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 $\pi: \mathfrak \to \mathfrak(V)$ 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 $\pi(\mathfrak)$ with $\operatorname V_i = i$, meaning that $\pi(X)(V_i) \subseteq V_i$ for each $X \in \mathfrak$ and ''i''.

Put in another way, the theorem says there is a basis for ''V'' such that all linear transformations in $\pi(\mathfrak)$ 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 $\pi(\mathfrak)$ is contained in some Borel subalgebra of $\mathfrak(V)$.

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

Step 2: Find an ideal $\mathfrak$ of codimension one in $\mathfrak$.

Let D\mathfrak = mathfrak, \mathfrak/math> be the derived algebra. Since $\mathfrak$ is solvable and has positive dimension, $D\mathfrak \ne \mathfrak$ and so the quotient $\mathfrak/D\mathfrak$ 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 $\mathfrak$.

Step 3: There exists some linear functional $\lambda$ in $\mathfrak^*$ such that
:$V_ = \$
is nonzero. This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional).

Step 4: $V_$ is a $\mathfrak$-invariant subspace. (Note this step proves a general fact and does not involve solvability.)

Let $Y \in \mathfrak$, $v \in V_$, then we need to prove $Y \cdot v \in V_$. If $v = 0$ then it's obvious, so assume $v \ne 0$ and set recursively $v_0 = v, \, v_ = Y \cdot v_i$. Let $U = \operatorname \$ and $\ell \in \mathbb_0$ 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 $m \in \mathbb_0$ be the smallest such that $v_m \notin \langle v_0,\ldots,v_\ell\rangle$, then obviously $m \ge \ell + 1$. Since $v_0,\ldots,v_$ are linearly dependent, $v_$ is a linear combination of $v_0,\ldots,v_\ell$. Applying the map $Y^$ it follows that $v_m$ is a linear combination of $v_,\ldots,v_$. 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 $n \in \mathbb_0$ and $X \in \mathfrak$ there exist elements $a_,\ldots,a_$ of the base field such that $a_=\lambda(X)$ and
:$X \cdot v_n = \sum_^ a_v_i.$
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 $n \in \mathbb_0$ and all elements of $\mathfrak$ and let $X \in \mathfrak$. Since $\mathfrak$ is an ideal, it's ,Y\in \mathfrak, and thus
:$X \cdot v_ = Y \cdot (X \cdot v_n) +$, Y\cdot v_n = Y \cdot \sum_^ a_v_i + \sum_^ a_v_i = a_v_0 + \sum_^ (a_ + a_)v_i + \lambda(X)v_,
and the induction step follows. This implies that for every $X \in \mathfrak$ 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 $\operatorname(\pi(X), _U) = \dim(U) \lambda(X)$. Applying this with ,Y\in \mathfrak instead of ''X'' gives \operatorname(\pi( ,Y, _U) = \dim(U) \lambda( ,Y. On the other hand, ''U'' is also obviously an invariant subspace of ''Y'', and so
:\operatorname(\pi( ,Y, _U) = \operatorname( pi(X),\pi(Y)_U]) = \operatorname( pi(X), _U, \pi(Y), _U = 0
since commutators have zero trace, and thus \dim(U) \lambda( ,Y = 0. Since $\dim(U) > 0$ is invertible (because of the assumption on the characteristic of the base field), $\lambda($, Y = 0 and
:$X \cdot (Y \cdot v) = Y \cdot (X \cdot v) +$, Y\cdot v = Y \cdot (\lambda(X) v) + \lambda(, Y v = \lambda(X) (Y \cdot v),
and so $Y \cdot v \in V_$.

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

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

Consequences

The theorem applies in particular to the adjoint representation $\operatorname: \mathfrak \to \mathfrak(\mathfrak)$ of a (finite-dimensional) solvable Lie algebra $\mathfrak$ over an algebraically closed field of characteristic zero; thus, one can choose a basis on $\mathfrak$ with respect to which $\operatorname(\mathfrak)$ consists of upper triangular matrices. It follows easily that for each $x, y \in \mathfrak$, $\operatorname($, y = operatorname(x), \operatorname(y)/math> has diagonal consisting of zeros; i.e., $\operatorname($, y is a strictly upper triangular matrix. This implies that mathfrak g, \mathfrak g/math> 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/math> 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 $\mathfrak \subseteq \mathfrak(V)$ a Lie subalgebra, then $\mathfrak$ is solvable if and only if $\operatorname(XY) = 0$ for every $X \in \mathfrak$ and Y \in mathfrak, \mathfrak/math>.''

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

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

See also

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

References

Sources

*
*.
*
*{{Citation , first = Jean-Pierre , last = Serre, title = Complex Semisimple Lie Algebras, series=Springer Monographs in Mathematics , publisher = Springer, location = Berlin, year = 2001, isbn = 3-5406-7827-1, doi = 10.1007/978-3-642-56884-8, mr = 1808366

Theorems about algebras