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I am working on a system of complex equations The question is the following:

Let $a_1,a_2,\ldots,a_N\in \mathbb{C}$ such that $$\sum_{j=1}^N \sum_{q=0}^{N-1-k} {N-1 \choose q} {N-1 \choose k+q} |a_j|^{2N-2k-2q-2} a_j^{k}=0,$$ for all $k=1,2,\ldots,N-1.$ Prove that $$e_1(a_1,a_2,\ldots,a_N)=e_2(a_1,a_2,\ldots,a_N)=\ldots=e_{N-1}(a_1,a_2,\ldots,a_N)=0,$$ where $e_k$ is an elementary symmetric polynomial of degree $k.$

I do not have much knowledge on Newton's identities. Could somebody help me prove it or give me some helpful references to read? I will be very appreciated. Thank you so much.

Suracha.

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  • $\begingroup$ Why do you think that these equations hold? $\endgroup$ – Ben McKay Mar 19 '16 at 20:14
  • $\begingroup$ Thank you for your interest in this question. I solved this question for small $N,$ for example, N=3 and $N=4.$ I believe that it is true in general. $\endgroup$ – Suracha Bosunoi Mar 19 '16 at 20:21
  • $\begingroup$ These equations appear when I solve another geometric problem. The expected solution of this geometric problem is that $a_1,a_2,\ldots,a_N$ are vertices of a regular $N$ gon centered at $0.$ $\endgroup$ – Suracha Bosunoi Mar 21 '16 at 12:34
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First it's more convenient to write these in the generating function form. Multiply by $s^k$, where $s$ is a new variable and sum over all $k$. Using the binomial identity two times we arrive at $F(s)=\sum_{j=1}^N (s+\bar{a}_j)^{N-1}(1/s+a_j)^{N-1}$. And the assumptions mean that the coefficients of $F$ at positive powers of $s$ vanish. However using the symmetry between $a_j$ and $\bar{a_j}$ the coefficients at negative powers of $s$ should also vanish. So the assumption is equivalent to $F(s)$ being constant $F(s)=\mathrm{const}$.

I believe this is the identity you had in the first place. Now we want to study it algebraically. This means that we should treat $a_j$ and $\bar{a_j}$ as independent variables. So we substitute $a_j=x_j$ and $\bar{a_j}=y_j$. We have $F(s)=\sum_{j=1}^N (1/s+x_j)^{N-1} (s+y_j)^{N-1}=\mathrm{const}$ and your question is: does it imply

\begin{equation}(*)\quad e_i(x_1,\ldots,x_N)=0 \quad (i=1,2,\ldots,N-1)? \end{equation} By symmetry it would also imply \begin{equation}(**)\quad e_i(y_1,\ldots,y_N)=0 \quad (i=1,2,\ldots,N-1). \end{equation}

Let's do dimension analysis first. The condition $F(s)=\mathrm{const}$ is $2N-2$ equations in $2N$ variables. So the set of solutions (call it $X$) is an algebraic set of dimension $\geq 2$. On the other hand the conditions $e_i(x_1,\ldots,x_N)=0$ are equivalent to $x_i$ being roots of the polynomial $t^N-A$ for some $A$. Let's replace $A$ by $A^N$. So we have $x_k=\zeta^{\pi_k} A$ for some permutation $\pi$ ($\zeta$ is the primitive $N$-th root of unity). Now for the $y$s we have $y_k=\zeta^{\sigma_k} B$ for a permutation $\sigma$. So the set of solutions to the conditions (*), (**) (call it $Y$) is a union of $2$-dimensional algebraic sets: one set for each pair of permutations $\pi, \sigma$.

It is not true that $X\subset Y$. Even for $N=3$ there is an irreducible component of $X$ that is not in $Y$, which you can find using sage, for instance.

On the other hand it is true that some components of $X$ and $Y$ coincide. For instance $x_k=\zeta^k A$ and $y_k=\zeta^{-k} B$ ($A$, $B\in\mathbb{C}$) defines a $2$-dimensional algebraic set that belongs to both $X$ and $Y$. This is the set of solutions to the geometric problem you mentioned.

So I believe the picture is as follows: The algebraic set $X$ has several irreducible components, some of which belong to $Y$, the others don't. It may happen that the other components do not contain points satisfying $x_i=\bar y_i$, so they don't produce solutions to the geometric problem. But from the purely algebraic point of view the statement is wrong, so we do not obtain some kind of generalized Newton identities.

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  • $\begingroup$ Thank you very much for your answer. I am not sure if I understand you correctly. Do you mean this question is not true even $N=3$? $\endgroup$ – Suracha Bosunoi Mar 28 '16 at 6:11
  • $\begingroup$ I'm not saying that your statement is not true for $N=3$. Newton identities have purely algebraic proof, i.e. using only operations like +, -, *, /, and I am saying that your statement, if it is true, cannot have purely algebraic proof, because in algebraic geometry the analogous statement is false. $\endgroup$ – Anton Mellit Mar 29 '16 at 21:39

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