[New release HaskellForMaths-0.1.8 - contains bugfix to Fractional instance for ExtensionField, plus documentation improvements.]

Over the last two weeks, we have looked at the finite affine geometries AG(n,Fq). Hopefully, this has all been fairly straightforward - affine geometry is just the familiar Euclidean geometry of points and lines, but without angles or distances, and all we have done is replace the reals R by the finite fields Fq.

This week we're going to look at finite projective geometries. Now, I know from my own experience that it can take a little while to "get" projective geometry. There are probably several reasons for this, but one of them is that mathematicians define projective geometry one way, but then most of the time think about it in a different way. So, I'll do my best to explain it, but don't worry if you don't get it at first, one day it will all click into place.

Okay, so let's start with the points. The points of PG(n,Fq) are defined to be the lines through the origin (ie the one-dimensional subspaces) in Fq^(n+1). That should sound a bit strange at first reading - how can the points (in PG(n,Fq)) be lines (in Fq^(n+1))? Bear with me - we'll see why it makes sense to think of them as points.

For example, the points of PG(2,F3) are the lines through 0 in F3^3. So the line {(0,0,0), (0,1,0), (0,2,0)} is a point of PG(2,F3), as is the line {(0,0,0), (1,0,2), (2,0,1)}. (Recall that arithmetic in F3 is modulo 3.) As a line in F3^3 is a one-dimensional subspace, it is generated by any non-zero point on the line, and consists of all scalar multiples of such a point. So we could represent the two lines we just mentioned as <(0,1,0)> and <(1,0,2)>, with angle brackets meaning "the line generated by". Any non-zero point on the line will do to represent the line, but it would be good to have a way of choosing a canonical representative. For the moment, let's say that we will choose the point whose last non-zero coordinate is 1. Here, then, are the points of PG(2,F3):

So the blue points are the points of the form (x,y,1). Every point of the form (x,y,1) represents a line through 0 in F3^3. However, this is not all the lines through 0, for it misses out those lines which are in the plane z = 0. The green points are the points of the form (x,1,0). Each point of the form (x,1,0) represents a line through 0 in the plane z = 0. However, this is still not all, for we are missing the line y = z = 0. This is represented by the red point (1,0,0). And that's it - these are all the "points" of PG(2,F3). (Confirm for yourself that every line through zero in F3^3 is represented by one of these "points".)

Since any scalar multiple of a point in F3^3 represents the same line through 0, and hence the same point of PG(2,F3), mathematicians often write the representatives as (x:y:z), with the colons indicating that it is only the ratios we are interested in. Thus (0:1:0) and (0:2:0) represent the same line in F3^3, and hence the

*same*point of PG(2,F3). (Note that (0:0:0) is*not*a point of PG(2,F3).)Okay, so I mentioned that mathematicians define PG(n,Fq) one way, but then think about it another way. So how do mathematicians think about it?

Well, the blue points in PG(2,F3) look just like a copy of AG(2,F3), don't they. Then the green points are called "the line at infinity". Finally, the red point is called "the point at infinity". So mathematicians think of PG(2,F3) as being like AG(2,F3), but with some additional points "at infinity". In particular, although PG(2,F3) was defined in terms of lines through 0 in F3^3, mathematicians actually think of it as consisting of points (not lines).

The idea that the additional points are "at infinity" comes from perspective drawing. Imagine an artist sitting in front of a canvas, looking along the z-axis. Their eye is the origin. The canvas is the plane {(x,y,1)}. To draw what they see, the artist needs to project a line from their eye, through the canvas, until it hits something. Given a large enough canvas, the artist can draw anything that is in front. However, things that are above or to the sides would have to be drawn "at infinity".

One thing I should emphasize is that, contrary to appearance, the line and point at infinity are no different from the other points. Indeed, I could have had my artist looking along the x-axis instead of the z-axis. In that case, the embedded affine plane would have been the points {(1,y,z)}, the line at infinity would have been {(0,1,z)}, and the point at infinity would have been (0,0,1). If we revert to thinking about lines in F3^3 for a moment, it should be clear that the points at infinity are not distinguished in any way from other lines in F3^3. Their appearance of being distinguished has arisen solely from our choice of coordinate system.

Okay, time for some code. In the above, we said that the canonical representative for a line through 0 would be the point with 1 as its last non-zero coordinate - corresponding to looking along the z-axis. That made things easier to explain. However, it turns out that it's usually more convenient to choose the point with 1 as its

*first*non-zero coordinate (ie, looking along the x-axis). Here is the code to list the points of PG(n,Fq):

`ptsPG 0 _ = [[1]]ptsPG n fq = map (0:) (ptsPG (n-1) fq) ++ map (1:) (ptsAG n fq)`

For example:

> ptsPG 2 f3[[0,0,1],[0,1,0],[0,1,1],[0,1,2],[1,0,0],[1,0,1],[1,0,2],[1,1,0],[1,1,1],[1,1,2],[1,2,0],[1,2,1],[1,2,2]]> map reverse it[[1,0,0],[0,1,0],[1,1,0],[2,1,0],[0,0,1],[1,0,1],[2,0,1],[0,1,1],[1,1,1],[2,1,1],[0,2,1],[1,2,1],[2,2,1]]

If we reverse the coordinates, you can see that we have the point at infinity, the line at infinity, and the embedded copy of AG(2,F3), as before.

That's probably enough to take in at one sitting. Next time, we'll look at lines in PG(n,Fq).

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