Exercise 1.7. Show that for the Thue-Morse substitution, Ωσ =
ΩT .
(2) The Fibonacci substitution is σ(a) = b, σ(b) = ab. The substitution
matrix is
0 1
1 1
, whose Perron-Frobenius eigenvalue equals the golden
mean τ = (1 +

5)/2. The corresponding left- and right-eigenvectors
are L = (1, τ ) and R =
. As with the Thue-Morse sequence, there
is a fixed point of
built from the seed b.a. To get a tiling, we take
the a-tile to have length 1 and the b-tile to have length τ . On average,
there are τ b-tiles for every a-tile. This shows that there are no periodic
tilings in Ωσ, since the ratio of a to b-tiles in a periodic tiling would
have to be rational.
(3) The period-doubling substitution has σ(a) = bb, σ(b) = ab. The matrix
0 1
2 1
, with λP
= 2, L = (1, 1) and R =
. Once again, a fixed
point of
can be built from the seed b.a
There is a map from the Thue-Morse tiling space to the period-
doubling tiling space. If a Thue-Morse tile is preceded by a tile of the
same type, replace it with a period-doubling a tile. If it is preceded by
a tile of the opposite type, replace it with a period-doubling b tile.
Exercise 1.8. Let ΩT
and ΩP
denote the Thue-Morse and period-
doubling substitution tiling spaces, respectively, and let f : ΩT
as above. Show that f intertwines the two substitutions. That is, σP
f =
f σT
. Use this fact to show that f is a 2:1 cover of ΩP
by ΩT
A brief digression into history. Substitution sequences were studied at
length long before aperiodic tilings became fashionable. Thue invented the
Thue-Morse sequence in the late 1800s, and Morse reinvented it in the 1930s
to prove properties of geodesics on Riemann surfaces. Traditionally, the
central object of study wasn’t the space of sequences, but rather a particular
sequence that was fixed by some power of the substitution. The sequence
space could then be recovered from this fixed point by taking its orbit under
a shift map, and then taking the completion of that orbit in a metric that
is similar to our tiling metric.
Substitution tilings in higher dimensions. A substitution in one
dimension is combinatorial replace each letter with a word. In higher di-
mensions, we must also consider the geometry of the tiles. Given a stretching
factor λ 1, a substitution is an operation that
(1) Stretches each tile by a linear factor λ, and
(2) Replaces each stretched tile by a cluster of (ordinary-sized) tiles, as in
Figure 1.10. As before, these clusters are called supertiles (of order 1).
The substitution matrix is as before, and Mij gives the number of i-tiles
in a substituted j-tile. The left-eigenvector L = (L1, . . . , Lk) specifies their
volumes of the different tile types, but has no information about their shapes.
Previous Page Next Page