## Cauchy’s Identity

In this article, our primary focus is the ring of symmetric polynomials in

Theorem (Cauchy’s Identity). Consider polynomials over all partitions [Recall that if ] We have an equality of formal power series:

**Note**.

For convenience, we will use and to denote and respectively. Also means

Note that the LHS sum is well-defined since for each , there are only finitely many partitions of so we get a finite sum in . The eventual sum lies in , a formal power series in

**Proof**

If , we have . Thus we need to show:

The proof is combinatorial: each term on the RHS is

Now we compute the coefficient of in the expanded product. E.g. for and , to obtain the term we can multiply:

Hence each occurrence of in the expansion corresponds to a matrix of non-negative integers with row sum and column sum , and the coefficient of on the RHS is exactly This means RHS = LHS. ♦

### Exercise

Simplify the following sum:

## Hall Inner Product for

Now and are dual bases under the Hall inner product and they satisfy Cauchy’s identity. We would like to prove the converse that Cauchy’s identity imply dual bases. However, the problem is that our results for the Hall inner product were proven for and not To obtain similar results for , we need to be a bit more careful.

We already know that

form a basis of . It turns out we also have:

Proposition. Let Then the setsare -bases of

**Proof**

Consider the surjective map which takes for all Its kernel has a basis given by the monomial symmetric polynomials Now if and we have as well. Thus:

Since is a basis of the set thus forms a basis of as desired.

Finally since and the matrix is invertible, we see that also forms a basis of ♦

Summary. The following are -bases of .

For example, when *d*=4 and *n*=3, we obtain the following, where each row gives a -basis of the abelian group.

Note: from this, it is clear that when we are confined to E.g. in the above diagram

Now we define the Hall inner product for by letting and to be dual bases. As before, we have the following.

Proposition.

- is an orthonormal basis.
- The Hall inner product on is symmetric.

**Proof**. Same as before (check carefully).

However, the involution is no longer unitary. This can be checked for *n*=2, *d*=3. Here, we get and A bit of computation then shows that which is not a unit vector. The reader should examine the proof to see why it does not carry over.

## Criterion for Orthonormal Basis

Now let us prove the desired result.

Theorem. Suppose we have a ollection indexed by for all and If Cauchy’s identity holds for them:then and are dual bases with respect to the Hall inner product.

**Proof**

We know that and both satisfy Cauchy’s identity and are dual. Taking the degree-*d* component in *x* or *y*:

Now, let us first show that is a basis of ; comparing dimensions, it suffices to show that the set spans this space.

- Suppose not; then there exists such that for all
- Write , where the are not all zero.
- The two Cauchy identities give us (note: is computed term-wise by taking as a power series in
*x*with coefficients in ):

- Hence as desired.

Now it remains to show that , where the dual is taken over We have:

for any ; by linearity we can replace with any In particular

which is what we wanted. ♦