Some aspects of rigidity for self-affine tilings
Boris Solomyak
University of Washington

December 21, 2011, RIMS

Boris Solomyak (University of Washington)

Rigidity for self-affine tilings

December 21, 2011, RIMS

1 / 69What is rigidity?

A structure is ”rigid” if it doesn’t allow a small perturbation
This often means some algebraic constraints
Another aspect: weak form of equivalence implies strong form ofequivalence
Substitution (self-affine) tilings exhibit both kinds of rigidity

Boris Solomyak (University of Washington)

Rigidity for self-affine tilings

December 21, 2011, RIMS

2 / 69Plan of the talk1

Preliminaries

2

Which expansions are possible for self-affine tilings? (Thurston,
Kenyon, Kenyon & S.)

3

When does the tiling have large discrete spectrum (diffraction or
dynamical)? (J.-Y. Lee & S.)

4

Topological and dynamical rigidity (J. Kwapisz)

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I. Preliminaries on tilings

Prototile set: A = {A1 , . . . , AN }, compact sets in Rd , which areclosures of its interior; interior is connected.
May have “colors” or “labels”
A tiling of Rd with the prototile set A: collection of tiles whose union
is Rd and interiors are disjoint. All tiles are translates of the prototiles.
A patch is a finite set of tiles. A+ denotes the set of patches withtiles from A.
Assume translational finite patch complexity (or finite patterncondition) (FPC).

Boris Solomyak (University of Washington)

Rigidity for self-affine tilings

December 21, 2011, RIMS

4 / 69Tile-substitutions in Rd
Let φ : Rd → Rd be an expanding linear map, that is, all its eigenvalues
are greater than 1 in modulus.
Definition. Let {A1 , . . . , Am } be a finite prototile set. A tile-substitution
with expansion φ is a map ω : A → A+ , where each ω(Ai ) is a patch
made of translates of Aj , such thatsupp(ω(Ai )) = φ(Ai ), i ≤ m.
The substitution is extended to all translates of prototiles by
ω(x + Aj ) = φx + ω(Aj ), and to patches and tilings by[
ω(P) = {ω(T ) : T ∈ P}.

Boris Solomyak (University of Washington)

Rigidity for self-affine tilings

December 21, 2011, RIMS

5 / 69Self-similar and self-affine tilings

Substitution matrix: S = (Si,j )i,j≤m , where S(i, j) = # tiles of type iin ω(Aj )Tile-substitution is primitive if S is primitive, that is, some power ofS has only positive entries (equivalently, ∃ k ∈ N, ∀ i ≤ m, the patchω k (Ai ) contains tiles of all types).
We say that T is a fixed point of the tile-substitution ω if
ω(T ) = T . Such tilings are called self-affine.
If φ is a similarity map (i.e. |φ(x)| = r |x| for some r > 1 and allx ∈ Rd ), then T is called self-similar.

Boris Solomyak (University of Washington)

Rigidity for self-affine tilings

December 21, 2011, RIMS

6 / 69Self-similar tilings: special cases

φ is a pure dilation: φ(x) = λx, for λ > 1, in Rd
λ is the real expansion constantφ(z) = λz for λ ∈ C, for |λ| > 1, in C ≈ R2
λ is the complex expansion constantBoris Solomyak (University of Washington)

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Example: chair tiling
Examples are shamelessly taken from ”Tiling Encyclopedia”, see
http://tilings.math.uni-bielefeld.de/Figure: tile-substitution, real expansion constant λ = 2

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Example: chair tiling

Figure: patch of the tiling

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December 21, 2011, RIMS

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Example: non-FLC tiling (Kenyon)

Figure: tile-substitution, real expansion constant λ = 3

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Example: non-FLC tiling (Kenyon)

Figure: patch of the tiling

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December 21, 2011, RIMS

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Example: non-FLC tiling (Kenyon)
To get the ”self-similar” one, need to iterate and rescale...

Figure: tile-substitution for the ”pre-self-similar” tiling

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Rigidity for self-affine tilings

December 21, 2011, RIMS

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Example: non-FLC tiling (Kenyon)

Figure: patch of the ”pre-self-similar” tiling (note optical illusion!)

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December 21, 2011, RIMS

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Example: self-similar tiling of the plane with fractal
boundary

Figure: Self-similar tiling with complex expansion constant λ, λ4 + λ + 1 = 0 (R.
Kenyon)Boris Solomyak (University of Washington)

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December 21, 2011, RIMS

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Example: self-similar tiling of the plane with fractal
boundary

Figure: Substitution rule

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Example: self-affine of the plane

Figure: A self-affine tiling in the plane with diagonal expansion matrix Diag[x1 , x2 ]where x1 ≈ 2.19869, x2 ≈ −1.91223 are roots of x 3 − x 2 − 4x + 3 = 0.

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December 21, 2011, RIMS

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Example: self-affine of the plane (cont.)

Figure: Subdivision rule: 1 → {3, 2}, 2 → {3, 2, 2, 2, 2}, 3 → {1, 1, 1}.
Construction uses free group F (a, b, c), with a, b, c corresponding to vectors(1, 1), (x1 − 1, x2 − 1), (x12 − x1 , x22 − x2 ) in R2 , the endomorphsm
ψ(a) = ab, ψ(b) = c, ψ(c) = ab 4 and tiles [b, a], [b, c], [a, c].

Boris Solomyak (University of Washington)

Rigidity for self-affine tilings

December 21, 2011, RIMS

17 / 69(Some) constructions of tilings with fractal boundaryFree group endomorphisms, iterating and rescaling, substitutions ofhigher-dimensional faces: [Dekking 1982], [Ito & Kimura 1991], [Ito& Ohtsuki 1991, 1993], [Kenyon 1990, 1996], [Arnoux & Ito 2001],[Sano, Arnoux & Ito 2001], [Ei & Ito 2005], [Ito & Rao 2006], [Ei, Ito& Rao 2006], [Furukado 2006], [Furukado, Ito & Robinson 2006], . . .
Finite state automata and numeration systems with a complex Pisot
base:[Thurston 1989], [Petronio 1994], [Akiyama & Sadahiro 1998],[Akiyama 2002], [Berthé & Siegel 2005], . . .
Connections with Markov partitions:
Many of the same papers + [Praggastis 1992, 1999], [Kenyon &Vershik 1998], [Enomoto 2008], . . .

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Self-affine tilings
Repetitivity

Lemma. For any primitive tile-substitution ω, there exists n ∈ N such ω n
has a fixed point.
Definition A tiling T of Rd is repetitive if every patch P ⊂ T appears
relatively dense in Rd ; more precisely, there exixts R = R(P) such thatevery ball of radius R contains a translated copy of P.
Lemma. Let T be an FLC fixed point of a tile-substitution with
substitution matrix S.
T is repetitive if and only if S is primitive and a T -patch, containing theorigin in the interior of its support, occurs in ω n (T ) for some T ∈ T andsome n ∈ N (in particular, if the origin lies in the interior of a tile).

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Self-affine tilings
Aperiodicity

Definition A tiling T of Rd is aperiodic if
T +x =T

Boris Solomyak (University of Washington)⇒ x = 0.

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Self-affine tilings
Boundary

Lemma. Let T be a self-affine tiling with expansion map φ whoserotational part is of infinite order (in R2 this means that the rotation isirrational modulo π). Then the tiles cannot be polyhedral, or evenpiecewise smooth.
Lemma. For any tile T of a self-affine tiling, Vol(∂T ) = 0.
Corollary. The PF eigenvalue of the substitution matrix of a self-affine
tiling with expansion map φ is | det φ|. The vector (Vol(Aj ))m
1 is a left PF
eigenvector.
Corollary. | det φ| is a Perron number, i.e. an algebraic integer θ > 1
whose Galois conjugates are strictly less than θ in modulus.

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December 21, 2011, RIMS

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Self-affine tilings and iterated function systems

There exist finite sets Dij ⊂ Rd , i, j ≤ m:
ω(Aj ) = {u + Ai : u ∈ Dij , i = 1, . . . , m}, j ≤ m,
with
φAj =

m
[

(Dij + Ai ), j ≤ m.

(1)

i=1

Here all the sets in the right-hand side must have disjoint interiors; it ispossible for some of the Dij to be empty. Note that S(i, j) = #Dij .

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Self-affine tilings and iterated function systems (cont.)

Rewrite the system of set equations (1):
Aj =

m
[

(φ−1 Ai + φ−1 Dij ), j ≤ m.

i=1

φ−1 is a contraction, so there is always a unique nonempty compact
solution {A1 , . . . , Am } (attractor of a graph-directed IFS). The difficulty is
to have Aj with nonempty interiors.

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II. Characterization of expansions I

Theorem. There is a self-similar tiling of the line R with expansion λ if
and only if |λ| is a Perron number.
We already know the necessity. Sufficiency follows from Lind’s Theorem:
every Perron number is the PF eigenvalue of a primitive matrix.
Theorem [Lind ’84] If λ > 1 is a Perron number, then there is a primitive
non-negative integral matrix M with the PF eigenvalue
P equal to λ.
Moreover, M can be chosen so that M11 > 0 and m
i=1 Mi1 ≥ 3.

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Characterization of expansions II
Theorem (?) [Thurston ’89] In R2 ≈ C, φ(z) = λz with complex λ, is an
expansion of a self-similar tiling if and only if λ is a complex Perron
number, i.e. an algebraic integer of modulus > 1 whose Galois conjugates,
except the complex conjugate, λ, are strictly less than |λ| in modulus.
Sufficiency in the theorem above was claimed by Thurston and proved by
Kenyon in his GAFA 1996 paper, but there still seems to be a gap... In
this talk I will focus on the proof of necessity.
Remark. |λ|2 Perronp
does not imply that λ is complex Perron. For√√
example: λ = −1 + i 1 + 5 has |λ|2 =p
2 + 5, which is Perron (even√Pisot), but λ has conjugates λ and −1 ± −1 + 5, one of which is≈ −2.11179, whereas |λ| ≈ 2.05817.

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December 21, 2011, RIMS

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Characterization of expansions III
Theorem [Kenyon-S. 2010] Let φ be an expanding linear map on Rd ,which is diagonalizable over C, and suppose there exists a self-affine tilingof Rd with expansion φ. Let λ be an eigenvalue of φ, and let γ be a Galois
conjugate of λ. Then either |γ| < |λ|, or γ is also an eigenvalue of φ, andits multiplicity is ≥ multiplicity of λ.
The case of non-diagonalizable maps (Jordan blocks) is open.
Similarity maps are diagonalizable over C, so this covers all
self-similar tilings.
I will present a sketch of the proof, skipping (many!) technical details; itwill be more complete in the self-similar case.

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Control points

Definition. For each T -tile T , fix a tile γT in the patch ω(T ); choose
γT with the same relative position for all tiles of the same type. Thisdefines a map γ : T → T called the tile map. Then define the control
point for a tile T ∈ T by{c(T )} =∞\φ−n (γ n T ).

n=0

There are finitely many choices for control points, depending on γ.
Let C = C(T ) = {c(T ) : T ∈ T }.

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Properties of control points

1

If T 0 ≈ T in T , that is, T 0 = T + x, then c(T 0 ) = c(T ) + x2

φ(c(T )) = c(γT ), for T ∈ T

3

φ(C) ⊂ C

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Rigidity for self-affine tilings

December 21, 2011, RIMS

28 / 69Proof of algebraicity of eigenvalues
Lemma. Let T be a self-affine tiling of Rd with expansion φ. Then all the
eigenvalues of φ are algebraic integers.
Proof: J := hCi, the subgroup of Rd generated by C = C(T ). It is a
finitely generated abelian group (by FLC).
By the Structure Theorem for Free Abelian Groups, there exist freegenerators v1 , . . . , vN ∈ Rd , i.e.,∀ ξ ∈ J, ∃ ! a1 , . . . , aN ∈ Z : ξ =NXaj vj .

j=1

Let V = [v1 . . . vN ], a d × N matrix, and a(ξ) = [a1 , . . . , aN ]T ∈ ZN .
Then ξ = Va(ξ).

Boris Solomyak (University of Washington)

Rigidity for self-affine tilings

December 21, 2011, RIMS

29 / 69Proof of algebraicity of eigenvalues (cont.)rank(V ) = d since C spans Rd , hence Ker(V T ) = {0}.
φ(C) ⊂ C ⇒ φJ ⊂ J, hence there exists an integer N × N matrix M
such thatφV = VM.
Every eigenvalue of φ is an eigenvalue of M: let eγ be an eigenvectorof φT :
φT eγ = γeγ ⇒ M T V T eγ = V T φT eγ = γV T eγ .
Thus, V T eγ 6= 0 is an eigenvector for M T corresponding to γ, hence
γ is an algebraic integer.

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December 21, 2011, RIMS

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Address map

We call ξ 7→ a(ξ) the address map. Note that a : J → ZN .
i

M

i

i

φ

i

ZN −−−−→ RN −−−−→ RN ←−−−− ZN
x


x




Vy
a
yV
a
J −−−−→ Rd −−−−→ Rd ←−−−− J

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Rigidity for self-affine tilings

December 21, 2011, RIMS

31 / 69Proof sketch of the Perron condition
Lemma. The matrix M is diagonalizable over C.
Proof in the case of a self-similar tiling of the plane, with a complex
expansion constant λ.
J = hCi is a finitely-generated Z-module. Then Q · J is a vector spaceover Q, on which φ acts. Note that {y1 , . . . , yN } is a basis, and thematrix of φ in this basis is M.
Note that Q · J is also a vector space over Q(λ) (the field). Let{ζ1 , . . . , ζr } be a basis of Q · J over Q(λ) and let n be the degree of
λ. Then{λs ζk : 0 ≤ s ≤ n − 1, 1 ≤ k ≤ r }is a basis for Q · J over Q.

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December 21, 2011, RIMS

32 / 69Proof of the lemma (cont.)

In this basis, φ = multiplication by λ has a matrix which is a direct
sum of r copies of the companion matrix of λ.
Each of them is diagonalizable over C, since the minimal polynomialof λ has no repeated roots. Thus, the linear operator given by M is
diagonalizable.

Remark. In fact, r = 1, which implies that C ⊂ Z[λ]ζ for some ζ. This isalso a kind of “rigidity”. This is a special case of Structure Theorem forcontrol points [Kenyon 1994,1996], [J.-Y. Lee & S. 2012].

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Rigidity for self-affine tilings

December 21, 2011, RIMS

33 / 69Proof sketch of the Perron condition (cont.)

Suppose that γ is a conjugate of λ and |γ| ≥ |λ| > 1. Then γ is aneigenvalue of M. We want to show that γ is an eigenvalue of φ.
Let Uγ be the (real) eigenspace for M corresponding to γ. The only
eigenvalues of Mγ := M|Uγ are γ and γ (if γ is nonreal).
Since M is diagonalizable over C, there is a projection πγ from RN to
Uγ commuting with M.
Consider the mapping fγ : C → Uγ given byfγ (ξ) = πγ a(ξ),

Boris Solomyak (University of Washington)

Rigidity for self-affine tilingsξ ∈ C.

December 21, 2011, RIMS

34 / 69Proof sketch of the Perron condition (cont.)

We have a commutative diagram:φ

J −−−−→


ay

J


ya

M

ZN −−−−→ ZN


π
πγ 
y
y γ
Mγ

Uγ −−−−→ Uγ

Boris Solomyak (University of Washington)

Rigidity for self-affine tilings

December 21, 2011, RIMS

35 / 69Proof sketch of the Perron condition (cont.)Letfγ (φ−k ξ) = M −k fγ (ξ),ξ ∈ C.

This is well-defined.
Now we have fγ defined on a dense set
C∞ :=

∞
[

φ−k C.

k=0

Boris Solomyak (University of Washington)

Rigidity for self-affine tilings

December 21, 2011, RIMS

36 / 69Proof sketch of the Perron condition (cont.)
Lemma The map fγ is uniformly continuous on C∞ , and thus it extends
by continuity to Rd , satisfying
fγ ◦ φ = M ◦ fγ .
Proof sketch. Step 1: the address map a, and hence fγ is uniformly
Lipschitz on C (large-scale):ka(ξ) − a(ξ 0 )k ≤ L1 kξ − ξ 0 k, ξ, ξ 0 ∈ C.
This is Thurston’s argument: one can get “quasi-efficiently” from acontrol point to a distant control point by moving from neighbor toneighbor, and then use FLC.
Remark. The address map is usually not continuous on J = hCi, since J is
usually dense, and the map is into ZN .
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December 21, 2011, RIMS

37 / 69Proof sketch of the Perron condition (cont.)

Step 2: The map fγ is Hölder continuous on C∞ : there exist r > 0 and
L2 > 0 such that for all ξ, ξ 0 ∈ C∞ with kξ − ξ 0 k < r ,kfγ (ξ) − fγ (ξ 0 )k ≤ L2 kξ − ξ 0 kα ,
where
α=

log |γ|, λmax = largest eigenvalue of φ.
log |λmax |Note: if φ is a similarity map, then |λmax | = |λ|, and |γ| ≥ |λ| byassumption. We then have that |γ| = |λ| (otherwise fγ ≡ const), and fγ is
Lipschitz on Rd .

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Rigidity for self-affine tilings

December 21, 2011, RIMS

38 / 69Proof sketch of the Perron condition (cont.)

Lemma. The function fγ (now defined on all Rd ) depends only on the tile
type in T up to an additive constant: if T , T + x ∈ T and ξ ∈ T , thenfγ (ξ + x) = fγ (ξ) + πγ a(x).
Proof sketch. It is enough to check this on the dense set C∞ and then it isa straightforward verification.

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Rigidity for self-affine tilings

December 21, 2011, RIMS

39 / 69Proof sketch of the Perron condition (cont.)
Conclusion in the self-similar caseWe havefγ ◦ φ = M ◦ fγ on Rd .
We want to show that fγ is linear.
fγ is Lipschitz ⇒ it is is differentiable a.e.
fγ is “almost flat” on a small neighborhood of some x ∈ Rd . Applyφk and notice that fγ gets flatter and flatter near φk x.
Now find a patch in some fixed BR (0) where T has the same pattern
as near φk x for all k (possible by repetitivity).

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Rigidity for self-affine tilings

December 21, 2011, RIMS

40 / 69Proof sketch of the Perron condition (conclusion!)

By the last lemma, there are points xk ∈ BR (0) near which fγ is thesame as near φk x (up to an additive constant), hence it is almost flat
there. By compactness, it must be exactly flat somewhere.
Using the expansiveness of φ and conjugation again, conclude that fγ
is flat everywhere.
fγ (0) = 0 by construction ⇒ fγ is linear. It is a surjection onto Uγ ,hence Mγ = M|Uγ is isomorphic, as a linear map, to a restriction ofφ, therefore, γ is an eigenvalue of φ.

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III. Discrete spectrum of self-affine tilings

Diffraction spectrum and Dynamical spectrum
Modern definition of a crystal: material with pure point diffraction
Dworkin (1993) showed that diffraction spectrum is a “part” ofdynamical spectrum, so we will be concerned with the latter.
Question: when is there “large” discrete component of the spectrum?

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Partial answer

Theorem [Jeong-Yup Lee & S., 2008, 2012] Let T be self-affine with adiagonalizable over C expansion map φ. Suppose that all the eigenvaluesof φ are algebraic conjugates with the same multiplicity. Then thefollowing are equivalent:(i) the set of eigenvalues of the tiling dynamical system associated with T(defined below) is relatively dense in Rd ;(ii) the spectrum of φ is a Pisot family: for every eigenvalue λ of φ and its
conjugate γ, either |γ| < 1, or γ is also an eigenvalue of φ;(iii) the set of control points C = C(T ) is a Meyer set, i.e. C − C is
uniformly discrete.

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Remarks
1(i) ⇒ (iii) is proved in [J.-Y. Lee & S. 2008], whereas (iii) ⇒ (i)follows from [Strungaru 2005] and [Dworkin 1993].

2(i) ⇒ (ii) was proved by [E. A. Robinson 2004], using the criterion for
eigenvalues in [S. 1997].

3(ii) ⇒ (i), the most technically difficult part, is proved in [J.-Y. Lee &S. 2012].

4

Examples show that the condition of having conjugates with the same
multiplicity cannot be omitted, but it is an open question how tohandle non-diagonalizable cases, or cases when not all eigenvalues of
φ are conjugates.

We will sketch the proof of (i) ⇔ (ii), but first we need to define the tilingdynamical system.
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Tiling space

Tiling space, or hull, generated by T :XT = {−g + T : g ∈ Rd },where the closure is in the “local” topology: two tilings are close if after asmall translation they agree on a large ball around the origin.
More precisely:%e(T1 , T2 ) := inf{r ∈ (0, 2−1/2 ) : ∃ g ∈ Br :T1 − g and T2 agree on B1/r }.
Then %(T1 , T2 ) := min{2−1/2 , %(T1 , T2 )} is a metric.

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Tiling dynamical system

Theorem. FLC ⇐⇒ XT is compact.
Rd acts by translations: T t (S) = S − t. Topological dynamical system(action of Rd by homeomorphisms):(XT , T t )t∈Rd = (XT , Rd )Definition. A topological dynamical system is minimal if every orbit isdense (equivalently, if it has no nontrivial closed invariant subsets).
Theorem. T is repetitive ⇐⇒ (XT , Rd ) is minimal.

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Uniform patch frequencies

For a patch P ⊂ T let LP (T , A) :=#{t ∈ Rd : −t + P ⊂ T , −t + supp(P) ⊂ A},
the number of T -patches equivalent to P that are contained in A.
Definition. A tiling T has uniform patch frequencies (UPF) if for any
non-empty patch P, the limit
freq(P, T ) := limr →∞LP (T , t + Qr )≥0rdexists uniformly in t ∈ Rd . Here Qr = [− 2r , 2r ]d .

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Unique ergodicity for tiling systems

Theorem. Let T be a tiling with FLC. Then the dynamical system(XT , Rd ) is uniquely ergodic, i.e. has a unique invariant probability
measure, if and only if T has UPF.
Theorem. Let T be a self-affine tiling. Then the dynamical system(XT , Rd ) is uniquely ergodic.
Denote by µ the unique invariant measure.

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Eigenvalues and Eigenfunctions

Definition. α ∈ Rd is an eigenvalue for the measure-preserving Rd -action(X , T t , µ)t∈Rd if ∃ eigenfunction fα ∈ L2 (X , µ), i.e., fα is not 0 in L2 andfor µ-a.e. x ∈ X
fα (T t x) = e 2πiht,αi fα (x), t ∈ Rd .
Here h·, ·i is the scalar product in Rd .
Warning: eigenvalue is a vector! (like “wave vector” in physics)
Theorem. If T is a self-affine tiling, then every measurable eigenfunction
for the system (XT , Rd , µ) coincides with a continuous function µ-a.e.

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Characterization of eigenvalues

Return vectors for the tiling:
Z(T ) := {z ∈ Rd : ∃ T , T 0 ∈ T , T 0 = T + z}.
Theorem [S. 1997] Let T be an aperiodic self-affine tiling with expansion
map φ. Then the following are equivalent for α ∈ Rd :(i) α is an eigenvalue for the topological dynamical system (XT , Rd );(ii) α is an eigenvalue for the measure-preserving system (XT , Rd , µ);(iii) α satisfies the condition:n

lim e 2πihφ z,αi = 1 for all z ∈ Z(T ).

n→∞Boris Solomyak (University of Washington)

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(2)

50 / 69Proof of (i) ⇒ (iii)Let z ∈ Z(T ), i.e., for some T ∈ T we have T + z ∈ T . Let ξ beany point in the interior of T . Then T − ξ and T − z − ξ agree onsome Bε .
Applying ω n we obtain that
ω n (T − ξ) = T − φn ξand
ω n (T − z − ξ) = T − φn z − φn ξ
agree on φn Bε .
By the definition of tiling metric,%(T − φn ξ, T − φn z − φn ξ) → 0, as n → ∞.

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51 / 69Proof (cont.)

A continuous function on a compact metric space is uniformly
continuous, hence|f (T − φn ξ) − f (T − φn z − φn ξ)| → 0, as n → ∞.
Using the eigenfunction equation we obtainn

1 − e 2πihφ z,αi → 0, n → ∞,proving (2).

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Relatively dense set of eigenvalues implies Pisot family

This uses a generalization of Pisot Theorem, due to I. Körneyi (1986) (asimilar result was independently obtained by C. Mauduit (1989)).
Theorem [Körneyi] (partial statement in a special case). Let λ1 , . . . , λr
be distinct complex algebraic integers, with |λj | ≥ 1, and there exist
nonzero αj , j ≤ r , such thatrX
dist 
αj λnj , Z → 0, n → ∞.
j=1
p(λ )

Then {λ1 , . . . , λr } is a Pisot family and αj = q(λjj ) for some p, q ∈ Z[x].

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Relatively dense set of eigenvalues implies Pisot family(cont.)
From (2) we obtain
dist(hφn z, αi, Z) → 0, n → ∞for a return vector z ∈ Rd and an eigenvalue α ∈ Rd .
We assumed φ is diagonalizable over C, hence there is a basis {ei } of
eigenvectors (need to be a bit careful with complex eigenvalues).
The set of return vectors is relatively dense, hence we can choose z sothat all of its coordinates w.r.t. to {ei } are nonzero. And we canmake sure that hei , αi =6 0 for all i.
Now application of Körneyi’s Theorem yields the result.

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Pisot family implies relatively dense set of eigenvalues
To outline the proof, assume for simplicity that1

φ has eigenvalues of multiplicity one (hence the characteristic
polynomial is irreducible);2

all these eigenvalues are real;3

the tiling T is aperiodic.

Without loss of generality, we can assume that
λ1 · · · 0
φ =  ... . . . ...  ,
0

···

λd

and |λ1 | ≤ |λ2 | ≤ . . . ≤ |λd |.
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Pisot family implies relatively dense set of eigenvalues(cont.)
Let
Q[φ] := {p(φ) : p ∈ Q[x]}, Z[φ] := {p(φ) : p ∈ Z[x]}.
The crucial step is to determine the structure of the control point set
C = C(T ).
Structure Theorem Let T be a self-affine tiling of Rd with adiagonalizable expansion map φ whose eigenvalues are all algebraicallyconjugate. Suppose that the additional assumptions (1)-(3) are satisfied.
Then there exists a vector α ∈ Rd such thatC ⊂ Z[φ]α.

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Pisot family implies relatively dense set of eigenvalues(cont.)First we use the Structure Theorem to finish the proof.
Observe that the set of return vectors satisfies Z ⊂ C − C ⊂ Z[φ]αThe set of control points C is relatively dense, hence the vector
α := [a1 , . . . , ad ]T

has all aj 6= 0

Considerβ := [a1−1 , . . . , ad−1 ]T .
We claim that the set {φj β}d−1
j=0 is contained in the set of
eigenvalues. This set is a basis of Rd (over R), and since the set of
eigenvalues forms an additive group, the proof will be complete.
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Pisot family implies relatively dense set of eigenvalues(cont.)We have for x = φi α, γ = φj β:
hφn x, γi = hφn+i α, φj βi =dXλn+i+j→ 0 (mod Z).
k

k=1

The convergence follows from the Pisot family property.

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58 / 69Proof sketch of Structure Theorem
Consider hCiQ , the linear space over Q generated by the set of control
points C. The FLC property implies that this space isfinite-dimensional over Q.
An easy argument shows that Q[φ] is a field. Since φ(C) ⊂ C, it
follows that
H := hCiQ = hCiQ[φ] .
Our result will follow if we show that H has dimension one as a linear
space over Q[φ].
Choose any control point ξ ∈ C with all non-zero coordinates. We
can define σ : H → Q[φ]ξ as a Q[φ]-module homomorphism which is
identical on Q[φ]ξ (basically, a projection commutingφ), and
S with−kthen let σ 0 be the restriction of σ to the set C∞ = ∞φC.
k=0Boris Solomyak (University of Washington)

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59 / 69Proof sketch of Structure Theorem (cont.)

The function σ 0 is uniformly continuous on the dense set C∞ , and so
can be extended by continuity to Rd . The extension commutes with
φ. This proceeds essentially following Thurston’s and Kenyon’s
arguments (see part II).
Next we show that the extension of σ 0 to Rd , which we also denote
σ 0 , is linear over R. This will be sufficient, since σ 0 is the identity on arelatively dense set, hence it is the identity on all of Rd whenceC ⊂ Q[φ]ξ, as desired.
Establishing linearity is the hardest part of the proof; it follows thescheme worked out in [Kenyon & S. 2010].

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60 / 69Proof sketch of Structure Theorem (end)σ 0 is Lipschitz on all lines parallel to the eigenvector e1 of φ with thesmallest in modulus eigenvalue λ1 .
By Rademacher’s Theorem, this implies that σ 0 is differentiable
almost everywhere in the e1 direction.
Another useful property of σ 0 is that it depends only on the tile type,
up to an additive constant.
Taking points of differentiability, “blowing up” by the expansion φand using the last item we prove that σ 0 is affine linear on all lines
parallel to e1 .
Projections of C to the coordinate axes cannot be discrete; this yields
linearity on the entire Rd .

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Remark. The last “non-discreteness” claim may seem surprising: why
can’t we have the tiling T as a direct product of, say, two tilings
corresponding to a partition of the set of eigenvalues of φ? The answer isthat this is prohibited by the Characterization of Expansions III from
[Kenyon & S. 2010]: it is impossible to split the set of eigenvalues of φ insuch a way that both parts form a Perron family.

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IV. Topological rigidity (after J. Kwapisz)
It turns out that the topology of the tiling space already determines
dynamics!
Theorem [J. Kwapisz, ETDS, published online 2011] Suppose that T0 ande
Te0 are self-similar aperiodic tilings of Rd , with expansions φ and φ,erepsectively. Let X = XT0 and X = XTe0 be the corresponding tiling spaces.
e , then there is a linear
If there is a homeomorphism h0 : X → X
dde conjugating
isomorphism A : R → R and a homeomorphism h : X → Xe:the translation action on X to the rescaling of the translation action on Xh(T − g ) = h(T ) − Ag , T ∈ X , g ∈ Rd .
Open Question: is the same true for self-affine tilings?

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Global linearization

Theorem [J. Kwapisz] In the context of the last theorem, if additionallyeh0 (T0 ) = Te0 , then h0 is homotopic to some homeomorphism hlin : X → X
such that hlin (T0 ) = Te0 and hlin is linear, i.e. for some linear map L,
hlin (T − g ) = hlin (T ) − Lg , g ∈ Rd .

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Rough sketch of the proof

Aperiodic tiling spaces are locally: Rd × Cantor set.
e has to take orbits to orbits:
A homeomorphism h0 : X → X
h0 (T − g ) = h0 (T ) − α(T , g ),where α(T , g ) is a cocycle over the Rd -action on X :α(T , g1 + g2 ) = α(T , g1 ) + α(T − g1 , g2 ).

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Step 1: Averaging of cocycles

Idea: cocycles are linear on large scales.
Lemma There exists a linear isomorphism Aα : Rd → Rd such that
lim

s→∞

α(T , sv )= Aα v , ∀ T ∈ X , v ∈ Rd .
s

The limit is uniform in the sense that
lim

sup

s→∞ |v |=1, T ∈X

Boris Solomyak (University of Washington)α(T , sv )− Aα v = 0.
s

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Step 2: Using that φ, φe are similarities (conformality)

Technical lemma. For any linear isomorphism Aα , if each of φ, φe issimilar to an orthogonal transformation of Rd , then there are sequences
mk , nk → ∞ of integers and a linear isomorphism A such thatkφe−nk φmk − I k → 0,kφe−nk Aα φmk − Ak → 0,
sup kφmk kkφe−nk k < ∞ and sup kφ−mk kkφenk k < ∞.
kkThis follows from the compactness of the orthogonal group.

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Step 3: ”Ironing” homeomorphisms to conjugacies
Aperiodicity of the tiling T0 implies that the substitution (“inflate andsubdivide”) action ω : X → X is invertible (“recognizability”); it ise →Xe.
hyperbolic in the Smale space sense; similarly for ω
e: X
The next idea goes back to linearization results in hyperbolic
dynamics (compare with Thurston’s argument!)
Step 1 (averaging) tells us that h0 is approximately linear on a large
scale, and we attempt to bring this linearity to the microscopic scaleby renormalizing h0 with the aid of high iterates ω mk and ω
e −nk .
Lemma. The family of homeomorphisms
hk := ωe −nk ◦ h0 ◦ ω mk

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is equicontinuous.

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Conclusion of proof sketch

Interchanging the roles of ω and ωe we see that hk−1 is also anequicontinuous family.
By passing to a subsequence we can ensure that hk and hk−1 converge
uniformly. Thus they must converge to h and h−1 respectively, where
h is a homeomorphism.
Then one shows thath(T − g ) = h(T ) − Ag , T ∈ X , g ∈ Rd .
where A is from the Technical Lemma.

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