2D Symmetry (1.5 weeks)

From previous lecture, we know that, in 2D, there are 3 basics symmetry elements: Translation,mirror (reflection),and rotation. What would happen to lattices that fulfill the requirement of more than one symmetry element (i.e. when these symmetry elements are combined!).

Start with the translation Add a rotation    lattice point lattice point T: scalar A translation vector connecting two lattice points! It must be some integer of or we contradicted the basic Assumption of our construction. p: integer Therefore,  is not arbitrary! The basic constrain has to be met! Combination of translation with rotation: 

 T T T tcos  b To be consistent with the original translation t: p must be integer p cos   n (= 2  /  ) b  2 3T  /3 3 2T 0  /2 4 T 0.5  /  (1) -T p > 4 or P < -2: no solution T TT AA’ B’B Allowable rotational symmetries are 1, 2, 3, 4 and 6.

Look at the case of p = 2  = 120 o  angle Look at the case of p = 1 n = 3; 3-fold n = 4; 4-fold  = 90 o  3-fold lattice. 4-fold lattice.

Look at the case of p = 0  = 60 o  n = 6; 6-fold Look at the case of p = 3n = 2; 2-fold Look at the case of p = -1n = 1; 1-fold 12 Exactly the same as 3-fold lattice.

1-fold 2-fold 3-fold 4-fold 6-fold Parallelogram Hexagonal Net Can accommodate 1- and 2-fold rotational symmetries Can accommodate 3- and 6-fold rotational symmetries Square Net Can accommodate 4-fold rotational Symmetry!

Combination of mirror line with translation: m Unless 0.5T centered rectangular constrain Or Primitive cell Rectangular m

Lattice + symmetries of motif (point group) = plane group (5)(1, 2, 3, 4, 6, m, etc) Parallelogram Hexagonal Net Square Net (1) (2) (3) Double cell (2 lattice points) Centered rectangular(4) Primitive cell Rectangular(5)

Oblique Rectangular Centered rectangular Square Hexagonal 1, 2 m 4 3,6 Five kinds of lattice The symmetry that the lattice point can accommodate Plane group 3D: space group. Group theory We will show the concept of group!

Group theory: set of elements (things) for a law of combination is defined and satisfies 3 postulates. (1) the combination of any two elements is also a member of the group; (2) “Identity” (doing nothing) is also a member of the group. “I” aI=Ia=a (a : an element) (3) for element, an inverse exists. a; a -1 a. a -1 = I a -1. a = I Example: Group {1, -1}; rank 2 rank (order) of the group = number of elements contained in a set Another Example: Group {1, -1, i, -i}; rank 4 We will show examples for point groups later!

n12346n12346 m [ ] In a point, there is no translation symmetry! Therefore, consider 2D point group, we only consider rotation and mirror! Put rotation symmetry and mirror together  ?

Example: m1m1 m2m2 R RL L {1,  1,  2, A  } group of rank 4 1 11 22 AA 1 11 22 AA 1 11 22 AA 1 11 22 AA 1 11 22 AA 1 11 22 AA Abelian group: a.b=b.a 2mm: point group 2 + m2 + m (1) (2) (3) (4) 1  1: 1 1  2:  2 1  3: A  1  4:  1

6-fold is a subset of 2-fold axis subgroup 3-fold axis

11 22 L L R Chirality not changed: Rotation is the right choice! ||       Combination theorem (1)(2) (3) (4) if 2mm

Show it is a group 1 11 22 AA 1 11 22 AA 1 11 22 AA 1 11 22 AA 1 11 22 AA 1 11 22 AA Satisfy 3 postulates? Rank 4 The number of motif in the pattern is exactly the same as the rank (order) of the group!

Hermann and Mauguin International notation Rotation axis n 1, 2, 3, 4, 6 Schonllies notation CnCn C 1, C 2, C 3, C 4, C 6 C: cyclic group – all elements are “powers” of some basic Operation e.g. Notation:

Hermann and Mauguin International notation Mirror plane m Schonllies notation CSCS C nv : Rotational symmetry with mirror plane vertical to the rotation axis. E.g. 2mm – C 2v.

11 (1) (2) (3) 22 S:C 4v HM: 4mmmm m m m m Only independent symmetry elements. The rank of this group is ? R L L 4 + m

11 22  /6 S:C 6v HM: 6mm The rank of this group is 12! 11 (1)(2) (3) (1) L (2) R (3) R 22 S:C 3v HM: 3mm (correct?) The rank of this group is 6!  2 is not independent of  1. HM (international notation): 3m

So far we have shown 10 point group or specifically 10 2-D crystallographic point group. HM notation,,,,,,,,, ; Schonllies notation,,,,,,,,, D crystallographic point group 5 2-D lattices 2-D crystallographic space group 12346m2mm3m4mm6mm C1C1 C2C2 C3C3 C4C4 C6C6 CsCs C 2v C 3v C 4v C 6v

Oblique Primitive Rectangular Centered rectangular Square Hexagonal 1, 2 m 4 3,6 Compatible with Compatibility: 2mm, 3m, 4mm, 6mm

2mm m m Put mirror planes along the edge of the cell. m m Primitive Rectangular Centered rectangular m, 2mmCompatible with Square4, 4mmCompatible with

30 o HexagonalCompatible with

Red ones Blue ones HexagonalCompatible with6mm

Oblique Primitive Rectangular Centered rectangular Square Hexagonal,,,, Compatible with 12 m2mm m 44mm 363m6mm

General oblique net. atoms Type of lattice Point group Symbol used to describe the space group P (for primitive) 1 Space group: p1 Upper case P for 3D lower case p for 2D

Primitive oblique net + 2 = p2 B (1) (2) (3) plane group: p2

p2p2 positions with symmetry  the lattice point!

p2p2

General relation between new symmetry position generated by combining rotation with translation at along the  -bisector of  /2 A (1) (2) (3) x  /2 B Question: what kind of symmetry operation is required in order for motif (1) get to motif (3)? A  : (1)  (2); : (2)  (3);

 /2 12 (1) (2) Could we always rotate  /2 respect to the dashed line  T! You can always define it that way!   /2

4 + lattice 4 1 || Correct? Combination of A  /2 with p4p4 at

p + 3 = p3 120 o 3 Combination of A 2  /3 with along the  -bisector of at T X 30 o mass center X/3

60 o (1) (2) (3) (1)  (2): A 2  /3 ; (2)  (3): Translation (1)  (3): B 2  /3 ; 60 o

p + 6 = p6 p has to be hexagonal net as well! 6 3-fold 2-fold From 2-fold rotation From 3-fold rotation Combination of A  /3 and A -  /3 with at

Combination of mirror symmetry with the translation! m + p + c p + m = pm  (1) R (2) L (3)  Independent mirror plane  is defined with respect to mirror line (plane)

c + m = cm  not an independent mirror plane! (lattice point!) (1)  (3) (1) R (2) L (3) L Glide plane with glide component  Two-step operation mmm cm gg

p + g = pg possible? gg (1) R (2) L (3) L (1)  (3)? General form: Remind:

c + g = cg possible? gg m gg m cg = cm rectangular net: pm, pg, cm!

p + 2mm = p2mm   c + 2mm = c2mm  

p (square) + 4mm Red: p4. Blue: pm. m p4mm Special case of a rectangular.

p (Hexagonal net) + 3m p3p3 60 o two ways  centered rectangular net m  edge m || edge

p3p3  Cell edge || Cell edge p31mp3m1

p31mp3m1 3m 3 Not yet done! Glide plane (or line).

p (Hexagonal net) + 6mm = p6 + p3m1 + p31m RedBlue Mirror line Glide line p6mm

2mm compatible with Rectangular! mirror plane? p2mm What if the mirror line is not passing through the rotation axis?

For example this way? Why not? How about this way? Why not? Leave all the two fold rotation axes maintain undisturbed!  OK

Center rectangular net (c2mm)? (m ok? )(g ok? ) p2mg XX Two fold rotation symmetries + offset mirror line

p2gg Two fold rotation symmetries + offset glide line

Three different ways: OK? X

p4gm The same results This is not C4gm! Because center position is not a lattice!

System (4)Lattice (5)Point group (10)Plane group (17) Oblique a  b  general Rectangular a  b,  = 90 o Square a = b,  = 90 o Hexagonal a = b,  = 120 o Primitive parallelogram Primitive or centered rectangular Square Hexagonal equilateral 2 pm pg cm 3 3m 6 6mm p3 p3m1 p31m p6 p6mm 4 4mm p4 p4mm p4gm m 2mm p2mm p2mg p2gg c2mm 1 p2 p1

Hermann-Mauguin Notation: pnab or cnab (1) First letter: p for primitive cell, c for centered cell (2) n: highest order of of rotational symmetry (1, 2, 3, 4, 6) (3) Next two symbols indicate symmetries relative to one translation axis. The first letter (a) is m (mirror), g (glide), or 1 (none). The axis of the mirror or glide reflection  main axis. The second letter (b) is m (mirror), g (glide), or 1 (none). The axis of the mirror or glide reflection is either || or tilted 180 o /n (when n>2) from the main axis. a b 1, 2 a b 3 60 o b 4 45 o a b 6 30 o a Old notes

 The short notation drops digits or an m that can be deduced, so long as that leaves no confusion with another group. E.g. p2 (p211): Primitive cell, 2-fold rotationp2 symmetry, no mirrors or glide reflections. p4g (p4gm): Primitive cell, 4-fold rotation, glidep4g reflection perpendicular to main axis, mirror axis at 45°. cmm (c2mm): Centred cell, 2-fold rotation,cmm mirror axes both perpendicular and parallel to main axis. p31m (p31m): Primitive cell, 3-fold rotation,p31m mirror axis at 60°. Short full pm p1m1 pg p1g1 cm c1m1 pmm p2mm pmg p2mg pgg p2gg p4m p4mm p6m p6mm p1: p111 p3: p311 p4: p411 p6: p611 p3m1 Old notes

Symbol for the plane group # of the particular plane group in the set Symbol for the group in 3D Point group Crystal system p2 No. 2 p211 2 oblique The information of the international X-ray table Diagram symmetry elements in the net. First line

origin at 2 (0 0) x y (x y)(x y) (1-x 1-y) General position (Unique for every Plane or space group) = Number/cell (rank of position) 2 e 1 Site symmetry Always 1 for general position Special position (on a symmetry Element) 1 d 2 1 c 2 1 b 2 1 a 2 x Wyckoff symbol by convention

 Notation for asymmetric used to represent point group symmetry: (a) : Asymmetric unit in the plane of the page (b) : Asymmetric unit above the plane of the page (c) : Asymmetric unit below the plane of the page (d) : Apostrophe indicating a left-handed asymmetric unit. Clear circle indicating right-handedness. (e) : Two asymmetric units on top of each other (f) : Two asymmetric units on top of one another, one left-handed and the other right-handed. + , + , and are mirror images of each other., Old notes

Another example pmm No. 6 p2mm mm Rectangular,,,,,,,, Origin at 2mm (x y)(x y) 2 m 1 2mm 4 1 ihgfedcbaihgfedcba

； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ；

； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ； ；

pmg No. 7 p2mg mm Rectangular,, (x y)(x y) Origin at 2,, Not an independent special position (mirror) 2 m 2 An independent special position a b c d How about glide plane? Atoms do not coincide! Glide is never a candidate for a special position!

rank Symmetry of the equipoints designation Condition limiting possible reflection (structure factor) Old notes

0, 0 1, 0 0, 1 x, y 1-y, x 1-x, 1-y y, 1-x 4 d 1 0, 0 1, 0 0, 1 1/2, 0 2 c 2 0, 1/2 1, 1/2 01/2, 1 = 4  1/2 0, 0 1, 0 0, 1 1/2, 1/2 1 b 4 0, 0 1, 0 0, 1 1, 1 1 a 4 Old notes 4  1/4 = 1

Supplement

Does the crystallographic group abelian? Some yes, some no! m m m m m m m (1) 11 (2) (3) (1) (2) (3) (1)(2) (3) (1) (2) (3) Commutative: a.b=b.a Noncommutative group a.b  b.a

1 11 22 A  /2 1 11 22 AA Group: 4mm AA A 3  /2 A  /2 A 3  /2 11 22 33 44 33 44 33 44 1 11 22 AA A  /2 A 3  /2 33 44 11 22 A  /2 AA A 3  /2 33 44 1 22 44 A  /2 AA A 3  /2 22 33 44 1A  /2 AA 33 44 11 AA A 3  /2 1A  /2 11 AA A 3  /2 1 11 22 33 44 11 22 33 1 AA 33 44 11 22 1A  /2 22 33 44 11 1 AA (1) Ask yourself how to get (1) to the rest of position?