Möbius strips and differential equations

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Möbius strips and differential equations

August 8, 2026

Algebraic topology is a field of mathematics which is usually presented as the study of certain amusing shapes, like the Klein bottle or Möbius strip. But this presentation perhaps fails to convey just how vital algebraic topology is to almost all areas of modern mathematics.

To give the reader a taste of this importance, in today's post we will start by exploring a funny property of the Möbius strip, and then show how the same funny phenomenon creeps up when you try to solve certain differential equations.

So, what is the Möbius strip, and what funny property does it have?

The Möbius strip

Turning math problems into geometry problems sometimes leaves you with very strange shapes. It turns out that the shape relevant to our problem will be the Möbius strip! So, before getting into our differential equation, let's take some time to explore the Möbius strip.

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Take a sheet of paper. If you tape two opposite edges together, then you will produce a cylinder. If you instead twist one of the edges before gluing, you get a Möbius strip.<br>twist one of the edges upside down. After this twist, then tape them together. -->

Möbius Strip

drag to rotate

The Möbius strip doesn't seem so different from a cylinder at first, but there is one very strange property of it: it only has one side!

But what does that mean, exactly? Specifying a side of the Möbius strip is the same as deciding which side of the paper is `up' and which side is `down.' These decisions need to be locally consistent: nearby points should agree on which way is up and which is down.<br>In other words, if I'm an ant on one side of the Möbius strip, if I walk a couple steps I should still be on the same side.

This suggests it should be pretty easy to find two sides of any surface: pick one point to start with, and arbitrarily label one side `up' and one side `down.' Then just move around, copying your meaning of up.

Here's a little animation of this; we start with a notion of up at one point (illustrated by an upwards pointing vector), and then move the vector around to nearby points, to give them a consistent notion of up.

Local Orientability

▶ Play<br>Press play to animate

Locally, the normal always comes back pointing the same way

Unfortunately, this procedure has a small hiccup. If we take a long journey, in a loop going around the entire Möbius strip, the notion of up flips when we get back to where we started!

We made a small animation to see this; after the animation plays, you can drag the Möbius strip to rotate it around, to better see the full path.

Non-Orientability

▶ Play<br>Press play to animate

After one full loop, the vector flips — same surface, opposite orientation

This is why we cannot define a notion of up or down on the Möbius strip; while in small patches it makes sense, if you travel all around the Möbius strip you'll end up reversing your notion of up and down! This is why people say a Möbius strip has only one side; while at every point on the strip, it looks like there's a `top' and a `bottom,' if you go around the entire strip you'll find that the top and bottom switch roles. If you have a Möbius strip made out of paper, you can observe this in the real world: trace your finger around a Möbius strip, and you'll find that if your finger started off on `top' of the strip, it will end on the `bottom.'

A problem in differential equations

With the Möbius strip introduced, let's turn to a problem of calculus.

The great mathematician Riemann was interested in solving differential equations over the complex numbers. As an example, consider the equation<br>\[\frac{df}{dz} = \frac{1}{2z}f(z).\]

Riemann was interested in finding complex solutions to this differential equation. It's easy to solve differential equations of this type by trying to find the Taylor series of \(f(z).\) Let's explain how. First, observe that the right hand side of the equation involves division by \(2z\); division by 0 is scary, so let's look for a solution near \(z=1.\) Then we can write the Taylor series of \(f(z)\) as<br>\[f(z) = a_0 + a_1(z-1) + a_2(z-1)^2 + a_3(z-1)^3 + \cdots,\]<br>for some coefficients \(a_0, a_1, a_2, ...\) which we need to solve for.

I'll remark now that this is a first order differential equation, so it has one degree of freedom. Thus we can actually make \(a_0\) take any value we want; to simplify the below computations, we'll set \(a_0 = 1.\) In other words, we're giving our differential equation the initial condition that \(f(1) = 1.\)

The equation<br>\[\frac{df}{dz} = \frac{1}{2z}f(z)\]<br>is equivalent to<br>\[2z\frac{df}{dz} = f(z).\]<br>If we term-by-term differentiate our Taylor series, then we find<br>\[\frac{df}{dz} = a_1 + 2a_2(z-1) + 3a_3(z-1)^2 + \cdots.\]

Multiplying the above expression by \(2z\) is a little tricky, because \(2z\) is a power series in the variable \(z,\) but \(df/dz\) is a power series centered at...