Cohomology groups for spaces of 12-fold tilings
CCohomology groups for spaces of 12-fold tilings
Nicolas B´edaride ∗ Franz G¨ahler † Ana G. Lecuona ‡ Abstract
We consider tilings of the plane with 12-fold symmetry obtained bythe cut and projection method. We compute their cohomology groupsusing the techniques introduced in [5]. To do this we completely describethe window, the orbits of lines under the group action and the orbitsof 0-singularities. The complete family of generalized 12-fold tilings canbe described using 2-parameters and it presents a surprisingly rich co-homological structure. To put this finding into perspective, one shouldcompare our results with the cohomology of the generalized 5-fold tilings(more commonly known as generalized Penrose tilings). In this case thetilings form a 1-parameter family, which fits in simply one of two types ofcohomology.
This paper deals with tilings of the plane R . Given a tiling, the group oftranslations of the plane acts on it, allowing us to associate to each tiling aspace, the hull of the tiling, defined as the closure of the orbit of the tilingunder the group action. This space, which has been thoroughly studied forthe last 20 years, has many interesting properties. We suggest [11] for a goodintroduction.Cohomology has been an essential tool of algebraic topology. It is a topo-logical invariant that associates groups (or a more complex ring structure) tospaces and can be used, among other things, to tell spaces apart. The compu-tation of the cohomology of the hull of a tiling emerged at the beginning of the2000’s. It has been successfully used to obtain dynamical results on tilings, seefor examples [8, 9] or [2, 12]. In this article we will compute the cohomologyof some tiling spaces, enhancing our understanding of them and providing thecommunity with some complete calculations.There are several classes of tilings with nice properties. Worth mentioningare the class of substitution tilings and the class of cut and project tilings. ∗ Aix Marseille Universit´e, CNRS, Centrale Marseille, I2M, UMR 7373, 13453 Marseille,France. Email: [email protected] † Faculty of Mathematics, Bielefeld University, 33615 Bielefeld, Germany. Email: [email protected] ‡ School of Mathematics and Statistics, University of Glasgow, Glasgow, UK. Email: [email protected] a r X i v : . [ m a t h . K T ] D ec or the former one, there is a well developed technique that allows to computethe cohomology groups of these tilings [1, 7]. For the latter ones, in spite ofthe general methods described by G¨ahler, Hunton and Kellendonk to computethese groups [5], very few examples have been treated in detail. Some examplescan be found in [6, 1]. However, most of these examples are also substitutiontilings; thus, the theoretical method concerning the cut and project tilings hasnever been optimized to be applied in actual computations.In this article we are going to focus on a two parameter family of cut andprojection tiling spaces: the generalized 12-fold tilings. The hull of these tilingswill be denoted by Ω E γ with γ ∈ R (the notation and construction method arefurther explained in Sections 1.1 and 1.2). Our goal is double: first, we wantto obtain a deeper understanding of this space of tilings using cohomology, andsecond we want to exploit the techniques in [5] to carry out the computations.Our computational efforts crystallize in the following theorem. Theorem 1.
The cohomology groups of Ω E γ satisfy:1. For all parameters γ , all the cohomology groups are torsion free.2. For all parameters γ , H (Ω E γ ) = Z .3. The rank of the group H (Ω E γ ) depends on γ and takes values in the set { , , , , , , } . The precise values of γ associated to the differentranks are explicit in Figure 8.4. The rank of the group H (Ω E γ ) depends on γ . Moreover, if the rank of H (Ω E γ ) ≤ , then the rank of H (Ω E γ ) is presented in Propositions 32to 39. As mentioned before, the proof of this theorem is based on the techniquesdescribed in [5], which we briefly summarize in the next subsection. A detailedplan of the proof is given in Subsection 1.4.In this article, the computation of the groups H (Ω E γ ) is completely ex-plicit; on the other hand, the results concerning the groups H (Ω E γ ) are com-puter assisted if γ (cid:54)∈ Z [ √ × Z [ √ γ . However, the information gathered in Proposition 42, particularly in Ta-bles 1 to 4, makes it very easy to compute the rank of H (Ω E γ ) for any fixedvalue of γ ∈ R . For a ‘generic’ value of γ we propose the following conjecture,supported by computer calculations. Conjecture 2.
The maximal cohomology attained among all the generalized12-fold tilings is H (Ω E γ ) = Z , H (Ω E γ ) = Z and H (Ω E γ ) = Z . .1 Background on cut-and-projection tilings cohomology Given a 2-plane E in the vector space R n containing no integer line (a linedirected by an element of Z n ) we want to study the tiling T of the plane E obtained by the cut and projection method. We recall this construction: tostart with we need to consider the following orthogonal decomposition R n = E ⊕ E ⊥ . Let π be the projection on E and π ⊥ the projection on E ⊥ . We are now readyto describe the tiling T we want to study: Its vertices are the images by π ofthe points in Z n ∩ ( E + [0 , n ). The tiles of the tiling are rhombi obtained bythe projection π of the 2-faces of the complex Z n which are inside the “band” E + [0 , n . Let us denote Ω E the hull of this tiling. In our case it is the closureof T under the action of the group of translations of the plane E .Remark that we can also consider an affine plane E = E + γ for some vector γ .We can associate a tiling to this affine plane by considering the band E + [0 , n and doing the same construction. We obtain a tiling of E + γ and can defineits hull, Ω γE . Obviously if γ is in Z n or in E we have the same hull. Otherwisethe tilings can be different. This is in particular the case for Penrose tilings,where one obtains for a generic value γ what is known under the name of a“generalized Penrose tiling”.During the last decade, starting with the work of [1] and [4], a lot of efforthas been put into understanding the topology of Ω E . We are particularly inter-ested in computing the cohomology groups with integer coefficients, H k (Ω E , Z )for k = 0 , ,
2, of the hull of the tiling. A general description of these coho-mology groups is given by G¨ahler-Hunton-Kellendonk in [5]. In order to statetheir results, we need to introduce quite a bit of notation and establish someconventions, which we will follow throughout the paper. • The group π ⊥ ( Z n ) will be called Γ. The closure of this group is theproduct of a free abelian group and a continuous group since it has notorsion part: Γ ∼ Z k × R (cid:96) . We call ∆ the discrete part of Γ. • Denote by F the vector space generated by ∆ ⊂ E ⊥ . We will decomposethe space E ⊥ as E ⊥ = F ⊕ F ⊥ , with F ⊥ parallel to Γ, and denote by ∆ the stabilizer of F ⊥ under the group Γ. • Assuming that (cid:96) = dim F ⊥ = 2, we let P be the collection of planesparallel to F ⊥ , defined as P = (cid:91) δ ∈ ∆ F ⊥ + δ. • The polytope obtained by projecting the cube [0 , n onto E ⊥ will be calledthe window and is denoted W . We need to assume that the intersectionbetween W and P is a union of polygons, segments and points. Theboundaries of the polygons and the segments are contained in lines directedby the vectors f , . . . , f n . 3 We consider the action of Γ on the lines described in the last point. Theset of orbits is denoted I and each orbit is called a 1-singularity. Thecardinality of the set I will be denoted L = | I | . • The intersection of any two 1-singularities will be called a 0-singularity.The set of all orbits of 0-singularities is denoted I . The cardinality of theset I will be denoted L = | I | . • Let Γ i ≤ Γ be the stabilizer under the action of Γ of the vector spacespanned by f i . Every 1-singularity α ∈ I is generated by some direction f i . We denote by L α the set of orbits of 0-singular points on α under theaction of Γ i , and we set Γ α := Γ i , if the 1-singularity α is in direction f i . • Since Γ α ⊂ ∆ for all α ∈ I , we can define the natural inclusion map β : (cid:77) α ∈ I Λ Γ α → Λ ∆ , (1)where Λ A denotes the exterior product of two copies of the free abeliangroup A . In our case, each group Γ α is of rank two, thus generated by twovectors, and Λ Γ α represents the exterior product of the two generators. • We define the numbers R and e as: R = rk β and e = − L + (cid:88) α ∈ I L α .e will turn out to be the Euler characteristic of the tiling space Ω E .With all this notation and conventions in place, we are now ready to statetwo general results about the cohomology groups H ∗ (Ω E , Z ).Firstly, we note (compare [5, Thm. 2.10]) that the cohomology is finitelygenerated, iff there exists a natural number ν such that rk ∆ = ν dim F ⊥ (or,equivalently, d = ( ν −
1) dim F ⊥ ), and ν = rk Γ α for all α ∈ I . In our case,these conditions are met for ν = 2, and the cohomologies of all dodecagonaltilings considered here are finitely generated.Secondly, with the notation introduced above, the cohomology groups cannow be expressed explicitly. Theorem 3 (Thm. 5.3 in [5]) . The free abelian part of H k (Ω E , Z ) is given bythe following isomorphisms: H (Ω E , Z ) H (Ω E , Z ) H (Ω E , Z ) Z Z L − R Z L + e − R The torsion part of H (Ω E , Z ) is isomorphic to the torsion in coker β :=Λ ∆ / (cid:104) Λ Γ α (cid:105) α ∈ I , whereas the other cohomology groups are torsion free.
4e note that R = rk β = rk (cid:104) Λ Γ α (cid:105) α ∈ I depends only on the directions ofthe 1-singularities α , not on their positions, and the same holds true for coker β .Hence, these quantities are the same for all dodecagonal tilings considered here.As we shall see, we have R = 3, and the torsion of coker β vanishes, so that thecohomology of all dodecagonal tilings is free. n -fold tilings In this section we recall the definition of the n -fold tiling and summarize someknown results on its cohomology groups. Following the description in the pre-vious section, to produce a tiling by the cut and projection method we need tostart with a 2-plane in a vector space. In this case the plane we intend to tile, E n , is defined as follows: • If n = 2 p + 1, then E n is the 2-plane in R p +1 generated by π p +1 ...cos pπ p +1 and π p +1 ...sin pπ p +1 . • If n = 2 p , then E n is the 2-plane in R p generated by πp ...cos ( p − πp and πp ...sin ( p − πp . The plane E n yields a decomposition of the ambient space as R k = E n ⊕ E ⊥ n ,where the value k depends on the parity of n . Furthermore, the space E ⊥ n decomposes as E ⊥ n = F n ⊕ F ⊥ n where F n is the vector space spanned by thediscrete part of π ⊥ ( Z k ) ⊂ E ⊥ n . The space E ⊥ n allows to define a family of“related” tilings by considering the translate planes E n + γ , γ ∈ E ⊥ n . If γ ∈ F ⊥ n ,then the tiling of the plane E n and of E n + γ coincide, so the interesting casesoccur when the translation vector γ belongs to F n . Moreover, if γ, γ (cid:48) ∈ F n and γ − γ (cid:48) ∈ ∆, then the tilings of E n + γ and E n + γ (cid:48) coincide. Notice that translating E n by γ and looking at integer points in the band is the same as keeping E n and looking at points of the form − γ + Z n in the band. Finally, remark that allthe results on the cohomology groups only depend on the geometry of W andthe action of Γ on the lines supporting W . We intersect W by planes F ⊥ n + δ and try to compute numbers involved in Theorem 3. Here if we translate E by γ we will study these tilings by cutting the window W , which does not dependon the parameter γ , with translates of the family of planes P of the form P γ = (cid:91) δ ∈ ∆ F ⊥ n + δ + γ, with γ ∈ F n . E γn . We refer to [3] for further references on these tilings.The following statement resumes all the known values of the cohomologygroups of the (undecorated) n -fold tilings (cf. [5, Table 5.1]). These results hadmostly been announced without proof already in [6], but that paper containssome mistakes in the cohomology of the generalized Penrose tilings, which werecorrected later by Kalugin [10]. Theorem 4.
For small values of n , the n -fold tilings satisfy:1. If n = 8 , these tilings are also known as Ammann–Beenker tilings. In thiscase F = 0 . The cohomology groups with integer coefficients of the -foldtilings are: H (Ω E ) H (Ω E ) H (Ω E ) Z Z Z
2. If n = 5 , these tilings are also known as (Generalized) Penrose tilings.In this case F is of dimension one and we can understand the translationparameter γ as a real number. The cohomology groups with integer coef-ficients of the -fold tilings depend on whether γ is in Z [ ϕ ] or not, where ϕ is the golden mean. The groups are as follows: γ H (Ω E γ ) H (Ω E γ ) H (Ω E γ ) Z [ ϕ ] Z Z Z R \ Z [ ϕ ] Z Z Z
3. If n = 12 , then F is of dimension two. For all γ ∈ ∆ , it holds Ω E γ =Ω E . In this case the cohomology groups with integer coefficients of the -fold tilings are: H (Ω E ) H (Ω E ) H (Ω E ) Z Z Z
4. If n = 7 , the cohomology groups with integer coefficients of the -foldtilings are not finitely generated. To give some perspective to the results in this paper, we will make a fast com-parison between the computation of the cohomology groups of the generalizedPenrose tilings and the 12-fold tilings.In the Penrose case, we start from the space R and we consider the cut andproject method onto a two dimensional plane. The main character in the con-struction, the polytope W , is in this case a 3-dimensional rhombic-icosahedronwith vertices in five parallel planes depicted in Figure 1. The intersection of6hese planes and W is a series of points and polygons. The boundaries of thepolygons are contained in lines directed by 5 different vectors. To compute thecohomology of the standard Penrose tiling, we need to study the orbits of theselines under the action of a certain group Γ.In the classic 12-fold tiling, we obtain a 4-dimensional polytope W , describedin Proposition 12. Its vertices are contained in 16 parallel planes. This time theboundaries of the polygons determined by these vertices are contained in linesdirected by 6 different vectors (see Figure 3).The main difference between these two settings though is not simply thatthere is one more direction in the 12-fold tiling. The main difference is apparentwhen we consider the generalized tilings. Indeed, to compute the cohomologyof the generalized tilings we need to shift the planes cutting the polytope. Inthe Penrose case, the shifting occurs along the vertical axis of the rhombic-icosahedron (see Figure 1), and it can be encoded by the real number γ in thestatement of Theorem 4.2. When we shift slightly the planes cutting the Penroserhombic-icosahedron we obtain some polygons with 10 sides (contained still in5 directions). Now, depending on whether or not the translation parameter γ isin Z [ ϕ ] (where ϕ is the golden mean) the parallel opposite lines on the decagonwill belong or not to the same orbit under the action of Γ.In the generalized 12-fold tiling, the situation is much richer. The shiftingof the planes that cut the polytope is encoded by a 2-dimensional parameter γ = ( γ , γ ). In Figure 6 one can see the number of sides that a polygoncut by the shifted planes will have. The 10 lines obtained in the generalizedPenrose become 24 in the generalized 12-fold tiling (see Proposition 24). Eachone of the original 6 directions can have at most 4 representatives in differentorbits. The game is now to understand when, depending on the value of γ ,these representatives are in the same orbit under the action of Γ. The result,surprisingly complicated, is the content of Proposition 27 and Figure 8. Thisinformation is enough to compute completely the first cohomology group of the12-fold generalized tilings. When trying to compute the second cohomologygroups, the intricacy grows further as explored in Section 4. The proof of the main theorem of this paper is a long and involved computationof the quantities e, L and R which will allow us, via Theorem 3, to computethe cohomology groups of the 12–fold tilings (as discussed after that theorem,these groups are all free). Roughly, the steps we will follow are:1. We intersect W with P γ : we obtain several polygons. We compute theequations of the lines which support the edges of the polygons. The case γ ∈ ∆ is presented in Section 2.2 while the general case can be found inSection 3.1.2. For each line α obtained in the preceding point, we consider its orbitunder the action of Γ. The number of orbits is the quantity L . The7igure 1: Window for the Penrose tilingsrelevant computations can be found in Section 3.2 and culminate withProposition 27 and Figure 8.The lines α are in one of 6 possible directions. The smallest value of L isthen six, with just one Γ-orbit per direction; and the largest is L = 24,corresponding to the case of four Γ-orbits per direction.3. We compute the stabilizer Γ α of each line α and obtain the quantity R , which is the rank of (cid:104) Λ Γ α (cid:105) α ∈ I . This computation takes place inSection 3.3 and allows us to compute the first cohomology groups for the12-fold tiling in Proposition 30.4. We count the number of intersections, up to the action, between a 1-singularity α and the other 1-singularities. This quantity is L α . We havenot completed the computation of L α for all the possible values of theparameter γ . The study breaks down into too many subcases which didnot seem worth looking into in full detail. However: • In Section 4.1 we present a complete calculation of the second coho-mology group when γ = 0. In this case we have actually computedthe values L and L α (Lemma 31), which is all that is needed to de-termine the cohomology groups of Ω E . These groups were known(Theorem 4.3) and are restated in Proposition 32.The case γ = 0 is the simplest analyzed and corresponds to the caseof L = 6 described above. 8 In Section 4.2 we have further computed the second cohomologygroups, and thus the full cohomology, for all the cases in which underthe action of Γ there are at most two representatives per line, thatis, the cases L ≤
12. This computation is computer assisted and wecollect the results in Propositions 32 to 39. • Finally, in Section 4.3 we analyze the general case and summarize theresults in Section 4.4. We consider the largest set of 1-singularities,which has 4 orbits per direction and thus 24 1-singularities. We inter-sect each of these with the translates by the action of Γ of the 20 nonparallel 1-singularities. This yields a complete list of 0-singularitiesin each 1-singularity. Our results are collected in Proposition 42,which provides an upper bound on the quantity L α . The completelists of 0-singularities are displayed in Tables 1 to 4. -fold tilings The 12-fold tiling space corresponding to the plane E γ with γ ∈ ∆ was intro-duced by Socolar in [13], where it was shown to be also a substitution tiling.Thus, the classical method exposed in [11] allows us to compute its cohomologygroups. Our goal now is to extend this computation to all parameters γ ∈ F .Since we will be only concerned with the case n = 12, from now on we willdenote simply by E, F etc. the spaces E , F etc. We start by establishingsome preferred bases for the spaces involved in the decompositions R = E ⊕ E ⊥ and E ⊥ = F ⊥ ⊕ F. The plane E will have the fixed orthonormal basis { u, v } given by u = 1 √ √ / / − / −√ / , v = 1 √ / √ / √ / / . The plane F will have the following fixed orthogonal basis: A = 13 − , B = 13 − . The basis { A, B } is not orthonormal, but it has been chosen to facilitate thecomputations. Finally, the plane F ⊥ will be chosen to be the algebraic conjugate9f E and we will denote { u (cid:48) , v (cid:48) } its orthonormal basis. (Recall that to obtain u (cid:48) , v (cid:48) we simply need to replace √ u and v .)The projection π ⊥ onto E ⊥ is described in the following lemma. The simpleproof is left to the reader. Lemma 5.
The image of the canonical basis of R by the projection π ⊥ onto F ⊥ ⊕ F defines six vectors g i , i = 1 , . . . , . We have g i = f i + δ i with f i ∈ F ⊥ , δ i ∈ F and: f = ( √ / , f = ( − / , √ / f = ( √ / , − / f = (0 , √ / f = ( −√ / , − / f = (1 / , √ /
6) and δ = (1 , δ = (0 , δ = ( − , δ = (0 , − δ = (1 , δ = (0 , . The vectors f i are expressed in the basis { u (cid:48) , v (cid:48) } and the δ i in the basis { A, B } . In this section we collect some technical lemmas that will be useful in the forth-coming computations. Many of them are straightforward and the proofs are leftto the reader.
Lemma 6.
The vectors f i ∈ F ⊥ , i = 1 , . . . , , satisfy the following two linearrelations over Z : f = f − f and f = f − f . Moreover, we have f − f = √ f f − f = −√ f f i + f i +2 = −√ f i +1 ∀ i = 1 , . . . , Corollary 7.
The following group isomorphisms hold π ⊥ ( Z ) ∼ Z A ⊕ Z B ⊕ (cid:104) f , . . . , f (cid:105) Z and π ⊥ ( Z ) ∼ Z ⊕ R . Proof.
Every element in π ⊥ ( Z ) is of the form (cid:80) n i g i . By Lemma 5, we canrewrite this expression as (cid:88) i =1 n i f i + (cid:88) i =1 n i δ i = (cid:88) i =1 n i f i + ( n − n + n ) A + ( n − n + n ) B. It is straightforward that Z is isomorphic to the abelian group generated by A, B , while Lemma 6 implies that the closure of the set { (cid:80) i =1 n i f i : n i ∈ Z } isisomorphic to R . 10 f f f f f Figure 2: The vectors f , . . . , f in F ⊥ and the directions of the 12 th roots ofunity in C . Corollary 8.
If we identify the plane F ⊥ with C , then the vectors f i correspondto some roots of the equation z = 3 − . The correspondence, illustrated inFigure 2, can be written as follows, where x denotes the complex number e i π : f i = 1 √ x i − , i = 1 , . . . , . (2) Lemma 9.
The complex number x = e i π is a simple root of the polynomial X −√ X +1 . Moreover, every element of Z [ x ] can be expressed as a polynomialin x of degree at most one with coefficients in Z [ √ . The precise expressionsare collected in the following chart for k ∈ N : x x x x x x x k x x √ − x − √ √ x − x − √ − − x k Notation.
Since the group Z [ √
3] will play a major role in this article, forsimplicity we will denote it G .When trying to determine the cohomology groups of the 12-fold tilings, wewill need to understand when certain R -linear combinations of powers of x belong to Z [ x ]. In some sense, the next lemma generalizes the equation √ f = f − f . Lemma 10.
Let α, β ∈ R and i < j two integers in [0 , .1. αx i ∈ Z [ x ] holds if and only if α belongs to G . . αx i + βx j ∈ Z [ x ] holds if and only if j − i = 1 , α, β ∈ G.j − i = 3 and (cid:40) β = ( β + β √ , α = ( α + α √ , α i , β i ∈ Z ,α − β , α − β ∈ Z .j − i = 2 and (cid:40) β = √ ( β + β √ , α = √ ( α + α √ , α i , β i ∈ Z ,α − β ∈ Z .j − i = 4 and (cid:40) β = √ ( β + β √ , α = √ ( α + α √ , α i , β i ∈ Z ,α − β ∈ Z . Proof.
We start with the first claim. By Lemma 9 it is clear that if α belongs to Z [ √
3] then αx i belongs to Z [ x ]. We prove the other implication. By assumption, αx i = (cid:80) j n j x j , n j ∈ Z , which implies α = x − i (cid:80) j n j x j . Again by Lemma 9,we know we can write α = ax + b , with a, b ∈ G . Since { x, } is a basis of C asan R vector space, we conclude that α = b ∈ G .We now proceed to argue the second claim in the statement. We needto understand which conditions on α, β imply αx i + βx j ∈ Z [ x ] . Since x i isinvertible in Z [ x ], this last expression is equivalent to α + βx j − i ∈ Z [ x ] , which,by Lemma 9 can be rewritten as α + βx j − i = ax + b, a, b ∈ G. Thus, dependingon the value of j − i and again by Lemma 9, we are looking to one of the followingequations βx + α = ax + b j − i = 1 .β √ x − β + α = ax + b j − i = 2 . βx − β √ α = ax + b j − i = 3 .β √ x − β + α = ax + b j − i = 4 .βx − β √ α = ax + b j − i = 5 . Now, since { x, } is an R basis for C , we deduce- If j − β = a, α = b ⇐⇒ β, α ∈ G .- If j − β = a, − β √ α = b ⇐⇒ β, α ∈ G .- If j − β = a, − β √ α = b ⇐⇒ ∃ β , β , α , α ∈ Z , β = 12 ( β + β √ ,α = 12 ( α + α √
3) and b = α − β − √ α − β ⇐⇒ β, α ∈ G, α − β , α − β ∈ Z . For the two remaining cases, an analogous computation to the precedingone, and using the same notations, yields- If j − β √ a, − β + α = b ⇐⇒ α, β ∈ √ G and α − β ∈ Z .
12 If j − β √ a, − β + α = b ⇐⇒ α, β ∈ √ G and α − β ∈ Z . We end this section with a last technical lemma which will not be neededuntil the reader is confronted with Figure 9.
Lemma 11.
Let S := { ( γ , γ ) (cid:54)∈ √ G × G } and consider its two subsets: A = { ( γ , γ ) ∈ S | γ √ γ ∈ G } and B = { ( γ , γ ) ∈ S | γ √ γ ∈ G } . Moreover, interchanging the roles of γ and γ , we obtain the analogous sets S (cid:48) := { ( γ , γ ) (cid:54)∈ G × √ G } , D = { ( γ , γ ) ∈ S (cid:48) | γ √ γ ∈ G } and E = { ( γ , γ ) ∈ S (cid:48) | γ √ γ ∈ G } . Finally, set C = S \ ( A ∪ B ) and F = S (cid:48) \ ( D ∪ E ) . Then,1. A ∩ B = D ∩ E = ∅ , which implies S = A (cid:116) B (cid:116) C and S (cid:48) = D (cid:116) E (cid:116) F .2. B ∩ S (cid:48) ⊂ F and E ∩ S ⊂ C .3. A ∩ D is the set of couples ( γ , γ ) subject to the following conditions:(a) ( γ , γ ) (cid:54)∈ √ ( G × G ) .(b) ( γ , γ ) ∈ ( G × G ) , i.e. γ = a + b √ , γ = c + d √ , a, b, c, d ∈ Z .(c) The integers a, b, c, d satisfy one of the following parity conditions:i. a, d even and c, b odd.ii. a, d odd and c, b even.iii. a, b, c, d odd.Proof. By the symmetry in the statement, it suffices to show the first claimfor the sets
A, B : if ( x, y ) ∈ A ∩ B then 2 x √ y and x √ y are in G .Subtracting these two expressions we obtain that x √ ∈ G which contradicts( x, y ) ∈ S . We conclude A ∩ B = ∅ and by an analogous argument D ∩ E = ∅ .Now, we proceed to show that B ∩ S (cid:48) ⊂ F and E ∩ S ⊂ C . This amounts toshow that B ∩ E = B ∩ D = A ∩ E = ∅ . • B ∩ E = ∅ : if there was an element ( x, y ) ∈ B ∩ E ⊂ S ∩ S (cid:48) , then x, y (cid:54)∈ √ G . However, since ( x, y ) ∈ B ∩ E we arrive to the followingcontradiction. (cid:40) x √ y ∈ G x + y √ ∈ G ⇒ (cid:40) x + 4 √ y ∈ G x + 3 √ y ∈ G ⇒ √ y ∈ G. B ∩ D = ∅ : if there was an element ( x, y ) ∈ B ∩ D ⊂ S ∩ S (cid:48) , then x, y (cid:54)∈ √ G . However, since ( x, y ) ∈ B ∩ D we arrive to the followingcontradiction. (cid:40) x √ y ∈ G x + 2 y √ ∈ G ⇒ (cid:40) x + 2 √ y ∈ G x + 2 √ y ∈ G ⇒ x ∈ G. • A ∩ E = ∅ : the analysis in this case is completely analogous to B ∩ D = ∅ swapping the two variables.Finally, we will completely determine the set A ∩ D ⊂ S ∩ S (cid:48) . If ( x, y ) ∈ A ∩ D ⊂ S ∩ S (cid:48) , then x, y (cid:54)∈ √ G and the following conditions need to besatistfied: (cid:40) x √ y ∈ G x + 2 y √ ∈ G ⇒ (cid:40) x √ y ∈ G x √ y ∈ G ⇒ y ∈ G. By the symmetry of the variables we must also have 4 x ∈ G . So, ( x, y ) ∈ ( G × G ) and there exist a, b, c, d ∈ Z such that 4 x = a + b √ y = c + d √ x, y ) ∈ A we have:2 √ a + b √
3) + 2 14 ( c + d √
3) = 3 b + c a + d √ ∈ G = Z [ √ . For this last condition to hold we need a + d, b + c ∈ Z . Analyzing theconstraints imposed by ( x, y ) ∈ D , we obtain c + b, d + a ∈ Z . It follows that a and d and that c and b have the same parity. Moreover, since x, y (cid:54)∈ √ G , a and b cannot be both even and likewise c and d . The parity conditions in thestatement follow. In this section we describe the polytope W = π ⊥ ([0 , ) and its intersectionwith the family of planes P . By definition, the window W is the convex hull ofthe 2 = 64 points (cid:80) i =1 n i g i , where each g i ∈ E ⊥ is the projection of the i th element of the canonical basis of R and n i is either 0 or 1. We end this sectionestablishing a convention regarding the description of the three dimensionalfaces of W, which will be thoroughly studied in the next section. The mainresult of this part is the following proposition. Proposition 12. • The polytope W ⊂ E ⊥ = F ⊥ ⊕ F has vertices, edges, faces ofdimension two and faces of dimension three. • All edges have the same length. The vertices are distributed in affine planes parallel to F ⊥ . Each ofthese planes intersects – F in a point (see Figure 3); and – W in either a point, a triangle or a hexagon. The proof of this proposition will be splitted in several lemmas.
Lemma 13.
The intersection of the window W and the family of translates of F ⊥ given by P = (cid:83) δ ∈ ∆ F ⊥ + δ consists of points, triangles and hexagons.Proof. Let us denote by P δ the affine plane F ⊥ + δ . Throughout this proof wewill follow the conventions established in Sections 1.1, 1.2, and 2.By Corollary 7, ∆ is isomorphic to Z with generators { A, B } . So, the family P consists of all translates of F ⊥ by an integer linear combination of A and B .The vertices of W are of the form n g + · · · + n g = n f + · · · + n f + ( n − n + n ) A + ( n − n + n ) B, (3)where n i ∈ { , } . Since the expressions n + n − n and n − n + n takeonly 4 different values, the numbers {− , , , } , it follows that the vertices of W project onto F onto 16 different points of the lattice ∆. The first step willbe to understand how many of these vertices are projected onto each of these 16lattice points. To this end, in the next array we collect in how many differentways the above mentioned numbers are obtained: n i + n i +4 − n i +2 − i = 1 1 3 3 1 i = 2 1 3 3 1It follows that on the plane P ( n + n − n ,n − n + n ) the number of vertices of W is at most the product of the corresponding entries on the second and third linesin the above array. (Notice that not all the 64 projections of the vertices of theunit cube in R need to be vertices of W .) After a straightforward computation,we obtain: • In each of P ( − , − , P ( − , , P (2 , − , P (2 , there is one single vertex of W . • In each of P ( − , , P ( − , , P (0 , − , P (0 , , P (1 , − , P (1 , , P (2 , , P (2 , thereare at most 3 vertices of W . It follows that the intersection of W withany of these planes is the convex hull of the 3 points obtained projectingthe canonical basis of R . These points turn out to not be aligned in anyof the 8 planes, so in this case the intersection of W and these planes is acollection of 8 triangles. • In each of P (0 , , P (0 , , P (1 , , P (1 , there are at most 9 vertices of W . Welist the 9 points in each plane and compute their convex hull. It turns outthat in each case the convex hull is a hexagon drawn in Figure 4. In eachof the 4 planes, the vertices of the corresponding hexagon in cyclic orderare given by: 15 , ,
1) (2 , , − − ,
1) (1 , AB Figure 3: The polytope W ⊂ E ⊥ = F ∪ F ⊥ has its vertices on 16 planesparallel to F ⊥ . The projection of each of these planes onto F , with fixed basis { A, B } , is a point. In this figure the number of vertices of W in each of thedifferent 16 planes is encoded by the following color code: Black=isolated point,red=triangle, green=hexagon.Figure 4: The four hexagons which appear in the planes P (0 , , P (1 , , P (0 , , P (1 , . – Hexagon in the plane P (0 , : f f f f f f f f f f f f f f f f – Hexagon in the plane P (1 , : f f f f f f f f f f f f f f f f f f – Hexagon in the plane P (0 , f f f f f f f f f f f f f f f f f f – Hexagon in the plane P (1 , : f f f f f f f f f f f f f f f f f f f f From this last lemma we deduce several results on the structure of W . Westate them here without proof. They are all the result of simple observations ordirect computations. In what follows, we will refer to the points of intersectionbetween W and the planes P ( − , − , P ( − , , P (2 , − , P (2 , as isolated points or isolated vertices . 16 orollary 14. The edges of the polygons in the statement of Lemma 13 areparallel to the directions f i ⊂ F ⊥ , i = 1 , . . . , . Corollary 15.
The 1–dimensional complex of W , that is, the set of projectionsof the edges of the unit cube in R onto F ⊥ ⊕ F , has the following properties: • No two vertices in one of the planes P δ are connected by an edge. • Every isolated vertex has valency . • Every vertex in a triangle has valency 5. The edges join the vertex inthe triangle with vertices on the hexagons, with vertices on the closesttriangle, and with the closest isolated vertex. • In every hexagon, vertices of valency alternate with vertices of valency4. The latter vertices connect to vertices in the planes at distance 1, theformer ones exhibit the same pattern plus two edges connecting to verticesin other hexagons. The following corollary is the last one in this section. In it we introduce thenotion of a cube , which is our shortcut notation for a “3-dimensional face ofthe polytope W ”. These cubes will play an essential role in this article and thenext subsection is devoted to their explicit description. Corollary 16.
The window W has faces of dimension . Every three di-mensional face has the combinatorics of a cube and comprises vertices. Wewill refer to these faces as “cubes”. Their disposition with respect to the verticesof W is as follows. • Vertices of valency belong to cubes. • Vertices of valency belong to cubes. • Vertices of valency belong to cubes. In what follows we will need a more precise description of these cubes. Noticethat, in order to completely determine a cube it suffices to know one of itsvertices and the three edges of the cube intersecting at it. A vertex of a cubeof W is a point in F ⊥ ⊕ F and we will express it as a linear combination of thevectors g , . . . , g . To describe the edges of the cube we start by remarking thefollowing: in the plane F there is a “rightwards” direction, given by the vector A and an “upwards” direction, given by the vector B (see Figure 3). Fromequation (3) it follows that an edge of a cube directed by • the vector g or g : is horizontal and rightwards when projected onto F . • the vector g or g : is vertical and upwards when projected onto F . • the vector g : is horizontal and leftwards when projected onto F . • the vector g : is vertical and downwards when projected onto F .17herefore, it is enough to recall the projections onto F ⊥ , given by vectors f i ,of the edges of the cube we want to describe, since it is immediate from thisinformation and the above remarks how to recover the cube. Furthermore,instead of the f i , since we are only interested at this point in the directions of theedges, we can use the complex vector x = e iπ and its multiples x i , i = 0 , . . . , • the vectors x i with i even represent edges of the cube which are horizon-tal in F . More specifically, x i with i = 0 , , F while x i with i = 2 , ,
10 projectonto leftwards vectors. • the vectors x i with i odd represent edges of the cube which are vertical in F . More specifically, x i with i = 1 , , F while x i with i = 3 , ,
11 project onto downwardsvectors.
Convention 17.
In the rest of the paper: A cube is given by a vertex g and threevectors x i , x j , x k which support the edges of the cubes containing the vertex. Wedenote the cube by { g, x i , x j , x k } . Example 18.
As an example, the set { g + g , x , x , x } determines a cubein W with a vertex in g + g ∈ P ( − , − and 3 edges which go upwards in theplane F and thus connect the vertex g + g with 3 vertices in the plane P ( − , .Each of these vertices is itself connected to 2 vertices in the plane P ( − , withedges that can be again encoded by the vectors x , x , x . Finally, the 3 verticesin P ( − , (each of which connects to 2 vertices in P ( − , ) are also connectedwith the single vertex in P ( − , . This is a complete description of the 6 verticesand 12 edges of a cube in W . With the notation developed above, we now proceed to describe all the cubesin W . The details of the computation will not be needed in the rest of thepaper; however, the last results of this section Proposition 21 and Corollary 22,summarizing the findings, are essential.According to Corollary 15 each of the 4 isolated vertices has valency 6 andfrom Corollary 16 we learn that these 6 edges belong to 8 different cubes. Afteran easy computation we find out that if we fix any one of the isolated vertices,precisely two of the 8 cubes which contain it contain also another isolated vertex.It follows that the number of different cubes containing isolated vertices is 8 +7 + 7 + 6 = 28. The combinatorics of these cubes is as follows: • The cubes containing two isolated vertices contain also the 6 vertices of thetwo triangles on a line (or column) in Figure 3. There are 4 such cubes.Following our convention, the two vertical cubes are given by the sets18 g + g , x , x , x } and { g + g + g , x , x , x } while the two horizontalones correspond to the sets { g + g , x , x , x } and { g + g + g , x , x , x } . • The remaining cubes containing one isolated vertex, which are 24 in num-ber, have all two edges going in the same direction on F and a third edgegoing in the perpendicular direction. The complete list follows: – Cubes containing v = g + g (bottom left isolated vertex): { v , x i , x i +3 , x i +4 } , i = 1 , , { v , x i , x i +1 , x i +4 } , i = 0 , , . – Cubes containing v = g + g + g (top left isolated vertex): { v , x i , x i +1 , x i +4 } , i = 3 , ,
11 and { v , x i , x i +3 , x i +4 } , i = 0 , , . – Cubes containing v = g + g + g + g (top right isolated vertex): { v , x i , x i +3 , x i +4 } , i = 3 , ,
11 and { v , x i , x i +1 , x i +4 } , i = 2 , , . – Cubes containing v = g + g + g (bottom right isolated vertex): { v , x i , x i +1 , x i +4 } , i = 1 , , { v , x i , x i +3 , x i +4 } , i = 2 , , . Example 19.
If we focus on the isolated vertex in the plane P ( − , − it belongsto 8 cubes which, when projected to F , can be described as: a cube developingcompletely to the right, with a vertex in the isolated point in P (2 , − ; a cubedeveloping completely vertically with a vertex in the isolated point P ( − , ; 3cubes that have two edges towards the right direction, with their furthermostvertices in the planes P (1 , − and P (0 , − and one edge upwards; and finally 3cubes with two upward edges, leaving the furthermost vertices in this directionon the planes P ( − , and P (0 , , and one rightward edge. The cubes around theother 3 isolated points have a completely analogous symmetric configuration. The cubes which contain no isolated vertex have vertices on the triangles andhexagons described in Lemma 13. Moreover, by Corollaries 15 and 16 we knowthat each vertex on a triangle has valency 5 and thus it belongs to 6 differentcubes. From the description above we have that 5 of these cubes contain anisolated vertex. The cube that we are missing connects the vertex in the triangleto two vertices in the closest hexagon and to one vertex in the closest triangle.There is a total of 12 such cubes, 3 for each pair of nearby triangles in Figure 3.We will not give a description as precise as the ones above for these cubes sincefor our purposes the precise vertex on the triangle that these cubes contain willbe irrelevant. The information we need from these cubes is the following: • Each vertex v , v , v in the triangle in the plane P ( − , belongs to a cubethat contains no isolated vertex of W . These three cubes are given by { v j , x i , x i +5 , x i +4 } , i = 0 , ,
8. 19
Each vertex v , v , v in the triangle in the plane P (0 , − belongs to a cubethat contains no isolated vertex of W . These three cubes are given by { v j , x i , x i − , x i +4 } , i = 1 , , • Each vertex v , v , v in the triangle in the plane P (2 , belongs to a cubethat contains no isolated vertex of W . These three cubes are given by { v j , x i , x i +5 , x i +4 } , i = 2 , , • Each vertex v , v , v in the triangle in the plane P (1 , belongs to a cubethat contains no isolated vertex of W . These three cubes are given by { v j , x i , x i − , x i +4 } , i = 3 , , W : Definition 20. • Standard cubes , which when projected onto F have two edges in the samedirection and a third edge in the perpendicular one; • Long cubes , whose 3 defining edges project onto the same direction on F .These two types of cubes can be horizontal or vertical , depending on which istheir longest direction when projected onto F , see Figure 5. We summarize our findings in the next proposition.
Proposition 21.
The polytope W has cubes. • vertical and 2 horizontal long cubes each containing two isolated points. • standard cubes each containing precisely one isolated point. • standard cubes containing only points in the triangles and hexagons. Corollary 22.
The sets of directions of the edges of the cubes in W are: • the long cubes yield the sets { x , x , x } and { x , x , x } . • The standard cubes with one isolated point yield { x i , x i +4 , x i +1 } and { x i , x i +4 , x i +3 } with i = 0 , . . . , . • The remaining standard cubes yield { x i , x i +4 , x i +5 } , i = 0 , , , , ,
10 and { x i , x i +4 , x i − } , i = 1 , , , , , . H (Ω E γ ) We want to apply the method described in Section 1.4 to compute the coho-mology groups of the generalized 12-fold tilings. The first step is to cut thepolytope W by the families of parallel planes P γ = (cid:83) δ ∈ ∆ F ⊥ + γ + δ with γ = ( γ , γ ) ∈ F (expressed with respect to the fixed orthogonal basis { A, B } ).Due to the periodicity of ∆, restricting γ , γ to vary in the range [0 , × [0 , P γ . The intersection of W with P γ when γ = (0 ,
0) has been treated in Proposition 12 and we now proceed to studythe intersection when γ (cid:54) = (0 , W if and only if it intersects one of its 3 dimensional faces. Indeed,since the 4-dimensional polytope W has a boundary consisting of 3–dimensionalfaces and it sits in the 4–dimensional space F ⊥ ⊕ F , a plane cannot intersectit without intersecting its boundary. (Notice that a plane could intersect W without intersecting an edge or a 2 dimensional face.) Thus, our next goal is tointersect the 40 cubes described in Proposition 21 with the planes in the families P γ = (cid:83) δ ∈ ∆ E + γ + δ and γ ∈ (0 , × (0 , P γ very few of them do actually intersect thewindow W . Indeed, it is evident from Figure 6 that if γ i (cid:54) = 0 for i = 1 , P γ with non trivial intersection: the onesthat project into the squares delimited by the colored dots in the figure. Onthe other hand, if γ (cid:54) = (0 ,
0) but γ = 0 or γ = 0, then there are 12 suchplanes. If we want to individuate one plane in the family P γ we will use thesame notation as in the preceding section, that is, P γδ is defined as the plane F ⊥ + γ + δ . Notice that in what follows, when we refer to a cube we will alwaysconsider it ‘parametrized’ as in the explicit list of the 40 cubes which constitutethe boundary of W described in Section 2.3. Lemma 23.
Consider a plane P γδ = F ⊥ + γ + δ intersecting a cube of W . • If the cube is standard and horizontal, then the intersection is a segmentdirected by the difference of the two horizontal vectors in the cube. If the cube is standard and vertical, then the intersection is a segmentdirected by the difference of the two vertical vectors. • If the cube is long, then P γδ is orthogonal to a diagonal of the cube andthus P γδ ∩ W is either a triangle or a hexagon.Proof. An arbitrary point p in a cube determined by the tuple { v, x i , x j , x k } has coordinates in E ⊥ = F ⊥ ⊕ F given by p = v + α g r + α g s + α g t where α (cid:96) ∈ [0 ,
1] and g r , g s and g t are the projections of the basis vectors in R onto F ⊥ ⊕ F or their opposites which are determined from x i , x j and x k as previouslyexplained. We are interested in the projection onto F of this point, which wewill denote π F ( p ) := π ⊥ ( p ) ∩ F .If the cube is horizontal, then two of the exponents defining it will be even,say they are the two first ones, and therefore π F ( p ) = π F ( v ) + ( α + α , α ). Onthe other hand, if the cube were vertical, we might assume that again the twofirst exponents defining it are odd and therefore in this case π F ( p ) = π F ( v ) +( α , α + α ). It follows that if p ∈ P γδ then, • in the horizontal case we have π F ( v ) + ( α + α , α ) = ( δ + γ , δ + γ ); • in the vertical case π F ( v ) + ( α , α + α ) = ( δ + γ , δ + γ ) must hold.So, for a point in a horizontal cube to be also in P γδ its α component, thevertical one in this case, is completely determined by the above equality while α and α are subject to the linear condition α + α = some constant. As α and α vary in [0 , P γδ . Indeed, theprojection of p onto this plane can be written as π P γδ ( v ) + α f r + α f s + α f t and the above conditions found on the α i ’s allow us to rewrite the expressionas: π P γδ ( v ) + ( δ + γ − π F ( v ) ) f r − α f r + α f s + ( δ + γ − π F ( v ) ) f t = π P γδ ( v ) + ( δ + γ − π F ( v ) ) f r + α ( f s − f r ) + ( δ + γ − π F ( v ) ) f t (4)Notice that in this last expressions all parameters are fixed except for α andas it varies it defines the segment of intersection between P γδ and the cube { v, x i , x j , x k } . As claimed in the statement, this segment is directed by thevector f s − f r . This vector is a multiple of x i − x j , the difference of the twoeven powers of x defining the cube. The argument is completely analogous inthe vertical case. See Figure 7.We now proceed to examine the intersection of P γδ with a long cube. For thesake of concreteness we will assume that the cube is horizontal, which meansthat it is directed by three even powers of x (the vertical case is completelyanalogous with three odd powers instead). Just as before, a point in this cubeis given by p = v + α g r + α g s + α g t and this time its projection onto F reads π F ( p ) = π F ( v ) + ( α + α + α , p ∈ P γδ we obtain theequality π F ( v ) + ( α + α + α ,
0) = ( δ + γ , δ + γ ). This means that theplane and the cube are in the same affine space of dimension 3. In this space222 999 966 66 6 9 66 9 63 6 33 6 3 363 363696 696Figure 6: Each of these colored grid points represent the projection on F of aplane F ⊥ containing some vertices of the polytope W . Recall that the plane F has basis { A, B } highlighted on the left-most grid. The numbers in these gridsrepresent the number of cubes in W that the planes P ( γ ,γ ) meet. The firstgrid collects the numbers for γ , γ (cid:54) = 0, the second for γ = 0 and the last onefor γ = 0.the equation of the plane is α + α + α = δ + γ − π F ( v ) and the cube is[0 , . Thus the plane is orthogonal to a diagonal of the cube. The intersectionis a polygon, more precisely, a triangle or a hexagon depending on the value of δ + γ − π F ( v ) . See Figure 7.With the information we have gathered about the planes P γ and the cubesin W we could now describe every segment in their intersection. Notice that, for γ with γ i (cid:54) = 0 for i = 1 ,
2, this would imply to explicit 72 segments. Indeed, wewould be looking at 9 different planes in the family P γ that effectively intersectthe cubes in W and the number of cubes intersecting each plane varies dependingon where in F, see Figure 6, these planes are projected to. For example in thecenter square a plane intersects 12 cubes of W whereas in a corner square itmeets only 6 cubes.Thankfully, for the computation we are interested in, we do not need to gointo this level of detail. In fact, the segments of intersection define lines in thespace E ⊥ = F ⊥ ⊕ F and we are interested in them only up to the action ofΓ = π ( Z ). In Lemma 23 we have described these segments/lines as points ofthe form v ± α g i ± α g j ± α g k where v is a vertex of a cube and the α i ’sare real numbers in the interval [0 ,
1] subject to some constraints. Since v is avertex of W it is also an element of the group Γ and therefore by a translation ofvector − v we might simply consider the expression α ( ± g i )+ α ( ± g j )+ α ( ± g k ).Geometrically, this amounts to look at all the lines we are interested in togetherin the single plane P γ (0 , = P γ . At this point, since all the information aboutthese lines on the F component is summarized in the point γ , we might look attheir expressions in the plane P γ , which we recall is parallel to F ⊥ , obtainingthe expressions α ( ± f i ) + α ( ± f j ) + α ( ± f k ). Our description of the cubes in W is in terms of the powers of the complex vector x = e iπ instead of the vectors f i . In this language, the lines we are interested in are of the form α ( 1 √ x r ) + α ( 1 √ x s ) + α ( 1 √ x t )23igure 7: The left figure represents the intersection between a plane and a longcube. The plane and the cube live in the same 3-dimensional space and theintersection is either a hexagon (pink) or a triangle (green). The two otherfigures depict the intersection of the planes and the standard cubes. Each planeintersects a given cube in one line whose direction is determined by the cube.In the figure dotted lines are meant to be inside the cubes and solid ones on thefaces.with the necessary constraints on the parameters α i .As explained in Section 1.4, to compute the cohomology groups of the hullof the tiling, we need to know the quantity L , which is the number of orbits oflines in the intersection between P γ and W under the action of Γ. With all theprevious conventions in place, we are now ready to give a convenient descriptionof these lines which will be handy to later compute L . We will use the standardconvention denoting by A + λu the line of direction u passing through the point A with λ ∈ R . Proposition 24.
The intersection of the family of planes P γ with the cubes in W yields a family of segments and polygons. Each of these segments and eachside of a polygon belongs to a line that, when translated to the plane P γ via avector in ∆ ⊂ F , can be described by one of the following equations: ± ( γ √ x i x i +2 x i +4 + γ √ x i +1 ) + λx i for i ∈ { , , }± ( γ √ x i +2 x i +4 x i +6 + γ √ x i +1 ) + λx i for i ∈ { , , } . Proof.
The list of the 40 cubes which constitute the boundary of the window isexplicit in the discussion before Proposition 21. Since we are working modulothe translations of ∆ we might ignore the vertices of the cubes and use only theinformation in Corollary 22.We start considering standard cubes, which are all of the form { x i , x i +4 , x k } for i = 1 , · · · ,
11 and k as in Corollary 22. By Lemma 23 the intersection of astandard horizontal cube and a single plane in P γ is a segment as in Equation (4)24irected by the vector x i +4 − x i , which is a multiple of the vector x i − . If thecube were vertical, the roles of γ and γ are swapped. For the sake of clarity,let us consider horizontal cubes. Since we are now working modulo the actionof ∆ and are interested in the whole line defined by the segment, the relevantpart of the expression (4) which we need reads γ f r + α ( f s − f r ) + γ f t where α ∈ R . This expression, rewritten in terms of the powers of x turns into γ √ x i + γ √ x k + α √ x i − . Finally, rewriting α √ as λ ∈ R we obtain the general expression of the lineswe are looking for. To compile the list in the statement of the proposition wesimply need to write down all the different lines obtained by this procedurefrom the standard cubes in Corollary 22. To ease future computations we haveadopted the convention that the lines will be written in such a way that thedirection is given by the i -th power of x yielding the expressions:horizontal cubes : γ √ x i +1 + γ √ x k + λx i for (cid:40) i odd mod 12 k = i + 2 , i + 4 , i + 6vertical cubes : γ √ x i +1 + γ √ x k + λx i for (cid:40) i even mod 12 k = i, i + 2 , i + 4Note that we can make the notation a bit more compact taking into accountthat x i = − x i +6 mod 12 . Adjusting the signs of γ and γ we obtain the firstpoint in the statement of the proposition. Indeed, for γ , γ (cid:54) = 0 all and onlystandard cubes are intersected by the planes in P γ since the long cubes areprojected onto lines where one γ i is constantly equal to zero (see Figure 5).We now consider the case of γ = (0 , γ ). The intersection of any plane in P γ with a standard horizontal cube will be either reduced to a point or a segment asdescribed in Lemma 23. Furthermore, the intersection with a standard verticalcube will be a segment as described in Lemma 23. The reader can “visualize”this in Figure 5: for γ = (0 , γ ) the planes intersecting standard horizontal cubesmight find their intersection on an edge of the cube or in the “middle”, whilethe intersection with a vertical cube for these values of γ happens on a “lateralface”. We obtain thus all the lines in the first point of this proposition with γ = 0. Furthermore, for these values of γ the intersection with the two verticallong cubes is non trivial. By Lemma 23 the intersection consists of triangles orhexagons perpendicular to the diagonal of the cube. Since we are interested onthe lines defined by the sides of the polygones only up to the action of ∆, it isenough to consider the intersection with just one of the two cubes. We choose { g + g , x , x , x } . The three planes in the family P (0 ,γ ) which intersect thislong vertical cube are P (0 ,γ )( − ,j ) with j = − , , j = − P γ , can be described as γ √ x i + λx i − for i = 1 , , λ ∈ R . (Recall that the vector x i +4 − x i is a multiple of x i − .) Following the conventionof writing the direction of the line as the i -th power we obtain the lines γ √ x i +1 + λx i for i = 0 , , λ ∈ R . Now, if j = 0 the intersection is a hexagon with parallel opposite sides. Thelines they define have the same directions as the ones obtained for j = − A + λv we have that A is a point of coordinates g + g plus some integermultiples of some basis vectors g i (yielding the coordinates of another vertexof the cube) plus γ √ x i with i = 1 , v is given by adifference of the form x i +4 − x i . A completely analogous argument can be madeto justify that modulo ∆ we do not obtain new lines when considering the case j = 1. Finally, notice that the 3 lines we have found for j = − γ and γ and of the evenand odd powers of x .Finally, the intersection P γ ∩ W when γ = (0 ,
0) was studied in Lemma 13.The lines we obtain in this case are supported by the sides of the trianglesand hexagons found in this intersection. According to Corollary 14 these lineshave all direction x i for some i = 0 , . . . ,
5. Moreover, since they all go throughvertices of W , it is evident that they can be translated to lines going through(0 , ∈ P via an element in ∆. Now that we have a complete list of lines explicit in Proposition 24, we need tocount the number of orbits of these lines under the action of Γ = ∆ ⊕(cid:104) f , . . . , f (cid:105) Z (see Corollary 7). From Proposition 24 we know that the lines appearing in W ∩ P γ have direction x i with i = 0 , . . . , γ ,for each fixed direction x i there are a certain number of lines when we considerthem translated to the plane P γ via an element in ∆. Since we are working inthe plane P γ the action of Γ is simplified to the action of ∆ := (cid:104) f , . . . , f (cid:105) Z .For computational reasons, it is better to use the powers of x than the f i ,so we remind the reader that Equation (2) tells us that ∆ ∼ √ Z [ x ]. Thenumber of orbits will depend on the value of γ , however we can do a first overallsimplification. 26 emma 25. The two lines in each of the following accolades γ √ (cid:40) x i +2 x i +4 + γ √ x i +1 + λx i , − ( γ √ (cid:40) x i +2 x i +4 + γ √ x i +1 ) + λx i , i ∈ { , , } γ √ (cid:40) x i +2 x i +4 + γ √ x i +1 + λx i , − ( γ √ (cid:40) x i +2 x i +4 + γ √ x i +1 ) + λx i , i ∈ { , , } . are in the same ∆ orbit.Proof. Since the argument is completely analogous for all the cases, in order toease the notation we will write the details of the proof only for the lines: γ √ x i +2 + γ √ x i +1 + λx i and γ √ x i +4 + γ √ x i +1 + λx i for i ∈ { , , } . The difference of two arbitrary points in these lines can be expressed as ± γ √ x i +2 − x i +4 ) + µx i , for some µ ∈ R . Now, it holds that ± ( x i +2 − x i +4 ) = ± x i , and therefore ± γ √ x i +2 − x i +4 ) + µx i ∈ ∆ ⇐⇒ ( ± γ √ µ ) x i ∈ Z [ x ] . By Lemma 10 this last condition holds if and only if ± γ √ + µ ∈ G , which isalways satisfied for, say, µ = ∓ γ √ and thus the two lines we started with are inthe same ∆ orbit. Corollary 26.
The lines in Proposition 24 in different ∆ orbits are at mostthe following : ± ( γ √ (cid:40) x i x i +2 + γ √ x i +1 ) + λx i for i ∈ { , , }± ( γ √ (cid:40) x i +4 x i +6 + γ √ x i +1 ) + λx i for i ∈ { , , } . Proposition 27.
The number L of different orbits of lines under the actionof Γ depends on γ = ( γ , γ ) and belongs to the set { , , , , , , } . Theprecise value depends on various constraints on γ and it can be found in Figure 8.When L = 24 we have that the 1-singularities have 6 different directionsand each direction has representatives under the action of Γ . The other valuesof L are obtained when some of the parallel representatives of the 1-singularitiesare identified under the action of Γ . The other extreme case, L = 6 , arriveswhen γ , γ ∈ G and there is only one orbit per direction. G √ G √ G G G √ G √ G γ γ
12 99 61518 912 1591218 2421 15 18 15 1224 18 21241821241821181512 181218 1521 18 2118Figure 8: The number of 1-singularities depending on γ = ( γ , γ ) ∈ R (recallthat G = Z [ √ A, B, C, D, E and F in Figure 9. Moreover,adding 1 to each of the values on the chart we obtain the rank of H (Ω E γ ) (cf.Proposition 30). Finally, the regions enclosed in blue are the ones for which thesecond cohomology groups have been completely determined in Section 4.228 roof. To start with, recall that we are working with the simplifying assumptionthat γ belongs to [0 , (see beginning of Section 3.1). Moreover, the orbits oflines under the action of Γ will be computed via de action of the group ∆ ontranslates by ∆ of the lines to the plane P γ (see beginning of Section 3.2). • If the value of γ = (0 , x i , and these lines are evidently indifferent ∆ orbits. • If precisely one between γ and γ is zero then, by Proposition 24, thereare 18 lines to be considered. We start considering the case γ = 0. In this casethere are six lines with 3 different directions x i , i ∈ { , , } and 12 lines withdirection x i , i ∈ { , , } . We want to understand under which conditions on γ two parallel lines are on the same orbit under the action of ∆ . For the evendirections, the equations of the lines under consideration are γ √ x i +1 + λx i and − γ √ x i +1 + λ (cid:48) x i and they will be in the same ∆ orbit if and only if there exist λ, λ (cid:48) ∈ R suchthat γ √ x i +1 + λx i − ( − γ √ x i +1 + λ (cid:48) x i ) ∈ ∆ . Reorganizing we obtain that this condition is equivalent to γ √ x i +1 + x i ( λ − λ (cid:48) ) ∈ ∆ for some λ, λ (cid:48) ∈ R . Finally, relabelling adequately the real parameters, this last condition reads2 γ x i +1 + µx i ∈ Z [ x ] for some µ ∈ R , and by Lemma 10 it will be fulfilled if and only if 2 γ , µ ∈ G .For the odd directions, i ∈ { , , } , the equations of the lines under consid-eration are γ √ x i +4 + λ x i , − γ √ x i +4 + λ x i , γ √ x i +6 + λ x i and − γ √ x i +6 + λ x i . Following the same arguments as before, we conclude that the two first lineswill be in the same ∆ orbit if2 γ x i +4 + µx i ∈ Z [ x ] for some µ ∈ R , which holds, via Lemma 10, if 2 γ ∈ √ G . Noticing that x i +6 = − x i weconclude that the last two lines are coincident and in fact one only line.Finally we need to analyze the possibility of the lines passing through thepoints ± γ √ x i +4 and γ √ x i being in the same ∆ orbit. This will be the case if ± γ √ x i +4 + x i ( λ − λ (cid:48) − γ √ ∈ ∆ for some λ, λ (cid:48) ∈ R ,
29r equivalently, if ± γ x i +4 + µx i ∈ Z [ x ] for some µ ∈ R . By Lemma 10 we conclude this is the case if γ ∈ √ G . Summing up, since √ G ⊂ √ G we have: ∗ For each even direction there is one ∆ orbit if γ ∈ G and two if γ (cid:54)∈ G . ∗ For each odd direction we can have: − three ∆ orbits if γ (cid:54)∈ √ G . − two ∆ orbits if γ ∈ √ G \ √ G with representatives passingthrough the points γ √ x i +4 and γ √ x i . − one ∆ orbit if γ ∈ √ G .The total number of ∆ orbits is presented in Figure 9, where the righthand side has the even directions with values 3 and 6 and the left hand sidethe odd directions, with values 3,6 and 9. The quantity L is computed bycombining together the information on the lines with even and odd directions.This information can be read in Figure 8.The arguments and calculations needed to arrive to the computation of thenumber of L when γ = ( γ ,
0) are analogous to the ones presented for thecase γ = (0 , γ ). Indeed, in the former case there will be 3 lines for each evendirection since two of the ones in Corollary 26 will be coincident and the usageof Lemma 10 will yield the same conclusions. All the relevant results are againcontained in Figures 8 and 9. • We now proceed to analyze the case γ i (cid:54) = 0 for i = 1 ,
2. To start withnotice that a point in any of these lines is of the form A + λv where A is a pointin P γ , λ ∈ R and v is the direction of the line. If two lines (cid:96) and (cid:96) (cid:48) are in thesame ∆ -orbit we have that A + λv − ( A (cid:48) + λ (cid:48) v ) ∈ ∆ for some λ, λ (cid:48) ∈ R , which implies ( A − A (cid:48) ) + ( λ − λ (cid:48) ) v ∈ ∆ , and relabeling the real parameter it is equivalent to A − A (cid:48) + µv ∈ ∆ for some µ ∈ R . The vector v = x i and the power i might be even or odd. We will write downthe full details for the i even case and the results for the odd case will be givenat the end, without details. For each even power there are 4 different lines andour aim is to understand the action of ∆ on them.30rom Corollary 26, we know that for a fixed even direction x i the points A, A (cid:48) we need to consider are of the form ± ( γ √ (cid:40) x i x i +2 + γ √ x i +1 )and, depending on the choices of signs, the difference A − A (cid:48) can be written asCase 1. ± √ γ ( x j − x k ) j, k ∈ { i , i + 2 } , j (cid:54) = k, orCase 2. ± √ (cid:0) γ ( x j + x k ) + 2 γ x i +1 (cid:1) j, k ∈ { i , i + 2 } . We start considering the first of these two cases. Under these circumstances A − A (cid:48) + µx i ∈ ∆ if and only if ± γ ( x j − x k ) + µ (cid:48) x i ∈ Z [ x ] for some µ (cid:48) ∈ R .Due to the ± sign in front of this expression, we can assume that j > k , so k = i and j = i + 2, which implies x j − x k = x i +4 . It follows that in this firstcase the condition we need to determine on γ is when does γ x i +4 + µ (cid:48) x i ∈ Z [ x ] for some µ (cid:48) ∈ R hold. By Lemma 10 we conclude that this is the case if and only if √ γ ∈ G. (5)We move on to consider Case 2 above. Now A − A (cid:48) + µx i ∈ ∆ if and onlyif γ ( x j + x k ) + 2 γ x i +1 + µ (cid:48) x i ∈ Z [ x ] , for some µ (cid:48) ∈ R . With the appropriatesubstitutions of x j + x k , depending on whether j = k or j (cid:54) = k , what we needto understand is for which values of γ i do the conditions2( γ x i + γ x i +1 ) + µ (cid:48) x i ∈ Z [ x ] ⇐⇒ (2 γ + µ (cid:48) ) x i + 2 γ x i +1 ∈ Z [ x ] (6)2( γ x i +2 + γ x i +1 ) + µ (cid:48) x i ∈ Z [ x ] ⇐⇒ γ x + γ x ) + µ (cid:48) ∈ Z [ x ] (7) γ √ x i +1 + 2 γ x i +1 + µ (cid:48) x i ∈ Z [ x ] ⇐⇒ ( γ √ γ ) x i +1 + µ (cid:48) x i ∈ Z [ x ](8)With the help of Lemma 10 we conclude that condition (6) is fulfilled if andonly if γ ∈ G , while condition (8) is fulfilled if and only if γ √ γ ∈ G .Now, condition (7) is equivalent to (2 √ γ + 2 γ ) x + µ − γ ∈ Z [ x ]. Thus weobtain the condition 2( γ √ γ ) ∈ G .To help the reader visualize the results obtained so far, we propose thefollowing figure: i − i i + 2 − ( i + 2) √ γ γ γ √ γ ) γ √ γ i, i + 2 , − i, − ( i + 2) in thefigure represents one of the four parallel lines in this direction. The labellingcorresponds to the power of x with coefficient γ in the description of the linein Corollary 26. The conditions we have studied above identify parallel lines asfollows: • Condition 5 identifies lines labelled ± i and ± ( i + 2) if √ γ ∈ G . • Condition 6 identifies lines labelled i and − i if 2 γ ∈ G . • Condition 7 identifies lines labelled i + 2 and − ( i + 2) if 2( γ √ γ ) ∈ G . • Condition 8 identifies lines labelled ± i and ∓ ( i + 2) if γ √ γ ∈ G .To finish the analysis and obtain the number of orbits of lines under theaction of Γ, we need to understand when do the above conditions arrive simul-taneously. The results follow. • If √ γ ∈ G and – γ ∈ G , then there is 1 orbit. – γ (cid:54)∈ G , then there are 2 orbits with representatives i and − i . • If √ γ ∈ G \ G and – γ ∈ G , then there are 2 orbits with representatives i and i + 2. – γ (cid:54)∈ G , then there are 4 different orbits. • If √ γ (cid:54)∈ G and – γ ∈ G , then there are 3 orbits with representatives i, ± ( i + 2). – γ (cid:54)∈ G and(A) 2 γ √ γ ∈ G , then 3 orbits with repr. ± i and i + 2.(B) γ √ γ ∈ G , then 2 orbits with repr. i and i + 2.(C) 2 γ √ γ , γ √ γ (cid:54)∈ G , then 4 orbits.Since the above results do not depend on the even direction fixed, to obtainthe total number of orbits for the even directions we need to simply multiply theabove numbers by 3. The labelling A, B, C in the last case corresponds to thesets and notation in Lemma 11. The results up to here have been summarizedin Figure 9.Next, we must consider the case of lines directed by v = x i with i an oddnumber. Following the same arguments as in the even case, we obtain theanalogous 4 conditions: • Lines labelled ± ( i + 4) and ± ( i + 6) are in the same ∆ -orbit if and onlyif for some µ (cid:48) ∈ R , it holds − γ x i +2 + µ (cid:48) x i ∈ Z [ x ] ⇐⇒ √ γ ∈ G γ R \ √ G ED F G G √ G √ G γ γ G √ G G √ G G R \ G C AB
Figure 9: The number of different orbits of lines depending on ( γ , γ ). On theleft picture we consider lines directed by x i with i odd. The right picture showsthe result for i even. • Lines labelled i + 4 and − ( i + 4) are in the same ∆ -orbit if and only iffor some µ (cid:48) ∈ R , it holds2( γ x i +4 + γ x i +1 )+ µ (cid:48) x i ∈ Z [ x ] ⇐⇒ (2 √ γ +2 γ ) x − (4 γ + µ (cid:48) ) ∈ Z [ x ]and this last condition is equivalent to 2( √ γ + γ ) ∈ G . This should becompared with set D in Lemma 11. • Lines labelled i + 6 and − ( i + 6) are in the same ∆ -orbit if and only iffor some µ (cid:48) ∈ R , it holds2( − γ x i + γ x i +1 ) + µ (cid:48) x i ∈ Z [ x ] ⇐⇒ γ x i +1 + ( − γ + µ (cid:48) ) x i ∈ Z [ x ]and this last condition is equivalent to 2 γ ∈ G . • Lines labelled ± ( i + 4) and ∓ ( i + 6) are in the same ∆ -orbit if and onlyif for some µ (cid:48) ∈ R , it holds γ √ x i +5 +2 γ x i +1 + µ (cid:48) x i ∈ Z [ x ] ⇐⇒ ( γ √ γ ) x +( µ (cid:48) − γ ) ∈ Z [ x ]and this last condition is equivalent to γ √ γ ∈ G . This should becompared with set E in Lemma 11.In the case of lines directed by an odd power of x the ‘square’ we obtain is:33 + 4 − ( i + 4) i + 6 − ( i + 6) √ γ √ γ + γ ) 2 γ γ √ γ A summary of our findings on the number of orbits of lines directed by oddpowers of x is summarized in Figure 9.Finally, to obtain L , the total number of orbits of lines under the actionof Γ, we need to combine the information on the lines directed by odd andeven powers of x . The considerations in Lemma 11 together with the analysiscarried out in this proof yield the final result, which we have encapsulated inFigure 8. H (Ω E γ ) In order to compute the rank of the group H (Ω E γ ) via Theorem 3 we need onemore ingredient: the value of R , which is the rank of the Z -module generatedby Λ Γ i , i = 1 , . . . , n . Recall that Γ i ≤ Γ is the stabilizer under the action of Γof the vector space spanned by f i . Lemma 28.
For all i = 1 , . . . , , the rank 2 groups Γ i are explicit in the fol-lowing chart: Γ Γ Γ (cid:104) f , f − f (cid:105) (cid:104) f , f + f (cid:105) (cid:104) f , f + f (cid:105) Γ Γ Γ (cid:104) f , f + f (cid:105) (cid:104) f , f + f (cid:105) (cid:104) f , f − f (cid:105) Proof.
Consider a line directed by the vector f i . The group Γ i is the subgroup { h ∈ Γ | f i + h = µf i , µ ∈ R } . Thus we need to solve an equation of the type λf i = h = n f + n f + n f + n f + n f + n f , where λ ∈ R and ( n , . . . , n ) ∈ Z . Using Equation (2) the above expressioncan be rewritten in terms of roots of unity and our task is to understand which λ ∈ R can be written as λ = (cid:80) k n k x k . By Lemma 10 we deduce λ ∈ Z [ √ f i which yield elements of the form Z [ √ f i arecollected in Lemma 6. The results in the table in the statement follow. Lemma 29.
Let β be the map from Eq. (1). For all parameters γ we have R = rk β = 3 , and coker β ∼ = Z . roof. By definition, the image of the map β depends only on the stabilizers Γ α of the 1-singularities α , that is, on their directions f i , and not on their positions.Let us denote Λ i = Λ Γ i for i = 1 , . . . ,
6, so that im β is generated by the Λ i .Using the explicit description of the groups Γ i in Lemma 28 and the relationsin Lemma 6, we compute the relevant exterior products: Λ = f ∧ f − f ∧ f = − f ∧ f + f ∧ f , Λ = − f ∧ f + f ∧ f , Λ = − f ∧ f + f ∧ f , Λ = f ∧ f − f ∧ f , Λ = 2 f ∧ f − f ∧ f = f ∧ f − f ∧ f + f ∧ f + 2 f ∧ f , Λ = 2 f ∧ f − f ∧ f = 2 f ∧ f − f ∧ f + f ∧ f + f ∧ f . In order to compute the rank of the Z -module im β generated by the Λ Γ i , werepresent each of these exterior products, with respect to the basis f i ∧ f j givenby the lexicographic order, as a column in the following matrix: − − − −
20 1 − − . The quantity R we seek is the rank of this matrix. It is evident that the firstthree columns are linearly independent. We have Λ = Λ − − , Λ = − + 3Λ + 2Λ , Λ = − + 2Λ and therefore R = 3.It is also easy to see that − (Λ + Λ ), Λ − + Λ ) and − Λ span a Z -module of rank 3 with free complement in Z . As the latter equals Λ ∆ inour basis, coker β is free of rank 3. Proposition 30.
The rank of H (Ω E γ ) depends on γ and equals the valuesdepicted in Figure 8 plus 1.Proof. By Theorem 3 the rank of H can be computed as 4 + L − R . The valueof L as a function of γ was computed in Lemma 27. Finally, by the previouslemma we know that for all γ we have R = 3 and the result follows. We are now left with the task of computing the rank of the second cohomologygroups. To this end we need to compute the quantity e in Theorem 3. It is35efined as a linear combination of the quantities L α and L which depend on γ .We recall the reader that the latter quantity is the cardinality of the set of allorbits of 0-singularities, while the former is the cardinality of the set of orbitsof 0-singularities on the specific 1-singularity α .The 0-singularities are obtained intersecting the representatives of orbits oflines studied in Section 3.2. From Proposition 27 we know that the numberof orbits depends on the value of the parameter γ . In the next subsection wecompute L α and and L in the case γ = (0 , L having value 6, 9 or 12. The corresponding values of γ are theones in the regions enclosed by a blue line in Figure 8. These results come froma computed aided calculation. Finally, in the last subsection we address thegeneral case. We have not obtained a complete description of the second coho-mology groups in this case. However, again with a computed aided calculation,we present an example of the extremal case in Proposition 27 when we have 24different line orbits to deal with. γ = (0 , Lemma 31. If γ = (0 , , then L = 14 and for each 1-singularity α we have L α = 6 .Proof. First of all we show that L α = 6 for each 1-singularity α . For a fixed line α = R f i , with i = 1 , . . . ,
6, we are interested in the intersections of this line withthe orbits of the other lines under the action of Γ. Using the bijection describedin Equation 2 we can write the relevant equations in terms of powers of the rootof unity x instead of the vectors f i . The generic formula we have to deal withsets a point in a line of direction x i , or rather points in a Γ i -orbit, equal a pointin a line of direction x j . In symbols we have, for i (cid:54) = j ∈ { , . . . , } : µx i + (cid:88) n k x k = λx j , n k ∈ Z , µ, λ ∈ R ⇐⇒ λ = µx i − j + (cid:88) k n k x k . It follows that λ is of the form cx + d + µ ( ax + b ) with a, b, c, d ∈ G . This givesus the conditions (cid:40) λ = d + µb − c = µa = ⇒ λ = d − ba c. (9)Inside the infinite collection of points of intersection between a line of direction x j and the translates by Γ of a line of direction x i , determined in this settingby the possible values of λ , we need to count how many are in different orbits.That is, the points λ and λ (cid:48) will be identified if λ − λ (cid:48) ∈ G .36ince x i − j = a + xb , by Lemma 9 when i − j varies we obtain five possiblevalues for − ba which are collected in the following set: S = { , − √ , −√ , − √ , −√ } . Keeping the notation λ to refer to its class modulo G , from (9) we see that weneed to understand how many different equivalence classes are there for numbersof the form λ = ba ( c + c √
3) where c , c ∈ Z and ba ∈ S . The complete list isthe following T = { , √ , √ , , √ , √ } , from which we deduce that L α = 6 independently of the value of α , that is, ofthe direction of the 1-singularity.We have identified on each line 6 different equivalence classes of points.However, some of these classes in different lines can be related by an element ofΓ. For example, the class of 0 belongs to every 1-singularity so there are at most31 different classes. The final quantity we need to compute, L , is precisely thistotal number of classes of 0-singularities under the action of Γ.Notice that the classes of 0-singularities in the line x i which are in the sameequivalence classes of 0-singularities in the line x j only depend on the difference i − j . Indeed, we have to solve an equation of the form λ x i − λ x j ∈ Γ ⇐⇒ λ x i − λ x j = a + bx with a, b ∈ G and λ , λ ∈ T , which is equivalent to λ x i − j − λ = a (cid:48) + b (cid:48) x forsome a (cid:48) , b (cid:48) ∈ G . With the help of Lemma 9 we see that the line of direction x and the line directed by x only share one equivalence class of 0-singularities,that of 0. So we have that L ≥
11. Comparing the equivalence classes betweenthe lines directed by x and x we obtain: λ x − λ = a + bx ⇐⇒ λ ( x √ − − λ = a + bx = ⇒ (cid:40) λ ∈ √ Gλ ∈ − λ + G .
It follows that the class of √ and the class of √ are in the same Γ orbit whenconsidered in lines directed by x i and x i − . From here we conclude that L ≥ x and x plus three more in the line directed by x .An equivalent computation to the one shown above, comparing the linesdirected by x and x and by x and x shows that there are no more orbitsof 0-singularities and gives a complete description. Indeed, it turns out thatthe line directed by x has four orbits in common with the one directed by x ;while it has three orbits in common with the line directed by x . Figure 10summarizes the the result: on each line of direction x i there are six markedpoints and points in different lines with the same color are in the same orbitunder the action of Γ. We conclude that L = 14.37 x x x x x √ √
32 1+ √
32 2 √ AA CC BB
Figure 10: Orbits of points if γ ∈ ∆ In this section, we summarize the results of some of the simpler cases, namely allthose with up to two 1-singularities per direction. For this, we need to determineall values of γ giving rise to these cases, and then determine for each of themthe values of L , L , and L α for all line orbits α . As R = 3 and coker β is freeindependently of γ , we have by Theorem 3: H (Ω E γ ) = Z L + e , H (Ω E γ ) = Z L and H (Ω E γ ) = Z , where e = − L + (cid:80) α ∈ I L α .For each value of γ involved, we have determined the combinatorics of lineintersections with the help of a computer program. In order to be able to connectthe number of orbits L α of points on the individual lines α to the total numberof point orbits L , we have split the numbers L α as L α = (cid:80) p L α ,p , where L α ,p is the number of those orbits whose points are intersections of exactly p L = (cid:80) α ∈ I L α ,p /p .These values are tabulated in each case. In these tables, we list L ,p for each type of line . For example, the symbol L ,p indicates lines with combinatoricsof “type 1”, the symbol L ,p lines with combinatorics of “type 2” etc. Thesuperscripts do not stand for directions or any other property of the lines, butthey collect lines with the same type of combinatorics. The column n in thetable indicates the number of lines of a certain type. The column dir specifieswhether lines of this type have even or odd directions, or both. In two further38ines in the tables, labelled (cid:80) L α ,p and L ,p , we give the sum of these data overall line types, and the corresponding data for the total orbits. The last columncontains the sum over all values of p . Case 1.
We start analyzing γ , γ ∈ G , which is equivalent to γ = (0 , L and L α in this case were carried out in Section 4.1. Theintersection combinatorics in the notation introduced in this section are givenin the following table: n dir p = 2 p = 3 p = 4 p = 5 p = 6 tot L ,p e, o (cid:80) L α ,p L ,p Proposition 32 (cf. Theorem 4.3) . The ranks of the cohomology groups of the12-fold tiling, together with the quantities L α , L , L and e , are given by: (cid:80) L α L L e rk H (Ω E ) rk H (Ω E ) rk H (Ω E )36 14 6 22 28 7 1 Case 2.
We now determine the intersection combinatorics under the as-sumption that there is only one line in each even direction and two lines perodd direction. From the proof of Proposition 27 (see particularly Figure 9) wesee that this situation can arise in two ways. • The first possibility is γ ∈ G and γ ∈ G \ G . All γ -values in this setgive the same intersection combinatorics, which are given in the followingtable: 39 dir p = 2 p = 3 p = 4 p = 5 p = 6 tot L ,p e
10 0 0 2 0 12 L ,p o L ,p o (cid:80) L α ,p L ,p
24 9 0 3 0 36We see here that we have two line types in each odd direction, with differ-ent intersection combinatorics, but only one line type in even directions.Putting these data together, we obtain:
Proposition 33.
The ranks of the cohomology groups of the generalized12-fold tilings with parameter γ = ( γ , γ ) satisfying γ ∈ G and γ ∈ G \ G , together with the quantities L α , L , L and e , are given by: (cid:80) L α L L e rk H (Ω E γ ) rk H (Ω E γ ) rk H (Ω E γ )90 36 9 54 63 10 1 • The second possibility occurs for γ ∈ √ G \ G and γ ∈ G . For all γ -values in this set, we obtain: n dir p = 2 p = 3 p = 4 p = 5 p = 6 tot L ,p e L ,p o (cid:80) L α ,p L ,p
36 5 0 0 2 43Here, for each odd direction, we have two lines of the same type. Thesetwo lines are related by symmetry - they are mirror images of each other.The cohomology of this case is slightly different:
Proposition 34.
The ranks of the cohomology groups of the generalized12-fold tilings with parameter γ = ( γ , γ ) satisfying γ ∈ √ G \ G and γ ∈ G , together with the quantities L α , L , L and e , are given by: (cid:80) L α L L e rk H (Ω E γ ) rk H (Ω E γ ) rk H (Ω E γ )99 43 9 56 65 10 1Of course, there is also the possibility of one line per odd direction, and twolines per even direction, but this is completely analogous to the above two cases,40nd we do not list them separately. This symmetry is reflected in Figure 8 bythe symbolic green diagonal.We now turn to the situation where we have two lines in each even and eachodd direction. Case 3.
We start with the case where the two lines in even directions aremirror images of each other. There are then three subcases. • In the first subcase, both γ and γ are contained in √ G \ G (cf. proof ofProposition 27 and Figure 8), in which case also the lines in odd directionsform mirror pairs. The corresponding γ -values then all lead to the sameintersection combinatorics as follows: n dir p = 2 p = 3 p = 4 p = 5 p = 6 tot L ,p e, o
12 1 0 0 2 15 (cid:80) L α ,p
12 144 12 0 0 24 180 L ,p
72 4 0 0 4 80We see here that in this case, all even and odd lines are of the same type,and are in fact all symmetry equivalent. This intersection combinatoricsresults in the following cohomology:
Proposition 35.
The ranks of the cohomology groups of the generalized12-fold tilings with parameter γ = ( γ , γ ) satisfying γ , γ ∈ √ G \ G ,together with the quantities L α , L , L and e , are given by: (cid:80) L α L L e rk H (Ω E γ ) rk H (Ω E γ ) rk H (Ω E γ )180 80 12 100 112 13 1 • In the second subcase, we have γ ∈ √ G \ G , but now γ (cid:54)∈ √ G . Moreprecisely, γ depends on γ , and must satisfy 2 γ + √ γ ∈ G , whichcorresponds to region E in Figure 9 (cf. proofs of Proposition 27 andLemma 11). All γ -values satisfying the conditions in this second subcaselead to the following intersection combinatorics: n dir p = 2 p = 3 p = 4 p = 5 p = 6 tot L ,p e
12 0 0 3 0 15 L ,p o
10 3 0 2 0 15 (cid:80) L α ,p
12 132 18 0 30 0 180 L ,p
66 6 0 6 0 78Here, the lines in even and odd directions have different intersection com-binatorics, but in each direction they form mirror pairs. This intersectioncombinatorics results in the following cohomology:41 roposition 36.
The ranks of the cohomology groups of the generalized12-fold tilings with parameter γ = ( γ , γ ) satisfying γ ∈ √ G \ G , γ (cid:54)∈ √ G and γ + √ γ ∈ G , together with the quantities L α , L , L and e ,are given by: (cid:80) L α L L e rk H (Ω E γ ) rk H (Ω E γ ) rk H (Ω E γ )180 78 12 102 114 13 1We note that this case also occurs with the roles of even and odd directionsinterchanged, which we do not list separately. • In the third subcase, the two lines in an odd direction are not symmetryequivalent, which is the case if γ ∈ G and γ ∈ √ G \ ( √ G ∪ G )(again this characterization of the parameters comes from the proof ofProposition 27). For all these γ -values, we obtain: n dir p = 2 p = 3 p = 4 p = 5 p = 6 tot L ,p e
16 0 0 2 0 18 L ,p o L ,p o (cid:80) L α ,p
12 132 18 0 30 0 180 L ,p
66 6 0 6 0 78Here, we have two types of odd lines with different intersection combina-torics, but overall, this results in the same cohomology ranks as above:
Proposition 37.
The ranks of the cohomology groups of the generalized12-fold tilings with parameter γ = ( γ , γ ) satisfying γ ∈ G and γ ∈ √ G \ ( √ G ∪ G ) , together with the quantities L α , L , L and e , aregiven by: (cid:80) L α L L e rk H (Ω E γ ) rk H (Ω E γ ) rk H (Ω E γ )180 78 12 102 114 13 1Of course, this latter case can also occur with the roles of even and odddirections interchanged, but we do not list it separately. Case 4:
Finally, we can have two line types in both even and odd direc-tions, which occurs if both γ and γ are contained in G \ G (cf. proof ofProposition 27 and Figure 8). This is the first time in our analysis in whichthe G -action on the set of 0-singularities, that is points of intersection betweenrepresentatives of lines, has more than one orbit. Indeed, the set of parame-ters G \ G splits in three G -cosets, so that altogether we need to analyze thebehavior of γ -values ( γ , γ ) from nine different ( G × G )-cosets. The analysis42f these 9 sets gives rise to only two different intersection combinatorics. Thesimpler combinatorics is obtained for the γ -values in the cosets with represen-tatives { (1 , √ , ( √ , , (1 + √ , √ } . These can be characterised asthose cosets for which γ + γ ∈ √ G . The intersection combinatorics thenis: n dir p = 2 p = 3 p = 4 p = 5 p = 6 tot L ,p e, o (cid:80) L α ,p
12 72 48 0 0 24 144 L ,p
36 16 0 0 4 56Here, we have again a single type of lines, and the cohomology becomes:
Proposition 38.
The ranks of the cohomology groups of the generalized 12-foldtilings with parameter γ = ( γ , γ ) satisfying γ , γ ∈ G \ G and γ + γ ∈ √ G , together with the quantities L α , L , L and e , are given by: (cid:80) L α L L e rk H (Ω E γ ) rk H (Ω E γ ) rk H (Ω E γ )144 56 12 88 100 13 1For the remaining six cosets of γ -values, the combinatorics is: n dir p = 2 p = 3 p = 4 p = 5 p = 6 tot L ,p e, o
18 0 2 0 0 20 L ,p e, o
12 0 4 0 0 16 (cid:80) L α ,p
12 180 0 36 0 0 216 L ,p
90 0 9 0 0 99Here, we have two line types, each of which occurs both in even and odddirections. This results in the following cohomology:
Proposition 39.
The ranks of the cohomology groups of the generalized 12-foldtilings with parameter γ = ( γ , γ ) satisfying γ , γ ∈ G \ G and γ + γ (cid:54)∈ √ G , together with the quantities L α , L , L and e , are given by: (cid:80) L α L L e rk H (Ω E γ ) rk H (Ω E γ ) rk H (Ω E γ )216 99 12 117 129 13 1 We now turn our efforts into computing the numbers L and L α in terms of γ = ( γ , γ ). The strategy is the same as the one described for the case γ =430 , , x i ) with i > F ⊥ (see Figure 2). For future reference, we collect in the following table theexpressions of the powers of x in these basis (cf. Lemma 9):(1 , x ) (1 , x ) (1 , x ) (1 , x ) x √ (1 + x ) x + √
32 2 √ + √ x √ x x + x √ x √ x x √ (2 x − √ + √ x √ x x x − − + x √ √ x x √ ( x − x − √ − √ + √ x Definition 40.
In order to simplify the following results we define the sets A = { , √ , √ } and A = { , , √ , √ } . The notation x + A stands forthe three values x + y with y ∈ A . We consider two generic lines given in Corollary 26 of directions x i , x k with k > i . There are three possible choices and we will deal with them separately: i, k are odd, i, k are even or i, k have different parity. We pick two lines directed by x i and compute their intersection with the trans-lates by Γ of the four different lines directed by x k when i, k are both even: γ √ (cid:40) x i x i +2 + γ √ x i +1 + λx i = ± ( γ √ (cid:40) x k x k +2 + γ √ x k +1 ) + µx k + (cid:88) n p x p . We simplify by x i and obtain γ √ (cid:40) x + γ √ x + λ = ± ( γ √ (cid:40) x k − i x k +2 − i + γ √ x k +1 − i ) + µx k − i + (cid:88) n p x p − i . (10)Recall that, by Lemma 9, the number (cid:80) n p x p − i is equal to ax + b with a, b ∈ G = Z [ √ , x k − i ) for F ⊥ and we compute the valueof λ using the array at the beginning of the section. Notice that because of ourchoice of basis we can compute directly λ without worrying about µ . There aretwo subcases to consider: 44 if k − i = 2, then subbing the information in (10) we obtain, consideringonly the first coordinate in the basis (1 , x ): γ √ (cid:40)
10 + γ √ √ / λ = ± ( γ √ (cid:40) − γ √ − √
33 )) + a √
33 + b, which implies λ = ± ( γ √ (cid:40) − − γ a √
33 + b − γ √ (cid:40) − γ . The 8 families of values for λ , each corresponding to the intersection ofone of the two lines directed by x i with one of the four lines directedby x k with k − i = 2, are collected in the following accolades. On theleft hand side we have the intersection with the line passing through thepoint γ √ x i + γ √ x i +1 and on the right hand side passing through the point γ √ x i +2 + γ √ x i +1 . λ = − γ + a √ − γ √ − γ + a √ − γ √ a √ − γ √ a √ and λ = − γ + a √ − γ + a √ − γ √ a √ a √ + γ √ The final step is to consider the above values modulo G . Since a, b ∈ G , wecan get rid of b in all of the above expressions and moreover, expressing a as a + a √ a , a ∈ Z we see that modulo G the numbers a √ splitin three different equivalence classes with representatives in { , √ , √ } .We conclude that the orbits of points on the two lines γ √ (cid:40) x i x i +2 + γ √ x i +1 + λx i obtained by intersecting them with the 4 lines directed by x i +2 are deter-mined by the following two sets of 12 values of λ . – Intersection with γ √ x i + γ √ x i +1 + λx i for the values of λ : − γ √ − γ + A − γ − γ √ + A − γ √ + A A (11) – Intersection with γ √ x i +2 + γ √ x i +1 + λx i for the values of λ : − γ + A − γ − γ √ + A A γ √ + A (12)45 emark 41. Before we analyze the next case, we remark that if insteadof the two lines γ √ (cid:40) x i x i +2 + γ √ x i +1 + λx i , we had considered the intersections with the two lines − (cid:32) γ √ (cid:40) x i x i +2 + γ √ x i +1 (cid:33) + λx i , then we would have obtained the same values as in the above two tables with the opposite signs . • if k − i = 4, we perform the same computations as in the preceding point.The details are left to the reader. The equations to consider, ignoring theparameter b ∈ G and written in the basis (1 , x ) are γ √ (cid:40)
11 + 2 γ λ = ± (cid:32) γ √ (cid:40) − − γ (cid:33) + a √ . These yield 4 different families of values for λ : λ = − γ − γ √ + a √ − γ − γ √ + a √ − γ + a √ − γ − γ √ + a √ Notice that in this case, the two parallel lines of direction x i under con-sideration yield the same values of λ which, modulo de action of G , arecollected in the following table: − γ − γ √ + A − γ − γ √ + A − γ + A − γ − γ √ + A (13) If i, k are both odd numbers then we obtain, as in the preceding case, thefollowing equations from Corollary 26 (with k > i ): ± (cid:32) γ (cid:40) x k +4 x k +6 + γ x k +1 (cid:33) + µ √ x k + √ (cid:88) p n p x p = λ √ x i + γ (cid:40) x i +4 x i +6 + γ x i +1 . We simplify, multiplying the equation by x − i : ± (cid:32) γ (cid:40) x k +4 − i x k +6 − i + γ x k +1 − i (cid:33) + µ √ x k − i + √ (cid:88) p n p x p − i = λ √ γ (cid:40) x x + γ x, (cid:80) p n p x p − i by ax + b using Lemma 9: ± (cid:32) γ (cid:40) x k +4 − i x k +6 − i + γ x k +1 − i (cid:33) + µ √ x k − i + √ ax + b ) = λ √ γ (cid:40) x x + γ x. (14)We now consider the above equation rewritten in the basis (1 , x k − i ), whichallows us to easily deduce the value of λ . To this end, we use the array atthe beginning of the section. We split the analysis in two cases, depending onwhether k − i is 2 or 4. We deduce • If k − i = 2, we use the basis (1 , x ) and we further split the computationsdepending on the sign on the right hand side of (14): – Considering Equation (14) with sign +: λ √ γ (cid:40) − − γ √
33 = b √ a − γ √ γ (cid:40) − ⇐⇒ λ √ a + b √ − γ √
33 + γ (cid:40) − − γ (cid:40) − − ⇐⇒ λ √ a + b √ − γ √
33 + γ (cid:40) ⇐⇒ λ = b + a √ − γ γ √ (cid:40) – Considering Equation (14) with sign − : λ √ γ (cid:40) − − γ √
33 = b √ a + γ √ − γ (cid:40) − ⇐⇒ λ √ b √ a − γ (cid:40) − − ⇐⇒ λ = b + a/ √ − γ √ (cid:40) − − λ modulo G . Remarkthat in this case the two parallel lines of direction x i yield the same valuesof λ . − γ + A − γ + γ √ + A γ √ + A γ √ + A (15)47 If k − i = 4, we use the basis (1 , x ) in (14) and perform a computationanalogous to the one in the preceding case: λ √ γ (cid:40) − γ √ b √ a ± (cid:32) − γ √
33 + γ (cid:40) − (cid:33) ⇐⇒ λ √ b √ a − γ √ + γ (cid:40) −
10 + γ (cid:40) − γ √ b √ a + γ √ − γ (cid:40) − − γ (cid:40) − − γ √ ⇐⇒ λ = b + 2 a √ − γ + γ √ −
101 or λ = b + 2 a √ − γ γ √ i odd, the orbits of points on the two lines γ √ (cid:40) x i +4 x i +6 + γ √ x i +1 + λx i obtained by intersecting them with the 4 lines directed by x i +4 are deter-mined by the following 2 sets of 12 values of λ . – Intersection with γ √ x i +4 + γ √ x i +1 + λx i for the values of λ : − γ + A − γ + A − γ − γ √ + A − γ + γ √ + A (16) – Intersection with γ √ x i +6 + γ √ x i +1 + λx i for the values of λ : − γ + A − γ + γ √ + A − γ + γ √ + A − γ + γ √ + A (17) k odd If i, k are of different parity and k is odd, then we use Corollary 26 to understandthe intersection of two lines of direction x i with the translates of 4 lines ofdirection x k . We obtain ± (cid:32) γ (cid:40) x k +4 x k +6 + γ x k +1 (cid:33) + µ √ x k + √ (cid:88) p n p x p = λ √ x i + γ (cid:40) x i x i +2 + γ x i +1 . (18)48e isolate λ and divide the whole expression by x i , λ √ ± (cid:32) γ (cid:40) x k − i +4 x k − i +6 + γ x k +1 − i (cid:33) + µ √ x k − i + √ (cid:88) p n p x p − i − γ (cid:40) x x − γ x. We use Lemma 9 to substitute the expression (cid:80) p n p x p − i : λ √ ± (cid:32) γ (cid:40) x k − i +4 x k − i +6 + γ x k +1 − i (cid:33) + µ √ x k − i + √ ax + b ) − γ (cid:40) x − γ x. We finally divide by √ i, k of different parity, k odd: λ = ± (cid:32) γ √ (cid:40) x k − i +4 x k − i +6 + γ √ x k +1 − i (cid:33) + µx k − i + ax + b − γ √ (cid:40) x − γ √ x. (19)The expression k − i can take any of the following values {− , − , , , } .Notice that since x n = − x n +6 , the cases − − λ . Indeed, in Equation 19, when written with respect tothe bases (1 , x ) or (1 , x − ) the only change is an overall sign change in the firstparenthesis, which leaves unchanged the set of possible values of λ . The sameoccurs when comparing the values k − i ∈ {− , } with respect to the bases(1 , x − ) or (1 , x ). It follows that we will have a complete answer by studyingthe cases k − i = − , − ,
1. To simplify the notation we will denote by (cid:60) [ · ] thelinear map that returns the first coordinate of an arbitrary expression on thebasis (1 , x k − i ). • If k − i = −
3, then Equation (19) yields, ignoring the monomial withcoefficient µ , λ = (cid:60) (cid:34) ± (cid:32) γ √ (cid:40) xx + γ √ x − (cid:33) + ax + b − γ √ (cid:40) x − γ √ x (cid:35) . Now, ignoring the parameter b , since ultimately we are interested in λ upto the action of G , and using the chart at the beginning of this section weobtain: λ = ± (cid:32) γ √ (cid:40) √ γ √ (cid:33) + a √ − γ √ (cid:40) − γ i even, the orbits of points on the two lines γ √ (cid:40) x i x i +2 + γ √ x i +1 + λx i , x k = x i − are determined by the following 2 sets of 16 values of λ modulo G . To compile the tables we used the fact that a √ with a ∈ G can be expressed as a √
32 = ( a + a √ √
32 = 12 ( a √ a ) with a , a ∈ Z . It follows that modulo G , the expression a √ is equivalent to an elementin the set { , , √ , √ } . – Intersection with γ √ x i + γ √ x i +1 + λx i for the values of λ : − γ √ + A − γ − γ √ + A − γ − γ √ + A − γ − γ √ + A (20) – Intersection with γ √ x i +2 + γ √ x i +1 + λx i for the values of λ : A − γ − γ √ + A − γ − γ √ + A − γ + A (21) • If k − i = 1 we follow the same steps as in the preceding case to obtain: λ = (cid:60) (cid:34) ± (cid:32) γ √ (cid:40) x x + γ √ (cid:33) + µx − + ax + b − γ √ (cid:40) x − γ √ x (cid:35) , which turns into λ = ± (cid:32) γ √ (cid:40) √
30 + γ √ (cid:33) + √ a − γ √ (cid:40) − γ . Notice that in this case, modulo de action of G , there is only one orbitof values of λ for each of the 4 lines of direction x i − . The values of λ obtained are: – Intersection with γ √ x i + γ √ x i +1 + λx i for the values of: λ ∈ { , − γ , − γ − γ √ , − γ − γ √ } . (22) – Intersection with γ √ x i +2 + γ √ x i +1 + λx i for the values of: λ ∈ {− γ √ , − γ − γ √ , − γ − √ γ , − γ − √ γ } . (23) • If k − i = 1, we follow the same steps as in the preceding cases to obtain: λ = (cid:60) (cid:34) ± (cid:32) γ √ (cid:40) x x + γ √ x (cid:33) + µx + ax + b − γ √ (cid:40) x − γ √ x (cid:35) , λ = ± (cid:32) γ √ (cid:40) −√ − γ √ (cid:33) + b − γ √ (cid:40) − , yielding, – Intersection with γ √ x i + γ √ x i +1 + λx i for the values of: λ ∈ {− γ √ , − γ √ − γ , , γ } . (24) – Intersection with γ √ x i +2 + γ √ x i +1 + λx i for the values of: λ ∈ { , − γ , γ √ , γ √ γ } . (25) k even The last case we need to analyze is k , i of different parity and k even. So,once again we fix two lines directed by x i and we intersect them with the fourdifferent lines directed by x k . That is, we are looking for the values of λ in thefollowing equation: ± (cid:32) γ √ (cid:40) x k x k +2 + γ √ x k +1 (cid:33) + µx k + (cid:88) p n p x p = γ √ (cid:40) x i +4 x i +6 + γ √ x i +1 + λx i , which we divide by x i to obtain: ± (cid:32) γ √ (cid:40) x k − i x k − i +2 + γ √ x k − i +1 (cid:33) + µx k − i + (cid:88) p n p x p = γ √ (cid:40) x x + γ √ x + λ. (26)This time the expression k − i can take any of the following values {− , − , − , , } ,and just like in the case k odd, i even, the analysis will be complete if we studythe cases k − i = − , − , • If k − i = −
3, then working with the basis (1 , x − ) we have: λ = ± (cid:32) γ √ (cid:40) − x − x − γ √ x (cid:33) − µx − + (cid:88) p n p x p − γ √ (cid:40) x x − γ √ x λ = ± (cid:32) γ √ (cid:40) −(cid:60) ( x ) − γ √ (cid:60) ( x ) (cid:33) + (cid:60) ( ax + b ) − γ √ (cid:40) (cid:60) ( x ) − − γ √ (cid:60) ( x ) λ = ± (cid:32) γ √ (cid:40) √ + γ √ (cid:33) + a √
32 + b − γ √ (cid:40) − − − γ Intersection with γ √ x i +1 + γ √ x i +4 + λx i for the values of: λ = ± (cid:32) γ √ (cid:40) √ + γ √ (cid:33) + a √
32 + b + γ √ − γ − γ + γ √ + A γ √ + A − γ + A − γ + A (27) – Intersection with γ √ x i +1 + γ √ x i +6 + λx i for the values of: λ = ± (cid:32) γ √ (cid:40) √ + γ √ (cid:33) + a √
32 + b + γ √ − γ − γ + γ √ + A − γ + γ √ + A − γ + γ √ + A γ √ + A (28) • If k − i = −
1, then remark that x − i = − x − i and we obtain λ = ± (cid:32) γ √ (cid:40) − x x + γ √ (cid:33) − µx + (cid:88) p n p x p − γ √ (cid:40) x x − γ √ x λ = ± (cid:32) γ √ (cid:40) (cid:60) ( x ) + γ √ (cid:33) + (cid:60) ( ax + b ) − γ √ (cid:40) (cid:60) ( x ) − − γ √ (cid:60) ( x ) λ = ± (cid:32) γ √ (cid:40) (cid:60) ( x ) + γ √ (cid:33) + a (cid:60) ( x ) + b − γ √ (cid:40) (cid:60) ( x ) − − γ √ (cid:60) ( x )which turns into λ = ± (cid:32) γ √ (cid:40) √ γ √ (cid:33) + a √ b − γ √ (cid:40) − − γ √ √ – Intersection with γ √ x i +1 + γ √ x i +4 + λx i for the values of: λ = ± (cid:32) γ √ (cid:40) √ γ √ (cid:33) + a √ b − γ √ − γ , that is λ ∈ {− γ − γ √ , − γ , − γ − γ √ , } . (29)52 Intersection with γ √ x i +1 + γ √ x i +6 + λx i for the values of: λ = ± (cid:32) γ √ (cid:40) √ γ √ (cid:33) + a √ b + γ √ − γ , that is, λ ∈ { γ √ , − γ + 2 γ √ , − γ , − γ } . (30) • If k − i = 1, then working with the basis (1 , x ) and Lemma 9 we have: λ = ± (cid:32) γ √ (cid:40) xx + γ √ x (cid:33) + µx + (cid:88) p n p x p − γ √ (cid:40) x x − γ √ x λ = ± (cid:32) γ √ (cid:40) −√ − γ √ (cid:33) − γ √ (cid:40) − − – Intersection with γ √ x i +1 + γ √ x i +4 + λx i for the values of: λ = ± (cid:32) γ √ (cid:40) −√ − γ √ (cid:33) + 2 γ √ λ ∈ {− γ + γ √ , γ √ , γ √ , γ √ γ } . (31) – Intersection with γ √ x i +1 + γ √ x i +6 + λx i for the values of: λ = ± (cid:32) γ √ (cid:40) −√ − γ √ (cid:33) + γ √ λ ∈ {− γ , , γ √ , γ √ γ } . (32) In the previous section we have computed for each line of direction x i all theintersection points with each of the Γ-translates of the other 20 lines of differentdirections. Moreover, we have collected these points in orbits under the Γ-action.The first step in order to compute the numbers L and L α , needed to deter-mine H (Ω E γ ), is to organize the information gathered so far, paying specialattention to repetitions in the above tables. This is done in Proposition 42.53e have not computed explicitly the values L and L α for all different pa-rameters γ . The number of subcases is too large to be considered of interest byitself. We finish our study with Proposition 42, which gives an upper bound on L α and moreover it describes explicitly the representatives of the 0-singularitieson each 1-singularity in terms of the parameter γ .Before we present the general picture in the next proposition, we show herethe intersection combinatorics, obtained with computer assistance, for the case γ = 1 / √ / γ = 1 /
13 + √ /
17. For technical reasons, our programrequires γ i ∈ Q ( √ γ , γ ) is taken in the region inFigure 8 labelled with a red 24. The intersection combinatorics for this caseare: n dir p = 2 p = 3 p = 4 p = 5 p = 6 tot L ,p e, o
42 0 2 0 0 44 (cid:80) L α ,p
24 1008 0 48 0 0 1056 L ,p
504 0 12 0 0 516We see that we have here a single line type occurring in both even andodd directions, and that the value of L α really attains the maximal value of44 admitted by Proposition 42. It is worth pointing out that in this genericcase, not all line intersections are generic in the sense that only two lines mayintersect in a single point (notice the entry in the table under p = 4). Thereason is that the positions of the singular lines do not move independently ofeach other when the γ -values vary. In this generic case, we obtain the followingcohomology: (cid:80) L α L L e rk H (Ω E γ ) rk H (Ω E γ ) rk H (Ω E γ )1056 516 24 540 564 25 1We conjecture this to be the maximal cohomology attained among all thegeneralized 12-fold tilings studied in this paper.Turning now our attention back to the general case, we present in the nextproposition our findings on the quantity L α . The statement refers to the fol-lowing explicit list of the 1-singularities α :(a) α = ± √ ( γ x i + γ x i +1 ) + λx i , λ ∈ R , i ∈ { , , } .(b) α = ± √ ( γ x i +2 + γ x i +1 ) + λx i , λ ∈ R , i ∈ { , , } .(c) α = ± √ ( γ x i +1 + γ x i +4 ) + λx i , λ ∈ R i ∈ { , , } .(d) α = ± √ ( γ x i +1 + γ x i +6 ) + λx i , λ ∈ R i ∈ { , , } .54 roposition 42. All the 1-singularities α have an associated value L α boundedby . Moreover, the values of λ identifying representatives of points of inter-section between α (in the above (a), (b), (c) and (d) cases) and the translatesof the 20 non-parallel 1–singularities are collected (respectively) in Table 1, Ta-ble 2, Table 3 and Table 4. (The values in the tables are for α with a positive √ factor; if the factor is negative, one should consider minus the entries in thetables.)Proof. The proof of this statement is nothing but organizing the informationcollected in the previous computations. We fix a direction x i and we look atthe 4 parallel lines given by Corollary 26.1. If i is even, then to understand the 0–singularities on:(a) α = √ ( γ x i + γ x i +1 )+ λx i , we collect the values of λ from the sets 11and 13 (for the intersections with lines directed by x i +2 and x i +4 respectively) and from the sets 20, 22 and 24 (for the intersectionswith lines directed by x i +3 , x i +5 and x i +1 respectively).Notice that the value − γ √ − γ appears in the sets 13, 22 and 24;the value 0 is in both sets 22 and 24 . The collection of values withno repetitions is the content of Table 1. Depending on the parameter γ , some of the values in the table could be in the same orbit underthe action of Γ. We conclude then that the total number of differentvalues in the table provides an upper bound for L α , which in thiscase is easily checked to be 44.If α = − √ ( γ x i + γ x i +1 ) + λx i we use Remark 41 to concludethat the values we are interested in are minus the ones displayed inTable 1.(b) α = √ ( γ x i +2 + γ x i +1 ) + λx i , we proceed just as before extractingthe values from the sets 12, 13, 21, 23 and 25. The summary of allthe values, with one repetition, is presented in Table 2.Again, if we consider α = − √ ( γ x i +2 + γ x i +1 )+ λx i , by Remark 41,we obtain minus the values in Table 2. This time, the upper boundfor L α we obtain is 44.2. If i is odd, we need to discuss the 0-singularities on:(c) α = √ ( γ x i +4 + γ x i +1 ) + λx i , for which we extract the values of λ from the sets 15, 16, 27, 29 and 31. Notice that in these sets the value − γ appears three times, − γ + γ √ twice and γ √ also twice. Thecollection of all these values with only one repetition is the content ofTable 3. The upper bound for L α in this case is found to be 44. Thecase of α = − √ ( γ x i +4 + γ x i +1 ) + λx i is dealt with via Remark 41as in the previous cases. 55d) α = √ ( γ x i +6 + γ x i +1 ) + λx i we use the values of λ in sets 15,17, 28, 30, and 32. This time the value − γ appears three timesand the values − γ + γ √ + A , − γ + γ √ and γ √ each appearstwice. The complete set of values with no repetitions is collectedin Table 4. This time the bound obtained is L α ≤
44. The case α = − √ ( γ x i +6 + γ x i +1 ) + λx i is dealt with via Remark 41. Remark 43. • If we replace γ by in the tables, the reader can remark that we obtain L α = 6 as in Lemma 31. • For all γ we have L ≤ (44 + 44) · · · . Of course in the case γ = 0 thesame upper bound gives · , while the actual result is . Thus wecan imagine that L is always strictly less than this value. − γ √ + A − γ √ − γ √ + A − γ √ − γ + A A − γ √ − γ + A − γ − γ √ + A − γ + A − γ − γ √ + A − γ √ + A − γ − γ √ + A − γ − γ √ + A − γ − γ √ + A − γ − γ √ − γ √ γ Table 1: Each value of λ is of the form aγ + bγ + c . The value aγ + bγ identifiesthe intersection point between the two 1–singularities, while c ∈ A ∪ A encodesthe different representatives of the equivalence classes under the action of G .The data on the first 2 lines is from the sets 11, 13 respectively while the lastthree lines have values from the sets 20, 22 and 24.56 √ + A − γ + A − γ − γ √ + A A − γ √ − γ + A − γ √ − γ + A − γ + A − γ √ − γ + A A − γ √ − γ + A − γ √ − γ + A − γ + A − γ √ − γ − √ γ − γ − √ γ γ √ γ √ + γ Table 2: Each value of λ is of the form aγ + bγ + c . The value aγ + bγ identifiesthe intersection point between the two 1–singularities, while c ∈ A ∪ A , encodesthe different representatives of the equivalence classes under the action of G .There is one value repeated in the table, namely − γ √ − γ . The data on thefirst 3 lines is from Tables 12, 13 and 21 respectively while the last line hasvalues from tables 23 and 25. − γ + A − γ + γ √ + A γ √ + A γ √ + A − γ + A − γ + A − γ − γ √ + A − γ + γ √ + A − γ + γ √ + A γ √ + A − γ + A − γ + A − γ − γ √ − γ − γ √ γ √ γ √ γ Table 3: Each value of λ is of the form aγ + bγ + c . The value aγ + bγ identifiesthe intersection point between the two 1–singularities, while c ∈ A ∪ A , encodesthe different representatives of the equivalence classes under the action of G .The data on the first 2 lines is from the sets 15, 16 respectively while the lastthree lines have values from the sets 27, 29 and 31. Notice the repetition of thevalue − γ . 57 γ + A − γ + γ √ + A γ √ + A γ √ + A − γ + A − γ + γ √ + A − γ + γ √ + A − γ + γ √ + A − γ + γ √ + A − γ + γ √ + A − γ + γ √ + A γ √ + A γ √ − γ γ + γ √ Table 4: Each value of λ is of the form aγ + bγ + c . The value aγ + bγ identifiesthe intersection point between the two 1–singularities, while c ∈ A ∪ A , encodesthe different representatives of the equivalence classes under the action of G .The data on the first 2 lines is from the sets 15, 17 respectively while the lastlines have values from the sets 28, 30 and 32.58 eferences [1] J. E. Anderson and I. F. Putnam. Topological invariants for substitutiontilings and their associated C ∗ -algebras. Ergodic Theory Dynam. Systems ,18(3):509–537, 1998.[2] M. Barge and J.M. Gambaudo. Geometric realization for substitutiontilings.
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