1.
Stability of quasigeodesics
This is a solution to the exercise on Sheet 2 about stability of quasigeodesics, which asked
you to prove the following theorem over several steps.
The ideas of Lemma 1.2 and the proof of Theorem 1.1 are important, so these proofs are
examinable (i.e. you might be asked to prove something similar using the same techniques).
Lemma 1.3 has an examinable statement, but not proof, mostly just because I nd it less
interesting than the other two.
Theorem 1.1. Let (X, d) be a δ hyperbolic space. Then for any K there exists κ so that if
σ : [0, D] → X is a (K, K)quasigeodesic, then there exists a geodesic γ so that im σ ⊆ Nκ (γ)
and γ ⊆ Nκ (im σ).
Let σ : [0, D] → X be any continuous rectiable path and let γ be a geodesic
joining its endpoints. Then for any x ∈ σ , we have d(x, γ) ≤ δ| log2 |α|| + 1.
Lemma 1.2.
Proof. Since α is continuous and rectiable, there exists t0 ∈ [0, D] so that m0 = α(t0 ) satises
d(α(0), m0 ) = d(m0 , α(D)).
Note that α|[0,t0 ] and α|[t0 ,D] are again continuous rectiable paths, so, proceeding inductively,
we can construct nite sets Mn of points in im α so that:
• M0 = {α(0), α(D)};
• M1 = {α(0), α(D), m0 };
• Mn = {mn0 , . . . , mn2n }, where mn0 = α(0), mn2n = α(D);
• for all n, and i, we have mn2i = min−1 and mn2i+1 is the midpoint of the subpath of α
from mn2i to mn2i+2 (midpoint is in the sense dened above, using lengths of subpaths).
Fix n. For each m ≤ n, consider geodesics γin joining mni to mni+1 , so that γ00 = γ . Then
n , γn
n
0
γ2i
2i+1 form a geodesic triangle ∆i , which is δ thin. If x ∈ γ0 = γ , then x lies δ close
1
1
to a point x1 in γ0 or γ1 . In turn, x1 lies δ close to a point x2 in one of two geodesics of the
form γj2 , γj20 . We thus have a sequence x, x1 , . . . , xn , where consecutive points are at distance
≤ δ , and xn lies at distance |α|/2n from some point in Mn ⊂ im α. If we choose n so that
|α|/2n+1 < 1 ≤ |α|/2n , we thus get d(x, α) ≤ δ| log2 |α|| + 1, as required.
γin−1 ,
Let σ : [0, D] → X be a (K, K)quasigeodesic. Then there exists K 0 = K 0 (K) and
a continuous rectiable (K 0 , K 0 )quasigeodesic α : [0, D] → X with:
(1) σ(0) = α(0) and σ(D) = α(D);
(2) each of α and σ is contained in the 10K 0 neighborhood of the other;
(3) for any s < t ∈ [0, D], the length of the subpath of α from α(s) to α(t) is at most
K d(α(s), α(t)) + K .
Lemma 1.3.
Proof. First we construct α. For i ∈ {0, 1, 2, . . . , bDc, D}, let xi = σ(i). For each i ≤ bDc − 1,
let αi be a geodesic from xi to xi+1 , and let αD be a geodesic from xbDc to xD . Let α =
α0 · · · αbDc · αD be the concatenation.
Each αi has length ≤ K d(xi , xi+1 ) + K ≤ 2K . Hence every point of α lies 2K close to
some point of σ (i.e. one of the xi ), so α ⊆ N2K (σ). On the other hand, every point of σ
lies 2K close to some xi , since σ is a (K, K)quasigeodesic, so σ ⊆ N2K (α). So, provided we
eventually choose K 0 ≥ K/5, the second assertion holds.
Now let's check that α is a quasigeodesic. For any t ∈ [0, D], let t̄ be the closest point in
{0, 1, 2, . . . , bDc, D} to t. Then for all s < t ∈ [0, D], we have
d(α(t), α(s)) ≤ d(α(t̄), α(s̄)) + 2K,
which is bounded by K|t̄−s̄|+3K , since α restricts to a (K, K)quasigeodesic on {0, 1, 2, . . . , bDc, D}
(because σ does and these restrictions coincide). But then we get d(α(t), α(s)) ≤ K|s − t| + 5K .
1
2
A similar computation gives the lower bound, so α is a (10K, 10K)quasigeodesic. The length
statement follows from a similar computation.
Now we can prove Theorem 1.1:
Proof of Theorem 1.1. In view of Lemma 1.3, we can assume that σ is a continuous, rectiable
(K, K)quasigeodesic with the property that for all s < t ∈ [0, D], the length of the subpath of
σ from σ(s) to σ(t) is ≤ K d(σ(s), σ(t)) + K . Also, we may assume K ≥ 1.
Let γ be a geodesic joining the endpoints of σ . Let E be the maximal distance from a point
in γ to a point in σ , and let x0 be a point in γ with d(x0 , σ) = E .
Choose y ∈ γ so that y lies between x0 and σ(0) and either d(y, x0 ) > 2E or, if that is
impossible, y = σ(0). Choose z ∈ γ analogously, with σ(0) replaced by σ(D). Let y 0 , z 0 ∈ im σ
be, respectively, closest points to y and z . Let σ 0 be the subpath of σ joining y 0 to z 0 , and let P
be the path [y, y 0 ]σ 0 [z 0 , z]. Since d(x0 , y), d(x0 , z) ≥ E (by the denition of E and the fact that
σ(0) ∈ im σ ), we have that P cannot enter the open E ball about x0 . Indeed, σ 0 does not enter
that ball, by the denition of E . If [y, y 0 ] enters the E ball about x0 , then one of two things
happens:
• if d(x0 , y) > 2E , then we can get from x0 to a point in [y, y 0 ] in distance < E , and then
to y in distance ≤ |[y, y 0 ]| ≤ E , so d(x0 , y) ≤ 2E , a contradiction;
• if y = σ(0), then y 0 = σ(0), and thus [y, y 0 ] is a single point in im σ , which can't lie in
the open E ball about x0 .
Similarly, [z, z 0 ] can't enter the forbidden ball.
Since d(y 0 , z 0 ) ≤ 6E (go to y in E , to x0 in 2E , to z in 2E , then to z 0 in E ), and the property
of σ from the beginning of the proof gives |P | ≤ 2E + 6EK + K . Lemma 1.2 provides
E ≤ δ log2 |2E + 6EK + K| + 1.
Since log2 |2E + 6EK + K| grows more slowly than E as E → ∞ (remember K is xed), there
exists C = C(K, δ) ≥ 1 so that E ≤ C . Hence γ ⊆ NC (im σ), as required.
Figure 1.
Proof that im σ lies in a uniform neighborhood of γ .
Let [s, t] ⊂ [0, D] be a largest subinterval so that σ([s, t]) lies entirely outside of NC (γ). Now,
we saw that every point of γ lies C close to somebody in im σ , and thus C close either to
somebody in im σ([0, s]) or im σ([t, D]). Since γ is connected, we therefore have x ∈ γ which is
C close to σ(s0 ) and to σ(t0 ) for some s0 ∈ [0, s] and some t0 ∈ [t, D]. Hence d(σ(s0 ), σ(t0 )) ≤ 2C .
Thus |σ([s, t])| ≤ |σ([s0 , t0 ])| ≤ 2CK + K . Hence im σ lies (2CK + K + C)close to γ . By
enlarging C (only depending on K ), we see that im σ ⊆ NC (γ), as required. (The second part
of the argument is illustrated in Figure 1.)