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.

Engel’s Theorem

To start with, choose a Lie algebra ${L \subset \mathfrak{gl} (V)}$ for some finite-dimensional ${k}$-vector space ${V}$; recall that ${\mathfrak{gl} (V)}$ is the Lie algebra of linear transformations ${V \rightarrow V}$ with the bracket ${[A,B] := AB - BA}$. 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 ${L \subset \mathfrak{gl} (V)}$ consist of nilpotent operators. Then there exists a vector ${v \in V}$, ${v \neq 0}$, such that ${Xv = 0}$ for all ${X \in L}$.

The theorem can also be stated in the form where ${V}$ is a representation for ${L}$ where the elemetns of ${L}$ act via nilpotent operators; in that case there is a vector annihilated by ${L}$. 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 ${V}$ with which each element of ${L}$ is represented as a strictly upper-triangular matrix.

The proof of the theorem goes by induction on ${\dim L}$ (and not on ${n}$!). When ${\dim L=0}$, the result is immediate. Otherwise, if ${\dim L \geq 1}$, we have a proper Lie subalgebra ${L' \subset L}$. Choose ${L' \subset L}$ to be a maximal subalgebra.

Claim 1 ${L'}$ has codimension one in ${L}$ and ${L'}$ is an ideal.

Proof: [Proof of the claim] Well, ${L'}$ acts via ${\mathrm{ad}}$ on both ${L'}$ and ${L'}$. In the latter case, we know by Engel’s theorem for ${L'}$, and the inductive hypothesis, that there exists a nonzero ${\bar{r} \in L/L'}$ such that ${[l', \bar{r}] = \bar{0}}$ for ${l' \in L'}$; lifting ${\bar{r}}$ to ${r \in L - L'}$, it follows that ${[L' + kr, L'] \subset L' + L' = L'}$. Moreover ${[L' + kr, L' + kr] \subset L' + L' + L' + 0 = L'}$. Both these imply ${L' + kr}$ is a Lie subalgebra of ${L}$, and contains ${L'}$ as an ideal. By maximality of ${L'}$, it follows that ${L' + kr=L}$, so we’re done. $\Box$

Now, to prove the theorem, choose ${L'}$ as before, and use the inductive hypothesis again to see that the vector space

$\displaystyle W := \{ w \in V, L' w = 0 \}$

is nonzero.

Claim 2 ${W}$ is stable under ${L}$.

In other words, if ${w \in W}$, ${l \in L}$, then ${l'( lw ) = 0}$ for any ${l' \in L'}$. But

$\displaystyle l' (lw) = l (l'w) + [l', l] w = 0 + [l', l]w,$

and ${[l',l] \in L'}$ since ${L'}$ is an ideal, so ${[l',l]w=0}$. This is essentially a new version of the “fundamental calculation” in discussing representations of ${\mathfrak{sl}_2}$.

Now, we have ${L = k r + L'}$ for some ${r \in L}$; we know that ${r}$ is a linear operator on ${W}$ by the previous claim, and is by assumption nilpotent. So ${\ker(r) \cap W \neq 0}$. Choose ${v}$ nonzero in that intersection; then both ${L', r}$ annihilate ${v}$, 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 ${L}$-stable spaces ${0 = V_0 \subset V_1 \subset \dots \subset V_n = V}$ such that ${\dim V_i/V_{i+1} = 1}$ for each ${i}$, and ${L}$ acts trivially on the quotients ${V_i/V_{i+1}}$. To construct the flag, proceed as follows. Choose ${V_1}$ to be spanned by a vector ${v}$ as in Engel’s theorem. Then ${L}$ acts as a family of nilpotent operators on ${V/V_1}$. Choose ${V_2/V_1 \subset V/V_1}$ such that ${V_2/V_1}$ is generated by an element ${v'}$ annihilated by ${L}$, by Engel again. Repeat until the process terminates, since ${\dim V}$ is finite. $\Box$

Nilpotence

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.

Formally:

Definition 3 The lower central series of a Lie algebra ${L}$ is defined by ${L_0 := L, L_1 := [L, L_0], \dots, L_n := [L, L_{n-1}], \dots}$. A Lie algebra is said to be nilpotent if its lower central series eventually becomes zero.

If a Lie algebra is nilpotent, then ${\mathrm{ad} \ x: L \rightarrow L}$ must be nilpotent for any ${x \in L}$. Indeed, ${\mathrm{ad} \ x}$ maps ${L_n \rightarrow L_{n+1}}$ since ${\mathrm{ad} \ x(y) = [x,y]}$.

The converse is also true:

Theorem 4 Suppose ${L}$ is a Lie algebra such that each ${\mathrm{ad} \ x}$ is nilpotent. Then ${L}$ is nilpotent.

Induction on ${\dim L}$, as usual. The image ${N := \mathrm{ad} (L) \subset \mathfrak{gl}(L)}$ is a Lie subalgebra (since ${\mathrm{ad} }$ is a homomorphism of Lie algebras). Each element of ${N}$ consists of a nilpotent operator on ${L}$ by assumption. So by Engel, there exists ${l \in L}$ such that ${[L, kl]=0}$. In other words, ${l}$ “commutes” with all of ${L}$, or lies in the center:

Definition 5 The center of a Lie algebra ${L}$ is the set of all ${x \in L}$ such that ${[x,l]=0}$ if ${l \in L}$.

The center is actually a Lie ideal of ${L}$. We have just shown that under the hypotheses of the theorem, the center ${Z}$ of ${L}$ is nonzero. Now ${L/Z}$ still satisfies the conditions of the theorem, and by the inductive hypothesis is nilpotent. So we are reduced to:

Lemma 6 Suppose ${L}$ is a Lie algebra, ${Z}$ its center, and ${L/Z}$ is nilpotent. Then ${L}$ is nilpotent.

Indeed, it follows by the nilpotence of ${L/Z}$ that some ${L_n \subset Z}$, which implies ${L_{n+1} = [L, L_n] \subset [L, Z]=0}$.