From now onwards, we will assume the base field *K* has characteristic 0.

## Example: *d*=3

Following the previous article, we examine the case of . We get 3 partitions: , and Let us compute for all From the previous article, we have:

Since , is absolutely irreducible (i.e. irreducible even over the algebraic closure of *K*). Since

each of and contains exactly one copy of Thus we write

where does not contain , i.e. . Hence

Taking , we obtain:

Thus so is absolutely irreducible. Since , contains two copies of Thus we write:

where does not contain or . We have

Since has 3 irreducible representations, we have found all of them: , , and:

Furthermore, we have shown that the 3 irreps of are in fact defined over .

## General Case

We will show that the above computations can be generalised to any Indeed, the representations satisfy:

Recall that the matrix satisfies where is the number of SSYT of shape and type and we have:

- for all ;
- if , then and so lexicographically.

Thus:

Theorem. Suppose is a finite partially ordered set and is a collection of representations. And suppose there are non-negative integers satisfying and:Then there are non-isomorphic absolutely irreducible defined over

Ksuch that:

**Proof**

Choose a maximal element ; then

So is absolutely irreducible. The number of copies of contained in each is:

so we can write: where does not contain This then gives us:

so the collection of for with satisfy the given conditions and we may proceed recursively with this reduced set. ♦

Applying the above theorem to the set and the representations of we get:

Corollary. There are absolutely irreducible representations of defined over such that:

## Consequences

If we apply this to the case *d*=3, we get irreps such that:

as we had computed earlier.

**Note 1**

Recall that is the number of SYT of shape , i.e. where Now is the regular representation and is the number of copies of contained in it. From character theory we thus have

- and
- which we had proven earlier with the RSK correspondence.

**Note 2: Frobenius Map**

If we let be the character for the irrep , then:

where the second equation is the vectorial form of the first.

Hence, if each corresponds to the complete symmetric polynomial then corresponds to the Schur polynomial and the Hall inner product corresponds to

Thus we have a group isomorphism between and the group of *virtual* characters of called the **Frobenius characteristic map**. [Recall that virtual characters are formal differences of the form where are group characters. The set of virtual characters of a finite group *G* forms a ring *R*(*G*), under addition and multiplication of characters as functions. *R*(*G*) is a finitely generated free abelian group, with a basis given by the irreps of *G*.]

Taken over the rational field, we obtain a linear isomorphism between:

- , the space of symmetric polynomials of degree
*d*with rational coefficients; - the space of class functions .