---- Engel's Theorem roughly says that nilpotency of a Lie algebra is the same thing as nilpotency of the operators comprising its [[adjoint representation]]. > [!theorem] Theorem. ([[Engel's Theorem]]) > 1. Let $\mathfrak{gl}(V)$ be the [[general linear Lie algebra]] over an $n$-dimensional [[vector space]] $V$, and let $\mathfrak{g}$ be a [[Lie subalgebra]]. Suppose that $x$ is a [[nilpotent linear operator]] for all $x \in \mathfrak{g}$. Then the elements of $\mathfrak{g}$ are [[simultaneously diagonalizable|simultaneously]] [[strictly triangular matrix|strictly upper triangularizable]], i.e., there exists a [[basis]] of $V$ such that $\mathfrak{g} \subset \mathfrak{n}_{n}$ and hence $\mathfrak{g}$ is a [[derived and central series of a Lie algebra|nilpotent]] [[Lie algebra]]. > >2. Suppose $\mathfrak{g}$ is any [[Lie algebra]] over $V$. Then $\mathfrak{g}$ is a [[derived and central series of a Lie algebra|nilpotent Lie algebra]] if and only if $\text{ad}(x) \in \mathfrak{gl}(V)$ is a [[nilpotent linear operator]] for all $x \in \mathfrak{g}$, where $\text{ad}(x)=[x,-]$ denotes the image of $x$ under the [[adjoint representation]] of $\mathfrak{g}$.[^1] ^theorem > [!basicnonexample] Warning. > It is *definitely not true* that a [[Lie subalgebra|subalgebra]] $\mathfrak{g} \subset \mathfrak{gl}(V)$ is nilpotent iff $x$ is a [[nilpotent linear operator]] for all $x \in \mathfrak{g}$. For example, take $\mathfrak{g}$ to be the [[diagonal matrix|diagonal matrices]] — $\mathfrak{g}$ is nilpotent (it is [[abelian Lie algebra|abelian]]), but its elements $x \in \mathfrak{g}$ are obviously not. ^784a3c We use the following intermediate results. 1. If $x \in \mathfrak{gl}(V)$ is nilpotent, then $\text{ad }x$ is nilpotent. 2. If $\mathfrak{g} \subset \mathfrak{gl}(V)$ is a [[Lie subalgebra]] and every $x \in \mathfrak{g}$ is nilpotent, then the $x \in \mathfrak{g}$ have a simultaneous zero [[eigenvector]] $v$ *(this part takes the most work)* 3. If $\mathfrak{g} \subset \mathfrak{gl}(V)$ is. [[Lie subalgebra]] and every $x \in \mathfrak{g}$ is nilpotent, then Induction with $(2)$ gives a [[basis]] of $V$ witnessing $\mathfrak{g} \subset \mathfrak{n}_{n}$. (From this follows **(1)** in the theorem statement.) > [!proof]- Proof. ([[Engel's Theorem]]) > We first prove the intermediate results $(1)$, $(2)$, $(3)$, where the intermediate result $(3)$ gives **(1)** in the theorem statement. Then we'll prove **(2)** in the theorem statement. > > **Proof of intermediate result $(1)$.** > Let $L_{x}:\mathfrak{gl}(V) \to \mathfrak{gl}(V)$ be the [[linear map]] $y \mapsto xy$, $R_{x}$ the map $y \mapsto yx$. So $\text{ad }x=L_{x}-R_{x}$. $L_{x}, R_{x}$ nilpotent since $x$ is. They also commute. Hence $\text{ad }x$ is nilpotent as a sum of commuting nilpotent linear operators. (See [[nilpotent element of a ring#^properties|here]].) > > **Proof of intermediate result $(2)$.** > We induct on $\text{dim }\mathfrak{g}$. > > *Base case.* When $\text{dim }\mathfrak{g}=1$, $\mathfrak{g}$ is spanned by one nilpotent $x \in \mathfrak{g}$. There is $n \geq 1$ such that $x^{n} \neq0$ but $x^{n+1}=0$. Any nonzero $v$ in the image of $x^{n}$ satisfies $xv=0$. > > *Induction step.* Suppose the result holds for all subdimensions. In particular, it holds for any maximal proper subalgebra $\mathfrak{k}$ of $\mathfrak{g}$. > > If $x \in \mathfrak{k}$, then since $x$ is nilpotent, $\text{ad }x:\mathfrak{g} \to \mathfrak{g}$ is nilpotent by intermediate result $(1)$, hence [[quotient representation|the induced adjoint action]] on the [[vector space]] $\mathfrak{g} / \mathfrak{k}$ consists of nilpotent linear operators. > > Using the induction hypothesis, there exists a nonzero element $\overline{y}$ in $\mathfrak{g} / \mathfrak{k}$ with $[x, \overline{y}]=0$ for all $x \in \mathfrak{k}$. Lifting $\overline{y}$ to an element $y \in \mathfrak{g} - \mathfrak{k}$, we get $[x,y] \in \mathfrak{k}$ for all $x \in \mathfrak{k}$. So $\mathfrak{k}$ has a 'simultaneously stabilizing element' $y$. > > Now, $\mathfrak{k}+\text{span}(y)$ is a subalgebra of $\mathfrak{g}$ strictly containing $\mathfrak{k}$; by maximality, this means $\mathfrak{g}=\mathfrak{\mathfrak{k}}+\text{span}(y)$. In addition, $\mathfrak{k}$ is an [[ideal]] of $\mathfrak{g}$, because it is stable under bracketing with both $\mathfrak{k}$ (duh) and $y$. > > The simultaneous zero-eigenspace $W=\{ v \in V: k \cdot v=0 \text{ for all } k \in \mathfrak{k} \} \subset V$ > is nontrivial by the induction hypothesis. It is $\mathfrak{k}$-stable by definition, but also $y$-stable[^1], meaning that $y \in \mathfrak{gl}(V)$ restricts to a nilpotent endomorphism of $W$. $W$ is already killed by everything in $\mathfrak{k}$... we want to find a nonzero element $w \in W$ killed also by $y$, then $w$ will satisfy the claim and we're done. Such $w$ exists by the $\text{dim }\mathfrak{g}=1$ case. > > > > > > **Proof of intermediate result $(3)$, and hence (1) in the theorem statement.** > > Bootstrap from $(2)$, pretty much > > **Proof of (2) in the theorem statement.** If $\mathfrak{g}$ is nilpotent, then obviously $\text{ad }x$ is nilpotent for all $x \in \mathfrak{g}$. Conversely, assume $\text{ad }x$ is nilpotent for all $x \in \mathfrak{g}$. By intermediate result $(3)$, the image of $\text{ad}:\mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ then lives in $\mathfrak{n}_{n}$, i.e., is a [[derived and central series of a Lie algebra|nilpotent Lie algebra]]. > > The [[kernel of a module homomorphism|kernel]] of $\text{ad}:\mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ is by definition the [[center of a Lie algebra|center]] $Z(\mathfrak{g})=\{ x \in \mathfrak{g}: [x,y]=0 \text{ for all } y \in \mathfrak{g} \}.$ [[first isomorphism theorem|Therefore]] $\mathfrak{g} / Z(\mathfrak{g})$ is [[isomorphism|isomorphic]] to the image of $\text{ad}$ in $\mathfrak{gl}(\mathfrak{g})$, and is hence a nilpotent Lie algebra. > > It follows that $\mathfrak{g}$ is nilpotent. Indeed, for some $n$ we have $(\mathfrak{g} / Z(\mathfrak{g}))^{n}=0$, [[formula for derived and central series of quotient Lie algebra|so]] $\mathfrak{g}^{n}+ Z(\mathfrak{g})=Z(\mathfrak{g})$, hence $\mathfrak{g}^{n} \subset Z(\mathfrak{g})$, hence $\mathfrak{g}^{n+1}=[\mathfrak{g}, \mathfrak{g}^{n}] \subset [\mathfrak{g}, Z(\mathfrak{g})]=0$. > > > ---- #### [^1]: One direction is straightforward; the other direction is tricky and substantiates calling this a 'Theorem'. ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```