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Engel’s theorem∗
rmilson†
2013-03-21 14:13:46
Before proceeding, it will be useful to recall the definition of a nilpotent Lie
algebra. Let g be a Lie algebra. The lower central series of g is defined to be
the filtration of ideals
D0 g ⊃ D1 g ⊃ D2 g ⊃ . . . ,
where
D0 g = g,
Dk+1 g = [g, Dk g],
k ∈ N.
To say that g is nilpotent is to say that the lower central series has a trivial
termination, i.e. that there exists a k such that
Dk g = 0,
or equivalently, that k nested bracket operations always vanish.
Theorem 1 (Engel) Let g ⊂ End V be a Lie algebra of endomorphisms of a
finite-dimensional vector space V . Suppose that all elements of g are nilpotent
transformations. Then, g is a nilpotent Lie algebra.
Lemma 1 Let X : V → V be a nilpotent endomorphism of a vector space V .
Then, the adjoint action
ad(X) : End V → End V
is also a nilpotent endomorphism.
Proof.
Suppose that
Xk = 0
for some k ∈ N. We will show that
ad(X)2k−1 = 0.
∗ hEngelsTheoremi created: h2013-03-21i by: hrmilsoni version: h32991i Privacy setting:
h1i hTheoremi h17B30i h15A57i
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Note that
ad(X) = l(X) − r(X),
where
l(X), r(X) : End V → End V,
are the endomorphisms corresponding, respectively, to left and right multiplication by X. These two endomorphisms commute, and hence we can use the
binomial formula to write
ad(X)2k−1 =
2k−1
X
(−1)i l(X)2k−1−i r(X)i .
i=0
Each of terms in the above sum vanishes because
l(X)k = r(X)k = 0.
QED
Lemma 2 Let g be as in the theorem, and suppose, in addition, that g is a
nilpotent Lie algebra. Then the joint kernel,
\
ker g =
ker a,
a∈g
is non-trivial.
Proof. We proceed by induction on the dimension of g. The claim is true for
dimension 1, because then g is generated by a single nilpotent transformation,
and all nilpotent transformations are singular.
Suppose then that the claim is true for all Lie algebras of dimension less
than n = dim g. We note that D1 g fits the hypotheses of the lemma, and
has dimension less than n, because g is nilpotent. Hence, by the induction
hypothesis
V0 = ker D1 g
is non-trivial. Now, if we restrict all actions to V0 , we obtain a representation
of g by abelian transformations. This is because for all a, b ∈ g and v ∈ V0 we
have
abv − bav = [a, b]v = 0.
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 g/D1 g acting on V0 , and the desired conclusion follows. QED
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Proof of the theorem. We proceed by induction on the dimension of g. The
theorem is true in dimension 1, because in that circumstance D1 g is trivial.
Next, suppose that the theorem holds for all Lie algebras of dimension less
than n = dim g. Let h ⊂ g be a properly contained subalgebra of minimum
codimension. We claim that there exists an a ∈ g but not in h such that
[a, h] ⊂ h.
By the induction hypothesis, h is nilpotent. To prove the claim consider the
isotropy representation of h on g/h. By Lemma 1, the action of each a ∈ h on
g/h 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 a ∈ g but
not in h such that
[b, a] ≡ 0 mod h,
for all b ∈ h. Equivalently, [h, a] ⊂ h and the claim is proved.
Evidently then, the span of a and h is a subalgebra of g. Since h has minimum
codimension, we infer that h and a span all of g, and that
D1 g ⊂ h.
(1)
Next, we claim that all the Dk h are ideals of g. It is enough to show that
[a, Dk h] ⊂ Dk h.
We argue by induction on k. Suppose the claim is true for some k. Let b ∈
h, c ∈ Dk h be given. By the Jacobi identity
[a, [b, c]] = [[a, b], c] + [b, [a, c]].
The first term on the right hand-side in Dk+1 h because [a, b] ∈ h. The second term is in Dk+1 h by the induction hypothesis. In this way the claim is
established.
Now a is nilpotent, and hence by Lemma 1,
ad(a)n = 0
for some n ∈ N. We now claim that
Dn+1 g ⊂ D1 h.
By (??) it suffices to show that
n times
z }| {
[g, [. . . [g, h] . . .]] ⊂ D1 h.
Putting
g1 = g/D1 h,
h1 = h/D1 h,
this is equivalent to
n times
z }| {
[g1 , [. . . [g1 , h1 ] . . .]] = 0.
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(2)
However, h1 is abelian, and hence, the above follows directly from (??).
Adapting this argument in the obvious fashion we can show that
Dkn+1 g ⊂ Dk h.
Since h is nilpotent, g must be nilpotent as well. QED
Historical remark. In the traditional formulation of Engel’s theorem, the
hypotheses are the same, but the conclusion is that there exists a basis B of V ,
such that all elements of g are represented by nilpotent matrices relative to B.
Let us put this another way. The vector space of nilpotent matrices Nil,
is a nilpotent Lie algebra, and indeed all subalgebras of Nil are nilpotent Lie
algebras. Engel’s theorem asserts that the converse holds, i.e. if all elements of a
Lie algebra g are nilpotent transformations, then g is isomorphic to a subalgebra
of Nil.
The classical result follows straightforwardly from our version of the Theorem and from Lemma 2. Indeed, let V1 be the joint kernel g. We then let U2 be
the joint kernel of g acting on V /V0 , and let V2 ⊂ V be the subspace obtained
by pulling U2 x back to V . We do this a finite number of times and obtain a flag
of subspaces
0 = V0 ⊂ V1 ⊂ V2 ⊂ . . . ⊂ Vn = V,
such that
gVk+1 = Vk
for all k. The choose an adapted basis relative to this flag, and we’re done.
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