We have established that all irreps of are defined over and hence any field of characteristic 0. For convenience we will fix .
For any group G and representation over if is a group homomorphism, we can twist as follows:
Sometimes, we also write for the twist; this gives an action on the set of all irreducible representations of G. Note that so the dimension of V is invariant among the equivalence classes of twists.
For example in , the character table is:
so we have 3 equivalence classes of irreps if we twist by .
When we let be the alternating character, i.e. Now we consider the modules
Question. What is the following:
Note that the underlying vector space for is the same as though the action has a twist. Recall that, as -modules,
and we wish to find all G-invariant elements in this space.
Each element on the RHS is a linear combination of elements Let us unwind the definition for to be invariant under G. Note that:
and takes it to . Now if any has more than one element, then the transposition which swaps them would flip the sign while leaving , . Thus we have shown:
- If for some , then the coefficient of in is zero.
Conversely, if for all , then no takes back to itself, so as it ranges over an orbit, the coefficient of in is uniquely determined.
Suppose d=3 and so we get
Then has 9 elements for . Any invariant element in must be a multiple of:
Conclusion. We have:
the number of binary matrices such that and
We saw that satisfies:
where J is the permutation matrix which swaps with its transpose Writing we get:
If then writing in matrix form gives Hence and we have Thus:
Since the matrix is invertible we have proven:
In particular, if and only if is its own transpose, e.g. (5, 3, 2, 1, 1).
Hence under the Frobenius map, the corresponding symmetric polynomial for is
Also, the involution which maps corresponds to This gives an algebraic interpretation for