WEBVTT
00:00:01.040 --> 00:00:06.600
I’ve got several fun things for you this video.
00:00:07.080 --> 00:00:14.520
An unsolved problem, a very elegant solution to a weaker version of the problem, and a little bit about what topology is and why people care.
00:00:15.400 --> 00:00:20.560
But before we jump into it, it’s worth saying a few words on why I’m excited to share the solution.
00:00:20.920 --> 00:00:24.520
When I was a kid, since I loved math and sought out various mathy things.
00:00:24.880 --> 00:00:32.360
I would occasionally find myself in some talk or a seminar where people wanted to get the youth excited about things that mathematicians care about.
00:00:33.240 --> 00:00:37.400
A very common go-to topic to excite our imaginations was topology.
00:00:38.160 --> 00:00:45.080
We might be shown something like a Mobius strip, maybe building it out of construction paper by twisting a rectangle and gluing its ends.
00:00:45.640 --> 00:00:46.080
“Look!”
00:00:46.080 --> 00:00:51.320
We’d be told as we were asked to draw a line along the surface, “It’s a surface with just one side!”
00:00:52.720 --> 00:00:58.480
Or we might be told that topologists view coffee mugs and donuts as the same thing, since each has just one hole.
00:00:59.560 --> 00:01:02.360
But these kinds of demos always left a lurking question.
00:01:03.000 --> 00:01:03.960
How is this math?
00:01:04.240 --> 00:01:06.760
How does any of this actually help to solve problems?
00:01:07.800 --> 00:01:18.320
It wasn’t until I saw the problem that I’m about to show you with its elegant and surprising solution that I started to understand why mathematicians actually care about some of these shapes and the properties they have.
00:01:21.200 --> 00:01:24.600
So there’s this unsolved problem called the inscribed square problem.
00:01:25.400 --> 00:01:31.640
If you have a closed loop, meaning you squiggle some line through space in a potentially crazy way and you end up back where you started.
00:01:32.400 --> 00:01:37.480
The question is whether or not you’ll always be able to find four points on this loop that make up a square.
00:01:38.600 --> 00:01:44.400
If your closed loop was a circle, for example, it’s quite easy to find an inscribed square, infinitely many, in fact.
00:01:49.360 --> 00:01:53.520
If your loop was, instead, an ellipse, it’s still pretty easy to find an inscribed square.
00:01:54.360 --> 00:02:00.760
The question is whether or not every possible closed loop, no matter how crazy, has at least one inscribed square.
00:02:01.760 --> 00:02:02.680
Pretty interesting, right?
00:02:03.320 --> 00:02:06.440
I mean, just the fact that this is unsolved is interesting.
00:02:07.000 --> 00:02:12.680
That the current tools of math can neither confirm nor deny that there’s some loop with no inscribed square in it.
00:02:14.080 --> 00:02:21.200
Now, if we weaken the question a bit and ask about inscribed rectangles instead of inscribed squares, it’s still pretty hard.
00:02:21.480 --> 00:02:27.000
But there is a beautiful video-worthy solution that might actually be my favorite piece of math.
00:02:28.320 --> 00:02:33.800
The idea is to shift the focus away from individual points on the loop and, instead, onto pairs of points.
00:02:34.960 --> 00:02:37.000
We’ll use the following fact about rectangles.
00:02:37.600 --> 00:02:40.920
Let’s label the vertices of some rectangle 𝑎, 𝑏, 𝑐, 𝑑.
00:02:41.760 --> 00:02:46.440
Then the pair of points 𝑎, 𝑐 has a few things in common with the pair of points 𝑏, 𝑑.
00:02:47.360 --> 00:02:55.400
The distance between 𝑎 and 𝑐 equals the distance between 𝑏 and 𝑑, and the midpoint of 𝑎 and 𝑐 is the same as the midpoint of 𝑏 and 𝑑.
00:02:56.480 --> 00:03:02.560
In fact, anytime you have two separate pairs of points in space, 𝑎, 𝑐 and 𝑏, 𝑑.
00:03:02.560 --> 00:03:08.480
If you can guarantee that they share a midpoint and that the distance between 𝑎, 𝑐 equals the distance between 𝑏 and 𝑑.
00:03:09.960 --> 00:03:13.280
It’s enough to guarantee that those four points make up a rectangle.
00:03:14.560 --> 00:03:17.640
So what we’re gonna do is try to prove that for any closed loop.
00:03:18.120 --> 00:03:24.760
It’s always possible to find two distinct pairs of points on that loop that share a midpoint and which are the same distance apart.
00:03:25.800 --> 00:03:27.120
Take a moment to make sure that’s clear.
00:03:27.680 --> 00:03:33.480
We’re finding two distinct pairs of points that share a common midpoint and which are the same distance apart.
00:03:38.400 --> 00:03:45.240
The way we’ll go about this is to define a function that takes in pairs of points on the loop and outputs a single point in 3D space.
00:03:45.640 --> 00:03:48.720
Which kind of encodes the midpoint and distance information.
00:03:49.320 --> 00:03:50.440
It will be sort of like a graph.
00:03:52.720 --> 00:03:56.600
Consider the closed loop to be sitting on the 𝑥𝑦-plane in 3D space.
00:03:57.360 --> 00:04:06.880
For a given pair of points, label their midpoint 𝑀, which will be some point on the 𝑥𝑦-plane, and label the distance between them 𝑑.
00:04:06.880 --> 00:04:11.720
Plot the point, which is exactly 𝑑 units above that midpoint 𝑀, in the 𝑧-direction.
00:04:15.160 --> 00:04:20.640
As you do this for many possible pairs of points, you’ll effectively be drawing through 3D space.
00:04:21.400 --> 00:04:27.240
And if you do it for all possible pairs of points on the loop, you’ll draw out some kind of surface above the plane.
00:04:28.760 --> 00:04:32.200
Now look at the surface and notice how it seems to hug the loop itself.
00:04:33.080 --> 00:04:34.720
This is actually gonna be important later.
00:04:34.720 --> 00:04:36.360
So let’s think about why it happens.
00:04:38.600 --> 00:04:47.320
As the pair of points on the loop gets closer and closer, the plotted point gets lower since its height is, by definition, equal to the distance between the points.
00:04:47.920 --> 00:04:52.040
Also the midpoint gets closer and closer to the loop as the points approach each other.
00:04:53.160 --> 00:04:59.200
Once the pair of points coincides, meaning the input of our function looks like 𝑥, 𝑥 for some point 𝑥 on the loop.
00:04:59.760 --> 00:05:04.240
The plotted point of the surface will be exactly on the loop at the point 𝑥.
00:05:05.520 --> 00:05:06.560
Okay, so remember that.
00:05:07.240 --> 00:05:10.200
Another important fact is that this function is continuous.
00:05:10.760 --> 00:05:14.200
And all that really means is that if you slightly adjust a given pair of points.
00:05:14.600 --> 00:05:18.680
Then the corresponding output in 3D space is only slightly adjusted as well.
00:05:19.240 --> 00:05:21.360
There’s never a sudden discontinuous jump.
00:05:22.560 --> 00:05:25.880
Our goal, then, is to show that this function has a collision.
00:05:26.440 --> 00:05:30.800
That two distinct pairs of points each map to the same spot in 3D space.
00:05:31.760 --> 00:05:35.800
Because the only way for that to happen is if they share a common midpoint.
00:05:36.080 --> 00:05:38.600
And if their distance, 𝑑, apart from each other is the same.
00:05:40.200 --> 00:05:46.720
So in some sense, finding an inscribed rectangle comes down to showing that this surface has to intersect itself.
00:05:51.920 --> 00:05:56.840
To move forward from here, we need to build up a relationship with the idea of pairs of points on a loop.
00:05:58.680 --> 00:06:03.160
Think about how we represent pairs of real numbers using a two-dimensional coordinate plane.
00:06:08.080 --> 00:06:14.360
Analogous to this, we’re gonna seek out a certain 2D surface, which naturally represents all pairs of points on the loop.
00:06:15.400 --> 00:06:21.920
Understanding the properties of this surface will help to show why the graph that we just defined has to intersect itself.
00:06:23.040 --> 00:06:26.760
Now, when I say pair of points, there are two things that I could be talking about.
00:06:27.480 --> 00:06:33.880
The first is ordered pairs of points, which would mean a pair like 𝑎, 𝑏 would be considered distinct from the pair 𝑏, 𝑎.
00:06:34.400 --> 00:06:37.160
That is, there’s some notion of which point is the first one.
00:06:39.480 --> 00:06:44.720
The second idea is unordered points, where 𝑎, 𝑏 and 𝑏, 𝑎 would be considered the same thing.
00:06:45.360 --> 00:06:50.120
Where all that really matters is what the points are and there’s no meaning to which one is first.
00:06:50.960 --> 00:06:53.880
Ultimately, we wanna understand unordered pairs of points.
00:06:54.320 --> 00:06:57.560
But to get there, we need to take a path of thought through ordered pairs.
00:06:59.800 --> 00:07:04.920
We’ll start out by straightening out the loop, cutting it at some point and deforming it into an interval.
00:07:05.720 --> 00:07:10.360
For the sake of having some labels, let’s say that this is the interval on the number line from zero to one.
00:07:11.560 --> 00:07:17.760
By following where each point ends up, every point on the loop corresponds with a unique number on this interval.
00:07:19.880 --> 00:07:27.640
Except for the point where the cut happened, which corresponds simultaneously to both endpoints of the interval, meaning the numbers zero and one.
00:07:29.040 --> 00:07:36.000
Now, the benefit of straightening out this loop like this is that we can start thinking about pairs of points the same way we think about pairs of numbers.
00:07:39.000 --> 00:07:48.400
Make a 𝑦-axis using a second interval then associate each pair of values on the interval with a single point in this one-by-one square that they span out.
00:07:49.800 --> 00:07:54.800
Every individual point of the square naturally corresponds to a pair of points on the loop.
00:07:55.320 --> 00:07:59.000
Since its 𝑥- and 𝑦-coordinates are each numbers between zero and one.
00:07:59.760 --> 00:08:02.480
Which are, in turn, associated to some unique point on the loop.
00:08:03.480 --> 00:08:08.760
Remember, we’re trying to find a surface that naturally represents the set of all pairs of points on the loop.
00:08:09.360 --> 00:08:11.440
And this square is the first step to doing that.
00:08:12.720 --> 00:08:16.560
The problem is that there’s some ambiguity when it comes to the edges of the square.
00:08:17.640 --> 00:08:23.200
Remember, the endpoints zero and one on the interval really correspond to the same point of the loop.
00:08:23.800 --> 00:08:28.920
As if to say that those endpoints need to be glued together if we’re gonna faithfully map back to the loop.
00:08:30.200 --> 00:08:42.160
So, all of the points on the left edge of the square like zero, 0.1; zero, 0.2; on and on and on really represent the same pair of points on the loop as the corresponding coordinates on the right edge of the square.
00:08:42.600 --> 00:08:45.960
One, 0.1; one, 0.2; on and on and on.
00:08:47.080 --> 00:08:54.240
So for this square to represent the pairs of points on the loop in a unique way, we need to glue this left edge to the right edge.
00:08:55.600 --> 00:08:59.480
I’ll mark each edge with some arrows to remember how the edges need to be lined up.
00:09:00.560 --> 00:09:03.480
Likewise, the bottom edge needs to be glued to the top edge.
00:09:03.840 --> 00:09:10.000
Since 𝑦-coordinates of zero and one really represent the same second point in a given pair of points on the loop.
00:09:10.000 --> 00:09:16.320
If you bend the square to perform the gluing.
00:09:16.760 --> 00:09:19.840
First rolling it into a cylinder to glue the left and right edges.
00:09:20.280 --> 00:09:24.240
Then gluing the ends of that cylinder, which represent the top and bottom edges.
00:09:24.840 --> 00:09:28.000
We get a torus, better known as the surface of a donut.
00:09:29.360 --> 00:09:34.320
Every individual point on this torus corresponds to a unique pair of points on the loop.
00:09:34.920 --> 00:09:39.800
And likewise, every pair of points on the loop corresponds to some unique point on this torus.
00:09:40.560 --> 00:09:46.240
The torus is to pairs of points on the loop what the 𝑥𝑦-plane is to pairs of points on the real number line.
00:09:50.120 --> 00:09:54.000
The key property of this association is that it’s continuous both ways.
00:09:54.480 --> 00:10:02.720
Meaning, if you nudge any point on the torus by just a tiny amount, it corresponds to only a very slight nudge to the pair of points on the loop and vice versa.
00:10:05.160 --> 00:10:11.560
So if the torus is the natural shape for ordered pairs of points on the loop, what’s the natural shape for unordered pairs?
00:10:12.240 --> 00:10:20.280
After all, the whole reason we’re doing this is to show that two distinct pairs of points on the loop share a midpoint and are the same distance apart.
00:10:22.200 --> 00:10:31.360
But if we consider a pair 𝑎, 𝑏 to be distinct from 𝑏, 𝑎, then that would trivially give us two separate pairs, which have the same midpoint and distance apart.
00:10:32.360 --> 00:10:37.600
That’s like saying you can always find a rectangle so long as you consider any pair of points to be a rectangle.
00:10:38.240 --> 00:10:38.840
Not helpful!
00:10:40.320 --> 00:10:41.560
So let’s think about this.
00:10:41.560 --> 00:10:46.080
Let’s think about how to represent unordered pairs of points looking back at our unit square.
00:10:46.880 --> 00:10:53.520
We need to say that the coordinates 0.2, 0.3 represent the same pair as 0.3, 0.2.
00:10:54.920 --> 00:11:00.560
Or that 0.5, 0.7 really represents the same thing as 0.7, 0.5.
00:11:02.600 --> 00:11:13.280
And in general, any coordinates 𝑥, 𝑦 has to represent the same thing as 𝑦, 𝑥.
00:11:13.280 --> 00:11:16.960
Once again, we capture this idea by gluing points together when they’re supposed to represent the same pair.
00:11:17.760 --> 00:11:21.400
Which, in this case, requires folding the square over diagonally.
00:11:23.680 --> 00:11:26.440
Now, notice this diagonal line, the crease of the fold.
00:11:27.240 --> 00:11:30.280
It represents all pairs of points that look like 𝑥, 𝑥.
00:11:30.520 --> 00:11:33.880
Meaning, the pairs which are really just a single point written twice.
00:11:34.800 --> 00:11:37.800
Right now, it’s marked with a red line, and you should remember it.
00:11:38.320 --> 00:11:43.360
It will become important to know where all of these pairs like 𝑥, 𝑥 live.
00:11:43.360 --> 00:11:45.360
But we still have some arrows to glue together here.
00:11:45.640 --> 00:11:47.920
We need to glue that bottom edge to the right edge.
00:11:48.960 --> 00:11:51.640
And the orientation with which we do this is gonna be important.
00:11:52.360 --> 00:11:56.880
Points towards the left of the bottom edge have to be glued to points towards the bottom of the right edge.
00:11:57.360 --> 00:12:01.760
And points towards the right of the bottom edge have to be glued to points towards the top of the right edge.
00:12:02.440 --> 00:12:04.000
It’s weird to think about, right?
00:12:04.720 --> 00:12:05.000
Go ahead.
00:12:05.200 --> 00:12:06.480
Pause and ponder this for a moment.
00:12:09.720 --> 00:12:15.000
The trick, which is kind of clever, is to make a diagonal cut, which we need to remember to glue back in just a moment.
00:12:15.840 --> 00:12:18.680
After that, we can glue the bottom on the right like so.
00:12:21.960 --> 00:12:24.080
But notice the orientation of the arrows here.
00:12:24.720 --> 00:12:29.600
To glue back what we just cut, we don’t simply connect the edges of this rectangle to get a cylinder.
00:12:30.280 --> 00:12:31.240
We have to make a twist.
00:12:32.480 --> 00:12:36.120
Doing this in 3D space, the shape we get is a Mobius strip.
00:12:36.720 --> 00:12:37.400
Isn’t that awesome?!
00:12:38.120 --> 00:12:43.720
Evidently, the surface which represents all pairs of unordered points on the loop is the Mobius strip.
00:12:44.720 --> 00:12:50.600
And notice, the edge of the strip, shown here in red, represents the pairs of points that look like 𝑥, 𝑥.
00:12:51.000 --> 00:12:53.760
Those which are really just a single point listed twice.
00:12:56.560 --> 00:13:02.280
The Mobius strip is to unordered pairs of points on the loop what the 𝑥𝑦-plane is to pairs of real numbers.
00:13:02.920 --> 00:13:05.480
That totally blew my mind when I first saw it.
00:13:09.000 --> 00:13:20.800
Now, with this fact that there is a continuous one-to-one association between unordered pairs of points on the loop and individual points on this Mobius strip, we can solve the inscribed rectangle problem.
00:13:22.480 --> 00:13:30.000
Remember, we had defined this special kind of graph in 3D space, where the loop was sitting in the 𝑥𝑦-plane.
00:13:30.000 --> 00:13:35.760
For each pair of points, you consider their midpoint 𝑀 which lives on the 𝑥𝑦-plane and their distance 𝑑 apart.
00:13:36.280 --> 00:13:39.640
And you plot a point which is exactly 𝑑 units above 𝑀.
00:13:41.200 --> 00:13:46.200
Because of the continuous one-to-one association between pairs of points on the loop and the Mobius strip.
00:13:46.960 --> 00:13:52.720
This gives us a natural map from the Mobius strip onto this surface in 3D space.
00:13:53.920 --> 00:14:02.120
For every point on the Mobius strip, consider the pair of points on the loop that it represents then plug that pair of points into the special function.
00:14:06.520 --> 00:14:07.360
And here’s the key point.
00:14:07.800 --> 00:14:13.560
When pairs of points on the loop are extremely close together, the output of the function is right above the loop itself.
00:14:14.000 --> 00:14:19.640
And in the extreme case of pairs of points like 𝑥, 𝑥, the output of the function is exactly on the loop.
00:14:21.880 --> 00:14:28.160
Since points on this red edge of the Mobius strip correspond to pairs like 𝑥, 𝑥.
00:14:28.160 --> 00:14:36.560
When the Mobius strip is mapped onto the surface, it must be done in such a way that the edge of the strip gets mapped right onto that loop in the 𝑥𝑦-plane.
00:14:39.200 --> 00:14:41.520
But if you stand back and think about it for a moment.
00:14:42.040 --> 00:14:51.120
Considering the strange shape of the Mobius strip, there is no way to glue its edge to something two-dimensional without forcing the strip to intersect itself.
00:14:53.160 --> 00:14:56.200
Since points of the Mobius strip represent pairs of points on the loop.
00:14:56.840 --> 00:15:11.600
If the strip intersects itself during this mapping, it means that there are at least two distinct pairs of points that correspond to the same output on this surface.
00:15:12.920 --> 00:15:16.000
Which means they share a midpoint and are the same distance apart.
00:15:17.480 --> 00:15:19.600
Which, in turn, means that they form a rectangle.
00:15:21.280 --> 00:15:22.040
And that’s the proof!
00:15:22.680 --> 00:15:29.200
Or, at least, if you’re willing to trust me in saying that you can’t glue the edge of a Mobius strip to a plane without forcing it to intersect itself.
00:15:29.640 --> 00:15:30.640
Then, that’s the proof.
00:15:33.400 --> 00:15:36.080
This fact is intuitively clear looking at the Mobius strip.
00:15:36.400 --> 00:15:41.120
But in order to make it rigorous, you basically need to start developing the field of topology.
00:15:42.040 --> 00:15:44.920
In fact, for any of you who have a topology class in your future.
00:15:45.520 --> 00:15:53.800
Going through the exercise of trying to justify this is a good way to gain an appreciation for why topologists choose to make certain definitions.
00:15:54.880 --> 00:15:56.480
And I want you to take note of something here.
00:15:57.040 --> 00:16:03.120
The reason for talking about the torus and the Mobius strip was not because we were playing around with construction paper.
00:16:03.480 --> 00:16:06.240
Or because we were daydreaming about deforming a coffee mug.
00:16:07.480 --> 00:16:11.280
They came up as a natural way to understand pairs of points on a loop.
00:16:11.720 --> 00:16:19.520
And that’s something that we needed to solve a concrete problem.