Möbius strips and differential equations

← back 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 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. 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. 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. 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. 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 \[\frac{df}{dz} = \frac{1}{2z}f(z).\] 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 \[f(z) = a_0 + a_1(z-1) + a_2(z-1)^2 + a_3(z-1)^3 + \cdots,\] 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 \[\frac{df}{dz} = \frac{1}{2z}f(z)\] is equivalent to \[2z\frac{df}{dz} = f(z).\] If we term-by-term differentiate our Taylor series, then we find \[\frac{df}{dz} = a_1 + 2a_2(z-1) + 3a_3(z-1)^2 + \cdots.\] 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 \(z=1.\) So, before doing the multiplication, we rewrite \[2z = 2 + 2(z-1).\] Thus \[2z\frac{df}{dz} = (2 + 2(z-1)) \cdot (a_1 + 2a_2(z-1) + 3a_3(z-1)^2 + \cdots),\] so that, when we distribute this product out, \[2z\frac{df}{dz} = 2a_1 + (2a_1 + 4a_2) \cdot (z-1) + (4a_2 + 6a_3) \cdot (z-1)^2 + \cdots\] Thus our equation \[2z\frac{df}{dz} = f(z)\] becomes \[\begin{align*} &\hspace{5mm}2a_1 + (2a_1 + 4a_2) \cdot (z-1) + (4a_2 + 6a_3) \cdot (z-1)^2 + \cdots \\ &= 1 + \hspace{17mm}a_1 \cdot (z-1) + \hspace{14mm}a_2 \cdot (z-1)^2 + \cdots.\end{align*}\] Equating the coefficients of the \((z-1)^n\) terms on both sides, we get the equations \[2a_1 = 1,\] \[2a_1 + 4a_2 = a_1,\] \[4a_2 + 6a_3 = a_2,\] and so on. From the first equation \(2a_1 = 1,\) we get \(a_1 = 1/2.\) But now that we know \(a_1 = 1/2,\) we can substitute this value of \(a_1\) into the second equation \[2a_1 + 4a_2 = a_1,\] to rewrite it as \[1 + 4a_2 = \frac{1}{2},\] implying \[a_2 = -\frac{1}{8}.\] Similarly, now that we know \(a_2 = -1/8,\) we can plug it into the third equation \[4a_2 + 6a_3 = a_2\] to get \[-\frac{4}{8} + 6a_3 = -\frac{1}{8},\] giving \[a_3 = \frac{1}{16}.\] We can solve for the Taylor coefficients of \(f(z)\) recursively, by just following our string of equations. Thus you can get as many Taylor coefficients for the solution \(f(z)\) to \[\frac{df}{dz} = \frac{1}{2z}f(z)\] as you'd like! While this method is great at finding a solution for \(z=1,\) the downside to Taylor series is that they don't always converge. It turns out that this particular Taylor series has radius of convergence 1, which means for any complex number \(z\) closer than 1 unit to \(z=1,\) the formal power series solution we just found will converge. You can play around, evaluating a few values of the power series, here. Click a point inside the disk below to evaluate the series there. z — plane Unfortunately, for values of \(z\) outside of the highlighted blue circle, the formal power series solution we just found will not always converge. As an example, if we plug in \(z = -1\) to our series we get \[f(-1)=1 - 1 - \frac{1}{2} - \frac{1}{2} - \frac{5}{8} - \frac{7}{8} - \frac{21}{16} - \cdots.\] Observe how, after the initial \(1-1,\) the terms just get more and more negative; it turns out that this continues to happen, and so \[f(-1) = -\infty,\] as we keep subtracting larger and larger numbers. This is annoying! We would love to solve our differential equation on the entire complex plane, but so far we can only solve it in a tiny disk. However, note that we found this power series by doing term-by-term expansion centered around the point \(z=1.\) There's no reason we couldn't just do term-by-term expansion centered around a different point. For example, let's take a point \(\alpha = 0.6 + 0.8i,\) which is near the edge of our blue circle. Using our first power series solution, we find that \(f(\alpha) \approx 0.8944 + 0.4472i.\) Using this as the initial condition, we can use the recurrence method from before to find a formal power series solution to \[\frac{df}{dz} = \frac{1}{2z}f(z)\] centered at \(\alpha.\) In the widget below, you can click on points to plug in values. expansion around 1 (same one as above) expansion around a = 0.6 + 0.8i z — plane disk around a = 1 disk around a; = 0.6+0.8i In the widget, we display two different formal power series solutions to our ODE, each centered around different points. The first one is the same one from before, and converges inside the blue circle; the second one is new, and converges inside the purple circle. If you click on a point \(z\) in the blue circle, your computer will use the formal power series to calculate \(f(z)\) using our power series from before; and if you click on a point \(z\) in the purple circle, your computer will use the new power series to calculate \(f(z).\) Click on different points of the intersection of the two circles, and compare the values of the two different power series. You should find that they are exactly the same, up to rounding errors (these are infinite sums, but your computer can only do finitely many terms!). So now we have a solution to \[\frac{df}{dz} = \frac{1}{2z}f(z)\] which obeys \(f(1) = 1,\) and makes sense in the union of the blue and purple circles: in the purple circle use the new formula, in the blue circle use the old formula, and on the overlap the two formulas give the same answer, so it doesn't matter which you use! Monodromy Let's recap what just happened: we found a solution to \[\frac{df}{dz} = \frac{1}{2z}f(z)\] inside a small blue circle, by expanding \[f(z) = a_0 + a_1(z-1) + a_2(z-1)^2 + \cdots\] as a formal power series around \(z=1\); the differential equation became a recurrence on the coefficients \(a_n,\) and the initial condition \(f(1) = 1\) just became the initial condition \(a_0 = 1\) of our recurrence. The resulting power series only made sense on a small blue disk, but we wanted a solution over the entire complex plane -- so, we repeated our recurrence procedure, but starting with the different center value \(0.6 + 0.8i\) instead. The differential equation again turns into a recurrence relation, and we can get the initial condition of our recurrence using the original formula to calcualte \(f(0.6 + 0.8i).\) This extended our solution from the blue circle to the union of the blue circle and the purple circle. Since this worked so well, let's go further: pick a point near the edge of the purple circle, and find a power series solution centered at that point! And then do this again and again and again, until we've covered the entire complex plane with circles!