MATH 534. SOLUTIONS TO PROBLEM SET 4
Problem 1. (5pts) (a) Define the following linear map:
Φ:V →V∗
v 7→ B(v, −)
It is injective since Ker(Φ) = V ⊥ = {0} (non-degenerate). It is also surjective, since V and V ∗ has the same dimension.
Now let r : V ∗ → W ∗ be the map obtained by restricting a linear function
l : V → C to W . Note that l is surjective (check!). The composition
l ◦ Φ: V ∗ W∗
gives a surjective map V → W ∗ . Explicitly, v 7→ B(v, −)|W . The kernel of
this map is precisely W ⊥ . Thus dim V = dim W ∗ + dim W ⊥ = dim W +
dim W ⊥ .
(b) By definition, W ⊂ (W ⊥ )⊥ . Note that by part (a), dim(W ⊥ )⊥ =
dim(V ) − dim(W ⊥ ) = dim(W ). Thus W ⊂ (W ⊥ )⊥ .
Problem 2. (5pts)
(“Only if”) Suppose y lies in the image of ad(x). That is, y = [x, v] for some
v ∈ L. Then ∀z ∈ Ker(ad(x)),
B(y, z) = −B([v, x], z) = −B(v, [x, z]) = −B(v, 0) = 0,
as desired.
(“If”) Conversely, assume B(y, z) = 0 ∀z ∈ Ker(ad(x)). Then
y ∈ Ker(ad(x))⊥ .
The “only if” part shows that Im(ad(x)) ⊂ (Ker(ad(x)))⊥ . Moreover,
dim Im(ad(x)) = dim L − dim Ker(ad(x)) = dim(Ker(ad(x)))⊥ .
Hence Im(ad(x)) = Ker(ad(x))⊥ and y ∈ Im(ad(x)).
Problem 3. (5pts)
Denote the image of v ⊗ w ∈ V ⊗ W in Hom(W ∗ , V ) under the isomorphism
φ : V ⊗ W and Hom(W ∗ , V ) described in the statement of this problem by
φv⊗w :
V ⊗ W ' Hom(W ∗ , V )
v ⊗ w 7→ φv⊗w .
We need to check that
a · φv⊗w = φa·v⊗w + φv⊗a·w , in Hom(W ∗ , V ).
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MATH 534. SOLUTIONS TO PROBLEM SET 4
Indeed, ∀l ∈ W ∗ , we have
[a · φv⊗w ](l) = a · (φv⊗w (l)) − φv⊗w (a · l)
= a · [l(w)v] − (a · l)(w)v
= a · [l(w)v] − [−l(a · w)v]
= l(w)a · v + l(a · w)v
= φa·v⊗w (l) + φv⊗a·w (l).
Problem 4 (Exercises 5a,d on page 30-31). (15pts)
(a) (6pts) L is a representation of the Lie algebra ad(L) ' L/Z(L) =
L/Rad(L) which is semisimple.
Z(L) is an ad(L)-submodule in L. By Weyl’s Theorem, it has a complement submodule L0 (ideal), i.e. L = Z(L) ⊕ L0 . Then L0 ' L/Z(L)
is semisimple. Hence L0 = [L0 , L0 ] ⊂ [L, L]. On the other hand [L, L] =
[Z(L) ⊕ L0 , Z(L) ⊕ L0 ] ⊂ [L0 , L0 ] ⊂ L0 . Thus L0 = [L, L].
(d) (9pts) By (a) L = Z(L) ⊕ [L, L]. Let (ρ, V ) be an L-module such
that Z(L) acts by semisimple elements of gl(V ). Then elements in ρ(Z(L))
can be diagonalized simultaneously, and V decomposes as a direct sum of
weight spaces, as a Z(L)-module:
M
V =
Vλ
λ
where λ ranges over Z(L)∗ and Vλ = {v ∈ V : ρ(z)v = λ(z)v, ∀z ∈ Z(L)}.
Now each Vλ is an L-submodule. Indeed, ∀v ∈ Vλ , ∀x ∈ [L, L] we have
ρ(x)v ∈ Vλ ∈ Vλ since ρ(z)ρ(x)v = ρ(x)ρ(z)v = λ(z)ρ(x)v, ∀z ∈ Z(L).
Since [L, L] is a semisimple Lie algebra, each Vλ decomposes as a direct
sum of irreducible [L, L] submodules:
Vλ =
mλ
M
Vλi .
i=1
Here each Vλi is irreducible as an [L, L]-modules. Since Z(L) acts on each
of these irreducible submodules by scalar multiplication, each Vλi is also
an L-module. Moreover, each Vλi is irreducible as an L-module, since it is
irreducible as an [L, L]-module. This shows that V is completely reducible,
V =
mλ
MM
λ
i=1
Vλi .
MATH 534. SOLUTIONS TO PROBLEM SET 4
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Problem 5 (Exercise 6 on page 31). (5pts)
β, γ give us two linear maps L → L∗ :
L → L∗
x 7→ β(x, −)
x 7→ γ(x, −)
Both maps are homomorphisms of L-modules, since β, γ are associative.
Both are isomorphisms of vector spaces, since β, γ are non-degenerate. As
L-modules, L ' L∗ . Hence the above two maps are in HomL (L, L∗ ) =
HomL (L, L). Since L is a simple Lie algebra (i.e., an irreducible L-module)
Schur’s lemma tells us that HomL (L, L) = C. Thus β and γ are scalar
multiples of each other.
Problem 6 (Exercise 7 on page 31). (5pts)
It is easy to check that the bilinear form (x, y) 7→ Tr(xy) is non-zero, symmetric and associative. We shall refer to it as ”the trace form”. As we
saw in class, the radical of an associative symmetric bilinear form on a Lie
algebra L is an ideal of L. Since sln is simple, we conclude that the radical
of trace form is zero. In other words, the trace form is non-degenerate. By
the previous problem, the Killing form is a scalar multiple of the trace form.
In other words, κ(x, y) = cTr(xy), ∀x, y ∈ sln for some c ∈ C.
To compute the scalar c, set x = y = E11 −E22 . xy = diag(1, −1, 0, . . . , 0)2 =
diag(1, 1, 0, . . . , 0) and Tr(xy) = 2. On the other hand, let us examine ad(x)
in the natural basis of sln consisting of the n2 − n elementary matrices Eij ,
i 6= j, and the n − 1 diagonal matrices Eii − Ei+1 i+1 , i = 1, . . . , n − 1. We
have
ad(x)(Eii − Ei+1,i+1 ) = 0
∀i = 1, 2, . . . , n − 1
ad(x)(E12 ) = 2E12
ad(x)(E1k ) = E1k
k = 3, . . . , , n
ad(x)(E2k ) = −E2k
k = 3, . . . , n
ad(x)(E21 ) = −2E21
ad(x)(Ek1 ) = −Ek1
ad(x)(Ek2 ) = Ek2
ad(x)(Eij ) = 0
k = 3, . . . , , n
k = 3, . . . , n
for other i, j
Hence ad(x) is diagonal in our basis of sln , with diagonal entries −2, 2, −1
and 1, appearing with multiplicities 1, 1, 2(n − 2) and 2(n − 2), respectively.
Thus
κ(x, y) = Tr(ad(x)ad(y)) = Tr(ad(x)2 ) = 22 +(−2)2 +12 ·2(n−2)+(−1)2 ·2(n−2) = 4n
and c =
4n
κ(x, y)
=
= 2n.
Tr(xy)
2