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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?

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    $\begingroup$ 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. $\endgroup$ – Berci Nov 9 '20 at 7:55
  • $\begingroup$ 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"? $\endgroup$ – user140161 Nov 9 '20 at 8:04
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    $\begingroup$ 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. $\endgroup$ – Arctic Char Nov 9 '20 at 9:10
  • $\begingroup$ 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. $\endgroup$ – Arctic Char Nov 9 '20 at 9:12
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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 $x=0,\ y=0,\ z=0$?

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  • $\begingroup$ the plane y-x=0. Is that right? $\endgroup$ – user140161 Nov 9 '20 at 19:21
  • $\begingroup$ Not quite. The three planes not mentioning $z$ intersect in a line. $\endgroup$ – Berci Nov 9 '20 at 19:50
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    $\begingroup$ 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? $\endgroup$ – user140161 Nov 9 '20 at 20:14
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    $\begingroup$ 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. $\endgroup$ – Berci Nov 9 '20 at 20:50
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    $\begingroup$ To add: if you randomly choose the 4 lines in the (projective) plane, they will almost surely satisfy the condition. $\endgroup$ – Berci Nov 9 '20 at 20:53

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