TOPICS IN NON COMMUTATIVE ALGEBRAIC GEOMETRY AND CONGRUENT ZETA FUNCTIONS

(PART VI). LIE ALGEBRAS AND THEIR ENVELOPING ALGEBRAS

YOSHIFUMI TSUCHIMOTO

1. Lie algebras

Definition 1.1. Let K be a commutative ring. Then a Lie algebra 𝔤 over K is a K-module with a bilinear (non associative) bracket product (“Lie bracket”)

[•, •] : 𝔤 × 𝔤 → 𝔤

which satisfies the following axioms:

(1) [X, X] = 0 for all X ∈ 𝔤.

(2) (“Jacobi identity”)

[X, [Y, Z]] = [[X, Y], Z]] + [Y, [X, Z]] (∀X, Y, Z ∈ 𝔤).

Example 1.2. Any associative algebra A over k may be regarded as a Lie algebra with the “commutator” as a Lie bracket.

In this talk, we always regard associative algebra as a Lie algebra equipped with the commutator product unless otherwise specified.

Lemma 1.3. Let 𝔤 be a Lie algebra over a ring k. Then there exists an associative unital algebra U(𝔤) with a Lie algebra homomorphism

ι_{𝔤} : 𝔤 → U(𝔤)

with the following universal property:

For any associative unital algebra A with a Lie algebra homomorphism ϕ : 𝔤 → A, there exists a unique algebra homomorphism

ψ : U(𝔤) → A

such that ψ ∘ ι_{𝔤} = ϕ holds.

The pair (U(𝔤), ι_{𝔤}) is unique up to an isomorphism.

Definition 1.4. Under the assumption of the previous Lemma, The pair (U(𝔤), ι_{𝔤}) is called the universal enveloping algebra the Lie algebra 𝔤.

Universal enveloping algebras of Lie algebras form an important class of non commutative associative algebras. Our task in this Part is to describe these algebras in our language.

2. Representations of a Lie algebra

Definition 2.1. Let k be a field. A finite dimensional representation of a Lie algebra 𝔤 over k is a Lie algebra homomorphism

ρ : 𝔤 → M_{n}(k).

Note: The full matrix algebra M_{n}(k), when regarded as a Lie algebra equipped with the commutator product, is commonly denoted as 𝔤𝔩_{n}(k).

Example 2.2. Let k be a field. Let 𝔤 be a finite dimensional Lie algebra over k. We then have an adjoint representation

𝔤 ∋ X ↦ ad(X) = (Y ↦ [X, Y]) ∈ End_{k−linear}(𝔤).

3. Poincaré-Birkoff-Witt Theorem

In this section we prove the Poincaré-Birkoff-Witt Theorem. The treatment here essentially follows [1]. Let 𝔤 be a Lie algebra over a field k. To prove the theorem we consider S_{k}(𝔤), the symmetric algebra of 𝔤 over k. Let us denote the multiplication of S_{k}(𝔤) by (x, y) ↦ x ∘ y. We note that each element x of S_{k}(𝔤) has its degree deg(x). (as a polynomial in elements of 𝔤.)

Lemma 3.1. We choose a ordered basis (x_{λ}; λ ∈ Ω). (That means, a basis with a totally ordered index set Ω.) Then there exists a linear action of 𝔤 on S_{k}(𝔤) which obeys the following rules:

(1) For any x ∈ 𝔤 and for any y ∈ S_{k}(𝔤),

deg(x.y − x ∘ y) ≤ deg(y)

(2) If λ_{0} ≤ λ_{1} ≤ λ_{2} ≤ ⋯ ≤ λ_{n}, then we have

x_{λ_{0}}.(x_{λ_{1}} ∘ x_{λ_{2}} ∘ x_{λ_{3}} ∘ ⋯ ∘ x_{λ_{n}}) = x_{λ_{0}} ∘ x_{λ_{1}} ∘ x_{λ_{2}} ∘ x_{λ_{3}} ∘ ⋯ ∘ x_{λ_{n}}.

(3) For any x, y ∈ 𝔤 and for any z ∈ S_{k}(𝔤), we have

x.(y.z) − y.(x.z) = [x, y].z

The proof is done by a careful use of induction. Namely,

Sublemma 3.2. We employ the same assumption of the above Lemma. Then for each m ∈ ℤ_{>0}, there exists a unique k-bilinear map

f_{m} : 𝔤 × S_{k}(𝔤)_{≤m} → S_{k}(𝔤)_{≤m+1}

which obeys the following rules:

(1) For any x ∈ 𝔤 and for any y ∈ S_{k}(𝔤)_{≤m},

deg(f_{m}(x, y) − x ∘ y) ≤ deg(y)

(2) If λ_{0} ≤ λ_{1} ≤ λ_{2} ≤ ⋯ ≤ λ_{n} and n ≤ m, then we have

f_{m}(x_{λ_{0}}, x_{λ_{1}} ∘ x_{λ_{2}} ∘ x_{λ_{3}} ∘ ⋯ ∘ x_{λ_{n}}) = x_{λ_{0}} ∘ x_{λ_{1}} ∘ x_{λ_{2}} ∘ x_{λ_{3}} ∘ ⋯ ∘ x_{λ_{n}}.

(3) For any x, y ∈ 𝔤 and for any z ∈ S_{k}(𝔤)_{≤m−1}, we have

f_{m}(x, f_{m}(y, z)) = f_{m}(y, f_{m}(x, z)) + f_{m}([x, y], z)

Proof. We note first that S_{k}(𝔤) has the set of monomials

{x_{λ_{1}} ∘ x_{λ_{2}} ∘ x_{λ_{3}} ∘ ⋯ ∘ x_{λ_{n}}; λ_{1} ≤ λ_{2} ≤ λ_{3} ≤ ⋯ ≤ λ_{n}}

as a k-basis. For monomial w = x_{λ_{1}} ∘ x_{λ_{2}} ∘ x_{λ_{3}} ∘ ⋯ ∘ x_{λ_{n}} such that λ_{1} ≤ λ_{2} ≤ λ_{3} ≤ ⋯ ≤ λ_{n}, we put z = λ_{2} ≤ λ_{3} ≤ ⋯ ≤ λ_{n}. Then

w = f_{m−1}(x_{λ_{1}}, z).

We define inductively the action of x_{λ_{0}} on it by the following equations.

f_{m}(x_{λ_{0}}, x_{λ_{1}}∘z) = \begin{cases}x_{λ_{0}} ∘ x_{λ_{1}} ∘ z & (if λ_{0} ≤ λ_{1}) \\ ( \begin{matrix}x_{λ_{1}} ∘ x_{λ_{0}} ∘ z \\ + f_{m−1}(x_{λ_{1}}, f_{m−1}(x_{λ_{0}}, z) − x_{λ_{0}} ∘ z) \\ + f_{m−1}([x_{λ_{0}}, x_{λ_{1}}], z) \end{matrix}) & (if λ_{0} > λ_{1}) \end{cases}

We first note that the above definition is necessary to meet our conditions. Indeed, by (2) we necessarily define as above for λ_{0} ≤ λ_{1}. When λ_{0} > λ_{1}, we compute

x_{λ_{0}}.(x_{λ_{1}} ∘ z)
=^{(3)}x_{λ_{1}}.x_{λ_{0}}.z + [x_{λ_{0}}, x_{λ_{1}}].z
=x_{λ_{1}}.(x_{λ_{0}}.z − x_{λ_{0}} ∘ z) + x_{λ_{1}}.(x_{λ_{0}} ∘ z) + [x_{λ_{0}}, x_{λ_{1}}].z
=^{(2)}x_{λ_{1}}.(x_{λ_{0}}.z − x_{λ_{0}} ∘ z) + x_{λ_{1}} ∘ x_{λ_{0}} ∘ z + [x_{λ_{0}}, x_{λ_{1}}].z

and take a careful look at degrees of each monomials using (1). From this argument we see in particular that the action is uniquely determined by conditions (1),(2),(3).

It is easy to see that the conditions (1),(2) are satisfied by f_{m} defined as above.. Let us proceed to verify that the f_{m} so defined also satisfies (3). Let us consider x_{λ}, x_{μ} z = x_{μ_{1}} ∘ x_{μ_{2}} ∘ ⋯ ∘ x_{μ_{n}} with μ_{1} ≤ μ_{2} ≤ ⋯ ≤ μ_{n}, n ≤ m − 1. We need to prove

(♭)  x_{λ}.x_{μ}.z − x_{μ}.x_{λ}.z = [x_{λ}, x_{μ}].z.

Since the equation above is antisymmetric in μ, ν, we may assume that λ ≤ μ.

(i) Case where λ ≤ μ_{1}.

x_{λ}.x_{μ}.z
=x_{λ}.(x_{μ} ∘ z) + x_{λ}.(x_{μ}.z − x_{μ} ∘ z)
=^{(1)}x_{λ} ∘ x_{μ} ∘ z + x_{λ}.(x_{μ}.z − x_{μ} ∘ z)

In other words,

f_{m}(x_{λ}, f_{m}(x_{μ}, z)) = x_{λ} ∘ x_{μ} ∘ z + f_{m−1}(x_{λ}, (f_{m−1}(x_{λ}, z) − x_{μ} ∘ z)).

On the other hand we have

x_{μ}.x_{λ}.z
=x_{μ}.(x_{λ} ∘ z)
=^{by def}x_{λ} ∘ x_{μ} ∘ z + f_{m−1}(x_{λ}, f_{m−1}(x_{μ}, z) − x_{μ} ∘ z) + f_{m−1}([x_{μ}, x_{λ}], z)

So the equation ♭ surely holds in this case. (ii) Case where λ, μ > μ_{1}.

In this case we need to “decompose” z further:

z = x_{ν}.w.

We first forget about the hypothesis λ ≤ μ and prove

x_{λ}.(x_{μ}.(x_{ν}.w)) (♡)

=x_{ν}.(x_{λ}.(x_{μ}.w)) + [x_{λ}, x_{ν}].(x_{μ}.w) + [x_{μ}, x_{ν}].(x_{λ}.w) + [x_{λ}, [x_{μ}, x_{ν}]].w

(Since we are doing induction, we need to pay a special attention on degrees on operands. That means, we should use f_{m}’s rather than the above “lazy” notation. But that is fairly cumbersome, so we keep on being lazy here.)

Let us now admit that the above equation ♡ is true and prove the rest of the equation (3). By interchanging λ and μ in the equation (♡), we obtain

x_{μ}.(x_{λ}.(x_{ν}.w)) (♢)

=x_{ν}.(x_{μ}.(x_{λ}.w)) + [x_{μ}, x_{ν}].(x_{λ}.w) + [x_{λ}, x_{ν}].(x_{μ}.w) + [x_{μ}, [x_{λ}, x_{ν}]].w

Then by subtracting (♢) from (♡), we obtain

x_{λ}.(x_{μ}.(x_{ν}.w)) − x_{μ}.(x_{λ}.(x_{ν}.w))
=x_{ν}.(x_{λ}.(x_{μ}.w) − x_{μ}.(x_{λ}.w))
+ ([x_{λ}, [x_{μ}, x_{ν}]] − [x_{μ}, [x_{λ}, x_{ν}]]).w.

Since deg(w) is smaller than deg(z), by induction hypothesis the first term in the right hand side may be replaced by x_{ν}.([x_{λ}, x_{μ}].w). The second term may be replaced, by the Jacobian identity, by [[x_{λ}, x_{μ}], x_{ν}]. So the equation (♭) holds in this case too.

It remains to prove the equation (♡). By the induction hypothesis we have

x_{μ}.(x_{ν}.w) = x_{ν}.(x_{μ}.w) + [x_{μ}, x_{ν}].w.

Also by the induction hypothesis we have

x_{λ}.([x_{μ}, x_{ν}].w) = [x_{μ}, x_{ν}].(x_{λ}.w) + [x_{λ}, [x_{μ}, x_{ν}]].w

Lastly, we decompose x_{μ}.w as

x_{μ}.w = (x_{μ} ∘ w) + (x_{μ}.w − x_{μ} ∘ w). = (x_{μ} ∘ w) + y

Then the second term y has degree smaller than deg(z) = deg(w) + 1. The case (i) applies to the first term and we obtain:

x_{λ}.(x_{ν}.(x_{μ}.w)) = x_{ν}.(x_{λ}.(x_{μ}.w)) + [x_{λ}, x_{ν}].(x_{μ}.w).

These altogether complete the proof. □

Theorem 3.3 (Poincaré, Birkoff, Witt(PBW)). Let 𝔤 be a Lie algebra over a field k. Then we have a k-algebra isomorphism

Ψ : Gr(U(𝔤)) ≅ S(𝔤).

Proof. Let

ι_{0} : 𝔤 → Gr(U(𝔤))

be the obvious k-linear map.

Using the universality of symmetric algebra, there exists a unique k-algebra homomorphism

Φ : S(𝔤) → Gr(U(𝔤))

which extends ι_{0}. On the other hand the action defined in the Lemma 1.3 gives us a linear map

Ψ_{0} : U(𝔤) ∋ x ↦ x.1 ∈ S(𝔤)

which is clearly degree-decreasing. So it defines a k-linear map

Ψ : Gr(U(𝔤)) → Gr(S(𝔤)) ≅ S(𝔤).

Now the composition we obtain

Ψ ∘ Φ : S(𝔤) →^{Φ} Gr(U(𝔤)) →^{Ψ} S(𝔤)

coincides with the identity map. Indeed, it coincides with the identity on monomials of the form

x_{λ_{1}} ∘ x_{λ_{2}} ∘ x_{λ_{3}} ∘ ⋯ ∘ x_{λ_{n−1}} ∘ x_{λ_{n}}.

The map Φ is easily verified to be surjective. So we conclude that Φ and Ψ are both bijective and are inverse to each other. □

4. Jordan-Chevalley decomposition of a square matrix

4.1. Existence and uniqueness of Jordan-Chevalley decomposition.

Definition 4.1. Let A be a square matrix over a field k. A Jordan-Chevalley decomposition (also called SN-decomposition) of A is a decomposition of A

A = S + N

which satisfies the following conditions.

(1) S is semisimple (that means, the minimal polynomial of S has only simple roots.)

(2) N is nilpotent.

(3) SN = NS

(4) S, N ∈ \overline{k}[A]

A main objective of this section is to prove the following proposition.

Proposition 4.2. For any square matrix A over a field k, there exists a unique Jordan-Chevalley decomposition.

To prove it, we need some basic facts from linear algebra.

Lemma 4.3. Let A be a square matrix over a field k. Let m_{A}(X) be the minimal polynomial of A over k. If m_{A} is decomposed into two coprime polynomials, that means,

m_{A} = m_{1}m_{2} (m_{1}, m_{2}) = 1,

then A is similar to a direct sum

A_{1} ⊕ A_{2} = (\begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix})

where m_{1}(A_{1}) = 0, m_{2}(A_{2}) = 0.

Proof. Since k[X] is an Euclidean domain, it is a principal ideal domain. thus we see that there exists a polynomial l_{1}(X), l_{2}(X) ∈ k[X] such that

l_{1}(X)m_{1}(X) + l_{2}(X)m_{2}(X) = 1

holds. We put

E_{j} = l_{j}(A)m_{j}(A) (j = 1, 2).

Then we see easily that E_{1}, E_{2} are mutually orthogonal projection. That means, we have

E_{1}^{2} = E_{1}, E_{2}^{2} = E_{2}, E_{1} + E_{2} = 1, E_{1}E_{2} = 0.

It is also easy to see that both E_{1} and E_{2} commute with A. Now putting A_{1} = A|_{Range E_{2}} and A_{2} = A|_{Range E_{1}} we see that

A = AE_{2} + AE_{1} = A_{1} ⊕ A_{2}.

with A_{1} and A_{2} satisfying the required property. □

Corollary 4.4. Every square matrix A over a field k is similar to a direct sum of square matrices A_{1}, A_{2}, …, A_{s} with each minimal polynomial m_{A_{j}} equals to a power f_{j}^{e_{j}} of a irreducible polynomial f_{j} over k.

□

Corollary 4.5. When k is algebraically closed, every square matrix A over a field k is similar to a direct sum of square matrices A_{1}, A_{2}, …, A_{s} with each minimal polynomial m_{A_{j}} equals to a power (X − c_{j})^{e_{j}} of a polynomial of degree 1 over k.

□

Corollary 4.6. A square matrix over a field k is semisimple if and only if it is diagonalizable (similar to a diagonal matrix) over \overline{k}.

□

Corollary 4.7. Let S_{1}, S_{2} be semisimple square matrices of the same size over k. if S_{1} and S_{2} commute, then both S_{1} + S_{2} and S_{1}S_{2} are also semisimple.

Proof. Using commutativity of S_{1} and S_{2}, we may easily see that S_{1} and S_{2} are simultaneously diagonalizable over \overline{k}. □

Corollary 4.8. Let k be a field. Let C be a commutative subalgebra of M_{n}(k). If C is generated by semisimple elements, then every element of C is also semisimple.

□

On the other hand we have

Lemma 4.9. Let k be a field. Let C be a commutative subalgebra of M_{n}(k). If C is generated by nilpotent elements, then every element of C is also nilpotent.

Proof. Easy. □

Corollary 4.10. A Jordan-Chevalley decomposition (if there exists) of a square matrix A is unique.

Proof. Let

A = S + N = S′ + N′

be two Jordan-Chevalley decompositions. Then S − S′ = N′ − N is a semisimple nilpotent element. Thus S − S′ = N′ − N = 0. □

Proof. (of Proposition 4.2.) It now remains to prove that Jordan-Chevalley decomposition of a square matrix exists. By definition we may assume that k is algebraically closed. In view of Corollary 4.5, we may then assume that the minimal polynomial m_{A} of A is of the form (X − c)^{e} for some c ∈ k and e ∈ ℤ_{>0}. Then

A = c + (A − c)

gives the required Jordan-Chevalley decomposition.

□

Definition 4.11. Let k be a field. For any square matrix x ∈ M_{n}(k), we denote by x_{s} (respectively, x_{n}) the semisimple (respectively, nilpotent) part of x in the Jordan-Chevalley decomposition of x.

Lemma 4.12. Let k be a field. Let x ∈ M_{n}(k) be a square matrix. then we have

(ad(x))_{s} = ad(x_{s}), (ad(x))_{n} = ad(x_{n})

Proof. Follows easily from the uniqueness of the Jordan-Chevalley decomposition. □

4.2. k-rationality.

Proposition 4.13. Let A be a square matrix over a field k. Let

A = S + N

be the Jordan-Chevalley decomposition of A.

Let m_{A} be the minimal polynomial of A. If all of the roots of m_{A} are separable over k, then S and N are defined over k. (That means, they are matrices over k).

Proof. In view of Corollary , we may assume that m_{A} is of the form f^{e} for some irreducible polynomial f and a positive integer e. By assumption, f has only simple roots.

f(X) = ∏_{j=1}^{d}(X − c_{j})

Let us define a polynomial χ_{s}(X) ∈ \overline{k}[X] as follows.

χ_{s}(X) = \frac{∏_{{1≤j≤d \atop j≠s}} (X − c_{j})}{∏_{{1≤j≤d \atop j≠s}} (c_{s} − c_{j})} (s = 1, 2, 3, …, d)

These polynomials are designed to satisfy the following property.

χ_{s}(c_{j}) = \begin{cases}1 & if j = s \\ 0 & if j ≠ s \end{cases}

Then we further define

ϕ_{s}^{(e)}(X) = 1 − (1 − χ_{s}^{e})^{e}

and

ψ(X) = ∑_{s=1}^{d} c_{s}ϕ_{s}^{(e)}(X).

It is fairly easy to see that

S = ψ(A)

holds.

The function ψ is symmetric with respect to roots {c_{s}} and thus ψ is a polynomial with coefficients in k. Thus S (hence also N) is defined over k.

□

The following example shows that the k-rationality of S does not necessarily hold when we drop off the assumption on A.

Example 4.14. Let A ∈ M_{p}(𝔽_{p}(x)) be a matrix of the following form.

A = (\begin{matrix}0 & 1 \\ & ... & ... \\ & & ... & 1 \\ x & & & 0 \end{matrix})

Then the minimal polynomial of A is given by X^{p} − x. The Jordan-Chevalley decomposition of A is given by

A = x^{1/p} + (A − x^{1/p}).

Thus the decomposition is not defined over 𝔽_{p}(x).

5. generalities in finite dimensional Lie algebras

5.1. Ideals of Lie algebras.

Definition 5.1. For a linear subspace S, T of a Lie algebra L, let us define denote by [S, T] the following linear subspace of L.

(1)

[S, T] = (linear span of {[x, y]; x ∈ S, y ∈ T })

Definition 5.2. Let L be a Lie algebra over a field k. A k-linear subspace 𝔞 of L is said to be an ideal of L if

[x, y] ∈ 𝔞 (∀x ∈ L, ∀y ∈ 𝔞)

holds. This clearly is equivalent to saying that

[L, 𝔞] ⊂ L

holds.

Proposition 5.3. Let 𝔞 be an ideal of a Lie algebra L. Then

(1) 𝔞 is a sub L-module (sub representation) of L.

(2) L/𝔞 caries a natural structure of a Lie algebra.

Proof. As usual. □

5.2. Simple, semisimple, solvable, and nilpotent Lie algebras: definition.

Definition 5.4. For a Lie algebra L, let us define the following ideals of L.

(1) Comm(L) = [L, L], and inductively,

Comm^{j}(L) = Comm(Comm^{j−1}(L)).

(2) ad(L)(L) = [L, L], and inductively,

ad^{j}(L)(L) = ad(ad^{j−1}(L)).

Lemma 5.5. For any Lie algebra L and for any positive integer j, we have

Comm^{j}(L) ⊂ ad^{j}(L).

Proof. Inductively, we have

Comm^{j}(L) = [Comm^{j−1}(L), Comm^{j−1}(L)] ⊂ [L, ad^{j−1}(L)] = ad^{j}(L).

□

Definition 5.6. A Lie algebra L over a field k is said to be

(1) semisimple if it has no abelian ideals.

(2) simple if it has no non trivial ideals and dim(L) > 1.

(3) solvable if Comm^{N}(L) = 0 for some N ∈ ℤ_{>0}.

(4) nilpotent if ad(L)^{N}(L) = 0 for some N ∈ ℤ_{>0}.

Proposition 5.7. We have the following implications.

(1) Simple Lie algebras are semisimple.

(2) Nilpotent Lie algebras are solvable.

Proof. (1)is Easy. (2) follows from Lemma 5.5. □

Semisimple algebras and solvable ones are “orthogonal”. For now we only mention the following

Lemma 5.8. Non zero solvable algebra L cannot be semisimple.

Proof. Let N_{0} be a positive integer such that

Comm_{0}^{N}(L) ≠ 0, Comm^{N_{0}+1}(L) = 0.

Then Comm^{N_{0}}(L) is a non-zero abelian ideal of L. □

5.3. The radicals of Lie algebras.

Definition 5.9. A radical of a Lie algebra L is a maximal solvable ideal of L.

Lemma 5.10. Let 𝔞 be an ideal of a Lie algebra L. If L/𝔞 and 𝔞 are both solvable Lie algebras, then L is also solvable.

Proof. Since L/𝔞 is solvable, there exists a positive integer N_{1} such that

Comm^{N_{1}}(L/𝔞) = 0.

Then we obviously have

Comm^{N_{1}}(L) ⊂ 𝔞.

On the other hand, since 𝔞 is solvable, there exists a positive integer N_{2} such that

Comm^{N_{2}}(𝔞) = 0.

We thus have

Comm^{N_{1}+N_{2}}(L) = Comm^{N_{2}}(Comm^{N_{1}}(L)) ⊂ Comm^{N_{2}}(𝔞) = 0.

□

Lemma 5.11. Every Lie subalgebras and quotients of solvable Lie algebras are solvable.

Proof. Obvious. □

Lemma 5.12. Let 𝔞, 𝔟 be ideals of a Lie algebra L. If 𝔞, 𝔟 are both solvable (as Lie algebras), then 𝔞 + 𝔟 is also solvable.

Proof.

𝔞 + 𝔟/𝔟 ≅ 𝔞/𝔞 ∩ 𝔟
□

Proposition 5.13. For a finite dimensional Lie algebra L over a field k, there exists a unique maximal solvable ideal of L. So we may call it the radical of L.

Proof. Let 𝔞_{0} be a solvable ideal of L which has the maximal dimension among solvable ideals. Then for any solvable ideal 𝔟 of L, 𝔞_{0} + 𝔟 is also solvable. Thus by the choice of 𝔞_{0} we see that

𝔞_{0} + 𝔟 = 𝔞_{0}. (That is, 𝔞_{0} ⊃ 𝔟.)

Thus we see that 𝔞_{0} is the largest solvable ideal of L.

□

Corollary 5.14. Let L be a finite dimensional Lie algebra over a field k. Let 𝔯 be its radical. Then:

(1) L/𝔯 is semisimple.

(2) L is semisimple if and only if 𝔯 = 0.

(3) A quotient L/𝔞 is semisimple if and only if 𝔯 ⊂ 𝔞.

Proof. (1) follows immediately from the definition and Lemma 5.8.

(2) is also easy. (3): L/𝔞 contains

(𝔯 + 𝔞)/𝔞(≅ 𝔯/(𝔯 ∩ 𝔞))

as a solvable ideal. □

5.4. Theorem of Engel.

Lemma 5.15. Let k be a commutative ring. Let x ∈ M_{n}(k) be a nilpotent matrix. Then ad(x) is also nilpotent.

Proof. Assume x^{N} = 0. We decompose ad(x) into left and right multiplication. Namely,

ad(x) = λ(x) − ρ(x)

Then λ(x) and ρ(x) commute with each other.

ad(x)^{2N−1} = ∑_{j=0}^{2N−1} λ(x)^{j}(−ρ(x))^{2N−1−j} = ∑_{j=0}^{2N−1} λ(x^{j})ρ((−x)^{2N−1−j}) = 0.

□

Theorem 5.16 (Engel). Let V be a finite dimensional vector space over a field k. Let L be a Lie subalgebra of 𝔤𝔩(V) such that each member of L is a nilpotent matrix. Then:

(1) If dim(L) ≥ 1, then L has an ideal of codimension 1.

(2) If dim(V) ≥ 1, then L has a simultaneous 0-eigen vector v. (That is, x.v = 0 (∀x ∈ L), v ≠ 0.)

Proof. If dim(L) = 0, then there is nothing to do. We proceed by induction on dim(L). Let L_{1} be a maximal among

{(Lie subalgebras of L which is not equal to L)}.

(The set above has a {0} as a member, so it is not empty.) In view of the lemma above, we see

∀x ∈ L_{1}∃N ∈ ℤ_{>0}(ad(x)^{N}(L) = 0).

We note that an vector space L/L_{1} admits adjoint actions by L_{1}. Thus

∀x ∈ L_{1}∃N ∈ ℤ_{>0}(ad(x)^{N}(L/L_{1}) = 0).

That means, ad_{L/L_{1}}(L_{1}) ⊂ 𝔤𝔩(L/L_{1}) also satisfy the assumption of the theorem. By the induction hypothesis, we see that conclusion (2) is applicable to this case. Namely, there exists an element y_{0} ∈ L \ L_{1} such that

ad(x)(y_{0})(= [x, y_{0}]) ∈ L_{1} (∀x ∈ L_{1})

holds. Now a vector subset

L_{2} = k.y_{0} + L_{1}(⊋ L_{1})

of L is closed under Lie bracket and therefore it is a Lie subalgebra of L. By the maximality of L_{1}, L_{2} should equal to L. It is then also easy to verify that L_{1} is an ideal of L(= L_{2}). This proves (1).

To prove (2), we note that (L_{1}, V) satisfies the assumption of the theorem. So again by induction we see that L_{1} has a simultaneous 0-eigen vector. In other words,

V_{1} := ⋂_{x∈L_{1}}{v ∈ V; x.v = 0} ⊋ {0}.

Let us then consider the action of y_{0}.

v ∈ V_{1} ⟹ x.(y_{0}.v) = y_{0}.(x.v) + [x, y_{0}].v = 0 ⟹ y_{0}.v ∈ V_{1}

Thus V_{1} admits an action of y_{0}. Since y_{0} is nilpotent on V by the assumption, we see that y_{0} has at least one 0-eigen vector v_{0}(≠ 0) in V_{1}. Then v_{0} surely is a simultaneous 0-eigen vector of L.

□

Theorem 5.17 (Engel). Let V be a finite dimensional vector space over a field k. Let L be a Lie subalgebra of 𝔤𝔩(V) such that each member of L is a nilpotent matrix. Then there exists a basis e_{1}, e_{2}, …, e_{n} of V such that

L ⊂ 𝔫_{n}(k) = \{(\begin{matrix}0 & ∗ & ∗ & … & ∗ \\ 0 & 0 & ∗ & … & ∗ \\ & & ... & ∗ & ∗ \\ 0 & 0 & … & 0 & ∗ \\ 0 & 0 & … & 0 & 0 \end{matrix})\}

holds with respect to this basis. In particular, L is nilpotent.

Proof. We apply the above theorem inductively to a vector space

L, L/(k.e_{1}), L/(k.e_{1} + k.e_{2}) …

and obtain the desired basis {e_{i}}. Since the Lie algebra 𝔫_{n}(k) is nilpotent, L is also nilpotent.

□

5.5. Ideals of 𝔤𝔩_{n}(k). Recall that 𝔫_{n}(k) denotes the Lie algebra of strictly upper triangular matrices. In this subsection we denote by e_{ij} the elementary matrices. (as we have done so without even mentioning...)

Lemma 5.18. Let k be a field of characteristic p (possibly 0). Let n ∈ ℤ_{>1}. We assume that (p, n) ≠ (2, 2). Then we have

{x ∈ 𝔤𝔩_{n}(k); [x, y] ∈ k.1_{n} (∀y ∈ 𝔫_{n}(k))} = k.1_{n} + k.e_{1n}.

Proof. Let us denote by L the left hand side of the lemma. Then we trivially have L ⊃ k.1_{n}. Furthermore, for all x ∈ 𝔫_{n}(k), we see easily that

xe_{1n} = 0, e_{1n}x = 0

holds. So we have

L ⊃ k.1_{n} + k.e_{1n}

Let us prove the opposite inclusion. We take an arbitrary element x ∈ L.

For any (i, j) satisfying i < j, we have e_{ij} ∈ 𝔫_{n}(k) and thus

[e_{i,j}, x] = c1_{n} (∃c ∈ k).

The rank of the left hand side is at most 2. So c must be equal to 0 when n ≥ 3. Otherwise (n = 2), we compare the trace of the both hand sides. The trace of the left hand side is clearly zero. The trace of a scalar matrix c.1_{2} is equal to 2c. Thus c = 0 by our assumption ((p, n) ≠ (2, 2)). In either case, we have [e_{i,j}, x] = 0. Then we compute some of special cases. First, let us examine the case where i = 1, j ≥ 2. Then

0 = [e_{1j}, x] = ∑_{st}[e_{1j}, x_{st}e_{st}] = ∑_{t} x_{jt}e_{1t} − ∑_{s} x_{s1}e_{sj}

By looking at (1, u) entry of the above equation, we conclude that equations in entries

∀j∀u((j ≥ 2, j ≠ u) ⟹ x_{ju} = 0)
∀j((j ≥ 2) ⟹ x_{jj} = x_{11})

hold. Similarly, by looking at the (u, n) entry of [e_{in}, x], we conclude that equations

∀i∀u((i ≤ n − 1, i ≠ u) ⟹ x_{ui} = 0)

hold. Putting the equations all together, we conclude that x is in the right hand side of the lemma. □

As an application of the Engel’s theorem, we prove the following proposition.

Proposition 5.19. Let k be a field of characteristic p (possibly 0). Let n be a positive integer. We assume that (p, n) ≠ (2, 2). Then each ideal I of 𝔤𝔩_{n}(k) is equal to the one in the following list.

(1) 0.

(2) k.1_{n}.

(3) 𝔰𝔩_{n}(k).

(4) 𝔤𝔩_{n}(k).

Proof. The case n = 1 is trivial. So let us assume n ≥ 2.

If I ⊂ k.1_{n}, then dim_{k}(I) ≤ 1 and hence I = 0 or I = k.1_{n}. Assume now I ⊄ k.1_{n}. Let us consider the Lie algebra 𝔫_{n}(k) of strictly upper triangular matrices. Then

(L, V) = (𝔫_{n}(k), (I + k.1_{n})/k.1_{n})

satisfies the assumption of the Engel’s theorem. So there exists a nonconstant element x ∈ I such that

[x, y] ∈ k.1_{n} (∀y ∈ 𝔫_{n}(k)).

holds. By using the previous lemma, we see that x may be presented as

x = c_{0}1_{n} + c_{1}e_{1n} (∃c_{0}, c_{1} ∈ k).

Since x is non-constant, we have c_{1} ≠ 0.

I ∋ [e_{11}, x] = c_{1}e_{1n}

Thus e_{1n} belongs to I. By changing the order of the base and repeating the above argument, we conclude that

∀i∀j((i ≠ j) ⟹ e_{ij} ∈ I.)

In addition we have

I ∋ [e_{ij}, e_{ji}] = e_{ii} − e_{jj}

This clearly proves I ⊃ 𝔰𝔩_{n}(k). Since the codimension of 𝔰𝔩_{n}(k) in 𝔤𝔩_{n}(k) is 1, we have either I = 𝔰𝔩_{n}(k) or I = 𝔤𝔩_{n}(k).

□

For the sake of completeness, we deal with the case (p, n) = (2, 2). In this case, situation is a bit different.

Lemma 5.20. Let k be a field of characteristic 2. Then:

(1) Any two-dimensional vector subspace V of 𝔰𝔩_{2}(k) with V ⊃ k.1_{2} is equal to a vector space L_{[b:c]} given by an element [b : c] ∈ ℙ^{1}(k) which is defined as

L_{[b:c]} = k.1_{n} + k. (\begin{matrix}0 & b \\ c & 0 \end{matrix}) .

(2) For any element [b : c] ∈ ℙ^{1}(k), the vector space L_{[b:c]} is an ideal of 𝔤𝔩_{2}(k).

(3) In particular, 𝔱_{2}(k) = k.1_{2} + 𝔫_{2}(k) is an ideal of 𝔤𝔩_{2}(k).

Proof. (1) There exists a traceless non constant matrix x ∈ V such that

I = k.1_{n} + k.x

holds. By subtracting a constant matrix, one may easily replace x by a matrix with zero diagonals.

(2) By a direct computation we see

[(\begin{matrix}x & y \\ z & w \end{matrix}) , (\begin{matrix}0 & b \\ c & 0 \end{matrix})] = (x−w) (\begin{matrix}0 & b \\ c & 0 \end{matrix})+(bz−cy)1_{2} (x, y, z, w, b, c, ∈ k)

(Note that char(k) = 0.)

(3) 𝔱_{2}(k) = L_{[1:0]}. □

Proposition 5.21. Let k be a field of characteristic 2. Then each ideal I of 𝔤𝔩_{2}(k) is equal to the one in the following list.

(1) 0.

(2) k.1_{n}.

(3) A two-dimensional Lie algebra L_{[b:c]} defined as in the lemma above.

(4) 𝔰𝔩_{n}(k).

(5) 𝔤𝔩_{n}(k).

Proof. We divide into several cases.

(i) Case where I ⊄ 𝔰𝔩_{2}(k). In this case there exists an element x ∈ I with tr(x) ≠ 0. Putting

x = (\begin{matrix}a & b \\ c & d \end{matrix}) ,

we compute as follows

I ∋ [e_{12}, x] = [(\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) , (\begin{matrix}a & b \\ c & d \end{matrix})] = (\begin{matrix}c & a + d \\ 0 & c \end{matrix}) .

(Note char(k) = 2.) Then we have

I ∋ [e_{11}, [e_{12}, x]] = [(\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}) , (\begin{matrix}c & a + d \\ 0 & c \end{matrix})] = (\begin{matrix}0 & a + d \\ 0 & 0 \end{matrix})

Thus we see that e_{12} ∈ I. In a same way (by changing the order of the base), we obtain, e_{21} ∈ I.

1_{2} = [e_{21}, e_{12}] ∈ I.

Since tr(x) ≠ 0, we see that {1_{2}, e_{12}, e_{21}, x} spans the 𝔤𝔩_{2}(k). thus I = 𝔤𝔩_{2}(k) in this case. (ii) Case where I ⊂ 𝔰𝔩_{2}(k) and I ∩ k.1_{2} = 0. Let x be arbitrary element of I and put

x = (\begin{matrix}a & b \\ c & d \end{matrix}) .

Then by computing [e_{12}, x] as in the case (i) above, we know that c = 0. Similarly, we know b = 0. Since x is traceless, a = d also holds. So the only possibility in this case is I = 0. (iii) Case where k.1_{2} ⊊ I ⊊ 𝔰𝔩_{2}(k). By a dimension consideration, we see that dim I = 2. Then we use the above lemma. (iv) The case I = k.1_{2} or I = 𝔰𝔩_{2}(k). Excellent. There is nothing to in this case.

□

5.6. Ideals of 𝔰𝔩_{n}(k).

Proposition 5.22. Let k be a field of characteristic p (possibly 0.) Let n be a positive integer.

(1) If p |̸n, then 𝔰𝔩_{n}(k) is a simple Lie algebra.

(2) If p|n, then 𝔰𝔩_{n}(k) has a unique nontrivial ideal k.1_{n}.

Proof. Let I be an ideal of 𝔰𝔩_{n}(k). By taking trace we see immediately that

1_{n} ∈ 𝔰𝔩_{n}(k) ⟺ p|n.

Thus if I ⊂ k.1_{n}, then The only nontrivial possibility is that p|n and I = k.1_{n}.

Assume now that I ⊄ k.1_{n}. Then by an argument similar to that in Proposition 5.19, we see that

x = c_{0}1_{n} + c_{1}e_{1n} ∈ I, (∃c_{0}, c_{1} ∈ k, c_{1} ≠ 0)

holds.

(1) If p |̸n, then by taking trace we see that e_{1n} ∈ I. By permuting the basis, we see that e_{ij} ∈ I whenever i ≠ j. Thus I = 𝔰𝔩_{n}(k) in this case.

(2) If p|n, then by assumption on (p, n) we have n ≥ 3. Thus

I ∋ [e_{21}, x] = c_{1}e_{2n} ∴ e_{1n} = [e_{12}, e_{2n}] ∈ I.

So in this case also we see that I = 𝔰𝔩_{n}(k).

□

Proposition 5.23. Let k be a field of characteristic 2. Then any two dimensional Lie algebra L_{[b:c]} as in Lemma 5.20 is an ideal of 𝔰𝔩_{2}(k). Thus each ideal I of 𝔰𝔩_{2}(k) is equal to the one in the following list.

(1) 0.

(2) k.1_{n}.

(3) L_{[b:c]}

(4) 𝔰𝔩_{n}(k)

Proof. Easy exercise. □

5.7. Invariant bilinear forms and Killing forms.

Definition 5.24. A symmetric bilinear form B : L × L → k of a Lie algebra over a field k is said to be invariant if it satisfies

B([Y, X], Z) + B(X, [Y, Z]) = 0 (∀X, Y, Z ∈ L)

(which means that “the Lie derivative of B is zero”), or, equivalently,

B([X, Y], Z) = B(X, [Y, Z]) (∀X, Y, Z ∈ L)

(which means that B is “balanced”.)

Lemma 5.25. Let L be a Lie algebra over a field k. Let B be an invariant bilinear form on L. Then for any ideal I of L,

L^{⊥} = {x ∈ L; B(x, y) = 0(∀y ∈ l)}

is an ideal of L.

Proof. Easy. □

Note: We need to be a bit careful when we use the notation •^{⊥}. It is safer to clarify the “container” (L) and bilinear form ρ. So the lemma above we should have written L^{⊥_{ρ,L}} (eek) in stead of L^{⊥}.

Definition 5.26. Let (ρ, V) be a finite dimensional representation of a Lie algebra L over a field k. Then the Killing form with respect to (ρ, V) is a bilinear form on L defined by

Tr_{ρ,V} (XY) = tr_{V} (ρ(X)ρ(Y)).

The ordinary(usual) Killing formκ_{L} of L is a bilinear form on L defined as the Killing form of the adjoint representation. That is,

κ_{L}(X, Y) = Tr_{ad,L}(XY) = tr_{ad,L}(ad(X) ad(Y)).

It is easy to verify that the Killing forms defined as above are invariant.

5.7.1. functoriality of Killing forms.

Lemma 5.27. Let L be a Lie algebra over a field k. Then the followings are true.

(1) Let V be a finite dimensional representation of L. Let W be a subrepresentation of V. Then we have

Tr_{V} (xy) = Tr_{W} (xy) + Tr_{V/W} (xy).

(2) Let I be an ideal of L. Assume L is finite dimensional. Then we have

κ_{L}(x, y) = Tr_{ad,L}(xy) = Tr_{ad,I}(xy)+Tr_{ad,L/I}(xy) = Tr_{ad,I}(xy)+κ_{L/I}(x̄, ȳ)

(where ¯• denotes the class of • in L/I.) In particular, for any x, y ∈ I, we have

κ_{L}(x, y) = κ_{I}(x, y)

Proof. (2): We choose a basis B = B_{1} ∐ B_{2} of L such that B_{2} forms a basis of I. Then B̄_{1} forms a basis of L/I. Under the basis B, ad(x) may be represented by a matrix

ad_{L}(x) = (\begin{matrix}ad_{L/I}(x̄) & ∗ \\ 0 & ad_{I} x \end{matrix}) .

We obtain the result easily from this. (1): may be proved in a same manner. □

5.8. Theorem of Iwasawa.

Theorem 5.28. Let L be a Lie algebra over a field k of characteristic p ≠ 0. Then L has a finite dimensional faithful representation. More precisely, there exists a two-sided ideal I of the universal enveloping algebra U(L) such that L acts faithfully on U(L)/I.

Before proving the above theorem, we first prove the next lemma.

Lemma 5.29. Under the hypothesis of the theorem, for any x ∈ L, there exists a monic non constant polynomial f_{x}(X) ∈ k[X] such that

f_{x}(x) ∈ Z(U(L))

holds.

Proof. Let us put s = dim(L). The linear transformation ad(x) on L is represented by a matrix of size s and has therefore its minimal polynomial m_{x}: Namely, m_{x} is a monic polynomial of degree no more than s such that

m_{x}(ad(x)) = 0

holds. Let us divide X, X^{p}, X^{p^{2}}, … X^{p^{s+1}} by m_{x}(X).

X^{p^{j}} = m_{x}(X)q_{j}(X)+r_{j}(X) (deg(r_{j}) < s) (j = 1, 2, 3, …, s+1)

Then s+1 polynomials {r_{j}(X)}_{j=1}^{s+1} of degree ≤ s−1 should be linearly dependent. That means, there exists a non trivial vector (c_{j}) ∈ k^{s+1} such that

∑_{j} c_{j}X^{p^{j}} ∈ m_{x}(X)k[X]

holds. Then we have

∑_{j} c_{j}(ad(x))^{p^{j}} = 0.

Thus we conclude

ad(∑_{j} c_{j}x^{p^{j}}) = 0.

By dividing ∑_{j} c_{j}X^{p^{j}} by leading coefficient, we obtain the required polynomial f_{x}.

□

Proof. of the Theorem Let {e_{1}, e_{2}, …, e_{s}} be a basis of L. Then by the above lemma we know that there exists a set of monic non constant polynomials {f_{1}, f_{2}, …, f_{s}} such that each h_{j} = f_{j}(e_{j}) belongs to the center of U(L). Let us put d_{j} = deg(f_{j}). Then using PWB theorem we may easily see that

\{h_{1}^{c_{1}}h_{2}^{c_{2}}h_{3}^{c_{3}} … h_{s}^{c_{s}}e_{1}^{l_{1}}e_{2}^{l_{2}}e_{3}^{l_{3}} … e_{s}^{l_{s}}; \begin{matrix}c_{1}, c_{2}, c_{3}, …, c_{s} ∈ ℕ, \\ l_{1}, l_{2}, l_{3}, …, l_{s} ∈ ℕ, \\ l_{j} < d_{j}(∀j) \end{matrix}\}

forms a basis of U(L). Let us now put

I = U(L)(h_{1}, …, h_{s})

Then A = U(L)/I is a finite dimensional vector space with the base

\{e_{1}^{l_{1}}e_{2}^{l_{2}}e_{3}^{l_{3}} … e_{s}^{l_{s}}; \begin{matrix}l_{1}, l_{2}, l_{3}, …, l_{s} ∈ ℕ, \\ l_{j} < d_{j}(∀j) \end{matrix}\}

The representation ρ_{A} of L on A is faithful. Indeed, for any x ∈ L, we have

ρ_{A}(x) = 0 ⟹ x.1 = 0 in A ⟹ x ∈ I ⟹ x = 0.
□

Definition 5.30. Let L be a Lie algebra.

(1) A representation V of L is called completely reducible if it is a direct sum of reducible sub representations.

(2) L is called completely reducible if every representation of L is completely reducible.

The following remark is (at least) in the Book of Bourbaki.

Proposition 5.31. Let L be a non zero finite dimensional Lie algebra over a field k of characteristic p ≠ 0. Then L can never be completely reducible.

Proof. Let us follow the proof of the theorem of Iwasawa. By taking f_{1}^{2} instead of f_{1} in the proof, we obtain a representation A_{1} = U(L)/I with a non trivial central nilpotent z = f_{1}(e_{1}). Then we see that zA_{1} cannot have a direct complementary L-module X. For if it existed, then X should necessarily a left ideal of A_{1}. On the other hand, by decomposing 1 ∈ A_{1} we obtain

1 = x + za (∃x ∈ X∃a ∈ A_{1}).

Then x = 1 − za has an inverse (1 + za). This implies that

X ⊃ A_{1}x ⊃ A_{1}.

which is a contradiction. □

5.9. Cartan’s criterion for solvability(Ccs). Cartan’s criterion relates several properties (semi simplicity, solvability) of Lie algebras with properties of their invariant bilinear forms.

To prove it we need some study on bilinear forms.

Definition 5.32 (in this subsection only). For any free module R^{t} over a ring R, We denote by ⟨⟩_{R} the “usual inner product”. That is,

⟨v, w⟩_{R} = ∑_{i} v_{i}w_{i}.

The first thing we do is to observe the property of this inner product when the base ring R is a “real field”. (Since we only need it for the case R = ℚ, we omit the definition of a real field and describe the following lemma only when R = ℚ.)

Lemma 5.33. Let W = ℚ^{t} be a vector space. Let b_{1}, b_{2}, … b_{s} ∈ W. Let B be a s × s matrix defined by

B = (⟨b_{i}, b_{j}⟩_{ℚ}).

Then the determinant of B is equal to 0 if and only if {b_{j}}_{j=1}^{s} are linearly dependent over ℚ.

Proof. Assume {b_{j}}_{j=1}^{s} are linearly dependent over ℚ. Then there exists a non trivial vector (c_{1}, c_{2}, … c_{s}) ∈ ℚ^{s} such that

∑_{j=1}^{s} c_{j}b_{j} = 0

holds. Thus

(c_{1}, c_{2} …, c_{s})B = 0

So B is a degenerate matrix which implies that det(B) = 0.

Let us now prove the opposite implication. Assume det(B) = 0. Then there exists a non trivial vector (c_{1}, c_{2}, …, c_{s}) such that

(c_{1}, c_{2} …, c_{s})B = 0

holds. Let us put

v = ∑_{j} c_{j}b_{j}.

Then we see that ⟨v, v⟩_{ℚ} = 0 and hence v = 0. (Note that for this implication we have used the fact that ℚ is a “real field”.) Thus {b_{j}} are linearly dependent over ℚ. □

The next task is to compare ℚ with other field.

Definition 5.34. For any subset S of a ℤ-module W_{ℤ}, Let us put

MG_{S} = max{| det_{l×l}(⟨b_{i}, b_{j}⟩_{ℤ})|b_{1}, … b_{l} ∈ S}.

(“The maximum modulus of Gram determinants”.) We denote by S_{k} the subset of W_{k} = W_{ℤ} ⊗_{ℤ} ⊗k defined by

S_{k} = {x ⊗ 1 ∈ W ⊗_{ℤ} k; x ∈ S}.

Lemma 5.35. Let S be a finite subset of a free module

W_{ℤ} = ℤ^{t} = {(v_{1}, v_{2}, …, v_{t}); v_{j} ∈ ℤ(∀j)}

over ℤ. Let k be a field of characteristic p. We assume either p = 0 or p > MG_{S} holds. Then we have

(a ∈ W_{k}, (a^{⊥} ∩ S_{k})^{⊥⊥} ∋ a) ⟹ a = 0

Proof. Assume a ≠ 0. Since the inner product ⟨•, •⟩_{k} is non degenerate on W_{k}, we see that (a^{⊥} ∩ S_{k})^{⊥⊥} is equal to the k-vector space spanned by (a^{⊥} ∩ S_{k}). Thus there exists a set of linearly independent vectors {b_{j}} ⊂ S_{k} so that we may write down a as

a = ∑_{i} a_{i}b_{i} (a_{i} ∈ k).

Then by the assumption on a, we see that

∑ a_{i}⟨b_{i}, b_{j}⟩_{k} = 0 (∀j)

Thus

det(⟨b_{i}, b_{j}⟩_{k}) = 0

which is equivalent to

(1)  p| det(⟨b_{i}, b_{j}⟩_{ℤ}).

Note on the other hand that b_{j} are linearly independent over ℤ. Thus

det(⟨b_{i}, b_{j}⟩_{ℤ}) ≠ 0

By the definition of MG_{S}, we see that

(2)  0 < | det(⟨b_{i}, b_{j}⟩_{ℤ})| ≤ MG_{S}

which contradicts to the condition (1). □

Definition 5.36. For any positive integer n, and for any ring k, we denote by Diag_{n}(k) the set of diagonal matrices in M_{n}(k). For any vector a = (a_{i}) ∈ k^{n}, we denote by diag(a) the diagonal matrix diag(a) = diag(a_{1}, …, a_{n}). Note that for any ring k, the restriction of the Killing form of 𝔤𝔩_{n} coincides with the “usual” inner product with this identification. That is,

tr(diag(a) diag(b)) = ⟨a, b⟩_{k}.

We define the following subset of Diag_{n}(ℤ).

S^{[n]} = {(diag((e_{i} − e_{j}) − (e_{m} − e_{l})); i, j, m, l ∈ {1, 2, 3, …, n}}.

(where the vectors {e_{i}}_{i=1}^{n} are elementary vectors.) We note that an obvious estimate

MG_{S^{[n]}} ≤ 4^{n}n!

holds.

Lemma 5.37. Let n be a positive integer. Let a = (a_{1}, …, a_{n}), b = (b_{1}, …, b_{n}) ∈ k^{n}.

If b satisfies b ∈ (a^{⊥}∩S^{[n]})^{⊥}, then there exist polynomials f, g ∈ k[X] such that

f(diag(a)) = diag(b), g(ad(diag(a))) = ad(diag(b))

holds.

Proof. Let us denote by ϵ_{ijlm} the vector

ϵ_{ijlm} = (e_{i} − e_{j}) − (e_{l} − e_{m}).

We first find a map f_{0} from Λ_{a} = {a_{i}; i = 1, 2, …, n} to Λ_{b} = {b_{i}; i = 1, 2, …, n} such that

f_{0}(a_{i}) = b_{i}.

Such a thing exists (is “well defined”) if and only if

∀i∀j(a_{i} = a_{j} ⟹ b_{i} = b_{j})

holds. This condition is equivalent to the condition

∀i∀j(a ⊥ ϵ_{ij11} ⟹ b ⊥ ϵ_{ij11})

which holds by the assumption on b. Thus we see that f_{0} exists. On the other hand, by using Lagrange interpolation formula we see that there exists a polynomial f ∈ k[X] such that f|_{Λ_{a}} = f_{0}. Then we have

f(diag(a)) = diag(b).

The adjoint action of a diagonal matrix diag(a) is represented by a diagonal matrix (a_{i} −a_{j})_{i,j}. Thus an argument similar to the one above proves the existence of g.

□

Proposition 5.38 (Cartan). Let V be an n dimensional vector space over a field k of characteristic p. We assume that either p = 0 or p > MG_{S^{[n]}} holds. Let L be a Lie subalgebra of 𝔤𝔩(V). If the Killing form of L with respect to V is identically equal to 0, then L is solvable.

Proof. We may assume that k is algebraically closed. Let us take an element x ∈ [L, L]. Then we have

(ad x_{s})(L) = (ad x)_{s}(L) ⊂ L

Let us now diagonalize x_{s} and write x_{s} = diag(a). Let us take arbitrary b ∈ (a^{⊥} ∩ S^{[n]})^{⊥}. By the lemma above we see that there exist polynomials f, g ∈ k[X] such that

f(diag(a)) = diag(b), g(ad(diag(a))) = ad(diag(b))

holds. For any w = ∑_{l}[y_{l}z_{l}] ∈ [L, L], we have:

tr(diag(b)[∑_{l} y_{l}z_{l}]) = ∑_{l} tr([diag(b), y_{l}]z_{l})

= ∑_{l} tr(ad(diag(b)).y_{l} z_{l}) = ∑_{l} tr(g(ad(diag(a))).y_{l} z_{l})

= ∑_{l} tr(g(ad(x_{s})).y_{l} z_{l}) ∈ ∑_{l} tr(LL)

= 0

That means, tr(diag(b)w) = 0. In particular, we have

tr(diag(b)x) = 0.

Since diag(b) = f(diag(a)) = f(x_{s}) is a polynomial in x, it commutes with x_{s} and with x_{n}. thus

diag(b)x = (diag(b)x_{s}) + (diag(b)x_{n})

gives the Jordan-Chevalley decomposition of diag(b)x. Therefore,

0 = tr(diag(b)x) = tr(diag(b)x_{s}) = tr(diag(b) diag(a)) = ⟨b, a⟩_{k}

thus b ⊥ a.

To sum up, we have shown

(a^{⊥} ∩ S)^{⊥} ∋ b ⟹ b ∈ a^{⊥}.

In other words,

(a^{⊥} ∩ S)^{⊥⊥} ∋ a

which is equivalent to saying that a is a linear combination of elements in (a^{⊥} ∩ S).

In view of Lemma 5.35, we see that a = 0. So x = x_{s} + x_{n} = x_{n} is a nilpotent element.

By the theorem of Engel, we conclude that [L, L] is nilpotent. Thus L is solvable (since we have shown that L/[L, L] and [L, L] are solvable).

□

Definition 5.39. We say that the Cartan’s criterion for solvability (Ccs) holds for a linear Lie algebra L ⊂ (𝔤𝔩_{n}(k)) over a field k if it satisfies the following condition. (Ccs) If the Killing form on L associated to k^{n} is identically zero, L is solvable.

Let n be a positive integer. We denote by Ccs(n) the set of p such that Ccs holds for any Lie algebra L of dimension less than or equal to n for any field k of characteristic p.

Ccs(n) = {p; Ccs holds for any(L ⊂ 𝔤𝔩_{n}(k)) provided char(k) = p, }

Corollary 5.40 (of Proposition). Let n be a positive integer. Then:

(1) 0 ∈ Ccs(n)

(2) For any prime p which is larger than MG_{S^{[n]}}, we have p ∈ Ccs(n).

(3) In particular, for any prime p which is larger than 4^{n}n!, we have p ∈ Ccs(n).

Note:

The estimate given in the above corollary is presumably far from the best one.

Proposition 5.41. Let n be a positive integer. Let k be a field of characteristic p ∈ Ccs(n). Let L be a Lie algebra over k whose dimension is less than or equal to n. If the usual Killing form κ = Tr_{ad,L} of L is identically equal to zero, then L is solvable. In particular, if p > 4^{n}n! or p = 0, then L is solvable if its usual Killing form is identically equal to zero.

Proof. Apply the definition to

L/(center ofL) ↪ 𝔤𝔩_{n}(L).
□

5.10. Cartan’s criterion for semisimplicity.

Definition 5.42. We call a Lie algebra L over k non degenerate if the Killing form κ_{L} of L is non degenerate.

Lemma 5.43. Every non degenerate Lie algebra L over a field k is semisimple.

Proof. Assume that there exists a non trivial abelian ideal A of L. Let y_{0} be a non zero element of A. Then for any x ∈ L, z = ad(x) ad(y_{0}) is nilpotent. Indeed,

z(L) = ad(x) ad(y_{0})(L) = ad(x)([y_{0}, L]) ⊂ ad(x)(A) ⊂ A,

z^{2}(L) = ad(x) ad(y_{0})(z(L)) ⊂ ad(x) ad(y_{0})(A) = ad(x)[y_{0}, A] = 0. Thus κ(x, y_{0}) = tr(ad(x) ad(y_{0})) = 0 for for any x ∈ L. That means, y_{0} ∈ L^{⊥}. This is contrary to the assumption on L. □

Proposition 5.44. Let n be a positive integer. Let k be a field of characteristic p ∈ Ccs(n). Let L be a Lie algebra of dimension n. Then the following conditions are equivalent:

(1) L is semisimple.

(2) L is non degenerate.

(3) L is a direct sum of simple ideals.

Proof. ((1) ⟹ (2)): Assume L is semisimple. Let us take an ideal I = L^{⊥} of L. Then the Killing form on I is identically equal to zero. since dim(I) ≤ n, I is a solvable algebra. Since L is semisimple, this implies I = 0. ((2) ⟹ (1)): holds (regardless of the base field) in view of the previous lemma. ((3) ⟹ (2)): We see that simple algebras are non degenerate in view of the argument above. Thus L is also non degenerate. ((2) ⟹ (3)): Let H be a nontrivial ideal of L. Then H ∩ H^{⊥} is an abelian ideal of L. Indeed, for any y, z ∈ H ∩ H^{⊥} and for any x ∈ L, we have

κ(x, [y, z]) = κ([x, y], z) ∈ κ(H, H^{⊥}) = 0

So that [y, x] ∈ L^{⊥} = 0. On the other hand, by the previous lemma we see that H ∩H^{⊥} is semisimple and so we have H ∩H^{⊥} = 0. Accordingly we have L = H ⊕ H^{⊥}.

□

5.11. examples.

Example 5.45. Let k be a field of characteristic p (possibly 0).

𝔤𝔩_{n}(k)

is a Lie algebra with the Killing form

κ_{𝔤𝔩_{n}(k)}(x, y) = tr((λ(x) − ρ(x))(λ(y) − ρ(y)))
= tr(λ(xy)) + tr(ρ(xy)) − tr(λ(x)ρ(y) − tr(λ(y)ρ(x))
=2n tr(xy) − 2 tr(x) tr(y).

𝔰𝔩_{n}(k) is an ideal of 𝔤𝔩_{n}(k) and so its Killing form is given by

κ_{𝔰𝔩_{n}(k)}(x, y) = 2n tr_{k^{n}}(xy).

If p |̸2n, then the Killing form is easily seen to be non-degenerate. so 𝔰𝔩_{n}(k) is a non-degenerated Lie algebra in this case. In this way we see that it is a semisimple Lie algebra. The Lie algebra is actually simple as we have shown in Proposition 5.22.

Example 5.46. Let p be an odd prime. Let k be a field of characteristic p. Then we have shown in Proposition 5.19 that the only non trivial ideals of 𝔤𝔩_{p}(k) are 𝔰𝔩_{p}(k) and k.1_{p}. So we see that

L = 𝔤𝔩_{p}(k)/k.1_{p}

is a semisimple Lie algebra (as it has no proper abelian ideals). It has a unique nontrivial ideal

M = 𝔰𝔩_{p}(k)/k.1_{p}.

Thus L cannot be a direct sum of simple Lie algebras.

5.12. Weyl’s theorem on complete reducibility.

Definition 5.47. Let L be a finite dimensional Lie algebra. Let B be a non-degenerate invariant bilinear form on L. Then we define the Casimir element C_{B} ∈ U(L) with respect to B by

C_{B} = ∑_{i} x_{i}x^{(i)}

where {x_{i}} is a basis of L, and {x^{(i)}} is the dual basis of the basis {x_{i}} with respect to B.

Proposition 5.48. Under the same assumption of the definition above, we have the following facts.

(1) The Casimir operator C_{B} is independent of the choice of the basis {x_{i}} of L.

(2) C_{B} commutes with L. So it is in the center of U(L).

Proof. (1): easy exercise in linear algebra. (2): For any a ∈ L, let us write the adjoint action of a on L by using the basis {x_{i}}. Namely,

[a, x_{i}] = ∑_{j} c_{i}^{(j)}(a)x_{j} (c_{i}^{(j)}(a) ∈ k).

Then the constants {c_{i}^{(l)}(a)} (“structure constants”) are expressed in terms of B as follows.

B(x^{(l)}, [a, x_{i}]) = ∑_{j} c_{i}^{(j)}(a)B(x^{(l)}, x_{j}) = c_{i}^{(l)}(a)

We note that from the invariance of B, we have

B([x^{(l)}, a], x_{i}) = c_{i}^{(l)}(a),

so that we have a dual expression

[x^{(l)}, a] = ∑_{i} c_{i}^{(l)}(a)x^{(i)}.

Then we compute as follows.

[a, C_{B}] = ∑_{i}[a, x_{i}]x^{(i)} + ∑_{i} x_{i}[a, x^{(i)}]
= ∑_{i} ∑_{j} c_{i}^{(j)}(a)x_{j}x^{(i)} − ∑_{i} ∑_{j} c_{j}^{(i)}(a)x_{i}x^{(j)} = 0.
□

Definition 5.49. Let L be a finite dimensional Lie algebra. Let V be a finite dimensional L-module. Tr_{V} (••) with respect to V. We assume that the Killing form Tr_{V} (••) with respect to V is non degenerate. Then we define the Casimir element with respect to V by

C_{V} = C_{Tr_{V}}

where {x_{i}} is a basis of L, and {x^{i}} is the dual basis of the basis {x_{i}} with respect to the Killing form Tr_{V}.

Lemma 5.50. Let k be a field of characteristic p. Let L be a ndimensional semisimple Lie algebra over a field k. Let V be a mdimensional L-module. Let I be the kernel of the representation ρ_{V} associated to V. We assume p ∈ Ccs(n) ∩ Ccs(m). Then the Killing form Tr_{V} on L/I is non degenerate.

Proof. L is semisimple and p ∈ Ccs(n) so L is semisimple. L/I is also non-degenerate so L/I is semisimple. We may thus assume I = 0. An ideal

J = L^{⊥_{Tr_{V}}}

of L is a solvable ideal. Since L is semisimple and p ∈ Ccs(m), we have by J = 0. That means, Tr_{V} is non-degenerate on L.

□

Lemma 5.51. Let k be a field of characteristic p (which may be 0). Let (L, W ⊂ V) be a triple which satisfies the following conditions.

(1) L is a semisimple Lie algebra over k.

(2) V is a finite dimensional irreducible L-module.

(3) W is an L-submodule of V of codimension 1.

(4) p ∈ Ccs(dim(L)) ∩ Ccs(dim(V)).

Then the exact sequence

0 → W → V → V/W → 0

splits. In other words, there exists a 1-dimensional L-submodule X of V which is complementary to W.

Proof. Since the question of existence of X is described in terms of existence of a solution of a set of linear equations, we may assume that k is algebraically closed. Let us denote by ρ_{V} the representation of L associated to V. Then by replacing L by L/ ker(ρ_{V}) if necessary, we may assume that the representation ρ_{V} is faithful.

Note that since L is semisimple, it acts on V/W trivially.

Let us first treat the case where W is irreducible. Let c = c_{V} be a Casimir element with respect to V. Since L is acts on V/W trivially, c|_{V/W} is equal to zero. Thus

dim(V) = tr_{V} (c) = tr_{W} (c|_{W}) + tr_{V/W} (c|_{V/W}) = tr_{W} (c|_{W}).

In particular, c|_{W} is not equal to zero. On the other hand, by Schur’s lemma, c|_{W} is equal to a scalar λ ∈ k. Thus X = Ker(c) is a required object in this case.

We next come to general case. Let W_{1} be the maximal proper Lsubmodule of W. Then we see that (L, W/W_{1} ⊂ V/W_{1}) satisfies the assumption of the lemma with W/W_{1} irreducible. By the argument above, we therefore see that there exists an L-submodule Y which contains W_{1} as a submodule of codimension 1 such that

V/W_{1} = Y/W_{1} ⊕ W/W_{1}

holds. Since (L, W_{1} ⊂ Y) also satisfies the assumption of the lemma with dim(Y) < dim(V), we deduce by induction that the lemma holds in general.

□

Lemma 5.52. Let L be a Lie algebra over a commutative ring k. Then for any L-modules V, W, each of the vector spaces

Hom_{k -linear}(V, W)

and

V ⊗_{k} W

admits a structure of L-module. Namely,

(x.f)(v) = x.(f(v))−f(x.v) (∀x ∈ L, ∀f ∈ Hom_{k -linear}(V, W)∀v ∈ V),

(x.v ⊗_{k} w) = (x.v) ⊗ w + v ⊗_{k} (x.w) (∀x ∈ L, ∀v ∈ V, ∀w ∈ W).

Proof. Easy. □

Theorem 5.53 (Weyl). Let k be a field of characteristic p (which may be 0). Let L be a non degenerate Lie algebra over k. Let V be a finite dimensional irreducible L-module. Assume p ∈ Ccs(dim(V)^{2}). Then V is completely reducible.

Proof. Let us define the following L-modules.

V_{1} = {f ∈ Hom_{k -linear}(V, W); f|_{W} ∈ k.1_{W} }
W_{1} = {f ∈ Hom_{k -linear}(V, W); f|_{W} = 0}

Then it is easy to see that the triple (L, W_{1} ⊂ V_{1}) satisfies the assumption of Lemma 5.51. We therefore have an element f ∈ Hom_{k -linear}(V, W) which satisfies the following conditions.

(1) f|_{W} ∈ k.1_{W}.

(2) f|_{W} ≠ 0.

(3) x.f = 0 (∀x ∈ L) (In other words, f is a L-linear homomor-

phism). by dividing by a suitable element in k, we may assume f|_{W} = 1_{W}. Then f gives a splitting of the exact sequence

0 → W → V → V/W → 0.
□

5.13. Semi direct products of Lie algebras.

Definition 5.54. Let L be a Lie algebra over a commutative ring k. Then:

(1) A (k-linear) derivation of L is a k-linear map D : L → L such that it obeys the following “Leibniz rule”.

D([x, y]) = [Dx, y] + [x, Dy] (∀x, y ∈ L).

(2) We denote by Der_{k}(L) the set of all derivations of L.

Lemma 5.55. (1) Any derivation D of a Lie algebra L is lifted

to a derivation on the universal enveloping algebra U(L).

(2) Der_{k}(L) forms a Lie algebra under the usual k-linear structure and the usual commutator as the bracket product.

Definition 5.56. Let L_{1}, L_{2} be Lie algebras over a commutative ring k. we say “L_{1} acts on L_{2} as a derivation” if there is given a Lie algebra homomorphism

π : L_{1} → Der_{k}(L_{2}).

If the action π is obvious in context, we shall simply denote x.y instead of π(x).y.

Definition 5.57. Let L_{1}, L_{2} be Lie algebras over a commutative ring k. Assume there is given an action π of L_{1} on L_{2}. Then we define a semi direct product L_{1} ⋉_{π} L_{2} of L_{1} and L_{2} by introducing the k-module L_{1} ⊕ L_{2} with the following bracket product.

[(x_{1}, x_{2}), (y_{1}, y_{2})] = ([x_{1}, y_{1}], [x_{2}, y_{2}] + x_{1}.y_{2} − y_{1}.x_{2}).

Note that (1) L_{1} and L_{2} are (identified with) subalgebras of L_{1} ⋉_{π} L_{2}.

(2) Further more, L_{2} is an ideal of L_{1} ⋉_{π} L_{2}.

(3) For x ∈ L_{1} and y ∈ L_{2}, we have

[x, y]_{L_{1}⋉L_{2}} = x.y.

5.14. Levi decomposition.

Definition 5.58. Let L be a Lie algebra over a field k. Let R be the radical of L. A Levi-subalgebra of L is a subalgebra L_{1} of L such that L is a direct sum of L_{1} and R as a vector space over k.

We have the following obvious lemma.

Lemma 5.59. Let L be a Lie algebra over a field k. Let R be the radical of L. Let L_{1} be a Levi-subalgebra of L. Then:

(1) L ≅ L_{1} ⋉ R.

(2) L_{1} ≅ L/R. In particular the isomorphism class of L_{1} is unique.

□

Lemma 5.60. Let n be a positive integer. Let L be an n-dimensional Lie algebra over a field k of characteristic p ∈ Ccs(n^{2}). Assume the radical R of L is abelian. Then:

(1) R = L^{⊥}. (We equip L with the usual Killing form.)

(2)

Hom_{k -linear}(L, R)

admits an action α of L. Namely,

(α(x).φ)(y) = [x, φ(y)] − φ([x, y]) (x ∈ L, y ∈ L).

(3) For any x ∈ L, we have (α(x).φ))|_{R} = 0.

(4) For any x ∈ R, α(x) is nilpotent.

(5)

V_{1} = {φ ∈ Hom_{k -linear}(L, R); φ|_{R} ∈ k · id_{R}}

is an R-submodule of Hom_{k -linear}(L, R).

(6) If R is not equal to zero, then

V_{2} = {φ ∈ Hom_{R -module}(L, R); φ|_{R} ∈ k · id_{R}} ≠ 0.

(7) There exists a Levi subalgebra L_{1} of L.

Proof. (1) Since L^{⊥} has a trivial Killing form, it is solvable. (Cartan’s criterion.) Thus, by the maximality of R, we have

R ⊃ L^{⊥}.

On the other hand, let us take an arbitrary x ∈ R, then for any y, z ∈ L, we have

ad(x) ad(y)z = [x, [y, z]] ∈ [x, R] ⊂ R,
(ad(x) ad(y))^{2}z ∈ [x, [y, R]] ⊂ [x, R] = 0 (Since R is abelian.)

So κ_{L}(x, y) = tr_{L}(ad(x) ad(y)) = 0 (as a trace of a nilpotent element.) Thus

R ⊂ L^{⊥}

also holds. (2):follows from the general theory. (3):follows easily from the definition of α. (4):Let us take an arbitrary x ∈ R. For any y ∈ L, we have by using(2)

((α(x)^{2}).φ)(y) = [x, (x.φ)(y)] − (x.φ)([x, y]) ∈ [R, R] − (x.φ)(R) = 0.

Thus α(x)^{2} = 0. (5): follows clearly from (3). (6): follows from (4),(5) and Engel’s theorem. (We need to note that

V_{2} = {φ ∈ V_{1}; ∀x ∈ R(α(x).φ = 0)}

holds.) (7): The action of R on V_{2} is equal to zero. So V_{2} admits an action by L/R, which is semisimple. Now consider the following exact sequence of L/R-modules.

0 → V_{0} → V_{2} →^{|_{R}} k → 0

where V_{0} is the kernel of the restriction map. By a special case of Weyl’s theorem on complete reducibility (Lemma 5.51), We see that the sequence splits. (Since we assumed p ∈ Ccs(n^{2})). This implies that there exists an element φ_{0} ∈ V_{2} such that

α(x).φ_{0} = 0, φ_{0}|_{R} = id_{R} .

Thus φ_{0} gives a splitting of the injection R ⊂ L. A Levi subalgebra of L is obtained by putting

L_{1} = {x − φ_{0}(x); x ∈ L}.
□

Theorem 5.61 (Levi decomposition of a Lie algebra). Let n be a positive integer. Let p ∈ Ccs(n^{2}). Let L be a n-dimensional Lie algebra over a field of characteristic p. Then L has a Levi subalgebra L_{1}. In other words, L may be expressed as a semi direct product

L = L_{1} ⋉ R

where L_{1} is a semisimple (Levi) subalgebra of L, and R is a solvable (radical) ideal of L.

Proof. If R = 0, then we only need to set L_{1} = L. So let us assume R ≠ 0. Let us put

R_{1} = [R, R].

Then from the definition, we R/R_{1} is an abelian Lie algebra. It is also easy to verify that R_{1} is an ideal of L. (R_{1} is a characteristic ideal of R). We apply the preceding lemma for R/R_{1} ⊂ L/R_{1} to obtain a Levi subalgebra M/R_{1} of L/R_{1}. Then M satisfies the following relations.

M + R = L, M ∩ R = R_{1}.

Since R is solvable (and we have assumed R ≠ 0), we see that dim(M) is strictly smaller than dim(L). By induction M have a Levi subalgebra M_{1}. Then it is clear that M_{1} is a Levi subalgebra of L.

□

5.15. Abstract Jordan Chevalley decomposition.

Proposition 5.62. Let n be a positive integer. Let L be a ndimensional semisimple Lie algebra over a field k of characteristic p ∈ Ccs(n^{4}). Then any derivation D ∈ Der_{k}(L) of L is inner. That is, there exists an element x = x_{D} such that

D(y) = [x_{D}, y] = ad(x_{D})(y).

Proof. Der_{k}(L) is itself a Lie algebra. Sending each element x of L to its “inner derivation” ad(x), we obtain a Lie algebra homomorphism

ad : L → Der_{k}(L)

We note that dim(Der_{k}(L)) ≤ dim(L)^{2}, and that ad may be viewed as a homomorphism of L-modules. (L acts on Der_{k}(L) via ad. Namely,

x.D = ad(x).D = [ad(x), D] = [x, D•] − D([x, •]) = − ad(D.x)

holds for any x ∈ L and for any D ∈ Der_{k}(L).) By the Weyl’s theorem on complete reducibility, we see that there exists a direct sum decomposition

Der_{k}(L) = ad(L) ⊕ X

of L-modules. Then for any D ∈ X and for any x ∈ L, we see that

x.D(= − ad(D.x)) ∈ X ∩ ad(L) = 0.

So D = 0. That means, X = 0.

□

Proposition 5.63. Let n be a positive number Let k be a separably closed field of characteristic p ∈ Ccs(n^{4}). We assume further that n is invertible in k. (This assumption is provided just in case: it probably is not necessary because the assumption p ∈ Ccs(n^{4}) is presumably much stronger.) Let L ⊂ 𝔤𝔩_{n}(k) be a linear semisimple Lie algebra. We assume that the representation L^{↷}k^{n} is irreducible. Then for any element x ∈ L, its semisimple part x_{s} and its nilpotent part x_{n} in 𝔤𝔩_{n}(k) lies in L.

Proof. We may assume k is algebraically closed. Let x ∈ L It is enough to prove x_{n} ∈ L. There exists a polynomial f ∈ k[X] such that x_{n} = f(x). Thus we see

ad x_{n}(L) ⊂ L.

Thus ad x_{n} is a derivation of L. By the preceding lemma we see that there exists an element y ∈ L such that

ad x_{n} = ad y

By Schur’s lemma, we see that there exists a constant c ∈ k such that

x_{n} = y + c · 1_{n}.

Let us compute traces of both hand sides. Since L = [L, L] (L has no non-trivial ideals.), we have tr(y) = 0. Since x_{n} is nilpotent, we have tr(x_{n}) = 0. Thus we conclude c = 0 (as we assumed n is invertible in k.) □

Proposition 5.64. Let n be a positive integer. Let L be a semisimple Lie algebra over a separably closed field k of characteristic p ∈ Ccs(n^{4}). Let V_{1} = (V_{1}, π_{1}), V_{2} = (V_{2}, π_{2}) be faithful irreducible representations of L with dimensions less than or equal to n. Then for any x ∈ L, the Jordan Chevalley decomposition of x

x = x_{s}^{(1)} + x_{n}^{(1)}

with respect to V_{1} and that

x = x_{s}^{(2)} + x_{n}^{(2)}

with respect to V_{2} coincides.

Proof. We consider a faithful representation (V, π) = (V_{1} ⊕V_{2}, π_{1} ⊕ π_{2}). For any x ∈ L,

π(x) = (\begin{matrix}π_{1}(x) & 0 \\ 0 & π_{2}(x) \end{matrix}) = (\begin{matrix}π_{1}(x_{s}^{(1)}) & 0 \\ 0 & π_{2}(x_{s}^{(2)}) \end{matrix})+(\begin{matrix}π_{1}(x_{n}^{(1)}) & 0 \\ 0 & π_{2}(x_{n}^{(2)}) \end{matrix})

satisfies the requirement for the Jordan Chevalley decomposition so by the uniqueness we see

π(x)_{s} = (\begin{matrix}π_{1}(x_{s}^{(1)}) & 0 \\ 0 & π_{2}(x_{s}^{(2)}) \end{matrix}) , π(x)_{n} = (\begin{matrix}π_{1}(x_{n}^{(1)}) & 0 \\ 0 & π_{2}(x_{n}^{(2)}) \end{matrix}) .

Now we argue in a same way as in the proof of the previous proposition and see that there exists a unique element y ∈ L such that

ad(x_{n}) = ad(y)

holds. By comparing entries, we obtain

ad(π_{1}(y)) = ad(π_{1}(x_{n}^{(1)})), ad(π_{2}(y)) = ad(π_{2}(x_{n}^{(2)})).

Since L has trivial center, we have

π_{1}(y) = π_{1}(x_{n}^{(1)}), π_{2}(y) = π_{2}(x_{n}^{(2)}).

Thus y = x_{n}^{(1)} = x_{n}^{(2)}. □

Definition 5.65. Let n be a positive integer. Let L be an ndimensional semisimple Lie algebra over a separably closed field k of characteristic p ∈ Ccs(n^{4}). Then the abstract Jordan Chevalley decomposition of x is an decomposition

x = x_{s} + x_{n} (x_{s}, x_{n} ∈ L)

such that

ad(x) = ad(x_{s}) + ad(x_{n})

is the Jordan Chevalley decomposition.

Proposition 5.66. Let n be a positive integer. Let L be an ndimensional Lie algebra over a separably closed field k of characteristic p ∈ Ccs(n^{4}) Then the abstract Jordan Chevalley decomposition of x exists. If furthermore there is given a m-dimensional representation (V, π) of L and p ∈ Ccs(m^{4}), then

π(x) = π(x_{s}) + π(x_{n})

gives the Jordan Chevalley decomposition of x.

Proof. Easy exercise. (Be sure to use Weyl’s theorem of complete reducibility. By taking quotient by a certain ideals (kernels of representations) one may reduce the proposition to a case where L is semisimple and π is faithful and irreducible. )

□

References

[1] J. E. Humphreys, Introduction to Lie algebras and representation theory, Springer-Verlag, 1972.