4 improved presentation of 2.1 2.2, undated the link of scilab code.

If we let L be the set of $v$'s such that three of the roots of $f_v$ lie on a line, we get that $Hull(P)=\bigcap_{v\in L}Conv(f_v)$ if $deg f≥3$.

• If for some fixed complex number $t\neq 0,1$,the holomorphic function $v\mapsto t a(v) + (1-t)b(v)$ is a constant $c$, then the polynomial has a rotational symmetry about $c$.And $c$ is a critical point of $f$. f$and$f$has a rotational symmetry about$c$. • If the segment$[a_0, b_0]$is a boundary edge of the polygon$Conv(f_{v_0})$(in particular and no other point in the line through$a_0, b_0$ismapped to$v_0$), v_0$, then

• As a consequence Comparing the coefficients we conclude that $\dfrac t{t-1}$ is a root of unity and $f'(0)=\dfrac{t}{t-1} f'(0)$. points of $f^{-1}(v_0)$ (and there is at least one ) due to the assumptionon the degree of $f$ and on the simplicity of $a_0, b_0$ as roots of $f_{v_0}$) are on one side of the line through $a_0, b_0$.

$z_0$ is on the right of the segment $[a(v(s)), b(v(s))]$ passes through $z(v(s))$ and remains almost vertical for sufficiently small $s$, so by has $z_0$ on its right side. By 2.1 we have may conclude
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I. First I want to share some computer experiments of H.H. Rugh. The following image supports the positive answer of the

QUESTION: is this hexagon equal to $Hull(X^3−1)$?

See the image

triangle (as a new user I was not allowed to use image tags).

A scilab program testing this problem can be found at roots-dancing. In particular an example $z^6-3z^3+z$ similar to the starting example of Zeb, showing that $Hull(P)$ can be strictly smaller then $\bigcap_vConv(f_v)$ for $v$ ranging the critical values of $f$ (shown in blue polygons). See smaller.

II. Here is a proof (communication of Rugh) of the same statement of Zeb (with $f=\Pi$):

So if

If we let L be the set of $v$'s such that three of the roots of $f_v$ lie on a line, we get that $Hull(P)=\bigcap_{v\in L}Conv(f_v)$ if $deg f≥3$.

The underlying idea is very similar to that of Zeb.

Statements: Let $f$ be a polynomial of degree at least 3. Assume that $a_0$ and $b_0$ are two distinct simple roots of $f(z)-v_0$. Then for $v$ in a small neighborhood of $v_0$, there are two simple roots $a(v)$, $b(v)$ of $f(z)-v$ with $a(v_0)=a_0$ and $b(v_0)=b_0$. In this case,

1. If for some complex $t\neq 0,1$, the holomorphic function $v\mapsto t a(v) + (1-t)b(v)$ is constant $c$, then the polynomial has a rotational symmetry about $c$. And $c$ is a critical point of $f$.

2. If the segment $[a_0, b_0]$ is a boundary edge of the polygon $Conv(f_{v_0})$ (in particular no other point in the line through $a_0, b_0$ is mapped to $v_0$), then

2.1 for $v$ sufficiently close to $v_0$, the segment $[a(v),b(v)]$ is a boundary edge of the polygon $Conv(f_{v})$, and for any $t\in ]0,1[$, the map $v\mapsto ta(v)+(1-t) b(v)$ is an open mapping.

2.2. The line through $a_0, b_0$ is outside $\bigcap_v Conv(f_{v})$.

Proof. 1. Replacing $f(z)$ by $f(z-c)$ if necessary, we may assume $c=0$. Note that $a\mapsto b(f(a))$ is defined and holomorphic in a neighborhood of $a_0$, satisfying that $b(f(a_0))=b_0$ and $f(b(f(a)))=f(a)$. It follows that $ta+(1-t)b(f(a))\equiv 0$ so $b(f(a))=\dfrac{ta}{t-1}$. Therefore $f(a)=f(\dfrac {ta}{t-1})$ in a neighborhood of $a$, thus in the entire complex plane. As a consequence $\dfrac t{t-1}$ is a root of unity and $f'(0)=\dfrac{t}{t-1} f'(0)$. Using $\dfrac t{t-1}\ne 1$ we get $f'(0)=0$. An example is $z^6+3z^4-5z^2$.

2.1. The condition means that all the other points in $f^{-1}(v_0)$ are contained in one of the open half planes delimited by the line through $a_0, b_0$. This is clearly an open condition.

Now if $ta(v)+(1-t) b(v)\equiv c$, by Point 1 $c$ must be a critical point and a center of symmetry and $Conv(f_{v_0})$ would have been symmetric with respect to $c$. This is not possible by our assumption that all the other points of $f^{-1}(v_0)$ (and there is at least one) are one side of the line through $a_0, b_0$.

2.2. We may look at the open set $W=\bigcup_v Outer(f_v)$ where $Outer(f_v)$ is the complement of $Conv(f_v)$. We may assume $a_0, b_0$ are on the imaginary axis and all the other points of $f^{-1}(v_0)$ are on the left half plane. We know already $i{\mathbb R}-[a_0, b_0]\subset Outer(f_{v_0})\subset W$.

Fix some $t\in [0,1]$ and let $z_0=ta_0+(1-t)b_0$. We may assume $z_0=0$. Now $v\mapsto z(v)=t a(v)+(1-t) b(v)$ is open. Choose a path $v(s)$ such that $z(v(s))$ is negative real. Then $z_0$ is on the right of the segment $[a(v(s)), b(v(s))]$ for sufficiently small $s$, so by 2.1 we have $z_0\in Outer(f_{v(s)})\subset W$. qed.

III. Finally I want to share some numerical experiments (with the help of Jos Leys) illustrating a refinement (communication by Thurston) of Gauss-Lucas property.

Consider a polynomial $f$ as a branched covering of the complex plane. Denote by ${\cal C}$ the convex hull of the critical points of $f$. It is called {\em the critical convex} of $f$. The following statements are equivalent: (1) For any $v\in \mathbb C$, we have $Conv(f_v)\supset {\cal C}$.

(2) The map $f$ is surjective on any closed half plane $H$ intersecting ${\cal C}$.

(1)$\Longrightarrow$(2). Assume $f(H)\not\ni v$. Then $f^{-1}(v)$ is contained in $\mathbb C-H$ which is an open half plane, in particular convex. Then $Conv(f|_v)$ is also contained in $\mathbb C - H$. So $Conv(f|_v)\not\supset {\cal C}$, contradicting (1).

(2)$\Longrightarrow$(1). Assume that $Conv(f|_v)$ does not contain the entire set ${\cal C}$. Then there is a closed half plane $H$ intersecting ${\cal C}$ but disjoint from $Conv(f|_v)$. Then $f(H)\not\ni v$, that is, $f$ is not surjective on $H$, contradicting (2).

Now the refinement (I'll leave the proof to Thurston if somebody requires) is that if one takes a supporting line $L$ of $\cal C$ there is a region on the outer half space of $L$ on which $f$ is a bijection onto $\mathbb C$ (injective in the interior and bijective on the union of the interior with half of the boundary arc). In fact this region is bounded by the two geodesic rays of the conformal metric $|f'(z)|\cdot |dz|$ tangent to $L$ at a critical point (if the critical point is simple, otherwise the region is even smaller). This means that we use the euclidean length in the range to measure tangent vectors in the domain. Geodesics in this metric are the pullbacks by $f$ of straight lines.

The following movie bijectivity is made by Jos Leys to illustrate this result.bijectivity

2 added 13 characters in body; deleted 1 characters in body

I. First I want to share some computer experiments of H.H. Rugh. The following image supports the positive answer of the

QUESTION: is this hexagon equal to $Hull(X^3−1)$?

See the image triangle (as a new user I was not allowed to use image tags).

A scilab program testing this problem can be found at roots-dancing. In particular an example $z^6-3z^3+z$ similar to the starting example of Zeb, showing that $Hull(P)$ can be strictly smaller then $\bigcap_vConv(f_v)$ for $v$ ranging the critical values of $f$ (shown in blues blue polygons). See smaller

II. Here is a proof (communication of Rugh) of the same statement of Zeb (with $f=\Pi$):

So if we let L be the set of $v$'s such that three of the roots of $f_v$ lie on a line, we get that $Hull(P)=\bigcap_{v\in L}Conv(f_v)$ if $deg f≥3$.

The underlying idea is very similar to that of Zeb.

Statements: Let $f$ be a polynomial of degree at least 3. Assume that $a_0$ and $b_0$ are two distinct simple roots of $f(z)-v_0$. Then for $v$ in a small neighborhood of $v_0$, there are two simple roots $a(v)$, $b(v)$ of $f(z)-v$ with $a(v_0)=a_0$ and $b(v_0)=b_0$. In this case,

1. If for some complex $t\neq 0,1$, the holomorphic function $v\mapsto t a(v) + (1-t)b(v)$ is constant $c$, then the polynomial has a rotational symmetry about $c$. And $c$ is a critical point of $f$.

2. If the segment $[a_0, b_0]$ is a boundary edge of the polygon $Conv(f_{v_0})$ (in particular no other point in the line through $a_0, b_0$ is mapped to $v_0$), then

2.1 for $v$ sufficiently close to $v_0$, the segment $[a(v),b(v)]$ is a boundary edge of the polygon $Conv(f_{v})$, and for any $t\in ]0,1[$, the map $v\mapsto ta(v)+(1-t) b(v)$ is an open mapping.

2.2. The line through $a_0, b_0$ is outside $\bigcap_v Conv(f_{v})$.

Proof. 1. Replacing $f(z)$ by $f(z-c)$ if necessary, we may assume $c=0$. Note that $a\mapsto b(f(a))$ is defined and holomorphic in a neighborhood of $a_0$, satisfying that $b(f(a_0))=b_0$ and $f(b(f(a)))=f(a)$. It follows that $ta+(1-t)b(f(a))\equiv 0$ so $b(f(a))=\dfrac{ta}{t-1}$. Therefore $f(a)=f(\dfrac {ta}{t-1})$ in a neighborhood of $a$, thus in the entire complex plane. As a consequence $\dfrac t{t-1}$ is a root of unity and $f'(0)=\dfrac{t}{t-1} f'(0)$. Using $\dfrac t{t-1}\ne 1$ we get $f'(0)=0$. An example is $z^6+3z^4-5z^2$.

2.1. The condition means that all the other points in $f^{-1}(v_0)$ are contained in one of the open half planes delimited by the line through $a_0, b_0$. This is clearly an open condition.

Now if $ta(v)+(1-t) b(v)\equiv c$, by Point 1 $c$ must be a critical point and a center of symmetry and $Conv(f_{v_0})$ would have been symmetric with respect to $c$. This is not possible by our assumption that all the other points of $f^{-1}(v_0)$ (and there is at least one) are one side of the line through $a_0, b_0$.

2.2. We may look at the open set $W=\bigcup_v Outer(f_v)$ where $Outer(f_v)$ is the complement of $Conv(f_v)$. We may assume $a_0, b_0$ are on the imaginary axis and all the other points of $f^{-1}(v_0)$ are on the left half plane. We know already $i{\mathbb R}-[a_0, b_0]\subset Outer(f_{v_0})\subset W$.

Fix some $t\in [0,1]$ and let $z_0=ta_0+(1-t)b_0$. We may assume $z_0=0$. Now $v\mapsto z(v)=t a(v)+(1-t) b(v)$ is open. Choose a path $v(s)$ such that $z(v(s))$ is negative real. Then $z_0$ is on the right of the segment $[a(v(s)), b(v(s))]$ for sufficiently small $s$ s$, so by 2.1 we have$z_0\in Outer(f_{v(s)})\subset W$. qed. III. Finally I want to share some numerical experiments (with the help of Jos Leys) illustrating a refinement (communication by Thurston) of Gauss-Lucas property. Consider a polynomial$f$as a branched covering of the complex plane. Denote by${\cal C}$the convex hull of the critical points of$f$. It is called {\em the critical convex} of$f$. The following statements are equivalent: (1) For any$v\in \mathbb C$, the we have$Conv(f_v)\supset {\cal C}$.\ C}$.

(2) The map $f$ is surjective on any closed half plane $H$ intersecting ${\cal C}$.

(1)$\Longrightarrow$(2). Assume $f(H)\not\ni v$. Then $f^{-1}(v)$ is contained in $\mathbb C-H$ which is an open half plane, in particular convex. Then $Conv(f|_v)$ is also contained in $\mathbb C - H$. So $Conv(f|_v)\not\supset {\cal C}$, contradicting (1).

(2)$\Longrightarrow$(1). Assume that $Conv(f|_v)$ does not contain the entire set ${\cal C}$. Then there is a closed half plane $H$ intersecting ${\cal C}$ but disjoint from $Conv(f|_v)$. Then $f(H)\not\ni v$, that is, $f$ is not surjective on $H$, contradicting (2).

Now the refinement (I'll leave the proof to Thurston if somebody requires) is that if one takes a supporting line $L$ of $\cal C$ there is a region on the outer half space of $L$ on which $f$ is a bijection onto $\mathbb C$ (injective in the interior and bijective on the union of the interior with half of the boundary arc). In fact this region is bounded by the geodesic of the conformal metric $|f'(z)|\cdot |dz|$ tangent to $L$ at a critical point. This means that we use the euclidean length in the range to measure tangent vectors in the domain. Geodesics in this metric are the pullbacks by $f$ of straight lines.

The following movie is made by Jos Leys to illustrate this result.bijectivity