The Frobenius-Schur Indicator and the Finite Rotation Groups
Daniel Penner Contents
1 Representation theory of finite groups 3 1.1 Basics of linear representations ...... 3 1.1.1 Definitions and theory ...... 3 1.1.2 Examples ...... 4 1.2 Character theory ...... 5 1.2.1 Basic definitions and lemmas ...... 5 1.2.2 Irreducible characters and the orthogonality relations . . . . . 8 1.2.3 The group algebra ...... 12 1.2.4 Character tables: some more theory ...... 14 1.2.5 Some examples of character tables ...... 17 1.2.6 Motivation: the character table of the quaternion group . . . . 19
2 Real representations and the Frobenius-Schur indicator 21 2.1 Real, complex, and quaternionic representations ...... 21 2.2 TheFrobenius-Schurindicator ...... 22 2.2.1 Relation to G-invariant bilinear forms ...... 22 2.2.2 The symmetric and alternating squares ...... 25 2.2.3 Bringing it all together ...... 27 2.3 Decomposing the group algebra R[G]...... 28 2.4 Some simple examples ...... 30
3 The finite subgroups of SO3 32 3.1 The cyclic groups Cn ...... 32 3.2 The dihedral groups Dn ...... 33 3.3 The tetrahedral group T ...... 34 3.4 The octahedral group O ...... 35 3.5 The icosahedral group I ...... 38
2 Chapter 1
Representation theory of finite groups
1.1 Basics of linear representations
In this first chapter, we will build the theory of group representations and their characters from the bottom up. In the first section, we present the basic defini- tions, prove the fundamental decomposition theorem of group representations, and introduce a couple simple examples. In the second section, we will present a broad overview of the theory of group characters, beginning with basic definitions and ele- mentary lemmas, and proceeding to prove the orthogonality relations of irreducible characters, and a theorem which reduces the study of representations to the study of their characters. Along the way we will briefly discuss the group algebra as it relates to this paper, and will conclude with orthogonality and divisibility theorems for character tables, and some simple examples of character tables. Much of the standard theory follows the first half of J.P. Serre’s classic book on the subject (see [7]).
1.1.1 Definitions and theory Let G be a finite group and V be a complex finite-dimensional vector space (note: the infinite-dimensional case will not be mentioned in this paper). A linear repre- sentation of G in V is a homomorphism ⇢ : G GL(V ). Here GL(V )denotesthe group of isomorphisms of V ,whichisisomorphictothegroupofinvertible! n n complex matrices, where n =dim(V ), and we call n the degree of the representa-⇥ tion. More concretely, for all s, t G there are complex n n matrices ⇢s,⇢t,⇢st such that ⇢ = ⇢ ⇢ .Inparticular,thismeansthat2 ⇢ = I⇥,where1 G is the st s t 1 n 2 identity element and In is the n n identity matrix. Let ⇢ : G GL(V )bearepresentationofdegree⇥ n,andW V be a proper subspace (that! is, neither the zero space nor the whole space).⇢ We say W is G- invariant if ⇢sw W for all w W and all s G.IfW is a G-invariant subspace of V ,thenforeach2 s G,therestriction2 ⇢ 2 is an endomorphism of W ,and 2 s W ⇢ = ⇢ ⇢ for all s, t G,so⇢ : G GL(W )isahomomorphism,and st W s W t W W thus a linear representation. We2 call it a subrepresentation! of ⇢.IfV = W W 1 2 3 for two G-invariant subspaces W1,W2 of V ,thenwesaythat⇢ = ⇢ ⇢ ,orthat W1 W2 ⇢ is a direct sum of its subrepresentations. Finally, we say that two representations 1 2 ⇢ : G GL(V1)and⇢ : G GL(V2) isomorphic if there is a vector space ! ! 1 1 2 isomorphism f : V1 V2 such that, for all s G, we have f ⇢s f = ⇢s. The following lemma! allows us to decompose2 a representation into a direct sum of subrepresentations.
Lemma 1.1.1. Let ⇢ : G GL(V ) be a representation, and let ⇢ W : G GL(W ) be a subrepresentation to! a G-invariant subspace W V . Then there! is a G- ⇢ invariant complement W 0 V such that V = W W 0, and thus ⇢ = ⇢ ⇢ . ⇢ W W 0 Proof. Let p be a projection of V onto W .Thenweconstructanewprojectiononto W by averaging over all the elements of G: 1 p0 = ⇢ p ⇢ 1 . G s · · s s G | | X2
Since W is G-invariant, and the Im(p) W ,weobtainthatIm(p0) W as well. Since p is a projection, it restricts to the⇢ identity on W . Therefore if ⇢w W ,then 2 for all s G,since⇢ 1 w W , we have 2 s 2
p0w = ⇢ p ⇢ 1 w = ⇢ ⇢ 1 w = w. s · · s s · s
So p0 is a projection onto W .LetW 0 =Ker(p0). Since the sum in p0 runs over all elements of G,wehavethatforanys G, 2 ⇢ p0 ⇢ 1 = p0,andthus⇢ p0 = p0 ⇢ .Soifw0 W 0,wehavep0 ⇢ w0 = ⇢ p0w0 = s · · s s · · s 2 · s s · ⇢ 0=0.So⇢ w0 W 0,andthusW 0 is G-invariant. So we have two G-invariant s · s 2 subspaces W, W 0 of V such that V = W W 0,andtherefore⇢ = ⇢ ⇢ . W W 0 We say that a representation ⇢ : G GL(V )isirreducible if V has no proper G-invariant subspaces. The following theorem! allows us to focus all our attention on understanding irreducible representations.
Theorem 1.1.2 (Maschke’s Theorem). Every representation ⇢ : G GL(V ) is a direct sum of irreducible subrepresentations. !
Proof. We go by induction on n =dim(V ). If n =1,thenV has no proper subspaces at all, and thus ⇢ is irreducible. Now suppose n>1. If V has no G-invariant subspace, then it is irreducible. Otherwise, let W V be a G-invariant subspace. The preceding lemma gives us that ⇢ V = W W ?, where W ? is also G-invariant, and thus ⇢ = ⇢ ⇢ .Butthese W W ? are both representations of degree strictly less than n,andsotheyarecoveredby our inductive hypothesis, and are thus direct sums of irreducible subrepresentations. So ⇢ is a direct sum of irreducible subrepresentations.
1.1.2 Examples
To show some of the theory in action, let’s examine the representations of S3,the 1 symmetric group on three letters. Every group has a representation ⇢s =1for
4 all s G, called the trivial representation.Thisisanexampleofadegree-one 2 representation, all of which are simply homomorphisms from G C. ! S3 has another degree-one representation, given by sending an permutation s S3 to its sign, sgn(s) 1 . Note that this is a representation since sign of2 a permutation respects products2{± } of permutations, and hence it is a homomor- phism G 1 .Thisrepresentationappearstobedi↵erentfromthetrivial representation!{± (that} is, nonisomorphic), but it is not immediately obvious how to prove that there are no satisfactory endomorphisms of C.Theirnon-isomorphism will follow from basic character theory a bit later. However, as we saw in the proof of Maschke’s theorem, both the trivial and sign representations are irreducible, since they have degree one. Next, since S3 acts on the set X = 1, 2, 3 ,eachs S3 defines a permutation x sx,where1x = x and s(tx)=({st)x for} any t 2 S .LetV be a X = 7! 2 3 | | 3-dimensional vector space with a basis ex x X indexed by the elements of X. { } 2 Then the elements of S3 permute the bases of this vector space, which induces a representation of S as a set of 3 3 permutation matrices. Hence we call this the 3 ⇥ permutation representation of S3.Wecanformthistypeofrepresentationfor any group G which acts on a finite set X.Itisnotclearatthemomentwhetheror not this representation, which has degree three, is irreducible. As before, we defer our curiosity to character theory, which will make everything clear. A particular and very important permutation representation is that induced by G acting on itself by left-multiplication. We call this the regular representation of G.InthiscaseletV be a G = n-dimensional vector space with basis es s G | | { } 2 indexed by the elements of G. Then for s, t G,welet⇢s send et est,and this induces a set of n n permutation matrices.2 We can already see7! that this (degree n)representationisnotirreducible,however,sincethevectorspaceelement⇥
s G es generates a 1-dimensional subspace on which every basis vector et acts as the2 identity (since it simply permutes the terms in the sum), and is thus G-invariant. WeP will return to the regular representation in the context of characters later.
1.2 Character theory
1.2.1 Basic definitions and lemmas
If ⇢ : G GL(V )isarepresentation,defineafunction ⇢ : G C as ⇢(s)= ! ! Tr(⇢s). We call this function the character of the representation ⇢ (we will drop the subscript when the representation is clear from context). Proposition 1.2.1. If is the character of a degree n representation ⇢ : G GL(V ), then if s, t G, we have the following: ! 2 1. (1) = n
1 2. (s )= (s)
1 3. (tst )= (s)
Proof. 1. ⇢ = I ,then n identity matrix, whose trace is n. 1 n ⇥ 5 m m 2. Since G is finite, there is some m>0suchthats =1.Therefore⇢s = In, since ⇢ is a homomorphism. So the eigenvalues i of ⇢s are roots of unity, whose
complex conjugates are their multiplicative inverses. Since Tr(⇢s)= i i,it follows that (s)=Tr(⇢ )=Tr(⇢ 1)= (s 1). s s P
3. For any square matrices A, B,Tr(AB)=Tr(BA). So if A = ⇢ts = ⇢t⇢s and 1 1 1 1 1 B = ⇢t ,weseethat (tst )=Tr(⇢tst )=Tr(⇢ts⇢t )=Tr(⇢t ⇢ts)= Tr(⇢s)= (s).
Proposition 1.2.2. If ⇢ : G GL(V ) and ⇢ : G GL(V ) are representations 1 ! 1 2 ! 2 of a group G with characters 1 and 2, respectively, then the character of the representation ⇢ = ⇢ ⇢ is = . 1 2 1 2 Proof. Since every matrix ⇢s of the representation ⇢ is in block diagonal form with upper block ⇢1(s)andlowerblock⇢2(s), the trace of ⇢s is simply the sum of the traces of its blocks. The following lemma and its corollaries facilitates much of the subsequent theory. Lemma 1.2.3 (Schur’s Lemma). Let ⇢1 : G GL(V ) and ⇢2 : G GL(V ) be ! 1 ! 2 irreducible representations of G, and let f : V1 V2 be a linear transformation for which ⇢2f = f⇢1 for all s G. Then the following! statements hold: s s 2 1. If ⇢1,⇢2 are not isomorphic representations, f =0.
1 2 2. If V1 V2 and ⇢ ⇢ , then f = I for some C. ' ' 2 Proof. 1. Suppose f =0.LetW1 =Ker(f) V1,andW2 =Im(f) V2. 6 ⇢ 2 1 ⇢ If w W1,thenforanys G we have 0 = ⇢sfw = f⇢sw,andhence 1 2 2 ⇢sw Ker(f)=W1.SoW1 is G-invariant (since f =0andthusitskernelis 2 6 2 2 nontrivial). Similarly, if w0 = fw W2,foranys G we have ⇢sw0 = ⇢sfw = f⇢1w Im(f)=W .SoW is G-invariant2 as well.2 Since ⇢1,⇢2 are irreducible, s 2 2 2 this means that either W1 =0andW2 = V2,orthatW1 = V1 and W2 =0. The latter is impossible since f =0,sowehavethatf is both injective and 6 2 1 surjective, and hence an isomorphism of vector spaces. Since ⇢sf = f⇢s for any s G, ⇢1,⇢2 are isomorphic representations of G. 2 1 2 2. Suppose V1 = V2 = V and ⇢ = ⇢ = ⇢.Weknowf has at least one eigenvalue, since V is a vector space over an algebraically closed field (C). Denote one such eigenvalue by ,andnotethatthelineartransformationg = f I has nontrivial kernel W . So if w W and s G,weseethat0=⇢ gw = g⇢ w, 2 2 s s so that ⇢sw Ker(g)=W ,andthereforeW is G-invariant. Since V is irreducible, W2is trivial or the whole space. But we know it is not trivial. So W = V ,andthereforeg =0,sothatf = I.
Corollary 1.2.4. Let h : V V be a linear map from a vector space V of 1 ! 2 1 dimension n to a vector space V2 of dimension m, and define
1 2 1 h0 = ⇢ 1 h⇢ , G s s s G | | X2 6 where ⇢i : G GL(V ) are representations. Then the following hold: ! i 1 2 1. If ⇢ ,⇢ are non-isomorphic representations, then h0 =0.
1 2 2. If V1 = V2 = V and ⇢ = ⇢ = ⇢, then h0 =(Tr(h)/dim(V ))I, where I is the identity on V .
1 2 3. Writing ⇢s =(ri1j1 (s))i1,j1 1,...,n (an n n matrix), ⇢s =(ri2j2 (s))i2,j2 1,...,m 2{ } 2{ } (an m m matrix), and h =(h ) (an⇥ m n matrix), so that we write ⇥ i2i1 ⇥ h0 =(hi02i1 ) where
1 1 h0 = r (s )h r (s), i2i1 G i2j2 j2j1 j1i1 s G,j1,j2 | | 2X 1 2 then if ⇢ ,⇢ are non-isomorphic representations, and i1,i2,j1,j2 are arbitrary indices, 1 1 r (s )r (s)=0. G i2j2 j1i1 s G | | X2 1 2 4. Given the same notation as above, if V1 = V2 = V and ⇢ = ⇢ = ⇢, we have that
1 1 i2i1 j2j1 1/dim(V ) if i1 = i2 and j1 = j2 r (s )r (s)= = G i2j2 j1i1 dim(V ) s G (0 otherwise. | | X2
Proof. 1. Since h is linear, and h0 is a linear combination of linear transforma- 2 1 tions, it is itself linear. Also, it satisfies ⇢sh0 = h0⇢s for all s G,sincethesum is symmetric in the elements of G.SobythefirstresultofSchur’sLemma,if2 1 2 ⇢ ,⇢ are non-isomorphic, then h0 =0.
1 2 2. If V1 = V2 = V and ⇢ = ⇢ = ⇢,thesecondresultofSchur’sLemmatellsus 1 that h0 = I for some C. Hence since Tr(⇢s h⇢s)=Tr(h)foranys G, we have that 2 2 1 1 Tr(h0)= Tr(⇢ 1 h⇢ )= Tr(h)=Tr(h). G s s G s G s G | | X2 | | X2
But since h0 = I,weseethat =Tr(h0)/dim(V )=Tr(h)/dim(V ).
1 2 3. By the first result of this corollary, if ⇢ ,⇢ are non-isomorphic, h0 =0,and
thus hi02i1 =0forallsuchindices.Sinceitisalinearcombinationofthe
entries hj2j1 ,andvanishesforanychoiceofh,weseethatthecoe cients 1 in this linear combination, s G ri2j2 (s )rj1i1 (s) for all j1 1,...,n and 2 2{ } j2 1,...,m , vanish as well. Since this holds for each choice of (i1,i2), we obtain2{ the result.} P
1 2 4. The second result of this corollary gives us that if ⇢ = ⇢ = ⇢ and V1 = V2 =
V ,thenh0 =(Tr(h)/dim(V ))I, and thus hi02i1 =(Tr(h)/dim(V )) i2i1 ,where =1ifi = i and is zero otherwise, for i ,i 1,...,n ,andwenote i2i1 2 1 1 2 2{ } 7 that Tr(h)= x ,sincetheo↵-diagonalentriesofh do not j1,j2 1,...,n j2j1 j2j1 contribute to the trace.2{ Therefore:} P 1 1 r (s )x r (s)=h0 G i2j2 j2j1 i1j1 i2i1 s G,j1,j2 1,...,n | | 2 X2{ } 1 = x . dim(V ) i2i1 j2j1 j2j1 j1,j2 1,...,n 2X{ } Therefore i2i1 j2j1 1 1 = r (s )r (s) dim(V ) G i2j2 i1j1 s G | | X2 for all choices of j1,j2 1,...,n .Sincethisholdsforallchoicesofi1,i2 1,...,n ,weobtaintheresult.2{ } 2 { }
1.2.2 Irreducible characters and the orthogonality relations We will now use the above-deduced tools to prove a number of surprising facts about irreducible characters (that is, the characters of irreducible representations). We will see that most of the interesting information about a given irreducible represen- tation is in a sense stored in its character, and also that the irreducible characters of a group must follow a strict set of guidelines, making them exceedingly pleasant to work with. First we define a Hermitian product on functions ', : G C: ! 1 (', )= '(s) (s), (1.1) G s G | | X2 and a similar symmetric product:
1 1 ', = '(s) 0(s ), h i G s G | | X2 1 and note that if we define the function 0(t)= (t ), then (', )= ', .Recall 1 h i that if 1, 2 are characters of a representation, then 2(s )= 2(s)(Proposition 1.2.1). Therefore, ( 1, 2)= 1, 2 ,sowecanregardtheseinnerproductsas identical. h i
Theorem 1.2.5 (Orthogonality Relations). If , 0 are irreducible characters of a group G, then they are orthonormal:
1 , 0 are characters of isomorphic representations ( , 0)= (0 otherwise.
Proof. Let ⇢,⇢0 be the representations whose characters are , 0,respectively.If
⇢s =(rij(s))i,j 1,...,n and ⇢s0 =(rkl0 (s))k,l 1,...,m ,then 2{ } 2{ }
1 1 ( , 0)= , 0 = r ,r0 = r (s)r0 k(s ). h i h ii kki G ii k i 1,...,n k 1,...,m i,k s G 2{X } 2{X } X | | X2 8 By the third result of Corollary 1.2.4, if ⇢,⇢0 are non-isomorphic representations, then each term of the outer sum is equal to zero, and hence ( , 0)=0.If⇢,⇢0 are isomorphic, then = 0,sothefourthresultofCorollary1.2.4revealsthefinalsum to be ( , 0)= , = r ,r h i h ii jji i,j 1,...,n 2X{ } 1 1 ij = r (s)r (s )= =1. G ii jj dim(V ) i,j 1,...,n s G i,j 1,...,n 2X{ } | | X2 2X{ }
The condition of orthonormality of a group’s irreducible characters allows us now to see that any group representation has a unique direct sum decomposition into irreducible subrepresentations (to up isomorphism of said subrepresentations):
Theorem 1.2.6. If ⇢ : G GL(V ) is a linear representation with character such that ! ⇢ = ⇢ ⇢ 1 ··· n for irreducible representations ⇢i, i =1,...,n, then for any irreducible representa- tion ⇢0 : G V with character 0, if n is the number of ⇢ isomorphic to ⇢0, then ! i i ni =( , 0).
n n Proof. By Proposition 1.2.2, = i=1 i,andtherefore( , 0)= i=1( i, 0). th Theorem 1.2.5 gives us that the i summand is nonzero if and only if ⇢i is isomorphic P P to ⇢0,andsothesumgivesuspreciselythenumberni of ⇢i’s isomorphic to ⇢0. Corollary 1.2.7. Two representations are isomorphic if and only if they have the same character.
n Proof. If ⇢,⇢0 are two representations with the same character = i=1 i,which decompose as ⇢ = ⇢1 ⇢n and ⇢0 = ⇢0 ⇢0 ,thenifnij is the number ··· 1 ··· n0 P of ⇢i isomorphic to ⇢j0 ,wehavebyTheorem1.2.6thatnij =( , j0 ), which is ex- actly the number of times j0 occurs in the sum decomposition of ,since⇢,⇢0 have the same character. So the two representations have each irreducible subrepresen- tation occur the same number of times in their direct sum decompositions (up to isomorphism of the subrepresentations), and are therefore isomorphic (note that in Theorem 1.2.6, the numbers ni are indepdendent of the particular decomposition into irreducible subrepresentations ⇢i). The converse is clear since trace is preserved under isomorphisms. With the above results, we can restrict our attention entirely on the characters of representations, since they uniquely determine the isomorphism class of their rep- resentations. To this end, we derive some further qualities of irreducible characters that allow us to work freely with them. Before we proceed, we note a useful criterion for irreducibility of a representation which follows directly from our preceding results.
9 Proposition 1.2.8. Let ⇢ : G GL(V ) be a representation with character . Then ( , ) is an integer, and is! equal to one if and only if ⇢ is an irreducible representation of G.
Proof. By Theorem 1.2.6, ⇢ = n1⇢1 nk⇢k for irreducible representaitons ⇢k. ··· k 2 By the orthogonality relations, we have that ( , )= i=1 nk,whichisaninteger. Furthermore, it equals one precisely if ⇢ is equal to one copy of one of the irreducible representations, in other words that ⇢ is irreducible. P Recall the regular representation ⇢G from section 1.1.2. The following proposition examines its character to see that every irreducible representation of G is contained within it as a subrepresentation, which will in turn yield some of our nice facts about irreducible characters. Proposition 1.2.9. Let G be a finite group, and G be the character of its regular representation. Then for s G, 2 G s =1 G(s)= | | (0 otherwise.
Furthermore, if ⇢i is an irreducible representation of G with character i, then ⇢i is G contained in the direct sum decomposition of ⇢ exactly ni = i(1) times. That is, we can write ⇢G = n ⇢ n ⇢ . 1 1 ··· m m Proof. The regular representation has degree G ,sincethebasis et t G of the 2 target vector space is indexed by the elements| of |G.So G(1) = G{ (Proposition} G | | 1.3). Then if s =1,since⇢s et = est for any t G,weseethatthediagonalof G 6 2 the matrix ⇢s must have all zeros, since it does not fix any of the basis elements es.Thereforeitstraceiszero.ThesecondclaimfollowsfromTheorem1.2.6,since G G ni =( , i)= i(1), using the form for we just proved.
Corollary 1.2.10. If G is a finite group with irreducible characters i, i =1,...,n, then we have the following: 1. n 2(1) = G i=1 i | | 2.P For any s G other than the identity, n (s) (1) = 0. 2 i=1 i i G n Proof. By Proposition 1.2.8, for any s G,wehaveP (s)= i=1 i(1) i(s). If s = 1, then we obtain that 2 P n G = G(1) = 2(1), | | i i=1 X and if s = 1, then we obtain that 6 n G 0= (s)= i(1) i(s). i=1 X
10 We now determine the number of isomorphism classes of irreducible representa- tions for a given group G.RecallthatwecanpartitionagroupG into conjugacy 1 classes C ,...,C ,whichsharethepropertythatifx C ,thensxs C for 1 m 2 i 2 i any s G.Ifafunctionf : G C is constant on conjugacy classes, then we call it 2 ! 1 a class function.Thatis,f is a class function if f(sxs )=f(x)forallx, s G. 2 Note that the set G of class functions of G forms a vector space over C,andthat H the irreducible characters 1,..., n of G are class functions. We will show that, if we equip G with the Hermitian inner product (1.1), the irreducible characters are an orthonormalH basis for . HG Lemma 1.2.11. If f G and ⇢ : G GL(V ) is an irreducible representation of G with character , define2H⇢ : V V !by f !
⇢f = f(s)⇢s. s G X2 Then ⇢f = I, where 1 G = f(s) (s)= | | (f, ). dim(V ) dim(V ) s G X2 1 1 1 Proof. For any s G,weseethat⇢s ⇢f ⇢s = t G f(t)⇢s ts = u G f(sus )⇢u = 2 2 1 2 u G f(u)⇢(u)=⇢f ,wherewesubstitutedu = s ts.So⇢f ⇢s = ⇢s⇢f ,andwe satisfy2 the conditions for the second statementP of Schur’s LemmaP (Lemma 1.2.3). P Therefore ⇢f = I for some C.But 2
dim(V ) =Tr( I)=Tr(⇢f )= f(s)Tr(⇢s)= f(s) (s). s G s G X2 X2 1 Therefore = dim(V ) s G f(s) (s), as desired. 2 Theorem 1.2.12. TheP irreducible characters 1,..., n are an orthonormal basis of , the space of class functions of G. HG Proof. By Theorem 1.2.5, we have that the irreducible characters are pairwise or- thonormal. We proceed to show that they span the space of class functions HG of G.LetW =Span( ,..., ). Then = W W ?.Letf W ?,sothat 1 n HG 2 (f, i)=0foralli =1,...,n. Now for a given representation ⇢ : G GL(V ), we can decompose it into a sum of irreducible subrepresentations: ⇢ = !⇢1 ⇢n. i i i i ··· Then if ⇢f = s G f(s)⇢s as in Lemma 1.2.10, we have that ⇢f = I =0(since 1 2 (s)= (s ), and f was assumed to be orthogonal to all the irreducible characters), P and therefore ⇢f = s G f(s)⇢s =0aswell. In particular, applying2 this to the regular representation ⇢G of G gives us that P 0=⇢f e1 = s G f(s)⇢se1 = s G f(s)es,sothatthecoe cientsf(t)inthe 2 2 linear combination are zero. Therefore f =0,sothatW ? = 0 ,andfinally P P { } G = W .Sothespanoftheirreduciblecharactersgeneratesthewholespaceof classH functions. Corollary 1.2.13. The number of irreducible representations of G is equal to the number of conjugacy classes of G.
11 Proof. By Theorem 1.2.11, the space G of class functions of G has dimension equal to n,thenumberofirreduciblecharactersofH G.Ontheotherhand,since class functions are constant on each of the conjugacy classes C1,...,Cm of G,we see that the space of class functions has dimension m,sincewecanrepresenteach class function by an m-vector of its values on the m conjugacy classes. So n = dim( )=m. HG 1.2.3 The group algebra In this section, we discuss an alternate and equivalent way to view group represen- tations, which will assist us in deriving some facts about character tables in the following section, and also enrich our analysis in the next chapter. If G is a finite group and K is a commutative ring, we define the group algebra of G over K to the set K[G]oflinearcombinationsx = s G ass,foras K,where 2 2 1 if x0 = s G bss K[G]thenx + x0 = s G(as + bs)s and xx0 = s,t G(btat s)s, so that addition2 and2 multiplication are directly2 inducedP by that in K and2 in G.We see thatP it is an algebra, since it is a ringP as well as a (finitely-generated)P K-module with basis G (therefore of dimension G ). | | For now, let us take K = C.SinceC[G] is a ring, we can define modules on it. In particular, if we have a linear representation ⇢ : G GL(V ), where V is ! acomplexvectorspace,thenforanyx = s G ass C[G]andv V ,wecan 2 2 2 let xv = s G as⇢sv,andseethatthisdefinitionofscalarmultiplicationdefinesa 2 P C[G]-module over V .WeseethatthismoduleV is finitely-generated, since if we choose aP basis = v ,...,v of the original V (when thought of as a complex { 1 n} vector space) every element of (the C[G]-module) V is a linear combination of the basis elements vi by elements of the group algebra C[G]. On the other hand, if we’re given an arbitrary finitely-generated C[G]-module V ,let = v1,...,vn be a basis of V ,andnotethatforanys, t G,wehave { } 1 1 2 that [st] =[s] [t] ,andthereforethat[s] [s ] =[ss ] =[1] ,then n identity matrix. This shows that the matrix of s under the basis ,forevery⇥s G,is invertible, so that by choosing this basis we have fixed a linear representation2 of G of degree dim V . Therefore, we see that we can speak equivalently of finitely-generated C[G]- modules and linear representations of G in C-vector spaces. In fact, all of the theory developed thus far can be expressed in the language of C[G]-modules, and though we will not reprove most things (see [4] for a thorough treatment of the general theory), we will make use of this language when it provides clarity or an additional perspective. We call an algebra A semisimple if every left A-module V is semisimple, which simply means that V can be written as a finite direct sum of irreducible submodules V = V1 Vn (in alignment with our definitions for representations, a module is irreducible ··· if it has no proper nonzero submodules). The following lemma is essentially a generalization of Maschke’s theorem (Theorem 1.1.2) in the language of K[G]-modules.
Lemma 1.2.14. If K has characteristic zero, then the group algebra K[G] is semisim- ple.
12 Proof. The proof is essentially the same as our proof of Lemma 1.1.1 (which fa- cilitated our original version of Maschke’s theorem): if W is a submodule of a K[G]-module V , then let p be a projection of V onto W which is K-linear. Then we can create a new projection 1 p0 = ⇢ p ⇢ 1 G s · · s s G | | X2 by averaging the conjugated projection p under all the elements of G.Thisprojection is K[G]linear,andthereforewecanfactorizeV into a direct sum of W and the kernel of p0. Using induction on dimension, as we did for Maschke’s theorem, we see that every K[G]-module is semisimple, or in other words that K[G]isasemisimple algebra.
1 2 Now, if we have any representations ⇢ : G GL(V1)and⇢ : G GL(V2)over afieldK,weletthespaceofintertwining operators! be defined as!
Hom (V ,V )= f Hom (V,W):⇢1 f = f ⇢2,s G . G 1 2 { 2 K s s 2 } 1 2 In the case where V1 = V2 and ⇢ = ⇢ we write HomG(V,V )=EndG(V ), the set of G-covariant endomorphisms of V .Equivalently,formodulesM,N over a group algebra K[G], we can define Hom (M,N)= f Hom (M,N):⇢1f = f⇢2,s G { 2 K s s 2 G ,andEndG(M)isdefinedsimilarly. } AringK is a division ring (sometimes called a skew field)ifeveryelement of K has a multiplicative inverse. Note that it di↵ers from a field only in that it needn’t be a commutative ring (but clearly every field is also a division ring). Now note that the statement of Schur’s lemma is equivalent to saying that if ⇢1,⇢2 are irreducible representations, then f HomG(V1,V2)iseither0oranisomorphism. It follows immediately from the invertibility2 of every nonzero intertwining operator that EndG(V )isadivisionringifourchosen⇢ : G GL(V )isirreducible. We take for granted the following corollary of the! structure theorem of semisimple algebras, which decomposes semisimple algebras into a product of matrix rings over the endomorphism rings of the components of the semisimple algebra:
Theorem 1.2.15 (Artin-Wedderburn). If K[G] V1 Vn is a semisimple group algebra over a division ring K, then K[G] can' be decomposed··· into a product of matrix rings Mn(Ki), where the matrices are n n and their entries come from ⇥ the division rings Ki = EndG(Vi):
N
A = Mni (Ki). i=1 Y In particular, we might be interested in the case K = C,whichcorresponds to the theory of finite group representations we have developed. First note that since C[G]isanalgebra,wecanconsideritasa G -dimensional module over itself. | | Then we can index a basis = es s G of C[G]bytheelementsofG,andseethat 2 the set of G G matrices of{ the} elements of G under this basis are is in fact the regular| representation.|⇥| | So in the correspondence we developed earlier between
13 finitely-generated C[G]-modules and representations of G over C, this tells us that C[G](asamoduleoveritself)correspondstotheregularrepresentation⇢G. Now note the following fact.
Proposition 1.2.16. C is the only division ring of finite degree over C.
Proof. Let V be a division ring of finite dimension over C,andletv V .Then 2 there is some complex polynomial in v with coe cients in C which evaluates to zero. But since C is algebraically closed, this polynomial factors completely, and therefore one of its degree-one factors (v zi)isequaltozeroforsomezi C.Therefore 2 v C,sothatV = C. 2 Therefore C[G]decomposesintoafiniteproductofmatrixrings
N
C[G]= Mni (C). i=1 Y Furthermore, the number N of factors in the decomposition is the number of dis- tinct irreducible representations of G over C,andthedimensionni is equal to th the dimension of the i irreducible representation ⇢i of G.Toseethis,welet G ⇢ = n1⇢1 nN ⇢N be the decomposition of the regular representation of G into irreducible subrepresentations,··· where ⇢ : G GL(V ). The ring of endomorphisms i ! i of each Vi is isomorphic to Mmi (C), where mi =dimVi,andsotherepresentations ⇢i,bylinearlyextendingthemandtakingthemalltogetherasahomomorphism from C[G]totheCartesianproductoftheendomorphismringsoftheVi gives us the same factorization for C[G]asabove,exceptweknowthatN is the number of irreducible representations, and ni is the degree of the irreducible representation ⇢i. Note that this factorization of the group algebra is directly analogous to the de- composition of the regular representation of G into the direct sum of all irreducible representations of G,asseeninProposition1.2.8. In chapter two, we will consider the somewhat more intricate case in which we take K = R.SinceourgeneralizedMaschke’stheorem,Lemma1.2.13,holdsfor any commutative ring of zero characteristic, we see that R[G]issemisimple,and therefore that Theorem 1.2.14 allows us to deconstruct this group algebra into a product of matrix rings over division rings of finite degree over R.Wewillseelater that there are only three such division rings, and that this decomposition provides us with new information about the nature of a group’s irreducible representations.
1.2.4 Character tables: some more theory Following our result in Corollary 1.2.12 that the number of irreducible characters is equal to the number of conjugacy classes, we introduce the notion of a character table,asquarematrix G whose ijth entry G is the value that the irreducible C Cij character i takes on the conjugacy class Cj of G. As Corollary 1.2.7 reduced our study of representations of G to the study of its irreducible characters, our task has been reduced to filling out the character table, and investigating what sort of information we can deduce about G from knowing its values.
14 To the first end, we have already found a number of useful tools. By Corollary 1.2.9, the squares of the degrees of the representations sum to G .ByTheorem 1.2.5, the rows of G are pairwise orthonormal. We will now proceed| | to show that the columns of theC character table are pairwise orthogonal, and that the degrees of the representations divide G . | | Proposition 1.2.17. The columns of G are pairwise orthogonal. C Proof. Consider a modified character table T where we multiply each entry in col- umn k (the values of the characters on conjugacy class k)by Ck / G .Soif g C is a representative element of each conjugacy class k =1,...,n| | |,thenwe| k k p see2 that from our orthogonality relations on the rows,
1 1 n n = (s) (s)= C (g ) (g )= T T . ij G i j G | k| i k j k ik jk s G k=1 k=1 | | X2 | | X X This shows that the rows of T are pairwise orthonormal under the standard Hermi- n tian dot product C .Itfollows,then,thatT ⇤T = I = TT⇤,i.e.thatT is a unitary matrix and thus its columns are also pairwise orthonormal. This means that for any two conjugacy classes Ci,Cj, we have
n n C C (g ) (g ) = T T = | i|| j| k i k j , ij ki kj G k=1 k=1 p X X | | and therefore that n G k(gi) k(gj)= | | ij. Ci Cj Xk=1 | || | p We will now prove that the degree of an irreducible representation divides the size of the group (adapting the proofs from [7, chapter 6] and [4, chapter 22]). This will require some facts about the center of the group algebra, as well as some elementary facts about algebraic integers. A complex number is an algebraic integer if it is the root of a polynomial with integer coe cients.
Lemma 1.2.18. We have the following facts about algebraic integers.
1. The set of algebraic integers forms a subring of C. 2. If is a character of a finite group G, then for every element s G, the number (s) is an algebraic number. 2
3. If a number is both an algebraic integer and a rational number, then it is an integer.
Proof. 1. We will first show that a complex number x is an algebraic integer if and only if the subring Z[x]ofC consisting of polynomials in x with integer coe cients is a finitely-generated Z-module.
15 Now, for any two algebraic integers x, y,weknowthattheringsZ[x], Z[y]are finitely-generated as Z-modules. But then clearly the tensor product Z[x] ⌦ Z[y]isfinitely-generated,sinceitcanbegeneratedbythetensorproducts of the generators of its factors. Therefore the ring Z[x, y]=Z[x][y]isalso finitely-generated, since we can map the tensor product into C surjective onto this ring by sending its generators to the products of the powers of x, y.But then every element z Z[x, y]isinafinitely-generatedZ-module contained 2 in C which itself contains both Z[z]. Since Z is a noetherian ring (i.e. every ideal is finitely-generated - in the case of Z,eachidealcanbegeneratedby one number), this is equivalent to saying that Z[z]isafinitely-generatedZ- module. By our first result here, this means that z is an algebraic integer. Since Z[x, y]containsx + y as well as the product xy,thisshowsthatthe algebraic integers form a ring.
2. First note that if is the character of a representation ⇢,then (s)isthesum of the eigenvalues 1,..., k of the matrix ⇢s.Letms be the degree of s in m G;thatis,theleastintegersuchthats =1.Thensince i is an eigenvalue ms ms of ⇢s,wehavethat i is an eigenvalue of ⇢s = I,theidentitymatrix.So ms i =1,andtherefore i is a root of unity. So the value (s)isasumof roots of unity. Clearly the roots of unity are algebraic integers, since they are roots of polynomials of the form xp 1. But since the algebraic integers form a ring, a sum of roots of unity is also an algebraic integer, and therefore (s) is an algebraic integer.
3. Let x Q be an algebraic integer. Suppose that x/Z.Thenwecanwrite x as a2 reduced fraction x = p/q,wherep and q are relatively2 prime integers. Since x is an algebraic number we have some a0,...,an 1 Z such that n n 1 n 2 x + an 1x + + a0 =0,or,mutiplyingthroughbyq ,that ··· n n 1 n 1 n p + an 1qp + + a1q p + a0q =0. ··· But since p =0(weassumedx/Z), this means that q divides pn,contradict- 6 2 ing that they are relatively prime. So x Z. 2
Lemma 1.2.19. If : G C is an irreducible character of a finite group G with conjugacy classes C ,...,C!, and g C for each i =1,...,m, then 1 m i 2 i (g ) = C i i | i| (1) is an algebraic integer.
Proof. For each conjugacy class Ci,let i = s C s,anelementofthegroupalgebra 2 i C[G]. Then for the irreducible C[G]-module U (which we know is a submodule of P C[G]byourdiscussionoftheregularrepresentationinsection1.2.3)withcharacter (that is, the module corresponding to the representation whose character is ), we have that u = u for every element u U. i i 2
16 To see why, we first note that i is in the center of the group algebra C[G]for each i (since conjugation by any group element fixes the element i). Then for any element t C[G], ut i = u it,andthereforef : U U defined by f(u)=u i is a homomorphism2 such that ⇢ f = f⇢ for any s G.ThereforeSchur’slemmatellsus! s s 2 that it is a multiple of the identity transformation. Thus u i = ⌘u for some ⌘ C. 2 But picking a basis of U and taking its trace gives us that s C (s)=⌘ (1), so 2 i that ⌘ = Ci (gi)/ (1) = ,asdesired. | | P In general, if we enumerate the elements of G as s1,...,sn,andwehavean n element r = i=1 aisi C[G]forintegersai,thenitsproductwiththegroup n 2 1 elements si, sir = j=1 bijsj with bij = al,wheresl = si sj,isaninteger. P 1 Our elements are of this form, where b =1whens s C ,andb =0 Pk ij i j 2 k ij otherwise. Therefore we see that the statement that u k = u implies that u is an n eigenvector of eigenvalue of the matrix (bij)ofthebij for which si k = j=1 bijsj, which are integers. So is an eigenvalue of an integer matrix, i.e. a root of its characteristic polynomial, which has all integer coe cients (since its coe Pcients are products and sums of the matrix entries). Therefore is an algebraic integer.
Proposition 1.2.20. Let : G C be an irreducible character of a finite group G. Then (1) divides G . ! | | Proof. Enumerate the conjugacy classes of G as C ,...,C ,andletg C for 1 m i 2 i each i =1,...,m (that is, any element of the conjugacy class Ci). Then since the irreducible characters are orthonormal and the conjugacy classes partition G,we have that m G = C (g ) (g ), | | | i| i i i=1 X and therefore that G m C (g ) (g ) | | = | i| i i . (1) (1) i=1 X 1 Since Ci (gi)/ (1) (Lemma 1.2.18) and (gi)= (gi )(Lemma1.2.17(2))are algebraic| | integers for i =1,...,m,thewholesumisanalgebraicinteger,sincethe algebraic integers form a ring (Lemma 1.2.17(1)). Therefore G / (1) is an algebraic integer. Since it is also a rational number, it is an integer| (Lemma| 1.2.17(3)). So (1) divides G . | | 1.2.5 Some examples of character tables
To get started, we return to the example of S3 D3,thesmallestnon-abeliangroup, which we began to explore in section 1.1.2. Let' : S 1 be the character of the 1 3 !{ } trivial representation. Recall the sign representation ⇢sgn : S 1 ,whichsends a permutation to its sign. We can now conclude that it is not!{± isomorphic} to the trivial representation, since their characters di↵er on the conjugacy classes of odd permutations (that is, sign 1). Furthermore, ⇢ is irreducible since it has degree sgn one. Denote its character be 2.Sowehaveobtainedtwoirreduciblecharactersof degree one. S3 has three conjugacy classes: C1 = id , C2 = (123), (132) ,and C = (12), (13), (23) .ByCorollary1.2.12,weknowthat{ } S has{ three irreducible} 3 { } 3 17 representations. By Corollary 1.2.9, 3 2(1) = G =6,andtherefore 2(1) = i=1 i | | 3 4, so that 3(1) = 2, and we obtain that S3 has one irreducible representation of degree two. Using Proposition 1.2.13,P we can obtain the entries in the third row (corresponding to 3)bynotingthatthesecondandthirdcolumnsareboth orthogonal to the first. Therefore, we have obtained the full character table:
S3 C1 C2 C3 1 111 2 11-1 3 2-10
Next we will find the character tables of D4, the dihedral group of 8 elements, and the quaternion group Q8,thesmallestnon-abeliangroupslargerthanS3. First, 4 2 2 D4 = x, y x = y = yxyx =1 has five conjugacy classes: C1 = 1 , C2 = x , C = h y, x| 2y , C = x, x3 ,andi C = xy, x3y ,andthereforebyCorollary{ } { } 3 { } 4 { } 5 { } 1.2.12, we know it has five irreducible representations, where we let 1 =Tr(⇢1)be 5 2 the trivial representation as usual. By Corollary 1.2.9, i=1 i (1) = G =8,so that 5 2(1) = 7. The only way to add four squares to make seven| is| to have i=2 i P 2(1) = 3(1) = 4(1) = 1, and 5(1) = 2. So we have filled out the first row and first columnP of the table. Note that the elements of C2,C3,C5 have order 2, and those of C4 have order 4. Therefore their values under the representations must have orders dividing those orders. So our degree one representations must map x 1, 1, i, i ,sincethese are the fourth roots of unity, and y 1 , since these7! { are the only} complex numbers of order 2. If x 1, then if y 7! {±1, we} obtain a very simple representation 7! 7! ⇢2 which is not isomorphic to ⇢1,sinceitscharactervaluesdi↵er.Sowehavefilled out the second row as D4 C1 C2 C3 C4 C5 2 11-11-1. 2 Now if ⇢3(x)= 1, then ⇢3(x )=1.If⇢3(y)=1,weobtainanotherrepresentation whose character sends C ,C to 1, and C ,C to 1. Note that this is not isomorphic 2 3 4 5 to ⇢1,⇢2,sincetheircharactersdi↵er:
D4 C1 C2 C3 C4 C5 3 111-1-1.
If ⇢4(x)= 1and⇢4(y)= 1, we obtain yet another representation whose character sends C ,C to 1 and C ,C to 1: 2 5 3 4
D4 C1 C2 C3 C4 C5 4 11-1-11.
Next if ⇢(x)= i for some degree one representation ⇢, then we would have that ⇢(x2)=⇢(x)2 =± 1, which is impossible since if ⇢(y) = 1 then we would have 1 = (y)= (x2y) = (x2) (y)= 1, and if ⇢(y)= 1thenwewouldsimilarlyhave 1= (y)= 6 (x2y) = (x2) (y)=1.Sothissituationisimpossible,confirming 6
18 that we have found all the right characters of degree one:
D4 C1 C2 C3 C4 C5 1 11111 2 11-11-1 3 111-1-1 4 11-1-11 5 2
This leaves our degree two character 5, whose values can be found by column orthogonality to the first column using Proposition 1.2.13. By comparison with column one, we easily see that (C )= 2, and (C )=0forj =3, 4, 5: 5 2 5 j
D4 C1 C2 C3 C4 C5 1 11111 2 11-11-1 3 111-1-1 4 11-1-11 5 2-20 0 0
1.2.6 Motivation: the character table of the quaternion group
In this section, we compute the character table of the quaternion group Q8 = 1, i, j, k ,wherei2 = j2 = k2 = 1, ij = k, jk = i,andki = j.Weseethat {± ± ± ± } Q8 has five conjugacy classes: C1 = 1 , C2 = 1 , C3 = i , C4 = j ,and C = k ,andthusfivedistinctirreduciblecharacters,with{ } { } {± } again{± the} trivial 5 {± } 1 character. Applying Corollary 1.2.9 exactly as with D4,weobtainthatthereare four degree one characters 1,...,4,andonedegreetwocharacter 5.Since i, j, k have order 4, they must all be mapped into 1, i by the degree one± represen-± ± tations, and this choice will determine the character{± ± } (since i2 = j2 = k2 = 1). If (i)=i (resp. (i)= i), then since i( i)=1,wemusthave ( i)= i (resp. ( i)=i). But this is impossible since characters are class functions. So we must have (i) 1 for all degree one characters (and the same holds for (j), (k), by symmetry).2{± If (}i)=1,then ( 1) = (i2)=1aswell,andtoavoidbeingthetriv- ial character let us put (j)= 1. Then we have (k)= (ij)= (i) (j)= 1. So if one of the i, j, k are sent to one, the other two are sent to 1 (again by symmetry). But this determines three distinct non-trivial degree one characters:
Q8 C1 C2 C3 C4 C5 1 11111 2 11-11-1 3 111-1-1 4 11-1-11 5 2
Note that we cannot have (i)= (j)= (k)= 1, since then ( 1) = (ijk)= 1, which implies that 1= (i) = ( i)= ( 1) (i)=1,whichisacon- tradiction since characters are class functions.6 This confirms that our degree one
19 characters are correct. At this stage, we can fill in the bottom row by column orthogonality:
Q8 C1 C2 C3 C4 C5 1 11111 2 11-11-1 3 111-1-1 4 11-1-11 5 2-20 0 0
Note that this character table is identical to the one we obtained for D4 at the end of the previous chapter. However, it is clear that D4 and Q8 are non-isomorphic 2 2 groups: D4 has three elements of order two (x ,y,x y), while Q8 only has one ( 1; the rest of the non-identity elements have order 4). Thus this example shows us that while character tables do uniquely determine the isomorphism classes of a group’s irreducible representations (Corollary 1.2.7), they are not in general strong enough to distinguish between isomorphism classes of groups. We see that any representation of either D4 or Q8 constructed from their degree one representations will be the same, since the degree one characters are themselves representations. This reduces the representation-theoretic distinction between these two groups to the di↵erence between their irreducible degree two representations. Just how di↵erent are they? The answer to this question will lead us to explore fur- ther how much information we can obtain about a representation simply by knowing its character.
20 Chapter 2
Real representations and the Frobenius-Schur indicator
In this chapter, we will further examine the depth of information one can derive about a group and its representations from its irreducible characters. As we saw in the previous chapter, representations of a group are isomorphic precisely if their characters are equal, but that in general non-isomorphic groups can have identical sets of characters. We will see that there are other ways to distinguish groups by their representations that may not carry over to the groups’ characters. In particular, we can decide, given a group’s character, whether it is possible to find a representation of the group over the real numbers which takes this character. Interestingly, we will also see that this criterion can be computed simply by looking at the groups’ character tables and conjugacy class structure.
2.1 Real, complex, and quaternionic representa- tions
Let ⇢ : G GL(V ) be a representation of a finite group G in a complex vector space V .Onewaytofurtherclassify! ⇢ is to determine whether there is a basis of V such that, for all s G,thematrix⇢s under this basis has only real-valued entries. If this is the case, we2 call ⇢ a real representation. Note that if ⇢ is a real representation, then its character is certainly real-valued, since trace of a linear transformation is independent of the choice of basis, and therefore if there is some real-valued matrix A = ⇢s,then s =Tr(A) R.If⇢ has a complex-valued character, then we call ⇢ a complex representation2 .Clearlyacomplex-valuedcharactercannotbethe trace of a matrix with real-valued entries, so our definition is consistent (i.e. no representation can be both real and complex). We can ask if the converse is true: if we have a real-valued character of some finite group G,isitnecessarilythecharacterofarealrepresentation⇢?Theanswer, as we shall see, is negative; there are groups with real-valued characters whose representation cannot be taken as real-valued. If ⇢ has real-valued character but is not a real representation, we say that it is a quaternionic representation,for reasons that will be made clear shortly. In particular, the theory we develop below
21 will give us the distinction we were looking for at the end of the previous chapter: the degree two irreducible representation of D4 is real, while that of Q8 is quaternionic. In the following section we will prove a simple computational criterion for classi- fying complex group representations as real, complex, or quaternionic, which solely depends on the group’s characters. Afterwards, we will see how this classification informs the relationship between the decomposition of the complex group algebra C[G] into irreducible sub-modules (i.e. the decomposition of the regular represen- tation into irreducible subrepresentations), and the corresponding decomposition of the real group algebra R[G] into irreducible submodules. Finally, we will work through some simple examples to illustrate the theory.
2.2 The Frobenius-Schur indicator
In this section, we will prove the following theorem, which allows us to classify a complex character as real, complex, or quaternionic by a simple computation. The proof vaguely follows that by Serre in [7, chapter 13.2], with certain details adapted from [3]. First, if : G C is a character of a complex representation ⇢ : G GL(V ), we define the Frobenius-Schur! indicator of to be ! 1 ◆( )= (s2). G s G | | X2 Theorem 2.2.1 (Frobenius-Schur). Let : G C be a character of a complex irreducible representation ⇢ : G GL(V ). Then! the Frobenius-Schur indicator ◆( ) satisfies the following statements:!
1. ◆( )=0if and only if ⇢ is a complex representation.
2. ◆( )=1if and only if ⇢ is a real representation.
3. ◆( )= 1 if and only if ⇢ is a quaternionic representation. 2.2.1 Relation to G-invariant bilinear forms Recall that if V is a complex, finite-dimensional vector space, then B : V V C is a bilinear form if B is linear in each coordinate separately. We say⇥ that! B is symmetric if B(v, w)=B(w, v)forallv, w V ,andthatB is alternat- ing or skew-symmetric if B(v, w)= B(w, v)forall2 v, w V (or equivalently, if B(v, v)=0foreveryv V ). Given a representation ⇢2: G GL(V )ofa finite group G over V ,wesaythat2 B is G-invariant if for every !s G we have 2 B(⇢sv, ⇢sw)=B(v, w). Finally, we say that B is non-degenerate if for each v V , we have that if B(v, w)=0forallw V ,thenv =0. 2 2 Recall that for any vector space V over C (or over any field), there is a dual space we denote as V ⇤,thespaceoflinearmapsf : V C.Forarepresentation ! ⇢ : G GL(V ), we can define its dual representation ⇢⇤ : G GL(V ⇤)overthe ! ! 1 dual space of V by setting ⇢s⇤' = '⇢s ,for' V ⇤ and s G.Thereforethe 2 2 1 character ⇤ of ⇢⇤ can be computed by the character of ⇢: ⇤(s)= (s )= (s).
22 T 1 Therefore ⇤ = ,sothat⇢s⇤ = ⇢s , the conjugate transpose. It follows then that ⇤(s)= (s). The following facts will allow us to reduce the classification of a representation ⇢ as complex, real, or quaternionic to determining of the existence of certain types of bilinear forms on the representation vector space V . Afterwards, we will see that the existence of such bilinear forms is determined by the indicator function defined at the beginning of the section.
Lemma 2.2.2. Let : G C be a character of a complex irreducible represen- tation ⇢ : G GL(V ). Then! is real-valued if and only if V has a G-invariant non-degenerate! bilinear form B. Furthermore, if this holds, B is unique up to mul- tiplication by a scalar; in other words, the space of such forms is one-dimensional.
Proof. First, note that is real-valued on all of G if and only if = = ⇤ on all of G,where ⇤ is the character of the dual representation ⇢⇤. Since two representations of G are isomorphic precisely when their characters are equal (Corollary 1.2.7), we conclude that is real-valued if and only if ⇢ an equivalent representation to its dual; that is, if and only if there is a vector space isomorphism f : V V ⇤ such ! that for every s G, f ⇢ = ⇢⇤ f. 2 s s But then we can define a bilinear form B : V V C as B(v, w)=fv(w), ⇥ ! which is G-invariant since B(⇢sv, ⇢sw)=f⇢sv(⇢sw)=⇢s⇤ fv ⇢sw = fv ⇢⇤ 1 ⇢sw = s fv(w)=B(v, w). Furthermore, B is non-degenerate since Schur’s Lemma gives us that f is an isomorphism (and is in fact multiplication by a scalar), and f being injective is equivalent to B being non-degenerate. Conversely, given a G-invariant non-degenerate bilinear form B(v, w), we can set f : V V ⇤ so that f (w)=B(v, w). Since B is non-degenerate, f is the zero ! v v map if and only if v = 0. Hence f is injective, and since dim V =dimV ⇤,itis an isomorphism. Also, (⇢⇤ f )(⇢ 1 w)=B(⇢ v, ⇢ w)=B(v, w)=f (w), so that s v s s s v f is an isomorphism of representations. Therefore, ⇢ ⇢⇤ if and only if there is a G-invariant non-degenerate bilinear form on V . ' Furthermore, let B0 : V V C be another non-degenerate G-invariant bilinear ⇥ ! form with corresponding isomorphism g : V V ⇤,whereg commutes with ⇢.Then 1 ! g f is an automorphism of V which commutes with ⇢.BySchur’slemma, 1 g f = I for some C,whichmeansthatf = g,andthereforethatB,B0 are identical up to scalar multiplication.2 This gives us that B is essentially unique, that that the space of G-invariant non-degenerate bilinear forms has dimension 1.
Proposition 2.2.3. Let : G C be a character of a complex irreducible repre- sentation ⇢ : G GL(V ). Then! the following statements hold: ! 1. ⇢ is a complex representation if and only if V has no G-invariant non-degenerate bilinear forms.
2. ⇢ is a real representation if and only if V has a (unique, up to scalar multipli- cation) G-invariant non-degenerate symmetric bilinear form.
3. ⇢ is a quaternionic representation if and only if V has a (unique, up to scalar multiplication) G-invariant non-degenerate alternating bilinear form.
23 Proof. Since ⇢ is a complex representation if and only if at least one value of is non-real, we have by Lemma 2.2.2 that ⇢ is a complex representation if and only if V has no G-invariant non-degenerate bilinear form. This establishes our first claim. Now suppose is real-valued, and let B : V V C be the G-invariant non-degenerate bilinear form. We claim that B is⇥ either! symmetric or alternat- + ing. To do this, we see that we can write B(v, w)=B (v, w)+B (v, w), where + 1 1 B (v, w)= 2 (B(v, w)+B(w, v)) and B (v, w)= 2 (B(v, w) B(w, v)). Note that + B is symmetric, and B is alternating, and that both are G-invariant and non- degenerate since they are linear combinations of such forms. Recall from the proof of Lemma 2.2.3 that there is a bijection between G-invariant non-degenerate bilinear forms B on V and isomorphisms f : V V ⇤ such that f ⇢s = ⇢s⇤ f for all s G. By Schur’s Lemma (Lemma 1.2.3), such! an f is a multiple of the identity map.2 As we defined the bijection in the proof of Lemma 2.2.2, we see that all such bilinear + forms are multiples of each other. Since B ,B are G-invariant, non-degenerate bi- + linear maps on V ,wehavethatB = B for some C.Butthisisonlypossible + 2 if either B =0orB =0,sinceoneissymmetricandtheotherisalternating(and not both, since then B =0,contradictingitsnondegeneracy).Sowehaveeither + B = B or B = B . IclaimthatB = B+ if and only if ⇢ is real. First suppose that ⇢ : G GL(V ) ! is a real representation. Then there is a basis under which the matrices ⇢s have real entries. This means that there is a G-invariant R-subspace V0 V such that ⇢ V = C V0.Inparticular,weseethatdim(V )=2dim (V0), so that if we have ⌦R R R an R-basis x1,...,xn of V0,wecanextendittoanR-basis x1,...,xn,y1,...,yn { } { } of V by letting yi = ixi. Then we can construct a G-invariant symmetric non-degenerate bilinear form as follows. Let (, ) be a positive-definite symmetric inner product on V0,anddefine 1 B0 = G s G(⇢sv, ⇢s,w). Then clearly B0 is positive-definite, symmetric, and G- 2 invariant| | (since the sum is simply permuted when we act on it by G). From here P we simply extend B0 to V as follows. Define B : V V C as B0(a + ib, c + id)= (B (a, c) B (b, d)) + i(B (a, d)+B (b, c)). B inherits⇥ symmetry! and G-invariance 0 0 0 0 from B0.ToseethatB is non-degenerate, we note that if we have some a + ib for which B(a + ib, c)=0forallc V ,thenbylinearindependenceinV we have 2 0 0=B(a + ib, c)=B(a, c)+iB(b, c), so that B(a, c)=B(b, c)=0forallc V0, and therefore a = b =0,sothata + ib =0.SoB is non-degenerate. Therefore2 there exists a G-invariant symmetric non-degenerate bilinear form on V ,andsince we saw that B is unique up to scalar multiplication, we have that B = B+. Conversely, suppose that B = B+;thatis,thereexistsaG-invariant non- degenerate symmetric bilinear form on V .Asbeforewenotethatwecantake it to be an isomorphism B : V V ⇤ sending v Bv(w)=B(v, w). Then let (, )be a positive-definite Hermitian inner! product on 7!V ,whichwetaketobeG-invariant as well (we can simply average a non-G-invariant one to obtain it, as before). Sim- ilarly, we can let B0 : V V ⇤ be the isomorphism defined by v (v, w)(taking ! 7! 1 the inner product as a function of w with fixed v). Then let T = B0 B be an R-automorphism of V ,wherewearethinkingofV as a 2n-dimensional R-vector space as before (n being its dimension as a complex one). Then T (v)istheele- ment of V for which B(v, w)=(f(v),w)forallw V .Furthermore,weseethat 2
24 1 1 (T (⇢sv),w)=B(⇢sv, w)=B(v, ⇢s ,w)=(T (v),⇢s w)=(⇢sT (v),w), so that ⇢ T = T ⇢ . Since this latter property holds for T 2 as well and ⇢ is an irreducible s s representation, Schur’s lemma tells us that T 2 = I for some C.Butsince(, ) is Hermitian and B is symmetric, we have that 2
( v, w)=(T 2(v),w)=B(T (v),w)=B(w, T(v))
=(T (w),T(v)) = (T (v),T(w)) = (v, T 2(w)) = (v, w), and therefore that R.Thensince(T (v),w)=v, T(w), we have then that (T (v),T(v)) = (v, T 2(w2)) = (v, v), so that is positive, because (, )ispositive- definite. Rescaling so that =1,wehavethatT 2 = I,andthereforecanonlyhave eigenvalues 1. ± Now let V ± = v V : T (v)= v be the eigenspaces corresponding to the eigenvalues 1. Note{ now2 that ± } ± (T (iv),w)=B(iv, w)=B(v, iw)=(T (v), iw)=( iT (v),w), so that T (iv)= iT (v), and therefore if v V ,thenT (iv)= iT (v)=v,so that iv V +.Butthenif v V +,wehaveT2(⇢ v)=⇢ T (v)=⇢ v ,sothatV + is 2 2 s s s G-invariant. Therefore V has a real G-invariant subspace of dimension n (over R -wheren is the dimension of V over C;thiscanbeseensincemultiplicationbyi + swaps V with V ). This shows that ⇢ is a real representation. It follows then that, if there exists such a B at all, B = B if and only if ⇢ is non-real and non-complex, i.e. ⇢ is quaternionic.
2.2.2 The symmetric and alternating squares Before we can prove Theorem 2.2.1, we need to take a short detour to discuss tensor products of linear representations and the decomposition of their characters.
Definition 2.2.4. Let V ,V be vector spaces with bases e , e , respectively. 1 2 { i1 } { i2 } Then we can define their tensor product V = V1 V2 to be the set of pairs v1 v2, for v V and v V , such that ⌦ ⌦ 1 2 1 2 2 2 1. The product v v is linear in both arguments. 1 ⌦ 2
2. The set of products ei1 ei2 forms a basis for the space V , and therefore dim V = dim V dim{ V .⌦ } 1 · 2 Accordingly, given two linear representations ⇢1 : G GL(V ) and ⇢2 : G ! 1 ! GL(V2), we can define their tensor product ⇢ : G GL(V1 V2)bysetting⇢s(v1 1 2 ! ⌦ 1 2 ⌦ v2)=⇢s(v1) ⇢s(v2)foreachs G.Inthiscasewewritethat⇢ = ⇢ ⇢ . Note that ⌦ 2 ⌦ 1 2 the character of ⇢ is equal to the product of the characters 1 and 2 of ⇢ ,⇢ , respectively: = 1 2. To see this we simply note that entries of the matrix · 1 2 representation of ⇢s are the entrywise products of the entries of ⇢s,⇢s,sothatthe trace is the sum of all products of the diagonal entries of the factor matrices. Given a representation ⇢ : G GL(V )ofdegreen with character ,weare interested in the tensor product space! V V ,withcorrespondingrepresentation ⌦
25 2 n ⇢ ⇢ and character .Let ei i=1 be a basis of V ,anddefine✓ : V V V ⌦ V to be the automorphism{ which} switches the basis elements in the coordinates:⌦ ! ⌦ ✓(ei ej)=ej ei for each pair (i, j). Since each element in V V can be written x ⌦y =( n ⌦a e ) ( n b e )= a b (e e ), we see⇥ that ✓(x y)= ⌦ i=1 i i ⌦ j=1 j j (i,j) i j i ⌦ j ⌦ a b (e e )=y x. (i,j) i j Pj ⌦ i ⌦P P Since ✓2 = I,theidentitytransformation,weobtainthattheeigenvaluesof P ✓ are 1. Let Sym2(V )denotethesetofelementsfixedby✓ (the eigenspace of the eigenvalue± 1), and let Alt2(V )denotethesetelementsv V V such that ✓(v)= v (the eigenspace of eigenvalue 1). We refer to these2 as the⌦ symmetric square and the alternating square,respectively.Clearly,thesetwosubsetsare disjoint. Furthermore, the set of elements 1 = (ei + ej) (ej + ei) i j is a linearly 2 { ⌦ { independent subset of Sym (V ), while the set 2 = (ei ej) (ej ei) i V V =Sym2(V ) Alt2(V ). ⌦ Note that since representations on V V are linear in each coordinate as well, that the symmetric and alternating squares⌦ are in fact G-invariant subspaces of V V , ⌦ 2 so that we can restrict a representation ⇢ : G GL(V V )torepresentations⇢ : 2 2 2 ! ⇥ G GL(Sym (V )) and ⇢↵ : G GL(Alt (V )). The following lemma decomposes their! characters. ! 2 Lemma 2.2.5. Let ⇢ : G GL(V ) be a representation with character . Let 2 ! and ↵ be the characters of the symmetric and alternating squares of V , respectively. Then if s G, we have 2 1 2 (s)= ( (s)2 + (s2)), 2 1 2 (s)= ( (s)2 (s2)). ↵ 2 Proof. First recall that we can construct an inner product (v1,v2)onV which is G-invariant (in the sense that (⇢tv1,⇢tv2)=(v1,v2)foreacht G, v1,v2 V ), and n 2 2 therefore we can pick an orthonormal basis ei i=1 for V of eigenvectors of ⇢s,for our given (and fixed) s G.Thereforeforeach{ i}=1,...,n,thereissomeeigenvalue 2 2 i C such that ⇢sei = iei.Therefore (s)= i i,andsince⇢s2 =(⇢s) ,the 2 2 2 2 2 eigenvalues of ⇢s2 are 1,..., n,sothat (s )= i i . Next we see that the eigenvalues of ⇢2 and ⇢2P,thesymmetricandalternating P↵ squares of ⇢,aretheproducts i j: (⇢ ⇢ )((e e ) (e e )) = ((e e ) (e e )), s ⌦ s i ± j ⌦ j ± i i j i ± j ⌦ j ± i so that 2 n 1 n n 1 2 = = 2 + = + 2 = ( (s)2 + (s2)) i j i i j 2 0 i i 1 2 i j i=1 i 2 2 2 Note that Lemma 2.2.3 implies that = (s)+ ↵(s), which we already knew, 2 2 2 2 2 since V V =Sym(V ) Alt (V ), and thus the character of ⇢ ⇢ = ⇢ ⇢↵ can be written⌦ as the sum of the characters of its direct summands (Proposition⌦ 1.2.2). 2.2.3 Bringing it all together We now explicate the relationship between the symmetric and alternating squares of a representation into a vector space V and the space of bilinear forms on V . First, note that there is a bijective correspondence between non-degenerate bi- linear forms B : V V C and linear functionals f : V V C of the tensor product: since tensor⇥ products! were defined such that they are⌦ bilinear,! these are in fact the same structures. In other words, there is a bijection between such bilinear forms and elements of the dual space (V V )⇤ V ⇤ V ⇤.Furthermore,sincethe ⌦ ' ⌦ space SB of bilinear forms B : V V C decomposes as SB = S↵ S ,whereS↵ is ⇥ ! the subspace of alternating forms and S the space of symmetric forms; and similarly 2 2 V V =Sym(V ) Alt (V )asseenbefore;wenotethatthebijectionSB V ⇤ V ⇤ respects⌦ this sum: it restricts to a bijection between the space of symmetric⇠ (resp.,⌦ 2 2 alternating) bilinear forms and elements of Sym (V ⇤) (resp., Alt (V ⇤)). In particular, the space of non-degenerate symmetric (resp., alternating) G- invariant bilinear forms on V is in bijective correspondence with the elements of 2 2 Sym (V ⇤) (resp., Alt (V ⇤)). Now, with all the pieces in place, we can prove our main theorem about the Frobenius-Schur indicator. 2 2 Proof of Theorem 2.2.1. First, recall from Lemma 2.2.5 that if and ↵ are the 2 characters of the symmetric and alternating squares, respectively, then (s)= 1 2 2 2 1 2 2 2 ( (s) + (s )), and that ↵(s)= 2 ( (s) (s )). Note that we can decompose the Frobenius-Schur indicator as follows: 1 1 ◆( )= (s2)= 2 (s) 2 (s)=( , 2 2 )=( , 2 ) ( , 2 ). G G ↵ 1 ↵ 1 1 ↵ s G s G | | X2 | | X2 Furthermore, we note that ( 1, )=( 1, ⇤)foranycharacter with dual 1 character ⇤,since ⇤(s)= (s ), and in the inner product we sum over all s G. 2 Now, we have seen that the dimension d (respectively, d↵)ofthespaceofG- invariant non-degenerate symmetric (resp., alternating) bilinear forms on V is equal to the number of times the trivial representation occurs in the decomposition of the 2 2 dual of the symmetric square ⇢ (resp., alternating square ⇢↵). But then we can rewrite the indicator function as ◆( )=d d . ↵ By Lemma 2.2.3, we have three distinct cases: 27 1. ⇢ is a complex representation; the space of G-invariant non-degenerate bilinear forms has dimension zero; d = d↵ =0;andthus◆( )=0. 2. ⇢ is a real representation; the space of G-invariant non-degenerate symmetric bilinear forms has dimension one, and the space of G-invariant non-degenerate alternating bilinear forms has dimension zero; d =1,d↵ =0;andthus ◆( )=1. 3. ⇢ is a quaternionic representation; the space of G-invariant non-degenerate symmetric bilinear forms has dimension zero, and the space of G-invariant non-degenerate alternating bilinear forms has dimension one; d =0,d↵ =1; and thus ◆( )= 1. 2.3 Decomposing the group algebra R[G] In section 1.2.3, we defined the group algebra of a finite group G over a commutative ring K,andnotedthatifwetakeK = C,thenthereisabijectivecorrespondence between finitely-generated modules over the algebra C[G]andrepresentationsofG in C-vector spaces. In discovering this correspondence, we did not use any partic- ular properties of C that do not also hold of R,andsoweseethatthereisalso abijectivecorrespondencebetweenfinitely-generatedmodulesovertherealgroup algebra R[G]andrepresentationsofG in R-vector spaces. Furthermore, since R has characteristic zero, Lemma 1.2.13 tells us that the algebra R[G]issemisimple, which, when combined with Theorem 1.2.14 (the Artin-Wedderburn decomposition theorem), implies that R[G]canbefactorizedintoafiniteproductofmatrixrings with coe cients in division rings of finite degree of R.Thefollowingresult(the slick, elementary proof proceeds according to [5]) corresponds to Proposition 1.2.15, in which we classified the division rings of finite degree over C. Theorem 2.3.1 (Frobenius’ Theorem). Let D be a division ring of finite degree over R. Then D is isomorphic to R, C, or the ring H of quaternions. Proof. Suppose that D is not isomorphic to R.Thenthereissomeelementx D R. 2 \ Let Rx be the two-dimensional R-vector space with basis 1,x . First we note that { } Rx is a commutative subset of D,ofdegreetwooverR,andthereforeisisomorphic to C.Furthermore,itcontainsalltheelementsofD which commute with x,andit has maximal degree over R among such subsets (this phrase makes sense since we took D to be finite-degree over R in the first place). To see this, let F be a commutative subspace of D containing Rx such that it has maximal degree (over R)ofallsuchspaces.Thenifsomex0 D commutes 2 with the elements of F ,thevectorspace a + rx0 : a F, r R is commutative, { 2 2 } so that it is actually equal to F (by our maximality assumption). Therefore x0 F , and F is maximal as a commutative subset of D.Thenforanyf F ,thesetof2 2 1 elements which commute with f is equal to the set which commutes with f ,so 1 that f F and therefore F is a field. Since it is commutative and contains Rx,it 2 is isomorphic to C and therefore equal to Rx. 28 Therefore we can let i denote some element of D such that i2 = 1(itexistssince Rx D is isomorphic to C). Then D becomes a C-vector space (not a commutative one)⇢ with left scalar multiplication. Let g : D D be the linear transformation defined by g(x)=xi.Theng2(x)= x,sothattheeigenvaluesof! g can only be i (note that they needn’t both be eigenvalues necessarily). Then if we let ± D± = x D : xi = xi be the eigenspaces of g in D corresponding of to i,then { 2 ± } 1 1 ± since for any x D we can write x = 2 (x ixi)+ 2 (x + ixi), where x ixi D±, + 2 + ⌥ 2 and clearly D D = , D = D D . \ ; Then since D+ is a maximal set which commutes with some element i D R, + 2+ \ we see that, as before, D C.ThereforeifD = 0, we have that D = D C. ' ' Otherwise, suppose that there is some nonzero y D .Thenforanyotherz D + 2 + 2 we have yzi = yiz = iyz,sothatyz D .Thenifh : D D is given by 1 2 ! 1 h(x)=xy,weseethath is defined by right-multiplication by y (recall that + D is a division ring), and therefore D D C,andinparticularbothhave ' ' dimension one as C-vector spaces. 2 Next note that, as before, Ry is isomorphic to C and thus y Ry.Butsince 2 + 2 + 2 2 y = h(y) D C, y D Ry = R.Furthermore,sincey is a square root of y , 2 ' 2 \ and y/R D+, y2 cannot have two real square roots, and hence y2 < 0. Therefore 2 ⇢ 2 we can define some j D for which j = 1(inparticular,itisapositivemultiple 2 of y). Finally let k = ij D ,andnotethatj, k are linearly independent over R. 2 + Therefore j, k is a basis for D over R,and 1,i is a basis for D over R,so { } { } that 1, i, j, k is a basis for D over R.Furthermore,k2 = ijij = i2j2 = 1, and ijk ={k2,sothat} i2 = j2 = k2 = ijk = 1, the defining relations of the quaternions. So D H,thequaternions. ' Corollary 2.3.2. If G is a finite group, then the group algebra of G over R factorizes into a product of matrix rings as follows: N1 N2 N3 R[G]= Mnj (R) Mnk (C) Mnl (H). j=1 Y kY=1 Yl=1 Proof. This follows directly from Frobenius’ theorem (2.3.1) and the Artin-Wedderburn decomposition theorem (Theorem 1.2.14). In the previous section, we inquried as to whether irreducible representations over the complex numbers could be realized over the reals. In this section, we dig deeper, and see that when we transform an irreducible representation over the complex numbers into a representation over the real numbers, we can obtain the irreducible representations over the reals, and furthermore determine their character and their corresponding simple component of the real group algebra R[G]. First we note that if ⇢ : G GL(V )isanirreduciblerepresentationoverthe complex numbers with character! ,whereV has dimension n and basis x ,...,x , { 1 n} then we can transform it into a 2n-dimensional vector space over R by letting yi = ixi for i =1,...,n,andtakingtheR-basis x1,...,xn,y1,...,yn .Thisturns⇢ into a 2n-degree representation ⇢R over the real{ numbers with character} + (since we multiply the basis elements by i in the “non-real” half of the vector space). The question, then, is whether or not ⇢ remains irreducible. We take note of the following criterion first. If is the character of a representation ⇢0 : G GL(V ) 0 ! 0 29 over the real numbers, then the inner product ( 0, 0)isequaltothenumberof irreducible representations contained in the decomposition of ⇢0 over the complex numbers. Since every G-invariant endomorphism of V0 is a representation of G over V0, we see that the dimension over R of the endomorphism ring EndG(V0)isequal to ( 0, 0). This will allow us to determine the simple component of the real group algebra R[G]whichcorrespondstorepresentation⇢. There are three scenarios: 1. If ⇢ is a real representation, then we know that there is a basis of V under which ⇢ has all real entries and still has character ,andthereforetherepresentation of character + =2 is reducible over R,andisinfactasumoftwocopies R of ⇢ .ThereforeitcorrespondstooneofthesimplecomponentsMni (R)in the decomposition of R[G]. 2. If ⇢ is a complex representation, then = ,andthereforethis⇢ corresponds to an irreducible real representation of degree6 2n and character + . Note that since ( + , + )=2,weseethatitsendomorphismringhasdimension 2overR and is thus equal to C. Therefore it corresponds to one of the simple components Mni (C)inthedecompositionofR[G]. 3. If ⇢ is a quaternionic representation, then = ,andso⇢ corresponds to a real representation of degree 2n and character 2 ,whichisnotreducibleover the real numbers. This is because, if it were, we would have a realization of ⇢ over the real numbers, contradicting that it is quaternionic. Furthermore, (2 , 2 ) = 4, and therefore the endomorphism ring of the representation has dimension 4 over R and is thus equal to H.Soitcorrespondstooneofthe simple components Mni (H)inthedecompositionofR[G]. 2.4 Some simple examples First, we return to the groups whose character tables we have computed. As a simplest example, we compute the Frobenius-Schur indicator of the degree two representation of S3.Recallthatitscharacterisgivenby: S3 C1 C2 C3 3 2-10 where C1 = id , C2 = (123), (132) and C3 = (12), (13), (23) .Thensince (123)2 =(132),and(12){ } 2 =(13){ 2 =(23)} 2 = id, we have{ that } 1 1 1 ◆( )= (s2)= (4 (C )+2 (C )) = (8 2) = 1. 3 G 3 6 1 3 2 6 s G | | X2 Therefore 3 is the character of a real representation. Now, we return to the identical character tables of D4 and Q8 to see that their degree two representations are indeed not the same. Recall that the degree two character of D4 is given by D4 C1 C2 C3 C4 C5 5 2-20 0 0 30 2 2 3 3 where C1 = 1 , C2 = x , C3 = y, x y , C4 = x, x and C5 = xy, x y . { } { } { } 2 { } { } Note that the elements of C4 square to x ;andthatthoseofC1,C2,C3,andC5 square to 1. Then we can compute the indicator: 1 1 ◆( )= (s2)= (6 (C )+2 (C )) = 1. 5 G 5 8 5 1 5 2 s G | | X2 Therefore 5 is also the character of a real representation. Now recall that the degree two character of Q8 is Q8 C1 C2 C3 C4 C5 5 2-20 0 0 where C = 1 , C = 1 , C = i , C = j ,andC = k .Inthiscase 1 { } 2 { } 3 {± } 4 {± } 5 {± } the elements of C3,C4,andC5 square to 1 C2,andthoseofC1 and C2 square to 1 C .Then 2 2 1 1 1 ◆( )= (s)= (2 (C )+6 (C )) = 1. 5 G 5 8 5 1 5 2 s G | | X2 Indeed, the degree two character of Q8 is quaternionic; it cannot be realized over R. In the final chapter, we will examine the finite subgroups of the special orthogonal group SO3 for more examples of character tables and Frobenius-Schur indicators. We will be led to compute the character tables of the symmetric group S4,andthe alternating groups A4 and A5, which have more high-degree irreducible represen- tations than those seen above. However, we will see that they for the most part turn out to be realizable over R. As it turns out, quaternionic representations are somewhat rare. For an example of a slightly larger group which has a quaternionic representation, we note that the special linear group SL2(F3)of2 2matricesof ⇥ determinant one with entries in the finite field F3 has a degree two quaternionic irreducible representation. 31 Chapter 3 The finite subgroups of SO3 In this section, we examine as a source of examples the finite subgroups of the special orthogonal group SO3,definedeitherasthespaceof3 3matricesofdeterminant1, or as the space of rotations in three dimensions. The classification⇥ of these subgroups is standard, and not particularly relevant to the contents of this paper. We defer to [1, chapter 6] or [2, chapter 3] for the details of the classification proof. We will take the following theorem as given: Theorem 3.0.1 (Classification of the finite subgroup sof SO3). Every finite sub- group of the special orthogonal group SO3 is isomorphic to one of the following: 1. A cyclic group Cn of order n 2. A dihedral group Dn of order 2n 3. The tetrahedral group T , defined as the group of symmetries of a regular tetra- hedron inscribed in the unit sphere. The group T is isomorphic to the alter- nating group A4. 4. The octahedral group O of symmetries of a regular octahedron (cube) inscribed in the unit sphere. It is isomorphic to the symmetric group S4. 5. The icosahedral group I of symmetries of a regular icosahedron inscribed in the unit sphere. It is isomorphic to the alternating group A5. In the remaining sections, we will determine the character tables of each class of groups listed above, and determine the Frobenius-Schur indicators of their repre- sentations of degree greater than one (if any). 3.1 The cyclic groups Cn Since Cn = x is abelian, it has n distinct conjugacy classes, and thus n nonisomor- phic irreducibleh i representations (since the sum of the squares of their dimensions must equal n,theyallhavedegreeone).SinceeveryelementofCn has order dividing j n and all representations are one-dimensional, the values i(x )mustbepowersof ! = e2⇡i/n,theprimitiventh root of unity. Of course, the value of a character on all 32 of Cn is determined by its value on a generator x,whichwecansendtoanyofthe n distinct powers of ! (the nth power, which is 1, accounts for the trivial character). Taking powers of i(x) generates the rest of the the values, since the characters are one-dimensional, and thus homomorphisms. From this, we know we needn’t bother calculating the Frobenius-Schur indicators of the characters, since unless the gener- ator x is sent to 1 (the trivial character) or to 1(whichwillnecessarilydetermine arealrepresentationthattakeson1onevenpowersofthegeneratorand 1onodd powers), it must be sent to a non-real number (as all other powers of roots of unity are), and hence the representations are complex. Note that, in any case, a character of degree one is not interesting from the perspective of real, complex, and quaternionic representations, since if the value of the character is always real, the representation is necessarily real, anyway (the character is itself the representation). 3.2 The dihedral groups Dn n 2 We write Dn = x, y x = y = xyxy =1 in its usual presentation. First, we note that the cases inh which| n is even are di↵erenti than those in which n is odd, as can be seen by the di↵erence in conjugacy class structures. n 1 If n is odd, then Dn has 2 +2 conjugacyclasses:theclassoftheidentity; the single class (of size n)containingtheelementsxky for all k =1,...,n;andthe k k (n 1)/2classeswhicheachcontain x ,x for each 1 k n/2 .Inthiscase, { } b c we can easily identify two irreducible (degree one) characters as the trivial one 1 and the representation 2 sending x 1andy 1. Since this is a homomorphism and is distinct from the trivial representation,7! 7! we see that it is irreducible. n Contrarily, if n is even, then Dn has 2 +3conjugacyclasses:theclassofthe identity; the class of elements xky for k even; the class of elements xky for k odd; k k n and the classes containing x ,x for each 1 k 1. Here we can easily { } 2 identify four degree one representations: 1,thetrivialcharacter; 2 which sends x 1andy 1; which sends x 1andy 1; and which sends x 1 7! 7! 3 7! 7! 4 7! and y 1. Again, these are all distinct homomorphisms from Dn C,andare thus irreducible7! representations. ! To obtain the remaining representations, in either case, we can construct the necessary amount of distinct two-dimensional representations as follows: let n = e2⇡i/n.Thenwecandefine k k n 0 k 01 ⇢x = k , and ⇢y = 0 10 n k 2 k n k k k k k so that (⇢y) = I,(⇢x) = I,and⇢y⇢x⇢y⇢x = I,sothat⇢ is a representation of Dn for each k =1,..., (n 1)/2 .Weseethattheyaredistinctsincetheircharacters take di↵erent valuesb on x forc each k.Toseethattheyareirreducible,wesimply take the inner product of their characters with themselves, and see that it comes out to one. In the case of n odd, this gives us our remaining (n 1)/2representations;andin the case of n even, this gives us our remaining n 1representations,andtherefore 2 33 we have found all the irreducible characters of Dn for all n. Note that our two- dimensional characters are all real, as defined, but the representations we gave are not. So in this case computing the Frobenius-Schur indicators wouldn’t be a bad idea. First note that every element of the form xky has order 2, and that if n is even, xn/2 does as well. So if n is odd, there are n+1 elements whose square is the identity; and if n is even, there are n +2.Fortheotherelements,ifn is odd, then squaring is an automorphism of the subgroup x (since n and 2 are relatively prime), and so k k h i each conjugacy class x ,x contains two squares. Therefore the Frobenius-Schur { } k indicator, where k is the character of ⇢ as defined earlier, is 1 1 ◆( k)= k(s)= (2(n +1) 2) = 1, Dn 2n s Dn | | X2 since the sum of the characters of the remaining conjugacy classes add to 2(since the character is orthogonal to the trivial character). Thus all the irreducible repre- sentations of Dn for n odd are real representations. If n is even, then first recall that there are n +2elementswhichsquaretothe identity. The remaining n 2 elements are the n 1powersofx excluding xn/2. Now since x2j = x2k if and only if k = j mod n/2, we see that each even power of k k x has two square roots. Since n is even, the conjugacy classes x ,x are either even or odd classes. So each even class contains four squares,{ except that} of xn/2 (which contains two), and each odd one contains none. Therefore the indicator is computed as follows: n 1 2 1 1 n/2 n/2 k/2 k/2 ◆( k)= k(s)= 2(n +2) 2( n + n )+4 ( n + n ) Dn 2n 0 1 s Dn k=1 | | X2 X @ A 1 = (2n +2 2+0)=1. 2n As in the odd case, we see that Dn has all real irreducible representations for even n,aswell. 3.3 The tetrahedral group T To find the character table of the tetrahedral group, we enumerate the vertices of aregulartetrahedronasv1,v2,v3,v4,andnotethatT simply permutes the set V of these vertices with orientation-preserving (that is, even) permutations. So T is isomorphic to A4,andwecaninsteadfinditscharactertable.Itselementsarethe identity, the eight three-cycles, and the three double-transpositions. A little fiddling reveals that the double-transpositions form a single conjugacy class 2,andthat the three-cycles split into two conjugacy classes in the alternating groupC A : : 4 C3 (123), (134), (142), (243), and 4 :(132), (143), (124), (234) (the identity is conjugacy class ). So by the main theoremC of characters, there are four nonisomorphic C1 irreducible characters of T ,where 1 is the trivial character. The main theorem also gives us that 4 (1)2 = T =12,sothat 4 (1)2 =11,forwhich i=1 i | | i=2 i P P 34 2 2 2 the only solution is 1 +1 +3 =11.So 2, 3 are one-dimensional, and 4 is three-dimensional. If ! is a primitive third root of unity, we see that for 2 to be non-trivial, it 2 must take on either ! or ! on conjugacy classes 3 and 4,sinceotherwiseitsvalue on both would be 1 (since these conjugacy classesC haveC elements of order 3, and so the order of their character value on the one-dimensional characters must have order dividing 3), and thus the value on would also be 1, since we can obtain C2 double-transpositions as products of certain three-cycles. So say that 2( 3)=!, 2 C and 2( 4)=! ,whichimpliesthat 2( 2)=1,sinceagainitselementscanbe obtainedC as products of those from the otherC two classes. By the same argument, 2 3 must take the values ! and ! on the final two classes, and since it is distinct it must reverse them. This fills out the first three rows of the character table: T 1(12)(34)(123)(132) 1 11 1 1 2 2 11 !! 2 3 11 ! ! 4 3 Finally we can use orthogonality of columns to fill out the rest: since the first and second column are orthogonal, 1+1+1+3 4( 2)=0,sothat 4( 2)= 1. Now for C C 2 the third and fourth columns, we have that if i =3, 4, then 1+(!+! )+3 4( i)=0, and hence ( )=0.Sothecompletetableisasbelow. C 4 Ci T 1(12)(34)(123)(132) 1 11 1 1 2 2 11 !! 2 3 11 ! ! 4 3-10 0 Now, we calculcate the Frobenius-Schur indicator of the three-dimensional irre- ducible character 4 of T .Weseethat 1 1 ◆( )= (g2)= (3 + 3 + 3 + 3) = 1. 4 G 4 12 g G | | X2 Hence 4 is realizable as a representation over R. As in our subsequent examples, this is completely natural, since the tetrahedral group is in fact a finite subgroup of SO3,thespaceof3 3 orthogonal (and thus real-valued) matrices of determinant one. ⇥ 3.4 The octahedral group O Recall that the octahedral group O is isomorphic to the symmetric group S4.Its elements are the identity, the eight three-cycles, the three double-transpositions, the six four-cycles, and the six transpositions. In the symmetric group, conjugacy class is determined by cycle type, so we have obtained our five conjugacy classes: of the C1 35 identity; of the transpositions; of the double-transpositions; of the three- C2 C3 C4 cycles; and 5 of the four-cycles. Therefore, there are five irreducible characters, C 5 with 1 being the trivial character. We must have that i=2 i(1) = O 1=23. If any of the squares were equal to 16, this could not add up; hence they| | must all be at most 9. But we must have at least two equal to 9P for it to add to 23, so we have that 2(1) = 1, 3(1) = 2, and 4(1) = 5(1) = 3. To obtain a non-trivial degree-one representation, we need look no further than the sign map, which sends a permutation to its sign. The transpositions and four-cycles are odd, and hence are sent to 1(thoseare 2 and 5); while the three-cycles and double-transposititons are sent to 1 ( and C ). So weC have filled out the first two rows: C3 C4 O C1 C2 C3 C4 C5 1 11111 2 1-11 1-1 3 2 4 3 5 3 To fill out another row, we consider the permutation representation of O op- erating on the set of pairs of opposite vertices of the octahedron inscribed in the unit sphere. Recall that the transpositions correspond to the rotations of angle C2 ⇡ about the line through opposite edges; the double-transpositions 3 correspond to the rotations of angle ⇡ about lines going through the center of oppositeC faces; the three-cycles 4 correspond to the rotations of angle 2⇡/3aboutthelinethrough opposite pairs ofC vertices; and the four-cycles correspond to the to the rotations C5 of angle ⇡/2aboutthesamelinesas 2. We see that fixes all four pairs;C fixes the two pairs adjacent to the edges C1 C2 through which the axis of rotation goes; 3 and 5 do not fix any pairs; and 4 fixes the pair of vertices through which theC axisC of rotation goes. So we haveC a representation A of character O C1 C2 C3 C4 C5 A 42010 Now note that if we subtract the trivial character from A,weobtainacharacter with values as follows: O C1 C2 C3 C4 C5 31-10-1 A 1 This character is irreducible, since its inner product with itself is equal to one. Call it 4.Thenwehaveobtainedthefourthrowofourcharactertable: O C1 C2 C3 C4 C5 1 11111 2 1-11 1-1 3 2 4 31-10-1 5 3 36 Since the product of two characters is a character (in particular, its representation is the tensor product of the two component representations, as we saw in section 2.2); we take the product 2 4, and see that its character is equal to O C1 C2 C3 C4 C5 4 2 3-1-101 In fact, this character is also irreducible, which we can see by taking its inner product with itself. We let 5 = 4 2, and see that we have filled out most of the table: O C1 C2 C3 C4 C5 1 11111 2 1-11 1-1 3 2 4 31-10-1 5 3-1-101 Now we can obtain 3 by column orthogonality. It is easily seen this way that 3( 2)= 3( 5)=0,that 3( 3)=2,andthat 3( 4)= 1. This completes the characterC table:C C C O C1 C2 C3 C4 C5 1 11111 2 1-11 1-1 3 202-10 4 31-10-1 5 3-1-101 Now we may determine whether the degree three and two representations of O are real or quaternionic by computing their Frobenius-Schur indicators. We see that the elements of 1, 2 and 4 square to the identity, so that we have 2 C C C s 1 for 10 elements g O.Theelementsof 5 square to other elements of 5, 2C 2 2 C C so there are 8 elements s 5. Finally, the elements of 3 square to those in 2,so there are 6 elements s2 2.Fromthisweobtainthat:C C C 2C2 1 1 (s2)= (10 2+6 2+8 1) = 1. G 3 24 ⇥ ⇥ ⇥ s G | | X2 1 1 (s2)= (10 3+6 1+8 0) = 1, G 4 24 ⇥ ⇥ ⇥ s G | | X2 1 1 (s2)= (10 3+6 1+8 0) = 1, G 5 24 ⇥ ⇥ ⇥ s G | | X2 so that indeed all three multi-degree representations of O can be realized over the real numbers. In fact this can be inferred at least in the case of the degree three representations, since O,asafinitesubgroupofSO3,hasarepresentationasthree- dimensional orthogonal matrices of determinant one; and we obtained the other degree three character by taking its tensor product with the sign representation. However, in the case of the degree two character, we do not have any obvious corresponding representation over the reals, and so this computation is at least somewhat enlightening. 37 3.5 The icosahedral group I To find the character table of I A ,firstwedeterminetheconjugacyclasses ' 5 of A5. Now, the conjugacy classes of the symmetric group S5 are equivalent to partitioning the group by cycle type. In general, for an alternating group A S , n ⇢ n each conjugacy class (that is, cycle type) that remains in An -theevenones-either remains an intact conjugacy class, or splits in two. In general, it doesn’t split if its cycle decomposition contains either an even cycle or two cycles of the same length. In the case of A5,thefive-cyclessplit,whilethedouble-transpositionsandthree- cycles do not. So we have five conjugacy classes: of the identity (size 1); of the C1 C2 double-transpositions (size 15); 3 of the three-cycles (size 20); 4 of the five-cycles generated by (12345) (size 12); andC of the five-cycles generatedC by (13524) (size C5 12). Therefore we have five irreducible representations, with 1 again the trivial 5 2 representation. This means that i=2 i (1) = 59, which we can easily see is only possible if the values are 5, 4, 3 and 3. Following [1], we use geometricP aspects of the rotation groups to fill out our table. We obtain one three-dimensional representation as the set of orthogonal matrices corresponding to the rotations in the icosahedral group I.Thetraceofarotation matrix which rotates by angle ✓ is 1 + 2 cos ✓,sincewecanpickabasisunder which the rotation matrix takes a block form with a 1 in one corner, and the block cos ✓ sin ✓ in the other. Since these angles are ⇡ for ,wegetthat ( )= 1; sin ✓ cos ✓ C2 2 C2 for , ✓ =2⇡ /3, so ( )=0;for , ✓ =2⇡/5and ( )=↵ =1+2cos(2⇡/5); C3 2 C3 C4 2 C4 similarly for 5, ✓ =4⇡/5sothat 2( 5)= =1+2cos(4⇡/5). Next, weC consider the action of CI A on 1,...,5 ;apermutationrepre- ' 5 { } sentation of dimension 5. The character P at each group element is equal to the number of indices fixed by that element. We see, then, that ( )=5; ( )=1; P C1 P C2 P ( 3)=2; P ( 4)= P ( 5)=0.Weknowthatthetrivialrepresentationappears in theC decompositionC of everyC permutation representation, so we take the inner prod- uct ( P 1, P 1)=1,andseethat 4 = P 1 is irreducible. So we have filled out three rows of the character table: I C1 C2 C3 C4 C5 1 11111 2 3-10↵ 3 3 4 401-1-1 5 5 Now recall that I is the group of symmetries of a regular dodecahedron inscribed in the unit circle, and consider the operation of I on the pairs of opposite faces of this dodecahedron. This gives us a degree six character with values ( )=6; D C1 D( 2)=2; D( 3)=0; D( 4)= D( 5)=1.Thisisbecausetheedgerotationsof angleC ⇡ fix two pairsC of faces (thoseC on eitherC side of the edges); the vertex rotatiions fix no faces; and the face rotations of both angles 2⇡/5and4⇡/5onlyfixthefaces they rotate about. We know, as before, that the trivial representation appears in this character’s decomposition. So we compute the inner product ( D 1, D 1)=1, and therefore = is irreducible. This fills out the bottom row of our 5 D 1 38 character table. We can fill out the third row by column orthogonality. Hence we have the completed character table of I A : ' 5 I C1 C2 C3 C4 C5 1 11111 2 3-10↵ 3 3-10 ↵ 4 401-1-1 5 51-100 1+p5 1 p5 where again ↵ = 2 and = 2 .Sincetheelementsof 1 and 2 square to the identity; the elements of square to other elements of C,andtheelementsC C3 C3 of 4 square to those in 5 (and vice versa), we can calculate the indicators of the non-trivialC characters : C 1 1 ◆( )= (s2)= (16 ( )+20 ( )+12( ( )+ ( ))). G 60 C1 C3 C4 C5 s G | | X2 Therefore 1 ◆( )=◆( )= (48 + 0 + 12) = 1, 2 3 60 1 ◆( )= (64 + 20 24) = 1, 4 60 1 ◆( )= (80 20) = 1. 5 60 As before, we see that all non-trivial irreducible representations of I A5 are indeed real. ' 39 Works Cited [1] M. Artin. (1991) Algebra. [2] J.H. Conway, D.A. Smith (2003). On Quaternions and Octonions: Their Geom- etry, Arithmetic, and Symmetry. [3] A. Deloro (2012). A First Encounter with Representations of Finite Groups. [4] Gordon James, Martin Biebeck (1993). Representations and Characters of Groups. [5] R.S. Palais (1968). The American Mathematical Monthly, Vol. 75, No. 4. “The Classification of Real Division Algebras.” [6] V. Serganova (2011). Representation Theory, Lecture Notes. [7] J.P. Serre (1977). Linear Representations of Finite Groups. 40