2024-08-22

Bar resolutions

We write for . is a -algebra and is an -module. The bar resolution tensored with is

Let us see why this is exact. We have

Claim: This is a chain complex of -modules. We claim that this complex is exact.

Approach

We will find such that

i.e., our goal is to show that and are homotopic.

Note that we don’t care if is an -module map since we are only measuring exactness. So it suffices to work in to check exactness.

Proof

We define by

We have

So we see that . Since and are homotopic, they induce the same map on homology. So on homology, we have . The only way this happen is if all homology groups of this complex are . So we’ve proved exactness.

Finite groups

Let be a finite group and be a -module. Recall that we have:

Theorem ( is -torsion)

Let be a finite group and be a -module. Then is -torsion, i.e., for all , we have .

Proof

We sum the condition

in over all to find

Now writing , we have

This matches the condition in . Hence, for all , we have .

Corollary

is -torsion for all .

Proof

Modulo the fact that: we can embed any module into an injective one. (Dimension shifting argument)

There exists a short exact sequence for an injective module

Because is a derived functor, we get a long exact sequence in cohomology

Now, if , we have because is injective. Therefore exactness forces

We have shifted the dimension down by . By induction on , we find that is also -torsion.

Let be a field. Let’s only work with finite-dimensional -modules.

Extra structure on the category of -modules

Tensor product

Let and be -modules. Then is a -module via

Warning

If is an algebra,

does not define an -module on . For example, if we let , then we would find that

In general, if is an -module and is a -module, then is a -module.

Duals

If is a -module, the dual has the same dimension (if finite-dimensional) as and can be turned into a -module via

Hom

If and are two -modules, we have the vector space . It can be turned into a -module by defining for

We have an isomorphism of -modules (requires finite-dimensionality)

i.e., the isomorphism is compatible with the -actions.

A priori, . However, we in fact have

To see this, the condition that translates into

i.e., is a -module homomorphism.

Adjunction

Let , , be -modules. Then

Tensoring with is adjoint to tensoring with . In fact, both sides are .

Lemma

  1. If is a projective -module and is any -module then is projective.
  2. If is an injective -module and is any -module then is injective.

General factor: functors with exact adjoints.

Proof

1.

We want to find a lift :

This is the same data as

by the adjunction property. In particular, is still a surjection by exactness of the functor .

Theorem

Denote by the -dimensional trivial module. Then

Proof

We take a projective resolution of :

We can then take . The resulting chain complex is

Alternatively, we can tensor with to get

This is still exact because we are working over a field and is a vector space (exactness doesn’t care about the additional -modules structure). By the Lemma, this is a projective resolution of . We can then take . The resulting chain complex is

By adjunction, this is chain complex is isomorphic to .