introduction to representation theory (etingof)

7/14/2019 Introduction to Representation Theory (Etingof)

1/106

IntroductiontorepresentationtheoryPavelEtingof,OlegGolberg,SebastianHensel,

TiankaiLiu,AlexSchwendner,ElenaUdovinaandDmitryVaintrobJuly13,2010

Contents1 Basicnotionsofrepresentationtheory 5

1.1 What is representation theory? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.5 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.6 Algebras defined by generators and relations . . . . . . . . . . . . . . . . . . . . . . . 111.7 Examples of algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.8 Quivers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.9 Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.10 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.11 The tensor algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.12 Hilb erts third problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.13 TensorproductsanddualsofrepresentationsofLiealgebras . . . . . . . . . . . . . . 201.14 Representationsofsl(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.15 Problems on Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Generalresultsofrepresentationtheory 232.1 Subrepresentationsinsemisimplerepresentations . . . . . . . . . . . . . . . . . . . . 232.2 The density theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.3 Representations of direct sums of matrix algebras . . . . . . . . . . . . . . . . . . . . 242.4 Filtrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.5 Finite dimensional algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1


2/106

2.6 Characters of representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.7 The Jordan-Holder theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.8 The Krull-Schmidt theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.10 Representations of tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 Representationsoffinitegroups: basicresults 333.1 Maschkes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.4 Duals and tensor products of representations . . . . . . . . . . . . . . . . . . . . . . 363.5 Orthogonality of characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.6 Unitaryrepresentations. AnotherproofofMaschkestheoremforcomplexrepresen

tations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.7 Orthogonality of matrix elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.8 Character tables, examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.9 Computingtensorproductmultiplicitiesusingcharactertables . . . . . . . . . . . . 423.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4 Representationsoffinitegroups: furtherresults 474.1 Frobenius-Schur indicator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.2 Frobenius determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.3 Algebraic numbers and algebraic integers . . . . . . . . . . . . . . . . . . . . . . . . 494.4 Frobenius divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.5 Burnsides Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.6 Representations of products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.7 Virtual representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.8 Induced Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.9 The Mackey formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.10 Frobenius reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.11 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.12 RepresentationsofSn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.13 Proof of Theorem 4.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.14 InducedrepresentationsforSn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

2


3/106

4.15 The Frobenius character formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.16 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.17 Th e h ook le n gth f or mu la . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.18 Schur-Weyldualityforgl(V) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.19 Schur-WeyldualityforGL(V) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.20 Schur polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.21 ThecharactersofL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.22 PolynomialrepresentationsofGL(V) . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.23 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.24 RepresentationsofGL2(Fq) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.24.1 Conjugacyclasses inGL2(Fq) . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.24.2 1-dimensional representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.24.3 Principal series representations . . . . . . . . . . . . . . . . . . . . . . . . . . 694.24.4 Complementary series representations . . . . . . . . . . . . . . . . . . . . . . 72

4.25 Artins theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.26 Representations of semidirect products . . . . . . . . . . . . . . . . . . . . . . . . . . 74

5 QuiverRepresentations 765.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 765.2 IndecomposablerepresentationsofthequiversA1, A2, A3 . . . . . . . . . . . . . . . . 795.3 IndecomposablerepresentationsofthequiverD4 . . . . . . . . . . . . . . . . . . . . 815.4 Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.5 Gabriels theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.6 Reflection Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885.7 Coxeter elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.8 Proof of Gabriels theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 925.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

6 Introduction tocategories 966.1 The definition of a category . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 966.2 Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 976.3 Morphisms of functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 986.4 Equivalence of categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 986.5 Representable functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

3


4/106

6.6 Adjoint functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1006.7 Abelian categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016.8 Exact functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

7 Structureoffinitedimensionalalgebras 1037.1 Projective modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1037.2 Lifting of idempotents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1037.3 Projective covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

INTRODUCTIONVeryroughlyspeaking,representationtheorystudiessymmetryinlinearspaces. Itisabeautiful

mathematicalsubjectwhichhasmanyapplications,rangingfromnumbertheoryandcombinatoricstogeometry,probabilitytheory,quantummechanicsandquantumfieldtheory.

Representation theory was born in 1896 in the work of the German mathematician F. G.Frobenius. ThisworkwastriggeredbyalettertoFrobeniusbyR.Dedekind. InthisletterDedekindmadethefollowingobservation: takethemultiplicationtableofafinitegroupGandturnitintoamatrixXG byreplacingeveryentryg ofthistablebyavariablexg. ThenthedeterminantofXGfactors into a product of irreducible polynomials in

{xg

}, each of which occurs with multiplicity

equaltoitsdegree. Dedekindcheckedthissurprisingfactinafewspecialcases,butcouldnotproveit in general. So he gave this problem to Frobenius. In order to find a solution of this problem(whichwewillexplainbelow),Frobeniuscreatedrepresentationtheoryoffinitegroups. 1

ThepresentlecturenotesarosefromarepresentationtheorycoursegivenbythefirstauthortotheremainingsixauthorsinMarch2004withintheframeworkoftheClayMathematicsInstituteResearch Academy forhighschool students, and its extendedversiongiven bythefirstauthor toMIT undergraduate math students in the Fall of 2008. The lectures are supplemented by manyproblemsandexercises,whichcontaina lotofadditionalmaterial;themoredifficultexercisesareprovidedwithhints.

Thenotescoveranumberofstandardtopicsinrepresentationtheoryofgroups,Liealgebras,andquivers. Wemostlyfollow[FH],withtheexceptionofthesectionsdiscussingquivers,whichfollow[BGP]. We alsorecommend thecomprehensive textbook [CR]. Thenotesshouldbeaccessible tostudentswithastrongbackground inlinearalgebraandabasicknowledgeofabstractalgebra.Acknowledgements. TheauthorsaregratefultotheClayMathematicsInstituteforhosting

thefirstversionofthiscourse. Thefirstauthor isvery indebtedtoVictorOstrik forhelpinghimpreparethiscourse, andthanksJoshNichols-Barrer andThomasLam for helpingrunthe coursein2004andforusefulcomments. HeisalsoverygratefultoDarijGrinbergforverycarefulreadingofthetext,formanyusefulcommentsandcorrections,andforsuggestingtheExercisesinSections1.10,2.3,3.5,4.9,4.26,and6.8.

Formoreonthehistoryofrepresentationtheory,see [Cu].4

1


5/106

1 Basicnotionsofrepresentation theory1.1 What isrepresentationtheory?Intechnicalterms,representationtheorystudiesrepresentationsofassociativealgebras. Itsgeneralcontentcanbeverybrieflysummarizedasfollows.

Anassociativealgebraoverafieldk isavectorspaceAoverkequippedwithanassociativebilinearmultiplication a, b ab,a, bA. Wewillalways considerassociative algebraswithunit,i.e., withanelement1suchthat1 a=a 1 =aforallaA. Abasic exampleofanassociative algebra isthe algebra EndV of linear operators froma vector spaceV to itself. Other importantexamplesincludealgebrasdefinedbygeneratorsandrelations,suchasgroupalgebrasanduniversalenvelopingalgebrasofLiealgebras.

Arepresentation of an associative algebra A (also called a left A-module) is a vector spaceV equippedwithahomomorphism:AEndV, i.e.,a linearmappreservingthemultiplicationandunit.

AsubrepresentationofarepresentationV isasubspaceU V which is invariantunderalloperators (a), aA. Also, if V1, V2 are two representations of A thenthedirect sum V1V2hasanobviousstructureofarepresentationofA.

A nonzero representation V of A is said to be irreducible if its only subrepresentations are0 and V itself, and indecomposable if it cannot be written as a direct sum of two nonzerosubrepresentations. Obviously, irreducible implies indecomposable,butnotviceversa.

Typicalproblemsofrepresentationtheoryareasfollows:1. Classify irreduciblerepresentationsofagivenalgebraA.2. Classify indecomposablerepresentationsofA.3. Do1and2restrictingtofinitedimensionalrepresentations.As mentioned above, the algebra A is often given to us by generators and relations. For

example, the universal enveloping algebra U of the Lie algebrasl(2) is generated by h,e,f withdefiningrelations

heeh= 2e, hff h=2f, eff e=h. (1)Thismeansthattheproblemoffinding,say,N-dimensionalrepresentationsofAreducestosolvinga bunch of nonlinear algebraic equations with respect to a bunch of unknown N by N matrices,forexamplesystem(1)withrespecttounknownmatricesh,e,f.

It isreally strikingthatsuch, atfirstglance hopelesslycomplicated, systemsofequationscanin factbesolvedcompletely bymethodsofrepresentationtheory! Forexample,wewillprovethefollowingtheorem.Theorem1.1. Letk=Cbe thefieldofcomplexnumbers. Then:

(i)ThealgebraU hasexactlyoneirreduciblerepresentation Vd ofeachdimension,uptoequivalence;thisrepresentation isrealized inthespaceofhomogeneouspolynomialsoftwovariablesx, yofdegree d1,anddefinedby theformulas

(h) =xxy y, (e) =x y, (f) =y x.

(ii)AnyindecomposablefinitedimensionalrepresentationofU isirreducible.Thatis,anyfinite5


6/106

|

|

|

dimensional representationofU isadirect sumof irreducible representations.Asanotherexampleconsidertherepresentationtheoryofquivers.Aquiver isafiniteorientedgraphQ. ArepresentationofQoverafieldk isanassignment

ofak-vectorspaceVi toeveryvertexiofQ,andofalinearoperatorAh :Vi Vj toeverydirectededgehgoingfromitoj(loopsandmultipleedgesareallowed). WewillshowthatarepresentationofaquiverQisthesamethingasarepresentationofacertainalgebraPQ calledthepathalgebraofQ. Thusonemayask: whatarethe indecomposablefinitedimensionalrepresentationsofQ?

More specifically, let us say that Q isof finite type if it has finitely many indecomposablerepresentations.

Wewillprovethefollowingstrikingtheorem,provedbyP.Gabrielabout35yearsago:Theorem 1.2. Thefinite type property of Q does not depend on the orientation of edges. Theconnectedgraphs thatyieldquiversoffinitetypearegivenby thefollowing list:

An :

Dn:

E6 :

E7 :

E8 : |

Thegraphs listed in thetheoremare called (simply laced)Dynkindiagrams. Thesegraphsarise ina multitudeof classification problems inmathematics, suchasclassification ofsimpleLiealgebras,singularities,platonicsolids,reflectiongroups,etc. Infact, ifweneededtomakecontactwith an alien civilization and show them how sophisticated our civilization is, perhaps showingthemDynkindiagramswouldbethebestchoice!

Asafinalexampleconsidertherepresentationtheoryoffinitegroups,whichisoneofthemostfascinating chapters of representation theory. In this theory, one considers representations of thegroupalgebraA=C[G] ofafinitegroupGthealgebra withbasisag, gGandmultiplicationlaw agah =agh. We will show that any finite dimensional representation of A is a direct sum ofirreducible representations, i.e., the notions of an irreducible and indecomposable representationarethesameforA(Maschkestheorem). Anotherstrikingresultdiscussedbelow istheFrobeniusdivisibility theorem: the dimensionof any irreduciblerepresentation of A divides the order of G.Finally,wewillshowhowtouserepresentationtheoryoffinitegroupstoproveBurnsidestheorem:

aanyfinitegroupoforderp qb,wherep, q areprimes, issolvable. Note thatthistheoremdoesnotmentionrepresentations,whichareusedonly in itsproof;apurelygroup-theoretical proofofthistheorem(notusingrepresentations)existsbut ismuchmoredifficult!

6


7/106

1.2 AlgebrasLetusnowbeginasystematicdiscussionofrepresentationtheory.

Letk beafield. Unlessstatedotherwise,wewillalwaysassumethatk isalgebraically closed,i.e., any nonconstant polynomial with coefficients in k has a root in k. The main example is thefieldofcomplexnumbersC,butwewillalsoconsiderfieldsofcharacteristicp,suchasthealgebraicclosureFp ofthefinitefieldFp ofpelements.Definition1.3. Anassociativealgebraoverk isavectorspaceAoverk togetherwithabilinearmapAAA, (a, b)ab,suchthat(ab)c=a(bc).Definition1.4. AunitinanassociativealgebraA isanelement1Asuchthat1a=a1 =a.Proposition1.5. Ifaunitexists, it isunique.Proof. Let1,1 betwounits. Then1=11 = 1.

From now on, by an algebra A we will mean an associative algebra witha unit. We will alsoassumethatA=0.Example1.6. Herearesomeexamplesofalgebrasoverk:

1. A=k.2. A=k[x1,...,xn]thealgebraofpolynomials invariablesx1,...,xn.3. A=EndV thealgebraofendomorphismsofavectorspaceV overk (i.e., linearmaps,or

operators,fromV to itself). Themultiplication isgivenbycompositionofoperators.4. The free algebra A = k x1,...,xn . A basis of this algebra consists of words in letters

x1,...,xn,andmultiplication inthisbasis issimplyconcatenationofwords.5. ThegroupalgebraA=k[G]ofagroupG. Itsbasis is{ag, gG},withmultiplication law

agah =agh.Definition1.7. AnalgebraAiscommutative ifab=ba foralla, bA.

Forinstance,intheaboveexamples,Aiscommutativeincases1and2,butnotcommutativeincases3(ifdimV >1),and4(ifn >1). Incase5,AiscommutativeifandonlyifGiscommutative.Definition 1.8. A homomorphism of algebras f : A B is a linear map such that f(xy) =f(x)f(y)forallx, yA,andf(1)=1.1.3 RepresentationsDefinition 1.9. Arepresentation ofan algebra A (also called a left A-module) isa vector spaceV togetherwithahomomorphismofalgebras:AEndV.

Similarly,arightA-module isaspaceV equippedwithanantihomomorphism:A

EndV;i.e.,satisfies(ab) =(b)(a)and(1)=1.

Theusualabbreviatednotation for(a)v isav fora leftmoduleandva forthe right module.Thenthepropertythatisan(anti)homomorphismcanbewrittenasakindofassociativity law:(ab)v=a(bv) forleftmodules,and(va)b=v(ab)forrightmodules.

Herearesomeexamplesofrepresentations.7


8/106

V

Example1.10. 1. V =0.2. V =A,and:AEndAisdefinedasfollows: (a)istheoperatorofleftmultiplicationby

a,sothat(a)b=ab(theusualproduct). Thisrepresentation iscalledtheregular representationofA. Similarly,onecanequipAwithastructureofarightA-modulebysetting(a)b:=ba.

3. A=k. ThenarepresentationofAissimplyavectorspaceoverk.4. A=k x1,...,xn . ThenarepresentationofAisjustavectorspaceV overkwithacollection

ofarbitrary linearoperators(x1),...,(xn) :V V (explainwhy!).Definition1.11. AsubrepresentationofarepresentationV ofanalgebraAisasubspaceW

which is invariantunderalltheoperators(a) :V V,aA.For instance,0andV arealwayssubrepresentations.

Definition1.12. ArepresentationV =0ofAisirreducible(orsimple)iftheonlysubrepresentationsofV are0andV.Definition 1.13. Let V1, V2 be two representations of an algebra A. A homomorphism (or intertwining operator) :V1 V2 is a linear operator which commutes with the action of A, i.e.,(av) =a(v)foranyvV1. Ahomomorphismissaidtobeanisomorphismofrepresentationsif it isan isomorphismof vector spaces. Theset(space)of all homomorphismsof representationsV1

V2 isdenotedbyHomA(V1, V2).

Note that if a linear operator:V1 V2 isan isomorphismofrepresentations thenso isthelinearoperator1 :V2 V1 (check it!).

Tworepresentationsbetweenwhichthereexistsanisomorphismaresaidtobeisomorphic. Forpracticalpurposes,twoisomorphicrepresentationsmayberegardedasthesame,althoughtherecould besubtleties related to the fact that an isomorphismbetween two representations, when itexists, isnotunique.Definition 1.14. Let V1, V2 be representations of an algebra A. Then the space V1V2 has anobviousstructureofarepresentationofA,givenbya(v1v2) =av1av2.Definition1.15. AnonzerorepresentationV ofanalgebraAissaidtobeindecomposableifitisnot isomorphictoadirectsumoftwononzerorepresentations.

It isobviousthatan irreduciblerepresentation is indecomposable. Ontheotherhand,wewillseebelowthattheconversestatement is false ingeneral.

Oneofthemainproblemsofrepresentationtheoryistoclassifyirreducibleandindecomposablerepresentationsofagivenalgebraupto isomorphism. Thisproblem isusuallyhardandoftencanbe solved onlypartially (say, for finite dimensionalrepresentations). Below we will see a numberofexamples inwhichthisproblemispartiallyorfullysolved forspecificalgebras.

We will now prove our first result Schurs lemma. Although it is very easy to prove, it isfundamental inthewholesubjectofrepresentationtheory.Proposition 1.16. (Schurs lemma)Let V1, V2 be representations ofanalgebra Aover anyfieldF (which need not be algebraically closed). Let : V1 V2 be a nonzero homomorphism ofrepresentations. Then:

(i)IfV1 is irreducible, is injective;8


9/106

(ii)IfV2 is irreducible, issurjective.Thusm, ifboth V1 andV2 are irreducible, isan isomorphism.

Proof. (i) The kernel K of is a subrepresentation of V1. Since = 0, this subrepresentationcannotbeV1. Soby irreducibilityofV1 wehaveK=0.

(ii) The imageI of isasubrepresentationofV2. Since=0, thissubrepresentationcannotbe0. SobyirreducibilityofV2 wehaveI=V2.Corollary 1.17. (Schurs lemmafor algebraically closed fields) Let V be afinite dimensionalirreduciblerepresentationofanalgebraAoveranalgebraicallyclosedfieldk,and:V V isanintertwiningoperator. Then= Idforsome k (ascalaroperator).Remark. Note that this Corollary is false over the field of real numbers: it suffices to take

A=C(regardedasanR-algebra),andV =A.Proof. Letbeaneigenvalueof(arootofthecharacteristic polynomialof). Itexistssincekisanalgebraicallyclosedfield. ThentheoperatorIdisan intertwiningoperatorV V,whichis not an isomorphism (since its determinant is zero). Thus by Proposition 1.16 this operator iszero,hencetheresult.Corollary

1.18.

Let

A

be

a

commutative

algebra.

Then

every

irreducible

finite

dimensional

representation V ofA is1-dimensional.

Remark. Notethata1-dimensionalrepresentationofanyalgebraisautomaticallyirreducible.Proof. LetV beirreducible. ForanyelementaA,theoperator(a) :V V isan intertwiningoperator. Indeed,

(a)(b)v=(ab)v=(ba)v=(b)(a)v(the second equality is true since the algebra is commutative). Thus, by Schurs lemma, (a) isa scalar operator for any a A. Hence every subspace of V is a subrepresentation. But V isirreducible,so0andV aretheonlysubspacesofV. ThismeansthatdimV =1(sinceV =0).Example1.19. 1. A=k. SincerepresentationsofAaresimplyvectorspaces,V =A istheonlyirreducibleandtheonly indecomposablerepresentation.

2. A = k[x]. Since this algebra is commutative, the irreducible representations of A are its1-dimensional representations. Aswediscussedabove, theyaredefinedbyasingleoperator(x).Inthe1-dimensionalcase,thisisjustanumberfromk. SoalltheirreduciblerepresentationsofAareV =k,k, inwhichtheactionofAdefinedby(x) =. Clearly,theserepresentationsarepairwisenon-isomorphic.

The classification of indecomposable representations of k[x] is more interesting. To obtain it,recall that any linear operator on a finite dimensional vector space V can be brought to Jordannormal form. More specifically, recall that the Jordan block J

,n is the operator on kn which in

thestandardbasis isgivenbythe formulasJ,nei =ei +ei1 fori >1,andJ,ne1 =e1. ThenforanylinearoperatorB:V V thereexistsabasisofV suchthatthematrixofB inthisbasisisadirectsumofJordanblocks. ThisimpliesthatalltheindecomposablerepresentationsofAareV,n =kn, k, with (x) =J,n. The fact that these representations are indecomposable andpairwise non-isomorphic follows from the Jordan normal form theorem (which in particular saysthattheJordannormalformofanoperator isuniqueuptopermutationofblocks).

9


10/106

Thisexampleshowsthatanindecomposablerepresentationofanalgebraneednotbeirreducible.3. ThegroupalgebraA=k[G],whereGisagroup. ArepresentationofAisthesamethingas

arepresentationofG,i.e.,avectorspaceV togetherwithagrouphomomorphism:GAut(V),whreAut(V) =GL(V)denotesthegroupof invertible linearmapsfromthespaceV to itself.Problem1.20. Let V beanonzerofinitedimensional representation ofanalgebra A. Show thatithas an irreducible subrepresentation. Then show by example that thisdoesnotalways holdforinfinitedimensionalrepresentations.Problem1.21.LetAbeanalgebraoverafieldk.ThecenterZ(A)ofAisthesetofallelementsz

AwhichcommutewithallelementsofA. Forexample, ifA iscommutative thenZ(A) =A.(a) Show that if V is an irreduciblefinite dimensional representation of A then any element

z Z(A) acts in V bymultiplication by some scalar V(z). Show that V : Z(A) k is ahomomorphism. It iscalled thecentralcharacterofV.

(b) Show that if V is an indecomposablefinite dimensional representation of A thenfor anyz Z(A), the operator (z) bywhich z acts in V has only one eigenvalue V(z), equal to thescalar by which z acts on some irreducible subrepresentation of V. Thus V : Z(A) k is ahomomorphism,which isagaincalled thecentralcharacterof V.

(c)Does (z) in(b)have tobeascalaroperator?Problem 1.22. Let A beanassociative algebra, and V a representation of A. By EndA(V)onedenotes thealgebraofallhomomorphisms of representations V V. Show that EndA(A) =Aop,thealgebra Awithoppositemultiplication.Problem1.23. Prove thefollowingInfinitedimensionalSchurs lemma (due toDixmier): LetAbeanalgebra over Cand V bean irreducible representation of Awithatmost countablebasis.Thenanyhomomorphism ofrepresentations :V V isascalaroperator.

Hint. By the usual Schurs lemma, the algebra D := EndA(V) is an algebra with division.ShowthatD isatmostcountablydimensional. Supposeisnotascalar,andconsiderthesubfieldC() D. Show that C() is a transcendental extension of C. Derivefrom this that C() isuncountablydimensionalandobtainacontradiction.1.4 IdealsA left ideal ofanalgebraA isasubspaceIAsuchthataII forallaA. Similarly,arightideal of analgebra A is a subspace I A suchthat IaI forall aA. A two-sided ideal is asubspacethat isbotha leftandaright ideal.

LeftidealsarethesameassubrepresentationsoftheregularrepresentationA. RightidealsarethesameassubrepresentationsoftheregularrepresentationoftheoppositealgebraAop.

Belowaresomeexamplesof ideals:

If A is any algebra, 0 andA are two-sided ideals. Analgebra A is called simple if 0 and A

are itsonlytwo-sided ideals. If:AB isahomomorphismofalgebras,thenker isatwo-sided idealofA.

IfS isanysubsetofanalgebraA,thenthetwo-sidedidealgenerated byS isdenoted S and is the span of elements of the form asb, where a, bA andsS. Similarly we can defineS =span{as}andSr =span{sb},the left,respectivelyright, idealgeneratedbyS.

10


11/106

1.5 QuotientsLet A be an algebra and I a two-sided ideal in A. Then A/I is the set of (additive) cosets of I.Let:AA/I bethequotientmap. WecandefinemultiplicationinA/I by(a) (b):=(ab).Thisiswelldefinedbecause if(a) =(a)then

(ab) =(ab+ (aa)b) =(ab) +((aa)b) =(ab)because(aa)bIbI=ker,asI isaright ideal;similarly, if(b) =(b)then

(ab) =(ab+a(b

b))=(ab) +(a(b

b))=(ab)becausea(bb)aII=ker,asI isalsoa left ideal. Thus,A/I isanalgebra.

Similarly, ifV isarepresentationofA,andW V isasubrepresentation,thenV /W isalsoarepresentation. Indeed,let:V V /W bethequotientmap,andsetV /W(a)(x):=(V(a)x).

Abovewenotedthat left idealsofAaresubrepresentationsoftheregularrepresentationofA,andviceversa. Thus,ifI isa left ideal inA,thenA/I isarepresentationofA.Problem 1.24. Let A = k[x1,...,xn] and I = A be any ideal in A containing all homogeneouspolynomialsofdegreeN. Show that A/I isan indecomposable representationofA.Problem 1.25. Let V

= 0 be a representation of A. We say that a vector v

V is cyclic if it

generates V, i.e., Av =V. A representation admitting a cyclic vector is said to be cyclic. Showthat

(a) V is irreducible ifandonly ifallnonzerovectorsofV arecyclic.(b) V iscyclic ifandonly if it is isomorphic toA/I,where I isa left ideal inA.(c)Giveanexampleofan indecomposable representationwhich isnotcyclic.Hint. Let A=C[x, y]/I2,where I2 is the ideal spanned byhomogeneous polynomials ofdegree

2(soAhasabasis 1, x , y). Let V =A be thespaceof linearfunctionalson A,with theactionofAgivenby((a)f)(b) =f(ba). Show that V providessuchanexample.1.6 AlgebrasdefinedbygeneratorsandrelationsIf f1, . . . , f m are elements of the free algebra k x1, . . . , xn , we say that the algebra A:=kx1, . . . , xn/{f1, . . . , f m}isgeneratedbyx1, . . . , xnwithdefiningrelationsf1 = 0, . . . , fm =0.1.7 Examplesofalgebras

1. TheWeylalgebra,k x,y / yxxy1 . 2. Theq-Weylalgebra,generatedbyx, x1, y , y1 withdefiningrelationsyx=qxyandxx1 =x1x=yy1 =y1y=1.Proposition. (i)AbasisfortheWeylalgebraA is{xiyj, i , j0}.(ii)Abasisfortheq-WeylalgebraAq is{xiyj, i , jZ}.

11


12/106

Proof. (i)Firstletusshowthattheelementsxiyj areaspanningsetforA. Todothis,notethatanyword inx, y canbeorderedtohaveallthexonthe leftofthey,atthecost of interchangingsome x and y. Since yxxy =1, this will lead to error terms, but these terms will be sums ofmonomialsthathaveasmallernumberoflettersx, ythantheoriginalword. Therefore,continuingthisprocess,wecanordereverythingandrepresentanywordasa linearcombinationofxiyj.

Theproofthatxiyj arelinearlyindependentisbasedonrepresentationtheory. Namely,letabeavariable,andE=tak[a][t, t1](hereta isjustaformalsymbol,soreallyE=k[a][t, t1]). ThenE

df d(ta+n)isarepresentationofAwithactiongivenbyxf =tf andyf = dt (where dt :=(a+n)ta+n1).iSupposenowthatwehaveanontrivial linearrelation

cijx yj =0. Thentheoperator

idjL= cijtdt

actsbyzero inE. LetuswriteLasr

djL= Qj(t) ,

dtj=0

whereQr =0. Thenwehaver

Lta =

Qj(t)a(a1)...(aj+1)taj.

j=0Thismust bezero, so we haver Qj(t)a(a1)...(aj+1)tj =0 in k[a][t, t1]. Taking thej=0leadingterm ina,wegetQr(t)=0,acontradiction.

(ii)Anywordinx,y,x1, y1 canbeorderedatthecostofmultiplyingitbyapowerofq. Thiseasily impliesboththespanningpropertyandthe linear independence.Remark. The proof of (i) shows that the Weyl algebra A can be viewed as the algebra of

polynomialdifferentialoperators inonevariablet.Theproofof(i)alsobringsupthenotionofafaithfulrepresentation.Definition. Arepresentation:AEndV isfaithful if isinjective.For example, k[t] is a faithful representation of the Weyl algebra, if k has characteristic zero

(check it!), but not in characteristicp, where (d/dt)pQ=0 for any polynomial Q. However, therepresentationE=tak[a][t, t1],asweveseen, isfaithful inanycharacteristic.Problem1.26. LetAbe theWeylalgebra,generatedbytwoelementsx, ywith therelation

yxxy1 = 0.(a)Ifchark= 0,whatare thefinitedimensionalrepresentationsofA?Whatare thetwo-sided

ideals inA?Hint.Forthefirstquestion,usethefactthatfortwosquarematricesB, C,Tr(BC) =Tr(CB ).

Forthesecondquestion,showthatanynonzerotwo-sidedidealinAcontainsanonzeropolynomialinx,anduse this tocharacterize this ideal.

Supposefor therestof theproblem that chark=p.(b)What is thecenterofA?

12


13/106

Hint. Show that xp and yp arecentralelements.(c)Findall irreduciblefinitedimensional representationsofA.Hint. Let V bean irreduciblefinitedimensional representation of A,and v bean eigenvector

ofy inV. Show that{v,xv,x2v,...,xp1v} isabasisofV.Problem1.27.Letqbeanonzerocomplexnumber,andAbetheq-WeylalgebraoverCgeneratedby x1 and y1 withdefiningrelations xx1 =x1x= 1, yy1 =y1y= 1,andxy=qyx.

(a)What is thecenterof Afordifferent q? If q isnota rootofunity,whatare the two-sidedideals inA?

(b)Forwhichqdoes thisalgebrahavefinitedimensional representations?Hint.Usedeterminants.(c)Findallfinitedimensional irreduciblerepresentationsofAforsuchq.Hint.This issimilar topart (c)of thepreviousproblem.

1.8 QuiversDefinition1.28. AquiverQ isadirectedgraph,possiblywithself-loopsand/ormultipleedgesbetween

two

vertices.

Example1.29.

WedenotethesetofverticesofthequiverQasI,andthesetofedgesasE. ForanedgehE,

leth,h denotethesourceandtargetofh,respectively:h h h

Definition1.30. ArepresentationofaquiverQisanassignmenttoeachvertexiI ofavectorspaceVi andtoeachedgehE ofalinearmapxh :Vh Vh.

Itturnsoutthatthetheoryofrepresentationsofquiversisapartofthetheoryofrepresentationsofalgebras inthe sensethat foreach quiverQ, thereexists acertain algebra PQ,called thepathalgebra of Q, such that a representation ofthe quiver Q is the same as a representation ofthealgebra PQ. We shall first define the path algebra of a quiver and thenjustify our claim thatrepresentationsofthesetwoobjectsarethesame.Definition 1.31. Thepath algebra PQ of a quiver Q is the algebra whose basis is formed byoriented paths in Q, including the trivial pathspi, iI, correspondingto the vertices of Q, andmultiplication isconcatenation ofpaths: ab isthepath obtained byfirsttracing bandthena. Iftwopathscannotbeconcatenated,theproductisdefinedtobezero.Remark1.32. Itiseasytoseethat forafinitequiverpi =1,soPQ isanalgebrawithunit.

iIProblem1.33. Show that thealgebra PQ isgenerated bypifor iI and ahfor hEwith thedefiningrelations:

13


14/106

1.p2i =pi,pipj = 0for i= j2. ahph =ah,ahpj = 0forj=h3.phah =ah,piah = 0for i=hWenowjustifyourstatementthatarepresentationofaquiveristhesamethingasarepresen

tationofthepathalgebraofaquiver.LetVbearepresentationofthepathalgebraPQ. Fromthisrepresentation,wecanconstructa

representationofQasfollows: letVi =piV,andforanyedgeh, letxh =ah|phV :phVphVbetheoperatorcorrespondingtotheone-edgepathh.

Similarly, let (Vi, xh) be a representation of a quiver Q. From this representation, we canconstruct a representation of the path algebra PQ: letV =iVi, letpi :V Vi V be theprojectionontoVi,andforanypathp=h1...hm letap =xh1...xhm :Vh Vh bethecompositionm 1oftheoperatorscorrespondingtotheedgesoccurring inp.

It isclearthattheaboveassignmentsV (piV)and(Vi) Vi are inversesofeachother.iThus,wehaveabijectionbetweenisomorphismclassesofrepresentationsofthealgebraPQ andofthequiverQ.Remark 1.34. In practice, it is generally easier to consider a representation of a quiver as inDefinition1.30.

We lastlydefineseveralpreviousconcepts inthecontextofquiversrepresentations.Definition1.35. Asubrepresentationofarepresentation(Vi, xh)ofaquiverQisarepresentation(Wi, xh

)whereWi Vi foralliI andwherexh(Wh)Wh andxh =xh|Wh :Wh Wh forallhE.Definition1.36. Thedirectsumoftworepresentations(Vi, xh)and(Wi, yh)istherepresentation(ViWi, xhyh).

As with representations of algebras, a nonzero representation (Vi) of a quiver Q is said to beirreducible if its only subrepresentations are (0) and (Vi) itself, and indecomposable if it is notisomorphictoadirectsumoftwononzerorepresentations.Definition1.37. Let(Vi, xh)and(Wi, yh)berepresentationsofthequiverQ. Ahomomorphism : (Vi) (Wi) of quiver representations is a collection of maps i : Vi Wi such thatyhh =h xh forallhE.Problem1.38. Let Abea Z+-graded algebra, i.e., A=n0A[n],and A[n] A[m] A[n+m].If A[n] isfinitedimensional, it isuseful to consider theHilbert series hA(t) =

dim

A[n]tn (the

generatingfunctionofdimensionsofA[n]).Oftenthisseriesconverges toarationalfunction,andtheanswer iswritten intheformofsuchfunction.Forexample, ifA=k[x]anddeg(xn) =nthen

hA(t) = 1 +t+t2+...+tn+...= 11t

Find theHilbertseriesof:(a) A=k[x1,...,xm](where thegrading isbydegreeofpolynomials);(b) A=k < x1,...,xm >(thegrading isby lengthofwords);

14


15/106

(c) A is the exterior (=Grassmann) algebra k[x1,...,xm], generated over some field k byx1,...,xm with the defining relations xixj +xjxi = 0 and x2i = 0for all i, j (the grading is bydegree).

(d)A isthepathalgebraPQ ofaquiverQ(thegrading isdefinedbydeg(pi) = 0,deg(ah) = 1).Hint.Theclosedanswer iswritten in termsof theadjacencymatrix MQ ofQ.

1.9 LiealgebrasLet

g

be

avector

space

over

afield

k,

and

let

[,] :

gggbeaskew-symmetricbilinearmap.(That is, [a, a]=0,andhence [a, b] =[b, a]).

Definition1.39. (g,[,]) isaLiealgebraif[,]satisfiestheJacobi identity[a, b], c + [b, c], a + [c, a], b = 0. (2)

Example1.40. SomeexamplesofLiealgebrasare:1. Anyspacegwith [,]=0(abelianLiealgebra).2. AnyassociativealgebraAwith [a, b] =abba .3. AnysubspaceU ofanassociative algebraAsuchthat [a, b]U foralla, bU.4. Thespace Der(A) of derivations of analgebra A, i.e. linearmaps D :AAwhich satisfy

theLeibnizrule:D(ab) =D(a)b+aD(b).

Remark 1.41. Derivations are important because they are the infinitesimal version of automorphisms. For example, assume that g(t) is a differentiable family of automorphisms of a finite dimensional algebra A over R or C parametrized by t (, ) such that g(0) = Id. ThenD :=g(0):AA isa derivation(check it!). Conversely, ifD :AA isaderivation, thenetDisa1-parameter familyofautomorphisms(giveaproof!).

This provides a motivation for the notion of a Lie algebra. Namely, we see that Lie algebrasariseasspacesof infinitesimalautomorphisms(=derivations)ofassociativealgebras. Infact,theysimilarlyariseasspacesofderivationsofanykindoflinearalgebraicstructures,suchasLiealgebras,Hopfalgebras,etc.,andforthisreasonplayavery importantrole inalgebra.

HereareafewmoreconcreteexamplesofLiealgebras:1. R3 with[u, v] =uv,thecross-productofuandv.2. sl(n),thesetofnnmatriceswithtrace0.

Forexample,sl(2)hasthebasis0 1 0 0 1 0

e= f = h=0 0 1 0 0 1

withrelations[h, e] = 2e, [h, f] =2f, [e, f] =h.

15


16/106

3. TheHeisenbergLiealgebraHofmatrices0 000 0 0

Ithasthebasis 0 0 0 0 1 0 0 0 1

x=0 0 1 y=0 0 0 c=0 0 00 0 0 0 0 0 0 0 0

withrelations [y, x] =cand [y, c] = [x, c]=0.4. Thealgebraaff(1)ofmatrices( 0 0)

ItsbasisconsistsofX= (1 0)andY = (0 1),with [X, Y] =Y.0 0 0 05. so(n),thespaceofskew-symmetricnnmatrices,with [a, b] =abba.Exercise. ShowthatExample1isaspecialcaseofExample5(forn=3).

Definition1.42. Letg1,g2 beLie algebras. Ahomomorphism :g1 g2 of Lie algebras is alinearmapsuchthat([a, b])=[(a), (b)].Definition 1.43. A representation of a Lie algebrag is a vector space V with a homomorphismofLiealgebras:gEndV.Example1.44. SomeexamplesofrepresentationsofLiealgebrasare:

1. V =0.2. AnyvectorspaceV with=0(thetrivialrepresentation).3. TheadjointrepresentationV =gwith(a)(b):=[a, b].

Thatthis isarepresentation followsfromEquation(2).ItturnsoutthatarepresentationofaLiealgebrag isthesamethingasarepresentationofa

certainassociativealgebraU(g). Thus,aswithquivers,wecanviewthetheoryofrepresentationsofLiealgebrasasapartofthetheoryofrepresentationsofassociativealgebras.

kDefinition1.45. LetgbeaLiealgebrawithbasisxi and [,]definedby [xi, xj] =kcijxk. Theuniversal enveloping algebraU(g) is the associative algebra generated by the xis with thedefiningrelationsxixjxjxi =kcijkxk.Remark. Thisisnotaverygooddefinitionsinceitdependsonthechoiceofabasis. Laterwe

willgiveanequivalentdefinitionwhichwillbebasis-independent.Example1.46. TheassociativealgebraU(sl(2))isthealgebrageneratedbye,f,hwithrelations

heeh= 2e hff h=2f eff e=h.Example1.47. Thealgebra

U(

H),where

HistheHeisenbergLiealgebra,isthealgebragenerated

byx,y,cwiththerelationsyxxy=c yccy= 0 xccx= 0.

NotethattheWeylalgebra isthequotientofU(H)bytherelationc=1.

16


17/106

1.10 TensorproductsInthissubsectionwerecallthenotionoftensorproductofvectorspaces,whichwillbeextensivelyusedbelow.Definition1.48. ThetensorproductVW ofvectorspacesV andW overafieldkisthequotientof the space V W whose basis is given by formal symbols vw, v V, w W (called puretensors),bythesubspacespannedbytheelements(v1+v2)wv1wv2w, v(w1+w2)vw1vw2, avwa(vw), vawa(vw),wherevV, wW, ak.Exercise. Show that V W can be equivalently defined as the quotient of the free abelian

groupV W generatedbyvw,vV, wW bythesubgroupgeneratedby(v1+v2)wv1wv2w, v(w1 +w2)vw1vw2, avwvaw,

wherevV, wW, ak.Thisallowsonetodefinethetensorproductofanynumberofvectorspaces,V1...Vn. Note

thatthistensorproductisassociative, inthesensethat(V1V2)V3 canbenaturally identifiedwithV1

(V2

V3).

Inparticular,peopleoftenconsidertensorproductsoftheformVn =V...V (ntimes)foragivenvectorspaceV,and,moregenerally,E :=Vn(V)m. Thisspaceiscalledthespaceoftensorsoftype(m, n)onV. For instance, tensorsoftype (0,1) arevectors, oftype(1,0) - linearfunctionals(covectors),oftype(1,1)- linearoperators,oftype(2,0)- bilinearforms,oftype(2,1)- algebrastructures,etc.

IfV isfinitedimensionalwithbasisei,i= 1,...,N,andei isthedualbasisofV,thenabasisofE isthesetofvectors

ei1 ...ein ej1 ...ejm,andatypicalelementofE is

N Ti1...in j1 jmj1...jmei1 ...ein e ...e ,

i1,...,in,j1,...,jm=1whereT isamultidimensionaltableofnumbers.

Physicistsdefineatensorasacollection ofsuchmultidimensionaltablesTB attachedtoeverybasis B in V, which change according to a certain rule when the basis B is changed. Here it isimportant to distinguish upper and lower indices, since lower indices of T correspond to V andupper ones to V. The physicists dont write the sum sign, but remember that one should sumover indices that repeat twice - once as an upper index and once as lower. This convention iscalled theEinstein summation, and it also stipulates that ifan indexappearsonce, thenthere isnosummationover it,whileno index issupposedtoappearmorethanonceasanupperindexormorethanonceasa lower index.

Onecanalsodefinethetensorproductoflinearmaps. Namely,ifA:V V andB :W Warelinearmaps,thenonecandefinethelinearmapAB:VW VW givenbytheformula(AB)(vw) =AvBw(checkthatthisiswelldefined!)

Themost importantpropertiesoftensorproductsaresummarized inthefollowingproblem.17


18/106

Problem 1.49. (a)Let U be any k-vector space. Construct a natural bijection between bilinearmaps V W U and linearmaps V W U.

(b) Show that if {vi} is a basis of V and {wj} is a basis of W then {viwj} is a basis ofV W.

(c) Construct a natural isomorphism V W Hom(V, W) in the case when V isfinitedimensional (naturalmeans that the isomorphism isdefinedwithoutchoosingbases).

(d)LetV beavectorspaceoverafieldk.LetSnV bethequotientofVn (n-foldtensorproductofV)bythesubspacespannedbythetensorsTs(T)whereT Vn,andsissometransposition.Also let

nV be thethequotientofV

n subspacespannedbythe tensorsT suchthat s(T) =Tfor

some transposition s. These spaces are called then-th symmetric, respectively exterior, power ofV. If{vi} isa basisof V, canyou constructa basisof SnV,nV? If dimV =m,what are theirdimensions?

(e) Ifkhascharacteristiczero,findanatural identificationof SnV with thespaceofT VnsuchthatT =sTforalltranspositionss,andofnV withthespaceofT Vn suchthatT =sTforall transpositions s.

(f)Let A :V W bea linearoperator. Thenwehaveanoperator An :Vn Wn,andits symmetric and exteriorpowers SnA :SnV SnW,nA :nV nW whichare defined inanobviousway. Suppose V =W andhasdimensionN,andassume that theeigenvaluesofAare1,...,N. FindT r(SnA), T r(

nA).

(g)Show that NA=det(A)Id, anduse this equality to givea one-lineproof of thefact thatdet(AB)=det(A)det(B).Remark. Notethatasimilardefinitiontotheabovecanbeusedtodefinethetensorproduct

V AW,whereAisanyring,V isarightA-module,andW isaleftA-module. Namely,VAWistheabeliangroupwhichisthequotientofthegroupV W freelygeneratedbyformalsymbolsvw,vV,wW,modulotherelations

(v1+v2)wv1wv2w, v(w1 +w2)vw1vw2, vawvaw,aA.

Exercise.

Throughoutthis

exercise,

we

let

k

be

an

arbitrary

field

(not

necessarily

of

characteristiczero,andnotnecessarilyalgebraicallyclosed).

If A and B are two k-algebras, then an (A, B)-bimodule will mean a k-vector space V withboth a left A-module structure and a right B-module structure which satisfy (av)b = a(vb) forany v V, a A and b B. Note that both the notions of left A-module and right A-moduleareparticularcasesofthenotionofbimodules;namely,aleftA-moduleisthesameasan(A, k)-bimodule,andarightA-module isthesameasa(k, A)-bimodule.

LetBbeak-algebra,W aleftB-moduleandV arightB-module. WedenotebyV BW thek-vector space(V k W)/vbwvbw|vV, wW, bB. We denotetheprojection ofapuretensorvw(withvV andwW)ontothespaceV B W byvB w. (NotethatthistensorproductV

B W istheonedefinedintheRemarkafterProblem1.49.)

If,additionally,A isanotherk-algebra,andiftherightB-modulestructureonV ispartofan(A, B)-bimodule structure, then V B W becomes a left A-module by a(vB w) =avB w foranyaA,vV andwW.

Similarly,ifC isanotherk-algebra,andiftheleftB-modulestructureonW ispartofa(B, C)bimodule structure, then V B W becomes a right C-module by (vB w)c = vB wc for any

18
http:///reader/full/Problem1.49.)http:///reader/full/Problem1.49.)


19/106

cC,vV andwW.IfV isan(A, B)-bimoduleandW isa(B, C)-bimodule,thenthesetwostructuresonV BW

canbecombined intoone(A, C)-bimodulestructureonV B W.(a)LetA,B,C,D befouralgebras. LetV bean(A, B)-bimodule,W bea(B, C)-bimodule,

andX a (C, D)-bimodule. Prove that (V B W)C X V B (WC X) as (A, D)-bimodules.=The isomorphism (from left to right) is given by (vB w)C x vB (wC x) for all vV,wW andxX.

(b)IfA, B,C arethreealgebras, and ifV isan(A, B)-bimoduleandW an(A, C)-bimodule,thenthevector space Hom

A(V, W)(the space ofall left A-linear homomorphisms from V to W)

canonicallybecomesa(B, C)-bimodulebysetting(bf) (v) =f(vb)forallbB,fHomA(V, W)andvV and(f c) (v) =f(v)cforallcC,fHomA(V, W)andvV.

LetA,B,C,Dbefouralgebras. LetV be a(B, A)-bimodule,W be a(C, B)-bimodule,andXa(C, D)-bimodule. ProvethatHomB(V,HomC(W, X))=HomC(WB V, X)as (A, D)-bimodules.The isomorphism (from left to right) is given by f (wB v f(v)w) for all v V, wWandfHomB(V,HomC(W, X)).1.11 The tensoralgebraThe notion of tensor product allows us to give more conceptual (i.e., coordinate free) definitionsofthefreealgebra,polynomialalgebra,exterioralgebra,anduniversalenvelopingalgebraofaLiealgebra.

Namely,givenavectorspaceV,defineitstensoralgebraT V overafieldktobeT V =n0Vn,withmultiplication definedbya b:=ab,aVn,bVm. Observethatachoice ofabasisx1,...,xN inV definesanisomorphismofT V withthefreealgebrak < x 1,...,xn >.

Also,onecanmakethe followingdefinition.Definition1.50. (i)ThesymmetricalgebraSV ofV isthequotientofT V bytheidealgeneratedbyvwwv,v, wV.

(ii)Theexterioralgebra

V ofV isthequotientofT V bytheidealgeneratedbyv

v,v

V.(iii)IfV isaLiealgebra,theuniversalenvelopingalgebraU(V)ofV isthequotientofT V by

the idealgeneratedbyvwwv[v, w],v, wV.It is easy to see that a choice of a basis x1,...,xN in V identifies SV with the polynomial

algebra k[x1,...,xN], V with the exterior algebra k(x1,...,xN), and the universal envelopingalgebraU(V)withonedefinedpreviously.

Also, itiseasytoseethatwehavedecompositionsSV =n0SnV,V =n0nV.1.12 Hilberts thirdproblemProblem1.51. It isknown that ifAand B are twopolygonsof thesamearea thenAcanbecutbyfinitelymanystraightcuts intopiecesfromwhichonecanmakeB.DavidHilbertasked in1900whether it is trueforpolyhedra in3dimensions. Inparticular, is it trueforacubeanda regulartetrahedronof thesamevolume?

19


20/106

The answer isno, aswasfound byDehn in 1901. The proof is very beautiful. Namely, toanypolyhedron A letusattach itsDehn invariant D(A) in V =R(R/Q) (the tensorproductofQ-vectorspaces).Namely, (a)

D(A) = l(a) ,

awhere arunsoveredgesof A,and l(a), (a)are the lengthofaand theangleata.

(a)Show that ifyoucutA intoB andC byastraightcut, then D(A) =D(B) +D(C).(b)Show that =arccos(1/3)/ isnotarationalnumber.Hint.Assumethat= 2m/n,forintegersm, n.Deducethatrootsoftheequationx+x1 = 2/3

arerootsofunityofdegreen.Concludethatxk+xk hasdenominator3k andgetacontradiction.(c)Using (a) and (b), show that the answer toHilberts question is negative. (Compute the

Dehn invariantof theregular tetrahedronand thecube).1.13 TensorproductsanddualsofrepresentationsofLiealgebrasDefinition1.52. Thetensorproductoftwo representationsV, W ofaLie algebrag isthespaceV W withVW(x) =V(x)Id+IdW(x).Definition

1.53.

The

dual

representation

V

to

arepresentation

V

of

aLie

algebra

gis

the

dual

spaceV toV withV(x) =V(x).

It iseasytocheckthattheseare indeedrepresentations.Problem 1.54. Let V,W,U befinite dimensional representations of aLie algebra g. Show thatthespaceHomg(V W, U) is isomorphic toHomg(V, UW). (HereHomg :=HomU(g)).1.14 Representationsofsl(2)Thissubsection isdevotedtotherepresentationtheoryofsl(2),which isofcentral importance inmany

areas

of

mathematics.

It

is

useful

to

study

this

topic

by

solving

the

following

sequence

of

exercises,whicheverymathematicianshoulddo,inoneformoranother.Problem 1.55. According to the above, a representation ofsl(2) isjust a vector space V with atriple of operators E ,F,H such that HEEH = 2E ,HF F H = 2F,EF F E = H (thecorrespondingmap isgivenby(e) =E, (f) =F,(h) =H).

LetV beafinitedimensional representationofsl(2)(thegroundfield in thisproblem is C).(a)Takeeigenvaluesof H andpickonewith the biggest real part. Call it . Let V() be the

generalizedeigenspacecorresponding to. Show that E|V() = 0.(b)Let W be any representation ofsl(2) and wW be a nonzero vector such that Ew = 0.

For any k >0find apolynomial Pk

(x) ofdegree k such that EkFkw =Pk

(H)w. (First computeEFkw, thenuse induction ink).

(c)Let v V() be a generalized eigenvector of H with eigenvalue . Show that there existsN >0such that FNv= 0.

(d) Show that H is diagonalizable on V(). (Take N to be such that FN = 0 on V(), andcompute ENFNv,vV(),by(b).Usethefact that Pk(x)doesnothavemultipleroots).

20


21/106

(e)LetNv be thesmallest N satisfying(c). Show that =Nv1.(f)ShowthatforeachN >0,thereexistsauniqueuptoisomorphismirreduciblerepresentation

ofsl(2)ofdimension N. Compute thematrices E ,F,H in this representationusingaconvenientbasis. (For V finite dimensional irreducible take as in (a) and v V() an eigenvector of H.Show thatv,Fv,...,Fv isabasisofV,andcompute thematricesoftheoperators E ,F,H in thisbasis.)

Denote the + 1-dimensional irreducible representationfrom (f) by V. Below youwill showthatanyfinitedimensionalrepresentation isadirectsumofV.

(g)Show that theoperator C =EF +F E+H2/2 (the so-calledCasimiroperator) commuteswith E ,F,H andequals (2+2)IdonV.

Now itwill beeasy toprove thedirect sumdecomposition. Namely, assume thecontrary, andletV beareduciblerepresentationof thesmallestdimension,which isnotadirectsumofsmallerrepresentations.

(h)Show thatC hasonlyoneeigenvalueonV,namely (2+2) forsomenonnegative integer.(usethatthegeneralizedeigenspacedecompositionofCmustbeadecompositionofrepresentations).

(i)Show that V has a subrepresentation W =V such that V /W =nVfor some n (use (h)and thefact thatV is thesmallestwhichcannotbedecomposed).

(j)Deducefrom (i) that the eigenspace V() of H is n+ 1-dimensional. If v1,...,vn+1 is itsbasis,show thatFjvi,1in+ 1,0jare linearly independentand thereforeformabasisofV (establish that ifF x= 0and Hx=x thenCx= (2)xandhence =).2

(k)DefineWi =span(vi, F vi,...,Fvi). Show that Vi aresubrepresentations ofV andderiveacontradictionwith thefact that V cannotbedecomposed.

(l)(Jacobson-MorozovLemma)LetV beafinitedimensionalcomplexvectorspaceandA:V V anilpotent operator. Show that there existsaunique,up toan isomorphism, representation ofsl(2)on V such that E=A. (Use theclassificationof therepresentations and theJordannormalform theorem)

(m) (Clebsch-Gordan decomposition)Find thedecomposition into irreduciblesof the representation

VV ofsl(2).

Hint. Forafinitedimensional representation V ofsl(2) it isuseful to introduce thecharacterV(x) =T r(exH), xC. Show that VW(x) =V(x) +W(x) and VW(x) =V(x)W(x).ThencomputethecharacterofV andofVV andderivethedecomposition. Thisdecompositionisoffundamental importance inquantummechanics.

(n)LetV =CM CN,andA=JM(0)IdN +IdM JN(0),whereJn(0) istheJordanblockofsizenwitheigenvaluezero(i.e.,Jn(0)ei =ei1,i= 2,...,n,andJn(0)e1 = 0). Find theJordannormalformof Ausing(l),(m).1.15 ProblemsonLiealgebrasProblem 1.56. (Lies Theorem) The commutant K(g) of a Lie algebra g is the linear spanof elements [x, y], x, y g. This is an ideal in g (i.e., it is a subrepresentation of the adjointrepresentation). Afinite dimensional Lie algebra g over afield k is said to be solvable if thereexists n such that Kn(g) = 0. Prove the Lie theorem: if k = C and V is afinite dimensionalirreduciblerepresentationofasolvableLiealgebrag thenV is1-dimensional.

21


22/106

Hint. Prove the result by induction in dimension. By the induction assumption, K(g) has acommon eigenvector v in V, that is there isa linearfunction :K(g)Csuch that av=(a)vfor any a K(g). Show that gpreserves common eigenspaces of K(g) (for this youwillneed toshowthat([x, a])=0forxgandaK(g).Toprovethis,considerthesmallestvectorsubspaceU containing v and invariantunder x. This subspace is invariantunder K(g)andany aK(g)actswith trace dim(U)(a) in thissubspace. Inparticular 0=Tr([x, a])=dim(U)([x, a]).).Problem1.57. Classify irreduciblefinitedimensionalrepresentationsof the twodimensionalLiealgebra with basis X, Y and commutation relation [X, Y] = Y. Consider the cases of zero andpositivecharacteristic. Is theLietheorem true inpositivecharacteristic?Problem 1.58. (hard!) Forany element xofaLiealgebra g let ad(x) denote theoperator gg, y [x, y].ConsidertheLiealgebragn generatedbytwoelementsx, ywiththedefiningrelationsad(x)2(y) =ad(y)n+1(x) = 0.

(a)Show that theLiealgebrasg1,g2,g3 arefinitedimensionalandfind theirdimensions.(b) (harder!) Show that theLiealgebrag4 has infinitedimension. Constructexplicitlyabasis

of thisalgebra.

22


23/106

2 Generalresultsofrepresentation theory2.1 Subrepresentations in semisimple representationsLetAbeanalgebra.Definition 2.1. A semisimple (or completely reducible) representation of A is a direct sum ofirreduciblerepresentations.Example. Let V be an irreducible representation of A of dimension n. Then Y = End(V),

withactionofAbyleftmultiplication, isasemisimplerepresentationofA, isomorphictonV (thedirect sum of n copies of V). Indeed, any basis v1,...,vn of V gives rise to an isomorphism ofrepresentationsEnd(V)nV,givenbyx(xv1,...,xvn).Remark. Note that by Schurs lemma, any semisimple representation V of A is canonically

identifiedwithXHomA(X, V)X,whereXrunsoverallirreduciblerepresentationsofA. Indeed,wehaveanaturalmapf :XHom(X, V)XV,givenbygxg(x),xX,gHom(X, V),and it iseasytoverifythatthismap isanisomorphism.

WellseenowhowSchurslemmaallowsustoclassifysubrepresentations infinitedimensionalsemisimplerepresentations.Proposition 2.2. Let Vi,1 i m be irreduciblefinite dimensional pairwise nonisomorphic

mrepresentations of A, and W be a subrepresentation of V = i=1niVi. Then W is isomorphic tom i=1riVi,ri ni,andtheinclusion:W V isadirectsumofinclusionsi :riVi niVi given

bymultiplication of a row vector of elements of Vi (of length ri) by a certain ri-by-ni matrix Xiwith linearly independentrows: (v1,...,vri) = (v1,...,vri)Xi.

mProof. Theproofisbyinductioninn:= i=1ni. Thebaseofinduction(n=1)isclear. Toperformthe induction step, let us assume that W is nonzero, and fix an irreducible subrepresentationP W. SuchP exists(Problem1.20). 2 Now,bySchurslemma,P isisomorphictoVi forsomei,andthe inclusion|P :P V factorsthroughniVi,andupon identificationofP withVi isgivenbytheformulav

(vq1,...,vqni),whereql

k arenotallzero.

Now note that the group Gi = GLni(k) of invertible ni-by-ni matrices over k acts on niViby (v1,...,vni) (v1,...,vni)gi (and by the identity on njVj,j = i), and therefore acts on theset of subrepresentations of V, preserving the property we need to establish: namely, under theaction of gi, the matrix Xi goes to Xigi, while Xj, j = i dont change. Take gi Gi such that(q1,...,qni)gi = (1,0,...,0). Then W gi contains the first summand Vi of niVi (namely, it is P gi),henceW gi =ViW,whereW n1V1...(ni1)Vi...nmVm isthekerneloftheprojectionof W gi to the firstsummandVi along the other summands. Thusthe requiredstatement followsfromthe inductionassumption.Remark2.3. InProposition2.2, it isnot importantthatk isalgebraically closed,noritmattersthatV isfinitedimensional. Iftheseassumptionsaredropped,theonlychangeneededisthattheentriesofthematrixXi arenolonger inkbutinDi =EndA(Vi),whichis,asweknow,adivisionalgebra. TheproofofthisgeneralizedversionofProposition2.2 isthesameasbefore(check it!).

2Another proof of the existence of P, which does not use the finite dimensionality of V, is by induction in n.Namely, if W itself is not irreducible, let K be the kernel of the projection of W to the first summand V1. ThenK isasubrepresentation of (n11)V1...nmVm,which isnonzerosince W isnot irreducible,so K containsanirreduciblesubrepresentation bytheinductionassumption.

23


24/106

2.2 Thedensity theoremLetAbeanalgebraoveranalgebraicallyclosedfieldk.Corollary2.4. Let V bean irreduciblefinitedimensional representation of A,and v1,...,vn Vbe any linearly independent vectors. Thenfor any w1,...,wn V there exists an element a Asuch that avi =wi.Proof. Assume the contrary. Then the image of the map A nV given by a (av1,...,avn)is a proper subrepresentation, so by Proposition 2.2 it corresponds to an r-by-n matrix X, r


25/106

by(a1, . . . , an) =a1y1+ +anyn

where{yi} isabasisofX. isclearlysurjective,ask A. Thus,thedualmap :XAnisinjective. ButAn =An asrepresentationsofA(check it!). Hence,Im =X isasubrepresentationofAn. Next,Matdi(k) =diVi, soA=ri=1diVi,An =ri=1ndiVi,asarepresentationofA.HencebyProposition2.2,X=ir=1miVi,asdesired.Exercise. The goal of this exercise is to give an alternative proof of Theorem 2.6, not using

anyofthepreviousresultsofChapter2.LetA1,A2,...,An benalgebraswithunits11, 12,..., 1n,respectively. LetA=A1A2...An.

Clearly,1i1j =ij1i,andtheunitofA is1 = 11+ 12+...+ 1n.For every representation V of A, it is easy to see that 1iV is a representation of Ai for every

i {1,2,...,n}. Conversely, if V1, V2, ..., Vn are representations of A1, A2, ..., An, respectively,thenV1V2...Vn canonically becomesarepresentationofA(with(a1, a2,...,an)AactingonV1V2...Vn as(v1, v2,...,vn)(a1v1, a2v2,...,anvn)).(a)ShowthatarepresentationV ofA is irreducible ifandonly if1iV isan irreduciblerepre

sentationofAi forexactlyonei {1,2,...,n},while1iV =0foralltheotheri. Thus,classifytheirreduciblerepresentationsofA intermsofthoseofA1,A2,...,An.(b)Letd

N. ShowthattheonlyirreduciblerepresentationofMatd(k) iskd,andeveryfinite

dimensionalrepresentationofMatd(k) isadirectsumofcopiesofkd.Hint: Forevery(i, j) {1,2,...,d}2,letEij Matd(k)bethematrixwith1intheithrowofthe

jthcolumnand0severywhereelse. LetV beafinitedimensionalrepresentationofMatd(k). ShowthatV =E11V E22V ...EddV,andthati :E11V EiiV,v Ei1v isanisomorphismforevery i {1,2,...,d}.Forevery vE11V,denote S(v) =E11v, E21v,...,Ed1v. Prove thatS(v)isasubrepresentationofV isomorphictokd (asarepresentationofMatd(k)),andthatvS(v).ConcludethatV =S(v1)S(v2)...S(vk),where{v1, v2,...,vk} isabasisofE11V.(c)ConcludeTheorem2.6.

2.4 FiltrationsLet A be an algebra. Let V be a representation of A. A (finite)filtration of V is a sequence ofsubrepresentations0=V0 V1 ... Vn =V. Lemma 2.8. Anyfinite dimensional representation V of an algebra A admits afinitefiltration0 =V0 V1 ... Vn =V such that thesuccessivequotientsVi/Vi1 are irreducible. Proof. Theproof isby induction indim(V). Thebase isclear, andonlythe inductionstepneedsto bejustified. Pick an irreducible subrepresentation V1 V, and consider the representationU =V/V1. Thenby the induction assumption U hasa filtration 0=U0 U1 ... Un1 =U such that U

i/U

i1 are irreducible. Define V

i for i

2 to be the preimages of U

i1 under the

tautological projectionV V /V1 =U. Then0=V0 V1 V2 ... Vn =V isafiltrationofV withthedesiredproperty.

25


26/106

2.5 FinitedimensionalalgebrasDefinition2.9. TheradicalofafinitedimensionalalgebraAisthesetofallelementsofAwhichactby0 inall irreduciblerepresentationsofA. ItisdenotedRad(A).Proposition2.10. Rad(A) isa two-sided ideal.Proof. Easy.Proposition2.11. Let Abeafinitedimensionalalgebra.

(i)LetI beanilpotent two-sided ideal inA, i.e.,In = 0forsomen.ThenI Rad(A).(ii)Rad(A) isanilpotent ideal. Thus,Rad(A) isthe largestnilpotent two-sided ideal inA.

Proof. (i)LetV bean irreduciblerepresentationofA. LetvV. ThenIv V isasubrepresenn

tation. IfIv=0 then Iv =V sothere is xI suchthat xv=v. Thenx =0, a contradiction.ThusIv=0,soI actsby0 inV andhenceI Rad(A).

(ii) Let 0 = A0 A1 ... An = A be a filtration of the regular representation of A by subrepresentations such that Ai+1/Ai are irreducible. It exists by Lemma 2.8. Let x Rad(A).Then x acts on Ai+1/Ai by zero, so x maps Ai+1 to Ai. This implies that Rad(A)n = 0, asdesired.Theorem2.12.Afinitedimensionalalgebra Ahasonlyfinitelymany irreduciblerepresentationsVi up to isomorphism, theserepresentationsarefinitedimensional,and

A/Rad(A) EndVi.=i

Proof. First, for any irreducible representation V of A, and for any nonzero vV, AvV is afinitedimensionalsubrepresentationofV. (It isfinitedimensionalasA isfinitedimensional.) AsV is irreducibleandAv=0,V =Av andV isfinitedimensional.

Next, supposewehave non-isomorphic irreduciblerepresentationsV1, V2, . . . , V r. ByTheorem2.5,

the

homomorphism

i :A EndVi

i iis surjective. So r idim EndVi dimA. Thus, A has only finitely many non-isomorphicirreduciblerepresentations(atmostdimA).

Now, let V1, V2, . . . , V r be all non-isomorphic irreducible finite dimensional representations ofA. ByTheorem2.5,thehomomorphism

i :A EndVii i

issurjective.

The

kernel

of

this

map,

by

definition,

is

exactly

Rad(A).

Corollary2.13. i(dimVi)2 dimA,where theVisare the irreduciblerepresentationsofA.Proof. AsdimEndVi =(dimVi)2,Theorem2.12impliesthatdimAdimRad(A) = dimEndVi =i

(dimVi)2. AsdimRad(A)0, (dimVi)2 dimA.i i

26


27/106

Example2.14. 1. LetA=k[x]/(xn). Thisalgebrahasauniqueirreduciblerepresentation,whichisa1-dimensionalspacek, inwhichxactsbyzero. SotheradicalRad(A) isthe ideal(x).

2. Let A be the algebra of upper triangular n by n matrices. It is easy to check that theirreduciblerepresentationsofAareVi,i= 1,...,n,whichare1-dimensional,andanymatrixxactsbyxii. SotheradicalRad(A) istheidealofstrictlyuppertriangularmatrices(as it isanilpotentidealandcontainstheradical). Asimilarresultholdsforblock-triangularmatrices.Definition2.15. AfinitedimensionalalgebraA issaidtobesemisimple ifRad(A)=0.Proposition2.16. ForafinitedimensionalalgebraA, thefollowingareequivalent:1. A issemisimple.2. i(dimVi)2 =dimA,where theVisare the irreduciblerepresentationsofA.3. A= Matdi(k)forsome di.i4. Anyfinite dimensional representation of A is completely reducible (that is, isomorphic to adirectsumof irreduciblerepresentations).

5. A isacompletelyreduciblerepresentationofA.Proof. AsdimA

dimRad(A) =

(dimV

i)2,clearlydimA=

(dimV

i)2ifandonlyifRad(A) =

i i0. Thus,(1)(2).Next, by Theorem 2.12, ifRad(A) = =iMatdi(k) fordi =dimVi.0, then clearly A Thus,

(1)(3). Conversely, ifA= Matdi(k),thenbyTheorem2.6,Rad(A) = 0,soAissemisimple.iThus(3)(1).

Next, (3)(4) by Theorem 2.6. Clearly (4)(5). To see that (5)(3), let A=iniVi.ConsiderEndA(A) (endomorphismsofAas a representation ofA). Asthe Visarepairwisenon-isomorphic,bySchurs lemma,nocopyofVi inAcanbemappedtoadistinctVj. Also,againbySchurs lemma,EndA(Vi) =k. Thus,EndA(A) Matni =Aop byProblem= (k). ButEndA(A)i

=(

(k))op 1.22,soAop = Matni(k). Thus,A Matni = Matni(k),asdesired.i i i2.6 CharactersofrepresentationsLet A be an algebra and V a finite-dimensional representation of A with action . Then thecharacter ofV isthelinear functionV :Ak givenby

V(a)=tr V((a)).|If[A, A]isthespanofcommutators [x, y]:=xyyxoverallx, yA,then[A, A]kerV. Thus,wemayviewthecharacterasamappingV :A/[A, A]k.Exercise. Showthat ifW V arefinite dimensional representationsofA,thenV =W +

V /W.Theorem2.17. (i)Charactersof(distinct)irreduciblefinite-dimensionalrepresentationsofAarelinearly independent.

(ii) If A is a finite-dimensional semisimple algebra, then these characters form a basis of(A/[A, A]).

27


28/106

Proof. (i)IfV1, . . . , V r arenonisomorphicirreduciblefinite-dimensionalrepresentationsofA,thenV1 Vr :AEndV1 EndVr issurjectivebythedensitytheorem,soV1, . . . , Vr arelinearly independent. (Indeed, ifiVi(a)= 0 forallaA,theniTr(Mi)= 0 forall Mi EndkVi. Buteachtr(Mi)canrangeindependentlyoverk,so itmustbethat1 = =r =0.)

(ii) First we prove that [Matd(k),Matd(k)]=sld(k), theset of all matrices with trace 0. It isclearthat[Matd(k),Matd(k)]sld(k). IfwedenotebyEij thematrixwith1intheithrowofthejthcolumnand0severywhereelse,wehave[Eij, Ejm ] =Eim fori= m,and[Ei,i+1, Ei+1,i] =EiiEi+1,i+1.Now{Eim}{EiiEi+1,i+1}formsabasisinsld(k),soindeed[Matd(k),Matd(k)]=sld(k),asclaimed.

Bysemisimplicity, wecanwriteA=Matd1 (k).Then[A, A] =sld1(k) Matdr (k) sldr(k), andA/[A, A] kr.= By Theorem 2.6, there are exactly r irreduciblerepresentations of A(isomorphic to kd1, . . . , kdr, respectively), and therefore r linearly independent characters on ther-dimensionalvectorspaceA/[A, A]. Thus,thecharacters formabasis.2.7 TheJordan-Holder theoremWe will now state and prove two important theorems aboutrepresentations of finite dimensionalalgebras- theJordan-HoldertheoremandtheKrull-Schmidttheorem.Theorem 2.18. (Jordan-Holder theorem). Let V be afinite dimensional representation of A,and 0 = V0 V1 ... Vn = V, 0 = V0 ... Vm = V befiltrations of V, such that the representations Wi :=Vi/Vi1 andW:=Vi/Vi1 are irreducibleforalli.Thenn=m,and thereiexistsapermutation of1,...,nsuchthat W(i) is isomorphic toWi.Proof. First proof (for k of characteristic zero). The character of V obviously equals the sumof characters of Wi, and also the sum of characters of Wi. But by Theorem 2.17, the charactersofirreduciblerepresentationsare linearly independent,sothemultiplicityofevery irreduciblerepresentationW ofAamongWi andamongWi arethesame. Thisimpliesthetheorem. 3Secondproof(general). Theproof isby inductionondimV. Thebaseof induction isclear,

soletusprovetheinductionstep. IfW1 =W1 (assubspaces),wearedone,sincebytheinductionassumption the theorem holds for V /W1. So assume W1 =

W1

. In this case W1

W1 = 0 (asW1, W1 are irreducible),sowehaveanembeddingf :W1W1V. LetU =V/(W1W1),and0 =U0 U1 ... Up =U beafiltrationofU withsimplequotientsZi =Ui/Ui1 (it existsby Lemma2.8). Thenweseethat:

1) V /W1 has a filtration with successive quotients W1, Z1,...,Zp, and another filtration withsuccessivequotientsW2,....,Wn.

2) V /W1 has a filtration with successive quotients W1, Z1,...,Zp, and another filtration withsuccessivequotientsW2 ,....,Wn .

Bytheinductionassumption,thismeansthatthecollectionofirreduciblerepresentationswithmultiplicitiesW1, W1 , Z1,...,Zp coincidesononehandwithW1,...,Wn,andontheotherhand,withW

1

,...,W . Wearedone.m

The Jordan-Holder theorem shows that the number n of terms in a filtration of V with irreducible successive quotients does not depend on the choice of a filtration, and depends only on

3Thisproofdoesnotworkincharacteristic pbecauseitonlyimpliesthatthemultiplicities ofWi andWi arethesamemodulop,whichisnotsufficient. Infact,thecharacteroftherepresentationpV,whereV isanyrepresentation,iszero.

28


29/106

V. Thisnumber iscalled the length ofV. It is easytoseethatn isalso themaximal lengthofafiltrationofV inwhichallthe inclusionsarestrict.2.8 TheKrull-Schmidt theoremTheorem2.19. (Krull-Schmidttheorem)AnyfinitedimensionalrepresentationofAcanbeuniquely(up toorderofsummands)decomposed intoadirectsumof indecomposable representations.Proof. It is clear that a decomposition of V into a direct sum of indecomposable representationsexists, so wejust need to prove uniqueness. We will prove it by induction on dimV. Let V =V1...Vm =V1 ...Vn. Letis :Vs V,is :VsV,ps :V Vs,ps :V Vs bethenaturalmapsassociatedtothesedecompositions. Lets =p1ispsi1 :V1 V1. Wehavens=1s =1. Nowweneedthe following lemma.Lemma2.20. LetW beafinitedimensional indecomposable representationofA. Then

(i)Anyhomomorphism :WW iseitheran isomorphismornilpotent;(ii)Ifs :W W,s= 1,...,narenilpotenthomomorphisms, thenso is:=1+...+n.

Proof. (i) Generalized eigenspaces of are subrepresentations of W, and W is their direct sum.Thus,canhaveonlyoneeigenvalue. Ifiszero, isnilpotent,otherwiseitisanisomorphism.

(ii)Theproofisbyinductioninn. Thebaseisclear. Tomaketheinductionstep(n1ton),assumethatisnotnilpotent. Thenby(i)isanisomorphism,soni=11i =1. Themorphisms1i are not isomorphisms,so they are nilpotent. Thus11n =11 +...+1n1 is anisomorphism,which isacontradictionwiththeinductionassumption.

By the lemma, we find that for some s, s must be an isomorphism; we may assume thats = 1. In this case, V1 = Im(p1i1)Ker(p1i1), so since V1 is indecomposable, we get thatf :=p1 i1 :V1 V1 andg:=p1i1 :V1V1 are isomorphisms.

Let B = j>1Vj, B = j>1Vj; then we have V = V1 B = V1B. Consider the maph : B

B defined as a composition of the natural maps B

V

B attached to these

decompositions. Weclaimthathisanisomorphism. Toshowthis,itsufficestoshowthatKerh= 0(ash isamapbetweenspacesofthesamedimension). AssumethatvKerh B. ThenvV1.Ontheotherhand,theprojectionofv toV1 iszero,sogv=0. Sinceg isan isomorphism,wegetv=0,asdesired.

Now by the induction assumption, m=n, and Vj V for some permutation of 2,...,n.= (j)Thetheorem isproved.Exercise. Let A be the algebra of real-valued continuous functions on R which are periodic

withperiod1. LetM betheA-moduleofcontinuousfunctionsf onRwhichareantiperiodicwithperiod1, i.e.,f(x+ 1)=f(x).

(i)ShowthatAandM are indecomposableA-modules.(ii)ShowthatA isnot isomorphictoM butAAis isomorphictoMM.Remark. Thus,weseethatingeneral,theKrull-Schmidttheoremfailsforinfinitedimensional

modules. However,itstillholdsformodulesoffinitelength,i.e.,modulesM suchthatanyfiltrationofM haslengthboundedabovebyacertainconstant l=l(M).

29


30/106

2.9 ProblemsProblem 2.21. Extensions of representations. Let A be an algebra, and V, W be a pair ofrepresentationsof A. Wewould like toclassify representations U of A such that V isa subrepresentationofU,and U/V =W.Ofcourse, there isanobviousexample U =V W,butare thereanyothers?

Suppose we have a representation U as above. As a vector space, it can be (non-uniquely)identifiedwithV W,sothatforanyaAthecorrespondingoperator U(a)hasblocktriangularform

V(a) f(a)U(a) = ,0 W(a)

where f :AHomk(W, V) isa linearmap.(a) What is the necessary and sufficient condition on f(a) under which U(a) is a repre

sentation? Maps f satisfying this condition are called (1-)cocycles (of A with coefficients inHomk(W, V)). Theyformavectorspacedenoted Z1(W, V).

(b)LetX :W V bealinearmap.ThecoboundaryofX,dX,isdefinedtobethefunctionAHomk(W, V)givenbydX(a) =V(a)XXW(a). ShowthatdX isacocycle,whichvanishesifandonly ifX isahomomorphism ofrepresentations. Thuscoboundariesformasubspace B1(W, V)Z1(W, V),whichisisomorphictoHomk(W, V)/HomA(W, V).ThequotientZ1(W, V)/B1(W, V)

is

denotedExt

1

(W, V).

(c) Show that if f, f Z1(W, V) and ff B1(W, V) then the corresponding extensions

U, U are isomorphicrepresentationsofA.Conversely, if:UU isan isomorphism such that(a) = 1V

1W0

thenffB1(V, W). Thus, thespace Ext1(W, V)classifiesextensionsofW byV.(d)Assume that W, V arefinite dimensional irreducible representations of A. For any f

Ext1(W, V), let Uf be the corresponding extension. Show that Uf is isomorphic to Uf as representations ifandonly iff andf areproportional. Thus isomorphismclasses(asrepresentations)ofnontrivialextensionsof W byV (i.e.,thosenot isomorphic toWV)areparametrized bytheprojectivespacePExt1(W, V). Inparticular,everyextensionistrivialifandonlyifExt1(W, V) = 0.Problem2.22. (a)LetA=C[x1,...,xn],and Va, Vb beone-dimensionalrepresentations inwhichxi act by ai and bi, respectively (ai, bi C). Find Ext1(Va, Vb)and classify2-dimensional representationsofA.

(b)Let B be thealgebraoverCgeneratedby x1,...,xn with thedefining relations xixj = 0forall i, j. Show thatfor n > 1 the algebra B has infinitelymany non-isomorphic indecomposablerepresentations.Problem 2.23. Let Q be a quiverwithout oriented cycles, and PQ the path algebra of Q. Findirreducible

representations

of

PQ

and

compute

Ext

1between

them.

Classify

2-dimensional

representationsofPQ.

Problem2.24. Let Abeanalgebra,and V arepresentation of A. Let :AEndV. AformaldeformationofV isaformalseries

=0+t1+...+tnn+...,30


31/106

where i :AEnd(V)are linearmaps, 0 =,and (ab) =(a)(b).Ifb(t)=1 +b1t+b2t2+...,wherebi End(V),and b1isaformaldeformationof,thenb

isalsoadeformationof ,which issaid tobe isomorphic to .(a)Show that ifExt1(V, V) = 0, thenanydeformationof is trivial, i.e., isomorphic to.(b) Is theconverse to(a) true? (consider thealgebraofdualnumbersA=k[x]/x2).

Problem 2.25. The Clifford algebra. Let V be a finite dimensional complex vector spaceequipped with a symmetric bilinearform (,). The Clifford algebra Cl(V) is the quotient of thetensoralgebraT V bythe idealgeneratedbytheelementsvv(v, v)1, vV.Moreexplicitly, ifxi,1iN isabasisofV and(xi, xj) =aij thenCl(V)isgeneratedbyxiwithdefiningrelations

xixj +xjxi = 2aij, xi2 =aii.Thus, if(,) = 0,Cl(V) =V.

(i)Show that if (,) isnondegenerate then Cl(V) is semisimple,andhasone irreducible representationofdimension2n ifdimV = 2n(so inthiscaseCl(V) isamatrixalgebra),and twosuchrepresentations ifdim(V) = 2n+ 1(i.e.,inthiscaseCl(V)isadirectsumoftwomatrixalgebras).

Hint. In the even case, pick a basis a1,...,an, b1,...,bn of V inwhich (ai, aj) = (bi, bj) = 0,(ai, bj) =ij/2, and construct a representation of Cl(V) on S :=(a1,...,an) inwhich bi actsasdifferentiationwith respect to ai. Show that S is irreducible. In the odd case the situation issimilar, except there should beanadditional basis vector c such that (c, ai) = (c, bi) = 0, (c, c) =1, and the action of c on S may be defined either by (1)degree or by (1)degree+1, giving tworepresentations S+, S (why are they non-isomorphic?). Show that there is no other irreduciblerepresentationsbyfindingaspanningsetofCl(V)with 2dimV elements.

(ii)ShowthatCl(V)issemisimpleifandonlyif(,)isnondegenerate. If(,)isdegenerate,whatisCl(V)/Rad(Cl(V))?2.10 RepresentationsoftensorproductsLet A, B be algebras. Then AB is also an algebra, with multiplication (a1 b1)(a2 b2) =a1a2b1b2. ThefollowingtheoremdescribesirreduciblefinitedimensionalrepresentationsofABintermsof irreduciblefinitedimensionalrepresentationsofAandthoseofB.Theorem 2.26. (i) Let V be an irreduciblefinite dimensional representation of A and W anirreduciblefinitedimensionalrepresentationofB.ThenV W isan irreduciblerepresentationofAB.

(ii)Any irreduciblefinitedimensional representation M of AB has theform (i)foruniqueV and W.Remark 2.27. Part (ii) of the theorem typically fails for infinite dimensional representations;e.g. it fails when A is the Weyl algebra in characteristic zero. Part (i) also may fail. E.g. letA=B=V =W =C(x). Then(i)fails,asAB isnotafield.Proof. (i)Bythedensitytheorem,themapsAEndV andBEndW aresurjective. Therefore,themapABEndV EndW =End(V W) issurjective. Thus,V W is irreducible.

(ii)FirstweshowtheexistenceofV andW. LetA, B betheimagesofA, B inEndM. ThenA, B are finite dimensional algebras, and M is a representation of A B, so we may assumewithout lossofgeneralitythatAandB arefinitedimensional.

31


32/106

Inthiscase,weclaimthatRad(AB)=Rad(A)B+ARad(B). Indeed,denotethelatterby J. ThenJ is a nilpotent ideal in AB, as Rad(A) andRad(B) are nilpotent. Onthe otherhand, (AB)/J = (A/Rad(A))(B/Rad(B)), which is a product of two semisimple algebras,hence semisimple. This implies J Rad(AB). Altogether, by Proposition 2.11, we see thatJ =Rad(AB),provingtheclaim.

Thus,weseethat(AB)/Rad(AB) =A/Rad(A)B/Rad(B).

Now, M is an irreducible representation of (A

B)/Rad(A

B), so it is clearly of the formM = V W, where V is an irreducible representation of A/Rad(A) and W is an irreduciblerepresentation of B/Rad(B), and V, W are uniquely determined by M (as all of the algebrasinvolvedaredirectsumsofmatrixalgebras).

32


33/106

3 Representationsoffinitegroups: basicresultsRecall that arepresentation of a group G over a field k is a k-vector space V together with agrouphomomorphism:GGL(V). Aswehaveexplainedabove,arepresentationofagroupGoverk isthesamethingasarepresentationof itsgroupalgebrak[G].

Inthissection,webeginasystematicdevelopmentofrepresentationtheoryoffinitegroups.3.1 MaschkesTheoremTheorem3.1. (Maschke)LetGbeafinitegroupandkafieldwhosecharacteristicdoesnotdivide|G|.Then:

(i)Thealgebra k[G] issemisimple.(ii)There isan isomorphism ofalgebras :k[G] iEndVi definedby g ig| ,where ViVi

are the irreducible representations of G. In particular, this is an isomorphism of representationsof G (where G acts on both sides by leftmultiplication). Hence, the regular representation k[G]decomposes into irreduciblesasidim(Vi)Vi,andonehas

G =dim(Vi)2.| |i

(thesumofsquaresformula).Proof. By Proposition 2.16, (i) implies (ii), and to prove (i), it is sufficient to show that if V isafinite-dimensionalrepresentationofGandW V isanysubrepresentation, thenthereexistsasubrepresentationW V suchthatV =WW asrepresentations.

ChooseanycomplementW ofW inV. (ThusV =WW asvectorspaces,butnotnecessarilyasrepresentations.) LetP betheprojectionalong W ontoW, i.e., theoperator onV definedbyP W =IdandP W =0. Let| |

P := 1 (g)P (g1),|G

|g

Gwhere(g) istheactionofg onV,and let

W =kerP .2

NowP|W =IdandP(V)W,soP =P,soP isaprojectionalongW. Thus,V =WW asvectorspaces.

Moreover, foranyhGandanyyW,P (h)y= 1 (g)P (g1h)y= 1 (h)P (1)y=(h)P y= 0,|G|

gG |G|Gso (h)ykerP =W. Thus, W is invariant undertheaction of G and istherefore a subrepresentationofV. Thus,V =WW isthedesireddecomposition intosubrepresentations.

TheconversetoTheorem3.1(i)alsoholds.Proposition3.2. Ifk[G] issemisimple, then thecharacteristicofkdoesnotdivide G.| |

33


34/106

| |

Proof. Write k[G] =ir=1EndVi, wheretheVi are irreduciblerepresentations andV1 =k isthetrivialone-dimensionalrepresentation. Then

r rk[G] =k EndVi =k diVi,

i=2 i=2wheredi =dimVi. BySchursLemma,

Homk[G](k, k[G])=kHomk[G](k[G], k) =k,

fornonzerohomomorphismsofrepresentations:k[G]kand:kk[G]uniqueuptoscaling.Wecantakesuchthat(g) = 1forallgG,andsuchthat(1)=gGg. Then

(1)= g = 1 =|G|.gG gG

If|G|=0,thenhasnoleft inverse,as(a)(1)=0 foranyak. Thisisacontradiction.Example3.3. IfG=Z/pZandkhascharacteristicp,thenevery irreduciblerepresentationofGoverk istrivial(sok[Z/pZ]indeedisnotsemisimple). Indeed,anirreduciblerepresentationofthisgroupisa1-dimensionalspace,onwhichthegeneratoractsbyap-throotofunity,andeveryp-throotofunity inkequals1,asxp1 = (x1)p overk.Problem3.4. LetGbeagroupoforderpn. Show thatevery irreduciblerepresentationofGoverafieldkofcharacteristicp istrivial.3.2 CharactersIf V is a finite-dimensional representation of a finite group G, then its character V : G kis defined by the formula V(g) = tr V((g)). Obviously, V(g) is simply the restriction of the|characterV(a)ofV asarepresentationofthealgebraA=k[G]tothebasisG A,so itcarriesexactlythesameinformation. Thecharacter isacentral orclassfunction: V(g)dependsonlyontheconjugacyclassofg; i.e.,V(hgh1) =V(g).Theorem3.5. If thecharacteristicof k doesnotdivide G,charactersof irreducible representa| |tionsofGformabasis in thespace Fc(G, k)ofclassfunctionson G.Proof. BytheMaschketheorem,k[G]issemisimple,sobyTheorem2.17,thecharactersarelinearlyindependent and are a basis of (A/[A, A]), where A = k[G]. It suffices to note that, as vectorspacesoverk,

(A/[A, A]) |ghhgkerg, hG}={Homk(k[G], k)={fFun(G, k) f(gh) =f(hg)g, hG}, |

which ispreciselyFc(G, k).Corollary3.6.Thenumberof isomorphismclassesof irreduciblerepresentationsofGequals thenumberofconjugacyclassesofG(if G = 0 ink).

34


35/106

Exercise. Show that if G = 0 in k then the number of isomorphism classes of irreducible| |representationsofGoverk isstrictly lessthanthenumberofconjugacyclasses inG.

Hint. LetP =gGgk[G]. ThenP2 =0. SoP haszero trace inevery finitedimensionalrepresentationofGoverk.Corollary 3.7. Any representation of G isdetermined by its character if k has characteristic 0;namely, V =W impliesV =W.3.3 ExamplesThefollowingareexamplesofrepresentationsoffinitegroupsoverC.

1. Finite abeliangroupsG=Zn1 Znk. LetG bethesetof irreduciblerepresentationsofG. EveryelementofG formsaconjugacyclass,so G = G. Recallthatall irreducible| | | |representationsoverC(andalgebraicallyclosedfieldsingeneral)ofcommutativealgebrasandgroupsareone-dimensional. Thus,G isanabeliangroup: if1, 2 :GC areirreduciblerepresentationsthensoare1(g)2(g)and1(g)1. G iscalledthedual orcharactergroupofG.Forgivenn1,define:Zn C by(m) =e2im/n. ThenZn ={k :k= 0, . . . , n1},soZ =Zn. Ingeneral,n

(G1G2 Gn) =G1 G2 Gn,

so G = G for any finite abelian group G. This isomorphism is, however, noncanonical:theparticulardecompositionofGasZn1 isnotuniqueas faraswhichelements Znkof G correspond to Zn1, etc. is concerned. On the other hand, G (G) is a canonical=isomorphism,givenby:G(G),where(g)() =(g).

2. The symmetric group S3. In Sn, conjugacy classes are determined by cycle decompositionsizes: two permutations are conjugate if and only if they have the same number of cyclesof each length. For S3, there are 3 conjugacy classes, so there are 3 different irreduciblerepresentationsoverC. Iftheirdimensionsared1, d2, d3,thend12+d22+d32 =6,soS3 musthavetwo1-dimensionalandone2-dimensionalrepresentations. The1-dimensionalrepresentationsarethetrivialrepresentationC+ givenby()=1andthesignrepresentationC givenby() = (1). The 2-dimensional representation can be visualized as representing the symmetries of theequilateraltrianglewithvertices1,2,3atthepoints(cos120,sin 120),(cos240,sin 240),(1,0)ofthecoordinateplane,respectively. Thus,forexample,

((12)) = 1 0 , ((123))= cos120 sin 120 .0 1 sin120 cos 120

Toshowthatthisrepresentation isirreducible,consideranysubrepresentationV. V mustbethe spanofasubsetof theeigenvectors of((12)), whicharethenonzeromultiples of(1,0)and (0,1). V must also be the span of a subset of the eigenvectors of ((123)), which aredifferentvectors. Thus,V mustbeeitherC2 or0.

3. ThequaterniongroupQ8 ={1,i,j,k},withdefiningrelationsi=jk=kj, j=ki=ik, k=ij=ji, 1 =i2 =j2 =k2.

35


36/106

The5conjugacyclasses are{1},{1},{i},{j},{k}, sothereare5different irreduciblerepresentations, the sum of the squares of whose dimensions is 8, so their dimensions mustbe1,1,1,1,and2.The center Z(Q8) is {1}, and Q8/Z(Q8) Z2 Z2. The four 1-dimensional irreducible=representationsofZ2Z2 canbepulledbacktoQ8. Thatis, ifq:Q8 Q8/Z(Q8) isthequotientmap,andanyrepresentationofQ8/Z(

introduction to representation theory (etingof)

Documents