Misplaced Pages

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 there's a flag ${\displaystyle V=V_{0}\supset V_{1}\supset \cdots \supset V_{n}=0}$ of invariant subspaces of ${\displaystyle \pi ({\mathfrak {g}})}$ with ${\displaystyle \operatorname {codim} V_{i}=i}$, meaning that ${\displaystyle \pi (X)(V_{i})\subseteq V_{i}}$ 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 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 ${\displaystyle \pi ({\mathfrak {g}})}$ is contained in some Borel subalgebra of ${\displaystyle {\mathfrak {gl}}(V)}$.

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), 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 ${\displaystyle {\mathfrak {g}}}$ 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 be 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 ${\displaystyle D{\mathfrak {g}}=}$ 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 ${\displaystyle {\mathfrak {g}}}$-invariant subspace. (Note this step proves a general fact and does not involve solvability.)

Let ${\displaystyle Y\in {\mathfrak {g}}}$, ${\displaystyle v\in V_{\lambda }}$, then we need to prove ${\displaystyle Y\cdot v\in V_{\lambda }}$. If ${\displaystyle v=0}$ then it's obvious, so assume ${\displaystyle v\neq 0}$ and set recursively ${\displaystyle v_{0}=v,\,v_{i+1}=Y\cdot v_{i}}$. Let ${\displaystyle U=\operatorname {span} \{v_{i}|i\geq 0\}}$ and ${\displaystyle \ell \in \mathbb {N} _{0}}$ be the largest such that ${\displaystyle v_{0},\ldots ,v_{\ell }}$ are linearly independent. Then we'll prove that they generate U and thus ${\displaystyle \alpha =(v_{0},\ldots ,v_{\ell })}$ is a basis of U. Indeed, assume by contradiction that it's not the case and let ${\displaystyle m\in \mathbb {N} _{0}}$ be the smallest such that ${\displaystyle v_{m}\notin \langle v_{0},\ldots ,v_{\ell }\rangle }$, then obviously ${\displaystyle m\geq \ell +1}$. Since ${\displaystyle v_{0},\ldots ,v_{\ell +1}}$ are linearly dependent, ${\displaystyle v_{\ell +1}}$ is a linear combination of ${\displaystyle v_{0},\ldots ,v_{\ell }}$. Applying the map ${\displaystyle Y^{m-\ell -1}}$ it follows that ${\displaystyle v_{m}}$ is a linear combination of ${\displaystyle v_{m-\ell -1},\ldots ,v_{m-1}}$. Since by the minimality of m each of these vectors is a linear combination of ${\displaystyle v_{0},\ldots ,v_{\ell }}$, so is ${\displaystyle v_{m}}$, and we get the desired contradiction. We'll prove by induction that for every ${\displaystyle n\in \mathbb {N} _{0}}$ and ${\displaystyle X\in {\mathfrak {h}}}$ there exist elements ${\displaystyle a_{0,n,X},\ldots ,a_{n,n,X}}$ of the base field such that ${\displaystyle a_{n,n,X}=\lambda (X)}$ and

${\displaystyle X\cdot v_{n}=\sum _{i=0}^{n}a_{i,n,X}v_{i}.}$

The ${\displaystyle n=0}$ case is straightforward since ${\displaystyle X\cdot v_{0}=\lambda (X)v_{0}}$. Now assume that we have proved the claim for some ${\displaystyle n\in \mathbb {N} _{0}}$ and all elements of ${\displaystyle {\mathfrak {h}}}$ and let ${\displaystyle X\in {\mathfrak {h}}}$. Since ${\displaystyle {\mathfrak {h}}}$ is an ideal, it's ${\displaystyle \in {\mathfrak {h}}}$, and thus

${\displaystyle X\cdot v_{n+1}=Y\cdot (X\cdot v_{n})+\cdot v_{n}=Y\cdot \sum _{i=0}^{n}a_{i,n,X}v_{i}+\sum _{i=0}^{n}a_{i,n,}v_{i}=a_{0,n,}v_{0}+\sum _{i=1}^{n}(a_{i-1,n,X}+a_{i,n,})v_{i}+\lambda (X)v_{n+1},}$

and the induction step follows. This implies that for every ${\displaystyle X\in {\mathfrak {h}}}$ the subspace U is an invariant subspace of X and the matrix of the restricted map ${\displaystyle \pi (X)|_{U}}$ in the basis ${\displaystyle \alpha }$ is upper triangular with diagonal elements equal to ${\displaystyle \lambda (X)}$, hence ${\displaystyle \operatorname {tr} (\pi (X)|_{U})=\dim(U)\lambda (X)}$. Applying this with ${\displaystyle \in {\mathfrak {h}}}$ instead of X gives ${\displaystyle \operatorname {tr} (\pi ()|_{U})=\dim(U)\lambda ()}$. On the other hand, U is also obviously an invariant subspace of Y, and so

${\displaystyle \operatorname {tr} (\pi ()|_{U})=\operatorname {tr} (|_{U}])=\operatorname {tr} ()=0}$

since commutators have zero trace, and thus ${\displaystyle \dim(U)\lambda ()=0}$. Since ${\displaystyle \dim(U)>0}$ is invertible (because of the assumption on the characteristic of the base field), ${\displaystyle \lambda ()=0}$ and

${\displaystyle X\cdot (Y\cdot v)=Y\cdot (X\cdot v)+\cdot v=Y\cdot (\lambda (X)v)+\lambda ()v=\lambda (X)(Y\cdot v),}$

and so ${\displaystyle Y\cdot v\in V_{\lambda }}$.

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 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. ${\displaystyle \square }$

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 ${\displaystyle {\mathfrak {g}}}$ over an algebraically closed field of characteristic zero; 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 ${\displaystyle x,y\in {\mathfrak {g}}}$, ${\displaystyle \operatorname {ad} ()=}$ has diagonal consisting of zeros; i.e., ${\displaystyle \operatorname {ad} ()}$ is a strictly upper triangular matrix. This implies that ${\displaystyle }$ 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 ${\displaystyle {\mathfrak {g}}}$ over a field of characteristic zero is solvable if and only if the derived algebra ${\displaystyle D{\mathfrak {g}}=}$ 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 ${\displaystyle {\mathfrak {g}}\subseteq {\mathfrak {gl}}(V)}$ 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 ${\displaystyle Y\in }$.

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 ${\displaystyle {\mathfrak {g}}}$ over an algebraically closed field of characteristic zero, 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 over any base field since in this case every vector subspace is a Lie subalgebra.

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

See also

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

References

1. ^ Serre 2001, Theorem 3
2. Humphreys 1972, Ch. II, § 4.1., Corollary A.
3. Serre 2001, Theorem 3″
4. Humphreys 1972, Ch. II, § 4.1., Corollary C.
5. Serre 2001, Theorem 4
6. Serre 2001, Theorem 3'
7. Jacobson 1979, Ch. II, § 6, Lemma 5.
8. Fulton & Harris 1991, Proposition 9.17.

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

• Theorems about algebras