# Penrose tilings as a cross section of a $5$-dimensional regular tiling

Could somebody explain to me how a penrose tiling , which is not periodic, can be a cross section of a regular tiling in $5$ dimensions, which is periodic? It does not make sense to me how a periodic tiling can produce an aperiodic cross-section.

Also, are there any examples of periodic $3$-dimensional tilings that can produce an aperiodic $2$-dimensional cross section? That would greatly help to visualize the previous question. Thanks!

The key to this is that the projection is onto a 2 dimensional space which sits at an angle of "irrational slope" with respect to the period lattice of the tiling

The following picture from the Tiling Encyclopedia should emphasize well what happens, it is a 1-dimensional projection of a lattice in 2 dimensions. Note that the key is the fact that since the line on which you project has an irrational slope with respect to $$\mathbb Z^2$$, it is impossible for the orange strip to have any period. Any period of the lattice will take some points in the orange strip, outside of the strip. And this leads to the aperiodicity of the Fibonacci tiling. Exactly the same thing happens with the Penrose.

P.S. The picture shows for me, if it doesn't load, here is the link to the Tiling Encyclopedia:

Fibonacci Pic

• Where's the picture? May 12, 2014 at 16:15
• @AidanF.Pierce It shows for me, I added a link to Tiling Enciclopedia. May 12, 2014 at 16:19
• Actualy Penrose tiles have fixed geometry, so there is only a finite set of cutting plane orientations in 5D that will yield true Penrose tilings. Most arbitrary orientations will yield tilings consisting of more than two distinct parallelogram shapes (not necessarily rhombi). Jun 22, 2020 at 8:38
• @Szczepan Hołyszewski I see you know this question. Do you have some references to give, in particular about possible implementations ? Oct 17, 2020 at 18:20
• @N. S. Unfortunately, 6 years later, the links to the Tiling Encyclopedia are broken... Dec 10, 2020 at 0:43

To give a 3-dimensional example: Let $$v$$ be the vector $$\left(1,\frac{\sqrt{2}}2,\frac{\pi}4\right)$$ and $$P$$ be the plane through the origin orthogonal to $$v$$. Take the points in $$\mathbb{Z}^3$$ within distance $$0.5$$ of $$P$$, and project them down onto $$P$$. Then we get the following planar arrangement of points, where the edges are those of $$\mathbb{Z}^3$$ colored according to their direction:

Here is a GIF of this projection as the thickness of the "slab" of points we project grows from $$0$$ to $$2.5$$. (The file was too large to embed within this post, so I have linked to Imgur directly.)

• In both examples, one projects the points from the lattice that are in a strip (in the first example, the strip is in gold, and in the second example, the strip includes all points within 0.5 of P). How does one determine, in general, what the width of the strip should be? Jul 23, 2021 at 14:42
• There’s no particular width that it “should” be, you’ll get tilings for a whole range of widths as shown in the GIF. For the Penrose tiling specifically, there would presumably be some width above which additional edges and vertices appear with epsilon frequency to produce configurations that don’t match the tiling rules (and below which vertices disappear, leading to too-large connected regions), but I’d need to know what specific projection was used to say what that might be. Jul 23, 2021 at 15:19
• Gratefully understood, RavenclawPrefect. One more question, a bit vaguer: the OP refers to the example of a difference in the number of dimensions between surface and lattice equal to three, and asks for (and gets) examples where the difference is one. Does the difference in the number of dimensions have a determinate effect upon the projection? Jul 24, 2021 at 9:07
• PS. When one indicates an irrational slope with a difference of more than one dimension (as in the Penrose tiling), would the slope need to be irrational with respect to every pair of axes of the lattice? Jul 24, 2021 at 10:22