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The question we are considering concerns a sum over all permutation $\sigma \in S_n$ (symmetric group) of a certain rational function:

$$\sum_{\sigma \in S_n} \frac{\sigma_{j1} \sigma_{j2}... \sigma_{jk}}{\prod_{i=1}^{n-1} (\sigma_i - \sigma_{i+1})} \stackrel{?}{=} 0,$$ with $k < n-1$ and $jm \in \{1,2,...,n\} \ \forall m \in \{1,..,k\}$.

It is easy to see that the sum is equal to zero if there is a constant in the numerator by adding a term in the denominator, such that it becomes invariant under cyclic permutations, i.e.$$\sum_{\sigma \in S_n} \frac{1}{\prod_{i=1}^{n-1} (\sigma_i - \sigma_{i+1})} = \sum_{\sigma \in S_n} \frac{(\sigma_n - \sigma_{1})}{\prod_{i=1}^{n-1} (\sigma_i - \sigma_{i+1}) (\sigma_n - \sigma_{1})} = 0 \, .$$ The question is whether this sum is still zero if we include the product of at most $n-2$ elements of the form $\sigma_{jm}$. The sum is unequal to zero for $k > n-2$.

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  • $\begingroup$ $jm$ being a way to say $j_m$, or an actual product? $\endgroup$ Aug 25, 2016 at 21:01
  • $\begingroup$ sorry for confusion, it should be $j_m$. $\endgroup$
    – Kathi
    Aug 26, 2016 at 7:43

2 Answers 2

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Perhaps you wish to look at the paper

Tewodros Amdeberhan, Explicit computations with the Divided Symmetrization operator, Proc. Amer. Math. Soc., 144 no 7 (2016) pp 2799-2810, doi:10.1090/proc/12931 arXiv:1406.0447

In particular, Proposition 0 is what you seek for.

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  • $\begingroup$ Equivalently (I assume), arXiv:1406.0447v1. I'm not sure how long the IP stays around :) $\endgroup$ Aug 25, 2016 at 21:00
  • $\begingroup$ @darijgrinberg: good point. :-) $\endgroup$ Aug 26, 2016 at 14:04
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    $\begingroup$ Well, since it hasn't been edited in three years, I thought I might as well do it, since the direct IP address link no longer works.... $\endgroup$
    – David Roberts
    Aug 15, 2019 at 7:52
  • $\begingroup$ @DavidRoberts Thank you! $\endgroup$ Aug 15, 2019 at 20:31
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Yes, it equals to 0 if $k<n-1$. Actually it is always a polynomial in $\sigma$'s (any singularity $\sigma_m-\sigma_k$ in the denominator cancels out when we couple summands in appropriate way). It's degree is (at most) $k-n+1$, thus the result.

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