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I have the following definitions:

Given a vector space $V$ over a field $k$, we can define the projective space $\mathbb P V = (V \backslash \{0\}) / \sim $ where $\sim$ identifies all points that lie on the same line through the origin.

A projective subspace $\mathbb P W$ of $\mathbb P V$ is of the form $\pi(W \backslash \{0\})$, where $\pi$ is the residue class map and $W$ is a vector subspace of $V$. Define $\mathrm{dim} (\mathbb P V) = \mathrm{dim}( V )- 1$. A line in $\mathbb P V$ is a $1$-dimensional projective subspace.

Now I'm finding it difficult to visualise what a line in projective space actually is. I can understand why any two lines in a projective plane intersect. Suppose I'm in $\mathbb P^3$ and want to write 'an equation' for the line that goes through the points $ p = (1:0:0:0)$ and $q = (a:b:c:d)$. How could I do that? Does my question even make sense? I'm concerned because $\mathbb P V$ isn't actually a vector space, so can I think of points inside it as vectors?

Thanks

Jonathan
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2 Answers2

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If you have two distinct points $A=[a_0:\ldots :a_n], B=[b_0:\ldots:b_n]\in \mathbb P^n$, they correspond to two vectors $a=(a_0,\ldots ,a_n), b= (b_0,\ldots,b_n)\in k^{n+1}$.
These vectors span a plane $\Lambda \subset k^{n+1}$ whose vectors are the $ua+vb, \; (u,v\in k)$.
The corresponding line $\overline {AB}=\mathbb P(\Lambda)\subset \mathbb P^n$ has its points of the form $[ua_0+vb_0:\ldots :ua_n+vb_n] \quad (u,v \in k, $ not both zero ).

In your particular case the projective line $\overline {pq}$ joining $p=[1:0:0:0]$ and $q=[a:b:c:d]$ has points with coordinates $[u+va:vb:vc:vd]$

Georges Elencwajg
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    Thanks a lot for your response. I was rather hoping it'd be you who responded; this ties in well with your answer to my previous question! – Jonathan Apr 30 '12 at 22:40
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    Yes, customers here deserve a good after-sales service :-) – Georges Elencwajg Apr 30 '12 at 22:45
  • So a line in projective space is actually a plane? – The Coding Wombat Mar 27 '19 at 20:34
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    @The Coding Wombat: Yes, a line in the projective space associated to a vector space is a plane in that vector space. – Georges Elencwajg Mar 28 '19 at 09:14
  • @GeorgesElencwajg So does that mean that the dimension of a line in $\mathbb{P}^n$ is equal to two? I do not understand that, since I read that $\dim \mathbb{P}^n=n$, but also that it is associated with $\mathbb{R}^5$ which is $\dim = 5$ – The Coding Wombat Mar 28 '19 at 14:30
  • No, the dimension of a line in $\mathbb P^n$ is one, even though the corresponding plane in the vector space has dimension two. Careful there :-) – Georges Elencwajg Mar 28 '19 at 14:58
  • I guess that's maybe because the dimension as a subspace of a vector space and dimension as a variety (which in case of it being a line in Projective space) are two completely different things? – Shreya Sep 09 '19 at 06:02
  • This is because the corresponding plane in the vector space has dimension two but after projection into P^n we get a one-dimensional line. – J.abdou Jul 21 '22 at 09:54
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Every linear subspace in $V$ is intersection of hyperplanes. Projecting down to $\Bbb P(V)$ you obtain that every linear subspace in $\Bbb P(V)$ is intersection of hyperplanes.

If $\dim V=4$, i.e. $\dim\Bbb P(V)=3$, a line $\ell\subseteq\Bbb P(V)$ corresponds to a $2$-dimensional subspace $U\subset V$ which can always be obtained as intesection of two $3$-dimensional subspaces, $U=W_1\cap W_2$, obviously not uniquely determined. Hence $\ell=\pi(W_1)\cap\pi(W_2)$.

If $\ell$ is given as the line through two points $A$ and $B$ you can always find some $W_1$ and $W_2$ as above knowing that an hyperplane in $V$ has equation $$ a_1X_1+a_2X_2+a_3X_3+a_4X_4=0\qquad\qquad(\ast) $$ and imposing that the coordinates of $A$ and $B$ satisfy this equation. You thus obtain a homogeneous linear system of two equations in 4 variables (the coefficients of $(\star)$) wich you can certainly solve.

Andrea Mori
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