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Engel's Theorem

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Engel’s Theorem

ad-Nilpotent

An element is ad-nilpotent if the map is a nilpotent endomorphism, i.e., for some .

Lemma

If is a nilpotent endomorphism, then is also nilpotent.

Proof Let be the left and right multiplication maps by . Then . Since and commute, we can use the binomial expansion: . If , then for , in each term of the sum either or , so either or . Thus .

Engel's Theorem

A finite-dimensional Lie algebra is nilpotent if and only if every element is ad-nilpotent.

A key lemma in the proof is another form of the theorem:

Engel's Theorem (Alternative Form)

Let be a non-zero finite-dimensional vector space. If is a Lie subalgebra consisting entirely of nilpotent endomorphisms, then there exists a non-zero vector such that for all .

Lie’s Theorem

Full Flag

Let be a finite dimensional vector space over a field , and let . A full flag in is a sequence of subspaces where .

e.g. If with a basis , then the standard full flag is given by for .

Lie's Theorem

Let be a solvable Lie algebra over an algebraically closed field of characteristic . Let be a finite-dimensional vector space over . If is a representation, then there exists a non-zero vector which is a common eigenvector for all .

Proof We proceed by induction on . For the base case, if , then for . Since is algebraically closed, has at least one eigenvalue, and therefore at least one eigenvector in . That is, there exists some nonzero such that . It follows that . Now, assume , and assume the theorem holds for all Lie algebras such that . The proof follows the following steps:

  1. Find an ideal with (i.e. ).
  2. Find a common eigenvector for .
  3. Show that , where is the subspace of common eigenvectors for .
  4. Show that for and find a nonzero eigenvector for .

Step 1. By assumption is solvable, so is a proper subspace of (otherwise for every , and so can never terminate). Then is abelian, meaning every subspace is an ideal. Let be a co-dimension one subspace, and hence an ideal, in . By the correspondence theorem, there exists an ideal with . Then , and . Step 2. By the inductive hypothesis, there exists some such that for every . It is easy to check that is a linear functional. So, is a common eigenvector for . Step 3. Let We note that by Step 2. We claim that . To show this, let , . We want to show that - that is, for every . Since is an ideal, , and then It is therefore sufficient to show that . Let be the largest integer such that are all linearly independent. Let for . By definition, for , and for all . It follows that , namely, restricts to an endomorphism of . We now show that, for all , we have for every . It is sufficient to show for every . We proceed by induction: if then . Now, assume for every . Then, for any . It only remains to check : since as it is an ideal. So, , meaning . If we let , then the matrix of with respect to is upper triangular: since preserves for each . But recall that when is applied to , the coefficient of in the result is exactly . So, this tells us thatThis applies to every , in particular to , henceBut as the trace of a commutator is zero, and the field has characteristic zero, the latter equation implies that . It follows that for any , we have , meaning . Step 4. Since , we have that for any . Fix such a ; by Step 3 , we know that . Since is algebraically closed, has an eigenvector in . That is, there exists some nonzero such that . Thus, , and . So, .

e.g.

  • Consider the Lie algebra of all upper triangular complex matrices:
  • This theorem is not true if the field is not algebraically closed. Consider Then for any arbitrary matrix such that , the characteristic polynomial is , which has no real roots. Thus, there are no eigenvectors in for such matrices.

Corollary

Let be a solvable Lie algebra, over an algebraically closed field of characteristic . Then, stabilises a full flag , i.e., for all .

Proof We’ll prove this inductively on the dimension of . In the base case where 0 , then is a full flag. Trivially, , so the statement is true for dimension . Suppose that the statement were true over all vector spaces of dimension . Then in of dimension , from the above theorem we conclude that there exists some such that ; that is, . So both and are -modules, so is also a -module. has dimension , so there exists a flag in that is stabilised by . By the correspondence theorem, we deduce that each for some subalgebra , with , and each stabilised by . Defining , then is a full flag that is stabilised by , as required.

There exists a similar result for nilpotent Lie algebras:

Corollary

Let be a nilpotent Lie algebra, over an algebraically closed field of characteristic . Then, for all .

Characterizations of Solvable Lie Algebras

Corollary

Let be a Lie algebra over an algebraically closed field of characteristic . Then the following are equivalent:

  1. is solvable.
  2. There exists a chain of ideals such that is abelian.
  3. is nilpotent.
  4. is solvable.

Proof : Suppose is soluble. Then there exists such that . Let . Then we get a chain of ideals and each is abelian because by definition is the commutator. : Suppose is a chain of ideals and each is abelian, then . So for all , that is, . Since is a representation preserving the Lie brackets, we have Hence by induction using the infusion above, In particular, , so is solvable. : Suppose is solvable. By Engel’s theorem, to show is nilpotent, it suffices to show is nilpotent for all . Since is spanned by for , it is enough to check is nilpotent for all . Let . By Lie’s theorem, there exists a common eigenvector for all endomorphisms in . Suppose , , then Therefore . Now let , which is of dimension strictly less than . We can invoke Lie’s theorem again to get another vector space . Since is finite dimensional, repeating the above process will finally terminate at some . That is, is the zero map. Thus is nilpotent. : Since is nilpotent, is soluble, for some . Thus , so is solvable.

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