Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\dots$

share|cite|improve this question
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. – Zhen Lin Jan 5 '12 at 12:49
@ZhenLin: The elementary expression is exactly what I'm looking for. – draks ... Jan 5 '12 at 12:59
Where is the question? – Marc van Leeuwen Jan 17 '12 at 14:25
The question is still not clear, since the scalar product seems to be defined in terms of itself. It looks like you want the monomial symmetric functions $m_\lambda$ to be mutually orthogonal; it would be clearer if you just said that. The normalization constants apparently could be anything, so the scalar product is not uniquely defined. There exist scalar products for which the $m_\lambda$ are mutually orthogonal, even orthonormal, but what is the interest? – Marc van Leeuwen Jan 18 '12 at 14:32
@MarcvanLeeuwen: When you say "There exist scalar products ..." does that imply that you can answer the question? My interest is to get a "procedure" to get the coefficients $c_k$? I thought the term scalar product would fit, but if this is what annoys you, I don't cling to that. Would projector be better? I thought I could learn something beyond the scope of my special interest. – draks ... Jan 27 '12 at 13:41
up vote 1 down vote accepted

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.

share|cite|improve this answer
In fact, you're not wrong. I just thought one could do something more mathematical. Thanks. – draks ... Mar 13 '12 at 9:13
Maybe you can tell me what's wrong with my question/answer? I would be happy to hear your opinion... – draks ... Mar 14 '12 at 19:33
A similar scalar product was also used here. How do you think is it related? – draks ... Mar 30 '12 at 18:49

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.

share|cite|improve this answer
In answer to you comment at my answer, this answer doesn't look wrong (though one might object that you forget to transform the constant function obtained after differentiation into its value). But it seems unnecessarily complicated. For one thing, if your functions are already symmetric, there is no need to symmetrize the differential operator as well, just $(\nabla\vec x)^\mu$ will do just fine, and pick out the same value as any permutation of $\mu$ would. But I never regard symmetric functions as functions, just expressions; if you want the coefficient of $X^\mu$, you can just look for it. – Marc van Leeuwen Mar 14 '12 at 22:58
@MarcvanLeeuwen What do you mean by "forget to transform the constant function obtained after differentiation into its value" ? And yes I know that my expression is complicated, but I like the use of the monomials themselves to calculate the scalar product. – draks ... Mar 16 '12 at 14:38
If $f(x)=x^3$, then $f'''(x)=6$ for all $x$, so $f'''$ is the constant function with value $6$, but writing $f'''=6$ is not quite correct, and writing $\langle f\mid g\rangle=f'''$ would be confusing. Similarly your equation $\nabla_\mu\cdot P(\vec X)=\frac1N\langle m_\mu\mid P\rangle$ has a (constant) function as LHS and a number as RHS. You can repair this by evaluating the LHS at any point, writing for instance $\nabla_\mu\cdot P(\vec X)|_{\vec X=0}$. – Marc van Leeuwen Mar 16 '12 at 15:16
@MarcvanLeeuwen 1. I don't see why $f'''(x)=6$ should not be correct. 2. I would rather write $f'''=\langle f| x^3 \rangle/3!$. 3. Since $\mu$ is the same as the total degree of any $P(\vec X)$ only constants terms stay and therefore setting $\vec X =0 $ is not necessary. I agree, that it feels a bit strange to have a function LHS and number RHS... – draks ... Mar 20 '12 at 23:09
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. – Marc van Leeuwen Mar 21 '12 at 8:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.