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It is known that A δ-hyperbolic space is a geodesic metric space in which every geodesic triangle is δ-thin. (http://en.wikipedia.org/wiki/%CE%94-hyperbolic_space)

The question is that if we remove the word "geodesic" from the above definition, is it still acceptable? Or does the sentence a δ-hyperbolic space is a metric space in which every geodesic triangle is δ-thin still make sense?

@Seirios @mhull @YCor:

This question actually comes from applications. It is observed that the network models of many real world systems, such biological networks and social networks, have small hyperbolicity[http://www.stat.berkeley.edu/~mmahoney/pubs/treelike-icdm13.pdf]. This property enables more efficient algorithm design for some network problems.

However, sometimes the chosen path in reality between a pairs of nodes in the network is not necessarily the shortest path(geodesic distance). But it is possible that the actual distance function is satisfying the four conditions of metric. Then the question is that does it still make sense to measure the hyperbolicty of graph with the new distance function? And if the obtained hyperbolicty is small, does the algorithm leveraging the small hyperbolicty of graphs still work efficiently?

Many thanks!

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  • $\begingroup$ In a discrete metric space, any geodesic triangle is degenerated, so every such spaces would be hyperbolic. On the other hand, it can be convenient to consider groups with a word distance, and such spaces are discrete. $\endgroup$ – Seirios Oct 10 '14 at 6:48
  • $\begingroup$ "In a discrete metric space, any geodesic triangle is degenerated, so every such spaces would be hyperbolic." Is it saying that as long as the space is discrete metric space then the δ value of every triangle under the distance function is always a finite number? And δ is always very small? $\endgroup$ – hulun Oct 10 '14 at 7:52
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$\delta$-hyperbolicity can be defined for non-geodesic metric spaces using the Gromov-product (this definition is given in the wikipedia article you cited). However, considering only the geodesic triangles in non-geodesic metric spaces does not work; consider $\mathbb Z\times\mathbb Z$ with the standard euclidean metric.

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  • $\begingroup$ Thanks for your reply! But I didn't quite understand the sentence "The geodesic triangles in non-geodesic metric spaces does not work". Could you please illustrate a little bit? $\endgroup$ – hulun Oct 10 '14 at 7:38
  • $\begingroup$ @hulun: as an exercise, you could observe that there are metric spaces with no geodesic segment at all, so that the geodesic hyperbolicity axiom becomes an empty condition $\endgroup$ – YCor Oct 10 '14 at 7:50

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