M5P46 - Lie Algebras [Lecture notes]

66 Pages·2015·0.302 MB·English

by  Ed Segal, I. Udrea

M5P46 - Lie Algebras
Lectures by Ed Segal
Notes by Iulian Udrea
Imperial College London, Spring 2015
Lecture 1 (15.01.2015) 1 Lecture 12 (10.02.2015) 33
Lecture 2 (20.01.2015) 3 Lecture 13 (12.02.2015) 35
Lecture 3 (20.01.2015) 6 Lecture 14 (17.02.2015) 38
Lecture 4 (22.01.2015) 9 Lecture 15 (17.02.2015) 41
Lecture 5 (27.01.2015) 13 Lecture 16 (19.02.2015) 43
Lecture 6 (27.01.2015) 15 Lecture 17 (24.02.2015) 45
Lecture 7 (29.01.2015) 20 Lecture 18 (26.02.2015) 48
Lecture 8 (03.02.2015) 22 Lecture 19 (03.03.2015) 51
Lecture 9 (03.02.2015) 24 Lecture 20 (05.03.2015) 55
Lecture 10 (05.02.2015) 26 Lecture 21 (10.03.2015) 58
Lecture 11 (10.02.2015) 29 Lecture 22 (12.03.2015) 62
Introduction
Iamresponsibleforallfaultsinthisdocument,mathematicalorotherwise. Anymeritsofthematerial
hereshouldbecreditedtothelecturer,nottome.
[e-mail].
ThelayoutofthisdocumentisbasedonZevChonoles’coursenotestemplate.
Lecture 1 (15.01.2015)
The term “Lie algebra”, named after Sophus Lie was introduced by Hermann Weyl in the 1930s. Lie
algebraswerealsoknownas“infinitesimalgroups”.
Lie algebras are algebraic objects. In this course we try to turn every abstract description into a
matrix description. Consider Matn×n((cid:67)), the (cid:67)-vector space of dimension n2 of n×n matrices over (cid:67).
Wehavematrixmultiplication: Matn×n((cid:67))×Matn×n((cid:67))→Matn×n((cid:67))sendingthepair(M,N)totheir
product, MN. The Lie bracket or commutator of M and N is denoted by [M,N]. We can send (M,N)
to MN −NM and set MN −NM = [M,N]. We can see that this map is bilinear, anti-symmetric (i.e.
[M,N]=−[N,M])andsatisfiestheJacobiidentity (i.e. [[M,N],L]+[[N,L],M]+[[L,M],N]=0).
To prove the Jacobi identity, one just needs to expand the left hand side and see that everything
cancelsout.
One particular thing worth mentioning is that [M,M]=0 if and only if [M,N]=−[N,M]. To see
this, we just need to check that [M +N,M +N]=[M,M]+[M,N]+[N,M]+[N,N]=0. But this is
clearsince[M,M]=0and[N,N]=0,andso[M,N]=−[N,M].
Definition. ALiealgebraisavectorspace L togetherwitha“bracket”(calledtheLiebracket)operation
[−,−]: L×L→ L
whichisbilinear,anti-symmetricandsatisfiestheJacobiidentity.
We can play with Lie algebras that are over any field, finite or infinite. It turns out that the theory
of infinite dimensional Lie algebras is not very easy. Throughout this course, we think of L as a finite
dimensional(cid:67)-vectorspace.
WenowgivesomeexamplesofLiealgebras.
Example. Matn×n((cid:67)) with Lie bracket [M,N]= MN −NM is a Lie algebra, denoted by gln((cid:67)) or just
gl . ThisisrelatedtothegroupGL of n×ninvertiblematricesasweshallseeinthenextlecture.
n n
Example. Wenowgiveanabstractversionoftheaboveexample. LetV beavectorspace. Considerthe
set of all endomorphisms of V, Hom(V,V) with Lie bracket [f,g]= f ◦g−g◦ f. This is a Lie algebra
denotedbygl(V).
Example. Set L to be any vector space and declare [x,y]=0, for any x,y ∈ L. This is an abelian Lie
algebra. TheseLiealgebrasarerelatedtoabeliangroups.
Example. Considersln={M ∈gln:Tr(M)=0}. RecallthatthetraceisalinearmapTr:Matn×n((cid:67))→(cid:67)
such that Tr(M) = (cid:80)na . And so, we have that sl is a subspace of gl . Also, Tr(MN) = Tr(NM).
i ii n n
Therefore,Tr([M,N])=Tr(MN)−Tr(NM)=0. Sowecanviewtheusualbracketongl asafunction
n
[−,−]:sl ×sl →sl .
n n n
AlltheaxiomsofaLiealgebraareautomaticallysatisfied,sincetheyholdingl .
n
(cid:26)(cid:18) (cid:19) (cid:27)
a b
Example. sl = : a,b,c∈(cid:67) . ThisisathreedimensionalLiealgebrawithbasis:
2 c −a
(cid:26) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)(cid:27)
1 0 0 1 0 0
h= , e= , f = .
0 −1 0 0 1 0
1
LetuscomputetheLiebracketonthisbasis:
(cid:18)0 1(cid:19) (cid:18)0 −1(cid:19)
[h,e]=he−eh= − =2e
0 0 0 0
(cid:18) (cid:19) (cid:18) (cid:19)
0 0 0 0
[h,f]=hf − fh= − =−2f
−1 0 1 0
(cid:18) (cid:19) (cid:18) (cid:19)
1 0 0 0
[e,f]=ef − fe= − =h
0 0 0 1
Wealsohave[e,h]=−[h,e]and[h,h]=0.
Remark. sl isthemostimportantLiealgebra!
2
Remark. What we have given above is a complete description of sl , a three dimensional Lie algebra
2
withbasis{h,e,f}andtheabovebrackets. Notethattheabovedataentirelydefinessl .
2
GivenanyLiealgebra L,wecanpickabasis{v ,...,v }for L andcomputethebrackets
1 n
[v ,v ]=λ1v +···+λnv
i j ij 1 ij n
withλk ∈(cid:67). Theλk arecalledstructureconstants. Thesestructureconstantsdependonthebasis!
ij ij
Thestructureconstantsdeterminethebracket,bybilinearity. Theysatisfy
λk =−λk
ij ji
forall i,j,k andalsotheJacobiidentity.
Letusnowcomputethestructureconstantsofgl . gl hasabasis:
2 2
(cid:26) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)(cid:27)
1 0 0 0 0 1 0 0
a= ,b= ,e= ,f = .
0 0 0 1 0 0 1 0
Thestructureconstantsare:
[a,e]=e, [a,f]=−f, [e,f]=a−b
[b,e]=−e, [b,f]= f, [a,b]=0.
Thisisnotthebestbasisforgl . Abetterbasisisthefollowingone:
2
(cid:26) (cid:18) (cid:19) (cid:18) (cid:19)(cid:27)
1 0 1 0
h= ,e,f,z= .
0 −1 0 1
Noticethat[z,−]=0. So,theLiealgebrasgl andsl areveryclose.
2 2
A really good way to find Lie algebras is to look for subspaces L of gl satisfying the following
n
condition. If M,N ∈ L,then[M,N]∈ L.
Definition. Let L be a Lie algebra. A Lie subalgebra is a subspace K ⊂ L such that if x,y ∈ K, then
[x,y]∈K. Hence(K,[−,−])isaLiealgebrainitsownright.
Wealwayshavetwotrivialsubalgebras,namely K = L and K ={0}.
Example. sl ⊂gl isaLiesubalgebra.
n n
Example. b ={M ∈gl : M isuppertriangular}. Infact,b isclosedundermatrixmultiplication(un-
n n n
likesl ),andsob isclosedunder[−,−]. Inotherwords,b isaLiesubalgebraofgl .
n n n n
Example. n ={M ∈gl : M isstrictlyuppertriangular}. Thisisclosedundermultiplication,andthus
n n
closedunder[−,−]. Son isaLiesubalgebraofgl andalsoofb .
n n n
2
Lecture 2 (20.01.2015)
Example. so ={M ∈gl : M(cid:62) =−M}. This is the set of all antisymmetric matrices. Let us show that
n n
so isaLiesubalgebraofgl . Suppose M,N ∈so ,then
n n n
[M,N](cid:62)=(MN −NM)(cid:62)=N(cid:62)M(cid:62)−M(cid:62)N(cid:62)=NM −MN =−[M,N].
Thus,[M,N]∈so . Therefore,so isaLiesubalgebraofgl .
n n n
Example. d = {M ∈ gl : M isdiagonal}. Note that diagonal matrices commute with each other. So
n n
weseethatif M,N ∈d ,then[M,N]=0andthusd isanabelianLiesubalgebraofgl .
n n n
Digression on Lie Groups
We have said in the first lecture that Lie algebras are related to groups. In the following section we
will talk a bit about this relationship. We shall however not give a rigorous exposition; we will do this
mainly for motivation. Roughly speaking, we can think of the elements of the Lie algebra as elements
of a group that are “infinitesimally close” to the identity. Let us first start with GL , the group of all
n
invertible n×n matrices. What is the connection between GL and the Lie algebra gl ? Let us think
n n
aboutmatriceswhichare“near” I ∈GL ,thatis,weconsideraperturbationoftheidentitymatrixbya
n
smallamountε>0,
I+εM,
for M anymatrixinMatn×n((cid:67))=gln andεisasmallcomplexnumber. Notethatwedonotrequire M
tobeinvertible. Ifεissmallenough,then
det(I+εM)≈det(I)=1,
sodet(I+εM)(cid:54)=0. Hence,(I+εM)∈GL . Letusnowtake(I+εM),(I+εN)∈GL ,thentakingtheir
n n
productgives
(I+εM)(I+εN)=I+ε(M +N)+ε2MN.
Soweseethatthefirstordermultiplicationisactuallyturnedintoaddition. Wealsohave
(I+εM)−1=I−εM +ε2M2+ (higherorderterms).
In any group G, given g,h ∈ G, we can consider their commutator ghg−1h−1 ∈ G. Note that
ghg−1h−1=e ifandonlyif gh=hg,where e istheidentityelementof G.
Letuscomputethecommutatorof(I+εM)and(I+εN)inGL :
n
(I+εM)(I+εN)(I+εM)−1(I+εN)−1=(I+εM)(I+εN)(I−εM)(I−εN)+O(ε2)
=I+ε(M +N −M −N)+O(ε2)
=I+O(ε2).
Letusnowcomputetheε2 term:
(I+εM)(I+εN)(I−εM +ε2M2)(I−εN +ε2N2)+O(ε3)=I+ε2(MN −NM)+O(ε3)
=I+ε2[M,N]+O(ε3).
SotheLiebracketistheleadingorderterminthemultiplicativecommutatorinthegroupGL . We
n
nowseetherelationshipbetweenthegroupGL andtheLiealgebragl .
n n
3
NowhowaboutotherLiealgebras,suchassl ,so ,b ,...?
n n n
Roughly speaking, a Lie group is a (nice) subgroup of GL , for some n. More precisely, a Lie group
n
isagroupthatisalsoadifferentiablemanifoldwiththeadditionalpropertythatthegroupoperationis
compatiblewiththesmoothstructure. Letusseeanexampleofsuchagroup.
Example. SL ={A∈GL :det(A)=1}. SL actuallysitsinsidetheLiegroupGL asthedeterminantis
n n n n
multiplicative,thatis,det(AB)=det(A)det(B).
ThisisaLiegroup. OurgoalistogetanassociatedLiealgebra. Asbefore,consider I+εM. When
doesthislieinSLn? Letusdothe n=2casecalculation. Let M =(cid:0)ac db(cid:1)∈Matn×n((cid:67)),then
(cid:18)1+εa εb (cid:19)
det(I+εM)=det
εc 1+εd
=(1+εa)(1+εd)−bcε2
=1+ε(a+d)+O(ε2)
=1+εTr(M)+O(ε2).
So,det(I+εM)=1ifandonlyifTr(M)=0. Inotherwords,(I+εM)∈SL ifandonlyif M ∈sl .
2 2
Thisistrueuptoorderε2.
The case n>2 is still true. That is, (I +εM)∈SL if and only if M ∈sl . Now suppose M,N ∈sl
n n n
andthecorrespondingperturbations(I+εM),(I+εN)∈SL . Hence,
n
(I+εM)(I+εN)(I+εM)−1(I+εN)−1=I+ε2[M,N]+O(ε3).
Therefore, [M,N] ∈ sl . So the fact that SL ⊂ GL is a subgroup implies that sl ⊂ gl is a Lie
n n n n n
subalgebra.
Ingeneral,supposewehaveaLiegroup G⊂GL . ThenthereisanassociatedLiealgebra
n
g={M ∈gl :(I+εM)∈G, uptoorderO(ε2)}.
n
ThisisautomaticallyaLiealgebra,becauseif M,N ∈g,then
(I+εM)(I+εN)(I+εM)−1(I+εN)−1=I+ε2[M,N]+(higherorderterms)
isanelementof G. Thus,[M,N]∈g,sogisaLiesubalgebraofgl .
n
Thisishowthegeneraltheorygoes,nowletusseesomemoreexamples.
Example. B ⊂GL isthegroupofinvertibleuppertriangularmatrices. Then(I+εM)∈B ifandonly
n n n
if M isuppertriangular,thatis, M ∈b .
n
Example. N = {A ∈ GL : Ais“uni-upper-triangular”}. Then (I +εM) ∈ N if and only if M ∈ n
n n n n
(strictlyuppertriangularmatrices).
Example. SupposeG⊂GL isabelian,soforallA,B∈G,wehaveABA−1B−1=I. Letgbetheassociated
n
Liealgebraand M,N ∈g. Then
(I+εM)(I+εN)(I+εM)−1(I+εN)−1=I+ε2[M,N]+O(ε3)=I
Thus,[M,N]=0. SogisanabelianLiealgebra.
Forexample,D ={A∈GL :Aisdiagonal}hasLiealgebrad ={M ∈gl : M isdiagonal}.
n n n n
4
Example. O ={A∈GL :AA(cid:62) = I}. This is the orthogonal group. Do not confuse this group with the
n n
(cid:62)
unitary group, U ={A∈GL ((cid:67)): AA = I}. Note that O can be over any field, or ring. On the other
n n n
hand,U isoverthecomplexnumbers,otherwisetheconjugateofAwouldnotmakeanysense.
n
Wealsohavethespecialorthogonalgroup,SO ={A∈O :det(A)=1}=O ∩SL .
n n n n
Notice that if A ∈ O , then det(AA(cid:62)) = (det(A))2 = 1. Hence, det(A) = I. SO is the kernel of
n n
det:O →(cid:90)/2,soSO hasindex2.
n n
LetuslookattheLiealgebra. Whendoes(I+εM)lieinO ? Wehave
n
(I+εM)(I+εM)(cid:62)=I+ε(M +M(cid:62))+(higherorderterms)=I.
So,(I+εM)∈O ifandonlyif M +M(cid:62)=0,thatis, M =−M(cid:62),thatis, M ∈so .
n n
SotheLiealgebraassociatedtotheLiegroupO isso ⊂gl . Itisnotsuprisingthatif(I+εM)∈O ,
n n n n
theninfact(I+εM)∈SO .
n
So,O andSO havethesameLiealgebrawhichbringsustothefollowingfactthat
n n
{Liegroups}=⇒{Liealgebras}
isnotaninjection.
The issue is that finite/discrete groups are “sent” to the zero Lie algebra. Apart from the discrete
groupissue,itisafactthattheLiealgebragtellsuseverythingabouttheassociatedLiegroup G.
Remark. WestudyLiegroupsusingLiealgebrasbecauseitiseasierandverylittleinformationislost.
5
Lecture 3 (20.01.2015)
Homomorphisms and Ideals
Definition. A function f : L → K between two Lie algebras L and K is called a homomorphism if f is
linearandpreservestheLiebracket,thatis,
[f(x),f(y)]= f([x,y])
forall x,y ∈ L.
The homomorphism f is called an isomorphism if f is bijective, or equivalently (thinking in cate-
goricalterms)ifthereexistsahomomorphism f−1: K → L whichisa2-sidedinverseof f. Thesecond
definitionisthe“correctone”.
If there exists an isomorphism f : L → K, we say that L and K are isomorphic. The composition of
twoLiealgebrahomomorphismsisclearlyaLiealgebrahomomorphism.
Example. Set L=gl and K =(cid:67),the1-dimensionalLiealgebra. Set
n
f =Tr:gl →(cid:67).
n
Thisislinear. Wealsohave
[Tr(M),Tr(N)]=0
bydefinition((cid:67)is1-dimensional,henceabelian,hencethebracketvanishes)andTr([M,N])=Tr(MN)−
Tr(NM)=0.
Example. Suppose L and K arebothabelian. Thenanylinearmap
f : L→K
isahomomorphism.
Example. Suppose L is 1-dimensional. Pick a basis x ∈ L, then [x,x]=0, and so L is abelian. There-
fore, any two 1-dimensional Lie algebras L and K must be isomorphic (just pick any non-zero linear
map f : L→K). So,
(cid:26)(cid:18) (cid:19)(cid:27)
0 b
n =
2 0 0
is1-dimensionalandabelian. Also,
(cid:26)(cid:18) (cid:19)(cid:27)
0 b
so =
2 −b 0
∼
is1-dimensionalandabelian. Hence,n =so .
2 2
Suppose f : L→K isahomomorphism. Theimageof f isaLiesubalgebra,since
[f(x),f(y)]= f([x,y])∈im(f).
We can view any Lie subalgebra as the image of a homomorphism. Conversely, if L ⊆ K is a Lie subal-
gebra,thentheinclusion
ι: L(cid:44)→K
x (cid:55)→ x
6
isaLiealgebrahomomorphismandim(ι)= L.
The kernel of a homomorphism f : L → K is a Lie subalgebra of L. It also has a stronger property.
If x ∈ker(f)and y isanyelementof L,then
f([x,y])=[f(x),f(y)]=[0,f(y)]=0,
so[x,y]∈ker(f).
Definition. A subspace I ⊂ L is called an ideal if for all x ∈ I and for all y ∈ L, we have [x,y] ∈ I.
(Equivalently,[y,x]∈I,byskew-symmetrysowejustneedtokeeptrackofminussigns.)
This is very similar to normal subgroups in group theory, also very similar to ideals of rings in ring
theory.
So,ker(f)isanidealof L foranyhomomorphism f : L→K.
Example. Consider
Tr:gl →(cid:67).
n
ThekernelofTrissl ⊂gl . Thus,sl isanideal.
n n n
Example.
(cid:26)(cid:18) (cid:19) (cid:27)
a b
b = ∈gl
2 0 c 2
(cid:28) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)(cid:29)
1 0 1 0 0 1
= z= ,h= ,e=
0 1 0 −1 0 0
where[h,e]=2e,andallotherbracketsvanish.
Thesubspace
(cid:28)(cid:18) (cid:19)(cid:29)
0 b
〈e〉= =n ⊂b
0 0 2 2
isanideal.
Thesubspace
(cid:28)(cid:18) (cid:19)(cid:29)
a 0
〈z,h〉= =d ⊂b
0 d 2 2
isaLiesubalgebra(andabelian),butitisnotanidealbecause[h,e]=2e∈/ d .
2
Quotient Lie algebras
Suppose I ⊂ L isanysubspaceofaLiealgebra L. Wecanformthequotientvectorspace
L/I ={x+I: x ∈ L},
where x+I ={x+ y: y ∈I}.
Letusdefine
[x+I,y+I]:=[x,y]+I.
Thisiswell-defined. Replace x by x+z forsomez∈I. Then[x+z,y]=[x,y]+[z,y],so[x,y]+I =
[x+z,y]+I because[z,y]getsabsorbedinto I. Similarlyfor y.
In fact, the bracket is well-defined on L/I if and only if I is an ideal. The Lie algebra axioms hold
automaticallyfor L/I becausetheyholdin L. Wecall L/I thequotientLiealgebraof L by I.
7
Thereisafunction
q: L→ L/I
x (cid:55)→ x+I
whichisautomaticallyaLiealgebrahomomorphismanditissurjectiveandker(q)= I. Soanyidealis
thekernelofahomomorphism.
Proposition (First isomorphism theorem). Suppose f : L → K is a surjective Lie algebra homomor-
∼
phism,andlet I =ker(f). Then K = L/I.
Proof. Wegetaninducedfunction
f(cid:101): L/I →K
x+I (cid:55)→ f(x)
which is well-defined and a Lie algebra homomorphism and an isomorphism by the first isomorphism
theoremforvectorspaces.
∼
If f : L→K isnotsurjective,thenwegetasurjection f : L→im(f)⊂K andim(f)= L/ker(f).
8