In mathematics, specifically the theory of Lie algebras, Lie's theorem states that,[1] over an algebraically closed field of characteristic zero, if
is a finite-dimensional representation of a solvable Lie algebra, then there's a flag
of invariant subspaces of
with
, meaning that
for each
and i.
Put in another way, the theorem says there is a basis for V such that all linear transformations in
are represented by upper triangular matrices.[2] 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
is contained in some Borel subalgebra of
.[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.
Proof
The proof is by induction on the dimension of
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
is positive. We also assume V is not zero. For simplicity, we write
.
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
.
Indeed, the theorem says in particular that a nonzero vector spanning
is a common eigenvector for all the linear transformations in
. Conversely, if v is a common eigenvector, take
to its span and then
admits a common eigenvector in the quotient
; repeat the argument.
Step 2: Find an ideal
of codimension one in
.
Let
be the derived algebra. Since
is solvable and has positive dimension,
and so the quotient
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
.
Step 3: There exists some linear functional
in
such that
![{\displaystyle V_{\lambda }=\{v\in V|X\cdot v=\lambda (X)v,X\in {\mathfrak {h}}\}}](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
is nonzero. This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional).
Step 4:
is a
-invariant subspace. (Note this step proves a general fact and does not involve solvability.)
Let
,
, then we need to prove
. If
then it's obvious, so assume
and set recursively
. Let
and
be the largest such that
are linearly independent. Then we'll prove that they generate U and thus
is a basis of U. Indeed, assume by contradiction that it's not the case and let
be the smallest such that
, then obviously
. Since
are linearly dependent,
is a linear combination of
. Applying the map
it follows that
is a linear combination of
. Since by the minimality of m each of these vectors is a linear combination of
, so is
, and we get the desired contradiction. We'll prove by induction that for every
and
there exist elements
of the base field such that
and
![{\displaystyle X\cdot v_{n}=\sum _{i=0}^{n}a_{i,n,X}v_{i}.}](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
The
case is straightforward since
. Now assume that we have proved the claim for some
and all elements of
and let
. Since
is an ideal, it's
, and thus
![{\displaystyle X\cdot v_{n+1}=Y\cdot (X\cdot v_{n})+[X,Y]\cdot v_{n}=Y\cdot \sum _{i=0}^{n}a_{i,n,X}v_{i}+\sum _{i=0}^{n}a_{i,n,[X,Y]}v_{i}=a_{0,n,[X,Y]}v_{0}+\sum _{i=1}^{n}(a_{i-1,n,X}+a_{i,n,[X,Y]})v_{i}+\lambda (X)v_{n+1},}](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
and the induction step follows. This implies that for every
the subspace U is an invariant subspace of X and the matrix of the restricted map
in the basis
is upper triangular with diagonal elements equal to
, hence
. Applying this with
instead of X gives
. On the other hand, U is also obviously an invariant subspace of Y, and so
![{\displaystyle \operatorname {tr} (\pi ([X,Y])|_{U})=\operatorname {tr} ([\pi (X),\pi (Y)]|_{U}])=\operatorname {tr} ([\pi (X)|_{U},\pi (Y)|_{U}])=0}](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
since commutators have zero trace, and thus
. Since
is invertible (because of the assumption on the characteristic of the base field),
and
![{\displaystyle X\cdot (Y\cdot v)=Y\cdot (X\cdot v)+[X,Y]\cdot v=Y\cdot (\lambda (X)v)+\lambda ([X,Y])v=\lambda (X)(Y\cdot v),}](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
and so
.
Step 5: Finish up the proof by finding a common eigenvector.
Write
where L is a one-dimensional vector subspace. Since the base field is algebraically closed, there exists an eigenvector in
for some (thus every) nonzero element of L. Since that vector is also eigenvector for each element of
, the proof is complete. ![{\displaystyle \square }](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Consequences
The theorem applies in particular to the adjoint representation
of a (finite-dimensional) solvable Lie algebra
over an algebraically closed field of characteristic zero; thus, one can choose a basis on
with respect to which
consists of upper triangular matrices. It follows easily that for each
,
has diagonal consisting of zeros; i.e.,
is a strictly upper triangular matrix. This implies that
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):[4]
- A finite-dimensional Lie algebra
over a field of characteristic zero is solvable if and only if the derived algebra
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
a Lie subalgebra, then
is solvable if and only if
for every
and
.[5]
Indeed, as above, after extending the base field, the implication
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
over an algebraically closed field of characteristic zero, each finite-dimensional simple
-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
-module V, let
be a maximal
-submodule (which exists by finiteness of the dimension). Then, by maximality,
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.[7]
Here is another quite useful application:[8]
- Let
be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero with radical
. Then each finite-dimensional simple representation
is the tensor product of a simple representation of
with a one-dimensional representation of
(i.e., a linear functional vanishing on Lie brackets).
By Lie's theorem, we can find a linear functional
of
so that there is the weight space
of
. By Step 4 of the proof of Lie's theorem,
is also a
-module; so
. In particular, for each
,
. Extend
to a linear functional on
that vanishes on
;
is then a one-dimensional representation of
. Now,
. Since
coincides with
on
, we have that
is trivial on
and thus is the restriction of a (simple) representation of
. ![{\displaystyle \square }](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
See also
References
- ^ a b Serre 2001, Theorem 3
- ^ Humphreys 1972, Ch. II, § 4.1., Corollary A.
- ^ Serre 2001, Theorem 3″
- ^ Humphreys 1972, Ch. II, § 4.1., Corollary C.
- ^ Serre 2001, Theorem 4
- ^ Serre 2001, Theorem 3'
- ^ Jacobson 1979, Ch. II, § 6, Lemma 5.
- ^ 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