I have the following definitions:
Given a vector space $V$ over a field $k$, we can define the projective space $\mathbb P V = (V \backslash \{0\}) / \sim $ where $\sim$ identifies all points that lie on the same line through the origin.
A projective subspace $\mathbb P W$ of $\mathbb P V$ is of the form $\pi(W \backslash \{0\})$, where $\pi$ is the residue class map and $W$ is a vector subspace of $V$. Define $\mathrm{dim} (\mathbb P V) = \mathrm{dim}( V )- 1$. A line in $\mathbb P V$ is a $1$-dimensional projective subspace.
Now I'm finding it difficult to visualise what a line in projective space actually is. I can understand why any two lines in a projective plane intersect. Suppose I'm in $\mathbb P^3$ and want to write 'an equation' for the line that goes through the points $ p = (1:0:0:0)$ and $q = (a:b:c:d)$. How could I do that? Does my question even make sense? I'm concerned because $\mathbb P V$ isn't actually a vector space, so can I think of points inside it as vectors?
Thanks
Best Answer
If you have two distinct points $A=[a_0:\ldots :a_n], B=[b_0:\ldots:b_n]\in \mathbb P^n$, they correspond to two vectors $a=(a_0,\ldots ,a_n), b= (b_0,\ldots,b_n)\in k^{n+1}$.
These vectors span a plane $\Lambda \subset k^{n+1}$ whose vectors are the $ua+vb, \; (u,v\in k)$.
The corresponding line $\overline {AB}=\mathbb P(\Lambda)\subset \mathbb P^n$ has its points of the form $[ua_0+vb_0:\ldots :ua_n+vb_n] \quad (u,v \in k, $ not both zero ).
In your particular case the projective line $\overline {pq}$ joining $p=[1:0:0:0]$ and $q=[a:b:c:d]$ has points with coordinates $[u+va:vb:vc:vd]$