In mathematics, specifically the theory of Lie algebras, Lie's theorem states that,[1] over an algebraically closed field of characteristic zero, if
is a finite-dimensional representation of a solvable Lie algebra, then there's a flag
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.
[2] 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 #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
[1] 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]/(xp), which has no eigenvectors.
(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
Conversely, if v is a common eigenvector, take
Step 3: There exists some linear functional
This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional).
(Note this step proves a general fact and does not involve solvability.)
Since by the minimality of m each of these vectors is a linear combination of
is upper triangular with diagonal elements equal to
On the other hand, U is also obviously an invariant subspace of Y, and so since commutators have zero trace, and thus
is invertible (because of the assumption on the characteristic of the base field),
Step 5: Finish up the proof by finding a common eigenvector.
Since the base field is algebraically closed, there exists an eigenvector in
The theorem applies in particular to the adjoint representation
of a (finite-dimensional) solvable Lie algebra
over an algebraically closed field of characteristic zero; thus, one can choose a basis on
consists of upper triangular matrices.
is a strictly upper triangular matrix.
Hence, one concludes the statement (the other implication is obvious):[4] Lie's theorem also establishes one direction in Cartan's criterion for solvability: Indeed, as above, after extending the base field, the implication
Conversely, assume the statement is true.
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.
[7] Here is another quite useful application:[8] By Lie's theorem, we can find a linear functional
By Step 4 of the proof of Lie's theorem,