 Today's lecture will be moving beyond classical material. I want to show first a somewhat straightforward application of Donaldson's theorem, which took quite a while to arrive. And then I'm going to discuss some enhancements to Donaldson's theorem, which come from Hagar fluorhomology, although I'm not going to really describe how that works. But I'm going to give a couple of applications of those enhancements. So the emphasis is going to be on applications today. And the protagonist in today's lecture is the class of lens spaces. And they're just a great class of examples for the different results I want to mention. But in every case, you can say more than, more beyond just the case of lens spaces. But somehow they're just perfect, perfect set of examples to run through this lecture. So by definition, if I have a pair of positive integers, p bigger than q greater than 0, or relatively prime, I can make a lens space. I'm going to denote it this way, lpq. And this is just going to be defined to be the result of doing p over q frame drain surgery, p over q slope drain surgery along the unknot. And if the denominator q is equal to 1, then this is the trace of surgery along a very simple four manifold, something with second Betty number 1. And I'm going to be interested in the class of fillings of this lens space when q is not necessarily equal to 1. So if you know your Kirby calculus, this space is homomorphic to surgery along this link that I'm writing down, the same link where I've been lazy about the crossings. This link is framed by integers. These are integers, and they're all greater than or equal to 2. And the relationship between these integers and pair pq is given by a continued fraction, the slightly unusual continued fraction, if you've not seen it before. This is the so-called Hertzberg Young Continued Fraction expansion. And what I do is I take p over q. I'm writing it with minus signs, where you might have expected plus signs, which are a little more classical. So any p over q can be uniquely expressed in this way. And I get this nice integer framed link presentation of my lens space. I'm going to, as a shorthand, well, I could call this link BlackboardLP comma q. So it's a framed link. It has n components, a1, up to an, or the labels. And continuing some abbreviations, four manifold, which this specifies the trace of surgery along this framed link. I'm just going to denote by this shorthand. And this has an intersection lattice, which also uses shorthand lambda pq to denote. So in the natural basis, the matrix which represents the gram matrix for the basis that you see described last time. It's just this matrix which records these coefficients on the diagonal. And it has minus ones along the sub-diagonal and minus ones along the super-diagonal and zeros elsewhere. And there's a shorthand for notating a form like this, which looks a lot like the link that I wrote down. I just write down a linear path and I put labels at the vertices in order. And that, to me, encodes a lattice which is freely generated by n classes whose self-pairings are given by the integers above them. And where two classes will pair 0 if they're not joined by an edge and will pair minus 1 if they are joined by an edge. So this is kind of a graphic representation. OK, so those are kind of lens-based basics. All right, so the application that I want to describe was theorem of Pala-Lisko. So if you go back to that list of problems from the course description, we knocked off the first two and we're going in order. The third was a question about two-bridge knots and being smoothly sliced. And I'm going to use this question I'm about to pose as a stand-in for that. The two are very, very closely related, but in the interest of time, I'm just going to pass over talking about two-bridge knots and vice-discs. So the reformulation of that question that I want to ask is simply, which lens space is bound rational homology balls? For which pairs PQ does a lens-based LPQ bound a rational homology ball, not notation I've used yet, but it's equivalent to asking that this is the boundary of a four-manifold W with vanishing second-bedding number. And I'm always going to ask these manifolds I'm considering to be smooth. W should be smooth. I do not care about the fundamental group. Well, if I have a four-manifold filling a lens space, I don't know anything about B1 necessarily. Well, OK, I'll confuse myself. But if I know that B2 of W is equal to 0, then some long exact sequence will tell me that B1 equals to 0 as well. So it's equivalent to being a rational homology ball. So it's the question implied by the answer. OK, absolutely. Sorry, I'm not thinking. So we'll just leave it at that then. This is not really a shorthand. OK, yeah, of course, yeah. So I made a mistake in asserting that this is the same thing as having vanishing second-bedding number, the rookie mistake. And the problem is, of course, if I have a lens space that bounds a rational ball, I can always just sum on S1 cross S3. And it'll continue to bound that space, but that thing's bedding number has gone up. First-bedding number is 1, but the second-bedding number is still 0. OK, this is the question I'm really interested in answering. So I'll give an example. So actually, on the homework, I saw a construction to show that the lens space L41 is the boundary of a rational homology ball. A somewhat more interesting example would be the lens space L95. This is also the boundary of a rational homology ball. In fact, it's the boundary of a four-manifold, which is built by taking a single 0 handle, a single 1 handle, and a single 2 handle. And I'll make it an exercise to work through a Kirby calculus description. I'll give for that later. But I wanted to point something interesting out, which is, let's call this four-manifold W. You see, given this, I can take that trace manifold whose boundary is L95. And I can glue it. I can close it up, like we did in that example at the end of last class. I can glue it up by plugging on this rational ball. But to actually close it up to something orientable, I'd better reverse the orientation on W. So I can glue the boundaries of these by an orientation reversing diphthalmorphism. And this four-manifold I've formed, which I'll denote x, is now closed, smooth. It has second-bedding number equal to the second-bedding number of that trace. And therefore, it inherits positive definiteness from the positive definiteness of this trace. Because as an exercise, you can prove that this lattice, this linear lattice, is definite. Lambda pq, positive definite. So as a sanity check, the lattice, L95, how do I sort out what that is? I take 9 fifths, and I have to write it as a continued fraction. So I think, how many times does 5 go into 9? It goes in once, but I have to overshoot. And I get 2 minus a fifth as my Heursebrook Young continued fraction. So my lattice graphically represented this way. And again, that means that I have a class x and a class y, whose self-pairings are given by the numbers above them, and whose pairing together is minus 1. So this steady number is 2. And now that I have a closed, smooth, positive definite four-manifold, I know what its intersection lattice is going to be. By Donaldson's theorem, it has to be isometric to this lattice we denoted I2 last time. This is just the standard Euclidean lattice, which I can emphasize by writing an orthonormal basis for it. Just like you usually write orthonormal basis vectors, E1 up to En. So what we'd be saying then is that the lattice on this guy must embed into this lattice. And we can check that that indeed is the case. So I could map x to, say, E1 plus E2. And I could map y to, well, does somebody want to help me? I need a class whose self-pairing is 5 and has pairing minus 1 with this class. I had to prepare this in advance, but I'll do it now. Oh, the question is, if I'm trying to embed this lattice as a sub-lattice of the standard Euclidean lattice, which I'm going to think of as the integer points in the process of trying to draw up. And so I know x has to map to an element in this lattice whose self-pairing is 2. And really up to change of basis, there's a single choice for what x could map to. The comment was that everybody failed and has to come back next year. Of course, the inner product of these two is plus 1, not minus 1. So you just wanted to negate that. That would be the embedding you really wanted. Hope I've given myself enough space. So the grid points are the points of I2. And I just embedded this lattice. This was my embedding of lambda 9, 5. So I have this vector E1 plus E2. And I have this vector E1 minus 2E2. And then the lattice that they generate. Did I get it? Oh, wait, E1 minus 2E2. So here? OK, thank you. So I get all multiples of this one, so the origin. So I'm getting all of these points. This is the sub lattice embedding. And I got that one. And I'm going to get everything on this diagonal line by adding that other vector to it. And then its negative is in here. So I get that right. Similarly, another night move takes me here. And a night move takes me here. So those would be the points that I get. OK, so that's a kind of a confirmation. Let's sanity check. This thing ought to embed. A non-example, though, would be the case of 2513. I'm going to argue that this is not the boundary of a rational homology ball. I haven't confirmed that this is the boundary of a rational homology ball. But I'm telling you that it is. And as a sanity check, I'm saying I could glue it on to this trace and get a closed, smooth, positive, definite four-manifold. I know what its intersection lattice is. So I know that the intersection lattice on this one has to embed. And I confirm that it doesn't, but feeling good. And I'm going to try to reverse that trick to show that this lens space is not the boundary of a rational homology ball. Just to move a little bit more quickly, you see if it did. So suppose, by way of contradiction, it did. OK, then I would make this gluing as before. I would take now the trace of 2513 and glue it on to minus W. And the same story would hold. I'd be getting a closed, smooth, positive, definite four-manifold. Well, let's check what the intersection lattice is on this. To do that, I have to write 2513 as an infraction. Fortunately, again, it's pretty short. If this is my closed, smooth, definite four-manifold, by Donaldson's theorem, this is isometric to I2. And so what I'm looking to do is to embed this lattice 213 into this Euclidean lattice. As before, there's really only one choice for where X could map. It has to map to E1 plus E2. But is there any place for Y to map? I'd like to say not to say that this thing does not embed. And that will contradict being the boundary of a W. I believe that's correct. The question is, isn't 139 plus 4? It might be a problem, yes. Which is, what could Y map to? So it's a morning of mistakes. That embedding would look like lambda 2513. I've got the same set of points embedding here. But now, 2E1 minus 3E2 is in my lattice. Similarly, minus 2E1 plus 3E2 is in my lattice. I'd be getting these circled points. Yes, a remark. Well, I know that X, the question is, why does X map to E1 plus E2? I know that X has to map to something in this lattice that has self-pairing 2. And so I'm trying to write 2 as a sum of squares. So it has to be plus minus 1 and plus minus 1. And after changing bases, I could just assume it's like that. That would appear to be a problem as far as ruling this thing out from bounding a rational ball. But if I take this lens space and I reverse its orientation, I've got another lens space. And all I do is I take the Q and I subtract it from P. And if this lens space bounds a rational ball W, then this lens space would have to bound the rational ball with reverse orientation. To show this statement, yes, there is the way to see this. Quickly, in general, I could add, well, I mean, it's a Kirby calculus exercise. Let me say that really quickly. I'll be in better shape to answer it more completely later. So the good news is that I can try to play this game again. So let's instead take this trace and glue on, I guess it's going to be plus W now. And I play the game I did before. I write out this continued fraction, 25-12ths. And what you find is that it's 3 minus 1 over 2, minus 1 over 2, minus et cetera, minus 1 over 2. And there are 11 of these 2s. The linear lattice would be of this form. And I'd be asking this to embed in the standard diagonal lattice of rank 12. And again, you can kind of argue where these classes must map. This element of self-pairing 2 would have to map to, say, e1 plus e2. And then the one before it would have to map to something like minus e2 plus e3, and so forth and so on. And by the time you get to here, you basically force how all of these two classes have to map in here. And what you'll discover is that there's just simply no element of self-pairing 3, which has pairing minus 1 with this guy, and 0 with all of its predecessors. So in fact, this embedding does not exist. Embedding does not exist. It's supposed to be i, standard diagonal lattice, rank 12. Yes? That shows it. This doesn't work, but. If this didn't work to rule out this thing from bounding a rational ball, this approach, but there was a trick to reverse orientation and get a lens-based of opposite orientation, which also bounds this nice positive definite trace manifold. By gluing the opposite rational ball onto it, we got a contradiction. So we concluded then that this does not indeed arise. We cannot, in fact, fill this lens-based 2513 by a rational ball. And the reason we know that it has to embed an i12 is, again, from Donaldson's theorem. And this is exactly what Liesch approved. He showed that the lens-based LPQ is the boundary of a rational homology ball, if and only if. The lattice lambda PQ embeds with full rank into the standard diagonal lattice and the linear lattice for the orientation reversal embeds with full rank into a Euclidean lattice, diagonal lattice. So if you have both of these embeddings, then it turns out that this lens-based does indeed bound a rational ball. And furthermore, although we're not really going to use it, but he works out the numerical criteria for this to occur. So this happens if and only if there are integers m and k, such that P is perfect square, m squared. And either q or P minus q is one of two forms. So it could look like mk plus 1, where the GCD of k and m is 1 or 2. Or it could look like k times m minus 1, where k is an odd divisor of either m minus 1 or 2m plus 1. So like I said, it's not so important to know that. It's what comes out of the combinatorial analysis of analyzing which of these linear lattices embed. And so you can imagine how this goes is he checks for which pairs PQ we get these embeddings and then finds that the numbers have to be of this form. And then he has some recipe for given values of this form producing rational balls, which fill that lens space. Was that lens space LPQ bounds a positive definite negative definite. I mean, we took the orientation reversal of LPQ and said that bounds a positive definite for manifold. That's how we're using the second bit. And there are many corollaries to this, surprisingly. So a corollary, which I will put as an exercise, this was a result of a Chateau and Larson from a few years ago, is that if you have a lens space and it's rational homology co-bordant to an integer homology sphere, then in fact, it bounds a rational ball. So it's rational homology co-bordant to the three sphere. I didn't really expect it, but it really is the source of a lot of refinements and other applications. This one, I think, is an especially nice one that you can state quickly and actually figure out a proof of with a hint. Yes, Paola? Full rank meaning that, what does it mean that this embedding is full rank? It means that the sublattice that you have has rank equal to the rank of the lattice that it embeds inside of. OK, the remark was, N1 is just the rank of this lattice and N2 is just the rank of this lattice. Say it one more time. OK, the question is roughly what goes into the construction of this rational homology ball, right? And the answer is that they correspond to the branch double covers of two-bridge links, which bound Euler characteristic one surfaces in the four ball. So you take the branch double cover of this surface, which in the case of a knot is going to be a sliced disk. And that's going to be a rational homology ball by an argument of Casson and Gordon. That was the part that I was kind of skipping over. But it's a great point that the construction of these rational homology balls really ties in with a construction in knot theory. So there's an enhancement of this. I use this made-up word to describe cubicity. So here's a theorem. Let's suppose I have a free manifold, which bounds a rational homology ball. And in addition, it bounds a really nice four manifold. So it's the boundary of a four manifold x, which is not only positive definite like we've been doing in these examples, but is furthermore sharp. And I'm not going to say what that is. But this is a condition. I think I'll describe it in the next lecture. But this is a condition coming from Hagar flow homology. Example would be if x was one of these traces that we're talking about associated to a lens space. Those are examples of not just positive definite four manifolds, but sharp ones as well. Then the intersection lattice on x has to embed to the diagonal lattice of rank n, where n is the rank of lambda x. That's exactly the obstruction that we've been using coming from Donaldson's theorem. But something more is true. For any point that I choose, I'll call q. For any point I choose in this lattice, I n, I take the unit cube based at q. This cube will necessarily contain a point of the lattice. Well, I'll write it like lambda. And lambda is the image of the embedding. So not only does the lattice embed in this diagonal lattice, but it's relatively dense. It hits every single unit cube with integer coordinates, with integer vertices. So if I look back at these examples, this was the case of the lens space L95. It bounds a rational homology ball. And it's lattice embedded with these yellow points. I'll try to shade them in. And if you stare at this, you do notice that every single unit square in this case that you pick in this picture hits the lattice in at least one point. Sometimes more than one point. On the other hand, for this lattice 2513, there was really just this one embedding up to automorphism. And you notice there are empty squares. Here's an empty square. So it failed to hit that unit square. And so simply on the basis of this embedding, this unique embedding, failing this cubicity condition, we can rule this lens space out from bounding a rational ball without playing Liska's trick of looking at the mirror and looking at the orientation reversal. Repeat the argument is the question. I'm saying that I can use this theorem to rule out this lens space from bounding a rational homology ball. And my argument is I look at this sharp four-manifold, which fills my lens space. And I'm going to argue that it doesn't have an embedding. And if we go back to our previous analysis, we saw there was really just one embedding of this lattice into I2, up to automorphism. And it looks like this, but these are the points you want your eye to draw attention to. So there's really just this one embedding, but it fails this cube condition. Here's a cube that doesn't contain any point of that sublattice. So this lens space couldn't bound a rational ball. It's a bit of an odd condition. I don't know why you would ever come up with this condition before having this application for low-dimensional topology. It just comes out naturally from Donaldson's theorem in tandem with some more information coming from Hagar'd Fleur Homology. The next application I'll describe is to Dane Surgery. And again, I'll let lens spaces play a antagonist here. So the question I want to ask here is which knots in the three-sphere have a Dane Surgery to a lens space? Well, obviously, the unknot does, because that was how we generated the class of lens spaces. Any of these are lens spaces. A more interesting example some people might have looked at on the problem set yesterday was if I take the right-hand trefoil and do plus-5 surgery along it. This ends up being homeomorphic to the result of minus-5 surgery along the unknot, which, thinking of it as 5 over negative 1, and I can, by the thing I did that didn't really answer that one question, I can change this to 5 over 4. This is a Kirby calculus description of a lens space. So that's a bit of a surprise that some non-trivial knots do have lens space surgeries. And actually, all torus knots, torus knots have lens space surgeries, all torus knots do, and many more. And the production of examples culminated in a conjecture due to John Berge from about 1990. And if this is the question, then Berge gave a conjectural answer. And the answer is the Berge knots, which I have a more intrinsic description. Berge knots, by a little theorem, are the same thing as so-called doubly primitive knots. Suffice it to say, it's a pretty exceptional class of knots. So these are knots that you can draw on the boundary of a genus to Hagard splitting. The standardly embedded genus to handle body in the threesphere, they're knots that you can draw on that in a particular way so that they're not included into the handle body in the inside or the handle body on the outside. This knot represents a primitive element of the fundamental group. So it's a primitive element. Hence they're called doubly primitive inside and out. It's not so important you know what that is. But he made a very influential conjecture, which is still open. Another piece of information I'd like you to know is that torus knots have lens-based surgeries. But it's a major theorem that if k is a knot and it has a lens-based surgery and k is not equal to a torus knot, then the surgery that you do has to be an integer. So this is a consequence of the famous cyclic surgery theorem of color Gordon-Lucy and Shailen. Yes, it has to stay on. So these knots stay on the boundary of a genus to handle body. So well, all torus knots will have this property, in fact. Sorry, repeat. They're knots which can be drawn, the doubly primitive knots, burgy knots, are knots which can be drawn on this genus to Hagard surface. But there is an additional condition I haven't told you, or I did, but not well. Yes, Paul? Picture, please. So the idea is that I have a lens space now, and it's being expressed as surgery along a knot. So think of this as the boundary of the trace of surgery along this knot. So the picture that I would draw would be the trace of surgery along some putative knot whose boundary is this lens space LPQ. This is a four-manifold with second-beddy number one. It's just trace of surgery along a single knot. Ah, OK. The picture we want is a picture of a burgy knot. So burgy is from Wisconsin. Here's this genus to handle body, standardly embedded in the three-sphere amount of practice. But am I that out of practice, though? So imagine these strands run around here in kind of like a cable of a torus knot. And the other strands come over to this side like so. So I might have a knot which is drawn on this genus to Hagard surface in this way. And what's special is that there's a compressing disc for the outer handle body, sorry, which meets this curve in a single point. And there's also a compressing curve for the inner handle body, which meets this curve in a single point. And it seems as though I failed to do that. Yes, would anybody like to correct this picture? See? Not as easy as it looks. Not as easy as it looks. Any loop on this surface, which happens to cross a compressing disc for the outer handle body in a single point, and which happens to also cross some cutting disc for the inner handle body in a single point. One disc for the outside and some other disc for the inside. As long as you can find those special discs, that gives you doubly primitive knot. But where this is leading is I could actually draw an example. I could build the trace of surgery perhaps as well. Again, close this thing off with a piece that I know well, this trace, P, P minus Q. So what I'm writing here is I'm supposing that my lens space is a knot surgery. And I'm saying if I have that, then I can glue up and get this big four manifold, which for I think the third time now, this is a closed, smooth, positive definite four manifold. So the intersection lattice on X is going to be diagonal. And the rank of this lattice is now what I'll write as N plus 1, where N is the rank of this lattice. I have all of these classes in here. I have these N classes that I understand how they pair. And then there's this special surface over here generating the homology in this piece. And together those classes rationally span the homology of this four manifold. So what I learned then is that the lattice, lambda P, P minus Q, must embed to this diagonal lattice. And its orthogonal complement is one dimensional. So I'm conflating this lattice with its image under the embedding. But this is going to be spanned by a single element. I'm going to write as sigma, where sigma you should think of as being the homology class that capital sigma represents. So an example would be we saw that L5, 4, 5 surgery along this knot. And so it means that the lattice lambda 5, 1 is just represented by this picture, should embed into the standard diagonal lattice of rank one more. And if I write my orthonormal basis vector as E1 and the other one is E2, we have some practice with this already. This class X maps to, say, 2E1 minus E2. That's an embedding. And the orthogonal complement, I'm just finding a primitive vector orthogonal to this one. And E1 plus 2E2 works, which I'll just write as a vector 1, 2. Another example would be L33, 2. I'm going to ask whether this could be 33 surgery along some knot. Well, if so, let me change this to 31. If so, then lambda of 33, 2. I have to do my continued fraction expansion. Then I get this lattice, because it's 17 minus 1 half is 33 halves. But this would have to embed in the diagonal lattice of rank 3, which is spanned by some classes, say E1, E2, and E3. And now back to the audience. Is it possible for me to map in classes X and Y to this lattice, that those earrings could take, say, again, 4E1 minus E2. And this one could be E2 plus minus E3. And what's the orthogonal complement? E1 plus 4E2 plus 4E3, which in vector notation is 1, 4, 4. So this lattice embeds. So it stands a chance, then, of being a knot surgery. Except yet again, there is an improvement on what Donaldson's theorem tells you. So it's a theorem. The length space is a knot surgery. Then the lattice lambda p minus q has to embed with co-dimension 1 into the standard diagonal lattice. I record the co-dimension 1 property by saying that the orthogonal complement is spanned by a single class sigma. And up to change of basis, if I write sigma as a vector in terms of entries sigma 1 up to sigma n plus 1, this vector is a special kind of a vector called a change maker. And again, this is a condition that I don't know why you would have studied, apart from it popping out of some Donaldson-Hagard-Fluer magic. But what this means is you, since we're out of time, you could do a demonstration. But you have to imagine these as values of coins. So imagine these as values of coins. And you want to board a bus. And the bus demands you pay an exact change. And you have enough money, but you're wondering whether you have exact change in any amount up to the total value of your coins. So what we're asking is that for all values 1 up to the total sum of these values. These are all positive values. I can make exact change in that amount. There exists a subset such that if I just add up my coins indexed by the elements of that subset, I get that amount. That's the fee to board the bus. And so when we look back at this example, if you have a one cent coin and a two cent coin, you can make exact change in any amount up to three cents. But if you have a one cent, a four cent, and a four cent coin, you have nine cents. But it's impossible to make two cents as change or three cents as change. So this is not a change maker. So that forbids this lens space from being a not surgery. And since we're over time, I'll wrap up the conclusion next time. But long story short, this is a necessary and sufficient condition for lens spaces to arise by not surgery. So that gives some evidence for the Berge conjecture. I'll put some polish on it next time. Thank you. The question is whether there is an easy way to see that the Berge knots admit a lens space surgery. And I'd say it's a quite interesting exercise. The answer is yes. It's a very pleasant exercise. But it involves taking the stained surgery solid torus and the space that you're surgerying it into and cutting them up in a creative way and being able to decompose it into two solid tori, which tells you that you're in a lens space. So there are nice details to work through there. Another question. You should look for a paper I wrote with Slavin. Oh, I'm sorry. So the question had to do with where to learn about cubicity. So a little argument on top of what's in that paper gives this cubicity condition. And there should hopefully be a paper with cubicity explicitly in it possibly this very month on the archive. That is also the question is whether this change maker condition is anywhere. And yes, this is in a paper from 2012 or 13. I bet if you googled my name and change maker, you'd a different embedding. OK, so to answer the question, to repeat the question, that's what I was going to do, the question is how did we actually show that this lens space is not equal to 33 surgery along some knot? Because what we did was we wrote down one embedding and we checked that the orthogonal complement's not a change maker. But that doesn't say there couldn't have been some other embedding of this where the orthogonal complement actually is a change maker. And there are only so many possibilities you can really check. I mean, why? It has self-pairing too. So it really has to look like a vector like that. And I think you end up finding that that's really the only possibility for X can map. So no, we didn't complete the argument, but we just did. Question is whether there are non-trivial knots with different surgeries to lens spaces. And the short answer is yes. In fact, all torus knots have that property. The PQ torus knot has the property that P times Q plus or minus 1 gives a lens space. But there are also hyperbolic knots with that property, like the Fintechel's third minus 2, 3, 7 pretzel knot, or 18 and 19 surgery, 18 and 18 plus or minus 1 surgery along it to the job. So those are interesting examples. 0, 1, and 2, yes. OK, the question is about the complexity of the rational balls filling lens spaces in the case of Liska's work. And the answer is yes. With a careful reading, first done by Jake Rasmussen, you can confirm that all of his lens spaces bound a rational ball built from a single 0, 1, and 2 handle. I mean, it's really direct. It's in that paper. But to think of it that way, I learned from him. So that's an entry into a very fascinating question, which is complexity of rational ball fillings. And so anytime a lens space bounds a rational ball, it bounds this one with the simplest handle decomposition you could have. Draw the surface, but thank you, Jake. Thank you.