projective plane in projective space I read in an article that the collection of lines going through a point in RP3 form a projective plane. I can't understand why .
we know that a point in RP3 is a line going through the origin of the vector space R4 and a line going through this point is actually a plane containing the line and going through origin.
so how can we think of the collection of lines going through a point in RP3 as a projective plane?
 A: You can embed $\mathbb R^3$ into $\mathbb R\mathrm P^3$. In this case, every affine line becomes a projective line. All the lines passing through a given point become projective lines, which means they gain one additional point but otherwise stay the same set of lines.
You can also embed $\mathbb R\mathrm P^2$ into $\mathbb R^3$. All the lines passing through the origin of $\mathbb R^3$ represent points of $\mathbb R\mathrm P^2$. Your question indicated that you have a good understanding of this.
Now you can combine these embeddings:


*

*You start with a given point in $\mathbb R\mathrm P^3$. For the sake of simplicity, you apply a projective transformation (or a change of basis, if you prefer to not move “actual objects” around) which change the coordinates of that point to $(0,0,0,1)$, i.e. the “origin” of $\mathbb R\mathrm P^3$. The set of lines through that point is a set of projective lines which meet in that point.

*Next you strip away infinity. Every line looses one point, but otherwise stays the same line. All your finite points get dehomogenized, so your common point now has coordinates $(0,0,0)$. You end up describing all one-dimensionsl subspaces of $\mathbb R^3$.

*Then you interpret this $\mathbb R^3$ as $\mathbb R\mathrm P^2$. Every linear subspace represents a point. You can intersect these with a drawing plane not passing through the origin.


So each line through the point in the original 3-space gets mapped to a point in the projective plane. The mapping is bijective.
A: Fix a point $P\in\Bbb P^3$ and a plane $\pi\subset\Bbb P^3$ such that $P\notin\Bbb \pi$. Also, let $\cal L$ be the set of lines through $P$.
Then there is a bijection
$$
\cal L\longleftrightarrow\pi
$$
as follows. To each $L\in\cal L$ associate the unique point $P_L\in\pi$ given by $P_L=L\cap\pi$. In the other direction, for each $Q\in\pi$ consider the unique $L\in\cal L$ such that $\{P,Q\}\subset L$. The two constructions are inverse of each other, obviously.
But $\pi$ is a $\Bbb P^2$, so you're basically done.
