Collection of lines through $P$ in projective space

Question

Let $P$ be a point in $\mathbb{P}^3$. Show that the collection of lines through $P$ has one-to-one correspondence with $\mathbb{P}^2$.

I'm a bit confused here. I thought projective lines can be determined by two points in $\mathbb{P}^3$. By picking a point $Q \neq P \in \mathbb{P^3}$, I'm giving the line passing through $P$ and $Q$, so wouldn't the collection of lines through $P$ be viewed as $\mathbb{P}^3$ itself?

Answer 1

Lines in $\mathbb{P}^3$ are the images of planes (2 dimensional) in $\mathbb{A}^4$. Given the point $P\in \mathbb{A}^4$, the planes containing $P$ are determined by $P$ and a point $Q\in\mathbb{A}^4$ linearly independent of $P$. Moreover, $P, Q$ span the same plane as $\lambda P, \mu Q$ for nonzero $\lambda, \mu$.

So, roughly speaking, you have a three dimensional space $\mathbb{A}^3$ from which to choose $Q$ (so it's independent of $P$), and then you have to quotient by the nonzero scalars, so there would be a correspondence with points in $\mathbb{P}^2$. I have left out some details, hope this helps.

Answer 2

If you take a plane $\pi \simeq \mathbb{P}^{2}$ embedded in $\mathbb{P}^{3}$, such that $P \notin \pi$, then each line through $P$ meets $\pi$ in a unique point. This gives your 1-1 correspondence (thinking of this in terms of vector spaces, you are taking the quotient space of your 4-dimensional vector space by the 1-dimensional subspace corresponding to $P$ to obtain a 3-dimensional vector space that we can associate with $\pi$).

The lines through $P$ cover the points of $\mathbb{P}^{3}$, and in fact partition the points of $\mathbb{P}^{3}\setminus \{P\}$ but that is different from what is being talked about here. There is no natural way to identify the lines through $P$ with $\mathbb{P}^{3}$.

Answer 3

By fixing and removing an hyperplane $\pi\subset\mathbf{P}^3$ "away" from $P$, i.e. such that $P\notin\pi$, you are in fact just looking at the set of lines through $P$ in an affine space $\mathbf{A}^3$.

Without loss of generality you can thus assume that $P=O=(0,0,0)$ under some affine coordinate frame that you can fix once for all. A line through $0$ has parametric equations ($t$ is the parameter) $$ x=mt\qquad y=nt\qquad z=pt $$ for some triple $(m,n,p)\neq(0,0,0)$. Since the lines corresponding to triples $(m,n,p)$ and $(\lambda m,\lambda n,\lambda p)$ for $\lambda\neq0$ are the same, you readily see that these are projective coordinates for points in $\mathbf{P}^2$, hence the identification.