2 Points at infinity
If you are already familiar with the basics of projective geometry, feel free to skip this section.
OK, time to come clean: proving Pappus’ theorem as stated previously is going to be tricky. Not because it’s a particularly hard theorem to prove, but because in its current form it isn’t strictly speaking true. Well, maybe that’s a bit harsh. The problem is that the theorem’s hypothesis assumes the existence of three points: \(AY \cap BX\), \(AZ \cap CX\) and \(BZ \cap CY\). Of course all these lines exist, but who’s to say that they intersect? Indeed, there are instances of the Pappus configuration where two of those lines are parallel. If we assume that they all intersect then the theorem works fine, but perhaps a more elegant solution exists.
This is where things get projective. When presenting projective geometry, people tend to focus on certain applications. They bring up perspective, projections and even the Italian Renaissance. Sometimes such introductions can feel a little overwrought. As with many parts of maths, my preferred approach revolves around one simple idea: convenience. It’s really annoying that lines don’t always intersect! We should change that. How? We just need to add a few new points.
The initial idea is to notice that, if we imagine the plane \(\mathbb{R}^{2}\) embedded into 3-dimensional space \(\mathbb{R}^{3}\) at some fixed non-zero height (for example the \(z=1\) plane), then to every point in that plane we can associate the line which passes though both that point and the origin.
While it is possible to assign an origin-intersecting line in \(\mathbb{R}^{3}\) to every point in our embedded \(\mathbb{R}^{2}\), there are a few extra origin-intersecting lines that will never be assigned to a point, namely the lines contained in the \(z=0\) plane. But that’s exactly what we’re after: some extra points! So let’s just rush optimistically ahead and make an extended plane called the projective plane, denoted \(\mathbb{RP}^{2}\), whose points are exactly the lines in \(\mathbb{R}^{3}\) that pass through the origin.
A point in \(\mathbb{RP}^{2}\) is a line in \(\mathbb{R}^{3}\) which passes through the origin. To visualise points in \(\mathbb{RP}^{2}\) we will often draw their intersection with the plane \(z=1\) (when it exists), as illustrated by the figure above.
So, if points in \(\mathbb{RP}^{2}\) correspond to a lines passing through the origin, what does a line in \(\mathbb{RP}^{2}\) correspond to? Pleasingly, the answer is a plane passing through the origin.
A line in \(\mathbb{RP}^{2}\) is a plane in \(\mathbb{R}^{3}\) which passes through the origin. In accordance with this terminology, three points in \(\mathbb{RP}^{2}\) are said to be collinear if they all lie within the same plane. To visualise lines in \(\mathbb{RP}^{2}\) we will often draw their intersection with the plane \(z=1\) (when it exists), as illustrated by the figure above.
The first thing to note is that this new space perfectly solves our original problem. Two origin-intersecting planes always intersect in an origin-intersecting line so we automatically know that two lines in \(\mathbb{RP}^{2}\) intersect in a point in \(\mathbb{RP}^{2}\). How does this work when the two lines are parallel lines in \(\mathbb{R}^{2}\)? Let’s have a look.
This is exactly what we want: the projective lines intersect in one of our extra projective points i.e. they intersect in an origin-intersecting line in \(\mathbb{R}^{3}\) that doesn’t intersect our old embedded \(\mathbb{R}^{2}\). Such points are called points at infinity, a name that seems fitting for the intersection points of parallel lines. The set of all points at infinity gives a projective line, the unique line at infinity.
We now know what a point in \(\mathbb{RP}^{2}\) is, but practically speaking, how are we going to write one down? Well there’s a surprisingly straightforward system for this called homogeneous coordinates. The idea is to simply specify the coordinates of a single non-zero point in \(\mathbb{R}^{3}\) that lies on the line.
In the figure above you can move both the light blue dot (which changes the point in \(\mathbb{RP}^{2}\)) and the pink dot (which gives possible homogeneous coordinates for that point). This figure illustrates how homogeneous coordinates are only defined up to scale.2 Despite this, they are extremely useful as the following proposition demonstrates.
Let \(A\), \(B\) and \(C\) be points in \(\mathbb{RP}^{2}\) with corresponding homogeneous coordinates \((a,b,c)\), \((d,e,f)\) and \((g,h,i)\). Then the following are equivalent,
- \(A\), \(B\) and \(C\) are collinear
- \(\det M = 0\)
where \(M\) is the matrix \({\small \begin{pmatrix} a&b&c \\[-2ex] d&e&f \\[-2ex] g&h&i \end{pmatrix}}\).
Proof. By definition, \(A\), \(B\) and \(C\) are collinear if and only if the vectors \((a,b,c)\), \((d,e,f)\) and \((g,h,i)\) lie within a common plane. This is equivalent to requiring that there exists a linear relation3 between these three vectors, which is equivalent to requiring that \(\det M = 0\). \(\square\)
Now that we have the basics of projective geometry in hand, there remains one technical result we should go over before proceeding with Pappus’ theorem. As the focus of this essay is uncovering cool facts hidden behind Pappus’ theorem, rather than establishing the foundations of projective geometry, we won’t see a proof of this result (although a reference will be provided).
A good way of thinking about projective geometry is that it’s just another attempt to solve geometric problems with algebra (like Cartesian coordinates). What distinguishes projective geometry is that we insist that the algebra be linear, which is why everything is represented by lines and planes that pass through the origin. A big part of the subject is therefore simply bringing the tools of linear algebra into play. If you’ve ever studied any linear algebra you’ll know that bases4 play an important role. The analogous notion in projective geometry is a collection of points in general position.
Four distinct points in \(\mathbb{RP}^{2}\) are said to be in general position when no three of them are collinear.
So how do we actually use such a collection of points? Once again, we bring it back to linear algebra; having a collection of points in general position allows us to choose a special basis.
Let \(A,B,C,D \in \mathbb{RP}^{2}\) be four distinct points in general position. Then there exists a basis \(\mathcal{B}\) of \(\mathbb{R}^{3}\) such that the homogeneous coordinates of \(A,B,C\) and \(D\) are \((1,0,0)\), \((0,1,0)\), \((0,0,1)\) and \((1,1,1)\) with respect to that basis.
Proof. See Section 2.2.5 in Bobenko (2020), publicly available here.
This proposition highlights how homogeneous coordinates aren’t a fundamental property of a point in \(\mathbb{RP}^{2}\) and should only be thought of with respect to a preferred basis (just like regular coordinates in linear algebra). With this final technicality in place, it’s time to move on.
In other words, the homogeneous coordinates \((a,b,c)\) define the same point in \(\mathbb{RP}^{2}\) as the homogeneous coordinates \(\lambda (a,b,c)\) for any non-zero \(\lambda\) in \(\mathbb{R}\).↩︎
I.e. a way of adding together scalar multiples of two of them to give the third.↩︎
A basis is a set of vectors that is both linearly independent and spanning. For more, see here.↩︎