Let’s work out some explicit examples of modular characters. First, we have a summary of the main results.
- Let be the modular characters of the simple k[G]-modules; they form a basis of
- Let be those of the projective indecomposable k[G]-modules; they form a basis of
- We have , the number of p-regular conjugancy classes of G.
- The and form a dual basis under the inner product so that
These relate to ordinary characters as follows: let be the standard irreducible characters of K[G], so they form an orthonormal basis of
- The map satisfies: for each the function is zero on the p-singular conjugancy classes of G.
- The map is the transpose of d and is injective.
- Thus, c = de is symmetric and positive definite.
Group S4 with p=2.
Let’s consider the usual character table for S4.
For the ring , we leave only the columns for e and (1,2,3), since the remaining conjugancy classes are 2-singular. Immediately, we obtain some linear relations:
So it remains to consider if d(χ2) is simple. If it weren’t it would be the sum of two 1-dimensional representations. But these are easily classifiable for Sn.
Lemma. There are at most two 1-dimensional representations of Sn over any field: the trivial and the alternating.
Indeed, these correspond to and since the image is abelian, it factors through the commutator and we obtain So we are left with the trivial and alternating representations. ♦
So d(χ2) is simple since otherwise it would be d(2χ1) which it clearly is not. Hence:
The basis elements of and are thus:
Note that we do have as expected. E.g.
Also and take all the 2-singular classes to 0.
Let φ be the regular representation. Find the multiplicity of and in the decomposition of φ, and the multiplicty of and among its composition factors.
Group S4 with p=3.
We remove the column (1,2,3) and keep the remaining four. Clearly, and are simple since they’re of dimension 1. Next, we have It remains to see if and are simple. Consider If it weren’t simple it must contain a submodule of dimension 1, which we saw is either or
- In the former case, where Since this means both eigenvalues for (1,2,3,4) are -1, and so those for its square (1,3)(2,4) are +1. But this contradicts
- In the latter case, where so both eigenvalues for (a, b) are +1. This means all elements of S4 have both eigenvalues equal to +1, which is absurd.
Thus and are simple and we have:
The basis elements of and are thus:
Group S5 with p=2.
First, we look at the usual character table for S5.
Removing the 2-singular conjugancy classes leaves us with the columns for e, 3-cyc and 5-cyc. Note that for any character χ so we are left with 4 rows. Next so we are left with which are clearly linearly independent. Writing them as linear combinations of simple modular characters:
where the matrix entries are all non-negative integers. It is not hard, albeit rather tedious, to list all possible 2 × 3 matrices. After solving for we are further reduced to 12 possibilities (corresponding to their values at e, the 3-cycle and 5-cycle):
((2, 5, 2), (2, -4, -3)), ((2, -1, -3), (2, -1, 2)), ((2, -1, -3), (2, -4, -3)),
((2, -1, -3), (3, 0, 3)), ((2, -1, -3), (3, -3, -2)), ((2, -1, -3), (4, -2, -1)),
((2, -1, -3), (5, -1, 0)), ((3, 0, -2), (3, -3, -2)), ((3, 0, -2), (4, -2, -1)),
((3, 0, -2), (5, -1, 0)), ((4, 1, -1), (4, -2, -1)), ((4, 1, -1), (5, -1, 0)).
Since for any g, this immediately removes the first 7 possibilities. Next is invalid since for we must have 3 fifth roots of unity summing up to -2, which is impossible. So we’re left with two choices. It turns out is the right choice, which we shall show below.
We need to show that modulo 2, the modular representation contains the trivial representation. Recall that is found in the representation where V is a 4-dimensional representation given by:
and acts on V by permuting the coordinates. Another way of expressing this is: where and g∈G acts on by taking Now is spanned by where multiplication commutes. A basis of W is given by where and
In the semisimple case, W contains exactly one copy of the trivial representation but modulo 2, we can find two copies.
- First, take the subspace X spanned by , which is G-invariant. Note that this vector is non-zero (simplify via ).
- Next, consider the map f : W→k which takes for all 1≤i≤j≤4. Note that and and so we see that f is G-equivariant, where G acts trivially on k. Note that f(X)=0 so we have at least two copies of the trivial representation among the composition factors.
Thus, has at least one copy of the trivial representation, and we get:
An explicit representation of φ3‘ is given by:
To calculate the modular character values at (1,2,3) and (1,2,3,4,5), we compute their characteristic polynomials, giving and respectively. Lifting the roots of unity to K gives us for the first case and for the second, where and so the values are and respectively.
Group S5 with p=3
From the character table of S5, we remove the two columns for 3-cyc and (2+3)cyc. The resulting modular characters satisfy the following:
The remaining 5 modular characters are linearly independent:
On the other hand, denote the 5 simple modular characters of by:
Let us say is even if Note that this holds if and only if φ is zero on the odd permutations: 2-cyc and 4-cyc (we’re ignoring the (2+3)cyc column for modular characters mod 3). Since φ is simple if and only if is, we see that are either all even, or exactly one of them is even. The former case is impossible since that would imply all take the same values for 2-cyc and 4-cyc. Hence, we may assume and that is even.
- First, cannot contain or . E.g. if we would have φ(e) = 3 and φ(5-cyc) = -2, which is impossible since we cannot have 3 fifth roots of unity summing to -2.
- Clearly, is impossible.
- Now suppose . This means are all of dimension 2. Since is even, it must be Hence we have which takes 2-cyc to -2. Hence takes 2-cyc to +2 and so it must be the identity (contradiction).
Hence, we have shown that and are both simple. Finally is either simple, or contains The latter would imply where Hence the eigenvalues for (2+2)cyc are all -1 and since its order is coprime to p=3, the matrix for (2+2)cyc is –I. But then we have (1,3)(2,4)·(1,2,3,4,5) = (1,4,5,3,2) and we cannot have both and
Thus we may write and we have: