Engel's theorem
Template:Short description In representation theory, a branch of mathematics, Engel's theorem states that a finite-dimensional Lie algebra <math>\mathfrak g</math> is a nilpotent Lie algebra if and only if for each <math>X \in \mathfrak g</math>, the adjoint map
- <math>\operatorname{ad}(X)\colon \mathfrak{g} \to \mathfrak{g},</math>
given by <math>\operatorname{ad}(X)(Y) = [X, Y]</math>, is a nilpotent endomorphism on <math>\mathfrak{g}</math>; i.e., <math>\operatorname{ad}(X)^k = 0</math> 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.
StatementsEdit
Let <math>\mathfrak{gl}(V)</math> be the Lie algebra of the endomorphisms of a finite-dimensional vector space V and <math>\mathfrak g \subset \mathfrak{gl}(V)</math> a subalgebra. Then Engel's theorem states the following are equivalent:
- Each <math>X \in \mathfrak{g}</math> is a nilpotent endomorphism on V.
- There exists a flag <math>V = V_0 \supset V_1 \supset \cdots \supset V_n =
0, \, \operatorname{codim} V_i = i</math> such that <math>\mathfrak g \cdot V_i \subset V_{i+1}</math>; i.e., the elements of <math>\mathfrak g</math> are simultaneously strictly upper-triangulizable.
Note that no assumption on the underlying base field is required.
We note that Statement 2. for various <math>\mathfrak g</math> and V is equivalent to the statement
- For each nonzero finite-dimensional vector space V and a subalgebra <math>\mathfrak g \subset \mathfrak{gl}(V)</math>, there exists a nonzero vector v in V such that <math>X(v) = 0</math> for every <math>X \in \mathfrak g.</math>
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 <math>\mathfrak g</math> is said to be nilpotent if the lower central series of it vanishes in a finite step; i.e., for <math>C^0 \mathfrak g = \mathfrak g, C^i \mathfrak g = [\mathfrak g, C^{i-1} \mathfrak g]</math> = (i+1)-th power of <math>\mathfrak g</math>, there is some k such that <math>C^k \mathfrak g = 0</math>. Then Engel's theorem implies the following theorem (also called Engel's theorem): when <math>\mathfrak g</math> has finite dimension,
- <math>\mathfrak g</math> is nilpotent if and only if <math>\operatorname{ad}(X)</math> is nilpotent for each <math>X \in \mathfrak g</math>.
Indeed, if <math>\operatorname{ad}(\mathfrak g)</math> consists of nilpotent operators, then by 1. <math>\Leftrightarrow</math> 2. applied to the algebra <math>\operatorname{ad}(\mathfrak g) \subset \mathfrak{gl}(\mathfrak g)</math>, there exists a flag <math>\mathfrak g = \mathfrak{g}_0 \supset \mathfrak{g}_1 \supset \cdots \supset \mathfrak{g}_n = 0</math> such that <math>[\mathfrak g, \mathfrak g_i] \subset \mathfrak g_{i+1}</math>. Since <math>C^i \mathfrak g\subset \mathfrak g_i</math>, this implies <math>\mathfrak g</math> is nilpotent. (The converse follows straightforwardly from the definition.)
ProofEdit
We prove the following form of the theorem:Template:Sfn if <math>\mathfrak{g} \subset \mathfrak{gl}(V)</math> is a Lie subalgebra such that every <math>X \in \mathfrak{g}</math> is a nilpotent endomorphism and if V has positive dimension, then there exists a nonzero vector v in V such that <math>X(v) = 0</math> for each X in <math>\mathfrak{g}</math>.
The proof is by induction on the dimension of <math>\mathfrak{g}</math> 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 <math>\mathfrak{g}</math> is positive.
Step 1: Find an ideal <math>\mathfrak{h}</math> of codimension one in <math>\mathfrak{g}</math>.
- This is the most difficult step. Let <math>\mathfrak{h}</math> be a maximal (proper) subalgebra of <math>\mathfrak{g}</math>, which exists by finite-dimensionality. We claim it is an ideal of codimension one. For each <math>X \in \mathfrak h</math>, it is easy to check that (1) <math>\operatorname{ad}(X)</math> induces a linear endomorphism <math>\mathfrak{g}/\mathfrak{h} \to \mathfrak{g}/\mathfrak{h}</math> and (2) this induced map is nilpotent (in fact, <math>\operatorname{ad}(X)</math> is nilpotent as <math>X</math> is nilpotent; see Jordan decomposition in Lie algebras). Thus, by inductive hypothesis applied to the Lie subalgebra of <math>\mathfrak{gl}(\mathfrak{g}/\mathfrak{h})</math> generated by <math>\operatorname{ad}(\mathfrak{h})</math>, there exists a nonzero vector v in <math>\mathfrak{g}/\mathfrak{h}</math> such that <math>\operatorname{ad}(X)(v) = 0</math> for each <math>X \in \mathfrak{h}</math>. That is to say, if <math>v = [Y]</math> for some Y in <math>\mathfrak{g}</math> but not in <math>\mathfrak h</math>, then <math>[X, Y] = \operatorname{ad}(X)(Y) \in \mathfrak{h}</math> for every <math>X \in \mathfrak{h}</math>. But then the subspace <math>\mathfrak{h}' \subset \mathfrak{g}</math> spanned by <math>\mathfrak{h}</math> and Y is a Lie subalgebra in which <math>\mathfrak{h}</math> is an ideal of codimension one. Hence, by maximality, <math>\mathfrak{h}' = \mathfrak g</math>. This proves the claim.
Step 2: Let <math>W = \{ v \in V | X(v) = 0, X \in \mathfrak{h} \}</math>. Then <math>\mathfrak{g}</math> stabilizes W; i.e., <math>X (v) \in W</math> for each <math>X \in \mathfrak{g}, v \in W</math>.
- Indeed, for <math>Y</math> in <math>\mathfrak{g}</math> and <math>X</math> in <math>\mathfrak{h}</math>, we have: <math>X(Y(v)) = Y(X(v)) + [X, Y](v) = 0</math> since <math>\mathfrak{h}</math> is an ideal and so <math>[X, Y] \in \mathfrak{h}</math>. Thus, <math>Y(v)</math> is in W.
Step 3: Finish up the proof by finding a nonzero vector that gets killed by <math>\mathfrak{g}</math>.
- Write <math>\mathfrak{g} = \mathfrak{h} + L</math> where L is a one-dimensional vector subspace. Let Y be a nonzero vector in L and v a nonzero vector in W. Now, <math>Y</math> is a nilpotent endomorphism (by hypothesis) and so <math>Y^k(v) \ne 0, Y^{k+1}(v) = 0</math> for some k. Then <math>Y^k(v)</math> is a required vector as the vector lies in W by Step 2. <math>\square</math>