Lie algebras [Lecture notes]

37 Pages·2015·0.473 MB·English

by  Kevin McGerty, Balász Szendroi, Dan Ciubotaru

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.