# Nested Tori

A blog dedicated to finding AWESOME visualizations in mathematics and other scientific disciplines. Even if you're not a mathematician or scientist, enjoy the view!

## Thursday, December 10, 2020

### Some Loxodromes

## Saturday, November 28, 2020

### The Uniform-Distribution-on-the-Sphere's Nemesis: Isothermal Coordinates

*Introduction to Riemann Surfaces*, section 1-3). We'll derive this another day, but the summary is, we end up deriving that $u$ has to satisfy the spherical Laplace equation \[\frac{1}{\sin\varphi} \frac{\partial}{\partial \varphi}\left(\sin \varphi \frac{\partial u}{\partial \varphi}\right) + \frac{1}{\sin \varphi} \frac{\partial^2 u}{\partial \theta^2} = 0,\] or really, it only has to satisfy this on an open subset of the sphere, in this case, a sphere with two points deleted. This fact often gets glossed over by people talking about geometry; we know that on a sphere, the only harmonic functions that are actually defined on the sphere everywhere are constant functions—so we can only find "interesting" harmonic functions on a sphere if they blow up at certain points. Using the very reasonable assumption that such a function $u$ should be independent of $\theta$, we derive $\frac{d}{d \varphi} \left(\sin\varphi \frac{d u}{d \varphi}\right) = 0$, making $\sin\varphi \frac{d u}{d \varphi} = C_1$ a constant, and $\frac{du}{d\varphi} = C_1 \csc \varphi$ to finally get $u = C_1 \ln(\csc \varphi + \cot \varphi) + C_2$, as before (just with two degrees of freedom, namely $C_1$ controls what the initial spacing of the grid lines, and $C_2$ controls what the value is at the equator). This function blows up at the poles, a fact which manifests itself visually by the grid curves $u =$ constant become more closely spaced together at the poles (and become infinitely dense there, meaning, this is only a valid coordinate system in the sphere outside the poles). We just set $C_1 = 1$ and $C_2 = 0$, which assigns the equator to $u=0$.

*isothermal coordinates*, and knowing what a big fan I am of stat mech, I'd like to fully understand the origin of the term someday. Finally, one might ask if there are coordinate systems (conformal or otherwise) that will cover up all the sphere without weird coordinate singularities like that. The answer is

*no,*and the short explanation of why is because a smooth vector field on the sphere must vanish somewhere (the

*hairy ball theorem*), and since coordinates always set up vector fields (take the vector in the direction of a coordinate curve) and aren't supposed to collapse at good points. This is a long topic for another day (we'll show some pretty cool coordinate systems on the sphere, though!)

## Saturday, October 31, 2020

### Happy Halloween from Nested Tori!

*principal branch*of the logarithm/root/power" function to be... and you don't get into the nuance with the details of differences between different branches of functions. This example here is exploring the mapping $z \mapsto \sqrt{z^2 + 1}$. This is already interesting from the Riemann surface point of view, because neither the function nor its inverse is definable in its most interesting form by mapping (subsets of) the complex plane alone.

*Formally*, to work on $\mathbb{C}$, we have to define things by taking a continuous choice of square root on $\mathbb{C}$ minus some ray from $0$ to $\infty$. The choice here is between a square root and its opposite (which is also a square root of the same complex number!) is not natural in the complex plane (unlike in the real numbers where you can always take the positive square root of a positive number, which is in some sense, a defining characteristic of real numbers, cementing why it is the unique complete ordered field), and being forced to choose forces a discontinuity somewhere. Riemann surface theory is, in some sense, about never having to make the choice. To quote Sir Roger Penrose (congrats on that Nobel Prize!):

*In particular, the domain of
the logarithm function would be ‘cut’ in some arbitrary way, by a line out
from the origin to infi**nity. To my way of thinking, this was a brutal
mutilation of a sublime mathematical structure*. *(*The Road to Reality*, p. 136)*

*Visual Complex Analysis*) suggest the most common branch cut for this is function is two horizontal lines to the left of $\mathsf{i}$ and $-\mathsf{i}$. I wrangled with this a bit, but I believe a subtlety is left out. To study $\sqrt{z^2 + 1}$ as truly the composition of a square root function, and the mapping $z^2 + 1$, your branch cuts will have to be the corresponding inverse image of the branch cut of the square root, under $z^2 + 1$. Using the negative real axis, this is the ray above $\mathsf{i}$ and the ray below $-\mathsf{i}$. Two rays to the left of $\pm \mathsf{i}$ cannot be realized as such an inverse image, for any ray to infinity (the image under $z^2 + 1$ is a whole parabola to infinity). Of course, the essential definition for complex analysis is that you define the function in this set by taking paths from an origin point and

*analytically continuing,*and entrusting your result to algebraic topology. But this is not very effective, computationally. This is basically saying that for some selections of branch cuts, the function composition $\sqrt{z^2 + 1}$ is a lie: you cannot get it by assembling square root, squaring, and plus one. What turns out to work, in terms of assembling functions rather than continuing along curves, is not $\sqrt{z^2 + 1}$, but $\sqrt{z+\mathsf{i}} \sqrt{z-\mathsf{i}}$, that is, if you take the branch of square root, along the negative axis, subtract and add $\mathsf{i}$, take the square root of each, and then multiply, you

*do*get something that reproduces what you theoretically get by paths. One might wonder what is the difference between $\sqrt{z^2 + 1} = \sqrt{(z+\mathsf{i})(z-\mathsf{i})}$ and $\sqrt{z+\mathsf{i}}\sqrt{z- \mathsf{i}}$; these are not the same because $\sqrt{ab} \neq \sqrt{a}\sqrt{b}$ in the complex plane unless you regard it as a multivalued set equality.

## Sunday, October 11, 2020

### Chaos Still Reigns!!

*The Fractal Geometry of Nature,*the set of complex numbers named after its author, Benoit Mandelbrot. I've spent a lot of time exploring the Mandelbrot set since I've been able to use computers, so you'd think I'd have plenty to say about it. And I do, but not this week. Indeed it is this, more than anything else, that started me on the visualization track. Instead, I'll leave you with the real reason why the thing is so damn captivating in the first place, with, what else, a visualization.

*Amygdala*back in the '90s. Unfortunately, I don't know where he is now and what the state of the newsletter is!

## Saturday, October 3, 2020

### Chaos Reigns!!!

Chaos and fractals were all the rage when I was growing up, and was probably one of the major factors in getting me interested in math, programming, and differential equations, since much of this is ultimately about how systems evolve over time. Oh, and getting me into TOTALLY AWESOME visualizations. This is a graph of the various possible final states of the quadratic iterator: $x_{n+1} = ax_n (1-x_n)$. What this means is, for a given parameter $a$, and some start value in the range $(0, 1)$, we consider iterating the map $F(a, x) = ax(1-x)$, namely, the sequence $F(a, x)$, $F(a, F(a, x))$, $F(a, F(a, F(a, x)))\dots$. For some ridiculous history here, one of my favorite things to do as a child was to repeatedly press buttons on a calculator over and over again to see what happens. Little did I know that this is basically, in math, *the only thing that actually exists*: everything is built on the foundation of recursion. Of course, I don't actually believe recursion is the only thing that exists, rather, there is interesting conceptual structure that is built up upon fundamental components, and it is damn amazing how something so simple can ultimately create something so complex. And here is one example in which this is true.

Anyway, to get back to it: if we keep iterating $F$ in the variable $x$, we will eventually fall into some stable pattern. For small $a$, iterating $F$ in the variable $x$ eventually gives convergence to a fixed point, namely $1-\frac{1}{a}$. How did we find this? If we have a fixed point $x = F(a, x)$, then $x = ax(1-x)$ or $-ax^2 + (a-1)x = 0$. This gives two solutions $x = 0$, and $-ax + (a-1) = 0$, or $x = (a-1)/a$.

Now, for each such $a$ (which we take to be the horizontal axis), we plot this fixed point (on the vertical axis). And as you move up to $a=3$, all is peaceful, as all iterations eventually settle on one and only one point. But at $a = 3$, something strange happens. You stop getting a single fixed point, but rather, you start bouncing between two values. We can actually solve for what these two values are using none other than ... the cubic equation (that's the small extent of continuity here with the previous posts!). The way to see this: if there's a period-2 sequence, $x = F(a, y)$ and $y = F(a, x)$ for the two values, which means $x = F(a, F(a, x))$ and therefore $x = aF(a, x)(1-F(a, x)) = a(ax(1-x))(1- ax(1-x))$. This is a priori a quartic equation, since the number of $x$'s eventually multiply out to degree $4$, but once you realize there's a trivial solution $x = 0$ here (because there's a factor of $x$ multiplying on both side), it's a cubic equation $a^2(1-x)(1-ax(1-x)) -1 = 0$. I'm not going to try solving this, despite having had much fun wrangling with Cardano's formula in the past few weeks, because that distracts from the ultimate insight (and indeed, as fun as that formula is, and the beautiful theory that eventually comes from trying to solve polynomial equations, Galois theory), you quickly enter territory in which you stop being able to solve for things in terms of roots. To visualize we instead proceed numerically, directly iterating from a start value and waiting for things to stabilize around a set of values. One sees that as $a$ grows higher and higher, the number of final points starts doubling and doubling, faster and faster (in fact, a factor of the Feigenbaum constant faster), until all chaos breaks loose and the iterator never settles on any fixed point. It's very weird how just varying the parameter $a$, one can go from very nice fixed point behavior to wild chaos.

Ok, you can't go very far analytically, but we can still try anyway:

To explain it, we'll have to back up a little. Note that I said the $2$-periodic solution involves the solution to a *cubic* equation. Where is the third one? (actually, as previously mentioned, it's really quartic, and $x=0$ is a solution). Is the cubic always one that has a zero discriminant? Probably not. We mentioned that these "fixed" points come about from iterating the mappings, i.e., there is dynamics involved. Solving the cubic, you will actually get $1 - 1/a$, the original fixed point, as one of the solutions (in fact, here is again where numerical analysis helps: you can get around using Cardano's formula if you realize from the picture that the original fixed point continues; then you guess the solution $1-1/a$, factor it out with synthetic division, and solve the remaining *quadratic* equation — this is exactly how I taught college algebra students to solve cubics). Of course, this has to satisfy $x = F(a, F(a, x)) = G(a, x)$, and the two solutions to this equation fixed points of $G$, and thus when $F$ is evaluated on it, it alternates between the two values. What goes wrong here is that this point (as well as the point $x=0$) is *unstable*, namely, if you start iterating from some point even slightly away from this fixed point, iterating the map will run away from that point $1 - 1/a$. What is significant about the value $a = 3$ is that the discriminant of the cubic forces two of the roots to become complex conjugates, so there is only one real solution, and the discriminant *also* is the switchover point from when the iteration is stable there versus anywhere else. A fixed point $x$ is stable whenever $|\partial G/\partial x| < 1$ at that point. This can be roughly seen by linearizing about the fixed point, a standard technique:

\[G(a, x+h) \approx G(a, x) + (\partial G/\partial x(a, x))h + O(h^2) = x + (\partial G/\partial x) h + O(h^2).\]

In other words, $G(a, x+h)$ is closer to $x$ than $x + h$ is, because $h$ is being multiplied by something of magnitude $< 1$. So iterating brings things close to the fixed point even closer. It can be seen that at least for a little bit after $a = 3$, the central $1-1/a$ continues, but the other two solutions are the stable fixed points. So if you plot the curves $x = G(a, x))$, you'll get something that divides into $3$ branches at $a = 3$ (referred to as a pitchfork bifurcation). Similarly, if you plot $x = F(a, F(a, F(a, F(a, x)))) = H(a, x)$, the 4th order iterate, it will divide into pitchforks twice. This is given in the above plot. However, to figure out which fixed points will actually happen if you iterate from a random starting point, you can figure out where $|\partial H/\partial x| < 1$. This gives the regions in the plane where solutions are stable. You can alternatively plot the complement, where it is $\geq 1$, which can be made to cover up the parts of the pitchforks that are not stable solutions. The weird gloopy visualization of these regions is just cool to look at, even where there aren't any stable points. The curves you see inside there are still the correct initial ones before it branches out into a complete mess. It of course quickly becomes ferociously hard to compute for higher degrees, so a numerical study is still the way to go. But this is conceptually fun, though!

I leave you with a similar implicit function graph, but now color-coded:

The red and blue correspond to $|\partial G/\partial x| \geq 1$ and the greens are where it is $< 1$.

## Saturday, September 5, 2020

### More Cubic Fun

*configuration space*of cubics. More explorations on this forthcoming, but I'm testing out a new graphing calculator interface called Desmos. Let's see how well it can be shared.

*discriminant*(that word is just so annoyingly hard to say), of the quadratic: $b^2 - 4ac$ in $ax^2 + bx+c$: when it passes through 0, the nature of the solution changes). One day I may make a post detailing the connection between this kind visualization and topological understanding of the concept of solvability by radicals. It might be far in the future, though!

## Saturday, August 22, 2020

### I need a half-tablespoon. So I have absolutely no choice but to accept complex numbers.

*height*point is going to give you less than half the

*volume*. Realizing the volume as a stack of disks, the answer is

*solve*for the height. It is easier to measure from the top of the spoon ($xy$-plane) than from the bottom, so we solve

*depressed*cubic (insert many jokes/puns here and catch flack from the sensitivity police for making light of serious, real mental illnesses). Of course, using Mathematician's Privilege (dimensional consistency is such a ... physicist ... thing), we can just set our units to $1$ and be done with it; this gives us the polynomial

*casus irreducibilis*. It being a depressed cubic, I don't have to deal with the Germans, and can skip directly to Italians with fancy cars (perhaps I'll "retcon" my story and say I was cooking a nice

*risotto*and needed to use half a hemispherical teacup), the solution to this is

$\zeta\sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} - 1}} + \bar\zeta\sqrt[3]{-\frac{1}{2} - \sqrt{\frac{1}{4} - 1}}$,

$\bar \zeta\sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} - 1}} + \zeta \sqrt[3]{-\frac{1}{2} -\sqrt{\frac{1}{4} - 1}}$, where $\zeta$ is a cube root of unity (please print these formulas out onto posters and hold them up to enforce social distancing). It happens that $\zeta = -\frac{1}{2} + \frac{\sqrt{3}}{2}\mathsf i$, which is exactly what is under the cube root, so we're dealing with

*ninth*roots of unity here! (who would have thought??). The way to see that these formulas work, generally from the depressed cubic, is basically to substitute $x = w - \frac{p}{3w}$, which results in a quadratic for $w^3$. That it's quadratic explains the nearly identical features of the stuff under the cube roots, but for a sign, and that it's $w^3$ that gets solved for is responsible for the cube roots outside, as well as mutiplying by roots of unity.

The cubic $y = x^3 - 3x + 1$. It has three real roots, but they only can be expressed algebraically by the sum of complex numbers! |

*possible*solutions are

*from the top*. Here's another picture and zoomed in:

A cross section of the hemisphere in question, together with a zoomed in view of the cubic curve (green). The vertical line (cyan) drawn through the cubic at its root will intersect the circle (red) where it needs to be filled (here you'd fill everything to the right of this line) |

*has*to go through the complex numbers is why complex numbers were finally accepted. It is not, as commonly believed, due to trying to solve for the quadratic, since in that case, complex numbers

*only*appear when the parabola fails to intersect the $x$-axis, meaning the complex solutions are of the impossible, nonexistent intersection points, so, the various Renaissance mathematicians thought, obviously should not be taken seriously. But the cubic forces it to happen no matter what, because a cubic curve

*always*crosses the $x$-axis. So there it is: half a tablespoon? You'll need complex numbers for that. Some more views (the title image sort of is looking from above, so might obscure the fact you must fill it

*more*than halfway from the bottom):

View from below the tablespoon. The little extra disk sticking out is the fill line. |

It's not taught because the solutions you get may not be easily reducible to something you can recognize, even in simple cases. Conceptually, however, it is a real adventure, and solving it by breaking it into a 2-step process makes it tractable and interesting. Our tablespoon example already skips step 1 ... all that about Germans and Italians.

Enjoy and stay safe!