1
$\begingroup$

Let $k$ be an algebraically closed field. From calculus, I learned to think of the line through two points $p,q \in k^n$ as the span of $q-p$, plus a displacement vector (either $p$ or $q$). $$ L_{pq} = \{ p + \lambda(q-p) : \lambda \in k\} $$ I'm taking a course in algebraic geometry, but I get confused by talk of lines in projective space $\mathbb{P}^n$ (over $k$). If we have two points $x,y \in \mathbb{P}^n$, is there any analogous way to describe the line containing $x,y$?

For example, let's start with an equation $x_0 + 2x_1 + 3x_2 = 0$. If $[x_0:x_1:x_2] \in \mathbb{P}^2$ lies on this line, then I know that $$ x_2 = - \frac 23 x_1 - \frac 13 x_0 $$ so the solutions to this equation should be the set $$ \{[3x_0 : 3x_1 : -2x_1 - x_0] | x_0, x_1 \in k, \text{not both zero}\} $$ I know that this line contains the points $[3:0:-1]$ and $[0:3:-2]$. Is there any way to write the solution set as some sort of "span" of these projective points? If we started with these two points and didn't have the linear equation, how could we reconstruct the linear equation?

$\endgroup$
  • $\begingroup$ A line in $\mathbb{P}^n$ is equivalent to a two-dimensional subspace of $\mathbb{R}^{n+1}$. $\endgroup$ – Daniel Schepler Sep 14 '17 at 23:09
  • $\begingroup$ I said that $k$ is an arbitrary algebraically closed field, which, unfortunately, does not include $\mathbb{R}$. $\endgroup$ – Joshua Ruiter Sep 14 '17 at 23:14
  • $\begingroup$ OK, then, a line in $\mathbb{P}^n_k$ is equivalent to a two-dimensional subspace of $k^{n+1}$. $\endgroup$ – Daniel Schepler Sep 14 '17 at 23:15
  • $\begingroup$ An $m$-dimensional affine subspace of $k^n$ is of the form $S= \{ x_0+Mv, v \in k^m\}$ for some base point $x_0$ and some matrix $M \in k^{n \times m}$. To embed it into $\mathbb{P}_k^n$ we need to make $S$ invariant under multiplication by constants, ie. $S^{proj} = k^* S = \{ a(x_0+Mv), v \in k^m, a \in k^*\}$ which is a plane passing through the origin in $k^n$ if $S$ was an affline line not containing the origin. $\endgroup$ – reuns Sep 14 '17 at 23:38
  • $\begingroup$ The line through a pair of points $\mathbf p$ and $\mathbf q$ (in homogeneous coordinates) can be described as their join, which is simply the set of all linear combinations $\lambda\mathbf p+\mu\mathbf q$. $\endgroup$ – amd Sep 15 '17 at 21:40
2
$\begingroup$

For this particular example, note that if $-x_0 - 2x_1 \ne 0$, then $$[3x_0 : 3x_1 : -x_0 - 2 x_1] = \left[\frac{3x_0}{-x_0 - 2x_1} : \frac{3x_1}{-x_0 - 2 x_1} : 1\right].$$ For the corresponding point in $k^2$, we have: $$\left(\frac{3x_0}{-x_0 - 2x_1}, \frac{3x_1}{-x_0 - 2 x_1} \right) = \lambda (-3, 0) + (1 - \lambda) (0, -3/2)$$ for $\lambda = \frac{x_0}{x_0 + 2 x_1}$. (The exceptional case generates the "point at infinity" $[2 : -1 : 0]$ on the projective line - and in the case $k = \mathbb{C}$ this could be thought of as the limit as $|\lambda| \to \infty$.)

For the final question: given the points $[3 : 0 : -1]$ and $[0 : 3 : -2]$, the points $(3, 0, -1)$ and $(0, 3, -2)$ of $k^3$ span a two-dimensional subspace of $k^3$, which corresponds to a line in $\mathbb{P}_k^2$. Finding the equation of this plane, you will get a homogeneous equation (in fact, a degree 1 equation) which will also give the equation of the projective line.

$\endgroup$

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.