The Unapologetic Mathematician

Mathematics for the interested outsider

Engel’s Theorem

When we say that a Lie algebra L is nilpotent, another way of putting it is that for any sufficiently long sequence \{x_i\} of elements of L the nested adjoint

\mathrm{ad}(x_n)\left[\dots\mathrm{ad}(x_2)\left[\mathrm{ad}(x_1)[y]\right]\right]

is zero for all y\in L. In particular, applying \mathrm{ad}(x) enough times will eventually kill any element of L. That is, each x\in L is ad-nilpotent. It turns out that the converse is also true, which is the content of Engel’s theorem.

But first we prove this lemma: if L\subseteq\mathfrak{gl}(V) is a linear Lie algebra on a finite-dimensional, nonzero vector space V that consists of nilpotent endomorphisms, then there is some nonzero v\in V for which l(v)=0 for all l\in L.

If \dim(L)=1 then L is spanned by a single nilpotent endomorphism, which has only the eigenvalue zero, and must have an eigenvector v, proving the lemma in this case.

If K is any nontrivial subalgebra of L then \mathrm{ad}(k)\in\mathrm{End}(L) is nilpotent for all k\in K. We also get an everywhere-nilpotent action on the quotient vector space L/K. But since \dim(K)<\dim(L), the induction hypothesis gives us a nonzero vector x+K\in L/K that gets killed by every k\in K. But this means that [k,x]\in K for all k\in K, while x\notin K. That is, K is strictly contained in the normalizer N_L(K).

Now instead of just taking any subalgebra, let K be a maximal proper subalgebra in L. Since K is properly contained in N_L(K), we must have N_L(K)=L, and thus K is actually an ideal of L. If \dim(L/K)>1 then we could find an even larger subalgebra of L containing K, in contradiction to our assumption, so as vector spaces we can write L\cong K+\mathbb{F}z for any z\in L\setminus K.

Finally, let W\subseteq V consist of those vectors killed by all \in K, which the inductive hypothesis tells us is a nonempty collection. Since K is an ideal, L sends W back into itself: k(l(w))=l(k(w))-[l,k](w)=0. Picking a z\in L\setminus K as above, its action on W is nilpotent, so it must have an eigenvector w with z(w)=0. Thus l(w)=0 for all l\in L=K+\mathbb{F}z.

So, now, to Engel’s theorem. We take a Lie algebra L consisting of ad-nilpotent elements. Thus the algebra \mathrm{ad}(L)\subseteq\mathfrak{gl}(L) consists of nilpotent endomorphisms on the vector space L, and there is thus some nonzero z\in L for which [L,z]=0. That is, L has a nontrivial center — z\in Z(L).

The quotient L/Z(L) thus has a lower dimension than L, and it also consists of ad-nilpotent elements. By induction on the dimension of L we assume that L/Z(L) is actually nilpotent, which proves that L itself is nilpotent.

August 22, 2012 - Posted by | Algebra, Lie Algebras

3 Comments »

  1. […] lemma leading to Engel’s theorem boils down to the assertion that there is some common eigenvector for all the endomorphisms in a […]

    Pingback by Lie’s Theorem « The Unapologetic Mathematician | August 25, 2012 | Reply

  2. […] like to have matrix-oriented versions of Engel’s theorem and Lie’s theorem, and to do that we’ll need flags. I’ve actually referred to […]

    Pingback by Flags « The Unapologetic Mathematician | August 25, 2012 | Reply

  3. […] obvious that if is nilpotent then will be solvable. And Engel’s theorem tells us that if each is ad-nilpotent, then is itself nilpotent. We can now combine this with our […]

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