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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 $C_{k<N}$ and some few other matrices?

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:

$$ \left( \begin{matrix} m_{4,0,0,0}\\ m_{3,1,0,0}\\ m_{2,2,0,0}\\ m_{2,1,1,0}\\ m_{1,1,1,1}\\ \end{matrix} \right)= \left( \begin{matrix} 0& 0& 0& 0& 1\\ 0& 0& 0& 1 &-1\\ 0& 0& 1/2& 0& -1/2\\ 0& 1/2& -1/2& -1& 1\\ 1/24& -1/4 &+1/8 &1/3 &-1/4\\ \end{matrix} \right) \left( \begin{matrix} p_1^4\\ p_2p_1^2\\ p_2^2\\ p_3p_1\\ p_4^1\\ \end{matrix} \right) $$

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Have you checked Stanley's Enumerative Combinatorics Vol. II? –  Qiaochu Yuan Nov 17 '11 at 23:46
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Or Cox/Little/O'Shea? –  J. M. Nov 18 '11 at 2:22
    
The section I linked to in Google Books describes the algorithm for the conversion in terms of elementary symmetrics. It doesn't show up for you? –  J. M. Nov 18 '11 at 13:10
    
@J.M.: I tried several times to find the solution in your reference, but either I'm not mathematician enough or the answer is not there. Do you know if there is any structure in the entries of the matrices I added to my original question? –  draks ... Jan 3 '12 at 19:44
    
Your formula for monomial symmetric functions in terms of power sums has a very nice combinatorial interpretation as multiset cycle indices. These cycle indices represent multiset operators using the Polya Counting formalism. The cycle index of a multiset operator makes it possible to compute the ordinary generating function of a multiset drawn from a set of objects enumerated by some other source generating function. Read more at this MSE link. –  Marko Riedel Jun 14 at 0:45

2 Answers 2

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+50

Probably a simple general formula for your matrix $C_n=(C_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$, where $\mathcal P_n$ denotes the partitions of $n$ (ordered in decreasing lexicographic ordering) does not exist. However a number of things can be said, notably your above guesses can be confirmed. One thing that your examples suggest but which is false is that the matrix is "upper-left triangular", which fails from $n=6$ on; the reason is that the true triangularity is given by $ C_{\lambda,\mu}\neq0\implies \lambda^{tr}\leq\mu$ in the dominance order where $\lambda^{tr}$ is the transpose (conjugate) partition of $\lambda$, and lexicographic order for $n\geq6$ does not assign complementary positions to $\lambda$ and $\lambda^{tr}$ (nor does any ordering for $n$ sufficiently large).

As you guessed it is easier to study the inverse matrix $C_n^{-1}=(M_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$, whose entry $M_{\lambda,\mu}$ gives the coefficient of $m_\mu$ in $e_\lambda$. This nonnegative number equals number of $0$-$1$ matrices with row sums $\lambda_1,\lambda_2,\ldots$ and column sums $\mu_1,\mu_2,\ldots$, and in particular $M$ is a symmetric matrix (Stanley EC2, Proposition 7.4.1, Corollary 7.4.2). The argument for this combinatorial interpretation is basically the balls-into-boxes you suggested: each term in $e_\lambda$ comes from choosing a monomial in each factor $e_{\lambda_i}$, and this choice can be represented by putting in row $i$ a $1$ (ball) in the selected columns and zeros elsewhere; the product monomial is found by summing up the columns (and by symmetry only those monomials with weakly decreasing exponents (column sums) need to be considered). Since $M_n$ is symmetric and $C_n$ is its inverse, an easy argument shows that $C_n$ is also symmetric.

You suggested that the sum of all entries of $C_n$ is $0$. This fails for $n=0,1$, but is true for all larger values, which can be seen as follows. By the definition of $C_n$, if one takes its column sums (equivalently, if one left-multiplies by $(1~1\cdots1)$), then one gets the coefficients the $e_\mu$ in $h_n=\sum_{\lambda\in\mathcal P_n}m_\lambda$, the $n$-th complete homogeneous symmetric function (sum of all distinct monomials of degree $n$). One would like to know the sum of those coefficients. Now the relation between elementary and complete homogeneous symmetric functions is given by the generating series identity $$ (1-e_1X+e_2X^2-e_3X^3+\cdots)(1+h_1X+h_2X^2+\cdots)=1 $$ or equivalently $\sum_{i=0}^n(-1)^ne_ih_{n-1}=0^n$. You can prove that the mentioned sum is $0$ for $n\geq2$ by induction from the latter equation. A more conceptual way is to use the generating series identity and the algebraic independence of the $e_i$ for $i\geq1$, which means there is a ring morphism sending all of them to $1$, and the obvious fact that $$ (1-X+X^2-X^3+\cdots)^{-1}=1+X $$ This shows that the mentioned morphism sends $h_0$ and $h_1$ to $1$ and all other $h_i$ to $0$; this value is precisely the sum of coefficients of all $e_\lambda$ we were after.

Finally for the computation of the matrix $C_n$, it also seems best to view it as the inverse of the matrix $(M_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$, which becomes unitriangular after permuting its columns according to the transposition of partitions. However $M_{\lambda,\mu}$ is best computed not by counting $0$-$1$ matrices (their numbers grow up to $n!$ in the bottom-right corner), but rather via $$ M_{\lambda,\mu}= \sum_{\lambda\leq\nu\leq\mu^{tr}}K_{\nu,\lambda,}K_{\nu^{tr},\mu} $$ where $K_{\nu,\lambda}$ designates a Kostka number, the number of semi-standard Young tableaux of shape $\nu$ and weight (content) $\lambda$. This identity is proved bijectively by the asymmetric RSK-correspondence, a bijective correspondence between $0$-$1$ matrices and pairs of semi-standard tableaux of transpose shapes, their weights being given respectively by the column sums and the row sums of the matrix. The Kostka numbers involved are generally quite a bit smaller than $M_{\lambda,\mu}$, and moreover there are ways to compute them without counting; one method is to interpret them as weight multiplicities for $GL_n$ and use a formula like Freudenthal's recurrence for them. (The LiE program which I maintain does so, and will do this computation easily; one can get results online up to $n=9$ from this site, if one takes into account the fact that a partition is represented by the sequence of differences of successive parts: $[4,2,2,1,0,0,0,0,0]$ is represented as $[2,0,1,1,0,0,0,0]$.)

One could either compute $M_{\lambda,\mu}$ this way and invert the result, or invert the matrix of Kostka numbers and deduce from above formula, which can be interpreted as a matrix product, the formula $$ C_{\lambda,\mu}= \sum_{\lambda^{tr}\leq\nu\leq\mu}K'_{\lambda,\nu^{tr}}K'_{\mu,\nu} $$ where $(K'_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$ is the inverse of the Kostka matrix $(K_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$. I don't know very much about these "inverse Kostka numbers", but you will find some information about them in the answers to this MO question; I'm not sure this allows them to be computed more efficiently than by inverting the Kostka matrix.

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Maybe you could also answer this question: math.stackexchange.com/q/17891/19341 –  draks ... Feb 2 '12 at 7:45

Beside the very nice answer by Marc, I found this Introduction to Symmetric Functions by Mike Zabrocki:

$$ e_\mu = \sum_\lambda B_{\lambda \mu} m_\lambda\tag{2.28} $$ where $B_{\lambda \mu}$ is the number of matricies with entries in $\{0, 1\}$ whose column sum is $\mu$ and row sum is equal to $\lambda$. $$\dots$$ $B_{\lambda \mu}$is the number of ways of filling the the Young diagram for the partition with $\mu_1$ 1s, $\mu_2$ 2s, etc. that are strictly increasing in the rows and there is no restriction on the relationship between the values in the columns.

In a table summarising all possible transitions at the end of chapter 2.1 they also mention what I was previously looking for $$ m_\mu = \sum_\lambda G_{\lambda \mu} e_\lambda, $$ but leave it as an exercise to determine some sort of formula for these ($G_{\lambda \mu}$) coefficients.

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