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Hyperbolic plane has a beautiful triangulation by congruent hyperbolic triangles where all the vertices of the triangulation have degree 7, this is of course not possible in the euclidean plane, even more, I think it is not possible to triangulate the euclidean plane with triangles that are $C$-bilipschitz with a unit triangulation for some constant $C$, because that would imply there is a bilipschitz map from $\mathbb{H^2}$ to the euclidean $\mathbb{R^2}$ and this is not possible, is that correct?

So my question is: is it possible to triangulate the euclidean plane with any kind of euclidean triangles such that at every vertex has degree 7. I've been trying to do it by hand but always a lot of little triangles start appearing and I'm not certain if I can continue triangulating in such a way that the whole plane is covered.

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    $\begingroup$ Take a conformal map of the hyperbolic plane to the disc and triangulate it in a way reminiscent of an M.C. Escher painting. Map the vertices from the disc to the plane with some stupid map - the inverse of $(x,y) \to (x/\sqrt{1+x^2+y^2},y/\sqrt{1+x^2+y^2)$, say. Then draw a straight line between every pair of vertices that had a straight line drawn between them before. If none of the lines cross and none of the triangles start overlapping, you're good, but I'm not sure how to figure out whether this happens. $\endgroup$
    – Will Sawin
    Commented Aug 8, 2012 at 18:11

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Unless I am missing something: start with the image of a degree-seven triangulation of the hyperbolic plane (coming from the (2, 3, 7) triangle group -- that is the triangulation both you and @Will seem to be alluding to) in the Beltrami-Klein model (this will be a triangulation of the disk with Euclidean straight edges). Then lift the vertices of which are distance $r$ from the origin by some monotoically increasing convex function of $r,$ such that $f(1) < \infty, f(0) = 0$ This will move your triangulation to a straight-line triangulation of a convex "cup." Now centrally project this from the point $(0, 0, f(1)).$ The image will be a combinatorially equivalent triangulation of the plane.

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    $\begingroup$ @Igor: I think, the easier way to describe what you are doing is to realize (2,3,7)-triangulation of hyperbolic plane on the hyperboloid $x^2+y^2-z^2=-1, z>0$, then straighten it replacing hyperbolic geodesic segments with Euclidean ones. Now, project the result orthogonally to the xy-plane and you are done. $\endgroup$
    – Misha
    Commented Aug 8, 2012 at 20:16
  • $\begingroup$ @Misha: yes, of course you are right. What I am describing is slightly more general, but also I was conjecturing that the OP might have been less familiar with the hyperboloid model... $\endgroup$
    – Igor Rivin
    Commented Aug 9, 2012 at 13:19
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I am not sure what you are asking in your first question (what is a "unit triangulation" ?) but one can prove the following:

There is no geodesic triangulation of the plane with all vertices of degree >6 and such that the following conditions holds:

b) the angles of triangles are bounded from below.

Condition b) means that each triangle is bi-Lipshitz to an equilateral triangle, so this is probably what you ask. And the reason you mention is valid.

Examples of geodesic triangulation with vertices of degree 7 are given in the previous answer, and it follows from, b) that some triangles in these examples are very thin.

Edited later. My first answer also had condition

a) all triangles are bounded.

I claimed that b) can be replaced with a).

This condition CANNOT be used to replace b). Here is an example of a geodesic trianglation of the plane with all vertices of arbitrarily high degree, and bounded triangles.

Begin with an equilateral triange inscribed in the unit circle. From each vertex draw segments from the unit circle to the concentric circle of radius 2, so that the degree of each vertex on the unit circle becomes large. Add some chords of the circle of radius 2, connecting the endpoints of these segmnts. Draw additional segments to make this a triangulation of some polygon inscribed in the circle of radius 2. All inner vertices of this triangulation have large degree.

Now repeat this construction. We obtain a sequence of larger and larger polygons $P_n$ each inscribed in a circle of radius $n$, and each polygon is triangulated so that the degrees of inner vertices is large. Continuing this indefinitely, we obtain a triangulation of the plane.

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  • $\begingroup$ 9/9/12 should be 8/9/12 in the US, 9.8.12 in Europe, or just omitted, since MathOverflow automatically keeps track of editing times. $\endgroup$
    – Andreas Blass
    Commented Aug 9, 2012 at 14:56

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