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Icosahedral quasicrystals with continuous phasons
L.S. Levitov
To cite this version:
L.S. Levitov. Icosahedral quasicrystals with continuous phasons. Journal de Physique, 1989, 50 (21),
pp.3181-3190. �10.1051/jphys:0198900500210318100�. �jpa-00211136�
3181
Icosahedral quasicrystals with continuous phasons
L. S. Levitov
L. D. Landau Institute for Theoretical Physics, Moscow (Reçu le 5 juin 1989, accepté le 6 juillet 1989)
Résumé.
2014Des familles de modèles de quasi cristaux icosaédriques sont construites. Dans ces
modèles, les atomes peuvent se déplacer continûment sous l’action des déplacements dans l’espace des phasons R3* leurs distances mutuelles restant toujours plus grandes qu’un diamètre
de « coeur dur » h > 0. Ces modèles possèdent la symétrie de l’icosaèdre et leurs propriétés de
diffraction sont celles de structures parfaitement quasi périodiques.
Abstract.
2014A variety of models for icosahedral quasicrystals is constructed such that atoms move
continuously under displacements along the phason space R3*, while their positions are separated by distances larger than some h > 0, a « hard core » diameter. The models exhibit icosahedral symmetry and the diffraction properties of perfect quasiperiodic structures.
J. Phys. France 50 (1989) 3181-3190 1er NOVEMBRE 1989,
Classification
Physics Abstracts
61.50E
1. Introduction.
For one-dimensional quasiperiodic structures it is known that phasons can be either
continuous or discontinuous, depending on the strength of the coupling [1]. These two regimes are separated by a line of analyticity breaking transition. In almost all incommensu- rate structures studied so far (Hg3 _ sAsF6, TTF-TCNQ, etc.) phasons are continuous. Are continuous phasons possible for quasicrystals ? All models constructed for an « ideal »
quasicrystal structure [2-4] lead to discontinuous phasons, i.e. each atom jumps at certain
value of phason shift. It was also suggested [5] that large symmetry groups of quasicrystals always forbid continuous phasons (theorems 1, 2 of Ref. [5]). However, a counterexample to
these theorems was reported recently : an icosahedrally symmetric structure with continuous phason transformations [6].
Here 1 return to this problem and construct many other structures having continuous phasons. Results of reference [6] seem to be not completely satisfactory : a) The structure suggested in [6] looks as a very artificially designed, so one might suspect it is unique. b) The
model [6] is constructed for only one of 11 possible icosahedral space groups [7-9], but the
situation for other 10 groups is also of interest. c) The method of reference [6] refers strongly
Article published online by EDP Sciences and available at http://dx.doi.org/10.1051/jphys:0198900500210318100
to the reader’s geometrical intuition, so for some people a more algebraic approach is
desirable.
Let us describe shortly the structure of the paper. Main results of this work are concentrated in sections 4, 5, 6. Models with continuous phasons are defined in section 2 as
periodic families of non-intersecting continuous atomic surfaces. The contents of section 2 is not new, it almost entirely overlaps with references [5, 6]. Icosahedral symmetry groups are discussed in section 3. It is shown that for constructing continuous atomic surfaces for six of these groups one needs to do this for only one of them. In sections 4, 5 we derive conditions which are necessary for the existence of continuous atomic surfaces. A wide class of solutions of these conditions is found. In section 6 the solutions are used to construct continuous atomic surfaces.
An important distinction of the structures constructed here and those described in reference [6] is that here we get straight linear atomic surfaces (hyperplanes in R 6) and
periodic interpenetrating lattices of atoms in 1R3, while in reference [6] the atomic surfaces are
curved and, thus, atomic positions in (iB3 form quasiperiodically modulated lattices.
Notation : trying to make the matter more transparent 1 almost everywhere use Gothic
letters for algebraic objects (groups, (sub-)lattices, subspaces) in order to distinguish them
from geometic ones (surfaces, hyperplanes, vectors). By the word « hyperplane » 1 denote
linear manifolds not necessarily containing the origin, in contrast with « subspaces » that always contain 0-vector.
2. Atomic surfaces, lattices and subspaces.
Atomic surfaces describe the dependence of atoms’ positions on phason variables. For a d- dimensional quasiperiodic structure with D = d + d’ incommensurate frequencies atomic
surfaces live in the space IRD = V 0153 V * (V = IR d represents the physical space, V * = R d’ is the phason space). The surfaces have dimension d’. We denote them by si (i labels atoms of
the structure). Three characteristic properties of atomic surfaces should be mentioned :
(i) (periodicity) The family of the surfaces si is periodic and has D independent periods generating a D-dimensional lattice 2 in RD. This periodicity in RD reflects quasiperiodicity in
the physical space V.
(ii) (conservation of atoms) Each atomic surface si has one-to-one projection on the phason
space V *. Thus si can be described by a function fi: V * -+ V as a set of points ( f (y ), y) e RD @ where standard coordinates in RD are used : (x, y), x e V, y e V *.
(iii) (hard core condition) Atoms cannot be closer to each other than by some constant
h >- o. In terms of functions fi this restriction reads : 1 fi (y ) - f j (y ) 1 >- h for all i,
je Z, y eV*.
Atomic surfaces define positions of atoms in the physical space as follows. Choose some a e V * (phason parameter) and consider a d-dimensional hyperplane Va consisting of all points (x, a) e IRD, @ i. e. the space V shifted by the vector (0, a). The hyperplane Va represents the physical space. Atoms’ positions are given by intersection points of the surfaces si and Va.
A family of atomic surfaces satisfying conditions (i), (ii), (iii) together with the rule of
finding atoms’ positions constitute the most general way of describing models for quasiperio-
dic arrangements of atoms. By varying the phason parameter a in the above given definition
on obtains different configurations of atoms. If the space V and the lattice 2 are
incommensurate ( [incommensurability ] - [2 n V = 0 ]), then all configurations of atoms
produced by varying a are physically equivalent. In this case phason shifts are included in the
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group of Goldstone equivalence transformations of the structure (like translations for
ordinary crystals).
It is important that atom’s positions as functions of phason shifts can be either continuous
or discontinuous. Since here we are interested in continuous phasons, let us introduce the
condition
(iv) (continuity) All atomic surfaces si (and corresponding functions fi) are continuous.
The continuity condition enables one to introduce lattices of periods of single atomic
surfaces. For an atomic surface si its lattice li is defined as a sublattice ouf 2 consisting of all
translations transforming si into itself. Simple analysis (presented in Ref. [5]) shows that
dim (l i) = d’. Besides i we need to introduce d’-dimensional subspaces ri : ri is a real linear
envelope of vectors of li. The subspace ri is « parallel » to si: si remains within a fixed distance of ri. Both lattices 1 i and subspaces ri will be used below.
3. Icosahedral symmetry.
Icosahedral group acts in (R6 = V (f) V* (V == (R3, V* = (R3*) according to its 3 + 3 representation (see [2-4]). Beginning from here we are interested only in icosahedrally symmetric structures, i. e. we demand that the symmetry group of the family of the atomic surfaces si coincides with one o f 11 icosahedral space groups. All possible icosahedral groups
were found in [7-9] (our notation of symmetry groups and related objects coincides with that of Ref. [8]). There are 2 point groups ffi : Y (60 elements, inversion is not included) and YI (120 elements, inversion is present). Their 6-dimensional representations 3 + 3 are point symmetry groups of the structures considered (thus, d = d’ = 3). Three Bravais lattices are
possible : DSC, £Fcc, 2BCC. For each pair [0, 2], ((5 = Y, Y,, 2 = 2sc, Spcc. £Bcc) two
space groups exist
-symmorphic and non-symmorphic
-except for the case (5 = YI, .2 = Spcc which yields only a symmorphic group. Thus get 11 groups.
For two groups H1, h2 we say that h1 includes H2 if H2 is isomorphic to a subgroup of
h1.
Proposition 1 : If two space symmetry groups H1, H2 are such that
a) -51 includes H2 ;
b) the group -51 allows for continuous atomic surfaces, then the group H2 also allows for continuous atomic surfaces.
Proo f : Given a family of atomic surfaces with the space symmetry group H1, one easily gets
a family of surfaces with the symmetry group H2 by lowering the symmetry from
H1 to H2 (this can be done by a proper continuous small deformation of the surfaces).
In what follows we work only with the symmorphic group .!5: (fj = Y, L = £sc, and
construct continuous atomic surfaces for it. One can check that this group includes all groups
(both symmorphic and non-symmorphic) with 6) = Y, L = .2sc, .2Fcc, 2BCC (totally,
6 groups). According to Proposition 1, the existence of continuous atomic surfaces for these groups is guaranteed by the existence of such surfaces for the group H.
Other 5 icosahedral groups (based on (fi = YI) will not be discussed here. Due to the presence of inversion not all of them allow for continuous phasons (proof will be reported elsewhere).
4. Non-transversality condition and its solutions.
Consider two atomic surfaces sl, s2 and find their lattices ll, l2 and corresponding subspaces
rl, r2. Calculate the dimension of r1 p r2, the linear envelope of rl and r2. Note that
dim ( rl ) = dim ( r2 ) = d’ = 3, so if dim ( ri q) r2 ) = d + d’ = 6, then the subspaces r, and r2 are transversal, i.e. they have only 0-vector in common. In this case the surfaces si, S2 are also transversal : they intersect and, moreover, their intersection is topologically
unremovable (see Refs. [6, 10]). But intersections of atomic surfaces are unphysical
-they
cannot exist due to the « hard core » condition. Thus, we obtain
Proposition 2 : For every two atomic surfaces si, sj the corresponding subspaces ri’ rj are not transversal :
For moving further we need to write the non-transversality condition (4.1) in the
coordinates (x, y), x e V, y E V *.
But first let us see how the subspaces ri can be described using these coordinates. Since for
any i e Z dim ( ri ) = 3 and ri has one-to-one projection on V*, one can find a linear
transformation Âi : V * - V such that all points of ri are given by (Âi (y), y), y e V *. After bases are chosen in V and V *, each subspace ri is given by 3 equations : xa = ¿ At/3 y/3
03B2
(03B1, f3 = 1, 2, 3 ), where At/3 is a matrix corresponding to Âi. Note that x and y live in
different spaces (V and V * respectively). Consequently, transformation rules for the matrix
At{3 are non-standard : if coordinates are changed by (x’, y’ ) _ (RI (x), R2 (y»), then
(here R1, R2 are arbitrary 3 x 3 matrices).
Non-transversality condition (4.1) means that ri and rj have a common nonzero vector. Let
it be (Âi (y), y ) _ (Âj(Y’), y’). We obtain y = y’, Âi (y) = Âj (y), that can be finally rewritten
as
This is an algebraic form of the geometric restriction (4.1).
Now we turn to the analysis of the action of the symmetry group -5 on the subspaces
ri. Beginning from here we use only orthonormal bases in V and V *. Operations of the point
group (5 are given by matrices 6 x 6:
where G and Gare 3 x 3 orthogonal matrices of the representations 3 and 3 respectively, Ô is 0-matrix.
The set of subspaces ri is symmetric under the point group (4.4), since the surfaces si are symmetric under the space group .5 . Let Âi be a 3 x 3 matrix connected with ri as explained above. Symmetry implies that not only ri but also other subspaces of the orbit
{Rg[ri]} are present. After applying transformation (4.4) to Âi and using (4.2) get
R9 [Âi] = GÂi G-1. Non-transversality condition (4.3) thus reads
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Sixty equations (4.5) impose strong constraints on possible matrices Âi. One solution of
equations (4.5) was found in reference [6]. Here we get many other solutions.
Proposition 3 : The orthogonal matrices  e SO (3 ) solve non-transversality equations (4.5).
Proof : If  E SO (3 ) then det - det 1 since an orthog-
onal 3 x 3 matrix with det = 1 always has at least one eigenvalue 1 (here Ê is unit matrix).
One could be interested whether there are other solutions of equations (4.5). Although the
whole set of solutions is not known, one can mention a) matrices proportional to orthogonal : À À, where À E R, Â E SO (3 ) ; b) matrices corresponding to the lattices laf3’Y introduced in reference [6]. Probably, other solutions also exist. We are not looking for them here, since orthogonal matrices suffice for our needs.
5. Rationality of the subspaces ri.
We have constructed a wide class of solutions of non-transversality equations (4.5). However,
besides non-transversality there exists another important restriction on the subspaces
ri
-their rationality. Indeed, r 1 is a linear envelope of li, a sublattice of 2. Thus,
ri is rational, i.e. it is commensurate with 2 ( [rationality ] - [commensurability ] - [dim (ri n Ë) = dim ( ri ) ] ). Let us study which of orthogonal matrices generate rational subspaces.
For this study a special choice of bases in V and V * will be crucial. It turns convenient to work with a cubic coordinate system in which the axes are aligned along 3 perpendicular 2-
fold symmetry axes of the icosahedral group (for details see Ref. [11] where this system was elaborated upon). The bases in V and V* are chosen so that vectors of the lattice
2sc projected on V and V * have the form
Here h, h’, k, k’,l, f’ are integers such that h + k’, k + l’ and + h’ are even numbers ;
T = (J5 + 1 )/2. Vectors commensurate with £sc are given by (5.1) with arbitrary rational h, h’, k, k’, f, f’. It turns convenient to modify representation (5.1) and write vectors of R6 commensurate with Ssc as
where a = (ai , a2, a3 ), b = (bl, b2, b3 ) E R3 have rational components : aa, ba e Q (a = 1, 2, 3 ). The connection of (5.2) and (5.1) is given by bi = h, b, = k, bi = l,
Let us explore rationality of ri using representation (5.2). Rationality of a vector
a = (E1 (y ), y) means that equations
can be solved simultaneously. From (5.3) find a = R (b ), where
Rationality of the subspace implies that R is a rational matrix (all its matrix
elements are rational numbers). Another important observation is that R is an orthogonal
matrix : R E SO (3). This is a direct consequence of two facts :
a) Rand  commute, hence they have common eigenvectors ;
b) Function z - (T 2 Z + 1 )/ (z + T 2) transforms the circle [ z 1 = 1 in the complex z-plane
into itself.
Thus, we come to
Proposition 4 : Orthogonal matrices  generating rational subspaces are parametrized by
rational orthogonal matrices R :
In order to see how large is the set (5.5) of allowed matrices we prove
Proposition 5 : Matrices  of the form (5.5) are dense in the group SO (3 ).
For the proof of Proposition 5 we need to prove
Lemma : Rational orthogonal matrices so (3) f1 GL3(Q) are dense in the group
SO (3 ).
Proof of Lemma : Each matrix R E SO(3) can be written using Euler parametrization
Here TX (a ) (T y ({3 ), T z ( l’ » is a matrix of rotation around the axis x (y, z ) by an angle a (f3, y ). The matrix elements of Tx( a), Ty(f3), Tz( 1’) are 1, 0, cos (a ), ± sin (a ),
cos (13 ), ± sin (f3), cos (y ), ± sin (y ). It is well known that the angles a for which
cos (a ) and sin (a ) are rational numbers are dense in the interval [0, 2 7T] (this is an easy consequence of the existence of infinitely many Pithagoras’ triplets a, b, c E Z such that az+ bz = C2). Since the matrix product (5.6) is continuous as a function of a, /3, y, rational matrices can be found in any vicinity of any matrix of SO (3 ).
Proof o f Proposition 5 : Consider a transformation p : SO (3) --+ SO (3 ), p (Â) =
Applying the method used above to prove orthogonality of R we find that p is a continuous one-to-one transformation. From Lemma we know that rational matrices are dense in SO (3 ). But continuous one-to-one image of a dense set is a
dense set.
Thus we found a wide class (5.5) of orthogonal matrices  yielding orbits of non-transversal rational subspaces. Matrices (5.5) will be used in section 6 for constructing a family of icosahedrally symmetric continuous atomic surfaces.
6. Continuous atomic surfaces.
The atomic surfaces discussed here are always linear, i.e. they are hyperplanes. Thus we use
the words « surface » and « hyperplane » without distinguishing between them. We begin with describing our Basic construction. Choose an orthogonal matrix  satisfying (5.5) and a vector
a = (xo, yo ) E V # V *. We consider an atomic surface
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and apply all operations of the space group .5 = ffi Q) 2 = Y ffi t sc to s. We get a family of
surfaces
The main result of this paper can be formulated as
Theorem : If the matrix  and the vector a = (xo, yo) are appropriately chosen, then an icosahedrally symmetric family of surfaces (6.2) satisfies conditions (i), (ii), (iii), (iv) of
section 2.
Proof : Properties (i), (ii), (iv) need no attention, since for all  and a the surfaces (6.2) are obviously periodic, continuous and have one-to-one projection on V*. Hence, only the
« hard core » condition (iii) needs special investigation. We perform it in two steps.
Step 1 (parallel surfaces). For this part of the proof we have to make use of a simple result
on commensurate subspaces :
Lemma : Given a subspace r commensurate with the lattice E, one can find a positive
constant ho 0 such that for any q E 2 the distance from the point q to the subspace r is either
0 or > h.
We leave this easy Lemma without proof.
Consider a subgroup of -5 consisting of pure translations belonging to Esc. Apply these
translations to the hyperplane s and get a family of parallel hyperplanes
We have to prove that the hyperplanes (6.3) either coincide or are separated by a constant
minimal distance.
Notation : Denote by so the hyperplane (6.1) with xo = 0, yo = 0. Since so contains 0, it is
not only a hyperplane but also a subspace of R6.
Two hyperplanes s + ql, s + q2 coincide if and only if ql - q2 E so. The lattice 1 [s ] = so n S has dimension 3, since so is commensurate with 2. Elements of factor-group £/ l [s] label
non-identical elements of the family (6.3).
Separation of parallel hyperplanes s + ql, s + q2 is equal to the distance between the point
ql - q2 and the subspace so. Using the Lemma we find that the « hard core » condition is true for the surfaces (6.3).
Step 2 (elimination of intersections). The family .5 [s 1 (6.2) can be viewèd as a 60-element
orbit generated by the group Y applied to the family of parallel surfaces (6.3) :
(here R. has the form (4.4)). For each g e Y the family i?[R9(s)] satisfies the « hard core »
condition, whatever  and a are chosen in (6.1) (see above). Thus, our goal is to provide non-
zero separation of different elements of the orbit (6.4) :
However, due to the periodicity of the family-5 [s ] condition (6.5) can be replaced by a weaker
one :
One checks easily that (6.6) combined with periodicity yields (6.5). Thus we have to choose  and a so that the intersections of hyperplanes h (s ) of the family (6.2) are absent.
For some  satisfying (5.5) we take a = 0. Obviously, in this case intersections are present.
Try to remove them by varying a. Let two different surfaces si , so belonging to
intersect : Then where, according to the
above discussion, gl =F g2. Change a, then The intersection cannot be removed if
We will see that condition (6.7) has a more convenient representation in terms of the subspace Wl , orthogonal complement to W12. We find
But,
Hence, W 2 has dimension -- 1, it is spanned by all vectors
where vectors u E V are such that always has a
solution since is an orthogonal 3 x 3 matrix). Rewritten in terms of vectors au of the form (6.10), condition (6.7) reads : or
for all au E W 2 given by (6.10). Since for all a condition (6.11) is true (see (b.7)), we obtain Taking au from (6.10) and R9n1 from (4.4) get Gi 1 (u ) - GZ 1 (u ) = 0
and
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