Continuing our discussion of modular representation theory, we will now discuss **block theory**. Previously, we saw that in any ring *R*, there is at most one way to write where is a set of orthogonal and centrally primitive idempotents. If such an expression exists, the are called **block idempotents** of *R*. For example, block idempotents exist when *R* is artinian.

We need the following refinement:

Lemma. Let be block idempotents of R. Suppose where the are orthogonal central idempotents. Then there exists a unique map such that:Note: if the sum is zero

Thus after a suitable renumbering of terms, we have:

**Proof**

For each we have , where are orthogonal central idempotents. Since is centrally primitive, we must have for some unique *j* and for all *k*≠*j*. The map then gives and for all And so

♦

**Decomposition of R-Modules**

Let *M* be an *R*-module and suppose where the are block idempotents of *R*. Then

- Indeed, since each
- On the other hand, if then we have and thus

Furthermore, since commutes with every *r*∈*R*, is in fact an *R*-submodule. The central idempotent acts as the identity on and zero on for *j*≠*i*. One can thus imagine:

where each is an -module.

## Block Idempotents of Semisimple *R*

Recall that a semisimple ring *R* is isomorphic to for some division ring where is the *n* × *n* matrix ring with entries in *D*. Since each matrix ring is a simple ring, we immediately obtain the central idempotents: for *i*=1,…,*m*, corresponds to the element whose component in is the identity matrix, and whose component in is the zero matrix.

As a module over itself, *R* is a direct sum of the spaces of column vectors, so where runs through a complete collection of simple *R*-modules, and the component gives the maximal decomposition as a direct sum of ideals.

In particular, this holds for the group ring *K*[*G*]. We have:

, where the direct sum is over all simple *V*.

Lemma. The block idempotent for is given by the following formula:where is the character of V.

**Proof**

Fix *V*; it suffices to show that induces the identity map on and zero on all other components. Now the coefficients of are constant over each conjugancy class, so and is *K*[*G*]-linear. If *W* is simple, induces a scalar map on it, say To compute we take the trace:

But and so the above sum is Thus as desired. ♦

## Block Idempotents of *R*[*G*]

Since *k*[*G*] is artinian, block idempotents are guaranteed to exist. Furthermore these can be lifted to block idempotents of *R*[*G*], which is a nice result since *R*[*G*] is not artinian.

Lemma. Suppose are orthogonal central idempotents of Z(k[G]). Then we can find orthogonal central idempotents of R[G] such that

**Proof**

The proof is conceptually similar to an earlier lemma. The main step is to show:

**Claim**: if *S* is a commutative ring with ideal *I* such that , then any orthogonal idempotents summing to 1 can be lifted to orthogonal idempotents summing to 1.

[ If we can show this, then idempotents can be lifted to , and in turn to etc. Since *R* is complete, this gives idempotents in *Z*(*R*[*G*]). ]

**Proof of Claim**.

Pick any such that . Thus:

As in the earlier proof, let and this gives Also for all *i*≠*j* since it is divisible by (since *S* is commutative). Finally, we claim that Indeed, from the factorisation we obtain:

where the first equality follows from and the second follows from for *i*≠*j*. Note that so Thus is zero. ♦

Conversely, if are orthogonal central idempotents summing to 1, then so are their images Finally, we have:

Lemma.If are central idempotents with the same image in k[G], then they are equal.

**Proof**.

Let , which is a central idempotent in π*R*[*G*]. But so we must have and so ♦

Thus, we have shown:

Summary.There is a 1-1 correspondence between:

- orthogonal central idempotents of k[G] summing to 1, and
- orthogonal central idempotents of R[G] summing to 1.

In particular, block idempotents for *k*[*G*] lift to those for *R*[*G*]:

Taking each , we can write it as a sum of the block idempotents of *K*[*G*]. Thus, we can partition the set of simple *K*[*G*]-modules as a disjoint union , one for each , such that:

For convenience, we also denote by where χ is the character of *V*. This gives the formula

## Example: *S*_{3}.

Let’s compute the central idempotents for *K*[*S*_{3}] using the above formula. Let *a* = (1,2) + (2,3) + (3,1) and *b* = (1,2,3) + (1,3,2). We recover the example at the end of the previous article:

Now let’s consider the case *p*=2. We have as block idempotents of *R*[*G*] (and also of *k*[*G*], after reduction mod 2), so the blocks are {*e*_{1}, *e*_{2}}, {*e*_{3}}. For *p*=3, {*e*_{1}, *e*_{2}, *e*_{3}} all belong to the same block.

If *M* is an indecomposable *k*[*G*]-module, then and thus there is exactly one *i* for which , and for all *j*≠*i*. As a result, the basis elements and can be classified into blocks, where *P* (resp. *M*) belongs to block if and only if (resp. ). Note that if *P* belongs to block *e _{i}*, then the idempotent

*e*acts as the identity on

_{i}*e*

_{i}*P*. The same holds for

*M*.

Similarly, for a simple *K*[*G*]-module *V*, there is a unique *i* for which In summary, we can think of a block as a collection of:

- indecomposable finitely-generated projective
*k*[*G*]-modules*P*; - simple
*k*[*G*]-modules*M*; - simple
*K*[*G*]-modules*V*.

Lemma.Suppose basis elements belong to distinct blocks e_{i}and e_{j}, where i≠j. Then the matrix entry of corresponding to [P], [V] is zero.

**Proof**

Indeed, we have and If is the lift of [*P*], we have and thus since is an indecomposable *R*[*G*]-module. Hence and we must have for any irreducible component *W* of This shows that *V* cannot be a component of ♦

Corollary. Let be basis elements.

- If [M] and [V] belong to different blocks, the matrix entry of is zero.
- If [P] and [V] belong to different blocks, the matrix entry of is zero.

**Proof**

The first statement follows from the fact that the matrix for *d* is the transpose of that for *e*; the second statement follows from *c* = *de*. ♦

Thus, the matrices for can be broken up as block matrices, one for each block idempotent of *k*[*G*].

## Example: *S*_{4}.

The character table of *S*_{4} gives us: letting *a* = (sum of 2-cycles), *b* = (sum of 3-cycles), *c* = (sum of 4-cycles), *d* = (sum of (2+2)-cycles), the central idempotents of *K*[*S*_{4}] are:

For *p* = 2, all five simple characters belong to the same block. For *p* = 3, we have:

Finally, we can check when two characters lie within the same block. First, we need the following lemma:

Lemma. Let R be a commutative k-algebra of finite dimension over k and be its block idempotents. Then any k-algebra homomorphism is uniquely determined by the image of the block idempotents

**Proof**

Corresponding to the block idempotents, write as a product of commutative *k*-algebras. Since is artinian, is semisimple and hence a product of matrix algebras. Since has no idempotent except 0 and 1, itself is a matrix algebra, i.e. for some division ring *D*/*k*. But *R* is commutative, so *n*=1 and *D* is a field extension *k’* of *k*.

On the other hand, let Since and the are orthogonal idempotents, exactly one is a *k*-algebra homomorphism while the remaining are zero maps for *j*≠*i*. Now must factor through the nilpotent ideal and so it is determined by which is either the zero map or the identity, depending on whether is 0 or 1. ♦

Theorem. Let V, W be simple K[G]-modules; the following are equivalent.

- V and W are in the same block.
- For any conjugancy class C⊆G and g∈C, we have:

**Note**

In the course of the proof, we will see that for any simple *V*, so the congruence is well-defined.

**Proof**

**Step 1**. For each simple *V*, we will define a ring homomorphism

Given any conjugancy class *C*⊆*G*, define as a *K*-linear map *V*→*V*. Since commutes with all *g*∈*G*, it is a *K*[*G*]-linear map *V*→*V*. But *V* is simple, hence such a map is a scalar; thus we get a ring homomorphism

**Step 2**. Show that is the LHS of the congruence and that this lies in *R*.

Taking the trace of , we get:

So for any representative To prove that this lies in *R*, recall that we can pick an *R*-lattice *M*⊂*V* which is an *R*[*G*]-module. Thus Picking a basis for *M*, we have

**Step 3**. Compute where is the *K*[*G*] block idempotent for *W*.

Let us rewrite

and so

**Step 4**. Complete the proof.

Since , step 3 tells us *V* and *W* belong to the same block if and only if for all *i*. But this value is either 0 or 1, so it holds if and only if

for all *i*.

On the other hand, so we also obtain a ring homomorphism Reduction mod π then gives us a *k*-algebra homomorphism By the above, *V* and *W* belong to the same block if and only if for all *i*. Since *Z*(*k*[*G*]) is a *commutative* *k*-algebra of finite dimension over *k*, the above lemma says this holds if and only if which is equivalent to for all *C*. ♦

## Example: *S*_{5}.

Modular 2, the blocks of the character table are labeled by the circles on the right:

The corresponding matrices are:

Modular 3, the table becomes:

The corresponding matrices are: