OK, now we’ve gotten some of the basic facts about the root space decomposition down. So, as usual is a semisimple Lie algebra and
a Cartan subalgebra; we have the decomposition
, where
is the root system. For each
, we can choose a pair of vectors
. Then
generate a subalgebra
which is isomorphic to
. Here
and
is a multiple of
, which in turn is the dual to
under the Killing form that identifies
.
That was a lightning review of where we are; if you’ve missed something, check back at this post.
The notation suggests that the algebra should only depend on
and not on the particular choice of
(but
is uniquely determined from
and
). Indeed, this is the case, and it follows from
Proposition 1 When
,
is one-dimensional.
Choose any coming from suitable
and
. We have a representation of
on
(recall ) and we can apply the representation theory of
to it.
Now belongs to the representation
—that is, the adjoint representation of
, which is irreducible because
is simple! So by Weyl’s theorem, we have a decomposition of
-modules
The space is also a
submodule of
, disjoint from
; indeed it is the zero representation. We can expand this decomposition
I claim . Indeed, all weights of the action of
on
are even because
, so any irreducible submodule of
(and a fortiori of
) must have an even highest weight, and thus a nonzero vector with weight zero. But all the vectors of weight zero have been accounted for in
. So
.
In particular,
The proposition thus follows.
From this we see in particular that the only integral multiples of that are roots are
. More strongly:
Proposition 2 If
, the only multiples of
that are roots are
.
Indeed, we can consider the representation of on
. On this
must act with integral weights only by the representation theory of
, so in particular, only multiples of
can occur. But if
were a root, then
as an integral multiple of
would not allowed to be a root. If
with
odd occurred, then
would be a weight for
—and so would
, meaning
, which we’ve just proved can’t happen.
We saw a trick in the previous arguments: consider the action of on a suitable subalgebra of
and apply the representation theory of
to it.
Now let . We want to see for which
is
a weight.
Let range over an interval
such that
for
, but
. Consider the submodule
Now preserves this. Moreover,
acts on this space with trace
This trace is also zero though—because preserve
and
. As a result:
since .
Now define a bilinear form on
by
. With this, we find
Proposition 3
If,
In particular, it is equal to
where
are above.
This will be important in the sequel, as this is one of the axioms of a root system. We will show that the roots form a root system.
If we had any two distinct sets as above (Helgason calls them “maximal strings”), one would have to be disjoint from the other, which contradicts
. So:
Corollary 4 The set of
such that
is an interval.
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