Given the elementary symmetric polynomials $e_k(X_1,X_2,...,X_N)$ generated via $ \prod_{k=1}^{N} (t+X_k) = e_0t^N + e_1t^{N-1} + \cdots + e_N. $ How can one get the monomial symmetric functions $m_\lambda(X_1,X_2,...,X_N)$ as products and sums in $e_k$? For example: $N=4$ $ m_{(2,1,1,0)}=X_1^2X_2X_3 + \text{all permutations}= e_3\cdot e_1 - 4 e_4 , $ $ m_{(2,2,0,0)}=X_1^2X_2^2 + ... = e_2^2-6e_4 - 2m_{(2,1,1,0)}=e_2^2 - 2e_3\cdot e_1 +2e_4 $ It seems clear that the products on the RHS run over all partitions $\mu$ of $N$. For each $\lambda$ there should be a set of integers $C_{\lambda\mu}$ such that $ m_\lambda = \sum_\mu C_{\lambda \mu} \prod_j e_{\mu_j} $ is true. Putting all values of $c_{\lambda \mu}$ together, you get the following equation: $ \left( \begin{matrix} m_{4,0,0,0}\\ m_{3,1,0,0}\\ m_{2,2,0,0}\\ m_{2,1,1,0}\\ m_{1,1,1,1}\\ \end{matrix} \right)= \left( \begin{matrix} -4&+4&+2&-4&+1\\ +4&-1&-2&+1&0\\ +2&-2&+1&0&0\\ -4&+1&0&0&0\\ +1&0&0&0&0\\ \end{matrix} \right)\left( \begin{matrix} e_{4}\\ e_3e_1\\ e_2^2\\ e_2e_1^2\\ e_1^4\\ \end{matrix} \right) . $ The transition matrix $C_4$ is symmetric. This also holds for $N=3$, where $C_3$ is $ \left( \begin{matrix} +3&-3&+1\\ -3&+1&0\\ +1&0&0\\ \end{matrix} \right) $ and for $N=2$ we get $ \left( \begin{matrix} -2&+1\\ +1&0\\ \end{matrix} \right). $
The main question is, if there exists a general formula for the entries of the matrices? For a given $N$, is it a kind of composition of matrices $C_{k
Beside that, the following is also of interest: The matrices so far are symmetric and their entries sum up to $0$. Is this true in general? Is this related to (conjugate) Young Tableaux?
Balls&Boxes If we bring the matrix on the LHS we get: $ \left( \begin{matrix} 0 &0&0&0&1\\ 0&0&0&1&4\\ 0&0&1&2&6\\ 0&1&2&5&12\\ 1&4&6&12&24\\ \end{matrix} \right) \left( \begin{matrix} m_{4,0,0,0}\\ m_{3,1,0,0}\\ m_{2,2,0,0}\\ m_{2,1,1,0}\\ m_{1,1,1,1}\\ \end{matrix} \right)= \left( \begin{matrix} e_{4}\\ e_3e_1\\ e_2^2\\ e_2e_1^2\\ e_1^4\\ \end{matrix} \right)\tag{*} . $
Maybe it's easier to interpret these values, since they are all positive, in terms of balls, that have to be put into boxes, according to the following rules :
You are given $N$ balls. Your balls are now divided into parts (and bless god that, this is a math forum :-) according to a partition $\mu$. These are the products of $e_k$'s. Now you are asked to put the balls partition-by-partition into a $N$ boxes. It is not allowed to put more than 1 ball in a box for the current partition.
The goal is to achieve a certain distribution of balls among the boxes, given by $\lambda$. These are the $m_\lambda$.
Power Sums Or would it help to express $e_k$ as power sums via Newton-Girard formulas? Here is the worked out example:
$ \left( \begin{matrix} m_{4,0,0,0}\\ m_{3,1,0,0}\\ m_{2,2,0,0}\\ m_{2,1,1,0}\\ m_{1,1,1,1}\\ \end{matrix} \right)= \left( \begin{matrix} 0& 0& 0& 0& 1\\ 0& 0& 0& 1 &-1\\ 0& 0& 1/2& 0& -1/2\\ 0& 1/2& -1/2& -1& 1\\ 1/24& -1/4 &+1/8 &1/3 &-1/4\\ \end{matrix} \right) \left( \begin{matrix} p_1^4\\ p_2p_1^2\\ p_2^2\\ p_3p_1\\ p_4^1\\ \end{matrix} \right) $