∫ Knotting is math. (So is everything.)

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There's a nice knot trick on the Internet where you can untie a knot without moving either ends across the bar. Here by World of Engineeringhttps://www.youtube.com/@world_of_engineering (fair use).

It's mesmerizing. And there are tons of videos about this famous knot. (Matt Parkerhttps://www.youtube.com/@standupmaths and Steve Mouldhttps://www.youtube.com/@SteveMould also have a nice video on the topic.) Despite that we have seen how it is untied, in continuous motion and indisputable, it's still not that satisfying.

My goal here is to prove it using mathematics.

∂ From knotting

First off, it's common for mathematicians to use symbols to represent problems. This step is optional. But working with symbols allow us to remove unnecessary details and focus only on the important bits. This is called "abstraction"https://en.wikipedia.org/wiki/Abstraction_(mathematics).

Interact with the graphic below to see one abstraction I suggest for this content:

What I propose here is a special representation for the string and bar situation. To elaborate the steps,

1. I trace the string from one end to the other.
2. When I stump upon a crossing, I jot down what type of crossing it is.
3. "o" if the string overcrosses itself or the bar
4. "u" if the string undercrosses itself or the bar
5. If the string crosses itself, and it's the cross that I jotted down before, I assign a running number for this one and the original.

Do you grasp it? Here is another example:

This representation works for any one string and one bar situation. The important bit that should be done is to show that the proposed representation system can bijectively represent the original problem. We call that this representation is "isomorphic"https://en.wikipedia.org/wiki/Isomorphism to the original problem. (For the sake of simplicty, I may have to skip the proof.In fact, there are situation where different knots may emit the same expression. Look for the comment in html. But it can be juggled without changing the equivalency. In a follow up post, I'll elaborate on this. My apologies.)

I've mentioned that abstraction is optional. But for mathematicians, learning about it the most fundamental step. Mathematicians are obsessed with generalization, they want to find a way to represent problems in a way that can be applied to a wide range of domains. So that everything they do or discover, in the problems that they are working on, can be applied to the other problems as well. This is the reason for all the numbers, the symbols, the curly curves, that terrorize many people. They are parts of the systematic effort to abstract the world as generalized as possible.

Let's see how we can use this representation to solve the original problem. I want to show that these two configurations are equivalent: Allow me to remind you about the two important rules of this game:

1. We are not allowed to move the bottom end across anything because it goes out of the area of interest.
2. And the top end is too bulky to cross under the bar.

We call them axioms. An axiom is a rule that we accept to be true without proof from start. And because of these axioms, it's not so obvious why the statement is true.

Perhaps we'd better start off with some things that are easier to follow. Interact with figures below to see various of equivalent configurations:

• Disentangle
  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, *, 0, 0, 0, 0; 0, 0, r, -, %, -, 7, 0, 0; 0, 0, |, 0, |, 0, |, 0, 0; =, =, %, =, +, =, %, =, =; 0, 0, |, 0, |, 0, |, 0, 0; 0, 0, |, 0, L, -, J, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;

  ["-1m", 0, 1, "-1m", 2]

  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, r, -, -, -, 7, 0, 0; 0, 0, |, 0, *, 0, |, 0, 0; =, =, %, =, +, =, %, =, =; 0, 0, |, 0, |, 0, |, 0, 0; 0, 0, |, 0, L, -, J, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;
• Slide through the barOnly guarantee to work when it's enclosed in a loop, or at the end. A loop is where there's a pair of enclosing crossing guarding the region.
  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, *, 0, 0; 0, 0, r, -, 7, 0, |, 0, 0; 0, 0, |, 0, |, 0, |, 0, 0; =, =, %, =, %, =, +, =, =; 0, 0, |, 0, |, 0, |, 0, 0; 0, 0, |, 0, L, -, J, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;

  [0, "-1m", "-1m"]

  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, *, 0, 0; 0, 0, 0, 0, 0, 0, |, 0, 0; 0, 0, 0, 0, 0, 0, |, 0, 0; =, =, =, =, =, =, +, =, =; 0, 0, 0, 0, 0, 0, |, 0, 0; 0, 0, r, -, -, -, J, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;
• Free cross
  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, *, 0, 0; 0, 0, r, -, 7, 0, |, 0, 0; 0, 0, |, 0, |, 0, |, 0, 0; =, =, %, =, +, =, +, =, =; 0, 0, |, 0, |, 0, |, 0, 0; 0, 0, |, 0, L, -, J, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;

  ["-1m", 0, 1]

  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, r, -, 7, 0, 0, 0, 0; 0, 0, |, 0, |, 0, 0, 0, 0; =, =, %, =, +, =, =, =, =; 0, 0, |, 0, |, 0, *, 0, 0; 0, 0, |, 0, L, -, J, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;
• Twist
  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, *, 0, 0, 0, 0, 0; 0, 0, 0, |, 0, 0, 0, 0, 0; 0, 0, 0, |, 0, 0, 0, 0, 0; =, =, =, %, =, =, =, =, =; 0, 0, r, %, 7, 0, 0, 0, 0; 0, 0, |, L, J, 0, 0, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;

  [0, "-1m", "-1m"]

  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, *, 0, 0, 0, 0, 0; 0, 0, 0, |, 0, 0, 0, 0, 0; 0, 0, 0, |, 0, 0, 0, 0, 0; =, =, =, %, =, =, =, =, =; 0, 0, r, J, 0, 0, 0, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;
• Flip end (required matching context)
  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, *, 0, 0, 0, 0, 0; 0, 0, 0, |, 0, 0, 0, 0, 0; 0, 0, 0, |, r, 7, 0, 0, 0; =, =, =, %, +, %, =, =, =; 0, 0, r, %, J, |, 0, 0, 0; 0, 0, |, L, -, J, 0, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;

  ["0p", "-1m", 1, 2, "-1m"]

  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, *, 0, 0; 0, 0, 0, 0, 0, 0, |, 0, 0; 0, 0, 0, 0, r, 7, |, 0, 0; =, =, =, =, +, %, %, =, =; 0, 0, r, -, J, |, |, 0, 0; 0, 0, |, 0, 0, L, J, 0, 0; 0, 0, |, 0, 0, 0, 0, 0, 0;
• Flip (check context)
  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, *, 0, 0, 0, 0, 0; 0, 0, r, %, -, 7, 0, 0, 0; 0, 0, |, |, r, %, 7, 0, 0; =, =, %, +, %, %, +, =, =; 0, 0, L, J, |, |, |, 0, 0; 0, 0, 0, 0, |, L, J, 0, 0; 0, 0, 0, 0, |, 0, 0, 0, 0;

  [0, 1, 2, 3, "-1m", "4p", "5p", "-1m", 6]

  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, *, 0, 0, 0, 0, 0; 0, 0, r, %, -, -, 7, 0, 0; 0, 0, |, |, r, 7, |, 0, 0; =, =, %, +, %, +, %, =, =; 0, 0, L, J, |, |, |, 0, 0; 0, 0, 0, 0, |, L, J, 0, 0; 0, 0, 0, 0, |, 0, 0, 0, 0;
• Separation
  0, 0, 0, 0, 0, 0, 0, 0; 0, 0, r, 7, 0, *, 0, 0; =, =, %, +, =, +, =, =; 0, 0, |, |, 0, |, 0, 0; 0, 0, L, %, 7, |, 0, 0; 0, 0, 0, L, +, J, 0, 0; 0, 0, 0, 0, |, 0, 0, 0; 0, 0, 0, 0, |, 0, 0, 0;

  [0, "-1m", "-1m", 1, 2, "-1m", "-1m"]

  0, 0, 0, 0, 0, 0, 0, 0; 0, 0, r, 7, 0, *, 0, 0; =, =, %, +, =, +, =, =; 0, 0, |, |, 0, |, 0, 0; 0, 0, |, L, -, J, 0, 0; 0, 0, L, -, 7, 0, 0, 0; 0, 0, 0, 0, |, 0, 0, 0; 0, 0, 0, 0, |, 0, 0, 0;
• Swap (check context)
  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, *, 0, 0, 0, 0; 0, r, -, -, +, 7, 0, 0, 0; 0, |, 0, 0, |, |, 0, 0, 0; =, +, =, =, %, +, =, =, =; 0, |, 0, r, %, J, 0, 0, 0; 0, |, 0, L, +, 7, 0, 0, 0; 0, L, -, -, J, |, 0, 0, 0;

  ["0p", "2m", "1m", 3, 4, "5p", "7m", "6m", 8]

  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, *, 0, 0, 0, 0; 0, r, -, -, +, 7, 0, 0, 0; 0, |, 0, r, %, J, 0, 0, 0; =, +, =, +, %, =, =, =, =; 0, |, 0, |, |, 0, 0, 0, 0; 0, |, 0, L, +, 7, 0, 0, 0; 0, L, -, -, J, |, 0, 0, 0;
• Pull across the barAgain, always work if it's in a loop. (check context)
  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, *, 0, 0, r, 7, 0, 0, 0; 0, |, r, -, %, +, -, 7, 0; =, %, +, =, %, %, =, +, =; 0, |, |, 0, |, |, 0, |, 0; 0, L, J, 0, |, |, 0, |, 0; 0, r, -, -, J, L, -, J, 0; 0, |, 0, 0, 0, 0, 0, 0, 0;

  [0, 1, "3p", "4p", 6, "2m", "7p", "8p", "5m"]

  0, 0, 0, 0, 0, 0, 0, 0, 0; 0, *, 0, 0, 0, 0, 0, 0, 0; 0, |, r, 7, 0, 0, r, 7, 0; =, %, +, %, =, =, %, +, =; 0, |, |, |, r, 7, |, |, 0; 0, L, J, L, %, +, J, |, 0; 0, r, -, -, J, L, -, J, 0; 0, |, 0, 0, 0, 0, 0, 0, 0;
• Transfer twist (check context)
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; 0, *, 0, 0, r, -, 7, 0, 0, 0, 0; 0, L, -, 7, |, r, %, -, -, -, -; 0, r, -, +, J, |, |, 0, 0, 0, 0; 0, |, 0, L, 7, L, +, -, -, 7, 0; 0, |, 0, 0, L, -, J, 0, 0, |, 0; =, +, =, =, =, =, =, =, =, %, =; 0, L, -, -, -, -, -, -, -, J, 0;

  ["4m", "0p", "1p", "7m", 2, 3, "5p", "6p", 8]

  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; 0, *, 0, 0, r, -, 7, 0, 0, 0, 0; 0, L, -, -, J, r, %, -, -, 7, 0; 0, 0, 0, 0, 0, |, |, 0, 0, |, 0; 0, r, -, -, 7, L, +, -, -, +, -; 0, |, 0, 0, L, -, J, 0, 0, |, 0; =, +, =, =, =, =, =, =, =, %, =; 0, L, -, -, -, -, -, -, -, J, 0;

These are called lemmas. A lemma is not only a small transformation which is relatively (but not always) simpler and more straightforward, it is also (most of the time) useful in other contexts. They are the building blocks of theorems.

Using these lemmas, I can show that both configurations above are equal in a few steps: The statement has now become a theorem! A theorem is a statement that has been (or can be) proven to be true. Unlike a lemma, it's common for a theorem to be something that is not so straightforward.

So once again we've reached the same conclusion as in the video. But this time it is through the power of math! We also get some useful things along the way.

1. We have the abstract representation of the string and bar situation.
2. We have lemmas that can be combinded to prove other knots too.

For example, we can also show that these knots are equivalent: using the lemmas alone this time, without the help from drawing: This well known knot can also be untied without moving the end under the bar: And we can untie this big one: Now try solve this one on your own before looking at the solution, it's really satisfying.

In fact, Parker has theorized that if net down and up directions of the string crossing under the bar is zero. We can untie the knot without moving the ends across the barhttps://www.youtube.com/watch?v=g3R_tc7YrFI. This will be a harder proof that involves more lemmas. (Again I would like to apologize that I will not prove it here.For example, pull across the bar can be done over sub-knots, or a sequence of crossing that pack together. This will be the crux of the follow up post I mentioned.)

There's a whole area that studies Knots beyond a string and a bar. It's called Knot Theoryhttps://en.wikipedia.org/wiki/Knot_theory. Its humble origin began with the study of knots in the real world, but it ends up being a powerful tool to study the structure of the building blocks of living things. Knot theory is used to match two configuration of DNAs and opens the door for us to perform genome sequencingBates, A. D., & Maxwell, A. (2005). DNA topology. Oxford University Press, USA..

∂ To everything

Whether you're trying to solve a knot or understand any problem in nature, the basic idea is the same. We start by abstracting a problem it into symbols. We prove theorems on the symbols. Then we combine the theorems to prove other theorems. Oftentimes these theorems find their way into the real world. And before anyone can imagine, they allow us to discover new truths. Truths those we have yet observe directly in the nature.

For a long time we can't observe black holes, but we can infer their existence from the general relativity. Only until recently that we have the means to verify their presence. Thanks to a collaborative effort of astronomers from around the world that we have a system of telescopes sensitive enough to capture themAkiyama, K. et al.[Event Horizon Telescope Collaboration]. (2019). First M87 event horizon telescope results. I. The shadow of the supermassive black hole. Astrophys. J. Lett., 875, L1.. This is just one of the examples that illustrate the power of mathematics.

A scottish mathematician once said, The ability to predict truth before we can observe them is where mathematics, as well as other disciplines of science, put conspiracy theories into shadow...

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