In the real projective plane $\mathbb{R}\mathbb{P}^2$ where points are represented by homogenous coordinates $[x,y,z]$, there is a degenerate conic consisting of a real double line. This conic is of equation $x^2=0$. I am looking to determine its corresponding conic envelope.
In Jurgen Richter-Gebert's Perspectives on Projective
Geometry, he says there are two such envelopes, one consiting of all lines passing through one point, and the other consisting of all lines passing through two points. Is this correct, or is it a misunderstanding?
Best Answer
I think you might have misunderstood the text here. A double line doesn’t have a unique dual: it has an infinite number of them, each consisting of the lines through each of two points on the line (which do not have to be distinct). Each of these dual conics can be obtained as the limiting case of the duals of a particular family of nondegenerate conics. For instance, given any two distinct points on the line one can obtain the corresponding two-point dual conic as the limiting case of the duals of the family of ellipses that have those points as common foci. Richter-Gebert spends much of section 9.6 on this. Indeed, it’s a major motivation for
Note that $\lambda$ can be zero, which is in fact the case for a double line. For your canonical double line $x^2=0$, then, we have $$A=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}$$ and any nonzero matrix of the form $$B=\begin{bmatrix}0&0&0\\0&a&b\\0&b&c\end{bmatrix}$$ is dual to it. A rank-one $B$ (which occurs when $b^2=ac$) represents a double point, while a rank-two $B$ represents two distinct points. In the latter case, you can recover the two points by splitting the conic, using the line $(1,0,0)^T$ as the “intersection point” for the algorithm.