# Conne's indexing construction for Penrose tilings

1. Jun 4, 2014

in Appendix D, pp. 179 ff, of Alain Conne's "Noncommutative Geometry", www.alainconnes.org/docs/book94bigpdf.pdf, he outlines a process. He looks at Penrose tilings of the plane which are composed of two types of triangles

L [with 2 longer sides of length $\phi$n+1, and one shorter side of $\phi$n) for integers n, whereby $\phi$ is the golden ratio (1+$\sqrt{5}$)/2] and

S [with 2 shorter sides of length $\phi$m, and one longer side of $\phi$m+1) for integers m]

put together by certain rules (among which include that a common side to two triangles must have the same endpoints for both sides). [The familiar darts and kites are combinations of these triangles that help in putting the triangles together, but are not important for this question.]
The process [the purpose of which is to code the tiling into a series of 0's and 1's, but that is also not important for this question] is to transform a tiling into another tiling by eliminating each side which fulfills two conditions:

(a) constitutes a border between two different kinds of triangles
(b) is a longer side of L and a shorter side of S.

This works fine, as long as n≠m (specifically, if n+1=m). But he himself gives an example (Figure 3, p. 94) of a tiling where n=m. In this case, I don't see how this process could work. Can anyone please explain this?

In alternative but similar process [which is to achieve the same set of codings] in pp. 412-413 of Appendix F, in the book "In Search of the Riemann Zeros" by Michel L. Lapidus, which uses the same triangles (labeling them L and R instead of L and S, resp.), the author eliminates sides that fulfill (a) and
(c) is the longer side of S (=R) and the shorter side of L (that is, the reverse of (b)).

Again, unless m+1=n, I don't see how this could work either. (Lapidus gives the example of a tiling with m=n in Figure 2 on page 413.)

(Lapidus' method is a bit nicer in that it is more definite: in Conne's method, you have to choose between two sides of each triangle to look at, whereas Lapidus has only one side of each triangle that has to be considered. That is seen in the difference of the indefinite and definite articles in conditions (b) and (c) resp.)

I would be grateful for help on this step.

2. Jul 2, 2014

### Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. Jul 3, 2014

To make this question perhaps more manageable, I would be happy for an answer for the case of n=m=0. Let me completely restate the question:

Let $\phi$ equal to the Golden Ratio
Make two triangles

L=$\phi$,$\phi$,1
S= $\phi$, 1,1

Tile the plane with these with a kite-and-darts Penrose tiling http://mathworld.wolfram.com/PenroseTiles.html Vertices of one triangle always touch vertices of another.

To make a new tiling further composed of these triangles, there is the possibility of "deflation" and "inflation". "Deflation" consists of cutting up each L into two smaller versions of L and one smaller version of S, and each S into one smaller version of L and one smaller version of L. OK, this I see.

"Inflation" is supposedly the opposite: combining triangles by eliminating borders in order to come up with enlargements of L and S.

But this is easier said than done. One method is allegedly to borders of length $\phi$. But when I follow this, it doesn't seem to work out.

The question: can someone explain, or at least direct me to a good site which does, a deflation process which starts from L and S? The sites I find either hand-wave over inflation (deflation is easier to explain) or use L and an enlarged version of S. (That is, with S* =S enlarged by $\phi$)

Motivation for the question: Connes and by Lapidus (and certainly others) code the inflation process (0 and 1 correspond to whether a triangle ends up in an enlarged L or an enlarged (by a factor of S) on the space of all possible tilings with the terniary Cantor set.