Polynomials and Representations III

Complete Symmetric Polynomials

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

h_0=1, \qquad h_k = \sum_{1 \le i_1 \le i_2 \le \ldots \le i_k \le n} x_{i_1} x_{i_2} \ldots x_{i_k} = \sum_{a_1 + \ldots + a_n = k} x_1^{a_1} x_2^{a_2} \ldots x_n^{a_n}.

For example when n=3, we have:

\begin{aligned} h_3 &= x_1^3 + x_2^3 + x_3^3 + x_1^2 x_2 + x_1 x_2^2 + x_2^2 x_3 + x_2 x_3^2 + x_1^2 x_3 + x_1 x_3^2 + x_1 x_2 x_3\\ &= m_3 + m_{21} + m_{111}. \end{aligned}

Thus, written as a sum of monomial symmetric polynomials, we have h_k = \sum_{\lambda\vdash k} m_\lambda. Note that while the elementary symmetric polynomials only go up to e_n, the complete symmetric polynomial h_k is defined for all k. Finally, we define as before:

Definition. If \lambda is any partition, we define:

h_\lambda := h_{\lambda_1} h_{\lambda_2} \ldots h_{\lambda_l},

assuming \lambda_{l+1} = 0.

Proceeding as before, let us write h_\lambda as in terms of  the monomial symmetric polynomials m_\mu.

Theorem. We have:

h_\lambda = \sum_{\mu} M_{\lambda\mu} m_\mu

where M_{\lambda\mu} is the number of matrices a_{ij} with non-negative integer entries such that \sum_i a_{ij} = \lambda_j for each j and \sum_j a_{ij} = \mu_i for each i.


The proof proceeds as earlier. Let us take the example of \lambda = (2, 2, 2) and \mu = (3, 2, 1). Multiplying h_2 h_2 h_2, we pick the following terms to obtain the product x_1^3 x_2^2 x_3.

\begin{aligned} h_2 &= \boxed{x_1^2} + x_2^2 + \ldots + x_1 x_2 + x_1 x_3 + x_2 x_3 + \ldots \\ h_2 &= x_1^2 + x_2^2 + \ldots + \boxed{x_1 x_2} + x_1 x_3 + x_2 x_3 + \ldots \\ h_2 &= x_1^2 + x_2^2 + \ldots + x_1 x_2 + x_1 x_3 + \boxed{x_2 x_3} + \ldots \\ \end{aligned} \implies \begin{array}{|c|ccc|}\hline 2 & 2 & 0 & 0 \\2 & 1 & 1 & 0 \\2 & 0 & 1 & 1 \\ \hline & 3 & 2 & 1 \\ \hline\end{array}.

Thus each matrix corresponds to a way of obtaining x^\mu := \prod_i x_i^{\mu_i} by taking terms from h_{\lambda_1}, h_{\lambda_2}, etc. ♦


Suppose we take partitions \lambda = (2, 2), \mu = (2, 1, 1). Then M_{\lambda\mu} = 4 since we have the following matrices:



Compute M_{\lambda\mu} for all partitions \lambda, \mu of 4. Calculate the resulting 5 × 5 matrix, by ordering the partitions reverse lexicographically.


Generating Functions

The elementary symmetric polynomials satisfy the following:

e_0 + e_1 t + e_2 t^2 + \ldots + e_n t^n = (1 + x_1 t)(1 + x_2 t) \ldots (1 + x_n t).

Thus their generating function is given by E(t) := \prod_{i=1}^n (1 + x_i t). Next, the generating function for the $h_k$’s is given by:

\begin{aligned}H(t) &:= h_0 + h_1 t + h_2 t^2 + \ldots \\ &= (1 + x_1 t + x_1^2 t^2 + \ldots) (1 + x_2 t + x_2^2 t^2 + \ldots ) \ldots (1 + x_n t + x_n^2 t^2 + \ldots)\\ &= \prod_{i=1}^n \frac 1 {1 - x_n t}.\end{aligned}

From H(t)E(-t) = 1, we obtain the following relation:

e_0 h_k - e_1 h_{k-1} + \ldots + (-1)^k e_k h_0 = 0, \qquad \text{ for } k=1, 2, \ldots.

Note that e_i = 0 for i > n.

From this recurrence relation, we can express each h_k as a polynomial in e_1, \ldots, e_n. E.g.

\begin{aligned} h_1 &= e_1, \\h_2 &= e_1^2 - e_2,\\h_3 &= e_1^3 - 2 e_1 e_2 + e_3,\\&\vdots\end{aligned}


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. e_3 = h_1^3 - 2 h_1 h_2 + h_3. As another example, if n=3, we have e_4 = h_1^4 - 3h_1^2 h_2 + 2h_1 h_3 + h_2^2 - h_4 = 0.

Definition. Since \Lambda_n \cong \mathbb{Z}[e_1, \ldots, e_n] is a free commutative ring, we can define a graded ring homomorphism

\omega: \Lambda_n \to \Lambda_n, \qquad e_i \mapsto h_i

for 1\le i \le n.

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

Proposition. \omega is an involution, i.e. \omega^2 is the identity on \Lambda_n.


We will prove by induction on k that w(h_k) = e_k for 0\le k\le n. For k=0 this is obvious; suppose 0<k\le n. Apply \omega to the above recurrence relation; since \omega(e_i) = h_i for 0\le i\le k we have:

h_0 \omega(h_k) - h_1 \omega(h_{k-1}) + \ldots + (-1)^k h_k \omega(h_0) = 0.

By induction hypothesis \omega(h_i) = e_i for i=0, \ldots, k-1; since h_0 = 1 we have

\omega(h_k) = h_1 e_{k-1} - h_2 e_{k-2} + \ldots + (-1)^{k-1} h_k e_0 = h_0 e_k = e_k.

Hence \omega^2(e_k) = e_k for all k; since e_1,\ldots, e_n generate \Lambda_n we are done. ♦

Now suppose |\lambda| = d; write h_\lambda \in \Lambda_n^{(d)} as an integer linear combination of the e_\mu for |\mu| = d, \mu_1 \le n. Applying \omega, this gives e_\lambda in terms of h_\mu for |\mu| = d, \mu_1 \le n. In particular, we get:

Corollary. The following gives a \mathbb{Z}-basis of \Lambda_n^{(d)}:

\{h_\lambda : |\lambda| = d, \lambda_1 \le n\}.

Hence we also have \Lambda_n \cong \mathbb{Z}[h_1, \ldots, h_n] as a free commutative ring; the isomorphism preserves the grading, where \deg(h_i)=i.


Consider the matrix \mathbf M = (M_{\lambda\mu}), \mathbf N = (N_{\lambda\mu}), where \lambda, \mu run through all partitions of d. Using the involution \omega, prove that

\left(\mathbf M\mathbf N^{-1}\right)^2 = \mathbf I.

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


Since h_{n+1}, h_{n+2}, \ldots \in \Lambda_n = \mathbb{Z}[h_1, \ldots, h_n], each h_k can be uniquely expressed as a polynomial in h_1, \ldots, h_n. For n=3, express h_4, h_5 in terms of h_1, h_2, h_3.

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