## Twisting

From the previous article, any irreducible polynomial representation of is of the form for some such that is the Schur polynomial .

Now given any analytic representation *V* of *G*, we can **twist** it by taking for an integer *k*. Then:

Twisting the irrep with gives us another irrep, necessarily of the form . What is this ? Note that from we can recover the partition by taking the minimal partition (with respect to ). Hence from we must have Thus:

Proposition. Any irreducible analytic representation of G can be uniquely written as:where is the polynomial representation satisfying:

Since , is not divisible by so the representation is polynomial if and only if Its degree is

## Dual of Irrep

The dual of the irrep is also a rational irrep, so it is of the form for some partition with and integer *k*. From:

we take the term with the smallest exponent for in lexicographical order. For large *N*, denoting for the reverse of a sequence , we have:

Hence and Pictorially we have:

## Weight Space Decomposition

By definition is the character of when acted upon by the torus group *S*. Since this polynomial is , as vector spaces we have:

where:

- runs through all partitions with and ;
- runs through all permutations of without repetition, e.g. if we get 3 terms: (5, 3, 3), (3, 5, 3) and (3, 3, 5);
- is the space of all for which
*S*acts with character , i.e.

and the dimension of is . This is called the **weight space decomposition** of We will go through some explicit examples later.

### Foreshadowing: SSYTs as a Basis

As noted above, the dimension of is exactly the number of SSYT with shape and type . Thus in a somewhat ambiguous way, we can take, as a basis of , elements of the form over all SSYT *T* of shape and entries from [*n*]={1,2,…,*n*}; each lies in the space

However, such a description does not distinguish between distinct SSYT of the same type. For that, one needs a construction like the determinant modules (to be described later).

## Example: *n*=2

Consider By the above proposition, each irreducible representation is given by where *m*, *k* are integers and To compute *V*(*m*), we need to find a polynomial representation of *G* such that

corresponding to the SSYT with shape (*m*) and entries comprising of only 1’s and 2’s. E.g. from:

Such a *V*(*m*) is easy to construct: take ; if {*e*, *f*} is a basis of *V*, then a corresponding basis of is given by . If , then the diagonal matrix *D*(*a*, *b*) takes so its character is as desired.

The weight space decomposition thus gives:

where each is 1-dimensional and spanned by

## Example: *d*=2

Consider . We have:

where each component is *G*-invariant. As shown earlier, we have:

Since the Schur polynomials are and , both and are irreps of *G*. The weight space decomposition of the two spaces are:

Hence in their weight space decompositions, all components have dimension 1.

## Example: *n*=3

Now let us take and compute where and To find we need to compute for all and We will work in the plane since partitions lie in the region , we only consider the coloured region:

The point has . Assuming , the condition then reduces to a single inequality Hence, lies in the brown region below:

To fix ideas, consider the case Calculating the Kostka coefficients gives us:

Taking into account all coefficients then gives us a rather nice diagram for the weight space decomposition.

E.g. we have and . These correspond to the following SSYT of shape (5,3):