Chapter 1
1.1
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.
1.2
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.
1.3
A quadrilateral is a set of four lines, no three concurrent.