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Course notes by Arjeh Cohen, Rosane Ushirobira and Jan Draisma, dated October 18, 2007, apparently a copy kept in the archive rather than Phil's own writing. Chapters cover symmetry and basic notions, permutation groups, cosets and Lagrange, Loyd's puzzle, group actions, finite subgroups of O(2) and O(3), representation theory, character tables with vibration symmetry, and compact Lie groups such as SU(2) and SO(3). Includes exercises.

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Group theory for Maths, Physics and Chemistry students Arjeh Cohen Rosane Ushirobira Jan Draisma October 18, 2007 2 Contents 1 Introduction 5 1.1 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Basic notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2 Permutation Groups 17 2.1 Cosets and Lagrange . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.2 Quotient groups and the homomorphism theorem . . . . . . . . . 18 2.3 Loyd’s puzzle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.4 Action of groups on sets . . . . . . . . . . . . . . . . . . . . . . . 28 3 Symmetry Groups in Euclidean Space 33 3.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.2 Isometries of n-space . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.3 The Finite Subgroups of O(2 ,R) . . . . . . . . . . . . . . . . . . 40 3.4 The Finite Subgroups of O(3 ,R) . . . . . . . . . . . . . . . . . . 42 4 Representation Theory 51 4.1 Linear representations of groups . . . . . . . . . . . . . . . . . . 51 4.2 Decomposing Displacements . . . . . . . . . . . . . . . . . . . . . 59 5 Character Tables 63 5.1 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 5.2 Orthogonality of irreducible characters . . . . . . . . . . . . . . . 64 5.3 Character tables . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 5.4 Application: symmetry of vibrations . . . . . . . . . . . . . . . . 77 5.5 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 6 Some compact Lie groups and their representations 81 6.1 Some examples of Lie groups . . . . . . . . . . . . . . . . . . . . 81 6.2 Representation theory of compact Lie groups . . . . . . . . . . . 84 6.3 The 2-dimensional unitary group . . . . . . . . . . . . . . . . . . 86 6.4 The 3-dimensional orthogonal group . . . . . . . . . . . . . . . . 88 3 4 CONTENTS Chapter 1 Introduction 1.1 Symmetry Group theory is an abstraction of symmetry Symmetry is the notion that an object of study may look the same from diﬀerent points of view. For instance, the chair in Figure 1.1 looks the same as its reﬂection in a mirror that would be placed in front of it, and our view on the wheel depicted next to the chair doesn’t change if we rotate our point of view overπ/6 around the shaft. But rather than changing viewpoint ourselves, we think of an object’s sym- metry as transformations of space that map the object ‘into itself’. What do we mean when we say that an object is symmetric ? To answer this question, consider once more the chair in Figure 1.1. In this picture we see a planeVcutting the chair into two parts. Consider the transformation rof three- dimensional space that maps each point pto the point p/primeconstructed as follows: Figure 1.1: Bilateral and rotational symmetry. From: [8]. 5 6 CHAPTER 1. INTRODUCTION letlbe the line through pand perpendicular to V. Denoting the distance of a pointxtoVbyd(x,V),p/primeis the unique point on lwithd(p/prime,V) =d(p,V) and p/negationslash=p/prime. The map ris called the reﬂection in V. Now apply rto all points in the space occupied by the chair. The result is a new set of points, but because of the choice of Vit equals the old space occupied by the chair. For this reason, we call the chair invariant underr, or say that ris asymmetry (transformation) of the chair. Transforming any point in space to its reﬂection in the mirror Wand next rotating it over an angle of πaround the axis V∩W, gives the same result as the reﬂectionrinV. This is an illustration of what we observed about symmetry: the change of view on the chair from the original chair to its mirror image, was ‘neutralized’ to the original look, but then on a transformed chair (transformed by means of r). Symmetry then is the phenomenon that the image chair is indistinguishable from the original chair. Next, consider the wheel depicted in Figure 1.1. It also has a symmetry, but of a diﬀerent type: the rotation saround the wheel’s axis mover 2π/6 moves all points occupied by the wheel to other points occupied by it. For this reason we call sa symmetry of the wheel. This type of symmetry is called rotational symmetry . Note that the rotations around mover the angles 2 kπ/6 are symmetries of the wheel as well, for k= 1,2,3,4,5. The wheel also has reﬂection symmetry: the reﬂection tin the plane Wperpendicular to mand cutting the wheel into two thinner wheels of the same size also maps the wheel to itself. Symmetry registers regularity, and thus records beauty. But it does more than that, observing symmetry is useful. Symmetry considerations lead to eﬃ- ciency in the study of symmetric objects. For instance, •when designing a chair, the requirement that the object be symmetric may reduce drawings to half the chair; •when studying functions on the plane that respect rotational symmetry around a ‘centre’ pin the plane, we can reduce the number of arguments from the (usual) two for an arbitrary function on the plane to one, the distance to p. Symmetry conditions A few more observations can be made regarding the previous cases. 1. A symmetry of an object or ﬁgure in space is a transformation of space that maps the object to itself. We only considered the space occupied by the chair and the wheel, respectively, and not the materials involved. For example, if the left legs of the chair are made of wood and the right ones of iron, one could argue that the chair is notsymmetric under reﬂection inV. Usually, we shall speak about symmetries of sets of points, but this subtlety should be remembered. 1.1. SYMMETRY 7 O Figure 1.2: The Hexagon 2. Symmetry transformations are bijective, i.e., they have an inverse trans- formation. In the case of r, it isritself. In the case of s, it is rotation aboutmover−2π/6, or, which amounts to the same, 10 π/6. 3. Two transformations can be composed to obtain another transformation; if both translations are symmetries of some object, then so is their com- position. In the case of s, we see that s◦s(sapplied twice) is a symmetry of the wheel as well; it is rotation over the angle 2 ·(2π/6). Also,t◦s (ﬁrst apply s, thent) is a symmetry of the wheel. Usually, we will leave out the composition sign ◦and writetsinstead oft◦s, ands2forss. 4. There is one trivial transformation which is a symmetry of any object, namely the identity, sending each point to itself. It is usually denoted by e(but also by 0 or 1, see below). Groups The transformations under which a given object is invariant, form a group . Group theory was inspired by these types of group. However, as we shall see, ‘group’ is a more general concept. To get a feeling for groups, let us consider some more examples. Planar groups The hexagon, as depicted in Figure 1.2, is a two-dimensional object, lying in the plane. There are lots of transformations of the plane leaving it invariant. For example, rotation raround 0 over 2 π/6 is one of them, as is reﬂection sin the vertical line lthrough the barycentre. Therefore, the hexagon has at least 8 CHAPTER 1. INTRODUCTION Figure 1.3: A frieze pattern. From: [11]. the following symmetries: e,r,r2,r3,r4,r5,s,sr,sr2,sr3,sr4,sr5. In fact, these are all symmetries of the hexagon. Thus, each is a (multiple) product of rands. These two elements are said to generate the group of symmetries. Exercise 1.1.1. Verify that the elements listed are indeed distinct and that r6=s2=e,andsrs=r5. Exercise 1.1.2. Show thatsris a reﬂection. In which line? Figure 1.3 shows the palmette motif, which is very frequently used as a dec- oration on the upper border of wallpaper. Imagine this pattern to be repeated inﬁnitely to the left and right. Then the translation tover the vector pointing from the top of one palmette to the top of the second palmette to its right is a symmetry of the pattern, and so is the reﬂection rin the vertical line lthrough the heart of a given palmette. With these two, all symmetries of the frieze pattern can be listed. They are: e,t,t2,t3,... andr,rt,rt2,rt3,.... We ﬁnd that the symmetry group of the pattern is inﬁnite. However, when ignoring the translational part of the symmetries, we are left with essentially only two symmetries: eandr. Exercise 1.1.3. Find frieze patterns that have essentially diﬀerent symmetry groups from the one of Figure 1.3. A pattern with a more complicated symmetry group is the wallpaper pattern of Figure 1.4, which Escher designed for ‘Ontmoeting’ (Dutch for ‘Encounter’). It should be thought of as inﬁnitely stretched out vertically, as well as hori- zontally. The translations sandtover the vectors a(vertical from one nose 1.1. SYMMETRY 9 Figure 1.4: Wallpaper pattern for Ontmoeting. From: M.C.Escher, 1944. to the next one straight above it) and b(horizontal, from one nose to the next one to its right) leave the pattern invariant, but this is not what makes the picture so special. It has another symmetry, namely a glide-reﬂection g, which is described as: ﬁrst apply translation over1 2a, and then reﬂect in the vertical linelequidistant to a left oriented and a neighbouring right oriented nose. Note thatg2=s. Exercise 1.1.4. List all symmetries of the pattern in Figure 1.4. Space groups Having seen symmetries of some 2-dimensional ﬁgures, we go over to 3-dimen- sional ones. The cube in Figure 1.5 has lots of symmetries. Its symmetry group is gen- erated by the rotations r3,r4,r/prime 4whose axes are the coordinate axes, and the reﬂectiontthrough the horizontal plane dividing the cube in two equal parts. Exercise 1.1.5. What is the size of the symmetry group of the cube? Closely related is the methane molecule, depicted in Figure 1.6. Not all of the symmetries of the cube are symmetries of the molecule; in fact, only those that map the hydrogen atoms to other such atoms are; these atoms form a regular tetrahedron . Exercise 1.1.6. Which of the symmetries of the cube are symmetries of the methane molecule? Finally consider the cubic grid in Figure 1.7. Considering the black balls to be natrium atoms, and the white ones to be chlorine atoms, it can be seen as the structure of a salt-crystal. It has translational symmetry in three perpendicular 10 CHAPTER 1. INTRODUCTION Figure 1.5: The cube. Figure 1.6: The methane molecule. From: [7]. 1.1. SYMMETRY 11 Figure 1.7: Structure of a crystal directions along vectors of the same size. To understand its further symmetry, consider all those symmetries of the crystal that leave a given natrium atom ﬁxed. Such a symmetry must permute the 6 chlorine atoms closest to the given natrium atom. These atoms form a regular octahedron, as depicted in Figure 1.8, and therefore we may say that the symmetry group of salt crystals is generated by that of the regular octahedron and three translations. Exercise 1.1.7. This exercise concerns symmetries of one-dimensional ﬁgures. Considere therefore the real line Rwith the Euclidean metric, in which the distance between x,y∈Ris|x−y|. A mapgfromRto itself is called an isometry if|g(x)−g(y)|=|x−y|for allx,y∈R. 1. Show that any isometry of Ris either of the form sa:x/mapsto→2a−x(reﬂection ina) or of the form ta:x/mapsto→x+a(translation over a). 2. Prove that {e}and{e,s0}, whereedenotes the identity map on R, are the only ﬁnite groups of isometries of R. To describe symmetries of discrete subsets of R, we restrict our attention to groupsGwhose translational part {a∈R|ta∈G}equals Z. 3. Show that if such a group Gcontainss0, then G={ta|a∈Z} ∪ {ta 2|a∈Z}. 4. Conclude that there are only two ‘essentially diﬀerent’ groups describing the symmetries of discrete subsets of Rhaving translational symmetry. 12 CHAPTER 1. INTRODUCTION Figure 1.8: The octahedron. Exercise 1.1.8. Analyse the symmetry groups of discrete subsets of the strip R×[−1,1] in a manner similar to that of Exercise 1.1.7. Compare the result with your answer to Exercise 1.1.3. 1.2 Basic notions Formal deﬁnition of a group Having seen some examples of groups, albeit from the narrow point of view of symmetry groups of ﬁgures in the plane or in three-dimensional space, we are ready for some abstract deﬁnitions that should reﬂect the experiences we had in those examples. Deﬁnition 1.2.1. Agroup is a setG, together with an operation ·, which maps an ordered pair ( g,h) of elements of Gto another element g·hofG, satisfying the following axioms. 1. The operation is associative , i.e., for all g,h,k ∈Gwe haveg·(h·k) = (g·h)·k. 2.Gcontains an identity element , i.e., an element ethat satisﬁes e·g= g·e=gfor allg∈G. 3. Each element of Ghas an inverse , i.e., for each g∈Gthere is an h∈G such thatg·h=h·g=e. This element is denoted by g−1. The cardinality of Gis called the order of the group, and often denoted by |G|; it may be inﬁnite. If the operation is not only associative but commutative as well (meaning g·h=h·gfor allg,h∈G), thenGis called an Abelian group. 1.2. BASIC NOTIONS 13 Exercise 1.2.2. Prove that a group Gcannot have more than one identity. Also, the notation g−1for the inverse of gseems to indicate the uniqueness of that inverse; prove this. Let us give some examples of groups. Example 1.2.3. 1. All sets of transformations found in §1.1 form groups; in each case, the composition ◦serves as operation ·. Some of them are Abelian; like the the chair, and some aren’t, like those of the cube. 2. The real numbers Rform a group with respect to addition +, the unit element being 0. They do notform a group with respect to ordinary multiplication, as 0 does not have an inverse. Leaving out 0 we do obtain an Abelian group ( R\ {0},·), the unit element being 1. 3. Also QandZform additive groups. Again Q\ {0}is a multiplicative group. However, Zcannot be turned into a multiplicative group, the only invertible elements being {±1}. 4. LetXbe a set. The bijections of Xform a group. Composition is again the group operation. The group inverse coincides with the ‘functional’ inverse. The group is often denoted by Sym( X). IfXis ﬁnite, say |X|=n, then |Sym(X)|=n!. If two sets XandYhave the same cardinality, then the groups Sym( X) and Sym( Y) are essentially the same. For ﬁnite sets of cardinality n, the representative Sym( {1,...,n}) of this class of groups is denoted by Sym( n) orSn. Its element are also called permutations. For X={1,2,3}, we have Sym(X) ={e,(1,2),(2,3),(1,3),(1,2,3),(3,2,1)}. This notation is explained in Section 2.3. 5. LetVbe a vector space. The set of all linear transformations of V, often denoted by GL( V), together with composition of maps, is a group. By ﬁxing a basis of V, its element can be written as invertible matrices. Let us make notation easier. Deﬁnition 1.2.4. Usually, we leave out the operation ·if it does not cause confusion. However, if the operation is written as +, it is never left out. For g in a groupGwe writegnfor the product g·g·...·g/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright n. Our ﬁrst lemma concerns the inverse of a product. Lemma 1.2.5. For two elements g,hin a groupGwe have (gh)−1=h−1g−1, and(gn)−1= (g−1)n. For the latter we also write g−n. 14 CHAPTER 1. INTRODUCTION This is very natural, as Weyl points out in [11]: ‘When you dress, it is not immaterial in which order you perform the operations; and when in dressing you start with the shirt and end up with the coat, then in undressing you observe the opposite order; ﬁrst take oﬀ the coat and the shirt comes last.’ Exercise 1.2.6. Show that the notation for fractions which is usual for integers, does not work here:h gcould stand for g−1has well as for hg−1, and the latter two may diﬀer. With the methane molecule in Figure 1.6, some of the symmetries of the cube form a smaller group, namely the symmetry group of the tetrahedron. We found an example of a subgroup. Deﬁnition 1.2.7. Asubgroup of a group Gis a subset HofGsatisfying the following conditions. 1.His non-empty. 2.His closed under the operation, i.e., for all g,h∈Hwe havegh∈H. 3.His closed under taking inverses. This means that, for all h∈H, we have h−1∈H. Exercise 1.2.8. Prove that a subgroup of a group contains the group’s identity. Prove that for a subset HofGto be a subgroup of Git is necessary and suﬃcient thatHbe non-empty and that gh−1∈Hfor allg,h∈H. Exercise 1.2.9. Prove that for a subset Hof a ﬁnite group Gto be a sub- group, it is necessary and suﬃcient that Hbe non-empty and closed under the operation of G. Generation The following lemma is needed to formalize the concept of a subgroup generated by some elements. Lemma 1.2.10. The intersection of a family of subgroups of a group Gis a subgroup of G. Deﬁnition 1.2.11. As a consequence of Lemma 1.2.10, for a subset SofG the intersection of all groups containing Sis a subgroup. It is denoted by /angbracketleftS/angbracketright and called the subgroup generated by S. Instead of /angbracketleft{g1,...,g n}/angbracketrightwe also write /angbracketleftg1,...,g n/angbracketright. The elements g1,...,g nare called generators of this group. Some groups are generated by a single element. For example, Zis gener- ated by 1, and the symmetry group of the triquetrum depicted in Figure 1.9 is generated by a rotation around its center over 2 π/3. 1.2. BASIC NOTIONS 15 Figure 1.9: The triquetrum. (0,1) (1,1) (0,0) (1,0) Figure 1.10: The ﬂat torus. Deﬁnition 1.2.12. If there exists an element g∈Gfor whichG=/angbracketleftg/angbracketright, thenG is called a cyclic group. Consider the integers modulo n, wherenis a positive integer. In this set, which is denoted by Z/nZfor reasons that will become clear later, two integers are identiﬁed whenever their diﬀerence is a multiple of n. This set forms a group, addition modulo nbeing the operation. The order of Z/nZisn, and it is a cyclic group. In fact, one could say that it is thecyclic group of this order, as all such groups are the same in a sense to be made precise later. Thus the symmetry group of the triquetrum (Figure 1.9) is ‘the same’ as Z/3Z. Deﬁnition 1.2.13. Letgbe an element of a group G. The order of /angbracketleftg/angbracketrightis called theorder ofg. It may be inﬁnite. Exercise 1.2.14. Prove that a cyclic group is Abelian. Exercise 1.2.15. Consider the ﬂat torus T. It can be seen as the square S= [0,1]×[0,1] with opposite edges glued together, such that the arrows in Figure 1.10 match. (Thus, the set is really in bijection with [0 ,1)×[0,1).) The torus has the following group structure: given two points p= (p1,p2) and q= (q1,q2) on the torus, their sum corresponds to the point ronSwith the property that r−(p1+q1,p2+q2), the latter term of which may not be in S, has integral entries. Determine which elements of Thave ﬁnite order, and which elements have inﬁnite order. 16 CHAPTER 1. INTRODUCTION Chapter 2 Permutation Groups The realization of a group by means of symmetries (more generally, bijections of a set) is closely related to the structure of its subgroups. Let us ﬁrst focus on the latter aspect. 2.1 Cosets and Lagrange Recall the notion of subgroup from Deﬁnition 1.2.7. Deﬁnition 2.1.1. For subsets SandTof a groupG, we writeSTfor the set {st|s∈S,t∈T}. Fors∈GandT⊂Gwe also write sTfor{s}TandTsforT{s}. Deﬁnition 2.1.2. For a subgroup HofGthe sets of the form gHwithg∈G are called left cosets ofHinG, and those of the form Hgare called right cosets ofHinG. The set of all left cosets of HinGis denoted by G/H . It would be logical to denote the set of right cosets of HinGbyH\G. Indeed, this is sometimes done, but this meaning of \should not be confused with the usual set-theoretic one. We shall only use G/H in these notes. Example 2.1.3. 1.G=Z, the additive group, and H=/angbracketleftm/angbracketright=mZ, the subgroup generated bym, for somem∈Z. The set of left cosets is {H,1 +H,2 +H,...,m − 1 +H}. Sincex+H=H+x, left cosets coincide with right cosets. 2.G=S3={(1),(12),(13),(23),(123),(132)}, the symmetry group of an equilateral triangle (check). Take the subgroup H=/angbracketleft(123)/angbracketright. Then the set of left cosets is {H,(12)H}, where (12) H={(12),(13),(23)}. Observe that (12)H=H(12) =G\H, so left and right cosets coincide. 17 18 CHAPTER 2. PERMUTATION GROUPS 3.G=Rn, the (additive) translation group of a real vector space and His a linear subspace of G. Then the left cosets are the aﬃne subspaces parallel toH. 4. A left coset need not be a right coset. Take for instance G=S3and Hthe cyclic subgroup /angbracketleft(12)/angbracketright. Then (13) H={(13),(123)}andH(13) = {(13),(132)}. It is easy to verify that left cosets are equivalence classes of the equivalence relation g∼h⇔g−1h∈H, and similarly for right cosets. So we have the following lemma. Lemma 2.1.4. The left cosets of a subgroup Hin the group Gform a partition of the group G. Forg∈G, the maph/mapsto→ghis a bijection between Hand the left cosetgH. A similar statement holds for right cosets. Exercise 2.1.5. Show that (13) /angbracketleft(12)/angbracketrightis not a right coset of /angbracketleft(12)/angbracketrightinS3, and that (12) /angbracketleft(123)/angbracketrightis a right coset of /angbracketleft(123)/angbracketrightinS3. Deﬁnition 2.1.6. If the number of left cosets of a subgroup Hin a groupGis ﬁnite, this number is called the index ofHinGand denoted by [ G:H]. Example 2.1.7. In the previous examples, we have [ Z:/angbracketleftm/angbracketright] =mand [S3:/angbracketleft(123)/angbracketright] = 2 . Theorem 2.1.8 (Lagrange’s Theorem) .IfGis a ﬁnite group and Ha subgroup, then|H|divides |G|. Corollary 2.1.9. IfGis a ﬁnite group and g∈G, then the order of gdivides that ofG. Exercise 2.1.10. IfHandKare subgroups of a ﬁnite group G, thenH∩K is also a subgroup of G. Prove that |HK|=|H||K| |H∩K|. 2.2 Quotient groups and the homomorphism the- orem If two groups GandHand some of their elements are in the picture, we shall always make it clear to which of the two groups each element belongs. This enables us to leave out the operation symbol throughout, as the product of two elements of His automatically understood to denote the product in Hand not inG. Also, we shall use efor the unit element of both groups. Next, we will see the functions between two groups that preserve the struc- ture of groups. 2.2. QUOTIENT GROUPS AND THE HOMOMORPHISM THEOREM 19 Deﬁnition 2.2.1. LetGandHbe groups. A homomorphism fromGtoHis a mapφ:G→Hsatisfying φ(g1g2) =φ(g1)φ(g2) for allg1,g2∈G, andφ(eG) =eH. If, in addition, φis a bijection, then φis called an isomorphism . Anautomorphism ofGis an isomorphism from Gto itself. By Aut( G) we denote the set of all automorphisms of G. If there exists an isomorphism from a group Gto a group H, thenGis called isomorphic toH. this fact is denoted by G∼=H, and it is clear that ∼= is an equivalence relation. We claimed earlier that the symmetry group of the triquetrum (Figure 1.9) is ‘the same as’ Z/3Z. The precise formulation is, that both groups are isomorphic. Deﬁnition 2.2.2. For a group homomorphism φ:G→Hthe subset {g∈G|φ(g) =e} ofGis called the kernel ofφ, and denoted by ker( φ). The set φ(G) ={φ(g)|g∈G} ⊆H is called the image ofφ, and denoted by im( φ). Exercise 2.2.3. Letφ:G→Hbe a homomorphism. Then φis injective if and only if ker φ={e}. Example 2.2.4. Forg∈G, we deﬁne φg:G→Gbyφg(h) =ghg−1. Then φgis an element of Aut( G), called conjugation with g. An automorphism of G is called an inner automorphism if it is of the form φgform someg; it is called outer otherwise. Exercise 2.2.5. Letφbe a group homomorphism from GtoH. Prove that ker(φ) and im(φ) are subgroups of GandH, respectively. Show that for every g∈Gwe havegker(φ)g−1⊆ker(φ). Exercise 2.2.6. Prove: 1. Aut(G) is a group with respect to composition as maps; 2. the set Inn( G) ={φg|g∈G}is a subgroup of Aut( G). Hereφgdenotes conjugation with g. Example 2.2.7. ForG= GL(V), the group of invertible linear transformations of a complex vector space V, the mapg/mapsto→det(g) is a homomorphism G→C∗. Its kernel is the group SL( V). Example 2.2.8. The map Z→Z/mZ,x/mapsto→x+mZ, is a homomorphism of additive groups. Its kernel is mZ. 20 CHAPTER 2. PERMUTATION GROUPS Example 2.2.9. There is an interesting homomorphism between SL(2 ,C) and the Lorentz group L, which is the group of all linear transformations of the vector space R4that preserve the Lorentz metric |x|:=x2 0−x2 1−x2 2−x2 3. Tox∈R4we associate a 2 ×2-matrixψ(x) as follows: ψ(x) =/parenleftbigg x0+x3x1−ix2 x1+ix2x0−x3/parenrightbigg , so that |x|= det(ψ(x)). Now, the map ϕ: SL(2,C)→Lgiven byϕ(A)(x) = ψ−1(Aψ(x)A∗) is a group homomorphism. Here A∗= (¯aji)ijifA= (aij)ij. Indeed,ψis a linear isomorphism from R4onto ψ(R4) =/braceleftbigg/parenleftbigg x y z u/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglex,y,z,u ∈Cand ¯x=x,¯y=z,¯u=u/bracerightbigg , and one checks that this space is invariant under M/mapsto→AMA∗for anyA∈ GL(2,C). Also, for A∈SL(2,C),ϕ(A) preserves the metric due to the multi- plicative properties of the determinant: |ϕ(A)(x)|= det(Aψ(x)A∗) = det(A) det(ψ(x)) det(A∗) = det(ψ(x)) =|x|. Exercise 2.2.10. Show thatϕin Example 2.2.9 satisﬁes ϕ(AB) =ϕ(A)ϕ(B), and thus ﬁnish the proof that ϕis a homomorphism SL(2 ,C)→L. Exercise 2.2.11. Determine the subgroups of ( Z/2Z)⊕(Z/2Z). Is (Z/2Z)⊕ (Z/2Z) isomorphic to Z/4Z? Exercise 2.2.12. Take the group Z/4Z, the subgroup /angbracketleft(1234) /angbracketrightofS4and the multiplicative subgroup generated by iinC. Verify that they are all isomorphic. As we have seen before, it is not always the case that left cosets and right cosets of a subgroup Hare the same. When that happens, we say that His normal. Deﬁnition 2.2.13. A subgroup HofGis called normal if every left coset of HinGis a right coset of HinG, or, equivalently, if gHg−1⊂Hfor allg∈G. We denote it by H/triangleleftequalG. IfHis normal, then we can deﬁne an operation on the set G/H of all left cosets ofHinH, as follows: (gH)·(kH) = (gk)H. This operation turns G/H into a group, called the quotient group of GbyH. The mapG/mapsto→G/H,g /mapsto→gHis a homomorphism G→G/H . 2.2. QUOTIENT GROUPS AND THE HOMOMORPHISM THEOREM 21 Exercise 2.2.14. LetHbe a normal subgroup of the group G(cf. Exercise 2.2.13) Verify that the operation on G/H is well-deﬁned, i.e., does not depend on the choice of representatives gandk. Verify that G/H does indeed become a group with this operation. Note that one really needs normality of H. Exercise 2.2.15. Prove that Inn( G) (cf. Example 2.2.4) is a normal subgroup of Aut(G) (cf. Deﬁnition 2.2.1). Example 2.2.16. 1. The subgroup /angbracketleft(12)(34),(13)(24) /angbracketrightis normal in S4. 2. LetGbe the group of Euclidean motions in R3(see Chapter 3). Write an element ofGas /parenleftbigg a v 0 1/parenrightbigg witha∈SO(3) and v∈R3. This presentation is convenient as the oper- ation inGis now given by the multiplication of 4 ×4 matrices. Consider the subgroup Nof all translations; thus, Nconsists of all matrices /parenleftbigg I v 0 1/parenrightbigg . ThenNis a normal subgroup of G. Lemma 2.2.17. The kernel of a group homomorphism φ:G→His a normal subgroup of G. See Exercise 2.2.5 for the proof. Exercise 2.2.18. Prove that any subgroup of index 2 is normal. Theorem 2.2.19. Letφ:G→Hbe a group homomorphism. Then G/kerφ∼= imφ. Proof. LetN:= kerφ. We claim that π:G/N →imφ,gN /mapsto→φ(g) is an isomorphism. Indeed, it is well-deﬁned: if g/primeN=gN, theng−1g/prime∈Nso that φ(g−1g/prime) =e, henceφ(g/prime) =φ(g). Also,πis a homomorphism: π((gN)(kN)) =π((gk)N) =φ(gk) =φ(g)φ(k) =π(gN)π(kN). Finally, ifπ(gN) =e, thenφ(g) =e, hencegN=eN; it follows that ker( π) = {eN}, and Exercise 2.2.3 shows that πis injective. It is also surjective onto its image, hence it is an isomorphism. Theorem 2.2.20 (First Isomorphism Theorem) .LetGbe a group, and H,K normal subgroups of Gsuch thatH⊇K. ThenKis a normal sugroup of H, H/K a normal subgroup of G/K , andG/H ∼=(G/K )/(H/K ). 22 CHAPTER 2. PERMUTATION GROUPS ABCD EFGH IJKL MNO Figure 2.1: Initial state of Sam Loyd’s 15-puzzle. Proof. For anyg∈G, we have gKg−1⊆K, so a fortiori this holds for g∈H. Hence,Kis normal in H. For any g∈Gandh∈H, we have (gK)(hK)(gK)−1= (ghg−1)K∈H/K asHis normal in G. Hence,H/K is normal inG/K . Consider the map π:G→(G/K )/(H/K ),g/mapsto→(gK)(H/K ). It is clearly a surjective homomorphism, and its kernel equals {g∈G|gK∈H/K}=H. By Theorem 2.2.19, G/H ∼=(G/K )/(H/K ) as claimed. Theorem 2.2.21 (Second Isomorphism Theorem) .LetGbe a group, Ha subgroup of G, andKa normal subgroup of G. ThenHK is a subgroup of G, Ka normal subgroup of HK,H∩Kis a normal subgroup of H, andHK/K ∼= H/(H∩K). Exercise 2.2.22. Prove the second Isomorphism Theorem. 2.3 Loyd’s puzzle A Messed-Up Puzzle Suppose that a friend borrows your 15-puzzle just after you solved it, so that it is in the initial state depicted in Figure 2.1. You leave the room to make some coﬀee and when you come back your friend returns the puzzle to you, sighing that it is too diﬃcult for him, and asking you to show him how to solve it. At this time, the puzzle is in the state of Figure 2.2. Conﬁdently you start moving around the plastic squares; for readers unfamiliar with the puzzle a possible sequence of moves is shown in Figure 2.3. As time passes, you start getting more and more nervous, for your friend is watching you and you don’t seem to be able to transform the puzzle back into its initial state. What is wrong? To answer this question, we shall look at some basic properties of permuta- tions. 2.3. LOYD’S PUZZLE 23 ABCD EFGH IJKL MON Figure 2.2: Messed up state of the 15-puzzle. ABCD EFGH IJKL MONABCD EFGH IJK MONLABCD EFGH IJ MONLK Figure 2.3: Possible moves of the 15-puzzle. 24 CHAPTER 2. PERMUTATION GROUPS Permutations and their Signs Recall the deﬁnition of a permutation and the symmetric groups Sn. Permuta- tions can be represented in several ways, three of which we shall discuss here. Two-row notation: for a permutation π:{1,...,n } → { 1,...,n }one writes a 2×nmatrix, in which both rows contain the numbers 1 ,...,n . The image ofiis the number that appears below iin the matrix. An example is π=/parenleftbigg 1 2 3 4 5 5 4 1 2 3/parenrightbigg ∈S5. Disjoint cycle notation: To explain this notation, consider the permutation πfrom the previous example. We have π(1) = 5,π(5) = 3 and π(3) equals 1 again. The permutation that satisﬁes these equations and that ﬁxes the other elements is denoted by (1,5,3) = (5,3,1) = (3,1,5). A permutation of this form is called a 3- cycle . Butπdoes not ﬁx 2 or 4; indeed,π(2) = 4 and π(4) = 2. We ﬁnd that π= (1,5,3)(2,4) = (2,4)(1,5,3), i.e.,πis the product of the 2-cycle (2 ,6) and the 3-cycle (3 ,5,1). These are called disjoint , because each ﬁxes the elements occurring in the other, and this fact makes them commute. Any permutation can be written as a product of disjoint cycles, and this factorization is unique up to changes in the order. In this notation, it is usual not to write down the 1-cycles. Leaving these out, it is not always clear to which Snthe permutation belongs, but this abuse of notation turns out to be convenient rather than confusing. Permutation matrix notation: consider once again the permutation π∈S5. With it we associate the 5 ×5-matrixAπ= (aij) with aij=/braceleftBigg 1 ifπ(j) =i, 0,otherwise. It can be checked that Aπσ=AπAσ, where we take the normal matrix product in the right-hand side. Hence Sncan be seen as a matrix group, multiplication being the ordinary matrix multiplication. Diagram notation: in this notation, a permutation is represented by a dia- gram as follows: draw two rows of dots labeled 1 ,...,n , one under the other. Then draw a line from vertex iin the upper row to vertex π(i) in the lower. In our example, we get the picture in Figure 2.4. 2.3. LOYD’S PUZZLE 25 12345 12345 Figure 2.4: The diagram notation. Exercise 2.3.1. Show that the 2-cycles in Sngenerate that whole group. Can you give a smaller set of 2-cycles that generates the whole group? How many do you need? Exercise 2.3.2. Determine the order of a k-cycle. Determine the order of π= (2,6)(1,5,3). Give and prove a general formula for the order of a permutation, given in disjoint cycle notation, in terms of the lengths of the cycles. Related to permutation matrices we have the following important deﬁnition. Deﬁnition 2.3.3. Forπ∈Snwe deﬁne the signofπby sgnπ:= detAπ. The sign of a permutation πis either 1 or −1. In the former case, πis called an even permutation, in the latter an oddpermutation. Exercise 2.3.4. Show that sgn : Sn→ {± 1}is a group homomorphism. The sign of a permutation can be interpreted in many ways. For example, in the diagram notation, the number of intersections of two lines is even if and only if the permutation is even. Exercise 2.3.5. Prove that the sign of a k-cycle equals ( −1)k+1. Exercise 2.3.6. Prove that the sign of a permutation equals ( −1)NwhereN is the number of crossings of the lines drawn in the diagram notation of the permutation. The even permutations in Snform a subgroup, called the alternating group and denoted by An. It is the kernel of the homomorphism sgn : Sn→ {± 1}of Exercise 2.3.4. 26 CHAPTER 2. PERMUTATION GROUPS 1 5 12 109 13141516112 63 74 8 Figure 2.5: Numbering of the positions. No Return Possible Let us return to Sam Loyd’s 15-puzzle. It can be described in terms of states and moves. The Figures 2.1 and 2.2 depict states of the puzzle. Formally, a state can be described as a function s:{1,..., 16} → { A,B,..., O,∗}. Here,s(p) is the square occupying position pin the state sand∗stands for the empty square. The positions are numbered as in Figure 2.5. Forp∈ {1,..., 16}, we denote by M(p)⊆S16the set of possible moves in a stateswiths(p) =∗. A movem∈M(p) should be interpreted as moving the square from position jto position m(j), forj= 1,..., 16. For example M(1) = {(1,2),(1,5)}, M(2) = {(1,2),(2,3),(2,6)},and M(6) = {(2,6),(5,6),(6,7),(6,10)}. Moves can be combined to form more general transformations . The set of all possible transformations starting in a state with the empty square in position pis denoted by T(p), and it is deﬁned as the set of all products (in S16) of the formmkmk−1...m 1, where mj+1∈M(mj◦mj−1◦...◦m1(p)) for allj≥0. This reﬂects the condition that mj+1should be a possible move, after applying the ﬁrst jmoves tos. In particular, in taking k= 0, we see that the identity is a possible transformation in any state. Also, if t∈T(p) and u∈T(t(p)), thenu◦t∈T(p). Finally, we have t−1∈T(t(p)). Moves and, more generally, transformations change the state of the puzzle. Lets∈Swiths(p) =∗and lett∈T(p). Thentchanges the state sto the 2.3. LOYD’S PUZZLE 27 Figure 2.6: Ordering of the non-empty squares in each state. statet·s, which is deﬁned by (t·s)(q) =s(t−1(q)). In words, the square at position qafter applying tis the same as the square at positiont−1(q) before that transformation. Now that we captured Sam Loyd’s 15-puzzle in a comfortable notation, we can formalize our question as follows: is there a t0∈T(16) that transforms the messed up state s1of Figure 2.2 into the initial state s0of Figure 2.1, i.e., that satisﬁes t0·s1=s0? The answer is no, and we shall presently make clear why. In each state, order the non-empty squares according to the path in Figure 2.6. That is, the j-th non-empty square one encounters when following this path in a particular state sis called the j-th square in s, forj= 1,..., 15. Now consider any position p, anyt∈T(p) and some s∈Swiths(p) =∗. Then we deﬁne a permutation πt∈S15by πt(i) =j⇔thei-th square in state sis thej-th square in state t·s. Note that this is independent of the choice of s. We have that πu◦t=πu◦πt (2.1) whenever the left-hand side makes sense, i.e., whenever u∈T(t(p)). We see that the t0we would like to ﬁnd equals (14 ,15), and then πt0also equals (14,15), considered as a permutation in S15. This is an odd permutation. Now consider the move mdepicted in Figure 2.7. It interchanges the squares in position 6 and 10 in Figure 2.5, so the move equals (6 ,10). The corresponding permutation πmis the cycle (10 ,9,8,7,6). A moment’s thought leads to the 28 CHAPTER 2. PERMUTATION GROUPS 678 910 Figure 2.7: Moves correspond to odd cycles. conclusion that πmis an odd cycle for anymovem, hence an even permutation. As a consequence of this and equation (2.3), a sequence of such moves will also be mapped to an even permutation by π. Therefore, such a transformation cannot possibly be (14 ,15). You conclude that your friend must have cheated while you were away mak- ing coﬀee, by lifting some of the squares out of the game and putting them back in a diﬀerent way. Exercise 2.3.7. Show that the set {πt|t∈T(p) for somep}forms a subgroup ofS15. Exercise 2.3.8. Prove that the set {t∈T(16)|t(16) = 16 } is a subgroup of A16. Use this to give another argument why your friend must have cheated. 2.4 Action of groups on sets Although the transformations in the 15-puzzle do not quite form a group, their ‘acting’ on the puzzle states reﬂects an important phenomenon in group theory, namely that of an action. Deﬁnition 2.4.1. LetGbe a group and Ma set. An action ofGonMis a mapα:G×M→Msatisfying 1.α(e,m) =mfor allm∈M, and 2.α(g,α(h,m)) =α(gh,m ) for allm∈Mandg,h∈G. 2.4. ACTION OF GROUPS ON SETS 29 If the action αis obvious from the context, we leave αout and write gmfor α(g,m). Also, for subsets S⊂GandT⊂Mwe writeSTfor{α(g,m)|g∈ S,m∈T}. For {g}Twe writegTand forS{m}we writeSm. Remark 2.4.2. Given an action α, we have a homomorphism G→Sym(M) given byg/mapsto→(m/mapsto→α(g,m)). Conversely, given a homomorphism φ:G→ Sym(M), we have an action α:G×M→Mgiven by α(g,m) =φ(g)m. In other words, an action on Mis nothing but a homomorphism G→Sym(M). Let us give some examples of a group acting on a set. Example 2.4.3. 1. The group Snacts on {1,...,n }byπ·i=π(i), and on the set of all two-element subsets of {1,...,n }byπ({i,j}) ={π(i),π(j)}. 2. The group GL(3 ,R) acts on R3by matrix-vector multiplication. 3. Consider the group of all motions of the plane, generated by all trans- lations (x,y)/mapsto→(x+α,y) forα∈Rinx-direction and all inﬂations (x,y)/mapsto→(x,βy ) forβ∈R∗. This acts on the solutions of the diﬀerential equation (d dx)2(y) =−y. Example 2.4.4. ConsiderG=S4and puta={12,34},b={13,24},c= {14,23}. Here 12 stands for {1,2}, etc. There is a natural action (as above) on the set of pairs, and similarly on the set {a,b,c}of all partitions of {1,2,3,4} into two parts of size two. Thus, we ﬁnd a homomorphism S4→Sym({a,b,c}). It is surjective, and has kernel a group of order 4. (Write down its nontrivial elements!) Example 2.4.5. The group SL(2 ,C) acts on the complex projective line P1(C) as follows. Denote by π:C2\ {0} →P1(C) the map sending a pair of homoge- neous coordinates to the corresponding point. The linear action of SL(2 ,C) on C2by left multiplication permutes the ﬁbers of π, hence induces an action on P1(C). On the aﬃne part of P1(C) where the second homogeneous coordinate is non-zero, we may normalize it to 1, and the action on the ﬁrst coordinate x is given by/parenleftbigg a b c d/parenrightbigg x=ax+b cx+d. Note, however, that this aﬃne part is not invariant under the group. Lemma 2.4.6. LetGbe a group. Then Gacts on itself by conjugation, that is, the map α:G×G→Gdeﬁned byα(h,g) =hgh−1is an action. 30 CHAPTER 2. PERMUTATION GROUPS Proof. We check the two conditions of Deﬁnition 2.4.1. First, α(e,g) =ege−1=g for allg∈G, and second, α(h,α(k,g)) =h(kgk−1)h−1= (hk)g(hk)−1=α(hk,g ). The following lemma allows us to partition a set Mon which a group Gacts into orbits. Lemma 2.4.7. LetGbe a group acting on a set M. Then the sets Gm={gm| g∈G}form∈Mpartition the set M. The setGm is called the orbit ofm underG. Deﬁnition 2.4.8. An action of a group Gon a setMis said to be transitive if it has only one orbit, or, stated diﬀerently, Gm=Mfor anym∈M. To deﬁne symmetry in a general context, we look at sets that are mapped to themselves by a given group. Deﬁnition 2.4.9. LetGbe a group acting on M. A subset T⊆Mis called invariant underS⊆GifST⊆T. In this case, the elements of Sare called symmetries ofT. Deﬁnition 2.4.10. LetGbe a group acting on Mand letm∈M. Then the set Gm:={g∈G|gm=m} is called the stabilizer ofm. Lagrange’s theorem has an important consequence for problems, in which one has to count the number of symmetries of a given set. It can be formulated as follows. Theorem 2.4.11. LetGbe a group acting on a set Mand letm∈M. Then Gmis a subgroup of G, and the map f:Gm→G/G m, g·m/mapsto→gGm is well deﬁned and bijective. As a consequence, if Gis ﬁnite, then |G|=|Gm||Gm|. Remark 2.4.12. One must distinguish the elements of Gfrom their actions on M. More precisely, the map sending g∈Gto the map α(g,·) :M→Mis a homomorphism, but need not be an isomorphism. We now want to make clear how the action of groups can help in classiﬁcation and counting problems, and we shall do so by means of the following problem. 2.4. ACTION OF GROUPS ON SETS 31 1 2 4 31 2 341 2 3 4 5 Figure 2.8: Three graphs, two of which are isomorphic. Given is a ﬁnite set Xwith|X|=v. What is the number of non- isomorphic simple graphs with vertex set X? This natural, but seemingly merely theoretical, question has strong analogue in counting all isomers with a given chemical formula, say C 4H10(see [6]). Of course, the question is formulated rather vaguely, and the ﬁrst step in solving it is to formalize the meanings of the words graph andnon-isomorphic . A simple graph is a function f:/parenleftbiggX 2/parenrightbigg → {0,1}. Here/parenleftbigX 2/parenrightbig stands for the set of all unordered pairs from X. If the function f takes the value 1 on such a pair {x,y}, this means that the pair is an edge in the graph. Often, graphs are associated with a picture like the ones in Figure 2.8. The set of all graphs on Xis denoted by Graphs( X). The ﬁrst two graphs in Figure 2.8 are essentially the same; to make them exactly the same one has to relabel the vertices. Such a relabeling is in fact a permutation of X. Formally, this leads to an action of Sym( X) on Graphs( X) in the following way. First of all Sym( X) acts on/parenleftbigX 2/parenrightbig by π· {x,y}={π(x),π(y)}. Furthermore, we deﬁne π·fby the commutativity of the following diagram. /parenleftbigX 2/parenrightbig f− − − − → { 0,1} π·/arrowbt/arrowbtId /parenleftbigX 2/parenrightbig − − − − → π·f{0,1} This diagram tells us that e∈/parenleftbigX 2/parenrightbig is an edge in the graph fif and only if π·e is an edge in the graph π·f. In formula, π·f(x) =f(π−1x). 32 CHAPTER 2. PERMUTATION GROUPS Graphs that can be obtained from each other by the action of some π∈ Sym(X) are called isomorphic . Now our original question can be rephrased very elegantly as follows. How many orbits does the action of Sym( X) have on Graphs( X)? The following lemma answers this question in general. Lemma 2.4.13 (Cauchy-Frobenius) .LetGbe a ﬁnite group acting on a ﬁnite setM. Then the number of orbits of GonMequals 1 |G|/summationdisplay g∈Gﬁx(g). Here ﬁx (g)denotes the number of m∈Mwithg·m=m. Exercise 2.4.14. Prove Lemma 2.4.13. To apply this lemma to counting graphs, we need to know for some π∈ Sym(X), the number of f∈Graphs(X) that are ﬁxed by π. The function fis ﬁxed byπif and only if fis constant on the orbits of /angbracketleftπ/angbracketrightin/parenleftbigX 2/parenrightbig . This reﬂects the condition that πshould map edges to edges and non-edges to non-edges. Denoting the number of orbits of /angbracketleftπ/angbracketrighton/parenleftbigX 2/parenrightbig byo(π), we ﬁnd that the number of non-isomorphic graphs on Xequals 1 v!/summationdisplay π∈Sym( X)2o(π), wherev=|X|. Exercise 2.4.15. Consider the permutation π= (1,2)(3,4,5) on {1,2,3,4,5}. Calculate the number o(π). Try to ﬁnd a general formula in terms of the cycle lengths ofπand the number v. Observe that o(π) only depends on the cycle type. Example 2.4.16. Forv= 4, we have o(e) = 6,o((12)) = 4, o((123)) = 2, o((1,2,3,4)) = 2 and o((12)(34)) = 4; the conjugacy clases of each of these permutations have sizes 1 ,6,8,6 and 3, respectively. Hence, the number of non-isomorphic graphs on 4 vertices equals (1·26+ 6·24+ 8·22+ 6·22+ 3·24)/24 = 11. Can you ﬁnd all? Chapter 3 Symmetry Groups in Euclidean Space This chapter deals with groups of isometries of n-dimensional Euclidean space. We set up a framework of results that are valid for general n, and then we restrict our attention to n= 2 or 3. In these two cases, we classify the ﬁnite groups of isometries, and point out which of these are crystallographic point groups , i.e., preserve a 2- or 3-dimensional lattice, respectively. First, however, we consider an example that motivates the development of this theory. 3.1 Motivation In low dimensions, there are few ﬁnite groups of isometries and this enables us, when given a two- or three-dimensional object, to determine its symmetry group simply by enumerating all axes of rotation and investigating if the object admits improper isometries such as inversion and reﬂections. As an example, suppose that we want to describe the symmetry group Gof the CH 4molecule depicted in Figure 3.1. The hydrogen atoms are arranged in a regular tetrahedron with theC-atom in the center. As a start, let us compute |G|. Consider a ﬁxed H-atom a. Its orbit Ga consists of all four H-atoms, as can be seen using the rotations of order 3 around axes through the C-atom and an H-atom. Applying Lagrange’s theorem we ﬁnd |G|= 4|Ga|. The stabilizer of aconsists of symmetries of the remaining 4 atoms. Consider one of the other H atoms, say b. Its orbit under the stabilizer Gahas cardinality 3, and the stabilizer ( Ga)bconsists of the identity and the reﬂection in the plane through a,band theC-atom, so that we ﬁnd |G|=|Ga||Ga|=|Ga||Gab||(Ga)b|= 4·3·2 = 24. Only half of these isometries are orientation-preserving. To gain more information on the group G, we try to ﬁnd all rotation axes: 33 34 CHAPTER 3. SYMMETRY GROUPS IN EUCLIDEAN SPACE HCH HH Figure 3.1: The CH 4molecule. Figure 3.2: A regular tetrahedron inside a cube. 1. There are 4 three-fold axes, namely those through an H-atom and the center of the opposite side of the tetrahedron. 2. There are three two-fold axes, namely those through the centers of two perpendicular sides. Later on, we shall see that this information determines the orientation pre- serving part of Guniquely; it is denoted by T. From the fact that Gdoes not contain the inversion, but does contain other improper isometries, we may conclude that GisT∪i(W\T), a group that can be described by Figure 3.2. Here we see a regular tetrahedron inside a cube. Half of the proper isometries of the cube are isometries of the tetrahedron, and the improper isometries of the tetrahedron may be obtained as follows: take any proper isometry of the cube which is nota isometry of the tetrahedron, and compose it with the in- versioni. Note that T∪i(W\) andWare isomorphic as abstract groups, but their realizations in terms of isometries are diﬀerent in the sense that there is no isometry mapping one to the other. This point will also be clariﬁed later. 3.2. ISOMETRIES OF N-SPACE 35 Exercise 3.1.1. Consider the reﬂection in a plane through two H-atoms and the C-atom. How can it be written as the composition of the inversion and a proper isometry of the cube? The ﬁnite symmetry group of the tetrahedron and the translations along each of three mutually orthogonal edges of the cube in Figure 3.2 generate the symmetry group of a crystal-like structure. For this reason, the group Gis called a crystallographic point group . 3.2 Isometries of n-space Consider the n-dimensional Euclidean space Rnwith the distance function d(x,y) =/bardblx−y/bardbl= (x−y,x−y)1/2= (/summationdisplay i(xi−yi)2)1/2. Deﬁnition 3.2.1. Anisometry is a mapg:Rn→Rnsuch that d(g(x),g(y)) =d(x,y) for allx,y∈Rn. The set of all isometries is a subgroup of Sym( Rn), which we denote by AO( n,R). The following deﬁnition and lemmas will explain the name of this group. Deﬁnition 3.2.2. For a ﬁxed a∈Rn, the isometry ta:x/mapsto→x+ais called the translation over a. Lemma 3.2.3. For anyg∈AO(n,R), there exists a unique pair (a,r)where a∈Rnandr∈AO(n,R)ﬁxes 0, such that g=tar. Proof. Seta:=g(0). Thent−1 ag(0) = 0, so t−1 agis an isometry ﬁxing 0, and it is clear that a=g(0) is the only value for which this is the case. Lemma 3.2.4. Letr∈AO(n,R)be an isometry ﬁxing 0. Thenris an orthog- onalR-linear map. Proof. Letx,y∈Rn, and compute (rx,ry ) = ( −d(rx,ry )2+d(rx,0)2+d(ry,0)2)/2 = (−d(x,y)2+d(x,0)2+d(y,0)2)/2 = (x,y), where we use that ris an isometry ﬁxing 0. Hence, rleaves the inner product invariant. This proves the second statement, if we take the ﬁrst statement for granted. For any z∈Rn, we have (z,rx +ry) = (z,rx) + (z,ry) = (r−1z,x+y) = (z,r(x+y)), so thatrx+ry−r(x+y) is perpendicular to all of Rn, hence zero. Similarly, one ﬁndsr(αx) =αrxforα∈R. This proves the ﬁrst statement. 36 CHAPTER 3. SYMMETRY GROUPS IN EUCLIDEAN SPACE Exercise 3.2.5. Prove that an isometry is bijective. (The assertion is used in the second part of Deﬁnition 3.2.1.) Hint: If ris an isometry, then t−r0r is an isometry ﬁxing 0, so it suﬃces to prove the assertion for rwithr0 = 0. Now the proof of Lemma 3.2.4 uses r−1, so we need another proof. Set v= r(λx+µy)−λr(x)−µr(y) and consider ( v,v). Using bilinearity of the inner product and orthogonality of r, it can be shown that ( v,v) = 0. The group of all orthogonal linear maps Rn→Rnis denoted by O( n,R). An isometry is thus an orthogonal linear map, followed by a translation; hence the name aﬃne orthogonal transformation . For a group G, we write T(G) :={a∈Rn|ta∈G}; this is an additive subgroup of Rn, which we shall identify with the group of all translations in G. Using the linearity of isometries ﬁxing 0, one can prove the following. Proposition 3.2.6. For a subgroup G⊆AO(n,R), the mapR:G→O(n,R) deﬁned by g=taR(g)for somea∈Rn is well-deﬁned, and a group homomorphism. Its kernel is T(G), and this set is invariant under R(G). Proof. Lemma 3.2.3 and 3.2.4 show that Ris well-deﬁned. Let r1,r2∈O(n,R) anda1,a2∈Rn, and check that ta1r1ta2r2=ta1+r1a2r1r2. This proves that Ris a homomorphism. Its kernel is obviously T(G). Let r∈R(G) anda1∈Rnsuch thatta1r1∈G, and leta2∈T(G). Then, for x∈Rn, ta1r1ta2r−1 1t−a1x=ta1r1ta2(r−1 1x−r−1 1a1) =ta1(x−a1+r1a2) =tr1a2x. This shows shows that T(G) is invariant under R(G). We are interested in certain groups that act discretely. Deﬁnition 3.2.7. A subsetV⊆Rnis called discrete if for every point p∈V there is an open neighbourhood U⊆Rnsuch that U∩V={p}. Deﬁnition 3.2.8. A discrete subgroup of Rnis called a lattice , and the dimen- sion of its R-linear span is called its rank. 3.2. ISOMETRIES OF N-SPACE 37 Lemma 3.2.9. LetL⊆Rnbe a lattice. Then Lis a closed set. Proof. As 0 is an isolated point, there is an /epsilon1 >0 such that /bardblv/bardbl ≥/epsilon1for all v∈L\{0}. But then /bardblv−w/bardbl ≥/epsilon1for all distinct v,w∈L. Hence, any sequence (vn)ninLwithvn−vn+1converging to 0, is eventually constant. Proposition 3.2.10. Leta1,...,a k∈Rnbe linearly independent over R. Then the set L=/summationdisplay iZai is a lattice. Conversely, every lattice is of this form. Proof. To prove that an additive subgroup of Rnis a lattice, it suﬃces to prove that 0 is an isolated point. For Las in the lemma, let mbe the minimum of the continuous function x→ /bardbl/summationdisplay ixiai/bardbl on the unit sphere {x∈Rk| /bardblx/bardbl= 1}. As theaiare linearly independent, m> 0. For any non-zero x∈Zk, we ﬁnd /bardbl/summationdisplay ixiai/bardbl=/bardblx/bardbl/bardbl/summationdisplay ixi /bardblx/bardblai/bardbl ≥ /bardblx/bardblm≥m, proving the ﬁrst statement. For the converse, we proceed by induction on the rank of the lattice. Clearly, a lattice of rank 0 is of the desired form. Now suppose that every lattice of rank k−1 is of that form, and let L⊆Rnbe a lattice of rank k. First let us show thatLis a closed subset of Rn. Being discrete and closed, Lhas only ﬁnitely many points in any compact set. In particular, one can choose a v1∈Lof minimal non-zero norm. Let π:Rn→v⊥ 1denote the orthogonal projection along v1. Thenπ(L) is an additive subgroup of v⊥ 1. Letv∈Lbe such that π(v)/negationslash= 0. After subtracting an integer multiple of v1fromv, we may assume that /bardblv−π(v)/bardbl ≤ /bardblv1/bardbl/2. Using /bardblv/bardbl ≥ /bardblv1/bardbland Pythagoras, we ﬁnd /bardblπ(v)/bardbl ≥√ 3/2/bardblv1/bardbl. This proves that π(L) is discrete, hence a lattice. It has rank k−1, so there exist v2,...,v k∈Lwhose images under πare linearly independent, and such that π(L) =k/summationdisplay i=2Zπ(vi). Nowv1,...,v kare linearly independent. For v∈Lwe can write π(v) =k/summationdisplay i=2ciπ(vi) 38 CHAPTER 3. SYMMETRY GROUPS IN EUCLIDEAN SPACE withci∈Z. Thenv−/summationtextk i=2civk∈Lis a scalar multiple of v1, and by minimality of/bardblv/bardblthe scalar is an integer. Hence, L=k/summationdisplay i=1Zvi. Deﬁnition 3.2.11. A subgroup G⊆AO(n,R) is said to act discretely if all its orbits in Rnare discrete. IfG⊆AO(n,R) is a group that acts discretely, then T(G) does so, too. In general, this does not hold for R(G), as the following example shows. Example 3.2.12. Letα∈Rbe an irrational multiple of π, and deﬁne r∈ O(3,R) by r= cosα sinα0 −sinαcosα0 0 0 1 . Consider the group Ggenerated by g=te3r. Fromgn=tne3rnit is clear that Gacts discretely, and that R(G) ={rn|n∈Z}. Asrhas inﬁnite order, we haveT(G) ={0}, andR(G) does not act discretely: the R(G)-orbit of (1 ,0,0)T is an inﬁnite set in the circle {x∈R3|x3= 0,/bardblx/bardbl= 1}. This phenomenon does not occur if we require T(G) to have the maximal possible rank. Theorem 3.2.13. LetGbe a subgroup of O(n,R)leaving invariant a lattice L of rankn. ThenGis ﬁnite. Proof. Leta1,...,a n∈Lbe linearly independent. As Lis discrete, it contains only ﬁnitely many vectors of norm /bardblai/bardbl, for eachi. Hence, the G-orbitsGai are ﬁnite. As the aispanRn, the permutation representation of Gon the ﬁnite setX:=/uniontextn i=1Gaiis faithful. Hence, Gis embedded into the ﬁnite group Sym(X). Deﬁnition 3.2.14. A subgroup G⊆AO(n,R) for which T(G) is a lattice of rankn, is called a crystallographic group inndimensions. A group G⊆O(n,R) leaving invariant a lattice of rank n, is called a crystallographic point group in ndimensions. Corollary 3.2.15. LetG⊆AO(n,R)be a crystallographic group. Then R(G) is a crystallographic point group. Any crystallographic point group is ﬁnite. Proof. By Proposition 3.2.6, T(G) is invariant under R(G). Now apply Theorem 3.2.13. 3.2. ISOMETRIES OF N-SPACE 39 This corollary allows for the following approach to the classiﬁcation of crys- tallographic groups in O( n,R): ﬁrst classify all ﬁnite subgroups of O( n,R), then investigate for each of them whether it leaves invariant a lattice of rank n. Fi- nally, for each pair ( R,L) of a crystallographic point group Rand a lattice Lof ranknleft invariant by R, there may be several non-equivalent groups Gwith T(G) =LandR(G) =R. Among them is always the semi-direct product L/multicloserightR: the subgroup of AO( n,R) generated by LandR. In this setting, ‘classiﬁcation’ means ‘up to equivalence under AO( n,R)’, i.e., the groups GandhGh−1for someh∈AO(n,R) are considered the same. This is a ﬁner equivalence relation than isomorphism, as is shown in the following example. Example 3.2.16. The groups C2andD1(see Section 3.3) are isomorphic as abstract groups, but not equivalent under AO(2 ,R). The same holds for the groupsWandT∪i(W\T) (see Section 3.4). Exercise 3.2.17. Show that the symmetry group Gof Figure 1.4 is not the semi-direct product of R(G) andT(G). In the next sections, we shall carry out this classiﬁcation in the cases n= 2 and 3, up to the point of determining the crystallographic point groups. The determinant of an orthogonal map turns out to be a useful tool there. Deﬁnition 3.2.18. An isometry r∈O(n,R) is called proper ororientation preserving if detr= 1. The set of all proper isometries in O( n,R) is denoted by SO(n,R). An isometry g∈AO(n,R) is called proper if R(g) is proper. Otherwise, it is called improper . Exercise 3.2.19. Prove that, in any subgroup G⊆AO(n,R) containing im- proper isometries, the proper ones form a subgroup of index 2. Exercise 3.2.20. Consider the following map: ra:v/mapsto→v−2(v,a)a, wherea∈Rnhas norm 1 (that is, ||a||= 1). Prove that 1. it belongs to O(n,R); 2. ﬁxes each vector in the hyperplane a⊥; 3.ra(a) =−a; 4. det(ra) =−1; 5.rahas order 2; Such an element is called a reﬂection . Exercise 3.2.21. Show that any ﬁnite subgroup G⊆AO(n,R) ﬁxes a point inRn. Hint: consider the ‘average’ of an arbitrary orbit. 40 CHAPTER 3. SYMMETRY GROUPS IN EUCLIDEAN SPACE Exercise 3.2.22. LetGbe a ﬁnite subgroup of the group GL( n,R) of all invert- iblen×n-matrices. Show that Gleaves invariant a positive deﬁnite symmetric bilinear form, and that it is hence conjugate to a subgroup of O( n,R). Hint: let (.,.) denote the standard inner product, and deﬁne /angbracketleftx,y/angbracketright:=1 |G|/summationdisplay g∈G(gx,gy ) forx,y∈Rn. Show that /angbracketleft.,./angbracketrightis symmetric and positive deﬁnite. Let Abe the matrix with entries aij=/angbracketleftei,ej/angbracketright, andB∈GL(n,R) such that BTB=A. Show thatBGB−1⊆O(n,R). 3.3 The Finite Subgroups of O(2 ,R) In nature, as well as in art and many other areas, many plane ﬁgures with certain symmetries exist. One may think of ﬂowers (seen from above), frieze patterns on picture frames, Escher’s plane tilings and decorative wall papers. In order to describe these symmetries, we investigate the ﬁnite subgroups of O(2,R). In two dimensions, we have two diﬀerent types of isometries. Lemma 3.3.1. The linear map rφ:R2→R2with matrix /parenleftbigg cosφ−sinφ sinφcosφ/parenrightbigg with respect to the standard basis is an element of SO(2,R), called rotation around 0 over φ. Conversely, any element of this group equals some rφ. Any improper isometry in O(2,R)is a reﬂection in a line, see Exercise 3.2.20. The proof of this lemma is straightforward. Let us describe two classes of ﬁnite subgroups of O(2 ,R). 1. The cyclic group of ordern, denoted by Cn, and generated by r2π/n. 2. The dihedral group of order 2n, denoted by Dn, and generated by r2π/n and the reﬂection in a ﬁxed line. In fact, these are the only two classes, as we shall now prove. Let Gbe a ﬁnite subgroup of O(2 ,R). There are two possibilities: 1.G⊆SO(2,R). By Lemma 3.3.1, it consists of rotations rφ. Letφ > 0 be minimal such that rφ∈G. AsGis ﬁnite,φ= 2π/nfor some integer n>0. Ifθis any other angle such that rθ∈G, thenθ=kφ+σfor some integerkand angleσ≥0 and<φ. By minimality of φ,σ= 0. Hence G is generated by r2π/n, and equal to Cn. 3.3. THE FINITE SUBGROUPS OF O(2,R) 41 2.Gcontains a reﬂection gin a linel. ThenG∩SO(2,R) is a subgroup of index 2 in G, and by the ﬁrst case generated by r2π/nfor some integer n>0. NowGis generated by the gandr2π/n, and conjugate to Dn. We have thus proved the following theorem. Theorem 3.3.2. LetGbe a ﬁnite subgroup of O(2,R). ThenGis conjugate to one of the groups CnorDnfor some integer n≥1. Exercise 3.3.3. Show that O( n,R) is generated by reﬂections. Hint: show ﬁrst that it is generated by elements that ﬁx an ( n−2)-dimensional subspace of Rn pointwise, and that O(2 ,R) is generated by reﬂections. Using this classiﬁcation, we can ﬁnd which of the ﬁnite groups are crystal- lographic point groups. Theorem 3.3.4. The following groups are crystallographic point groups: C1,C2,C3,C4,C6, D1,D2,D3,D4,D6. Moreover, any crystallographic point group Gin two dimensions is conjugate to one of the above. Proof. For the ﬁrst part, one only needs to construct invariant lattices for D4 andD6(see 3.3.5), as the other groups are contained in one of these see. For the second statement, suppose that Gis a crystallographic point group. According to Theorem 3.2.13, Gmust be ﬁnite. The lattice Lis invariant under the subgroup Kof O(2,R) generated by G∪ {i}, whereidenotes the inversion. As the inversion commutes with all elements of G,Kis ﬁnite (either equal to Gor twice as large). Therefore, according to Theorem 3.3.2, Kis conjugate to CnorDnfor somen; asKcontains the inversion, nmust be even. So either n≤6, in which case Gis of one of the types mentioned in the theorem, or n≥8. Suppose that the latter were the case and choose a vector cin the lattice of shortest possible length. Now r2π/nc∈Landc−r2π/nc∈L. A little calculus shows that the latter vector is shorter than cbecause the rotation angle too small (for n= 8, this can be seen from Figure 3.3), and we arrive at a contradiction. Exercise 3.3.5. Show thatD4andD6are crystallographic point groups. Note that we have only classiﬁed all crystallographic point groups, and not the crystallographic groups. There are 17 of the latter, and only 10 of the former; see [2]. 42 CHAPTER 3. SYMMETRY GROUPS IN EUCLIDEAN SPACE chcc-hc Figure 3.3: Why ncannot not be 8. 3.4 The Finite Subgroups of O(3 ,R) In this section, we shall repeat the classiﬁcation of §3.3, but now for three dimensions. Lemma 3.4.1. Any element g∈SO(3,R)ﬁxes a line, i.e., with respect to some orthonormal basis ghas the matrix /parenleftbigg 1 0 0M/parenrightbigg , whereMis the matrix of an element of SO(2,R). Proof. Letλ1,λ2,λ3be the eigenvalues of g. Asgleaves the norm invariant, all have norm 1. As gis proper, we have λ1λ2λ3= 1. Now either all are real, in which case at least one of them is 1, or we may assume that λ1is real and λ3=λ2. But then λ2λ3=|λ2|2= 1, soλ1= 1. An element of SO(3 ,R) is called a rotation , the line ﬁxed by it its axisand the angle corresponding to Mitsrotation angle . We present some classes of ﬁnite subgroups of SO(3 ,R). 1. Fix an axis land consider the proper rotations around lover the angles 2kπ/n fork= 0,...,n −1. They form a group, denoted by Cn, due to its analogy to the class denoted by Cnin the two-dimensional setting. 2. Start with the group Cnof rotations around land choose a second axis m perpendicular to l. Consider the group generated by Cnand the rotation aroundmoverπ. It is twice as large as Cn, and denoted by D/prime n. Note that it contains only proper isometries. The proper symmetries of an n-prism form a group of type D/prime n. Note that D/prime 1is conjugate to C2. 3.4. THE FINITE SUBGROUPS OF O(3,R) 43 3. The ﬁve platonic solids (tetrahedron, cube, octahedron, dodecahedron and icosahedron) with vertices on the unit sphere yield examples of ﬁnite subgroups of SO(3 ,R). However, only three distinct classes arise: •T, the group of all proper rotations leaving a tetrahedron invariant. •W, the group of all proper rotations that are symmetries of a cube. •P, the group of all proper rotations leaving a dodecahedron invariant. The reason is that the octahedron and the cube are polar ﬁgures: it is easily seen that the proper symmetries of a cube are exactly those of the octahedron which has its vertices in the middles of the faces of the cube. A similar statement holds for the dodecahedron and the icosahedron. Theorem 3.4.2. Any ﬁnite subgroup of SO(3,R)is conjugate to one of the following. Cn forn≥1, D/prime n forn≥2, T,W,P. The following proof is due to Euler. Proof. LetGbe a ﬁnite subgroup of SO(3 ,R). Consider the set Sdeﬁned by S:={p∈R3| ||p||= 1,there exists g∈Gsuch thatg/negationslash=eandg(p) =p}. Count the number Nof pairs (g,p), withe/negationslash=g∈Gandpon the unit sphere, that satisfy g(p) =p. On one hand, each g/negationslash=eﬁxes exactly two anti-podal points, hence this number is N= 2(|G| −1). On the other hand, for each p∈S, the number of g/negationslash=eﬁxingpequals |Gp|−1. Hence N=/summationdisplay p∈S(|Gp| −1). Ifgp=pandg/negationslash=e, andh∈G, then (hgh−1)hp=hpandhgh−1/negationslash=e, soSis invariant under G. Partition Sinto orbits of G. The value of |Gp|is constant on such an orbit o, namely |G|/|o|, due to Lagrange’s theorem. Hence we ﬁnd 2(|G| −1) =/summationdisplay o|o|(|G|/|o| −1), where the sum is taken over all orbits of GinS. Hence, dividing both sides by |G|, we get 2−2 |G|=/summationdisplay o(1−|o| |G|). 44 CHAPTER 3. SYMMETRY GROUPS IN EUCLIDEAN SPACE Group type Order Number of orbits on STheir sizes rotation axes Cn n 2 1 ,1 1 ×n D/prime n 2n 3 2 ,n,n 1×n,n×2 T 12 3 4 ,4,6 4 ×3,3×2 W 24 3 8 ,6,12 4 ×3,3×4,6×2 P 60 3 12 ,20,30 6 ×5,10×3,15×2 Table 3.1: The ﬁnite groups of proper rotations in space. Hence, if we set ao:=|G|/|o|for each orbit oonS, then we have /summationdisplay o(1−1/ao)<2. As eachs∈Shas non-trivial stabilizer, each orbit in Shas size at most |G|/2; hence each aois at least 2. Now Exercise 3.4.3 leads to Table 3.1. That is, it provides the numerical data for it; it remains to check that this data determines the group up to conjugation. Exercise 3.4.3. Consider the following function in GAP. extend:=function(a,s) #Pre-condition: a is a non-empty non-decreasing list of integers >=2, #and s equals (sum i: (1-1/a[i])). This procedure ‘extends’ a in all #possible ways, under the condition that this sum remains smaller than #2. local b,g,o; if s >= 2 then return false; else g:=2/(2-s); o:=List(a,x->g/x); Print("Orbit sizes:",o," |G|:",g,"\n"); b:=a[Length(a)]; while extend(Concatenation(a,[b]),s+(1-1/b)) do b:=b+1; od; return true; fi; end; Explain why the function call extend([2,2],1) results in an endless loop. On the other hand, when called as extend([2,3],1/2+2/3) , the function does halt, and prints Orbit sizes:[ 6/5, 4/5 ] |G|:12/5 Orbit sizes:[ 6, 4, 4 ] |G|:12 Orbit sizes:[ 12, 8, 6 ] |G|:24 Orbit sizes:[ 30, 20, 12 ] |G|:60 3.4. THE FINITE SUBGROUPS OF O(3,R) 45 Figure 3.4: The orbits of Won the points on rotation axes. Compare this with Table 3.1. Exercise 3.4.4. In Figure 3.4 we see a cube with vertices on the unit sphere. Its group of proper symmetries contains non-trivial rotation axes through the points drawn in that picture; they form the set Sin the proof of Theorem 3.4.2. The points in the diﬀerent orbits are given diﬀerent shades of grey. Make similar pictures for the group types Cn,D/prime nandT. Describe the symmetry axes for the icosahedron. Now let us try to include improper rotations. Let Gbe a ﬁnite subgroup of O(3,R) containing improper rotations. Then H:=G∩SO(3,R) has index 2 in G. IfGcontains the inversion i, then we have a disjoint union G=H∪iH. Note that this is isomorphic to H×C2as an abstract group. Now assume that Gdoes not contain i. Then the set G/prime:=H∪i(G\H) is a subgroup of SO(3 ,R) in whichHhas index 2, and one can recover Gby G=H∪i(G/prime\H). Conversely, if G/prime,H⊆SO(3,R) are ﬁnite subgroups, and Hhas index 2 in G/prime, then this equation deﬁnes a ﬁnite subgroup of O(3 ,R), which is isomorphic to G/primeas an abstract group, but not equivalent to it. This enables us to classify all ﬁnite subgroups of O(3 ,R). Theorem 3.4.5. Any ﬁnite subgroup of O(3,R)is conjugate to one of the following: Cn,Cn∪iCn,Cn∪i(C2n\Cn), forn≥1, D/prime n,D/prime n∪iD/prime n,Dn∪i(D/prime 2n\D/prime n),Cn∪i(D/prime n\Cn), forn≥2, T,W,P,T ∪iT,W ∪iW,P ∪iP,T∪i(W\T). 46 CHAPTER 3. SYMMETRY GROUPS IN EUCLIDEAN SPACE We do not give all details of the proof, but restrict ourselves to the following remarks. 1. Note that D/prime 1andC2describe the same class; for this reason the second list starts with n= 2. 2. The last group in the list is made possible by the fact that a group of typeWhas a subgroup of type Tof index 2. This can be seen as follows: consider the group of proper symmetries of a cube with center 0, and look at the lines between anti-podal vertices of the cube; they are axes of rotations of order 3 leaving the cube invariant; these proper rotations form the symmetry group of a regular tetrahedron with center 0 and points on those axes. In applications, one wants to determine the (ﬁnite) symmetry group of a given object. The following procedure, based on Table 3.1 and the list in The- orem 3.4.5, can then be helpful. 1. List all rotation axes and their orders. They determine the subgroup H ofGof all proper rotations. 2. Check whether Gcontains any improper rotations. If not, then G=H. 3. Suppose that Gcontains improper rotations. Check whether i∈G. If so, thenG=H∪iH. 4. Suppose that Gcontains improper rotations, but not the inversion. Then, according to the list of Theorem 3.4.5, there is only one possibility for G, unlessHis of typeCn. 5. Suppose that H=Cnand thatGcontains improper rotations, but not the inversion. Then if Gis cyclic, it is of type Cn∪i(C2n\Cn). Otherwise, it is of type Cn∪i(D/prime n\Cn). Exercise 3.4.6. Consider the molecule depicted in Figure 3.5. Identify the class of its symmetry group in the list of Theorem 3.4.5. Exercise 3.4.7. Consider the full symmetry group of the tetrahedron, including its improper rotations. Which class in the list of Theorem 3.4.5 does it belong to? Exercise 3.4.8. Determine the symmetry group of the CH 4molecule depicted in Figure 3.1. Exercise 3.4.9. Consider the ethane molecule in staggered conﬁguration as depicted in Figure 3.6. What is the type of its symmetry group? Exercise 3.4.10. Determine the symmetry group of the N 4P4-molecule, de- picted in Figure 3.7. 3.4. THE FINITE SUBGROUPS OF O(3,R) 47 HH HN Figure 3.5: The NH 3- molecule. H HCH CH HH Figure 3.6: The ethane molecule in staggered conﬁguration. 48 CHAPTER 3. SYMMETRY GROUPS IN EUCLIDEAN SPACE NN NN PPP P Figure 3.7: The ethane molecule in staggered conﬁguration. Most of the above examples are taken from [2]. Finally, the question arises which of the ﬁnite subgroups of O(3 ,R) are crys- tallographic point groups. Theorem 3.4.11. LetGbe a crystallographic point group in 3dimensions. ThenGis conjugate to one of the following groups: Cm,Cm∪iCm m= 1,2,3,4,6 D/prime m,D/prime m∪iD/prime m m= 2,3,4,6 Cm∪i(C2m\Cm) m= 1,2,3 D/prime m∪i(D/prime 2m\D/prime m) m= 2,3 Cm∪i(D/prime m\Cm) m= 2,3,4,6 T,W,T ∪iT,W ∪iW,T ∪i(W\T) Like in the two-dimensional case, a crystallographic point group may leave several lattices invariant. Indeed, there are 230 mutually non-equivalent crys- tallographic groups in 3 dimensions, while there are only 32 crystallographic point groups, see [2]. The proof runs like that of Theorem 3.3.4. For the sake of compatibility with other texts on crystallographic groups, especially texts on the applications in physics and chemistry, we include table 3.4. It is organized as follows: in the ﬁrst column, one ﬁnds lattices classiﬁed up to their symmetry groups. The second column contains the groups leaving that lattice invariant (but no lattice with a smaller symmetry group), in our notation. The last row for each lattice is the full symmetry group of that lattice. There is only one exception to this rule: a trigonal lattice is really hexagonal. The third and fourth column contain two commonly used sets of symbols for them. 3.4. THE FINITE SUBGROUPS OF O(3,R) 49 Type of Lattice Our notation Schoenﬂies International Triclinic C1 C1 1 C1∪iC1Ci=S2 1 Monoclinic C2 C2 2 C1∪i(C2\C1)Cs=C1h m C2∪iC2C2h 2/m Orthorhombic D/prime 2 D2=V 2 2 2 C2∪i(D/prime 2\C2)C2v m m 2 D/prime 2∪iD/prime 2D2h=Vh m m m Trigonal C3 C3 3 D/prime 3 D3 3 2 C3∪iC3C3i=S6 3 C3∪i(D/prime 3\C3)C3v 3 m D/prime 3∪iD/prime 3D3d 3 m Tetragonal C4 C4 4 D/prime 4 D4 4 2 2 C4∪iC4C4h 4/m C2∪i(C4\C2)S4 4 C4∪i(D/prime 4\C4)C4v 4 m m D/prime 4∪i(D/prime 4\D/prime 2)D2d=Vd 4 2 m D/prime 4∪iD/prime 4D4h 4/m m m Hexagonal C6 C6 6 D/prime 6 D6 6 2 2 C6∪iC6C6h 6/m C3∪i(C6\C3)C3h 6 C6∪i(D/prime 6\C6)C6v 6 m m D3∪i(D/prime 6\D/prime 3)D3h 6 m 2 D/prime 6∪iD/prime 6D6h 6/m m m Isometric T T 2 3 W O 4 3 2 T∪iT T h m 3 T∪i(W\T)Td 4 3 m W∪iW W h m 3 m Table 3.2: A Dictionary of Crystallographic Group Names. 50 CHAPTER 3. SYMMETRY GROUPS IN EUCLIDEAN SPACE Chapter 4 Representation Theory 4.1 Linear representations of groups An abstract group can be represented in various ways. In Chapter 2, we en- countered permutation representations; in the present chapter we will start the theory of linear representations. In this chapter, Gdenotes a group and Ka ﬁeld of characteristic zero. All vector spaces will be over K. The notion We introduce the concept of a linear representation of G. Deﬁnition 4.1.1. A(linear) representation of a groupGon a vector space V is a homomomorphism ρ:G→GL(V). Ifρis injective, it is called faithful . After a choice of basis of V, we can view a representation as a homomorphism G→GL(n,K), where GL(n,K) is the group of invertible matrices with entries in K; such a map is called matrix representation of G. Ifρ:G→GL(V) is a representation of GonV, then we often write gv instead ofρ(g)v, if no confusion is possible. Also, Gis said to act linearly onV, andVis called aG-module . The dimension of the representation is by deﬁnition the dimension of V. Note that ( g,v)/mapsto→ρ(g)vis indeed an action of GonVin the sense of Deﬁnition 2.4.1. Exercise 4.1.2. To experience the use of matrix representations of groups, try to guess which orthogonal linear map is the product ρxρyρz. Hereρxstands for rotation over πabout thex-axis, and similarly for yandz. To check your guess, represent the rotations by matrices and compute the product. 51 52 CHAPTER 4. REPRESENTATION THEORY Example 4.1.3. LetGbe the cyclic group Cnof ordernwith generator c. It has one-dimensional complex representations rl(l∈Z) deﬁned by rl:Cn→GL(1,C), ck/mapsto→(ekl2πi/n). Note thatrlis faithful if and only if nandlare co-prime. Example 4.1.4. Recall (2.3) the permutation matrix presentation of the sym- metric group S5given in Chapter 2: to each permutation π∈S5, we associate the 5 ×5 matrixMπwithMπei=eπ(i). This is a matrix representation of S5. The same thing works for S3. The permutation (1 ,2) can be represented by the matrix M(1,2)= 0 1 0 1 0 0 0 0 1 , and (1,2,3) can be represented by M(1,2,3)= 0 0 1 1 0 0 0 1 0 . We know that (1 ,2)(1,2,3) = (2,3) and this can be seen from the matrices as well: M(1,2)M(1,2,3)= 1 0 0 0 0 1 0 1 0 . Exercise 4.1.5. Prove that the correspondence given above between S3and GL(n,K) is a group homomorphism, that is, a matrix representation of S3. Example 4.1.6. Recall from 3.3 the dihedral group Dnof order 2n. It contains the cyclic group Cnof ordernwith generator cas a subgroup of index 2 and has a reﬂection a(of order 2) with aca=c−1. The group Dnhas a two-dimensional real representation r:Dn→GL(2,R) determined by ck/mapsto→/parenleftbigg cos(2kπ/n )−sin(2kπ/n ) sin(2kπ/n ) cos(2kπ/n )/parenrightbigg , a/mapsto→/parenleftbigg 0 1 1 0/parenrightbigg . Example 4.1.7. An important example of a linear representation of a ﬁnite groupGis the regular representation . LetG={g1,...,g n}and takeVto be the vector space of dimension nand basis {eg1,...,e gn}. Deﬁneρ:G→GL(V) byρ(gi)(egj) =egigj, extended linearly to all of V. Example 4.1.8. The above examples can be summarized in a more general construction. Let π:G→Snbe a permutation representation. Then πgives rise to a linear representation Mπ:G→GL(V), whereVis a complex vector space with (formal) basis ei(i= 1,...,n ), in the following way: Mπ(g)ej=eπ(g)j(g∈G;j∈ {1,...,n }). 4.1. LINEAR REPRESENTATIONS OF GROUPS 53 The regular representation of Example 4.1.7 is Mπwithπ:G→Sym(G) the left regular permutation representation of G. The matrix representation of Example 4.1.4 is Mπ, whereπis the natural permutation representation of Sn(actually, for n= 5,3, respectively). Example 4.1.9. An important class of constructions of new representations from known ones arises from function space constructions. Let r:G→GL(V) be a linear representation of GonV. Consider the vector space F(V), of all mapsV→Kwith pointwise addition ( f+g)x=f(x) +g(x) and pointwise scalar multiplication ( λf)x=λf(x) (whereλ∈K,x∈V,f,f/prime:V→K). Then rinduces the following representation r∗:G→GL(F(V)): r∗(g)(f)x=f(r(g)−1x) (x∈V,f∈F(V),g∈G). As done before, we usually unwind the heavy notation by dropping rwherever possible, and write ( gf)(x) =f(g−1x) instead. Note thatF(V) has the dual space V∗ofK-linear functions as an invariant subspace. Exercise 4.1.10. LetR[x]≤3be the linear space of all polynomials in xof degree ≤3. Fora∈Rwe consider the R-linear representation ρa:R→GL(R[x]≤3) deﬁned asρa(r)p=p(x−ar) (r∈R,p∈R[x]≤3). Find a corresponding matrix representation. Equivalence and Subrepresentations Just as with groups, we are mainly interested in representations up to isomor- phism. Deﬁnition 4.1.11. Letrandsbe representations of a group Gon vector spacesVandW, respectively. A linear map T:V→Wis said to intertwiner andsifs(g)T=Tr(g) for allg∈G. In this case, Tis called a homomorphism ofG-modules . The space of all homomorphisms of G-modules is denoted by Hom G(V,W ). The representations randsare called equivalent if there exists a linear isomorphism T:V→Wintertwining them. In this case, Tis an isomorphism ofG-modules . It is straightforward to prove that this really deﬁnes an equivalence relation ∼on representations. Exercise 4.1.12. Consider the linear representations rlofCndeﬁned in Exam- ple 4.1.3. For which pairs ( l,m) are the representations rlandrmequivalent? Exercise 4.1.13. Consider the linear representations ρaof Exercise 4.1.10. Show thatρa∼ρbif and only if ab/negationslash= 0. In Example 4.1.9, we saw that V∗is a invariant under the action of Gon F(V). This gives rise to the following deﬁnition. 54 CHAPTER 4. REPRESENTATION THEORY Deﬁnition 4.1.14. Letrbe a representation of a group GonV. A subspace Wisinvariant ifr(g)W⊂Wfor allg∈G. In this case, r|W:G→GL(W) given byr|W(g) =r(g)|W, is called a subrepresentation ofr, andWis called a sub-G-module ofV. IfGis clear from the context, we leave it out and call Wa submodule of V. Example 4.1.15. Suppose that T:V→Wis aG-module homomorphism ofGonVandW. Then ker Tis aG-invariant subspace of Vand imTan invariant subspace of W. Example 4.1.16. Consider the two-dimensional matrix representation ρ:R→ GL(R2) deﬁned by Mt=/parenleftbigg 1t 0 1/parenrightbigg . Then it is not hard to see that the 1 dimensional subspace spanned by (1 ,0) is an invariant subspace. Example 4.1.17. LetVbe a ﬁnite-dimensional vector space over K, and let ρ:G→GL(V) be a representation of a group g. LetP(V) be the set of all poly- nomial functions on V, i.e., the set of all elements of F(V) (see Example 4.1.9) that can be written as a K-linear combination of products of elements of V∗. The subspace P(V) ofF(V) is invariant under G, and so are its homogeneous components P(V)d, deﬁned by P(V)d={f∈P(V)|f(λx) =λdf(x) for allx∈Vandλ∈K}. We ﬁnd that P(V)0is the one-dimensional space of constant functions, and P(V)1=V∗has dimension n:= dimV. Choosing a basis x1,...,x nofV∗, we ﬁnd thatP(V)dis precisely the K-linear span of monomials xa1 1· · ·xannwhere (a1,...,a n)∈Nnsatisﬁesa1+...+an=d. This implies dim P(V)d=/parenleftbign+d−1 d/parenrightbig ; prove this! To make all this more explicit, suppose that n= 2, and consider an element A∈GL(V) whose matrix is/parenleftbigg a b c d/parenrightbigg with respect to the basis e1,e2dual tox1,x2. We compute the matrix of AonP(V)1with respect to x1,x2. We have A−1= (ad−bc)−1/parenleftbigg d−b −c a/parenrightbigg , so Ax1(e1) =x1(A−1e1) = (ad−bc)−1x1(de1−ce2) = (ad−bc)−1d Ax1(e2) =x1(A−1e2) = (ad−bc)−1x1(−be1+ae2) =−(ad−bc)−1b Ax2(e1) =x2(A−1e1) = (ad−bc)−1x2(de1−ce2) =−(ad−bc)−1c Ax2(e2) =x2(A−1e2) = (ad−bc)−1x2(−be1+ae2) = (ad−bc)−1a 4.1. LINEAR REPRESENTATIONS OF GROUPS 55 from which we conclude Ax1= (ad−bc)−1(dx1−bx2) andAx2= (ad− bc)−1(−cx1+ax2). HenceAhas matrix (ad−bc)−1/parenleftbigg d−c −b a/parenrightbigg on the basis x1,x2ofP(V)1. Note that this is the tranposed inverse of A. Finally, let us compute the matrix of AonP(V)2with respect to the basis x2 1,x1x2,x2 2. Now Ax2 1 = (Ax1)2= ( ad−bc)−2(dx1−bx2)2 = (ad−bc)−2(d2x2 1−2bdx 1x2+b2x2 2) A(x1x2) = (Ax1)(Ax2) = ( ad−bc)−2(dx1−bx2)(−cx1+ax2) = (ad−bc)−2(−cdx2 1+ (ad+bc)x1x2−abx2 2) Ax2 2 = (Ax2)2= ( ad−bc)−2(−cx1+ax2)2 = (ad−bc)−2(c2x2 1−2acx 1x2+a2x2 2), so the matrix sought for is (ad−bc)−2 d2−cd c2 −2bd ad +bc−2ac b2−ab a2 . In conclusion, we have found a matrix representation of GL(2 ,K) onK3 sendingAto the 3 ×3 matrix just computed. Can you determine its kernel? Exercise 4.1.18. Find the invariant subspaces of the matrix representation ρ:R→GL(R2) deﬁned by Mt=/parenleftbigg etet−1 0 1/parenrightbigg . Lemma 4.1.19. Suppose that Gis ﬁnite, and that ρis a representation of Gon a vector space V. LetUbe a sub-G-module of V. Then there exists a sub-G-moduleWofVsuch thatV=U⊕W. Proof. LetW/primebe any vector space complement of UinV, and letπ/primebe the projection of VontoUwith kernel W/prime. Deﬁne a new map π:=1 |G|/summationdisplay g∈Gρ(g)π/primeρ(g−1) onV. We claim that πis a projection commuting with all ρ(h), h∈G. To see that it does indeed commute with ρ(h), consider πρ(h) =1 |G|/summationdisplay g∈Gρ(g)π/primeρ(g−1h); 56 CHAPTER 4. REPRESENTATION THEORY by the dummy transformation g/mapsto→hgthis equals 1 |G|/summationdisplay g∈Gρ(hg)π/primeρ((hg)−1h) =1 |G|/summationdisplay g∈Gρ(h)ρ(g)π/primeρ(g−1h−1h) =ρ(h)1 |G|/summationdisplay g∈Gρ(g)π/primeρ(g−1) =ρ(h)π, as claimed. To verify that πis a projection, we ﬁrst note that ππ/prime=π/prime; indeed, both are zero on W/primeand the identity on U. Compute π2=π1 |G|/summationdisplay g∈Gρ(g)π/primeρ(g−1) =1 |G|/summationdisplay g∈Gρ(g)ππ/primeρ(g−1) =1 |G|/summationdisplay g∈Gρ(g)π/primeρ(g−1) =π, where the second equality follows from the fact that πcommutes with each ρ(g). This concludes the proof that πis a projection. This implies that V= im(π)⊕ker(π), where both im( π) and ker(π) are sub-G-modules of Vby Exercise 4.1.15. Now note that im( π) = im(π/prime) =U, so that we may take W= ker(π) to conclude the proof of the lemma. Remark 4.1.20. Letρbe a linear representation of Gon a ﬁnite-dimensional complex vector space V. Suppose that Vis endowed with a Hermitian inner product (.,.) satisfying (ρ(g)(v),ρ(g)(w)) = (v,w)∀v,w∈V, g∈G. Now ifUis aG-invariant subspace of V, then the orthogonal complement U⊥ ofUinVis an invariant subspace of Vcomplementary to U. It is important note that, if Gis ﬁnite, one can always build an invariant Hermitian inner product ( .|.) from an arbitrary one ( .,.) as was done in Exercise 3.2.22 for positive deﬁnite inner products: (v|w) =/summationdisplay g∈G(ρ(g)(v),ρ(g)(w)). Thus, the above argument gives an alternative proof of the lemma in the case whereVis a ﬁnite-dimensional complex vector space. 4.1. LINEAR REPRESENTATIONS OF GROUPS 57 Exercise 4.1.21. Show that in Exercises 4.1.10 and 4.1.18, there are invariant subspaces that have no invariant complements. Deﬁnition 4.1.22. AG-moduleVis called reducible if it has sub- G-modules other than 0 and V; it is called irreducible otherwise. If Vis the direct sum of irreducible sub- G-modules, then Vis called completely reducible . This terminology is also used for the corresponding representations. By def- inition, each irreducible representation is completely reducible. Exercise 4.1.18 gives an example of a reducible completely reducible representation. In Exer- cise 4.1.27 we shall see that not every reducible representation is completely reducible. Exercise 4.1.23. Consider the cyclic group Cnof ordernwith generator c. It has a two-dimensional real representation deﬁned by r:Cn→GL(2,R), ck/mapsto→/parenleftbigg cos(2kπ/n )−sin(2kπ/n ) sin(2kπ/n ) cos(2kπ/n )/parenrightbigg . Show that it is irreducible for n>2. If we replace RbyC, we obtain a complex representation. Show that this representation is completely reducible. Exercise 4.1.24. Show that any irreducible representation of an Abelian group Gover an algebraically closed ﬁeld is one-dimensional. For ﬁnite groups, irreducible representations are the building blocks of all representations. Theorem 4.1.25. For a ﬁnite group G, every ﬁnite-dimensional G-module is completely reducible. Proof. Proceed by induction. Suppose that the stament holds for all representa- tions of dimension smaller than n, and letVbe ann-dimensional representation. IfVis irreducible, then we are done. Otherwise there exists a sub- G-moduleU with 0 /subsetnoteqlU/subsetnoteqlV. By Lemma 4.1.19, Uhas a invariant complementary sub- G- moduleW. As both dim Uand dimWare smaller than n, we may apply the induction hypothesis, and ﬁnd that Uis the direct sum of irreducible sub- G- modulesU1,...,U kofUandWis the direct sum of irreducible sub- G-modules W1,...,W l. But then V=U1⊕...⊕Uk⊕W1⊕...⊕Wl. Remark 4.1.26. The representation of Rin Example 4.1.16 is reducible, but it is not the direct sum of irreducibles. Exercise 4.1.21 also gives examples of this phenomenon. This shows that we need at least some condition on G(like ﬁniteness) for Theorem 4.1.25 to hold. 58 CHAPTER 4. REPRESENTATION THEORY Exercise 4.1.27. Letpbe a prime, and consider the cyclic group Cpwith generatorc. The map Cp→GL2(Z/pZ) deﬁned by ck/mapsto→/parenleftbigg 1k 0 1/parenrightbigg is a matrix representation of CpoverZ/pZ. The one-dimensional subspace spanned by (1 ,0)Tis invariant, but it has no invariant complement. In general, representation theory of a group Gover a ﬁeld of characteristic ptends to be much harder if pdivides |G|than otherwise. An important result about irreducible representations known as Schur’s Lemma. Lemma 4.1.28 (Schur’s Lemma) .Suppose that Kis algebraically closed, and letVbe an irreducible ﬁnite-dimensional G-module over K. Then any homo- morphism of G-modulesV→Vis a scalar. In other words, a matrix commuting with all matrices of an irreducible ma- trix representation is a multiple of the identity matrix. Proof. LetT:V→Vbe a homomorphism of G-modules, and let λ∈Kbe an eigenvalue of T. ThenT−λIis also a homomorphism of G-modules. By Example 4.1.15, its kernel is a sub- G-module of V, and it is non-zero as λis an eigenvalue of T. AsVis irreducible, ker( T−λI) equalsV. Exercise 4.1.29. Compute the set of all 2 ×2 matrices commuting with Cn in one of the 2-dimensional complex representations given in Example 4.1.6. Conclude that the representation is reducible. Exercise 4.1.30. Prove that the following converse of Schur’s lemma holds: Letr:G→GL(V) be a completely reducible representation of a group G. For rto be irreducible, it suﬃces that any linear map A:V→Vthat commutes with allr(g),g∈G, is a scalar. Exercise 4.1.31. SupposeVis a vector space of dimension nwith a subspace Wof dimension kwhere 1< k < n , and letGbe the group of elements in GL(V) that leave Winvariant. Then the embedding r:G→GL(V) is clearly a faithful reducible representation of GonV. 1. Establish that ris not completely reducible. 2. Show that the only G-module homomorphisms A:V→Vare scalar multiplications. 3. Conclude that the converse of Schur’s lemma does not necessarily hold if Gis not completely reducible, cf. Exercise 4.1.30. 4.2. DECOMPOSING DISPLACEMENTS 59 3 2 1 Figure 4.1: A system with 3 point masses and springs. 3 2 1 Figure 4.2: A displacement. 4.2 Decomposing Displacements Consider the system of Figure 4.1. It depicts three point masses (labelled with P={1,2,3}) at positions ( −1 2,−1 2√ 3),(−1 2,1 2√ 3),(0,1) in two-dimensional spaceE=R2, and these masses are connected with springs of equal lengths. The system can be thought of as vibrating (in two dimensions) about its equi- librium state, in which the point masses form an equilateral triangle. Adisplacement of the system is a function f:P→R2which attaches to each point mass a velocity vector. Displacements can be depicted as in Figure 4.2. The set of all displacements is an R-linear space, which we denote by Γ. Its dimension is clearly 3 ×2 = 6, i.e., there are 6 degrees of freedom. We would like to decompose Γ into subspaces in a manner that is consis- tent with the symmetry group D3that acts on the system. To be precise, we 60 CHAPTER 4. REPRESENTATION THEORY 3 2 11 23 (1,2,3) Figure 4.3: The action of (1 ,2,3) on a displacement. have a permutation representation π:D3→Sym(P). The map is in fact an isomorphism, and we shall identify D3with Sym(P) =S3. For example, the permutation (1 ,2) corresponds to reﬂection in the line bisecting the edge 12 and meeting point 3. The groupD3also acts linearly on E. The corresponding linear representa- tionD3→GL(E) is given by (1,2)/mapsto→/parenleftbigg −1 0 0 1/parenrightbigg and (1,2,3)/mapsto→1 2/parenleftbigg −1√ 3 −√ 3−1/parenrightbigg . The space Γ becomes a D3-module by setting (gf)(i) =gf(g−1i), i∈P,f∈Γ,g∈G. Our goal can now be formalized as follows: try to write Γ as a direct sum of irreducible submodules. Theorem 4.1.25 tells us that this is possible. Here, we shall try to solve this problem ‘by hand’, without using too much representation theory. In this way, the reader will appreciate the elegant char- acter arguments which are to be given in Example 5.2.23. As a ﬁrst step, we can write each displacement in a unique way as a purely ‘radial’ displacement and a purely ‘tangental’ displacement (see Figure 4.4), and the spaces Γ radand Γ tanof radial and tangental displacements, respectively, are three-dimensional submodules of Γ. The submodule Γ radcontains a one-dimensional submodule Γ1 rad, consisting of displacements with the ‘same’ velocity vector in all three directions, such as the one depicted in Figure 4.5. To be precise: this space consists of all radial displacements ffor which (1 ,i)(f(1)) =f(i). 4.2. DECOMPOSING DISPLACEMENTS 61 3 2 13 2 1 Figure 4.4: A radial (left) and a tangental (right) displacement. 3 2 1 Figure 4.5: A special radial displacement. 62 CHAPTER 4. REPRESENTATION THEORY 3 2 13 2 1 Figure 4.6: Two more special radial displacements. The space Γ1 radmust have an invariant complement in Γ rad, and looking at the pictures we ﬁnd that the space Γ2 radof radial displacements spanned by the two in Figure 4.6 is also invariant. Note that this space consists of all radial displacements for which (1 ,3)f(1) + (1,2)f(1) +f(1) = 0. One may wonder if the two-dimensional module Γ2 radcan be decomposed further, but this is not the case, as we shall see in Example 5.2.23. The rep- resentation Γ tancan be decomposed in a similar fashion. However, instead of going into details rightaway, we will ﬁrst treat more representation theory, and come back to this application in Example 5.2.23. Notes This section is based on a similar (but more diﬃcult) example in [10]. If your interest is aroused by this example, you are encouraged to read the pages in that book concerned with the displacemements of a tetrahedral molecule. Chapter 5 Character Tables In this chapter, Gdenotes a ﬁnite group and Kan algebraically closed ﬁeld of characteristic 0, e.g. C. 5.1 Characters An extremely useful notion regarding matrices is the trace . The trace of a matrix A= (aij)n×nis tr(A) =a11+· · ·+ann. Exercise 5.1.1. Prove that for two matrices AandB, one has tr( AB) = tr(BA). Exercise 5.1.2. Show that in the case of a permutation representation π:G→ Sn, the number tr( Mπ(g)) (see Example 4.1.8) is equal to the number of ﬁxed points of the permutation π(g). Deﬁnition 5.1.3. Letrbe a representation of a group Gon a ﬁnite-dimensional vector space V. The character ofris the function χrdeﬁned onGby χr(g) = tr(r(g)) (g∈G). Ifris irreducible, then the character χris also called irreducible. Thus, the character of rongis the sum of the eigenvalues of r(g) counted with multiplicities. Remark 5.1.4. IfGhas ﬁnite order n, andρ:G→GL(V) is a ﬁnite- dimensional representation, then ρ(g)n=Ifor allg∈G. Hence, over the algebraic closure of K, eachρ(g) is diagonalizable, and all its eigenvalues are n-th roots of unity. If K=C, then all eigenvalues have absolute value 1. We next give some properties of complex characters. Lemma 5.1.5. For any complex representation r:G→GL(V)of a ﬁnite group, we have: 63 64 CHAPTER 5. CHARACTER TABLES 1.χr(e) = tr(I) = dimV; 2.χr(g−1) =χr(g); 3.χr(ghg−1) =χr(h)for allg,h∈G. Proof. The ﬁrst part is easy. As for the second part, we have seen that all the eigenvalues λ1,...,λ nofr(g) have absolute value 1. Therefore: χr(g−1) = tr(r(g−1)) = tr(r(g)−1) =/summationdisplay λ−1 i=/summationdisplay λi=tr(r(g)) =χr(g). The last part follows directly from Exercise 5.1.1. The last part states that characters are constant on conjugacy classes of G. Example 5.1.6. TakeG= GL(2,K). The trace of an arbitrary element A=/parenleftbigg a b c d/parenrightbigg isa+d. This is the character of the identity, which in this case is a representation (often called the natural representation ). But, as we have seen in Example 4.1.17, there are also linear actions on P(V)1and onP(V)2. We read oﬀ the corresponding characters as traces of the matrices found in 4.1.17. They are ( a+d)/(ad−bc) onP(V)1, and (a2+ad+ bc+d2)/(ad−bc)2onP(V)2. Lemma 5.1.7. LetV1,...,V sbe representations of Gwith characters respec- tively,χ1,...,χ s. Then the character of the representation V1⊕ · · · ⊕Vsis the sum of characters χ1+· · ·+χs. 5.2 Orthogonality of irreducible characters The following theorem is very important in the representation theory of ﬁnite groups. Theorem 5.2.1. LetGbe a ﬁnite group. The irreducible complex characters of Gform an orthonormal basis for the space of class functions on G, with respect to the Hermitian inner product (· | ·). We need a deﬁnition to make clear what this theorem states. Deﬁnition 5.2.2. Aclass function on a groupGis a function f:G→Cthat is constant on conjugacy classes, or, equivalently, that is ﬁxed by the linear action ofGonCGdeﬁned by (g·f)(h) :=f(ghg−1). The space of all class functions is denoted by H. The Hermitian inner product is given by (χ|ψ) =1 |G|/summationdisplay g∈Gχ(g)ψ(g) forχ,ψ∈ H. 5.2. ORTHOGONALITY OF IRREDUCIBLE CHARACTERS 65 We have noticed earlier that characters are class functions. We divide the proof of Theorem 5.2.1 into two parts: ﬁrst, we will show the orthonormality of the irreducible characters, and then their being a full basis of the space of class functions. For our ﬁrst purpose, we introduce the tensor product of two representations. Deﬁnition 5.2.3. LetVandWbe vector spaces over K. ThenV⊗W, called thetensor product of VandW, is the quotient of the free vector space over K with basis {v⊗w|v∈V, w ∈W}by its subspace spanned by all elements of the following forms (for v,v1,v2∈V,w,w 1,w2∈Wandλ∈K): (v1+v2)⊗w−v1⊗w−v2⊗w, v⊗(w1+w2)−v⊗w1−v⊗w2, (λv)⊗w−λ(v⊗w),and v⊗(λw)−λ(v⊗w). Exercise 5.2.4. Show that if v1,...,v mandw1,...,w nare bases of VandW, respectively, then {vi⊗wj|i= 1,...,m, j = 1,...,n } is a basis of V⊗W. It is not hard to see that if U,V, andWare vector spaces over K, then the spaces ( U⊗V)⊗WandU⊗(V⊗W) are canonically isomorphic. We will identify them, leave out parentheses, and write Td(V) for thed-fold tensor product of Vwith itself. The direct sum of all Td(V) ford∈N(including 0, for whichTd(V) is isomorphic to K) forms an associative algebra with respect to the operator ⊗; this algebra is called the tensor algebra on Vand denoted byT(V). Deﬁnition 5.2.5. LetVbe a vector space over Kand letdbe a natural number. Then the quotient of Td(V) by the subspace spanned by the set {v1⊗v2⊗...⊗vd−vπ(1)⊗vπ(2)⊗...⊗vπ(d)|π∈Sd} is denoted by Sd(V), and called the d-th symmetric power ofV. Just likeT(V), the direct sum of all Sd(V) ford∈Nforms an associative algebra, called the symmetric algebra on Vand denoted by S(V). This algebra is commutative, and to stress this fact we usually leave out the operator ⊗from expressions in S(V). Exercise 5.2.6. Show that if v1,...,v nis a basis of V, then the set {vm1 1· · ·vmn n|m1,...,m n∈N, m 1+...+mn=d} is a basis for Sd(V). Prove that Sd(V∗) is canonically isomorphic to P(V)d(see Example 4.1.17). 66 CHAPTER 5. CHARACTER TABLES Deﬁnition 5.2.7. Letr:G→GL(V) ands:G→GL(W) be linear represen- tations ofG. Then we deﬁne a linear representation r⊗s:G→GL(V⊗W) of Gby (r⊗s)(g)(v⊗w) := (r(g)v)⊗(s(g)w), extended linearly to all of V⊗W. This representation is denoted by r⊗sand called the tensor product of rands. Lemma 5.2.8. LetG,r,V,s, andWbe as in Deﬁnition 5.2.7. Then the char- acter ofr⊗ssatisﬁesχr⊗s=χrχs. Proof. Fix basesv1,...,v mandw1,...,w nofVandW, respectively. Consider, for a ﬁxedg∈G, the matrix A= (aij,kl) of (r⊗s)(g) with respect to the basis vi⊗wjofV⊗W; We have (r⊗s)(g)(vk⊗wl) =/summationdisplay i,jaij,klvi⊗wj. (5.1) On the other hand, by the deﬁnition of the tensor representation, the left-hand side equals (r(g)vk)⊗(s(g)wl) = (/summationdisplay irikvi)⊗(/summationdisplay jsjlwj) =/summationdisplay i,jriksjlvi⊗wj. (5.2) Comparing the right-hand sides of equations (5.1) and (5.2) we obtain aij,kl=riksjl, and therefore χs⊗r(g) =/summationdisplay i,jaij,ij=/summationdisplay i,jriisjj=χr(g)χs(g), concluding the proof. Exercise 5.2.9. LetVbe aG-module, and consider the linear isomorphism σ:V⊗V→V⊗Vdetermined by σ(x⊗y) =y⊗xforx,y∈V. Show that σ has precisely two eigenspaces, with eigenvalues 1 and −1, and that both spaces areG-invariant. Here Gacts onV⊗Vvia the tensor product representation of the linear representation of GonV(given by the G-module structure of V) with itself. Another important ingredient in our proof of Theorem 5.2.1 is the so-called dual representation. Deﬁnition 5.2.10. Letr:G→GL(V) be a representation of the group G. Then we write V∗for the dual of V(in the notation of Example 4.1.17, this is P(V)1), andr∗for the corresponding representation, given by (r∗(g)f)(v) :=f(r(g)−1v). 5.2. ORTHOGONALITY OF IRREDUCIBLE CHARACTERS 67 Exercise 5.2.11. Show thatχr∗=χr. One ﬁnal lemma before proceeding with the proof of Theorem 5.2.1. Lemma 5.2.12. Letr,V,s,W,G be as in Deﬁnition 5.2.7. Deﬁne a repre- sentationtofGinHom(V,W ), the space of all C-linear maps from VtoW by t(g)H:=s(g)Hr(g)−1, H ∈Hom(V,W ). This representation is equivalent to s⊗r∗. Proof. DeﬁneT:W⊗V∗→Hom(V,W ) by T(w⊗f)(v) =f(v)·w, v ∈V,f∈V∗,w∈W, and extend it linearly. It is a matter of elementary linear algebra to see that T is a linear isomorphism. We claim that it intertwines the representations tand s⊗r∗. The best way to see this is to do the computation yourself, so we leave it as an exercise for the reader. Exercise 5.2.13. Prove that the map Tdeﬁned in the proof of Lemma 5.2.12 is really an intertwining map. If we choose a basis ( vj)jofVwith dual basis ( vj)jofV∗, and a basis ( wi)i ofW, then the map Tsendswi⊗vjto the linear map V→Whaving the matrix Eij(the matrix with 1 on position ( i,j), and zeroes elsewhere) with respect to the bases (vj)jand (wi)i. Having completed all preparations, of which the tensor product is a very useful tool in its own right, we proceed with the ﬁrst part of the proof of this section’s main theorem. Proof of Theorem 5.2.1, ﬁrst part. Letr,V,s,W,G be as in Deﬁnition 5.2.7 and tas in Lemma 5.2.12, and moreover, assume that both representations rand sare irreducible. We wish to compute the inner product ( χs|χr). It can be written in a particularly nice form, using tensor products. Deﬁne the linear map P: Hom(V,W )→Hom(V,W ) by P=1 |G|/summationdisplay g∈Gt(g). The trace of Pequals 1 |G|/summationdisplay g∈Gχt(g) =1 |G|/summationdisplay g∈Gχs(g)χr∗(g) =1 |G|/summationdisplay g∈Gχs(g)χr(g)) = (χs|χr). 68 CHAPTER 5. CHARACTER TABLES Here we used Lemmas 5.2.12, 5.2.8 and Exercise 5.2.11. We claim that Pis a projection from Hom( V,W ) onto the space Hom G(V,W ). To show that it is indeed a projection, let H∈Hom(V,W ), and compute P2H=1 |G|/summationdisplay g1 |G|/summationdisplay hs(g)s(h)Hr(h−1)r(g−1) =1 |G|/summationdisplay g1 |G|/summationdisplay hs(gh)Hr((gh)−1) =1 |G|/summationdisplay g1 |G|/summationdisplay hs(h)Hr(h−1) =1 |G|/summationdisplay gPH =PH, where in third equality the dummy his replaced by g−1h. This proves that P is a projection. To see that im( P) is contained in Hom G(V,W ), letg0∈G, H∈Hom(V,W ), andv∈V, and compute (PH)(g0v) =1 |G|/summationdisplay gs(g)Hr(g−1g0)v =1 |G|/summationdisplay gs(g0g)Hr(g−1g−1 0g0)v =s(g0)1 |G|/summationdisplay gs(g)Hr(g−1)v =g0(PHv ), where in the second equality, the dummy gis replaced by g0g. Finally, if H∈ Hom G(V,W ), then PH=1 |G|/summationdisplay gs(g)Hr(g−1) =1 |G|/summationdisplay gs(g)s(g−1)H =H, so that im( P) = Hom G(V,W ). This proves the claim. We now distinguish two cases. First, suppose that randsare not equivalent. Then, by Schur’s lemma, the space Hom G(V,W ) is zero-dimensional, so that Pand, a fortiori , its trace are 0. In this case ( χs|χr) = 0. Alternatively, suppose that randsare equivalent. As we only wish to compute the inner product of their characters, we may assume that they are 5.2. ORTHOGONALITY OF IRREDUCIBLE CHARACTERS 69 equal:W=Vands=r. In this case, the image of Pis the linear span of the identity map IonV. Taking any basis of Hom( V,V) that starts with I, we see that the matrix of Pwith respect to that basis has only non-zero elements on the ﬁrst row. The only contribution to the trace of Pis therefore the element in the upper left corner. Since PI=I, that element is 1. Hence ( χr|χr) = 1. This proves that the characters are an orthonormal system in the space of class functions on G. Corollary 5.2.14. The number of non-isomorphic irreducible representations ofGis ﬁnite. Proof. The dimension of His the number of conjugacy classes of G. The irre- ducible characters form an orthonormal, whence independent, set in this ﬁnite- dimensional space by the above. Lemma 5.2.15. Letrbe a complex representation of Gon a space V, and write V=V1⊕ · · · ⊕Vt, where theViare irreducible submodules. Let sbe an irreducible representation of Gon a spaceW. Then the number of ifor whichViis isomorphic to Wis equal to(χr|χs). In particular, this number does not depend on the decomposition in irreducibles. Proof. Ifχiis the character of Vithen the character χrofVcan be written as χr=χ1+· · ·+χt. Hence (χr|χs) = (χ1|χs) +· · ·+ (χt|χs), and by the above ( χi|χs) is one ifVi∼=W, and zero otherwise. Corollary 5.2.16. Two representations of a ﬁnite group are equivalent if and only if they have the same character. Recall that the regular representation ρ:G→GL(V) of the group Gis deﬁned byρ(g)(eh) =egh, withVthe vector space of dimension |G|and basis {eg|g∈G}. Proposition 5.2.17. The character of the regular representation ρofGsatis- ﬁesρ(g) =δg,e|G|(g∈G). Proof. Sinceeg=g, the transformation ρ(e) ﬁxes each element gofG. There- fore, tr(ρ(e)) =n=|G|. Ifg/negationslash=e, theneh/negationslash=eghfor eachh∈G. This implies that tr(ρ(g)) = 0. Lemma 5.2.18. Every irreducible representation WofGis a sub-representation of the regular representation with multiplicity dimW. 70 CHAPTER 5. CHARACTER TABLES Proof. From Proposition 5.2.17 we have (χρ|χs) =1 |G|/summationdisplay h∈Gχρ(h)χs(h) =1 |G||G|χs(e) = dim(W), and the result follows from Lemma 5.2.15. Corollary 5.2.19. Letr1,r2,...,r sbe the irreducible representations of G. Then |G|=/summationdisplay n2 i whereniis the dimension of the representation ri. Proof. We saw in the lemma that the regular representation ρcan be written as a sum of all irreducible representations riofG. Moreover, each of these representations appears nitimes in the decomposition. So ρ=n1times/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright r1⊕...⊕r1⊕...⊕nstimes/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright rs⊕ · · · ⊕rs. Therefore |G|=χρ(e) =/summationtexts i=1niχri(e) =/summationtexts i=1n2 i. Suppose that we manage to construct some non-isomorphic irreducible rep- resentations of G. Then Corollary 5.2.19 can be used to decide whether we have already ﬁnd all. In order to ﬁnish the proof of Theorem 5.2.1, we still need to show that the irreducible characters span H. Lemma 5.2.20. Suppose that r:G→GL(V)is an irreducible n-dimensional complex representation of G, and letf∈ H. Then the linear map H:=1 |G|/summationdisplay g∈Gf(g)r(g) equals 1 n(f|χr)I. Proof. First of all, note that Hveriﬁesr(g)H=Hr(g) for allg∈G. Hence, as ris irreducible, Hmust be multiplication by a scalar λI. Its trace equals nλ. On the other hand, it equals 1 |G|/summationdisplay g∈Gf(g)χr(g) = (f|χr). Consequently, λ= (f|χr)/n. 5.2. ORTHOGONALITY OF IRREDUCIBLE CHARACTERS 71 Proof of Theorem 5.2.1, second part. In order to prove that the irreducible char- acters form a basis of H, it suﬃces to prove that the orthoplement of their linear span in His zero. Hence, let f∈ H be orthogonal to all irreducible characters. According to Lemma 5.2.20, the linear map H=1 |G|/summationdisplay g∈Gf(g)r(g) is zero for all irreducible representations r. But, writing an arbitrary represen- tation as a direct sum of irreducible ones, we see that this linear map is zero for anyrepresentation r. Let us apply this knowledge to the regular representation. Compute Heh=1 |G|/summationdisplay g∈Gf(g)r(g)eh=1 |G|/summationdisplay g∈Gf(g)egh. Thus, forHto be zero,fmust be identically zero. This concludes the proof. Theorem 5.2.21. The number of irreducible characters equals the number of conjugacy classes of G. Proof. The dimension of Hequals the number of conjugacy classes of G, and the irreducible characters form a basis of this linear space. Half of the following proposition is Exercise 4.1.24, and can be done without character theory. Proposition 5.2.22. The groupGis Abelian if and only if all of its irreducible representations are one dimensional. Proof. Denote bysthe number of irreducible characters of Gand denote their dimensions, as before by ni. IfGis Abelian, all conjugacy classes consist of a single element of G. Then by the theorem above, there are |G|irreducible representations of G. Hence by Corollary 5.2.19, all ni’s are equal to 1. This settles one implication. As for the converse, suppose that ni= 1 for alli. Then |G|=sand so there are|G|distinct conjugacy classes of G. Therefore, Gis Abelian. Example 5.2.23. Recall from Section 4.2 the D3-module Γ of displacements. Let us employ character theory to ﬁnd the decomposition of Γ into irreducible modules. First of all, note that D3has three conjugacy classes, with represen- tatives (1),(12) and (123). Hence D3has three distinct irreducible characters by Theorem 5.2.21. Let V=R3, and lets:D3→GL(V) be the linear rep- resentation corresponding to the permutation representation of D3on{1,2,3} (see Example 4.1.8). Then χs((1)) = 3, χs((1,2)) = 1,andχs((1,2,3)) = 0. 72 CHAPTER 5. CHARACTER TABLES We have (χs|χs) = (9 + 3 ∗1)/6 = 2, so that Vis reducible. Indeed, the line spanned by (1 ,1,1)∈Vis invariant. Let r1be the representation of D3on this line, and let r2be the representation of D3on an invariant complement. Then χr1((1)) = 1, χr1((1,2)) = 1,andχr1((1,2,3)) = 1 and χr2((1)) = 2, χr2((1,2)) = 0,andχr2((1,2,3)) = −1. Bothχr1andχr2have squared norm 1, so they are irreducible. The represen- tation ofD3onEis equivalent to r2, as can be read of from the traces of the matrices on page 4.2. An element of Γ can be extended in a unique way to a linear map V→E, so that we may identify Γ with the space Hom K(V,E) ofK-linear maps from Vto E. By Lemma 5.2.12, Hom K(V,E) is isomorphic to V∗⊗E; letr:D3→V∗⊗E be the corresponding representation. Lemma 5.2.8 shows that χr=χsχr2. Hence, χr((1)) = 6, χr((1,2)) = 0,andχr((1,2,3)) = 0. First, we can decompose χr= (χr1+χr2)χr2=χr1χr2+ (χr2)2=χr2+ (χr2)2. The character ( χr2)2has squared norm 3, so it is reducible. We have (χr1|(χr2)2) = 1 and also ( χr2|(χr2)2) = 1, so that ( χr2)2−χr1−χr2is the char- acter of a third irreducible representation, say r3, of dimension 1. Its character equals χr3((1)) = 1, χr3((12)) = −1,andχ(r3)(123) = 1, that is,χr3= sgn . By Theorem 5.2.21, the characters r1,sgn,r2exhaust all irreducible characters of D3. We conclude that the character χrofD3on Γ decomposes into χr1+ 2χr2+χr3. 5.3 Character tables Letχ1,χ2,...,χ sbe the characters of the distinct irreducible representations of G. LetC1,C2,...,C sbe the distinct conjugacy classes of G, with representatives g1,...,g s, respectively. Construct a table with rows labelled by χ1,χ2,...,χ sand columns labelled byg1,g2,...,g s; under each giwe write the size of Ci. The (i,j)-entry of the table isχi(gj). It is common to take C1={e}={g1}, so that the ﬁrst column contains the dimensions of the respective characters, and also to take χ1to be the trivial representation, so that the ﬁrst row contains only ones. This table is called the character table ofG. Example 5.3.1. Let us check the character table of S3. First notice that there are three conjugacy classes: C1=e,C2={(12),(13),(23)}consisting of all transpositions and C3={(123),(132)}consisting of all 3-cycles. By Theorem 5.2.21, we know that there are three irreducible representations. By convention, χ1is the trivial character. For the second one we take the sign representation. Recall this is the one that assigns to each cycle its sign 5.3. CHARACTER TABLES 73 (2-cycles are odd, 3-cycles are even). Since it is a 1-dimensional representation, it is equal to its character. Finally for the third representation, we proceed as follows. The permutation representation πofS3onC3has an invariant subspace Uspanned by (1 ,1,1). Its orthogonal complement is also an invariant subspace, say V; it consists of all (z1,z2,z3)∈C3such thatz1+z2+z3= 0. The latter representation has dimension 2 and is irreducible; it is called the standard representation of S3. To ﬁnd its character, we observe that χV=χπ−χU. ButUis just the trivial representation and χπis equal to 3 on C1, 1 onC2and 0 onC3(why? see Exercise 5.1.2) We have found the complete character table. S3 (1) (12) (123) 1 3 2 trivialU 1 1 1 sign 1−1 1 standardV 2 0 −1 Proposition 5.3.2. The standard representation of Snis irreducible for all n>1. Proof. Letχbe the character of the n-dimensional linear representation corre- sponding to the permutation representation of Snon{1,...,n }(see Example 4.1.8), and let χ1be the trivial character of Sn. We wish to show that χ−χ1is irreducible; to this eﬀect, it suﬃces to prove that ( χ|χ) = 2. Denoting by ﬁx( σ) the set of ﬁxed points of σ∈Snon{1,...,n }, we haveχ(σ) =|ﬁx(σ)|. Hence, we ﬁnd (χ|χ) =1 n!/summationdisplay σ∈Sn|ﬁx(σ)|2. To evaluate the right-hand side of this equation we count the number of per- mutations having precisely kﬁxed points (0 ≤k≤n). This number is clearly equal to /parenleftbiggn k/parenrightbigg F(n−k), whereF(m) denotes the number of ﬁxed-point-free permutations in Sm. A standard inclusion-exclusion argument shows that F(m) =m/summationdisplay i=0(−1)im! i!. 74 CHAPTER 5. CHARACTER TABLES Combining the above considerations we ﬁnd (χ|χ) =1 n!n/summationdisplay k=0k2/parenleftbiggn k/parenrightbiggn−k/summationdisplay i=0(−1)i(n−k)! i! =n/summationdisplay k=0k21 k!n−k/summationdisplay i=0(−1)i i! =:Mn. It is readily seen that M2= 2. To prove that Mn= 2 for all n≥2, we note that forn≥3 we have: Mn−Mn−1=n/summationdisplay i=0(n−i)2(−1)i 1 (n−i)!i!, and it is not hard to prove that the right-hand side is zero for n≥3. Example 5.3.3. Consider the cyclic group Cnwith generator c, and letζ∈C be a primitive n-th root of unity. Deﬁne, for j= 0,...,n −1, the map χj: Cn→Cbyχj(ck) :=ζjk. Then the χjare characters of Cn, and irreducible as they are one-dimensional. By Theorem 5.2.21 there are no other characters, so that the character table of Cnis as follows. Cne c c2... cn−1 1 1 1 ... 1 χ11 1 1 ... 1 χ21ζ ζ2... ζn−1 χ31ζ2ζ4... ζ2(n−1) ·· · · ... · ·· · · ... · ·· · · ... · χn1ζn−1ζ2(n−1)... ζ(n−1)2 Example 5.3.4. To compute the character table of S4, we ﬁrst count the number of conjugacy classes. They are the identity, the 2-cycles, the 3-cycles, the 4-cycles and the product of two distinct 2-cycles; ﬁve in total. Hence, by Theorem 5.2.21, S4has ﬁve distinct irreducible characters. It has three of them in common with S3(or, indeed, with any symmetric group; see Proposition 5.3.2), namely the trivial character, the sign and the standard character. They are listed in the following partial character table. S4 (1) (12) (123) (1234) (12)(34) 1 6 8 6 3 χ1(trivial) 1 1 1 1 1 χ2(sign) 1−1 1 −1 1 χ4(standard) 3 1 0 −1 −1 5.3. CHARACTER TABLES 75 By Lemma 5.2.19, the squares of the dimensions of the irreducible characters sum up to |S4|= 24, and we only have 1 + 1 + 9 = 11 so far. We conclude that the two remaining irreducible characters have dimensions 2 and 3. The latter is the product χ5:=χ2χ4, which is irreducible as it has norm 1. The remaining 2- dimensional character χ3can be found using the orthogonality relations among characters. We then ﬁnd the following character table of S4. S4 (1) (12) (123) (1234) (12)(34) 1 6 8 6 3 χ1 1 1 1 1 1 χ2 1−1 1 −1 1 χ3 2 0 −1 0 2 χ4 3 1 0 −1 −1 χ5=χ2χ4 3−1 0 1 −1 Exercise 5.3.5. Compute the number of elements in each conjugacy class of S4. Exercise 5.3.6. Consider the permutation representation of S4on partitions of{1,2,3,4}into two sets of size 2; that is, the action of S4on the set {{A,B} | |A|=|B|= 2, A∪B={1,2,3,4}, A∩B=∅}. Show that the character of the corresponding linear representation is χ1+χ3; this explains the character χ3. A more or less explicit description of the irreducible representations of Snis known for general n. It involves the beautiful combinatorics of Young tableaux, a good textbook on which is Fulton’s book [5]. Exercise 5.3.7. Construct the character table of D4. Exercise 5.3.8. Compute the character table of the Klein group ( Z/2Z)⊕ (Z/2Z). Example 5.3.9. Consider the subgroup Tof the orthogonal group SO3(R) consisting of the proper rotations of the tetrahedron. We wish to compute its character table. The conjugacy classes are those of (1) ,(234),(243) and (13)(24) (make a picture that corresponds to this numbering), and their cardinalities are 1, 4, 4 and 3, respectively. Thus, there must be 4 irreducible characters, of representations with dimensions m1≤m2≤m3≤m4. The sum of their squares should equal |T|= 12. Their turns out to be only one solution, namely m1=m2=m3= 1 andm4= 3. Hence, apart from the trivial character, there should be two more characters of degree 1. Let rbe one of them. As ris nothing but a homomorphism from TtoC∗, the complex number r((234)) should have multiplicative order 3; hence it is either ω=e2πi/3orω2. The image of (243) is ﬁxed in both cases, and so is the image of (13)(24), as χrshould be orthogonal to the trivial representation. This explains the ﬁrst three lines in Table 5.1. The last line can be obtained from the orthonormality conditions. 76 CHAPTER 5. CHARACTER TABLES T(1) (234) (243) (13)(24) 1 4 4 3 χ1 1 1 1 1 χ2 1ω ω21 χ3 1ω2ω 1 χ4 3 0 0 −1 Table 5.1: The character table of the tetrahedral group Exercise 5.3.10. Find a representation corresponding to the last line of Table 5.1. Exercise 5.3.11. LetGbe a group and let Hbe a normal subgroup of G. Letrbe a representation of Gwith the property that H⊆kerr. Show that r induces a representation of G/H . Conversely, any representation of G/H can be lifted to one on G. Apply this principle to the group T, which has the Klein 4-group as a normal subgroup. Exercise 5.3.12. Prove the column orthogonality of the character table: for g,h∈G, andχ1,...,χ sa complete set of irreducible characters of G, s/summationdisplay i=1χi(g)χi(h) =|CG(h)|δg,h. This formula is also useful for the construction of character tables. Exercise 5.3.13. Letχbe the character of a complex linear representation of GonV. Show that the character of Gon the symmetric square S2(V) ofV equals g/mapsto→χ(g)2+χ(g2) 2. What is the character of GonS3(V)? Use these results to ﬁnd new characters ofA5from the irreducible 4-dimensional character occurring in the standard permutation representation of degree 5. Exercise 5.3.14. Consider the alternating group A5on 5 letters. 1. Write down representatives for each of its conjugacy classes. 2. Determine the sizes of the corresponding conjugacy classes. 3. Show that A5has only one irreducible character of degree 1. (Hint: if there were one more, then there would have to be a normal subgroup N such thatA5/Nis abelian.) 5.4. APPLICATION: SYMMETRY OF VIBRATIONS 77 4. The rotation group of the icosahedron is A5. This gives an irreducible representation ρofA5of degree 3. Write down its character. 5. Verify that the composition of ρwith conjugation by (1 ,2) gives a non- equivalent representation and determine its character. 6. Apply the symmetric square formula of Exercise 5.3.13 to ρto ﬁnd a 6-dimensional character. Show that this character splits into the trivial character and an irreducible 5-dimensional character. 7. Conclude from the orthogonality of the character table that there is one more irreducible character. Compute this character, and compare it with the character of the restriction of the standard representation of S5toA5. Exercise 5.3.15. Up to isomorphism, there are precisely two groups of order 8. They have identical character tables. Exhibit these groups and their character tables. 5.4 Application: symmetry of vibrations There exist many spectroscopic techniques in chemistry that can be used to mea- sure symmetry properties of molecules of a given substance. In this section, we describe two of them: infrared (IR) spectroscopy and Raman spectroscopy. In the former method, the substance is exposed to radiation of various frequencies, and the intensity of the light leaving the substance is measured. If the radiation excites the molecules from the ground state into vibration, then it is absorbed, giving rise to a (downward) peak in the intensity (plotted as a function of the frequency). As we have seen in Section 4.2, there are several essentially diﬀerent vibrations, corresponding to irreducible submodules of the representation of the symmetry group of the molecule on the space of all displacements. As with Ra- man spectroscopy—another technique whose physical details we do not discuss here—only actual vibrations may contribute a peak to the intensity spectrum, i.e., the (inﬁnitesimal) isometries of three-dimensional space do not contribute. However, not even all irreducible modules in the module of vibrations can be observed by IR spectroscopy; due to technical properties of this technique, only those irreducibles occurring also in the natural 3-dimensional representation do actually contribute a peak in the spectrum. This distinguishes IR from Ra- man spectroscopy, which allows to observe all irreducibles that occur in the symmetric square of the aforementioned natural representation. Hence, to predict the number of peaks using either of the above methods, for a given molecule with nmolecules, we proceed as follows. 1. Determine the symmetry group Gof the molecule at hand, and compute or look up its character table. 2. Determine the character χdisofGon the 3n-dimensional module of dis- placements of the molecule, as well as its character χisoon the 6-dimensional 78 CHAPTER 5. CHARACTER TABLES Cl ClClCl P Cl Figure 5.1: The PCl 5molecule submodule of (inﬁnitesimal) isometries of R3. Then χvib:=χdis−χiso is the character of Gon vibrations of the molecule. 3. Compute the character χ1ofGonR3, and the character χ2ofGon the symmetric square of R3. 4. Then the number of peaks in IR spectroscopy is the number of irreducible characters in χvib(counted with multiplicity) that also occur in χ1, and and the number of peaks in Raman spectroscopy is the number of irre- ducible characters in χvib(counted with multiplicity) that also occur in χ2. Example 5.4.1. We want to predict the number of peaks in IR and Raman spectroscopy when applied to the molecule PCl 5of Figure 5.1. In these chemical applications, it is common to use the Schoenﬂies notation (see Table 3.2), and we will adhere to this tradition in the present example. 1. The symmetry group of this molecule is D3h, whose character table is as follows (Eis the identity, Cistands for a rotation of order i,σhandσv are reﬂections in a horizontal and vertical plane, respectively, and S3is the composition of a C3and aσh). 5.4. APPLICATION: SYMMETRY OF VIBRATIONS 79 D3hE C 33C2σhS3σv 1 2 3 1 2 3 A/prime 1 1 1 1 1 1 1 A/prime 2 1 1 -1 1 1 -1 E/prime2 -1 0 2 -1 0 A/prime/prime 1 1 1 1 -1 -1 -1 A/prime/prime 2 1 1 -1 -1 -1 1 E/prime/prime2 -1 0 -2 1 0 2. Letχ1denote the character of D3hin its natural representation on R3, and letχ3denote the character of D3hcorresponding to its permutation representation on the 6 molecules of PCl 5(see Example 4.1.8). Their values are easily calculated to be as follows. D3hE C 33C2σhS3σv χ1 3 0 -1 1 -2 1 χ3 6 3 2 4 1 4 As in Example 5.2.23, the character of the module of displacements equals the dual of χ3timesχ1. Having only real values, χ3is self-dual, so that we ﬁnd the following values for χdis. D3hE C 33C2σhS3σv χdis 18 0 -2 4 -2 4 This character can be decomposed into irreducible ones as follows: χdis= 2A/prime 1+A/prime 2+ 4E/prime+ 3A/prime/prime 2+ 2E/prime/prime The character χisois the sum of the characters of D3hon translations and on rotations. The character of D3hon translations equals χ3=E/prime+A/prime/prime 2, and the character χ4on rotations turns out to be as follows. D3hE C 33C2σhS3σv χ4 3 0 -1 -1 2 -1 We ﬁnd that χ4=A/prime 2+E/prime/prime. We conclude that χvib= 2A/prime 1+ 3E/prime+ 2A/prime/prime 2+E/prime/prime. 3. We have already computed χ1. Using Exercise 5.3.13 we ﬁnd the following values for its symmetric square χ2. D3hE C 33C2σhS3σv χ2 6 0 2 2 2 2 We ﬁnd that χ2= 2A/prime 1+E/prime+E/prime/prime, whileχ1=E/prime+A/prime/prime 2. 4. From the above, we conclude that only vibrations with characters E/primeor A/prime/prime 2are observable with IR spectroscopy; the number of such irreducible characters in χvibis 5, so that is the predicted number of peaks when using this technique. Using Raman spectroscopy, only vibrations with charactersA/prime 1,E/primeandE/prime/primecan be observed; this leads to 6 peaks. 80 CHAPTER 5. CHARACTER TABLES PdCl (NH ) 232PdCl (NH ) 232N N Cl ClN Cl NCl cis− trans−Pd Pd Figure 5.2: Two conﬁgurations of PdCl 2(NH 3)2 The following exercises shows how spectroscopic techniques may or may not be used to recognize molecules. Exercise 5.4.2. The square planar coordination compound PdCl 2(NH 3)2can be prepared with two diﬀerent conﬁgurations; see Figure 5.2. Is it possible to discern the cis- and the trans-conﬁgurations using information concerning Pd−Cl stretch vibrations in IR spectroscopy? Exercise 5.4.3. Is it possible to deduce the symmetry of Ni(CO) 4(tetrahedron or square plane) from the IR and/or Raman spectra of the CO-stretches? Exercise 5.4.4. LetTbe the octahedron whose 6 vertices are joined by strings. Determine the decomposition of the 18-dimensional space of vibrations of irre- ducibles of the group of rotational symmetries of T(of order 24). Interpret the result. Which spaces fuse into irreducibles for the group of all (proper or improper) symmetries of T(of order 48)? 5.5 Notes Most of the content of this chapter can be found in Serre’s book [9]. The last section is entirely based on Theo Beelen’s lectures in the course for which these notes were prepared. Chapter 6 Some compact Lie groups and their representations In the previous two chapters we got acquainted with the representation theory ofﬁnite groups. In this chapter we will treat some of the representation theory ofcompact Lie groups , without going into the details of deﬁning a Lie group and showing, for example, the existence of an invariant measure. A good reference, on which in fact most of this chapter is based, is [1]. Some general, less detailed remarks can be found in [9]. 6.1 Some examples of Lie groups One can think of a Lie group Gas a smooth surface embedded in some Euclidean space. Moreover, Gcarries the structure of a group in such a way that the multiplication map ( g,h)/mapsto→g·h,G×G→Gand the inverse map g/mapsto→g−1,G→ Gare smooth. Examples, some of which we already saw, are: 1. GL(n,R), 2. SL(n,R), 3. O(n,R), 4. SO(n,R), 5. U(n) := {g∈Mn(C)|g∗g=I}, whereg∗stands for the conjugate transpose of g, i.e.,g∗= (gji)ijifg= (gij)ij. 6. SU(n) :={g∈U(n)|det(g) = 1}. The latter two groups are called the unitary group and the special unitary group . A more intrinsic deﬁnition is the following: let ( .,.) be a Hermitian inner product on a ﬁnite-dimensional complex vector space V. Then the corresponding unitary 81 82CHAPTER 6. SOME COMPACT LIE GROUPS AND THEIR REPRESENTATIONS group is the group of all linear maps on Vwith (gv,gw ) = (v,w) for allg∈G andv,w∈V. Note that each of the real (complex) matrix groups above is the zero set of some smooth map FfromMm(R) (Mm(C)) to some Euclidean space Rdsuch that the rank of the Jacobian of Fat every point of the zero set ofFisd. If, in this situation, the zero set of Fis a group with respect to matrix multiplication—as it is in the above cases—then it is automatically a Lie group. Such Lie groups are called linear because they are given by a linear representation. In the present chapter, this class of Lie groups suﬃces for our needs. Exercise 6.1.1. How can GL( n,R) be constructed from Mn+1(R) via the above construction? What about ( R,+)? Exercise 6.1.2. Show that O( n,R)∼=SO(n,R)/multicloseright(Z/2Z). Here /multicloserightmeans that the subgroup at the left is a normal subgroup and the subgroup at the right is a complementary subgroup in the sense that their product is the whole group and their intersection is the trivial group. Such a product is called semi-direct . Ifnis odd, then it is a direct product. Exercise 6.1.3. Show that, in SU(2), any element is conjugate to a diagonal matrix of the form e(t) :=/parenleftbigg eit0 0e−it/parenrightbigg for certaint∈R. The theory of Lie groups is a blend of group theory and topology (surfaces, continuity, diﬀerentiability, etc.), as might be expected. For example, one can ask whether a given Lie group is (path-wise) connected, or whether it is compact. A typical lemma from the theory is the statement that the connected component of the identity element is a normal subgroup. Exercise 6.1.4. Show that O( n,R) is not connected. Proposition 6.1.5. The Lie group SO(n,R)is connected. Proof. We will connect an arbitrary matrix g= (aij)ijin SO(n,R) to the iden- tity via a continuous and piecewise smooth path in SO( n,R). The ﬁrst step is to connect it to a diagonal matrix; to this end, suppose ﬁrst that there exists a position (i,j) such that i > j andaij/negationslash= 0. Then one can choose this position such that, in addition, ai/primej/prime= 0 for all ( i/prime,j/prime) withi/prime> j/primeand eitherj/prime< jor j/prime=jandi/prime>i. Schematically, ghas the following form: g= T∗ ∗ ∗ ∗ 0ajj∗aji∗ 0∗ ∗ ∗ ∗ 0aij∗aii∗ 0 0 ∗ ∗ ∗ , 6.1. SOME EXAMPLES OF LIE GROUPS 83 whereTis a (j−1)×(j−1) upper triangular matrix. Now let r(t), t∈Rbe the realn×nmatrix given by r(t)i/primej/prime=  cos(t),ifi/prime=j/prime=jori/prime=j/prime=i, sin(t),ifi/prime=iandj/prime=j, −sin(t),ifi/prime=jandj/prime=i, 1,ifi/negationslash=i/prime=j/prime/negationslash=j,and 0 otherwise . Thenr(t) has the following block form: r(t) = I 0 0 0 0 0 cos(t) 0 −sin(t) 0 0 0 I 0 0 0 sin(t) 0 cos( t) 0 0 0 0 0 I ; note thatr(t)∈SO(n,R) for allt∈R. Now we have c(t) :=r(t)g= T ∗ ∗ ∗ ∗ 0ajjcos(t)−aijsin(t)∗ajicos(t)−aiisin(t)∗ 0 ∗ ∗ ∗ ∗ 0ajjsin(t) +aijcos(t)∗ajisin(t) +aiicos(t)∗ 0 0 ∗ ∗ ∗  NowR(sin(t),cos(t)) runs through all 1-dimensional subspaces of R2astruns from 0 toπ, so that there exists a t0∈[0,π) for which (sin( t0),cos(t0)) is perpendicular to ( ajj,aij) in the standard inner product; we ﬁnd that c(t0)ij= 0, in addition to c(t0)i/primej/prime= 0 for all ( i/prime,j/prime) withi/prime>j/primeand eitherj/prime<jorj/prime=j andi/prime> i. Moreover, the path c(t), t∈[0,t0] lies in SO( n,R) and connects g toc(t0). Repeating the above construction, we ﬁnd a path connecting gto an upper triangular matrix h. Being an element of O( n,R),his in fact diagonal and has only entries ±1 on the diagonal. Moreover, as det( h) = 1, the number of −1s is even, and they can be partitioned into pairs. Each such pair can be smoothly transformed to 1s by multiplication from the left with an appropriate r(t), t∈[0,π] as above. This proposition implies that SO( n,R) is the connected component of the identity in O( n,R). The following exercise will be used later on, for the repre- sentation theory of SO(3 ,R). Exercise 6.1.6. Show that SU(2) is connected. Hint: ﬁnd a continuous path from any given element of SU(2) to some e(t) as in Exercise 6.1.3. 84CHAPTER 6. SOME COMPACT LIE GROUPS AND THEIR REPRESENTATIONS 6.2 Representation theory of compact Lie groups Compact Lie groups (i.e., with the property that their embedding in Euclidean space is closed and bounded) are particularly nice, because much of the repre- sentation theory of ﬁnite groups works just as well for these. The main tool is the so-called Haar measure , which enables us to take ‘the average over a group’ just as we did in ﬁnite group theory. Let us state the existence and uniqueness of this measure in a theorem. Theorem 6.2.1. LetGbe a compact real Lie group, and denote by C(G)the linear space of all continuous functions G→RonG. There is a unique map f/mapsto→/integraltext Gf(g)dg,C(G)→Rwhich is 1.linear , i.e., for all e,f∈C(G)andα∈R, we have /integraldisplay G(e+f)(g)dg=/integraldisplay Gf(g)dg+/integraldisplay Ge(g)dgand/integraldisplay G(αf)(g)dg=α/integraldisplay Gf(g)dg, 2.monotonous , i.e., ife(g)≤f(g)for allg∈G, then also /integraldisplay Ge(g)dg≤/integraldisplay Gf(g)dg, 3.left-invariant , i.e., for the action of Ginduced on C(G)by the left multi- plication, we have for all h∈Gandf∈C(G)that /integraldisplay Gh·f(g)dg=/integraldisplay Gf(g)dg, whereh·fis the function g/mapsto→f(h−1g) (h∈G). 4. and normalized , i.e., /integraldisplay G1dg= 1. This linear map is called the invariant ( Haar )-integral. Exercise 6.2.2. A ﬁnite group is a compact Lie group. What is the invariant integral? In order to ﬁnd an invariant Haar integral on the compact Lie group T= {e(t)|t∈R}, wheree(t) is deﬁned as in Exercise 6.1.3, we search for invariant measures one(t), that is, diﬀerential forms invariant under left multiplication. In general, given a representation r:G→GL(V), we can make invariant measures on Gby taking certain entries of the matrix r(g)−1dr(g). For the groupTand the natural representation, this is e(t)−1de(t). The 1,1 entry of this matrix is e−itdeit=idt. Since/integraltext2π 0idt= 2πi, the invariant Haar-integral, viewed as a map assigning a real number to a function f:R/2πZ→R, is 1 2π/integraldisplay2π 0f(t)dt. 6.2. REPRESENTATION THEORY OF COMPACT LIE GROUPS 85 A similar but more laborious computation yields that the invariant Haar mea- sure forSU2(C) is 1 2π2sin2θsinψdθdψdφ, where /parenleftbigg cosθ−isinθcosψ −sinθsinψeiφ sinθsinψe−iφcosθ+isinθcosψ/parenrightbigg is a general element of SU2(C). The result of the following exercise will be used in proving that certain irreducible representations of SU 2(C) exhaust all irreducible representations. Exercise 6.2.3. Show that the map sending a class function f: SU(2) →C to (f◦e) :R→C, whereeis the one-parameter subgroup of SU(2) deﬁned in Exercise 6.1.3, is a linear bijection between the space Hof continuous class functions on SU(2) and the linear space of all even, 2π-periodic continuous functions R→C. Moreover, the norm corresponding to the invariant inner product (6.1) below corresponds, under this linear bijection, to the normalized L2inner product on the space of continuous even 2 π-periodic functions on R. One could deﬁne representations of compact Lie groups in Hilbert spaces, but here we shall restrict our attention to ﬁnite-dimensional representations. Also, we shall only consider complex representations. Deﬁnition 6.2.4. A ﬁnite-dimensional complex representation of the Lie group Gis a smooth homomorphism G→GL(V), whereVis a ﬁnite-dimensional complex vector space. In fact, one can prove that continuous homomorphisms from a Lie group G are automatically smooth. For ﬁnite-dimensional complex representations of compact Lie groups, al- most all theorems that we proved in the preceding chapters for ﬁnite groups hold, be it that at some places a sum must be replaced by an integral. To give an idea how things work, one can try the following exercise. Exercise 6.2.5. Letr:G→GL(V) be a ﬁnite-dimensional representation of the compact Lie group G, and letW⊂Vbe an invariant subspace. Show that Whas an invariant complement in V. We will study the representations of the groups SU(2) and SO(3 ,R), so it is good to verify that these are indeed compact Lie groups. Exercise 6.2.6. In this exercise we prove that U( n), SU(n), O(n,R) and SO(n,R) are compact Lie groups. To this end, endow the space Mn(C) with the Hermitian inner product given by ((aij)ij,(bij)ij) :=/summationdisplay i,jaijbij. 1. Show that ( a,b) = tr(ab∗) for alla,b∈Mn(C). 86CHAPTER 6. SOME COMPACT LIE GROUPS AND THEIR REPRESENTATIONS 2. Show that U( n) is bounded with respect to the norm corresponding to this inner product. 3. Show that U( n) is the zero set of a smooth map F:Mn(C)→Rn2, whose Jacobian has rank n2at each element of U(n). Thus, U(n) is a Lie group by the construction of the beginning of this chapter, and compact as it is a closed and bounded set in Mn(C). Find similar arguments for the other groups above. Show, on the contrary, that the group O( n,C) of complex matrices gsatisfyinggTg=Iis not compact for n≥2. A ﬁnal remark concerns the compact analogon of the inner product on the space of complex-valued functions on a ﬁnite group: in the case of a compact Lie group, one should consider the space of Hilbert space of continuous class functions with Hermitian inner product deﬁned by (f|h) :=/integraldisplay Gf(g)h(g)dg. (6.1) Here, the irreducible characters form an orthonormal Hilbert basis in the space of continuous class functions. 6.3 The 2-dimensional unitary group This section is based on section II.5 of [1]. Letr: SU(2) →GL(V) be the standard representation of SU(2) on V=C2, the action being given by matrix-column multiplication (the elements of Vare understood to be columns). As we have seen before, this action induces linear actions onP(V)d, the space of homogeneous polynomials of degree ddeﬁned on V, by letting (gP)(ξ) =P(g−1ξ) forξ∈V,P∈P(V)d,g∈SU(2). Letrd: SU(2) →GL(P(V)d) be the corresponding representation and χdits character. We are ready to state the main theorem of this section. Theorem 6.3.1. The representations rdare irreducible for all d= 1,2,.... Proof of theorem 6.3.1. We will use Exercise 4.1.30. Let A∈End(P(V)d) be a linear map that commutes with all rd(g) forg∈SU(2). The spaceP(V)dis spanned by the polynomials Pk:ξ/mapsto→ξk 1ξd−k 2fork= 0,...,d , and is therefore ( d+ 1)-dimensional. Consider the diagonal subgroup of SU(2). It consists of all matrices of the form ga:=/parenleftbigg a 0 0a−1/parenrightbigg 6.3. THE 2-DIMENSIONAL UNITARY GROUP 87 Note that the element a∈Cmust necessarily have norm 1. We compute (gaPk)(ξ) =Pk(g1 aξ) = (ξ1/a)k(aξ2)d−k =ad−2kPk(ξ). Hence we conclude that Pkis an eigenfunction of rd(ga) with eigenvalue ad−2k. Choosea0such that none of these eigenvalues coincide. If Acommutes with rd(ga0), then the eigenspaces of the latter map are invariant under A. Hence, as these eigenspaces are all one-dimensional, there are complex numbers ckfor k= 0,...,d such that APk=ckPk. It remains to show that all ckare equal. For this purpose, consider the real rotationsstdeﬁned by st:=/parenleftbigg cost−sint sintcost/parenrightbigg and compute (stPn)(ξ) =Pn(s−tξ) = (ξ1cost+ξ2sint)n =/summationdisplay k/parenleftbiggn k/parenrightbigg coskt·sinn−kt·Pk(ξ) ApplyingAto the result, we get A(stPn) =/summationdisplay k/parenleftbiggn k/parenrightbigg coskt·sinn−kt·ckPk. First applying Aand thenrd(st), and using the fact that the two commute, we ﬁnd that also A(stPn) =/summationdisplay k/parenleftbiggn k/parenrightbigg coskt·sinn−kt·cnPk. Takingtsuch that cos tsint/negationslash= 0, it follows from the fact that the Pkare linearly independent that cn=ckfor allk= 1,...,n , and hence that A=cnI. We wish to prove that these are all irreducible characters. To this end we shall use some Fourier theory, and the result of Exercise 6.2.3. Theorem 6.3.2. The representations rdare, up to equivalence, all irreducible representations of the group SU(2) . Proof. In view of Exercise 6.2.3, we can identify Hwith the space of even 2 π- periodic complex-valued functions on R. The matrix of rd(e(t)) with respect to the basisP0,...,P dis diagonal, and its trace is κd(t) =d/summationdisplay k=0ei(d−2k)t. 88CHAPTER 6. SOME COMPACT LIE GROUPS AND THEIR REPRESENTATIONS It is readily veriﬁed that κ0(t) = 1 (this corresponds to the trivial representation of the group), and that κ1(t) = 2 cost. Indeed, one has κd+2(t) =κd(t) + 2 cos((d+ 2)t). Hence, the functions κdford= 0,1,2,...generate the same space as the functions cos dt. From Fourier theory we know that this subspace is dense in the Hilbert space of continuous even 2 π-periodic functions. This implies, using the isomorphism from Exercise 6.2.3, that the characters χdspan a dense subspace in the Hilbert space of continuous class functions on SU(2). Any irreducible character diﬀerent from all rdwould have inner product 0 with allrd, and hence be zero. This concludes the proof that the rdexhaust the set of irreducible characters of SU(2). 6.4 The 3-dimensional orthogonal group We will obtain the irreducible representations of SO(3 ,R) from those of SU(2) by means of the following lemma, which appeared earlier as Exercise 5.3.11. Lemma 6.4.1. LetGandHbe groups,φ:G→Ha surjective homomorphism. Then the (irreducible) representations of Hare in bijective correspondence with the (irreducible) representations rofGwith kerr⊇kerφ. Thus, we wish to ﬁnd a surjective homomorphism of SU(2) onto SO(3) and to determine its kernel. Recall the homomorphism φ: SL(2,C)→Lof Example 2.2.9. Here, Lis the group of Lorentz-norm preserving linear maps from Mto itself, where Mis theR-linear span of the following four complex 2 ×2-matrices. e0=I,e1=/parenleftbigg 0 1 1 0/parenrightbigg ,e2=/parenleftbigg 0−i i0/parenrightbigg ,e3=/parenleftbigg 1 0 0−1/parenrightbigg . It can be checked that in this way det(x0e0+x1e1+x2e2+x3e3) =x2 0−x2 1−x2 2−x2 3. SoLis the group {B∈GL(M)|det(Bx) = det(x),for allx∈M}. The homomorphism φwas deﬁned by φ(A)(x) =AxA∗,for allx∈M,A ∈SL(2,C). From now on, φwill denote the restriction of this homomorphism to SU(2). We claim that the image is the group G:={B∈L|Be0=e0and detB= 1}, which is, by its action on the invariant 3-space e⊥ 0, easily seen to be isomorphic to SO(3,R). First of allφ(SU(2)) ⊆G. Indeed, if A∈SU(2), then A∗=A−1. Therefore, φ(A)(e0) =e0, sinceeois just the 2 ×2 identity matrix. Also, the function 6.4. THE 3-DIMENSIONAL ORTHOGONAL GROUP 89 det◦φis a continuous function SU(2) →R, which can apparently take only values in {−1,1}. As SU(2) is connected, and det( φ(I)) = 1, the determinant of any element in the image φ(SU(2)) is 1. Now let us prove that G⊆φ(SU(2)). We follow the approach given in [4]. First we prove the following two lemmas. Lemma 6.4.2. LetFbe a subgroup of SO(3,R)with the following two proper- ties. 1.Facts transitively on the unit sphere. 2. There is an axis such that Fcontains all rotations around that axis. ThenF= SO(3,R). Exercise 6.4.3. Prove this lemma. Lemma 6.4.4. For every non-zero x∈e⊥ 0there is an A∈SU(2) such that φ(A)x=ce3for somec∈Randc > 0. Heree⊥ 0denotes the orthogonal complement of e0with respect to the Lorentz inner product, i.e., e⊥ 0={x1e1+x2e2+x3e3|xi∈R} ⊆M. Proof. Any suchxis in fact a Hermitian matrix with trace 0. As such it has two real eigenvalues c,−c, wherec > 0, (note that xis non-zero). Let two (necessarily orthogonal) eigencolumns of norm 1 corresponding to cand−c be the ﬁrst and second column of a matrix B, respectively. Then the usual coordinate transformation rule gives x=B−1(cH)B. We have that B∈U(2). Now take A=dBwhered2= 1/(detB); thenAis still unitary, since |d|= 1. Furthermore, we have AxA−1= (dB)x(dB)−1=dd−1BxB−1=cH, and detA=d2detB= 1. Now let us continue with the proof that F:=φ(SU(2)) coincides with G. The group Facts on the unit sphere Sine⊥ 0deﬁned by x2 1+x2 2+x2 3, and by Lemma 6.4.4 this action is transitive, since any vector in the sphere is mapped toe3by some group element. Hence, to apply Lemma 6.4.2 to F, we need only check thatFcontains all rotations around some axis. We leave this as an exercise to the reader: Exercise 6.4.5. Show thatφ(e(t)) is the rotation around /angbracketlefte3/angbracketrightthrough an angle of 2t. Heree:R→SU(2) is the one-parameter group deﬁned earlier. We conclude that φ: SU(2) →Gis a surjective homomorphism. Its kernel contains {I,−I}, as is easily checked, and the following exercise shows that this is the whole kernel. 90CHAPTER 6. SOME COMPACT LIE GROUPS AND THEIR REPRESENTATIONS Exercise 6.4.6. Show that ker φ={I,−I}. Hint: the elements of this kernel must commute with all of M, hence in particular with e1,e2, ande3. Finally we can apply lemma 6.4.1: the irreducible representations of SO(3) correspond to those of SU(2) for which −Iis mapped to the identity transfor- mation. More precisely, we wish to know for which dwe haverd(−I) =I. For this to be the case, it is necessary and suﬃcient that ((−I)P)(ξ) =P(ξ) for allξ∈V,P∈P(V)d. This holds if and only if dis even. Hence the irreducible representations are Wd:=P(V)2d. Note that the dimension of Wdis 2d+ 1. In the following example, we will see another interpretation of the irreducible representations of SO(3 ,R). Example 6.4.7. ConsiderP(V)dthe space of homogeneous polynomials of degreeddeﬁned on the vector space Vwhich will be of dimension three here. As we have seen before, the group GL(3 ,R) acts on this space P(V)dby (A·p)(x) = p(A−1x), for allA∈GL(3,R),p∈P(V)dandx∈R3. So does the subgroup SO(3). Take the subspace UofP(V)2generated by the polynomial x2 1+x2 2+x2 3. This is an invariant subspace for the action of SO(3). Hence P(V)2is not irreducible. Actually the spaces P(V)dare not irreducible for all d≥2. To ﬁnd SO(3)- invariant subspaces of P(V)d, we need to deﬁne the Laplace operator ∆ on R3: ∆ =∂2 ∂x2 1+∂2 ∂x2 2+∂2 ∂x2 3. The space of harmonic polynomials hdis the vector space: hd={p∈P(V)d|∆p= 0}. As was noticed in Example 4.1.17, the dimension of P(V)dis dimP(V)d=/parenleftbigd+2 d/parenrightbig . We now want to prove that the dimension of hdis 2d+ 1. A homogeneous polynomial pof degreedcan be written as a sum of homogeneous polynomials pkinx2andx3with power of x1’s as coeﬃcients: p(x1,x2,x3) =p0(x2,x3) +x1p1(x2,x3) +x2 1 2p2(x2,x3) +· · ·+xd 1 d!pd(x2,x3) =d/summationdisplay k=0xk 1 k!pk(x2,x3) where thepk’s have degree d−k. If one applies the Laplace operator to p(you see why we took factorials): ∆(p) =d−2/summationdisplay k=0xk 1 k!pk+2(x2,x3) +d/summationdisplay k=0xk 1 k!/parenleftbigg∂2 ∂x2 2+∂2 ∂x2 3/parenrightbigg pk(x2,x3) 6.4. THE 3-DIMENSIONAL ORTHOGONAL GROUP 91 Hence an element p∈P(V)dis in hdif and only if for all k= 0,...,d −2: pk+2(x2,x3) =−/parenleftbigg∂2 ∂x2 2+∂2 ∂x2 3/parenrightbigg pk(x2,x3) This formula implies that pinhddepends only on the ﬁrst two homogeneous polynomials p0andp1. Therefore we conclude that the dimension of hdis precisely the number of homogeneous polynomials in two variables of degree d andd−1. That is, dim hd= 2d+ 1. Of course, the degree of hdreminds us of the irreducible representations Wd of SO(3) we met before. Indeed, we have the following. Theorem 6.4.8. The space hdof harmonic polynomials of degree dis an irre- ducible representation of SO(3) . To see that what we just stated is true, we should ﬁrst check that hdis a representation of SO(3). We state here the following fact without proof. Fact: The action of the Laplace operator on P(V)dcommutes with the action of SO(3). From this we conclude that hdis a representation of SO(3). To prove that hdis irreducible, we claim that hd=Wd. Setpa harmonic polynomial in hd,p(x1,x2,x3) = (x2+ix3)d. LetR(t) be the matrix of SO(3) deﬁned by R(t) = 1 0 0 0 cost−sint 0 sintcost  ApplyR(t) top. That gives us R(t)p= exp( −idt)p. Sopgenerates an invariant subspace Wofhdunder the action of R(t). Any element in SO(3) is conjugate to R(t). So in order to ﬁnd the character ofWdonR(t) it suﬃces to compute it on R(t). That is just the value of the character χ2dofrd(the irreducible representation of SU(2)) at et/2, i.e.,/summationtext2d k=0ei(d−k)t. By the decomposition of hdin irreducible Ws’s, we know that its character is a linear combination of exp( ist). So what we saw above about the action of R(t) onWleads us to conclude that hd=Wd. 92CHAPTER 6. SOME COMPACT LIE GROUPS AND THEIR REPRESENTATIONS Bibliography [1] T. Br¨ ocker and T. tom Dieck. Representations of Compact Lie Groups . Springer Verlag, New York, 1995. [2] F. Albert Cotton. Chemical applications of group theory . 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Princeton University Press, Princeton, 1989. 93