 All right. Good morning to be the last lecture in this series about lattices and low-dimensional topology. And I don't have a title for the lecture. But we'll do more applications and discuss a little bit techniques beyond those which go into proving and using Donaldson's theorem. So I just wanted to recap a couple of applications pointed out in the last lecture. Two natural problems. When does a manifold bound a rational homology ball? When does a manifold arise by integer-dain surgery along a knot in the three-sphere? And we can specialize both questions to the case of lens spaces. And these are cases where now we have complete answers. And we have them, thanks in large part, to advances in gauge theory and flow homology where lattices play a role. So the first theorem is not exactly what I had stated in the last lecture, but it is also true. So the condition that a lens space bounds a rational ball is equivalent to an associated lattice, this linear lattice, coefficients corresponding to the continued fraction expansion of P over Q. If and only if that lattice has an embedding into Euclidean space in which it hits every unit cube with integer vertices. So it's ubiquitous in the sense that it hits every cube. That's the reason for that compound. Related theorem is that for a lens space, being integer-dain surgery along a knot in S3 is, and again, I don't think I've stated the equivalence, but it is equivalent to that lattice embedding into, again, a diagonal lattice with co-dimension 1 and so that the orthogonal complement is a very kind of a vector, so-called change maker, kind of a curious combinatorial condition. And related to that, I wanted to show the picture of one in the family of doubly primitive knots. These are the knots conjectured to be precisely the knots in the three-sphere with lens-based surgeries. So you see here a curve drawn on a genus-2 surface, and that genus-2 surface is embedded as a Higard surface in the three-sphere. And the special property about this curve is that I can pick a loop on this surface which bounds a disc in the exterior handle body, and it meets the curve in a single point. And I can also draw a disc, a loop, which bounds a disc, the interior handle body, and it also meets the curve in a single point of intersection. Yes, question? Right, the question was, is it necessarily the case that the disc chosen for the outside and for the inside will, as in this picture, meet in a single point? And the answer is no, that's not part of the definition. And for other examples of doubly primitive knots, these discs look a lot more complicated with respect to one another. So a big part of Berge's project was actually enumerating all of the different classes of doubly primitive knots. And they fall into various families. This is a cable of a torus knot. So certain cables of torus knots, two cables of torus knots fall into this family. But others as well, curves you can draw on the fiber of a genus one, fibered knot in the three sphere, and some others, including some sporadic families that don't really fit into, well, they're just sporadic families. So maybe a remarkable outcome of the theorem is that I'm asserting a perfect obstruction. And what actually gets sorted out is a rather complicated list of lens spaces. I mean, I told you what the pairs of PQR that Liska found correspond to lens spaces bounding rational balls. And it's already a somewhat complicated list, and this one even more so. So it's remarkable that the information from Donaldson's theorem, from Lattice's, from Higart-Fleur homology, is exactly picking out a really complicated set. It's just kind of a recurring miracle that these lattice theoretic techniques seem to pick out exactly complicated phenomena. And I wanted to give one more example. This is to do with unknotting. At first, it doesn't look anything like what we've been discussing. So I'll draw a picture of an alternating diagram of a two-bridge knot. Two bridges referring to the fact that this has a diagram where you see two maxima and two minima, two local maxima and two local minima. And alternating in that, if you trace the diagram with your finger as you go along it and you encounter crossings, you alternately encounter them over, under, over, under, et cetera. So it's a fact that any two-bridge knot admits an alternating two-bridge knot diagram and alternating knots greatly generalize the class of two-bridge knots. This is a non-trivial knot. If I circle this crossing, I'm going to call my diagram D. And if I make a crossing change, so I'm going to even change the color here. So if I just keep the knot the same outside of that window, but I change that crossing, quickly draw the rest of it. OK, well, it's no longer alternating. It's almost alternating. And you can check that actually making that crossing change unknot this knot. So you can do an isotope which exhibits that this diagram represents the unknot. It's always kind of interesting, fun to see knot diagrams simplify. So I promise it works here. So alternating diagrams, alternating knots, they have a diagrammatic definition. They have these alternating diagrams. And there is a question as to which alternating knots have a knotting number one. Which alternating knots can you unknot by changing the single crossing? And note that's different from asking which alternating diagrams have an unknotting crossing in them. Because a priori, you might have a knot with an alternating diagram, but you have to put it into a very complicated non-alternating diagram before you're able to spot a crossing you can change to unknot it. But it was conjectured that alternating knots are so special that if you had one with unknotting number one, it will display an unknotting crossing in an actual alternating diagram. And that is indeed a theorem of Duncan McCoy. So if k is an alternating knot with unknotting number one, meaning it can be unknotted by changing a single crossing, then any alternating diagram be as an unknotting crossing, which had been, I think, a 25-year-old conjecture at the time he had proven it. And moreover, D prime. So I'm just defining D prime implicitly here. It's the diagram that you get by changing that on knotting crossing in D. So D prime simplifies through the round unknot diagram via a sequence of moves. And I'll show those to you. One move is what's called flight. Flight is where I see a bit of knot diagram, which I conceal by a disk labeled with T for tangle. And out of T come a couple of strands to the left. And out of T come a couple of strands with a cross in them to the right. And what I'm allowed to do is to, you can picture doing this physically, grabbing this tangle here and rotating the top towards you 180 degrees round. That'll have the effect of basically moving that crossing over to the left side and turning the tangle upside down. And these strands join up to the rest of the knot diagram. It is the same knot. So I shouldn't be changing the knot type in any of these moves. So Flights were introduced by Tate. It's a Scottish word meaning roughly to turn something inside out. And it's the subject of a famous conjecture of Tates from the 1800s, which is now a theorem of Manasco and Thistleplate. So this is a well-known move. Notice it doesn't change the complexity of the diagram, at least as far as crossing numbers concerned. There are as many crossings in the diagram before as after. Question, Dominic. It means that I can change a single crossing in some diagram of the knot and thereby unknot it, get an unknotted curve. So in this diagram, I changed this crossing. It was just one crossing change. And I got a new diagram, which is actually a diagram of the unknot. I guess the question is about what is the content of the theorem? The content is that if you have a knot, which you know you can change by unknotting a single crossing in some diagram, you might not know that you can necessarily change a crossing in an alternating diagram of that knot. Evidence for the claim that alternating diagrams of alternating knots are the best diagrams. They already manifest all of the subtle properties a knot could possess. Are we ready for the questions? We're ready for the next moves. OK, that's not all. There is a move where I see something that looks like a righto meister 2 move, except that there is another strand of the knot behind it. And you see, I'm going to highlight this crossing. That's the one that if I were to change it, I would be looking at an alternating diagram. And what I can do is I can affect that righto meister 2 move in the presence of that other strand. This is called a tsukamoto untunging. And that move you see simplifies it. This diagram on the left had two more crossings than the one on the right. Lastly, or unultimately, you can do a simplifying righto meister 2 move. And you can do a sequence of simplifying righto meister 1 moves. So this gives an easy test then to see whether an alternating knot has a knotting number 1, the question is whether the moves are directional. And yes, the moves are all unidirectional. So these are all monotonically decreasing and crossing number. So flyping is the only move that keeps the crossing number the same. So you might look at a diagram, look at all the possible ways to flyp and there might be many. But then once you spot a tsukamoto untunging or one of the righto meister 2 or 1 possibilities, you do that move and you reduce the number of crossings. The question is, can you do any sequence of these moves and is it guaranteed to a knot the knot or do you have to do them in a special order? If ever you see a possibility of doing a tsukamoto, any one of these last three moves, you do it. But if you don't see a way to do one of these last three moves, you might have to flyp around a bunch first. If you flyp around and you're unable to do any of these moves, then it's telling you you weren't looking at an almost alternating diagram of the unknot. But if you do a sequence of flypes and suddenly you do spot one of these moves, you do it and then keep going. If you don't spot one of these moves and you keep flapping, so the question is in another diagram, so I'm stressing any alternating diagram of this one knot that you're looking at, well, I did a process, that's my fault. Do you want to try it one more time? Yes. OK, it's a slightly subtle question about getting stuck doing this. And I think the most optimistic interpretation of what I've written works. It's almost impossible to get stuck. I think it's impossible to get stuck. The question is, why are these better than just righto-meister moves? The reason is that these moves are always monotonically simplifying the diagram down. Whereas if you, in general, are trying to simplify a diagram of the unknot down to the round unknot diagram, you might have to increase the number of crossings. You might have to do a righto-meister one or two move that increases the number of crossings and things can get out of control. So this is a monotonic procedure. And a consequence is that if you have, if D is an alternating diagram with, say, n crossings, then there exists an algorithm to either locate an unknotting crossing or else certify the not, does not have unknotting number one. Unknotting number one. And the running time of this algorithm is a very small polynomial. I don't actually know. But I think maybe, I think it's quadratic to test a given crossing and then, therefore, cubic. But don't hold me to that. But it's very small. And as a human being, it's a pleasant algorithm to undertake. Question. Right. The question is whether there are alternating knots with unknotting number one. But you look at a nonalternating diagram of it, and it does not have an unknotting crossing. I get that right. And I want to say that the answer is yes. The answer is yes. There are examples amongst two-bridge knots. So you can give an example of a two-bridge knot with a knotting number one and give it a nonalternating two-bridge diagram where you don't actually spot any unknotting crossing. So how does this result tie in with the whole business we've been discussing about lattice embeddings and these other pair of results? Well, the intermediary is this. So there's what's called the Montesinos trick. What the Montesinos trick says is that if k is any knot, not necessarily an alternating knot, I have any knot in the free sphere. And it has unknotting number one. So it can be unknotted by a single crossing change. Then there's an affiliated space, which we haven't really discussed very much. But if we had more time, we would have talked about the branched double cover some more. So this space sigma k, which is defined to be the double cover of S3, branched along k. This is the way to make it an English sentence. This space is homeomorphic to some half integer chain surgery along some other knot called k prime free sphere. The picture then is that whatever knot I'm studying, I'm looking at its branched double cover. And if, indeed, this knot has unknotting number one, then there's some other knot, k prime, that of k prime hanging off to prove that it's a knot. And I'm doing a half integer surgery along it, d over 2. d's an odd number. I'll write it like this. 2d minus 1 over 2. That's the surgery coefficient. So if you want to obstruct a knot from having unknotting number one, a way to do it is to prove that the branched double cover is not half integer surgery along a knot in the free sphere. And that doesn't necessarily sound like you've made life any easier for yourself. But we have a precedent. We have, in these lectures, already results about spaces arising by dain surgery along knots in the free sphere. The more ingredients is that if k is alternating, then it turns out that the branched double cover of k has a lot in common with lend spaces. So the first thing it has in common is that it bounds a sharp four manifold, moreover with either orientation. But never mind. The key quality that we're going to use, or that McCoy uses, is that this space bounds a sharp four manifold. And another key property of these spaces, so I'm assuming something about, I'm assuming my knot is alternating. I'm not assuming it has a knotting number one for these statements. This is really, I think, two separate things. The other fact is that this is what's called an L space. And both of these are conditions coming from Hagar floor homology. And so the machine that's behind this second theorem about lend spaces arising by integer dain surgery, that whole machine applies to this more general class of spaces, branched double covers of alternating knots. For the reason that these spaces are L spaces and they're sharp. So you get very nice lattice theoretic information. What feeds into McCoy's theorem then? So this would be kind of the more technical version of what McCoy does. It proves that if K is alternating, then the branch double cover of K is indeed, oh, I missed a question. Questions. OK, first question is whether there's a relationship between bounding a sharp four manifold and being an L space. And the short answer is no. That there are wide classes of manifolds. In fact, all cypher fibered spaces bound sharp four manifolds. However, they're typically not L spaces. They manifest very complicated Hagar floor homology. So there are spaces that bound sharp four manifolds, which are not L spaces. And conversely, there are L spaces which do not bound sharp four manifolds. Moreover, there are L spaces which don't even bound definite manifolds. Moreover, there are L spaces which don't even bound definite manifolds of either sign, which is a few week old theorem of Daimie and then, depending on how you are supposed to alphabetize the name Miller-Isemeyer, that one, and then Lidman. So there are independent conditions. And then the second question had to do with why branch double covers of alternating links are L spaces. And there are a couple of ways to do that. So in fact, these spaces have Hagar diagrams where the number of generators is equal to the order of first homology. Generators of the Hagar floor chain complex. So the differential vanishes. And you get for free that these are L spaces. And the original proof of Osvath and Sabo that these are L spaces makes use of the surgery exact triangle and crossing resolution, or crossing changes and crossing resolutions in alternating diagrams. So I was just going behind the scenes of what McCoy does. He proves that if you have an alternating knot, then its branch double cover is half integer surgery along some other knot, if and only if the lattice associated to this sharp four manifold I never told you about, so called Gordon Litherland manifold. This space is half integer surgery along a knot, if and only if this is a so-called, for this lecture only, a two-step change maker, meaning that it's a step away from being a change maker lattice. So it's much in the spirit of the Lens-based Dain surgery theorem. And what I just find sensational is that it gives effectively a diagram structure theorem for the knot, the alternating knots with a knotting number one. So you could instead think about starting with the knot and applying any sequence of these moves that always, so you start with a round knot, you do a bunch of these Reitermeister moves keeping you alternating, you do a single Reitermeister two move to make you almost alternating, and then you do a bunch of reverse untonggings and flips and make as complicated diagram as you like and those hit all of the alternating knots with a knotting number one. So it's a great structure theorem with echoes of, you know, graph structure theorems. A question. Well, I'll probably say a little bit more then. What is a two-step change maker lattice? So it means that the lattice admits an embedding once gained into a diagonal lattice. Now it's ranked too larger and the orthogonal complement is spanned by a pair of vectors. I'll call sigma and tau. And tau is a very simple vector. Let's say it's the vector one minus one and a bunch of zeros, n zeros. Sigma is a vector zero, one, I don't know how to index, but let's say sigma one up to sigma n. And this is all a change maker. Well, I guess the whole thing is a change maker, but you have a worthless coin. And there are theorems about other, you know, third quarter integral surgeries and so forth. But I just wanted to, this one has the nicest statement. OK. Now the rest of the lecture is entitled Miscellany. So I just want to, I think at this point, was there a question? Yeah. Oh, sorry, yeah. So what was the relationship between this theorem and the one which I just erased? Right. So if and only if k has a knotting number one. So maybe the way to read this theorem is start with an alternating knot whose branch double cover is a half integer surgery. Then it's a two-step change maker lattice. And then McCoy's real work is looking at when these lattices are two-step change maker lattices. And he analyzes their structure. And he can classify them all. And he can figure out that any time you have one, the diagram that you're looking at of your alternating knot must manifest an unknotting crossing. So it kind of completes the loop. I mean, this was going from the back to the front is the Montecino's trick. And then this is really the hard. This is kind of the floor homology Donaldson step. And this is the combinatorial analysis step. And this is the kind of classic topology step. I mean, it's all unidirectional. So I'll write that in. So this is kind of the lattice magic. Let's just say Hagar floor Donaldson. This is a combinatorial analysis. Sorry for the mess. Make it messier. And this is just a Montecino's trick. So kind of classic topology. So two questions. First, have sort of this lattice theory machine that operates well whenever we have a manifold which bounds sharp four manifolds with both orientations. One question would be, you could ask simply which three manifolds definite four manifolds with both orientations. I think that's an intractable problem. But a version which plays into this story a little bit better would be which instead bound sharp four manifolds with both orientations. They're two multi-part questions. And to refine that even more, to really get this set up, you might want to know which Lenspaces, excuse me, which L spaces bound, well, let me see if I can just squeeze it in here. So which L spaces bound sharp four manifolds with both orientations? And on the basis of kind of the lattice theory, I had conjectured that the only ones, so this is, I had conjectured that the answer to 1c is just branch double covers of alternating links. There was reason from the lattice theory to expect this, which would be kind of a cool classification of the set of branch covers of alternating links. But this is false. So there's a counter example. I'll just say the answer to this is no. Well, the counter example due to William Ballinger. And they have to do with embedded graphs in two complexes in S3. And these counter examples are pretty, to me, exciting because they suggest a wider class of manifolds than we've ever really thought about before to which all our favorite tricks apply and should be investigated. And which of those are integer surgeries along knots? Which of those bound rational balls? It's open terrain. The next question would be, remember, if you have a lens space and it's integer surgery along a knot, then this lens space is the trace of that surgery. And this manifold is simply connected. And it has second-beddy number one. And so you could conjecture, as I did, that if you have a lens space and it bounds a four-manifold w smoothly, which is simply connected and has second-beddy number equal to one, then the reason is it's because that lens space is integer surgery along a knot. How could you give any other examples? But there are counter examples. And as before, there are counter examples due to Ballinger, like get a life, come on. So for example, the lens space 398.75 does bound a four-manifold, which is simply connected and has first-beddy number one. But it's not an integer knot surgery. So the way Ballinger knows that it bounds such a four-manifold is he works in a, he takes a configuration of lines in CP2, three lines in CP2, kind of like we did in the first lecture. And he does a sequence of blow-ups, much like we did in the first lecture. And he gets to a very interesting configuration of spheres and, well, it's not doing a, it's going to be in a connected sum of CP2s. So not actually blow-up, but you know, topologists positive blow-up. So he gets an interesting configuration of spheres in a connected sum of CP2s. And then by removing one of these spheres and doing some smoothings, he gets a configuration of spheres whose neighborhood boundary is a lens space. And what's left over is a simply connected bit with buddy number one. So he has a way of constructing examples. It's a wide class of examples. This is just the smallest one I could read off from his list. And the way you know that it's not knot surgery is, you know, you compare it to Berge's list. Or, you know, you look at all the ways this thing could possibly embed into a Euclidean lattice of rank one more, and you check that the orthogonal complement is not a changemaker. So that's how it's ruled out. So it would be wonderful to know more examples of this. It would be wonderful to know more counter examples. So there are two things I think are very interesting stemming from Ballinger's discoveries. The conjecture after one C is that I had conjectured that the only L space is bounding sharp four manifolds with both orientations are branched double covers of alternating links. And that's false. There are even more examples. Finally, I want to talk about bounding other kinds of definite forms. OK? So we've seen this example of the Poincare sphere a number of times already. We know that it's plus one surgery along the right hand trefoil. And we know that it's also surgery along the E8 link. So it bounds the traces of surgeries of these manifolds. So then what I would say is that P3 bounds both this definite form I1 corresponding to the intersection lattice of the trace of surgery along this knot. And it also bounds the E8 lattice, which is the intersection lattice of the trace of surgery along this link. And by taking as many connect sums as you like, you can get all of these forms, IN plus 1 and E8 plus IN for all N greater than or equal to 0. So while we're not obstructing this space from bounding definite forms, it bounds some. But then we ask, which forms can we actually get? And these are the ones that occur to us from these examples. And it turns out that that's all. So this is the theorem of Kim Freudschoff. I want to say it's from the late 90s that these are all. So P3 only bounds these forms. And so I guess I won't say very much about the proof. But there's a modern proof. Well, his proof was also modern. It wasn't that long ago. But let me put it like this. You can give a proof that uses complementary input about lattices coming from, on the one hand, Hagard floor homology. And on the other hand, instanton floor homology. So the way that in rough sketch, Part A works, so you use what's called the D invariant in Hagard floor homology, which is really a theft of an invariant Freudschoff defined in Cyberg-Witton floor homology, his H invariant, and got renamed. So you don't really need to do this in Hagard floor. You could do it in Cyberg-Witton floor just like Freudschoff was thinking. But you use the D invariant in Hagard floor homology. And this has something to do with self-pairings of so-called characteristic elements in lattices. And if you know enough about D invariants of lattices, what you find is that from this bit of the argument alone, P3 could possibly bound IN direct some lattice lambda chosen from the following list. So it could be, so here N is greater than or equal to 0, could be I1. It could be E8, like we've seen. Or it could be a more complicated lattice. It could be the lattice D12 plus. It could be the so-called short-leach lattice, O23. The list of 15 examples. And this is kind of a lattice theoretic theorem. So here, there's a theorem of Noam-Elkies working to get you this list. But if that's all you knew, there are some more exotic forms here, which could arise. But the input from instanton floor homology, like I said, it gives complementary input. So the instanton floor input has to do instead with self-pairings, not of characteristic elements, but of irreducible elements, lattices. And long story short, you're able to rule out these 13 possibilities. Because these 13 possibilities, each of them, these all contain an irreducible element of self-pairing free. So it violates some bound that instanton floor is giving you. And so using instanton floor, you can cross these other possibilities off of this list and get Vorschoff's theorem. That's sort of the proof sketch. It's not exactly the proof sketch which Vorschoff followed. But it's one which was written down by Chris Gaduto a few years ago. And he showed how to get interesting proofs on it. So for example, if you do this punk or a sphere is plus 1 surgery along the right-hand trefoil. If instead you do plus 1 surgery along the next simplest torus knot, T25, and ask which forms this bounds. It bounds exactly the forms of a diagonal piece, direct sum with a lattice lambda, where lambda is either standard 1 of rank 1. That's what comes from this knot surgery description. It also bounds E8. And it also bounds this lattice D12 plus. And maybe just for show, here's a graphic depiction of E12 plus. What I think would be really interesting to do is to go beyond the case of integer homology spheres, which has been looked at. And try to explore this beyond the case of integer homology spheres to the case of rational homology spheres, which include a lot more examples. Finally, in the miscellaneous category, you can ask which lens spaces and you get by doing surgery along a knot in the Poincare homology sphere. And you can change this to anything you like, but this fits very naturally into this program of exceptional Dane surgery of looking at surgeries between non-hyperbolic manifolds. So this is sort of like the next simplest case, in a sense, beyond which lens spaces can you get by surgery along a knot in S3, which is solved. And as it happens, I'll call this a theorem of Caudal. If this occurs, so modulo a condition on the knot genus, cipher genus of the knot that you're talking about, it turns out that lambda pq has to embed into the direct sum of E8 plus a diagonal form, In. And it's what you might call Jacob calls a hard 8. You could call it an E8 changemaker. So there's some analog to a changemaker vector inside E8 direct sum diagonal lattice. There's an analog to that. And the statement is that if your lens space occurs by surgery along a knot in P3, then that linear lattice associated that lens space embeds as the orthogonal complement to 1. And it's expected, and it's work in progress, that this is a complete obstruction. So expected to be complete obstruction. And so you would get then a great big list of the lens spaces, which are surgery along a knot in the Poincare sphere, which is sort of an analog of the Berge conjecture, which was written down by Mototongue some years ago. So that's an expected outcome. OK, so that ends the miscellaneous section. And I'll wrap up here. The question is, how does the de-invariant count the self-pairings? There's a formula. So there's just, I'll write it really quickly. So the quick answer is, for ease, you assume that you have a definite unimodular lattice. And now you're looking at self-pairings of some vectors. So this is an element in the lattice, but it's a so-called characteristic element. And I subtract the rank of the lattice and divide by 4 for good luck. I look at this set, and I take the minimum value. And I declare that to be the de-invariant of this lattice. So it's a purely lattice-theoretic quantity. And then it turns out if y is an integer homology sphere and it bounds a four-manifold x, which is definite, then the de-invariant of the lattice attached to x bounds the Hagarred-Fleur de-invariant of y, which has just a single spin-C structure because I'm assuming that it's an integer homology sphere. And these are typically negative values by a theorem of Elkies, and you have an inequality like this. So that's a hint of where it comes into play.