Engel’s Theorem and Nilpotent Lie Algebras July 23, 2009Posted by Akhil Mathew in algebra, representation theory.
Tags: algebra, Engel's theorem, Lie algebras, linear algebra, nilpotent
Now that I’ve discussed some of the basic definitions in the theory of Lie algebras, it’s time to look at specific subclasses: nilpotent, solvable, and eventually semisimple Lie algebras. Today, I want to focus on nilpotence and its applications.
To start with, choose a Lie algebra for some finite-dimensional -vector space ; recall that is the Lie algebra of linear transformations with the bracket . The previous definition was in terms of matrices, but here it is more natural to think in terms of linear transformations without initially fixing a basis.
Engel’s theorem is somewhat similar in its statement to the fact that commuting diagonalizable operators can be simultaneously diagonalized.
Theorem 1 (Engel) Let consist of nilpotent operators. Then there exists a vector , , such that for all .
The theorem can also be stated in the form where is a representation for where the elemetns of act via nilpotent operators; in that case there is a vector annihilated by . I will tacitly use this equivalence below.
The corollary makes the analogy between commuting diagonalizable operators clearer:
Corollary 2 Under the same hypothesis, there exists a basis of with which each element of is represented as a strictly upper-triangular matrix.
The proof of the theorem goes by induction on (and not on !). When , the result is immediate. Otherwise, if , we have a proper Lie subalgebra . Choose to be a maximal subalgebra.
Claim 1 has codimension one in and is an ideal.
Proof: [Proof of the claim] Well, acts via on both and . In the latter case, we know by Engel’s theorem for , and the inductive hypothesis, that there exists a nonzero such that for ; lifting to , it follows that . Moreover . Both these imply is a Lie subalgebra of , and contains as an ideal. By maximality of , it follows that , so we’re done.
Now, to prove the theorem, choose as before, and use the inductive hypothesis again to see that the vector space
Claim 2 is stable under .
In other words, if , , then for any . But
and since is an ideal, so . This is essentially a new version of the “fundamental calculation” in discussing representations of .
Now, we have for some ; we know that is a linear operator on by the previous claim, and is by assumption nilpotent. So . Choose nonzero in that intersection; then both annihilate , proving Engel’s theorem.
Proof: [Proof of the corollary] By linear algebra, the statement of the corollary is equivalent to the statement: There exists a flag of -stable spaces such that for each , and acts trivially on the quotients . To construct the flag, proceed as follows. Choose to be spanned by a vector as in Engel’s theorem. Then acts as a family of nilpotent operators on . Choose such that is generated by an element annihilated by , by Engel again. Repeat until the process terminates, since is finite.
The notion of nilpotence in Lie algebras is slightly more complicated than simply considering strictly upper-triangular matrices. The idea instead is to say that a Lie algebra is nilpotent if it is at least “almost” commutative, in that taking successive brackets eventually yields zero.
Definition 3 The lower central series of a Lie algebra is defined by . A Lie algebra is said to be nilpotent if its lower central series eventually becomes zero.
If a Lie algebra is nilpotent, then must be nilpotent for any . Indeed, maps since .
The converse is also true:
Theorem 4 Suppose is a Lie algebra such that each is nilpotent. Then is nilpotent.
Induction on , as usual. The image is a Lie subalgebra (since is a homomorphism of Lie algebras). Each element of consists of a nilpotent operator on by assumption. So by Engel, there exists such that . In other words, “commutes” with all of , or lies in the center:
Definition 5 The center of a Lie algebra is the set of all such that if .
The center is actually a Lie ideal of . We have just shown that under the hypotheses of the theorem, the center of is nonzero. Now still satisfies the conditions of the theorem, and by the inductive hypothesis is nilpotent. So we are reduced to:
Lemma 6 Suppose is a Lie algebra, its center, and is nilpotent. Then is nilpotent.
Indeed, it follows by the nilpotence of that some , which implies .