Duality (projective geometry)

Duality in the projective plane refers to the interchangeability between points and lines which preserves incidence properties.

Notice that both points and lines can be represented (on a plane) by means of ordered pairs. A point is represented by the ordered pair (x,y), where x is the abscissa and y is the ordinate, which together are coordinates of the point. A line can likewise be represented by an ordered pair (m,b) where m is the slope and b is the y-intercept.

Given three points

$P_{1}:(x_{1},y_{1}),$

$P_{2}:(x_{2},y_{2}),$

$P_{3}:(x_{3},y_{3});$

these three points are collinear iff their coordinates satisfy the equation

${y_{2}-y_{1} \over x_{2}-x_{1}}={y_{3}-y_{2} \over x_{3}-x_{2}}={y_{3}-y_{1} \over x_{3}-x_{1}}\qquad \qquad (1)$.

Likewise, given three lines

$L_{1}:(m_{1},b_{1}),$

$L_{2}:(m_{2},b_{2}),$

$L_{3}:(m_{3},b_{3});$

one can verify that these three lines are concurrent iff their parameters satisfy the equation

${b_{2}-b_{1} \over m_{2}-m_{1}}={b_{3}-b_{2} \over m_{3}-m_{2}}={b_{3}-b_{1} \over m_{3}-m_{1}}.\qquad \qquad (2)$

Equations (1) and (2) are equivalent to each other up to an exchange of x with m and y with b. Therefore there exists is a way to exchange lines with points in such a way that concurrency is exchanged with collinearity.

It is possible to distinguish lines from points by conjugating ordered pairs. That is, let line (m,b) be represented instead by its conjugate $(m,-b)^{\star }.$ Then it can be verified that the intersection L₁.L₂ of a pair of lines L₁ and L₂ is

$(m_{1},b_{1})^{\star }.(m_{2},b_{2})^{\star }=\left({b_{2}-b_{1} \over m_{2}-m_{1}},{m_{1}b_{2}-m_{2}b_{1} \over m_{2}-m_{1}}\right)\qquad \qquad (3)$

where b₁ and b₂ are negative y-intercepts. Also, the common line P₁.P₂ passing through a pair of points P₁ and P₂ is

$(x_{1},y_{1}).(x_{2},y_{2})=\left({y_{2}-y_{1} \over x_{2}-x_{1}},{x_{1}y_{2}-x_{2}y_{1} \over x_{2}-x_{1}}\right)^{\star }.\qquad \qquad (4)$

Equation (4) can be seen to be the same as equation (3), after exchanging m with x and b with y, and applying the following rules of conjugation:

$A^{\star \star }=A,\qquad \qquad (5)$

$A=B\rightarrow A^{\star }=B^{\star },\qquad \qquad (6)$

$(A.B)^{\star }=A^{\star }.B^{\star }=B^{\star }.A^{\star }.\qquad \qquad (7)$

Indeed, if equation (3) is represented as

$A^{\star }.B^{\star }=C$

then applying rule (6) yields

$(A^{\star }.B^{\star })^{\star }=C^{\star }.$

Applying rule (7) then yields

$A^{\star \star }.B^{\star \star }=C^{\star }$

and applying rule (5) finally yields

$A.B=C^{\star }.$