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:
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 j and for each i.
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. ♦
Suppose we take partitions , . Then since we have the following matrices:
Compute for all partitions of 4. Calculate the resulting 5 × 5 matrix, by ordering the partitions reverse lexicographically.
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 homomorphism
From what we have seen, the following comes as no surprise.
Proposition. is an involution, i.e. is the identity on
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
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.
Since each can be uniquely expressed as a polynomial in For n=3, express in terms of