# The Geometry of Rank Drop in a Class of
Face-Splitting Matrix Products

Erin Connelly  Department of Mathematics, University of Washington, Seattle
<EMAIL_ADDRESS>,  Sameer Agarwal  Google Inc., Seattle
<EMAIL_ADDRESS>,  Alperen Ergur  Department of Mathematics,
Univerity of Texas at San Antonio<EMAIL_ADDRESS>and  Rekha R. Thomas
Department of Mathematics, University of Washington, Seattle<EMAIL_ADDRESS>
###### Abstract.

Given $k\leq 6$ points $(x_{i},y_{i})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$,
we characterize rank deficiency of the $k\times 9$ matrix $Z_{k}$ with rows
$x_{i}^{\top}\otimes y_{i}^{\top}$ in terms of the geometry of the point
configurations $\\{x_{i}\\}$ and $\\{y_{i}\\}$. While this question comes from
computer vision the answer relies on tools from classical algebraic geometry:
For $k\leq 5$, the geometry of the rank-drop locus is characterized by cross-
ratios and basic (projective) geometry of point configurations. For the case
$k=6$ the rank-drop locus is captured by the classical theory of cubic
surfaces.

##  1\. Introduction

Let $\otimes$ denote the Kronecker product [LOAN200085]. We are interested in
solving the following problem:

######  Problem 1.1.

Given $k$ points $(x_{i},y_{i})\in\mathbb{R}^{3}\times\mathbb{R}^{3}$, $k\leq
9$, consider the $k\times 9$ matrix $Z_{k}$ whose rows are
$x^{\top}_{i}\otimes y^{\top}_{i}$ for $i=1,\ldots,k$, i.e.,

$Z_{k}=\begin{bmatrix}x^{\top}_{1}\otimes y^{\top}_{1}\\\ \vdots\\\
x^{\top}_{k}\otimes y^{\top}_{k}\end{bmatrix}$

Delineate the geometry of point configurations $\\{x_{i}\\}$ and $\\{y_{i}\\}$
for which $\operatorname{rank}(Z_{k})<k$.

In signal processing, the matrix $Z_{k}$ is known as the face-splitting
product [Slyusar] of the two matrices $X_{k}\in\mathbb{R}^{k\times 3}$ with
rows $x^{\top}_{1},\ldots,x^{\top}_{k}$ and $Y_{k}\in\mathbb{R}^{k\times 3}$
with rows $y^{\top}_{1},\ldots,y^{\top}_{k}$. The face-splitting product is a
special case of the Khatri-Rao matrix product in linear algebra and statistics
[khatri-rao-1968]. However, our interest 1.1 comes from a pair of estimation
problems in $3$-d computer vision known as the 7 and 5 point problems
respectively [hartley-zisserman-2004, kruppa1913ermittlung,
nister2004efficient].

######  Problem 1.2 (7-point problem).

Given 7 points
$(\hat{x}_{i},\hat{y}_{i})\in\mathbb{R}^{2}\times\mathbb{R}^{2}$ find a matrix
$F\in\mathbb{R}^{3\times 3}$ such that

$\displaystyle y^{\top}_{i}Fx_{i}$ $\displaystyle=0,\ \forall
i=1,\dots,7~{}\text{ and }~{}\det(F)=0.$ (1)

where $x_{i}=\begin{bmatrix}\hat{x}_{i}\\\ 1\end{bmatrix}$ and
$y_{i}=\begin{bmatrix}\hat{y}_{i}\\\ 1\end{bmatrix}$.

######  Problem 1.3 (5-point problem).

Given 5 points
$(\hat{x}_{i},\hat{y}_{i})\in\mathbb{R}^{2}\times\mathbb{R}^{2}$ find a matrix
$E\in\mathbb{R}^{3\times 3}$ such that

$\displaystyle y^{\top}_{i}Ex_{i}$ $\displaystyle=0,\ \forall
i=1,\dots,5~{}\text{ and }~{}2EE^{\top}E-\operatorname{Tr}(EE^{\top})E=0.$ (2)

where $x_{i}=\begin{bmatrix}\hat{x}_{i}\\\ 1\end{bmatrix}$ and
$y_{i}=\begin{bmatrix}\hat{y}_{i}\\\ 1\end{bmatrix}$.

A solution $F$ to the $7$ point algorithm that has rank $2$ is known as a
fundamental matrix, and a solution $E$ to the $5$ point algorithm is known as
an essential matrix [hartley-zisserman-2004]. Observe that in both problems,
the quantity of interest is a $3\times 3$ matrix $M$ that satisfies equations
of the form

$y_{i}^{\top}Mx_{i}=0,\ \forall i=1,\dots,k,$

which can be re-written as 111Here $\operatorname{vec}(\cdot)$ is the so
called vectorization operator that concatenates the columns of a matrix into a
vector.

$(x_{i}^{\top}\otimes y_{i}^{\top})\operatorname{vec}(M)=0,\ \forall
i=1,\dots,k$

where $\textup{vec}(M)$ is the $9$-dimensional vector obtained by
concatenating the columns of $M$. Thus the $7$ point algorithm involves
computing the intersection of the null space of a $7\times 9$ face splitting
product with a determinantal variety. Similarly, the $5$ point algorithm
involves finding the intersection of the null space of a $5\times 9$ face
splitting product with the so called essential variety [demazure1988deux].

We wish to understand the point configurations $\\{x_{i}\\}$ and $\\{y_{i}\\}$
for which 1.2 and 1.3 are ill-posed. In both cases, it can be shown that a
sufficient condition for ill-posedness is that the corresponding face-
splitting product matrix be rank deficient and thus our interest in 1.1

A first answer to 1.1 is that $\operatorname{rank}(Z_{k})<k$ if and only if
all the maximal minors of $Z_{k}$ are zero. These minors are bi-homogeneous
polynomials in $x_{i}$ and $y_{i}$, and typically do not shed much light on
the geometry of the point configurations $\\{x_{i}\\}$ and $\\{y_{i}\\}$ that
cause $Z_{k}$ to drop rank. Our goal is to characterize the rank deficiency of
$Z_{k}$ in terms of the geometry of the $x$ and $y$ points.

To this end, we begin by observing that the rank of $Z_{k}$ is a projective
invariant. First, note that multiplying the $x_{i}$’s and $y_{i}$’s by non-
zero scalars scales the rows of $Z_{k}$ and does not change its rank. Second,
if $H_{1},H_{2}$ are invertible matrices, then from the mixed-product property
of Kronecker products we have that:

$\displaystyle
Z^{\prime}_{k}=\begin{bmatrix}(H_{1}x_{1})^{\top}\otimes(H_{2}y_{1})^{\top}\\\
\vdots\\\
(H_{1}x_{k})^{\top}\otimes(H_{2}y_{k})^{\top}\end{bmatrix}=\begin{bmatrix}(x_{1}^{\top}\otimes
y_{1}^{\top})(H^{\top}_{1}\otimes H^{\top}_{2})\\\ \vdots\\\
(x_{k}^{\top}\otimes y_{k}^{\top})(H^{\top}_{1}\otimes
H^{\top}_{2})\end{bmatrix}=Z_{k}(H^{\top}_{1}\otimes H^{\top}_{2}).$ (3)

The matrix $H_{1}^{\top}\otimes H_{2}^{\top}$ has full rank since $H_{1}$ and
$H_{2}$ are invertible, and hence,
$\operatorname{rank}(Z^{\prime}_{k})=\operatorname{rank}(Z_{k})$. These
observations imply that $\operatorname{rank}(Z_{k})$ is independent of scaling
and choice of coordinates in $\mathbb{R}^{3}\times\mathbb{R}^{3}$. Therefore,
we should be studying the problem over projective space. While our computer
vision applications care about real input, mathematically there is no loss in
assuming that the input points $(x_{i},y_{i})$ are complex because
$\mathbb{R}\subset\mathbb{C}$. Therefore, letting $\mathbb{P}$ denote
projective space over $\mathbb{C}$ we pass to the following formulation of 1.1
which will be the main subject of study in this paper.

######  Problem 1.4.

Given $k$ points $(x_{i},y_{i})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$, $k\leq
9$, consider the $k\times 9$ matrix $Z_{k}$ whose rows are
$x^{\top}_{i}\otimes y^{\top}_{i}$ for $i=1,\ldots,k$, i.e.,

$Z_{k}=\begin{bmatrix}x^{\top}_{1}\otimes y^{\top}_{1}\\\ \vdots\\\
x^{\top}_{k}\otimes y^{\top}_{k}\end{bmatrix}.$

Delineate the geometry of point configurations $\\{x_{i}\\}$ and $\\{y_{i}\\}$
for which $\operatorname{rank}(Z_{k})<k$.

Let $Z_{k}\\{l\\}$ denote a submatrix of $Z_{k}$ with $l$ of the $k$ rows. If
$\operatorname{rank}(Z_{k}\\{l\\})<l$, then $\operatorname{rank}(Z_{k})<k$,
and any condition on the $l$ points $(x_{i},y_{i})$ that contribute the rows
of $Z_{k}\\{l\\}$ and cause $Z_{k}\\{l\\}$ to drop rank will be a condition on
$\\{(x_{i},y_{i}),i=1,\ldots,k\\}$ such that $Z_{k}$ drops rank. We will call
such a condition an inherited condition for rank drop. For each value of $k$
we will be most interested in the non-inherited conditions on the input that
make $Z_{k}$ rank deficient.

As an illustration of the type of answers we will provide, consider the
version of Problem 1.4 in $\mathbb{P}^{1}\times\mathbb{P}^{1}$. Suppose
$\\{x_{i}\\}$ and $\\{y_{i}\\}$ are each sets of distinct points in
$\mathbb{P}^{1}$. In Theorem 4.1 we prove that the $4\times 4$ matrix $Z_{4}$
is rank deficient if and only if the cross-ratio of $x_{1},x_{2},x_{3},x_{4}$
equals that of $y_{1},y_{2},y_{3},y_{4}$. The cross-ratio is the fundamental
projective invariant of four points in $\mathbb{P}^{1}$.

It is a classical fact that the geometry of a finite set of points in
$\mathbb{P}^{2}$, that remains invariant under the action of
$\textup{PGL}(3)$, can be expressed via polynomials in the $3\times 3$
determinants (brackets) of the matrices whose rows are the points. Many of our
answers to Problem 1.4 will be phrased in terms of these bracket expressions.
The results have analogs in terms of cross-ratios when the points are in
general position.

Even though the motivation for this paper are the 5 and 7-point problems, we
will only address the cases $2\leq k\leq 6$. This is because a great deal of
classical geometry emerges for $k=6$, and we expect a similarly rich story for
the cases $k\geq 7$. In the interest of keeping the paper at a reasonable
length we will address the cases $7\leq k\leq 9$ in future work.

### Summary of results and organization of the paper.

In Section 2 we show that one can make some simplifying assumptions on the
input to Problem 1.4 without any loss of generality. While these assumptions
are not needed for the theory, they are helpful and sometimes critical for
computations. In Section 3 we review the tools from invariant theory that are
needed in this paper.

In Section 4 we present our results for $k\leq 4$. Theorem 4.2 solves Problem
1.4 for $k=2,3,4$. Section 5 considers $k=5$ and presents two theorems.
Theorem 5.1 characterizes the new conditions for rank drop in $Z_{5}$ in terms
of bracket equations (24). A consequence is that when $k=5$, a necessary
condition for rank drop is that one set of points (say the $y_{i}$’s) is on a
line $\ell$. The second result is Theorem 5.2 which gives a geometric
characterization of rank deficiency: when the $y_{i}$’s on the line $\ell$ are
distinct and the $x_{i}$’s are in general position, then the rank drop of
$Z_{5}$ is equivalent to the existence of an isomorphism from the conic
$\omega$ through the $x_{i}$’s to the line $\ell$, taking $x_{i}$ to $y_{i}$.

Finally, we consider the case $k=6$ in Section 6 where we will see that
Problem 1.4 is intimately related with the theory of cubic surfaces in
$\mathbb{P}^{3}$. An important new fact here is that given $5$ points
$(x_{i},y_{i})$ in general position there is a unique $6$th point
$(x_{6},y_{6})$ that make $Z_{6}$ rank deficient. This $6$th point admits a
rational expression in terms of the $5$ points (Lemma 6.1).

The results in this section are organized in two parts. In Section 6.1 we
characterize the rank deficiency of $Z_{6}$ using brackets; Theorem 6.4 covers
the case of neither the $x_{i}$’s nor the $y_{i}$’s being in a line, and
Theorem 6.7 addresses the situation in which one set of points is in a line.
When the points are in general position, we identify a simple algebraic check
(Theorem 6.6) for rank deficiency in terms of a vector of classical invariants
(46) of $6$ points in $\mathbb{P}^{2}$.

In Section 6.2, we explain the origins of the algebraic theorems in Section
6.1 using geometry. The central result of this section is Theorem 6.9 which
says that when the points are in general position,
$\operatorname{rank}(Z_{6})<6$ if and only if there is a smooth cubic surface
$\mathcal{S}$ that arises as the blow up of $\mathbb{P}^{2}$ in both the $x$
points and the $y$ points such that the two sets of exceptional curves form a
Schläfli double six on $\mathcal{S}$. This result allows for a simple
determinantal representation of $\mathcal{S}$ using the null space of $Z_{6}$.

We then interpret Theorem 6.9 in two different ways. The first is Theorem 6.20
that identifies the unique $6$th point, given $5$ points in general position,
that forces $Z_{6}$ to drop rank. The construction in this theorem relies on a
recipe due to Sturm via conics in $\mathbb{P}^{2}\times\mathbb{P}^{2}$, and
this result proves the algebraic Theorem 6.4.

Theorem 6.9 can be rephrased in yet another way using the Cremona hexahedral
form of a cubic surface (Theorem 6.21 ). This result explains the algebraic
Theorem 6.6. Using the invariants needed in this theorem along with the
classical Joubert invariants of $6$ points in a line, we complete the answer
to Problem 1.4 when $k=6$ by proving Theorem 6.7.

### A note about computations.

All of the algebraic statements in this paper are proven using the
computational algebra software package Macaulay2 [M2]. The codes that we used
can be found at

github.com/rekharthomas/rankdrop

A basic strategy we employ to understand rank drop is to use Macaulay2 to
decompose the ideal of maximal minors of $Z_{k}$ over $\mathbb{Q}$ into its
associated primes (over $\mathbb{Q}$). Then we interpret these ideals
geometrically to state the conditions for rank drop. These geometric
conditions will hold over $\mathbb{R}$ and $\mathbb{C}$ as well since the
prime decomposition over $\mathbb{Q}$ can only refine over the larger fields.

### Acknowledgments

We thank Jarod Alper, Timothy Duff, and Giorgio Ottaviani for helpful
discussions. A.E. is supported by NSF CCF 2110075.

##  2\. Preprocessing the input

We begin by showing that one can make simplifying assumptions on the input to
Problem 1.4 which is a collection of $k\leq 6$ points
$(x_{i},y_{i})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$.

We sometimes refer to $(x_{i},y_{i})$ as a point pair, especially when we need
to decouple the components and think of the configurations $x_{i}$ and $y_{i}$
separately. It will also be convenient to refer to the $\mathbb{P}^{2}$
containing $\\{x_{i}\\}$ as $\mathbb{P}^{2}_{x}$ and the $\mathbb{P}^{2}$
containing $\\{y_{i}\\}$ as $\mathbb{P}^{2}_{y}$. Let $\textup{PGL}(3)$ be the
group of invertible $3\times 3$ matrices over $\mathbb{C}$ up to scale. An
element $H\in\textup{PGL}(3)$ induces a homography (projective isomorphism) on
$\mathbb{P}^{2}$.

In the introduction we informally proved that the rank of
$Z_{k}=(x_{i}^{\top}\otimes y_{i}^{\top})_{i=1}^{k}$ is invariant under
$\textup{PGL}(3)\times\textup{PGL}(3)$ acting on
$\mathbb{P}^{2}_{x}\times\mathbb{P}^{2}_{y}$ by left multiplication. This
allows us to change bases in $\mathbb{P}^{2}_{x}$ and $\mathbb{P}^{2}_{y}$
independently. We state the result formally below:

######  Lemma 2.1.

Let $(x_{i},y_{i})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$ for $i=1,\ldots,k$
and let $H_{1},H_{2}\in\textup{PGL}(3)$ be homographies acting on
$\mathbb{P}_{x}^{2}$ and $\mathbb{P}^{2}_{y}$ respectively (i.e.,
$(x,y)\mapsto(H_{1}x,H_{2}y)$). Then $Z_{k}=\left(x^{\top}_{i}\otimes
y^{\top}_{i}\right)_{i=1}^{k}$ has the same rank as
$Z^{\prime}_{k}=\left((H_{1}x_{i})^{\top}\otimes(H_{2}y_{i})^{\top}\right)_{i=1}^{k}$.

Lemma 2.1 allows us to make certain assumptions on the configurations
$\\{x_{i}\\}$ and $\\{y_{i}\\}$ which we phrase as corollaries to the lemma.
While our proofs do not need these assumptions, they become critical for
computations in Macaulay2. The first assumption is that the $x_{i}$’s and
$y_{i}$’s in the input to Problem 1.4 are finite points. With a view towards
computations, in the following lemma and in several parts of the paper, we fix
representatives of points in $\mathbb{P}^{2}$ and write them out as vectors in
$\mathbb{C}^{3}$.

######  Corollary 2.2.

Without loss of generality, we can assume that the input to Problem 1.4 is of
the form $x_{i}=(x_{i1},x_{i2},1)^{\top}$ and $y_{i}=(y_{i1},y_{i2},1)^{\top}$
for all $i=1,\ldots,k$, where $x_{i1},x_{i2},y_{i1},y_{i2}\in\mathbb{C}$.
Hence

$\displaystyle x_{i}^{\top}\otimes
y_{i}^{\top}=(x_{i1}y_{i1},x_{i1}y_{i2},x_{i1},x_{i2}y_{i1},x_{i2}y_{i2},x_{i2},y_{i1},y_{i2},1).$
(4)

###### Proof.

Let $x_{i}\in\mathbb{P}^{2}_{x}$ and $y_{i}\in\mathbb{P}^{2}_{y}$ be $k$ pairs
of points where $k$ is finite, and suppose that some of these points lie on
the line at infinity in their $\mathbb{P}^{2}$. Since we have only finitely
many points, there exists a line $\ell$, common to both $\mathbb{P}^{2}$,
which contains none of these points. Let $H$ be a homography that sends $\ell$
to the line at infinity in both $\mathbb{P}^{2}$. Then $Hx_{i}$ and $Hy_{i}$
are finite points for all $i$. By Lemma 2.1, the rank of
$Z_{k}=(x_{i}^{\top}\otimes y_{i}^{\top})_{i=1}^{k}$ equals the rank of
$Z^{\prime}_{k}=((Hx_{i})^{\top}\otimes(Hy_{i})^{\top})_{i=1}^{k}$ and hence
we can work with the transformed points to understand the rank deficiency of
$Z_{k}$. ∎

The other assumption we can make is that sufficiently generic configurations
allow some of the points to be fixed, which will prove useful for
computations.

######  Corollary 2.3.

When no three of the $x_{i}$’s or $y_{i}$’s are collinear in their respective
$\mathbb{P}^{2}$, we may fix the first four in each set to be:

$\displaystyle\begin{split}x_{1}=y_{1}=(0,0,1),\,\,\,x_{2}=y_{2}=(1,0,1),\\\
x_{3}=y_{3}=(0,1,1),\,\,\,x_{4}=y_{4}=(1,1,1).\end{split}$ (5)

At certain times, for the sake of symmetry, we will instead fix the first four
points as:

$\displaystyle\begin{split}x_{1}=y_{1}=(1,0,0),\,\,\,x_{2}=y_{2}=(0,1,0),\\\
x_{3}=y_{3}=(0,0,1),\,\,\,x_{4}=y_{4}=(1,1,1).\end{split}$ (6)

Since the two fixings are related by a homography, results from one fixing
transfer to the other.

##  3\. Cross-ratios and brackets

By Lemma 2.1, the rank deficiency of $Z_{k}$ is invariant under the action of
$\textup{PGL(3)}\times\textup{PGL}(3)$ on $\mathbb{P}^{2}\times\mathbb{P}^{2}$
given by $(H_{1},H_{2})\cdot(x,y)\mapsto(H_{1}x,H_{2}y)$. Therefore, the input
to Problem 1.4 can be thought of as the $\textup{PGL}(3)$-orbits of $k$ points
in each $\mathbb{P}^{2}$. The intrinsic geometry of $k$ points
$p_{1},\ldots,p_{k}\in\mathbb{P}^{2}$ are the properties of the configuration
that remain intact/invariant under the action of $\textup{PGL}(3)$. Since an
element of $\textup{PGL}(3)$ can be assumed to have determinant $1$, the above
geometry is precisely the invariant properties, under the action of
$\textup{SL}(3)$, of the configuration obtained by replacing each
$p_{i}\in\mathbb{P}^{2}$ with a representative in $\mathbb{C}^{3}$. We now
introduce the basic tools from the invariant theory of point sets needed to
formulate our results. There are numerous references for this material such as
[coble], [derksen_kemper], [dolgachev_ortland], [dolgachev] and
[SturmfelsInvariant].

Let $\mathbb{C}[{\bf p}]:=\mathbb{C}[p_{i1},p_{i2},p_{i3},i=1,\ldots,k]$ be
the polynomial ring in the variables $p_{ij}$ for $i=1,\ldots,k$, $j=1,2,3$,
that represent the coordinates of $k$ points $p_{1},\ldots,p_{k}$ in
$\mathbb{P}^{2}$ written via representatives in $\mathbb{C}^{3}$. A polynomial
$f\in\mathbb{C}[{\bf p}]$ is invariant under $\textup{SL}(3)$ if
$f(p_{1},\ldots,p_{k})=f(Hp_{1},\ldots,Hp_{k})$ for all $H\in\textup{SL}(3)$.
It is a well-known classical result that the invariant ring $\mathbb{C}[{\bf
p}]^{\textup{SL}(3)}$, which is the collection of all polynomials in
$\mathbb{C}[{\bf p}]$ that are invariant under $\textup{SL}(3)$, is generated
by the $3\times 3$ minors of the matrix whose rows are the symbolic
$p_{1}^{\top},\ldots,p_{k}^{\top}$. In other words, $\mathbb{C}[{\bf
p}]^{\textup{SL}(3)}$ is generated as an algebra by the bracket polynomials

$[ijl]:=\det\begin{bmatrix}p_{i1}&p_{i2}&p_{i3}\\\ p_{j1}&p_{j2}&p_{j3}\\\
p_{l1}&p_{l2}&p_{l3}\end{bmatrix}$ (7)

for all distinct choices of $i,j,l\in\\{1,\ldots,k\\}$. In our situation, we
will write $[ijl]_{x}$ (respectively, $[ijl]_{y}$) when the input points come
from $\\{x_{i}\\}$ (respectively, $\\{y_{i}\\}$). Equations in brackets can be
used to express geometry. For example, three points
$p_{i},p_{j},p_{l}\in\mathbb{P}^{2}$ are collinear if and only if $[ijl]=0$,
and $6$ points lie on a conic if and only if they satisfy the bracket equation
(17) in Lemma 3.5.

Often we consider points in a line $\ell$ in $\mathbb{P}^{2}$ as points in
$\mathbb{P}^{1}$ where we can compute the bracket $[ij]$ for
$p_{i},p_{j}\in\ell$ by identifying $\ell$ with the line parameterized as say
$(*,*,0)$, and then dropping the $0$ coordinate to get

$[ij]:=\det\begin{bmatrix}p_{i1}&p_{i2}\\\ p_{j1}&p_{j2}\end{bmatrix}.$ (8)

When we have many points $p_{1},\ldots,p_{k}$ on a line and a point $u$ not on
that line, the $\mathbb{P}^{2}$ brackets $[iju]$ will be closely related to
the $\mathbb{P}^{1}$ brackets $[ij]$.

######  Lemma 3.1.

For points $p_{1},\ldots,p_{k}$ on a line and a point $u$ not on the line,
there exists a nonzero scalar $\lambda$ depending only on the choice of
coordinate representative of $u$ such that $[iju]=\lambda[ij]$ for all $i,j$.

###### Proof.

We can assume without loss of generality that $p_{i}=(p_{i1},p_{i2},0)$ for
all $i$. We then calculate

$[iju]=\text{det}\begin{bmatrix}p_{i1}&p_{i2}&0\\\ p_{j1}&p_{j2}&0\\\
u_{1}&u_{2}&u_{3}\end{bmatrix}=u_{3}[ij].$ (9)

∎

######  Definition 3.2.

The cross-ratio of an ordered list of $4$ points
$p_{1},\ldots,p_{4}\in\mathbb{P}^{1}$ is

$(p_{1},p_{2};p_{3},p_{4}):=\frac{[13][24]}{[14][23]}.$ (10)

The cross-ratio is the only projective invariant of $4$ points in
$\mathbb{P}^{1}$ in the following sense.

######  Lemma 3.3.

If $p_{1},\ldots,p_{4}\in\mathbb{P}^{1}$ and
$q_{1},\ldots,q_{4}\in\mathbb{P}^{1}$ are two sets of points then
$(p_{1},p_{2};p_{3},p_{4})=(q_{1},q_{2};q_{3},q_{4})$ if and only if there
exists a homography $H:\mathbb{P}^{1}\to\mathbb{P}^{1}$ such that
$Hp_{i}=q_{i}$ for all $i$.

Permuting the points on the line changes the cross-ratio in a systematic way,
i.e., if

$(p_{1},p_{2};p_{3},p_{4})=(q_{1},q_{2};q_{3},q_{4})$ (11)

then

$(p_{\sigma(1)},p_{\sigma(2)};p_{\sigma(3)},p_{\sigma(4)})=(q_{\sigma(1)},q_{\sigma(2)};q_{\sigma(3)},q_{\sigma(4)})$
(12)

for all $\sigma\in S_{4}$. Thus we do not need to consider multiple orderings.
The cross-ratio is $0,1,\infty$, or $\frac{0}{0}$ if and only if the points
are not distinct. See [semple-kneebone]*III, §4-5.

######  Example 1.

We now give an example to illustrate that equations in brackets can be more
robust than equalities of cross-ratios. Consider the following points in
$\mathbb{P}^{1}$ where the first set has repetition:

$\displaystyle x_{1}$ $\displaystyle=(1,1)\quad x_{2}=(1,1)$
$\displaystyle\quad\quad y_{1}$ $\displaystyle=(8,1)\quad y_{2}=(4,1)$
$\displaystyle x_{3}$ $\displaystyle=(1,1)\quad x_{4}=(3,1)$
$\displaystyle\quad\quad y_{3}$ $\displaystyle=(2,1)\quad y_{4}=(5,1).$

Then

$(x_{1},x_{2};x_{3},x_{4})=\frac{0}{0},\quad\quad(y_{1},y_{2};y_{3},y_{4})=-1$
(13)

and therefore the cross-ratios are not equal. However, if we write the cross-
ratio equality

$(x_{1},x_{2};x_{3},x_{4})=(y_{1},y_{2};y_{3},y_{4})$ (14)

in bracket form as

$[13]_{x}[24]_{x}[14]_{y}[23]_{y}=[13]_{y}[24]_{y}[14]_{x}[23]_{x}$ (15)

then equality does hold.

In Example 1, the points $(x_{i},y_{i})$ lie on the variety defined by the
bracket equation (15) but do not satisfy the cross ratio equality (14). The
discrepancy is because cross-ratios are well-defined only on an open set,
while the corresponding bracket equation holds on the Zariski closure of this
open set. Since our interest is in describing the algebraic variety of input
points that make $Z_{4}$ rank deficient, the bracket expressions are the
correct choice. However, whenever the points are in sufficiently general
position, we can pass to cross-ratio expressions which, besides being elegant,
have the advantage that they can be visualized easily.

We will also need planar and conic cross-ratios which we now define.

######  Definition 3.4.

The $5$ point cross-ratio of $p_{1},\ldots,p_{5}\in\mathbb{P}^{2}$ is

$(p_{1},p_{2};p_{3},p_{4};p_{5}):=\frac{[135][245]}{[145][235]}.$ (16)

This is also called a planar cross-ratio or the cross-ratio around $p_{5}$.

Note that planar cross-ratios are preserved under a homography of
$\mathbb{P}^{2}$. For $5$ distinct points, the cross-ratio around $x_{5}$ can
be obtained geometrically by drawing the $4$ lines $\overline{p_{i}p_{5}}$ for
$i=1,2,3,4$, cutting them with a transversal, and then computing the cross-
ratio of the intersection points. See Figure 1. Also, the planar cross-ratio
is transformed by permutations of $p_{1},\ldots,p_{4}$ similarly to the line
cross-ratio.

$p_{5}$$p_{1}$$p_{2}$$p_{3}$$p_{4}$$a_{2}$$a_{1}$$a_{4}$$a_{3}$ .

Figure 1. For distinct points $p_{1},\ldots,p_{5}\in\mathbb{P}^{2}$, the
planar cross-ratio $(p_{1},p_{2};p_{3},p_{4};p_{5})$ equals the $4$ point
cross-ratio $(a_{1},a_{2};a_{3},a_{4})$

######  Lemma 3.5.

A collection of $6$ points $p_{1},\ldots,p_{6}\in\mathbb{P}^{2}$ lie on a
conic if and only if

$[135][245][146][236]=[136][246][145][235],$ (17)

or generically, the following cross-ratio equality holds:

$(p_{1},p_{2};p_{3},p_{4};p_{5})=(p_{1},p_{2};p_{3},p_{4};p_{6}).$ (18)

Lemma 3.5 makes the following definition of a conic cross-ratio well-defined.

######  Definition 3.6.

The conic cross-ratio of $4$ points $p_{1},\ldots,p_{4}$ on a (non-degenerate)
conic $\omega$ is

$(p_{1},p_{2};p_{3},p_{4})_{\omega}:=(p_{1},p_{2};p_{3},p_{4};p)$ (19)

where $p\in\omega$ is any new point.

Any non-degenerate conic $\omega$ is isomorphic to $\mathbb{P}^{1}$. If
$\iota:\omega\to\mathbb{P}^{1}$ is one such isomorphism then

$(p_{1},p_{2};p_{3},p_{4})_{\omega}=(\iota(p_{1}),\iota(p_{2});\iota(p_{3}),\iota(p_{4})).$
(20)

##  4\. $k\leq 4$

We will now answer Problem 1.4 in the cases $k\leq 4$. As a warm up to Problem
1.4 we first consider the analogous question in
$\mathbb{P}^{1}\times\mathbb{P}^{1}$. Recall that any condition on a proper
subset of $\\{(x_{i},y_{i})\\}$ that make the corresponding rows of $Z_{k}$
dependent is called an inherited condition for the rank drop of $Z_{k}$.

######  Theorem 4.1.

Consider the $k\times 4$ matrix $Z_{k}=(x_{i}\otimes y_{i})_{i=1}^{k}$ where
$(x_{i},y_{i})\in\mathbb{P}^{1}\times\mathbb{P}^{1}$ and $k\leq 4$. Then,

  1. (1)

$\operatorname{rank}(Z_{2})<2$ if and only if $x_{1}=x_{2}$ and $y_{1}=y_{2}$,

  2. (2)

$\operatorname{rank}(Z_{3})<3$ if and only if either an inherited condition
holds or $x_{1}=x_{2}=x_{3}$ or $y_{1}=y_{2}=y_{3}$,

  3. (3)

$\operatorname{rank}(Z_{4})<4$ if and only if the following bracket equation
holds:

$[13]_{x}[24]_{x}[14]_{y}[23]_{y}=[14]_{x}[23]_{x}[13]_{y}[24]_{y}.$ (21)

Generically, $x_{1},\ldots,x_{4}$ and $y_{1},\ldots,y_{4}$ are distinct and
then (21) holds if and only if there is a homography $H\in\textup{PGL}(2)$
such that $Hx_{i}=y_{i}$ for $i=1,\ldots,4$.

###### Proof.

In each of the above cases, $Z_{k}$ will be rank deficient if and only if all
its maximal minors vanish. Hence we compute the ideal of $k\times k$ minors of
the symbolic $Z_{k}$ in each case and decompose it to understand rank
deficiency. While computations in Macaulay2 are done over $\mathbb{Q}$, our
results hold over $\mathbb{C}$ as explained at the end of the Introduction. By
Corollary 2.2 we can fix the last coordinates of $x_{i}$ and $y_{i}$ to $1$.

If $k=2$ the ideal of $2\times 2$ minors of $Z_{2}$ is the prime ideal
$\langle x_{11}-x_{21},y_{11}-y_{21}\rangle$ which says that rank drops if and
only if $x_{1}=x_{2}$ and $y_{1}=y_{2}$. We will call this the repeated point
condition.

If $k=3$, the ideal of $3\times 3$ minors of the symbolic $Z_{3}$ matrix is
radical and decomposes into $5$ prime ideals. Three of them correspond to
repeated points $x_{i}=x_{j}$ and $y_{i}=y_{j}$ for $i\neq j$. These are
inherited for rank drop from the $k=2$ case. The remaining $2$ prime ideals
say that $x_{1}=x_{2}=x_{3}$ and $y_{1}=y_{2}=y_{3}$ respectively.

When $k=4$, we can use Macaulay2 to check that

$\det(Z_{4})=[14]_{x}[23]_{x}[13]_{y}[24]_{y}-[13]_{x}[24]_{x}[14]_{y}[23]_{y}$

which yields (21). In this case, the ideal of maximal minors is principal and
prime, generated by $\det(Z_{4})$, and does not decompose. However, note that
the bracket expression for $\det(Z_{4})$ vanishes under all of the conditions
inherited from $k=2$ and $k=3$. Furthermore, when the $x_{i}$’s and $y_{i}$’s
are distinct then (21) is equivalent to
$(x_{1},x_{2};x_{3},x_{4})=(y_{1},y_{2};y_{3},y_{4})$ and the homography
follows by Lemma 3.3. ∎

We now consider Problem 1.4 for $k\leq 4$ points
$(x_{i},y_{i})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$. Again all of the
statements can be checked in Macaulay2 by following the same general strategy
as in the proof of Theorem 4.1.

######  Theorem 4.2.

Suppose $(x_{i},y_{i})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$ and
$Z_{k}=(x_{i}^{\top}\otimes y_{i}^{\top})_{i=1}^{k}$ for $k\leq 4$. Then

  1. (1)

$\operatorname{rank}(Z_{2})<2$ if and only if $x_{1}=x_{2}$ and $y_{1}=y_{2}$.

  2. (2)

$\operatorname{rank}(Z_{3})<3$ if and only if either an inherited condition
holds or $x_{1}=x_{2}=x_{3}$ and $y_{1},y_{2},y_{3}$ are on a line, or
$x_{1},x_{2},x_{3}$ are on a line and $y_{1}=y_{2}=y_{3}$.

  3. (3)

$\operatorname{rank}(Z_{4})<4$ if and only if an inherited condition holds or
one of the following is true:

    1. (a)

$x_{1}=x_{2}=x_{3}=x_{4}$ or $y_{1}=y_{2}=y_{3}=y_{4}$.

    2. (b)

The points $x_{1},\ldots,x_{4}\in\mathbb{P}^{2}_{x}$ are in a line and
$y_{1},\ldots,y_{4}\in\mathbb{P}^{2}_{y}$ are in a line, and

$[13]_{x}[24]_{x}[14]_{y}[23]_{y}=[14]_{x}[23]_{x}[13]_{y}[24]_{y}.$ (22)

Generically, the collinear points $x_{1},\ldots,x_{4}$ and
$y_{1},\ldots,y_{4}$ are distinct and then (22) holds if and only if there is
a homography $H\in\textup{PGL}(3)$ such that $Hx_{i}=y_{i}$ for
$i=1,\ldots,4$.

###### Proof.

The proof of (1) and (2) can be done using Macaulay2 by computing the ideal of
maximal minors of $Z_{k}$ and then decomposing the radical of the ideal to get
all possible conditions for rank drop. The matrix $Z_{k}$ is written
symbolically as in (4) where, by Corollary 2.2 we assume that the last
coordinate of all $x_{i}$’s and $y_{i}$’s are $1$. For instance, the ideal of
$3\times 3$ minors of $Z_{3}$ is radical and of dimension $8$ and degree $3$.
It is the intersection of $5$ prime ideals of which $3$ correspond to the
repeated point conditions inherited from $k=2$. The fourth ideal provides the
new condition that $x_{1}=x_{2}=x_{3}$ and $y_{1},y_{2},y_{3}$ are in a line.
The fifth ideal provides the analogous condition with the roles of $x$ and $y$
switched. These ideals appear in the form:

$\langle
x_{11}-x_{21},x_{12}-x_{22},x_{11}-x_{31},x_{12}-x_{32},[123]_{y}\rangle\quad\textup{
and }\quad\langle
y_{11}-y_{21},y_{12}-y_{22},y_{11}-y_{31},y_{12}-y_{32},[123]_{x}\rangle.$

The ideal of maximal minors of $Z_{4}$ is radical of dimension $12$ and degree
$6$. It decomposes into $17$ primes ideals:

  * •

$6$ of the primes correspond to repeated points (inherited from $k=2$). All of
these have dimension $12$ and degree $1$.

  * •

$8$ of the primes correspond to $3$ points in one side coinciding and the
corresponding points on the other side on a line (inherited from $k=3$). All
of these ideals have dimension $11$ and degree $2$.

  * •

$2$ of the primes say $x_{1}=x_{2}=x_{3}=x_{4}$ and $y_{1}=y_{2}=y_{3}=y_{4}$
respectively (new conditions for $k=4$). These ideals have dimension $10$ and
degree $1$.

  * •

The last prime component, call it $J$, has dimension $11$ and degree $24$. It
implies that the $x_{i}$ and $y_{i}$ are in a line (new condition for $k=4$).
One way to verify this claim is to check that the ideal generated by the
brackets $[123]_{x},[124]_{x}$ and $[123]_{y},[124]_{y}$ is contained in $J$.
These brackets being $0$ encode the collinearity of $x_{i}$’s and the
collinearity of $y_{i}$’s.

One can understand $J$ more precisely by making the following calculation. If
the $x$ and $y$ points are both in a line, then we can use homographies to fix
these lines, and assume that $x_{i}=(a_{i},0,1)$ and $y_{i}=(b_{i},0,1)$ for
$i=1,\ldots,4$. This means that we can identify $x_{i}$ with
$(a_{i},1)\in\mathbb{P}^{1}$ and $y_{i}$ with $(b_{i},1)\in\mathbb{P}^{1}$.
Under this substitution, $Z_{4}$ has exactly one non-zero minor

$\det(M)=\det\begin{bmatrix}a_{1}b_{1}&a_{1}&b_{1}&1\\\
a_{2}b_{2}&a_{2}&b_{2}&1\\\ a_{3}b_{3}&a_{3}&b_{3}&1\\\
a_{4}b_{4}&a_{4}&b_{4}&1\end{bmatrix}=[14]_{x}[23]_{x}[13]_{y}[24]_{y}-[13]_{x}[24]_{x}[14]_{y}[23]_{y}.$
(23)

The second equality in (23) can be checked in Macaulay2. The bracket
expression set to $0$ is the equality of cross-ratios:

$\displaystyle(x_{1},x_{2};x_{3},x_{4})=(y_{1},y_{2};y_{3},y_{4})$

which becomes well-defined when the $x_{i}$’s are distinct and the $y_{i}$’s
are distinct. ∎

##  5\. $k=5$

We now consider the rank deficiency of $Z_{5}$. In this case, we present two
theorems. 5.1 is a characterization of rank deficiency in terms of brackets.
We will see that a necessary condition for rank drop is that one set of points
(say $\\{y_{i}\\}$) is on a line $\ell$. 5.2 gives a geometric
characterization of rank deficiency when the points are sufficiently generic,
i.e., when the $y_{i}$’s on the line $\ell$ are distinct and the $x_{i}$’s are
in general position. In this case the rank drop of $Z_{5}$ is equivalent to
the existence of an isomorphism from the conic $\omega$ through the $x_{i}$’s
to the line $\ell$ taking $x_{i}$ to $y_{i}$. This is analogous to Theorem 4.2
(3)(b) where there is a homography taking $x_{i}$ to $y_{i}$ under sufficient
genericity. We will illustrate both results using Example 2.

######  Theorem 5.1.

Given a configuration $\\{(x_{i},y_{i})\\}_{i=1}^{5}$ in
$\mathbb{P}^{2}\times\mathbb{P}^{2}$, the matrix $Z_{5}$ is rank deficient if
and only if either, one of the inherited conditions hold, or the points in one
$\mathbb{P}^{2}$ (say in $\mathbb{P}^{2}_{y}$) are on a line and the following
bracket equations hold for all distinct
$i_{1},i_{2},i_{3},i_{4},j\in\\{1,2,3,4,5\\}$:

$[i_{1}i_{3}j]_{x}[i_{2}i_{4}j]_{x}[i_{1}i_{4}]_{y}[i_{2}i_{3}]_{y}=[i_{1}i_{4}j]_{x}[i_{2}i_{3}j]_{x}[i_{1}i_{3}]_{y}[i_{2}i_{4}]_{y}.$
(24)

By the action of permutations, there are only $5$ equations to consider in
(24), one for each choice of $j\in\\{1,2,3,4,5\\}$. In the case that both the
$x_{i}$’s and the $y_{i}$’s are contained in lines then (24) holds
automatically. Generically, (24) is equivalent to the cross-ratio equalities:

$(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}};x_{j})=(y_{i_{1}},y_{i_{2}};y_{i_{3}},y_{i_{4}})$
(25)

for all distinct $i_{1},i_{2},i_{3},i_{4},j\in\\{1,2,3,4,5\\}$.

###### Proof.

The proof splits into two cases. For the first case, we assume that the points
in neither side are entirely in a line and show that rank will drop only under
inherited conditions. For the second case, we are then free to assume that one
side is contained in a line, which allows us to simplify the Macaulay2
calculations.

  1. (1)

Suppose neither the $x_{i}$’s nor the $y_{i}$’s are in a line, and none of the
inherited conditions hold. Then we can assume without loss of generality that
$x_{1}\neq x_{2}$ and $y_{1}\neq y_{2}$ because no side has four coincident
points (due to the absence of inherited conditions). Since neither side is
contained entirely in a line, we can also assume that $x_{1},x_{2},x_{3}$ are
not collinear. Then either $y_{1},y_{2},y_{3}$ are also not collinear, or
$y_{1},y_{2},y_{3}$ are in a line which means that we can assume
$y_{1},y_{2},y_{4}$ are not collinear. Applying homographies, we therefore
need to work with two different fixings of points:

$x_{1}=(1,0,0)=y_{1}\quad\quad\textup{ and }\quad\quad
x_{2}=(0,1,0)=y_{2}\quad\quad\textup{ and }\quad\quad x_{3}=(0,0,1)=y_{3}$
(26)

and

$x_{1}=(1,0,0)=y_{1}\quad\quad\textup{ and }\quad\quad
x_{2}=(0,1,0)=y_{2}\quad\quad\textup{ and }\quad\quad x_{3}=(0,0,1)=y_{4}.$
(27)

Note that these fixings are not to finite points as we have done so far, but
are equivalent to the fixing in (5) as remarked after Corollary 2.2. They
allow for faster computations in Macaulay2.

In (26), the ideal generated by the $5\times 5$ minors of $Z_{5}$ is the
intersection of $20$ prime ideals. Of these, $4$ correspond to invalid inputs
$x_{i},y_{i}=(0,0,0)$; $7$ correspond to the inherited repeated point
condition from Theorem 4.2(1); $6$ correspond to the inherited condition from
Theorem 4.2(2); and the remaining $3$ correspond to the inherited
bracket/cross-ratio condition from Theorem 4.2(3)(b).

In (27), the ideal generated by the $5\times 5$ minors of $Z_{5}$ is the
intersection of $17$ prime ideals. Of these, $4$ correspond to invalid inputs
$x_{i},y_{i}=(0,0,0)$; $7$ correspond to repeated points; $4$ correspond to
the inherited condition from Theorem 4.2(2); and the remaining $2$ correspond
to the inherited bracket/cross-ratio condition from Theorem 4.2(3)(b).

Thus, there are no new conditions for rank drop if neither the $x_{i}$’s nor
the $y_{i}$’s are in a line.

  2. (2)

Suppose $y_{1},\ldots,y_{5}$ are on a line. Applying homographies, we may
assume that $y_{i}=(m_{i},0,1)$ for $i=1,\ldots,5$ and all $x_{i}$ are finite.
We may further assume that $x_{1}=(0,0,1)$ and that $y_{1}=(0,0,1)$. Then
rearranging the columns of $Z_{5}$ we get

$Z^{\prime}_{5}=\begin{bmatrix}0&0&0&0&0&1&0&0&0\\\
x_{21}m_{2}&x_{22}m_{2}&x_{21}&x_{22}&m_{2}&1&0&0&0\\\
x_{31}m_{3}&x_{32}m_{3}&x_{31}&x_{32}&m_{3}&1&0&0&0\\\
x_{41}m_{4}&x_{42}m_{4}&x_{41}&x_{42}&m_{4}&1&0&0&0\\\
x_{51}m_{5}&x_{52}m_{5}&x_{51}&x_{52}&m_{5}&1&0&0&0\end{bmatrix}.$ (28)

Let $I$ be the ideal generated by the $6$ non-zero $5\times 5$ minors of
$Z^{\prime}_{5}$ and let $J$ be the ideal generated by the $5$ differences of
left and right hand sides in the bracket equations (24) under the same fixings
of points. Then we can verify using Macaulay2 that $I=\sqrt{J}$. A symmetric
argument will switch the roles of the $x$ and $y$ points.

∎

######  Example 2.

The following example illustrates Theorem 5.1 and motivates Theorem 5.2.
Consider

$\displaystyle x_{1}=(0,0,1)\quad$ $\displaystyle y_{1}=(0,0,1)\quad\quad$
$\displaystyle x_{4}=(1,1,1)\quad$ $\displaystyle y_{4}=(-4,0,1)$
$\displaystyle x_{2}=(1,0,1)\quad$ $\displaystyle y_{2}=(1,0,1)\quad\quad$
$\displaystyle x_{5}=(50,98,113)\quad$ $\displaystyle y_{5}=(8,0,1)$
$\displaystyle x_{3}=(0,1,1)\quad$ $\displaystyle y_{3}=(3,0,1)$

Here the $y_{i}$’s are on the line $y_{i}=(m_{i},0,1)$ and we can verify that
(24) is satisfied. The matrix $Z_{5}$ shown below, has rank $4$.

$Z_{5}=\begin{bmatrix}0&0&0&0&0&0&0&0&1\\\ 1&0&1&0&0&0&1&0&1\\\
0&0&0&3&0&1&3&0&1\\\ -4&0&1&-4&0&1&-4&0&1\\\
400&0&50&784&0&98&904&0&113\end{bmatrix}.$

Note that in this example, the $y_{i}$’s (which are on a line) are distinct,
and that no $3$ of the $x_{i}$’s are in a line. In this case, there is a
$1$-parameter family of rank $2$ projective transformations
$\\{T_{\lambda}\\}$ such that $T_{\lambda}(x_{i})=y_{i}$ for $i=1,\ldots,5$
except for $5$ unique values of $\lambda$, one for each of
$x_{1},\ldots,x_{5}$. In this example, this is the family

$T_{\lambda}=\begin{bmatrix}0&12&0\\\
-7&-3&7\end{bmatrix}+\lambda\begin{bmatrix}7&-15&0\\\ 7&-5&0\end{bmatrix}.$
(29)

For each $i=1,\ldots,5$ there exists unique $\lambda_{i}$ such that
$T_{\lambda_{i}}$ has nullvector (center) $x_{i}$. These are

$\lambda_{1}=\infty\quad\lambda_{2}=0\quad\lambda_{3}=\frac{4}{5}\quad\lambda_{4}=\frac{3}{2}\quad\lambda_{5}=\frac{21}{20}.$
(30)

Since $T_{\lambda_{i}}(x_{i})=(0,0,0)\not\sim y_{i}$, these $5$
transformations can be seen as degenerate members of the family.

$x_{1}$$x_{2}$$x_{3}$$x_{4}$$x_{5}$$c$$a_{1}$$a_{2}$$a_{3}$$a_{4}$$a_{5}$$\ell$

Figure 2. We construct the projection with center $c$.

We now formalize the second half of Example 2 as a theorem. As seen in Theorem
5.1, for $Z_{5}$ to drop rank, either the $x_{i}$’s or the $y_{i}$’s have to
be in a line. Suppose the $y_{i}$’s are distinct and in a line, and the
$x_{i}$’s are in general position. Then the $x_{i}$’s lie on a conic $\omega$
and we will see that there is an isomorphism taking $\omega$ to the line
containing the $y_{i}$’s so that each $x_{i}$ maps to $y_{i}$.

######  Theorem 5.2.

Suppose we have a configuration $\\{(x_{i},y_{i})\\}_{i=1}^{5}$ such that no
$3$ of the $x_{i}$’s are collinear and the $y_{i}$’s are distinct and on a
line $\ell_{y}$. Let $\omega$ be the conic through $x_{1},\ldots,x_{5}$.
Suppose further that no inherited condition for rank drop holds. Then the
following are equivalent:

  1. (1)

The matrix $Z_{5}$ is rank deficient.

  2. (2)

For all $c\in\omega\setminus\\{x_{1},\ldots,x_{5}\\}$ there exists a
projective transformation $T_{c}\,:\,\mathbb{P}^{2}\rightarrow\mathbb{P}^{1}$
centered at $c$ such that $T_{c}(x_{i})\sim y_{i}$ for all $i$.

  3. (3)

For some $c\in\omega\setminus\\{x_{1},\ldots,x_{5}\\}$ there exists a
projective transformation $T_{c}\,:\,\mathbb{P}^{2}\rightarrow\mathbb{P}^{1}$
centered at $c$ such that $T_{c}(x_{i})\sim y_{i}$ for all $i$.

  4. (4)

There exists a cross-ratio preserving isomorphism $F:\omega\to\ell_{y}$ such
that $F(x_{i})=y_{i}$ for all $i$.

###### Proof.

To prove (1) implies (2), let $c\in\omega\setminus\\{x_{1},\ldots,x_{5}\\}$.
Then for all distinct $i_{1},i_{2},i_{3},i_{4},j\in\\{1,\ldots,5\\}$

$(y_{i_{1}},y_{i_{2}};y_{i_{3}},y_{i_{4}})=(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}};x_{j})=(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}})_{\omega}=(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}};c)$
(31)

for any $c\in\omega$. The first equation is from (25) in Theorem 5.1 and the
other two equalities follow by Definition 3.6. Let $\overline{cx_{i}}$ be the
line through $c$ and $x_{i}$ and let $\ell$ be any line such that
$c\not\in\ell$. Then if we define $a_{i}$ to be the unique intersection of
$\ell$ and $\overline{cx_{i}}$ it follows that

$(a_{i_{1}},a_{i_{2}};a_{i_{3}},a_{i_{4}})=(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}};c)=(y_{i_{1}},y_{i_{2}};y_{i_{3}},y_{i_{4}})$
(32)

for all $i_{1},i_{2},i_{3},i_{4}\in\\{1,\ldots,5\\}$. See Figure 2. It follows
that there exists homography $H$ such that $Ha_{i}\sim y_{i}$ for all $i$.
Therefore if $T^{\prime}_{c}$ is the projection in $\mathbb{P}^{2}_{x}$,
centered at $c$, onto the line $\ell$ then $T_{c}:=HT^{\prime}_{c}$ is the
desired projective transformation with $T_{c}(x_{i})\sim y_{i}$ for all $i$.

Clearly (2) implies (3). To show that (3) implies (1), suppose such a
projection exists and is centered at $c$. Then

$(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}};x_{j})=(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}})_{\omega}=(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}};c)=(y_{i_{1}},y_{i_{2}};y_{i_{3}},y_{i_{4}})$
(33)

for all distinct $i_{1},i_{2},i_{3},i_{4},j\in\\{1,\ldots,5\\}$. The first two
equalities follow by Definition 3.6 and the last equality follows by
hypothesis. It immediately follows that $Z_{5}$ is rank deficient by Theorem
5.1.

To show that (1) implies (4) we may assume that the cross-ratio equalities
(25) hold. Then define an isomorphism $F:\omega\to\ell_{y}$ by
$F(x_{1})=y_{1},F(x_{2})=y_{2},F(x_{3})=y_{3}$. Then for all
$p\in\omega\setminus\\{x_{1},x_{2},x_{3}\\}$, $F(p)=q$ is such that
$(x_{1},x_{2};x_{3},p)_{\omega}=(y_{1},y_{2};y_{3},q)$. It then follows that
$F(x_{4})=y_{4}$ and $F(x_{5})=y_{5}$ because

$(x_{1},x_{2};x_{3},x_{4})_{\omega}=(y_{1},y_{2};y_{3},y_{4})\quad\quad(x_{1},x_{2};x_{3},x_{5})_{\omega}=(y_{1},y_{2},y_{3};y_{5})$
(34)

by hypothesis (Theorem 5.1).

Finally, to show that (4) implies (1), suppose such an isomorphism $F$ exists.
Then

$(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}};x_{j})=(x_{i_{1}},x_{i_{2}};x_{i_{3}},x_{i_{4}})_{\omega}=(F(x_{i_{1}}),F(x_{i_{2}});F(x_{i_{3}}),F(x_{i_{4}}))=(y_{i_{1}},y_{i_{2}};y_{i_{3}},y_{i_{4}})$
(35)

for all $i_{1},i_{2},i_{3},i_{4},j\in\\{1,\ldots,5\\}$. The first equality
follows by Definition 3.6 and the second and third equalities follow by
hypothesis. It then follows that $Z_{5}$ is rank deficient by Theorem 5.1. ∎

##  6\. $k=6$

The final case we will consider in this paper is a set of $6$ inputs
$\\{(x_{i},y_{i})\\}_{i=1}^{6}\in\mathbb{P}^{2}\times\mathbb{P}^{2}$. We
organize our results in two parts. In the first part we state algebraic
characterizations of the rank deficiency of $Z_{6}$. These are analogous to
the results in the previous sections in that the results are in terms of
invariant theory. In the second part, we explain the geometry that leads to
these algebraic statements. This relies on the classical theory of cubic
surfaces and its combinatorics. The geometry works in an open region where the
input is sufficiently generic. However, as we saw in previous sections, the
algebra that it translates to describes the Zariski closure of these open
regions and hence the set of all inputs that make $Z_{6}$ rank deficient.

We say that $p_{1},\ldots,p_{6}\in\mathbb{P}^{2}$ are in general position if
at most $2$ points are in a line and at most $5$ points are on a conic. In
several results below we will need that the $x$ points and/or the $y$ points
are in general position. We will also need that a collection of point pairs
$\\{(x_{i},y_{i})\\}\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ is in general
position, by which we will mean that

  1. (1)

the $x$ points and $y$ points are in general position, and

  2. (2)

no more than $4$ pairs $(x_{i},y_{i})$ are related by a homography
$Hx_{i}=y_{i}$.

Note that if (1) holds then $Z_{5}$ has full rank because all of our previous
conditions for rank drop require at least the $x_{i}$’s or the $y_{i}$’s to be
in a line.

The variety $\mathbb{P}^{2}\times\mathbb{P}^{2}$ can be embedded in
$\mathbb{P}^{8}$ as the image of the Segre map

$\displaystyle\mathbb{P}^{2}\times\mathbb{P}^{2}\rightarrow\mathbb{P}^{8}\,\,\,\textup{
such that }\,\,\,(x,y)\mapsto x^{\top}\otimes y^{\top}.$ (36)

In what follows it will be helpful to identify $\mathbb{P}^{8}$ with the set
of all $3\times 3$ complex matrices up to scale via the bijection
$M\leftrightarrow\textup{vec}(M)$. Then the Segre embedding of
$\mathbb{P}^{2}\times\mathbb{P}^{2}$ in $\mathbb{P}^{8}$ (which we will also
call $\mathbb{P}^{2}\times\mathbb{P}^{2}$), can be identified with the set of
rank one $3\times 3$ matrices up to scale since

$\displaystyle x^{\top}\otimes y^{\top}=(\textup{vec}(yx^{\top}))^{\top}.$
(37)

The Segre variety $\mathbb{P}^{2}\times\mathbb{P}^{2}\subset\mathbb{P}^{8}$
has dimension $4$ and degree $6$. It is the zero locus of the determinantal
ideal generated by the $2\times 2$ minors of a symbolic $3\times 3$ matrix
whose entries denote the coordinates of $\mathbb{P}^{8}$.

We first show that given $5$ points $\\{(x_{i},y_{i})\\}_{i=1}^{5}$ in general
position, there exists a unique $6$th point
$(x_{6},y_{6})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$ such that
$\\{(x_{i},y_{i})\\}_{i=1}^{6}$ is rank deficient.

######  Lemma 6.1.

Let $\\{(x_{i},y_{i})\\}_{i=1}^{5}$ be in general position. Then there is a
unique new $6$th point $(x_{6},y_{6})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$
such that the matrix $Z_{6}$ is rank deficient. This $6$th point is a rational
function of the first $5$; if we use homographies to fix the first $4$ points
as

$\displaystyle x_{1}=(1,0,0)=y_{1}\quad\quad x_{2}=(0,1,0)=y_{2}$
$\displaystyle x_{3}=(0,0,1)=y_{3}\quad\quad x_{4}=(1,1,1)=y_{4},$

then the $6$th point has the formula:

$\begin{split}x_{6}=&(\frac{y_{53}-y_{52}}{y_{53}x_{52}-x_{53}y_{52}},\frac{y_{53}-y_{51}}{y_{53}x_{51}-x_{53}y_{51}},\frac{y_{51}-y_{52}}{y_{51}x_{52}-x_{51}y_{52}})\\\
y_{6}=&(\frac{x_{53}-x_{52}}{y_{53}x_{52}-x_{53}y_{52}},\frac{x_{53}-x_{51}}{y_{53}x_{51}-x_{53}y_{51}},\frac{x_{51}-x_{52}}{y_{51}x_{52}-x_{51}y_{52}}).\end{split}$
(38)

###### Proof.

Let $\mathcal{R}(Z_{5})$ denote the row span of $Z_{5}=(x_{i}\otimes
y_{i})_{i=1}^{5}$. By general position, $\operatorname{rank}(Z_{5})=5$ which
means that $\mathcal{R}(Z_{5})\cong\mathbb{P}^{4}$. Therefore, generically,
$\mathcal{R}(Z_{5})$ intersects the $4$-dimensional, degree $6$ Segre variety
$\mathbb{P}^{2}\times\mathbb{P}^{2}$ transversally in $6$ distinct points.
Each of these points corresponds to a rank one matrix $yx^{\top}$ which we
have identified with $x^{\top}\otimes y^{\top}$. Five of these intersection
points are the matrices $x_{i}^{\top}\otimes y_{i}^{\top}$ for $i=1,\ldots,5$
and the $6$th point $x_{6}^{\top}\otimes y_{6}^{\top}$ corresponds to a unique
point $(x_{6},y_{6})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$.

To prove the second part of the Lemma, we first use homographies to fix the
first $4$ points as mentioned. We can then use Macaulay2 to verify that we can
solve for $(x_{6},y_{6})$ in terms of $(x_{5},y_{5})$ as in equations (38). ∎

The existence of, and constructions for, this $6$th point were noted in both
[chiral] and [wernerfivepoints], but not the rational formula. The rational
formula shows that if $(x_{i},y_{i})_{i=1}^{5}$ are real then $(x_{6},y_{6})$
will be real as well. Note that if $\\{x_{i}\\}_{i=1}^{5}$ and
$\\{y_{i}\\}_{i=1}^{5}$ were related by homography then the denominators in
(38) would all be $0$. Here is an illustration of Lemma 6.1.

######  Example 3.

We begin with $5$ point pairs

$\displaystyle x_{1}$ $\displaystyle=(1,0,0)\quad y_{1}=(1,0,0)$
$\displaystyle\quad\quad x_{2}$ $\displaystyle=(0,1,0)\quad y_{2}=(0,1,0)$
$\displaystyle x_{3}$ $\displaystyle=(0,0,1)\quad y_{3}=(0,0,1)$
$\displaystyle\quad\quad x_{4}$ $\displaystyle=(1,1,1)\quad y_{4}=(1,1,1)$
$\displaystyle x_{5}$ $\displaystyle=(3,5,1)\quad y_{5}=(8,2,1)$

If we leave $(x_{6},y_{6})$ symbolic and construct the matrix $Z_{6}$, we can
verify with Macaulay2 that the unique new point pair that will result in rank
deficiency is
$x_{6}\sim(-\frac{1}{3},\frac{7}{5},\frac{3}{17}),~{}y_{6}\sim(-\frac{4}{3},\frac{2}{5},-\frac{1}{20})$,
which is exactly what we get from the formula above.

In Lemma 6.15 we will prove a counterpart to the above result, where we
instead begin with $x_{1},\ldots,x_{6}$ and calculate (up to homography) the
points $y_{1},\ldots,y_{6}$ that will result in rank deficiency.

###  6.1. Algebraic Characterizations of Rank Deficiency

In this section we will write down various algebraic characterizations of rank
deficiency. While these statements are by no means obvious or intuitive, once
stated, they are straightforward to check using Macaulay2. In the next section
we will explain the geometric origins of these statements.

Our first result is a characterization of rank drop using brackets and cross-
ratios as in the previous sections.

######  Definition 6.2.

Given $6$ input points $(x_{i},y_{i})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$,
let $\mathcal{B}_{rackets}(Z_{6})$ be the ideal generated by the differences
of the two sides in the following bracket equations:

$[ikp]_{x}[jrp]_{x}[irs]_{y}[jks]_{y}=[irp]_{x}[jkp]_{x}[iks]_{y}[jrs]_{y}$
(39)

for all distinct indices $i,j,k,r,p,s\in\\{1,2,3,4,5,6\\}$.

Up to permutations, there are $30$ equalities in (39): one for each choice of
$p$ and $s$. Under sufficient genericity, the bracket equations are the planar
cross-ratio equalities:

$(x_{i},x_{j};x_{k},x_{r};x_{p})=(y_{i},y_{j};y_{k},y_{r};y_{s}).$ (40)

Next we define a specific kind of highly degenerate configuration for which
$Z_{6}$ does not drop rank.

######  Definition 6.3.

We say that $\\{(x_{i},y_{i})\\}_{i=1}^{6}$ is an asymmetric double triangle
if the $\\{x_{i}\\}$ and $\\{y_{i}\\}$ each consist of $3$ distinct points
such that, up to reordering, $y_{1}=y_{2}$, $y_{3}=y_{4}$, $y_{5}=y_{6}$,
$x_{1}=x_{4}$, $x_{2}=x_{5}$, and $x_{3}=x_{6}$.

Let $\mathcal{M}_{inors}(Z_{6})$ denote the ideal of maximal minors of
$Z_{6}$. The following statement can be checked using Macaulay2.

######  Theorem 6.4.

Let $\\{(x_{i},y_{i})\\}_{i=1}^{6}$ be such that neither the $x_{i}$’s nor the
$y_{i}$’s are in a line. Then $Z_{6}$ is rank deficient if and only if one of
the inherited conditions hold or the $30$ bracket equations (39) hold and the
configuration is not in an asymmetric double triangle.

###### Proof.

As in the proof of Theorem 5.1 we split into two cases. Applying homographies
we can assume either that

$\displaystyle x_{1}=(1,0,0)=y_{1}\quad\quad x_{2}=(0,1,0)=y_{2}\quad\quad
x_{3}=(0,0,1)=y_{3}.$

or

$\displaystyle x_{1}=(1,0,0)=y_{1}\quad\quad x_{2}=(0,1,0)=y_{2}\quad\quad
x_{3}=(0,0,1)=y_{4}.$

In the first case the ideal $\mathcal{M}_{inors}(Z_{6})$ decomposes into $19$
components. The first $6$ correspond to invalid inputs $x_{i},y_{i}=(0,0,0)$,
the next $12$ correspond to repeated point pairs, and the $19$th component we
refer to as $M_{19}$. The ideal $\mathcal{B}_{rackets}(Z_{6})$ decomposes into
$13$ components. One component is $M_{19}$ and the other $12$ components
correspond to asymmetric double triangles.

In the second case the ideal $\mathcal{M}_{inors}(Z_{6})$ decomposes into $17$
components. The first $6$ correspond to invalid inputs $x_{i},y_{i}=(0,0,0)$,
the next $10$ correspond to repeated point pairs, and the $17$th component we
refer to as $M_{17}$. The ideal $\mathcal{B}_{rackets}(Z_{6})$ decomposes into
$21$ components. One component is $M_{17}$ and the other $20$ components
correspond to asymmetric double triangles. ∎

In Theorem 6.4 the only assumption is that neither the $x_{i}$’s nor the
$y_{i}$’s are in a line. In Theorem 6.7 we will tackle the case of one side in
a line, and together the two theorems cover all possibilities.

Before we can do that, we consider an important subcase of 6.4, namely the
generic case. The main result here is 6.6 which uses tools from invariant
theory that we have not encountered thus far. Besides being a key result of
this paper, it will also lead to a proof of 6.7, completing the story.

Suppose $x_{1},\ldots,x_{6}\in\mathbb{P}^{2}_{x}$ and
$y_{1},\ldots,y_{6}\in\mathbb{P}^{2}_{y}$ are in general position. Then we may
fix

$\displaystyle\begin{split}x_{1}=y_{1}=(1,0,0),\quad\quad
x_{2}=y_{2}=(0,1,0),\\\ x_{3}=y_{3}=(0,0,1),\quad\quad
x_{4}=y_{4}=(1,1,1)\end{split}$ (41)

and consider the rank deficiency of $Z_{6}$ whose $5$th and $6$th rows are
left symbolic as:

$\displaystyle\begin{split}(x_{51},x_{52},x_{53})\otimes(y_{51},y_{52},y_{53}),(x_{61},x_{62},x_{63})\otimes(y_{61},y_{62},y_{63}).\end{split}$
(42)

Recall that the general position assumption guarantees that no submatrix of
$Z_{6}$ with $5$ rows is rank deficient. Under these assumptions, one can use
Macaulay2 to check that $\mathcal{M}_{inors}(Z_{6})$ is a radical ideal of
dimension $9$ and degree $4$. Its prime decomposition consists of $14$
components. The first $4$ of these components correspond to invalid inputs
$x_{i},y_{i}=(0,0,0)$, the next $9$ correspond to repeated points, and the
$14$th component, denoted as $M_{14}$, encodes a new set of conditions, and
has dimension $8$ and degree $35$. Since $Z_{6}$ is rank deficient if and only
if all its maximal minors vanish, we conclude that the new conditions for rank
drop are exactly that $(x_{5},y_{5})$ and $(x_{6},y_{6})$ lie in the variety
of $M_{14}$.

We will now interpret $M_{14}$ in terms of the invariants of
$x_{1},\ldots,x_{6}$ and $y_{1},\ldots,y_{6}$. For a set of $6$ points
$p_{1},\ldots,p_{6}\in\mathbb{P}^{2}$ define

$[(ij)(kl)(rs)]:=[ijr][kls]-[ijs][klr].$ (43)

This is a classical invariant of points in $\mathbb{P}^{2}$ whose vanishing
expresses that the three lines $\overline{p_{i}p_{j}}$,
$\overline{p_{k}p_{l}}$ and $\overline{p_{r}p_{s}}$ meet in a point [coble]*pp
169. Using these invariants, Coble defines the following $6$ scalars
[coble]*pp 170:

$\displaystyle\begin{split}\bar{a}&=[(25)(13)(46)]+[(51)(42)(36)]+[(14)(35)(26)]+[(43)(21)(56)]+[(32)(54)(16)]\\\
\bar{b}&=[(53)(12)(46)]+[(14)(23)(56)]+[(25)(34)(16)]+[(31)(45)(26)]+[(42)(51)(36)]\\\
\bar{c}&=[(53)(41)(26)]+[(34)(25)(16)]+[(42)(13)(56)]+[(21)(54)(36)]+[(15)(32)(46)]\\\
\bar{d}&=[(45)(31)(26)]+[(53)(24)(16)]+[(41)(25)(36)]+[(32)(15)(46)]+[(21)(43)(56)]\\\
\bar{e}&=[(31)(24)(56)]+[(12)(53)(46)]+[(25)(41)(36)]+[(54)(32)(16)]+[(43)(15)(26)]\\\
\bar{f}&=[(42)(35)(16)]+[(23)(14)(56)]+[(31)(52)(46)]+[(15)(43)(26)]+[(54)(21)(36)]\end{split}$
(44)

Let $\bar{a}_{x},\bar{b}_{x},\bar{c}_{x},\bar{d}_{x},\bar{e}_{x},\bar{f}_{x}$
be the polynomials obtained by computing the expressions in (44) using $6$
symbolic vectors $x_{1},\ldots,x_{6}$ where each
$x_{i}=(x_{i1},x_{i2},x_{i3})^{\top}$. Similarly, let
$\bar{a}_{y},\bar{b}_{y},\bar{c}_{y},\bar{d}_{y},\bar{e}_{y},\bar{f}_{y}$ be
obtained from symbolic $y_{1},\ldots,y_{6}$ where each
$y_{i}=(y_{i1},y_{i2},y_{i3})^{\top}$.

######  Definition 6.5.

Let $\mathcal{I}_{nvariants}(Z_{6})$ denote the ideal of $2\times 2$ minors of

$\displaystyle\begin{bmatrix}\bar{a}_{x}&\bar{b}_{x}&\bar{c}_{x}&\bar{d}_{x}&\bar{e}_{x}&\bar{f}_{x}\\\
\bar{a}_{y}&\bar{b}_{y}&\bar{c}_{y}&\bar{d}_{y}&\bar{e}_{y}&\bar{f}_{y}\end{bmatrix}$
(45)

after specializing the first $4$ points $(x_{i},y_{i})$ as in (41).

The ideal $\mathcal{I}_{nvariants}(Z_{6})$ is radical of dimension $9$ and
degree $4$. Its prime decomposition consists of $14$ components. The first $4$
of these components correspond to invalid inputs $x_{i},y_{i}=(0,0,0)$, the
next $8$ correspond to $3$ repeated points in either $\mathbb{P}^{2}_{x}$ or
$\mathbb{P}^{2}_{y}$, the $13$th component corresponds to $x_{5}\sim y_{5}$
and $x_{6}\sim y_{6}$ (which encodes that $\\{x_{i}\\}$ and $\\{y_{i}\\}$ are
related by a homography). The $14$th component, denoted as $I_{14}$, provides
a new condition. This ideal has dimension $8$ and degree $35$.

######  Theorem 6.6.

Given a configuration $\\{(x_{i},y_{i})\\}_{i=1}^{6}$ such that the
$\\{x_{i}\\}$ and $\\{y_{i}\\}$ are in general position,

$\displaystyle\begin{pmatrix}\bar{a}_{x}&\bar{b}_{x}&\bar{c}_{x}&\bar{d}_{x}&\bar{e}_{x}&\bar{f}_{x}\end{pmatrix}\sim\begin{pmatrix}\bar{a}_{y}&\bar{b}_{y}&\bar{c}_{y}&\bar{d}_{y}&\bar{e}_{y}&\bar{f}_{y}\end{pmatrix}$
(46)

if and only if either the matrix $Z_{6}$ is rank deficient or $y_{i}=Hx_{i}$
for some homography $H\in\textup{PGL}(3)$.

###### Proof.

It can be checked using Macaulay2 that $M_{14}=I_{14}$. ∎

The statement of Theorem 6.6 begs the question of how homography ties in with
rank drop, if at all. We give an example to show that the existence of a
homography such that $y_{i}=Hx_{i}$ for $6$ points $(x_{i},y_{i})$ is not
sufficient for $Z_{6}$ to be rank deficient. In contrast, it does suffice for
$7$ point pairs.

######  Example 4.

By Lemma 2.1, if there is a $H\in\textup{PGL}(3)$ such that $y_{i}=Hx_{i}$ for
$i=1,\ldots,6$, then we may assume that $x_{i}=y_{i}$ for $i=1,\ldots,6$. Then
the null space of $Z_{6}$ consists exactly of the vectors $\textup{vec}(F)$
for the $3\times 3$ matrices $F$ such that

$x_{i}^{\top}Fx_{i}=0,\,\,\forall\,\,i=1,\ldots,6.$ (47)

Generically, this is exactly the set of $3\times 3$ skew-symmetric matrices,
$\text{Skew}_{3}\cong\mathbb{P}^{2}$. Hence $\operatorname{rank}(Z_{6})=6$.
For example, consider

$\displaystyle x_{1}=y_{1}=(1,0,0)\quad\quad x_{2}=y_{2}=(0,1,0)\quad\quad
x_{3}=y_{3}=(0,0,1)$ $\displaystyle x_{4}=y_{4}=(1,1,1)\quad\quad
x_{5}=y_{5}=(2,3,1)\quad\quad x_{6}=y_{6}=(3,7,1)$

for which the nullspace of $Z_{6}$ is $\text{Skew}_{3}$. In general, if the
homography is given by the matrix $H$ then the null space of $Z_{6}$ is
obtained by left action of $H^{\top}$ on $\text{Skew}_{3}$. We note that this
null space is unusual in that will always consist entirely of rank $2$
matrices.

To finish our algebraic discussion, we consider the case of one
$\mathbb{P}^{2}$ having all input points in a line. In order to state the
result, we introduce the classical Joubert invariants of $6$ points in
$\mathbb{P}^{1}$ [coble] which are:

$\displaystyle\begin{split}A_{y}&=[25][13][46]+[51][42][36]+[14][35][26]+[43][21][56]+[32][54][16]\\\
B_{y}&=[53][12][46]+[14][23][56]+[25][34][16]+[31][45][26]+[42][51][36]\\\
C_{y}&=[53][41][26]+[34][25][16]+[42][13][56]+[21][54][36]+[15][32][46]\\\
D_{y}&=[45][31][26]+[53][24][16]+[41][25][36]+[32][15][46]+[21][43][56]\\\
E_{y}&=[31][24][56]+[12][53][46]+[25][41][36]+[54][32][16]+[43][15][26]\\\
F_{y}&=[42][35][16]+[23][14][56]+[31][52][46]+[15][43][26]+[54][21][36].\end{split}$
(48)

######  Theorem 6.7.

Suppose $y_{1},\ldots,y_{6}\in\mathbb{P}^{2}_{y}$ are in a line. Then $Z_{6}$
is rank deficient if and only if

$\bar{a}_{x}A_{y}+\bar{b}_{x}B_{y}+\bar{c}_{x}C_{y}+\bar{d}_{x}D_{y}+\bar{e}_{x}E_{y}+\bar{f}_{x}F_{y}=0$
(49)

Furthermore, if $x_{1},\ldots,x_{6}$ are in general position and
$y_{1},\ldots,y_{6}$ are distinct, then (49) holds if and only if there exists
a projection $T:\mathbb{P}^{2}_{x}\to\mathbb{P}^{1}_{y}$ such that
$T(x_{i})\sim y_{i}$ for all $i$.

This theorem is analogous to Theorems 4.2 (3)(b) and 5.2. The proof of this
theorem will follow from the geometry that we develop in the next section.

###  6.2. Geometric Characterizations of Rank Deficiency

We next explain where the algebraic results (and ideals) from the previous
section come from by providing a geometric characterization of rank deficiency
of $Z_{6}$. This relies on the classical theory of smooth cubic surfaces in
$\mathbb{P}^{3}$. For the convenience of the reader, we have collected the
needed facts about cubic surfaces in Appendix A.

####  6.2.1. Cubic surfaces and Schäfli double sixes

A cubic surface in $\mathbb{P}^{3}$ is a hypersurface defined by a single
homogenous polynomial of degree $3$ in $\mathbb{C}[z_{0},z_{1},z_{2},z_{3}]$
where $(z_{0},\ldots,z_{3})$ are the coordinates of $\mathbb{P}^{3}$.

######  Definition 6.8.

A pair of 6 lines $\\{\ell_{1},\ldots,\ell_{6}\\}$ and
$\\{\ell_{1}^{\prime},\ldots,\ell_{6}^{\prime}\\}$ in $\mathbb{P}^{3}$ is
called a Schläfli double six if

  1. (1)

the $6$ lines in each set are mutually skew (i.e., they don’t intersect
pairwise), and

  2. (2)

$\ell_{i}$ intersects $\ell_{j}^{\prime}$ if and only if $i\neq j$.

A smooth cubic surface has $36$ Schläfli double sixes. The rich and highly
symmetric properties of these line configurations are listed in Appendix A and
lie at the heart of all of our major results in this section. Our main
(geometric) theorem in this section is the following.

######  Theorem 6.9.

Suppose
$\\{(x_{i},y_{i})\\}_{i=1}^{6}\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ is in
general position. Then $Z_{6}$ is rank deficient if and only if there exists a
cubic surface $\mathcal{S}$ that can be obtained as the blow up of both
$\mathbb{P}^{2}_{x}$ in $x_{1},\ldots,x_{6}$ and $\mathbb{P}^{2}_{y}$ in
$y_{1},\ldots,y_{6}$ such that their respective exceptional curves
$\\{\ell_{1},\ldots,\ell_{6}\\},\\{\ell_{1}^{\prime},\ldots,\ell_{6}^{\prime}\\}$
form a Schläfli double six on $\mathcal{S}$. When $Z_{6}$ is rank deficient,
this cubic surface has the determinantal representation
$\mathcal{N}(Z_{6})\cap\mathcal{D}$, where $\mathcal{N}(Z_{6})$ denotes the
right null space of $Z_{6}$ and $\mathcal{D}$ is the cubic hypersurface of all
rank deficient $3\times 3$ matrices.

To prepare for the proof of this theorem, we begin by constructing a cubic
surface from $5$ points $\\{(x_{i},y_{i})\\}_{i=1}^{5}$ in general position.
Recall that we can identify a point in $\mathbb{P}^{8}$ with a $3\times 3$
matrix $M$ via its vectorization $\textup{vec}(M)$. Under this identification,
$\mathcal{D}\subset\mathbb{P}^{8}$ is the cubic hypersurface
$\mathcal{D}=\\{M\,:\,\det(M)=0\\}$. Let
$\mathcal{N}(Z_{5})\subset\mathbb{P}^{8}$ be the null space of $Z_{5}$. Since
$\\{(x_{i},y_{i})\\}_{i=1}^{5}$ is in general position,
$\operatorname{rank}(Z_{5})=5$ and so,
$\mathcal{N}(Z_{5})\cong\mathbb{P}^{3}$. Therefore,

$\displaystyle S:=\mathcal{N}(Z_{5})\cap\mathcal{D}$ (50)

is a cubic surface in
$\mathcal{N}(Z_{5})\cong\mathbb{P}^{3}\subset\mathbb{P}^{8}$. This surface is
smooth by the general position assumption. (We note that if $\\{x_{i}\\}$ and
$\\{y_{i}\\}$ were related by a homography then $S$ would be degenerate (see
Example 4).)

The surface $S$ has a natural determinantal representation (see Section A for
the definition): Pick a basis $M_{0},\ldots,M_{3}$ of $\mathcal{N}(Z_{5})$ and
set

$\displaystyle M(z)=z_{0}M_{0}+\cdots+z_{3}M_{3}$ (51)

where $z_{i}\in\mathbb{C}$. If $(x_{i},y_{i}),i=1,\ldots,5$ are real, we can
choose $M_{0},\ldots,M_{3}$ to be real matrices. Since $M(z)$ is a
parameterization of $\mathcal{N}(Z_{5})$, $S$ is cut out by $\det(M(z))=0$,
and $M(z)$ is a determinantal representation of $S$. Generically, the
$3$-dimensional $\mathcal{N}(Z_{5})$ does not intersect the $4$-dimensional
$\mathbb{P}^{2}\times\mathbb{P}^{2}$ of rank one matrices, and so a point
$M(z)$ on $S$ has rank exactly $2$. This also follows from $S$ being smooth.

Next we construct a Schläfli double six on the cubic surface
$S=\mathcal{N}(Z_{5})\cap\mathcal{D}$, from $\\{(x_{i},y_{i})\\}_{i=1}^{5}$.

######  Definition 6.10.

Given $\\{(x_{i},y_{i})\\}_{i=1}^{5}$ in general position, define

$\displaystyle\ell_{x_{i}}:=$ $\displaystyle\\{M(z)\,:\,M(z)x_{i}=0\\}\textup{
for }i=1,\ldots,5,\textup{ and }$ (52) $\displaystyle\ell_{y_{j}}^{\prime}:=$
$\displaystyle\\{M(z)\,:\,y_{j}^{\top}M(z)=0\\}\textup{ for }j=1,\ldots,5,$
(53)

where $M(z)$ is of the form (51).

For each $i=1,\ldots,5$, $\ell_{x_{i}}\subset\mathcal{N}(Z_{5})$ by the
definition of $M(z)$ in (51). Further, since $M(z)x_{i}=0$, $M(z)$ has a non-
trivial right null space and hence, $\det(M(z))=0$. Therefore,
$\ell_{x_{i}}\subset S=\mathcal{N}(Z_{5})\cap\mathcal{D}$. The condition that
$M(z)\in\mathcal{N}(Z_{5})$ is equivalent to $M(z)$ satisfying the $5$ linear
constraints

$\displaystyle\langle x_{i}^{\top}\otimes
y^{\top}_{i},M(z)\rangle=y_{i}^{\top}M(z)x_{i}=0\textup{ for }i=1,\ldots,5.$
(54)

Therefore, $\ell_{x_{i}}$ is cut out by $8$ linear constraints including the
$3$ constraints given by $M(z)x_{i}=0$. However one of them is redundant since
$M(z)x_{i}=0$ implies that $y_{i}^{\top}M(z)x_{i}=0$. Therefore,
$\ell_{x_{i}}$ is cut out by $7$ independent linear constraints and is hence a
line on the cubic surface $S$. Similarly, each
$\ell_{y_{j}}^{\prime},j=1,\ldots,5$ is a line on $S$.

######  Lemma 6.11.

The lines $\ell_{x_{i}}$ and $\ell_{y_{j}}^{\prime}$ satisfying the following
incidence relations:

  1. (1)

The lines $\\{\ell_{x_{i}},i=1,\ldots,5\\}$ are mutually skew.

  2. (2)

The lines $\\{\ell_{y_{j}}^{\prime},j=1,\ldots,5\\}$ are mutually skew.

  3. (3)

$\ell_{x_{i}}$ intersects $\ell_{y_{j}}^{\prime}$ if and only if $i\neq j$.

Therefore, $\\{\ell_{x_{i}},i=1,\ldots,5\\}$ and
$\\{\ell_{y_{j}}^{\prime},j=1,\ldots,5\\}$ form part of a unique Schläfli
double six on $S$.

###### Proof.

The proof of the incidence relations follows from dimension counting and can
be found in [chiral, Lemma 6.1]. The two sets of lines are part of a unique
Schläfli double six on $S$ since each double six is already uniquely defined
by three of its line pairs, and we have $5$ line pairs here with the correct
incidences (see Lemma A.1 (3)). ∎

By Lemma A.1, given a Schläfli double six on $S$, there are two sets of $6$
points $\\{p_{i}\\}\subset\mathbb{P}^{2}_{p}$ and
$\\{q_{j}\\}\subset\mathbb{P}^{2}_{q}$ such that $S$ is the blow up of
$\mathbb{P}^{2}_{p}$ at $\\{p_{i}\\}$ with blow up morphism
$\pi_{p}\,:\,S\rightarrow\mathbb{P}^{2}_{p}$, and also the blow up of
$\mathbb{P}^{2}_{q}$ at $\\{q_{j}\\}$ with blow up morphism
$\pi^{\prime}_{q}\,:\,S\mapsto\mathbb{P}^{2}_{q}$. Each set of skew lines in
the given double six is the set of exceptional curves of these point sets
under $\pi_{p}$ and $\pi^{\prime}_{q}$. The blow up morphism
$\pi_{p}\,:\,S\rightarrow\mathbb{P}^{2}_{p}$ sends $M(z)\in S$ to the unique
$p\in\mathbb{P}^{2}$ in its right null space and the blow up morphism
$\pi^{\prime}_{q}\,:\,S\rightarrow\mathbb{P}^{2}_{q}$ sends $M(z)\in S$ to the
unique $q\in\mathbb{P}^{2}$ in its left null space. See the second half of
Appendix A.

We are now ready to prove the forward direction of Theorem 6.9.

######  Definition 6.12.

Let $\ell_{6},\ell_{6}^{\prime}$ denote the $6$th pair of lines in the
Schläfli double six on $S$ consisting of $\\{\ell_{x_{i}}\\}_{i=1}^{5}$ and
$\\{\ell_{y_{j}}^{\prime}\\}_{j=1}^{5}$ from Lemma 6.11.

######  Lemma 6.13.

Suppose $\\{(x_{i},y_{i})\\}_{i=1}^{5}$ is in general position. Then the
addition of a new row $x^{\top}_{6}\otimes y^{\top}_{6}$ makes $Z_{6}$ rank
deficient if and only if $\\{\ell_{x_{i}}\\}_{i=1}^{6}$ and
$\\{\ell_{y_{j}}^{\prime}\\}_{j=1}^{6}$ form a Schläfli double six on $S$
where

$\displaystyle\ell_{6}=\ell_{x_{6}}=\\{M(z)\,:\,M(z)x_{6}=0\\}\quad\textup{
and
}\quad\ell^{\prime}_{6}=\ell_{y_{6}}^{\prime}=\\{M(z)\,:\,y_{6}^{\top}M(z)=0\\}.$
(55)

###### Proof.

To prove the forward direction, suppose $Z_{6}$ is rank deficient. Then
$\mathcal{N}(Z_{5})=\mathcal{N}(Z_{6})$ and
$S=\mathcal{N}(Z_{6})\cap\mathcal{D}$. Check that $\ell_{x_{6}}$ is a line on
$S$ by the same argument as for $\ell_{x_{i}},i\leq 5$. Similarly,
$\ell_{y_{6}}^{\prime}$ is a line on $S$. One can also check that the
collection $\\{\ell_{x_{i}}\\}_{i=1}^{6}$ and
$\\{\ell_{y_{j}}^{\prime}\\}_{j=1}^{6}$ satisfies the incidence relations of a
Schläfli double six. This must mean that $\ell_{6}=\ell_{x_{6}}$ and
$\ell_{6}^{\prime}=\ell_{y_{6}}^{\prime}$ since the first $5$ line pairs
determine a unique double six on $S$.

To prove the reverse direction, suppose we have a Schläfli double six on $S$
as in the statement of the lemma. Then by the dicussion before Definition
6.12, $S$ is the blow up of $\mathbb{P}^{2}_{x}$ at $\\{x_{i}\\}$ under a
morphism $\pi_{x}$ and $S$ is also the blow up of $\mathbb{P}^{2}_{y}$ at
$\\{y_{i}\\}$ under a morphism $\pi_{y}^{\prime}$. Further,
$\pi_{x}(\ell_{x_{i}})=x_{i}$ for $i=1,\ldots,6$ and
$\pi^{\prime}_{y}(\ell_{y_{j}}^{\prime})=y_{j}$ for $j=1,\ldots,6$. We need to
argue that $Z_{6}=(x^{\top}_{i}\otimes y^{\top}_{i})_{i=1}^{6}$ is rank
deficient. Recall Lemma 6.1 and let $(x_{0},y_{0})$ be the unique new point
pair such that $Z^{\prime}$ with rows $x_{i}^{\top}\otimes y_{i}^{\top}$ for
$i=0,\ldots,5$ is rank deficient. Then, by the forwards direction, we must
have $\ell_{6}$ be the exceptional curve of $x_{0}$ under $\pi_{x}$ and hence
$x_{0}=\pi_{x}(\ell_{6})=x_{6}$. Similarly,
$y_{0}=\pi_{y}(\ell^{\prime}_{6})=y_{6}$. We conclude that
$(x_{0},y_{0})=(x_{6},y_{6})$ and therefore, $Z_{6}$ is rank deficient. ∎

The following corollary to Lemma 6.13 and Theorem A.5 concludes the proof of
the forward direction of Theorem 6.9.

######  Corollary 6.14.

If $\\{(x_{i},y_{i})\\}_{i=1}^{6}$ is in general position and $Z_{6}$ is rank
deficient then the cubic surface $S=\mathcal{N}(Z_{6})\cap\mathcal{D}$ is the
blow up of $\mathbb{P}^{2}_{x}$ at $x_{1},\ldots,x_{6}$ with exceptional
curves $\ell_{x_{i}},i=1,\ldots,6$, and also the blow up of
$\mathbb{P}^{2}_{y}$ at $y_{1},\ldots,y_{6}$ with exceptional curves
$\ell_{y_{j}}^{\prime},j=1,\ldots,6$. The two sets of lines form a Schläfli
double six on $S$.

Note that Lemma 6.13 gives a second construction for the unique point pair
$(x_{6},y_{6})$ in Lemma 6.1. Namely, as the images under the blow up
morphisms, of the final pair of lines in the unique Schläfli double six on $S$
determined by the lines $\\{\ell_{x_{i}}\\}_{i=1}^{5}$ and
$\\{\ell_{y_{j}}^{\prime}\\}_{j=1}^{5}$. This was also noted in [chiral].

To prove the backward direction of Theorem 6.9, we will first prove a
counterpart to Lemma 6.1.

######  Lemma 6.15.

If $x_{1},\ldots,x_{6}$ are in general position then there is a unique set of
$6$ points $y_{1},\ldots,y_{6}$ in general position (up to homography) such
that $Z_{6}$ is rank deficient. If we assume that

$\displaystyle x_{1}=(1,0,0)=y_{1}\quad\quad x_{2}=(0,1,0)=y_{2}$
$\displaystyle x_{3}=(0,0,1)=y_{3}\quad\quad x_{4}=(1,1,1)=y_{4}.$

then we can obtain $y_{5},y_{6}$ via the rational function

$\displaystyle y_{5}=$
$\displaystyle(\frac{x_{63}-x_{62}}{x_{53}x_{62}-x_{63}x_{52}},\frac{x_{63}-x_{61}}{x_{53}x_{61}-x_{53}x_{61}},\frac{x_{61}-x_{62}}{x_{61}x_{52}-x_{51}x_{62}})$
(56) $\displaystyle y_{6}=$
$\displaystyle(\frac{x_{53}-x_{52}}{x_{53}x_{62}-x_{63}x_{52}},\frac{x_{53}-x_{51}}{x_{53}x_{61}-x_{53}x_{61}},\frac{x_{51}-x_{52}}{x_{61}x_{52}-x_{51}x_{62}}).$
(57)

###### Proof.

We first prove that this construction will result in rank deficiency; the
uniqueness will come afterwards as a consequence of Corollary 6.14. We
construct the matrix $Z_{6}=(x^{\top}_{i}\otimes y^{\top}_{i})_{i=1}^{6}$ and
fix $x_{1},\ldots x_{4},y_{1},\ldots y_{6}$ as in the statement of the Lemma.
We can then verify with Macaulay2 that all $6\times 6$ minors of $Z_{6}$ are
$0$ and therefore $Z_{6}$ is rank deficient.

To see that this choice is unique, construct
$S=\mathcal{N}(Z_{6})\cap\mathcal{D}$. Then by Corollary 6.14, $S$ is both the
blow up of $\mathbb{P}^{2}_{x}$ at $x_{1},\ldots,x_{6}$ and the blow up of
$\mathbb{P}^{2}_{y}$ at $y_{1},\ldots,y_{6}$ and the two sets of exceptional
curves
$\\{\ell_{x_{1}},\ldots,\ell_{x_{6}}\\},\\{\ell^{\prime}_{y_{1}},\ldots,\ell^{\prime}_{y_{6}}\\}$
form a Schläfli double six. Now suppose there was some other
$y_{1}^{\prime},\ldots,y_{6}^{\prime}$ in general position such that
$Z^{\prime}_{6}=(x_{i}^{\top}\otimes{y^{\prime}_{i}}^{\top})_{i=1}^{6}$ was
rank deficient. Then $S^{\prime}=\mathcal{N}(Z^{\prime}_{6})\cap\mathcal{D}$
is also the blow up of $\mathbb{P}^{2}_{x}$ at $x_{1},\ldots,x_{6}$. Moreover,
it is also the blow up of $\mathbb{P}^{2}_{y}$ at
$y_{1}^{\prime},\ldots,y_{6}^{\prime}$ and the exceptional curves
$\ell_{x_{1}},\ldots,\ell_{x_{6}},\ell^{\prime}_{y^{\prime}_{1}},\ldots,\ell^{\prime}_{y^{\prime}_{6}}$
form a Schläfli double six. It follows by Lemma A.2 that
$\ell^{\prime}_{y_{i}}=\ell^{\prime}_{y^{\prime}_{i}}$ for all $i$. If we
compose the blow up morphisms
$\pi_{y^{\prime}}\circ\pi_{y}^{-1}:\mathbb{P}^{2}_{y}\to\mathbb{P}^{2}_{y}$ we
obtain an invertible projective transformation; thus this composition is a
homography taking $y_{i}\mapsto y_{i}^{\prime}$. This concludes the proof. ∎

The formula shows that if $x_{1},\ldots,x_{6}$ are real then, up to
homography, $y_{1},\ldots,y_{6}$ will also be real.

######  Example 5.

Consider Example 3 again and now fix the points

$\displaystyle x_{1}$ $\displaystyle=(1,0,0)\quad x_{2}=(0,1,0)$
$\displaystyle\quad\quad y_{1}$ $\displaystyle=(1,0,0)\quad y_{2}=(0,1,0)$
$\displaystyle x_{3}$ $\displaystyle=(0,0,1)\quad x_{4}=(1,1,1)$
$\displaystyle\quad\quad y_{3}$ $\displaystyle=(0,0,1)\quad y_{4}=(1,1,1)$
$\displaystyle x_{5}$ $\displaystyle=(3,5,1)\quad
x_{6}=(-\frac{1}{3},\frac{7}{5},\frac{3}{17}).$

If we leave $y_{5},y_{6}$ symbolic and construct the matrix $Z_{6}$, we can
verify with Macaulay2 that the unique new points that will result in rank
deficiency are
$y_{5}\sim(8,2,1),~{}y_{6}\sim(-\frac{4}{3},\frac{2}{5},-\frac{1}{20})$, which
is exactly what we get from the formula above. Moreover, check that this
matches up exactly with the points in Example 3.

The following corollary now completes the reverse direction of Theorem 6.9.

######  Corollary 6.16.

Given a smooth cubic surface $\mathcal{S}$ with a Schläfli double six
$\\{\ell_{i}\\}_{i=1}^{6}$ and $\\{\ell_{j}^{\prime}\\}_{i=1}^{6}$ that arises
as the exceptional lines of $x_{1},\ldots,x_{6}\in\mathbb{P}^{2}_{x}$ and
$y_{1},\ldots y_{6}\in\mathbb{P}^{2}_{y}$, the matrix
$Z_{6}=(x_{i}^{\top}\otimes y_{i}^{\top})_{i=1}^{6}$ is rank deficient.

###### Proof.

Lemma 6.15 provides us with unique (up to homography) points
$y_{1}^{\prime},\ldots,y_{6}^{\prime}\in\mathbb{P}^{2}_{y}$ such that
$Z^{\prime}_{6}=(x_{i}^{\top}\otimes{y^{\prime}}^{\top}_{i})_{i=1}^{6}$ is
rank deficient. Then, following along the same lines as the proof of Lemma
6.15, $S=\mathcal{N}(Z^{\prime}_{6})\cap\mathcal{D}$ and the points
$\\{y_{i}\\}$ and $\\{y_{i}^{\prime}\\}$ must be related by homography. It
follows by Lemma 2.1 that $Z_{6}$ is rank deficient. ∎

Combining Corollaries 6.14 and 6.16 we obtain a method to compute a
determinantal representation of a smooth cubic surface starting with a
Schläfli double six on it.

######  Corollary 6.17.

Let $\mathcal{S}$ be a smooth cubic surface with a Schläfli double six,
$\\{\ell_{i}\\}$ and $\\{\ell_{j}^{\prime}\\}$, and let
$\\{(x_{i},y_{i})\\}_{i=1}^{6}\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$ be
such that $\ell_{i}$ are the exceptional lines of the blow up of
$\mathbb{P}^{2}_{x}$ at $x_{1},\ldots,x_{6}$ and $\ell_{6}^{\prime}$ are the
exceptional lines of the blow up of $\mathbb{P}^{2}_{y}$ at $y_{1},\ldots
y_{6}$. Then a determinantal representation of $\mathcal{S}$ is given by
$M(z)=\sum z_{i}M_{i}$ where $M_{0},\ldots,M_{3}$ is a basis of the null space
of $Z_{6}=(x^{\top}_{i}\otimes y^{\top}_{i})_{i=1}^{6}$.

This gives us the final statement of Theorem 6.9, concluding the proof.

######  Remark 6.18.

In the classical literature one finds a determinantal representation of a
cubic surface from $6$ points in a $\mathbb{P}^{2}$ (whose blow up at these
points is the surface), via the Hilbert-Burch theorem (see Appendix A). It is
worthwhile to contrast that method with that in Corollary 6.17 where the
starting point is $6$ point pairs in $\mathbb{P}^{2}\times\mathbb{P}^{2}$.
Then a determinantal representation (51) of the surface can be obtained from a
basis of $\mathcal{N}(Z_{6})$ using simple linear algebra.

####  6.2.2. Conic intersections and the bracket ideal
$\mathcal{B}_{rackets}(Z_{6})$

We now use Theorem 6.9 to explain the ideal $\mathcal{B}_{rackets}(Z_{6})$
used in the algebraic characterization of rank defeciency in Theorem 6.4.

Consider a Schläfli double six
$\\{\ell_{i}\\}_{i=1}^{6},\\{\ell_{j}^{\prime}\\}_{j=1}^{6}$ on a cubic
surface $\mathcal{S}$ where the lines $\ell_{i}$ are the exceptional lines
under the blow up $\pi_{x}$ of points $x_{i}\in\mathbb{P}^{2}_{x}$ and the
lines $\ell^{\prime}_{j}$ are the exceptional lines under the blow up
$\pi^{\prime}_{y}$ of points $y_{j}\in\mathbb{P}^{2}_{y}$. Let $C_{i}$ be the
unique conic passing through $\\{x_{1},\ldots,x_{6}\\}\backslash\\{x_{i}\\}$
and $C^{\prime}_{i}$ the unique conic passing through
$\\{y_{1},\ldots,y_{6}\\}\backslash\\{y_{i}\\}$. By Lemma A.1, the lines
$\ell_{i}$ are the strict transforms of the conics
$C_{i}^{\prime}\subset\mathbb{P}^{2}_{y}$ and and the lines
$\ell_{j}^{\prime}$ are the strict transforms of the conics
$C_{j}\subset\mathbb{P}^{2}_{x}$. There is then a birational automorphism
$\mathbb{P}^{2}_{x}\to S\to\mathbb{P}^{2}_{y}$ such that $x_{i}\mapsto
C_{i}^{\prime}$ and $C_{i}\mapsto y_{i}$. This map is a degree $5$ Cremona
transformation and can be formalized using the following recipe due to Rudolf
Sturm [sturm], mentioned by Werner in [wernerfivepoints]. These conics can be
used to give a third construction of the unique $6$th point pair
$(x_{6},y_{6})$ that can be added to $\\{(x_{i},y_{i})\\}_{i=1}^{5}$ in
general position so that $Z_{6}$ is rank deficient. We state the result
without proof in Lemma 6.19 and illustrate it in Figure 3. The data for this
figure was chosen to make the conics easily visible.

######  Lemma 6.19.

[sturm] Let $C$ be the unique conic through $x_{1},\ldots,x_{5}$ in general
position, and let $H_{i}$ be the unique homography such that
$y_{j}=H_{i}x_{j}$ for all $j\neq i$. Define $C_{i}^{\prime}:=H_{i}(C)$ for
$i=1,\ldots,5$. Then $y_{6}\in\bigcap_{i=1}^{5}C_{i}^{\prime}$ is the unique
point in the intersection of the conics $C_{i}^{\prime}$.

Similarly, let $C^{\prime}$ be the unique conic through $y_{1},\ldots,y_{5}$,
and let $G_{j}$ be the unique homography such that $x_{i}=G_{j}y_{i}$ for all
$i\neq j$. Define $C_{j}:=H_{j}(C^{\prime})$ for $j=1,\ldots,5$. Then
$x_{6}\in\bigcap_{j=1}^{5}C_{j}$ is the unique point in the intersection of
the conics $C_{j}$.

Figure 3. The point $y_{6}$ is the unique intersection point of the $5$ conics
$C_{i}^{\prime}$ in $\mathbb{P}^{2}_{y}$.

We can now restate Theorem 6.9 using Lemma 6.19 to yield another
characterization of minimal rank deficiency of $Z_{6}$.

######  Theorem 6.20.

Let $\\{(x_{i},y_{i})\\}_{i=1}^{5}$ be in general position. Then
$(x_{6},y_{6})$ is the unique new point under which $Z_{6}$ is rank deficient
if and only if $y_{6}$ is the unique intersection point
$\cap_{j=1}^{5}C_{i}^{\prime}$ and $x_{6}$ is the unique intersection point
$\cap_{j=1}^{5}C_{j}$.

Theorem 6.20 explains the bracket and cross-ratio conditions that appear in
Theorem 6.4. Let $C$ be the conic through $x_{1},\ldots,x_{5}$ and let $H$ be
the homography such that $Hx_{i}=y_{i}$ for all $i=1,2,3,4$. Then
$C^{\prime}:=HC$ is a conic that passes through
$y_{1}=Hx_{1},y_{2}=Hx_{2},y_{3}=Hx_{3},y_{4}=Hx_{4}$ and $Hx_{5}$. By Lemma
3.5, $y_{6}\in C^{\prime}$ if and only if

$\det(H)[135]_{x}\det(H)[245]_{x}[146]_{y}[236]_{y}=[136]_{y}[246]_{y}\det(H)[145]_{x}\det(H)[235]_{x}.$
(58)

This can be rewritten as the bracket equation (39) by eliminating $\det(H)$:

$\displaystyle[135]_{x}[245]_{x}[146]_{y}[236]_{y}$
$\displaystyle=[136]_{y}[246]_{y}[145]_{x}[235]_{x}$ (59)

The other 29 bracket equations follow by selecting different conics and
homographies.

####  6.2.3. The Cremona hexahedral form of a cubic surface

We now give a third characterization of minimal rank deficiency of $Z_{6}$ via
the Cremona hexahedral form of a cubic surface. This will explain the
statement of Theorem 6.6.

Recall that every smooth cubic surface is the blow up of $\mathbb{P}^{2}$ at
$6$ points in general position. The Cremona hexahedral form of the surface
provides explicit equations for it in terms of the points being blown up. We
briefly describe these equations and the properties we need. More details can
be found in [coble, Section 4], [dolgachev, Section 9.4.3],
[Ottaviani2013FiveLO, Section 5.6].

Given $6$ points $p_{1},\ldots,p_{6}\in\mathbb{P}^{2}$, consider the following
list of $6$ cubic polynomials in $u=(u_{0},u_{1},u_{2})$, the coordinates of
$\mathbb{P}^{2}$, that vanish on them.

$\displaystyle\begin{split}a(u)&=[25u][13u][46u]+[51u][42u][36u]+[14u][35u][26u]+[43u][21u][56u]+[32u][54u][16u]\\\
b(u)&=[53u][12u][46u]+[14u][23u][56u]+[25u][34u][16u]+[31u][45u][26u]+[42u][51u][36u]\\\
c(u)&=[53u][41u][26u]+[34u][25u][16u]+[42u][13u][56u]+[21u][54u][36u]+[15u][32u][46u]\\\
d(u)&=[45u][31u][26u]+[53u][24u][16u]+[41u][25u][36u]+[32u][15u][46u]+[21u][43u][56u]\\\
e(u)&=[31u][24u][56u]+[12u][53u][46u]+[25u][41u][36u]+[54u][32u][16u]+[43u][15u][26u]\\\
f(u)&=[42u][35u][16u]+[23u][14u][56u]+[31u][52u][46u]+[15u][43u][26u]+[54u][21u][36u]\end{split}$
(60)

In fact, each summand in each cubic is the product of the lines through $3$
disjoint pairs of the $6$ points and already vanishes on the points. Formally
these expressions are similar to the scalars in (44). These cubic polynomials
are covariants of $p_{1},\ldots,p_{6}$ under the action of $\textup{PGL}(3)$,
and appear in the work of Coble [coble].

The $6$ cubics in (60) span the $4$-dimensional vector space of cubics that
vanish on $p_{1},\ldots,p_{6}$ and can be used to blow up $\mathbb{P}^{2}$ at
$p_{1},\ldots,p_{6}$ to obtain a cubic surface ${S}$. This surface is the
closure of

$\displaystyle\\{(a(u),b(u),c(u),d(u),e(u),f(u))~{}:~{}u\in\mathbb{P}^{2}\\})\subset\mathbb{P}^{5},$
(61)

and the zero set of the Cremona hexahedral equations:

$\displaystyle\begin{split}z_{1}^{3}+z_{2}^{3}+z_{3}^{3}+z_{4}^{3}+z_{5}^{3}+z_{6}^{3}&=0\\\
z_{1}+z_{2}+z_{3}+z_{4}+z_{5}+z_{6}&=0\\\
\bar{a}z_{1}+\bar{b}z_{2}+\bar{c}z_{3}+\bar{d}z_{4}+\bar{e}z_{5}+\bar{f}z_{6}&=0\end{split}$
(62)

where the scalars $\bar{a},\ldots,\bar{f}$ are as in (44). Here,
$z_{1},\ldots,z_{6}$ as the coordinates of $\mathbb{P}^{5}$.

The Cremona hexahedral form of $S$ can be used to find the $27$ lines on $S$
explicitly via expressions for the $45$ tritangent planes of $S$, see [coble,
Section 4]. In particular the $15$ lines that are not part of the Schläfli
double six corresponding to the blow up are obtained by permuting the
equations:

$z_{1}+z_{4}=z_{2}+z_{5}=z_{3}+z_{6}=0.$ (63)

Under the blow up morphism $\pi:S\to\mathbb{P}^{2}$, these lines correspond to
specific lines $\overline{p_{i}p_{j}}$; the one above corresponds to the line
between $p_{1}$ and $p_{2}$.

We can characterize the rank deficiency of $Z_{6}$ via the Cremona hexahedral
form.

######  Theorem 6.21.

Let $x_{1},\ldots,x_{6}\in\mathbb{P}^{2}_{x}$ and
$y_{1},\ldots,y_{6}\in\mathbb{P}^{2}_{y}$ be in general position and let
$S_{x}$ (respectively, $S_{y}$) be the blow up of $\mathbb{P}^{2}_{x}$
(respectively, $\mathbb{P}^{2}_{y}$) at $x_{1},\ldots,x_{6}$ (respectively,
$y_{1},\ldots,y_{6}$) in Cremona hexahedral form. Then $S_{x}=S_{y}$ if and
only if either there exists homography $H$ such that $Hx_{i}=y_{i}$ for all
$i=1,\ldots,6$ or the matrix $Z_{6}$ is rank deficient.

###### Proof.

If $S_{x}=S_{y}$, then they both contain the $15$ lines in equation (63), none
of which correspond to the proper transforms of either set of $6$ points under
their respective blow up morphisms. It follows that either the exceptional
lines of $x_{1},\ldots,x_{6}$ are exactly the exceptional lines of
$y_{1},\ldots,y_{6}$ or that the two sets are entirely distinct. Moreover,
since each of the $15$ lines in (63) have specific intersection conditions,
either $\ell_{x_{i}}=\ell_{y_{i}}$ for all $i$, or
$\\{\ell_{x_{1}},\ldots,\ell_{x_{6}}\\}$ and
$\\{\ell_{y_{1}},\ldots,\ell_{y_{6}}\\}$ form a Schläfli double six.

In the first case, the exceptional lines of $x_{1},\ldots,x_{6}$ are exactly
the exceptional lines of $y_{1},\ldots,y_{6}$. Then we can define an
invertible projective transformation
$\pi_{y}\circ\pi_{x}^{-1}:\mathbb{P}^{2}_{x}\to\mathbb{P}^{2}_{y}$, where
$\pi_{x}$ and $\pi_{y}$ again refer to the blow up morphisms, such that
$x_{i}\mapsto y_{i}$ for all $i$. This means that for each $i=1,\ldots,6$,
$x_{i}$ and $y_{i}$ are related by a homography $H$.

In the second case the exceptional lines of $x_{1},\ldots,x_{6}$ are distinct
from the exceptional of $y_{1},\ldots,y_{6}$. In this case they form a
Schläfli double six and by Theorem 6.9, $Z_{6}$ is rank deficient.

To prove the reverse direction, suppose $Hx_{i}=y_{i}$ for some homography
$H$. Then $S_{x}=S_{y}$ because
$(\bar{a}_{x},\ldots\bar{f}_{x})\sim(\bar{a}_{y},\ldots,\bar{f}_{y})$ by the
formula in (44) which means that $S_{x}$ and $S_{y}$ have the same Cremona
hexahedral form. Similarly, if $Z_{6}$ is rank deficient then by Theorem 6.9,
$\mathbb{P}^{2}_{x}$ blown up at $\\{x_{i}\\}$ agrees with
$\mathbb{P}^{2}_{y}$ blown up at $\\{y_{i}\\}$ and hence $S_{x}=S_{y}$. ∎

Theorem 6.21 provides a geometric proof of the algebraic characterization of
rank deficiency in Theorem 6.6. As before, we subscript the cubics in (60)
with $x$ or $y$ depending on whether they come from $\\{x_{i}\\}$ or
$\\{y_{i}\\}$. The Cremona hexahedral equations (62) imply that the two cubic
surfaces $S_{x}$ and $S_{y}$ are equal if and only if the following
polynomials in $u$ (coordinates on $\mathbb{P}^{2}_{x}$) and $v$ (coordinates
on $\mathbb{P}^{2}_{y}$) are identically zero.

$\begin{split}\bar{a}_{x}a_{y}(v)+\bar{b}_{x}b_{y}(v)+\bar{c}_{x}c_{y}(v)+\bar{d}_{x}d_{y}(v)+\bar{e}_{x}e_{y}(v)+\bar{f}_{x}f_{y}(v)\equiv
0\\\
\bar{a}_{y}a_{x}(u)+\bar{b}_{y}b_{x}(u)+\bar{c}_{y}c_{x}(u)+\bar{d}_{y}d_{x}(u)+\bar{e}_{y}e_{x}(u)+\bar{f}_{y}f_{x}(u)\equiv
0.\end{split}$ (64)

In other words,

$\begin{split}\bar{a}_{x}z_{1}+\bar{b}_{x}z_{2}+\bar{c}_{x}z_{3}+\bar{d}_{x}z_{4}+\bar{e}_{x}z_{5}+\bar{f}_{x}z_{6}=0&\quad\quad\forall
z\in S_{y}\\\
\bar{a}_{y}z_{1}+\bar{b}_{y}z_{2}+\bar{c}_{y}z_{3}+\bar{d}_{y}z_{4}+\bar{e}_{y}z_{5}+\bar{f}_{y}z_{6}=0&\quad\quad\forall
z\in S_{x}\end{split}$ (65)

or equivalently, the matrix

$\displaystyle\begin{bmatrix}1&1&1&1&1&1\\\
\bar{a}_{x}&\bar{b}_{x}&\bar{c}_{x}&\bar{d}_{x}&\bar{e}_{x}&\bar{f}_{x}\\\
\bar{a}_{y}&\bar{b}_{y}&\bar{c}_{y}&\bar{d}_{y}&\bar{e}_{y}&\bar{f}_{y}\end{bmatrix}$
(66)

is rank deficient. One can check using Macaulay2 that the ideal of $3\times 3$
minors of (66) coincides with the ideal $\mathcal{I}_{nvariants}(Z_{6})$ of
$2\times 2$ minors of (45). Therefore, we conclude that $S_{x}$ and $S_{y}$
coincide if and only if (46) holds, which is the same as

$\displaystyle\begin{pmatrix}\bar{a}_{x}&\bar{b}_{x}&\bar{c}_{x}&\bar{d}_{x}&\bar{e}_{x}&\bar{f}_{x}\end{pmatrix}=\lambda\begin{pmatrix}\bar{a}_{y}&\bar{b}_{y}&\bar{c}_{y}&\bar{d}_{y}&\bar{e}_{y}&\bar{f}_{y}\end{pmatrix}$

for some non-zero scalar $\lambda$.

This concludes the discussion about $6$ points $\\{(x_{i},y_{i})\\}_{i=1}^{6}$
in general position.

####  6.2.4. The Proof of Theorem 6.7

We can now use the tools developed in the previous subsection to prove Theorem
6.7 which is the last statement left to prove. This is the case of one
$\mathbb{P}^{2}$ having all input points in a line; suppose this is
$\mathbb{P}^{2}_{y}$. We first show that under this assumption, (64) can be
simplified using Lemma 3.1. Note that if $y_{1},\ldots,y_{6}$ are in a line
$\ell$ then for all points $v\in\ell$

$(a_{y},b_{y},c_{y},d_{y},e_{y},f_{y})(v)=(0,0,0,0,0,0),$ (67)

and for all points $v\in\mathbb{P}^{2}_{y}\backslash\ell$, using Lemma 3.1,

$(a_{y},b_{y},c_{y},d_{y},e_{y},f_{y})(v)\sim(A_{y},B_{y},C_{y},D_{y},E_{y},F_{y})$
(68)

where $A_{y},\ldots,F_{y}$ are the Joubert Invariants introduced in (48). The
Joubert invariants are known to satisfy

$\begin{split}A+B+C+D+E+F=0\\\
A^{3}+B^{3}+C^{3}+D^{3}+E^{3}+F^{3}=0\end{split}$ (69)

(see [dolgachev, Theorem 9.4.10]), and any $5$ of them give a complete basis
for the space of invariants for $6$ ordered points in $\mathbb{P}^{1}$. In
particular, if $p_{1},\ldots,p_{6}\in\mathbb{P}^{1}$ and
$q_{1},\ldots,q_{6}\in\mathbb{P}^{1}$ are each sets of $6$ distinct points,
then there exists a homography $H$ sending $p_{i}\mapsto q_{i}$ if and only if

$(A_{p},\ldots,F_{p})\sim(A_{q},\ldots,F_{q}).$ (70)

We further note that if $\\{y_{i}\\}_{i=1}^{6}$ are on a line then
$\bar{a}_{y}=\cdots=\bar{f}_{y}=0$ by looking at the definition of
$[(ij)(kl)(rs)]$. We can then simplify (64) to the single equation (49) which
was:

$\bar{a}_{x}A_{y}+\bar{b}_{x}B_{y}+\bar{c}_{x}C_{y}+\bar{d}_{x}D_{y}+\bar{e}_{x}E_{y}+\bar{f}_{x}F_{y}=0.$

The following lemma proves the first part of Theorem 6.7.

######  Lemma 6.22.

Let $\\{(x_{i},y_{i})\\}_{i=1}^{6}\subset\mathbb{P}^{2}\times\mathbb{P}^{2}$
be such that the points in one $\mathbb{P}^{2}$ are in a line. Then $Z_{6}$ is
rank deficient if and only if (49) holds.

###### Proof.

The proof is relatively straightforward. Suppose $y_{1},\ldots,y_{6}$ are in a
line. Applying homographies to each side we may assume that the $x_{i}$’s are
all finite and that $y_{i}=(y_{i1},1,0)$. Then

$Z_{6}=\begin{bmatrix}x_{11}y_{11}&x_{11}&0&x_{12}y_{11}&x_{12}&0&y_{11}&1&0\\\
x_{21}y_{21}&x_{21}&0&x_{22}y_{21}&x_{22}&0&y_{21}&1&0\\\
x_{31}y_{31}&x_{31}&0&x_{32}y_{31}&x_{32}&0&y_{31}&1&0\\\
x_{41}y_{41}&x_{41}&0&x_{42}y_{41}&x_{42}&0&y_{41}&1&0\\\
x_{51}y_{51}&x_{51}&0&x_{52}y_{51}&x_{52}&0&y_{51}&1&0\\\
x_{61}y_{61}&x_{61}&0&x_{62}y_{61}&x_{62}&0&y_{61}&1&0\end{bmatrix}$ (71)

and $Z_{6}$ is rank deficient if and only if its only obviously non-zero
$6\times 6$ minor $\det(M)=0$. Furthermore, a quick computation will give

$\bar{a}_{x}A_{y}+\bar{b}_{x}B_{y}+\bar{c}_{x}C_{y}+\bar{d}_{x}D_{y}+\bar{e}_{x}E_{y}+\bar{f}_{x}F_{y}=24\det(M)$
(72)

yielding the desired result. ∎

Before we prove the second part of Theorem 6.7, we compute an example to
illustrate the statement.

######  Example 6.

Consider the following points in which the $x$ points are in general position
and the $y$ points are all distinct and contained in a line. This can be seen
as a configuration in $\mathbb{P}^{2}\times\mathbb{P}^{1}$ so we fix

$\displaystyle x_{1}=(0,0,1)\quad$ $\displaystyle y_{1}=(0,1)\quad\quad$
$\displaystyle x_{4}=(1,1,1)\quad$ $\displaystyle y_{4}=(-4,1)$ $\displaystyle
x_{2}=(1,0,1)\quad$ $\displaystyle y_{2}=(1,1)\quad\quad$ $\displaystyle
x_{5}=(3,5,1)\quad$ $\displaystyle y_{5}=(8,1)$ $\displaystyle
x_{3}=(0,1,1)\quad$ $\displaystyle y_{3}=(3,1)$

and leave $x_{6},y_{6}$ as symbolic finite points. This creates a $6\times 6$
matrix

$Z_{6}=\begin{bmatrix}0&0&0&0&0&1\\\ 1&1&0&0&1&1\\\ 0&0&3&1&3&1\\\
-4&1&-4&1&-4&1\\\ 24&3&40&5&8&1\\\
x_{61}y_{61}&x_{61}&x_{62}y_{61}&x_{62}&y_{61}&1\end{bmatrix}.$ (73)

(The reader may also view the $y_{i}$ as points on the line at infinity in
$\mathbb{P}^{2}$ where the third coordinate is $0$; the non-zero part of the
corresponding $6\times 9$ matrix $Z_{6}$ is (73).)

The matrix $Z_{6}$ in (73) is rank deficient exactly when its determinant, a
quadratic polynomial in the coordinates of $x_{6}$ and $y_{6}$, is $0$. If we
arbitrarily fix $x_{6}=(2,11,1)$, then $\det(Z_{6})=-918y_{61}+2942$, and
$y_{6}=(2942,918)$ is the unique point $y_{6}$ such that the configuration is
rank deficient.

If we consider only the first $5$ point pairs we fixed initially, there exists
a unique projective transformation

$T=\begin{bmatrix}146&-294&0\\\ 135&-109&11\end{bmatrix}$ (74)

such that $Tx_{i}\sim y_{i}$ for all $i=1,\ldots,5$. Moreover, we can check
that $Tx_{6}=(-2942,-918)\sim y_{6}$. This transformation $T$ has center
$u=(-1617,-803,11888)$. This point actually has some further significance,
because we can check that

$(a_{x}(u),\ldots,f_{x}(u))\sim(48079,-55599,-88559,-17265,22529,90815)\sim(A_{y},\ldots,F_{y})$
(75)

∎

The following now finishes the proof of Theorem 6.7.

###### Proof of Theorem 6.7 (second part).

Let $S$ be the blow up of $\mathbb{P}^{2}_{x}$ in $x_{1},\ldots,x_{6}$ in
Cremona hexahedral form, and recall (62) and (69). It follows that (49) holds
if and only if $(A_{y},\ldots,F_{y})\in S$, which holds if and only if there
exists $u\in\mathbb{P}^{2}_{x}$ such that

$(a_{x}(u),\ldots,f_{x}(u))\sim(A_{y},\ldots,F_{y}).$ (76)

We first show that if $Z_{6}$ is rank deficient then a projection exists. Let
$\ell\subset\mathbb{P}^{2}_{x}$ be a line such that $u\not\in\ell$. We then
construct a projection $T^{\prime}:\mathbb{P}^{2}_{x}\to\ell$ with center $u$
as in the proof of Theorem 5.2 and Figure 2. This projection has
$T^{\prime}(x_{i})=a_{i}$ for all $i$, where the $a_{i}$ are defined as in
Figure 2. Then

$(A_{a},\ldots,F_{a})\sim(a_{x}(u),\ldots,f_{x}(u))\sim(A_{y},\ldots,F_{y})$
(77)

Since the Joubert Invariants generate the space of all invariants for $6$
ordered points on a line it follows that there exists a homography
$H:\ell\to\mathbb{P}^{1}_{y}$ such that $Ha_{i}=y_{i}$. We conclude that
$T=HT^{\prime}$ is the desired projection
$T:\mathbb{P}^{2}_{x}\to\mathbb{P}^{1}_{y}$.

To prove the reverse direction, suppose such a projection $T$ exists. We can
use homography on each side to fix $u=(0,0,1)$ and to fix the image line at
infinity. Equivalently, pick homographies on each side such that

$T=\begin{bmatrix}1&0&0\\\ 0&1&0\end{bmatrix}.$ (78)

It then follows that $x_{i}=(y_{i1},y_{i2},b_{i})$ for all $i$, where $b_{i}$
are unknown scalars. The matrix $Z_{6}$ then becomes

$\begin{bmatrix}y_{11}^{2}&y_{11}y_{12}&y_{11}y_{12}&y_{12}^{2}&y_{11}b_{1}&y_{12}b_{1}\\\
y_{21}^{2}&y_{21}y_{22}&y_{21}y_{22}&y_{22}^{2}&y_{21}b_{2}&y_{22}b_{2}\\\
y_{31}^{2}&y_{31}y_{32}&y_{31}y_{32}&y_{32}^{2}&y_{31}b_{3}&y_{32}b_{3}\\\
y_{41}^{2}&y_{41}y_{42}&y_{41}y_{42}&y_{42}^{2}&y_{41}b_{4}&y_{42}b_{4}\\\
y_{51}^{2}&y_{51}y_{52}&y_{51}y_{52}&y_{52}^{2}&y_{51}b_{5}&y_{52}b_{5}\\\
y_{61}^{2}&y_{61}y_{62}&y_{61}y_{62}&y_{62}^{2}&y_{61}b_{6}&y_{62}b_{6}\end{bmatrix}$
(79)

which is clearly rank deficient. This concludes the proof of Theorem 6.7. ∎

## References

##  Appendix A Facts about cubic surfaces

This is a collection of classical facts about cubic surfaces needed in Section
6.2. See [Hartshorne, V. Chapter 4] for a modern reference and [coble] or
[Hilbert-Cohn-Vossen] for classical references.

######  Lemma A.1.

Let $\mathcal{S}\subset\mathbb{P}^{3}$ be a smooth cubic surface. Then,

  1. (1)

$\mathcal{S}$ can be obtained by blowing up $6$ points
$p_{1},\ldots,p_{6}\in\mathbb{P}^{2}$ in general position. The exceptional
curves of $p_{1},\ldots,p_{6}$ in the blow up morphism
$\pi\,:\,\mathcal{S}\rightarrow\mathbb{P}^{2}$ are $6$ lines
$\ell_{1},\ldots,\ell_{6}$ on $\mathcal{S}$.

  2. (2)

The cubic surface $\mathcal{S}$ has exactly $27$ (complex) lines. They are:

    1. (a)

the exceptional lines $\ell_{1},\ldots,\ell_{6}$,

    2. (b)

the strict transforms $\ell_{ij}$ of the lines in $\mathbb{P}^{2}$ passing
through $p_{i},p_{j}$ for $1\leq i<j\leq 6$ ($15$ of them), and

    3. (c)

the strict transforms $\ell_{j}^{\prime}$ of the conics in $\mathbb{P}^{2}$
passing through five of the $p_{i}$ for $i\neq j,j=1,\ldots,6$ ($6$ of them).

  3. (3)

The two sets of lines $\\{\ell_{1},\ldots,\ell_{6}\\}$ and
$\\{\ell_{1}^{\prime},\ldots,\ell_{6}^{\prime}\\}$ form a Schäfli double six
on $\mathcal{S}$. There are $36$ Schläfli double sixes on $\mathcal{S}$. Each
double six is uniquely determined by three of the line pairs
$(\ell_{i},\ell_{i}^{\prime})$.

  4. (4)

Any set of $6$ mutually skew lines among the $27$ lines on $\mathcal{S}$ can
play the role of $\ell_{1},\ldots,\ell_{6}$. Precisely, if
$\\{g_{1},\ldots,g_{6}\\}$ is a set of $6$ mutually skew lines among the $27$
lines, then there is another morphism
$\pi^{\prime}\,:\,\mathcal{S}\rightarrow\mathbb{P}^{2}$, making $\mathcal{S}$
isomorphic to $\mathbb{P}^{2}$ with six points $q_{1},\ldots,q_{6}$ in general
position blown up such that $g_{1},\ldots,g_{6}$ are the exceptional curves of
$q_{1},\ldots,q_{6}$ under $\pi^{\prime}$.

We now note several facts about double sixes in $\mathbb{P}^{3}$ (or more
generally line configurations in $\mathbb{P}^{3}$).

######  Lemma A.2.

  1. (1)

Each double six in $\mathbb{P}^{3}$ is contained in a unique non-singular
cubic surface and thus forms part of the set of $27$ lines on that surface.

  2. (2)

Any set of $6$ mutually skew lines among $27$ lines that satisfy the incidence
relations is one half of a unique Schläfli double six and determines the other
lines and the cubic surface.

Next we talk about determinantal representations of a cubic surface.

######  Definition A.3.

A determinantal representation of a cubic surface $\mathcal{S}$ defined by a
cubic equation $F(z)=F(z_{0},z_{1},z_{2},z_{3})=0$ is a $3\times 3$ matrix of
linear forms

$\displaystyle
M(z)=M(z_{0},z_{1},z_{2},z_{3})=z_{0}M_{0}+z_{1}M_{1}+z_{2}M_{2}+z_{3}M_{3}$
(80)

such that $\det(M(z))=\lambda F(z)$ for some $\lambda\neq 0$. Two
representations $M(z)$ and $M^{\prime}(z)$ are equivalent if
$M^{\prime}(z)=AM(z)B$ for $A,B\in\textup{GL}(3)$.

Note that if $M(z)$ is a determinantal representation of $\mathcal{S}$, then
so is $M(z)^{\top}$. We present the following facts about determinantal
representations of a cubic surface [buckley_kosir].

######  Lemma A.4.

  1. (1)

Every smooth cubic surface $\mathcal{S}$ has a determinantal representation.

  2. (2)

The equivalence classes of determinantal representations of $\mathcal{S}$ are
in bijection with sets of $6$ mutually skew lines on $\mathcal{S}$ that are
the exceptional curves of the blow up of $\mathbb{P}^{2}$ at the $6$ points
that gives $\mathcal{S}$.

  3. (3)

If the determinantal representation $M$ of $\mathcal{S}$ corresponds to the
skew lines $\ell_{1},\ldots,\ell_{6}$ on $\mathcal{S}$, then $M^{\top}$
corresponds to another $6$ skew lines
$\ell_{1}^{\prime},\ldots,\ell_{6}^{\prime}$ on $\mathcal{S}$ such that the
pair of $6$ lines form a Schläfli double six on $\mathcal{S}$. Thus there are
$72$ equivalence classes of determinantal representations of $\mathcal{S}$ one
for each half of the $36$ Schläfli double sixes on $\mathcal{S}$.

  4. (4)

The choice of a Schlafli double six
$\ell_{1},\ldots,\ell_{6},\ell_{1}^{\prime},\ldots,\ell_{6}^{\prime}$ on
$\mathcal{S}$ is equivalent to the choice of a pair of equivalence classes of
determinantal representations of $\mathcal{S}$, namely$[M]$ and $[M^{\top}]$.
(A smooth cubic surface cannot have a symmetric determinantal representation
and hence $[M]\neq[M^{\top}]$.)

Here are various recipes to explicitly realize some of the above assertions.

  1. (1)

If $\mathcal{S}$ is the blow up of $\mathbb{P}^{2}$ at the $6$ points
$p_{1},\ldots,p_{6}$, then $\mathcal{S}$ can be constructed as follows. Pick a
basis $\\{f_{0},f_{1},f_{2},f_{3}\\}$ of the $4$-dimensional vector space of
cubics through $p_{1},\ldots,p_{6}$ and map

$\displaystyle
u\in\mathbb{P}^{2}\mapsto(f_{0}(u):f_{1}(u):f_{2}(u):f_{3}(u))\in\mathbb{P}^{3}.$
(81)

  2. (2)

The variety of $\langle f_{0},\ldots,f_{3}\rangle$ is
$\\{p_{0},\ldots,p_{6}\\}$. By the Hilbert-Burch theorem [eisenbud-book,
Section 20.4] there is a $3\times 4$ matrix $L(u)$ of linear forms in
$u_{0},u_{1},u_{2}$ whose maximal minors are $f_{0},\ldots,f_{3}$. The matrix
$L(u)$ generically has rank $3$ and has rank $2$ exactly when $u=p_{i}$ for
some $i=1,\ldots,6$. Define a $3\times 3$ matrix $M(z)$ of linear forms in
$z_{0},z_{1},z_{2},z_{3}$ by the relation

$\displaystyle M(z)\cdot\begin{pmatrix}u_{0}\\\ u_{1}\\\
u_{2}\end{pmatrix}=L(u)\cdot\begin{pmatrix}z_{0}\\\ z_{1}\\\ z_{2}\\\
z_{3}\end{pmatrix}.$ (82)

Then $M(z)$ is a determinantal representation of $\mathcal{S}$.

  3. (3)

The blow up of $\mathbb{P}^{2}$ at $p_{1},\ldots,p_{6}$ can be seen via (82).
If $\bar{u}\in\mathbb{P}^{2}$ is not one of $p_{1},\ldots,p_{6}$, then some
$f_{i}(\bar{u})\neq 0$ and $\operatorname{rank}(L(\bar{u}))=3$. Therefore, the
null space of $L(\bar{u})$ is a unique point $\bar{z}\in\mathbb{P}^{3}$ and
$M(\bar{z})\bar{u}=L(\bar{u})\bar{z}=0$. This means that $\det(M(\bar{z}))=0$
and $\bar{z}\in\mathcal{S}$. If $\bar{z}$ is a generic point on $\mathcal{S}$,
then $\operatorname{rank}(M(\bar{z}))=2$, and the blow up morphism $\pi$ sends
$\bar{z}$ to the unique point $\bar{u}\in\mathbb{P}^{2}$ that generates the
right null space of $M(\bar{z})$.

  4. (4)

The $6$ exceptional curves of the blow up are precisely

$\displaystyle\ell_{i}:=\\{z\in\mathbb{P}^{3}\,:\,M(z)p_{i}=0\\}\textup{ for
}i=1,\ldots,6.$ (83)

By what is stated above, $\operatorname{rank}(L(u))=2$ if and only if
$u=p_{i}$ for some $i=1,\ldots,6$ if and only if (via (82)) $M(z)u=L(u)z=0$ is
a system of $2$ linearly independent equations in $z$ (i.e., defines a line in
$\mathbb{P}^{3}$). This line is entirely on $\mathcal{S}$ since $\det(M(z))=0$
if $M(z)p_{i}=0$.

  5. (5)

Corresponding results hold for the determinantal representation $M(z)^{\top}$
of $\mathcal{S}$.

    1. (a)

To obtain the points $q_{1},\ldots,q_{6}\in\mathbb{P}^{2}$ whose blow up is
$\mathcal{S}$, compute $L^{\prime}(u)$ via

$\displaystyle M(z)^{\top}u=L^{\prime}(u)z.$ (84)

The variety of the maximal minors of $L^{\prime}(u)$ is
$\\{q_{1},\ldots,q_{6}\\}$.

    2. (b)

A point $\bar{y}\in\mathbb{P}^{2}$ not equal to any $q_{i}$ is sent to the
unique point $\bar{z}$ in $\mathbb{P}^{3}$ in the right nullspace of
$L^{\prime}(\bar{y})$. This means
$L^{\prime}(\bar{y})\bar{z}=M(\bar{z})^{\top}\bar{y}=0$ and
$\bar{z}\in\mathcal{S}$ since $\det(M(\bar{z}))^{\top}=0$.

    3. (c)

The morphism $\pi^{\prime}\,:\,\mathcal{S}\rightarrow\mathbb{P}^{2}$ sends
$\bar{z}$ to the unique point $\bar{y}\in\mathbb{P}^{2}$ that generates the
right nullspace of $M(\bar{z})^{\top}$, or equivalently, the left nullspace of
$M(\bar{z})$.

    4. (d)

The exceptional curves of the blow up are the lines

$\displaystyle\ell_{j}^{\prime}:=\\{z\in\mathbb{P}^{3}\,:\,q_{j}^{\top}M(z)=0\\}\textup{
for }j=1,\ldots,6.$

The following theorem follows from the above facts.

######  Theorem A.5.

Given a determinantal representation $M(z)$ of a smooth cubic surface
$\mathcal{S}\subset\mathbb{P}^{3}$, there exists $6$ points
$(p_{i},q_{i})\in\mathbb{P}^{2}\times\mathbb{P}^{2}$ such that $\mathcal{S}$
is the blow up of the first $\mathbb{P}^{2}$ at $\\{p_{i}\\}_{i=1}^{6}$ and
also the blow up of the second $\mathbb{P}^{2}$ at $\\{q_{j}\\}_{j=1}^{6}$.
The exceptional curves of the two blow ups form a Schläfli double six on
$\mathcal{S}$:

$\displaystyle\ell_{i}:=\\{z\in\mathbb{P}^{3}\,:\,M(z)p_{i}=0\\}\textup{ for
}i=1,\ldots,6,$ (85)
$\displaystyle\ell_{j}^{\prime}:=\\{z\in\mathbb{P}^{3}\,:\,q_{j}^{\top}M(z)=0\\}\textup{
for }j=1,\ldots,6.$ (86)