Lie's theorem explained

In mathematics, specifically the theory of Lie algebras, Lie's theorem states that, over an algebraically closed field of characteristic zero, if

\pi: ak{g} \to ak{gl}(V) is a finite-dimensional representation of a solvable Lie algebra, then there's a flag V = V _{0} \supset V _{1} \supset … \supset V _{n} = 0 of invariant subspaces of \pi( ak{g}) with \operatorname{codim} V _{i} = i , meaning that \pi(X)(V _{i)} \subseteq V _{i} for each X \in ak{g} and i.

Put in another way, the theorem says there is a basis for V such that all linear transformations in

\pi( ak{g}) are represented by upper triangular matrices. This is a generalization of the result of Frobenius that commuting matrices are simultaneously upper triangularizable, as commuting matrices generate an abelian Lie algebra, which is a fortiori solvable.

A consequence of Lie's theorem is that any finite dimensional solvable Lie algebra over a field of characteristic 0 has a nilpotent derived algebra (see

1. Consequences

). Also, to each flag in a finite-dimensional vector space V, there correspond a Borel subalgebra (that consist of linear transformations stabilizing the flag); thus, the theorem says that

\pi( ak{g}) is contained in some Borel subalgebra of ak{gl}(V) .

Counter-example

For algebraically closed fields of characteristic p>0 Lie's theorem holds provided the dimension of the representation is less than p (see the proof below), but can fail for representations of dimension p. An example is given by the 3-dimensional nilpotent Lie algebra spanned by 1, x, and d/dx acting on the p-dimensional vector space k[''x'']/(x^p), which has no eigenvectors. Taking the semidirect product of this 3-dimensional Lie algebra by the p-dimensional representation (considered as an abelian Lie algebra) gives a solvable Lie algebra whose derived algebra is not nilpotent.

Proof

The proof is by induction on the dimension of

ak{g} and consists of several steps. (Note: the structure of the proof is very similar to that for Engel's theorem.) The basic case is trivial and we assume the dimension of ak{g} is positive. We also assume V is not zero. For simplicity, we write X ⋅ v = \pi(X)(v) .

Step 1: Observe that the theorem is equivalent to the statement:

• There exists a vector in V that is an eigenvector for each linear transformation in

\pi( ak{g}) .Indeed, the theorem says in particular that a nonzero vector spanning V _{n-1} is a common eigenvector for all the linear transformations in \pi( ak{g}) . Conversely, if v is a common eigenvector, take V _{n-1} to its span and then \pi( ak{g}) admits a common eigenvector in the quotient V/V _{n-1} ; repeat the argument.

Step 2: Find an ideal

ak{h} of codimension one in ak{g} .

Let

D ak{g} = [ ak{g}, ak{g}] be the derived algebra. Since ak{g} is solvable and has positive dimension, D ak{g} \ne ak{g} and so the quotient ak{g}/D ak{g} is a nonzero abelian Lie algebra, which certainly contains an ideal of codimension one and by the ideal correspondence, it corresponds to an ideal of codimension one in ak{g} .

Step 3: There exists some linear functional

λ in ak{h} ^{*} such thatV _{λ} = \{ v \in V | X ⋅ v = λ (X) v, X \in ak{h} \} is nonzero. This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional).

Step 4:

V _{λ} is a ak{g} -invariant subspace. (Note this step proves a general fact and does not involve solvability.)

Let

Y \in ak{g} , v \in V _{λ} , then we need to prove Y ⋅ v \in V _{λ} . If v = 0 then it's obvious, so assume v \ne 0 and set recursively v _{0} = v, v _{i+1} = Y ⋅ v _{i} . Let U = \operatorname{span} \{ v _{i} | i \ge 0 \} and \ell \in N _{0} be the largest such that v _{0,\ldots,v} _{\ell} are linearly independent. Then we'll prove that they generate U and thus \alpha = (v _{0,\ldots,v} _{\ell)} is a basis of U. Indeed, assume by contradiction that it's not the case and let m \in N _{0} be the smallest such that v _{m} \notin \langle v _{0,\ldots,v} _{\ell\rangle} , then obviously m \ge \ell + 1 . Since v _{0,\ldots,v} _{\ell+1} are linearly dependent, v _{\ell+1} is a linear combination of v _{0,\ldots,v} _{\ell} . Applying the map Y ^{m-\ell-1} it follows that v _{m} is a linear combination of v _{m-\ell-1} ,\ldots,v _{m-1} . Since by the minimality of m each of these vectors is a linear combination of v _{0,\ldots,v} _{\ell} , so is v _{m} , and we get the desired contradiction. We'll prove by induction that for every n \in N _{0} and X \in ak{h} there exist elements a _{0,n,X} ,\ldots,a _{n,n,X} of the base field such that a _{n,n,X} =λ (X) andX ⋅ v _{n} = \sum ^{n} _{i =0} a _{i,n,X} v _{i.} The n =0 case is straightforward since X ⋅ v _{0} = λ (X) v _{0} . Now assume that we have proved the claim for some n \in N _{0} and all elements of ak{h} and let X \in ak{h} . Since ak{h} is an ideal, it's [X,Y] \in ak{h} , and thusX ⋅ v _{n+1} = Y ⋅ (X ⋅ v _{n)} + [X, Y] ⋅ v _{n} = Y ⋅ \sum ^{n} _{i =0} a _{i,n,X} v _{i} + \sum ^{n} _{i =0} a _{i,n,[X,Y]} v _{i} = a _{0,n,[X,Y]} v _{0} + \sum ^{n} _{i =1} (a _{i-1,n,X} + a _{i,n,[X,Y]} )v _{i} + λ (X)v _{n+1} , and the induction step follows. This implies that for every X \in ak{h} the subspace U is an invariant subspace of X and the matrix of the restricted map \pi(X)| _{U} in the basis \alpha is upper triangular with diagonal elements equal to λ (X) , hence \operatorname{tr}(\pi(X)| _{U)} = \dim(U) λ (X) . Applying this with [X,Y] \in ak{h} instead of X gives \operatorname{tr}(\pi([X,Y])| _{U)} = \dim(U) λ ([X,Y]) . On the other hand, U is also obviously an invariant subspace of Y, and so\operatorname{tr}(\pi([X,Y])| _{U)} = \operatorname{tr}([\pi(X),\pi(Y)]| _{U])} = \operatorname{tr}([\pi(X)| _{U,} \pi(Y)| _{U])} = 0 since commutators have zero trace, and thus \dim(U) λ ([X,Y]) = 0 . Since \dim(U) > 0 is invertible (because of the assumption on the characteristic of the base field), λ ([X, Y]) = 0 andX ⋅ (Y ⋅ v) = Y ⋅ (X ⋅ v) + [X, Y] ⋅ v = Y ⋅ ( λ (X) v) + λ ([X, Y]) v = λ (X) (Y ⋅ v), and so Y ⋅ v \in V _{λ} .

Step 5: Finish up the proof by finding a common eigenvector.

Write

ak{g} = ak{h} + L where L is a one-dimensional vector subspace. Since the base field is algebraically closed, there exists an eigenvector in V _{λ} for some (thus every) nonzero element of L. Since that vector is also eigenvector for each element of ak{h} , the proof is complete. \square

Consequences

\operatorname{ad}: ak{g} \to ak{gl}( ak{g}) of a (finite-dimensional) solvable Lie algebra ak{g} over an algebraically closed field of characteristic zero; thus, one can choose a basis on ak{g} with respect to which \operatorname{ad}( ak{g}) consists of upper triangular matrices. It follows easily that for each x, y \in ak{g} , \operatorname{ad}([x, y]) = [\operatorname{ad}(x), \operatorname{ad}(y)] has diagonal consisting of zeros; i.e., \operatorname{ad}([x, y]) is a strictly upper triangular matrix. This implies that [ ak g, ak g] is a nilpotent Lie algebra. Moreover, if the base field is not algebraically closed then solvability and nilpotency of a Lie algebra is unaffected by extending the base field to its algebraic closure. Hence, one concludes the statement (the other implication is obvious):

A finite-dimensional Lie algebra

ak g over a field of characteristic zero is solvable if and only if the derived algebra D ak g = [ ak g, ak g] is nilpotent.

Lie's theorem also establishes one direction in Cartan's criterion for solvability:

If V is a finite-dimensional vector space over a field of characteristic zero and

ak{g} \subseteq ak{gl}(V) a Lie subalgebra, then ak{g} is solvable if and only if \operatorname{tr}(XY) = 0 for every X \in ak{g} and Y \in [ ak{g}, ak{g}] .

Indeed, as above, after extending the base field, the implication

⇒ is seen easily. (The converse is more difficult to prove.)

Lie's theorem (for various V) is equivalent to the statement:

For a solvable Lie algebra

ak g over an algebraically closed field of characteristic zero, each finite-dimensional simple ak{g} -module (i.e., irreducible as a representation) has dimension one.Indeed, Lie's theorem clearly implies this statement. Conversely, assume the statement is true. Given a finite-dimensional ak g -module V, let V _{1} be a maximal ak g -submodule (which exists by finiteness of the dimension). Then, by maximality, V/V _{1} is simple; thus, is one-dimensional. The induction now finishes the proof.

The statement says in particular that a finite-dimensional simple module over an abelian Lie algebra is one-dimensional; this fact remains true over any base field since in this case every vector subspace is a Lie subalgebra.

Here is another quite useful application:

Let

ak{g} be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero with radical \operatorname{rad}( ak{g}) . Then each finite-dimensional simple representation \pi: ak{g} \to ak{gl}(V) is the tensor product of a simple representation of ak{g}/\operatorname{rad}( ak{g}) with a one-dimensional representation of ak{g} (i.e., a linear functional vanishing on Lie brackets).

By Lie's theorem, we can find a linear functional

λ of \operatorname{rad}( ak{g}) so that there is the weight space V _{λ} of \operatorname{rad}( ak{g}) . By Step 4 of the proof of Lie's theorem, V _{λ} is also a ak{g} -module; so V = V _{λ} . In particular, for each X \in \operatorname{rad}( ak{g}) , \operatorname{tr}(\pi(X)) = \dim(V) λ (X) . Extend λ to a linear functional on ak{g} that vanishes on [ ak g, ak g] ; λ is then a one-dimensional representation of ak{g} . Now, (\pi, V) \simeq (\pi, V) ⊗ ( -λ ) ⊗ λ . Since \pi coincides with λ on \operatorname{rad}( ak{g}) , we have that V ⊗ ( -λ ) is trivial on \operatorname{rad}( ak{g}) and thus is the restriction of a (simple) representation of ak{g}/\operatorname{rad}( ak{g}) . \square

See also

• Engel's theorem, which concerns a nilpotent Lie algebra.
• Lie–Kolchin theorem, which is about a (connected) solvable linear algebraic group.

Sources

• • .