Projective transformation in $\mathbb{P}^1$

Question

Let $A=\{p_1,p_2,p_3\}$ and $B=\{q_1,q_2,q_3\}$ be two sets of 3 points in the (complex) projective line $\mathbb{P}^1$. My question is:

Is it true that there always exist $A \in PGL_2(\mathbb{C})$ inducing a projectivity (projective transformation) that sends $A$ to $B$?

Thank you in advance!

Answer 1

Hint In any affine chart, the projective transformations $\Bbb P^1 \to \Bbb P^1$ are given by the linear fractional transformations, $$z \mapsto \frac{a z + b}{c z + d} , \qquad a d - b c \neq 0.$$ (NB that scaling $a, b, c, d$ by the same nonzero factor does not change the transformation, giving rise to the $P$ in $PGL_2(\Bbb C)$.)

Now, can you show that for any (if you like, ordered) triple $A = (p_1, p_2, p_3)$ of distinct points that there is a linear fractional transformation $\phi$ such that $\phi(0) = p_1, \phi(1) = p_2, \phi(\infty) = p_3$?

---

Remark 1 The choice of the reference triple $(0, 1, \infty)$ here is not essentially, but it does make a computational approach using the explicit form of the f.l.t.s particularly nice. In fact, one can even do better by choosing the affine chart at which the given point $p_3$ is at infinity, so that $\phi$ fixes $\infty$. But the linear transformations that do this are precisely the affine transformations $z \mapsto a z + b$, and it's easy to find such a $\phi$ such that $\phi(0) = p_1$ and $\phi(1) = p_2$.

Remark 2 This line of reasoning shows that the answer to the question is yes (at least provided the triples are of distinct points), but NB the answer to the analogous question for four points is no, and this leads to the important notion of cross-ratio, which is an especially simple example of a geometry invariant.

Answer 2

To give another formulation of Travis's perfect answer...

You can certainly find an element $g$ of $GL_2$ to take arbitrary 'basis lines ' $p_1, p_2$ to $q_2, q_2$. So WLOG, $p_1=q_1 = (1:0)$, and $p_2= q_2 = (0:1)$. Any such $g$ (i.e., that preserves those two lines) is of the form diag$(z,w)$, with $z$ and $w$ non-zero. If $ q_3 = (a:b)$, and $p_3= (c:d)$, with all entries non-zero, one can certainly solve for non-zero $z$ and $w$ so that $a = zc$, and $b= w d$.