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Suppose a multinomial $P(X_1, X_2,\ldots, X_n)$, that is given as a sum of monomials $m_\lambda$ with coefficients $c_k$: $ P(\vec{X})=P(X_1, X_2,\ldots, X_n) = \sum_k c_k m_{\lambda_k} . $

Since the monomials form a basis of the vector space of multinomials, there is also a scalar product $ c_k=\frac{1}{N}\langle m_{\lambda_k} \mid P\rangle, $ where $N=\langle m_{\lambda_k}\mid m_{\lambda_k}\rangle$ would be a normalization constant.

My question is: Does the Scalar Product, such that the $m_λ$ are mutually orthogonal or better orthonormal, have an elementary expression?

An application could allow calculation of Kostka number, since $ s_{\lambda} = \sum K_{\lambda\mu} m_\mu, $ where $s_\lambda$ is a Schur polynomial. If this is an efficient way or not, is a different question. First I thought that I had to deal with something like square integrable functions, but then I found what I posted below $\dots$

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    Calculating the Kostka numbe$r$s was also mentioned [here](http://math.stackexchange.com/questions/47805/schur-skew-functions#comment106508_47805).2012-03-30

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It seems like the real question is what one can say about (the computation of) the coordinate functions for the basis of monomial symmetric functions; this does not involve a scalar product for which the monomial symmetric functions are orthonormal. (Of course once one knows the coordinate functions, the scalar product making the monomial symmetric functions an orthonormal basis can be computed as usual by multiplying corresponding coordinates and then summing them all). Whether computing these coordinate functions is difficult depends on how the symmetric polynomials are given (for instance they could be given as expressions in the elementary symmetric polynomials, or as combinations of Schur functions). For the most obvious way to give a symmetric polynomial, namely as an expression expanded in the monomials in the $X$s, the problem is almost trivial: the coefficient of $m_\lambda$ is equal to the coefficient of any monomial in the $X$s that occurs in $m_\lambda$, for instance the monomial $X^\lambda=X_1^{\lambda_1}\ldots X_n^{\lambda_n}$ (all these coefficients are equal by definition of a symmetric function). For other ways to specify symmetric functions, the problem therefore reduces to that of expanding them into monomials, which may be tedious, but is entirely straightforward.

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    A similar scalar product was also used [here](http://math.stackexchange.com/questions/47805/schur-skew-functions#comment106508_47805). How do you think is it related?2012-03-30
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I think the following could do:

Let $\mu=(\mu_1,\dots\mu_k,\dots\mu_{m\le n})$ be a given partition of $n$. Further let $\triangledown_{\mu}$ be an operator defined as follows: $ \nabla_{\mu}:=\frac{1}{|m_\mu|}\sum_\sigma \prod_{k} \frac{d^{\mu_k}}{\mu_k!d(X_k)^{\mu_k}}=\frac{m_{\mu}(\nabla\vec{X} )}{|m_\mu|\prod_k \mu_k!}, $ where the sum runs over all permutations $\sigma(\mu)=(\mu_{\sigma(1)},\mu_{\sigma(2)},\dots,\mu_{\sigma(m)} )$ and $|m_\mu|$ means the number of elements of the monomial (which can be calculated by $m_\mu(1,1,\dots,1)\;$).

Now we apply $\nabla_{\mu}$ on $P(\vec{X})$. Since $P$ is homogenous, all terms in $P$ have the same total degree $n$. Therefore only those $m_{\lambda_k}$ will contribute, that match the pattern $\mu$. So it should give the coefficient $c_k$, so I conclude $ \nabla_{\mu}\cdot P(\vec{X})=c_k=\frac{1}{N}\langle m_{\mu} \mid P\rangle. $ EDIT: But can a differential be used as part of a scalar product? And further, referring to Zhen's comment

Since the monomials form an honest basis for your vector space, there is a (unique) inner product for which it is an orthonormal basis. The question is whether this inner product is convenient in some sense: Is it continuous? Does it have an elementary expression? etc.

is this the elementary expression of the unique inner product ?


An example: $P=r(X_1^3+X_2^3+X_3^3) +s(X_1^2X_2+...) + tX_1X_2X_3$ and $\mu=(2,1,0)$. Then $ \nabla_{(2,1,0)}=\frac{1}{6}\left( \frac{d^{2}}{2!dX_1^2}\frac{d^{\phantom{1}}}{1!dX_2^{\phantom{1}}}+...\right). $ The only terms that survive, are those, that fit exactly to the given pattern $c_{(2,1,0)}=s$.


Joriki is doing something similar in his answer here. He additionally sets $z=0$, since the polynomial there is not homogenous.

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    I never wrote that $f'''(x)=6$ is not correct (in fact I said it holds for all $x$), I said that $f'''=6$ is not quite correct (since the LHS is a function and the RHS is a number). If you cannot understand that a constant function is not the same as the value it takes everywhere, then don't bother. But mathematically they are not the same thing, and one shouldn't write equations where the LHS can _only_ be viewed as a function and the RHS _only_ as a number, as is the case with $f'''=\langle f\mid x^3 \rangle/3!$. Setting the function argument to any value makes the conversion needed.2012-03-21