###### Abstract

Let $P$ be a set of $n$ points in real projective $d$-space, not all contained
in a hyperplane, such that any $d$ points span a hyperplane. An ordinary
hyperplane of $P$ is a hyperplane containing exactly $d$ points of $P$. We
show that if $d\geqslant 4$, the number of ordinary hyperplanes of $P$ is at
least $\binom{n-1}{d-1}-O_{d}(n^{\lfloor(d-1)/2\rfloor})$ if $n$ is
sufficiently large depending on $d$. This bound is tight, and given $d$, we
can calculate the exact minimum number for sufficiently large $n$. This is a
consequence of a structure theorem for sets with few ordinary hyperplanes: For
any $d\geqslant 4$ and $K>0$, if $n\geqslant C_{d}K^{8}$ for some constant
$C_{d}>0$ depending on $d$, and $P$ spans at most $K\binom{n-1}{d-1}$ ordinary
hyperplanes, then all but at most $O_{d}(K)$ points of $P$ lie on a
hyperplane, an elliptic normal curve, or a rational acnodal curve. We also
find the maximum number of $(d+1)$-point hyperplanes, solving a
$d$-dimensional analogue of the orchard problem. Our proofs rely on Green and
Tao’s results on ordinary lines, our earlier work on the $3$-dimensional case,
as well as results from classical algebraic geometry.

title = On sets defining few ordinary hyperplanes, author = Aaron Lin and
Konrad Swanepoel, plaintextauthor = Aaron Lin, Konrad Swanepoel, year=2020,
number=4, received=26 April 2019, revised=17 January 2020, published=24 April
2020, doi=10.19086/da.11949,

[classification=text]

##  1 Introduction

An _ordinary line_ of a set of points in the plane is a line passing through
exactly two points of the set. The classical Sylvester–Gallai theorem states
that every finite non-collinear point set in the plane spans at least one
ordinary line. In fact, for sufficiently large $n$, an $n$-point non-collinear
set in the plane spans at least $n/2$ ordinary lines, and this bound is tight
if $n$ is even. This was shown by Green and Tao [GT13] via a structure theorem
characterising all finite point sets with few ordinary lines.

It is then natural to consider higher dimensional analogues. Motzkin [M51]
noted that there are finite non-coplanar point sets in $3$-space that span no
plane containing exactly three points of the set. He proposed considering
instead hyperplanes $\Pi$ in $d$-space such that all but one point contained
in $\Pi$ is contained in a $(d-2)$-dimensional flat of $\Pi$. The existence of
such hyperplanes was shown by Motzkin [M51] for $3$-space and by Hansen [H65]
in higher dimensions.

Purdy and Smith [PS10] considered instead finite non-coplanar point sets in
$3$-space with no three points collinear, and provided a lower bound on the
number of planes containing exactly three points of the set. Referring to such
a plane as an _ordinary plane_ , Ball [B18] proved a $3$-dimensional analogue
of Green and Tao’s [GT13] structure theorem, and found the exact minimum
number of ordinary planes spanned by sufficiently large non-coplanar point
sets in real projective $3$-space with no three points collinear. Using an
alternative method, we [LS18] were able to prove a more detailed structure
theorem but with a stronger condition; see Theorem 4.1 in Section 4.

Ball and Monserrat [BM17] made the following definition, generalising ordinary
planes to higher dimensions.

######  Definition.

An _ordinary hyperplane_ of a set of points in real projective $d$-space,
where every $d$ points span a hyperplane, is a hyperplane passing through
exactly $d$ points of the set.

They [BM17] also proved bounds on the minimum number of ordinary hyperplanes
spanned by such sets (see also [M15]). Our first main result is a structure
theorem for sets with few ordinary hyperplanes. The elliptic normal curves and
rational acnodal curves mentioned in the theorem and their group structure
will be described in Section 3. Our methods extend those in our earlier paper
[LS18], and we detail them in Section 2.

######  Theorem 1.1.

Let $d\geqslant 4$, $K>0$, and suppose $n\geqslant
C\max\\{(dK)^{8},d^{3}2^{d}K\\}$ for some sufficiently large absolute constant
$C>0$. Let $P$ be a set of $n$ points in $\mathbb{R}\mathbb{P}^{d}$ where
every $d$ points span a hyperplane. If $P$ spans at most $K\binom{n-1}{d-1}$
ordinary hyperplanes, then $P$ differs in at most $O(d2^{d}K)$ points from a
configuration of one of the following types:

  1. ( )

A subset of a hyperplane;

  2. ( )

A coset $H\oplus x$ of a subgroup $H$ of an elliptic normal curve or the
smooth points of a rational acnodal curve of degree $d+1$, for some $x$ such
that $(d+1)x\in H$.

It is easy to show that conversely, a set of $n$ points where every $d$ span a
hyperplane and differing from ( ) ‣ 1.1 or ( ) ‣ 1.1 by $O(K)$ points, spans
$O(K\binom{n-1}{d-1})$ ordinary hyperplanes. By [BM17]*Theorem 2.4, if a set
of $n$ points where every $d$ points span a hyperplane itself spans
$K\binom{n-1}{d-1}$ ordinary hyperplanes, and is not contained in a
hyperplane, then $K=\Omega(1/d)$. Theorem 1.2 below implies that $K\geqslant
1$ for sufficiently large $n$ depending on $d$.

For a similar structure theorem in dimension $4$ but with $K=o(n^{1/7})$, see
Ball and Jimenez [BJ18], who show that $P$ lies on the intersection of five
quadrics. Theorem 1.1 proves [BJ18]*Conjecture 12, noting that elliptic normal
curves and rational acnodal curves lie on $\binom{d}{2}-1$ linearly
independent quadrics [Klein]*p. 365[Fi08]*Proposition 5.3. We also mention
that Monserrat [M15]*Theorem 2.10 proved a structure theorem stating that
almost all points of the set lie on the intersection of $d-1$ hypersurfaces of
degree at most $3$.

Our second main result is a tight bound on the minimum number of ordinary
hyperplanes, proving [BM17]*Conjecture 3. Note that our result holds only for
sufficiently large $n$; see [BM17][M15][J18] for estimates when $d$ is small
or $n$ is not much larger than $d$.

######  Theorem 1.2.

Let $d\geqslant 4$ and let $n\geqslant Cd^{3}2^{d}$ for some sufficiently
large absolute constant $C>0$. The minimum number of ordinary hyperplanes
spanned by a set of $n$ points in $\mathbb{R}\mathbb{P}^{d}$, not contained in
a hyperplane and where every $d$ points span a hyperplane, is

$\binom{n-1}{d-1}-O\left(d2^{-d/2}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right).$

This minimum is attained by a coset of a subgroup of an elliptic normal curve
or the smooth points of a rational acnodal curve of degree $d+1$, and when
$d+1$ and $n$ are coprime, by $n-1$ points in a hyperplane together with a
point not in the hyperplane.

Green and Tao [GT13] also used their structure theorem to solve the classical
orchard problem of finding the maximum number of $3$-point lines spanned by a
set of $n$ points in the plane, for $n$ sufficiently large. We solved the
$3$-dimensional analogue in [LS18]. Our third main result is the
$d$-dimensional analogue. We define a _$(d+1)$ -point hyperplane_ to be a
hyperplane through exactly $d+1$ points of a given set.

######  Theorem 1.3.

Let $d\geqslant 4$ and let $n\geqslant Cd^{3}2^{d}$ for some sufficiently
large absolute constant $C>0$. The maximum number of $(d+1)$-point hyperplanes
spanned by a set of $n$ points in $\mathbb{R}\mathbb{P}^{d}$ where every $d$
points span a hyperplane is

$\frac{1}{d+1}\binom{n-1}{d}+O\left(2^{-d/2}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right).$

This maximum is attained by a coset of a subgroup of an elliptic normal curve
or the smooth points of a rational acnodal curve of degree $d+1$.

While the bounds in Theorems 1.2 and 1.3 are asymptotic, we provide a
recursive method (as part of our proofs) to calculate the exact extremal
values for a given $d$ and $n$ sufficiently large in Section 5. In principle,
the exact values can be calculated for any given $d$ and turns out to be a
quasi-polynomial in $n$ with a period of $d+1$. We present the values for
$d=4,5,6$ at the end of Section 5.

### Relation to previous work

The main idea in our proof of Theorem 1.1 is to induct on the dimension $d$,
with the base case $d=3$ being our earlier structure theorem for sets defining
few ordinary planes [LS18], which in turn is based on Green and Tao’s
Intermediate Structure Theorem for sets defining few ordinary lines
[GT13]*Proposition 5.3.

Roughly, the structure theorem in $3$-space states that if a finite set of
points is in general position (no three points collinear) and spans few
ordinary planes, then most of the points must lie on a plane, two disjoint
conics, or an elliptic or acnodal space quartic curve. In fact, we can define
a group structure on these curves encoding when four points are coplanar, in
which case our point set must be very close to a coset of the curve. (See
Theorem 4.1 for a more precise statement.)

As originally observed by Ball [B18] in $3$-space, the general position
condition allows the use of projection to leverage Green and Tao’s
Intermediate Structure Theorem [GT13]*Proposition 5.3. This avoids having to
apply their Full Structure Theorem [GT13]*Theorem 1.5, which has a much worse
lower bound on $n$, as it avoids the technical Section 6 of [GT13], dealing
with the case in the plane when most of the points lie on a large, though
bounded, number of lines. On the other hand, to get to the precise coset
structure, we used additive-combinatorial results from [GT13]*Section 7,
specifically [GT13]*Propositions A.5, Lemmas 7.2, 7.4, 7.7, and Corollary 7.6.
In this paper, the only result of Green and Tao [GT13] we explicitly use is
[GT13]*Proposition A.5, which we extend in Proposition 4.3, while all other
results are subsumed in the structure theorem in $3$-space. In dimensions
$d>3$, the general position condition also allows the use of projections from
a point to a hyperplane (see also Ball and Monserrat [BM17]). In Section 2.2
we detail various technical results about the behaviour of curves under such
projections, which are extensions of $3$-dimensional results in [LS18].

While the group structure on elliptic or singular space quartic curves are
well studied (see for instance [Muntingh]), we could not find references to
the group structure on singular rational curves in higher dimensions. This is
our main focus in Section 3, which in a way extends [LS18]*Section 3. In
particular, we look at Sylvester’s theorem on when a binary form can be
written as a sum of perfect powers, which has its roots in classical invariant
theory. In extending the results of [LS18]*Section 3, we have to consider how
to generalise the catalecticant (of a binary quartic form), which leads us to
the secant variety of the rational normal curve as a determinantal variety.

Green and Tao’s Intermediate Structure Theorem in $2$-space has a slightly
different flavour to their Full Structure Theorem, the structure theorem in
$3$-space, and Theorem 1.1. However, this is not the only reason why we start
our induction at $d=3$. A more substantial reason is that there are no smooth
rational cubic curves in $2$-space; as is well known, all rational planar
cubic curves are singular. Thus, both smooth and singular rational quartics in
$3$-space project onto rational cubics, and we need some way to tell them
apart. In higher dimensions, we have Lemma 3.7 to help us, but since this is
false when $d=3$, the induction from the plane to $3$-space [LS18] is more
technical. This is despite the superficial similarity between the $2$\- and
$3$-dimensional situations where there are two almost-extremal cases while
there is essentially only one case when $d>3$.

Proving Theorem 1.1, which covers the $d>3$ cases, is thus in some sense less
complicated, since not only are we leveraging a more detailed structure
theorem (Theorems 1.1 and 4.1 as opposed to [GT13]*Proposition 5.3), we also
lose a case. However, there are complications that arise in how to generalise
and extend results from $2$\- and $3$-space to higher dimensions.

##  2 Notation and tools

By $A=O(B)$, we mean there exists an absolute constant $C>0$ such that
$0\leqslant A\leqslant CB$. Thus, $A=-O(B)$ means there exists an absolute
constant $C>0$ such that $-CB\leqslant A\leqslant 0$. We also write
$A=\Omega(B)$ for $B=O(A)$. None of the $O(\cdot)$ and $\Omega(\cdot)$
statements in this paper have implicit dependence on the dimension $d$.

We write $A\mathbin{\triangle}B$ for the symmetric difference of the sets $A$
and $B$.

Let $\mathbb{F}$ denote the field of real or complex numbers, let
$\mathbb{F}^{*}=\mathbb{F}\setminus{\\{0\\}}$, and let
$\mathbb{F}\mathbb{P}^{d}$ denote the $d$-dimensional projective space over
$\mathbb{F}$. We denote the homogeneous coordinates of a point in
$d$-dimensional projective space by a $(d+1)$-dimensional vector
$[x_{0},x_{1},\dots,x_{d}]$. We call a linear subspace of dimension $k$ in
$\mathbb{F}\mathbb{P}^{d}$ a _$k$ -flat_; thus a point is a $0$-flat, a line
is a $1$-flat, a plane is a $2$-flat, and a hyperplane is a $(d-1)$-flat. We
denote by $Z_{\mathbb{F}}(f)$ the set of $\mathbb{F}$-points of the algebraic
hypersurface defined by the vanishing of a homogeneous polynomial
$f\in\mathbb{F}[x_{0},x_{1},\dots,x_{d}]$. More generally, we consider a
(closed, projective) _variety_ to be any intersection of algebraic
hypersurfaces. We say that a variety is pure-dimensional if each of its
irreducible components has the same dimension. We consider a _curve_ of degree
$e$ in $\mathbb{C}\mathbb{P}^{d}$ to be a variety $\delta$ of pure dimension
$1$ such that a generic hyperplane in $\mathbb{C}\mathbb{P}^{d}$ intersects
$\delta$ in $e$ distinct points. More generally, the degree of a variety
$X\subset\mathbb{C}\mathbb{P}^{d}$ of dimension $r$ is

$\deg(X):=\max\left\\{|\Pi\cap X|:\text{$\Pi$ is a $(d-r)$-flat such that
$\Pi\cap X$ is finite}\right\\}.$

We say that a curve is _non-degenerate_ if it is not contained in a
hyperplane, and _non-planar_ if it is not contained in a $2$-flat. We call a
curve _real_ if each of its irreducible components contains infinitely many
points of $\mathbb{R}\mathbb{P}^{d}$. Whenever we consider a curve in
$\mathbb{R}\mathbb{P}^{d}$, we implicitly assume that its Zariski closure is a
real curve.

We denote the Zariski closure of a set $S\subseteq\mathbb{C}\mathbb{P}^{d}$ by
$\overline{S}$. We will use the _secant variety
$\operatorname{Sec}_{\mathbb{C}}(\delta)$_ of a curve $\delta$, which is the
Zariski closure of the set of points in $\mathbb{C}\mathbb{P}^{d}$ that lie on
a line through some two points of $\delta$.

###  2.1 Bézout’s theorem

Bézout’s theorem gives the degree of an intersection of varieties. While it is
often formulated as an equality, in this paper we only need the weaker form
that ignores multiplicity and gives an upper bound. The (set-theoretical)
intersection $X\cap Y$ of two varieties is just the variety defined by
$P_{X}\cup P_{Y}$, where $X$ and $Y$ are defined by the collections of
homogeneous polynomials $P_{X}$ and $P_{Y}$ respectively.

######  Theorem 2.1 (Bézout [Fu84]*Section 2.3).

Let $X$ and $Y$ be varieties in $\mathbb{C}\mathbb{P}^{d}$ with no common
irreducible component. Then $\deg(X\cap Y)\leqslant\deg(X)\deg(Y)$.

###  2.2 Projections

Given $p\in\mathbb{F}\mathbb{P}^{d}$, the _projection from $p$_,
$\pi_{p}\colon\mathbb{F}\mathbb{P}^{d}\setminus\\{p\\}\to\mathbb{F}\mathbb{P}^{d-1}$,
is defined by identifying $\mathbb{F}\mathbb{P}^{d-1}$ with any hyperplane
$\Pi$ of $\mathbb{F}\mathbb{P}^{d}$ not passing through $p$, and then letting
$\pi_{p}(x)$ be the point where the line $px$ intersects $\Pi$ [H92]*Example
3.4. Equivalently, $\pi_{p}$ is induced by a surjective linear transformation
$\mathbb{F}^{d+1}\to\mathbb{F}^{d}$ where the kernel is spanned by the vector
$p$.

As in our previous paper [LS18], we have to consider projections of curves
where we do not have complete freedom in choosing a generic projection point
$p$.

Let $\delta\subset\mathbb{C}\mathbb{P}^{d}$ be an irreducible non-planar curve
of degree $e$, and let $p$ be a point in $\mathbb{C}\mathbb{P}^{d}$. We call
$\pi_{p}$ _generically one-to-one on $\delta$_ if there is a finite subset $S$
of $\delta$ such that $\pi_{p}$ restricted to $\delta\setminus S$ is one-to-
one. (This is equivalent to the birationality of $\pi_{p}$ restricted to
$\delta\setminus\\{p\\}$ [H92]*p. 77.) If $\pi_{p}$ is generically one-to-one,
the degree of the curve $\overline{\pi_{p}(\delta\setminus\\{p\\})}$ is $e-1$
if $p$ is a smooth point on $\delta$, and is $e$ if $p$ does not lie on
$\delta$; if $\pi_{p}$ is not generically one-to-one, then the degree of
$\overline{\pi_{p}(\delta\setminus\\{p\\})}$ is at most $(e-1)/2$ if $p$ lies
on $\delta$, and is at most $e/2$ if $p$ does not lie on $\delta$
[H92]*Example 18.16, [Kollar]*Section 1.15.

The following three lemmas on projections are proved in [LS18] in the case
$d=3$. They all state that most projections behave well and can be considered
to be quantitative versions of the trisecant lemma [KKT08]. The proofs of
Lemmas 2.3 and 2.4 are almost word-for-word the same as the proofs of the
$3$-dimensional cases in [LS18]. All three lemmas can also be proved by
induction on the dimension $d\geqslant 3$ from the $3$-dimensional case. We
illustrate this by proving Lemma 2.2.

######  Lemma 2.2.

Let $\delta$ be an irreducible non-planar curve of degree $e$ in
$\mathbb{C}\mathbb{P}^{d}$, $d\geqslant 3$. Then there are at most $O(e^{4})$
points $p$ on $\delta$ such that $\pi_{p}$ restricted to
$\delta\setminus\\{p\\}$ is not generically one-to-one.

###### Proof.

The case $d=3$ was shown in [LS18], based on the work of Furukawa [Fu11]. We
next assume that $d\geqslant 4$ and that the lemma holds in dimension $d-1$.
Since $d>3$ and the dimension of $\operatorname{Sec}_{\mathbb{C}}(\delta)$ is
at most $3$ [H92]*Proposition 11.24, there exists a point
$p\in\mathbb{C}\mathbb{P}^{d}$ such that all lines through $p$ have
intersection multiplicity at most $1$ with $\delta$. It follows that the
projection $\delta^{\prime}:=\overline{\pi_{p}(\delta)}$ of $\delta$ is a non-
planar curve of degree $e$ in $\mathbb{C}\mathbb{P}^{d-1}$. Consider any line
$\ell$ not through $p$ that intersects $\delta$ in at least three distinct
points $p_{1},p_{2},p_{3}$. Then $\pi_{p}(\ell)$ is a line in
$\mathbb{C}\mathbb{P}^{d-1}$ that intersects $\delta^{\prime}$ in three points
$\pi_{p}(p_{1}),\pi_{p}(p_{2}),\pi_{p}(p_{3})$. It follows that if
$x\in\delta$ is a point such that for all but finitely many points
$y\in\delta$, the line $xy$ intersects $\delta$ in a point other than $x$ or
$y$, then $x^{\prime}:=\pi_{p}(x)$ is a point such that for all but finitely
many points $y^{\prime}:=\pi_{p}(y)\in\delta^{\prime}$, the line
$x^{\prime}y^{\prime}$ intersects $\delta^{\prime}$ in a third point. That is,
if $\pi_{x}$ restricted to $\delta$ is not generically one-to-one, then the
projection map $\pi_{x^{\prime}}$ in $\mathbb{C}\mathbb{P}^{d-1}$ restricted
to $\delta^{\prime}$ is not generically one-to-one. By the induction
hypothesis, there are at most $O(e^{4})$ such points and we are done. ∎

######  Lemma 2.3.

Let $\delta$ be an irreducible non-planar curve of degree $e$ in
$\mathbb{C}\mathbb{P}^{d}$, $d\geqslant 3$. Then there are at most $O(e^{3})$
points $x\in\mathbb{C}\mathbb{P}^{d}\setminus\delta$ such that $\pi_{x}$
restricted to $\delta$ is not generically one-to-one.

######  Lemma 2.4.

Let $\delta_{1}$ and $\delta_{2}$ be two irreducible non-planar curves in
$\mathbb{C}\mathbb{P}^{d}$, $d\geqslant 3$, of degree $e_{1}$ and $e_{2}$
respectively. Then there are at most $O(e_{1}e_{2})$ points $p$ on
$\delta_{1}$ such that $\overline{\pi_{p}(\delta_{1}\setminus\\{p\\})}$ and
$\overline{\pi_{p}(\delta_{2}\setminus\\{p\\})}$ coincide.

##  3 Curves of degree $d+1$d+1

In this paper, irreducible non-degenerate curves of degree $d+1$ in
$\mathbb{C}\mathbb{P}^{d}$ play a fundamental role. Indeed, the elliptic
normal curve and rational acnodal curve mentioned in Theorem 1.1 are both such
curves. In this section, we describe their properties that we need. These
properties are all classical, but we did not find a reference for the group
structure on singular rational curves of degree $d+1$, and therefore consider
this in detail.

It is well-known in the plane that there is a group structure on any smooth
cubic curve or the set of smooth points of a singular cubic. This group has
the property that three points sum to the identity if and only if they are
collinear. Over the complex numbers, the group on a smooth cubic is isomorphic
to the torus $(\mathbb{R}/\mathbb{Z})^{2}$, and the group on the smooth points
of a singular cubic is isomorphic to $(\mathbb{C},+)$ or
$(\mathbb{C}^{*},\cdot)$ depending on whether the singularity is a cusp or a
node. Over the real numbers, the group on a smooth cubic is isomorphic to
$\mathbb{R}/\mathbb{Z}$ or $\mathbb{R}/\mathbb{Z}\times\mathbb{Z}_{2}$
depending on whether the real curve has one or two semi-algebraically
connected components, and the group on the smooth points of a singular cubic
is isomorphic to $(\mathbb{R},+)$, $(\mathbb{R},+)\times\mathbb{Z}_{2}$, or
$\mathbb{R}/\mathbb{Z}$ depending on whether the singularity is a cusp, a
crunode, or an acnode. See for instance [GT13] for a more detailed
description.

In higher dimensions, it turns out that an irreducible non-degenerate curve of
degree $d+1$ does not necessarily have a natural group structure, but if it
has, the behaviour is similar to the planar case. For instance, in
$\mathbb{C}\mathbb{P}^{3}$, an irreducible non-degenerate quartic curve is
either an elliptic quartic, with a group isomorphic to an elliptic curve such
that four points on the curve are coplanar if and only if they sum to the
identity, or a rational curve. There are two types, or species, of rational
quartics. The rational quartic curves of the first species are intersections
of two quadrics (as are elliptic quartics), they are always singular, and
there is a group on the smooth points such that four points on the curve are
coplanar if and only if they sum to the identity. Those of the second species
lie on a unique quadric, are smooth, and there is no natural group structure
analogous to the other cases. See [LS18] for a more detailed account. The
picture is similar in higher dimensions.

######  Definition (Clifford [Clifford], Klein [Klein]).

An _elliptic normal curve_ is an irreducible non-degenerate smooth curve of
degree $d+1$ in $\mathbb{C}\mathbb{P}^{d}$ isomorphic to an elliptic curve in
the plane.

######  Proposition 3.1 ([S09]*Exercise 3.11 and Corollary 5.1.1,
[S94]*Corollary 2.3.1).

An elliptic normal curve $\delta$ in $\mathbb{C}\mathbb{P}^{d}$, $d\geqslant
2$, has a natural group structure such that $d+1$ points in $\delta$ lie on a
hyperplane if and only if they sum to the identity. This group is isomorphic
to $(\mathbb{R}/\mathbb{Z})^{2}$.

If the curve is real, then the group is isomorphic to $\mathbb{R}/\mathbb{Z}$
or $\mathbb{R}/\mathbb{Z}\times\mathbb{Z}_{2}$ depending on whether the real
curve has one or two semi-algebraically connected components.

A similar result holds for singular rational curves of degree $d+1$. Since we
need to work with such curves and a description of their group structure is
not easily found in the literature, we give a detailed discussion of their
properties in the remainder of this section.

A _rational curve_ $\delta$ in $\mathbb{F}\mathbb{P}^{d}$ of degree $e$ is a
curve that can be parametrised by the projective line,

$\delta\colon\mathbb{F}\mathbb{P}^{1}\to\mathbb{F}\mathbb{P}^{d},\quad[x,y]\mapsto[q_{0}(x,y),\dots,q_{d}(x,y)],$

where each $q_{i}$ is a homogeneous polynomial of degree $e$ in the variables
$x$ and $y$. The following lemma is well known (see for example [SR85]*p. 38,
Theorem VIII), and can be proved by induction from the planar case using
projection.

######  Proposition 3.2.

An irreducible non-degenerate curve of degree $d+1$ in
$\mathbb{C}\mathbb{P}^{d}$, $d\geqslant 2$, is either an elliptic normal curve
or rational.

We next describe when an irreducible non-degenerate rational curve of degree
$d+1$ in $\mathbb{C}\mathbb{P}^{d}$ has a natural group structure. It turns
out that this happens if and only if the curve is singular.

We write $\nu_{d+1}$ for the _rational normal curve_ in
$\mathbb{C}\mathbb{P}^{d+1}$ [H92]*Example 1.14, which we parametrise as

$\nu_{d+1}:[x,y]\mapsto[y^{d+1},-xy^{d},x^{2}y^{d-1},\dotsc,(-x)^{d-1}y^{2},(-x)^{d}y,(-x)^{d+1}].$

Any irreducible non-degenerate rational curve $\delta$ of degree $d+1$ in
$\mathbb{C}\mathbb{P}^{d}$ is the projection of the rational normal curve, and
we have

$\delta[x,y]=[y^{d+1},-xy^{d},x^{2}y^{d-1},\dotsc,(-x)^{d-1}y^{2},(-x)^{d}y,(-x)^{d+1}]A,$

where $A$ is a $(d+2)\times(d+1)$ matrix of rank $d+1$ (since $\delta$ is non-
degenerate) with entries derived from the coefficients of the polynomials
$q_{i}$ of degree $d+1$ in the parametrisation of the curve (with suitable
alternating signs). Thus $\delta\subset\mathbb{C}\mathbb{P}^{d}$ is the image
of $\nu_{d+1}$ under the projection map $\pi_{p}$ defined by $A$. In
particular, the point of projection
$p=[p_{0},p_{1},\dots,p_{d+1}]\in\mathbb{C}\mathbb{P}^{d+1}$ is the
($1$-dimensional) kernel of $A$. If we project $\nu_{d+1}$ from a point
$p\in\nu_{d+1}$, then we obtain a rational normal curve in
$\mathbb{C}\mathbb{P}^{d}$. However, since $\delta$ is of degree $d+1$,
necessarily $p\notin\nu_{d+1}$. Conversely, it can easily be checked that for
any $p\notin\nu_{d+1}$, the projection of $\nu_{d+1}$ from $p$ is a rational
curve of degree $d+1$ in $\mathbb{C}\mathbb{P}^{d}$. We will use the notation
$\delta_{p}$ for this curve. We summarise the above discussion in the
following proposition that will be implicitly used in the remainder of the
paper.

######  Proposition 3.3.

An irreducible non-degenerate rational curve of degree $d+1$ in
$\mathbb{C}\mathbb{P}^{d}$ is projectively equivalent to $\delta_{p}$ for some
$p\in\mathbb{C}\mathbb{P}^{d+1}\setminus\nu_{d+1}$.

We use the projection point $p$ to define a binary form and a multilinear form
associated to $\delta_{p}$. The _fundamental binary form_ associated to
$\delta_{p}$ is the homogeneous polynomial of degree $d+1$ in two variables
$f_{p}(x,y):=\sum_{i=0}^{d+1}p_{i}\binom{d+1}{i}x^{d+1-i}y^{i}$. Its
_polarisation_ is the multilinear form
$F_{p}\colon(\mathbb{F}^{2})^{d+1}\to\mathbb{F}$ [D03]*Section 1.2 defined by

$F_{p}(x_{0},y_{0},x_{1},y_{1},\dots,x_{d},y_{d}):=\frac{1}{(d+1)!}\sum_{I\subseteq\\{0,1,\dots,d\\}}(-1)^{d+1-|I|}f_{p}\left(\sum_{i\in
I}x_{i},\sum_{i\in I}y_{i}\right).$

Consider the multilinear form
$G_{p}(x_{0},y_{0},\dots,x_{d},y_{d})=\sum_{i=0}^{d+1}p_{i}P_{i}$, where

$P_{i}(x_{0},y_{0},x_{1},y_{1},\dots,x_{d},y_{d}):=\sum_{I\in\binom{\\{0,1,\dots,d\\}}{i}}\prod_{j\in\overline{I}}x_{j}\prod_{j\in
I}y_{j}$ (1)

for each $i=0,\dots,d+1$. Here the sum is taken over all subsets $I$ of
$\\{0,1,\dots,d\\}$ of size $i$, and $\overline{I}$ denotes the complement of
$I$ in $\\{0,1,\dots,d\\}$. It is easy to see that the binary form $f_{p}$ is
the _restitution_ of $G_{p}$, namely [D03]*Section 1.2

$f_{p}(x,y)=G_{p}(x,y,x,y,\dots,x,y).$

Since the polarisation of the restitution of a multilinear form is itself
[D03]*Section 1.2, we must thus have $F_{p}=G_{p}$. (This can also be checked
directly.)

######  Lemma 3.4.

Let $\delta_{p}$ be an irreducible non-degenerate rational curve of degree
$d+1$ in $\mathbb{C}\mathbb{P}^{d}$, $d\geqslant 2$, where
$p\in\mathbb{C}\mathbb{P}^{d+1}\setminus\nu_{d+1}$. A hyperplane intersects
$\delta_{p}$ in $d+1$ points $\delta_{p}[x_{i},y_{i}]$, $i=0,\dots,d$,
counting multiplicity, if and only if
$F_{p}(x_{0},y_{0},x_{1},y_{1},\dots,x_{d},y_{d})=0$.

###### Proof.

We first prove the statement for distinct points
$[x_{i},y_{i}]\in\mathbb{C}\mathbb{P}^{1}$. Then the points
$\delta_{p}[x_{i},y_{i}]$ are all on a hyperplane if and only if the
hyperplane in $\mathbb{C}\mathbb{P}^{d+1}$ through the points
$\nu_{d+1}[x_{i},y_{i}]$ passes through $p$. It will be sufficient to prove
the identity

$D:=\det\begin{pmatrix}\nu_{d+1}[x_{0},y_{0}]\\\ \vdots\\\
\nu_{d+1}[x_{d},y_{d}]\\\
p\end{pmatrix}=F_{p}(x_{0},y_{0},x_{1},y_{1},\dots,x_{d},y_{d})\prod_{0\leqslant
j<k\leqslant d}\begin{vmatrix}x_{j}&x_{k}\\\ y_{j}&y_{k}\end{vmatrix},$ (2)

since the second factor on the right-hand side does not vanish because the
points $[x_{i},y_{i}]$ are distinct. We first note that

$\displaystyle D$
$\displaystyle=\begin{vmatrix}y_{0}^{d+1}&-x_{0}y_{0}^{d}&x_{0}^{2}y_{0}^{d-1}&\dotsc&(-x_{0})^{d}y_{0}&(-x_{0})^{d+1}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
y_{d}^{d+1}&-x_{d}y_{d}^{d}&x_{d}^{2}y_{d}^{d-1}&\dotsc&(-x_{d})^{d}y_{d}&(-x_{d})^{d+1}\\\
p_{0}&p_{1}&p_{2}&\dotsc&p_{d}&p_{d+1}\end{vmatrix}$
$\displaystyle=(-1)^{\left\lfloor\frac{d+2}{2}\right\rfloor}\begin{vmatrix}y_{0}^{d+1}&x_{0}y_{0}^{d}&x_{0}^{2}y_{0}^{d-1}&\dotsc&x_{0}^{d}y_{0}&x_{0}^{d+1}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
y_{d}^{d+1}&x_{d}y_{d}^{d}&x_{d}^{2}y_{d}^{d-1}&\dotsc&x_{d}^{d}y_{d}&x_{d}^{d+1}\\\\[3.0pt]
p_{0}&-p_{1}&p_{2}&\dotsc&(-1)^{d}p_{d}&(-1)^{d+1}p_{d+1}\end{vmatrix}.$ (3)

We next replace $(-1)^{i}p_{i}$ by $x^{i}y^{d+1-i}$ for each $i=0,\dots,d+1$
in the last row of the determinant in (3) and obtain the Vandermonde
determinant

$\displaystyle\mathrel{\phantom{=}}(-1)^{\left\lfloor\frac{d+2}{2}\right\rfloor}\begin{vmatrix}y_{0}^{d+1}&x_{0}y_{0}^{d}&x_{0}^{2}y_{0}^{d-1}&\dotsc&x_{0}^{d}y_{0}&x_{0}^{d+1}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
y_{d}^{d+1}&x_{d}y_{d}^{d}&x_{d}^{2}y_{d}^{d-1}&\dotsc&x_{d}^{d}y_{d}&x_{d}^{d+1}\\\\[3.0pt]
y^{d+1}&xy^{d}&x^{2}y^{d-1}&\dotsc&x^{d}y&x^{d+1}\end{vmatrix}$
$\displaystyle=(-1)^{\left\lfloor\frac{d+2}{2}\right\rfloor}\prod_{0\leqslant
j<k\leqslant d}\begin{vmatrix}y_{j}&y_{k}\\\
x_{j}&x_{k}\end{vmatrix}\prod_{0\leqslant j\leqslant
d}\begin{vmatrix}y_{j}&y\\\ x_{j}&x\end{vmatrix}$
$\displaystyle=(-1)^{\left\lfloor\frac{d+2}{2}\right\rfloor}(-1)^{\binom{d+2}{2}}\prod_{0\leqslant
j<k\leqslant d}\begin{vmatrix}x_{j}&x_{k}\\\
y_{j}&y_{k}\end{vmatrix}\prod_{0\leqslant j\leqslant
d}\begin{vmatrix}x_{j}&x\\\ y_{j}&y\end{vmatrix}.$

Finally, note that $(-1)^{\lfloor(d+2)/2\rfloor}(-1)^{\binom{d+2}{2}}=1$ and
that the coefficient of $x^{i}y^{d+1-i}$ in $\prod_{0\leqslant j\leqslant
d}\begin{vmatrix}x_{j}&x\\\ y_{j}&y\end{vmatrix}$ is

$\sum_{I\subseteq\binom{\\{0,\dots,d\\}}{i}}\prod_{j\in
I}(-y_{j})\prod_{j\in\overline{I}}x_{j}=(-1)^{i}P_{i},$

where $P_{i}$ is as defined in (1). It follows that the coefficient of $p_{i}$
in (3) is $P_{i}$, and (2) follows.

We next complete the argument for the case when the points $[x_{i},y_{i}]$ are
not all distinct. First suppose that a hyperplane $\Pi$ intersects
$\delta_{p}$ in $\delta_{p}[x_{i},y_{i}]$, $i=0,\dots,d$. By Bertini’s theorem
[H77]*Theorem II.8.18 and Remark II.8.18.1, there is an arbitrarily close
perturbation $\Pi^{\prime}$ of $\Pi$ that intersects $\delta_{p}$ in distinct
points $\delta_{p}[x_{i}^{\prime},y_{i}^{\prime}]$. By what has already been
proved,
$F_{p}(x_{0}^{\prime},y_{0}^{\prime},\dots,x_{d}^{\prime},y_{d}^{\prime})=0$.
Since $\Pi^{\prime}$ is arbitrarily close and $F_{p}$ is continuous,
$F_{p}[x_{0},y_{0},\dots,x_{d},y_{d}]=0$.

Conversely, suppose that $F_{p}(x_{0},y_{0},\dots,x_{d},y_{d})=0$ where the
$[x_{i},y_{i}]$ are not all distinct. Perturb the points
$[x_{0},y_{0}],\dots,[x_{d-1},y_{d-1}]$ by an arbitrarily small amount to
$[x_{0}^{\prime},y_{0}^{\prime}],\dots,[x_{d-1}^{\prime},y_{d-1}^{\prime}]$
respectively, so as to make
$\delta_{p}[x_{0}^{\prime},y_{0}^{\prime}],\dots,\delta_{p}[x_{d-1}^{\prime},y_{d-1}^{\prime}]$
span a hyperplane $\Pi^{\prime}$ that intersects $\delta_{p}$ again in
$\delta_{p}[x_{d}^{\prime},y_{d}^{\prime}]$, say, and so that
$[x_{0}^{\prime},y_{0}^{\prime}],\dots,[x_{d}^{\prime},y_{d}^{\prime}]$ are
all distinct. If we take the limit as
$[x_{i}^{\prime},y_{i}^{\prime}]\to[x_{i},y_{i}]$ for each $i=0,\dots,d-1$, we
obtain a hyperplane $\Pi$ intersecting $\delta_{p}$ in
$\delta_{p}[x_{0},y_{0}],\dots,\delta_{p}[x_{d-1},y_{d-1}],\delta_{p}[x_{d}^{\prime\prime},y_{d}^{\prime\prime}]$,
say. Then
$F_{p}(x_{0},y_{0},\dots,x_{d-1},y_{d-1},x_{d}^{\prime\prime},y_{d}^{\prime\prime})=0$.
Since the multilinear form $F_{p}$ is non-trivial, it follows that
$[x_{d},y_{d}]=[x_{d}^{\prime\prime},y_{d}^{\prime\prime}]$. Therefore, $\Pi$
is a hyperplane that intersects $\delta_{p}$ in $\delta_{p}[x_{i},y_{i}]$,
$i=0,\dots,d$. ∎

The secant variety $\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})$ of the
rational normal curve $\nu_{d+1}$ in $\mathbb{C}\mathbb{P}^{d+1}$ is equal to
the set of points that lie on a proper secant or tangent line of $\nu_{d+1}$,
that is, on a line with intersection multiplicity at least $2$ with
$\nu_{d+1}$. We also define the real secant variety of $\nu_{d+1}$ to be the
set $\operatorname{Sec}_{\mathbb{R}}(\nu_{d+1})$ of points in
$\mathbb{R}\mathbb{P}^{d+1}$ that lie on a line that either intersects
$\nu_{d+1}$ in two distinct real points or is a tangent line of $\nu_{d+1}$.
The _tangent variety_ $\operatorname{Tan}_{\mathbb{F}}(\nu_{d+1})$ of
$\nu_{d+1}$ is defined to be the set of points in $\mathbb{F}\mathbb{P}^{d+1}$
that lie on a tangent line of $\nu_{d+1}$. We note that although
$\operatorname{Tan}_{\mathbb{R}}(\nu_{d+1})=\operatorname{Tan}_{\mathbb{C}}(\nu_{d+1})\cap\mathbb{R}\mathbb{P}^{d+1}$,
we only have a proper inclusion
$\operatorname{Sec}_{\mathbb{R}}(\nu_{d+1})\subset\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})\cap\mathbb{R}\mathbb{P}^{d+1}$
for $d\geqslant 2$.

We will need a concrete description of
$\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})$ and its relation to the
smoothness of the curves $\delta_{p}$. For any
$p\in\mathbb{F}\mathbb{P}^{d+1}$ and $k=2,\dots,d-1$, define the
$(k+1)\times(d-k+2)$ matrix

$M_{k}(p):=\begin{pmatrix}p_{0}&p_{1}&p_{2}&\dots&p_{d-k+1}\\\
p_{1}&p_{2}&p_{3}&\dots&p_{d-k+2}\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
p_{k}&p_{k+1}&p_{k+2}&\dots&p_{d+1}\end{pmatrix}.$

Suppose that $\delta_{p}$ has a double point, say
$\delta_{p}[x_{0},y_{0}]=\delta_{p}[x_{1},y_{1}]$. This is equivalent to $p$,
$\nu_{d+1}[x_{0},y_{0}]$, and $\nu_{d+1}[x_{1},y_{1}]$ being collinear, which
is equivalent to $p$ being on the secant variety of $\nu_{d+1}$. (In the
degenerate case where $[x_{0},y_{0}]=[x_{1},y_{1}]$, we have that
$p\in\operatorname{Tan}_{\mathbb{F}}(\nu_{d+1})$.) Then
$\delta_{p}[x_{0},y_{0}]$, $\delta_{p}[x_{1},y_{1}]$,
$\delta_{p}[x_{2},y_{2}]$,…, $\delta_{p}[x_{d},y_{d}]$ are on a hyperplane in
$\mathbb{F}\mathbb{P}^{d}$ for all
$[x_{2},y_{2}],\dots,[x_{d},y_{d}]\in\mathbb{F}\mathbb{P}^{1}$. It follows
that the coefficients of
$F_{p}(x_{0},y_{0},x_{1},y_{1},x_{2},y_{2},\dots,x_{d},y_{d})$ as a polynomial
in $x_{2},y_{2},\dots,x_{d},y_{d}$ all vanish, that is,

$p_{i}x_{0}x_{1}+p_{i+1}(x_{0}y_{1}+y_{0}x_{1})+p_{i+2}y_{0}y_{1}=0$

for all $i=0,\dots,d-1$. This can be written as
$[x_{0}x_{1},x_{0}y_{1}+y_{0}x_{1},y_{0}y_{1}]M_{2}(p)=0$. Conversely, if
$M_{2}(p)$ has rank $2$ with say $[c_{0},2c_{1},c_{2}]M_{2}(p)=0$, then there
is a non-trivial solution to the linear system with $c_{0}=x_{0}x_{1}$,
$c_{1}=x_{0}y_{1}+y_{0}x_{1}$, $c_{2}=y_{0}y_{1}$, and we have
$c_{0}x^{2}+2c_{1}xy+c_{2}y^{2}=(x_{0}x+y_{0}y)(x_{1}x+y_{1}y)$. In the
degenerate case where $[x_{0},y_{0}]=[x_{1},y_{1}]$, we have that the
quadratic form has repeated roots.

It follows that $M_{2}(p)$ has rank at most $2$ if and only if
$p\in\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})$ (also note that $M_{2}(p)$
has rank $1$ if and only if $p\in\nu_{d+1}$). We note for later use that since
the null space of $M_{2}(p)$ is $1$-dimensional if it has rank $2$, it follows
that each $p\in\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})$ lies on a unique
secant (which might degenerate to a tangent). This implies that $\delta_{p}$
has a unique singularity when
$p\in\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})\setminus{\nu_{d+1}}$, which is
a node if
$p\in\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})\setminus\operatorname{Tan}_{\mathbb{C}}(\nu_{d+1})$
and a cusp if
$p\in\operatorname{Tan}_{\mathbb{C}}(\nu_{d+1})\setminus{\nu_{d+1}}$. In the
real case there are two types of nodes. If
$p\in\operatorname{Sec}_{\mathbb{R}}(\nu_{d+1})\setminus\nu_{d+1}$, then the
roots $[x_{0},y_{0}],[x_{1},y_{1}]$ are real, and $\delta_{p}$ has either a
cusp when $p\in\operatorname{Tan}_{\mathbb{R}}(\nu_{d+1})\setminus\nu_{d+1}$
and $[x_{0},y_{0}]=[x_{1},y_{1}]$, or a crunode when
$p\in\operatorname{Sec}_{\mathbb{R}}(\nu_{d+1})\setminus\operatorname{Tan}_{\mathbb{R}}(\nu_{d+1})$
and $[x_{0},y_{0}]$ and $[x_{1},y_{1}]$ are distinct roots of the real binary
quadratic form $c_{0}x^{2}+2c_{1}xy+c_{2}y^{2}$. If
$p\in\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})\setminus\operatorname{Sec}_{\mathbb{R}}(\nu_{d+1})\cap\mathbb{R}\mathbb{P}^{d+1}$
then the quadratic form has conjugate roots
$[x_{0},y_{0}]=[\overline{x_{1}},\overline{y_{1}}]$ and $\delta_{p}$ has an
acnode.

If $p\notin\operatorname{Sec}(\nu_{d+1})$, then $\delta_{p}$ is a smooth curve
of degree $d+1$. It follows that $\delta_{p}$ is singular if and only if
$p\in\operatorname{Sec}(\nu_{d+1})\setminus{\nu_{d+1}}$. For the purposes of
this paper, we make the following definitions.

######  Definition.

A _rational singular curve_ is an irreducible non-degenerate singular rational
curve of degree $d+1$ in $\mathbb{C}\mathbb{P}^{d}$. In the real case, a
_rational cuspidal curve_ , _rational crunodal curve_ , or _rational acnodal
curve_ is a rational singular curve isomorphic to a singular planar cubic with
a cusp, crunode, or acnode respectively.

In particular, we have shown the case $k=2$ of the following well-known
result.

######  Proposition 3.5 ([H92]*Proposition 9.7).

Let $d\geqslant 3$. For any $k=2,\dots,d-1$, the secant variety of $\nu_{d+1}$
is equal to the locus of all $[p_{0},p_{1},\dots,p_{d+1}]$ such that
$M_{k}(p)$ has rank at most $2$.

######  Corollary 3.6.

Let $d\geqslant 3$. For any $k=2,\dots,d-1$ and
$p\in\mathbb{C}\mathbb{P}^{d+1}\setminus\nu_{d+1}$, the curve $\delta_{p}$ of
degree $d+1$ in $\mathbb{C}\mathbb{P}^{d}$ is singular if and only if
$\operatorname{rank}M_{k}(p)\leqslant 2$.

We next use Corollary 3.6 to show that the projection of a smooth rational
curve of degree $d+1$ in $\mathbb{C}\mathbb{P}^{d}$ from a generic point on
the curve is again smooth when $d\geqslant 4$. This is not true for $d=3$, as
there is a trisecant through each point of a quartic curve of the second
species in $3$-space. (The union of the trisecants form the unique quadric on
which the curve lies [H92]*Exercise 8.13.)

######  Lemma 3.7.

Let $\delta_{p}$ be a smooth rational curve of degree $d+1$ in
$\mathbb{C}\mathbb{P}^{d}$, $d\geqslant 4$. Then for all but at most three
points $q\in\delta_{p}$, the projection
$\overline{\pi_{q}(\delta_{p}\setminus\\{q\\})}$ is a smooth rational curve of
degree $d$ in $\mathbb{C}\mathbb{P}^{d-1}$.

###### Proof.

Let $q=\delta_{p}[x_{0},y_{0}]$. Suppose that
$\overline{\pi_{q}(\delta_{p}\setminus\\{q\\})}$ is singular. Then there exist
$[x_{1},y_{1}]$ and $[x_{2},y_{2}]$ such that
$\pi_{q}(\delta_{p}[x_{1},y_{1}])=\pi_{q}(\delta_{p}[x_{2},y_{2}])$ and the
points $\delta_{p}[x_{0},y_{0}]$, $\delta_{p}[x_{1},y_{1}]$, and
$\delta_{p}[x_{2},y_{2}]$ are collinear. Then for arbitrary
$[x_{3},y_{3}],\dots,[x_{d},y_{d}]\in\mathbb{C}\mathbb{P}^{1}$, the points
$\delta_{p}[x_{i},y_{i}]$, $i=0,\dots,d$ are on a hyperplane, so by Lemma 3.4,
$F_{p}(x_{0},y_{0},\dots,x_{d},y_{d})$ is identically $0$ as a polynomial in
$x_{3},y_{3},\dots,x_{d},y_{d}$. The coefficients of this polynomial are of
the form

$p_{i}x_{0}x_{1}x_{2}+p_{i+1}(x_{0}x_{1}y_{2}+x_{0}y_{1}x_{2}+y_{0}x_{1}x_{2})+p_{i+2}(x_{0}y_{1}y_{2}+y_{0}x_{1}y_{2}+y_{0}y_{1}x_{2})+p_{i+3}y_{0}y_{1}y_{2}$

for $i=0,\dots,d-2$. This means that the linear system
$[c_{0},3c_{1},3c_{2},c_{3}]M_{3}(p)=0$ has a non-trivial solution
$c_{0}=x_{0}x_{1}x_{2}$,
$3c_{1}=x_{0}x_{1}y_{2}+x_{0}y_{1}x_{2}+y_{0}x_{1}x_{2}$,
$3c_{2}=x_{0}y_{1}y_{2}+y_{0}x_{1}y_{2}+y_{0}y_{1}x_{2}$,
$c_{3}=y_{0}y_{1}y_{2}$. The binary cubic form
$c_{0}x^{3}+3c_{1}x^{2}y+c_{2}xy^{2}+c_{3}y^{3}$ then has the factorisation
$(x_{0}x+y_{0}y)(x_{1}x+y_{1}y)(x_{2}x+y_{2}y)$, hence its roots give the
collinear points on $\delta_{p}$. Since $\delta_{p}$ is smooth, $M_{3}(p)$ has
rank at least $3$ by Corollary 3.6, and so the cubic form is unique up to
scalar multiples. It follows that there are at most three points $q$ such that
the projection $\overline{\pi_{q}(\delta_{p}\setminus\\{q\\})}$ is not smooth.
∎

We need the following theorem on the fundamental binary form $f_{p}$ that is
essentially due to Sylvester [S51] to determine the natural group structure on
rational singular curves. Reznick [Rez2013] gives an elementary proof of the
generic case where $p$ does not lie on the tangent variety. (See also Kanev
[K99]*Lemma 3.1 and Iarrobino and Kanev [IK99]*Section 1.3.) We provide a very
elementary proof that includes the non-generic case.

######  Theorem 3.8 (Sylvester [S51]).

Let $d\geqslant 2$.

  1. ( )

If $p\in\operatorname{Tan}_{\mathbb{C}}(\nu_{d+1})$, then there exist binary
linear forms $L_{1},L_{2}$ such that $f_{p}(x,y)=L_{1}(x,y)^{d}L_{2}(x,y)$.
Moreover, if $p\notin\nu_{d+1}$ then $L_{1}$ and $L_{2}$ are linearly
independent, and if $p\in\mathbb{R}\mathbb{P}^{d+1}$ then $L_{1}$ and $L_{2}$
are both real.

  2. ( )

If
$p\in\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})\setminus\operatorname{Tan}_{\mathbb{C}}(\nu_{d+1})$,
then there exist linearly independent binary linear forms $L_{1},L_{2}$ such
that $f_{p}(x,y)=L_{1}(x,y)^{d+1}-L_{2}(x,y)^{d+1}$. Moreover, if
$p\in\mathbb{R}\mathbb{P}^{d+1}\setminus\operatorname{Sec}_{\mathbb{R}}(\nu_{d+1})$
then $L_{1}$ and $L_{2}$ are complex conjugates, while if
$p\in\operatorname{Sec}_{\mathbb{R}}(\nu_{d+1})$ then there exist linearly
independent real binary linear forms $L_{1},L_{2}$ such that
$f_{p}(x,y)=L_{1}(x,y)^{d+1}\pm L_{2}(x,y)^{d+1}$, where we can always choose
the lower sign when $d$ is even, and otherwise depends on $p$.

###### Proof.

( ) ‣ 3.8: We work over $\mathbb{F}\in\\{\mathbb{R},\mathbb{C}\\}$. Let
$p=[p_{0},p_{1},\dots,p_{d+1}]\in\operatorname{Tan}_{\mathbb{F}}(\nu_{d+1})$.
Let $p_{*}=\nu_{d+1}[\alpha_{1},\alpha_{2}]$ be the point on $\nu_{d+1}$ such
that the line $pp_{*}$ is tangent to $\nu_{d+1}$ (if $p\in\nu_{d+1}$, we let
$p_{*}=p$). We will show that

$f_{p}(x,y)=\sum_{i=0}^{d+1}p_{i}\binom{d+1}{i}x^{d+1-i}y^{i}=(\alpha_{2}x-\alpha_{1}y)^{d}(\beta_{2}x-\beta_{1}y)$
(4)

for some $[\beta_{1},\beta_{2}]\in\mathbb{F}\mathbb{P}^{1}$.

First consider the special case $\alpha_{1}=0$. Then $p_{*}=[1,0,\dots,0]$ and
the tangent to $\nu_{d+1}$ at $p_{*}$ is the line
$x_{2}=x_{3}=\dots=x_{d+1}=0$. It follows that
$f_{p}(x,y)=p_{0}x^{d+1}+p_{1}(d+1)x^{d}y=(1x-0y)^{d}(p_{0}x+p_{1}(d+1)y)$. If
$p_{1}=0$, then $p=p_{*}\in\nu_{d+1}$. Thus, if $p\notin\nu_{d+1}$, then
$p_{1}\neq 0$, and $x$ and $p_{0}x+p_{1}(d+1)y$ are linearly independent.

We next consider the general case $\alpha_{1}\neq 0$. Equating coefficients in
(4), we see that we need to find $[\beta_{1},\beta_{2}]$ such that

$p_{i}\binom{d+1}{i}=\binom{d}{i}\alpha_{2}^{d-i}(-\alpha_{1})^{i}\beta_{2}-\binom{d}{i-1}\alpha_{2}^{d-i+1}(-\alpha_{1})^{i-1}\beta_{1}$

for each $i=0,\dots,d+1$, where we use the convention
$\binom{d}{-1}=\binom{d}{d+1}=0$. This can be simplified to

$p_{i}=\left(1-\frac{i}{d+1}\right)\alpha_{2}^{d-i}(-\alpha_{1})^{i}\beta_{2}-\frac{i}{d+1}\alpha_{2}^{d-i+1}(-\alpha_{1})^{i-1}\beta_{1}.$
(5)

Since we are working projectively, we can fix the value of $\beta_{1}$ from
the instance $i=d+1$ of (5) to get

$p_{d+1}=-(-\alpha_{1})^{d}\beta_{1}.$ (6)

If $p_{d+1}\neq 0$, we can divide (5) by (6). After setting
$\alpha=\alpha_{2}/\alpha_{1}$, $\beta=\beta_{2}/\beta_{1}$, and
$a_{i}=p_{i}/p_{d+1}$, we then have to show that for some
$\beta\in\mathbb{F}$,

$a_{i}=-\left(1-\frac{i}{d+1}\right)(-\alpha)^{d-i}\beta+\frac{i}{d+1}(-\alpha)^{d-i+1}$
(7)

for each $i=0,\dots,d$. We next calculate in the affine chart $x_{d+1}=1$
where the rational normal curve becomes
$\nu_{d+1}(t)=((-t)^{d+1},(-t)^{d},\dots,-t)$, $p=(a_{0},\dots,a_{d})$, and
$p_{*}=\nu_{d+1}(\alpha)$. The tangency condition means that $p_{*}-p$ is a
scalar multiple of

$\nu_{d+1}^{\prime}(\alpha)=((d+1)(-\alpha)^{d},d(-\alpha)^{d-1},\dots,2\alpha,-1),$

that is, we have for some $\lambda\in\mathbb{F}$ that
$(-\alpha)^{d+1-i}-a_{i}=\lambda(d+1-i)(-\alpha)^{d-i}$ for all $i=0,\dots,d$.
Set $\beta=\alpha+\lambda(d+1)$. Then
$(-\alpha)^{d+1-i}-a_{i}=(\beta-\alpha)(1-\frac{i}{d+1})(-\alpha)^{d-i}$, and
we have

$\displaystyle a_{i}$
$\displaystyle=(-\alpha)^{d+1-i}-(\beta-\alpha)\left(1-\frac{i}{d+1}\right)(-\alpha)^{d-i}$
$\displaystyle=-\left(1-\frac{i}{d+1}\right)(-\alpha)^{d-i}\beta+\frac{i}{d+1}(-\alpha)^{d-i+1},$

giving (7) as required. If $\alpha=\beta$, then $\lambda=0$ and
$p=p_{*}\in\nu_{d+1}$. Thus, if $p\notin\nu_{d+1}$, then $\alpha\neq\beta$,
and $\alpha_{2}x-\alpha_{1}y$ and $\beta_{2}x-\beta_{1}y$ are linearly
independent.

We still have to consider the case $p_{d+1}=0$. Then $\beta_{1}=0$ and we need
to find $\beta_{2}$ such that

$p_{i}=\left(1-\frac{i}{d+1}\right)\alpha_{2}^{d-i}(-\alpha_{1})^{i}\beta_{2}$
(8)

for all $i=0,\dots,d$. Since $p_{d+1}=0$, we have that
$\nu_{d+1}^{\prime}(\alpha)$ is parallel to $(p_{0},\dots,p_{d})$, that is,

$p_{i}=\lambda(d+1-i)(-\alpha)^{d-i}$

for some $\lambda\in\mathbb{F}^{*}$. Set
$\beta_{2}=\lambda(d+1)/(-\alpha_{1})^{d}$. Then

$p_{i}=\frac{(-\alpha_{1})^{d}\beta_{2}}{d+1}(d+1-i)\left(\frac{\alpha_{2}}{-\alpha_{1}}\right)^{d-i}=\left(1-\frac{i}{d+1}\right)\alpha_{2}^{d-i}(-\alpha_{1})^{i}\beta_{2},$

again giving (8) as required. Note that since $\alpha_{1}\neq 0$ but
$\beta_{1}=0$, $\alpha_{2}x-\alpha_{1}y$ and $\beta_{2}x-\beta_{1}y$ are
linearly independent. Note also that since $\lambda\neq 0$, we have
$\beta_{2}\neq 0$ and $p\neq[1,0,\dotsc,0]$, hence $p\notin\nu_{d+1}$.

( ) ‣ 3.8: Let
$p=[p_{0},\dots,p_{d+1}]\in\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})\setminus\operatorname{Tan}_{\mathbb{C}}(\nu_{d+1})$,
and suppose that $p$ lies on the secant line through the distinct points
$p_{1}:=\nu_{d+1}[\alpha_{1},\alpha_{2}]$ and
$p_{2}:=\nu_{d+1}[\beta_{1},\beta_{2}]$. Since $p,p_{1},p_{2}$ are distinct
and collinear, there exist $\mu_{1},\mu_{2}\in\mathbb{C}^{*}$ such that
$p=\mu_{1}p_{1}+\mu_{2}p_{2}$. This means that for $i=0,\dotsc,d+1$, we have

$p_{i}=\mu_{1}(-\alpha_{1})^{i}\alpha_{2}^{d+1-i}+\mu_{2}(-\beta_{1})^{i}\beta_{2}^{d+1-i}.$

Then

$\displaystyle f_{p}(x,y)$
$\displaystyle=\sum_{i=0}^{d+1}p_{i}\binom{d+1}{i}x^{d+1-i}y^{i}$
$\displaystyle=\mu_{1}\sum_{i=0}^{d+1}\binom{d+1}{i}(\alpha_{2}x)^{d+1-i}(-\alpha_{1}y)^{i}+\mu_{2}\sum_{i=0}^{d+1}\binom{d+1}{i}(\beta_{2}x)^{d+1-i}(-\beta_{1}y)^{i}$
$\displaystyle=\mu_{1}(\alpha_{2}x-\alpha_{1}y)^{d+1}+\mu_{2}(\beta_{2}x-\beta_{1}y)^{d+1}$
$\displaystyle=L_{1}(x,y)^{d+1}-L_{2}(x,y)^{d+1}$

where the linear forms $L_{1},L_{2}$ are linearly independent.

If
$p\in\mathbb{R}\mathbb{P}^{d+1}\setminus{\operatorname{Sec}_{\mathbb{R}}(\nu_{d+1})}$,
then $f_{p}$ is real and $p_{1}$ and $p_{2}$ are non-real points. Taking
conjugates, we have

$p=\overline{\mu_{1}}\nu_{d+1}[\overline{\alpha_{1}},\overline{\alpha_{2}}]+\overline{\mu_{2}}\nu_{d+1}[\overline{\beta_{1}},\overline{\beta_{2}}]$

as vectors, and because of the uniqueness of secants of the rational normal
curve through a given point, we obtain $\overline{\mu_{1}}=\mu_{2}$ and
$\nu_{d+1}[\overline{\alpha_{1}},\overline{\alpha_{2}}]=\nu_{d+1}[\beta_{1},\beta_{2}]$,
hence $\overline{\alpha_{1}}=\beta_{1}$ and $\overline{\alpha_{2}}=\beta_{2}$.
It follows that $\overline{L_{1}(x,y)}=L_{2}(\overline{x},\overline{y})$.

If $p\in\operatorname{Sec}_{\mathbb{R}}(\nu_{d+1})$, then $p_{1}$ and $p_{2}$
are real, so
$[\mu_{1},\mu_{2}],[\alpha_{1},\alpha_{2}],[\beta_{1},\beta_{2}]\in\mathbb{R}\mathbb{P}^{1}$,
and we obtain $f_{p}(x,y)=L_{1}^{d+1}\pm L_{2}^{d+1}$ for some linearly
independent $L_{1},L_{2}$ over $\mathbb{R}$, where the choice of sign depends
on $p$. ∎

We are now in a position to describe the group laws on rational singular
curves. We first note the effect of a change of coordinates on the
parametrisation of $\delta_{p}$. Let
$\varphi\colon\mathbb{F}\mathbb{P}^{1}\to\mathbb{F}\mathbb{P}^{1}$ be a
projective transformation. Then $\nu_{d+1}\circ\varphi$ is a reparametrisation
of the rational normal curve. It is not difficult to see that there exists a
projective transformation
$\psi\colon\mathbb{F}\mathbb{P}^{d+1}\to\mathbb{F}\mathbb{P}^{d+1}$ such that
$\nu_{d+1}\circ\varphi=\psi\circ\nu_{d+1}$. It follows that if we
reparametrise $\delta_{p}$ using $\varphi$, we obtain

$\delta_{p}\circ\varphi=\pi_{p}\circ\nu_{d+1}\circ\varphi=\pi_{p}\circ\psi\circ\nu_{d+1}=\psi^{\prime}\circ\pi_{\psi^{-1}(p)}\circ\nu_{d+1}\cong\delta_{\psi^{-1}(p)},$

where $\psi^{\prime}\colon\mathbb{F}\mathbb{P}^{d}\to\mathbb{F}\mathbb{P}^{d}$
is an appropriate projective transformation such that first transforming
$\mathbb{F}\mathbb{P}^{d+1}$ with $\psi$ and then projecting from $p$ is the
same as projecting from $\psi^{-1}(p)$ and then transforming
$\mathbb{F}\mathbb{P}^{d}$ with $\psi^{\prime}$. So by reparametrising
$\delta_{p}$, we obtain $\delta_{p^{\prime}}$ for some other point
$p^{\prime}$ that is in the orbit of $p$ under the action of projective
transformations that fix $\nu_{d+1}$.

Since
$\delta_{p}\circ\varphi[x_{0},y_{0}],\dots,\delta_{p}\circ\varphi[x_{d},y_{d}]$
lie on a hyperplane if and only if the $\delta_{\psi^{-1}(p)}[x_{i},y_{i}]$’s
are on a hyperplane, it follows from Lemma 3.4 that
$F_{p}(\varphi(x_{0},y_{0}),\dots,\varphi(x_{d},y_{d}))$ is a scalar multiple
of $F_{\psi^{-1}(p)}(x_{0},y_{0},\dots,x_{d},y_{d})$, in which case
$f_{p}\circ\varphi=f_{\psi^{-1}(p)}$ up to a scalar multiple. Thus, we obtain
the same reparametrisation of the fundamental binary form $f_{p}$.

######  Proposition 3.9.

A rational singular curve $\delta_{p}$ in $\mathbb{C}\mathbb{P}^{d}$ has a
natural group structure on its subset of smooth points $\delta_{p}^{*}$ such
that $d+1$ points in $\delta_{p}^{*}$ lie on a hyperplane if and only if they
sum to the identity. This group is isomorphic to $(\mathbb{C},+)$ if the
singularity of $\delta_{p}$ is a cusp and isomorphic to
$(\mathbb{C}^{*},\cdot)$ if the singularity is a node.

If the curve is real and cuspidal or acnodal, then it has a group isomorphic
to $(\mathbb{R},+)$ or $\mathbb{R}/\mathbb{Z}$ depending on whether the
singularity is a cusp or an acnode, such that $d+1$ points in $\delta_{p}^{*}$
lie on a hyperplane if and only if they sum to the identity. If the curve is
real and the singularity is a crunode, then the group is isomorphic to
$(\mathbb{R},+)\times\mathbb{Z}_{2}$, but $d+1$ points in $\delta_{p}^{*}$ lie
on a hyperplane if and only if they sum to $(0,0)$ or $(0,1)$, depending on
$p$.

###### Proof.

First suppose $\delta_{p}$ is cuspidal and
$\mathbb{F}\in\\{\mathbb{R},\mathbb{C}\\}$, so that
$p\in\operatorname{Tan}_{\mathbb{F}}(\nu_{d+1})\setminus{\nu_{d+1}}$. By
Theorem 3.8, $f_{p}=L_{1}^{d}L_{2}$ for some linearly independent linear forms
$L_{1}$ and $L_{2}$. By choosing $\varphi$ appropriately, we may assume
without loss of generality that $L_{1}(x,y)=x$ and $L_{2}(x,y)=(d+1)y$, so
that $f_{p}(x,y)=(d+1)x^{d}y$ and $p=[0,1,0,\dots,0]$, with the cusp of
$\delta_{p}$ at $\delta_{p}[0,1]$. It follows that the polarisation of $f_{p}$
is $F_{p}(x_{0},y_{0},\dotsc,x_{d},y_{d})=P_{1}=x_{0}x_{1}\dotsb
x_{d}\sum_{i=0}^{d}y_{i}/x_{i}$. For $[x_{i},y_{i}]\neq[0,1]$, $i=0,\dots,d$,
the points $\delta_{p}[x_{i},y_{i}]$ are on a hyperplane if and only if
$\sum_{i=0}^{d}y_{i}/x_{i}=0$. Thus we identify
$\delta_{p}[x,y]\in\delta_{p}^{*}$ with $y/x\in\mathbb{F}$, and the group is
$(\mathbb{F},+)$.

Next suppose $\delta_{p}$ is nodal, so that
$p\in\operatorname{Sec}_{\mathbb{C}}(\nu_{d+1})\setminus\operatorname{Tan}_{\mathbb{C}}(\nu_{d+1})$.
By Theorem 3.8, $f_{p}=L_{1}^{d+1}-L_{2}^{d+1}$ for some linearly independent
linear forms $L_{1}$ and $L_{2}$. Again by choosing $\varphi$ appropriately,
we may assume without loss of generality that $L_{1}(x,y)=x$ and
$L_{2}(x,y)=y$, so that $f_{p}(x,y)=x^{d+1}-y^{d+1}$ and $p=[1,0,\dots,0,-1]$,
with the node of $\delta_{p}$ at $\delta_{p}[0,1]=\delta_{p}[1,0]$. The
polarisation of $f_{p}$ is
$F_{p}(x_{0},y_{0},\dots,x_{d},y_{d})=P_{0}-P_{d+1}=x_{0}x_{1}\dotsb
x_{d}-y_{0}y_{1}\dotsb y_{d}$. Therefore, $\delta_{p}[x_{i},y_{i}]$,
$i=0,\dotsc,d$, are on a hyperplane if and only if
$\prod_{i=0}^{d}y_{i}/x_{i}=1$. Thus we identify
$\delta_{p}[x,y]\in\delta_{p}^{*}$ with $y/x\in\mathbb{C}^{*}$, and the group
is $(\mathbb{C}^{*},\cdot)$.

Now suppose $\delta_{p}$ is real and the node is an acnode. Then the linearly
independent linear forms $L_{1}$ and $L_{2}$ given by Theorem 3.8 are
$L_{1}(x,y)=\alpha x+\beta y$ and
$L_{2}(x,y)=\overline{\alpha}x+\overline{\beta}y$ for some
$\alpha,\beta\in\mathbb{C}\setminus\mathbb{R}$. There exists
$\varphi\colon\mathbb{R}\mathbb{P}^{1}\to\mathbb{R}\mathbb{P}^{1}$ such that
$L_{1}\circ\varphi=x+iy$ and $L_{2}\circ\varphi=x-iy$, hence we may assume
after such a reparametrisation that $f_{p}(x,y)=(x+iy)^{d+1}-(x-iy)^{d+1}$ and
that the node is at $\delta_{p}[i,1]=\delta_{p}[-i,1]$. The polarisation of
$f_{p}$ is
$F_{p}(x_{0},y_{0},\dots,x_{d},y_{d})=\prod_{j=0}^{d}(x_{j}+iy_{j})-\prod_{j=0}^{d}(x_{j}-iy_{j})$,
and it follows that $\delta_{p}[x_{0},y_{0}],\dotsc,\delta_{p}[x_{d},y_{d}]$
are collinear if and only if
$\prod_{j=0}^{d}\frac{x_{j}+iy_{j}}{x_{j}-iy_{j}}=1$. We now identify
$\mathbb{R}\mathbb{P}^{1}$ with the circle
$\mathbb{R}/\mathbb{Z}\cong\left\\{z\in\mathbb{C}:|z|=1\right\\}$ using the
Möbius transformation $[x,y]\to\frac{x+iy}{x-iy}$.

It remains to consider the crunodal case. Then, similar to the complex nodal
case, we obtain after a reparametrisation that $\delta_{p}[x_{i},y_{i}]$,
$i=0,\dotsc,d$, are on a hyperplane if and only if
$\prod_{i=0}^{d}y_{i}/x_{i}=\pm 1$, where the sign depends on $p$. Thus we
identify $\delta_{p}[x,y]\in\delta_{p}^{*}$ with $y/x\in\mathbb{R}^{*}$, and
the group is $(\mathbb{R}^{*},\cdot)\cong\mathbb{R}\times\mathbb{Z}_{2}$,
where $\pm 1\in\mathbb{R}^{*}$ corresponds to
$(0,0),(0,1)\in\mathbb{R}\times\mathbb{Z}_{2}$ respectively. ∎

The group on an elliptic normal curve or a rational singular curve of degree
$d+1$ as described in Propositions 3.1 and 3.9 is not uniquely determined by
the property that $d+1$ points lie on a hyperplane if and only if they sum to
some fixed element $c$. Indeed, for any $t\in(\delta^{*},\oplus)$, $x\boxplus
y:=x\oplus y\oplus t$ defines another abelian group on $\delta^{*}$ with the
property that $d+1$ points lie on a hyperplane if and only if they sum to
$c\oplus dt$. However, these two groups are isomorphic in a natural way with
an isomorphism given by the translation map $x\mapsto x\ominus t$. The next
proposition show that we always get uniqueness up to some translation. It will
be used in Section 5.

######  Proposition 3.10.

Let $(G,\oplus,0)$ and $(G,\boxplus,0^{\prime})$ be abelian groups on the same
ground set, such that for some $d\geqslant 2$ and some $c,c^{\prime}\in G$,

$x_{1}\oplus\dotsb\oplus x_{d+1}=c\iff x_{1}\boxplus\dotsb\boxplus
x_{d+1}=c^{\prime}\quad\text{for all }x_{1},\dots,x_{d+1}\in G.$

Then $(G,\oplus,0)\to(G,\boxplus,0^{\prime}),x\mapsto x\boxminus 0=x\oplus
0^{\prime}$ is an isomorphism, and

$c^{\prime}=c\boxplus\underbrace{0\boxplus\dotsb\boxplus 0}_{\text{$d$
times}}=c\ominus(\underbrace{0^{\prime}\oplus\dotsb\oplus
0^{\prime}}_{\text{$d$ times}}).$

###### Proof.

It is clear that the cases $d\geqslant 3$ follow from the case $d=2$, which we
now show. First note that for any $x,y\in G$, $x\boxplus y\boxplus(c\ominus
x\ominus y)=c^{\prime}$ and $(x\oplus y)\boxplus 0\boxplus(c\ominus x\ominus
y)=c^{\prime}$, since $x\oplus y\oplus(c\ominus x\ominus y)=(x\oplus y)\oplus
0\oplus(c\ominus x\ominus y)=c$. Thus we have $x\boxplus y=(x\oplus y)\boxplus
0$, hence $(x\oplus y)\boxminus 0=x\boxplus y\boxminus 0\boxminus
0=(x\boxminus 0)\boxplus(y\boxminus 0)$. Similarly we have $x\oplus
y=(x\boxplus y)\oplus 0^{\prime}$, hence $x\boxplus y=x\oplus y\ominus
0^{\prime}$, so in particular $0^{\prime}=0\boxminus 0=0\oplus(\boxminus
0)\ominus 0^{\prime}$, and $\boxminus 0=0^{\prime}\oplus 0^{\prime}$. So we
also have $x\boxminus 0=x\oplus(\boxminus 0)\ominus 0^{\prime}=x\oplus
0^{\prime}$, and $(G,\oplus,0)\to(G,\boxplus,0^{\prime}),x\mapsto x\boxminus
0=x\oplus 0^{\prime}$ is an isomorphism. ∎

##  4 Structure theorem

We prove Theorem 1.1 in this section. The main idea is to induct on the
dimension $d$ via projection. We start with the following statement of the
slightly different case $d=3$, which is [LS18]*Theorem 1.1. Note that it
contains one more type that does not occur when $d\geqslant 4$.

######  Theorem 4.1.

Let $K>0$ and suppose $n\geqslant C\max\\{K^{8},1\\}$ for some sufficiently
large absolute constant $C>0$. Let $P$ be a set of $n$ points in
$\mathbb{R}\mathbb{P}^{3}$ with no $3$ points collinear. If $P$ spans at most
$Kn^{2}$ ordinary planes, then up to projective transformations, $P$ differs
in at most $O(K)$ points from a configuration of one of the following types:

  1. ( )

A subset of a plane;

  2. ( )

A subset of two disjoint conics lying on the same quadric with $\frac{n}{2}\pm
O(K)$ points of $P$ on each of the two conics;

  3. ( )

A coset of a subgroup of the smooth points of an elliptic or acnodal space
quartic curve.

We first prove the following weaker lemma using results from Section 2.

######  Lemma 4.2.

Let $d\geqslant 4$, $K>0$, and suppose $n\geqslant
C\max\\{d^{3}2^{d}K,(dK)^{8}\\}$ for some sufficiently large absolute constant
$C>0$. Let $P$ be a set of $n$ points in $\mathbb{R}\mathbb{P}^{d}$ where
every $d$ points span a hyperplane. If $P$ spans at most $K\binom{n-1}{d-1}$
ordinary hyperplanes, then all but at most $O(d2^{d}K)$ points of $P$ are
contained in a hyperplane or an irreducible non-degenerate curve of degree
$d+1$ that is either elliptic or rational and singular.

###### Proof.

We use induction on $d\geqslant 4$ to show that for all $K>0$ and all
$n\geqslant f(d,K)$, for all sets $P$ of $n$ points in
$\mathbb{R}\mathbb{P}^{d}$ with any $d$ points spanning a hyperplane, if $P$
has at most $K\binom{n-1}{d-1}$ ordinary hyperplanes, then all but at most
$g(d,K)$ points of $P$ are contained in a hyperplane or an irreducible non-
degenerate curve of degree $d+1$, and that if the curve is rational then it
has to be singular, where

$g(d,K):=\sum_{k=0}^{d}k^{3}2^{d-k}+C_{1}2^{d}(d-1)K$

and

$f(d,K):=d^{2}(g(d,K)+C_{2}d^{10})+C(d-1)^{8}K^{8}$

for appropriate $C_{1},C_{2}>0$ to be determined later and $C$ from Theorem
4.1. We assume that this holds in $\mathbb{R}\mathbb{P}^{d-1}$ if $d\geqslant
5$, while Theorem 4.1 takes the place of the induction hypothesis when $d=4$.

Let $P^{\prime}$ denote the set of points $p\in P$ such that there are at most
$\frac{d-1}{d-2}K\binom{n-2}{d-2}$ ordinary hyperplanes through $p$. By
counting incident point-ordinary-hyperplane pairs, we obtain

$dK\binom{n-1}{d-1}>(n-|P^{\prime}|)\frac{d-1}{d-2}K\binom{n-2}{d-2},$

which gives $|P^{\prime}|>n/(d-1)^{2}$. For any $p\in P^{\prime}$, the
projected set $\pi_{p}(P\setminus\\{p\\})$ has $n-1$ points and spans at most
$\frac{d-1}{d-2}K\binom{n-2}{d-2}$ ordinary $(d-2)$-flats in
$\mathbb{R}\mathbb{P}^{d-1}$, and any $d-1$ points of
$\pi_{p}(P\setminus\\{p\\})$ span a $(d-2)$-flat. To apply the induction
hypothesis, we need

$f(d,K)\geqslant 1+f(d-1,\tfrac{d-1}{d-2}K),$

as well as $f(3,K)\geqslant C\max\\{K^{8},1\\}$, both of which easily follow
from the definition of $f(d,K)$. Then all except $g(d-1,\frac{d-1}{d-2}K)$
points of $\pi_{p}(P\setminus\\{p\\})$ are contained in a $(d-2)$-flat or a
non-degenerate curve $\gamma_{p}$ of degree $d$ in
$\mathbb{R}\mathbb{P}^{d-1}$, which is either irreducible or possibly two
conics with $\frac{n}{2}\pm O(K)$ points on each when $d=4$.

If there exists a $p\in P^{\prime}$ such that all but at most
$g(d-1,\frac{d-1}{d-2}K)$ points of $\pi_{p}(P\setminus\\{p\\})$ are contained
in a $(d-2)$-flat, then we are done, since $g(d,K)>g(d-1,\frac{d-1}{d-2}K)$.
Thus we may assume without loss of generality that for all $p\in P^{\prime}$
we obtain a curve $\gamma_{p}$.

Let $p$ and $p^{\prime}$ be two distinct points of $P^{\prime}$. Then all but
at most $2g(d-1,\frac{d-1}{d-2}K)$ points of $P$ lie on the intersection
$\delta$ of the two cones $\overline{\pi^{-1}_{p}(\gamma_{p})}$ and
$\overline{\pi^{-1}_{p^{\prime}}(\gamma_{p^{\prime}})}$. Since the curves
$\gamma_{p}$ and $\gamma_{p^{\prime}}$ are $1$-dimensional, the two cones are
$2$-dimensional. Since their vertices $p$ and $p^{\prime}$ are distinct, the
cones do not have a common irreducible component, so their intersection is a
variety of dimension at most $1$. By Bézout’s theorem (Theorem 2.1), $\delta$
has total degree at most $d^{2}$, so has to have at least one $1$-dimensional
irreducible component. Let $\delta_{1},\dotsc,\delta_{k}$ be the
$1$-dimensional components of $\delta$, where $1\leqslant k\leqslant d^{2}$.
Let $\delta_{1}$ be the component with the most points of $P^{\prime}$ amongst
all the $\delta_{i}$, so that

$|P^{\prime}\cap\delta_{1}|\geqslant\frac{|P^{\prime}|-2g(d-1,\frac{d-1}{d-2}K)}{d^{2}}.$

Choose a $q\in P^{\prime}\cap\delta_{1}$ such that $\pi_{q}$ is generically
one-to-one on $\delta_{1}$. By Lemma 2.2 there are at most
$O(\deg(\delta_{1})^{4})=O(d^{8})$ exceptional points, so we need

$|P^{\prime}\cap\delta_{1}|>C_{2}d^{8}.$ (9)

Since $|P^{\prime}|>n/(d-1)^{2}$, we need

$\frac{\frac{n}{(d-1)^{2}}-2g(d-1,\frac{d-1}{d-2}K)}{d^{2}}>C_{2}d^{8},$

or equivalently, $n>(d-1)^{2}(2g(d-1,\frac{d-1}{d-2}K)+C_{2}d^{10})$. However,
this follows from the definition of $f(d,K)$. If $\pi_{q}$ does not map
$\delta_{1}\setminus\\{q\\}$ into $\gamma_{q}$, then by Bézout’s theorem
(Theorem 2.1), $n-1-g(d-1,\binom{d-1}{d-2}K)\leqslant d^{3}$. However, this
does not occur since $f(d,K)>g(d-1,\binom{d-1}{d-2}K)+d^{3}+1$. Thus,
$\pi_{q}$ maps $\delta_{1}\setminus\\{q\\}$ into $\gamma_{q}$, hence
$\delta_{1}$ is an irreducible curve of degree $d+1$ (or, when $d=4$, possibly
a twisted cubic containing at most $n/2+O(K)$ points of $P$).

We first consider the case where $\delta_{1}$ has degree $d+1$. We apply Lemma
2.4 to $\delta_{1}$ and each $\delta_{i}$, $i=2,\dots,k$, and for this we need
$|P^{\prime}\cap\delta_{1}|>C^{\prime\prime}d^{4}$, since
$\deg(\delta_{1})\leqslant d^{2}$ and $\sum_{i=2}^{d}\deg(\delta_{i})\leqslant
d^{2}$. However, this condition is implied by (9). Thus we find a
$q^{\prime}\in P^{\prime}\cap\delta_{1}$ such that
$\overline{\pi_{q^{\prime}}(\delta_{1}\setminus\\{q^{\prime}\\})}=\gamma_{q^{\prime}}$
as before, and in addition, the cone
$\overline{\pi_{q^{\prime}}^{-1}(\gamma_{q^{\prime}})}$ does not contain any
other $\delta_{i}$, $i=2,\dots,k$. Since all points of $P$ except
$2g(d-1,\frac{d-1}{d-2}K)+d^{2}$ lie on $\delta_{1}\cup\dots\cup\delta_{k}$,
we obtain by Bézout’s theorem (Theorem 2.1) that

$|P\setminus{\delta_{1}}|\leqslant
d(d^{2}-d-1)+d^{2}+2g(d-1,\tfrac{d-1}{d-2}K)<g(d,K).$

We next dismiss the case where $d=4$ and $\delta_{1}$ is a twisted cubic. We
redefine $P^{\prime}$ to be the set of points $p\in P$ such that there are at
most $12Kn^{2}$ ordinary hyperplanes through $p$. Then $|P^{\prime}|\geqslant
2n/3$. Since we have $|P\cap\delta_{1}|\leqslant n/2+O(K)$, by Lemma 2.3 there
exists $q^{\prime}\in P^{\prime}\setminus\delta_{1}$ such that the projection
from $q^{\prime}$ will map $\delta_{1}$ onto a twisted cubic in
$\mathbb{R}\mathbb{P}^{3}$. However, by Bézout’s theorem (Theorem 2.1) and
Theorem 4.1, $\pi_{q^{\prime}}(\delta_{1}\setminus\\{q^{\prime}\\})$ has to be
mapped onto a conic, which gives a contradiction.

Note that $g(d,K)=O(d2^{d}K)$ since $K=\Omega(1/d)$ by [BM17]*Theorem 2.4. We
have shown that all but $O(d2^{d}K)$ points of $P$ are contained in a
hyperplane or an irreducible non-degenerate curve $\delta$ of degree $d+1$. By
Proposition 3.2, this curve is either elliptic or rational. It remains to show
that if $\delta$ is rational, then it has to be singular. Similar to what was
shown above, we can find more than $3$ points $p\in\delta$ for which the
projection $\overline{\pi_{p}(\delta\setminus\\{p\\})}$ is a rational curve of
degree $d$ that is singular by the induction hypothesis. Lemma 3.7 now implies
that $\delta$ is singular. ∎

To get the coset structure on the curves as stated in Theorem 1.1, we use a
simple generalisation of an additive combinatorial result used by Green and
Tao [GT13]*Proposition A.5. This captures the principle that if a finite
subset of a group is almost closed, then it is close to a subgroup. The case
$d=3$ was shown in [LMMSSZ18].

######  Lemma 4.3.

Let $d\geqslant 2$. Let $A_{1},A_{2},\dotsc,A_{d+1}$ be $d+1$ subsets of some
abelian group $(G,\oplus)$, all of size within $K$ of $n$, where $K\leqslant
cn/d^{2}$ for some sufficiently small absolute constant $c>0$. Suppose there
are at most $Kn^{d-1}$ $d$-tuples $(a_{1},a_{2},\dotsc,a_{d})\in A_{1}\times
A_{2}\times\dotsb\times A_{d}$ for which $a_{1}\oplus a_{2}\oplus\dotsb\oplus
a_{d}\notin A_{d+1}$. Then there is a subgroup $H$ of $G$ and cosets $H\oplus
x_{i}$ for $i=1,\dotsc,d$ such that

$|A_{i}\mathbin{\triangle}(H\oplus
x_{i})|,\left|A_{d+1}\mathbin{\triangle}\left(H\oplus\bigoplus_{i=1}^{d}x_{i}\right)\right|=O(K).$

###### Proof.

We use induction on $d\geqslant 2$ to show that the symmetric differences in
the conclusion of the lemma have size at most
$C\prod_{i=1}^{d}(1+\frac{1}{i^{2}})K$ for some sufficiently large absolute
constant $C>0$. The base case $d=2$ is [GT13]*Proposition A.5.

Fix a $d\geqslant 3$. By the pigeonhole principle, there exists $b_{1}\in
A_{1}$ such that there are at most

$\frac{1}{n-K}Kn^{d-1}\leqslant\frac{1}{1-\frac{c}{d^{2}}}Kn^{d-2}$

$(d-1)$-tuples $(a_{2},\dotsc,a_{d})\in A_{2}\times\dotsb\times A_{d}$ for
which $b_{1}\oplus a_{2}\oplus\dotsb\oplus a_{d}\notin A_{d+1}$, or
equivalently $a_{2}\oplus\dotsb\oplus a_{d}\notin A_{d+1}\ominus b_{1}$. Since

$\frac{1}{1-\frac{c}{d^{2}}}K\leqslant\frac{c}{d^{2}-c}n\leqslant\frac{c}{(d-1)^{2}}n,$

we can use induction to get a subgroup $H$ of $G$ and $x_{2},\dotsc,x_{d}\in
G$ such that for $j=2,\dotsc,d$ we have

$|A_{j}\mathbin{\triangle}(H\oplus x_{j})|,\left|(A_{d+1}\ominus
b_{1})\mathbin{\triangle}\left(H\oplus\bigoplus_{j=2}^{d}x_{j}\right)\right|\leqslant
C\prod_{i=1}^{d-1}\left(1+\frac{1}{i^{2}}\right)\frac{1}{1-\frac{c}{d^{2}}}K.$

Since $|A_{d}\cap(H\oplus x_{d})|\geqslant
n-K-C\prod_{i=1}^{d-1}(1+\frac{1}{i^{2}})\frac{1}{1-\frac{c}{d^{2}}}K$, we
repeat the same pigeonhole argument on $A_{d}\cap(H\oplus x_{d})$ to find a
$b_{d}\in A_{d}\cap(H\oplus x_{d})$ such that there are at most

$\displaystyle\frac{1}{n-K-C\prod_{i=1}^{d-1}\left(1+\frac{1}{i^{2}}\right)\frac{1}{1-\frac{c}{d^{2}}}K}Kn^{d-1}$
$\displaystyle\leqslant\frac{1}{1-\frac{c}{d^{2}}-C\prod_{i=1}^{d-1}\left(1+\frac{1}{i^{2}}\right)\frac{c}{d^{2}-c}}Kn^{d-2}$
$\displaystyle\leqslant\frac{1}{1-C_{1}\frac{c}{d^{2}-c}}Kn^{d-2}$
$\displaystyle\leqslant\left(1+\frac{C_{2}c}{d^{2}-c}\right)Kn^{d-2}$
$\displaystyle\leqslant\left(1+\frac{1}{d^{2}}\right)Kn^{d-2}$

$(d-1)$-tuples $(a_{1},\dotsc,a_{d-1})\in A_{1}\times\dotsb A_{d-1}$ with
$a_{1}\oplus\dotsb\oplus a_{d-1}\oplus b_{d}\notin A_{d+1}$, for some absolute
constants $C_{1},C_{2}>0$ depending on $C$, by making $c$ sufficiently small.
Now $(1+\frac{1}{d^{2}})K\leqslant cn/(d-1)^{2}$, so by induction again, there
exist a subgroup $H^{\prime}$ of $G$ and elements
$x_{1},x_{2}^{\prime},\dotsc,x_{d-1}^{\prime}\in G$ such that for
$k=2,\dotsc,d-1$ we have

$|A_{1}\mathbin{\triangle}(H^{\prime}\oplus
x_{1})|,|A_{k}\mathbin{\triangle}(H^{\prime}\oplus
x_{k}^{\prime})|,\left|(A_{d+1}\ominus
b_{d})\mathbin{\triangle}\left(H^{\prime}\oplus
x_{1}\oplus\bigoplus_{k=2}^{d-1}x_{k}^{\prime}\right)\right|\leqslant
C\prod_{i=1}^{d-1}\left(1+\frac{1}{i^{2}}\right)\left(1+\frac{1}{d^{2}}\right)K.$

From this, it follows that $|(H\oplus x_{k})\cap(H^{\prime}\oplus
x_{k}^{\prime})|\geqslant n-K-2C\prod_{i=1}^{d}(1+\frac{1}{i^{2}})K=n-O(K)$.
Since $(H\oplus x_{k})\cap(H^{\prime}\oplus x_{k}^{\prime})$ is non-empty, it
has to be a coset of $H^{\prime}\cap H$. If $H^{\prime}\neq H$, then
$|H^{\prime}\cap H|\leqslant n/2+O(K)$, a contradiction since $c$ is
sufficiently small. Therefore, $H=H^{\prime}$, and $H\oplus
x_{k}=H^{\prime}\oplus x_{k}^{\prime}$. So we have

$|A_{i}\mathbin{\triangle}(H\oplus
x_{i})|,\left|A_{d+1}\mathbin{\triangle}\left(H\oplus\bigoplus_{\ell=1}^{d-1}x_{\ell}\oplus
b_{d}\right)\right|\leqslant C\prod_{i=1}^{d}\left(1+\frac{1}{i^{2}}\right)K.$

Since $b_{d}\in H\oplus x_{d}$, we also obtain

$\left|A_{d+1}\mathbin{\triangle}\left(H\oplus\bigoplus_{i=1}^{d}x_{i}\right)\right|\leqslant
C\prod_{i=1}^{d}\left(1+\frac{1}{i^{2}}\right)K.\qed$

To apply Lemma 4.3, we first need to know that removing $K$ points from a set
does not change the number of ordinary hyperplanes it spans by too much.

######  Lemma 4.4.

Let $P$ be a set of $n$ points in $\mathbb{R}\mathbb{P}^{d}$, $d\geqslant 2$,
where every $d$ points span a hyperplane. Let $P^{\prime}$ be a subset that is
obtained from $P$ by removing at most $K$ points. If $P$ spans $m$ ordinary
hyperplanes, then $P^{\prime}$ spans at most $m+\frac{1}{d}K\binom{n-1}{d-1}$
ordinary hyperplanes.

###### Proof.

Fix a point $p\in P$. Since every $d$ points span a hyperplane, there are at
most $\binom{n-1}{d-1}$ sets of $d$ points from $P$ containing $p$ that span a
hyperplane through $p$. Thus, the number of $(d+1)$-point hyperplanes through
$p$ is at most $\frac{1}{d}\binom{n-1}{d-1}$, since a set of $d+1$ points that
contains $p$ has $d$ subsets of size $d$ that contain $p$. If we remove points
of $P$ one-by-one to obtain $P^{\prime}$, we thus create at most
$\frac{1}{d}K\binom{n-1}{d-1}$ ordinary hyperplanes. ∎

The following lemma then translates the additive combinatorial Lemma 4.3 to
our geometric setting.

######  Lemma 4.5.

Let $d\geqslant 4$, $K>0$, and suppose $n\geqslant C(d^{3}K+d^{4})$ for some
sufficiently large absolute constant $C>0$. Let $P$ be a set of $n$ points in
$\mathbb{R}\mathbb{P}^{d}$ where every $d$ points span a hyperplane. Suppose
$P$ spans at most $K\binom{n-1}{d-1}$ ordinary hyperplanes, and all but at
most $dK$ points of $P$ lie on an elliptic normal curve or a rational singular
curve $\delta$. Then $P$ differs in at most $O(dK+d^{2})$ points from a coset
$H\oplus x$ of a subgroup $H$ of $\delta^{*}$, the smooth points of $\delta$,
for some $x$ such that $(d+1)x\in H$. In particular, $\delta$ is either an
elliptic normal curve or a rational acnodal curve.

###### Proof.

Let $P^{\prime}=P\cap\delta^{*}$. Then by Lemma 4.4, $P^{\prime}$ spans at
most $K\binom{n-1}{d-1}+d\frac{1}{d}K\binom{n-1}{d-1}=2K\binom{n-1}{d-1}$
ordinary hyperplanes.

First suppose $\delta$ is an elliptic normal curve or a rational cuspidal or
acnodal curve. If $a_{1},\dotsc,a_{d}\in\delta^{*}$ are distinct, then by
Propositions 3.1 and 3.9, the hyperplane through $a_{1},\dotsc,a_{d}$ meets
$\delta$ again in the unique point $a_{d+1}=\ominus(a_{1}\oplus\dotsb\oplus
a_{d})$. This implies that $a_{d+1}\in P^{\prime}$ for all but at most
$d!O(K\binom{n-1}{d-1})$ $d$-tuples $(a_{1},\dotsc,a_{d})\in(P^{\prime})^{d}$
with all $a_{i}$ distinct. There are also at most $\binom{d}{2}n^{d-1}$
$d$-tuples $(a_{1},\dotsc,a_{d})\in(P^{\prime})^{d}$ for which the $a_{i}$ are
not all distinct. Thus, $a_{1}\oplus\dotsb\oplus a_{d}\in\ominus P^{\prime}$
for all but at most $O((dK+d^{2})n^{d-1})$ $d$-tuples
$(a_{1},\dotsc,a_{d})\in(P^{\prime})^{d}$. Applying Lemma 4.3 with
$A_{1}=\dotsb=A_{d}=P^{\prime}$ and $A_{d+1}=\ominus P^{\prime}$, we obtain a
finite subgroup $H$ of $\delta^{*}$ and a coset $H\oplus x$ such that
$|P^{\prime}\mathbin{\triangle}(H\oplus x)|=O(dK+d^{2})$ and $|\ominus
P^{\prime}\mathbin{\triangle}(H\oplus dx)|=O(dK+d^{2})$, the latter being
equivalent to $|P^{\prime}\mathbin{\triangle}(H\ominus dx)|=O(dK+d^{2})$. Thus
we have $|(H\oplus x)\mathbin{\triangle}(H\ominus dx)|=O(dK+d^{2})$, which
implies $(d+1)x\in H$. Also, $\delta$ cannot be cuspidal, otherwise by
Proposition 3.9 we have $\delta^{*}\cong(\mathbb{R},+)$, which has no finite
subgroup of order greater than $1$.

Now suppose $\delta$ is a rational crunodal curve. By Proposition 3.9, there
is a bijective map
$\varphi:(\mathbb{R},+)\times\mathbb{Z}_{2}\rightarrow\delta^{*}$ such that
$d+1$ points in $\delta^{*}$ lie in a hyperplane if and only if they sum to
$h$, where $h=\varphi(0,0)$ or $\varphi(0,1)$ depending on the curve $\delta$.
If $h=\varphi(0,0)$ then the above argument follows through, and we obtain a
contradiction as we have by Proposition 3.9 that
$\delta^{*}\cong(\mathbb{R},+)\times\mathbb{Z}_{2}$, which has no finite
subgroup of order greater than $2$. Otherwise, the hyperplane through distinct
$a_{1},\dotsc,a_{d}\in\delta^{*}$ meets $\delta$ again in the unique point
$a_{d+1}=\varphi(0,1)\ominus(a_{1}\oplus\dotsb\oplus a_{d})$. As before, this
implies that $a_{d+1}\in P^{\prime}$ for all but at most
$O((dK+d^{2})n^{d-1})$ $d$-tuples $(a_{1},\dotsc,a_{d})\in(P^{\prime})^{d}$,
or equivalently $a_{1}\oplus\dotsb\oplus a_{d}\in\varphi(0,1)\ominus
P^{\prime}$. Applying Lemma 4.3 with $A_{1}=\dotsb=A_{d}=P^{\prime}$ and
$A_{d+1}=\varphi(0,1)\ominus P^{\prime}$, we obtain a finite subgroup $H$ of
$\delta^{*}$, giving a contradiction as before. ∎

We can now prove Theorem 1.1.

###### Proof of Theorem 1.1.

By Lemma 4.2, all but at most $O(d2^{d}K)$ points of $P$ are contained in a
hyperplane or an irreducible curve $\delta$ of degree $d+1$ that is either
elliptic or rational and singular. In the prior case, we get Case ( ) ‣ 1.1
of the theorem, so suppose we are in the latter case. We then apply Lemma 4.5
to obtain Case ( ) ‣ 1.1 of the theorem, completing the proof. ∎

##  5 Extremal configurations

We prove Theorems 1.2 and 1.3 in this section. It will turn out that
minimising the number of ordinary hyperplanes spanned by a set is equivalent
to maximising the number of $(d+1)$-point planes, thus we can apply Theorem
1.1 in both theorems. Then we only have two cases to consider, where most of
our point set is contained either in a hyperplane or a coset of a subgroup of
an elliptic normal curve or the smooth points of a rational acnodal curve.

The first case is easy, and we get the following lower bound.

######  Lemma 5.1.

Let $d\geqslant 4$, $K\geqslant 1$, and let $n\geqslant 2dK$. Let $P$ be a set
of $n$ points in $\mathbb{R}\mathbb{P}^{d}$ where every $d$ points span a
hyperplane. If all but $K$ points of $P$ lie on a hyperplane, then $P$ spans
at least $\binom{n-1}{d-1}$ ordinary hyperplanes, with equality if and only if
$K=1$.

###### Proof.

Let $\Pi$ be a hyperplane with $|P\cap\Pi|=n-K$. Since $n-K>d$, any ordinary
hyperplane spanned by $P$ must contain at least one point not in $\Pi$. Let
$m_{i}$ be the number of hyperplanes containing exactly $d-1$ points of
$P\cap\Pi$ and exactly $i$ points of $P\setminus\Pi$, $i=1,\dots,K$. Then the
number of unordered $d$-tuples of elements from $P$ with exactly $d-1$
elements in $\Pi$ is

$K\binom{n-K}{d-1}=m_{1}+2m_{2}+3m_{3}+\dots+Km_{K}.$

Now consider the number of unordered $d$-tuples of elements from $P$ with
exactly $d-2$ elements in $\Pi$, which equals $\binom{K}{2}\binom{n-K}{d-2}$.
One way to generate such a $d$-tuple is to take one of the $m_{i}$ hyperplanes
containing $i$ points of $P\setminus\Pi$ and $d-1$ points of $P\cap\Pi$,
choose two of the $i$ points, and remove one of the $d-1$ points. Since any
$d$ points span a hyperplane, there is no overcounting. This gives

$\displaystyle\binom{K}{2}\binom{n-K}{d-2}$
$\displaystyle\geqslant(d-1)\left(\binom{2}{2}m_{2}+\binom{3}{2}m_{3}+\binom{4}{2}m_{4}+\dotsb\right)$
$\displaystyle\geqslant\frac{d-1}{2}(2m_{2}+3m_{3}+4m_{4}+\dotsb).$

Hence the number of ordinary hyperplanes is at least

$m_{1}\geqslant
K\binom{n-K}{d-1}-\frac{K(K-1)}{d-1}\binom{n-K}{d-2}=K\binom{n-K}{d-1}\frac{n-2K-d+3}{n-K-d+2}.$

We next show that for all $K\geqslant 2$, if $n\geqslant 2dK$ then

$K\binom{n-K}{d-1}\frac{n-2K-d+3}{n-K-d+2}>\binom{n-1}{d-1}.$

This is equivalent to

$K>\frac{n-K+1}{n-2K-d+3}\prod_{i=1}^{K-2}\frac{n-i}{n-d-i+1}.$ (10)

Note that

$\frac{n-K+1}{n-2K-d+3}<2$ (11)

if $n>3K+2d-5$ and

$\frac{n-i}{n-d-i+1}<\frac{i+2}{i+1}$ (12)

if $n\geqslant(i+2)d$ for each $i=1,\dots,K-2$. However, since $2dK>(i+2)d$
and also $2dK>4K+2d-5$, the inequality (10) now follows from (11) and (12). ∎

The second case needs more work. We first consider the number of ordinary
hyperplanes spanned by a coset of a subgroup of the smooth points $\delta^{*}$
of an elliptic normal curve or a rational acnodal curve. By Propositions 3.1
and 3.9, we can consider $\delta^{*}$ as a group isomorphic to either
$\mathbb{R}/\mathbb{Z}$ or $\mathbb{R}/\mathbb{Z}\times\mathbb{Z}_{2}$. Let
$H\oplus x$ be a coset of a subgroup $H$ of $\delta^{*}$ of order $n$ where
$(d+1)x=\ominus c\in H$. Since $H$ is a subgroup of order $n$ of
$\mathbb{R}/\mathbb{Z}$ or $\mathbb{R}/\mathbb{Z}\times\mathbb{Z}_{2}$, we
have that either $H$ is cyclic, or $\mathbb{Z}_{n/2}\times\mathbb{Z}_{2}$ when
$n$ is divisible by $4$. The exact group will matter only when we make exact
calculations.

Note that it follows from the group property that any $d$ points on
$\delta^{*}$ span a hyperplane. Also, since any hyperplane intersects
$\delta^{*}$ in $d+1$ points, counting multiplicity, it follows that an
ordinary hyperplane of $H\oplus x$ intersects $\delta^{*}$ in $d$ points, of
which exactly one of them has multiplicity $2$, and the others multiplicity
$1$. Denote the number of ordered $k$-tuples $(a_{1},\dotsc,a_{k})$ with
distinct $a_{i}\in H$ that satisfy $m_{1}a_{1}\oplus\dotsb\oplus m_{k}a_{k}=c$
by $[m_{1},\dotsc,m_{k};c]$. Then the number of ordinary hyperplanes spanned
by $H\oplus x$ is

$\frac{1}{(d-1)!}[2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-1$ times}}\\!;c].$
(13)

We show that we can always find a value of $c$ for which (13) is at most
$\binom{n-1}{d-1}$.

######  Lemma 5.2.

Let $\delta^{*}$ be an elliptic normal curve or the smooth points of a
rational acnodal curve in $\mathbb{R}\mathbb{P}^{d}$, $d\geqslant 2$. Then any
finite subgroup $H$ of $\delta^{*}$ of order $n$ has a coset $H\oplus x$ with
$(d+1)x\in H$, that spans at most $\binom{n-1}{d-1}$ ordinary hyperplanes.
Furthermore, if $d+1$ and $n$ are coprime, then any such coset spans exactly
$\binom{n-1}{d-1}$ ordinary hyperplanes.

###### Proof.

It suffices to show that there exists $c\in H$ such that the number of
solutions $(a_{1},\dotsc,a_{d})\in H^{d}$ of the equation $2a_{1}\oplus
a_{2}\oplus\dotsb\oplus a_{d}=c$, where $c=\ominus(d+1)x$, is at most
$(d-1)!\binom{n-1}{d-1}$.

Fix $a_{1}$ and consider the substitution $b_{i}=a_{i}-a_{1}$ for
$i=2,\dotsc,d$. Note that $2a_{1}\oplus\dotsb\oplus a_{d}=c$ and
$a_{1},\dots,a_{d}$ are distinct if and only if $b_{2}\oplus\dotsb\oplus
b_{d}=c\ominus(d+1)a_{1}$ and $b_{2},\dots,b_{d}$ are distinct and non-zero.
Let

$A_{c,j}=\left\\{(j,a_{2},\dotsc,a_{d}):2j\oplus a_{2}\oplus\dotsb\oplus
a_{d}=c,\text{$a_{2},\dotsc,a_{d}\in H\setminus\\{j\\}$ distinct}\right\\},$

and let

$B_{k}=\left\\{(b_{2},\dotsc,b_{d}):b_{2}\oplus\dotsb\oplus
b_{d}=k,\text{$b_{2},\dotsc,b_{d}\in H\setminus\\{0\\}$ distinct}\right\\}.$

Then $|A_{c,j}|=|B_{c\ominus(d+1)j}|$, and the number of ordinary hyperplanes
spanned by $H\oplus x$ is

$\frac{1}{(d-1)!}\sum_{j\in H}|A_{c,j}|.$

If $d+1$ is coprime to $n$, then $c\ominus(d+1)j$ runs through all elements of
$H$ as $j$ varies. So we have
$\sum_{j}|B_{c\ominus(d+1)j}|=(n-1)\dotsb(n-d+1)$, hence for all $c$,

$\frac{1}{(d-1)!}\sum_{j\in H}|A_{c,j}|=\binom{n-1}{d-1}.$

If $d+1$ is not coprime to $n$, then $c\ominus(d+1)j$ runs through a coset of
a subgroup of $H$ of size $n/\gcd(d+1,n)$ as $j$ varies. We now have

$\sum_{j\in H}|B_{c\ominus(d+1)j}|=\gcd(d+1,n)\sum_{k\in
c\ominus(d+1)H}|B_{k}|.$

Summing over $c$ gives

$\displaystyle\sum_{c\in H}\sum_{j\in H}|A_{c,j}|$
$\displaystyle=\gcd(d+1,n)\sum_{c\in H}\sum_{k\in c\ominus(d+1)H}|B_{k}|$
$\displaystyle=\gcd(d+1,n)\frac{n}{\gcd(d+1,n)}(n-1)\dotsb(n-d+1)$
$\displaystyle=n(n-1)\dotsb(n-d+1).$

By the pigeonhole principle, there must then exist a $c$ such that

$\frac{1}{(d-1)!}\sum_{j\in H}|A_{c,j}|\leqslant\binom{n-1}{d-1}.\qed$

We next want to show that $[2,\\!\\!\overbrace{1,\dotsc,1}^{\text{$d-1$
times}}\\!\\!;c]$ is always very close to $(d-1)!\binom{n-1}{d-1}$,
independent of $c$ or the group $H$. Before that, we prove two simple
properties of $[m_{1},\dotsc,m_{k};c]$.

######  Lemma 5.3.

$[m_{1},\dots,m_{k};c]\leqslant 2m_{k}(k-1)!\binom{n}{k-1}$.

###### Proof.

Consider a solution $(a_{1},\dotsc,a_{k})$ of $m_{1}a_{1}\oplus\dotsb\oplus
m_{k}a_{k}=c$ where all the $a_{i}$ are distinct. We can choose
$a_{1},\dotsc,a_{k-1}$ arbitrarily in $(k-1)!\binom{n}{k-1}$ ways, and $a_{k}$
satisfies the equation $m_{k}a_{k}=c\ominus m_{1}a_{1}\ominus\dotsb\ominus
m_{k-1}a_{k-1}$, which has at most $m_{k}$ solutions if $H=\mathbb{Z}_{n}$ and
at most $2m_{k}$ solutions if $H=\mathbb{Z}_{2}\times\mathbb{Z}_{n/2}$. ∎

######  Lemma 5.4.

We have the recurrence relation

$\displaystyle[m_{1},\dots,m_{k-1},1;c]=(k-1)!\binom{n}{k-1}$
$\displaystyle-[m_{1}+1,m_{2},\dots,m_{k-1};c]$
$\displaystyle-[m_{1},m_{2}+1,m_{3},\dots,m_{k-1};c]$ $\displaystyle-\dotsb$
$\displaystyle-[m_{1},\dots,m_{k-2},m_{k-1}+1;c].$

###### Proof.

We can arbitrarily choose distinct values from $H$ for $a_{1},\dots,a_{k-1}$,
which determines $a_{k}$, and then we have to subtract the number of
$k$-tuples where $a_{k}$ is equal to one of the other $a_{i}$,
$i=1,\dots,k-1$. ∎

######  Lemma 5.5.

$[2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-1$
times}}\\!\\!;c]=(d-1)!\left(\binom{n-1}{d-1}+\varepsilon(d,n)\right),$

where

$|\varepsilon(d,n)|=\begin{cases}O\left(2^{-d/2}\binom{n}{(d-1)/2}+\binom{n}{(d-3)/2}\right)&\text{if
$d$ is odd,}\\\
O\left(d2^{-d/2}\binom{n}{d/2-1}+\binom{n}{d/2-2}\right)&\text{if $d$ is
even.}\end{cases}$

###### Proof.

Applying Lemma 5.4 once, we obtain

$[2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-1$
times}}\\!\\!;c]=(d-1)!\binom{n}{d-1}-[3,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-2$
times}}\\!\\!;c]-(d-2)[2,2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-3$
times}}\\!\\!;c].$

Note that at each stage of the recurrence in Lemma 5.4 (as long as it
applies), there are $(d-1)(d-2)\dotsb(d-k)$ terms of length $d-k$, where we
define the _length_ of $[m_{1},\dotsc,m_{k};c]$ to be $k$.

If $d$ is odd, we can continue this recurrence until we reach

$\displaystyle[2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-1$ times}}\\!\\!;c]$
$\displaystyle=(d-1)!\left(\binom{n}{d-1}-\binom{n}{d-2}+\dotsb+(-1)^{(d+1)/2}\binom{n}{(d+1)/2}\right)$
$\displaystyle\qquad+(-1)^{(d-1)/2}R,$

where $R$ is the sum of $(d-1)(d-2)\dotsb(d-(d-1)/2)$ terms of length
$(d+1)/2$. Among these there are

$\frac{\binom{d-1}{2}\binom{d-3}{2}\dotsb\binom{2}{2}}{(\frac{d-1}{2})!}=(d-2)(d-4)\dotsb
3\cdot 1$

terms of the form $[2,\dotsc,2;c]$. We now write $R=A+B$, where $A$ is the
same sum as $R$, except that we replace each occurrence of $[2,\dots,2;c]$ by
$[1,\dots,1;c]$, and

$B:=(d-2)(d-4)\dotsb 3\cdot 1([\underbrace{2,\dotsc,2}_{\text{$\frac{d+1}{2}$
times}};c]-[\\!\underbrace{1,\dotsc,1}_{\text{$\frac{d+1}{2}$ times}}\\!;c]).$

We next bound $A$ and $B$. We apply Lemma 5.4 to each term in $A$, after which
we obtain $(d-1)(d-2)\dotsb(d-(d+1)/2)$ terms of length $(d-1)/2$. Then using
the bound in Lemma 5.3, we obtain

$\displaystyle A$
$\displaystyle=(d-1)!\binom{n}{(d-1)/2}-O\left((d-1)(d-2)\dotsb(d-(d+1)/2)\left(\tfrac{d-3}{2}\right)!\binom{n}{(d-3)/2}\right)$
$\displaystyle=(d-1)!\left(\binom{n}{(d-1)/2}-O\left(\binom{n}{(d-3)/2}\right)\right).$

For $B$, we again use Lemma 5.3 to get

$\displaystyle|B|$ $\displaystyle=O\left((d-2)(d-4)\dotsb 3\cdot
1\left(\frac{d-1}{2}\right)!\binom{n}{(d-1)/2}\right)$
$\displaystyle=O\left((d-2)(d-4)\dotsb 3\cdot 1\cdot
2^{-\frac{d-1}{2}}(d-1)(d-3)\dotsb 4\cdot 2\binom{n}{(d-1)/2}\right)$
$\displaystyle=O\left((d-1)!2^{-\frac{d-1}{2}}\binom{n}{(d-1)/2}\right).$

Thus we obtain

$[2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-1$
times}}\\!\\!;c]=(d-1)!\left(\binom{n}{d-1}-\binom{n}{d-2}+\dotsb+(-1)^{\frac{d+1}{2}}\binom{n}{(d+1)/2}\right)\\\
+(-1)^{\frac{d-1}{2}}(d-1)!\left(\binom{n}{(d-1)/2}-O\left(\binom{n}{(d-3)/2}\right)\right)+(-1)^{\frac{d-1}{2}}B\\\
=(d-1)!\left(\binom{n-1}{d-1}+(-1)^{\frac{d+1}{2}}O\left(\binom{n}{(d-3)/2}\right)\pm
O\left(2^{-\frac{d-1}{2}}\binom{n}{(d-1)/2}\right)\right),$

which finishes the proof for odd $d$.

If $d$ is even, we obtain

$[2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-1$
times}}\\!\\!;c]=(d-1)!\left(\binom{n}{d-1}-\binom{n}{d-2}+\dotsb+(-1)^{\frac{d}{2}+1}\binom{n}{d/2}\right)+(-1)^{d/2}R,$

where $R$ now is the sum of $(d-1)(d-2)\dotsb(d-d/2)$ terms of length $d/2$.
Among these there are

$\frac{(d-1)\binom{d-2}{2}\binom{d-4}{2}\dotsb\binom{2}{2}}{(\frac{d-2}{2})!}+\frac{2\binom{d-1}{3}\binom{d-4}{2}\dotsb\binom{2}{2}}{(\frac{d-4}{2})!}=(d+1)(d-1)\dotsb
7\cdot 5$

terms of the form $[3,2,\dots,2;c]$. Again we write $R=A+B$, where $A$ is the
same sum as $R$, except that each occurrence of $[3,2,\dots,2;c]$ is replaced
by $[1,\dots,1;c]$, and

$B:=(d+1)(d-1)\dotsb 7\cdot
5([3,\\!\\!\underbrace{2,\dotsc,2}_{\text{$\frac{d}{2}-1$
times}}\\!\\!;c]-[\underbrace{1,\dotsc,1}_{\text{$\frac{d}{2}$ times}};c]).$

Similar to the previous case, we obtain

$A=(d-1)!\left(\binom{n}{d/2-1}-O\left(\binom{n}{d/2-2}\right)\right)$

and

$|B|=O\left((d+1)(d-1)\dotsb 7\cdot
5(\tfrac{d}{2}-1)!\binom{n}{d/2-1}\right)=O\left(2^{-d/2}d!\binom{n}{d/2-1}\right),$

which finishes the proof for even $d$. ∎

Computing $[2,\dotsc,2;c]$ and $[3,2,\dotsc,2;c]$ exactly is more subtle and
depends on $c$ and the group $H$. We do not need this for the asymptotic
Theorems 1.2 and 1.3, and will only need to do so when computing exact
extremal values.

To show that a coset is indeed extremal, we first consider the effect of
adding a single point. The case where the point is on the curve is done in
Lemma 5.6, while Lemma 5.7 covers the case where the point is off the curve.
We then obtain a more general lower bound in Lemma 5.8.

######  Lemma 5.6.

Let $\delta^{*}$ be an elliptic normal curve or the smooth points of a
rational acnodal curve in $\mathbb{R}\mathbb{P}^{d}$, $d\geqslant 2$. Suppose
$H\oplus x$ is a coset of a finite subgroup $H$ of $\delta^{*}$ of order $n$,
with $(d+1)x\in H$. Let $p\in\delta^{*}\setminus(H\oplus x)$. Then there are
at least $\binom{n}{d-1}$ hyperplanes through $p$ that meet $H\oplus x$ in
exactly $d-1$ points.

###### Proof.

Take any $d-1$ points $p_{1},\dotsc,p_{d-1}\in H\oplus x$. Suppose that the
(unique) hyperplane through $p,p_{1},\dots,p_{d-1}$ contains another point
$p^{\prime}\in H\oplus x$. Since $p\oplus p_{1}\oplus\dots\oplus p_{d-1}\oplus
p^{\prime}=0$ by Propositions 3.1 and 3.9, we obtain that $p\in H\ominus dx$.
Since $(d+1)x\in H$, we obtain $p\in H\oplus x$, a contradiction. Therefore,
the hyperplane through $p,p_{1},\dots,p_{d-1}$ does not contain any other
point of $H\oplus x$.

It remains to show that if
$\\{p_{1},\dots,p_{d-1}\\}\neq\\{p_{1}^{\prime},\dots,p_{d-1}^{\prime}\\}$
where also $p_{1}^{\prime},\dots,p_{d-1}^{\prime}\in H\oplus x$, then the two
sets span different hyperplanes with $p$. Suppose they span the same
hyperplane. Then $\ominus(p\oplus p_{1}\oplus\dotsb\oplus p_{d-1})$ also lies
on this hyperplane, but not in $H\oplus x$, as shown above. Also,
$p_{i}^{\prime}\notin\\{p_{1},\dots,p_{d-1}\\}$ for some $i$, and then
$p_{1},\dots,p_{d-1},p_{i}^{\prime}$, and $\ominus(p\oplus
p_{1}\oplus\dotsb\oplus p_{d-1})$ are $d+1$ distinct points on a hyperplane,
so their sum is $0$, which implies $p=p_{i}^{\prime}$, a contradiction.

So there are $\binom{n}{d-1}$ hyperplanes through $p$ meeting $H\oplus x$ in
exactly $d-1$ points. ∎

The following Lemma generalises [GT13]*Lemma 7.7, which states that if
$\delta^{*}$ is an elliptic curve or the smooth points of an acnodal cubic
curve in the plane, $H\oplus x$ is a coset of a finite subgroup of order
$n>10^{4}$, and if $p\notin\delta^{*}$, then there are at least $n/1000$ lines
through $p$ that pass through exactly one element of $H\oplus x$. A naive
generalisation to dimension $3$ would state that if $\delta^{*}$ is an
elliptic or acnodal space quartic curve with a finite subgroup $H$ of
sufficiently large order $n$, and $x\in\delta^{*}$ and $p\notin\delta^{*}$,
then there are $\Omega(n^{2})$ planes through $p$ and exactly two elements of
$H\oplus x$. This statement is false, even if we assume that $4x\in H$ (the
analogous assumption $3x\in H$ is not made in [GT13]), as can be seen from the
following example.

Let $\delta$ be an elliptic quartic curve obtained from the intersection of a
circular cylinder in $\mathbb{R}^{3}$ with a sphere which has centre $c$ on
the axis $\ell$ of the cylinder. Then $\delta$ is symmetric in the plane
through $c$ perpendicular to $\ell$, and we can find a finite subgroup $H$ of
any even order $n$ such that the line through any element of $H$ parallel to
$\ell$ intersects $H$ in two points. If we now choose $p$ to be the point at
infinity on $\ell$, then we obtain that any plane spanned by $p$ and two
points of $H$ not collinear with $p$, intersects $H$ in two more points. Note
that the projection $\pi_{p}$ maps $\delta$ to a conic, so is not generically
one-to-one. The number of such $p$ is bounded by the trisecant lemma (Lemma
2.3). However, as the next lemma shows, a generalisation of [GT13]*Lemma 7.7
holds except that in dimension 3 we have to exclude such points $p$.

######  Lemma 5.7.

Let $\delta$ be an elliptic normal curve or a rational acnodal curve in
$\mathbb{R}\mathbb{P}^{d}$, $d\geqslant 2$, and let $\delta^{*}$ be its set of
smooth points. Let $H$ be a finite subgroup of $\delta^{*}$ of order $n$,
where $n\geqslant Cd^{4}$ for some sufficiently large absolute constant $C>0$.
Let $x\in\delta^{*}$ satisfy $(d+1)x\in H$. Let
$p\in\mathbb{R}\mathbb{P}^{d}\setminus\delta^{*}$. If $d=3$, assume
furthermore that $\delta$ is not contained in a quadric cone with vertex $p$.
Then there are at least $c\binom{n}{d-1}$ hyperplanes through $p$ that meet
the coset $H\oplus x$ in exactly $d-1$ points, for some sufficiently small
absolute constant $c>0$.

###### Proof.

We prove by induction on $d$ that under the given hypotheses there are at
least $c^{\prime}\prod_{i=2}^{d}(1-\frac{1}{i^{2}})\binom{n}{d-1}$ such
hyperplanes for some sufficiently small absolute constant $c^{\prime}>0$. The
base case $d=2$ is given by [GT13]*Lemma 7.7.

Next assume that $d\geqslant 3$, and that the statement holds for $d-1$. Fix a
$q\in H\oplus x$, and consider the projection $\pi_{q}$. Since $q$ is a smooth
point of $\delta$, $\overline{\pi_{q}(\delta\setminus\\{q\\})}$ is a non-
degenerate curve of degree $d$ in $\mathbb{R}\mathbb{P}^{d-1}$ (otherwise its
degree would be at most $d/2$, but a non-degenerate curve has degree at least
$d-1$). The projection $\pi_{q}$ can be naturally extended to have a value at
$q$, by setting $\pi_{q}(q)$ to be the point where the tangent line of
$\delta$ at $q$ intersects the hyperplane onto which $\delta$ is projected.
(This point is the single point in
$\overline{\pi_{q}(\delta\setminus\\{q\\})}\setminus\pi_{q}(\delta\setminus\\{q\\})$.)
The curve $\pi_{q}(\delta)$ has degree $d$ and is either elliptic or rational
and acnodal, hence it has a group operation $\boxplus$ such that $d$ points
are on a hyperplane in $\mathbb{R}\mathbb{P}^{d-1}$ if and only if they sum to
the identity.

Observe that any $d$ points
$\pi_{q}(p_{1}),\dots,\pi_{q}(p_{d})\in\pi_{q}(\delta^{*})$ lie on a
hyperplane in $\mathbb{R}\mathbb{P}^{d-1}$ if and only if
$p_{1}\oplus\dots\oplus p_{d}\oplus q=0$. By Proposition 3.10 it follows that
the group on $\pi_{q}(\delta^{*})$ obtained by transferring the group
$(\delta^{*},\oplus)$ by $\pi_{q}$ is a translation of
$(\pi_{q}(\delta^{*}),\boxplus)$. In particular, $\pi_{q}(H\oplus
x)=H^{\prime}\boxplus x^{\prime}$ for some subgroup $H^{\prime}$ of
$(\pi_{q}(\delta^{*}),\boxplus)$ of order $n$, and $(d+1)x^{\prime}\in
H^{\prime}$.

We would like to apply the induction hypothesis, but we can only do that if
$\pi_{q}(p)\notin\pi_{q}(\delta^{*})$, and when $d=4$, if $\pi_{q}(p)$ is not
the vertex of a quadric cone containing $\pi_{q}(\delta)$. We next show that
there are only $O(d^{2})$ exceptional points $q$ to which we cannot apply
induction.

Note that $\pi_{q}(p)\in\pi_{q}(\delta^{*})$ if and only if the line $pq$
intersects $\delta$ with multiplicity $2$, which means we have to bound the
number of these lines through $p$. To this end, we consider the projection of
$\delta$ from the point $p$. Suppose that $\pi_{p}$ does not project $\delta$
generically one-to-one to a degree $d+1$ curve in
$\mathbb{R}\mathbb{P}^{d-1}$. Then $\pi_{p}(\delta)$ has degree at most
$(d+1)/2$. However, its degree is at least $d-1$ because it is non-degenerate.
It follows that $d=3$, and that $\pi_{p}(\delta)$ has degree $2$ and is
irreducible, so $\delta$ is contained in a quadric cone with vertex $p$, which
we ruled out by assumption.

Therefore, $\pi_{p}$ projects $\delta$ generically one-to-one onto the curve
$\pi_{p}(\delta)$, which has degree $d+1$ and has at most $\binom{d}{2}$
double points (this follows from the Plücker formulas after projecting to the
plane [W78]*Chapter III, Theorem 4.4). We thus have that an arbitrary point
$p\in\mathbb{R}\mathbb{P}^{d}\setminus\delta$ lies on at most $O(d^{2})$
secants or tangents of $\delta$ (or lines through two points of $\delta^{*}$
if $p$ is the acnode of $\delta$).

If $d=4$, we also have to avoid $q$ such that $\pi_{q}(p)$ is the vertex of a
cone on which $\pi_{q}(\delta)$ lies. Such $q$ have the property that if we
first project $\delta$ from $q$ and then $\pi_{q}(\delta)$ from $\pi_{q}(p)$,
then the composition of these two projections is not generically one-to-one.
Another way to do these to successive projections is to first project $\delta$
from $p$ and then $\pi_{p}(\delta)$ from $\pi_{p}(q)$. Thus, we have that
$\pi_{p}(q)$ is a point on the quintic $\pi_{p}(\delta)$ in
$\mathbb{R}\mathbb{P}^{3}$ such that the projection of $\pi_{p}(\delta)$ from
$\pi_{p}(q)$ onto $\mathbb{R}\mathbb{P}^{2}$ is not generically one-to-one.
However, there are only $O(1)$ such points by Lemma 2.3. Thus there are at
most $Cd^{2}$ points $q\in H\oplus x$ to which we cannot apply the induction
hypothesis.

For all remaining $q\in H\oplus x$, we obtain by the induction hypothesis that
there are at least
$c^{\prime}\prod_{i=2}^{d-1}(1-\frac{1}{i^{2}})\binom{n}{d-2}$ hyperplanes
$\Pi$ in $\mathbb{R}\mathbb{P}^{d-1}$ through $\pi_{q}(p)$ and exactly $d-2$
points of $H^{\prime}\boxplus x^{\prime}$. If none of these $d-2$ points equal
$\pi_{q}(q)$, then $\pi_{q}^{-1}(\Pi)$ is a hyperplane in
$\mathbb{R}\mathbb{P}^{d}$ through $p$ and $d-1$ points of $H\oplus x$, one of
which is $q$. There are at most $\binom{n-1}{d-3}$ such hyperplanes in
$\mathbb{R}\mathbb{P}^{d-1}$ through $\pi_{q}(q)$. Therefore, there are at
least
$c^{\prime}\prod_{i=2}^{d-1}(1-\frac{1}{i^{2}})\binom{n}{d-2}-\binom{n-1}{d-3}$
hyperplanes in $\mathbb{R}\mathbb{P}^{d}$ that pass through $p$ and exactly
$d-1$ points of $H\oplus x$, one of them being $q$. If we sum over all
$n-Cd^{2}$ points $q$, we count each hyperplane $d-1$ times, and we obtain
that the total number of such hyperplanes is at least

$\frac{n-Cd^{2}}{d-1}\left(c^{\prime}\prod_{i=2}^{d-1}\left(1-\frac{1}{i^{2}}\right)\binom{n}{d-2}-\binom{n-1}{d-3}\right).$
(14)

It can easily be checked that

$\frac{n-Cd^{2}}{d-1}\binom{n}{d-2}\geqslant\left(1-\frac{1}{2d^{2}}\right)\binom{n}{d-1}$
(15)

if $n>2Cd^{4}$, and that

$c^{\prime}\prod_{i=2}^{d-1}\left(1-\frac{1}{i^{2}}\right)\frac{1}{2d^{2}}\binom{n}{d-1}\geqslant\frac{n-Cd^{2}}{d-1}\binom{n-1}{d-3}$
(16)

if $n>4d^{3}/c^{\prime}$. It now follows from (15) and (16) that the
expression (14) is at least

$c^{\prime}\prod_{i=2}^{d}\left(1-\frac{1}{i^{2}}\right)\binom{n}{d-1},$

which finishes the induction. ∎

######  Lemma 5.8.

Let $\delta^{*}$ be an elliptic normal curve or the smooth points of a
rational acnodal curve in $\mathbb{R}\mathbb{P}^{d}$, $d\geqslant 4$, and let
$H\oplus x$ be a coset of a finite subgroup $H$ of $\delta^{*}$, with
$(d+1)x\in H$. Let $A\subseteq H\oplus x$ and
$B\subset\mathbb{R}\mathbb{P}^{d}\setminus(H\oplus x)$ with $|A|=a$ and
$|B|=b$. Let $P=(H\oplus x\setminus A)\cup B$ with $|P|=n$ be such that every
$d$ points of $P$ span a hyperplane. If $A$ and $B$ are not both empty and
$n\geqslant C(a+b+d^{2})d$ for some sufficiently large absolute constant
$C>0$, then $P$ spans at least $(1+c)\binom{n-1}{d-1}$ ordinary hyperplanes
for some sufficiently small absolute constant $c>0$.

###### Proof.

We first bound from below the number of ordinary hyperplanes of $(H\oplus
x)\setminus A$ that do not pass through a point of $B$.

The number of ordinary hyperplanes of $(H\oplus x)\setminus A$ that are
disjoint from $A$ is

$\frac{1}{(d-1)!}\left|\left\\{(a_{1},\dots,a_{d})\in(H\setminus(A\ominus
x))^{d}:\begin{array}[]{c}2a_{1}\oplus a_{2}\oplus\dotsb\oplus
a_{d}=\ominus(d+1)x,\\\ \text{$a_{1},\dots,a_{d}$ are
distinct}\end{array}\right\\}\right|.$

If we denote by by $[m_{1},\dotsc,m_{k}]^{\prime}$ the number of ordered
$k$-tuples $(a_{1},\dotsc,a_{k})$ with distinct $a_{i}\in H\setminus(A\ominus
x)$ that satisfy $m_{1}a_{1}\oplus\dotsb\oplus m_{k}a_{k}=\ominus(d+1)x$, then
we obtain, similar to the proofs of Lemmas 5.3 and 5.4, that

$\displaystyle[2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-1$
times}}\\!]^{\prime}$
$\displaystyle=(d-1)!\binom{n-b}{d-1}-[3,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-2$
times}}\\!]^{\prime}-(d-2)[2,2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-3$
times}}\\!]^{\prime}$
$\displaystyle\geqslant(d-1)!\binom{n-b}{d-1}-2(d-2)!\binom{n-b}{d-2}-2(d-2)(d-2)!\binom{n-b}{d-2}$
$\displaystyle=(d-1)!\binom{n-b}{d-1}-2(d-1)!\binom{n-b}{d-2},$

and it follows that the number of ordinary hyperplanes of $(H\oplus
x)\setminus A$ disjoint from $A$ is at least
$\binom{n-b}{d-1}-2\binom{n-b}{2}$.

Next, we obtain an upper bound on the number of these hyperplanes that pass
through a point $q\in B$. Let the ordinary hyperplane $\Pi$ pass through
$p_{1},p_{2},\dots,p_{d}\in(H\oplus x)\setminus A$, with $p_{1}$ being the
double point. Since $q\in\Pi$ and any $d$ points determine a hyperplane, $\Pi$
is still spanned by $q,p_{1},\dots,p_{d-1}$, after a relabelling of
$p_{2},\dots,p_{d}$. Let $S$ be a minimal subset of
$\\{p_{2},\dots,p_{d-1}\\}$ such that the tangent line $\ell$ of $\delta$ at
$p_{1}$ lies in the flat spanned by $S\cup\\{q,p_{1}\\}$.

If $S$ is empty, then $\ell$ is a tangent from $q$ to $\delta$, of which there
are at most $d(d+1)$ (this follows again from projection and the Plücker
formulas [W78]*Chapter IV, p. 117[NZ]*Corollary 2.5). Therefore, the number of
ordinary hyperplanes through $p_{1},p_{2},\dots,p_{d}\in(H\oplus x)\setminus
A$ with the tangent of $\delta$ at $p_{1}$ passing through $q$ is at most
$d(d+1)\binom{n-b}{d-2}$.

If on the other hand $S$ is non-empty, then there is some $p_{i}$, say
$p_{d-1}$, such that $q,p_{1},\dots,p_{d-2}$ together with $\ell$ generate
$\Pi$. Therefore, $\Pi$ is determined by $p_{1}$, the tangent through $p_{1}$,
and some $d-3$ more points $p_{i}$. There are at most
$(n-b)\binom{n-b-1}{d-3}=(d-2)\binom{n-b}{d-2}$ ordinary hyperplanes through
$q$ in this case.

The number of ordinary hyperplanes of $(H\oplus x)\setminus A$ that contain a
point from $A$ is at least

$a\left(\binom{n-b}{d-1}-a\binom{n-b}{d-2}-(n-b)\binom{n-b-1}{d-3}\right)=a\binom{n-b}{d-1}-(a^{2}+a(d-2))\binom{n-b}{d-2},$

since we can find such a hyperplane by choosing a point $p\in A$ and $d-1$
points $p_{1},\dots,p_{d-1}\in(H\oplus x)\setminus A$, and then the remaining
point $\ominus(p\oplus p_{1}\oplus\dots\oplus p_{d-1})$ might not be a new
point in $(H\oplus x)\setminus A$ by either being in $A$ (possibly equal to
$p$) or being equal to one of the $p_{i}$. The number of these hyperplanes
that also pass through some point of $B$ is at most $ab\binom{n-b}{d-2}$.

Therefore, the number of ordinary hyperplanes of $(H\oplus x)\setminus A$ that
miss $B$ is at least

$(1+a)\binom{n-b}{d-1}-\left(2+b(d(d+1)+d-2)+a^{2}+a(d-2)+ab\right)\binom{n-b}{d-2}.$
(17)

Next, assuming that $B\neq\emptyset$, we find a lower bound to the number of
ordinary hyperplanes through exactly one point of $B$ and exactly $d-1$ points
of $(H\oplus x)\setminus A$. The number of hyperplanes through at least one
point of $B$ and exactly $d-1$ points of $(H\oplus x)\setminus A$ is at least
$bc^{\prime}\binom{n-b}{d-1}-ab\binom{n-b}{d-2}$ by Lemmas 5.6 and 5.7 for
some sufficiently small absolute constant $c^{\prime}>0$. The number of
hyperplanes through at least two points of $B$ and exactly $d-1$ points of
$(H\oplus x)\setminus A$ is at most $\binom{b}{2}\binom{n-b}{d-2}$. It follows
that there are at least
$bc^{\prime}\binom{n-b}{d-1}-\bigl{(}ab+\binom{b}{2}\bigr{)}\binom{n-b}{d-2}$
ordinary hyperplanes passing though a point of $B$.

Combining this with (17), $P$ spans at least

$(1+a+bc^{\prime})\binom{n-b}{d-1}-\left(2+b(d(d+1)+d-2)+a^{2}+a(d-2)+2ab+\binom{b}{2}\right)\binom{n-b}{d-2}=:f(a,b)$

ordinary hyperplanes. Since

$f(a+1,b)-f(a,b)=\binom{n-b}{d-1}-(2a+2b+d-1)\binom{n-b}{d-2}$

is easily seen to be positive for all $a\geqslant 0$ as long as
$n>(2a+2b+d-1)(d-1)+b+d-2$, we have without loss of generality that $a=0$ in
the case that $b\geqslant 1$. Then $f(0,b+1)-f(0,b)$ is easily seen to be at
least

$c^{\prime}\binom{n-b-1}{d-1}-(d^{2}+d-2+b)\binom{n-b-1}{d-2},$

which is positive for all $b\geqslant 1$ if $n\geqslant C(b+d^{2})d$ for $C$
sufficiently large. Also,
$f(0,1)=(1+c^{\prime})\binom{n-1}{d-1}-(d^{2}+2d)\binom{n-1}{d-2})\geqslant(1+c)\binom{n-1}{d-1}$
if $n\geqslant Cd^{3}$. This completes the proof in the case where $B$ is non-
empty.

If $B$ is empty, then we can bound the number of ordinary hyperplanes from
below by setting $b=0$ in (17), and checking that the resulting expression

$(1+a)\binom{n}{d-1}-\left(d+a^{2}+a(d-2)\right)\binom{n}{d-2}$

is increasing in $a$ if $n>(2a+d-1)(d-1)+d-2$, and larger than
$\frac{3}{2}\binom{n-1}{d-1}$ if $n>Cd^{3}$. ∎

We are now ready to prove Theorems 1.2 and 1.3.

###### Proof of Theorem 1.2.

Let $P$ be the set of $n$ points. By Lemma 5.2, we may assume that $P$ has at
most $\binom{n-1}{d-1}$ ordinary hyperplanes. Since $n\geqslant Cd^{3}2^{d}$,
we may apply Theorem 1.1 to obtain that up to $O(d2^{d})$ points, $P$ lies in
a hyperplane or is a coset of a subgroup of an elliptic normal curve or the
smooth points of a rational acnodal curve.

In the first case, by Lemma 5.1, since $n\geqslant Cd^{3}2^{d}$, the minimum
number of ordinary hyperplanes is attained when all but one point is contained
in a hyperplane and we get exactly $\binom{n-1}{d-1}$ ordinary hyperplanes.

In the second case, by Lemma 5.8, again since $n\geqslant Cd^{3}2^{d}$, the
minimum number of ordinary hyperplanes is attained by a coset of an elliptic
normal curve or the smooth points of a rational acnodal curve. Lemmas 5.2 and
5.5 then complete the proof. Note that the second term in the error term of
Lemma 5.5 is dominated by the first term because of the lower bound on $n$,
and that the error term here is negative by Lemma 5.2. ∎

Note that if we want to find the exact minimum number of ordinary hyperplanes
spanned by a set of $n$ points in $\mathbb{R}\mathbb{P}^{d}$, $d\geqslant 4$,
not contained in a hyperplane and where every $d$ points span a hyperplane, we
can continue with the calculation of $[2,1,\dotsc,1;c]$ in the proof of Lemma
5.5. As seen in the proof of Lemma 5.2, this depends on $\gcd(d+1,n)$. We also
have to minimise over different values of $c\in H$, and if $n\equiv
0\pmod{4}$, consider both cases $H\cong\mathbb{Z}_{n}$ and
$H\cong\mathbb{Z}_{n/2}\times\mathbb{Z}_{2}$.

For example, it can be shown that if $d=4$, the minimum number is

$\begin{cases}\binom{n-1}{3}-4&\text{if }n\equiv 0\pmod{5},\\\
\binom{n-1}{3}&\text{otherwise},\end{cases}$

if $d=5$, the minimum number is

$\begin{cases}\binom{n-1}{4}-\frac{1}{8}n^{2}+\frac{1}{12}n-1&\text{if
}n\equiv 0\pmod{6},\\\ \binom{n-1}{4}&\text{if }n\equiv 1,5\pmod{6},\\\
\binom{n-1}{4}-\frac{1}{8}n^{2}+\frac{3}{4}n-1&\text{if }n\equiv
2,4\pmod{6},\\\ \binom{n-1}{4}-\frac{2}{3}n+2&\text{if }n\equiv
3\pmod{6},\end{cases}$

and if $d=6$, the minimum number is

$\begin{cases}\binom{n-1}{5}-6&\text{if }n\equiv 0\pmod{7},\\\
\binom{n-1}{5}&\text{otherwise.}\end{cases}$

###### Proof of Theorem 1.3.

We first show that there exist sets of $n$ points, with every $d$ points
spanning a hyperplane, spanning at least
$\frac{1}{d+1}\binom{n-1}{d}+O\left(2^{-d/2}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right)$
$(d+1)$-point hyperplanes. Let $\delta^{*}$ be an elliptic normal curve or the
smooth points of a rational acnodal curve. By Propositions 3.1 and 3.9, the
number of $(d+1)$-point hyperplanes spanned by a coset $H\oplus x$ of
$\delta^{*}$ is

$\frac{1}{(d+1)!}[\\!\underbrace{1,\dotsc,1}_{\text{$d+1$ times}}\\!;c]$

for some $c\in\delta^{*}$. Note that

$[\\!\underbrace{1,\dotsc,1}_{\text{$d+1$
times}}\\!;c]=d!\binom{n}{d}-d[2,\\!\\!\underbrace{1,\dotsc,1}_{\text{$d-1$
times}}\\!\\!;c],$

so if we take $H\oplus x$ to be a coset minimising the number of ordinary
hyperplanes, then by Theorem 1.2, there are

$\displaystyle\mathrel{\phantom{=}}\frac{1}{d+1}\left(\binom{n}{d}-\binom{n-1}{d-1}\right)+O\left(2^{-\frac{d}{2}}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right)$
$\displaystyle=\frac{1}{d+1}\binom{n-1}{d}+O\left(2^{-\frac{d}{2}}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right)$
(18)

$(d+1)$-point hyperplanes.

Next let $P$ be an arbitrary set of $n$ points in $\mathbb{R}\mathbb{P}^{d}$,
$d\geqslant 4$, where every $d$ points span a hyperplane. Suppose $P$ spans
the maximum number of $(d+1)$-point hyperplanes. Without loss of generality,
we can thus assume $P$ spans at least
$\frac{1}{d+1}\binom{n-1}{d}+O\left(2^{-d/2}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right)$
$(d+1)$-point hyperplanes.

Let $m_{i}$ denote the number of $i$-point hyperplanes spanned by $P$.
Counting the number of unordered $d$-tuples, we get

$\binom{n}{d}=\sum_{i\geqslant d}\binom{i}{d}m_{i}\geqslant
m_{d}+(d+1)m_{d+1},$

hence we have

$m_{d}\leqslant\binom{n}{d}-\binom{n-1}{d}-O\left(d2^{-\frac{d}{2}}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right)=O\left(\binom{n-1}{d-1}\right),$

and we can apply Theorem 1.1.

In the case where all but $O(d2^{d})$ points of $P$ are contained in a
hyperplane, it is easy to see that $P$ spans $O(d2^{d}\binom{n}{d-1})$
$(d+1)$-point planes, contradicting the assumption.

So all but $O(d2^{d})$ points of $P$ are contained in a coset $H\oplus x$ of a
subgroup $H$ of $\delta^{*}$. Consider the identity

$(d+1)m_{d+1}=\binom{n}{d}-m_{d}-\sum_{i\geqslant d+2}\binom{i}{d}m_{i}.$

By Theorem 1.2 and Lemma 5.8, we know that
$m_{d}\geqslant\binom{n-1}{d-1}-O\left(d2^{-d/2}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right)$
and any deviation of $P$ from the coset $H\oplus x$ adds at least
$c\binom{n-1}{d-1}$ ordinary hyperplanes for some sufficiently small absolute
constant $c>0$. Since we also have

$\displaystyle\sum_{i\geqslant d+2}\binom{i}{d}m_{i}$
$\displaystyle=\binom{n}{d}-m_{d}-(d+1)m_{d+1}$
$\displaystyle=\binom{n}{d}-\binom{n-1}{d-1}-\binom{n-1}{d}+O\left(d2^{-\frac{d}{2}}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right)$
$\displaystyle=O\left(d2^{-\frac{d}{2}}\binom{n}{\lfloor\frac{d-1}{2}\rfloor}\right),$

we can conclude that $m_{d+1}$ is maximised when $P$ is exactly a coset of a
subgroup of $\delta^{*}$, in which case (18) completes the proof. ∎

Knowing the exact minimum number of ordinary hyperplanes spanned by a set of
$n$ points in $\mathbb{R}\mathbb{P}^{d}$, $d\geqslant 4$, not contained in a
hyperplane and where every $d$ points span a hyperplane then also gives the
exact maximum number of $(d+1)$-point hyperplanes.

Continuing the above examples, for $d=4$, the maximum number is

$\begin{cases}\frac{1}{5}\binom{n-1}{4}+\frac{4}{5}&\text{if }n\equiv
0\pmod{5},\\\ \frac{1}{5}\binom{n-1}{4}&\text{otherwise},\end{cases}$

for $d=5$, the maximum number is

$\begin{cases}\frac{1}{6}\binom{n-1}{5}+\frac{1}{48}n^{2}-\frac{1}{72}n+\frac{1}{6}&\text{if
}n\equiv 0\pmod{6},\\\ \frac{1}{6}\binom{n-1}{5}&\text{if }n\equiv
1,5\pmod{6},\\\
\frac{1}{6}\binom{n-1}{5}+\frac{1}{48}n^{2}-\frac{1}{8}n+\frac{1}{6}&\text{if
}n\equiv 2,4\pmod{6},\\\
\frac{1}{6}\binom{n-1}{5}+\frac{1}{9}n-\frac{1}{3}&\text{if }n\equiv
3\pmod{6},\end{cases}$

and for $d=6$, the maximum number is

$\begin{cases}\frac{1}{7}\binom{n-1}{6}+\frac{6}{7}&\text{if }n\equiv
0\pmod{7},\\\ \frac{1}{7}\binom{n-1}{6}&\text{otherwise}.\end{cases}$

## Acknowledgments

We thank Peter Allen, Alex Fink, Misha Rudnev, and an anonymous referee for
helpful remarks and for pointing out errors in a previous version.

## References

[AL] Aaron Lin
Department of Mathematics
London School of Economics and Political Science
United Kingdom
aaronlinhkgmailcom [KS] Konrad Swanepoel
Department of Mathematics
London School of Economics and Political Science
United Kingdom
kswanepoellseacuk
http://personal.lse.ac.uk/swanepoe/