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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