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 $\pi :\mathfrak{𝔤}\to \mathfrak{𝔤}\mathfrak{𝔩}(V)$ 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 $\mathrm{codim}{V}_i=i$, 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 #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 $\pi (\mathfrak{𝔤})$ is contained in some Borel subalgebra of $\mathfrak{𝔤}\mathfrak{𝔩}(V)$.

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 $\mathfrak{𝔤}$ 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:

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

Indeed, the theorem says in particular that a nonzero vector spanning $V_{n-1}$ is a common eigenvector for all the linear transformations in $\pi (\mathfrak{𝔤})$. Conversely, if v is a common eigenvector, take $V_{n-1}$ to be its span and then $\pi (\mathfrak{𝔤})$ admits a common eigenvector in the quotient $V/V_{n-1}$; repeat the argument. 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 $U=\mathrm{span}\{v_i|i\ge 0\}$ and $\ell \in {\mathbb{\mathbb{N}}}_0$ 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 $m\in {\mathbb{\mathbb{N}}}_0$ 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 $n\in {\mathbb{\mathbb{N}}}_0$ and $X\in \mathfrak{𝔥}$ 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 $n\in {\mathbb{\mathbb{N}}}_0$ 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 $\mathrm{tr}(\pi (X){|}_U)=\dim (U)\lambda (X)$. Applying this with $[X,Y]\in \mathfrak{𝔥}$ 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 $\mathfrak{𝔤}=\mathfrak{𝔥}+L$ 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. $◻$

Consequences

The theorem applies in particular to the adjoint representation $\mathrm{ad}:\mathfrak{𝔤}\to \mathfrak{𝔤}\mathfrak{𝔩}(\mathfrak{𝔤})$ 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 $\mathrm{ad}(\mathfrak{𝔤})$ consists of upper triangular matrices. It follows easily that for each $x,y\in \mathfrak{𝔤}$, $\mathrm{ad}([x,y])=[\mathrm{ad}(x),\mathrm{ad}(y)]$ has diagonal consisting of zeros; i.e., $\mathrm{ad}([x,y])$ is a strictly upper triangular matrix. This implies that $[\mathfrak{𝔤},\mathfrak{𝔤}]$ 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 $\mathfrak{𝔤}\subseteq \mathfrak{𝔤}\mathfrak{𝔩}(V)$ a Lie subalgebra, then $\mathfrak{𝔤}$ is solvable if and only if $\mathrm{tr}(XY)=0$ 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 $\mathrm{rad}(\mathfrak{𝔤})$. Then each finite-dimensional simple representation $\pi :\mathfrak{𝔤}\to \mathfrak{𝔤}\mathfrak{𝔩}(V)$ is the tensor product of a simple representation of $\mathfrak{𝔤}/\mathrm{rad}(\mathfrak{𝔤})$ 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 $\mathrm{rad}(\mathfrak{𝔤})$ so that there is the weight space $V_{\lambda }$ of $\mathrm{rad}(\mathfrak{𝔤})$. 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 $X\in \mathrm{rad}(\mathfrak{𝔤})$, $\mathrm{tr}(\pi (X))=\dim (V)\lambda (X)$. 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 $\mathrm{rad}(\mathfrak{𝔤})$, we have that $V\otimes (-\lambda )$ is trivial on $\mathrm{rad}(\mathfrak{𝔤})$ and thus is the restriction of a (simple) representation of $\mathfrak{𝔤}/\mathrm{rad}(\mathfrak{𝔤})$. $◻$

See also

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

Sources

• Fulton, William; Harris, Joe (1991). Representation theory. A first course. Graduate Texts in Mathematics, Readings in Mathematics. Vol. 129. New York: Springer-Verlag. doi:10.1007/978-1-4612-0979-9. ISBN 978-0-387-97495-8. MR 1153249. OCLC 246650103.
• Humphreys, James E. (1972), Introduction to Lie Algebras and Representation Theory, Berlin, New York: Springer-Verlag, ISBN 978-0-387-90053-7.
• Jacobson, Nathan (1979), Lie algebras (Republication of the 1962 original ed.), New York: Dover Publications, Inc., ISBN 0-486-63832-4, MR 0559927
• Serre, Jean-Pierre (2001), Complex Semisimple Lie Algebras, Berlin: Springer, doi:10.1007/978-3-642-56884-8, ISBN 3-5406-7827-1, MR 1808366