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Consider a $(p,q)$ regular tiling of the hyperbolic plane projected on the Poincare disc (that is, a tiling of q p-gons joining at each vertex).

Obviously the area of all tilings converge to $\pi$, but what about the total perimeter? that is, the sum of all the lengths of the tiling edges?

Is the total perimeter infinite? and if it is finite, is there a known formula for the total length?

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  • $\begingroup$ "Obviously the area of all tilings converge to $ \pi $ " Why do you think that? $\endgroup$ – Willemien Nov 29 '14 at 10:07
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    $\begingroup$ because the tilings cover the Poincare disc completely, which has area $\pi$ $\endgroup$ – lurscher Dec 2 '14 at 18:56
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    $\begingroup$ I thought you would say something like that , but it is wrong , the (hyperbolic) area of the Poincare disk is infinite,(the length contracts towards the edge), The maximum area of a triangle is $\pi$ (the triangle in an $ ( 3, \infty)$ tiling ) $\endgroup$ – Willemien Dec 2 '14 at 20:12
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    $\begingroup$ Which is why I made the comment of $\pi$ as the total area, to make it clear that I was referring to the lengths of the projected (euclidean) perimeters on the Poincare disc coordinates, not on the underlying hyperbolic plane $\endgroup$ – lurscher Dec 3 '14 at 5:22
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The total perimeter will be infinite.

You can see that like this: start with one polygon in the center of the Poincaré disk. Now add to that every adjacent polygon, i.e. every polygon which has at least one vertex in common with the central one. The outer boundary will be a sequence of hyperbolic line segments which goes once around your central polygon. So the length of all together will be greater than the perimeter of the Euclidean convex hull of the central polygon. Furthermore, none of these line segments is an edge of the central polygon, so the two sets of edges are disjoint.

Repeat this, adding all the polygons which have at least one vertex in common with your union from the previous step. At each step you'll be considering a new set of edges, which sum up to more than the central convex hull. You can repeat this process an infinite number of times. And the sum of an infinite number of lengths with a common positive lower bound must be infinite. This isn't even accounting for the edges between polygons of the same step.

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  • $\begingroup$ to make it clear, it seems that this argument applies both for the projected Euclidean metric perimeters and the underlying hyperbolic lengths $\endgroup$ – lurscher Dec 3 '14 at 5:25
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    $\begingroup$ @lurscher: Well, to see that the underlying hyperbolic lengths have an infinite sum is far easier: you have an infinite number of polygons with a finite number of edges, each of which has the same positive length. Even if you divide by two to account for each edge belonging to two polygons, this is obviously infinite. The Euclidean length inside the model is therefore far more interesting, even though the argument I used for that could indeed be made for the hyperbolic lengths as well. $\endgroup$ – MvG Dec 3 '14 at 6:22
  • $\begingroup$ Cool, I just wanted to clarify because Willemien commented on my question and he assumed I meant distances under the underlying hyperbolic metric, wanted to make sure this applied to the Euclidean projection $\endgroup$ – lurscher Dec 3 '14 at 21:04
  • $\begingroup$ Great answer! This leaves the interesting question of whether the outer perimeter (of, say, all tiles distance $\leq n$ from a center tile) converges or not. $\endgroup$ – Steven Stadnicki Feb 21 '15 at 4:08
  • $\begingroup$ @StevenStadnicki: I guess that should converge to the unit circle. Any structure which might cause you to “go back and forth” along the boundary of the disk is bound to vanish after a few iterations. Sure, only to be replaced by many more smaller structures of the same kind, but none of them persists, so the limit should be a single cover of the smooth circle. The argument here is a lot of hand-weaving, though, so if you want a proper proof you may want to ask that as a separate question. $\endgroup$ – MvG Feb 21 '15 at 9:38

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