Pop maths: A sliver of algebraic geometry – Part 3

This is the final post in this series, where I give a layperson’s explanation of a result from algebraic geometry, using only high school–level algebra as a starting point. If you’ve not read the first two in the series, you might want to start here.

In this post, we’re going to need a couple of facts from high-school algebra. The first is a technique for solving quadratic equations called “completing the square”. This refers to the fact that

    \[ ax^2 + bx = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a}.\]

You can verify this for yourself by expanding the brackets using the FOIL method on the right-hand side. The second algebra fact is that

    \[ (a - b)(a + b) = a^2 - b^2,\]

commonly referred to as the “difference of two squares”, and is another fact you can check for yourself using FOIL.

The claim

Recall that a conic in \P^2 is a curve given by an equation that is homogeneous of degree 2. That is, an equation of the kind

(1)   \begin{equation*} \alpha_1 X^2 + \alpha_2 XY + \alpha_3 XZ + \alpha_4 Y^2 + \alpha_5 YZ + \alpha_6 Z^2 = 0,\end{equation*}

where the \alpha_i‘s are any number — we only exclude the possibility that all six \alpha_i‘s are zero. This covers every possible equation you can make with 3 letters multiplied together in pairs, once all term have been moved to the left hand side (there’s no reason whatever to think of X^2 = Y^2 and X^2 - Y^2 = 0 as separate equations when all we want to study is the geometry of the solutions).

In the first article, I talked about the affine plane \A^2, and how curves there are equivalent if one can be transformed into the other by a substitution

    \begin{align*} x' &= ax + by \\ y' &= cx + dy, \end{align*}

where ad - bc \neq 0, which we referred to as non-degeneracy. These substitutions can equivalently be thought of as the transformation of one figure into another equivalent figure, or as simply a change of the coordinate system used to describe \A^2; they are two sides of the same coin, and you can choose your preferred interpretation. The same thing is true in \P^2, but using the letters X, Y, and Z rather than x and y. This suggests that some conics might be equivalent to others. Can we classify the equivalent types of conics in \P^2? And what relation will they have to Apollonius’ classification of the Euclidean/affine conic sections?

I claim that up to equivalence, there are really only five conics in \P^2, despite the infinitely many equations of the type (1) that are possible.

Proving the claim

One way to show this is to employ a technique called the Lagrange diagonalization algorithm. It uses coordinate transformations like the kind above to remove the “mixed” terms XY, XZ, and YZ from the picture. I’ll now explain the algorithm and give an example. This is definitely the most formula-heavy part of the whole series, so feel free to just skim-read it and pick up again at the conclusion, although the techniques used are nothing more than the two facts from basic algebra outlined in the intro.

  1. Look at the left hand side of your equation, and look for a squared term like X^2, Y^2, or Z^2. When you find one, look for a mixed term containing the squared letter. So for example, if you find X^2, look to see if there is a term such as XY or XZ. If there is no such mixed term, nothing to do here, move on to the next squared term, such as Y^2. If there is, then “complete the square” (as in the introduction) on the squared and mixed term you have found. For example, if the terms are X^2 and XY, rewrite them like this

        \[aX^2 + bXY = a\left(X + \frac{bY}{2a}\right)^2 - \frac{b^2Y^2}{4a}.\]

    Then just make the non-degenerate transformation.

        \begin{align*} X' &= X + \frac{b}{2a}Y \\ Y' &= Y \end{align*}

    This means that

        \[aX^2 + bXY = a\left(X + \frac{bY}{2a}\right)^2 - \frac{b^2}{4a}Y^2 = a{X'}^2 - \frac{b^2}{4a}{Y'}^2.\]

    Notice how we have essentially “killed off” a mixed term using the coordinate transformation. With a little re-arranging of the two substitutions above, this also implies that X =  X' - \frac{b}{2a}Y', so if there is another mixed term, say XZ, we can swap the X for X' - \frac{b}{2a}Y'. This means there are no more Xs and Ys in the equation, only X' and Y'. We can then repeat the process outlined about on the resulting {X'}^2 and X'Z term. Once this is done, look for more squared terms you can use to eliminate mixed terms with further transformations—these may now be terms like {Y'}^2 and so on.

  2. But what if there are no squared terms at some stage in the algorithm? What if we are stuck with only mixed terms? In that case, we can take a mixed term like XY and make the non-degenerate transformation

        \begin{align*} X &= X' + Y' \\ Y &= X' - Y', \end{align*}

    so that XY = (X' + Y')(X' - Y') = {X'}^2 - {Y'}^2 by the “difference of two squares” formula above. Now we have some squared terms to work with, but we may have created some more mixed terms in the process. So go back to step 1. and use the newly created squared terms to eliminate more mixed terms.

  3. Repeat steps 1 and 2 until there are no more mixed terms left. The resulting equation is not uniquely defined: it depends on the choices you made along the way about what order to do things in. You and I could each perform the algorithm and get different results. We’ll deal with that after.Let’s do a fully worked example of this part of the algorithm to make sure that it is clear. Our conic is defined by an equation, say, 2XY - 3YZ = 0. There are no squared terms, so let’s pick XY and make the transformation in step 2: X = X' + Y', and Y = X' - Y'. Thus the equation becomes

        \[2(X' + Y')(X' - Y') - 3(X' - Y')Z = 2{X'}^2 - 2{Y'}^2 - 3X'Z + 3Y'Z = 0.\]

    Remember: this equation defines an equivalent projective curve to the original equation, because all we did was make a valid non-degenerate change of coordinates. Now we can use the {X'}^2 term to kill off the X'Z term. Just write 2{X'}^2 - 3X'Z = 2(X' - \frac{3}{4}Z)^2 - \frac{9}{8}Z^2. Then we can make the change of coordinates X'' = X' - \frac{3}{4}Z, Z' = Z, so the equation becomes:

        \[2{X''}^2 - 2{Y'}^2 - \frac{9}{8}{Z'}^2 + 3Y'Z' = 0.\]

    Finally, we can use {Y'}^2 to eliminate the Y'Z' term. I can write

        \[-2{Y'}^2 + 3Y'Z'  = \frac{9}{8}{Z'}^2 - 2\left(Y' - \frac{3}{4}Z'\right)^2,\]

    and then make the transformation Z'' = Z' and Y'' = Y' - \frac{3}{4}Z'. So the equation becomes

        \[2{X''}^2 - \frac{9}{8}{Z'}^2 - 2{Y''}^2 + \frac{9}{8}{Z'}^2 = 2{X''}^2 - 2{Y''}^2 = 0.\]

    Now there are no more mixed terms, so we are done.

  4. If we relabel all our symbols so that the 's are hidden, we are left with an equation that looks like

        \[aX^2 + bY^2 + cZ^2 = 0, \]

    where a, b, and c are some combination of the \alpha_i‘s in the original equation, and may or may not be zero. The final step is to make the non-dengerate transformation X' = \sqrt{a}X, and then again for Y' = \sqrt{b}Y and Z' = \sqrt{c}Z. Thus we have the following situation. If a is positive, then aX^2 = (\sqrt{a}X)^2 = {X'}^2. If a is negative, then aX^2 = -(\sqrt{a}X)^2 = -{X'}^2. If a = 0, there’s nothing to do. The same is true analogously for b and Y, and c and Z. After making these changes, we are left with a new equation of the form a' {X'}^2 + b'{Y'}^2 + c'{Z'}^2 = 0, where a', b' and c' are either 1, -1, or 0, and the number of 1s, -1s, and 0s appearing in this form is uniquely defined. You and I could do the algorithm, or indeed any algorithm using valid coordinate changes, to get the equation to a form with only 1s, -1s and 0s as coefficients, but we’d always end up with the same number — the only possible difference is the names of the variables, but we can always swap these around anyway. This is a result called (for some mysterious reason) “Sylvester’s law of inertia”1, but it’s a bit more advanced so I won’t prove2 it here.

Therefore, if we swap the names of the variables if need be, and perhaps replace an equation with its negative (since aX^2 + bY^2 + cZ^2 = 0 and -(aX^2 + bY^2 + cZ^2) = 0 clearly have the same solutions), then every conic in \P^2 is equivalent to one of the following five conics:

    \begin{align*} X^2 &= 0 \\ X^2 - Y^2 &= 0 \\ X^2 + Y^2 &= 0 \\ X^2 + Y^2 - Z^2 &= 0 \\ X^2 + Y^2 + Z^2 &= 0. \end{align*}

That’s it for the algebra. Now what about the geometry?


In the previous article, I wrote that we could obtain an affine curve from a projective curve by simply writing (X : Y :Z) = (X/Z : Y/Z : 1), and then studying the points in \A^2 corresponding to (x, y)= (X/Z, Y/Z). These are just the points of the projective curve where Z \neq 0, and is commonly called an “affine piece” of the curve. The “shortcut” method for doing this calculation is just to rewrite the equation, but replacing Z with 1, and replace X by x and Y by y. Geometrically, we are intersecting the curve with a copy of \A^2, and studying the intersection. Hence we will do that for each of the five conics to see what these affine pieces of the curve look like, and then briefly examine the points at infinity with respect to the affine pieces.

First, X^2 = 0. Wherever Z \neq 0, its points are equivalent to (X/Z)^2 = x^2 = 0. This implies that x = 0 on these points. However, y could be anything. So the affine part of this curve contains all points (0, y) — it’s just a line running along the y-axis in our coordinate system. Is this a conic section? Well, actually yes: it’s a particularly ugly conic section obtained by intersecting a cone with a plane that just glances the edge of the cone:

The point (0 : 1 : 0) also satisfies the equation. This point does not lies in the affine piece where Z \neq 0, but is the point at infinity where the opposite ends of the affine line meet. Projective lines are like circles — they loop back on themselves.

Now for X^2 - Y^2 = 0. Repeating the process, we have obtain the affine curve x^2 - y^2 = 0. But that’s a difference of two squares, so this can also be written as (x - y)(x + y) =0. This equation is true whenever (x - y) is zero, or (x + y) is zero. Therefore this curve is comprised of two straight lines, described by x - y = 0 and x + y = 0, which intersect at the point (0,0). It’s a cross shape, and is also a degenerate conic section:

The points where Z = 0, are (1 : 1 : 0) and (1 : -1 : 0). These are the points the two lines disappear off to in the distance.

Then we have X^2 + Y^2 = 0. In the projective plane, because (0 : 0 : 0) does not refer to a projective point, this equation holds only at (0 : 0 : 1). However, this point does lie in the affine plane where Z \neq 0, so when we take it to the affine piece, we find that x^2 + y^2 = 0 also defines a single point, sitting at (0,0). It corresponds to another degenerate conic section—when the plane only cuts at the point of the cone.

Now we come to the two most interesting cases. Let’s rewrite the curve equation X^2 + Y^2 - Z^2 = 0 to X^2 + Y^2 = Z^2 and study the affine part Z \neq 0. The affine curve we obtain here is x^2 + y^2 = 1: an affine ellipse, one of Apollonius’s conic sections. On the other hand, if Z = 0, then we have X^2 + Y^2 = 0 which has no projective solutions, and so there are no points at infinity. But that’s exactly what we would expect, because ellipses do not extend infinitely far.

But what about the other conic sections of Apollonius? If we relabel variables to X = X', Y = Z' and Z = Y', we have {X'}^2 + {Z'}^2 - {Y'}^2 = 0, and then write this as {X'}^2 - {Y'}^2 = {Z'}^2, we can study the part where Z' \neq 0. But that’s just the affine curve {x'}^2 - {y'}^2 = 1: precisely the points of a hyperbola. If we check the points at infinity by examining the curve where Z' = 0, we have {X'}^2 - {Y'}^2 = 0, with solution (1 : 1 : 0) and (1 : -1 : 0). These are the same points at infinity as with the cross above, and with a picture, we can see why:

The arms of the hyperbola meet with the cross at the same points at infinity.

What about the parabola, the third conic section? Returning to X^2 + Y^2 - Z^2 = 0, we can write this using the difference of two squares formula Y^2 - (Z- X)(Z + X) = 0, and then making the non-degenerate change of coordinates X' =Z - X and Z' = Z + X. Hence we have Y^2 - X'Z'= 0. Then at the part of the curve where Z' \neq 0, we have y^2 = x', the parabola. There is a single point infinitely far from this parabola, at (1 : 0 : 0) in (X' : Y : Z') coordinates.

These considerations lead us to the inescapable conclusion that the three conic sections identified by Apollonius are just three pieces of a single projective curve X^2 + Y^2 - Z^2 = 0. The projective plane unifies our vision of all three curves, which can now be studied as a single geometric entity. It is an example of how even ideas many hundreds of years old can be given a new perspective by the abstract power of more modern mathematics.

Now we come to the final projective conic, given by the equation X^2 +Y^2 + Z^2 = 0. This one presents an unfortunate problem, because it doesn’t seem to describe a projective curve at all. It could only be solved if X, Y, and Z were all zero, but (0 : 0 : 0) is not a point in projective space. This equation doesn’t have any geometry.

That troublesome fifth conic

But perhaps we’re being too hasty. There is a way out, but it involves moving to an even more abstract space. The complex numbers are the number system obtained by adding square roots to negative numbers, and are a foundational part of much of modern mathematics and physics. And we can transfer everything we have done so far to the complex world. We can study equations where the numbers involved are complex numbers, form an affine plane \A_\C^2 that has points given in coordinates as pairs (a, b) where a and b are each complex numbers, and form the complex projective plane \P_\C^2 by dividing three-dimensional complex affine space into complex lines through the origin, and has coordinates as ratios (A: B : C) where each entry is complex. By this point, there is no way that any human can imagine what a complex projective space “looks” like3. We can work with it perfectly well mathematically, but any visualization we do is via analogy, which with training and intuition can be surprisingly accurate guides to navigating this strange world.

We can study conics given by equations

    \[\alpha_1 X^2 + \alpha_2 XY + \alpha_3 XZ + \alpha_4 Y^2 + \alpha_5 YZ + \alpha_6 Z^2 = 0,\]

where now the \alpha_is are complex numbers, and even repeat steps 1-3 of the algorithm above. In step 4, however, we can even go a step further. In the complex world, there is no distinction between positive and negative numbers. Once we have arrived at the form aX^2 + bY^2 + cZ^2 = 0, where a, b, and c, are complex we can just make the change of coordinates X' = \sqrt{a}X, Y' = \sqrt{b}Y, and Z' = \sqrt{c}Z. These square roots always exist in the complex numbers, no matter what a, b and c are.4  Now the curve can only possibly be in one of three forms (after possibly swapping the names of the variables):

    \begin{align*} X^2 &= 0 \\ X^2 + Y^2 &= 0 \\ X^2 + Y^2 + Z^2 &= 0, \end{align*}

simply depending on how many of a, b, and c were zero.

Classical — and much of modern — algebraic geometry is really the study of curves, surfaces, and higher dimensional geometric objects living in complex projective spaces. In the complex world, every algebraic equation has a geometric meaning. The theory is more unified, and frankly easier, owing to a number of powerful results about complex geometry that make the correspondence between geometry and algebra as rich as can be.

Let’s take one last look at that final curve X^2 + Y^2 + Z^2 = 0. In the complex case, this definitely has solutions, for example (1 : \sqrt{1} : 0). In fact, it has infinitely many. And as a final nod to our friend Apollonius, we can make another non-degenerate change of coordinates Z = \sqrt{-1}Z', so that the equation becomes

    \[X^2 + Y^2 - {Z'}^2 =0.\]

In the complex projective plane, that fifth conic, the most degenerate and problematic of them all, is one and the same with the fourth conic, which was the most beautiful and interesting. Apollonius’s conic sections live inside it, as the complex numbers contain the ordinary “real” numbers, and so this complex curve contains all the original solutions to the fourth equation. The whole family of conic sections, studied by the ancient Greeks, and later by Islamic and Renaissance mathematicians, are still there despite the long journey they have been on, through increasingly abstract spaces, from \mathbf{E}^2, to \P^2, to \P_\C^2.


  1. Link
  2. See, for example Roman’s Advanced Linear Algebra, page 269, or the lecture notes here. The theorem is usually stated using matrices whose entries correspond to the \alpha_i‘s in our equation.
  3. Actually that’s not quite true. The complex projective line \P_\C^1 looks exactly like a sphere. But a complex projective plane? No chance
  4. Actually every non-zero complex number has two square roots, so to make this change of coordinates we have to choose one.