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 1When , 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 2If , 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 3If ,

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 4The set of such that is an interval.

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