# 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 is 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 #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 ${\mathfrak {gl}}(V)$.

## 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 ${\mathfrak {g}}$ 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, $D{\mathfrak {g}}\neq {\mathfrak {g}}$ 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 $v\neq 0$ and set recursively $v_{0}=v,\,v_{i+1}=Y\cdot v_{i}$. Let $U=\operatorname {span} \{v_{i}|i\geq 0\}$ 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 $m\geq \ell +1$. 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 ${\mathfrak {g}}/\operatorname {rad} ({\mathfrak {g}})$. $\square$