At the end of the 19th century, David Hilbert, who was a famous mathematician at the time,
Listed 20 problems that he thought were going to direct the future of 20th Century mathematics.
But one of them, Hilbert's 3rd problem, was so easy that
Hilbert's student solved it 2 years after he posed it.
So that's the one I want to explain today.
Yeah, I would call it Hilbert's 3rd problem,
but it's maybe, maybe let's call it the equidecomposability problem.
If you have two polygons 𝑃₁ and 𝑃₂--
Brady Haran: So no rules about convex or concave? Daniel Litt: No rules about convex or concave.
All you need is a finite number of straight lines which meet and form a closed figure.
So here are 2 polygons, 𝑃₁ and 𝑃₂, and they have the same area,
then you can cut off 𝑃₁ with finitely many straight lines and rearrange the pieces to get 𝑃₂.
So here's an example of a polygon which I've already pre-cut.
It's a parallelogram and here's how we can rearrange it into a rectangle with the same area.
Just move this triangle over here. So that was pretty easy.
So the proof actually gives you an algorithm for cutting up a polygon
and decomposing it into another one.
So Hilbert's 3rd question,
which was in 1898, I think, was, "Is the same thing true in three dimensions?"
So what do I mean by that?
I mean if you have two polyhedra, so three-dimensional figures
which are made out of, you know, flat faces edges, and vertices,
and if they have the same volume,
can you cut one up into pieces with finitely many cuts and rearrange the pieces to make the other?
Brady Haran: Well, in my intuition, would be yes.
It's because if you put two polyhedra with the same volume into, like, a bath,
they'll displace the same amount of water.
Daniel Litt: So, yeah, so if you were allowed to melt the polyhedra, you could definitely do it.
but if you're only allowed to use straight cuts, it turns out you can't,
and this is what Dehn proved,
and I think Hilbert probably expected this,
so I think he probably tried with the cube, tried to cut it up into pieces,
and rearrange it into a tetrahedron,
so that's a figure with 4 faces all a bunch of triangles, which are the same, and he couldn't do it.
and so the actual question was:
can you cut up a cube with finitely many knife cuts and rearrange the pieces to get a tetrahedron?
And Dehn ended up proving that you can't do that. Brady Haran: You can't do it?
Daniel Litt: You can't do it.
Brady Haran: What changes between 2 and 3 dimensions that suddenly makes it impossible?
Daniel Litt: Yeah, so I can explain at least the idea behind Dehn's proof.
Dehn was the student of Hilbert, and it was his thesis problem to solve Hilbert's third question,
it's one of the greatest theses of all time, and so what was his idea?
Well, his idea was that, well, let's say you have 2 polygons, which have a different area.
Why can't you cut one up and rearrange it to the other?
Well, the area is invariant under this action of cutting it and then moving the pieces around;
the area doesn't change.
So Dehn's idea was that you should find some invariant of a polyhedron
that doesn't change when you cut it up and move the pieces around,
and then he wanted to compute that invariant for the cube and for the tetrahedron and see if they're different.
So what that means is that, you know,
no matter how you take the cube, cut it up, and move the pieces around,
if these invariants are different, you can't get a tetrahedron.
Brady Haran: The deal breaker, the thing that can't change--
Daniel Litt: The obstruction, yeah, to move, changing one piece
and one polyhedron into another polyhedron. Brady Haran: What was it? Are you going to tell me?
Daniel Litt: I'll tell you, it's kind of complicated.
So we should maybe we should build up to it.
Let's prove this 1833 theorem.
So this is equidecomposability for polygons.
All right, so there are several steps. So what's the goal?
Let me first make the problem a little bit easier.
Let's try to cut up a polygon of area 𝐴,
so 𝐴 could be 5,
and rearrange the pieces into a 1 by 𝐴 rectangle.
And I claim that's enough.
Why is that? It's because cutting something up and rearranging the pieces is a reversible operation,
so if you have a polygon of area 𝐴,
you can make it into 1 by 𝐴 rectangle
and another polygon with area 𝐴; you can make it into the same 1 by 𝐴 rectangle,
and then, well, how do you get from your first polygon to your second one?
you just pass through the 1 by 𝐴 rectangle that you are able to make both of them into,
so it's enough to do this.
so we can assume one of 𝑃₁ and 𝑃₂ is just a rectangle.
What's the first step? It's easy.
You just cut it into triangles.
Step 2 is a little trickier.
We're going to turn triangles into parallelograms
All right, so we have our triangle. What do we do?
We take the midpoints of these two sides,
we draw the line through them so that this has the same length as this,
and then, well, we just move this triangle over here.
So, here's a triangle,
And now you can cut it up and turn it into a parallelogram.
Brady Haran: So any triangle can be parallel, er, parallelogram-ized.
Daniel Litt: Yeah, that's exactly right.
All right, now we're gonna turn parallelograms into rectangles.
So here what do you do?
You just drop a perpendicular like this and then move this triangle over here.
Here's my parallelogram and I'm going to turn it into a rectangle.
That was easy.
All right, so now, we have some rectangle, which has the same area as our original polygon.
Well, we do this for every triangle we got. And we get a bunch of rectangles,
and now, we're gonna make it into a 1 by 𝐴 rectangle.
So here's some rectangle,
and what we do is we measure out a distance 1,
let's say, 1 inch or so on the bottom,
and draw a rectangle with the same area.
Now we connect these two vertices,
and it turns out that the triangle 𝐴𝐵𝐶 is the same--
it's congruent to the triangle 𝐷𝐸𝐹,
So we can start by moving this triangle and just sliding it down here.
Now, there's a little bit left over,
it's these two triangles, and it turns out they're also the same;
it turns out the triangle 𝐴𝐷𝐺 is the same as the triangle 𝐶𝐹𝐻.
How do we cut this rectangle up into this one which has the desired property
to 1 by Area rectangle? Just move this triangle over here and this triangle over here.
Okay, and that finishes the proof.
We first cut into triangles, then we cut into parallelograms,
then we cut into rectangles, then we cut it special rectangles.
And then we undo all of that to get into the other polygon that we wanted to aim at at the beginning.
Brady Haran: Cool. Daniel Litt: Okay.
So, why did this work?
It worked because we had this one invariant that we knew to search for the area,
where we were gonna search for a 1 by 𝐴 rectangle,
and it turns out that there's a secret hidden invariant in three dimensions
that will stop us from running this algorithm.
All right, so let's see what Dehn did.
So what he wanted to do is he wanted to take a cube and a tetrahedron,
which have the same volume,
and he wanted to show you couldn't cut this up in any way
with a finite number of slices and rearrange the pieces to get this one,
so he had to come up with some way of distinguishing the two,
something that is different about these guys,
and it doesn't change when you cut them up and rearrange the pieces.
Brady Haran: Daniel, you're saying he wanted to show you couldn't do it
That's right. Brady Haran: Wouldn't he rather have shown
you could do it? Did he already know-- Brady Haran: He probably tried, I mean,
yeah, so why do you conjecture something's impossible?
you, like, to try really hard and try really hard and you just fail,
and you get really frustrated and you're like, "I'm gonna prove this can't be done."
and in Dehn's case he was able to do it, right?
So here's the basic observation.
so let's zoom in on some edge of a polyhedron
and there's two invariants of this edge.
You can look at-- you can look at the length,
So 𝓁 is the length of an edge,
and there's another invariant which I'll call 𝜃. That's the dihedral angles.
It's the angle it takes to get from this face to this one,
so it's the angle swept out by my hands; I moved from one face to the other
It's the angle it takes to get from this face to this one,
so that's an angle between 0 and 2𝜋
So what's the observation? The observation is there's two ways to cut,
so one is you can cut like this.
so you can cut with a plane that kind of passes through the edge.
It cuts this edge in half, right?
Brady Haran: It changes the length.
But it leaves the dihedral angle the same, right?
And the lengths of the two edges you get are the-- they sum to the original length.
So if you cut, let's say, vertically, you get (𝓁₁, 𝜃)
and (𝓁₂, 𝜃), so the two lengths of these two pieces, 𝓁₁ and that's 𝓁₂,
the 𝜃's are the same and 𝓁₁+𝓁₂ is the original length.
Right, so the other thing you can do is cut along the edge
so the other thing you can do is cut like this,
and then you get, well, two edges there, on these two pieces.
They have the same length and angles, summed together,
so you get (𝓁, 𝜃₁) and (𝓁, 𝜃₂),
and here, 𝜃₁+𝜃₂ is the original 𝜃.
so the two dihedral angles add together to give you what you started with.
And, so, Dehn's idea was to stare at this
and just make it into an invariant in the most brutal way possible.
So how did he do it?
So here's the invariant: so you take the sum of the following symbols 𝓁ᵢ⊗𝜃ᵢ
We're here-- this runs over the edges of the polyhedron.
so what you do is you take the length of each edge and the angle at each edge
and write down just this sum, which is right now just a sum of symbols.
It doesn't mean anything.
So, let's figure out what it means. Brady Haran: How you add an angle to a length?
That's right, it doesn't. Well, so we're sort of we're formally multiplying them
and then we're adding these symbols together,
so let me let me say where this lives.
So this lives in something, which is called the Real numbers, tensored with Real numbers mod 2π
so let me explain what this set is.
Brady Haran: You're cracking out the new symbols on me now?
Daniel Litt: That's right, yeah.
So, let's say what this is: it's an element of this thing.
This weird object is just a sum 𝑎₁⊗𝑏₁+𝑎₂⊗𝑏₂.
these are just symbols so far, all the way up to 𝑎_𝑛⊗𝑏_𝑛 for some 𝑛,
where these 𝑎ᵢ and 𝑏ᵢ are just real numbers,
but we say two symbols like this are the same if certain rules are satisfied.
So what rules?
Well, first of all, 𝑎⊗𝑏 is the same as 𝑎⊗(𝑏+2𝜋). What is this expressing?
It's just expressing that if you take an angle and you add 2𝜋 to it, you get the same angle.
All right, now we have some other rules.
We say that 𝑎₁⊗𝑏+𝑎₂⊗𝑏=(𝑎₁+𝑎₂)⊗𝑏
So you're allowed to add on the left if the 𝑏's on the right are the same.
So what does this express?
It expresses that if you've 2𝜃's which are the same, if you cut, you can add two lengths.
That's just expressing this.
All right, there's one more rule:
It's just the same thing on the other side.
So it says that if you have 𝑎⊗𝑏₁+𝑎⊗𝑏₂,
that's the same as 𝑎⊗(𝑏₁+𝑏₂).
And that's just expressing this rule.
So this is just a set of symbols which are allowed to add and subtract,
but they follow some funny rules when you do that,
and Dehn's Invariant is this object inside this set,
And he's saying, "Well,
no matter how you cut off your shape,
When you compute this invariant, you get the same thing in this set."
That's called the Dehn Invariant.
Brady Haran: I'm not gonna pretend to totally understand it,
but I believe you. What's wrong with having an invariant though?
What does that mean? I can't just, you know,
be strategic and make millions of cuts and still be able to build these things--
Yes, and the point of this invariant is that
it doesn't change when you make a cut, right?
You get more edges and you get more angles,
but because of these rules over here and equivalently these rules over here,
the invariant in the set just doesn't change
So now what you have to do is give into a polyhedron.
You have to compute this invariant and sometimes, they're different, right?
So Dehn's observation
Showed that the invariant of the cube is not the same as the invariant of the tetrahedron.
Brady Haran: So this invariant is almost like a fingerprint or DNA.
Daniel Litt: That's exactly right. Brady Haran: It's unique to the polyhedron.
Yeah, so it turns out and Dehn didn't know this,
but it was proven about 30 years later
that if you know the volume of a polyhedron and you know the Dehn invariant,
then that determines the polyhedron up to this operation of cutting and moving the pieces around.
So in fact, it's true that
if you have two polyhedra with the same volume and the same Dehn invariant,
then you can cut one up and get the other one by rearranging the pieces.
But here, Dehn, what Dehn did is just showed that if you have two polyhedra with different invariants,
You just can't do that. You're stuck.
There's no way you can change the Dehn there.
Brady Haran: Because once you've got that Dehn invariant,
you're stuck with it for life. Daniel Litt: That's right. Yeah.
There's nothing you can do, at least with this operation of cutting with a knife
and moving the pieces around that changes that invariant. That's why it's called an invariant.
Brady Haran: What's the Dehn invariant of a cube? Is it like--
Daniel Litt: Yes, we can do the Dehn Invariant of a cube. That's a good example.
So let's do the cube. So luckily, all the edges are the same.
So we only have to compute with one edge.
So let's say we have a cube of side length 1,
and then what do we have to do?
We have to compute the angle. Well, that's just 90°, or 𝜋/2.
So for each edge contributes 1⊗(𝜋/2), and then how many edges are there?
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Well because of this rule that if the second guys are the same,
we can add the first guys.
This is just the same as 12⊗(𝜋/2), if you'd like.
That's the Dehn Invariant of the cube.
Now, the tetrahedron is a little harder while you compute this angle
which I'll just call 𝜃, a little bit of a pain to compute it, so I won't do it here.
You have to compute what side length gives you a volume of 1,
so let's call it 𝓁.
And how many edges are there?
There's 1, 2, 3, 4, 5, 6, so you get 6𝓁⊗𝜃.
and once you compute what these actually are, you'll see that it isn't the same as this.
Brady Haran: So if I gave you a cube and said, "Turn this into a tetrahedra,"
Daniel Litt: Yeah. Brady Haran: You're gonna say, "Yeah, I can do that,"
but unfortunately, I'm gonna have to change it's volume.
No, unfortunately you have to do both with the volume and the Dehn invariant.
Oh you have to change both
Because, of course, the volume doesn't change when you cut and move things around either,
So those are exactly the two invariants of a polyhedron--
the volume and the Dehn Invariant.
Dehn's big contribution was to discover that this is there is the second one. Nobody knew that before.
Brady Haran: How come I can do it when I melt them?
Daniel Litt: Yeah, that's a good question.
That's because melting them doesn't change the volume, right?
but it does change this invariant; you've destroyed all the edges,
so how would you even compute the Dehn invariant?
In fact, it turns out in higher dimensions, there are also even more invariants.
We don't know what they are in general,
So that's an open problem.
If you want to think about, I don't know, 12 dimensional polyhedra,
and one you can cut one up and get the other one.
It's actually a really beautiful and exciting part of mathematics.
Brady Haran: There's a little bit of complicated mathematics there are,
Daniel Litt: That's right. Brady Haran: But,
I can still appreciate what Dehn did--
the cleverness of it-- that he found it-- that he realized those edges were the problem?
Daniel Litt: Yeah, that was clever.
Well, I think the really amazing thing is
he knew, well, the lengths of the edges don't stay the same
and the dihedral angles don't stay the same when you cut things up,
but there's some funny way of combining them all together. That does stay the same.
That's really the miracle of the Dehn Invariant.
Brady Haran: Clever. Daniel Litt: Yeah.
Brady Haran: Good thesis. Daniel Litt: Yeah.
Brady Haran: I bet there was no questioning about giving him his doctorate.
Daniel Litt: Yeah, I mean, he went on to do a lot of amazing stuff to the rest of mathematics too.
But yeah, this was an amazing start.
[Brady Haran laughs] Brady Haran: Cool!
Because the hotel's full, but the manager is clever, and here's what he does:
He shifts the person in Room 1 to Room 2,
and the person in Room 2 to Room 3,
and the person in Room 3 to Room 4,
and so on.
Everyone gets shifted forward one room,
and because there are infinitely many rooms,
you never run out of rooms to put people in.