 So welcome to our final day. Just to remind everybody that the abbreviated or shortened program for the day. So talks 9, 10, 11, 15, and 12, 15. There's tea after the 10 o'clock talk. And then, so anyway, all four packed into the morning. And action packed, so then, I guess, dash up for lunch. So anyway, you could all check the schedule. But that's for today. Now, I'm delighted to introduce Lisa Piccarello, who will tell us about four manifolds with boundary and fundamental group Z. Thanks, Denny. Yeah, thanks, everybody, for getting up early on the last day. I'm happy to be here. It's been a really wonderful conference in place to be. I also should take the opportunity to say whatever I want about Tom, because he's conspicuously not there. But maybe I'm still at MIT. So I should not do that. And so let me tell you what this talk is going to be about, which it's going to be entirely about some work that's joint with Anthony Conway and Mark Powell. And what we're going to study is some objects, which, well, it's sort of not very original. I'm going to try to study a little bit about manifolds and a little bit about knots in them. But the context I'm going to be working in is that the dimension will be four. And so in order to have a talk that contains me convincing you that we know really anything at all, I need to put some really strong restrictions on what kind of four manifolds I'm talking about. So I'm going to assume that pi 1 of either the manifold or the surface is either trivial or Z. So some comments about what knots mean. Dave made a compelling case yesterday that codimension 1 is an interesting place to study knotting, but again, being less original, knots will be codimension 2. So you might want to think about this as surfaces. And then just like in classical knot theory, when I talk about the group of a knot, I want to talk about its complement. So all of my surfaces are going to live in simply connected spaces. And when I say the group of the surface, I mean the group of the complement. OK, and then now what I want to try to do is tell you what we know about these objects in a few categories. So I'll tell you what we know about them topologically, what we know about them for the smooth closed setting, and in the smooth with boundary setting. Topologically, we know quite a lot. I'll state them more precisely. But for these groups, both for four manifolds and for surfaces in them, we have classification theorems. And the main theorem that I'm going to talk about today is contributes towards this classification. OK, so we know quite a lot in the topological setting. For the setting with smooth closed manifolds, we also have started to make some progress on the baby's case of this, which is we've taken kind of the simplest classification problem you could think of, and we've given it a name. So this is the Pongray conjecture. And for surfaces, there's something called the unknotting conjecture. And these are apparently hard problems. I'm not going to talk much about them today because I don't have much to say. But what I will talk about is the fact that appropriate analogs of these problems are a lot more approachable if you ask them about smooth manifolds with boundary. So I'll talk a little bit about an analog of the Pongray conjecture here, which is known to be false, and an analog of the unknotting conjecture here, which is also known to be false. So this is the plan. Before I get started, let me make one very important disclaimer, which is something you might already know, and it's that other groups exist. And we know a little bit about manifolds with other groups or surfaces with other groups. In the topological category, for example, for finite cyclic groups and something called solvable bomb-solitaire groups, there's some classification work due to people like Hamilton and Crack-Tichner. For surfaces or manifolds, we have some exotic results for other groups as well. But these are the two that we have the most systematic understanding of. So I'm just going to restrict attention to these groups. Any questions before I get going? I'm going to start by, in this category, stating the classifications. So I'm going to try to state four theorems all at the same time because, as you might expect, classification of manifolds with or without boundary and with one of these two groups looks very similar. So starting with all the attributions that the theorem that you know is that Friedman in 82 gave us this classification of manifolds where I'm going to start by talking about the topological category and talking about manifolds. So Friedman classified simply connected closed four manifolds. A bit later, Boyer classified simply connected manifolds with boundary. For pi 1z, it's work of Friedman and Quinn in the setting with no boundary. And the theorem that I'm going to tell you about is a combination of work of my collaborators independently. And then the three of us, we're going to give a classification for pi 1z and with non-empty boundary. So the way I'm going to do the statement is that when I need to modify the classification a little bit for the setting where the group is z, I'll add a modification in green. So the statement is going to be that our set of objects, which is topological four manifolds, let me say with boundary, but I'm allowing the boundary to be the empty set. And pi 1g, or g is one of our two groups, these manifolds. And I'll add a couple of hypotheses in a second. These manifolds classified up to homeomorphism. I'm going to say rel boundary. There's no boundary. That doesn't mean anything. This set of manifolds is in bijection with a set of, in the most generality, five invariants. So the statement I'm going to write here is not completely precise. If I really wanted to state this as a precise bijection, I'd need to give you a list of five purely algebraic thingies over here that you can define without referencing the manifolds. That statement's kind of a mess. So the way I'm going to state this now is sort of more intuitively. If you start with a four manifold with these properties, what are the pieces of data you need to write down to completely capture the manifolds? But we do also have a realization statement. So if you give me those invariants, then I can give you a memo. So the invariant, well, the first one is probably something you're expecting. It's the homeomorphism type of the boundary. That's certainly something you need to record. If there is no boundary, there's not a lot of data there. The second thing you need to record is also kind of very simple object. You need to know the inclusion map from pi 1 of the boundary into pi 1 of the filling. So if you have no boundary, neither of these are very interesting. The first invariant that shows up all the time you're expecting, you know, Freedmen's classification is in terms of that intersection form. That is indeed what you want in the setting with no fundamental group. If you have fundamental group, you want a modification of this called the equivariant intersection form, which I'll describe in just a moment. The fourth invariant is going to be kind of the most mysterious one. I think most people here won't have seen it before. It's something we're calling the automorphism invariant. I'll describe that as well in just a second. This invariant also doesn't exist in the no boundary setting. And then the final invariant is a classical thing called the Kirby-Seabaman invariant. Really loosely speaking, topologically, you can think of this as measuring whether the manifold has a smooth structure. So this is the shape of all of the theorems. If you have no boundary, you really just see these two invariants. If you have no fundamental group, then there's no data here either. So Boyer would maybe just have four. And we have five bits of data. Any questions about the statement? Yeah, Marco? Yes, we do. But then, yeah, we do for simplicity. I think you could get away from that. There's a couple of hypotheses that I need to add here. Is that I want to be working with boundary irrational homology sphere or some appropriate analog of that. So h1 of the boundary of x, it needs to be torsion. A bit different if when pi1 and z, I'll explain that. And I also want to know that the inclusion map from pi1 of the boundary into pi1 of the manifold is surjective. So the first things I want to do is tell you what these things are. Any more questions before I start that? What you're familiar with is the plain old intersection form, which pairs elements of h2 and spits integers. And generally, I think of this as this form that tells you what is the h2 of the manifold and how does it interact with itself. That's an invariant you can define for manifolds with pi1 z as well. But it doesn't quite capture everything you need to know. Because if you have interesting pi1, you might also want to record how does the h2 of the manifold interact with the pi1. And a really elegant way to capture that is instead of looking at the homology of the space, you can look at the homology of the universal cover, or in this case, the infinite cyclic cover. So the equivariant form pairs these homology groups, and it's going to spit Laurent polynomials. And I'll tell you in just a second exactly what it does. But before I do, let me maybe get a running example going. So maybe you want to think about your manifold as being s1 times b3 for starters. So I'm imagining this tube is, there's an s1 like this. I haven't identified the ends, but you can. And then I want to attach a two-handle to this, maybe like this. So this is a perfectly nice, very simple, four-manifold with pi1 z. And to compute its equivariant intersection form, we'll need to look at the cover. So that's also very easy to understand. We see r times b3. And it now has a whole bunch of two-handles stuck to it. And they happen to become three-framed, but you shouldn't worry about that. So OK, let me now tell you exactly what this form does. Just like the regular intersection form asks how do surfaces representing the classes intersect, we're going to do the same thing. So if you have a pair of classes, let's say represented by sigma and gamma. Well, again, you want to ask how sigma intersects, in this case not just gamma, but all of the translates of gamma. So we'll see a sum. So in this case, well, at first, if you look at the homology of this cover, as a module over z, it looks infinitely generated. I don't like to write down forms over infinitely generated modules. So you want to think about this, in fact, as being a ZTT inverse module, at which point it becomes finitely generated. Let's say it's generated by this two-handle here. Well, OK, we just have to ask ourselves if I take a class running once over this handle, how does it interact with itself? So we'll see self-intersection 3 from the framing. But then we also have to record how does this class intersect this? And we see a minus 1 from the link. And then if i is larger than 1, well, we don't have any intersections between these. For i minus 1, we get a t, and that's it. So this is what the equilibrium form looks like in this case. If the boundary's not empty, so the other more interesting invariant on this list is the automorphism invariant. And to tell you geometrically what this is measuring, let me remind you that when you have this H1 torsion hypothesis that we have, there's a short exact sequence, which looks like this. This is what you see in that simply connected case. This should be very familiar. And if you're working with a manifold with pi 1z, you also have a sequence like this on the covers. And what the automorphism invariant is doing morally is capturing this map right here. So in a picture, if you have your boundary manifold and you have some H1 class in there, well, what the automorphism invariant is telling you is which relative class should it be the boundary of? I'm not going to say anything more precise about what this invariant is. You can make it precise algebraically. Let me just try to say something about why you should be expecting to need to record something like this. So if you think about maybe trying to prove the surge activity result here, somebody's going to give you some data and you're going to need to build them a four manifold with boundary and pi 1z. Well, as you can sort of imagine what data you're going to want to be told. So you know you have to have pi 1z. They're going to tell you what you want the boundary manifold to be. But we know that there's a relationship between pi 1 and the boundary manifold and pi 1 of the filling. So you have to ask them, don't just tell me the boundary. Tell me what that relationship should be. And that's this right here. And then the next thing we specify is what they want the H2 to be. Now it should interact with itself. That's great. That's this. But then you know there's a relationship between H1 of the boundary and H2 of the filling. And that's what the automorphism. So I want to sketch how the proof of this result goes in the surge activity setting. The breakdown here is that we proved surge activity. My collaborators had proved injectivity a bit before. And the proof is the outline is very straightforward. It follows sort of in four steps. So the first step is just to begin with the boundary that has been insisted upon. So start with m cross i. And what you want to do is you want to attach two handles to it in a way sort of that's instructed by the intersection form, the aquavariant form, and the automorphism invariant. Somebody has insisted on what the H2 should be and how it should line up with the boundary. You're just going to force that by putting two handles here in sort of the way that you were told to do it. A couple of comments about this. This so far, what we've built is smooth. And this step here is the main technical work. Once you've attached all your two handles, you sort of have some new m prime. And what you have to show is that the homology of the cover of this upper boundary, the one that is associated to the inclusion map on pi1 down here, the cover of this has no H1 as a ZTT inverse module. So we have this hypothesis here where sort of H1 of the boundary was torsion. That's something like being a rational homology sphere. And what you show here is that after you finish adding your two handles, what you get is something like an analog of being an integer. And the reason you want to know that is because you can show that such a manifold, such an m prime, bounds a topological homotopy S1. So this step here is where the proof is just topological. So this maybe seems very strong and very surprising. But in fact, you already know a theorem like this. In the simply connected case, you know that Friedman proved that integer homology sphere is all bound contractable. So this is a pi1 Z analog. OK, good. So once you have a homotopy S1 filling of m prime, you can stick it up here and define this whole thing to be your x. And finally, you just verify the x has the right invariance. No, sorry, g is one of my two permitted groups here. That would be a really great theorem that we definitely did not prove. More questions? Yeah. Oh, sorry. Where does the restriction that h1 of boundary is torsion come in? I mean, in fact, a few places you, but maybe one that you can see is that this sequence wouldn't necessarily be short without it. But it also, you want it to get a little bit of control over what that covariant form can be as well. Boyer, by the way, does not need it. His theorem, he can remove it, and the theorem will look very different, but he can remove it. We did not pursue removing it. This is what I wanted to say about manifolds. I'm going to move on to talking about surfaces. Anything lingering before? So then, again, I'll start in the topological category. And here, as I said in the beginning, we again have classification theorems. So Boyer's work gives a classification of, oh, sorry, let me give a whole bunch of setup before I get going. So the settings we're always going to be working in when I talk about surfaces is that x is going to be a topological manifold with pi 1 trivial. The boundary of x is either going to be empty or s3. All of my surfaces are going to be proper. And what I'm going to use for an equivalence here is that I'm going to say that in either category, a pair of surfaces are the same if there's a homeomorphism or diffeomorphism of pairs. You can, there are results about everything I'll say in this section. If instead you say isotope, but it's a bit messier. And then the groups we're really interested in are the not groups. And again, that's not going to be everything. It's going to be either one or two. So in this context, we have some classifications. So Boyer gives us a classification when pi 1 of the complement is 1 and the surface is closed. And our work gives a classification when pi 1 of the complement is z. And the surface can have boundary or not. And I'm not going to state the whole classification here. It looks rather similar to this. So let me just say that classifications exist. A comment is that it's not written down in the literature for simply connected complements for surfaces with boundary, but it's recoverable from this. So in the smooth category for surfaces, there's this motivating conjecture, which is called the unnotting conjecture. And what the unnotting conjecture says is that there is supposed to be a unique s2 and s4 with pi 1z. So in the topological category, this is a theorem for most genera. So Friedman proved this for genus zero surfaces and Conway and Powell proved this for genus greater than two. Before you ask, it's open in genus one and two. It's probably true, but their technique's kind of... So this conjecture appears to be rather difficult for closed surfaces. But let me try to tell you a little bit about what we do and don't know about them. So before I do that, let me make a definition. We'll say that a surface is exotic if there exists another surface, which is topologically but not smoothly equivalent to it. If you can ask, or the unnotting conjecture is asking, is there an exotic pi 1z sphere in s4? And what we know is that there exist exotic surfaces in some closed-form manifolds with no pi 1, that's work of Fintuchel and Stern and it's work of Kim and Ruberman with pi 1z. We don't have, so the unnotting conjecture wants you to find exotic surfaces in the simplest manifolds you can work in. We're not great at finding exotic surfaces and in particularly simple manifolds, but there is sort of one very good thing that we can do, which is this work of Hoffman and Sennigtschian, which says there are exotic surfaces in the trivial homology class in some manifolds. And this is important because if you're gonna try to produce exotic surfaces in s4, then they're certainly going to have to live in them. Yeah. They are genus 1. I think that there are genus G surfaces in some big manifolds. In particular, exotic, sorry, exotic genus 0 surfaces in bigger manifolds. In particular, it's certain that exotic manifolds stabilize with S2 times S2. But there's sort of still a lot we don't know here, so we don't know whether there exists exotic in s4 of any genus. We don't know whether there is, for example, an exotic CP1 in CP2. So this is sort of a Pi 1 Z setting. This is a no Pi 1 setting. And we also don't know, like, I don't know, more globally, like, given, if you're given some random four manifold, can you find an exotic closed surface? So these problems are, especially these apparently very hard for closed surfaces, but something that happens a lot, or sort of a common theme for smooth manifolds is that if you work with boundary, things get easier. So what I want to do with the rest of the time is tell you that we know answers to questions like these for smooth manifolds with boundary and try to sketch a proof there. Questions before I do. The main result that I sort of want to tell you, and in fact it's the result that I'll sketch, is this really wonderful work of Kyle Hayden in 2019, and he showed that there exists exotic D2, embedded in B4 with Pi 1 Z. And he doesn't write it down, but his work also shows that there are exotic D2 in punctured CP2 with no Pi 1. So that's kind of an answer to these two questions in the setting where you have boundary. And with Mark and Anthony, we answered sort of a version of this question in the setting with boundary. So let's say for any two-handle body X with S3 boundary, there exists exotic D2. And all of these theorems can also be sort of kicked up to surfaces of any genus. This is kind of simple. The rest of the time, I want to outline, in fact, a proof of Kyle's result. Our proof relies on what Kyle does and then you sort of need to do more. But the techniques Kyle uses are, I think, really beautiful and also sort of very common in the literature about exotic manifolds or surfaces with boundary. So let me sort of try to give you a flavor of what that looks like. Questions before I do. So let me start off by reminding you of a couple of facts. This notation, the dotted circle, this is used to denote S1 times B3. And if we have a dotted circle and let me just say zero frame two-handle, linking it as the half-link, well, this is B4. And the boundary of this is exactly the zero surgery on this half-link, which is S3 as you know. So you know this, but Kyle's argument is really just this sort of beautiful little trick with this sort of very standard stuff. So Kyle's argument is very straightforward. What we need to do is, one, build a pair of disks with pi1z and such that they are topologically equivalent. And then we're going to show that their complements are not defiomorphic. And this will in particular show that their can have been a defiomorphism of pairs and so they can't be smoothed in. So I'm gonna build some disks for you explicitly. Let me start off by drawing a picture. So what we have here is a four-manifold built out of a one-handle and a two-handle. And while I've drawn a circle in it and I'll explain that in a second, if you squint at this picture, you'll see that while it looks messy right now, in fact, the one and two-handle can be isotoped just to look like this. So this is really just a picture of the four ball. So this green curve that I've drawn, that's some knot in S3. And in fact, we can see in this picture that that knot in S3 bounds a disk in the four ball because you can just take this little disk and push it a bit below the two-handle. So there's some disk here. So in fact, instead of thinking about this as a knot, let's think about it as a disk. And well, let me leave it like that for a moment. So this is gonna be one of our disks, D. To describe the other disk, let me again draw a picture quickly. I'll ask you that question in a second. Okay, so this should look like exactly the same picture I had before. And again, I'm gonna think of it as a zero and a one-handle, but I'm gonna order them in the other direction because in your head, you checked that this was a hop-link before, I'm allowed to put the dot and the zero in the other order. And that will again give me some sort of convoluted looking diagram of the four ball. So again, this green curve, this is some knot in the boundary of the four ball. And in fact, it's exactly the same knot as we had before, right? Because the boundary of this four ball is like bang on identified with the boundary of this four ball. They have exactly the same surgery diagram. So these are the same two knots. And then something that's a little bit less obvious in this picture, but still true, is that this knot here in sort of this four ball, this also bounds a disk. And the way you see that is that if you close this up right here, you would be able to drag the two-handle around so that this only had intersections with the two-handle and then it'd be able to bound a disk for the same. So this is this pair of disks in the four ball. They have the same boundary. And it's not too difficult to check that indeed their complements have Pi 1z. We know how to describe their complements. It's actually just to think of these as themselves dotted circles. And well, in this case, for example, you can then just read off Pi 1 and n. It turns out to be z. So these two disks, they have the same boundary. They're both in the four ball. They both have Pi 1z. And then you can just appeal to the classification to the side that they are indeed topologically equivalent. Let me break for questions there about the construction. So too. Now I just have to show that these complements here, which are in fact that the manifolds whose handle diagrams are drawn here, I just have to show that these are not defiant morphic. The first thing you want to observe, and this is something you can just sort of check using Snappy, because we know a whole lot about three manifolds. You check that the math and class group of the boundary of, well, either of these surface complements, check that this is trivial. So what this tells me is that if I did have a defiant morphism, it would have to restrict to only this, it would have to restrict to this one map on the boundary. And in fact, I can reasonably call this map the identity map because I have this, exactly the same surgery diagram of these two. So schematically what we see going on here is that if we had a defiant morphism on the boundary, it would have to be the identity. So what that tells me is that if I had some homology class in the boundary, then I could take slightly bigger manifolds, which are built by attaching two handles along that homology class. Let's say this is W and this is W prime. And these would be defiant morphic because I can just use the identity map on a handle. So this seems a little bit gnarly. We wanted to show that these manifolds weren't defiant morphic, and what I've done here is I've made them bigger. That maybe seems unnecessary, but what's really great about this step is that these manifolds now have some H2 if you add enough two handles. And in particular, maybe you wanna add two handles here and here. Now, as Andres told you the other day, the genus function steps in as a really great way of showing that these manifolds aren't defiant morphic. So for example, you can ask, let's say any constructive four-manifold topologists to show you that there's a sphere-generating homology here, and you can ask Paolo to show you that there's not a minus one sphere-generating homology here. So I'm gonna stop with this. This is this sort of very classic argument, and Andres is kind of an extension of this. Well, I guess I mean, I have surgery diagrams of both discomplements, which are literally the same as surgery diagrams. So if you feel very strongly that I not call it the identity, that's fine. I'll call it the point-wise fixed map surgery diagram of the boundary. Yeah, it's because, oh, sorry, I'm supposed to repeat the question. Why did I draw this looking like a front projection? And the reason is because once you've added some two handles to this discomplement, it's easy to check that this is a Stein manifold, and therefore it's not allowed to have a minus one sphere. I guess it's just kind of an augmentation of the standard Mazer cork. Maybe we need to get a little more creative in this field, but something we use all the time to produce exotica is this little contractible guy. And what Kyle has essentially done is said, instead of this clasp, I'll think about kind of removing my disc right there. Now you don't want to use this because it turns out I don't have to help pi 1z. And it also has the mapping class group of this complement once you appropriately surgery is not trivial, but then you just wiggle this example to try to have pi 1z and trivial mapping class group and that's a solution. So the question is how do you improve this argument to prove the more general theorem for any two-handle body? So what you do is you start with a handle diagram of your two-handle body and then off to the side, you just drill out a trivial disc for the unknot. Now what I'm gonna try to do is modify that picture until I have something where I have more than one disc. So the first thing you do is you kind of start running the two-handles through that trivial one-handle to sort of set up whatever algebraic topology you're looking for. And then you try to get a picture that looks relatively stined from that by also adding some stuff which has large TB but doesn't contribute to anything else. And then maybe you use a trick like Kyle's to add something where you have a pair of discs and then an argument like this goes.