Lie Algebra
January 30, 2017
Jakelic
From before:
Ex 1:
L = M2 (F)
T2 (F) 5 L, [E11 , E21 = −E21 ∈
/ T]
St2 (F) 5 L
Let L be a Lie Algebra over F and gl(L) = {T : L → L|T is a linear transformation} .
Denition 0.1. Linear transformation ∂ ∈ gl(L) is a derivation if ∂ [x, y] =
[∂(x), y] + [x, ∂(y)] , ∀x, y ∈ L1 Der L := {∂ : L → L| ∂ is a derivation}
Proposition 1.
Proof.
Der(L) is a Lie subalgebra of
gl(L) under the commutator bracket.
1 Often called Leibniz's rule for
its similarity to the product rule
of dierentiation.
Let ∂1 , ∂2 ∈ Der(L), α, β ∈ F.
Closure under linear combinations: Let x, ∈ L. Then (α∂1 + β∂2 ) [x, y] = . . . [(α∂1 +
β∂2 )].
Closure under bracket: [∂1 , ∂2 ] [x, y] = . . . ∈ Der(L)
Fix x ∈ L Dene ad(x) : L → L by y → [x, y].2
2 called the adjoint of x
Proposition 2.
1.
adx ∈ gl(L) 3
3 ad is a linear map
x
2.
adx ∈ Der(L) 4
4 ad
Note:
Jacobi identity
follows product rule.
x
Hint: use Jacobi rule.
⇔ adx ∈ Der(L) (product rule applies)
Proposition 3.
1.
αadx + βady = adαx+βy ∀x, y ∈ L and ∀α, β ∈ F
2.
[∂, adx ] = ad∂x , ∀x ∈ L and ∀∂ ∈ Der(L)
Proof.
Apply to arbitrary element.
Denition 0.2. Let
L be a Lie Algebra. Then ad(L) = {adx |x ∈ L}. ad(L) ⊆
Der(L)
Corollary 0.1.
Proof.
ad(L) E Der(L)
By Proposition.
Denition 0.3. Let L be a Lie Algebra over a eld
F and let X ⊆ L. Dene
CL (X) = {y ∈ L| [y, x] = 0, ∀x ∈ X} . Then CL (X) is the centralizer of L in X.
Lemma 1.
Lie Algebra
January 30, 2017
1.
CL (X) ≤ L
2.
CL (L) = Z(L)
Denition 0.4. Let
normalizer of K in L.
K be a subspace of L. Then NL (K) = {[x, K] ⊆ K} is the
Proposition 4.
1.
NL (K) ≤ L
2.
K ≤ L ⇔ K ⊆ NL (K)
3.
K E L ⇔ L = NL (K)
Proof. Subspace. Let x1 , x2 ∈ NL (K). Need to show bracket of x1 and x2 is in
normalizer of k. Use Jacobi.
Proposition 5.
Let
M, N E L. Then
1.
M + N := {m + n|m ∈ M, n ∈ N }
2.
[M, N ] E L
3.
M ∩N EL
Proof.
Do number 1. Number 3: Let x ∈ M ∩ N. Remember M , N are ideal.
Number 2:
Jakelic