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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

PeterM
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2 Answers2

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  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.

John Hughes
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    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. – PeterM Jul 15 '14 at 22:11
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    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 $P^{n+1}$. – John Hughes Jul 15 '14 at 23:51
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    For (2), it's Theorem 3.9 that you want to look at. – John Hughes Jul 15 '14 at 23:54
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This is a rewritten of the accepted answer above. I wish this is right. First, it is easier to see the picture from the affine plane where a point is an array starting from the origin.

Let $(x_i)$ and $(y_i)$ be two distinct points in $\mathbb P^n$. Let $(u_0,u_1)$ be a point of $\mathbb P^1$ and consider the point $(z_i)$ with $z_i=(u_1-u_0)x_i+u_0y_i$.

This is a linear variety since the above can be written as $r_{ij}z_k=r_{ik}z_j+r_{kj}z_i$ where $r_{ij}=y_ix_j-y_jx_i$.

Since a line has dimension 1, it is easy to show its uniqueness. To see it is a line, choose a pair $(i,j)$ such that $r_{ij}$ is nonzero, which exists since $(x_i) $ and $(y_i)$ are distinct points, and then we have a matrix $A_{(n-1)\times (n+1)}=(r_{ij}I_{n-1} *)$ which is of full rank. Since it has two linearly independent solutions, it has dimension 1.

Goulag
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