# Projective plane

1. Dec 9, 2013

### gentsagree

Can somebody expand on the fact that the projective plane is defined as the "set of equivalence classes of $R^{3}/ (0,0,0)$"?

Why do we need to "delete the origin"???

I understand, I think, other definitions of it such as the set of lines in $R^{3}$ passing through the origin (0,0,0) like Wikipedia says; however, I can neither understand the first statement nor make a connection between the two.

Thank you

2. Dec 9, 2013

### economicsnerd

Recall: For any topological space $X$ and any equivalence relation $\sim$ on $X$, we have a topological space called $X/\sim$. We're just looking at the special case of $X=\mathbb R^3\setminus \{\vec 0\}$ and the relation $\sim$ with $$\vec x \sim\vec y \iff \vec y=c\vec x \text{ for some } c\in\mathbb R\setminus\{0\}.$$ Do we "need to" delete the origin? No... It's our topological space, and we can define it however we like.

3. Dec 9, 2013

### Office_Shredder

Staff Emeritus
The purpose of deleting the origin is that if I asked you which line it belongs to, the correct answer is all of them. Since I can't decide which line to put it in, it's easiest to just delete it. You could arbitrarily assign it to a line but then we'd need to keep specifying which line it belongs with.

4. Dec 11, 2013

### R136a1

Of course, we can leave the origin in there if we wish. We can then proceed like usual and get a perfectly fine topological space. So we don't "need" to delete anything.

The reason we do delete the origin is not for topological reasons, but for other reasons, and it touches on what the projective plane is actually used for.

The idea behind projective planes and such is the following. You want to project a 3D space on a 2D plane. For example, you want to paint a 3D figure on a painting as follows:
http://xahlee.info/3d/i2/Albrecht_Durer_Man_Drawing_a_Lute.png

So you take a fixed point. Then with any point on the 3D-figure, you connect the fixed point with that point using a tight rope. This rope will intersect some canvas. Where this intersection happens, you place a dot. Then you take another point on the 3D figure, and so on.

This gives rise to some weird geometry. For example, two parallel lines in the 3D world might easily be projected to two intersecting lines. For example:

This weird geometry is called projective geometry. The projective plane is used to study projective geometry.

We give the "fixed point" the coordinates $(0,0,0)$. The canvas is the plane $z=1$. The goal is for any point $(x,y,z)$ to see where it projects on the canvas. This point is exactly $(x/z,y/z,1)$. In particular, we see that two points $(x,y,z)$ and $(\alpha x, \alpha y, \alpha z)$ project on the same point on the canvas. And we see that $(0,0,0)$ can't project at all (in principle, a point of the form $(x,y,0)$ can't project at all, but this can be solved using points on infinity which we describe later, however not being able to project $(0,0,0)$ can't be solved at all!). So this is the reason for the weird equivalence relation and leaving out $(0,0,0)$.

Another description of the projective plane is when we add to the ordinary plane $\mathbb{R}^2$ certain "points on infinity". The idea is the following: in a plane, two lines either intersect or they don't (or they coincide). Now, if they are parallel (and don't coincide), then they don't have a point in common. The idea behind the projective plane is to "add" a point on infinity representing the intersection of these two parallel lines. In this way, every two distinct lines always intersect in one point (possibly at infinity).

This can be made analytic using homogeneous coordinates. The idea is this: given two parallel lines $ax + by = c$ and $ax + by = d$, we represent the point on infinity by the coordinate $(-b,a)$ (which is just a nonzero point on the line $ax+ by=0$, thus the line through the origin parallel to these two lines). This leaves us with a nasty situation because we need to distinguish between ordinary coordinates and points on infinity. The solution is to add a new coordinate number. So given an ordinary point $(x,y)$ on $\mathbb{R}^2$, we add a $1$ in the end to denote that it's an ordinary point, so we get $[x:y:1]$. On the other hand, given a point on infinity $(x,y)$, we add a $0$ to denote that it's a fictional point: $[x:y:0]$.

So, given lines of the form $ax +by = c$, we mentioned how to represent the point on infinity. Just take a nonzero point on the line $ax + by = 0$. For example $(-b,a)$ is such a point. This gives us $[-b:a:0]$. But $(-2b,2a)$ is also such a point, and this gives us $[-2b:2a:0]$. So multiplying with a constant should give the same point. And we force this to be true indeed: we say that $[x:y:0]$ is the same as $[\alpha x: \alpha y :0]$. And for convenience, we say the same thing for "ordinary points". So we allow $[x:y:1]$ to be the same as $[\alpha x: \alpha y : \alpha]$.

Now, given a line $ax + by = c$, we can easily find which points are on the line by making the equation homogeneous. This gives us $ax + by = cz$. A point $[x:y:z]$ is on this line iff $ax + by = cz$.

Back to your topological space. The $\mathbb{R}^3$ in your OP is just the set of homogeneous coordinates, and the equivalence relation merely is the same equivalence relation I've just put on the homogeneous coordinates.

Now, why don't we allow $[0:0:0]$? Well, it has a $0$ in the end, so that would indicate that it's a point at infinity. But it's a point of infinity that would lie on every line. Indeed, every equation $ax + by = cz$ would allow $[0:0:0]$. This is behaviour that is unwanted.

A more mathematically advanced view of the projective plane is the following. You take $\mathbb{R}^3$ and you take the set of all one-dimensional subspaces through the origin. This also corresponds to the projective plane. Indeed, a point $[x:y:z]$ on the projective plane corresponds to the subspace $\textrm{span}\{(x,y,z)\}$. The point $[0:0:0]$ corresponds to $\textrm{span}\{(0,0,0)\}$, which is not one-dimensional, so we remove it.

So, now you have seen three approaches to projective geometry. These approaches are more or less equivalent, but I won't bother doing this here. I hope you see now where the projective plane comes from.

5. Dec 11, 2013

### jgens

Not quite. Including the origin would induce a slightly pathological topology on the projective space. Unless of course you topologize the resulting space with something other than the quotient topology. Granted this is certainly not the only, nor the most illuminating, reason to delete the origin.

Edit: While there are compelling reasons to omit the origin based on the geometrical notions projective space ought to capture, the topological viewpoint illustrates why the naive approach of just assigning the origin to an arbitrary line fails. That is really the point of my aside.

Last edited: Dec 11, 2013
6. Dec 11, 2013

### R136a1

That's exactly what I said.

7. Dec 11, 2013

### jgens

Whoops! I stopped reading after the "not for topological reasons" because it is precisely for topological reasons that adding in the origin can never work! My bad for not reading further.