 Before introducing the next speaker, let me make a very short announcement. So for speakers, please, in case of questions, repeat the question for the benefit of the online participants. With that said, it's my pleasure to introduce our next speaker, Josh Green, who will guide us from ball feelings to cube timings. Please. Thank you, Andras. Thank you all for being here. I think I've been more excited to attend this conference than any I can remember, at least in my recent memory. So I'm very glad to be here and to participate in honor of Tom. So the question I'll begin with is a well-known open-ended problem posed by Cassin and Herrera, I believe, from 81. And the question is, which rational homology spheres found in here and throughout the talk, I'm always referring to smooth manifolds. Which rational homology spheres found smooth rational homology for ball? It's pretty open-ended. The problem to take a rational homology sphere and determine whether or not it does or it does not, I don't know if that's a decidable problem. And directions usually given to this problem by focusing on a special class of three manifolds and attempting to answer a problem within that class. And starting point for the work I'll talk about today is a partial answer which Palleliska gave. Well, he gave a complete answer. So he solved the problem for lens spaces. And I'll quickly sketch what went into Palleliska's argument. If I have a lens space, it's the boundary of a plumbing manifold. So I take a linear chain like this and I put some integers greater than or equal to 2. And I'm taking the disk bundle of that Euler number over the two sphere and I'm plumbing a bunch of them together in a chain like that. And I get a compact, smooth, four manifold whose boundary, if I chose these according to the Hertzberg Young continued fraction expansion of P over Q, gives me a manifold whose boundary is this lens space manifold x. And if this thing happened to bound a smooth rational homology ball, then I could reverse its orientation and glue it on. And the result then is a closed, smooth, definite four manifold. So by Donaldson's theorem, I know what its intersection lattice is. Its intersection lattice is just a standard diagonal lattice. And it will have rank n, all of the rational homology is accounted for by the space x sitting within. And the intersection lattice on x, which I'm going to denote lambda x, therefore has an embedding into this standard diagonal lattice. So that places a constraint on what PQ can be. This lattice has to embed into the diagonal lattice. And if it doesn't, then you know that your lens space doesn't bound a rational ball. And that's practically a complete obstruction to a lens space bounding a rational ball. So this result, as far as I'm aware, was the starting point of a theme, which is that we have a problem at hand, say, to determine which lens spaces bound a rational ball. And there's a complete obstruction. Well, there's an obstruction which one proves is complete, which comes from an application of Donaldson's theorem and usually comes with a little twist. And the twist in the case of lens spaces bounding rational balls is that, although that's not a complete test, you have this trick of being able to reverse orientation on the lens space. And that'll bound a different plumbing. And then if it bounds a rational ball, then you're going to get a pair of lattice embeddings into diagonal lattices, a full rank. And it turns out that possessing those two embeddings is a complete obstruction to the lens space bounding a rational ball. So if it passes that test, then in fact, there is a construction to show that that lens space bounds a smooth rational homology ball. So that's a theme. And another aspect of the theme is that the examples that pass the obstruction, they tend to fall to a fairly simple iterative scheme or maybe not always so simple, but sometimes simple. And this was sort of the first instance of this I'm aware of. But it's a similar thing that comes up when you study which lens spaces arise by integer-dane surgery along a knot in the three sphere. And there the twist is different. It's coming from some information from the de-invariant floor homology. Another instance would be McCoy's work on alternating knots with a knotting number one. It gives a complete classification of those knots and a sort of a structure theorem for those knots. And it's again coming from an obstruction using Donaldson's theorem plus a little twist. And the result I'll describe today is a continuation of this theme. And in particular, I'm going to be interested in a broader class of three manifolds, branched double covers of alternating links. And I'd like to know which of those bound to smooth rational homology ball. So on to alternating links. Here's my main example for the talk. There I've drawn an alternating diagram of a link, alternating link L. And I can, this is supposed to be red, giving the diagram a checkerboard coloring. And I'm thinking of the red regions with the half twists between them as giving a spanning surface for this link. And I'm going to form the double cover of the foreball branched along a copy of this surface where I've let its interior sink a little bit into that foreball. So this is the branch double. I'll just think of it as the branch double cover of F. And what do I want to say? So that's a four manifold. And it fills the branch double cover of the three sphere branched along the link. And it's a very important four manifold. It plays for these links the role that linear plumbings do for lens spaces. And in fact, you can show, if you start with a two bridge link diagram, the four manifold that you get back actually is diffimorphic to one of these plumbings. So it is a strict generalization. And I want to understand this four manifold a bit, because it's such a key tool. So before telling you more about that, I need to tell you about the take graph. So the take graph, I remind you, I am putting a vertex into each region of the surface. And I'm connecting an edge two vertices if there's a crossing joining those two regions. I could have drawn on the surface itself. It's a spine of the surface. So this is g. This is the take graph of this diagram. And I'll tell you a two part theorem. First, if I'm interested in the intersection lattice on this four manifold, which I am, this is isometric to the lattice of integer flows on the take graph. So let me explain that by way of example. So the flow lattice is just the cycle space of this graph, which in this case, I see it's two dimensional, generated by the classes of these two cycles. And this space inherits a natural inner product sitting inside of the chain group, first chain group of the graph with integer coefficients, thinking of edges as forming an orthonormal basis. So in this example, the flow lattice of g I can think about as being generated by these two classes, a and b. And a, paired with itself, is 5. It's a cycle of length 5. b, paired with itself, is 4. It's a cycle of length 4. And a, paired with b is minus 2 because they overlap in two edges, but they point oppositely around the two cycles. So that's a complete description of the flow lattice of this graph. So this, and a generalization of it, is a theorem of Gordon and Litherland, I think, from 78. That's the first theorem I want to tell you. And the second theorem is more recent. It's the statement that this four manifold is what is called sharp. So that's a definition. And also, this is a theorem due to Ohspath and Sabo. I can't remember when exactly, maybe 2005. But I'll tell you more about sharp manifolds in a moment. But the defining feature is that whenever you have a definite filling of a three manifold that gives you inequality of the de-invariance of the three manifold in terms of the intersection lattice on the filling four manifold, you have an inequality. And the condition to be sharp is saying that those inequalities are all attained by some spin C structures on the four manifold. So you can really read off the de-invariant of the boundary directly from the intersection form on the four manifold. But I'm not going to go any deeper into that. What's nice about this is that it feeds into a theorem from some years ago, which states that if ever you have a rational homology sphere and it bounds a rational homology ball and it also bounds a sharp four manifold, then one thing you know is that the intersection lattice on x embeds into the standard diagonal lattice with full rank. So that's just coming from the argument we gave before, if that's Donaldson's theorem. But something more is true, namely that every unit cube in I n, maybe to stress, I'll say, because of the pictures I'll draw, with integer vertices, really just a unit cube, whose points belong to this lattice, contains at least one point of the embedded lattice. So as an example, actually this link that I'm carrying around, the branch double cover of this example link is the boundary of a rational homology ball. And let me convince you of that by quickly drawing the picture. So there's my link. I'm going to attach a band to this link right here. When I make that band sum, what happens is that that ring becomes free. Then depending on whether you're looking up or down, you can cap that off with a zero handle after having done your one handle attachment. It goes away. And after I do write a Meister move and an isotope, this is the 2, 4 torus link. And I can do that trick again of adding a band. And I'll get then a two-component unlink. And I can cap those off with zero handles. And by virtue of this procedure, what I've shown you is that L is actually the boundary of a ribbon surface, R. So R is a ribbon surface with Euler characteristic 1. OK? And when you have a link which bounds a ribbon surface with Euler characteristic 1, that is sometimes what's called chi-slice. And it's a fact that if you have a chi-slice link, you take the branch double cover of that ribbon surface, then what you're getting is a rational ball, which fills the branch double cover of your link. Good. So therefore, our theorem applies. And we should be able then to embed the flows on this graph thanks to this part of this theorem into the standard integer lattice of rank 2. So here it is. I want to embed flows on my graph into the integer lattice I2, which I'm taking an orthonormal basis for, labeled E1 and E2. So it labels the axes this way. And A has to map to some class of self-pairing 5. And B has to map to some class of self-pairing 4, which then together have pairing minus 2. And one choice that works is to map A to 2 E1 minus E2 and to map B to 2 E2. And so I'll map where those points go. So that's where B goes. A goes here. That was the image of A. That's the image of B. And then I can look at the sublattice that they span. So I'm always allowed to go 2 up or 2 down from any point that I've already marked. And I can also make a night move in any direction. Got one wrong? 2 down, 2 down, 2 down. And I think that's everything that would fit onto this. So there's my embedding. It turns out that this is a perfect obstruction as well for obstructing lens spaces from bounding rational balls. So you can show that for the case of a lens space, having your linear lattice embed in Euclidean space subject to this condition is equivalent to having both the lattice and the dual, so to speak, linear lattice embedding into Euclidean lattice with full rank. But I don't know an argument which goes directly from this criterion to a reproof of Liska's theorem. So I don't see a simplification. This condition is going to continue with us. We call this condition cubiquity for the reason that the lattice embeds in a way that's sort of ubiquitous from the point of all the unit cubes. You hit each unit cube in at least one point. Pause for questions and to collect my thoughts. Yes, the question is what is the idea of the proof? Josh, the idea is that, OK, so you're going to get a bunch of spin C structures on this four-manifold, when you do C1 squared minus B2 over 4, you're getting V invariant 0. And this four-manifold's intersection lattice embeds into the Euclidean lattice. So you're thinking, which characteristic elements in the Euclidean lattice have self-pairing squared minus rank over 4 equal to 0? And those are precisely plus minus 1 vectors. And so the way the criterion comes out is you find that every coset, if you look at characteristic elements in this lattice, modulo 2 times this embedded sub-lattice, each one of those cosets has to contain one of those plus minus 1 vectors. And then you massage that into saying that actually every coset of the sub-lattice in here contains a point of the unit cube based at the origin. And then that ends up, I'm actually going to want to use that in a moment. And then that ends up being shown to be equivalent to having every unit cube possess a point of your embedded lattice. OK. So on the basis of that, the fact that it does what we want for Lenspace is we have a conjecture. So L is alternating. And if I don't say it, it would always be non-split. And if I'm ever talking about a diagram, I'm always coloring it according to a convention that the red surface at every crossing looks the way it does in this diagram. So I'm suppressing that. But the conjecture is that we have a complete obstruction. So the branch double cover of the link bounds a rational homology ball if and only if the lattice attached to the link which I never really defined. But it's this lattice on the Gordon-Lutherland lattice, on the Gordon-Lutherland manifold. Conjecture is that cubiquity is a perfect obstruction to bounding a rational ball, meaning admitting a cubiquitous embedding. But we don't know how to solve it yet. And what I want to do with the talk then is to give evidence in a special case. One special case is solved. Those are lens spaces. And I'm going to go to kind of an opposite extreme in a sense. So to get there, note, if I have a lattice which is already embedded in IM, is a sublattice of IM, if this is cubiquitous, then as I was explaining, every coset inside of IM has to hit the unit cube at the origin. So that means then the number of cosets is bounded by the number of points in the unit cube, which is 2 to the n. And this quantity is almost a quantity called the determinant of the lattice. The determinant of the lattice would be the index of the lattice in its dual lattice. And being a unimodular lattice, IM would sit halfway between. So this index is actually going to be the square root of this determinant. So that means if I take the determinant of lambda and I divide by 4 to the power of the rank, that's less than or equal to 1. It has to be a somewhat dense embedding. So this is a quantity that just kind of naturally pops up for us. And it probably needs a better name. But for purposes of this talk, I'm just going to write it like this and think of it as kind of a renormalized determinant of this lattice. And yes, what is the index in that example? Anybody care to answer? How many points per unit volume are there? There are a couple of ways of proceeding. You see in every unit cube, there's actually a unique point. So that's suggesting maybe the answer is 4. And you can confirm that by taking the gram matrix. The way you calculate the determinant of a lattice is you take a basis for it. You make the matrix of inner products of the basis elements. And then you take the absolute value of the determinant of that matrix. And that's 16, which is indeed 4 squared. So this is kind of an extremal example, which will where we're going. And so based on this, if I have an alternating link L, I'm just going to define the normalized determinant of the link to be, well, just the normalized determinant of its associated lattice. And in terms of the picture of the link, well, so this is the link determinant, but then it's divided by something which is kind of not really a good invariant of knots or links. This is going to be the number of red regions. I'll try to write it a little bit bigger. It's the number of red regions minus 1, 4 to that power. So it's very sensitive to the coloration of the link used. If you take the mirror link, you'll get something different because there's no relationship between number of red and black regions in general. So there's a little corollary, then, that if you looked in some link tables and you wanted to know which branch double covers of some low crossing alternating links bound rational balls, one thing you could do is to calculate this normalized determinant. And you see that if that normalized determinant is bigger than 1, then there's no way that that branch double cover could bound a rational ball. So it's kind of a simple numerical test. Not a complete test. So for example, the normalized determinant of our running example link is equal to 1. Also, the normalized determinant of the pretzel, 1, 2, 4, 4 is equal to 1. And in general, it can be practically anything. It can be less than 1. It can be bigger than 1. It can be equal 1. And notice that lambda, if we have a cubiquitous lattice, so if lambda is cubiquitous and it has normalized determinant 1, this happens if and only if lambda is equal to the set of center points of a tiling of Euclidean space by cubes of side length 2, or 2 cubes for short. So I'll highlight those cubes. I can't stop drawing. There, you see the cube tiling staring at you now. And wouldn't you know? But there's some interesting results about tiling of Euclidean space by equal sized cubes. So it had been conjectured by Minkowski, 1897, and then proved him by Hayosh, 1942, that in every lattice tiling of Euclidean space by, it's a redundant to say this, but congruent, cubes. There always exists a pair of cubes which touch along a full facet of each. And since it's a lattice tiling, any cube really looks the same as any other to translating. So it suffices to look at the cube centered at the origin. And of course, you see the one just above it touches it along a full facet. So the separation in dates and Hayosh being one of the giants of geometry of an era suggests that there's some depth to this result. In the end, an elementary argument involving bases of finite Abelian groups, whatever that is. But it's a hard argument to come up with. It's an easy argument to read. And then it's an easy argument to forget. And Hayosh himself apparently was, so this was his second attempt at writing a PhD thesis. The first one apparently wasn't good enough for Bayer, was it? Yeah, so OK. So he had to go back and work for three more years before he got it. And he was like, OK, he has a 45-year-old conjecture of Minkowski. That could be a thesis. And he never felt comfortable with the proof of this result. Well, anyway, there are other stories you can tell, but I'll just stop with that. Great, so we can actually buy something from this result. And I guess it's time for me to tell you what the main theorem is, which is going to use it. Well, the slightly darkly humorous thing is that he finally decided later in life he would give a seminar where he'd exposed the argument at last. But then he passed away shortly before he could do that. The theorem, the L is alternating. That's one of these extremal examples. And D is going to be alternating diagram of it, which is colored correctly to make. So the number of red regions is raised to the valve that fits this formula. Then the following are equivalent. One is that this link L is chi ribbon, which I showed you by example earlier, what that is. Chi means Euler characteristic 1. Two is the link is chi slice. So if you know about slice ribbon, this is a slice surface with no closed components with Euler characteristic 1. So if you understand the definitions, this is supposed to be obvious. Third is that the branch double cover of L bounds a rational homology of all. This is more or less a classic result of Cassin and Gordon, as I understand. The fourth is the result I told you before, which is that the lattice attached to L is chi decodis. All of these things we've seen. And then the final thing is where the meat of the theorem lies. And that is that this diagram you've represented your link by belongs to a certain explicit class of diagrams called denote curly D, in which I'll now describe. And by the way, once I've described it, it'll be fairly clear that being a diagram in that set implies that the associated link is chi ribbon. So you complete the chain of implications. So it certifies the conjecture that I'm erasing in the extremal case where normalized determinant is as large as it could be. So here's D. Again, it's this kind of simple iterative scheme like I was alluding to earlier in the lecture of the kind I was talking about earlier in the lecture. So it's the smallest set. And they'll necessarily be colored, alternating link diagrams, such that first the round unknot diagram. And I hope this is the right convention. This crossing, one crossing diagram of the unknot line D. And I guess I do need to tell you how the colors work, at least in this one. I think it should be like that, if not the other one, which is also closed under a certain composition. So here's the composition. I take a pair of diagrams, D1 and D2. I expose a crossing from each one. Maybe I won't put in the coloration because it gets a little noisy. But the coloration on this won't really match the one here. It's not that there's going to be a consistent color on this region, because the crossings, it wouldn't look right. So the composition is you take those diagrams and you join a cross in the way that I'm doing. You take that little composition circle and you enlarge it and put it right here. So that's the composition law. And that's the set D. Yeah, you almost see direct some of the intersection forms. But then it gets a little screwed up. Well, there's gluing. But I don't think that that's what that is. It's sort of vaguely reminiscent of two or three sum and graph theory. But maybe there should. OK, I'd like to know the word. OK, so about as clean as structured theorem for link diagrams as you might hope for. And as far as five going back to one, that proceeds basically like as in the example. You pick a diagram from this set. And if it's not one of your seed examples, then it looks like this. You find your decomposing circle is what I call this circle. And I put in a band and it pops off. And then it simplifies down to a connected sum of things that were already in D and induction. You prove that the link represented as chi-ribbon. So examples are a running example of a diagram is certainly in this set. You can check that directly. And the standard pretzel diagram, although this has the right normalized determinant, this diagram's not in this. I have five minutes. All right, so all the work is in now showing that four implies five. So it issues quickly from Hayosh's theorem. So you suppose that lambda is cubicotis. So you suppose that lambda is a two cubed tiling lattice. And then what Hayosh will tell you is that there's a nice presentation for this lattice that find a lower triangular matrix with two's on the diagonal, zeros and ones below it, so that the column vectors form a basis for lambda. We call it the Hayosh basis. And it's a very nice basis. So if you take any one of these basis elements, or if you take the difference of any two of these basis elements, these are what are called irreducible elements in this lattice. Meaning you can't write it as a sum of vectors of non-zero vectors with non-negative inner product. Parse that. Kind of like root vectors in a root lattice. Sometimes they're called simple elements. OK, so you've got a two cubed tiling lattice on your hands. And you're wondering, could lambda be actually isometric to the lattice of flows on a planar graph G? Well, there's a nice fact, which is that the irreducible elements on the lattice of flows on the graph, these are one-to-one correspondence with oriented cycles in the graph. So you know that these standard basis elements must match to some cycles in the graph. And the fact that differences are irreducible also tells you that cycles that touch each other can only do so in a very simple way, in a way that their sine sum also gives an oriented cycle. So there are a lot of tricks like this going back and forth. And in particular, this last column, which is a vector of self-pairing four, what that's going to find for you in your graph is a four cycle. And you're going to find that this graph decomposes. So I have a bit of graph going from one vertex to the other in my planar drawing. And I have a bit more graph going between the opposite pair of vertices. I'm not telling you what's happening in these looms. But that's the kind of structure that the graph has. And that four cycle is the take graph echo of this decomposing cycle circle here. So what you can do is you can do a decomposition and see that the graphs that you get, again, have lattices of flows, which are two cubed tiling lattices. So you kind of decompose and you induct. And then what you find is that G has to be the take graph of a diagram D inside of this set script D. So that's how the proof that goes. I'll leave that up, but I'll just close with a conjecture. So maybe a remark and then a conjecture. We can go a little further. So we've got this whole class of links now. We know their branch double covers bound rational balls. And we can study the relation of ribbon rational homology co-boardism between elements of this set now. And we have a complete understanding of how that works. So if I have two of these links with a ribbon rational homology co-boardism between their branch double covers, then start with one and do some decompositions until you get down to a diagram of the other. And it leads to some conjectures. So I'll make two, then end with a proof of a special case of one. One is that if there exists a ribbon rational homology co-boardism from a three-manifold to the branch double cover of an alternating link L2, then L2 is alternating, then the y itself is actually the amorphic to the branch double cover of an alternating link. And a special case of this would be, well, on the level of links of related conjectures that if there exists a ribbon concordance from a knot k1 to a knot k2, where k2 is alternating, then we conjecture that k1 is alternating. OK, but let me give you the evidence and then I heard the bell rings and make the quick joke that lots of people are studying ribbon concordances these days from E and A to E and Z. So evidence for two, I could say a lot of things about lattices and support of it, but let me just bring it full circle. The case that L2 is a two-bridge link, and either of these conjectures implies that L1 or k1, however I'm notating it, is two-bridge. I'll give you the proof really fast and then we'll quit. So here's why, and here's my ribbon rational homology co-boardism to the double cover of L2, which is a lens space. And what Gordon tells us is that the fundamental group of Y has to inject into the fundamental group of the co-boardism, which is surjected onto by the fundamental group of this lens space. So this is finite cyclic, so this is finite cyclic, so this is finite cyclic. So geometrization then tells you why is itself a lens space, and then if you're trying to do the conjecture at the level of links, then Hodgson and Rubenstein tell you that this thing had to be the branch double cover of a two-bridge link, and only that. That's a little bit of evidence. I'll leave it at that, thank you. So the question is why are we using diagrams to look at alternating links as a, right, thank you. I guess he didn't know better. Yeah, I mean, to give a kind of a surfacey characterization of what's going on, yes. It's monotonic, maybe, but it is monotonic, yeah. Then any time what could be a decomposing circle, you can decompose and get into smaller elements of your family.