as is well known, we can put a metric on the upper half plane $\mathbb{R}^+ \times \mathbb{R}$ by setting $$ d\left((x,t);(x',t')\right):=\log\left(\frac{1 + \delta}{1 - \delta}\right)^{1/2}, $$ where $$ \delta := \left(\frac{(x'-x)^2 + (t'-t)^2}{(x'-x)^2+(t'+t)^2}\right)^{1/2}. $$

My question is: what is the most elementary proof that this is indeed a metric, meaning that the triangle inequality is satisfied?

Elementary means that no geometric arguments can be used (e.g. from hyperbolic geometry).

Thanks for your answers!

Edit: I am looking for computational solutions which do not make use of any invariance property.

  • 6
    $\begingroup$ "Elementary means that no geometric arguments can be used"??? $\endgroup$
    – Igor Rivin
    Nov 25 '11 at 11:12
  • 9
    $\begingroup$ You are looking for a computational argument, but even that is helped by geometric insight. For example, instead of just charging ahead, we can first show that the metric is invariant under the Mobius group, and that means that in your triangle you can put one vertex at $i = (0, 1)$, another vertex at $ki = (0, k).$ The computation is then fairly trivial. $\endgroup$
    – Igor Rivin
    Nov 25 '11 at 12:23
  • 2
    $\begingroup$ I agree with Igor. The way I prove it when I teach the subject, is by showing that the formula you give is the length metric associated with the standard Riemannian structure on $\mathbb{H}^2$; this in turn uses te M\"obius group; see A.F. Beardon, "The geometry of discrete groups", Springer-Verlag 1983. $\endgroup$ Nov 25 '11 at 17:17
  • 1
    $\begingroup$ are you sure there is no mistake in your definition above? I am getting a violation of the triangle inequality using the above definition. $\endgroup$
    – Suvrit
    Nov 25 '11 at 21:13
  • 2
    $\begingroup$ Why would you not want to make use of the invariance property? $\endgroup$
    – Yemon Choi
    Nov 28 '11 at 8:09

The question is not obvious because it is stated in the upper half-plane, while it is much easier if you translate it in terms of the Klein projective model of the hyperbolic plane (in a ball of radius $1$). Then I believe that the formula is equivalent to the one given by the Hilbert metric of the disk: $$ d_H(U,V)=-\frac{1}2 \log[U,V;A,B]~, $$ where $A$ and $B$ are the intersections with the boundary of the ball of the line through $U$ and $V$, and $[U,V;A,B]$ is the cross-ratio of the four points. (The minus sign might depend on the convention for the cross-ratio).

Now in this form there is a beautiful and quite simple proof of the triangle inequality, originally due to Hilbert but which can be found in "Metric spaces, convexity and nonpositive curvature" by Papadopoulos, pp153-154, here: http://books.google.com/books?id=JrwzXZB0YrIC&lpg=PA153&ots=V5xkvJE6rO&dq=hilbert%20metric%20triangle%20inequality%20convex&pg=PA153#v=onepage&q=hilbert%20metric%20triangle%20inequality%20convex&f=false

In other terms: the harder part is to check that the distance you wrote is the same as the expression given by Hilbert in the ball, then you can use his simple and nice proof.


If you replace the formulas above by the correct formula for $d$, then you can probably prove the triangle inequality by identifying the minimum value of the function $(y,t) \mapsto d((x,s), (y,t)) + d((y,t), (z,u))$ by finding the critical points of this function.

According to wikipedia, the correct formula is $$ d((x,s), (y,t)) = \operatorname{arccosh}(\delta + \sqrt{\delta^2-1}), $$ where $$ \delta = 1 + \frac{(x-y)^2 + (s-t)^2}{st}. $$


It is worth pointing out, from the calculus of variations, the parametrized curves that achieve your distance are these: with some constant $A,$ and another constant $R > 0,$ either

$$(A, e^t) $$


$$(A + R \tanh t, \; R \, \mbox{sech} \, t). $$

That is, your two points are along one of these curves, the distance is the absolute value of the difference in values of $t.$

Maybe you can do something with that.


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