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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 :)2012-12-03
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    It looks like it may be a typo, and should rather say "adjust $p_i$ by $1/c_i.$"2012-12-03
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    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.$2012-12-03
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    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.2012-12-03
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    Right, but it doesn't matter, since adjusting the points $p_i\in\Bbb P^2$ by scalar factors has no effect. We only see an effect when we consider the affine coordinates lifting the projective coordinates of these points. So the point is that by a judicious choice of affine coordinates that lift the projective coordinates of the $p_i$'s, we will have $q=p_0+p_1+p_2$ in affine coordinates.2012-12-03
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    But adjusting scalar factors *after* taking a linear combination? For example, consider $p=(0:1)$ and $q=(1:0)$ in $\Bbb P^1$. We have $p+2q=(1:2)$, but if make such an adjustment, then we are left with $p+q=(1:1)$, which is not the same point in $\Bbb P^1$.2012-12-03
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    We have $(0,1)+2(1,0)=(2,1).$ Suppose however, that $(0:1)$ and $(1:0)$ were initially given with affine coordinates $(0,3)$ and $(-5,0).$ He says, we know there are constants $a,b$ such that $a(0,3)+b(-5,0)=(2,1)$ (namely, $a=1/3,b=-2/5.$) Thus, by choosing the affine coordinates $(0,1)$ and $(2,0)$ to begin with, we will get $(0,1)+(2,0)=(2,1)$, i.e. a linear combination resulting in $(2,1)$ without multiplying the representatives for $(0:1)$ and $(1:0)$ by scalars.2012-12-03
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    I see; we are absorbing the coefficients into the basis, thereby making a change of basis which is invisible under the scaling action.2012-12-03
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    @TabesBridges: your statement is false in general even for $n=2$: if the $p_i$'s are colinear, then their images by $\phi$ are also colinear, so you can't reach arbitrary $n+2$-uple.2012-12-03
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    Thank you, I forgot to include the general position hypothesis; editing now.2012-12-03

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