On the Algebraic Geometry of Multiview

Abstract

We study the multiviews of algebraic space curves X from n pin-hole cameras of a real or complex projective space. We assume the pin-hole centers to be known, i.e., we do not reconstruct them. Our tools are algebro-geometric. We give some general theorems, e.g., we prove that a projective curve (over complex or real numbers) may be reconstructed using four general cameras. Several examples show that no number of badly placed cameras can make a reconstruction possible. The tools are powerful, but we warn the reader (with examples) that over real numbers, just using them correctly, but in a bad way, may give ghosts: real curves which are images of the emptyset. We prove that ghosts do not occur if the cameras are general. Most of this paper is devoted to three important cases of space curves: unions of a prescribed number of lines (using the Grassmannian of all lines in a 3-dimensional projective space), plane curves, and curves of low degree. In these cases, we also see when two cameras may reconstruct the curve, but different curves need different pairs of cameras.

1. Introduction

Computer vision is an essential guest in our life [1,2,3,4,5,6,7,8,9]. In the last 10 years many authors have written important contributions to mathematical problems related to computer vision using algebraic geometry. They addressed one of its key problems: the reconstruction of algebro-geometric objects (lines, conics, and even more general algebraic curves) from their linear projections into a plane. As usual in algebraic geometry, first one proves a theorem over an algebraically closed field, here the field $\mathbb{C}$ of the complex numbers, and then, explains why/how this theorem over $\mathbb{C}$ helps to prove something interesting about a picture over $\mathbb{R}$, i.e., a real curve of a projective plane ${\mathbb{P}}^2\left(\mathbb{R}\right)$. We are in the setup of pin-hole cameras. Take as $\mathbb{K}$ either $\mathbb{C}$ or $\mathbb{R}$. For a single picture there is a projective plane $M\left(\mathbb{K}\right)\subset {\mathbb{P}}^3\left(\mathbb{K}\right)$ and we always project on $M\left(\mathbb{K}\right)$ from a point of ${\mathbb{P}}^3\left(\mathbb{K}\right)\setminus M\left(\mathbb{K}\right)$. In the abstract algebro-geometric setting there is no specific screen, just the notion of a projection from one or more points to an abstract plane [10,11,12,13,14,15,16,17]. We are aware that in most textbooks there is a screen for each camera and the reconstruction of each screen is an important task [3].

As in [10,11,12,13,15,16,17], we assume we know the projection points $o_1,\dots ,o_n$ and call $X_{o_i}$ the minimal cone with vertex $o_i$ containing X. The cone $X_{o_i}$ is often called the back-projection of the image of X by the camera $o_i$. We are aware that a very important task is the recovering of the pin-hole centers $o_1,\dots ,o_n$ [1,2,3,6]. However, this part is not considered in our paper.

We say that X is $\alpha$-reconstructible using the pin-holes $o_1,\dots o_n$ if $X=X_{o_1}\cap \dots \cap X_{o_n}$. Now, assume that X is defined over $\mathbb{R}$ and that $o_1,\dots ,o_n\in {\mathbb{P}}^3\left(\mathbb{R}\right)$. We say that X is $\alpha$-reconstructible over $\mathbb{R}$ if $X\left(\mathbb{C}\right)=X_{o_1}\left(\mathbb{C}\right)\cap \dots \cap X_{o_n}\left(\mathbb{C}\right)$ and each $X_{o_i}\left(\mathbb{C}\right)$ is uniquely determined by the images by $o_i$ of the real part, $X\left(\mathbb{R}\right)$, of the complex curve $X\left(\mathbb{C}\right)$. More precisely, we say that X is set-theoretically (resp., scheme-theoretically) $\alpha$-reconstructible by $o_1,\dots ,o_n$ if $X\left(\mathbb{C}\right)=X_{o_1}\left(\mathbb{C}\right)\cap \dots \cap X_{o_n}\left(\mathbb{C}\right)$ as sets (resp., as algebraic schemes). Over $\mathbb{R}$ for scheme-theoretic $\alpha$-reconstruction we add the condition that the set-theoretic images of $X\left(\mathbb{R}\right)$ determine the schemes $X_{o_1}\left(\mathbb{C}\right)$. In all our examples, the latter condition is always satisfied.

In one short section, we prove the following results.

Theorem 1. 

Take a positive integer s. Let $X_i\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$, $1\le i\le s$, be integral curves such that $X_i\ne X_j$ for all $i\ne j$. Set $X:=X_1\cup \dots \cup X_s$. There is a non-empty Zariski open subset U of ${({\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X)}^4$ such that X is set-theoretically α-reconstructible by all cameras $(c_1,c_2,c_3,c_4)\in U$.

We recall that Zariski open subsets U of ${({\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X)}^4$ are very large. Indeed, the set ${\mathbb{P}}^3{\left(\mathbb{C}\right)}^4\setminus U$ has complex dimension at most 11, it is a finite union of differentiable manifolds of dimension at most 22, and hence, it has measure zero for any measure on ${\mathbb{P}}^3{\left(\mathbb{C}\right)}^4$ locally equivalent to the Lebesgue measure. Thus, “random” choices of the centers should give centers in U. Anyway, given any $(o_1,o_2,o_3,o_4)$ it is easy to obtain $X_{o_1}\cap X_{o_2}\cap X_{o_3}\cap X_{o_4}$, and then, choose four other centers if $X_{o_1}\cap X_{o_2}\cap X_{o_3}\cap X_{o_4}$ seems not to be the correct answer or to independently check the answer $X_{o_1}\cap X_{o_2}\cap X_{o_3}\cap X_{o_4}$.

Question 1: Give conditions on X assuring that $\alpha$-reconstruction is not possible only using three general centers or three non-birational centers.

Theorem 2. 

Take a positive integer s. Let $X_i\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$, $1\le i\le s$, be integral curves such that $X_i\ne X_j$ for all $i\ne j$, each $X_i$ is defined over $\mathbb{R}$ and each $X_i\left(\mathbb{R}\right)$ is infinite. Set $X:=X_1\cup \dots \cup X_s$. There is a non-empty Zariski open subset U of ${({\mathbb{P}}^3\left(\mathbb{R}\right)\setminus X\left(\mathbb{R}\right))}^4$ such that X is set-theoretically α-reconstructible over $\mathbb{R}$ by all cameras $(c_1,c_2,c_3,c_4)\in U$.

Theorem 3. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be a smooth and connected curve. Then, X is scheme-theoretically α-reconstructed by four general cameras.

By §7 in [18] there are smooth curves $X\subset {\mathbb{P}}^3$ which are not scheme-theoretically the intersection of three surfaces. Hence, in many cases scheme-theoretical $\alpha$-reconstruction is not possible using only three cameras. Every space curve (even singular and reducible) is set-theoretically the intersection of three surfaces by a theorem of Kneser or its higher dimensional generalizations due to Eisenbud and Evans [19,20]. D. Jaffe gave a non-tautological description of the space curves which are the intersection of a cone with another surface [21]. We do not have a non-tautological description of all curves which are complete intersections of two cones, i.e., of the curves which scheme-theoretically may be $\alpha$-reconstructed using two cameras.

A large part of the paper is devoted to the study of three classes of curves X to be projected:

• finite unions of lines;
• plane curves;
• irreducible curves of degree at most 4.

Refs. [11,14] study the algebraic geometry of n projections of a single curve. F. Rydell and I. Sundeliu study plane curves and rational normal curves [17]. The latter are exactly the irreducible space curves of degree 3 not contained in a plane. See Example 5 for our discussion of these curves.

Recently, the classical trifocal tensor (Ch. 14 in [3]) was deeply studied and extended to linear projections of linear subspace of any dimension [22,23,24,25,26,27,28,29].

In Section 2, we fix the notation, collect from the references several results we need, and recall (Remark 5): how to obtain from a planar picture the equation of the plane curve. We will need this step several times.

In Section 3, we recall the known facts about real algebraic curves we use and see how to use a real “arc” (diffeomorphic to an open interval) to $\alpha$-reconstruct a curve over $\mathbb{R}$. We show that with the official algebro-geometric definition of a real curve there are real curves X and $o\in {\mathbb{P}}^3\left(\mathbb{R}\right)$ such that $X\left(\mathbb{R}\right)=\varnothing$, but the image of $X\left(\mathbb{C}\right)$ by o intersects ${\mathbb{P}}^2\left(\mathbb{R}\right)$ in a circle. We prove that this is not the case for a general pin-hole (Theorem 4).

In Section 4, we prove Theorems 1, 2, and 3 and discuss them.

In Section 5, we study the images by n cameras of the union of m lines, $m>1$.

In Section 6, we study plane curves. The aim is to reconstruct them using only two cameras: one general and one special. We perform a test on the images which implies that we started with a smooth plane curve (Proposition 4). We give a criterion to obtain the plane curve using two cameras (not for $\alpha$-reconstruction, but to identify the curve from the two cones $X_{o_1}$ and $X_{o_2}$) if the curve has no nontrivial automorphism (Proposition 5).

In Section 7, we discuss space curves lying on an irreducible quadric surface. This section is used for curves of low degree and for the reconstruction of an irreducible curve whose genus is high with respect to its degree.

In Section 8, we discuss the linear projection of curves of degree at most 4 and give parameter spaces for finite unions of easy objects, for instance, smooth conics. In Section 11 (Conclusions), we recall the key points obtained in this paper.

Many thanks to the reviewers for useful suggestions. These suggestions improved the exposition of the paper and, perhaps, future papers.

2. Notation and Preliminary Results

Notation 1. 

For each $p\in {\mathbb{P}}^3$ let ${\mathit{\ell }}_p:{\mathbb{P}}^3\setminus \left\{p\right\}\to {\mathbb{P}}^2$ denote the linear projection from p.

Note that if $p\in {\mathbb{P}}^3\left(\mathbb{R}\right)$, then ${\mathit{\ell }}_{p|{\mathbb{P}}^3\left(\mathbb{R}\right)}$ induces a surjection whose fibers are diffeomorphic to $\mathbb{R}$.

Let $\sigma :{\mathbb{P}}^m\left(\mathbb{C}\right)\to {\mathbb{P}}^m\left(\mathbb{C}\right)$, $m\ge 1$, denote the complex conjugation. We have ${\mathbb{P}}^m\left(\mathbb{R}\right)=\{p\in {\mathbb{P}}^m\left(\mathbb{C}\right)\mid \sigma \left(p\right)=p\}$.

For any subset S of a projective space, let $\langle S\rangle$ denote its linear span.

For any curve $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ and any $o\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X$ the set ${\mathit{\ell }}_o\left(X\right)\subset {\mathbb{P}}^2\left(\mathbb{C}\right)$ is well-defined. Now, assume $o\in X$ and let $X^{\prime }$ be the union of all irreducible components of X which are not lines containing o. The set ${\mathit{\ell }}_o(X^{\prime }\setminus \left\{o\right\})$ is well-defined and we call ${\tilde{\mathit{\ell }}}_o\left(X\right)$ the closure of ${\mathit{\ell }}_o(X^{\prime }\setminus \left\{o\right\})$ in ${\mathbb{P}}^2\left(\mathbb{C}\right)$. Here, closure in the Euclidean sense or in the sense of the Zariski topology gives the same set ${\tilde{\mathit{\ell }}}_o\left(X\right)$ by a theorem of Chevalley. If $o\notin X$, set ${\tilde{\mathit{\ell }}}_o\left(X\right):={\mathit{\ell }}_o\left(X\right)$. If either $o\notin X$ or $o\in X$ and $X^{\prime }\ne \varnothing$, then ${\tilde{\mathit{\ell }}}_o\left(X\right)$ is a curve and we call $X_o$ the cone with vertex o and ${\tilde{\mathit{\ell }}}_o\left(X\right)$ as its base. Note that $deg\left(X_o\right)=deg\left({\tilde{\mathit{\ell }}}_o\left(X\right)\right)$. To obtain the integer $deg\left({\tilde{\mathit{\ell }}}_o\left(X\right)\right)$ we apply the next observation to the irreducible components of X.

Remark 1. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an integral projective curve. Fix $o\in {\mathbb{P}}^3\left(\mathbb{C}\right)$ and assume that X is not a line containing o. Set $d:=deg\left(X\right)$ and let μ be the multiplicity of X at o, with the convention $\mu =0$ if and only if $o\notin X$. We have $\mu =1$ if and only if o is a smooth point of X. Set $k:=deg\left({\tilde{\mathit{\ell }}}_o\left(X\right)\right)$. Let $X^{\prime }$ be the normalization of X and $Y^{\prime }$ the normalization of ${\tilde{\mathit{\ell }}}_o\left(X\right)$. $X^{\prime }$ and $Y^{\prime }$ are smooth and connected projective curves and ${\mathit{\ell }}_o$ induces a surjective map $f:X^{\prime }\to Y^{\prime }$. We have $d=m+kdeg\left(f\right)$.

Remark 2. 

Let $X\left(\mathbb{C}\right)\subset {\mathbb{P}}^3\left(C\right)$ be an integral and non-degenerate curve. The set of all $p\in {\mathbb{P}}^3\left(\mathbb{C}\right)$ such that ${\mathit{\ell }}_{p\left|X\right(\mathbb{C})\setminus \{p\}}:X\left(\mathbb{C}\right)\setminus \left\{p\right\}\to {\mathbb{P}}^2\left(\mathbb{C}\right)$ is not birational onto its image and is finite [30,31]. The recent paper [31] contains an upper bound for the number of these exceptional points. Set $d:=deg\left(X\right(\mathbb{C}\left)\right)$. By Th. 1.3 in [31], there are at most ${(d+1)}^3-d(d+1)$ such points. Note that this is not true if $X\left(\mathbb{C}\right)$ spans a plane M, because for each $p\in M\setminus X\left(\mathbb{C}\right)$ the curve ${\mathit{\ell }}_p\left(X\left(\mathbb{C}\right)\right)$ is a line.

Remark 3. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an integral curve. Set $d:=deg\left(X\right)$. Assume the existence of $c_1,c_2\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X\left(\mathbb{C}\right)$ such that $X_{c_1}\cap X_{c_2}=X$ scheme-theoretically. The theorem of Bezout gives $deg\left(X_{c_1}\right)deg\left(X_{c_2}\right)=d$, and hence, either $d=1$, i.e., X is a line, or at least one among ${\mathit{\ell }}_{c_1|X}$ and ${\mathit{\ell }}_{c_2|X}$ is not birational onto its image. Moreover, exactly one among ${\mathit{\ell }}_{c_1|X}$ and ${\mathit{\ell }}_{c_2|X}$ is not birational onto its image if and only if X is a plane curve. Let X be a plane curve, but not a line, and let $\langle X\rangle$ be the plane spanned by X. We have $X=X_{c_1}\cap X_{c_2}$ if we take $c_1\in (\langle X\rangle \setminus X)$ and $c_2\in ({\mathbb{P}}^3\left(\mathbb{C}\right)\setminus \langle X\rangle )$. Here, there is a non-planar example. Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be a smooth degree 4 elliptic curve. By case 2 of Th.1 in [32], X is contained in four quadric cones and none of their vertices is a point of X. Thus, the theorem of Bezout gives $X=X_{c_1}\cap X_{c_2}$ if $c_1$ and $c_2$ are the vertices of two different quadric cones containing X.

Remark 4. 

Let $Y\subset {\mathbb{P}}^2$ be an integral degree d curve. Let $u:X\to Y$ denote the normalization map. Let q be the genus of X. If Y is the image of a degree d smooth curve $X\subset {\mathbb{P}}^3$, then $0\le q\le \lfloor d/2\rfloor \lceil d/2\rceil -d+1$ (Th. 3.7 in [33]). Moreover, if $q=\lfloor d/2\rfloor \lceil d/2\rceil -d+1$, then X is contained in an integral quadric surface (Th. 3.11 in [33]). This quadric surface is unique, unless $d=4$ and $q=1$. Assume $d\ge 7$ and set ${\pi }_1(d,3):=d^2/6-d/2+1$ if $d\equiv 0mod3$ and ${\pi }_1(d.3):=d^2/6-d/2+1/3$ if $d\equiv 1,2mod3$. If $q<\lfloor d/2\rfloor \lceil d/2\rceil -d+1$, then $q\le {\pi }_1(d,3)$ (Th. 3.13 in [33]).

Remark 5. 

Let $Y\subset {\mathbb{P}}^2\left(\mathbb{C}\right)$ be a plane curve without multiple components. At several places it would be essential to obtain the equation of Y from some of its points. If Y is defined over $\mathbb{R}$ we will only use real points of Y. In the next section, we show why we can use the objects usually called open arcs. Take homogeneous coordinates $z_0,z_1,z_2$ of ${\mathbb{P}}^2\left(\mathbb{C}\right)$. For each integer $k\ge 0$ let $\mathbb{C}{[z_0,z_1,z_2]}_k$ (resp., $\mathbb{R}[z_0,z_1,z_2]$) denote the set of all homogeneous polynomials of degree k with complex (resp., real) coefficients, i.e., the linear span of all monomials $z_0^a{z}_1^b{z}_2^c$ with a, b, and c non-negative integers and $a+b+c=k$. Since these monomials are linearly independent, ${dim}_{\mathbb{C}}\mathbb{C}{[z_0,z_1,z_2]}_k={dim}_{\mathbb{R}}\mathbb{R}{[z_0,z_1,z_2]}_k=\left(\genfrac{}{}{0pt}{}{k+2}{2}\right)$. To obtain a homogeneous equation of Y (unique up to a non-zero multiplicative constant), we assume we have a set $A\subset Y\left(\mathbb{C}\right)$ such that each irreducible component of $Y\left(\mathbb{C}\right)$ contains infinitely many points of A. These assumptions are equivalent to assuming that A is Zariski dense in $Y\left(\mathbb{C}\right)$. Thus, if $k:=deg\left(Y\right)$, up to a non-zero constant, there is a unique $f\in \mathbb{C}{[z_0,z_1,z_2]}_k$ such that $Y\left(\mathbb{C}\right)=\{f=0\}$; just find a non-zero solution of a linear homogeneous system with $\left(\genfrac{}{}{0pt}{}{k+2}{2}\right)$ equations. If $A\subset {\mathbb{P}}^2\left(\mathbb{R}\right)$, then we find f with real coefficients.

From an equation f of Y, we easily obtain the set of singular points of Y and other geometric information which will be important in Section 6, with X a plane curve.

Almost always we use the case in which $deg\left(Y\right)$ is known, often the integer $deg\left(X\right)$. In other cases, we use Remark 1 and test the possible integers $deg\left(Y\right)$.

3. Over the Real Numbers

Let X be a smooth and connected curve of genus g defined over $\mathbb{R}$. Let $X\left(\mathbb{C}\right)$ be the set of its complex points and let $\sigma :X\left(\mathbb{C}\right)\to X\left(\mathbb{C}\right)$ denote the complex conjugation. Each connected component of $X\left(\mathbb{R}\right)$ is a one-dimension connected and compact manifold, and hence, it is diffeomorphic to a circle.

Proposition 1 

(Prop. 3.1 in [34]). Let $n\left(X\right)$ be the number of connected components of $X\left(\mathbb{R}\right)$. The topological space $X\left(\mathbb{C}\right)\setminus X\left(\mathbb{R}\right)$ has at most two connected components. Set $a\left(X\right)=1$ if $X\left(\mathbb{C}\right)\setminus X\left(\mathbb{R}\right)$ is connected and $a\left(X\right)=0$ if $X\left(\mathbb{C}\right)\setminus X\left(\mathbb{R}\right)$ has two connected components. Then:

1. 

$0\le n\left(X\right)\le g+1$.

2. 

If $n\left(X\right)=0$, then $a\left(X\right)=1$. If $n\left(X\right)=g+1$, then $a\left(X\right)=1$.

3. 

If $a\left(X\right)=1$, then $n\left(X\right)\equiv g+1mod2$.

F. Klein proved that all pairs $\left(n\right(X),a(X\left)\right)$ allowed by Proposition 1 are realized by some genus g smooth curve defined over $\mathbb{R}$.

We recall that the real projective space ${\mathbb{P}}^2\left(\mathbb{R}\right)$ is a compact 2-dimensional connected but not simply connected differential manifold with the 2-dimensional sphere as its universal cover and with a degree 2 covering map $S^2\to {\mathbb{P}}^2\left(\mathbb{R}\right)$. Let $\sigma :{\mathbb{P}}^2\left(\mathbb{C}\right)\to {\mathbb{P}}^2\left(\mathbb{C}\right)$ denote the complex conjugation. For any algebraic curve $X\subset {\mathbb{P}}^2$ (or $X\left(\mathbb{C}\right)\subset {\mathbb{P}}^2\left(\mathbb{C}\right)$) defined over $\mathbb{R}$ set $X\left(\mathbb{R}\right):=X\left(\mathbb{C}\right)\cap {\mathbb{P}}^2\left(\mathbb{R}\right)=\{p\in X\left(\mathbb{C}\right)\mid \sigma \left(p\right)=p\}$. Set $d:=deg\left(X\right)$. We recall that any degree d plane curve has arithmetic genus $(d-1)(d-2)/2$. Note that if d is odd, then $X\left(\mathbb{R}\right)\ne \varnothing$ and each real line meets $X\left(\mathbb{R}\right)$. For each even d there is a smooth planar curve $X\left(\mathbb{C}\right)$ with $X\left(\mathbb{R}\right)=\varnothing$, e.g., take the Fermat curve $x_0^d+x_1^d+x_3^d=0$.

Remark 6. 

Let $Y\subset {\mathbb{P}}^2\left(\mathbb{C}\right)$ be a complex degree d plane curve. Take $f\in \mathbb{C}{[z_0,z_1,z_2]}_d\setminus \left\{0\right\}$ such that $Y=\{f=0\}$. Let $\sigma \left(f\right)\in \mathbb{C}{[z_0,z_1,z_2]}_d$ denote the polynomial obtained applying the complex conjugation to all coefficients of f. The curve $V\left(\sigma \right(f\left)\right)$ is well-defined, i.e., for all $c\in \mathbb{C}\setminus \left\{0\right\}$ we have $V\left(\sigma \right(f\left)\right)=V\left(\sigma \right(cf\left)\right)$. Note that $V\left(\sigma \right(f\left)\right)=\sigma \left(Y\right)$. Since ${\mathbb{P}}^2\left(\mathbb{R}\right)$ is the fixed-point set of the complex conjugation $\sigma :{\mathbb{P}}^2\left(\mathbb{C}\right)\to {\mathbb{P}}^2\left(\mathbb{C}\right)$, $Y\cap \sigma \left(Y\right)\subset {\mathbb{P}}^2\left(\mathbb{R}\right)$, the curve Y is defined over $\mathbb{R}$ if and only if $Y=\sigma \left(Y\right)$.

Definition 1. 

Take ${\mathbb{P}}^2\left(\mathbb{R}\right)$ with the Euclidean topology. An open arc $A\subset {\mathbb{P}}^2\left(\mathbb{R}\right)$ is a locally closed set $A\subset T$ such that there is a homeomorphism u from $\overline{A}$ to the closed interval $[a,b]\subset \mathbb{R}$, $a<b$, such that $u_{|A}$ induces a homeomorphism $A\to (a,b)$. An open arc $A\subset {\mathbb{P}}^2\left(\mathbb{R}\right)$ is said to be an open arc of degree d or an open arc of a degree d algebraic curve if the Zariski closure of A in ${\mathbb{P}}^2\left(\mathbb{C}\right)$ is a degree d plane curve without multiple components.

Circles appear as a connected component of $X\left(\mathbb{R}\right)$ in Proposition 1. Each such circle is the union of two open arcs.

Remark 7. 

Let A be an open arc of ${\mathbb{P}}^2\left(\mathbb{R}\right)$ and let Y be the Zariski closure of A in ${\mathbb{P}}^2\left(\mathbb{C}\right)$. Since A is an infinite set, the definition of the Zariski topology of ${\mathbb{P}}^2\left(\mathbb{C}\right)$ shows that either $Y={\mathbb{P}}^2\left(\mathbb{C}\right)$ or Y the union of finitely many points, say a finite set S, and a curve $Y^{\prime }$ without multiple components. Since a finite subset of ${\mathbb{P}}^2\left(\mathbb{C}\right)$ is closed and it has the discrete topology, then $Y^{\prime }=Y$, i.e., it is a curve with no multiple component. Note that if Y is a curve without multiple components, then the integer $deg\left(Y\right)$ is uniquely determined by A.

Lemma 1. 

Let A be a degree d open arc of ${\mathbb{P}}^2\left(\mathbb{R}\right)$ and let Y be the Zariski closure of A in ${\mathbb{P}}^2\left(\mathbb{C}\right)$. Then:

1. 

Y is an irreducible curve of degree d;

2. 

Y is defined over $\mathbb{R}$ and $A\subset Y\left(\mathbb{R}\right)$.

Proof. 

Remark 7 gives that Y is a curve. By Remark 6, the degree d complex curve $\sigma \left(Y\right)$ is well-defined. Since $\sigma \left(A\right)=A$, the degree d curve $\sigma \left(Y\right)$ is the Zariski closure of A. Thus, $Y=\sigma \left(Y\right)$. By Remark 6, Y is defined over $\mathbb{R}$. Since ${\mathbb{P}}^2\left(\mathbb{R}\right)$ is the fixed-point set of the complex conjugation. Assume that Y is not irreducible and let $Y=Y_1\cup \dots \cup Y_s$, $s\ge 2$, be the decomposition over $\mathbb{C}$ of Y into its irreducible components, i.e., each $Y_i$ is irreducible and $Y_i⊈Y_j$ for all $i\ne j$. Since $\sigma \left(Y\right)=Y$, either $\sigma \left(Y_1\right)=Y_1$ or there is an integer $j>1$ such that $\sigma \left(Y_1\right)=Y_j$. Assume the existence of $j>1$ such that $\sigma \left(Y_1\right)=Y_j$ and let $Y^{\prime }$ be the closure of $Y\setminus (Y_1\cup Y_j)$. Since $\sigma$ is continuous for the Euclidean topology and two different irreducible curves have only finitely many common points, $\sigma \left(Y^{\prime }\right)=Y^{\prime }$. Note that either $Y^{\prime }=\varnothing$ (case $s=2$) or $Y^{\prime }$ is a curve of degree $<d$. Since ${\mathbb{P}}^2\left(\mathbb{R}\right)\cap (Y_1\cup Y_j)\subseteq Y_1\cap Y_j$ is finite, A is contained in $Y^{\prime }$, contradicting the definition of degree of an open arc. □

In the next example, we see that if we use the standard definitions of algebraic geometry sometimes ghosts appear.

Example 1. 

Take a plane $M\subset {\mathbb{P}}^3$ defined over $\mathbb{R}$. Let $D\subset M$ be a smooth conic defined over $\mathbb{R}$. We also assume $D\left(\mathbb{R}\right)\ne \varnothing$, so that $D\left(\mathbb{R}\right)$ is diffeomorphic to a circle. Thus, $D\left(\mathbb{R}\right)$ is a smooth conic. Fix $c\in {\mathbb{P}}^3\left(\mathbb{R}\right)\setminus M\left(\mathbb{R}\right)$ and let $T\subset {\mathbb{P}}^2\left(\mathbb{C}\right)$ be the cone with vertex c and $D\left(\mathbb{C}\right)$ as a base. The quadric cone T is defined over $\mathbb{R}$ and $T\left(\mathbb{R}\right)$ is the cone with vertex c and base $D\left(\mathbb{R}\right)$. Take a general plane $H\left(\mathbb{C}\right)\subset {\mathbb{P}}^2\left(\mathbb{C}\right)$, i.e., a plane with an equation with general complex coefficients. We have $c\notin H\left(\mathbb{C}\right)$ and $H\left(\mathbb{C}\right)$ is not defined over $\mathbb{R}$. To be sure that $H\left(\mathbb{C}\right)$ is not defined over $\mathbb{R}$ it is sufficient that one of its equations has 1 as a coefficient and an element of $\mathbb{C}\setminus \mathbb{R}$ as another coefficient, i.e., that an equation of $H\left(\mathbb{C}\right)$ has two non-zero coefficient whose ratio is not real. Thus, $L\left(\mathbb{C}\right):=H\left(\mathbb{C}\right)\cap \sigma \left(H\right(\mathbb{C}\left)\right)$ is a line. The degree 4 curve $W:=\left(H\right(\mathbb{C})\cap T(\mathbb{C}\left)\right)\cup \sigma \left(H\right(\mathbb{C})\cap T(\mathbb{C}\left)\right)$ is defined over $\mathbb{R}$. Since $\sigma \left(L\right(\mathbb{C}\left)\right)=L\left(\mathbb{C}\right)$, $L\left(\mathbb{C}\right)$ is defined over $\mathbb{R}$. Note that $L\left(\mathbb{R}\right)\cap T\left(\mathbb{R}\right)$ is formed by two points. Thus, $W\left(\mathbb{R}\right)$ is finite. The real camera with c as its pin-hole has the circle $D\left(\mathbb{R}\right)$ as the real part of the image of $W\left(\mathbb{C}\right)$.

Lemma 2. 

Take $c\in {\mathbb{P}}^3\left(\mathbb{R}\right)$. Let $X\subset {\mathbb{P}}^3$ be a reduced curve defined over $\mathbb{R}$ and such that $c\notin X\left(\mathbb{C}\right)$. Assume $deg\left(X\left(\mathbb{C}\right)\right)=deg({\mathit{\ell }}_c\left(X\left(\mathbb{C}\right)\right)$, i.e., assume that ${\mathit{\ell }}_{c\left|X\right(\mathbb{C})}$ is birational onto its image. Then, ${\mathit{\ell }}_o\left(X\left(\mathbb{C}\right)\right)\cap {\mathbb{P}}^2\left(\mathbb{R}\right)\setminus {\mathit{\ell }}_o\left(X\left(\mathbb{R}\right)\right)$ is a finite set.

Proof. 

Set $d:=deg\left(X\right)$. Let T be the cone with vertex c and ${\mathit{\ell }}_c\left(X\left(\mathbb{C}\right)\right)$ as a base. T is defined over $\mathbb{R}$ and $deg\left(T\left(\mathbb{C}\right)\right)=deg({\mathit{\ell }}_c\left(X\left(\mathbb{C}\right)\right)$. By assumption $deg\left(T\right(\mathbb{C}\left)\right)=d$. Set $E:={\mathit{\ell }}_c\left(X\left(\mathbb{C}\right)\right)\cap {\mathbb{P}}^2\left(\mathbb{R}\right)\setminus {\mathit{\ell }}_c\left(X\left(\mathbb{R}\right)\right)$ and assume $E\ne \varnothing$. Fix $p\in E$. The set ${\mathit{\ell }}_c^{-1}\left(p\right)$ is the unique complex line $L\left(\mathbb{C}\right)$ containing c and with ${\mathit{\ell }}_c(L\left(\mathbb{C}\right)\setminus \left\{c\right\})=p$. Since both p and c are defined over $\mathbb{R}$, $L\left(\mathbb{C}\right)$ is defined over $\mathbb{R}$. Thus, $L\left(\mathbb{C}\right)\cap X\left(\mathbb{C}\right)$ is $\sigma$-invariant. Since $\mathbb{C}$ is algebraically closed, $L\left(\mathbb{C}\right)\cap X\left(\mathbb{C}\right)\ne \varnothing$. By the definition of E, we have $L\left(\mathbb{R}\right)\cap X\left(\mathbb{R}\right)=\varnothing$. Since $L\left(\mathbb{C}\right)\cap X\left(\mathbb{C}\right)$ is $\sigma$-invariant, $deg\left(L\right(\mathbb{C})\cap X(\mathbb{C}\left)\right)>1$. Since ${\mathit{\ell }}_{c\left|X\right(\mathbb{C})}$ is birational onto its image, E is finite. □

Example 1 shows that the birationality of ${\mathit{\ell }}_{c\left|X\right(\mathbb{C})}$ is required in Lemma 2. In Example 2, X is irreducible over $\mathbb{R}$, but not over $\mathbb{C}$. The following example works with X irreducible and smooth over $\mathbb{C}$, $o\notin X\left(\mathbb{C}\right)$, $X\left(\mathbb{R}\right)=\varnothing$, and ${\mathit{\ell }}_o\left(X\right)\left(\mathbb{R}\right)$ diffeomorphic to a circle.

Example 2. 

Take $c\in {\mathbb{P}}^3\left(\mathbb{R}\right)$ and a smooth real conic $D\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ such that $D\left(\mathbb{R}\right)\ne \varnothing$ and $c\notin D$. The set $D\left(\mathbb{R}\right)$ is diffeomorphic to a circle. Let $T\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be the cone with base D and c as its vertex. Take as X the complete intersection of D with a general smooth quadric surface Q defined over $\mathbb{R}$, but with $Q\left(\mathbb{R}\right)=\varnothing$.

Proposition 2. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an integral curve. Let $Y\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an integral curve such that $X\ne Y$. There is a non-empty Zariski open subset $U\subset {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus (X\cup Y)$ such that ${\mathit{\ell }}_c\left(X\right)\ne {\mathit{\ell }}_c\left(Y\right)$ for all $c\in U$.

Proof. 

Set $d:=deg\left(X\right)$. Since $X\ne Y$ and X, Y are irreducible curves, $X\cap Y$ is finite. By Remark 2, there is a finite set $S\subset {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus (X\cup Y)$ such that ${\mathit{\ell }}_{c|X}$ and ${\mathit{\ell }}_{c|Y}$ are birational onto their images for all $c\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus (X\cup Y\cup S)$. Fix $c\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus (X\cup Y\cup S)$ and assume ${\mathit{\ell }}_c\left(X\right)={\mathit{\ell }}_c\left(Y\right)$. We have $deg\left({\mathit{\ell }}_c\left(X\right)\right)=deg\left(X\right)$ and $deg\left({\mathit{\ell }}_c\left(Y\right)\right)=deg\left(Y\right)$. Thus, $deg\left(X\right)=deg\left(Y\right)$. Fix a general $p\in Y$. Since Y has only finitely many singular points, p is a smooth point of Y. Since $X\cap Y$ is finite, $p\notin X$. Remark 2 gives $deg\left({\mathit{\ell }}_p\left(X\right)\right)=deg\left(X\right)$ and $deg\left({\tilde{\mathit{\ell }}}_p\left(Y\right)\right)=deg\left(Y\right)-1\ne deg\left(X\right)$. Thus, ${\mathit{\ell }}_p\left(X\right)$ and ${\tilde{\mathit{\ell }}}_p\left(Y\right)$ have only finitely many common points. There is a Zariski open neighborhood U of p in ${\mathbb{P}}^3\left(\mathbb{C}\right)$ such that ${\mathit{\ell }}_c\left(X\right)\cap {\mathit{\ell }}_c\left(Y\right)$ is finite for all $c\in (U\setminus U\cap Y)$. □

Proposition 3. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an integral curve defined over $\mathbb{R}$ and such that $X\left(\mathbb{R}\right)$ is infinite. Let $Y\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an integral curve such that $X\ne Y$, and Y is defined over $\mathbb{R}$. There is a non-empty Zariski open subset $U\subset {\mathbb{P}}^3\left(\mathbb{R}\right)\setminus (X\cup Y)$ such that ${\mathit{\ell }}_c\left(X\left(\mathbb{R}\right)\right)\ne {\mathit{\ell }}_c\left(Y\left(\mathbb{R}\right)\right)$ for all $c\in U$.

Proof. 

The proposition is obvious if $Y\left(\mathbb{R}\right)$ is finite. Thus, we may assume that $Y\left(\mathbb{R}\right)$ is infinite. Thus, $X\left(\mathbb{R}\right)$ (resp., $Y\left(\mathbb{R}\right)$) is Zariski dense in $X\left(\mathbb{C}\right)$ (resp., $Y\left(\mathbb{C}\right)$). By Remark 2, there is a finite set $S\subset {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus (X\cup Y)$ such that $deg\left({\mathit{\ell }}_c\left(X\right)\right)=deg\left(X\right)$ and $deg\left({\mathit{\ell }}_c\left(Y\right)\right)=deg\left(Y\right)$ for all $c\notin (X\cup Y\cup S)$. Thus, $deg\left(Y\right)=deg\left(X\right)$. Since $Y\left(\mathbb{R}\right)$ is infinite, there is $p\in Y\left(\mathbb{R}\right)$ such that $p\notin X\left(\mathbb{R}\right)$ and p is contained in an open arc of $Y\left(\mathbb{R}\right)$. We use this point p to adapt the proof of Proposition 2. □

Theorem 4. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be a reduced curve defined over $\mathbb{R}$. Then, there is a non-empty Zariski open subset V of ${\mathbb{P}}^3\left(\mathbb{R}\right)\setminus X\left(\mathbb{R}\right)$ such that ${\mathit{\ell }}_o\left(X\right)\left(\mathbb{R}\right)\setminus {\mathit{\ell }}_o\left(X\left(\mathbb{R}\right)\right)$ is finite for all $o\in V$.

Proof. 

Let $X\left(\mathbb{C}\right)=X_1\left(\mathbb{C}\right)\cup \dots \cup X_s\left(\mathbb{C}\right)$, $s\ge 1$, be the decomposition of X into its irreducible components. Since X is real, $\sigma \left(X\right(\mathbb{C}\left)\right)=X\left(\mathbb{C}\right)$, i.e., up to a permutation of the indices, there is an integer e such that $0\le e\le s$, $s-e$ is even, $\sigma \left(X_i\left(\mathbb{C}\right)\right)=X_i\left(\mathbb{C}\right)$, i.e., $X_i$ is defined over $\mathbb{R}$, for $i=1,\dots ,e$, and for $i=e+1,e+3,\dots ,s-1$, the complex conjugation $\sigma$ permutes $X_i\left(\mathbb{C}\right)$ and $X_{i+1}\left(\mathbb{C}\right)$, i.e., neither $X_i\left(\mathbb{C}\right)$ not $X_{i+1}\left(\mathbb{C}\right)$ is defined over $\mathbb{R}$, while $X_i\left(\mathbb{C}\right)\cup X_{i+1}\left(\mathbb{C}\right)$ is defined over $\mathbb{R}$. Apply Remark 1 to each irreducible component of X, and then, apply Proposition 2 to any pair of irreducible components of X. □

4. Proofs of Theorems 1, 2 and 3

Proof of Theorem 1: 

Set $d:=deg\left(X\right)$. By Remark 2, there is a finite set $S\subset {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X$ such that ${\mathit{\ell }}_{c|X_i}$ is birational onto its image for all $i=1,\dots ,s$ and all $c\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus (X\cup S)$. By Proposition 2, there is a non-empty Zariski open subset V of ${\mathbb{P}}^3\left(\mathbb{C}\right)\setminus (X\cup S)$ such that ${\mathit{\ell }}_{c|X}$ is birational onto its image for all $c\in V$, i.e., $deg\left({\mathit{\ell }}_c\left(X\right)\right)=d$ for all $c\in V$. Fix a general $(c_1,c_2)\in V^2$ and let $X_{c_1}$ and $X_{c_2}$ be the corresponding cones. The scheme $X_{c_1}\cap X_{c_2}$ is scheme-theoretically a complete intersection of degree $d^2$, perhaps with multiple components, containing X. □

Claim 1: The scheme $X_{c_1}\cap X_{c_2}$ contains each $X_i$ with multiplicity 1.

Proof of Claim 1: 

Fix a general $p\in X_i$. Hence, $X_i$ is smooth at p. Since ${\mathit{\ell }}_{c_h|X}$, $h=1,2$, is birational onto its image and p is general in $X_i$, ${\mathit{\ell }}_{c_h}\left(a\right)={\mathit{\ell }}_{c_h}\left(p\right)$ for some $a\in X$ implies $a=p$. If $c_h\notin T_p{X}_i$, we also obtain that ${\mathit{\ell }}_{c_h}\left(X\right)$ is smooth at p. Thus, the cones $X_{c_1}$ and $X_{c_2}$ are smooth at p. The tangent space of $X_{c_i}$ at p is the plane spanned by $T_p{X}_i$ and $c_i$. Thus, $X_{c_1}$ and $X_{c_2}$ are transversal at p for a general $(c_1,c_2)$.

Applying Proposition 2 to the irreducible components of $X_{c_1}\cap X_{c_2}$ not contained in X we obtain that $X_{c_1}\cap X_{c_2}\cap X_{c_3}$ is the union of X (counted with multiplicity one) and a finite set E. Taking $c_4\in V$ not contained in one of the finitely many cones with base X and vertex a point of E, we obtain $X=X_{c_1}\cap X_{c_2}\cap X_{c_3}\cap X_{c_3}$ (set-theoretically). □

Proof of Theorem 2: 

Mimic the proof of Theorem 1 quoting Proposition 3 instead of Proposition 2. □

Proof of Theorem 3: 

Take a general $(c_1,c_2,c_3,c_4)$ as in the proof of Theorem 1. With the assumption of Theorem 1 we prove that $X_{c_1}\cap X_{c_2}\cap X_{c_3}\cap X_{c_4}$ are scheme-theoretically the same in a neighborhood of each point of X, which would obviously prove Theorem 3. For a general $(c_1,c_2)$ the scheme $X_{c_1}\cap X_{c_2}$ is singular at only finitely many points of X. For a general $c_3$ the scheme $X_{c_1}\cap X_{c_2}\cap X_{c_3}$ is smooth at all points of $X_{\mathrm{reg}}\setminus S$ with S finite and at each $p\in S$ the Zariski tangent space of $X_{c_1}\cap X_{c_2}\cap X_{c_3}$ has dimension 2. Since $S\subset X_{\mathrm{reg}}$, for a general $c_4$ the cone $X_{c_4}$ at no $p\in S$ has a tangent plane containing the plane $T_p(X_{c_1}\cap X_{c_2}\cap X_{c_3})$. □

Remark 8. 

If $s\ge 2$ in Theorem 2 it is not sufficient to assume that each $X_i$ is defined over $\mathbb{R}$ and that $X\left(\mathbb{R}\right)$ is infinite. For instance, take $X_1\left(\mathbb{R}\right)$ infinite and $X_i\left(\mathbb{R}\right)=\varnothing$ for all $i>1$. A real and smooth projective curve $X_i$ has either $X_i\left(\mathbb{R}\right)=\varnothing$ or $X\left(\mathbb{R}\right)$ infinite and a union of finitely many disjoint circles, each of them being the union of two open arcs (Proposition 1). Odd-degree curves $X_i$ have $X_i\left(\mathbb{R}\right)$ infinite, but there is a degree 4 singular rational curve Y with a unique singular point, p, which is an ordinary node and $Y\left(\mathbb{R}\right)=\left\{p\right\}$. Reconstruction over $\mathbb{R}$ by any number of real cameras is also impossible if X is defined over $\mathbb{R}$, but some of its irreducible components are not, e.g., if $\sigma \left(X_s\right)=X_{s-1}$ and $\sigma \left(X_{s-1}\right)=X_s$, because the real part of the real curve $X_{s-1}\cup X_s$ is contained in $X_{s-1}\cap X_s$, and hence, it is finite.

Example 3. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be a smooth conic and let $\langle X\rangle$ be the plane spanned by X. Take $c\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X$. The map ${\mathit{\ell }}_{c|X}$ is birational onto its image if and only if $c\notin \langle X\rangle$. Take $c_1,c_2\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus \langle X\rangle$ such that $c_1\ne c_2$. The scheme $X_{c_1}\cap X_{c_2}$ is a degree 4 curve which is the union of X and another conic Y with $deg(Y\cap X)=2$. Indeed, $Y\ne X$, i.e., $X_{c_1}\cap X_{c_2}$ is not a multiplicity 2 structure on X, because both $X_{c_1}$ and $X_{c_2}$ are smooth along X and their tangent planes along C are not always the same. Take $c_3\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus (\langle X\rangle \cup \{c_1,c_2\})$. Since $deg(Y\cap X_{c_2})=4$, while $deg(X\cap Y)=2$, $X_{c_1}\cap X_{c_2}\cap X_{c_3}$ is not scheme-theoretically equal to X.

5. Multiview of m Distinct Lines

In this section, we consider the case of m lines and n cameras.

Let $\mathbb{G}$ be the Grassmannian of all lines of a 3-dimensional projective space. The algebraic variety $\mathbb{G}$ is irreducible and projective and $dim\mathbb{G}=4$. The compact complex manifold $\mathbb{G}\left(\mathbb{C}\right)$ is isomorphic to a smooth quadric hypersurface of ${\mathbb{P}}^5\left(\mathbb{C}\right)$ and $\mathbb{G}\left(\mathbb{R}\right)$ is a compact and connected 4-dimensional manifold. See [11,14] for a detailed description of $\mathbb{G}$ and how to obtain from a line as an element of $\mathbb{G}$ its image in the screen ${\mathbb{P}}^2$ using the camera o. See Ch. 10 in [35] for a detailed study.

Call $\mathcal{C}$ the $3n\times 4$ matrix obtained stacking the $3\times 4$ matrices. Given homogeneous coordinates $x_0,x_1,x_2,x_3$ the matrix $\mathcal{C}$ gives a morphism ${\mathbb{P}}^3\left(\mathbb{C}\right)\setminus \{o_1,\dots ,o_n\}\to {\mathbb{P}}^2{\left(\mathbb{C}\right)}^n$. In only using $\mathcal{C}$ there is an explicit way to obtain o, a rational map ${\gamma }_{\mathcal{C}}:\mathbb{G}⤏{\mathbb{P}}^2{\left(\mathbb{K}\right)}^n$ which is not defined exactly at the lines $L\in \mathbb{G}$ containing at least one point $o_i$ [11]. The closure of the image of ${\gamma }_{\mathcal{C}}$ is called the line multiview variety associated to $\mathcal{C}$. This variety is deeply studied in [11,14]. The former paper also gives easy conditions on $o_1,\dots ,o_n$, say $n\ge 3$, to say if a general fiber of ${\gamma }_{\mathbb{C}}$ is a single point, i.e., “generic reconstruction” holds (prop. 2.4 in [11]) and which line L in the domain of $\gamma$ we have ${\gamma }_{\mathcal{C}}^{-1}\left({\gamma }_{\mathcal{C}}\left(L\right)\right)=\left\{L\right\}$, i.e., L maybe reconstructed (in their sense) from its m pictures (Th. 2.6 in [11]).

Remark 9. 

Fix $o\in {\mathbb{P}}^3\left(\mathbb{C}\right)$ and take lines $L,R\in \mathbb{G}$ such that $L\ne R$ and $o\notin L\cup R$. We have ${\mathit{\ell }}_o\left(L\right)={\mathit{\ell }}_o\left(R\right)$ if and only if the planes $\langle L\cup \left\{o\right\}\rangle$ and $\langle L\cup \left\{o\right\}\rangle$ are the same. Note that if $\langle L\cup \left\{o\right\}\rangle =\langle R\cup \left\{o\right\}\rangle$, then $L\cap R\ne \varnothing$. Thus, this problem (less lines in a screen even if no line contains the center of one camera) does not arise if we start with m pairwise disjoint lines.

Lemma 3. 

Fix $o_1,o_2\in {\mathbb{P}}^3\left(\mathbb{C}\right)$ such that $o_1\ne o_2$. Let A be the set of all $L\in \mathbb{G}$ such that $L\cap \{o_1,o_2\}=\varnothing$. Let $\gamma :A\to {\mathbb{P}}^2\left(\mathbb{C}\right)\times {\mathbb{P}}^2\left(\mathbb{C}\right)$ and ${\gamma }_{\mathbb{R}}:A\cap \mathbb{G}\left(\mathbb{R}\right)\to {\mathbb{P}}^2\left(\mathbb{R}\right)$ denote the maps associated to the cameras $o_1,o_2$, with ${\gamma }_{\mathbb{R}}$ being defined only if $\{o_1,o_2\}\subset {\mathbb{P}}^3\left(\mathbb{R}\right)$. Then:

(a) γ and ${\gamma }_{\mathbb{R}}$ are surjective.

(b) Take $L\in A$. If $\langle L\cup \{o_1,o_2\}\rangle ={\mathbb{P}}^3\left(\mathbb{C}\right)$, then ${\gamma }^{-1}\left(\gamma \left(L\right)\right)=\left\{L\right\}$. If $\langle L\cup \{o_1,o_2\}\rangle$ is a plane, then ${\gamma }^{-1}\left(\gamma \left(L\right)\right)$ (resp., ${\gamma }_{\mathbb{R}}^{-1}\left(\gamma \left(R\right)\right)$ is the set of all lines (resp., real lines) in $\langle L\cup \{o_1,o_2\}\rangle$ containing neither $o_1$ nor $o_2$.

(c) Take $L,R\in A$ such that $L\cap R=\varnothing$. Then, $\gamma \left(L\right)\ne \gamma \left(R\right)$.

Proof. 

Fix $(R_1,R_2)\in {\mathbb{P}}^2\left(\mathbb{C}\right)\times {\mathbb{P}}^2\left(\mathbb{C}\right)$ and let $C_i$ denote the back-projection of $R_i$ with respect to $o_i$. Either $C_1=C_2$ or $C_1\cap C_2$ is a line. In the latter case, $C_1\cap C_2={\gamma }^{-1}\left(\gamma \left(C_1\right)\right)$. Now, assume $C_1=C_2$ and take any line R such that $R\cap \{o_1,o_2\}=\varnothing$ and $R\subset C_1$. By Remark 9, we have $\gamma \left(R\right)=(R_1,R_2)$, concluding the proof of part (a).

Remark 9 gives which lines V satisfies $\gamma \left(V\right)=(R_1,R_2)$, proving part (b).

Part (b) implies part (c). □

To discuss the reconstruction of $m>1$ lines we need to discuss several options. We start with the union $A:=L_1\cup \dots \cup L_m$ of n lines. To determine the domain of the multiple-view map for m lines we distinguish if we just start with A or the ordered set $(L_1,\dots ,L_n)$. In the latter case, the natural parameter space is the Zariski open subset ${\mathcal{U}}_m$ of ${\mathbb{G}}^m$ formed by all $(R_1,\dots ,R_m)$ such that $R_i\ne R_j$. The quasi-projective variety ${\mathcal{U}}_m$ has ${\mathbb{G}}^m$ as its natural compactification. In the next section, we discuss and solve the easiest case: m ordered lines in which we may also distinguish their images, i.e., m distinct lines with m distinct colors. Of course, in this setup it is not sufficient to reconstruct A, for a perfect reconstruction we also need to reconstruct the name of each of the m irreducible components of A. In this setup the natural domain ${\mathcal{V}}_m$ is the quotient of ${\mathcal{U}}_m$ by the action of the symmetric group with m elements. The set ${\mathcal{V}}_m$ is a smooth irreducible quasi-projective variety of dimension $4m$ defined over $\mathbb{R}$.

Remark 10. 

Assume $m>1$ and set $e:=\lfloor m/2\rfloor$. Both ${\mathcal{U}}_m$ and ${\mathcal{V}}_m$ are defined over $\mathbb{R}$ and so their real part ${\mathcal{U}}_m\left(\mathbb{R}\right)$ and ${\mathcal{V}}_m\left(\mathbb{R}\right)$ are well-defined. As expected, ${\mathcal{U}}_m\left(\mathbb{R}\right)$ is the set of all m distinct lines of ${\mathbb{P}}^3\left(\mathbb{R}\right)$, which is a Zariski open subset of the compact $4m$-dimensional differential manifold $\mathbb{G}{\left(\mathbb{R}\right)}^m$. However, in algebraic geometry ${\mathcal{V}}_m\left(\mathbb{R}\right)$ has an official definition. With this official scheme-theoretic definition ${\mathcal{V}}_m\left(\mathbb{R}\right)$ is the set of all $A\in {\mathcal{V}}_n\left(\mathbb{C}\right)$ such that $\sigma \left(A\right)=A$. With this definition ${\mathcal{V}}_n\left(\mathbb{R}\right)$ has $e+1$ connected components, ${\mathcal{V}}_{n,i}\left(\mathbb{R}\right)$, $i=0,\dots ,e$, each of them a differentiable manifold of dimension $4m$, with ${\mathcal{V}}_{m,i}\left(\mathbb{R}\right)$ parametrizing the set of all $A\in {\mathcal{V}}_m\left(\mathbb{R}\right)$ with A union of $m-2i$ real lines and i pairs of lines L and $\sigma \left(L\right)$ with L not real. Note that $L\cup \sigma \left(L\right)$ has a real point if and only if L and $\sigma \left(L\right)$ are coplanar, and hence, $L\cap \sigma \left(L\right)$ is a real point. Note that a perfect real screen, say ${\mathit{\ell }}_o$, see $A\in {\mathcal{V}}_{n,i}\left(\mathbb{R}\right)$ as the union of at most $m-2i$ real lines (exactly $m-2i$ real lines if A is formed by m pairwise disjoint lines) and i real points. The same image would appear using a perfect camera from $B=R_1\dots \cup R_m$ with each $R_j$ a real line and exactly $2e$ lines containing o. With at least three non-collinear pin-hole cameras $o_1,\dots ,o_n$ we always see if $A\in {\mathcal{V}}_m\left(\mathbb{R}\right)$ is an element of the unique connected component, ${\mathcal{V}}_{m,0}\left(\mathbb{R}\right)$, we are interested in.

In the next observation, we collect several useful practical suggestions coming from [11].

Remark 11. 

Call $\mathcal{C}$ n cameras $o_1,\dots ,o_n$, $n\ge 3$, with, say, at least three not collinear cameras. Call M the associated $3n\times 4$ matrix. There is a complete description of the closed variety $\overline{{\gamma }_{\mathcal{C}}}\subset {\left({\mathbb{P}}^2\right)}^n$ of line views of $\mathcal{C}$ in terms of the matrix M and the Plücker coordinates of $\mathbb{G}$ (Th. 2.5 in [11]). The explicit equations of $\overline{{\gamma }_{\mathcal{C}}}$ inside ${\left({\mathbb{P}}^2\right)}^n$ give an easy test to say if $(R_1,\dots ,R_n)\in \overline{{\gamma }_{\mathcal{C}}}$ or not. The explicit description of $\mathrm{Im}\left({\gamma }_{\mathcal{C}}\left(\mathbb{G}\right)\right)$ also gives if $(R_1,\dots ,R_n)$ comes from $L\in \mathbb{G}$ not containing any $o_1,\dots ,o_n$. The subvariety $\overline{{\gamma }_{\mathcal{C}}}$ of ${\left({\mathbb{P}}^2\right)}^n$ has dimension 4 and so its has measure 0 in ${\left({\mathbb{P}}^2\right)}^n$. Thus, a small modification of $(R_1,\dots ,R_n)\in \overline{{\gamma }_{\mathcal{C}}}$ is not an element of $\overline{{\gamma }_{\mathcal{C}}}$. We should take as image some nearby $(R_1^{\prime },\dots ,R_n^{\prime })\in \mathrm{Im}\left({\gamma }_{\mathcal{C}}\left(\mathbb{G}\right)\right)$.

5.1. Two Compactifications of the Set of m Lines in ${\mathbb{P}}^2\left(\mathbb{C}\right)$

We explain two possible compactifications of the set of unions of m distinct lines of ${\mathbb{P}}^2\left(\mathbb{C}\right)$, i.e., of the subset of the dual complex projective plane ${\mathbb{P}}^{2\vee }$ with cardinality m. Then, if we have n screens we just take the product of n copies of the same compactification. We prefer the second one, often called the Hilbert scheme compactification.

(a) The symmetric product.

We do not need to take ${\mathbb{P}}^2\left(\mathbb{C}\right)$. Let W be a smooth and connected complex projective variety. Take the quotient ${\mathrm{Sym}}^m\left(X\right)$ of $W^m$ by the symmetric group $S_m$ permuting the factors. The set ${\mathrm{Sym}}^m\left(W\right)$ is an irreducible projective variety of dimension $mdimW$. If W is defined over $\mathbb{R}$, then ${\mathrm{Sym}}^m\left(W\right)$ is defined over $\mathbb{R}$. If $dimW>1$ (as in our cases with ${\mathbb{P}}^2$ and ${\mathbb{P}}^3$), for any $m>1$ the variety ${\mathrm{Sym}}^m\left(W\right)$ is singular. This is the main reason for not using it for algebro-geometric calculations, as in §4 in [11].

(b) The Hilbert scheme compactification.

This compactification exists for all W, but it is nice, i.e., it smooth and irreducible, only if $dimW\le 2$. If W is a smooth and connected projective curve, then the Hilbert scheme compactification is ${\mathrm{Sym}}^m\left(W\right)$. Now, assume that W is a smooth and connected projective surface. This is the only case we need. Let $W_0^m$ denote the set of all subset $A\subset W$ such that $\#A=m$. The Hilbert scheme compactification $W^{\left[m\right]}$ of $W_0^m$ parametrizes the set of all degree m zero-dimensional schemes of W. The set $W^{\left[m\right]}$ is irreducible and smooth [36], $dim{W}^{\left[m\right]}=2m$ and, if $m>1$, $W^{\left[m\right]}\setminus W_0^m$ is an irreducible hypersurface of $W^{\left[m\right]}$. If $m>1$ we have $\mathrm{Pic}\left(W^{\left[m\right]}\right)\cong \mathrm{Pic}\left(W\right)\oplus \mathbb{Z}$ [37]. Thus, for $W={\mathbb{P}}^2$ we have $\mathrm{Pic}\left(W^{\left[m\right]}\right)\cong {\mathbb{Z}}^2$.

5.2. The Practical Suggestion

Fix $m\ge 2$. Take cameras $o_1,\dots ,o_n$ and assume for the moment that all pictures $(A_1,\dots ,A_m)$ are formed by m distinct lines, say $A_i=R_{i,j}$, $1\le i\le n$, $1\le j\le m$. Fix $R_{1,1}$ and $R_{2,1}$. Lemma 3 gives a family B of lines not containing $o_1,o_2$ and mapping to $R_{1,1}$ and $R_{2,1}$. In almost all cases B is a singleton. Take $L\in B$. See if the cameras $o_3,\dots ,o_n$ give components of $A_3,\dots ,A_n$. If all these tests work, L is an irreducible component and we reduce to the case of $m-1$ lines. If at least one of these test fails, use $R_{2,2}$ instead of $R_{2,1}$ and continue.

The main problem is when B is not a singleton.

We use the setup with screen ${\mathsf{\Pi }}_1,\dots ,{\mathsf{\Pi }}_n$ with $o_i$ projecting on the plane ${\mathsf{\Pi }}_i$. We assume ${\mathsf{\Pi }}_i\cap \{o_1,\dots ,o_n\}=\varnothing$, which is not a strong restriction from the abstract point of view. With this assumption for all $(i,j,h)\in {\{1,\dots ,n\}}^3$ such that $j\ne h$, the set $\langle \{o_j,o_h\}\rangle \cap {\mathsf{\Pi }}_i$ is a single point, $p_{i,j,h}$. If $p_{1,1,2}\notin R_{1,1}$, then $\#B=1$ for any choice of $R_{2,1}$ (Lemma 3). To conclude, it is sufficient that $p_{1,1,2}\in R_{1,x}$ for at most $m-1$ indices.

5.3. m-Lines of Different Colors

We assume that we start with m ordered lines, say $(L_1,\dots ,L_m)\in {\mathcal{U}}_m$, and that these lines have different colors (m lines of m distinct colors) and that these different colors are distinguished by our m screens. Here, it is the same over $\mathbb{R}$ or over $\mathbb{C}$ and we use $\mathbb{K}$ to denote that both possibilities occur. We also assume that if in one screen two lines map to the same line we see that the image gives both colors, so this is not a problem for the reconstruction of the m ordered lines. So, if no $L_1,\dots ,L_m$ pass through $o_i$ the screen associated to $o_i$ gives m lines, each of them associated (by its color) to a unique $L_i$. In this extreme situation, one has essentially m times the single line case. When we have $a<m$ colors and we know how many lines have a given color, then we reduce to a times the situation described in the first part of this section.

6. Plane Curves

Smooth plane curves $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ of degree d are one of the main characters in [17]. In this section, we consider the reconstruction of a smooth plane curve X over the complex numbers, leaving the case of real plane curves to a subsection in which we may use everything we said over $\mathbb{C}$ to obtain useful results over $\mathbb{R}$. We always assume that X has no multiple component, because the set $X\left(\mathbb{C}\right)$ cannot distinguish multiple components.

Remark 12. 

Let $C\subset {\mathbb{P}}^2\left(\mathbb{C}\right)$ be a degree d plane curve without multiple components. We recall that in Remark 5 we explained how to obtained an equation of C and that if C is irreducible and defined over $\mathbb{R}$ and $C\left(\mathbb{R}\right)$ is infinite, then an equation of C with real coefficients may be obtained just by one open arc contained in $C\left(\mathbb{R}\right)$ (Remark 1).

Several times we use the following strategy to reconstruct X.

Remark 13. 

Suppose we know that the degree $d>1$ curve X is a plane curve. If in some way we identify the plane $\langle X\rangle$, then one camera $o\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus \langle X\rangle$ is sufficient to reconstruct X as a scheme, because $X=\langle X\rangle \cap {\mathit{\ell }}_o^{-1}\left({\mathit{\ell }}_o\left(X\right)\right)$. To identify a plane M it is sufficient to know three non-collinear points of M. Thus, there is the following test to check if X is a plane curve, if we know how to produce some of its points. Take three non-collinear points of X and obtain its linear span, N. If $X⊈N$, then X is not a plane curve. Using cameras would be sufficient to reconstruct at least three non-collinear points of X.

Remark 14. 

Suppose X is a degree $d>1$ plane curve. If $o\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus \langle X\rangle$, then ${\mathit{\ell }}_o\left(X\right)\cong X$. If $o\in \langle X\rangle$ and X is not the union of d lines passing through o, then ${\tilde{\mathit{\ell }}}_o\left(X\right)$ is a line. If X is the union of d lines through o, then ${\mathit{\ell }}_o(X\setminus \left\{o\right\})$ is the union of d collinear points, the line containing them being ${\mathit{\ell }}_o(\langle X\rangle \setminus \left\{o\right\})$.

Remark 14 shows that it is very useful if one camera sends X to a line. No number of cameras would be sufficient if their position is “bad”, i.e., all of them map the plane to a line.

Remark 15. 

Take a plane curve $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$. For any $o_1,\dots ,o_n\in \langle X\rangle \setminus X$ each ${\mathit{\ell }}_{o_i}\left(X\right)$ is a line and their back-projected images would only give $\langle X\rangle$, information we obtained just using camera $o_1$.

6.1. Using a Unique Camera to Prove That X Is a Plane Curve

Remark 16. 

Fix X and o with the only restriction that no line containing o is an irreducible component of X. If ${\tilde{\mathit{\ell }}}_o\left(X\right)$ is a line, then X is contained in the back-projection of ${\tilde{\mathit{\ell }}}_o\left(X\right)$, which is a plane.

Proposition 4. 

Take o and X. Set $d:=deg\left(X\right)$ and assume that ${\tilde{\mathit{\ell }}}_o\left(X\right)$ is a smooth degree d plane curve. Then, X is contained in a plane.

Proof. 

Since ${\tilde{\mathit{\ell }}}_o\left(X\right)$ has degree $deg\left(X\right)$, $o\notin X$ (Remark 1), and hence, ${\tilde{\mathit{\ell }}}_o\left(X\right)={\mathit{\ell }}_o\left(X\right)$.

First, assume $d=2$. If X is a conic, then it is contained in a plane. If $X=L\cup R$ with $L,R$ lines, then ${\mathit{\ell }}_o\left(X\right)\cap {\mathit{\ell }}_o\left(R\right)\ne \varnothing$, and hence, ${\mathit{\ell }}_o(L\cup R)$ is not a smooth conic.

From now on, we assume $d\ge 3$. By Remark 1, X is smooth and connected and X is isomorphic as an abstract curve to ${\mathit{\ell }}_o\left(X\right)$. The genus formula for degree smooth plane curves gives that X has genus $(d-1)(d-2)/2$. Remark 4 gives that X is a plane curve. □

6.2. Reconstructing X from Two Cameras

We assume we have $o\in {\mathbb{P}}^3\left(\mathbb{C}\right)$ such that $o\notin X$. We assume that we know that X is a plane curve. If $deg\left({\mathit{\ell }}_o\left(X\right)\right)<d$, then ${\mathit{\ell }}_o\left(X\right)$ is a line and the back-projection of this line is the plane M containing X (Remark 13). Thus, to reconstruct X in this case it is sufficient to use that $X=X_o\cap X_{o_2}$ with $o_2$ any point of ${\mathbb{P}}^3\left(\mathbb{C}\right)\setminus M$.

Remark 17. 

In Section 5, we made a practical suggestion, given the pictures of m lines in two screens. The same suggestion works for recovering a set $A\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ with $\#A=m$ from its image $A_1$ and $A_2$ in two screens.

We first show the case $d=2$.

Remark 18. 

Take a smooth conic and $o_1,o_2\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus \langle X\rangle$ such that $o_1\ne o_2$. The set $X_{o_1}\cap X_{o_2}$ is the union of X and another conic $X^{\prime }$. $X^{\prime }$ is smooth if and only if $o_2\notin X_{o_1}$. If $X^{\prime }$ is smooth, to distinguish between X and $X^{\prime }$ we need another camera.

From now on we assume $d\ge 3$. We explain how to use $o_1$ and $o_2$ to find a finite $S\subset X$ such that $\langle S\rangle =\langle X\rangle$, and then, use that $X=X_{o_1}\cap \langle S\rangle$.

Remark 19. 

Let $C\subset {\mathbb{P}}^2\left(\mathbb{C}\right)$ be a degree $d>1$ plane curve. From C we obtain an equation, f, of C (Remark 5). Take homogeneous coordinates $z_0,z_1,z_2$ of ${\mathbb{P}}^2\left(\mathbb{C}\right)$. The set $\mathrm{Sing}\left(C\right)$ is the solution set of the system $\frac{\partial }{{\partial }_{z_0}}\left(f\right)=\frac{\partial }{{\partial }_{z_0}}\left(f\right)=\frac{\partial }{{\partial }_{z_0}}\left(f\right)$ of three homogeneous degree $d-1$ equations with one of the partial derivatives, say $\frac{\partial }{{\partial }_{z_0}}\left(f\right)$, being identically zero if and only if C is a union of lines containing the point $[1:0:0]$.

Remark 20. 

Let $C\subset {\mathbb{P}}^2\left(\mathbb{C}\right)$ be an irreducible curve of degree $d\ge 3$. From C we obtain an equation, f, of C (Remark 17). Take homogeneous coordinates $z_0,z_1,z_2$ of ${\mathbb{P}}^2\left(\mathbb{C}\right)$. We recall that a smooth point $p\in C_{\mathrm{reg}}$ is a flex if and only if the tangent line $T_{p}C$ of C at p has order of contact at least 3. Let $\mathrm{Hess}\left(f\right)$ denote the Hessian of f, i.e., the determinant of the symmetric $3\times 3$ matrix whose $(i,j)$ entry is the polynomial $\frac{{\partial }^2}{{\partial }_{z_i}{\partial }_{z_j}}\left(f\right)$. The Hessian has degree $3(d-2)$. Since C is irreducible, $\mathrm{Hess}\left(f\right)$ is not divisible by f ([35] p. 16) (here we only need that C is not the union of d lines through one point). Since C is irreducible, the system $\mathrm{Hess}\left(f\right)=f=0$ has only finitely many solutions and the set of all solution is the union of the set $\mathrm{Sing}\left(C\right)$ and the set of all flexes of $C_{\mathrm{reg}}$.

Remark 21. 

Assume X is smooth, so that $C_i:={\mathit{\ell }}_{o_i}\left(X\right)$ is a smooth degree d curve. Let A be the set of all flexes of X. Set $a:=\#A$. If $d=3$, then $a=9$, but for $d>4$ the integer a depends on the smooth degree d curve X we need to identify. Since A is the common zero of an equation of X and its Hessian, $a\le 3d(d-2)$. On each ${\mathit{\ell }}_{o_i}\left(X\right)$, there is a set $A_i:={\mathit{\ell }}_{o_i}\left(A\right)$ of a points of $C_i$. For any $s>2$ let $a_s$ be the number of flexes of X with the flex line with order of contact s. We have ${\sum }_{3\le s\le d}(s-2)a_s=3d(d-2)$. By Remark 17, to reconstruct X from $A_1$ and $A_2$ it is sufficient to observe that $a\ge 3$ and that A is not formed by collinear points, which is true by the formula just given. For a general X we have $a=3d(d-2)$, and this is the worst information we can obtain from the point of view of reconstruction.

Example 4. 

Take an integral singular degree $d>2$ plane curve X. Set $B:=\mathrm{Sing}\left(X\right)$ and let A be the flexes of $X_{\mathrm{reg}}$. In a screen we may distinguish between a singular point and a smooth flex. Thus, using two screens $o_1,o_2\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus \langle X\rangle$ and Remark 17 we may reconstruct $\langle X\rangle$, and hence, X if and only if $\langle A\cup B\rangle$ is a plane. This is not always true, but the only example we know in which $\langle A\cup B\rangle$ is not a plane is the following curve: a cuspidal cubic curve. Now, we explain this case and show how we are able to overcome the fact that $\langle A\cup B\rangle$ is a line and reconstruct the curve using its geometry. An irreducible cuspidal curve has a unique singular point and a unique flex. Indeed, up to a projective equivalence we may take $X=\{f=0\}$ with $f=z_2{z}_1^2-z_0^3$. Since $\mathrm{Hess}\left(f\right)=24z_0{z}_1^2$, $[0:0:1]$ is the singular point of X and $[0:1:0]$ its smooth flex point. It is possible to reconstruct $\langle X\rangle$ for the following reason. The cuspidal tangent line $\{x_1=0\}$ is uniquely determined by X and it does not contain the flex $[0:1:0]$. Hence, from ${\mathit{\ell }}_{o_1}\left(X\right)$ and ${\mathit{\ell }}_{o_2}\left(X\right)$ we identify a line of $\langle X\rangle$ and another point of $\langle X\rangle$ not on this line, and hence, we reconstruct the plane $\langle X\rangle$, and hence, X. An irreducible nodal curve has one singular point and three flexes. By the theorem of Bezout, these three flexes are not contained in a line containing the node of X.

Remark 22. 

Let $X=\{f=0\}$ be a degree $d>2$ reducible curve without multiple components. The zero-locus of the Hessian curve of X contains all lines which are irreducible components of X and no other irreducible components of X. We do not see how to fully reconstruct X with two cameras not contained in the plane $\langle X\rangle$ just with the use of the singular points and the Hessian curve only in the following two cases:

1. 

d points through a common point;

2. 

$d=3$ and X is the union of a smooth conic and of one of its tangent lines.

Question 2: Is the cuspidal plane cubic the only case among the irreducible curves for which the union of the singular points and the flexes does not span the plane? Are the two cases described in Remark 22 the only ones among the reducible plane curves?

Proposition 5. 

Take a plane curve $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ such that no element of $\mathrm{Aut}(\langle X\rangle )$, except the identity, induces an automorphism of X. Take $o_1,o_2\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus \langle X\rangle$ such that $o_1\ne o_2$. Then, X is the only plane curve of degree $>1$ contained in $X_{o_1}\cap X_{o_2}$.

Proof. 

Assume the existence of another plane curve $Y\subset X_{o_1}\cap X_{o_2}$ which is not a line. Since Y is not a line, ${\mathit{\ell }}_{o_i}$, $i=1,2$, induces an isomorphism $f_i:Y\to X$. Since $o_1\ne o_2$, $f_2\circ f_1^{-1}$ is not the identity map, this is a contradiction. □

A general smooth plane curve of degree at least 4 has no nontrivial automorphism, and hence, we may apply Proposition 5 to this plane curve. The case $d=2$ shows that some assumptions are necessary.

6.3. Real Plane Curves

Now, assume that the degree d plane curve X is defined over $\mathbb{R}$. We take $o\in {\mathbb{P}}^3\left(\mathbb{R}\right)$ and we assume that ${\tilde{\mathit{\ell }}}_o\left(X\left(\mathbb{C}\right)\right)$ is not a line, and hence, it has degree d. Since its real part $X\left(\mathbb{R}\right)$ may be finite or empty (even if X is smooth) by Proposition 1, ${\mathit{\ell }}_o\left(X\right)\left(\mathbb{R}\right)$ may be larger than ${\mathit{\ell }}_o(X\left(\mathbb{R}\right)$, but ${\mathit{\ell }}_o\left(X\right)\left(\mathbb{R}\right)\setminus {\mathit{\ell }}_o\left(X\left(\mathbb{R}\right)\right)$ is always finite (Proposition 3). We assume that $X\left(\mathbb{R}\right)$ is infinite, and hence, it contains an open arc. We showed how to obtain the curve ${\mathit{\ell }}_o\left(X\right)\left(\mathbb{C}\right)$ from the open arc (Remark 5). Having two different points $o_1,o_2$ not in $\langle X\left(\mathbb{C}\right)\rangle$ we may use all suggestions from the reconstruction of $X\left(\mathbb{C}\right)$. If we know $X\left(\mathbb{C}\right)$, then we know $X\left(\mathbb{R}\right)=X\left(\mathbb{C}\right)\cap {\mathbb{P}}^3\left(\mathbb{R}\right)$.

7. Curves in a Smooth Quadric Surface and a Quadric Cone

We use this section to consider the $\alpha$-reconstruction of curves of degree at most 4 (see Section 8). However, as explained in Remark 4, this section may also be used to study the $\alpha$-reconstruction of irreducible curves whose arithmetic genus is very high with respect to their degree, i.e., the next step after plane curves.

Let Q be a smooth quadric surface defined over $\mathbb{C}$. We have $Q\left(\mathbb{C}\right)\cong {\mathbb{P}}^1\left(\mathbb{C}\right)\times {\mathbb{P}}^1\left(\mathbb{C}\right)$ and we fix homogeneous coordinates $x_0,x_1$ on the first factor of Q and homogeneous coordinates $y_0,y_1$ on the second factor of Q. Fix $(a,b)\in {\mathbb{N}}^2$. Let $\mathbb{C}{[x_0,x_1,y_0,y_1]}_{a,b}$ denote the set of all bihomogeneous polynomials of bidegree $(a,b)$, i.e., the set of all $\mathbb{C}$-linear combinations of the monomials $x_0^c{x}_1^d{y}_0^e{y}_1^f$ with $c\ge 0$, $d\ge 0$, $e\ge 0$, $f\ge 0$, $c+d=a$, and $e+f=b$. These monomials are linearly independent, and hence, $\mathbb{C}{[x_0,x_1,y_0,y_1]}_{a,b}$ is a $\mathbb{C}$-vector space of dimension $(a+1)(b+1)$. The important result is that all curves $X\subset Q$ have a bidegree and that the set of all curves $X\subset Q$ of bidegree $(a,b)$ is parametrized one to one by the projective space $\mathbb{P}\mathbb{C}{[x_0,x_1,y_0,y_1]}_{a,b}$ (case $g=e=0$ of Ch. V.2 in [38]).

Now, assume that Q is defined over $\mathbb{R}$ and that each of its two rulings is defined over $\mathbb{R}$. Under these assumptions we have $Q\left(\mathbb{R}\right)\cong {\mathbb{P}}^1\left(\mathbb{R}\right)\times {\mathbb{P}}^1\left(\mathbb{R}\right)$ and for all $(a,b)\in {\mathbb{N}}^2$ the complex conjugation $\sigma$ acts on the $\mathbb{C}$ vector space $\mathbb{C}{[x_0,x_1,y_0,y_1]}_{a,b}$. For each $f\in \mathbb{C}{[x_0,x_1,y_0,y_1]}_{a,b}$ the bihomogeneous polynomial $\sigma \left(f\right)$ is obtained from f, taking the complex conjugate of each coefficient of f. Thus, $\sigma \left(f\right)=f$ if and only if each coefficient of f is real, i.e., if and only if $f\in \mathbb{R}{[x_0,x_1,y_0,y_1]}_{a,b}$. A curve $\{f=0\}\subset Q$, $f\in \mathbb{C}{[x_0,x_1,y_0,y_1]}_{a,b}$, $f\ne 0$, is defined over $\mathbb{R}$ if and only if a non-zero multiple of f is in $\mathbb{R}{[x_0,x_1,y_0,y_1]}_{a,b}$. Thus, $\mathbb{P}\left(\mathbb{R}{[x_0,x_1,y_0,y_1]}_{a,b}\right)$ parametrizes all curves defined over $\mathbb{R}$ and of bidegree $(a,b)$.

Now, assume that X is defined over $\mathbb{R}$, i.e., $\sigma \left(X\right(\mathbb{C}\left)\right)=X\left(\mathbb{C}\right)$, where $\sigma :{\mathbb{P}}^3\left(\mathbb{C}\right)\to {\mathbb{P}}^3\left(\mathbb{C}\right)$ is the complex conjugation. Since Q is the unique quadric surface containing X, $\sigma \left(Q\right)=Q$, i.e., Q is a rank 4 quadric surface defined over Q. There are two types of such quadrics: either $Q\left(\mathbb{R}\right)=\varnothing$ or $Q\left(\mathbb{R}\right)\cong {\mathbb{P}}^1\left(\mathbb{R}\right)\times {\mathbb{P}}^1\left(\mathbb{R}\right)$, which is diffeomorphic to the product of two circles. The two cases are distinguished by the action of $\sigma$ on the two rulings of $Q\left(\mathbb{C}\right)$. We have $Q\left(\mathbb{R}\right)=\varnothing$ if and only if $\sigma$ exchanges the two rulings of $Q\left(\mathbb{C}\right)$, while $Q\left(\mathbb{R}\right)\cong {\mathbb{P}}^1\left(\mathbb{R}\right)\times {\mathbb{P}}^1\left(\mathbb{R}\right)$ if and only if $\sigma$ preserves the two rulings, i.e., it sends each line of one ruling to a line of the same ruling. Take an irreducible curve $Y\in |{\mathcal{O}}_Q(a,b)|$. In the first (resp., second case) $\sigma \left(Y\right)\in |{\mathcal{O}}_Q(b,a)|$ (resp., $\sigma \left(Y\right)\in |{\mathcal{O}}_Q(a,b)|$. Since $\sigma \left(X\right)=X$ and $1\ne 3$, we obtain $Q\left(\mathbb{R}\right)\cong {\mathbb{P}}^1\left(\mathbb{R}\right)\times {\mathbb{P}}^1\left(\mathbb{R}\right)$. Since $X\in |{\mathcal{O}}_Q(1,3)|$, the second projection of Q induces an isomorphism $f:X\to {\mathbb{P}}^1$. Since $\sigma$ preserves each ruling, f is defined over $\mathbb{R}$, and hence, $X\left(\mathbb{R}\right)\cong {\mathbb{P}}^1\left(\mathbb{R}\right)$ is diffeomorphic to a circle.

Let $T\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ denote an irreducible quadric cone. Call o the vertex of T. Take a curve $X\subset T$ and set $d:=deg\left(X\right)$. Let $\mu$ be the multiplicity of X at o. Note that $\mu =0$ if and only if $o\notin X$ and $\mu =1$ if and only if o is a smooth point of X. All lines of T contain o and there is a pencil of lines through p. From now on we assume $d>1$.

We first describe how to use Ex. V.2.9 in [38] to describe all curves with $\mu \le 1$. First, assume $\mu =0$, i.e., $o\notin X$. In this case, d is even and X is the complete intersection of T and a degree $d/2$ surface not containing X. Now, assume $\mu =1$. Fix a line $L\subset T$. In this case, d is odd and the curve $X\cup L$ is the complete intersection of T and a surface of degree $(d+1)/2$. To solve the quoted exerciseEx. V.2.9 in [38], and hence, to prove all the claims we just gave, we need a setup which works for an arbitrary $\mu$.

Let $\pi :F_2\to T$ be a minimal desingularization of T. The smooth surface $F_2$ is one of the Hirzebruch surfaces described in Ch. V, §2 in [38]. Set $h:={\pi }^{-1}\left(o\right)$ and let f be the class of an element of a ruling of $F_2$. We have $\mathrm{Pic}\left(F_2\right)\cong \mathbb{Z}h\oplus \mathbb{Z}f$ (Prop. 2.3 in [38]) and we write ${\mathcal{O}}_{F_2}(ah+bf)$, $a,b\in \mathbb{Z}$, for the line bundles on X. We have $h^0\left({\mathcal{O}}_{F_2}(ah+bf)\right)>0$ if and only if $a\ge 0$ and $b\ge 0$. Assume $a\ge 0$. The linear system $|{\mathcal{O}}_{F_2}(ah+bf)|$ is base-point-free if and only if $b\ge 2a$ (Th. V.2.17 in [38]). The images by $\pi$ of the elements of $|{\mathcal{O}}_{F_2}\left(bf\right)|$, $b>0$, are the unions of b lines contained in T. Now, assume $a>0$. If $0\le b<2a$, then $|{\mathcal{O}}_{F_2}(ah+bf)|$ is formed by the union of h with multiplicity $\lceil (2a-b)/2\rceil$ and an element of $|{\mathcal{O}}_{F_2}(a-\lceil (2a-b)/2\rceil )h+bf\left)\right|$. Now, assume $b\ge 2a$. We have $h^0\left({\mathcal{O}}_{F_2}(ah+bf)\right)=(a+1)(b+1)-a(a+1)$. By the adjunction formula, all $C\in |{\mathcal{O}}_{F_2}(ah+bf)|$ have arithmetic genus q with $2q-2=-2(a-2)a+(a-2)b$. By a theorem of Bertini and Th. V.2.17 in [38] and in general, $C\in |{\mathcal{O}}_{F_2}(ah+bf)|$ is smooth and (if $a>0$) it is connected. Any $C\in |{\mathcal{O}}_{F_2}(ah+bf)|$ without h as a component has image $\pi \left(C\right)$ with $\mu =b-2a$ and degree ${\mathcal{O}}_{F_2}(h+2f)·{\mathcal{O}}_{F_2}(ah+bf)=b+2a$. Moreover, C may be described as a single equation in a weighted projective plane. Now, assume that T is defined over $\mathbb{R}$. In this case, $o\in T\left(\mathbb{R}\right)$. Take an irreducible $X\subset T$ defined over $\mathbb{R}$, with X not a line, and let $Y\in |{\mathcal{O}}_{F_2}(ah+bf)|$ be the strict transform of X. We have $deg\left(X\right)=b+2a$, $\mu =b-2a$ and Y is obtained as the zero-locus of a single equation in a weighted projective plane. Note that if we project X from the vertex o we only obtain a smooth conic.

8. Curves of Degree 3 and 4

Irreducible and non-planar degree 3 space curves are the rational normal curves considered in [17]. In the next example we consider them from the point of view of $\alpha$-reconstruction.

Example 5. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an integral degree 3 curve not contained in a plane. It is well known that X is unique, up to a projective transformation, and that it is smooth and rational, i.e., isomorphic to ${\mathbb{P}}^1$, and all possible Xs are projectively equivalent. Moreover, some Xs may be defined over $\mathbb{R}$, all of them are projectively equivalent over $\mathbb{R}$ and $X\left(\mathbb{R}\right)$ is a circle. Since 3 is a prime number and X is not contained in a plane, $X\ne X_{o_1}\cap X_{o_2}$ (not even set-theoretically), but we may reconstruct X from two cameras with a center in $X\left(\mathbb{C}\right)$ in the following way. Fix $o_1,o_2\in X\left(\mathbb{C}\right)$ such that $o_1\ne o_2$. Each ${\tilde{\mathit{\ell }}}_{o_i}(X\left(\mathbb{C}\right)$ is a smooth conic and $X_{o_1}\left(\mathbb{C}\right)\cap X_{o_2}\left(\mathbb{C}\right)$ is the union of $X\left(\mathbb{C}\right)$ and the line $\langle \{o_1,o_2\}\rangle$. If we know $o_1$ and $o_2$, we know the line $\langle \{o_1,o_2\}\rangle$ and we subtract it from $X_{o_1}\left(\mathbb{C}\right)\cap X_{o_2}\left(\mathbb{C}\right)$.

Now, assume that X is defined over $\mathbb{R}$. Since 3 is odd, intersecting $X\left(\mathbb{C}\right)$ with a real plane gives that $X\left(\mathbb{R}\right)\ne \varnothing$. Proposition 1 gives that $X\left(\mathbb{R}\right)$ is topologically a circle. We take $o_1,o_2\in X\left(\mathbb{R}\right)$ and obtain that $X_{o_1}\cap X_{o_2}$ is the union of X and the real line $\langle \{o_1,o_2\}\rangle$.

The pictures one obtains projecting from $o\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X\left(\mathbb{C}\right)$ are listed in Figure 2 in [17].

Note that to use the recipe in Example 5 we do not need to be clever: any choice of points $o_1$ and $o_2$ of $X\left(\mathbb{C}\right)$ or $X\left(\mathbb{R}\right)$, $o_1\ne o_2$, works.

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an integral and non-degenerate projective curve. By Remark 4 and the classification of curves on a quadric surface, the curve X is one of the following ones:

• X is an elliptic curve, i.e., it is smooth and it has genus 1;
• X is singular;
• X is smooth and rational, i.e., it is smooth and of genus 0.

We first consider the case in which X is an elliptic curve and prove that $X=X_{o_1}\cap X_{o_2}$ for some $o_1,o_2\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X\left(\mathbb{C}\right)$. We also see that X is contained in exactly four quadric cones.

Example 6. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an elliptic curve. The Riemann–Roch theorem gives that X is the complete intersection of two quadric surfaces. By case 2 of the proof of Th. 1 in [32], X is contained in exactly four quadric cones, and hence, there is a set $A\subset {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X\left(\mathbb{C}\right)$ such that $\#A=4$ and $X=X_{o_1}\cap X_{o_2}$ if and only if $o_1$ and $o_2$ are distinct elements of A.

Now, assume that X is defined over $\mathbb{R}$. By Proposition 1, either $X\left(\mathbb{R}\right)=\varnothing$ or $X\left(\mathbb{R}\right)$ is topologically a circle or $X\left(\mathbb{R}\right)$ is topologically the union of two disjoint circles. We claim that each of these three cases occurs for some X. For the first one, just take a general complete intersection of two quadric surfaces $Q_1$ and $Q_2$ with $Q_2\left(\mathbb{R}\right)=\varnothing$, i.e., associated to a definite quadratic form. This case is described in [39] p. 21. For the second and third cases, just take an elliptic curve C defined over $\mathbb{R}$, with $C\left(\mathbb{R}\right)$ prescribed (one or two circles) and embed it in ${\mathbb{P}}^3$ by the complete linear system $4p$ with $p\in C\left(\mathbb{R}\right)$.

In the next example, we prove that every degree 4 singular and irreducible space curve is of the form $X=X_{o_1}\cap X_{o_2}$ for some $o_1,o_2\in {\mathbb{P}}^3\left(\mathbb{C}\right)\setminus X\left(\mathbb{C}\right)$. The general set up at the beginning of Example 7 allows us to describe which reducible Xs are complete intersections of two quadric cones. The union of a rational normal curve and one of its tangent lines is not the intersection of two quadric cones.

Example 7. 

Pencils of quadrics over $\mathbb{C}$ or over $\mathbb{R}$ are uniquely determined by their Segre symbol (§8.6.1 in [35]). See Ex. 3.1 in [40] for a full list in the case of pencils of quadrics in ${\mathbb{P}}^3\left(\mathbb{C}\right)$ with, for each pencil, the associated degree 4 complete intersection. Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be an irreducible and non-degenerate degree 4 curve with at least one singular point, o. By Remark 4, o is the unique singular point of X, X has either an ordinary node or an ordinary cusp at o, X is rational, and X is the complete intersection of two quadric surfaces. To prove that X may be reconstructed by two (very particular) linear projections it is sufficient to prove that X is contained in at least two quadric cones. Since X has multiplicity 2 at o, $deg\left(X\right)=4$ and X is non-degenerate, $\tilde{\mathit{\ell }}\left(X_o\right)$ is a smooth conic. Hence, X is contained in a unique quadric cone with vertex o. Up to a projective transformation of ${\mathbb{P}}^3\left(\mathbb{C}\right)$ there are only two possible Xs, the one with a node at o and the one with a cusp at o. We will call the former the nodal degree 4 curve and the latter the cuspidal degree 4 curve. Thus, it is sufficient to find two pairs, $(X_1,C_1)$ and $(X_2,C_2)$, such that $X_1$ is a nodal degree 4 curve, $X_2$ is a cuspidal degree 4 curve, and $C_i$ is a quadric cone containing $X_i$ but with vertex $p_i$ not equal to the singular point, $o_i$, of $X_i$. Fix $p\in {\mathbb{P}}^3\left(\mathbb{C}\right)$ and let C be any irreducible quadric cone with vertex p. Fix $o\in C\setminus \left\{p\right\}$. The plane $T_{o}C$ is a plane containing the line $\langle \{o,p\}\rangle$. Let $2o$ denote the closed subscheme of C with ${\left({\mathcal{I}}_{o,C}\right)}^2$ as its ideal sheaf. The scheme $2o$ is zero-dimensional, $deg\left(2o\right)=3$, and $\langle 2o\rangle =T_{o}C$. Let $|{\mathcal{I}}_{2o}\left(2\right)|$ denote the set of all quadric surfaces containing $2o$. The set $|{\mathcal{I}}_{2o}\left(2\right)|$ is a projective space of dimension 6. We have $C\in |{\mathcal{I}}_{2o}\left(2\right)|$ and for any $T\in |{\mathcal{I}}_{2o}\left(2\right)|$ with $T\ne C$, the scheme $T\cap C$ is a degree 4 connected curve of arithmetic genus 1 with a singular point at o. Each line contained in C contains o. Taking T with $p\notin T$, which is easy, we obtain a curve $T\cap C$ containing no line. The set of all unions $X^{\prime }\subset C$ of two conics singular at p is parametrized by the pairs of planes containing o, and hence, they form a 4-dimensional family. Thus, there are T such that $C\cap T$ is a degree 4 irreducible curve with o as its singular point. We saw that any such $T\cap C$ is the complete intersection of C and a quadric cone with vertex o. A general T has as $T\cap C$ a nodal degree 4 curve. For a general $T\in |{\mathcal{I}}_{2o}\left(2\right)|$ the curve $T\cap C$ is nodal at o. To obtain that the cuspidal degree 4 curves are complete intersection of two cones it is sufficient to observe that their pencil of quadric surfaces contains exactly two quadric cones (case 4 on p. 325 in [40]).

Now, we work over $\mathbb{R}$ and assume that X is defined over $\mathbb{R}$. Since o is the unique singular point of $X\left(\mathbb{C}\right)$ and $\sigma \left(X\right(\mathbb{C}\left)\right)=X\left(\mathbb{C}\right)$, $\sigma \left(o\right)=o$, i.e., $o\in X\left(\mathbb{R}\right)$. There are nodal X with $X\left(\mathbb{R}\right)=\left\{o\right\}$ (and so nothing can come from a real picture). These curves are obtained in the following way. Let $U\subset {\mathbb{P}}^4\left(\mathbb{C}\right)$ be a degree 4 rational normal curve defined over $\mathbb{R}$ and such that $U\left(\mathbb{R}\right)=\varnothing$ (such a real rational normal curve exists for all even degrees). Take $a\in U\left(\mathbb{C}\right)$. Projecting from $\langle \{a,\sigma (a\left)\right\}\rangle \setminus \{a,\sigma (a\left)\right\}$ we obtain a nodal real degree curve with its singular point as the only real point. There are two isomorphism classes as real algebraic curves of nodal degree 4 real curves X with $X\left(\mathbb{R}\right)$ infinite. Both types are obtained in the following way. Let $V\subset {\mathbb{P}}^4$ be a degree 4 rational normal curve defined over $\mathbb{R}$ and with $V\left(\mathbb{R}\right)\ne \varnothing$. $V\left(\mathbb{R}\right)$ is topologically a circle. Fix $b_1,b_2\in V\left(\mathbb{R}\right)$ such that $b_1\ne b_2$ and $c\in V\left(\mathbb{C}\right)\setminus V\left(\mathbb{R}\right)$. Consider the linear projection of V from a point of $\langle \{b_1,b_2\}\rangle$. Thus, the real nodal degree 4 curve X has $X\left(\mathbb{R}\right)$ connected and homeomorphic to a circle with two points identified, i.e., it is homeomorphic to the union of two circles with a common point. The complex line $\langle \{c,\sigma (c)\rangle$ is σ-invariant, and hence, it is defined over $\mathbb{R}$ with $\langle \{c,\sigma (c)\rangle (\mathbb{R})$ homeomorphic to a circle. Since V is a rational normal curve, $\langle \{c,\sigma (c\left)\right\}\rangle \cap V=\{c,\sigma (c\left)\right\}$, and hence, $\langle \{c,\sigma (c\left)\right\}\rangle \left(\mathbb{R}\right)\cap V\left(\mathbb{R}\right)=\varnothing$. The image of V by a point of $\langle \left\{c\sigma \right(c\left)\right\}\rangle \left(\mathbb{R}\right)$ is a real nodal degree 4 curve W such that $W\left(\mathbb{R}\right)$ has two connected components, one of them being homeomorphic to a circle and the other one being the isolated singular point. There is a unique isomorphism class of real degree 4 cuspidal curves. They are obtained projecting V by a point of $T_{p_1}V\setminus \left\{b_1\right\}$. For these curves the real part is homeomorphic to $V\left(\mathbb{R}\right)$, i.e., to a circle.

The following observation is the converse part of Examples 6 and 7:

Remark 23. 

A pencil of quadric surfaces not having a plane as a fixed component of all its element has as its common intersection Y a degree 4 connected space curve spanning ${\mathbb{P}}^3$. Y is either integral (as the default X here) or the union of a line and a degree 3 rational normal curve (described in [17]) or the union of conics and lines with total degree 4.

Example 8. 

Let $X\subset {\mathbb{P}}^3\left(\mathbb{C}\right)$ be a smooth and rational degree 4 curve. The Riemann–Roch theorem gives that X is contained in a unique quadric surface Q. By ex. V.2.9 in [38], the quadric Q is smooth. Hence, X is contained in no quadric cone. The best we can achieve with low-degree cones is to take sufficiently general $o_1,o_2,o_3\in X$ and obtain that $X_{o_1}\cap X_{o_2}\cap X_{o_3}$ is the union of X and a finite set with each $X_{o_i}$ of degree 3.

Now, assume that X is defined over $\mathbb{R}$. By Proposition 1, either $X\left(\mathbb{R}\right)=\varnothing$ or $X\left(\mathbb{R}\right)$ is topologically a circle. We claim that both cases occur. Indeed, the curve X is a linear projection of a rational normal curve C of ${\mathbb{P}}^4$ from a point o not contained in the secant variety of $C\left(\mathbb{C}\right)$. Since X is defined over $\mathbb{R}$, C is defined over $\mathbb{R}$ and $o\in {\mathbb{P}}^4\left(\mathbb{R}\right)$. Moreover, we claim that the converse holds. To prove this claim it is sufficient to find rational normal curves $C_1$ and $C_2$ of ${\mathbb{P}}^4$ defined over $\mathbb{R}$ with $C_1\left(\mathbb{R}\right)=\varnothing$ and $C_2\left(\mathbb{R}\right)$ a circle, and then, take a linear projection of $C_i$ from a general point of ${\mathbb{P}}^4\left(\mathbb{R}\right)$. We fix real models $U_1,U_2$ of ${\mathbb{P}}^1$ with $U_1\left(\mathbb{R}\right)=\varnothing$ and $U_2\left(\mathbb{R}\right)$ a circle (Proposition 1) and embed them in ${\mathbb{P}}^4$ using a degree 4 σ-invariant effective divisor.

Finite Unions of Low-Degree Curves

It would be nice to extend the multiview by n cameras of the union of m lines to the case of finite unions of a prescribed number of low-degree curves, say $m_1$ lines, $m_2$ smooth conics, and $m_3$ rational normal curves. It is a natural assumption that with cameras we are able to distinguish between the image of a line, the image of a smooth conic, and the image of a rational normal curve. We saw that this is the case for general cameras (Theorem 4). It is a natural assumption that in each picture we may separate the $m_1$ lines, $m_2$ smooth conics, and $m_3$ rational normal curves.

In algebraic geometry, one usually starts with a compactification of the data, because in the compact case, i.e., the projective case, we may count the number of solutions with algebro-geometric tools. In the case of multiviews of a single line in ${\mathbb{P}}^3$, this is the Grassmannian $\mathbb{G}$, and the explicit way to use it to obtain the image as the first key step in [11]. An explicit compactification of degree 3 rational normal curves is known [41] and it was quoted in [17]. For a single smooth conic (or a single curve of any degree) in ${\mathbb{P}}^3$ one option is to use the Hilbert scheme compactification. An easy method for conics is to look at pairs $(M,C)$, where M is a plane and C is a conic of M (perhaps reducible or a double line). The set of all planes of ${\mathbb{P}}^3$ is isomorphic to ${\mathbb{P}}^3$. Thus, this compactification is smooth and it is a fiber bundle over ${\mathbb{P}}^3$ with ${\mathbb{P}}^5$ as fibers. This compactification works well over $\mathbb{R}$. If instead of conics we take degree d plane curves, we obtain a compactification of the set of all degree d plane curves (with the plane not fixed in ${\mathbb{P}}^3$), which is a fiber bundle over ${\mathbb{P}}^3$ with as fibers a projective space of dimension $\left(\genfrac{}{}{0pt}{}{d+2}{2}\right)-1$.

Then, we need the compactification for finite unions of m objects, say m conics. There are two natural compactifications, the symmetric product compactification and the Hilbert scheme compactification (Section 6).

Then, we need to compactify the images. The best method is the Hilbert scheme. For instance, as a compactification of all unions $A\subset {\mathbb{P}}^2$ of m distinct smooth conics we obtain the projective space of all degree $2m$ homogeneous polynomials whose irreducible factors have degree at most 2.

9. Materials and Methods

All the results of this paper are given with a mathematical proof.

10. Discussion

In this paper, we consider the reconstruction of algebro-geometric curves in a projective space (real or complex) using n pin-hole cameras whose location is known. We prove several general theorem in which reconstruction is proved using only four general cameras. The objects considered here are finite union of lines, plane curves, and space curves of degree 3 and 4.

11. Conclusions

In this paper, we study the reconstruction of algebro-geometric objects (finite union of lines, plane curves, low-degree curves) using a low number of pin-hole cameras with known centers. We prove that they can be reconstructed (even over the real numbers) by four general cameras, and hence, hopefully, by four random cameras. We show that sometimes no number of badly put cameras are sufficient. We found examples over the real numbers in which, using in a bad way the algebro-geometric definitions, ghosts occur, i.e., the emptyset may have as its image a circle. We show that ghosts do not occur for general cameras. We suggest the extension of one shown here (union of m lines using n cameras) to the case of m objects, e.g., m conics or unions of $m_1$ lines, $m_2$ conics, and $m_3$ rational normal curves. In some cases, we see how two very particular cameras are sufficient or sufficient up to subtracting one specific line which we know will appear in the back-projection of the two images. We gave a few suggestions for further study at the end of Section 8 and two open questions in earlier sections.