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  • 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 ?₁ and ?₂--

  • 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, ?₁ and ?₂, and they have the same area,

  • then you can cut off ?₁ with finitely many straight lines and rearrange the pieces to get ?₂.

  • 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 ?,

  • so ? could be 5,

  • and rearrange the pieces into a 1 by ? 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 ?,

  • you can make it into 1 by ? rectangle

  • and another polygon with area ?; you can make it into the same 1 by ? rectangle,

  • and then, well, how do you get from your first polygon to your second one?

  • you just pass through the 1 by ? rectangle that you are able to make both of them into,

  • so it's enough to do this.

  • so we can assume one of ?₁ and ?₂ 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 ? 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 ??? is the same--

  • it's congruent to the triangle ???,

  • 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 ??? is the same as the triangle ???.

  • 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 ? 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 ? is the length of an edge,

  • and there's another invariant which I'll call ?. 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 2?

  • 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 (?₁, ?)

  • and (?₂, ?), so the two lengths of these two pieces, ?₁ and that's ?₂,

  • the ?'s are the same and ?₁+?₂ 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 (?, ?₁) and (?, ?₂),

  • and here, ?₁+?₂ is the original ?.

  • 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 ?ᵢ⊗?ᵢ

  • 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

  • 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 ?₁⊗?₁+?₂⊗?₂.

  • these are just symbols so far, all the way up to ?_?⊗?_? for some ?,

  • where these ?ᵢ and ?ᵢ 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, ?⊗? is the same as ?⊗(?+2?). What is this expressing?

  • It's just expressing that if you take an angle and you add 2? to it, you get the same angle.

  • All right, now we have some other rules.

  • We say that ?₁⊗?+?₂⊗?=(?₁+?₂)⊗?

  • So you're allowed to add on the left if the ?'s on the right are the same.

  • So what does this express?

  • It expresses that if you've 2?'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 ?⊗?₁+?⊗?₂,

  • that's the same as ?⊗(?₁+?₂).

  • 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