The main theme of the last chapter has been the analogy between Lie algebras and groups. In this chapter we pursue another idea, namely, relations between Lie algebras and associative algebras. We consider an associative algebra 
—usually the algebra of linear transformations in a finite dimensional vector space—and a subalgebra 
of L. We are interested in studying relations between the structure of and of the subalgebra * of generated by . We study this particularly for solvable or nilpotent.
The results we obtain in this way include the classical theorems of Lie and Engel on solvable Lie algebras of linear transformations and a criterion for complete reducibility of a Lie algebra of linear transformations. We introduce the notion of weight spaces and we establish a decomposition into weight spaces for the vector space of a “split” nilpotent Lie algebra of linear transformations. These results will play an important role in the structure theory of the next chapter.
Our first results can be established for subsets of an associative algebra which are more general than Lie algebras. It is not much more difficult to treat these more general systems. Moreover, occasionally these are useful in the study of Lie algebras themselves.
DEFINITION 1. A subset 
of an associative algebra over a field Φ is called weakly closed if for every ordered pair (a, b), a, b ∈ , there is defined an element γ(a, b) ∈ Φ such that ab + γ(a, b)ba ∈ . We assume the mapping (a, b) → γ(a, b) is fixed and write a × b = ab + γ(a, b)ba. A subset 
of is called a subsystem if c × d ∈ for every c, d ∈ and is a left ideal (ideal) if a × c ∈ (a × c and c × a ∈ ) for every a ∈ , c ∈ .
(1) Any subalgebra of L is weakly closed in relative to γ(a, b) ≡ – 1.
(2) If 
is the Lie algebra with basis (e, f, h) such that [ef] = h, [eh] = 2e, [fh] = –2f then = Φe ∪ ρf ∪ Φh is a subsystem of .
(3) The set of symmetric matrices is weakly closed in the algebra Φn of n × n matrices if we take γ(a, b) = 1.
(4) Let = 
∪ 
where is the set of symmetric matrices and is the set of skew matrices. Define γ(a, b) = 1 if a and b are symmetric and γ(a, b) = –1 otherwise. Then is weakly closed and is an ideal in .
(5) If γ(a, b) ≡ 0 we have a multiplicative semigroup in .
DEFINITION 2. If is a subset of an associative algebra (an algebra = associative algebra with 1) we denote by *(†) the subalgebra of (subalgebra containing 1) generated by We call *(†) the enveloping associative algebra (enveloping algebra) of (in ).
We shall now note some properties of weakly closed systems which will be needed in the proof of our main theorem on such sets.
I. If W is an element of a weakly closed system , then 
= {W}* 
is a subsystem of such that * = {W}*.
Proof: The enveloping associative algebra {W}* is the algebra of polynomials in W with constant terms 0. If W1 and W2 are two such polynomials then Wx × W2 is a polynomial. Hence = {W}* is a subsystem. Since ∋ W, * ⊇ {W}*. Since ⊆ {W}* and the latter is a subalgebra, * ⊆ {W}*. Hence * = {W}*.
II. If 
is a subsystem of and W is an element of such that B × W ∈ * for every B ∈ then
Proof : The elements of * are linear combinations of monomials B1 B2 … Br, Bi ∈ . If B ∈ then BW = –γ(B W)WB + B × W ∈ W* + *. Induction on r now shows that if Bi ∈ then B1, … Br W ∈ * + *. This proves (1).
III. Let be a subsystem of such that * is nilpotent and * ≠ *. Then there exists a W ∈ such that W 
* but B × W ∈ * for every B in .
Proof: The assumption * ≠ * implies that there exists a W1 ∈ , W1 *. If B × W1 ∈ * for all B ∈ then we take W = W1. Otherwise we have a W2 = B1 × W1 ∈ , *. We repeat the argument with W2 in place of W1. Either this can be taken to be the W of the statement or we obtain W3 = B2 × W2 = B2 × (B, × W1)e , *. This procedure leads in a finite number of steps to the required element W or else we obtain an infinite sequence W1 W2, …, Wi = Bi–1 × Wi–1, where Wi ∈ but W *. We shall show that this last possibility cannot occur and this will complete the proof. We note that Wk is a linear combination of products of k – 1 B’s belonging to and W1. Since * is nilpotent there exists a positive integer n such that any product of n elements of is 0. Now W2n is a linear combination of terms C1 … CjW1Di … Dk where the Cr and D8 ∈ and j + k = 2n – 1. Either j 
n or k n so that we have C1 … CjW1D1 ··· Dk = 0. Hence W2n = 0 and W2n € * contrary to assumption
The main result we shall obtain on weakly closed systems is the following
THEOREM 1. Let be a weakly closed subset of the associative algebra 
of linear transformations of a finite-dimensional vector space 
over Φ. Assume every W ∈ is associative nilpotent, that is, Wk = 0 for some positive integer k. Then the enveloping associative algebra * of is nilpotent.
Proof: We shall prove the result by induction on dim The result is clear if dim = 0 or if = {0}. Hence we assume dim > 0, ≠ {0}. Let Ω be the collection of subsystems of such that * is nilpotent and let be an element of Ω such that dim * is maximal (for the elements of Ω). We shall show that * = * and this will imply the theorem. We note first that * ≠ 0. Thus let W be a non-zero element of . Then by I, = {W}* is a subsystem and * = {w}*. Since {W}* is the set of polynomials in W with constant term 0, {W}* is nilpotent. Hence ∈ Ω. Since ≠ 0 it follows that * ≠ 0. This implies that the subspace 
spanned by all the vectors xB*, x ∈ , B* ∈ , is not 0. Also ≠ . For otherwise, any 
. 
If we use similar expressions for the xi we obtain 
in *. A repetition of this process gives 
in *. Since * is nilpotent this implies x = 0, and = 0 contrary to assumption. Hence we have ⊃ ⊃ 0. Let be the subset of of elements C such that C ⊆ . Then it is clear that is a subsystem containing . Moreover induces weakly closed systems of nilpotent linear transformations in and in the factor space /. Since dim , dim / ≥ dim we may assume that the induced systems have nilpotent enveloping associative algebras. The nilpotency of the algebra in / implies that there exists a positive integer p such that if x is any element of and C1, C2, …, Cp are any Ci in 
then xC1C2 … Cp ∈ 
. Also there exists an integer q such that if D1 D2, …, Dq ∈ and y ∈ then yDi … Dq = 0. This implies that if C1 …, Cp + q ∈ then C1C2 … Cp + q = 0. Thus * is nilpotent and ∈ Ω. We can now conclude that = *. Otherwise, by III., there exists W ∈ , ∈ * such that B × W ∈ * for all B in . By II., we have for any x ∈ , B* ∈ *, 
Hence W 
so W ∈ . Since W *, dim * > dim * and since ∈ Ω this contradicts the choice of . Hence * = * is nilpotent.
If is a set of nil triangular matrices, that is, triangular matrices with 0 diagonal elements, then * is contained in the associative algebra of nil triangular matrices. The latter is nilpotent; hence * is nilpotent. The following result is therefore essentially just a slightly different formulation of Theorem 1.
THEOREM 1′. Let be as in Theorem 1. Then there exists a basis for such that the matrices of all the W ∈ have nil triangular form relative to this basis.
Proof. We may assume ≠ 0. Then the proof of Theorem 1 shows that ⊃ * where * is the space spanned by the vectors xW*, x ∈ , W* ∈ *. In general, if is a subspace and 
is a set of linear transformations, then we shall write for the subspace spanned by all the vectors yS, y ∈ , S ∈ . Then it is immediate that (*)2 = (*)*. Also if * ≠ 0 the argument used before shows that * ⊃ 2. Hence we have a chain ⊃ * ⊃ ()2 ⊃ ()3 ⊃ … ⊃ (*)N – 1 ⊃ (*)N = 0, if (*)N = 0, if (*)N = 0 and (*)N – 1 ≠ 0. We now choose a basis (e1 …, en1) for such that (e1 … eni) is a basis for (*)N – 1, (e1, … en1, …, en1 + n2) is a basis for (*)N – 2, …, (e1, …, en1 + n2 + … + nk is a basis for (*)N – k Then the relations (*)N – k ∈ (*)N – k + 1 imply that the matrix of any W ∈ has the form
Theorem 1 can be generalized to a theorem about ideals in weakly closed sets. For this purpose we shall derive another general property of enveloping associative algebras of weakly closed sets:
IV. Let be a weakly closed subset of an associative algebra and let be an ideal in , *, * the enveloping associative algebras of and , respectively. Then
Proof : By II. of § 1, the condition that B × W ∈ for B ∈ , W ∈ , implies that *W W * + *. It follows that if W1 W2, …, Wr ∈ , then * W1W2 … Wr 
** + *. Hence ** ** + *. Induction now proves (*)k* *(*)k + (*)k. Similarly the condition that is a left ideal implies that *(*)k (*)k* + (*)k. Hence (2) holds. Now (3) is clear for k = 1 and if it holds for some k, then
Similarly, the second part of (3) holds.
We can now prove
THEOREM 2, Let be a weakly closed set of linear transformations acting in a finite-dimensional vector space and let be an ideal in such that every element of is nilpotent. Then *(hence ) is contained in the radical of .*
Proof: ** + * is an ideal since *(** + *) ** and (** + *)* *(** + *) + * * + * = * * + *. Also (** + *)k * * + (*)k. By Theorem 1, * is nilpotent. Hence k can be chosen so that (* * + *)k **. On the other hand (* *)l *(*)l and l can be taken so that (**)l = 0. Thus ** + * is a nilpotent ideal and so it is contained in . It follows that * and are contained in .
Theorem 1 applies in particular to Lie algebras. In this case it is known as Engel’s theorem on Lie algebras of linear transformations: If is a Lie algebra of linear transformations in a finitedimensional vector space and every L ∈ is nilpotent, then * is nilpotent. The conclusion can also be stated that a basis exists for the underlying space so that all the matrices are nil triangular. These results can be applied via the adjoint representation to arbitrary finite-dimensional Lie algebras. Thus we have
Engel’s theorem on abstract Lie algebras. If is a finite-dimensional Lie algebra, then is nilpotent if and only if ad a is nilpotent for every a ∈ .
Proof : A Lie algebra is nilpotent if and only if there exists an integer N such that [… [a1a2]a3…aN] = 0 for ai ∈ . This implies that […. [xa][a] = 0 if the product contains N – 1 a’s. Hence (ad a)N – 1 = 0. Conversely, let be finite-dimensional and assume that ad a is nilpotent for every a ∈ . The set ad of linear transformations ad a (acting in ) is a Lie algebra of nil- potent linear transformations. Hence (ad )* is nilpotent. This means that there is an integer N such that a1 ad a2 … ad aN = 0 for ai ∈ . Thus [… [aa1]a2] … aN] = 0 and so N + 1 = 0.
We can apply Theorem 2 of the last section to obtain two characterizations of the nil radical of a Lie algebra.
THEOREM 3. Let be a finite-dimensional Lie algebra. Then the nil radical of can be characterized in the following two ways: (1) For every a ∈ , ad a (acting in ) is nilpotent and if is any ideal such that ad b (in ) is nilpotent for every b ∈ , then .
(2) is the set of elements b ∈ such that ad b ∈ the radical of (ad )*.
Proof:
(1) If b ∈ and a ∈ , then [ab] ∈ and [… [ab]b] … b] = 0 for enough b’s. Hence ad b is nilpotent. Next let be an ideal such that ad b is nilpotent for every b ∈ . Then the restriction adb of ad b to is nilpotent and is nilpotent by Engel’s theorem. Hence .
(2) The image of under the adjoint mapping is an ideal in ad consisting of nilpotent transformations. Hence, by Theorem 2, ad . It is clear that ad is an ideal in ad ; hence its inverse image 1 under the adjoint representation is an ideal in . But 1 is the set of elements b such that ad b ∈. Every ad1b, b ∈ 1, is nilpotent. Hence 1 is nilpotent and . We saw before that ad so that 1. Hence = 1.
For a linear transformation A in a finite-dimensional vector space ft the well-known Fitting’s lemma asserts that = 0A ⊕ 1A where the i A are invariant relative to A and the induced transformation in 0 A is nilpotent and in 1 A is an isomorphism. We shall call 0 A and 1 A, respectively, the Fitting null and one component of relative to A. The space 
and 0 A = {z | zAi = 0 for some i}. The proof of the decomposition runs as follows. We have 
A A2 …. Hence there exists an r such that Ar = Ar + 1 = … = 1A. Let i = {Zi | Z Ai = 0}. Then 1 2 …., so we have an s such that s = s + 1 = … = 0A. Let t = max(r, s). Then 0A = 1 and 1A = At. Let x ∈ 0 A. Then x At = y A2t for some y since At = A2t. Thus x = (x – y At) + y At and y At ∈ 1 A, while (x – y At)At = 0 so x – y At ∈ 0 A Hence = 0 A + 1 A. Let z ∈ 0 A 1A. Then z = wAt and 0 = zAt = w A2t. Since wA2t = 0, w ∈ 0A = l and wAt = 0. Hence z = 0 and 0A ∩ 1A = 0; hence = 0A ⊕ 1A. Since m0 A = l, At = 0 in 0 A. Since 1A = Ar = Ar + 1 = 1 AA, A is surjective in 1A. Hence A is an isomorphism of 1 A.
We recall also another type of decomposition of relative to A namely, the decomposition into primary components. Here = 
where 
are the irreducible factors with leading coefficients 1 of the minimum (or of the characteristic) polynomial of A, and if μ(λ) is any polynomial, then
(cf. Jacobson, [2], vol. II, p. 130).
If μ(λ) = π(λ) is irreducible with leading coefficient one, then π A = 0 unless π = πi for some i. The space μ A is evidently invariant relative to A. The characteristic polynomial of the restriction of A to πiA has the form πi(λ)ei and dim πiA = eirj where ri = deg πi(λ). If πi(λ) = λ, then the characteristic polynomial of A in λ A is λe, so Ar = 0 in λ A and λ A 0 A If πi(λ) ≠ λ, the characteristic polynomial of the restriction of A to πiA is not divisible by λ, so this restriction is an isomorphism. Hence πiA = πiAA = …, so πiA 1 A. Thus Σπi ≠ λ πiA 1A. Since = 0 A ⊕ 1A = 
it follows that 
. In particular, we see that the Fitting null component is the characteristic space of the characteristic root 0 of A and dim 0 A is the multiplicity of the root 0 in the characteristic polynomial of A.
We shall now extend these results to nilpotent Lie algebras of linear transformations. It would suffice to consider the primary decomposition since the result on the Fitting decomposition could be deduced from it. However, the Fitting decomposition is applicable in other situations (e.g., vector spaces over division rings), so it seems to be of interest to treat this separately. We shall need two important commutation formulas. Let be an associative algebra, a an element of and consider the inner derivation Da: x → x′ = [xa] in . We write x(k) = (x(k – 1))′, x(0) = x. Then one proves directly by induction the following formulas:
We apply this to linear transformations to prove
LEMMA 1. Let A and B be linear transformations in a finite-dimensional vector space. Suppose there exists a positive integer N such that 
Then the Fitting components 0 A, 1 A of relative to A are invariant under B.
Proof: Suppose xAm = 0. Then for k = N + m – 1
Hence xB ∈ 0A. Next let x ∈ 1A. If t is the integer used in the proof of Fitting’s lemma, then we can write x = yAt + N – 1. Then
Hence 1AB 1A.
THEOREM 4. Let be a nilpotent Lie algebra of linear transformations in a finite-dimensional vector space and let 
, 
Then the i are invariant under (that is under every B ∈ ) and = 0 ⊕ 1. Moreover, 1 = ΣA ∈ 1A.
Proof: Suppose first that = 0A for every A ∈ . Then 1A = 0 for all A and Σ1 A = 0. Also by Engel’s theorem (*)N = 0 for some N. Hence 
Thus the result holds in this case. We shall now use induction on dim and we may assume 0A ≠ for some A. By Lemma 1, the B ∈ map 0 A into itself; hence induces a nilpotent Lie algebra of linear transformations in = 0 A ⊂ . We can write = 0 ⊕ 1 where 0 = 
and 0 and 1 are invariant under . Then = 0 ⊕ 1 ⊕ 1 A. It is clear from the definitions that 
and 
. On the other hand, by Engel’s theorem the algebra induced by * in 0 is nilpotent so that we have an N such that 0(*)N = 0. Then
Hence 
and the theorem is proved.
Our discussion of the primary decomposition will be based on a criterion for multiple factors of polynomials. Let Φ[λ] denote the polynomial ring in the indeterminate λ with coefficients in Φ. We define a sequence of linear mappings D0 = 1, D1, D2, … in Φ[λ] as follows: Di is the linear mapping in Φ[λ] whose effect on the element λ3 in the basis (1, λ, λ2, …) for Φ[λ] over Φ is given by
where we take 
Thus we have 
so that D1 is the usual formal derivation in Φ[λ]. Also 
Hence if the characteristic is zero, then 
we shall write Φi = ΦDi. Then we have the Leibniz rule:
It suffices to check this for 
in the basis (1, λ, λ2, …). Then
Since 
the above reduces to 
Hence (8) is valid.
We can now prove
Proof: This is true for k = 0 since μ0(λ) = μ(λ). Write μ(λ) = Φ(λ)ψ(λ) where 
Then we may assume that 
, 
and 
. Then 
by (8).
Let 
and multiply (5) by αk and sum on k. This gives
which we shall use to establish
LEMMA 3. Let A, B be linear transformations in a finite-dimensional vector space satisfying 
for some N. Let μ(λ) be a polynomial and let 
for some m}. Then μA is invariant under B.
Proof: Let y ∈ μ A and suppose yμ(A)m = 0. Set 
, 
. Then, by (9),
By Lemma 2, 
Hence 
and 
Hence y B ∈ μ A.
THEOREM 5. Let be a nilpotent Lie algebra of linear transformations in a finite-dimensional vector space . Then we can decompose = 1 ⊕ 2 ⊕ … ⊕ r where i is invariant under and the minimum polynomial of the restriction of every A ∈ to i is a prime power.
Proof. If the minimum polynomial of every A ∈ is a prime power, then there is nothing to prove. Otherwise, we have an A ∈ such that = π1A ⊕ … ⊕ πSA, πi = πi(λ) irreducible, πiA ≠ 0, s > 1. By Lemma 3, πlAB πiA. for every B ∈ . This permits us to complete the proof by induction on dim .
If the base field is algebraically closed, then the minimum polynomials in the subspaces i are of the form (λ – α(A))kiA, A ∈ . Setting Zi(A) = A – α(A) for the space i we see that Zi(A) is a nilpotent transformation in i. Hence every A ∈ is a scalar plus a nilpotent in i. We therefore have the following
COROLLARY (Zassenhaus). If is a nilpotent Lie algebra of linear transformations in a finite-dimensional vector space over an algebraically closed field, then the space can be decomposed as 1 ⊕ 2 ⊕ … ⊕ r where the * are invariant subspaces such that the restriction of every A ∈ to i is a scalar plus a nilpotent linear transformation.
Consider a decomposition of as in Theorem 5. For each i and each A we may associate the prime polynomial πi A(λ) such that πi A(λ)ki A is the minimum polynomial of A restricted to i. Then the mapping π2: A → πi A(λ) is a primary function for in the sense of the following
DEFINITION 3. Let be a Lie algebra of linear transformations in . Then a mapping π: A → πA(λ), A in , πA(λ) a prime polynomial with leading coefficient one, is called a primary function of for if there exists a non-zero vector x such that x πA(A)mx, A = 0 for every A ∈ . The set of vectors (zero included) satisfying this condition is a subspace called the primary component π corresponding to π.
Using this terminology we may say that the space i in Theorem 5 is contained in the primary component πi. By adding together certain of the i, if necessary, we may suppose that if i ≠ j, then πi: A → πi A(λ) and πj: A → πjA(λ) determined by i and j are distinct. We shall now show that in this case i coincides with πi and the πi are the only primary functions. Thus let π: A →A(λ) be a primary function and let x ∈ π. Assume π ≠ πi so that there exists Ai ∈ such that πAi(λ) ≠ πi Ai(λ) Write x = x1 + … + xr, xi ∈ i. Since x ∈ π there exists a positive integer m such that
Since the decomposition = 1 ⊕ … ⊕ r is direct, we have xiπAi(Ai)m = 0 and since 
this implies xi = 0. It follows that if π ≠ πi for i = 1, 2, …, r, then x = 0 and this contradicts the definition of a primary function. Thus the are the only primary functions. The same argument shows that if x ∈ πi, then x ∈ i; hence i = πi.
The argument just given was based on the following two properties of the decomposition: i πi. where πi is a primary function and πi ≠ πj if i ≠ j. The existence of such a decomposition is a consequence of Theorem 5. Hence we have the following
THEOREM 6. If is a nilpotent Lie algebra of linear transformations in a finite-dimensional vector space , then has only a finite number of primary functions. The corresponding primary components are submodules and is a direct sum of these. Moreover, if = 1 ⊕ 2 ⊕ … ⊕ r is any decomposition of as a direct sum of subspaces i ≠ 0 invariant under such that (1) for each i the minimum polynomial of the restriction of every A ∈ to i is a prime power πi A(λ)miA and (2) if i ≠ j, there exists an A such that πi A(λ) ≠ πj A(λ), then the mappings A → πi A(λ), i = 1, 2, ···, r, are the primary functions and the i are the corresponding primary components.
It is easy to establish the relation between the primary decomposition and the Fitting decomposition: 0 is the primary component λ if A → λ is a primary function and is 0 otherwise, 1 is the sum of the primary components π, π ≠ λ. We leave it to the reader to verify this.
We shall now assume that the nilpotent Lie algebra of linear transformations has the property that the characteristic roots of every A ∈ are in the base field. A Lie algebra of linear transformations having this property will be called split. Evidently, if the base field Φ is algebraically closed, then any Lie algebra of linear transformations in a finite-dimensional vector space over Φ is split. Now let be nilpotent and split. It is clear that the characteristic polynomial of the restriction of A to the primary component corresponding to A → λA(λ) is of the form πA(λ)r. Since this is a factor of the characteristic polynomial of A and πA(λ) is irreducible, our hypothesis implies that πA(λ) = λ – α(A), α(A) ∈ Φ. Under these circumstances it is natural to replace the mapping A → πA(λ) by the mapping A → α(A) of into Φ and to formulate the following
DEFINITION 4. Let be a Lie algebra of linear transformations in . Then a mapping α: A → α(A) of into the base field Φ is called a weight of for if there exists a non-zero vector x such that x(A – α(A)1)mx, A = 0 for all A ∈ . The set of vectors (zero included) satisfying this condition is a subspace α called the weight space of corresponding to the weight a.
Theorem 6 specializes to the following result on weights and weight spaces.
THEOREM 7. Let be a split nilpotent Lie algebra of linear transformations in a finite-dimensional vector space. Then has only a finite number of distinct weights, the weight spaces are submodules, and is a direct sum of these. Moreover, let = 1 ⊕ 2 ⊕ … ⊕ r be any decomposition of into subspaces i ≠ 0 invariant under such that (1) for each i, the restriction of any A ∈ has only one characteristic root αi(A) (with a certain multilicity) in i and (2) if i ≠ j, then there exists an A ∈ such that αi(A) ≠ αj(A). Then the mappings A → αi(A) are the weights and the spaces i are the weight spaces.
Our main result in this section gives the structure of a Lie algebra of linear transformations whose enveloping associative algebra * is semi-simple. In the next section the proof of Lie’s theorems will also be based on this result. For all of this we shall have to assume that the characteristic is 0. We recall that the trace of a linear transformation A in a finite-dimensional vector space, which is defined to be 
for any matrix (αij) of A, is the sum of the characteristic roots ρi i = 1, …, n of A. Also tr 
If A is nilpotent all the ρi are 0, so tr Ak = 0, k = 1, 2, ···. If the characteristic of Φ is 0, the converse holds: If tr Ak = 0, k = 1, 2, …, then A is nilpotent. Thus we have 
The formulas (Newton’s identities; cf. Jacobson, [2], vol. I, p. 110) expressing 
in terms of the elementary symmetric functions of the pi show that if the characteristic is 0 our conditions imply that all the elementary symmetric functions of the Pi are 0, so all the pi are 0. Hence A is nilpotent. We use this result in the following
LEMMA 4. Let C ∈ , the algebra of linear transformations in a finite-dimensional vector space over a field of characteristic 0 and suppose 
and [CAi] = 0, i = 1, 2, …, r. Then C is nilpotent.
Proof: We have [Ck – 1 Ai] = 0, k = 1, 2, … (C0 = 1). Hence Ck = 
Since the trace of any commutator is 0, this gives tr Ck = 0 for k = 1, 2, …. Hence C is nilpotent.
The key result for the present considerations is the following
THEOREM 8. Let be a Lie algebra of linear transformations in a finite-dimensional vector space over a field of characteristic 0. Assume that the enveloping associative algebra * is semi-simple. Then = 1 ⊕ where is the center of and 1 is an ideal of which is semi-simple (as a Lie algebra).
Proof: Let be the radical of . We show first that = the center of . Otherwise, 1 = [] is a non-zero solvable ideal. Suppose 
and set 
If C ∈ 3, C = Σ[AiBi], Ai ∈ 2, Bi ∈ , and [CAi] = 0 since 3 2 and 2 is abelian. Hence, by Lemma 4, C is nilpotent. Thus every element of the ideal 3 of is nilpotent; hence, by Theorem 2, 3 is in the radical of *. Since * is semi-simple, 3 = 0. Since 3 = [2] this shows that 2 Since 2 [ ] ; ′ every element C of 2 has the form Σ[AiBi], Ai Bi ∈ , and we have [CAi] = 0, since 2 . The argument just used implies that 2 = 0, which contradicts our original assumption that 1 = [] ≠ 0. Hence we have [] = 0, and = . The argument we have used twice can be applied to conclude also that ′ = 0. Hence we can find a subspace 1 ′ such that = 1 ⊕ . Since 1 contains ′, 1 is an ideal. Also 1 ≅ / = / is semi-simple. This concludes the proof.
Remark’. In the next chapter we shall show that 1 = ′, so we shall have = ′ ⊕ , ′, semi-simple.
COROLLARY 1. Let be as in the theorem, * semi-simple. Then is solvable if and only if is abelian. More generally, if is solvable and is the radical of *, then */ is a commutative algebra.
Proof: If is abelian it is solvable, and if is solvable and * is semi-simple, then = ⊕ 1 where 1 is semi-simple. Since the only Lie algebra which is solvable and semi-simple is 0 we have 1 = 0 and = is abelian. To prove the second statement it is convenient to change the point of view slightly and consider as a subalgebra of L, a finite-dimensional associative algebra over a field of characteristic 0. Since any such can be considered as a subalgebra of an associative algebra of linear transformations in a finite-dimensional vector space, Theorem 8 is applicable to and . We now assume solvable, so ( +)/, which is a homomorphic image of , is solvable. Moreover, the enveloping associative algebra of this Lie algebra is the semi-simple associative algebra */. Hence ( + )/ is abelian. This implies that (a + ) (b + ) = (b + ) (a + ) for any a, b ∈ . Since the cosets a + generate */, it follows that */ is commutative.
COROLLARY 2. Let be a Lie algebra of linear transformations in a finite-dimensional vector space over a field of characteristic 0, let be the radical of and the radical of *. Then is the totality of nilpotent elements of and [] .
Proof: Since is associative nilpotent it is Lie nilpotent. Hence . Moreover, the elements of are nilpotent so 0 the set of nilpotent elements of . If 0 denotes the radical of the enveloping associative algebra * then, by Corollary 1, */0 is commutative. Now any nilpotent element of a commutative algebra generates a nilpotent ideal and so belongs to the radical. Since */0 is semi-simple, it has no non-zero nilpotent elements. It follows that 0 is the set of nilpotent elements of * and so 0 = 0 is a subspace of . Next consider ( + )/. The enveloping associative algebra of this Lie algebra is */ which is semi-simple. Hence the radical of ( + ) is contained in its center. Since ( + )/ is a solvable ideal in ( + ) we must have [( + )/, ( + )/] = 0, which means that [] . Hence [] 0. Then [0] 0 is an ideal in . Since its elements are nilpotent, 0 by Theorem 2. Hence 0 and 0 = which completes the proof.
There is a more useful formulation of Theorem 8 in which the hypothesis on the structure of * is replaced by one on the action of on . In order to give this we need to recall some standard notions on sets of linear transformations. Let Σ be a set of linear transformations in a finite-dimensional vector space over a field Φ. We recall that the collection L(Σ) of subspaces which are invariant under Σ(A , A ∈ Σ) is a sublattice of the lattice of subspaces of . We refer to the elements of L(Σ) as Σ-subspaces of . If is a subspace then the collection of A ∈ such that A is a subalgebra of . It follows that if ∈ L(Σ), then is invariant under every element of the enveloping associative algebra Σ* of Σ and under every element of the enveloping algebra Σ†. Thus we see that L(Σ) = L(Σ*) = L(Σ†)
The set Σ is called an irreducible set of linear transformations and is called Σ-irreducible if L(Σ) = {, 0} and ≠ 0. Σ is indecomposable and is Σ-indecomposable if there exists no decomposition = 1 ⊕ 2 where the i are non-zero elements of L(Σ). Of course, irreducibility implies indecomposability. Σ (and relative to Σ) is called completely reducible if = Σα, α ∈ L(Σ) and α irreducible. We recall the following well-known result.
THEOREM 9. Σ is completely reducible if and only if L(Σ) is complemented, that is, for every ∈ L(Σ) there exists an ′ ∈ L(Σ) such that = ⊕ ′. If the condition holds then = 1 ⊕ 2 ⊕ … ⊕ r where the i ∈ L(Σ) and are irreducible.
Proof : Assume = Σα, α irreducible in L(Σ) and let ∈ L(Σ) If dim = dim , = and = ⊕ 0. We now suppose dim < dim and we may assume the theorem for subspaces 1 such that dim 1 > dim . Since = Σα there is an α such that α 
. Consider the subspace α . This is a Σ-subspace of the irreducible Σ-space α. Hence either * = α or α = 0. If α = α, α contrary to assumption. Hence α = 0 and 1 = + α = ⊕ α. We can now apply the induction hypothesis to conclude that = 
Then 
where ′ = 
Conversely, assume L(Σ) is complemented. Let 1 be a minimal element ≠ 0 of L(Σ). (Such elements exist since dim is finite.) Then we have = 1 ⊕ where ∈ L(Σ) We note now that the condition assumed for carries over to . Thus let 
be a Σ-subspace of . Then we can write = ⊕ ′, ′ ∈ L(Σ). Then, by Dedekind’s modular law, = = + (′ ) since . If we set ′ = ′ then ′ = ′ = 0. Hence = ⊕ ″ where ″ ∈L(Σ) and ″ . We can now repeat for the step taken for ; that is, we can write = 2 ⊕ where 2, ∈ L(Σ) and 2 is irreducible. Continuing in this way we obtain, because of the finiteness of dimensionality, that = 1 ⊕ 2 ⊕ … ⊕ r, i irreducible, i ∈ L(Σ). This completes the proof.
We suppose next that Σ = is a subalgebra of the associative algebra (possibly not containing 1), and we obtain the following necessary condition for complete reducibility.
THEOREM 10. If is an associative algebra of linear transformations in a finite-dimensional vector space, then completely reducible implies that is semi-simple.
Proof : Let be the radical of and suppose = Σα, α ∈ L() and irreducible. Consider the subspace α* spanned by the vectors of the form yN, y ∈ *α, N ∈ . This is an -subspace contained in *. Since k = 0 for some k we must have α *α (cf. the proof of Theorem 1). Since *α is irreducible we can conclude that α = 0 for every α. Since = Σα this implies that = 0, that is, = 0. Hence is semi-simple.
Since Σ is completely reducible if and only if Σ* and Σ† are completely reducible, we have the
COROLLARY. If Σ is completely reducible, then Σ* and Σ† are semisimple.
We shall say that a single linear transformation A is semi-simple if the minimum polynomial μ(λ) of A is a product of distinct prime polynomials. This condition is equivalent to the condition that {A}† has no nilpotent elements ≠ 0. Thus if 
then 
is nilpotent and Z ≠ 0 if some ei > 1. Conversely, suppose the πi are distinct primes and all ei = 1. Let Z = Φ(A) be nilpotent. If Zr = 0, Φ(A)r and Φ(λ)r is divisible by μ(λ) Hence Φ(λ) is divisible by μ(λ) and Z = Φ(A) = 0. We can now prove
THEOREM 11. Let be a completely reducible Lie algebra of linear transformations in a finite-dimensional vector space over a field of characteristic 0. Then = ⊕ 1 where is the center and 1 is a semi-simple ideal. Moreover, the elements of are semisimple.
Proof: If is completely reducible, * is semi-simple. Hence the statement about = ⊕ 1 follows from Theorem 8. Now suppose that contains a C which is not semi-simple. Then {C}† contains a non-zero nilpotent N. Moreover, {C}† is the set of polynomials in C, so that N is a polynomial in C. Hence N is in the center of †. Since this is the case, N† = †N is an ideal in †. Moreover, (N†)k Nk† = 0 if k is sufficiently large. Since N ∈ N†, N† is a non-zero nilpotent ideal in † contrary to the semi-simplicity of †.
We shall show later (Theorem 3.10) that the converse of this result holds, so the conditions given here are necessary and sufficient for complete reducibility of a Lie algebra of linear transformations in the characteristic zero case. Nothing like this can hold for characteristic p ≠ 0 (cf. § 6.3). We remark also that the converse of Theorem 10 is valid too. For the associative case one therefore has a simple necessary and sufficient condition for complete reducibility, valid for any characteristic of the field. The converse of Theorem 10 is considerably deeper than the theorem itself. This will not play an essential role in the sequel.
We need to recall the notion of a composition series for a set Σ of linear transformations and its meaning in terms of matrices. We recall that a chain = 1 = 1 ⊃ 2 ⊃ … ⊃ s ⊃ s+1 = 0 of Σ-subspaces is a composition series for relative to Σ if for every i there exists no ′ ∈ L(Σ) such that i ⊃ ′ ⊃ i + 1. If is a Σ-subspace of , then Σ induces a set 
of linear transformations in / As is well known (and easy to see) for groups with operators the -subspaces of = / have the form / where is a Σ-subspace of containing . It follows that there exists no ∈ L(Σ) with ⊃ ⊃ if and only if / is -irreducible. Hence = 1 ⊃ 2 ⊃> … ⊃ s + 1 = 0, i ∈ L(Σ), is a composition series if and only if every i/i + 1 is i-irreducible, i the set of induced transformations in i/i + 1 determined by the A ∈ Σ.
The finiteness of dimensionality of assures the existence of a composition series. Thus let 2 be a maximal Σ-subspace properly contained in 1 = . Then 1 ⊃ 2 and 1/2 is irreducible. Next let 3 be a maximal invariant subspace of 2, ≠ 2, etc. This leads to a composition series 1 ⊃ 2 ⊃ 3 ⊃ … ⊃ s ⊃ s + 1 = 0 for .
Let 1 ⊃ 2 ⊃ 3 ⊃ … ⊃ s + 1 = 0 be a descending chain of Σ-subspaces and let (e1, …, en) be a basis for such that (e1 …, eni) is a basis for s, (e1 …, en1 + n2) is a basis for s – 1, etc. Then it is clear that if A ∈ Σ the matrix of A relative to (ei) is of the form
Thus we have
where x is a vector in s – j + 2. Hence we have (10) with Mj = 
The cosets 
form a basis for the factor space s – j + 1/s – j + 2 and (11) gives
which shows that the matrix of the linear transformation A induced in s – j + 1/s – j + 2 is Mj.
We shall now see what can be said about solvable Lie algebras of linear transformations in a finite-dimensional vector space over an algebraically closed field of characteristic zero. For this we need the following
LEMMA 5. Let be an abelian Lie algebra of linear transformations in the finite-dimensional space over Φ, Φ algebraically closed. Suppose is -irreducible. Then is one-dimensional.
Proof. If A ∈, , A has a non-zero characteristic vector x. Thus xA = αx, α ∈ Φ. Now let α be the set of vectors y satisfying this equation. Then if B ∈ and y ∈ α (yB) A = yAB = αyB. Hence yB ∈ α This shows that α is invariant under every B ∈ . Since is irreducible, = α, which means that A = α1 in . Now this holds for every A ∈ . It follows that any subspace of is an -subspace. Since is -irreducible, has no subspaces other than itself and 0. Hence is one-dimensional.
Lie’s theorem. If i is a solvable Lie algebra of linear transformations in a finite-dimensional vector space over an algebraically closed field of characteristic 0, then the matrices of can be taken in simultaneous triangular form.
Proof: Let = 1 ⊃ 2 ⊃ … s + 1 = 0 be a composition series for relative to . Let 
denote the set of induced linear transformations in the irreducible space i/i + 1. Then is a solvable Lie algebra of linear transformations since it is a homomorphic image of . Also is irreducible, hence completely reducible in i/i + 1. Hence is abelian by Theorem 11. It follows from the lemma that dim i/i + 1 = 1. This means that if we use a basis corresponding to the composition series then the matrices Mi in (10) are one-rowed. Hence every M corresponding to the A ∈ is triangular.
If is nilpotent, it is solvable, so Lie’s theorem is applicable. We observe also that Lemma 5 and, consequently, Lie’s theorem are valid for any field of characteristic 0, provided the characteristic roots of the linear transformations belong to the field. We have called such Lie algebras of linear transformations split. If we combine this extension of Lie’s theorem with Theorem 7 we obtain
THEOREM 12. Let be a split nilpotent Lie algebra of linear transformations in a finite-dimensional vector space over a field of characteristic 0. Then is a direct sum of its weight spaces α, and the matrices in the weight space α can be taken simultaneously in the from
Proof: The fact that is a direct sum of the weight spaces * has been proved before. We have xα(A – α(A)1)m = 0 for every xα in α, which implies that the only characteristic root of A in α is α(A). Since the diagonal elements in (10) are characteristic roots it follows that the Mi in (10) for A are α(A). Hence we have the form (13) for the matrix of A in α.
The proof of this result (or the form of (13)) shows that in α there exists a non-zero vector x such that xA = α(A)x, A ∈ . If A, B ∈ we have 
and 
. The first two of these equations imply that α is a linear function on . Since every element of the derived algebra ′ is a linear combination of elements [AB], the linearity and the last condition imply that α(C) = 0, C ∈ ′. We therefore have the following important consequence of Theorem 12.
COROLLARY. Under the same assumptions as in Theorem 12, the weights α: A → α(A) are linear functions on which vanish on ′.
As usual, the results we have derived for Lie algebras of linear transformations apply to abstract Lie algebras on considering the set of linear transformations ad . We state two of the results which can be obtained in this way.
THEOREM 13. Let be a finite-dimensional Lie algebra over a field of characteristic 0, the radical, and the nil radical. Then [] .
Proof. By Corollary 2 to Theorem 8, [ad , ad ] is contained in the radical of (ad )*. This implies that there exists an integer N such that for any N transformations of the form [ad ai, ad si], 
, we have [ad a1 ad si] [ad a2, ad s2] … [ad aN ad sN] = 0. Hence ad [a1s1] ad [a2s2] … ad [aNsN] = 0. Thus for any x ∈ we have [… [[x[a1s1]] [a2s2]] …, [aNsN]] = 0. This implies that []N + 1 = 0, so [] is nilpotent. Since this is an ideal, [] .
COROLLARY 1. The derived algebra of any finite-dimensional solvable Lie algebra of characteristic 0 is nilpotent.
Proof: Take = in the theorem.
COROLLARY 2. Let . be solvable finite-dimensional of characteristic 0, the nil radical of . Then D for any derivation D of .
Proof: Let = ⊕ ΦD the split extension of ΦD by (cf. § 1.5). Here is an ideal and [s, D] = sD for s ∈ . Since / is one dimensional, it is solvable. Hence ′ is a nilpotent ideal in . On the other hand, ′ = ′ + D, so ′ + D is a nilpotent ideal in and ′ + D . Hence D .
THEOREM 14 (Lie). Let be a finite-dimensional solvable Lie algebra over an algebraically closed field of characteristic 0. Then there exists a chain of ideals = n ⊃ n – 1 ⊃ … ⊃ 1 ⊃ 0 such that dim i = i.
Proof: Since is solvable, ad is a solvable Lie algebra of linear transformations acting in the vector space . Let = 1 2 ⊃ … ⊃ s + 1 = 0 be a composition series of relative to ad . Then i ad i is equivalent to the statement that i is an ideal. By the proof of Lie’s theorem we have that dim i/i + 1 = 1. Hence the composition series provides a chain of ideals of the type required.
We shall now show that the assumption that the characteristic is 0 is essential in the results of §§ 5 and 6. We begin our construction of counter examples in the characteristic p ≠ 0 case with a p-dimensional vector space over Φ of characteristic p. Let E and F be the linear transformations in whose effect on a basis (e1 e2, … ep) is given by
Then eiEF = iei + 1, i 
P – 1, epEF = 0 and eiFE = (i – 1)ei + 1, epFE = – e1. Hence
so that we have [EF] = E. Hence = ΦE + ΦF is a two-dimensional solvable Lie algebra. We assert that acts irreducibly in . Thus let be an -subspace ≠ 0 and let x = Σξiei ≠ 0 be in . Then
The Vandermonde determinant
If we multiply xFj – 1 by the cofactor of the (j, i) element of V and sum on j we obtain that Vξiei ∈ and so ξiei ∈ . It follows from this that contains one of the ei. If we operate on this ei by the powers of E we obtain every ej. Hence = , which proves irreducibility and shows that Theorem 11 fails for characteristic p. Also since complete reducibility implies semi-simplicity (actually * = is simple), Theorem 8 also fails for characteristic p.
Next let denote the two-dimensional non-abelian Lie algebra Φe + Φf, [ef] = e, Φ of characteristic p. Then e → E, f → F given by (14), defines a representation of acting in and is an module with basis (e1, e2, …, eP). Let 
be the split extension = ⊕ . Then is an abelian ideal in and / ≅ is solvable. Hence is solvable. The derived algebra ′ = Φe + [] = Φe + . Since [e] = E = , (′)2 = (′)3 = … = . Hence ′ is not nilpotent. This shows that also Theorem 13 fails for characteristic p.
It is still conceivable that Lie’s theorem might hold for characteristic p if one replaces the word “solvable” by “nilpotent.” This would imply that we have the nice canonical’ form (13) for nilpotent Lie algebras. However, this is not the case either. Thus let E and F be as before. Then EF – FE = E implies that E(FE– 1) – (FE– 1)E = 1. Set G = FE–1, H = 1. Then we have
Hence = ΦE + ΦG + ΦH is a nilpotent Lie algebra. On the other hand, * = *( = ΦE + ΦF) is irreducible even if the base field is algebraically closed. Hence the matrices cannot be put in simultaneous triangular form. If X = ξE + ηG + ζ1 one can verify that (X – (ξ + ζ)1)p = 0 if p ≠ 2. Hence the mapping X → ξ + ζ is the only weight for . Note that this is linear. (Cf. Exercise 24, Chap. V.)
In these exercises all algebras and vector spaces are finite-dimensional.
1. Let Φ be algebraically closed of characteristic 0 and let be a solvable subalgebra of m dimensions of the Lie algebra properly containing . Show that is contained in a subalgebra of m + 1 dimensions.
2. Φ as in Exercise 1, solvable. Show that has a basis (e1, e2, …, en) with a multiplication table 
3. Φ as in Exercise 1. Let be a solvable subalgebra of ΦnL. Prove that dim n(n + 1)/2 and that can be imbedded in a solvable subalgebra of n(n + 1)/2 dimensions. Show that if 1 and 2 are solvable subalgebras of n(n + 1)/2 dimensions, then there exists a non-singular matrix A such that 2 = A– 11A.
4. Show that if is not nilpotent over an algebraically closed field then contains a two-dimensional non-abelian subalgebra.
5. Φ algebraically closed, any characteristic. If a is a characteristic root of a linear transformation A, the characteristic space a = {xα | xα(A – α1)k = 0 for some k}. We have = Σ ⊕ α be an automorphism in a non-associative algebra and let = Σ ⊕ α be the decomposition of into characteristic spaces relative to A. Show that
Let S be the linear transformation in such that S = α1 in α for every α. Show that S is an automorphism which commutes with A and that U = S–1A is an automorphism of the form 1 + Z, Z nilpotent.
6. Φ as in Exercise 5, a Lie algebra over Φ, A an automorphism in Let = Σ ⊕ α be the decomposition of into characteristic subspaces relative to A. Let = ∪ ad α where ad α = {ad xα | xα ∈ α}. Show that is weakly closed relative to A × B = [AB].
7. Φ, , A, as in Exercise 6. Assume A is of prime order. Show that if the only x ∈ such that xA = x is x = 0 then every element of is nilpotent. Hence prove that if a Lie algebra over an arbitrary field has an automorphism of prime order without fixed points except 0, then is nilpotent.
8. Let D be a derivation in a non-associative algebra and let = Σ ⊕ α be the decomposition of into characteristic spaces relative to D. Show that
Let S be the linear transformation in such that S = α1 in α for every α. Show that S is a derivation which commutes with D and that D – S = Z is a nilpotent derivation.
9. Prove that if is a Lie algebra over a field of characteristic 0 and has a derivation without non-zero constants, then is nilpotent.
10. (Dixmier-Lister). Show that if is the nilpotent Lie algebra of Exercise 1.7 and the characteristic is not 2 or 3, then every derivation D satisfies D ′ and hence is nilpotent. Show that this Lie algebra cannot be the derived algebra of any Lie algebra.
11. Let be a Lie algebra of linear transformations such that every A ∈ has the form α(A)1 + Z where α(A) ∈ Φ and Z is nilpotent. Prove that is nilpotent.
The next group of exercises is designed to prove the finiteness of the derivation tower of any Lie algebra with 0 center. The corresponding result for finite groups is due to Wielandt. The proof in the Lie algebra case is due to Schenkman [1]. It follows fairly closely Wielandt’s proof of the group result but makes use of several results which are peculiar to the Lie algebra case. These effect a substantial simplification and give a final result which is sharper than that of the group case.
Let be a Lie algebra with 0 center. Then is isomorphic to the ideal 1 of inner derivations in the derivation algebra 2 = 
(). We can identify with 1 and so consider as an ideal in 2 We have seen (Exercise 1.4) that 2 has 0 center. Hence the process can be repeated and we obtain 1 2 3 where 3 is the derivation algebra of (), ()()), and 2 is identified with the ideal of inner derivations of 2. This process leads to a chain = 1 2 3 …, where each i is invariant in i + 1. Hence 1 = is sub-invariant in i (Exercise 1.8) and every i has 0 center. The tower theorem states that from a certain point on we must have r = r + 1 = ….
This means that if has 0 center, then for a suitable 
is a complete Lie algebra.
Exercise 1.4, 1.8, and 1.9 will be needed in the present discussion. Besides these all that is required are elementary results and Engel’s theorem that if ad a is nilpotent for every α then is nilpotent.
12. Prove that every Lie algebra has a decomposition = w + 
where 
and is a nilpotent subalgebra.
13. Let be a Lie algebra and let be the centralizer of w Show that if has 0 center, then w.
Sketch of proof:$ is an ideal. Write 
a nilpotent subalgebra. Set = + , which is a subalgebra since is an ideal. Write 
a nilpotent subalgebra. Then 
and = w + 1. If is the center of 1, 1 is contained in the center of so that 1 = 0. This implies that 1 = 0 since 1 is an ideal in 1. Then = 
.
14. Let be subinvariant in and assume that the centralizer of in is 0. Prove that the centralizer of ω in is contained in ω.
Sketch of proof. Assume ω. Then, by Exercise 13, . is an ideal and = + is a subalgebra; ⊃ . The normalizer of in properly contains and contains a z ∈ , ∉ . Then = Φz + is a subalgebra such that ω = ω. The centralizer of ω in contains z ∉ ω. Hence, by Exercise 13, has a non-zero center and so has non-zero centralizer in 
contrary to assumption.
15. Let be a Lie algebra, a subinvariant subalgebra, an ideal in . Show that if the centralizer of in is 0 and the centralizer of in is 0, then the centralizer of S in is 0.
16. Let = 1 2 3 …, i = 
(i – 1) be the tower of derivation algebras for a Lie algebra with 0 center. Show that the centralizer of 1 in i is 0 and that the centralizer of 
in i is contained in 
Let 
be the center of 
and 
the derivation algebra of Prove that
Hence prove that there exists an m such that m = m + 1 = ···. This is Schenkman’s derivation tower theorem.
17. Let be a Lie algebra of linear transformations in a vector space over a field of characteristic 0. Let be a subinvariant subalgebra of such that every K ∈ is nilpotent. Prove that is contained in the radical of * (Hint: Use Exercise 1.22.)
18. Let be a nilpotent Lie algebra of linear transformations in a vector space such that is a sum of finite-dimensional subspaces invariant under . Show that = Σ ⊕ π where the π are the primary components corresponding to the primary functions π: A → π(A). Show that the π are invariant. Show that if is any invariant subspace, then = Σ ⊕ π where π = π.
then 
.
