On pg. 400 of the book Metric Spaces of Non-Positive Curvature by Bridson and Haefliger, it is mentioned that:

The hyperbolic plane $\mathbf H^2=\{(a, b)\in \mathbf R^2:\ b>0\}$, (with Riemannian metric $(dx^2+dy^2)/y^2$) is $\delta$-hyperbolic for some $\delta>0$.

(A complete Riemannian manifold is said to be $\delta$-hyperbolic if each side of a geodesic triangle is contained in the union of the $\delta$-neighborhood of the other two. If such a delta exists, we say the manifold is hyperbolic.)

Towards the proof of the above mentioned statement, the authors write

To see that $\mathbf H^2$ is hyperbolic, note that since the area of geodesic triangles in $\mathbf H^2$ is bounded by $\pi$, there is a bound on the radius of semi circles that can be inscribed in a geodesic triangle.

I can see that the area of any goedesic triangle is bounded by $\pi$. This is immediate by applying the Gauss-Bonnet formula and using the fact that $\mathbf H^2$ has constant curvature $-1$, since the sum of exterior angles of any triangle cannot exceed $3\pi$.

But I am unable to see how this gives us $\delta$-hyperbolicity of $\mathbf H^2$. I do not follow what the authors mean by "inscribing a semi circle in a geodesic triangle."

  • $\begingroup$ en.wikipedia.org/wiki/Ideal_triangle "Because the ideal triangle is the largest possible triangle in hyperbolic geometry, the measures above are maxima possible for any hyperbolic triangle, this fact is important in the study of δ-hyperbolic space." en.wikipedia.org/wiki/… $\endgroup$
    – Will Jagy
    Jul 19, 2017 at 18:36
  • $\begingroup$ You misquote the book on page 399 definition 1.1 : " a geodesics triangle is said to be $\delta $ -slim if each of its sides is contained in the delta neighbourhood of the union of the other two sides. And so on . Important is the union bit that is absent in your definition , each point on a side needs to be max delta away from a point on any of the other two sides $\endgroup$
    – Willemien
    Jul 21, 2017 at 21:36
  • $\begingroup$ @Willemien Right. I missed the 'union' in my definition. I am editing it. $\endgroup$ Jul 22, 2017 at 10:46
  • $\begingroup$ Hope you did not find me complaining to mutch but it is those small mistakes that make statements true or false, also I noticed you wrote "Since the sum of exterior angles of any triangle cannot exceed 3pi " depending how you define the exterior angle ( is the max of an exterior angle pi or 2pi) this statement can be true or false) $\endgroup$
    – Willemien
    Jul 22, 2017 at 12:14
  • $\begingroup$ @Willemien I did not mind, of course. That was a small mistake in the typographical sense but a crucial one in the mathematical sense. Someone not familiar with the definition from before could easily get confused. But I want to point out, you misspelled 'much' :P $\endgroup$ Jul 22, 2017 at 12:25

1 Answer 1


The definition of $\delta$-hyperbolic space says: any point on a side of a geodesic triangle is within the distance $\le \delta$ of some other side.

If the above fails, then some side (say, $AB$) of some geodesic triangle contains a point $P$ at large distance $R$ from other two sides $BC, AC$. Draw the circle of radius $R$ (in the hyperbolic metric) centered at $P$. The side $AB$ divides it evenly in two halves -- it's a diameter of the circle, since it is a geodesic passing through the center. So, one half of the circle is within the triangle. This implies that the triangle has large area, a contradiction.

  • $\begingroup$ I am unable to see why a geodesic passing through the center of a geodesic ball divides the area of the disc into two equal halves. I am trying to use the Gauss-Bonnet formula. The problem is that the term which involves the integral of the unsigned curvature has to be shown equal for the two halves, for the exterior angles are the same. Can you please explain how to get that the area of the two halves are indeed equal. Thanks. $\endgroup$ Jul 20, 2017 at 9:55
  • 3
    $\begingroup$ Apply an isometry that maps that geodesic to a straight line. $\endgroup$
    – user357151
    Jul 20, 2017 at 13:06
  • $\begingroup$ I am trying to understand the arguments. I don't understand at the last line where you claim that the area of triangle is greater than the circle due to that half circle lie inside the triangle. If you kindly address my doubt will be appreciated. Thank you in advance $\endgroup$ Nov 1, 2022 at 11:46

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