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 $\pi ({\mathfrak {g}})$ stabilizes a flag $V=V_{0}\supset \cdots \supset V_{i}\supset \cdots \supset V_{n}=0,\operatorname {codim} V_{i}=i$; "stabilizes" means $\pi (X)V_{i}\subset V_{i}$ 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 form 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.

Counter-example

For algebraically closed fields of characteristic p>0 Lie's theorem holds provided the dimension of the representation is less than p, 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. The basic case is trivial and we assume the dimension of ${\mathfrak {g}}$ is positive. 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,

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, $D{\mathfrak {g}}\neq {\mathfrak {g}}$ 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 ${\mathfrak {g}}$-module.

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

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

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

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

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

Write ${\mathfrak {g}}={\mathfrak {h}}+L$ where L is one-dimensional vector subspace. Since the base field k 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. $\square$

See also

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

References

• 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.
• Jean-Pierre Serre: Complex Semisimple Lie Algebras, Springer, Berlin, 2001. ISBN 3-5406-7827-1