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 ![[L,L]](https://s-ssl.wordpress.com/latex.php?latex=%5BL%2CL%5D&bg=e6e6e6&fg=333333&s=0 "[L,L]")
, or else the derived series wouldn’t be able to even start heading towards 
. The quotient ![L/[L,L]](https://s-ssl.wordpress.com/latex.php?latex=L%2F%5BL%2CL%5D&bg=e6e6e6&fg=333333&s=0 "L/[L,L]")
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
\\&=l(\lambda(k)w)-\lambda([l,k])w\\&=\lambda(k)l(w)-\lambda([l,k])w\end{aligned}](https://s-ssl.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7Dk%28l%28w%29%29%26%3Dl%28k%28w%29%29-%5Bl%2Ck%5D%28w%29%5C%5C%26%3Dl%28%5Clambda%28k%29w%29-%5Clambda%28%5Bl%2Ck%5D%29w%5C%5C%26%3D%5Clambda%28k%29l%28w%29-%5Clambda%28%5Bl%2Ck%5D%29w%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}k(l(w))&=l(k(w))-[l,k](w)\\&=l(\lambda(k)w)-\lambda([l,k])w\\&=\lambda(k)l(w)-\lambda([l,k])w\end{aligned}")
But if 
, then we’d have 
; we need to verify that ![\lambda([l,k])=0](https://s-ssl.wordpress.com/latex.php?latex=%5Clambda%28%5Bl%2Ck%5D%29%3D0&bg=e6e6e6&fg=333333&s=0 "\lambda([l,k])=0")
. 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
)\\&=\lambda(k)l^i(w)+l(w')-\lambda([l,k])l^{i-1}(w)-w''\end{aligned}](https://s-ssl.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7Dk%28l%5Ei%28w%29%29%26%3Dk%28l%28l%5E%7Bi-1%7D%28w%29%29%29%5C%5C%26%3Dl%28k%28l%5E%7Bi-1%7D%28w%29%29%29-%5Bl%2Ck%5D%28l%5E%7Bi-1%7D%28w%29%29%5C%5C%26%3D%5Clambda%28k%29l%5Ei%28w%29%2Bl%28w%27%29-%5Clambda%28%5Bl%2Ck%5D%29l%5E%7Bi-1%7D%28w%29-w%27%27%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}k(l^i(w))&=k(l(l^{i-1}(w)))\\&=l(k(l^{i-1}(w)))-[l,k](l^{i-1}(w))\\&=\lambda(k)l^i(w)+l(w')-\lambda([l,k])l^{i-1}(w)-w''\end{aligned}")
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 ![[l,k]](https://s-ssl.wordpress.com/latex.php?latex=%5Bl%2Ck%5D&bg=e6e6e6&fg=333333&s=0 "[l,k]")
on is ![n\lambda([l,k])](https://s-ssl.wordpress.com/latex.php?latex=n%5Clambda%28%5Bl%2Ck%5D%29&bg=e6e6e6&fg=333333&s=0 "n\lambda([l,k])")
. 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 
.
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August 25, 2012 -
Posted by John Armstrong |
Algebra, Lie Algebras
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