| layout |
page |
| usemath |
true |
| title |
Chapter 1 |
Consider the following correspondence from one and two-dimensional subspaces of $\mathbb{R}^3$ (i.e., the points and lines of $\mathsf{PG}(2,\mathbb{R})$) with the elements of $(\mathcal{P},\mathcal{L})$, induced by intersection with $S^2$:
|
$\mathsf{PG}(2,\mathbb{R})$ |
$(\mathcal{P},\mathcal{L})$ |
| points |
1-dim. space $\langle v\rangle$
|
$\langle v\rangle\cap S^2$ |
| lines |
2-dim. space $\langle u, v\rangle$
|
$\langle u,v\rangle\cap S^2$ |
First of all, this mapping is well-defined since a one-dimensional
subspace of $\mathbb{R}^3$ intersects $S^2$ in a pair of antipodal
points, and a two-dimensional subspace of $\mathbb{R}^3$ intersects
$S^2$ in a great circle. Moreover, it is a bijection because the inverse
map takes antipodal points to the subspace they span, and a great circle
spans a plane of $\mathbb{R}^3$ passing through the origin. The
incidence relation for $(\mathcal{P},\mathcal{L})$ is inherited from the
natural incidence relation for subspaces, so we indeed have an
isomorphism of geometries.
Solving for $(x_1,y_1,z_1)\cdot [a,b,c]=0$ and
$(x_2,y_2,z_2)\cdot [a,b,c]=0$ is the same problem as finding the normal
to two vectors in $\mathbb{R}^3$, and hence we can find $[a,b,c]$ by
computing $(x_1,y_1,z_1)\times (x_2,y_2,z_2)$ and replacing round
brackets with square ones.
A quadrilateral is a set of four lines, no three concurrent.