# The Unapologetic Mathematician

## 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.

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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 […]

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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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