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 πR[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: S3.
Let’s compute the central idempotents for K[S3] 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 {e1, e2}, {e3}. For p=3, {e1, e2, e3} 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 ei, then the idempotent ei acts as the identity on eiP. 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 ei and ej, 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: S4.
The character table of S4 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[S4] 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: S5.
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: