## The Submodule of Invariants

If is a module of a Lie algebra , there is one submodule that turns out to be rather interesting: the submodule of vectors such that for all . We call these vectors “invariants” of .

As an illustration of how interesting these are, consider the modules we looked at last time. What are the invariant linear maps from one module to another ? We consider the action of on a linear map :

Or, in other words:

That is, a linear map is invariant if and only if it intertwines the actions on and . That is, .

Next, consider the bilinear forms on . Here we calculate

That is, a bilinear form is invariant if and only if it is associative, in the sense that the Killing form is:

## More New Modules from Old

There are a few constructions we can make, starting with the ones from last time and applying them in certain special cases.

First off, if and are two finite-dimensional -modules, then I say we can put an -module structure on the space of linear maps from to . Indeed, we can identify with : if is a basis for and is a basis for , then we can set up the dual basis of , such that . Then the elements form a basis for , and each one can be identified with the linear map sending to and all the other basis elements of to . Thus we have an inclusion , and a simple dimension-counting argument suffices to show that this is an isomorphism.

Now, since we have an action of on we get a dual action on . And because we have actions on and we get one on . What does this look like, explicitly? Well, we can write any such tensor as the sum of tensors of the form for some and . We calculate the action of on a vector :

In general we see that . In particular, the space of linear endomorphisms on is , and so it get an -module structure like this.

The other case of interest is the space of bilinear forms on a module . A bilinear form on is, of course, a linear functional on . And thus this space can be identified with . How does act on a bilinear form ? Well, we can calculate:

In particular, we can consider the case of bilinear forms on itself, where acts on itself by . Here we read

## New Modules from Old

There are a few standard techniques we can use to generate new modules for a Lie algebra from old ones. We’ve seen direct sums already, but here are a few more.

One way is to start with a module and then consider its dual space . I say that this can be made into an -module by setting

for all , , and . Bilinearity should be clear, so we just check the defining property of a module. That is, we take two Lie algebra elements and check

so for all , as desired.

Another way is to start with modules and and form their tensor product . Now we define a module structure on this space by

We check the defining property again. Calculate:

while

These are useful, and they’re only just the beginning.

## Reducible Modules

As might be surmised from irreducible modules, a reducible module for a Lie algebra is one that contains a nontrivial proper submodule — one other than or itself.

Now obviously if is a submodule we can form the quotient . This is the basic setup of a short exact sequence:

The question is, does this sequence split? That is, can we write as the direct sum of and some other submodule isomorphic to ?

First of all, let’s be clear that direct sums of modules do make sense. Indeed, if and are -modules then we can form an action on by defining it on each summand separately

Clearly the usual subspace inclusions and projections between , , and intertwine these actions, so they’re the required module morphisms. Further, it’s clear that .

So, do all short exact sequences of representations split? no. Indeed, let be the algebra of upper-triangular matrices, along with the obvious -dimensional representation. If we let be the basic column vector with a in the th row and elsewhere, then the one-dimensional space spanned by forms a one-dimensional submodule. Indeed, all upper-triangular matrices will send this column vector back to a multiple of itself.

On the other hand, it may be the case for a module that any nontrivial proper submodule has a complementary submodule with . In this case, is either irreducible or it’s not; if not, then any proper nontrivial submodule of will also be a proper nontrivial submodule of , and we can continue taking smaller submodules until we get to an irreducible one, so we may as well assume that is irreducible. Now the same sort of argument works for , showing that if it’s not irreducible it can be decomposed into the direct sum of some irreducible submodule and some complement, which is another nontrivial proper submodule of . At each step, the complement gets smaller and smaller, until we have decomposed entirely into a direct sum of irreducible submodules.

If is decomposable into a direct sum of irreducible submodules, we say that is “completely reducible”, or “decomposable”, as we did when we were working with groups. Any module where any nontrivial proper submodule has a complement is thus completely reducible; conversely, complete reducibility implies that any nontrivial proper submodule has a complement. Indeed, such a submodule must consist of some, but not all, of the summands of , and the complement will consist of the rest.

## Irreducible Modules

Sorry for the delay; it’s getting crowded around here again.

Anyway, an irreducible module for a Lie algebra is a pretty straightforward concept: it’s a module such that its only submodules are and . As usual, Schur’s lemma tells us that any morphism between two irreducible modules is either or an isomorphism. And, as we’ve seen in other examples involving linear transformations, all automorphisms of an irreducible module are scalars times the identity transformation. This, of course, doesn’t depend on any choice of basis.

A one-dimensional module will always be irreducible, if it exists. And a unique — up to isomorphism, of course — one-dimensional module will always exist for simple Lie algebras. Indeed, if is simple then we know that . Any one-dimensional representation must have its image in . But the only traceless matrix is the zero matrix. Setting for all does indeed give a valid representation of .

## Lie Algebra Modules

It should be little surprise that we’re interested in concrete actions of Lie algebras on vector spaces, like we were for groups. Given a Lie algebra we define an -module to be a vector space equipped with a bilinear function — often written satisfying the relation

Of course, this is the same thing as a representation . Indeed, given a representation we can define ; given an action we can define a representation by . The above relation is exactly the statement that the bracket in corresponds to the bracket in .

Of course, the modules of a Lie algebra form a category. A homomorphism of -modules is a linear map satisfying

We automatically get the concept of a submodule — a subspace sent back into itself by each — and a quotient module. In the latter case, we can see that if is any submodule then we can define . This is well-defined, since if is any other representative of then , and , so and both represent the same element of .

Thus, every submodule can be seen as the kernel of some homomorphism: the projection . It should be clear that every homomorphism has a kernel, and a cokernel can be defined simply as the quotient of the range by the image. All we need to see that the category of -modules is abelian is to show that every epimorphism is actually a quotient, but we know this is already true for the underlying vector spaces. Since the (vector space) kernel of an -module map is an -submodule, this is also true for -modules.

## More Kostka Numbers

First let’s mention a few more general results about Kostka numbers.

Among all the tableaux that partition , it should be clear that . Thus the Kostka number is not automatically zero. In fact, I say that it’s always . Indeed, the shape is a single row with entries, and the content gives us a list of numbers, possibly with some repeats. There’s exactly one way to arrange this list into weakly increasing order along the single row, giving .

On the other extreme, , so might be nonzero. The shape is given by , and the content gives one entry of each value from to . There are no possible entries to repeat, and so any semistandard tableau with content is actually standard. Thus — the number of standard tableaux of shape .

This means that we can decompose the module :

But , which means each irreducible -module shows up here with a multiplicity equal to its dimension. That is, is always the left regular representation.

Okay, now let’s look at a full example for a single choice of . Specifically, let . That is, we’re looking for semistandard tableaux of various shapes, all with two entries of value , two of value , and one of value . There are five shapes with . For each one, we will look for all the ways of filling it with the required content.

Counting the semistandard tableaux on each row, we find the Kostka numbers. Thus we get the decomposition

## Kostka Numbers

Now we’ve finished our proof that the intertwinors coming from semistandard tableauxspan the space of all intertwinors from the Specht module to the Young tabloid module . We also know that they’re linearly independent, and so they form a basis of the space of intertwinors — one for each semistandard generalized tableau.

Since the Specht modules are irreducible, we know that the dimension of this space is the multiplicity of in . And the dimension, of course, is the number of basis elements, which is the number of semistandard generalized tableaux of shape and content . This number we call the “Kostka number” . We’ve seen that there is a decomposition

Now we know that the Kostka numbers give these multiplicities, so we can write

We saw before that when , the multiplicity is one. In terms of the Kostka numbers, this tells us that . Is this true? Well, the only way to fit entries with value , with value , and so on into a semistandard tableau of shape is to put all the entries on the th row.

In fact, we can extend the direct sum by removing the restriction on :

This is because when we have . Indeed, we must eventually have , and so we can't fit all the entries with values through on the first rows of . We must at the very least have a repeated entry in some column, if not a descent. There are thus no semistandard generalized tableaux with shape and content in this case.

## Intertwinors from Semistandard Tableaux Span, part 3

Now we are ready to finish our proof that the intertwinors coming from semistandard generalized tableaux span the space of all intertwinors between these modules.

As usual, pick any intertwinor and write

Now define the set to consist of those semistandard generalized tableaux so that for some appearing in this sum with a nonzero coefficient. This is called the “lower order ideal” generated by the in the sum. We will prove our assertion by induction on the size of this order ideal.

If is empty, then must be the zero map. Indeed, our lemmas showed that if is not the zero map, then at least one semistandard shows up in the above sum, and this would itself belong to . And of course the zero map is contained in any span.

Now, if is not empty, then there is at least some semistandard with in the sum. Our lemmas even show that we can pick one so that is maximal among all the tableaux in the sum. Let’s do that and define a new intertwinor:

I say that is with removed.

Every appearing in has , since if is semistandard then is the largest column equivalence class in . Thus must be a subset of since we can’t be introducing any new nonzero coefficients.

Our lemmas show that if , then must appear with the same coefficient in both and . That is, they must be cancelled off by the subtraction. Since is maximal there’s nothing above it that might keep it inside the ideal, and so .

So by induction we conclude that is contained within the span of the generated by semistandard tableaux, and thus must be as well.

## Intertwinors from Semistandard Tableaux Span, part 2

We continue our proof that the intertwinors that come from semistandard tableaux span the space of all such intertwinors. This time I assert that if is not the zero map, then there is some semistandard with .

Obviously there are some nonzero coefficients; if , then

which would make the zero map. So among the nonzero , there are some with maximal in the column dominance order. I say that we can find a semistandard among them.

By the results yesterday we know that the entries in the columns of these are all distinct, so in the column tabloids we can arrange them to be strictly increasing down the columns. What we must show is that we can find one with the rows weakly increasing.

Well, let’s pick a maximal and suppose that it does have a row descent, which would keep it from being semistandard. Just like the last time we saw row descents, we get a chain of distinct elements running up the two columns:

We choose the sets and and the Garnir element just like before. We find

The generalized tableau must appear in with unit coefficient, so to cancel it off there must be some other generalized tableau with for some that shows up in . But since this just interchanges some and entries, we can see that , which contradicts the maximality of our choice of .

Thus there can be no row descents in , and is in fact semistandard.