Solutions to some exercises in the book "J. E. Humphreys, An Introduction to Lie Algebras and Representation Theory"

50 Pages·2013·0.381 MB·English

by  Zhihua Chang

Solutions to some exercises in the book “J. E. Humphreys, An
Introduction to Lie Algebras and Representation Theory”
July20,2013
Contents
1 DefinitionsandFirstExamples 3
2 IdealsandHomomorphisms 9
3 SolvableandNilpotentLieAlgebras 13
4 TheoremsofLieandCartan 16
5 KillingForm 17
6 CompleteReducibilityofRepresentations 20
7 Representationsofsl(2,F) 24
8 RootSpaceDecomposition 30
9 Axiomatics 33
10 SimpleRootsandWeylGroup 36
11 Classification 39
12 ConstructionofRootSystemsandAutomorphisms 39
13 AbstractTheoryofWeights 40
14 IsomorphismTheorem 41
15 CartanSubalgebras 41
16 ConjugacyTheorems 41
17 UniversalEnvelopingAlgebras 43
18 GeneratorsandRelations 43
19 TheSimpleAlgebras 43
20 WeightsandMaximalVectors 45
21 FiniteDimensionalModules 46
1
22 MultiplicityFormula 47
23 Characters 47
24 FormulasofWeyl,Kostant,andSteinberg 48
25 ChevalleybasisofL 48
26 Kostant’sTheorem 49
27 AdmissibleLattices 50
Page2
1 Definitions and First Examples
1. LetLbetherealvectorspaceR3. Define[xy]=x×y(crossproductofvectors)forx,y ∈L,andverifythatL
isaLiealgebra. WritedownthestructureconstantsrelativetotheusualbasisofR3.
Solution: Clearly,[,]isbilinearandanti-commutative,itneedonlytochecktheJacobiIdentity:
[[x,y],z] = (x×y)×z
= (x.z)y−(y.z)x
= (z.x)y−(y.x)z+(x.y)z−(z.y)x
= [[z,y],x]+[[x,z],y]
where(.)istheinnerproductofR3.
TakethestandardbasisofR3: e = (1,0,0),e = (0,1,0),e = (0,0,1). Wecanwritedownthestructure
1 2 3
equationsofL:
[e ,e ] = e
1 2 3
[e ,e ] = e
2 3 1
[e ,e ] = e
3 1 2
2. Verify that the following equations and those implied by (L1)(L2) define a Lie algebra structure on a three
dimensionalspacewithbasis(x,y,z):[xy]=z,[xz]=y,[yz]=0.
Solution:
(L1)(L2)aresatisfied,itissufficienttoshowtheJacobiIdentityholdforthebasis:
[[x,y],z] = [z,z]=0
[[y,z],x] = [0,x]=0
[[x,z],y] = [y,y]=0
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
0 1 1 0 0 0
3. Let x = ,h = ,y = be an ordered basis for sl(2,F). Compute the matrices of
0 0 0 −1 1 0
adx,adh,adyrelativetothisbasis.
Solution:
Bythestructureequationsofsl(2,F):
adx(y) = −ady(x)=h
adx(h) = −adh(x)=−2x
ady(h) = −adh(y)=2y
Wecanwritedownthematricesofadx,adh,adyrelativetothisbasiseasily:
     
0 −2 0 2 0 0 0 0 0
adx∼0 0 1 adh∼0 0 0  ady ∼−1 0 0
0 0 0 0 0 −2 0 2 0
Page3
4. FindalinearLiealgebraisomorphictothenonabeliantwodimensionalalgebraconstructedin(1.4).
Solution:
TwodimensionalLiealgebraconstructedin(1.4)isgivenbybasis(x,y)withcommutation[x,y]=x.
Ingl(F2),let
(cid:18) (cid:19) (cid:18) (cid:19)
0 1 −1 0
x(cid:55)→ , y (cid:55)→
0 0 0 0
Thisisaisomorphism.
5. Verifytheassertionsmadein(1.2)aboutt(n,F),d(n,F),n(n,F),andcomputethedimensionofeachalgebra,
byexhibitingbases.
Solution:
Assertionsmadein(1.2):
t(n,F)=d(n,F)+n(n,F) vectorspacedirectsum (1)
[d(n,F),n(n,F)]=n(n,F) (2)
[t(n,F),t(n,F)]=n(n,F) (3)
Evidently, (1)holdsand[d(n,F),n(n,F)] ⊆ n(n,F). Sowejustneedtoshowtheconverseconclusionis
alsotrue.
Lete denotesthematrixwith(i,j)-elementis1,and0otherwise.
ij
n(n,F)=span {e |i<j}
F ij
Butweknow
e =e e −e e =[e ,e ]⊆[d(n,F),n(n,F)], i<j
ij ii ij ij ii ii ij
So(2)iscorrect.
(3)followsfrom(1)and(2):
[t(n,F),t(n,F)] = [d(n,F)+n(n,F),d(n,F)+n(n,F)]
⊆ [d(n,F),d(n,F)]+[n(n,F),n(n,F)]+[d(n,F),n(n,F)]
⊆ n(n,F)
Conversely,n(n,F)=[d(n,F),n(n,F)]⊆[t(n,F),t(n,F)].
6. Letx∈gl(n,F)havendistincteigenvaluesa ,··· ,a inF. Provethattheeigenvaluesofadxarepreciselythe
1 n
n2scalarsa −a (1(cid:54)i,j (cid:54)n),whichofcourseneednotbedistinct.
i j
Solution:
 
v
1i
Let v =  ..  ∈ Fn,1 (cid:54) i (cid:54) n are eigenvectors of x respect to eigenvalue a respectively. Then
i  .  i
v
ni
v ,1 (cid:54) i (cid:54) narelinearindependent. LetA = (v ) ,E denotesthen×nmatrixwith(i,j)-elementis
i ij n×n ij
Page4
1,and0otherwise,e =AE A−1,ThenwehaveAisanonsingularmatrixand
ij ij
e v =δ v
ij k jk i
adx(e )v =x.e .v −e .x.v =(a −a )δ v =(a −a )e .v
ij k ij k ij k i k jk i i j ij k
So
adx(e )=(a −a )e
ij i j ij
i.e.,a −a areeigenvaluesofadx,theeigenvectorsaree respectively. Henceadxisdiagonalizable. So
i j ij
wecanconcludethattheeigenvaluesofadxarepreciselythen2scalarsa −a (1(cid:54)i,j (cid:54)n).
i j
7. Lets(n,F)denotethescalarmatrices(=scalarmultiplesoftheidentity)ingl(n,F).IfcharF is0orelseaprime
notdividingn,provethatgl(n,F)=sl(n,F)+s(n,F)(directsumofvectorspaces),with[s(n,F),gl(n,F)]=
0.
Solution:
∀A=(a )∈gl(n,F),Iftr(A)=0,A∈sl(n,F);elseA= tr(A)I+(A−tr(A)I)withtr(A−tr(A)I)=0,
ij n n n
i.e.,A− tr(A)I ∈sl(n,F).(charF is0orelseaprimenotdividingn.) sl(n,F)∩s(n,F)={0}isclearly.
n
Hencegl(n,F)=sl(n,F)+s(n,F)(directsumofvectorspaces).
ForaI ∈s(n,F),∀A∈gl(n,F),[aI,A]=aA−aA=0,So
[s(n,F),gl(n,F)]=0.
8. VerifythestateddimensionofD .
l
Solution:
(cid:18) (cid:19) (cid:18) (cid:19)
m n 0 I
Supposex= (m,n,p,q ∈gl(n,F),s= l . Bysx=−xts,wehave
p q I 0
l
m=−qt,n=−nt,p=−pt,q =−mt
WecanenumerateabasisofD :
l
e −e ,1(cid:54)i,j (cid:54)l; e −e ,e −e ,1(cid:54)i<j (cid:54)l
i,j l+j,l+i i,l+j j,l+i l+i,j l+j,i
wheree isthematrixhaving1inthe(i,j)positionand0elsewhere. Hence
ij
1 1
dimD =l2+ l(l−1)+ l(l−1)=2l2−l
l 2 2
9. When charF = 0, show thateach classical algebra L = A ,B ,C , or D is equal to [LL]. (Thisshows again
l l l l
thateachalgebraconsistsoftrace0matrices.)
Solution:
[L,L]⊆Lisevident. ItissufficienttoshowL⊆[L,L].
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• A :
1
e = 1[h,e ]
12 2 12
e = 1[e ,h]
21 2 21
h = [e ,e ]
12 21
• A (l(cid:62)2):
l
e = [e ,e ], k (cid:54)=i,j,i(cid:54)=j
ij ik kj
h = [e ,e ]
i i,i+1 i+1,i
• B (l(cid:62)2):
l
e −e = [e −e ,e −e ]
1,l+i+1 i+1,1 1,j+1 l+j+1,1 j+1,l+i+1 i+1,l+j+1
e −e = [e −e ,e −e ]
1,i+1 l+i+1,1 1,l+j+1 j+1,1 l+j+1,i+1 l+i+1,j+1
e −e = [e −e ,e −e ]
i+1,i+1 l+i+1,l+i+1 i+1,1 1,l+i+1 1,i+1 l+i+1,1
e −e = [e −e ,e −e ]
i+1,j+1 l+i+1,l+j+1 i+1,1 1,l+i+1 1,j+1 l+j+1,1
e −e = [e −e ,e −e ]
i+1,l+j+1 j+1,l+i+1 i+1,i+1 l+i+1,l+i+1 i+1,l+j+1 j+1,l+i+1
e −e = [e −e ,e −e ]
l+i+1,j+1 j+l+1,i+1 l+i+1,l+i+1 i+1,i+1 l+i+1,j+1 j+l+1,i+1
where1(cid:54)i(cid:54)=j (cid:54)l.
• C (l(cid:62)3):
l
e −e = [e ,e ]
ii l+i,l+i i,l+i l+i,i
e −e = [e −e ,e −e ] i(cid:54)=j
ij l+j,l+i ii l+i,l+i ij l+j,l+i
e +e = [e −e ,e +e ]
i,l+j j,l+i ii l+i,l+i i,l+j j,l+i
e +e = [e −e ,e +e ]
l+i,j l+j,i l+i,l+i ii l+i,j l+j,i
• D (l(cid:62)2):
l
e −e = 1[e −e ,e −e ]
ii l+i,l+i 2 ij l+j,l+i ji l+i,l+j
+1[e −e ,e −e ]
2 i,l+j j,l+i l+j,i l+i,j
e −e = [e −e ,e −e ]
ij l+j,l+i ii l+i,l+i ij l+j,l+i
e −e = [e −e ,e −e ]
i,l+j j,l+i ii l+i,l+i i,l+j j,l+i
e −e = [e −e ,e −e ]
l+i,j l+j,i l+i,l+i ii l+i,j l+j,i
wherei(cid:54)=j.
10. Forsmallvaluesofl,isomorphismsoccuramongcertainoftheclassicalalgebras. ShowthatA ,B ,C areall
1 1 1
isomorphic,whileD istheonedimensionalLiealgebra. ShowthatB isisomorphictoC ,D toA . Whatcan
1 2 2 3 3
yousayaboutD ?
2
Solution:
TheisomorphismofA ,B ,C isgivenasfollows:
1 1 1
A → B → C
1 1 1
e −e (cid:55)→ 2(e −e ) (cid:55)→ e −e
11 22 22 33 11 22
e (cid:55)→ 2(e −e ) (cid:55)→ e
12 13 21 12
e (cid:55)→ 2(e −e ) (cid:55)→ e
21 12 31 21
ForB ,C wefirstcalculatetheeigenvectorsforh =e −e ,h =e −e andh(cid:48) =e −e ,h(cid:48) =
2 2 1 22 44 2 33 55 1 11 33 2
e −e respectively. We denote λ = (λ(h ),λ(h )) for the eigenvalue of h ,h , λ(cid:48) is similar. See the
22 44 1 2 1 2
followingtable:
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B C
2 2
α=(1,0) e −e α(cid:48) =(−1,1) e −e
21 14 21 34
−α=(−1,0) e −e −α(cid:48) =(1,−1) e −e
12 41 12 43
β =(−1,1) e −e β(cid:48) =(2,0) e
32 45 13
−β =(1,−1) e −e −β(cid:48) =(−2,0) e
23 54 31
α+β =(0,1) e −e α(cid:48)+β(cid:48) =(1,1) e +e
15 31 14 23
−(α+β)=(0,−1) e −e −(α(cid:48)+β(cid:48))=(−1,−1) e +e
13 51 41 32
2α+β =(1,1) e −e 2α(cid:48)+β(cid:48) =(0,2) e
25 34 24
−(2α+β)=(−1,−1) e −e −(2α(cid:48)+β(cid:48))=(0,−2) e
43 52 42
Wemakealineartransformation
h˜(cid:48) =−1h(cid:48) + 1h(cid:48),h˜(cid:48) = 1h + 1h
1 2 1 2 2 2 2 1 2 2
Thenα(h )=α(cid:48)(h˜(cid:48)),α(h )=α(cid:48)(h˜(cid:48)),β(h )=β(cid:48)(h˜(cid:48)),β(h )=β(cid:48)(h˜(cid:48)). SotheisomorphismofB ,C
1 1 2 2 1 1 2 2 2 2
isgivenasfollows:
B → C
2 2
e −e (cid:55)→ −1(e −e )+ 1(e −e )
22 44 2 11 33 2 22 44
e −e (cid:55)→ 1(e −e )+ 1(e −e )
33 55 2 11 √ 33 2 22 44
e −e (cid:55)→ 2(e −e )
12 41 √2 12 43
e −e (cid:55)→ 2(e −e )
21 14 2 21 34
e −e (cid:55)→ e
32 45 13
e −e (cid:55)→ e
23 54 √ 31
e −e (cid:55)→ 2(e +e )
15 31 √2 14 23
e −e (cid:55)→ 2(e +e )
13 51 2 32 41
e −e (cid:55)→ e
25 34 24
e −e (cid:55)→ e
43 52 42
For A and D , we calculate the eigenvalues and eigenvectors for h = e −e ,h = e −e ,h =
3 3 1 11 22 2 22 33 3
e −e andh(cid:48) =e −e ,h(cid:48) =e −e ,h(cid:48) =e −e respectively.
33 44 1 11 44 2 22 55 3 33 66
A D
3 3
α=(1,1,−1) e α(cid:48) =(0,1,1) e −e
13 26 35
−α=(−1,−1,1) e −α(cid:48) =(0,−1,−1) e −e
31 62 53
β =(−1,1,1) e β(cid:48) =(0,1,−1) e −e
24 23 65
−β =(1,−1,−1) e −β(cid:48) =(0,−1,1) e −e
42 32 56
γ =(−1,0,−1) e γ(cid:48) =(1,−1,0) e −e
41 12 54
−γ =(1,0,1) e −γ(cid:48) =(−1,1,0) e −e
14 21 45
α+γ =(0,1,−2) e α(cid:48)+γ(cid:48) =(1,0,1) e −e
43 16 34
−(α+γ)=(0,−1,2) e −(α(cid:48)+γ(cid:48))=(−1,0,−1) e −e
34 61 43
β+γ =(−2,1,0) e β(cid:48)+γ(cid:48) =(1,0,−1) e −e
21 13 64
−(β+γ)=(2,−1,0) e −(β(cid:48)+γ(cid:48))=(−1,0,1) e −e
12 31 46
α+β+γ =(−1,2,−1) e α(cid:48)+β(cid:48)+γ(cid:48) =(1,1,0) e −e
23 15 24
−(α+β+γ)=(1,−2,1) e −(α(cid:48)+β(cid:48)+γ(cid:48))=(−1,−1,0) e −e
32 51 42
Wetakealineartransformation
h˜(cid:48) =−h(cid:48) +h(cid:48),h˜(cid:48) =h(cid:48) +h(cid:48),h˜(cid:48) =−h(cid:48) −h(cid:48)
1 1 3 2 1 2 3 1 3
Page7
thenα(h ) = α(cid:48)(h˜(cid:48)),β(h ) = β(cid:48)(h˜(cid:48)),γ(h ) = γ(cid:48)(h˜(cid:48)),i = 1,2,3;TheisomorphismofA andD canbe
i i i i i i 3 3
givenasfollows:
A (cid:55)→ D
3 3
e −e (cid:55)→ −(e −e )+e −e
11 22 11 44 33 66
e −e (cid:55)→ (e −e )+(e −e )
22 33 11 44 22 55
e −e (cid:55)→ −(e −e )−(e −e )
33 44 11 44 33 66
e (cid:55)→ e −e
13 26 35
e (cid:55)→ e −e
31 62 53
e (cid:55)→ e −e
24 23 65
e (cid:55)→ e −e
42 32 56
e (cid:55)→ e −e
41 12 54
e (cid:55)→ e −e
14 21 45
e (cid:55)→ e −e
43 16 34
e (cid:55)→ e −e
34 61 43
e (cid:55)→ e −e
21 13 64
e (cid:55)→ e −e
12 31 46
e (cid:55)→ e −e
23 15 24
e (cid:55)→ e −e
32 51 42
11. VerifythatthecommutatoroftwoderivationsofanF-algebraisagainaderivation,whereastheordinaryproduct
neednotbe.
Solution:
AisaF-algebra,δ,δ(cid:48) ∈Der(A),a,b∈A
[δ,δ(cid:48)](ab) = δδ(cid:48)(ab)−δ(cid:48)δ(ab)
= δ(δ(cid:48)(a)b+aδ(cid:48)(b))−δ(cid:48)(δ(a)b+aδ(b))
= δ(δ(cid:48)(a))b+δ(cid:48)(a)δ(b)+δ(a)δ(cid:48)(b)+aδ(δ(cid:48)(b))
−δ(cid:48)(δ(a))b−δ(a)δ(cid:48)(b)−δ(cid:48)(a)δ(b)−aδ(cid:48)(δ(b))
= ([δ,δ(cid:48)](a))b−a[δ,δ(cid:48)](b)
∴[δ,δ(cid:48)] ∈ Der(A)
12. LetLbeaLiealgebraoveranalgebraicallyclosedfieldandletx ∈ L. ProvethatthesubspaceofLspannedby
theeigenvectorsofadxisasubalgebra.
Solution:
Supposey,zareeigenvectorsofadxrespecttoeigenvaluesλ,µ,i.e.,[x,y]=λy,[x,z]=µzthen
adx[y,z] = [x,[y,z]]
= [y,[x,z]]−[z,[x,y]]
= (λ+µ)[y,z]
So [y,z] is also a eigenvector of adx. i.e., the subspace of L spanned by the eigenvectors of adx is a
subalgebra.
Page8
2 Ideals and Homomorphisms
1. Provethatthesetofallinnerderivationsadx,x∈L,isanidealofDerL.
Solution:
∀δ ∈DerL,x,y ∈L
[δ,adx](y) = δ([x,y])−[x,δ(y)]
= [δ(x),y]+[x,δ(y)]−[x,δ(y)]
= ad(δ(x))(y)
∴[δ,adx] = adδ(x)isainnerderivations.
2. Showthatsl(n,F)ispreciselythederivedalgebraofgl(n,F)(cf. Exercise1.9).
Solution:
∀x,y ∈gl(n,F),tr[x,y]=tr(xy)−tr(yx)=0. Wehave
[gl(n,F),gl(n,F)]⊆sl(n,F)
Conversely,byexercise1.9,
sl(n,F)=[sl(n,F),sl(n,F)]⊆[gl(n,F),gl(n,F)]
3. Prove that the center of gl(n,F) equals s(n,F) (the scalar matrices). Prove that sl(n,F) has center 0, unless
charF dividesn,inwhichcasethecenteriss(n,F).
Solution:
(cid:80)
Clearly, we have s(n,F) ⊆ Z(gl(n,F)). Conversely, Let a = a e ∈ Z(gl(n,F)), then for each
ij ij
i,j
e ∈gl(n,F),
kl
(cid:88)
[a,e ] = a [e ,e ]
kl ij ij kl
i,j
(cid:88)
= a (δ e −δ e )
ij jk il li kj
i,j
n n
(cid:88) (cid:88)
= a e − a e
ik il lj kj
i=1 j=1
n n
(cid:88) (cid:88)
= (a −a )e + a e − a e
kk ll kl ik il lj kj
i=1 j=1
i(cid:54)=k j(cid:54)=l
So
a =a ,a =0,i(cid:54)=j
kk ll ij
i.e.
a∈s(n,F)
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Forsl(n,F),ifc∈Z(sl(n,F)),∀x∈sl(n,F),[x,c]=0. Butweknowgl(n,F)=sl(n,F)+s(n,F)and
s(n,F)isthecenterofgl(n,F). Hencec∈Z(gl(n,F))=s(n,F). Wehave
Z(sl(n,F))=sl(n,F)∩s(n,F)
If charF does not divide n, each aI ∈ s(n,F) has trace na (cid:54)= 0, so aI (cid:54)∈ sl(n,F). i.e., Z(sl(n,F)) =
sl(n,F) ∩ s(n,F) = 0. Else if charF divides n, each aI ∈ s(n,F) has trace na = 0, in this case
Z(sl(n,F))=sl(n,F)∩s(n,F)=s(n,F)
4. Show that (up to isomorphism) there is a unique Lie algebra over F of dimension 3 whose derived algebra has
dimension1andliesinZ(L).
Solution:
LetL bethe3-dimensionalliealgebraoverF withbasis(x ,y ,z )andcommutation:
0 0 0 0
[x ,y ]=z ,[x ,z ]=[y ,z ]=0
0 0 0 0 0 0 0
.
SupposeLbeany3-dimensionalliealgebraoverF whosederivedalgebrahasdimension1andliesinZ(L).
Wecantakeabasis(x,y,z)ofLsuchthatz ∈ [LL] ⊆ Z(L). Byhypothesis,[x,y] = λz,[x,z] = [y,z] =
0,λ∈F. ThenL→L ,x(cid:55)→x ,y (cid:55)→y ,z (cid:55)→λz isaisomorphism.
0 0 0 0
5. SupposedimL = 3,L = [LL]. ProvethatLmustbesimple. [ObservefirstthatanyhomomorphicimageofL
alsoequalsitsderivedalgebra.] Recoverthesimplicityofsl(2,F),charF (cid:54)=2.
Solution:
LetI isanidealofL,then[L/I,L/I]=[L,L]/I =L/I.
Suppose L has a proper ideal I (cid:54)= 0, then I has dimension 1 or 2. If I has dimension 2, then L/I is a
1-dimensionalalgebra, [L/I,L/I] = 0 (cid:54)= L/I. ElseI hasdimension1, wecantakeabasis(x,y,z)ofL
suchthatzisabasisofI,so
[x,z]∈I,[y,z]∈I
Hence[LL]iscontainedinthesubspaceofLspannedby[x,y],z. Itsdimensionisatmost2,thiscontradict
with[LL]=L.
Now,weconcludethatLhasnopropernonzeroideal,i.e.,LisasimpleLiealgebra.
6. Provethatsl(3,F)issimple, unlesscharF = 3(cf. Exercise3). [Usethestandardbasish ,h ,e (i (cid:54)= j). If
1 2 ij
I (cid:54)=0isanideal,thenIisthedirectsumofeigenspacesforadh oradh ;comparetheeigenvaluesofadh ,adh
1 2 1 2
actingonthee .]
ij
Solution:
7. Prove that t(n,F) and d(n,F) are self-normalizing subalgebras of gl(n,F), whereas n(n,F) has normalizer
t(n,F).
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