## Complete Symmetric Polynomials

Corresponding to the elementary symmetric polynomial, we define the **complete symmetric polynomials** in to be:

For example when , we have:

Thus, written as a sum of monomial symmetric polynomials, we have Note that while the elementary symmetric polynomials only go up to , the complete symmetric polynomial is defined for all Finally, we define as before:

Definition. If is any partition, we define:assuming

Proceeding as before, let us write as in terms of the monomial symmetric polynomials

Theorem. We have:where is the number of matrices with non-negative integer entries such that for each

jand for eachi.

**Proof**

The proof proceeds as earlier. Let us take the example of and . Multiplying , we pick the following terms to obtain the product

Thus each matrix corresponds to a way of obtaining by taking terms from etc. ♦

**Example**

Suppose we take partitions , . Then since we have the following matrices:

**Exercise**

Compute for all partitions of 4. Calculate the resulting 5 × 5 matrix, by ordering the partitions reverse lexicographically.

## Generating Functions

The elementary symmetric polynomials satisfy the following:

Thus their generating function is given by Next, the generating function for the $h_k$’s is given by:

From , we obtain the following relation:

Note that for

From this recurrence relation, we can express each as a polynomial in . E.g.

## Duality Between *e* and *h*

From the symmetry of the recurrence relation, we can swap the *h*‘s and *e*‘s and the expressions are still correct, e.g. As another example, if *n*=3, we have

Definition. Since is a free commutative ring, we can define a graded ring homomorphismfor

From what we have seen, the following comes as no surprise.

Proposition. is an involution, i.e. is the identity on

**Proof**.

We will prove by induction on that for For this is obvious; suppose . Apply to the above recurrence relation; since for we have:

By induction hypothesis for ; since we have

Hence for all ; since generate we are done. ♦

Now suppose ; write as an integer linear combination of the for Applying , this gives in terms of for In particular, we get:

Corollary. The following gives a -basis of :Hence we also have as a free commutative ring; the isomorphism preserves the grading, where

**Exercise**

Consider the matrix , where run through all partitions of *d*. Using the involution prove that

In particular, **M** is invertible; this is not obvious from its definition.

**Exercise**

Since each can be uniquely expressed as a polynomial in For *n*=3, express in terms of