# Is there a set-theoretic definition of Projective Space?

I posted this on mathOverflow previously which was the wrong place to post it and I was asked to try this forum instead. Can anyone explain this in simple terms:

I met projective space via a recent class on perspective drawing, believe it or not, but I didn't know that this was the "space" we were using. I came across a more detailed description trawling the net. In a book on point-set topology that I bought, it describes Euclidean n-space as a field made of (sorry I don't know how to write mathematical symbols yet):

[ {n-tuples of reals}, Op("+"), Op(".") ]

So what is the equivalent set-theoretic description for projective space? I haven't been able to find one anywhere. All I've found is that basically it is constructed by taking a regular plane and adding the 'horizon' line but I want to understand mathematically what it is.

EDIT(2): I guess it's the fact of being on the sphere that matters. I think I get it... The whole antipodal points minus origin thing is how you would describe the surface of a sphere, isn't it? Parallel lines in R3 would hit the spherical boundary and then NECESSARILY converge as they follow that new curved surface... Yay! :o)

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See math.stackexchange.com/questions/13763/… for a description of the projective plane. –  Arturo Magidin Feb 27 '11 at 20:04

Albraically a projective space is $\frac{V \setminus \underline{0}}{\sim}$, where $\sim$ is an equivalence relation defined as $\underline{x} \sim \underline{y} \Leftrightarrow \underline{x} = \rho \underline{y} \; \rho \in \mathbb{R}^{*}$, $V$ is a vector space, $\underline{0}$ is the null vector and $\mathbb{R}^* = \mathbb{R} \setminus \{ 0 \}$.

Basically, what the definition says is that a projective space is a vector space without the zero vector and where all the vectors that multiples of each other are identified. This leads to topological representations of projective spaces such as the projective line, which can be identified as a circle (if you add the "point at infinity" on a line, it closes).

As for your intuition, it is right. What happens in a projective space is that you "add" some "points at infinity" which identify a direction (all parallel lines "end" in the same point).

Like aaron points out, since we qutient out all multiples, the projective space has a dimension less than the vector space we construct it from.

EDIT: re-reading the OP's comment, I see I answered only partially. The point that the horizon is a circle is derived from a topological model of the 2 dimensional projective space (projective plane). You can see it as your "normal" cartesian plane, where you add a "circle" at infinity, where all the antipodal points are identified. This means that the points lying on the circumference on the first quadrant are the same as the points on the circumference on the third (same for the second and fourth quadrant). This is just a description of what aaron said: you're taking all lines from the origin and identifying their directions (their "points at infinity") so that all parallel lines end up at the same point. (I could draw something on the computer, but it would look really silly, because I can't program anything that would do it for me automatically, so I'd have to use MS Paint) So, yes, the horizon is actually a circle and "in perspective" parallel lines meet "at infinity", which was the motivation for artists to start and think mathematically of perspective (this is, I think, one of the historical origins to the projective plane).

As for the projective line, you should try to think to add one single point to a regular line, let's say "at infinity". What happens is that the line starts from this point very far away and ends in this same point (intuitively, a line has only "one infinity" to go to) so it closes in a circle.

Is this more explanatory? :)

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I'm getting on a wee bit better with this description. 2 things: 1) I presume the null vector is the "origin" (R \ {0})? 2) Are you saying the "horizon line" is actually a circle, and that effectively an artist is zoomed in to just a segment of it so that it appears as a line? –  mathmoggy Feb 27 '11 at 18:42
@mathmoggy; if you don't believe that the horizon is a circle, just stand up in an open field, look to the horizon and rotate yourself 360 degrees :) –  ypercube Feb 27 '11 at 21:20
@mathmoggy, exactly, the null vector is the vector in which all entries are zeros. Also, the horizon you see is a semi-circle (you can actually see a little more than 180 degrees from both eyes) and if you turn around and look the other way, you complete the circle, like @ypercube said. –  Andy Feb 27 '11 at 23:44
@Aaron, thanks so much. I must be the most exasperating person to try and answer. @ypercube... gosh, i know THAT. I feel so silly now. OK so I get vector space minus origin, all the antipodean lines shooting through the origin and ray-tracing the sky at infinity... which sky is the horizon. I think I understand my error: y'all were imagining a sphere in R^3 when offering explanations; I was literally visualizing a flat piece of paper that you draw on, or at best R^2. I think I'm caught up dimensionally now... –  mathmoggy Feb 28 '11 at 1:39

See my answer from MO. The point is that a line through the origin is determined by a point (actually a pair of points) on the sphere, so if you want an $n$-dimensional space then you need to take lines through 0 in $R^{n+1}$.

Think of a line through 0 as a "direction." For example the space of lines through 0 in $R^2$ is 1-dimensional. (In fact it is a circle.) In general projective space is not just the same as an $n$-dimensional sphere because two antipodal points (opposite ends of a diameter) represent the same line through 0, hence the same point of projective space.

Thus another way to define $n$-dimensional projective space is as a quotient of an $n$-dimensional sphere by identifying antipodal points. The $n$-dimensional sphere is defined as a subset of $R^{n+1}$.

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sorry... the "n+1" concept is still completely lost on me. Do you mean something like: a dimensionless point (dim 0) lies on a line (dim 0+1 = 1), or line (dim 1) exists on a surface (dim 1+1 = 2)? –  mathmoggy Feb 27 '11 at 17:06
A 2-dimensional sphere naturally lives in 3-dimensional space. A 1-dimensional sphere (i.e. a circle) naturally lives in 2-dimensional space (i.e. a plane). In general an $n$-dimensional sphere is the set of points in $n+1$-dimensional space that are a fixed distance (radius) from a given point (the center). –  aaron Feb 27 '11 at 17:13
OK so it's as I thought... you're saying n-dimensional things live in an n+1-dimensional space. I am still incapable of extrapolating from this to a mathematical understanding of what projective space is. –  mathmoggy Feb 27 '11 at 18:27
I FINALLY understand your answer over on MO and also the last paragraph in your post here; but just cos Andy did all the heavy lifting to bring me round, I'll select his answer. Thanks! –  mathmoggy Feb 28 '11 at 2:18
No problem. The more people there are thinking about projective space, the better! –  aaron Feb 28 '11 at 2:24

The projective plane, also called two-dimensional projective space, is by definition the set of all straight lines through the origin in three-dimensional Euclidean space. So a "point" in projective space is actually a line through the origin $O$ in the ordinary 3-space.

If you fix some plane in 3-space which doesn't pass through the origin (take the plane $z=1$ to be concrete), then each point $P=(x,y,1)$ in that plane corresponds to a unique element in the projective plane, namely the line in 3-space passing through $O$ and $P$. These lines (those that intersect the plane $z=1$) are the "finite points" in the projective plane, and they can be said to have finite coordinates $(x,y)$, so they look just like points in an ordinary Euclidean plane. But there are additional elements in the projective plane, namely the lines through $O$ which lie along the $xy$-plane (so that they never intersect the plane $z=1$). It's these lines that are called the "points at infinity" in the projective plane (and they, in turn, form a one-dimensional projective space, the "projective line at infinity", also known as the "horizon").

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Thanks Hans! I have learned a lot today, that's for sure... –  mathmoggy Feb 28 '11 at 2:20
You're welcome! I might add that the points at infinity are not really any different from the finite points, since every line through the origin in 3-space looks just like any other. The finite/horizon splitting is just an artifact coming from our choice of a particular plane (say $z=1$) to project upon. –  Hans Lundmark Feb 28 '11 at 7:52