Exploring Exotic 4-Manifolds and Topology

It's my pleasure to introduce the second speaker for today, Lisa Piccirillo from MIT and UT Austin. Lisa Piccirillo got her PhD from UT Austin after an undergrad at Boston College, and after that she has continued to alternate between Boston and Texas. She is most famous for solving the Conway knot problem and she's been awarded the Miriam Merzakany New Frontiers Prize, the Clay Research Fellowship, Sloan Fellowship and other awards. So it's my pleasure to introduce her and her talk on Exotic Phenomena in Dimension 4. Thank you. Hi, yeah thanks so much for the introduction and for coming. I was very pleased to be invited to speak here. I was also a little bit surprised because every time I've spoken at Harvard before, I've really just given an awful talk. So hopefully today we're going to break the curse. But also you've been warned. So, okay, so I'm a low-dimensional topologist. So I'm primarily interested in studying manifolds. And if you want to study manifolds, maybe a particularly obtuse goal that you might have is to classify. Okay, but if you're going to do that, you probably first need to specify a little bit more carefully what you mean by manifolds. So you probably should declare what dimension you're going to be working in, but you also probably should declare what kind of manifold you want to study. Historically, there's kind of three types of manifolds that people are interested in. The one you're thinking of, which is just a... a nice topological space, local atlas to Rn. That's what we're going to call a topological manifold today. So that's kind of no extras. But we also study smooth manifolds, where we're going to just require that, in addition, the transition functions are C infinity. And also, historically, people are interested in something called a PL manifold, but I'm not going to tell you what that is today. Reason being is that in dimension 4 it's the same as smooth so it won't really come up. Okay, let me also just make a couple of convenience assumptions today. All of my manifolds are going to be oriented and unless I explicitly say otherwise, compact and without boundary. So these are the kind of manifolds, maybe, that I'll talk about what we know about the classification of today. I'm going to be primarily talking about dimension 4, but let me just very quickly tell you what we know in other dimensions. So in the sort of classical low dimensions, dimensions 1, 2, and 3, maybe, I don't know, check. We have classification theorems. In dimension 3, that was not easy, but that's a story you've heard before and not the one I'm going to tell today. Let me also just tell you that there's a theorem in dimension 3 of Moise that there's no difference between smooth and topological manifolds in low dimensions. Okay. The classical high dimensions, so greater than or equal to 5, in high dimensions, there's something called the surgery program or surgery theory. And this is a theory that was developed in the 60s. by a whole bunch of folks, so let me call out Browder, Novikov, Sullivan, and Wall in particular. And in a slogan, what you can think is that surgery theory is going to reduce the classification of manifolds to problems in algebraic topology. I want to comment here that I think there is a misconception, I had this misconception for a long time, that because surgery theory exists, the classification of high dimensional manifolds is kind of done. That's not true. It's kind of not my problem anymore, but it's not known in general. So there are things that we know, for example, simply connected smooth 5 manifolds are classified, but there's lots of things we don't know. For example, the smooth Poincare conjecture is open in dimension 126. Okay. And in high dimensions, as perhaps you know, we can have a difference between the smooth and topological categories. That's originally work of Milner. Okay, so that brings us to dimension four. Let me start off by talking about the topological category, where a groundbreaking work of Friedman from 82 tells us, one way to phrase it, is that surgery theory works topologically. It works when the fundamental group is what he calls good. So sometimes this machinery will let you get classifications in dimension 4 for topological manifolds. And he carries this out. For example, he gives a classification of simply connected topological 4-manifolds. So the set of x4 top with no pi 1. Friedman tells us that this is in bijection with some other set of algebraic doodads that are more well understood. So it's in bijection with this set of pairs. There's a form from Zn to Z, which is whatever unimodular, bilinear, symmetric. And there's an integer in Z mod. So, right, the theorem. And I wanted to kind of state this to see what classifications in high dimensions have a tendency to look like. There's usually a big algebraic set that you maybe do or don't think is any nicer than this, but anyway, there's a big algebraic set, and you say that all of these correspond to a unique manifold. So let me interject here with just a couple definitions for people who don't do topology or geometry all the time. This pi 1 thing I keep referencing, this is the fundamental group. It's not important that you know what that is today. Just know that it's a group that you can associate to a manifold. It's kind of a primary invariant for them. And this thing right here, I'll mention a lot as well. This is called the intersection form. Again, it doesn't matter that you know what that is, but it's a form which is an invariant of a manifold. And Friedman tells us that for simply connected topological manifolds, it's like the invariant, basically. If you know what this means, the intersection form is the cup product on H upper 2. Okay, any questions so far? This is kind of a board on everything I'm not going to talk about. Let me also say that I really like to get a lot of questions. So what I really want to try to tell you about today is what we know about smooth 4-manifolds, which is nothing. We have no classification theorems. We have no conjectured classification theorems. So far the field is, in a sense, really still in its infancy. One of the primary things we study is just the extent to which whatever is true topologically fails to hold smoothly. Let me make a definition to make that a little more precise. We're going to say that smooth 4-manifold x is exotic. if there exists another smooth four-manifold, x prime, which is homeomorphic but not diphyomorphic to x. In this language, the smooth four-dimensional Poincaré conjecture says that S4 is not exotic. But in fact, we know from Donaldson's work in the early 80s that there do exist exotic four-manifolds. And these two statements together serve to motivate the question that will sort of guide the talk today, which is which smooth four manifolds are exotic? And since we're maybe particularly interested in one day eventually understanding the Poincare conjecture, maybe we're particularly interested in understanding which small four manifolds are exotic. Where small will maybe be taken to mean today that the fundamental group is trivial. Donaldson's examples have that. And maybe I also am going to ask that B2 be fairly small. So this is as an integer, and I want it to be small. So b2 is the final invariant that I'm going to sort of impose upon you today. This is the second Betty number. It doesn't, again, matter that you know what it is. It's an integer. It's an invariant of a manifold. It's really crudely measuring how complicated it is. b2 of the four-sphere is zero. So this is the question that I'm going to use to guide the talk today. The plan... is to tell you classically what are the techniques and what do we know. Then I'm supposed to, I guess, tell you about current developments. So what do we know and how do we know it? But then what I really want to talk about, and what I'll spend a lot of the second talk on, is starting to try to say something about the structure of where exotica comes from, and maybe a little bit about quantifying it. So asking how you could measure how different a pair of smooth structures are. Questions? Okay, so before I move on, let me just first give a little bit of a disclaimer. This is a question that gets a lot of airtime. I'm going to give it more airtime today. But it's really not the only question in smooth four-manifold topology. There's a lot of things we want to know about four-manifolds, and we're also interested in things other than the manifolds themselves. For example, sub-manifolds, particularly in co-dimension 2, that's kind of higher knot theory. We're also interested in diffeomorphisms and manifolds. And for these two types of objects, submanifolds and diffeomorphisms, the story I'm going to tell today, by and large, has a parallel story. And in fact, both of those stories are very active right now as well. And maybe the final comment is that from now on, unless I explicitly say otherwise, all manifolds are smooth and simply connected. Okay, so classically, how do you build Exotica? So this is really kind of a three-step process. The first step is to give yourself some candidates. So build a pair of smooth... smooth four manifolds x and x prime, which you think reasonably might be homeomorphic, but not diffeomorphic. Step two, show they're homeomorphic. This is really riveting stuff. Step three, you know where this is going. Okay. But I wanted to point this out because I'm going to do it a lot. It'll give me a little structure. And I also wanted to sort of point out that this is basically under control by Friedman. So really what you need to do is you build yourself some manifolds and you just get the intersection form right and then Friedman says you're good to go here and now you just need to distinguish them. Okay so then the techniques I need to tell you about are really how do you build four manifolds and how do you tell them apart. So let me give a list of some ways that classically we get four manifolds to study. And it's also good to have these kind of, I don't know, in mind, have some examples to be thinking about as we go through the talk. Any questions before I do? I'm going to keep stopping and asking for questions. Okay, well, eventually. So the first sort of place we get for manifolds is some sort of very basic small examples. For example, there's the four sphere. That's great. There's complex projective space. There's its orientation reverse. And you can take connected sums of these things. That maybe seems very basic, but in fact, sums of CP2 and CP2 bars play a huge role in the exotica literature. Okay, you can also try to build yourself some four manifolds out of lower dimensional manifolds by taking products or bundles. For example, S2 cross S2, or more generally, products of surfaces, are very interesting examples, or maybe you want to take a 3-manifold cross S1. Another way you can get some interesting formantless folds to study is you can ask a geometer to give you some. And in particular, complex surfaces and symplectic manifolds have played a large role. And those are actually kind of, for the most part, those are our root examples. I'll say a little bit more about mangling them, but this is kind of the natural sources for the most part. studied classically. So we also like to take them and kind of Frankenstein them together using cut and paste operations, which is what it sounds like. You have some manifold x, which contains some codimension 0 submanifold z, and you're going to cut it out and replace it with something else. Okay, and then the final construction of manifolds, which I just want to sort of mention here. This is a way most people work with manifolds on a day-to-day basis, but it actually doesn't really play a role in the classical exotica literature. But you can build manifolds out of these really basic building blocks called handles. And I'll say what those are in the second talk. So that's how we build manifolds. We also need to be able to tell them apart. And classically, if you want to distinguish manifolds, you have one option. It's gauge theory. So I'm going to say very little about this today. There are like four or more world-class gauge theorists in this room. None of them is me. Very crudely, this is an invariant that counts solutions to a PDE on the manifold. It's a very powerful invariant theory. It's not an exaggeration to say that everything we know about exotica, we know thanks to gauge theory. So this is how we're going to tell our manifolds apart, for the most part, at least classically. But I just wanted to give you a couple of warnings about gauge theory. The first is that, in general, it's not possible to compute it explicitly. And, you know, if you want to build some manifolds and then distinguish them, you need to be able to actually literally make computations and then say, well, they don't match. So, well, people are able to sort of get around this, but it's largely by using formal properties. You can sometimes get your hands on things if you have some geometric structure, and we have some understanding of how these invariants behave under gluing. Okay, so that's the first warning. The second warning is that gauge theory doesn't work when you have no B2. The invariant just isn't defined. invariants. So in particular, while this theory has taught us a lot and there's still a lot it can do for us, you're not going to disprove the Poincare conjecture using a gauge theoretic invariant, at least as we know today. Yeah, good. If you're in the ask a geometer category, does that influence it? The methods are the same if you're working with a manifold that has a little bit more structure. In general, it's just that the structure is going to help you maybe get your hands on this. Sometimes we can say more about classifying manifolds that have more structure among those manifolds which have more structure, but that's a different question. More questions? Okay, so that's what I wanted to say about classical techniques. So now I'm meant to tell you what we know. Which is quite a lot. This board is deliberately too small to read. There's a lot of exotic illiterature. We would be here for a long time if I tried to kind of state everything carefully and give you all the names. So let me instead try to give you, like, tell you what the patterns are on this board. So how's the board working? There'll be kind of three eras I'll discuss. In each era, I'll tell you how small we can get. I'm assuming manifolds are simply connected, unless otherwise. I'll tell you how many smooth structures we can build, who did it. And then these are the two sort of... important columns, I'll try to say something about what the big major developments are, both in constructions and in obstructions. So, so the first era in, in development of, of exotica is, is the Donaldson era in the 80s. It's kicked off with Donaldson's development of, of Yang-Mills gauge theory. The examples in this time, in this era are, for the most part complex, so we start to see non-complex examples computed using gluing formula and cut and paste operations towards the end. That's kind of era one. Era two is kicked off by the development of the Seiberg-Witten equations in 1995. And there's an incredible amount of work then. We don't actually see any increase or any decreasing in the size of the manifolds we can get during this era. But I sort of want to emphasize that this chart is problematic in that that question is not the only question you want to know about form manifolds. So I'm only presenting sort of smallness here. There isn't smallness developments. But in this, there's a ton of new results about smooth form manifolds. For example, questions about minimal genus surfaces representing homology classes and a lot of other things. So this maybe makes. makes it look, well, this is an important theory. What is the second column? Oh, sorry. This one? No, no, I'm sorry. The first column. The first one. Yeah, great. So this is B2. So this is the size of the manifold. So Donaldson's first invariance of B2 equals 10. In the 80s, we got 9, and that's about it. In this second Seiberg-Witten era, no development, nothing smaller. Good. Yes, other questions? Other things you want to be said out loud because you can't see them? Yeah, that's a great question. Yeah, it's, I think it's because, but one of the reasons is because of a lack of small examples. When you look at this list and you say you're only interested in simply connected, so you probably don't want products of surfaces or things like that, well, you immediately end up here, and these tend to have a lot of B2. And so if you want to have less B2, Well, you have to kind of get your scissors out and start trying to put them together in ways where, like, one of them eats some B2 out of the other one, and you end up with very messy objects. Yeah, thanks. More questions? Yeah? Yeah? Oh yeah, sorry Tom. That's Tom. But nobody can see that anyway. Thanks. Okay, so this is the second era, the cyber-grinton era in the mid-90s. Big development in techniques, both constructively and obstructively, but actually no movement on the smallness of the exotica we can produce. More questions about that era? Anything? Okay, and then the third era is kicked off in 2005 by a result of General Park, who uses rational blowdown, a technique that was developed in the cyber-Witten era, to get the first improvement in the size of exotica in a long time. So he gives us exotic simply connected manifolds with b2 equals 8. And that, like, kicked off an arms race. Everybody works on this. There's a lot of results by these people and probably others in all sorts of permutations that I'm not going to write down. We start getting smaller and smaller b2, but nobody cares about 5, but that's fine. Until the end of the era is this result of Achmedev and Park. So they give simply connected B2 equals three exotic manifolds. And that's the smallest exotica we have for a long, long time. There are a lot of candidates. If you... Just kind of forget about ever being able to compute a gauge theoretic invariant for it. You got lots of options. More questions? It doesn't. We just, by and large, don't have a more combinatorial invariant, but I'm about to say something about that. Yeah, more questions. Okay, so let me try to tell you about a couple of more recent results. I'm going to use the same table, but I'm going to write a lot bigger. Okay, so... There's three kind of camps of results recently that I want to say something about. The first is some work from 2021, which really doesn't make things smaller. It's like 23, the biggest thing on the table yet. No development in techniques, the development is in the obstruction. So in 2021, we get the first example of a pair of exotic manifolds which are distinguished with something called the Slice-Nuts Approach. which is this method of distinguishing manifolds, which is a little more convoluted than just saying, like, compute invariance for both. Hey, look, they don't match. And this approach, I'll tell you what it is in just a minute, but let me sort of tell you now that the thing that's exciting about it is that versions of this approach could possibly work with no v2. So it was good to know that this approach can work ever, since we're maybe hopeful that it could work in cool settings later. And that's joint work with Chakrian Manalescu and Marco Marangon. The second set of results that I want to tell you about is from last year. Uh... And again, the examples in these results are not particularly small. There's not really movement here. What's exciting about this result is the development in the techniques. So classically, we build exotic manifolds by starting with some big geometric stuff and making a hot mess. But we wanted to be a lot more juvenile about our construction. So we're really just going to start with handles, with these basic building blocks. We're going to make a stack of them. and that's how we're going to get our manifolds, so explicit Handel constructions. And the reason that we can get away with this is because we're prepared to make explicit computations of something called the Haggard-Floer mixed invariant. So you should not really think that we are getting away from gauge theory here. Conjecturally, the Hager-Fleur mixed invariant is equivalent to the Seiberg-Witten invariance. So this probably is a gauge theoretic proof. But we're accessing these invariants via a pretty different method that allows us to do computations much more explicitly. This is joint work with Adam Levine and Ty Ledman. Because our constructions are much more explicit, we can monkey around and make other stuff happen. So we're also able to produce exotic 4-manifolds with b2 equals 4, so not the smallest, but pretty small, at the expense of having a little bit of pi1, z mod 2. But in fact, what's interesting about these is that their intersection form is definite. Okay, so that maybe doesn't mean too much to you. An intersection form, as you know, can be definite or indefinite, and let me sort of assert that all the exotica previously was indefinite. And, you know, there's other things we ask about manifolds, whether then is there small exotica, and for some of those other questions, you see a difference between the behavior of definite and indefinite manifolds. So it was maybe reasonable to wonder whether definite manifolds were more rigid. We're saying here that they're not. Okay, and then the most exciting result I'm going to tell you about today, which I wish I could tell you was due to me, but it's not, is work from last year from Stipschitz and Sabo. And they're going to give you exotic manifolds with B2 equals 1, pi 1, z mod 2, and again definite. And this I think you should think of as the first movement in actually getting smaller exotica in a long time. Their techniques are going to be a mix from the good old days. Oh, I lost the green. And from us. And they can compute the Seiberg-Ritten invariance. And then the final result that I kind of want to mention, in terms of progress in exotica, is very, very recent, from earlier this year. This is a result about exotic manifolds with boundary, which in a sense doesn't even belong on this table. The with boundary setting is very different. It doesn't make sense to compare. I'll talk about that a bit more later on. This is work of Kiu Ren, who is a graduate student, and Mike Willis. And what's so exciting about their work is that their obstruction is combinatorial. So they use something called a skein lasagna module, which comes from Khovanov homology lasagna. And this is the first time that compact exotic manifolds have been produced without gauge theory. Questions? Yeah. No. More questions? You can put a description of a manifold into my collaborators and maybe get something here. Yeah? Yep. Yep. You look at the universal cover. If the manifolds have a... non-diffeomorphic universal covers and they're non-diffeomorphic and the universal covers have B2+. Good question. Other questions? Oh, yeah. Right. So, okay, what I'd want at the end of the day is a set of smooth manifolds with pi1 trivial. is in bijection with anything. Just give me a reasonable other set that you can relate them to. That would be awesome. For now, the question that seems to motivate us more, or the question that we can sometimes work on, apparently, is the one where we say, where can I find Exotica? Sort of a geography question. More questions? Yeah? No. It's bad. The situation is bad. I'm going to try to say something a bit deeper in the talk about, you know, maybe if you could try to start saying where do all smooth structures come from, but it'll be not at all ready to see what kind of set. looking for so the plan for the rest of the first half is um to tell you what the slice approach is This is an argument that's due to Kassin in the 70s. And here's how Kassin says you can maybe try to distinguish manifolds. So let's suppose... you have a pair of candidates for exotica. So you have a pair of homeomorphic smooth 4-manifolds. Here they are. That's x, and this is x prime, and they're homeomorphic. All right, so what Kasson would like you to do is remove an open neighborhood of a point from both of them. These manifolds now have boundary. They have an S3 boundary component. From that ball, you removed. They're still homeomorphic. And now here's sort of what Kastner wants you to do. He says, suppose you could find a not K in S3. Such that two things happen. One, there is a D2 embedded smoothly in here, in XCirc, such that the boundary of the disk is this knot. So what you should have in mind is that your knot, it lives in S3, so you could think of it as being in this boundary component right here. That's K. And we're asking that there's a smooth disk that it bounds somewhere in here. And then also, you should want that there does not exist such a disk in x prime where the boundary is k. If you have both of these things, then that implies that the manifolds are not diffeomorphic. It's easy to see that the punctured manifold can't be diffeomorphic. If you had a diffeomorphism, you push this disc across it, and the knot would bound over here, and it extends to the closed things. So that's the argument. It's, you know, it's not quite as direct as, like, compute the things. They don't match, but it's also not too bad. So why should, or any questions about the approach or anything else? Thank you. More questions, more mistakes on the board. Okay. So why should you want to do something like this? Well, for one, this argument can be used to show that R4 is exotic. So this is the non-compact analog of the Poincaré conjecture. And it's false. So this argument has teeth sometimes in a slightly different setting. Okay, so that sounds kind of good. Then we'd like to know whether you can use this argument to show that S4 is exotic. And, well, we don't know, but there is an invariant, or there's a few now, actually. There exist invariants in the literature which could... due to, they could provide the obstruction here, even when x prime is S4. So unlike using gauge theory, we could conceivably disprove the Poincare conjecture like this using tools we have today. So all of that sounds kind of good. But here's what's not so good. It's not easy. It's not at all clear how to sort of set this up, how to find a good knot, okay? So if you have even two, even a known exotic pair, it's not clear how to find a knot which bounds a smooth disc in one of them but not the other. And that's what we did here. Now it was the first time that this argument had sort of been run in a compact setting. Okay, question? So, okay, this is an argument. It works sometimes. It could maybe disprove the Poincare conjecture. Maybe that's something we should be trying to do. So in follow-up work with Chibri and Manolescu, well, we just sort of tried to think about how you would try to get this running. So we gave a very systematic method for producing... four manifolds X, which are homeomorphic to S4, and which come with a K as in the first step. So it's this really pretty straightforward construction. It gives you candidates for counterexamples of the Poincaré conjecture, and those candidates come equipped with a knot, which we don't know that it doesn't bound a disc in B4, but it doesn't, you know, seem to have a good reason to. And the method is, is really systematic. I personally made a computer spit 3K examples, and I personally am terrible at coding. So we produced a whole bunch of examples, and some of them were kind of interesting. Interesting meaning we, we, you know, we have, we get some knots and there's something like if the knots have blah property, then the Poincare conjecture is false. And we just sort of don't know. We fail to show that the knots don't have that property. Um, I don't think this was ever meant to be like, aha, these ones are going to disprove the Poincare conjecture. It's more meant to, to mean somehow the method isn't kind of trivial out of the gates. Like it does spit some kind of subtle looking stuff. Um, okay. Since then, um. It's been proven that these examples are actually not particularly interesting by my younger sibling, Kai Nakamura. But what Kai did, it really just ruled out the examples we wrote down. It doesn't kill the method. You can build more. There's a bazillion ways you can build more, but if you want to see some that have been written down, Chhapriyan student Chahakya has done so. And you can start to see how, like, hopefully there's going to be a little bit of a conversation here then. Build examples, rule them out, maybe that tunes the parameters on the construction, maybe eventually you'll show the construction's not going to disprove the Poincaré conjecture, that would be fine. So that's sort of the idea. But of course, given that it's whatever year it is, not only are we having that conversation, ML is having that conversation. So there's work of Gukov, Halverson, Chiprian. and, uh, who's name I really can't spell, sorry, uh, who are using ML to both build, so to run our construction, and study the examples it produces. So, uh, the neural net that kind of evaluates them is out, that came out last year, and this one is in progress. And maybe a comment about these is, um, if the, if the neural net, if, if, if the neural nets win, it will really be a proof. Um, they are spitting explicit constructions, and the way they study examples, it, it does spit something verified. But that's, um, that's what I wanted to say for the first half. Um, questions? Yeah. Oh, well, I don't know because, yeah, they don't, they haven't produced. They claim they're producing. But they're going to be homotopy studios. How big are the knots? I don't know. Big. So, right, good. So there's actually a little bit of a convolution, and this is why I kind of hid what I meant by some of them interesting. There's a flip in the logic. What they're actually going to build is they're going to spit knots where if you could prove the knot did bound a disk, then the Park-Wright construction is false. So what the net actually does is looks for disks. Oh, they definitely are detected by... Oh, yeah, definitely are. It's a standard... Yeah, it's the Maser manifold, which we'll all learn what that is next time. So far, all of the new approaches are really in proof-of-concept stage. No new actual... Phenomena, just like new argument. So, welcome back for Lisa Piccirillo's second talk on exotic phenomena in four dimensions. Hey, thanks for coming back. So the plan for the second talk is what I said I was going to do, but not in the order I said I was going to do it. So I'll spend at least half of the talk talking about where Exotica comes from. I can give you explicit pictures of how you can get exotic manifolds in a pretty high level of generality, and I want to do that because it's very nice. It doesn't have to be sort of mystical. Then I'll talk a little bit about how you might try to... Say how distinct a pair of smooth structures are, and then hopefully I'll run out of time. But if I don't, I'll tell you a little bit about the work with Ty and Adam. Okay, so starting off with kind of where exotica comes from. So let me start off with an observation. Let's suppose you have a pair of, a candidate exotic pair. So you have a pair of smooth form manifolds and they're homeomorphic. The observation is that if x and x prime co-bounded. a 5-manifold W. And so what you should be thinking is that it looks something like this. Like maybe that's x, and here's x prime. And I'm asking that there's some 5-manifold which has the two of them as its boundary components. So if they co-bound a 5-manifold like this, and in fact, that 5-manifold is diffeomorphic to a product. It's actually just diffeomorphic to x cross i. Well, then that's going to imply that my manifolds are themselves diffeomorphic. If this cobordism is really just a product, then I can kind of think about flowing x along the interval direction, and it'll land very nicely on x prime. That'll be a diffeomorphism. So manifolds are diffeomorphic if they co-bound products. And in fact, candidates for exotica do co-bound something that's very close to a product. So this is the theorem of Wall. He tells us that such four manifolds are going to co-bound something called an H-cobordism. So an H-cabordism is not literally a product, but you should think it's a 5-manifold that has the algebraic topology of a product. So you have your candidate exotic pair. You have this cobordism, which algebraically is pretty boring. By the way, more formally, what this means is that the boundary inclusions are homotopy equivalences. But you have your two manifolds. They co-bound this, like, algebraically very nice five manifolds. And if that five manifold were actually just literally a product, then they would be the same. So there's a sense in which that five manifold is capturing how different the manifolds are. Like the failure of that thing to be a product is what's generating. what's generating the difference in these smooth structures. Um, and we can actually make that even kind of more, uh, explicit and a little more four-dimensional. Uh, so, so let me kind of get a picture going. So what's our H cobordism look like? It's like there's x, here's x prime. Uh, we have something running between them, which is maybe like a bit weird. Okay? So, in fact, Wall tells us that not only do we have this nice manifold running between them, but actually most of it's boring. There's a sub-H cobordism, W prime, such that W minus W prime, that part is actually a product. So there's this sub-mat5 manifold here, where all the interesting non-product stuff is happening here. And this is very boring. And in particular, like, the fact that we have a product here tells me kind of that this part of x is actually the same as this part of x prime. The difference is, like, these parts, whatever that means. Okay? And the theorem I actually really kind of want to spend most of the rest, or this section talking about, is something called the Korg theorem, which is going to improve this even more. So the quark theorem was proven by everybody in around about 1997. So let me give you some names but not write them. Curtis, Tseng, Friedman, Stong, Matveyev, Matveyev, Bizaka, maybe we should say Kirby. And Casim proved a non-compact version of it, like, before all of them were born. OK. So, so this kind of was in, came out of, I don't know, anyway, it's due to a lot of people. What it says is that, even better, W prime, this bit that has all the interesting stuff going on, can be taken. such that if you look at how W prime intersects X, so let's call that C. That's right here. And if you look at where W prime intersects X prime, let's call that C prime. That's here. We can take W prime so that these are contractible. So contractible means that their algebraic topology is as simple as possible. We have the alge top of B4. So kind of all of this put together is telling me that if I have two homeomorphic smooth 4-manifolds, then what's different between them is all packed in this piece, and this piece doesn't have any algebraic topology. Let me just write down, to be explicit, a corollary. What we learn here is that if you take X, you remove C, you glue in C prime, what you get is diffeomorphic to X prime, right? We start here, lose this, that's the same as this, glue that back. So these two doohickeys, rats, are called quarks. And this operation of cutting one of them out and gluing in the other is called cork twisting. Any questions about this statement? Or anything? So there's several things I think are, is really, really surprising about this result. One of them is, is that, well, these quarks are pretty interesting objects in their own right. You know that they can't be diffeomorphic because, well, Well, if they were, then all of my four manifolds would be diffeomorphic, which they're not. So let me write that down kind of explicitly. This is a result that was originally proven a little bit earlier in a slightly different form. Let me also quote Achboli-Ruberman for the form I want. What we get from this is that there exists exotic contractible. for manifolds. Now, let me emphasize that contractible manifolds have boundary. So this is not quite the setting I've been talking about kind of all along. But contractible manifolds are very simple. They have essentially no algebraic topology. b2 is 0. Everything's trivial. So this is, very reasonably, the width boundary. analog of the Poincare conjecture, and again, it's false. And another thing I think is really surprising about this theorem is that, you know, over here, we're having this really hard time producing exotica with no boundary when there's very little algebraic topology. But this is telling me that that That issue is technical. Like, I can pack the exoticness between, like, the difference between the two manifolds into something that has no algebraic topology. So somehow it's not the lack of algebraic topology that's giving us the issue here. So I'm going to... give you fairly explicitly examples of exotic contractible manifolds. Any questions before I do? Yeah, they'll be integer homology three spheres. They can be as simple as surgery on a knot. More questions? Before I try to set this up, try to explicitly build some of these, let me just comment that this is the theorem that Rann and Willis reproved. Technically, they kind of reproved that version of it. All right, so to set this up, I have to tell you about handles. And I also have to remember not to write in the shadow. So help me out with that if it doesn't go well. So I'm going to try to do one. We can see this part of the board. I'm going to try for a definition by picture here. So please bother me if it's not clear. So handles are a set of building blocks for building manifolds. And you should really think that they're kind of a... manifold-y version of a cell complex. So I think most of us know how to build a, a cell complex. And, and we're going to, to do the, basically the same thing. But all of our cells, we're just gonna thicken them up so that everything is consistently dimensioned whatever I want. in my case, four. So let me start off with a 3D example of building a manifold out of handles. Let's suppose I want to build a manifold that kind of has one zero cell and one one cell. So, okay, I might start with a zero cell. There it is. It's a point. Okay, but that's, like, not great as a three manifold. So my zero handle will be this three-dimensional thickening of that. Okay. And okay, now I guess I wanted to add a one cell to this. So, you know, what would that look like from a cell complex perspective? It looks something like this. Not a three manifold, but we can get around that by, again, thickening it up in the set case in two more dimensions. That's a one cell. This is a one handle. In dimension four. works the same. Let me try to give something of an example. So let's suppose I want to build a, a, a four manifold out of a zero handle and a two handle, okay? So, so what do I do? I start with, you know, my generalization of zero cells, so there's my zero cell. I need to thicken this up in, in four dimensions, so I, I can't draw you a picture of that, but let me just draw you this picture again and declare that that's B4. B4, there it is. That's my zero handle. Okay. And now I said I want to glue, I want to use a two handle. So that's going to be this kind of thickening of a two cell. So all right, here's my two cell. And how do you glue a 2-cell onto a cell complex? You have to glue its boundary on to whatever you are doing. So, okay, here's the boundary. That's an S1 here. So I need to glue this S, I need, yeah, this S1 needs to get glued on here. The boundary of this B4, well, that's the 3-sphere, so I'm looking for an S1 in the 3-sphere, so I'm looking for a knot. Okay, pick a knot, fine, glue it. And then, well, all right, we need to turn the whole thing into a four-manifold by crossing this with another D2. Technically, you need a framing here. We're not going to worry about that. And I'm only going to build four-manifolds out of two handles today. So that's what you need. Questions about, you know, handles in the kind of broad sense? Okay, let me give you two more quick facts about handles. One is just an even kind of cheaper picture of this. So a schematic or an even more schematic of a four-dimensional two-handle that you'll see me use a bunch of times is you draw your zero handle like this. So this is my zero handle. It's B4 and here's its S3 boundary. And then I'm going to attach my 2-handle to, OK, we have some not in that boundary. My 2-handle is, you know, fundamentally there's kind of the D2. And my 2-handle, you know, looks something like this. So I'll use that representation of a 4-ball with a 2-handle. And then maybe the final thing I wanted to say is that it's the theorem of Morse that all smooth manifolds can be built in this way. Thank you. Questions? Anything wrong on the board? To build a contractible exotic manifold, I need two handles, and I need one more operation, which is something called carving. So building manifolds by carving. So let's suppose I'm given a 4-manifold X, smooth, with a disk, a D2, embedded smoothly in X, such that the boundary of the disk goes into the boundary of the 4-manifold. So you should be picturing something very similar schematic, something like this. Here's x, here's its boundary, and we have, uh, a disk with boundary. Um, okay. So, uh, I wanna build a slightly more interesting four manifold out of this. Um, and, well, not very exciting, I suppose. What we're gonna do is cut it out. So we're going to remove an open neighborhood of that disk. And we'll get something that looks in the schematic like, I don't know, like that. This is x minus y. If I have submanifolds, I can cut them out. Last thing we need is a lemma. We need to check that. We're kind of all on board. Attaching a two-handle and carving out a disk are, in a sense, kind of dual operations. Here we're adding a thickened-up disk to the outside. Here I'm removing a thickened-up disk from the inside. And the lemma relates them sort of even further. So let's suppose we're given a disk embedded smoothly, let's just say, in the four-ball. And, okay, I want the boundary of the disk, again, to land in the boundary of the four-ball. then there's two manifolds that I can build out of this data. I can consider B4 minus the neighborhood of that disk. And I can also consider B4 union a two-handle glued along the boundary of that disk. Right? So this and that. And the statement is not that these manifolds are the same, but that they have the same boundary. And the proof of this is super cute. Here's S4, which I can decompose along an S3 into two 4-balls. Let's suppose that my disk lives in this one. Now I'm just going to repartition S4. where the neighborhood of this disc goes with this. That's before union a two-handle along the boundary of the disc, and the rest is the four-ball minus the neighborhood of that disc, and their boundaries are just identified by setup. I'm going to go. Okay, so now we're ready. But really, let me check in. Questions or anything or unhappiness with the construction before I do? Yeah. Yes, let's say it's smooth. It turns out for three manifolds there's no difference, but it comes to you looking smooth. You don't have to appeal to anything. More questions? All right, so this method I'm going to give for building contractible exotica is, in fact, due to Barry Mazur, the number theorist. It was his undergraduate thesis here. But the development of this owes a lot to Salman. And here's the construction. So first off, if we're going to build exotica, you know, build the thing, show they're homeomorphic, show they're not to be amorphic, for the homeomorphism, we're always going to ask Friedman what he can do for us. And what Friedman can do in this setting is he says that if you have any contractible C and C prime that have the same boundary, are automatically homeomorphic. So if I want to build candidates for exotic contractibles, I'll just build contractibles and make them have the same boundary and that'll work. So I'm really in the market for manifolds with the same boundary and we just learned how to build manifolds with the same boundary, which is good except that these manifolds are super duper not the same or contractible. This one has B2, this one has B1. Okay, not the same. But But Mazur gives us this really beautiful kind of way to use that lemma anyway. So what he says you should do is consider a link L, which has two components, and this link lives in S3. And the link has the following properties. So one, both components are individually unknots. So maybe what you have in mind is something like this. Maybe that's L1, and maybe L2 is like that. We can still see. Getting borderline. OK. And I won't go any lower. It's the size of it. OK. Thanks. I will not go smaller. This is a link. This one's kind of obviously an unknot. If you squint at this one and you can see it, it's also an unknot. OK. And unknots, they have this nice property, which is that they bound disks. They just bound disks in S3. There's one. You can push that disc into the four ball a little bit, so unknots are also going to mount discs in B4. So what we can do is construct a pair of manifolds in the following way. We'll start with B4. Our link, L1, L2, lives in its boundary. And we can build a manifold here by removing a disk for L2 and attaching a 2-handle along L1. Let's call that C. And we can build a C-prime by doing the same thing in the other order. So we'll remove a disk for L1 and attach a disk for L2. One interpretation of the lemma is that the boundary 3 manifold can't tell if you add a disk or remove a disk. So the boundary just can't see the difference between these two setups, so these guys have the same boundary. And as long as you assume that the linking number of L is 1, then they're both contractible. So I'm not now going to go on to prove that contractible manifolds you build like this aren't diffeomorphic. Let me tell you that you can. And in fact, for this literal link that you may or may not be able to see, the contractible manifolds you get are not diffeomorphic. And in fact, this literal link is the one that was used. here, and it's also the one that these guys redetect. So, you know, this kind of... these two manifolds, which are defined very simply from a very straightforward link like this, well, they give you a pair of contractible exotic manifolds. Okay, and the theorem, the thing... The theorem, the quark theorem maybe gets, like, even better. So this is the last kind of leg of the quark theorem. In fact, you can prove that... Any corks can be built using a generalization of this construction. So any exotic pair are related by cutting out and re-gluing a contractible manifold, and you can assume that those manifolds are built using something like this. Uh, no. You might need a bunch of components and then a bunch of components. And maybe you're not using the four ball either. Maybe you're using a fixed other contractible. More questions? Yeah. Yeah, there's some really easy ways to check it. They factor through a little bit of contact or symplectic topology, so they're gauge theoretic somewhere in there. But, like, if you can draw a Legendrian diagram of them where some Thurston-Bennican number is positive, then you win, or something like this. I'm going to talk about quantifying exoticness. Anything before I do? Yeah, so I just erased the theorem that had to... attributions here, and it was one relative one, absolutely. Yes. Yeah, it's not too bad. Let me at least... It's not too bad. So it turns out, this isn't clear, but when you remove a disk for the unknot, what you get is S1 cross D3, so you get this thing's homotopy equivalent to a circle. And so if you want that dead, you better make that, that sort of two cell go around once. And somehow that turns into the linking condition. Okay, so... All exotica comes from cork twisting. Corks can always be taken to come from this construction. So if you want to know how different two smooth structures are, one way you can try to measure that is by saying, how bad does the link that gives the cork have to be? So a couple of definitions of how bad a link like that might be are, well, okay, let's suppose we have a candidate for an exotic pair. So x is homeomorphic to x prime. Then we'll say that. the complexity of the pair is something like the minimum of the amount of geometric linking of L taken over all quarks, which gets you from x to x prime. Uh, and, uh, we're also-we also study a-a version of-of distance, which is called, uh, the stabilization number. That's gonna be the minimum size of the link you need. Again, over, uh, all quarks. Maybe a normalization for both. A comment, perhaps, for people who come from a more high-dimensional perspective, both of these notions of complexity can be defined in terms of the H-cobordism, but I'm defining them kind of discreetly here. So these definitions are set up so that for a homeomorphic pair, if the complexity or... or the stabilization distance, stabilization number, r0. Well, that's the same thing as the manifolds being diffeomorphic. All right, so we certainly know that there are pairs of manifolds where this complexity is positive. But in fact, it's a major open question whether it's ever bigger than one. Do there exist x, x prime, what's above complexity or, and this one especially is kind of a very famous and apparently very hard question. No. Probably of this bigger, there's like a bipartite link. I definitely want to subtract one. Yeah, yeah, there's normalization. Sure. That's right. Yeah, subtract one, divide by two. But, you know, something that's counting that and then appropriately being what it should be. I think you probably want to divide this by two, this by two also. More questions. Okay. So we are apparently pretty far away from being able to address this. But there have been current developments. So for manifolds with boundary, we actually can answer this question. So we start with the complexity. This is the theorem. The heavy lifting of this is really from Morgan and Szabo in 1998. We're going to need something that's due to Akboli-Ruberman. And if you want to see it written down, that was done very recently by Robert de la Du. So the statement is that there exist contractibles, so with boundary, where the complexity of that contractible pair is big. And even more surprising and exciting is a result from Sun Kyung Kang a couple years ago who said that there are such contractibles where the stabilization distance equals 2. And this, I think, yeah, this was a very surprising result. Let me maybe say one more thing that's a little bit outside the purview of the talk before I kind of move on from complexity, which is another very recent result from Junpeng Lin, 2021, who said that for, let me say, there exist exotic diffeomorphisms, so non-manifolds, with stabilization of of the diffeomorphism and the identity, whatever. Let me just say stabilization number two. So I'm not going to make this precise, but I said at the beginning there's a largely parallel story to the one I'm telling for submanifolds and for diffeomorphisms with manifolds, and Zhen Feng solved this problem for diffeomorphisms, and what's really exciting about it is he's working with closed manifolds, so he solved the real one. That's what I'm going to say about distance or quantification. Questions here? You should ask questions. Yeah, if you have a spin C structure... if you have different Seiberg-Witten invariants in spin C structures with a large expected dimension, then you can't have small complexity, which is going to run you smack into the simple type conjecture, but that's a bound. More questions? So, let me tell you something about the techniques that Tai and Adam and I use. To start, let me give you a statement that I'll talk about. So, one of the things we show is that there exist exotic four-manifolds, which are homeomorphic. but not diffeomorphic. There exists four manifolds that are homeomorphic but not diffeomorphic. To S1 cross S3, connect some two copies of CP2 bar. Nope, two copies of CP2 and nine copies of CP2 bar. Okay, so if you're kind of keeping track of stats here, this is b2 is 11, so not particularly small, and pi1 is z. That's the biggest pi1 you've seen all day. So this is maybe not... not a good theorem from the perspective of the table. This is the one I'm gonna tell you about anyway. And the reason I wanna tell you about this one is because I think, well, because the proof is really easy, and it still captures, I think, most of the kind of ideas and developments that we use with other results. To prove this, we develop a new four-manifold invariant. Alpha, it's an integer. It's only defined for four manifolds that have some B3, that have some second homology. Let me just say, for convenience, B3 is 1 for today. And this invariant really is new. It's provably different from Donaldson, Seiberg-Witten, Heger-Flermix, Baffaruta invariants. We can prove some other things with it that I won't write down because I'm not really going to motivate them. Maybe just give me three sentences out loud, and then I'll come back to saying things that I've backed up or defined. So some things we can prove with alpha are that we can build manifolds with non-vanishing alpha invariant that have, that contain square zero embedded homologically essential spheres. That's something usually gauged theoretic invariants vanish on. And we can also distinguish four manifolds that are related. by a functional sort of not surgery on an Alexander polynomial 1, not using this invariant. Okay, end unjustified sentences. So, I'm going to define alpha for you and then show you a little bit about how you can expect to actually literally compute it sometimes. To do that, I need to give you, like, the world's shortest crash course in Heger-Fleur homology. Any questions before we do that? If you filibuster, then you don't have to do Heger-Fleur homology. All right, rats. So what is this? This is an invariant package defined by Osh-Voth-Sabo-rats. I have it backwards? Yes. Oh, okay, great. Thanks. Around, I don't know, 2002, everything I'm going to write down is due to them. For experts, I'm thinking about HFRED, and I'm summing over spin C structures. Okay, so what's the invariant? Out of the box, it's an invariant of a 3-manifold. And it's a, let's just say it's a finitely presented module over Z mod 2. It satisfies a TQFT structure, so cobordisms. Z from a 3-manifold to some other 3-manifold. which, you know, that's something like this. You have your two three-manifolds, and there's some four-manifold running between them. These cobordisms are going to induce maps between HF of Y and HF of Y prime. In this way, you can think that it gives you a four-manifold invariant. This map is an invariant of the cobordism. Okay? And these maps can be written down sometimes. And one really powerful tool for writing them down is the existence of an exact triangle. So there exists a, so let's see, for z, two-handle cobordism. So if this is pretty simple, just start with y cross i, add a two-handle, that's it. Then you have this exact triangle. which has the map you want in it. So it's going to go from hf of y to hf of y prime. That's the map you care about. And then, okay, well, there's something else. Let me call it y sub k. It doesn't matter what it is. I'll just tell you that it's some explicit ancillary 3-manifold. And if you built the cobordism, then you know what this 3-metafold is. And that's it. That's your crash course. So this alpha invariant for a 4-manifold with a little bit of B3 is defined to be the minimum dimension of H of red of y for y generating H3. This is a very common way in general of getting an invariant of a manifold. We're frequently building invariants by saying, I don't know, what's the minimum genus of a surface that represents some class in H2? This is an analog of that. Because you take a minimum, you get invariants for free, but usually there's a cost, and it's that you can't compute your thingy. Let me try to convince you that you can sometimes compute your thingy in this case. State dilemma, let me get a little bit of a picture going. If your manifold has some B3, then it also has B1, so it's got a little bit of a kind of circle in there somewhere. So maybe that's my picture of X. And my generator of H3 maybe looks like that. Okay, so something I can do, this thing's going to have to be non-separating, so I can cut along it, and I get a cobordism from Y to itself. And the lemma is that if that cobordism, if the map associated to it, which goes from HFY to itself, if that's an isomorphism, then Y was minimal. So you can find minimizers if you can find cuts where you get nice ones. So let me just conclude by saying then, how should you build? Four manifolds where alpha of x is whatever you want. Well, we're going to work backwards. You're going to take a 3-manifold y. where the rank of HF of Y is N. It's whatever you want. Three-manifold invariants are reasonably computable, especially if you ask somebody who knows what they're doing. So, okay, you can find something like this. And now we're going to build a cobordism Z from Y to itself. So, something like this. Such that, two things. I want the map f to be an isomorphism, and I want pi to be trivial. Then you define x to be z with the n's glued together. And by setup, you get. That alpha of x is whatever you want it. So you can kind of vary, well, this is a pretty straightforward way of writing down a four manifold that has whatever alpha invariant you want, supposing that you can do this. So let me just say one thing maybe in words. even, about why you can do this. In particular, how are you gonna get your hands on this? These maps are, you know, you can't necessarily always just write down any cobordism map. And, well, you can, you can keep control of this by keeping a good handle on this. So if you build your Z by only using two-handle cobordisms, and you only attach them in ways where this ancillary manifold has no HF, then you just force an isomorphism across the top at every step. So as long as you build the thing and you're pretty careful, then you get this. And I will stop there. Thanks.

I was very pleased to be invited to speak here. I was also a little bit surprised because every time I've spoken at Harvard before, I've really just given an awful talk. So hopefully today we're going to break the curse. But also you've been warned.

So, okay, so I'm a low-dimensional topologist. So I'm primarily interested in studying manifolds. And if you want to study manifolds, maybe a particularly obtuse goal that you might have is to classify. Okay, but if you're going to do that, you probably first need to specify a little bit more carefully what you mean by manifolds.

So you probably should declare what dimension you're going to be working in, but you also probably should declare what kind of manifold you want to study. Historically, there's kind of three types of manifolds that people are interested in. The one you're thinking of, which is just a... a nice topological space, local atlas to Rn.

That's what we're going to call a topological manifold today. So that's kind of no extras. But we also study smooth manifolds, where we're going to just require that, in addition, the transition functions are C infinity. And also, historically, people are interested in something called a PL manifold, but I'm not going to tell you what that is today.

Reason being is that in dimension 4 it's the same as smooth so it won't really come up. Okay, let me also just make a couple of convenience assumptions today. All of my manifolds are going to be oriented and unless I explicitly say otherwise, compact and without boundary. So these are the kind of manifolds, maybe, that I'll talk about what we know about the classification of today. I'm going to be primarily talking about dimension 4, but let me just very quickly tell you what we know in other dimensions.

So in the sort of classical low dimensions, dimensions 1, 2, and 3, maybe, I don't know, check. We have classification theorems. In dimension 3, that was not easy, but that's a story you've heard before and not the one I'm going to tell today. Let me also just tell you that there's a theorem in dimension 3 of Moise that there's no difference between smooth and topological manifolds in low dimensions.

Okay. The classical high dimensions, so greater than or equal to 5, in high dimensions, there's something called the surgery program or surgery theory. And this is a theory that was developed in the 60s. by a whole bunch of folks, so let me call out Browder, Novikov, Sullivan, and Wall in particular.

And in a slogan, what you can think is that surgery theory is going to reduce the classification of manifolds to problems in algebraic topology. I want to comment here that I think there is a misconception, I had this misconception for a long time, that because surgery theory exists, the classification of high dimensional manifolds is kind of done. That's not true.

It's kind of not my problem anymore, but it's not known in general. So there are things that we know, for example, simply connected smooth 5 manifolds are classified, but there's lots of things we don't know. For example, the smooth Poincare conjecture is open in dimension 126. Okay. And in high dimensions, as perhaps you know, we can have a difference between the smooth and topological categories. That's originally work of Milner.

Okay, so that brings us to dimension four. Let me start off by talking about the topological category, where a groundbreaking work of Friedman from 82 tells us, one way to phrase it, is that surgery theory works topologically. It works when the fundamental group is what he calls good.

So sometimes this machinery will let you get classifications in dimension 4 for topological manifolds. And he carries this out. For example, he gives a classification of simply connected topological 4-manifolds. So the set of x4 top with no pi 1. Friedman tells us that this is in bijection with some other set of algebraic doodads that are more well understood. So it's in bijection with this set of pairs.

There's a form from Zn to Z, which is whatever unimodular, bilinear, symmetric. And there's an integer in Z mod. So, right, the theorem.

And I wanted to kind of state this to see what classifications in high dimensions have a tendency to look like. There's usually a big algebraic set that you maybe do or don't think is any nicer than this, but anyway, there's a big algebraic set, and you say that all of these correspond to a unique manifold. So let me interject here with just a couple definitions for people who don't do topology or geometry all the time. This pi 1 thing I keep referencing, this is the fundamental group. It's not important that you know what that is today.

Just know that it's a group that you can associate to a manifold. It's kind of a primary invariant for them. And this thing right here, I'll mention a lot as well. This is called the intersection form. Again, it doesn't matter that you know what that is, but it's a form which is an invariant of a manifold.

And Friedman tells us that for simply connected topological manifolds, it's like the invariant, basically. If you know what this means, the intersection form is the cup product on H upper 2. Okay, any questions so far? This is kind of a board on everything I'm not going to talk about. Let me also say that I really like to get a lot of questions. So what I really want to try to tell you about today is what we know about smooth 4-manifolds, which is nothing.

We have no classification theorems. We have no conjectured classification theorems. So far the field is, in a sense, really still in its infancy. One of the primary things we study is just the extent to which whatever is true topologically fails to hold smoothly. Let me make a definition to make that a little more precise.

We're going to say that smooth 4-manifold x is exotic. if there exists another smooth four-manifold, x prime, which is homeomorphic but not diphyomorphic to x. In this language, the smooth four-dimensional Poincaré conjecture says that S4 is not exotic.

But in fact, we know from Donaldson's work in the early 80s that there do exist exotic four-manifolds. And these two statements together serve to motivate the question that will sort of guide the talk today, which is which smooth four manifolds are exotic? And since we're maybe particularly interested in one day eventually understanding the Poincare conjecture, maybe we're particularly interested in understanding which small four manifolds are exotic. Where small will maybe be taken to mean today that the fundamental group is trivial. Donaldson's examples have that.

And maybe I also am going to ask that B2 be fairly small. So this is as an integer, and I want it to be small. So b2 is the final invariant that I'm going to sort of impose upon you today. This is the second Betty number. It doesn't, again, matter that you know what it is.

It's an integer. It's an invariant of a manifold. It's really crudely measuring how complicated it is. b2 of the four-sphere is zero. So this is the question that I'm going to use to guide the talk today.

The plan... is to tell you classically what are the techniques and what do we know. Then I'm supposed to, I guess, tell you about current developments.

So what do we know and how do we know it? But then what I really want to talk about, and what I'll spend a lot of the second talk on, is starting to try to say something about the structure of where exotica comes from, and maybe a little bit about quantifying it. So asking how you could measure how different a pair of smooth structures are.

Questions? Okay, so before I move on, let me just first give a little bit of a disclaimer. This is a question that gets a lot of airtime.

I'm going to give it more airtime today. But it's really not the only question in smooth four-manifold topology. There's a lot of things we want to know about four-manifolds, and we're also interested in things other than the manifolds themselves. For example, sub-manifolds, particularly in co-dimension 2, that's kind of higher knot theory. We're also interested in diffeomorphisms and manifolds.

And for these two types of objects, submanifolds and diffeomorphisms, the story I'm going to tell today, by and large, has a parallel story. And in fact, both of those stories are very active right now as well. And maybe the final comment is that from now on, unless I explicitly say otherwise, all manifolds are smooth and simply connected. Okay, so classically, how do you build Exotica?

So this is really kind of a three-step process. The first step is to give yourself some candidates. So build a pair of smooth... smooth four manifolds x and x prime, which you think reasonably might be homeomorphic, but not diffeomorphic.

Step two, show they're homeomorphic. This is really riveting stuff. Step three, you know where this is going. Okay.

But I wanted to point this out because I'm going to do it a lot. It'll give me a little structure. And I also wanted to sort of point out that this is basically under control by Friedman.

So really what you need to do is you build yourself some manifolds and you just get the intersection form right and then Friedman says you're good to go here and now you just need to distinguish them. Okay so then the techniques I need to tell you about are really how do you build four manifolds and how do you tell them apart. So let me give a list of some ways that classically we get four manifolds to study. And it's also good to have these kind of, I don't know, in mind, have some examples to be thinking about as we go through the talk.

Any questions before I do? I'm going to keep stopping and asking for questions. Okay, well, eventually.

So the first sort of place we get for manifolds is some sort of very basic small examples. For example, there's the four sphere. That's great. There's complex projective space. There's its orientation reverse.

And you can take connected sums of these things. That maybe seems very basic, but in fact, sums of CP2 and CP2 bars play a huge role in the exotica literature. Okay, you can also try to build yourself some four manifolds out of lower dimensional manifolds by taking products or bundles. For example, S2 cross S2, or more generally, products of surfaces, are very interesting examples, or maybe you want to take a 3-manifold cross S1.

Another way you can get some interesting formantless folds to study is you can ask a geometer to give you some. And in particular, complex surfaces and symplectic manifolds have played a large role. And those are actually kind of, for the most part, those are our root examples.

I'll say a little bit more about mangling them, but this is kind of the natural sources for the most part. studied classically. So we also like to take them and kind of Frankenstein them together using cut and paste operations, which is what it sounds like. You have some manifold x, which contains some codimension 0 submanifold z, and you're going to cut it out and replace it with something else.

Okay, and then the final construction of manifolds, which I just want to sort of mention here. This is a way most people work with manifolds on a day-to-day basis, but it actually doesn't really play a role in the classical exotica literature. But you can build manifolds out of these really basic building blocks called handles. And I'll say what those are in the second talk.

So that's how we build manifolds. We also need to be able to tell them apart. And classically, if you want to distinguish manifolds, you have one option. It's gauge theory. So I'm going to say very little about this today.

There are like four or more world-class gauge theorists in this room. None of them is me. Very crudely, this is an invariant that counts solutions to a PDE on the manifold. It's a very powerful invariant theory. It's not an exaggeration to say that everything we know about exotica, we know thanks to gauge theory.

So this is how we're going to tell our manifolds apart, for the most part, at least classically. But I just wanted to give you a couple of warnings about gauge theory. The first is that, in general, it's not possible to compute it explicitly. And, you know, if you want to build some manifolds and then distinguish them, you need to be able to actually literally make computations and then say, well, they don't match. So, well, people are able to sort of get around this, but it's largely by using formal properties.

You can sometimes get your hands on things if you have some geometric structure, and we have some understanding of how these invariants behave under gluing. Okay, so that's the first warning. The second warning is that gauge theory doesn't work when you have no B2. The invariant just isn't defined.

invariants. So in particular, while this theory has taught us a lot and there's still a lot it can do for us, you're not going to disprove the Poincare conjecture using a gauge theoretic invariant, at least as we know today. Yeah, good.

If you're in the ask a geometer category, does that influence it? The methods are the same if you're working with a manifold that has a little bit more structure. In general, it's just that the structure is going to help you maybe get your hands on this.

Sometimes we can say more about classifying manifolds that have more structure among those manifolds which have more structure, but that's a different question. More questions? Okay, so that's what I wanted to say about classical techniques.

So now I'm meant to tell you what we know. Which is quite a lot. This board is deliberately too small to read.

There's a lot of exotic illiterature. We would be here for a long time if I tried to kind of state everything carefully and give you all the names. So let me instead try to give you, like, tell you what the patterns are on this board.

So how's the board working? There'll be kind of three eras I'll discuss. In each era, I'll tell you how small we can get. I'm assuming manifolds are simply connected, unless otherwise.

I'll tell you how many smooth structures we can build, who did it. And then these are the two sort of... important columns, I'll try to say something about what the big major developments are, both in constructions and in obstructions. So, so the first era in, in development of, of exotica is, is the Donaldson era in the 80s.

It's kicked off with Donaldson's development of, of Yang-Mills gauge theory. The examples in this time, in this era are, for the most part complex, so we start to see non-complex examples computed using gluing formula and cut and paste operations towards the end. That's kind of era one.

Era two is kicked off by the development of the Seiberg-Witten equations in 1995. And there's an incredible amount of work then. We don't actually see any increase or any decreasing in the size of the manifolds we can get during this era. But I sort of want to emphasize that this chart is problematic in that that question is not the only question you want to know about form manifolds. So I'm only presenting sort of smallness here.

There isn't smallness developments. But in this, there's a ton of new results about smooth form manifolds. For example, questions about minimal genus surfaces representing homology classes and a lot of other things.

So this maybe makes. makes it look, well, this is an important theory. What is the second column? Oh, sorry. This one?

No, no, I'm sorry. The first column. The first one. Yeah, great.

So this is B2. So this is the size of the manifold. So Donaldson's first invariance of B2 equals 10. In the 80s, we got 9, and that's about it. In this second Seiberg-Witten era, no development, nothing smaller. Good.

Yes, other questions? Other things you want to be said out loud because you can't see them? Yeah, that's a great question.

Yeah, it's, I think it's because, but one of the reasons is because of a lack of small examples. When you look at this list and you say you're only interested in simply connected, so you probably don't want products of surfaces or things like that, well, you immediately end up here, and these tend to have a lot of B2. And so if you want to have less B2, Well, you have to kind of get your scissors out and start trying to put them together in ways where, like, one of them eats some B2 out of the other one, and you end up with very messy objects. Yeah, thanks. More questions?

Yeah? Yeah? Oh yeah, sorry Tom.

That's Tom. But nobody can see that anyway. Thanks.

Okay, so this is the second era, the cyber-grinton era in the mid-90s. Big development in techniques, both constructively and obstructively, but actually no movement on the smallness of the exotica we can produce. More questions about that era? Anything? Okay, and then the third era is kicked off in 2005 by a result of General Park, who uses rational blowdown, a technique that was developed in the cyber-Witten era, to get the first improvement in the size of exotica in a long time.

So he gives us exotic simply connected manifolds with b2 equals 8. And that, like, kicked off an arms race. Everybody works on this. There's a lot of results by these people and probably others in all sorts of permutations that I'm not going to write down. We start getting smaller and smaller b2, but nobody cares about 5, but that's fine. Until the end of the era is this result of Achmedev and Park.

So they give simply connected B2 equals three exotic manifolds. And that's the smallest exotica we have for a long, long time. There are a lot of candidates.

If you... Just kind of forget about ever being able to compute a gauge theoretic invariant for it. You got lots of options. More questions?

It doesn't. We just, by and large, don't have a more combinatorial invariant, but I'm about to say something about that. Yeah, more questions.

Okay, so let me try to tell you about a couple of more recent results. I'm going to use the same table, but I'm going to write a lot bigger. Okay, so...

There's three kind of camps of results recently that I want to say something about. The first is some work from 2021, which really doesn't make things smaller. It's like 23, the biggest thing on the table yet. No development in techniques, the development is in the obstruction. So in 2021, we get the first example of a pair of exotic manifolds which are distinguished with something called the Slice-Nuts Approach.

which is this method of distinguishing manifolds, which is a little more convoluted than just saying, like, compute invariance for both. Hey, look, they don't match. And this approach, I'll tell you what it is in just a minute, but let me sort of tell you now that the thing that's exciting about it is that versions of this approach could possibly work with no v2.

So it was good to know that this approach can work ever, since we're maybe hopeful that it could work in cool settings later. And that's joint work with Chakrian Manalescu and Marco Marangon. The second set of results that I want to tell you about is from last year. Uh...

And again, the examples in these results are not particularly small. There's not really movement here. What's exciting about this result is the development in the techniques.

So classically, we build exotic manifolds by starting with some big geometric stuff and making a hot mess. But we wanted to be a lot more juvenile about our construction. So we're really just going to start with handles, with these basic building blocks.

We're going to make a stack of them. and that's how we're going to get our manifolds, so explicit Handel constructions. And the reason that we can get away with this is because we're prepared to make explicit computations of something called the Haggard-Floer mixed invariant. So you should not really think that we are getting away from gauge theory here.

Conjecturally, the Hager-Fleur mixed invariant is equivalent to the Seiberg-Witten invariance. So this probably is a gauge theoretic proof. But we're accessing these invariants via a pretty different method that allows us to do computations much more explicitly. This is joint work with Adam Levine and Ty Ledman.

Because our constructions are much more explicit, we can monkey around and make other stuff happen. So we're also able to produce exotic 4-manifolds with b2 equals 4, so not the smallest, but pretty small, at the expense of having a little bit of pi1, z mod 2. But in fact, what's interesting about these is that their intersection form is definite. Okay, so that maybe doesn't mean too much to you.

An intersection form, as you know, can be definite or indefinite, and let me sort of assert that all the exotica previously was indefinite. And, you know, there's other things we ask about manifolds, whether then is there small exotica, and for some of those other questions, you see a difference between the behavior of definite and indefinite manifolds. So it was maybe reasonable to wonder whether definite manifolds were more rigid.

We're saying here that they're not. Okay, and then the most exciting result I'm going to tell you about today, which I wish I could tell you was due to me, but it's not, is work from last year from Stipschitz and Sabo. And they're going to give you exotic manifolds with B2 equals 1, pi 1, z mod 2, and again definite. And this I think you should think of as the first movement in actually getting smaller exotica in a long time.

Their techniques are going to be a mix from the good old days. Oh, I lost the green. And from us. And they can compute the Seiberg-Ritten invariance.

And then the final result that I kind of want to mention, in terms of progress in exotica, is very, very recent, from earlier this year. This is a result about exotic manifolds with boundary, which in a sense doesn't even belong on this table. The with boundary setting is very different. It doesn't make sense to compare.

I'll talk about that a bit more later on. This is work of Kiu Ren, who is a graduate student, and Mike Willis. And what's so exciting about their work is that their obstruction is combinatorial. So they use something called a skein lasagna module, which comes from Khovanov homology lasagna.

And this is the first time that compact exotic manifolds have been produced without gauge theory. Questions? Yeah.

No. More questions? You can put a description of a manifold into my collaborators and maybe get something here.

Yeah? Yep. Yep.

You look at the universal cover. If the manifolds have a... non-diffeomorphic universal covers and they're non-diffeomorphic and the universal covers have B2+.

Good question. Other questions? Oh, yeah. Right.

So, okay, what I'd want at the end of the day is a set of smooth manifolds with pi1 trivial. is in bijection with anything. Just give me a reasonable other set that you can relate them to. That would be awesome.

For now, the question that seems to motivate us more, or the question that we can sometimes work on, apparently, is the one where we say, where can I find Exotica? Sort of a geography question. More questions?

Yeah? No. It's bad.

The situation is bad. I'm going to try to say something a bit deeper in the talk about, you know, maybe if you could try to start saying where do all smooth structures come from, but it'll be not at all ready to see what kind of set. looking for so the plan for the rest of the first half is um to tell you what the slice approach is This is an argument that's due to Kassin in the 70s.

And here's how Kassin says you can maybe try to distinguish manifolds. So let's suppose... you have a pair of candidates for exotica.

So you have a pair of homeomorphic smooth 4-manifolds. Here they are. That's x, and this is x prime, and they're homeomorphic. All right, so what Kasson would like you to do is remove an open neighborhood of a point from both of them.

These manifolds now have boundary. They have an S3 boundary component. From that ball, you removed.

They're still homeomorphic. And now here's sort of what Kastner wants you to do. He says, suppose you could find a not K in S3.

Such that two things happen. One, there is a D2 embedded smoothly in here, in XCirc, such that the boundary of the disk is this knot. So what you should have in mind is that your knot, it lives in S3, so you could think of it as being in this boundary component right here.

That's K. And we're asking that there's a smooth disk that it bounds somewhere in here. And then also, you should want that there does not exist such a disk in x prime where the boundary is k. If you have both of these things, then that implies that the manifolds are not diffeomorphic.

It's easy to see that the punctured manifold can't be diffeomorphic. If you had a diffeomorphism, you push this disc across it, and the knot would bound over here, and it extends to the closed things. So that's the argument.

It's, you know, it's not quite as direct as, like, compute the things. They don't match, but it's also not too bad. So why should, or any questions about the approach or anything else?

Thank you. More questions, more mistakes on the board. Okay.

So why should you want to do something like this? Well, for one, this argument can be used to show that R4 is exotic. So this is the non-compact analog of the Poincaré conjecture.

And it's false. So this argument has teeth sometimes in a slightly different setting. Okay, so that sounds kind of good.

Then we'd like to know whether you can use this argument to show that S4 is exotic. And, well, we don't know, but there is an invariant, or there's a few now, actually. There exist invariants in the literature which could... due to, they could provide the obstruction here, even when x prime is S4.

So unlike using gauge theory, we could conceivably disprove the Poincare conjecture like this using tools we have today. So all of that sounds kind of good. But here's what's not so good.

It's not easy. It's not at all clear how to sort of set this up, how to find a good knot, okay? So if you have even two, even a known exotic pair, it's not clear how to find a knot which bounds a smooth disc in one of them but not the other.

And that's what we did here. Now it was the first time that this argument had sort of been run in a compact setting. Okay, question?

So, okay, this is an argument. It works sometimes. It could maybe disprove the Poincare conjecture. Maybe that's something we should be trying to do.

So in follow-up work with Chibri and Manolescu, well, we just sort of tried to think about how you would try to get this running. So we gave a very systematic method for producing... four manifolds X, which are homeomorphic to S4, and which come with a K as in the first step.

So it's this really pretty straightforward construction. It gives you candidates for counterexamples of the Poincaré conjecture, and those candidates come equipped with a knot, which we don't know that it doesn't bound a disc in B4, but it doesn't, you know, seem to have a good reason to. And the method is, is really systematic.

I personally made a computer spit 3K examples, and I personally am terrible at coding. So we produced a whole bunch of examples, and some of them were kind of interesting. Interesting meaning we, we, you know, we have, we get some knots and there's something like if the knots have blah property, then the Poincare conjecture is false. And we just sort of don't know. We fail to show that the knots don't have that property.

Um, I don't think this was ever meant to be like, aha, these ones are going to disprove the Poincare conjecture. It's more meant to, to mean somehow the method isn't kind of trivial out of the gates. Like it does spit some kind of subtle looking stuff.

Um, okay. Since then, um. It's been proven that these examples are actually not particularly interesting by my younger sibling, Kai Nakamura.

But what Kai did, it really just ruled out the examples we wrote down. It doesn't kill the method. You can build more. There's a bazillion ways you can build more, but if you want to see some that have been written down, Chhapriyan student Chahakya has done so. And you can start to see how, like, hopefully there's going to be a little bit of a conversation here then.

Build examples, rule them out, maybe that tunes the parameters on the construction, maybe eventually you'll show the construction's not going to disprove the Poincaré conjecture, that would be fine. So that's sort of the idea. But of course, given that it's whatever year it is, not only are we having that conversation, ML is having that conversation. So there's work of Gukov, Halverson, Chiprian.

and, uh, who's name I really can't spell, sorry, uh, who are using ML to both build, so to run our construction, and study the examples it produces. So, uh, the neural net that kind of evaluates them is out, that came out last year, and this one is in progress. And maybe a comment about these is, um, if the, if the neural net, if, if, if the neural nets win, it will really be a proof. Um, they are spitting explicit constructions, and the way they study examples, it, it does spit something verified. But that's, um, that's what I wanted to say for the first half.

Um, questions? Yeah. Oh, well, I don't know because, yeah, they don't, they haven't produced. They claim they're producing. But they're going to be homotopy studios.

How big are the knots? I don't know. Big. So, right, good.

So there's actually a little bit of a convolution, and this is why I kind of hid what I meant by some of them interesting. There's a flip in the logic. What they're actually going to build is they're going to spit knots where if you could prove the knot did bound a disk, then the Park-Wright construction is false. So what the net actually does is looks for disks. Oh, they definitely are detected by...

Oh, yeah, definitely are. It's a standard... Yeah, it's the Maser manifold, which we'll all learn what that is next time. So far, all of the new approaches are really in proof-of-concept stage.

No new actual... Phenomena, just like new argument. So, welcome back for Lisa Piccirillo's second talk on exotic phenomena in four dimensions.

Hey, thanks for coming back. So the plan for the second talk is what I said I was going to do, but not in the order I said I was going to do it. So I'll spend at least half of the talk talking about where Exotica comes from. I can give you explicit pictures of how you can get exotic manifolds in a pretty high level of generality, and I want to do that because it's very nice.

It doesn't have to be sort of mystical. Then I'll talk a little bit about how you might try to... Say how distinct a pair of smooth structures are, and then hopefully I'll run out of time.

But if I don't, I'll tell you a little bit about the work with Ty and Adam. Okay, so starting off with kind of where exotica comes from. So let me start off with an observation.

Let's suppose you have a pair of, a candidate exotic pair. So you have a pair of smooth form manifolds and they're homeomorphic. The observation is that if x and x prime co-bounded. a 5-manifold W. And so what you should be thinking is that it looks something like this.

Like maybe that's x, and here's x prime. And I'm asking that there's some 5-manifold which has the two of them as its boundary components. So if they co-bound a 5-manifold like this, and in fact, that 5-manifold is diffeomorphic to a product. It's actually just diffeomorphic to x cross i.

Well, then that's going to imply that my manifolds are themselves diffeomorphic. If this cobordism is really just a product, then I can kind of think about flowing x along the interval direction, and it'll land very nicely on x prime. That'll be a diffeomorphism.

So manifolds are diffeomorphic if they co-bound products. And in fact, candidates for exotica do co-bound something that's very close to a product. So this is the theorem of Wall.

He tells us that such four manifolds are going to co-bound something called an H-cobordism. So an H-cabordism is not literally a product, but you should think it's a 5-manifold that has the algebraic topology of a product. So you have your candidate exotic pair.

You have this cobordism, which algebraically is pretty boring. By the way, more formally, what this means is that the boundary inclusions are homotopy equivalences. But you have your two manifolds.

They co-bound this, like, algebraically very nice five manifolds. And if that five manifold were actually just literally a product, then they would be the same. So there's a sense in which that five manifold is capturing how different the manifolds are.

Like the failure of that thing to be a product is what's generating. what's generating the difference in these smooth structures. Um, and we can actually make that even kind of more, uh, explicit and a little more four-dimensional. Uh, so, so let me kind of get a picture going. So what's our H cobordism look like?

It's like there's x, here's x prime. Uh, we have something running between them, which is maybe like a bit weird. Okay?

So, in fact, Wall tells us that not only do we have this nice manifold running between them, but actually most of it's boring. There's a sub-H cobordism, W prime, such that W minus W prime, that part is actually a product. So there's this sub-mat5 manifold here, where all the interesting non-product stuff is happening here.

And this is very boring. And in particular, like, the fact that we have a product here tells me kind of that this part of x is actually the same as this part of x prime. The difference is, like, these parts, whatever that means. Okay? And the theorem I actually really kind of want to spend most of the rest, or this section talking about, is something called the Korg theorem, which is going to improve this even more.

So the quark theorem was proven by everybody in around about 1997. So let me give you some names but not write them. Curtis, Tseng, Friedman, Stong, Matveyev, Matveyev, Bizaka, maybe we should say Kirby. And Casim proved a non-compact version of it, like, before all of them were born. OK.

So, so this kind of was in, came out of, I don't know, anyway, it's due to a lot of people. What it says is that, even better, W prime, this bit that has all the interesting stuff going on, can be taken. such that if you look at how W prime intersects X, so let's call that C. That's right here. And if you look at where W prime intersects X prime, let's call that C prime.

That's here. We can take W prime so that these are contractible. So contractible means that their algebraic topology is as simple as possible. We have the alge top of B4. So kind of all of this put together is telling me that if I have two homeomorphic smooth 4-manifolds, then what's different between them is all packed in this piece, and this piece doesn't have any algebraic topology.

Let me just write down, to be explicit, a corollary. What we learn here is that if you take X, you remove C, you glue in C prime, what you get is diffeomorphic to X prime, right? We start here, lose this, that's the same as this, glue that back. So these two doohickeys, rats, are called quarks. And this operation of cutting one of them out and gluing in the other is called cork twisting.

Any questions about this statement? Or anything? So there's several things I think are, is really, really surprising about this result.

One of them is, is that, well, these quarks are pretty interesting objects in their own right. You know that they can't be diffeomorphic because, well, Well, if they were, then all of my four manifolds would be diffeomorphic, which they're not. So let me write that down kind of explicitly. This is a result that was originally proven a little bit earlier in a slightly different form. Let me also quote Achboli-Ruberman for the form I want.

What we get from this is that there exists exotic contractible. for manifolds. Now, let me emphasize that contractible manifolds have boundary. So this is not quite the setting I've been talking about kind of all along.

But contractible manifolds are very simple. They have essentially no algebraic topology. b2 is 0. Everything's trivial. So this is, very reasonably, the width boundary.

analog of the Poincare conjecture, and again, it's false. And another thing I think is really surprising about this theorem is that, you know, over here, we're having this really hard time producing exotica with no boundary when there's very little algebraic topology. But this is telling me that that That issue is technical. Like, I can pack the exoticness between, like, the difference between the two manifolds into something that has no algebraic topology. So somehow it's not the lack of algebraic topology that's giving us the issue here.

So I'm going to... give you fairly explicitly examples of exotic contractible manifolds. Any questions before I do? Yeah, they'll be integer homology three spheres. They can be as simple as surgery on a knot.

More questions? Before I try to set this up, try to explicitly build some of these, let me just comment that this is the theorem that Rann and Willis reproved. Technically, they kind of reproved that version of it. All right, so to set this up, I have to tell you about handles.

And I also have to remember not to write in the shadow. So help me out with that if it doesn't go well. So I'm going to try to do one. We can see this part of the board.

I'm going to try for a definition by picture here. So please bother me if it's not clear. So handles are a set of building blocks for building manifolds. And you should really think that they're kind of a... manifold-y version of a cell complex.

So I think most of us know how to build a, a cell complex. And, and we're going to, to do the, basically the same thing. But all of our cells, we're just gonna thicken them up so that everything is consistently dimensioned whatever I want. in my case, four.

So let me start off with a 3D example of building a manifold out of handles. Let's suppose I want to build a manifold that kind of has one zero cell and one one cell. So, okay, I might start with a zero cell.

There it is. It's a point. Okay, but that's, like, not great as a three manifold. So my zero handle will be this three-dimensional thickening of that. Okay.

And okay, now I guess I wanted to add a one cell to this. So, you know, what would that look like from a cell complex perspective? It looks something like this.

Not a three manifold, but we can get around that by, again, thickening it up in the set case in two more dimensions. That's a one cell. This is a one handle. In dimension four. works the same.

Let me try to give something of an example. So let's suppose I want to build a, a, a four manifold out of a zero handle and a two handle, okay? So, so what do I do? I start with, you know, my generalization of zero cells, so there's my zero cell.

I need to thicken this up in, in four dimensions, so I, I can't draw you a picture of that, but let me just draw you this picture again and declare that that's B4. B4, there it is. That's my zero handle.

Okay. And now I said I want to glue, I want to use a two handle. So that's going to be this kind of thickening of a two cell.

So all right, here's my two cell. And how do you glue a 2-cell onto a cell complex? You have to glue its boundary on to whatever you are doing.

So, okay, here's the boundary. That's an S1 here. So I need to glue this S, I need, yeah, this S1 needs to get glued on here. The boundary of this B4, well, that's the 3-sphere, so I'm looking for an S1 in the 3-sphere, so I'm looking for a knot. Okay, pick a knot, fine, glue it.

And then, well, all right, we need to turn the whole thing into a four-manifold by crossing this with another D2. Technically, you need a framing here. We're not going to worry about that. And I'm only going to build four-manifolds out of two handles today.

So that's what you need. Questions about, you know, handles in the kind of broad sense? Okay, let me give you two more quick facts about handles.

One is just an even kind of cheaper picture of this. So a schematic or an even more schematic of a four-dimensional two-handle that you'll see me use a bunch of times is you draw your zero handle like this. So this is my zero handle. It's B4 and here's its S3 boundary.

And then I'm going to attach my 2-handle to, OK, we have some not in that boundary. My 2-handle is, you know, fundamentally there's kind of the D2. And my 2-handle, you know, looks something like this. So I'll use that representation of a 4-ball with a 2-handle. And then maybe the final thing I wanted to say is that it's the theorem of Morse that all smooth manifolds can be built in this way.

Thank you. Questions? Anything wrong on the board?

To build a contractible exotic manifold, I need two handles, and I need one more operation, which is something called carving. So building manifolds by carving. So let's suppose I'm given a 4-manifold X, smooth, with a disk, a D2, embedded smoothly in X, such that the boundary of the disk goes into the boundary of the 4-manifold. So you should be picturing something very similar schematic, something like this. Here's x, here's its boundary, and we have, uh, a disk with boundary.

Um, okay. So, uh, I wanna build a slightly more interesting four manifold out of this. Um, and, well, not very exciting, I suppose. What we're gonna do is cut it out.

So we're going to remove an open neighborhood of that disk. And we'll get something that looks in the schematic like, I don't know, like that. This is x minus y. If I have submanifolds, I can cut them out.

Last thing we need is a lemma. We need to check that. We're kind of all on board. Attaching a two-handle and carving out a disk are, in a sense, kind of dual operations.

Here we're adding a thickened-up disk to the outside. Here I'm removing a thickened-up disk from the inside. And the lemma relates them sort of even further. So let's suppose we're given a disk embedded smoothly, let's just say, in the four-ball.

And, okay, I want the boundary of the disk, again, to land in the boundary of the four-ball. then there's two manifolds that I can build out of this data. I can consider B4 minus the neighborhood of that disk.

And I can also consider B4 union a two-handle glued along the boundary of that disk. Right? So this and that. And the statement is not that these manifolds are the same, but that they have the same boundary.

And the proof of this is super cute. Here's S4, which I can decompose along an S3 into two 4-balls. Let's suppose that my disk lives in this one.

Now I'm just going to repartition S4. where the neighborhood of this disc goes with this. That's before union a two-handle along the boundary of the disc, and the rest is the four-ball minus the neighborhood of that disc, and their boundaries are just identified by setup.

I'm going to go. Okay, so now we're ready. But really, let me check in.

Questions or anything or unhappiness with the construction before I do? Yeah. Yes, let's say it's smooth.

It turns out for three manifolds there's no difference, but it comes to you looking smooth. You don't have to appeal to anything. More questions?

All right, so this method I'm going to give for building contractible exotica is, in fact, due to Barry Mazur, the number theorist. It was his undergraduate thesis here. But the development of this owes a lot to Salman.

And here's the construction. So first off, if we're going to build exotica, you know, build the thing, show they're homeomorphic, show they're not to be amorphic, for the homeomorphism, we're always going to ask Friedman what he can do for us. And what Friedman can do in this setting is he says that if you have any contractible C and C prime that have the same boundary, are automatically homeomorphic. So if I want to build candidates for exotic contractibles, I'll just build contractibles and make them have the same boundary and that'll work.

So I'm really in the market for manifolds with the same boundary and we just learned how to build manifolds with the same boundary, which is good except that these manifolds are super duper not the same or contractible. This one has B2, this one has B1. Okay, not the same.

But But Mazur gives us this really beautiful kind of way to use that lemma anyway. So what he says you should do is consider a link L, which has two components, and this link lives in S3. And the link has the following properties. So one, both components are individually unknots. So maybe what you have in mind is something like this.

Maybe that's L1, and maybe L2 is like that. We can still see. Getting borderline.

OK. And I won't go any lower. It's the size of it.

OK. Thanks. I will not go smaller. This is a link.

This one's kind of obviously an unknot. If you squint at this one and you can see it, it's also an unknot. OK.

And unknots, they have this nice property, which is that they bound disks. They just bound disks in S3. There's one. You can push that disc into the four ball a little bit, so unknots are also going to mount discs in B4.

So what we can do is construct a pair of manifolds in the following way. We'll start with B4. Our link, L1, L2, lives in its boundary.

And we can build a manifold here by removing a disk for L2 and attaching a 2-handle along L1. Let's call that C. And we can build a C-prime by doing the same thing in the other order. So we'll remove a disk for L1 and attach a disk for L2.

One interpretation of the lemma is that the boundary 3 manifold can't tell if you add a disk or remove a disk. So the boundary just can't see the difference between these two setups, so these guys have the same boundary. And as long as you assume that the linking number of L is 1, then they're both contractible.

So I'm not now going to go on to prove that contractible manifolds you build like this aren't diffeomorphic. Let me tell you that you can. And in fact, for this literal link that you may or may not be able to see, the contractible manifolds you get are not diffeomorphic. And in fact, this literal link is the one that was used.

here, and it's also the one that these guys redetect. So, you know, this kind of... these two manifolds, which are defined very simply from a very straightforward link like this, well, they give you a pair of contractible exotic manifolds.

Okay, and the theorem, the thing... The theorem, the quark theorem maybe gets, like, even better. So this is the last kind of leg of the quark theorem.

In fact, you can prove that... Any corks can be built using a generalization of this construction. So any exotic pair are related by cutting out and re-gluing a contractible manifold, and you can assume that those manifolds are built using something like this. Uh, no.

You might need a bunch of components and then a bunch of components. And maybe you're not using the four ball either. Maybe you're using a fixed other contractible. More questions?

Yeah. Yeah, there's some really easy ways to check it. They factor through a little bit of contact or symplectic topology, so they're gauge theoretic somewhere in there.

But, like, if you can draw a Legendrian diagram of them where some Thurston-Bennican number is positive, then you win, or something like this. I'm going to talk about quantifying exoticness. Anything before I do?

Yeah, so I just erased the theorem that had to... attributions here, and it was one relative one, absolutely. Yes. Yeah, it's not too bad. Let me at least...

It's not too bad. So it turns out, this isn't clear, but when you remove a disk for the unknot, what you get is S1 cross D3, so you get this thing's homotopy equivalent to a circle. And so if you want that dead, you better make that, that sort of two cell go around once.

And somehow that turns into the linking condition. Okay, so... All exotica comes from cork twisting. Corks can always be taken to come from this construction. So if you want to know how different two smooth structures are, one way you can try to measure that is by saying, how bad does the link that gives the cork have to be?

So a couple of definitions of how bad a link like that might be are, well, okay, let's suppose we have a candidate for an exotic pair. So x is homeomorphic to x prime. Then we'll say that. the complexity of the pair is something like the minimum of the amount of geometric linking of L taken over all quarks, which gets you from x to x prime.

Uh, and, uh, we're also-we also study a-a version of-of distance, which is called, uh, the stabilization number. That's gonna be the minimum size of the link you need. Again, over, uh, all quarks.

Maybe a normalization for both. A comment, perhaps, for people who come from a more high-dimensional perspective, both of these notions of complexity can be defined in terms of the H-cobordism, but I'm defining them kind of discreetly here. So these definitions are set up so that for a homeomorphic pair, if the complexity or... or the stabilization distance, stabilization number, r0.

Well, that's the same thing as the manifolds being diffeomorphic. All right, so we certainly know that there are pairs of manifolds where this complexity is positive. But in fact, it's a major open question whether it's ever bigger than one. Do there exist x, x prime, what's above complexity or, and this one especially is kind of a very famous and apparently very hard question.

No. Probably of this bigger, there's like a bipartite link. I definitely want to subtract one.

Yeah, yeah, there's normalization. Sure. That's right. Yeah, subtract one, divide by two. But, you know, something that's counting that and then appropriately being what it should be.

I think you probably want to divide this by two, this by two also. More questions. Okay.

So we are apparently pretty far away from being able to address this. But there have been current developments. So for manifolds with boundary, we actually can answer this question. So we start with the complexity. This is the theorem.

The heavy lifting of this is really from Morgan and Szabo in 1998. We're going to need something that's due to Akboli-Ruberman. And if you want to see it written down, that was done very recently by Robert de la Du. So the statement is that there exist contractibles, so with boundary, where the complexity of that contractible pair is big. And even more surprising and exciting is a result from Sun Kyung Kang a couple years ago who said that there are such contractibles where the stabilization distance equals 2. And this, I think, yeah, this was a very surprising result. Let me maybe say one more thing that's a little bit outside the purview of the talk before I kind of move on from complexity, which is another very recent result from Junpeng Lin, 2021, who said that for, let me say, there exist exotic diffeomorphisms, so non-manifolds, with stabilization of of the diffeomorphism and the identity, whatever.

Let me just say stabilization number two. So I'm not going to make this precise, but I said at the beginning there's a largely parallel story to the one I'm telling for submanifolds and for diffeomorphisms with manifolds, and Zhen Feng solved this problem for diffeomorphisms, and what's really exciting about it is he's working with closed manifolds, so he solved the real one. That's what I'm going to say about distance or quantification.

Questions here? You should ask questions. Yeah, if you have a spin C structure...

if you have different Seiberg-Witten invariants in spin C structures with a large expected dimension, then you can't have small complexity, which is going to run you smack into the simple type conjecture, but that's a bound. More questions? So, let me tell you something about the techniques that Tai and Adam and I use. To start, let me give you a statement that I'll talk about.

So, one of the things we show is that there exist exotic four-manifolds, which are homeomorphic. but not diffeomorphic. There exists four manifolds that are homeomorphic but not diffeomorphic.

To S1 cross S3, connect some two copies of CP2 bar. Nope, two copies of CP2 and nine copies of CP2 bar. Okay, so if you're kind of keeping track of stats here, this is b2 is 11, so not particularly small, and pi1 is z. That's the biggest pi1 you've seen all day. So this is maybe not...

not a good theorem from the perspective of the table. This is the one I'm gonna tell you about anyway. And the reason I wanna tell you about this one is because I think, well, because the proof is really easy, and it still captures, I think, most of the kind of ideas and developments that we use with other results. To prove this, we develop a new four-manifold invariant.

Alpha, it's an integer. It's only defined for four manifolds that have some B3, that have some second homology. Let me just say, for convenience, B3 is 1 for today.

And this invariant really is new. It's provably different from Donaldson, Seiberg-Witten, Heger-Flermix, Baffaruta invariants. We can prove some other things with it that I won't write down because I'm not really going to motivate them.

Maybe just give me three sentences out loud, and then I'll come back to saying things that I've backed up or defined. So some things we can prove with alpha are that we can build manifolds with non-vanishing alpha invariant that have, that contain square zero embedded homologically essential spheres. That's something usually gauged theoretic invariants vanish on. And we can also distinguish four manifolds that are related.

by a functional sort of not surgery on an Alexander polynomial 1, not using this invariant. Okay, end unjustified sentences. So, I'm going to define alpha for you and then show you a little bit about how you can expect to actually literally compute it sometimes. To do that, I need to give you, like, the world's shortest crash course in Heger-Fleur homology. Any questions before we do that?

If you filibuster, then you don't have to do Heger-Fleur homology. All right, rats. So what is this? This is an invariant package defined by Osh-Voth-Sabo-rats.

I have it backwards? Yes. Oh, okay, great.

Thanks. Around, I don't know, 2002, everything I'm going to write down is due to them. For experts, I'm thinking about HFRED, and I'm summing over spin C structures.

Okay, so what's the invariant? Out of the box, it's an invariant of a 3-manifold. And it's a, let's just say it's a finitely presented module over Z mod 2. It satisfies a TQFT structure, so cobordisms. Z from a 3-manifold to some other 3-manifold.

which, you know, that's something like this. You have your two three-manifolds, and there's some four-manifold running between them. These cobordisms are going to induce maps between HF of Y and HF of Y prime.

In this way, you can think that it gives you a four-manifold invariant. This map is an invariant of the cobordism. Okay?

And these maps can be written down sometimes. And one really powerful tool for writing them down is the existence of an exact triangle. So there exists a, so let's see, for z, two-handle cobordism. So if this is pretty simple, just start with y cross i, add a two-handle, that's it.

Then you have this exact triangle. which has the map you want in it. So it's going to go from hf of y to hf of y prime.

That's the map you care about. And then, okay, well, there's something else. Let me call it y sub k.

It doesn't matter what it is. I'll just tell you that it's some explicit ancillary 3-manifold. And if you built the cobordism, then you know what this 3-metafold is.

And that's it. That's your crash course. So this alpha invariant for a 4-manifold with a little bit of B3 is defined to be the minimum dimension of H of red of y for y generating H3. This is a very common way in general of getting an invariant of a manifold. We're frequently building invariants by saying, I don't know, what's the minimum genus of a surface that represents some class in H2?

This is an analog of that. Because you take a minimum, you get invariants for free, but usually there's a cost, and it's that you can't compute your thingy. Let me try to convince you that you can sometimes compute your thingy in this case. State dilemma, let me get a little bit of a picture going.

If your manifold has some B3, then it also has B1, so it's got a little bit of a kind of circle in there somewhere. So maybe that's my picture of X. And my generator of H3 maybe looks like that. Okay, so something I can do, this thing's going to have to be non-separating, so I can cut along it, and I get a cobordism from Y to itself.

And the lemma is that if that cobordism, if the map associated to it, which goes from HFY to itself, if that's an isomorphism, then Y was minimal. So you can find minimizers if you can find cuts where you get nice ones. So let me just conclude by saying then, how should you build?

Four manifolds where alpha of x is whatever you want. Well, we're going to work backwards. You're going to take a 3-manifold y.

where the rank of HF of Y is N. It's whatever you want. Three-manifold invariants are reasonably computable, especially if you ask somebody who knows what they're doing. So, okay, you can find something like this.

And now we're going to build a cobordism Z from Y to itself. So, something like this. Such that, two things.

I want the map f to be an isomorphism, and I want pi to be trivial. Then you define x to be z with the n's glued together. And by setup, you get.

That alpha of x is whatever you want it. So you can kind of vary, well, this is a pretty straightforward way of writing down a four manifold that has whatever alpha invariant you want, supposing that you can do this. So let me just say one thing maybe in words.

even, about why you can do this. In particular, how are you gonna get your hands on this? These maps are, you know, you can't necessarily always just write down any cobordism map. And, well, you can, you can keep control of this by keeping a good handle on this.

So if you build your Z by only using two-handle cobordisms, and you only attach them in ways where this ancillary manifold has no HF, then you just force an isomorphism across the top at every step. So as long as you build the thing and you're pretty careful, then you get this. And I will stop there. Thanks.

Transcript for:Exploring Exotic 4-Manifolds and Topology

Transcript for:
Exploring Exotic 4-Manifolds and Topology