archive-nl.com » NL » M » METAFYSICA.NL Total: 972 Choose link from "Titles, links and description words view": Or switch to
"Titles and links view". |

- 3-D Crystals XVII

gd b So b becomes b gcg 1 ga c So c becomes c gdg 1 gb d So d becomes d gfg 1 gh f So f becomes f ggg 1 g So g becomes g ghg 1 gf h So h becomes h This amounts to the identity automorphism sixth inner automorphism When finally the transforming element is h we get the following h1h 1 hh 1 1 So 1 becomes 1 hah 1 haf hd c So a becomes c hbh 1 hbf ha d So b becomes d hch 1 hcf hb a So c becomes a hdh 1 hdf hc b So d becomes b hfh 1 hff hg f So f becomes f hgh 1 hgf hh g So g becomes g hhh 1 h So h becomes h This amounts to the seventh determined inner automorphism of D 4 and is represented by the permutation of the group elements and can be expressed by the cycle a c b d This automorphism was already encountered earlier So all in all we have found four different inner automorphisms of the group D 4 represented by the cycles 1 a b c d f g h identity automorphism b d f h a c f h a c b d So the inner automorphisms of D 4 constitute four permutations of the elements one of period 1 identity permutation and three of period 2 So they form the group D 2 And this group will turn out to be a normal subgroup of the full group of automorphisms inner and outer In addition to the four inner automorphisms D 4 also admits of four outer automorphisms To find these we look to all elements of period 2 except the element g which is unique because it commutes with all elements of the group To show this we repeat the group table of D 4 and highlight this special status of the element g 1 a b c d f g h period 1 1 a b c d f g h 1 a a 1 h g f d c b 2 b b f 1 h g a d c 2 c c g f 1 h b a d 2 d d h g f 1 c b a 2 f f b c d a g h 1 4 g g c d a b h 1 f 2 h h d a b c 1 f g 4 Table 17 6 Table of D 4 ab h ba f So neither a nor b commutes with every element of the group cb f bc h So c does not commute with every element of the group df c fd a So neither d nor f commutes with every element of the group ha d ah b So h does not commute with every element of the group Indeed apart from the identity only the element g commutes with every element of the group so the center of the group is 1 g This means that of the non identity elements g has a unique status and cannot be interchanged by any other element of the group and at the same time preserve the structure of the group table So with respect to period 2 elements we are left with the set a b c d Geometrically the elements of this set represent the four mirrors of the Square or of the Regular Tetragonal Pyramid In some of the inner automorphisms some of them were interchanged See for instance Figure 1 above There are still some possible interchanges left for these mirrors namely expressed as cycles a b c d and its inverse a d c b a b c d and a d b c According to BUDDEN p 421 these latter two have the effect of interchanging f and h I cannot figure out what he means by this because an interchange solely of f and h can according to me only be accomplished by interchanging b and d as Figure 1 illustrates There as a result of the interchange of b and d the figure is flipped over resulting in the two rotations of period 4 90 0 and 270 0 i e f and h which are each other s inverses being interchanged We will look into the matter shortly The final solution of this problem is stated HERE i e much further down and was found by systematically deriving all the automorphisms of the group D 4 by means of considering the substitution of generators This consideration starts a little further down a b c d renames the mirrors in anticlockwise direction while a d c b renames them in clockwise direction The effect of a b c d is illustrated in the next Figure Figure 2 Effect of a b c d expressing a permutation of the elements a b c and d of the group D 4 This permutation is one of D 4 s outer automorphisms The interchange is between the two types of mirrors Let us summarize the eight automorphisms of the group D 4 We ll give the cycle the corresponding permutation of the group elements and their periods The permutations will be given by their lower part only Inner automorphisms 1 a b c d f g h 1 a b c d f g h period 1 b d f h 1 a d c b h g f period 2 b d a c 1 c d a b f g h period 2 a c f h 1 c b a d h g f period 2 They accordingly form the group D 2 Outer automorphisms a b c d 1 b c d a f g h period 4 a d c b 1 d a b c f g h period 4 a b c d 1 b a d c f g h period 2 a d b c 1 d c b a f g h period 2 Together these inner and outer automorphisms of the group D 4 form the full automorphism group and this group has the structure of D 4 as the distribution of the periods indicates So we can write Aut D 4 D 4 The inner automorphisms form the group D 2 and this is a normal subgroup of the full automorphism group An alternative way of deriving the eight automorphisms of the group D 4 It is possible to express the elements of the group D 4 in terms of generators Two such generators independent generators are needed because D 4 contains a subgroup C 4 which can be generated by a single element but then we do not get outside this subgroup and consequently cannot generate the remaining elements of the full group So we need another generator in addition to one that generates the subgroup To generate the subgroup we can take the element f which has a period of 4 corresponding to a rotation of 90 0 However we prefer to call this element r The other generator we will call a Now in terms of these generators the group D 4 has as defining relations r 4 1 a 2 1 ar r 3 a BUDDEN p 290 The first relation implies that the period of r is 4 The second relation implies that the period of a is 2 The third relation says that a rotation r followed by a reflection a has the same result as a reflection a followed by three times a rotation r The next Figure shows the interpretation of the third defining relation in terms of the symmetries of the Square Figure 3 To show geometrically that ar r 3 a It is valid for all mirror lines of the Square An automorphism of the group D 4 can be found by choosing an alternative and equally valid pair of generators a pair because this group requires two generators and then generate the sequence of group elements We then obtain one permutation of the elements such that the table structure is preserved i e we have found an automorphism Then we choose another possible pair of generators and generate another sequence of the group elements and obtain yet another permutation representing a second automorphism We will now generate the group from the generators r and a and after that consider all other possible pairs of generators In order to do so we set up a group table with the element r generating the subgroup C 4 and the element a 1 r r 2 r 3 a period 1 1 r r 2 r 3 a 1 r r r 2 r 3 1 4 r 2 r 2 r 3 1 r 2 r 3 r 3 1 r r 2 4 a a 2 Partial table of D 4 But r and a generate a new element namely ra so this element must be added 1 r r 2 r 3 a ra period 1 1 r r 2 r 3 a ra 1 r r r 2 r 3 1 ra 4 r 2 r 2 r 3 1 r 2 r 3 r 3 1 r r 2 4 a a 2 ra ra 2 Partial table of D 4 further filled in But the elements r and ra generate a new element again namely r 2 a so this new element must be added 1 r r 2 r 3 a ra r 2 a period 1 1 r r 2 r 3 a ra r 2 a 1 r r r 2 r 3 1 ra r 2 a 4 r 2 r 2 r 3 1 r 2 r 3 r 3 1 r r 2 4 a a 2 ra ra 2 r 2 a r 2 a 2 Partial table of D 4 still further filled in Finally the element r and the element r 2 a produce again a new element namely r 3 a So we must add this element to the table We will see that we now have generated all elements of the group i e by subsequent multiplications no new elements will be created anymore 1 r r 2 r 3 a ra r 2 a r 3 a period 1 1 r r 2 r 3 a ra r 2 a r 3 a 1 r r r 2 r 3 1 ra r 2 a r 3 a 4 r 2 r 2 r 3 1 r 2 r 3 r 3 1 r r 2 4 a a 2 ra ra 2 r 2 a r 2 a 2 r 3 a r 3 a 2 Partial table of D 4 Last distinct element added We must now complete the group table of D 4 in terms of the generators r and a On the basis of the above given defining relations we can compute the remaining entries of the table Let us complete the first four rows 1 r r 2 r 3 a ra r 2 a r 3 a period 1 1 r r 2 r 3 a ra r 2 a r 3 a 1 r r r 2 r 3 1 ra r 2 a r 3 a a 4 r 2 r 2 r 3 1 r r 2 a r 3 a a ra 2 r 3 r 3 1 r r 2 r 3 a a ra r 2 a 4 a a 2 ra ra 2 r 2 a r 2 a 2 r 3 a r 3 a 2 Partial table of D 4 The first four rows being filled in Now we re going to compute the last four rows of the table still making use of the defining relations r 4 1 a 2 1 ar r 3 a The second element of the fifth row headed by a is ar But this is not a new element because according to the third defining relation we see that ar r 3 a And this element we have already found earlier We now generate the remaining elements of this fifth row ar 2 arr r 3 ar r 3 r 3 a r 6 a r 4 r 2 a r 2 a and this indeed is again not a new element We can put it in its right place in the table as the third element of the fifth row ar 3 arrr r 3 arr r 3 r 3 ar r 4 r 2 ar r 2 ar r 2 r 3 a r 5 a r 4 ra ra We place this result in the appropriate location of the table The next element is aa and this is equal to 1 Then the next element of the row is ara and this can be reduced as follows ara r 3 aa r 3 In the same way we compute all the remaining elements of the fifth row and then all the elements of the last three rows The resulting c o m p l e t e group table of D 4 in terms of the generator pair r and a is given below 1 r r 2 r 3 a ra r 2 a r 3 a period 1 1 r r 2 r 3 a ra r 2 a r 3 a 1 r r r 2 r 3 1 ra r 2 a r 3 a a 4 r 2 r 2 r 3 1 r r 2 a r 3 a a ra 2 r 3 r 3 1 r r 2 r 3 a a ra r 2 a 4 a a r 3 a r 2 a ra 1 r 3 r 2 r 2 ra ra a r 3 a r 2 a r 1 r 3 r 2 2 r 2 a r 2 a ra a r 3 a r 2 r 1 r 3 2 r 3 a r 3 a r 2 a ra a r 3 r 2 r 1 2 Table 17 7 Complete group table of D 4 in terms of the generator pair r and a In this group table the sequence of elements is 1 r r 2 r 3 a ra r 2 a r 3 a This sequence is generated by the generators r period 4 and a period 2 In order to find automorphisms we replace the original generators by two others also of periods 4 respectively 2 only elements of the same period can be interchanged with each other The elements of period 4 are r r 3 The elements of period 2 are r 2 a ra r 2 a r 3 a So the element r 3 of period 4 can replace r The elements r 2 ra r 2 a and r 3 a can replace a But when we replace a by r 2 period 2 we will duplicate this element because it already occurs at the third place of the row of group elements So we will not use this particular replacement The remaining elements of period 2 can be interchanged with a without duplicating elements On the basis of this we can now determine all the possible alternative pairs of generators r a r ra r r 2 a r r 3 a r 3 a r 3 ra r 3 r 2 a r 3 r 3 a We will now determine the possible sequences of group elements generated by the alternative pairs of generators as given above We begin with the original sequence of elements generated by the pair r and a and consider the generator pair r and ra In the original sequence we replace r by r wherever it occurs changing nothing of course and we replace a by ra wherever it occurs We then will obtain a new sequence of elements i e a permutation of them and thus we find an automorphism Next we consider the generator pair r and r 2 a and replace every r in the original sequence by r which doesn t change anything and we replace every a in the original sequence by r 2 a and so obtain yet another sequence of group elements i e another permutation and thus another automorphism We continue to do so with respect to the remaining possible pairs of generators and we then will obtain the following eight permutations of the group elements and thus all the eight automorphisms inner and outer of the group D 4 1 r r 2 r 3 a ra r 2 a r 3 a 1 r r 2 r 3 ra r 2 a r 3 a a 1 r r 2 r 3 r 2 a r 3 a a ra 1 r r 2 r 3 r 3 a a ra r 2 a 1 r 3 r 2 r a r 3 a r 2 a ra 1 r 3 r 2 r ra a r 3 a r 2 a 1 r 3 r 2 r r 2 a ra a r 3 a 1 r 3 r 2 r r 3 a r 2 a ra a Each row of the above array represents a permutation of the group elements of the group D 4 these elements being expressed in terms of r and a And these permutations must of course correspond to the permutations found earlier We found these permutations by interchanging pairs of independent generators and for example with respect to the second row we have let r the same while replacing a by ra We then substituted every a in the first row by ra and obtained the new sequence of elements This procedure can also be outlined

Original URL path: http://www.metafysica.nl/turing/d3_lattice_17.html (2016-02-01)

Open archived version from archive - 3-D Crystals XVIII

r 2 a ra 1 rar 2 aa 1 r 1 rar 2 r 1 rar 2 r 2 rar rr 2 a a With this we have the fifth automorphism 1 1 r 62 r 2 r 2 62 r a 62 r 2 a ra ra r 2 a 62 a And this corresponds to the fifth automorphism in the above list of automorphisms found by replacing generators Finally transformation of each non identity element by the transforming element r 2 a Earlier we saw that r 2 1 r and that a 1 a Before we proceed we see that r 2 a 1 a 1 r 2 1 a r 2 1 ar r 2 a r 2 a r r 2 a 1 r 2 ara 1 r 2 1 r 2 ar a 1 r r 2 r 2 aa 1 r r 2 r 2 r r 2 r 2 a r 2 r 2 a 1 r 2 ar 2 a 1 r 2 1 r 2 ar 2 a 1 r r 2 ar 2 ar r 2 ar 2 r 2 a r 2 ara r 2 r 2 aa r 2 r 2 r r 2 a a r 2 a 1 r 2 r 2 a 1 r 2 a 1 r 2 1 r 2 a 1 r r 2 ar r 2 r 2 a ra r 2 a ra r 2 a 1 r 2 ar aa 1 r 2 1 r 2 ar r 2 1 r 2 r 2 a r 2 1 ra r 2 1 rar rr 2 a a r 2 a r 2 a r 2 a 1 r 2 ar 2 aa 1 r 2 1 r 2 ar 2 r r 2 a This gives the sixth inner automorphism 1 1 r 62 r 2 r 2 62 r a 62 ra ra a r 2 a 62r 2 a And this corresponds to the sixth automorphism in the above list of automorphisms found by replacing generators So we have demonstrated that all automorphisms of the group D 3 are inner automorphisms We can now write down the above determined permutations representing automorphisms as cycles in order to determine their periods The permutation 1 1 r 62 r r 2 62 r 2 a 62 r 2 a ra a r 2 a 62 ra can be expressed as the cycle a r 2 a ra so the period is 3 The permutation 1 1 r 62 r r 2 62 r 2 a 62 ra ra r 2 a r 2 a 62 a can be expressed by the cycle a ra r 2 a so the period is 3 The permutation 1 1 r 62 r 2 r 2 62 r a 62 a ra r 2 a r 2 a 62 ra can be expressed by the cycle r r 2 ra r 2 a so the period is 2 The permutation 1 1 r 62 r 2 r 2 62 r a 62 r 2 a ra ra r 2 a 62 a can be expressed by the cycle r r 2 a r 2 a so the period is 2 Finally the permutation 1 1 r 62 r 2 r 2 62 r a 62 ra ra a r 2 a 62r 2 a can be expressed by the cycle r r 2 a ra so the period is 2 The group of automorphisms of D 3 therefore contains One element of period 1 identity automorphism Two elements of period 3 Three elements of period 2 This distribution of periods is the fingerprint of the group D 3 so the automorphism group of the group D 3 is isomorphic to D 3 Aut D 3 D 3 Now having seen several examples of automorphisms we characterize automorphisms once again in order to get more and more to grip with this rather difficult concept Two groups G and H are isomorphic to each other if there can be found a one to one correspondence between the elements of G and those of H such that products are preserved implying that the two groups have identical structure Of course a group G is always isomorphic to itself because one can always find a one to one correspondence between the elements of G and those of the same group G such that products are preserved namely the identity permutation of the group elements If moreover there exists in addition to this permutation yet another one to one correspondence between the elements of G and those of the same group G i e yet another permutation of G s elements such that products are preserved then we have an automorphism other than the identity automorphism of the group The set of all possible automorphisms of the group G itself forms a group under successive application of those permutations and is called the automorphism group of G i e Aut G Sometimes this automorphism group is isomorphic with the group itself Aut G G but very often it is a different group It can happen that the automorphism group of an Abelian group itself is non Abelian However the automorphisms of a cyclic group which is always Abelian do form an Abelian Group which however need not be cyclic For example Aut C 12 D 2 See later on If we have a cyclic group of prime order all the elements except the identity can serve as generator of the group and they all generate a different sequence of elements So the number of possible automorphisms is one less than the order of the group and moreover the automorphism group is in this case also cyclic Aut C p C p 1 when p is prime In a relevant section of Part IV we have given an example of such a cyclic group of prime order namely C 7 considered its automorphisms and spoke about automorphism generally We will continue to study the automorphisms of the Abelian groups C 12 and C 6 x C 2 The cyclic group of order 12 composite order i e the group C 12 can be represented by the set 0 1 2 3 4 5 6 7 8 9 10 11 under addition modulo 12 Because the order of this cyclic group is not prime but composite not all non identity elements can serve as generators Only those of period 12 can do so These are the elements 1 5 7 and 11 When we now first generate the group elements by the generator 1 then generate them with the generator 5 then do the same with respect to the generator 7 and finally do it with the generator 11 we get four different sequences of group elements which are thus permutations of the twelve group elements and because they generated sequences they are automorphisms of our group C 12 The first sequence we get by setting the element 0 identity element then setting the given element 1 generator and then repeat the generator till all the group elements are generated which here means that we keep adding modulo 12 1 till all elements are generated The second sequence is obtained when setting 0 then 5 generator and then adding 5 modulo 12 till all elements are generated In the same way the other two sequences are obtained All in all we get four sequences each consisting of the twelve group elements 0 1 2 3 4 5 6 7 8 9 10 11 0 5 10 3 8 1 6 11 4 9 2 7 0 7 2 9 4 11 6 1 8 3 10 5 0 11 10 9 8 7 6 5 4 3 2 1 These four sequences are thus permutations of the group elements and they correspond to the following cycles 1 5 2 10 4 8 7 11 1 7 3 9 5 11 1 11 2 10 3 9 5 7 As can be seen all these three cycles are of period 2 Together with the identity permutation they form the group D 2 so Aut C 12 D 2 Next we take the direct product group C 6 x C 2 defined by r 6 1 a 2 1 ar ra BUDDEN p 423 where r and a are independent generators both or alternative pairs are needed to generate all group elements of periods 6 and 2 respectively In terms of these generators the group Also discussed in HERE in Part IX contains Six elements of period 6 arranged here in inverse pairs r r 5 ar ar 5 ar 2 ar 4 Two elements of period 3 r 2 r 4 Three elements of period 2 r 3 a ar 3 To seek automorphisms we replace the generators r and a by other pairs of independent generators of period 6 and 2 Because there are six elements of period 6 and three of period 2 there must be 18 r a r r 3 r ar 3 r 5 a r 5 r 3 r 5 ar 3 ar a ar r 3 ar ar 3 ar 5 a ar 5 r 3 ar 5 ar 3 ar 2 a ar 2 r 3 ar 2 ar 3 ar 4 a ar 4 r 3 ar 4 ar 3 Among these possible pairs there are however six pairs of dependent generators i e two generators which can generate each other but cannot generate all elements of the group In the above listing these pairs are indicated by red coloring Let us consider these six pairs show that the members of those pairs are dependent on each other i e one can generate the other while in the calculations realizing that the group is Abelian that is all products commute r 5 r 3 r 5 r 5 r 10 r 4 and r 4 r 5 r 9 r 3 ar 5 ar 3 ar 5 ar 5 aar 10 r 10 r 4 and r 4 ar 5 ar 9 ar 3 ar 2 a ar 2 ar 2 r 4 and r 4 ar 2 a ar 4 a ar 4 ar 4 r 8 r 2 and r 2 ar 4 a ar ar 3 arar r 2 and r 2 ar ar 3 r r 3 rr r 2 and r 2 r r 3 The twelve remaining pairs are pairs of independent generators They generate twelve different sequences of group elements which are because they are generated and thus preserve products at the same time the twelve automorphisms of our group i e of the group C 6 x C 2 Before we tabulate the twelve element sequences automorphisms generated by the twelve pairs of alternative independent generators we show how as an example the pair of independent generators ar 2 ar 3 generates all elements of the group and as an automorphism results in a certain sequence of those elements But before that we generate the original sequence of group elements by the original pair of generators r and a The original sequence of group elements is generated by the generators r and a as follows First we have as being given the identity element 1 which does not partake in any swapping of group elements in order to generate an automorphism because its period which is 1 is always unique Secondly we have r which is given Then we have rr r 2 r 2 r r 3 r 3 r r 4 r 4 r r 5 r 5 r r 6 1 Further we have a which is given and then we obtain ar from multiplication with r arr ar 2 ar 2 r ar 3 ar 3 r ar 4 ar 4 r ar 5 ar 5 r ar 6 a So the initial sequence of group elements representing the identity permutation and thus the identity automorphism is 1 r r 2 r 3 r 4 r 5 a ar ar 2 ar 3 ar 4 ar 5 An alternative sequence of group elements can be generated by the independent pair of generators ar 2 ar 3 as follows First of course we have 1 And then ar 2 which is now given ar 2 ar 2 r 4 r 4 ar 2 a aar 2 r 2 r 2 ar 2 ar 4 ar 4 ar 2 1 ar 3 given ar 3 ar 2 r 5 r 5 ar 2 ar 7 ar arar 2 r 3 r 3 ar 2 ar 5 ar 5 ar 2 r 7 r So with the pair ar 2 ar 3 we have generated the alternative sequence of group elements 1 ar 2 r 4 a r 2 ar 4 ar 3 r 5 ar r 3 ar 5 r Seen as a permutation of group elements with respect to the original sequence of those same elements we can determine what cycle it represents 1 62 1 r 62 ar 2 r 2 62 r 4 r 3 62 a r 4 62 r 2 r 5 62 ar 4 a 62 ar 3 ar 62 r 5 ar 2 ar ar 3 62 r 3 ar 4 62 ar 5 ar 5 62 r This implies the cycle r ar 2 ar r 5 ar 4 ar 5 r 2 r 4 r 3 a ar 3 and from this we can see that the automorphism defined by this cycle is of period 6 In the same way we can determine the remaining ten alternative sequences permutations of the group elements by generating them by the remaining alternative pairs of independent generators and we can also determine their periods The result will be the twelve automorphisms of the group C 6 x C 2 Each row representing a permutation automorphism concludes with the period of that permutation 1 r r 2 r 3 r 4 r 5 a ar ar 2 ar 3 ar 4 ar 5 1 1 r r 2 r 3 r 4 r 5 ar 3 ar 4 ar 5 a ar ar 2 2 1 r 5 r 4 r 3 r 2 r a ar 5 ar 4 ar 3 ar 2 ar 2 1 r 5 r 4 r 3 r 2 r a 3 ar 2 ar a ar 5 ar 4 2 1 ar r 2 ar 3 r 4 ar 5 a r ar 2 r 3 ar 4 r 5 2 1 ar r 2 ar 3 r 4 ar 5 r 3 ar 4 r 5 a r ar 2 3 1 ar 5 r 4 ar 3 r 2 ar a r 5 ar 4 r 3 ar 2 r 2 1 ar 5 r 4 ar 3 r 2 ar r 3 ar 2 r a r 5 ar 4 6 1 ar 2 r 4 a r 2 ar 4 ar 3 r 5 ar r 3 ar 5 r 6 1 ar 2 r 4 a r 2 ar 4 r 3 ar 5 r ar 3 r 5 ar 2 1 ar 4 r 2 a r 4 ar 2 ar 3 r ar 5 r 3 ar r 5 3 1 ar 4 r 2 a r 4 ar 2 r 3 ar r 5 ar 3 r ar 5 2 These twelve automorphisms form a group From the distribution of the periods over these elements we can identify it as D 6 so Aut C 6 x C 2 D 6 We have noted that the group C 2 x C 2 x C 2 has 168 automorphisms The reason for this large number is that its seven elements og period 2 are so to speak indistinguishable Now C 3 x C 3 has eight elements of period 3 these being four pairs of inverses p p 1 q q 1 r r 1 s s 1 Evidently this group will also have a large number of automorphisms since any pair of the eight elements may be taken as generators provided they are not inverses Automorphisms of non Abelian groups In the case of non Abelian groups we may obtain all the inner automorphisms systematically by transforming all the elements of the group by each element in turn An interesting example is provided by Q 4 and we use the notation 1 r a s t u v w for its elements Its group table is We will now show that and how the group elements can be generated by the element pair r and t both of period 4 1 r a s t u v w period 1 1 r a s t u v w 1 r r a s 1 v t w u 4 a a s 1 r w v u t 2 s s 1 r a u w t v 4 t t u w v a s r 1 4 u u w v t r a 1 s 4 v v t u w s 1 a r 4 w w v t u 1 r s a 4 Table 18 1 Table of Q 4 A group table of a certain group G can be seen as the complete definition of the group and so is the above group table with respect to the group Q 4 We will now show that and how all the group elements can be generated by the element pair r and t both of period 4 The identity element 1 is given as also is the element r and also the element t When we now consider the product r r we see by checking the group table that we get the element a

Original URL path: http://www.metafysica.nl/turing/d3_lattice_18.html (2016-02-01)

Open archived version from archive - 3-D Crystals XIX

in the TABLE ABOVE and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt t t 1 r r 1 rt tr t w r s v u The resulting permutation accordingly is corresponding to the cycle r t s w u v which is of period 2 Next we replace the initial generator pair r t by the generator pair t s and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt t t 1 s s 1 st ts t w s r u v The resulting permutation accordingly is corresponding to the cycle r t s w which is of period 4 Next we replace the initial generator pair r t by the generator pair w r and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt w w 1 r r 1 rw wr w t r s u v The resulting permutation accordingly is coresponding to the cycle r w s t which is of period 4 Next we replace the initial generator pair r t by the generator pair w s and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt w w 1 s s 1 sw ws w t s r v u The resulting permutation accordingly is corresponding to the cycle r w s t u v which is of period 2 Next we replace the initial generator pair r t by the generator pair u r and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt u u 1 r r 1 ru ur u v r s t w The resulting permutation accordingly is corresponding to the cycle r u t s v w which has period 3 Next we replace the initial generator pair r t by the generator pair u s and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt u u 1 s s 1 su us u v s r w t The resulting permutation accordingly is corresponding to the cycle r u w s v t which is of period 3 Next we replace the initial generator pair r t by the generator pair u t and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt u u 1 t t 1 tu ut u v t w s r The resulting permutation accordingly is corresponding to the cycle r u s v which is of period 4 Next we replace the initial generator pair r t by the generator pair u w and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt u u 1 w w 1 wu uw u v w t r s The resulting permutation accordingly is corresponding to te cycle r u s v t w which is of period 2 Next we replace the initial generator pair r t by the generator pair v r and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt v v 1 r r 1 rv vr v u r s w t The resulting permutation accordingly is corresponding to the cycle r v t s u w which is of period 3 Next we replace the initial generator pair r t by the generator pair v s and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt v v 1 s s 1 sv vs v u s r t w The resulting permutation accordingly is corresponding to the cycle r v w s u t which is of period 3 Next we replace the initial generator pair r t by the generator pair v t and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt v v 1 t t 1 tv vt v u t w r s The resulting permutation accordingly is corresponding to te cycle r v s u which is of period 4 Finally we replace the initial generator pair r t by the generator pair v w and determine the implied permutation of the elements r s t w u v which corresponds to yet another automorphism of the group Q 4 r s t w u v r r 1 t t 1 tr rt v v 1 w w 1 wv vw v u w t s r The resulting permutation accordingly is corresponding to the cycle r v s u t w which is of period 2 We have now determined all 24 automorphisms of the group Q 4 Let s now summarize the total result We ve found 24 permutations of group elements Of each of these permutations we have determined the cycles they represent and by means of these cycles we could determine the periods of those permutations In the next table we will summarize these cycles and the determined periods in order to identify the group they form i e to identify the automorphism group of the group Q 4 The identity element of this automorphism group is of course the identity permutation having period 1 and is one of the 24 automorphisms cycle period identity 1 t w u v 2 r s u v 2 r s t w 2 r s t u w v 2 r s t v w u 2 r t s w u v 2 r w s t u v 2 r u s v t w 2 r v s u t w 2 r t u s w v 3 r t v s w u 3 r w v s t u 3 r w u s t v 3 r u t s v w 3 r u w s v t 3 r v t s u w 3 r v w s u t 3 t u w v 4 t v w u 4 r t s w 4 r w s t 4 r u s v 4 r v s u 4 So we have found one element of period 1 nine elements of period 2 eight elements of period 3 and six elements of period 4 And this leaves no doubt that the full group of automorphisms of Q 4 i e Aut Q 4 is S 4 the inner automorphisms being the normal subgroup with structure D 2 as we had found out in the previous document The group table of the group S 4 was depicted in Part XIII and can be seen by clicking HERE The first four cycles in the above list correspond to the inner automorphisms The way in which the group S 4 arises may be seen in another way BUDDEN p 426 For the three pairs of elements of period 4 may be likened to the three pairs of opposite faces of a cube If we label these faces r s t w u v in opposite pairs each of the twenty four rotations of the cube will correspond to exactly one of the permutations in the list of automorphisms of Q 4 These rotations are 3 x 3 90 0 180 0 270 0 9 rotatations with respect to the three crystallographic axes connecting centers of opposite faces 4 x 2 120 0 240 0 8 rotations with respect to the cube s diagonals 6 x 1 180 0 6 rotatations with respect to axes connecting opposite edges of the cube Identity 0 0 equivalent to 360 0 The inner automorphisms as established in the previous document are represented in the above list of cycles by the following identity t w u v r s u v r s t w And above we interpreted r s t w u v as pairs of opposite faces of a cube Well in this interpretation the three transpositions just given do in fact interchange opposite pairs of faces Each one of them interchanges two pairs of opposite faces so they must each correspond to a rotation of 180 0 about an axis connecting the two remaining opposite faces of the cube So we now have a geometrical interpretation of the four including the identity inner automorphisms of the group Q 4 Indeed there are three such rotations i e three rotations of 180 0 of the cube that interchange two pairs of opposite faces The interest in all this lies in the fact that though the group Q 4 is not the symmetry group of any geometrical figure yet it has this curious association with the Cube or Octahedron This concludes our discussion concerning the automorphism group of the group Q 4 Now Q 6 is also a group with six elements of period 4 and one might expect that its automorphisms would also number twenty four This is not so because one s freedom of choice is now more restricted in the following way See next table and what follows 1 a b c d f g h j k l m period 1 1 a b c d f g h j k l m 1 a a l 1 d f g h j c m k b 6 b b 1 m j c d f g h l a k 6 c c j d k l a 1 b m g h f 4 d d c f m k l a 1 b h j g 4 f f d g b m k l a 1 j c h 4 g g f h 1 b m k l a c d j 4 h h g j a 1 b m k l d f c 4 j j h c l a 1 b m k f g d 4 k k m l g h j c d f 1 b a 2 l l k a f g h j c d b m 1 3 m m b k h j c d f g a 1 l 3 Table 19 1 Table of Q 6 The group contains two elements of period 6 a and b in the notation used in the above group table of Q 6 and two of period 3 l and m It is essential that an automorphism shall either leave each of these pairs alone or else interchange them i e interchanges the members of such a pair Thus for example dg a while fc b So an automorphism which replaces d by f and g by c is feasible because then two elements of the same period a and b both of period 6 are interchanged And indeed one of the automorphisms of Q 6 is c g d f h j But since for example jh m we do not have an automorphism which replaces d by j and g by h because then elements of different periods a which is of period 6 and m which is of period 3 would be interchanged resulting obviously in products not being preserved Automorphisms of S 4 Finally we consider the automorphisms of S 4 and using the notation of the group table as it is given HERE in Part XIII we may take two of the elements of period 4 out of the subset j k n s t x to be generators of the whole group so long as they are not inverses We are therefore in a similar position to the one we were in when obtaining the automorphisms of Q 4 and the group of these is of order 24 Not unexpectedly its structure i e the structure of Aut S 4 is isomorphic to S 4 but what is surprising is that all the automorphisms are inner This may be seen by considering the effect of conjugation upon the subset j k n s t x Recall that the sequential order of elements of a set is immaterial as the next table shows Note that in every case the period of the transforming element is equal to the period of the induced permutation Recall that the conjugate or transform of an element p of a group G by an element q of that same group is qpq 1 where q is the transforming element So as we also did on other occasions in the following table we transform every element generally denoted by X of the mentioned set by each group element in turn And what we will see is that we get 24 inner automorphisms and knowing that the total number of automorphisms of the group S 4 is 24 we see that all the automorphisms of this group are inner automorphisms We show the resulting permutations representing the automorphisms their corresponding cycles and the implied period of those permutations evident from the cycles conjugation permutation of elements of period 4 cycle period 1X1 1 j t k n s x 1 aXa 1 k n j t x s j k t n s x 2 bXb 1 s x n k j t j s t x k n 2 cXc 1 x s t j k n j x n t s k 3 dXd 1 n k s x t j j n x t k s 3 fXf 1 t j x s n k j t k x n s 2 gXg 1 n k t j x s j n t k s x 2 hXh 1 t j n k s x j t k n 2 iXi 1 k n x s t j j k x t n s 3 jXj 1 j t s x n k k s n x 4 kXk 1 x s k n j t j x t s 4 l Xl 1 s x j t k n j s k t x n 3 mXm 1 x s j t n k j x k t s n 3 nXn 1 s x k n t j j s t x 4 pXp 1 t j s x k n j t k s n x 2 qXq 1 n k x s j t j n s t k x 3 rXr 1 j t n k x s k n s x 2 sXs 1 k n t j s x j k t n 4 tXt 1 j t x s k n k x n s 4 uXu 1 k n s x j t j k s t n x 3 vXv 1 s x t j n k j s n t x k 3 wXw 1 x s n k t j j x t s k n 2 xXx 1 n k j t s x j n t k 4 yXy 1 t j k n x s j t s x 2 All Automorphisms of S 4 In the above table we see that the permutations representing automorphisms are such that one of them is of period 1 nine of them are of period 2 eight are of period 3 and six are of period 4 And this indicates that they form a group isomorphic to S 4 so the automorphism group of the group S 4 is isomorphic to S 4 or for short Aut S 4 S 4 And all the automorphisms are inner automorphisms It will readily be seen that the six elements of period 4 may be associated with the six faces of a cube and that the automorphism may be thought of as a re labelling of the six quarter turns When we rotate the cube about one axis connecting two opposite faces we can rotate by 90 0 180 0 and 270 0 as being three symmetry operations Two of these rotations namely 90 0 and 270 0 are of period 4 and both can be denoted as quarter turns the former anti clockwise the latter clockwise And because the cube has three such axes there are six quarter turns These quarter turns can then be labelled j t k n s x i e corresponded to the six period 4 elements of the group S 4 The automorphisms then constitute swappings of these

Original URL path: http://www.metafysica.nl/turing/d3_lattice_19.html (2016-02-01)

Open archived version from archive - 3-D Crystals XX

We will now subject this motif to a translation t We have chosen this translation to be substantially smaller in length than the corresponding translation inherent in the pattern of Figure 1 The eventually resulting pattern will accordingly be more dense but this is of course theoretically immaterial Figure 3 A translation t and its reverse t 1 applied to the intial motif 1 So in the above Figure we generated two new group elements t and t 1 represented by two new locations of the basic motif It is clear that these two new elements do not have a finite period because however often they are repeated they will never end up to be the identity element represented by the position of the initial motif 1 so the period is equal to infinity Indeed when we repeat the action of the two elements one of them is a translation t to the right while the other is that same translation but directed to the left indicated by t 1 we get the following pattern Figure 4 repeated application of the translation t and t 1 generating new motifs representing new group elements By constantly reapeating the mentioned translations an infinite sequence of motifs will appear forming a one dimensional pattern of course still not being the pattern of Figure 1 The translation t can be seen as a generator of the group i e one of the generators of the group which will when fully generated underly the symmetry of the pattern of Figure 1 When we now introduce as a second generator a mirror line as indicated in the next Figure the pattern will develop further To begin with the mirror image of the initial motif 1 will be formed Figure 5 A second generator the mirror reflection a is added to be the next stage in the built up of the pattern of Figure 1 The initial motif is reflected in the mirror line a thereby creating yet another motif representing yet another group element a But at the same time a combines with t i e the motif representing the group element t is reflected in the mirror line a and creates yet another motif which represents the group element at which means first apply t and then apply to the result a See next Figure Figure 6 Also the motif representing the group element t is reflected in the mirror line a generating a new motif representing the group element at Of course the other elements t 3 t 2 t 1 t 2 t 3 are also reflected in the mirror line a thereby generating still more motifs representing still more group elements See next Figure Figure 7 The motifs t 4 t 3 t 2 also being reflected in a Figure 8 The motifs t 3 t 2 t 1 also being reflected in a where ta at 1 All this results in the following infinite pattern Figure 9 The total result of combining the translation t and the reflection a It is a one dimensional infinite pattern Not yet however the pattern of Figure 1 In this intermediary result we can already detect a subgroup namely 1 at Let s explain that at 2 atat And when we perform this operation on the intial motif we will end up again at the initial motif Figure 10 The operation at 2 t is a translation a is a reflection in the line a If we first apply the operation t on the intial motif 1 and then on the result apply a then on this result in turn t and finally on the latter result the operation a we will end up at the initial motif again All this clearly shows that at 2 1 which means that the period of the element at is 2 And together with the identity element which is represented by the initial motif denoted by 1 this element at forms a subgroup which has the structure of C 2 Figure 11 The subgroup 1 at as we find it in the infinite pattern sofar generated But the pattern of Figure 1 is not yet generated which means that we need yet another generator to be introduced This will be a horizontal mirror line denoted b placed just above the motifs such that the string of motifs as we have it in Figure 9 will be reflected Figure 12 A third generator the mirror line b is added The motifs will all be reflected in the newly added mirror line b Figure 13 The new mirror line b reflects the motifs and so creating as many new motifs which represent as many new group elements A second subgroup can now be detected 1 a b ba Let s check this The period of a is 2 because it is a reflection By the same token the element b has period 2 ba 2 baba it represents the following action Figure 14 The action of ba 2 It turns out that we end up at the initial motif again So ba 2 1 From the above Figure it is clear that the period of the element ba is 2 And this means that all the elements of the set 1 a b ba except the identity element 1 are of period 2 The next figure shows that ba ab Figure 15 ba ab Compare with ba in the previous Figure All in all we have the set 1 a b ba with defining relations a 2 b 2 ba 2 1 and ba ab From this we deduce aba aab b bba a baa b bab bba a So the structure of the set 1 a b ba is clearly that of the group D 2 The group D 2 as a subgroup of the group yet to be generated Along the same line we can detect yet another subgroup with D 2 structure namely 1 b at bat

Original URL path: http://www.metafysica.nl/turing/d3_lattice_20.html (2016-02-01)

Open archived version from archive - 3-D Crystals XXI

marks a two fold rotation axis to be present in the P2 pattern There are of course many such axes but this particular one is chosen to be one of the generators of the pattern from an initial motif the latter indicated by the numeral 1 The other generators needed are two translations The next Figure shows how the P2 pattern can be generated from the three chosen generators And this is equivalent to the generation of the group P2 Figure 4 The pattern and with it the group elements can be generated by two translations s and t and the half turn a The pattern must be viewed as becoming to extend indefinitely in 2 D space The pattern according to the Plane Group P3 Figure 5 Arranging motifs with point symmetry 3 in a hexagonal 2 D lattice yields a pattern that represents Plane Group P3 The pattern must be conceived as to be indefinitely extended in two dimensional space A unit cell outlined by the hexagonal net is indicated light blue Each motif consists of three motif units One such unit is considered as the initial motif unit and is indicated by the numeral 1 As generators for building up this pattern we choose a horizontal translation t and an anticlockwise rotation of 120 0 denoted by p about the point R So from the initial motif unit 1 representing the identity element of the group the motif unit p is generated which represents the group element p And when on the initial motif unit the rotation p is twice applied we get the motif unit p 2 and with it the corresponding group element p 2 Three times applying p yields the identity element so p 3 1 Wherever we have some already generated motif unit representing a group element we can generate a new motif unit representing a new group element by rotating it 120 0 about the point R Another new motif unit can be produced by shifting it according to the translation t See Figure 6 The total symmetry content of the Plane Group P3 is given in the next Figure Figure 5a Total symmetry content of the Plane Group P3 There are no mirror lines and also no glide lines The Plane Group only possesses 3 fold rotation axes and translations Figure 6 This Figure shows the tri radiate nature of the P3 pattern with respect to the point R That point is explicitly indicated in the previous Figure The whole pattern returns as it was before i e occupies the same space as it did before when we rotate it 120 0 about R The point R is not the only point with this property There are many such points But the rotation about the specific point R by 120 0 anticlockwise is chosen as a generator of the group Any already existing motif unit representing a group element of the pattern yields a new motif unit representing a new group element when it is subjected to rotation of 120 0 or of 240 0 about the point R In addition to this particular generator a second generator is needed which is chosen to be a horizontal translation to the right and implying also such a translation to the left By this translation the motif units representing group elements t 1 t t 2 etc are generated from the initial motif The next Figures show the generation of the P3 pattern A number of motif units are provided with the indication of how they can be generated by means of the two generators p rotation of 120 0 about the point R indicated in Figure 5 and t translation horizontally to the right Figure 7 From the element 1 the elements t 1 t t 2 etc are generated by the element t From the element 1 is generated the element p by applying the transformation p which is an anticlockwise rotation by 120 0 From the element p is generated the element p 2 by again applying p From the element p are generated the elements t 1 p tp t 2 p etc by applying t From the element p 2 are generated the elements t 1 p 2 tp 2 t 2 p 2 etc by the element t Figure 8 From the element p the element tp is generated by applying t and from tp the element ptp is generated by applying p i e an anticlockwise rotation of 120 0 about the point R as indicated in Figure 5 From the element ptp the elements tptp t 2 ptp etc are generated by applying t Figure 9 From the element p 2 the element tp 2 is generated by applying t and from tp 2 the element ptp 2 is generated by applying p rotation of 120 0 anticlockwise about R and from ptp 2 the elements tptp 2 t 2 ptp 2 etc are generated by t Figure 10 From the element p 2 the element tp 2 is generated by applying t and from the element tp 2 the element p 2 tp 2 is generated by applying p 2 A rotation of 240 0 anticlockwise about the point R As given in Figure 5 From p 2 tp 2 are generated the elements t 1 p 2 tp 2 tp 2 tp 2 etc by applying t Figure 11 From the element 1 the element t is generated by applying the translation t and from the element t the element pt is generated by applying the rotation p about the point R see Figure 5 and from the element pt the element tpt is generated by applying the translation t From the element tpt the element p 2 tpt is generated by applying the rotation p 2 about the point R Figure 12 From the element p 2 the element t 2 p 2 is generated by applying the translation t 2 and

Original URL path: http://www.metafysica.nl/turing/d3_lattice_21.html (2016-02-01)

Open archived version from archive - 3-D Crystals XXII

motif unit under the reflection a is chosen as a generator The motif unit b which is the image of the initial motif unit under the reflection b is chosen as a second generator The motif unit t which is the image of the initial motif unit under the translation t is chosen as a third generator A unit cell is indicated yellow The pattern must be conceived as extending indefinitely in two dimensional space We will now give the Pm pattern again and label some motif units such as representing generated group elements Figure 9 Generation of some group elements represented by motif units of the group Pm by the generators a b and t The pattern according to the Plane Group Pg Figure 10 Assymmetric motifs in fact motif units placed in a primitive rectangular net such that they alternate along the y edges horizontal of the meshes produce a pattern that represents the Plane Group Pg The only symmetry elements the Plane Group Pg possesses are simple translations in the x vertical and y horizontal directions and glide lines parallel to the y direction See Figure 11 Figure 11 Patterns representing Plane Group Pg have glide lines parallel to the y direction One of them is shown The total symmetry content of the Plane Group Pg is shown in the next Figure Figure 12 Total symmetry content of the Plane Group Pg Glide lines are indicated by red dashed lines For clarity the nodes of the net are indicated black dots Figure 13 The pattern representing Plane Group Pg consists of motif units commas They can be generated from a given initial motif unit denoted 1 by two genrators g a horizontal glide reflection and t a vertical translation The pattern according to the Plane Group P2mm Figure 14 Placing motifs with 2mm point symmetry in a primitive rectangular 2 D lattice creates a periodic pattern of these motives representing the Plane Group P2mm The pattern must be conceived as extending indefinitely in two dimensional space The total symmetry content of this Plane Group is given in Figure 15 Figure 15 The total symmetry content of the Plane Group P2mm Solid lines black and red indicate mirror lines Small red solid ellipses indicate 2 fold rotation axes perpendicular to the plane of the drawing The next Figure again gives the P2mm patttern and prepares for letting it be generated Figure 16 Each composed motif of our version of a P2mm pattern consists of four motif units commas partially overlapping such that the symmetry of the resulting composed motif is 2mm As before each motif unit represents a group element One such motif unit is chosen to be the initial motif unit representing the identity element of the group and denoted 1 A second motif unit denoted a is chosen as a first generator It is produced from 1 by the mirror line a A third motif unit denoted b is chosen as a second generator It is

Original URL path: http://www.metafysica.nl/turing/d3_lattice_22.html (2016-02-01)

Open archived version from archive - 3-D Crystals XXIII

cannot be further divided anymore And they indeed can represent the group elements To highlight these genuine motif units more clearly we can use colors provided we do not interpret the difference in color to represent an asymmetry one motif unit red or blue is asymmetric Two of them together one red and one blue make up a symmetric entity representing not a group element but a subgroup of the full group and twelve of them six red six blue make up the full composed hexagonal motif The next Figure has such motifs be placed in a hexagonal net point lattice Figure 14a A representation of a P6mm pattern in terms of genuine basic motif units red and blue Here the difference in color should not be interpreted as asymmetry it only serves to indicate the basic motif units making up the hexagonal composed motifs Each such composed motif is supposed to have 6mm symmetry i e it has D 6 structure not C 6 structure It has therefore six rotations and six reflections Each genuine motif unit either red or blue represents a group element of the Plane Group P6mm In order to generate the P6mm pattern which means to generate the Plane Group P6mm which in turn means to generate all group elements we choose one motif unit to be the initial motif unit and as such to represent the identity element denoted 1 and three others to represent generator elements A generator element m resulting from a reflection of the initial motif unit in a mirror line m A generator element p resulting from a 60 0 anticlockwise rotation of the initial motif unit about the point R See next Figure A generator element t resulting from a horizontal translation t of the initial motif unit Figure 14b Three generators m p and t represented by motif units and one initial motif unit representing the identity element are chosen The motif unit p results from a 60 0 anticlockwise rotation of the initial element 1 about the point R The motif unit t results from a horizontal shift of the initial motif unit to the right For clarity we enlarge the region around the point R Figure 14c The composed motif enlarged from the P6mm pattern of the previous Figure at the point R It consists of twelve true motif units red and blue representing group elements The difference in color should here not be interpreted as representing asymmetry The motif units 1 m p and the mirror line m are indicated The other motif units of the composed motif of Figure 14c are then automatically implied Figure 14d The composed motif and its twelve constituent basic motif units at the point R of Figure 14b These basic motif units represent the group elements 1 m p mp mp 2 mp 3 mp 4 mp 5 p 2 p 3 p 4 and p 5 They form the subgroup D 6 of the group P6mm The elements 1 p p 2 p 3 p 4 p 5 also form a subgroup of the group P6mm as well as of the subgroup D 6 with structure C 6 In order to further generate the pattern and with it the group elements we can subject the elements of the subgroup D 6 Figure 14d to a translation t resulting in 12 new group elements represented by 12 basic motif units A few of these new elements are indicated in the next Figure Figure 14e Some new group elements resulting from the translation t are indicated Also some points bearing composed motifs of the lattice are indicated R Q U W The next Figure shows the composed motif next to the right of the one at point R with the group elements represented by basic motif units indicated as they were produced by the translation t with respect to the composed motif at the point R Figure 14f The composed motif next to the right of the one at the point R The red motif units in Figure 14f together form the left coset of the subgroup 1 p p 2 p 3 p 4 p 5 with structure C 6 See Figure 14d by the element t All the twelve elements red and blue in Figure 14f together form the left coset of the D 6 subgroup See Figure 14d by the element t Next we re going to generate the composed motif at the point Q in Figure 14e It can be obtained by rotating the previously generated composed motif the motif next to the one at point R 300 0 anticlockwise or which is the same 60 0 clockwise about the point R which means that we subject all elements of the previously generated composed motif to the action of p 5 The next Figure gives this new composed motif Inside the image we have left the notations for the elements as they were in the motif next to R while the identities of the newly generated group elements are given at the perimeter of the image Figure 14g The newly generated composed motif at lattice point Q as indicated in Figure 14e The names of the new elements are given at the perimeter of the image i e outside the image All the elements in Figure 14g red and blue together form the left coset of the D 6 subgroup See Figure 14d by the element p 5 t The third row Figure 14e of composed motifs can now be completed by means of applying to this lastly obtained composed motif the translations t 3 t 2 t 1 t t 2 t 3 And along the same lines we can complete row 2 of the pattern In order to reach the fourth row we first determine the composed motif at the point U by applying t 2 2 times applying the translation t to the elements of the composed motif at the point R and

Original URL path: http://www.metafysica.nl/turing/d3_lattice_23.html (2016-02-01)

Open archived version from archive - 3-D Crystals XXIV

motifs differs by 30 0 from that of the motifs of the Plane Group P3m1 yields a periodic pattern representing Plane Group P31m See Figure 7 Figure 7 When motifs having point symmetry 3m i e having a 3 fold rotation axis and three equivalent mirror lines are inserted in a primitive hexagonal net in the way i e the orientation shown a pattern of repeated motifs will emerge that represents Plane Group P31m Each composed motif consists of three augmented motif units in such a way that the symmetry of the composed motif is 3m Each augmented motif unit represents a subgroup or a coset The pattern must be conceived as extending indefinitely in two dimensional space The total symmetry content of the Plane Group P31m is given in the next diagram Figure 8 Total symmetry content of Plane Group P31m Mirror lines are indicated by solid lines red and black Now we will show the generation of the group P31m by means of basic motif units that legitimately represent group elements i e elements of the full group P31m In Figure 7 we had motifs each consisting of three units making angles with each other of 120 0 and providing the composed motif with 3m symmetry Which is equivalent to a D 3 group structure But such a motif unit is still symmetric in itself so in fact it is composed of still more basic motif units It is an augmented motif unit And indeed the basic units as can be obtained by dividing the augmented motif units do not have and ought not to have any symmetry at all They re going to build up a symmetric pattern from scratch They can legitimately represent group elements i e elements of the full group P31m The next image shows a composed motif equivalent to the ones in Figure 7 but partitioned into six basic motif units This type of motif can perhaps be more clearly expressed as follows The two basic motif units composing an augmented motif unit of which three together make up the full composed motif can conveniently be distinguished by colors provided we do not interpret the difference between colors as expressing an asymmetry The two basic motif units red and blue in the next Figures are symmetrically related to each other The latter motifs we will now place in a hexagonal lattice That lattice having the same orientation as that used for depicting the pattern of the group P3m1 resulting in the periodic pattern according to the group P31m The effect is that both patterns P31m and P3m1 have the same type of composed motif and also of basic motif unit for that matter but in each case those motifs are differenly orientated with respect to the lattice lines i e the edges of the unit cell This difference in orientation is 30 0 Figure 8a Pattern according to the Plane Group P31m The composed motifs consist of six basic motif units red and blue each representing a group element i e an element of the group P31m The difference in color should not be interpreted as an asymmetry The pattern must be conceived as extending indefinitely in two dimensional space The next Figure gives this same pattern Some lattice points are marked R S X U W One basic motif unit of the composed motif at point R is chosen to be the initial motif unit representing the identity element 1 of the group Three other basic motif units are chosen as generators The basic motif unit p representing an anticlockwise rotation of 120 0 about the lattice point R The basic motif unit m representing a reflection in the mirror line m The basic motif unit t representing a horizontal translation t Figure 8b Pattern according to the Plane Group P31m The initial motif unit three generators and some lattice points are indicated The next Figure depicts an enlargement of the composed motif at the lattice point R The initial basic motif unit 1 the generators p and m and the mirror line m are indicated Figure 8c Composed motif consisting of six basic motif units of the P31m pattern at the point R in Figure 8b The identity of the remaining basic motif units of the composed motif at the lattice point R can now be determined p 3 1 Figure 8d The group elements of the composed motif at the point R Together they form the subgroup 1 p p 2 m mp mp 2 having the structure of D 3 The elements 1 p p 2 form a subgroup with the structure of C 3 The elements 1 m form the subgroup C 2 We will now produce the composed motif at the point S Figure 8b We ll do this by subjecting all the elements basic motif units of the composed motif at the point R to a translation t i e we will form the left coset of the subgroup D 3 materialized as the composed motif at the point R by the element t See next Figure Figure 8e The elements basic motif units of the composed motif at the lattice point S They form the left coset of the D 3 subgroup by the element t The notations i e identities of the newly generated elements are placed at the perimeter of the image i e outside the image Next we determine the elements of the composed motif at the lattice point U Figure 8b To generate those elements we must subject the elements of the composed motif at the point S to an anticlockwise rotation of 240 0 about the point R that is to say we must apply the transformation p 2 In this way we form the left coset of the subgroup D 3 by the element p 2 t See next Figure Figure 8f Generation of the basic motif units of the composed motif at the lattice point U The names

Original URL path: http://www.metafysica.nl/turing/d3_lattice_24.html (2016-02-01)

Open archived version from archive