Lie's theorem
In mathematics,[] specifically the theory of Lie algebras, Lie's theorem states that, over an algebraically closed field of characteristic zero, if is a finite-dimensional representation of a solvable Lie algebra, then there is a flag of invariant subspaces of with, meaning that for each and i.
Put in another way, the theorem says there is a basis for V such that all linear transformations in 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. Also, to each flag in a finite-dimensional vector space V, there correspond a Borel subalgebra ; thus, the theorem says that is contained in some Borel subalgebra of.
Counter-example
For algebraically closed fields of characteristic p>0 Lie's theorem holds provided the dimension of the representation is less than p, 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/, which has no eigenvectors. Taking the semidirect product of this 3-dimensional Lie algebra by the p-dimensional representation gives a solvable Lie algebra whose derived algebra is not nilpotent.Proof
The proof is by induction on the dimension of and consists of several steps. The basic case is trivial and we assume the dimension of is positive. We also assume V is not zero. For simplicity, we write.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.
Step 2: Find an ideal of codimension one in.
Let be the derived algebra. Since is solvable and has positive dimension, and so the quotient 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.
Step 3: There exists some linear functional in such that
is nonzero. This follows from the inductive hypothesis.
Step 4: is a -invariant subspace.
Let,, then we need to prove. If then it's obvious, so assume and set recursively. Let and be the largest such that are linearly independent. Then we'll prove that they generate U and thus is a basis of U. Indeed, assume by contradiction that it's not the case and let be the smallest such that, then obviously. Since are linearly dependent, is a linear combination of. Applying the map it follows that is a linear combination of. Since by the minimality of m each of these vectors is a linear combination of, so is, and we get the desired contradiction. We'll prove by induction that for every and there exist elements of the base field such that and
The case is straightforward since. Now assume that we have proved the claim for some and all elements of and let. Since is an ideal, it's, and thus
and the induction step follows. This implies that for every the subspace U is an invariant subspace of X and the matrix of the restricted map in the basis is upper triangular with diagonal elements equal to, hence. Applying this with instead of X gives. On the other hand, U is also obviously an invariant subspace of Y, and so
since commutators have zero trace, and thus. Since is invertible, and
and so.
Step 5: Finish up the proof by finding a common eigenvector.
Write where L is a one-dimensional vector subspace. Since the base field is algebraically closed, there exists an eigenvector in for some nonzero element of L. Since that vector is also eigenvector for each element of, the proof is complete.
Consequences
The theorem applies in particular to the adjoint representation of a solvable Lie algebra over an algebraically closed field of characteristic zero; thus, one can choose a basis on with respect to which consists of upper triangular matrices. It follows easily that for each, has diagonal consisting of zeros; i.e., is a strictly upper triangular matrix. This implies that 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 :Lie's theorem also establishes one direction in Cartan's criterion for solvability:
Indeed, as above, after extending the base field, the implication is seen easily.
Lie's theorem is equivalent to the statement:
Indeed, Lie's theorem clearly implies this statement. Conversely, assume the statement is true. Given a finite-dimensional -module V, let be a maximal -submodule. Then, by maximality, 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:
By Lie's theorem, we can find a linear functional of so that there is the weight space of. By Step 4 of the proof of Lie's theorem, is also a -module; so. In particular, for each,. Extend to a linear functional on that vanishes on ; is then a one-dimensional representation of. Now,. Since coincides with on, we have that is trivial on and thus is the restriction of a representation of.