We continue our discussion of modular representations; recall that all modules are finitely-generated even if we do not explicitly say so. First, we introduce a new notation: for each projective finitely-generated *k*[*G*]-module *P*, we have a unique projective finitely-generated *R*[*G*]-module denoted for which

First we have:

Proposition. The matrices for e and d are transpose of each other. Hence, c = de is positive-semidefinite and symmetric.

**Proof**

We need to show for all basis elements and , i.e. *P* is projective indecomposable and *M* is simple. As seen earlier:

- By definition where
- Write for some
*R*[*G*]-module*N*, free over*R*, so that*d*(*y*) =*N*/π*N*. - We need to show

Consider ; this is *R*-free since it is a submodule of a free *R*-module. Our desired result would follow if we could show:

But since *Q* is finitely-generated and *R*[*G*]-projective, we can write for some *R*[*G*]-module *Q’* and *n*>0. Since direct sum commutes with tensor product, it suffices to prove the above two isomorphisms for Now this is obvious since in this case, *X*=*N*, and we have

as desired. ♦

**Note**

In fact, we shall see later that *c* is positive definite, or equivalently, it’s injective. If *c*(*x*) = 0, then so *e*(*x*) = 0. But a basis of is given by [*P*] for indecomposable projective *P*, so it suffices to show that are linearly independent.

## Modular Characters

Recall that a finitely-generated *K*[*G*]-module *M* can be represented by its character where its trace as a *K*-linear map. Since *K*[*G*] is semisimple, standard character theory says uniquely determines *M*. Now, is a class function (i.e. it is constant on each conjugancy class) and in fact, forms an orthonormal basis of the space of class functions, where orthonormality follows from Schur’s lemma.

We would like to produce a similar theory for *k*[*G*]-modules. The naive approach of taking where does not lead to a satisfactory theory. The better approach is to lift the character to a function *G*→*R*.

Definition. An element g of G is said to bep-regularif its order is coprime to p; if it is not p-regular, we say it isp-singular. The collection of p-regular elements of G is denoted Note that this is a union of conjugancy classes.

Now assume *K* contains all *n*-th roots of 1, where *n* = |*G*|. It follows that *k* contains all *m*-th roots of 1, where *m* is the largest factor of *n* coprime to *p*. The *m*-th roots of 1 in *k* are all distinct and the canonical map *R* → *R*/π induces a bijection between the *m*-th roots of 1 in *K* and in *k*. Denote this bijection by λ.

Now we are ready to define modular characters.

Definition. Let M be a finitely-generated k[G]-module and Consider the k-linear map Let r be the order of g; since r|m, k contains all the eigenvalues of g, denotedNow the

modular characterof M is defined as follows:

We have the following properties:

- If
*N*⊆*M*is a submodule, then This is easily seen by taking a*k*-basis of*N*, then extending to give a*k*-basis of*M*. Hence is independent of our choice of [*M*] in the Grothendieck group - For any
*k*[*G*]-module*M*, consider its*k*-dual*M** with*G*acting on it via Then - For any two
*k*[*G*]-modules*M*,*N*, we have for all*p*-regular*g*. [*Note that the tensor product is over k and not k*[*G*]. ]

Next we would like to relate modular characters and standard ones, so suppose *N* is an *R*[*G*]-module. Then we have:

a *K*[*G*]-module and *k*[*G*]-module. The first one gives a standard character and the second one gives a modular character From the definition, we have:

Hence, in our *c*–*d*–*e* diagram:

we have:

On the other hand, let us compare and These clearly give the same values for all *p*-regular *g*. Furthermore, the following shows that when *g* is *p*-singular.

Proposition. Let M be a finitely-generated projective R[G]-module. If g is p-singular then

**Proof**

First, note that *M* is projective over *R*[*H*] for any subgroup *H* of *G* since *R*[*G*] is free over *R*[*H*]. Hence, replacing *G* with the subgroup generated by *g*, we may assume *G* is cyclic and generated by *g*. This gives:

where *m* is coprime to *p* and *r*>0 since *g* is *p*-singular. We need to show that multiplication-by-*T* has trace zero for every indecomposable projective *R*[*G*]-module *M*. But this corresponds to an indecomposable projective *k*[*G*]-module *N* := *M*/π*M* and we may classify all *N* by decomposing *k*[*G*]:

where each direct-sum factor is indecomposable since its radical is of codimension 1. The corresponding *R*[*G*]-module is then Since *r*>0, multiplication by *T* has trace zero in each : this is readily seen by taking the basis which maps to ♦

Summary. In the c-d-e diagram above, we have:where in the second equality, is extended to a function G→K by mapping all p-singular elements to zero.

## Computing Pairing over *k*

Our next job is to find a formula for the pairing:

for a projective *k*[*G*]-module *P* and general *k*[*G*]-module *M*. First some preliminaries:

Lemma. If M is a projective k[G]-module, then so are and for any k[G]-module N.

**Proof**

Since *M* is projective, it is a direct summand of some *k*[*G*]^{n}. Since dual and tensor product commute with finite direct sums, it suffices to prove this for *M* = *k*[*G*]. In this case, and are in fact free over *k*[*G*].

To check that a general *k*[*G*]-module *V* is free, it suffices to find a *k*-subspace *W* ⊆ *V* such that For the dual we pick where *f* : *k*[*G*]→*k* takes *e* to 1 and all other *g* to 0; and for the tensor product, we pick ♦

Now we compute the pairing:

The third equality follows from this:

- If
*Q*is a projective*k*[*G*]-module, then is a free*R*-module and - [ To prove this, note that is a
*R*-submodule of a free*R*-module, so it must be*R*-free. Now write for some*R*[*G*]-module and so On the other hand, we have so Comparing terms then gives ]

Since is *R*[*G*]-projective, its character vanishes on the *p*-singular elements and the above equals:

Thus, we define the following:

Definition. A class function is said to be ap-regular class function. An inner product is defined on the space of all such functions:

Hence, the basis and for projective indecomposable *P* and simple *M* give rise to dual bases and under the above inner product.

We thus have:

Theorem. The map is injective. Hence is the number of p-regular conjugancy classes of G.

**Proof**

We will show that for projective indecomposable *P*, the collection of is linearly independent over *K* and hence over **Z**. Indeed, if then taking the inner product with for each simple *M* gives:

since {[*M*]} forms a dual basis for {[*P*]}.

**Note**

This also shows that *d*, being the transpose of *e*, is surjective when extended to *K*. In fact, it is even true that is surjective but the proof of this is rather involved and we may revisit this again some day.

## Grothendieck Rings

Given finitely-generated *k*[*G*]-modules *M* and *N*, their tensor product over *k* is also finitely-generated. By linear algebra, tensor product over a field is always exact for both and So if is an exact sequence of *k*[*G*]-modules, then so is:

This gives a pairing which is bi-additive. Since tensor product is associative, we obtain a ring structure on Furthermore if *M* is projective, then so is as we saw above. Thus, identifying with its image we see that is an ideal of

**Example**

Let us revisit the example from the previous article. The character table of *S*_{3} is given by:

For *p*=3, we have and So the modular character tables for and are given by:

Note that and and both vanish at the 3-cycle as expected. Furthermore one can verify by direct computation that taking into account (1,2) has class size of 3.

The Grothendieck ring is isomorphic to where {1, *T*} maps to respectively. Then is the ideal generated by *T*+2 and