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 $\pi :𝔤\to 𝔤𝔩(V)$ 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 $\pi (𝔤)$ with $\mathrm{codim}{V}_i=i$, meaning that $\pi (X)(V_i)\subseteq V_i$ for each $X\in 𝔤$ and i.

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

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]/(x^p), 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 $𝔤$ 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 $𝔤$ 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:

• There exists a vector in V that is an eigenvector for each linear transformation in $\pi (𝔤)$.

Indeed, the theorem says in particular that a nonzero vector spanning $V_{n-1}$ is a common eigenvector for all the linear transformations in $\pi (𝔤)$. Conversely, if v is a common eigenvector, take $V_{n-1}$ to be its span and then $\pi (𝔤)$ admits a common eigenvector in the quotient $V/V_{n-1}$; repeat the argument.

Step 2: Find an ideal $𝔥$ of codimension one in $𝔤$.

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

Step 3: There exists some linear functional $\lambda$ in ${𝔥}^{*}$ such that

$V_{\lambda }=\{v\in V|X\cdot v=\lambda (X)v,X\in 𝔥\}$

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

Let $Y\in 𝔤$, $v\in V_{\lambda }$, 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 $U=\mathrm{span}\{v_i|i\ge 0\}$ and $\ell \in {\mathbb{N}}_0$ be the largest such that $v_0,\dots ,v_{\ell }$ are linearly independent. Then we'll prove that they generate U and thus $\alpha =(v_0,\dots ,v_{\ell })$ 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 $v_m\notin ⟨v_0,\dots ,v_{\ell }⟩$, then obviously $m\ge \ell +1$. Since $v_0,\dots ,v_{\ell +1}$ are linearly dependent, $v_{\ell +1}$ is a linear combination of $v_0,\dots ,v_{\ell }$. Applying the map $Y^{m-\ell -1}$ it follows that $v_m$ is a linear combination of $v_{m-\ell -1},\dots ,v_{m-1}$. Since by the minimality of m each of these vectors is a linear combination of $v_0,\dots ,v_{\ell }$, so is $v_m$, and we get the desired contradiction. We'll prove by induction that for every $n\in {\mathbb{N}}_0$ and $X\in 𝔥$ there exist elements $a_{0,n,X},\dots ,a_{n,n,X}$ 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 $𝔥$ and let $X\in 𝔥$. Since $𝔥$ is an ideal, it's $[X,Y]\in 𝔥$, 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 $X\in 𝔥$ 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 $\mathrm{tr}(\pi (X){|}_U)=\dim (U)\lambda (X)$. Applying this with $[X,Y]\in 𝔥$ instead of X gives $\mathrm{tr}(\pi ([X,Y]){|}_U)=\dim (U)\lambda ([X,Y])$. On the other hand, U is also obviously an invariant subspace of Y, and so

$\mathrm{tr}(\pi ([X,Y]){|}_U)=\mathrm{tr}([\pi (X),\pi (Y)]{|}_U])=\mathrm{tr}([\pi (X){|}_U,\pi (Y){|}_U])=0$

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 $𝔤=𝔥+L$ 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 $𝔥$, the proof is complete. $◻$

Consequences

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

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 $𝔤$ over an algebraically closed field of characteristic zero, each finite-dimensional simple $𝔤$-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 $𝔤$-module V, let $V_1$ be a maximal $𝔤$-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 $𝔤$ be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero with radical $\mathrm{rad}(𝔤)$. Then each finite-dimensional simple representation $\pi :𝔤\to 𝔤𝔩(V)$ is the tensor product of a simple representation of $𝔤/\mathrm{rad}(𝔤)$ with a one-dimensional representation of $𝔤$ (i.e., a linear functional vanishing on Lie brackets).

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

See also

• Engel's theorem, which concerns a nilpotent Lie algebra.
• Lie–Kolchin theorem, which is about a (connected) solvable linear algebraic group.

Sources

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