Hyperbolic Distance

We have already indiated that there is a concept of distance in
hyperbolic geometry, and that this is related to the function D.

Definition
If G is a group of transformations of a set S, then a distance function
for the geometry on S is a function d from SxS to R such that,

     for A, B, C in S, and g in G
D1 d(A,B) ≥ 0,
with equality if and only if A = B.
D2 d(A,B) = d(B,A).
D3 d(A,B) ≤ d(A,C) + d(C,B),
with equality if and only if C lies between A and B.
D4 d(g(A),g(B)) = d(A,B).

In the hyperbolic circles pages, we introduced the function D.
We observed that it satisfies D1, D2 and D4. A direct calculation
shows that it does not satisfy D3.

 For z, w in D0
D(z,w) = |z-w|/|w*z-1|
Definition
For z, w in D0, the hyperbolic distance between z and w given by
d(z,w) = 2arctanh(D(z,w)).

Note that arctanh is an increasing bijection from [0,1) to [0,∞).
It follows that d satisfies D1, D2 and D4.

To verify that it satisfies D4, we begin by looking at the case
where C lies between A and B.

Lemma 1
If C lies between A and B, then d(A,B) = d(A,C) + d(C,B).

proof of lemma 1

Lemma 2
If C does not lie between A and B, then d(A,B) < d(A,C) + d(C,B).

proof of lemma 2

Together, these establish D4, so d is a distance function on the disk.

The hyperbolic radius of a hyperbolic circle.

The h-circle K(w,r) consists of points z with D(z,w) = r.
In our new notation, it consists of the points z with
d(z,w) = 2arctanh(r), so we say that K(w,r) has h-radius
2arctanh(r). Note that as r tends to 1, the h-radius
tends to ∞, so we have arbirarily large hyperbolic
distances in the disk.

The CabriJava window shows an h-segment AB with D(A,B) = 1/2.
Thus d(A,B) = 2arctanh(1/2) = ???.
If you drag A, then B moves so that d(A,B) always has this value.
You should observe that the representation appears smaller as it
approaches the boundary.
return to main hyperbolic menu