Pop maths: A sliver of algebraic geometry – Part 2

You might want to check out part 1 before reading this.

A little motivation

Last week, we took the Euclidean plane \mathbf{E}^2 and forgot what angles and distances were, while retaining the concept of parallel lines. This gave us a new space called the affine plane, \A^2. Our mission today is to describe an extension of the affine plane called the projective plane, much beloved by algebraic geometers.

Why do algebraic geometers like projective space? One reason is that as well as studying the geometry of equations, they like to study the geometry of systems of equations. Consider the pair of lines:

    \begin{align*} x + y &= 1 \\ x - y &= 2. \end{align*}

A pair (a, b) that satisfies both equations is a point lying on both lines—the only one is (\frac{3}{2}, -\frac{1}{2}). Now consider

    \begin{align*} x + y &= 1 \\ x + y &= 2. \end{align*}

It is obvious that these two equations have no common solutions, and therefore, there’s no geometry lying behind this system. Each equation individually describes a line parallel to the other. The system as a whole describes the intersection of two parallel lines: nowhere.

When standing in the middle of a road it seems like the parallel edges of the road do meet in the distance. What if there were a way to make this idea of parallel lines meeting at an infinitely far away point a mathematical reality?

The projective plane is a way of extending the affine plane so that parallel lines really do meet at infinity. Rather than being some “hack”, this really is where figures defined by algebraic equations “want” to live. It gives us a more complete, unified picture of algebraic geometry. We don’t lose any of the geometry of the affine situation: wherever you’re standing in projective space, everything “looks” affine from your perspective. Nonetheless, there is a broader perspective that allows us to see the infinitely far-away horizon, just like how on planet Earth everything looks flat, until we go up in a space shuttle and see the true geometry from space. So what is the projective plane?

How to make a projective plane

When introducing a new concept in mathematics, it is best if it can be fully described using objects and concepts that are already well understood and defined, to guarantee the new concept fits in with the rest of mathematics. One common way to do this is via an equivalence relation: take an existing structure, divide it into distinct parts called equivalence classes, and then use the classes as single entities in a new structure, retaining all the properties of the old structure except those that were destroyed in the partitioning process. We’ll use a space we already know — the affine space — to construct the projective plane. However, once the construction is complete, the projective plane will stand as a mathematical entity on its own; rarely will we have to lift the hood to see how it was originally built.

The first step is to introduce the affine space \A^3. There’s nothing mysterious here. You’ve met the 2d affine plane \A^2. So the 3d affine space is just like ordinary 3d space but with no concept of angle or length, but parallel lines (and now, planes) are to be respected. The next step is to choose a privileged point, which we’ll called O for now — any point will do for this construction.

The equivalence relation I’m going to introduce on the affine space is to say that two points P and Q are in the same class if there is a line through P and Q that also passes through O, the privileged point, in which case we write P \sim Q to show they are now “equivalent”, but not equal. There’s a slight problem with this though: under this definition every point P would be in the same class as O, since for any P there is certainly a line passing through P and O that also passes through O. The easy fix is to “puncture” the space, removing O from all further considerations.

The points P and Q are equivalent because they lie on the same point through O, here drawn at the base of the axes

Now every point R in \A^3 (with O missing) is equivalent to every other point on an infinitely long line through R and where O used to be. These lines through O are the equivalence classes of our equivalence relation. And now for the punchline: the points of the projective plane \P^2 are the equivalence classes of the relation \sim described above.

The projective plane is the set of all lines through O.

In other words, \P^2 is a geometric space whose points are lines through a chosen point in \A^3. Or in fancy mathematical notation1:

    \[ \P^2 =( \A^3 - \{O\}) / \sim.\]

If you find this hard to get your head around, don’t worry: it is. We can’t even properly draw a projective curve; it is too far removed from ordinary 2d space. It isn’t even clear why this is a 2d plane at this stage.

Getting our bearings

Let’s add some coordinate to make this plane easier to navigate. If I briefly go back to \A^3, I can choose a coordinate system that has O = (0,0,0). What this means for the equivalence relation is that P \sim Q if and only if, supposing P = (\alpha, \beta, \gamma), then Q = (\lambda \alpha, \lambda \beta, \lambda \gamma) where \lambda is a non-zero number. So for example, (1, 1, 2) \sim (2, 2, 4). Since these two points lie on the same line through O in \A^3, they represent the same point in \P^2.

Now we just let \P^2 inherit this coordinate system: a point P in \P^2 is given coordinates by taking the coordinates of any point in the line that compromises it in \A^3. The coordinates for a point in \P^2 are therefore not unique. However, the point is uniquely determined by the ratios between the coordinates, so we write

    \[P = (\alpha : \beta : \gamma) = (\lambda \alpha : \lambda \beta : \lambda \gamma)\]

for any non-zero \lambda, where the colons remind us that it is ratios that count. Of course, (0 : 0 : 0) does not refer to any point in \P^2, because we excluded it from \A^3 when constructing the space. And that’s it: we no longer have to think about \A^3 at all. We are just working on a strange plane, where points are uniquely identified by a ratio of 3 coordinates. Let’s do some geometry and see if we can get more of a feel for this.

Using these special coordinates I can now explain why \P^2 is an extension of \A^2. I somehow have to provide a method for embedding the entirety of \A^2 into \P^2 without losing any information—each point in \A^2 should be correspond to a unique point in \P^2. The trick here is as follows: by the rule that it is coordinate ratios that matter, any point (X : Y : Z) in \P^2 where Z \neq 0 is the same as (\frac{X}{Z}: \frac{Y}{Z}: 1). If we then just look at the first two coordinates, then (\frac{X}{Z}, \frac{Y}{Z}) is point on the affine plane. By a reverse argument, we can choose any Z we like, and assign a point on the affine plane by

(1)   \begin{equation*} (x, y) = \left(\frac{X}{Z}, \frac{Y}{Z}\right) \mapsto \left(\frac{X}{Z} : \frac{Y}{Z} : 1\right) = (X : Y : Z). \end{equation*}

For simplicity, we can just choose Z = 1, so that (x, y) \mapsto (x : y : 1). 2. The important thing is that each pair (x, y) is of a different ratio to Z, and so corresponds to a different point in \P^2. Therefore \P^2 contains \A^2. In fact there are many ways to contain \A^2 inside \P^2, for instance by choosing a different value for Z, or by sending (x, y) to (x : 1 : y), or however you please, so that wherever we are in \P^2, there is a local neighbourhood that looks just like \A^2. This means that to study just some region of a figure in \P^2, we can just use the comparatively simpler methods of affine geometry. For this post we’ll stick with the simple (x, y) \mapsto (x : y : 1) embedding as much as possible.

The line at infinity

Now let’s see what this means for the parallel lines x + y = 1 and x + y = 2 we met above. We can embed the points as we described above by (1), so that these affine lines now lives in \P^2. But we can go a step further, and continue the lines in \P^2 by applying this same transformation to the equations themselves:

    \begin{align*} x + y = 1 \quad &\mapsto  \quad\frac{X}{Z} + \frac{Y}{Z} = 1 \quad \text{ and } \\ x + y = 2 \quad&\mapsto \quad \frac{X}{Z} + \frac{Y}{Z} = 2. \end{align*}

It is not hard to see that, after choosing an embedding, say by setting Z = 1, then the solutions to these equations in X and Y correspond to solutions of the corresponding affine lines. But then if we multiply through by Z, we obtain the following relations between projective coordinates that lines satisfy:

    \begin{align*} X + Y &= Z \\ X + Y &= 2Z. \end{align*}

These new “projectivized” equations do have a solution in projective coordinates. Namely, the point (1 : -1 : 0). This point does not lie on our affine plane at all, because we chose our affine plane to specifically be that subset of \P^2 where Z \neq 0. Rather, it is “infinitely far away”.

The set of all points (X : Y : 0) are therefore the “points at infinity” with respect to this copy of \A^2 that lives inside \P^2, and there aren’t that many of them: there’s point (X: 1 :0) for each number X, forming an (affine) line, plus the point (1: 0: 0). Together, we call this “line at infinity”3. Thus \P^2 is simply \A^2 with one additional line added. This is an intuitive reason that \P^2 deserves to be called a 2d plane—It is much more like \A^2 than the original \A^3 that it was constructed from.

Projective algebraic curves

Finally, for this rather long and challenging post, we’ll consider solutions of equations in the projective plane. There’s a problem. Suppose I have an equation, like X^2 + Z = 1. A triple of coordinates such as (1 , 1 , 0) satsifies this equation. However, (2, 2, 0), which when considered as projective coordinates is the same point as (1 : 1 : 0), does not. The fix for this is to consider the solutions of homogeneous equations only. These are equations in which every term is of the same degree. So for example, X^2 + YZ = 0 is homogeneous, but X^3 + 3Z = 1 is not. Since we’re studying conics, which we defined to be algebraic curves of degree 2, we require every term to be of degree 2. This means that if we have the general conic equation

    \[ a_0X^2 + a_1 XY + a_2XZ + a_3Y^2 + a_4 YZ + a_5 Z^2 = 0,\]

then any coordinate triple representing the same point in \P^2 will satisfy the equation, since if we have two coordinates of the same point (\alpha :  \beta : \gamma) = (\lambda \alpha : \lambda \beta : \lambda \gamma) (with \lambda \neq 0, of course), then

    \[ a_0\alpha^2 + a_1 \alpha\beta + a_2\alpha\gamma + a_3\beta^2 + a_4 \beta\gamma +a_5 \gamma^2 = 0,\]

if and only if

    \begin{align*} &a_0(\lambda\alpha)^2 + a_1 (\lambda\alpha)(\lambda\beta) + a_2(\lambda\alpha)(\lambda\gamma) + a_3(\lambda\beta)^2 + a_4 (\lambda\beta)(\lambda\gamma) +a_5 (\lambda\gamma)^2\\ {}={}&\lambda^2 ( a_0\alpha^2 + a_1 \alpha\beta + a_2\alpha\gamma + a_3\beta^2 + a_4 \beta\gamma + a_5 \gamma^2)\\{}={}& 0. \end{align*}

This insistence on studying only homogeneous equations may seem like a limitation. However, we have already seen that have a way of switching seamlessly between non-homogeneous equations for plane curves in \A^2 and into homogeneous equations in \P^2, and switching back is just as easy. For example the hyperbola:

    \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \rightsquigarrow \frac{(X/Z)^2}{a^2}- \frac{(Y/Z)^2}{b^2} = 1 \rightsquigarrow \frac{X^2}{a^2} - \frac{Y^2}{b^2} = Z^2,\]

and this projective curve contains a perfect copy of the original affine curve.

That covers all I’d like to say about the projective plane. But where is all this going? What are the equivalents of Apollonius’ conic sections in the projective plane? How many are there, and how do they relate to the original conic sections from the first post? Is there anything to be gained from migrating to an even weirder geometric space? Find out in the final post of the series.

  1. This is included just as a point of interest; you don’t need to understand this notation to understand the article. See if you can guess how the notation describes what we have done, by giving new meanings to familiar arithmetic symbols like - and /.
  2. Any other choice of Z just stretches/shrinks the plane before embedding it, but remember, shrinking and stretching don’t matter on the affine plane!
  3. Roughly, if all possible (X:Y:Z) spans a projective plane, then (X:Y:0) spans a projective line, which itself decomposes into an affine line and a point. Ask me in the comments for more details if you like.