## Representations of the Symmetric Group

Let [*d*] be the set {1,…,*d*}, and *S _{d }*be the group of bijections From here on, we shall look at the representations of Note that this requires a good understanding of representation theory (character theory) of finite groups.

To start, let us first find *some* representations.

Definition. Given a partition , let be the set of partitions of into disjoint subsets such that for each The group then acts on via:

**Note**

If , the partitioning is considered different when we swap and For example if *d* = 4 and , then has 6 elements:

In general,

As is a *G*-set, we see that is a -module for any field *K*. Two special cases are of interest.

- :
*X*has a single element so is the*trivial representation*; - :
*X*is just the set of permutations of [*d*] so is the*regular representation*.

**Exercise**

Show that for , is the standard representation of on

Let us study such representations in a more general setting.

## Representations From *G*-Sets

Let *G* be a group and *X* a *G*-set; then *K*[*X*] is naturally a *K*[*G*]-module. We then have the following isomorphisms of *K*[*G*]-modules:

Lemma.

- (i.e. is isomorphic to its dual);
- ;
- ;
- has dimension , the number of orbits of
X.

**Note**

In general, a representation is not isomorphic to its dual.

**Proof**

Take the bilinear map induced linearly from the map:

This induces an isomorphism of vector spaces. It is also *G*-equivariant since the bilinear map is *G*-equivariant, which proves the first statement.

For the second, linear algebra gives as vector spaces, taking One easily checks that it is *G*-equivariant.

The third statement follows from the first two since generically, we have for any *K*[*G*]-modules *V* and *W*. Thus:

Finally, if is invariant under every , then for all and Thus is constant when runs through an orbit. ♦

Now assume that char(*K*) = 0.

Question. Given what is the following value:where is the algebraic closure of ?

**Note**

The reader may assume to simplify matters, in which case *L* = *K* and when we mention “absolutely irreducible”, just replace it with “irreducible”.

**Answer**

By the above lemma, we have:

which has dimension . Writing and let us pick and Then the orbit of the pair satisfies:

so the collection of these numbers is invariant over all .

Conversely, suppose and are partitions of [*d*] satisfying:

for all . Then we can find mapping the pairwise intersection sets to so the pairs and are in the same orbit. Hence the number of orbits is the number of solutions to

in non-negative integers

In particular, the above holds when we replace *K* by *L* so:

Conclusion. We have