what does project away mean?

Question

I realise I should know this but I have no idea what people mean when they say "we project away from this point" (or replace point with line, plane or whatever in projective space). What does this mean?

Answer 1

If $L\subset \mathbb P^n$ is a linear subspace of dimension $l$ and $M\subset \mathbb P^n$ is a linear subspace of dimension $m=n-l-1$ such that $L\cap M=\emptyset$, the projection from $L$ onto $M$ is the morphism $\mathbb P^n\setminus L\to M$ sending $p$ to the unique point of intersection $M\cap \overline {pL}$ of $M$ with the linear subspace generated by $p$ and $L$.

This is quite close to the way Renaissance painters invented projective geometry: they projected the three-dimensional world from their eye, the zero-dimensional $L$, onto their two-dimensional canvas $M$.
How not to be nostalgic about a time when projective geometry meant the art of the likes of Della Francesca, Brunelleschi, Donatello or Alberti?
[Please, excuse the lyric tone: I'm just back from a holiday in Florence ...]

Edit: a concrete computation
As an attempt to answer carmelo's question in his comment, let me make an explicit computation. Consider $l=[0:0:1]\in \mathbb P^2$ (with coordinates $x:y:z$) and let us project onto the line $M$ of equation $z=0$ .
We get the map $f:\mathbb P^2\setminus \{l\}\to M:[x:y:z]\mapsto [x:y:0]$, a rational map from $\mathbb P^2$ to $M$ which cannot be extended regularly through $l$.

However if we blow up $l$ we get the variety $\tilde {\mathbb P^2}\subset \mathbb P^2\times \mathbb P^1$ defined by the condition $([x:y:z][u:v])\in \tilde {\mathbb P^2}$ iff $xv=yu$ (where $\mathbb P^1$ has coordinates $u:v$).
The rational map $f$ now extends to the regular morphism $$\tilde f:\tilde {\mathbb P^2} \to M:([x:y:z],[u:v])\mapsto [u:v:0]$$
The indeterminacy of $f$ at $l$ has been resolved by blowing up the point $l$ to the copy of $\mathbb P^1$ formed by all the pairs $([0:0:1],[u:v])\in \tilde {\mathbb P^2}$