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\title{Arcs with large conical subsets\\ in Desarguesian planes of even order}
\author{Kris Coolsaet\qquad Heide Sticker\\
  \small Department of Applied Mathematics, Computer Science and Statistics, \\[-0.8ex]
  \small Ghent University, \\[-0.8ex]
  \small Krijgslaan 281--S9, B--9000 Gent, Belgium\\[-0.8ex]
  \small \texttt{\{Kris.Coolsaet,Heide.Sticker\}@UGent.be}}
\date{\dateline{Jan 13, 2014}{Jan 13, 2014}\\
\small Mathematics Subject Classification: 51E21}

\begin{document}
\maketitle
\begin{abstract}
  We give an explicit classification of the arcs in $\PG(2,q)$ ($q$
  even) with a large conical subset and excess 2, i.e., that consist
  of $q/2+1$ points of a conic and two points not on that conic.
  Apart from the initial setup, the methods used are similar to those
  for the case of odd $q$, published earlier (Electronic Journal of
  Combinatorics, 17, \#R112).
\end{abstract}
%
%
%-------------
% Introduction
%-------------
\section{Introduction}
Consider the Desarguesian projective plane $\PG(2,q)$ over the finite
field of order $q$. An \emph{arc} %of $\PG(2,q)$ 
is a set of points of
$\PG(2, q)$ with the property that no three points of this set are
collinear.  An arc is called \emph{complete} if it
is not contained in a bigger arc.

When $q$ is odd it is well known that an arc can be of size at most
$q+1$ and in that case always coincides with the set of points of some
conic. When $q$ is even, the maximum size of an
arc is $q+2$ and again examples are known.  (For instance, one can
always add the nucleus to the conic.) For further
information on the geometrical and combinatorial properties of arcs
we refer to \cite[Chapters 9--10]{Hirschfeld}.

%It is natural
%to ask what is the second biggest size for a complete arc in $\PG(2,q)$.
Removing some points from a conic again yields an arc, but this arc is
obviously not complete. However, after removing a sufficient number of
points it may be possible to extend the set thus obtained to a
nonconical arc by adding a point that does not belong to the original
conic. This new arc might not be complete, but can be made complete by
adding yet more points. 


We shall be interested in arcs for which the
conical part is as large as possible
and that contain (at least) two points not on that conic.
A simple result by B.\ Segre \cite{Segre} and L.\ Lombado-Radice
\cite{L-R} shows that the size of the conical part can be at most
$(q+3)/2$ when $q$ is odd, and $(q+2)/2$ when $q$ is even.
%
% OK TO REMOVE (by editor)
%
%The case of odd $q$ was investigated in detail by G.\
%Pellegrino \cite{Pelle-a,Pelle-2}, and later by G.~Korchm{\'a}ros and
%A.~Sonnino \cite{Korch-1,Korch-2}. 

In \cite{Coolsaet} we classified all arcs of this type with two points
not on the conic, for odd $q$ only.  Our method made use
of an initial step that needs an element %$\beta$
of the field $\GF(q)$ that is not a square. We could therefore not
immediately extend this technique to the case where $q$ is even.
% (and every field element is a square).

In this paper we show that this problem can be overcome by starting with a
different setup, described in Section \ref{sec-prelim}. The end result
for $q$ even turns out to be much simpler than that for $q$ odd. For
instance, instead of three different types of arcs for the odd case,
there is only one case to consider when $q$ is even (cf.\ Section
\ref{sec-arcs}). This is a consequence of the fact that there is no
distinction between internal and external points of a conic in the
even case.




%

\section{Notation and preliminary definitions}
\label{sec-prelim}
(Where possible we have chosen notations in such a way that they
conform to the notations for similar concepts in \cite{Coolsaet}.)


Let $S$ denote an arc of $\PG(2,q)$. We define a \emph{conical subset} of $S$ to
be any subset $R$ of $S$ of the form $R = S \cap C$ where $C$ is a
conic. (Most of the time the conic $C$ and the conical
subset $R$ will be clear from context.  We will therefore usually
leave out the reference to $C$ when talking about tangents and secants
of $C$ and lines external to $C$.)
%
The points of $S$ that do not belong to $R$ will be called
\emph{supplementary} points. The number $e=|S\setminus R|$ will be
called the \emph{excess} of the arc. We shall always assume that the arc $S$
has at least one supplementary point, i.e., that $S$ is not fully
contained in a conic.

When $q$ is even, all tangents to $C$ go through a common point $N$
(not on $C$) which is called the \emph{nucleus} of $C$. Consider a
point $S$ of the plane. If $Q$ does
not belong to $C$ and $Q\ne N$, then there is exactly one line through
$Q$ which is tangent to $C$, and there are $q/2$ lines that intersect
$C$ in two points. Each of these may intersect $R$ in at most one
point, and hence $|R|\leq q/2+1$. A conical subset that attains this bound will be
called \emph{large}. Arcs with large conical
subsets are the subject of this paper.

Henceforth let $K$ denote the finite field of order $q$, with $q$ even.

Without loss of generality we may fix $C$ to be the conic with
equation $XZ = Y^2$. Mapping $t\in K$ to the point with (homogeneous)
coordinates $(1:t:t^2)$ and $\infty$ to the point with coordinates
$(0:0:1)$ defines a one--to--one relation between $K \cup\{\infty\}$ and
$C$. The nucleus $N$ of $C$ has coordinates $(0:1:0)$.
The tangent through the point $(1:t:t^2)$ has equation $Z = t^2X$. The
tangent through the point $(0:0:1)$ has equation $X=0$.

The subgroup of $\PGL(3,q)$ that stabilizes $C$ is isomorphic to
$\PGL(2,q)$. The element of $\PGL(2,q)$ associated to the 
matrix $\begin{matrix}a&b\\c&d\end{matrix}$ maps the point $(1:t)$ of
the projective line $\PG(1,q)$ onto the point
$(1:\displaystyle{\frac{b + dt}{a + ct}})$. This yields a
corresponding action on $C$ that can be
extended to an action on any point $(x:y:z)$ of the whole plane $\PG(2,q)$, which
can be written as follows:
\begin{equation}\label{eq-PGL3}
\begin{matrix}a&b\\c&d\end{matrix}
\quad\leftrightarrow\quad
(x\ y\ z) \mapsto (x\ y\ z)
\begin{matrix}
a^2 & ab & b^2 \\
2ac & ad+bc & 2bd \\
c^2 & cd & d^2 \\
\end{matrix}
\end{equation}
where 2 is obviously zero in even characteristic.
% (which makes $\PGL(2,q)$ fix the nucleus $N$).

Let $L$ denote the quadratic extension field of $K$ and let $N(\cdot)$
and $\mathrm{Tr}(\cdot)$ be the norm and trace function for this
extension.
%
Consider $\phi\in L$ such that $\phi$ has multiplicative order $q+1$
in $L$.  Then $N(\phi)=\phi^{q+1} = 1$ and $\mathrm{Tr}(\phi)=\phi +
\phi^q = T$, with $T\in K\setminus\{0\}$. The characteristic equation of
$\phi$ is $X^2 + TX + 1 = 0$. In particular $\phi^2 + T\phi + 1 = 0$.
Note that $1,\phi,\phi^2,\ldots$ runs through all elements of $L$ with
norm equal to 1. Since $q$ is even, the only element of $K$ with
norm equal to 1 is 1 itself.

For any integer $i$ define
\[
f_i \isdef \frac1T \mathrm{Tr}(\phi^i) = \frac1T (\phi^i + \phi^{iq}) = \frac1T (\phi^i + \phi^{-i}).
\]
We have the following properties:
\begin{equation}\label{eq-1}
f_0=0,\ f_1=1,\ f_2= T,\ f_{i+2} = T f_{i+1} + f_i,
\end{equation}
\begin{equation}\label{eq-2}
f_{i+q+1}=f_i,\ f_{-i} = f_i,\ f_{i+1}+f_{i-1} = Tf_i,\ 
f_{i+j} + f_{i-j} = T f_if_j.
\end{equation}
(Note that the index $i$ of $f_i$ can be treated as an element of
$\Z_{q+1}$.) 

From (\ref{eq-2}) we may easily compute 
\begin{equation}\label{eq-3}
f_{i+j} = f_{i-1}f_j + f_if_{j+1} = f_if_{j-1} + f_{i+1}f_j,
\end{equation}
and hence
\begin{equation}\label{eq-4}
\begin{array}{rcl}
f_{2i} &=& (f_{i-1}+f_{i+1}) f_i = Tf_i^2\\
f_{2i-1} &=& f_i^2 + f_{i-1}^2
= (f_i + f_{i-1})^2,\\
 f_{2i+1} &=& f_i^2 + f_{i+1}^2
= (f_i + f_{i+1})^2.
\end{array}
\end{equation}

% The first few values of $f_0,f_1,\ldots$ are as follows~:
% \[
% 0, 1, T, T^2 + 1, T^3, T^4 + T^2 + 1, T^5 + T, T^6 + T^4 + 1, 
% T^7, \ldots
% \]
Consider the following element of $\PGL(2,q)$~: 
\begin{equation}\label{eq-M'}
M'_{i} \isdef \begin{matrix}f_{i-1} & f_i \\ f_i & f_{i+1}
\end{matrix}.
\end{equation}
It follows from (\ref{eq-1}--\ref{eq-3}) that
\[
M'_0 = \begin{matrix}1&0\\0&1\end{matrix},\ M'_{i+j} = M'_i M'_j,\ M'_i = M_1^{\prime i},\ \det M'_i = 1.
\]
The equivalent action of $\PGL(3,q)$ has the following form~:
\[
\begin{matrix}
f_{i-1}^2 & f_{i-1}f_i & f_{i}^2 \\
0 & 1 & 0 \\
f_{i}^2 & f_{i+1}f_i & f_{i+1}^2 \\
\end{matrix}.
\]
Note that $(1:0:1)$ is mapped to $(f_{2i-1}:f_{2i}:f_{2i+1})$. 
Moreover, all of these points lie on the line
$\ell$ with equation $x + Ty + z = 0$, which is an external line to
the conic $C$.

We shall number the points of $\ell$ as $Q_0, Q_1, \ldots, Q_q$ such
that $Q_i$ has coordinates $(f_{i-1}:f_i:f_{i+1})$. Note that all
these points are different and hence $\ell =
\{Q_0,Q_1,\ldots,Q_{q}\}$. The index $i$ of $Q_i$ will be called the
\emph{orbital index} of $Q_i$. Orbital indices can be treated as
elements of $\Z_{q+1}$.

The tangent through $Q_0(1:0:1)$ is the line with equation $X + Z = 0$ (i.e.,
$X=Z$) which intersects $C$ in $P_0(1:1:1)$. Define $P_i$ to be the
image of $P_0$ under $M'_i$. Then $P_i$ has coordinates 
\[
(f_{2i-1}: f_{2i} + 1: f_{2i+1}) \approx (1: t_i: t_i^2),
\]
by (\ref{eq-4}), with
\[
t_i \isdef \frac{f_{i}+f_{i+1}}{f_{i-1} + f_{i}}.
\]
(The `$\approx$'-sign denotes equality up to a scalar factor.)

We have
\[t_{q/2}=0,\ t_{q/2+1}=\infty,\ t_0 = 1,\ t_{-i} = 1/t_i.
\]
Again, the index $i$ of $P_i$ will be called its orbital index, and
again it can be treated as an element of $\Z_{q+1}$.

% The following lemma illustrates that orbital indices are a useful
% concept in this context.
\begin{lemma}
\label{lemma-1}
Let $i,j,k\in\Z_{q+1}$. Then 
\begin{itemize}
\item $P_i,P_j,Q_k$ are collinear if and only if $k \equiv i+j\pmod{q+1}$.
\item $P_iQ_k$ is a tangent to $C$ if and only if $k\equiv 2i\pmod{q+1}$.
\end{itemize}
\end{lemma}

\begin{proof}
 Indeed
\begin{eqnarray*}
\begin{determ}
f_{2i-1} & f_{2i} + 1 & f_{2i+1} \\
f_{2j-1} & f_{2j} + 1 & f_{2j+1} \\
f_{i+j-1} & f_{i+j} & f_{i+j+1}
\end{determ}
&=&
\begin{determ}
f_{2i-1}  & f_{2i} + 1 & Tf_{2i} \\
 f_{2j-1} & f_{2j} + 1 & Tf_{2j}\\
f_{i+j-1} & f_{i+j} & Tf_{i+j} 
\end{determ}
=
T
\begin{determ}
f_{2i-1} & 1 & f_{2i} \\
f_{2j-1} & 1 & f_{2j} \\
f_{i+j-1} & 0  & f_{i+j}
\end{determ}\\
&=&
T
\begin{determ}
f_{2i-1} & f_{2i} \\
f_{i+j-1} & f_{i+j}
\end{determ}
+
T
\begin{determ}
f_{2j-1} & f_{2j} \\
f_{i+j-1} & f_{i+j}
\end{determ}
\end{eqnarray*}
and this can be expanded to
\[
f_{3i+j-1} + f_{i-j+1} + f_{3i+j-1} + f_{i-j-1} +
f_{3j+i-1} +f_{j-i+1} + f_{3j+i-1} + f_{j-i+1} = 0.
\]
This proves that $P_i,P_j,Q_k$ are collinear when $k \equiv
i+j\pmod{q+1}$. Because the line $P_iP_j$ intersects the line $\ell$
in exactly one point, and the orbital index of that point is uniquely
determined (mod $q+1$), $P_i,P_j,Q_k$ cannot be collinear for other
values of $k$.

The tangent through $P_i$ has equation $Z=t_i^2 X$ which is
clearly satisfied by the point $Q_{2i}$. Again, the intersection of
this tangent with $\ell$ is a single point, and hence no
other point $Q_k$ can satisfy this property.
\end{proof}

As in \cite{Coolsaet}, apart from the `rotations' $M'_i$ defined in
(\ref{eq-M'}) we also introduce the `symmetries' $M_i$, as follows:
\[
M_{i} \isdef \begin{matrix}f_i & f_{i+1}\\ f_{i-1} & f_i
\end{matrix},
\]
or represented as elements of $\PGL(3,q)$~:
\[
\begin{matrix}
f_{i}^2 & f_{i+1}f_{i} & f_{i+1}^2 \\
0 & 1 & 0 \\
f_{i-1}^2 & f_{i-1}f_i & f_{i}^2 \\
\end{matrix}.
\]
Note that $M_0$ interchanges $X$- and $Z$-coordinates, that
$M_i=M_0M'_i$ and $M'_iM_0=M_{-i}$.
\goodbreak
\begin{lemma}
\label{lemma-2}
Let $H$ denote the subgroup of $\PGL(3,q)$ that leaves both the conic $C$ and its
external line $\ell$ invariant (and hence is a subgroup of $\PGL(2,q)$). Then $H$
consists of the elements $M_i,M'_i$, $i\in\Z_{q+1}$
and  is a dihedral group of order
$2(q+1)$.

The action of $H$ on $C$ and $\ell$ is given by
\begin{equation}\label{eq-HPQ}
\begin{array}{ccccccccccccccc}
M_i &: & P_j &\mapsto & P_{i-j},&\quad&Q_j&\mapsto&Q_{2i-j},\\
M'_i &: & P_j &\mapsto & P_{j+i},&\quad&Q_j&\mapsto&Q_{j+2i}.\\
\end{array}
\end{equation}
\end{lemma}
\begin{proof}
We leave it to the reader to prove that the action of $H$ on the
points of $C$ and $\ell$ is indeed as given in (\ref{eq-HPQ}).

Consider the action of a general element of $\PGL(2,q)$, as presented
in (\ref{eq-PGL3}), on the line $\ell$ (represented by a column
vector)~:
\begin{equation}\label{eq-abcd}
\begin{matrix}
a^2 & ab & b^2 \\
0 & ad+bc & 0 \\
c^2 & cd & d^2 \\
\end{matrix}^{-1}
\begin{matrix}1\\T\\1
\end{matrix}
=
\begin{matrix}
d^2 & bd & b^2 \\
0 & ad+bc & 0 \\
c^2 & ac & a^2 \\
\end{matrix}
\begin{matrix}1\\T\\1
\end{matrix}
=
\begin{matrix}
b^2 + Tbd + d^2 \\ T(ad+bc) \\ a^2 + Tac + c^2
\end{matrix}
\end{equation}
It follows that $\ell$ is stabilized by this action if and only if 
\begin{equation}\label{eq-cond1}
a^2 + Tac + c^2 = b^2 + Tbd + d^2 = ad+bc.
\end{equation}
Note that
\[
f_i^2 + Tf_{i+1}f_{i} + f_{i+1}^2 =
f_i^2 + Tf_{i-1}f_{i} + f_{i-1}^2 =
f_i^2 + f_{i-1}f_{i+1} = 1,
\]
by (\ref{eq-1}--\ref{eq-4}). Hence $M_i$ and $M'_i$ satisfy this condition. It is easily seen
that the set of all elements $M_i,M'_i$ with $i\in\Z_{q+1}$ forms a
group isomorphic to the dihedral group of order $2(q+1)$. It is
therefore sufficient to show that no other elements leave both $C$ and
$\ell$ invariant, or equivalently, that the stabilizer of $H$ of the
point $Q_0\in\ell$ has size at most 2.

The fact that $Q_0(1:0:1)$ is stabilized is expressed as follows:
\begin{equation}\label{eq-cond2}
a^2 + c^2 = b^2 + d^2,\quad ab + cd = 0.
\end{equation}
This yields $a+b+c+d=0$ and $ab=cd$, and therefore 
\[\begin{array}{cccccccccc}
0=a(a+b+c+d) &=& a^2+ab+ac+ad&=& a^2+cd+ac+ad &=& (a+c)(a+d), \\
0=b(a+b+c+d) &=& ab+b^2+bc+bd&=& cd+b^2+bc+bd &=& (b+c)(b+d), \\
0=c(a+b+c+d) &=& ac+bc+c^2+cd&=& ac+bc+c^2+ab &=& (a+c)(b+c), \\
0=d(a+b+c+d) &=& ad+bd+cd+d^2&=& ad+bd+ab+d^2 &=& (a+d)(b+d), \\
\end{array}
\]
and hence $a=c$ or $a=d$, $b=c$ or $b=d$, $a=c$ or $b=c$ and $a=d$ or
$b=d$. And therefore $a=c$ and $b=d$, or $a=d$ and $b=c$.

Together with (\ref{eq-cond1}), the first case leads to
\[
a^2 + Ta^2 + a^2 = b^2 + Tb^2 + b^2 = ab+ba,
\]
and then $Ta^2=Tb^2=0$, implying  $a=b=c=d=0$, which is not allowed.
%
The second case implies
\[
a^2 + Tab + b^2 = b^2 + Tab + a^2 = a^2+b^2,
\]
and then $ab=0$. When $a=0$ this yields transformation $M_0$, when
$b=0$ this yields $M'_0$.
\end{proof}
%
\section{Classification up to projective equivalence}
\label{sec-arcs}
After these preliminaries we can continue with the classification up
to projective equivalence of arcs with large conical subsets much in
the same way as when $q$ is odd. Because of the similarity with
\cite{Coolsaet} we shall try to be brief.

First $q$ must be large enough to ensure that the conic $C$ is completely
determined by the arc $S$. Let $S$ be an arc with conical subset $R$
and excess $|S\setminus R|=e$. If $C'$ is another conic giving rise to the conical
subset $R'=S\cap C'$ with $|S\setminus R'|=e'$. Then by
\cite[Lemma 3]{Coolsaet}, we know that $e'\ge |R| - 4$. As an immediate
consequence we obtain the following result.
\begin{lemma}
\label{lemma-16}
Let $\PG(2,q)$ be of even order.
If $q \ge 16$ then an arc of $\PG(2,q)$  with a large conical subset
of excess $e < 5$ can contain at most one such conical subset.
\end{lemma}
\begin{proof}
If $S$ has a large conical subset of excess $e$, then by the above
$e'\ge q/2 - 3$ and hence $e' \ge 5$ when $q\ge 16$. Hence, any other
conical subset will have an excess that is too large.
\end{proof}

In this section we shall consider an arc $S$ with a large conical
subset of excess 2. First note that the nucleus of the corresponding conic
cannot be one of the supplementary points: otherwise the line joining
the supplementary points would be a tangent to $C$ and then 3 points
of that tangent would belong to $S$.

Hence we may assume that the line joining the supplementary points is
external to $C$, and without loss of generality we may assume that
this line is the line $\ell$ introduced in the previous section. In
fact, we may even assume that one of the supplementary points is
$Q_0$, and then the other is of the form $Q_{2a}$ for some
$a\in\Z_{q+1}$.

Consider the following graph $\Gamma$:
\begin{itemize}
\item Vertices are the elements of $\Z_{q+1}$,
\item Two different vertices $i,j$ are adjacent if and only if the
  line $P_iP_j$ contains either $Q_0$ or $Q_{2a}$, i.e., if and only
  if $i+j\equiv0$ or $2a\pmod{q+1}$. (Cf.\ Lemma \ref{lemma-1}.)
\end{itemize}
Note that the degree of a vertex of $\Gamma$ is 2 in most cases,
except for the vertices $0$ and $a$ which are of degree 1, because
they lie on a tangent through $Q_0$ or $Q_{2a}$.
%
 It
follows that $\Gamma$ is the disjoint union of one path and a number
(possibily zero) of cycles. We may enumerate the vertices of the cycle or path that
contains $i$ as follows:
\begin{equation}
\label{HL eq cycle E}
\ldots,i,-i,2a+i,-2a-i,4a+i,-4a-i,\ldots
\end{equation}
For a cycle, this sequence eventually starts to repeat. For the path
this sequence stops at $0$ and $a$.

Define $n$ to be the order of $2a\pmod{q+1}$ and let $d
=\gcd(a,q+1)=(q+1)/n$. ($n$ and $d$ will always be odd.)  Note that
each of the vertices in (\ref{HL eq cycle E}) is equal to $\pm
i\pmod{d}$. Also note that $0$ and $a$ are both divisible by
$d$. Hence, if $i\not\equiv 0\pmod d$, then (\ref{HL eq cycle E}) denotes a
cycle, and not a path.  The cycle either has length $2n$ with $i\equiv
2na+i \pmod{q+1}$, or length $2n+1$ with $i\equiv
-2na-i\pmod{q+1}$. The latter case is impossible as the graph only has
two types of edges and two consecutive edges never are of the same
type, i.e. the number of edges in the cycle must be even. Hence the
cycle has length $2n$ and must contain all vertices that are equal to
$\pm i\pmod{d}$.
%
The path has endpoints $0$ and $a$ and contains all points that are multiple of $2a$. This implies that the path has size $n$.

This proves the following lemma.
\begin{lemma}
  $\Gamma$ is the disjoint union of $(d-1)/2$ cycles of length $2n$
  and one path of $n$ vertices, where $n$ is the order of
  $2a\pmod{q+1}$ and $d=(q+1)/n$, i.e. $d=\gcd(a,q+1)$.
\end{lemma}

Define the set $Z_k \isdef k+2a\Z_{q+1}=k+d\Z_{q+1}$. We call $Z_k$ a
\emph{half cycle} of $\Gamma$.  The cycles of $\Gamma$ are precisely
the sets $Z_k\cup Z_{-k}$, with $k$ in the range $1,\ldots,(d-1)/2$.

Denote by $N(R)$ the set of orbital indices of vertices of the large
conical subset $R$.  Since $S$ is an arc, $N(R)$ must be an
\textsl{independent set} of $\Gamma$.  To have an independent set of
size $q/2+1$ we need to take the largest possible independent
set for each component of $\Gamma$.  

When $n$ is odd, as is the case
here, the largest independent set in a
path with $n$ vertices is unique and has size $(n+1)/2$. The largest
independent set in a cycle is one of its two half cycles and has length $n$.  This
proves the following result.
\begin{theorem}
\label{HL theo extern}
Let $a\in \{1,\ldots,q\}$. Let $d=\gcd(a,q+1)$, $n=(q+1)/d$. 
Let $S = R \cup\{Q_0,Q_{2a}\}$, with $R\subset C$ and $|R|=q/2+1$.
Then $S$ is an arc of $\PG(2, q)$ if and only if $N(R)$ can be written as the union of pairwise disjoint sets, of the form
\[N(R)= \Pi \cup Z_{\pm 1} \cup \ldots\cup Z_{\pm(d-1)/2}, \]
with independent choices of sign, and
\begin{eqnarray*}
\Pi  &=& \{0,-2a,-4a,\cdots,a \}.
\end{eqnarray*}
\end{theorem}

Every arc listed in Theorem \ref{HL theo extern} can be denoted by its
\emph{signature} $(a;\epsilon_1,\ldots,\epsilon_{(d-1)/2})$, where
$\epsilon_k=\pm 1$ depending on the choice made for the half cycle
$Z_{\pm k}$.

\begin{theorem}
\label{HL theo extern2}
Let $q$ be even,  $a\in \{1,\ldots,q\}$, $d=\gcd(a,q+1)$ and
$n=(q+1)/d$. Let $H_a$ denote the subgroup of $\PGL(3,q)$ that leaves the conic $C$
invariant and fixes the pair $\{Q_0,Q_{2a}\}$. Then
$H_a=\{M'_0,M_a\}$ and acts as follows on the components of $\Gamma$
and the arc with signature
$(a;\epsilon_1,\ldots,\epsilon_{(d-1)/2})$
\[\def\arraystretch{1.2}
\begin{array}{|l|c|c|c|}
\hline
\mbox{Element of $H_a$} &\mbox{Image of $Z_k$}&\mbox{Image of}\\
& &(a;\epsilon_1,\ldots,\epsilon_{\frac{d-1}{2}}) \\
\hline
M'_0 \mbox{ (the identity)}& Z_{k}&(a;\epsilon_1,\ldots,\epsilon_{\frac{d-1}{2}})\\ 
M_{a}   &   Z_{-k} &(a;-\epsilon_{1},\ldots,-\epsilon_{\frac{d-1}{2}})\\  
\hline 
\end{array}\]
\end{theorem}
\begin{proof}
  Note that $H_a$ must fix the line joining $Q_0$ and $Q_{2a}$, i.e.,
  $\ell$, and hence is a subgroup of $H$. From (\ref{eq-HPQ}) it
  follows that the identity is the only transformation that fixes both
  $Q_0$ and $Q_{2a}$ and that $M_a$ is the only element of $H$ that
  interchanges $Q_0$ and $Q_{2a}$.

By (\ref{eq-HPQ}), $M_a$ maps $Z_k$ onto $Z_{a-k}$. Half cycle indices
are determined modulo $d$, and $a$ is a multiple of $d$. Hence $Z_{a-k}=Z_{-k}$
\end{proof}

\begin{corollary}
  Let $q\ge 16$, $q$ even. Let $H_S$ denote the subgroup of
  $\PG(3,q)$ that leaves invariant the arc $S$ with signature
  $(a;\epsilon_1,\ldots,\epsilon_{d-1})$.  If $d=1$, then $H_S =
  H_a$. Otherwise, $H_S$ is trivial.
\end{corollary}

\begin{lemma}
\label{lem L_q(a)}
Let $L_q(a)$ denote the number of projectively  inequivalent arcs
$S$ with a signature of the form
$(a;\epsilon_1,\ldots,\epsilon_{(d-1)/2})$, with $d=\gcd(a,q+1)$. Then
\begin{equation}
\label{HL eq E_q(a)}
L_q(a) = 
\left\{\begin{array}{ll}
 1, &\mbox{when $d=1$},\\
2^{\frac{d-3}{2}},\quad&\mbox{when $d>1$.}
\end{array}\right.
\end{equation}
\end{lemma}
\begin{proof}
If $d=1$, then there is clearly the one signature $(a)$. 
Otherwise, when $d>1$, the number of signatures is equal to $2^{(d-1)/2}$ and the value of $|S^{H_a}| =
|H_a|/|H_S|$ is equal to $2$ for each such arc. Hence, $L_q(a)=\frac{2^{(d-1)/2}}{2}$ for $d>1$.
 \end{proof}
 
 
 %
\begin{theorem}
\label{HL number}
Let $q\ge 16$, $q$ even. The number $L_q$ of projectively inequivalent arcs $S$ in
$\PG(2,q)$ of size $|S|=q/2+3$, with a conical subset $R=S\cap C$ of
size $|R|=q/2+1$, is given by
\begin{equation}\label{eq-HL number}
L_q =
 \frac12 
{\sum_{d}}'
\phi\left(\frac{q+1}{d}\right) L_q(d)
\end{equation}
where the sum is restricted to all proper divisors $d$ of $q+1$,
$\phi$ denotes Eulers totient function, and $L_q(d)$ is as given
in Lemma \ref{lem L_q(a)}.
\end{theorem}
\begin{proof}
Note that $M_0$ interchanges $Q_{2a}$ and $Q_{-2a}$ and fixes $Q_0$,
and hence to enumerate all arcs up to
isomorphism, we should consider only one of $a$
and $q+1-a$. In other words, we may take $a$ in the range
$1,\ldots,q/2$.

The total number of inequivalent arcs is then given by
$\sum_{a=1}^{q/2} L_q(a)$.  Note that $L_q(a)$ does not directly
depend on $a$, but only on $d=\gcd(a,q+1)$.  The number of integers
$a$, $1\le a \le q/2$ such that $d=\gcd(a,q+1)$ is equal to
$\frac12 \phi((q+1)/d)$.
\end{proof}

The table below lists the values of $L_q$ for small
values of $q$.
\[
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
q & 16 & 32 & 64 & 128 & 256 \\
\hline
L_q & 8 & 31 & 100 & 1048639 & 128 \\
\hline
\end{array}
\]
For larger values of $q$ the value of $L_q$ becomes very large, unless
$q+1$ is a (Fermat) prime, in which case $L_q=q/2$.

For $32\le q\le 256$ we have checked by computer that all of these
arcs are complete. For $q=16$ four of the eight arcs are complete and
there exist three inequivalent arcs of excess 3, all of them complete.

% For $q=8$, where we can no longer rely on
% Lemma \ref{lemma-16}, there is one complete arc of size $6$ (i.e., of
% excess one), and one complete arc of size $10$, the conic with added nucleus.



%
\subsection*{Acknowledgements}

We would like to thank the anonymous referees for helpful comments and
suggestions and for correcting an error in the original proof of
Lemma \ref{lemma-2}.
%
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%
\end{document}
