Margulis Lemma

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