Convex hulls of surfaces in fourspace

Abstract

This is a case study of the algebraic boundary of convex hulls of varieties. We focus on surfaces in fourspace to showcase new geometric phenomena that neither curves nor hypersurfaces exhibit. Our method is a detailed analysis of a general purpose formula by Ranestad and Sturmfels in the case of smooth real algebraic surfaces of low degree (that are rational over the complex numbers). We study both the complex and the real features of the algebraic boundary of Veronese and Del Pezzo surfaces. The main difficulties and the possible approaches to the case of general surfaces are discussed for and complemented by the example of Bordiga surfaces.

1 Introduction

Convex algebraic geometry is the study of convex semi-algebraic sets. The Zariski closure of the Euclidean boundary of such a semi-algebraic convex body is an algebraic hypersurface, called its algebraic boundary. This hypersurface is a central object, generalizing the hyperplane arrangement associated to a polytope, see [22]. Information regarding the algebraic boundary of a convex body can help in the study of the convex body itself. As for polytopes, it is a hard problem to describe the algebraic boundary given the extreme points of the convex body. This is the problem that we study in this paper. There is a general formula for the algebraic boundary of a convex hull of a variety in terms of projective duality, due to Ranestad and Sturmfels [18], inspired by their previous work on convex hulls of space curves [19]. The formula gives an inclusion of the algebraic boundary of the convex hull of a variety, when it is a full dimensional convex compact set, into the union of some varieties $X^{[k]}$ that we will define below. Here, we take this work further into higher dimension: we consider convex hulls of surfaces in four-dimensional space and determine irreducible components that actually contribute to their algebraic boundaries, and their degrees. Our goal is to provide a case study of the relevant phenomena for smooth surfaces of low degree. We give a complete answer in the case of Veronese and Del Pezzo surfaces, and discuss the main issues for general surfaces using Bordiga surfaces as an example.

Our case study illustrates the following three steps in the general recipe described by Ranestad and Sturmfels in [18]. The first step is mostly solved by previous work of Kazarian’s, so that our work has a particular focus on the last two steps. Our starting point is a smooth real algebraic surface $X\subset \mathbb {P}^4$. In the first step, we want to characterize subvarieties denoted by $X^{[k]}$ of the projective dual variety of X. These subvarieties are given as families of hyperplanes in $\mathbb {P}^4$ satisfying certain tangency conditions to X and are in general reducible. The computation of the degrees of the $X^{[k]}$ relies on various results in (enumerative) complex algebraic geometry. Most important are the formulae by Kazarian from [11] and Vainsencher from [23] counting singular plane algebraic curves with fixed degree and prescribed types of singularities. We also want to determine the irreducible components of these varieties.

The second step is to determine the invariants of the dual variety of each $X^{[k]}$, preferably irreducible component by irreducible component. Here, we aim for dimension and degree. For nonsingular varieties, there is a general formula to compute the degree of the dual hypersurface in Fulton’s book [9, Example 3.2.21]. There is also a formula for singular hypersurfaces [9, Example 4.4.3]. However, in our case of a singular variety of higher codimension, this is a notoriously difficult problem and requires a good understanding of the singularities of the variety. With an ad hoc approach we succeed in determining all dimensions and degrees in the case of Del Pezzo surfaces in Sect. 3. The case of general surfaces is discussed in Section 4. We showcase the main difficulties and, in the special case of the Bordiga surfaces, suitable methods to overcome them. Here, we use previous work of Pandharipande [16] combined with the Riemann-Hurwitz formula to compute the degree of the projective dual variety of the curve $X^{[3]}$. For the variety $X^{[2]}$, which is a surface, the characterization of its singularities remains open. More generally, it is an open problem to compute dimension and degree of the dual variety (of an irreducible component) of $X^{[k]}$.

The first two steps are in the realm of complex algebraic geometry. The third step is the transition to real algebraic geometry. The general recipe provides a list of candidates for the algebraic boundary of the convex hull of X in the first two steps. At this point, we need to select those hypersurfaces in this list that can actually occur in the algebraic boundary of the convex hull. We completely work out the case of Del Pezzo surfaces, which clearly illustrates the main difficulties in this step. Firstly, the answer depends on the real form of the complex algebraic surface. The real forms of Del Pezzo surfaces are classified and the list is fairly short so that we can do a case analysis. In general, the classification of real forms and their embeddings is open for surfaces, including Bordiga surfaces. Secondly, the answer might also depend on the choice of affine chart, in which we take the convex hull. Indeed, there is a real form of a Del Pezzo surface in $\mathbb {P}^4$, where different pairs of quadric cones form the algebraic boundary of the convex hull depending on the choice of hyperplane at infinity. To illustrate the subtleties on a higher level, consider the variety $X^{[4]}$ which is a 0-dimensional real projective variety in all our examples so that its dual variety is a union of hyperplanes. Among them are real hyperplanes and non-real hyperplanes that come in complex conjugate pairs. It is a problem in real enumerative geometry to determine the number of real points of $X^{[4]}$. Additionally, from convexity considerations, among the real hyperplanes in this finite set, we want to select the supporting hyperplanes to the convex body, which is a nonnegativity constraint on the variety X and this notion of nonnegativity is relative to the hyperplane at infinity (since a linear form does not have a well-defined sign on real projective space). Furthermore, to be an irreducible component of the algebraic boundary, the intersection of the hyperplane and the convex body also has to be full-dimensional (equivalently Zariski dense) in the hyperplane. This is the condition that fails for all points in $X^{[4]}$ in the example of Del Pezzo surfaces.

We briefly describe the results of our case study, which focuses on smooth surfaces in $\mathbb {P}^4$ of even degree that are rational as complex algebraic varieties. Those are the Veronese surfaces, the Del Pezzo surfaces (both of degree 4), and the Bordiga surfaces (of degree 6). Our restriction to even degrees comes from real considerations: we are interested only in those cases in which the convex hull of the surface is a full-dimensional convex body, hence compact. For a compact real Veronese surface, the algebraic boundary of the convex hull is irreducible and of degree 6, see Theorem 2.1. The algebraic boundary of the convex hull of a compact real Del Pezzo surface is the union of two quadric cones over a nonsingular quadric in $\mathbb {P}^3$, see Theorem 3.7. For generic Bordiga surfaces, the picture becomes rather involved: we expect that the algebraic boundary can have at most two irreducible components. We were not able to determine the degree exactly. However, one irreducible component has degree 384. Our results in this case are summarized in Theorem 4.6, only for real forms arising as blow ups of the real projective plane in a set of 5 pairs of complex conjugate points.

1.1 Setting

Here, we begin with the setting in complex algebraic geometry. The discussion of the real geometry will be done for each surface case separately. Let $X \subset \mathbb {C}^d$ be a complex algebraic variety with a compact real part. We consider the convex body ${{\,\textrm{conv}\,}}(X) \subset \mathbb {R}^d$, the convex hull of the real part of X. This is a semi-algebraic set by quantifier elimination. The (projective) Zariski closure of its topological boundary $\partial ( {{\,\textrm{conv}\,}}(X))$ is thus a hypersurface, called the algebraic boundary of ${{\,\textrm{conv}\,}}(X)$, see [22]. We denote it by $\partial _a ({{\,\textrm{conv}\,}}(X))\subset \mathbb {P}^d$. When X is smooth, [18, Theorem 1.1] provides all the irreducible components of $\partial _a ({{\,\textrm{conv}\,}}(X))$, that we now recall.

Let $X\subset \mathbb {P}^d$ be a projective variety, not contained in any hyperplane. Fix an integer $k\le d$. Define $X^{[k]}$ as the Zariski closure of the set

$$\begin{aligned} \{ u \in (\mathbb {P}^d)^* \,|\, u^\perp \hbox { is tangent to } X \hbox { at } k \hbox { linearly independent points }\} \subset (\mathbb {P}^d)^* \end{aligned}$$

of hyperplanes tangent to our variety at k linearly independent points. These k points are singularities in the intersection $u^\perp \cap X$, where $u^\perp = \{[x_0,\ldots ,x_d]\in \mathbb {P}^d \,|\, u_0 x_0 + \ldots + u_d x_d = 0\}$.

We write ${{\,\textrm{conv}\,}}(X)$ for the convex hull of the real points of X in an affine chart $\mathbb {R}^d\subset \mathbb {P}^d(\mathbb {R})$. We assume that this real algebraic set is compact. Then by [18, Theorem 1.1]

$$\begin{aligned} \partial _a ({{\,\textrm{conv}\,}}(X)) \subset \bigcup _{k = 1}^d \left( X^{[k]} \right) ^*. \end{aligned}$$

(1.1)

There are various ways in which one can improve the inclusion in (1.1). For instance, since $\partial _a ({{\,\textrm{conv}\,}}(X))$ is a hypersurface, we can get rid of the irreducible components of $\left( X^{[k]} \right) ^*$ that have higher codimension. A necessary condition (based on Terracini’s Lemma, see [18, Section 1]) for $\left( X^{[k]} \right) ^*$ to be a hypersurface is that the k-th secant variety of X has at most codimension 1, i.e., that $k \ge \left\lceil {\frac{n}{{\dim X + 1}}} \right\rceil$. Another issue could be that some of the components in the right hand side of (1.1) do not have real points. In this case they do not contribute to the algebraic boundary either. Our goal is to understand which are the relevant irreducible components of the algebraic boundary in the particular case of smooth surfaces in fourspace. We approach this question by analyzing smooth (or even generic) surfaces of fixed low degree and sectional genus. Since we want to compute convex hulls, we need surfaces with a compact real locus, and this can happen only if $\deg X$ is even. Degree 2 surfaces are embeddings of $\mathbb {P}^1\times \mathbb {P}^1$ in $\mathbb {P}^3\subset \mathbb {P}^4$, therefore they have been already investigated in [18, Section 2.3]. The first interesting degree is then 4; here we find two families of smooth surfaces: the Veronese and the Del Pezzo surfaces. A Veronese surface has sectional genus 0 and it is a projection into $\mathbb {P}^4$ of the image of the Veronese map $\nu _2(\mathbb {P}^2)\subset \mathbb {P}^5$. A Del Pezzo surface has sectional genus 1 and it is the complete intersection of two quadrics or, equivalently, the blow up of $\mathbb {P}^2$ in five generic points. In degree 6 we find Bordiga and K3 surfaces, with sectional genus 3 and 4 respectively. Bordiga surfaces are the blow up of $\mathbb {P}^2$ in ten generic points, whereas K3 surfaces in fourspace are the complete intersection of a quadric and a cubic hypersurface. This is a classification as complex algebraic varieties. These surfaces can have other real forms. For Del Pezzo surfaces, we discuss the complete classification of the real forms. In the case of Bordiga surfaces, we only discuss the blow ups of the real projective plane in a real set of 10 points because of the absence of a complete classification of their real forms.

Using [23] or [11], we can compute the degrees of the varieties $X^{[k]}$, as shown in Table 1. The two papers provide formulae to compute the number of singular curves on a surface, having certain types of singularities. The formulae cannot be applied to the case of the Veronese surface; this requires an ad hoc geometric study of the varieties $X^{[k]}$.

Table 1 The degrees of the $X^{[k]}$’s for low degree surfaces in $\mathbb {P}^4$

A general fact is that $X^{[1]}$ is the dual variety $X^*$, therefore by the biduality theorem [10, Theorem 15.24], $\left( X^{[1]} \right) ^* = X$ is not a hypersurface in our framework. The values of k for which $X^{[k]}$ might be relevant for us, are thus $k=2,3,4$. K3 surfaces are the only irrational surfaces in the range that we are considering. There is a whole literature regarding the associated varieties $X^{[k]}$ when X is a K3 surface (see for instance [2]), but not enough for us to compute the degrees of $\left( X^{[k]} \right) ^*$. Hence, we focus here on the rational cases of the Veronese, Del Pezzo, and Bordiga surfaces. The strategy is to partition the numbers in Table 1 with one summand for each irreducible component. We then investigate each irreducible component separately, in order to determine what will be relevant after dualization. When we are dealing with blow ups of the plane from a complex point of view, studying the family of hyperplanes tangent to X at a certain number of points, translates into studying the family of curves in $\mathbb {P}^2$ passing through the points that we blow up, and having prescribed singularities.

1.1.1 Kazarian and Vainsencher’s numbers

Let X be a surface and consider a linear system |D| of divisors on X. Then [11, Section 10] and [23, Section 5] provide the recipes for computing the number of singular curves in the linear system |D| with prescribed singularities, and going through the correct number of points, in such a way that the answer is finite. For instance, consider the surface $X=\mathbb {P}^2$ and suppose we want to compute the number of rational quartics in X. Denoting by L the class of a line in $\mathbb {P}^2$, we have $D=4L$. Plane quartics constitute a $\mathbb {P}^{14}$. In order for the quartics to be rational, they need to have generically three nodes: these are our prescribed singularities. Each node condition drops the dimension of this variety by 1, thus the variety of rational quartics has codimension 3 in $\mathbb {P}^{14}$. We want to compute the degree of this variety, so we impose 11 linear conditions, namely that these curves pass through 11 points in $\mathbb {P}^2$ in general position. We obtain in this way a finite set, whose cardinality is computed by Kazarian and Vainsencher’s formulae. Vainsencher computes singular curves with a certain number of nodes, whereas Kazarian allows also other types of singularities. In particular, $A_1$ denotes a node and $A_2$ denotes a cusp. These will be our cases of interest. We will mainly talk about Kazarian’s formulae, to avoid jumping between one paper and the other when we need to count cusps. There are four parameters to be considered

$$\begin{aligned} \begin{aligned}&\textsf{d} = D.D,{} & {} \textsf{k} = D.K_X, \\&\textsf{s} = K_X.K_X,{} & {} \textsf{x} = \chi (X), \end{aligned} \end{aligned}$$

(1.2)

where $K_X$ is the canonical divisor on X and $\chi$ denotes the topological Euler characteristic. Here ‘ . ’ denotes the intersection product of two divisors, defined on the Picard group ${{\,\textrm{Pic}\,}}(X)$ of the surface X [1, Chapter 1]. Then, by plugging such quantities in [11, Theorem 10.1] one gets the desired numbers. The linear system |D| is given by the embedding of X as the linear system of hyperplane sections. We explain the notation $k \pi ^*(L) - \sum E_i$ below. Table 2 computes (1.2) for our surfaces. Using these numbers one can get the degrees in Table 1:

$$\begin{aligned} \deg X^{[k]} = N_{A_1^k}, \end{aligned}$$

using the notation of [11, Theorem 10.1] in which $N_{A_1^k}$ is the number of singular curves of |D| with k nodes.

Table 2 The quantities required for Kazarian’s formulae

2 Veronese

Consider the Veronese surface Y in $\mathbb {P}^5$, i.e., the image of the second Veronese map

$$\begin{aligned} \nu _2 : \mathbb {P}^2&\rightarrow \mathbb {P}^5 \\ [ t_0, t_1, t_2]&\mapsto [ t_0^2, t_0 t_1, t_0 t_2, t_1^2, t_1 t_2, t_2^2]. \end{aligned}$$

Its secant variety is a cubic hypersurface in $\mathbb {P}^5$ [10, p. 144f]. Therefore, by projecting Y onto $\mathbb {P}^4$ from a point that does not belong to the secant variety, we obtain a smooth irreducible surface $X\subset \mathbb {P}^4$. We only consider the real surfaces obtained as the projection from a real point in $\mathbb {P}^5$ and no other real forms of this complex algebraic surface. We now make this precise.

2.1 The $\mathbb {C}$omplex picture

We treat $\mathbb {P}^5$ as the projective space of the space of $3\times 3$ symmetric matrices. Using this point of view, one can prove that the Veronese surface Y is the variety of rank one matrices. We will first study Y and its associated varieties $Y^{[k]}$, and then we will take care of the projection. In this setting, we have the pairing of two matrices consisting of the trace of their product; we will generally denote by A matrices in the primal space and by B matrices in the dual space. Given a symmetric matrix $B\in (\mathbb {P}^5)^*$, the corresponding hyperplane in the primal space is

$$\begin{aligned} H_B = \{ A \in \mathbb {P}^5 \,|\, {{\,\textrm{trace}\,}}(BA) = 0 \}. \end{aligned}$$

On the other hand, every $3\times 3$ symmetric matrix B corresponds to the plane conic $C_B = \{ t\in \mathbb {P}^2 \,|\, t^T B t = 0\}\subset \mathbb {P}^2$. This plane curve is isomorphic via $\nu _2$ to the hyperplane section $H_B\cap Y$, since $t^T B t = {{\,\textrm{trace}\,}}(B ( t t^T))$ and clearly $t t^T \in Y$. So in order to study hyperplane sections of the Veronese surface, we can study plane conics. There are three distinct cases: if ${{\,\textrm{rank}\,}}B = 3$, then $C_B$ is a smooth conic; if ${{\,\textrm{rank}\,}}B = 2$, then $C_B$ is the union of two lines; if ${{\,\textrm{rank}\,}}B = 1$, then $C_B$ is a double line. Hence, the number of singular points of $C_B$, and therefore of the corresponding hyperplane section of the Veronese, can be 0, 1, or infinity. In the last case, the hyperplane section is a double conic. This shows that

$$\begin{aligned} Y^{[2]} = Y^{[3]} = \{ B\in (\mathbb {P}^5)^* \,|\, {{\,\textrm{rank}\,}}B = 1\} = \nu _2((\mathbb {P}^2)^*) \end{aligned}$$

is again a Veronese surface (in the dual space). Since a conic is always planar, $Y^{[4]} = \emptyset$.

We now focus on the projection onto $\mathbb {P}^4$. The secant variety of Y is the set of $3\times 3$ symmetric matrices of rank 2. Thus, fix a matrix $A \in \mathbb {P}^5$ of full rank and let $\pi _A : \mathbb {P}^5 \dashrightarrow \mathbb {P}^4$ be the projection centered at A. Denote by $X_A = \pi _A (Y)$ our surface of interest in $\mathbb {P}^4$. The projection $\pi _A$ corresponds, in the dual space, to the intersection with the hyperplane $H_A$. Therefore, the variety $X_A^{[2]} = X_A^{[3]}$ is given by $Y^{[2]} \cap H_A = \nu _2( (\mathbb {P}^2)^* ) \cap H_A$. By the reasoning above, it is isomorphic to the plane conic $C_A \subset (\mathbb {P}^2)^*$. Since A is full rank, the variety $\nu _2(C_A) \subset (\mathbb {P}^4)^*$ is the rational normal curve. Therefore,

$$\begin{aligned} \dim X_A^{[2]} = 1, \qquad \deg X_A^{[2]} = 4. \end{aligned}$$

The dual variety $\left( X_A^{[2]} \right) ^* \!\!\!=\! \left( X_A^{[3]} \right) ^* \!\!\!\subset \!\mathbb {P}^4$ is a sextic hypersurface. This completes the analysis of the complex structure.

2.2 The $\mathbb {R}$eal picture

Regarding the real parts of these varieties and the possible contribution to the algebraic boundary of ${{\,\textrm{conv}\,}}(X_A)$, we need to distinguish two cases in terms of the signature of the projection center A. Since A must be of rank 3 in order for $X_A\subset \mathbb {P}^4$ to be smooth, its signature can be either (3, 0) or (2, 1), up to a global sign. Here (p, n) specifies the number p of positive eigenvalues of A and the number n of negative eigenvalues.

Firstly, we assume that A has signature (3, 0), i.e., it is positive definite. Pick any point $\overline{M}\in \mathbb {P}^4$; then $\pi _A^{-1}(\overline{M}) = \{\mu M + \lambda A | [\mu ,\lambda ]\in \mathbb {P}^1\}$. For $\mu = 1$ and a sufficiently large real $\lambda$, the matrix $M+\lambda A$ is positive definite as well, and therefore it is a combination of three rank one matrices: $M+\lambda A = \frac{1}{3} (v_1 v_1^T + v_2 v_2^T + v_3 v_3^T)$. Hence $\overline{M}\in {{\,\textrm{conv}\,}}(X_A)$, i.e., ${{\,\textrm{conv}\,}}(X_A) = \mathbb {R}^4$. In this case, the conic $C_A$ and the rational normal curve $X_A^{[2]}$ are real varieties with empty real locus. The real points of $\left( X_A^{[2]} \right) ^*$ must then be singular, so they do not form a hypersurface. Therefore, there are no hypersurfaces that can describe $\partial _a ({{\,\textrm{conv}\,}}(X_A))$. The geometric picture in this case is that ${{\,\textrm{conv}\,}}(X_A)$ is the projection of the cone of positive semidefinite matrices from an interior point and therefore $\mathbb {R}^4$.

Assume now that the center of the projection A has signature (2, 1). In this case the rational normal curve $X_A^{[2]}$ has a non-empty real locus, and its dual hypersurface has smooth real points. Being the only irreducible variety that can contribute to the algebraic boundary of ${{\,\textrm{conv}\,}}(X)$, it coincides with it. Geometrically, ${{\,\textrm{conv}\,}}(X)$ is the projection of the cone of positive semidefinite matrices, from a point outside it. We summarize our findings in the following theorem.

Theorem 2.1

Let $X_A\subset \mathbb {P}^4$ be the projection of $\nu _2(\mathbb {P}^2)\subset \mathbb {P}^5$ with center the full-rank $3\times 3$ symmetric matrix A. Then,

• if A has signature (3, 0), then ${{\,\textrm{conv}\,}}(X_A) = \mathbb {R}^4$;

• if A has signature (2, 1), then $\partial _a ({{\,\textrm{conv}\,}}(X_A))$ is a hypersurface of degree 6.

Example 2.2

As an explicit example of the $A=(2,1)$ case, consider the projection of the Veronese surface onto $\mathbb {P}^4$ centered at $A = [0,0,-1,2,0,0]\in \mathbb {P}^5$, which corresponds to the matrix

$$\begin{aligned} \begin{bmatrix} 0 &{} 0 &{} -\frac{1}{2} \\ 0 &{} 2 &{} 0 \\ -\frac{1}{2} &{} 0 &{} 0 \end{bmatrix} \end{aligned}$$

having signature (2, 1). Then, $X_A$ is the image of the map

$$\begin{aligned} \nu _2 : \mathbb {P}^2&\rightarrow \mathbb {P}^4 \\ [t_0,t_1,t_2]&\mapsto [t_0^2, -2 t_0 t_1, 2 t_0 t_2 + t_1^2, -2 t_1 t_2, t_2^2]. \end{aligned}$$

Its ideal is generated by the following 7 cubic polynomials:

$$\begin{aligned}&x_3^3-4 x_2 x_3 x_4+8 x_1 x_4^2,\\&x_2 x_3^2-4 x_2^2 x_4+2 x_1 x_3 x_4+16 x_0 x_4^2,\\&x_1 x_3^2-4 x_1 x_2 x_4+8 x_0 x_3 x_4,\\&x_0 x_3^2-x_1^2 x_4,\\&x_1^2 x_3-4 x_0 x_2 x_3+8 x_0 x_1 x_4,\\&x_1^2 x_2-4 x_0 x_2^2+2 x_0 x_1 x_3+16 x_0^2 x_4,\\&x_1^3-4 x_0 x_1 x_2+8 x_0^2 x_3. \end{aligned}$$

We identify $\mathbb {R}^4\subset \mathbb {P}^4(\mathbb {R})$ with the affine chart whose hyperplane at infinity is given by the equation $2 x_0 + x_1 + 2 x_2 + 4 x_3 = 0$. Since

$$\begin{aligned} 2 t_0^2 - 2 t_0 t_1 + 2 ( 2 t_0 t_2 + t_1^2 ) + 4 t_2^2 = ( t_0 - t_1 )^2 + (t_0 + 2 t_2)^2 + t_1^2 \end{aligned}$$

is a sum of three squares, then the Veronese surface in this affine chart does not have real points at infinity, hence it is compact. This implies that its convex hull is a convex body and by Theorem 2.1 its algebraic boundary is the variety $\left( X_A^{[2]} \right) ^*$ of degree 6 defined as the zero locus of

$$\begin{aligned}&x_1^2 x_2^2 x_3^2-4 x_0 x_2^3 x_3^2-4 x_1^3 x_3^3+18 x_0 x_1 x_2 x_3^3-27 x_0^2 x_3^4-4 x_1^2 x_2^3 x_4\\&+16 x_0 x_2^4 x_4+18 x_1^3 x_2 x_3 x_4-80 x_0 x_1 x_2^2 x_3 x_4-6 x_0 x_1^2 x_3^2 x_4+144 x_0^2 x_2 x_3^2 x_4\\&-27 x_1^4 x_4^2+144 x_0 x_1^2 x_2 x_4^2-128 x_0^2 x_2^2 x_4^2-192 x_0^2 x_1 x_3 x_4^2+256 x_0^3 x_4^3 \end{aligned}$$

which is dual to the standard quartic rational normal curve in $(\mathbb {P}^4)^*$ parametrized by the map $[s_0,s_1]\mapsto [s_0^4,s_0^3s_1,s_0^2s_1^2,s_0s_1^3,s_1^4]$, for $[s_0,s_1]\in \mathbb {P}^1$.

Example 2.3

For the $A=(3,0)$ case, let us pick the identity matrix, namely the point $[1,0,0,1,0,1]\in \mathbb {P}^5$, and consider the projection of the Veronese surface given by

$$\begin{aligned} \nu _2 : \mathbb {P}^2&\rightarrow \mathbb {P}^4 \\ [t_0,t_1,t_2]&\mapsto [t_0^2 - t_1^2, t_0^2 - t_2^2, t_0 t_1, t_0t_2, t_1t_2]. \end{aligned}$$

The image of $\nu _2$ is our surface of interest $X_A$, namely the projection of the standard Veronese surface to $\mathbb {P}^4$. The ideal of $X_A$ is generated by

$$\begin{aligned}&x_2 x_3^2-x_0 x_3 x_4-x_2 x_4^2,\\&x_2^2 x_3-x_1 x_2 x_4-x_3 x_4^2,\\&x_1 x_2 x_3-x_0 x_1 x_4-x_2^2 x_4+x_4^3,\\&x_0 x_2 x_3-x_0 x_1 x_4-x_3^2 x_4+x_4^3,\\&x_0^2 x_3-x_0 x_1 x_3-x_3^3+x_0 x_2 x_4+x_3 x_4^2,\\&x_0 x_1 x_2-x_1^2 x_2+x_2^3-x_1 x_3 x_4-x_2 x_4^2,\\&x_0^2 x_1-x_0 x_1^2+x_0 x_2^2-x_1 x_3^2-x_0 x_4^2+x_1 x_4^2. \end{aligned}$$

The curve $X_A^{[2]}$ is a rational normal quartic with no real points. It is the zero locus of the ideal with generators

$$\begin{aligned}&x_2 x_3-2 x_0 x_4-2 x_1 x_4,\\&2 x_0 x_3+x_2 x_4,\\&2 x_1 x_2+x_3 x_4,\\&4 x_1^2+x_3^2+x_4^2,\\&4 x_0 x_1-x_4^2,\\&4 x_0^2+x_2^2+x_4^2. \end{aligned}$$

The corresponding dual hypersurface $\left( X_A^{[2]} \right) ^*$, is defined by the sextic:

$$\begin{aligned} s(x_0,\ldots ,x_4) =&\; x_0^4 x_1^2-2 x_0^3 x_1^3+x_0^2 x_1^4+2 x_0^3 x_1 x_2^2+2 x_0^2 x_1^2 x_2^2-8 x_0 x_1^3 x_2^2+4 x_1^4 x_2^2+x_0^2 x_2^4 \\ {}&+8 x_0 x_1 x_2^4 -8 x_1^2 x_2^4+4 x_2^6+4 x_0^4 x_3^2-8 x_0^3 x_1 x_3^2+2 x_0^2 x_1^2 x_3^2+2 x_0 x_1^3 x_3^2\\&+20 x_0^2 x_2^2 x_3^2 -38 x_0 x_1 x_2^2 x_3^2 +20 x_1^2 x_2^2 x_3^2+12 x_2^4 x_3^2-8 x_0^2 x_3^4+8 x_0 x_1 x_3^4+x_1^2 x_3^4\\&+12 x_2^2 x_3^4+4 x_3^6+8 x_0^3 x_2 x_3 x_4-12 x_0^2 x_1 x_2 x_3 x_4-12 x_0 x_1^2 x_2 x_3 x_4+8 x_1^3 x_2 x_3 x_4\\&+36 x_0 x_2^3 x_3 x_4-72 x_1 x_2^3 x_3 x_4-72 x_0 x_2 x_3^3 x_4 +36 x_1 x_2 x_3^3 x_4-2 x_0^3 x_1 x_4^2\\&+8 x_0^2 x_1^2 x_4^2-2 x_0 x_1^3 x_4^2+2 x_0^2 x_2^2 x_4^2-2 x_0 x_1 x_2^2 x_4^2+20 x_1^2 x_2^2 x_4^2 +12 x_2^4 x_4^2\\&+20 x_0^2 x_3^2 x_4^2-2 x_0 x_1 x_3^2 x_4^2+2 x_1^2 x_3^2 x_4^2-84 x_2^2 x_3^2 x_4^2+12 x_3^4 x_4^2+36 x_0 x_2 x_3 x_4^3 \\&+36 x_1 x_2 x_3 x_4^3+x_0^2 x_4^4-10 x_0 x_1 x_4^4+x_1^2 x_4^4+12 x_2^2 x_4^4+12 x_3^2 x_4^4+4 x_4^6. \end{aligned}$$

A computation in Julia shows that this polynomial is a sum of squares. For instance:

$$\begin{aligned} s(x_0,\ldots ,x_4) =&\;\, 4 (-x_1^2 x_2+x_0 x_1 x_2-2 x_2 x_3^2+x_2^3-x_1 x_3 x_4+2 x_0 x_3 x_4+x_2 x_4^2)^2 \\&+ 4 (x_0^2 x_3-x_0 x_1 x_3+2 x_2^2 x_3-x_3^3+x_0 x_2 x_4-2 x_1 x_2 x_4-x_3 x_4^2)^2 \\&+ (2 x_0 x_1 x_4 - x_0 x_2 x_3 - x_1 x_2 x_3 + x_2^2 x_4 + x_3^2 x_4 - 2 x_4^3)^2 \\&+ (x_0^2 x_1 - x_0 x_1^2 + x_0 x_2^2 - x_0 x_4^2 - x_1 x_3^2 + x_1 x_4^2)^2 \\&+ 3 (x_0 x_2 x_3-x_1 x_2 x_3+x_2^2 x_4-x_3^2 x_4)^2 \\&+ 12 (x_2 x_3^2-x_0 x_3 x_4-x_2 x_4^2)^2 \\&+ 12 (x_2^2 x_3-x_1 x_2 x_4-x_3 x_4^2)^2. \end{aligned}$$

The seven cubic equations in this sum-of-squares decomposition generate the ideal of the Veronese surface $X_A \in \mathbb {P}^4$. This surface has smooth real points and is the (Zariski closure of the) real zero locus of s. Therefore, the real points of $\left( X_A^{[2]} \right) ^*$ do not form a hypersurface. Moreover, it is one of the two irreducible components of the singular locus of the sextic hypersurface defined by s in $\mathbb {P}^4$.

3 Del Pezzo

In this section we study the case in which $X\subset \mathbb {P}^4$ is a Del Pezzo surface of degree 4. Over $\mathbb {C}$ this type of surfaces arises as a complete intersection of two quadric threefolds. Alternatively, X can be realized as a complex algebraic surface as the blow up $\textrm{Bl}_{p_1,\ldots ,p_5} \mathbb {P}^2$ of five points in the (complex) projective plane, such that no three points lie on a line [7, Proposition 8.1.25]. Such a surface can then be embedded in $\mathbb {P}^4$ via its anticanonical linear system of divisors

$$\begin{aligned} |D| = |3\pi ^*(L) - \sum _{i=1}^5 E_i|, \end{aligned}$$

where L is the class of a line in $\mathbb {P}^2$, $\pi ^*(L)$ is its pullback on $\textrm{Bl}_{p_1,\ldots ,p_5} \mathbb {P}^2$, and $E_i$ is the exceptional divisor corresponding to $p_i$ (see [7, Chapter 8] for more). We study the real picture for all possible real forms of Del Pezzo surfaces of degree 4 in Subsection 3.2 below. The real forms have been classified for instance in [20].

3.1 The $\mathbb {C}$omplex picture

The embedding of $\textrm{Bl}_{p_1,\ldots ,p_5} \mathbb {P}^2$ into $\mathbb {P}^4$ defined by the linear system $|D| = |3\pi ^*(L) - \sum _{i=1}^5 E_i|$ fits into the following diagram:

The hyperplane sections $C_u$, $u\in (\mathbb {P}^4)^*$, of X are curves in |D|, i.e., strict transforms of cubic curves $Q_u \subset \mathbb {P}^2$ that pass through the points $p_1,...,p_5$. In a hyperplane section $C_u$ that contains an exceptional line $E_i$, the curve $C_u-E_i$, residual to the line in the hyperplane section, is the strict transform of a cubic curve $Q_u$ with multiplicity 2 at the point $p_i$.

If a hyperplane $u^\perp \subset \mathbb {P}^4$ is tangent to X, then the hyperplane section $C_u$, and hence also the plane cubic curve $Q_u$, is singular. Therefore, instead of studying hyperplanes tangent to X at a certain number of points, we can focus on curves in $\mathbb {P}^2$ passing through $p_1,\ldots ,p_5$, with prescribed singularities. The images of those points under the embedding of the surface to $\mathbb {P}^4$ are always linearly independent. Using this point of view, we want to analyze the irreducible components of the varieties $X^{[k]}\subset (\mathbb {P}^4)^*$ whose duals are hypersurfaces. First, we discuss the various irreducible components. The findings in the cases $k=2,3,4$ are summarized in Propositions 3.1, 3.2, 3.3.

Since a node at a base point corresponds to two singularities of a hyperplane section, whereas a node outside the base locus gives one singularity, our following case analysis for $X^{[k]}$ is based on partitions of k into parts that are restricted to be 1 or 2.

3.1.1 The irreducible components for k=2

Points of $X^{[2]}\subset (\mathbb {P}^4)^*$ are hyperplanes in $\mathbb {P}^4$ tangent to X at 2 points. So if $u\in X^{[2]}$, then $Q_u$ is a singular cubic curve. Let $q_1,q_2$ be the two singular points on $C_u$. There are two possibilities: either the points $q_1$ and $q_2$ are mapped to the same point in $\mathbb {P}^2$, or they are mapped to distinct points. We describe the two cases in the following paragraphs. We point out that by conic we always mean a smooth quadric curve in $\mathbb {P}^2$.

(A): irreducible cubic. If the two singularities $q_1$ and $q_2$ on $C_u$ are mapped to the same point in $\mathbb {P}^2$, they must lie on one of the exceptional lines $E_i$ so that they are mapped to $p_i$. The curve $Q_u$ is then singular at $p_i$ and has no other singularities. It is an irreducible cubic curve with multiplicity 2 at $p_i$, see Figure 1. For each exceptional line $E_i$ these hyperplane sections form a $\mathbb {P}^2\subset (\mathbb {P}^4)^*$. Therefore, the contribution of these irreducible components to the degree of $X^{[2]}$ is 1 each and so $5\cdot 1 = 5$ in total.

Fig. 1

Irreducible cubic curve through five fixed points, with a node at $p_i$

(B): conic + line. If the two singularities $q_1$ and $q_2$ on $C_u$ are mapped to distinct points in $\mathbb {P}^2$, the cubic curve $Q_u$ must be smooth at the points $p_i$ and have two singular points. An irreducible cubic has at most one singular point by Bézout’s Theorem, so $Q_u$ must be reducible: the union of a conic and a line. Five, four or three of the points $p_i$ must lie on the conic, see Fig. 2. In the first case, the conic is fixed, while there are no restrictions on the line, so the irreducible component of $X^{[2]}$ whose points identify this type of plane cubics is a $\mathbb {P}^2\subset (\mathbb {P}^4)^*$ and therefore of degree 1.

Fig. 2

Left: a conic through five points and a line. Center: a conic through three points and a line through two points. Right: a conic through four points and a line through one point

(C): conic + line. If three of the points $p_i$ lie on the conic, then the remaining two lie on and fix the line. (Figure 2, center). Again, for every choice of the two points, the associated component of $X^{[2]}$ is a $\mathbb {P}^2 \subset (\mathbb {P}^4)^*$, hence the sum of their degrees is $\left( {\begin{array}{c}5\\ 2\end{array}}\right) \cdot 1 = 10$.

(D): conic + line. The last possible case is obtained by imposing that the conic goes through four points and the line through one, as in Fig. 2, right. This can be done in 5 different ways, and every time we get a copy of $\mathbb {P}^1\times \mathbb {P}^1$. The degree of this type of components is $5\cdot 2 = 10$.

Summing up over all the irreducible components that we have just described, we get that $\deg X^{[2]} = 5+1+10+10 = 26$, as predicted in Table 1.

We now determine for which of these irreducible components, the dual varieties are hypersurfaces: this does not happen when the component is $\mathbb {P}^2\subset (\mathbb {P}^4)^*$. Therefore, the only relevant case is (D). The dual of a smooth quadric surface, for us $\mathbb {P}^1\times \mathbb {P}^1$ naturally embedded in a hyperplane of $(\mathbb {P}^4)^*$, has degree 2 [8, Proposition 2.9]: it is a quadric cone with vertex the hyperplane that contains $\mathbb {P}^1\times \mathbb {P}^1$. Hence, by dualizing these $\mathbb {P}^1\times \mathbb {P}^1$, we get the union of the five singular quadrics of the pencil defined by X. More precisely, X being the complete intersection of two quadric hypersurfaces $V_0,V_\infty \subset \mathbb {P}^4$, we can construct the associated pencil of quadrics that they generate:

$$\begin{aligned} \mathscr {L} = \{\lambda V_0 + \mu V_\infty \,|\, [\lambda ,\mu ] \in \mathbb {P}^1 \}. \end{aligned}$$

Every element of $\mathscr {L}$ is a quadric in $\mathbb {P}^4$, which is therefore represented by a $5\times 5$ matrix with linear homogeneous entries in $\lambda , \mu$. The determinant of this matrix has five zeros, which are isolated exactly when X is smooth. These values of $[\lambda ,\mu ]$ define the five singular quadrics $V_1,\ldots ,V_5$ of the pencil $\mathscr {L}$. To see that these constitute $\left( X^{[2]}_{(D)} \right) ^*$, note that $\mathbb {P}^1\times \mathbb {P}^1$ is contained in a $\mathbb {P}^3$ inside $(\mathbb {P}^4)^*$. Hence, its dual variety is a cone of degree 2 containing X: this is by definition one of the $V_i$’s. We summarize our findings on $X^{[2]}$ and $\left( X^{[2]} \right) ^*$.

Proposition 3.1

Let $X\subset \mathbb {P}^4$ be a smooth Del Pezzo surface, arising as the complete intersection of two generic quadric hypersurfaces $V_0, V_\infty$. Then, $X^{[2]}\subset (\mathbb {P}^4)^*$ is the union of 21 irreducible components, 16 of which are copies of $\mathbb {P}^2$. The remaining 5 are the singular quadrics in the pencil generated by $V_0, V_{\infty }$. Therefore, $\deg X^{[2]} = 26$.

The dual variety $\left( X^{[2]} \right) ^*\subset \mathbb {P}^4$ is the union of 16 copies of $\mathbb {P}^1$ and 5 quadric hypersurfaces, namely the singular quadrics of the pencil.

3.1.2 The irreducible components for k=3

The points of $X^{[3]}\subset (\mathbb {P}^4)^*$ are hyperplanes in $\mathbb {P}^4$ tangent to X at 3 points. Every such $u\in X^{[3]}$ identifies a hyperplane section $C_u$ with three singular points $q_1,q_2,q_3$, and a plane cubic curve $Q_u$ with singularities at the images of $q_1,q_2,q_3$ in $\mathbb {P}^2$. As in the case of $k=2$, either two of these three points are mapped to the same point $p_i$, or the three points are mapped to three different points distinct from the $p_i$ in $\mathbb {P}^2$. In the first case, $Q_u$ has multiplicity two both at a point $p_i$ and at some point in $\mathbb {P}^2$ distinct from the $p_i$. In the second case, $Q_u$ is smooth at the $p_i$ and has three points of multiplicity 2 distinct from the $p_i$. In the first case, $Q_u$ must be the union of a line and a conic, and in the second case, $Q_u$ must be the union of three lines, as in Fig. 3.

Fig. 3

Left: a conic through five points and a line through one point: they intersect at $p_i$. Center: a conic through four points and a line through two points: they intersect at $p_i$. Right: three lines through five points

(A): a node at $p_i$. When $Q_u$ is the union of a line and a conic, one or two of the points $p_i$ lie on the line as in Fig. 3, left or center. The curve of $u\in (\mathbb {P}^4)^*$, that identify plane curves $Q_u$ of this type, has degree 5 for any choice of the node $p_i$. The curve in $(\mathbb {P}^4)^*$ is indeed the union of five lines $\mathbb {P}^1\subset (\mathbb {P}^4)^*$. One of them arises by fixing the conic through $p_1,\ldots ,p_5$ and moving the line through $p_i$; the other four arise by fixing the line through $p_i$ and another point $p_j$ ($j\ne i)$. In total, the degree of these components sums to $5\cdot 5 = 25$. To compute the degree 5 we can also use Kazarian’s formulae for the case :

$$\begin{aligned} D = 3\pi ^*(L) - 2E_i - \sum _{k\ne i} E_k \qquad \quad K_X = -3\pi ^*(L) + \sum _{i=1}^5 E_i \end{aligned}$$

which implies, in Kazarian’s notation, that $\textsf{d} = 1, \textsf{k} = -3, \textsf{s} = 4, \textsf{x} = 8$. Substituting these quantities in the expression for $N_{A_1}$ gives 5.

(B): three lines. When the cubic $Q_u$ splits into three lines, two of the lines contain two points $p_i$, while the third line contains the last point (see Figure 3, right). There are $5\cdot \frac{1}{2}\left( {\begin{array}{c}4\\ 2\end{array}}\right) = 15$ ways in which one can choose the roles of the five points on the three lines. Each choice determines a line $\mathbb {P}^1\subset X^{[3]}$. So these components contribute with degree $15 \cdot 1 = 15$.

Proposition 3.2

Let $X\subset \mathbb {P}^4$ be a smooth Del Pezzo surface. Then $X^{[3]}\subset (\mathbb {P}^4)^*$ is the union of 40 lines $\mathbb {P}^1\subset (\mathbb {P}^4)^*$. Therefore, $\deg X^{[3]} = 40$.

The dual variety $\left( X^{[3]} \right) ^*\subset \mathbb {P}^4$ is the union of 40 projective planes $\mathbb {P}^2 \subset \mathbb {P}^4$.

3.1.3 The irreducible components for k=4

The variety $X^{[4]}$ is a union of points u such that $C_u$ has 4 singular points $q_1,q_2,q_3,q_4$. The cubic curve $Q_u$ can have at most three singular points so at least one of the singular points on $Q_u$ must be a point $p_i$. Hence, either one or two pairs of the points $q_i$ are mapped to the same point. If two pairs are mapped to a the same points, the cubic $Q_u$ has two singularities, at two of the points $p_i$, and it is the union of a line and a conic, as in Figure 4, left. If only one pair of points is mapped to the same point, the cubic $Q_u$ has three singular points and it is the union of three lines, as in Figure 4, right.

Fig. 4

Left: a conic through five points and a line through two points. Right: three lines through five points, two of them intersecting at $p_i$

(A): conic + line. If $Q_u$ is the union of a conic and a line, the only way for obtaining four singularities on $C_u$ is that the 2 intersection points of the two components are among the base points $p_1,\ldots ,p_5$, as in Figure 4, left. So the conic must pass through all five points and is therefore unique. This leaves $\left( {\begin{array}{c}5\\ 2\end{array}}\right)$ choices for the line which is uniquely determined by any two points $p_i$, $p_j$. These choices give therefore 10 points of $X^{[4]}$.

(B): three lines. Consider now plane cubics that split into three lines. Since they determine only three intersection points in $\mathbb {P}^2$, two of these lines must intersect in one of the base points $p_i$, as in Figure 4, right. To count them, we can first choose the point containing two lines and then choose the two points that determine the line that does not go through our first choice. Then, all three lines are determined so that this gives precisely $5\cdot \left( {\begin{array}{c}4\\ 2\end{array}}\right) = 30$ points in $X^{[4]}$.

There are no other possible configurations of plane cubics $Q_u$ such that $C_u$ has four singularities. Indeed, the cubic must be reducible again, so that it is either a conic and a line, or three lines. The conic and the line intersect in two points, hence these are forced to be two of the $p_i$’s, in order for $C_u$ to have four singularities. Then the conic is forced to pass also through the remaining three points, so it is fixed. In the case of three lines, two lines are forced to go through two of the base points each. The remaining line must go through the fifth point; to obtain in the end 4 singularities, this line must also intersect one of the other two lines in one of the $p_i$’s. These are therefore the only possibilities and they are the two cases discussed above. We summarize:

Proposition 3.3

Let $X\subset \mathbb {P}^4$ be a generic Del Pezzo surface. Then $X^{[4]}\subset (\mathbb {P}^4)^*$ is the union of 40 points. Therefore, $\left( X^{[4]}\right) ^*\subset \mathbb {P}^4$ is the union of 40 hyperplanes.

It turns out to be useful for the real picture in Sect. 3.2 to relate these 40 points to the quadrics of $X^{[2]}_{(D)}$.

Proposition 3.4

Let $Y_i$ be the five irreducible components of $X_D^{[2]}$ in some order. (They are the dual varieties to the five singular quadrics in the pencil defining X.) The set $X^{[4]}$ is the union of the pairwise intersections of the $Y_i$:

$$\begin{aligned} X^{[4]} = \bigcup _{i\ne j} Y_i\cap Y_j. \end{aligned}$$

Each intersection $Y_i\cap Y_j$ consists of four points and so the union is disjoint.

Proof

Let u be a point of $X^{[4]}$ so that the hyperplane section $C_u = u^\perp \cap X$ is a curve with four singularities. The case analysis above shows that this hyperplane section is a union of four lines in $\mathbb {P}^4$ (with either two exceptional divisors or one among them). Their intersection graph is a four cycle. The ideal of the curve $C_u = u^\perp \cap X$ in $u^\perp \cong \mathbb {P}^3$ is generated by two quadrics, namely those in the pencil defining X modulo the linear form given by u. This pencil contains two products of planes, where one of the planes is spanned by a pair of intersecting lines. The two planes of each pair intersect in a line spanned by two singular points of $C_u$, giving the missing two lines between the four points. These two singular quadrics in the pencil associated to $u^\perp \cap X$ come from singular quadrics in the pencil associated to X, say $V_i$ and $V_j$. Since $V_i$ and $V_j$ contain a plane $\mathbb {P}^2$, they must be singular. The line in $u^\perp$ in which the two planes intersect must pass through the cone point of the appropriate $V_k$ containing both planes. This shows that the line, in which the two planes intersect, is a line of the ruling of $V_k$. Now $u^\perp$ is the tangent hyperplane to $V_k$ along this line, which implies that u is in $Y_k$. Since this argument works both for $V_i$ and $V_j$, it shows $u\in Y_i\cap Y_j$ as desired. Since the varieties $Y_1,\ldots ,Y_5$ are surfaces of degree 2, the intersection $Y_i\cap Y_j$ contains 4 points. There are 10 such pairs and the union $X^{[4]}$ has 40 points, so that the union must be disjoint. $\square$

3.2 The $\mathbb {R}$eal picture

Up to now, we investigated the varieties $X^{[k]}$ and their duals for the Del Pezzo surfaces $X\subset \mathbb {P}^4$ from a point of view of complex projective geometry. Now we want to dive in the realm of convex and real algebraic geometry. In what follows, we want to address the question:

Which of the irreducible hypersurfaces in $\left( X^{[k]} \right) ^*$ can contribute to $\partial _a ({{\,\textrm{conv}\,}}(X))$?

This question is equivalent to ask:

Are the points of $X^{[k]}\subset (\mathbb {P}^4)^*$ supporting hyperplanes for X?

Recall that a hyperplane $u^\perp$ supports ${{\,\textrm{conv}\,}}(X)$ if $X(\mathbb {R})\cap u^\perp \ne \emptyset$ and <u, x> has the same sign for all $x\in X(\mathbb {R})$ (in affine coordinates on a chosen chart where $X(\mathbb {R})$ is compact).

Since we are dealing with convex bodies, we want our smooth real algebraic Del Pezzo surface X to have a compact real locus on some affine chart of $\mathbb {P}^4$. By the compactness of $\mathbb {P}^4(\mathbb {R})$, this is equivalent to the fact that the curve $H\cap X$ has no real points. The blow ups of the real projective plane in a real points and b pairs of complex conjugate points (with $a+2b = 5$) always contain a real line in $\mathbb {P}^4$ and hence do not satisfy this requirement. However, there is a full classification of the real forms of Del Pezzo surfaces, see [20]. There are three relevant topological types: the 4 real forms in [20, Corollary 3.2] that are not blow ups of the real projective plane do not contain real lines. The second to last case $(Q^{3,0} \times Q^{3,0})(0,4)$ in Russo’s list has no real points at all, though. So the interesting cases for us are the three types $Q^{3,1}(0,4)$, $Q^{2,2}(0,4)$ and $\mathbb {D}_4$. Here, the notation V(a, 2b) for a real algebraic variety V denotes the blow up of V in a real points and b complex conjugate pairs of points. The notation $Q^{r,s}$ stands for the quadric in $\mathbb {P}^{r+s-1}$ with signature (r, s, 0). The type $\mathbb {D}_4$ is a non-standard real structure (called de Jonquières involution) of the blow up of the real projective plane in five real points described in [20, Example 2 for $n=3$]. The structure of the algebraic boundary of the convex hull of X depends only on the topological type of our Del Pezzo surface X.

The complex algebraic discussion of the varieties $X^{[k]}$ in the previous subsections is relevant for all these real types: a real algebraic Del Pezzo surface X of degree 4 is embedded in $\mathbb {P}^4$ by its anticanonical linear system, so there is a complex projective automorphism $A\in {\mathrm PGL}_4(\mathbb {C})$ that maps it to a (complex algebraic) surface that is the blow up of $\mathbb {P}^2$ in 5 points as a complex algebraic variety. Since A is an automorphism of $\mathbb {P}^4$, tangency of X with hyperplanes $H\subset \mathbb {P}^4$ is preserved and the above discussion of the irreducible components of subvarieties $X^{[k]}$ of $X^*$ for surfaces that are blow ups is still relevant here. For $k=2$, the only candidates for irreducible components of the algebraic boundary of ${{\,\textrm{conv}\,}}(X)$ are the singular quadrics in the pencil of quadrics vanishing on X. Here we need to distinguish cases, depending on the signature of the real singular quadrics of the pencil.

So what are the signatures in each pencil? Every topological type of a smooth real Del Pezzo surface forms a connected family, see [13, Corollary 5.6]. Since a Del Pezzo surface of degree 4 in $\mathbb {P}^4$ is smooth if and only if there are precisely 5 distinct singular quadrics in its defining pencil, the connectedness of the topological types implies that the number of real singular quadrics in the pencil of a real Del Pezzo surface $X\subset \mathbb {P}^4$ and their signature is constant for every type. They can therefore be computed from an example, as we do in Examples 3.8, 3.9, 3.10. Table 3 shows the cases that are relevant for us.

Table 3 Topological types of Del Pezzo surfaces with no real lines, and the signatures of the real singular quadrics of the associated pencil

We now show that each topological type that does not contain a line has a compact real locus on a suitable affine chart of $\mathbb {P}^4$. In order to find a hyperplane H that contains no real points of X, we can reason as follows. Fix one of the real singular quadrics Q of the pencil associated to X having signature (3, 1, 1) and consider the projection $p:\mathbb {P}^4\dashrightarrow \mathbb {P}^3$ from its singular point. The image of the quadric itself under this projection is a quadric with signature (3, 1, 0), hence its real points form a sphere $S^2$. This projection, restricted to the Del Pezzo surface X, is a 2 : 1 cover. Let $H'\subset \mathbb {P}^3$ be a hyperplane that contains no real points of p(Q). Then, the preimage $H = p^{-1}(H')\subset \mathbb {P}^4$ is a hyperplane that contains no real points of X. The center of the projection does not lie on X since X is smooth and a complete intersection. In conclusion, the Del Pezzo surface X has a compact real locus in the affine chart $\mathbb {P}^4\setminus H$.

For dimensional reasons, we already know that $\left( X^{[3]}\right) ^*$ is not part of the algebraic boundary of ${{\,\textrm{conv}\,}}(X)$. On the other hand, $\left( X^{[4]}\right) ^*$ is a union of 40 hyperplanes, but none of them is a supporting hyperplane for $X(\mathbb {R})$. Indeed, according to Proposition 3.4 and its proof, if $u\in X^{[4]}$ is a real supporting hyperplane, then it supports ${{\,\textrm{conv}\,}}(X)$ at an edge, namely one of the lines in a ruling of two singular quadrics $V_i,V_j$. Since the dimension of the supported face is 1, its Zariski closure is not equal to $u^\perp$, so the latter cannot be an irreducible component of the algebraic boundary. In fact, such face is already captured by the varieties in $X^{[2]}_{(D)}$.

Let us focus now on $\left( X^{[2]}\right) ^*$. The convexity of ${{\,\textrm{conv}\,}}(X)$ imposes curvature conditions on the hypersurfaces in its algebraic boundary. A quadric with signature (2, 2, 1) is a cone over an hyperboloid in $\mathbb {P}^3$ which has negative Gaussian curvature at every point, and therefore a positive and a negative principal curvatures. However, the principal curvatures at a point in the boundary of a convex body must be, when defined, nonnegative [21, Section 2.5]. Hence, these quadrics cannot bound a convex set. Therefore, in the $Q^{3,1}(0,4)$ and $Q^{2,2}(0,4)$ cases there can be at most two of the varieties in $\left( X^{[2]} \right) ^*$, namely the quadrics with signature (3, 1, 1), that contribute to the algebraic boundary of the convex hull. In the case when X is of type $\mathbb {D}_4$, there are at most 4 quadric cones that can contribute to the algebraic boundary of the convex hull. For any of these four quadrics, there exists an affine chart in which it contributes to the algebraic boundary. To show this, fix one of these real singular quadrics, say V. Consider the projection p to $\mathbb {P}^3$ from the singular point of V. As we noticed above, this gives a 2 : 1 cover when restricted to X. The image p(X) is a quadric with signature (3, 1, 0), namely $S^2$. The preimage of each tangent hyperplane to a point on the sphere is a hyperplane in $\mathbb {P}^4$. By choosing an affine chart for instance with p at infinity, this hyperplane is a supporting hyperplane for X, because it was a supporting hyperplane in the projection. The same reasoning holds for all the singular quadrics in the pencil with signature (3, 1, 1).

In fact, not all four singular quadrics can show up in the boundary of ${{\,\textrm{conv}\,}}(X)$ in the same affine chart: they come in two pairs, and only one pair can be part of the algebraic boundary of ${{\,\textrm{conv}\,}}(X)$ in a given affine chart. We make this rigorous in the following proposition.

Proposition 3.5

Let $X\subset \mathbb {P}^4$ be a smooth real Del Pezzo surface of degree 4 and type $\mathbb {D}_4$. Let $V_1,V_2,V_3,V_4$ be the four singular quadrics of the pencil having signature (3, 1, 1). Then, there are exactly two pairs of quadrics such that in a given affine chart (where $X(\mathbb {R})$ is compact) only one pair can contribute to $\partial _a ({{\,\textrm{conv}\,}}(X))$.

Proof

The signatures of the matrices in the pencil associated to X follow from Table 3 and they are shown in Fig. 5.

Let us consider an affine chart on this $\mathbb {P}^1$ by assuming that the singular quadric with signature (2, 2, 1) is the point at infinity.

Fix now the quadric $V_1$. Up to a change of coordinates in $\mathbb {P}^4$, it has equation

$$\begin{aligned} x_0^2+x_1^2+x_2^2-x_3^2=0. \end{aligned}$$

Denote by $p_i$ the singular point of $V_i$. With this choice of coordinates $p_1 = [0,0,0,0,1]$. Consider the quadric $Q_i = V_1 + \varepsilon V_i$, for $i=2,3,4$ and $\varepsilon >0$. For $\varepsilon$ small enough, because of the diagram in Fig. 5 and our choice of affine chart in the pencil, the quadric $Q_i$ has signature (4, 1, 0). The determinant of the matrix associated to $Q_i$ will then be

$$\begin{aligned} -\varepsilon c_i + \varepsilon ^2 (\ldots ), \end{aligned}$$

where $c_i$ is the coefficient of $x_4^2$ in $V_i$. The signature of $Q_i$ implies that the determinant is negative. Hence, it determines the sign of $c_i$: in this case, for all i we have $c_i>0$. Notice that $V_i(p_1)=c_i$ for $i=2,3,4$, hence $V_i(p_1)>0$. Here we are using the notation $V_i$ for both the quadrics and its defining polynomial. We can repeat the same argument for $V_2,V_3,V_4$ and we get the sign conditions shown in Table 4.

Let us consider now an affine chart in which $V_2$ contributes to $\partial _a ({{\,\textrm{conv}\,}}(X))$. Since $V_2$ has signature (3, 1, 1), the convex hull of the Del Pezzo surface X in this chart must be contained in one of the regions in which $V_2<0$. Consider the quadric $V_3$ in this affine chart. The cone point $p_3$ belongs to the locus where $V_2$ is negative, therefore a real line of $V_3$ through $p_3$ will intersect X in two real points. The segment connecting these two points will contain $p_3$ in its interior. Since the discriminant of $V_3$ with respect to the projection from $p_2$ is definite (and hence has no real points), the two extrema of this segment belong to two distinct connected components. Assume by contradiction that this segment belongs to the boundary of ${{\,\textrm{conv}\,}}(X)$. Then the hyperplane H exposing that segment is the tangent hyperplane to $V_3$ at the points of the segment (they all have the same tangent hyperplane, since they lie on a line of a quadric cone through its vertex). Then H has a different sign on the two real connected components of X. Indeed, consider a line through $p_3$ such that $V_3(x)\le 0$ for every x on that line. Then H changes sign on that line in $p_3$. So H cannot be a supporting hyperplane.

This shows that if in the chosen affine chart $V_2$ is part of the algebraic boundary of ${{\,\textrm{conv}\,}}(X)$, then $V_3$ cannot be part of it. Repeating the same argument for all quadrics we find that in the same affine chart the following pairs of quadrics can bound simultaneously ${{\,\textrm{conv}\,}}(X)$:

$$\begin{aligned} V_1, V_2 \qquad \hbox {or} \qquad V_3, V_4 \qquad \hbox {or} \qquad V_1, V_4. \end{aligned}$$

In the next step, we exclude the pair $V_1, V_4$. Table 4 shows that $V_3(p_1)>0$ and $V_1(p_3)<0$. Geometrically, each $V_i$ is a cone over a quadric with signature (3, 1, 0) in $\mathbb {P}^3$, namely, a sphere. By the choice of the signature, in every affine chart $\mathbb {R}^4$ of $\mathbb {P}^4$ the region that contains the points inside the sphere is convex and it is given by a connected component of $V_i\le 0$. Therefore, $p_3$ is inside the sphere of $V_1$ and $p_1$ is outside the sphere of $V_3$. This implies that the two real intersection points of a line on $V_3$ through $p_3$ lie on the two connected components of $X(\mathbb {R})$. Since $p_1$ is outside of the sphere of $V_3$ and $p_3$ inside the sphere of $V_1$, two real intersection points with a line on $V_1$ through $p_1$ lie on the same connected component of $X(\mathbb {R})$. In particular, the projection from $p_1$ of the real points of X has two connected components. Of course, the same argument works symmetrically for $V_4$ instead of $V_1$. Regarding $V_2$ and $V_3$, the argument above with the hyperplane H shows that the projection of $X(\mathbb {R})$ from $p_2$ or $p_3$ is surjective.

Now fix an affine chart where $X(\mathbb {R})$ is compact and such that $V_1$ is an irreducible component of $\partial _a ({{\,\textrm{conv}\,}}(X))$. Assume by contradiction that the other boundary component of $\partial _a ({{\,\textrm{conv}\,}}(X))$ is $V_4$. The quadric $V_2$ is a cone over a sphere, with singular point $p_2\in \{x \in \mathbb {P}^4(\mathbb {R}):V_1(x)<0\}$. Since the set $\{x \in \mathbb {P}^4(\mathbb {R}):V_2(x)<0\}$ in any affine chart $\mathbb {R}^4$ containing $p_2$ is the union of two convex cones $C_1, C_2$, with $C_1\cap C_2 = p_2$, there exists an affine linear function $\ell$ on $\mathbb {R}^4$ that separates the cones, i.e., $\ell (p_2)=0$, $\ell (C_1{\setminus } p_2)>0$ and $\ell (C_2{\setminus } p_2)<0$. Since $p_1$ belongs to the region where $V_2$ is positive, we can also assume that $\ell (p_1) = 0$. Let $S_1$ and $S_2$ be the two connected components of the real locus $X(\mathbb {R})$ on our affine chart $\mathbb {R}^4$. Since $X(\mathbb {R})$ is compact on that chart, the reasoning above shows that $S_1 \subset C_1$ and $S_2 \subset C_2$, and therefore $\ell$ separates the two connected components of $X(\mathbb {R})$.

Let $p = \lambda y_1 + (1-\lambda ) y_2$ for some $\lambda \in (0,1)$, $y_i\in S_i$, such that $\ell (p)=0$. Consider then the line L spanned by p and $p_1$, which satisfies $\ell (L)=0$. Since $p_1$ does not belong to ${{\,\textrm{conv}\,}}(X)$, there exists a real point $q\in L$ such that $q\in \partial ({{\,\textrm{conv}\,}}(X))$. Since $\partial _a ({{\,\textrm{conv}\,}}(X))\subset (X^{[2]})^*$, we can write q as a convex combination of real points $x_1,x_2 \in X(\mathbb {R})$ where these two points belong to two different connected components of $X(\mathbb {R})$. Since $\partial _a ({{\,\textrm{conv}\,}}(X)) = V_1 \cup V_4$, then $q\in V_i$ for $i=1$ or 4. Therefore, $x_1, x_2, q \in V_i$ and since this is a quadric then the line spanned by the three points is in the ruling of $V_i$. By the previous part of the proof, for $i=1,4$ if a line of the ruling of $V_i$ meets X in two real points, then the latter are forced to lie on the same connected component of X. This gives a contradiction, since $x_1, x_2$ must lie on two different connected components. In conclusion, we have that

$$\begin{aligned} \partial _a ({{\,\textrm{conv}\,}}(X)) = V_1 \cup V_2 \qquad \text {or} \qquad \partial _a ({{\,\textrm{conv}\,}}(X)) = V_3 \cup V_4. \end{aligned}$$

$\square$

Fig. 5

The pencil of quadrics in the $\mathbb {D}_4$ type of Del Pezzo surface, with the respective signatures

Table 4 Signs of the singular quadrics with signature (3, 1, 1) evaluated at the singular points of the other quadrics

Remark 3.6

We point out that the pairs of quadrics in Proposition 3.5 are formed by consecutive quadrics in the pencil, such that the segment connecting them (the one that does not have other singular quadrics) is given by quadrics with signature (4, 1, 0). In both pairs we have one quadric whose ruling connects points of the same connected component of X (namely $V_1$ and $V_4$), and one quadric whose ruling connects points of different connected components of X (namely $V_2$ and $V_3$).

In conclusion, the algebraic boundary of a Del Pezzo surface consists of two singular quadrics of the associated pencil, with signature (3, 1, 1). We summarize our results for a Del Pezzo surface in the following theorem.

Theorem 3.7

Let $X\subset \mathbb {P}^4$ be a smooth real Del Pezzo surface of degree 4. Then there is an affine chart $\,\mathbb {P}^4\setminus H$ of $\,\mathbb {P}^4$ such that $X(\mathbb {R})\cap (\mathbb {P}^4\setminus H)$ is compact if and only if X does not contain any real line. There are three such topological types with $X(\mathbb {R})$ homeomorphic to $S^2$, $S^1\times S^1$ or the disjoint union $S^2\sqcup S^2$ of two copies of $S^2$, namely $Q^{3,1}(0,4)$, $Q^{2,2}(0,4)$, and $\mathbb {D}_4$. Then

$$\begin{aligned} \partial _a ({{\,\textrm{conv}\,}}(X)) = V_1 \cup V_2, \qquad \deg \partial _a ({{\,\textrm{conv}\,}}(X)) = 4. \end{aligned}$$

Here $V_1, V_2$ are two singular quadrics of the pencil having signature (3, 1, 1). For the types $Q^{3,1}(0,4)$, $Q^{2,2}(0,4)$ there is a unique choice; for the type $\mathbb {D}_4$ they are one of the two possible pairs, according to Proposition 3.5.

We want to point out the big improvement that we obtained with this case-by-case analysis: Table 1 gives a variety of degree $12+26+40+40$ in $\mathbb {P}^4$ to be dualized in order to find the algebraic boundary of X. In the end, taking into account real and convex aspects, only 4 of the 118 degrees and 2 of the 113 irreducible components are relevant.

Example 3.8

Consider the two quadrics $V_0,V_{\infty }$ defined respectively by the polynomials

$$\begin{aligned} f_0&= x_0^2-x_1^2-x_2^2-x_3^2-x_4^2,\\ f_{\infty }&= 2 x_2^2-2 x_1 x_3+2 x_0 x_4. \end{aligned}$$

Let $X\subset \mathbb {P}^4$ be the smooth Del Pezzo surface with ideal $I = \langle f_0, f_{\infty } \rangle$.

Fig. 6

A hyperplane section of the Del Pezzo surface in Example 3.8: here the green curve. The blue and the red quadric are (3.1) and (3.2) respectively

The associated pencil of quadrics $\mathscr {L}$ is defined by the pencil of matrices

$$\begin{aligned} \begin{bmatrix} -\lambda &{} 0 &{} 0 &{} 0 &{} \mu \\ 0 &{} \lambda &{} 0 &{} -\mu &{} 0 \\ 0 &{} 0 &{} \lambda +2\mu &{} 0 &{} 0 \\ 0 &{} -\mu &{} 0 &{} \lambda &{} 0 \\ \mu &{} 0 &{} 0 &{} 0 &{} \lambda \end{bmatrix} \end{aligned}$$

for $[\lambda ,\mu ] \in \mathbb {P}^1$. This Del Pezzo surface is of type $Q^{3,1}(0,4)$, hence it is topologically $S^2$ and among the five singular quadrics of the pencil, three are real. The signature of their matrices is either (3, 1, 1) or (2, 2, 1). These quadrics are:

$$\begin{aligned} (3,1,1)&:{} & {} x_0^2-x_1^2-3 x_2^2+2 x_1 x_3-x_3^2-2 x_0 x_4-x_4^2 = 0, \end{aligned}$$

(3.1)

$$\begin{aligned}{} & {} {}&2 x_0^2-2 x_1^2-2 x_1 x_3-2 x_3^2+2 x_0 x_4-2 x_4^2 = 0,\\ (2,2,1)&:{} & {} x_0^2-x_1^2+x_2^2-2 x_1 x_3-x_3^2+2 x_0 x_4-x_4^2 = 0,\nonumber \end{aligned}$$

(3.2)

and can be obtained for $[\lambda ,\mu ] = [1,1], [1,-1], [2,-1]$ respectively. Figure 6 shows the hyperplane section $2 x_1 - 1 = 0$ of the Del Pezzo surface and the two (3, 1, 1) quadrics, in the affine chart $\{x_0 \ne 0\}$. Notice that the hyperplane section of the convex hull of X is not the convex hull of the hyperplane section of X.

Example 3.9

Consider the two quadrics $V_0,V_{\infty }$ defined respectively by the polynomials:

$$\begin{aligned} f_0&= 2 x_0^2 - 3 x_1^2 - x_2^2 + x_4^2,\\ f_{\infty }&= 3 x_0^2 - 2 x_1^2 - x_3^2 - x_4^2. \end{aligned}$$

Let $X\subset \mathbb {P}^4$ be the smooth Del Pezzo surface with ideal $I = \langle f_0, f_{\infty } \rangle$.

Fig. 7

The hyperplane section $\{x_2=1\}$ of the Del Pezzo surface in Example 3.9, in the affine chart $\{x_0\ne 0\}$: here the green curve. The red and the blue cylinders are the zero loci of $f_{\infty }$ and the other (3, 1, 1) quadric, respectively

The associated pencil of quadrics $\mathscr {L}$ is defined by the pencil of matrices

$$\begin{aligned} \begin{bmatrix} 2 \lambda + 3 \mu &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} - 3 \lambda - 2 \mu &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} - \lambda - \mu &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} -\mu &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} \lambda - \mu \end{bmatrix} \end{aligned}$$

for $[\lambda ,\mu ] \in \mathbb {P}^1$. This Del Pezzo surface is of type $Q^{2,2}(0,4)$, hence it is topologically $S^1\times S^1$ and all the five singular quadrics of the pencil are real. There are three matrices with signature (2, 2, 1) and two with signature (3, 1, 1). These quadrics are:

$$\begin{aligned} (3,1,1)&:{} & {} 5 x_0^2 - 5 x_1^2 - x_2^2 - x_3^2 = 0, \qquad \qquad \\{} & {} {}&3 x_0^2 - 2 x_1^2 - x_3^2 - x_4^2 = 0, \qquad \qquad \\ (2,2,1)&:{} & {} 2 x_0^2 - 3 x_1^2 - x_2^2 + x_4^2 = 0, \qquad \qquad \\{} & {} {}&5 x_1^2 + 3 x_2^2 - 2 x_3^2 - 5 x_4^2 = 0, \qquad \qquad \\{} & {} {}&5 x_0^2 + 2 x_2^2 - 3 x_3^2 - 5 x_4^2 = 0, \qquad \qquad \end{aligned}$$

and can be obtained for $[\lambda ,\mu ] = [1,1], [0,1] , [1,0], [-3,2], [-2,3]$ respectively. The real part of X is compact in the affine chart $\{x_0=1\}$. Figure 7 shows X and the two (3, 1, 1) quadrics in the hyperplane section $\{x_2=1\}$ of the affine chart $\{x_0\ne 0\}$.

Example 3.10

Consider the two quadrics $V_0,V_{\infty }$ defined respectively by the polynomials:

$$\begin{aligned} f_0&= 2 x_0^2 - x_1^2 - x_3^2 - x_4^2,\\ f_{\infty }&= x_0^2 - 2 x_1^2 - x_2^2 - x_4^2. \end{aligned}$$

Let $X\subset \mathbb {P}^4$ be the smooth Del Pezzo surface with ideal $I = \langle f_0, f_{\infty } \rangle$.

Fig. 8

The hyperplane section $\{x_1=1\}$ of the Del Pezzo surface in Example 3.10, in the affine charts $\{x_0\ne 0\}$ (left) and $\{x_3\ne 0\}$ (right). The section of the Del Pezzo surface is here the green curve. The blue and red as well as the orange and purple (3, 1, 1) quadrics are the two pairs of possible algebraic boundaries of X, according to Proposition 3.5

The associated pencil of quadrics $\mathscr {L}$ is defined by the pencil of matrices

$$\begin{aligned} \begin{bmatrix} 2 \lambda + \mu &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} - \lambda - 2 \mu &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} - \mu &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} - \lambda &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} - \lambda - \mu \end{bmatrix} \end{aligned}$$

for $[\lambda ,\mu ] \in \mathbb {P}^1$. This Del Pezzo surface is of type $\mathbb {D}_4$, hence it is topologically $S^2\sqcup S^2$ and all the five singular quadrics of the pencil are real. The signature of one of their matrices is (2, 2, 1), and it is (3, 1, 1) for the other four. These quadrics are:

and can be obtained for $[\lambda ,\mu ] = [0,1] , [1,0], [1,-2], [1,-1], [2,-1]$ respectively. The teal labels refer to Fig. 5. The real part of X is compact in the affine chart $\{x_0=1\}$. Figure 8 shows X and the four (3, 1, 1) quadrics in the two possible configurations.

4 Beyond

We discuss in this section criteria and difficulties of generalizing the previous discussion for any surface $X \subset \mathbb {P}^4$. In particular, the Bordiga surface, isomorphic to the blow up of $\mathbb {P}^2$ in 10 generic points $p_1,...,p_{10}$, will be used as a recurrent example. Assuming that no three points are collinear, no six points on a conic, and no ten points on a cubic, this is a smooth irreducible surface, also relevant in Computer Vision [3]. The embedding of $\textrm{Bl}_{p_1,\ldots ,p_{10}} \mathbb {P}^2$ into $\mathbb {P}^4$ is defined by the linear system of divisors

$$\begin{aligned}|D|=|4\pi ^*(L) - \sum _{i=1}^{10} E_i|,\end{aligned}$$

and fits into the diagram:

For more details on the geometry of this surface we refer to [4, 7].

4.1 The $\mathbb {C}$omplex picture

As already pointed out in the course of the paper, to find the algebraic boundary of the convex hull, we first describe $X^{[k]}$. Their degrees can be computed using Kazarian’s formulae [11]. The next step is for each k, and each irreducible component of $X^{[k]}$, to find the dual variety $\left( X^{[k]} \right) ^*$, its dimension, degree, and irreducible components. Finally, one has to select the hypersurfaces, among the $\left( X^{[k]} \right) ^*$, that are actually real and support the convex hull of X. This is the algorithm that allows to compute the algebraic boundary of the convex hull of X. In practice, however, the above steps are not trivial at all and require a very good understanding of all the varieties involved. The starting point, as we saw in the Del Pezzo case, is to investigate the geometry of the $X^{[k]}$ themselves. We already observed that $X^{[1]}$ is not important, since $\left( X^{[1]} \right) ^*=X$ is not a hypersurface. On the other hand, we know that $\left( X^{[4]} \right) ^*$ is a union of finitely many hyperplanes. We are therefore left to study the components of $X^{[2]}$, $X^{[3]}$, and their duals. The dualization process is done irreducible component by irreducible component. Depending on the surface X we are dealing with, the number of components can be very different. For instance, if X is a K3 surface in $\mathbb {P}^4$, then by [5, 6] the varieties $X^{[k]}$ are irreducible. If we consider instead a rational surface X, with a larger Picard group, there may be various irreducible components of $X^{[2]}$ and $X^{[3]}$, corresponding to families of plane curves, as in the Del Pezzo case. When studying a Bordiga surface X, we get that

$$\begin{aligned} X^{[2]}&= X^{[2]}_{(A)} \cup X^{[2]}_{(B)} = \bigcup _{i=1}^{10} \mathbb {P}^2 \cup X^{[2]}_{(B)}{} & {} \subset (\mathbb {P}^4)^*,\nonumber \\ X^{[3]}&= X^{[3]}_{(A)} \cup X^{[3]}_{(B)} \cup X^{[3]}_{(C)} = \bigcup _{i=1}^{55} \mathbb {P}^1 \cup C_4 \cup \bigcup _{i=1}^{10} Y_i{} & {} \subset (\mathbb {P}^4)^*. \end{aligned}$$

(4.1)

A 2-nodal hyperplane section in $X^{[2]}_{(A)}$ has an exceptional line as component, so $X^{[2]}_{(A)}$ is the union of ten planes, one for each exceptional line. The 2-nodal hyperplane sections in $X^{[2]}_{(B)}$ correspond to plane quartics through $p_1,...,p_{10}$, with two nodes. So $X^{[2]}_{(B)}$ is a linear surface section of the variety of 2-nodal plane curves, a surface of degree 225.

The 3-nodal hyperplane sections in $X^{[3]}_{(A)}$ correspond to the 45 pencils of curves that consists of the union of a plane cubic curve through eight $p_i$ and the line through the remaining two $p_i$, and 10 pencils of curves that consist of the union of a cubic curve through nine $p_i$ and a line through the tenth $p_i$. The 3-nodal hyperplane sections in $C_4$ correspond to plane quartic curves through the points $p_i$ of genus zero, while the 3-nodal hyperplane sections $Y_i$ are hyperplane sections that contain an exceptional line and have an additional node outside this line. For each exceptional line, these form a plane curve of degree 20 in $(\mathbb {P}^4)^*$. For details see [14, Section 4.1.2]. At this stage, one can get rid of the linear components, since their dual linear spaces are not hyperplanes, and focus on the remaining ones.

In general, $X^{[k+1]}$ will be contained in the singular locus of $X^{[k]}$. However, $X^{[k+1]}$ detects only nodal singularities, and it is reasonable to expect that the singular locus of $X^{[k]}$ will have also other components, see Remark 4.5. The expected dimensions of $X^{[2]}$ and $X^{[3]}$ are 2 and 1 respectively. Because they are neither hypersurfaces nor smooth varieties, there are no general formulae for computing the degree of their duals. In fact, both the singular locus and the topological invariants of the desingularization are needed to compute the dual of such a singular variety of codimension larger than one. In the special case of a Bordiga surface we are still able to compute the degree of $\left( X^{[3]}\right) ^*$.

4.1.1 Computations for the Bordiga surface

Consider a Bordiga surface realized as a complex algebraic surface as the blow up $\textrm{Bl}_{p_1,\ldots ,p_{10}} \mathbb {P}^2$ of ten generic points in the complex projective plane, embedded in $\mathbb {P}^4$ via the linear system of divisors $|D| = |4\pi ^*(L) - \sum _{i=1}^{10} E_i|$. The associated varieties $X^{[2]}$, $X^{[3]}$ are described in (4.1). Regarding $X^{[2]}$, it is not clear to the authors how to find the degree of the dual of the surface $X^{[2]}_{(B)}$. On the other hand, using some ad hoc strategies, we are able to compute the degrees of the components $C_4^*$ and $Y_i^*$ of $\left( X^{[3]}\right) ^*$. Since the methods might be useful for other cases, we discuss them here.

The general points on the curve $C_4$ correspond, under the blow up map, to plane quartic curves through $p_1,\ldots ,p_{10}$ with three nodes. The curve $C_4$ was studied extensively by Pandharipande: it is an irreducible curve of degree 620 [12], with arithmetic genus 5447 and geometric genus 725, which were computed (and denoted as $g_4$ and $\tilde{g}_4$ respectively) in [16, Section 3]. We will compute the degree of the hypersurface dual to $C_4$. We reduce the computation of this degree to the Riemann-Hurwitz formula with the following well-known observation.

Proposition 4.1

Let $C\subset \mathbb {P}^n$ be an irreducible curve and let $\mathcal {T}$ be its tangent developable (that is the Zariski closure of the union of all tangent lines to C at smooth points). Then, the degree of $\mathcal {T}$ and the degree of the dual variety $C^*$ agree.

Proof

The tangent developable $\mathcal {T}$ of C is a surface in $\mathbb {P}^n$. Its degree is obtained by intersecting it with a generic subspace $U\subset \mathbb {P}^n$ of codimension 2. On the other hand, $C^*$ is a hypersurface and its degree is computed by intersecting it with a generic line $L\subset (\mathbb {P}^n)^*$. A generic subspace $U\subset \mathbb {P}^n$ of codimension 2 is dual to a generic line $L\subset (\mathbb {P}^n)^*$. An intersection point of L with $C^*$ corresponds to a hyperplane $H\subset \mathbb {P}^n$ that is tangent to C at a smooth point x. Since this hyperplane H moves in the pencil of hyperplanes containing $L^* = U$, the tangent line to C at x intersects U. Conversely, an intersection point x of U with the tangent developable $\mathcal {T}$ of C gives rise to a tangent hyperplane H to C, namely the span of U and the tangent line to C, on which x lies. So this H is an intersection point of L with $C^*$. This one-to-one correspondence between intersection points of U with $\mathcal {T}$ and intersection points of L with $C^*$ shows the claim that $\deg \mathcal {T} = \deg C^*$. $\square$

Remark 4.2

The tangent developable $\mathcal {T}$ gives rise to a curve $C_{\mathcal {T}}$ in $\textbf{Gr}(2,n+1)$, and $C^*$ to a curve $C^*_G$ in $\textbf{Gr}(n-1,n+1)$. Duality sets up a natural isomorphism $\textbf{Gr}(2,n+1) \rightarrow \textbf{Gr}(n-1,n+1)$ that restricts to an isomorphism $C_{\mathcal {T}}\rightarrow C^*_G$.

Now we compute $\deg \mathcal {T}$ via the Riemann-Hurwitz formula: consider the projection of $C_4$ from a generic projective plane to $\mathbb {P}^1$; each tangent line that intersects the chosen plane gives a ramification point of the projection. The degree of the projection is $\deg C_4 = 620$. Pandharipande’s work provides an accurate description of the singularities of $C_4$. We deduce that the ramification points of the map arise either from points of $C_4$ at which the tangent line intersects the given plane, or as projections of the cusps of $C_4$. The cusps of this curve correspond exactly to plane quartics $Q_u$ with two nodes and a cusp [16, Lemma 3]. There are $N_{A_1^2 A_2}=2304$ curves with that property that go through $p_1,\ldots ,p_{10}$. We now put together all these information in the Riemann-Hurwitz formula:

$$\begin{aligned} 2\cdot 725 - 2 = 620 \cdot (-2) + (2304 + \deg \mathcal {T}), \end{aligned}$$

from which we obtain that $\deg C_4^* = \deg \mathcal {T} = 384$.

In order to conclude the algebraic study of the degree of $\left( X^{[3]}\right) ^*$, we are left with the computation of $\deg Y_i$. Fix $i\in \{1,\ldots ,10\}$. The points of $Y_i\subset (\mathbb {P}^4)^*$ correspond, via the blow up map, to plane curves with a node at $p_i$ and another node in the plane. Every i gives rise to one such irreducible component. We write $Y=Y_i$ in what follows, to simplify the notation.

Theorem 4.3

The curve Y is a plane curve of degree 20 and geometric genus 9 with 114 nodes and 48 cusps. A point $u\in Y$ is a node if and only if $Q_u$ has a node at $p_1$ and two other nodes in $\mathbb {P}^2$. A point $u\in Y$ is a cusp if and only if $Q_u$ has a node at $p_1$ and a cusp in $\mathbb {P}^2$. The plane curve dual to Y has degree 8 and 12 cusps and no nodes.

Proof

It will be convenient to change the point of view and introduce an appropriate incidence variety. Let us first define the ambient space. Notice that the curve $C_u$ always contains the exceptional line $E_1$, since $Q_u$ has a node at $p_1$. Hence, the u’s that belong to Y must satisfy $u^\perp \supset E_1$. This condition on the points cuts out a projective plane in $(\mathbb {P}^4)^*$ that we denote by $\mathcal {H}_{E_1}$. Therefore, Y is a plane curve in $\mathcal {H}_{E_1}\subset (\mathbb {P}^4)^*$.

The curves $C_u$ for $u\in \mathcal {H}_{E_1}$ form the (base-point free) linear system associated to the divisor $D = 4\,L - 2E_1 - \sum _{i=2}^{10} E_i$ on X. The associated line bundle has 3 global sections and the associated morphism to $\mathbb {P}^2$ has degree $D^2 = 16 - 4 - 9 = 3$. We denote this 3 : 1 cover of $\mathbb {P}^2$ by f. Our curve Y turns out to be the dual curve to the branch divisor of f, which is described in detail in [15, Section 10].

The Tschirnhausen bundle associated to f in [15] is the bundle $E = \mathcal {O}_{\mathbb {P}^2}(-2)\oplus \mathcal {O}_{\mathbb {P}^2}(-2)$ of rank 2 on $\mathbb {P}^2$, compare ibid., Table 10.5. Indeed, the associated surface S is the blow up of $\mathbb {P}^1 \times \mathbb {P}^1$ in 9 points. The pullback of a line in $\mathbb {P}^2$ corresponds to a curve of bidegree (2, 3) in $\mathbb {P}^1\times \mathbb {P}^1$ through the nine base points. This surface S is isomorphic to our Bordiga surface via the isomorphism of the blow up of $\mathbb {P}^1\times \mathbb {P}^1$ in one point and the blow up of $\mathbb {P}^2$ in two points since $2(L-E_2) + 3(L-E_1) - (L-E_1-E_2) - \sum _{i=3}^{10} E_i = 4\,L - 2E_1 - \sum _{i=2}^{10}E_i$. This decomposition is meaningful because the $(-1)$-curve $L-E_1-E_2$ in the blow up of $\mathbb {P}^2$ in two points corresponds to the exceptional divisor of the blow up of $\mathbb {P}^1\times \mathbb {P}^1$ in one point. Miranda’s results give us the numbers in Theorem 4.3 via the branch divisor B of f. First, the degree of the branch divisor $B\subset \mathbb {P}^2$ is 8 by [15, Proposition 4.7]. Secondly, its singularities are determined by ibid., Lemma 4.8, which says that every singularity of the branch divisor B of f is a point of total ramification, and ibid., Corollary 5.8(ii), which shows that such a total ramification point is a double point of B with one tangent and therefore generally an ordinary cusp. The number of such cusps is then counted in ibid., Lemma 10.1 as $3c_2(E)$ in terms of the second Chern class of the Tschirnhausen bundle E. Since this bundle is split in our case, we can compute the Chern classes of E as the elementary symmetric polynomials in the Chern classes of the line bundle $\mathcal {O}_{\mathbb {P}^2}(-2)$ so that $c_1(E) = -4L$ and $c_2(E) = 4 L^2$ in the Chow ring of $\mathbb {P}^2$, see [8, Corollary 5.4]. So in summary, the curve B has degree 8, 12 cusps, and genus 9.

We show next that Y is indeed the dual curve to the branch divisor B of f. The pullback of a line L in $\mathbb {P}^2$ via f is the curve $C_u$ for some appropriate $u\in \mathcal {H}_{E_1}$ and the restriction of f to $Q_u$ is a triple cover of $\mathbb {P}^1$. Since $Q_u$ has genus 2 and the branch divisor has degree 8, the curve $Q_u$ is smooth if the intersection of the line L and the branch divisor B is transversal by Riemann-Hurwitz. If the line L is tangent to B at a smooth point $p\in B$, then we choose a local coordinate x on L so that $p=0$. By Miranda’s results, the fiber $f^{-1}(p)$ contains two points, one of them double. Locally (in an analytic sense) around the double point, we get a 2 : 1 cover of an affine line that we can write as $z^2 = a_2x^2 + a_3x^3 + \ldots$. Here, the term on the right hand side starts with a quadratic term because L is tangent to B at p. If $a_2\ne 0$, this shows that the curve $Q_u$ has a node at the double point in $f^{-1}(p)$. If the line L is actually a flex line, then $a_3$ is also 0 and the double point in $f^{-1}(p)$ is a cusp of $Q_u$. This shows that Y is the dual curve of B. Moreover, it provides the characterization of the singularities of Y in the theorem. Their numbers are determined by Plücker’s formulae in terms of the degree and the singularities of B. If d, g, $\delta$, and $\kappa$ denote the degree, geometric genus, number of nodes, and number of cusps for a plane curve C and dually $d^*$, $g^*$, $\delta ^*$, and $\kappa ^*$ the invariants of $C^*$ then

$$\begin{aligned}&d^* = d(d-1) - 2\delta - 3\kappa \\&d = d^*(d^*-1) - 2\delta ^* - 3 \kappa ^*\\&g = g^* =\frac{1}{2} (d^*-1)(d^*-2) - \delta ^* - \kappa ^* \end{aligned}$$

holds, see [17]. So with $Y = B^*$ as well as $d=8$, $\delta =0$, and $\kappa = 12$, we find $d^* = 20$, $\delta ^* = 114$, and $\kappa ^* = 48$. $\square$

The dual of the curve Y as a curve in $\mathbb {P}^4$ is a hypersurface that is a cone, with vertex a line, over the plane curve $Y^*$. So we get the following result.

Corollary 4.4

The variety $\left( X_{(C)}^{[3]} \right) ^* \subset \mathbb {P}^4$ is the union of 10 hypersurfaces, arising as cones over plane curves, each of degree 8.

Remark 4.5

We can also count the singularities of Y by counting plane quartics $Q_u$ that are more singular. More precisely, the number of nodes of Y is the number of plane quartics with a node at $p_1$ and two other nodes on $\mathbb {P}^2$; the number of cusps of Y is the number of plane quartics with a node at $p_1$ and a cusp somewhere in $\mathbb {P}^2$. They are Kazarian’s numbers $N_{A_1^2}=114$ and $N_{A_2}=48$ respectively.

4.2 The $\mathbb {R}$eal picture

Once the complex study of the $\left( X^{[k]}\right) ^*$ is done, we can focus on the real geometry. The first question regards the variety X directly: does there exist an affine chart in $\mathbb {P}^4$ such that the real points of X in this chart form a compact set? If X is the blow up of $\mathbb {P}^2$ in a certain number of points, then we can reason as follows. If a base point were real, then the associated exceptional curve would be real as well. In our cases of interest (Del Pezzo and Bordiga surfaces), these $(-1)$-curves will be embedded as lines and since a line and a hyperplane in $\mathbb {P}^4$ always intersect, there cannot be any affine chart where the real part of X is compact. Therefore, the base points must be all complex. Since we want X to be defined over the reals though, the base points must come in pairs of complex conjugates. This implies for instance that a necessary condition is that there is an even number of base points. In general however, an exceptional curve on a smooth surface need not be a line and may have compact real locus. In such a case, there could be real points in the base locus of the blow up.

As for $\left( X^{[k]}\right) ^*$, we can deduce from their real geometry which irreducible components support the convex hull of X. In fact, if $Y\subset X^{[k]}$ is an irreducible component, such that its dual variety $Y^* \subset \left( X^{[k]}\right) ^*$ is a hypersurface, we want that for a full dimensional subset $U\subset Y$ the following holds: for all hyperplanes $u^\perp \in U$, X is contained in one of the halfspaces bounded by $u^\perp$. In the case in which X is the blow up of $\mathbb {P}^2$ in some points, this boils down to study the the sign of a polynomial. In the notation of the blow up that we used so far, we have that $u^\perp$ is a supporting hyperplane if and only if the polynomial $q_u$ defining the quartic $Q_u\subset \mathbb {P}^2$ has constant sign on $\mathbb {P}^2$. We exhibit the computation for the case in which X is a Bordiga surface.

4.2.1 Computations for the Bordiga surface

Continuing Sect. 4.1.1, we are going to examine the points of the $X^{[k]}$ to understand if they are realizable over the reals and if they can give rise to supporting hyperplanes, in the various cases. We will focus only on those components of the $X^{[k]}$ whose duals are hypersurfaces and so have a chance to be part of the algebraic boundary. This analysis will affect the choices of the ten base points $p_1,\ldots , p_{10}$. The results will be summarized in Theorem 4.6.

We start with $X^{[2]}$. The only candidate here is $X^{[2]}_{(B)}$, whose points u correspond to plane quartic curves through $p_1,\ldots ,p_{10}$, with two nodes in $\mathbb {P}^2$. In order to show that this configuration can produce supporting hyperplanes, we will construct first the hyperplane section, and then choose the Bordiga surface by selecting suitable 10 points to be blown up. Define the quartic polynomial $q_u$ in the following way. Pick two points x, y in the plane and let $c_1, c_2, c_3$ be homogeneous quadratic polynomials in three variables defining three conics that intersect in x, y. Define $q_u = c_1^2+c_2^2+c_3^2$. Since it is a sum of squares, $q_u$ is non-negative on the whole $\mathbb {P}^2$, and zero only at the two singular points x, y. Hence $u^\perp$ is a separating hyperplane for X, where X is the blow up of $\mathbb {P}^2$ in ten points that are five pairs of complex conjugate zeros of $q_u$.

Regarding $X^{[3]}$, we have to analyze both $C_4$ and $Y_i$. Points of $C_4$ correspond to rational quartics through $p_1,\ldots ,p_{10}$, which have three nodes x, y, z in $\mathbb {P}^2$. We can repeat the same procedure as for $X^{[2]}_{(B)}$ to construct a separating hyperplane; the only difference is that now the three conics will have to intersect in the three nodes x, y, z. The points in $Y_i$ represent plane quartics $Q_u$ with a node at $p_i$ and another node $x\in \mathbb {P}^2$. Since we want the hyperplanes coming from $X^{[k]}$ to support X in k real points, then in this case $p_i$ is forced to have real coordinates. This implies that there is also another base point, say $p_j$ which is real. However, $p_j$ is a smooth point for the quartic $Q_u$. Therefore, in a neighborhood of $p_j$ the polynomial $q_u$ changes sign. This implies that these $u^\perp$ cannot be supporting hyperplanes for X, independently of the choice of the base points.

Finally, $X^{[4]}$ is subdivided in three cases. In the first case (which counts 666 points), the corresponding quartics $Q_u$ are reducible, thus they are union of either two conics or a line and a cubic. In both cases, there is no way to make just the four singular points real, hence these reducible quartics change sign in $\mathbb {P}^2$. So the associated hyperplanes cannot support X, independently of the choice of the $p_i$. The second case (counting 1050) is similar to the discussion on $Y_i$, since we have a node at $p_i$ and therefore some $p_j$ must be real as well. However, it is a smooth point of $Q_u$, and therefore there is a sign change that prevent $u^\perp$ from being a supporting hyperplane, independently of the choice of the base points. The points of the last type are 45 and correspond to quartics with two nodes at, say, $p_1,p_2$. Therefore, both of them must be real. In this setting, we can consider three conics $c_1,c_2,c_3$ that intersect exactly at $p_1,p_2$. Then, define $q_u = c_1^2+c_2^2+c_3^2$. This provides a non-negative quartic on $\mathbb {P}^2$ which is zero precisely at $p_1,p_2$. Hence $u^\perp$ is a separating hyperplane for X, where X is the blow up of $\mathbb {P}^2$ in ten points that are $p_1,p_2$ and four pairs of complex conjugate zeros of $q_u$. In particular, this shows that just one hyperplane among these 45 can be a supporting hyperplane of X. However, since in this case two base points need to be real, there are no affine charts in which X has a compact real locus.

The following statement summarizes the whole discussion on the Bordiga case. One can construct explicit examples which prove that the inclusion and the inequality in Theorem 4.6 below can be equalities.

Theorem 4.6

Let $X\subset \mathbb {P}^4$ be a smooth Bordiga surface with compact real locus in an appropriate affine chart, arising as the blow up of $\mathbb {P}^2$ at ten base points $p_1,\ldots ,p_{10}$ which are five generic pairs of complex conjugates. Then

$$\begin{aligned} \partial _a ({{\,\textrm{conv}\,}}(X)) \subset \left( X^{[2]}_{(B)} \right) ^* \cup C_4^*, \qquad \deg \partial _a ({{\,\textrm{conv}\,}}(X)) \le \deg \left( X^{[2]}_{(B)} \right) ^* + 384. \end{aligned}$$

It is natural to ask whether equality in the formulae in Theorem 4.6 can be achieved. The answer is positive, as shown in the following example. We believe there are cases in which the inequality is strict, but we don’t have an example.

Example 4.7

Consider the two homogeneous quartic polynomials

$$\begin{aligned} \begin{aligned} f_2 =&(2 x_0^2 + x_0 x_1 + 2 x_0 x_2 - x_1^2)^2 \!+\! (x_0^2 - x_1^2 - x_2^2)^2 \!+\! (x_0^2 - 2 x_0 x_2 - x_1^2 - x_1 x_2 - 3 x_2^2)^2,\\ f_3 =&(x_0 x_1 + x_0 x_2 - x_1^2 - x_2^2)^2 \!+\! (x_0 x_1 + 4 x_0 x_2 - x_1^2 - 4 x_1 x_2 - 4 x_2^2)^2\\&+ (2 x_0 x_1 - 2 x_0 x_2 - 2 x_1^2 - x_1 x_2 + 2 x_2^2)^2. \end{aligned} \end{aligned}$$

Fig. 9

Left: the three conics summands of $f_2$, intersecting at $(-1,0), (0,-1)$. Right: the three conics summands of $f_3$, intersecting at (0, 0), (1, 0), (0, 1)

They are the sum of the squares of three quadratic polynomials, defining plane conics. The three conics of $f_i$ intersect in i real points, as shown in Fig. 9. Therefore, the real locus of the quartic $\{f_i=0\}$ consists of i nodes, the green dots in Figure 9. One can check that the set $\{f_2=f_3=0\}$ consists of 8 pairs of complex conjugate points. Let $p_1,\ldots ,p_{10}$ be five such pairs, and let $X = \text {Bl}_{p_1, \ldots ,p_{10}} \mathbb {P}^2 \subset \mathbb {P}^4$ be the associated Bordiga surface. Then there exist $u_2,u_3 \in (\mathbb {P}^4)^*$ such that $Q_{u_i} = \{f_i = 0\}$. By construction, this implies that $u_2 \in X^{[2]}_{(B)}$ and $u_3 \in X^{[3]}_{(B)}$. Moreover, since the quartics $f_i$ are sums of squares, the corresponding hyperplanes separate X and therefore belong to the algebraic boundary. Then, the algebraic boundary consists of two irreducible components:

$$\begin{aligned} \partial _a({{\,\textrm{conv}\,}}X) = \left( X^{[2]}_{(B)} \right) ^* \cup C_4^*. \end{aligned}$$