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
admits a common eigenvector in the quotient
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
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,