Representation Theory - James Lingard

1 Representation TheoryRepresentationsLetGbe a group andVa vector space over a fieldk. ArepresentationofGonVis a grouphomomorphism :G Aut(V). Thedegree(ordimension) of is just representationsLet :G Aut(V) and :G Aut(V ) be two representations ofG. Then aG-linear mapfrom to is a linear map :V V such that ( (g)) = ( (g)) for allg G, or equivalently such that the following diagram commutes:V (g)// V V (g)//V If additionally is an isomorphism of vector spaces then we say that is anisomorphismfrom to , and that isisomorphicto . Notice that :V V is an isomorphism from to iff 1is an isomorphism from to , so isomorphism is an equivalence Gandv V, we often writegvinstead of (g)v. In this notation, :V V is anisomorphism iffg (v) = (gv)for allg Gand allv onV, andWis a subspace ofVsuch thatg(W) Wfor allg G, then we say thatWis subrepresentation ofVistrivialif it is 0 orV, and indecomposible representationsA Representation is calledirreducibleif it has no non-trivial adirect sumofWandW , writtenV=W W , ifWandW are subrepresentations ofVandV=W W as vector spaces.

2 Given a representationV, we want to break it up intosmaller pieces, that is, write is asV=W1 W2 Wkwhere eachWidoes not break up into smaller say that a Representation isindecomposibleif it is not a direct sum of smaller indecomposible then it is irreducible, but the converse does not follow in representationsLetGact on a setX. Then thepermutation representationofGwith respect to this action,k[X], is a|X|-dimensional vector space overkwith basis{ex|x X}. The action ofGon thisvector space is defined bygex=egxfor allg Gand allx [X] is never irreducible, for the 1-dimensional subspace spannedby x Xexis invariant representationsIf :G Aut(V) is a Representation then thekernelof is ker ={g G| (g) = id}.A Representation ofGisfaithfulif ker ={1}; in this case we say thatGactsfaithfullyonV,andGis isomorphic to a subgroup of Aut(V). Note that ker CG, so ifGis simple then everynon-trivial Representation is a finite group then it posesses a faithful finite-dimensional Representation .

3 ForGacts onitself faithfully by left-multiplication; thus the permuation representationk[G] for this action reducibilityLetGbe a finite group andVa Representation ofGover a field of characteristic zero. Vis aG-invariant subspace then there exists aG-invariant complement irreducible Vis be any vector space complement toW. Let :V Wbe the projection ofVontoWdefined by (w+w ) =wfor allw Wandw W , and define (v) =|G| 1 g Gg (g 1v).Then(a)Ifv Vthen (v) W, and ifw Wthen (w) =w, so is a projection ontoW.(b)im =Wand |W= id, soker im =V.(c)For allv Vh (v) =|G| 1 g Ghg (g 1v)=|G| 1 g G(hg) ((hg) 1hv) = (hv)so isG-linear.(d)If (v) = 0 thenh (v) = (hv) = 0, so ker ker is aG-invariant complement follows easily from (1).2 CharactersFor the whole of this sections, all groups will be finite and all representations will be on finite-dimensional vector spaces :G Aut(V) is a represention, thecharacterof is the function :G Cg7 tr( (g)).

4 Properties of the character1. does not depend on a choice of basis and are isomorphic representations then (g) = (g) for allg (1) = (g) = (hgh 1) and so is constant on conjugacy classes (g) = (g 1).6. (g) = (g) + (g).The space of class functionsAclass functiononGis a functionf:G Cwhich is constant on conjugacy classes ofG. SoifVis a Representation ofGoverCthen is a class function onG. We writeCGfor the set ofall class functions onG. This is a complex vector space, with a basis O:G Cg7 {1 ifg O0 ifg / OwhereOranges over the conjugacy classes can define a Hermitian inner product onCGby f, f =|G| 1 g Gf(g)f (g).If and are irreducible representations, then , ={1 if is isomorphic to 0 if is not isomorphic to .Thus the irreducible characters form part of an orthonormal basis forCG, and so the number ofdistinct irreducible reresentations is at most the number of conjugacy classes ofG. In fact, the3irreducible characters form an orthonormal basis forCG, and hence there are precisely as manydistinct irreducible representations as there are conjugacy classes of orthogonalityIf is an arbitrary Representation ofGwith character , and 1.}}

5 , kare the distinct irre-ducible characters, then by complete reducibility =n1 1+ +nk someni N. Therefore , i =niby orthogonality, and so = ni iwhereni= , i . any decomposition of into a sum of irreducible representations, each irreducible rep-resentation occurs the same number of and are representations ofGwith the same character, then = . as above, , = n2iand so is irreducible iff , = regular representationAny groupGacts on itself by left-multiplication. The permutation Representation of this actionis called theregular representationofG. If is the character of the regular Representation ofGthen (g) ={|G|ifg= [G] = (dim 1) 1 (dim k) k,and in particular|G|= (dim i) orthogonalityFixgandh G. Then irreducible (g) (h) ={|CG(h)|ifgis conjugate toh0ifgis not conjugate is a formal consequence of the orthogonality of of orthogonalityFirst we need the following (Schur s Lemma)Suppose that ( , V) and ( , V ) are irreducible representations ofGoverC, with characters and respectively.}}

6 Suppose that :V V is aG-linear map. is an isomorphism or = :V Vis an isomorphism then is multiplication by a scalar C, and soHomG(V, V ) ={Cif is isomorphic to 0 if is not isomorphic to . that ker is a subrepresentation ofVand im is a subrepresentation ofV . SinceVandV are both irreducible, ker = 0 orVand im = 0 orV . The result algebraically closed, has an eigenvalue and an eigenvectorvfor . Then = Iis also aG-linear mapV V. But (v) = 0 and so ker 6= 0. But then sinceVis irreducible, ker =Vand so = let ( , V) and ( , V ) be as above and let :V V beanylinear map. DefineAv =|G| 1 g Gg 1 Av is aG-linear map. Furthermore, tr(Av ) = tr , so in particular if tr 6= 0 thenAv 6= on to the main part of the proof. Choose bases forVandV and write (g) and (g) asmatrices with respect to these bases. Then , =|G| 1 g G (g) (g)=|G| 1 g Gtr( (g))tr( (g))=|G| 1 g Gi,j (g)ii (g 1) be in passing that a consequence of Schur s Lemma and the orthogonality of characters isthat if ( , V) and ( , V ) are two representations ofGwith characters and , thendim HomG(V, V ) =.}

7 5 Operations on charactersMotivationLetf, f CG. Then the following operations are defined: inner product: f, f sum: (f+f )(g) =f(g) +f (g) involution:f (g) =f(g 1) product: (ff )(g) =f(g)f (g).Note that = + , so that the sum of characters has a Representation theoretic inter-pretation but what about the others? We shall define thedualof a Representation and thetensor productof two representations dual of a representationIf is a Representation ofGonV, we can makeGact on the dual vector spaceV . Define thedual Representation of by( (g) )(v) = ( (g 1)v).By considering the matrices of and with respect to a pair of dual bases, we see that thematrix of (g) is the transpose of the matrix of (g 1), and so (g) = (g 1) = (g).Note that in general there is noG-linear isomorphism betweenVandV . In fact,V =V iff V= V, that is, if V(g) Rfor allg tenor productLetVandWbe vector spaces overk, with basesv1, .. , vmandw1, .. , wnrespectively. Thenwe define thetensor productofVandWto be the vector spaceV Wwith basisvi wjfor1 i mand 1 j n.

8 So dim(V W) = dim(V) dim(W).We define a bilinear mapV W V Wby defining it on a basis as(vi, wj)7 vi wjand extending linearly, so( ivi, jwj)7 i j(vi wj).We denote the image of (v, w) V Wbyv important property of tensor products is that for any vector spaceUoverk, there exists abijection{bilinear mapsV W U} {linear mapsV W U},andV Wis the unique vector space with this property. We say that the tensor product isuniversal for bilinear mappings .6 Iff:V Vandf :W Ware linear maps, we can define a linear map (f f ) asV W V Wvi wj7 f(vi) f (wj). w7 f(v) f(w) for allv Vand allw (f f ) = tr(f) tr(f ).Symmetric and exterior powersWe may define a linear map :V V V Vbyvi vj7 vj vi,sov v 7 v vfor allv, v V. Then 2= 1, and so ( 1)( + 1) = 0. Hence has thetwo eigenvalues 1 onV V. DefineS2V={a V V| a=a} 2V={a V V| a= a}to be the two eigenspaces of . These are called thesecond symmetric powerand thesecondexterior the basisvivj=vi vj+vj vi(1 i j d)and so dim(S2V) =d(d+ 1)/2, whered= dimV.

9 2 Vhas the basisvi vj=vi vj vj vj(1 i < j d)and so dim( 2V) = (d 1)d/2. HenceV V=S2V general, we writeV nto meanV V, and for each Snwe define a linear map :V n V nbyvi1 vin7 vi (1) vi (n).Then we defineSnV={a V n| a=afor all Sn}={a V n| a=afor all = (i(i+ 1))} nV={a V n| a= sgn( )afor all Sn}={a V n| a= afor all = (i(i+ 1))},anddim(SnV) =(d+ 1n)dim( nV) =(dn).7 The tensor product of two representationsLet ( , V) and ( , V ) be two representations ofG. Then thetensor product of and is defined by( )(g) = (g) (g). is a Representation ofGonV ,SnVand nVare subrepresentations ofV n. In the casen= 2,V V=S2V 2 Vand the characters ofS2 Vand 2 Vare S2V=12[( (g))2+ (g2)] V2V=12[( (g))2 (g2)].8 Induction and RestrictionInductionLetGbe a finite group and letH G. Given a representationVofHwe define theinducedrepresentationasIndGHV= HomH(CG, V)={f:G V|f(hg) =hf(g) for allh H, g G},the space ofH-linear maps on IndGHVin the following manner: Ifx Gandf:G V HomH(CG, V) then(x f)(g) =f(gx).

10 Properties of the induced representationLetVbe a Representation ofHand let be its character. IndGHV=|H\G| dimV= (|G|/|H|) is the character of IndGHVthen (x) =|H| 1 g Ggxg 1 H (gxg 1) = Hg H\GHgx=Hg (gxg 1).RestrictionLetGbe a finite group and letH G. Given a representationWofGwe get a representationResGHWofHjust by resricting the domain of the Representation obenius reciprocityLetVbe a Representation ofHwith character and letWbe a Representation ofGwithcharacter . Then ,IndGH G= ResGH , HandHomG(W,IndGHV) = HomH(ResGHW, V).The Mackey formulaSee lecture GroupsAtopological groupis a group which is also a topological space, and where the group operationsare continuous maps with respect to this topology. Acompact groupis a topological group whichis compact as a topological a topological group is acontinuousgroup homomorphism :G Aut(V)for some finite dimensional vector spaceV. (In fact every Representation of a compact group isalso differentiable but we won t prove this.)