Engel's theorem

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

$\mathrm{ad}(X):𝔤\to 𝔤,$

given by $\mathrm{ad}(X)(Y)=[X,Y]$, is a nilpotent endomorphism on $𝔤$; i.e., $\mathrm{ad}(X{)}^k=0$ for some k.Template:Sfn 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 Template:Harv. Engel's student K.A. Umlauf gave a complete proof in his 1891 dissertation, reprinted as Template:Harv.

Statements

Let $𝔤𝔩(V)$ be the Lie algebra of the endomorphisms of a finite-dimensional vector space V and $𝔤\subset 𝔤𝔩(V)$ a subalgebra. Then Engel's theorem states the following are equivalent:

1. Each $X\in 𝔤$ is a nilpotent endomorphism on V.
2. There exists a flag $V=V_0\supset V_1\supset \cdots \supset V_n=0,\mathrm{codim}{V}_i=i$ such that $𝔤\cdot V_i\subset V_{i+1}$; i.e., the elements of $𝔤$ are simultaneously strictly upper-triangulizable.

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

We note that Statement 2. for various $𝔤$ and V is equivalent to the statement

• For each nonzero finite-dimensional vector space V and a subalgebra $𝔤\subset 𝔤𝔩(V)$, there exists a nonzero vector v in V such that $X(v)=0$ for every $X\in 𝔤.$

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 $𝔤$ is said to be nilpotent if the lower central series of it vanishes in a finite step; i.e., for $C^0𝔤=𝔤,C^i𝔤=[𝔤,C^{i-1}𝔤]$ = (i+1)-th power of $𝔤$, there is some k such that $C^k𝔤=0$. Then Engel's theorem implies the following theorem (also called Engel's theorem): when $𝔤$ has finite dimension,

• $𝔤$ is nilpotent if and only if $\mathrm{ad}(X)$ is nilpotent for each $X\in 𝔤$.

Indeed, if $\mathrm{ad}(𝔤)$ consists of nilpotent operators, then by 1. $\iff$ 2. applied to the algebra $\mathrm{ad}(𝔤)\subset 𝔤𝔩(𝔤)$, there exists a flag $𝔤={𝔤}_0\supset {𝔤}_1\supset \cdots \supset {𝔤}_n=0$ such that $[𝔤,{𝔤}_i]\subset {𝔤}_{i+1}$. Since $C^i𝔤\subset {𝔤}_i$, this implies $𝔤$ is nilpotent. (The converse follows straightforwardly from the definition.)

Proof

We prove the following form of the theorem:Template:Sfn if $𝔤\subset 𝔤𝔩(V)$ is a Lie subalgebra such that every $X\in 𝔤$ 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 $𝔤$.

The proof is by induction on the dimension of $𝔤$ 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 $𝔤$ is positive.

Step 1: Find an ideal $𝔥$ of codimension one in $𝔤$.

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

Step 2: Let $W=\{v\in V|X(v)=0,X\in 𝔥\}$. Then $𝔤$ stabilizes W; i.e., $X(v)\in W$ for each $X\in 𝔤,v\in W$.

Indeed, for $Y$ in $𝔤$ and $X$ in $𝔥$, we have: $X(Y(v))=Y(X(v))+[X,Y](v)=0$ since $𝔥$ is an ideal and so $[X,Y]\in 𝔥$. Thus, $Y(v)$ is in W.

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

Write $𝔤=𝔥+L$ 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 $Y^k(v)\ne 0,Y^{k+1}(v)=0$ for some k. Then $Y^k(v)$ is a required vector as the vector lies in W by Step 2. $◻$

See also

• Lie's theorem
• Heisenberg group

Works cited

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