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Revision difference : finite plane |
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Version 9 |
| Let $\mathcal{H} = (V, \mathcal{E})$ be a linear space. A \emph{finite plane} is an intersecting linear space. That is to say, a linear space in which any two edges in $\mathcal{E}$ have a nonempty intersection. |
Let $\mathcal{H} = (V, \mathcal{E})$ be a linear space. A \emph{finite plane} is an intersecting linear space. That is to say, a linear space in which any two edges in $\mathcal{E}$ have a nonempty intersection. |
| Finite planes are rather restrictive hypergraphs, and the following holds. |
Finite planes are rather restrictive hypergraphs, and the following holds. |
| \begin{thm*} |
\begin{thm*} |
| Let $\mathcal{H} = (V, \mathcal{E})$ be a finite plane. Then for some positive integer $k$, $\mathcal{H}$ is $(k+1)-$regular, $(k+1)-$uniform, and $|\mathcal{E}| = |V| = k^2 + k + 1$. |
Let $\mathcal{H} = (V, \mathcal{E})$ be a finite plane. Then for some positive integer $k$, $\mathcal{H}$ is $(k+1)-$regular, $(k+1)-$uniform, and $|\mathcal{E}| = |V| = k^2 + k + 1$. |
| \end{thm*} |
\end{thm*} |
| The above $k$ is the \emph{\PMLinkEscape{order}} of the finite plane. It is not known in general if finite planes exist of \PMLinkEscape{order} other than $k$ a power of a prime. The terminology "finite plane" is suggestive, as we can think of the edges as a finite collection of lines in Euclidean space in general position so that they all intersect pairwise in exactly one point. The added restriction that all pairs of vertices determine an edge (i.e. "any two points determine a line"), however, makes it impossible to depict a finite plane in the Euclidean plane by means of straight lines, except for the trivial case $k = 1$. The finite plane of \PMLinkEscape{order} $2$ is known as the \emph{Fano plane}. The following is a diagrammatic representation: |
The above $k$ is the \emph{\PMLinkEscape{order}} of the finite plane. It is not known in general if finite planes exist of \PMLinkEscape{order} other than $k$ a power of a prime. The terminology "finite plane" is suggestive, as we can think of the edges as a finite collection of lines in Euclidean space in general position so that they all intersect pairwise in exactly one point. The added restriction that all pairs of vertices determine an edge (i.e. "any two points determine a line"), however, makes it impossible to depict a finite plane in the Euclidean plane by means of straight lines, except for the trivial case $k = 1$. The finite plane of \PMLinkEscape{order} $2$ is known as the \emph{Fano plane}. The following is a diagrammatic representation: |
| \begin{center} |
\begin{center} |
| \includegraphics{fano.ps} |
\includegraphics{fano.ps} |
| \end{center} |
\end{center} |
| An edge here is represented by a straight line, and the inscribed circle is also an edge. In other words, for a vertex set $\{1, 2, 3, 4, 5, 6, 7 \}$, the edges of the Fano plane are |
An edge here is represented by a straight line, and the inscribed circle is also an edge. In other words, for a vertex set $\{1, 2, 3, 4, 5, 6, 7 \}$, the edges of the Fano plane are |
| \begin{center} |
\begin{center} |
| \begin{array}{c} |
\begin{array}{c} |
| \{1, 2, 4 \} \\ |
\{1, 2, 4 \} \\ |
| \{2, 3, 5 \} \\ |
\{2, 3, 5 \} \\ |
| \{3, 4, 6 \} \\ |
\{3, 4, 6 \} \\ |
| \{4, 5, 7 \} \\ |
\{4, 5, 7 \} \\ |
| \{5, 6, 1 \} \\ |
\{5, 6, 1 \} \\ |
| \{6, 7, 2 \} \\ |
\{6, 7, 2 \} \\ |
| \{7, 1, 3 \} |
\{7, 1, 3 \} |
| \end{array} |
\end{array} |
| \end{center} |
\end{center} |
| Notice that the Fano plane is generated by the ordered |
Notice that the Fano plane is generated by the ordered triplet $(1, 2, 4)$ and adding $1$ to each entry, modulo $7$. The generating triplet has the property that the differences of any two elements, in either \PMLinkEscape{order}, are all pairwise different modulo $7$. In general, if we can find a set of $k+1$ elements of $\ZZ_{k^2 + k + 1}$ (the integers modulo $k^2 + k + 1$) with all pairwise differences distinct, then this gives a cyclic representation of the finite plane of order $k$. |
| lows because every pair of vertices is contained in precisely one edge. For a fixed edge $e$, ${e \choose 2}$ counts all the pairs of vertices in $e$, and this summed over all edges gives all the possible pairs of vertices, which is ${n \choose 2}$. The second one holds because given any vertex, this vertex forms exactly one edge with every other vertex, so $|e| - 1$ counts the number of vertices $v$ shares in vertex $e$, summed over all edges gives all the vertices except $v$. |
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