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Zheng, H. Lie's Theorem. Encyclopedia. Available online: https://encyclopedia.pub/entry/31063 (accessed on 03 December 2023).
Zheng H. Lie's Theorem. Encyclopedia. Available at: https://encyclopedia.pub/entry/31063. Accessed December 03, 2023.
Zheng, Handwiki. "Lie's Theorem" Encyclopedia, https://encyclopedia.pub/entry/31063 (accessed December 03, 2023).
Zheng, H.(2022, October 25). Lie's Theorem. In Encyclopedia. https://encyclopedia.pub/entry/31063
Zheng, Handwiki. "Lie's Theorem." Encyclopedia. Web. 25 October, 2022.
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 $\displaystyle{ \pi: \mathfrak{g} \to \mathfrak{gl}(V) }$ is a finite-dimensional representation of a solvable Lie algebra, then $\displaystyle{ \pi(\mathfrak{g}) }$ stabilizes a flag $\displaystyle{ V = V_0 \supset V_1 \supset \cdots \supset V_n = 0, \operatorname{codim} V_i = i }$; "stabilizes" means $\displaystyle{ \pi(X) V_i \subset V_i }$ for each $\displaystyle{ X \in \mathfrak{g} }$ and i. Put in another way, the theorem says there is a basis for V such that all linear transformations in $\displaystyle{ \pi(\mathfrak{g}) }$ 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 (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 $\displaystyle{ \pi(\mathfrak{g}) }$ is contained in some Borel subalgebra of $\displaystyle{ \mathfrak{gl}(V) }$.

linear transformations triangularizable subalgebra

## 1. 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]/(xp), 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.

## 2. Proof

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:

• 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, $\displaystyle{ D\mathfrak{g} \ne \mathfrak{g} }$ 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 $\displaystyle{ U = \operatorname{span} \{ v_i | i \ge 0 \} }$ 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. $\displaystyle{ \square }$

## 3. Consequences

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:

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 $\displaystyle{ Y \in [\mathfrak{g}, \mathfrak{g}] }$.

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:

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.

Here is another quite useful application:

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 $\displaystyle{ \mathfrak{g}/\operatorname{rad}(\mathfrak{g}) }$. $\displaystyle{ \square }$

### References

1. Serre, Theorem 3″
2. Humphreys, Ch. II, § 4.1., Corollary C.
3. Serre, Theorem 4
4. Serre, Theorem 3'
5. Jacobson, Ch. II, § 6, Lemma 5.
6. Fulton & Harris, Proposition 9.17.
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