# Substitutions with Vanishing Rotationally Invariant First Cohomology

Authors: Juan García Escudero
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

The cohomology groups of tiling spaces with three-fold and nine-fold symmetries
are obtained. The substitution tilings are characterized by the fact that they have vanishing
first cohomology group in the space of tilings modulo a rotation. The rank of the rational first
cohomology, in the tiling space formed by the closure of a translational orbit, equals the Euler
totient function evaluated atN if the underlying rotation group is ZN. When the symmetries
are of crystallographic type, the cohomologies are infinitely generated.


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## Body

## 1. Introduction

The computation of topological invariants of tiling spacesΩ is a subject of increasing research activity, mainly since the work of Anderson and Putnam [1].We use substitution rules to define collections of tilings. The expanding hierarchical structures that arise are essentially the same at each level, because they are described by a single substitution map. More general formalisms for handling general spaces of hierarchical tilings are studied in [2].The simplest invariants of tiling spaces are the Čech cohomology groups. A relevant fact, also from the point of view of applications, is that the Čech cohomology is related to the gap distribution in the spectrum of the Schroedinger operator with a potential associated to a particular tiling. The Čech cohomology groups may be interpreted also in terms of certain tiling properties. For projection tilings, there is in the first cohomology at least a subgroup isomorphic to the reciprocal lattice of the tiling [3], and hence for a tiling with N-fold symmetry its minimal dimension is given by the Euler totient function Φ(N). If the stretching factor of a one-dimensional substitution tiling is a Pisot number of algebraic degree k and the rank of the first rational Čech cohomology is also k, then the number of appearances of a patch in a return word is determined by the Euclidean length of the return word [4]. Based on this results, it has been shown in [5] that the top-dimensional cohomology for a tiling in Rd governs patch frequencies and the number of appearances of a patch in a finite region.A class of tiling spaces with fivefold symmetry has been analyzed in [6]. The rank of the first cohomology coincides with Φ(5) and its restriction in the space of tilings modulo a fivefold rotation is zero. In this paper we study several examples of tiling spaces with crystallographic and noncrystallographic symmetries with analogous properties. They have vanishing first cohomology in the space of tilings modulo a rotation Ω̅. When the integer first cohomology in Ω is finitely generated, as in the tilings with ninefold and pentagonal symmetries, its rank coincides with the Eulers totient function. In the cases with crystallographic symmetries, it is the rank of the rational first cohomology which equals the Eulers totient function.


## 2. Geometric Constructions of Substitution Tilings withd-Fold Symmetry and Algebraic Surfaces

An extension of the methods introduced in [7] for tilings with odd symmetries nondivisible by three to all the symmetries has been given in [8]. There are four types of geometric constructions consisting in straight lines in d orientations: (A) for d=2m, (B) for d=2m+1, (C) for d=6m, and (D) for d=6m+3. They produce simplicial arrangements SX,X=A,B,C,D, which are the basis for the generation of planar tilings with arbitrarily high symmetry. The systems with d odd are included in the systems with d even: (B) is included as a subsystem of (A), and (D) is included in (C). The fact that, in general, a system with d lines is included in another with 2d lines allows to introduce in a systematic way arrows on the edges, which are necessary for the definition of the inflation rules generating face-to-face tilings [6]. We give an alternative formulation which simplifies the construction. Systems of lines, equivalent up to scaling, translation, and rotation, to those obtained with the constructions (A) and (B) are {Λk,d(x,y,0)=0},k=1,2,…d, where(2.1)Λk,d(x,y,τ):=y-(cos⁡(k2πd)-x)tan(kπd+τ)-sin⁡(k2πd)The simplicial arrangementsSC,SD and their mirror images S̅C,S̅D can be obtained also by using Λk,d(x,y,τ) for certain values of τ. For fixed d we consider the subset formed by the lines Λ2k-1,2d(x,y,τ) with k=1,2,…d and τ=nπ/4d. When d=3m, we get, for n=0,1,2,…8d, successive rotations of two types of simple arrangements of lines (no more than two lines intersect) Σ1,Σ2 and the two types of simplicial arrangements SA,SC and SB,SD for d even and odd, respectively. In addition to the nontriangular shapes, Σ1,Σ2, have two sets of triangular prototiles: Σ1 has the prototiles of SA,SB while Σ2 and its mirror image Σ̅2 have the prototiles of SC,SD and S̅C,S̅D. In Figure 1(a) we show the case d=5 with (from top to bottom) n=0,1,2,…6. Two rotated versions of SB correspond to n=2,6, the others are simple arrangements of lines Σ1 obtained when n=0,4. The case d=6 with n=0,1,2,…6 can be seen in Figure 1(b). Now we have two simplicial arrangements: SC with 3 prototiles when n=2 and SA with one prototile for n=6. When n=0, the simple arrangement Σ1 has 6 copies of the prototile appearing in SA, and when n=4, Σ2 has seven triangles with 3 different shapes (prototiles in SC). In both cases the prototiles have edges with three different lengths as in Section 4.Line configurations described by (2.1) for (from top to bottom) (a) d=5;n=0(Σ1),1,2(SB),3,4(Σ1),5,6(SB) and (b) d=6;n=0(Σ1),1,2(SC),3,4(Σ2),5,6(SA).
(a)(b)The line configurations ford=9 can be seen in Figure 2(a) when n=0,1,2,3 and Figure 2(b) for n=4,5,6. We get Σ1 when n=0,12, while Σ2 and Σ̅2 are obtained with n=4,8. The simplicial arrangement SB corresponds to n=6 and we get SD,S̅D when n=2,10. By using the notation in [9], the prototiles appearing in Σ1 and SB are a,b,c (we use capital letters in Section 3 to describe the same tiles), while in Σ1 and SD there are seven prototiles a,b,c,d,e,f,g. All the prototiles now have edges with four different lengths as in Section 3.Simple and simplicial arrangements ford=9 (a) n=0(Σ1),1,2(SD),3, (b) n=4(Σ2),5,6(SB).
(a)(b)The simple arrangementsΣ1,Σ2 are on the basis of a construction of algebraic surfaces with many real nodes proposed recently [10]. The surfaces have affine equations S(x,y,z):=F(x,y)+λF(z,0)+μ=0,λ,μ∈R, where the plane curve F(x,y) is a product of lines corresponding to a simple arrangement. Real variants of Chmutov surfaces [11] are based on Σ1 as shown in [12]. On the other hand Σ2, and its mirror image Σ̅2, have one more triangle than Σ1, and this property can be used to construct surfaces with more singularities. The dynamical formulation of the line configurations given by (2.1) can then be used to get deformations of the surfaces by varying τ, where some singularities disappear. Other deformations of interest, giving smooth surfaces, are obtained by varying μ.


## 3. Cohomology Groups of Tiling Spaces with Ninefold Symmetry

The tiling spacesΩX are formed by the closure of the translational orbit of one tiling X. A substitution σ is said to force the border [13] if there is a positive integer n such that if t1 and t2 are two tiles of the same type, then the two level-n supertiles σn(t1) and σn(t2) have, up to translation, the same pattern of neighboring tiles. If a substitution forces the border, then the tiling space is the inverse limit of the Anderson-Putnam complex Γ0, which contains one copy of every kind of tile that is allowed with some edges identified [1]. For 2D tiling spaces the cohomology classes are generated by the vertices, edges, and faces of Γ0.In addition toΩX we consider also another type of tiling spaces Ω̅X. The finite rotation group ZN associated to the tiling acts on ΩX, and the quotient ΩX/ZN yields the space Ω̅X of tilings modulo a rotation about the origin [14].We study first the cohomology of a ninefold symmetry tiling spaceΩN. The substitution rules for this and other composite patterns were studied in [9]. They can be obtained by using SB in Figure 2(b).Up to mirror reflection and rotation, the tiling has three triangular prototiles𝒯 with inflation or substitution rules (level-1 supertiles) given in Figure 3. Iteration of the inflation rules shows that the tiling has 18 vertex configurations. After three inflation steps, they are transformed into just two vertex configurations that we call a and b representing a threefold and a ninefold star, respectively (Figure 4). The substitution forces the border in four steps. The analysis of the vertex sequences appearing in the level-4 superedges shows that there are four edge types αb,a,βa,b,γa,a,δb,a, where the subindexes denote the vertices appearing on the edge borders. The lengths of α,β,γ,δ are s1,s2,s3,s4, with sν≡sin(νπ/9).Figure 3
Inflation rules for the ninefold symmetry tiling.Figure 4
Level-3 supertiles in the ninefold symmetry tiling.The rotation groupZ9 acts freely on edges and tiles. The two vertices satisfy a=r3a,b=rb with r9=1. We have six tile types A=𝒯(r4γ,r4β,r5α),Ã=𝒯(r4γ,r3α,r5β),B=𝒯(δ,r8γ,rα),B̃=𝒯(δ,r7α,γ),C=𝒯(δ,r3γ,r6β),C̃=𝒯(δ,r3β,r5γ), and each appears in 9 orientations. The Anderson-Putnam complex Γ0 has Euler characteristic χ=22.The three irreducible representations ofZ9 over the integers are the one-dimensional scalar (r=1), a two-dimensional and a six-dimensional representation. The 2D and the 6D representations have r acting by multiplication on the rings ℛ1=Z[r]/(r2+r+1) and ℛ2=Z[r]/(r6+r3+1), respectively. In this case the vertex b appears only in the scalar representation, the vertex a in the scalar and 2D representations, while the edges and faces appear in all representations. Vertices, edges, and faces form a basis for the spaces of chains Ck with k=0,1,2. The boundary maps ∂k:Ck+1↦Ck for k=0,1 are(3.1)∂1=(r5-r3-rr700r4-r500r6-r3-r4r4-r81r3-r5001-11-1)
in all representations and
(3.2)∂0=(-r210-11-101)
in the scalar and 2D representations, but with appropriate modification having in mind that only the vertex a appears in the 2D representation. If the cochain groups are denoted by Ck, then, for k=0,1, the coboundary maps δk:Ck↦Ck+1 are the transposes of the matrices ∂k given above with r replaced with rT=r-1.Scalar Representationr=1
In the scalar representation,C0=Z2,C1=Z4,C2=Z6. We have rankδ0=1 and rankδ1=3. The cohomologies of Γ0 in this representation are ℋ0=Kerδ0=Z,ℋ1=Kerδ1/Imδ0=0, and ℋ2=Z6/Imδ1=Z3.2D Representation
There is one vertex in the 2D representation, andC0,C1,C2 are free modules of dimensions 1,4,6 over the ring ℛ1. The rank of δ0 over ℛ1 is 1, the matrix δ1 has rank 3 over ℛ1, and, as abelian groups, we have ℋ0=ℋ1=0,ℋ2=Z6.6D Representation
There are no vertices in the 6D representation, andC1,C2 are free modules of dimensions 1 and 3 over the ring ℛ2. The map δ1 has rank 3 over ℛ2 and ℋ0=0,ℋ1=Z6,ℋ2=Z18.We have obtained the cohomology of the complexΓ0 which is enough because the tiling forces the border [1]. To get the cohomology of the tiling space ΩN, we need to compute the direct limit of the cohomologies under the substitution.The substitution matrix on vertices is(3.3)σ0=(r2001),
while the matrix on 1-chains is(3.4)σ1=(000100rr501r51rr5rr5),
and the substitution on 2-chains(3.5)σ2=(0r0r1r10101r0r50r5+r8r4r5+r8r50r2+r50r2+r5r6r6rr6r+r41+r3+r6r+r4+r71r41+r6r41+r3+r6r+r4+r7).The induced matrices on cochainsσk* are obtained by transposition of the matrices σk with r replaced with r-1. They are isomorphisms, and the direct limit of each ℋk is simply ℋk. By taking into account all the irreducible representations, we get the following(3.6)Ȟ0(ΩN)=Z,Ȟ1(ΩN)=Z6,Ȟ2(ΩN)=Z27.The cohomology ofΩ̅X is the rotationally invariant part of the cohomology of ΩX, and we have(3.7)Ȟ0(Ω̅N)=Z,Ȟ1(Ω̅N)=0,Ȟ2(Ω̅N)=Z3.Another tiling space with noncrystallographic symmetries and vanishing first cohomology inΩ̅X was analyzed in [6]. The pentagonal tiling space ΩΞ+ has the same prototiles as the Robinson decomposition of the Penrose tilings: A,B with edges α,β and lengths s1,s2, with sν≡sin(νπ/5). The inflation rules can be obtained from SB in Figure 1(a) with arrows as indicated in [6]. The cohomology groups are(3.8)Ȟ0(ΩΞ+)=Z,Ȟ1(ΩΞ+)=Z4,Ȟ2(ΩΞ+)=Z14,
and the rotationally invariant part is ℋ̌0(Ω̅Ξ+)=Z,ℋ̌1(Ω̅Ξ+)=0,ℋ̌2(Ω̅Ξ+)=Z2.Fractal tilings related with Penrose patterns have been studied in [15, 16]. In addition to the computation of the cohomologies, the analysis of the patterns of vertex configurations, appearing in Ξ+ on the level-6 superedges, can be used also to construct tilings whose edges and tiles have irregular shapes. In Figure 5(a) we have shown the vertex sequences arising in the A-type level-6 supertile (see Figure  12 in [6]). The borders of the new prototile shapes are formed by concatenation of scaled copies of oriented α and β. The substitution rules for the corresponding pattern can be seen in Figure 5(b).Pentagonal tiling with irregular shapes: (a) construction of the boundaries, (b) inflation rules.
(a)(b)Tiling spaces with crystallographic symmetries also may have zero first cohomology. For instance the chair tiling spaceΩCH has [14](3.9)Ȟ0(ΩCH)=Z,Ȟ1(ΩCH)=Z[12]2,Ȟ2(ΩCH)=Z[14]⊕Z[12]2
with ℋ̌0(Ω̅CH)=Z,ℋ̌1(Ω̅CH)=0,ℋ̌2(Ω̅CH)=Z[1/4]. In the next section we study other tiling spaces with crystallographic symmetries, which also have ℋ̌1(Ω̅T)=0.


## 4. Tiling Spaces with Threefold Symmetry

### 4.1. Equithirds Tiling

The equithirds tiling was obtained independently by L. Danzer and B. Kalahurka ([5] and references therein). It is based on the substitution of Figure 6. The prototiles are an equilateral triangle of side length 1 and an isosceles triangle with sides of length 1,1, and 3.Figure 6
Inflation rules for the equithirds tiling.The analysis of the five vertex configurations shows that, after one inflation step, they are transformed into one: vertexa. The substitution forces the border in two steps. We denote the edges by α,β with lengths 1,3, and the rotation group Z3 acts freely on them.We have four tile typesA=𝒯(α,rα,r2α),A̅=𝒯(α,rα,r2α),B=𝒯(α,β,rα),B̅=𝒯(α,r2α,r2β), with r3=1. The tiles B,B̅ appear in three orientations and A,A̅ in one. The Anderson-Putnam complex Γ0 has therefore Euler characteristic χ=3.The two irreducible representations ofZ3 over the integers are the one-dimensional scalar (r=1) and a two-dimensional vector representation. The vector representation has r acting by multiplication on the ring ℛ1=Z[r]/(r2+r+1). The edges and the tiles B,B̅,D appear in all representations, while the vertex and the tiles A,A̅ appear only in the scalar representation. The boundary maps are ∂0=0 and
(4.1)∂1=(1-11-r1-r200-1r2).Scalar Representationr=1
In the scalar representation,C0=Z,C1=Z2,C2=Z4. In this case rankδ1=2 and we have ℋ0=Kerδ0=Z,ℋ1=Kerδ1/Imδ0=0, and ℋ2=Z4/Imδ1=Z2.Vector Representation
There are two tiles in the 2D representation, andC1,C2 are free modules of dimensions 2 over the ring ℛ1. The rank of δ1 over ℛ1 is 1 and we have ℋ0=0,ℋ1=Z2 and ℋ2=Z2.The cohomology of the complexΓ0 is therefore ℋ0(Γ0)=Z,ℋ1(Γ0)=Z2 and ℋ2(Γ0)=Z4. Now we compute the direct limit of the cohomologies under the substitution.The substitution on the vertex is the identity. The substitution on edges is represented by(4.2)σ1=(3003),
and the substitution on 2-chains is(4.3)σ2=(301103111+r+r21+r+r23+r+r21+r1+r+r21+r+r21+r23+r+r2).In the scalar representation the direct limits of eachℋk, under the action of the induced matrices on cochains σk*, are ℋk for k=0,1. We can take a basis for ℋ2 in terms of the canonical basis of Z4 formed by e1,e1+e3 with the classes of v3=e4-e3,v4=e2-e1 cohomologous to zero. The eigenvalues of σ2* are 9,3,1 with eigenvectors v1=(1,1,1,1)T,v2,v3,v4=(-3,-3,1,1)T. Its action on ℋ2 gives the direct limit Z⊕Z[1/3].In the vector representation the direct limit ofℋ1 is Z[1/3]2. A base for ℋ2 in this representation is given by the vector (-1,0)T~(r,1)T which is the eigenvector of σ2* with eigenvalue 1. The direct limit is now Z2. The cohomology groups for the equithirds tiling space ΩEQ are (see also [5])(4.4)Ȟ0(ΩEQ)=Z,Ȟ1(ΩEQ)=Z[13]2,Ȟ2(ΩEQ)=Z3⊕Z[13].The rotationally invariant part of the cohomology is(4.5)Ȟ0(Ω̅EQ)=Z,Ȟ1(Ω̅EQ)=0,Ȟ2(Ω̅EQ)=Z⊕Z[13].


### 4.2. The Tiling SpacesΩT

Although the paper [7] deals with tilings having odd symmetries nondivisible by three, the authors consider a case which can be derived with a set of six lines in the plane. The tiling spaces ΩT that we treat now are based on such system of lines. In the construction discussed in Section 2, it corresponds to the simplicial arrangements  SA; SC in Figure 1(b).The substitution rules forT are represented in Figure 7, and its seven vertex configurations can be seen in Figure 8. After two inflation steps the vertex configurations are transformed into the configurations 1 and 2 which are denoted by a,b. The substitution forces the border also in two steps: the vertex sequences in the level-2 superedges (Figure 9) are α:53,β:676̅,γ:2354 with lengths s1,s2,s3, sν≡sin(νπ/6), and the vertices in the edge borders are α:ab,β:aa,γ:ab. The rotation group Z3 acts freely on the edges and a=ra,b=rb with r3=1.Figure 7
Inflation rules forT.Figure 8
Vertex configurations inT.Figure 9
Level-2 supertiles inT.The four tile types areA=𝒯(α,r2α,β),D=𝒯(β,rα,γ),D̅=𝒯(γ,α,r2β),andE=𝒯(β,rβ,r2β), and each appears in 3 orientations, except E which appears in only one. The Euler characteristic of Γ0 is χ=3.In this case the edges and the tilesA,D,D̅ appear in all representations of Z3, while the vertices a,b, and the tile E appear only in the scalar representation. The boundary maps are
(4.6)∂0=(-10-1101),∂1=(1-r2r-10-11r2-10-110).Scalar Representation  r=1
In this representation,C0=Z2,C1=Z3,C2=Z4. Here rankδ0=1 and rankδ1=2 with cohomologies ℋ0=Kerδ0=Z,ℋ1=Kerδ1/Imδ0=0 and ℋ2=Z4/Imδ1=Z2.Vector Representation
There are three edges and three tiles in the 2D representation, andC1,C2 are free modules of dimension 3 over the ring ℛ1. The rank of δ1 over ℛ1 is 2, and, as abelian groups, we have ℋ0=0, ℋ1=Z2 and ℋ2=Z2.Adding up the contributions of each representations, we get the cohomology of the complexΓ0: ℋ0(Γ0)=Z, ℋ1(Γ0)=Z2, and ℋ2(Γ0)=Z4.The substitution on the vertices is the2×2 identity matrix. The substitution on 1-chains is
(4.7)σ1=(002r02r2010r2),
and the substitution on 2-chains is(4.8)σ2=(0r2r0r201+r211r+r2010110).In the scalar representation the direct limits ofℋk for k=0,1 are ℋk. A base for ℋ2 in this representation is given by the vectors u3=(0,0,1,0)T,u4=(0,0,0,1)T, and the classes of uk,k=1,2, with u1=(0,1,-1,0)T,u2=(-1,0,2,-1)T, are cohomologous to zero. The action of σ2* applied twice to u3 gives 4u3~4u4, and the direct limit yields a contribution of Z[1/2].In the vector representationv3=(1,1-r,-2r)T forms a base for ℋ1. The matrix σ1* has eigenvalues -r,2r,2r with eigenvectors (-r2,0,1)T,(r2,0,2)T,(0,1,0)T, and the direct sum of the eigenspaces is not isomorphic to all Z6. The action of σ1* gives a contribution of Z[1/2]2 to the first cohomology. A base for ℋ2 is given by u3=(0,1,0)T with u1=(1-r,r2,1)T,u2=(-1,-r2,0)T cohomologous to zero. We have σ1*3u3~u3, and the direct limit of ℋ2 is Z2. The cohomology groups are then(4.9)Ȟ0(ΩT)=Z,Ȟ1(ΩT)=Z[12]2,Ȟ2(ΩT)=Z2⊕Z[12].The rotationally invariant part of the cohomology is(4.10)Ȟ0(Ω̅T)=Z,Ȟ1(Ω̅T)=0,Ȟ2(Ω̅T)=Z[12].With the purpose of studying the meaning of the topological invariants in tiling spaces, Sadun shows that there are patches in the equithirds tiling that play a role analogous to return words in one dimension [5]. The control patches are the equilateral triangle and a rhombus, formed by two isosceles triangles, in three different orientations. This case illustrates the more general property that if the rank of the second rational cohomology is k, then the number of control patches is also k. The number of appearances of any patch in a sufficiently substituted rhombus in the equithirds tiling is then determined by the number of appearances of the four control patches.For the tilingT we may find four patches that generate the tilings, because they obey certain substitution rules induced by the inflation on the triangular prototiles. The patches Y,Z, and their substitution rules are represented in Figure 10. The patch content in the inflation rules is(4.11)Y⟼YZZ̅Z̿Z⟼Z̅YYY,
where Z̅ and Z̿ represent rotated versions of Z by 120 and 240 degrees.Figure 10
Inflation rules for the patchesY and Z in T.Although we have studied in detail only some particular cases, we can see that the general constructions of substitutions proposed in [7, 8] are a rich source of tiling spaces with unusual properties. The tilings with noncrystallographic symmetries studied in this paper, derived with such constructions, have vanishing rotationally invariant first cohomology, but this is not true in general. For instance, the analysis of a tiling T* obtained with the same basic construction as T (which has inflation factor s3/s1) but with scaling factor s2/s1 shows that the associated tiling space has in ℋ̌1 a nontrivial rotationally invariant piece. The tiling T* has six prototiles with the same three shapes A,D,E as in T. The mirror images of A,D, and E shape level-1 supertiles are different in T*, while in T we have two mirror inflation rules only for D (Figure 7). The first cohomology for the tiling space ΩT* in the scalar representation of Z3 is now Z. An open question is, for general tiling spaces, which property distinguishes them in relation to the existence of a nontrivial rotationally invariant component in the first cohomology group.


## 4.1. Equithirds Tiling

The equithirds tiling was obtained independently by L. Danzer and B. Kalahurka ([5] and references therein). It is based on the substitution of Figure 6. The prototiles are an equilateral triangle of side length 1 and an isosceles triangle with sides of length 1,1, and 3.Figure 6
Inflation rules for the equithirds tiling.The analysis of the five vertex configurations shows that, after one inflation step, they are transformed into one: vertexa. The substitution forces the border in two steps. We denote the edges by α,β with lengths 1,3, and the rotation group Z3 acts freely on them.We have four tile typesA=𝒯(α,rα,r2α),A̅=𝒯(α,rα,r2α),B=𝒯(α,β,rα),B̅=𝒯(α,r2α,r2β), with r3=1. The tiles B,B̅ appear in three orientations and A,A̅ in one. The Anderson-Putnam complex Γ0 has therefore Euler characteristic χ=3.The two irreducible representations ofZ3 over the integers are the one-dimensional scalar (r=1) and a two-dimensional vector representation. The vector representation has r acting by multiplication on the ring ℛ1=Z[r]/(r2+r+1). The edges and the tiles B,B̅,D appear in all representations, while the vertex and the tiles A,A̅ appear only in the scalar representation. The boundary maps are ∂0=0 and
(4.1)∂1=(1-11-r1-r200-1r2).Scalar Representationr=1
In the scalar representation,C0=Z,C1=Z2,C2=Z4. In this case rankδ1=2 and we have ℋ0=Kerδ0=Z,ℋ1=Kerδ1/Imδ0=0, and ℋ2=Z4/Imδ1=Z2.Vector Representation
There are two tiles in the 2D representation, andC1,C2 are free modules of dimensions 2 over the ring ℛ1. The rank of δ1 over ℛ1 is 1 and we have ℋ0=0,ℋ1=Z2 and ℋ2=Z2.The cohomology of the complexΓ0 is therefore ℋ0(Γ0)=Z,ℋ1(Γ0)=Z2 and ℋ2(Γ0)=Z4. Now we compute the direct limit of the cohomologies under the substitution.The substitution on the vertex is the identity. The substitution on edges is represented by(4.2)σ1=(3003),
and the substitution on 2-chains is(4.3)σ2=(301103111+r+r21+r+r23+r+r21+r1+r+r21+r+r21+r23+r+r2).In the scalar representation the direct limits of eachℋk, under the action of the induced matrices on cochains σk*, are ℋk for k=0,1. We can take a basis for ℋ2 in terms of the canonical basis of Z4 formed by e1,e1+e3 with the classes of v3=e4-e3,v4=e2-e1 cohomologous to zero. The eigenvalues of σ2* are 9,3,1 with eigenvectors v1=(1,1,1,1)T,v2,v3,v4=(-3,-3,1,1)T. Its action on ℋ2 gives the direct limit Z⊕Z[1/3].In the vector representation the direct limit ofℋ1 is Z[1/3]2. A base for ℋ2 in this representation is given by the vector (-1,0)T~(r,1)T which is the eigenvector of σ2* with eigenvalue 1. The direct limit is now Z2. The cohomology groups for the equithirds tiling space ΩEQ are (see also [5])(4.4)Ȟ0(ΩEQ)=Z,Ȟ1(ΩEQ)=Z[13]2,Ȟ2(ΩEQ)=Z3⊕Z[13].The rotationally invariant part of the cohomology is(4.5)Ȟ0(Ω̅EQ)=Z,Ȟ1(Ω̅EQ)=0,Ȟ2(Ω̅EQ)=Z⊕Z[13].


## 4.2. The Tiling SpacesΩT

Although the paper [7] deals with tilings having odd symmetries nondivisible by three, the authors consider a case which can be derived with a set of six lines in the plane. The tiling spaces ΩT that we treat now are based on such system of lines. In the construction discussed in Section 2, it corresponds to the simplicial arrangements  SA; SC in Figure 1(b).The substitution rules forT are represented in Figure 7, and its seven vertex configurations can be seen in Figure 8. After two inflation steps the vertex configurations are transformed into the configurations 1 and 2 which are denoted by a,b. The substitution forces the border also in two steps: the vertex sequences in the level-2 superedges (Figure 9) are α:53,β:676̅,γ:2354 with lengths s1,s2,s3, sν≡sin(νπ/6), and the vertices in the edge borders are α:ab,β:aa,γ:ab. The rotation group Z3 acts freely on the edges and a=ra,b=rb with r3=1.Figure 7
Inflation rules forT.Figure 8
Vertex configurations inT.Figure 9
Level-2 supertiles inT.The four tile types areA=𝒯(α,r2α,β),D=𝒯(β,rα,γ),D̅=𝒯(γ,α,r2β),andE=𝒯(β,rβ,r2β), and each appears in 3 orientations, except E which appears in only one. The Euler characteristic of Γ0 is χ=3.In this case the edges and the tilesA,D,D̅ appear in all representations of Z3, while the vertices a,b, and the tile E appear only in the scalar representation. The boundary maps are
(4.6)∂0=(-10-1101),∂1=(1-r2r-10-11r2-10-110).Scalar Representation  r=1
In this representation,C0=Z2,C1=Z3,C2=Z4. Here rankδ0=1 and rankδ1=2 with cohomologies ℋ0=Kerδ0=Z,ℋ1=Kerδ1/Imδ0=0 and ℋ2=Z4/Imδ1=Z2.Vector Representation
There are three edges and three tiles in the 2D representation, andC1,C2 are free modules of dimension 3 over the ring ℛ1. The rank of δ1 over ℛ1 is 2, and, as abelian groups, we have ℋ0=0, ℋ1=Z2 and ℋ2=Z2.Adding up the contributions of each representations, we get the cohomology of the complexΓ0: ℋ0(Γ0)=Z, ℋ1(Γ0)=Z2, and ℋ2(Γ0)=Z4.The substitution on the vertices is the2×2 identity matrix. The substitution on 1-chains is
(4.7)σ1=(002r02r2010r2),
and the substitution on 2-chains is(4.8)σ2=(0r2r0r201+r211r+r2010110).In the scalar representation the direct limits ofℋk for k=0,1 are ℋk. A base for ℋ2 in this representation is given by the vectors u3=(0,0,1,0)T,u4=(0,0,0,1)T, and the classes of uk,k=1,2, with u1=(0,1,-1,0)T,u2=(-1,0,2,-1)T, are cohomologous to zero. The action of σ2* applied twice to u3 gives 4u3~4u4, and the direct limit yields a contribution of Z[1/2].In the vector representationv3=(1,1-r,-2r)T forms a base for ℋ1. The matrix σ1* has eigenvalues -r,2r,2r with eigenvectors (-r2,0,1)T,(r2,0,2)T,(0,1,0)T, and the direct sum of the eigenspaces is not isomorphic to all Z6. The action of σ1* gives a contribution of Z[1/2]2 to the first cohomology. A base for ℋ2 is given by u3=(0,1,0)T with u1=(1-r,r2,1)T,u2=(-1,-r2,0)T cohomologous to zero. We have σ1*3u3~u3, and the direct limit of ℋ2 is Z2. The cohomology groups are then(4.9)Ȟ0(ΩT)=Z,Ȟ1(ΩT)=Z[12]2,Ȟ2(ΩT)=Z2⊕Z[12].The rotationally invariant part of the cohomology is(4.10)Ȟ0(Ω̅T)=Z,Ȟ1(Ω̅T)=0,Ȟ2(Ω̅T)=Z[12].With the purpose of studying the meaning of the topological invariants in tiling spaces, Sadun shows that there are patches in the equithirds tiling that play a role analogous to return words in one dimension [5]. The control patches are the equilateral triangle and a rhombus, formed by two isosceles triangles, in three different orientations. This case illustrates the more general property that if the rank of the second rational cohomology is k, then the number of control patches is also k. The number of appearances of any patch in a sufficiently substituted rhombus in the equithirds tiling is then determined by the number of appearances of the four control patches.For the tilingT we may find four patches that generate the tilings, because they obey certain substitution rules induced by the inflation on the triangular prototiles. The patches Y,Z, and their substitution rules are represented in Figure 10. The patch content in the inflation rules is(4.11)Y⟼YZZ̅Z̿Z⟼Z̅YYY,
where Z̅ and Z̿ represent rotated versions of Z by 120 and 240 degrees.Figure 10
Inflation rules for the patchesY and Z in T.Although we have studied in detail only some particular cases, we can see that the general constructions of substitutions proposed in [7, 8] are a rich source of tiling spaces with unusual properties. The tilings with noncrystallographic symmetries studied in this paper, derived with such constructions, have vanishing rotationally invariant first cohomology, but this is not true in general. For instance, the analysis of a tiling T* obtained with the same basic construction as T (which has inflation factor s3/s1) but with scaling factor s2/s1 shows that the associated tiling space has in ℋ̌1 a nontrivial rotationally invariant piece. The tiling T* has six prototiles with the same three shapes A,D,E as in T. The mirror images of A,D, and E shape level-1 supertiles are different in T*, while in T we have two mirror inflation rules only for D (Figure 7). The first cohomology for the tiling space ΩT* in the scalar representation of Z3 is now Z. An open question is, for general tiling spaces, which property distinguishes them in relation to the existence of a nontrivial rotationally invariant component in the first cohomology group.



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*Source: 818549-2012-02-28.xml*