Well, basics of the theory is here:

http://bendwavy.org/klitzing/explain/tiling-conf.htm

Some pictures are at http://superliminal.com/geometry/tyler/gallery/hyperbolic/6664-all.html, http://superliminal.com/geometry/tyler/gallery/hyperbolic/666p.html and http://superliminal.com/geometry/tyler/gallery/hyperbolic/marek3.html

Lately, I tried using HyperbolicApplet (http://aleph0.clarku.edu/~djoyce/poincare/PoincareApplet.html), but it requires some image editing to get a proper colored picture.

]]>Is there a place I can see pictures of these?

]]>For example, a tiling with three hexagons and one triangle around each vertex has three different forms. With more complex vertex configurations, there can be many distinct tilings.

]]>as close to 1 as possible, while still less than 1, the very best you can do is It comes within of equalling 1, and this has amazing consequences.

]]>Thanks!

I noticed yesterday that we can easily tile a sphere with just two heptagons: take one to be the northern hemisphere, one to be the southern hemisphere, and divide the equator into 7 segments.

But in this case we only have two heptagons meeting at each vertex: that’s a bit degenerate.

So let me show that **you can’t subdivide a sphere into just polygons with 6 or more sides, if at least 3 polygons meet at each vertex**. By the reasoning above, if we could do this we’d have

So, expressing everything in terms of edges, we get

so

so

which contradicts

This argument also gives a bit more. For a torus we must have

There are lots of ways to subdivide a torus into hexagons that meet 3 at a vertex: just take a hexagonal lattice and curl it up. But if we try to stick in one or more polygons with more than 6 edges, or try to have more than 3 hexagons meet somewhere, the argument above shows we’re doomed: then we’ve tipped the situation over from

to

A similar calculation shows that **if you subdivide a sphere into just hexagons and pentagons, with 3 meeting at each vertex, there have to be exactly 12 pentagons**. People working on fullerenes know this well! The first interesting example is this:

but people go further, like this:

]]>That’s a great proof, thank you!

]]>I still don’t really understand what you’re saying, so I’ll just blather at random.

I’m pretty sure it’s impossible to take a sphere and subdivide its surface into just hexagons, or indeed any mix of just hexagons, heptagons, octagons, nonagons and so on. You need some polygons with fewer sides! This should be a purely topological result, nothing to do with the geometry of the sphere or the polygons: their edges don’t need to be ‘straight’ in any sense. It should follow from Euler’s formula

where is the genus, or number of holes in our doughnut; here .

It’s easiest for me to explain this in the case where we imagine subdividing a sphere into heptagons with meeting at each vertex. You can easily generalize what I’m about to say.

In this special case we have

since each face has 7 edges but each edge is shared by 2 faces. We also have

since each vertex has 3 edges coming out of it, but each edge ends at 2 vertices.

So

and thus

so the problem is, there’s no way to obtain

with a positive number of faces. (Please don’t invent polyhedra with a negative number of faces to get around this, at least not today.)

So, we can’t ’tile’ a 2-sphere, no matter how irregularly, with heptagons meeting three at a vertex. Nor can we do a torus, since that would require . If we try a 2-holed torus we get

so maybe we can do that with 12 heptagons—I don’t know. The case I’m most familiar with is the 3-holed torus, which gives

or . That’s why we can tile a 3-holed torus with 24 heptagons meeting 3 at each vertex! Or dually we can tile the 3-holed torus with triangles, 7 meeting at each of 24 vertices, as shown in this picture by Greg Egan:

If you allow more than 3 heptagons to meet at some vertices we instead have

and I think you can show it’s still impossible to tile a sphere, which would require

I get the impression you’re trying to violate such laws of topology.

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