Lie’s Theorem
The lemma leading to Engel’s theorem boils down to the assertion that there is some common eigenvector for all the endomorphisms in a nilpotent linear Lie algebra on a finite-dimensional nonzero vector space
. Lie’s theorem says that the same is true of solvable linear Lie algebras. Of course, in the nilpotent case the only possible eigenvalue was zero, so we may find things a little more complicated now. We will, however, have to assume that
is algebraically closed and that no multiple of the unit in
is zero.
We will proceed by induction on the dimension of using the same four basic steps as in the lemma: find an ideal
of codimension one, so we can write
for some
; find common eigenvectors for
; find a subspace of such common eigenvectors stabilized by
; find in that space an eigenvector for
.
First, solvability says that properly includes
, or else the derived series wouldn’t be able to even start heading towards
. The quotient
must be abelian, with all brackets zero, so we can pick any subspace of this quotient with codimension one and it will be an ideal. The preimage of this subspace under the quotient projection will then be an ideal
of codimension one.
Now, is a subalgebra of
, so we know it’s also solvable, so induction tells us that there’s a common eigenvector
for the action of
. If
is zero, then
must be one-dimensional abelian, in which case the proof is obvious. Otherwise there is some linear functional
defined by
Of course, is not the only such eigenvector; we define the (nonzero) subspace
by
Next we must show that sends
back into itself. To see this, pick
and
and check that
But if , then we’d have
; we need to verify that
. In the nilpotent case — Engel’s theorem — the functional
was constantly zero, so this was easy, but it’s a bit harder here.
Fixing and
we pick
to be the first index where the collection
is linearly independent — the first one where we can express
as the linear combination of all the previous
. If we write
for the subspace spanned by the first
of these vectors, then the dimension of
grows one-by-one until we get to
, and
from then on.
I say that each of the are invariant under each
. Indeed, we can prove the congruence
that is, acts on
by multiplication by
, plus some “lower-order terms”. For
this is the definition of
; in general we have
for some .
And so we conclude that, using the obvious basis of the action of
on this subspace is in the form of an upper-triangular matrix with
down the diagonal. The trace of this matrix is
. And in particular, the trace of the action of
on
is
. But
and
both act as endomorphisms of
— the one by design and the other by the above proof — and the trace of any commutator is zero! Since
must have an inverse we conclude that
.
Okay so that checks out that the action of sends
back into itself. We finish up by picking some eigenvector
of
, which we know must exist because we’re working over an algebraically closed field. Incidentally, we can then extend
to all of
by using
.
[…] like to have matrix-oriented versions of Engel’s theorem and Lie’s theorem, and to do that we’ll need flags. I’ve actually referred to flags long, long ago, but […]
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