#compactification

Okay so something I’ve been kind of obsessed with since learning topology this semester is the fact that the real number line is homeomorphic to the open interval (-1,1) (and to any open interval really but y’know). I liked the idea of taking any real function and squishing it into a 2x2 square. Turns out this totally works and is great, so I’ll put some screenshots of Desmos graphs and such here. Feel free to join in with any cool things about this concept :)

The homeomorphism I chose is ψ: (-1,1) → ℝ given by

This one was just the first one I thought of, but it turns out to have some very interesting properties that others (maybe? should check this) don’t have. We’ll get to that. 

Now, for any function f: ℝ → ℝ we can compose ψ with f followed by ψ⁻¹ so that we can squish the entire graph of f over all of the real numbers into a 2x2 square around the origin. Concretely we transform f(x) into

As an example, this is what the sine and exponential functions look like when squishificated:

Representing functions like this obviously has its downsides: the details of the sine (like its translational symmetry) are lost for inputs further from 0. An upside is that behaviour of the function as inputs near infinity are immediately clear: e^x goes to infinity for increasing x, and to 0 for decreasing x.

Speaking of infinity, wouldn’t it be handy if we could say that e^x literally approaches ∞ as x → ∞. Well, it seems quite obvious how this would be done: define f not only on ℝ, but on ℝ ∪ {∞}, where f(∞) = lim{x → ∞} f(x), should this limit exist. We can also say that f(a) = ∞ if lim{x → a} f(x) = ∞. This would allow us to say for example that 1/0 = ∞, and 1/∞ = 0. The squishificated graph of f(x) = 1/x looks like this:

Two things to remark about this: firstly 1/x, unlike sin(x) and e^x, is well-defined at infinity, as it approaches the same value from both sides. It’s also more clear why we choose to add an unsigned infinity, rather than defining the functions on ℝ ∪ {∞,-∞}. Secondly, this graph is suspiciously linear looking. This is because of a property that the tangent function has: we have 1/ψ(x) = -ψ(x - 1). In fact: taking the reciprocal of any function transforms its squishificated graph by reflecting all values above the x-axis through the line y=1/2, and all values below through y=-1/2. This might not sound like much of a symmetry, but that just depends on how we look at our square.

Recall that we took ∞ = -∞. This makes it so that the top and bottom edges of our square are really the same edge. Same for the left and right. ‘Pasting’ a square together like this yields a donut (or torus if you’re fancy like that)! Consider then the donut on which we draw our graphs. Take the line y=0 to be the outer perimeter, and y=∞ to be the perimeter of the hole. Then y=±1 are the circles on which the donut would rest if you laid it on a flat surface. Every line of the form x=a looks like a small circle perpendicular to these earlier circles, a line through which you’d slice the donut. x=0, x=1,x=∞, and x=-1 would each be a quarter turn away from each other.

Now we can consider certain reflections in our donut, and how a symmetry of a squishificated graph under these reflections would translate to a certain property that the given function has.

Reflection through the plane through y=0 and y=∞ corresponds to multiplying the output by -1, therefore a function whose squishificated graph is symmetrical under this reflection would satisfy the identity f(x) = -f(x). The only functions that satisfy this are the ones that only take values in 0 or ∞.

Another obvious reflection is through the plane that goes through x=0 and x=∞. This would correspond with the identity f(x) = f(-x). There’s many functions that satisfy this identity, so-called even functions.

These two reflections are also just reflections on the regular real plane, not our weird graphing donut, so they’re not that riveting. Functions that are symmetric under the composition of these two reflections satisfy f(x) = -f(-x), the uneven functions.

Here’s where it gets interesting: reflecting through the plane going through x=1 and x=-1 (thereby exchanging inputs of 0 and ∞) corresponds with the identity f(x) = f(x⁻¹) in exactly the same way as the previous two (given that we consider ∞ to be a valid value and input). There’s not really any obvious nice examples, but constant functions certainly satisfy it.

Lastly we have the identity f(x) = (f(x))⁻¹, which corresponds to the same kind of transformation on the square as the previous, but less obviously to a donut reflection. Here’s where it kind of breaks down to consider the donut as an actual physical round thing, where distance is defined like in regular 3D space. If we reflect through the cylinder going through y=1 and y=-1 (and thus exchanging outputs of 0 and ∞), the distances along the donut’s surface are conserved, and this reflection is therefore a valid symmetry of the donut. Anyway, this corresponds to the last identity.

conformal compactification of two copies of minkowski space along their conformal infinities into S¹ × Sⁿ⁻¹

Sarah Koch and John Hubbard, An Analytic Construcion of the Deligne-Mumford Compactification

ℂ² ℂ³

someone from my cohort bought this book for like a dollar and gave it to me because I kept using this story as an analogy to my research when giving talks