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 $$\pi: \mathfrak{g} \to \mathfrak{gl}(V)$$ is a finite-dimensional representation of a solvable Lie algebra, then there's a flag $$V = V_0 \supset V_1 \supset \cdots \supset V_n = 0$$ of invariant subspaces of $$\pi(\mathfrak{g})$$ with $$\operatorname{codim} V_i = i$$, meaning that $$\pi(X)(V_i) \subseteq V_i$$ for each $$X \in \mathfrak{g}$$ and i.

Put in another way, the theorem says there is a basis for V such that all linear transformations in $$\pi(\mathfrak{g})$$ 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 ). 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 $$\pi(\mathfrak{g})$$ is contained in some Borel subalgebra of $$\mathfrak{gl}(V)$$.

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]/(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 $$\mathfrak{g}$$ 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 $$\mathfrak{g}$$ is positive. We also assume V is not zero. For simplicity, we write $$X \cdot v = \pi(X)(v)$$.

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 $$\pi(\mathfrak{g})$$. Conversely, if v is a common eigenvector, take $$V_{n-1}$$ to its span and then $$\pi(\mathfrak{g})$$ admits a common eigenvector in the quotient $$V/V_{n-1}$$; repeat the argument.
 * There exists a vector in V that is an eigenvector for each linear transformation in $$\pi(\mathfrak{g})$$.

Step 2: Find an ideal $$\mathfrak{h}$$ of codimension one in $$\mathfrak{g}$$.

Let $$D\mathfrak{g} = [\mathfrak{g}, \mathfrak{g}]$$ be the derived algebra. Since $$\mathfrak{g}$$ is solvable and has positive dimension, $$D\mathfrak{g} \ne \mathfrak{g}$$ and so the quotient $$\mathfrak{g}/D\mathfrak{g}$$ 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 $$\mathfrak{g}$$.

Step 3: There exists some linear functional $$\lambda$$ in $$\mathfrak{h}^*$$ such that
 * $$V_{\lambda} = \{ v \in V | X \cdot v = \lambda(X) v, X \in \mathfrak{h} \}$$

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 $$\mathfrak{g}$$-invariant subspace. (Note this step proves a general fact and does not involve solvability.)

Let $$Y \in \mathfrak{g}$$, $$v \in V_{\lambda}$$, then we need to prove $$Y \cdot v \in V_{\lambda}$$. If $$v = 0$$ then it's obvious, so assume $$v \ne 0$$ and set recursively $$v_0 = v, \, v_{i+1} = Y \cdot v_i$$. Let $$U = \operatorname{span} \{ v_i | i \ge 0 \}$$ and $$\ell \in \mathbb{N}_0$$ 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 $$m \in \mathbb{N}_0$$ 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 $$n \in \mathbb{N}_0$$ and $$X \in \mathfrak{h}$$ there exist elements $$a_{0,n,X},\ldots,a_{n,n,X}$$ of the base field such that $$a_{n,n,X}=\lambda(X)$$ and
 * $$X \cdot v_n = \sum_{i=0}^{n} a_{i,n,X}v_i.$$

The $$n=0$$ case is straightforward since $$X \cdot v_0 = \lambda(X) v_0$$. Now assume that we have proved the claim for some $$n \in \mathbb{N}_0$$ and all elements of $$\mathfrak{h}$$ and let $$X \in \mathfrak{h}$$. Since $$\mathfrak{h}$$ is an ideal, it's $$[X,Y] \in \mathfrak{h}$$, and thus
 * $$X \cdot v_{n+1} = Y \cdot (X \cdot v_n) + [X, Y] \cdot v_n = Y \cdot \sum_{i=0}^{n} a_{i,n,X}v_i + \sum_{i=0}^{n} a_{i,n,[X,Y]}v_i = a_{0,n,[X,Y]}v_0 + \sum_{i=1}^{n} (a_{i-1,n,X} + a_{i,n,[X,Y]})v_i + \lambda(X)v_{n+1},$$

and the induction step follows. This implies that for every $$X \in \mathfrak{h}$$ 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 $$\lambda(X)$$, hence $$\operatorname{tr}(\pi(X)|_U) = \dim(U) \lambda(X)$$. Applying this with $$[X,Y] \in \mathfrak{h}$$ instead of X gives $$\operatorname{tr}(\pi([X,Y])|_U) = \dim(U) \lambda([X,Y])$$. 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 $$\dim(U) \lambda([X,Y]) = 0$$. Since $$\dim(U) > 0$$ is invertible (because of the assumption on the characteristic of the base field), $$\lambda([X, Y]) = 0$$ and
 * $$X \cdot (Y \cdot v) = Y \cdot (X \cdot v) + [X, Y] \cdot v = Y \cdot (\lambda(X) v) + \lambda([X, Y]) v = \lambda(X) (Y \cdot v),$$

and so $$Y \cdot v \in V_{\lambda}$$.

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

Write $$\mathfrak{g} = \mathfrak{h} + L$$ 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 $$\mathfrak{h}$$, the proof is complete. $$\square$$

Consequences
The theorem applies in particular to the adjoint representation $$\operatorname{ad}: \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$$ of a (finite-dimensional) solvable Lie algebra $$\mathfrak{g}$$ over an algebraically closed field of characteristic zero; thus, one can choose a basis on $$\mathfrak{g}$$ with respect to which $$\operatorname{ad}(\mathfrak{g})$$ consists of upper triangular matrices. It follows easily that for each $$x, y \in \mathfrak{g}$$, $$\operatorname{ad}([x, y]) = [\operatorname{ad}(x), \operatorname{ad}(y)]$$ has diagonal consisting of zeros; i.e., $$\operatorname{ad}([x, y])$$ is a strictly upper triangular matrix. This implies that $$[\mathfrak g, \mathfrak g]$$ 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 $$\mathfrak g$$ over a field of characteristic zero is solvable if and only if the derived algebra $$D \mathfrak g = [\mathfrak g, \mathfrak g]$$ 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 $$\mathfrak{g} \subseteq \mathfrak{gl}(V)$$ a Lie subalgebra, then $$\mathfrak{g}$$ is solvable if and only if $$\operatorname{tr}(XY) = 0$$ for every $$X \in \mathfrak{g}$$ and $$Y \in [\mathfrak{g}, \mathfrak{g}]$$.

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

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