13
$\begingroup$

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.

$\endgroup$
14
  • 2
    $\begingroup$ 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". $\endgroup$ Apr 15, 2015 at 20:56
  • 1
    $\begingroup$ $S_k(x+y)=\sum_{i=0}^k S_i(x)S_{k-i}(y)$, where $S_0=1$ by definition. This is trivial from a combinatorial point of view, and also the best you'll get. $\endgroup$ Apr 15, 2015 at 21:52
  • 1
    $\begingroup$ Hi @AlexanderWoo, this formula is not valid for the example in question or i am sleepy. $\endgroup$
    – vudu vucu
    Apr 15, 2015 at 22:53
  • 2
    $\begingroup$ .@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. $\endgroup$ Apr 16, 2015 at 4:45
  • 3
    $\begingroup$ $S_k$ for the elementary symmetric functions, what a horrible notation... $\endgroup$ May 7, 2015 at 17:58

4 Answers 4

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ May 7, 2015 at 19:58
  • $\begingroup$ Sigh! And maybe you can't peel because the whole thing is not symmetric any more. $\endgroup$ May 7, 2015 at 20:06
2
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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 $S_1$.

$\endgroup$
8
  • $\begingroup$ 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, 2015 at 19:00
  • $\begingroup$ Yes, I did. Also why can't both solutions be correct? $\endgroup$
    – vuur
    May 9, 2015 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, 2015 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, 2015 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, 2015 at 20:16
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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