Show that $\mathbb{RP}^2$ has four "lines", no three of which have a common "point"

365 Views Asked by At

I'm going through Stillwells The Four Pillars of Geometry and this is one of the questions from an exercise. I tried searching for this both here and other places online but all of them contain language that I am unable to comprehend. I am a complete beginner to projective geometry and i'm finding it hard to wrap my head around these concepts.

Here's what i've done so far: I know that any "point" in $\mathbb{RP}^2$ is a line in $\mathbb{R^3}$, and any "line" in $\mathbb{RP}^2$ is a plane in $\mathbb{R}^3$. This reduces the problem to one in $\mathbb{R}^3$: given that I have 4 planes that pass through $\text{O}$, i need to show that no three of them intersect in a line. Is my approach correct? How do I go forward with this?

1

There are 1 best solutions below

6
On BEST ANSWER

The problem asks to

Show that $\Bbb{RP}^2$ has four lines such that no three of them has a common point.

You are right about the correspondence of lines in $\Bbb{RP}^2$ with planes in $\Bbb R^3$ going through the origin.

Thus, to solve the problem, it's enough to specify 4 planes in $\Bbb R^3$ such that the intersection of any 3 of them is trivial.

Can you find a 4th plane for the 3 coordinate planes $x=0,\ y=0,\ z=0$?