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:

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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

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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.

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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.

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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 .

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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.

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If we set and , we can see that . Indeed, if is any derivation and , then we can check that

This makes an ideal, so the Killing form of is the restriction of of the Killing form of . Then we can define to be the subspace orthogonal (with respect to ) to , and the fact that the Killing form is nondegenerate tells us that , and thus .

Now, if is an outer derivation — one not in — we can assume that it is orthogonal to , since otherwise we just have to use to project onto and subtract off that much to get another outer derivation that is orthogonal. But then we find that

since this bracket is contained in . But the fact that is injective means that for all , and thus . We conclude that and that , and thus that is onto, as asserted.

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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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which lets us write out matrices for the adjoint action:

and from here it’s easy to calculate the Killing form. For example:

We can similarly calculate all the other values of the Killing form on basis elements.

So we can write down the matrix of :

And we can test this for degeneracy by taking its determinant to find . Since this is nonzero, we conclude that is nondegenerate, which we know means that is semisimple — at least in fields where .

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Now, the radical of the Killing form is more than just a subspace of ; the associative property tells us that it’s an ideal. Indeed, if is in the radical and are any other two Lie algebra elements, then we find that

thus is in the radical as well.

We recall that there was another “radical” we’ve mentioned: the radical of a Lie algebra is its maximal solvable ideal. This is not necessarily the same as the radical of the Killing form, but we can see that the radical of the form is contained in the radical of the algebra. By definition, if is in the radical of and is any other Lie algebra element we have

Cartan’s criterion then tells us that the radical of is solvable, and is thus contained in , the radical of the algebra. Immediately we conclude that if is semisimple — if — then the Killing form must be nondegenerate.

It turns out that the converse is also true. In fact, the radical of contains all abelian ideals . Indeed, if and then , and the square of this map sends into . Thus is nilpotent, and thus has trace zero, proving that , and that is contained in the radical of . So if the Killing form is nondegenerate its radical is zero, and there can be no abelian ideals of . But the derived series of eventually hits zero, and its last nonzero term is an abelian ideal of . This can only work out if is already zero, and thus is semisimple.

So we have a nice condition for semisimplicity: calculate the Killing form and check that it’s nondegenerate.

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