User peter erskin - MathOverflow

Expressing power sum symmetric polynomials in terms of lower degree power sums

2010-08-26T15:11:47Z

Is there an explicit formula expressing the power sum symmetric polynomials $$p_k(x_1,\ldots,x_N)=\sum\nolimits_{i=1}^N x_i^k = x_1^k+\cdots+x_N^k$$ of degree $k$ in $N < k$ variables entirely through the power sum symmetric polynomials $p_j(x_1,\ldots,x_N)$ of degrees $ j \le N $?

Examples: $$N=1,\ k=2: \quad p_2=x^2=x\times x=p_1^2$$

$$N=2,\ k=3: \quad p_3 = x^3 + y^3 = [3(x^2+y^2)(x+y)-(x+y)^3]/2 = (3 p_2 p_1-p_1^3)/2$$

What is the general formula?

I am looking for a formula similar to that for the expansion of the Schur functions $s_\lambda$ in terms of the symmetric power sums:

$$ s_\lambda=\sum_{\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)}\chi^\lambda_\rho \prod_j \frac{p^{r_j}_j}{r_j!},$$ where the coefficients $\chi^\lambda_\rho$ are the characters of the representation of the symmetric group indexed by the partition $\lambda$ evaluated at elements of cycle type indexed by the partition $\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)$, which contains $ r_j $ parts of length $j$.

Clearly, the power sums of degree higher than $N$ can be expanded in a similar manner: $$ p_k=\sum_{\rho}a_{k;\rho}\prod_{j=1}^N p_j^{r_j}, $$ where $\rho=(1^{r_1},2^{r_2},\dots,N^{r_N})$ is the partition of $k$ such that $k=r_1+2r_2+3r_3+...+Nr_N$.

In the above example for $N=2,\ k=3$ one has $a_{3;\ (1^{1},2^{1}) }=3/2$ and $a_{3;\ (1^{3},2^{0})}=-1/2$.

My question can be thus reformulated as follows: given $r_1,...,r_N$ what is the explicit formula for $a_{k;\rho}$?

Note Added

Actually, Wikipedia tells us how to construct a certain explicit formula for $p_k$. It gives the following expressions for $p_n$ with $n=N$ in terms of $ e_j, $

$$ p_n = \begin{vmatrix} e_1 & 1 & 0 & \cdots & \\ 2e_2 & e_1 & 1 & 0 & \cdots & \\ 3e_3 & e_2 & e_1 & 1 & \cdots & \\ \vdots &&& \ddots & \ddots & \\ ne_n & e_{n-1} & \cdots & & e_1 & \end{vmatrix}, $$

and for $e_n$ with $n=N$ in terms of $ p_j, $

$$ e_n=\frac1{n!} \begin{vmatrix}p_1 & 1 & 0 & \cdots\\ p_2 & p_1 & 2 & 0 & \cdots \\ \vdots&& \ddots & \ddots \\ p_{n-1} & p_{n-2} & \cdots & p_1 & n-1 \\ p_n & p_{n-1} & \cdots & p_2 & p_1 \end{vmatrix}. $$

As far as I can see from the derivation described in Wikipedia, these determinant expressions are also valid for $p_n$ with $ n > N $ and for $e_n$ with $ n < N $.

For $p_n$ with $n>N$ one should take into account that all $ e_k=0 $ for $ k > N $, so that the resulting matrix has zeros in both right-upper and left-lower corners.

Substituting the determinants for $e_j$ into the determinant for $p_k$, one gets the explicit formula which seems to solve the problem.

However, I still don't know how to obtain the coefficients $a_{k;\rho}$ in the expansion of $ p_k $ in terms of the first $N$ power sums which would be the desired (really explicit) formula.

Comment by Peter Erskin

2010-08-27T15:54:06Z

Thank you, Robin! Your formula relates $ p_k $ to lower degree power sums including $ p_m $ with $ m>N, $ right? However, I wanted to express $ p_k $ entirely through $ p_1,...,p_N. $ This is formally done in the next answer by Gjergji Zaimi, as well as in the "Note Added" above (probably, the original formulation of my question was not particularly clear, sorry). Nevertheless, I think your contribution to the discussion is really very important. Thanks.

Comment by Peter Erskin

2010-08-27T15:40:11Z

Thank you very much for the explicit (in contrast to previous suggestions) formula and for the clarification! The formula really works and it looks simpler than the determinant-of-determinants that I proposed in the "Note Added". Unfortunately, I do not see how to present your expression for $ p_k$ in the form similar to your formula for $ e_n $. Is it possible to find the combinatorial coefficients $ a_{k;\rho} $ as discussed in my question?

Comment by Peter Erskin

2010-08-27T09:42:54Z

I am not sure that the formula for the Schur functions which I quoted from Wikipedia (see my question) includes only $p_j$ with $j\le N$ when the degree of the Schur function is higher than $N$. Apparently, in this case one has $p_k$ with $k > N$ on the r.h.s. However, the corresponding coefficients might be identically zero --- I did not check. If this is not the case, it is interesting to learn what is the expansion of the Schur function $s_\lambda$ of degree higher than $N$ in terms of only the first $N$ power sums $p_1,...,p_N$.

Comment by Peter Erskin

2010-08-27T09:00:21Z

Thanks! If I understand correctly, this yields a nice relation between the power sum polynomials $p_1,\ldots,p_k$ for $k > N$. However, as you mentioned, this is not an explicit formula. Indeed, in order to expand $$p_k(x_1,\ldots,x_N)$$ in terms of only $N$ power sums $$p_1(x_1,\ldots,x_N),\ldots,p_N(x_1,\ldots,x_N)$$ that form a complete basis for symmetric polynomials in $N$ variables, one should further apply your formula iteratively for each of the lower degree power sums $p_{k-1},\ p_{k-2},\ldots, p_{N+1}$ that appear in your formula for $p_k$.