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Let the Poincaré Half Plane be the set $\{(x, y) \in \mathbb{R}^2 : y > 0\}$. It is a known result that the the metric

$ds^2 = \frac{dx^2 + dy^2}{y^2}$

yields a distance function $f$ such that its output is the length of the geodesic between two points on the Poincaré Half Plane. Through some process that I do not understand, it is possible to prove that

$f((x_1, y_1), (x_2, y_2)) = \operatorname{arcosh} \left( 1 + \frac{ {(x_2 - x_1)}^2 + {(y_2 - y_1)}^2 }{ 2 y_1 y_2 } \right)$.

How does one prove the previous equality from the given metric?

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  • $\begingroup$ Have you tried computing it directly from the equation of he geodesics? $\endgroup$
    – Sak
    Mar 4, 2016 at 6:00
  • $\begingroup$ No, nor would I know how to do this. If there is a resource (preferably one that is digital that I can browse for myself without either paying or having to wait 1-2 weeks for a physical copy) that would explain how to do this, that would be a boon. $\endgroup$
    – Fomalhaut
    Mar 4, 2016 at 20:23
  • $\begingroup$ Recall that in a Riemannian manifold the metric (in the usual metric spaces sense) induced by the Riemannian metric is given by taking the infimum of the lengths of curves joining the two points in question. It can be proved that there is always a curve actually attaining this distance ( the geodesics). These geodesics satisfy a differential equation given in terms of certain coefficients which are in turn in terms of the Riemannian metric. If ou solve this equation you solve your question. (see here: en.wikipedia.org/wiki/Solving_the_geodesic_equations) $\endgroup$
    – Sak
    Mar 4, 2016 at 21:01

1 Answer 1

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One way is to take advantage of symmetry.

  • Prove that vertical lines $y=\text{constant}$ are geodesics, and that your formula holds for any pair of points lying on a vertical line.
  • Prove that the metric $ds^2$ is invariant under the group of fractional linear transformations $x+iy \mapsto \frac{a(x+iy) + b}{c(x+iy)+d}$.
  • Prove that your formula is invariant under the group of fractional linear transformations.
  • Prove that the set of images of the vertical lines, under the action of the group of fractional linear transformations, are the semicircles with diameter on the $x$-axis. Hence, your formula holds for any two points lying on such a semicircle.
  • Prove that any two points lie on such a semicircle.
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  • $\begingroup$ This is not logically sound, because it only shows that $f$ holds all of the necessary properties to be a candidate for a distance function compatible with hyperbolic geometry (as evidenced by your appeal to the Möbius functions). But what I am asking is, given the metric, why $f$ must be the function given in the equality. $\endgroup$
    – Fomalhaut
    Mar 4, 2016 at 20:21
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    $\begingroup$ I'm a little unsure what additional thing you are asking. Perhaps what you are missing is that geodesics in this metric are unique? (which is true because this is a complete Riemannian metric of negative curvature on a simply connected manifold). Thus, by the symmetry argument I have given, the geodesics must be as described in my answer. Also, $f$ must be obtained by integration along the geodesics described, which as my argument shows agrees with the formula in your question. And therefore $f$ must be given by that formula. $\endgroup$
    – Lee Mosher
    Mar 4, 2016 at 20:48
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    $\begingroup$ @LeeMosher I think the OP (original poster) wants to know how do you get from $(1) \ ds^2 = \frac{dx^2 + dy^2}{y^2}$ to $(2) \operatorname{arcosh} \left( 1 + \frac{ {(x_2 - x_1)}^2 + {(y_2 - y_1)}^2 }{ 2 y_1 y_2 } \right)$. and back. I think (1) is the square of the derrivative of (2) but am not able to produce the deduction of that. $\endgroup$
    – Willemien
    Mar 5, 2016 at 10:03
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    $\begingroup$ @Willemien: While it may be that he wants an algebraic/analytic solution to his problem, it can be useful to have other solutions. $\endgroup$
    – Lee Mosher
    Mar 5, 2016 at 14:00
  • $\begingroup$ @LeeMosher Is the very best to have both solutions, I did not manage the algebraic/analytic solution and it would be nice to be able to study that one as well. Hope you can add it to your answer. $\endgroup$
    – Willemien
    Mar 5, 2016 at 17:02

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