HYPERBOLIC GEOMETRY: SOLUTION SHEET 3
JIMMY TSENG
Exercise (36). For s > t > 0, let Bs,t be the (Euclidean) box in H bounded by the following four
curves
• The curve in H := {z := x + iy : y > 0} defined by the equation x = −1 for all y > 0,
• The curve in H := {z := x + iy : y > 0} defined by the equation x = 1 for all y > 0,
• The curve in H := {z := x + iy : y > 0} defined by the equation y = t.
• The curve in H := {z := x + iy : y > 0} defined by the equation y = s.
Calculate the hyperbolic area of Bs,t . Justify your steps. How does the hyperbolic area of B3,2
compare to that of B2,1 ? To that of B11,10 ?
What happens to the hyperbolic area of Bt+1,t as t → ∞?
Solution. We need to use the hyperbolic area formula and, in particular, the information we are
given allows us to determine the integration bounds (to see this, draw a picture of the Bs,t ):
s
Z Z
Z sZ 1
1
1
1 1 1
AreaH (Bs,t ) :=
dx dy =
dx dy = −2 = 2
−
.
2
2
y t
t
s
Bs,t y
t
−1 y
Thus AreaH (B2,1 ) = 2(1−1/2) = 1 > AreaH (B3,2 ) = 2(1/2−1/3) = 1/3 ≈ 0.33 > AreaH (B11,10 ) =
2(1/10 − 1/11) = 1/55 ≈ 0.02.
Finally, we have that
1
1
1
−
lim AreaH (Bt+1,t ) = lim 2
= lim 2
= 0.
t→∞
t→∞
t→∞
t t+1
t(t + 1)
Exercise (38). Let a be a positive real number. Prove that the mapping z 7→ z + ia does not
preserve the hyperbolic area.
Solution. Let us call the mapping ϕ. To show that ϕ does not preserve the hyperbolic area, it
suffices to find a subset S ⊂ H such that AreaH (S) 6= AreaH (ϕ(S)). Consider the set B2,1 from the
previous exercise (i.e. Exercise 36). Note that ϕ(B2,1 ) = B2+a,1+a (draw this!). Thus Exercise 36
tells us that
1
2
1 1
1
AreaH (ϕ(B2,1 )) = 2
−
=2
−
=
t
s
1+a 2+a
(2 + a)(1 + a)
and that AreaH (B2,1 ) = 1. These areas are equal only when
1=
2
(2 + a)(1 + a)
holds and this equation holds only when a = 0 or a = −3. Since a > 0, we have that the areas are
not equal: AreaH (B2,1 ) 6= AreaH (ϕ(B2,1 )).
Exercise (39). Let (X, d) be a metric space. Prove that a point y is not isolated if there exist
{xn }∞
n=1 ⊂ X such that xn → y and all the points xn are pairwise distinct.
c 2016 Jimmy Tseng All Rights Reserved.
Date: 1
J. Tseng
Hyperbolic Geometry Solution Sheet 3
2
Solution. Assume the conclusion is false—i.e. assume that y is isolated. Then for some ε > 0,
the closed ball B(y, ε) of radius ε around y does not contain any point of X other than y itself.
But, xn → y implies that there exists some N ∈ N for which d(xN , y) ≤ ε. There are two cases. If
xN 6= y, then we have arrived at a contradiction.
If xN = y, then we claim that there exists some M ∈ N such that M > N for which xM ∈ B(y, ε).
If this were not true, then, for all m > N , we have that d(xm , y) > ε, which contradicts the
condition that xn → y. Thus xM ∈ B(y, ε). Since the points {xn } are pairwise distinct, we must
have that xN 6= XM . Hence, xM 6= y and xM ∈ B(y, ε) and we have arrived at a contradiction in
this case also. This shows the desired result.
Remark. Note that the converse is also true: If y is not isolated, then there exist {xn }∞
n=1 ⊂ X
such that xn → y and all the points xn are pairwise distinct.
To see the converse, we will construct such a sequence. Pick a r1 > 0 and consider the closed ball
B(y, r1 ) around y of radius r1 > 0. Since y is not isolated, there exists a point x1 ∈ B(y, r1 ) such
that x1 6= y. Let x1 be the first element of the sequence we are constructing. Let r2 = 12 d(y, x1 ) > 0.
Consider the closed ball B(y, r2 ). For the same reason, there exists x2 ∈ B(y, r2 ) such that x2 6= y.
/ B(y, r2 ). Recursively defining elements xn in this manner
Also, by choice of the ball B(y, r2 ), x1 ∈
yields the desired sequence.
Exercise 1 (40). Prove that set {1/n : n ∈ N} is a discrete subset of R. (Note that N :=
{1, 2, 3, · · · }.)
Solution. Let S := {1/n : n ∈ N}. The subset S is discrete if every point in S is isolated. Pick
an arbitrary point 1/n ∈ S. We claim that there is no element of S lying strictly between 1/n and
1/(n + 1). To prove the claim, we will assume that the conclusion is false: there exists 1/m ∈ S
such that
1
1
1
<
< ,
n+1
m
n
which implies that
1
1
1
1
−
>0
−
> 0.
n m
m n+1
This, in turn, implies that
n+1−m
m−n
>0
> 0.
mn
m(n + 1)
As the denominators are both > 0, we have that n + 1 > m > n. This gives a contradiction
because there is no integer strictly between n and n + 1. This proves the claim.
Now the claim also shows that there is no element of S lying strictly between 1/(n − 1) and 1/n
when n ≥ 2. As
1
1
>
,
n(n − 1)
n(n + 1)
1
the closed ball around 1/n of radius 21 n(n+1)
contains only the point 1/n from the set S. This
shows that 1/n is isolated. As 1/n is an arbitrary point of S, this shows that all points in S are
isolated and thus S is a discrete subset of R.
Exercise (48). Let (X, d) be a metric space and d a proper metric. Let Γ be a group of homeomorphisms of X. Prove that if Γ acts properly discontinuously on X, then, for all x ∈ X, the
orbit Γ(x) is a discrete subset of X.
Solution. We give a proof by contradiction. Assume that Γ(x) is not discrete. Then (by the
Remark from the solution of Exercise 39) there exist γn ∈ Γ and x0 ∈ X such that γn (x) → x0
and all the points γn (x) are pairwise distinct. Let ε > 0 and Bε (x0 ) denote the open ball of radius
J. Tseng
Hyperbolic Geometry Solution Sheet 3
3
ε around the point x0 . Since d is a proper metric, Bε (x0 ) is compact. Now as γn (x) → x0 , there
exists N ∈ N such that for all n ≥ N we have that γn (x) ∈ Bε (x0 ). Since the points γn (x) are
pairwise distinct, it follows that the transformations γn are pairwise distinct. Thus, there are
infinitely many γn ∈ Γ such that γn (x) ∈ Bε (x0 ). This contradicts the fact that Γ acts properly
discontinuously and gives the desired result.
J.T.: School of Mathematics, University of Bristol, University Walk, Bristol, BS8 1TW UK
E-mail address: [email protected]