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A quick way to define the hyperbolic metric in the Poincare disc is via the cross ratio: Given points a,b in the disc, let p,q be the endpoints of the hyperbolic line (halfcircle/line perpendicular to the circle) through a and b. Then the hyperbolic metric is given by

d(a,b) = log [a:b:p:q].

(Depending on the precise definition of cross ratio, the order of the four entries might vary.)

This definitions has many advantages, e.g. it is easy to show that the metric is invariant under Möbius transformations preserving the circle and that the corresponding geodesics are precisely the hyperbolic lines. However, the disadvantage seems to be that the triangle inequality is not quite obvious.

Does anybody know a quick and elegant proof of the triangle inequality for the above definition of the hyperbolic metric?

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It helps if you know that for a,b and a',b' with $d(a,b) = d(a',b')$ there is a Möbius transformation $A$ such that $a' = Aa$ and $b'=Ab$. You can then assume that $a = 0$, $0 \lt b \lt 1$ and compare the distances to a third point $c$ in the Poincaré disc. Not extremely quick and elegant, but simpler than the generic case. –  commenter Oct 28 '12 at 11:27
    
You can even do better if you know that the action on the boundary circle is 3-transitive (which is easy). But even then, the computation remains ugly. I was hoping for a smarter way. –  Tom Hase Oct 28 '12 at 12:37

1 Answer 1

I would suggest doing this computation in the Klein model rather than the Poincare model. Since there's an equivalence between them by an appropriate projection sending lines to circles, and preserving cross-ratios, the computation should be equivalent.

The hyperbolic metric in the Klein model is the Hilbert metric. A nice exposition of the triangle inequality for the Hilbert metric is given in Section 2 of a paper of McMullen.

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