Let $S_k$ be the $k$-th symmetric polynomial of $n$-variable. How can I rewrite $$S_k(x+y)-S_k(x)-S_k(y)$$

by just using $x,y,S_1,S_2,\cdots S_{k-1}$ where $x=(x_1,x_2,\cdots,x_n)$ and $y=(y_1,y_2,\cdots,y_n)$.

Example: Let $x=(x_1,x_2)$ and $y=(y_1,y_2)$

$$S_2(x+y)-S_2(x)-S_2(y)=(x_1+y_1)(x_2+y_2)-x_1x_2-y_1y_2$$ $$=x_1y_2+x_2y_1=(x_1+x_2)(y_1+y_2)-(x_1y_1+x_2y_2)$$ $$=S_1(x)S_1(y)-S_1(xy).$$

How can we generalize this for any $n$ and $k$?

I believe somebody found this before but my research area is far to symmetric polynomials. References are also accepted. Thanks.

  • $\begingroup$ Can you include what do you mean by $S_k$? I mean the definition of $S_k$ $\endgroup$
    – Elaqqad
    Commented Apr 15, 2015 at 20:27
  • $\begingroup$ elementary symmetric polynomials. en.wikipedia.org/wiki/Elementary_symmetric_polynomial $\endgroup$
    – vudu vucu
    Commented Apr 15, 2015 at 20:30
  • $\begingroup$ I guess you only have to do it for $k=n$, right? Since in any other case, $S_k(x_1,\cdots,x_n)$ decomposes into $\binom{n}{k}$-many terms $S_k(y_1,\cdots,y_k)$... $\endgroup$ Commented May 7, 2015 at 16:26
  • $\begingroup$ Hi @Theo, I saw your comment now. Yes, with this trick, it is enough. $\endgroup$
    – vudu vucu
    Commented May 21, 2015 at 8:32

1 Answer 1


This is just an observation, not an answer. Too long for a comment.
See https://mathoverflow.net/questions/203033. The first (symmetric) part $S_k(x+y)$ can be generalised to any number of variable types x, y, z, .. by observing that the elementary summetric functions never produce a variable to a power >1. Hence we can 'encode' the variable types x, y, z, .. by $x^1, x^2, x^3, ..$ and obtain using 4 variables (=subscripts) of 3 types

$S_2(x+x^2+x^3)=x_2^3 x_1^3+x_3^3 x_1^3+x_4^3 x_1^3+x_2^2 x_1^3+x_3^2 x_1^3+x_4^2 x_1^3+x_2 x_1^3+x_3 x_1^3+x_4 x_1^3+x_2^3 x_1^2+x_3^3 x_1^2+x_4^3 x_1^2+x_2^2 x_1^2+x_3^2 x_1^2+x_4^2 x_1^2+x_2 x_1^2+x_3 x_1^2+x_4 x_1^2+x_2^3 x_1+x_3^3 x_1+x_4^3 x_1+x_2^2 x_1+x_3^2 x_1+x_4^2 x_1+x_2 x_1+x_3 x_1+x_4 x_1+x_2^3 x_3^3+x_2^2 x_3^3+x_2 x_3^3+x_2^3 x_4^3+x_3^3 x_4^3+x_2^2 x_4^3+x_3^2 x_4^3+x_2 x_4^3+x_3 x_4^3+x_2^3 x_3^2+x_2^2 x_3^2+x_2 x_3^2+x_2^3 x_4^2+x_3^3 x_4^2+x_2^2 x_4^2+x_3^2 x_4^2+x_2 x_4^2+x_3 x_4^2+x_2^3 x_3+x_2^2 x_3+x_2 x_3+x_2^3 x_4+x_3^3 x_4+x_2^2 x_4+x_3^2 x_4+x_2 x_4+x_3 x_4$
which is just this simple expression in monomial symmetric functions:
'Decoding' $x^1 -> x, x^2 -> y, x^3 -> z, ..$ recovers the original $S_2(x+y+z)$.

Generally, the following conjecture seems to hold: (using e for the elementary symm. pol.)
$e_k(v,t) = \sum m_\lambda$
where v gives the number of variables (=subscripts) and t gives the number of types used in the argument (x,y,z,..); the sum goes over all partitions $\lambda$ in k parts <= v, thus partitions of k up to k*v.


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