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In some work on QFT the following identity has come up: $$ \sum_{\sigma \in S_n}\sum_{j=1}^n \left(\sum_{l=1}^j w_{\sigma_l}\right)\prod_{i=1,i\neq j}^{n}\frac{1}{\sum_{l=1}^j z_{\sigma_l}-\sum_{l=1}^i z_{\sigma_l}}=0 $$ for $2<n$ where $w_1,...,w_n$ are arbitrary numbers and $z_1,...,z_n$ are such that $$ \sum_{z\in S}z\neq 0,\ \ \forall S\subsetneq \{z_1,...,z_n\} $$ I have verified this symbolically for $2<n<6$ and using specific values for $w, z$ I have verified it up to $n=8$. Do you have any ideas on how to approach a proof?

An idea I've thought of so far: We could write a function $$ f(z)=\sum_{\sigma\in S_n}\prod_{i=1}^{n}\frac{1}{z-\sum_{l=1}^i z_{\sigma_l}} $$ The integral of $f(z)$ around $\infty$ is then clearly 0 since it dies off fast enough. We can also express this integral as a sum over residues and we find almost the above identity but without the weights $w_l$. In this case the identity is true also without the sum over permutations but somehow the sum over permutations allows for the weights $w_l$.

For convenience, here is a mathematica expression of the identity:

Sum[
    Sum[
        Sum[w[p[[l]]], {l, 1, j}]
            Product[
                If[i == j,
                    1, 
                    1/(Sum[z[p[[l]]], {l, 1, j}] - Sum[z[p[[l]]], {l, 1, i}])],
                {i,n}],
            {j, n}],
        {p, Permutations[Range[n]]}]
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1 Answer 1

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June 18, 2019 Edit: The identity is now proved and generalized. See below.

I did not have time to look at your identity for very long but I am pretty sure it follows from Lemma II.2 in my article "Trees, forests and jungles: a botanical garden for cluster expansions" with Vincent Rivasseau.

A starting point. Make a set of $n+1$ elements by adding a root numbered say $0$. Define the edge variables $u_{ij}$ $(i<j)$ to be zero except $u_{0,i}=z_i$. An ordered forest of length $\tau$, i.e., $(\{i_1,j_1\},\{i_2,j_2\},\ldots,\{i_{\tau},j_{\tau}\})$ which contributes in Lemma II.2 namely for which $$ \prod_{\nu=1}^{\tau}u_{i_{\nu},j_{\nu}}\neq 0 $$ basically is sequence of distinct elements $(j_1,\ldots,j_{\tau})$ in $\{1,2,\ldots,n\}$. And for maximal length $\tau=n$, this is just a permutation $\sigma$.


Not quite there yet, although it looks promising.

I use formula (II.6) in Lemma II.2 of my article for the set $\{0,1,\ldots,n\}$ instead of $\{1,\ldots,n\}$. In the notations of the paper, I define the edge variables $u_{\{ij\}}$ by $u_{\{0i\}}=z_i$ for $1\le i\le n$ and zero otherwise. I also define the edge variables $v_{\{ij\}}$ by $v_{\{0i\}}=y_i$ for $1\le i\le n$ and $1$ otherwise. I will rename $\nu$ from the paper and now call it $j$. I will also rename $\mu$ from the paper and now call it $i$. An ordered forest of length $\tau$ is a sequence of edges $(l_1,\ldots,l_{\tau})$. In order to contribute it has to be of the form $l_a=\{0,\sigma(a)\}$ for $\sigma=(\sigma(1),\ldots,\sigma(\tau))$ a sequence of distinct elements of $\{1,2,\ldots,n\}$ of length $\tau$. With this harmonization of notations, Lemma II.2 from the paper gives $$ y_1\cdots y_n=\sum_{\tau=0}^{n}\sum_{\sigma}z_{\sigma(1)}\cdots z_{\sigma(\tau)} \sum_{j=0}^{\tau}\frac{y_{\sigma(1)}\cdots y_{\sigma(j)}}{\prod_{i=0,i\neq j}^{\tau} (Z_{j}^{\sigma}-Z_{i}^{\sigma})} $$ where for $k=0,1,\ldots,\tau$, $Z_{k}^{\sigma}=z_{\sigma(1)}+\cdots+z_{\sigma(k)}$. Might as well define also for $k=0,1,\ldots,\tau$, $W_{k}^{\sigma}=w_{\sigma(1)}+\cdots+w_{\sigma(k)}$.

I will then carefully separate the cases where $i$ or $j$ are zero from the rest.

$$ y_1\cdots y_n=\sum_{\tau=0}^{n}\sum_{\sigma}z_{\sigma(1)}\cdots z_{\sigma(\tau)} \sum_{j=1}^{\tau}\frac{y_{\sigma(1)}\cdots y_{\sigma(j)}}{Z_{j}^{\sigma}\prod_{i=1,i\neq j}^{\tau} (Z_{j}^{\sigma}-Z_{i}^{\sigma})} $$ $$ +\sum_{\tau=0}^{n}\sum_{\sigma}z_{\sigma(1)}\cdots z_{\sigma(\tau)} \frac{1}{\prod_{i=1}^{\tau} (-Z_{i}^{\sigma})} $$ Now I set $y_i=e^{sz_i+tw_i}$. I take $\frac{\partial^2}{\partial s\partial t}$ and set $s=t=0$. This gives the identity $$ (w_1+\cdots+w_n)(z_1+\cdots+z_n)= \sum_{\tau=0}^{n}\sum_{\sigma}z_{\sigma(1)}\cdots z_{\sigma(\tau)} \sum_{j=1}^{\tau}\frac{W_{j}^{\sigma}}{\prod_{i=1,i\neq j}^{\tau} (Z_{j}^{\sigma}-Z_{i}^{\sigma})} $$ Now I am stuck. I would like to "take the coefficient" of $z_1\cdots z_n$ which would somehow force $\tau=n$ and $\sigma$ to be a permutation but I don't see it yet.


It's a bit late for me so can't give details but I think I know how to peal off $\tau=n$. Use Möbius inversion in the (Boolean) poset of subsets $I=\{i_1<\cdots<i_k\}$ of $[n]=\{1,\ldots,n\}$. Namely apply the identity I just proved not just for the whole set $\{1,\ldots,n\}$ and corresponding variables, but also for all the $I$'s.

Really really have to go but basically the wanted result amounts to showing $$ \sum_{I\subset [n]} (-1)^{n-|I|}\left(\sum_{i\in I} w_i\right) \left(\sum_{i\in I} z_i\right)=0 $$ when $n\ge 3$.


OK here is the endgame.

I will denote by $|\cdot|$ the cardinality of a finite set. For $I\subset[n]$, let $$ B_I=\sum_{\sigma}z_{\sigma(1)}\cdots z_{\sigma(|I|)} \sum_{j=1}^{|I|}\frac{W_{j}^{\sigma}}{\prod_{i=1,i\neq j}^{|I|} (Z_{j}^{\sigma}-Z_{i}^{\sigma})} $$ where the sum is over sequences $\sigma=(\sigma(1),\ldots,\sigma(|I|))$ which form a permutation of the elements of $I$. Also define $$ A_I=\sum_{J\subset I}B_J\ . $$ Then the previous identity applied to the variables in $I$ instead of the whole set $[n]$, give $$ A_I=\left(\sum_{a\in I} w_a\right) \left(\sum_{b\in I} z_b\right)\ . $$ By Möbius inversion in the Boolean lattice, we get $$ B_I=\sum_{J\subset I}(-1)^{|I|-|J|}A_J\ . $$ The LHS in the OP's identity is $B_{[n]}$ divided by $z_1\cdots z_n$. However, $$ B_{[n]}=\sum_{J\subset[n]}(-1)^{n-|J|} \left(\sum_{a\in J} w_a\right) \left(\sum_{b\in J} z_b\right) $$ $$ =\sum_{a,b\in [n]}w_a z_b\sum_{J,\{a,b\}\subset J\subset[n]} (-1)^{n-|J|}=\sum_{a,b\in [n]}w_a z_b(1-1)^{n-|\{a,b\}|}=0 $$ by Newton's binomial theorem and the hypothesis $n\ge 3$. QED

Note that the OP's identity admits a trivial generalization to families of weights $w^{(1)},\ldots,w^{(k)}$, namely $$ \sum_{\sigma\in\mathfrak{S}_n} \sum_{j=1}^{n}\frac{W_{j}^{(1),\sigma}\cdots W_{j}^{(k),\sigma}}{\prod_{i=1,i\neq j}^{n} (Z_{j}^{\sigma}-Z_{i}^{\sigma})}\ =\ 0 $$ as soon as $n\ge k+2$. Here I used the generalized notation $W_{j}^{(r),\sigma}=w_{\sigma(1)}^{(r)}+\cdots+w_{\sigma(j)}^{(r)}$.

Finally more information about identities like in Lemma II.2 of my article with Rivasseau can be found in my dedicated webpage (in dire need of updates though).

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  • $\begingroup$ Great thanks! I think I see what you mean, but will need a bit more time to process it. For the $v_{ij}$ I make a similar choice $v_{0i}=w_i+X$ and the rest 0, then I can look at the $X^{n-1}$ term on both sides which is 0 on the LHS and $\sum_{l=1}^j w_{\sigma{l}}$ on the RHS. I'll just go over all the details and make sure it is correct then accept this as the answer. $\endgroup$
    – Petter
    Jun 17, 2019 at 18:11
  • $\begingroup$ I am working on it too. Your identity is quite interesting. Which part of QFT did it come from? Scattering equations? MHV type computations? $\endgroup$ Jun 17, 2019 at 18:13
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    $\begingroup$ Ah, I initially misunderstood formula II.6, thinking that it was true for a fixed $\tau$, though now I see $\tau$ is also summed over and it is more difficult to use. Yes, this is indeed close. Why is the LHS $y_1...y_n$ and not 0 here? The product is over all pairs and if I understand correctly many $v_{ij}$ are 0. Perhaps it can now be solved by induction in $n$? $\endgroup$
    – Petter
    Jun 17, 2019 at 23:08
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    $\begingroup$ I'm considering symmetrized fermion loop diagrams for a Fermi surface in 2 dimensions as studied here: math.ubc.ca/~feldman/papers/loops.pdf A cancellation at small momenta was noted here arxiv.org/pdf/1509.07783.pdf journals.aps.org/prb/abstract/10.1103/PhysRevB.96.155125 I've now generalized this a bit and then this identity showed up. $\endgroup$
    – Petter
    Jun 17, 2019 at 23:21
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    $\begingroup$ I've now accepted this as answer. Sorry about the delay, it took me some time to understand the proof. The generalization is nice. $\endgroup$
    – Petter
    Jun 19, 2019 at 13:55

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