 I just want to welcome, and I will just say the practical things first. So for the Zoom attendees, I don't know, I hope you can hear me. We will take your questions at the end of the lecture, but then during the talk you can also post your questions, but we'll only take your questions after the lecture. But then for those of you who are at the lecture room right now with us, you may ask questions during the lecture as well, if you want. Okay, so I'm very honored to kind of reopen this presidential lecture series, which did not happen for a while because of the coronavirus situation, but I'm even more delighted to introduce you, Professor Andrew Lobb, who is a mathematician in the field of topology. Andrew has studied at Oxford for his bachelor's degree, later obtained his PhD at Harvard University, and also spent some postdoc research years at Imperial College London, Stony Brook, and Berkeley. Now he has been a professor at Durham University in the UK, and he actually has been visiting us since October last year as an excellence chair in the Oist Month Visitors Program, which in fact just only last year we launched this program, and so Andrew actually is the first visitor in this Oist Month Visitors Program. And today he will tell us about this kind of, at least a part of his remarkable solution to this big long-standing conjectures in math called the rectangular peg problem. And actually what is nice is that he has started working on this problem while he was at Oist in back in February, and then he finished solving this problem in May, three months later only, but still when he was at Oist. And there's, if I can say, a funny twist to this, which is that he was supposed to leave at the end of March to Oist to go back to the UK, but because of the coronavirus pandemic situation he couldn't really, and so he stayed here as extended stay, and that's when he actually finished solving this problem. So, and to me it was kind of interesting to observe, sorry, observe Andrew working on this problem. And one of the things that he told me was that when he kind of had this kind of, I guess, breakthroughish idea, he let it sit for a while, one hour or so, to kind of enjoy fully, enjoy this kind of sensation. And without, you know, before actually computing, he started computing it to check whether it's correct or possibly incorrect in that case. It ruins his kind of sensation. And later I was watching documentary of one of the, another honorable mathematician, fields metalist, Maryam Mirzakani. And she also basically mentioned the similar habit that she has, so it's interesting. And anyhow, I'm very excited that we created this opportunity to kind of, for you to also have a sneak peek at how Andrew views, you know, maybe some particular problem and kind of take it into, kind of recast it into a different, like view it in a different way in this particular problem, I mean topology. And then so that you can see it in a rather simpler and more beautiful manner. Okay, please join me to welcome Andrew and let us enjoy his lecture today. So thanks very much, lovely introduction. It's a pleasure to be here at OIST and pleasure to be given the opportunity to give this lecture. That's a picture of you and you're looking particularly happy in this picture because you've just won a second prize in a beauty contest, two contestants. And you've won, you've won this factory. There it is, square peg factory. And it's the pegs that are square, not the factory. And what is a, well, so a peg is something that goes in a hole, but like a nail goes in a hole in order to hold something down. And a square peg is a peg that has a square cross section, right? Okay, so we're on the same page. So if somebody comes to you once you've got your square peg factory and says I've got a hole and it's this shape, square. Do you have anything that'll fit the hole? The answer is yes, because you make square pegs. And you can make square pegs of all sizes just long as they're square. So you can make a peg that fits this hole kind of exactly, well, not even kind of exactly, just exactly. There it is, so the hole was in white and the peg is in pink and it fits the square hole. Exactly, all right, but maybe you're a bit cheeky and somebody comes to you with a square hole and says, can you make a square peg to fit my square hole? Holes for us are always going to be these sort of closed shapes in the plane, like this. Somebody else, can you make a square peg to fit the hole? You can say yes and make a slightly smaller peg. And following the rules of the game that we're going to use today, this is also a square peg that fits this hole, right? The point is I've got my hole as just some shape and I've got to try and find four points on that hole that form the vertices of a square. And then I say I can find a square peg to fit the hole. So even if somebody comes to you with a circular hole, get the phrase square peg in a round hole, you can actually make a square peg to fit this round hole. Because you can find four points on the circle that form the vertices of a square. And there are many choices, of course. Okay, fantastic. What about if somebody comes to you with this shape? So it's a bit like a circle, but it's got this slice taken out of it. So this is the shape of the hole. Can you find a square peg to fit this hole? The answer is still yes, and it's basically the same square peg. Now the interior of the peg isn't fully in the interior of the hole. But the rules of the game that we're playing today don't really require that. What we're trying to look for is four points on the boundary of the hole that form the vertices of a square. And then we say we found a square peg that fits that shape of a hole. Does that make sense? All right, cool. So this is a game we're playing. There's a very old conjecture, older conjectures go, these open conjections, due to turplets. In 1912, maybe he formulated it a little bit earlier than that, which is the following. Any shape hole, a square peg that fits it. Meaning I can draw any closed loopy thing in the xy plane, then I can always find four points on that loop that form the vertices of a square. This is known as the square peg conjecture or the turplets conjecture. And it's completely open in full generality, even to this day. So we're going to be talking about rectangles instead of just squares. And thinking about the problem of sticking rectangular pegs inside holes. So there's a rectangle that lives inside this circular hole. Here's a rectangle that lives inside that hole. I've just got to, again, find four points on the white hole curve that I'm given that form the vertices of a rectangle. Rectangles come in different shapes. And the shape of a rectangle is given by its aspect ratio, which is the ratio of the long side to the small side, which is a number greater than or equal to one. So square is nothing more than an aspect ratio one rectangle. It's a rectangle. And the result that Reiko introduction alluded to, that proved earlier this year with a collaborator, Josh Green is the following theorem. Any smoothly shaped hole has rectangular pegs fitted aspect ratios. So you can see, I guess, just from these pictures, that you can find rectangles of any aspect ratio that live inside the circle. You can just shrink this one, make it thinner. And you can find rectangular pegs of any aspect ratio inside the square as well. So it turns out, when you've got a smooth hole, this is always true. This isn't just a special property of the circle or the rectangle. And that's the result that we proved earlier. And that collaboration started here at OIST. And this is known as a rectangular peg problem. So you might think, if you're reading it quickly, that corollary is this conjecture of turplets. So it's not quite the case. And the key is this word here, smoothly, which doesn't appear in turplets's conjecture. So although we know that there is going to be a square in every smoothly shaped hole, we don't know it for every possible hole. Smoothly, what does it mean? I'm not going to tell you precisely. Some of you, I mean, means differentiable, basically. And that rules out things like fractals. OK. Good. So that's the thing we proved. And we were pleased and surprised. I guess more surprised than pleased, probably. But it was picked up by the media, and there are a few articles about it here and there. I'm not going to tell you about this theorem today. I'm going to tell you about a theorem that uses, has lots of the same ideas in it that go into the proof of this theorem. This is older. This is due to Vaughn in 1981. And he didn't write it up. He left it to a friend to write up. So hard to track down that reference. But the theorem is this. So any shaped, any hole has a rectangular peg that fits it. So Vaughn has dropped the smooth condition, but he doesn't get all possible rectangles. He just tells you there's one rectangle. If he could tell you that rectangle is always a square, then you would have proved the Turplitz conjecture, but he hasn't. He's just said, given any hole that you give me, I can find four points in it that form the vertices of a rectangle. And this is the thing I'm going to talk about today. I'm not going to talk about the proof of that theorem. I'm going to talk about a proof of this theorem. Vaughn's theorem. So apologies if you're expecting a nice easy presidential lecture just about the history of mathematics. I'm actually going to prove something. But I promise you, you're going to be able to follow most of this for sure. All right, so how are we going to prove it? So if we're going to start by collecting three facts, and then the three facts are going to fit together and give us the proof. Sounds easy, right? OK, so let's do that. The first facts are going to come from a game called cut-and-paste topology. And this is a game you can indulge in if you have some rubber sheets, some scissors, and some glue. Show you what I mean. So I suppose you've got this square sheet of rubber, and then you're given some assembly instructions that take the form of these two arrows. So what you've got to do is glue the two edges with the arrows together, like respecting the direction of the arrows. So you've got to line those arrows up. And then you ask, what do you get? That's part of the game. So that's quite easy to see what one gets. I can even see it in ghost outline on the board. You get this, which is a cylinder. Right, because you just take these two vertical edges, stretch them out, and then identify them around the back. And you get a cylinder. OK, cool. So that's the easiest part of the game. What about this? So I've got a rubber sheet, again, and I've got these two vertical edges with arrows on. I've got to line up the arrows. So I can do that, stretch it out, but I've got to put a half twist in it now so the arrows line up and then join it together. So what do you get? Don't all shout out at once or at all. You get a Mobius band. So this is, Mobius band is going to play big role in our later arguments. This is a non-oriental surface, and you've got one side. And whereas the cylinder's got these two circles that it's bound to, the top circle and the bottom circle, the Mobius band has just got a single circle. You trace it around. All right, it's a Mobius band. Let's get a bit more complicated. So I start with a right-angled triangle sheet of rubber. And I want to identify this bottom edge with the vertical edge, following these arrows and gluing them together. So if this arrow were pointing in the other direction, say, that would be kind of an easy thing to do. You just fold them together. You get like a cone. And because everything's a rubber, that's just like a disk. But these arrows point in different directions. So the question is, what do you get? And this is quite hard to see. I don't, yeah, I think you can't see this automatically. I think one needs to make use of the cut part of the cut-and-paste topology in order to figure out what this is. OK, so we're going to figure out what this is, and we're going to do it by making a cut. So we're going to cut this shape along there. What do we get? Get this. Get these two triangles. We still want to identify this edge with this edge. And we've got to remember we made the cut because we've got to glue that back together if we want to get back to the original shape, all right? And now the trick is what we're going to do is now we're going to glue the single arrows to the single arrows. Great, so I'm going to take this triangle and just lives here. And then this triangle is going to come around and stick on to that triangle. In order to make the arrows line up, of course, I've got to flip this triangle over. When I do that, I get this. So you get this shape. You get a square like that. Well, let's lose that edge in the middle that's a bit distracting. So it's just a square with opposite edges identified like this. And this is nothing more than a Mobius band because these two arrows go in different directions. Saw that already. Killer Mobius band. Amazing. Good. Two more shapes. This one. So now I want to identify the vertical edges with each other and then the horizontal edges with each other. Let's start with the vertical edges. That's just going to give us a cylinder. We saw that before. So we get a cylinder. It is stuck together there. And this is wrapping in that direction. So now I've got to identify the top circle with the bottom circle, glue those things together, and see what I get. So let's do it like this. So what I've done is I've made out rubber, remember? So I just grabbed hold of this top circle and stretched it round like this. And this kind of happened to the arrows. And you see, if I keep going, I can line up these arrows with those arrows and glue them together. Everything will be hunky-dory and I'll have my shape. And my shape will be, of course, a torus, otherwise known as the frosting on a donut. There it is. This is a torus. Another example. So I'm going to do the same thing. Glue the vertical edges together and the horizontal edges together. But the thing that's changed is the direction of these double-headed arrows on the top. Now they don't match with the ones on the bottom. So I'll get a cylinder as before and I can do this stretching out business again. So let's just do that all at once. But now the direction of these arrows has changed because the direction of those arrows that I started with at the top wasn't in the other direction. So what I can't do is continue to make a torus again because these arrows aren't going to line up with those arrows. And so I won't be following the rules of the construction set that I've been given. So what do I do? Well, I have to do something drastic. And I'll draw a picture then we'll talk about it. What I'm drawing is a Klein bottle. It's hard to draw a Klein bottle because a Klein bottle always has to self-intersect itself in three-dimensional space. So I'm going to do my best. There. This is a picture of a Klein bottle. So what I've done is I've shrunk this circle down, made it into this sort of neck, and then it's come in and it's like punctured the other side of this tube. And then it's able to join up with this circle sort of at the bottom and they match. But the problem is it had to kind of come in and puncture the surface at the side. Klein bottles in three-dimensional space always have to intersect themselves. That's either a weakness or a feature of Klein bottles in three-dimensional space. Because there's no way of matching up this circle with that circle without the surface puncturing itself. That's our first fact. Fact one, a Klein bottle in 3D space always self-intersects, always has to puncture itself in three-dimensional space. All right, that's our first fact. I want to do another fact. Let me draw that other fact somewhere. Maybe I can raise this statement of this theorem. I'm not going to prove. So the other fact concerns the Klein bottle. So here's my Klein bottle. So it's another fact about the Klein bottle. So I've got a Klein bottle. And what I'm going to do is cut it up and make this cut from that corner to the midpoint and then from this midpoint to the corner like that and see what I get. So I get two right-angled triangles and one parallelogram. Right, so it looks like that. But there's some identifications that go on this edge. It's got to be glued to that edge. The front part of these double-headed arrows gets identified with the front part of these double-headed arrows like that. That's how that gluing takes place. So that means that this up here gets identified with that down there, right? And the back of this double-headed arrow gets identified with the back of that double-headed arrow. So that means that this here gets identified with this here. So I made that cut and that's the result. So what is the result? Well, this piece is on its own. It's just glued to itself and it's a Mobius band, right? Because the top and the bottom get identified with a twist. This is a Mobius band. The other two pieces, well, they get glued together in a couple of ways by the white arrow and by the orange double-headed arrow. So let's glue them together by the white arrow. Well, once you do that, you see it's also nothing but a Mobius band. It's also a Mobius band. So what we've done is we've taken the Klein bottle and we've managed to cut it into two Mobius bands. Is there a question? Yeah, I mean, so this is a slightly mathy question, sorry guys, but the point is this is going to be one single cut. And so this point has to line up with that point and that sort of forces things to happen. All right. Anyway, this is two Mobius bands. So what we've done is we've taken a Klein bottle and we've managed to cut it into two Mobius bands. This is the form of the second fact. Klein bottle equals Mobius band plus Mobius band. Short-hand way of writing down the fact that you can cut a Klein bottle into two Mobius bands, or in other words, if you've got two Mobius bands, you can stick them together and make a Klein bottle. That's our second fact. That's our last fact, I think, from cut and paste apology that I'm going to use. Our third fact is in the form of a question and answer, fact three. Here's the question. When does ABCD form a rectangle? What do I mean by ABC and D? So ABCD, it just can be the vertices of some quadrilateral inside the XY plane. And I want to know when that forms a rectangle. Let me draw your quadrilateral that looks very rectangle-like. ABCD, I want to know when is that quadrilateral rectangle? That's the question. There are several good answers to this, several correct answers as well. And the standard answer, of course, is that this quadrilateral is a rectangle when these four angles are right angles. That's the standard answer and it's a good answer. But it's not the answer that's going to be useful for us. The answer that's going to be useful for us is actually in terms of the diagonals of this rectangle. Let's put in some diagonals. There we are, flag. So you'll notice the diagonals of a rectangle have the same length as each other. Now, that doesn't guarantee that your quadrilateral is definitely going to be a rectangle. You can make, say, a kite shape where you've got the two diagonals having the same length. But another property that the diagonals have is they share a common midpoint. The midpoint of AC is the midpoint of BD. And when the midpoints line up and the lengths are the same, you see that you've got to have the diagonals of a rectangle. So this is the characterization of rectangles that's going to be important for us. So here's the answer. When AC and BD, the same midpoints and the same length. So that's the answer. It's very easy to convince yourself of this fact. And these are the three facts that we're going to put together and use to deduce this result that any shaped hole has a rectangular peg sitting inside it. And I don't think it should be at all obvious to any of you how one goes about assembling these three facts to give a proof of this theorem. But I'm going to give you a plan of the proof. And then we're going to do the proof. So it's going to happen. Plan of the proof. And it's going to be a proof by contradiction, otherwise known as a reductio ad absurdum. Your pretentious or Roman. And what a proof by contradiction means is, well, we're going to assume that we don't have the thing we want and then go on to deduce an impossibility. Therefore, we must have had the thing that we really did want. So what we're going to assume is that there is a hole that has no rectangular peg. In other words, there's going to be some whole H where you can't find four points that form the vertices of a rectangle. So assume H is a hole that has no rectangular pegs fitting. All right? So we're going to assume that form theorem isn't true because there's some hole that doesn't have a rectangular peg. And then we're going to go on and deduce something that's false, and that'll tell us that our assumption must have been false. Therefore, all holes must have rectangular pegs. It's going to be the idea. So what are we going to do? We'll use H to put the Mobius band in 3D space. So what we're going to do is first thing, doubling the Mobius band. And I'm going to tell you precisely what doubling is yet. But doubling the Mobius band will give a Klein bottle because two Mobius bands make a Klein bottle in 3D space with no self-intersection. Klein bottles in 3D space always have self-intersection. So that's the contradiction that we wanted. But it shouldn't be obvious how we do any of these things yet because this is the plan of the proof. How are we going to use H to put a Mobius band in three-dimensional space? And what does doubling mean? And how do we see that the Klein bottle has no self-intersection? How does H play into this strategy at all? All right, so that's the plan. So here's H. H is just some hole shape in the XY plane. Going to make it look particularly blobby like this. This is H. It's living in the XY plane. And I want to be able to refer to points on H. This thing we're assuming exists in our plan of proof. We're assuming this is a hole that has no rectangular pegs that fit it. I want to be able to refer to points on H easily. So I'm just going to start a particular point, call that 0, and travel around and measure the distance. So this curve looks to me like it's got length 8, 6, 7, and then back to there. Don't take that 8 equals 0 expression too seriously. This is H. All right. And I'm going to use H to put a Mobius band inside three-dimensional space. So where's my Mobius band? Here is my Mobius band. We're going to use this picture of the Mobius band. The one where it starts with a triangle and you identify the bottom edge with the top edge. You remember we showed that as a Mobius band by cutting and pasting. So what I've got to do is, given any point inside this Mobius band, I've got to give you an address for that point in three-dimensional space. I've got to give you three coordinates for it, X, Y, and Z. I do that for all the points inside the Mobius band. That'll tell you how the Mobius band sits inside three-dimensional space. That's the plan. All right. So how are we going to do it? Well, let me draw three-dimensional space. How about that? X, Y, Z. Now, when I stick this Mobius band in three-dimensional space, I'll tell you this now, it's not going to lie just anywhere in three-dimensional space. It's going to lie in the part where the Z coordinate is positive. So if I think of this floor as being the X, Y plane, then the Mobius band is going to live in level E and higher, and none of it's going to live in level D. So it's just going to live there. Good. Everybody with me? So let's go ahead and do this. So what I'm going to do is put some little numbers here. 8, 4, 0, 8. I use the same length as the curve to put this thing just in a grid like this. And remember, this is the Mobius band. I've got to tell you, I've got to give you a three-dimensional address for each point in the Mobius band. So let's do that with a particular point that I pick at random or as far as you know at random. So this is the point 7, 1. Now I've got to tell you where do this point in the Mobius band live in three-dimensional space? What do I do? This is what I do. I look at the coordinates 7, 1. I travel over to H. I find 7 and 1. I draw a line between them. The midpoint gives me the X and Y coordinates because this is just in the XY plane. The midpoint gives me the X and Y coordinates. And the length gives me the Z coordinate. So it seems a little bit esoteric, but that's what we're going to do. Say again, let's take my point in the Mobius band. This is the point 7, 1. That gives me two points on this whole H. I connect them, and then I use the midpoint and the length to tell me where this point in the Mobius band lives in three-dimensional space. All right, so that's a process that isn't super obvious. So this is, I don't know where it goes. Somewhere in three-dimensional space, with those coordinates XYZ. Doing this a lot, XYZ. OK, fantastic. Good, so a few things to note. This is what I really was explaining was how I tell you about a point in this right-angled triangle where it goes in three-dimensional space. But in the Mobius band, this edge is identified with this edge. So it would be good if when I stuck this triangle in three-dimensional space, this edge matched up with that edge, and then I'd really know I had a Mobius band. So is that happening? Let's show you what I mean. If I look at this point, 5, 0, that's really the same point as this point up here, which is 8, 5. So I'd like them to get sent to the same point in three-dimensional space so that these two edges get glued together. When I follow the recipe for 5, 0, what do I do? I join 0 and 5. I look at the length and the midpoint, and that gives me the point in three-dimensional space. If I follow the recipe for 8, 5, what do I do? I join 8 and 5. I just get the same line. So it's got the same midpoint and the same length. So these two points indeed go to the same point in three-dimensional space. So this is really telling me how to put my Mobius band in three-dimensional space. So first point, there's going to be two other points. Next point is I'm going to worry about points on this line. So this line is the boundary of the Mobius band. It's kind of the circle boundary of the Mobius band. So where do points on this line go? Let's look at this point at 4, 4. What happens? So I've got to follow my recipe. 4, 4, that determines two points on this curve. But they're both the same point. They're both the point 4. So it's there and there. Just the same point. The midpoint of those two points is just the same point itself because it's a zero-length little interval. And, well, the length is zero, so the z-coordinate is zero. So where 4, 4 ends up going in the xy-plane is just exactly to this point with z-coordinate zero. And if I think about where all these points go, it's going to be the same story. They're just going to trace out this curve in the xy-plane. So this boundary of the Mobius band, where is it getting sent to? It's getting sent here, the curve h in the xy-plane. And every other point on the Mobius band is living kind of above that with positive z-coordinate. All right, so I've got this picture. I don't want to take that big. So that's my picture of the Mobius band. And this is the boundary of the Mobius band on the xy-plane. All right. And there's going to be one more point about the Mobius band. And this last point is that this Mobius band doesn't intersect itself anywhere. Now, you should be expecting another point about the Mobius band because we haven't used our assumption, which is that h has no rectangular pegs. The assumption that h has no rectangular pegs tells us that this Mobius band doesn't self-intersect. Why is that? Well, suppose it did self-intersect. What that would mean would be there would be two points here that correspond to two arcs of this curve h that have the same midpoint, given the same xy-coordinates and the same length, given the same z-coordinate. But then those arcs would be diagonals of a rectangle that live on h. But h doesn't have any rectangles. So all right, so what have we learned? This is a Mobius band with no self-intersection. Great. And it meets the xy-plane in its boundary curve like this. So what do I do? I double it. And doubling just means I reflect an xy-plane. Get this Mobius band down here. This is a reflected Mobius band. And so what I've got is two Mobius bands that are just being glued together. And that, of course, we know is a Klein bottle. So this thing is a Klein bottle. And the top Mobius band has no self-intersection. And the bottom Mobius band has no self-intersection. Of course, don't meet each other because the top one lies in positive z and bottom one lies in negative z. So it's a Klein bottle without self-intersection. And that's the contradiction that we wanted. And so that tells us that our assumption that H has no rectangles must have been false. So therefore, H must have some rectangular peg that fits it. And that's the end of that proof. And I guess the end of that talk. So thanks very much. Thank you for your talk. That was really fun. I had a small maybe technical question. When you cut up the Klein bottle right behind you and you sort of glued it back together into two Mobius bands, and then you use this to say that, oh, Klein bottle is then to glue together. And so I'm understanding that you're gluing them on sort of since the Mobius band only has one sort of circle as a boundary, you're gluing the two circles. But do you have any issues of chirality as in sort of are there like two ways to glue like two Mobius bands, depending on how you would do that half twist in 3D space? So Mobius bands can definitely live in 3D space in different ways. But however you glue together two Mobius bands, you'll get a Klein bottle. It just might be living in space in different ways. Any more questions from the lecture room? Yeah, is this OK? I had a question about the result that you didn't prove today, which is about the regularity of the curve. So what kind of smoothness assumption do you make? And where does that quantitatively enter into the estimate? So we make the, I think you can get away with C1, just differentiable, continuously differentiable. But we make the smooth assumptions infinitely differentiable. And so where that enters in is we end up working not with a Klein bottle in three dimensional space, but a Klein bottle in four dimensional space. And that extra degree of freedom allows us to sort of parameterize all the aspect ratios. So we get all the aspect ratios. However, we need to use the geometry of four dimensional space rather than the topology as is used here. And it's rather important for the geometry that we have smoothness so that we can talk about differential forms and stuff. And I guess that at some point there's no way to remove like, because this is a, as first just a qualitative assumption, there's actually a quantitative place that it enters so that after you obtain the result, then you can kind of remove these assumptions later. There's some place where you can't remove it later. Yeah, I mean, one can't, yeah, basically you can't do simplistic geometry things on smooth. I mean, that's somehow the question. I mean, the problem with getting some, so our result shows, and it was already known, that if you've got a smooth curve, then you can always find a square inside it. So you can maybe try this if you want to prove the original Turplitz conjecture. You take some weird looking hole that isn't smooth and then you approximate it by smooth things. You get a sequence of squares and all the smooth things and then you sort of use compactness or something to show you the limit square in the non-smooth thing. Problem that can happen is that the square might decrease in area and vanish away to nothing in the limit, but nobody's really found a way to overcome that problem yet. Okay, thank you. Anybody else? I'm going to be very brave and ask a very, I'm not a mathematician at all, but I think to me, the magic of this proof came when there is an assumption made and there is the relationship you established between how you transformed the triangle into that shape. I think I didn't quite understand how that assumption was the basis for this transformation. Which assumption? What, yes, assume H is a hole which has no rectangular pegs. Oh, right, yes, so this is, where we use that is here, really. So the fact that H has no rectangular pegs is telling us that the Mobius band has no self-intersection. And why is that? Well, if this Mobius band has self-intersection, what it means is there are two points in this Mobius band that get mapped to the same place in three-dimensional space. And if you follow through this process, see what it means for two points in the Mobius band to get mapped to three-dimensional space, it's going to tell you that there are two arcs of this curve that might look something like this that have the same midpoint, giving you the same X, Y coordinates three-dimensional space, and the same length as each other, giving you the same Z coordinates three-dimensional space. And that's going to tell you that there was a rectangle. But because there isn't a rectangle, you know that can't happen, so that's what's telling the Mobius band has no self-intersection. Thank you for a great lecture. So I'm sorry for this question, because I'm not a mathematician, but part of being this kind of fundamental research, does it have any applied idea here? Like, I don't know. Could it be helpful in understanding the nature of black hole or time traveling or something like that? Thank you. Well, the assumption right at the beginning of the lecture was that you had inherited a square peg factory. And so, yeah, I mean, the result quite useful from that point of view, because you know you can fulfill all of society's needs. But in terms of applications, I don't really think there are any. I mean, this was 1912, this is back before relativity and so on, it entered the picture of math and physics that had not really meshed in this way that they did in the next few years. No, this is just a problem that's intriguing because it's so simple to state and yet apparently so hard to make any traction with. So, no, we don't, I don't think there are necessarily any applications for this. One thing it suggests is that one should maybe be thinking about applying modern day techniques to these old problems because our proof of the rectangular peg problem was really very quick once we'd had the idea to use symplectic geometry and the main body of the proof really only took like a page. And so it might be an opportunity for enterprising mathematicians to go back and look at some of these old conjectures and see where the up-to-date techniques can really be applied. I'll open the questions to the Zoom attendees, no? For those of you who are attending the Zoom, do you have questions? Could you also type your question in the Zoom somehow? Please click the Q&A button at the bottom center of your screen. Okay, so the question is, do we assume that the curve does not have self-intersections? Yeah, I think, yeah, that's, yeah. Yes, we do. Follow-up questions are welcome too, so. If there are any other questions also from this room, no? If not, then if not on Zoom, then let's thank Andrea again for this wonderful lecture.