 The speaker of the morning, Larry Gooth, who will speak about Lipschitz's constant and degree of. OK, thanks for having me. I was Tom's grad student 15 or 20 years ago, and he was a great advisor. And I wanted to come and say thank you. The winner of my first year, I started doing a reading class with Tom. And that winter, my mom needed an operation. And I was kind of nervous, and I didn't do my reading. And I came to talk with Tom about it. And he was very down-to-earth and real. He shared some of his experiences. And he made me feel comfortable to talk to him about anything. And in that spring, I joined his group. And I was part of Tom's student reading group. And that was my, oh, by the way, my mom did well. She's well, and she'll probably watch the video with this. So then I joined Tom's reading group, which was my most important learning experience in grad school. And it's kind of delicate to run a student reading group, because we're all learning to explain math. And so the explanations are a little rough sometimes. And the advisor has to kind of help everybody understand and help us learn to explain better. And at the same time, to be kind of careful of everybody's feelings in the group. And he did a really good job. He somehow created a sense that we're all trying to understand the math together. And so I have tried to carry this sense, now that I'm an advisor and I have my reading seminar. So to carry this sense, and also when I teach and when I give talks and things like that. Okay, which maybe is a good moment to mention that this talk is in a little bit different area than most of the talks in the conference. And I hope it will be clear to everybody. And so please stop me any time if you have questions. The notation isn't clear. If the main idea isn't clear, or whatever. So as I talk about the relationship between some geometry and some topology. And the geometry is the Lipschitz constant. So I have to start by reminding everybody, what is the Lipschitz constant of a map? Okay, so suppose I have two Romanian manifolds. M and N are Romanian manifolds. And I have a map between them. Then the Lipschitz constant of this map is the supremum over all of the curves in the domain. So I wrote gamma one, because gamma is a one dimensional curve. Of the length of the image divided by the length of the curve. The Lipschitz constant is like a stretch factor. It measures how much this mapping stretches curves. Here are a couple facts about it. It's also related to how much it stretches other things. So the infinitesimal version of this is that if you look at the length of Df of a tangent vector, that is at most the Lipschitz constant times the length of the tangent vector. And if you look at what f does to surfaces of higher dimensions, then you can say that if you have a k dimensional surface in your manifold, then the k dimensional volume of the image is at most L to the k times the k dimensional volume of the surface. So hopefully that gives some sense of what is a Lipschitz constant. And what the talk is about is how does this geometric property of a map relate to its topology? And there are fancier topological questions, but we're just gonna talk about the degree of the map. How does the Lipschitz constant relate to the degree? Okay, now there's an old sort of fundamental proposition about this. It says that if m and n are closed and orientable, so we can talk about the degree, and f is a map between them, then the norm of the degree of f is at most the volume of the domain over the volume of the range times the Lipschitz constant to the power n. What is n is the dimension of these manifolds. Okay, and I'm gonna sketch for you a proof of this. There are several reasonable proofs, but the one I'm gonna sketch for you uses differential forms, because differential forms are going to be one of our main characters going forward. Okay, so let me say a little bit about differential forms. So suppose I have a k form on a manifold, then I'm gonna talk about a notion of how big it is, which is called its L infinity norm. So the L infinity norm of the k form is the supremum over any point in the manifold and any k tangent vectors. Of what happens if I take my form and I evaluate it on those tangent vectors, and I normalize it by dividing by the norms of the tangent vectors. Or if you like, I could insist that I take unit tangent vectors and I ask how big the form could be. So that's the L infinity norm of a differential form. Okay, and this L infinity norm, it also behaves nicely in terms of mappings and Lipschitz constants. So I'll put that on my list of facts. So if I have any omega, which is a k form on the target, and I pull it back, here's what happens. The L infinity norm of the pullback is bounded by L to the k times the L infinity norm of the form. So this one is on N and this one is on N. Okay, so this isn't hard to prove. I won't do the proof, but if you take this fact about tangent vectors and you plug it into the definition of L infinity, it pops out. And another way to think about it is that k forms are sort of dual to k manifolds, and this fact is dual to this fact. Let's see, there's one more basic fact about differential forms. Suppose I integrate a differential form over a k-chain or a k manifold. The norm of that is at most the volume of the k manifold times the L infinity norm of the form. Questions or comments so far? Yeah, good. Why it's not what? Oh, it is, yeah, good question. Okay, it is also, create, the L infinity norm of the differential of the map. Yeah, thank you. And so this is like the infinitesimal version of this. This is the soup of how much the mapping stretches tangent vectors, and that's the same as the soup of how much it stretches macroscopic curves. Yeah, good, any other questions? Okay, so now we have the tools to prove this fundamental proposition. So here's the proof. I let omega be a normalized volume form on the target. So it's one over the volume of the target times d volume of the target, and so it's normalized, so it's integral is one. And also I can think about its norm, its L infinity norm, and the L infinity norm of the volume form would just be one, but because of the normalization, we divide by the volume of N. So now how does this relate to the degree? The norm of the degree of F is the norm of the integral over the domain of the pullback of omega, okay? So now that's now bounded by the volume of the domain times the L infinity norm of the pullback form. Then this norm is controlled by that. So that's bounded by the volume of the manifold times the Lipschitz constant of the map to the N times the L infinity norm of the original form. And then we plug in what that is. That's one over the volume of the target, and that's the proof. Okay, so the question that I wanna talk with you about today is how accurate, how sharp is this basic upper bound on the degree? And this question was raised by Gromov in the 70s, and he worked out a few interesting examples, and then later he came up with some questions. So for example, what happens if we're thinking about maps from SN to itself? So let's say these are the unit N sphere. How accurate is this bound? So these have the same volume, so the volume factors just cancel, and the proposition says the degree is at most Lipschitz constant to the N. And that's basically, that's accurate up to a constant factor. So for every L at least one, there exists a map F with the following properties. It's Lipschitz constant is at most L, and its degree is at most some dimensional constant times L to the N. Here's what the maps look like. So these were discovered by Gromov in the 70s. So here's the domain, and here's the target, and we're going to try to build this map F. What I do is I find a bunch of disjoint small balls in the domain. So those balls, they are right over here. So we have BJ, our 10 over L balls in the domain. Okay, how many are there? How many 10 over L balls can I fit in here? Well, they have volume, something like one over L to the N, and the total volume is one. So the number of these balls is around L to the N. Okay, now what do I do? What does my mapping do? Well, for each ball, it maps BJ comma boundary of BJ to SN comma base point degree one. So let's draw a base point on this SN. So over there is the base point. And actually, let me put it at the bottom. Pretend it's at the south pole. Okay, so what happens? I take one of these balls. It's small, so it's basically Euclidean. So it's basically a Euclidean ball of radius 10 over L. And now I'm allowed to stretch it by a factor of L. So after I do that, I have basically a Euclidean ball of radius 10. And I take that Euclidean ball of radius 10, and I wrap it over this sphere, and I put the whole boundary at the south pole to make a degree one map. So that's the mapping on each of these balls. And notice that it takes the whole bound, all of the boundaries, they all go to the base point. And that makes it easy to extend the map. So in between these balls, there's this like blue interstitial region. And what F does is it maps the rest of SN colored in blue and it just sends that all to the base point. So it has Lipschitz constant L because each of these had Lipschitz constant L. And this has Lipschitz constant zero. And it has the desired degree because the degree, there's one degree for each ball. The number of balls is around L to the N. This one should be equals. Yeah, this one could be equals. Yeah, you're right, I did. Any other questions or comments? So for spheres, this proposition is sharp up to a constant factor. And you do actually need the constant factor. So Gromov showed that you can't have degree exactly L to the N. And you can't have degree 0.99 L to the N. You need some modest constant. Okay. And then he raised the question, how often does it happen that this proposition is sharp up to a constant factor? So here are some examples. So proposition one is sharp up to a constant factor. Okay, so we just saw it's true for maps from the sphere to itself. It's true for maps from a product of several spheres to itself just by taking the products of the maps we've constructed. It's sharp for maps from CPN to itself. Takes a little more work. Anyway, there are a bunch of examples where it's sharp. But there are many examples that were not so easy to understand. And about 20 years later, Gromov proposed a problem, a particular manifold that he couldn't figure out what the answer was, and he thought it could be interesting. And what it is is a connect sum of two of these things. More generally, I'm gonna have XK be S2 cross S2, connect sum to the K. And this problem maybe is in the Russian tradition of problem solving, where the idea is to give the simplest problem that goes beyond current ideas and points towards maybe a general theory. Okay, so the problem Gromov asked about is what if you take a self map of one of these guys? So let me give it a name. D sub K of L. So that's gonna be the maximum of the degree of F, where F is a self map of X to the K and the Lipschitz constant of F is at most L. Okay, so Proposition 1 says that this is at most L to the fourth. And the question Gromov asked is, is it approximately L to the fourth, like for these other manifolds, or is it actually much smaller than that? And trying just a little bit to build examples, the examples had much smaller degree, maybe like L cubed. Okay, so that question was open for 20 some years and my graduate student, Sasha Bairdnikov, made a cool construction for S2 cross S2, connect sum with itself. He proved, so Bairdnikov, for K equals two, D sub K of L is at least a constant times L to the fourth. And his construction also works when K is three. And he described this to me in my office and then he said, but I think K equals four will be different and he was right and he worked together with Fegermanine and they proved it. So theorem, Bairdnikov and Manin. So for any K which is at least four, D sub K of L is little O of L to the fourth. So remember that little O means that it's less than any small constant times L to the fourth, asymptotically as L goes to infinity. Cool, so there's a phase transition between three and four. Okay, and then we worked together on understanding better, more quantitatively, what this little O is like. And we pinned it down decently. The three of us together proved that D K of L is in some range. So it's less than a big constant L to the fourth over log of L to the half and it is more than a little constant L to the fourth over log of L to the fourth. So it gives the, is that legible from? Okay, great. So it gives the general shape that this thing behaves like L to the fourth divided by some power of log L and we don't really know what the right power is. So my goal for today is to try to give some intuition why there is a transition between three and four to describe something about what these maps are like and something about why these are different. Any questions or comments about the plan? Right, right. So the question is, is there any connection with simplicial volume? There might potentially be some connection but there's not any very direct connection with simplicial volume. Right, thanks for this question. Okay, so the question is basically, what is the metric on X K? So to define the Lipschitz constant I need to have a reminding manifold with a metric. And for all these phases there were canonical metrics which is what I've had in mind. But for this space there is no super canonical metric. So we pick a metric. So G is some metric. Now, this also depends on G. So okay, I'll stick a little G over here. But these results are independent of G up to the constants. So if I stick a little G over here the only thing that changes in this result is that that constant depends on little G. And the only thing that changes in this result is that those constants depend on little G. Oh, is there any hope that the constant doesn't depend on the metric? No, I don't think so. But conversely, it's a nice feature of, if you ask about the exact value of something like this it depends on the metric almost always. But if you ask about the asymptotics up to constant factors then it's independent of the metric. So in some sense it's a question about the topology of this space. And FEDGA has developed this into a fairly systematic theory of how to compute these things for many different manifolds. And it's based on the rational homotopy properties of the manifold, which I don't know super well so I'm gonna focus on this example. So now we're ready to look at this picture that I pre-drew. This is a picture of the manifold S2 cross S2 connects some with itself. So on the left here I have a copy of S2 cross S2. This is like the connect some part and here's a second copy of S2 cross S2. So this picture corresponds to K equals two but you can imagine bigger case. Okay, and I drew some stuff on it. The main characters in the story are going to be the differential forms on these manifolds. And so I drew something that hopefully helps suggest kind of standard basis of differential forms. So what is for example A1? So the AIs and BIs are two forms. It is Poincare dual to an S2. So actually each of these forms is Poincare dual to an S2. So A1 is this, so okay. So this here is an S2 in my S2 cross S2. And then I tried to draw like a blue neighborhood of it. And A1 is a differential form that lives on this blue neighborhood and it is Poincare dual to this S2. And so on for all the other letters. Okay, so I'm gonna go over here I think and I'm gonna explain our plan. Okay, so I'll start with an observation. If I take the integral on XK of A1 wedge B1, what do I get? Well this was Poincare dual to an S2. That was Poincare dual to another S2. So the wedge product is Poincare dual to their intersection which is a point. So this integral is one. And therefore if I have a map from anybody to XK, then the degree of the map is the integral over the domain of the pullback of A1 wedge the pullback of B1. So this is the, we're gonna try to study that thing. Okay, now our main lemma is that if I just have a map from a ball, that this is a standard Euclidean ball to XK, and if K is at least four, then when I look at the integral over the ball of the pullback of A1 wedge the pullback of B1, then that is smaller than L to the fourth over log of L to the half. And this implies the upper bound down there pretty easily. So if I have actually any manifold mapping to XK, I cover it with charts which are balls. And if I restrict the map to each of these balls, then I can apply this lemma to bound this thing. And then I put together the balls and I bound that thing which is the degree. So everything can take place on a single chart which is a ball and this is the main thing we have to understand. And on the other hand, if K is two or three, this inequality is false. This integral could be as big as L to the fourth, constant times L to the fourth if K is two or three. So here's our plan to try to understand this. So I'm gonna suppose that the integral pullback of A1 wedge pullback of B1 is around L to the fourth. And if K is two or three, that can happen. And we're gonna try to sort of visualize how it happens. In particular, the thing we're gonna try to visualize are the pullback forms. We're gonna try to visualize the pullback of A1 and so on. Okay, and if K is at least four, then in our visualization, we're gonna see something that starts to look like it can't happen. So we're gonna try to move towards getting a contradiction. And that would sort of, so if we really got a contradiction here, this would be a proof by contradiction that says this cannot be as big as L to the fourth. We won't give a full proof, but I'm gonna try to give a sense of what happens there, of what changes when K is four. All right, so let's begin with the case K equals one. So here is my four ball. And inside of there, I'm gonna try to visualize the pullback of A1 and the pullback of B1. And it might look as follows. The pullback of A1 might look like some parallel two planes. And then the pullback of B1 might look like some parallel two planes going in a transverse direction. Okay, so let me do some writing to go with this picture. The pullback of A1 is supported on the horizontal slabs. So it lives here. And what are they? They're thick planes and they go in the x1, x2 direction. And now, I don't wanna write anything else about that. Okay, and I guess one other thing we can say about them is we know how big they are. So the L infinity norm of this should be around L squared, which we thought about earlier. Okay, and then the pullback of B1, everything is the same, except it's on the neighborhood of these vertical planes. And let me also put a little scaling scale in this picture. So that distance there is one over L. And the thickness of one of these guys is also approximately one over L. So that's maybe one over 10. Is all the description of K equals one. So this two plane, which has thickness one over 10 L, it got stretched out. So it's thickness got increased by a factor of L to be as thick as this thing. And then each bit like that goes here and wraps around that two sphere. And so if you go down this two plane, it wraps around the two sphere many times. So that's what the map looks like. And then, all right, let me pause there. How is your image coming of these pullback forms? So then let's think about the integral of pullback of A1 wedge pullback of B1. So A1 is a two form. Yeah, right, that's right. The picture is confusing. So this is a two sphere and A1 is a two form. Okay, so what happens when we take the integral of the pullback of A1 wedge, the pullback of B1? Well, the pullback of A1 wedge, the pullback of B1 lives on these intersections here and here and here. And there are L to the fourth of them. You can count by thinking about it. This is like a four dimensional grid with spacing one over L. So there are L to the fourth of these intersections. And each one of them is just a pullback of this. And so when we integrate that one, it's like integrating that one, which is one. Okay, so this is around L to the fourth. So, okay, so we did the case K equals one. The case K equals one isn't a hard case. It's just S2 cross S2 map to itself. It's just the product of two of the mappings that we constructed earlier. Let's go to K equals two. K equals two, something tricky happens. So now we have to think about the pullback of all four of these forms. So maybe the pullback of A1 and B1 that might be as above as before. But then what do we do with the pullback of A2 and B2? Okay, here's a first attempt. Here's my B4. I'll draw it a little stretched out. And the pullback of A1 and B1 look just like before. And I think to myself, well, that worked well. So maybe I should make the pullback of A2 and B2 look similar and maybe I'll stick them over here. Okay, so this is not possible. And let me tell you what the problem is. So this attempt does not work. Here's what the issue is. A1 wedge B1 is the same cohomology class as A2 wedge B2. So these are equal in H2 Dram. And that means that there exists three forms, C, where DC is A1 wedge B1 minus A2 wedge B2. Let me draw C in the picture here. So geometrically, C lives on a neighborhood of a curve that goes from A1 wedge B1, which lives on this ball, to A2 wedge B2 that lives on that ball. Okay, this picture is now really confusing from the point of view of the dimension of everything that came up before. So that's a two-sphere and that's a curve. So let me try to label it a little bit. So C is a three-form on XK. And C is supported on, lives on, a three-disc cross a curve. And the curve goes from here to here. So if our pullbacks of A1 and B1 and A2 and B2 look like this, what would the pullback of C look like? Well, there would be one strand of it connecting each disc of A1, pullback A1 wedge, pullback B1 to each disc of pullback A2 wedge, pullback B2, have all of these strands. And the problem is that if we think about the geometry of these strands, there is not enough space to put them all in the four ball. So geometrically, the purple strands, thickness, radius, one over L, they have some length that I'll call A, which we don't know what it is, but in the picture, in attempt one, A would be one. And the number of purple strands is L to the fourth. Okay, let's digest each of those and see if they make sense. So why do they have radius one over L? Well, if I take a purple strand over here and I map it forward by F, it needs to cover this purple tube over here. And the thickness of the purple tube over here is some fixed number. It may not be huge, but maybe a 10th or a 100th. And this purple strand can only stretch by a factor of L. So its thickness needs to be at least around one over L so that it can cover the strand in the target. Okay, so that's why this is true. Its length is some number called A. That's just the definition of A, so we don't have to check it. And the number of strands is L to the fourth. That's because the degree of our map is L to the fourth. So there are L to the fourth, there are L to the fourth pre-images of this guy and there are L to the fourth pre-images of that guy. We have to connect them all. So we need L to the fourth strands. Okay, so what's the total volume of all that? The total volume is L to the fourth, one over L cubed times A, which is L times A. And that should be less than one because it fits in the unit ball. So the conclusion is that A is actually smaller than one over L. So these strands that connect a blue vertex to a red vertex, they can't have length one. They need to have a much smaller length scale, one over L, like that length scale. So let's try again, now that we know where A2 wedge B2, now that we know that we need to have these short purple strands. So attempt two of our four ball. We have pullback of A1 and pullback of B1, just like before. And we know that we should put the vertices, pullback of A2 wedge B2 close by so that they're close enough that we can connect them. So what are the labels in this picture? So the blue stuff is the same as before. This red dot is the pullback of A2 wedge, the pullback of B2, and this purple thing is the pullback of C. Okay, but now we have a new issue, which is, I know I've picked what that should look like, but what does the pullback of A2 look like? And here's what's tricky about it. The support of the pullback of A2 is disjoint, the support of the pullback of A1 or B1, just because the support of A2 itself is disjoint from the support of A1 or B1. Okay, so I would sort of like to make the pullback of A2 to be some kind of two-plane like I did before, but that two-plane is not allowed to hit the blue two-plane, so that hopefully it looks sort of like this, and we'll have to take out our four-dimensional visualization skills. Think if this makes sense. I would like to have the pullback of A2 and the pullback of B2 be some parallel planes like before, but I want this plane there to somehow be going behind those planes there and not hit them. So here's how to do that. At this point, so we're gonna try to do some visualization in 4D, so in that sense, I'm in a good place. But at this point, I think it's best to describe in coordinates by algebra what all these things are. So let's remind ourselves. The pullback of A1 is supported on some thick, thickened two-planes in the x1, x2 direction, and the pullback of B1 similarly supported on some thickened two-planes in the x3, x4 direction. Now what are we gonna do with the pullback of A2? It's gonna be supported on some thickened two-planes in a different direction, say the x1, x3 direction, and then the pullback of B2 will be supported on some thickened two-planes in the x2, x4 direction. So what do I wanna know? I wanna know that this two-plane doesn't intersect either one of those. So let me explain why this two-plane doesn't intersect that two-plane. So I brought some two-planes with me, two-plane, two-plane. Okay, so one of them is going in the x1, x2 direction, so that's this two-plane, and one of them is going in the x1, x3 direction, that's that two-plane. I brought a stiffer one, and they appear to be intersecting along a line. But what I have right now, and what I'm holding up in front of you, are some two-planes in R3, and these two-planes live in R4. So if I want to get them to not intersect, all I have to do is take this two-plane in my right hand and move it a little bit in the fourth dimension, and then it will become disjoint from the two-plane in my left hand. So as long as I may put these sort of in general position, these ones won't intersect that one, and for the same reason these ones won't intersect that one, and for the same reason these ones won't intersect these two. But those two will intersect each other transversely, just like we'd like them to. So that's the trick of how to arrange all these things to build the map from a ball to X2 that has a really big pullback of this differential form. And secretly we have almost constructed the mapping. If you look at this picture, like various parts of this picture, I've told you should be mapped to various parts of X2. So I know who's mapped here, I know who's mapped there, I know who's mapped to C, and so on. And to finish the job, I just have to fill in the map to send the in-between stuff here, which admittedly looks like a mess, to the in-between stuff over here. And the in-between stuff over here is a four ball. It's not very topologically, it's trivial to map something to a four ball, and geometrically it's not that hard either. So that's how we do it for K equals two. Can think for a moment, but I don't have too much time. So this is for K equals two. Then if I had to go further and do it for K equals three, what would I do? Now I need to have a pullback of A3 and a pullback of B3. Okay, so I need another plane. And there's something I haven't done yet. I haven't yet done the X1, X4 plane. And this one is the X2, X3 plane. So that's how we map to X3. Okay, and then if I try to map to X4, I have a problem. So first of all, I have run out of coordinate two planes. There are not any more of them. And then I say, well, maybe I'll try a plane which is not a coordinate two plane. But if I take a plane that's not a coordinate two plane, some funny angle, it will intersect some of these two planes. So it will hit the things that I've already put there a lot. And even if I'm allowed to curve it or something, it will hit the things that I've already put there a lot. And that was the intuition that Sasha explained to me in my office like three, four years ago. Why he thought that the case K equals four should be different from all the other cases. Okay, now that's not a proof yet. The main issue is, you know, in this special example, I assume that the pullback of A1 was supported on a bunch of parallel planes. And that doesn't have to be the case. So we need to make this argument more general. And just briefly say, there's some analysis involved in doing this, especially in getting this quantitative bound, is based on doing Fourier analysis on these differential forms. We break them into low-frequency parts that are simple and those low-frequency parts look sort of like this. And we can tell a version of this story that why K equals four is different. And high-frequency parts, and Fourier analysis lets us prove some bounds that the high-frequency parts don't contribute very much to the degree. But that'll be a story for another day. Okay, thank you, Tom, and thank you all for having me. Questions? Yeah, that's correct. So the function that we defined, if you change the metric, it only affects this function, dK of L by a constant factor. Yes. That will also only change the constant by a constant factor. And that's because our question about degree is a homotopy question. So if two spaces are homotopy equivalent and the maps in the homotopy equivalents have some Lipschitz constant, then all the homotopy properties will agree up to a constant factor. If we were to ask about the Lipschitz constant of some diffeomorphisms or something like that, then maybe we could get maybe the smooth structure. Except I haven't come up with what is an interesting question to ask. I would involve that. Right, the question is have I thought about a more interesting format of fold, like a K3 surface? I personally haven't. Based on this, FEDJA has been trying to develop like a systematic classification. You're given any manifold of any dimension and you try to compute these things. Which is going pretty well, but I don't know if it's like completely and totally covers all the cases. And topologically what it involves is the rational homotopy theory, the like minimal, Sullivan minimal model about which I wish I knew more. Yeah, everything we said adapts to a connected sum in higher dimensions. Right, that's right. So the number three is special to S2 cross S2. There's a different number, which I think we've even computed what that transition number is in each dimension. For S n cross S n, it probably, it is that dimension, yeah. So Tom, building on David's comments at the beginning of his talk, Tom just answered Francesco's question. Sure, the answer is if you look at S n cross, if you have maps from S n cross n connect some K times to itself. The transitional K is you take the dimension of lambda n of R2n. And you divide it by two. Any more questions? Yeah, you get the same, you get the same kinds of answers. Yeah, yeah. So I think if we look at connect sum of S k cross S n minus K some number of times that we can say where the transition is and you could connect some with different with various Ks, it's a little bit messier. But in, I mean, I think in principle, we could reduce the question to some purely topological question. Any more questions? Okay, if not, let's thank Larry again.