# Polynomials with roots in convex position

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).

The set $$\mathcal P_n$$ is clearly an open subset of all monic polynomials of degree $$n$$ over $$\mathbb R$$ or $$\mathbb C$$.

What is the geometry and topology of $$\mathcal P_n$$?

Facts:

(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 $$\alpha_0 > \alpha_1 > \cdots$$ and a sequence of signs $$\epsilon_0,\epsilon_1,\ldots,\epsilon_{n-2}\in\lbrace \pm 1\rbrace$$. Then
$$x^n+\sum_{k=0}^{n-2}\epsilon_k\alpha_k x^k\in \mathcal P_n(\mathbb R)$$ and different choices of signs correspond to different connected components. (Proof: Up to translation, we can assume that all roots sum up to $$0$$. If two polynomials of the above form are connected by a continuous path of real polynomials with roots summing up to $$0$$ , then all derivatives except the very last one do not have a root at $$0$$ (since it is otherwise in the interior of the convex hull of all roots). All coefficients up to degree $$d-2$$ are thus necessarily non-zero and different signs correspond to different connected components). Are there other connected components?

(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 $$\mathcal C_n=\mathcal P_n(\mathbb C)$$?

(3) We have the equalities $$\mathcal P_{n-1}=\lbrace P'/n \ |\ P\in\mathcal P_n\rbrace$$.

• Ī̲ have some ideas how to prove $\mathcal C_n=\mathcal P_n(\mathbb C)$ and, similarly, to make a retraction of the real case to components described by the OP. Still searching for a suitable compactification of $\mathcal P_n$ because Ī̲ detest to construct ε–δ proofs. – Incnis Mrsi Aug 24 '15 at 22:36
• I am not sure of why all these real polynomials are in different CCs. Do you pretend that if a coefficient of degree $\le n-2$ is zero, then $P$ is not in ${\mathcal P)_n$ ? That seems wrong for the constant coefficient. – Denis Serre Jul 5 '16 at 12:57
• I have sketched a proof in order to adress the the comment of Denis Serre. – Roland Bacher Jul 30 at 13:41