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汉弗莱斯李代数和表示理论导学小黄书的课后习题1-7章,包括原来的题目,参考解答,解答不一定对的,仅仅是我个人理解做出的解答,如果有问题请及时反馈,谢谢
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Introduction to Lie Algebras and Respresentation Theory
ZENG,Bowen
2016.09.15
Contents
1 Preface 1
2 Ch1 Basic Concept 1
3 Ch2 Ideals and homomorphisms 4
4 Ch3 Solvable and nilpotent Lie algebra 6
5 Ch4 Theorem of Lie and Cartan 8
6 Ch5 Killing form 10
7 Ch6 Complete reducibility of representations 19
8 Ch7 Representations of sl(2, F ) 20
1 Preface
2 Ch1 Basic Concept
1.Let L be the real vector space R
3
. Define [xy] = x × y (cross product of vectors) for x, y ∈ L, and
verify that L is a Lie algebra. Write down the structure constants relative to the usual basis of R
3
.
Verify L is a Lie algebra
(L1) Cross product is bilinear
(L2) [xx] = x × x = 0
(L3) [x[yz]] = [x, y × z] = x × (y × z) = (x, z)y − (x, y)z
[y[zx]] = y × (z × x) = (y, x)z − (y, z)x
[z[xy]] = z × (x × y) = (z, y)x − (z, x)y
[x[yz]]+[y[zx]]+[z[xy]] = (x, z)y −(x, y)z +(y, x)z−(y, z)x+(z, y)x−(z, x)y = [(x, z)−(z, x)]y+
[(y, x) − (x, y)]z + [(z, y) − (y, z)]x = 0
So L is a Lie algebra
and the multiplication table as follows
1
[ ] x y z
x 0 z -y
y -z 0 x
z y -x 0
2.Verify that the following equations and those implied by (L1)(L2) define a Lie algebra structure on a
three dimensional vector space with basis (x, y, z) : [xy] = z, [xz] = y, [yz] = 0. We only to verify
Jacobi identity [x[yz]] + [y[zx]] + [z[xy]] = [x0] + [y, −y] + [zz] = 0
3.Let x =
0 1
0 0
, h =
1 0
0 −1
, y =
0 0
1 0
be an ordered basis for sl(2, F ). Computer the matri-
ces of adx, adh, ady relative to this basis.
∵ [x, h] = xh − hx = −2x
[x, y] = xy − yx = h
[y, h] = yh − hy = 2y
∴ adx(x, h, y) = (adx(x), adx(h), adx(y)) = ([xx], [xh], [xy]) = (0, −2x, h) = (x, h, y)
0 −2 0
0 0 1
0 0 0
ady(x, h, y) = (ady(x), ady(h), ady(y)) = ([yx], [yh], [yy]) = (−h, 2y, 0) = (x, h, y)
0 0 0
−1 0 0
0 2 0
adh(x, h, y) = (adh(x), adh(h), adh(y)) = ([hx], [hh], [hy]) = (2x, 0, −2y) = (x, h, y)
2 0 0
0 0 0
0 0 −2
4.Find a linear Lie algebra isomorphic to the nonabelian two dimensional algebra constructed in (1.4).
Define a map ϕ : L → adL by x 7→ adx, y 7→ ady, [xy] 7→ ad[xy] give an operation ad[xy] =
[adxady]in adL, then adL
∼
=
L because of [adxady] = ad[xy] = adx
5.Verify the assertions made in (1.2) about t(n, F ), δ(n, F ), n(n, F ), and compute the dimension of each
algebra, by exhibiting bases.
(i) t(n, F )n(n, F )and δ(n, F ) are closed under the bracket.
If A ∈ t(n, F ), B ∈ t(n, F ), then [A, B] = AB − BA also lies in t(n, F )
If A ∈ n(n, F ), B ∈ n(n, F ), then [A, B] = AB − BA also lies in n(n, F )
If A ∈ δ(n, F ),B ∈ δ(n, F ), then [A, B] = AB − BA also lies in δ(n, F )
(ii) A ∈ δ(n, F ), B ∈ n(n, F ), [AB] = AB − BA is also a strictly upper matrix
(iii) If A
i
∈ t(n, F ), i=1,2 A
i
= B
i
+C
i
, where B
i
∈ δ(n, F ), C
i
∈ n(n, F ) [A
1
, A
2
] = [B
1
+C
1
, B
2
+
C
2
] = [B
1
B
2
] + [B
1
C
2
] + [C
1
B
2
] + [C
1
C
2
] = B
1
B
2
− B
2
B
1
+ B
1
C
2
− C
2
B
1
+ C
1
B
2
− B
2
C
1
+
C
1
C
2
− C
2
C
1
∈ n(n, F )
for δ(n, F ), we take all the diagonal matrices e
ii
, 1 6 i 6 n n in number
for n(n, F ), we take all the matrices e
ij
, 1 6 i ≤ j 6 n,
n(n−1)
2
in number
and t(n, F ) is the direct sum of δ(n, F ) and n(n, F ) as space, so dim(t(n, F )) = dim(δ(n, F )) +
dim(n(n, F )) = n +
n(n−1)
2
6.Let x ∈ gl(n, F ) have n distinct eigenvalues a
1
, . . . a
n
in F . Prove that the eigenvalues of adx are
precisely the n
2
scalars a
i
− a
j
(1 6 i, j 6 n), which of course need not be distinct.
Proof. We can find a suitable basis such that the matrix of x on this basis is diag(a
1
, . . . a
n
), and
let e
ij
, 1 6 i, j 6 n be the standard basis of gl(n, F ), and adx(e
ij
) = [x, e
ij
] = xe
ij
− e
ij
x =
a
1
.
.
.
a
n
1
−
1
a
1
.
.
.
a
n
=
.
.
.
a
i
.
.
.
−
.
.
.
a
j
.
.
.
=
2
.
.
.
a
i
− a
j
.
.
.
= (a
i
− a
j
)e
ij
7. Let s(n, F ) denote the scalar matrices(=scalar multiples of the identity) in gl(n, F ). If char F is 0 or
else a prime not diving n, prove that gl(n, F ) = sl(n, F ) + s(n, F ) (direct sum of vector spaces), with
[s(n, F ), gl(n, F )] = 0.
Proof. A ∈ sl(n, F ), B = aE ∈ s(n, F ), [A, B] = AB − BA = aA − aA = 0, a ∈ F , what‘s
more dim(sl(n, F )) + dim(s(n, F )) = n
2
− 1 + 1 = n
2
= dim(gl(n, F )), so gl(n, F ) = sl(n, F ) +
s(n, F )
8.Verify the stated deimension of D
l
Since sx = −x
t
s, where x =
m n
p q
, s =
0 I
l
I
l
0
, we have
0 I
l
I
l
0
m n
p q
=−
m
t
p
t
n
t
q
t
0 I
l
I
l
0
,
which induces p = −p
t
, q = −m
t
, n = −n
t
, since m = −q
t
, we take e
ii
− e
l+i l+i
, l in number;
e
i j
− e
t
i+l j+l
, l
2
− l in number; sincen = −n
t
, we take e
i j+l
− e
j i+l
, 1 6 i 6= j 6 l, 1/2l(l − 1) in
number, similar p = −p
t
, 1/2l(l − 1) in number, so the total number of basis elements is l + l
2
− l +
1/2l(l − 1) + 1/2l(l − 1) = 2l
2
− l.
9.When char F = 0, show that each classical algebra L = A
l
, B
l
, C
l
, orD
l
is equal to [LL].
Proof. Let x
1
, . . . x
l
be the basis of L(A
l
, B
l
, C
l
, orD
l
), [LL] = {[x
i
x
j
] =
l
P
k=1
a
ij
x
k
, x
i
, x
j
∈ L},
dim[LL] = l
10.For small values of `, isomorphisms occur among certain of the classical algebra. Show thatA
l
, B
l
, C
l
are isomorphic, while D
1
is the one dimensional Lie algebra. Show that B
2
is isomorphic to C
2
, D
3
to
A
3
, What can you say about D
2
?
Since dim A
l
= (l+1)
2
−1, dim B
l
= 2l
2
+l, dim C
l
= 2l
2
−l, dim D
l
= 2l
2
−l, dim A
1
=dimB
1
=dim
C
1
= 3, dim D
1
= 1, dim B
2
=dimC
2
= 10, dim D
3
=dimA
3
= 15, while DimD
2
= 6
11.Verify that the commutator of two derivations of an F -algebra is again a derivation, whereas the ordi-
nary product need not be.
Proof. If δ, δ
0
∈ DerU , a, b ∈ U , [δ, δ
0
](ab) = (δ δ
0
− δ
0
δ)(ab) = δ δ
0
(ab) − δ
0
δ(ab) = aδ δ
0
(b) +
δ δ
0
(a)b − a δ
0
δ(b) − δ
0
δ(a)b = a(δδ
0
(b) − δ
0
δ(b)) + (δδ
0
(a) − δ
0
δ(a))b = a[δδ
0
](b) + [δδ
0
](a)b, while
δ δ
0
(ab) = δ(a δ
0
(b) + δ
0
(a)b) = δ(a δ
0
(b)) + δ(δ
0
(a)b) = aδ δ
0
(b) + δ(a) δ
0
(b) + δ
0
(a)δ(b) + δ δ
0
(a)b 6=
aδ δ
0
(b) + δ δ
0
(a)b
12.LetL be a Lie algebra and let x ∈ L. Prove that the subspace of L spanned by the eigenvectors of adx
is a subalgebra.
Proof. Let v
1
, . . . v
l
be the eigenvectors of adx belong to eigenvalues a
i
, . . . a
l
, i.e. adx(v
i
) = a
i
v
i
,
1 6 i 6 l, adx([v
i
v
j
]) = [x, [v
i
v
j
]] = [v
i
[xv
j
]] − [v
j
[xv
i
]] = [v
i
, a
j
v
j
] − [v
j
, a
i
v
i
] = a
j
[v
i
v
j
] +
a
i
[v
i
v
j
] = (a
i
+ a
j
)[v
i
v
j
]
3
3 Ch2 Ideals and homomorphisms
1.Prove that the set of all inner derivations adx, x ∈ L, is an ideal of DerL
Proof. DerL = {δ ∈ EndL|δ(xy) = xδ(y) + δ(x)y, x, y ∈ L}, and IntL = {δ
0
∈ DerL| δ
0
[yz] =
[δ
0
(y)z]+[y δ
0
(z)]}, [δ δ
0
]([yz]) = (δ δ
0
− δ
0
δ)([yz]) = δ δ
0
([yz])−δ
0
δ([yz]) = δ([δ
0
(y)z]+[y δ
0
(z)])−
δ
0
δ(yz − zy) = δ(δ
0
(y)z − z δ
0
(y) + y δ
0
(z) − δ
0
(z)y) − δ
0
(δ(yz) − δ(zy)) = δ δ
0
(y)z + δ
0
(y)δ(z) −
δ(z) δ
0
(y)−zδ δ
0
(y)+δ(y) δ
0
(z)+yδ δ
0
(z)−δ δ
0
(z)y−δ
0
(z)δ(y)−δ
0
(δ(y)z +yδ(z)−δ(z)y)−zδ(y)) =
δ δ
0
(y)z + δ
0
(y)δ(z) − δ(z) δ
0
(y) − zδ δ
0
(y) + δ(y) δ
0
(z) + yδ δ
0
(z) − δ δ
0
(z)y − δ
0
(z)δ(y) − δ
0
([δ(y)z] +
[yδ(z)]) = δ δ
0
(y)z + δ
0
(y)δ(z) − δ(z) δ
0
(y) − zδ δ
0
(y) + δ(y) δ
0
(z) + yδ δ
0
(z) − δ δ
0
(z)y − δ
0
(z)δ(y) −
([δ
0
δ(y)z] + [δ(y) δ
0
(z)] + [δ
0
(y)δ(z)] + [y δ
0
δ(z)]) = δ δ
0
(y)z + δ
0
(y)δ(z) − δ(z) δ
0
(y) − zδ δ
0
(y) +
δ(y) δ
0
(z) + yδ δ
0
(z) − δ δ
0
(z)y − δ
0
(z)δ(y) − δ
0
δ(y)z + z δ
0
δ(y) − δ(y) δ
0
(z) + δ
0
(z)δ(y) − δ
0
(y)δ(z) +
δ(z) δ
0
(y) − y δ
0
δ(z)+ δ
0
δ(z)y, and[[δ δ
0
](y)z]+ [y[δ δ
0
](z)] = [(δ δ
0
− δ
0
δ)(y)z] + [y(δ δ
0
− δ
0
δ)(z)] =
[δ δ
0
(y)−δ
0
δ(y), z]+[y, δ δ
0
(z)−δ
0
δ(z)] = [δ δ
0
(y)z]− [δ
0
δ(y)z]+[yδ δ
0
(z)]−[y δ
0
δ(z)] = δ δ
0
(y)z −
zδ δ
0
(y) − δ
0
δ(y)z + z δ
0
δ(y) + yδ δ
0
(z) − δ δ
0
(z)y − y δ
0
δ(z) + δ
0
δ(z)y
2.Show that sl(n, F ) is precisely the derived algebra of gl(n, F ).
Proof. It’s easy to prove sl(n, F ) is an ideal of gl(n, F ) because of T r([xz]) = T r(xz−zx) = Tr(xz)−
T r(zx) = 0, x ∈ sl(n, F ), z ∈ gl(n, F ) and sl(n, F ) = [gl(n, F ), gl(n, F )].
3.Prove that the center of gl(n, F ) equals s(n, F ). Prove that sl(n, F ) has center 0, unless char F divides
n, in which case the center is s(n, F ).
Proof. Z
gl(n,F )
= {z ∈ gl(n, F )|[zx] = 0, forallx ∈ gl(n, F )}, we only need to notice that only scalar
matrices can commute with any matrix, when charF - n, set s(n, F ) = {aE|a ∈ F }, T r(aE) = na =
0, it induces a = 0, when charF |n, T r(aE) = na = 0.
4.Show that(up to isomorphism) there is a unique Lie algebra over F of dimension 3 whose derived alge-
bra has dimension 1 and lies in Z(L)
Proof. Let x, y, z be the standard basis of L, if [LL]={x}({y},{z}), it’s obvious that x ∈ Z(L) when we
write down the multiplication table.
5.Suppose dim L=3, L=[LL]. Prove that L must be simple
Proof. Let x, y, z be the standard basis of L, I is an ideal of L, take arbitrary element ax+by+cz of I,
because [x,ax+by+cz]=b[xy]+c[xz],[y,ax+by+cz]=-a[xy]+c[yz], [z,ax+by+cz]=-a[xz]-b[yz] all lie in I, if
a6= 0, then [xy]∈ I, [xz]∈ I, dim I> 2, if dim I=2, then b or c cannot be 0, if b6= 0, then [yz]∈ I, dim I=3,
contradiction, so dim I= 3, I=L, similarly, b6= 0, I=L; c6= 0, I=L; otherwise a=b=c=0.
6.Prove that the sl(n,F) is simple, unless char F=3.
4
Proof. Take the standard basis h
1
= e
11
−e
22
, h
2
= e
22
−e
33
, e
ij
(1 6 i 6= j 6 3), If I6= 0 is an ideal of s-
l(3,F), adh
1
(h
1
, h
2
, e
12
, e
13
, e
21
, e
23
, e
31
, e
32
) = (h
1
, h
2
, e
12
, e
13
, e
21
, e
23
, e
31
, e
32
)diag(0, 0, 2, 1, −2, −1, −1, 1),
adh
2
((h
1
, h
2
, e
12
, e
13
, e
21
, e
23
, e
31
, e
32
)) = (h
1
, h
2
, e
12
, e
13
, e
21
, e
23
, e
31
, e
32
)diag(0, 0, −1, 1, 1, 2, −1, −2),
If h
i
∈ I, then[h
i
e
ij
] = λe
ij
∈ I; if e
ij
(i 6= j) ∈ I, [e
ij
e
ij
] = e
ii
− e
jj
= h
i
∈ I, it means that I=L
unless I is vacant. If char F=3, then the centre of sl(3,F) is s(3,F), what’s more, s(3,F) is an ideal of
sl(3,F).
7.Prove that t(n, F ) and δ(n, F ), are self-normalizing subalgebra of gl(n, F ), whereas n(n, F ) has
normalizert(n, F ).
Proof. We firstly write down the normalizers N
gl(n,F )
(t(n, F )) = {x ∈ gl(n, F )|[x, t(n, F )] ⊂ t(n, F )},
N
gl(n,F )
(δ(n, F )) = {x ∈ gl(n, F )|[x, δ(n, F )] ⊂ δ(n, F )}, N
gl(gl(n,F ))
(n(n, F )) = {x ∈ gl(n, F )|[x, n(n, F )] ⊂
n(n, F )}, take the standard basis of t(n,F)e
ij
i < j, e
kl
∈ N
gl(n,F )
(t(n, F )), 1 ≤ k, l ≤ n, [e
kl
e
ij
] =
e
kl
e
ij
− e
ij
e
kl
= δ
li
e
kj
− δ
jk
e
il
∈ t(n, F )whenk > j, e
kj
= 0andi > l, e
il
= 0, hencek > j > i >
l, e
kl
= 0. Similarly, take the standard basis of δ(n, F ) e
ii
, 1 ≤ i ≤ n, e
kl
∈ N
gl(n,F )
(δ(n, F )), 1 ≤
k, l ≤ n, [e
kl
e
ii
] = e
kl
e
ii
− e
ii
e
kl
= δ
li
e
kl
− δ
ik
e
il
, when k 6= i, e
ki
= 0andi 6= le
il
= 0, only when
k=i=l, δ
li
e
kl
− δ
ik
e
il
∈ δ(n, F ), hence e
kl
∈ δ(n, F ) ,we notice that t(n, F ) = δ(n, F ) + n(n, F ),
[x + y, n(n, F )] = [x, n(n, F )] + [y, n(n, F )] ⊂ n(n, F ), where x ∈ δ(n, F ), y ∈ n(n, F ), hence
t(n, F ) ⊂ N
gl(n,F )
(n(n, F )), to the contrary, if y ∈ N
gl(n,F )
(n(n, F )), to prove y ∈ t(n, F ), set
y = e
kl
, for any e
ij
∈ n(n, F ) i < j, [e
kl
e
ij
] = e
kl
e
ij
− e
ij
e
kl
= δ
li
e
kj
− δ
jk
e
il
, when kleqj, e
kj
=
0, i ≤ l, e
il
= 0, when k ≤ j ≤ i ≤ l[e
lk
e
ij
] = 0, so e
kl
= 0
8.Prove that in each classical linear Lie algebra, the set of diagonal matrices is a self-normalizing subal-
gebra, when char F=0.
Proof. If D is the set of diagonal matrices of A
l
, {h
i
= e
ii
− e
i+1i+1
}, 1 ≤ i ≤ l are the standard
basis of D, N
A
l
(D) = {x ∈ A
l
|[x, D] ⊂ D}, if e
kl
∈ N
A
l
(D), [e
klh
i
] = [e
kl
e
ii
] − [e
kl
e
i+1 i+1
] =
δ
li
e
ki
− δ
ik
e
il
− δ
l i+1
e
k i+1
+ δ
i+1 k
e
i+1 l
lies in D, so k=l=i, or k=l=i+1, that e
kl
∈ D.
9.Prove Proposition 2.2
Proof. (i) Define a map ψ : L/Kerφ → Imφ by x + Kerφ 7→ φ(x), since ψ([x + Kerφ, y + Kerφ]) =
ψ([xy] + Kerφ) = φ([xy]) = [φ(x)φ(y)], and set ψ(x + Kerφ) = φ(x), ψ(x
0
+ Kerφ) = φ(x
0
),
ifφ(x) = φ(x
0
), then φ(x) − φ(x
0
) = φ(x − x
0
) = 0, it induces x − x
0
∈ Kerφ, x + Kerφ =
x
0
+ Kerφ, now we prove there exists a unique ψ : L/I → L
0
, making the following diagram commute
L
π
φ
//
L
0
ψ
L/I
define ψ : L/I → L
0
by x + I 7→ φ(x), if φ( x
1
) = φ(x
2
), it will induce that x
1
− x
2
∈ I, so ψ is
well-defined. if there exists another φ
0
such that ψ ◦ π = φ
0
, then (φ − φ
0
)(x) = (ψ ◦ π − ψ ◦ π)(x) = 0;
(ii) it’s easy to prove that J/I is an ideal of L/I, and define epic homomorphism π:L/I → L/J by
x+I 7→ x+J, x ∈ L, if and only if x ∈ J, π(x+I) = 0, Kerπ = J/I, from the standard homomorphism
theorem(i), we have L/I/J/I
∼
=
L/J, what’s more π is homomorphic because of x, y ∈ L, π([x +
I, y + I]) = π([xy] + I) = [xy] + J = [π(x + I), π(y + I)]; (iii) it’s easy to prove that I ∩ J is
an ideal of I, let π be the canonical map L → L/J, f be the restriction to π on I, i.e. f = π|
I
I →
I/J, f(I)
∼
=
I/Kerf, π(I + J)
∼
=
(I + J)/J, f(I) = π(J + I) = π(J) + π(I) = 0 + π|
I
(I = f(I),
5
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