Lie Algebras and Representation Theory

Homework 2

Debbie Matthews

2015-04-13
HW2 Debbie Matthews Spring 2015

Problem 3.1
Let I be an ideal of L. Then each member of the derived series or descending
central series of I is also an ideal of L.

To begin with, we note that bilinearity of of the bracket ensures that each
member of the derived series and descending central series of I is in fact a sub-
space of L. So we proceed to check that it closed under the action of adL .

(Derived Series) First note that I (0) = I is an ideal. Assume for induction
that I (n) is an ideal. Cosider I (n+1) = [I (n) , I (n) ]. Let x ∈ L and y, z ∈ I (n) .
By the Jacobi relation

[x, [y, z]] = −[y, [z, x]] − [z, [x, y]]

⊆ [y, I (n) ] + [z, I (n) ]
⊆ [I (n) , I (n) ] + [I (n) , I (n) ]
= I (n+1) + I (n+1)
= I (n+1) .

Therefore, I (n+1) is an ideal and we are done by induction.

(Descending Central Series) Again start by observing I 0 = I is an ideal.

Assume for induction that I n is an ideal. Consider I n+1 = [I, I n ]. Let x ∈ L,
y ∈ I, and z ∈ I n . By the Jacobi relation

[x, [y, z]] = −[y, [z, x]] − [z, [x, y]]

⊆ [y, I n ] + [z, I]
⊆ [I, I n ] + [I n , I]
= I n+1 + I n+1
= I n+1

Therefore, I n+1 is an ideal and we are done by induction.

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Problem 3.2
Prove that L is solvable if and only if there exists a chain of subalgebras L =
L0 ⊇ L1 ⊇ L2 ⊇ . . . ⊇ Lk = 0 such that Li+1 is an ideal of Li and such that
each quotient Li /Li+1 is abelian.

(⇒) Suppose L is solvable. Consider the chain of subalgebras formed by the

derived series: Li = L(i) . Since L is an ideal of itself, we know from problem
3.1 that L(i) ⊇ L(i+1) where L(i+1) is an ideal in L. Thus L(i+1) is an ideal in
L(i) ⊆ L. Finally, each quotient L(i) /L(i+1) is abelian by definition since L(i+1)
consists of all brackets of elements in L(i) .

(⇐) Suppose there exists a chain of subalgebras L = L0 ⊇ L1 ⊇ L2 ⊇ . . . ⊇

Lk = 0 such that Li+1 is an ideal of Li and such that each quotient Li /Li+1
is abelian. Note that Lk = 0 is solvable and Lk−1 /Lk is abelian and hence
solvable. Therefore Lk−1 is solvable. Proceed down the chain. In the final step
we have L1 is solvable and L0 /L1 is solvable hence L0 is solvable. Therefore
L = L0 is solvable and we are done.

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Problem 3.3
Let char F = 2. Prove that sl(2, F) is nilpotent.

Since sl(2, F) is the set of all trace zero matrices, consider the standard basis
     
1 0 0 1 0 0
x= ,y = ,z = .
0 −1 0 0 1 0

With this basis we have the relations

[x, y] = 2y, [x, z] = −2z, [y, z] = x.

Considering that char F = 2, these relations can be modified to read

[x, y] = 0, [x, z] = 0, [y, z] = x.

With these relations, we observe that [L, L] = span{x}. Then since [x, x] =
[x, y] = [x, z] = 0, we conclude

[L, [L, L]] = 0.

Therefore, sl(2, F) is nilpotent.

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Problem 3.4
Prove that L is solvable (resp. nilpotent) if and only if ad L is solvable (resp.
nilpotent).

(Solvable) First we show that adL(n) = (adL )(n) by inducting on n. We use

the basic fact that ad[x,y] = [adx , ady ]. For n = 0 we have L = L(0) and hence
(adL )(0) = adL = adL(0) . Assume for induction that adL(n) = (adL )(n) . Then
adL(n+1) = ad[L(n) ,L(n) ] = [adL(n) , adL(n) ] = [(adL )(n) , (adL )(n) ] = (adL )(n+1) .

(⇒) Armed with the fact above, assume L is solvable. Then there exists
some n ∈ N such that L(n) = 0. Then adL(n) = (adL )(n) = 0 and hence adL is
solvable.

(⇐) Meanwhile if there is some n such that (adL )(n) = 0, then adL(n) = 0.
This means L(n) ⊆ Z(L). Then both L/Z(L) and Z(L) are solvable. Therefore
L is solvable.

(Nilpotent) A similar argument as above shows that adLn = (adL )n , where

this time we use adLn+1 = ad[L,Ln ] = [adL , adLn ] = [adL , (adL )n ] = (adL )n+1 .
The rest of the argument is the same as above, replacing solvable with nilpotent
and replacing the superscripts (i) with i.

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Problem 3.5
Prove that the nonabelian two dimensional algebra constructed in (1.4) is solv-
able but not nilpotent. Do the same for the algebra in Exercise 1.2.

(1.4) : L = span{v, w} such that [v, w] = v

(1.2) : K = span{x, y, z} such that [x, y] = z, [x, z] = y, [y, z] = 0

(1.4 : L) Consider L(1) = [L, L] = span{v}. Since L(1) is one-dimensional,

it is abelian. Therefore L(2) = [L(1) , L(1) ] = 0 and thus L is solvable. However,
[−w, v] = v, we have that (ad−w )n v = v for all n ≥ 0. Thus Ln 6= 0 for all
n ≥ 0 which means L is not nilpotent.

(1.2 : K) Now we have K (1) = [K, K] = span{y, z}. But since [y, z] = 0, we
again have that K (1) is abelian. Therefore, K (2) = [K (1) , K (1) ] = 0 and thus K
is solvable. However, since [x, y] = z and [x, z] = y we have that (adx )2n y = y
and (adx )2n+1 = z for all n ≥ 0. Thus K n 6= 0 for all n ≥ 0 which means K is
not nilpotent.

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HW2 Debbie Matthews Spring 2015

Problem 3.6
Prove that the sum of two nilpotent ideals of a Lie algebra L is again a nilpotent
ideal. Therefore, L possesses a unique maximal nilpotent ideal. Determine this
ideal for each algebra in Exercise 5.

Let I and J be nilpotent ideals in L. Then there exists n, m ∈ N such

that I n = J m = 0. Assume without loss of generality that n ≥ m and thus
I n = J n = 0. We will show that all elements of I + J are ad-nilpotent and then
apply Engel’s theorem. Consider x + y ∈ I + J, where x ∈ I and y ∈ J.

We begin by noting that adL has associative multiplication where

adx (ady adz )w = adx [y[z, w]]

= [x, [y, [z, w]]]
= [x, [y, adz w]]
= (adx ady ) adz w.

We make one additional note that adx+y z = [x + y, z] = [x, z] + [y, z] =

(adx )z + (ady )z = (adx + ady )z. Finally, since the multiplicative structure of
adL is not commutative, we can’t apply the binomial theorem directly. How-
ever, because of associativity and anticommutativity, we know that (adx+y )2n =
P2n
(adx + ady )2n will expand to a sum of the form k=0 ak (adx )k (ady )2n−k where
ak ∈ F. In every term of this expression, either k ≥ n or 2n − k ≥ n. Therefore,
this sum is zero and we conclude that all elements of I + J are ad-nilpotent. By
Engel’s theorem, we then know I + J is itself nilpotent.

Note that 0 is a nilpotent ideal. For any finite dimensional L, it is then

clear that L contains a maximal nilpotent ideal. If L has two maximal nilpotent
ideals I and J, then I + J is nilpotent and by maximality we get I = J = I + J.
Hence, L possesses a unique maximal nilpotent ideal.

(1.4 : L) Consider span{v} = L1 . We have established that this is an ideal.

It is nilpotent since it is one-dimensional. Since all of L is two-dimensional and
not nilpotent, we conclude that the one-dimensional ideal span{v} is the unique
maximal nilpotent ideal.

(1.2 : K) Consider span{y, z} = L1 . Again, we have established that this is

an ideal. It is nilpotent since it is abelian. Since all of L is three-dimensional
and not nilpotent, we conclude that the two-dimensional ideal span{y, z} is the
unique maximal nilpotent ideal.

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Problem 3.7
Let L be nilpotent, K a proper subalgebra of L. Prove that NL (K) includes K
properly.

Consider the descending central series L = L0 ⊇ L1 ⊇ L2 ⊇ . . . ⊇ Ln = 0.

Choose m maximal such that there exists x ∈ Lm \K. Since K is properly
contained in L, there is some y ∈ L0 \K = L\K. Furthermore Ln = 0 ⊆ K.
Hence 0 ≤ m < n. Consider [x, K]. Since x ∈ Lm we know [x, K] ⊆ Lm+1 . By
maximality of m we conclude [x, K] ⊆ K. Therefore x ∈ NL (K). Since x ∈ / K,
we have that K is properly contained in NL (K).

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Problem 3.8
Let L be nilpotent. Prove that L has an ideal of codimension 1.

Consider the descending central series L = L0 ⊇ L1 ⊇ L2 ⊇ . . . ⊇ Ln = 0,

where n is minimal. Choose x0 ∈ L0 \L1 . Then complete a basis for L with
x1 , x2 , . . . xm . Consider K = span{x1 , x2 , . . . , xm }. In particular L1 ⊆ K so K
is an ideal. By construction, K has codimension 1.