i )) ≤<br />
C 3 vol(B R/2 (p i )), we get that<br />
∫ ∫ t0<br />
∫<br />
−<br />
Mx(ψ i )(φ t (p)) ≤ C 3 t 0 − Mx(ψ)<br />
B R/2 (p i)<br />
0<br />
It is now easy to find R i → ∞ and δ i → 0 with<br />
B R (p i)<br />
≤ C 4 (t 0 , R, n) √ ε i .<br />
∫ ∫ t0<br />
− Mx(ψ i )(φ t (p)) ≤ δ i for all R ∈ [1, R i ].<br />
B R (p i) 0