On a question of supports

Abstract

We give a sufficient condition in order that n closed connected subsets in the n-dimensional real projective space admit a common multitangent hyperplane.

1 Introduction

The motivation for the present note is a step in the proof of the following statements [2, Corollary 5.5 and Theorem 6.1] or [3, 4, Section 5.3]:

Theorem 1.1

Let X be a real del Pezzo surface of degree 2 such that $X({\mathbb {R}})$ is homeomorphic to the disjoint union of four spheres. Then a smooth map $f:X({\mathbb {R}})\rightarrow {\mathbb {S}}^2$ can be approximated by regular maps if and only if its topological degree is even.

Theorem 1.2

Let X be a real del Pezzo surface of degree 1 such that $X({\mathbb {R}})$ is homeomorphic to the disjoint union of four spheres and a projective plane. Then every smooth map $f:X({\mathbb {R}})\rightarrow {\mathbb {S}}^2$ can be approximated by regular maps.

In the statements above ${\mathbb {S}}^2\subset {\mathbb {R}}^{3}$ is the real locus of the quadric $x_1^2+x_2^2+x_3^2=1$ and a regular map is only regular on real algebraic loci, see [3, 4, Definitions 1.2.54 and 1.3.4] for details.

One key point in the proof of the former statements was the existence of a bitangent line to any pair of connected components of a plane quartic and the existence of a tritangent conic to any triple of connected components of certain space sextic. To be precise we need the following:

Proposition 1.3

Let $n=2,3$ and $X\subset {\mathbb {P}}^{\,n}$ be a smooth real algebraic curve of degree 2n whose real locus $X({\mathbb {R}})$ has at least $n+1$ connected components. If $n=3$, assume furthermore that X lies on a singular quadric.

Choose n connected components $\Omega _1,\dots ,\Omega _n$ of $X({\mathbb {R}})$. Then there exists a hyperplane of  ${\mathbb {P}}^{\,n}({\mathbb {R}})$ which is tangent to $\Omega _i$ for all $1\leqslant i \leqslant n$.

Given a pair of embedded circles in the plane, it seems rather clear that a line tangent to each of them exists provided the circles are unnested. Anyway, finding a rigorous proof of this is not straightforward and we did not find proper reference in the literature. It is less obvious to find a tritangent conic to three embedded circles in a cone. More generally, we can wonder how to generalize the obvious necessary condition to be unnested in a more general setting and, even better we can seek for a necessary and sufficient condition. We find a sufficient (but still not necessary) condition in a rather general setting. This is the main result of this short note (Theorem 3.3) from which we derive easily Proposition 1.3 as a particular case. Sections 2 and 3 are devoted to the proof of this theorem. In Sect. 3, we prove Proposition 1.3 and propose a conjecture with a sufficient condition weaker than Theorem 3.3. We refer to the cited references for the proofs of Theorems 1.1 and 1.2.

2 Some reminders

We start with some well-known definitions from convex geometry.

Definition 2.1

(Convex hull) Let E be a Euclidean space of dimension n. A subset $A\subset E$ is called convex in E if and only if for all $x,y\in A$ and every $t\in [0,1]$ we have

i.e. the line segment joining x and y is contained in A. The convex hull of a subset $A\subset E$ is the smallest (in the inclusion sense) convex subset of E containing A.

Definition 2.2

(Extremal point) Let E be a Euclidean space of dimension n and $A\subset E$ be a subset. We say that a point $x\in A$ is an extremal point of A if the convex hull of is still convex.

Theorem 2.3

(Krein–Milman, see for instance [1, Chapter II.4, Theorem 1]) Every non-empty compact convex subset of a Euclidean space admits an extremal point.

Corollary 2.4

Every non-empty compact subset of a Euclidean space admits an extremal point.

Proof

Let A be a non-empty compact subset of a Euclidean space. Let $A_{\mathrm{c}}$ be the convex hull of A. By Krein–Milman, there exists an extremal point $x\in A_{\mathrm{c}}$. If $x\notin A$, then the convex set contains A and it is a strict subset of $A_{\mathrm{c}}$, which contradicts $A_{\mathrm{c}}$ being the convex hull of A. Therefore, $x\in A$. $\square$

3 n-supporting hyperplanes

Definition 3.1

(Supporting hyperplane) Let H be a hyperplane of a Euclidean space E given by the equation $l(x)=a$, where l is a linear form and $a\in {\mathbb {R}}$. We denote by $H^{+}$ and $H^-$ the half-spaces

Let $A\subset E$ be a subset of E and $x\in A$. We say that H is a supporting hyperplane of A in x (or that H leans on A in x) if and only if the following hold:

• (a) $x\in A\cap H$,

• (b) $A\subset H^+$ or $A\subset H^-$.

If A is a subset of ${\mathbb {P}}^{\,n}({\mathbb {R}})$ and $x\in A$, we say that H leans on A in x if and only if there exists an affine chart E of ${\mathbb {P}}^{\,n}({\mathbb {R}})$ such that $x\in E$ and H leans on A in x inside E.

Definition 3.2

(r-supporting hyperplane) Let $A_1,\dots ,A_r$ be subsets of ${\mathbb {P}}^{\,n}({\mathbb {R}})$. We say that H is a hyperplane of r-support of $A_1,\dots ,A_r$ if there exist points $x_1\in A_1,x_2\in A_2,\dots , x_r\in A_r$ such that H is a supporting hyperplane of $A_i$ in $x_i$ for all $1\leqslant i\leqslant r$.

Theorem 3.3

Let $n\in {\mathbb {N}}$ and let $A_1,\dots ,A_n\subset {\mathbb {P}}^{\,n}({\mathbb {R}})$ be closed connected subsets of ${\mathbb {P}}^{\,n}({\mathbb {R}})$. Suppose that there exists a point $p\in {\mathbb {P}}^{\,n}({\mathbb {R}})$ such that no hyperplane passing through p meets all the $A_i$. Then there exists an n-supporting hyperplane of $A_1,\dots ,A_n$.

Proof

We write ${\mathbb {P}}={\mathbb {P}}^{\,n}({\mathbb {R}})$ and ${\mathbb {P}}^*=({\mathbb {P}}^{\,n}({\mathbb {R}}))^{*}$ for the dual projective space. To each hyperplane $H\subset {\mathbb {P}}$ given by an equation $\sum \lambda _kx_k=0$, we associate the point in ${\mathbb {P}}^*$ To each point $q\in {\mathbb {P}}$ we associate the dual hyperplane in ${\mathbb {P}}^*$

The hypothesis that there exists a point $p\in {\mathbb {P}}$ such that no hyperplane passing through p meets all the $A_i$ implies that the $A_i$ are pairwise disjoint. Let ${\mathscr {H}}$ be the set of hyperplanes in ${\mathbb {P}}$ that meet all the $A_i$. Since there is a hyperplane through n points in ${\mathbb {P}}$, we see that ${\mathscr {H}}$ is non-empty. Let ${\mathscr {H}}^{*}$ be the image of ${\mathscr {H}}$ in the dual space ${\mathbb {P}}^*$ via the above correspondence. Since $p^{*}$ corresponds to the set of hyperplanes in ${\mathbb {P}}$ passing through p, the set ${\mathscr {H}}^{*}$ is contained in the complement of the hyperplane $p^{*}$ in ${\mathbb {P}}^*$ Let $U_p$ be the open affine complement of $p^{*}$ in ${\mathbb {P}}^*$

Lemma 3.4

The set  ${\mathscr {H}}^{*}$ is compact in $U_p$.

Proof

For each $1\leqslant i\leqslant n$, let ${\mathscr {H}}_i$ be the set of hyperplanes that meet $A_i$. We have ${\mathscr {H}}^{*}=\bigcap _{\,i=1}^n ({\mathscr {H}}_i)^{*}$ The set $A_i$ being closed implies that $({\mathscr {H}}_i)^{*}$ is closed. We start by showing that the complement of ${\mathscr {H}}^{*}$ in $U_p$ is open.

Indeed, the natural map induces a continuous double cover ${\mathbb {S}}^n\rightarrow {\mathbb {P}}$. The inverse image $B_i$ of $A_i$ through this map is a closed subset in the unit sphere of ${\mathbb {R}}^{n+1}$ If H is an hyperplane in ${\mathbb {P}}$ that does not meet $A_i$, then its preimage $H'$ is a hyperplane in ${\mathbb {R}}^{n+1}$ which does not meet $B_i$. The intersection $H'\!\cap {\mathbb {S}}^n$ is the unit sphere of dimension $n-1$ in $H'$ and in particular is closed in ${\mathbb {S}}^n$.

If $d>0$ is the distance between the two compacts $B_i$ and $H'$, we can take $U_i$ the subset of ${\mathbb {P}}^*$ formed by the duals of hyperplanes whose traces on ${\mathbb {S}}^n$ are at distance less than 1/2 of $B_i$. Then is open in $U_p$.

This shows that the complement of $({\mathscr {H}}_i)^{*}$ in ${\mathbb {P}}^*$ is open. It follows that ${\mathscr {H}}^{*}$ is closed in ${\mathbb {P}}^*$ Moreover, the set ${\mathscr {H}}^{*}$ is bounded in $U_p$ because it is closed and ${\mathscr {H}}^{*}\!\cap p^*=\varnothing$. Hence ${\mathscr {H}}^{*}$ is compact in $U_p$.$\blacksquare$

By Corollary 2.4 of Krein–Milman and Lemma 3.4, the set ${\mathscr {H}}^{*}$ admits an extremal point $H^*$ Let us show that H is an n-supporting hyperplane of $A_1,\dots ,A_n$.

We proceed by contradiction and without loss of generality, we can suppose that H does not support $A_1$. Since $H\in {\mathscr {H}}$, there exists for each $i=2,\dots ,n$ a point $y_i\in A_i\cap H$. Let $P_1$ be a hyperplane passing through p and $y_2,\dots ,y_n$ and recall that $P_1$ does not meet $A_1$ by hypothesis. Since H does not lean on $A_1$, it does not lean on $A_1$ in the aff ine chart . We place ourselves inside E. The hyperplane $H\cap E$ defines two half-spaces $H^+$ and $H^-$ in E and there exists and .

Let S be the closed segment $[x_1,x_2]$ in E. It intersects H. Let us show that

$$\begin{aligned} \text { any hyperplane in } E \text { that meets } S \text { also meets } A_1. \end{aligned}$$

(1)

Let P be a hyperplane of E meeting S. If it meets S in $x_1$ or $x_2$, we are finished. Suppose that $P\cap S\subset ]x_1,x_2[$ and $A_1\cap P=\varnothing$. Let and . The sets $O^+$ and $O^-$ are open subsets of E and $A_1\subset O^+\cup O^-$. The subspace $A_1$ being connected in E, we have $A_1\subset O^+$ or $A_1\subset O^-$ This is impossible because $x_1\in O^+$ and $x_2\in O^-$ (or the other way around). this ends the proof of (1).

Let $y\in S$. Since $y_2,\dots ,y_n$ are pairwise distinct and are not contained in E (remember that $y_i\in A_i\cap P_1$ for $i\in \{2,\dots ,n\}$ by definition of $P_1$) and $S\subset E$, there exists a hyperplane $H_y\subset {\mathbb {P}}$ through $y,y_2,\dots ,y_n$. The hyperplane $H_y$ is contained in ${\mathscr {H}}$ because it meets $A_1$ by property (1).

The points $y_2,\dots ,y_n$ define a line D in ${\mathbb {P}}^*$ and we have $(H_y)^{*}\!\in D$. Therefore, the set of $(H_y)^{*}$, $y\in S$, is a closed segment $S^{*}$ It is contained in $U_p$, because $p\notin H_y$, and $S^*$ is contained in ${\mathscr {H}}^{*}$ as a consequence of (1). Let $y_0=S\cap H$, where $H^*$ is the extremal point of ${\mathscr {H}}^*$ from above. Then $H^{*}=(H_{y_0})^{*}$ is a point in the interior of $S^{*}$ It is therefore contained in the convex hull of ${\mathscr {H}}^{*}$ and cannot be an extremal point, because we lose convexity if we take it away. Hence the contradiction. $\square$

4 Conclusion

Proof of Proposition 1.3

First recall that any hyperplane meets any connected component of $X({\mathbb {R}})$ in an even number of intersection points, counted with multiplicity, see e.g. [3, 4, Lemma 2.7.8]. Let p be a point of . By definition of the degree, a hyperplane passing through p cannot meet n other components of $X({\mathbb {R}})$ because X has degree 2n in ${\mathbb {P}}^{\,n}$

The conclusion follows from Theorem 3.3. $\square$

Theorem 3.3 is enough to prove Proposition 1.3, but it is easy to see that the existence of a point p such that no hyperplane passing through p meets all the $A_i$ is not necessary. Take for example two intersecting circles in the plane: as in Theorem 3.3, these are two subsets in the 2-dimensional plane, but by any point p, there is a line meeting the two circles. Anyway, there is clearly a line tangent to them.

We propose the following conjecture using a weaker sufficient condition (which can be applied to the former example):

Conjecture 4.1

Let $\{A_i\}_{1\leqslant i\leqslant n}$ be closed connected subsets contained in an affine subset of ${\mathbb {P}}^{\,n}({\mathbb {R}})$. Let $C_i$ be the union of all $(n\,{-}\,2)$-dimensional linear subspaces $P\subset {\mathbb {P}}^{\,n}({\mathbb {R}})$ such that for all $j \ne i$, $1 \leqslant j \leqslant n$, P meets the convex hull of $A_j$. Assume that for all $1 \leqslant i \leqslant n$, $A_i$ is not included in interior of $C_i$, then there exists an n-supporting hyperplane of $A_1,\dots ,A_n$.

Remark that this new sufficient condition is still unnecessary: consider three disjoint spheres $A_1$, $A_2$ and $A_3$ with the same radius and whose centers are on the same line. If $A_1$ is not the sphere in the middle it is in the interior of the union of all lines meeting $A_2$ and $A_3$.

We can see that the sufficient condition of the conjecture is weaker than the one of Theorem 3.3, by contraposition. If the condition of the conjecture is not satisfied, then there exists i such that $A_i$ is included in the interior of the union of the $(n\,{-}\,2)$-dimensional linear subspaces meeting each convex hull of $A_j$, $j\ne i$. Then there exists an $(n\,{-}\,2)$-dimensional linear subspace P meeting each convex hull of $A_i$. Let $p\in {\mathbb {P}}$, then the hyperplane generated by p and P meets each convex hull of $A_i$, hence each $A_i$ as they are connected, which contradicts the condition of the theorem.

We could also ask about the number of multi-tangent planes.

Proposition 4.2

Under the conditions of Theorem 3.3, if each $A_i$ contains a non-empty open subset, then there is at least $n+1$ distinct n-supporting hyperplanes of $A_1,\dots ,A_n$.

Proof

If each $A_i$ contains a non-empty open subset, so does ${\mathscr {H}}^*$ This implies that there are at least $n+1$ distinct extremal points for ${\mathscr {H}}^*$ Indeed, if ${\mathscr {H}}^*$ has less than $n+1$ extremal points, it is the convex hull of its extremal points and therefore it is a hyperplane of dimension at most $n-1$ hence it does not contain any open set. Then, the proof of Theorem 3.3 establishes that each extremal point for ${\mathscr {H}}^*$ corresponds to distinct n-supporting hyperplanes. $\square$

However, it seems that the conditions of this theorem imply that we have $2^n$ extremal points (in dimension 2: 4 bitangent lines, 8 in dimension 3, etc.) by going either below or above each $A_i$. This suggests that ${\mathscr {H}}^*$ reassembles to a cube. Moreover, all the examples we studied lead us to propose the following conjecture.

Conjecture 4.3

The main condition of Theorem 3.3 is sufficient and necessary to have $2^n$ multi-tangent planes when the $A_i$ are not thin (i.e. contain an open subset).