User peter erskin - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:27:05Z http://mathoverflow.net/feeds/user/8802 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36771/expressing-power-sum-symmetric-polynomials-in-terms-of-lower-degree-power-sums Expressing power sum symmetric polynomials in terms of lower degree power sums Peter Erskin 2010-08-26T15:11:47Z 2010-09-01T07:35:04Z <p>Is there an explicit formula expressing the <a href="http://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial" rel="nofollow">power sum symmetric polynomials</a> $$p_k(x_1,\ldots,x_N)=\sum\nolimits_{i=1}^N x_i^k = x_1^k+\cdots+x_N^k$$ of degree $k$ in $N &lt; k$ variables entirely through the power sum symmetric polynomials $p_j(x_1,\ldots,x_N)$ of degrees $j \le N$? </p> <p>Examples: $$N=1,\ k=2: \quad p_2=x^2=x\times x=p_1^2$$</p> <p>$$N=2,\ k=3: \quad p_3 = x^3 + y^3 = [3(x^2+y^2)(x+y)-(x+y)^3]/2 = (3 p_2 p_1-p_1^3)/2$$</p> <blockquote> <p>What is the general formula?</p> </blockquote> <p>I am looking for a formula similar to that for the expansion of the <a href="http://en.wikipedia.org/wiki/Schur_polynomial#Relation_to_representation_theory" rel="nofollow">Schur functions</a> $s_\lambda$ in terms of the symmetric power sums:</p> <p>$$s_\lambda=\sum_{\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)}\chi^\lambda_\rho \prod_j \frac{p^{r_j}_j}{r_j!},$$ 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 $j$. </p> <p>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 $k=r_1+2r_2+3r_3+...+Nr_N$. </p> <p>In the above example for $N=2,\ k=3$ one has $a_{3;\ (1^{1},2^{1}) }=3/2$ and $a_{3;\ (1^{3},2^{0})}=-1/2$. </p> <blockquote> <p>My question can be thus reformulated as follows: given $r_1,...,r_N$ what is the explicit formula for $a_{k;\rho}$?</p> </blockquote> <hr> <p><strong>Note Added</strong></p> <p>Actually, <a href="http://en.wikipedia.org/wiki/Newton%27s_identities#Expressions_as_determinants" rel="nofollow">Wikipedia</a> 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> <p>$$p_n = \begin{vmatrix} e_1 &amp; 1 &amp; 0 &amp; \cdots &amp; \\ 2e_2 &amp; e_1 &amp; 1 &amp; 0 &amp; \cdots &amp; \\ 3e_3 &amp; e_2 &amp; e_1 &amp; 1 &amp; \cdots &amp; \\ \vdots &amp;&amp;&amp; \ddots &amp; \ddots &amp; \\ ne_n &amp; e_{n-1} &amp; \cdots &amp; &amp; e_1 &amp; \end{vmatrix},$$</p> <p>and for $e_n$ with $n=N$ in terms of $p_j,$</p> <p>$$e_n=\frac1{n!} \begin{vmatrix}p_1 &amp; 1 &amp; 0 &amp; \cdots\\ p_2 &amp; p_1 &amp; 2 &amp; 0 &amp; \cdots \\ \vdots&amp;&amp; \ddots &amp; \ddots \\ p_{n-1} &amp; p_{n-2} &amp; \cdots &amp; p_1 &amp; n-1 \\ p_n &amp; p_{n-1} &amp; \cdots &amp; p_2 &amp; p_1 \end{vmatrix}.$$</p> <p>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 $n &lt; N$.</p> <p>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. </p> <p>Substituting the determinants for $e_j$ into the determinant for $p_k$, one gets the explicit formula which seems to solve the problem. </p> <p>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. </p> http://mathoverflow.net/questions/36771/expressing-power-sum-symmetric-polynomials-in-terms-of-lower-degree-power-sums/36779#36779 Comment by Peter Erskin Peter Erskin 2010-08-27T15:54:06Z 2010-08-27T15:54:06Z Thank you, Robin! Your formula relates $p_k$ to lower degree power sums including $p_m$ with $m&gt;N,$ 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 &quot;Note Added&quot; 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. http://mathoverflow.net/questions/36771/expressing-power-sum-symmetric-polynomials-in-terms-of-lower-degree-power-sums/36873#36873 Comment by Peter Erskin Peter Erskin 2010-08-27T15:40:11Z 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 &quot;Note Added&quot;. 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? http://mathoverflow.net/questions/36771/expressing-power-sum-symmetric-polynomials-in-terms-of-lower-degree-power-sums Comment by Peter Erskin Peter Erskin 2010-08-27T09:42:54Z 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 $k &gt; N$ 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 $p_1,...,p_N$. http://mathoverflow.net/questions/36771/expressing-power-sum-symmetric-polynomials-in-terms-of-lower-degree-power-sums Comment by Peter Erskin Peter Erskin 2010-08-27T09:00:21Z 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 $k &gt; N$. 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 $p_k$.