I'm trying to solve exercise I.3.15 in Hartshorne's Algebraic Geometry. The question starts as follows:

Projection from a point: Let $ \mathbb{P}^{n } $ be a hyperplane in $ \mathbb{P}^{n+1 } $ and let $ P \in \mathbb{P}^{n +1 } - \mathbb{P}^{n }$. Define a mapping $ \varphi : \mathbb{ P}^{n+1 } - \{P \} \rightarrow \mathbb{P}^{n } $ by $\varphi(Q) =$ the intersection of the unique line containing $ P$ and $Q $ with $ \mathbb {P}^ {n }$.

I haven't been exposed to projective geometry prior to this. The question assumes that there is a unique line and the line intersects with the hyperplane uniquely. I'm trying to show this, but I inevitably end up with $ n$ linear equations that are difficult to deal with. Furthermore, some notes online suggest that a transformation can make the hyperplane $x_0 = 0 $ and the point $(1:0 : \cdots 0) $. I can see how a transformation can move the hyperplane, but I can't come up with a transformation that simultaneously moves both the point and the hyperplane.

Could you please help me with the following questions:

  • How to show that a unique line passes through the two points and intersects the hyperplane in one point.
  • How to transform the projective space so that the hyperplane is $x_{ 0} =0$ and the point is $(1:0 : \cdots 0) $.

Thank you

  1. If you have two distinct points $P$ and $Q$, then the parametric line $s(t) = (1-t)P + tQ$ passes through them.

  2. Pick $n+1$ points in some plane and one point outside the plane. You want to send these to $n+1$ points of "the standard plane" and one more point. That's a system of $n+2$ equations in the unknown entries of the matrix, but the "scale factors" are also unknown, alas. Harthshorne's little book on Project Geometry has a nice reduction of this problem to a sequence of two linear-equation-solving problems. I suggest you take a look at that. The rest of the book may also provide you with a useful intro to Projective Geometry -- I recommend it. I believe that someone has LaTeX-ed it and put a version online.

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    $\begingroup$ Thanks. On 1) Is this actually well defined? It looks like I get different points if I pick another representative for $ P$. Also why is this line unique? On 2) I found the book. Trying to locate this sequence now. Thanks again. $\endgroup$ – PeterM Jul 15 '14 at 22:11
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    $\begingroup$ For (1), the answer is "yes, but it's complicated." The image of $s$ (i.e., the set $\{s(t) | t \in \mathbb R \}$ is independent of the representative you pick for $P$ or $Q$, but $s(t)$ depends on the choice. If $P_0$ and $Q_0$ are representatives of the classes $P$ and $Q$, then I should have said $s_0(t) = (1-t)P_0 + tQ_0$. You could then pick other reps, say $P_1$ and $Q_1$, and define $s_1(t)$. Then $image(s_0) = image(s_1)$, but in general $s_0(t) \ne s_1(t)$. By writing $s(t) = \frac{1-t}{t} P + Q$, you can make $s(\infty)$ make sense; then $s$ becomes a map from $P^1$ into $P^{n+1}$. $\endgroup$ – John Hughes Jul 15 '14 at 23:51
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    $\begingroup$ For (2), it's Theorem 3.9 that you want to look at. $\endgroup$ – John Hughes Jul 15 '14 at 23:54

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