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.

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Duality

Duality is a transformation that maps lines and points into points and lines, respectively, while preserving certain geometric properties. It involves interchanging the roles of points and lines in an incidence structure.

Let \( \sigma = (\mathscr{P}, \mathscr{L}, \mathcal{I}) \) be an incidence structure. The dual of \( \sigma \), denoted by \( \sigma^d \), is defined as:

\( \sigma^d = (\mathscr{L}, \mathscr{P}, \mathcal{I}^{-1}) \)

where points of \( \sigma^d \) are the lines of \( \sigma \), and the lines of \( \sigma^d \) are the points of \( \sigma \).

Points are collinear

The dual statement of "Points are collinear" is "Lines are concurrent"

Lines are concurrent

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Principle of Duality

In projective geometry, any true statement expressed in terms of points, lines, and incidence has a dual statement obtained by interchanging the words “point” and “line”. If the original statement is valid, then so is its dual — without requiring a separate proof. This symmetry is known as the Principle of Duality.

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Show that the principle of duality holds in the class of projective planes.

Let \( \pi = (\mathscr{P}, \mathscr{L}, \mathcal{I}) \) be a projective plane, and let \( \pi^d \) be its dual, where:

1. \(P1^d\):If \(L\) and \(M\) are two distinct lines, there is exactly one point incident with both.
2. \(P2^d\): If \(p\) and \(q\) are two distinct points, there is at least one line incident with both.
3. \(P3^d\):Through any point \(p\), there are at least three lines.
4. \(P4^d\):For any point \(p\), there exists at least one line not passing through \(p\).
5. \(P5^d\): There exists at least one point.

We now verify that \( \pi^d \) satisfies the axioms of a projective plane.

1. \(P_1\): Two points in \( \pi^d \) determine a unique line.
   By \(P2^d\), two points have at least one common line. Suppose two lines pass through them — this would contradict \(P1^d\). Hence, exactly one line passes through two points. ✓
2. \(P_2\): Any two lines meet in at least one point.
   By \(P1^d\), two lines intersect in exactly one point. ✓
3. \(P_5\): There exists at least one line.
   From \(P5^d\), there is a point \(p\); from \(P4^d\), there’s a line not through \(p\). So a line exists. ✓
4. \(P_4\): For any line \(L\), there exists a point not on it.
   By \(P_5^d\), there exist a point, say \(p\).
   Case 1:
   If \(p \notin L\), we are done.
   Case 2:
   If \(p \in L\), the we use \(P3^d\) to get another line \(N\) through \(p\), and \(P4^d\) to get line \(M\) not through \(p\).
   Then intersection \(q = M \cap N\) is not on \(L\). ✓
For any line \(L\), there exists a point \(q\) not on it
5. P3: Every line contains at least three points.
   Let \(L\) be any line.
   By \(P4\), there exists a point \(p \notin L\).
   By \(P3^d\), three lines \(A, B, C\) pass through \(p\).
   Each intersects \(L\) at a distinct point: \(A \cap L\), \(B \cap L\), \(C \cap L\).
   Thus, \(L\) has at least three points. ✓

Hence, \( \pi^d \) satisfies all axioms of a projective plane.

NOTE

Therefore, the dual of a projective plane is also a projective plane. This proves the Principle of Duality

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Isomorphism

[isomorphism ! definition]

Let \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two planes then \( \sigma \) and \( \sigma '\) are isomorphic if there exist a bijections
\( f:\mathscr{P} \to \mathscr{P}', F:\mathscr{L} \to \mathscr{L}'\) provided \( (p,L) \in \mathscr{I} \) iff \( (p',L') \in \mathscr{I'} \) .
Isomorphic planes \( \sigma \) and \( \sigma '\) are denoted by \( \sigma \sim \sigma '\).

Note:
Let \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two isomorphic planes, then

• \( f:\mathscr{P} \to \mathscr{P}'\) is bijection
• \( F:\mathscr{L} \to \mathscr{L}'\) is bijection
• \( f \) preserves the collinearity

Theorem: If \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two planes and \( f:\mathscr{P} \to \mathscr{P}'\) is bijection such that points \( a_1,a_2,a_3, \cdots \in P \) are collinear if and only if \( f(a_1), f(a_2), f(a_3),\cdots \in P'\) are collinear then \( \sigma \sim \sigma '\).

[isomorphism ! theorem]

Proof
Given, \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two planes with a bijection \( f:\mathscr{P} \to \mathscr{P}'\) and preserves collinearity.
Now, we show \( F:\mathscr{L} \to \mathscr{L}'\) is a bijection.

• F is well defined and one-one
  Let \(L_1 \in \mathscr{L}\), and \( L_2 \in \mathscr{L'}\), then then \( x_1,y_1 \in L_1 \) and \( x_2,y_2 \in L_2\).
  Suppose, \(L_1=L_2 \), then, \(x_1,y_1, x_2,y_2 \) are collinear. This implies that
  \(f(x_1),f(y_1),f(x_2),f(y_2)\) are collinear, so \(F(L_1)=F(L_2)\).
  Hence, \( F:\mathscr{L} \to \mathscr{L}'\) is well defined and one-one.
• F is onto
  For each \( L' \in \mathscr{L'} \) there exists \( x',y' \in L' \) such that
  \(f(x)=x', f(y)=y'\)
  Since, \(f(x),f(y)\) are collinear, therefore \(x,y\) are collinear
  Thus, there exists a line \( L \in \mathscr{L}\) such that \(F(L)=L'\), where \(L=xy\)
  Hence, \(F\) is onto

This completes the proof.

© Geometry Notes. All rights reserved.

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Real Affine and Projective Plane

Real Affine Plane

Definition: Real affine plane

The Real Affine Plane is an incidence structure satisfying the following axioms:

1. Point is an ordered pair \((x,y)\) where \(x, y \in \mathbb{R}\).
2. Line is the set of points \((x,y)\) satisfying \(ax + by + c = 0\), where \(a, b, c \in \mathbb{R}\), and not both \(a\) and \(b\) are zero.
3. A point \((x_0, y_0)\) lies on the line \(ax + by + c = 0\) if and only if \(ax_0 + by_0 + c = 0\).

We denote the Real Affine Plane (also known as the Euclidean Plane) by \(\alpha_R\).

Theorem: Show that \(\alpha_R\) is an affine plane.

Proof: We verify the axioms:

1. Two points determine a unique line:
   Let \((x_1, y_1)\), \((x_2, y_2)\) be two points. The line through them is: \[ (y_2 - y_1)x + (x_1 - x_2)y + (x_2 y_1 - x_1 y_2) = 0 \quad \text{(A)} \] Suppose another line \(ax + by + c = 0\) passes through both. Then: \[ ax_1 + by_1 + c = 0, \quad ax_2 + by_2 + c = 0 \] Solving gives \(a\) and \(b\) in terms of \(c\), leading back to (A). So the line is unique.
2. Parallel line through a point:
   Given line \(L: ax + by + c = 0\) and point \(P = (x_0, y_0)\) not on \(L\), the line: \[ ax + by = ax_0 + by_0 \] passes through \(P\) and is parallel to \(L\). Uniqueness follows from slope and constant comparison.
3. Three non-collinear points exist:
   \((0,0)\), \((1,0)\), \((0,1)\) are non-collinear.

Hence, \(\alpha_R\) is an affine plane. \(\blacksquare\)

Real Projective Plane

Definition: Real projective plane

In the projective plane, points are homogeneous coordinates: \([x_1, x_2, x_3]\), where scaling by \(k \ne 0\) gives the same point: \([x_1,x_2,x_3] = [kx_1,kx_2,kx_3]\).

If \(x_3 \ne 0\), we identify \([x_1,x_2,x_3]\) with \((x_1/x_3, x_2/x_3)\) in the affine plane.

Example: \([1,2,3] = [2,4,6]\), but as triples they are different.

Definition: The Real Projective Plane \(\pi_R\) is an incidence structure with:

1. Point: Proportionality class \([x_1,x_2,x_3]\), not all zero.
2. Line: Proportionality class \(\langle l_1,l_2,l_3 \rangle\), not all zero.
3. Incidence: \([x_i]\) lies on \(\langle l_j \rangle\) iff \(l_1x_1 + l_2x_2 + l_3x_3 = 0\).

Theorem: \(\pi_R\) is a projective plane.

Proof: We show:

1. Two points in \(\pi_R\) determine a line:
   Let \(x=[x_1,x_2,x_3]\) and \(y=[y_1,y_2,y_3]\) be two points, then \(\begin{vmatrix}x_1 & x_2\\y_1 & y_2\end{vmatrix} \neq 0\).
   Now, there exists a line: \[ \left\langle \begin{vmatrix}x_2 & x_3\\y_2 & y_3\end{vmatrix}, \begin{vmatrix}x_3 & x_1\\y_3 & y_1\end{vmatrix}, \begin{vmatrix}x_1 & x_2\\y_1 & y_2\end{vmatrix} \right\rangle \tag{A} \] Suppose there exists another line \(\langle l_1,l_2,l_3\rangle\) (i) passing through \(x\) and \(y\), then \[ l_1 x_1 + l_2 x_2 + l_3 x_3 = 0 \quad (ii), \quad l_1 y_1 + l_2 y_2 + l_3 y_3 = 0 \quad (iii) \] Solving (ii) and (iii), we get \[ l_1 = \frac{\begin{vmatrix}x_2 & x_3\\y_2 & y_3\end{vmatrix}}{\begin{vmatrix}x_1 & x_2\\y_1 & y_2\end{vmatrix}} l_3, \quad l_2 = \frac{\begin{vmatrix}x_3 & x_1\\y_3 & y_1\end{vmatrix}}{\begin{vmatrix}x_1 & x_2\\y_1 & y_2\end{vmatrix}} l_3 \] Hence, (i) and (A) are the same. Therefore, two points in \(\pi_R\) determine a line.
2. Two lines always meet:
   Let \(L=\langle l_1,l_2,l_3\rangle\) and \(M=\langle m_1,m_2,m_3\rangle\) be two lines. Then \[ \left[\begin{vmatrix}l_2 & l_3\\m_2 & m_3\end{vmatrix}, \begin{vmatrix}l_3 & l_1\\m_3 & m_1\end{vmatrix}, \begin{vmatrix}l_1 & l_2\\m_1 & m_2\end{vmatrix}\right] \] is a point on both.
3. There is a four-point:
   By inspection, \([1,1,1],[1,0,0],[0,1,0],[0,0,1]\) is a four-point.

Hence \(\pi_R\) is a projective plane. \(\blacksquare\)

Summary: From Affine to Projective

• \((x,y) \in \alpha_R \mapsto [x,y,1] \in \pi_R\)
• Line \(ax+by+c=0 \mapsto \langle a,b,c \rangle\)
• Points \([a,b,0]\) are points at infinity
• Line \(\langle 0,0,1 \rangle\) is the line at infinity

Important: These results connect algebra and geometry. Determinants characterize collinearity and concurrency.

Theorem: Points \([x_1,x_2,x_3]\), \([y_1,y_2,y_3]\), \([z_1,z_2,z_3]\) are collinear iff \[ \begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3 \end{vmatrix} = 0 \]

Proof: The point \(z\) lies on the line through \(x\) and \(y\) iff the incidence condition holds, which expands to the determinant being zero. \(\blacksquare\)

Corollaries

1. Lines \(\langle l_i \rangle\), \(\langle m_i \rangle\), \(\langle n_i \rangle\) concurrent iff determinant of coefficients is zero.
2. If \(x, y\) lie on a line, then \(z\) lies on it iff \(z_i = \lambda x_i + \mu y_i\) for some \(\lambda,\mu\).
3. Three points collinear iff there exist \(\alpha,\beta,\gamma\) not all zero such that \(\alpha x_i + \beta y_i + \gamma z_i = 0\).

Projective Plane over a Division Ring

Let \(D\) be a division ring. Define:

• \(\mathscr{P}_D = \{[x_1,x_2,x_3] \mid (x_i) \in D^3 \setminus \{0\}\}\)
• \(\mathscr{L}_D = \{\langle l_1,l_2,l_3 \rangle \mid (l_i) \in D^3 \setminus \{0\}\}\)
• Incidence: \(l_1x_1 + l_2x_2 + l_3x_3 = 0\)

Theorem: Prove that \(\pi_D=(\mathscr{P,L,I}) \) is projective plane

Proof:

Let \(D\) be a division ring and \(T = D^3 \setminus \{(0,0,0)\}\) be the set of all non-zero triples of \(D\). Now we show:

1. We define \(\mathscr{P}_D= \{[x_1,x_2,x_3] : (x_1,x_2,x_3) \in T \}\), where each point is a proportionality class of triples, not all zero.
2. We define \(\mathscr{L}_D= \{ \langle l_1,l_2,l_3\rangle : (l_1,l_2,l_3) \in T \}\), where each line is a proportionality class of triples, not all zero.
3. We define that a point \([x_1,x_2,x_3]\) lies on a line \(\langle l_1,l_2,l_3 \rangle\) iff \(l_1 x_1 + l_2 x_2 + l_3 x_3 = 0\).

Here, \(\pi_D=( \mathscr{P}_D, \mathscr{L}_D, I)\) is an incidence structure. Using Theorem \ref{pitheorem}, we conclude that \(\pi_D\) is a projective plane. Hence the theorem. \(\blacksquare\)

Theorem. If \(F\) is a field, then \(\pi_F=(\mathscr{P,L,I})\) is a projective plane.
Proof. The proof of the theorem is similar to the theorem for \(\pi_D\), so left for the reader.

Important Note.
If \(q\) is a power of a prime, then there exists a projective plane of order \(q\). If \(q\) is congruent to 1 or 2 modulo 4 and is not the sum of the squares of two integers, then there exists no projective plane of order \(q\).

Isomorphism of Planes

Planes \(\sigma = (\mathscr{P},\mathscr{L},\mathscr{I})\) and \(\sigma'\) are isomorphic (\(\sigma \sim \sigma'\)) if there are bijections: \[ f: \mathscr{P} \to \mathscr{P'}, \quad F: \mathscr{L} \to \mathscr{L'} \] preserving incidence: \((P,L) \in \mathscr{I} \iff (f(P), F(L)) \in \mathscr{I'}\).

Theorem: If \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two planes and \( f:\mathscr{P} \to \mathscr{P}'\) is bijection such that points \( a_1,a_2,a_3, \cdots \in P \) are collinear if and only if \( f(a_1), f(a_2), f(a_3),\cdots \in P'\) are collinear then \( \sigma \sim \sigma '\).

Proof: Given, \( \sigma=(\mathscr{P,L,I}) \) and \( \sigma '=(\mathscr{P',L',I'}) \) are two planes with a bijection \( f:\mathscr{P} \to \mathscr{P}'\) and preserves collinearity. Now, we show \( F:\mathscr{L} \to \mathscr{L}'\) is a bijection.

1. \(F\) is well defined and one-one:
   Let \(L_1 \in \mathscr{L}\), and \( L_2 \in \mathscr{L'}\), then \( x_1,y_1 \in L_1 \) and \( x_2,y_2 \in L_2\). Suppose \(L_1=L_2 \), then \(x_1,y_1, x_2,y_2 \) are collinear. This implies that \(f(x_1),f(y_1),f(x_2),f(y_2)\) are collinear, so \(F(L_1)=F(L_2)\). Hence, \( F:\mathscr{L} \to \mathscr{L}'\) is well defined and one-one.
2. \(F\) is onto:
   For each \( L' \in \mathscr{L'} \) there exists \( x',y' \in L' \) such that \(f(x)=x', f(y)=y'\). Since \(f(x),f(y)\) are collinear, therefore \(x,y\) are collinear. Thus, there exists a line \( L \in \mathscr{L}\) such that \(F(L)=L'\), where \(L=xy\). Hence, \(F\) is onto.

This completes the proof. \(\blacksquare\)

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