Characterizing intersection of zero sets of elementary symmetric polynomials on R^n - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:11:03Z http://mathoverflow.net/feeds/question/57859 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57859/characterizing-intersection-of-zero-sets-of-elementary-symmetric-polynomials-on-r Characterizing intersection of zero sets of elementary symmetric polynomials on R^n Nick 2011-03-08T17:08:25Z 2011-03-08T22:57:44Z <h2>Stated simply, the question is:</h2> <blockquote> <p>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 <code>$\{x\in\mathbb{R}^{n}: x_{i_{1}} = \dotsb = x_{i_{j}} = 0\}$</code> and $W_{n-k+1} = \cup_{i_{1} &lt; i_{2} &lt; \dotsb &lt; i_{n-k+1}}V_{i_{1}\dotsb i_{n-k+1}}$, the set of all points in $\mathbb{R}^{n}$ with at least $n-k+1$ coordinates equal to zero. Does $$(*) \qquad U_{k} \cap U_{k+1} = W_{n-k+1}$$ ? Why?</p> </blockquote> <p>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</p> <blockquote> <p>$$(**) \qquad \overline{\Gamma_{k}^{+}} \cap \overline{\Gamma_{k+1}^{+}} \subset W_{n-k+1}$$</p> </blockquote> <p>? It is well known (see "An Inequality for Hyperbolic Polynomials", Lars Garding) that $$(***)\qquad \Gamma_{k}^{+} \supset \Gamma_{k+1}^{+}$$ and that <code>$$(****)\qquad \{\sigma_{k+1} &gt; 0\} \cap \Gamma_{k}^{+} = \Gamma_{k+1}^{+}$$</code> 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.</p> <h2>Background and motivation:</h2> <p>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</p> <p>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 $x \in W_{n}$.</p> <p>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 $x \in W_{n-1}$.</p> <p>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</p> <p>$$\sigma_{k}(x^{2}) = \sigma_{k}^{2}(x)-2\sigma_{k-1}(x)\sigma_{k+1}(x)+2\sigma_{k-2}(x)\sigma_{k+2}(x)+\dotsb+(-1)^{k}2\sigma_{0}(x)\sigma_{2k}(x)$$</p> <p>for other values of $k$, the above approach fails in general.</p> <hr> <p>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.</p> http://mathoverflow.net/questions/57859/characterizing-intersection-of-zero-sets-of-elementary-symmetric-polynomials-on-r/57887#57887 Answer by Emmanuel Briand for Characterizing intersection of zero sets of elementary symmetric polynomials on R^n Emmanuel Briand 2011-03-08T22:16:48Z 2011-03-08T22:57:44Z <p>To establish that $U_k \cap U_{k+1} \subset W_{n-k+1}$:</p> <p>Consider the polynomial $P(t)=\prod (t-x_i)$. The elementary symmetric polynomials $\sigma_i$ are its coefficients, up to the sign.</p> <p>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:</p> <blockquote> <p>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 $Q$.</p> </blockquote>