Perron Number Tiles

There is a mathematical result from Thurston and Kenyon that the mapping of a self-similar tiling in ℝ2 to itself is always the multiplication of each point of ℝ2 by an integral power of a Perron number and that there is at least one self-similar tiling for every Perron number.

This suggests that there are self-similar tiles whose construction involves Perron numbers. However it may not imply that there is a self-similar tile for every Perron number as some of these tilings involve tiles which are the attractors of graph directed IFSs (Rauzy tiles, sensu lato, as studied by Cantorini, Siegel and Arnoux), rather than "simple" (stateless) IFSs. However self-similar tiles are known for some Perron numbers.

As noted the transforms of any IFS generating a self-similar figure are of the form aep + x. Let re be a Perron number. Intuition suggests that the inverse transformation represented by r-1e-iα is involved in the transformations of the IFS whose attractor is a self-similar figure. As the complex conjugate of a Perron number is also a Perron number we can use r-1e instead. (The choice between these two expressions represents the choice between two mirror image tiles.) I conjecture that, when c = r-1e and P(c) is a polynomial in c, the transforms of the IFS for many Perron number tiles are of the form ±cnp + sP(c), where s is a scaling factor which may arbitrarily be set to 1. Thus the IFS for a Perron number tile is

{ Ti: p → uicni p + Pi(c) }

where ui is a unit..

The next step is to determine ni, ui and Pi(c). If A denotes the attractor then the ratio of the areas of A and Ti.A is 1:|c2ni|. From this it can be deduced that there is a "dissection polynomial" D(x) such that D(|c|2) = 1 which specifies the values of ni.

It is obvious that the coefficients of D(x) are non-negative integers, that a0 = 0, and that the sum of the coefficients is the number of transforms of the IFS and the number of elements in the self-similar dissection of the tile. This leads to a strategy for finding candidate values of ni. A dissection polynomial obviously has a single root in the range (0 .. 1), which can be found numerically by bisection. Thus, given a list of Perron numbers, those roots of a set of candidate dissection polynomials can be calculated, to find Perron numbers with a magnitude equal to the reciprocal of the square root of the calculated root. Finding Perron numbers is a harder task, as all roots (real and complex) of a candidate polynomial have to be calculated, and surveys run the risk of hitting a combination of co-efficients for which the numerical calculation of the roots does not converge. I have found this to be a problem with some polynomial solving code available online, but the PolyLib C# library on CodeProject.org appears to work reliably.

In the general case ui may be +1 or -1, and the IFS is

{ Ti: p → ±cni p + Pi(c) }

I conjecture that the coefficients of Pi(c) are rational, which means that they can be scaled so that they are all integers. (But see notes below about pletals and hextals.)

P0(c) and P1(c) can be arbitrarily be set to 0 and 1. (It is sometimes more convenient to set them to -1 and 1, or 0 and n). This means that checking candidate two element tiles is relatively easy, though there is the problem of cryptic (not visually obvious) tiles such as Levy's Curve. A non-exhaustive survey of 3 element tiles is computationally practicable, but larger numbers of elements are impracticable, and tiles with larger numbers of elements are more readily found by using techniques to programmatically derive one tile from another. (For example, from 8 order 3 6th unit cubic Pisot tiles found by a survey I derived 58 order 5 tiles and 256 order 7 tiles.)

There are 3 sets of Perron numbers for which I know that there are related tiles.

1) Those Perron numbers which are roots of quadratic equations, and which are either complex or integer. Each such number is satisifed by, inter alia, a dissection polynomial nx = 1, corresponding to tiles with n elements all of the same size, which are known as rep-tiles. I have a construction which generates tiles for all such Perron numbers, but can only formally prove the existence of a tile for Perron numbers of the form i√n, for which a geometrical construction is available to produce the rep-n-rectangle. I am however confident in the existence of "linear" and "caterpillar" rep-tiles for all such Perron numbers.

Some rep-tiles are associated with a square grid. In this case the allowed units are the unit Gaussian integers (+1, -1, +i and -i), and the co-effecients of the polynomial become Gaussian rationals, i.e. numbers of form a + ib, where a and b are rational. These form a richer set of tiles, and I coin the term pletals (polyPLET fractALs) to refer to them. The units can also be written eimπ/2. Thus the IFS for a pletal is

{ Ti: p → cni eimiπ/2 p + Pi(c) }

Some rep-tiles are associated with a hexagonal grid. In this case the allowed units are the unit Eisenstein integers ( (+1, +e11, +e01, -1, -e11 and -e01)), and the co-efficients of the polynomial take the form a + e01b, where a and b are rational, and e01 is the Eisenstein integer (√3-i)/2. These also form a rich set of tiles, and I coin the term hextals (polyHEX fractALs) to refer to them. The units can also be written eimπ/3. Thus the IFS for a hextal is

{ Ti: p →cni eimiπ/3 p +Pi(c) }

2) Perron numbers whose magnitude is the square root of a unit cubic Pisot number. I have found tiles for the 1st (plastic number), 2nd, 3rd (Douady number), 4th (Tribonacci number) and 6th unit cubic Pisots, and have a construction for the dissection polynomial nx + x2 + x3, which gives some additional such numbers, including the 8th and 12th unit cubic Pisots. (The corresponding Pisot polynomials are x3 - nx2 - x - 1.) An attempt to find a tile for the 5th unit cubic Pisot, which has a minimal number of 4 elements was unsuccessful.

3) Perron numbers of the form i√m, where m is a metallic number. There is a construction for m:1 rectangles, so there is a tile for each such number.

© 2016 Stewart R. Hinsley