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We define $H=\{u \in \mathbb{C}: \text{im}(u)>0\}$ to be the open upper half plane. For $n \in \mathbb{N}$ and $k = 1, \ldots n$ let $e_{n,k}(x_1,\ldots,x_n)=\sum\limits_{1 \leq j_1 < \ldots < j_k \leq n}~\prod\limits_{r=1}^k~x_{j_r}$ be the elementary symmetric polynomial of degree $k$ in $n$ variables. Show that any $u \in \mathbb{C}^n$ with $e_{n,k}(u_1,\ldots,u_n)=0$ does not lie in $H^n$.

This statement is easy to see for $e_{n,n}=\prod\limits_{r=1}^n x_r$, as any root $u$ of $e_{n,n}$ must have at least one $u_i=0$, and $0 \notin H$, thus $u \notin H^n$. Similarly for $e_{n,1}=\sum\limits_{r=1}^n x_r~$ and any $u \in H^n$ we have $e_{n,1}(u) \in H$ as $H$ is closed under addition. Thus $e_{n,1}(u) \neq 0$ as $0 \notin H^n$. It is also easy to see that the statement holds for $e_{n,k}$ iff it holds for $e_{n,n-k}$, as $H$ is closed under inversion $u \mapsto -\frac{1}{u}$.

But I still miss an argument to cover the remaining cases.

leon.fuchsler
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  • Atleast we can say difference between max angle and min angle of ${x_i : i \in [n]}$ must be greater than $\pi/k$. –  Aug 13 '23 at 05:34
  • A hypothesis/conjecture : If we assume $|x_i| = 1$ for all $i \in [n]$ then $e_{n,k} = 0 \iff$ ${\prod_{j=1}^k x_{i_j}}$ undergo antipodal cancellation in which case if the angle of $\prod_{j=1}^k x_{i_j}$ is $\theta$ then there is a product with angle $\theta \pm \pi$. –  Aug 13 '23 at 05:45

1 Answers1

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Let's do a proof by contradiction.

Your question is equivalent to showing that if a polynomial of degree n has a coefficient of zero then at least one of the roots does not lie in the upper half plane. This is because the coefficients of the powers of x in an nth degree polynomial are given by the elementary symmetric polynomial of the roots by Vieta's formulas (although possibly multiplied by a constant).

For example, for $n=3$: $(x-a)(x-b)(x-c) = x^3 - (a+b+c) x^2 + (ab+bc+ac) x - abc$

So, in order to show the contradiction we need:

(1) The Gauss-Lucas Theorem states that: If P is a (nonconstant) polynomial with complex coefficients, all zeros of P' (derivative of P) belong to the convex hull of the set of zeros of P. (From Wikipedia)

and

(2) Obviously, if all the roots of P are in the UHP (upper half plane) then the convex hull of the roots is in the UHP.

So, finally, if we repeatedly take derivatives of our polynomial we will eventually get to the term with coefficient equal to 0 so that our polynomial has no constant term. At this point it is obvious that 0 is a root to this polynomial because we can factor out an x. Since 0 is not in the UHP this provides the contradiction.

TheJack
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