4Hyperbolic geometry
IB Geometry
4.6 Hyperboloid model
Recall we said there is no way to view the hyperbolic plane as a subset of
R
3
,
and hence we need to mess with Riemannian metrics. However, it turns out we
can indeed embed the hyperbolic plane in R
3
, if we give R
3
a different metric!
Definition (Lorentzian inner product). The Lorentzian inner product on
R
3
has the matrix
1 0 0
0 1 0
0 0 −1
This is less arbitrary as it seems. Recall from IB Linear Algebra that we can
always pick a basis where a non-degenerate symmetric bilinear form has diagonal
made of 1 and
−
1. If we further identify
A
and
−A
as the “same” symmetric
bilinear form, then this is the only other possibility left.
Thus, we obtain the quadratic form given by
q(x) = hx, xi = x
2
+ y
2
− z
2
.
We now define the 2-sheet hyperboloid as
{x ∈ R
2
: q(x) = −1}.
This is given explicitly by the formula
x
2
+ y
2
= z
2
− 1.
We don’t actually need to two sheets. So we define
S
+
= S ∩ {z > 0}.
We let
π
:
S
+
→ D ⊆ C
=
R
2
be the stereographic projection from (0, 0, -1) by
π(x, y, z) =
x + iy
1 + z
= u + iv.
P
π(P )
We put r
2
= u
2
+ v
2
. Doing some calculations, we show that
(i) We always have r < 1, as promised.
(ii) The stereographic projection π is invertible with
σ(u, v) = π
−1
(u, v) =
1
1 − r
2
(2u, 2v, 1 + r
2
) ∈ S
+
.
(iii) The tangent plane to S
+
at P is spanned by
σ
u
=
∂σ
σu
, σ
v
=
∂σ
∂v
.
We can explicitly compute these to be
σ
u
=
2
(1 − r
2
)
2
(1 + u
2
− v
2
, 2uv, 2u),
σ
v
=
2
(1 − r
2
)
2
(2uv, 1 + v
2
− u
2
, 2v).
We restrict the inner product
h·, ·i
to the span of
σ
u
, σ
v
, and we get a
symmetric bilinear form assigned to each u, v ∈ D given by
E du
2
+ 2F du dv + G dv
2
,
where
E = hσ
u
, σ
u
i =
4
(1 − r
2
)
2
,
F = hσ
u
, σ
v
i = 0,
G = hσ
v
, σ
v
i =
4
(1 − r
2
)
2
.
We have thus recovered the Poincare disk model of the hyperbolic plane.