  # 3.3 The Wallpaper Groups

Definition 3.3.1. A subgroup K of DE2 is the wallpaper group if it's translational subgroup is generated by two translations.

A wallpaper group will be denoted by K, it's corresponding translational and point group with TK and OK, respectively.

The lattice consists of points ma+nb, m,nÎZ, where translations by independent vectors a and b generate TK.

From the condition that K is discrete, it follows: there are such a and b, that their module is smaller or equal to the module of any other translation in TK. In the further exposition, we will consider TK = áa, bñ, with |a| and |b| as minimal. Depending on the proportion of vectors a and b, we will make the difference between the lattices. The classification of wallpaper groups we begin by the classification of TK, that is by their lattices. Namely, the lattice will be 'built' of parallelograms (with two exceptions). Regarding the form of 'bricks', the lattice will get it's name.

Without loosing generality, let:

(1) |a|£|b| (otherwise, a « b);

(2) |a-b|£|a+b| (otherwise a ® - b, while because of |b| = |-b|, we have the condition (1) satisfied);

From the conditions (1) and (2) result the possible relations between |a|, |b|, |a-b| and |a+b|:

1) |a| < |b| < |a-b| < |a+b|

- oblique lattice;

2) |a| < |b| < |a-b| = |a+b|

- rectangular lattice;

3) (i) |a| < |b| = |a-b| < |a+b|

(ii) |a| = |b| < |a-b| < |a+b|, (by the replacement of b and a-b we obtain (i))

- centered rectangular lattice;

4) |a| = |b| < |a-b| = |a+b|

- square lattice;

5) |a| = |b| = |a-b| < |a+b|

- hexagonal lattice.

So there are 5 types of lattices:

- oblique;

- rectangular;

- centered rectangular;

- square;

- hexagonal. All the lattices except centered will be called simple lattices.

The Theorem 3.1 and the discreteness condition restrict the possibilities for OK to a finite number of different rotations that may belong to OK for any wallpaper group K. By proving the Crystallographic Restriction Theorem, we will show that there are allowed only four non-trivial rotations through the angles [(2p)/ 6], [(2p)/ 4], [(2p)/ 3], [(2p)/ 2], i.e. the rotations of the order 6, 4, 3, 2, respectively.

Every orthogonal group of a wallpaper group is finite, because the wallpaper group is discrete, so it contains the rotations through the angles [(2p)/ k], k Î Z. Since the vectors a and b that generate TK are independent, they represent the basis of R2. In this basis, according to Theorem 3.1, the matrix from OK preserves the lattice (this matrix represents the rotations about the origin), and it will contain the integer values (as lattice nodes). It's trace will be integer to. Because a matrix trace is invariant under the change of basis, and in standard basis the trace of

M æ
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is

 tr(M) = 2cosq,
all the terms 2cosq will be integers. Therefore,
cosq æ
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We proved:

Crystallographic Restriction Theorem. In a wallpaper group, the point group of rotations must be of order 1, 2, 3, 4 or 6.

This theorem is stated by Barlow in 1893 .

By using the proved existence of five lattice forms, four possible non-trivial rotations (Crystallographic Restriction Theorem) and the fact that these rotation preserve a lattice (Theorem 3.1), we will demonstrate the existence of 17 non-isomorphic wallpaper groups. Their classification will be done by exhausting all possible cases: for every combination of a lattice and a orthogonal group we will construct the corresponding group, or show that such a group does not exist. At the end, we will prove that all groups derived are mutually non-isomorphic.

We will use the standard notation, where the letters p or c denote a primitive or centered lattice, respectively. The number following the letter gives the order of rotation. It is followed by the next letter: m for reflection, g for glide reflection. This scheme has some exceptions: the groups p3m1 and p31m. That disagreement comes from the older notation that used four symbols.

In that classification, we will consider the vector a as collinear to x-axis, and b in the first quadrant. By Mq will be denoted the matrix of rotation about the origin, through the angle q, and Sf will be a reflection in the line passing through the origin, and of the slope f/2 .

Let`s mention that for every TK and it's orthogonal group OK, the wallpaper group will be TKÈXiÎOK(t,X), where TK and (t,Xi) represents the realization of Xi from OK in K.

We begin the classification from the oblique lattice:

(1) As for the oblique lattice R, |a| < |b| < |a-b| < |a+b|,

the only element of OD that preserves R is the rotation through f about the origin, so

 OK Ì {±I}.

(i) If the only rotation in K is the identity I, we get the simplest case: the wallpaper group generated by translations, K = {(ma+nb,I) | m,nÎZ}.

According to the notation accepted, this is the wallpaper group p1. It's orthogonal group is trivial. (ii) OK contains -I. We get K = TK{(0,I),(0,-I)}, that is the union of two neighboring classes of TK. The elements of K, not belonging to TK are (ma+nb,I)(0,-I) = (ma+nb,-I),    m,nÎ Z (that means that the elements are translations and half-turns about the points (1/2ma+1/2nb)).

Here, K is the wallpaper group p2. The orthogonal group is C2. (2) Rectangular lattice:

|a| < |b| < |a-b| = |a+b|,

Here, except the coincidence, we have the transformations -I, S0 and Sp preserving the lattice:

a) S0 is realized in K as a reflection (0,S0). K = TKÈTK(S0), and we have the wallpaper group pm. It's orthogonal group is D1. b) S0 is realized in K as a glide reflection, so K contains (0,S0). This is the glide reflection (1/2a,S0).

Let that glide reflection be given in the form (aa+bb,S0). Then (aa+bb,S0)2=( aa +bb+aa-bb,I)=((2aa,I) ÎTKÞ a = m/2, m Î Z. For m even, we will have (1/2ka,I)(1/2ka,S0) ÎTK Þ (0,S0) Î TK, and S0Ï OK, so this is the contradiction. Hence, m is odd, and the glide reflection is of the form (ma+nb,I)(1/2a,S0). The group K we obtain as TKÈTK(1/2a,S0).

This is the wallpaper group pg. It's orthogonal group is D1. (ii) Orthogonal group

OK is {I,-I, S0, Sp }

For OK = { I,Sp} we get the group isomorphic to a). Therefore, OK Ì { S0,Sp}. As

 (0,S0)(0,Sp) = (0,S0Sp),
S0Sp æ
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= -I.

We have -I Î OK. According to that, there are three wallpaper group corresponding to this orthogonal group:

a) If S0 and Sp are realized as reflections, K = TKÈXÎOK(0,X). This is the wallpaper group pmm. It's orthogonal group is D1×D1.

b) The only element of orthogonal group realized as a reflection is (0,Sp). Now K must contain the glide reflection parallel to the y-axis. By analogy with (1), we obtain that the glide reflections are of the form (1/2,S0). The transformation -I from OK is realized as (1/2,S0)(0,Sp) = (1/2a,-I). K = TKÈTK(0,S0)ÈTK(0,Sp)ÈTK(1/2a,-I). This is the wallpaper group pmg. The orthogonal group is D1×D1.

c) The third case occurs when K does not contain the reflections. So,  S0 is realized as (1/2a,S0), and Sp as (1/2b,Sp). -I is realized as (1/2a,S0)(1/2b,Sp) = (1/2(a-b),-I).

K = TKÈTK(1/2a,S0)ÈTK(1/2b,Sp)ÈTK(1/2(a-b),-I).

This is the wallpaper group pgg. It's orthogonal group is D1×D1. (3) Centered lattice

 |a| < |b| = |a-b| < |a+b|

The elements of OK as in (2) are I, -I, S0, Sp.

Since |b| = |a-b|, for a = (xa,ya) and b = (xb,yb) holds:

 xb2+yb2 = (xa-xb)2+(ya-zb)2 Û 2xbxa-xa2+2ybya-ya2 = 0 Û xa(2xb-xa)+ya(2yb-ya) = 0
 Û áa,2b-añ = 0 Û 2b-a ^a Þ 2b-a

so we obtain again the rectangular lattice.

To avoid the groups isomorphic to the already derived ones (pm, pg, pmm, pmg), we take that K, where the reflections from OK are realized as both reflections and glide reflections.

Namely such OK are characteristic for the centered lattice, in order to make a difference between the centered and the rectangular lattice.

(i) In the case that OK = { I,S0}, S0 is realized as a reflection (2b-a,S0), and glide reflection as (1/2(2b-a)+a/2,S0) = (b,S0), we have K = TKÈTK(b,S0).

This is the wallpaper group cm. The structure of OK for this group is D1. (ii) The case when OK={ I,-I,S0,Sp } is analogous to (i) (by mutually exchanging x- and y-axis).

Hence, we have K = TKÈTK(b,S0)ÈTK(a,Sp ).

The wallpaper group derived is cmm. It's orthogonal group is D1×D1. 4) Square lattice

It is preserved by the dihedral group D4. Therefore,

 |a| = |b| < |a-b| = |a+b|

To get the wallpaper groups characteristic for this lattice (non-isomorphic to already derived ones) OK must contain M[(p)/ 2].

(i) In the case OK = áM[(p)/ 2]ñ we have

K = TKÈTK(0,M[(p)/ 2])ÈTK(0,Mp)ÈTK(0,M[(3p)/ 2]).

The result is the wallpaper group p4. The structure of it's orthogonal group is C4. (ii) In the case S0 ÎOK, OK = {I,M[(p)/2],-I,M[(3p)/ 2], S0,S[(p)/ 2],Sp,S[(3p)/ 2]},

because of

S0M[(p)/ 2] æ
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= S[(p)/ 2]

a) S0 is realized as a reflection.

We get the wallpaper group p4m. It's orthogonal group is D4. b) S0 is not realized as a reflection. Let S0 be realized as (aa+bb,S0). So

 (aa+bb,S0)2 = (2a,S0) ÎK

It follows that 2a ÎZ. If a is an integer, we have (-aa,I)(aa+bb,S0) = (bb,S0) ÎK, and that is the contradiction. Therefore, S0 is realized as (1/2 a+bb,S0). By analogy, Sp is realized as (a1 a+1/2b,S0). From

 ( 1 2 a+bb,S0)(a1a+ 1 2 b,Sp) = ((a1 + 1 2 )a+(b- 1 2 )b,Mp)

it follows that a1 = 1/2+k and b = 1/2+l, where k, l Î Z.

 K = TKÈTKá(0,M[(p)/ 2])ñÈTK( 1 2 a+ 1 2 b,S0)ÈTK( 1 2 a+ 1 2 b,S[(p)/2])ÈTK( 1 2 a+ 1 2 b,Sp)ÈTK( 1 2 a+ 1 2 b,S[(3p)/ 2])

This is the wallpaper group p4g. It's orthogonal group is D4. (6) Hexagonal lattice

There we have the lattice:

 |a| = |b| = |a-b| < |a+b|,

and it is preserved by the dihedral group D6, generated by M[(p)/ 3] and S0 from OK. In the subgroup of direct transformations from D6, we have áM[(2p)/3]ñ£áM[(p)/ 3]ñ. Except S0, the indirect transformations in D6 are S[(kp)/ 3], k = 1,¼,5 .

Since S0M[(kp)/ 3] = S[(kp)/ 3]. Each of them is realized as a reflection in K.

It is sufficient to prove that S0 is realized as the reflection (for (0,S0)(0,M[(kp)/ 3]) = (0,S[(kp)/ 3])). Let S0 be realized as (aa+bb,S0). Then

(0,M[(2p)/ 3])(aa+bb,S0) = (a(b-a)-ba,S[(2p)/ 3]) Î K.

(a(b-a)-ba,S[(2p)/ 3])2 = ((a-b)b,I) ÎKÞ a- Z Þ (aa+bb,S0)((b-a)a,I) = (b(a+b),I) ÎKÞ a, Z Þ (aa+bb,S0)(-aa,I) = (bb,S0) ÎK, i.e. S0 is realized as the reflection in K.

When OK = áM[(2p)/ 3],S[(p)/ 3]ñ, S0 is not in OK.

Now we see that the wallpaper groups corresponding to the hexagonal lattice are:

(i) OK is generated by M[(p)/ 3]. Here

 K = TK È k = 1,¼,5 (0,M[(kp)/ 3])

This is the wallpaper group p6. It's orthogonal group is C6. (ii) The orthogonal group is D6, OK = áM[(p)/3],S0ñ.

 K = TK È k = 1,¼,5 (0,M[(kp)/ 3]) È k = 1,¼,5 (0,S[(kp)/ 3])

The wallpaper group K is p6m, and it's orthogonal group is D6. (iii) OK is generated by M[(2p)/ 3].

 K = TK È k = 1,¼,3 (0,M[(2kp)/ 3])

The obtained wallpaper group is p3, and the orthogonal group is isomorphic to C3. (iv) When OK is generated by M[(2p)/ 3] and S0:

 K = TK È k = 1,¼,3 (0,M[(2kp)/ 3]) È k = 1,¼,3 (0,S[(2kp)/ 3])

This wallpaper group is p3m1, and it's orthogonal group is isomorphic to D3. (v) OK = áM[(2p)/ 3],S[(p)/ 3] ñ.

 K = TK È (0,M-[(2p)/ 3]) È TK(0,M-[(2p)/ 3]) È TK(0,S[(p)/ 3]) È TK(0,S[(5p)/ 3]) È TK(0,Sp).

This is the group p31m, and the orthogonal group is D3. In the presented analysis we distinguished 17 different cases for the wallpaper groups: 2 for the oblique, 6 for rectangular, 7 for centered, 11 for square, and 11 for hexagonal lattice. For completing the classification, we need to show that these groups are not mutually isomorphic.

The only possible isomorphic groups can be those with isomorphic orthogonal groups (Theorem 3.2, Corollary 3.1). So, the groups of the following classes (p2, pm, pg, cm), (pmm, pmg, pgg, cmm), (p4m, p4g; p3m1, p31m) might be mutually isomorphic.

In the first class, the only group possessing a rotation is p2, so it is not isomorphic to any other group of that class (Theorem 3.2). The group pg is the unique group of that class which does not posses reflections, so it is not isomorphic to pm or cm. The remaining groups pm and cm are not mutually isomorphic, because every glide reflection from pm can be decomposed as the product of translations and reflections from pm (Theorem 3.2, Corollary 3.2), which is not true for cm.

The group pgg does not contain the reflections, so it is not isomorphic with pmm, pmg or cmm. The groups pmg and cmm are not isomorphic, because the product of every two reflections from pmg is a translation, while in cmm this product can give -I (Theorem 3.2, Corollary 3.3). Finally, the group pmm is not isomorphic with pmg or cmm, because every glide reflection from pmm can be decomposed as the product of translations and reflections belonging to that group (Theorem 3.2, Corollary 2), which not holds for pmg or cmm.

Every rotation from p4m can be decomposed as the product of two reflections from p4g, but this is not true for p4g. Therefore, p4m and p4g are not isomorphic (Theorem 3.2, Corollary 3.2).

In the same way, the previous consideration holds for p31m and p31m.

Thus, the classification of 17 wallpaper groups is completed.

Theorem 3.3.2. There are 17 different wallpaper groups.

The first attempt to derive and classify the wallpaper groups we may find in the works of C. Jordan (1867/1869), but he omitted the group pgg, recognized by L. Sohncke in 1874. The first complete classification of wallpaper groups was done by E.S. Fedorov in 1890 .