§ Proof of projective duality

In projective geometry, we can interchange any statement with "points" and "lines" and continue to get a true statement. For example, we have the dual statements:
  • Two non-equal lines intersect in a unique point (including a point at infintyfor a parallel line).
  • Two non-equal points define a unique line.
The proof of the duality principle is simple. Recall that any point in projective geometry is of the from [a:b:c](b/a,c/a)[a : b : c] \simeq (b/a, c/a). A projective equation is of the form px+qy+rz=0px + qy + rz = 0 for coefficients p,q,rCp, q, r \in \mathbb C.
  • if we have a fixed point [a:b:c][a : b : c], we can trade this to get aline ax+by+cz=0ax + by + cz = 0.
  • If we have a line ax+by+cz=0ax + by + cz = 0, we can trade this to get a point [a:b:c][a:b:c].
  • The reason we need projectivity is because this correspondence is only welldefined upto scaling: the line x+2y+3x + 2y + 3 is the same as the line 2x+3y+62x + 3y + 6.
  • In using our dictionary, we would get [1:2:3][1:2:3], [2:4:6][2:4:6]. Luckily for us,projectivity, these two points are the same! (2/1,3/1)=(4/2,6/2)(2/1, 3/1) = (4/2, 6/2).
  • The "projective" condition allows us to set points and lines on equal footing:lines can be scaled, as can points in this setting.