 Thanks for that introduction. It's a great honor to be here. So I wanted to talk about this topic, building multiple embedded curves and quantifying non-orientability. I wanted to talk about this for a couple of reasons. First, this is a project that I started here. Several years ago, Larry Gooth and I were both visiting IHES. And we started talking about this project. And eventually, I completed this. And then second, because the tools that show up here end up having a very systolic feel to them. And since this is a conference in honor of Marcel Becher, I wanted to talk about something a little bit systolic. So let me start by explaining the title. So the title is, I want to talk about filling multiples of curves and quantifying non-orientability. So let me start with a question. So suppose you have a closed curve in Rn. Then there's a minimal surface whose boundary is this curve. So what happens if we go around this curve twice? What happens if we take two copies of this curve? What happens to the minimal surface? Well, the obvious thing to happen would be that if you take two copies of this curve, then you should get two copies of the minimal surface. But it turns out that there are examples where that doesn't happen. LC Young's found some examples in the 60s of curves where if you take two copies of the curve, then the area of the corresponding minimal surface only increases by a factor of about 1 and 1 half. So the topology of the minimal surface changes completely when you take two copies of the curve. And so that's the phenomenon that I wanted to talk about today. It turns out that this phenomenon is closely related to non-orientability. And so what I want to do today is I want to start by giving an example of one of LC Young's constructions. I want to connect this to non-orientability. I want to define a quantitative version of non-orientability, a way of measuring how non-orientable a surface in Rn is. And then I want to show that this quantitative non-orientability is bounded in terms of the area of the surface. And that consequently, we can get bounds on the filling areas of multiple curves. OK? So let me start with the definition. So if T is an integral one-cycle, so suppose I take a simple closed curve or a union of oriented closed curves, in Rn, we define the filling area of T to be the minimal area of an integral two-chain. Just think about a surface whose boundary is that curve. That is boundary equal to T. And so the question I want to start with today is this. How is the filling area of T related to the filling area of 2T? How is the filling area of the curve related to the filling area of two copies of the curve? One direction is clear. If I have a filling of T, if I have a surface whose boundary is T, then I can take two copies of that surface to get a surface whose boundary is 2T. So that means that the area of a minimal filling of 2T is at most 2 times the filling area of T. And in low dimensions, this is going to be sharp. So for instance, when n is equal to 2, when we're dealing with curves in the plane, then the filling area of 2T is exactly equal to 2 times the filling area of T. And you can show this by looking at winding numbers. If I have a curve, say, like this in the plane, then it winds around this region once, this region once, this region once, this region twice. And so the minimal area of a surface whose boundary is this curve is this area plus this area plus this area plus 2 times this area. And if I go around that same curve twice, then what happens to the winding numbers? Well, they all double. Now it winds around this area twice, twice, twice, four times. And so the integral of the winding numbers, the area of a minimal surface whose boundary is this curve exactly doubles. The inequality is also sharp in R3. If you have a curve in R3, then the filling area of two copies of that curve is twice the filling area of the curve. But now this takes a little more work to do because it doesn't make sense to talk about winding numbers in R3. So now this is a theorem of Federer. But if we go up one more dimension, if we go to R4, then there are examples due to LC young of curves in R4. So the filling area of two copies of the curve is only about one and a half times the filling area of the original curve. And so why does this happen? And why does this happen in R4 rather than R3? Well, like I said at the beginning, this has to do with non-orientability. So if we want to construct an example like this, we need to start with a non-orientable surface. Say we take a Klein bottle embedded in R4 like this. Then we can construct a curve that lies on this Klein bottle. What we're going to do is we're going to construct our curve to be the sum of 2k plus 1 loop, where k is some large number. So a large odd number of loops around the Klein bottle, all going in alternating directions. And because we're on a Klein bottle, so if this one goes this way, then the next one goes the other way, and the next one goes the other way. Since we're on a Klein bottle, if we change directions an odd number of times and go around the Klein bottle, then we end up back where we started. So I can have all of these in all of these, having alternating orientations like this. And then the filling area of this curve, if I want to find a minimal surface where the boundary is this curve, a natural thing to do is to take pairs of these loops, one goes one way, the next goes the other way, and fill them in with annuli. And if we do that, we get a picture like this. So pairs of adjacent loops get filled with annuli. But because we started with an odd number of loops, we're going to have one loop left over here, and we have to fill that loop with a disk. So the filling area of T, this is approximately minimal. The filling area of T is going to be about half the area of the Klein bottle plus the area of the disk. OK? So now what happens if we take two copies of the curve? If we take two copies of the curve, now we have an even number of loops going around the Klein bottle. And because we have an even number now, we can pair all of them up. We pair them all up going all the way around the Klein bottle. And the surface we get is almost exactly the surface of the Klein bottle, except the orientations alternate. So here, we're oriented one way. Here, we're oriented the other way. Here, we're oriented one way. Here, we're oriented the other way. But we have a surface which is made up of the surface of the Klein bottle with alternating orientations in the area of that surface is the area of the Klein bottle, or is approximately the area of the Klein bottle. And so what that means is that we've saved two copies of the disk. The filling area of 2T, the minimal surface filling two copies of the curve, has area less than twice the area of the minimal surface filling one copy of the curve by exactly two copies of that disk. So is this OK so far? Is the example clear? We're going to be coming back to this a couple of times during the talk. So I just want to make sure this is clear. That's a three-dimensional example. Well, no, this is a four-dimensional example because we need to start with an embedded Klein bottle. Yeah, it needs to be embedded. We'll see later what happens if you only have an immersed Klein bottle. So yeah, so what does that mean? So that means that the obvious relationship that you would expect between the filling area of T and the filling area of 2T doesn't hold. It's not always going to be twice the filling area of 2T. It's not always going to be twice the filling area of T. And so the natural question is, OK, is there any relationship between the filling area of 2T and the filling area of T? And it turns out the answer is yes. It turns out that for any dimension of d and n, there's a C greater than 0. Such that if T is a d-cycle in Rn, then the filling area of 2T is at least C times the filling area of T. So this is what I want to talk about today. So how do we prove something like this? What we want to do is we want to show that the filling area of T is bounded from above by the filling area of 2T. And so the way we're going to do this is we're going to take, our strategy is going to be to try to take a filling of 2T and then try to take half of it. So for example, if we start with something simple, we start with a zero-cycle. So a zero-cycle is just going to be a collection of points in Rn with positive and negative orientations. And then a filling of 2T, that's going to be a one-chain. So it's going to be a collection of curves, oriented curves. So there we have two curves coming out of each minus, and then two curves going into each plus. So this is the filling of 2T. If I want to bound the filling area of T from above, then I want to find a way to take half of this filling. What does it mean to take half of this? Well, one thing we could do is we could just take all of the curves that go, say, clockwise. So we take all the curves that go in this direction. That's half of B, half of the filling. That's a filling of T. And the area of that filling should be about half the area of our filling of 2T. So what are we actually doing here? What are we actually doing when we say, when I say half take half of B? Well, one way to look at it is that this is about orientations. If I take this filling of 2T and I forget the orientation on it, if I forget the directions of all the curves, what I end up with is a cycle. I have two edges going into each minus, and I've got two edges going into each plus. And so I get a cycle like this, a mod 2 cycle. And because this is one-dimensional, it's orientable. It's a sum of closed curves. So I can pick an orientation, say like this, call that orientation P. And then when I say take half of B, one way of formalizing that is to say, I take all the parts of B that match up with the orientation. I take the sum of B plus P. These have the same curves, but with different orientations. And so whenever I have a piece of B that goes parallel to the orientation, I get two copies of that in the sum. Whenever I have a piece of B that goes opposite to the orientation, it gets canceled out in the sum. And so what I end up with is two copies of each edge that goes parallel to the orientation, no copies of each edge that goes opposite to the orientation. And so that's a filling of 2T still. And I can divide it by 2, because now all the coefficients are even to get a filling of T. And the same thing almost works in higher dimensions. The same thing almost works for the Klein bottle example. So what are we doing? We take a filling of 2T, we forget the orientations, we orient that, we orient that, the resulting mod 2 cycle, we take a sum, and we get half of it, and we get two times, and we get some chain with even coefficients. And we can almost do that for the Klein bottle example. What happens if we try this for the Klein bottle example? Well, here's our curve, here's our filling of two copies of the curve. If we forget the orientations on this filling, we end up with a mod 2 cycle. This is the fundamental class of the Klein bottle. Unfortunately, we can't take the next step because we can't orient this. Our next step would be to find an orientation of this cycle. What we can do is we can say, OK, we can't quite orient this. But if you cut the Klein bottle, if you cut the Klein bottle right there, you're going to get a cylinder. That cylinder is orientable. And if you cap off both ends of the cylinder, you get a sphere, and that sphere is orientable. So what you end up with, if you do that, is what I'm going to call a pseudo orientation. So what is this? So this is an orientation on, this is the cycle with integer coefficients. And the cycle is you cut the Klein bottle, you cap off both ends of the resulting cylinder, and you take the fundamental class of that sphere. And so what you end up with is a cycle which has, well, it's either plus or minus 1 on the surface of the Klein bottle. And then in that place where you cut, that disk where you cut, you have two copies of a disk. So this is an integral cycle. And it's congruent to b mod 2. We have 1 plus or minus 1 on the surface of the Klein bottle and then 2 on this disk in between, and that's going to cancel out when we take it to cycle mod 2. And we call this a pseudo orientation of b. It's a cycle with integer coefficients that's congruent to b mod 2. And this we can do nearly the same thing that we did before. We can take the sum of b plus this cycle. That's going to be congruent to 0 mod 2 because b is congruent to p. Because that's congruent to 0 mod 2, that means that all the coefficients are even. That means we can divide it by 2 to get an integer cycle, a cycle with integer coefficients. And the boundary of b plus p over 2 is going to be 2t plus 0 over 2, which is t. And the picture looks like this. Here is our original filling of 2t. Here is our pseudo orientation. If we sum these together, we're going to get two copies of each of the annuali that run parallel to the orientation. We're going to get zero copies of each annuali that run opposite to the orientation. And then we're going to get two copies of that extra disk we added here. And so we end up with exactly what we saw before. We have alternating bands going around the climb bottle and then one disk to fill that last piece of the climb bottle. And so we can do nearly the same thing for in higher dimensions. It's just that instead of taking an orientation, we have to take a pseudo orientation. Is this OK so far? So to summarize, what we end up with is this. So let's define an invariant. Let's say that if A is a mod 2 cycle, I'm going to define the non-orientability of A. By NO of A is the infimal mass of an integral cycle, which is congruent to A mod 2. This is the smallest possible mass of a pseudo orientation. If A is, in fact, orientable, then we can know that if A is, in fact, the fundamental class of an orientable manifold, we can just take P to be the integral fundamental class of that manifold. And then NO of A is just the mass of A. If A is non-orientable, then we're going to have to do some cutting and gluing like we saw before in order to get a pseudo orientation. And NO of A is going to be bigger than A. So this measures how hard it is to lift A to an integer cycle. And what we just saw was that if the boundary of B is 2T, then we can debound the filling volume of T in terms of the mass of that filling of 2T plus the non-orientability of that filling of 2T. And so in order to prove that the filling volume of T is bounded by a constant times the filling volume of 2T, it suffices to show this. It suffices to show that if A is a mod 2 cycle in Rn, then the non-orientability of A is bounded by the mass of A, bounded by the area of A. Is this OK so far? OK. So before I get to the proof of this theorem, I want to mention a couple of corollaries. So I've been motivating this in terms of filling areas because I think it's the easiest way to motivate this. But this is also important in geometric measure theory. So part of geometric measure theory involves various completions of the space of chains. And so if we can take these quantitative bounds on filling areas of chains and on the difficulty of lifting chains to lifting mod 2 chains to integral chains to show various facts about completions of the space of chains, about currents and flat chains. So for instance, and these are going to take a little bit of explaining because these are going to look very trivial, but these have been open questions for a long time. So for instance, we can show that if K is a positive integer, then the map which sends T to KT is an embedding with closed image on the space of integral flat chains. The integral flat chains are a certain completion of the space of chains based on the flat norm. The flat norm is closely related to filling areas. And so the fact that these examples that show that filling area is not multiplicative also show that the flat norm is not multiplicative. And so these examples tell you that the flat norm is not multiplicative. But we can use this theorem to show that in fact, the flat norm is not too far from being multiplicative. That if you multiply a chain by T, then the flat norm can only go down by a constant factor. Consequently, this tells us that the space of multiples of K is actually reasonably well behaved. And so this is important if you want to define, for instance, currents mod K or flat chains mod K. This lets you show that if T is a mod K current, then you can lift it to an integral current. Again, this is something that looks kind of obvious, but it actually takes a theorem like this to prove. The problem is that any mod K chain can be lifted to a chain with integer coefficients, but typically a current is going to be a limit of mod K chains. And it's not clear that if you take a sequence of mod K chains that converge, it's not clear that the lifts to integral chains are also going to converge. In order to show that those lifts all converge, you need some bound like this. You need some bound that says that if I have a chain with mass A, then I can lift it to an integer chain with mass comparable to the mass of A. And if you do that, you can prove something like this. You can prove that any mod K current can be lifted to an integral current. And consequently, you can show that the mod K currents are actually a quotient of the integral currents, just like the mod K chains are a quotient of the integral chains. But this is a diversion. I'm not going to talk about currents and chains, currents of flat chains for the rest of the talk. I just wanted to mention these corollaries. OK, so here's the, so here's the. So we want to prove this theorem. We want to prove that if A is a mod 2 d cycle in Rn, say some non-orientable surface in Rn, then its non-orientability is bounded by the mass of A, that we can lift A to an integral cycle without too much larger mass. So how do we prove this? The basic idea is not too far different. The basic idea is not too far removed from what we did for the Klein bottle. For the Klein bottle, we said, OK, if you take the Klein bottle and you cut the Klein bottle, you can cut it into a cylinder, you can cut it into an orientable surface, and then cap that off to get a pseudo-orientation. We can do kind of the same thing in general. So our strategy in general is going to be this. If I have some mod 2 d cycle in Rn, Rn is contractable. So there's a mod 2 d plus 1 chain F so that the boundary of F is equal to A. In the case of the Klein bottle, you can take this to be the solid torus whose boundary is the twisted solid torus whose boundary is the Klein bottle. This d plus 1 chain can be cut into orientable pieces. It might be non-orientable, but we can cut it into orientable pieces. And we can construct a lift F sub z of F that has integer coefficients. Basically, we take our d plus 1 chain, we cut it into orientable pieces. We just pick an orientation on each of those orientable pieces so that we can turn it into a chain with integer coefficients. Then the boundary of that lift is going to be a pseudo-orientation of A. And if we are lucky, if we've made good choices of how to cut, and if we've made good choices of orientations for the pieces that we cut it into, then that pseudo-orientation isn't going to be too much larger than the original. Then the mass of P shouldn't be too much larger than the mass of A. And in some cases, we can do this very well. So just like before, I'm going to start by showing what happens in a particularly simple example. In this case, where our example is going to be co-dimension 1 rather than dimension 0, if we start off with a cycle A, which is co-dimension 1. So for instance, let's take this figure 8 curve in the plane, then we can find it. If we can find an orientation on that cycle by doing this, we write this as the boundary of a top-dimensional chain F. So here, F is this gray area in here. Right now, A and F, these are both chains with mod 2 coefficients. But because F is top-dimensional, we can put an orientation on Rn and then inherit that orientation to F. In this case, I've chosen the clockwise orientation. And then the boundary of F is going to be A again. But now, the boundary of F is going to be an oriented version of A. So F is orientable. So A is orientable. And the non-orientability of A is equal to the mass of A. We can do the same thing in higher dimensions, just as long as we keep this co-dimension 1 assumption. And so in particular, this is why the Klein bottle example had to be done in R4 rather than R3. Because in R3, if we look at a Klein bottle that's immersed in R3 rather than embedded in R4, then a Klein bottle immersed in R3 has an inside and an outside. So you can see, here's my Klein bottle again, except now I've got a self-intersection here where the neck intersects the bulb. And if I have a Klein bottle immersed in R3, then you can imagine filling it with smoke. You can imagine piping smoke into the Klein bottle. And the smoke can't escape because it can't escape out that way because the bulb is solid. It can't go through the bulb. And it can't escape through the neck because if you go out around the neck, you're going to run into the outside of the bulb right there. So if you fill this up with smoke, then the smoke is going to take a shape like this. You're going to have a solid torus of smoke and then a neck coming out. And then that neck is going to be capped off like that. And so in particular, the Klein bottle immersed in R3 is actually kind of orientable. We can put an orientation on that like this. And this is the boundary of a solid torus. So this is topologically a torus. A torus is orientable. And so we can put an orientation on it like this. And that's going to be an integer cycle, which is congruent to my original mod 2 cycle. And it has the same mass as my original cycle. So we can think of the Klein bottle immersed in R3 is orientable. And so we can do exactly this construction. In general, we can think of a co-dimension one cycle in any Rn as orientable. And if we do that, we get this proposition that every n minus 1 cycle in Rn is orientable. The non-orientability is equal to the mass. Consequently, as one consequence, we get this theorem of Federer that if T is an integral n minus 2 cycle in Rn, then the filling volume of 2T is equal to twice the filling volume of T. So this is essentially why this phenomenon happens in R4, but not in R3. Because in R3, when you do this construction, when you start with this curve and you take this filling of 2T and then you take it mod 2, you end up with a mod 2 cycle. But that mod 2 cycle is orientable. And so you end up with a sharp bound on the filling area of 2T in terms of the filling area of T. Is this OK so far? Are there any questions so far? OK. OK, so this is a low-co-dimension case. What happens in higher-co-dimensions? So in higher-co-dimensions, things get more difficult. Because now, in higher-co-dimensions, you don't have this natural orientation of the interior of A. And so you have to do something more complicated. So what I'm going to do in higher-co-dimensions is I'm going to start with a very simple argument that gives a very bad bound on the non-orientability. And then I'm going to progressively improve that argument, try and make it more and more complicated to get better and better bounds. So here's the basic argument. Let's suppose that A is a mod 2 d-cycle in V, mod 2 d-cycle in Rn. Let's say it has mass V. And let's make it a cellular cycle. By a cellular, I mean that it's a cycle in the unit grid. We take the unit grid and there's a sum of d-cells in the unit grid. The reason I want to make it cellular is because that way I can do this. I can say, oops. OK, good. OK, what was I saying? So let's suppose that A is a mod 2 cellular d-cycle of mass V. But what I want to do is this. Let's fill A with a mod 2 chain F. Because we started with a cellular cycle. We can make F cellular as well. So F is a sum of d plus 1 cells in the unit grid. In particular, by the isoparametric inequality, it's a sum of V to the d plus 1 over d cells in the unit grid, each of which is a unit cube. Consequently, it's a sum of, well, this many orientable pieces. Each of those cubes is orientable. So let's just pick an orientation on each of those cubes, essentially at random. The problem with choosing an orientation on those cubes at random is that if I have two adjacent cubes, then I have a 1 half probability that I chose a different orientation on each side of those two adjacent cubes. And so if I take the boundary of Fz to get a pseudo-orientation, then odds are that the area of that pseudo-orientation, the mass of that pseudo-orientation, is going to be approximately the total boundary of all of those cubes, about half of the total boundary of all those cubes, which is to say, it's going to be on the order of V to the d plus 1 over d. And so we get a very simple argument, but unfortunately, it gives us a very bad bound on the non-orientability. It tells us that the non-orientability is at most V to the d plus 1 over d. Rather than what we want, which is that the non-orientability should be on the order of V. So how can we improve this? Well, the first thing we can do is we can say, OK, well, it's not so efficient to take a bunch of little cubes like this, let's try to make these cubes bigger. Let's say instead of taking a bunch of little cells like that, let's try to construct something like this. If we can replace some of these cubes by bigger cubes, then the total boundary is going to decrease by a lot. And so the area of the resulting pseudo-orientation should decrease by a lot. And how do we do this? How do we construct a filling like this? What we can do is we can construct a filling by using a sequence of approximations. So here is, say, our original curve. Let's call this A. And this is a curve in the unit grid. If we approximate this in the 2 by 2 grid, we get a curve like this. If we approximate in the 4 by 4 grid, we get a curve like this. And we can keep on approximating the 8 by 8, 16 by 16, in larger and larger and larger grids. Furthermore, the difference between two consecutive approximations between the original curve and the approximation in the 2 by 2 grid or between the 2 by 2 grid and the 4 by 4 grid and so on, these are going to be relatively small. So for instance, the difference between the original curve and the approximation in the 2 by 2 grid is just about this. And this is made up of roughly v many squares, each of them unit squares, so each of them with perimeter on the order of 1. If we take the approximation in the 2 by 2 grid and the 4 by 4 grid, these are two approximations of the same curve. And so again, these are going to be close together. These are going to be about distance to a part, which means that I can fill in the annulus between this approximation and this approximation with roughly half as many squares, v over 2 squares. But now each of the squares is twice as big, v over 2 squares, each with perimeter on the order of 2. And we can, yeah? Sorry, which is the rule for the approximation choice? We use the Federer Fleming deformation lemma. And so we can fill in this annulus with v squares with perimeter 1. We can fill in this annulus with v over 2 squares with perimeter 2. And we can keep on going and going and going. This is eventually going to stop. Where does it stop? Well, eventually this grid is going to get larger than my original surface. Eventually, when my grid is, say, side length v, then it's going to be bigger than my original surface. And the approximation is going to be 0. And so eventually, if we add up log many steps like this, log many annuli like this, we're going to get a filling of A. And that's going to be a filling of A made up of, well, v squares with perimeter 1 or v unit squares, v over 2, 2 by 2 squares, v over 4, 4 by 4 squares, and so on and so forth. And so now when we take the total boundary of all the squares that we used, in the first step, we used squares with total perimeter on the order v. In the second step, we used total squares with total perimeter on the order of v. We have log many steps. And so we end up with the following proposition that if A is a cellular mod 2 cycle with volume v, then it has a pseudo-orientation p, where the mass of that pseudo-orientation is only on the order of v log v. Because we have log many steps and each of the steps we used about perimeter v. So how do we do better than this? Because now we want to get rid of this log factor. And this is where it starts to get hard. Actually, I guess they're OK. Because this is where things start to get difficult. So in order to understand how to get rid of this log factor, we should try to understand where this log factor comes from. The reason we have this extra log factor is basically because this procedure is very wasteful in cases where this procedure is very wasteful in places where the surface is already close to a plane. So if I have part of my surface that's close to a plane, if I have a chunk like this, then when we think about what we're doing, we're taking an approximation of this in the 1 by 1 grid. And then we're taking an approximation in the 2 by 2 grid. And then we're taking an approximation in the 4 by 4 grid and the 8 by 8 grid, and so on and so forth. We're taking log many approximations of this same piece of A. And each of those log many approximations is contributing about whatever that area is to the non-orientability. And really, this is kind of unnecessary. Because if we can find some piece of A, if we can find some piece of our surface that starts out close to a plane, then we should be able to just pick an orientation on that piece of the surface, say I orient it that way. And if I pick an orientation on that piece of the surface, then that's going to give me orientations on each of the approximations. I have an approximation of this 1 by 1 approximation going this way and an approximation of the 2 by 2 grid going the same way, approximation of the 4 by 4 grid going the same way. I should be able to get at least local orientations on each of those approximations. And if I get local approximations on each of those approximations, then I can choose consistent orientations on all of the squares that result from this process. And if I choose consistent orientations on all those squares, I get cancellation when I take the boundary. And so I don't need to use all the boundary that I drew here. I can just end up with something where the boundary is just one of those approximations. So really, what we want to do to get rid of the log factor is we want to take advantage of the fact that choosing these orientations of squares randomly is very wasteful when our surface is already close to a plane. The problem we run into is that there's no particular reason that A should ever be close to a plane. You might have started with a surface that was never close to a plane. So here's an example. So say I want to construct a surface that has a very large non-orientability. Say I want to construct a surface that's as non-orientable as possible. So what I can do is I can start with, say, a cube. And this is going to be a cube. So my surface is going to be two-dimensional. It's going to be embedded in R4. So this is a three-dimensional cube in R4. And I'm going to make the side length on the order of v so that the area of this cube is v. So this is orientable. And if I want to make it non-orientable, one thing I can do is that I can start adding cross-handles. I can take, say, a disk here and a disk here, cut those out and replace them with a twisted cylinder here, a cylinder here glued with a twist. That makes it non-orientable. But what I really want is I want to make this as non-orientable as possible. So if I want to make this more non-orientable, I can do the same thing on this face. I can do the same thing on this face. I can add, I guess I have six faces, so I can add six cross-handles of scale root v. But once I've done that, I've kind of run out of space to add more cross-handles. I've kind of run out of space to add more handles of this size. Instead, what I can do is I can go one scale down. Instead of trying to add another big cross-handle, I can try to add a bunch of little cross-handles. Say I take something 1 tenth the size, I cut out two circles like that, and add it in a little cross-handle there, and add a little cross-handle there, and maybe on this previous handle and over here. And if I do this 600 times, if I go down by a factor of 10, if I scale down by a factor of 10, now I have 100 times more space to add handles. And so I can add, say, 600 cross-handles of scale root v over 10. That's going to increase the non-orientability, because now in order to make this orientable, I have to cut all six big handles, and then I need to cut all 600 small handles. But now I'm out of room to add the small handles. So the next thing I do, well, I go down by another factor of 10. Now I add on some very tiny handles, 1 100th the scale of my originals. And again, I add so many that they fill up the entire surface. That's going to take about 60,000 handles of scale root v over 100. And I can keep on going. I can just, every time I run out of space, I go down to the next scale and add more. Where does this stop? Well, eventually this has to stop, because I wanted my surface to be cellular to start with. I wanted my surface to be made out of unit cubes. So if I'm building a surface out of unit cubes, that means I have to stop when I have to stop when this is smaller than 1. I have to stop when I hit scale smaller than 1. That's going to take about log many steps. And so what I also, so, so, so, so, so now I can try to count how much non-orientability I've added. The total amount of non-orientability for this surface is going to be, well, let's see. In order to make this orientable, I need to cut each of the cross handles, each of the big cross handles. That's going to be six of those, each contributing on the order of, that's going to be six big handles, each of which contributes on the order of v to the non-orientability. Then I need to cut these 600 smaller handles. That's going to be about 600 small handles, each of which is going to contribute on the order of v over 100, because they're 10 times smaller than what I started with. Same thing for these tiny ones. That's going to be 60,000 handles that contribute each area on the order of v over 10,000, and so on. And there are log many steps. And each of the log v many steps, each of those steps contributes on the order of v. And so when I add all these together, I end up with a surface with non-orientability on the order of v log v. But there has to be some problem here, because my theorem said that a surface with area v should have non-orientability on the order of v. And this is a surface with non-orientability on the order of v log v. What's the problem? Well, the problem is that I can't add all these handles for free. Every time I add a handle, it adds to the area of the surface. So I added these six big handles at the beginning. This is going to contribute area v. I added the 600 smaller handles on top of those. That's going to contribute another v to the area. And same for the next ones and the next ones each of these scales contributes another v to the area. There are log many scales. So by the time I've done this construction, I started with a cube with area v, but I end up with a surface with area on the order of v log v. And so the reason that the theorem still holds for the surface is that in order to add all this non-orientability to the surface, I needed to add a lot of area. And so what we need to do in order to get rid of this log factor is to show that that happens in general. We need to be able to show how do we prove the proposition for sets like this, for sets that are. This is not exactly a fractal because we only iterated log v many times. We only iterated down to scale 1, but this is close to a fractal. And so the way that we prove this proposition for sets that are close to fractals, we need to show that adding topological complexity to our surface also adds extra area. And then we need to prove the theorem for surfaces with low complexity. And I'm putting this in quotes because the notion that I'm actually going to use is something called uniform rectifiability. So is this OK so far? Is the picture clear so far? OK. So what I'm actually going to use is a notion called uniform rectifiability. This is developed by David and Sems to sort of quantify and sort of quantitative version of rectifiability. Remember, the surface is rectifiable if it has a tangent plane almost everywhere. And so for instance, well, a line is rectifiable, a square is rectifiable because you have a tangent plane everywhere except the corners. And even a surface like this where we have something close to a fractal, even a surface like this is rectifiable. Because for almost every point, if you zoom in small enough, if you zoom into scale 1, then you're going to see something that looks like a plane. You're going to see something that has a tangent plane. This is a surface that's very ugly on scale as bigger than 1, but as long as it's nice on sufficiently small scales, then it's rectifiable. And that is really the problem with rectifiability, that you can have surfaces that are rectifiable but are still very poorly behaved. And so the idea of uniform rectifiability is to exclude that. We say that a surface is uniformly rectifiable, or a set is uniformly rectifiable. And I apologize for this. We say that a set is uniformly rectifiable if and only if it has what's called a coronary decomposition. And I'm not going to give a rigorous definition of what a coronary decomposition is, but very roughly speaking, what this means is that for all but a few balls centered on the set and for all but a few balls centered on the set, the intersection of the ball and the set is close to the graph of a Lipschitz function with small Lipschitz constant. So in particular, if I have a line, this is going to be uniformly rectifiable, because no matter where I am on the line, if I draw a ball like this, I'm going to see something that's close to a Lipschitz graph with small Lipschitz constant. If I have a square like this, I can do nearly the same thing. I can say, OK, I'm going to take these four Lipschitz graphs here, one for each side. And then if I take a point on the square and I take a ball around that point, then most of the time, unless I'm unlucky, unless I pick to the corner or unless I pick something very close to the corner, when I take a ball around that point, I get something that's close to one of these Lipschitz graphs. But surfaces like this, where you have structure, where you have topological complexity at every scale or at many scales, these are not uniformly rectifiable, because no matter what ball I take, if I take a ball like that, then it sees that cross handle. If I take a ball like that, it sees that cross handle. Most of the balls are going to see a cross handle until you get down to scale one. But any ball of scale bigger than one is going to see a cross handle. And so uniformly rectifiable sets are essentially the opposite of these close to fractal sets. You can have complexity some of the places, some of the time, but you can't have complexity everywhere at every scale. In a uniformly rectifiable set, you can have complexity at some scales and some points, but you can't have complexity at every scale and every point. So I apologize for not being able to give a rigorous definition, but the rigorous definition is complicated. But is this OK so far? Is the basic idea clear? So what we do is we take a complicated set and we break it down into it. So the way we prove the theorem is we show that every surface can be broken down into a sum of uniformly rectifiable surfaces. That every mod 2 cellular d cycle a can be written as the sum a is the sum of a sub i's of mod 2 cellular d cycles with uniformly rectifiable support so that the total mass of each of those pieces is bounded by a constant times the mass of the original cycle. So we can cut it into pieces so that all the pieces are uniformly rectifiable and the total area of the pieces is not too much bigger than the total area of what we started with. In this case, in this example of the cube with all these cross handles, what this decomposition looks like is like this. We can write, let me call this a, we can write a as a sum of the original cube and then we can add on each of these cross handles. And what do I mean by this? I mean I want the, so to add each of these cross handles we cut out these disks and we added on this twisted cylinder what I want to add here is this twisted cylinder plus these disks so that the sum of this plus this mod 2 is going to be the cube with the cross handle added on top. And similarly, we have a cross handle on the front and a cross handle on the side and then three more on the back and the bottom and the other side. And so we can write, so this first generation we can write as the cube plus six of these cross handles. This second generation, we added 600 more of these so we can add 600 of these smaller cross handles and then we can add 60,000 of these tiny cross handles and so on and so forth. And on one hand you can check that the sum of all these pieces is going to be my surface here. And furthermore, all these pieces that we added are essentially rotations and scalings of one particular piece. They're all essentially rotations and scalings of this. And a piece like this is simple enough that it's uniformly rectifiable. As long as I pick a point on the surface and I take a ball around it, as long as I don't take a ball that sees one of these corners, I'm going to get something that's reasonably smooth. I'm going to get something that's reasonably close to a Lipschitz graph. So each of these, so this cube is uniformly rectifiable. Each of these pieces use uniformly rectifiable and the sum of everything is this surface here. And the total area of the pieces in this sum, well, this contributes area V. These all contribute another factor of V. These all contribute another factor of V and so on and so forth. And so the total area of all these pieces is on the order of the area of A. In general, the proof of this theorem uses a result of David and Sam's that any quasi-minimizing set is uniformly rectifiable. So in general, the way we prove this theorem is that we take some arbitrary surface here and then we say, OK, well, if this surface is too complicated in one region, if this surface is very poorly behaved in one region, then we can just cut that out and that region becomes one of our pieces. And if we're careful about doing this, if we do this starting with small regions and then going to bigger and bigger regions, then the result is going to be a decomposition like this. So the result is going to be a decomposition into uniformly rectifiable pieces so that the sum of the areas of the pieces is comparable to the area of the whole. Then the second part of the theorem is that the second part of the proof is that the theorem holds for uniformly rectifiable pieces. We show that if you have a mod 2 cellular d-cycle with uniformly rectifiable support, then it has a pseudo-orientation p with a massive p bounded by a constant times the mass of a. And this goes back to, and the idea of this is basically this idea about choosing the orientations of squares correctly, this idea that if we have a piece of the surface that looks like a plane, then we can choose all the approximations of that piece of the surface to have orientations that go the same way, and so you can get some cancellation there. This uses the fact that each of these pieces are uniformly rectifiable because a uniformly rectifiable surface has to be close to a plane at most balls, at all but a few balls. So a mod 2 cellular d-cycle with uniformly rectifiable support has a pseudo-orientation with comparable area. And then when you put these two together, you get the proof of the theorem. Is this OK so far? OK. So let me finish by talking about a couple of open questions. So one natural, so what this shows is this shows that in general, if I have a cycle T, oops, there's supposed to be a 2 in there somewhere. So there's supposed to be a 2 on the left. It's supposed to ask, is the filling volume of 2T greater than or equal to the filling volume of T? We'll just say, does the filling volume always increase when you take a multiple? And this isn't clear. What we show is that the filling volume of 2T can't go down by too much when compared to the filling volume of T. What we show is that the filling volume of 2T is greater than or equal to some constant times the filling volume of T, but it's not at all clear whether that constant is greater than or less than 1. Conceivably, you could try to estimate that constant from the proof, but the proof is quite long. And so, so, so, so, so, so, so, so, so, so my guess is that if you tried to estimate the constant from the proof, you would end up with something extremely small. It would be very natural for the filling volume of 2T to always be greater than or equal to the filling volume of T, but it's not at all clear. Another question we can ask. More generally, this shows that the filling volume of KT is at least a constant times the filling volume of T for any K. We can ask whether we can choose these constants C sub K uniformly. And then finally, a more general question. So this is sort of a, this is sort of a systolic, this is sort of a, this non-orientability is sort of a systolic invariant. Because one way we can think about non-orientability is it's a question about how much cutting you need to do in order to cut a non-orientable surface into an orientable one. We started with this climb bottle example where you take the climb bottle and you cut along a particular curve to make it into a cylinder and you cap off that curve to get a pseudo-orientation. In general, the picture for any surface is roughly the same. If you have some large complicated non-orientable surface, you want to cut by a curve that makes it orientable and then you want to take, then you want to fill in all the curves that you cut, all the curves you cut. So this is a sort of a systolic question. You have the, in order for a curve to cut a non-orientable surface into an orientable one, it has to be in a particular homology class. So we're asking about the smallest representative of that homology class, not necessarily in terms of length but in terms of area. And so the unusual thing about this theorem is that this is a systolic estimate for surfaces that's independent of genus. This bound that the non-orientability of A is at most some constant times the area of A is completely independent of genus. Whereas usually when you look at systoles of a surface, they get larger and larger as the genus increases. Part of the reason for that is because we're making this assumption that our surface is embedded in RN. Because it's embedded in RN, you can't have some of the strange geometry that you usually see in a surface of large genus. So one natural question is this. Can we say something more general? If I have a surface embedded in RN by a bilipjits map and by a bilipjits map, I mean, I don't mean like a Nash embedding. I mean that two points that are some distance apart in the surface map the points in RN that are comparable distance apart. If I have a surface embedded in RN by a bilipjits map, are there systolic inequalities for that surface that are better than the inequalities for an arbitrary surface? And this suggests that at least some of the time there should be some inequalities that are better for an embedded surface than an arbitrary surface. But I don't have many examples beyond this. I don't have a general picture of the situation. Okay, thank you. Are there any questions? Can you hear me? At some point, you mentioned Mock K cycles. The rest of the time Mock two. So, does K differ from two? It's essentially the same story, it's harder to draw pictures. Other questions. I actually have one myself. You talked in the beginning about integrating the winding number. But if you integrate the square of the winding number, that's also an interesting invariant. And that actually works for surface curves in three space because instead of integrating the stricter points, you integrate with respect to lines. Is that anything you've looked at or is that anything that comes up in this? Inequality of Bill Pohl, that was. I haven't thought about that. I'll have to look at the inequality. Yes, it's in Santolo's book. Okay. So, other questions? Yes. So, can you say something about what happens when N becomes large and the surface? No. N being the dimension or? Oh, the dimension of the ambient space? No. So, a lot of the steps we use are, you use the federal filming deformation lemma. And that's dependent on the dimension of the ambient space. It's possible that in Gromov's filling inequality, or he's independent of the ambient space. So, it's certainly possible that there are ways to do this that are independent of the dimension of the ambient space, but what I'm using is not. Any other questions? If not, what do we do? We take a small break, five minutes. Five minute break, you can get that and come back and then cross the river. Thank you very much. Thank you.