Decomposition of Semisimple Lie Algebras
We say that a Lie algebra is the direct sum of a collection of ideals
if it’s the direct sum as a vector space. In particular, this implies that
, meaning that the bracket of any two elements from different ideals is zero.
Now, if is semisimple then there is a collection of ideals, each of which is simple as a Lie algebra in its own right, such that
is the direct sum of these simple ideals. Further, every such simple ideal of
is one in the collection — there’s no way to spread another simple ideal across two or more summands in this decomposition. And the Killing form of a summand is the restriction of the Killing form of
to that summand, as we expect for any ideal of a Lie algebra.
If is any ideal then we can define the subspace
of vectors in
that are “orthogonal” to all the vectors in
with respect to the Killing form
. The associativity of
shows that
is also an ideal, just as we saw for the radical. Indeed, the radical of
is just
. Anyhow, Cartan’s criterion again shows that the intersection
is solvable, but since
is semisimple this means
, and we can write
.
So now we can use an induction on the dimension of ; if
has no nonzero proper ideal, it’s already simple. Otherwise we can pick some proper ideal
to get
, where each summand has a lower dimension than
. Any ideal of
is an ideal of
— the bracket with anything from
is zero — so
and
must be semisimple as well, or else there would be a nonzero solvable ideal of
. By induction, each one can be decomposed into simple ideals, so
can as well.
Now, if is any simple ideal of
, then
is an ideal of
. It can’t be zero, since if it were
would be contained in
, which is zero. Thus, since
is simple, we must have
. But the direct-sum decomposition tells us that
, so all but one of these brackets
must be zero, and that bracket must be
itself. But this means
for this simple summand, and — by the simplicity of
—
.
From this decomposition we conclude that for all semisimple we have
. Every ideal and every quotient of
must also be semisimple, since each must consist of some collection of the summands of
.
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