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's 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](https://bliptext.com/articles/derived-algebra-of-a-lie-algebra) (see ). 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 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 (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. 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 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 is positive. We also assume V is not zero. For simplicity, we write. Step 1: Observe that the theorem is equivalent to the statement: Indeed, the theorem says in particular that a nonzero vector spanning V_{n-1} is a common eigenvector for all the linear transformations in. Conversely, if v is a common eigenvector, take V_{n-1} to be its span and then admits a common eigenvector in the quotient V/V_{n-1}; repeat the argument. 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 \lambda in such that is nonzero. This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional). Step 4: V_{\lambda} is a -invariant subspace. (Note this step proves a general fact and does not involve solvability.) Let, , then we need to prove. If v = 0 then it's obvious, so assume v \ne 0 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, v_{\ell+1} is a linear combination of. Applying the map it follows that v_m is a linear combination of. Since by the minimality of m each of these vectors is a linear combination of, so is v_m, 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 n=0 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 \pi(X)|U in the basis \alpha is upper triangular with diagonal elements equal to \lambda(X), 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 \dim(U) > 0 is invertible (because of the assumption on the characteristic of the base field), 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 V{\lambda} for some (thus every) nonzero element of L. Since that vector is also eigenvector for each element of, the proof is complete. \square

Consequences

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 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 (the other implication is obvious): Lie's theorem also establishes one direction in Cartan's criterion for solvability: Indeed, as above, after extending the base field, the implication \Rightarrow is seen easily. (The converse is more difficult to prove.) Lie's theorem (for various V) is equivalent to the statement: Indeed, Lie's theorem clearly implies this statement. Conversely, assume the statement is true. Given a finite-dimensional \mathfrak g-module V, let V_1 be a maximal \mathfrak 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: By Lie's theorem, we can find a linear functional \lambda of so that there is the weight space V_{\lambda} of. By Step 4 of the proof of Lie's theorem, V_{\lambda} is also a -module; so. In particular, for each,. Extend \lambda to a linear functional on that vanishes on ; \lambda is then a one-dimensional representation of. Now,. Since \pi coincides with \lambda on, we have that is trivial on and thus is the restriction of a (simple) representation of. \square

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