Mathematical Crystallography 9781501508912, 9780939950195

Citation preview

MATHEMATICAL CRYSTALLOGRAPHY A N INTRODUCTION TO THE MATHEMATICAL FOUNDATIONS OF CRYSTALLOGRAPHY T H E REVISED EDITION OF VOLUME 1 5 R E V I E W S IN M I N E R A L O G Y

AUTHORS:

M . B. BOISEN, JR. DEPARTMENT OF MATHEMATICS

G . V . GIBBS DEPARTMENT OF GEOLOGICAL SCIENCES

Virginia Polytechnic Institute & State University Blacksburg, Virginia 24061

SERIES EDITOR: PAUL H . RIBBE DEPARTMENT OF GEOLOGICAL SCIENCES

Virginia Polytechnic Institute & State University Blacksburg, Virginia 24061 MINERALOGICAL SOCIETY OF AMERICA WASHINGTON, D.C.

COPYRIGHT: FIRST EDITION 1985 REVISED EDITION 1990 MINERALOGICAL

SOCIETY

of

AMERICA

Printed by BookCrafters, Inc., Chelsea, Michigan 48118 REVIEWS in MINERALOGY (Formerly: SHORT COURSE NOTES) ISSN 0275-0279 VOLUME 15, Revised Edition MATHEMATICAL CRYSTALLOGRAPHY ISBN 0-939950-26-X ADDITIONAL COPIES of this volume as well as those listed below may be obtained from the MlNERALOGICAL SOCIETY OF AMERICA 1625 I Street N.W., Suite 414, Washington, D.C. 20006 U.S.A. Volume 1: Sulfide Mineralogy, 1974; P. H. Ribbe, Ed. 284 pp. Six chapters on the structures of sulfides and sulfosatts; the crystal chemistry and chemical bonding of sulfides, synthesis, phase equilibria, and petrology. I S B N # 0-939950-01-4. Volume 2: Feldspar Mineralogy, 2nd Edition, 1983; P. H. Ribbe, Ed. 362 pp. $17. Thirteen chapters on feldspar chemistry, structure and nomenclature; Al.Si order/disorder in relation to domain textures, diffraction patterns, lattice parameters and optical properties; determinative methods; subsolidus phase relations, microstructures, kinetics and mechanisms of exsolution, and diffusion; color and interference colors; chemical properties; deformation. I S B N # 0-939950-14-6. Volume 4: Mineralogy and Geology of Natural Zeolites, 1977; F. A. Mumpton, Ed. 232 pp. Ten chapters on the crystal chemistry and structure of natural zeolites, their occurrence in sedimentary and low-grade metamorphic rocks and closed hydrologic systems, their commercial properties and utilization. I S B N * 0-939950-04-9. Volume 5: Orttiosiiicates, 2nd Edition, 1982; P. H. Ribbe, Ed. 450 pp. Liebau's "Classification of Silicates" plus 12 chapters on silicate garnets, olivines, spinels and humites; zircon and the actinide orthosilicates; titanite (sphene), chloritoid, staurolite, the aluminum silicates, topaz, and scores of miscellaneous orthosilicates. Indexed. I S B N # 0-939950-13-8. Volume 8: Marine Minerals, 1979; R. G. Bums, Ed. 380 pp. Ten chapters on manganese and iron oxides, the silica polymorphs, zeolites, day minerals, marine phosphorites, barites and placer minerals; evaporite mineralogy and chemistry. I S B N # 0-939950-06-5. Volume 7: Pyroxenes, 1980; C. T. Prewrtt, Ed. 525 pp. Nine chapters on pyroxene crystal chemistry, spectroscopy, phase equilibria, subsolidus phenomena and thermodynamics; composition and mineralogy of terrestrial, lunar, and meteoritic pyroxenes. I S 8 N # 0-939950-07-3. Volume 8: Kinetics of Geochemlcal Processes, 1981; A. C. Lasaga and R. J. Kirkpatrlck, Eds. 398 pp. Eight chapters on transition state theory and the rate laws of chemical reactions; kinetics of weathering, diagenesis, igneous crystallization and geochemical cycles; diffusion in electrolytes; irreversible thermodynamics. I S B N # 0-939950-06-1. Volume 9A: Amphibotes and Other Hydrous Pyriboies—Mineralogy, 1981; D. R. Veblen, Ed. 372 pp. Seven chapters on biopyribole mineralogy and polysomatism; the crystal chemistry, structures and spectroscopy of amphiboles; subsolidus relations; amphibole and serpentine asbestos—mineralogy, occurrences, and health hazards. I S B N # 0-939950-10-3. Volume 9B: Amphiboles: Petrology and Experimental Phaae Relations, 1982; D. R. Veblen and P. H. Ribbe, Eds. 390 pp. Three chapters on phase relations of metamorphic amphiboles (occurrences and theory); igneous amphiboles; experimental studies. I S B N # 0-939950-11-1. Volume 10: Characterization of Metamorphism through Mineral Equilibria, 1982; J. M. Ferry, Ed. 397 pp. Nine chapters on an algebraic approach to composition and reaction spaces and their manipulation; the Gibbs' formulation of phase equilibria; geologic thermobarometry; buffering, infiltration, isotope fractionation, compositional zoning and inclusions; characterization of metamorphic fluids. I S B N # 0-939950-12-X. Volume 11: Carbonates: Mineralogy and Chemistry, 1983; R. J. Reader, Ed. 394 pp. Nine chapters on crystal chemistry, polymorphism, microstructures and phase relations of the rbombohedral and orthorhombic carbonates; the kinetics of CaCO, dissolution and precipitation; trace elements and isotopes in sedimentary carbonates; the occurrence, solubility and solid solution behavior of Mg-caldtes; geologic thermobarometry using metamorphic carbonates. I S B N # 0-939950-15-4.

Volume 12: Fluid Inclusions, 1984; by E. Roedder. 844 pp. Nineteen chapters providing an introduction to studies of aN types of fluid inclusions, gas, liquid or melt, trapped in materials from the earth and space, and their application to the understanding of geological processes. I S B N # 0-939950-16-2. Volume 13: Micas, 1984; S. W. Bailey, Ed. 584 pp. Thirteen chapters on structures, crystal chemistry, spectroscopic and optical properties, occurrences, paragenesis, geochemistry and petrology of micas. I S B N # 0-93995^17-0. Volume 14: Microscopic to Macroscopic: Atomic Environments to Mineral Thermodynamics, 1985; S. W. Kleffer and A. Navrotsky, Eds. 428 pp. Eleven chapters attempt to answer the question, "What minerals exist under given constraints of pressure, temperature, and composition, and why?" Includes worked examples at the end of some chapters. I S B N # 0-93995016-9. Volume 15: Mathematical Crystallography, 1985; by M. B. Bolsen, Jr. and G. V. Glbbs. 408 pp.4 p A matrix and group theoretic treatment of the point groups, Bravais lattices, and space groups presented with numerous examples and problem sets, Including solutions to common crystallography problems involving the geometry and symmetry of crystal structures. I S B N # 0-93995019-7. Volume 18: Stable Isotopes in High Temperature Geological Processes, 1988; J. W. Valley, H. P. Taylor, Jr., and J. R. O'Neil, Eds. 570 pp. Starting with the theoretical, kinetic and experimental aspects of isotopic fractionation, 14 chapters deal with stable isotopes in the early solar system, in the mantle, and in the igneous and metamorphic rocks and ore deposits, as well as in magmatic volatiles, natural water, seawater, and in meteoric-hydrothermal systems. ISBN #0-939950-20-0. Volume 17: Thermodynamic Modelling of Geological Materials: Minerals, Fluids, Melts, 1987; H. P. Eugster and 1. S E. Carmlchael, Eds. 500 pp. Thermodynamic analysis of phase equilibria in simple and multi-component mineral systems, and thermodynamic models of crystalline solutions, igneous gases and fluid, ore fluid, metamorphic fluids, and silicate melts, are the subjects of this 14-chapter volume. ISBN # 0-939950-21-9. Volume 18: Spectroscopic Methods in Mineralogy and Geology, 1988; F. C. Hawthorne, Ed. 898 pp. Detailed explanations and encyclopedic discussion of applications of spectroscopies of major importance to earth sciences. Included are IR, optical, Raman, Mossbauer, M A S NMR, EXAFS, XANES, EPR, Auger, XPS, luminescence, XRF, PIXE, R B S and EELS. ISBN # ->-939f 0-939950-22-7. Volume 19: Hydrous Phyllosillcates (exclusive of micas), 1988; S. W. Bailey, Ed. 898 pp. Seventeen chapters covering the crystal structures, crystal chemistry, serpentine, kaolin, talc, pyrophyllite, chlorite, vermiculite, smectite, mixed-layer, sepiolite, palygorskite, and modulated type hydrous phyllosilicate minerals. Volume 20: Modem Powder Diffraction, 1989; D. L Blah and J. E. Post, Eds. The principles, instrumentation, experimental procedures, and computer analysis of X-ray (synchrotron) and neutron powder diffraction, with chapters on sample preparation, quantitative analysis, profile fitting, and Rletveld refinement of crystal structures. ISBN # 6-939950-24-3. Volume 21: Geochemistry and Mineralogy of Rare Earth Elements, 1989; B. R. Upln and G. A. McKay, Eds.

REVIEWS IN MINERALOGY Foreword t o Volume 15 M A T H E M A T I C A L C R Y S T A L L O G R A P H Y represents a new direction for the Reviews in Mineralogy series. This text book is not a review volume in any sense of the term, but in fact it is, as its subtitle suggests, "An Introduction to the Mathematical Foundations of Crystallography." Written by a mathematician, M. B. Boisen, Jr., and a mineralogist, G. V. Gibbs, Volume 15 was carefully prepared and illustrated over a period of several years. It contains numerous worked examples, in addition to problem sets (many with answers) for the reader to solve. The book was first introduced at a Short Course of the same title in conjunction with the annual meetings of the Mineralogical Society of America and the Geological Society of America at Orlando, Florida, October 24-27, 1985. Boisen and Gibbs instructed 35 participants with the assistance of E. Patrick Meagher, (University of British Columbia), James W. Downs (Ohio State University), and Bryan C. Chakoumakos (University of New Mexico), who led the computer-based laboratory sessions. Paul H. Ribbe Series Editor Blacksburg, VA 9/13/85 - a Friday

Preface to t h e R e v i s e d Edition of M a t h e m a t i c a l Crystallography In the Revised Edition we have corrected the errors, misprints and omissions that we have found and our students and other users have kindly pointed out to us. In particular, we are pleased to thank our bright and talented students K.M. Whalen, K.L. Bartelmehs, L.A. Buterakos, V.K Chapman, B.C. Chakoumakos, J.W Downs and R.T. Downs for bringing some of these errors to our attention. The Revised Edition also includes a more comprehensive index and a set of solutions for all of the problems presented in the book. We are especially pleased to thank V.K. Chapman and K.M Whalen for their care and efforts in working out the solutions to these problems in detail. We also thank Sharon Chiang for providing us with her drafting skills. M. B. Boisen, Jr. and G. V. Gibbs Blacksburg, VA 2/28/90

to o u r w i v e s H E L E N and

NANCY

whose patience a n d s u p p o r t is s i n c e r e l y appreciated

MATHEMATICAL

CRYSTALLOGRAPHY PREFACE

This book is written with two goals in mind.

The first is to derive

the 32 crystallographic point groups, the 14 Bravais lattice types and the 230 crystallographic space group types.

The second is to develop the

mathematical tools necessary for these derivations in such a manner as to lay the mathematical foundation needed to solve numerous basic problems in crystallography and to avoid extraneous discourses.

To

demonstrate

how these tools can be employed, a large number of examples are solved and problems are given.

The book is, by and large, self-contained.

In

particular, topics usually omitted from the traditional courses in mathematics that are essential to the study of crystallography are discussed. For

example,

the techniques needed to work in vector spaces with non-

cartesian bases are developed.

Unlike the traditional group-theoretical

approach, isomorphism is not the essential ingredient in crystallographic classification schemes.

Because alternative classification schemes must

be used, the notions of equivalence relations and classes which are fundamental to such schemes are defined, discussed and illustrated. example, we will

For

find that the classification of the crystallographic

space groups into the traditional 230 types is defined in terms of their matrix representations.

Therefore, the derivation of these groups from

the point groups will be conducted using the 37 distinct matrix groups rather than the 32 point groups they represent. We have been greatly influenced by two beautiful books. Weyl's book entitled Symmetry

Hermann

based on his lectures at Princeton Uni-

versity gives a wonderful development of the point groups as well as an elegant

exposition

Zachariasen's

book

of

symmetry

entitled

in

art

Theory

of

and nature. X~ray

Fredrik W. H.

Diffraction

in

Crystals

presents important insights on the derivation of the Bravais lattice types and the crystallographic space groups. These two books provided the basis for many of the ideas developed in this book. The

theorems,

examples,

definitions

and corollaries are labelled

sequentially as a group whereas the problems are labelled separately as a group as are the equations. self-explanatory.

The manner in which these are labelled is

For example, T4.15 refers to Theorem (T) 15 in Chapter

4 while DA1.1 refers to Definition (D) 1 in Appendix (A) 1. v

We have strived to write this book so that it is self-teaching. reader is encouraged to attempt to solve the examples before

The

appealing

to the solution presented and to work all of the problems.

ACKNOWLEDGEMENTS

We wish to thank Virginia Chapman, a lady of exceptional talent and ingenuity, project.

for

her

enthusiastic

and

tireless

contribution

this

In particular, her preparation of the GML files used to produce

this book is greatly appreciated.

We thank John C. Groen for his dedi-

cation and his meticulous preparation of the many illustrations book.

to

in this

Margie Strickler is gratefully acknowledged for her preparation

of the GML files for the appendices.

It is also a pleasure to thank Karen

L. Geisinger for painstakingly preparing the stereoscopic pair diagrams of the C-equivalent ellipsoids for the 32 crystallographic point groups G.

We thank Don Bloss and Hans Wondratschek for beneficial discussions,

Sharon Chang for important technical advice on the drafting of the figures and her assistant, Melody L. Watson, for her draftwork.

We also thank

Bryan C. Chakoumakos, Department of Geology, University of New

Mexico,

Albuquerque, New Mexico; James W. Downs, Department of Geology and Mineralogy,

Columbus, Ohio; Karen L. Geisinger, Department of Geoscience,

The Pennsylvania State University, University Park, Pennsylvania and Neil E.

Johnson,

Department

of

Geological

Sciences,

VPI&SU,

Blacksburg,

Virginia and David R. Veblen, Department of Earth and Planetary Sciences, The Johns Hopkins University, Baltimore, Maryland for their reading the earlier drafts and useful remarks.

However, they are neither

of re-

sponsible for any errors that may be present in the book nor for the point of view we have taken in this project.

Finally, we gratefully acknowledge

the Series Editor Paul H. Ribbe for his helpful comments and criticisms and the National Science Foundation Grant EAR-8218743 for partial support of this project.

vi

EXPLANATION OF SYMBOLS SYMBOL

DESCRIPTION

S

Geometric three-dimensional space.

P

A primitive lattice or the basis for a primitive lattice.

[r]p

The triple representation of r with respect to the basis

Lp *

D

*

*

*

= {a ,b ,c }

D.

The lattice generated by D . The reciprocal lattice of

D.

Mp(a)

The 3x3 matrix representation with respect to the basis D of a when a is a point isometry and of the linear component of a when o is an isometry.

M

The set of all MpCtx) where a t G .

D(C)

I

The set of all isometries.

T

The set of all translations.

D{O)

The basis {T (O),T (O),T (O)} of S where D = {t ,T ,T } is a basis fori. x ^ y ^' z x y i

j

R

(C)

The set of all R c,( 0 )(°) where a t G .

{M | t}

The Seitz notation for the 4x4 matrix representation of an isometry.

orb-y-(o)

The orbit of O under the translations of 7".

A Q (a)

The linear component of a.

Aq(C)

The set of A Q ( o ) where a e

T{G)

The set of all translations in C .

T^

The set of translations associated with the lattice L.

tr(M)

The trace of the matrix M.

det(M) = |M|

The determinant of the matrix M.

#(H)

The number elements in H .

o(a)

The order of the isometry a.

[uvw)n

An nth-turn whose axis is along ua + v b + w c where D = {a,b,c} is a given basis.

[uvw]n

m

G.

An nth-turn screw about the vector ua + v b + w c with a translation of m r / n where r is the shortest nonzero vector in the lattice in the [uvw] direction.

/'

The inversion.

Basic conventions

Points in S are denoted by lower case letters, vectors and their endpoints by bold-faced lower case letters, lengths of vectors by italics, sets by capital italics and matrices by capital letters. vii

MATHEMATICAL CRYSTALLOGRAPHY CONTENTS

Page

COPYRIGHT; LIST OF PUBLICATIONS

ii

DEDICATION

iii

FOREWORD

iv

PREFACE

v

ACKNOWLEDGMENTS

vi

EXPLANATION OF SYMBOLS

C h a p t e r 1.

vii

MODELING SYMMETRICAL PATTERNS AND OF MOLECULES A N D C R Y S T A L S

GEOMETRIES

INTRODUCTION

1

Symmetrical patterns in molecular structures Symmetrical patterns in crystals

1 3

A MATHEMATICAL DESCRIPTION OF THE GEOMETRIES OF MOLECULES AND CRYSTALS

8

Geometric three-dimensional space Vector addition and scalar multiplication Triples Space lattices Vector spaces Vector space bases The one-to-one correspondence between S and R 3 Coordinate axes LENGTHS AND ANGLES

25

Inner product Metrical matrix Cross product Triple scalar product

Chapter 2.

9 10 11 12 16 18 21 23

25 26 34 38

S O M E G E O M E T R I C A L A S P E C T S OF

CRYSTALS

INTRODUCTION

41

EQUATION OF PLANES AND LATTICE PLANES

42

Lattice planes The equation of a plane Miller indices cf-spacings

42 42 44 46 viii

RECIPROCAL BASIS VECTORS

47

Direct and reciprocal lattices D and D* compared

51 51

CHANGE OF BASIS

56

Zones

63

APPLICATIONS

65

A DESCRIPTION OF THE GEOMETRY OF A CRYSTAL IN TERMS OF A CARTESIAN BASIS

72

Calculation of angular coordinates from crystallographic data DRAWING CRYSTAL STRUCTURES C h a p t e r 3.

75 83

POINT ISOMETRIES - VEHICLES DESCRIBING SYMMETRY

FOR

INTRODUCTION

91

ISOMETRIES

91

Rotations Orientation symbols Compositions of isometries Rotoinversions

92 95 95 96

SYMMETRY ELEMENTS

99

DEFINING SYMMETRY

100

LINEAR MAPPINGS

101

Matrix representations of linear mappings Matrix representations of compositions of linear mappings Algebraic properties of 322 THE CONSTRUCTION OF A SET OF MATRICES DEFINING THE ROTATIONS OF 322

Chapter 4.

THE MONAXIAL C R Y S T A L L O G R A P H I C

105 108 117 110

POINT

GROUPS

INTRODUCTION

123

ALGEBRAIC CONCEPTS

123

Binary operations Groups Symmetry groups

123 125 127 ix

CRYSTALLOGRAPHIC RESTRICTIONS

129

MONAXIAL ROTATION GROUPS

134

Matrix representations and basis vectors Equivalent points and planes

C h a p t e r 5.

THE POLYAXIAL CRYSTALLOGRAPHIC

141 144

POINT

GROUPS

INTRODUCTION

157

PROPER POLYAXIAL POINT GROUPS

157

CONSTRUCTION OF THE DIHEDRAL GROUPS

168

CONSTRUCTION OF THE CUBIC AXIAL GROUPS

173

CONSTRUCTION OF THE IMPROPER CRYSTALLOGRAPHIC POINT GROUPS

. . . .

180

THE CRYSTAL SYSTEMS

183

SCHOENFLIES SYMBOLS

191

THE ICOSAHEDRAL POINT GROUPS

192

C h a p t e r 6.

THE BRAVAIS

LATTICE

TYPES

INTRODUCTION

199

LATTICES

199

A DERIVATION OF THE 14 BRAVAIS LATTICE TYPES

207

Lattices Lattices Lattices Lattices Lattices Lattices Lattices Lattices Lattices Lattices Lattices

invariant invariant invariant invariant invariant invariant invariant invariant invariant invariant invariant

under under under under under under under under under under under

7 2 3 4 6 222 322 U22 622 23 432

208 208 213 217 220 220 222 222 222 222 222

THE 14 BRAVAIS LATTICE TYPES

223

MATRIX GROUPS REPRESENTING THE CRYSTALLOGRAPHIC POINT GROUPS

x

...

225

Chapter 7.

THE CRYSTALLOGRAPHIC

SPACE GROUPS

INTRODUCTION

229

TRANSLATIONS

229

ISOMETRIES

237

CRYSTALLOGRAPHIC SPACE GROUPS

249

CRYSTALLOGRAPHIC SPACE GROUP OPERATIONS

258

THE CRYSTALLOGRAPHIC SPACE GROUP TYPES DERIVED FROM THE ONE-GENERATOR POINT GROUPS

262

THE CRYSTALLOGRAPHIC SPACE GROUP TYPES DERIVED FROM THE TWO-GENERATOR POINT GROUPS

276

THE CRYSTALLOGRAPHIC SPACE GROUP TYPES DERIVED FROM THE

THREE-GENERATOR POINT GROUPS

295

Appendix 1.

MAPPINGS

303

Appendix 2.

M A T R I X METHODS

309

OPERATIONS

310

SOLVING SYSTEMS OF LINEAR EQUATIONS

312

Reduced row echelon matrices

317

DETERMINANTS

326

INVERSES Appendix 3.

330 C O N S T R U C T I O N A N D I N T E R P R E T A T I O N OF M A T R I C E S REPRESENTING POINT ISOMETRIES

INTRODUCTION

339

Cartesian bases General bases

339 340

INTERPRETATION OF MATRICES REPRESENTING POINT ISOMETRIES Cartesian bases General bases

342 342 345

PROOFS OF MAIN RESULTS

352

xi

Appendix 4.

POTPOURRI

HANDEDNESS OF BASES

357

DISCUSSION AND PROOF OF T6.15

358

Appendix 5.

SOME P R O P E R T I E S OF LATTICE PLANES

361

Appendix 6.

INTERSECTION ANGLES BETWEEN ROTATION A X E S

DIHEDRAL GROUPS

371

CUBIC AXIAL GROUPS

373

Appendix 7.

EQUIVALENCE RELATIONS, COSETS AND FACTOR GROUPS

EQUIVALENCE RELATIONS

379

EQUIVALENCE CLASSES

382

COSETS

385

FACTOR GROUPS

389

Appendix 8.

ISOMORPHISMS

395

REFERENCES

399

NEW to the REVISED EDITION SOLUTIONS TO PROBLEMS

402

INDEX

456

xii

CHAPTER 1

M O D E L I N G S Y M M E T R I C A L P A T T E R N S A N D GEOMETRIES OF MOLECULES AND

CRYSTALS

"All things are made of atoms - little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repeling upon being squeezed into one anotherRichard Feynman INTRODUCTION

Symmetrical principle

Patterns

in

Molecular

Structures:

The

minimum

an arrangement wherein the total energy of the resulting is minimized. gregate,

energy

states that the atoms in an aggregate of matter strive to adopt configuration

When such a condition is realized, the atoms in the ag-

whether

large

or small in number, are characteristically

peated at regular intervals in a symmetrical pattern.

re-

In recent years,

molecular orbital methods have had great success in finding minimum total energy structures for small aggregates

(molecules), using various algo-

rithms for optimizing molecular geometry (cf. Hehre et at., references therein).

1986

and

Not only do these calculations reproduce known mo-

lecular structures within the experimental error, but they also show that when the total energy is minimized certain atoms in the molecule are repeated at regular intervals in a symmetrical pattern about a point, line or a plane.

( E 1 . 1 ) Example - Repetition of a p a t t e r n at r e g u l a r i n t e r v a l s about a point and a line:

Monosilicic acid, H^SiO»,

is a small molecule whose atoms

are repeated at regular intervals in a symmetrical pattern about a point and a line (Figure 1.1).

An optimization of the geometry of this molecule

using molecular orbital methods shows that its total energy is minimized when an OH group of the molecule is repeated at regular intervals of 90° about a point at the center of the Si atom and at regular intervals of 180° about a line to give the molecular structure displayed in Figure 1.1. The repetition of the group at regular intervals about a line is called rotational symmetry, whereas that about a point is called rotoinversion symmetry.

Rotations and rotoinversions and how they can be used to define

the symmetry of this molecule will be examined in Chapters 3 and 4.

F i g u r e 1.1 ( t o the l e f t ) : as

determined

in

near

A drawing of the molecular structure of monosilicic acid, Hj.SiOj,, Hartree-Fock molecular orbital calculations

O'Keeffe and Gibbs, 1984).

(Gibbs et al., 1981;

The intermediate-sized sphere centering the molecule represents

Si, the four largest spheres represent 0, and the smallest spheres represent H.

The sizes

of the spheres in this drawing or in any other drawing in this book are not intended to mimic the actual sizes of atoms in molecules or crystals.

F i g u r e 1.2 ( t o the r i g h t ) :

A drawing of the structure of a tricyclosiloxane molecule com-

posed of three "tetrahedral" Si(0H) 2 groups bonded together into a 6-membered ring.

The

large spheres represent 0, the intermediate-sized spheres Si, and the small ones H.

The

structure of the molecule was determined by O'Keeffe and Gibbs (1984), using molecular orbital methods.

( E 1 . 2 ) Example - Repetition of a p a t t e r n at r e g u l a r i n t e r v a l s about lines and across planes:

Tricyclosiloxane, H 6 S i 3 0 3 , is an example of a mole-

cule whose atoms are repeated at regular intervals in a symmetric pattern about

lines and across planes (Figure 1.2).

The total energy of this

molecule is minimized when a "tetrahedral" Si0 2 H 2 group is repeated at regular intervals about a line to form a planar 6-membered ring of three Si and three 0 atoms.

Imagine a line drawn perpendicular to the plane

of the ring and passing through its center.

Note that the atoms of the

molecule are repeated in a symmetrical pattern about this line at regular intervals of 120°. repeated

Also note that the hydrogen atoms of the molecule are

across the plane of the ring.

The regular repetition of the

structure across a plane is called reflection symmetry.

There are other

lines and planes in the molecule about which the atoms of the molecule are repeated, but a study of these elements will be deferred to Chapter

2

Symmetrical Patterns in C r y s t a l s :

When a large aggregate of atoms (typ-

21

ically "lO ") strives to adopt a configuration wherein the total energy is minimized, we may again find that some atomic pattern of the aggregate is similarly

repeated

across planes.

at

regular

intervals about points and lines or

But, unlike a molecule, the pattern may also be repeated

at regular intervals along straight lines to produce a periodic pattern of atoms in three dimensions. is said to be a crystal

Such a three-dimensional aggregate of atoms

or a crystalline

solid,

whereas the actual atom

arrangement of the solid is referred to as its crystal book, we shall only be concerned with ideal crystals.

structure.

which contain a variety of flaws and irregularities, envisaged as being perfect in every respect.

In this

Unlike real crystals ideal crystals are

Also the atoms in such a

solid are envisaged to repeat indefinitely at regular intervals in three dimensions.

In addition, the positions of the atoms in such a crystal

are envisaged to be static and to be specified exactly by a set of atomic coordinates.

Thus, when we make reference to a crystal or a crystalline

solid, we shall be referring to an ideal crystal. structure

at

regular

intervals

along

The repetition of a

straight

lines

is

called

translational symmetry, a subject discussed in Chapters 6 and 7. (El.3)

Example - Repetition of a pattern at regular

straight line:

intervals

along

a

As stated above, a characteristic feature of a crystalline

solid is that some atomic pattern in a large aggregate of atoms is repeated at regular intervals along straight lines.

Consider the string

of silicate tetrahedra in Figure 1.3 isolated from the crystal structure of

a-quartz, a common mineral of Si02 composition comprising about 12

percent of the continental crust.

Fix your attention on any two adjacent

tetrahedra in the string and imagine that this pair of tetrahedra is repeated at regular intervals indefinitely in the direction of the string. Whenever some atomic pattern in a crystal (like the pair of tetrahedra in the string) is repeated along some line at regular intervals, it is convenient to represent such a periodic pattern by a directed line segment (a vector) parallel to the string and whose length equals the repeat unit of the pattern.

Such a vector (labelled a in this case), drawn next to

the string of tetrahedra in Figure 1.3, is called a translation vector. (El.4)

Example - The c r y s t a l s t r u c t u r e of a-quartz:

a-quartz crystal structure is presented in Figure 1.4.

3

A drawing of the As observed for

Figure 1.3: A string of silicate tetrahedra isolated from the crystal structure of aquartz. Each silicon atom (small sphere) in the string is bonded to four oxygen atoms (large spheres) disposed at the corners of a SiCU silicate tetrahedron. The lines connecting the atoms in the string represent the bonds between Si and 0. Direct your attention on any two adjacent tetrahedra in the string and note that this pair of tetrahedra is repeated at regular intervals so that the repeat unit is given, for example, by the distance between Si atoms in alternate silicate tetrahedra. The repeat along the string is represented by a parallel vector a whose magnitude» cr, equals the repeat unit along the string. The ellipses, (...), at both ends of the string indicates that this sequence of silicate tetrahedra repeats indefinitely in the direction of the string, even though the sequence is terminated in the figure.

? •CTi J—O

cS

w? W W

w?

Figure 1.4: The crystal structure of a-quartz projected down a direction in the crystal along which the pattern of atoms displayed in the drawing is repeated. The SiOfc groups in the structure share corners and form spirals of tetrahedra that advance toward the reader and that are linked laterally to form a continuous framework of corner silicate tetrahedra. The vector a represents the repeat unit of the structure in the direction of a; b represents the repeat unit in the direction of b.

4

F i g u r e 1.5: A view of the a - q u a r t z structure tilted about 2° off the viewing d i r e c t i o n in Figure 1 . 4 . The repeat unit along the viewing d i r e c t i o n is represented by the vector c w h o s e m a g n i t u d e equals the separation b e t w e e n equivalent atoms in the lines of atoms paralleling c. Because crystals consist o f periodic three-dimensional patterns of atoms disp o s e d along w e l l - d e f i n e d lines, w h e n e v e r a crystal like a - q u a r t z is v i e w e d along one of these d i r e c t i o n s , the arrangement w i l l b e simplified by the fact that only the atoms in the repeating u n i t will b e seen as in F i g u r e 1 . 4 . Also, the shorter the repeat u n i t along such a v i e w i n g d i r e c t i o n , t h e simplier the arrangement because the repeat unit must involve fewer atoms. W h e n the crystal is t i l t e d off this direction, t h e n the view becomes m u c h more c o m p l i c a t e d w i t h the u n c o v e r i n g of the m a n y atoms that lie b e n e a t h the atoms in the repeating unit. A l t h o u g h the repeat p a t t e r n of the structure d i s p l a y e d in Figure 1 . 4 and 1 . 5 is finite, the p a t t e r n is assumed to continue uninterrupted in the d i r e c t i o n of a , b a n d c indefinitely in a ideal crystal.

monosilicic acid, each Si atom in the structure is bonded to four nearest neighbor 0 atoms disposed at the corners of a tetrahedron.

In addition,

as each 0 atom is bonded to two nearest neighbor Si atoms, the structure can

be

viewed

as

a

framework

structure

of

corner

sharing

silicate

tetrahedra.

As the structure in Figure 1.4 is viewed down one of the lines along which an atomic pattern

is

repeated, the atoms along this line are one

on top of the other so that only the atoms in one repeat unit along the line

are visible.

Thus beneath each Si atom displayed in the

figure,

there exists a line of Si atoms equally spaced at regular intervals that extends indefinitely.

Likewise beneath each oxygen atom in the

figure,

there exists a comparable line of oxygen atoms also equally spaced at the same

regular

intervals.

However, when the structure

in Figure

1.4 is

tilted off the viewing direction by about 2°, the repeating pattern of atoms along this direction is exposed as in Figure 1.5.

5

By convention,

the repeat unit along each of these lines is represented by a parallel vector c whose length, c, is equal to the separation

(the repeat unit)

between adjacent atoms in the lines of equivalent Si atoms and 0 atoms that parallel the viewing direction. Returning to Figure 1.4, note that the tetrahedra running across the figure in strings from left to right are exact replicas of the ones comprising the string in Figure 1.3.

In fact, several such parallel strings

are displayed in the figure running left to right. another

such

set

of

strings

of

tetrahedra

In addition, note that

runs

across

the

figure

diagonally from the lower right to the upper left and another runs from the lower left to the upper right, both intersecting the first set strings at exactly 120°.

of

As shown in Figure 1.4, the unit repeat along

the left-to-right trending set of strings is represented by the translation vector a ,

whereas that along the diagonal set of strings running

from the lower right to the upper left is represented by the translation vector b.

In addition to defining the

repeat

unit

along

strings

of

tetrahedra, the pattern at both ends of these vectors is exactly the same regardless of their location in the structure provided the vectors have not been rotated from their original orientations.

This property can be

illustrated by laying a sheet of tracing paper on Figure 1.4, tracing a and

b

on

it and then sliding

(translating) the sheet over the figure

making sure that the vectors on the sheet are kept parallel with those on the drawing.

Note that the pattern at the end points of both a and

b is an exact copy of that at their origin regardless of the placement of the vectors on the drawing, provided that the vectors on the sheet are kept parallel with those on the figure. The lengths

(magnitudes) a, b and c of the translation vectors D =

{a,b,c} and the interaxial angles between them denoted a =

CN

( c )

D

"-1

-1

1

1

0

0

1

0

t

"l" (b)

D

=

0

o' 1

»

0

D

=

-1 0

20

0

>

0

1

"-21

1 ( d )

"0

y

2 0

0' >

0 1

T h e one-to-one correspondence between S a n d R3:

Let D = {a,b,c} de-

note a basis for S and let W denote any vector in S.

Then

Wi w2

[W],

w

where

the

unique

representation

of

vectors of D is W = W j a + w 2 b + w 3 C . R

as a linear combination of the

Hence corresponding to each vector

there is a unique triple [ W ] ^ in R3.

in S, 3

w

>.

Conversely, each vector in

is the triple of some linear combination of D and hence gives rise

to a unique vector in S.

Consequently, we have the one-to-one

corre-

spondence between S and R3 given by

w

[w].

or written another way: Wi

Wia + w 2 b + WjC

w2 w3

This one-to-one correspondence is the fundamental reason why R3 can be used to model three-dimensional geometric space S. We have already seen that S and R3 are both three-dimensional vector spaces.

As such, each has a vector addition and scalar product defined

on it.

One important property of the W

it

"preserves"

these

vector

space

< +

-

[ w ] ^ correspondence is that

operations.

V £ S, then u + v corresponds to [u]^ + [v]^.

For

example,

if

U,

Hence, to add two vectors

in S we first find their corresponding triples, add the triples and then convert

that sum into its corresponding vector in S.

This eliminates

the necessity of using the clumsy parallelogram rule for adding geometric vectors. x[u]Q.

Similarly,

if

U z S

and

x e R,

then

xu

corresponds

to

To formalize and justify these remarks we have the following

theorem. (T1.9) T h e o r e m :

Let D = {a,b,c> denote a basis of S.

lowing two statements are true:

21

Then the fol-

[u + V]Q = [U] D + [v]Q

(1) (2)

Proof: u,,

t

x u

]/)

=

f o r

Let u, V e S.

U2,

U3,

VL

V2

V,a + v 2 b + V 3 C.

a 1 1

for all u, v x

z

R

and

u

e

e

S;

S.

Since D is a basis of S, there exist real numbers and

v3

such

that

U = u ^A + u2 b + u, c and V =

Hence

u + v = (ui a + u2b + u3c) + (x^a + y2b + v3c) +

= In R

3

+ (u2 + v2)b + (u2 + v3)c .

(1.6

we have

[u]D =

U1

VI

u2

Vi

and

[u]D + lv]D =

Ui u2 U

"3.

But according to equation (1.6), this

last triple

+ Vi +

3 + V3

is also

[U + V]^.

Hence [u + v ] ^ = [u] D + [v]^.

(PI.7) Problem:

•

Prove part (2) of T1.9.

(PI.8) Problem - Calculating vectors in the lattice of a-quartz: In El.5 we found the triple corresponding to several vectors labeled in Figure 1.14.

Estimate r + U using the parallelogram rule and then determine

[r +

from the figure.

and compare your answers.

Now calculate [r + U]^ using Theorem

T1.9

Use the theorem to calculate each of the fol-

lowing: (a)

[-r] D

(b)

[6r + 2u]0

(c)

[3r - 5 u ] d .

Theorem T1.9 shows that, as vector spaces, S and

are identical.

The mathematical statement of this fact is that S and R 3

are isomorphic

and that the mapping of S to R ! that takes w to [wj^ is an isomorphism. If we choose an origin

0

and a basis

D = {a,b,c}

in

S,

then as ob-

served earlier the vectors 0, a, b and C are fundamental to establishing a frame of reference on S.

To find the triples that correspond to these

vectors,

0 =

we

observe

that

0a + Ob + 0c,

0a + lb + 0c and c = 0a + Ob + lc.

Hence

22

a =

la + Ob + 0c,

b =

[0 ]

D

[a].

=

1

[c]D

D

=

In particular, the basis of R 3 that corresponds to D is

"l 0 0

"0 »

1 0

0 y

0

(1.7)

_1

Consequently, no matter how peculiar the geometric relationship between a, b and c may be, the corresponding triples in R 3 are very simple. This simplicity is a great help in conveniently describing features in S with respect to the established frame of reference.

For example, the

set of all vectors in the space lattice defined by D is represented in R3

as the set Z 3 of all triples with integer entries.

However, caution

must be exercised when lengths of vectors and angles between vectors are considered since, for example, in spite of the great similarity of the three vectors in (1.7), their corresponding vectors a, b and C may vary greatly in length.

We shall discuss how to overcome this problem later

in the chapter. Vectors in S that lie in the direction of a can be written in the simple form xa for some x z R. written

in the

form y b

Similarly, vectors along b and C can be

and ZC, respectively, where y , z E R.

Since

D = {a,b,c} forms a basis, any vector V E S can be written as a linear combination V = xa + y b + ZC.

Hence, geometrically v can be

pictured

as the sum of three vectors each in the direction of a basis vector (see As the correspondence between S and R3

Figure 1.15(a)).

can decompose a vector in R3

[xa]

D

in a similar manner.

[yb]D =

suggests, we

Since

[ZC].

and since these are scalar multiples of the basis vectors in (1.7), we have the correspondence pictured in Figure 1.15(b). Coordinate A x e s :

Let D = {a,b,c} denote a basis for S.

As in the case

of a-quartz, a, b and c will usually be chosen so that each lies in an 23

Z=

\rsc\r,eR\

v = >-a+/b+zc

\rt€f\

Figure 1.15.

A graphical representation of the basis vectors and coordinate axes in (a)

geometric space 5 and (b) R1 .

In (a), basis vectors a, b and C are directed along coor-

dinate axes X, Y and Z, respectively. are displayed with [ a ] [ b ] p there

exists

xa + yb + zc.

three

real

In (b), the triple representatives of these vectors

and [c]^ placed along X, Y and Z.

numbers

X,

y

and

z

such

that

V

For each point V t S, can

be written

as

V =

Likewise, for each vector V t S, there exists a triple representative of

v, [v]^ E R3 such that [v] D = x[a\D The

purpose of this

V E 5 and [V]^ t R1.

figure

is to show

+ y[b)D

+ z[c]D

the correspondence that exists between vectors

No geometrical significance should be attached to Figure

other than as a model for Figure 1.15(a).

1.15(b)

The vectors in 5 have length and direction and

an angle can be defined between any two nonzero vectors in S.

On the other hand, as R J

is just a set of triples, there is no intrinsic meaning to the notions of the length of a vector or the angles between vectors in RJ.

24

important direction relative to the structure of the crystal under study. Since these directions will be important to us, we call the set of all points

lying

along them coordinate axes.

For example, the X-axis is

defined to be the set of points lying on a line passing through 0 including a. and

is

shown

| r2

{r 2 b

and

That is, the X-axis is the set of vectors {rja | Ti s in

E R)

Figure

1.15(a).

The

V-axis

is

defined

to

and the Z-axis is defined to be {r 3 c | r 3 E /?}.

X-axis consists of the vectors r = r ^

+ r 2 b + r 3 c where r2

the triples for the points on the X-axis in R

3

R) be

As the

= r3

=

0,

are (see Figure 1.15(b)):

ri X =

{

0

I r1

e

R)

0 Similarly, the triples for the points on the Y-

V =

respectively.

{

I r2

e R]

and

Z =

{

and Z-axes in R3

I r3

e

are

R)

By convention, the interaxial angles between these coor-

dinate axes are denoted a = b")b + ((a")'* c )c

53

.

(2.16)

Since ( D )

is the reciprocal basis with respect to D

, equations

of

(2.10) yield a"' (a")" = 1 and b"* (a")" = c"* (a")" = 0

Since the inner product is commutative, (2.16) becomes (a )

= a.

Sim-

ilarly (b")" = b and (c")" = c. Hence (D")" = {a,b,c}.

In

o

view of T2.6 and the definition of a reciprocal basis we have

shown that a = (b x c")/ [a"'» (b"x c*)] b = (c*x a")/[a** (b"x c"")] c = (a"x b' )/ [ a ( b " x

Let H denote the metrical matrix for D

H[r]D*

for all r s S.

=

But by T2.6, (O")" = D.

H[r]c*

for all r z S. is

[r]

=

Hence

( D

By (2.15),

y

Hence

[r] D

Also, by (2.15), G [ r ] D =

the metrical matrix for D .

.

c")]

[r] D * for all r E S where G

[r]^ = G "^[r]^* (see Appendix 2

for a discussion on the inverse of a matrix and why, since det(G) i 0, G

1

exists).

Therefore, H[r]^- = G

equality holds for all r E S, H = G

1

.

for all r e S.

Since the

Hence we have proved the following

result.

(C2.7) Corollary:

If G is the metrical matrix for D,

then G

1

is the

metrical matrix for D .

The fact that the metrical matrix of D

is the "reciprocal" of the

metrical matrix of D is consistent with calling D of D. of D

The metrical matrix of D

is denoted G .

the reciprocal basis The complete geometry

is described in terms of that of D in the following theorem.

54

(T2.8)

Theorem:

Let D = {a,b,c} denote a basis

{a ,b ,C } denote its reciprocal basis. written in terms of that of D

of

S

and let

Then the geometry of D

D

=

can be

in the following way:

a

=

(fec/y)sina

;

b

=

(ac/v)sin|3

;

c

=

(ab/v)sin!f

;

cosa

=

(cosgcosi - cosa)/sinPsini

cos (5

=

(cosacos? - cosf3)/sinasini

cosJf

=

(cosacosfJ - cosi)/sinasin3

where y = a • (b X c) .

Proof:

Since

||b x c | = facsina and

a • (b x c) =

v

(see

(1.15)

and

(1-17)), a

= (||b x c||)/(a • (b x c)) =

The expressions for b cosa , we

shall

(foc/v)sina

and c can be similarly established.

use the fact that b

• c

= b c cosa

To calculate

and the vector

identity (White, 1960, p. 76) x • z

x • w

y

y • w

(2.17)

(x x y ) • (2 x w ) = • z

Hence cosa" = (b" •

C")/(b"c") =

( v 2 b " • c")/( o 2 t>csingsini)

(2.18)

But b" • c"

=

(1/v 2 ) (c x a) • (a x b )

=

(1/v 2 )

c • a a • a =

(a 2 bc/v2)(cos&cosi

c • b a • b

occosp =

Q/V2)

- cosa)

2

a

cbcosa

abcosZ (2.19)

Substituting (2.19) into (2.18) we obtain

cosa

= (cosfScos? - cosa)/sin(5siny .

The other parts of the theorem are proved in a perfectly similar manner.•

55

Because of the * duality between D and D

*

discussed in T2.6, any ex-

= (cosScosi - cosoO/sinPsini can be used to write the

pression like cosa equality cosa

=

(cosfi* cosi * — cosa *)/sin3 * sini *

cosfS

=

(cosa* cosi * — cosP * )/sina * sini *

cosï

=

(cosa* cosft* — cosï * )/sina * sinp *

Similarly, and

CHANGE OF BASIS We learned in Chapter 1 that there are innumerable directions in a crystal along which a representative pattern of the structure is repeated at regular intervals.

Furthermore, not only do these directions define

a

system along which three non-coplanar coordinate

natural

coordinate

axes can be placed, but the repeat unit along each of these directions defines a natural set of basis vectors D = {a,b,c}.

Thus, for a given

crystal, innumerable sets of axes, basis vectors and planes can be chosen, depending on the problem to be studied.

In addition, as there are innu-

merable ways of describing such a crystal, it is essential to be able to change its geometric framework with respect to the observations made.

Moreover,

being

if we wish to study crystal structure relationships

(Smith, 1982), polymorphism, phase transformations

(Hazen and Finger,

1982) or defects (Lasaga and Kirkpatrick, 1981), it is imperative that these features be related to a single set of basis vectors.

In partic-

ular, if data are provided by one investigator describing some feature of a crystal with respect to one basis and if another investigator provides a similar set of data on another such crystal but defined with respect to a second basis, it is important to express both data sets with respect to a common basis in order to make comparisons and to draw conclusions.

In other words, we must be able to make a change of basis.

In the last section we saw how, given a basis D = {a,b,c} and its reciprocal

basis D

*

*

*

*

= {a ,b ,C }, we can easily translate geometrical *

information given in terms of D into the language of D .

For example,

if r t S and if we know [i*]^ then we can find [r]^* by, see (2.15), G[r] 0 = [r] D *

56

,

where G is the metrical matrix of D.

In this section, we will see how

to translate geometrical information from a basis D j = { a ^ b ^ C ] }

to a

- {a2,b2,c2}.

basis D2

Just as in the special case when be found that will convert

[r]n

= D and D2

into [ r ] n

basis D2

, a matrix will

for all r E S and this matrix

will be used to calculate the metrical matrix of D2 (D2.9) Definition:

- D

from that of

D^.

The change of basis matrix from the basis D i to the

is the matrix T such that

T [ r ] D i = [r] for all r

t

D ,

S.

Since we know what job T is supposed to perform, we can find T by observing that

112

' 11

t23

0

,'il

t,2

f

0_

M l

112

tl3

"o"

*2 1

t2 2

t2 3

1 0

3 3.

.'¡1

tl2

* 3 3.

'll

til

tl3

"o"

'2 1

t2 2

Î23

0

.'3 1

t 32

=

= 1st

column

'til' =

t21

\

T

of

1st

column

of T

=

2nd

column

of T

=

3rd

column

of T

>12 t

=

2 2

(2.

Î32.

tl3 =

Î23

.

.tu.

t [010] t , and [ C l ] D i = [001]

=

bi

[bi]0l

=

.tn.

1

[100]

T [ a 1 ] 0 i = [a 1 ] £ ? 2 , u

~1

1

Di

[ai]n

tl3

t22

=

T,

[bl]

[b^p. u

2

and

D2

T[Cll

D1=

[Ci]

u

basis,

exists

rlt

r2,

have

= 2nd column of T and [Cj],,

Now let r denote an arbitrary vector in S .

there

we

2

column of T.

r, E R

By T1.9, we have [r]^ = r j a j ^

D2

such that r = r ^

+ ^[bj^

57

+ ^[c^^.

= 3rd 2

Since D i

is a

+ r 2 b j + r3Cj. But then

T[r]Di=

T(r1[a1]0i + M b , ] ^

+

c , ^ ^

=

r 1 T [ a 1 ] D i + r 2 T [ b , ] D i + r,T[c,]0i

=

r1[*i]Di + r2[b,]Di + r3[Cl]D2

=

[r^i J

+ r2b! + r,Ci]n D,

•

Therefore, the fact that T "works" for [a,]- , [ b ^ n U

that it "works" for all [ r ] n , r e S.

L>1

x

and [ c ^ , ,

L>1

implies

Consequently, w e have established

the following theorem.

(T2.10) T h e o r e m :

D2

= {a2,b2,c2>

The change of basis matrix T from

= {a^b^Cx}

to

is

T =

t a i] r>

If T is the change of basis

[biln

matrix

from

f c i]/

Di

to

D2,

then

it

is

straightforward to show that T ^ exists (since the columns of T are linearly T[r] nu

independent). 1

= [r]_. u

i

by T

Furthermore,

-1

by

multiplying

both

sides

of

, we have

Therefore, T ^ is the change of basis matrix from D2 T2.10 w i t h the roles of D , and D2

to

Applying

switched, we have the following cor-

ollary.

(C2.11) Corollary:

If T is the change of basis matrix from D^

then T * is the change of basis matrix from D2

T_1

=

[a«]Di

58

[bi Id,

to D ] and

^ D ,

to

D2,

As we shall see in a later section on applications, it is often easy to find by inspection the change of basis matrix from Dz

to

in one direction, say

The change of basis matrix in the other direction is then

found by inverting the first matrix. Now that w e have a procedure for finding the change of basis matrix from Di of D2

to D2,

we need to discover how to find the metrical matrix

from the metrical matrix Gj of D x .

expression

for

T

D* = {at,b*,c*>.

using

the

reciprocal

G2

W e begin by obtaining a useful bases

= {a^b^Ci)

and

By T2.5,

b, [aj

0,

ai • b 2 .

bi

[b x ]D

2

*

•

b* ,

Ci • bi

[Ci

bi • Cj

Hence, J.

a

l

•

a2

bi

•

a

l

•

b2

b,

•

•

*

bx

c'i

•

*

bi *

Ci

Ci

•

Ci

•

"ft 32 *

b2

• C

Ci

*

2.

We now have the following equations:

^

^

=

^ d ,

G2[r]D2

=

[r]D*

Gltr]Di

= [r]D*

. -Ij.

Defining S to b e the change of basis matrix from Di

^L

to DZi

w e also have

These relationships are summarized in the following circuit diagram:

59

(2.

-

[••ID,

i

Gi

^D*

flo,

s

*

G

*

tr]

D*

>

According to this diagram,

S G x T ' V ] ^ = GAr]Di for all r E S. S.

By T2.6, D2

Hence, SGjT

-1

= G2 .

Consequently we need to calculate

is the reciprocal basis of D2 • Hence,

^D*

ai "V a.

=

(a 'zy

ai

a2

(b*)*

a'i

b2

(4)*

a'i

C2

Applying this observation to each of the columns of the change of basis matrix from D j to D 2 written in the form of (2.21), we obtain -v ai S =

a'i ai

•

•Jbx • a 2

a2

•

*

b2

•

bi

C

2

•

*

•

a

•

b

Ci

•

c

C2

•

bx

C2

•

.*

c'i

c

•

C

b2

bi

•

c2

b2

•

ai

b2

•

b2

•

Ci *

c'i *

Consider, *

•

t s =

a2

•

a2

•

a

2

ai

.* *

•

Ci

*

. *

2

bi *

l.

According to the analogous statement to (2.21), S*" is the change of basis matrix from D 2 to D x . (T

= (T^) \

That is, S 1 = T _ 1 .

we write T ^.

(T2.12) Theorem:

Since

Hence we have the following theorem.

Let D j and D2

matrix from D j to D2

Therefore, S = (T _1 ) t .

denote bases, T the change of basis

and let G x denote the metrical matrix for D x . 60

Then

the relationship between Gj and G 2 , the metrical matrix for D2,

is given

by Gi = T^G-J

Gj = T ' ^ h T " 1

and

.

We summarize these results in the following circuit diagram. T [r] T^GiT = Gj

[r]

I T

^DÏ (T2.13) Theorem: G.

Then the volume

^D*

Let D = {a,b,c} denote a basis with metrical matrix v of the parallelpiped outlined by D is such that

v 2 = det(G)

Proof:

(2.22)

G i = T" t G 1 T" 1

I _t

.

Let C denote a cartesian basis and let T denote the change of

basis matrix from C to D.

By (1.18) we know that v = det(M) where

M =

[a] r

I

[b],

Note that TM

Hence det(TM) = 1.

[T[a]c

=

=

[[a] D

=

I3

I T[b]c I [b]D

I T[c] c ]

I [c] D ]

•

Therefore,

1

= det(TM) = det(T)det(M) = det(T)v

Consequently, det(T) = 1/v.

Since the metrical matrix for C is I 3 ) by

T2.12, 13

(by (A2.8))

.

= T^T

61

.

Since det(T ) = det(T), 1

= det(T)det(G)det(T) = (l/v2)det(G)

.

v2

Therefore, det(G) =

We are now in a position to show how the cross product is calculated when the vectors are expressed with respect to an In

Chapter

arbitrary

basis

D.

1, we discussed how to calculate v x w with respect to a

cartesian basis. (T2.14) Theorem:

Let D = {a,b,c} denote a basis with metrical matrix

G and let r and S denote vectors in S.

Then a

det(G) -i

r x s

b

G[r]

c Proof:

D

(2.23)

G[S]

D

Let rj, r 2 , r 3 , Sj, s 2 and s 3 E R be such that JU

X

JL

r

=

rja

+ r2b

+ r3c

,

s

=

Sia

+ s2b

+ s3c

,

and

J-

where D

rV

= {a ,b ,C } is the reciprocal basis for D. r x s = =

(ria" + r 2 b" + r 3 c") x (ija" + 5 2 b" + s3c") Cr253 - r 3 s 2 )(b" x c") + ( r 3 s 1 - r^j)c" x a" + (r^j - s ^ ) a

where we have used Chapter 1.

Then

the properties

x b

,

of the cross product discussed in

By T2.6,

r x 5 = v"[(r 2 s 3 - r 3 5 2 )a + (r3Si - ri5 3 )b + (rjS2 - 5^2)0] where v

= a

• (b" X C ).

Hence,

62

,

r x s = v

By T2.13, (v")2 = det(G").

a

ri

Si

b

r2

s2

c

r3

s3

By C2.7, G" = G _ 1 and so =

(detCG*))*

=

(detCG"1))^ (det(G))"i

thus the theorem is established. Let D = {a,b,c} denote a basis with metrical matrix G.

V 2 = det(G) =

a2

abcosTS

abcosi

b2

becosa

accosg

bccosa

c2

By T2.13,

accosfl

and so we find that V =

obc(1

(P2.8) Problem:

- cos2a - cos2f5 - cos2Y + 2cosacosf5cosi)^ Show that (v")2 =

the parallelepiped outlined D V

= a b c (1 - cos a

det(G") where v" is the volume of

= {a ,b ,C } and that

— cos &

— cos i

+ 2cosa cosfS cosi• M )

-1 * By C2.7, G is the metrical matrix for D . This reciprocal relationship between the metrical matrices of D and D yields the reciprocal volume relationship V2 = det(G) = l/det(G_1) =

1/v*2

and therefore v = 1/v . Zones: indices

Suppose the nonparallel lattice planes P^ and P2 k^ 2.x) and (h2

k2

have Miller

H2). respectively, then the zone axis de-

fined by them is in the direction of a vector parallel to the intersection 63

of Pi and P 2 .

Hence if

si

=

h^a

+ /Cib

+ £iC

s2

=

h2a

+ Ar2b" + Z2c"

,

and ,

then the zone axis is in the direction of SI X S 2 . (E2.15)

Example

lattice v e c t o r s :

-

The

vector p r o d u c t of two non-collinear reciprocal i Show that det(G) 2 (S! X s 2 ) is in the direct lattice L p

where G is the metrical matrix for D. Solution:

By T2.14, a 2

det(G) (S! X S 2 )

b

=

G[ S l

'D

G[S2

'D

c a

hi

h2

b

ki

k2

c

n2

Hence det(G) ? (S! X s 2 ) = ua + vb + W C where

Hi

Since hi, k i y llt

k2

h2

hi

h2

«2

ii

ki

k2

h2,

k2

and l2 are all integers, so are u, v and w .

Hence det(G)^(S! X S 2 ) c L P .

c

The U, v and w in this example define the zone axis.

To distinguish

such a set of integers from a set defining a crystal plane (/?/(£), the integers for the zone are enclosed between a pair of brackets [t/vw] . (E2.16) by

two

Example - F i n d i n g non-collinear

a vector p e r p e n d i c u l a r to the plane

vectors

in

a

crystal:

The

lunar

defined mineral

pyroxferroite (Fe,Ca)Si03 is triclinic with cell dimensions a = 6.621A,

64

b = 7.551A, c = 17.381A, 1971).

a = 114.27°,

3 = 82.68°,

X = 94.58°

(Burnham,

Calculate the cross product of the following two vectors r3„

= -0.0963a + 0.1243b + 0.2018c

rs„

=

0.1084a - 0.0880b + 0.2947c

to find a vector perpendicular to their plane. Solution:

Calculating G with the cell dimension of pyroxferroite, we find 43.8376 G =

and (detG)2 = 7 8 5 . 3 4 6 6 .

r3„ x r 5 4

=

-3.9922

14.6624

-3.9922

57.0176

-53.9461

14.6624

-53.9461

302.0992

Hence,

(785.3466)

-1

a

-1.7589

9.4243

b

-3.4146

-21.3482

c

52.8461

95.3653

= 1.0219a + 0.8478b + 0.0888c is a vector perpendicular to the plane of r 3 1 ( and r 5 1 i . (P2.9)

Problem:

Calculate the cross product

k x i

for pyroxferroite

given that k

=

0.1166a + 0.0968b + 0.0101c

i

= -0.0286a + 0.0369b + 0.0599c

and .

APPLICATIONS

In the previous section, we introduced a method for finding the entries of a transformation matrix T.

In practice, however, the entries

of such a matrix may be found by making a careful drawing of a crystal structure or its lattice representation that shows how one set of basis vectors is related to another set.

Then by examining how the vectors of

one basis can be expressed as a linear combination of the other, the coefficients of the second basis vectors are used as columns in the ma-

65

X,

Figure

2.4:

A drawing of the oxygen atoms (stippled spheres) in the

kyanite structure idealized so that they are close-packed at the corners

19

of a cubic face-centered array (modified after Taylor and Jackson, 1928). The Di = ( a ^ b x , ^ ) basis vectors directed along Y i,

Z i,

outline

(a 2 > b 2 ,Ci) directed

a

unit

cell

of

along Xlt

the

kyanite

coordinate

axes

structure

and

outlines a cell of the

Xlt O2 = cubic

face-centered array of oxygen atoms.

trix.

This method will be illustrated with an example.

Another method

for finding T is illustrated in EA2.23. (E2.17) Example - Determination of the cell dimensions of a subcell of the kyanite structure:

The oxygen atoms comprising the structure of kyanite,

Al 2 Si0 5 , can be viewed as a slightly distorted cubic close-packed facecentered structure with Si occupying 10% of the available tetrahedral voids and A1 occupying 40% of the available octahedral voids (Taylor and Jackson, 1928).

Before the structure of this mineral was solved, the

close-packed nature of the structure was theorized because the dimensions of

the

unit

cell

of

kyanite

outlined

by

the

natural

basis

Di =

{aijb^Ci} are similar to those calculated from a set of basis vectors Di

=

i a 2>b2C2) defining a face-centered subcell of oxygen atoms. The relationship between the D2 -basis of the subcell and the

basis of the kyanite cell is displayed in Figure 2.4.

Dx-

Inspection of this

figure shows that the D^basis vectors can be written in terms of the D2-basis vectors as follows: ax

= 3/2a2 - l/2b2 + c 2

bx

= 2b 2

Cj

= - a 2 + c2

Cfi

.

A recent measurement of the cell dimensions of kyanite at 25°C

(Winter

and Ghose, 1979) yielded the values a x = 7.126A, b ! = 7.852A, c x = 5.572A, Oi = 89.99°, 3! = 101.11° and

= 106.03° for D1.

W i t h this information,

calculate the cell dimensions of the subcell defined by the oxygen atoms in kyanite.

Using T2.10, we can construct the change of basis matrix

Solution:

[ai]

3/2

0

-1

-1/2

2

0

1 0

from Dx

to

1

D2

Using the cell dimensions for D j , its metrical matrix Gj is

Gx

=

50.7799

-15.4510

-7.6511

-15.4510

61.6539

0.0076

-7.6511

0.0076

31.0472

To compute the cell dimensions of the face-centered cell, we need to solve for the metrical matrix G 2 of D 2 .

According to T2.12,

G2 = T"tG1T"1

Inverting T (see Appendix 2),

3/2 T"

1

=

-1/2 1

.

(2.

we find that

0

-1

2

0

0

1

-1 =

' 2/5

0

2/5

1/10

1/2

1/10

-2/5

0

3/5

Substituting T " t , Gj and t " 1 into (2.24) we have

67

2/5 0

2/5

1/10

-2/5

50.7799

-15.4510

-7.6511

0

-15.4510

61.6539

0.0076

3/5

-7.6511

0.0076

31.0472

1/2

1/10

2/5

0

2/5

1/10 1/2 -2/5

1/10

0

3/5

14.9205

-0.0090

-0.5580

a2

a2b2cosZ2

a2c2cosfl2

-0.0090

15.4135

-0.0052

o 2 b 2 cosi 2

2 b2

b2c2coscc2

-0.5580

-0.0052

15.0106

a 2 c 2 cosP 2

b 2 c 2 cosa 2

2 c2

Equating the entries of these two matrices and solving for the cell dimensions of the face-centered subcell, we find that a2 = 3.8627A, b2 = 3.9260A, c 2 = 3.8743A, o 2 = 90.02°, & 2 = 92.14° and Z2 = 90.03°.

As is

evident from these results, the departure of the oxygen atoms in kyanite from

an

ideal

cubic

close

packed

subcell

with

o 2 = b2 = c2

and

o 2 = P 2 = y 2 = 90° is small. (P2.10)

o

Problem: The crystal structure of the relatively rare mineral

sapphirine, (AlMg)h(Al2Si)202„, can also be described as a slightly distorted cubic close-packed array of oxygen atoms

(Moore,

1968).

The

equations that define the basis vectors of the sapphirine structure Dx = {a 1 ,b 1 I c 1 } in terms of the face-centered subcell D2 = {a 2 ,b 2 ,c 2 } are aj = 2a2 + 2b 2 bj = -5/2a2 + 5/2b z c x = -a 2 - b 2 + 2c2 (1)

Using

the

cell

Ci = 9.957A,

dimensions

.

o, = 11.286A,

bx = 14.438A,

= 125.4°, and o, = Ti = 90.0°, measured for the

mineral (Higgins and Ribbe, 1979), show that the metrical matrix defining the geometry of the face-centered subcell is

G2 =

(2)

16.2991

-0.3774

-0.1762

-0.3774

16.2991

-0.1762

-0.1762

-0.1762

16.4722

(2.25)

Evaluate the entries of (2.25) and show that the cell dimensions of

the

subcell

are

a2 = b2 = 4.037A,

o 2 = P 2 = 90.62 and Z2 = 91.33. 68

C2 = 4.059,

(E2.18) Example - Transforming indices of planes with change of basis: The close-packed monolayers of the kyanite structure parallel (111), (111) and (111) of its face-centered subcell.

(111),

Determine the

indices of these planes in terms of the kyanite basis. Solution:

We recall that perpendicular to each stack of planes

{h2k1l1)

there exists a vector S with the coordinates such that

D,

=

To find the coordinates of this vector in terms of the

basis, we use

the equation (see (2.22)) =

^ D *

'

(2

"

where the components of [ s ] n * are the indices of the same plane but defined in terms of the kyanite basis.

Transposing T and substituting

this result into (2.26), we obtain "3/2 -1/2

hi ki

=

Replacing (h2k2l2} we get

(220),

1

h2

0

2

0

k2

-1

0

1

-l2.

3h2/2 - k2/2 + l2 2k2

=

—Aï 2 + l2

in succession by (111), (111), (111) and (111),

(122), (022), and (320).

The integers in (220) and

(022) are reduced to (110) and (011), respectively, because both sets contain a common factor of 2.

With this reduction, we can conclude that

the close-packed monolayers of oxygen atoms in kyanite are parallel to planes with indices (110), (122), (011) and (320).

•

(P2.11) Problem: The close-packed monolayers of oxygen atoms of

the

sapphirine structure parallel (111), (111), (111) and (111) of its face-centered subcell.

Show that these monolayers parallel (100), (052),

(101) and (052) of the sapphirine structure.

69

(P2.12)

Problem:

The crystal structure of tremolite, a calcic amphibole

of C a 2 M g 5 ( S i ^ O ! i ) 2 ( 0 H ) 2 composition, bears a close structural resemblance w i t h that of the calcic pyroxene diopside, C a M g S i 2 0 s , w h e n viewed down [010] w i t h a , C, and (5 being nearly the same in both minerals.

However,

the cell dimensions of the two minerals differ in that the b - c e l l edge of tremolite is double that of diopside.

W h e n Warren

(1930) solved the

structure of tremolite, he took advantage of these relationships and derived the basic structural unit (a double chain) in the amphibole structure by reflecting the diopside structure over a plane perpendicular to [010].

In a paper on this choice of basis vectors, Whittaker and Zussman

(1961) have observed that although

Warren's

choice,

Dj =

{aijbj.Cj},

illustrates the relationship to the diopside structure, it does not conform w i t h the conventional choice of basis, D 2 = { a 2 , b 2 , c 2 } ,

(1)

»2

=

»1 ~

b2

=

b,

c2

=

Cj

defined by

Cl

.

Using the cell dimensions (Oj = 9.78A, b j = 17.8A, Ci = 5.26A. 0! = 73.97°, o! = y x = 90°) determined by Warren, show that the cell dimensions of the conventional cell are o 2 = 9.74A, b2

=

17.8A, c 2 = 5.26A, P 2 = 105.23°, o 2 = Zz = 90°.

(2)

A Ca-rich amphibole like tremolite Ca-poor amphibole adopted by Warren.

is commonly exsolved

lamallae developed along

(3)

Warren's

(001) of the cell

Show that these lamellae parallel

of the conventional cell (cf. Ross et at.,

(101)

1969).

structural analysis of tremolite showed that one of

the Si atoms in the double chain has coordinates 0.01) t .

with

(0.29,

0.08,

Show that the coordinates of this Si atom in the con-

ventional cell are (0.29, 0.08, 0.30)*.

(E2.19) Example - A calculation of the cell dimensions of kyanite assuming its o x y g e n

atoms are c u b i c c l o s e - p a c k e d .

If we assume that the

face-

centered subcell displayed in Figure 2.4 for kyanite has cubic geometry

70

(i.e., the oxygen atoms in kyanite are ideally cubic close-packed so that a

2 ~ b2 = c2, a2 = B2 = 2"2 = 90°) and the oxygen atoms are in

contact

along the face diagonals of the cell, then a2 = 2/2r0 where r 0 is the effective nonbonded radius of an oxygen atom.

If we equate the volume

3

of this cell, v = 16/2r0, with that obtained by evaluating the determi3

nant of G 2 , then 16/2r0 = 58.719A3.

Solving for r„, we get r, = 1.374A.

Our problem is to determine the cell dimensions of kyanite assuming that its oxygen atoms are ideally cubic closest packed each with an effective radius of 1.374A.

As the cell edge of such a cube is 2/2r 0 , then its cell

edges are given by cr2 = 2/2" x 1.374A = 3.886A.

The metrical matrix for

this ideal structure with a2 = b2 = c2 = 3.886A and ot2 = 62

=

=

90.0°

is obtained by evaluating (2.22):

Gi = T t G 2 T 3/2

-1/2

1

15.1010

0.0

0

2

0

0.0

-1

0

1

0.0

0.0

52.8535

-15.1010

-15.1010

60.4040

15.1010

-78.5505 Solving G! for the cell dimensions

-1

0.0

-1/2

2

0

1

0

1

15.1010

0.0

-7.5505 0.0 30.2020

of the

ideal

structure, we obtain t^ = 7.270A, bx = 7.772A, Pi = 100.89° and ïj = 105.50°.

0

0.0

close-packed

kyanite

= 5.496A, c^ = 90.0°,

These values show a close correspondence

with those measured for kyanite (E2.17), indicating the centers of the oxygen atoms in the structure define rather well a face-centered cubic lattice. (P2.13)

(1)

• Problem:

Evaluate the determinant of the metrical slightly

distorted

subcell of

sapphirine

matrix and

G2

for

the

show that its

volume is v2 = 66.125A3. (2)

Equate the volume obtained in (1) with that of a cubic face3

centered cell, 16/2r0, and show that r 0 = 1.430A.

71

2.5: >cal

A set of direct * * * * D

-

Lng from a k

D -

(a,b,c) and

(a ,b ,c } basis vectors each common

origin,

0.

The

unit

is directed along I so that c = ck, i

:ed in the (a.c)-plane perpendicular to C it a = osinPi + ocos&j and j is directed * * * b so that b = D J. No significance be attached to the relative lengths of the * basis vectors.

D

(3) Assuming that the cell edge of the cubic cell is 2/2r0 and that r 0 = 1.430A, show that the metrical matrix for an ideal closepacked sapphirine structure is 130.896

0.0

0.0

204.525

-65.448

0.0

Gi =

-65.448 0.0 98.172

(4) Calculate the cell dimensions of the sapphirine assuming that it consists of a cubic close-packed array of oxygen atoms of radius 11.44A,

1.430A.

Compare

these calculated dimensions

= 14.30A, cx = 9.91A,

= 125.3°, a1 =

(Oj = = 90°)

with those given in P2.10. A DESCRIPTION OF THE GEOMETRY OF A CRYSTAL IN TERMS OF A CARTESIAN BASIS It is often useful to define a natural basis D = {a,b,c} of a crystal in terms of a cartesian basis.

We shall choose the cartesian basis C =

{i,j,k} such that k is in the direction of C, j is in the direction of C X a and i is set perpendicular to the plane of k and j so as to complete a right-handed cartesian basis set (Figure 2.5).

For this setup of the

cartesian basis, there must exist real numbers a., such that the natural basis D for a crystal can be expressed as

72

a = Q u i + o 3 jk (2.27)

b = a 1 2 i + cr22j + o 3 2 k c = oJ3k

Note that by the way in which the C-basis (in particular in regard to the handedness of D

and C )

is defined in relation to the

° n s °2 2 and o 3 3 must all be positive.

D-basis,

Thus, the change of basis matrix

A from D to C is given by

A =

[a],

[e],

[b],

=

On

oi2

0

0

o22

0

o3i

O3 2

033.

(2

To find the entries of this matrix, we use the following circuit diagram (compare with 2.22)

[r ] D G

[r]c \

t \ I iJ A" T " [r]r

[r] D and obtain

G = AtI3A = AtA

(2.29)

Expanding (2.29) we have

a2 G =

b2

occosP

becosa

2 ^ 2

»11 + Oji

alia12

abcost ac cosfi

abcos?

+ o31o32

= AfcA

becosa c2

o,,a,, + a 3,o3 2

2

2

o12 + o22 + a32

1O33 (2.30)

o32o33 2 °33

By equating corresponding elements of (2.30), expressions are found from which the a., entries of A shall be deduced.

We begin by observing that

2 a33 = c 2 and so o 3 3 = c.

Next, replacing o 3 3 in both o 3 2 o 3 3

by c, we find that 0 3 1 = ocosB and o 3 2 =

73

facosa.

and O31O33

Replacing o 3 1 in

2 °n

+

cr32

2 °31 are

= ai

by acosf, we find that O n = osinP.

replaced

by

crsinfJ,

acos(5

and

When a l l t

b cosa,

o 3 1 and

respectively,

(cosi - cosacos|3)/sinfS is replaced by -cosi sina (T2.8), we *

o 12

=

by

-fasinacosJT

-bsinacosi .

2

and

find 2

that 2

Finally, replacing a 1 2 and o 3 2 in o 1 2 + o 2 2 + o 3 2

and facosa, respectively, and simplifying, we find that

o 2 2 = fasinasini .

When these results are substituted into (2.28), the

A matrix becomes denoted a(i is defined

to be the mapping cc3(r) = 1, then ni is denoted

We denote 1/ simply as /'. The names and symbols for some represen-

tative rotoinversions are given in Table 3.2. (1976), we

Following Boisen and Gibbs

assign to n, the turn angle and rotation axis of n.

orientation symbol for n is taken to be that of n. a quarter-turn inversion with the axis of

its

directed along a and with a turn angle of 90°.

Thus, associated

denotes quarter-turn

Likewise, 3 denotes a

third-turn inversion whose rotation axis parallels C.

96

The

Figure 3.3: The action of a rotoinversion a/ on an arbitrary point r e S where 0 denotes the origin, a is a rotation about I with turn angle p and / is the inversion isometry. The vector r is mapped onto —r by / (i.e., /(r) = - r ) and a maps (-r) onto S = a(-r) - a 0 ( f ) ) ~ ai(r). Hence, the rotoinversion isometry ai maps r onto s. The turn angle and the rotation axis of o/ are inherited from a.

Figure 3.4: The action of a reflection isometry m = /2 on an arbitrary vector r t S where 0 is the origin and m is the mirror plane perpendicular to the rotation axis £ of a half-turn 2 over which the vector r is reflected. The vector r is mapped onto s by 2, s is mapped onto -S by / so that /2(r) = -s. Hence ¡2 is equivalent to a reflection m of the vector r directly over the plane m onto -S such that m(r) = -s.

Figure 3.3 shows the effect of a rotoinversion o/ on a point r e S where a has a turn angle of p°.

As the inversion sends r to —r, we observe

that o/(r) = o(/(r)) = a(-r).

Then a rotates —r through a turn angle of

p mapping —r onto s as shown in Figure 3.3. So far we have only considered rotoinversions of the form a/.

This

does not result in loss of generality since, as we shall now show, ia = ai for any rotation a.

Let V denote a vector in 5 and let a denote a

rotation such that a(V) = W.

As -v lies on the same line as V, a(—V)

must lie on the same line as W but point in the opposite direction such that

a(-V) = —W.

—ot(v) = a(—V). /a(V) .

Next, multiplying a(v) = W by -1, we observe that

Hence, a/(V) = a(/(v)) = a(-V) = -a(v) = /(a(V)) =

Therefore, a/ and ia are the same mapping.

that a and / commute.

Since a/ = /a, we say

Hence / commutes with all rotations.

As we shall

see later, isometries do not in general commute and the fact that / commutes with each rotation is an important property of /. All rotoinversions, with the exception of 2, leave exactly one point 97

/(c) W(b) /{aï

/'(c)

Figure 3.5: A collection of vectors that are images of an arbitrary vector r t_S under the point isometries that comprise a 6 inversion axis paralleling £.

in S fixed.

Figure 3.6: An illustration that an inversion isometry i changes the handedness of a basis from right to left. (a) A right-handed basis D = {a,b,c} emanating from an origin 0. (b) The images of a, b and c, {/'(a) ,/(b) ,/(c)} under /. (c) The figure in (b) rotated so that /'(a) and /(b) are parallel to a and b, respectively, in (a). Observe that the basis vectors in D = {/'(a) ,/(b) ,/(c)) are left-handed illustrating that an inversion isometry changes the handedness of a set of basis vectors.

The 2 operation leaves a plane of points fixed, as illus-

trated in Figure 3.4. Rather than viewing 2 as the composition of 2 and /', it is simpler to view the operation as a reflection about the plane of points left fixed by 2.

By a reflection we mean that each point and

its image lie on a line perpendicular to the plane such that the plane bisects the line segment between them. Because of this fact, 2 is almost always referred to as a reflection

isometry and is denoted by m.

thermore, the plane of fixed points is called the mirror

plane m.

FurThe

orientation symbol assigned to m is inherited from that of the axis of the perpendicular half-turn part of the operation.

For example,

^^^m

denotes a reflection of space over a mirror plane that is perpendicular to 2a + b , whereas m denotes a reflection of space about the plane that is perpendicular to C. (P3.1) Problem:

Determine the names and symbols of the point isometries

o that map the vector r in Figure 3.5 onto r , s, t , u, v and W , assuming that the rotation axis £ of a parallels C. 98

Consider the right-handed set of basis vectors D = {a,b,c} displayed in Figure 3.6(a).

The images of these vectors under the inversion is the

set D = {/'(a), /(b), /'(C)} (Figure 3.6(b)).

If we orient D with /'(a) on

the left and /'(b) on the right, then /'(c) will be directed downward (Figure 3.6(c)).

Consequently, the inversion transforms a right-handed basis set

to a left-handed one (see Appendix 4) .

Since rotations do not change the

handedness of a basis, the net effect of a rotoinversion ai is to change the

handedness.

basis.

Hence, all rotoinversions

change the handedness of a

One important consequence of this is that no isometry

is both a

rotation and a rotoinversion.

(E3.3) E x a m p l e

- The

inversion m a p p i n g

the composition of two successive

is its o w n

inversions

about

inverse:

Show that

a

point

common

is

equivalent to the identity.

Solution:

Since /'(r) = - r for all r E S , then /'(/'(r)) = /'(-r) = 1(r).

Hence /'/' = i2

= 1.

•

SYMMETRY

(D3.4) Definition:

The symmetry

ELEMENT

element

of a point isometry a is defined

to be the set of all points in space left fixed by a, i.e., a point p e S

is on the symmetry element of a if a ( p ) = p. Let a denote a rotation other than the identity with a rotation axis

I.

Since each point p along i is left fixed by a and every other point

in S is moved, i is the symmetry element of a. by the identity, and so S

Every point p E 5 is fixed

is the symmetry element of the identity.

As

a plane of points is left fixed by a reflection, this plane, referred to as the mirror plane, is the symmetry element of the reflection.

Because

all other rotoinversions

element

each fix a single point, the symmetry

of these isometries is a point.

It should be noted that our

of symmetry element differs somewhat from that used

definition

in the ITFC (Hahn,

1983). In isometry

general,

the

type

of

symmetry

a reveals what type of point

a ( p ) = p for all p £ S,

element

then a is the identity.

of a is a line I, then a is a rotation. a plane, then a is a reflection.

possessed

isometry a is.

by

a

point

For example, if

If the symmetry element

If the symmetry element of a is

Finally, if the symmetry element of a

is a single point, then a is a rotoinversion other than 2. 99

DEFINING

The

symmetry

SYMMETRY

of an object is often described intuitively by such

phrases as "the object is well-balanced or well-proportioned" or that "it consists of a pattern that is repeated at

regular

point, line or plane or along a direction in space." of phrases are too ambiguous for our purposes.

intervals

around

a

However, these types

By using isometries, we

can define symmetry with sufficient rigor so as to bring powerful mathematical tools to bear on the subject. Consider an object in space occupying the subset of points B in S. Let a denote an isometry.

By a(B) we mean the set consisting of all of

the images of the points in B .

That is,

a(B) = (o(b) | b

If b2,C 2 } to be the set of vectors emanating from p such that a 2 parallel to aj and cr2 = Oi, b 2 is parallel to bj and b2 parallel to Cj and c 2 = c t .

= bx

is

and c 2 is

Then the set of end points of the vectors

in

LQ

= {uai

is described in terms of D2 {q + ua2

+ ybj + wCi

| u,v,w

t Z}

as + vb 2 + w c 2

| u,v,w

E

Z} ,

where q is the vector emanating from p with terminus 0. If we let Ln = {ua2 + v b 2 + w c 2 I u,v,w E Z}, then the set of points in L n U Ux equals 2 {q + l2

|

E

LD

To show that a maps our new lattice Ln U2 show that if SL2

e

L n , then a(H 2 ) L>2

E

Lo • u2

)

.

into self-coincidence, we shall By adding and subtracting q,

we have a(d 2 ) = a(q + i2

- q) .

Since the origin p is fixed by a, a is a linear mapping with respect to the vectors emanating from p, and so a(q + l 2 - q) = a(q + l 2 ) - a(q) . Since end points q + l2 end points in L n . •J i

and q are both in Ln

Therefore, t

"(q +

, a(q + H 2 ) and a(q) have

- q +

tt and a(q) = q + i2 131

,

o(H 2 )

= o ( q + H2) - o ( q ) t t! = [q + n 2 ] - [q - ii] t2

-

H2

Since a is an isometry, the fact that a(Ln that a(Ln

) = Ln

U 2

L/ 2

E

L,

"D 2

L>2

and so Ln is mapped I/;

•

) is a subset of Z_n implies U2

into self-coincidence

by a.

Thus, if a lattice is mapped into self-coincidence by a, then the origin of the lattice can always be chosen as a fixed point of a. (T4.15)

Theorem:

•

If C is a crystallographic point group, then C has

a finite number of elements. Proof:

Let LQ

denote the lattice that is mapped into self-coincidence

by each of the isometries in C.

By T4.14, without loss of generality,

the origin 0 of L ^ may be taken to be a common point left fixed by each of the isometries in C.

Consider a ball centered at 0 with large enough

radius so that it contains all three basis vectors a, b , c e LQ

and let

a be an element of C.

Since a does not stretch vectors, and

a(Lp)

and a(c) must also be vectors in L^

= Lp,

a(a),

within the ball.

a(b)

(Newman, 1972, p.90).

lattice points

of L ^

in the ball

Consequently, there are only a finite number of

choices for ot(a), a ( b ) and a ( c ) . a(b),

contained

Moreover, inasmuch as the radius of the ball is finite,

there are only a finite number of

{a(a),

since

Since a is completely determined by

a ( c ) } , we conclude that there is only a finite number of

distinct point isometries in C that leave L p invariant. (D4.16) Definition:

•

When a group C is finite, the order of C , denoted

#(C), is the number of distinct elements in C. For example, since 322 consists of 6 elements, we write #(322) = 6. (P4.4) Problem: (T4.17) Theorem:

Determine #(422). Let a denote a crystallographic point isometry.

Then

the turn angle p associated with a is a multiple of either 60° or 90°. 132

Proof:

Since a is a crystallographic point

into self-coincidence.

Let D = { a , b , c }

Since a(Lp) = L Q , a ( a ) , a ( b ) ,

and a ( c ) are in

lattice LQ such that o maps LQ denote a basis for L LQ and so [ a ( a ) ] p ,

D-

isometry, there exists a

[ a ( b ) ] ^ j and [ a ( c ) ] ^ have integer coordinates.

these triples are the columns of M^(a), all of the entries of integers.

Consequently, the trace of

is an integer.

Since are

By TA3.6,

tr(M n (o)) = ±(1 + 2cosp)

(4. sign when a

where the "+" sign is used when a is a rotation and the is a rotoinversion.

The solutions to (4.1) given that tr(M 0 (a)) = N

for

some

integer N

are enumerated

in Table 4.1.

Note that

since

|cosp| < 1, we have the inequalities -1 < 1 + 2cosp < 3

and

-3 < -(1 + 2cosp) < 1

.

An examination of Table 4.1, where those values of N that satisfy these inequalities are listed and the corresponding p values are found, reveals that p is always either a multiple of 60° or 90°.

The point

isometries associated with each p value is also shown in Table 4.1. Table 4.1:

•

Solutions sets of the turn angles p for any crystallographic point isometry a. —(1 + 2cp) Solutions

(1 + 2cp) Solutions

N

cp = (/V - l)/2

p

o

-1

-1

180°

0

-i

±120°

1

0

±90°

2

i

±60°

3

1

0°

N

a

2

cp = -(/V + 1 ) / 2

p

0

a

1

-3

1

0°

-2

4

±60°

6,6*1

4,4-1

-1

0

±90°

4,4-1

6,6*1

0

-i

±120°

3,3*1

1

1

-1

180°

133

m

The 16 isometries found in Table 4.1 will serve as the elements for all of the crystallographic point groups. that such a group G is a bona

fida

However, before we can conclude

crystallographic point group, we must

show that there is some lattice that every element of G maps into selfcoincidence.

In this chapter and the next we will find all of the finite

groups that can be constructed from these isometries.

In Chapter 6, we

will find a lattice that is left invariant for each of these groups. MONAXIAL ROTATION The

structure

of

a group C

GROUPS

can be better understood by studying

certain special subsets of C called subgroups. (D4.18) Definition:

A subset H of a group C is a subgroup

of C if, under

the binary operation of C , H is a group. For example, in Chapter 3, we observed that 322 has several subgroups including 3 = {

1

,

2 - {1,1-^^2}.

3

,

3

difficult

to

"/0>2 =

{ l , ^ } ,

O.t110^}

and

We confirmed that these were subgroups by creating

their multiplication tables. are

=

see

However, some of the rules defining a group

directly

from

such

a

table.

The

rule

of

associativity is not readily verifiable from a simple inspection of the table.

The determination of which element is the identity and which el-

ement is the inverse of a given element can be seen from a table, but it is somewhat tedious.

Furthermore, if a subset of a group is very large,

then creating its multiplication table can be impractical. theorem

eliminates

most

of

these

difficulties

The following

in the case of finite

groups.

(T4.19) Theorem:

If H is a finite nonempty subset of a group C ,

then

H is a subgroup of C if and only if H is closed under the binary operation of C . We

will

not

prove

T4.19

here

but its proof can be found in any

standard modern alegbra text (e.g., Herstein, 1964). (E4.20) Example - 222 is a s u b g r o u p of 422:

Consider the subset

O,2V00hW°h}

222 =

134

Figur*

4.2:

The multiplication table for the

group 222 under composition.

[100]2

2

222

1

1

2

2

2

1

[100]2 [010]2

[100] 2

[010]j

[010]2

[100]2

1 2

2 1

[°10l2 [ioo]2

[10012 [010]2

[010]2

and show that it is a subgroup of H22 (see P3.15). Solution

To determine whether 222 is a subgroup of H22, we construct the

multiplication table shown in Figure 4.2. of elements in 222,

we know that 222

Since we have a nonempty list

is a nonempty set.

entry in the table belongs to the set 222,

222

is closed under the binary

operation (composition) of 422 and so is a subgroup. H

Also, since each

In other words, if

is a nonempty subset of a group (G,*), then (H,*) is a subgroup of G

if each entry in the multiplication table of H is an element of H.

(E4.21)

Example:

•

Let K denote a crystallographic point group and let

D denote a basis.

Then the set of matrices C = M n ( K ) = {M n (a) | a E K) D D

forms a finite group under matrix multiplication. set (T4.15), then M Q ( K ) must be a finite group. of

C

Since K is a finite Let H denote the subset

consisting of all matrices M in C such that det(M) = 1, i.e., M

represents a rotation isometry.

Show that H is closed and therefore a

group under the binary operation of C . Solution: I3 £ C . Mj,

Since 1 t K

(it is the identity element of K) , then M p O ) =

But det(I 3 ) = 1 and so I 3 c H.

M 2 £ H.

Then

MIM 2 E C

det (Mj )det (M 2 ) = 1 • 1 = 1. the binary operations of C .

since

C

Therefore, H is nonempty. is

a

group

Hence, MiM 2 £ H and so H

and

Let

detiM^j) =

is closed under

This result also shows that the composition

of two proper crystallographic operations is a proper

crystallographic

operation.

(P4.5) Problem: K

•

Let a and B be elements of a crystallographic point group

(E4.21). 135

(1)

Show that a(5 is a proper operation when a and 3 are both improper operations.

(2)

Show that ct$ is an improper operation when a is a proper and B is an improper operation.

(D4.22) Definition:

Let g denote an element of the group G

binary operation *.

The nth

is a positive

the

power of g , denoted g " is defined to be

g " = g * 9 * • •* g where n

under

(n g's)

When n = 0, g ^

integer.

is defined

to

be

identity element of C and when n < 0, g " is defined to be (g ^)

the that

is g" = o " 1 when n < 0.

* g"1*

•* g " 1 )

9's)

The usual rules of exponents hold true for the powers of

an element g e C

including the following: n

(1)

g

(2)

(g )

(E4.23) Example:

Let C

* g

m

= g

n + m

= g

r

for all r>, m

for all n,

m

E

E

1

,

1 .

denote a crystallographic point

group.

Let

g E C and define H = {gn

| n s Z)

,

then show that H is a subgroup of C. Solution:

According to T4.19, we need only show that H is closed under

composition.

Let X , y E H.

n and m such that x = g1

By the definition of H, and y = gm.

there exists integers

Hence,

n m n + m xy = g g = g

Since n + m is an integer, x y E H.

(D4.24) Defi nition: subgroup

Hence, H is a subgroup of C .

Let (G,*) denote a group and let g E G .

of G generated by g, denoted , is defined to be

136

The

c

cyclic

= ig" I n z Z) We will now justify calling a subgroup. that when C

In (E4.23) we showed

is a crystallographic point group, then is a subgroup.

In the next problem, you shall be asked to prove this fact in general.

(P4.6) Problem: subgroup of C

Show that if g E C where C is a group, then is a

(do not assume that C is finite).

(D4.25) Definition:

Let C denote a group and let g z C.

If g' = e for

some positive integer /' where e is the identity of C , then the smallest such positive integer is called the order of g and is denoted o(g).

Let g z C where C is a group such that there is a positive integer /' such that g' = e, the identity element of C.

= {g,g 2 .g 3 ,•••,g

Let k = o(g).

Then

= e} .

This means that any element of {g" | n z Z} can be written as g' where 1 < j < k.

For example, g ^ = g ^

gg

j

and g

k +1

= g since g

k - 1

k + 1

o(g).

= g

k

= g g

Furthermore, it is straightforward distinct.

^ since

k

= e

= eg =

to

show

,

g.

that

g,g 2 ,...,g

are


That is, if g' = e for some positive integer /', then # ( 9 )

=

A cyclic subgroup of a crystallographic point group is a set of

isometries having a common axis.

These form such symmetry elements as

an n-fold rotation axis or rotoinversion axis.

(E4.26) Example - T h e elements of = 4:

Let C denote the group of

all point isometries leaving the origin fixed and consider a quarter-turn 4 z C.

Since 41* = 1 and 1 is the identity element of C , = {4,4 2 ,4 3 ,4* = 1} .

We give this cyclic group the symbol 4 and note that the elements of this group define a 4-fold axis.

Since all of these proper rotations fix the 137

same axis, t is an example of a proper monaxial group. 43

is usually written in the equivalent form 4 ^ and 4 2

By convention, is written as 2

to conform with the convention adopted with respect to the turn angle in Chapter 3.

Hence, 4 = = ( M ^ ^ "

(D4.27) Definition:

1

}

.

A group consisting of rotations that fix a

axis is called a proper

monaxial

common

group.

A finite proper monaxial group C is cyclic and is generated by the rotation

in

C with the least non-negative turn angle.

generator for U.

Thus 4 is the

If C is a proper monaxial crystallographic point group,

then by T4.17, a generator for C can be found with a turn angle 0°, 60°, 90°, 120° or 180°.

This can be shown by observing that rotations with

turn angles greater than 180° generate groups that include rotations with turn angles less than 180°. a3

is a rotation of 90°.

For example, if a is a rotation of 270°, then A complete list of all of the possible proper

monaxial crystallographic point groups is given below:

1 =

=

{1 V

V

£

z}

V

=

{1}

2 =

=

(2

V

E

Z} =

{1,2}

3 =

=

{3 V

V

£

Z} =

{1,3,3 - 1 }

4

=

=

{4 V

V

E

Z} =

{1,4,2,4'h

6 =

=

{6 V

V

E

Z} =

{i.e.a^.s"1^

We observe that each of these proper monaxial groups define an

n-fold

axis where rt = 1, 2, 3, 4 and 6. Now that we have all of the possible proper monaxial crystallographic point groups, we can use the following theorem to obtain all of the possible improper monaxial crystallographic point groups.

(T4.28) T h e o r e m :

( T h e improper

point

group

generating

theorem.)

If

/ is an improper crystallographic point group, then there exists a proper crystallographic point group C such that either

(1)

(2)

I =

C

{gi

| g

I = H U

U E

Gi,

where

is

i

the

inversion

(here

C } ) , or

(C \ H)i

where H is a subgroup of C such that

138

Gi =

#(C)/#(W) = 2 (here (C \ «)/ = {gi | g e C and g | H)). Furthermore, all of the sets constructed from a proper crystallographic group C as in (1) or (2) are groups and hence improper

crystallographic

point groups. By the symbol "U" used in I = C U Ci, we mean the union of the two sets C

and Ci where the union of two sets A

and B,

A U B is defined

to be the set consisting of all of the elements of A

together with all

of the elements of B . not in H. from C . group

By (C \ H ) we mean the set of elements in G but

The set (C \ W ) is obtained by deleting each element in H A subgroup H of C such that #(G)/#(H) = 2 is called a halving

of C

(since it has half of the elements of C).

The proof of T4.28

is given by Boisen and Gibbs (1976) .

(E4.29) Example:

Use T4.28 to find the improper crystallographic point

groups for the case where C = t. Solution:

Applying part (1) of T4.28, we obtain

C U Ci = 4 U ti = {1,4,2,4" 1 } U O ,4,2,4~^}i = n,4,2,4" 1 ,1/,4/,2/,4" 1 /} = {1 ) 4,2,4~ 1 ,/,4,m,4~ 1 } . This group contains a 4-fold rotation axis, 4 = {1,4,2,4

perpendic-

ular to a mirror plane, m = {1,m}, and is consequently designated 4 / m read "4 upon m".

Applying part (2) of

{1,2} is the only halving group of of C .

T4.28,

we

note

that

=2

2 =

since both 4 and 4 ^ generate all

Then

H U (C \ H)i

H =

U (4 \ 2)/ = {1,2} U ({1,4,2,4^} \ {1,2})/

= {1,2} U {4,4"1}/ = {1,2,4M"1/} = £ 1,2,4,4 _1 > .

139

This crystallographic point group is designated 4 because it is the cyclic group generated by 4.

We also

observe that T4.28 states the groups

constructed using (1) and (2) are improper crystallographic point groups and so there is no need to examine whether these groups obey the rules of a group.

•

(E4.30) Example - Derivation of the possible improper point groups based on 3:

Use T4.28 to find the improper crystallographic point groups for

the case where G = 3.

Solution:

Applying part (1) of the theorem, we obtain = {1.3.3" 1 } U

3 U 3i

{/.M"1}

= {1,3,3~\i,3Xh •

Since this is a cyclic group generated by 3, the group is denoted by 3 Since 3 has an odd number of elements, it has no halving

read "three bar".

groups and therefore (2) does not apply.

Note that #(C U Gi) #(G).

if

C

is

a proper

•

crystallographic

point

= 2#(C) and if H is a halving group of G , Furthermore,

the

nature

of

G U Ci

differs

group,

then

U (G\W)/') = from

that

H U (G\W)/ and G in the sense that G U G/' contains the inversion. reason C\H.

H U (G\H)/

does

not

The

contain /' is that 1 is not an element of

A point group containing i is called a centrosymmetric Note that when C

of

group.

is monaxial, then G U Gi and H U (G \ H)i

are

also monaxial with the same axis.

In the case of C U Ci, we observe that

if a E G U Ci,

g for some g E C or a = gi

g £ G.

In

then

either

either case

a

a = and

g

have

the

same

for some

axis.In the case of

H U (C \ H)/, if a E H U (G \ H)/', either a = g for some g E H, plying that a and g have the same axis, or a = gi where g E C \ H, implying that a and g have the same axis.

imagain

Therefore, we see that the

set of axes in any improper crystallographic point group is the same as that of the proper group from which it is derived.

(P4.7) Problem:

Use T4.28 to find the improper crystallographic point

groups for the case where G = 2.

Show that

140

(2 U 2/ is denoted by 2/m).

Show that H - {1} is a halving group of

2 = {1,2} and that / U

(2 \ 7)/ = {1 ,m} .

This group is called m because it is a cyclic group generated by m. (P4.8) Problem:

Use T4.28 to find the improper crystallographic point

groups for the case C = 7.

Show that

7 U 7/ = {1,/}

(7 U 7/ is denoted 7).

(P4.9)

Problem:

Explain why 7 has no halving group.

Use T4.28 to find the improper crystallographic point

group for the case C = 6.

6 U 6i =

(6 U 6i is called 6/m reflection plane).

Show that

{l,6,3,2,3" 1 ) 6" 1 J /,6 ) 3,m J 3" 1 ,6" 1 }

because it has a 6-fold axis perpendicular to

a

Show that 3 is a halving group of 6 and that 3 U (5 \ 3)i = O ^ S " 1 , ^ , « ; " 1 }

•

This group is denoted 6 since it is ). We have now derived all 13 of the possible monaxial crystallographic point groups.

These are listed and their derivations summarized in Table

4.2. Matrix

representations

and basis v e c t o r s :

In order to obtain a matrix

representation for each of the point isometries for a given group, we need to select a basis for each group. To accomplish this, we will define, for each crystallographic point group C , a set of bases such that

the

matrix

representations

of

the

isometries is the same for any basis in the set and such that the matrices are of a simple form.

In fact, each entry of the resulting matrices will

be either 1, 0 or —1.

In chapter 6 we will show that every lattice left

invariant by a point group C will contain a sublattice that has a basis

141

Table 4 . 2 :

T h e 13 monaxial c r y s t a l l o g r a p h i c point g r o u p s and t h e i r orders d e r i v e d from the p r o p e r monaxial c r y s t a l l o g r a p h i c point group

Improper Monaxial Point Groups

Proper Monaxial

Halving

Point Groups

Groups"

Containing / (centrosymmetrie)

C

HC)

1

l

2

2

3

3

6

C U CI

H

none

1 none

#(C U

Not Containing /'

H U (C \

Ci)

H)i

1KH U (C '

1

2

none

2/m

4

m

3

6

none

8

Ü

4

12

6

6

4

2

t/m

6

3

6/m

of the type we define for C in this chapter and the next.

2 —

Since such a

sublattice will be called a primitive lattice, we will use the letter P to denote these bases. We

begin

by

considering

the

nth-turns

crystallographic point groups in Table 4.2.

of

the

monaxial

In all cases we will assume

that the bases described are right-handed.

1 (identity):

Since 1 is represented by the identity matrix for

every

basis, any basis can be chosen.

2 (half-turn):

For n > 1, we shall choose C to be any nonzero vector in

the positive direction of the rotation axis.

The a and b vectors

are

chosen to be any non-collinear, nonzero vectors in the plane perpendicular to c .

Since any vector perpendicular to the rotation axis of 2 will be

mapped to its negative, '-1 Mp(2) =

0 0

142

0 -

o' 1

0

0 1

.

3 (third-turn):

As in the case of a half-turn, c is chosen to be a nonzero

vector in the positive direction of the rotation axis. As 3 ( a )

to be any nonzero vector perpendicular to C. with a , we choose b = 3 ( a ) .

We then choose a is not collinear

With this choice of basis, we recall from

our discussion of 322 in Chapter 3 that 3 ( b ) = - a — b and so

0 - 1 0 1 - 1 0

M p (3) =

0

4 (quarter-turn):

(4.1

0

1

As before we choose C to be a nonzero vector in the Also, we let a be any nonzero

positive direction of the rotation axis.

vector perpendicular to C and let b = 4 ( a ) .

For this choice of basis,

we obtain 0 - 1 0 Mp(4) =

6 (sixth-turn):

To keep the number of basis types small, we choose as

the basis for 6 that which is used for 3 , where C is chosen as a nonzero vector in the positive direction of the rotation axis of 6 . a + b and 6 ( b ) = - a .

Then 6 ( a ) =

Hence, "l

-1

o"

Mp(6) =

(E4.31)

Example:

Find M p (3) = { M p ( 3

of Hp (3) and recognizing that Hp(3k)

)

| k e Z)

by finding the powers

(.Hp(3))k

After finding Mp(3),

=

find M p (3) U M p (3)M p (/).

Mp(3) Solution :

By (4.1),

Mp(3 )

II

2

"o

-1

o'

1

-1

0

0

0

1

2 =

143

-1

1

o"

-1

0

0

0

0

1

=

Mp(3"1)

;

II

Mp(3 ) =

0

-1

o'

1

-1

0

0

0

1

3

"l

0

o"

0

1

0

0

0

1

=

Mp(1)

Hence, 1 Mp(3) =

0

0

0

-1

0

0

1

0

0

0

1

-1

0

1

0

0

1

>

,

-1

1

0

-1

0

0

0

0

1

Mp(a)Mp(/) = -Mp (a) for any point isometry

Mp(3)M p(/)

{"Mp

=

(4.2)

a

(1),-Mp(3),- Mp(3"

Hence "-1

0

0

-1

0

0

0

-1

Mp(3)Mp(/) =

Then

Mp(3)

consists

of

o'

1

0

-1 y

1

0

0

"l

-1

0

1

0

0

0

0

-1

0 -1

]

o" (4.3)

the matrices in (4.2) together with those in

(4.3).

(P4.10)

o

Problem:

Then find M p ( n )

Find M p ( n ) = { M p ( n U Mp(n)Mp(/') .

)

| k t Z) for n = 1, 2, 4,

6.

For each C = n and each halving group H

of C that exists, find

Mp(tt U (C \ H)i) = M p ( W ) U (Mp(C) \ Mp(W))Mp(/).

Equivalent

Points and Planes:

We defined the symmetry group G of an

object to be the set of all isometries that map the object into coincidence.

self-

Utilizing G we can impose an "equivalence" on the points

in S.

( D 4 . 3 2 ) Definition: and y

Let G denote a group of isometries.

in S are said to be

C - e q u i v a l e n t

if there exists

a

The points X e C such that

a(X) = y .

In the case where C is the symmetry group of an object 6 , then the intuitive interpretation of C-equivalent points is that if x and y are C-quivalent, then the object B

appears the same in every respect to a

person viewing the object from either point x or point y .

144

The notion of C-equivalent points is a generalization of the notion of equality.

That is, while X and y may be d istinct points, the statement

that they are C-equivalent expresses the idea that in some specific sense they are equal.

This is similar to our usual handling of fractions in

arithmetic where we think of the fractions 2/3 and 4/6 as being "equal" even though they are clearly not identical.

If C-equivalence is to mimic

equality, we would hope that the basic properties of equality hold. (T4.33)

Theorem:

Let C denote a group of isometries and let x ~ y

denote "x is C-equivalent to y " .

Then

(1)

X ~ X for all X E S

(2)

For all X, y E S,

(3)

For all X, y, Z E S, if X ~ y and y ~ Z, then X ~ z (transi-

(reflexive property) ;

if X ~ y then y ~ X (symmetric property) ;

tive property) . Proof:

Since C is a group of isometries, 1 t C.

all X E 5 , we have X ~ x for all X E S . in S such that X ~ y .

Let X and y denote elements

Then there exists a E C such that a(X) = y .

C is a group, a ^ E C.

Since

Then a'VtX))

=

a" 1 (y)

(a_1a)(x)

=

a-1(y)

1(X) =

a_1(y)

X

Hence y ~ X.

Since 1 (X) = x for

=

a_1(y) .

Let X,y,Z E S such that X ~ y and y ~ Z. Then there exists

a, 3 E C such that a ( x ) = y and £(y) = Z.

Then

ga(X) = B ( o ( x ) ) = B(y) = Z. Since C is a group, it is closed and so &a E C.

Therefore X ~ Z.

Any relation defined on the elements of a set satisfying the three parts of T4.33 is called an equivalence lations are fundamental principles of concept.

relation.

Since equivalence re-

to any true understanding of the mathematical

crystallography, we have devoted much of Appendix 7 to this

Those who are unfamiliar with this concept should

appendix carefully. 145

read this

•

Equivalence relations organize the set on which they are defined into subsets of related elements called equivalence classes.

In the case of

C-equivalent points, if X t S, then the equivalence

of x, denoted

class

[X] is

[x]

= {y E 5

I x ~ y}

= {y e S

| there exists g e C such that g ( x ) = y>

= (g(x) | g E C).

Since, in the case of C-quivalence, [X] is the set of images of X under C , [x] is called the orbit

of x under C and is designated orb^(x).

When

the number of points in orb^(x) equals #(C), then x is called a point of

general

position.

x is a point

Otherwise,

of

special

position.

When

x

is a point of special position, there exists one or more nonidentity elements g t C such that g ( x ) = x.

Let D denote a basis of S, x be an

element of S and C be a point group.

Then we define

o r b C ) D ( [ x ] D ) = {[g(x)]D

| g E C) .

(E4.34) Example - T h e symmetry of H4SiO,, is 4: of HtSiO,,, are given in Table 1.1. to C

The atomic coordinates

Show that the coordinates with respect

of the oxygen atoms designated Oi, 0 2 , 0 3 and 0» are mapped into

self-coincidence by the matrix representations of the monaxial group 4 where M c ( 4 ) = {M c (1), M c ( 4 ) , M c ( 2 ) , M c ( 4 " 1 ) } .

That is, the four oxygen atoms of the molecule are 4-equivalent.

Solution:

This will be done by showing that 0j is 4-equivalent to each

of 0 2 , 0 3 , and 0».

Since the coordinates of the vector r emanating from

the origin to 0j given in Table 1.1 are

[r] c = [1.281, 0.466, 0.877] t

,

and since

Mc(1)[r]c =

"l

0

0

"l 281

0

1

0

0 466

0

0

1

0 877

146

"l 281 =

0 466 0 877

Mc(4)[r]c =

' 0

1

0

"l 281

-1

0

0

0 466

0

0

-1

0 877

-1

0

0

1 281

0

-1

0

0 466

0

0

1

0 877

Mc(2)[r]c =

Mc(4"1)[r]c =

0.466' -1.281

=

-0.877 -1.281" -0.466

=

0.877

0

-1

o'

"l 281"

1

0

0

0 466

0

0

-1

0 877

"-0 466" 1 281

=

- 0 877

we see that the orbit of r is orb ?

c

(r)

= {Mc(o)[r]c

| « £ 4)

= {Mc(1)[r]c, Mc(4)[r]c, Mc(2)[r]c>

But these

Mc(4"1)[r]c>

"l 281

" 0 466"

- 1 281

0 466

- 1 281

- 0 466

1 281

0 877

- 0 877

0 877

-0 877

>

are the coordinates

of Oi, 0 2 )

03

^-equivalent to each of the other oxygen atoms.

- 0 466"

and 0,,.

Hence Oj is

The reader should repeat

this process for each of the remaining oxygens in the molecule and observe that a similar result is obtained. (P4.11) Problem:

•

Show that the set of hydrogen atoms designated H 1 } H 2 ,

H 3 and H 4 in H^SiO« is mapped into self-coincidence by the elements of M^C^)-

Do this by forming the orbit of [rj^, where [r]^ is triple of

the coordinates of one of the hydrogen atoms in Table 1.1. (P4.12)

Problem:

Show that the silicon atom in H»SiOi, is on a point of

special position and that its orbit consists of a single point. (E4.35) Example:

Show that the faces on the crystal in Figure 4.3 are 147

Figure 4.3: A set of faces comprising a trigonal dipyramid that are i-equivalent to (321).

mapped into self-coincidence by the isometries of the monaxial group 6. The natural basis earlier for 6.

for the crystal is the basis P = {a,b,c} described "ii ~v ?v

Recall that the face poles s = ha

crystal are defined in terms of its basis P . that

Solution:

+ kb

+ He

of a

In Appendix 3 it is shown

-t M p *(a) = M p (a) This problem will be solved by showing that the image under

6 of each face pole of the crystal is another of its face poles.

That

is, if [S]p* represents a face pole, then we must show that M p *(cO[s] p * also represents a face pole for each a t 6. face pole [s]p* = (327)

For example, consider the

and note that 1

Mp*(1)[»]p* =

o

0 0

1 0

2

1

1

0

0 M p *(6)[s] p *

-1

0

1

0

0-1

-1

-1

0

1 0 0

M p *(m)[s] p * =

2

=

1 0

-1

M p *(3)[S] p *

3

o' "3

0 0

1

"l

0

o'

3

0

1

0

2

0

0

-1

1

148

3 =

2 -1_

Figure 4.4:

A dioptase crystal with a (110) face

of a hexagonal prism labelled (o), and (021) and (131) faces

of

rhombohedra

labelled

5

and

X,

respectively.

,-1 Mp*(3 ')[»]p*

o

1

o

-1

-1

0

0

1

0

MpM^'lIslp* =

-1

-1

0

1

0

0

0

0 - 1

When these triples are collected together into a set, we have the orbit of face poles that are 6-equivalent to (321):

=

{M p *(a)

| a e

6

}

This collection of equivalent faces is called a crystal face form and is designated by placing the indices (hkl) of the representative plane of the orbit between braces, {hkl}. class

Thus, {321} denotes the

equivalence

of faces on the crystal in Figure 4.3 that are 6-equivalent to

(321).

• 149

(P4.13)

Problem:

Figure 4.4 is a drawing of an idealized

crystal

of

dioptase, CuSi0 3 *H 2 0, showing a (110) face of an hexagonal prism labelled

a and (021) and (131) faces of rhombohedra labelled s and x,

respectively.

Assuming that the point symmetry of the dioptase crystal is 3, find the indices

of each of the faces on the crystal that are 3-equivalent to

(110), (021) and (131).

Then assign indices to each of the 3-equivalent

faces on the crystal. Table 4.3:

Coordinates of the atoms comprising an SiO,,

group in narsarsukite (Peacor and B u e r g e r ,

Atom

X

1962).

z

y 0 3085

Si

0 0118

0,

- 0 0400

0 3024

0

0,

0 0488

0 1754

0 2684

0s

0 1324

0 4023

0 1934

0«

- 0 0977

0 3676

0 3062

(P4.14) Problem:

0 1921

The symmetry of the atoms about the origin chosen for

narsarsukite (a = b = 10.727A, c = 7.948A, a = 3 = TS = 90°) is

U/m.

(1) Determine the elements of M p ^ / m ) and the coordinates of the atoms in narsarsukite that are 4/m-equivalent to those in Table 4.3. (2) With atomic coordinates obtained in (1), prepare a drawing of the atoms in narsarsukite that are 4/m-equivalent to those in Table 4.3 viewed down the Z-axis. (3) Calculate the SiO bond lengths and the OSiO angles for the SiO* group described in Table 4.3.

Then calculate the bond lengths and angles

for an SiO» group that is the image under 4 of the one described in the table and observe that the geometries of the two are identical, demonstrating that 4 is an isometry. (P4.15)

Problem:

presented

on

the

Stereoscopic drawings of C-equivalent ellipsoids are next

four

pages

crystallographic point groups G .

for

each

of

the

13

monaxial

Examine these drawings in Figure 4.5

and confirm their point symmetries. 150

F i g u r a 4 . 5 (on this a n d tha following p a g a * ) : Stereoscopic pair plots of C-equivalent ellipsoids for each of the 13 monaxial crystallographic point groups G.

1

2

152

V] 4/m

4 /

6/m

\i

\i "Ì

m

154

4

6

155

CHAPTER

5

THE POLYAXIAL CRYSTALLOGRAPHIC

"Much

of

the

symmetry.

importance

Just

as

of measurement

numbers

has

metry.

With each

terizes

the symmetry

of

been figure

groups can

be

chosen'),

comes used

groups

w e associate

of the figure."

POINT

from

their

to measure can

GROUPS

connection size

be used

a group,

and

this

with

(once a

to measure group

unit sym-

charac-

-- J. R. Durbin

INTRODUCTION

In Chapter 4 we derived all of the possible monaxial crystallographic In this chapter we shall learn how all of the

possible

proper polyaxial point groups are constructed from the proper

point groups.

monaxial

point groups. (T4.28),

we

Then, using the Improper Point Group Generating Theorem shall

derive

all

crystallographic point groups.

of

the

possible

improper

polyaxial

The interaxial angles between the rota-

tion axes of the rotations participating in any one of these groups are then determined.

As in Chapter 4, a special basis P will be chosen for

each point group.

PROPER

( D 5 . 1 ) Definition:

POLYAXIAL POINT

GROUPS

Let C denote a proper point group.

If C has more

than one axis associated with its nonidentity rotations, then we call C a proper

polyaxial

group.

In our investigation of the polyaxial groups, we will be examining combinations of monaxial groups.

The task of finding the possible proper

polyaxial point groups will be considerably more difficult than that of finding the monaxial point groups.

Our first goal is to establish the

inequality which will state that three monaxial groups associated

with

nonequivalent pole points with orders Vi, v 2 and v 3 , respectively, can be used to form a proper polyaxial point group only if

1/Vi + l/v2 + l/v3 > 1

157

.

Figure 5.1: 322

The set of all pole points belonging to the nonidentity rotations of group

(see Figure 3.13 for a drawing of rotation axes of the group).

The basis vectors a ,

b and c coincide with the coordinate axes X , Y and 1, respectively, where ct = B = 90° and 1 = 120°. those in in [OlOJj

To

The nonidentity rotations in 3 leave the antipodal points p ! 1 and p 1 2 2 leave p 2 i leave

paj

and p ) 2

fixed, those in

fixed,

leave p , 2 and p 2 2 fixed and those

fixed.

and

facilitate our proof that will establish this inequality, we shall

consider the surface B of a unit ball centered at 0. Any point isometry a acting on B maps B onto itself.

In fact, a is completely determined

by its action on B because the effect of a on B determines the images {a(i) ,a(j) ,a(k)}, where C = {¡,j,k} is a cartesian basis, which in turn completely determines a.

A nonidentity proper rotation h about the ro-

tation axis t leaves exactly two antipodal points on B unmoved.

These

points are precisely the points, on opposite sides of the ball, at which i and B intersect and are called the pole points belonging to h.

If these

two points are labelled p and q, then they are the only points x on B that satisfy the equality /?(X) = x. The set of all pole points belonging to the nonidentity rotations in C will be denoted by P(C) and will be called the set of pole points belonging to C. is an equivalence relation on P(C).

Note that C-equivalence

Applying C-equivalence to P(C),

we partition P(C) into its set of equivalence classes.

For example, the

pole points of 322 P{322)

=

{Pll,Pl2,P21>Pl2>P23>P31>Pl2>P33}

158

are denoted in Figure 5.1 by p.. class p..

where / indicates to which equivalence

We denote the ith

belongs.

belonging to C by

equivalence class of pole

points

C.(C).

( E 5 . 2 ) Example - The 322-equivalence class of p class under 322-equivalence of p

u

:

Find the equivalence

shown in Figure 5.1.

u

That is, find

0^322).

Solution:

By definition of an equivalence class,

[p 11 ] = { q E PCC)

I

P l l

~ q}

.

By definition of 322-equivalence, [Pu]

= { q e P(322) = ig(Pii)

=

| there exists g e 322 such that g ( P n ) = q }

I 9 e 322}

i1(Pxx).3(pll),3-1(pM),I1«>]2Cp11),I01012(p11),I110l2(pll)>

= {Pii.Pizl Hence Ct(322) (P5.1)

=

Problem:

(P3i>P32>P33)

as

{p„,p

n

) Cz(322)

Verify that

=

{p2j,p22,p23}

and

C3(322)

=

shown in Figure 5.1.

Note that {C1 (322) ,C 2 (322), P(322)

= Cx(322)

C3 (322)} partitions P(322). U C2(322)

U

That is

C,(322)

and C.(322) where 0 is the empty set.

n

C .(322)

= 0 when /' / /

,

Note that # ( C 1 ( 5 2 2 ) ) = 2, #(CZ(322))

= 3 and

# ( C 3 ( 3 2 2 ) ) = 3. ( P 5 . 2 ) Problem:

Draw a diagram showing the rotation axes of 422.

P!! denote a pole point of

p 2 1 a pole point of

159

Let

and p 3 1 a pole

point of

Find ^ ( 4 2 2 ) ,

C2(422)

and C 3 (022).

Note that every

pole point of 022 is in one of these equivalence classes.

Now w e shall turn our attention to polyaxial rotation groups G

and

the subgroups of C associated with its pole points.

(D5.3)

Definition:

Let p denote a pole point belonging to C .

collection of all rotations of C that have p the stabilizer

as a pole point

of p in C and is denoted by C p .

C

= {g e C

p

Then the is called

That is

| g ( p ) = p}

.

The stabilizer of a pole point p., is denoted by C...

(T5.4) T h e o r e m :

Let p denote a pole point of C .

Then C ^ is a subgroup

of C .

Proof:

Since G

9i> 92 e G

•

is finite, we need only show that C ^

is closed.

Let

Then

9i9i(P) = 9i(92(P)) = 9l(P) = P Since g i g 2 ( P ) = P and g i g 2 nonempty, C p

E C

(C is closed) g i g 2 e C p .

Since

is

is a subgroup.

Since the rotations of

c

leave p fixed, they all have the line I

containing 0 and p as their rotation axis.

Consequently, C p is a proper

monaxial group and therefore isomorphic to one of the cyclic groups listed in Table 4.2. In the case of

and

322,

(322) 2 1

=

(322) 3 2 =

(322), x

=

(322) 2 ,

(322)„

=

(322) 3 , =

J

(322) X 1

=

(322) 1 2 =

{3,3-1,1}

=

(HOOl^j {[iio]2>1}

0 1 0

^,!}

We observe in this example that two pole points are associated w i t h each

160

monaxial group (Figure 5.1).

It is also evident that the pole

p 21 j p 2 2 , p 2 3 j P3i, p 3 2 j and p J 3

where the rotation axes have different orientations. points p

n

points

are each associated with the group 2 The remaining pole

and p l 2 are similarly associated with group 3.

( T 5 . 5 ) Theorem:

Let C denote a proper crystallographic point

Let p denote a pole point and let q be G-equivalent to p .

group.

Then C ^ is

isomorphic to C ^ and hence they are both isomorphic to the same proper monaxial point group of Table 4.2. Proof:

Since q is C-equivalent to p , there exists a rotation g E G such

that g ( p ) = q. g

1

0;

(q) = p and so g h g \ q )

all h £ C p . an

Consider

C p •+ C q defined by 8 ( h ) = g h g 1 . = g h ( p ) = g ( p ) = q.

By EA8.5, 8 is an isomorphism.

Note that

Hence 8 ( h ) e C q for

Since isomorphism defines

equivalence relation, C ^ and C ^ are isomorphic to the same proper

monaxial point group of Table 4.2.

( E 5 . 6 ) Example:

Solution:

e

Show that (322)22 = { 3 h 3 _ 1

| h E (322) 21 ).

As shown above

(322)21

=

(522)22

=

[ ^ ^ { l . t

1 0 0

^ }

and [07°]2={1,[010]2) .

Then {3h3_1

| h E (322)21)

=

{313" 1 . S ^ 1 0 0 ^ " 1 }

=

{1,

= (P5.3) Problem:

[010]

(322)22

2}

Confirm the results in E5.6 with matrices showing that

M 0 ((322) 2 Z ) = { M D ( 3 ) M D ( h ) M D ( 3 " 1 ) Note that since M^(3 for MQ((322)21)

D

•

1

) = M^(3)

| h E (322) 21 )

the matrices described in P5.3

are similar (see Appendix 7) to the matrices in

161

.

(322)22). equivalence

Hence while p 2 i relation

of

and p 2 2

D4.32,

the

are 322-equivalent under the

matrices

in

(322) 2 1)

and

M^((322) 2 2) are equivalent under the equivalence relation of DA3.3 (see also EA7.4).

A relation corresponding to similarity can be defined on

the subgroups of G. (T5.7) Theorem:

This relation is called

Let C denote a group.

conjugation.

The relation defined on the set

of subgroups of C defined by

Hi ~ H 2 where H x

and H 2

relation

~

conjugate

to H 2 .

there exists a g E C such that g ^ g

^ = H2

,

are subgroups of C , is an equivalence relation. The

is called conjugation

and if Hi ~ H2,

H¡, is said to be

The proof of T5.7 is essentially the same as that of EA7.4 where we demonstrated

that

similarity

of matrices

is an equivalence relation.

Conjugation and C-equivalence are very closely related concepts.

Con-

sider T

| p e P(G)}

= {Cp

.

Conjugation is an equivalence relation on T while C-equivalence is an equivalence relation on P(C). p ~ q

Furthermore, we have «-»• C p ~ C q

where C-equivalence is used on the left and conjugation on the

right.

Consequently, if p is a pole point, then an isomorphic image of C ^ occurs about any axis having g(p) as a pole point for each g E G.

Suppose we

have located the position of the axis of one of the rotations of G.

Then

we can find others by mapping the axis under the operations of G.

Fur-

thermore, the axes found in this manner will be associated with cyclic groups of the same order as the original axis.

This observation will be

extremely important in our construction of the rotation groups.

(L5.8) Lemma:

Let C denote a proper point group such that #(G) > 1.

Then 2(/V - 1) =

t I

".(v. - 1)

/'=1 162

where N = #(G), t is the number of equivalence classes of pole points, n. = #(C.(C)) and v. = Proof:

iKC..).

The basic strategy for establishing this theorem will be to find

two distinct ways of counting the nonidentity rotations of C. will be two expressions each equaling twice the number of rotations of C.

The result nonidentity

This will establish the result since these two equal

expressions will be precisely those appearing in the equation. by taking each pole point p.. in

P(C)

We begin

one at a time and counting the

number of nonidentity rotations of C leaving p.. fixed.

The sum of these

numbers taken over all the pole points in P(C) will equal twice the number of nonidentity

rotations

in C because each of these rotations leaves

exactly two pole points fixed and hence is counted twice.

The number of

nonidentity rotations leaving p.. fixed is if(C..) - 1 = v. — 1. Thus, for the pole points in C.(C) we have ¿((nonidentity rotations leaving p.

fixed) = v. — 1

\

#(nonidentity rotations leaving p.^ fixed) = v. — 1

I \

//(nonidentity rotations leaving p.

fixed) = v. - 1

Summing up these numbers we find that the contribution n. (\>. - 1) for each 1 < i < t.

n. equations

from

C.(C)

is

Adding the contribution from each of the

t equivalence classes, we find that the sum is

t Z n.(v. - 1). /=1

Since

N - 1 is the number of nonidentity rotations in C, it follows that twice the number of nonidentity rotations in C is 2{N - 1) and so we have established that t I

2(/V - 1) =

n.(v. - 1)

.

•

/=1 From

E5.2

#(C2 (322)) =

3

and

P5.1,

we

see

that

nx =

and

n3 =

#(C 3 (322)) = 3.

#(C t (322)) = 2 ,

Also

Vl

n2 =

= 1K3) = 3, v 2 =

# ( [ 7 0 0 ] 2 ) = 2 and v 3 = #( i 7 7 ° ! 2) = 2. Since N = #(.322) = 6, we can verify L5.8 in this case by observing that 2(6 - 1) = 2(3 - 1) + 3(2 - 1) + 3(2 - 1) 163

.

(P5.4)

Problem:

Using the information you have developed

about

422

(including the solution to P5.2) verify the equation in L5.8 for 422. (T5.9)

Theorem:

Let p and q denote C-equivalent pole points and let

T denote the set of all elements of G that map p to q.

Then T is a left

coset of G .

P

Proof:

A discussion of cosets and related topics can be found in Appendix

Since p is C-equivalent to q ,

7.

there exists g E C such that g ( p ) Let t E T.

We shall show that T = 9 C p .

q.

t ( p ) = q.

=

By definition of T ,

Since g ( p ) = q , we have

g_1t(p)

Hence g ' t E C

Conversely, suppose h E gC

Therefore t E g C

Then h = g k where k E C

(q)

g

p

Hence

gk(p)

h(p)

g(p) q

Hence h E T.

c

Consequently T = g^p-

One consequence of T5.9 is that if G is a finite proper point group such that #(C) > 1 and p is a pole point of G, then there is a one-to-one correspondence between the cosets of

and the pole points that are

C-equivalent to p .

then n,,

Hence, if p £ C.(C),

be the number of pole points in C .(C) C p in C .

which is defined to

, equals the number of cosets of

Recall (see the proof of TA7.13) that each coset of C ^ has

the same number of elements as does C

, that is v. elements. P' '

Since the

cosets of C p partition C into n. cosets each having \>. elements we have established that

#(C) In the case of 322, v

i

=

3 >

v

2

=

v

3

= 2

N = n.v, for each 1 < i < t i i

(5.

we observe that since /7j = 2, n 2 = n 3 = 3 and

and that N = 6

n.v. in all three cases. ; /

164

(P5.5)

Problem:

Show that n.v. = N

for each 1 < / < 3 in the case

of

422.

(T5.10) Theorem:

Let f denote the number of equivalence classes of pole

points of a finite proper point group C where N = # ( G ) > 1.

Then t = 2

or 3 and (1)

if t = 2, C is a monaxial group and

(2)

if t = 3, C

is a polyaxial group such that

1/vi + l/v 2 + 1/v3 > 1 and # ( C ) = 2/(1/vi + l/v 2 + l/v3 - 1) Proof:

.

By L5.8, we have

2(N - 1) =

and by (5.1), we have /V = n.v..

I n (v. - 1) ;=1

.

(5.2)

Dividing the left side of

(5.2) by N

n v

and the right side of (5.2) by j j> we obtain

2 - Z/N =

t I (1 - 1/v.) /=1

.

(5.3)

Since v. is the order of the stabilizer of a pole point, v. > 2. t t I (1 - (1/v.)) ^ Z (1 - i ) /=1 /'= 1

=

t I (i) = t/2 /'=1

Hence

.

Also, since N S 2, 2 > 2 - 2//V

.

Hence, from (5.3) 2 > t/2

Hence 4 > t.

.

Therefore, t can only equal 1, 2 or 3.

We can thus conclude

that there are no rotation groups having more than 3 equivalence classes of pole points.

We now examine each of these three cases for the value

of t. C a s e w h e r e t = 1:

2 - 2/N =

In this case, Equation (5.3) becomes t 1 I (1 - 1/v.) = Z (1 - 1/v,) /'=1 /=1 165

or 1 - 2/N

The

=

1/vi

left member of this equation is always nonnegative because N ^ 2,

but the right member is always negative because Vj > 2, which is a conTherefore, t cannot equal 1, from which we conclude that

tradiction.

P ( C ) must contain more than one equivalence class of pole points.

Case where t = 2:

In this case Equation (5.3) becomes

2 2 - 2/N =

I (1 - 1/v.) = (1 - 1 / v O + (1 - l/v 2 ) /=1

.

By a little algebraic manipultion we find that

2 = A//V! + /V/v2

.

From Equation (5.1) we have that N/\>. = n., and so the above expression simplifies to 2 = n1

Because n^

and n 2

+ n2

are positive integers, we conclude that n ± = n2

is the only possible solution.

- 1

Hence, for a rotation group with t = 2,

we have two equivalence classes consisting of one pole point each.

Al-

together C has a total of two pole points, which defines one and only one rotation axis. of

Therefore, those groups with two equivalence classes

pole points must be the proper monaxial groups given in Table 4.2.

The number of elements in each of these possible monaxial groups is equal to the order of the rotation axis, # ( G ) = Vj = v 2 •

Case whe r© t — 31

In this case Equation (5.3) expands to

2 - 2/N = (1 - 1/v,) + (1 - l/v 2 ) + (1 - 1/v,)

.

Rewriting this result we see that

1 + 2/N = 1 / v i + 1 / v2 + 1/v,

Since N > 2, it follows that 1 + 2/N

> 1 and so

166

.

(5.

#(C,(C)) =

1,KC,(C))

N/Vi

groups.

#(C,(C))

Group Name

c

N

#(C) =

C - vlv2vj

II

Symbol for

Possible finite proper polyaxial point

11
6. The group 332

is usually designated by 23 and the group 532

designated by

l/Vj

S o l v i n g for N

+

l/v2

+

l/v3

>

1

.

in (5.4) we o b t a i n

#(C) =

Us ing part Vj > v 2 S v 3 ,

is usually

235.

we

N =

2/ (1/v! + l/v 2 + l/v 3 - 1)

(2) of T5.10 w h e r e , shall

polyaxial point groups.

construct

for

all

Note that if

.

convenience,

of

the

we

possible

assume finite

> 2, then e a c h of the

that proper

fractions

1/v. w o u l d be less t h a n or equal to 1/3 for each i and so (5.5) w o u l d not be satisfied. or

equal

to

v 2 = 2 or 3. (5.5). groups

H e n c e v 3 = 2. -J and so

If v 2

> 3, then

(5.5) w o u l d

Suppose v 2 = v 3 = 2.

T h e s e groups, denoted n22

1/vi + l/v 2

again not be satisfied.

is

less

than

Therefore,

T h e n any value of Vj > 1 w o u l d satisfy w h e n n = Vj,

of w h i c h t h e r e are an infinite number.

#{n22)

= In

are c a l l e d the

.

If v 2 = 3 and v 3 = 2, then if v t > 5, (5.5) w o u l d not be satisfied. the only groups of this type are 332,

432

dihedral

Using (T5.10) w e see that

and 532.

finite proper point groups are r e c o r d e d T a b l e 5.1.

Hence

A l l of t h e s e p o s s i b l e N o t e that w e h a v e

The orientation of the three mutually

Figure 5 . 2 :

dicular 2-fold axes in 222 ordinate ^ ^ 2

system

with

along b .

2

perpen

These axes define a natural co-

lying

along c

Each 2-fold axis

[ 100]2

aiong

a and

is represented by a

diad

symbol.

Figure 5 . 3 :

Multiplication table f o r

Hp(222)

Mp(l)

Mp(1)

Mp(2)

Mp(2)

V

[10

°>2)

Mp([°10]2)

10

M„(I °]2, V

[010]

2)

Mp(222).

Mp(2)

V[100]2)

Mp([°10l2)

Mp(2)

V(,00)2>

V[010]2)

Mp(l)

v[010]2>

M„CI,00I2)

V

[010]

V

[100]

Mp(2)

2) 2)

Mp(2)

Mp(1)

yet to show that each of these possibilities actually occurs as a point group. CONSTRUCTION OF THE DIHEDRAL GROUPS In Appendix 6 we proved, for

r>22,

that the n-fold axis is perpen-

dicular to each 2-fold symmetry axis and adjacent 2-fold axes group must intersect at an angle of 180/n (TA6.1).

in this

To confirm that each

n22 is actually a group, we shall define a basis of S for each, write the elements of

r>22

as described in TA6.1 with respect to this basis and

then form the multiplication table to check closure. The construction of 222:

Since the 2-fold axes are mutually perpendic-

ular, we define P = { a , b , c } to be a basis where a, b and c are also mutually perpendicular such that each lies along an axis (see Figure 5.2). Hence the metrical matrix G for P is 168

G

g ii

0

0

0

g 22

0

0

0

g 33

=

[100] With respect to this choice of basis, the half-turns are denoted 2, [010] 2, 2 along a, b , C, respectively. By Table 5.1, 222 can only have 4 elements and so we conjecture that 222 = {1 ) 2 ) is a group.

[ioo] 2 ) [ oio] 2 }

The matrix representation of each of these can be found era-

ploying the approach used for 322 in Chapter 3. Mp(222)

Hence

= {Mp(1),Mp(2),MpC[100]2),Mp([010]2)}

.

To confirm that 222

These matrices are described in Table 5.2.

is

a

group, we form the multiplication table of Mp(222) shown in Figure 5.3. Since no new entries resulted in the formation of the table, Mp(222) is closed under matrix multiplication and, since it is finite, it is a group. Since the mapping from Mp(222) to 222 that maps Mp(ot) to a for each a preserves the operation, the multiplication table of 222 under composition can be obtained by deleting the Mp(

) for each element (Figure 4.2).

Hence, 222 is a group. T h e construction of 322:

As in Chapter 3 (see Figure 3.10), we choose

P = {a,b,c} where C coincides with the 3-fold axis, a coincides with one of the two-fold axes and b = 3 ( a ) (Figure 5.4).

Hence the metrical matrix

G of P is 9

G =

il

-g

n/2

o

-g n/2

g w

0

o

o

g 33

Since 60 = 180/3, there are two-fold axes at 60° intervals starting with the one coinciding with a.

Hence, a two-fold axis lies along b.

Recall

that this is the same basis that was used for the monaxial groups 3, 3, 6, 6 and 6/m.

With respect to P, 169

Table 5 . 2 :

T h e nonzero e n t r i e s of the matrix r e p r e s e n t a t i o n s M ^ ( a ) f o r

rotation isometries o g r o u p s 1, 2, 4, 222, 422, 23 a n d 432 f o r P -

(a,b,c).

Mp(1) : In= «22 = t,, = 1

«22 = 1; «11 = «12 -1 =

Mp(2) : I,,= ii2 = -l; in = l

= -1 : «12 - 1; «21 = «12

V

[010I

2)

Mp([101]2)

»n = i; in = «n = -i

M / " 1 ^ ) : «21 = i; »12 ' «11 = -1

in = i; tn = in = -i

M/^V

«n = «n = l; in =-1

M / " 1 ^ ) : «ii~ i; «12 = «21

1

) • »12 = i; «11

=

«21= -1 -1

[ l] 1 = -1 i.j = ti. = i; t.i = -l Mp( " 3- ) : «11 = 1; «21 = «12

V

ti01]

Mp(

2)

[0il]

2)

V[110]2) V

ni0]

2)

V [ 1 1 1 ]3)

in = in = in = -l

Mp(4) : «ai =«.. = l; «12 = -1

«11 = In = «21 = -l

Hp(4"1) : «12= «ii = l; «21 = -l

«21 = «12 = 1; «1. = -1

V

1

=

&X2

=

=

[100]

= = 4 ) = «11 " «12 i; «21 -1

v 1 1 0 0 1 «- 1 )

:

«ii = «21

=

«21 = «12 = «11 = 1

Mp(I01014) :«22

: «11 = «12 = «21 = 1

V 1 0 1 0 1 « - 1 ) '• «ii ~ «22

=

i; «12= -1

In ~ 1; «11 -1 =

= -1 i; «11

522 = {1,3,3-1,I100]2 [110]2 [010]2} In Figures 322,

3.14 and 3.15, the multiplication tables for M^j(522) and

respectively, are displayed.

From these tables, we observed that

322 is a group. The construction of 422:

As in Chapter 4, we choose P = { a , b , c } where

c coincides with the 4-fold axis, a 4(a).

is along a 2-fold axis and b =

The metrical matrix G for P is

011 G =

0 0

0

0

Sii

0

0

Su

Since 45 = 180/4, there are two-fold axes at 45° intervals starting with the one coinciding with a.

Hence, there is a two-fold axis along b .

The elements of 422 with respect to P, 5.5) 170

given in P3.14 are (see Figure

Table 5.3:

N o n - z e r o e n t r i e s of t h e m a t r i x r e p r e s e n t a t i o n s M ^ ( a ) a n d

Mp*(a)

f o r t h e r o t a t i o n ¡sometries a of p o i n t g r o u p s 3 , 6 , 3 2 2 , a n d 622 f o r P = {a,b,c} a n d P* = {a*,b*,c*>. MpC«)

Hp(l) : t n = «22 = «3 3 = 1

V

[ 0 1 0 1

2)

: i22 = 1; 1.1 = «21 =

Mp(2) : i n = 1 2 2 = - 1 ; i 3 3 = 1

V

[ 1 1 0 ]

2)

: «21 = i,2 = ill = - 1

V

[ 1 0 0 ]

Mp([

21

2,

in

= i; «12 = i 2 2 = «33 = - 1

Mp(3)

°]2)

in

= in

= l; i 2 2 = «33 = "I

Mp(3_1)

= i; l „

Hp(6)

«21 = i n M^C[120]2>

i12

=

i2 2

=

= "I

: t2, = l „

= 1; t 1 2 = l 2 2 = - 1

: i, 2 = t,, = 1; I,, = i 2 , = - 1

: 1,, = t 2 , =

Mp(6_1)

1; ill = «33 = "I

= -1

: tn

= 1; «i 2 = - 1

= til = «33 = 1; «21 = - 1

Mp-(a)

(1) : t,i = «22 = tjj = 1

[010] 2) Hp' (

: i22 = 1 ;

Hp* (2) : «,, = «22 = - 1 ; «i. = 1

[1l0] 2) Hp- (

: t2i = « . 2

v

v

Hp*

Mp*

Hp*

([100]2)

([210]2)

([110]2)

([120]2)

: «1, = 1; «21 = «22

=

«13 = "I

Hp- (3) : l 2 , = i „

l„

= i, 2 = «3. = - 1

=

= 1; I,, = «i 2 = -1

: «11 = «12 = 1; «22 = «33 = -1

1 Hp" (3" ) : l l 2 = I,, = 1; t 2 , = «22 = - 1

: 12, = 1 „

Hp- (6) : i 2 1 = H 2 2 = I,, = 1; l 1 2 = -1

= 1; «33 = "I

: «21 = «21 = l; «11 = «3 3 = - 1

Figure

1 Hp- (6" ) : I,, = l,i = t „

5.4:

= 1; «1, = - 1

The orientat ion of the rotation axes in

322.

The 3-fold axis of the group is perpendicular to a plane of three

2-fold

axes

with

adjacent

intersecting at an angle of 60°.

2-fold axes in the plane The basis vector C is de-

fined to lie along the 3-fold axis, a is defined to lie along one of the 2-fold axes and b is defined to lie along another at

120°

so that b = 3(a).

parallels a, ^ ^ ^

Thus, 3 parallels c,

parallels a + b and

parallels

b.

The 3-fold axis is represented by a triad and each 2-fold by a diad svmbol.

171

Figure

5.5

The

orientât ion of the rotation axes

The 4-fold axis of the group is perpendicular four

2-fold

axes

with

intersecting at 45°.

adjacent

2-fold

in 422.

to a plane

axes

in

the

of

plane

The basis vector a is defined to

lie

along one of the 2-folds and b is defined to lie along another at 90° to a so that b = 4(a). and

1

2 are disposed as

j [110]-/

and

n

.

parallels - a

b.

Thus, 4 parallels c,

in 222,

'2

parallels

a + b

The 4-fold axis is represented

by a tetrad and each 2-fold by a diad symbol.

Figure 5.6: 622.

The orientation of the rotation axes of group

The 6-fold axis

is perpendicular

2-fold axes w i t h adjacent at 30°.

2-folds

to

a

plane

in the plane

of

six

intersecting

The vector C is defined to lie along the 6-fold axes,

a is defined to lie along one of the 2-folds and b is defined at 120° to a so that b = 3(a). Hence, 6 parallels [100].2, 1/ H O '2 /. . [010J. . . ,,, . /270/J. c, 1 and 1 '2 are disposed as in 322 and 1 '2

— | — y

as in 322

parallels 2a + b , lels - a + b .

l120h

parallels a + 2 b

and

l11°h

w

a

paral-

The 6-fold axis is represented by a hexad and

the 2-folds by diad symbols.

The multiplication table for M p ( 4 2 2 ) and (422) were found in P3.15.

An

examination of the multiplication table of 422 shows that 422 is closed under composition and hence is a group. T h e construction of 622:

We choose the basis used for 322 for 622.

Since

30 = 180/6, there are two-fold axes at intervals of 30° starting with the one

coinciding with a.

elements of 622

622 =

Thus, there is a two-fold axis along b.

The

(see Figure 5.6) are

{1,6,3,2,a'Ve"1,^100^,

(P5.6) Problem:

Find all of the matrices in Hp(622).

Confirm your re-

sults with those given in Table 5.3.

(P5.7)

Problem:

that Mp(622)

Prepare a multiplication table for Mp(622) and observe

is closed under multiplication, demonstrating that 622 172

is

a group.

(P5.8) Problem:

For 522 no basis exists that gives all of the pole points

and matrices with integer entries. where k axes.

Therefore, a cartesian basis is used

is along the five-fold axis and i is along one of the two-fold Find the matrices in M£.(522) and form the multiplication

table

showing that M^(522) is closed and hence 522 is a group.

Note that 522 is not a crystallographic group and hence does not map a lattice into self-coincidence.

This is why no basis can be found such

that the representation of the pole points and the matrices of the mappings consist entirely of integers. crystallographic point groups. all of this type.

This fact

is

true

for

all

non-

In particular, n22 groups with n > 6 are

However, using a cartesian basis as in P5.8, each of

these can be shown to be groups. C O N S T R U C T I O N OF T H E C U B I C A X I A L

GROUPS

In the construction of these groups we shall need to map pole points using the matrix representation of the constituent rotations.

Since the

pole points are constrained to be on a unit ball, the triple representing a given pole point can be somewhat complicated. [/3/3,/3/3,/3/3]t.

the pole point

For example, consider

As this pole point lies on the zone

[111], we shall use [111]1" to represent the pole point to simplify the computations.

We shall see that each pole can be easily represented in

this manner.

(E5.11) of

222,

Example: form

the

Using the zone symbols to represent the pole points equivalence

classes

of

pole points with respect to

222-equivalence.

Solution: points.

Since 222 has three 2-fold rotation axes,

it

has

six

pole

Recall that by the way the basis P = { a , b , c } for 222 was defined,

the triples for these pole points on the unit ball B are

{a/a,-a/a,b/b,-b/b ,c/c,-c/c}

.

The representatives formed from the zone symbol associated with these pole 173

points are {[a]p,[-a]p,[b]p,[-b]p,[c]p,[-c]p]}

=

{[100]t,[100]tJ[010]1,[010]t,[001]t,[001I1}

The

equivalence

{g(p)

class

| g t 222}.

of

a

given

Starting w i t h

pole

.

point

p

is

[a] p , the equivalence class

the

set

[a] of

a

is [a]

=

{Mp(g)[a]p

| g t

[b]

=

{[b]p,[-b] p )

[c]

=

{[c]p,[-c]p}

222}

= {[a]p,[-a]p}

.

Similarly,

and

This

.

is consistent with the information given in Table 5.1 w h e r e w e ob-

serve that the pole points of 222

are

partitioned

into

3

equivalence

classes with two pole points each.

(P5.9) 322,

Problem:

form

the

322-equivalence.

Construction

of

satisfied if 332

c

Using the zone symbols to represent the pole points of equivalence

classes

of

pole

points

with

respect

to

Confirm your answer by referring to Figure 5.1.

332(=23):

We

shall discover what conditions must be

is to be a group.

Once these conditions have been es-

tablished, we shall use them to determine the interaxial angles that must occur between the generating rotations it is a group.

and then construct 332

We begin by noting that if there

does

not

and show exist

two

third-turns whose composition is a half-turn, then the set of all third turns (recall that the inverse third-turn is a third-turn about the other end of the axis) together with the identity would be a closed set composition

which would violate part (2) of T5.10. in 332

under

and hence we would have a polyaxial group of the form

Hence, there exist two third-turns

whose composition is a half-turn.

In Appendix 6, we showed (PA6.2

and PA6.5) that when 33 = 2, then

• • • >9 n ) denote the generators of C. under C

If L is invariant

then it is invariant under each of {g^ .g^, . . . >9^) • N°w suppose

L is invariant under each of {g l ,g„, . . . ,g } . Let g t C. 205

Then

g is a

finite product

(composition)

}, say

g

Then

h 1 h 2 -. .hf where h(. £ { g ^ g ^ ... ,g n ). g(Z.) =

of

h 1 h 2 ...h f (Z.)

=

h 1 h 2 ...h f _

=

h ¿L)

=

L

^L)

.

•

Given a point isometry a and a lattice LQ we need a strategy for determining whether L ^ is left invariant under a or not.

Note that

a{LQ)

is the lattice generated by a(£>) = (a(a),a(b),a(c)} since a(L/a + vb + W C ) = t/a(a) + Va(b) + wa(c) Hence L^

= a ( L i f

and only if D and ot(D) generate the same lattice.

Using T6.1, to show that L ^ is left invariant under a, we need only show that each vector in D can be expressed as an

integral

combination

of

a(D) and vice versa. (T6.13) Theorem: a basis of S.

Let C denote a point group and let D = {a,b,c} denote

Then L^

is invariant under G if and only if M^(a) is an

integral matrix for each generator a of C. Proof:

Recall that

M 0 (cO

[a(a)]

D

[«(b)]D

[o(c)]D

which is the change of basis matrix from a(D) to D.

Since we know (see

CA3.8) that det(M_.(oO) = ±1, T6.3 yields the result that /.„ = /. if U U a\U) and only if M^(a) is integral. • (E6.14) Example - A lattice left invariant under a point group C: a and f5 are generators for some point group C,

206

Suppose

D = {a,b,C} denotes

a

basis for S and a(a) = b , a ( b ) = - a + c, a(c) = c, 3(a) = - b , 3(b) = - a , and 3(c) = -C. Solution:

Show that LQ

is invariant under G.

To show that the lattice L p is invariant under C , we must show

that M n (a) and M n (3) are both unimodular.

M D ( then T 2 = TJT * and p2 = 1 0 ! where T is the translation such

=

that t(p!(o 2 )) = o 2 . Proof:

Note that TPI(O2) = T ( P I ( o 2 ) ) = o2

and so TPJ is a point isometry.

Since T X T ' is a translation and iPt is

a point isometry such that a = (lit "*")(tPI), P2

=

,

and

TPJ.

Suppose the origin O is fixed and a is an isometry. there exists a unique P(o)

1

by T/.10, t 2 =

= o and a = TP.

the linear

component

with respect to O .

translation

D = {a,b,c}

and

point

isometry

We call t the translational

component

P such

that

of a and P

of a (since a point isometry is a linear mapping) Since a = TP, a(r)

where t = T ( o )

T

Then by T7. 10

=

t(P (r))

=

P(r) + t

.

is the translational vector for T.

is chosen.

NOW suppose a basis

Then P can be represented by the 3 x 3

matrix

M^j(P) and the translational vector t of T can be represented as the triple [t]p.

Creating a 4 x 4 matrix, denoted

= {M^(P) | [t]^}, from

these in the.following manner,

RN(A) = {MN(P)

MD(P)

| [t]D

0 0 0

I I

| [t]n} =

238

1

we

obtain a matrix representation for a.

In order to accomplish this

representation, we use 4-tuples, elements of R1*, to represent the vectors, r in S .

The 4-tuples are constructed by putting a dummy 1 in the

fourth position below [r]£j-

(E7.12) Example:

Verify that R^Ca) = {M^CP) | [t]^} represents a = iß

where t = t(o) in the sense that

[«(r)] D

" [r] D RD(a) 1

Solution:

Let "il MD(S)

=

i2l

^12

^13

^22

^2 3

£39

[t] D =

£33

tr] 0 =

Then fc 1 2

£ 13

tl"

X

12 1

^2 2

^2 3

tz

y

2-3

t. 1

_1

"Uli

[r] D { M D ( ß ) I [t]D' n}

=

1

d3 2

£3 3

0

0

0

« h *

+

Î12 y

+

1Î13Z

z

+

ti

12 iX + l22y + l22z + 12 i31x

+ n 3 2 y + 1"

t2

13 0

[t] D

MD(3)[r]D + 1

0

[t] D

[P(r)] D +

0

1

[fHr)] 0 + [t] D

1

[P(r) + t ] D

Kr)]

D (by (7.8)) •

In E7.12 we showed that the 4 x 4 a where a = xfi.

matrix given in (7.9) represents

The notation {M^(B) I [tj^}

(Seitz, 1935) for a with respect to o and

D.

(D7.13)

an

Definition:

Let

a =

if5 denote

is

called the Seitz

isometry

where

notation

t

is

the

translational component and 3 is the linear component of a with respect to

the

origin

O.

Let

{M^(3) I [ t (o) ] i s respect

to

Let

D

denote

a

basis.

called the 4 x 4

matrix

Then the matrix Rp(a) representation

of

a

= with

D.

O

denote

a choice of origin and D

isometries, then so is aTS and since

240

a basis.

If a and Z are

aï(r)

=

a(ï(r))

,

[r]D

[r]

R D ( a ï)

D

RD(a)|RD(ï)

Since matrix multiplication is associative, this equals

[r]

D

RD(a)RD(ï)

Hence R0(aï) = RD(a)RD(ï)

(7.10)

The composition aï can also be viewed in terms of the decomposition described in T7.9.

Suppose Ï = tiBi and a = T2fÎ2> then

(aï)(r)

Hence, a? = tf3 where 3

=

=

(T2MiPi)(r)

=

(TîPîtJCBiCr))

=

( t 2 e 2 ) ( M r ) + t i (o))

=

T 2 CB 2 CBiCr)) + P î d x C o ) ) )

=

B

2

M r ) + M t x C o ) ) + t2(o)

3 2 Bi and T is the translation whose translational

vector is P2(t1 Co)) + t 2 (o).

Therefore, in the Seitz notation, given an

origin o and a basis D,

R 0 ( a i ) = {M 2 M! I M 2 11 + t 2 }

,

(7.11)

where Mj = M p C M , M 2 = M ^ C M , t t = [Tjio)]^, t 2 = [t 2 (o)] 0 -

Note that

tt

R^(ai) =

and

t2

are

defined

here

to

be

vectors

in

R* .

Since

RD(a)RD(y),

{M 2

(P7.11) Problem: for any 3 x 3

I t 2 }{M! I t j = {M 2 Mi

I M2tx + t 2 }

.

Using matrix multiplication, confirm that (7.12) holds

matrices Mj and M 2 and any two vectors t x and t 2 in R3 .

241

(7.12)

Suppose that R

n(a)

=— {M ^LI | | t}. WJ . Then I IICII we W C can A N find I IIIU Rn(ot

by noting

that RpCa" 1 ) = (R^Ca))" 1 and observing that if R^Ca" 1 ) = {N | r} then

{M | t}{N | r} = {I3

I 0}

.

{MN I Mr + t} = {I3

I 0}

.

Hence, by (7.11)

Hence N = M

1

and r is such that Mr + t = 0.

That is,

r = -m'H Consequently, {M I t}"1 = {M" 1 I -M _ 1 t} (P7.12)

Problem:

Problem:

(7.13)

Use (7.11) to confirm (7.13) by showing that {M | t}{M _1

(P7.13)

.

| -M _ 1 t} = {I, | 0}

.

Show that {M I t)"1 = {M" 1 I -M _ 1 t}

by extending the procedure for calculating the inverse of a 3 x 3 matrix given in Appendix 2 to the 4 x 4 (T7.14)

Theorem:

case.

Let / denote the group of all isometries.

Then T is

a normal subgroup of /. Proof:

Let a E / and T E T.

According to the remark following DA7.11,

we need only show that ata ^ E T.

Using the Seitz notation with respect

to an origin O and a basis D, we can write R^j(a) = {M | tj} and Rpit) {I3

I t 2 }.

By (7.13), R^ia" 1 ) = {m" 1 | - m " 1 ^ } . R^axa"1)

=

{M | ^ > { 1 , | t j H M " 1 {M I Mt2 + t ^ i M " 1 {I, I Mt2 + ti {I, I Mt 2 }

242

.

=

Hence | -M"1^}

I -M"1^} M (7.14)

Hence

aia

1

is

a translation.

Therefore cua * e T for all a s I and

T £ T and so T is a normal subgroup of I.

This

result

isometries.

says

that

we

can

•

almost

commute

translations

and

Since ata ^ £ T where a e I and T E T, we have ata ^ = T'

for some T' E T.

Hence at = T ' a.

That is, if we start with ax we can

move the translation t to the other side of a if we trade x for a different translation T'.

This will be used to simplify some of our expressions.

(T7.15) Theorem:

Let a denote an isometry, let Oi and o 2 denote points

in S and let D = {X

x

,T } denote a basis of T. y' z

Then

where ¡5! and f$2 are the linear components of a with respect to O! and o 2 , respectively.

Proof:

Let { a ^ b ^ C i } denote the vectors in D{oj

in D{o2). column

The first column of M n ,

and {a 2 ,b 2 ,c 2 } those

*(Pi) is [Pi(al')]ri,

U(Oi)

of M n ,

,(P2)

t(Pi(o2)) = o 2 .

Hence

is

[P 2 (a 2 )] n .

g 2 (a 2 )

=

•

By

T7.ll

. and the first oj P2 =

TPJ

xPx(a 2 )

=

TP!(T X (O 2 ))

=

t(P 1 T x )(O 2 )

. t

Since T is a normal subgroup of / there exists a T

IT 1

and t

= P! T x &l^

£

r.

X

such that

= T P X

Hence

5 2 (a 2 )

=

TCT^PJCOJ)

=

T X (I(P 1 (O 2 ))

(R is abelian)

=

T x (°2)

(since T(PI(O2)) = o 2 )

=

P1xxP11(o2)

243

where

Also M ^ i ^ d )

=

MVß/coi)))

=

PI(t

=

Pi(«i)

Since PJT^B!^ c T, there exists p,q,r

M1

and so [ P I ( 3 I ) ] N , . l of M n .

1 PT = 1

.. = [pc/r]^

of

0(0l)(Pl)

Mn,

=M

.(PI)

equal

(since P I fixes Oj)

•

z R such that

, '

x y z

= [P2(a2)]n/-

. ( P I ) equals that of M n ,

columns M

x

(Oi))

i-

Hence the first column

..(P2).

Similarly, the second and third

those

of

Mn,

•

Therefore,

D(0l)(P2)-

Theorem T7.15 tells us a remarkable fact: the component

- (32)

of a is

the

same

regardless

nature

of the choice

of the

of origin.

linear For ex-

ample, if Pj and P 2 denote the linear components of a with respect to Oi and o 2 , respectively, and if f$j is, say, a half-turn about

an

axis

£

through Oi, then p 2 is a half-turn about an axis I' through o 2 where £' parallels I. Besides being important to our derivation of the

crystallographic

space groups, the next several results will enable us to transfer information about a crystal structure that is described in terms of a basis Di(o,) to a new basis and origin D 2 ( o 2 ) . ( T 7 . 1 6 ) Theorem:

Let a denote an isometry, let D = ^ T x ' T y > T z ^ denote

a basis for T and let O] and o 2 denote points in S.

R

o(0l)(a)

=

01

where M is a 3 x 3 matrix and t E R3,

1 t}

'

then

R D ( O j ) ( c 0 = {M | (M - I,)p + t} where p = Proof:

Then if

,

[o2]D(oi).

Let Pi and P 2 denote the linear components of a with respect to

0[ and o 2 .

Since M = M n .

. ( P I ) , T7.15 implies that M = M n , ,(P2). OjJ o2j 244

Let Tj and t 2 be the translational components of a with respect to Oj and o 2 , respectively.

Then

t2 =

x *(o 2 ) =

x iT *

where

Rp

Bi(o2).

translational vector for t 2 with translation such that t p C ° i )

T

Since

-1 t

Pi(o 2 ),

(oj) =

{M | [ t 2 ( o 2 ) ] > D

(a) =

We

respect

= o2.

to

shall

next

Oj

Let

Then

}

.

By

T7.ll,

determine T P

the

denote

the

-1

=

Tit

=

1,1

=

dlTp

-1

- 1

T T P P )(T

Tp)

(T is abelian)

^ ) represents

[ 5 x (o2 ) ]

the translational

to o,. The translational vector of T p 1 , P with respect F i respect to Oj, is represented by —[o 2 ] Since t represents

vector of I with

the translational vector of T j,

12(0!)

= = =

M

t°^D(0l)

l 3 [ 0 2 ]

-

(M- l3)[o2]0(oi)

+ t

D(0l)

+t

t

,

+ t is the translational vector of t 2 with respect to

CM - l3)[o 2 ] ri . D{Oi).

+

[ M ° 2 ) ] D ( 0 i ) - [°*]D(0l)

But the translational vector of t 2 with respect to D{o2) + t by (7.7).

also (M - I 3 ) [ o 2 ] n

R

where p = [o2^

D(o2)(o°

=

{M

1 (M

"

l3)P

( T 7 . 1 7 ) Theorem:

using the R

+

t}

'

D (oj' Let D j and D2 denote bases for T, o denote the origin,

T denote the change of basis matrix from D1 11

is then

Consequently,

to D2

and let r e S.

Then,

representation of r, we have

[r]

Dx (o)

D 2 CO)

{T | [OOO]1"}

The proof of T7.17 is straightforward and is left to the reader. 245

(T7.18)

Theorem:

points in S .

Let D denote a basis for T and let O; and o 2 denote

Let p =

[r]

Then if r t S,

[r]

0(o2) = (la

0(Ol)

I "P>

Again, the proof of this theorem is left to the reader. (P7.14) Problem:

Show that if we move the origin from o x to o 2 and if

we then change the basis from Di to D2,

[r]

D2

we have

[r]

( O J )

Dx (0l)

= {T | -Tp}

where T is the change of basis matrix from D j to D2, r

E

p = [o2]n,

. and

S.

(P7.15) Problem:

A structural analysis of coesite by Zoltai and Buerger

(1959) shows it to be monoclinic with space group symmetry B2/b.

The

coordinates of the atoms determined in the analysis (Table 7.1) define a framework structure of silicate tetrahedra with double-crankshaft chains like those in sanidine, KAlSi 3 O e , a monoclinic feldspar with space group symmetry C2/m.

The structures differ in that the chains in sanidine are

related by a mirror plane whereas those in coesite are related by a glide plane.

Figure 7.4 displays the sanidine structure between 0 and i.

The

origin Oj and basis vectors D j = { a ^ b ^ C j } define the setting of

the

coesite unit cell and 0 2 and D2 sanidine

cell.

= {a 2 ,b 2 ,C 2 ) define the setting of the

The connection between these cells was established by

Megaw (1970) who wrote a matrix for transforming the coesite coordinates to match those of sanidine.

Following a comparison of the resulting co-

ordinates with those of sanidine, she concluded that the coesite structure is impossible for KAlSi 3 0, because there are no enough to accommodate K.

cavities

in

it

large

She also concluded that the feldspar structure

is impossible for coesite because it would require a bridging SiOSi angle of 114°, which theoretical evidence indicates is too narrow for stability (Gibbs, 1982). 246

F i g u r e 7.4: of

the

A drawing of the tetrahedral

double —crankshaft

chains

spanning the structure along

in

[100].

nodes

sanidine

The origin

O x and the D x - b a s i s vectors (Cj is perpendicular to the plane forming a right-handed system) define the setting defined by Zoltai and

Buerger

(1959) for the coesite unit cell, and o a and the D2-basis

vectors

(b2

is

perpendicular

to the

plane forming a right-handed system) define the setting for the sanidine unit cell. d is the change of origin vector.

The vector This drawing

is modified after Megaw (1970).

T a b l e 7.1:

Zoltai

C o m p a r i s o n of the atomic c o o r d i n a t e s of coesite and s a n i d i n e .

&

Buerger

Atom

Sanidine

Megaw

(1959)

(1970)

yi

Ii

Atom

X

I

y

Si,

.1403

.0735

. 1084 -.3597

. 1084

.3168

T,

.7089

.1178

.3444

Si,

.5063

.5388

.1576

.1576

.2175

T,

.0097

.1850

.2233

. 3653

0

. 28S8

0

.1472

0

.1793

.1269

.4025

.1722

.1469

.2244

.0341

.3100

.2574

.0063

0,

0

0

0

0

1/4

0,

1/2

3/4

.1166

0

.1166

0

0,

.7306

.5595

.1256

.2306

.1256

.4211

0,

.3080

.3293

. 1030 - . 1 9 2 0

.1030

.2287

0,

.4877

.5274

.2878 - . 0 1 2 3

.2878

.2103

-1/2

247

°A(2)

"

°A(D °D °B °C

"

(1)

Examine the drawing of the double-crankshaft chains in Figure 7.4 and show that a 2 = aj + fc>! b2 = ca c 2 = -fc>! and [o 2 ] D i = [p] D i = [4i3/4,0] .

(2)

Formulate the 4 x 4 transformation matrix 1 0 {T | -Tp} =

(3)

0 -i

0

0

1

-1

0

i

0

0

0

1

1 0

Use the matrix {T | -Tp} to transform the atomic

coordinates

of coesite (Zoltai and Buerger, 1959) to match those in sanidine and compare your results with those obtained by Megaw

(Table

7.1, Column 3). Let D j and D2

(T7.19) Theorem:

denote bases of T, let Oj and o 2 denote

points in S and let a denote an isometry.

R

02(o2)(a)

=

{

™T~1

1 T[(M

"

Then

l 3 ) P + t]>

*

where R n . . (a) = {M | t}, T is the change of basis matrix from D x to Is 1 vO 1 ) D2 and p = [ ° 2 ] D i ( 0 i ) . Proof:

Consider the following circuit diagram where r e S:

[r] Di(oi)

{T | -Tp}

o2) 1

1 {M | t}

[«Cr)] Di(o,)

[r]

D.(o,)(B) {T | -Tp}

248

t«(r)]D ( O ) 2 2

It follows from the circuit diagram that Rn

. .(a) = {T | -Tp}{M | t}{T | -Tp}" 1

(7.

= {T | -Tp}{M | tHT" 1 | p} = {TMT"1 | -Tp + T[t + Mp]} = {TMT"1 | T[(M - I 3 )P + t]}

.

•

Note that (7.15) expresses the fact that R n . (a) and R n , . (a) U2KO2) i-'lKOl) are similar matrices in the realm of all 4 x 4 real matrices (see Appendix 7). CRYSTALLOGRAPHIC SPACE G R O U P S Not all groups of isometries are suitable for describing the symmetry of crystals.

In Chapter 4, for example, we found the crystallographic

restrictions that must be imposed on point isometries.

In this section

we study the restrictions that must be imposed on groups of isometries that leave the structure of some crystal invariant. a crystal implies that there structure.

is three-dimensional

The very nature of periodicity

in

the

This periodicity is described in terms of a crystallographic

translation group. (D7.20) Definition:

A translation group T

Let o denote a point in S.

is said to be a crystallographic

translation

group

if and only if orb^-(o)

is a three-dimensional lattice whose vectors emanate from o. Note that, by (7.6), if T

is a crystallographic translation group

and if {a,b,c} is a basis for the lattice orbT-(o), then D = { T ,T ,T } T ' x y z is a generating set for T where

T

= a

> T y(°)

v(°)

=

b

T

z (°)

= c

-

That

is, if orby-(o) = {ua + vb + wc | u ,v ,w e Z} then

orb-,- (o) = { T U T V ' T V V ' ( O )

T

and so

(D7.21) Definition:

'

x y z

t U V W , / = {r t t t x y z A

set

of

I u ,v ,w e Z)

,

,

u,y,weZ} ' ' translations D = (T 1 ,T 2 ,ti)

249

of

a

translation group T

crystallographic

T

is said to b e a generating set of

if and only if D is a basis for T and -r '

Let D = { X 1 , T 2 , T 3 }

=

,

U

( T I T

denote

V W 2

I J

a

. I

U,V,W

t

generating

translation group T and let o s S.

,,

Z}

set

.

for

the

crystallographic

T h e n D ( o ) is a basis for the lattice

orby-(o) as shown in the following problem.

(P7.16)

Problem:

Let T denote a crystallographic translation group and

let D = { T 1 j I 2 , T ) } be a generating orb-y-(o) w h e r e o e

set for T.

Show that Lp^^

S.

We shall now turn our attention to groups of

(P7.17)

Problem:

Show that if G

isometries.

denotes a group of isometries, then the

set T of all of the translational isometries in C

is a subgroup of C .

The subgroup T in P7.17 is called the translation (D7.22)

space

Definition:

group

is the

group

A group of isometries C is called a

if the translation group of C

of

C.

crystallographic

is a crystallographic

trans-

lation group.

(T7.23)

Theorem:

Let C

the translation group of C

(P7.18)

Problem:

denote a crystallographic

space group.

Then

is a normal subgroup of C .

Prove T7.23.

(Hint:

See the proof of T7.14 where

C

will play the role of / and the translation group of C will play the role of T.)

Let C

denote a crystallographic

translation group. as shown in P7.16,

space group and

If D =

a

i-s t h 6

the 3 x 3 to D.

n

'- S

let T set

f°

lattice equal to orb^-(o).

then, by T7.15, the matrix representation respect to D{o)

8

enerat:

for the

250

| a E C}

,

then,

with

Let M ^ ( a ) denote

m a t r i x representation of the linear component of a

M 0 ( C ) = {MD(a)

T,

its

If a s C ,

linear component

is the same for all choices of origin O.

Also, let M ^ ( C ) denote the set

denote r

w i t h respect

where C

(P7.19)

is any groups of

isometries.

Let

Problem:

C

denote

{ t , t , t } denote a basis bas for T. x' y' z See (7.12) and (7.13)).

a

group

of

isometries

Prove that M ^ ( C )

and

let D =

is a group.

(Hint:

H e n c e a s s o c i a t e d w i t h each c r y s t a l l o g r a p h i c space group C there is a group of 3 x 3 m a t r i c e s all possible

crystallographic

ciated group of

and a lattice L-,. .. U{ O)

U{u)

space groups

3 x 3 matrices

C

and lattice

that

these linear components (D7.24) origin.

of the

the same asso-

and

Before

Since

isometries

how

in C ,

M^(C)

we will

study

first.

Let a denote

Definition:

share

see

can be constructed.

w e can do this w e must learn more about M ^ ( C ) represents the linear components

We will

an isometry

let o e S denote the

and

T h e n w e d e n o t e the linear component of a by

AQ(a).

If C

A

and A

is

a

group of isometries, then we define A Q ( G ) to be

A Q ( C ) = {AQ(a)

(P7.20)

Prove that A Q ( C )

Problem:

By T7.15, note that

| a £ C}

.

is a group.

for any points

Oj, o 2 £ S,

o!

(C)

are isomorphic since they are b o t h represented by M ^ ( C ) . w e will discover the relationship b e t w e e n M ^ ( C )

(T7.25)

Theorem:

Let

C

denote

Using

o2

(C)

Ao(C),

and

a crystallographic

space group, D =

a g e n e r a t i n g set for its t r a n s l a t i o n group, and o the origin. Then AQ(C)

leaves

invariant.

crystallographic p o i n t

Proof: V e LQ^

Recall

that

Hence

AQ(C)

one

of

the

32

groups.

V

if

if and only if

is

a

[v]^^

vector E

in

% o )

( e )

=

{ M

D(B)

I

3

0 }

and D ( 0 )

(T)

= (I,

I

251

[V]

emanating

S

L e t

Let T denote the t r a n s l a t i o n s u c h that x ( o ) = V .

R

is

D ( 0 )

>

E

AQ(G) Then

from O,

and V £ L

then

Hence R

Since the

D(o)(ikfrl)

translation group

v

Hence [P( )]£)(0) of L p ^ y

z

an
« C or t E Z 3 + [i,i,0] t .

if and only if t2 = tx + t where t E V [-2.5,1.5,-S]1 = {MD([010]2

[-3,l,-3]t + [i,i,0]^ E

| [000]*} E

R

DD (( oO ))

Since

Z 3 + [i,i,0]t,

(C)

The next theorem yields more information on what types of elements t of R 3 can occur in matrices {M | t } in ( T 7 . 2 9 ) Theorem: HII.-jjh

Let C denote a crystallographic space group, let h 1 }

E A (C) be such that h 1 h 1 . . . h A f = h fc

Let D

denote

a basis

for

1hfc +

+

T and

{Mj | t j} , . . . > (M ^ | t ^ } E

R

...hn

2

.

(7.16)

let O E S. C

0(0)( )'

Let M. = M,-,(h.).

i

then

there

D r

If

{I3 | s} E

exists

such that

(MI

I TXHM,

{13

| T,}...^

I -HM

( P 7 . 2 1 ) Problem: (D7.30)

Definition:

f c + 1

| TK}

I tfc

+

= 1 HM f c +

2

I tfc

+

2

)...{Mn

| y

.

(7.17)

Use T7.26 to prove T7.29. Let H denote a point group, let T denote a trans-

lation group and let D denote a basis for T.

If H!,...,H

E H are such

that the relation

h l h I . . . h f c = hfc

+

1hfc +

254

2

...hn

(7.18)

z R3

holds and if M. = M,-.(h.)> then we say 1 that t, 1 , . . . ,t i D r ' ' ' n with

the

relation

{Mj | ^>,...,01 R^(7"). we

say

(7.18)

with

| t } satisfy

respect

condition

T

to

(7.17)

for

consistent

are D

and some

if

{I3 | s} E

In the case where the relation is in the form h 1 h 2 ---h^. = 1, that

t 1( t 2 ,...,t^

tj ,t2 , . . .

is

is consistent.

consistent

with

the

relation

if

That is, if

{Mi I t J t M j | t2)...{Mfc | t k ) = {I, | s}{I3

| 0} = {I, | s}

e

R

0

(D

.

In the terminology of D7.30, T7.29 says that if the relation (7.18) occurs

where

hi,..., h

E

A (C)

and

if

tu-.-.t

E

ft3

such

that

{M. 1I t.} E R „ . .(C) must be consistent J where M. = M~(h.), then t,,...,t 1 / r D(o) i D r ' ' n with the relation (7.18) with respect to T(C)

and D.

This result will

enable us to transfer relations from a point group to its corresponding space groups.

See Grossman and Magnus (1964) for a discussion of

the

generators and relations approach to group theory. By

T7.26,

we

see

that

each

matrix

M

such

that

{M |

TJ.} E

where C is a crystallographic space group, is associated with a collection of elements of C characterized by |{M I t 2 }

{I, | t 2 - tx>

{M I t 2 }

{I3

E

7(C)J =

| t}{M | t x } = {M | t 2 } where {I, | t> e T ( C ) |

= RD(T(C)){H

| t,}

,

the right coset of Rp(7~(C)) with respect to {M | t!). of A Q ( C ) corresponds to a right coset of

T(C).

Hence each element

This

correspondence

yields an isomorphism from A Q ( C ) to

G/T(C).

(P7.22) Problem:

to R D ( o ) (C)/R D (T (C)) defined by

Prove that e : A Q ( C )

6(&) = R 0 (7~(C)){M | t} where M = M ^ B ) and {M | t} z isomorphism

(you may

choose

to

skip

R0(o)(C)

is

an

this problem on first reading).

Conclude that A Q ( C ) is isomorphic to C/7"(C). We are now ready to give a clear statement of how we will construct all of the crystallographic space groups C.

Let H denote an oriented

point group and let L denote a lattice left invariant by H.

Let P denote

that basis for the primitive sublattice of L associated with H and let O denote the origin. sponds to L by

.

Then we denote the translation group that correThat is

255

TL

= {T E r | X(o) £ L}

Let M p ( H ) = {M,,Mj,...,M }. R

3

.

(7.20)

Then we seek all possible

T L ^ , . . . , !

E

such that

Rp(C) = RpC7'z_)[{M1 | t1},{M2 | t 1 },...,{M n | t n >]

(7.21)

for some crystallographic space group C such that Rp(7~(C)) = R p (7" L ) and

(7.22) Mp(C) = MPCH) By

finding all such sets

crystallographic

space groups

.

{^,...,1 } we will

find

all

of

the

(through their matrix representations).

The strategy that we will use to obtain these sets is to determine

a

sufficient number of relations on H so that a set {tj ,t2, . . . ,t } yields an Rp^ o ^(G) if and only if the set is consistent with all of these relations . Once all of the crystallographic space groups satisfying (7.22) for some choice of H and L are found, we will need to classify them into space group types. employed.

To do this, many different equivalence relations have been

Some of these are discussed by Hans Wondratschek in Chapter 8

of ITFC (Hahn, 1983).

We shall use an equivalence relation that gives

rise to the traditional 230 crystallographic space group types. idea behind this relation is relatively simple.

The basic

Suppose C j and C 2 are

two crystallographic space groups. If there exist points Oj and o 2 and generating sets D^ and D2

of T(GI)

and 7~(C2) such that Di(oJ

D 2 ( o 2 ) are both right handed and R n , , (C x ) and R n •JI(OI) same set of matrices, then C1

and C 2 are equivalent.

,

and

,(C 2 ) are the

Hence to show that

C j is equivalent to C 2 we show that, by a change of origin and generating sets, the matrix representation of one can be converted into that of the other. In view of T7.16 we have the following definition. (D7.31) Definition:

Let C j and C 2 denote crystallographic space groups.

Then Cj is equivalent to C 2 if and only if there exists a vector R

3

r E

and an integral matrix T with det(T) = 1 such that the matrix rep-

resentations {M | t} of the elements of Cx with respect to D, a generating set for 7"(C), become those of C 2 by the transformation 256

{TMT-1

01 I t}

I T[(M - I,)p + t]}

When discussing a space group C where orb-y-^^(o) is

a

centered-

lattice, we will usually write the matrices representing A ^ C C ) with respect to the basis P of the primitive lattice.

However, in order to test

for equivalence, D7.31 requires that {M | t} be written with respect to a basis D that generates T(C).

While this is a serious problem relative

to the change of basis matrix T, we need not change to the basis D in order to detect equivalence that occurs because of a change of origin as the following theorem shows.

Let a and (5 denote isometries, let D

(T7.32) T h e o r e m :

and P

denote

bases of T, 0[, o 2 t S and T be the change of basis matrix from D to

P.

Then R

D(oi)(a)=R0(o2)(fJ)

R

P(°i)(a)

if and only if

Proof:

R

P(O2)(P)

Suppose R D ( 0 i ) ( a ) = R D ( 0 2 ) ( P ) .

where V £ S,

R

=

D(o2)(^

If

R

D(o1))(a) =

{M

I

[v]

DC0l)}

then R n . (3) = {M | [ w ] n , . } where W is such that U(o1) L>{ o i j

=

{M

I

(M

- ^ ^ ^ ( o j

+

^ ( o ^

=

{M

I

[V]

D(0l)}

•

Hence R p ( o i ) ( a ) = {TMT" 1

| T[v]D(oi)>

and

R p ( 0 i ) ( P ) = iTMl" 1

| T[w]D(oi)>

Then R

P(o2)(e) = i ™ 1 " 1

I

(

™ T " 1 " l3)[o2]p(oi) + T[W]0(Oi)>

= {TMT" 1

| T(M - I 3 ) T " 1 [ o 2 ] p ( o i ) + T [ w ] D ( 0 i ) }

= {TMT" 1

| T[(M - l 3 ) [ o 2 ] 0 ( O i ) +

(Since T _ 1 [ o 2 ] D ( o i ) = = {TMT" 1 =

R

P(

0 l

[W]D(0i)]}

[o2]D(0i))

| T[V]D(oi)} )^

•

The converse is proved in a completely similar fashion.

257

By T7.32, we may use the basis P of the primitive lattice to detect equivalence due to change of origin.

However, one must change to the

basis of the lattice under consideration to detect equivalence due to a change of basis. C R Y S T A L L O G R A P H I C SPACE GROUP OPERATIONS When an oriented point group H and a lattice L are given, to find the space groups satisfying (7.22) we must search for sets of elements in R3 that are consistent with the relations in H. = 1 is one relation that is always present.

Let h E H.

Then

The next theorem gives

the constraint that is imposed by this relation. (T7.33) Theorem:

Let h be a point isometry, let T denote a translation

group and D be a basis for T.

Then t is consistent with the relation

ho(h)= 1 W i t h respect to T and D if and only if {I 3 | Nt} t RQ(.T)

where

N = M + M 2 +...+ M0(-h:) and M = M^Ch). Proof: to T.

Suppose t is consistent with the relation h 0 ^ ' ' = 1 with respect Then {M | t}°(h:) = {I3 | s} E Rp(7).

But note that

(M | t } 2

=

{ M 2 | Mt + t}

{M | t} 3

=

{ M 3 | M 2 t + Mt + t}

{M | t } ° ( h )

=

{M°(h)

| M ( ° ( h ) _ 1 ) t +...+ Mt + t} .

But {M | t } ° ( h ) = {I, | s}. Since M ° ( h ) = I 3 , t = M ° ( h ) t and so, by rearranging terms, we have S = (M + M 2 + ... + M ° ( h ) ) t = Nt . By reversing this argument the converse is also proved. The constraint described by T7.33 gives rise to two new types of isometries.

The screw and glide operations. 258

•

( E 7 . 3 4 )

Consider a space group

E x a m p l e :

R p ( T ( C ) )

=

Let P

R p C T p ) .

=

denote the basis chosen in Chapter

{ a , b , c }

6 with C along the axis of 4 ,

a and b perpendicular to C with Y = 90°

such that D is a right-handed system.

M =

Furthermore,

since

the

o r b - j - ^ (O)} is Z . V

where N = M + M

p

( 4 )

=

lattice

Then

0

- 1

0

1

0

0

0

0

1

type

is

{[v]^

primitive,

| V c

Then by T7.29 and T7.33, if {M | t} E C, then Nt £

3

2

M

such that M p ( G ) = M p { t ) and

C

+ M 3 + M*.

Calculating N we obtain

N =

0

0

0

0

0

0

0

0

4

Hence Nt E Z 3 implies that 41 3 E Z. Hence to be consistent with 4 " = 1 ,

t

must be chosen such that t3 = 0, i, | or 3/4.

•

The

operations

found

in E7.34 are {M | [000]*}, {M | [0,0,i]t},

{M | [O.O,!]1^} and {M | [0,0,3/4]*} where M = quarter-turn

represent

screw

operations.

Mp(4) .

These

matrices

The translational component

of each of these is directed along the axis of the rotation which in this case is along C.

Hence, under {M | [0,0,i]*} a point [xyz]t is rotated

a quarter turn about C and then displaced a distance of C/4 to the point [-y,x,i

+

2 ]

t

.

Consequently, this operation is called a quarter-turn of c/4 and will be denoted 4j.

screw with a screw translation

Since the

screw translation of {M | [000]*"} is 0, it is merely a quarter-turn and will be denoted by the usual symbol 4 . [xyz] t

to

[~y,x,i

+ z] 1 ".

The operation {M I [00*]L} maps

Since the screw translation

quarter-turn screw will be denoted 4j.

presents a quarter-turn screw denoted 4^. tional convention that

Tuvwl

n

m

is C/2, this

Similarly, {M | [0,0,3/4]t) reIn general, we make the nota-

represents an n

th

-turn screw about the

vector ua + v b + wc with a screw translation of mr/n

where

shortest (nonzero) vector in the lattice in the

direction.

symmetry

element of a screw

associated rotation.

operation

[t/vw]

r

is the The

will be taken to be that of its

Hence in the case of a primitive cubic lattice,

259

^ ^^

denotes a third-turn screw about [111] with a screw translation

of 2/3(a + b + c) since a + b + C is the shortest nonzero vector in the [111] direction.

However, in the body-centered cubic

represents a third-turn screw about

lattice,

[111] with a screw

translation

of

(2/3) (|a + 4-b + i c ) = (l/3)a + (l/3)b + (l/3)c since |a + +b + ic is the shortest nonzero lattice vector in the direction [111]. (P7.23)

Problem:

Show that with respect to the primitive cubic lattice

P, the Seitz notation for

is

{M | [2/3,2/3,2/3]^}

where M =

,

and P is the basis for the primitive lattice given

in Table 6.3.

(P7.24) Problem:

Find the Seitz notation for

with respect

to

the body-centered cubic lattice / described in Table 6.3.

(P7.25) Problem:

Write the full 4 x 4

matrix for the screw

operations

discussed in P7.23 and P7.24.

(P7.26)

Problem:

Find the Seitz notation for

with respect to a

primitive cubic lattice P .

(P7.27) Problem:

Find the Seitz notation for

body-centered cubic lattice /.

with respect to a

Note that your answer here should be the

same as in P7.26.

(P7.28) Problem:

Find the Seitz notation for

with respect to a

primitive hexagonal lattice.

(E7.35) Example:

Consider the space group C of E7.34 based on 4

but

with a body-centered lattice type I.

The basis D

in E7.34 and so M and N are the same.

However { [ V ] ^ | V E orb^-^^(o)}

is now Z 3 + Zfi.i.i] 1 .

Since Nt =

[0,0,4t 3

there must exist a

[ v'iv 2 v 3 ] T E Z 3 and an n z Z such that [0,0,4t 3 ] t = [ v 1 v 2 v 3 ] t +

260

is the same here as

i

nt

V =

Since V i + ni

= 0 and v

n e Z,

l t

n = 0.

Hence 4t 3 e Z.

as in E7.34, we obtain only the operation 4 ,

4t,

42

Consequently,

and 4 3

for a body-

centered lattice left invariant under 4.

(E7.36)

Example:

•

Consider the space group C based on the point group

m with lattice type P (in the second setting with Y as the unique axis). Then 1 M = Mp(

[010]

0

o"

0 - 1 0 .

m) =

0

0

1

Nt = [2t J, 0, 2t j ] E Z 3

and so t1

If we take ti = i and t3

= 0 or \ and t3 = 0 or

a point [x,y,z]^"

is reflected across the plane (010) and then translated

parallel to it a distance of a/2 along [100]. a glide glide

operation,

(010) is called the glide

translation.

Such an operation is called plane

and a/2 is called the

The reflection plane of the mirror part of the glide

operation will be taken to be its symmetry element. translation 1

t° °]a.

is

a

along

this

glide

is

called

an

Because the glide a-glide

denoted

by

In the case where fj = t3 = -J, the glide translation is |a + |c

which again parallels the (010) plane. glide

= 0, then

translation

called an n-glide.

is not

This type of glide (where

the

in the direction of a basis vector) will be

The glide planes that arise in our example are listed

below (note that a mirror operation is a glide with a 0 glide translation modulo the translation group).

Symbol t j 0 1 2 0 1 2 (P7.29) the

Problem:

lattice

axis).

t3 0 0 2

A.

i

I

Glide

of

Operation [010] m [010] a [010] c [010] n

I

Type

reflection a-glide c-glide n-glide

Consider the space group C described in E7.36 where

type P is taken to be in the first setting (Z the unique

In this case the mirror plane is perpendicular to c .

261

Show that

m, a , b and n (recall that

the following glide operations are possible:

when no orientation symbol is given, [001] is assumed) where the glide translation of the n-glide is ia + -Jb.

(P7.30)

Problem:

Show that the glide translations of the glide

oper-

ations in P7.29 are unchanged if an /4-centered or a S-centered lattice is used instead of a primitive lattice.

Show that, in E7.36, the result

is unchanged when P is replaced by an /4-centered or C-centered lattice.

For a beautifully illustrated discussion of space group operations, see Bloss (1971) Chapter 7.

THE CRYSTALLOGRAPHIC

SPACE GROUP T Y P E S D E R I V E D FROM T H E

ONE-GENERATOR

P O I N T GROUPS

Given an oriented point group H,

a lattice L left invariant under

H and the basis P for the primitive lattice, the results stated thus far give us conditions that must hold for an {M | t} to be in Rp^^iG), G

is a space group satisfying (7.22).

essary

and

Rp(T^)

sufficient

conditions

Theorem T7.37 below states nec-

for

generate Rp(C) for some C

generator point group.

where

a t

to be such that {M | t} and

satisfying (7.22) when H

is a one-

By a one-generator group we mean a cyclic group.

Such a group H has an element a such that H = { a , a

2

,...}.

These

groups are 1, 2, 3, 4 , 6, 7, m, 3, 4 and 6.

(T7.37)

Theorem:

Let L denote a lattice left invariant by

generator point group H

the

one-

generated by a and let P denote the basis for

the primitive sublattice of L that is associated with H.

Let M =Mp(a).

There exists a crystallographic space group C satisfying (7.22) such that each element of Rp^ o ^(G) can be written in the form

{I3 where { I 3

| v}

| r}'

,

E R p ( 7 " ^ ) and 1 < /' S 0(a) if and only if r

with the relation

Proof:

I vHM

= 1 with respect to T^

is consistent

and P.

If there is such a crystallographic space group C , then, since

262

{M | r> £ a°(

a

Rp^CG),

) _ i

reSpect

by

T7.29, r must be consistent with the relation

to T ^ and P.

Now suppose that r is consistent

with a0(-a^ = 1 with respect to T ^ and P.

By (D7.30),

{M | r} o ( c ° = {I, | s}

where {I3

| s} E Rp(7~^).

,

(7.23)

We will use (7.23) to show that the set

K = R p (7" L ){{M | r}' | / = 1, . . • ,o(a)} is closed.

Let A = {I3

| s^iM

| r}' and B = {I3

| S 2 }{M | r}' be ele-

ments of K where Sj and s 2 are triples representing lattice vectors with respect to P

and where 1 < /',/' < o(ot).

Then {M | r}' = {m' | p} for some

p E R 2 and so

{M | r}'{I3 | s 2 } = {f/ I p}{I 3

| sj}

= {M' I p + M'S2} = {I, | M'S2}{M' I p} = {I, I M'S 2 HM Since L

is invariant under H,

I r}'

.

(7.24)

M'S2 is a triple representing a

vector in L with respect to P.

lattice

Hence

AB = {I, | s j ) {M | r>'{ 1 3 | S 2 H N

= {I, | S L }{I 3

I M'S2HM

I r}'

I r)'{M | r}' =

= {I, | S L + H'S2){M I r}' + ' .

where

{I3

| SI + M'S2} E Rp(T^).

If /' + /'< o(a) , then {M | r}'

in {{M | r}' | /' = 1, . . . ,o(a)} and so then AB is in K. then setting k = i + j - o(a) {M | t}' + ' = {M | t}° ( a ) {M

By (7.23) , we have, 263

| t)k

.

+

1

is

If o(a) < / + /',

I s}{M I r )

AB = {I, I S l + M ^ H I ,

= {I,

where {I3 | S 3 } E R ( s i n c e

k

I r}

I S,}{M

k

,

R p ( T i s

closed)

1 < k < o(a).

and

Hence AB E K.

(P7.31)

Problem:

Rp(T.

Suppose that A = { I 3

| SjMM | r } '

where

{I3

A" 1 = {I3 1

by confirming that AA

| - M o ( a ) - '"(s + si)}{M | r > o ( a ) " ' | 0}.

= {I3

(Hint:

use (7.23) to help with the

Now show that A ^ is in K.

simplification).

Using P7.31, we have completed the proof that K is a group. K = Rp(C)

for some space group C .

satisfied.

By the way K was constructed, Mp(G) = Mp(AQ.

Rp(T(C)) = Rp(F^) {I3 | t } (P7.32)

| Sj} e

Show that

we

show

To show that

that the only elements of the form

are those where t represents a lattice vector in L.

in Rp(G) Problem:

1 < /' < o(a).

shall

Hence

Now we need to show that (7.22) is

Suppose

Show

that

m' =

| t} =

{I3 I3

and

| s^íM

{I3

so /' =

o(a).

| r}' Then

where show

that

{I3 | t> e Rp(7" L ) and that R p ( r ( C ) ) = R p ( T ¿ ) . Since we now have T(C)

isomorphic to L,

we have also shown that

C

is a crystallographic space group. (E7.38)

Example:

•

Find all of the crystallographic space group types C

such that Mp(C) = M p (6). Solution:

From Table 6.3, the only lattice type left invariant under 6

is the primitive lattice P with basis P such that

1 - 1 0 M = Mp(6) =

1 0 0

264

0 0

1

Consider {M | r}.

By T7.33, we need to find all vectors r such that Nr

represents a lattice vector.

Then by T7.37 we will have found all of the

crystallographic space groups C that satisfy (7.22) with H = 6 and L = P.

Since the lattice is primitive

R p i T p ) = ({la

I s} I s

Z3}.

E

Since N = M + M 2 + M 3 + M* + M 5 + M e , we have

0 N =

0

0

0

0

0

0

0

6

and so Nr = [0 0 6r 3 ] , which implies that 6r 3 must be an integer.

Hence,

we have the generators

{M | [ f , / , ^ ] 4 } . {M | [r 1 ,r2 , (1/6) ] t }

, {M | [ri

{M | [ r . / i J ] ' ) , {M | {r l ,r 2 ,(2/3)] t >, {M | [rltr2,

Since

and r2

are equivalent under D7.31. position to some vector p ,

If the origin is shifted from its current

then

becomes {M|

(M - I 3 ) P + r)

.

But "0 -1 (M - I,)p + r =

1 -1

0

P2

0

0

P3.

0

Pi - P2

i

a

nd Px = rt

+

- r 2 , then

265

r2 r

r i +

r2 r

0

r

'ri

0' Pi

-P2 =

of

However many of these

{M | r)

=

(5/6) ] t } .

can take on any values, we have an infinite number

crystallographic space groups associated with 6.

If we set p2

,r2 , (1/3) ]*} ,

3.

3.

0 (M - I 3 )p + r =

0 r3.

Hence by an appropriate choice of origin, each of the generators we have found

can

be

written

in

the

form

{M | [0,0,Tj]1"}.

For

example

{M | [ 0 , 0 , 1 / 6 ] i s the representative of the infinite number of space groups generated by the elements of the form {M | [Tj ,r2,1/6]*"}.

Hence

every space group C with Mp(C) = Mp(6) is equivalent to one generated by one of

{M | [0,0,0^}, {M | [0,0,1/6]*}, {M | [0,0,1/3]*}, {M I [0,0,i)]1}, {M | [0,0,2/3]*}, {M | [0,0,5/6]1} Since

.

(M - I 3 )p has a zero in the third component, a change of origin

does not affect the third component of r.

Hence the only way two of these

generators, say {M | [0,0,1/6]*}

and {M | [0,0,l/3]t>

can lead to equivalent space groups is if there exists a proper unimodular matrix T such that {TMT"1 | T[0,0,l/6]t} = {M | [0,0,l/3]t} . It can be shown that no such matrix T exists.

In fact all six of the

generators lead to nonequivalent space group types.

Hence we have found

six nonequivalent space groups generated by G, Gj, 6 2 , 6 3 , G s and respectively.

denoted P6,

P6lt

G5,

Since these all have the primitive lattice P, they are

P62,

P63,

P6,, and P6S.

•

A brute force technique for showing that the six generators found in (E7.38) are nonequivalent can be developed along the lines of the argument used at the end of Chapter b.

More elegant approaches require the

development of further algebraic tools. book to develop those tools.

We do not have the space in this

Hence we will leave this detail unexplored.

266

(E7.39) E x a m p l e :

Consider the space group P A c c o r d i n g

to E7.38

and

T7.37,

6 RpCfii)

=

U Rp(7 /'= 1

){M

=

Rp(7~p){M

I r}

U

where falls

M = Mp(6)

RpCTp){M

and

| r}'

| r}"

R

U

I [0,0,1/6]1}, R p ( r p ) { M 2

Rp(7"p){M"

|

[0,0,2/3]*},

R

p

(r

p

(r

p

| r}5

){M

5

r}3

|

Rp(7"p){M

U

each

element

| r}6

of

,

P6!

cosets:

| [0,0,1/3]1},

){M

Rp(7"p){M

U

Consequently

in e x a c t l y o n e o f t h e s i x r i g h t

Rp(7>){M

p

[0,0,l/6]t.

r =

I r}1

Rp(7"p){M

U

|

[0,0,5/6]1},

Rp(rp){M3

R

p

(r

p

){I

[0,0,i]t>,

|

I

3

[OOO]1},

where 1 - 1 0 M

1

=

0

0

W e shall now describe the elements in R p ( T p ) { M 2

an element

1

0

0

1

0

0

0

1

0

0

0

1

where u , v , w for

u

0

analyzing

-1

0

0 "

V

1

-1

0

0

w

0

0

1

0

0

0

since

the

we

find that

the

the

matrix

cussed by Wondratschek

The

points

of

"o

-1

0

1

-1

0

1/3

0

0

1

1

0

0

0

=

lattice

in

(7.25)

and

defines

v)a +

this

that appear

form

V 1/3 + W 1

Boisen

267

the

and

a third-turn

techniques

Gibbs

screw

dis-

(1978),

operation in

the

axis

in the u n i t cell are t h o s e

for

point

(1/3) (t/ + v ) b

which

Using

occurring about an

to c p a s s i n g t h r o u g h t h e

(1/3) (2u -

example

of a space group operation

(1967)

((1/3) + w ) c

For

form

u

is p r i m i t i v e .

representation

w i t h a screw t r a n s l a t i o n of direction parallel

in e a c h o f t h e s e c o s e t s .

and Neubuser

matrix

1

| [0 ,0, (1/3) ] t } is of t h e

0

e Z

0 0

.

0 < (1/3)(2u - v) < 1

and

0 < (1/3)(u + v) < 1

Adding the corresponding terms of these inequalities we have 0 < u < 2 .

Since u is an integer, u can only equal 0 or 1.

When u = 0, the ine-

qualities become 0 < —(1/3)v < 1 and 0 < (1/3)v < 1. that v < 0 and the second that v > 0.

The first implies

Hence, when u = 0, v must equal

0. When u = 1, a similar analysis shows that v can only be 0 or 1. when u = 1, V can equal either 0 or 1. located at the origin,

Thus, besides the

Hence,

operation

there are two others in the unit cell passing

through the points [1/3,2/3,0]^ and [2/3,1/3,0]t, respectively.

An ele-

t

ment in the coset R p C r p ) { M " | [0,0,2/3] > is of the form 1

0

0

u

-1

1

0

0

1

0

0

0

0

0

-1

0

V

-1

0

1

w

0

0

0

1

0

0

1

0

0

0

1

2/3

-1

0

0

0

0

1

0

1

0

0

0

u V 2/3 + W 1

An analysis of this matrix shows that it defines a negative third-turn screw operation (3

occurring about an axis passing through the point (1/3)(u + V)a + (1/3)(2v - u)b .

Consequently, besides the (3 others in the unit

cell passing

[2/3,1/3,0]^, respectively. (3

operation at the origin, there are two through the points

[1/3,2/3,0]t

and

This is as one would expect, since (3^)^ =

and this composition does not move the axis of the 3^ operation.

(P7.33) Problem:

Analyze the matrices in each of the remaining cosets

of P6i and show that the axes are positioned as shown in Figure 7.5. (P7.34)

Problem:

Let x denote the translational isometry defined by

t(o) = t where t = 2a + 3b + 5c and P = {a,b,c}.

268

Show that

Y

Figure 7.5:

A diagram of a unit cell containing the screw axes of space group P6i each

taking place about rotation axes paralleling c .

The 6;-screw axis is symbolized by a pronged

hexad, the 3!-screw axes by pronged triads and the 2i-screw axes by pronged diad symbols.

Figure 7.6:

A diagram showing the compos ition of the trans 1ational

isometry x where

T (o) = 2a + 3b + 5c = t with the rotational screw operations comprising a 6x-screw axis that parallels c and passes through the origin.

The resulting operations formed by this compo-

sition take place about rotation axes paralleling c and are located on the perpendicular bisector of t at a distance of t/2ctn(p/2) from the line from O to t where p is the turn angle of the rotational component of the screw operation.

269

1 0 RpOO =

Find

0

0

1 0

0

0

0

0

2 3 1 5

0

1

Rp(ta) for each of the six coset representatives a enumerated in

E7.39 (For example, find Rp(ia) where Rp(a) = {M | [0,0,1/6]t}).

Observe

that these axes are located at points along the perpendicular bisector of the line segment £ from O to t as shown in Figure 7.6.

The distance

of the axis of xa from £ is (i/2)ctn(p/2) where p is the turn angle of the linear component of a and t = ||t||.

This formula can be derived by

noting that xa maps the Z axis to the line parallel to the Z axis passing through the point t.

(P7.35) Problem:

Consider the space group P62 •

Then

6

Rp(P62) = U R„(rp){M | r}' /= 1

where M = Mp(6), r = [0,0, l/3]t. P62

Locate all of the axes of elements in

that pass through the unit cell.

in the ITFC (Hahn, 1983, p. 546).

Check your answer with that given

Note that your drawing should not in-

clude those shown in the table for [r 1 ,r 2 ,r 3 ] t where Tj or r 2 are equal to 1. (P7.36) Problem: (D7.40) Definition:

Do P7.35 for P6,.

(In this case r = [O.O.i]^.)

Let C denote a crystallographic space group and let

P denote the basis of the primitive lattice associated with A Q (G).

Then

each right coset of Rp(7~p) in Rp(C) has a representative of the

form

{M | r} where r = [ r 1 , r 2 , r i ] t sentative is called a principal

(P7.37) Problem:

and 0 < r. < 1 for each /. representative

of C with

Such a reprerespect

to

P.

Use (7.24) to prove that Rp(Tp) is a normal subgroup

of Rp(C). (P7.38) Problem:

Show that the principal representative defined in D7.40

exists for each coset of R p ( T p ) in R p ( C ) and that there is only one such

270

representative for each coset.

(P7.39) P6lt

Problem:

Using E7.39, find the principal

with respect to P,

representatives

of

the basis for the primitive lattice associated

with C.

( D 7 . 4 1 ) Definition:

Let C denote a crystallographic space group and let

[xyz]*" denote a point that is moved by each isometry of C .

Then the set

of images of [xyz] t under the principal representatives of C with respect to P is called the general

position

of C

in the set are called the general

(cf. Hahn, 1983).

equivalent

positions

The triples

of C

(Henry and

Lonsdale, 1952).

Suppose that we know the triple that is the image of [X,y,z]*" with respect to a basis P under an isometry a and wish to find the 4 x 4 matrix representation

Rp

of

a

with

t

[fi(.x,y,z),f2(x,y,z),f3(x,y,z)]

respect is

the

to

P.

image.

Suppose

Then /\(x,y,z)

V = is

a

linear polynomial for each 1 £ / £ 3 and

X

^ 11

r

y

r

r

z

—

21

r

1

!1

12

22

r

13

tl

r

23

y

32

f3 I

z

0

0

r

0

1

fi(.x,y,z)

X —

f3(x,y,z)

1

1

Hence we obtain the following set of polynomial identities:

r ii* + r12y

+ rl3z

r 21* + r22y

+ r 2 3 z + t2 =

+ t1 = ^ ( x , y , z )

r 3 1 x + r32y

+ r33z

f2(x,y,z)

+ t3 = f-,{x,y,z)

.

Therefore, r., , r.„, r.„ and t. are the coefficients of f.(x,y,z). ' /1' / 2* / 3 / /

For

example, if the image of [x,y,'Z]^ under a is [(3/4) + z,(l/4) - y,(3/4) - x]fc

,

then r

u

x + r12y

+ r 1 3 z + ti = Ox + Oy + lz + (3/4)

implying that r 1 1 = r 1 2 = 0 , r 1 3 = l and ti=3/4. 271

Continuing this reasoning,

we see that

R

Consequently, the 4 x 4

P(0)(a)

Example:

0

1

3/4

0

-1

0

1/4

-1

0

0

3/4

0

0

0

1

=

representative of a can be found by an inspection [x.yjZ] 1 ".

of the corresponding image of

(E7.42)

0

Find the general equivalent positions of P6

In P7.39, the principal

Solution:

representatives

of P61

with

respect

to P are {M | [ 0 ,0,1/6 ] t } , {M 2

| [0,0,1/3]*}, {M 3

| 0,0,i] 1 },

{M* | [0,0,2/3]*}, {M 5

| [0,0,5/6]1"}, {I 3

| [000] t >,

where "l

M =

The images of [x,y,z]

- 1

1

0

0

0

0"

0 . 1

under these representations are

[x - y , x , (1/6) + z] t ,

- y , (1/3) + z] t , [-x ,-y

[~y,x

[y - x ,-x, (2/3) + z] t , [y,y - x,(5/6) + z f ,

+ z]1,

[x.y.z]1

.

These triples are the general equivalent positions of P61 .

(P7.40) P63.

Problem:

the

general equivalent positions

for P62

and

Verify your answers with the results given in (Hahn, 1983).

(E7.43) C

Find

c

Example:

Find all of the crystallographic space groups

types

such that A (C) is isomorphic to 2.

From Table 6.3, the

Solution: P and C Hence

lattice types left invariant under 2 are

(where we have chosen the setting where Y

we

Mp(^'^2).

seek

crystal lographic

With respect to P,

space

we have

272

groups

C

is the unique such

that

axis).

Mp(G) =

- 1 0 10

M = M p (I° ^2) =

We will find the space groups C

such

0

0

1 0

0

0 - 1

that

Rp(7~(G)) = Rp(7~p)

1 P E Z

'} '

first.

By T7.37, since

V

=

V

{{l3

we seek those elements of r e R3

1 r}

-26)

such that Nr e Z 3 where

0 N = M + M

Hence Nr =

(7

=

[0,2r2,0] must be in Z 3 .

0

0

0

2 0

0

0

0

Therefore, modulo P, we have the

generators {M I [ r ^ O , ^ ] * } where rit

r 3 t R.

and

{M | [ r 1 , i , r 3 ] t >

(7.27)

By T7.37, the generators in (7.27) yield all of the

space groups C that satisfy (7.22) for H = To

,

and L = P.

find a list of the nonequivalent space groups from all of the

space groups based on (7.27), we determine the impact of moving the origin to p .

If this shift is made, {M | t } becomes {M | (M - I 3 )P + r} = {M | [-2p l ,0 ) -2p 3 ] t + r]

.

Choosing p ! and p 2 so that 2pj = Tj and 2p 3 = r2, we see that each of the generators in (7.27) is equivalent to either {M | [000]1"}

It

{M | [0,i,0] t )

or

can be shown that these are nonequivalent.

crystallographic

space

group

types

where

Thus we have found two Mp ( C ) =

M

and

R p ( T ( C ) ) = R p { T p ) . We call the space group generated by {M | [OOO]1"}, P2, and that generated by {M | [0,i,0]t}, (P7.41)

Problem:

Consider the space group P21.

273

Locate all of the sym-

metry axes in this space group that pass through the unit cell, find the principal representatives of P2i with respect to P, and calculate set of general equivalent positions.

its

Compare your results to those in

ITFC (Hahn, 1983).

Continuing with E7.43, we consider the case where the lattice left invariant under 2 is taken to be C .

Even though the basis of this lattice

is D = {-Ja + ^b,b,c), we continue to write all of our matrices in terms of the basis P.

Hence

w j[I3

I s}

=

1

where m E Z 3 and u = 0 or ] j

S = m + uti.i.O] "

Consequently, we seek those elements r =

[r 1 ,r 2 ,r 3 ]^ E R3

the condition that Nr represents a lattice vector.

[O^ri.O] 1 = m + u[i,i, 0]*

where u = 0 or 1.

(7.

that satisfy

That is, such that

,

Then 0 = m^ + ui where m , E Z.

Hence u = 0 and so,

as in the case of P, we have the generators

{M | [/• 1 ,0 J r 3 ] t }

and

{M | [ r ^ ^ r , ] * )

.

As before, these are equivalent to

{M | [000] t }

respectively.

In

and

particular

{N | [O.i.O]*} ,

{M | [0,^,0]*"}

1

is

(M I [i.ijO] "} where we have chosen Tj = -J and r3 = 0. {M I [i.i.O]*} = {I, | [¿,i,0]t}{M

and since {I3

Rp(7" c ){I 3

| [bi.O]1} E

R

P

| [000] t >

,

| [OOO] 1 } =

| [OOO]*) U R p ( 7 c ) { M

| U.i.O]1}

and so {M | [000]t} together with R p ( T q ) generates the same group of

274

to

Note that since

{ T ,

| [000]t} U R p ( f c ) { M Rp(7" c ){I 3

equivalent

matrices as that generated by {M | [i,i,0] } and Hence groups.

equivalent

space

Consequently we obtain only one space group denoted C2.

{M | [000] } and

Hence

there are three distinct

{M | [0,i,0] } generate

crystallographic

monoclinic point group (E7.44) Example:

space groups

based

on the

They are P2, P2j and C2.

•

Find the principal coset representatives of C2 with

respect to the P basis and the set of general equivalent positions

of

C2.

Solution:

By (7.26) and (7.28) we see that R p ( T c ) = Rp(7-p){I, I [000]t} U R p (7~ p ){I 3 I [i.i.O]*} = Kp(.Tp)

U

R p y( . T p ) P P

{I,

I [i.i.O]*)

.

But R p ( C 2 ) = R p ( r c ) { I , | [000]t}U R p ( 7 c ) { M | [OOO]1"} where M =

.

.

Hence | [U,0]t}j{M

R p ( C 2 ) = ^Rp(Tp) U R p C T p ) { I 3

| [000]t}

U [ R p ( r p ) U R p (7~ p ){I 3 | [i,i,0]t}]{I3 | [OOO]1") = R p ( r p ) { M | [OOO]1} U Rp(7"p){M | [i,i,0]t} U Rp(rp){i3 The

principal

coset

I [ooo]*} u R p ( r p ) { i 3

representatives

of

C2

with

| [i.i.on* .

respect

t

{M | [000] > {M | [i.i.O]*}, {I, | [000]*), {Ij | - 1 0

Mi

.

to

P

Since

0

0

1 0

0

0 - 1

the general equivalent positions for C2 are [-x.y.-z] 1 , [i - X,i + y,-z] t ,[x.y,*]* and [1 + x +

275

y.z]*.

are

(P7.42)

Problem:

Show that the set of all crystallographic space group

types C such that A Q ( G ) is isomorphic to 3 consist of P3,

P3 j,

P32

and

R3.

Applying the techniques illustrated in this section one can verify that the one-generator point groups give rise to the following list of 30 crystallographic space groups. 1:

PI P2,P21,C2

2:

3:

P3,P31,P3Z,R3

6:

P6,P61,P62,P63,P6h,P6i

1:

PI

m:

Pm,Pc,

3:

THE

Cm,Cc

P3,R3

4:

P4,14

6:

P6

CRYSTALLOGRAPHIC

S P A C E GROUP T Y P E S

DERIVED

FROM T H E T W O - G E N E R A T O R POINT G R O U P S The two generator point groups H contain two elements a and (5 such that each element of H

is expressed

1 < / < o(a) and 1 < j < o(P).

uniquely

in the

form a'fs' where

In order to do this, the elements a and

fi must be related in a special way.

For each point group, we will choose

a and 3 such that 3 = a ¡5a and such that D

= {1}.

A listing of the two-generator

(7.29 point

groups together with their generators is given in Table 7.2. (P7.43)

Problem:

Show

that the a and g listed for each of the two-

generator point groups satisfies (7.29). (P7.44) Problem:

Show that

n = {1} for each a and fS pair given in Table 7.2 where denotes the cyclic group generated by Z. 276

Table 7.2:

T h e two-generator oriented point

g r o u p s and their g e n e r a t o r s . Group

a

Group

a

2/m

i

[010]2

31m

3

1/m

i

4

122

4

6/m

i

222

2

mm2

2

321

3

312

3

3 ml

3

31m

3 3

3m1

(E7.45) Example:

(S

6

Qmm

4

[ioo]2

U2m

4

[100]m

4m2

4

622

6

6mm

6

[1C0] 2 [210]2 [100]m [210]m

62m

6

6m2

6

I [210] 2 [ioo]2 [ioo]m [100]2 [100]m [100] 2 i m

m

[100]2 noo]m

[100]2

Let a and (5 be the generators listed in Table 7.2 for

a two-generator point group.

Show that if

c P f

= 1,

then a" = 1 and f5m = 1 . Solution:

Suppose a n $ m = 1.

But g"7 £ and a B

m

n

t .

Then

Since tf" = -a", & m c