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projective line configurations
1 Introduction
1.1 Definition
A projective line configuration consists of a collections of $p$ points and $\ell$ lines in a projective space such that through each point of the configuration there pass a fixed number $\lambda$ of lines of the configuration and on each line of the configuration there are found a fixed number $\pi$ of points of the configuration. It is not required that the intersection of any two lines belonging to the configuration be a line of the configuration nor that through every two points of the configuration there pass a line of the configuration. (Indeed, this will usually not be the case except for the very simplest examples with small values of $\pi$ and $\lambda$.) Besides being interesting geometric objects in their own right, projective configurations arise naturally in geometric definitions, constructions, and theorems, and also in such contexts as collections of special points and lines associated to algebraic varieties.
1.2 Examples
A simple example of a projective configuration is a triangle:
$\xy,(0,0);(30,30)**@{},(14,17)*{b},(22,30);(52,0)**@{},(38,17)*{a},(0,5);(52% ,5)**@{},(26,2)*{c},(5,5)*{\bullet},(5,2)*{A},(47,5)*{\bullet},(46,2)*{B},(26% ,26)*{\bullet},,(30,26)*{C}$ 
Here we have 3 points, $A$,$B$,$C$ and 3 lines, $a$,$b$,$c$ with the property that each of the points lines on exactly 2 of the lines ($A$ lies on $b$ and $c$, $B$ lies on $a$ and $c$, and $C$ lies on $a$ and $b$) and that each of the lines passes through exactly two of the points ($a$ passes through $B$ and $C$, $b$ passes through $A$ and $C$, and $c$ passes through $A$ and $B$.)
Another example is a complete quadrangle. This configuration consists of 4 points $A$, $B$, $C$, $D$ and the 6 lines $a$, $b$, $c$, $d$, $e$, $f$ as illustrated below:
$\xy,(5,35)*{\bullet},(8,37)*{A},(35,35)*{\bullet},(32,37)*{B},(5,5)*{\bullet},% (8,2)*{C},(35,5)*{\bullet},(32,2)*{D},(0,5);(40,5)**@{},(20,2)*{c},(0,35);(40% ,35)**@{},(20,37)*{a},(5,0);(5,40)**@{},(2,20)*{d},(35,0);(35,40)**@{},(37,% 20)*{b},(0,0);(40,40)**@{},(27,30)*{f},(0,40);(40,0)**@{},(12,30)*{e}$ 
Since each of the points lies on exactly 3 of the lines and each the lines contains exactly two of the points, this is indeed a bona fide projective configuration. (Note that the intersection of the lines $e$ and $f$ is not highlighted as this is not one of the points of the configurations — as mentioned at top, not every intersection of two lines of the configuration need be a point of the configuration.)
1.3 Notation
When discussing line configurations, use is made of the notation $(p_{{\lambda}}\ell_{{\pi}})$ or $\begin{pmatrix}p&\lambda\\ \pi&\ell\end{pmatrix}$ to indicate that the configuration contains $p$ points and $\ell$ lines with $\lambda$ lines passing through each point and $\pi$ points on each line. Thus, we would say that the triangle is a configuration of type $(3_{2}3_{2})$ (or $\begin{pmatrix}3&2\\ 2&3\end{pmatrix}$) and the complete quadrangle is a configuration of type $(4_{3},6_{2})$.
It is worth pointing out that the four numbers $p$, $\ell$, $\pi$, $\lambda$ are linked by the relation $p\cdot\lambda=\pi\cdot\ell$. The reason for this is a counting argument — we could count pairs consisting of a point and a line passing through that point two ways. We could start with the $p$ points and count $\lambda$ lines passing through each point to arrive at $p\lambda$ pairs. Alternatively, we could start with the $\ell$ lines and count $\pi$ points on each line to arrive at $\pi\ell$ pairs. Since we are counting the same objects (pairs consisting of incident points and lines), we must arrive at the same number either way.
Finally, when dealing with cases where $p=\ell$ (and hence, by what was said above, $\pi=\lambda$), one may use the abbreviated notation $(p_{\lambda})$. Thus, one could say instead that the triangle is a configuration of type $(3_{2})$. Thus notation is most commonly encountered in the context of selfdual configurations (which will be defined in the next subsection).
1.4 Choice of Projective Space
So far, we have only spoken of projective configurations in a general manner. To discuss the matter in more detail, we need to take into account the projective space within which our configuration is situated.
The need for doing so is rather well illustrated by the fact that certain configurations may not exist in all projective spaces. As an example, we may consider the Hesse configuration, which is a configuration of type $(9_{4}12_{3})$. In this configuration, if we label the points by the letters $A$ through $I$ suitably, the lines pass through the following triplets of points: $ABC$, $DEF$, $GHI$, $ADG$, $BEH$, $CFI$, $AEI$, $BFG$, $CDH$, $AFH$, $BDI$, $CEG$. It is not possible to find 9 points and 12 lines in the real projective plane $\mathbb{R}\mathbb{P}^{2}$ (or, for that matter, any real projective space $\mathbb{R}\mathbb{P}^{n}$) which form such a configuration. However, such a configuration can be found in the complex projective plane $\mathbb{C}\mathbb{P}^{2}$ — for instance we could take the 9 points with homogeneous coordinates
$\begin{matrix}A:(1,r_{+},0)&B:(1,1,0)&C:(1,r_{},0)\\ D:(r_{+},0,1)&E:(1,0,1)&F:(r_{},0,1)\\ G:(0,1,r_{+})&H:(0,1,1)&I:(0,1,r_{})\end{matrix}$ 
and the 12 lines with equations
$\begin{matrix}\hbox to 50.0pt{$\hfil ABC:\;$}z=0\hfill&\hbox to 50.0pt{$\hfil DEF% :\;$}y=0\hfill&\hbox to 50.0pt{$\hfil GHI:\;$}x=0\hfill\\ \hbox to 50.0pt{$\hfil ADG:\;$}x+r_{}y+r_{+}z=0\hfill&\hbox to 50.0pt{$\hfil BEH% :\;$}x+y+z=0\hfill&\hbox to 50.0pt{$\hfil CFI:\;$}x+r_{+}y+r_{}z=0\hfill\\ \hbox to 50.0pt{$\hfil AEI:\;$}x+r_{}y+z=0\hfill&\hbox to 50.0pt{$\hfil BFG:% \;$}x+y+r_{}z=0\hfill&\hbox to 50.0pt{$\hfil CDH:\;$}r_{}x+y+z=0\hfill\\ \hbox to 50.0pt{$\hfil AFH:\;$}r_{+}x+y+z=0\hfill&\hbox to 50.0pt{$\hfil BDI:% \;$}x+y+r_{+}z=0\hfill&\hbox to 50.0pt{$\hfil CEG:\;$}x+r_{+}y+z=0\hfill,\end{matrix}$ 
where $r_{\pm}=(1\pm i\sqrt{3})/2$. That the appropriate points lie on the appropriate lines may be readily verified by a computation which is especially effortless if one makes use of the facts that $r_{+}^{3}=r_{}^{3}=1$ and $r_{+}r_{}=1$.
This notion of certain configurations being only found in certain spaces may be clarified by an intrinsic/extrinsic approach. Define an abstract line configuration of type $(p_{\lambda}\ell_{\pi})$ to be a triplet $\langle P,L,I\rangle$, where $P$ and $Q$ are sets and $I$ is a relation on $P\times L$ such that the following conditions hold:

The cardinality of $P$ is $p$.

The cardinality of $L$ is $\ell$.

For every $x\in P$, the cardinality of $\{y\in L\mid I(x,y)\}$ is $\lambda$.

For every $x\in L$, the cardinality of $\{y\in P\mid I(y,x)\}$ is $\pi$.

Given two distinct elements $x,y$ of $P$, there exists at most one element $z$ of $L$ such that $I(x,z)$ and $I(y,z)$.

Given two distinct elements $x,y$ of $L$, there exists at most one element $z$ of $P$ such that $I(z,x)$ and $I(z,y)$.
Given a projective space $S$, we may then define an embeding of an abstract line complex $\langle P,L,I\rangle$ to be an assignment of a point of $S$ to every element of $P$ and a line to every element of $L$ in such a way that the point assigned to an element $x$ of $P$ will lie on the line assigned to an element $y$ of $L$ if and only if $I(x,y)$. For instance, returning to our first example above, the abstract configuration of the triangle is
$\langle\{A,B,C\},\{a,b,c\},\{(A,b),(A,c),(B,a),(B,c),(C,a),(C,b)\rangle.$ 
Not only is this way of thinking useful conceptually, but, as we shall see in the next section, it is useful in practise because it lets us divide the work of finding configurations into a combinatorial task of determining abstract configurations and a geometric task of determining which abstract configurations may be embedded in which space.
When the space within which our configuration is embedded is two dimensional, i.e. happens to be a projective plane, then we can apply the duality operation to obtain another line configuration in which the role of lies and points has been interchanged. We call this new configuration the dual of the original configuration. If it happens that the dual configuration is projectively equivalent to the original configuration, we call it a selfdual configuration. For instance, the triangle is a selfdual configuration in $\mathbb{R}\mathbb{P}^{2}$, given two triangles, there will be a collineation which maps one into the other. In the example of the complete quadrangle, its dual is a configuration known as the complete quadrangle, which consists of four lines such that each pair of lines intersects in on of the six points of the configuration.
$\xy,(6,6)*{\bullet},(8,4)*{C},(26,66)*{\bullet},(29,66)*{E},(66,26)*{\bullet},% (66,29)*{F},(42,18)*{\bullet},(45,17)*{D},(18,42)*{\bullet},(16,45)*{B},(36,36% )*{\bullet},(39,39)*{A},(4,0);(28,72)**@{},(26,11)*{c},(0,4);(72,28)**@{},(1% 0,26)*{d},(72,24);(0,48)**@{},(52,34)*{b},(24,72);(48,0)**@{},(34,52)*{a}$ 
This notion of duality can be extended to abstract configurations. Given an abstract line configuration $\langle P,L,I\rangle$, its dual is $\langle L,P,I^{{\prime}}\rangle$ where $I^{{\prime}}(x,y)$ if and only if $I(y,x)$. For instance, the abstract configuration of the complete quadrangle is
$\displaystyle\langle$  $\displaystyle\{A,B,C,D\},\{a,b,c,d,e,f\},$  
$\displaystyle\{(A,a),(A,d),(A,e),(B,a),(B,b),(B,f),(C,c),(C,d),(C,f),(D,b),(D,% c),(D,e)\}\rangle$ 
so its dual is
$\displaystyle\langle$  $\displaystyle\{a,b,c,d,e,f\},\{A,B,C,D\},$  
$\displaystyle\{(a,A),(d,A),(e,A),(a,B),(b,B),(f,B),(c,C),(d,C),(f,C),(b,D),(c,% D),(e,D)\}\rangle$ 
which corresponds to the complete quadrilateral. Likewise, we can define a notion of selfduality at the abstract level. We will say that an abstract line configuration $\langle P,L,I\rangle$ is selfdual if there exists a onetoone correspondence $g\colon P\to L$ such that, for all $x,y\in P$, we have $I(x,g(y))$ if and only if $I(y,g(x))$.
1.5 Symmetries
Next, we consider the effect of collineations on configurations. Given a configuration in a projective space, a collineation of that space will map that configuration into some configuration. If it happens to be mapped into the same configuration, then we say that the collineation is a symmetry of the configuration in question.
To illustrate these ideas, let us consider a triangle in $\mathbb{R}\mathbb{P}^{2}$ consisting of the points $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ and the lines $x=0$, $y=0$, $z=0$. Firstly, consider the collineation
$\displaystyle x$  $\displaystyle\mapsto x+y$  
$\displaystyle y$  $\displaystyle\mapsto y$  
$\displaystyle z$  $\displaystyle\mapsto z$ 
Under this mapping, the lines and points of our configuration transform as follows:
$\begin{matrix}(1,0,0)\mapsto(1,0,0)\hfill&x=0\mapsto x+y=0\hfill\\ (0,1,0)\mapsto(1,1,0)\hfill&y=0\mapsto y=0\hfill\\ (0,0,1)\mapsto(0,0,1)\hfill&z=0\mapsto z=0\hfill\end{matrix}$ 
Since the point $(1,1,0)$ does not belong to the original configuration, this transform is not a symmetry of our triangle.
Secondly, consider the transformation
$\displaystyle x$  $\displaystyle\mapsto y$  
$\displaystyle y$  $\displaystyle\mapsto z$  
$\displaystyle z$  $\displaystyle\mapsto x$ 
Under this mapping, the lines and points of our configuration transform as follows:
$\begin{matrix}(1,0,0)\mapsto(0,1,0)&\hfil&x=0\mapsto y=0\hfill\\ (0,1,0)\mapsto(0,0,1)&\hfil&y=0\mapsto z=0\hfill\\ (0,0,1)\mapsto(1,0,0)&\hfil&z=0\mapsto z=0\hfill\end{matrix}$ 
Since the images of the points are points of the original configuration and the images of the lines also belong to the original configuration, this collineation is a symmetry of the configuration.
Thirdly, consider the collineation
$\displaystyle x$  $\displaystyle\mapsto 2x$  
$\displaystyle y$  $\displaystyle\mapsto y$  
$\displaystyle z$  $\displaystyle\mapsto z$ 
Under this mapping, the lines and points of our configuration transform as follows:
$\begin{matrix}(1,0,0)\mapsto(2,0,0)\hfill&x=0\mapsto 2x=0\hfill\\ (0,1,0)\mapsto(0,1,0)\hfill&y=0\mapsto y=0\hfill\\ (0,0,1)\mapsto(0,0,1)\hfill&z=0\mapsto z=0\hfill\end{matrix}$ 
Since the images of the points are points of the original configuration and the images of the lines also belong to the original configuration, this collineation is a symmetry of the configuration. (Remember that, since we are dealing with homogeneous coordinates on projective space, overall scalings do not matter, so $(1,0,0)$ and $(2,0,0)$ label the same point, likewise $x=0$ and $2x=0$ describe the same line.) Note that this symmetry differs from the one in the previous example because each point and line is individually left invariant as opposed to only having the set of all points and the set of all lines be left invariant.
We may also consider permuting the points and lines in abstract line configurations. Given an abstract line configuration $\langle P,L,I\rangle$, we will define a symmetry of this configuration to be a pair of permutations $f\colon P\to P$ and $g\colon L\to L$ such that, for all $x\in P$ and all $y\in L$, we have $I(f(x),g(y))$ if and only if $I(x,y)$.
We may relate these abstract and concrete symmetry groups as follows. Suppose that we have an abstract configuration $C$ which is embedded in a projective space $P$ as a configuration $C^{{\prime}}$. Let $G_{a}$ be the group of symmetries of $C$. Let $G_{e}$ be the group of collineations of $P$ which preserves $C^{{\prime}}$ and let $G_{f}$ be the group of collineations which leaves the points and lines of $C^{{\prime}}$ fixed individually. Then $G_{f}$ is a normal subgroup of $G_{e}$ and the quotient group $G_{e}/G_{f}$ is a subgroup of $G_{a}$.
To illustrate how this works, we will consider the symmetry groups associated to the example of the triangle studied above. We begin with $G_{e}$. Writing down the effect of a linear transform and asking that it preserve the configuration, we find that, in order for a transformation to preserve the configuration, it should have one of the following forms:
$\begin{matrix}\left\{\begin{matrix}x\mapsto\alpha x\\ y\mapsto\beta y\\ z\mapsto\gamma z\end{matrix}\right.&\left\{\begin{matrix}x\mapsto\alpha y\\ y\mapsto\beta z\\ z\mapsto\gamma x\end{matrix}\right.&\left\{\begin{matrix}x\mapsto\alpha z\\ y\mapsto\beta x\\ z\mapsto\gamma y\end{matrix}\right.&\left\{\begin{matrix}x\mapsto\alpha x\\ y\mapsto\beta z\\ z\mapsto\gamma x\end{matrix}\right.&\left\{\begin{matrix}x\mapsto\alpha z\\ y\mapsto\beta y\\ z\mapsto\gamma x\end{matrix}\right.&\left\{\begin{matrix}x\mapsto\alpha y\\ z\mapsto\beta y\\ z\mapsto\gamma z\end{matrix}\right.\end{matrix}$ 
Here, $\alpha,\beta,\gamma$ are arbitrary nonzero real numbers. If we instead ask that the points and lines of the triangle be fixed individually, only transforms of the form
$\left\{\begin{matrix}x\mapsto\alpha x\\ y\mapsto\beta y\\ z\mapsto\gamma z\end{matrix}\right.$ 
remain. These form the group $G_{f}$. Examining the abstract configuration of the triangle, we may verify that the following permutations are the ones which preserve incidence:
$\begin{matrix}\begin{tabular}[]{cccc}x&A&B&C\\ \hline f(x)&A&B&C\end{tabular}\quad\begin{tabular}[]{cccc}x&a&b&c\\ \hline g(x)&a&b&c\end{tabular}&\begin{tabular}[]{cccc}x&A&B&C\\ \hline f(x)&A&C&B\end{tabular}\quad\begin{tabular}[]{cccc}x&a&b&c\\ \hline g(x)&a&c&b\end{tabular}\\ \\ \begin{tabular}[]{cccc}x&A&B&C\\ \hline f(x)&B&C&A\end{tabular}\quad\begin{tabular}[]{cccc}x&a&b&c\\ \hline g(x)&b&c&a\end{tabular}&\begin{tabular}[]{cccc}x&A&B&C\\ \hline f(x)&C&B&A\end{tabular}\quad\begin{tabular}[]{cccc}x&a&b&c\\ \hline g(x)&c&b&a\end{tabular}\\ \\ \begin{tabular}[]{cccc}x&A&B&C\\ \hline f(x)&C&A&B\end{tabular}\quad\begin{tabular}[]{cccc}x&a&b&c\\ \hline g(x)&c&a&b\end{tabular}&\begin{tabular}[]{cccc}x&A&B&C\\ \hline f(x)&B&A&C\end{tabular}\quad\begin{tabular}[]{cccc}x&a&b&c\\ \hline g(x)&b&a&c\end{tabular}\end{matrix}$ 
The group presented above is isomorphic to $S_{3}$. From what was described earlier, one may also check that $G_{e}/G_{f}$ is also isomorphic to $S_{3}$ so, in this case, $G_{e}/G_{f}$ is isomorphic to the whole of $G_{a}$.
As another illustrative example of symmetry groups of configurations, we shall consider the configuration of type $(4_{1}1_{4})$ in $\mathbb{R}\mathbb{P}^{2}$ consisting of a line $a$ with equation $z=0$ and four points $A$, $B$, $C$,$D$ with coordinates $(1,0,0)$, $(1,1,0)$, $(1,2,0)$, $(1,3,0)$ respectively.
$\xy,(0,5);(50,5)**@{};(25,7)*{a},(10,5)*{\bullet},(10,2)*{A},(20,5)*{\bullet}% ,(20,2)*{B},(30,5)*{\bullet},(30,2)*{C},(40,5)*{\bullet},(40,2)*{D}$ 
In order to preserve this configuration, a collineation must have one of the following forms:
$\begin{matrix}\left\{\begin{matrix}x\mapsto\alpha x+\beta z\hfill\\ y\mapsto\alpha y+\gamma z\hfill\\ z\mapsto\delta z\hfill\end{matrix}\right.&\left\{\begin{matrix}x\mapsto\alpha x% +\beta z\hfill\\ y\mapsto 3\alpha x\alpha y+\gamma z\hfill\\ z\mapsto\delta z\hfill\end{matrix}\right.&\left\{\begin{matrix}x\mapsto 3% \alpha x2\alpha y+\beta z\hfill\\ y\mapsto 3\alpha x3\alpha y+\gamma z\hfill\\ z\mapsto\delta z\hfill\end{matrix}\right.&\left\{\begin{matrix}x\mapsto 3% \alpha x2\alpha y+\beta z\hfill\\ y\mapsto 6\alpha x3\alpha y+\gamma z\hfill\\ z\mapsto\delta z\hfill\end{matrix}\right.\end{matrix}$ 
Here $\alpha$, $\beta$, $\gamma$, $\delta$ are real numbers with neither $\alpha$ nor $\beta$ equal to zero. These transforms form the group $G_{e}$. Of these, the transforms
$\left\{\begin{matrix}x\mapsto\alpha x+\beta z\hfill\\ y\mapsto\alpha y+\gamma z\hfill\\ z\mapsto\delta z\hfill\end{matrix}\right.$ 
fix the points individually so form the normal subgroup $G_{f}$. As for symmetries of the abstract configuration, since there is only one line, $g$ is trivial whilst $f$ can be any permutation of $4$ objects because the only relation to be preserved is that all $4$ points line on the same line. Hence, $G_{a}$ is isomorphic to $S_{4}$. However, $G_{e}/G_{f}$ is isomorphic to the Klein viergruppe, so here we have a case in which $G_{e}/G_{f}$ is a proper subgroup of $G_{a}$.
1.6 Generalizations
In projective spaces of dimension higher than two, we can consider configurations consisting not only of points and lines but also of higherdimensional subspaces. For instance, in three or more dimensions, we can consider configurations consisting of points, lines, and planes. Specifically, such a configuration consists of a set of $n_{{00}}$ points, $n_{{11}}$ lines, and $n_{{22}}$ planes such that each point lies on exactly $n_{{01}}$ lines and $n_{{02}}$ planes, each line contains $n_{{10}}$ points and lies on $n_{{12}}$ planes, and each plane contains exactly $n_{{20}}$ points and $n_{{21}}$ planes, where $n_{{00}}$, $n_{{01}}$, $n_{{10}}$, $n_{{02}}$, $n_{{11}}$, $n_{{20}}$, $n_{{12}}$, $n_{{21}}$, and $n_{{22}}$ are positive integers. An example of such an object consists of the four points, six lines, and four planes which comprise the vertices, edges, and faces of a tetrahedron. Other than mentioning that there exists such a generalization, we shall not pursue this topic further here, but shall confine our attention to configurations consisting only of points and lines in this article.
2 Determination
2.1 Introduction
Having described the general theory of line configurations, we now turn our attention to the determination of configurations. Following the methodology described above, we will proceed in two steps, first determining abstract configurations, then studying their embeddings in projective spaces.
2.2 Restrictions on $p$, $\ell$, $\pi$, $\lambda$
We will begin by deriving some conditions which limit the possible values of $p$, $\ell$, $\pi$, $\lambda$ which can occur for a line configuration. Already, we have noted one such restriction above, namely $p\cdot\lambda=\pi\cdot\ell$.
Because there must be at least as many points as there are points on any line, we must have $p\geq\pi$. Likewise, because there must be at least as many lines as pass through any point, we must also have $\ell\geq\lambda$.
Let $P$ be any point of the configuration. Then there will be $\lambda$ lines passing through $P$, each of which will pass through $\lambda1$ points in addition to $P$. Since a line is determined by two points and all $\lambda$ lines have $P$ in common, no two of them will have any other point in common, hence there will be $\lambda(\pi1)$ distinct points located on these lines. Thus, we conclude that $p\geq\lambda(\pi1)$. By interchanging “point” and “line” in the argument just given, we conclude that $\ell\geq\pi(\lambda1)$.
Suppose that $\pi\geq 2$. Since at most one line of the configuration can pass through two points but every line of the configuration must pass through at least two points of the configuration, there can be no more lines in the configuration than there are pairs of points, so $\ell\leq{p\choose 2}$. Dually, we must have $p\leq{\ell\choose 2}$.
As an illustration of these conditions, we will now ask what limitations they impose on the types of configurations which have 12 points. From the inequality $p\geq\pi$, we see that $\pi$ is limited to the values 1,2,3,4,5,6,7,8,9,10,11,12.
If $\pi=1$, then we have $\ell=12\lambda$, hence the possible types are $(12_{\lambda}12\lambda_{1})$. All the other inequalities are satisfied or irrelevant and, as we shall see, for every choice of $\lambda$, there is a configuration of this type.
Likewise, if $\lambda=1$, then we have $\ell\pi=12$. Thus, we have the possibilities $(12_{1}12_{1})$, $(12_{1}6_{2})$, $(12_{1}4_{3})$, $(12_{1}3_{4})$, $(12_{1}2_{6})$, and $(12_{1}1_{1}2)$, all of which satisfy the remaining inequalities and which happen to occur as types of configurations.
We now focus our attention to the cases where $\pi\geq 2$ and $\lambda\geq 2$. Then we have the inequalities $\ell\leq{p\choose 2}=66$ and ${\ell\choose 2}\geq p=12$ to reckon with. These limit $\ell$ to the values $6\leq\ell\leq 66$. When $\lambda\geq 2$, the inequality $p\geq\lambda(\pi1)$ implies that $7\geq\pi$, so it turns out that the possible values 8,9,10,11,12 mentioned above are ruled out. Summarrizing, in this case we have the following restrictions on the ranges of our constants:
$\displaystyle 6\leq$  $\displaystyle\ell\leq 66$  
$\displaystyle 2\leq$  $\displaystyle\lambda\leq\ell$  
$\displaystyle 2\leq$  $\displaystyle\pi\leq 7$ 
To finish, we will consider the remaining possible values of $\pi$ one at a time. When $\pi=2$, we have $\ell=6\lambda$. Thus, the inequality $\ell\geq\pi(\lambda1)$ is automatically satisfied and the inequality $p\geq\lambda(\pi1)$ reduces to $\lambda\leq 12$. We have the following possibilities:
$\begin{matrix}(12_{2}12_{2})&(12_{3}18_{2})&(12_{4}24_{2})&(12_{5}30_{2})&(12_% {6}36_{2})&(12_{7}42_{2})\\ (12_{8}48_{2})&(12_{9}54_{2})&(12_{{10}}60_{2})&(12_{{11}}66_{2})\end{matrix}$ 
When $\pi=3$, we have $\ell=4\lambda$. Again, the inequality $\ell\geq\pi(\lambda1)$ is automatically satisfied. The inequality $p\geq\lambda(\pi1)$ reduces to $6\geq\lambda$, hence we have the following possibilities:
$\begin{matrix}(12_{2}8_{3})&(12_{3}12_{3})&(12_{4}16_{3})&(12_{5}20_{3})&(12_{% 6}24_{3})\end{matrix}$ 
When $\pi=4$, we have $\ell=3\lambda$. Then, the inequality $p\geq\lambda(\pi1)$ becomes $4\geq\lambda$ and the inequality $\ell\geq\pi(\lambda1)$ also becomes $4\geq\lambda$, hence we have the following possibilities:
$\begin{matrix}(12_{2}6_{4})&(12_{3}9_{4})&(12_{4}12_{4})\end{matrix}$ 
When $\pi=5$, we have $12\lambda=5\ell$. Since 5 and 12 are coprime, this implies that $\lambda=5n$ and $\ell=12n$ for some positive integer $n$. But then, the inequality $p\geq\lambda(\pi1)$ would become $12\geq 20n$, which is impossible because $n>0$, so we have no configurations with $p=12$ and $\pi=5$.
When $\pi=6$, we have $\ell=2\lambda$. Then, the inequality $p\geq\lambda(\pi1)$ becomes $12\geq 5\lambda$, so only $\lambda=2$ is possible. Thus, we must also have $\ell=4$. However, the inequality $\ell\geq\pi(\lambda1)$ is not satisfied when $\ell=4$, $\pi=6$, and $\lambda=2$, so we have no configurations with $p=12$ and $\pi=6$.
When $\pi=7$, we have $12\lambda=7\ell$. Since 7 and 12 are coprime, this implies that $\lambda=7n$ and $\ell=12n$ for some positive integer $n$. But then, the inequality $p\geq\lambda(\pi1)$ would become $12\geq 42n$, which is impossible because $n>0$, so we have no configurations with $p=12$ and $\pi=7$.
2.3 The Cases $\lambda=1$ and $\pi=1$
Having deduced and illustrated limitations on the four constants $p$, $\ell$, $\pi$, $\lambda$, we now turn our attention to the determining which sets of numbers describe actual configurations. We begin with the easy case where one or both of $\lambda$ and $\pi$ equals $1$.
Before proceeding further, it is worth pointing out that we are only interested in classifying abstract configurations up to equivalence by permutation. That is to say, we will consider two abstract configurations $\langle P,L,I\rangle$ and $\langle P^{{\prime}},L^{{\prime}},I^{{\prime}}\rangle$ equivalent if there exist onetoone correspondences $f\colon P\to P^{{\prime}}$ and $g\colon L\to L^{{\prime}}$ such that $I(x,y)$ if and only if $I^{{\prime}}(f(x),g(y))$. The reason for doing this is that, since it is easy enough to permute elements in a given configuration, listing only one configuration out of an equivalence class cuts down on the number.
Suppose that we have an abstract configuration with $\pi=1$. Then, to every element of $L$ we may associate exactly one element of $P$. Furthermore, we may define an equivalence relation on $L$ by equating the lines which pass through the same point. Thus, our lines are partitioned into $p$ partitions of $\lambda$ points each. Conversely, given two numbers $\lambda$ and $p$ and setting $\ell=p\lambda$, we can make a configuration by taking a set $L$ with $\ell$ elements and partitioning it into $p$ subsets of $\lambda$ elements each, then associating to each equivalence class an element of the set $P$.
3 Catalogue
[Under Construction]
Mathematics Subject Classification
14N20 no label found14N10 no label found05B99 no label found52C30 no label found51N15 no label found51E20 no label found51A45 no label found51A05 no label found51A20 no label found Forums
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