Short Segments and Long Segments

CRITERION OF INCOMPRESSIBILITY

2.5 Short Segments and Long Segments

We call a segmentPiPi+1short if its length is less thane⁻⁵, and we callPiPi+1long if it’s not short. The next lemma is a corollary of Lemma 3.2.1. It gives us the freedom to perturb the endpoints of a long segment.

Lemma 2.5.1. Let α and β be any two geodesics in Hⁿ, choose P,P⁰ ∈ α and Q,Q⁰ ∈ β. Let R > 0. We assume that d(P,Q) > ¹

2e⁻⁵, d(P,P⁰) = O(R⁻¹) and d(Q,Q⁰)=O(R⁻¹), then we have the following inequalities.

d(P⁰,Q⁰) d(P,Q) −1

=O(R⁻¹),

Θ(v(P⁰,Q⁰),v(P,Q)@P⁰)= O(R⁻¹) and Θ(v(Q⁰,P⁰),v(Q,P)@P⁰) =O(R⁻¹).

In particular, the oriented angle between v(P,Q) and α is O(R⁻¹)-close to the oriented angle betweenv(P⁰,Q⁰) andα. Similar results hold on the geodesic β.

For a long segmentPiPi+1, we will prove the following result.

Lemma 2.5.2. Assume that PiPi+1is a long segment. Then for any pointQj onC⁰_j that is ¹⁰_R-close to f(Pj), where j =i,i+1, we have

d(Qi,Qi+1) d(Pi,Pi+1) −1

=O(R⁻¹). (2.13)

Moreover, the angle betweenv(Qi,Qi+1)andC_i⁰isO(R⁻¹)-close to the angle between v(Pi,Pi+1)andCi. Similarly, the angle betweenv(Qi+1,Qi)andC_i⁰₊₁isO(R⁻¹-close to to the angle betweenv(Pi+1,Pi)andCi+1.

Proof. We first consider the case whenPiPi+1doesn’t cross any seams. In this case, Di is a seam inH² and D⁰_i is a seam inHⁿ. Denote A = Di ∩Ci, B = Di∩Ci+1, A⁰ = D⁰_i ∩C_i⁰ and B = D_i⁰∩ C_i⁰₊₁. We then choose point P_i⁰ on ray A⁰f(Pi) such that |A⁰P_i⁰| = |APi|. Similarly, we choose P_i⁰₊₁ on ray B⁰f(Pi+1) such that

|B⁰P_i⁰₊₁| = |BPi+1|. Denote d= d+iθ the complex distance betweenC_i⁰andC_i⁰₊₁. Denoted₀the real distance betweenCiandCi+1. Then by definition of good pants, we have

1− 1 R² ≤ d

d₀ ≤ 1+ 1

R², and θ d < 1

R².

So we can apply Proposition 2.3.1 and Proposition 2.3.4 to compare quadrangle B⁰A⁰P_i⁰P_i⁰₊₁withB APiPi+1. ForRsufficiently large, we have

1− 4

R² ≤ d(P_i⁰,P_i⁰₊₁)

d(Pi,Pi+1) ≤ 1+ 4

R², (2.14)

|∠A⁰P_i⁰P_i⁰₊₁−∠APiPi+1| ≤ 4

R². (2.15)

Moreover, the direction v(P_i⁰,P_i⁰₊₁) is almost in the plane that contains D_i⁰andC_i⁰. By our construction of f, we have

d(P⁰_j, f(Pj)) =O(R⁻²),

for j =i,i+1. So we getd(Qj,P⁰_j) =O(¹_R), for j =iori+1. By Equation (2.14), d(P_i⁰,P_i⁰₊₁) > ¹

2e⁻⁵when Rlarge. Then the statement of Lemma 2.5.2 just follows from Lemma 2.5.1.

Now we assume that PiPi+1 crosses some seams. Denote S₁,· · · ,Sm to be the intersection points betweenPiPi+1and seams. We denoteS₀= Pi andSm+1 = Pi+1. ThenSjSj+1 lies in a standard right angled hexagon and cuts the hexagon into two regions, where one of the regions can only be a triangle, a quadrangle or a pentagon.

Based on the shape of that region, we have three cases. In all cases, we will prove the following inequality

|f(Sj)f(Sj+1)|

|SjSj+1| −1

=O(R⁻²). (2.16)

Moreover, we will show that the bending angle at each point f(Sj) is O(R⁻²). To study the angles, we denote by θj the entering angle of SjSj+1 into the right angle hexagon and we denote by ϕj the exit angle of SjSj+1 leaving the right angled hexagon (see Figure 2.5). Similarly, we define θ⁰_j and ϕ⁰_j for the segment

f(Sj)f(Sj+1)inHⁿ.

Case 1 Triangle. Denote by∆the triangle that contains sideSjSj+1and denote∆⁰the triangle that is the image of∆mapped by f. Recall that f is(1+²⁰_R2)-bilipschitz when restricted on each side of a right angled hexagon, so Equation (2.16) follows from Corollary 2.3.2 by comparing triangle∆⁰with∆. Moreover, we have the following inequalities on angles:

θ⁰_j−θj

=O(R⁻²) and

ϕ⁰_j −ϕj

=O(R⁻²).

Case 2 Quadrangle. By the geometry of a right angled hexagon (cf. A.3), the length of a seam is on the order ofe^−R2. So the lengths ofSjSj+1and f(Sj)f(Sj+1)are bothO(R⁻²)-close to ^R₂. In this case, all anglesθj, θ⁰_j, ϕj, ϕ⁰_jare exponentially close to ^π₂.

θj

φ_j

S_j

Sj+1

Δ

Figure 2.5: An illustration of the anglesθj andϕjin the case of a triangle

Case 3 Pentagon. Without loss of the generality, we assume that Sj is on a cuff and Sj+1 is on a seam. Let point T be a foot of the seam that contains Sj+1, and denoteT⁰ = f(T). ThenT is exponentially close to Sj+1 and T⁰ is exponentially close to f(Sj+1). It is reduced to the quadrangle case, after replacingSj+1byTand replacing f(Sj+1)byT⁰. Then Equation (2.16) follows from Lemma 3.2.1.

SinceSj,Sj+1andSj+2are in a geodesic, so we haveϕj = θj+1. By the above angle estimates, we have |ϕ⁰_j −θ⁰_j+1| = O(R⁻²). However, it is not enough to show that the bending angle at P⁰_j+1 is O(R⁻²) due to the freedom in dimension. Since any pair of pants in the surfaceSis R-good. So one can easily check thatv(Sj+1,⁰S⁰_j+2) is almost in the hyperbolic plane determined by S⁰_j,S⁰_j+1and the seam that contains S⁰_j+1. Therefore, the bending angle atSjisO(R⁻²). Now we consider the piecewise geodesic αby concatenating f(S₀),· · ·, f(S_m+1). Since the distance between two seams is roughly ^R₂, a unit length subarc ofαcan cross at most one bending point whenRlarge. So the bending norm ofαis bounded byO(R⁻²). By Corollary 2.2.4, we have the following inequality:

d(f(Pi), f(Pi+1)) Pm

i=0

f(Sj)f(Sj+1)

−1

=O(R⁻²).

Combined with Equation (2.16), we proved

d(f(Pi, f(Pi+1)) d(Pi,Pi+1) −1

=O(R⁻²).

In particular, we haved(f(Pi), f(Pi+1)) ≥ ¹

2e⁻⁵whenRlarge. So Inequality (2.13) follows from Lemma 2.5.1. One can check that the angle relations are also satisfied.

The above lemma shows that "long is flexible". In the above proof, It’s important thatPiPi+1is long. It gives us the freedom to perturbP_i⁰andP_i⁰₊₁such that Inequality (2.13) still hold. When PiPi+1 is short, such a perturbation no longer works. We will use the earthquake technic whenPiPi+1is short.

Now we consider the short segments. Any component of ˜γ\ ∪ {long segments} is composed of short segments. Let PiPi+m be such a component. Then the twists on Ci+1,· · ·,Ci+m−1 are exactly 1, and the twists on C_i⁰₊₁,· · ·,C_i⁰_+m−

1 are all ¹⁰_R2-close to 1. We will perform a earthquake map E on the lamination λ. The measure µ= (h₀,· · ·,hk+1)associated to the earthquake is defined as follows: we set

hj = 







 d(D⁰_j−

1,D⁰_j)−1, ifi < j < i+m

0, otherwise.

Here, d(D⁰_j−

1,D⁰_j) denotes the real distance between D⁰_j−

1 and D⁰_j. Denote by E(Ci),· · · ,E(Ci+m) the images of geodesics Ci,· · ·,Ci+m under the earthquake E. By definition of a real earthquake, E(Dj) is the common orthogonal between E(Cj)andE(Cj+1). Then, by our construction ofµ, the distance betweenE(Dj)and E(D_j+1)is the same as the real distance betweenD⁰_jandD⁰_j+1forj =i,· · ·,i+m−2.

By Lemma 2.1.1, we havem < R. Denote by xthe earthquake norm, then x =

i+m−1

X

j=i+1

|d(D⁰_j−

1,D⁰_j)−1|< 10 R.

By Corollary 2.4.2, we can find a point Ri on E(Ci) and a point Ri+m on E(Ci+m) such that the following hold.

1. d(Ri,E(Pi)) < ¹⁰_R andd(Ri+m,E(Pi+m)) < ¹⁰_R. 2. ₁₊¹10

R

d(Pi,Pi+m) ≤ d(Ri,Ri+m) ≤ (1+ ¹⁰_R)d(Pi,Pi+m).

3. ∠Ri is ²⁰_R-close to∠Pi, and∠Ri+m is ²⁰_R-close to ∠Pi+m.

We connect Ri and Ri+m by a geodesic and denote by Rj the intersection between geodesic RiRi+m and E(Cj), for j = i +1,· · ·,i +m−1. Now we can construct those points Qi,· · ·,Qi+m in Claim 2.2.5. Since the distance between E(Dj) and E(Dj+1)is the same as the real distance between Dj and D⁰_j+1, then there is a map from

i+m

F

j=i

E(Cj)to

i+m

F

j=i

C⁰_j such that all the feet are mapped to the corresponding feet and the map is an isometry when restricted on each connected component. Next, we chooseQj to be the image of Rj under the map, for j = i,· · · ,i+m. Now we can apply Proposition 2.3.1 and Proposition 2.3.4 to estimate the length ofQjQj+1

and the bending angle atQj. Precisely, we get

d(Q_j,Qj+1) d(Rj,Rj+1) −1

≤ 4

R², (2.17)

for j =i,· · · ,i+m−1. Next, we consider the bending angle atQj. If the two feet on C⁰_jare in the exact opposite direction after parallel transporting, then by Proposition (2.3.1) and (2.4.1) the bending angle at Qj is bounded by _R⁴2. In general, the two feet onCj, after parallel transporting at a same base point, can form an angle at most

10

R². So the bending angle atQjis bounded by ¹⁴_R2. By Equations (2.17), we obtain

(1− 4 R²)

RiRi+m

≤

i+m−1

X

j=i

d(Qj,Qj+1) ≤ (1+ 4 R²)

RiRi+m .

Combined with the second property ofRj above, we get (1− 20

R)

PiPi+m

≤

i+m−1

X

j=i

d(Qj,Qj+1) ≤ (1+ 20 R)

PiPi+m

, (2.18)

when R large. It follows from d(E(Pi),Ri) < ¹⁰_R and our construction of Qi that d(Qi, f(Pi)) < ²⁰_R. By symmetry,d(Qi+m, f(Pi+m)) < ²⁰_R.

For each component of ˜γ\ ∪ {long segments}, we can define Qi in the above way.

If, for somei ∈ {0,· · · ,k+1},Qiis not defined yet, then eitherPi = Pk+1orPiPi+1

is a long segment. For thisi, we just letQi = f(Pi). Thus, we have definedQi for alli= 0,· · ·,k+1. For a long segmentPiPi+1, and from our constructionQi(resp.

Qi+1) is ²⁰_R-close to f(Pi) (resp. f(Pi+1)), so the result of Lemma 2.5.2 holds. In

the case of short segments, Inequality (2.18) holds and the bending angles at Qi

are bounded byO(_R¹₂). So the bending norm is bounded byO(_R¹). Therefore, the assumption of Claim 2.2.5 is satisfied and it completes the proof.

C h a p t e r 3