# Maximal number of Skew lines in the Fermat Surface

Sally Andria, Jacqueline Rojas and Wállace Mangueira

###### Abstract.

It is well-known that the Fermat surface of degree $d\geq 3$ has $3d^{2}$
lines. However, it has not yet been established what is the maximum number of
pairwise disjoint lines that it can have if $d\geq 4$. In this article we show
that the maximal number of skew lines on the Fermat surface of degree $d\geq
4$ is $3d$ either $d$ even or $d$ odd distinct of 5, otherwise ($d=5$) it
contains no more than 13 pairwise disjoint lines.

###### Key words and phrases:

Skew lines, Fermat surface

## Introduction

It is well-known that the Fermat surface of degree $d$ in the complex
projective space has $3d^{2}$ lines for $d\geq 3$, so it is an lower bound for
$\ell_{d}$, the maximum number of lines that a smooth surface of degree $d$ in
$\mathbb{P}^{3}$ can have. In fact, since 1882 has been know that the so
called Schur’s quartic contains exactly 64 lines ([12]). And only in 1943, B.
Segre proved that $\ell_{4}=64$ ([13])111 Even though a gap was discovered in
Segre’s proof by Rams-Schütt in 2015 ([11]), the claim is still correct., but
$\ell_{d}$ remains unknown for $d\geq 5$. In this direction the articles of
Caporaso-Harris-Mazur ([6]) and Boissière-Sarti ([5]) exhibited lower bounds
for these numbers, which leads us to infer that $3d^{2}$ not provide the
maximum number of lines on a smooth surface of degree $d\geq 4$ in
characteristic zero. On the other hand according the Bauer-Rams
$11d^{2}-30d+18$ is an upper bound for the maximum number of lines on a smooth
surface of degree $d\geq 3$ in $\mathbb{P}^{3}(k)$ being $k$ a field of
characteristic 0 or of characteristic $p>d$ ([4]). For example, the Fermat
surface defined by the vanishing of the polynomial
$x^{q+1}+y^{q+1}+z^{q+1}+w^{q+1}$ em $\mathbb{P}^{3}(k)$ being $k$ a field
extension of $\mathbb{F}_{q^{2}}$ where $q=p^{e}$ for a prime $p$ contains
$q^{4}+q^{3}+q+1$ lines, which exceed the Bauer-Rams’s upper bound and leads
the authors (cf. [2] and references there in) to conjecture that these Fermat
surfaces may provide the maximum number of lines possible on a surface of a
given degree in characteristic $p>0$.

Another problem related to this is to determine the maximum number,
${\mathfrak{s}}_{d}$, of pairwise disjoint lines (or skew lines) that a smooth
surface of degree $d$ can have. It is known that ${\mathfrak{s}}_{3}=6$,
${\mathfrak{s}}_{4}=16$ ([9]) and ${\mathfrak{s}}_{6}=48$ ([7]).222 In 1975,
Miyaoka gave the upper bound $r_{d}\leq 2d(d-2)$ if $d\geq 4$ ([8]). Some
lower bounds were given by Rams ([10]) and Boissiere-Sarti ([5]). However,
$r_{d}$ remain unknown for $d=5$ and $d\geq 7$.

To our knowledge, there is no place in the modern literature where the maximum
number of pairwise disjoint lines on the Fermat surfaces is explicitly stated.
For example, over the complex numbers in Rams article ([10]) he claims: “Let
us mention that the Fermat surface ${{\rm F}_{d}}$, i.e. the surface with
$3d^{2}$ lines (the largest number known so far for $d\neq 4,6,8,12,20$),
contains no family of $3d$ pairwise disjoint lines.” without any proof. For
Fermat surfaces over fields of characteristic $p\neq 0$, in [2] the authors
exhibit some bounds for $p=2,3$.

The aim of our work is to show in an elementary and self-contained way that
the maximum number of pairwise disjoint lines on Fermat surfaces of degree
$d\geq 3$ over the complex numbers is exactly $3d$ for any $d$ even and $d$
odd distinct from 3 and 5 (being such numbers 6 and 13, for $d=3,5$,
respectively), according to Theorem In order to do that, we first established
a notation for the set of lines in ${{\rm F}_{d}}$ (see (, in such a way that,
we obtain the stratification
${\mathcal{L}}^{0}\,\dot{\cup}\,{\mathcal{L}}^{1}\,\dot{\cup}\,{\mathcal{L}}^{2}$
with $\\#{\mathcal{L}}^{i}=d^{2},$ for $i=0,1,2$ of these lines in ${{\rm
F}_{d}}$ (cf. Proposition Moreover, the relations established in Proposition
ogether with Proposition ve us enough conditions to study the intersection
between the lines in the families ${\mathcal{L}}^{i}$ and ${\mathcal{L}}^{j}$
for $i\neq j$. Next, we check that the maximum number of pairwise disjoint
lines on the family ${\mathcal{L}}^{i}$ is $d$ for all $i$, which implies that
${\mathfrak{s}}({{\rm F}_{d}})\leq 3d$ (being ${\mathfrak{s}}({{\rm F}_{d}})$
the maximum number of pairwise disjoint lines that ${{\rm F}_{d}}$ can have).
In fact, if $d$ is even, then we easily get a family consisting of $3d$
pairwise disjoint lines on ${{\rm F}_{d}}$ (cf. Proposition otherwise we are
faced with a real/generalized ‘Sudoku game’ to find such maximal set of
pairwise disjoint lines on ${{\rm F}_{d}}$ (cf. Sections Secdgeq4). To our
surprise the case $d=5$ was the only one (for $d\geq 4$) for that there is no
family with $3d$ skew lines.

Finally, we note that to study the maximum number of rational curves (in
particular lines) which do not intersect one another on a surface is an
important tool to classify surfaces in the projective space (cf. [9], [3] and
[1]), as well as to determine all the lines on a smooth surface from some set
of its skew lines ([14]).

##  1\. Lines on Fermat’s surfaces

Let ${\rm F}_{d}$ be the degree $d$ Fermat surface in the projective complex
space defined as the zeros locus of

$x^{d}-y^{d}-z^{d}+w^{d}\in\mathbb{C}[x,y,z,w].$

Set $\Phi({{\rm F}_{d}})=\\{\ell\subset{{\rm F}_{d}}\ |\ \ell\textrm{ is a
line}\\}$. An easy verification allows us to conclude that
$\mathcal{L}^{j}=\\{L_{k,i}^{j}\\}\subset\Phi({{\rm F}_{d}})$ for $j=0,1,2$
being

$L^{0}_{k,i}:\left\\{\begin{array}[]{l}y=\eta^{i}x\\\
w=\eta^{k}z\end{array}\right.\\!\\!\\!\\!,\,\,\,\,L^{1}_{k,i}:\left\\{\begin{array}[]{l}x=\eta^{k+i}z\\\
y=\eta^{i}w\end{array}\right.\,\,\hbox{ and }\quad
L^{2}_{k,i}:\left\\{\begin{array}[]{l}x=v\eta^{i}w\\\
y=v\eta^{k+i}z\end{array}\right.$ (1.0.1)

where $\eta$ is a primitive $d$th root of the unity, $v$ is a complex number
such that $v^{d}=-1$ and $k,i\in\\{0,1,2,\ldots,d-1\\}$.333Here we use the
indices $i$ and $k+i$ to describe the lines in the families
${{\mathcal{L}}}^{1}$and ${{\mathcal{L}}}^{2}$ instead of simply $i,k$,
because this simplifies the writing of incidence relations between the lines
in ${{\rm F}_{d}}$, as we will see later. Moreover
$\\#(\mathcal{L}^{j})=d^{2}$ for $j=0,1,2$.

######  Proposition 1.1.

With the above notation $\Phi({{\rm
F}_{d}})={\mathcal{L}}^{0}\,\dot{\cup}\,{\mathcal{L}}^{1}\,\dot{\cup}\,{\mathcal{L}}^{2}$.
Thus $\\#(\Phi({{\rm F}_{d}}))=3d^{2}$.

###### Proof.

Let us consider the line $L=Z(x,y)$ in $\mathbb{P}^{3}$. Note that we can
stratified the lines in ${{\rm F}_{d}}$ studying their intersection with the
line $L$, i.e.

$\Phi({{\rm F}_{d}})=\Big{\\{}\ell\in\Phi({{\rm F}_{d}})\ |\ \ell\cap
L\neq\emptyset\Big{\\}}\,\,\dot{\cup}\,\,\Big{\\{}\ell\in\Phi({{\rm F}_{d}})\
|\ \ell\cap L=\emptyset\Big{\\}}.$

Let $\ell$ be a line in ${{\rm F}_{d}}$. Have in mind that ${{\rm F}_{d}}\cap
L=\Big{\\{}[0:0:1:\eta^{j}]\Big{\\}}_{j=0}^{d-1}$ where $\eta$ is a primitive
$d$th root of the unity. Therefore, according to $\ell\cap L\neq\emptyset$ or
$\ell\cap L=\emptyset$ we have, respectively:

  * •

$\ell$ is determined by the points ${\rm p}=[0:0:1:\eta^{k}]\in L$ for exactly
one value of $k\in\\{0,\ldots,d-1\\}$ (since $L\not\in\Phi({{\rm F}_{d}})$)
and ${\rm q}=[\alpha:\beta:0:\gamma]$ with $\alpha,\beta,\gamma\in\mathbb{C}$
not all zero. Thus

$\begin{array}[]{lcl}\ell\subset{{\rm
F}_{d}}&\Longleftrightarrow&\alpha^{d}v^{d}-\beta^{d}v^{d}-u^{d}+(\eta^{k}u+\gamma
v)^{d}=0\quad\forall\,[u:v]\in\mathbb{P}^{1}.\\\
&\Longleftrightarrow&\left\\{\begin{array}[]{rl}\alpha^{d}-\beta^{d}+\gamma^{d}=0&\\\
\eta^{k(d-j)}\gamma^{j}=0&\hbox{ for }j=1,\ldots,d-1.\end{array}\right.\\\
&\Longleftrightarrow&\alpha^{d}-\beta^{d}=0\,\,(\alpha\beta\neq 0)\,\,\hbox{
and }\,\,\gamma=0.\\\
&\Longleftrightarrow&\ell=Z(w-\eta^{k}z,y-\eta^{i}x)=L^{0}_{k,i}\in{\mathcal{L}}^{0}\,\,\hbox{
with }\,\,\alpha^{-1}\beta=\eta^{i}.\end{array}$

Therefore, ${\mathcal{L}}^{0}=\Big{\\{}\ell\in\Phi({{\rm F}_{d}})\ |\ \ell\cap
L\neq\emptyset\Big{\\}}.$

  * •

If $\ell\cap L=\emptyset$, then we can assume that $\ell$ is defined by

$x-\alpha z-\beta w\,\,\hbox{ and }\,\,y-\gamma z-\delta w\,\,\hbox{ with
}\,\,\alpha\delta-\beta\gamma\neq 0.$

Thus,

$\begin{array}[]{lcl}\ell\subset{{\rm F}_{d}}&\Longleftrightarrow&(\alpha
z+\beta w)^{d}-(\gamma z+\delta w)^{d}+z^{d}-w^{d}=0.\\\
&\Longleftrightarrow&\left\\{\begin{array}[]{rl}\alpha^{d}-\gamma^{d}+1=0&\\\
\beta^{d}-\delta^{d}-1=0&\\\
\alpha^{d-j}\beta^{j}-\gamma^{d-j}\delta^{j}=0&\hbox{ for
}j=1,\ldots,d-1.\end{array}\right.\end{array}$ (1.1.2)

From (or $j=1,2$ (as $d\geq 3$), we get $\alpha^{d-1}\beta=\gamma^{d-1}\delta$
and $\alpha^{d-2}\beta^{2}=\gamma^{d-2}\delta^{2}$, which implies that
$\gamma^{d-2}\delta(\gamma\beta-\alpha\delta)=0$. Therefore, $\gamma\delta=0$.
In fact, we have

$\left\\{\begin{array}[]{l}\gamma=0\Longrightarrow\beta=0\Longrightarrow\ell\in{\mathcal{L}}^{2}.\\\
\delta=0\Longrightarrow\alpha=0\Longrightarrow\ell\in{\mathcal{L}}^{1}.\end{array}\right.$

Finally, note that $[\eta^{k+i}:\eta^{i}:1:1]\in L^{1}_{k,i}\\--L^{2}_{t,j}$
for any $t,j$. Thus, ${\mathcal{L}}^{1}\cap{\mathcal{L}}^{2}=\emptyset$. ∎

### Studying the intersections between the lines on ${{\rm F}_{d}}$

In what follows we use the notation $a\equiv_{d}b$ instead of $a\equiv
b\,({\rm mod}\,d)$ to indicate that $a$ is congruent to $b$ modulo $d$.

######  Proposition 1.2.

With the notation as in
$(\hyperref@@iv{retasL0L1L2}{)}{$}{.}Forany$a,b,i,j,k,t∈{0,1,…,d-1}$holds\begin{enumerate}
\par\enumerate@item@[$(\rm a)$] $L_{a,b}^{0}\cap
L_{k,i}^{s}\not=\emptyset\iff\left\\{\begin{array}[]{ll}a=k\,\,\hbox{ or
}\,\,b=i&\hbox{ if }\,\,s=0,\\\ b-a\equiv_{d}k&\hbox{ if }\,\,s=1,\\\
b+a\equiv_{d}k&\hbox{ if }\,\,s=2.\end{array}\right.$
\par\enumerate@item@[$(\rm b)$] $L_{k,i}^{s}\cap
L_{t,j}^{s_{1}}\not=\emptyset\iff\left\\{\begin{array}[]{ll}k+i\equiv_{d}t+j\,\,\hbox{
or }\,\,i=j&\hbox{ if }\,\,s=s_{1}\in\\{1,2\\},\\\
v^{2}\eta^{t+2j}=\eta^{k+2i}&\hbox{ if }\,\,s=1,\,s_{1}=2.\end{array}\right.$
\par\enumerate@item@[$(\rm c)$] If $d$ is odd, then we can choose $v=-1$ and
follows that \begin{center} $L_{k,i}^{1}\cap L_{t,j}^{2}\not=\emptyset\iff
k+2i\equiv_{d}t+2j$. \end{center} \par\enumerate@item@[$(\rm d)$] If $d$ is
even, then \begin{center} $L_{k,i}^{1}\cap L_{k,j}^{2}=\emptyset$ for all
$i,j$. \end{center} \par\end{enumerate}$

###### Proof.

The statements (a) and (b) are straightforward verification (from the
definitions of the lines $L^{s}_{k,i}$ in (, and (c) follows from (b).

Now, let us consider $d\geq 4$ even and suppose that $L_{k,i}^{1}\cap
L_{k,j}^{2}\neq\emptyset$ for some $i,j$. Thus, follows from (b) that
$v^{2}\eta^{2j}=\eta^{2i}$, which implies that $v\eta^{j}=\pm\eta^{i}$. Here,
if we computed the $d$-th power of $v\eta^{j}=\pm\eta^{i}$, we lead to an
absurd result. ∎

######  Definition 1.3.

Let ${\mathfrak{s}}(X)$ be the maximum number of skew lines in
$X\subseteq{{\rm F}_{d}}$.

The relations in the above proposition allows us to show that.

######  Corollary 1.4.

${\mathfrak{s}}(\mathcal{L}^{s})=d,$ for $s=0,1,2$. In particular,
${\mathfrak{s}}({{\rm F}_{d}})\leq 3d$.

###### Proof.

From (a) in Proposition have that $L^{0}_{a,b}\cap L_{k,i}^{0}=\emptyset$ iff
$a\neq k$ and $b\neq i$. Thus, any subset ${\rm C}\subset{\mathcal{L}}^{0}$ of
pairwise disjoint lines is constituted by lines $L^{0}_{a,b}$, to whom the
indices $a$ are all distinct. Hence, ${\mathfrak{s}}({\mathcal{L}}^{0})\leq d$
(since $a\in\\{0,\ldots,d-1\\}$). For another hand
$\\{L^{0}_{a,a}\\}_{a=0}^{d-1}$ is a family of $d$ skew lines in
${\mathcal{L}}^{0}$. Therefore, ${\mathfrak{s}}(\mathcal{L}^{0})=d$.

One more time, from (b) in Proposition have that $L^{s}_{k,i}\cap
L_{t,j}^{s}=\emptyset$ iff $i\neq j$ and $k+i\not\equiv_{d}t+j$. Again, the
condition $i\neq j$ (with $i,j\in\\{0,1,...,d-1\\})$ implies that
${\mathfrak{s}}({\mathcal{L}}^{s})\leq d$ for $s=1,2$. However,
$\\{L^{s}_{k,i}\\}_{i=0}^{d-1}$ is constitute by $d$ skew lines in
${\mathcal{L}}^{s}$. Therefore, ${\mathfrak{s}}(\mathcal{L}^{s})=d$ for
$s=1,2$. Finally, note that ${\mathfrak{s}}({{\rm
F}_{d}})\leq{\mathfrak{s}}(\mathcal{L}^{0})+{\mathfrak{s}}(\mathcal{L}^{1})+{\mathfrak{s}}(\mathcal{L}^{2})=3d$.∎

##  2\. Characterizing sets of skew lines on ${{\rm F}_{d}}$

From Corollary have the upper bound $3d$ for ${\mathfrak{s}}({{\rm F}_{d}})$.
So we are invited to look for maximal subsets of skew lines in ${{\rm
F}_{d}}$. In this direction an important tool is the next proposition, which
will established some kind of sudoku’s rule for our game444 The game is: given
$d\geq 4$ find the maximum number of pairwise lines in the Fermat surface
${{\rm F}_{d}}$.. In fact, the lower bound $2d$ for ${\mathfrak{s}}({{\rm
F}_{d}})$ will be established in Corollary rom this point onward, we start
playing (pay attention to the rules!).

######  Proposition 2.1.

Let $R_{d}=\\{0,1,\ldots,d-1\\}$ and $r_{d}:\mathbb{Z}\longrightarrow R_{d}$
the remainder555If $a\in\mathbb{Z}$, then $r_{d}(a)=r$ where $r\in R_{d}$ and
$a\equiv_{d}r$. function by $d$. Consider the functions

$\begin{array}[]{ccccccccc}\psi_{d}:&R_{d}\times
R_{d}&\longrightarrow&R_{d}&\,\,\hbox{ and }&\varphi_{d,{\pm}}:&R_{d}\times
R_{d}&\longrightarrow&R_{d}\\\
&(k,i)&\longmapsto&r_{d}(k+2i)&&&(k,i)&\longmapsto&r_{d}(i\pm k).\end{array}$

For $u\in R_{d}$, $s\in\\{0,1,2\\}$ define

$\displaystyle{\rm D}_{u}^{s}$
$\displaystyle:=\big{\\{}\mathcal{L}_{k,i}^{s}\in\mathcal{L}^{s}\ \mid\
(k,i)\in\psi_{d}^{-1}(u)\big{\\}},\hbox{ for }s=1,2;$ $\displaystyle{\rm
D}_{u,\pm}^{0}$
$\displaystyle:=\big{\\{}\mathcal{L}_{k,i}^{s}\in\mathcal{L}^{0}\ \mid\
(k,i)\in\varphi_{d,\pm}^{-1}(u)\big{\\}}.$

It is verified that

  1. $(\rm a)$

the restriction of $\psi_{d}$ and $\varphi_{d,\pm}$ to $R_{d}\times\\{i\\}$ is
a bijection for all $i$;

  2. $(\rm b)$

${\left.\kern-1.2pt\varphi_{d,\pm}\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{\\{k\\}\times
R_{d}}}:\\{k\\}\times R_{d}\to R_{d}$ is a bijection for all $k$;

  3. $(\rm c)$

$\psi_{d}\mid_{\\{k\\}\times R_{d}}:\\{k\\}\times R_{d}\to R_{d}$ is a
bijection for all $k$, if $d$ is odd;

  4. $(\rm d)$

$\\#\psi_{d}^{-1}(u)=d$ and $\\#{\varphi_{d,\pm}}^{-1}(u)=d$ for all $u\in
R_{d}$;

  5. $(\rm e)$

${\rm D}_{u,\pm}^{0}\subset{\mathcal{L}}^{0}$ and ${\rm
D}_{u}^{s}\subset{\mathcal{L}}^{s}$ for $s=1,2,$ are families of $d$ skew
lines.

###### Proof.

It is left to the reader.∎

In the next corollary we find the lower bound $2d$ for ${\mathfrak{s}}({{\rm
F}_{d}})$.

######  Corollary 2.2.

${\mathfrak{s}}(\mathcal{L}^{0}\cup\mathcal{L}^{s})=2d$ for $s=1,2$ and
${\mathfrak{s}}(\mathcal{L}^{1}\cup\mathcal{L}^{2})=2d$. Thus
$2d\leq{\mathfrak{s}}({{\rm F}_{d}})\leq 3d$.

###### Proof.

From Corollary have that
${\mathfrak{s}}(\mathcal{L}^{0})={\mathfrak{s}}(\mathcal{L}^{1})={\mathfrak{s}}(\mathcal{L}^{2})=d$.
Which implies that

${\mathfrak{s}}(\mathcal{L}^{0}\cup\mathcal{L}^{s})\leq 2d,\textrm{ for
}s=1,2\hskip 28.45274pt\textrm{ and }\hskip
28.45274pt{\mathfrak{s}}(\mathcal{L}^{1}\cup\mathcal{L}^{2})\leq 2d.$

Thus, it is enough to find a family of $2d$ skew lines on
$\mathcal{L}^{0}\cup\mathcal{L}^{s}$ for $s=1,2$ and on
$\mathcal{L}^{1}\cup\mathcal{L}^{2}$, respectively. For the first statement,
from item (a) of Proposition e conclude that the following two sets are
constituted by $2d$ skew lines

$\\{L^{0}_{0,0},L^{0}_{1,1},\ldots,L^{0}_{d-1,d-1},L_{1,0}^{1},L_{1,1}^{1},\ldots,L_{1,d-1}^{1}\\};$
$\\{L^{0}_{0,0},L^{0}_{1,d-1},L^{0}_{2,d-2},\ldots,L^{0}_{d-1,1},L_{1,0}^{2},L_{1,1}^{2},\ldots,L_{1,d-1}^{2}\\}.$

Now, for the second statement, we have that $\\#{\rm D}_{0}^{1}=d$ and
$\\#{\rm D}_{1}^{2}=d$ in accordance with the item (e) of Proposition oreover,
by item (b) of Proposition have that $L_{k,i}^{1}\cap L_{m,n}^{2}=\emptyset$
for any $L^{1}_{k,i}\in{\rm D}_{0}^{1}$ and $L_{m,n}^{2}\in{\rm D}_{1}^{2}$.
Therefore, ${\rm D}_{0}^{1}\cup{\rm D}_{1}^{2}$ is a family of $2d$ skew lines
in ${\mathcal{L}}^{1}\cup{\mathcal{L}}^{2}$. ∎

From now on, we will focus on capturing maximal subsets of skew lines in
${{\rm F}_{d}}$, revisiting the conditions that must be satisfied by such
subsets.

####  2.0.1. Rewriting conditions for subsets of skew lines in ${{\rm F}_{d}}$

In order to find maximal sets of skew lines in ${{\rm F}_{d}}$, we started by
characterizing those subsets of skew lines in ${\mathcal{L}}^{s}$ for each
$s=0,1,2$ in terms of $\psi_{d}$ and $\varphi_{d,\pm}$ (cf. Proposition when
it comes to the case.

One more time from the Proposition obtain the following two corollaries.

######  Corollary 2.3.

Let ${\rm C}\subset\Phi({{\rm F}_{d}})$ and define ${\rm C}^{s}:={\rm
C}\cap{\mathcal{L}}^{s}$ for $s\in\\{0,1,2\\}$.

  * $({\rm a})$

${\rm C}^{0}$ is constituted by skew lines $\iff$
$\left\\{\textrm{\begin{tabular}[]{l}${\rm
C}^{0}=\\{L^{0}_{a_{1},b_{1}},\ldots,L^{0}_{a_{m},b_{m}}\\}$ with $\\#{\rm
C}^{0}=m$,\\\ $0\leq a_{1}<\cdots<a_{m}\leq d-1$ and there is\\\ a permutation
$\sigma$ of $R_{d}$ such that $\sigma(a_{i})=b_{i}$.\end{tabular}}\right.$

  * $({\rm b})$

${\rm C}^{1}$ is constituted by skew lines $\iff$
$\left\\{\textrm{\begin{tabular}[]{l}${\rm
C}^{1}=\\{L^{1}_{a_{1},b_{1}},\ldots,L^{1}_{a_{m},b_{m}}\\}$ with $\\#{\rm
C}^{1}=m$,\\\ $0\leq b_{1}<\cdots<b_{m}\leq d-1$ and $\varphi_{d,+}$\\\
restricted to $\\{(a_{i},b_{i})\\}_{i=1}^{m}$ is
injective.\end{tabular}}\right.$

  * $({\rm c})$

${\rm C}^{2}$ is constituted by skew lines $\iff$
$\left\\{\textrm{\begin{tabular}[]{l}${\rm
C}^{2}=\\{L^{2}_{a_{1},b_{1}},\ldots,L^{2}_{a_{m},b_{m}}\\}$ with $\\#{\rm
C}^{2}=m$,\\\ $0\leq b_{1}<\cdots<b_{m}\leq d-1$ and $\varphi_{d,+}$\\\
restricted to $\\{(a_{i},b_{i})\\}_{i=1}^{m}$ is
injective.\end{tabular}}\right.$

######  Remark 2.4.

Note that ${\mathcal{L}}^{s}_{k}=\\{L^{s}_{k,i}\in{\mathcal{L}}^{s}\mid i\in
R_{d}\\}$ for $k\in R_{d}$ and $s\in\\{0,1,2\\}$ is constituted by $d$ skew
lines if $s\in\\{1,2\\}$ (according to $({\rm b})$ in Proposition Moreover,

${\mathcal{L}}^{s}={\mathcal{L}}^{s}_{0}\,\,\dot{\cup}\,\cdots\,\dot{\cup}\,{\mathcal{L}}^{s}_{d-1}\quad\hbox{
for any }\,s\in\\{0,1,2\\}.$

Now, we will concentrate our attention on the description of those subsets
${\rm C}^{s}$ of ${\mathcal{L}}^{s}$ consisting of skew lines such that ${\rm
C}^{s}\cup{\rm C}^{s_{1}}$ is also formed by skew lines (for $0\leq
s<s_{1}\leq 2$).

In what follows, for any subset ${\rm X}\subseteq\Phi({{\rm F}_{d}})$ we may
identify the line ${\mathcal{L}}^{s}_{k,i}\in{\rm X}$ with the pair $(k,i)$
(which will be clear from the context). Having this in mind we will consider
$\psi_{d}({\rm X})$ and $\varphi_{d,\pm}({\rm X})$.

######  Corollary 2.5.

With the above notation. Assume that ${\rm C}^{0},{\rm C}^{1}$ and ${\rm
C}^{2}$ consist of skew lines. Then we have

  * $({\rm a})$

${\rm C}^{0}\,\cup\,{\rm C}^{1}$ is constituted by skew lines $\iff$ ${\rm
C}^{1}\cap{\mathcal{L}}^{1}_{k}=\emptyset$ for every $k\in\varphi_{d,-}({\rm
C}^{0})$.

  * $({\rm b})$

${\rm C}^{0}\,\cup\,{\rm C}^{2}$ is constituted by skew lines $\iff$ ${\rm
C}^{2}\cap{\mathcal{L}}^{2}_{k}=\emptyset$ for every $k\in\varphi_{d,+}({\rm
C}^{0})$.

  * $({\rm c})$

For $d$ odd holds ${\rm C}^{1}\,\cup\,{\rm C}^{2}$ is constituted by skew
lines $\iff$ $\psi_{d}({\rm C}^{1})\,\cap\,\psi_{d}({\rm C}^{2})=\emptyset$.

######  Remarks 2.6.

Assume $d$ odd. If ${\rm C}^{s}\subset{\mathcal{L}}^{s}$ consist of skew lines
for $s=0,1,2$, then Corollary lows to concluded that

  * •

If ${\rm C}^{1}={\mathcal{L}}^{1}_{k}$ $($resp. ${\rm
C}^{2}={\mathcal{L}}^{2}_{k})$ for some $k\in R_{d}$, then ${\rm
C}^{2}=\emptyset$ $($resp. ${\rm C}^{1}=\emptyset)$.

  * •

If $\varphi_{d,+}({\rm C}^{0})=R_{d}$ $($resp. $\varphi_{d,-}({\rm
C}^{0})=R_{d}$), then ${\rm C}^{2}=\emptyset$ $($resp. ${\rm
C}^{1}=\emptyset)$666It is also true for $d$ even..

  * •

$\\#\psi_{d}({\rm C}^{1})+\\#\psi_{d}({\rm C}^{2})\leq d.$ In particular, if
$\psi_{d}({\rm C}^{1})=R_{d}$, then ${\rm C}^{2}=\emptyset$ and vice versa.

Let’s see an example of a family of 13 skew lines in ${\rm F}_{5}$.

######  Example 2.7.

Note that ${\rm C}^{0}=\\{L^{0}_{0,4},L^{0}_{2,0},L^{0}_{3,1},L^{0}_{4,3}\\}$
consists of four skew lines (cf. (a) in Corollary Furthermore, in the rows of
the next table we register the values of $\varphi_{5,\pm}({\rm C}^{0})$,
respectively.

${\rm C}^{0}$ | $L^{0}_{0,4}$ | $L^{0}_{2,0}$ | $L^{0}_{3,1}$ | $L^{0}_{4,3}$
---|---|---|---|---
$\varphi_{5,-}$ | 4 | 3 | 3 | 4
$\varphi_{5,+}$ | 4 | 2 | 4 | 2

Now, having in mind Corollary r the choice of ${\rm
C}^{s}\subset{\mathcal{L}}^{s}$ such that ${\rm C}^{0}\cup{\rm C}^{s}$ to be
constituted by skew lines for $s=1,2$, it is necessary that

${\rm
C}^{1}\cap{\mathcal{L}}^{2}_{k}=\emptyset\quad\forall\,k\in\\{3,4\\}\quad\hbox{and}\quad{\rm
C}^{2}\cap{\mathcal{L}}^{2}_{k}=\emptyset\quad\forall\,k\in\\{2,4\\}.$

So, ${\rm C}^{1}=\\{L^{1}_{0,1},L^{1}_{0,4},L^{1}_{2,0},L^{1}_{2,3}\\}$ and
${\rm
C}^{2}=\\{L^{2}_{0,0},L^{2}_{1,2},L^{2}_{3,1},L^{2}_{3,3},L^{2}_{3,4}\\}$ are
admissible choices. As well as, according to the information on the rows in
the following two tables.

${\rm C}^{1}$ | $L^{1}_{0,1}$ | $L^{1}_{0,4}$ | $L^{1}_{2,0}$ | $L^{1}_{2,3}$
---|---|---|---|---
$\varphi_{5,+}$ | 1 | 4 | 2 | 0
$\psi_{5}$ | 2 | 3 | 2 | 3
${\rm C}^{2}$ | $L^{2}_{0,0}$ | $L^{2}_{1,2}$ | $L^{2}_{3,1}$ | $L^{2}_{3,3}$ | $L^{2}_{3,4}$
---|---|---|---|---|---
$\varphi_{5,+}$ | 0 | 3 | 4 | 1 | 2
$\psi_{5}$ | 0 | 0 | 0 | 4 | 1

We have that ${\rm C}^{s}$ consists of skew lines for $s=1,2$ (cf. (b) and (c)
in Corollary nd $\psi_{5}({\rm C}^{1})\cap\psi_{5}({\rm C}^{2})=\emptyset$,
which implies that ${\rm C}^{1}\cup{\rm C}^{2}$ also is formed by skew lines
(cf. (c) in Corollary Therefore, ${\rm C}:={\rm C}^{0}\cup{\rm C}^{1}\cup{\rm
C}^{2}$ consists of 13 skew lines in ${\rm F}_{5}$. In fact, in Theorem will
prove that ${\mathfrak{s}}({\rm F}_{5})=13$.

######  Remarks 2.8.

Note that the correspondence between lines in ${\mathcal{L}}^{s}$ and pairs in
$R_{d}\times R_{d}$ for each $s\in\\{0,1,2\\}$ allows us to associate to the
$d^{2}$ lines in ${\mathcal{L}}^{s}$ (for each $s\in\\{0,1,2\\}$) the
following $d\times d$ square matrix:

$\left(\begin{array}[]{cccc}(0,0)&(1,0)&\cdots&(d-1,0)\\\
(0,1)&(1,1)&\cdots&(d-1,1)\\\ \vdots&\vdots&\cdots&\vdots\\\
(0,d-1)&(1,d-1)&\cdots&(d-1,d-1)\end{array}\right).$ (2.8.3)

Now, let us investigate the families of lines identified by the entries in the
rows, columns, diagonals, and anti-diagonals of the matrix mentioned above.
But first of all, it is important to make clear that:

For each $r\in R_{d}$

  * •

$\varphi_{d,-}^{-1}(r)$ will be named a diagonal with remainder $r$ of the
matrix in ( • $\varphi_{d,+}^{-1}(r)$ will be named a anti-diagonal with
remainder $r$ of the matrix in (itemize

So, for example we say that $L^{s}_{a,b}$ and $L^{s}_{a_{1},b_{1}}$ are in the
same diagonal (resp. anti-diagonal) if
$\varphi_{d,-}(a,b)=\varphi_{d,-}(a_{1},b_{1})$ (resp.
$\varphi_{d,+}(a,b)=\varphi_{d,+}(a_{1},b_{1})$).

Note that

    * $({\rm i})$

The family ${\mathcal{L}}^{s}_{k}$ is labeled by the pairs in the $(k+1)$-th
column of the matrix in ( – [$({\rm ii})$] Any two lines labeled by pairs in
the same row of the matrix in (eet.

    * $({\rm iii})$

Each of such diagonals and anti-diagonals determine $d$ skew lines in
${\mathcal{L}}^{0}$.

    * $({\rm iv})$

Each diagonal (resp. anti-diagonal) with remainder $r$ meet the column in the
matrix (n exactly one pair (i.e. in exactly one line in
${\mathcal{L}}^{s}_{k}$ for $k\in R_{d}$).

    * $({\rm v})$

Let $L^{s}_{a,b},L^{s}_{a_{1},b_{1}}\in{\mathcal{L}}^{s}$ be disjoint and $d$
odd and $s\in\\{0,1,2\\}$. If
$\varphi_{d,\pm}(a,b)=\varphi_{d,\pm}(a_{1},b_{1})$, then
$\varphi_{d,\mp}(a,b)\neq\varphi_{d,\mp}(a_{1},b_{1})$. In other words, if
$L^{s}_{a,b},L^{s}_{a_{1},b_{1}}$ are lines in the same diagonal, then they
are in distinct anti-diagonal, and vice versa.

##  3\. Computing ${\mathfrak{s}}({{\rm F}_{d}})$ for $d\in\\{3,5\\}$

Next we will exhibit the only two Fermat surfaces ${{\rm F}_{d}}$ satisfying
${\mathfrak{s}}({{\rm F}_{d}})<3d$.

###  3.1. Showing that ${\mathfrak{s}}({\rm F}_{3})=6$

######  Proposition 3.1.

Let ${\rm C}$ be a set of skew lines in the Fermat cubic ${\rm F}_{3}$ and
consider ${\rm C}^{s}={\rm C}\cap\mathcal{L}^{s}$ for each $s=0,1,2$. If
$\\#{\rm C}^{s}=3$ then there exists $k\in\\{0,1,2\\}\setminus\\{s\\}$ such
that ${\rm C}^{k}=\emptyset$.

###### Proof.

Next, we will subdivide in the following three cases:

$\bullet$ Assume $\\#{\rm C}^{0}=3$ and let ${\rm
C}^{0}=\left\\{L^{0}_{0,b_{0}},L^{0}_{1,b_{1}},L^{0}_{2,b_{2}}\right\\}$ with
$b_{j}\in\\{0,1,2\\}$ and

$b_{0}\not=b_{1},\,\,b_{0}\not=b_{2},\,\,b_{1}\not=b_{2}.$ (3.1.4)

We claim that $\\#\varphi_{3,+}({\rm C}^{0})=1$ or $\\#\varphi_{3,-}({\rm
C}^{0})=1$. Note that $r_{3}(\\{b_{i},b_{i}+1,b_{i}+2\\})=R_{3}$, for any $i.$
Thus $b_{0}\equiv_{3}b_{1}+j$ for some $j=0,1,2.$ In fact,
$b_{0}\not\equiv_{3}b_{1}$, so we have

$\underbrace{b_{0}\equiv_{3}b_{1}+1}_{(i)}\quad\hbox{ or
}\quad\underbrace{b_{0}\equiv_{3}b_{1}+2.}_{(ii)}$

For $(i)$ have in mind that $b_{1}+1\equiv_{3}b_{2}+j$ for some $j=0,1,2.$ In
fact, by Equation (3.1.4) we have

$\displaystyle b_{1}+1\not\equiv_{3}b_{2}+1\quad\hbox{ and }\quad
b_{1}+1\not\equiv_{3}b_{2}.$

Thus $b_{0}\equiv_{3}b_{1}+1\equiv_{3}b_{2}+2$ and consequently
$\varphi_{3,+}({\rm C}^{0})=\\{b_{0}\\}$, so $\\#\varphi_{3,+}({\rm
C}^{0})=1$.

For $(ii)$ we used that $b_{1}+2\equiv_{3}b_{2}+j$ for some $j=0,1,2.$
However, by Equation (3.1.4) we have

$\displaystyle b_{1}+2\not\equiv_{3}b_{2}\quad\hbox{ and }\quad
b_{1}+2\not\equiv_{3}b_{2}+2\hbox{.}$

Thus $b_{0}\equiv_{3}b_{1}+2\equiv_{3}b_{2}+1$, that is,
$b_{0}\equiv_{3}b_{1}-1\equiv_{3}b_{2}-2$ and consequently $\varphi_{3,-}({\rm
C}^{0})=\\{b_{0}\\}$, so $\\#\varphi_{3,-}({\rm C}^{0})=1$.

Finally, if $\\#\varphi_{3,+}({\rm C}^{0})=1$ then $\\#\varphi_{3,-}({\rm
C}^{0})=3$ (cf. item (v) Remark Which implies that ${\rm C}^{1}=\emptyset$
(cf. Remark Analogously, if $\\#\varphi_{3,-}({\rm C}^{0})=1$ then ${\rm
C}^{2}=\emptyset$.

$\bullet$ Assume $\\#{\rm C}^{1}=3$ and let ${\rm
C}^{1}=\left\\{L^{1}_{a_{0},0},L^{1}_{a_{1},1},L^{1}_{a_{2},2}\right\\}$ with
$a_{i}\in\\{0,1,2\\}$ and

$a_{0}\not\equiv_{3}a_{1}+1,\,\,a_{0}\not\equiv_{3}a_{2}+2,\,\,a_{1}+1\not\equiv_{3}a_{2}+2.$
(3.1.5)

We will analyze the following two possibilities: $a_{0}\equiv_{3}a_{1}+2$ or
$a_{0}\not\equiv_{3}a_{1}+2.$

$a_{0}\equiv_{3}a_{1}+2$ One more time have in mind that
$a_{1}+2\equiv_{3}a_{2}+j$ for some $j=0,1,2.$ In fact, follows from Equation
(3.1.5)777Note that $a_{1}+2\equiv_{3}a_{2}\Longrightarrow
a_{1}+1\equiv_{3}a_{2}+2$. As well as,
$a_{1}+2\equiv_{3}a_{2}+2\Longrightarrow a_{0}\equiv_{3}a_{2}+2$.

$\displaystyle a_{1}+2\not\equiv_{3}a_{2}\quad\hbox{ and }\quad
a_{1}+2\not\equiv_{3}a_{2}+2.$

Thus $a_{0}\equiv_{3}a_{1}+2\equiv_{3}a_{2}+1$. This implies that
$\\#\psi_{3}({\rm C}^{1})=1$ and therefore ${\rm
C}^{0}=\emptyset$.888$\\#\psi_{3}({\rm C}^{1})=1\implies
a_{0}\equiv_{3}a_{1}+2\equiv_{3}a_{2}+1\implies\\{a_{0},a_{1},a_{2}\\}=R_{3}\implies{\rm
C}^{0}=\emptyset$.

$a_{0}\not\equiv_{3}a_{1}+2$ In this case we will show that $\\#\psi_{3}({\rm
C}^{1})=3$ (ie, $r_{3}(\\{a_{0},a_{1}+2,a_{2}+1\\})=R_{3}$).

Since $a_{0}\not\equiv_{3}a_{1}+1$ (cf. (3.1.5)) then necessarily
$a_{0}\equiv_{3}a_{1}$. On another hand, note that

$a_{1}+2\equiv_{3}a_{2}+1\Longrightarrow
a_{0}+2\equiv_{3}a_{2}+1\Longrightarrow a_{0}\equiv_{3}a_{2}+2$

and

$a_{0}\equiv_{3}a_{2}+1\implies a_{1}+1\equiv_{2}a_{2}+2$

which are both absurd (cf. (3.1.5)). Therefore
$a_{0}\not\equiv_{3}a_{1}+2,a_{0}\not\equiv_{3}a_{2}+1,a_{1}+2\not\equiv
a_{2}+1$ and this implies that $\psi_{3}({\rm C}^{1})=R_{3}$. Furthermore
${\rm C}^{2}=\emptyset$ (cf. Remarks The case where $\\#{\rm C}^{2}=3$ we left
for reader. ∎

######  Corollary 3.2.

${\mathfrak{s}}({\rm F}_{3})=6$.

###### Proof.

Let ${\rm C}\subset{\rm F}_{3}$ be a set of skew lines such that $\\#{\rm
C}>6$. Then ${\rm C}\cap\mathcal{L}^{i}={\rm C}^{i}\not=\emptyset,$ for each
$i=0,1,2$ (cf. Corollary By Proposition may conclude that $\\#{\rm C}^{i}\leq
2$ for each $i=0,1,2$ which is an absurd. This implies that
${\mathfrak{s}}({\rm F}_{3})\leq 6$. Now use Corollary proof

###  3.2. Showing that ${\mathfrak{s}}({\rm F}_{5})=13$

######  Lemma 3.3.

Let ${\rm C}^{0}\subset\mathcal{L}^{0}$ be a set of skew lines in ${\rm
F}_{5}$. If $\\#{\rm C}^{0}=5$ then $\\#\varphi_{5,+}({\rm C}^{0})\geq 3$ or
$\\#\varphi_{5,-}({\rm C}^{0})\geq 3$.

###### Proof.

Assume that ${\rm
C}^{0}=\\{L^{0}_{a_{0},b_{0}},\ldots,L^{0}_{a_{4},b_{4}}\\}$. If
$\\#\varphi_{5,+}({\rm C}^{0})\leq 2$, then without lost of generality we may
assume that $a_{0}+b_{0}\equiv_{5}a_{1}+b_{1}\equiv_{5}a_{2}+b_{2}.$ This
implies that $b_{0}-a_{0}\not\equiv_{5}b_{1}-a_{1}$,
$b_{0}-a_{0}\not\equiv_{5}b_{2}-a_{2}$ and
$b_{1}-a_{1}\not\equiv_{5}b_{2}-a_{2}$ (cf. (v) in Remarks Therefore
$\\#\varphi_{5,-}({\rm C}^{0})\geq 3$ as we desired. ∎

######  Lemma 3.4.

Let ${\rm
C}^{s}=\\{L^{s}_{a_{0},b_{0}},\ldots,L^{s}_{a_{4},b_{4}}\\}\subset\mathcal{L}^{s}$
be a set of skew lines in ${\rm F}_{5}$ such that $\\#{\rm C}^{s}=5$ for
$s=1,2$. If $\\#\\{a_{0},\ldots,a_{4}\\}\leq 3$ then $\\#\psi_{5}({\rm
C}^{s})\geq 3$.

###### Proof.

We will divide the proof in three cases according to the
$\\#\\{a_{0},\ldots,a_{4}\\}$. The first case is
$\\#\\{a_{0},\ldots,a_{4}\\}=1$. In this case follows that $\\#\psi_{5}({\rm
C}^{s})=5$ since ${\rm C}^{s}={\mathcal{L}}^{s}_{k}$ for some $k\in R_{5}$.
The second one is when $\\#\\{a_{0},\ldots,a_{4}\\}=2$ and which case at least
three are equal, so we may assume that $a_{0}=a_{1}=a_{2}$. This implies that
$\\#\psi_{5}({\rm C}^{s})\geq 3$.999Since,
$a_{0}+2b_{i}\equiv_{5}a_{0}+2b_{j}\Longleftrightarrow b_{i}\equiv_{5}b_{j}$
for $i\neq j$, and $i,j\in\\{0,1,2\\}$ The last one occurs when
$\\#\\{a_{0},\ldots,a_{4}\\}=3$. In this case we have two possibilities
(reordering indexes if necessary):
(i) $a_{0}=a_{1}=a_{2}$ and $\\#\\{a_{0},a_{3},a_{4}\\}=3$, which implies
$\\#\psi_{5}({\rm C}^{s})\geq 3$.
(ii) $a_{0}=a_{1}$, $a_{2}=a_{3}$ and $\\#\\{a_{0},a_{2},a_{4}\\}=3$. In this
case,

$a_{0}+2b_{0}\not\equiv_{5}a_{1}+2b_{1}\,\hbox{and
}\,a_{2}+2b_{2}\not\equiv_{5}a_{3}+2b_{3},$

which implies that $\\#\psi_{5}({\rm C}^{s})\geq 2$. Let us suppose by absurd
that $\\#\psi_{5}({\rm C}^{s})=2$. So we may assume that

$a_{0}+2b_{0}\equiv_{5}a_{2}+2b_{2}\equiv_{5}a_{4}+2b_{4}\,\textrm{ and
}\,a_{1}+2b_{1}\equiv_{5}a_{3}+2b_{3}.$

Now note that

$\displaystyle\sum_{i=0}^{4}(a_{i}+b_{i})\equiv_{5}$ $\displaystyle\
\sum_{i=0}^{4}a_{i}\equiv_{5}\sum_{i=0}^{4}(a_{i}+2b_{i})\equiv_{5}3(a_{0}+2b_{0})+2(a_{1}+2b_{1})$
$\displaystyle\equiv_{5}$ $\displaystyle\
(a_{0}+b_{0})+2a_{0}+(a_{1}+b_{1})+a_{1}+3b_{1}$

which implies that
$(a_{2}+b_{2})+(a_{3}+b_{3})+(a_{4}+b_{4})\equiv_{5}2a_{0}+a_{1}+3b_{1}$.
Having in mind that $r_{5}(\\{a_{i}+b_{i}\\}_{i=1}^{5})=R_{5}$ (since $\\#{\rm
C}^{s}=5$), we have that

$(a_{2}+b_{2})+(a_{3}+b_{3})+(a_{4}+b_{4})\equiv_{5}4(a_{0}+b_{0})+4(a_{1}+b_{1}).$

Thus,

$\displaystyle
4(a_{0}+b_{0})+4(a_{1}+b_{1})\equiv_{5}2a_{0}+a_{1}+3b_{1}\implies$
$\displaystyle 3a_{1}+b_{1}+2a_{0}+4b_{0}\equiv_{5}0$ $\displaystyle\implies$
$\displaystyle 2(a_{0}+2b_{0})\equiv_{5}2(a_{1}+2b_{1})$
$\displaystyle\implies$ $\displaystyle a_{0}+2b_{0}\equiv_{5}a_{1}+2b_{1}$

and this is an absurd. Therefore $\\#\psi_{5}({\rm C}^{s})\geq 3$ for
$s=1,2$.∎

######  Lemma 3.5.

Let ${\rm C}$ be a set of skew lines in ${\rm F}_{5}$, ${\rm C}^{i}={\rm
C}\cap\mathcal{L}^{i}$ for $i=0,1,2$ such that $\\#{\rm C}^{0}\geq 4$. If
${\rm C}^{s}=\\{L^{s}_{a_{0},b_{0}},\ldots,L^{s}_{a_{4},b_{4}}\\}$ with
$\\#{\rm C}^{s}=5$ and $\\#{\rm C}^{r}=4$ where $\\{s,r\\}=\\{1,2\\}$ then
$\\#\\{a_{0},\ldots,a_{4}\\}\geq 3$.

###### Proof.

Note that, if $\\#\\{a_{0},\ldots,a_{4}\\}=1$, then ${\rm
C}^{s}={\mathcal{L}}^{s}_{k}$ for some $k\in R_{5}$. And this implies that
${\rm C}^{r}=\emptyset$ which is an absurd (cf. Remarks Now, if
$\\#\\{a_{0},\ldots,a_{4}\\}=2$, then we have two possibilities (reordering
indexes if necessary):

(i) $a_{0}=a_{1}=a_{2}=a_{3}$ and $a_{0}\not=a_{4}\quad$ (ii)
$a_{0}=a_{1}=a_{2}$ and $a_{0}\not=a_{3}=a_{4}$.

In case (i) follows that $\\#\psi_{5}({\rm C}^{s})\geq 4$ and
$\\#\psi_{5}({\rm C}^{r})\in\\{0,1\\}$ (since $\psi_{5}({\rm
C}^{s})\cap\psi_{5}({\rm C}^{r})=\emptyset$). Hence, if $\\#\psi_{5}({\rm
C}^{r})=0$ then ${\rm C}^{r}=\emptyset$, else $\\#\psi_{5}({\rm C}^{r})=1$
which implies $\\#\\{a_{0},\ldots,a_{4}\\}\leq 1$ and this is an
absurd.101010In fact, assume that $s=1,r=2$, ${\rm
C}^{2}=\\{L^{2}_{a_{0}^{\prime},b_{0}^{\prime}},\ldots,L^{2}_{a_{3}^{\prime},b_{3}^{\prime}}\\}$.
As $\\#\psi_{5}(C^{2})=1$ then
$\\#\\{a_{0}^{\prime},\ldots,a_{3}^{\prime}\\}=4$ (if
$a_{0}^{\prime}=a_{1}^{\prime}$, then
$a_{0}^{\prime}+2b_{0}\not\equiv_{5}a_{1}^{\prime}+2b_{1}^{\prime}$ since
$b_{0}^{\prime}\not\equiv_{5}b_{1}^{\prime}$, which is an absurd). Therefore,
$\\#\varphi_{5,+}({\rm C}^{0})=1$ (cf. Corollary hich implies that
$\\#\varphi_{5,-}({\rm C}^{0})\geq 4$. Hence, $\\#\\{a_{0},\ldots,a_{4}\\}\leq
1$.

For (ii) note that

$\displaystyle
0\equiv_{5}\sum_{i=0}^{4}(a_{i}+b_{i})\equiv_{5}3a_{0}+2a_{3}\implies
3a_{0}\equiv_{5}3a_{3}\implies a_{0}\equiv_{5}a_{3}\implies a_{0}=a_{3}$

which is an absurd.∎

######  Theorem 3.6.

${\mathfrak{s}}({\rm F}_{5})=13$.

###### Proof.

Let ${\rm C}$ be a set of skew lines in ${\rm F}_{5}$. Let us suppose that
$\\#{\rm C}\geq 14$. In fact, it is enough to analyse the case $\\#{\rm
C}=14.$ Define ${\rm C}^{i}={\rm C}\cap\mathcal{L}^{i}$, with $i=0,1,2$. Note
that only one of the possibilities happen:

  1. $(a)$

$\\#{\rm C}^{0}=4$ and $\\#{\rm C}^{1}=\\#{\rm C}^{2}=5$;

  2. $(b)$

$\\#{\rm C}^{0}=5$, $\\#{\rm C}^{s}=5$ and $\\#{\rm C}^{r}=4$ for
$\\{r,s\\}=\\{1,2\\}$.

For $(a)$ let us consider ${\rm
C}^{1}=\\{L^{1}_{a_{0},b_{0}},\ldots,L^{1}_{a_{4},b_{4}}\\}$ and ${\rm
C}^{2}=\\{L^{2}_{a_{0}^{\prime},b_{0}^{\prime}},\ldots,L^{2}_{a_{4}^{\prime},b_{4}^{\prime}}\\}$.
Note that

$\\#\\{a_{0},\ldots,a_{4}\\}\leq 3\ \textrm{ or }\
\\#\\{a_{0},\ldots,a_{4}\\}>3.$

The last inequality can not occur because other way

  1. $(i)$

if $\\#\\{a_{0},\ldots,a_{4}\\}=5$ then $\\#\varphi_{5,-}(C^{0})=0$ and this
implies that ${\rm C}^{0}=\emptyset$;

  2. $(ii)$

if $\\#\\{a_{0},\ldots,a_{4}\\}=4$, then $\\#\varphi_{5,-}({\rm C}^{0})=1$.
Hence $\\#\varphi_{5,+}(C^{0})=4$. Therefore
$\\#\\{a_{0}^{\prime},\ldots,a_{4}^{\prime}\\}=1$, which implies that
$\\#\psi_{5}(C^{2})=5$. Furthermore $\\#\psi_{5}(C^{1})=0$ which is an absurd.

Therefore $\\#\\{a_{0},\ldots,a_{4}\\}\leq 3$. Analogously, we may conclude
that $\\#\\{a_{0}^{\prime},\ldots,a_{4}^{\prime}\\}\leq 3$. Follows from Lemma
at $\\#\psi_{5}({\rm C}^{1})\geq 3$ and $\\#\psi_{5}({\rm C}^{2})\geq 3$,
which is an absurd by Remark For $(b)$ let us assume that $\\#{\rm
C}^{0}=\\#{\rm C}^{1}=5$ and $\\#{\rm C}^{2}=4$ (the other case is analogous).
Let us consider ${\rm
C}^{0}=\\{L^{0}_{a_{0},b_{0}},\ldots,L^{0}_{a_{4},b_{4}}\\}$, ${\rm
C}^{1}=\\{L^{1}_{a_{0}^{\prime},b_{0}^{\prime}},\ldots,L^{1}_{a_{4}^{\prime},b_{4}^{\prime}}\\}$
and ${\rm
C}^{2}=\\{L^{2}_{a_{0}^{\prime\prime},b_{0}^{\prime\prime}},\ldots,L^{2}_{a_{3}^{\prime\prime},b_{3}^{\prime\prime}}\\}$.
Using arguments analogous to cases $(i)$ and $(ii)$ we may conclude111111If
$\\#\\{a_{0}^{\prime},...,a_{4}^{\prime}\\}=5$ then ${\rm C}^{0}=\emptyset$.
If $\\#\\{a_{0}^{\prime},...,a_{4}^{\prime}\\}=4$ then $\\#\varphi_{5,-}({\rm
C}^{0})=1$. Hence $\\#\varphi_{5,+}({\rm C}^{0})=5$ and this implies that
${\rm C}^{2}=\emptyset$, which is an absurd. that
$\\#\\{a_{0}^{\prime},\ldots,a_{4}^{\prime}\\}\leq 3$. Now, follows from Lemma
at $\\#\\{a_{0}^{\prime},\ldots,a_{4}^{\prime}\\}=3$. So
$\\#\big{(}\varphi_{5,-}({\rm C}^{0})\big{)}\leq 2$. Now we will analyze all
the three possibilities:

  1. $(iii)$

if $\\#\varphi_{5,-}({\rm C}^{0})=0$, then ${\rm C}^{0}=\emptyset$;

  2. $(iv)$

if $\\#\varphi_{5,-}({\rm C}^{0})=1$, then $\\#\varphi_{5,+}({\rm C}^{0})=5$
and this implies that ${\rm C}^{2}=\emptyset$;

  3. $(v)$

if $\\#\varphi_{5,-}({\rm C}^{0})=2$ and then we may assume that

$b_{0}-a_{0}\equiv_{5}b_{1}-a_{1}\equiv_{5}b_{2}-a_{2}\equiv_{5}b_{3}-a_{3}\quad\textrm{
and }\quad b_{0}-a_{0}\not\equiv_{5}b_{4}-a_{4}$ (3.6.6) or
$b_{0}-a_{0}\equiv_{5}b_{1}-a_{1}\equiv_{5}b_{2}-a_{2}\quad\textrm{ and }\quad
b_{0}-a_{0}\not\equiv_{5}b_{4}-a_{4}\equiv_{5}b_{3}-a_{3}$ (3.6.7)

By Equation (3.6.6) we may conclude that $\\#\big{(}\varphi_{5,+}({\rm
C}^{0})\big{)}\geq 4$. So
$\\#\\{a_{0}^{\prime\prime},\ldots,a_{3}^{\prime\prime}\\}=1$. This implies
that $\\#\psi_{5}({\rm C}^{2})=4$ and consequently $\\#\psi_{5}({\rm
C}^{1})=1$ by Remark nd this is an absurd by Lemma inally by Equation (3.6.7)
follows that

$\displaystyle\sum_{i=0}^{4}(b_{i}-a_{i})\equiv_{5}0\implies$ $\displaystyle
3(b_{0}-a_{0})+2(b_{4}-a_{4})\equiv_{5}0$ $\displaystyle\implies$
$\displaystyle 3(b_{0}-a_{0})\equiv_{5}3(b_{4}-a_{4})$ $\displaystyle\implies$
$\displaystyle b_{0}-a_{0}\equiv_{5}b_{4}-a_{4}$

which is an absurd. Therefore $\\#{\rm C}\leq 13$ for any set ${\rm C}$ of
skew lines in $F_{5}$. On other hand the Example ow an example with $13$ skew
lines, thus ${\mathfrak{s}}({\rm F}_{5})=13$. ∎

##  4\. Addressing the case $d\geq 4$ and $d\not=5$

From Corollary have that $2d\leq{\mathfrak{s}}({{\rm F}_{d}})\leq 3d$ for any
$d\geq 3$. For $d\geq 4$ even, we have that ${\mathfrak{s}}({{\rm
F}_{d}})=3d$, as we prove in the next proposition. However for $d\geq 7$ odd,
we will divide our study in two cases: $d\equiv_{4}1$ and $d\equiv_{4}3$ being
$d\geq 7$.

######  Proposition 4.1.

Let $d\geq 4$ even. If ${\rm C}^{0}=\Big{\\{}L^{0}_{a,a}\Big{\\}}_{a=0}^{d-1}$
and ${\rm C}^{s}=\Big{\\{}L^{s}_{1,i}\Big{\\}}_{i=0}^{d-1}$ for $s=1,2$, then
${\rm C}^{0}\cup{\rm C}^{1}\cup{\rm C}^{2}$ consists of $3d$ skew lines in
${{\rm F}_{d}}$.

###### Proof.

Follows from Corollary at ${\rm C}^{s}$ consists of $d$ skew lines for each
$s\in\\{0,1,2\\}$.121212Note that $\varphi_{d,+}({\rm
C}^{1})=R_{d}=\varphi_{d,+}({\rm C}^{2})$ and $\\#{\rm C}^{s}=d$ for all $s$.
On the other hand, $\varphi_{d,-}({\rm C}^{0})=\\{0\\}$ and
$\varphi_{d,+}({\rm C}^{0})=\\{0,2,\ldots,2d-2\\}$, which implies that ${\rm
C}^{0}\cup{\rm C}^{s}$ consists of skew lines for $s=1,2$, respectively (cf.
Corollary Finally, note that the statement (d) in Proposition sures us that
${\rm C}^{1}\cup{\rm C}^{2}$ also is formed by skew lines. Therefore, ${\rm
C}^{0}\cup{\rm C}^{1}\cup{\rm C}^{2}$ consists of $3d$ skew lines in ${{\rm
F}_{d}}$. ∎

Below we will discuss some more examples that led us to believe that
${\mathfrak{s}}({{\rm F}_{d}})=3d$ for $d\geq 7$ odd.

######  Example 4.2.

For $d=7$, let us consider ${\rm
C}^{0}=\\{L^{0}_{0,0},L^{0}_{1,3},L^{0}_{2,2},L^{0}_{3,5},L^{0}_{4,4},L^{0}_{5,6},L^{0}_{6,1}\\}$
which consists of seven skew lines (cf. (a) in Corollary Furthermore, in the
rows of the next table we register the values of $\varphi_{7,\pm}({\rm
C}^{0})$, respectively.

${\rm C}^{0}$ | $L^{0}_{0,0}$ | $L^{0}_{1,3}$ | $L^{0}_{2,2}$ | $L^{0}_{3,5}$ | $L^{0}_{4,4}$ | $L^{0}_{5,6}$ | $L^{0}_{6,1}$
---|---|---|---|---|---|---|---
$\varphi_{7,-}$ | 0 | 2 | 0 | 2 | 0 | 1 | 2
$\varphi_{7,+}$ | 0 | 4 | 4 | 1 | 1 | 4 | 0

Now, having in mind Corollary r the choice of ${\rm
C}^{s}\subset{\mathcal{L}}^{s}$ such that ${\rm C}^{0}\cup{\rm C}^{s}$ to be
constituted by skew lines for $s=1,2$, it is necessary that

${\rm
C}^{1}\cap{\mathcal{L}}^{2}_{k}=\emptyset\quad\forall\,k\in\\{0,1,2\\}\quad\hbox{and}\quad{\rm
C}^{2}\cap{\mathcal{L}}^{2}_{k}=\emptyset\quad\forall\,k\in\\{0,1,4\\}.$

So, ${\rm
C}^{1}=\\{L^{1}_{4,0},L^{1}_{4,2},L^{1}_{5,3},L^{1}_{5,4},L^{1}_{5,5},L^{1}_{6,1},L^{1}_{6,6}\\}$
and ${\rm
C}^{2}=\\{L^{2}_{2,0},L^{2}_{2,5},L^{2}_{2,6},L^{2}_{3,1},L^{2}_{3,2},L^{2}_{3,3},L^{2}_{6,4}\\}$
are admissible choices. As well us, according to the information on the rows
in the following two tables.

${\rm C}^{1}$ | $L^{1}_{4,0}$ | $L^{1}_{4,2}$ | $L^{1}_{5,3}$ | $L^{1}_{5,4}$ | $L^{1}_{5,5}$ | $L^{1}_{6,1}$ | $L^{1}_{6,6}$
---|---|---|---|---|---|---|---
$\varphi_{7,+}$ | 4 | 6 | 1 | 2 | 3 | 0 | 5
$\psi_{7}$ | 4 | 1 | 4 | 6 | 1 | 1 | 4

${\rm C}^{2}$ | $L^{2}_{2,0}$ | $L^{2}_{2,5}$ | $L^{2}_{2,6}$ | $L^{2}_{3,1}$ | $L^{2}_{3,2}$ | $L^{2}_{3,3}$ | $L^{2}_{6,4}$
---|---|---|---|---|---|---|---
$\varphi_{7,+}$ | 2 | 0 | 1 | 4 | 5 | 6 | 3
$\psi_{7}$ | 2 | 5 | 0 | 5 | 0 | 2 | 0

We have that ${\rm C}^{s}$ consists of skew lines for $s=1,2$ (cf. (b) and (c)
in Corollary nd $\psi_{7}({\rm C}^{1})\cap\psi_{7}({\rm C}^{2})=\emptyset$,
which implies that ${\rm C}^{1}\cup{\rm C}^{2}$ also is formed by skew lines
(cf. (c) in Corollary Therefore, ${\rm C}:={\rm C}^{0}\cup{\rm C}^{1}\cup{\rm
C}^{2}$ consists of 21 skew lines in ${\rm F}_{7}$. Thus ${\mathfrak{s}}({\rm
F}_{7})=21$ (since ${\mathfrak{s}}({\rm F}_{7})\leq 21$).

Let us go now to case $d=9$ and $d=11$.

######  Example 4.3.

The next tables contain the necessary information to concluded that those 27
lines (in ${\rm F}_{9}$) bellow are pairwise disjoint.

${\rm C}^{0}$ | $L^{0}_{4,1}$ | $L^{0}_{5,2}$ | $L^{0}_{6,3}$ | $L^{0}_{7,7}$ | $L^{0}_{8,8}$ | $L^{0}_{0,0}$ | $L^{0}_{1,4}$ | $L^{0}_{2,5}$ | $L^{0}_{3,6}$
---|---|---|---|---|---|---|---|---|---
$\varphi_{9,-}$ | 6 | 6 | 6 | 0 | 0 | 0 | 3 | 3 | 3
$\varphi_{9,+}$ | 5 | 7 | 0 | 5 | 7 | 0 | 5 | 7 | 0

${\rm C}^{1}$ | $L^{1}_{5,0}$ | $L^{1}_{5,1}$ | $L^{1}_{1,2}$ | $L^{1}_{1,3}$ | $L^{1}_{7,4}$ | $L^{1}_{5,5}$ | $L^{1}_{2,6}$ | $L^{1}_{2,7}$ | $L^{1}_{8,8}$
---|---|---|---|---|---|---|---|---|---
$\varphi_{9,+}$ | 5 | 6 | 3 | 4 | 2 | 1 | 8 | 0 | 7
$\psi_{9}$ | 5 | 7 | 5 | 7 | 6 | 6 | 5 | 7 | 6

${\rm C}^{2}$ | $L^{2}_{1,0}$ | $L^{2}_{1,1}$ | $L^{2}_{6,2}$ | $L^{2}_{6,3}$ | $L^{2}_{2,4}$ | $L^{2}_{2,5}$ | $L^{2}_{6,6}$ | $L^{2}_{6,7}$ | $L^{2}_{6,8}$
---|---|---|---|---|---|---|---|---|---
$\varphi_{9,+}$ | 1 | 2 | 8 | 0 | 6 | 7 | 3 | 4 | 5
$\psi_{9}$ | 1 | 3 | 1 | 3 | 1 | 3 | 0 | 2 | 4

Now, we show the tables for the lines in ${\rm F}_{11}$

${\rm C}^{0}$ | $L^{0}_{5,0}$ | $L^{0}_{6,1}$ | $L^{0}_{7,2}$ | $L^{0}_{8,3}$ | $L^{0}_{9,7}$ | $L^{0}_{10,8}$ | $L^{0}_{0,9}$ | $L^{0}_{1,4}$ | $L^{0}_{2,5}$ | $L^{0}_{3,6}$ | $L^{0}_{4,10}$
---|---|---|---|---|---|---|---|---|---|---|---
$\varphi_{11,-}$ | 6 | 6 | 6 | 6 | 9 | 9 | 9 | 3 | 3 | 3 | 6
$\varphi_{11,+}$ | 5 | 7 | 9 | 0 | 5 | 7 | 9 | 5 | 7 | 9 | 3
${\rm C}^{1}$ | $L^{1}_{8,0}$ | $L^{1}_{8,1}$ | $L^{1}_{4,2}$ | $L^{1}_{4,3}$ | $L^{1}_{0,4}$ | $L^{1}_{0,5}$ | $L^{1}_{7,6}$ | $L^{1}_{7,7}$ | $L^{1}_{4,8}$ | $L^{1}_{1,9}$ | $L^{1}_{1,10}$
---|---|---|---|---|---|---|---|---|---|---|---
$\varphi_{11,+}$ | 8 | 9 | 6 | 7 | 4 | 5 | 2 | 3 | 1 | 10 | 0
$\psi_{11}$ | 8 | 10 | 8 | 10 | 8 | 10 | 8 | 10 | 9 | 8 | 10

${\rm C}^{2}$ | $L^{2}_{1,0}$ | $L^{2}_{1,1}$ | $L^{2}_{1,2}$ | $L^{2}_{6,3}$ | $L^{2}_{6,4}$ | $L^{2}_{6,5}$ | $L^{2}_{2,6}$ | $L^{2}_{8,7}$ | $L^{2}_{8,8}$ | $L^{2}_{8,9}$ | $L^{2}_{8,10}$
---|---|---|---|---|---|---|---|---|---|---|---
$\varphi_{11,+}$ | 1 | 2 | 3 | 9 | 10 | 0 | 8 | 4 | 5 | 6 | 7
$\psi_{11}$ | 1 | 3 | 5 | 1 | 3 | 5 | 3 | 0 | 2 | 4 | 6

In Propositions d f it is written $L^{s}_{a,b}$ where $a$ and/or $b$ do not
belong to $R_{d}$, then assume that $L^{s}_{a,b}:=L^{s}_{a_{1},b_{1}}$ which
$a_{1},b_{1}\in R_{d}$ such that $a\equiv_{d}a_{1}$ and $b\equiv_{d}b_{1}$.

######  Proposition 4.4.

Let $d=2n+1$ with $n=2k$ and $k\geq 3$. Consider the families

${\rm
C}^{0}=\Big{\\{}L^{0}_{1+i,2k+i}\Big{\\}}_{i=0}^{k}\cup\Big{\\{}L^{0}_{k+i,k+i}\Big{\\}}_{i=2}^{k-1}\cup\Big{\\{}L^{0}_{2k+i,1+i}\Big{\\}}_{i=0}^{k}\cup\Big{\\{}L^{0}_{3k+i,3k+i}\Big{\\}}_{i=1}^{k+1},$
${\rm
C}^{1}=\Big{\\{}L^{1}_{2k+1,2k+i}\Big{\\}}_{i=1}^{k-1}\cup\Big{\\{}L^{1}_{3k+1,2k}\Big{\\}}\cup\Big{\\{}L^{1}_{1,k+i}\Big{\\}}_{i=0}^{k-1}\cup\Big{\\{}L^{1}_{2k+1,i}\Big{\\}}_{i=0}^{k-1}\cup\Big{\\{}L^{1}_{3k+2,4k}\Big{\\}}\cup\Big{\\{}L^{1}_{2,3k+i}\Big{\\}}_{i=0}^{k-1},$
${\rm
C}^{2}=\Big{\\{}L^{2}_{1,i}\Big{\\}}_{i=0}^{k-1}\cup\Big{\\{}L^{2}_{2k+2,3k+i}\Big{\\}}_{i=0}^{k}\cup\Big{\\{}L^{2}_{2,2k+i}\Big{\\}}_{i=0}^{k-1}\cup\Big{\\{}L^{2}_{2k+2,k+i}\Big{\\}}_{i=0}^{k-1}.$

It is verified that ${\rm C}^{0}\cup{\rm C}^{1}\cup{\rm C}^{2}$ consists of
$3d$ skew lines in ${\rm F}_{d}$.

###### Proof.

Let us divide the proof in four steps.
Step 1: ${\rm C}^{0}$ is constituted by $d$ skew lines.

First of all, note that ${\rm C}^{0}$ is defined by four strata below

${\rm C}^{0}=\underbrace{\Big{\\{}L^{0}_{1+i,2k+i}\Big{\\}}_{i=0}^{k}}_{({\rm
i})}\cup\underbrace{\Big{\\{}L^{0}_{k+i,k+i}\Big{\\}}_{i=2}^{k-1}}_{({\rm
ii})}\cup\underbrace{\Big{\\{}L^{0}_{2k+i,1+i}\Big{\\}}_{i=0}^{k}}_{({\rm
iii})}\cup\underbrace{\Big{\\{}L^{0}_{3k+i,3k+i}\Big{\\}}_{i=1}^{k+1}}_{({\rm
iv})},$

where the stratum (ii) is non-empty if and only if $k\geq 3$ (so, $d\neq 5$
and $d\neq 9$). Furthermore, we have that the label $t$ in each
$L^{0}_{t,j}\in{\rm C}^{0}$ is varying throughout the set

$\\{\underbrace{1,\ldots,k+1}_{({\rm i})},\underbrace{k+2,\ldots,2k-1}_{({\rm
ii})},\underbrace{2k,\ldots,3k}_{({\rm
iii})},\underbrace{3k+1,\ldots,4k,4k+1\equiv_{d}0}_{({\rm iv})}\\}.$ (4.4.8)

And the label $j$ throughout the set

$\\{\underbrace{1,\ldots,k+1}_{({\rm
iii})},\underbrace{k+2,\ldots,2k-1}_{({\rm
ii})},\underbrace{2k,\ldots,3k}_{({\rm
i})},\underbrace{3k+1,\ldots,4k,4k+1\equiv_{d}0}_{({\rm iv})}\\}.$ (4.4.9)

Since the sets in (nd (re equal to $R_{d}$ follows that ${\rm C}^{0}$ is
constituted by $d$ skew lines.
Step 2: ${\rm C}^{0}\cup{\rm C}^{s}$ is constituted by skew lines for $s=1,2$.

Next we display the values of $\varphi_{d,\pm}$ over ${\rm C}^{0}$ (using the
stratification (i),…,(iv) for ${\rm C}^{0}$).

${\rm C}^{0}$/ (i) | $L^{0}_{1,2k}$ | $L^{0}_{2,2k+1}$ | $L^{0}_{3,2k+2}$ | $\cdots$ | $L^{0}_{k-1,3k-2}$ | $L^{0}_{k,3k-1}$ | $L^{0}_{1+k,3k}$
---|---|---|---|---|---|---|---
$\varphi_{d,-}$ | $2k-1$ | $2k-1$ | $2k-1$ | $\cdots$ | $2k-1$ | $2k-1$ | $2k-1$
$\varphi_{d,+}$ | $2k+1$ | $2k+3$ | $2k+5$ | $\cdots$ | $4k-3$ | $4k-1$ | $4k+1\equiv_{d}0$

${\rm C}^{0}$/ (ii) | $L^{0}_{k+2,k+2}$ | $L^{0}_{k+3,k+3}$ | $L^{0}_{k+4,k+4}$ | $\cdots$ | $L^{0}_{2k-3,2k-3}$ | $L^{0}_{2k-2,2k-2}$ | $L^{0}_{2k-1,2k-1}$
---|---|---|---|---|---|---|---
$\varphi_{d,-}$ | $0$ | $0$ | $0$ | $\cdots$ | $0$ | $0$ | $0$
$\varphi_{d,+}$ | $2k+4$ | $2k+6$ | $2k+8$ | $\cdots$ | $4k-6$ | $4k-4$ | $4k-2$

${\rm C}^{0}$/ (iii) | $L^{0}_{2k,1}$ | $L^{0}_{2k+1,2}$ | $L^{0}_{2k+2,3}$ | $\cdots$ | $L^{0}_{3k-2,k-1}$ | $L^{0}_{3k-1,k}$ | $L^{0}_{3k,1+k}$
---|---|---|---|---|---|---|---
$\varphi_{d,-}$ | $2k+2$ | $2k+2$ | $2k+2$ | $\cdots$ | $2k+2$ | $2k+2$ | $2k+2$
$\varphi_{d,+}$ | $2k+1$ | $2k+3$ | $2k+5$ | $\cdots$ | $4k-3$ | $4k-1$ | $0$

${\rm C}^{0}$/ (iv) | $L^{0}_{3k+1,3k+1}$ | $L^{0}_{3k+2,3k+2}$ | $L^{0}_{3k+3,3k+3}$ | $\cdots$ | $L^{0}_{4k-1,4k-1}$ | $L^{0}_{4k,4k}$ | $L^{0}_{0,0}$
---|---|---|---|---|---|---|---
$\varphi_{d,-}$ | $0$ | $0$ | $0$ | $\cdots$ | $0$ | $0$ | $0$
$\varphi_{d,+}$ | $2k+1$ | $2k+3$ | $2k+5$ | $\cdots$ | $4k-3$ | $4k-1$ | $0$

Now, having in mind Corollary r the choice of ${\rm
C}^{s}\subset{\mathcal{L}}^{s}$ such that ${\rm C}^{0}\cup{\rm C}^{s}$ to be
constituted by skew lines for $s=1,2$ and the tables (involving ${\rm C}^{0}$)
above, it is necessary that

$\left\\{\begin{array}[]{l}{\rm
C}^{1}\cap{\mathcal{L}}^{1}_{t}=\emptyset\quad\forall\,\,t\in\\{0,2k-1,2k+2\\}\\\
\vspace{-0.4cm}\\\ {\rm
C}^{2}\cap{\mathcal{L}}^{2}_{t}=\emptyset\quad\forall\,\,t\in\\{2k+j\\}_{j=1}^{2k+1}\\--\\{2k+2,4k\\}.\end{array}\right.$
(4.4.10)

Now, it is a straightforward verification to see that the label $t$ in each
$L^{s}_{t,j}\in{\rm C}^{s}$ belong to the set

$\\{1,2,2k+1,3k+1,3k+2\\}\,\hbox{ for }\,s=1\,\hbox{ and
}\,\\{1,2,2k+2\\}\,\hbox{ for }\,s=2.$

Thus, using (e concluded that ${\rm C}^{0}\cup{\rm C}^{s}$ is constituted by
skew lines for $s=1,2$.
Step 3: ${\rm C}^{s}$ is constituted by $d$ skew lines for $s=1,2$.

Let us stratified ${\rm C}^{1}$ as follows: ${\rm C}^{1}={\rm
A}_{1}\,\dot{\cup}\,{\rm A}_{2}\,\dot{\cup}\,{\rm A}_{3}\,\dot{\cup}\,{\rm
A}_{4}\,\dot{\cup}\,{\rm A}_{5}\,\dot{\cup}\,{\rm A}_{6}$ where

$\begin{array}[]{c}{\rm
A}_{1}:=\Big{\\{}L^{1}_{2k+1,2k+i}\Big{\\}}_{i=1}^{k-1},\,\,{\rm
A}_{2}:=\Big{\\{}L^{1}_{3k+1,2k}\Big{\\}},\,\,{\rm
A}_{3}:=\Big{\\{}L^{1}_{1,k+i}\Big{\\}}_{i=0}^{k-1},\\\
\vspace{-0.3cm}\hfil\\\ {\rm
A}_{4}:=\Big{\\{}L^{1}_{2k+1,i}\Big{\\}}_{i=0}^{k-1},\,\,{\rm
A}_{5}:=\Big{\\{}L^{1}_{3k+2,4k}\Big{\\}},\,\,{\rm
A}_{6}:=\Big{\\{}L^{1}_{2,3k+i}\Big{\\}}_{i=0}^{k-1}.\end{array}$ (4.4.11)

Note that the label $j$ in each $L^{1}_{t,j}\in{\rm C}^{1}$ is varying
throughout the set

$\\{\underbrace{0,\ldots,k-1}_{{\rm A}_{4}},\underbrace{k,\ldots,2k-1}_{{\rm
A}_{3}},\underbrace{2k}_{{\rm A}_{2}},\underbrace{2k+1,\ldots,3k-1}_{{\rm
A}_{1}},\underbrace{3k,\ldots,4k-1}_{{\rm A}_{6}},\underbrace{4k}_{{\rm
A}_{5}}\\},$ (4.4.12)

which it is equal to $R_{d}$. Furthermore, $\varphi_{d,+}({\rm C}^{1})$ is
given by

${\rm A}_{1}$ | $L^{1}_{2k+1,2k+1}$ | $L^{1}_{2k+1,2k+2}$ | $L^{1}_{2k+1,2k+3}$ | $\cdots$ | $L^{1}_{2k+1,3k-3}$ | $L^{1}_{2k+1,3k-2}$ | $L^{1}_{2k+1,3k-1}$
---|---|---|---|---|---|---|---
$\varphi_{d,+}$ | $1$ | $2$ | $3$ | $\cdots$ | $k-3$ | $k-2$ | $k-1$

${\rm A}_{2}$ | $L^{1}_{3k+1,2k}$
---|---
$\varphi_{d,+}$ | $k$

${\rm A}_{3}$ | $L^{1}_{1,k}$ | $L^{1}_{1,k+1}$ | $L^{1}_{1,k+2}$ | $\cdots$ | $L^{1}_{1,2k-3}$ | $L^{1}_{1,2k-2}$ | $L^{1}_{1,2k-1}$
---|---|---|---|---|---|---|---
$\varphi_{d,+}$ | $k+1$ | $k+2$ | $k+3$ | $\cdots$ | $2k-2$ | $2k-1$ | $2k$

${\rm A}_{4}$ | $L^{1}_{2k+1,0}$ | $L^{1}_{2k+1,1}$ | $L^{1}_{2k+1,2}$ | $\cdots$ | $L^{1}_{2k+1,k-3}$ | $L^{1}_{2k+1,k-2}$ | $L^{1}_{2k+1,k-1}$
---|---|---|---|---|---|---|---
$\varphi_{d,+}$ | $2k+1$ | $2k+2$ | $2k+3$ | $\cdots$ | $3k-2$ | $3k-1$ | $3k$

${\rm A}_{5}$ | $L^{1}_{3k+2,4k}$
---|---
$\varphi_{d,+}$ | $3k+1$

${\rm A}_{6}$ | $L^{1}_{2,3k}$ | $L^{1}_{2,3k+1}$ | $L^{1}_{2,3k+2}$ | $\cdots$ | $L^{1}_{2,4k-3}$ | $L^{1}_{2,4k-2}$ | $L^{1}_{2,4k-1}$
---|---|---|---|---|---|---|---
$\varphi_{d,+}$ | $3k+2$ | $3k+3$ | $3k+4$ | $\cdots$ | $4k-1$ | $4k$ | $0$

Thus, $\varphi_{d,+}({\rm C}^{1})=R_{d}$. Taking into account the established
facts, we may use Corollary concluded that ${\rm C}^{1}$ is constituted by $d$
skew lines.

In a similar way, let us consider the following stratification for ${\rm
C}^{2}$: ${\rm C}^{2}={\rm B}_{1}\,\dot{\cup}\,{\rm B}_{2}\,\dot{\cup}\,{\rm
B}_{3}\,\dot{\cup}\,{\rm B}_{4}$ where

$\begin{array}[]{c}{\rm
B}_{1}:=\Big{\\{}L^{2}_{1,i}\Big{\\}}_{i=0}^{k-1},\,\,{\rm
B}_{2}:=\Big{\\{}L^{2}_{2k+2,3k+i}\Big{\\}}_{i=0}^{k},\\\
\vspace{-0.3cm}\hfil\\\ {\rm
B}_{3}:=\Big{\\{}L^{2}_{2,2k+i}\Big{\\}}_{i=0}^{k-1},\,\,{\rm
B}_{4}:=\Big{\\{}L^{2}_{2k+2,k+i}\Big{\\}}_{i=0}^{k-1}.\end{array}$ (4.4.13)

Note that the label $j$ in each $L^{2}_{t,j}\in{\rm C}^{2}$ is varying
throughout the set

$\\{\underbrace{0,\ldots,k-1}_{{\rm B}_{1}},\underbrace{k,\ldots,2k-1}_{{\rm
B}_{4}},\underbrace{2k,\ldots,3k-1}_{{\rm
B}_{3}},\underbrace{3k,\ldots,4k}_{{\rm B}_{2}}\\},$ (4.4.14)

which it is equal to $R_{d}$. Furthermore, $\varphi_{d,+}({\rm C}^{2})$ is
given by

${\rm B}_{1}$ | $L^{2}_{1,0}$ | $L^{2}_{1,1}$ | $L^{2}_{1,2}$ | $\cdots$ | $L^{2}_{1,k-3}$ | $L^{2}_{1,k-2}$ | $L^{2}_{1,k-1}$
---|---|---|---|---|---|---|---
$\varphi_{d,+}$ | $1$ | $2$ | $3$ | $\cdots$ | $k-2$ | $k-1$ | $k$

${\rm B}_{2}$ | $L^{2}_{2k+2,3k}$ | $L^{2}_{2k+2,3k+1}$ | $L^{2}_{2k+2,3k+2}$ | $\cdots$ | $L^{2}_{2k+2,4k-2}$ | $L^{2}_{2k+2,4k-1}$ | $L^{2}_{2k+2,4k}$
---|---|---|---|---|---|---|---
$\varphi_{d,+}$ | $k+1$ | $k+2$ | $k+3$ | $\cdots$ | $2k-1$ | $2k$ | $2k+1$

${\rm B}_{3}$ | $L^{2}_{2,2k}$ | $L^{2}_{2,2k+1}$ | $L^{2}_{2,2k+2}$ | $\cdots$ | $L^{2}_{2,3k-3}$ | $L^{2}_{2,3k-2}$ | $L^{2}_{2,3k-1}$
---|---|---|---|---|---|---|---
$\varphi_{d,+}$ | $2k+2$ | $2k+3$ | $2k+4$ | $\cdots$ | $3k-1$ | $3k$ | $3k+1$

${\rm B}_{4}$ | $L^{2}_{2k+2,k}$ | $L^{2}_{2k+2,k+1}$ | $L^{2}_{2k+2,k+2}$ | $\cdots$ | $L^{2}_{2k+2,2k-3}$ | $L^{2}_{2k+2,2k-2}$ | $L^{2}_{2k+2,2k-1}$
---|---|---|---|---|---|---|---
$\varphi_{d,+}$ | $3k+2$ | $3k+3$ | $3k+4$ | $\cdots$ | $4k-1$ | $4k$ | $0$

Thus, $\varphi_{d,+}({\rm C}^{2})=R_{d}$. Again, using Corollary concluded
that ${\rm C}^{2}$ is constituted by $d$ skew lines.
Step 4: ${\rm C}^{1}\cup{\rm C}^{2}$ is constituted by $2d$ skew lines.

Having in mind (c) in Corollary is enough to prove that $\psi_{d}({\rm
C}^{1})\,\cap\,\psi_{d}({\rm C}^{2})=\emptyset$. So, we will use again the
stratification for ${\rm C}^{1}$ in (nd ${\rm C}^{2}$ in (o display the
computation of $\psi_{d}({\rm C}^{1})$ and $\psi_{d}({\rm C}^{2})$ bellow.

${\rm A}_{1}$ | $L^{1}_{2k+1,2k+1}$ | $L^{1}_{2k+1,2k+2}$ | $L^{1}_{2k+1,2k+3}$ | $\cdots$ | $L^{1}_{2k+1,3k-3}$ | $L^{1}_{2k+1,3k-2}$ | $L^{1}_{2k+1,3k-1}$
---|---|---|---|---|---|---|---
$\psi_{d}$ | $2k+2$ | $2k+4$ | $2k+6$ | $\cdots$ | $4k-6$ | $4k-4$ | $4k-2$

${\rm A}_{2}$ | $L^{1}_{3k+1,2k}$
---|---
$\psi_{d}$ | $3k$

${\rm A}_{3}$ | $L^{1}_{1,k}$ | $L^{1}_{1,k+1}$ | $L^{1}_{1,k+2}$ | $\cdots$ | $L^{1}_{1,2k-3}$ | $L^{1}_{1,2k-2}$ | $L^{1}_{1,2k-1}$
---|---|---|---|---|---|---|---
$\psi_{d}$ | $2k+1$ | $2k+3$ | $2k+5$ | $\cdots$ | $4k-5$ | $4k-3$ | $4k-1$

${\rm A}_{4}$ | $L^{1}_{2k+1,0}$ | $L^{1}_{2k+1,1}$ | $L^{1}_{2k+1,2}$ | $\cdots$ | $L^{1}_{2k+1,k-3}$ | $L^{1}_{2k+1,k-2}$ | $L^{1}_{2k+1,k-1}$
---|---|---|---|---|---|---|---
$\psi_{d}$ | $2k+1$ | $2k+3$ | $2k+5$ | $\cdots$ | $4k-5$ | $4k-3$ | $4k-1$

${\rm A}_{5}$ | $L^{1}_{3k+2,4k}$
---|---
$\psi_{d}$ | $3k$

${\rm A}_{6}$ | $L^{1}_{2,3k}$ | $L^{1}_{2,3k+1}$ | $L^{1}_{2,3k+2}$ | $\cdots$ | $L^{1}_{2,4k-3}$ | $L^{1}_{2,4k-2}$ | $L^{1}_{2,4k-1}$
---|---|---|---|---|---|---|---
$\psi_{d}$ | $2k+1$ | $2k+3$ | $2k+5$ | $\cdots$ | $4k-5$ | $4k-3$ | $4k-1$

Thus,

$\psi_{d}({\rm C}^{1})=\\{2k+1,2k+2,\ldots,4k-2,4k-1\\}.$ (4.4.15)

${\rm B}_{1}$ | $L^{2}_{1,0}$ | $L^{2}_{1,1}$ | $L^{2}_{1,2}$ | $\cdots$ | $L^{2}_{1,k-3}$ | $L^{2}_{1,k-2}$ | $L^{2}_{1,k-1}$
---|---|---|---|---|---|---|---
$\psi_{d}$ | $1$ | $3$ | $5$ | $\cdots$ | $2k-5$ | $2k-3$ | $2k-1$

${\rm B}_{2}$ | $L^{2}_{2k+2,3k}$ | $L^{2}_{2k+2,3k+1}$ | $L^{2}_{2k+2,3k+2}$ | $\cdots$ | $L^{2}_{2k+2,4k-2}$ | $L^{2}_{2k+2,4k-1}$ | $L^{2}_{2k+2,4k}$
---|---|---|---|---|---|---|---
$\psi_{d}$ | $0$ | $2$ | $4$ | $\cdots$ | $2k-4$ | $2k-2$ | $2k$

${\rm B}_{3}$ | $L^{2}_{2,2k}$ | $L^{2}_{2,2k+1}$ | $L^{2}_{2,2k+2}$ | $\cdots$ | $L^{2}_{2,3k-3}$ | $L^{2}_{2,3k-2}$ | $L^{2}_{2,3k-1}$
---|---|---|---|---|---|---|---
$\psi_{d}$ | $1$ | $3$ | $5$ | $\cdots$ | $2k-5$ | $2k-3$ | $2k-1$

${\rm B}_{4}$ | $L^{2}_{2k+2,k}$ | $L^{2}_{2k+2,k+1}$ | $L^{2}_{2k+2,k+2}$ | $\cdots$ | $L^{2}_{2k+2,2k-3}$ | $L^{2}_{2k+2,2k-2}$ | $L^{2}_{2k+2,2k-1}$
---|---|---|---|---|---|---|---
$\psi_{d}$ | $1$ | $3$ | $5$ | $\cdots$ | $2k-5$ | $2k-3$ | $2k-1$

Thus,

$\psi_{d}({\rm C}^{2})=\\{0,1,2,\ldots,2k-1,2k\\}.$ (4.4.16)

Thus, from (nd e have that $\psi_{d}({\rm C}^{1})\,\cap\,\psi_{d}({\rm
C}^{2})=\emptyset$. ∎

######  Proposition 4.5.

Let $d=2n+1$ with $n=2k+1$ and $k\geq 3$. Consider the families

${\rm
C}^{0}=\big{\\{}L^{0}_{2k+1+i,0}\big{\\}}_{i=0}^{k+1}\cup\big{\\{}L^{0}_{k+1+i,3k+i}\big{\\}}_{i=1}^{k+2}\cup\big{\\{}L^{0}_{2+i,2k-1+i}\big{\\}}_{i=0}^{k+1}\cup\big{\\{}L^{0}_{k+4+i,k+2+i}\big{\\}}_{i=0}^{k-4},$
${\rm
C}^{1}=\big{\\{}L^{1}_{2k+1,2k+i}\big{\\}}_{i=1}^{k-1}\cup\big{\\{}L^{1}_{3k+1,2k}\big{\\}}\cup\big{\\{}L^{1}_{1,k+i}\big{\\}}_{i=0}^{k-1}\cup\big{\\{}L^{1}_{2k+1,i}\big{\\}}_{i=0}^{k-1}\cup\big{\\{}L^{1}_{3k+2,4k}\big{\\}}\cup\big{\\{}L^{1}_{2,3k+i}\big{\\}}_{i=0}^{k-1},$
${\rm C}^{2}=\
\big{\\{}L^{2}_{3,i}\big{\\}}_{i=0}^{k}\cup\big{\\{}L^{2}_{2k+4,k+1+i}\big{\\}}_{i=0}^{k}\cup\big{\\{}L^{2}_{2,2k+2+i}\big{\\}}_{i=0}^{k}\cup\big{\\{}L^{2}_{2k+4,3k+3+i}\big{\\}}_{i=0}^{k-1}.$

It is verified that ${\rm C}^{0}\cup{\rm C}^{1}\cup{\rm C}^{2}$ consists of
$3d$ skew lines in ${\rm F}_{d}$.

###### Proof.

It is analogous to the proof presented in the Proposition proof

It follows the from the previous Propositions Prop:dimparnimpar and the
Examples missingex:d9ed11 that

######  Corollary 4.6.

Assume $d\geq 7$ odd. Then ${\mathfrak{s}}(F_{d})=3d$.

######  Theorem 4.7.

Let ${{\rm F}_{d}}$ be the Fermat surface of degree $d\geq 3$. If
$\textrm{r}({{\rm F}_{d}})$ is the maximum number of skew lines in ${{\rm
F}_{d}}$, then $\textrm{r}({{\rm F}_{d}})=3d$ for all $d\not=3,5.$ Being
$\textrm{r}(\textrm{F}_{3})=6$ and $\textrm{r}(\textrm{F}_{5})=13.$

###### Proof.

For $d\in\\{3,5\\}$ see Section or the other cases, see the previous results
in this section.∎

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Sally ANDRIA | Jacqueline ROJAS | Wállace MANGUEIRA
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Dep. of Mathematics, UFF | Dep. of Mathematics, UFPB | Dep. of Mathematics, UFPB
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