# sum over permutations equals zero?

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 $k > n-2$.

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

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.