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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