18.745 Introduction to Lie algebra Victor Kac , Fall 2004 Lecture 6 , noted by Liu Ruochuan

Throughout this lecture 𝔽 will be assumed to be an algebraically closed field.

Jordan Decomposition

Now suppose V is a finite dimensional vector space over 𝔽 and A is a linear operator on V. Let {λ_{1}, ⋯, λ_{s}} denote the set of distinct eigenvalues of A. By the theorem of Jordan normal form, in some basic of V, the matrix of A is a direct sum of the Jordan blocks J_{λ_{i}} assigned to some eigenvalue λ_{i}. If we denote by V_{λ_{i}} the span of the vectors in the basis which correspond to all Jordan blocks assigned to λ_{i}, we obtain the following generalized eigenspace decomposition of V.

V = ⨁_{i=1}^{s} V_{λ_{i}},

where the generalized eigenspace V_{λ_{i}} can also be defined as

V_{λ_{i}} = {v ∈ V |(A − λ_{i}I)^{N}v = 0 for some N }

We take A_{s} as the diagonal part of the Jordan normal form of A, A_{n} = A − A_{s}. Then A_{s} is semisimple, i.e diagonalizable operator and A_{n} is a nilpotent operator. Moreover, A_{s}A_{n} = A_{n}A_{s}. This decomposition is called a Jordan decomposition of A.

Ex6.1 Show that there exist polynomials P (x) and Q(x) such that A_{s} = P (A) and A_{n} = Q(A) Solution. By Chinese remainder theorem, there exists a polynomial P (x) such that P (x) ≡ λ_{i}(mod(x − λ_{i})^{n}) for every λ_{i}, where n =dimV. Then for v ∈ V_{λ_{i}}, we have P (A)v = λ_{i}v. That means P (A) = A_{s} and (1 − P)(A) = A_{n}.

Ex6.2 If linear operators A and B commute, then any eigenspace and generalized eigenspace of A is B-invariant. Conclude that two commuting semi-simple operators can be diagonalized in the same basic. Solution. Let V_{λ_{i}} be the generalized eigenspace of A with eigenvalue λ_{i}. For every v ∈ V_{λ_{i}}, we have (A − λ_{i})^{N}v = 0 for some N. Since A commutes with B, we get (A − λ_{i})^{N}Bv = B(A − λ_{i})^{N}v = 0. Thus Bv is in V_{λ_{i}} by definition. That just means the generalized eigenspace of A is B-invariant. The method for eigenspace case is the same. Now suppose A and B are both semi-simple. Let V = ⨁_{i=1}^{s} V_{λ_{i}} be the eigenspace decomposition of A. Since B is semi-simple and V_{λ_{i}} is an invariant subspace w.r.t B, B is semi-simple on V_{λ_{i}}. Then B can be diagonalized in V_{λ_{i}} under some basis. Now under the basis, which is the union of these basis, A and B are all diagonal.

The Jordan decomposition of a linear operator A is unique in sense of Theorem 1. Let A = As^{′} + An^{′} where As^{′} and An^{′} are also linear operators which satisfy (1)As^{′} is diagonalizable (2)An^{′} is nilpotent

(3)As^{′}An^{′} = An^{′}As^{′}

Then As^{′} = A_{s}, An^{′} = A_{n}.

Proof. We first have As^{′} − A_{s} = A_{n} − An^{′}. Note that As^{′} and An^{′} commute with A. So by Ex 6.1, As^{′} and An^{′} commute with both A_{s} and A_{n}. Then by Ex 6.2, As^{′}−A_{s} is semisimple. But A_{n} − An^{′} is nilpotent by the binomial formula. Since the only nilpotent semisimple operator is 0, we conclude that As^{′} = A_{s} and An^{′} = A_{n} from the equation given above.

Generalized Weight Space Decomposition

Let 𝔤 be a finite dimensional Lie algebra over 𝔽 and π a representation of 𝔤 in a finite dimensional vector space V over 𝔽. Consider the generalized eigenspace decomposition of 𝔤 w.r.t ada and of V w.r.t π(a).

𝔤 = ⨁_{α} 𝔤_{α}^{a} where 𝔤_{α}^{a} = {g ∈ 𝔤|(ada − α)^{N}g = 0 for some N}
V = ⨁_{λ} V_{λ}^{a} where V_{λ}^{a} = {v ∈ V |(π(a) − λ)^{N}v = 0 for some N}

These two decompositions are related by

Proposition 1. π(𝔤_{α}^{a})V_{λ}^{a} ⊆ V_{λ+α}^{a}

We need the following lemma to finish the proof.

Lemma 1. Let A be a unital associate algebra over 𝔽. Let a, b ∈ A and α, λ ∈ 𝔽. Then we have the following identity

(a − α − λ)^{N}b = ∑_{j=0}^{N}(_{j}^{N})(ada − α)^{j}b(a − λ)^{N−j}

Proof. Let L_{x} and R_{x} denote the operators of left and right multiplication in A by x. Then associativity means that L_{x}R_{y} = R_{y}L_{x}. We have L_{a−α−λ} = (ada − α) + R_{a−λ}. Note that ada − α commute with R_{a−λ}, hence L_{a−α−λ}^{N} = ∑_{j=0}^{N}(_{j}^{N})(ada − α)^{j}R_{α−λ}^{N−j} by the binomial formula. Now apply both sides to b to obtain the identity.

P roof of the proposition. Suppose g ∈ 𝔤_{α}^{a}, v ∈ V_{λ}^{a}. Apply the lemma to A =End(V), a is π(a), b = π(g), hence

(π(a) − α − λ)^{N}π(g)v = ∑_{j=0}^{N}(_{j}^{N})(adπ(a) − α)^{j}π(g)(π(a) − λ)^{N−j}v

Take N >dim𝔤_{α}^{a}+dimV_{λ}^{a}, then each summand of the right hand side of the above equation is zero. That means the desired result.

Suppose 𝔥 is a finite dimensional Lie algebra over 𝔽. Let π be a representation of 𝔥 in a finite dimensional vector space V over 𝔽 and let λ ∈ 𝔥^{∗}. Then the generalized weight space attached to λ is defined as

V_{λ} = {v ∈ V |(π(h) − λ(h))^{N}v = 0 for every h ∈ 𝔥 and some N}.

Theorem 2. Notation as above and further assume char.𝔽 = 0 and 𝔥 is nilpotent. Then

V = ⨁_{λ∈𝔥^{∗}} V_{λ}

Ex 6.3 a)Deduce the theorem from Ex 6.2 in the case when 𝔥 is abelian. b)Consider the adjoint representation of the unique 2-dimensional non-abelian Lie algebra to show that theorem 2 fails if 𝔥 is not nilpotent. Solution. a) For any a ∈ 𝔥, π(𝔥)V_{λ}^{a} ⊂ V_{λ}^{a} by Ex 6.2 . So if there is at least one π(a) has distinct eigenvalues, then we can apply induction on dimV. Otherwise we just need to prove that the only eigenvalue of every π(a) is linear functional on 𝔥. Let a and b are elements of 𝔥. Suppose eigenvalues of π(a) and π(b) are λ and μ respectively. Then since 𝔥 is abelian, we have

(π(a + b) − λ − μ)^{N} = ∑_{j=0}^{N}(_{j}^{N})(π(a) − λ)^{j}(π(b) − μ)^{N−j}

Now choose N > 2n where n =dimV. Then apply the above equation to every v ∈ V to conclude that the eigenvalue of π(a + b) is λ + μ. b)Denote the unique 2-dimensional non-abelian Lie algebra by 𝔟 = 𝔽a+𝔽b which satisfies [a, b] = b. It is easy to see that 𝔽a is a generalized eigenspace of ada, but it is not a generalized eigenspace of b. Thus the theorem may fails if the given Lie algebra is not nilpotent.

Proof. Take any a ∈ 𝔥, then 𝔥 = 𝔥_{0}^{a}. Hence by the proposition which we just proved π(𝔥)V_{λ}^{a} ⊂ V_{λ}^{a} for all a ∈ 𝔥 and eigenvalue λ of a. So if there is at least one π(a) has distinct eigenvalues, then we can apply induction on dimV. Otherwise we just need to prove that the only eigenvalue of every π(a) is linear functional on 𝔥. Apply Lie’s theorem that all π(a) are upper triangular in some basis, then we deduce the desired result since the eigenvalue are just the numbers on diagonal.

Remark It is easy to see that we can still get the decomposition even if char.𝔽 ≠ 0. But λ may not be linear functional on 𝔥 in this case.

Ex 6.4 Consider the 2-dimensional representation of H_{1} for char.𝔽 = 2

V = 𝔽[x]/x^{2}𝔽[x], p = \frac{∂}{∂x}, q = x, c = 1.

Then V = V_{λ}, but λ is not a linear function. Solution. Note that p and q are all nilpotent. But p + q is not nilpotent since (p + q)^{2}(x) = x. So in this case λ is not a linear function.

Ex 6.5^{∗} If 𝔥^{p} = 0, then the theorem still holds for char.𝔽 = p.