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Let $n$ be a positive integer and $S_n$ be the symmetric group on $\{1,2,\ldots,n\}$. For any $w\in S_n$ and polynomial $f\in \mathbb{R}[x_1,x_2,\ldots,x_n]$, denote $w(f)=f(x_{w(1)},x_{w(2)},\ldots,x_{w(n)})$. Furthermore, define $$\alpha(f)=\sum\limits_{w\in S_n}\epsilon(w)w(f),$$ where the sum is over all the $w\in S_n$ and $\epsilon(w)$ denotes the sign of the permutation $w$.

It is easy to see that for any $f\in \mathbb{R}[x_1,x_2,\ldots,x_n]$, $\alpha(f)$ is divisible by $\prod\limits_{1\leq i<j\leq n}(x_i-x_j).$ So when $f$ is a homogeneous polynomial of degree $\frac{n(n-1)}{2}$ , we have $$\alpha(f)=a_f\prod\limits_{1\leq i<j\leq n}(x_i-x_j)$$ for some $a_f\in \mathbb{R}$ .

Suppose $f$ is a homogeneous polynomial of degree $\frac{n(n-1)}{2}$, how to know whether $a_f$ is $0$ or not when the expansion $$f(x_1,x_2,\ldots,x_n)=\sum\limits_{i_1+i_2+\cdots+i_n=\frac{n(n-1)}{2}}a_{i_1,i_2,\ldots,i_n}x_1^{i_1}x_2^{i_2}\cdots x_n^{i_n}$$ is not easy to get?

For example, let $n=5$ and $$f(x_1,x_2,x_3,x_4,x_5)=(x_1+x_2)^2(x_2+x_3)^2(x_3+x_4)^2(x_4+x_5)^2(x_5+x_1)^2,$$ then $f$ is homogeneous polynomial of degree $10$.

With the expansion of $f$ expanded by the computer, I can calculate that $a_f$ is nonzero. But how to get this without the expansion of $f$?

Maybe it is hard to answer my question in general, so I hope someone can answer my question for the specific polynomial I gave above.

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    $\begingroup$ I am afraid that this is hard in general. But you may try to ask for some specific polynomials. $\endgroup$
    – Fedor Petrov
    Commented Mar 22, 2016 at 9:57
  • $\begingroup$ Fedor Petrov: yes, I know it is hard to answer in general, so I gave the specific polynomial above and I hope someone can answer my question for this polynomial. $\endgroup$
    – user173856
    Commented Mar 22, 2016 at 13:54
  • $\begingroup$ Can you elaborate on what can be expected to be known about $f$? E.g: do you know a factorization, and expansions of the factors (as in your example)? Can you evaluate $f$ at (finite) sets of points? $\endgroup$
    – pinaki
    Commented Mar 22, 2016 at 15:11
  • $\begingroup$ auniket: Yes, I want to know whether $a_f$ is 0 or not for $f$ which can factor as a product of polynomials of degree $1$ as follow:$$f=\prod_{k=1}^{\frac{n(n-1)}{2}}(x_{i_k}+x_{j_k}).$$ $\endgroup$
    – user173856
    Commented Mar 22, 2016 at 15:44
  • $\begingroup$ just linking to this related MO question mathoverflow.net/questions/56627/… $\endgroup$
    – Abdelmalek Abdesselam
    Commented Mar 26, 2016 at 15:31

2 Answers 2

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As for your polynomial $f=\prod_{i=1}^5 (x_i+x_{i+1})^2$, you may do the following trick. At first, you replace your polynomial to $g=\prod_{i=1}^5 (x_i+x_{i+1}-4)(x_i+x_{i+1}-3)$. Antisymmetrizations of $f$ and $g$ are the same, since their difference has degree less than 10. Next, we use points 0,1,2,3,4 as $x$'s. I think, there is unique (up to dihedral group symmetry) permutation 10234 for which $g$ is non-zero. Hence all non-zero summands in the sum $\sum {\rm\,sign}\,(\pi) g(x_{\pi_1},x_{\pi_2},\dots,x_{\pi_5})$ are equal, so this sum is non-zero.

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  • $\begingroup$ Nice idea. I wish however bring a correction. After the remark that we may always restrict to permutations fixing $0$, because of the cyclic invariance of $f$, there remains two permutations $\pi$ for with $g^\pi$ does not vanish at the point $(0,1,2,3,4)$, namely the transposition $(24)$ and the $4$-cycle $(1,2,3,4)$. Fortunately, they give the same contribution, with the same signature. $\endgroup$
    – Denis Serre
    Commented Mar 22, 2016 at 15:56
  • $\begingroup$ of course, these are the same cycle 10234 in two orders, totally 10 summands $\endgroup$
    – Fedor Petrov
    Commented Mar 22, 2016 at 16:04
  • $\begingroup$ @Denis In other words, you mention cyclic invariance, but really there is dihedral $D_5$-invariance, and up to this invariance there is unique non-zero summand. $\endgroup$
    – Fedor Petrov
    Commented Mar 22, 2016 at 17:54
  • $\begingroup$ Fedor Petrov: Thanks for your answer! But I think your method is not suitable for big $n$. Do you still have any method to deal with similar problems when $n$ is big? By the way, is the conception "antisymmetrization" for a polynomial you mentioned above a defined mathematical concept? $\endgroup$
    – user173856
    Commented Mar 26, 2016 at 7:52
  • $\begingroup$ Sometimes it does work. I do not understand which specific polynomial do you consider for large $n$. Antisymmetrization is your $\alpha (f)$. $\endgroup$
    – Fedor Petrov
    Commented Mar 26, 2016 at 8:07
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Because $a_f$ is the coefficient of $x_1^{n-1}\cdots x_{n-2}^2x_{n-1}$ in $\alpha(f)$, you find that $$a_f=\sum_{\sigma\in S_n}\epsilon(\sigma)a_{\sigma(n)-1,\ldots,\sigma(1)-1}.$$ The other coefficients $a_{i_1,\ldots,i_n}$ do not contribute. Essentially because two exponents $i_k$ are repeated and therefore are cancelled in the summation, thanks to transpositions.

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  • $\begingroup$ Well, $x_2^{n-1} x_1^{n-2}x_3^{n-3}\cdots x_{n-1}$ also contributes $\endgroup$
    – Venkataramana
    Commented Mar 22, 2016 at 11:59
  • $\begingroup$ @Venkataramana. This is exactly what I say. Your monomial corresponds to the transposition $\tau=(n-1,n)$. $\endgroup$
    – Denis Serre
    Commented Mar 22, 2016 at 12:13
  • $\begingroup$ Denis Serre: Thank you for your answer. I knew what you said, but sometimes we can not obtain the expansion of $f$. So I asked how to know whether $a_f$ is zero or not without the expansion of $f$. Maybe it is hard to answer in general, so I gave a specific polynomial above and I hope someone can answer my question for this polynomial. $\endgroup$
    – user173856
    Commented Mar 22, 2016 at 14:10

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