crystallographic and quasicrystallographic groups · eary 1981: de bruijn found the "cut and...

CRYSTALLOGRAPHIC AND

QUASICRYSTALLOGRAPHIC

GROUPS

Valery A. Churkin

Sobolev Institute of Mathematics SB RAS

Erlagol, 2010

1

Table of contents

Ñîäåðæàíèå

1 A SHORT HISTORICAL REVIEW... 41.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 A short plan of lectures . . . . . . . . . . . . . . . . . . . . . . 91.3 Advisable sources . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 AFFINE TRANSFORMATIONS 112.1 De�nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Properties of compositions . . . . . . . . . . . . . . . . . . . . 112.3 Pseudo-orthogonal transformations . . . . . . . . . . . . . . . 12

3 CRYSTALLOGRAPHIC GROUPS: DEFINITIONS 13

4 THE WEAK BIEBERBACH THEOREM 144.1 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.2 Corollary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 FINITE SUBGROUPS OF GLn(Z) 165.1 Jordan's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 165.2 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 NUMBER OF CRYSTALLOGRAPHIC GROUPS UP TOISOMORPHISM 206.1 Bieberbach's theorem . . . . . . . . . . . . . . . . . . . . . . . 206.2 Cocycles and extensions . . . . . . . . . . . . . . . . . . . . . 206.3 Bound for cocycles . . . . . . . . . . . . . . . . . . . . . . . . 216.4 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226.5 Pictures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

7 ALGEBRAIC INTEGERS AND ROOTS OF UNITY 24

8 Angles in QC-groups 268.1 2D general case . . . . . . . . . . . . . . . . . . . . . . . . . . 268.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278.3 Rank 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2

9 ANGLES OF ROTATIONS FOR QUADRATIC QUASILATTICES 299.1 De�nition, examples . . . . . . . . . . . . . . . . . . . . . . . 299.2 Lemma, corollary . . . . . . . . . . . . . . . . . . . . . . . . . 309.3 Theorems 2D and 3D . . . . . . . . . . . . . . . . . . . . . . . 32

10 WILD 4D AND 3D QUASICRYSTALLOGRAPHIC GROUPS 3310.1 4D and quaternions . . . . . . . . . . . . . . . . . . . . . . . . 3310.2 3D example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 3610.3 3D example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

11 PSEUDOORTHOGONALITY OF THE ACTION OF APOINT GROUP ON A QUASILATTICE 3711.1 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3711.2 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

12 MAIN RESULTS 3912.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3912.2 Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

13 Normal Abelian Subgroups of PEC-Groups 4013.1 De�nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4013.2 Lemma 1 (on invariance of a form) . . . . . . . . . . . . . . . 4113.3 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4113.4 Remarks on Euclead and Minkovsky spaces . . . . . . . . . . . 4213.5 Realization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4313.6 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

14 Proof of Theorem 1 4314.1 De�ning relators for two lattice . . . . . . . . . . . . . . . . . 4314.2 The inequalities for Z . . . . . . . . . . . . . . . . . . . . . . . 4514.3 The inequalities for T . . . . . . . . . . . . . . . . . . . . . . . 4614.4 The end of proof theorem 1 . . . . . . . . . . . . . . . . . . . 4714.5 Lemma 2 (a matrix criterion for two lattices) . . . . . . . . . . 47

15 Proof of Theorem 2 48

16 Proof of Theorem 3 4916.1 A rotation symmetry cube for p = q . . . . . . . . . . . . . . . 4916.2 A cube for p = q . . . . . . . . . . . . . . . . . . . . . . . . . 51

3

16.3 A cube for p < q . . . . . . . . . . . . . . . . . . . . . . . . . 52

17 Proof of Theorem 4 53

18 Example with exactly two lattices 56

19 Problems 58

20 Survey by Moodi R.V. 5820.1 Properties of a quasicrystall . . . . . . . . . . . . . . . . . . . 5820.2 De�nition model set . . . . . . . . . . . . . . . . . . . . . . . 5920.3 Icosian model set . . . . . . . . . . . . . . . . . . . . . . . . . 61

21 References 62

1 A SHORT HISTORICAL REVIEW...

1.1 History

1890�91 years: E.S. Fedorov and A. Sch�on�ies classi�ed all possiblesymmetry groups of natural crystalls or crystallographic groups.

A crystall symmetry is a transformation (isometry) of a euclidean plane,which moves the atomic lattice of the crystall to itself. It turned out that thenumber of such groups and, consequently, di�erent atomic lattices is equalto 219.

If we distinguish a lattice from its mirror re�ection then this numberequals 230. Crystall always admits parallel shifts (translations) in threenon-coplanar directions. There are only �nitely many of rotation symmetriesmodulo translation subgroup.

Year 1900: D. Gilbert formulated his 18-th problem on �lling the spaceusing congruent polyhedra.

The �rst part of the problem is the question: are there only �nitely manyessentially di�erent groups posessing a fundamental region (crystallographicgroups) in the n-dimensional euclidean space?

The second part asks if there exist any polyhedra completely �lling thespace without gaps and which are not fundamental regions for any groups ofmotions.

4

Years 1911�1912: L. Bieberbach answered the �rst part of Gilbertproblem in the a�rmative: the translation subgroup of a crystallographicgroup has the �nite index, this index is bounded by Jordan theorem (1878,1880), there are only �nitely many of types of isomorphisms.

Year 1928: K. Reinhardt answered the second part of the 18-th Gilbertproblem in the a�rmative: such polyhedra exist.

Year 1947: H. Zassenhaus o�ered an algorithm for encountering crystallographicgroups and, together with his pupils, found the number of crystallographicgroups in 4-dimensional euclidean space. It turned out that this numberequals 4783.

Year 1961: B.N. Delaunay (Delone) proved that there are only�nitely many of types of such polyhedra, which are Dirichlet regions for somecrystallographic groups in the n-dimensional euclidean space. He o�ered analgorithm for encountering such polyhedra for given n.

Years 1960�1975: Hao Wang, R. Berger, D.E. Knuth, R.M.Robinson, R. Ammann, ... there were discovered many examples of aperiodictilings ("aperiodic" means "without any nontrivial translations along itself")of the plane by "bricks" of �nitely many types. The corresponding groups ofmotions, moving the tiling to itself, turned out to be trivial.

Year 1978: R. Penrose there were discovered aperiodic Penrose tilings.

5

Year 1981: de Bruijn found the "cut and project method" for Penrosetilings. He obtained Penrose tilings by projecting 2-dimensional faces of a

"cubic" neighbourhood of some appropriate plane in the cubic tiling of the5-dimensional space to this plane.

Year 1984: D. Schechtman and others there were discovered reallyexisting quasicrystalls with the atomic lattice of a new type. There were

experimentally discovered rotation symmetries of orders 5, 8, 10, 12, whichis impossible in Fedorov�Sch�oen�ies groups. It is possible to obtain Penrosetilings by intersecting a quasicrystall with a plane.

6

Year 1986: S.P. Novikov and A.P. Veselov introduced the notion ofa quasicrystallographic group containing a translation quasilattice.

The translation quasilattice is �nitely generated and contains a basis ofthe space.

Main problem Found a classi�cation of quasicrystallographic groups indimension 2 and 3.

7

Years 1990�92: there were investigated �at quasicrystallographic groupsby Novikov's school (S.A. Piunikhin, A.P. Veselov, A. Lazareva, A. Chalych,V.Roganova).Theory of such groups with quadratic quasilattices provides

rotations of �nite orders 5, 8, 10, 12.

The possibility of rotation symmetries of an in�nite order was discovered.There were investigated icosahedral models of quasicrystalls which admitphysical realizations.

There was found a connection between quasicrystallographic groups andcrystallographic groups in pseudo-euclidean spaces.

G.A. Margulis constructed examples of wild quasicrystallographic groupsin the 3-dimensional space.

It is possible that a rotation group is everywhere dense in the group SO3

of all rotations and, at the same time, the whole group is not contained in a�nitely generated quasicrystallographic group.

Years 2000�2003: R.M. Garipov found a new "algebraic method ofencountering crystallographical groups".

Algorithm is based on the weakened Bieberbach theorem and allows tocompute not only classical crystallographic groups, but also crystallographicgroups of motions of Minkowsky spaces with a �xed rotation group. Therewere performed corresponding computations for the Minkowsky spaces ofdimension 6 4.

But the problem of extending the weakened Bieberbach theorem to arbitrarypseudo-euclidean spaces wasn't solved.

1998-2008 year: R.V. Moody formulated the general de�nition of aquasicrystall in terms of cut and projection.

There were showed outreachs to geometry, discrete groups and arithmetics,analysis and dynamic systems theory.

2006-10 years: V.A. Artamonov and S. Sanchez introduced thenotion of the symmetry group of a quasicrystall in "cut and projectionmodel"terms.

Numbers 5, 8, 10, 12 appear again.

V.A. Churkin solved the problem of extending the weakened Bieberbachtheorem to arbitrary pseudo-euclidean spaces.

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1.2 A short plan of lectures

• Main de�nitions and formulas for a�ne transformations and groups ofmotions.

Di�erent de�nitions of crystalls and crystallographic groups.

• The weakened Bieberbach theorem in euclidean spaces.

• The Jordan theorem.

• A short proof of the Bieberbach theorem on the �niteness of the numberof isomorphism classes of n-dimensional crystallographic groups.

• Auxiliary information about algebraic integers.

• Characterization of rotation angles for 2-dimensional and 3-dimensionalquasicrystallographic groups and rotation angles for groups with quadraticquasilattices.

• Wild examples in dimensions 4 and 3.

• Pseudo-orthogonality of the rotation group of a quasicrystallographicgroup.

• The weakened Bieberbach theorem in pseudo-euclidean spaces of index6 2.

• Examples of crystallographic groups of motions of pseudo-euclideanspaces of index > 3, that contain several "lattices and also exceptionalautomorphisms, which permute the translation lattice with anotherabstract "lattices".

• Unsolved problems.

1.3 Advisable sources

• J.G. Ratcli�, Hyperbolic manifolds, Springer-Verlag, 1994.

There is a detailed elementary proof of Bieberbach theorem and thetheory of discrete cocompact groups of motions of euclidean, sphericaland hyperbolic spaces.

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• Â.À. Àðòàìîíîâ, Þ.Ë. Ñëîâîõîòîâ, Ãðóïïû è èõ ïðèìåíåíèå âôèçèêå, õèìèè è êðèñòàëëîãðàôèè, Ìîñêâà, Àêàäåìèÿ, 2005.

This is a detailed and interesting textbook, intended for chemists,physisists and applied mathematicians.

• Ð.Ì. Ãàðèïîâ, Àëãåáðàè÷åñêèé ìåòîä âû÷èñëåíèÿ êðèñòàëëîãðàôè-÷åñêèõ ãðóïï è ðåíòãåíîñòðóêòóðíûé àíàëèç êðèñòàëëîâ, Íîâîñè-áèðñê, ÍÃÓ, 2003.

A textbook with Garipov's algorithm and it's realization in the 3-dimensional space.

• Ñ.Ï. Íîâèêîâ, È.À. Òàéìàíîâ, Ñîâðåìåííûå ãåîìåòðè÷åñêèå ñòðóê-òóðû è ïîëÿ, Ìîñêâà, ÌÖÍÌÎ, 2005.

There is an introduction to crystallographic groups theory and it'sgeneralizations.

• Ëå Òû Êóîê Òõàíã, Ñ.À. Ïèóíèõèí, Â.À. Ñàäîâ, Ãåîìåòðèÿ êâà-çèêðèñòàëëîâ, Óñïåõè ìàòåìàòè÷åñêèõ íàóê, 1993, Ò. 48, N 1, Ñ.41�102.

There are results of Novikov's school on the quasicrystallographic groups.

• È.Ð. Øàôàðåâè÷, Îñíîâíûå ïîíÿòèÿ àëãåáðû, Èæåâñê, ÐÕÄ, 1999.

A beautiful outline of the classical theory of crystallographic groups(without proofs).

• R.V. Moody,Model sets: a survey, From Quasicrystals to More ComplexSystems, Springer-Verlag, Berlin, 1998, p.145�166. (Variant 2008 year:arXiv: math.MG/0002020.)

General de�nition of quasicrystall in the "cut and projection"model:review of results and emphasizing on relationships with geometry, discretegroups and arithmetics, analysis and dynamic systems theory.

• R.V. Moody,Model sets: a survey, From Quasicrystals to More ComplexSystems, Springer-Verlag, Berlin, 1998, p.145�166. (Variant 2008 year:arXiv: math.MG/0002020.)

10

General de�nition of quasicrystall in the "cut and projection"model:review of results and emphasizing on relationships with geometry, discretegroups and arithmetics, analysis and dynamic systems theory.

2 AFFINE TRANSFORMATIONS

2.1 De�nitions

Let us consider Rn as a vector space and as an a�ne space simultaneously.Elements of the vector space Rn are vectors and there is a special (distinguished)vector 0, automorphisms of this vector space are invertible linear transormationsx 7→ Ax, detA 6= 0.

The a�ne space Rn has no distinguished elements, its elements are called'points' and automorphisms are invertible a�ne transformations g : x 7→a+ Ax, where a ∈ Rn, detA 6= 0.

A linear transormationA is called 'the linear part' of an a�ne transformationg or 'the di�erential' dg of this transformation.

A map x 7→ a+ x is called 'a parallel translation' by the vector a.It is convenient to write g := a+A. Here + is a formal notation, but in the

formula g(x) = a+Ax the symbol + means, as usually, the addition of vectors.Translation has the form a + I where I is the identity map. Translation isde�ned uniquely by the corresponding vector and so it may be identifyedwith this vector.

2.2 Properties of compositions

Theorem1) g : x 7→ a+ Ax determines a and A uniquely.2) The composition law: if g = a+A, h = b+B then g ·h = a+Ab+AB.3) The inversion law: if g = a+ A then g−1 = −A−1a+ A−1.4) The conjugation law: if g = a + A is an a�ne transformation and

z = z + I is a translation then g · z · g−1 = Az is a translation.

Proof. 1) If a+Ax = b+Bx for all x then for x = 0 we obtain a = b andAx = Bx for all x, A = B.

2) g(h(x)) = a+ A(b+Bx) = a+ Ab+ ABx.3) If g ·h = 0+I then as a consequence of 2) and 1) we obtain a+Ab = 0,

AB = I, thus b = −A−1a, B = A−1.

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4) g ·z ·g−1 = (a+A)(z+ I)(−A−1a+A−1) = = (a+Az+A)(−A−1a+A−1) = a+ Az − a+ I = Az + I = Az.

The proof of the theorem is complete.

Corollary 1) The set Affn of all a�ne transformations Rn is a groupunder the operation of composition.

2) The map g 7→ dg is a group homomorphism Affn → GLn.3) The subgroup Tn, consisting of all translations, is normal in Affn.4) Let Γ 6 Affn and Z = Γ ∩ Tn. Then for all g = a+ A ∈ Γ and z ∈ Z

we have Az ∈ Z.

2.3 Pseudo-orthogonal transformations

De�nitionA non-degenerate symmetric bilinear formB(x, y) of the signature(p, q), de�ned on the vector space Rn, determines on Rn the structure ofthe pseudoeuclidean space Rp, q. In the canonical form the correspondingquadratic form looks like

x21 + . . .+ x2

p − x2p+1 − . . .− x2

p+q,

where p + q = n. The space Rn, 0 is called 'a euclidean space', Rn−1, 1 is aMinkowsky space.

De�nition A linear transformation preserving a form of the signature(p, q) is called 'pseudoorthogonal of type (p, q)'. Such transformations constitutethe pseudoorthogonal group Op, q. A�ne transformations g = a + A, whereA ∈ Op, q, are called 'motions' or 'isometries' of the pseudoeuclidean spaceRp, q. They constitute the group of motions Is (Rp, q).

Remark In the euclidean space the group Is (Rn, 0) indeed coincides withthe set of mappings preserving the distance between any two points (anexcercise). In general case a quadratic form of the signature (p, q) determinesonly pseudometric but not a metric: it is possible that B(x, x) = 0 for x 6= 0.

The group corresponding to the the 4-dimensional Minkowsky space iscalled 'the Poincare group', the corresponding motions preserve the Maxwell'sequations for the electromagnetic wave in vacuum.

12

3 CRYSTALLOGRAPHIC GROUPS: DEFINITIONS

There exist di�erent de�nitions of crystallographic groups and their generalizations.

Let X be a locally compact metric space with the metric |x− y|.A subset

D ⊂ X is a Delaunay set or a net in Xif there exist such numbers R > r > 0

that∀a, b ∈ D : a 6= b⇒ |a− b| > r,

∀x ∈ X ∃a ∈ D : |x− a| < R.

De�nition 1The group Γ of isometries of a metric space X is called crystallographic

if its orbit Γx is a Delaunay set in X.

'Crystall' is a union of a �nite number of orbits of a crystallographicgroup Γ. Usually, in applications the crystalls under investigation are ones ineuclidean spaces.

The following de�nition of the crystallographic group doesn't involve thenotion of metric. It is su�cient to have the topological structure, which canbe found, for example, in a�ne spaces.

Let Γ be a group of continuous transformations of a locally compacttopological space X. Then Γ acts on X properly discontinuously if

∀K ⊂ X : K is compact ⇒ set {g ∈ Γ | gK ∩K 6= �} is �nite.

A group Γ acts on X cocompactly

if∃K ⊂ X : K is compact and X =

⋃g∈Γ

gK.

De�nition 2 Crystallographic group is a properly discontinuous cocompactgroup of topological transformations.

For the groups of motions of euclidean spaces these two de�nitions of acrystallographical group are equivalent.

Because of Bieberbach theorem it is possible to give the third de�nitionof a crystallographic group.

De�nition 3

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The group of motions of a euclidean space is crystallographic if all of itstranslations form a lattice, that is, a free abelian group which is generatedby the basis of the space.

The hard part of the proof of Bieberbach theorem is how to deduce thethird de�nitions from the �rst and the second de�nition.

The third de�nition can be easily generalized to the Novikov-Veselovde�nition of a quasicrystallographic group of motions of a euclidean space.

De�nition 4The group of motions of a euclidean space is quasicrystallographic if all of

its translations form a quasilattice, that is, a �nitely generated abelian groupcontaining the basis of the space:

Γ < Is (Rn, 0), Z = Γ ∩ Tn = 〈e1, . . . , eN〉,

e1, . . . , en − basis of Rn, 0, n 6 N.

De�nition 5 Let bee Γ crystallographic or quasicrystallographic groupwith lattice Z of translations. The quotient group G = Γ/Z is called a groupof rotations or a point symmetry group. This relation is usually expressed bythe formula Γ = Γ(G,Z).

Further we will use the third de�nition of a crystallographic (and aquasicrystallographic) group in euclidean spaces. The third de�nition of acrystallographic group will also be used for pseudoeuclidean spaces.

Remark. The Auslender conjecture, stating that crystallographic groupsof an a�ne space are virtually solvable, in general case still is not proved. Itholds if the dimension of the space is 6 3.

4 THE WEAK BIEBERBACH THEOREM

4.1 Proof

Theorem A lattice of translations in a crystallographic group of motions ofa euclidean space is uniquely determined by the crystallographic group itselfas an abstract group.

More exactly, it isa) a normal free abelian subgroup of �nite rank,b) a maximal abelian subgroup,

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c) endowed with a positive de�nite symmetric bilinear form which isinvariant under the action of the group by conjugation.

Proof. Let Z be the lattice of translations of a crystallographic group Γ.

Let a subgroup T 6 Γ satisfy the conditions a), b), c). Then� T ∩ Z / TZ,� TZ/T ∩ Z ' (T/T ∩ Z)× (Z/T ∩ Z) is an abelian group,� T ∩ Z is the center of TZ.

Hence, TZ is a nilpotent group of the class 2. Then

∀g ∈ T, ∀z ∈ Z : [g, [g, z]] = 1.

If g = a+ A then

[g, z] = gzg−1z−1 = A(z)− z = (A− I)z.

Then [g, [g, z]] = (A− I)2z. Therefore (A− I)2z = 0, ∀z ∈ Z.As Z contains a basis of Rn it follows that (A− I)2 = 0. But A ∈ On(R).

Hence A − I is a normal transformation. If (A − I)2 = 0 then A − I = 0,A = I so g is a translation, T ⊂ Z. Due to the maximality T = Z.


4.2 Corollary

Remark The proof does not work for pseudoeuclidean spaces because nontrivialunipotent transformation can be a motion of a pseudoeuclidean space.

But for quasicrystallographic groups it work.

Corollary Let ϕ : Γ→ Γ′ be an abstract isomorphism of crystallographicgroups Γ(G,Z) and Γ′(G′, Z ′). If ϕ(Z) = Z ′ then after an appropriate conjugationin the group Affn we may assume that

G = G′ 6 GLn(Z).

Proof. Let

ϕ : g = a+ A 7→ g′ = a′ + A′, z 7→ f(z), z ∈ Z.

Here f is an isomorphism of abelian groups Z and Z ′ in Rn. Then f canbe represented by a linear transformation F of the space Rn.

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Due to the isomorphism

g · z · g−1 7→ g′ · f(z) · g′−1.

By the conjugation formula we have

A(z) 7→ A′(f(z)),

f(A(z)) = A′(f(z)), ∀z ∈ Z,

FA = A′F, A′ = FAF−1 = fAf−1.

Replacing Γ with a group which is conjugate to Γ by f we can assumethat G = G′. If we choose the basis of the space which is also a basis of thelattice Z, we get G < GLn(Z).

The proof of the corollary is complete.

5 FINITE SUBGROUPS OF GLn(Z)

5.1 Jordan's Theorem

It is easy to see, that the orders of �nite subgroups of GLn(Z) bounded bythe function depending on n. Thus there are only �nite number of �nitesubgroups up to isomorphism.

Indeed, ring homomorphism Z → Zm by the rule x 7→ x(modm) inducegroup homomorphism ϕ : GLn(Z) → GLn(Zm). If m > 3 then Kerϕ =Γn(m) has no torsion (exercise).

Therefore any �nite subgroup ofGLn(Z) embedded in �nite groupGLn(Z3)and it's number not greater then the number of subgroups of GLn(Z3).

It is important to know the number of conjugacy classes of �nite subgroupsof GLn(Z) for the classi�cation of crystallographic groups.

Theorem(Jordan)There are only �nitely many �nite subgroups of GLn(Z) up to conjugacy

in GLn(Z).

There are only a bounded number of possibilities for the action of �nitegroup on a rank-n free abelian group Z.

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5.2 Proof

Proof. The second statement follows from the �rst because the action of �nitegroup G on Zn is given by a homomorphism ϕ : G→ GLn(Z).

Two actions is isomorphic if and only if images Imϕ of homomorphism isconjugated in GLn(Z).

We now prove the �rst statement.Given a �nite subgroup G of GLn(Z), we will �nd a basis for Zn so that

elements of G are represented by matrices with entries whose absolute valuesare bounded by a constant C = C(n).

Consider the standard embedding Zn ⊂ Rn, so that G can be seen as asubgroup of GLn(R).

By averraging over G any positive de�nite quadratic form on Rn, �ndG-invariant positive de�nite quadratic form.

Scale so that the lattice points nearest the origin have length 1. Let V 6Rn be the vector subspace spanned by these lattice points.

Scale the subspace V ⊥ until the lattice points in Rn \V nearest the originhave length 1.

The scaled quadratic form is G-invariant because G preserves V and V ⊥.

Continue this way until the vector subspace spanned by lattice points ofminimum (unit) norm is all of Rn.

There may not be a basis for lattice consisting of elements of norm 1(for example, union of set all cubes vertices and centers for cubical lattice inR5). We will show that there is a basis consist of elements of norm at most(n+ 1)/2.

Lemma 1. Number N(r) of lattice points in ball with radius r and centerat origin Rn is at most (2r + 1)n.

Proof. Indeed, balls with radius 1/2 and centered at lattice points haveno common points and contains in ball B(0, r + (1/2)).

We can using formula from math. analysis for volume of ball with radiusr in Rn:

volBn(r) = cnrn,

constant cn = volSn−1/n depends only on n, Sn−1 is a sphere with unit

radius from Rn.

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Lets compare volumes:

N(r) · cn/2n 6 cn(r + (1/2))n, N(r) 6 (2r + 1)n.

Lemma 1 is proved.

Choose linearly independent elements a1, . . . , an of norm 1 from Zn.

Let Wk be the vector space spanned by a1, . . . , ak. Suppose inductivelythat for some k < n the lattice Zn ∩Wk has a basis b1, . . . , bk of vectors ofnorm 6 (k + 1)/2.

Lemma 2. The distance from (Zn ∩ Wk+1)\Wk to Wk is greater than(k + 3)−(k+1).

Proof. Let exist w = u + v, w ∈ Zn ∩ Wk+1, u ∈ Wk, v ∈ W⊥k , 0 <

‖v‖ 6 (k + 3)−(k+1). Then the distance from the vectors λw, where λ =0, ±1, ±2, . . . , ±(k + 3)k+1, to space Wk is 6 1.

Let P be a parallelepiped spanned by a1, . . . , ak with center at origin.By adding to λw appropriate Z-linear combination of elements a1, . . . , ak

we can get such vector (λw)′ ∈ Zn ∩Wk+1 that it's projection to Wk belongto P .

Parallelepiped P has diameter not greater than | a1 | + . . .+ | ak |=1+ . . .+1 = k. Thus P contained in a ball of radius k/2 with center at originand (λw)′ contained in a ball of radius r = (k/2) + 1 with center at origin.

The number of points (λw)′ is not greater than N(r), r = (k/2)+1. Thenby lemma 1 for a space Wk+1 = Rk+1 we have that

2(k + 3)k+1 + 1 6 (2r + 1)k+1 = (k + 3)k+1.

This inequality is false. Lemma 2 is proved.

A lattice is the additive group, then the in�mum of lenghts projection ofvectors w ∈ Zn∩Wk+1 on W

⊥k is a minimum for some vector w0 ∈ Zn∩Wk+1

by lemma 2. Analogously, the projection of vector ak+1 is proportional to theprojection of vector w0 with a coe�cient in Z. Therefore b1, . . . , bk, w0 is abasis of lattice Zn∩Wk+1. By adding to w0 appropriate Z-linear combinationof elements a1, . . . , ak or b1, . . . , bk we can get such a new basis vector bk+1

that it's projection to Wk belong to P . Then

| bk+1 |6√

(k/2)2 + 1 < (k + 2)/2.

18

This completes the inductive step.

We conclude by showing that matrices representing elements of G withrespect to basis b1, . . . , bn for lattice Zn have bounded entries.

Since G preserves length, it su�cient to show that the coordinates of anyvector v are bounded, if | v |6 (n+ 1)/2.

By the Cramer's formulas i-th coordinate of vector v with respect to basisZn has absolute value volPi/volQ, where Pi is a parallelepiped spanned byb1, . . . , bi−1, v, bi+1 . . . , bn, andQ is a parallelepiped spanned by b1, . . . , bi−1, bi, bi+1, . . . , bn.

Since volume of parallelepiped is not greater than product of lengths ofit's edges, we have volPi 6 ((n+ 1)/2)n.

Now we only need a bottom estimation for volQ.Open balls with radius

rn = (n+ 3)−(n+1)

and centered at the vertices of Q have no pairwise intersection.

It's intersections with Q give us a number of sectors, sum of volumes ofwhich is not less then volume of Q.

Using parallel transfer we can collect all sectors at the same vertex of Q,

in that case they form a ball B with radius rn and volQ > volB(rn).

As a result coordinates of vector v bounded by function

volPi/volQ 6 ((n+ 1)/2)n/(cnrnn)

depending only on n.


19

6 NUMBEROF CRYSTALLOGRAPHIC GROUPS

UP TO ISOMORPHISM

6.1 Bieberbach's theorem

Theorem(Bieberbach)There are only �nitely many isomorphic classes of n-dimensional euclidian

crystallographical groups for any natural n.Proof. Let Γ = Γ(G,Z) be a crystallographic group of motions of euclidian

space Rn with translation lattice Z and point group G.

6.2 Cocycles and extensions

By the weakened Bieberbach theorem group isomorphism induces latticeisomorphism and point group isomorphism. By the Jordan's theorem we

suppose that

• Z = Zn,

• G is �xed �nite subgroup of GLn(Z),

• action G on Z = Zn is given.

We will �nd the number of such groups Γ 6 Is (Rn) that

• Z / Γ,

• Z is a maximal abelian subgroup,

• Γ/Z = G.

Let g ∈ G = Γ/Z and ρ(g) is given representative of coset g = ρ(g)Z.Then

ρ(g)ρ(h) = ρ(gh)α(g, h),

here α(g, h) ∈ Z. Therefore α : G×G→ Z is mapping (�cocycle� or �systemof factors�).

If α is given, then multiplying of elements of Γ is also given. Indeed, any

element of Γ uniquely presented as ρ(g)a, there g ∈ Γ, a ∈ Z.

20

Also,ρ(g)a · ρ(h)b = ρ(g) ρ(h) ρ(h)−1a ρ(h) b =

= ρ(gh) · α(g, h) (ρ(h)−1 a ρ(h)) b.

By choosing of representatives ρ(g) we can bound function α dependingonly on n.

There are only �nitely many di�erent functions α if n is �xed. Thereforenumber of extension Z with G is �nite.

6.3 Bound for cocycles

We will use the proof of theorem above. Firstly we will construct such Γ-invariant euclidean metric of Rn as all nonzero shortest vectors of lattice Zhave unit length and contain basis Rn.

Let a1, . . . , an be n linearly independent shortest vectors of Z. Let P bea parallelepiped spanned by a1, . . . , an.

ThenRn =

⋃g∈Γ

gP.

Since all edges P have unit length, then diamP 6 n.

We can choose element ρ(g), which move center of P at minimal distance,as representative of coset g.

Since g = ρ(g)Z and Z-translations of parallelepiped P covers Rn thenthis distance is 6 n.

Therefore α(g, h) = ρ(gh)−1ρ(g)ρ(h) move the center of P at distance6 3n.

Number of such elements of Z is bounded because number N(r) of latticepoints Z in ball with radius r and center at origin is at most (2r+1)n (lemma1).

Therefore number of functions α bounded depending on |G| 6 f(n) andn = rkZ.


21

6.4 Remarks

Remark 1 Bieberbach proved more. He proved that abstract isomorphismof crystallographic groups can be realized as a�ne transformation.

Remark 2 System of factors is not arbitrary function. Since group operationis associative, some condition holds for that function. There is the origins ofextension theory for groups with abelian kernel and cohomological theory forgroups.

Remark 3 Exact number of isomorphic classes n-dimensional crystallographicgroups at n = 1, 2, 3, 4 is numbers 2, 17, 219, 4783 correspondingly.

6.5 Pictures

22

7 ALGEBRAIC INTEGERS AND ROOTS OF

UNITY

De�nition. Complex number λ is algebraic, if

∃a1, . . . , an ∈ Q : λn + a1λn−1 + . . .+ an = 0.

Polynomial xn + a1xn−1 + . . .+ an is annulator for λ.

If it's degree is minimal, then it is minimal (annulator) for λ. It's degreeis called degree of number λ.

If it is possible to choose a1, . . . , an ∈ Z then λ is called integer algebraicnumber.

Lemma 1) Minimal polynomial divide any annulator polynomial.2) Minimal polynomial is unique.3) The set of all algebraic integers is a ring Z.4) Minimal polynomial for algebraic integer has integer coe�cients.Lets prove, for example, 4). All roots of annulator polynomial with integer

coe�cients is algebraic integers.Using 1) we can see that minimal polynomial for one of them divide

annulator.Therefore all roots of minimal polynomial is integer algebraic numbers.

24

All coe�cients of this polynomial can be expressed through its roots byViet formulas. Thus, in view of 3), all coe�cients are algebraic integers.

Since it is also rational then it is just integer numbers.De�nition Number λ ∈ C is called primitive root of unity degree k if

λk = 1, λl 6= 1 ∀l : 0 < l < k.

Theorem Let K/Q be algebraic extension degree n. Let λ ∈ K beprimitive root degree k. Then

k 6 2n2.

Proof. All primitive roots degree k is exactly roots of k-th cyclotomicpolynomial Φk(x), deg Φk(x) = ϕ(k), there ϕ is phi-function of Euler.

If k = pα11 . . . pαs

s , where 2 6 p1 < p2 . . . < ps is prime numbers, then

ϕ(k) = k∏i

(1− 1

pi) =

=k

p1 . . . ps

∏i

(pi − 1) > (p1 − 1) . . . (ps − 1) > 2s−1.

By the conditions n > ϕ(k).Then 2n > 2ϕ(k) > 2s, log2 2n > s.Therefore

n > ϕ(k) > ks∏i=1

(1− 1

pi) > k(1− 1

2)log2 2n =

k

2n.

Hence, 2n2 > k. Proof is complete.Remark This estimation is raw. Lets calculate exactly, which primitive

roots k-th degree of unity is belong to extension K/Q degree 4. By theoremk 6 2 · 42 = 32.

Direct solution of inequality ϕ(k) 6 4 give values k = 1, 2, 3, 4, 5, 6, 8, 10, 12.Primitive roots degree k = 2, 3, 4, 6 belongs to quadratic extension of

�eld Q. Remain possibilities are 5, 8, 10, 12.Rotation symmetries of namely orders 5, 8, 10, 12 has been experimentally

observed of quasicrystalls.

25

8 ANGLES OF ROTATIONS IN 2-DIM

QUASICRYSTALLOGRAPHIC GROUP

8.1 2D general case

Theorem For a real number ϕ there exists a 2D quasicrystallographic groupcontaining a rotation by ϕ if and only if

λ = exp(iϕ) = cosϕ+ i sinϕ

is an algebraic integer.Proof. Let Γ = Γ(G,Z) be some quasicrystallographic group of R2 with

quasilattice Z. Let g ∈ Γ be a rotation of R2 by angle ϕ.

We can identify the center of the rotation and the origin. We will identifyR2 with the �eld C. Then g : z 7→ λz, z ∈ C. Since |λ| = 1, it follows thatλ = exp(iϕ).

Since Z ' Zn, it follows that g|Z ∈ AutZ ' GLn(Z).

Every integer unimodular matrix is a root of it's monic characteristicpolynomial p(x) with integer coe�cients. Thus p(g|Z) = 0, p(g)z = p(λ)z =0,∀z ∈ Z. Therefore p(λ) = 0 and λ is an integer algebraic number.

Conversely, let

λn + a1λn−1 + . . .+ an = 0, ai ∈ Z, n = min .

Then p(x) = xn + a1xn−1 + . . .+ an is the minimal polynomial for λ.

Since λ = λ−1 and p(λ) = 0, it follows that

1

λn+ a1

1

λn−1+ . . .+ an = 0,

anλn + an−1λ

n−1 + . . .+ 1 = 0,

λn +an−1

anλn−1 + . . .+

1

an= 0.

The uniqueness of the minimal polynomial yields

an = ±1, a1 = ±an−1, a2 = ±an−2, . . . .

26

Let's identify euclidean plane R2 and C. Consider quasilattice

Z = 〈1, λ, λ2, . . . , λn−1〉.

Let g : z 7→ λz, z ∈ C. Since |λ| = 1, g is a rotation.

Let G = 〈g〉 = {gk | k ∈ Z} is a cyclic group and Γ = Γ(G,Z) is thesemidirect product of Z and G.

Then Γ is a quasicrystallographic group, g ∈ Γ, g is a rotation of R2 byangle ϕ.

8.2 Example

Example Two-dimensional quasicrystallographic group with in�nite cyclicrotation group.

Let's consider irreducible polynomial of degree 4 with integer coe�cientssuch that two of its roots have absolute value equal to 1 and two others aremutually inverse positive real numbers. For example, let

p(x) = (x2 − (2−√

3)x+ 1)(x2 − (2 +√

3)x+ 1) = x4 − 4x3 + 3x2 − 4x+ 1.

Put A = 2 −√

3, B = 2 +√

3. Since numbers A and B are irrational,p(x) is irreducible over Q.

It's easy to see that roots λ, λ of polynomial x2 − Ax+ 1 have absolutevalue 1 and roots µ, µ−1 of polynomial x2 − Bx + 1 are real and mutuallyinverse.

Assume that µ > 1. If λk = 1 then since p(x) is the minimal polynomialfor λ it divides xk − 1. But then µk = 1; a contradiction.

Consequently, λk 6= 1 for all k 6= 0. Let's construct the quasicrystallographicgroup Γ = Γ(G,Z) using this λ as in the theorem above.

This group is a required example.Remark 1 All rotations in crystallographic groups have �nite orders.

There exist quasicrystallographic groups containing a rotation of in�niteorder.

Remark 2 Let A be a linear operator of R4 with unimodular matrix,whose characteristic polynomial is equal to p(x). Then R4 decomposes intoa direct sum of two A-invariant planes.

27

Furthermore, A preserves an euclidean structure and acts as a rotationin one of these planes, and acts as a hyperbolic rotation in the second plane.

Therefore, we conclude that operator A preserves a pseudo-euclideanstructure of type (3,1).

8.3 Rank 3

Corollary There are no two-dimensional quasicrystallographic groups within�nite rotation group and translation subgroup of rank 3.

Proof. Assume that there exists a quasicrystallographic group Γ(G,Z) ⊂Is C with quasilattice Z of rank 3.

Since rkZ = 3, if a rotation of G has �nite order then this order is lessthan 2 · 32 = 18.

The group generated by all rotations of �nite orders is a �nite cyclicgroup. Consequently, there exists a rotation g : z 7→ λz of in�nite order with|λ| = 1.

By the theorem λ is a root of an irreducible over Q minimal polynomialp(x), deg p(x) 6 rkZ = 3.

If deg p(x) = 1 then p(x) = x± 1, λ = ∓1 and λ2 = 1.

If deg p(x) = 2 then p(x) = x2 −mx + 1 with m ∈ Z, p(λ) = p(λ) = 0,λ+ λ = 2 cosϕ = m, |m| 6 2.

If m = 0 then p(x) = x2 ± 1 and λ4 = 1.

If m = −1 then p(x) = x2 + x+ 1 and λ3 = 1.

If m = 1 then p(x) = x2 − x+ 1 and λ6 = 1.

If m = ±2 then p(x) = (x± 1)2 is reducible.

Consequently, deg p(x) = 3 and by the theorem p(x) = x3 +mx2 +mx+1(or p(x) = x3 +mx2 −mx− 1).

Since λ 6= λ is a root of p(x), applying Viet formula to the last coe�cientwe infer that the third root of p(x) is equal to ±1.

This contradicts to irreducibility of p(x). The corollary is proved.

28

9 ANGLES OF ROTATIONS FOR QUADRATIC

QUASILATTICES

9.1 De�nition, examples

De�nition n-dimensional quasilattice Z ⊂ Rn is called quadratic if thereexists a real quadratic extension K of the �eld Q such that

Z ⊂ Ke1 + . . .+Ken

for some e1, . . . , en in Z.Example 1 Quasilattice generated in the �eld of complex numbers by the

roots of equation x5 = 1 is quadratic for K = Q(√

5).Indeed, if λ = cos(2π/5) + i sin(2π/5) then cos(2π/5) = (

√5 − 1)/4,

cos(4π/5) = 2−√

5.

Let us denote L = K · 1 +K · λ. Then L contains

• i sin(2π/5),

• i sin(4π/5),

• λ2,

• λ3 = λ2,

• λ4 = λ,

• Z = 〈1, λ, λ2, λ3, λ4〉.

Example 2 Quasilattice, generated in R3 by vectors corresponding to thevertices of the regular icosahedron (dodecahedron), is quadratic.

Indeed, let τ : τ 2 = τ + 1, τ = (√

5 + 1)/2.

Vertices of the icosahedron have the following coordinates

(0, ±τ, ±1), (±1, 0, ±τ), (±τ, ±1, 0).

Vertices of the dodecahedron have the coordinates

(±1, ±1, ±1), (0, ±τ−1, ±τ),

(τ, 0, ±τ−1), (±τ−1, ±τ, 0).

29

(see, for instance, M. Berger, Geometry, chapter 12, 12.5.5.3, 12.5.5.5).Remark From the viewpoint of applications in chemistry and physics it

is important to classify the angles of rotations in two- and three-dimensionalquasicrystallographic groups with quadratic quasilattice. For such groups therank of the quadratic quasilattice is equal to 4 and 6 correspondingly.

9.2 Lemma, corollary

Lemma Let λ ∈ C, |λ| = 1. The number λ is integer algebraic of degree 4if and only if the number λ+ λ is integer algebraic of degree 2.

More exactly,

λ+ λ =m±

√m2 − 4k

2,

with

m, k ∈ Z, |m±√m2 − 4k

2| < 2

for some choice of the sign and irrational number√m2 − 4k.

Proof. Let A = λ+ λ ∈ R. As λ = λ−1 we have that

λ2 − Aλ+ 1 = 0.

Minimal polynomials for λ and λ coincide, so A is an algebraic integer.

If A is rational then A is integer and λ is a number of degree 2, whichcontradicts our assumption. Therefore, A must be irrational.

Letp(x) = x4 −mx3 + nx2 − ax+ b

be the minimal polynomial for λ. Then

p(x) = (x2 − Ax+ 1)(x2 −Bx+ C).

Comparing the corresponding coe�cients we obtain the equations

m = A+B, n = C + 1 + AB, a = CA+B, b = C.

30

It is obvious that a−m = CA− A = (C − 1)A = (b− 1)A.

If b 6= 1 then A = (a−m)/(b− 1) ∈ Q, which contradicts irrationality ofA.

Therefore, b = 1 = C, a = m, n = 2 + AB.Denoting k = AB = n− 2 ∈ Z we obtain the equations

A+B = m, AB = k.

Solving the quadratic equation x2 − mx + k = 0 we get the requiredpresentation:

λ+ λ = A =m±

√m2 − 4k

2, B =

m∓√m2 − 4k

2.

Now the proof of su�ciency is an easy excercise.Corollary If the point group of a two-dimensional quasicrystallographic

group acts irreducible on Q-completed quasilattice, then the quasilattice isof the rank 4 if and only if it is quadratic.

Proof. If g : z 7→ λz, z ∈ C is a rotation element in the point group then|λ| = 1 and λ must be an algebraic integer of degree 4,

otherwise the action of the point group would be reducible because of thecommutativity of the group of rotations.

Let us put A = λ+ λ, K = Q(√m2 − 4k).

Then A ∈ K andλ2 − Aλ+ 1 = 0,

λ2 = Aλ− 1 ∈ L = K · 1 +K · λ,and λ3 = Aλ2 − λ ∈ L, Z ⊂ L.

On the contrary, if Z ⊂ Ke1 +Ke2 and |K : Q| = 2 then rkZ 6 4.

By the corollary on the quasilattices of rank 3 we have that

rkZ 6= 3

andrkZ 6= 2,

otherwise Z would not be a quasilattice.

Therefore, rkZ = 4. The proof of the corollary is complete.

31

9.3 Theorems 2D and 3D

TheoremAn angle ϕ is the angle of rotation in some 2-dimensional quasicrystallographicgroup with quadratic quasilattice if and only if

cosϕ =m+ n

√d

4, m, n ∈ Z, d ∈ N, 4 | (m2 − n2d).

Proof. Because of the lemma we have

cosϕ =m±

√m2 − 4k

4.

Let d = m2 − 4k, n = 1 if the sign of the square root is 'plus' andn = −1, if the sign of the square root is 'minus'. Then m2 − n2d = 4k.

Vice versa, if m2 − n2d = 4k then

cosϕ =m+ n

√d

4=m±

√m2 − 4k

4.

The proof is complete.Theorem An angle ϕ is the angle of rotation in some 3-dimensional

quasicrystallographic group with quadratic quasilattice if and only if

cosϕ =m+ n

√d

4, m, n ∈ Z, d ∈ N, 4 | (m2 − n2d).

Proof. Su�ciency. Let the condition on the cosine of the angle ϕ hold.Let us construct a 3-dimensional quasicrystallographic group with the givenangle of rotation.

Because of the theorem 1 there exists a 2-dimensional quasicrystallographicgroup Γ whose point group consists of rotations by the angles which aremultiples of ±ϕ.

Its subgroup of translations Z is a quadratic quasilattice: Q⊗ Z has thedimension 2 over the �eld K = Q(cosϕ).

Let us include R2 into R3 and choose a vector e ⊥ R2.

Let Z be the quasilattice in R3 generated by Z, e, e cosϕ.

Let G consists of rotations around the axis Re by angles which aremultiples of ±ϕ.

32

Then the subgroup Γ, generated by Z and G is crystallographic and thequasilattice Z is quadratic.

Necessity. Let Γ(G,Z) be a 3-dimensional quasicrystallographic groupwith quadratic quasilattice and let g ∈ Γ be a rotation of R3 by the angle ϕaround some axis.

Then its matrix A is de�ned over the �eld K = Q(√d).

As det(A− I) = 0, the subspace W = Im (A− I) is de�ned over the �eldK and it has the real dimension 2.

Then T = Z ∩W is a quadratic quasilattice in the plane W .

Therefore, the rotation g ∈ Γ induces the rotation by the angle ϕ in theplane W and it preserves the quadratic quasilattice T . Because of theorem 1we obtain the formula for cosϕ.


10 WILD 4D AND 3D QUASICRYSTALLOGRAPHIC

GROUPS

10.1 4D and quaternions

Wild 4D and 3D quasicrystallographic groupsOur goal now is to construct examples of 4-dimensional and 3-dimensional

quasicrystallographic groups with a quadratic quasilattice,

whose rotation group is similar in some sense to a free nonabelian group.

Let us identify R4 and the algebra of quaternions H with orthonormalbasis 1, i, j, k.

Let us put

g = exp(iϕ), h = exp(jψ), G = 〈g, h〉.

Consider the additional requirement

0 < ϕ < π, | cosϕ| < 1

2, cosϕ =

m+ n√d

4,

m, n ∈ Z, d ∈ N, 4 | (m2 − n2d);

33

0 < ψ < π, cosψ = 2 cosϕ.

Choose numbers d,m, n in order to make the angles to be irrational:ϕ, ψ /∈ Qπ.

It's easy to see that the group G is dense in SU2 = S3 = {q ∈ H | |q| =1}.

By Tits theorem the groupG contains a free nonabelian subgroup, otherwiseit would be virtually solvable,

so it would satisfy the corresponding identity and, therefore, its closureSU2 would be virtually solvable.

But there is well known fact that SU2 contains free nonabelian subroupsand, consequently, it cannot be a virtually solvable group.

Let now Z = Z[G] is a Z-module in the additive group of the algebra H,generated by all quaternions contained in G:

gl1hl2gl3 . . . gls−1hls .

Herel1, l2, . . . , ls ∈ Z.

Theorem In the previous notations the semidirect product Γ = Z h Gis a 4-dimensional quasicrystallographic group with

• the quadratic quasilattice Z ⊂ H of rank 8,

• the point group G which is dense in SU2 = {q ∈ H : |q| = 1}.

Proof. In fact all we need to prove is that Z is a free abelian group ofrank 8.

From requirements for cosϕ and cosψ it follows that g and h are algebraicintegers of degree 4 over Q.

Therefore we can assume that li = 0, 1, 2, 3 for all i.

Let us write the following quaternions in the scalar-vector form

g = α + u, α = cosϕ, u = i sinϕ,

h = β + v, β = cosψ = 2 cosϕ, v = j sinψ.

34

Thengh = αβ + αv + βu− (u, v) + [u, v],

hg = βα + αv − βu+ (v, u)− [v, u],

gh− hg = 2βu = i · 4 cosϕ · sinϕ =

= (g + g)(g − g) = g2 − g−2 ∈ Z[g].

Therefore,gh = hg + z, g = g−1, z ∈ Z[g].

Finally, the subgroup Z is a Z-hull of the set of products

hlgm, 0 6 l, m < 4,

andrkZ 6 16.

Let us show now that we can assume that 0 6 l < 2.

Indeed,

h2 + 1 = 2h · cosψ, cosψ = 2 cosϕ = g + g−1 = P (g),

where P (x) ∈ Z[x], degP (x) < 4.Then

• h2 = −1 + 2h · P (g),

• h3 = −h+ 2h2 · P (g) = −h+ 2(−1 + 2h · P (g)) · P (g).

It allows us to eliminate h2 and h3 in the products hlgm.

Consequently, rkZ 6 8.Prove now that Z is quadratic over K = Q(

√d).

As

g + g−1 = 2 cosϕ =m+ n

√d

2= A ∈ K,

we have that

g2 = −1 + Ag, g3 = −g + Ag2 = −A+ (−1 + A2)g.

From this we have

Z ⊂ K1 +Kg +Kh+Kgh.

The proof is complete.

35

10.2 3D example 1

Corollary 1 There exists a 3-dimensional quasicrystallographic group witha quadratic quasilattice of rank 6 and

an in�nite nonabelian rotation group which is dense in SO3.

Proof. Let R3 be the euclidean space of the pure imaginary quaternionswith the basis i, j, k from the algebra H.

Then the group SU2 of quaternions of the norm equal 1 acts on R3 byconjugation

g : v 7→ gvg−1, v ∈ R3.

In the previous example we had G ⊂ SU2 and Z = Z[G]. It is clear thatZ is invariant under conjugation by elements of G. Then T = Z ∩R3 is alsoG-invariant.

The group Z is generated as an additive subgroup by the elements

1, g, g2, g3, h, hg, hg2, hg3.

Then the generators for T = Z ∩ R3 are twice the vector parts of thegenerators for Z:

g − g = 2i sinϕ, g2 − g2 = 2i sin 2ϕ, . . . .

If g = α + u, h = β + v then the vector part of gh is of the formαv + βu+ [u, v].

As αβ ∈ K = Q(√d) and the coordinates u, v in the basis i, j, k are in

K, it follows that T ⊂ K · i+K · j+K ·k and the quasilattice T is quadraticof rank 6.

The group G acts on T by conjugation.Let H be the image of G in SO3 under the representation induced by the

conjugation action.

Then the semidirect product ∆ = ThH is a required quasicrystallographicgroup.

Note that H is dense in SO3

and by Tits theorem it contains a free nonabelian subgroup F of anin�nite rank.The proof of the corollary is complete.

36

10.3 3D example 2

Corollary 2 There exists a 3-dimensional quasicrystallographic group whichis not contained as a subgroup in a �nitely generated 3-dimensional quasicrystallographicgroup.

Proof. Let F be a free group of the countable rank from the Corollary 1with a basis gi, i ∈ N. Then F preserves a quadratic quasilattice T .

Let {αi}i∈N be a countable sequence of real numbers which are algebraicallyindependent over Q.

Choose a vector v ∈ R3, v 6= 0.

Lethi : x 7→ gi(x) + αiv, x ∈ R3.

Let H = 〈hi | i ∈ N〉.As hiTh

−1i = T then Γ = HT is a quasicrystallographic group.

Besides that, Γ is not contained in a �nitely generated quasicrystallographicgroup, otherwise all αi would belong to a �nite extension of the �eld Q.


11 PSEUDOORTHOGONALITY OF THE ACTION

OF A POINT GROUP ONAQUASILATTICE

11.1 Lemma

Lemma Let Γ(G,Z) be a n-dimensional quasicrystallographic group. LetV 6 Z ⊗Q be a minimal subspace invariant under the action of G on Z byconjugation.

Then there exists a symmetric bilinear form on the space Z ⊗ Q whichis non-degenerate on the subspace V and invariant under the action of thegroup G.

Proof. The set Z = Z ⊗Q ⊂ Rn is a subspace over the �eld Q.

By our assumption the group G preserves the scalar product B(x, y) inthe euclidean space Rn.

The image of Z ×Z in R under B(x, y) has the dimension 6 N2 over Q,where N = rkZ.

37

Let us choose a basis β1, . . . , βs.

Then for x, y ∈ Z we have

B(x, y) =∑i

βiBi(x, y),

where Bi(x, y) ∈ Q.As B(x, y) is a bilinear symmetric form on Z, all of the forms Bi(x, y)

are also bilinear symmetric.

As B(x, y) is invariant under the action of G, it follows that all of theforms Bi(x, y) are also invariant under this action.

Because of the positive de�niteness we have that B(x, y) 6= 0 on V .

Therefore, there exist rational numbers αi (αi is close to βi)

such that the form S(x, y) =∑αiBi(x, y) is not identically zero on V .

Besides that, we have that S(x, y) is invariant under the action of G.

Let KerS be the kernel of the symmetric bilinear form S(x, y). Then

0 6 V ∩KerS < V.

The subspace V ∩KerS is invariant under the action of G.

Therefore V ∩KerS = 0 because of the minimality of V .

Hence, S(x, y) is non-degenerate on V . The proof of the lemma is complete.

11.2 Theorem

Theorem Let Γ(G,Z) be a n-dimensional euclidean quasicrystallographicgroup.

Then there exists a non-degenerate symmetric bilinear form which isde�ned on the quasilattice Z,

• taking the integer values on this quasilattice and

• invariant under the action of the point group G on Z by conjugation.

Proof. Let V be a minimal G-invariant subspace in Z = Z⊗Q and S(x, y)be a G-invariant symmetric bilinear form which is non-degenerate on V .

38

Let W = V ⊥ be the orthogonal complement with respect to the formS(x, y). Then V ∩W = 0 because of the minimality of V and the fact thatS|V is non-degenerate.

Therefore, Z = V ⊕W and W is invariant under G.

Proceeding with this process of decomposition, �nally we obtain Z =⊕sj=1Vj,

where Vj is a minimalG-invariant subspace, endowed with a non-degenerateG-invariant symmetric bilinear form Sj(x, y), j = 1, . . . , s.

Then the direct sum of these forms gives us the required form on Z.Multiplying this form by an appropriate natural number we can make it

to be integer-valued.

The proof of the theorem is complete.Remark Garipov R.M. and Churkin V.A. (Siberian Math. J., 2009)

proved that in conditions of this theorem every quasicrystallographic groupΓ = Γ(G,Z) in euclidean space is isomorphic to a crystallographic group inpseudo-euclidean space.

Main problem Found a classi�cation of crystallographic groups on pseudo-Euclidean space of small dimension.

12 MAIN RESULTS

12.1 Introduction

Garipov R. M. (Algebra and Logic, 2003) proved that crystallographic groupson Minkowski spaces uniquely determine their translation lattices as abstractgroups. Basing on an analog of the weak Bieberbach theorem, he obtaineda full description of some crystallographic classes on Minkowski spaces ofsmall dimension up to isomorphism (Algebra and Logic, 2003; Dokl. Akad.Nauk, 2006; Algebra and Logic, 2008). The weak Bieberbach theorem �??? He also asked about the validity of the weak Bieberbach theorem forcrystallographic groups on arbitrary pseudo-Euclidean spaces.

12.2 Theorems

Here we prove (Churkin V.A., Siberian Mathematical Journal, 2010, Vol. 51,No. 3, P. 700�714)

39

Theorem 1 If the rotation group of a crystallographic group on a pseudo-Euclidean space includes no normal free abelian subgroup of �nite rankgreater than 2 which acts identically on a suitable isotropic subspace and onthe quotient by it, then the pseudo-Euclidean lattice in this crystallographicgroup is unique.

Theorem 2 Each crystallographic group on the pseudo-Euclidean spaceRp,q with min{p, q} 6 2 contains a unique pseudo-Euclidean lattice.

Theorem 3 If min{p, q} > 3 then the pseudo-Euclidean space Rp,q alwaysadmits a crystallographic group with at least two distinct pseudo-Euclideanlattices of signature (p, q) and an automorphism exchanging these lattices.

If min{p, q} = 3 or min{p, q} > 5 then we can choose these lattices sothat the corank of their intersection in each lattice is equal to min{p, q}.

If min{p, q} = 4 then we can choose these lattices so that the corank oftheir intersection in each lattice is equal to 3.

Theorem 4 In every crystallographic group with two distinct pseudo-Euclidean lattices of the same rank the corank of their intersection in eachlattice must di�er from 2 and 4. It can only take the values 3, 5, 6, 7, 8, 9, . . . .

13 Normal Abelian Subgroups of PEC-Groups

13.1 De�nitions

�1. Normal Abelian Subgroups of PEC-GroupsTake a crystallographic group Γ = Γ(G,Z) in Rp,q with a lattice Z of

rank p+ q and a rotation group G.Hence, the translation lattice

(a) is a �nite rank normal free abelian subgroup coinciding with itscentralizer (a maximal abelian subgroup);

(b) is a subgroup equipped with a nondegenerate real symmetric bilinearform invariant under the conjugation action of the group.

Certainly, abstract isomorphisms preserve these properties. We refer toeach subgroup of a crystallographic group with properties (a) and (b) as apseudo-Euclidean lattice. If the form is of signature (p, q) then we also speakof a lattice of signature (p, q).

Suppose that two crystallographic groups Γ = Γ(G,Z) and Γ′ = Γ′(G′, Z ′)are isomorphic and take an abstract isomorphism ϕ : Γ→ Γ′ of groups.

40

If ϕ(Z) 6= Z ′ then Z 6= ϕ−1(Z ′) = T . Since the lattice Z ′ enjoys properties (a)and (b) as a subgroup of Γ′, and ϕ : Γ → Γ′ is a group isomorphism, thesubgroup T = ϕ−1(Z ′) must enjoy properties (a) and (b) as a subgroup of Γ.Consequently, Γ contains the two distinct lattices Z and T .

13.2 Lemma 1 (on invariance of a form)

Lemma 1 (criterion for the invariance of a form) Given an a�ne crystallographicgroup Γ with the lattice Z and a bilinear form ζ : Z × Z → R on Z, takea �nite rank normal free abelian subgroup T of Γ with T * Z. Then

(1) TZ is a class 2 nilpotent group with center T ∩ Z and commutant[T, Z] ⊂ T ∩ Z;

(2) the invariance of ζ under the conjugation action of T on Z is equivalentto the ful�lment of two conditions:

(a) the commutant [T, Z] is isotropic with respect to ζ,

(b) the operators a− 1 : x 7→ axa−1x−1, x ∈ Z are skew-symmetric withrespect to ζ for all a ∈ T .

13.3 Proof

Proof(1) It is obvious that TZ and T ∩ Z are normal subgroups of Γ, and the

quotientTZ/(T ∩ Z) ' T/(T ∩ Z)× Z/(T ∩ Z)

is an abelian group. Consequently, the commutant [TZ, TZ] = [T, Z] isincluded into T ∩ Z. However, T ∩ Z is the center of TZ since Z is itsown centralizer. Therefore, TZ is a class 2 nilpotent group.

(2) Suppose that ζ is invariant under the conjugation action of T on Z.Denote by [a, b] = aba−1b−1 the group commutator of two elements a and b.

(a) Since ζ is bilinear and invariant under T , for all a ∈ T , b ∈ Z, andc ∈ T ∩ Z we have

ζ(aba−1b−1, c) = ζ(aba−1, c)− ζ(b, c) =

= ζ(aba−1, aca−1)− ζ(b, c) = ζ(b, c)− ζ(b, c) = 0.

41

This means that ζ([a, b], c) = 0 for all a ∈ T , b ∈ Z, and c ∈ T ∩ Z, or,equivalently,

ζ([T, Z], T ∩ Z) = ζ(T ∩ Z, [T, Z]) = 0. (1)

Since [T, Z] ⊆ T ∩ Z, it follows that ζ(x, y) = 0 for all x, y ∈ [T, Z].(b) For all a ∈ T and x, y ∈ Z the bilinearity of ζ yields

ζ(axa−1, aya−1) = ζ([a, x]x, [a, y]y) = ζ([a, x], [a, y])+

+ζ([a, x], y) + ζ(x, [a, y]) + ζ(x, y). (2)

If ζ(axa−1, aya−1) = ζ(x, y) then taking the isotropy of [T, Z] into accountwe deduce that

ζ([a, x], y) + ζ(x, [a, y]) = 0. (3)

This amounts to the skew symmetry of the operator a− 1 with respect to ζsince (a− 1)x = [a, x].

Suppose now that conditions 2(a) and 2(b) hold for a bilinear form ζ.Then (3) and (2) imply the invariance of ζ under the conjugation action ofa ∈ T on Z.

The proof of Lemma 1 is complete.

13.4 Remarks on Euclead and Minkovsky spaces

Remark 1 Condition 2(a) means that in Rn there is a subspace Σ isotropicwith respect to ζ which includes the ranges of the operators a − 1 for alla ∈ T . In particular, the ranks of a − 1 are bounded from above by thedimension of Σ. Thus, if the space is Euclidean then [T, Z] = 1 and T = Z.

Remark 2 In an arbitrary basis the matrix form of condition 2(b) is

((A− I)x)>Jy + x>J(A− I)y = 0,

with the matrix A of the operator a, the identity matrix I, and the matrix Jof the form ζ. Since J> = J , it follows that (J(A− I))> = −J(A− I), whichexpresses the skew symmetry of J(A−I). If the form is nondegenerate then Jis a nondegenerate matrix, and the ranks of J(A − I) and A − I coincide.Thus, if this is a Minkowski space then rk(A − I) 6 1 by Remark 1, andthen J(A − I) is a skew-symmetric matrix of rank 6 1. However, the rankof every skew-symmetric matrix is even; consequently, J(A− I) = 0, A = I,and T = Z.

42

13.5 Realization

Corollary Take a subspace Σ in Rp+q isotropic with respect to a symmetricbilinear form ζ of signature (p, q). Take a lattice Z in Rp,q such thatN = Z∩Σis a lattice in Σ. Take the group G consisting of the linear operators A whichkeep Z and N invariant, act identically on the quotients of the chain 0 <N < Z of subgroups, and moreover all operators A− I are skew-symmetricwith respect to ζ. Then the subgroup T = GN is a �nite rank normal freeabelian group in the crystallographic group Γ = GZ, and the group G 'T/T ∩Z under multiplication is isomorphic to the discrete group of matrices{A− I | A ∈ G} under addition.

13.6 Proof

Given A ∈ G, the hypotheses yield (A − I)Z ⊂ N and (A − I)N = 0.The operators A,B ∈ G commute when so do A − I and B − I. However,(A− I)(B− I) = 0 = (B− I)(A− I). Moreover, the operators in G preserve(centralize) the elements of N . Thus, T = GN is an abelian group. It has notorsion because Ak = (I + (A − I))k = I + k(A − I) 6= I for A 6= I. SinceG ⊂ AutZ and the abelian group of integer matrices is �nitely generated,it follows that T = GN is a �nite rank free abelian group. Its normality inΓ = GZ follows from the inclusion (A − I)Z ⊂ N . The group Γ = GZ =GNZ = TZ is a class 2 nilpotent by Lemma 1. The invariance of ζ underthe operators in G follows from Lemma 1.

The correspondenceA 7→ A−I is an isomorphism between the multiplicativegroup and the additive group since

0 = (A− I)(B − I) = AB − A−B + I,

AB − I = (A− I) + (B − I).


14 Proof of Theorem 1

14.1 De�ning relators for two lattice

� 2. Crystallographic Groups with a Unique LatticeHere we prove Theorems 1 and 2.

43

Take a crystallographic group Γ with two distinct pseudo-Euclidean lattices Zand T and �nd de�ning relations for TZ if this group Γ exists. Assume thatthe maximal dimension of an isotropic subspace is at least 1.

Choose a basis for Z as a free abelian group which agrees with the chainof subgroups

Z > T ∩ Z > [T, Z] > 0.

Take a basis b1, . . . , br for Z modulo T∩Z, a basis c1, . . . , cm for T∩Z modulothe isolator √

[T, Z] = {x ∈ T ∩ Z | ∃k : xk ∈ [T, Z]}of [T, Z] in T ∩ Z, and a basis e1, . . . , en for

√[T, Z]. It is clear that T ∩ Z

is isolated in Z because [ak, b] = [a, b]k for class 2 nilpotent groups, and Z istorsion-free. Thus,

b1, . . . , br, c1, . . . , cm, e1, . . . , en (4)

is a basis for Z, as required.Choose also a basis

a1, . . . , as, c1, . . . , cm, e1, . . . , en (5)

for T which agrees with the chain of subgroups

T > T ∩ Z > [T, Z] > 0.

Observe that the distance between the lattices is measured by the numbers rand s equal to the coranks of the intersection T ∩Z in Z and T respectively.

Write

[ai, bj] =∏k

eλk

ij

k , λkij ∈ Z. (6)

Together with the commutation relations between ai, bj, ck, el we can regardthese relations as de�ning relations for TZ.

On the other hand, λkij determine the conjugation action of T on Z.

Indeed, for the linear operator a : x 7→ axa−1 on Z we have a − 1 : x 7→axa−1x−1 = [a, x] by the multiplicative notation for the operation in Z.

If Ai is the matrix of ai in the basis (4) then by (6) the subdivision of (4)induces the block structure

Ai − I =

O O OO O OLi O O

, Li =(λkij).

Here k is the row index and j is the column index in the n× r matrix Li.

44

14.2 The inequalities for Z

The self-centralization of Z is equivalent to the linear independence of thematrices Li. Indeed, since TZ is class 2 nilpotent, it follows that[∏

i

aαii , bj

]=∏i

[ai, bj]αi =

=∏i

(∏k

eλk

ij

k

)αi

=∏k

e

∑iαiλ

kij

k = 1

for all j if and only if∑

i αiλkij = 0 for all j and k; i.e., when the matrices Li

are linearly dependent with coe�cients αi.Take now the Gram matrix J of ζ in the basis (4). By (1), the symmetry

of ζ, and the isotropy of [T, Z], it has the block form

J =

P Q XQ> R OX> O O

in accordance with the subdivision of (4), where the important block X hassize r × n. Observe that the columns of X are linearly independent since Jis nondegenerate. In particular,

rk [T, Z] = n 6 r = rk (Z/T ∩ Z). (7)

By Lemma 1 the invariance of ζ under the conjugation action of T on Zis equivalent to the skew symmetry of J(Ai− I) for all i. However, J(Ai− I)is equal P Q X

Q> R OX> O O

· O O O

O O OLi O O

=

XLi O OO O OO O O

.

Thus, the invariance of ζ under the conjugation action of T on Z is equivalentto the skew symmetry of XLi for all i:

(XLi)> = −XLi, i = 1, . . . , s. (8)

Conversely, the existence of X, of the indicated size r × n and maximalrank n, satisfying (8) guarantees the existence of a nondegenerate symmetric

45

form ζ on Z, which is invariant by Lemma 1 under the conjugation action of Ton Z since we can include this matrix X into a nondegenerate symmetricmatrix, for instance into

J =

I O XO I OX> O O

.

Indeed, the matrix identity I O OO I O−X> O I

I O XO I OX> O O

=

I O XO I OO O −X>X

yields det J = det(−X>X) 6= 0 since X>X is the Gram matrix of theindependent system of columns of X.

The matrix X has left inverse; thus, the real linear independence of Liis equivalent to the real linear independence of the skew-symmetric r × rmatrices XLi. Since the dimension of the space of skew-symmetric matricesis equal to r(r − 1)/2, we have

s 6r(r − 1)

2. (9)

14.3 The inequalities for T

Suppose now that the lattice T is also equipped with a nondegenerate symmetricbilinear form τ invariant under the conjugation action of Z on T . Then byLemma 1, the analog of (1), and the isotropy of [T, Z] with respect to τ , thematrix of τ in the basis (5) has the block form

J ′ =

U V YV > W OY > O O

in accordance with the subdivision of the basis, where Y is an s× n matrix.

If b ∈ Z then b : x 7→ bxb−1 is a linear operator on T . Then b − 1 : x 7→bxb−1x−1 = [b, x] by the multiplicative notation for the operation in T . If Bj

is the matrix of bj in the basis (5) then by (6) the subdivision of (5) induces

46

the block structure

Bj − I =

O O OO O O−Mj O O

, Mj =(λkij).

Here k is the row index, and i is the column index in the matrix Mj.By analogy with the above, the invariance of τ under the conjugation

action of Z on T is equivalent to the skew symmetry condition

(YMj)> = −YMj, j = 1, . . . , r. (10)

The columns of Y are linearly independent since J ′ is a nondegenerate matrix.Thus,

n 6 s, (11)

and Y has left inverse. Consequently, the linear independence ofMj is equivalentto the linear independence of the skew-symmetric s× s matrices YMj. Thedimension of the space of skew-symmetric matrices is equal to s(s − 1)/2;thus,

r 6s(s− 1)

2. (12)

14.4 The end of proof theorem 1

It follows from (9) and (12) that the coranks s and r of T ∩ Z in T and Zare at least 3. Indeed, if s 6 2 then r 6 1 by (12), and then s = 0 by (9),and T = Z. This means that if a crystallographic group in Rp,q includes twolattices, then its rotation group must contain an abelian normal subgroupT/(T ∩ Z) of rank > 3 acting identically on the quotients of the �ag ofsubspaces

Rp,q > Σ > O,

where Σ is the real linear span of the commutant [T, Z]. The proof ofTheorem 1 is complete.

14.5 Lemma 2 (a matrix criterion for two lattices)

Lemma 1 and its translation into the matrix language in the proof of Theorem 1imply Lemma 2 (criterion for the existence of two lattices) The existence oftwo pseudo-Euclidean lattices Z and T in a crystallographic group of the

47

form Γ = TZ with the coranks r and s of T ∩Z in Z and T respectively, andrank n of [T, Z], is equivalent to the existence of a �cube�

(λkij)of integers,

where 1 6 i 6 s, 1 6 j 6 r, 1 6 k 6 n, as well as a pair of matrices X and Yof size r × n and s× n respectively and the maximal rank n satisfying

(a) all face sections Li of the cube(λkij)determined by i = const are

linearly independent, and all lateral sectionsMj of the cube(λkij)determined

by j = const are linearly independent;

(b) all matrices XLi and YMj are skew-symmetric.


Take a crystallographic group Γ = Γ(G,Z) in Rp,q with a lattice Z of rankp+ q and a rotation group G. Take a lattice T in G with T 6= Z.

Firstly examine the case min{p, q} 6 1.For Euclidean spaces condition 2 in Lemma 1 means that T and Z

commute. Since Z is a maximal abelian subgroup, this is impossible forT 6= Z; i.e., the lattice is unique.

For Minkowski spaces Lemma 1 implies that the operators a−1 are of rankat most 1 and are skew-symmetric with respect to ζ. Since the rank of everyskew-symmetric matrix is even, all these operators are zero by Remark 3following Lemma 1. This means that T and Z commute, and T = Z.

Suppose now that the maximal dimension of an isotropic subspace withrespect to ζ is equal to 2. We may assume that in a suitable basis thequadratic form corresponding to ζ is

−x21 − x2

2 + x23 + x2

4 + · · ·+ x2n.

Fix in R2,q, with q > 2, a maximal isotropic subspace

Σ = {(α, β, α, β, 0, . . . , 0)> | α, β ∈ R}.

By Witt's theorem, two arbitrary maximal isotropic subspaces are isometric.Thus, there exists a transformation h of R2,q preserving ζ and satisfyingh([T, Z]) ⊂ Σ.

Since (h 00 1

)(G Z0 1

)(h−1 00 1

)=

(hGh−1 hZ

0 1

);

48

we can, if need be, replace Γ = Γ(G,Z) by a conjugate subgroup in theisometry group of the pseudo-Euclidean space, and assume that [T, Z] ⊂ Σ.

If A is the matrix of a and J is the matrix of ζ in the standard basisthen we �nd respectively that the columns of A− I are contained in Σ, andJ(A− I) must be an ordinary skew-symmetric matrix. However,

J(A− I) =

−α −γ −λ −ν . . .−β −δ −µ −ξ . . .α γ λ ν . . .β δ µ ξ . . .0 0 0 0 . . .

. . . . . . . . . . . . . . .

.

If this matrix is skew-symmetric then α = δ = λ = ξ = 0 and β = −γ =−µ = ν.

Consequently, the set of matrices

J(A− I) = β

0 −1 0 1 0 . . .1 0 −1 0 0 . . .0 −1 0 1 0 . . .1 0 −1 0 0 . . .0 0 0 0 0 . . .

. . . . . . . . . . . . . . . . . .

and then A− I constitutes a one-parameter family.

By the corollary to Lemma 1 it is discrete, and as an additive group it isisomorphic to the image of T acting on Z by conjugation.

Consequently, this family is additively generated by one matrix. Sincethe kernel of the homomorphism a 7→ a, for a ∈ T , coincides with T ∩ Z bythe self-centralization of Z, the group T/(T ∩ Z) is cyclic, and s = 1. From(12) it follows that Z = T , which contradicts our assumption. The proof ofTheorem 2 is complete.


16.1 A rotation symmetry cube for p = q

� 3. Crystallographic Groups with Two Lattices

49

Here we prove Theorems 3 and 4.Use Lemma 2. Construct the required matrices of the forms and a cube of

numbers(λkij)in the case p = q = r = s = n > 3, p 6= 4, m = 0. In this case

in the matrices J and J ′ the submatrices Q,R, V , and W are absent. PutP = U = O and X = Y = I. Then the bilinear forms ζ and τ are symmetricand of type (p, p). In this case in the p× p× p cube

(λkij)(for brevity simply

a size p cube) all face and lateral sections are skew-symmetric. The doubleskew symmetry

λkij = −λjik, λkij = −λikj ∀i, j, k

implies the symmetry of the cube with respect to the rotation i 7→ j 7→ k 7→ iabout the axis i = j = k through 2π/3:

λkij = −λjik = λijk,

as well as the skew symmetry of the horizontal sections k = const.Conversely, the skew symmetry of the face sections only and the symmetry

of the cube with respect to the reverse rotation k 7→ j 7→ i 7→ k (equivalently,the rotation i 7→ j 7→ k 7→ i) imply the skew symmetry of the other sections:

λkij = −λjik = −λikj, λkij = −λjik = −λkji.

Now express a cube of size p as the union of embedded cubes of increasingsize with the common diagonal i = j = k corresponding to the increasingchain

{p} ⊂ {p, p− 1} ⊂ {p, p− 1, p− 2} ⊂ · · · ⊂ {p, p− 1, p− 2, . . . , 1}

of indexing sets. Each cube results from the previous one by adding threeouter matrix sections of size increasing by 1. Construct them so that theseouter matrix sections result from one by the rotation about the axis i = j = kthrough 2π/3.

Start with the zero cube of size 2. Then, beginning with size 3, take eachof the subsequent outer matrix sections i = const so that it is an integer skew-symmetric matrix with the zero �rst row and column. Require in additionthat this matrix be of maximal rank and independent of the previous parallelsections. For instance, for p = 3 the outer sections of the cube may have the

50

form

0i↖ −1 1 ↗ j

0 0 00 0

1 0 −10 0

0 0 0−1 1

0

k ↓Its face sections

L1 =

0 0 00 0 −10 1 0

, L2 =

0 0 10 0 0−1 0 0

, L3 =

0 −1 01 0 00 0 0

are linearly independent. Since M1 = −L1, M2 = −L2, and M3 = −L3, thelateral sections are also independent.

16.2 A cube for p = q

In the general case the required cube exists for p > 2, p 6= 4. Let us provethis claim.

If p > 3 is odd then the submatrix of the outer face matrix sectionresulting by removing the �rst row and column has an even size p− 1 > 2.

Choose it so that the determinant of the submatrix be nonzero, andcolumns (and rows) be independent. Under the rotation through 2π/3 therows of the outer face matrix become the �rst column s of the lateral matrixsections.

Thus, the lateral sections, with the exception of the outer one, are linearlyindependent.

If all lateral sections are dependent then the outer section can be linearlyexpressed in terms of the others. However, its �rst row is zero; thus, allcoe�cients of the linear combination are zero, and the outer lateral section

51

is zero. We arrive at a contradiction since we choose all outer sections to benonzero.

If p > 4 is even then, removing the face, lateral, and upper outer sections,we obtain a cube of odd size p− 1 > 3.

Its sections appear as submatrices in the sections of the original size pcube, with the exception of the outer sections. By the previous argument wecan choose them linearly independent.

The dimension (p− 1)(p− 2)/2 of the space of skew-symmetric matricesis greater than p− 1 for p > 4. Choose the outer section of the size p cube sothat its submatrix obtained by removing the �rst row and column is skew-symmetric and independent of the corresponding submatrices of the otherparallel sections. Then all parallel sections are independent.

Observe that for p = 4 a cube with independent sections cannot exist. InTheorem 4 we prove an even stronger claim.

Let us �nd an automorphism exchanging the lattices. It is easy to seethat the correspondence

ai ←→ bi, ck ←→ ck, i, k = 1, . . . , p,

preserves the de�ning relations of Γ = TZ, and thus, it extends to an automorphismof Γ exchanging the subgroups T and Z. Indeed, (6) is carried into the validequality

[bi, aj] =∏k

eλk

ij

k

since λkij = −λkji by the skew symmetry of the k-sections in the constructed

cube(λkij).

16.3 A cube for p < q

By construction, the coranks of the intersection Z∩T in the lattices Z and Tare equal to p.

Suppose now that p < q and m = q−p. Decompose the pseudo-Euclideanspace Rp,q as the orthogonal sum of pseudo-Euclidean spaces

Rp,q = Rp,p ⊕ Rm,

where Rm is a Euclidean space. For p > 3, p 6= 4, choose in Rp,p one ofthe constructed crystallographic groups Γ with two lattices Z and T of type

52

(p, p). Take an abelian lattice R on the Euclidean space Rm and considerthe direct product Γ′ = Γ×R of groups. Then Γ′ is a crystallographic groupwith two distinct lattices Z×R and T ×R, which, obviously, we may assumeequipped with Γ′-invariant symmetric bilinear forms of type (p, q). Extendan automorphism of Γ exchanging Z and T by the identity on R. This yieldsan automorphism of the crystallographic group Γ′ exchanging Z × R andT × R. The condition on the coranks of the intersection of the lattices isobviously ful�lled.

It remains to examine the case p = 4 and q > p. Apply the previousdecomposition. Represent the pseudo-Euclidean space R4,q as the orthogonalsum of pseudo-Euclidean spaces

R4,q = R3,3 ⊕ R1,q−3.

Choose in R3,3 one of the constructed crystallographic groups Γ with twolattices Z and T of signature (3, 3). Take a lattice R in R1,q−3 and considerthe direct product Γ′ = Γ × R of groups. Extend an automorphism of Γexchanging Z and T by the identity on R. This yields an automorphism ofthe crystallographic group Γ′ exchanging Z ×R and T ×R. In this case thecoranks of the intersection of the lattices are equal to 3.

The proof of Theorem 3 is complete.


Proof of Theorem 4By Theorem 3 together with (9) and (12) it su�ces to show that the

coranks of the intersection of distinct lattices cannot be simultaneously equalto 4. Use Lemma 2, keep the previous notation, and assume on the contrarythat r = s = 4.

Split the subsequent argument into steps.

Step 1. The matrices XLi are linearly independent since Li are linearlyindependent and X has left inverse. Similarly, the matrices YMj are linearlyindependent as well.

Step 2. All matrices XLi, YMj, Li, and Mj are of rank 2. Indeed,multiplying every column of the cube

(λkij)by Y we obtain a new cube

with the face sections Y Li and skew-symmetric lateral sections YMj. Thenevery face section Y Li contains a zero row and rkY Li < 4. However, the

53

matrices X and Y have left inverses; thus, rkXLi = rkLi = rkY Li < 4.On the other hand, XLi is skew-symmetric, and its rank is even. Takingthe linear independence of XLi into account, we deduce that rkXLi = 2.Similarly we can prove that rkYMj = 2.

Step 3. Elimination of Y .The matrix Y has the left inverse Y −1 of size n × 4, while Y −1Y = I is

an n× n matrix. Then

XLi = X · I · Li = XY −1Y Li = X ′L′i,

where X ′ = XY −1, and L′i = Y Li is a face section of the new cube(νkij)

resulting from the multiplication of all columns of the old cube by Y . Itslateral sections M ′

j = YMj are skew-symmetric matrices.

Replacing the old cube with the new cube, we can eliminate Y andassume that there exist a cube (now not necessarily of integers) and a 4× 4matrix X of rank n such that all lateral sections Mj of the cube are skew-symmetric, as well as all matrices XLi for the face sections Li. Moreover, thematrices Mj are linearly independent and of rank 2. A similar claim holdsfor XLi, as well as Li.

Step 4. If the set of all rows of the concatenation of the matrices Li,i = 1, . . . , 4, is included into a proper subspace of R4 then all rows satisfya nontrivial linear equation

∑4j=1 λjxj = 0. Then

∑4j=1 λjMj = 0, which

contradicts the linear independence of the lateral sectionsMj. Consequently,the system of rows of the union of Li includes four independent rows. Denotethem by a, b, c, and d. We may assume that they lie below the main diagonalsof the skew-symmetric matrices Mj. Denote two other remaining rows by xand y.

Step 5. List the possible locations of the intersections of a, b, c, and dwith the skew-symmetric matrices Mj, considering that rkLi = 2:

0 −xj −aj −bjxj 0 −cj −djaj cj 0 −yjbj dj yj 0

,

0 −aj −xj −bjaj 0 −cj −yjxj cj 0 −djbj yj dj 0

,

0 −aj −bj −xjaj 0 −yj −cjbj yj 0 −djxj cj dj 0

.

54

It is easy to verify that there are no other intersections.Step 6. Verify that the rows x and y of Li are zero. Indeed, considering

that rkLi = 2, express them in terms of the independent a, b, c, or d in Liusing the occurrences of x, −x, y, and −y. We have

x = αa+ βb = γc+ δd, y = α′a+ β′c = γ′b+ δ′d.

The linear independence of a, b, c, and d yields x = y = 0. Consequently, allsections Mj are either of type I, or of type II, or of type III, where

I :

0 0 −aj −bj0 0 −cj −djaj cj 0 0bj dj 0 0

, II :

0 −aj 0 −bjaj 0 −cj 00 cj 0 −djbj 0 dj 0

,

III :

0 −aj −bj 0aj 0 0 −cjbj 0 0 −dj0 cj dj 0

Step 7. In a matrix Mj of either of these types the determinant of the

submatrices of the crossings with the rows a, b, ±c, ±d of size 2 is equalto zero, for otherwise detMj 6= 0 by Laplace's formula for expanding thedeterminant by two rows. Consequently, the columns of these submatrices arelinearly dependent, and then the corresponding columns ofMj are proportional.

Step 8. Suppose that Xu and Xv are linearly independent columnswith indices j and l of the skew-symmetric matrix XLi of rank 2. Thenthe submatrix of XLi with rows and columns j and l cannot be zero, forotherwise XLi would be a skew-symmetric matrix of rank 2, and argumentssimilar to step 7 imply that Xu and Xv would be linearly dependent. Hence,the entries (j, l) and (l, j) of XLi contain opposite nonzero numbers; denotethem by α and −α.

It is clear that u and v constitute a basis for the system of columnswith indices j and l of the matrices Li. Then Mj and Ml by step 7 containproportional columns λu and µv, which by step 6 lie in the common sectionLm, where m 6= i. Thus, the skew-symmetric matrix XLm includes thecolumns X(λu) = λ(Xu) and X(µv) = µ(Xv) with indices j and l, whichin the entries (j, l) and (l, j) contain λα and −µα. By the skew symmetryλα = µα, and so λ = µ. Consequently, the basis columns Xu and Xv for

55

XLi, as well as λXu and λXv for XLm with the same indices j and l, areproportional with the common coe�cient λ.

Step 9. Verify that XLm = λXLi, invoking the skew symmetry ofthe matrices and the condition on the rank. This will yield the requiredcontradiction with the claim of their independence in step 1.

Using arguments similar to steps 4�6, we may assume that all skew-symmetric matrices XLi are either of type I, or of type II, or of type III.Suppose that it is type I, and

XLi =

0 0 −α −β0 0 −γ −δα γ 0 0β δ 0 0

.

Pick columns 1 and 3 as a basis. If α = 0 then β 6= 0 and γ 6= 0 by theindependence of columns 1 and 3. However, then the rank of XLi is equalto 4 rather than 2. Hence, α 6= 0, and δ is uniquely determined in terms of α,β, and γ from the equality αδ−βγ = 0, which follows from the condition onthe rank of XLi. Thus, XLi is uniquely reconstructed from columns 1 and 3using the following conditions: the skew symmetry, type I, and the rank beingequal to 2. Multiplying columns 1 and 3 by λ, we obtain the correspondingcolumns of XLm. Since XLm is skew-symmetric, of type I, and of rank 2, itis completely reconstructed from columns 1 and 3, and then XLm = λXLi.

Similar arguments are valid if we replace the indices of the basis columnsor the type.

The proof of Theorem 4 is complete.

18 Example with exactly two lattices

Example with exactly two latticesLet Γ be the group in R3,3 with cube of numbers (λkij) constructed in proof

of theorem 3. Then Γ = TZ is nilpotent group of class 2, Z is normal freeabelian subgroup with basis b1, b2, b3, c1, c2, c3, T is normal free abeliansubgroup with basis a1, a2, a3, c1, c2, c3 and

[a1, b1] = 1, [a2, b1] = c−13 , [a3, b1] = c2,

[a1, b2] = c3, [a2, b2] = 1, [a3, b2] = c−11 ,

[a1, b3] = c−12 , [a2, b3] = c1, [a3, b3] = 1.

56

Brie�y,[ai, bi] = 1, [ai, bj] = c

ε(i j k)k if i 6= j.

Here [a, b] = aba−1b−1 is the commutator of elements a and b, ε(i j k) is thesignum of permutation (i j k).

The group Γ is crystallographic in R3,3 if

b1, b2, b3, c1, c2, c3

is a basis of lattice Z = Z6 of translations of R6,

the generators a1, a2, a3 induce the linear transformations

ai : x 7→ aixa−1i , x ∈ Z ' Z6

with block-matrices

Ai =

(I OLi I

), i = 1, 2, 3,

in the basis b1, b2, b3, c1, c2, c3. Here I is identity (3× 3)-matrix and

L1 =

0 0 00 0 −10 1 0

, L2 =

0 0 10 0 0−1 0 0

,

L3 =

0 −1 01 0 00 0 0

.

The nondegenerate symmetric bilinear form

ζ(x, y) = x1y4 + x4y1 + x2y5 + x5y2 + x3y6 + x6y3

on R6 has type (3, 3) and is invariant under the action of operators ai.For example, the associated quadratic form κ(x) = ζ(x, x) is A1-invariant:

κ(A1x) = κ(x1, x2, x3, x4,−x3 + x5, x2 + x6) =

= 2x1x4 + 2x2(−x3 + x5) + 2x3(x2 + x6) = 2x1x4 + 2x2x5 + 2x3x6 = κ(x).

The automorphism ai ↔ bi, ci ↔ ci, i = 1, 2, 3, permutes the subgroupsT and Z.

Therefore T has a Z-invariant nondegenerate symmetric bilinear form.

57

Theorem (Ch., 2009)1) The crystallographic group Γ, constructed above, is unique up to

isomorphizm under the conditionsa) Γ has two distinct pseudo-Euclidean lattices of rank 6 6,b) Γ is minimal: the derived subgroup and the center of Γ coincide.

2) Γ has exactly two pseudo-Euclidean lattices.

3) The group Γn is crystallographic and has exactly 2n pseudo-Euclideanlattices.

19 Problems

Problem 1 Conjecture: Every crystallographic group on pseudo-Euclideanspace has a �nite number of pseudo-Euclidean lattices.

Problem 2 Obtain a classi�cation of all crystallographic groups on pseudo-Euclidean space R2,2 for a �xed classic discrete group G of rotations.

Problem 3: Auslender's conjecture, 1964 Properly discontinuous cocompactgroup of a�ne transformations of Rn is virtually solvable.

A survey and new results: Abels H., Margulis G.A., Soifer G.A. Properlydiscontinuous groups of a�ne transformations with orthogonal linear part,C. R. Acad. Sci. Paris, 1997, T. 324 (Ser.I), P. 253�258.

20 Survey by Moodi R.V.

20.1 Properties of a quasicrystall

Robert V. Moody, Model set: A SurveyThis article surveys the mathematics of the cut and project method as

applied to point sets Λ, called here model sets. It covers the geometric,arithmetic, and analytical sides of this theory as well as di�raction and theconnection with dynamical systems.

Following properties may be considered as representative:

• discreteness

• extensiveness

• �niteness of local complexity

58

• repetitivity

• di�ractivity

• aperiodicity

• existence of exotic symmetry.

1) Discreteness and extensiveness of Λ means that Λ is Delone set.

2) Finiteness of local complexity of Λ means that for each r > 0 there are,up to translation, only �nitely many point sets (called patches of radius r) ofthe form Λ∩Br(v). Here Br(v) is the ball of radius r about the point v ∈ Rd.So, on each scale, there are only �nitely many di�erent patterns of points.This condition can be expressed topologically: Λ has �nite local complexityi� the closure of Λ− Λ is discrete.

3) Repetitivity of Λ means loosely that any �nite patch that appears,appears in�nitely often. More precisely, given any patch of radius r there isan R so that within each ball of radius R, no matter its position in Rd, thereis at least one translate of this patch.

4)Di�ractivity of Λ means that the Fourier transform of the autocorrelationdensity that arises by placing a delta peak on each point of Λ, should containa part that looks discrete and point-like.

5)Aperiodicity of Λ means that Λ has not nontrivial translation symmetry.

6) Existence of exotic symmetry: for example, Penrose tiling has a self-similarity with factor τ 4, τ = (1 +

√5)/2, Golden ratio.

20.2 De�nition model set

By de�nition, a cut and project scheme or model set consists of a collectionof spaces and mappings

Rn π1←− Rn ×G π2−→ G

∪

L

where

• Rn is 'physical' space � a real euclidean space

59

• G is 'internal' space � a locally compact abelian group, for example,Rd

• L is a lattice (discrete and cocompact subgroup) in 'hyperspace' Rn×G

• π1 is a projection map and π1 |L is injective

• π2 is a projection map and π2(L) is dense in G

Let L = π1(L) ⊂ Rn and the map ∗ : L→ G be de�ned by the rule

x∗ = π2(π−11 (x)), x ∈ L.

Let W ⊂ G be a 'window' with properties

• W is nonempty and W = int(W ) is compact

• ∂W ∩ π2(L) = ∅

• ∂W is of Haar measure 0.

Then setΛ(W ) = {u ∈ L |u∗ ∈ W}

orQ(W ) = {v ∈ L |π2(v) ∈ W}

is a 'quasicrystal'.De�nition (Artamonov, Sanches, 2008-10) Let G = Rd. The group of

all a�ne transformations g of hyperspace Rn × Rd with g(Q(W )) = Q(W )is called a proper group of symmetries of quasicrystal Q(W ) or Λ(W ) and isdenoted by Sym (Λ(W )).

The group of all a�ne transformations g of hyperspace Rn × Rd with

g(L) = L, g(Rn) = Rn, g(Rd) = Rd

is called a group of symmetries of a quasicrystal and is denoted by Sym (L).Our euclidean quasicrystallographic groups are subgroups of the group

symmetries of a quasicrystal.

60

20.3 Icosian model set

Icosian model set (Kramer, Baake et al , 1990)The elements of norm 1 of the usual quaternion ring

H = R + Ri+ Rj + Rk

form a group isomorphic to SU2. Since this group is a 2-fold cover of theorthogonal group SO3, in particular it contains 2-fold covers of the icosahedralgroup. One such example is the following list I of 120 vectors:

1

2(±1, ±1, ±1, ±1), (±1, 0, 0, 0) and all permutations,

(0, ±1, ±τ ′, ±τ) and all even permutations,

where τ = (1 +√

5)/2 is the Golden ratio and ' indicates the conjugationmap

√5 7→ −

√5.

The subring I generated by this group is called the icosian ring. Theform of the points of I makes it clear that I is a Z[τ ]-module. We let ∗

denote the mapping on I that conjugates each of the coordinates with respectto the unique Galois non-trivial automorphism on Z[τ ] (de�ned by sending√

5 7→ −√

5). Note that I∗ 6= I.The ring I is of rank 4 over Z[τ ] and rank 8 over Z. We make an explicit

embedding of I as a lattice I in R8 by the mapping x 7→ (x, x∗).This already provides the framework of a cut and project scheme:

R4 π1←− R4 × R4 π2−→ R4

∪I

with the projections being given by the �rst and second components of (x, x∗).Remarkably the lattice I has an entirely natural interpretation as the root

lattice of type E8. Now we wish to show that this cut and project schemerespects the symmetry that is inherent in its construction. Geometrically thepoints of I form the vertices of a regular polytope P in 4-space and also formthe vectors of a root system ∆4 of type H4. The Coxeter group H4 is noneother than the group of automorphisms of P (and also of ∆4), and is in factvery easily described: it is the set of all (14400) maps

x 7→ uxv; x 7→ uxv

61

where u, v ∈ I.The subgroup of these transformations in which v = u−1 is obviously a

copy of the icosahedral group and this subgroup stabilizes the 3-dimensionalspace Ri+Rj+Rk of pure quaternions. These maps provide automorphismsof the rings I and, via conjugation, on I∗ too, and thus give rise to an actionof I as automorphisms on the entire cut and project scheme. If the windowW is chosen to be invariant under I then the resulting model set is alsoI-invariant. Restricting everything to the pure quaternions we get a newcut and project scheme based on the 6-dimensional root lattice D6 and anicosahedral symmetry.

21 References

References

1. Bieberbach L. � �Uber die Bewegungsgruppen der Euklidischen R�aume.I,� Math. Ann., Bd 70, 297�336 (1911).

2. Bieberbach L. � �Uber die Bewegungsgruppen der Euklidischen R�aume. II,�Math. Ann., Bd 72, 400�412 (1912).

3. Novikov S. P. and Taimanov I. A. Modern Geometric Structures andFields, Providence, Amer. Math. Soc., 2006.

4. Le Thang Tu Quoc, Piunikhin S. A., and Sadov V. A. The geometryof quasicrystals, Russian Math. Surveys, 1993, Vol. 48, N 1, P. 37�100.

5. Piunikhin S. A.,Quasicrystallographic groups in the sense of Novikov,Math. Notes, 1990, Vol. 47, N. 5, P. 478�482.

6. Piunikhin S. A. Several new results on quasicrystallographic groupsin Novikov sense, Math. Notes, 1990, Vol. 48, N. 3, P. 944�949.

7. Abels H., Margulis G.A., Soifer G.A. Properly discontinuous groupsof a�ne transformations with orthogonal linear part, C. R. Acad. Sci. Paris,1997, T. 324 (Ser.I), P. 253�258.

8. Garipov R. M. and Churkin V. A. Quasicrystallographic groups onMinkowski spaces, Siberian Math. J., 2009, Vol. 50, N. 4, P. 616�631.

9. Garipov R. M. Ornament groups on a Minkowski plane, Algebra andLogic, 2003, Vol. 42, N. 6, P. 365�381.

10. Garipov R. M. Quasicrystallographic groups on the Minkowski spaceR1,2, Dokl. Akad. Nauk, 2006, Vol. 409, N. 3, P. 300�304.

62

11. Garipov R. M. Crystallographic classes in 4-dimensional Minkowskispace, Algebra and Logic, 2008, Vol. 47, N.1, P. 18�31.

12. Churkin V.A. The Weak Bieberbach Theorem for CrystallographicGroups on Pseudo-Euclidean Spaces, Siberian Math. J., 2009, Vol. 50, N. 4,P. 616�631.

13. Moody R.V. Model sets: a survey, From Quasicrystals to MoreComplex Systems, Springer-Verlag, Berlin, 1998, p.145�166. (Variant 2008year: arXiv: math.MG/0002020.)

14. Àðòàìîíîâ Â.À., Ñàí÷åñ Ñ. Î ãðóïïàõ ñèììåòðèé êâàçèêðè-ñòàëëîâ, Ìàòåì. çàìåòêè, 2010, Ò. 87, N. 3, Ñ. 323�329.

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crystallographic and quasicrystallographic groups · eary 1981: de bruijn found the "cut and...

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