One Complete Character Table (part 2)
Last time we wrote down the complete character table of :
which is all well and good except we haven’t actually seen a representation with the last line as its character!
So where did we get the last line? We had the equation , which involves the characters of the defining representation and the trivial representation . This equation should correspond to an isomorphism .
We know that there’s a copy of the the trivial representation as a submodule of the defining representation. If we use the standard basis of , this submodule is the line spanned by the vector . We even worked out the defining representation in terms of the basis to show that it’s reducible.
But what we want is a complementary subspace which is also -invariant. And we can find such a complement if we have a -invariant inner product on our space. And, luckily enough, permutation representations admit a very nice invariant inner product! Indeed, just take the inner product that arises by declaring the standard basis to be orthonormal; it’s easy to see that this is invariant under the action of .
So we need to take our basis and change the second and third members to make them orthogonal to the first one. Then they will span the orthogonal complement, which we will show to be -invariant. The easiest way to do this is to use . Then we can calculate the action of each permutation in terms of this basis. For example:
and write out all the representing matrices in terms of this basis:
These all have the required form:
where the in the upper-left is the trivial representation and the block in the lower right is exactly the other representation we’ve been looking for! Indeed, we can check the values of the character:
exactly as the character table predicted.