Question about how to use projective transformation to simplify condisions

Question

I am slightly confused about what you can do with projective transformations. For instance, say we have P, Q on a projective curve C, can we assume that after some projective transformation P is [1,0,0], Q is [0,1,0], AND their tangent lines are $\{x_1=0\}$, $\{x_0=0\}$ respectively? What is the maximum simplification a projective transformation can give? I think the above example is one. Thanks!

Answer 1

You can certainly assume the first part, as long as $P$ and $Q$ are distinct points - just take the line segment through $P$ and $Q$ and you can find a projective transformation mapping it to the line segment between $[1,0,0]$ and $[0,1,0]$.

The second part does not work out if $P$ and $Q$ share a tangent line - you can't have one line mapped to two different lines.

Otherwise, I believe such a projective transformation can be found:

By a known result, you can find a projective transformation taking four points to four points in $\mathbb R P^2$.

Let $R, S$ be points on the tangent lines at $P,Q$ respectively. Then choose the mappings

$$P \to [1,0,0], Q \to [0,1,0], R\to[1,0,1], S \to[0,0,1]. $$

Projective transformations preserve collinearity and incidence, so the tangent lines get mapped the way we want.

As for how much you can simplify, the theorem I mentioned above is probably useful. I will state it in full:

Let $P, Q, R, S$ and $P', Q', R', S'$ be points in $\mathbb R P^2. $

If $P, Q, R, S$ are in general position and $P', Q', R', S'$ are also in general position, there is a unique projective transformation $\pi$ such that

$\pi(P)=P'$, $\pi(Q)=Q'$, $\pi(R)=R'$, $\pi(S)=S'.$

Pappus' hexagon theorem is a nice example - you can transform four points to the unit square, which makes a large part of the proof trivially easy. Note that we can use projective transformations here because the theorem only talks about collinearity and incidence, two properties that are preserved under a projective transformation.

My intuition agrees with you that your example is probably the most you can assume. The more points you specify, the harder it is to find a suitable transformation. Four generic points define just a single transformation, so you probably can't add any more restrictions after that!