Introduction
In this post, we aim to glean as much as we cam about the characters of the symmetric groups (today we’ll be focusing on and ) using simple properties of characters.
First up, !
The most natural representation of (in fact, this hardly seems like a reprsentation at all!) would to be let be such that it sends to its by permutation matrix and . For instance, and
Unfortunately, this is not an irreducible one: all the ’s leave for a real alone. Hence, is an invariant subspace of (an invariant line). The projection operator onto is for all , and That is, is the orthogonal complement of under the dot product. By definition, .
According to Maschke’s theorem (or Theorem 1 from Serre’s 1977 book), is also an invariant subspace. We can do a quick spot check: for . Setting , we see that and , works and so , as we expect.
The subrepresentation is thus just the trivial representation: There just isn’t much freedom offered by a good ’ol line. However, the degree 2 reprsentation is more interesting. Notice that so can be indentified with by matrices. Fixing as a basis for , and then writing for as a matrix, gives us:
and Of course, the identity goes to as usual.
The corresponding character (recall that the trace does not depend on the choice of the basis of ) is just if the permutation is even and not the identity and is the permutation is odd. Taking into account the identity, we can say if is even and is is odd.
We call this irreducible character (why is this not reducible?) and the corrsponding representation the standard representation, .
Now, let that unknown third irreducible character be : Let for transpositions and for 3-cycles . By the orthogonality of irreducible characters, we know and utilizing the other character we have Putting these numbers toegther, we get and .
Thus, is simply the sign of ! We call it the sign character: .
All in all, we now have the character table of !
Onto !
Let’s continue our analysis of symmetric group with the next one: . As usual, we have the trivial character: , that returns for all . In much the same way as last time, we can construct a natural representation for , that assigns a to the corressponding by permutation matrix, as viewed as an element of . This won’t we irreducible however, as the vectors that have all coordinates equal in will be invariant under the action of the ’s. The dimensional complement of this invariant line will be invariant, and that is our standard representation, which character . Doing the computations, we get , , and .
It’s time to invoke the orthogonality! We still have two unknown characters: and . Using the sum of squares formula, we have , which implies , which forces and . Leting take on values , , and and using the three equations: we get
- ,
- ,
- .
Adding the first two equations, . Notice that the ’s must be integers, so this is linear diophantine equation. Upon solving, we get and , for . Adding the last two equations, we get and so , and substituting these expressions into the first equation yields .
We do the same drill with (which takes on the values to get that , , and for .
Lastly, we have an equation involving both the ’s and ’s as , which gives us . This implies and by the SFFT.
This completes our character table for - just using orthogonality!
Tensor Products
Recall that we have the notion of the tensor product of two representations - a tool that we can use to possibly build and from , and . In that direction, we will derrive an expression for the character of a tensor product. But Before that, we look at a slightly different expression for a character of a representation.
Let’s fix a basis for . Recall that we have a corresponding dual basis, for , the dual space of : Given a , is the coefficient of in the expansion of in terms of the basis. Thus, the matrix representation (with respect to ) for has entry . Taking the sum of the diagonal entries to get the trace, we have
Now we can work with the character of a tensor product better: Let be a vector space with basis . Then a basis for is . A corresponding dual basis for would be , where we define and extend linearly. That is, extracts the coefficient of in the expansion of the input in the basis . That gives us for an elementary tensor is , which is also just .
All in all, armed with this new formula for the trace, we have
So the tensor produt a representations just has the effect of multiplying the corresponding characters! In fact, going back to our character table for , we can see that - that’s another free character for us!