*K*-Representations and *G*-Representations

As mentioned at the end of the previous article, we shall attempt to construct analytic representations of from continuous representations of

Let . Consider , where is the group of diagonal matrices in *K* so

as a topological group. From our study of representations of the *n*-torus, we know that is a direct sum of 1-dimensional irreps of the form:

where . E.g. if then the representation is trivial; if the representation is obtained by taking the determinant.

Hence, the character of is a Laurent polynomial with non-negative integer coefficients, i.e.

Definition. For a continuous finite-dimensional representation , we will write for expressed as a Laurent polynomial in .The same holds for a complex analytic representation of G.

**Examples**

- If is the identity map, its Laurent polynomial is
- For , its Laurent polynomial is

The following are clear for any continuous representations *V*, *W* of *K*.

In summary, so far we have the following:

## Main Examples: Sym and Alt

For any vector space *V*, the group acts on by permuting the components. We denote:

where is the sign of *w*. Let be a fixed basis of *V*. The case *d*=2 is quite easy to describe for the above spaces, for we can just take the following bases:

- ;
- .

Denote these two types of elements by and respectively, so that and Note that ; this is not true for higher values of *d*.

Similarly, in general, we can pick the following as bases:

- ;
- .

where the components commute in Sym and anticommute in Alt (e.g. ).

Now suppose and acts on it; *G* and act on in the following manner:

These two actions commute, from which one easily shows that and are *G*-invariant subspaces of .

**Example**

Suppose we have acting on . This takes:

To compute the Laurent polynomials of these spaces, we let the diagonal matrix act on them, giving:

Hence we have:

## Some Lemmas

Lemma 1. For a representation V of K, the Laurent polynomial is symmetric.

**Proof**

For any , take the corresponding permutation matrix ; we have:

Thus for any ♦

Lemma 2. Given K-representations V, W, if , thenHence by the previous article, the same holds for analytic G-representations V, W.

**Proof**

Any , being unitary, is diagonalizable by a unitary matrix, i.e. there exists such that . Hence the given condition implies:

.

By character theory of compact groups, as *K*-reps. ♦

Now for the final piece of the puzzle.

Lemma 3. Let be a symmetric polynomial. There are polynomial representations:such that

**Proof**

Taking homogeneous parts, let us assume for some degree Write *f* as a linear sum of elementary symmetric polynomials with integer coefficients; separating terms we have , where *g*, *h* are both linear combinations of with non-negative integer coefficients. Hence, it suffices to show that for some polynomial representation

Since , we can just pick:

from the above. ♦

## Consequences

Immediately we have:

Corollary 1. For any symmetric Laurent polynomial , there exist rational representations:such that

**Proof**

Indeed, for some symmetric polynomial *g* in Applying lemma 3, we can find such that ; now we take and ♦

Finally we have:

Corollary 2. Any irreducibleK-module V can be lifted to a rational irreducibleG-module.

**Proof**

By corollary 1, where *W*, *W’* are rational *G*-modules. Then so as *K*-modules by lemma 2 above. By corollary 3 here, is a rational irreducible *G*-module. ♦

## Summary of Results

Thus we obtain:

Note the following.

- When we consider
*virtual*representations (recall that these are formal differences of two representations), this corresponds to all symmetric Laurent polynomials with integer coefficients. - Any rational
*G*-representation is of the form where is a polynomial*G*-representation.

**Moving Ahead**

Our next task is to identify the irreducible rational *G*-modules *V*. Tensoring by some power of det, we may assume *V* is polynomial, so is a symmetric polynomial in Studying these will lead us back to Schur polynomials.