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Let $S_k$ be the $k$-th elementary symmetric polynomial of $n$ variables. 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.

Edit: (or addition:) It can be closed form or an algorithm.. Useful comments.

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    $\begingroup$ This looks like $c_k(\eta\otimes\xi)-c_k(\eta)-c_k(\xi)$ for vector bundles. (Though, the $xy$ in your example is kind of cheating.) That would probably be useful in topology, but, as far as I know, no one has ever come up with a "closed formula". $\endgroup$ – Alex Degtyarev Apr 15 '15 at 20:56
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    $\begingroup$ $S_k(x+y)=\sum_{i=0}^k S_i(x)S_{k-i}(y)$, where $S_0=1$ by definition. This is trivial from a combinatorial point of view, and also the best you'll get. $\endgroup$ – Alexander Woo Apr 15 '15 at 21:52
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    $\begingroup$ Hi @AlexanderWoo, this formula is not valid for the example in question or i am sleepy. $\endgroup$ – vudu vucu Apr 15 '15 at 22:53
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    $\begingroup$ .@vudu vucu Your pairing of the two alphabets $X=\{x_i\}_{i\in I};\ Y=\{y_i\}_{i\in I}$ is not conventional (although you may need it), setting $X+Y=\{x_i+y_i\}_{i\in I}$ destroys the symmetry of the expression $S_k(X+Y)-S_k(X)-S_k(Y)$. Are you aware of this ? What is used as definition of the sum is $$X+Y=\{x_i+y_j\}_{i,j\in I}$$ In this case Alexander Woo's formula is right. $\endgroup$ – Duchamp Gérard H. E. Apr 16 '15 at 4:45
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    $\begingroup$ $S_k$ for the elementary symmetric functions, what a horrible notation... $\endgroup$ – Abdelmalek Abdesselam May 7 '15 at 17:58
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Express $S_k$ in power sums via the Newton formula. For power sums this is the binomial formula for each summand. Then express power sums again in elementary symmetric functions.

Added:

The expression $S_k(x+y)-S_k(x)-S_k(y)$ is still invariant under the diagonal action of the symmetric group $\mathfrak S(n)$ acting on $x$ and $y$ in the same way. So it can written as a polynomial in a basis for this diagonal representation. One has to determine such a basis. By Corollary 2.17 of here the algebra of invariant polynomials of the diagonal action is the integral closure of the the algebra of 2-polarizations of the algebra of symmetric polynomials. By Example 2.18, in the case of the permutation group, the algebra of 2-polizations is already integrally closed.

However, the expression $S_2(x+y)-S_2(x)-S_2(y)$ is a 2-polarization.

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  • $\begingroup$ It's a good start but one has sums $\sum_i x_i^a y_i^b$ say with $a>b$ and one has to peel off the $x_i^{a-b}$ from the $(x_i y_i)^b$ somehow. $\endgroup$ – Abdelmalek Abdesselam May 7 '15 at 19:58
  • $\begingroup$ Sigh! And maybe you can't peel because the whole thing is not symmetric any more. $\endgroup$ – Peter Michor May 7 '15 at 20:06
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Not quite an answer. The logarithmic derivative of the generating function trick (as described very well in Yakovlenko's notes would seem to give a reasonable approach to this (I am not quite up to working through it right this moment).

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This is just a suggestion how to proceed in the case $k=3$, which is too long for a comment.

In case of degree $=$ number of variables $=3$ we have this formula: \begin{align} & S_3(x+y+z)-S_3(x+y)-S_3(y+z)-S_3(z+y)+S_3(x)+S_3(y)+S_3(z) \\ =\;& S_1(x)S_1(y)S_1(z)-S_1(xy)S_1(z)-S_1(yz)S_1(x)-S_1(zx)S_1(y)+2S_1(xyz) \end{align}

Specializing to $z=-\frac{x+y}{2}$ yields a formula for $S_3(x+y)-S_3(x)-S_3(y)$ in terms of $S_1$.

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  • $\begingroup$ Have you worked it out? In my comment to the OP, there remain two $S_2$ terms. So a priori either mine or yours must be wrong. $\endgroup$ – Wolfgang May 9 '15 at 19:00
  • $\begingroup$ Yes, I did. Also why can't both solutions be correct? $\endgroup$ – vuur May 9 '15 at 19:14
  • $\begingroup$ well I think, the mean $e_2$ can only be expressed by means of $e_1$ if using reciprocals. Anyway, can you provide your final expression? $\endgroup$ – Wolfgang May 9 '15 at 19:20
  • $\begingroup$ It's minus the RHS with $z$ replaced by $-\frac{x+y}{2}$. If you think an expression such as $S_1(-\frac{x+y}{2}x)$ is weird, one can write $-S_1(xx)/2-S_1(xy)/2$ instead. $\endgroup$ – vuur May 9 '15 at 19:27
  • $\begingroup$ In your final expression, some coefficients $1/2$ remain, which is strange. I think I've found the flaw: your formula is correct, but the problem is $S_3(2z)=8S_3(z)$ not $2S_3(z)$. So the terms of the LHS don't cancel out as you would like... $\endgroup$ – Wolfgang May 9 '15 at 20:16
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I will use the standard notation $e_k$ for the $k$-th elementary symmetric polynomial of $n$ variables, and just $e$ instead of $e_1$ for better readability.
For $k\le n$, denote the set of the "base polynomials" of degrees $\leqslant k$ that are writable in terms of $x,y,e_1,e_2,\cdots,e_{k-1}$, possibly including coordinate-wise products as in the OP, by $P_{k-1}$.

For a partition $k=\ell+m$, define, in somewhat sloppy notation, the symmetric polynomial $$E_{(\ell,m)}:=\sum (x_1\cdots x_\ell y_{\ell+1}\cdots y_k +y_1\cdots y_\ell x_{\ell+1}\cdots x_k ).$$ So the sum is over all pairs of disjoint $\ell$- and $m$-tuples of indices. (If $\ell=m$, each monomial occurs twice, but we will count it only once.)
For instance, the following sum has $8$ monomials, and it can be checked that we can decompose it into polynomials of $ P_{3}$ as follows:

$$E_{(3,1)}=\sum (x_1x_2x_3y_4+y_1y_2y_3x_4)=\\ e_3(x)e(y)+e_3(y)e(x)-[e_2(x)+e_2(y)]e(xy) \\ +e(x)e(xxy)+e(y)e(yyx)-e(xxxy)-e(yyyx).$$ (BTW this splits of course into two non-symmetric identities, one which contains all terms featuring three instances of $x$ and one of $y$, and the other one switching $x\leftrightarrow y$.)

Now, taking $k=4$, it is easy to see that we have $$e_k(x+y)-e_k(x)-e_k(y)=E_{(3,1)}+E_{(2,2)}$$ but it seems to me that $$E_{(2,2)}=\sum x_1x_2y_3y_4$$ can not be decomposed into polynomials of $ P_{3}$ . Or am I wrong?

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