Lie's theorem

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.

The proof is by induction on the dimension of ${\displaystyle {\mathfrak {g}}}$ 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 ${\displaystyle {\mathfrak {g}}}$ is positive. We also assume V is not zero. For simplicity, we write ${\displaystyle X\cdot v=\pi (X)v}$.

Step 1: Observe that the theorem is equivalent to the statement:[3]

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

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

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

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

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

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

is nonzero.

This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional).

Step 4: ${\displaystyle V_{\lambda }}$ is a ${\displaystyle {\mathfrak {g}}}$-module.

(Note this step proves a general fact and does not involve solvability.)

Let ${\displaystyle Y}$ be in ${\displaystyle {\mathfrak {g}}}$, ${\displaystyle v\in V_{\lambda }}$ and set recursively ${\displaystyle v_{0}=v,\,v_{i+1}=Y\cdot v_{i}}$. For any ${\displaystyle X\in {\mathfrak {h}}}$, since ${\displaystyle {\mathfrak {h}}}$ is an ideal,

${\displaystyle X\cdot v_{i}=\lambda (X)v_{i}+\lambda ([X,Y])v_{i-1}}$.

This says that ${\displaystyle X}$ (that is ${\displaystyle \pi (X)}$) restricted to ${\displaystyle U=\operatorname {span} \{v_{i}|i\geq 0\}}$ is represented by a matrix whose diagonal is ${\displaystyle \lambda (X)}$ repeated. Hence, ${\displaystyle \dim(U)\lambda ([X,Y])=\operatorname {tr} ([\pi (X)|_{U},\pi (Y)|_{U}])=0}$. Since ${\displaystyle \dim(U)}$ is invertible, ${\displaystyle \lambda ([X,Y])=0}$ and ${\displaystyle Y\cdot v}$ is an eigenvector for X.

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

Write ${\displaystyle {\mathfrak {g}}={\mathfrak {h}}+L}$ where L is a one-dimensional vector subspace. Since the base field k is algebraically closed, there exists an eigenvector in ${\displaystyle V_{\lambda }}$ for some (thus every) nonzero element of L. Since that vector is also eigenvector for each element of ${\displaystyle {\mathfrak {h}}}$, the proof is complete. ${\displaystyle \square }$

The theorem applies in particular to the adjoint representation ${\displaystyle \operatorname {ad}$ :{\mathfrak {g}}\to {\mathfrak {gl}}({\mathfrak {g}})} of a (finite-dimensional) solvable Lie algebra ${\displaystyle {\mathfrak {g}}}$; thus, one can choose a basis on ${\displaystyle {\mathfrak {g}}}$ with respect to which ${\displaystyle \operatorname {ad} ({\mathfrak {g}})}$ consists of upper-triangular matrices. It follows easily that for each ${\displaystyle x,y\in {\mathfrak {g}}}$, ${\displaystyle \operatorname {ad} ([x,y])=[\operatorname {ad} (x),\operatorname {ad} (y)]}$ has diagonal consisting of zeros; i.e., ${\displaystyle \operatorname {ad} ([x,y])}$ is a nilpotent matrix. By Engel's theorem, this implies that ${\displaystyle [{\mathfrak {g}},{\mathfrak {g}}]}$ is a nilpotent Lie algebra; the converse is obviously true as well. Moreover, whether a linear transformation is nilpotent or not can be determined after extending the base field to its algebraic closure. Hence, one concludes the statement:[4]

A finite-dimensional Lie algebra ${\displaystyle {\mathfrak {g}}}$ over a field of characteristic zero is solvable if and only if the derived algebra ${\displaystyle D{\mathfrak {g}}=[{\mathfrak {g}},{\mathfrak {g}}]}$ is nilpotent.

Lie's theorem also establishes one direction in Cartan's criterion for solvability: if V is a finite-dimensional vector over a field of characteristic zero and ${\displaystyle {\mathfrak {g}}\subset {\mathfrak {gl}}(V)}$ a Lie subalgebra, then ${\displaystyle {\mathfrak {g}}}$ is solvable if and only if ${\displaystyle \operatorname {tr} (XY)=0}$ for every ${\displaystyle X\in {\mathfrak {g}}}$ and ${\displaystyle Y\in [{\mathfrak {g}},{\mathfrak {g}}]}$.[5]

Indeed, as above, after extending the base field, the implication ${\displaystyle \Rightarrow }$ is seen easily. (The converse is more difficult to prove.)

Lie's theorem (for various V) is equivalent to the statement:[6]

For a solvable Lie algebra ${\displaystyle {\mathfrak {g}}}$, each finite-dimensional simple ${\displaystyle {\mathfrak {g}}}$-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 ${\displaystyle {\mathfrak {g}}}$-module V, let ${\displaystyle V_{1}}$ be a maximal ${\displaystyle {\mathfrak {g}}}$-submodule (which exists by finiteness of the dimension). Then, by maximality, ${\displaystyle V/V_{1}}$ 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 without the assumption that the base field has characteristic zero.[7]

Here is another quite useful application:[8]

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

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

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

• Fulton, William; Harris, Joe (1991). Representation theory. A first course. Graduate Texts in Mathematics, Readings in Mathematics. 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, Lie algebras, Republication of the 1962 original. Dover Publications, Inc., New York, 1979. ISBN 0-486-63832-4
• Jean-Pierre Serre: Complex Semisimple Lie Algebras, Springer, Berlin, 2001. ISBN 3-5406-7827-1