# Projective Equivalence of $(n+2)$-tuples in $\mathbb{P}^n$

Let $p_1,...,p_{n+2}$ and $q_1,...,q_{n+2}$ be two sets of distinct points in general position in $\Bbb P^n$ (there may however be overlaps between the two sets.). Then there exists $\phi\in Aut(\mathbb{P}^n)$ taking $p_i\mapsto q_i$.

I've been having trouble with what should be the extremely simple proof of this fact. Michael Artin proves it for $n=2$ (though the same idea should work for any $n$) here http://math.mit.edu/classes/18.721/chpcurvev5.pdf following the line of reasoning I was trying to use. My confusion arises when he says to adjust the $p_i$ by a factor of $1/c_1$. This sounds like a perfectly fine thing to do in $\Bbb P^n$, but I don't see how it leads to the conclusion that $q=p_0+p_1+p_2$. If there is a typo, and he meant to write that we adjust each $p_i$ by $1/c_i$, then I understand how he reaches the conclusion, but I don't see how we are allowed to separately scale three tuples of coordinates before adding them together, i.e. Artin seems to be jumping back-and-forth between projective and affine coordinates without justification. Do I misunderstand the nature of $PGL(n)$ in some way?

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Gah, yes! Thanks :) – Tabes Bridges Dec 3 '12 at 2:08
It looks like it may be a typo, and should rather say "adjust $p_i$ by $1/c_i.$" – Andrew Dec 3 '12 at 2:10
Which makes sense, since he only needs to work with some coordinate representative of the $p_i$'s,i.e., any particular choice can be scaled without affecting the fact that the coordinates represent $p_i.$ – Andrew Dec 3 '12 at 2:12
Yes; the weird part is that he seems to make this adjustment and then add the coordinates together as one might do in an affine situation. – Tabes Bridges Dec 3 '12 at 2:16
I see; we are absorbing the coefficients into the basis, thereby making a change of basis which is invisible under the scaling action. – Tabes Bridges Dec 3 '12 at 3:08

Based on Andrew's comments, I'm going to answer myself. I was misunderstanding Artin's proof. On one side of the linear map he works in terms of arbitrary $p_0,p_1,p_2,q\in\Bbb P^2$ in general position; on the other side he fixes $e_0,e_1,e_2\in\Bbb P^2$ the projections of the standard basis vectors, and for the final point chooses $\epsilon:=e_0+e_1+e_2$.

Since the given quadruples are in general position, the lifts of $p_0,p_1,p_2$ to $\Bbb A^3$ form a basis, so we can write $q=c_0p_0 + c_1p_1 + c_2p_2$. At this point, Artin says to adjust each $p_i$ by a factor of $1/c_i$, which I misinterpreted as invoking the projective scaling action; rather, we simply absorb the $c_i$ into the chosen representative affine coordinates for the $p_i$, and we map $q\mapsto\epsilon=e_0+e_1+e_2=(1:1:1)$, regardless of the values of $c_i$.

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