3
$\begingroup$

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

$\endgroup$
2
$\begingroup$
  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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ For (2), it's Theorem 3.9 that you want to look at. $\endgroup$ – John Hughes Jul 15 '14 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.