Engel's theorem

In representation theory, a branch of mathematics, Engel's theorem states that a finite-dimensional Lie algebra $\mathfrak g$ is a nilpotent Lie algebra if and only if for each $X \in \mathfrak g$, the adjoint map

:$\operatorname(X)\colon \mathfrak \to \mathfrak,$
given by $\operatorname(X)(Y) =$, Y/math>, is a nilpotent endomorphism on $\mathfrak$; i.e., $\operatorname(X)^k = 0$ for some ''k''. It is a consequence of the theorem, also called Engel's theorem, which says that if a Lie algebra of matrices consists of nilpotent matrices, then the matrices can all be simultaneously brought to a strictly upper triangular form. Note that if we merely have a Lie algebra of matrices which is nilpotent ''as a Lie algebra'', then this conclusion does ''not'' follow (i.e. the naïve replacement in Lie's theorem of "solvable" with "nilpotent", and "upper triangular" with "strictly upper triangular", is false; this already fails for the one-dimensional Lie subalgebra of scalar matrices).

The theorem is named after the mathematician Friedrich Engel, who sketched a proof of it in a letter to Wilhelm Killing dated 20 July 1890 . Engel's student K.A. Umlauf gave a complete proof in his 1891 dissertation, reprinted as .

Statements

Let $\mathfrak(V)$ be the Lie algebra of the endomorphisms of a finite-dimensional vector space ''V'' and $\mathfrak g \subset \mathfrak(V)$ a subalgebra. Then Engel's theorem states the following are equivalent:
# Each $X \in \mathfrak$ is a nilpotent endomorphism on ''V''.
# There exists a flag $V = V_0 \supset V_1 \supset \cdots \supset V_n = 0, \, \operatorname V_i = i$ such that $\mathfrak g \cdot V_i \subset V_$; i.e., the elements of $\mathfrak g$ are simultaneously strictly upper-triangulizable.

Note that no assumption on the underlying base field is required.

We note that Statement 2. for various $\mathfrak g$ and ''V'' is equivalent to the statement
*For each nonzero finite-dimensional vector space ''V'' and a subalgebra $\mathfrak g \subset \mathfrak(V)$, there exists a nonzero vector ''v'' in ''V'' such that $X(v) = 0$ for every $X \in \mathfrak g.$

This is the form of the theorem proven in #Proof. (This statement is trivially equivalent to Statement 2 since it allows one to inductively construct a flag with the required property.)

In general, a Lie algebra $\mathfrak g$ is said to be nilpotent if the lower central series of it vanishes in a finite step; i.e., for C^0 \mathfrak g = \mathfrak g, C^i \mathfrak g = mathfrak g, C^ \mathfrak g/math> = (''i''+1)-th power of $\mathfrak g$, there is some ''k'' such that $C^k \mathfrak g = 0$. Then Engel's theorem implies the following theorem (also called Engel's theorem): when $\mathfrak g$ has finite dimension,
*$\mathfrak g$ is nilpotent if and only if $\operatorname(X)$ is nilpotent for each $X \in \mathfrak g$.
Indeed, if $\operatorname(\mathfrak g)$ consists of nilpotent operators, then by 1. $\Leftrightarrow$ 2. applied to the algebra $\operatorname(\mathfrak g) \subset \mathfrak(\mathfrak g)$, there exists a flag $\mathfrak g = \mathfrak_0 \supset \mathfrak_1 \supset \cdots \supset \mathfrak_n = 0$ such that mathfrak g, \mathfrak g_i\subset \mathfrak g_. Since $C^i \mathfrak g\subset \mathfrak g_i$, this implies $\mathfrak g$ is nilpotent. (The converse follows straightforwardly from the definition.)

Proof

We prove the following form of the theorem: ''if $\mathfrak \subset \mathfrak(V)$ is a Lie subalgebra such that every $X \in \mathfrak$ is a nilpotent endomorphism and if ''V'' has positive dimension, then there exists a nonzero vector ''v'' in ''V'' such that $X(v) = 0$ for each ''X'' in $\mathfrak$.''

The proof is by induction on the dimension of $\mathfrak$ and consists of a few steps. (Note the structure of the proof is very similar to that for Lie's theorem, which concerns a solvable algebra.) The basic case is trivial and we assume the dimension of $\mathfrak$ is positive.

Step 1: Find an ideal $\mathfrak$ of codimension one in $\mathfrak$.

:This is the most difficult step. Let $\mathfrak$ be a maximal (proper) subalgebra of $\mathfrak$, which exists by finite-dimensionality. We claim it is an ideal of codimension one. For each $X \in \mathfrak h$, it is easy to check that (1) $\operatorname(X)$ induces a linear endomorphism $\mathfrak/\mathfrak \to \mathfrak/\mathfrak$ and (2) this induced map is nilpotent (in fact, $\operatorname(X)$ is nilpotent as $X$ is nilpotent; see Jordan decomposition in Lie algebras). Thus, by inductive hypothesis applied to the Lie subalgebra of $\mathfrak(\mathfrak/\mathfrak)$ generated by $\operatorname(\mathfrak)$, there exists a nonzero vector ''v'' in $\mathfrak/\mathfrak$ such that $\operatorname(X)(v) = 0$ for each $X \in \mathfrak$. That is to say, if v = /math> for some ''Y'' in $\mathfrak$ but not in $\mathfrak h$, then $$, Y= \operatorname(X)(Y) \in \mathfrak for every $X \in \mathfrak$. But then the subspace $\mathfrak' \subset \mathfrak$ spanned by $\mathfrak$ and ''Y'' is a Lie subalgebra in which $\mathfrak$ is an ideal of codimension one. Hence, by maximality, $\mathfrak' = \mathfrak g$. This proves the claim.

Step 2: Let $W = \$. Then $\mathfrak$ stabilizes ''W''; i.e., $X (v) \in W$ for each $X \in \mathfrak, v \in W$.

:Indeed, for $Y$ in $\mathfrak$ and $X$ in $\mathfrak$, we have: $X(Y(v)) = Y(X(v)) +$, Yv) = 0 since $\mathfrak$ is an ideal and so $$, Y\in \mathfrak. Thus, $Y(v)$ is in ''W''.

Step 3: Finish up the proof by finding a nonzero vector that gets killed by $\mathfrak$.

:Write $\mathfrak = \mathfrak + L$ where ''L'' is a one-dimensional vector subspace. Let ''Y'' be a nonzero vector in ''L'' and ''v'' a nonzero vector in ''W''. Now, $Y$ is a nilpotent endomorphism (by hypothesis) and so $Y^k(v) \ne 0, Y^(v) = 0$ for some ''k''. Then $Y^k(v)$ is a required vector as the vector lies in ''W'' by Step 2. $\square$

See also

* Lie's theorem
* Heisenberg group

Notes

Citations

Works cited

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Representation theory of Lie algebras
Theorems in representation theory