Configuration

Configuration is a plane consisting finite points and finite lines. It is described by an incidence table in which points are listed down to left and lines are across the top in the table.

Example 1

A triangle is a configuration with a total of three points and three lines.

Triangle A B C a x x b x x c x x

The triangle is self-dual.

Example 2

A complete four point is a configuration with a total of four points and six lines.

Complete four point \(L_1\) \(L_1'\) \(L_2\) \(L_2'\) \(L_3\) \(L_3'\) \(a_0\) x x x \(a_1\) x x x \(a_2\) x x x \(a_3\) x x x

The complete four point is not self-dual.

Example 3

A fano-configuration is a configuration with a total of seven points and seven lines.

Fano \(L_1\) \(L_1'\) \(L_2\) \(L_2'\) \(L_3\) \(L_3'\) \(D\) \(a_0\) x x x \(a_1\) x x x \(a_2\) x x x \(a_3\) x x x \(d_1\) x x x \(d_2\) x x x \(d_3\) x x x

The fano-configuration is self-dual.

Example 4

Pappus configuration is a configuration with a total of nine points and nine points.

Papus \(L\) \(L'\) \(A\) \(B\) \(C\) \(A'\) \(B'\) \(C'\) \(P\) \(a\) x x x \(b\) x x x \(c\) x x x \(a'\) x x x \(b'\) x x x \(c'\) x x x \(a''\) x x x \(b''\) x x x \(c''\) x x x

The Pappus configuration is self-dual.

Example 5

A Desargues configuration is a configuration with a total of
ten points each on three lines and
ten lines each on three point.
The incidence of points and lines on Desargues configuration is given below with
A=(b,c,a''),..., A'=(b',c',a'')...,A''=(a,a',p) so on and L=(a'',b'',c'')

Desargues \(A''\) \(B''\) \(C''\) \(A\) \(B\) \(C\) \(A'\) \(B'\) \(C'\) \(L\) \(a\) x x x \(b\) x x x \(c\) x x x \(a'\) x x x \(b'\) x x x \(c'\) x x x \(a''\) x x x \(b''\) x x x \(c''\) x x x \(p\) x x x

The Desargues configuration is self-dual.

Tactical configuration

Let \( \sigma\) be a configuration and \( r,s \in \mathbb{z}^+\) then \(\sigma \) is called tactical configuration if and only if
each point is exactly on r lines and
each line is exactly on s points.
If such points and lines are respectively m and n in number, we denote the tactical configuration by
\( (m_r, n_s)\)
Such four numbers r,s,m,n in tactical configuration are independent but evidently satisfy the equation
mr=ns
Also the dual of
\((m_r, n_s)\) is \((n_s,m_r)\)
For instance dual of
\( (4_3,6_2)\) is \( (6_2,4_3)\)

Some examples of tactical configuration are given below.

1. Triangle is a tactical configuration with \((3_2,3_2)=(3_2)\)
2. Complete four point is tactical configuration with \((4_3,6_2)\)
3. Fano-configuration is tactical configuration:\((7_3,7_3)=(7_3)\)
4. Pappus configuration is tactical with \((9_3,9_3)=(9_3)\)
5. Desargues configuration is tactical with \((10_3,10_3)=(10_3)\)

The tactical configuration of points and lines satisfies:

1. The number of points is finite.
2. The number of lines is finite.
3. Each point is on the same number of lines (2 or greater)
4. Each line is on the same number of points (2 or greater)
5. Each pair of distinct points is on at most one line.
6. Each pair of distinct lines is on at most one point.
7. Not all points are on the same line.
8. There exists at least one line.

If \( \pi \) is tactical configuration with \( (m_n):m=n^2-n+1,n \geq 3\) then show that \( \pi \) is projective plane.

Given \( \pi \) is tactical configuration with \((m_n):m=n^2-n+1,n \geq 3\), we show P5 to P1.

1. P5: By definition of tactical configuration, there are m lines
   where \(m=n^2-n+1\) which is greater than 1
   This is sufficient to claim that there exist line.
2. P4: Let L is arbitrary line in \( \pi \) then there are exactly n-points on it.
   Also there are total of \(n^2-n+1\) points, which is greater than n
   This is sufficient to claim that there is at least one point not on L.
3. P3: Let L is arbitrary line in \( \pi \) then there are exactly n-points on it.
   Since \( n \geq 3 \)
   This is sufficient to claim that there are at least three points on L.
4. P2: Let L be arbitrary line in \( \pi \) then L lies exactly on n points and also each of these point are exactly on n lines, thus total number of lines including L is
   \( n(n-1)+1\) lines
   \(n^2-n+1\) lines
   \(m \) lines
   Since we have chosen arbitrary line L and shown that every remaining lines are incident to L, this sufficient to claim that two lines always meet.
5. P1: Let p be arbitrary point in \( \pi \) then p lies exactly on n lines and also each of these lines are exactly on n points, thus total number of points including p is
   \(n(n-1)+1\) points
   \(n^2-n+1\) points
   \(m\) points
   Since we have chosen arbitrary point p and shown that every remaining point are joined to p, this sufficient to claim that two points determine a line.

Thus \( \pi \) is projective plane. This completes the proof.

A finite projective plane is a tactical configuration and for some \(n\geq 3\), it has form \((m_n)\) with \(m=n^2-n+1\) .

Let \( \pi \) be a finite projective plane then
By P5- there is a line, say L
Being finite projective plane, L contain finite number of points, say n
By P3- \(n\geq3\)
Now we show

1. There are exactly m points for \( m=n^2-n+1\)
2. There are exactly m lines for \(m=n^2-n+1\)
3. Every point are exactly on \(n\) lines
4. Every line are exactly on \(n\) points

The proof are as follows

1. Claim 4 : Every line is exactly on n points
   We assumed that line L has n points, say \(p_1,p_2,p_3,p_4 \ldots, p_n\)
   Let M be arbitrary line in \( \pi \) other than L, then
   By P2- there is a point on both L and M, say p_1
   By P3- the line M has second point, say q
   By P1,P3- the line \({qp}_n\) has third point, say r
   By P2- each line \({rp}_i\) must meet M at n-2 distinct points other than \(p_1\) and \(p_n\), say \(q_i;i=2,, \ldots,n-1\).
   Hence, total number of points on M including \(p_1\) and \(p_n\) is
   \((n-2)+2=n\) points
   The possibility of some more point on M contradict that line L contains exactly n points.
2. Claim 2: Every point is exactly on n lines.
   Let p be arbitrary point in \( \pi \).
   By dual of P2- there is a line not on p, say L.
   Since the line L has n points, say \(q_1,q_2,q_3,q_4 \ldots,q_n\)
   By P2- each line \({pq}_i\) incident at p for \(i=1,\ 2,, \ldots,n\).
   The possibility of some more line on p contradict that L contains
   exactly n points.
   Hence p is exactly on n lines
3. Claim 3:There are exactly m points for \(m=n^2-n+1\)
   Let p be arbitrary point in \( \pi \) then p lies exactly on n lines and also each of these lines are exactly on n points, thus total number of points including p is
   \(n(n-1)+1\) points
   \(n^2-n+1\) points
   \(m\) points
4. Claim 4: There are exactly m lines for \(m=n^2-n+1\)
   Let L be arbitrary line in \( \pi \) then L lies exactly on n points and also each of these point are exactly on n lines, thus total number of lines including L is
   \( n(n-1)+1\) lines
   \(n^2-n+1\) lines
   \(m \) lines

This completes the proof

If \( \alpha\) is tactical configuration with then show that is affine plane.

Given is tactical configuration, we show A5 to A1.

1. A5: By definition of tactical configuration, there are lines
   which is greater than 1
   This sufficient to claim that there exists lines.
2. A4: Let L is arbitrary line in \( \alpha\) then there are exactly n-points on it.
   Also there are total of \(n^2\) points, which is greater than n.
   This sufficient to claim that there is at least one point not on L
3. A3: Let L is arbitrary line in \( \alpha\) then there are exactly n-points on it.
   Since \(n \ge 2\)
   This is sufficient to claim that there are at least two points on L.
4. A2: Let L be arbitrary line in \( \alpha\) and p be a point not on it.
   Then p lies exactly on n+1 lines, say \(M_1, M_2,..., M_n , M_{n+1}\)
   Since L contains only n points, and every two points determine a line,
   There is exactly one line on p that can not meet L.
   This sufficient to claim that, there is exactly one line on p parallel to L.
5. A1: Let p be arbitrary point in \( \alpha\) then p lies exactly on n+1 lines and also each of these lines are exactly on n points, thus total number of points including p is
   (n-1)(n+1)+1 points
   \(n^2-1+1 \) points
   \(n^2\) points
   Since we have chosen arbitrary point and shown that every remaining points are joined to , this sufficient to claim that two points determine a line.

Thus \( \alpha \) is affine plane.

A finite affine plane is a tactical configuration and for some \(n \ge 2\) , it has form \( \left (n^2_{n+1},n^2+n_n \right )\).

Let \( \alpha\) be a finite affine plane then
By A5- there is a line, say L.
Being finite affine plane, L contain finite number of points, say n
By A3-\(n \ge 2\)
Now we show

1. There are exactly \(n^2\) points
2. There are exactl \(n^2+n\) lines
3. Every point is exactly on \(n+1\) lines
4. Every line is exactly on\(n\) points

The proof are as below

1. Claim 4 : Every line is exactly on n points
   We assumed that line L has n points, say \(p_1,p_2,...,p_n\)
   Let M be arbitrary line other than L, then
   Case1: if L and M are intersecting lines, without loss of generality, we assume that \(p_1\) on both L and M.
   By A3- there is second point on M, say q.
   By A1- there is a line \(qp_n\), say N.
   By A2- there is exactly one line on each \(p_i\) ; i=2,3,...,n-1, parallel to N.
   All of these lines must intersect M at n-2 different points other than \(p_1\) and q. [otherwise it contradict A2]
   Hence, total number of points on M including \(p_1\) and q is (n-2)+2
   n points
   
   Case2- if L and M are parallel lines
   By A3- there is second point on M, say q.
   By A1- there is a line \(qp_1\), say N.
   By A2- there is exactly one line on each \(p_i;i=2,...,n\) parallel to N.
   All of these lines must intersect M at different points other than q. Otherwise it contradicts A2
   Hence, total number of points on M including q is
   (n-1)+1=n points
2. Claim 3: Every point is exactly on lines.
   Let p be arbitrary point, then
   By A5- there is a line, say L.
   Case 1
   
   If p is not on L
   By A1- there are n distinct lines on p corresponding to n points on L
   By A2- there is one more line on p that is parallel to L
   Hence p is exactly on (n+1) lines.
   Case 2
   
   If p is on L, without loss of generality we assume that \(p_1\)=p.
   By A4- there is a point not on L, say q
   By A3- there is second point on L, without loss of generality say r
   By A1- there is a line qr with n points on it.
   By A1- there are n lines on p corresponding to each n-points on qr
   By A2- there is one more line on p parallel to qr
   Hence p is exactly on (n+1)lines.
3. Claim 3: There are exactly points.
   Let p be arbitrary point in then p lies exactly on lines and also each of these lines are exactly on n points, thus total number of points on p is
   points
   points
   points
4. Claim 4: There are exactly lines
   Let L be arbitrary line in then
   Since there are n-points on L and also each of these points are exactly on lines, there are
   lines including L it self.
   Also, By A4- there is a point p not on L.
   And, By A2- there is a line on p parallel to L, say M.
   Now we choose a line, say N, joining points p and q, where q is on L.
   Then By A2, there are lines corresponding to each of the remaining n-1 points other than p on M that are parallel to L.
   Thus there are total of
   lines
   lines

This completes the solution.