Download Hyperbolic geometry quiz solutions

MATH32051/42051/62051 Hyperbolic Geometry
Quiz
Hyperbolic geometry quiz solutions
A1
(i) We say that γ1 , γ2 ∈ Möb(H) are conjugate if there exists g ∈ Möb(H) such that
γ1 = g−1 γ2 g.
[2 marks]
If you had g and g−1 the other way round, or γ1 and γ2 the other way
round, then that’s fine—all of these are equivalent.
(ii) Let γk (z) = kz. Writing γk as
√k z + 0
kz + 0
= k
0z + 1
0z + √1k
√
√
so that it is normalised form, we see that τ (γk ) = ( k+1/ k)2 .
[2 marks]
+ 1)2
k2 );
The trace of γk is not (k
(or
if you get something like this as
your√answer √
then you’ve forgotten to normalise γk . Leaving the answer
as ( k + 1/ k)2 is fine; some of you chose to expand out the bracket
to obtain τ (γk ) = k + k1 + 2, which is also fine.
(iii) There are several ways to do this. The quickest is to quote the result (stated in
the course) that if γ1 and γ2 are conjugate then τ (γ1 ) = τ (γ2 ).
From this it follows that γk , γℓ conjugate implies
√
√
√
√
( k + 1/ k)2 = ( ℓ + 1/ ℓ)2 ,
equivalently
1
1
=ℓ+ .
k
ℓ
Note that this does not implt that k = ℓ or k = 1/ℓ (in the same way that
1 + 4 = 2 + 3 does not imply that 1 = 2 or 1 = 3). Multiplying by kℓ gives
k2 ℓ + ℓ = kℓ2 + k which can be factorised as (k − ℓ)(kℓ − 1) = 0. Hence k = ℓ or
k = 1/ℓ.
[6 marks]
k+
Other methods include letting g(z) = (az + b)(cz + d) and working out
g(γk (z)) and γℓ (g(z)) and comparing coefficients.
B2
(i) This is asking you to prove the Gauss-Bonnet Theorem, given in the lecture
notes. Here’s the proof given in lectures.
Let ∆ be a hyperbolic triangle with internal angles α, β and γ.
We first study the case when at least one of the vertices of ∆ belongs to ∂H,
and hence the angle at this vertex is zero. By applying a Möbius transformation
of H, we can map this vertex to ∞ without altering the area or the angles. By
applying the Möbius transformation z 7→ z + b for a suitable b we can assume
that the circle joining the other two vertices is centred at the origin in C. By
applying the Möbius transformation z 7→ kz we can assume it has radius 1.
1
MATH32051/42051/62051 Hyperbolic Geometry
Quiz
The question asks you to state why the reduction to this special case
is valid. Möbius transformations are isometries, and so preserve distance. But we’re also using the fact that the Möbius transformation
we applied didn’t change the hyperbolic area of ∆ and didn’t change
the internal angles. This is because (as we stated in Lecture 5) Möbius
transformations are area-preserving and conformal (=angle-preserving).
Hence (see Figure 1.1)
AreaH (∆) =
=
=
=
=
1
dx dy
2
∆ y
Z b Z ∞
1
dy dx
√
2
1−x2 y
a
Z b
−1 ∞
dx
y √1−x2
a
Z b
1
√
dx
1 − x2
a
Z β
−1 dθ substituting x = cos θ
Z Z
π−α
= π − (α + β).
This proves (??) when one of the vertices of ∆ lies on ∂H.
∆
α
β
α
a
π−α
0
β
b
Figure 1.1: The Gauss-Bonnet Theorem with one vertex of ∆ at ∞
Now suppose that ∆ has no vertices in ∂H. Let the vertices of ∆ be A, B and
C, with internal angles α, β and γ, respectively. Apply a Möbius transformation
of H so that the side of ∆ between vertices A and C lies on a vertical geodesic.
Let δ be the angle at B between the side CB and the vertical. This allows us to
construct two triangles, each with one vertex at ∞: triangle AB∞ and triangle
CB∞. See Figure 1.2.
AreaH (∆) = AreaH (ABC) = AreaH (AB∞) − AreaH (BC∞).
2
MATH32051/42051/62051 Hyperbolic Geometry
Quiz
Now
AreaH (AB∞) = π − (α + (β + δ))
AreaH (BC∞) = π − ((π − γ) + δ).
Hence
AreaH (ABC) = π − (α + (β + δ)) − (π − ((π − γ) + δ))
= π − (α + β + γ).
C
π−γ
γ
β δ
α
A
B
Figure 1.2: The Gauss-Bonnet Theorem for the triangle ABC with no vertices on ∂H
[10 marks]
(ii) Let Q be a hyperbolic quadrilateral with vertices A, B, C, D (labelled, say, anticlockwise) and corresponding internal angles α, β, γ, δ. Construct the geodesic
from A to C, creating triangles ABC (with internal angles α1 , β, γ1 ) and CDA
(with internal angles γ2 , δ, α2 ), where α1 + α2 = α and γ1 + γ2 = γ. By the
Gauss-Bonnet Theorem
AreaH (Q) = AreaH (ABC) + AreaH (CDA)
= π − (α1 + β + γ1 ) + π − (α2 + β + γ2 )
= 2π − (α + β + γ + δ).
[4 marks]
A common mistake is to assume that the geodesic from A to C bisects
the angles, i.e. α1 = α2 = α/2, γ1 = γ2 = γ/2. In general, this doesn’t
happen.
(iii) This is a straightforward integration:
AreaD (Cr ) =
Z Z
2π
θ=0
3
Z
r
ρ=0
4
ρ dρ dθ
(1 − ρ2 )2
MATH32051/42051/62051 Hyperbolic Geometry
Quiz
1 r
= 4π
1 − ρ2 ρ=0
= 4π
r2
.
1 − r2
[4 marks]
(iv) Let r ∈ (0, 1) and consider the points r, ir, −r, −ir in D. Let Sr denote the
square with vertices at r, ir, −r, −ir.
Let α(r) be the internal angle at the vertex r. Let γ(z) = eiπ/2 z denote the
Möbius transformation given by rotation around 0 through angle π/2. Then γ
leaves Sr invariant and maps r to ir. As Möbius transformations are conformal,
it follows that the internal angle at ir is the same as the internal angle at r.
Rotating through angle π and 3π/2 shows that the internal angles at −r and
−ir are all equal to α(r), and all four sides have the same length. Hence Sr is a
regular polygon.
As r → 1, the vertices converge to 1, i, −1, −i, respectively. The internal angle
at a vertex on the boundary is 0.
Note that Sr ⊂ Cr . Hence
0 ≤ AreaD (Sr ) ≤ AreaD (Cr ) =
4πr 2
→0
1 − r2
as r → 0. Hence AreaD (Sr ) → 0 as r → 0. By part (ii), AreaD (Sr ) = 2π − 4α(r).
Hence 2π − 4α(r) → 0 as r → 0. Hence α(r) → π/2 as r → 0.
As α(r) varies continuously in r, it follows from the Intermediate Value Theorem
that for each α ∈ [0, π/2), there exists a regular square S with internal angle α.
[8 marks]
(v) The first statement is clear as AreaD (Sr ) = 2π − 4α(r) ≤ 2π.
The (hyperbolic) perimeter Pr of S satisfies
Pr = dD (r, ir) + dD (ir, −r) + dD (−r, −ir) + dD (−ir, r)
≥ dD (r, ir) + dD (ir, −r)
≥ dD (−r, r)
≥ dD (0, r)
1+r
= log
→∞
1−r
as r → 1.
[4 marks]
4