or equivalently, that nested bracket operations always vanish.
Theorem 1 (Engel)
Let be a nilpotent endomorphism of a vector space . Then, the adjoint action
is also a nilpotent endomorphism.
for some . We will show that
are the endomorphisms corresponding, respectively, to left and right multiplication by . These two endomorphisms commute, and hence we can use the binomial formula to write
Each of terms in the above sum vanishes because
Suppose then that the claim is true for all Lie algebras of dimension less than . We note that fits the hypotheses of the lemma, and has dimension less than , because is nilpotent. Hence, by the induction hypothesis
Now a finite number of mutually commuting linear endomorphisms admits a mutual eigenspace decomposition. In particular, if all of the commuting endomorphisms are singular, their joint kernel will be non-trivial. We apply this result to a basis of acting on , and the desired conclusion follows. QED
Proof of the theorem.
We proceed by induction on the dimension of . The theorem is true in dimension 1, because in that circumstance is trivial.
By the induction hypothesis, is nilpotent. To prove the claim consider the isotropy representation of on . By Lemma 1, the action of each on is a nilpotent endomorphism. Hence, we can apply Lemma 2 to deduce that the joint kernel of all these actions is non-trivial, i.e. there exists a but not in such that
for all . Equivalently, and the claim is proved.
Evidently then, the span of and is a subalgebra of . Since has minimum codimension, we infer that and span all of , and that
Next, we claim that all the are ideals of . It is enough to show that
We argue by induction on . Suppose the claim is true for some . Let be given. By the Jacobi identity
The first term on the right hand-side in because . The second term is in by the induction hypothesis. In this way the claim is established.
Now is nilpotent, and hence by Lemma 1,
for some . We now claim that
By (1) it suffices to show that
this is equivalent to
However, is abelian, and hence, the above follows directly from (2).
In the traditional formulation of Engel’s theorem, the hypotheses are the same, but the conclusion is that there exists a basis of , such that all elements of are represented by nilpotent matrices relative to .
Let us put this another way. The vector space of nilpotent matrices , is a nilpotent Lie algebra, and indeed all subalgebras of are nilpotent Lie algebras. Engel’s theorem asserts that the converse holds, i.e. if all elements of a Lie algebra are nilpotent transformations, then is isomorphic to a subalgebra of .
The classical result follows straightforwardly from our version of the Theorem and from Lemma 2. Indeed, let be the joint kernel . We then let be the joint kernel of acting on , and let be the subspace obtained by pulling back to . We do this a finite number of times and obtain a flag of subspaces
for all . The choose an adapted basis relative to this flag, and we’re done.
|Date of creation||2013-03-22 12:42:25|
|Last modified on||2013-03-22 12:42:25|
|Last modified by||rmilson (146)|