# Higher weight spectra of ternary codes associated to the quadratic Veronese
$3$-fold

Krishna Kaipa  and  Puspendu Pradhan  Department of Mathematics, Indian
Institute of Science Education and Research, Pune, Maharashtra, 411008 India.
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

The problem studied in this work is to determine the higher weight spectra of
the Projective Reed-Muller codes associated to the Veronese $3$-fold
$\mathcal{V}$ in $PG(9,q)$, which is the image of the quadratic Veronese
embedding of $PG(3,q)$ in $PG(9,q)$. We reduce the problem to the following
combinatorial problem in finite geometry: For each subset $S$ of
$\mathcal{V}$, determine the dimension of the linear subspace of $PG(9,q)$
generated by $S$. We develop a systematic method to solve the latter problem.
We implement the method for $q=3$, and use it to obtain the higher weight
spectra of the associated code. The case of a general finite field
$\mathbb{F}_{q}$ will be treated in a future work.

###### Key words and phrases:

Quadratic Veronese varieties, Veronese code, Linear system of quadrics,
Generalized weight enumerator polynomial, Extended weight enumerator
polynomial

###### 1991 Mathematics Subject Classification:

94B27, 51E20, 05B25

##  1\. Introduction

Let $PG(n,q)$ denote the $n$-dimensional projective space defined over a
finite field $\mathbb{F}_{q}$. Let $K=(n^{2}+3n+2)/2$. The quadratic Veronese
$n$-fold $\mathcal{V}_{n}$ in $PG(K-1,q)$ is the image of the quadratic
embedding

$\imath:PG(n,q)\hookrightarrow PG(K-1,q),$

defined with respect to coordinates $(x_{0},\dots,x_{n})$ on $PG(n,q)$ and
$(z_{0},\dots,z_{K-1})$ on $PG(K-1,q)$ by

$\imath(x_{0},\dots,x_{n})=(x_{0}^{2},x_{0}x_{1}\dots,x_{n-1}x_{n},x_{n}^{2}),$

for some fixed ordering of the $K$ monomials of degree $2$ in the variables
$x_{0},\dots,x_{n}$. Let $P_{1},\dots,P_{N}$ where $N=1+q+\dots+q^{n}$, be a
fixed ordering of the points of $PG(n,q)$ and let $\mathbb{G}_{n}$ denote the
matrix of size $K\times N$ defined by

$\mathbb{G}_{n}=\left[\begin{matrix}v_{1}&\dots&v_{N}\end{matrix}\right],$

where $v_{j}\in\mathbb{F}_{q}^{K}$ represents the point $\imath(P_{j})$ of
$PG(K-1,q)$. The linear code of length $N$ and dimension $K$ generated by the
matrix $\mathbb{G}_{n}$ is the quadratic Veronese code $C_{n}$, or the second
order Projective Reed Muller code of length $N$ ([5, §5]). A codeword of the
code $C_{n}$ is a quadratic form $f$ on $PG(n,q)$ and the weight of a codeword
is the number of points of $PG(n,q)$ not lying on the corresponding quadric
hypersurface defined by the vanishing of $f$. An $r$-dimensional subcode of
$C_{n}$ is a linear subspace of $C_{n}$ generated by some $r$ linearly
independent quadratic forms $f_{1},\dots,f_{r}$ on $PG(n,q)$. The weight of
this subcode is the number of points of $PG(n,q)$ which do not lie on the
common zero set of $f_{1},\dots,f_{r}$. Equivalently, $r$-dimensional subcodes
of $C_{n}$ correspond to codimension $r$ linear subspaces $\mathcal{L}$ of
$PG(K-1,q)$, with the weight of the subcode being the number of points of
$\mathcal{V}_{n}$ not lying on $\mathcal{L}$. The $r$-th generalized weight
enumerator (in short GWE) of $C_{n}$ is the polynomial

$W^{(r)}(Z)=\sum_{\mathcal{L}\subset PG(K-1,q)\text{ of codim
$r$}}Z^{\\#\mathcal{V}_{n}\setminus\mathcal{L}}=\sum_{i}A_{i}^{(r)}Z^{i},$

where $A_{i}^{(r)}$ is the number of codimension $r$ linear subspaces of
$PG(K-1,q)$ which intersect $\mathcal{V}_{n}$ in $N-i$ points.

We also explain the algebraic-geometric significance of the polynomial
$Z^{N}W^{(r)}(\frac{1}{Z})$. It gives the answer to the following geometric
problem:

######  Problem.

Given natural numbers $n$ and $r$, let $\mathcal{S}^{(r)}_{n}$ denote the
collection of varieties in $\mathbb{P}^{n}$ defined by $r$ linearly
independent quadratic forms
$f_{1}(X_{0},\dots,X_{n}),\dots,f_{r}(X_{0},\dots,X_{n})$ over
$\mathbb{F}_{q}$. For $V\in\mathcal{S}^{(r)}_{n}$, let $\\#V(\mathbb{F}_{q})$
denote the number of $\mathbb{F}_{q}$-rational points of $V$. Determine the
generating function

$\sum_{V\in\mathcal{S}^{(r)}_{n}}Z^{\\#V(\mathbb{F}_{q})}=\sum_{i\geq
0}c_{i}^{(r)}(n)Z^{i},$

where $c_{i}^{(r)}(n)$ is the number of varieties in $\mathcal{S}^{(r)}_{n}$
having $i$ rational points over $\mathbb{F}_{q}$.

The numbers $c_{i}^{(r)}(n)$ are just the numbers $A^{(r)}_{N-i}(C_{n})$.

The first GWE of $C_{n}$ is easily calculated (see Theorem 2.1). However, for
$r>1$, the $r$-th GWEs of $C_{n}$ are hard to determine. In the work [2],
Johnsen and Verdure determined the $r$-GWE’s of the code $C_{2}$ for all $r$
and all finite fields $\mathbb{F}_{q}$. In the work [3], they further obtained
the $r$-th GWEs of the code $C_{3}$ for all $r$ over the binary field
$\mathbb{F}_{2}$. In this work, we develop a method to determine all the
$r$-GWE’s of the code $C_{3}$ over a general finite field $\mathbb{F}_{q}$. We
now explain the gist of our method: Given a linear code $C$ of dimension $K$
with a generator matrix $M$, let $B_{j,i}$ denote the number of $K\times j$
submatrices of $M$ that have rank $i$. As shown by Jurrius-Pellikaan [4], the
knowledge of these quantities $B_{j,i}$ for all $(j,i)$ suffices to determine
the $r$-GWE’s of the code $C$ for all $r$. For the code $C_{3}$, the
quantities $B_{j,i}$ are the number of configurations of $j$ points in
$PG(3,q)$ whose images in $\mathcal{V}_{3}$ span a $(i-1)$-dimensional linear
subspace in $PG(9,q)$. We treat this problem from a finite geometry
perspective, and develop a systematic method to calculate the quantities
$B_{j,i}$ for all $(j,i)$, and hence obtain the $r$-th GWE of $C_{3}$ for all
$r$ and for all fields $\mathbb{F}_{q}$. We execute the method for $q=3$ in
the present work, and the case of general $q$ will be treated in a future
work.

######  Theorem 1.1.

For $1\leq r\leq 10$, let $W^{r}(Z)$ denote the $r$-th GWE of the code $C_{3}$
over the ternary field $\mathbb{F}_{3}$. We have:

$\displaystyle W^{(1)}(Z)=$ $\displaystyle
2Z^{18}(195Z^{18}+4212Z^{12}+4700Z^{9}+5265Z^{6}+390)$
$\displaystyle\begin{split}W^{(2)}(Z)={}&13Z^{24}(67797Z^{16}+243000Z^{15}+597780Z^{14}+933120Z^{13}+1151610Z^{12}+991440Z^{11}\\\
&+816480Z^{10}+428400Z^{9}+182250Z^{8}+116640Z^{7}+43200Z^{6}+14580Z^{4}+760Z^{3}+360).\end{split}$
$\displaystyle\begin{split}W^{(3)}(Z)={}&10Z^{26}(408114369Z^{14}+615347208Z^{13}+471425994Z^{12}+230935536Z^{11}+79511991Z^{10}\\\
&+20839572Z^{9}+5231304Z^{8}+968760Z^{7}+237978Z^{6}+50544Z^{5}+9360Z^{4}+52Z+108).\end{split}$
$\displaystyle\begin{split}W^{(4)}(Z)={}&Z^{27}(304888261329Z^{13}+152054622720Z^{12}+37402159860Z^{11}+5713746480Z^{10}\\\
&+641358900Z^{9}+69893928Z^{8}+7076160Z^{7}+627120Z^{6}+89505Z^{5}+40).\end{split}$
$\displaystyle\begin{split}W^{(5)}(Z)={}&26Z^{33}(49159423665Z^{7}+8099569140Z^{6}+649260360Z^{5}+31541400Z^{4}\\\
&+1366565Z^{3}+73548Z^{2}+2430Z+120).\end{split}$ $\displaystyle W^{(6)}(Z)=$
$\displaystyle
2Z^{35}(237163797936Z^{5}+12886036860Z^{4}+334019790Z^{3}+4956120Z^{2}+104975Z+2340).$
$\displaystyle W^{(7)}(Z)=$ $\displaystyle
10Z^{36}(1800649161Z^{4}+31769712Z^{3}+252954Z^{2}+936Z+13).$ $\displaystyle
W^{(8)}(Z)=$ $\displaystyle 13Z^{38}(5557197Z^{2}+30160Z+60).$ $\displaystyle
W^{(9)}(Z)=$ $\displaystyle 4Z^{39}(7371Z+10).$ $\displaystyle W^{(10)}(Z)=$
$\displaystyle Z^{40}.$

In particular, the generalized Hamming weights of the code $C_{3}$ over
$\mathbb{F}_{3}$ are given by

$d_{1}=18,\,d_{2}=24,\,d_{3}=26,\,d_{4}=27,\,d_{5}=33,\,d_{6}=35,\,d_{7}=36,\,d_{8}=38,\,d_{9}=39,\,d_{10}=40.$

The rest of this paper is organized as follows. In §2, we record the ordinary
weight enumerator of the code $C_{n}$. In §3, we explain how the quantities
$B_{j,i}$ determine the $r$-GWE’s of a code $C$, and we develop the geometric
method to determine $B_{j,i}$’s for the code $C_{3}$. In §4, we determine
$B_{j,1},B_{j,2},B_{j,3},B_{j,4}$ and $B_{j,5}$ for all $j$ and for a general
finite field $\mathbb{F}_{q}$. In §5, §6, §7 and §8, we determine the
quantities $B_{j,6},B_{j,7},B_{j,8}$ and $B_{j,9}$ respectively for all $j$
and $q=3$. Finally, in §9 we prove Theorem 1.1.

##  2\. Ordinary weight enumerator of the codes $C_{n}$

We will write down the first generalized weight enumerator
$W^{(1)}_{C_{n}}(Z)$ of $C_{n}$. The ordinary weight enumerator of $C_{n}$ is
just $1+(q-1)W^{(1)}_{C_{n}}(Z)$. We need some notation. Let $GL_{n}(q)$
denote the general linear group of invertible $n\times n$ matrices over
$\mathbb{F}_{q}$, The corresponding projective general linear group will be
denoted $PGL_{n}(q)$. Let

(1)
$\genfrac{[}{]}{0.0pt}{0}{r}{j}_{q}=\frac{\prod_{i=0}^{j-1}(q^{r}-q^{i})}{|GL_{j}(q)|}=\prod_{i=0}^{j-1}\frac{(q^{r}-1)(q^{r-1}-1)\dots(q^{r-j+1}-1)}{(q^{j}-1)(q^{j-1}-1)\dots(q-1)},$

denote the Gaussian binomial coefficient, which is the number of
$j$-dimensional subspaces in $\mathbb{F}_{q}^{r}$. Let $f_{0}(X,Y)$ denote a
fixed quadratic form in the variables $X,Y$ which is irreducible over
$\mathbb{F}_{q}$. We recall from [1, Chapter 1] that the non-degenerate
quadratic forms on $PG(2t,q)$ form a single orbit with representative
$X_{0}X_{1}+X_{2}X_{3}+\dots+X_{2t-2}X_{2t-1}+X_{2t}^{2}$. The corresponding
quadrics in $PG(2t,q)$ are known as parabolic quadrics, and have
$(q^{2t}-1)/(q-1)$ points. The size of this orbit is
$q^{t^{2}+t}\prod_{i=1}^{t}(q^{2i+1}-1)$. The non-degenerate quadratic forms
on $PG(2t-1,q)$ form two orbits represented by
$X_{0}X_{1}+X_{2}X_{3}+\dots+X_{2t-2}X_{2t-1}$ and
$X_{0}X_{1}+X_{2}X_{3}+\dots X_{2t-4}X_{2t-3}+f(X_{2t-2},X_{2t-1})$. These
orbits have sizes
$\tfrac{1}{2}q^{t^{2}}(q^{t}+1)\prod_{i=1}^{t-1}(q^{2i+1}-1)$ and
$\tfrac{1}{2}q^{t^{2}}(q^{t}-1)\prod_{i=1}^{t-1}(q^{2i+1}-1)$ respectively.
The corresponding quadrics in $PG(2t-1,q)$ are known as hyperbolic and
elliptic quadrics respectively, and they have $(q^{t-1}+1)(q^{t}-1)/(q-1)$ and
$(q^{t-1}-1)(q^{t}+1)/(q-1)$ points respectively. Given $m<n$ and a quadratic
form $f(X_{0},\dots,X_{m})$ such that the corresponding quadric in $PG(m,q)$
has $\nu$ points, then the quadric defined by $f$ in $PG(n,q)$ has
$1+q+\dots+q^{n-m-1}+\nu q^{n-m}$ points. Thus, the weight of $f$ viewed as a
codeword of $C_{n}$ is $q^{n-m}(1+q+\dots+q^{m}-\nu)$. Since there are
$\genfrac{[}{]}{0.0pt}{1}{n+1}{m+1}_{q}$ ways to pick a $m$-dimensional linear
subspace of $PG(n,q)$ we conclude that the values that weights of codewords of
$C_{n}$ can take are as listed below:

  1. (1)

$q^{n}+q^{n-t}$, where $1\leq t\leq\lfloor(n+1)/2\rfloor$. The number of
codewords with this weight are

$A_{q^{n}+q^{n-t}}^{(1)}=\tfrac{1}{2}\genfrac{[}{]}{0.0pt}{1}{n+1}{2t}_{q}\,q^{t^{2}}(q^{t}-1)\prod_{i=1}^{t-1}(q^{2i+1}-1).$

  2. (2)

$q^{n}$, and the number of codewords with this weight are

$A_{q^{n}}^{(1)}=\sum_{t=0}^{\lfloor
n/2\rfloor}\genfrac{[}{]}{0.0pt}{1}{n+1}{2t+1}_{q}\,q^{t^{2}+t}\prod_{i=1}^{t}(q^{2i+1}-1).$

  3. (3)

$q^{n}-q^{n-t}$, where $1\leq t\leq\lfloor(n+1)/2\rfloor$. The number of
codewords with this weight are

$A_{q^{n}-q^{n-t}}^{(1)}=\tfrac{1}{2}\genfrac{[}{]}{0.0pt}{1}{n+1}{2t}_{q}\,q^{t^{2}}(q^{t}+1)\prod_{i=1}^{t-1}(q^{2i+1}-1).$

We summarize this:

######  Theorem 2.1.

The ordinary weight enumerator of the code $C_{n}$ is given by
$1+(q-1)W^{(1)}(C_{n})$ where:

(2)
$W^{(1)}_{C_{n}}(Z)=A_{q^{n}}^{(1)}\,Z^{q^{n}}+\sum_{t=1}^{\lfloor(n+1)/2\rfloor}\left(A_{q^{n}-q^{n-t}}^{(1)}\,Z^{q^{n}-q^{n-t}}+A_{q^{n}+q^{n-t}}^{(1)}\,Z^{q^{n}+q^{n-t}}\right),$

and the quantities $A^{(1)}_{j}$ are as above.

In particular, for $n=3$ we have $6$ orbits of quadratic forms on $PG(3,q)$,
which we record in Table 1, including their sizes, and weights when regarded
as codewords of the code $C_{n}$.

Orbit | Representative | Size | weight
---|---|---|---
$\mathcal{O}_{1}$ | $X_{0}X_{1}$ | $\tbinom{q^{3}+q^{2}+q+1}{2}$ | $q^{3}-q^{2}$
$\mathcal{O}_{2}$ | $X_{0}X_{3}-X_{1}X_{2}$ | $\tfrac{1}{2}(q^{6}+q^{4})(q^{3}-1)$ | $q^{3}-q$
$\mathcal{O}_{3}$ | $X_{0}^{2}$ | $(1+q)(1+q^{2})$ | $q^{3}$
$\mathcal{O}_{4}$ | $X_{1}^{2}-X_{0}X_{2}$ | $(1+q)(1+q^{2})(q^{5}-q^{2})$ | $q^{3}$
$\mathcal{O}_{5}$ | $X_{2}X_{3}-f_{0}(X_{0},X_{1})$ | $\tfrac{1}{2}(q^{6}-q^{4})(q^{3}-1)$ | $q^{3}+q$
$\mathcal{O}_{6}$ | $f_{0}(X_{0},X_{1})$ | $\tfrac{1}{2}(q^{3}-1)(q^{3}+q)$ | $q^{3}+q^{2}$
Table 1. List of $G$-orbits of quadratic forms on $PG(3,q)$

Specializing (2) to $n=3$, we get:

(3)
$W^{(1)}_{C_{3}}(Z)=\tfrac{1}{2}(q^{3}+q^{2}+q+1)(q^{3}+q^{2}+q)\,Z^{q^{3}-q^{2}}+\tfrac{1}{2}(q^{6}+q^{4})(q^{3}-1)\,Z^{q^{3}-q}\\\
+(q^{5}-q^{2}+1)(q^{3}+q^{2}+q+1)\,Z^{q^{3}}\\\
+\tfrac{1}{2}(q^{6}-q^{4})(q^{3}-1)\,Z^{q^{3}+q}+\tfrac{1}{2}(q^{3}-1)(q^{3}+q)\,Z^{q^{3}+q^{2}}$

Further specializing this to $q=3$, we get

$2Z^{18}(195Z^{18}+4212Z^{12}+4700Z^{9}+5265Z^{6}+390),$

which matches with Theorem 1.1.

##  3\. Approach to the Problem

Let $C$ be a linear $[N,K,d]_{q}$ code over $\mathbb{F}_{q}$ of length $N$,
dimension $K$ and minimum distance $d$ with generator matrix $M_{K\times N}$.
For $J\subset\\{1,\dots,N\\}$, let $r_{J}$ be the rank of the submatrix of
$M_{K\times N}$ on the columns indexed by $J$. The higher weight spectrum of
$C$ can be obtained just by knowing $r_{J}$ for each
$J\subset\\{1,\dots,N\\}$. To see this, consider for each $1\leq j\leq N$, the
polynomials $B_{j}(T)\in\mathbb{Z}[T]$ defined by

(4) $\displaystyle B_{j}(T)=\sum_{J}(T^{K-r_{J}}-1),$

where the sum runs over all subsets $J\subset\\{1,\dots,N\\}$ of size $j$. We
define $B_{0}(T)=(T^{K}-1)$. The _extended weight enumerator_ polynomial (in
short EWE) is the polynomial $W_{C}(Z;T)\in\mathbb{Z}[Z,T]$ defined by

$\displaystyle W(Z;T)=$ $\displaystyle
1+\sum_{j=0}^{N}B_{j}(T)(1-Z)^{j}Z^{N-j}$ (5) $\displaystyle=$ $\displaystyle
1+(T^{K}-1)Z^{N}+\sum_{j=1}^{N}B_{j}(T)(1-Z)^{j}Z^{N-j}.$

Let $A_{j}(T)\in\mathbb{Z}[T]$ be polynomials defined by

(6) $\displaystyle W(Z;T)=1+\sum_{j=1}^{N}A_{j}(T)Z^{j}.$

The polynomial $W(Z;q^{m})$ is the weight enumerator polynomial of the
extended code $C\otimes\mathbb{F}_{q^{m}}$ (see [4, Proposition 5.23]). Thus,
$A_{j}(q^{m})$ is the number of weight $j$ codewords of
$C\otimes\mathbb{F}_{q^{m}}$. The $r$-th generalized weight enumerator
polynomial of $C$ can be obtained from extended weight enumerator polynomial:

######  Theorem 3.1.

[4, Theorem 5.7] Let $C$ be a linear code over $\mathbb{F}_{q}$ and $0\leq
r\leq K$. Then the $r$-th generalized weight enumerator polynomial of $C$ can
be obtained from the extended weight enumerator polynomial of $C$ by:

$W^{(r)}(Z)=\frac{1}{|GL_{r}(q)|}\sum_{j=0}^{r}{r\brack
j}_{q}(-1)^{r-j}q^{\tbinom{r-j}{2}}W(Z;q^{j}).$

From (4), (3) and Theorem 3.1, it follows that determining the quantities
$r_{J}$ for all $J\subset\\{1,\dots,N\\}$ allows us to obtain the $r$-th
generalized weight enumerator polynomial of $C$. It will be convenient to
introduce the numbers $B_{j,i}$ defined to be the number of
$J\subset\\{1,\dots,N\\}$ such that $|J|=j$ and rank$(G_{J})=i$. In terms of
$B_{j,i}$ we can write

$B_{j}(T)=\sum_{i\geq 1}B_{j,i}(T^{K-i}-1).$

We note that in the polynomial $B_{j}(T)$ the quantities $B_{j,K}$ are
multiplied by $(T^{K-K}-1)=0$. Therefore, there is no need to determine the
quantities $B_{j,K}$. In other words, we may re-define

(7) $B_{j}(T)=\sum_{i=1}^{K-1}B_{j,i}(T^{K-i}-1).$

By [4, Theorem 5.4], we have $A_{0}(T)=1$, and for $j>0$:

(8)
$A_{j}(T)=\sum_{i=N-j}^{N-d}(-1)^{N+j+i}\textstyle\tbinom{i}{N-j}B_{i}(T).$

It will be useful to note that :

(9) $B_{j,j}=\textstyle\tbinom{N}{j}-\sum_{i<j}B_{j,i},\quad\text{ if $j\leq
K$}.$

Although, in the expression (7) the quantities $B_{j,K}$ were not required,
for later use we note that

(10) $B_{j,K}=\textstyle\tbinom{N}{j}-\sum_{i<K}B_{j,i},\quad\text{ if
$j>K$.}$

We now return to the code $C_{3}$. Here the Veronese embedding
$\imath:PG(3,q)\to PG(9,q)$ is given by:

(11)
$\imath(x_{0},\dots,x_{3})=(x_{0}^{2},x_{0}x_{1},x_{1}^{2},x_{0}x_{2},x_{1}x_{2},x_{2}^{2},x_{0}x_{3},x_{1}x_{3},x_{2}x_{3},x_{3}^{2}).$

The generator matrix $\mathbb{G}_{3}$ of $C_{3}$ will be denoted as just
$\mathbb{G}$. We need some more notation and definitions.

  1. –

A _configuration_ $S$ is just a set of points of $PG(3,q)$.

  2. –

The group $PGL_{4}(q)$ will be denoted as $G$.

  3. –

For a homogeneous ideal $I$ of $\mathbb{F}_{q}[X_{0},X_{1},X_{2},X_{3}]$
generated by quadratic forms $f_{1},\dots,f_{r}$, let $V(I)$ denote the
intersection of the quadrics $f_{1}=\dots=f_{r}=0$ in $PG(3,q)$.
For $S\subset PG(3,q)$, let $I_{2}(S)$ denote the vector space of quadratic
forms $f(X_{0},X_{1},X_{2},X_{3})$ vanishing on $S$.

  4. –

The _rank_ of a configuration $S$ is defined by any one of the following
equivalent numbers:

    1. (a)

one more than the dimension of the linear subspace of $PG(3,q)$ spanned by
$\imath(S)$.

    2. (b)

the rank of the submatrix of $\mathbb{G}$ on the columns representing the
points of $\imath(S)$.

    3. (c)

$10-\text{dim}(I_{2}(S))$

######  Definition 3.2.

A configuration $S\subset PG(3,q)$ of rank $r$ is a maximal rank $r$
configuration – in short a $\mathcal{C}_{r}$– if there is no rank $r$
configuration properly containing $S$. Equivalently, $S$ is a
$\mathcal{C}_{r}$ if $V(I_{2}(S))=S$ and $\text{dim}(I_{2}(S))=10-r$.

For example, the line $\mathcal{L}$ given by $x_{2}=x_{3}=0$ has rank $3$
because $\imath(\mathcal{L})$ is the conic $z_{1}^{2}=z_{0}z_{2}$ in the plane
$z_{3}=\dots=z_{9}=0$ and the span of this conic is the whole plane, or in
other words the vector space $I_{2}(\mathcal{L})$ has basis
$X_{3}^{2},X_{3}X_{2},X_{3}X_{1},X_{3}X_{0},X_{2}^{2},X_{2}X_{1},X_{2}X_{0}$.
Clearly $V(I_{2}(\mathcal{L}))$ is the points of intersection of the
hyperplanes $X_{3}=0$ and $X_{2}=0$ which is the line $\mathcal{L}$. Thus,
$\mathcal{L}$ is a $\mathcal{C}_{3}$. Similarly, the plane $\mathcal{H}$ given
by $x_{3}=0$, has rank $6$ because $\imath(\mathcal{H})$ is a Veronese surface
in the $5$-dimensional linear subspace $PG(5,q)\subset PG(9,q)$ given by
$z_{6}=\dots=z_{9}=0$, and the linear span of this surface is $PG(5,q)$. Here
$I_{2}(\mathcal{H})$ has basis $X_{3}^{2},X_{3}X_{2},X_{3}X_{1},X_{3}X_{0}$
and clearly $V(I_{2}(\mathcal{H}))$ is just the hyperplane $X_{3}=0$ which is
$\mathcal{H}$. Thus, $\mathcal{H}$ is a $\mathcal{C}_{6}$.

We note that there is a monomorphism $\rho:G\hookrightarrow PGL_{10}(q)$ (see
[1, Theorem 4.15]) such that for each $g\in G$ and $v\in PG(3,q)$ we have

$\imath(g\cdot v)=\rho(g)\imath(v),$

and the image $\mathcal{G}(\mathcal{V}_{3})\subset PGL_{10}(q)$ of this
homomorphism is the automorphism group of $\mathcal{V}_{3}$ in $PGL_{10}(q)$.
Therefore, given a line $\mathcal{L}$ (resp. plane $\mathcal{H}$) of $PG(3,q)$
we may assume that up to $G$-equivalence, $\mathcal{L}$ is given by
$x_{2}=x_{3}=0$ (resp. $\mathcal{H}$ is given by $x_{3}=0$). Thus, each line
of $PG(3,q)$ is a $\mathcal{C}_{3}$ and each plane of $PG(3,q)$ is a
$\mathcal{C}_{6}$.

######  Lemma 3.3.

Given a rank $r$ configuration $S$, there is a unique maximal rank $r$
configuration $\bar{S}\supset S$. We have $\bar{S}=V(I_{2}(S))$.

###### Proof.

Let $S^{\prime}=V(I_{2}(S))$. We must show that i) $S^{\prime}$ has rank $r$,
and ii) any rank $r$ configuration $T$ containing $S$, is contained in
$S^{\prime}$.
As for i), it suffices to show $I_{2}(S^{\prime})=I_{2}(S)$. Since
$S^{\prime}\supset S$, we have $I_{2}(S^{\prime})\subset I_{2}(S)$. On the
other hand, if $f\in I_{2}(S)$ then by definition of $S^{\prime}$, it follows
that $f$ vanishes on $S^{\prime}$, and hence $f\in I_{2}(S^{\prime})$, i.e.
$I_{2}(S)\subset I_{2}(S^{\prime})$. Thus, $I_{2}(S^{\prime})=I_{2}(S)$.
As for ii), if $T\supset S$ has rank $r$, then the fact that
$\text{rank}(T)=\text{rank}(S)=r$ implies $I_{2}(S)=I_{2}(T)$, and hence
$T\subset V(I_{2}(T))=V(I_{2}(S))=S^{\prime}$. We define
$\bar{S}=V(I_{2}(S))$. ∎

The next lemma allows us to build $\mathcal{C}_{r}$’s by adding points to each
$\mathcal{C}_{r-1}$.

######  Lemma 3.4.

If $S$ is a $\mathcal{C}_{r}$, and $T\subset S$ then $\bar{T}\subset S$. In
particular, $S$ contains a $\mathcal{C}_{r-1}$ configuration $T$.

###### Proof.

Since $T\subset S$, we get $I_{2}(S)\subset I_{2}(T)$, and hence
$\bar{T}=V(I_{2}(T))\subset V(I_{2}(S))=S$.

We delete points of $S$ one at a time until the rank drops by one. If
$S^{\prime}$ is the resulting rank $(r-1)$ configuration, then
$T=\overline{S^{\prime}}$ is a $\mathcal{C}_{r-1}$ configuration contained in
$S$. ∎

In order to determine the rank $r$ configurations of a given size $j$, it
suffices to look for size $j$ and rank $r$ subsets of each $\mathcal{C}_{r}$.
There are two steps in our approach.

  1. Step 1 $\rangle$

Determine the set $\mathfrak{C}_{r}$ of all $\mathcal{C}_{r}$’s. i.e., the
collection of all maximal rank $r$ configurations. We recursively construct
all $\mathcal{C}_{r}$ configurations by adding points to $\mathcal{C}_{r-1}$
configurations. We will also obtain a decomposition,

$\mathfrak{C}_{r}=\cup_{\alpha\in\\{a,b,c,\dots\\}}\mathcal{M}_{r\alpha}.$

Each of the parts $\mathcal{M}_{r\alpha}$ will be a class of maximal
configurations consisting of a union of $G$-orbits of $\mathcal{C}_{r}$’s. We
will also determine the size of each class $\mathcal{M}_{r\alpha}$.

  2. Step 2 $\rangle$

For each class $M_{r\alpha}$ and each integer $j>r$, we will determine
$B_{j,r}(\mathcal{M}_{r\alpha})$, which is the number of rank $r$
configurations of size $j$, contained in a configuration $T$ representing the
class $\mathcal{M}_{r\alpha}$.

From Step 1 and Step 2, we calculate $B_{j,r}$ for $j>r$ by:

(12) $B_{j,r}=\sum_{\alpha\in\\{a,b,c,\dots\\}}|\mathcal{M}_{r\alpha}|\cdot
B_{j,r}(\mathcal{M}_{r\alpha}),\qquad j>r.$

The quantity $B_{r,r}$ is determined by the identity

$B_{r,r}=\tbinom{N}{r}-\sum_{i<r}B_{r,i}.$

We also give a systematic procedure to determine the quantities
$B_{j,r}(\mathcal{M}_{r\alpha})$ recursively with respect to $r$:

######  Lemma 3.5.

Let $T$ be a $\mathcal{M}_{r\alpha}$ configuration. For each $i<r$, let
$\mathfrak{C}_{i}(T)$ denote the collection of all $\mathcal{C}_{i}$
configurations that are contained in $T$, and that have at least $(r+1)$
points. We can write
$\mathfrak{C}_{i}(T)=\cup_{\beta\in\\{a,b,c,\dots\\}}\mathcal{M}_{i\beta}(T)$
where $\mathcal{M}_{i\beta}(T)$ denotes the class of configurations
$\mathcal{M}_{i\beta}$ that are contained in $T$ and have at least $r+1$
points. We have

(13)
$B_{j,r}(\mathcal{M}_{r\alpha})=\tbinom{|T|}{j}-\sum_{i<r}\sum_{\mathcal{M}_{i\beta}(T)\in\mathfrak{C}_{i}(T)}|\mathcal{M}_{i\beta}(T)|\cdot
B_{j,i}(\mathcal{M}_{i\beta}(T)).$

###### Proof.

In order to obtain $B_{j,r}(\mathcal{M}_{r\alpha})$, we must subtract from
$\tbinom{|T|}{j}$, the number of $j$ point configurations $S\subset T$ having
rank at most $r-1$. If $S\subset T$ is a $j$ point configuration of rank
$i<r$, then by Lemma 3.4 and Lemma 3.3, $\bar{S}$ is the unique
$\mathcal{M}_{i\beta}(T)$ configuration containing $S$. Thus, the second term
in the right hand side of (13) exactly counts all the $j$-point configurations
in $T$ that have rank strictly less than $r$.

∎

In §4, we study configurations of rank at most $5$ (for a general finite field
$\mathbb{F}_{q}$), and in §5, §6, §7 and §8, we study configurations of rank
$6$, $7$, $8$ and $9$ respectively over the field $\mathbb{F}_{3}$. Each of
the sections §4 – §8 is divided into two subsections dedicated to Step 1 and
Step 2 of the above method.

##  4\. Configurations of rank at most $5$ for a general finite field
$\mathbb{F}_{q}$

###  4.1. Classification of $\mathcal{C}_{r}$, for $r\leq 5$

The only configuration of rank $1$ is a single point of $PG(3,q)$, and the
only configuration of rank $2$ is a set of two distinct points of $PG(3,q)$.
We recall that maximal rank $r$ configurations can be built from maximal rank
$(r-1)$ configurations by adding points. There are two ways to add a point
$P_{3}$ to the $\mathcal{C}_{2}$ given by $\\{P_{0},P_{1}\\}$: if $P_{2}$ lies
on the line $L=\overline{P_{0},P_{1}}$, then by Lemma 3.3, the unique
$\mathcal{C}_{3}$ containing this configuration of $3$ points is the line $L$.
All such configurations form one orbit denoted
$\mathcal{M}_{3a}=G\cdot\overline{P_{0},P_{1}}$ of size
$\genfrac{[}{]}{0.0pt}{1}{4}{2}_{q}=(1+q^{2})(1+q+q^{2})$.

If $P_{2}\notin L$ then the resulting configuration
$S=\\{P_{0},P_{1},P_{2}\\}$ of $3$ points in general position in a plane $\Pi$
is also a $\mathcal{C}_{3}$: indeed we may assume $S=\\{e_{0},e_{1},e_{2}\\}$
so that $I_{2}(S)$ has basis

$\\{X_{0}X_{1},X_{0}X_{2},X_{1}X_{2},X_{0}X_{3},X_{1}X_{3},X_{2}X_{3},X_{3}^{2}\\}.$

It is easily checked that $V(I_{2}(S))=S$. All such $\mathcal{C}_{3}$’s form a
single $G$-orbit denoted $\mathcal{M}_{3b}=G\cdot\\{e_{0},e_{1},e_{2}\\}$, and
of size $q^{3}(1+q)^{2}(1+q^{2})(1+q+q^{2})/6$. In summary:

The $G$-orbits in $\mathfrak{C}_{r}$ for $r\in\\{1,2,3\\}$ are as follows:

  * There is one orbit of $\mathfrak{C}_{1}$ given by $\mathcal{M}_{1a}$ of size $(1+q)(1+q^{2})$. This orbit consists of points of $PG(3,q)$.

  * There is one orbit of $\mathfrak{C}_{2}$ given by $\mathcal{M}_{2a}$ of size $\tbinom{(1+q)(1+q^{2})}{2}$. This orbit consists of unordered pairs of points of $PG(3,q)$.

  * There are $2$ orbits in $\mathfrak{C}_{3}$:

    1. $\mathcal{M}_{3a}$

of size $|G(1,3)|$. This orbit consists of the lines of $PG(3,q)$.

    2. $\mathcal{M}_{3b}$

of size $q^{3}(1+q)^{2}(1+q^{2})(1+q+q^{2})/6$. This orbit consists of
unordered triples of non-collinear points of $PG(3,q)$.

####  4.1.1. Classification of $\mathcal{C}_{4}$’s

We now turn to determining the set $\mathfrak{C}_{4}$. Let $S$ be a
$\mathcal{C}_{4}$ containing $\mathcal{M}_{3a}$ configuration represented by
$\overline{e_{0},e_{1}}$. We may assume $S$ contains
$T=\\{e_{2}\\}\cup\overline{e_{0},e_{1}}$. Here $I_{2}(T)$ has basis

$\\{X_{0}X_{2},X_{1}X_{2},X_{0}X_{3},X_{1}X_{3},X_{2}X_{3},X_{3}^{2}\\},$

and $V(I_{2}(T))$ is clearly equal to $T$. Thus, $T$ is a $\mathcal{C}_{4}$.
The set of all such configurations forms a single orbit
$\mathcal{M}_{4a}=G\cdot T$ whose elements consist of a pair $(L,P)$ where $L$
is a line not containing the point $P$. Since there are
$\genfrac{[}{]}{0.0pt}{1}{4}{2}_{q}$ ways to pick a line of $PG(3,q)$ and
$(q^{2}+q^{3})$ ways to pick a point not on this line, we see that the size of
$\mathcal{M}_{4a}$ is $\genfrac{[}{]}{0.0pt}{1}{4}{2}_{q}\cdot(q^{2}+q^{3})$
which simplifies to $q^{2}(1+q)(1+q^{2})(1+q+q^{2})$.

Next, we consider $S$ which is a $\mathcal{C}_{4}$ containing a
$\mathcal{M}_{3b}$ configuration represented by $\\{e_{0},e_{1},e_{2}\\}$. If
$S$ contains a point $P$ which lies on one of the $3$ lines
$\overline{e_{i},e_{j}}$, then by maximality, $S$ contains a
$\mathcal{M}_{3a}$ configuration, which case has already been treated. We now
assume $S$ contains no other point on any line $\overline{e_{i},e_{j}}$,
$0\leq i<j\leq 2$. If $S$ contains another point $P$ of the plane $\Pi$
spanned by $\\{e_{0},e_{1},e_{2}\\}$ then we may assume $S$ contains
$T=\\{e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2}\\}$ which has rank $4$. Here
$I_{2}(T)$ has basis

$\\{X_{3}^{2},X_{3}X_{2},X_{3}X_{1},X_{3}X_{0},X_{0}X_{1}-X_{0}X_{2},X_{0}X_{1}-X_{1}X_{2}\\},$

and clearly $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{4}$. All
such configurations of $4$ points in general position in a plane of $PG(3,q)$
form a single orbit $\mathcal{M}_{4c}=G\cdot T$, because given a such a
$4$-tuple $(P_{1},\dots,P_{4})$, there are $(q-1)q^{3}$ elements $g\in
PGL_{4}(q)$ with the property that
$g\cdot(e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2})=(P_{1},P_{2},P_{3},P_{4})$. Thus,

$\mathcal{M}_{4c}=\tfrac{1}{4!}\tfrac{|PGL_{4}(q)|}{q^{3}(q-1)}=q^{3}(q^{4}-1)(q^{3}-1)(q+1)/24.$

On the other hand, if $S$ contains a point $P$ not lying on the plane
generated by $\\{e_{0},e_{1},e_{2}\\}$ then we may assume $S$ contains
$T=\\{e_{0},e_{1},e_{2},e_{3}\\}$. Here $I_{2}(T)$ has basis

$\\{X_{0}X_{1},X_{0}X_{2},X_{1}X_{2},X_{0}X_{3},X_{1}X_{3},X_{2}X_{3}\\},$

and clearly $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{4}$. Given
$4$ ordered points $(P_{1},\dots,P_{4})$ in general position in $PG(3,q)$,
there are $(q-1)^{3}$ elements $g\in PGL_{4}(q)$ with the property that
$G\cdot(e_{0},e_{1},e_{2},e_{3})=(P_{1},\dots,P_{4})$. Thus, all such
configurations form a single orbit which we denote
$\mathcal{M}_{4b}=G\cdot\\{e_{0},e_{1},e_{2},e_{3}\\}$ of size

$|\mathcal{M}_{4b}|=|PGL_{4}(q)|/(24(q-1)^{3})=q^{6}(1+q)^{2}(1+q^{2})(1+q+q^{2})/24.$

We summarize this:

The $G$-orbits of $\mathfrak{C}_{4}$ are:

  1. $\mathcal{M}_{4a}$

of size $q^{2}(1+q)(1+q^{2})(1+q+q^{2})$. This orbit consists of pairs $(L,P)$
of a line $L$ and a point $P$ of $PG(3,q)$ such that $P\notin L$.

  2. $\mathcal{M}_{4b}$

of size $q^{6}(1+q+q^{2}+q^{3})(1+q+q^{2})(q+1)/24$. This orbit consists of
unordered $4$-tuples of points in general position in $PG(3,q)$.

  3. $\mathcal{M}_{4c}$

of size $q^{3}(q^{4}-1)(q^{3}-1)(q+1)/24$. This orbit consists of unordered
$4$-tuples of points in general position in some plane of $PG(3,q)$.

####  4.1.2. Classification of $\mathcal{C}_{5}$’s

Next, we determine the set $\mathfrak{C}_{5}$. Let $S$ be a $\mathcal{C}_{5}$
containing a $\mathcal{M}_{4a}$ configuration represented by
$\\{e_{2}\\}\cup\overline{e_{0},e_{1}}$. If $S$ contains another point $P$ of
the plane $\Pi$ spanned by $\\{e_{0},e_{1},e_{2}\\}$, then the line
$L=\overline{P,e_{2}}$ meets the line $\overline{e_{0},e_{1}}$ in a third
point $Q$. The configuration $\\{e_{2},P,Q\\}$ is contained in the
$\mathcal{M}_{3a}$ configuration $L$, and hence by Lemma 3.4, $S$ must contain
$L$. Thus, $S$ contains the union of the lines $L$ and
$\overline{e_{0},e_{1}}$ which we denote $T$. We may take $Q=e_{0}$. Here
$I_{2}(T)$ has basis

$\\{X_{1}X_{2},X_{0}X_{3},X_{1}X_{3},X_{2}X_{3},X_{3}^{2}\\},$

and clearly $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{5}$. All
such configurations consisting of $2$ intersecting lines in $PG(3,q)$ form a
single orbit $G\cdot T$ which we denote $\mathcal{M}_{5a}$. Since
$\mathcal{M}_{5a}$ consists of a pair of lines in a plane, we see that the
size of $\mathcal{M}_{5a}$ is:

$|\mathcal{M}_{5a}|=(1+q+q^{2}+q^{3})\cdot\tbinom{1+q+q^{2}}{2}=q(1+q)^{2}(1+q^{2})(1+q+q^{2})/2.$

If $S$ contains no other point $P$ of the plane $\Pi$ spanned by
$\\{e_{0},e_{1},e_{2}\\}$, then we may assume $S$ contains
$T=\\{e_{2},e_{3}\\}\cup\overline{e_{0},e_{1}}$. Here $I_{2}(T)$ has basis

$\\{X_{0}X_{2},X_{0}X_{3},X_{1}X_{2},X_{1}X_{3},X_{3}X_{2}\\},$

and clearly $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{5}$. The
set of all configurations consisting of a line $L$ together with $2$ points on
a line $L^{\prime}$ skew to $L$ form the class $\mathcal{M}_{5b}$ of size

$|M_{4a}|\cdot q^{3}/2=q^{5}(1+q)(1+q^{2})(1+q+q^{2})/2.$

Next, let $S$ be a $\mathcal{C}_{5}$ containing a $\mathcal{M}_{4c}$
configuration represented by
$S_{0}=\\{e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2}\\}$. Suppose $S$ contains
another point $P$ of the plane $\Pi$ spanned by $\\{e_{0},e_{1},e_{2}\\}$. If
$P$ lies on one of the $6$ lines between the $4$ points of $S_{0}$, then by
maximality, the whole line is contained in $S$, and hence $S$ contains a
$\mathcal{M}_{4a}$ configuration, which case has already been treated. If $P$
does not lie on any of the $6$ lines of $S_{0}$ (this case is possible only if
$q>3$), then $S_{0}\cup\\{P\\}$ are $5$ points in general position in the
plane $\Pi$ (i.e., no $3$ of the points are collinear). For $q\geq 4$, given
any $5$ points in general position of $PG(2,q)$, there exists a unique non-
degenerate conic passing through these points. Moreover, all the non-
degenerate conics of $PG(2,q)$ are $PGL_{3}(q)$ equivalent. Let $T$ be the
conic $X_{1}^{2}-X_{0}X_{2}$ in the plane $X_{3}=0$. Here $I_{2}(T)$ has basis

$\\{X_{3}^{2},X_{3}X_{2},X_{3}X_{1},X_{3}X_{0},X_{1}^{2}-X_{0}X_{2}\\},$

and clearly $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{5}$. Let
$\mathcal{M}_{5e}$ denote this class of $\mathcal{C}_{5}$’s consisting of a
non-degenerate conic in a plane of $PG(3,q)$. Since there are
$|PGL_{3}(q)|/|PGL_{2}(q)|=(q^{5}-q^{2})$ non-degenerate conics in a plane,
and there are $(1+q)(1+q^{2})$ planes in $PG(3,q)$, we see that

$|\mathcal{M}_{5e}|=(1+q)(1+q^{2})\cdot|PGL_{3}(q)|/|PGL_{2}(q)|=(q^{5}-q^{2})(1+q^{2})(1+q).$

We now assume $S$ is a $\mathcal{C}_{5}$ containing a $\mathcal{M}_{4c}$
configuration represented by $\\{e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2}\\}$, and
$S$ contains no other point $P$ of the plane $\Pi$ spanned by
$e_{0},e_{1},e_{2}$. We may assume $S$ contains
$T=\\{e_{3},e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2}\\}$. Here $I_{2}(T)$ has basis

$\\{X_{3}X_{2},X_{3}X_{1},X_{3}X_{0},X_{0}X_{1}-X_{0}X_{2},X_{0}X_{1}-X_{1}X_{2}\\},$

and clearly $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{5}$. Let
$\mathcal{M}_{5c}$ denote this class of $\mathcal{C}_{5}$’s consisting of $5$
points $4$ of which are in general position in a plane (there are
$|\mathcal{M}_{4c}|$ ways to pick these), and the fifth point lies outside
this plane (there are $q^{3}$ ways to pick this point), Since there are
$|\mathcal{M}_{4c}|$ ways to pick the $4$ points, and $q^{3}$ ways to pick the
fifth point outside the plane, we see that

$|\mathcal{M}_{5c}|=|\mathcal{M}_{4c}|\cdot
q^{3}=q^{6}(q^{4}-1)(q^{3}-1)(q+1)/24.$

It remains to consider $S$ which is a $\mathcal{C}_{5}$ containing a
$\mathcal{M}_{4b}$ configuration represented by
$\\{e_{0},e_{1},e_{2},e_{3}\\}$. If $S$ contains a point $P$ which lies on one
of the $6$ lines $\overline{e_{i},e_{j}}$, say $\overline{e_{0},e_{1}}$, then
by maximality, $S$ contains $\mathcal{M}_{4a}$ which case has already been
treated. If $P$ does not lie on any of these $6$ lines, but it lies on one of
the $4$ planes formed by $\\{e_{0},e_{1},e_{2},e_{3}\\}\setminus\\{e_{i}\\}$,
$i=0,1,2,3$, $S$ will be a $\mathcal{M}_{5c}$ configuration. Therefore, we
assume any $P\in S\setminus\\{e_{0},e_{1},e_{2},e_{3}\\}$ has the property
that the points of $T=\\{e_{0},e_{1},e_{2},e_{3},P\\}$ are in general
position. Up to $G$-equivalence, we may assume $P=e_{0}+e_{1}+e_{2}+e_{3}$.
Here $I_{2}(T)$ has basis

$\\{X_{0}(X_{2}-X_{1}),X_{1}(X_{2}-X_{0}),X_{0}(X_{3}-X_{1}),X_{1}(X_{3}-X_{0}),X_{2}(X_{3}-X_{0})\\},$

and $V(I_{2}(T))=T$. Thus, $T$ is a $\mathcal{C}_{5}$. Since $5$-tuples
$(P_{1},\dots,P_{5})$ of points in general position in $PG(3,q)$ correspond
bijectively with $PGL_{4}(q)$, we see that the size of the class
$\mathcal{M}_{5d}$ consisting of unordered $5$-tuples of points in general
position in $PG(3,q)$ is

$|\mathcal{M}_{5d}|=|PGL_{4}(q)|/5!=q^{6}(q^{4}-1)(q^{3}-1)(q^{2}-1)/120.$

We summarize this:

The set $\mathfrak{C}_{5}$ can be partitioned into the following five classes:

  1. $\mathcal{M}_{5a}$

of size $q(1+q)^{2}(1+q^{2})(1+q+q^{2})/2$. This class consists of a pair of
intersecting lines of $PG(3,q)$.

  2. $\mathcal{M}_{5b}$

of size $q^{5}(1+q)(1+q^{2})(1+q+q^{2})/2$. This class consists of a line $L$
and two points on a line $L^{\prime}$ skew to $L$.

  3. $\mathcal{M}_{5c}$

of size $q^{6}(q^{4}-1)(q^{3}-1)(q+1)/24$. This class consists of four points
in general position in a plane, together with a point outside the plane.

  4. $\mathcal{M}_{5d}$

of size $q^{6}(q^{4}-1)(q^{3}-1)(q^{2}-1)/120$. This class consists of $5$
points of $PG(3,q)$ in general position.

  5. $\mathcal{M}_{5e}$

of size $(q^{5}-q^{2})(1+q^{2})(1+q)$. This class consists of a non-degenerate
conic in a plane of $PG(3,q)$.

###  4.2. Calculations of $B_{j,i}$ for $j>i$

Since a maximal configuration of rank $1$ and $2$ contains exactly $1$ and $2$
points of $PG(3,q)$ respectively. Therefore, in these cases we can see that
$B_{j,i}=0$ for $i\in\\{1,2\\}$ and $j>i$.

####  4.2.1. $B_{j,3}$ for $j>3$

Let $T$ be a $\mathcal{M}_{3a}$ configuration, i.e., a line in $PG(3,q)$. Then
$T$ has $(1+q)$ points of $PG(3,q)$. Since $j>3$ and any
$S\in\mathfrak{C}_{i}(T)$, $i\leq 2$ has at most $2$ points. By (13), we have

$B_{j,3}(\mathcal{M}_{3a})=\tbinom{1+q}{j}.$

A $\mathcal{M}_{3b}$ configuration has $3$ points, so
$B_{j,3}(\mathcal{M}_{3b})=0$. Therefore from (12), we have

(14)
$B_{j,3}=|\mathcal{M}_{3a}|\cdot\tbinom{1+q}{j}=(1+q^{2})(1+q+q^{2})\cdot\tbinom{1+q}{j}.$

####  4.2.2. $B_{j,4}$ for $j>4$

An $\mathcal{M}_{4a}$ configuration $T$ represented by a pair $(L,P)$ of a
line $L$ and a point $P$ of $PG(3,q)$, has size $(q+2)$ points. Thus:

$\mathfrak{C}_{1}(T)=\mathfrak{C}_{2}(T)=\varnothing,\;\mathfrak{C}_{3}(T)=\mathcal{M}_{3a}(T).$

Since there is only one line in $T$, we have $|\mathcal{M}_{3a}(T)|=1$. By
(13), we have,

$B_{j,4}(\mathcal{M}_{4a})=\tbinom{q+2}{j}-\tbinom{1+q}{j}.$

Since the $\mathcal{M}_{4c}$ and $\mathcal{M}_{4b}$ configurations have
exactly $4$ points of $PG(3,q)$ and $j>4$, we see that
$B_{j,4}(\mathcal{M}_{4b})=B_{j,4}(\mathcal{M}_{4c})=0$. Therefore for $j>4$,
by (12), we have

(15)
$B_{j,4}=q^{2}(1+q)(1+q^{2})(1+q+q^{2})\cdot(\tbinom{q+2}{j}-\tbinom{1+q}{j}).$

####  4.2.3. $B_{j,5}$ for $j>5$

A $\mathcal{M}_{5a}$ configuration $T$ represented by a pair of intersecting
lines of $PG(3,q)$ contains $(2q+1)$ points. Thus,

$\mathfrak{C}_{1}(T)=\mathfrak{C}_{2}(T)=\varnothing,\quad\mathfrak{C}_{3}(T)=\mathcal{M}_{3a}(T),\quad\mathfrak{C}_{4}(T)=\mathcal{M}_{4a}(T).$

We also note that $|\mathcal{M}_{3a}(T)|=2$ and $|\mathcal{M}_{4a}(T)|=2q$. By
(13), we have,

$B_{j,5}(\mathcal{M}_{5a})=\tbinom{2q+1}{j}-2q\cdot(\tbinom{q+2}{j}-\tbinom{1+q}{j})-2\tbinom{1+q}{j}=\tbinom{2q+1}{j}-2q\cdot\tbinom{q+2}{j}+2(q-1)\cdot\tbinom{1+q}{j}.$

####

A $\mathcal{M}_{5b}$ configuration $T$ represented by a line $L$ and two
points $\\{P_{1},P_{2}\\}$ on a line $L^{\prime}$ skew to $L$ contains $(q+3)$
points of $PG(3,q)$. Again,

$\mathfrak{C}_{1}(T)=\mathfrak{C}_{2}(T)=\varnothing,\quad\mathfrak{C}_{3}(T)=\mathcal{M}_{3a}(T),\quad\mathfrak{C}_{4}(T)=\mathcal{M}_{4a}(T).$

We also note that $|\mathcal{M}_{3a}(T)|=1$ and $|\mathcal{M}_{4a}(T)|=2$. By
(13), we have

$B_{j,5}(\mathcal{M}_{5b})=\tbinom{q+3}{j}-2\cdot(\tbinom{2+q}{j}-\tbinom{1+q}{j})-\tbinom{1+q}{j}=\tbinom{q+3}{j}-2\cdot\tbinom{2+q}{j}+\tbinom{1+q}{j}.$

####

A $\mathcal{M}_{5c}$ configuration $T$ represented by the $(q+1)$ points of a
non-degenerate conic in a plane of $PG(3,q)$. Here

$\mathfrak{C}_{1}(T)=\mathfrak{C}_{2}(T)=\mathfrak{C}_{3}(T)=\mathfrak{C}_{4}(T)=\varnothing,$

because $T$ does not contain a line which rules out
$\mathcal{M}_{3a},\mathcal{M}_{4a}$ and configurations in
$\mathfrak{C}_{i}(T)$ must have at least $6$ points which rules out
$\mathcal{M}_{3b},\mathcal{M}_{4b},\mathcal{M}_{4c}$. By (13), we have

$B_{j,5}(\mathcal{M}_{5c})=\tbinom{1+q}{j}.$

Since each of the $\mathcal{M}_{5d}$ or $\mathcal{M}_{5e}$ configurations
contains $5$ points $PG(3,3)$, we have for $j>5$,

$B_{j,5}(\mathcal{M}_{5d})=B_{j,5}(\mathcal{M}_{5e})=0.$

Therefore for $j>5$, we have

(16)
$\begin{split}B_{j,5}=&\tfrac{1}{2}q(1+q)^{2}(1+q^{2})(1+q+q^{2})\cdot\left(\tbinom{2q+1}{j}-2q\cdot\tbinom{q+2}{j}+2(q-1)\cdot\tbinom{1+q}{j}\right)\\\
&{}+\tfrac{1}{2}q^{5}(1+q)(1+q^{2})(1+q+q^{2})\cdot(\tbinom{q+3}{j}-2\cdot\tbinom{2+q}{j}+\tbinom{1+q}{j})\\\
&{}+(q^{5}-q^{2})(1+q^{2})(1+q)\cdot\tbinom{1+q}{j}.\end{split}$

##  5\. Configurations of rank $6$

For the rest of this paper, we assume $q=3$. The size of the group
$G=PGL_{4}(q)$ is

$|G|=12130560=2^{8}\cdot 3^{6}\cdot 5\cdot 13.$

Since an irreducible conic in a projective plane has only $4$ points when
$q=3$. We see that the class of configuration $\mathcal{M}_{5e}$ in
$\mathfrak{C}_{5}$ does not occur when $q=3$. Therefore, the set
$\mathfrak{C}_{5}$ is partitioned into the following four classes when $q=3$:

  1. $\mathcal{M}_{5a}$

of size $2^{4}\cdot 3\cdot 5\cdot 13$. This class consists of a pair of
intersecting lines of $PG(3,3)$.

  2. $\mathcal{M}_{5b}$

of size $2^{2}\cdot 3^{5}\cdot 5\cdot 13$ This class consists of a line $L$
and two points on a line $L^{\prime}$ skew to $L$.

  3. $\mathcal{M}_{5c}$

of size $2^{4}\cdot 3^{5}\cdot 5\cdot 13$. This class consists of four points
in general position in a plane, together with a point outside the plane.

  4. $\mathcal{M}_{5d}$

of size $2^{5}\cdot 3^{5}\cdot 13$. This class consists of $5$ points of
$PG(3,3)$ in general position.

As before, we build the $\mathcal{C}_{6}$’s by adding points to
$\mathcal{C}_{5}$’s. We carry this out systematically by running through the
classes in $\mathfrak{C}_{5}$.

######  Lemma 5.1.

The set $\mathfrak{C}_{6}$ can be partitioned into the following 9 classes:

  1. $\mathcal{M}_{6a}$

of size $2^{3}\cdot 5$. It consists of planes of $PG(3,3)$.

  2. $\mathcal{M}_{6b}$

of size $2^{4}\cdot 3^{4}\cdot 5\cdot 13$. It consists of configurations
$(\\{L_{1},L_{2}\\},\\{P\\})$ where $\\{L_{1},L_{2}\\}$ is a pair of
intersecting lines, and $P$ does not lie on the plane generated by the two
lines.

  3. $\mathcal{M}_{6c}$

of size $3^{4}\cdot 5\cdot 13$. It consists of a pair of skew lines in
$PG(3,3)$.

  4. $\mathcal{M}_{6d}$

of size $2^{6}\cdot 3^{4}\cdot 5\cdot 13$. It consists of configurations
$(L,\\{P_{1},\dots,P_{4}\\})$ where $\\{P_{1},\dots,P_{4}\\}$ are $4$ points
in general position in a plane $\Pi$ and the line $L$ intersects $\Pi$ in one
of these $4$ points.

  5. $\mathcal{M}_{6e}$

of size $2^{4}\cdot 3^{5}\cdot 5\cdot 13$. It consists of configurations
$\\{P_{1},\dots,P_{6}\\}$ where the $3$ lines
$L_{i}=\overline{P_{2i-1},P_{2i}}$ are non-coplanar lines meeting at a single
point different from $\\{P_{1},\dots,P_{6}\\}$.

  6. $\mathcal{M}_{6f}$

of size $2^{4}\cdot 3^{6}\cdot 5\cdot 13$, The elements of $\mathcal{M}_{6f}$
are configurations $\\{P_{1},\dots,P_{6}\\}$ where $P_{1},P_{2}$ are on a line
$L$, and $P_{3},P_{4}$ are on a line $L^{\prime}$, and $P_{5},P_{6}$ are on a
line $L^{\prime\prime}$ with $\\{L^{\prime},L^{\prime\prime}\\}$ being a pair
of skew lines which meet $L$ at distinct points $P,Q\notin\\{P_{1},P_{2}\\}$.

###### Proof.

Let $S$ be a $\mathfrak{C}_{6}$ configuration.

#### Classes $\mathcal{M}_{6a},\mathcal{M}_{6b}$

We begin with $\mathcal{C}_{6}$ configurations $S$ containing a
$\mathcal{M}_{5a}$ configuration represented by the pair of lines
$X_{1}X_{2}=0$ in the plane $X_{3}=0$. If $S$ contains another point $P$ of
the plane $X_{3}=0$, then by maximality of $\mathcal{M}_{5a}$, the
configuration $\\{P\\}\cup\mathcal{M}_{5a}$ has rank $6$. We have already
shown that a plane of $PG(3,3)$ is $\mathcal{C}_{6}$, therefore, $S$ is
represented by the plane $X_{3}=0$. Let $\mathcal{M}_{6a}$ denote the class of
configurations consisting of planes of $PG(3,3)$. The size of this class is
$40$.

If $S$ contains no other point of the plane $X_{3}=0$, then we may assume $S$
contains $T=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\\{e_{3}\\}$.
Here $I_{2}(T)$ has basis $\\{X_{1}X_{2},X_{0}X_{3},X_{1}X_{3},X_{2}X_{3}\\}$
and clearly $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{6}$. Let
$\mathcal{M}_{6b}$ denote the class of configurations consisting of a pair of
intersecting lines (which can be chosen in $|\mathcal{M}_{5a}|$ ways) together
with a point not on the plane generated by the pair of lines (which can be
chosen in $3^{3}$ ways). The size of this class is

$|\mathcal{M}_{6b}|=|\mathcal{M}_{5a}|\cdot
3^{3}=3^{4}(1+3)(1+3+3^{2})(1+3+3^{2}+3^{3})/2=84240.$

#### Classes $\mathcal{M}_{6c},\mathcal{M}_{6d}$

Next, we consider $\mathcal{C}_{6}$ configurations $S$ which contain the
$\mathcal{M}_{5b}$ configuration represented by
$S_{0}=\overline{e_{0},e_{1}}\cup\\{e_{2},e_{3}\\}$. Let
$L=\overline{e_{0},e_{1}}$ and let $L^{\prime}=\overline{e_{2},e_{3}}$. If
$P\in S$ lies on the line $L^{\prime}=\overline{e_{2},e_{3}}$, then by
maximality, $S_{0}\cup\\{P\\}$ has rank $6$, and is contained in the
configuration $T=L\cup
L^{\prime}=\overline{e_{0},e_{1}}\cup\overline{e_{2},e_{3}}$. Here $I_{2}(T)$
has basis $\\{X_{0}X_{2},X_{0}X_{3},X_{1}X_{2},X_{1}X_{3}\\}$ and clearly
$V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{6}$. Let
$\mathcal{M}_{6c}$ denote this class consisting of a pair of skew lines. This
class has size

$|\mathcal{M}_{6c}|=130\cdot 3^{4}/2=5265.$

We now assume $S\supset S_{0}$ does not contain any other point of $L\cup
L^{\prime}$. Given a point $P$ of $S\setminus S_{0}$, if one of the lines
$\overline{P,e_{2}}$ or $\overline{P,e_{3}}$ meets the line
$L=\overline{e_{0},e_{1}}$, then by maximality, $S$ contains a
$\mathcal{M}_{5a}$ configuration, which has already been treated. Thus, the
three points $P,e_{2},e_{3}$ are non-collinear and $L$ does not intersect any
of the $3$ lines generated by $P,e_{2},e_{3}$. Up to $G$-equivalence, we may
assume $T=\\{P\\}\cup S_{0}$ is represented by
$\\{e_{0},e_{1},e_{2}\\}\cup\overline{e_{3},e_{0}+e_{1}+e_{2}}$. Here
$I_{2}(T)$ has basis

$\\{X_{3}(X_{2}-X_{0}),X_{3}(X_{1}-X_{0}),X_{0}(X_{1}-X_{2}),X_{1}(X_{0}-X_{2})\\},$

and we verify that $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{6}$.
Let $\mathcal{M}_{6d}$ be the class consisting of $4$ points in general
position in a plane $\Pi$ (which can be chosen in $|\mathcal{M}_{4c}|$ ways),
together with a line which intersects $\Pi$ in one of these $4$ points (which
can be chosen in $4\cdot 3^{2}$ ways). Thus, the size of this class is

$|\mathcal{M}_{6d}|=4\cdot
3^{2}\cdot|\mathcal{M}_{4c}|=3^{5}(3^{4}-1)(3^{3}-1)(3+1)/6=336960.$

#### Classes $\mathcal{M}_{6e},\mathcal{M}_{6f}$

Next we consider $\mathcal{C}_{6}$ configurations $S$ which contain the
$\mathcal{M}_{5c}$ configuration
$S_{0}=\\{e_{3},e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2}\\}$. If $S$ contains a
point $P$ of the plane $X_{3}=0$, then by maximality, $S$ contains a
$\mathcal{M}_{5a}$, which case has already been treated. So we assume
$S\setminus S_{0}$ contains no other point of the plane $X_{3}=0$. For $P\in
S\setminus S_{0}$, let $T=\\{P,e_{3},e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2}\\}$.

Let $Q$ be the intersection point of the line $\overline{e_{3},P}$ with the
plane $X_{3}=0$. Consider the $6$ lines $L_{1},\dots,L_{6}$ in the plane
$X_{3}=0$ generated by
$\\{P_{1},\dots,P_{4}\\}=\\{e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2}\\}$. Since
$q=3$, the union of these $6$ lines is the whole plane $X_{3}=0$. If a point
$R$ lies on $3$ of these lines then $R\in\\{P_{1},\dots,P_{4}\\}$. There are
three points $R_{1}=e_{0}+e_{1},R_{2}=e_{1}+e_{2},R_{3}=e_{2}+e_{0}$ which lie
on exactly two of these $6$ lines. There are three possibilities for the point
$Q$ above: i) $Q\in\\{e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2}\\}$, ii)
$Q\in\\{R_{1},R_{2},R_{3}\\}$, iii) $Q$ lies on exactly one of the lines
$L_{1},\dots,L_{6}$. We consider each of these possibilities one by one.

i) If $Q\in\\{P_{1},\dots,P_{4}\\}$, then by maximality, $S$ contains a
$\mathcal{M}_{5b}$ configuration, which case has already been treated.

ii) If $Q\in\\{R_{1},R_{2},R_{3}\\}$ say $Q=e_{0}+e_{1}$, then we may assume
$P=e_{0}+e_{1}+e_{3}$ (by replacing $e_{3}$ with $ce_{3}$ for a suitable
scalar $c$). We note that each of the three lines
$\overline{Q,e_{0}},\overline{Q,e_{2}},\overline{Q,e_{3}}$ through $Q$ contain
exactly $2$ points of $T$, namely
$\\{e_{0},e_{1}\\},\\{e_{2},e_{0}+e_{1}+e_{2}\\},\\{e_{3},e_{0}+e_{1}+e_{3}\\}$.
Thus, $T$ is $G$-equivalent to the configuration
$\\{e_{0},e_{0}+e_{3},e_{1},e_{1}+e_{3},e_{2},e_{2}+e_{3}\\}$, with $e_{3}$
representing the point of concurrence of the $3$ lines
$\overline{e_{3},e_{i}}$ for $i=0,1,2$. Here $I_{2}(T)$ has basis

$\\{X_{0}X_{1},X_{1}X_{2},X_{0}X_{2},X_{3}(X_{0}+X_{1}+X_{2}-X_{3})\\},$

and clearly $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{6}$. Let
$\mathcal{M}_{6e}$ be this class of configurations consisting of $2$ points
each on three non-coplanar lines through a point $P$. The configuration
$\mathcal{M}_{6e}$ is obtained by i) picking a point $P$ and three non-
coplanar lines through $P$, and ii) picking $2$ points on each of these $3$
lines different from $P$. The size of this class is

$|\mathcal{M}_{6e}|=40\cdot\tfrac{1}{6}(1+3+3^{2})(1+3)3^{3}\cdot\tbinom{3}{2}^{3}=252720.$

For later use, we note that the stabilizer group in $G$ of the
$\mathcal{M}_{6e}$ configuration $T$ is group of order $48$ which is the
wreath product $Z_{2}\wr S_{3}$ where $S_{3}$ permutes the three lines through
$e_{3}$ (generated by $\left(\begin{smallmatrix}P&0\\\
0&1\end{smallmatrix}\right)$ where $P$ is $3\times 3$ permutation matrix) and
$Z_{2}$ interchanges the $2$ points on a line, (for example generated by
$g_{0}=\left(\begin{smallmatrix}-1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
-1&0&0&1\end{smallmatrix}\right)$ for the line through
$\overline{e_{0},e_{3}}$). In more detail, if $g_{0},g_{1},g_{2}$ are
involutions interchanging the points $\\{e_{i},e_{i}+e_{3}\\}$ for $i=0,1,2$
respectively, then $g_{0},g_{1},g_{2}$ commute with each other and hence
generate the group $Z_{2}\times Z_{2}\times Z_{2}$. The group $S_{3}$ above
acts on $Z_{2}\times Z_{2}\times Z_{2}$ by permutation action on the three
factors. As a corollary, we note that given a point
$Q=ae_{0}+be_{1}+ce_{2}+de_{3}$ with $abc\neq 0$, there exist automorphisms of
the $\mathcal{M}_{6e}$ configuration that allow us to independently change the
sign of $a,b$ and $c$ (the coordinate $d$ is not necessarily fixed). Thus, up
to $G$-equivalence we can assume $Q=e_{0}+e_{1}+e_{2}+d^{\prime}e_{3}$ for
some $d^{\prime}\in\mathbb{F}_{3}$.

iii) If $Q$ is on exactly one of the lines $L_{1},\dots,L_{6}$, say
$\overline{e_{0},e_{1}}$ then $Q\notin\\{R_{1},R_{2},R_{3}\\}$ and hence
$Q\neq e_{0}+e_{1}$. Thus, $P=e_{0}+\lambda e_{1}+e_{3}$ for some
$\lambda\in\mathbb{F}_{3}\setminus\\{0,1\\}$, i.e. $\lambda=-1$. Thus,

$T=\\{e_{0},e_{1},e_{2},e_{0}+e_{1}+e_{2},e_{3},e_{3}+e_{0}-e_{1}\\}.$

Here $I_{2}(T)$ has basis

$\\{X_{3}X_{2},X_{3}(X_{1}+X_{0}),X_{0}(X_{1}-X_{2}+X_{3}),X_{2}(X_{0}-X_{1})\\},$

and it is readily checked that $V(I_{2}(T))$ equals $T$. Thus, $T$ is a
$\mathcal{C}_{6}$. We will denote this class configuration by
$\mathcal{M}_{6f}$. We note that $6$ points of $T$ generate $14$ planes of
which only $2$ of them have $4$ points: these are the planes $X_{2}=0$ and
$X_{3}=0$. Thus, the class $\mathcal{M}_{6f}$ consists of configurations
$G$-equivalent to $T$. An element of $\mathcal{M}_{6f}$ can be constructed by
i) choosing $2$ points $P_{1},P_{2}$ on the line of intersection of two planes
$\Pi_{1},\Pi_{2}$ ii) $2$ points $P_{3},P_{4}$ on the plane $\Pi_{1}$ and iii)
$2$ points $P_{5},P_{6}$ with the property that if
$Q_{1}=\overline{P_{3},P_{4}}\cap L$ and $Q_{2}=\overline{P_{5},P_{6}}\cap L$
then the points $P_{1},P_{2},Q_{1},Q_{2}$ are distinct points of $L$. The pair
of planes can be chosen in $\tbinom{40}{2}$ ways. The points $P_{1},P_{2}$ can
be chosen in $\tbinom{3+1}{2}$ ways, and the points $Q_{1},Q_{2}$ can be
chosen in $(3-1)(3-2)$ ways. The points $P_{3},P_{4}$ and $P_{5},P_{6}$ can be
chosen in $(3\tbinom{3}{2})^{2}$ ways. Thus, the class $\mathcal{M}_{6f}$ has
size

$|\mathcal{M}_{6f}|=\tbinom{40}{2}\cdot\tbinom{3+1}{2}\cdot(3-1)(3-2)\cdot(3\tbinom{3}{2})^{2}=758160.$

For later use, we record the stabilizer group in $G$ of the $\mathcal{M}_{6f}$
configuration $T$. Consider the following four involutions which preserve $T$:

$g_{0}=\left(\begin{smallmatrix}&1&&\\\ 1&&&\\\ &&1&\\\
&&&-1\end{smallmatrix}\right),\,g_{1}=\left(\begin{smallmatrix}&1&-1&\\\
1&&-1&\\\ &&-1&\\\
&&&-1\end{smallmatrix}\right),\,g_{2}=\left(\begin{smallmatrix}&1&&1\\\
1&&&-1\\\ &&1&0\\\
&&&1\end{smallmatrix}\right),\,g_{3}=\left(\begin{smallmatrix}-1&&&\\\ &1&&\\\
&&&-1\\\ &&-1&\end{smallmatrix}\right).$

We note that i) $g_{0}$ interchanges $\\{P_{1},P_{2}\\}$, ii) $g_{1}$
interchanges $\\{P_{1},P_{2}\\}$, $\\{P_{3},P_{4}\\}$, iii) $g_{2}$
interchanges $\\{P_{1},P_{2}\\}$, $\\{P_{5},P_{6}\\}$, and iv) $g_{3}$
interchanges $\\{P_{3},P_{5}\\}$ and $\\{P_{4},P_{6}\\}$. The matrices
$g_{1},g_{2}$ commute and generate the group $Z_{2}\times Z_{2}$. The matrix
$g_{3}$ and satisfies $g_{3}g_{1}g_{3}=g_{2}$, $g_{3}g_{2}g_{1}=g_{2}$. Thus,
the group generated by $g_{1},g_{2},g_{3}$ is isomorphic to the dihedral
group. The element $g_{0}$ commutes with $g_{2},g_{3},g_{4}$. Thus, the
stabilizer group of $T$ is isomorphic to the direct product $D_{4}\times
Z_{2}$

$Z_{2}\times D_{4}:\langle
g_{0},g_{1},g_{2},g_{3}\;|\;g_{0}^{2},g_{1}^{2},g_{2}^{2},g_{3}^{2},(g_{1}g_{2})^{2},(g_{0}g_{1})^{2},(g_{0}g_{2})^{2},(g_{0}g_{3})^{2},g_{3}g_{1}g_{3}g_{2}\rangle.$

Finally, we come to the case when $S$ contains a $\mathcal{M}_{5d}$
represented by $5$ points $\\{P_{1},\dots,P_{5}\\}$ in general position in
$PG(3,3)$. We may take
$\\{P_{1},\dots,P_{5}\\}=\\{e_{0},e_{1},e_{2},e_{3},e_{0}+e_{1}+e_{2}+e_{3}\\}$.
If $P$ is another point of $S$, let $T=\\{P,P_{1},\dots,P_{5}\\}$. Since in
$PG(3,3)$ there is no collection of $6$ points in general position, there are
only two possibilities: i) some $3$ points of $T$ are collinear, or ii) no $3$
points of $T$ are collinear but some $4$ points of $T$ are coplanar. In case
(i), we may assume $P$ lies on the line $\overline{P_{4}P_{5}}$. In this case
by maximality, $S$ contains a $\mathcal{M}_{6d}$ configuration, and in case
(ii), $S$ contains a $\mathcal{M}_{6f}$ configuration. These cases have
already been treated above.

∎

###  5.1. $B_{j,6}$ for $j>6$

####

A $\mathcal{M}_{6a}$ configuration $T$ represented by a plane $\Pi$ contains
$13$ points of $PG(3,3)$. Since $j\geq 7$, we have

$\mathfrak{C}_{1}(T)=\dots=\mathfrak{C}_{4}(T)=\varnothing,\quad\mathfrak{C}_{5}(T)=\mathcal{M}_{5a}(T).$

We note that $|\mathcal{M}_{5a}(T)|=\tbinom{13}{2}=78$. By (13), we have,

$B_{j,6}(\mathcal{M}_{6a})=\tbinom{13}{j}-78\tbinom{7}{j}.$

####

A $\mathcal{M}_{6b}$ configuration $T$ of $8$ points of $PG(3,3)$ represented
by a pair of intersecting lines $L_{1},L_{2}$ and point $P$ outside the plane
$\langle L_{1},L_{2}\rangle$. Here, again:

$\mathfrak{C}_{1}(T)=\dots=\mathfrak{C}_{4}(T)=\varnothing,\quad\mathfrak{C}_{5}(T)=\mathcal{M}_{5a}(T),$

and $|\mathcal{M}_{5a}(T)|=1$. By (13), we have,

$B_{j,6}(\mathcal{M}_{6b})=\tbinom{8}{j}-\tbinom{7}{j}.$

####

A $\mathcal{M}_{6c}$ configuration $T$ of $8$ points of $PG(3,3)$ represented
by a pair of skew lines $L_{1},L_{2}$. Here $\mathfrak{C}_{i}(T)=\varnothing$
for all $1\leq i\leq 5$. By (13), we have,

$B_{j,6}(\mathcal{M}_{6c})=\tbinom{8}{j}.$

####

A $\mathcal{M}_{6d}$ configuration $T$ represented by
$(L,\\{P_{1},\dots,P_{4}\\})$ where $\\{P_{1},\dots,P_{4}\\}$ are $4$ points
in general position in a plane $\Pi$, and the line $L$ intersects $\Pi$ at
$P_{4}$. The configuration $T$ contains $7$ points of $PG(3,3)$. Here again,
$\mathfrak{C}_{i}(T)=\varnothing$ for all $1\leq i\leq 5$, and hence by (13),
we have,

$B_{j,6}(\mathcal{M}_{6d})=\tbinom{7}{j}.$

We have, $B_{j,6}(\mathcal{M}_{6e})=B_{j,6}(\mathcal{M}_{6f})=0$ as each of
$\mathcal{M}_{6e}$ or $\mathcal{M}_{6f}$ configurations contain exactly $6$
points of $PG(3,3)$. Therefore, by (12), for $j>6$, we have

$B_{j,6}=40\cdot(\tbinom{13}{j}-78\tbinom{7}{j})+84240\cdot(\tbinom{8}{j}-\tbinom{7}{j})+5265\cdot\tbinom{8}{j}+336960\cdot\tbinom{7}{j},$

i.e.

(17)
$B_{j,6}=40\cdot\tbinom{13}{j}+89505\cdot\tbinom{8}{j}+249600\cdot\tbinom{7}{j}.$

##  6\. Configurations of rank $7$

As before, we build the $\mathcal{C}_{7}$’s by adding points to
$\mathcal{C}_{6}$’s. We carry this out systematically by running through the
classes in $\mathfrak{C}_{6}$.

######  Lemma 6.1.

The set $\mathfrak{C}_{7}$ can be partitioned into the following 7 classes:

  1. $\mathcal{M}_{7a}$

of size $2^{3}\cdot 3^{3}\cdot 5$. This class consists of a plane of $PG(3,3)$
and a point outside that plane.

  2. $\mathcal{M}_{7b}$

of size $2^{4}\cdot 3^{2}\cdot 5\cdot 13$. This class consists of $3$
concurrent non coplanar lines of $PG(3,3)$.

  3. $\mathcal{M}_{7c}$

of size $2^{4}\cdot 3^{4}\cdot 5\cdot 13$. This class consists of three lines
$L_{1},L_{2},L_{3}$ of $PG(3,3)$, where $L_{2}$ and $L_{3}$ are skew to each
other and the line $L_{1}$ meets both the lines.

  4. $\mathcal{M}_{7d}$

of size $2^{5}\cdot 3^{5}\cdot 5\cdot 13$. This class consists of
configurations $(\\{L_{1},L_{2}\\},\\{P_{1},P_{2}\\})$, where
$\\{L_{1},L_{2}\\}$ is a pair of intersecting lines and $P_{1},P_{2}$ lie
outside the plane $\langle L_{1},L_{2}\rangle$. Also, the line
$\overline{P_{1},P_{2}}$ does not meet $L_{1}$ and $L_{2}$.

  5. $\mathcal{M}_{7e}$

of size $2^{5}\cdot 3^{6}\cdot 5\cdot 13$. This class consists of
configurations $(\\{P_{1},P_{2},P_{3},P_{4}\\},L)$, where the points
$P_{1},P_{2},P_{3},P_{4}$ are in general position in $PG(3,3)$ and the line
$L$ does not meet any of the $6$ lines $\overline{P_{i},P_{j}}$, where $1\leq
i<j\leq 4$.

  6. $\mathcal{M}_{7f}$

of size $2^{6}\cdot 3^{5}\cdot 5\cdot 13$. This class consists of
configurations $\\{P_{1},\dots,P_{6},P_{7}\\}$ where the $3$ lines
$L_{i}=\overline{P_{2i-1},P_{2i}}$, $i=1,2,3$, are non-coplanar lines meeting
at a single point $Q$ different from $\\{P_{1},\dots,P_{6}\\}$. Let $L_{4}$ be
a line through $Q$ such that no three of the four lines
$\\{L_{1},L_{2},L_{3},L_{4}\\}$ are coplanar. Of the three points
$\\{Q_{1},Q_{2},Q_{3}\\}$ on $L_{4}$ different from $Q$, there is only one
point $Q_{1}$ which lies on two of the planes spanned by three points of
$\\{P_{1},\dots,P_{6}\\}$, whereas the other two points $\\{Q_{2},Q_{3}\\}$
lie on three such planes. We take $P_{7}=Q_{1}$.

  7. $\mathcal{M}_{7g}$

of size $2^{2}\cdot 3^{5}\cdot 5\cdot 13$. This class consists of
configurations $\\{P_{1},\dots,P_{8}\\}$ where the four lines
$L_{i}=\overline{P_{2i-1},P_{2i}}$, $i=1,2,3,4$, meet at a single point $Q$
different from $\\{P_{1},\dots,P_{8}\\}$ and no three of these four lines are
coplanar. If we arbitrarily pick $\\{P_{2i-1},P_{2i}\\}$ on $L_{i}$ for
$i=1,\dots,3$, then the points $\\{P_{7},P_{8}\\}$ on $L_{4}$ are the pair of
points $\\{Q_{2},Q_{3}\\}$ in the notation of the description of
$\mathcal{M}_{7f}$ configuration above.

###### Proof.

Let $S$ be a $\mathcal{C}_{7}$ configuration.

#### Class $\mathcal{M}_{7a}$

If $S$ contains a $\mathcal{M}_{6a}$ configuration represented by the plane
$\Pi$ given by $X_{3}=0$, then we may assume $S$ contains
$T=\\{e_{3}\\}\cup\Pi$. Here $I_{2}(T)$ has basis
$\\{X_{3}X_{0},X_{3}X_{1},X_{3}X_{2}\\}$ and $V(I_{2}(T))$ clearly equals $T$.
Thus, $T$ is a $\mathcal{C}_{7}$ which we denote $\mathcal{M}_{7a}$. The size
of $\mathcal{M}_{7a}$ is $3^{3}\cdot|\mathcal{M}_{6a}|=1080$.

#### Classes $\mathcal{M}_{7b}$, $\mathcal{M}_{7c}$, $\mathcal{M}_{7d}$

If $S$ contains a $\mathcal{M}_{6b}$ configuration represented by
$S_{0}=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\\{e_{3}\\}$, and
$P$ is another point of $S$, there are $4$ possibilities: i) $P$ lies on the
plane $\Pi:X_{3}=0$, ii) $P$ lies outside $\Pi$, and the line
$\overline{P,e_{3}}$ meets $\Pi$ at $e_{0}$, iii) $P$ lies outside $\Pi$, and
the line $\overline{P,e_{3}}$ meets $\Pi$ at a point $Q$ of $S_{0}$ different
from $e_{0}$, iv) $P$ lies outside $\Pi$, and the line $\overline{P,e_{3}}$
meets $\Pi$ at a point $Q$ not lying on $S_{0}$.

In case i) by maximality, $S$ contains a $\mathcal{M}_{6a}$ configuration
which case has been treated above.

In case ii) by maximality, $S$ contains the configuration $T$ of three
concurrent lines
$\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\overline{e_{0},e_{3}}$.
Here $I_{2}(T)$ has basis $\\{X_{3}X_{2},X_{3}X_{1},X_{1}X_{2}\\}$ and
$V(I_{2}(T))$ clearly equals $T$. Thus, $T$ is a $\mathcal{C}_{7}$
configuration. We denote the class of configurations $G$-equivalent to $T$ by
$\mathcal{M}_{7b}$. The size of $\mathcal{M}_{7b}$ is
$|\mathcal{M}_{6e}|/\tbinom{3}{2}^{3}$, i.e. $9360$.

In case iii) by maximality of $S$, we may assume $S$ contains the
configuration $T$ of the three lines $L_{1}\cup L_{2}\cup L_{3}$ where
$L_{1}=\overline{e_{0},e_{1}}$, $L_{2}=\overline{e_{0},e_{2}}$ and
$L_{3}=\overline{e_{1},e_{3}}$. Here $I_{2}(T)$ has basis
$\\{X_{3}X_{0},X_{3}X_{2},X_{1}X_{2}\\}$ and $V(I_{2}(T))$ clearly equals $T$.
Thus, $T$ is a $\mathcal{C}_{7}$. We denote the class of configurations
$G$-equivalent to $T$ by $\mathcal{M}_{7c}$. The lines $L_{2}$ and $L_{3}$ are
skew to each other and the line $L_{1}$ meets both the lines. We can choose
two skew lines $L_{2}$ and $L_{3}$ in $PG(3,3)$ in $130\cdot 3^{4}/2$ ways and
pick the line $L_{1}$ in $(3+1)^{2}$ ways. Thus, the size of the class
$\mathcal{M}_{7c}$ is $65\cdot 3^{4}\cdot 4^{2}$, i.e. $84240$.

In case iv) we may assume $S$ contains the configuration $T$ consisting of the
lines $L_{1}=\overline{e_{0},e_{1}}$ and $L_{2}=\overline{e_{0},e_{2}}$ in the
plane $X_{3}=0$, and points $\\{e_{3},e_{3}+e_{1}+e_{2}\\}$. Here $I_{2}(T)$
has basis $\\{X_{1}(X_{3}-X_{2}),X_{2}(X_{3}-X_{1}),X_{3}X_{0}\\}$ and
$V(I_{2}(T))$ clearly equals $T$. We denote the class of configurations
$G$-equivalent to $T$ by $\mathcal{M}_{7d}$. The number of ways to choose $T$
is the number of ways to pick the configuration $\mathcal{M}_{6b}$ multiplied
by the number of ways to pick the point $P$ which is $(3-1)(3^{2}-3)$ (there
are $(3^{2}-3)$ ways to pick the point $Q$ and $(3-1)$ ways to pick the point
$P$ on the line $\overline{e_{3},Q}$). Thus,
$|\mathcal{M}_{7d}|=|\mathcal{M}_{6b}|\cdot\frac{3(3-1)^{2}}{2}$, i.e.
$505440$.

In case $S$ contains the $\mathcal{M}_{6c}$ configuration represented
$L_{1}=\overline{e_{0},e_{1}}$ and $L_{2}=\overline{e_{2},e_{3}}$. Given any
other point $P$ of $S$, we may assume $P=e_{0}+e_{1}+e_{2}+e_{3}$. In this
case the line $L_{3}=\overline{e_{0}+e_{1},e_{2}+e_{3}}$ also contains $P$,
and hence by maximality, $S$ contains $L_{1}\cup L_{2}\cup L_{3}$ which is the
configuration $\mathcal{M}_{7c}$ discussed above.

#### Class $\mathcal{M}_{7e}$

Next we consider the case when $S$ contains the $\mathcal{M}_{6d}$
configuration represented by $S_{0}=\\{P_{1},\dots,P_{4}\\}$ $\cup L$ where
$\\{P_{1},\dots,P_{4}\\}$ are in general position in a plane $\Pi$ and the
line $L$ intersects $\Pi$ in $P_{4}$. Let $P$ be another point of $S$ and let
$T=S_{0}\cup\\{P\\}$. If $P\in\Pi$, then by maximality, $S$ contains a
$\mathcal{M}_{6b}$ configuration, which cases have been treated above.
Similarly, if the plane $\Pi^{\prime}$ generated by $P$ and $L$ contains a
point of $\\{P_{1},P_{2},P_{3}\\}$, then by maximality, $S$ contains a
$\mathcal{M}_{6b}$ configuration, which case has been treated above. So we
assume $\Pi^{\prime}$ does not contain $P_{1},P_{2}$ or $P_{3}$. In this case
$\\{P_{1},P_{2},P_{3},P\\}$ is a $\mathcal{M}_{4b}$ configuration which we may
represent as $\\{e_{0},e_{1},e_{2},e_{3}\\}$. The line $L$ does not meet any
of the $6$ lines $\overline{e_{i},e_{j}}$, where $0\leq i<j\leq 3$. We may
assume $L=\overline{e_{0}+e_{1}+e_{2},e_{0}+c^{\prime}e_{1}+e_{3}}$ where
$c^{\prime}\notin\\{0,1\\}$, i.e. $c^{\prime}=-1$. Here $I_{2}(T)$ has basis

$\\{X_{3}(X_{1}+X_{0}+X_{2}),X_{0}(X_{1}-X_{2}+X_{3}),X_{1}(X_{0}-X_{2})+X_{3}(X_{2}+X_{0})\\},$

and $V(I_{2}(T))$ equals $T$. Thus, $T$ is a $\mathcal{C}_{7}$. We denote, by
$\mathcal{M}_{7e}$, the class of configurations which are $G$-equivalent to
$T$. There are $|\mathcal{M}_{4b}|$ ways to choose the $\mathcal{M}_{4b}$
configuration represented by $\\{e_{0},e_{1},e_{2},e_{3}\\}$. Let $\Pi_{i}$ be
the plane spanned by $\\{e_{0},e_{1},e_{2},e_{3}\\}\setminus\\{e_{i}\\}$ and
let $Q_{i}=\Pi_{i}\cap L$, $i=0,1,2,3$. Writing $Q_{3}=e_{0}+ae_{1}+be_{2}$
and $Q_{2}=e_{0}+ce_{1}+de_{3}$ the condition for the line
$L=\overline{Q_{1},Q_{2}}$ to not meet any of the $6$ lines
$\overline{e_{i},e_{j}}$ is $abcd\neq 0$ and $a\neq c$. Thus, there are
$(3-1)^{3}(3-2)$ ways to pick $L$. Thus,
$|\mathcal{M}_{7e}|=3^{6}(3^{4}-1)(3^{3}-1)(3^{2}-1)(3-2)/24$, i.e. $1516320$.

#### Classes $\mathcal{M}_{7f}$ and $\mathcal{M}_{7g}$

Next, we consider the case when $S$ contains the $\mathcal{M}_{6e}$
configuration represented by
$S_{0}=\\{e_{0},e_{0}+e_{3},e_{1},e_{1}+e_{3},e_{2},e_{2}+e_{3}\\}$. Let
$P_{2i+1}=e_{i}$ and $P_{2i+2}=e_{i}+e_{3}$ for $i\in\\{0,1,2\\}$. Let $P_{7}$
be another point of $S$ and let $T=\\{P,P_{1},\dots,P_{6}\\}$. There are $3$
possibilities for $P$: (i) $P_{7}$ lies on any of the lines
$\overline{P_{2i+1},P_{2i+2}}$, $i\in\\{0,1,2\\}$ , (ii) $P_{7}$ is contained
in any of the $3$ Planes $\Pi_{0}:\langle e_{0},e_{1},e_{3}\rangle$,
$\Pi_{1}:\langle e_{0},e_{2},e_{3}\rangle$ and $\Pi_{2}:\langle
e_{1},e_{2},e_{3}\rangle$ , (iii) $P_{7}$ lies in the complement of
$\Pi_{0}\cup\Pi_{1}\cup\Pi_{2}$.

In case (i) and (ii), by maximality, $S$ contains $\mathcal{M}_{6b}$, which
cases have already been treated.

In case (iii), we have $P_{7}=ae_{0}+be_{1}+ce_{2}+de_{3}$ with $abc\neq 0$.
Here $I_{2}(T)$ has basis

$\\{X_{1}(aX_{2}-cX_{0}),X_{0}(bX_{2}-cX_{1}),bX_{3}(X_{0}+X_{1}+X_{2}-X_{3})-d(a+b+c-d)X_{0}X_{1}\\}.$

A direct calculation shows that

$\bar{T}=V(I_{2}(T))=T\cup\\{P^{\prime}\\},\quad\text{where
}P^{\prime}=ae_{0}+be_{1}+ce_{2}+(a+b+c-d)e_{3}.$

Thus, $T$ is a $\mathcal{C}_{7}$ if and only if $2d=a+b+c$. Thus, there are
$4$ choices for $P_{7}$, namely { $(1,b,c,1+b+c-d):b,c\in\\{\pm 1\\}\\}$. We
denote the class of configuration $G$-equivalent to $T$ by $\mathcal{M}_{7f}$.
In this case, we claim that the $\mathcal{M}_{6e}$ sub-configuration of
$T=\mathcal{M}_{7f}$ is unique, in other words removing a point $Q$ from $T$
results in a $\mathcal{M}_{6e}$ configuration if and only if $Q=P_{7}$: We
note that the $6$ points $P_{1},\dots,P_{6}$ represented by
$e_{i}+\epsilon_{i}e_{3}$ where $i\in\\{0,1,2\\}$ and
$\epsilon_{i}\in\\{0,1\\}$ generate $11$ planes of $PG(3,q)$. Of these $11$
planes there are $3$ planes which contain $4$ of $\\{P_{1},\dots,P_{6}\\}$,
these planes denoted $\Pi^{i}$ are generated by
$\\{e_{3},e_{0},e_{1},e_{2}\\}\setminus\\{e_{i}\\}$ for $i=0,1,2$. The
remaining $8$ planes are denoted $\Pi_{\epsilon}$ where
$\epsilon=(\epsilon_{0},\epsilon_{1},\epsilon_{2})\in\\{0,1\\}^{3}$, and are
generated by
$\\{e_{0}+\epsilon_{0}e_{3},e_{1}+\epsilon_{1}e_{3},e_{2}+\epsilon_{2}e_{3}\\}$.
We recall from the discussion about the automorphism group of the
$\mathcal{M}_{6e}$ configuration, that we may assume
$P_{7}=e_{0}+e_{1}+e_{2}+de_{3}$ for some $d\in\mathbb{F}_{3}$. Since
$2d=1+1+1=0$, we must have $P_{7}=e_{0}+e_{1}+e_{2}$. Clearly, of the $11$
planes above, $P_{7}$ is contained only in the planes $\Pi_{\epsilon}$ for
$\epsilon=(0,0,0)$ and $\epsilon=(1,1,1)$ Thus, $P_{7}$ lies on exactly two
planes of the form $\Pi_{\epsilon}$ and $\Pi_{\epsilon}$. Therefore, none of
the $6$ configurations $\\{P_{1},\dots,P_{7}\\}\setminus P_{i}$ for
$i\in\\{1,\dots,6\\}$ are $\mathcal{M}_{6e}$ configurations, Thus, proving
that $\\{P_{1},\dots,P_{6}\\}$ is the unique $\mathcal{M}_{6e}$ configuration
in $T$. Therefore, the number of $\mathcal{M}_{7f}$ configurations is
$|\mathcal{M}_{6e}|\cdot 4$, i.e. $1010880$. We also note that the
automorphism group of $\mathcal{M}_{7f}$ configuration. We recall that the
automorphism group of the $\mathcal{M}_{6e}$ configuration is $(Z_{2}\times
Z_{2}\times Z_{2})\rtimes S_{3}$, where the $S_{3}$ factor permutes the lines
$\overline{e_{i},e_{3}}$ for $i\in\\{0,1,2\\}$ and the factor $Z_{2}\times
Z_{2}\times Z_{2}$ has generators $g_{0},g_{2},g_{2}$ where $g_{i}$
interchanges the points $\\{e_{i},e_{i}+e_{3}\\}$ for $i=0,1,2$ respectively.
Since $P_{7}=e_{0}+e_{1}+e_{2}$, the automorphism group of the
$\mathcal{M}_{7f}$ configuration is $Z_{2}\times S_{3}$, where the $Z_{2}$
factor is generated by the involution $g_{0}g_{1}g_{2}\in Z_{2}\times
Z_{2}\times Z_{2}$.

We now consider the case

$\bar{T}=V(I_{2}(T))=T\cup\\{P_{8}\\},\quad\text{where
}P_{8}=ae_{0}+be_{1}+ce_{2}+(a+b+c-d)e_{3}.$

The number of such pairs $\\{P_{7},P_{8}\\}$ is $4$. We denote the class of
configuration $G$-equivalent to $\bar{T}$ by $\mathcal{M}_{7g}$. As above, we
may assume $P,P^{\prime}$ are given by $e_{0}+e_{1}+e_{2}\pm e_{3}$. Thus,
$P_{7}$ is on the planes $\Pi_{\epsilon}$ for
$\epsilon=(1,0,0),(0,1,0),(0,0,1)$ and similarly $P_{8}$ is on the planes
$\Pi_{\epsilon}$ for $\epsilon=(0,1,1),(1,0,1),(1,1,0)$. In other words,
$P_{7}$ is on the planes generated by $\\{P_{2},P_{3},P_{5}\\}$,
$\\{P_{1},P_{4},P_{5}\\}$, and $\\{P_{1},P_{3},P_{6}\\}$, and similarly
$P_{8}$ is on the planes generated by $\\{P_{1},P_{4},P_{6}\\}$,
$\\{P_{2},P_{3},P_{6}\\}$, and $\\{P_{2},P_{4},P_{5}\\}$. Thus, a list of
planes containing $4$ points of the configuration $T^{\prime}$ is as below. We
use the notation $(ijkl)$ to mean that $P_{i},P_{j},P_{k},P_{l}$ are coplanar:

$(1234),(1256),(3456),(1278),(5678),(3478)(2357),(1457),(1367),(1468),(2368),(2458).$

Of the $\tbinom{8}{2}=28$ configurations
$T_{ij}=\\{P_{1},\dots,P_{8}\\}\setminus\\{P_{i},P_{j}\\}$ obtained from
$T^{\prime}$ by deleting $2$ of the points, we see that there are $16$
configurations of type $\mathcal{M}_{6e}$:

$T_{12},T_{14},T_{16},T_{17},T_{23},T_{25},T_{28},T_{34},T_{36},T_{37},T_{45},T_{48},T_{56},T_{57},T_{68},T_{78},$

and there are $12$ configurations of type $\mathcal{M}_{6f}$:

$T_{13},T_{15},T_{18},T_{24},T_{26},T_{27},T_{35},T_{38},T_{46},T_{47},T_{58},T_{67}.$

While there are $4$ ways (choices of $\\{P_{7},P_{8}\\}$) to extend a given
$\mathcal{M}_{6e}$ configuration to a $\mathcal{M}_{7g}$ configuration, we see
that there are $16$ choices for the base $\mathcal{M}_{6e}$ configuration that
yield the same $\mathcal{M}_{7e}$ configuration. Therefore, the number of
$\mathcal{M}_{7g}$ configurations is given by $\frac{|\mathcal{M}_{6e}|\cdot
4}{16}$, i.e. $63180$.

Next, we consider the case when $S$ contains the $\mathcal{M}_{6f}$
configuration represented by

$S_{0}=\\{P_{1}=e_{0},P_{2}=e_{1},P_{3}=e_{2},P_{4}=e_{0}+e_{1}+e_{2},P_{5}=e_{3},P_{6}=e_{3}+e_{0}-e_{1}\\}.$

We use the label $(ijk)$ for the plane generated by $P_{i},P_{j},P_{k}$. If
$P_{l}$ also lies on this plane, the label $(ijkl)$ will also denote the same
plane. The $6$ points of $S_{0}$ generate $14$ planes, of which only the
planes $(1234)$ and $(1256)$ contain $4$ points of $S_{0}$. These are the only
planes containing both $P_{1}$ and $P_{2}$. Of the remaining $12$ planes,
there are $8$ of them of the form $(ijk)$ for
$i\in\\{1,2\\},j\in\\{3,4\\},k\in\\{5,6\\}$, and there are $4$ planes
$(345),(346),(356),(456)$.

We can write $PG(3,3)$ as the union of the $4$ planes containing the line
$\overline{P_{1},P_{2}}$ . These are the planes $(1234)$, $(1256)$,
$\Pi_{1},\Pi_{-1}$ , where $\Pi_{1}$ is the plane generated by
$P_{1},P_{2},e_{2}+e_{3}$ and $\Pi_{-1}$ is the plane generated by
$P_{1},P_{2},e_{2}-e_{3}$. Let $P_{7}$ be another point of $S$ and let
$T=\\{P_{7},P_{1},\dots,P_{6}\\}$. There are $3$ possibilities for $P_{7}$:
(i) some $3$ points of $T$ are collinear, (ii) some $5$ points of $T$ are
coplanar, (iii) no $3$ points of $T$ are collinear and no $5$ points of $T$
are coplanar.

In case (i) and (ii), by maximality, $S$ contains a $\mathcal{M}_{6b}$
configuration, which cases which have already been treated. In case (iii) we
represent $P_{7}=ae_{0}+be_{1}+ce_{2}+de_{3}$. The condition that no $5$
points of $T$ are coplanar, implies that $P_{7}$ is not on the planes $(1234)$
and $(1256)$, i.e. $cd\neq 0$. Therefore, we take $d=1$ and $c\in\\{\pm 1\\}$.
We recall that the automorphism group of the $\mathcal{M}_{6f}$ configuration
is isomorphic to the group $Z_{2}\times D_{4}$. The element $g_{0}$ generating
the $Z_{2}$ factor interchanges the planes $\Pi_{1}$ and $\Pi_{-1}$, so that
we may assume $c=1$. We parametrize the points of the affine plane
$\pi_{1}:=\Pi_{1}\setminus\overline{P_{1}P_{2}}$ as
$xe_{0}+ye_{1}+(e_{2}+e_{3})$.

The $12$ planes $(ijk)$ (not containing both $P_{1},P_{2}$) meet $\pi_{1}$ in
the twelve lines:

$\begin{split}&L_{345}:y=x,\quad L_{346}:y=-x,\quad L_{356}:y=x+1,\quad
L_{456}:y=-x-1.\\\ &L_{135}=L_{146}:y=0,\;L_{145}:y=1,\;L_{136}:y=-1.\\\
&L_{235}:x=0,\;L_{245}=L_{236}:x=1,\;L_{246}:x=-1.\end{split}$

The $D_{4}$ factor generated by $g_{1},g_{2},g_{3}$ preserves $\Pi_{1}$ and
the $13$ points of $\Pi_{1}$ form orbits under this group. One of these orbits
consists of the $4$ points of $\overline{P_{1},P_{2}}$. The remaining $9$
points are the points of the affine plane $\pi_{1}$. The four lines
$L_{345},L_{346},L_{356},L_{456}$ in $\pi_{1}$ form a pair of parallel lines
of slope $1$ and another pair of parallel lines of slope $-1$, which we refer
to as a $\\#$ pattern. These $4$ lines account for $8$ of the $9$ points of
$\pi_{1}$. The remaining point is $(x,y)=(0,1)$ which forms a single
$D_{4}$-orbit. The four points of intersection of the pattern $\\#$ are:
$(x,y)\in\\{(0,0),(1,1),(1,-1),(-1,0)\\}$ and they form the second orbit.
Finally, the $4$ points of the $\\#$ pattern which lie on exactly one of the
four lines form the third orbit: $(x,y)\in\\{(0,1),(-1,-1),(0,-1),(-1,1)\\}$.
We summarize the three $D_{4}$ orbits on $\pi_{1}$:

  1. (1)

the four points with coordinates $(x,y)\in\\{(0,0),(1,1),(1,-1),(-1,0)\\}$
which lie on one of the $4$ lines $\overline{P_{i}P_{j}}$ for
$i\in\\{3,4\\},j\in\\{5,6\\}$,

  2. (2)

the four points with coordinates $(x,y)\in\\{(0,1),(-1,-1),(0,-1),(-1,1)\\}$,
which lie on the intersection of three planes

    1. (a)

one of which is in the $4$ planes $(345),(346),(356),(456)$

    2. (b)

one of which is in the $4$ planes $(135),(145),(136),(146)$

    3. (c)

one of which is in the $4$ planes $(235),(245),(236),(246)$

  3. (3)

the point with coordinates $(x,y)=(1,0)$ which is on the intersection of four
planes

    1. (a)

none of which is in the $4$ planes $(345),(346),(356),(456)$

    2. (b)

two of which are in the $4$ planes $(135),(145),(136),(146)$

    3. (c)

two of which are in the $4$ planes $(235),(245),(236),(246)$.

The condition that no $3$ points of $T$ are collinear implies that $P_{7}$
lies in $\pi_{1}$ and is either in the second or the third orbit above. Thus,
we may assume $P_{7}$ to be represented by $(x,y)=(0,1)$ or $(x,y)=(1,0)$. In
the former case, $P_{7}$ lies on the planes $(356),(145),(235)$. In this case,
the configuration $T\setminus\\{P_{6}\\}=\\{P_{1},\dots,P_{5},P_{7}\\}$ is a
$\mathcal{M}_{6e}$ configuration because it consists of the three non-coplanar
lines $\overline{P_{1}P_{4}},\overline{P_{2}P_{3}},\overline{P_{5}P_{7}}$
which are concurrent through the point $e_{1}+e_{2}$. Since we have already
accounted for the $\mathcal{C}_{7}$ configurations arising from
$\mathcal{M}_{6e}$, we skip this case. Therefore, we are left with $P_{7}$
represented by $e_{0}+e_{2}+e_{3}$.

Here $I_{2}(T)$ has basis

$\\{-X_{3}(X_{2}+X_{1}+X_{0}),X_{2}(X_{0}-X_{1}-X_{3}),X_{0}(X_{2}-X_{3}-X_{1})\\}.$

Next we calculate $\bar{T}=V(I_{2}(T))$. If $Q=(x,y,z,w)\in\bar{T}$ and
$zw=0$, then we just get $Q\in S_{0}$. So we take $Q=(x,y,z,1)$ with
$z\in\\{\pm 1\\}$. A direct calculation shows that i) if $z=1$ then $Q=P_{7}$
and ii) if $z=-1$ then the coordinates $(x,y,-1,1)$ of $Q$ satisfy the
conditions $(x,y)=(0,-1)$, i.e., $Q=P_{8}=-e_{1}-e_{2}+e_{3}=g_{0}P_{7}$. The
planes containing $4$ points of the configuration $\\{P_{1},\dots,P_{8}\\}$
are

$(1234),(1256),(1357),(1467),(2457),(2367),(1458),(1368),(2358),(2468),(3478),(5678).$

Of the $\tbinom{8}{2}=28$ configurations
$T_{ij}=\\{P_{1},\dots,P_{8}\\}\setminus\\{P_{i},P_{j}\\}$ we have $16$
$\mathcal{M}_{6e}$ configurations:

$T_{13},T_{14},T_{15},T_{16},T_{23},T_{24},T_{25},T_{26},T_{37},T_{38},T_{47},T_{48},T_{57},T_{58},T_{67},T_{68},$

and $12$ $\mathcal{M}_{6f}$ configurations:

$T_{12},T_{17},T_{18},T_{27},T_{28},T_{34},T_{35},T_{36},T_{45},T_{46},T_{56},T_{78}.$

Thus, $T^{\prime}$ extends a $\mathcal{M}_{6e}$ configuration, which has
already been treated above (namely $\mathcal{M}_{7g}$ configuration). ∎

###  6.1. $B_{j,7}$ for $j>7$

####

An $\mathcal{M}_{7a}$ configuration $T$ represented by a plane $\Pi$ and a
point $P$ outside $\Pi$, contains $14$ points of $PG(3,3)$. Since $j>7$, for
$S\in\mathfrak{C}_{i}(T)$, $i\leq 6$, the only possible cases are
$S=\mathcal{M}_{6a},\mathcal{M}_{6b}$, and $|\mathcal{M}_{6a}(T)|=1$,
$|\mathcal{M}_{6b}(T)|=\tbinom{13}{2}=78$. By (13), we have,

$B_{j,7}(\mathcal{M}_{7a})=\tbinom{14}{j}-\tbinom{13}{j}-78\tbinom{8}{j}=\tbinom{13}{j-1}-78\tbinom{8}{j}.$

####

An $\mathcal{M}_{7b}$ configuration $T$ represented by three concurrent lines
not contained in a plane contains $10$ points of $PG(3,3)$. Since $j>7$, for
$S\in\mathfrak{C}_{i}(T)$, $i\leq 6$, the only possible case is
$S=\mathcal{M}_{6b}$, and $|\mathcal{M}_{6b}(T)|=\tbinom{3}{2}\cdot 3=9$ (as
we have $\tbinom{3}{2}$ choices for two lines and $3$ choices for a point from
the remaining line). By (13), we have,

$B_{j,7}(\mathcal{M}_{7b})=\tbinom{10}{j}-9\tbinom{8}{j}.$

####

An $\mathcal{M}_{7c}$ configuration $T$ is represented by three lines
$L_{1},L_{2},L_{3}$ of $PG(3,3)$, where $L_{2}$ and $L_{3}$ are skew to each
other and the line $L_{1}$ meets both the lines. It contains $10$ points of
$PG(3,3)$. Since $j>7$, the only possible cases for $S\in\mathfrak{C}_{i}(T)$,
$i\leq 6$ are $S=\mathcal{M}_{6b},\mathcal{M}_{6c}$. As there are $2$ pairs of
intersecting lines ($\\{L_{1},L_{2}\\}$ and $\\{L_{2},L_{3}\\}$) in $T$, and a
point can be chosen from the remaining line in $3$ ways, we have
$|\mathcal{M}_{6b}(T)|=6$. It is clear that $|\mathcal{M}_{6c}(T)|=1$, as
there is only one configuration of a pair of skew lines and which is
$\\{L_{1},L_{3}\\}$. By (13), we have,

$B_{j,7}(\mathcal{M}_{7c})=\tbinom{10}{j}-6\cdot\tbinom{8}{j}-\tbinom{8}{j}=\tbinom{10}{j}-7\cdot\tbinom{8}{j}.$

####

A $\mathcal{M}_{7d}$ configuration $T$ represented by
$(\\{L_{1},L_{2}\\},\\{P_{1},P_{2}\\})$, where $\\{L_{1},L_{2}\\}$ is a pair
of intersecting lines and the line $\overline{P_{1},P_{2}}$ not contained in
the plane does not meet $L_{1}$ and $L_{2}$, contains $9$ points of $PG(3,3)$.
Since $j>7$, the only possible case for $S\in\mathfrak{C}_{i}(T)$, $i\leq 6$
is $S=\mathcal{M}_{6b}$. Clearly, $|\mathcal{M}_{6b}(T)|=2$, as there is one
pair of intersecting lines and $2$ choices for a point from
$\\{P_{1},P_{2}\\}$. By (13), we have,

$B_{j,7}(\mathcal{M}_{7d})=\tbinom{9}{j}-2\cdot\tbinom{8}{j}.$

####

Let $T$ be a $\mathcal{M}_{7e}$ configuration represented by
$(\\{P_{1},P_{2},P_{3},P_{4}\\},L)$, where the points
$P_{1},P_{2},P_{3},P_{4}$ are in general position in $PG(3,3)$ and the line
$L$ does not meet any of the $6$ lines $\overline{P_{i},P_{j}}$, where $1\leq
i<j\leq 4$. The configuration $T$ has $8$ points of $PG(3,3)$. In this case,
$\mathfrak{C}_{i}(T)=\varnothing$ for $i\leq 6$. By (13), we have

$B_{8,7}(\mathcal{M}_{7e})=1.$

####

Let $T$ be a $\mathcal{M}_{7g}$ configuration represented by
$\\{P_{1},\dots,P_{6},P_{7},P_{8}\\}$ where the $4$ lines
$L_{i}=\overline{P_{2i-1},P_{2i}}$, $i=1,2,3,4$, meet at a single point $Q$
different from $\\{P_{1},\dots,P_{6},P_{7},P_{8}\\}$ and no three of them are
contained in a plane. The configuration $T$ has $8$ points of $PG(3,3)$. Since
$j>7$, we have no such $S$ with $S\in\mathfrak{C}_{i}(T)$, $i\leq 6$. By (13),
we have,

$B_{8,7}(\mathcal{M}_{7g})=1.$

Since a $\mathcal{M}_{7f}$ configuration contains exactly $7$ points $PG(3,3)$
and $j>7$, we get $B_{j,7}(\mathcal{M}_{7f})=0$. Therefore, by (12), for
$j>7$, we have

$\begin{split}B_{8,7}&=1080\cdot(\tbinom{13}{7}-78\tbinom{8}{8})+9360\cdot(\tbinom{10}{8}-9\tbinom{8}{8})\\\
&+84240\cdot(\tbinom{10}{8}-7\cdot\tbinom{8}{8})+505440\cdot\tbinom{9}{8}-2\cdot\tbinom{8}{8}+1516320+63180,\end{split}$

and, if $j\geq 9$,

$B_{j,7}=1080\cdot(\tbinom{13}{j-1})+9360\cdot\tbinom{10}{j}+84240\cdot\tbinom{10}{j}+505440\cdot\tbinom{9}{j},$

i.e.

(18) $B_{8,7}=9413820,\text{ and }j\geq
9,B_{j,7}=1080\cdot\tbinom{13}{j-1}+93600\cdot\tbinom{10}{j}+505440\cdot\tbinom{9}{j}.$

##  7\.  Configurations of rank $8$

As before, we build the $\mathcal{C}_{8}$’s by adding points to
$\mathcal{C}_{7}$’s. We carry this out systematically by running through the
classes in $\mathfrak{C}_{7}$. If $T$ is a $\mathcal{C}_{8}$, then $I_{2}(T)$
is a $2$ dimensional vectors space, and its projectivization is a pencil of
quadratic forms on $PG(3,3)$. If $S$ denote the set of $(q+1)=4$ elements of
this pencil, let $\nu_{i}(T)=|S\cap\mathcal{O}_{i}|$, here
$\mathcal{O}_{1},\dots,\mathcal{O}_{6}$ are the $G$-orbits of quadratic forms
on $PG(3,3)$, listed in Table 1. We will write down the numbers
$\nu(T)=(\nu_{1}(T),\dots,\nu_{6}(T))$.

######  Lemma 7.1.

The set $\mathfrak{C}_{8}$ can be partitioned into the following $4$ classes:

  1. $\mathcal{M}_{8a}$

of size $2^{3}\cdot 3^{2}\cdot 5\cdot 13$. This class consists of a plane of
$PG(3,3)$ and a line not contained in that plane.

  2. $\mathcal{M}_{8b}$

of size $2^{4}\cdot 3^{2}\cdot 5\cdot 13$. This class consists of $4$
concurrent lines such that no three are contained in a plane.

  3. $\mathcal{M}_{8c}$

of size $2^{2}\cdot 3^{6}\cdot 5\cdot 13$. This class consists of $4$ lines
$L_{1},L_{2},L_{3},L_{4}$ such that $L_{i}$ and $L_{j}$ intersect if and only
if $i\neq j\mod 2$ (and $L_{i}$ and $L_{j}$ are skew if and only if $i=j\mod
2$).

  4. $\mathcal{M}_{8d}$

of size $2^{6}\cdot 3^{4}\cdot 5\cdot 13$. This class consists of
$(\\{L_{1},L_{2}\\},\\{P_{1},P_{2},P_{3}\\})$ where $\\{L_{1},L_{2}\\}$ is a
pair of lines in a plane $\pi_{1}$, and the points $P_{1},P_{2},P_{3}$ are in
a different plane $\pi_{2}$. The point $Q=L_{1}\cap L_{2}$ lies on the line
$L=\pi_{1}\cap\pi_{2}$, but none of the points $P_{1},P_{2},P_{3}$ lie on $L$.
The four points $\\{Q,P_{1},P_{2},P_{3}\\}$ are in general position in
$\pi_{2}$.

  5. $\mathcal{M}_{8e}$

of size $2^{3}\cdot 3^{6}\cdot 5\cdot 13$. This class consists of
configurations $\\{P_{1},\dots,P_{6},P_{7},P_{8}\\}$ where the $3$ lines
$L_{i}=\overline{P_{2i-1},P_{2i}}$, $i=1,2,3$, are non-coplanar lines meeting
at a single point $Q$ different from $\\{P_{1},\dots,P_{6}\\}$. There are four
lines $L$ passing through $Q$ with the property that no three of the four
lines $\\{L_{1},L_{2},L_{3},L\\}$ are coplanar. Let $L_{4},L_{5}$ be two such
lines. On each of these lines $\\{L_{4},L_{5}\\}$ there is a unique point
$\\{Q_{1}(L_{4}),Q_{1}(L_{5})\\}$ as given in the description of the
$\mathcal{M}_{7f}$ configuration (see Lemma 6.1). We take
$\\{P_{7},P_{8}\\}=\\{Q_{1}(L_{4}),Q_{1}(L_{5})\\}$.

###### Proof.

Let $S$ be a $\mathcal{C}_{8}$ configuration.

#### Class $\mathcal{M}_{8a}$

Let $S$ contain a $\mathcal{M}_{7a}$ configuration represented by the point
set $T=\\{e_{3}\\}\cup\Pi$, where the plane $\Pi$ given by $X_{3}=0$. If $S$
contains a point $P\notin T$, then $P$ will be on a line $\overline{e_{3},Q}$,
where $Q$ is a point of $\Pi$. Then by maximality, $\overline{e_{3},P}$ is
contained in $S$. We can assume $P=e_{0}+e_{1}+e_{3}$ and let
$T^{\prime}=\overline{e_{3},e_{0}+e_{1}+e_{3}}\cup\Pi$. Here
$I_{2}(T^{\prime})$ has basis $\\{X_{3}(X_{0}-X_{1}),X_{3}X_{2}\\}$ and
$V(I_{2}(T^{\prime}))$ clearly equals $T^{\prime}$. Thus, $T^{\prime}$ is a
$\mathcal{C}_{8}$ which we denote $\mathcal{M}_{8a}$. Clearly, all the
$(q+1)=4$ elements of the pencil $\mathbb{P}(I_{2}(T^{\prime}))$ are in
$\mathcal{O}_{1}$ (pairs of planes). Thus, $\nu(T^{\prime})=(4,0,0,0,0,0)$.
The number of ways to choose the plane $\Pi$ is $40$ and a line not contained
in $\Pi$ is $13\cdot 3^{2}$. Therefore, the size of $\mathcal{M}_{8a}$ is
$40\cdot 13\cdot 3^{2}=4680$.

#### Class $\mathcal{M}_{8b}$

Let $S$ contain a $\mathcal{M}_{7b}$ configuration represented by
$S_{0}=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\overline{e_{0},e_{3}}$.
Let $P$ be another point of $S$. There are two possibilities for $P$: (i) $P$
lies on one of the planes $\Pi_{1}:X_{1}=0$, $\Pi_{2}:X_{2}=0$,
$\Pi_{3}:X_{3}=0$, and (ii) $P$ lies outside $\Pi_{1},\Pi_{2}$ and $\Pi_{3}$.

In case (i) by maximality, $S$ contains a $\mathcal{M}_{7a}$ configuration
which case has been treated above.

In case (ii), if $P=ae_{0}+be_{1}+ce_{2}+de_{3}$, then the condition that $P$
does not lie on the planes $\Pi_{1},\Pi_{2}$ and $\Pi_{3}$ is that $bcd\neq
0$. So we can assume $d=1$ and $P=ae_{0}+be_{1}+ce_{2}+e_{3}$. Let
$T^{\prime}=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\overline{e_{0},e_{3}}\cup\\{P\\}$
. Here $I_{2}(T^{\prime})$ has basis
$\\{X_{3}(cX_{1}-bX_{2}),X_{2}(X_{1}-bX_{3})\\}$ and $V(I_{2}(T^{\prime}))$
equals
$T=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\overline{e_{0},e_{3}}\cup\overline{e_{0},P}$.
Thus, $T$ is a $\mathcal{C}_{8}$ which we denote $\mathcal{M}_{8b}$. In the
pencil $\mathbb{P}(I_{2}(T))$, apart from the mentioned generators, the
remaining $(q-1)=2$ elements are in $\mathcal{O}_{4}$. Thus,
$\nu(T)=(2,0,0,2,0,0)$.

The number of ways a $\mathcal{M}_{7b}$ configuration $S_{0}$ can be chosen is
$|\mathcal{M}_{7b}|$. Since no three lines of a $\mathcal{M}_{8b}$
configuration are contained in a plane, there are $4$ choices for the line
$\overline{e_{0},P}$. The configuration $T\setminus\overline{e_{0},e_{i}}$,
$i=1,2,3$ is $\mathcal{M}_{7b}$. Thus, there are $4$ different
$\mathcal{M}_{7b}$ configurations contained in a $\mathcal{M}_{8b}$
configuration. Therefore, the size of $\mathcal{M}_{8b}$ is
$\frac{|\mathcal{M}_{7b}|\cdot 4}{4}=|\mathcal{M}_{7b}|=9360$.

#### Class $\mathcal{M}_{8c}$

Let $S$ contain a $\mathcal{M}_{7c}$ configuration represented by
$S_{0}=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\overline{e_{1},e_{3}}$.
Let $P$ be another point of $S$. There are two possibilities for $P$: (i) $P$
lies on one of the planes $\Pi_{1}:X_{3}=0$, $\Pi_{2}:X_{2}=0$, and (ii) $P$
lies outside $\Pi_{1},\Pi_{2}$. In case (i) by maximality, $S$ contains a
$\mathcal{M}_{7a}$ configuration which case has been treated above. In case
(ii), if $P=ae_{0}+be_{1}+ce_{2}+de_{3}$, then the condition that $P$ does not
lie on the planes $\Pi_{1},\Pi_{2}$ is that $cd\neq 0$. Switching to the basis
$f_{0}=e_{0},f_{1}=e_{1},f_{2}=ce_{2}+ae_{0},f_{3}=de_{3}+be_{1}$ of
$PG(3,3)$, the configuration $S_{0}$ is again given by
$S_{0}=\overline{f_{0},f_{1}}\cup\overline{f_{0},f_{2}}\cup\overline{f_{1},f_{3}}$,
and $P=f_{2}+f_{3}$. Thus, we may take $T^{\prime}=S_{0}\cup\\{P\\}$ to be
$T^{\prime}=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\overline{e_{1},e_{3}}\cup\\{e_{2}+e_{3}\\}$
. Here $I_{2}(T^{\prime})$ has basis $\\{X_{3}X_{0},X_{2}X_{1}\\}$ and
$V(I_{2}(T^{\prime}))$ equals
$T=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\overline{e_{1},e_{3}}\cup\overline{e_{2},e_{3}}$.
Thus, $T$ is a $\mathcal{C}_{8}$ which we denote $\mathcal{M}_{8c}$. In the
pencil $\mathbb{P}(I_{2}(T))$, apart from the mentioned generators, the
remaining $(q-1)=2$ elements are in $\mathcal{O}_{2}$. Thus,
$\nu(T)=(2,2,0,0,0,0)$. The size of $\mathcal{M}_{8c}$ is
$3\cdot|\mathcal{M}_{4b}|=189540$, because there are $3$ ways to obtain the
configuration $T$ from four points in general position of $PG(3,3)$.

#### Class $\mathcal{M}_{8d}$

Let $S$ contain a $\mathcal{M}_{7d}$ configuration represented by
$S_{0}=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\\{e_{3},e_{3}+e_{1}+e_{2}\\}$.
If $P$ is another point of $S$, the there are $4$ possibilities for $P$: (i)
$P$ lies in the plane $\Pi_{0}:\langle e_{0},e_{1},e_{2}\rangle$, (ii) $P$
lies in at least one of the planes $\Pi_{1}:\langle e_{3},e_{0},e_{1}\rangle$,
$\Pi_{2}:\langle e_{3},e_{0},e_{2}\rangle$, $\Pi_{3}:\langle
e_{1}+e_{2}+e_{3},e_{0},e_{1}\rangle$, $\Pi_{4}:\langle
e_{1}+e_{2}+e_{3},e_{0},e_{2}\rangle$, (iii) $P$ lies on the line
$L:\overline{e_{3},e_{1}+e_{2}+e_{3}}$, (iv) $P$ lies in the complement of
$\Pi_{0}\cup\Pi_{1}\cup\Pi_{2}\cup\Pi_{3}\cup\Pi_{4}\cup L$ in $PG(3,3)$.

In the cases (i) and (iii) by maximality, $S$ contains a $\mathcal{M}_{7a}$
configuration, which case has been treated above.

In the case (ii) by maximality, $S$ contains a $\mathcal{M}_{7b}$ or
$\mathcal{M}_{7c}$ configuration, which case has been treated above.

In case (iv), let $P=ae_{0}+be_{1}+ce_{2}+de_{3}$. The condition for
$P\notin\Pi_{0}$ is that $d\neq 0$. So we can assume $d=1$. The conditions for
$P\notin\Pi_{1},\Pi_{2},\Pi_{3},\Pi_{4}$ are $c\neq 0,1$ and $b\neq 0,1$.
Therefore $b=c=-1$. The condition for $P$ to lie on the line $L$ is $a=b-c=0$
. Thus, we may take $P$ to be $ae_{0}-e_{1}-e_{2}+e_{3}$ with $a\in\\{\pm
1\\}$. Let
$T^{\prime}=\overline{e_{0},e_{1}}\cup\overline{e_{0},e_{2}}\cup\\{e_{3},e_{3}+e_{1}+e_{2},P\\}$.
Here $I_{2}(T^{\prime})$ has basis

$\\{aX_{1}(X_{3}-X_{2})-X_{0}X_{3},X_{3}X_{1}+X_{2}X_{1}+X_{3}X_{2}\\},$

and $V(I_{2}(T^{\prime}))$ equals $T^{\prime}$. Thus, $T^{\prime}$ is a
$\mathcal{C}_{8}$ which we denote $\mathcal{M}_{8d}$. Of the $(q+1)=4$ points
of the pencil $\mathbb{P}(I_{2}(T^{\prime}))$, except for the element
$X_{3}X_{1}+X_{2}X_{1}+X_{3}X_{2}$ which is in $\mathcal{O}_{4}$, the
remaining $q=3$ elements are in $\mathcal{O}_{2}$ . Thus,
$\nu(T^{\prime})=(0,3,0,1,0,0)$.

We will now count the size of $\mathcal{M}_{8d}$. This configuration consists
of a pair of intersecting lines $L_{1},L_{2}$ in a plane $\Pi$, and three non-
collinear points $P_{1},P_{2},P_{3}$ lying outside $\Pi$, such that the $3$
lines formed by $P_{1},P_{2},P_{3}$ do not meet $L_{1},L_{2}$. Since $q=3$,
with this condition, the plane $\langle P_{1},P_{2},P_{3}\rangle$ meets $\Pi$
in a line $M$ through $L_{1}\cap L_{2}$. A pair of intersecting lines
$L_{1},L_{2}$ in $PG(3,3)$ can be chosen in $\mathcal{M}_{5a}$ ways, the line
$M$ can be chosen in $\tbinom{2}{1}$ ways in the plane $\langle
L_{1},L_{2}\rangle$, a plane $\Pi^{\prime}$ through $M$ other than $\Pi$ can
be chosen in $\tbinom{3}{2}$ ways and on $\Pi^{\prime}$ a set of $3$
noncollinear points with the property no two are lying on a line through
$L_{1}\cap L_{2}$, can be chosen in $(3^{3}-3^{2})$ ways. Hence, the size of
$\mathcal{M}_{8d}$ is
$|\mathcal{M}_{5a}|\cdot\tbinom{2}{1}\cdot\tbinom{3}{2}\cdot(3^{3}-3^{2})$,
i.e. $336960$.

Let $S$ contain a $\mathcal{M}_{7e}$ configuration represented by
$S_{0}=\\{e_{0},e_{1},e_{2},e_{3}\\}\cup\overline{e_{0}+e_{1}+e_{2},e_{0}-e_{1}+e_{3}}$.
Let us denote $P_{1}=e_{0}+e_{1}+e_{2}$ and $P_{2}=e_{0}-e_{1}+e_{3}$. All the
points of $PG(3,q)$ are on the $4$ planes $\Pi_{0}:\langle
P_{1},P_{2},e_{0}\rangle$, $\Pi_{1}:\langle P_{1},P_{2},e_{1}\rangle$,
$\Pi_{2}:\langle P_{1},P_{2},e_{2}\rangle$, $\Pi_{3}:\langle
P_{1},P_{2},e_{3}\rangle$, through $\overline{P_{1},P_{2}}$. Let $P$ be
another point of $S$. Then $P$ will lie on one of these $4$ planes, say
$\Pi_{0}$. In this case, the plane $\Pi_{0}$ contains $5$ points of $S$, and
by maximality, $S$ contains a $\mathcal{M}_{7d}$ configuration, which case has
been treated above.

#### Class $\mathcal{M}_{8e}$

Let $S$ contain a $\mathcal{M}_{7f}$ configuration represented by
$S_{0}=\\{e_{0},e_{0}+e_{3},e_{1},e_{1}+e_{3},e_{2},e_{2}+e_{3},e_{0}+e_{1}+e_{2}\\}$.
Here, $I_{2}(S_{0})$ has a basis

$\\{X_{1}(X_{2}-X_{0}),X_{0}(X_{2}-X_{1}),X_{3}(X_{0}+X_{1}+X_{2}-X_{3})\\}.$

Let $P^{\prime}=ae_{0}+be_{1}+ce_{2}+de_{3}$ be another point of $S$. Then
there are three possibilities for $P^{\prime}$: (i) $P^{\prime}$ lies on one
of the planes $\Pi_{0}:\langle e_{0},e_{1},e_{3}\rangle$, $\Pi_{1}:\langle
e_{0},e_{2},e_{3}\rangle$ and $\Pi_{2}:\langle e_{1},e_{2},e_{3}\rangle$,
$\Pi_{3}:\langle e_{0},e_{1},e_{2}\rangle$, $\Pi_{4}:\langle
e_{0}+e_{3},e_{1}+e_{3},e_{2}+e_{3}\rangle$ containing $4$ points of $S_{0}$,
(ii) $P^{\prime}$ lies on one of the 21 lines containing two points of
$S_{0}$, (iii) $P^{\prime}$ lies on the line
$\overline{e_{3},e_{0}+e_{1}+e_{2}}$, (iv) $P^{\prime}$ lies in the complement
of these planes and lines in $PG(3,3)$.

In cases (i) and (ii), by maximality, $S$ contains a $\mathcal{M}_{6b}$
configuration. From a $\mathcal{M}_{6b}$ configuration, we have seen earlier
that a $\mathcal{M}_{7b}$, $\mathcal{M}_{7c}$ or $\mathcal{M}_{7d}$
configuration can be obtained, and these cases have been treated above.

In case (iii), if $P^{\prime}$ lies on the line
$\overline{e_{3},e_{0}+e_{1}+e_{2}}$, then $P^{\prime}$ is either
$e_{0}+e_{1}+e_{2}+e_{3}$ or $-e_{0}-e_{1}-e_{2}-e_{3}$. Thus, the $8$ points
of $S_{0}\cup\\{P^{\prime}\\}$ lies on $4$ concurrent lines
$\overline{e_{3}e_{0}}$, $\overline{e_{3}e_{1}}$, $\overline{e_{3}e_{2}}$,
$\overline{e_{3}e_{0}+e_{1}+e_{2}}$. But these $8$ points of
$S_{0}\cup\\{P^{\prime}\\}$ do not satisfy the condition to be a
$\mathcal{M}_{7g}$ configuration. So, the configuration consisting of points
of $S_{0}\cup\\{P^{\prime}\\}$ is of rank $8$ and contained in a
$\mathcal{M}_{8b}$ configuration represented by the $4$ concurrent lines
$\overline{e_{3}e_{0}}$, $\overline{e_{3}e_{1}}$, $\overline{e_{3}e_{2}}$,
$\overline{e_{3}e_{0}+e_{1}+e_{2}}$. By Lemma 3.3, $S_{0}\cup\\{P^{\prime}\\}$
is not a maximal configuration and $S$ is a $\mathcal{M}_{8b}$ configuration.

Now we will consider case (iv). The number of points contained in the union
$\Pi_{0}\cup\Pi_{1}\cup\Pi_{2}\cup\Pi_{3}\cup\Pi_{4}$ is $35$. All the $21$
lines formed by the $2$ points of $S_{0}$ are contained in these $5$ planes.
Also, there are two points on the line $\overline{e_{3},e_{0}+e_{1}+e_{2}}$
which are not in the union of these $5$ planes and $21$ lines. Thus, there are
total $37$ points of $PG(3,3)$ covered by union of the planes
$\Pi_{0},\Pi_{1},\Pi_{2},\Pi_{3},\Pi_{4}$, and
$\overline{e_{3},e_{0}+e_{1}+e_{2}}$. The remaining $3$ points
$\\{e_{0}-e_{1}-e_{2}+e_{3},-e_{0}+e_{1}-e_{2}+e_{3},-e_{0}-e_{1}+e_{2}+e_{3}\\}$
which form a single orbit under the automorphism group $Z_{2}\times S_{3}$ of
the $\mathcal{M}_{7f}$ configuration. So we may take
$P^{\prime}=e_{0}-e_{1}-e_{2}+e_{3}$. Let $T=S_{0}\cup\\{P^{\prime}\\}$. A
basis of the $I_{2}(T)$ is given by

$\\{X_{1}(X_{2}-X_{0})+X_{3}(X_{0}+X_{1}+X_{2}-X_{3}),X_{0}(X_{2}-X_{1})\\},$

and it is readily checked that $V(I_{2}(T))$ equals $T$. Thus, $T$ is a
maximal $\mathcal{C}_{8}$, which we denote as $\mathcal{M}_{8e}$. The
configuration $T_{i}=T\setminus\\{P_{i}\\}$ for $i=1,\dots,8$ are
$\mathcal{C}_{7}$ configurations with $7$ points and containing no lines. In
the list $\mathcal{M}_{7a}-\mathcal{M}_{7g}$, the only configuration with $7$
points and containing no lines is $\mathcal{M}_{7f}$. Therefore, the size of
the class of $\mathcal{M}_{8e}$ configurations is:

$|\mathcal{M}_{8e}|=\tfrac{1}{8}|\mathcal{M}_{7f}|\cdot 3=|G|/32=379080.$

In the pencil $\mathbb{P}(I_{2}(T))$, there is one element each in
$\mathcal{O}_{1}$ and $\mathcal{O}_{2}$ and the remaining $2$ elements are in
$\mathcal{O}_{6}$. Thus, $\nu(T)=(1,1,0,0,0,2)$.

Let $S$ contain a $\mathcal{M}_{7g}$ configuration represented by
$S_{0}=\\{e_{0},e_{0}+e_{3},e_{1},e_{1}+e_{3},e_{2},e_{2}+e_{3},e_{0}+e_{1}-e_{2},e_{0}+e_{1}-e_{2}+e_{3}\\}$.
Let us consider the six planes $\Pi_{0}:\langle e_{0},e_{1},e_{3}\rangle$,
$\Pi_{1}:\langle e_{2},e_{0}+e_{1}-e_{2},e_{3}\rangle$, $\Pi_{2}:\langle
e_{0},e_{2},e_{3}\rangle$, $\Pi_{3}:\langle
e_{0},e_{0}+e_{3},e_{0}+e_{1}-e_{2}\rangle$, $\Pi_{4}:\langle
e_{1},e_{2},e_{3}\rangle$, and $\Pi_{5}:\langle
e_{1},e_{1}+e_{3},e_{0}+e_{1}-e_{2}\rangle$. Each of the plane
$\Pi_{i},i\in\\{0,\dots,5\\}$, contain $4$ points of $S_{0}$. Now,

$\displaystyle|\cup_{i=0}^{5}\Pi_{i}|$
$\displaystyle=|\Pi_{0}|+|\Pi_{1}\setminus\Pi_{0}|+|\Pi_{2}\setminus(\Pi_{0}\cup\Pi_{1})|+|\Pi_{3}\setminus(\cup_{i=0}^{2}\Pi_{i})|+|\Pi_{4}\setminus(\cup_{i=0}^{3}\Pi_{i})|+|\Pi_{5}\setminus(\cup_{i=0}^{4}\Pi_{i})|$
$\displaystyle=13+9+6+6+3+3=40.$

This shows that the union $\cup_{i=0}^{5}\Pi_{i}$ contains all the points of
$PG(3,3)$. If $S$ contains another point $P$, then $P$ will be on at least one
of the planes $\Pi_{i}$ and since $\Pi_{i}$ contains $5$ points of $S$, by
maximality, it will contain a $\mathcal{M}_{5b}$ configuration. From a
$\mathcal{M}_{5b}$ configuration we only get a$\mathcal{M}_{6b}$ configuration
and from a $\mathcal{M}_{6b}$ configuration we can get a $\mathcal{M}_{7b}$, a
$\mathcal{M}_{7c}$ or a $\mathcal{M}_{7d}$ configuration. So, this case has
been treated above. ∎

###  7.1. $B_{j,8}$ for $j>8$

####

Let $T$ be a $\mathcal{M}_{8a}$ configuration of $16$ points represented by a
plane $\Pi$ of $PG(3,3)$ and a line $L$ not contained in $\Pi$. Let
$L\cap\Pi=\\{Q\\}$. Since $j>8$, the possible cases for
$S\in\mathfrak{C}_{i}(T)$, $i\leq 7$ are as follows:

  1. (i)

$S=\mathcal{M}_{7a}$ consisting of a plane $\pi$ and a point $P$ not on $\pi$.
In this case $|\mathcal{M}_{7a}(T)|=3$, as we must take $\pi=\Pi$ and there
are $3$ ways to pick $P$ from $L\setminus Q$.

  2. (ii)

$S=\mathcal{M}_{7b}$ consisting of $3$ non-coplanar lines
$\\{\ell_{1},\ell_{2},\ell_{3}\\}$ through a point $P$. In this case,
$|\mathcal{M}_{7b}(T)|=6$, as we must take $P=Q$, $\ell_{1}=L$, and there are
$\tbinom{4}{2}=6$ choices for $\\{\ell_{2},\ell_{3}\\}$ in $\Pi$.

  3. (iii)

$S=\mathcal{M}_{7c}$ consisting of three lines
$\\{\ell_{1},\ell_{2},\ell_{3}\\}$ where $\ell_{2}$ and $\ell_{3}$ are skew to
each other and the line $\ell_{1}$ meets both the lines. We must take
$\ell_{3}=L$, $\ell_{2}$ to be a line in $\Pi$ not passing through $Q$, and
$\ell_{1}$ a line in $\Pi$ passing through $P$. Thus, there are $1$, $q^{2}=9$
and $(q+1)=4$ choices for $\ell_{1},\ell_{2}$ and $\ell_{3}$ respectively.
Therefore, $|\mathcal{M}_{7c}(T)|=36$.

  4. (iv)

$S=\mathcal{M}_{7d}$ which consists of a pair of lines $\ell,\ell^{\prime}$ in
a plane $\pi$ , and a pair of points $P,P^{\prime}$ not lying on $\pi$ such
that the line $\overline{P,P^{\prime}}$ meets $\pi$ in a point not lying on
$\ell\cup\ell^{\prime}$. In this case $|\mathcal{M}_{7d}(T)|=108$, as there
are $\tbinom{9}{2}=36$ choices for the pair $\ell,\ell^{\prime}$ of lines not
passing through $Q$ and $\tbinom{3}{2}=3$ choices for the $2$ points
$P,P^{\prime}$ from $L\setminus Q$.

  5. (v)

$S=\mathcal{M}_{6a}$ consisting of a plane $\pi$. In this case we must take
$\pi=\Pi$ and hence $|\mathcal{M}_{6a}(T)|=1$.

By (13), we have,

$B_{j,8}(\mathcal{M}_{8a})=\tbinom{16}{j}-3\cdot(\tbinom{14}{j}-\tbinom{13}{j})-6\cdot\tbinom{10}{j}-36\cdot\tbinom{10}{j}-108\cdot\tbinom{9}{j}-\tbinom{13}{j},$

i.e.

$B_{j,8}(\mathcal{M}_{8a})=\tbinom{16}{j}-3\cdot\tbinom{13}{j-1}-42\cdot\tbinom{10}{j}-108\cdot\tbinom{9}{j}-\tbinom{13}{j}.$

####

A $\mathcal{M}_{8b}$ configuration $T$ of $13$ points represented by $4$
concurrent lines $L_{1},\dots,L_{4}$ through a point $Q$, such that no three
of these lines are coplanar. Since $j>8$, the possible cases for
$S\in\mathfrak{C}_{i}(T)$, $i\leq 7$ are listed as follows.

  1. (i)

$S=\mathcal{M}_{7b}$ consisting of three non-planar concurrent lines. In this
case $|\mathcal{M}_{7b}(T)|=4$, as there $\tbinom{4}{3}$ ways to choose $3$
concurrent lines in $T$.

  2. (ii)

$S=\mathcal{M}_{7d}$ which consists of a pair of lines $L,L^{\prime}$ in a
plane $\pi$, and a pair of points $P,P^{\prime}$ not lying on $\pi$ such that
the line $\overline{P,P^{\prime}}$ meets $\pi$ in a point not lying on $L\cup
L^{\prime}$. In this case $|\mathcal{M}_{7d}(T)|=\tbinom{4}{2}\cdot 3\cdot
3=54$, as there are $\tbinom{4}{2}$ ways to choose to pick $L,L^{\prime}$ from
$\\{L_{1},\dots,L_{4}\\}$, and $9$ ways to pick a point each (different from
$Q$) from the remaining two lines of
$\\{L_{1},\dots,L_{4}\\}\setminus\\{L,L^{\prime}\\}$.

By (13), we have,

$B_{j,8}(\mathcal{M}_{8b})=\tbinom{13}{j}-4\cdot\tbinom{10}{j}-54\cdot\tbinom{9}{j}.$

####

Let $T$ be a $\mathcal{M}_{8c}$ configuration of $13$ points represented by
$4$ lines $L_{1},L_{2},L_{3},L_{4}$ such that $L_{i}$ meets $L_{j}$ if and
only if $i\neq j\mod 2$ . Since $j>8$, the possible cases for
$S\in\mathfrak{C}_{i}(T)$, $i\leq 7$ are as follows.

  1. (i)

$S=\mathcal{M}_{7c}$ consisting of three lines
$\\{\ell_{1},\ell_{2},\ell_{3}\\}$ where $\ell_{2}$ and $\ell_{3}$ are skew to
each other and the line $\ell_{1}$ meets both the lines. In this case
$|\mathcal{M}_{7c}(T)|=4$, as we must take $S$ to be the union of any $3$ of
the $4$ lines $\\{L_{1},\dots,L_{4}\\}$ .

  2. (ii)

$S=\mathcal{M}_{7d}$ which consists of a pair of lines $L,L^{\prime}$ in a
plane $\pi$, and a pair of points $P,P^{\prime}$ not lying on $\pi$ such that
the line $\overline{P,P^{\prime}}$ meets $\pi$ in a point not lying on $L\cup
L^{\prime}$. In this case, we must take $L,L^{\prime}$ to a pair $L_{i},L_{j}$
with $i\neq j\mod 4$ (of which there are $4$ choices). If
$\\{L,L^{\prime}\\}=\\{L_{1},L_{2}\\}$ then there are $2$ choices of points on
each of the lines $L_{3},L_{4}$ for the pair $P,P^{\prime}$. Thus,
$|\mathcal{M}_{7d}(T)|=4\cdot 2\cdot 2=16$

By (13), we have,

$B_{j,8}(\mathcal{M}_{8c})=\tbinom{12}{j}-4\cdot\tbinom{10}{j}-16\cdot\tbinom{9}{j}.$

####

Let $T$ be a $\mathcal{M}_{8d}$ configuration of $10$ points, represented by
$(L_{1},L_{2},\\{P_{1},P_{2},P_{3}\\})$ where $\\{L_{1},L_{2}\\}$ is a pair of
lines in a plane $\pi_{1}$, and the points $P_{1},P_{2},P_{3}$ are in a
different plane $\pi_{2}$. The point $Q=L_{1}\cap L_{2}$ lies on the line
$L=\pi_{1}\cap\pi_{2}$, but none of the points $P_{1},P_{2},P_{3}$ lie on $L$.
The four points $\\{Q,P_{1},P_{2},P_{3}\\}$ are in general position in
$\pi_{2}$. Since $j>8$, the possible case for $S\in\mathfrak{C}_{i}(T)$,
$i\leq 7$ is $S=\mathcal{M}_{7d}$. In this case $|\mathcal{M}_{7d}(T)|=3$, as
we must take $S=T\setminus\\{P_{i}\\}$ for $i=1\dots 3$. By (13), we have,

$B_{j,8}(\mathcal{M}_{8d})=\tbinom{10}{j}-3\cdot\tbinom{9}{j}.$

We have, $B_{j,8}(\mathcal{M}_{8e})=0$, as an $\mathcal{M}_{8e}$ configuration
contains exactly $8$ points $PG(3,3)$. Therefore, by (12), for $j>8$,

$\begin{split}B_{j,8}&=4680\cdot(\tbinom{16}{j}-3\cdot\tbinom{13}{j-1}-42\cdot\tbinom{10}{j}-108\cdot\tbinom{9}{j}-\tbinom{13}{j})+9360\cdot(\tbinom{13}{j}-4\cdot\tbinom{10}{j}-54\cdot\tbinom{9}{j})\\\
&+189540\cdot(\tbinom{12}{j}-4\cdot\tbinom{10}{j}-16\cdot\tbinom{9}{j})+336960\cdot(\tbinom{10}{j}-3\cdot\tbinom{9}{j}),\end{split}$

i.e.,

$B_{j,8}=4680\cdot\tbinom{16}{j}+4680\cdot\tbinom{13}{j}-14040\cdot\tbinom{13}{j-1}-655200\cdot\tbinom{10}{j}-5054400\cdot\tbinom{9}{j}.$

##  8\. Configurations of rank $9$

A configuration $S$ of rank $9$ is, by definition, contained in the set of
points $\mathcal{Q}$ of a quadric of $PG(3,3)$ defined by some quadratic form
$f(X_{0},\dots,X_{3})$. The points of $\mathcal{Q}$ are a rank $9$
configuration if and only if $I_{2}(\mathcal{Q})$ is one dimensional. As seen
in Table 1, there are six $G$-orbits of quadratic forms on $PG(3,3)$
represented by

$\varphi_{1}=X_{0}X_{1},\,\varphi_{2}=X_{0}X_{1}-X_{2}X_{3},\,\varphi_{3}=X_{1}^{2}+X_{3}^{2}-X_{0}X_{2},\,\varphi_{4}=X_{0}X_{2}-X_{1}^{2},\,\varphi_{5}=X_{0}^{2},\,\varphi_{6}=X_{0}^{2}+X_{1}^{2}.$

The quadric $\mathcal{Q}_{i}=V(\varphi_{i})$ is a rank $9$ configuration if
and only if $I_{2}(V(\varphi_{i}))$ is the one dimensional vector space
generated by $\varphi_{i}$. This is not the case for i) $\varphi_{6}$, ii)
$\varphi_{5}$ and iii) $\varphi_{4}$ because: i) $V(\varphi_{6})$ is a line
which has rank $3$, ii) $V(\varphi_{5})$ is a plane which has rank $6$, and
iii) $I_{2}(V(\varphi_{4}))$ is $2$-dimensional vector space spanned by
$\varphi_{4}$ and $X_{0}(X_{2}-X_{1})$.

The quadrics $\mathcal{Q}_{1},\mathcal{Q}_{2}$ and $\mathcal{Q}_{3}$ have
$22,16$ and $10$ points respectively. The configurations
$\mathcal{M}_{8a},\mathcal{M}_{8c}$ and $\mathcal{M}_{8e}$ have $16,12$ and
$8$ points respectively. We recall from §7, that the pencil of quadrics
defining $\mathcal{M}_{8a},\mathcal{M}_{8c}$ and $\mathcal{M}_{8e}$
configurations contains an element of $G\cdot\varphi_{1}$ ,
$G\cdot\varphi_{2}$ and $G\cdot\varphi_{3}$ respectively. Therefore, i)
$\mathcal{Q}_{1}$ properly contains an $\mathcal{M}_{8a}$ configuration, ii)
$\mathcal{Q}_{2}$ properly contains an $\mathcal{M}_{8c}$ configuration, and
iii) $\mathcal{Q}_{3}$ properly contains an $\mathcal{M}_{8e}$ configuration.
Since the containment is proper in each case, and the configurations
$\mathcal{M}_{8a},\mathcal{M}_{8c},\mathcal{M}_{8e}$ are maximal, it follows
that the quadrics $\mathcal{Q}_{1},\mathcal{Q}_{2},\mathcal{Q}_{3}$ are
$\mathcal{C}_{9}$’s. We summarize this:

######  Lemma 8.1.

The set $\mathfrak{C}_{9}$ can be partitioned into the following 3 classes:

  1. $\mathcal{M}_{9a}$

of size $2^{2}\cdot 3\cdot 5\cdot 13$, consisting of a pair of hyperplanes of
$PG(3,3)$.

  2. $\mathcal{M}_{9b}$

of size $2\cdot 3^{4}\cdot 5\cdot 13$, consisting of hyperbolic quadrics of
$PG(3,3)$.

  3. $\mathcal{M}_{9c}$

of size $2^{3}\cdot 3^{4}\cdot 13$, consisting of elliptic quadrics of
$PG(3,3)$.

###  8.1. $B_{j,9}$ for $j>9$

####

Let $T$ be a $\mathcal{M}_{9a}$ configuration of $22$ points represented by a
pair of hyperplanes $\pi_{1},\pi_{2}$ intersecting in a line $L$. Since $j>9$,
the possible cases for $S\in\mathfrak{C}_{i}(T)$, $i\leq 8$ are listed as
follows.

  1. (i)

$S=\mathcal{M}_{8a}$ consisting of a plane $\pi$ and a line $\ell$ not
contained in $\pi$. In this case $|\mathcal{M}_{8a}(T)|=2\cdot 12=24$, as
there are $2$ choices for $\pi$, and $12$ choices for $\ell$. Therefore,

$|\mathcal{M}_{8a}(T)|\cdot
B_{j,9}(\mathcal{M}_{8a}(T))=24\cdot(\tbinom{16}{j}-3\cdot\tbinom{13}{j-1}-42\cdot\tbinom{10}{j}-\tbinom{13}{j}).$

  2. (ii)

$S=\mathcal{M}_{8b}$. For a $\mathcal{M}_{8b}(T)$ configuration of $4$
concurrent lines $L_{1},\dots,L_{4}$ through a point $Q$, no three of which
are coplanar, the point $Q$ has to be on the line $L$, and $\pi_{1}$,
$\pi_{2}$ each contain two of the lines $\\{L_{1},\dots,L_{4}\\}$. The point
$Q$ on $L$ can be chosen in $4$ ways, and the lines $L_{1},\dots,L_{4}$
through $Q$ can be chosen in $9$ ways. Thus, $|\mathcal{M}_{8b}(T)|=4\cdot
9=36$. Therefore,

$|\mathcal{M}_{8b}(T)|\cdot
B_{j,9}(\mathcal{M}_{8b}(T))=36\cdot(\tbinom{13}{j}-4\cdot\tbinom{10}{j}).$

  3. (iii)

$S=\mathcal{M}_{8c}$. The $\mathcal{M}_{8c}(T)$ configurations can be chosen
in the following way: pick a pair of points $\\{P,P^{\prime}\\}$ on the line
$L$. A pair of lines $L_{1},L_{2}$ through $P$, and another pair of lines
$L_{1}^{\prime},L_{2}^{\prime}$ through $P^{\prime}$ such that
$L_{1},L_{2},L_{1}^{\prime},L_{2}^{\prime}$ are distinct from $L$, and
$L_{i},L_{i}^{\prime}$ are in the plane $\pi_{i}$ for $i\in\\{1,2\\}$. There
are $\tbinom{4}{2}$ ways to pick $\\{P,P^{\prime}\\}$ on $L$, and there are
$3^{4}$ ways to choose the four lines
$\\{L_{1},L_{1}^{\prime},L_{2},L_{2}^{\prime}\\}$. Thus,
$|\mathcal{M}_{8c}(T)|=486$. Therefore,

$|\mathcal{M}_{8c}(T)|\cdot
B_{j,9}(\mathcal{M}_{8c}(T))=486\cdot(\tbinom{12}{j}-4\cdot\tbinom{10}{j}).$

  4. (iv)

$S=\mathcal{M}_{8d}$ consisting of
$(\\{L_{1},L_{2}\\},\\{P_{1},P_{2},P_{3}\\})$ as described in Lemma 7.1. Such
an $S$ can be chosen as follows: Pick a pair of lines $L_{1},L_{2}$ in one of
the planes $\pi_{1},\pi_{2}$ intersecting at a point $Q$ of $L$. There are
$2\cdot 4\cdot 3=24$ ways to pick these. In the other plane, there are $3$
lines $L_{1},L_{2},L_{3}$ through $Q$ apart from $L$. There are $18$ ways to
pick $P_{i}\in L_{i}\setminus\\{Q\\}$ for $i\in\\{1,2,3\\}$ such that
$P_{1},P_{2},P_{3}$ are non-collinear. Thus, $|\mathcal{M}_{8c}(T)|=24\cdot
18=432$. Therefore,

$|\mathcal{M}_{8d}(T)|\cdot
B_{j,9}(\mathcal{M}_{8d}(T))=432\cdot\tbinom{10}{j}.$

  5. (v)

$S=\mathcal{M}_{7a}$ consisting of a plane $\pi$ and a point $P$ not on $\pi$.
There are $2$ choices for $\pi$ namely $\pi_{1},\pi_{2}$. Any of the $9$
points of the other plane not lying on $L$ can be chosen as $P$. Thus,
$|\mathcal{M}_{7a}(T)|=2\cdot 9=18$.Therefore,

$|\mathcal{M}_{7a}(T)|\cdot
B_{j,9}(\mathcal{M}_{7a}(T))=18\cdot\tbinom{13}{j-1}.$

  6. (vi)

$S=\mathcal{M}_{7b}$. Let us consider a $\mathcal{M}_{7b}(T)$ configuration
represented by $3$ concurrent non-coplanar lines $L_{1},L_{2},L_{3}$ through a
point $Q$. There are two cases depending on whether
$L\in\\{L_{1},L_{2},L_{3}\\}$ or not.

    1. (a)

In the former case, we may assume $L=L_{1}$. The number of way to choose $Q$
on $L$ is $4$. Then the two other lines $L_{2},L_{3}$ through $Q$ lie in
different planes, and can be chosen in $\tbinom{3}{1}^{2}=9$ ways. This case
contributes $36$ to $|\mathcal{M}_{7b}(T)|$.

    2. (b)

If $L\notin\\{L_{1},L_{2},L_{3}\\}$, then two of the lines
$\\{L_{1},L_{2},L_{3}\\}$ lie in the same plane. The number of ways of
choosing $Q$ on $L$ is $4$, the number of ways choosing two lines through $Q$
in one of the plane $\pi_{1},\pi_{2}$ is $2\cdot\tbinom{3}{2}=6$. The
remaining line through $Q$ in the other plane can be chosen in $\tbinom{3}{1}$
ways. Thus, this case contributes $4\cdot 2\cdot 3\cdot 3=72$ to
$|\mathcal{M}_{7b}(T)|$.

Thus, $|\mathcal{M}_{7b}(T)|=36+72=108$. Therefore,

$|\mathcal{M}_{7b}(T)|\cdot
B_{j,9}(\mathcal{M}_{7b}(T))=108\cdot\tbinom{10}{j}.$

  7. (vii)

$S=\mathcal{M}_{7c}$ represented by three lines $L_{1},L_{2},L_{3}$, where
$L_{2}$ and $L_{3}$ are skew to each other and the line $L_{1}$ meets both the
lines. There are two cases depending on whether $L\in\\{L_{1},L_{2},L_{3}\\}$
or not.

    1. (a)

If $L\in\\{L_{1},L_{2},L_{3}\\}$, say $L=L_{3}$. let $P_{i}=L\cap L_{i}$,
$i=1,2$. The number of way to choose $P_{1}$ and $P_{2}$ is $4\cdot 3$. Then
the two lines $L_{1}$ and $L_{2}$ can be chosen in $\tbinom{3}{1}^{2}$. Thus
the number of ways of choosing $L_{1},L_{2},L_{3}$ is $4\cdot
3\cdot\tbinom{3}{1}^{2}=108$. Thus, this case contributes $108$ to
$|\mathcal{M}_{7c}(T)|$.

    2. (b)

If $L\notin\\{L_{1},L_{2},L_{3}\\}$, then exactly one of the two pairs of
lines $\\{L_{1},L_{2}\\}$ and $\\{L_{2},L_{3}\\}$ are coplanar, say
$\\{L_{1},L_{2}\\}$. In this case, the pair $\\{L_{2},L_{3}\\}$ can be chosen
$4\cdot 3\cdot 3$ ways, and $L_{1}$ can be chosen in $9$ ways. Thus, this case
contributes $2\cdot 4\cdot 3\cdot 3\cdot 9=648$ to $|\mathcal{M}_{7c}(T)|$.

Therefore, $|\mathcal{M}_{7c}(T)|=108+648=756$ and

$|\mathcal{M}_{7c}(T)|\cdot
B_{j,9}(\mathcal{M}_{7c}(T))=756\cdot\tbinom{10}{j}.$

  8. (viii)

$S=\mathcal{M}_{6a}$. In this case $|\mathcal{M}_{6a}(T)|=2$. Therefore,

$|\mathcal{M}_{6a}(T)|\cdot
B_{j,9}(\mathcal{M}_{6a}(T))=2\cdot\tbinom{13}{j}.$

By (13), for $j>9$, we have,

$\displaystyle\begin{split}B_{j,9}(\mathcal{M}_{9a})&=780\cdot(\tbinom{22}{j}-24\cdot(\tbinom{16}{j}-3\cdot\tbinom{13}{j-1}-42\cdot\tbinom{10}{j}-\tbinom{13}{j})-36\cdot(\tbinom{13}{j}-4\cdot\tbinom{10}{j})\\\
&-486\cdot(\tbinom{12}{j}-4\cdot\tbinom{10}{j})-432\cdot\tbinom{10}{j}-18\cdot\tbinom{13}{j-1}-108\cdot\tbinom{10}{j}-756\cdot\tbinom{10}{j}-2\cdot\tbinom{13}{j}),\end{split}$

i.e.

$\displaystyle
B_{j,9}(\mathcal{M}_{9a})=780\cdot(\tbinom{22}{j}-24\cdot\tbinom{16}{j}+1800\cdot\tbinom{10}{j}-486\cdot\tbinom{12}{j}+54\cdot\tbinom{13}{j-1}-62\cdot\tbinom{13}{j}).$

####

Let $T$ be a $\mathcal{M}_{9b}$ configuration represented by a hyperbolic
quadric $\mathcal{H}$. A hyperbolic quadric in $PG(3,3)$ can be written as a
disjoint union of $4$ skew lines $H_{1},\dots,H_{4}$ (known as a regulus), and
also as a disjoint union of a dual set of $4$ skew lines $V_{1},\dots,V_{4}$
(known as the opposite regulus). The set $T$ consists of the $16$ points
$H_{i}\cap V_{j}$ for $1\leq i,j\leq 4$. Since $j>9$, the possible cases for
$S\in\mathfrak{C}_{i}(T)$, $i\leq 8$ are as follows.

  1. (i)

$S=\mathcal{M}_{8c}$ consisting of $4$ lines $L_{1},\dots,L_{4}$ with
$L_{i},L_{j}$ being skew if and only if $i\equiv j\mod 2$. Here,
$|\mathcal{M}_{8c}(T)|=\tbinom{4}{2}^{2}=36$, as $\\{L_{1},\dots,L_{4}\\}$
consists of $2$ lines from $\\{H_{1},\dots,H_{4}\\}$ and $2$ lines from
$\\{V_{1},\dots,V_{4}\\}$.Therefore,

$|\mathcal{M}_{8c}(T)|\cdot
B_{j,9}(\mathcal{M}_{8c}(T))=36\cdot(\tbinom{12}{j}-4\cdot\tbinom{10}{j}).$

  2. (ii)

$S=\mathcal{M}_{8d}$ consisting of
$(\\{L_{1},L_{2}\\},\\{P_{1},P_{2},P_{3}\\})$ as described in Lemma 7.1. We
must have
$|\\{L_{1},L_{2}\\}\cap\\{H_{1},\dots,H_{4}\\}|=|\\{L_{1},L_{2}\\}\cap\\{V_{1},\dots,V_{4}\\}|=1$,
and hence there are $16$ ways to pick $\\{L_{1},L_{2}\\}$. If $L_{1}=H_{4}$
and $L_{2}=V_{4}$, the points $\\{P_{1},P_{2},P_{3}\\}$ must be $\\{H_{1}\cap
V_{\sigma(1)},H_{2}\cap V_{\sigma(2)},H_{3}\cap V_{\sigma(3)}\\}$ for some
permutation $\sigma$ of $\\{1,2,3\\}$. i.e., there are $6$ ways to pick
$\\{P_{1},P_{2},P_{3}\\}$. Thus, $|\mathcal{M}_{8d}(T)|=16\cdot 6=96$.
Therefore,

$|\mathcal{M}_{8d}(T)|\cdot
B_{j,9}(\mathcal{M}_{8d}(T))=96\cdot\tbinom{10}{j}.$

  3. (iii)

$S=\mathcal{M}_{7c}$ represented by three lines $L_{1},L_{2},L_{3}$, where
$L_{2}$ and $L_{3}$ are skew to each other and the line $L_{1}$ meets both the
lines. We must have $L_{1}\in\\{H_{1},\dots,H_{4}\\}$ and
$L_{2},L_{3}\in\\{V_{1},\dots,V_{4}\\}$ or $L_{1}\in\\{V_{1},\dots,V_{4}\\}$
and $L_{2},L_{3}\in\\{H_{1},\dots,H_{4}\\}$. Thus,
$|\mathcal{M}_{7c}(T)|=2\cdot\tbinom{4}{2}\cdot\tbinom{4}{1}=48$. Therefore,

$|\mathcal{M}_{7c}(T)|\cdot
B_{j,9}(\mathcal{M}_{7c}(T))=48\cdot\tbinom{10}{j}.$

By (13), for $j>9$, we have,

$B_{j,9}(\mathcal{M}_{9b})=\tbinom{16}{j}-36\cdot(\tbinom{12}{j}-4\cdot\tbinom{10}{j})-96\cdot\tbinom{10}{j}-48\cdot\tbinom{10}{j}=\tbinom{16}{j}-36\cdot\tbinom{12}{j}.$

####

Let $T$ be a $\mathcal{M}_{9c}$ configuration represented by an elliptic
quadric $\mathcal{Q}_{3}=V(\varphi_{3})$ consisting of $10$ points. The only
size $j$ subset of $T$ for $j>9$, is $T$ itself. Therefore,

$B_{j,9}(\mathcal{M}_{9c})=\tbinom{10}{j}.$

Finally, for $j>9$, we have by (12):

$\begin{split}B_{j,9}&=780\cdot(\tbinom{22}{j}-24\cdot\tbinom{16}{j}+1800\cdot\tbinom{10}{j}-486\cdot\tbinom{12}{j}+54\cdot\tbinom{13}{j-1}-62\cdot\tbinom{13}{j})\\\
&+10530\cdot(\tbinom{16}{j}-36\cdot\tbinom{12}{j})+8424\cdot\tbinom{10}{j},\end{split}$

i.e.,

(19)
$\begin{split}B_{j,9}=780\cdot\tbinom{22}{j}-8190\cdot\tbinom{16}{j}-48360\cdot\tbinom{13}{j}-14025960\cdot\tbinom{12}{j}+1412424\cdot\tbinom{10}{j}+42120\cdot\tbinom{13}{j-1}\end{split}$

##  9\. Proof of Theorem 1.1

We recall from §3, that we have defined $B_{0}(T)=(T^{K}-1)$. Using
(7),(9),(10),(14)-(19) we calculate the values of $B_{j}(T)$:

$\displaystyle B_{1}(T)=$ $\displaystyle 40(T^{9}-1),$ $\displaystyle
B_{2}(T)=$ $\displaystyle 780(T^{8}-1),$ $\displaystyle B_{3}(T)=$
$\displaystyle 9880(T^{7}-1),$ $\displaystyle B_{4}(T)=$ $\displaystyle
130(T^{7}+702T^{6}-703),$ $\displaystyle B_{5}(T)=$ $\displaystyle
936(5T^{6}+698T^{5}-703),$ $\displaystyle B_{6}(T)=$ $\displaystyle
780(109T^{5}+4812T^{4}-4921),$ $\displaystyle B_{7}(T)=$ $\displaystyle
1560(2T^{5}+663T^{4}+11286T^{3}-11951),$ $\displaystyle B_{8}(T)=$
$\displaystyle 585(T-1)(241T^{3}+16333T^{2}+131461T+131461),$ $\displaystyle
B_{9}(T)=$ $\displaystyle 520(T-1)(55T^{3}+5500T^{2}+138016T+525844),$
$\displaystyle B_{10}(T)=$ $\displaystyle
104(T-1)(110T^{3}+8435T^{2}+399125T+4410326),$ $\displaystyle B_{11}(T)=$
$\displaystyle 3120(T-1)(T^{3}+100T^{2}+6211T+171775),$ $\displaystyle
B_{12}(T)=$ $\displaystyle 260(T-1)(2T^{3}+326T^{2}+29837T+1921619),$
$\displaystyle B_{13}(T)=$ $\displaystyle
40(T-1)(T^{3}+352T^{2}+61426T+9659872),$ $\displaystyle B_{14}(T)=$
$\displaystyle 1080(T-1)(T^{2}+508T+230582),$ $\displaystyle B_{15}(T)=$
$\displaystyle 18720(T-1)(4T+7103),$ $\displaystyle B_{16}(T)=$ $\displaystyle
1170(4T+49739)(T-1),$ $\displaystyle B_{17}(T)=$ $\displaystyle
20540520(T-1),$ $\displaystyle B_{18}(T)=$ $\displaystyle 5705700(T-1),$
$\displaystyle B_{19}(T)=$ $\displaystyle 1201200(T-1),$ $\displaystyle
B_{20}(T)=$ $\displaystyle 180180(T-1),$ $\displaystyle B_{21}(T)=$
$\displaystyle 17160(T-1),$ $\displaystyle B_{22}(T)=$ $\displaystyle
780(T-1).$

Having calculated the polynomials $B_{j}(T)$, we use them in (8), to determine
the polynomials $A_{j}(T)$:

######  Theorem 9.1.

The extended weight enumerator of $C_{3}$ is given by

$W(X,Y,T)=X^{40}+(T-1)\sum_{i=1}^{40}a_{i}(T)X^{40-i}Y^{i},$

where the polynomials $a_{i}(T)$ are as follows:

$\displaystyle a_{18}(T)=$ $\displaystyle 780,$ $\displaystyle a_{24}(T)=$
$\displaystyle 1170(4T-3),$ $\displaystyle a_{26}(T)=$ $\displaystyle
1080(T-3)(T-9),$ $\displaystyle a_{27}(T)=$ $\displaystyle
40(T^{3}-26T^{2}+442T-884),$ $\displaystyle a_{28}(T)=$ $\displaystyle
189540(T-3),$ $\displaystyle a_{30}(T)=$ $\displaystyle
936(100T^{2}-600T+909,)$ $\displaystyle a_{31}(T)=$ $\displaystyle
505440(T-3)(T-6),$ $\displaystyle a_{32}(T)=$ $\displaystyle
5265(T-3)(17T^{2}-160T+513),$ $\displaystyle a_{33}(T)=$ $\displaystyle
3120(T-3)(T^{3}+84T^{2}-972T+3000),$ $\displaystyle a_{34}(T)=$ $\displaystyle
63180(T-3)(T^{3}-5T^{2}-45T+249),$ $\displaystyle a_{35}(T)=$ $\displaystyle
936(T-3)(5T^{4}+243T^{3}-6408T^{2}+51654T-142020),$ $\displaystyle a_{36}(T)=$
$\displaystyle
130(T^{6}+523T^{5}-15635T^{4}+199500T^{3}-1349544T^{2}+4639614T-6020826),$
$\displaystyle a_{37}(T)=$ $\displaystyle
9360(T-3)(T^{5}-30T^{4}+405T^{3}-3132T^{2}+13932T-27864),$ $\displaystyle
a_{38}(T)=$ $\displaystyle
780(T-3)(T^{6}-33T^{5}+507T^{4}-4698T^{3}+28242T^{2}-105678T+189054),$
$\displaystyle a_{39}(T)=$ $\displaystyle
40(T-3)(T^{7}-35T^{6}+585T^{5}-6096T^{4}+43320T^{3}-214344T^{2}+695448T-1128816),$
$\displaystyle\begin{split}a_{40}(T)=&(T-3)\left(T^{8}-36T^{7}+633T^{6}-7110T^{5}+56241T^{4}-325134T^{3}\right.\\\
&\qquad\qquad\left.+1371006T^{2}-3936114T+5856786\right).\end{split}$

Having obtained the extended weight enumerator $W(Z;T)$ of $C_{3}$ over
$\mathbb{F}_{3}$, we are in a position to compute the quantities $W(Z;q^{j})$
for $q=3$. We calculate the $r$-GWE of $C_{3}$ listed in Theorem 1.1, by using
the quantities $W(Z;q^{j})$ in the result of Theorem 3.1.

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