Crystallographic Point Groups
Crystallographic Point Groups
Crystallographic Point Groups
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INTERNATIONAL SCHOOLON FUNDAMENTALCRYSTALLOGRAPHY
CRYSTALLOGRAPHICPOINT GROUPS(short review)Mois I. AroyoUniversidad del Pais Vasco, Bilbao, Spain
GROUP THEORY(few basic facts)
<strong>Crystallographic</strong> symmetry operations inthe planeMirror symmetry operationMirror line my at 0,yMatrix representationdrawing: M.M. JulianFoundations of CrystallographycTaylor & Francis, 2008myxy= -x y= -1 1xyFixed pointsmy x fyf= x fyfdet -1 1= ? tr-11= ?
Symmetry operations in the planeMatrix representations2-fold rotation3-fold rotationMatrix representation2zx= -x= -1 y -x -1xy3 + x y= -yx-y= 0 -11 -1xydet -1 -1=?0 -1det =1 -1?tr-1-1= ?tr0 -1= ?1 -1
GROUP AXIOMS1. CLOSUREg1o g2=g12g1, g2, g12∈G2. IDENTITYg o e = e o g = g3. INVERSE ELEMENTg o g -1 = e4. ASSOCIATIVITY(g1o g2)o g3= g1o ( g2 o g3)= g1o g2 o g3
Group Properties4. How to define a groupMultiplication tableGroup generatorsa set of elements such that each element ofthe group can be obtained as a product of thegenerators
.. .g1g 2Isomorphic groups.Φ(g1). g 1g2Φ(g2)G G’.Φ(g1g2)G={g}Φ(g)=g’Φ -1 (g’)=gG’={g’}Φ: G G’Φ -1 : G’GhomomorphicconditionΦ(g1)Φ(g2)= Φ(g1g2)-groups with the same multiplication table
<strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong> in 2D<strong>Point</strong> group 2 = {1,2}Motif withsymmetry of 22 x 2 =-group axioms?-1-1x-1-1=11-order of 2?-multiplication tabledrawing: M.M. JulianFoundations of CrystallographycTaylor & Francis, 2008Where is the two-foldpoint?-generators of 2?
<strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong> in 2D<strong>Point</strong> group m = {1,m}Motif with-group axioms?symmetry of mm x m =-11x-11=11Whereis themirrorline?-order of m?-multiplication tabledrawing: M.M. JulianFoundations of CrystallographycTaylor & Francis, 2008-generators of m?
GIsomorphic groups. .Φ(g)=g’Φ -1 (g’)=gΦ(g1)Φ(g2)= Φ(g1 g2)G’<strong>Point</strong> group 2 = {1,2}<strong>Point</strong> group m = {1,m}-groups with the same multiplication table
<strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong> in 2D<strong>Point</strong> group 1 = {1}Motif with-group axioms?symmetry of 11 x 1 =11x11=11-order of 1?-multiplication tabledrawing: M.M. JulianFoundations of CrystallographycTaylor & Francis, 2008-generators of 1?
Problem 2.1Consider the model of the molecule of the organicsemiconductor pentacene (C22H14):Determine:-symmetry operations:matrix and (x,y) presentation-generators-multiplication table
Problem 2.2Consider the symmetry group of the square. Determine:-symmetry operations:matrix and (x,y)presentation-generators-multiplication table
Visualization of <strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong>- general position diagram- symmetry elements diagramStereographic ProjectionsP’’<strong>Point</strong>s P in theprojection plane
EXAMPLEStereographic Projections ofmm2<strong>Point</strong> group mm2 = {1,2z,mx,my}Molecule ofpentaceneStereographic projections diagramsgeneral positionsymmetry elements
Problem 2.2 (cont.)Stereographic Projections of4mmgeneral positiondiagramsymmetry elementsdiagram? ?
Problem 2.3 (additional)Consider the symmetry group of the equilateraltriangle. Determine:-symmetry operations:matrix and (x,y)presentation-general-position and symmetryelementsstereographicprojection diagrams;-generators-multiplication table
Conjugate elementsConjugate elementsgi ~ gk if ∃ g: g -1 gig = gk,where g, gi, gk, ∈ GClasses of conjugateelementsL(gi)={gj| g -1 gig = gj, g∈G}Conjugation-properties(i) L(gi) ∩ L(gj) = {∅}, if gi ∉ L(gj)(ii) |L(gi)| is a divisor of |G|(iii)L(e)={e}(iv) if gi, gj ∈ L, then (gi) k =(gj) k = e
Example (Problem 2.2):Classes of conjugate elementsThe group of the square 4mmClasses of conjugate elements:{1}, {2},{4,4 - },{mx,my}, {m+,m-}
EXERCISESProblem 2.1 (cont)Distribute the symmetry elements of the groupmm2 = {1,2z,mx,my} in classes of conjugate elements.multiplicationtablestereographicprojection
CRYSTALLOGRAPHICPOINT GROUPS(brief overview)
Symmetry operations in 3DRotations
Rotation <strong>Crystallographic</strong> <strong>Point</strong><strong>Groups</strong> in 3DCyclic: 1(C1), 2(C2), 3(C3), 4(C4), 6(C6)Dihedral: 222(D2), 32(D3), 422(D4), 622(D6)Cubic: 23 (T), 432 (O)
Dihedral <strong>Point</strong> <strong>Groups</strong>222422{e,2z, 2y,2x}{e,4z,4z, 2z,2y,2x,2+,2-}32622{e,3z,3z ,21,22,23}{e,6z,6z, 3z,3z, 2z21,22,23, 21,22,23} ´ ´ ´
Cubic Rotational <strong>Point</strong> <strong>Groups</strong>23 (T){e, 2x, 2y, 2z,31,31,32,32,33,33,34,34}{e, 2x, 2y, 2z,4x,4x,4y,4y,4z,4z432(O)31,31,32,32,33,33,34,3421,22,23,24,25,26}
Symmetry operations in 3DRotoinvertions
Symmetry operations in 3DRotoinvertions
Symmetry operations in 3DRotoinversions
Symmetry operations in 3D3 Roto-inversiongeneralviewdown thesymmetry axis3 + generalview3 + down thesymmetry axis
<strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong> in 3DProper rotations: det =+1: 1, 2, 3, 4, 6- 1Improper rotations: det =-1:- 2- 3- 4- 6Hermann-Mauguin symbolism (International Tables A)-symmetry elements along primary, secondary andternary symmetry directionsrotations: by the axes of rotationplanes: by the normals to the planes-symmetry elements in decreasing order of -symmetry (except for two cubic groups: 23 and m 3 )
Crystal systems and<strong>Crystallographic</strong> point groupsprimary secondary ternary
Crystal systems and<strong>Crystallographic</strong> point groupsprimary secondary ternary
<strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong> in 3D
Direct-product groupsLet G1 and G2 are two groups. The set of all pairs {(g1,g2),g1∈G1, g2∈G2} forms a group with respect toG1 x G2the product: (g1,g2) (g’1,g’2)= (g1g’1, g2g’2).The group G=G1 x G2is called a direct-product groupProperties of(i)(ii)G1 x G2G1 x ≅ G2 {(g1,e2), g1∈G1} G1G1 x ≅ G2 {(e1,g2), g2∈G2} G2{(g1,e2), g1∈G1} ∩ {(e1,g2), g2∈G1}= {(e1,e2)}(iii)∀ (g1,g2)∈ G1 x G2 : (g1,g2)= (g1,e2) (e1,g2)
Examples: Direct product groups<strong>Point</strong> group mm2 = {1,2z,mx,my}G1={1,2z} G2={1,mx}G1 x G2= {1.1, 2z.1, 1.mx, 2zmx=my}Centro-symmetrical groupsG1: rotational groupsG2={1,1}group of inversionG1 x {1,1}=G1+1.G1
<strong>Crystallographic</strong> point groups andabstract groupsCyclic groupsAbstractgroupsAbelian non-cyclic groupsNon-Abelian groupsDirect products of non-Abelian groups withcyclic groups of order 2CrystalclassesAffineequivalenceAbstractgroups32 18
<strong>Crystallographic</strong>point groupsasabstract groups
<strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong>G G+1G G(G’) G’+1(G-G’)1 (C1) 1+1.1=1 (Ci) ---- -----2 (C2) 2+1.2=2/m (C2h) 2(1) m (Cs)3 (C3) 3+1.3=3 (C3i or S6) ---- ----4 (C4) 4+1.4=4/m (C4h) 4(2) 4 (S4)6 (C6) 6+1.6=6/m (C6h) 6(3) 6 (C3h)
<strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong>G G+1G G(G’) G’+1(G-G’)222 (D2) 222+1.222=2/m2/m2/m 222(2) 2mm (C2v)mmm (D2h)32 (D3) 32+1.32=32/m 3m(D3d) 32(3) 3m (C3v)422 (D4) 422+1.422=4/m2/m2/m 422(4) 4mm (C4v)4/mmm(D4h) 422(222) 42m (D2d)622 (D6) 622+1.622=6/m2/m2/m 622(6) 6mm (C6v)6/mmm(D6h) 622(32) 62m (D3h)23 (T) 23+1.23=2/m3 m3 (Th) ---- -----432 (O) 432+1.432=4/m32/m 432(23) 43m (Td)m3m(Oh)
<strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong><strong>Groups</strong> isomorphic to 422422 e 4z 4z 2z 2x 2y 2+2-4mm42me 4z 4z 2z mx my m+m-e 4z 4z 2z 2x 2y m+m-4m2 e 4z 4z 2z mx my 2+2-<strong>Groups</strong> isomorphic to 622622 e 6z6z 3z3z 2z 212223 212223 ´ ´ ´6mm e 6z6z 3z3z 2z m1m2m3 m1m2m3 ´ ´ ´62m e 6z6z 3z3z mz 212223 m1m2m3 ´ ´ ´6m2 e 6z6z 3z3z mz m1m2m3 212223 ´ ´ ´
Problem 2.44mmConsider the following three pairs of stereographicprojections. Each of them correspond to a crystallographicpoint group isomorphic to 4mm:(i) Determine those point groups by indicating their symbols,symmetry operations and possible sets of generators;(ii) Construct the corresponding multiplication tables;(iii) For each of the isomorphic point groups indicate the one-toonecorrespondence with the symmetry operations of 4mm.
<strong>Crystallographic</strong> <strong>Point</strong> <strong>Groups</strong> in 3D
ExampleSymmetry groups of moleculesDetermine the symmetry group of the configuration:AB4 type moleculezyx?Hint: cubic crystal systemsymmetry directions: [100] [111][110]
Problem 2.5(a)Determine the symmetry elements and the corresponding pointgroups for each of the following models of molecules:
Problem 2.5(b)Determine the symmetry elements and the corresponding pointgroups for each of the following models of molecules:
GROUP-SUBGROUPRELATIONSI. Subgroups: index, coset decompositionand normal subgroupsII. Conjugate subgroupsIII. Group-subgroup graphs
Subgroups:Some basic results (summary)Subgroup H < G1. H={e,h1,h2,...,hk} ⊂ G2. H satisfies the group axioms of GProper subgroups H < G, andtrivial subgroup: {e}, GIndex of the subgroup H in G: [i]=|G|/|H|(order of G)/(order of H)Maximal subgroup H of GNO subgroup Z exists such that:H < Z < G
EXERCISESProblem 2.6Consider the group of the square anddetermine its subgroups
Problem 2.7 (additional)(i) Consider the group of the equilateral triangle anddetermine its subgroups;(ii) Distribute the subgroups into classes of conjugatesubgroups;(iii) Construct the maximal subgroup graph of 3mmxmyyxmxxMultiplication table of 3m
Coset decomposition G:HGroup-subgroup pair H < Gleft cosetdecompositionright cosetdecompositionG=H+g2H+...+gmH, gi∉H,m=index of H in GG=H+Hg2+...+Hgm, gi∉Hm=index of H in GCoset decomposition-properties(i) giH ∩ gjH = {∅}, if gi ∉ gjH(ii) |giH| = |H|(iii) giH = gjH, gi ∈ gjH
Coset decomposition G:HNormalsubgroupsHgj= gjH, for all gj=1, ..., [i]Theorem of Lagrangegroup G of order |G|subgroup H
EXERCISESProblem 2.8Consider the subgroup {e,2} of 4mm, of index 4:-Write down and compare the right and leftcoset decompositions of 4mm with respect to{e,2};-Are the right and left coset decompositions of4mm with respect to {e,2} equal or different? Canyou comment why?Problem 2.9Demonstrate that H is always anormal subgroup if |G:H|=2.
Conjugate subgroupsConjugate subgroupsLet H1
EXERCISESProblem 2.6 (cont)Distribute the subgroups of the groupof the square into classes of conjugatesubgroupsHint: The stereographic projections could be rather helpful
Complete and contractedgroup-subgroup graphsComplete graph ofmaximal subgroupsContracted graph ofmaximal subgroups
International Tables for Crystallography, Vol. A, Chapter 10Group-subgroup relations of point groupsHahn and Klapper
Factor groupproduct of sets:G={e, g2, ...,gp}{Kj={gj1,gj2,...,gjn}Kk={gk1,gk2,...,gkm}Kj Kk={ gjpgkq=gr | gjp ∈ Kj, gkq ∈Kk}Each element gr is takenonly once in the productKj Kkfactor group G/H: H GG=H+g2H+...+gmH, gi∉H,group axioms:G/H={H, g2H, ..., gmH}(i) (giH)(gjH) = gijH(ii) (giH)H =H(giH)= giH(iii) (giH) -1 = (gi -1 )H
Problem 2.8 (cont)Consider the normal subgroup {e,2} of 4mm, ofindex 4, and the coset decomposition 4mm: {e,2}:(3) Show that the cosets of the decomposition 4mm:{e,2}fulfil the group axioms and form a factor group(4) Multiplication table of the factor group(5) A crystallographic point group isomorphic to thefactor group?
GENERATION OFCRYSTALLOGRAPHIC POINTGROUPS
Generation of point groups<strong>Crystallographic</strong> groups are solvable groupsComposition series: 1 Z2 Z3 ... Gindex 2 or 3Set of generators of a group is a set of groupelements such that each element of the group can beobtained as an ordered product of the generatorskh kh-1 k2W=(gh) * (gh-1) * ... * (g2) * g1g1 - identityg2, g3, ... - generate the rest of elements
ExampleGeneration of the group of the square2z 4z mxComposition series: 1 2 4 4mmStep 1:[2] [2] [2]1 ={1}Step 2:2 = {1} + 2z {1}Step 3:4 ={1,2} + 4z {1,2}Step 4:4mm = 4 + mx 4
Generation of sub-cubic point groups
Composition series of cubic point groupsand their subgroups
Generation of sub-hexagonal point groups
Composition series of hexagonal pointgroups and their subgroups
Problem 2.10Generate the symmetry operations of thegroup 4/mmm following its compositionseries.Generate the symmetry operations of thegroup 3m following its composition series.
GENERAL AND SPECIALWYCKOFF POSITIONS
General and special Wyckoff positionsSite-symmetry group So={W} of a point XoWXo = Xoa b cx0x0d e fy0=y0g h iz0z0General position XoS= 1 ={1}Special position XoS> 1 ={1,...,}Site-symmetry groups: oriented symbols
ExampleGeneral and special Wyckoff positions<strong>Point</strong> group 2 = {1,2z}Site-symmetry group So={W} of a point Xo=(0,0,z)So = 22z:-1-100=00WXo = Xo1zz2 b 1 (x,y,z) (-x,-y,z)1 a 2 (0,0,z)
ExampleGeneral and special Wyckoff positions<strong>Point</strong> group mm2 = {1,2z,mx,my}Site-symmetry group So={W} of a point Xo=(0,0,0)So = mm22z:-1-100=00WXo = Xo1zz100my:-10=01zz4 d 1 (x,y,z) (-x,-y,z) (x,-y,z) (-x,y,z)2 c m.. (0,y,z) (0,-y,z)2 b .m. (x,0,z) (-x,0,z)1 a mm2 (0,0,z)
EXERCISESProblem 2.11Consider the symmetry group of the square 4mmand the point group 422 that is isomorphic to it.Determine the general and special Wyckoff positionsof the two groups.Hint: The stereographic projections could be rather helpful
EXAMPLEWyckoff positions splitting schemesGroup-subgroup pair mm2 >2, [i]=2mm224 d 1 (x,y,z)(-x,-y,z)(x,-y,z)(-x,y,z)x,y,z= x1,y1,z1 2 b 1-x,-y,z=-x1,-y1,z1x,-y,z=x2,y2,z2 2 b 1-x,y,z=-x2,-y2,z2
EXERCISESProblem 2.12Consider the general and special Wyckoffpositions of the symmetry group of the square4mm and those of its subgroup mm2 of index 2.Determine the splitting schemes of the general andspecial Wyckoff positions for 4mm > mm2.Hint: The stereographic projections could be rather helpful
GROUP-SUPERGROUPRELATIONS
Supergroups: Some basic results (summary)Supergroup G>HH={e,h1,h2,...,hk} ⊂ GProper supergroups G>H, andtrivial supergroup: HIndex of the group H in supergroup G: [i]=|G|/|H|(order of G)/(order of H)Minimal supergroups G of HNO subgroup Z exists such that:H < Z < G
The Supergroup ProblemGiven a group-subgrouppair G>H of index [i]Determine: all Gk>Hof index [i], Gi≃GG...G G2 G3 Gn[i][i]HHall Gk>H contain H as subgroupGk=H+g2H+...+gikH
Example: Supergroup problemGroup-subgroup pair422>222Supergroups 422 ofthe group 222[2]422422?222222How many arethe subgroups222 of 422?How many arethe supergroups422 of 222?
Example: Supergroup problemGroup-subgroup pair422>222Supergroups 422 ofthe group 2224224z224x224y22[2][2]2z2x2y2z2+2-2224z22= 2z2x2y +4z(2z2x2y)4z22= 2z2+2- +4z(2z2+2-)4z22=222+4z2224y22=222+4y2224x22=222+4x222
NORMALIZERS
Normalizer of H in GNormal subgroupHG, if g -1 H g = H, for ∀g∈GNormalizer of H in G, H
Problem 2.13Consider the group 4mm and its subgroups of index 4.Determine their normalizers in 4mm. Distribute thesubgroups into conjugacy classes with the help of theirnormalizers in 4mm.Hint: The stereographic projections could be rather helpful
Problem 2.10{e,2,4,4 -1 ,mx,my,m+,m-}SOLUTIONNormalizer of {1,my}in 4mm4mm{1,my}2mm={e,2,mx,my}Conjugate subgroups: 4mm=2mm+4(2mm){ }{1,my}{1,mx}