§ 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 infinty for 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 $[a : b : c] \simeq (b/a, c/a)$. A projective equation is of the form $px + qy + rz = 0$ for coefficients $p, q, r \in \mathbb C$.

• if we have a fixed point $[a : b : c]$, we can trade this to get a line $ax + by + cz = 0$.
• If we have a line $ax + by + cz = 0$, we can trade this to get a point $[a:b:c]$.
• The reason we need projectivity is because this correspondence is only well defined upto scaling: the line $x + 2y + 3$ is the same as the line $2x + 3y + 6$.
• In using our dictionary, we would get $[1:2:3]$, $[2:4:6]$. Luckily for us, projectivity, these two points are the same! $(2/1, 3/1) = (4/2, 6/2)$.
• The "projective" condition allows us to set points and lines on equal footing: lines can be scaled, as can points in this setting.