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

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?

• It's not true that the statement holds for any four lines (planes), you only need to prove existence: simply pick four specific planes in $\Bbb R^3$ with the given property. – Berci Nov 9 '20 at 7:55
• I'm not sure I understand. Why does it not hold for any 4 planes that pass through the origin? And how does picking four specific planes in $\mathbb{R}^3$ show that $\mathbb{RP}^2$ has exactly four "lines"? – user140161 Nov 9 '20 at 8:04
• As suggested you have a wrong translation of the problem. The question asks you to find specific 4 lines with some properties, while in your translation you wish to show this property for any chosen 4 lines. – Arctic Char Nov 9 '20 at 9:10
• Note that one can easily find 4 lines which passes through the same point. Indeed, given any point, there are infinitely many lines passing through it. – Arctic Char Nov 9 '20 at 9:12

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 $$x=0,\ y=0,\ z=0$$?

• the plane y-x=0. Is that right? – user140161 Nov 9 '20 at 19:21
• Not quite. The three planes not mentioning $z$ intersect in a line. – Berci Nov 9 '20 at 19:50
• I think i got it. x-y+z=0 should satisfy the condition. I did get to this by using an online 3d plotter though. How can I prove this mathematically? – user140161 Nov 9 '20 at 20:14
• Actually, for any nonzero $a,b,c$, the line (=plane) $ax+by+cz=0$ will work. It's simple algebra. The intersection of planes is the common solution of their equations, so it's clear that the three coordinate planes only intersect in $x=y=z=0$. Now, if $x=0,\ y=0,\ ax+by+cz=0$ then $z=0$ follows. – Berci Nov 9 '20 at 20:50
• To add: if you randomly choose the 4 lines in the (projective) plane, they will almost surely satisfy the condition. – Berci Nov 9 '20 at 20:53