Thoughts while drunk

Had a thought while I was drunk last night.

The rules for a finite projective plane are pretty straightforward: you have some finite number of points, and some finite number of lines. A line is simply some subset of your points. Every pair of lines must cross at one point, and every pair of points must have one line running through them.

If you rephrase these rules to “there must not exist a pair of points without a line running through them”, the second-simplest example is a space of one point p and one line {p}. But the first iteresting example is the space of seven points a-g and the seven lines {abc}{adf}{cef}{age}{cdg}{bfg}{bde}.

Now, aside from the pathological cases “one line, all the points on that one line” and “one point, all the lines through that one point”, there’s a rather pretty way of constructing a finite plane for any prime number p.

Build a p-by-p grid of points. These are our points px,y where x and y range over 0 to (p-1) – ie, the integers modulo p.

Through each point a0,y, construct p lines ly,z by starting the line at a0,y and then chosing a “slope” z. (of course, the line isn’t really sloped in projective space, just in the the way we draw it up as a grid). Thus the line ly,z consists of all points an,y+z*n (mod p).

Visually, its easy to confirm that for most points, there is a line running through them. Just grab your grid and shear it – moving each column down by one as you go across, and roll the points at the bottom of the column up to the top. This doesn’t disturb things all that much – it just alters the “slope” of the lines. Do this as many times as you like. If your two points are not in the same column, then after fewer than p clicks, there will be a horizontal line running through them. (if you prefer, you can dispense with the visual metaphor and just prove it from the definition of px,y and ly,z).

It is also easy to confirm that for most pairs of lines, they have one common point. To do this we observe that the grid height is a prime number, and the process of picking a sloped line (that wraps around) means that as each line ly,z goes across the columns of the grid, it visits each row once – that’s just a feature of multiplication modulo p. So pick one of your two lines and do the shear thing until it is horizontal. The other line must intersect it.

Now, obviously we have an issue. Two points ax,y in the same column y do not have a line running through them. Two ly,z lines that are parallel (have the same z) do not have a common point.

So we introduce the idea of there being points at infinity – which is where parallel lines meet.

For each set of lines an,z*n, we define a point a∞,z that all lines ly,z pass through.

Similarly (and more simply), for each column of points ax,n we define a line of “inifinite slope” – a vertical line lx,∞.

Awesome! But wait! Where do all these vertical lines intersect? There can be no point, because two different vertical lines have a disjoint set of points “on” them. There is an analagouse problem for the points at infinity – each belongs to a different family of parallel lines.

So we introduce a point a∞,∞ at which all the vertical lx,∞ intersect, and a line l∞,∞ joining all the points at infinity a∞,z.

One final problem: What line joins the points a∞,∞ and a∞,z? What point lies at the intersection of l∞,∞ and lx,∞? We address this by saying that the pointa t infinity is part of the line at infinity.

Problem solved. Every pair of points has one line, every pair of lines intersects at one point. Even better, all points have p+1 lines running through them, and all lines consist of p+1 points – including the ones at infinity.

Now, here’s the thing.

You can pick any point and any line running through it, and declare “These! These are now my point and line at infinity!” and rearrange all the other lines and points into a grid just like the original.

  1. Pick a point do be your point at infinity. There are p2+p+1 to choose from.
  2. choose one line on it to be your line at infinity. There are p+1 to choose from
  3. arrange the other lines into an order. There are p! ways to do this. Each line becomes a column in your grid.
  4. Choose, for column 0, an order in which the points appear. There are p! ways of doing this.
  5. Choose, for column 1, an order in which the points appear. There are p! ways of doing this.

And that suffices to determine a unique arrangement of the grid. If you don’t care about the arrangement of the grid, and only care about what points are at infinity or not, then there are just the (p2+p+1)(p+1) possible choices.

This process is much like choosing a point of view to paint a picture from, and the freaky thing is that infinity behaves just like any other point under these transformations.

Now … can we extend this to higher degree spaces? Can we construct a pn hypercube, with hyperplanes of all degrees 0 to n-1, with the rule that every pair of planes of degree x intersects at a distinct “plane” of degree x-1 and x+1? That it takes three “planes” to define a “point”, or n degree x objects to define a degree x+n object in general? It works ok in eucledian space – can it work with finite numbers of points?

I suspect it cannot, because the *only* projective planes (apart from the degenerate ones) are the planes constructed as above, and there’s a certain number of points involved.

But if it can be done, then observe that we can construct finite projective spaces based on any pn. And for all of them, we can rearrange them by choosing a point, line (etc) at infinity. It really feels suspiciously like the process of choosing a single polynomial with no root to define a finite field pn. Is there a link? The “zero” of the finite field – which is not part of the multiplication group – is a “special rules” singleton like a∞,∞. Can we construct a polynomial having no root by choosing a layout of a finite projective space, rather than having to hunt for one? Or is it just that there just a lot of problems of order pn that have no particular relationship with one another?


One Response to Thoughts while drunk

  1. rolawren says:

    Interesting, not sure I really understand it tho’. Maybe if I have another drink it might help.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: