Stated simply, the question is:
-------------------------------



> 
> Consider two elementary symmetric polynomials $\sigma\_{k}$ and $\sigma\_{k+1}$ on $\mathbb{R}^{n}$ with zero sets $U\_{k}$ and $U\_{k+1}$. Let $V\_{i\_{1}i\_{2}\dotsb i\_{j}}$ be the coordinate linear space $\{x\in\mathbb{R}^{n}: x\_{i\_{1}} = \dotsb = x\_{i\_{j}} = 0\}$ and $W\_{n-k+1} = \cup\_{i\_{1} < i\_{2} < \dotsb < i\_{n-k+1}}V\_{i\_{1}\dotsb i\_{n-k+1}}$, the set of all points in $\mathbb{R}^{n}$ with at least $n-k+1$ coordinates equal to zero. Does
>  \begin{equation}
> (\\*) \qquad U\_{k} \cap U\_{k+1} = W\_{n-k+1}
> \end{equation}
>  ? Why?
> 
> 
> 


Note: I am interested in a slightly more specific question. Namely, if $\Gamma\_{k}^{+}$ is the component of $\sigma\_{k} > 0$ containing the point $(1,1,\dotsb,1)$, then is



> 
> \begin{equation}
> (\\*\\*) \qquad \overline{\Gamma\_{k}^{+}} \cap \overline{\Gamma\_{k+1}^{+}} \subset W\_{n-k+1}
> \end{equation}
> 
> 
> 


? It is well known (see "An Inequality for Hyperbolic Polynomials", Lars Garding) that 
\begin{equation}
(\\*\\*\\*)\qquad \Gamma\_{k}^{+} \supset \Gamma\_{k+1}^{+}
\end{equation}
and that
`\begin{equation}
(****)\qquad \{\sigma_{k+1} > 0\} \cap \Gamma_{k}^{+} = \Gamma_{k+1}^{+}
\end{equation}`
If $(\\*)$ is true, then $(\\*\\*)$ follows. The equation $(\\*)$ is simpler to state and appears to be true. Of course $(\\*\\*)$ may hold with $(\\*)$ failing, but I think this is unlikely.


Background and motivation:
--------------------------


Numerical simulations in mathematica suggest that $(\\*)$ holds.
Certainly $W\_{n-k+1} \subset U\_{k} \cap U\_{k+1}$.
I can prove the reverse inclusion in certain cases. For example, denoting $x^{j} = (x\_{1}^{j},\dotsb,x\_{n}^{j})$ for $x\in\mathbb{R}^{n}$, notice that $W\_{n-k+1}$ is exactly the zero set of $\sigma\_{k}(x^{2})$. So we have


n arbitrary, k=1: From Newton's identities (see wikipedia), $\sigma\_{1}^{2}(x) = \sigma\_{1}(x^{2})+2\sigma\_{2}(x)$. So if $\sigma\_{1}(x) = \sigma\_{2}(x) = 0$, then $\sigma\_{1}(x^{2}) = 0$ which means $x = (0,\dotsb,0)$ so that $x \in W\_{n}$.


n arbitrary, k = 2, $(\\*\\*)$ only: can be proved similarly to $k=1$ by writing $0 \le \sigma\_{2}(x^{2}) = \sigma\_{2}^{2}(x) -2\sigma\_{1}(x)\sigma\_{3}(x)+2\sigma\_{4}(x)$. So if $\sigma\_{2}(x) = \sigma\_{3}(x) = 0$, then $\sigma\_{2}(x^{2}) = 2\sigma\_{4}(x)$. Now, $\sigma\_{4}(x) \le 0$ because of $(\\*\\*\\*)$ and $(\\*\\*\\*\\*)$. Hence $\sigma\_{2}(x^{2}) = 0$ so that $x \in W\_{n-1}$.


n arbitrary, $k=n-2$ and $k =n-1$ can be proved in a similar fashion. However, because of the additional terms in the equation


\begin{equation}
\sigma\_{k}(x^{2}) = \sigma\_{k}^{2}(x)-2\sigma\_{k-1}(x)\sigma\_{k+1}(x)+2\sigma\_{k-2}(x)\sigma\_{k+2}(x)+\dotsb+(-1)^{k}2\sigma\_{0}(x)\sigma\_{2k}(x)
\end{equation}


for other values of $k$, the above approach fails in general.




---


I'm actually trained as a differential geometer, so I may be approaching this problem in the wrong way (perhaps there's a technique in real algebraic geometry?). If you can suggest an alternative approach I would be grateful to hear it. This result seems rather elementary to state so I would be surprised if the result is not known. Thank you.



Characterizing intersection of zero sets of elementary symmetric polynomials on R^n
To establish that $U\_k \cap U\_{k+1} \subset W\_{n-k+1}$:


Consider the polynomial $P(t)=\prod (t-x\_i)$. The elementary symmetric polynomials $\sigma\_i$ are its coefficients, up to the sign.


Suppose that $\sigma\_k=\sigma\_{k+1}=0$. This means that $0$ is a multiple root of the derivative of order $(n-k-1)$ of $P$. Now you can conclude that $0$ is a root of $P$ of multiplicity at least $n-k+1$ (that is: at least $n-k+1$ of the numbers $x\_i$ are zero) by means of the following facts:



> 
> If a non-constant polynomial $Q$ has
>  all its roots real, then so does its
>  derivative $Q'$. Then the roots of
>  $Q'$ are the multiple roots of $Q$,
>  plus one root between each pair of
>  consecutive distinct roots of $Q$,
>  necessarily simple (otherwise the sum
>  of the multiplicities of the roots of
>  $Q'$ would exceed its degree). In
>  particular, if $x$ is a multiple root
>  of $Q'$ then it is also necessarily a
>  root of $Q$.
> 
> 
> 


