1.4. CONTRUCTING INTERESTING TILINGS 13

Figure 1.10. The chair substitution

We then define the tiling space

Ωσ = {Tilings T | every patch of T is found in a supertile of some order}

(1.1)

Theorem 1.4. If the substitution matrix is primitive, then there exists

a fixed point of

σn

for some n 0. (Such fixed points are called “self-similar

tilings”.) In particular, Ωσ is non-empty.

Proof. If σ is primitive, there exists an integer k 0 for which each

supertile of order k and type a contains a copy of the tile a. By taking k

large enough, we can assume that the supertile contains a copy of a in its

interior. Let t1 be the set of points in our supertile, and let t2 be the chosen

tile of type a inside t1. We will find a point x inside t2 in such a way that

(t1 − x) =

λk(t2

− x). This means that t2 − x can be used as a seed for a

self-similar tiling.

To find the point p, we subdivide the tile t2 into smaller subtiles along

the same pattern by which the supertile t1 is subdivided into tiles. Let t3 be

the subtile that sits inside t2 in the same way that t2 sits inside t1. Repeat

the process indefinitely, subdividing tn and letting tn+1 sit inside tn the way

that tn sits inside tn−1. The intersection of the nested sequence of subtiles

tn is nonempty and has diameter 0, hence is a single point, which we call

x.

This procedure is illustrated for the chair tiling in Figure 1.11. If a is the

chair in standard orientation (a square with the northeast corner missing),

then σ(a) contains two copies of a, but neither are in the interior. We have

to go to

σ2(a)

to get interior tiles, and there are two copies of a in the

interior of

σ2(a),

both of which are shaded. Using them as seed tiles, we can

get two different self-similar tilings, one of which is shown in Figure 1.12.

Why did we need interior tiles? If we had chosen t2 from σ(a), then the

point x would have been on the boundary of t2, as in Figure 1.13. Using

that as a seed would have yielded an infinite self-similar structure, but it

would have covered only part of the plane. To get the whole plane we must