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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