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The product of monomial symmetric polynomials can be expressed as $m_{\lambda} m_{\mu} = \Sigma c_{\lambda\mu}^{\nu}m_{\nu}$ for some constants $c_{\lambda\mu}^{\nu}$.

In the case of Schur polynomials, these constants are called the Littlewood-Richardson coefficients. What are they called for monomial symmetric polynomials, and how do I calculate them?

draks ...
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DavidA
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2 Answers2

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I found this reference, where the authors deal with the products you asked for.


EDIT The reference is

A MAPLE program for calculations with Schur functions by M.J. Carvalho, S. D’Agostino Computer Physics Communications 141 (2001) 282–295


From the paper (p.5 chap. 3.1 Multiplication and division of $m$-functions):

Let’s define the result of the addition and subtraction of two partitions $(\mu_1,\mu_2, . . .)$ and $(\nu_1, \nu_2, . . .)$ as being the partition whose parts are $(\mu_1 ± \nu_1,\mu_2 ± \nu_2, . . .)$. For these operations to be meaningful, it is necessary that both partitions have an equal number of parts; if they do not, then one increases the number of parts of the shortest one by adding enough zeros at the end. ... The multiplication (and division) of two m-functions are then defined as $$ m_{\alpha} m_{\beta} = \Sigma I_{\gamma}m_{\gamma} $$ and $$ m_{\alpha}/ m_{\beta} = \Sigma I_{\gamma'}m_{\gamma'} $$ where the partitions $\gamma$,$\gamma'$ result from adding to or subtracting, respectively, from $\alpha$ all distinct partitions obtained by permuting in all possible ways the parts of $\beta$. Clearly, all $m$-functions involved are functions of the same $r$ indeterminates, i.e. have the same number of total parts. The coefficient $I_\nu$, with $\nu = \gamma$ is given by $$ I_\nu=n_\nu \frac{\dim (m_\alpha)}{\dim (m_\nu)} $$ where $n_\nu$ is the number of times the same partition $\nu$ appears in the process of adding or subtracting partitions referred to above.

As far as I read, they don't give a special name to these coefficients.

draks ...
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  • Not mentioning the details of the reference in the answer, but only (supposedly) on the other end of a link is not a great idea. The link is dead now, and I have no idea where it used to be pointing. – Marc van Leeuwen May 19 '13 at 13:22
  • And is "subtracting" missing after "respectively"? Adding $\alpha$ from something else is bad syntax... – Marc van Leeuwen May 19 '13 at 13:27
  • @Marc thanks for reporting that. I try to fix the link (if possible) and provide more information about the paper in the body of the post...do you have any special interest in the topic? – draks ... May 19 '13 at 13:34
  • @marc sorry you were right. I stripped the reference a bit to fit to the question. Stand by... – draks ... May 19 '13 at 21:39
  • Reading the answer again, I have strong doubts about the division case. What do they mean by that? Note that the monomial function $m_{2,1}=e_2e_1-3e_3$ corresponds to an irreducible polynomial in the polynomial ring $\Bbb C[e_1,e_2,e_3,\ldots]$ and I think this is typical for the $m_\alpha$. But then they cannot be divisible by any other $m_\beta$, and the quotient $m_\alpha/m_\beta$ can only make sense as a formal rational fraction, not a polynomial as the RHS of the equation indicates. So either they are using a curious definition of division, or this is just plain false. – Marc van Leeuwen May 22 '13 at 06:23
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It is a matter of basic linear algebra to get the answer. Let me reformulate the question.

  1. Let $s_{\cdot}$ be the Schur polynomials and $c_{i,j}^k$ be the Littlewood-Richardson coefficients. Namely, with Einstein summation convention, $$s_{i}s_{j} = c_{i,j}^{k} s_{k}$$

  2. Let $m_{\cdot}$ be the symmetric monomials and $d_{a,b}^c$ be such that $$m_{a}m_{b} = d_{a,b}^{c} m_{c}.$$

We want an expression of the $d$'s in terms of the $c$'s.


Deduction

It is wellknown that both $s$ and $m$ forms a basis for the ring of symmetric function over $\mathbb{Q}$. So there are matrices $P, Q$ inverse to each other such that $m_a = P_a^i s_i$ and $s_i = Q_i^a m_a$ for any $a$ and $i$.

Then $m_a m_b$ equals $(P_a^i s_i)(P_b^j s_j)$, so $$P_a^i P_b^j c_{i,j}^k = d_{a,b}^c P_c^k.$$

Since $Q$ is the inverse of $P$, we have the formula

$$d_{a,b}^c = P_a^i P_b^j Q_k^c c_{i,j}^k.$$

Student
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