LIE ALGEBRAS KEVINMCGERTY2012,WITHMINORMODIFICATIONSBYBALA´ZSSZENDRO˝I2013,DANCIUBOTARU2015 1. BACKGROUND InthissectionIusesomematerial,likemultivariableanalysis,whichisnotnecessaryforthemainbodyofthe course,butifyouknowit(andhopefullyifyoudon’tbutarewillingtothinkimpreciselyatsomepoints)itwill helptoputthecourseincontext. Forthoseworriedaboutsuchthings,fearnot,itisnon-examinable. Inmathematics,groupactionsgiveawayofencodingthesymmetriesofaspaceorphysicalsystem. Formallythesearedefinedasfollows: anactionofagroupGonaspace1X isamapa: G X X, × → written(g.x) a(g,x)ormorecommonly(g,x) g.xwhichsatisfiestheproperties 7→ 7→ (1) e.x=x,forallx X,wheree Gistheidentity; ∈ ∈ (2) (g g ).x=g .(g .x)forallg ,g Gandx X. 1 2 1 2 1 2 ∈ ∈ Natural examples of actions are that of the group of rigid motions SO on the unit sphere x R3 : 3 { ∈ x =1 ,orthegenerallineargroupGL (R)onRn. n || || } WheneveragroupactsonaspaceX,thereisaresultinglinearaction(arepresentation)onthevector spaceoffunctionsonX.IndeedifFun(X)denotesthevectorspaceofreal-valuedfunctionsonX,then theformula g(f)(x)=f(g−1.x), (g G,f Fun(X),x X), ∈ ∈ ∈ defines a representation of G on Fun(X). If X and G have more structure. e.g. that of a topological space or smooth manifold, then this action may also preserve the subspaces of say continuous, or differentiablefunctions.Liealgebrasariseasthe“infinitesimalversion”ofgroupactions,whichloosely speakingmeanstheyarewhatwegetbytryingtodifferentiategroupactions. Example 1.1. Take for example the natural action of the circle S1 by rotations on the plane R2. This actioncanbewrittenexplicitlyusingmatrices: cos(t) sin(t) g(t)= − sin(t) cos(t) (cid:18) (cid:19) where we have smoothly parametrized the circle S1 using the trigonometric functions. Note that for thisparametrization,g(t)−1 = g( t). TheinducedactiononFun(R2)restrictstoanactionon ∞(R2) − C thespaceofsmooth(i.e. infinitelydifferentiable)functionsonR2. Usingourparametrization,itmakes sensetodifferentiatethisactionattheidentityelement(i.e. att = 0)togetanoperationν: ∞(R2) C → ∞(R2),givenby C d x f f(g( t). ) 7→dt − y |t=0 (cid:18) (cid:19) (cid:0) df df (cid:1) =y x . dx − dy Itisimmediatefromtheproductrulefordifferentiation,thattheoperatorνconstructedintheabove exampleobeysthe“Leibnizrule”: ν(f.g)=ν(f).g+f.ν(g). An operator on smooth functions which satisfies this property is called a derivation. It’s not hard to see that any such operator on ∞(R2) will be of the form a(x,y) d +b(x,y) d where a,b ∞(R2). C dx dy ∈ C Thus,heuristicallyfornow,wethinkoftheinfinitesimalversionofagroupactionasthecollectionof derivationsonsmoothfunctionsweobtainby“differentiatingthegroupactionattheidentityelement”. (Forthecirclethecollectionofvectorfieldswegetarejustthescalarmultiplesofthevectorfieldν,but for actions of larger group we would attach a larger space of derivations). It turns out this set of 1I’mbeingdeliberatelyvaguehereaboutwhata“space”is,X couldjustbeaset,butitcouldalsohaveamoregeometric nature,suchasatopologicalspaceorsubmanifoldofRn. 1 2 KEVINMCGERTY2012,WITHMINORMODIFICATIONSBYBALA´ZSSZENDRO˝I2013,DANCIUBOTARU2015 derivations forms a vector space, but it also has a kind of “product” which is a sort of infinitesimal remnantofthegroupmultiplication2. Let’ssetthisupalittlemoreformally. Definition1.2. AvectorfieldonX =Rn(or,withabitmorework,anymanifold)isa(smooth)function ν: Rn Rn,whichonecanthinkofasgivingtheinfinitesimaldirectionofaflow(e.g. ofafluid,oran → electricfieldsay). ThesetofvectorfieldsformsavectorspacewhichwedenotebyΘ . Suchfieldscan X bemadetoactonfunctionsf: X Rbydifferentiation. Ifν =(a ,a ,...,a )instandardcoordinates 1 2 n → (herea :Rn R),thenset i → n ∂f ν(f)= a . i ∂x i i=1 X ThisformulagivesanactionofΘ onthespaceofsmoothfunctions ∞(X),sinceiff ∞(X),then X C ∈C soisν(f). Thisactionislinear,andinteractswithmultiplicationoffunctionsaccordingtotheLeibniz rule,thatis,ifν isavectorfield,andf ,f ∞(X)then 1 2 ∈C ν(f f )=ν(f ).f +f .ν(f ), 1 2 1 2 1 2 inotherwords, vectorfieldsactasderivations onsmoothfunctionsintheabovesense. Notewecan talk about vector fields and derivations interchangeably, since the derivation given by a vector field completelydeterminesthevectorfield,andanyderivationcomesfromavectorfield(thisisanexercise thatisworthchecking). Note that if we compose two derivations ν ν we again get an operator on functions, but it is 1 2 ◦ not given by a vector field, since it involves second order differential operators. However, it is easy tocheckusingthesymmetryofmixedpartialderivativesthatifν ,ν arederivations, then[ν ,ν ] = 1 2 1 2 ν ν ν ν is again a derivation. Thus the space Θ of vector fields on X is equipped with a 1 2 2 1 X ◦ − ◦ naturalproduct3[.,.]whichiscalledaLiebracket. Thederivativesofagroupactiongivesubalgebrasof thealgebraΘ . X Example 1.3. Consider the action of SO (R) on R3. Using the fact that every element of SO (R) is a 3 3 rotation about some axis through the origin it is not too hard to find the space of vector fields on R3 whichcanbeassociatedtothisaction,andcheckthatitformsaLiealgebra. Indeedaswesawbefore, theactionofthecirclefixingthez-axisgivesthederivation y d +x d , andthederivationobtained − dx dy from rotation about any other axis will be obtained by an orthogonal change of coordinates. It can beshownthattheseformthe3-dimensionalspaceg = span x d y d ,y d z d ,z d x d ,and R{ dy − dx dz − dy dx − dz} moreoveritisthennothardtocheckthatgisclosedunderthebracketoperations[, ]. (Thisalsogives · · anon-trivialexampleofa3-dimensionalLiealgebra). 2. DEFINITIONSANDEXAMPLES ThedefinitionofaLiealgebraisanabstractionoftheaboveexampleoftheproductonvectorfields. Itispurelyalgebraic,soitmakessenseoveranyfieldk. Definition 2.1. A Lie algebra over a field k is a pair (g,[.,.] ) consisting of a k-vector space g, along g withabilinearoperation[.,.] : g g gtakingvaluesingknownasaLiebracket,whichsatisfiesthe g × → followingaxioms: (1) [.,.] isalternating,i.e. [x,x] =0forallx g. g g ∈ (2) TheLiebracketsatisfiestheJacobiIdentity: thatis,forallx,y,z gwehave: ∈ [x,[y,z] ] +[y,[z,x] ] +[z,[x,y] ] =0. g g g g g g Remark2.2. ItiseasytocheckdirectlyfromthedefinitionthattheLiebracketweputonthespaceof vectorfieldsΘ satisfiestheaboveconditions. X Note that by considering the bracket [x+y,x+y] it is easy to see that the alternating condition g impliesthatforallx,y Lwehave[x,y] = [y,x] ,thatis[.,.] isskew-symmetric. Ifchar(k) = 2, g g g ∈ − 6 the alternating condition is equivalent to skew-symmetry. Note that a Lie algebra is an algebra with a product which is neither commutative nor associative, and moreover it does not have an identity element4. Wewillnormallysimplywrite[.,.]andreserveusethedecoratedbracketonlyforemphasis orwherethereisthepotentialforconfusion. 2Tobeabitmoreprecise,itcomesfromtheconjugationactionofthegrouponitself. 3Thisisintheweakestsense,inthatitisabilinearmapΘX×ΘX →ΘX.Itisnotevenassociative–theaxiomitdoessatisfy isdiscussedshortly. 4Thismakesthemsoundawful.However,aswewillseethisisnotthewaytothinkaboutthem! LIEALGEBRAS 3 Example2.3. (1) IfV isanyvectorspacethensettingtheLiebracket[.,.]tobezerowegeta(not veryinteresting)Liealgebra. SuchLiealgebrasarecalledabelianLiealgebras. (2) Generalising the example of vector fields a bit, if A is a k-algebra and δ: A A is a k-linear map,thenwesayδisak-derivationifitsatisfiestheLeibnizrule,thatis,if: → δ(a.b)=δ(a).b+a.δ(b), a,b A. ∀ ∈ ItiseasytoseebyadirectcalculationthatifDerk(A)denotesthek-vectorspaceofk-derivations onA,thenDerk(A)isaLiealgebraunderthecommutatorproduct,thatis: [δ ,δ ]=δ δ δ δ . 1 2 1 2 2 1 ◦ − ◦ Indeedthealternatingpropertyisimmediate,sotheonlythingtocheckistheJacobiidentity, whichisaneasycomputation. (3) For a more down-to-earth example, take g = gl the k-vector space of n n matrices with entriesink. ItiseasytocheckthatthisisaLiealgnebraforthecommutatorp×roduct: [X,Y]=X.Y Y.X. − Slightly more abstractly, if V is a vector space, then we will write gl(V) for the Lie algebra End(V)equippedwiththecommutatorproductasformatrices. (4) IfgisaLiealgebraandN < gisak-subspaceofgonwhichtherestrictionoftheLiebracket takesvaluesinN,sothatitinducesabilinearform[.,.] : N N N,then(N,[.,.] )isclearly N N × → aLiealgebra,andwesayN isa(Lie)subalgebraofg. (5) Letsl = X gl :tr(X)=0 bethespaceofn nmatriceswithtracezero.Itiseasytocheck n { ∈ n } × thatsl isaLiesubalgebraofgl (eventhoughitisnotasubalgebraoftheassociativealgebra n n End(V)). MoregenerallywesayanyLiesubalgebraofgl(V)foravectorspaceV isalinearLie algebra. (6) IfAisanassociativek-algebra,thenifa Aletδ : A Abethelinearmapgivenby a ∈ → δ (b)=a.b b.a, b A. a − ∈ One can check that δ is a derivation on A, and that this is equivalent to the statement that a (A,[.,.] )isaLiealgebra,where[.,.] isthecommutatorbracketonA,thatis[a,b] =a.b b.a. A A A − ThusanyassociativealgebracanbegiventhestructureofaLiealgebra.(Thisisageneralisation ofthecaseofn nmatrices). × Remark 2.4. One could begin to try and classify all (say finite-dimensional) Lie algebras. In very low dimension this is actually possible. For dimension 1 clearly there is a unique (up to isomorphism5) Lie algebra since the alternating condition demands that the bracket is zero. In dimension two, one can again have an abelian Lie algebra, but there is another possibility: if g has a basis e,f then we { } may set [e,f] = f, and this completely determines the Lie algebra structure. All two-dimensional Liealgebraswhicharenotabelianareisomorphictothisone(checkthis). Itisalsopossibletoclassify three-dimensionalLiealgebras,butitbecomesrapidlyintractabletodothisingeneralasthedimension increases. In this course we will focus on a particular class of Lie algebras, known as semisimple Lie algebras,forwhichanelegantclassificationtheoremisknown. 3. HOMOMORPHISMSANDIDEALS We have already introduced the notion of a subalgebra of a Lie algebra in the examples above, but there are other standard constructions familiar from rings which make sense for Lie algebras. A homomorphism of Lie algebras (g,[.,.]g) and (g′,[.,.]g′) is a k-linear map φ: g g′ which respects the → Liebrackets,thatis: φ([a,b]g)=[φ(a),φ(b)]g′ a,b g. ∀ ∈ AnisomorphismofLiealgebrasisabijectivehomomorphism. AnidealinaLiealgebragisasubspace I such that for all a g and x I we have [a,x] I. It is easy to check that if φ: g g′ is a g ∈ ∈ ∈ → homomorphism,thenker(φ)= a g:φ(a)=0 isanidealofg. Conversely,ifI isanidealofgthen { ∈ } itiseasytocheckthatthequotientspaceg/I inheritsthestructureofaLiealgebra,andthecanonical quotientmapq: g g/I isaLiealgebrahomomorphismwithkernelI. → Remark3.1. NotethatbecausetheLiebracketisskew-symmetric,wedonotneedtoconsidernotions ofleft,rightandtwo-sidedideals,astheywillallcoincide. IfanontrivialLiealgebrahasnonontrivial idealswesayitissimple. 5OfcourseIhaven’tsaidwhatanisomorphismofLiealgebrasisyet(seebelow)butyouprobablyknow... 4 KEVINMCGERTY2012,WITHMINORMODIFICATIONSBYBALA´ZSSZENDRO˝I2013,DANCIUBOTARU2015 Just as for groups and rings, we have the normal stable of isomorphism theorems, and the proofs areidentical. Theorem3.2. (1) Let φ: g g′ be a homomorphism of Lie algebras. The subspace φ(g) = im(φ) is a subalgebraofg′andφind→ucesanisomorphismφ¯: g/ker(φ) im(φ). → (2) IfJ I gareidealsofgthenwehave: ⊂ ⊂ (g/J) (I/J)=g/J ∼ (3) IfI,J areidealsofgthenwehave (cid:14) (I +J)/J =I/(I J). ∼ ∩ 4. REPRESENTATIONSOFLIEALGEBRAS Justasforfinitegroups(orindeedgroupsingeneral)onewayofstudyingLiealgebrasistotryand understandhowtheycanactonotherobjects. ForLiealgebras,wewilluseactionsonlinearspaces,or inotherwords,“representations”. Formallywemakethefollowingdefinition. Definition4.1. ArepresentationofaLiealgebragisavectorspaceV equippedwithahomomorphism ofLiealgebrasρ: g gl(V). Inotherwords,ρisalinearmapsuchthat → ρ([x,y])=ρ(x) ρ(y) ρ(y) ρ(x) ◦ − ◦ where denotescompositionoflinearmaps. Wemayalsorefertoarepresentationofgasag-module. ◦ A representation is faithful if ker(ρ) = 0. When there is no danger of confusion we will normally suppressρinournotation,andwritex(v)ratherthanρ(x)(v),forx g,v V. ∈ ∈ WewillstudyrepresentationofvariousclassesofLiealgebrasinthiscourse,butforthemomentwe willjustgivesomebasicexamples. Example4.2. (1) Ifg = gl(V)forV avectorspace, thentheidentitymapgl(V) gl(V)isarep- → resentationofgl(V)onV,whichisknownasthevectorrepresentation. Clearlyanysubalgebra gofgl(V)alsoinheritsV asarepresentation,wherethenthemapρisjusttheinclusionmap. (2) GivenanarbitraryLiealgebrag,thereisanaturalrepresentationadofgongitselfknownas theadjointrepresentation. Thehomomorphismfromgtogl(g)isgivenby ad(x)(y)=[x,y], x,y g. ∀ ∈ Indeedthefactthatthisisarepresentationisjustarephrasing6oftheJacobiidentity. Notethat while the vector representation is clearly faithful, in general the adjoint representation is not. Indeedthekernelisknownasthecentreofg: z(g)= x g:[x,y]=0, y g . { ∈ ∀ ∈ } Note that if x z(g) then for any representation ρ: g gl(V) the endomorphism ρ(x) com- ∈ → muteswithalltheelementsρ(y) End(V)forally g. ∈ ∈ (3) IfgisanyLiealgebra,thenthezeromapg gl isaLiealgebrahomomorphism. Thecorres- → 1 ponding representation is called the trivial representation. It is the Lie algebra analogue of the trivialrepresentationforagroup(whichsendeverygroupelementtotheidentity). (4) If(V,ρ)isarepresentationofg,wesaythatasubspaceU <V isasubrepresentationifφ(x)(U) ⊆ U forallx g. Itfollowsimmediatelythatφrestrictstogiveahomomorphismfromgtogl(U), ∈ hence(U,φ )isagainarepresentationofg. Notealsothatif V : i I areacollectionofin- |U i { ∈ } variantsubspaces,theirsum V isclearlyalsoinvariant,andsoagainasubrepresentation. i∈I i There are a number of standard wPays of constructing new representations from old, all of which havetheiranalogueforgrouprepresentations. Forexample,recallthatifV isak-vectorspace,andU isasubspace,thenwemayformthequotientvectorspaceV/U.Ifφ: V V isanendomorphismofV whichpreservesU,thatisifφ(U) U,thenthereisaninducedmapφ¯:→V/U V/U. Applyingthisto ⊆ → representationsofaLiealgebragweseethatifV isarepresentationofgandU isasubrepresentation we may always form the quotient representation V/U. Next, if (V,ρ) and (W,σ) are representations of g, then clearly V W the direct sum of V and W becomes a g-representation via the obvious homo- ⊕ morphismρ σ. Moreinterestingly,thevectorspaceHom(V,W)oflinearmapsfromV toW hasthe ⊕ structureofag-representationvia (4.1) x(φ)=σ(x) φ φ ρ(x), x g,φ Hom(V,W). ◦ − ◦ ∀ ∈ ∈ 6Checkthis!It’salso(forsomepeople)ausefulwayofrememberingwhattheJacobiidentitysays. LIEALGEBRAS 5 Itisstraight-forwardtocheckthatthisgivesaLiealgebrahomomorphismfromgtogl(Hom(V,W)). One way to do this is simply to compute directly. Another, slightly quicker way, is to notice that Endk(V W)=Homk(V W,V W)containsEndk(V),Endk(W)andHomk(V,W)since ⊕ ⊕ ⊕ Homk(V W,V W)=Homk(V,V) Homk(V,W) Homk(W,V) Homk(W,W), ⊕ ⊕ ⊕ ⊕ ⊕ thuswemaycombineρandσtogiveahomomorphismτ: g gl(V W).SinceHomk(V,W)isclearly → ⊕ preservedbyτ(g),itisasubrepresentationofgl(V W),wherethelatterisag-representationviathe ⊕ compositionofτ: g gl(V W)withtheadjointrepresentationad: gl(V W) gl(gl(V W)). It → ⊕ ⊕ → ⊕ istheneasytoseethattheequation(4.1)describestheresultingactionofgonHomk(V,W). AnimportantspecialcaseofthisiswhereW = kisthetrivialrepresentation(asabove,sothatthe map σ: g gl(k) is the zero map). This allows us to give V∗ = Hom(V,k), the dual space of V, a → natural structure of g-representation where (since σ = 0) the action of x g on f V∗ is given by ∈ ∈ ρ∗: g gl(V∗)where → ρ∗(x)(f)= f ρ(x) (f V∗). − ◦ ∈ Ifα: V V isanylinearmap,recallthatthetransposemapαt: V∗ V∗isdefinedbyαt(f)=f α, → → ◦ thusourdefinitionoftheactionofx gonV∗isjust7 ρ(x)t.Thismakesitclearthattheactionofgon ∈ − V∗ iscompatiblewiththestandardconstructionsondualspaces,e.g. ifU isasubrepresentationofV, theU0 theannihilatorofU willbeasubrepresentationofV∗,andmoreover,thenaturalisomorphism ofV withV∗∗isanisomorphismofg-representations. Weendthissectionwithsometerminologywhichwillbeusefullater. Definition4.3. Arepresentationissaidtobeirreducibleifithasnopropernon-zerosubrepresentations, anditissaidtobecompletelyreducibleifitisisomorphictoadirectsumofirreduciblerepresentations. Example 4.4. Giving a representation of gl is equivalent to giving a vector space equipped with a 1 linear map. Indeed as a vector space gl = k, hence if (V,ρ) is a representation of gl we obtain a 1 1 linearendomorphismofV bytakingρ(1). Sinceeveryotherelementofgl isascalarmultipleof1this 1 completelydeterminestherepresentation,andthiscorrespondenceisclearlyreversible. Ifweassumekisalgebraicallyclosed,thenyouknowtheclassificationoflinearendomorphismsis given by the Jordan canonical form. From this you can see that the only irreducible representations ofgl aretheone-dimensionalones,whileindecomposablerepresentationscorrespondtolinearmaps 1 withasingleJordanblock. 5. NILPOTENTLIEALGEBRAS We now begin to study particular classes of Lie algebra. The first class we study are nilpotent Lie algebras,whicharesomewhatanalogoustonilpotentgroups. Weneedafewmoredefinitions. Definition 5.1. If V,W are subspaces of a Lie algebra g, then write [V,W] for the linear span of the elements [v,w]:v V,w W . NoticethatifI,J areidealsingthensois[I,J]. Indeedtocheckthis, { ∈ ∈ } notethatifi I,j J,x gwehave: ∈ ∈ ∈ [x,[i,j]]= [i,[j,x]] [j,[x,i]]=[i,[x,j]]+[[x,i],j] [I,J] − − ∈ usingtheJacobiidentityinthefirstequalityandskew-symmetryinthesecond. Definition5.2. ForgaLiealgebra,letC0(g) = g,andCi(g) = [g,Ci−1(g)]fori 1. Thissequenceof ≥ idealsofgiscalledthelowercentralseriesofg,andwesaygisnilpotentifCN(g)=0forsomeN >0. If N isthesmallestintegersuchthatCN(g)=0thenwesaythatgisanN-stepnilpotentLiealgebra. For example, a Lie algebra is 1-step nilpotent if and only if it is abelian. The definition can be rephrasedasfollows:thereisanN >0suchthatforanyN elementsx ,x ,...,x ofgtheiteratedLie 1 2 N bracket [x ,[x ,[...,[x ,x ]]...]=adx (adx (...adx (x ))...)=0. 1 2 N−1 N 1 2 N−1 N Inparticular,alltheelementsad(x) gl(g)forx garenilpotent. ∈ ∈ Remark5.3. TheidealC1(g) = [g,g]isknownasthederivedsubalgebra8ofgandisalsodenoted9D(g) andsometimesg′. 7NotethattheminussigniscrucialtoensurethisisaLiealgebrahomomorphism–concretelythisamountstonoticingthat A7→−Atpreservesthecommutatorbracketonn×nmatrices. 8Oddly,notasthederivedidealeventhoughitisanideal. 9Partlyjusttocauseconfusion,butalsobecauseitcomesupalot,playingslightlydifferentroles,whichleadstothedifferent notation.We’llseeitagainshortlyinaslightlydifferentguise. 6 KEVINMCGERTY2012,WITHMINORMODIFICATIONSBYBALA´ZSSZENDRO˝I2013,DANCIUBOTARU2015 Lemma5.4. LetgbeaLiealgebra. Then (1) Ifgisnilpotent,soisanysubalgebraorquotientofg. (2) Ifg/z(g)isnilpotent,thengisnilpotent. Proof. Thefirstpartisimmediatefromthedefinition. Indeedifh gisasubalgebraofgthenclearly ⊆ wehaveCi(h) Ci(g),sothatifCN(g)=0wealsohaveCN(h)=0. Similarly,aneasyinductionsay ⊆ showsthatifhisanideal,thenCi(g/h)=(Ci(g)+h)/h,andsoagainifCN(g)=0wemustalsohave CN(g/h)=0. Forthesecondclaim,takingh=z(g),anidealofg,weseefromthefirstpartthatifg/z(g)isnilpotent, thenthereissomeN with(CN(g)+z(g))/z(g) = CN(g/z(g)) = 0,andsoCN(g) z(g). Butthenitis clearthatCN+1(g)=0,andsogisnilpotentasrequired. ⊆ (cid:3) Remark5.5. NoticethatifI isanarbitraryidealing,andI andg/I arenilpotentitdoesnotfollowthat gisnilpotent. Indeedrecallthatifgisthenon-abelian2-dimensionalLiealgebra,thengcanbegivena basisx,ywith[x,y]=y.Hencek.yisa1-dimensionalidealing(whichisthusabelianandsonilpotent) andthequotientisagain1-dimensionalandsonilpotent. However,ad(x)hasyasaneigenvectorwith eigenvalue1,sogcannotbenilpotent,andindeedCi(g)=k.yforalli 1. ≥ Example5.6. LetV beavectorspace,and =(0=F F F ... F =V) 0 1 2 n F ⊂ ⊂ ⊂ ⊂ asequenceofsubspaceswithdim(F )=i(suchasequenceisknownasaflagorcompleteflaginV). Let i n = n bethesubalgebraofgl(V)consistingoflinearmapsX gl(V)suchthatX(F ) F forall F i i−1 ∈ ⊆ i 1. WeclaimtheLiealgebranisnilpotent. Toseethisweshowsomethingmoreprecise. Indeedfor ≥ eachpositiveintegerk,considerthesubspace nk = x gl(V):x(F ) F i i−k { ∈ ⊂ } (where we let 0 = F for all l 0). Then clearly nk n, and nk = 0 for any k n. We claim that l ≤ ⊂ ≥ Ck(n) nk+1,whichwillthereforeprovenisnilpotent. Theclaimisimmediatefork = 0,sosuppose ⊆ weknowbyinductionthatCk(n) nk+1. Thenifx nandy nk+1,wehavexy(F ) x(F ) i i−k−1 ⊆ ∈ ∈ ⊂ ⊂ F ,andsimilarlyyx(F ) F ,thuscertainly[x,y] nk+2andsoCk+1(n) nk+2asrequired. i−k−2 i i−k−2 ⊂ ∈ ⊆ In fact you can check that Ck(n) = nk+1, so that n is (n 1)-step nilpotent i.e. Cn−2(n) = 0, and − 6 Cn−1(n) = 0 (note that if dim(V) = 1 then n = 0). If we pick a basis e ,e ,...,e of V such that 1 2 n { } F = span(e ,e ,...,e thenthematrixArepresentinganelementx nwithrespecttothisbasisis i 1 2 i } ∈ strictlyuppertriangular,thatis,a = 0foralli j. Itfollowsthatdim(n) = n ,sowhenn = 2we ij ≥ 2 justgetthe1-dimensionalLiealgebra,thusthefirstnontrivialcaseiswhenn = 3andinthatcasewe (cid:0) (cid:1) geta3-dimensional2-stepnilpotentLiealgebra. Notethatthesubalgebrat gl ofdiagonalmatricesisnilpotent,sinceitisabelian,soanilpotent ⊂ n linearLiealgebraneednotconsistofnilpotentendomorphisms. Neverthelesswewillshowthereisa closeconnectionbetweenthetwonotions. Lemma5.7. Ifx gl(V)beanilpotentendomorphismthenad(x) End(gl(V))isalsonilpotent. ∈ ∈ Proof. Themapλ : gl(V) gl(V)givenbyy xy isclearlynilpotentifxisnilpotent,andsimilarly x → 7→ forthemapρ : gl(V) gl(V)givenbyy yx.Moreover,λ andρ clearlycommutewitheachother, x x x → 7→ sosincead(x)=λ ρ ,itisalsonilpotent. Indeedform 0wehave x x − ≥ m m (λ ρ )m = ( 1)i λm−iρi x− x − i x x i=0 (cid:18) (cid:19) X andsoifxn = 0,sothatλn = ρn = 0,thenifm 2n,everytermontheright-handsidemustbezero, andsoad(x)m =0asrequxired.x ≥ (cid:3) Forthenextpropositionweneedthenotionofthenormaliserofasubalgebra. Definition5.8. LetgbeaLiealgebraandletabeasubalgebra. Thesubspace N (a)= x g:[x,a] a, a a g { ∈ ∈ ∀ ∈ } is a subalgebra of g (check this using the Jacobi identity) which is called the normaliser of a. It is the largestsubalgebraofginwhichaisanideal. LIEALGEBRAS 7 Definition5.9. IfgisanyLiealgebraand(V,ρ)isarepresentationofg,thendefine Vg = v V :ρ(x)(v)=0, x g . { ∈ ∀ ∈ } Itiscalledthe(possiblyzeroingeneral)subrepresentationofinvariantsinV. Proposition5.10. LetnbeaLiealgebra,and(V,ρ)arepresentationofnsuchthatforeveryx n,thelinear ∈ mapρ(x)isnilpotent. Thentheinvariantsubspace Vn = v V :ρ(x)(v)=0, x n { ∈ ∀ ∈ } isnon-zero. Proof. Weuseinductionond=dim(n),thecased=1beingclear. Clearlythestatementofthepropos- ition is unchanged if we replace n by its image in gl(V), so we may assume that n is a subalgebra of gl(V). Nowconsidera ( napropersubalgebra. Sincetheelementsa aarenilpotentendomorphisms, ∈ thepreviouslemmashowsthatad(a) gl(n)isalsonilpotent,hencead(a)alsoactsnilpotentlyonthe ∈ quotient10n/a. Sincedim(a) < dim(n), byinduction(appliedwithV = n/aanda)wecanfindx / a ∈ suchthatad(a)(x)=[a,x] aforalla a. Itfollowsthatforanypropersubalgebraa,itsnormaliser ∈ ∈ N (a)= x n:[x,a] a, a a n { ∈ ∈ ∀ ∈ } strictlycontainsa. Thus if a is a proper subalgebra of n of maximal dimension, we must have N (a) = n, or in other n words,amustactuallybeanidealofn. Nowifn/aisnotone-dimensional,thepreimageofaone-dimensionalsubalgebrainn/awouldbe apropersubalgebraofnstrictlycontaininga,whichagaincontradictsthemaximalityofa. Thusn/ais one-dimensional,andwemayfindz nsothatk.z a=n. ∈ ⊕ Byinduction,weknowthatVa = v V : a(v) = 0, a a isanonzerosubspaceofV. Weclaim { ∈ ∀ ∈ } thatzpreservesVa. Indeed a(z(v))=[a,z](v)+z(a(v))=0, a a,v Va, ∀ ∈ ∈ since[a,z] a. Buttherestrictionofz toVa isnilpotent,sothesubspaceU = v Va : z(v) = 0 is nonzero. Si∈nceU =Vnwearedone. { ∈ }(cid:3) Definition5.11. IfgisaLiealgebraandx g,wesaythatanelementxisad-nilpotentifad(x) gl(g) ∈ ∈ isanilpotentendomorphism. Forconvenience11wesaythatgisad-nilpotentifallofitselementsare. Theorem5.12. (Engel’stheorem)ALiealgebragisnilpotentifandonlyifad(x)isnilpotentforeveryx g. ∈ Proof. In the terminology of the above definition, the theorem states that a Lie algebra is nilpotent if andonlyifitisad-nilpotent. Aswealreadynoted,itisimmediatefromthedefinitionthatanilpotent Liealgebraisad-nilpotent,soourtaskistoshowtheconverse. Forthis,notethatitisclearthatifgis ad-nilpotent, then any quotient of it is also ad-nilpotent. Hence using induction on dimension along withLemma5.4(2),itwillbeenoughtoshowthatz(g)isanon-zero. Now z(g)= z g:[z,x]=0, x g { ∈ ∀ ∈ } = z g: ad(x)(z)=0, x g { ∈ − ∀ ∈ } =gad(g), henceapplyingProposition5.10totheadjointrepresentationofg,itfollowsimmediatelythatz(g)=0 6 andwearedone. (cid:3) ThefollowingisanimportantconsequenceoftheProposition5.10. Corollary5.13. LetgbeaLiealgebraand(V,ρ)arepresentationofgsuchthatρ(x)isanilpotentendomorphism forallx g. Thenthereisacompleteflag =(0=F F ... F =V)suchthatρ(g) n . 0 1 n F ∈ F ⊂ ⊂ ⊂ ⊆ 10Thenotationherecanbeabitconfusing:theLiealgebranisana-representationbytherestrictionoftheadjointrepresent- ationofgtoa. Sinceaisasub-a-representationofg,thequotientg/aisana-representation. Notehoweverthatitisnotitselfa Liealgebraunlessaisanidealofg. 11Thoughbecauseofthefollowingtheoremthisterminologywon’tbeusedmuch. 8 KEVINMCGERTY2012,WITHMINORMODIFICATIONSBYBALA´ZSSZENDRO˝I2013,DANCIUBOTARU2015 Proof. Letussaythatgrespectsaflag ifρ(g) n . Useinductionondim(V). ByProposition5.10, F F ⊆ we see that the space Vg = 0. Thus by induction we may find a flag ′ in V/Vg which g respects. 6 F Takingitspreimageandextendingarbitrarily(bypickinganycompleteflaginVg),wegetacomplete flaginV thatgclearlyrespectsasrequired. (cid:3) 6. SOLVABLELIEALGEBRAS We now consider another class of Lie algebras which is slightly larger than the class of nilpotent algebras. ForaLiealgebrag, letD0g = g, andDi+1g = [Dig,Dig]. Digisthei-thderivedidealofg. NotethatC1(g)=D1gisDgthederivedsubalgebraofg. Definition6.1. ALiealgebragissaidtobesolvableifDNg=0forsomeN >0. Since it is clear from the definition that Dig Ci(g), any nilpotent Lie algebra is solvable, but as ⊂ onecanseebyconsideringthenon-abelian2-dimensionalLiealgebra,therearesolvableLiealgebras whicharenotnilpotent. Example 6.2. Let V be a finite dimensional vector space and = (0 = F < F < ... < F = V) a 0 1 n F completeflaginV. Let b = x gl(V):x(F ) F , F i i { ∈ ⊆ } thatis,b isthesubspaceofendomorphismswhichpreservethecompleteflag . Weclaimthatb is F F F solvable. SinceanynilpotentLiealgebraissolvable,andclearlyb issolvableifandonlyifD1b is, F F thesolvabilityofgwillfollowifwecanshowthatD1b n . Toseethis,supposefirstthatx,y b F F F ⊆ ∈ and consider [x,y]. We need to show that [x,y](F ) F for each i, 1 i n. Since x,y b , i i−1 F ⊂ ≤ ≤ ∈ certainly we have [x,y](F ) F for all i,1 i n, thus it is enough to show that the map [x,y] i i ⊆ ≤ ≤ inducedby[x,y]onF /F iszero. Butthismapisthecommutatorofthemapsinducedbyxandy i i−1 inEnd(F /F ),whichsinceF /F isone-dimensional,isabelian,sothatallcommutatorsarezero. i i−1 i i−1 Ifwepickabasis e ,e ,...,e ofV suchthatF =span(e ,...,e ),thengl(V)getsidentifiedwith 1 2 n i 1 i { } gl and b corresponds to the subalgebra b of upper triangular matrices. It is straight-forward to n F n showbyconsideringthesubalgebrat ofdiagonalmatricesthatb isnotnilpotent. n n Wewillseeshortlythat,incharacteristiczero,anysolvablelinearLiealgebrag gl(V),whereV is ⊂ finitedimensional,isasubalgebraofb forsomecompleteflag . Wenextnotesomebasicproperties F F ofsolvableLiealgebras. Lemma6.3. LetgbeaLiealgebra,φ: g hahomomorphismofLiealgebras. → (1) Wehaveφ(Dkg) = Dk(φ(g)). Inparticularφ(g)issolvableifgis,thusanyquotientofasolvableLie algebraissolvable. (2) Ifgissolvablethensoareallsubalgebrasofg. (3) Ifim(φ)andker(φ)aresolvablethensoisg. ThusifI isanidealandI andg/I aresolvable,soisg. Proof. Thefirsttwostatementsareimmediatefromthedefinitions. Forthethird,notethatifim(φ)is solvable,thenforsomeN wehaveDNim(φ)= 0 ,sothatbypart(1)wehaveDN(g) ker(φ),hence ifDMker(φ)= 0 wemusthaveDN+Mg= 0{a}srequired. ⊂ (cid:3) { } { } Notethat,asmentionedabove,thepart3)oftheaboveLemmaisfalsefornilpotentLiealgebras. For the rest of this section we will assume that our field k is algebraically closed of characteristic zero. Lemma6.4. (Lie’sLemma)LetgbeaLiealgebraandletI gbeanideal,andV afinitedimensionalrepresent- ation. Supposev V isavectorsuchthatx(v) = λ(x).vf⊂orallx I,whereλ: I gl (k). Thenλvanishes ∈ ∈ → 1 on[g,I] I. ⊂ Proof. Letx g. Foreachm N,letW =span v,x(v),...,xm(v) . TheW formanestedsequence m m ∈ ∈ { } ofsubspacesofV. Weclaimthathxm(v) λ(h)xmv+W forallh I andm 0. Usinginduction m−1 ∈ ∈ ≥ onm,theclaimbeingimmediateform=0,notethat hxm(v)=[h,x]xm−1(v)+xhxm−1(v) (λ([h,x])xm−1v+W )+x(λ(h)xm−1(v)+W ) m−2 m−2 ∈ λ(h)xm(v)+W , m−1 ∈ whereinthesecondequalityweuseinductiononmforbothh,[h,x] I. ∈ NowsinceV isfinitedimensional,thereisamaximalnsuchthatthevectors v,x(v),...,xn(v) are { } linearlyindependent,andsoW = W forallm n. ItthenfollowsthatW ispreservedbyx,and m n n ≥ LIEALGEBRAS 9 fromtheclaimitfollowsthatW isalsopreservedbyeveryh I. Moreover,theclaimalsoshowsthat n ∈ foranyh I thematrixof[x,h]withrespecttothebasis v,x(v)...,xn(v) ofW isuppertriangular n ∈ { } withdiagonalentriesallequaltoλ([x,h]). Itfollowsthattr([x,h])=(n+1)λ([x,h]). Sincethetraceof acommutatoriszero12,itfollowsthat(n+1)λ([x,h]) = 0,andsosincechar(k) = 0weconcludethat λ([x,h])=0. (cid:3) Theorem 6.5. (Lie’s theorem) Let g be a solvable Lie algebra and V is a g-representation. Then there is a homomorphismλ: g gl (k)andanonzerovectorv V suchthatx(v)=λ(x).vforallx g. → 1 ∈ ∈ Proof. We use induction ondim(g). Ifdim(g) = 1, then g = k.x forany nonzero x g, and since k is ∈ algebraicallyclosed,xhasaneigenvectorinV andwearedone. Fordim(g) > 1,considerthederived subalgebraD1g.Sincegissolvable,D1gisaproperidealofg.Thequotientg/D1gisabelian,andtaking thepreimageofanycodimensiononesubspaceofitgivesacodimension1idealIofg.Byinductionwe maypickahomomorphismλ: I gl (k)suchthatthesubspaceU = w V :h(w)=λ(h).w, h I → 1 { ∈ ∀ ∈ } isnonzero. Nowifx g,then ∈ h(x(w))=[h,x](w)+xh(w) =λ([h,x])(w)+λ(h).x(w) =λ(h).x(w). whereinthesecondequalityweusedLie’sLemma. ThusgpreservesU. NowsinceI iscodimension one in g, we may write g = kx I for some x g. Taking an eigenvector v U of x completes the proof. ⊕ ∈ ∈ (cid:3) TheanalogueofCorollary5.13forsolvableLiealgebrasisthefollowing: Corollary6.6. Letb gl(V)beasolvablesubalgebraofgl(V). Thenthereisacompleteflag ofV suchthat ⊂ F b b . F ⊆ Proof. Byinductionondim(V),whereLie’stheoremprovidestheinductionstep. (cid:3) NotethatthistheoremshowsthatifgisasolvableLiealgebra,thenanyirreduciblerepresentationof gisone-dimensional. Sinceconverselyanyone-dimensionalrepresentationofgisclearlyirreducible, we can rephrase Lie’s theorem as the statement that any irreducible representation of a solvable Lie algebraisone-dimensional. Moregenerally,itisnaturaltonotethefollowingsimpleLemma. Lemma 6.7. Let g be a Lie algebra. The one-dimensional representations of g are parametrized by the vector space(g/Dg)∗ asfollows: if(V,ρ)isaarepresentationofganddim(V) = 1,thenρ: g End(V) = k,and ρ(Dg)=0,sothatρinducesalinearfunctionalα: g/Dg k. → ∼ → Proof. Ifρ: g gl(V),thensincedim(V) = 1,theassociativealgebraEnd(V)iscommutative,sothat → gl(V)isabelian. Thus ρ([x,y])=ρ(x)ρ(y) ρ(y)ρ(x)=0, ( x,y g), − ∀ ∈ hence ρ(Dg) = 0, and ρ induces α (g/Dg)∗ as claimed. Conversely, given α: (g/Dg)∗, the map ρ: g gl (k)givenbyx α(x+Dg∈)isaLiealgebrahomomorphism,because → 1 7→ ρ([x,y])=0=α(x+Dg)α(y+Dg) α(y+Dg)α(x+Dg). − (cid:3) Anisomorphismclassesofone-dimensionalrepresentationsofgarethusgivenbythespaceoflinear functionals(g/Dg)∗. Wewillrefertoanelementα (g/Dg)∗ asaweightofg. Theyshouldbethought ∈ of as the generalization of the notion of an eigenvalue of a linear map. Lie’s theorem shows that if g is a solvable Lie algebra (hence in particular if g is nilpotent), then any irreducible representation is one-dimensional, sothat(g/Dg)∗ parametrizes theirreducible representations ofg. Forα (g/Dg)∗ letuswritek fortherepresentation(k,α)ofgonthefieldkgivenbyα. ∈ α 12Itisimportantherethatρ([x,h])isthecommutatorofρ(x)andρ(h)bothofwhichpreserve Wn–bytheclaiminthecaseof ρ(h),andbyourchoiceofninthecaseofρ(x)–inordertoconcludethetraceiszero. 10 KEVINMCGERTY2012,WITHMINORMODIFICATIONSBYBALA´ZSSZENDRO˝I2013,DANCIUBOTARU2015 7. REPRESENTATIONSOFNILPOTENTLIEALGEBRAS Inthissectionweassumethatkisanalgebraicallyclosedfieldofcharacteristiczero. A representation (ρ,V) of the one-dimensional Lie algebra gl (k) on a k-vector space V is given, 1 once we chose a basis vector e of gl (k), by a single linear map φ: V V via the correspondence φ = ρ(e). Thustheclassificationofre1presentationsofgl (k)isequivalen→ttotheclassificationoflinear 1 endomorphisms.13 ThisclassificationisofcoursegivenbytheJordannormalform(atleastoveranal- gebraicallyclosedfield). Inthissectionwewillseethata(slightlyweaker)versionofthisclassification holdsforrepresentationsofanynilpotentLiealgebra. Webeginbyreviewingsomelinearalgebra. Letx: V V bealinearmap. LetV bethegeneralized λ → eigenspacesforxwitheigenvalueλ: V = v V : N >0,(x λ)N(v)=0 . λ { ∈ ∃ − } (thusV iszerounlessλisaneigenvalueofx). λ Lemma7.1. Letx: V V bealinearmap. Thereisacanonicaldirectsumdecomposition → V = V , λ λ∈k M ofV intothegeneralizedeigenspacesofx. Moreover,foreachλ,theprojectiontoa : V V (withkernelthe λ λ → remaininggeneralizedeigenspaceofx)canbewrittenasapolynomialinx. Proof. Letm k[t]betheminimalpolynomialofx. Thenifφ: k[t] End(V)givenbyt xdenotes x ∈ → 7→ thenaturalmap, wehavek[t]/(mx) ∼= im(φ) ⊆ End(V). Ifmx = ki=1(t−λi)ni wheretheλi arethe distincteigenvaluesofx,thentheChineseRemainderTheoremshowsthat Q k k[t]/(m )= k[t]/(t λ )ni, x ∼ − i i=1 M Itfollowsthatwemaywrite1 k[t]/(m )as1 = e +...+e accordingtotheabovedecomposition. x 1 k ∈ Now clearly e e = 0 if i = j and e2 = e , so that if U = im(e ), then we have V = U . i j 6 i i i i 1≤i≤k i Moreover,eache canbewrittenaspolynomialsinxbypickinganyrepresentativeink[t]ofe (thought i i ofasanelementofk[t]/(m )). NoteinparticularthismeansthateachU isinvariantunderLim(φ). x i ItthusremainstocheckthatUi =Vλi. Since(t−λi)niei =0∈k[t]/(mx),itisclearthatUi ⊆Vλi. To seethereverseinclusion,supposethatv V sothat,say,(x λ )n(v)=0.Writev =v +...+v ,where ∈ λi − i 1 k v U . Nowsince(x λ )n(v)=0,andeachU isstableunderim(φ)itfollowsthat(x λ )n(v )=0 j j i j i j ∈ − − foreachj,(1 j k). If1 l kandl = k,thensince(t λi)n and(t λl)nl arecoprime,wemay finda,b k[t]≤such≤thata.(t≤ λ≤i)n+b.(t 6 λl)nl =1. − − ∈ − − Settingt=xinthisequation,andapplyingtheresulttothevectorv wefind: l v =1.v = a.(t λ )n+b.(t λ )nl (v ) l l i l l − − (7.1) =(cid:0)(t λi)n(vl)+(t λl)nl(v(cid:1)l) − − =0, where the first term is zero by the above, and the second term is zero since v U . It follows that l l ∈ v =v U andsoV U asrequired. i ∈ i λi ⊆ i (cid:3) Lemma7.2. i) Let V be a vector space and write A = Homk(V,V) for the associative algebra of linear mapsfromV toitself. Thenwehave: n n (x λ µ)ny = (ad(x) λ)i(y)(x µ)n−i, ( x,y A), − − i − − ∀ ∈ i=0(cid:18) (cid:19) X whereasusualad(x)(y)=[x,y]=xy yx. − ii) LetgbeaLiealgebraand(V,ρ)arepresentationofg. Thenforallx,y gwehave ∈ n n (7.2) (ρ(x) λ µ)nρ(y)= ρ((ad(x) λ)i(y))(ρ(x) µ)n−i. − − i − − i=0(cid:18) (cid:19) X 13EssentiallythesameistrueforrepresentationsoftheabeliangroupZ.