Polynomials with roots in convex position - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T20:46:22Zhttp://mathoverflow.net/feeds/question/26799http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/26799/polynomials-with-roots-in-convex-positionPolynomials with roots in convex positionRoland Bacher2010-06-02T08:21:03Z2010-06-02T10:18:29Z
<p>Let $\mathcal P_n$ denote the set of all monic polynomials of degree $n$ with real or complex coefficients such that $P\in\mathcal P_n$ if for all $k\in\lbrace 0,1,\dots,n-2\rbrace$ the $n-k$ roots of $P^{(k)}=\left(\frac{d}{dx}\right)^kP$ are in strictly convex position (ie are the $n-k$ vertices of a convex polygon with $n-k$ extremal vertices).</p>
<p>The set $\mathcal P_n$ is clearly an open subset of all monic polynomials of degree $n$ over $\mathbb R$ or $\mathbb C$.</p>
<p>What is the geometry and topology of $\mathcal P_n$?</p>
<p>Facts:</p>
<p>(1) Over the reals, $\mathcal P_n$ has at least $2^{n-1}$ connected components: Indeed, consider a very fast decreasing sequence (I guess $n\longmapsto 1/((1+n)^{(1+n)^{1+n}})!$ will probably work) of strictly positive reals $\alpha_0>\alpha_1\dots$ and a sequence of signs $\epsilon_0,\epsilon_1,\dots\in\lbrace \pm 1\rbrace^{\mathbb N}$. Then<br>
$$x^n+\sum_{k=0}^{n-2}\epsilon_k\alpha_k x^k\in \mathcal P_n(\mathbb R)$$
and different choices of signs correspond to different connected components. Are there other connected components?</p>
<p>(2) Over the complex numbers, all the polynomials described in (1) are in the same connected component. Choosing $\epsilon_i$ on the complex unit circle suggests however that $\pi_1(\mathcal C_n)$ might be $\mathbb Z^{n-1}$ where $\mathcal C_n$ denotes the connected component of $x^n+\sum_{k=0}^{n-2}\alpha_kx^k$ in $\mathcal P_n(\mathbb C)$. Do we have $\mathcal C_n=\mathcal P_n(\mathbb C)$?</p>
<p>(3) We have the equalities $\mathcal P_{n-1}=\lbrace P'/n|P\in\mathcal P_n\rbrace$.</p>