This post is the second in the series on and the third in the series on Lie algebras. I’m going to start where we left off yesterday on , and go straight from there to classification. Basically, it’s linear algebra.

** Classification **

We’ve covered all the preliminaries now and we can classify the -representations, the really interesting material here. By Weyl’s theorem, we can restrict ourselves to irreducible representations. Fix an irreducible .

So, we know that acts diagonalizably on , which means we can write

where for each , i.e. is the -eigenspace.

Definition 1The eigenvalues occuring nontrivially in the above decomposition are called theweightsof .

Here “weight” is just another word for “eigenvalue” but in the general case of semisimple Lie algebras, weights are actually linear functionals on an abelian subalgebra, which here just happens to be 1-dimensional (spanned by ).

I claim now:

This proposition says that, basically, increases the weight by 2, and decreases it by 2. Its generalization says a lot about how semisimple Lie algebras behave, but isn’t really any different in the proof.

The proof in our case runs as follows. Suppose . We must show , which is what the first assertion states. Indeed, by the definition of representations:

by the all-important relations between . For the story is similar:

The proposition itself didn’t use irreducibility, which we’ll soon have to invoke. First of all, is finite-dimensional, so the weights are finite, and there must be a *maximal* weight (a “highest weight”), with an associated eigenvector : the **highest weight vector**.

Claim 1. In other words, is annihilated by the Lie subalgebra of strictly upper-triangular matrices.

Indeed, otherwise by (1) would be an eigenvector of eigenvalue , though was maximal. This proves the claim.

So, we now want to see that repeated multiplication by the ‘s on leads to vectors that form a basis all of . Since the are weight vectors of weight by (1) (and induction), we will have a nice decomposition of . Thus, define the sequence ; the notation refers to repeated multiplication by on .

Claim 2Suppose . Then are linearly independent.

Well, we know already that () are eigenvectors of of eigenvalues , which are different for distinct . The assertion now follows from the general fact from linear algebra:

Let be a linear transformation on a finite-dimensional vector space . If the ‘s ) are nonzero eigenvectors for , say , with different eigenvalues (say if ), then the ‘s are linearly independent.

To prove this, suppose given a nontrivial linear relation between the ‘s, which we may assume has as *few variables as possible.* Renumbering the if necessary, write the relation in the form

with no ; then apply to (2) to get

and multiply (2) by to get

Subtracting the last two identties gives a nontrivial (as ) linear relation between the with fewer variables, contradiction.

So, back to . Our is finite-dimensional, and we can’t have infinitely many linearly independent vectors. Thus eventually we get ; assume moreover is the smallest integer for which this happens.

Our next goal is to show that the , , for a *basis* for . To do this, we just need to show they span , and we’ll be done if we show that their span is invariant under the action of : that will mean their span is a nontrivial -submodule of , hence all of by irreducibility.

The ‘s have decreasing weights, and increases weights. Thus the following makes sense:

Claim 3We have if .

Induction on . First of all, let’s take ; then

Here we used Claim 1.

Now assume the claim true for , and we prove it for :

So, both and map into itself, and we have proved most of the following:

Theorem 3 (Classification Theorem for Finite-Dimensional Irreducibles)The , form a basis for the irreducible -module ; if is the highest weight, we have

each factor in this sum is one-dimensional, spanned by ; and move the weight spaces up and down by respectively. The maps for eigenvalues are even isomorphisms at all but boundary values of . The compositions are scalar multiplication by nonzero scalars.

All that remains is the following sub-result:

Claim 4if .

Otherwise, and would both map into itself, and that would be a proper -submodule of , contradiction.

It’s actually true in the finite-dimensional case that the highest weight is actually a positive integer. This follows from making some of the earlier arguments with myseterious “constants” more precise; I’ll probably return to this later on sometime, in the context of general semisimple Lie algebras. I do want to go back to full generality rather than focusing only on specific examples, but we’ve now covered the key ideas for .

July 22, 2009 at 2:07 pm

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