duality-theoremsprojective-geometry
On duality, what would be the dual of "coplanar points"?
Let's say $p_1, p_2, p_3 \in \mathbb{P}^3$ are coplanar points, i.e. they all belong to a plane $\pi$. If I want to "translate" this statement to its dual, how would I do it?
I know that, in $\mathbb{P}^3$, duals of points are planes and duals of planes are points. Would it be that the three planes $p_1^*, p_2^*, p_3^*$ intersect at the point $\pi^*$? (I use $L^*$ to represent the dual of a projective linear variety $L$)
For a modern approach, see Richter-Gebert's
Perspectives on Projective Geometry: A Guided Tour Through Real and Complex Geometry.
The problem is a lot simpler than it looks and I was able to figure out the solution thanks to a tip given by the instructor. The tip is:
One can assume, via projective transformation, that $P_1=(1:0:0)$, $P_2=(0:1:0)$, $P_3=(0:0:1)$ and $Q=(1:1:1)$.
With this, one is able to express $S_1$, $S_2$ and $S_3$ simply in terms of $R$ and calculate directly the intersection between the three lines $\overline{P_1S_1}$, $\overline{P_2S_2}$ and $\overline{P_3S_3}$.
Best Answer
Yes. Just replace every "point" with "plane" and vice versa:
"The three points $a,b,c$ lie on the plane $d$."
becomes:
"The three planes $a,b,c$ $\color{red}{lie~on}$ the point $d$."
Then fix the incidence relation (marked in red), because saying that the three planes "lie on" the same point sounds off:
"The three planes $a,b,c$ $\color{red}{intersect}$ at the point $d$."