The geometry of rank drop in a class of face-splitting matrix products: Part II

Definition 2.1

A quadratic Cremona transformation of ℙ² is a birational automorphism f : ℙ² ⇢ ℙ² defined as f(x) = (f₁(x) : f₂(x) : f₃(x)) where f₁, f₂, f₃ are homogeneous quadratic polynomials in x = (x₁, x₂, x₃).

Lemma 2.2

Let g be a Cremona transformation and let f be the standard Cremona involution. Then there are projective transformations H₁, H₂ such that g = H₁ ∘ f ∘ H₂.

Proof

Let a₁, a₂, a₃ ∈ ℙ² denote the base points of g. The coordinates (g₁, g₂, g₃) of g form a basis for the three-dimensional vector space of quadratics vanishing on the points a₁, a₂, a₃. Another basis is h = (ℓ₂ ℓ₃, ℓ₁ℓ₃, ℓ₁ℓ₂) where ℓ_i ∈ ℂ[x, y, z]₁ defines the line joining a_j and a_k for every labeling {i, j, k} = {1, 2, 3}. Therefore there is some invertible linear transformation H₁ for which g = H₁h. Similarly, (ℓ₁, ℓ₂, ℓ₃) is a basis for ℂ[x, y, z]₁ and so there is a linear transformation H₂ for which H₂ (x, y, z) = (ℓ₁, ℓ₂, ℓ₃). The map h is given by f ∘ H₂ and so g = H₁ ∘ f ∘ H₂. □

Lemma 2.3

Let f : Px2⇢Py2 be a Cremona transformation. If f and f⁻¹ have base points e1x=e1y = (1:0:0), e2x=e2y=(0:1:0),e3x=e3y=(0:0:1) in the domain and codomain, then f has the form

where a, b, c ∈ ℂ ∖ {0}.

Proof

Suppose that f = (f₁, f₂, f₃) where f₁, f₂, f₃ are quadratic polynomials. Since f is undefined at the three base points in the domain, it follows that f₁, f₂, f₃ contain only the monomials x₁x₂, x₁x₃, x₂x₃. Moreover, we know that f(x₁, x₂, 0) = (0 : 0 : 1). It follows that f₁, f₂ do not contain the monomial x₁x₂. In examining the other two exceptional lines, we find that f₁, f₂, f₃ contain only one monomial each and that f has the desired form. □

Definition 2.4

Let f be a Cremona transformation with base points B(f). For a curve C ⊂ ℙ², define f(C) := f(C ∖ B(f)), and for a given point p, let ν_p(C) be the multiplicity of the curve C at the point p.

Lemma 2.5

(See [4]). Let C ⊂ ℙ² be a plane curve of degree n and let f be a Cremona transformation. Then

In particular, if C is a smooth cubic curve then f(C) is also a cubic curve if and only if the base points of f lie on C. In this case, f⁻¹(f(C)) = C implies that the base points of f⁻¹ lie on f(C).

Lemma 2.6

Let C be a smooth cubic curve and let f be a Cremona transformation with base points a₁, a₂, a₃ ∈ C in the domain and b₁, b₂, b₃ in the co-domain. Then f(C) is a smooth cubic curve and f̄ : C → f(C), defined by taking the closure of f|_C∖B(f), is an isomorphism.

Proof

By Lemma 2.5, f(C) is a cubic curve. Moreover, since f⁻¹(f(C)) = C is a cubic curve, it also follows that b₁, b₂, b₃ ∈ f(C). The fact that f̄ is an isomorphism follows from the corollary after [8, § 1.6, Theorem 2] which says that a birational map between nonsingular projective plane curves is regular at every point, and is a one-to-one correspondence. □

Lemma 2.7

Let f : C → C′ be an isomorphism of smooth cubic plane curves. Then there is a two-parameter family of Cremona transformations fσ′ : ℙ² ⇢ ℙ² such that fσ′|C = f. The base points of these Cremona transformations will lie on the cubic curves.

Proof

Since C and C′ are isomorphic, they have the same Weierstraß form C₀. There are therefore homographies H₁, H₂ ∈ PGL(3) such that H₁(C) = C₀ = H₂(C′) and therefore H1−1 H₂(C′) = C. Then H1−1 H₂ ∘ f : C → C is an automorphism of C and it follows by [4, Theorem 1.3] that this is induced by some two-parameter family of Cremona transformations g_σ; the members of this family are obtained by picking the first two base points arbitrarily on C and then letting the third base point be determined by the equation a₃ = a(3b − a₁ − a₂). Then fσ′:=H2−1H1∘gσ is the desired family of Cremona transformations. By Lemma 2.5 the base points of each of these Cremona transformations lie on the cubic curves. □

Theorem 3.2

(Trinity correspondence). Consider the following three sets:

1. 𝓛: the set of all generic lines ℓ in ℙ(ℂ^3×3),

2. 𝓠: the set (up to projective equivalence) of pairs of linear projections π₁, π₂ : ℙ³ ⇢ ℙ² with non-coincident centers c₁, c₂, along with a permissible quadric Q ⊂ ℙ³ through c₁, c₂,

3. 𝓒: the set of (non-degenerate) Cremona transformations from ℙ² ⇢ ℙ².

Then there is a 1 : 1 correspondence between 𝓛 and 𝓒, a 1 : 3 correspondence between 𝓛 and 𝓠, and a 3 : 1 correspondence between 𝓠 and 𝓒, such that the diagram (6) commutes:

(6)

Lemma 3.3

Fix a pair of linear projections π₁, π₂ : ℙ³ ⇢ ℙ² with non-coincident centers c₁, c₂ and let F be its fundamental matrix. There is a 1 : 1 correspondence between the quadrics Q ⊂ ℙ³ through c₁, c₂ and the lines ℓ ⊂ ℙ(ℂ^3×3) through F.

Proof

Applying projective transformations, we can assume that c₁ = (1 : 0 : 0 : 0), c₂ = (0:1:0:0), π₁(u₁ : u₂ : u₃ : u₄) = (u₂ : u₃ : u₄) and π₂(u₁ : u₂ : u₃ : u₄) = (u₁ : u₃ : u₄). If F = (F_ij) is the fundamental matrix of (π₁, π₂), then for all u ∈ ℙ³ we have

Since the entries in position (2, 3) and (3, 2) of π₂(u) π₁(u)^⊤ are the same, F is a scalar multiple of

and B_F(x, y) = x₃ y₂ − x₂ y₃. In particular, there exists some p ∈ ℙ³ with π₁(p) = x and π₂(p) = y if and only if x₃ y₂ = x₂ y₃.

Consider the image of φ : ℙ³ ⇢ ℙ(ℂ^3×3) where φ(u) = π₂(u) π₁(u)^⊤. By (7), φ(ℙ³) is contained in the hyperplane F^⊥ ⊂ ℙ(ℂ^3×3). Any matrix in ℙ(ℂ^3×3) can be written as sF + M for some scalar s and M ∈ F^⊥. Therefore,

since π₂(u)^⊤ F π₁(u) = 0, and any linear function on the image of φ can be identified with its image in F^⊥. On the other hand, a line ℓ in ℙ(ℂ^3×3) through F is of the form {sF + tM : (s : t) ∈ ℙ¹}, where M ∈ F^⊥. Therefore, lines through F are in bijection with linear functions on φ(ℙ³), up to scaling.

The monomials u1u2,u1u3,u1u4,u2u3,u2u4,u32,u3u4,u42 form a basis for the 7-dimensional vector space of homogeneous quadratic polynomials that vanish on c₁, c₂. Thus any quadratic polynomial in ℂ[u₁, u₂, u₃, u₄]₂ vanishing at c₁ and c₂ can be written as 〈M, π₂(u) π₁(u)^⊤ 〉 for a unique matrix M ∈ F^⊥. This gives a linear isomorphism between linear functions on the image of φ, up to global scaling (which have been identified with lines through F), and quadrics passing though c₁ and c₂. □

Corollary 3.4

Let π₁, π₂ : ℙ³ ⇢ ℙ² be two linear projections with centers c₁ ≠ c₂ and fundamental matrix F. Let ℓ_F be a line in ℙ(ℂ^3×3) through F. The correspondence ℓ_F ↦ Q, where Q ⊂ ℙ³ is a quadric passing through c₁, c₂, is as follows. Let M ∈ ℓ_F be any M ≠ F. Then Q is cut out by the bilinear form

Lemma 3.5

([1], [5], [6, Result 22.11]). Under the 1 : 1 correspondence in Lemma 3.3, the line ℓ corresponds to a permissible quadric Q through c₁, c₂ if and only if ℓ is a generic line.

Lemma 3.6

Fix π_i : ℙ³ ⇢ ℙ² to be linear projections with non-coincident centers c_i for i = 1, 2. A permissible quadric Q through c₁, c₂ defines a Cremona transformation f : ℙ² ⇢ ℙ² such that f(π₁(p)) = π₂(p) for any point p ∈ Q. The base points of f are π₁(c₂) and the image under π₁ of the two lines contained in Q passing through c₁. Similarly, the base points of f⁻¹ are π₂(c₁) and the image under π₂ of the two lines contained in Q passing through c₂.

Proof

Since c₁, c₂ ∈ Q, the restriction of π₁ (and π₂) to Q is generically 1 : 1. Therefore, π₁(Q) and π₂(Q) are each birational to a ℙ². The map f will be π₂ ∘ (π₁|_Q)⁻¹. Let us check that this is a quadratic Cremona transformation.

As before, we can take π₁(u) = (u₂ : u₃ : u₄) and π₂(u) = (u₁ : u₃ : u₄). Then c₁ = (1 : 0 : 0 : 0) is the kernel of π₁, and we are given that it lies on Q. As we saw already, these assumptions imply that Q is defined by the vanishing of a polynomial of the form q(u) = α u₁ u₂ + β u₁ + γ u₂ + δ where α ∈ ℂ is a scalar, β, γ ∈ ℂ[u₃, u₄] are of degree 1, and δ ∈ ℂ[u₃, u₄] is of degree 2. We can then write q as

where a = (α u₂ + β), b = (γ u₂ + δ) ∈ ℂ[u₂, u₃, u₄] with deg(a) = 1, deg(b) = 2. The map (π₁|_Q)⁻¹ is then given by

To verify this, first check that π₁(u) = a(x) ⋅ x where ⋅ denotes scalar multiplication. To see that u ∈ Q we compute

where the last equality comes from the homogeneity of a, b with deg(a) = 1, deg(b) = 2.

Composing with π₂ we have

whose coordinates are indeed quadratic. Since f = π₂ ∘ (π₁|_Q)⁻¹ is defined by quadratics and generically 1 : 1, it is a quadratic Cremona transformation.

To show that this transformation is non-degenerate, we must demonstrate that it has three unique base points. To understand the base points of f, recall that on a smooth quadric surface there are two distinct (possibly complex) lines passing through each point. The images of the two lines passing through c₁ under the projection π₁ will each be a single point. Therefore f is not well-defined on these image points in ℙ². Similarly, f is undefined on π₁(c₂) since π2(π1−1(π1(c2))) = π₂(c₂) = 0. Therefore these three points are exactly the base points of f in the domain. Finally, because c₁ c₂ ⊄ Q, these base points are all distinct. The base points in the codomain can be found symmetrically. □

Lemma 3.7

Given a generic line ℓ ⊂ ℙ(ℂ^3×3), the set of points (x, y) ∈ ℙ² × ℙ² satisfying y^T Mx = 0 for all M ∈ ℓ coincides with the closure of the graph {(x, f(x)) : x ∈ ℙ² ∖ B(f)} of a unique Cremona transformation f : ℙ² ⇢ ℙ². This gives a 1 : 1 correspondence between generic lines ℓ ⊂ ℙ(ℂ^3×3) and Cremona transformations f : ℙ² ⇢ ℙ². Moreover, when F ∈ ℓ has rank two, this Cremona transformation agrees with that induced by the maps 𝓛_F → 𝓠_F → 𝓒.

Proof

Since ℓ is generic, we may assume without loss of generality that ℓ = span{F, M} where F has rank two. This gives a pair of linear projections π₁, π₂ : ℙ³ ⇢ ℙ² with non-coincident centers c₁, c₂ with fundamental matrix F. In the 1 : 1 correspondence 𝓛_F ↔ 𝓠_F given in Corollary 3.4, the line ℓ corresponds to the permissible quadric Q given by the zero set of q(u) = π₂(u)^⊤ Mπ₁(u). By Lemma 3.6, the Cremona transformation f : ℙ² ⇢ ℙ² corresponding to q(u) in the correspondence 𝓠_F → 𝓒 satisfies f(π₁(p)) = π₂(p) for all p ∈ Q ∖ {c₁, c₂}. Since π₁(Q) is dense in ℙ², the graph of f and the set {(π₁(p), π₂(p)): p ∈ Q ∖ {c₁, c₂}} ⊂ ℙ² × ℙ² are both two-dimensional, as is their intersection. Each is the image of an irreducible variety under a rational map and so the Zariski-closures of these two sets are equal. By construction, this is contained in the zero sets of y^TFx and y^TMx, as π₂(p)^T Fπ₁(p) = 0 for all p ∈ ℙ³ and π₂(p)^T Mπ₁(p) = 0 for all p ∈ Q. Since F, M are linearly independent, the variety {(x, y) : y^T Fx = y^TMx = 0} in ℙ² × ℙ² is two-dimensional. It therefore coincides with the Zariski-closure of the graph of f.

Conversely, suppose that f : ℙ² ⇢ ℙ² is a Cremona transformation. We claim that {f(x)x^⊤ : x ∈ ℂ³} spans a 7-dimensional linear space V ⊂ ℂ^3×3. Up to projective transformations on Px2  and  Py2, we can take f to be the standard Cremona involution, giving

One can check explicitly that seven distinct monomials appear in this matrix and so the span of all such matrices is 7-dimensional. Projectively, the orthogonal complement gives a line ℓ = V^⊥ in ℙ(ℂ^3×3). By definition, ℓ is exactly the set of all matrices M such that y^⊤ Mx = 0 for all (x, y) in the graph of f. Under the assumption that f is the standard Cremona transformation, ℓ is the span of the diagonal matrices F₁ = diag(1, −1, 0) and F₂ = diag(0, 1, −1); in general ℓ will be projectively equivalent to this line. We can verify that this line contains exactly the three rank-two matrices F₁, F₂, F₁ + F₂, and is therefore generic. □

Remark 3.8

Given ℓ = span{F, M} we can solve for the coordinates of the corresponding Cremona transformation f : ℙ² ⇢ ℙ² as follows. Given x ∈ ℙ², the corresponding point y = f(x) will be the left kernel of the 3 × 2 matrix FxMx The coordinates of y can be written explicitly in terms of the 2 × 2 minors of this matrix, which are quadratic in x. Note that, up to scaling, this formula for y is independent of the choice of basis {F, M} for ℓ. Any point x ∈ ℙ² for which FxMx has rank at most 1 will be a base point of this Cremona transformation. In particular, if Fx = 0, then x is a base point of f. As we will see below, there are three such points when ranging over all rank-two matrices in ℓ.

Lemma 3.9

Let ℓ be a generic line in ℙ(ℂ^3×3), i.e., ℓ contains three rank-two matrices F₁, F₂, F₃.

1. Then ℓ gives rise to three permissible quadrics Q₁, Q₂, Q₃ ⊂ ℙ³, each containing the centers of a pair of linear projections with fundamental matrices F₁, F₂, F₃ respectively.

2. The quadrics Q₁, Q₂, Q₃, in conjunction with their distinguished linear projections, all induce the same Cremona transformation f. The base points of f are e1x,e2x,e3x in the domain and e1y,e2y,e3y in the codomain, where eix  and  eiy generate the right and left nullspaces of F_i respectively.

Proof

A generic line ℓ ⊂ ℙ(ℂ^3×3) intersects the determinantal variety 𝓓 cut out by det X = 0 in three rank-two matrices F₁, F₂, F₃. Each F_i is the fundamental matrix of a pair of linear projections ℙ³ ⇢ ℙ² with non-coincident centers, and by Lemma 3.3 and Lemma 3.5 there is a unique permissible quadric Q_i through these centers corresponding to the line ℓ. By Lemma 3.7, each of these quadrics induces the same Cremona transformation f : ℙ² ⇢ ℙ².

To conclude, we show that the base points of f and f⁻¹ are e1x,e2x,e3x and e1y,e2y,e3y , respectively. We show that e1x,e2x,e3x are the base points of f and the argument for the base points of f⁻¹ follows symmetrically. First, note that each eix is a base point of f. This follows from Remark 3.8, since each F_i ∈ ℓ has rank two. Since the Cremona transformation f has three base points, it only remains to show that these points are distinct. If e1x=e2x, then by linearity Fe1x = 0 for all F ∈ ℓ = span{F₁, F₂}. This would imply that rank(F) ≤ 2 for all F ∈ ℓ, contradicting the genericity of the line ℓ. □

Corollary 3.10

The correspondence 𝓠 → 𝓒 is 3 : 1.

Proof

Let Q = (Q, π₁, π₂) ∈ 𝓠 be a permissible quadric along with a pair of linear projections that correspond to f ∈ 𝓒. If F is the fundamental matrix associated to (π₁, π₂), then there exists a unique generic line ℓ through F corresponding to Q by Lemma 3.3 and Lemma 3.5. With the full trinity correspondence, this line ℓ contains three fundamental matrices F₁, F₂, F₃ corresponding to Q₁, Q₂, Q₃ ∈ 𝓠 that each produce the Cremona transformation f. Moreover, by Lemma 3.7 this line ℓ is the unique line in ℙ(ℂ^3×3) corresponding to f. Therefore if Q′ ∈ 𝓠 is such that Q′ ↦ f it follows that π1′,π2′ have one of F₁, F₂, F₃ as their fundamental matrix and that the quadric Q′ is produced by the line ℓ. We conclude that Q′ is, up to projective equivalence, one of Q₁, Q₂, Q₃. □

Theorem 3.11

Given a generic codimension-two subspace V ⊂ ℙ(ℂ^3×3), the intersection of V with R₁, the Segre embedding of ℙ² × ℙ², is a del Pezzo surface of degree six, and can be described explicitly via the trinity correspondence. Specifically, if g : ℙ² ⇢ ℙ² is the Cremona transformation corresponding to the line V^⊥, then

Proof

For convenience, we denote

To see that this is a degree-six del Pezzo surface, we show that V₁ can be obtained as the blowup of ℙ² in three non-collinear points, specifically, at the base points of g: e1x,e2x,e3x . Let π_x : V₁ ⇢ ℙ² be the morphism defined by π_x(vu^⊤) = u. Let the ℓiy be the exceptional lines of g−1(ℓiy)=eix. Then π_x is 1 : 1 except on three mutually skew lines {y(eix)⊤:y∈ℓiy} which are taken to the points {eix} Therefore V₁ is the blowup of ℙ² in three non-collinear points and is a del Pezzo surface of degree six.

In particular, V₁ must be Zariski closed and it follows by Lemma 3.7 that V ∩ R₁ = V₁. □

Lemma 3.12

Suppose that P = (xi,yi)i=1k admits a generic line ℓ ⊆ 𝓝_Z (passing through three rank-two matrices F₁, F₂, F₃). Then for all j = 1, 2, 3 there is no i such that yi⊤ F_j = 0 = F_jx_i.

Proof

Suppose, without loss of generality, y1⊤ F₁ = 0 = F₁ x₁. From the matrix F₁ and the line ℓ through it we obtain a pair of projections π₁, π₂ with centers c₁, c₂ and a smooth permissible quadric Q passing through them. Then π₂(c₁) and π₁(c₂) are the left and right epipoles of F₁, but since y1⊤ F₁ = 0 = F₁ x₁, it must be that y₁ ∼ π₂(c₁) and x₁ ∼ π₁(c₂). On the other hand, for any point p on the line connecting c₁, c₂, we have

since F₂ ∈ 𝓝_Z. Therefore, by Corollary 3.4, p ∈ Q and thus c₁ c₂ ⊂ Q, which is a contradiction since Q is permissible. □

Definition 3.13

A generic line ℓ ⊆ 𝓝_Z is P-degenerate if there exists a rank-two matrix F ∈ ℓ such that either Fx_i = 0 or yi⊤ F = 0 for some i. We call a generic line that is not P-degenerate a P-generic line.

Definition 3.14

A quadric Q ⊂ ℙ³ passes degenerately through a reconstruction c1,c2,{pi}i=1k of P if it passes through these k + 2 points and contains the line through a center point c_i and a reconstructed point p_j.

Definition 3.15

A Cremona transformation f : ℙ² ⇢ ℙ² maps x_i ↦ y_i degenerately if x_i is a base point of f and y_i lies on the corresponding exceptional line, or symmetrically, y_i is a base point of f⁻¹ and x_i lies on the corresponding exceptional line.

Theorem 3.16

Given a configuration of point pairs P = (xi,yi)i=1k and the matrix Z = (xi⊤⊗yi⊤)i=1k , define the following subsets of 𝓛, 𝓠, 𝓒:

1. 𝓛_P : the set of all P-generic lines ℓ ⊆ 𝓝_Z := nullspace(Z),

2. 𝓠_P : the set (up to projective equivalence) of all permissible quadrics passing non-degenerately through some reconstruction c₁, c₂, p₁, …, p_k of P,

3. 𝓒_P : the set of all Cremona transformations f : ℙ² ⇢ ℙ² mapping x_i ↦ y_i non-degenerately for all i = 1, …, k.

Then there is a 1 : 1 correspondence between the elements of 𝓛_P and 𝓒_P, a 1 : 3 correspondence between the elements of 𝓛_P and 𝓠_P, and a 3 : 1 correspondence between the elements of 𝓠_P and 𝓒_P as in the diagram

(16)

Proof

We need to show that the trinity correspondence (6) can be restricted to the sets 𝓛_P, 𝓠_P, 𝓒_P. We will therefore examine each leg of this diagram.

(𝓛_P → 𝓠_P) We begin by considering a P-generic line ℓ = span{F, M} ⊆ 𝓝_Z. Without loss of generality, we can take F to be one of the three fundamental matrices in ℓ with corresponding projections π₁, π₂ : ℙ³ ⇢ ℙ² with non-coincident centers c₁, c₂ that give reconstructions p₁, …, p_k ∈ ℙ³ of the point pairs P. By Lemma 3.3, the line ℓ corresponds to a smooth permissible quadric Q defined by the vanishing of q(u) = π₂(u)^T Mπ₁(u). For any point p_i in the reconstruction, we have

since M ∈ ℓ ⊂ 𝓝_Z. Therefore Q passes through the reconstruction c₁, c₂, p₁, …, p_k. It remains to show that it does so non-degenerately. By Lemmas 3.6 and 3.9, a reconstructed point p_i lies on one of the lines through c₁ (or symmetrically through c₂) if and only if there exists M ∈ ℓ such that Mx_i = 0 (or symmetrically yi⊤ M = 0). Since ℓ is P-generic there is no such M, implying that the quadric passes through the reconstruction non-degenerately.

(𝓠_P → 𝓒_P) Consider a permissible quadric Q passing through a reconstruction c₁, c₂, p₁, …, p_k of P with linear projections π₁, π₂. As in Theorem 3.2, the tuple (Q, π₁, π₂) induces a Cremona transformation f := π₂ ∘ (π₁|_Q)⁻¹. By Lemma 3.6, the base points of f are the images of the point c₂ and each of the lines in Q passing through c₁. Since p_i ≠ c₂ and does not belong to these lines, the point x_i = π₁(p_i) is not a base point of f. Similarly, the base points of f⁻¹ are the images of the point c₁ and the lines in Q passing through c₂ under π₂, so a symmetric argument shows that y_i = π₂(p_i) is not a base point of f⁻¹. Therefore f maps x_i = π₁(p_i) to y_i = π₂(p_i) non-degenerately.

(𝓒_P → 𝓛_P) Consider a Cremona transformation f : ℙ² ⇢ ℙ² such that x_i ↦ y_i non-degenerately for all i. As in Lemma 3.7, f corresponds to a unique line ℓ ⊂ ℙ(ℂ^3×3) defined by the property that f(x)^⊤ Mx = 0 for all M ∈ ℓ and x ∈ ℙ². In particular, yi⊤ Mx_i = 0 for all M ∈ ℓ and i = 1, …, k, implying that ℓ ⊆ 𝓝_Z. By assumption, no point x_i is a base point of f and no point y_i is a base point of f⁻¹. By Lemma 3.9, it then follows that Mx_i ≠ 0 and yi⊤ M ≠ 0 for all M ∈ ℓ. Therefore ℓ is not P-degenerate. □

Remark 3.17

The assumptions of non-degeneracy can be removed from the 1 : 1 correspondence between generic lines in 𝓝_Z and Cremona transformations mapping x_i ↦ y_i. Extending this to quadrics is more subtle, as some rank-two matrices F ∈ ℓ ⊂ 𝓝_Z may not give full reconstructions of the point pairs P.

Proof of the only-if direction of Theorem 3.1

For 8 semi-generic point pairs, the matrix Z = (xi⊤⊗yi⊤)i=18 is rank deficient exactly when 𝓝_Z =: ℓ is a line. This line ℓ is generic because it is also the nullspace of any submatrix of Z of size 7 × 9 and the corresponding seven point pairs are generic. Pick a subset of seven point pairs, say (xi,yi)i=17 , from the original eight pairs. Since these seven point pairs are generic, and ℓ is also generic, we can assume that Fx_i ≠ 0 and yi⊤ F ≠ 0 for any rank-two matrix F ∈ ℓ and all i = 1, …, 7. On the other hand, if we pick a different set of seven point pairs, say (xi,yi)i=28 , then ℓ is also the nullspace of the corresponding Z₇ and by the same argument as before, Fx_i ≠ 0 and yi⊤ F ≠ 0 for any rank-two matrix F ∈ ℓ and all i = 2, …, 8. Therefore, ℓ is P-generic.

Since ℓ is P-generic, by Theorem 3.16, ℓ gives rise to a Cremona transformation f : Px2⇢Py2 such that f(x_i) = y_i for i = 1, …, 8. This finishes the proof of Theorem 3.1. □

Example 3.18

Consider the quadric Q ⊂ ℙ³ defined by the equation x² + y² − z² − w² = 0 and the following 10 points p₁, …, p₈, c₁, c₂ ∈ Q:

The two projections (cameras) with centers c₁, c₂ have matrices

and we can calculate the image points and epipoles:

The point pairs (x_i, y_i) give us the matrix

which we can check is rank deficient and has nullspace spanned by the vectors

The reconstruction we started with has fundamental matrix

and if we take a different matrix

in the nullspace of Z₈ we can verify that A2⊤MA1 yields the original quadric Q:

The other two possible choices for fundamental matrices in the nullspace of Z₈ are

which have epipoles ex2=(0:1:1),ey2=(−1:0:1),ex3=(0:−1:1)  and  ey3=(1:0:1). Moreover, we can verify that there is a unique Cremona transformation

such that f(x_i) = y_i for all i. This Cremona transformation has base points exactly matching the epipoles. Finally, we can check that each camera center lies on two real lines on the quadric Q, parameterized by (a : b) ∈ ℙ¹ as

whose images are exactly the other two possible pairs of epipoles/base points (ex2,ey2)  and  (ex3,ey3).