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Could somebody explain to me how a penrose tiling 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!

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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.

Fibonacci

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

More About Fibonacci

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  • $\begingroup$ Where's the picture? $\endgroup$ – Aidan F. Pierce May 12 '14 at 16:15
  • $\begingroup$ @AidanF.Pierce It shows for me, I added a link to Tiling Enciclopedia. $\endgroup$ – N. S. May 12 '14 at 16:19
  • $\begingroup$ 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). $\endgroup$ – Szczepan Hołyszewski Jun 22 at 8:38

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