 All right. Can everyone hear me? The question was, everyone can hear me. Right. Well, so for this lecture series, I'm discussing lattices and low-dimensional topology. And while lattices have their own rich theory, it's not as though there's a theory of lattices and low-dimensional topology that I'm going to be developing. Rather, I just wanted to highlight their role in the service of low-dimensional topology. And I think if there are one or two takeaways you get from the lectures, I think they'd be that the techniques I'll describe are very accessible and easy to get into. And if you're lucky enough to be working on a problem where a lattice naturally arises, then that's an important structure you should try to exploit. Like if you're studying a problem with some symmetry, you have to use the symmetry. So I'd say if you see a lattice in your problem, you should think hard. It may conceal some rich algebraic combinatorial structure you didn't realize was there. And I'll show some examples of that as we go. This lecture is going to be, for the most part, classical material, meaning material at least 40 years old, up until maybe the very end of the lecture. And so for that reason, it might be well known to some people, but if you're earlier in graduate school, then maybe this is new stuff, and so I get the chance to expose some classics for you. So where this all begins is the intersection pairing. So to begin, x is going to denote closed, meaning compact, connected, and without boundary, oriented, formatful. And you know that the cut product, in that case, on second dimensional cohomology, and by the way, now here's a convention. Anytime I write a homology or a cohomology group without coefficients here, I'm suppressing integer coefficients. And other times I'll make the coefficient ring visible. This cut product pairing lands in the top dimensional cohomology. Because it's oriented, I have a fundamental class, which I can cap with, got an integer. And this is, as we're in even dimensions, it's a symmetric, and it's a bilinear integer valued, i.e., integral pairing. And Poincare duality, let's just say a couple of things. First, I can switch to looking at, I can view this as a pairing instead on second homology thanks to Poincare duality. And when I do that, I change the notation to qx. And q is to make you think of quadratic form. But this is going to the integers. And I'm going to get 0 if either of my entries into this pairing is torsion class. So to get something interesting, I'll kill the torsion. So I'm going to favor this point of view on the intersection pairing. And it'll explain why it's called the intersection pairing. And for this case of a closed manifold Poincare duality tells us that this pairing is unimodular. And there are many ways to say what that means. That is to say, maybe most concretely, that if you choose a basis, any basis for this free abelian, finitely generated free abelian group, I can look at this matrix of pairings. I get an n by n matrix. So if I have a basis for a lattice in general and I make this matrix, this is called the gram matrix, gram of Gram-Schmidt thing. This matrix has determinant plus or minus 1. So it's invertible over the integers. OK. Now the reason for calling it the intersection pairing is what happens in case this four manifold is smooth. So if x is smooth, then any pair of classes that I choose, second homology, mod torsion, by the miracle of small dimensions, I can find closed oriented surfaces. We'll call them sigma alpha and sigma beta in side effects, which represent these classes and which furthermore are transverse to one another. So the picture in half the dimension would be if this is my sigma alpha, this is my sigma beta. Because of dimension equaling co-dimension, these surfaces will meet in a finite number of intersection points. And the intersection points will come with signs. And so I can make the sign sum of these. I'm just going to denote it by a dot. That will denote the sign sum of the intersection numbers between sigma alpha and sigma beta. So in this example, I'm just getting plus 1. And this is the intersection pairing between the classes, alpha and beta. So it doesn't matter which surfaces you choose or how they're positioned, you'll get this canonical value. We still call it the intersection pairing, even if the manifold's not smooth. Although there, it's harder to give an interpretation. And this is exactly in line with the intersection pairing of curves on the surface. So I've seen you process Paul's note sitting in the back row if you want to be heard. So let's look at a slew of examples. So in no particular order, the first example to consider, well, maybe the 0th example to consider is the force sphere. In that case, the second homology vanishes. And the form I get is just identically 0. So let me sneak in a bit of notation up here. I'm going to write lambda x to denote the free Abelian group h2 of x mod torsion equipped with this pairing, qx. And I'll call this object a lattice. So this is the intersection lattice, x. And I'll probably be a bit sloppy at times about conflating qx and lambda x. But hopefully, it won't be confusion for you. So in this case, lambda x is just 0, hence its example, 0. Example 1 would be the case of CP2. And in this case, the second homology is infinite cyclic generated by a single generator. I'm calling h for summaries and maybe for hyperplane. And well, what is h? H is represented by any complex line that I choose inside CP2. So if I want to figure out the intersection lattice, all I need to think about is how this single basis element pairs with itself. So to do that, I represent my class by a surface which represents it. And then I have to take another representative, which is transverse to the original. So I just take another line. And any two lines in CP2 will meet in a positive point of intersection. So the intersection lattice in this case is z equipped with respect to this basis. I'm just getting the 1 by 1 matrix 1. That's the gram matrix. And I'll often abbreviate this i1. And what else do I want to point out? Well, simply that the rank of this lattice is 1. And the signature, meaning the signature of this form, sometimes called the signature of the four manifold, like in Francesco's lecture, is also 1. So it's a definite lattice, meaning that any class that I choose, I'll often just write the dot instead of qx. This is greater than 0 for any non-zero class that I choose. Or precisely, this is a positive definite lattice. Pretty boring lattice, but a lot to say about it. OK, moving right along. Next example would be the product of two copies of a two-sphere. In this case, the second homology is two-dimensional. And because I'm not using Engel brackets to denote an inner product, I'll keep using them to note generated by. This is generated by the class of a two-sphere across any point and any point across a two-sphere. So the picture is a perpendicular pair of two-spheres. They're already transverse to one another. They have intersection number plus 1, if I'm careful how I orient everything. And individual copies can be pushed off of themselves. So they'll have self-pairing 0. So in this basis, then, you get a rank two lattice. And its form is represented in this basis by the grand matrix 0, 1, 1, 0. Also sometimes annoyingly denoted h, which I might do. I'll try not to. So this lattice has rank two. Its signature 0, it has a positive and a negative eigenvalue if I diagonalize over the reals. So this is an indefinite form. And it's also even, meaning that any class that I choose, the self-pairing of that class is an even number. That suffices to check on a basis, which we've already done. This lattice doesn't have that property, so it's called odd. Odd, meaning it's not even. With these basic examples. Now I have two permutable examples. Let me just pick one. Here's a more interesting example of a closed smooth four-manifold. It's the zero locus of this polynomial, this homogeneous polynomial, side CP3. This is a quartic surface, complex surface, sometimes called the Fermat quartic because of the form it takes. And this is an example of a K3 surface Francesco mentioned in his lecture. So if you're good enough at algebraic topology and differential topology, you can calculate that all these examples, by the way, let me point out, are simply connected. So this is actually a smooth manifold. Simply connected. And if you make a characteristic class calculation, you can figure out that the second betting number of this manifold, which is after all the same thing as the rank of the intersection lattice, works out to 22. The signature of this lattice works out to minus 16. And so it's indefinite. And it's also even. All elements have even self-pairing. And you could go and hunt for a basis for this and actually write down a gram matrix for it. But as I'll describe in a little bit, if you know enough about lattices, this already tells you exactly what the lattice is isometric to. So that's going to be one of the first classic theorems I tell you. So we'll revisit this example and fill in the blank what it's actually corresponding to. So that ties in a little bit with, I think Francesco gave us that calculation of minus 16 yesterday. Now to tie in with Jen's lecture, here's a fourth example. So this example is going to be not closed. But nevertheless, we can talk about the intersection pairing on a compact for manifold. So this will be an example of manifolds. So this is going to be a family of manifolds, which are simply connected, smooth, compact, but not closed. So the construction is this. I'm going to let Blackboard L denote a framed link, free sphere. So I'll choose one particular example. Take a moment to draw. All right, now I have to count my example. OK. So that's a link in the three sphere. And to frame it means just to, I could have put more interesting knots in, by the way. I mean, I can actually draw more interesting knots, but I chose not to. This is a particular example I want to work with. And to frame it means just to place an integer at each component. I also know other numbers besides this one, but I choose to put twos on all of those. OK. So that's a framed link in the three sphere. And it has a linking matrix. I should orient it, actually. So my framed links should always come with an orientation. And the linking matrix is just going to, off the diagonal, it's going to record the linking between different link components. And on the diagonal, it's going to record the framing I put on that link component. So if I number them 1, 2, 3, 4, 5, 6, 7, 8, I'm going to get an 8 by 8 linking matrix, which looks kind of boring down the diagonal. It's minus 1's on the super and sub diagonals down to ultimate entry. And then, assuming things have lined up correctly, there should be a minus 1 in the 8, 3 position and also in the 3, 8 position. So that's the linking matrix of this link. Random example. And there's a four-manifold encoded by this, which Jen was denoting x. I've chosen to use truck for trace. This is the trace of surgery along this link. So you can view these as pain surgery instructions for producing a closed-oriented three-manifold. I do two surgery along each of these link components. And that's going to show up as the boundary of this trace of surgery along L. So just I'll reproduce the picture that Jen put up last time, but do a little bit more with it. So to build the trace of surgery, I start with a four-dimensional zero cell, a zero handle, something just diffeomorphic to a four-ball, OK? And I'm picturing my link as sitting in its boundary. So my link component looks like an S0 in this picture. Here's another link component. It's another S0. And what I'm to do is attach, pick up two cells. It's called two handles, four-dimensional two handles, along these link components. The core of one of these just looks like a disk that's had its boundary identified with that link component. So I'm drawing in the cores of these two handles. They've been thickened up by product with another two-dimensional disk, OK? So in words then, what I've taken is a zero handle. And I've unioned with the two handle, one for each link component. And without going into the details, when you attach a two-handle along a link component, there's essentially an integer's worth of different gluings that you could choose. Different meaning up to changing the topological type of the glued up four manifold. And that integer's worth is what's called the framing of that two-handle. And the framing that I choose for the two-handle is just chosen to match the framing I wrote down on the link component. I'm sorry? Yes. The question is whether I could repeat the last part. So when I attach a two-handle, I need to specify a framing. So the two-handle is diffeomorphic to d2 cross d2, which is a ball. And what I'm doing is I'm taking a part of its boundary, which is diffeomorphic to d2 cross s1 and identifying it with a neighborhood of a link component, which is also diffeomorphic to d2 cross s1. So how do I identify d2 cross s1 with d2 cross s1? Up to isotope, there is an integer's worth of choices corresponding to the mapping class group of d2 cross s1. Which one of those I choose is governed by the integer that I wrote down on that link component. OK. So if you're thinking in terms of homotopy, it looks like I just have a single zero cell and a number of two cells. And so the space that I've built has the homotopy type of a wedge of two spheres, one for each link component, which makes the computation of the fundamental group being trivial clear. And the second dimensional homology group is then going to be free abelian of rank equal to the number of link components. And in this picture, there's a very nice suggestion for a choice of representatives for the basis for second homology. So each link component bounds a ciphered surface. Just a surface embedded in the three sphere who's a compact oriented surface embedded in the three sphere whose boundary is that component. And look, it hit another link component, but I don't care. And what I can do for each of these link components is maybe just in the interest of making the picture clearer is to let that ciphered surface, which over here would kind of sit in the boundary, to just let its interior sink inside of the foreball a little bit. And in my example, the ciphered surface is a disk. So this is a disk. And it has the same boundary as this disk. So I glue them together, and I get a sphere. And I can do that for each of my link components. So what I would say is that we get a basis for the second homology of this space. I don't need to mod out by torsion. It shows torsion free examples. I mean, they're simply connected. So I get a basis for the second homology, represented by capped off. These cores are the caps. Cores are caps. By capped off ciphered surfaces. But I chose to push them in. So I'll call them pushed-in ciphered surfaces for the link components. The gram matrix, with respect to this basis, is none other than the linking matrix. This is the trace of surgery. So many remarks to make. The boundary of this trace of surgery, in general, well, I mean, it's always non-empty. It's non-empty. It's equal to what you might denote the result of surgery along this link. That's often written this way. There are lots of, as many people doing Dane surgery as notations for a Dane surgery. Dane surgeons. So in general, I mean, these are not closed. And nor can they be closed off in any kind of obvious way. If you lucked out and you happen to get a three-sphere boundary, then you could plug it up with a four-ball and you'd get a closed four-manifold. Whose intersection pairing will agree with that of this compact one. But in general, they're not closed and not easily closed up. And this choice that I made is particularly famous one. This space, which I built here, well, this link L, I'll call the E8 link. And the surgery along L is not called the E8 manifold. It's actually a representation of a famous space called the Poncare homology sphere, which is going to be a key player in these lectures. It's the first example discovered of a topological, well, of a three-manifold with the same homology groups as the three-sphere, but which is not defumorphic to the three-sphere. It's not simply connected. And this intersection lattice, lambda x, equivalently, the matrix is denoted E8. And you can see right away this is even because all the basis elements have even self-pairing, that's recorded on the diagonal. This is one of the most famous exceptional structures in all mathematics. And I think we all probably heard a lot about it last month when the fields metals were announced. And in particular, Marina Villazoska was honored for, amongst other things, showing that if you take the E8 lattice, which has a natural embedding into eight-dimensional Euclidean space, and you center balls at those lattice points and blow them up until they touch one another, that's the densest packing of balls in eight-dimensional Euclidean space. And so that's getting into some very deep properties about lattices and their theta series and modular forms and quasi-modular forms and so forth that I'm just giving lip service to in the lectures. But a remarkable structure, which is going to recur for us, it's not obvious from how I've written it down. This should be obviously even, should obviously also have rank A. Maybe what's not obvious is that it is indeed a definite form, definite, positive definite, and it's also unimodular. We have a question. Yes, the question is how to really define the intersection pairing for a four-manifold, which is compact with non-empty boundary. And the two answers are you could cap with the relative fundamental class, which is in relative H4. So that's one way to proceed. You could also use this very direct interpretation that I had written here before of representing homology classes by surfaces and taking their intersection numbers and that'll do the job just as well. Besides introducing several very useful constructions, the other thing this example shows, which you could perhaps think about in an office hour, is I can get rid of the force here. Sort of always there. If you take any lattice, so again, it's just a free abelian group equipped with a symmetric bilinear pairing, I can find a four-manifold x, which is simply connected, a compact and smooth, whose intersection lattice is isometric to lambda. And it follows from this construction. What you would do is pick whatever lattice you'd like to work with, pick any basis for it, write down the gram matrix, and then cleverly cook up a link, a framed link, whose linking matrix coincides with that gram matrix, and then you'll have built a link for which the trace of surgery is the desired manifold. This is via construction four. So lots of examples. Building more examples from old, just very quickly, we can make connected sum. So if I take a pair of four-manifolds and I connect some of them together, the intersection lattice is what you might expect. It's just the direct sum of the two. So for instance, if I connected some together a bunch of copies of CP2, I would be just getting n copies of the form I denoted i1 earlier. And I'll come up with an even shorter shorthand for that. That's in. So again, this is going to look like the free abelian group of rank n equipped with just this diagonal, all-ones matrix. That's going to be the most obvious form you could think of in n dimensions, and it's going to come up a whole lot for us. Another interesting example would be if you took CP2 and you blew it up at two points, you would be getting something whose form looks like 1 minus 1 minus 1. Or you could instead take CP2 and connect sum it with S2 cross S2. And this would look like 1 with this hyperbolic form tacked on. And these turn out to be equivalent over the integers. There's a change of basis taking one to the other. And a satisfying explanation of that fact is that, in fact, these four manifolds I wrote down are actually lithomorphic to one another. There's something, if you didn't know, good. And there's a very elegant proof of this using Kirby calculus, which again may be something to think about in the office hour if you're fresh to it. I guess I didn't say. I guess I snuck in here orientation reversal. So if I reverse orientation on a manifold, that means I'm replacing its fundamental class with the negative of it so that when I pair elements together, I'm going to be getting the negatives of their values. So if I reverse orientation, I'll denote that by putting a bar over my manifold. Its intersection lattice is just isometric to the intersection lattice of the original guy, but I negate all the pairings. And so that snuck in here. So it turns out these, so four is special because it's not a closed example. But the closed examples I put up in these connected sum and orientation reversal operations generate a whole family of lattices that are realized as intersection lattices of simply connected smooth four manifolds. And it's half a major theorem and half a major conjecture that the forms that you get are actually all the forms possible to get. So I'll elaborate on that statement in a little bit. But these are sort of canonical examples and operations for that reason. Question, comment? Yeah, so the E8 lattice is definite. Why don't I emphasize that? It's got a really funny history. It was kind of known to exist before it was constructed. People knew that there was this even eight-dimensional integral lattice, but didn't really have an explicit description of it for maybe 10 years or something. This is like in the late 1800s. And it was finally brought to life by Corkin and Zolotarov, I think are the names. So I mean, once you have this matrix, you can just go back to your 19th century computer and then put it in. It'll come out with all positive eigenvalues. And so that's definite. And by the way, check the determinant equals 1. But here's an explicit way of embedding it in eight-dimensional Euclidean space. So you can regard it as take eight tuples of points in the integers or all half integers. So it's kind of like the Z8 lattice. And then it's been shifted off of itself by the vector of 1 half. And the requirement is that the sum of these coordinates is even. So it's isometric to, well, that's what it is as a set. And I just take the standard Euclidean product on R8 and pair together elements accordingly. And it's isometric to E8 question. It is. The two is now changing to an 8 because it was a typo. That's a very concrete representation of E8. And the challenge would be to find a set of eight elements of that form so that their respective pairings, so that it gives a basis which matches this basis. I mean, whose grand matrix matches this, which might start by taking the, you know, if you call your, you might start by taking one minus one and a bunch of zeros for your first element. Zero one minus one zeros for your next set. You can kind of get the top row or the first seven in that way. And then you have to start using your half integer freeness to get the last guy. Well, that's fun. Go for it. OK, so from CP2, S2 cross S2, and this Fermat Quartek, if you take those three and you make all four manifolds you can by taking connect sums and orientation reversals thereof, generate a list of closed, simply connected, smooth, four manifolds, and a bunch of accompanying intersection lattices. And that is up to a great theorem and a great conjecture, the complete list of all intersection lattices of such manifolds. So we'll get back to that, I promise. But at this point I want to talk about some classic theorems which gets us in that direction. So the first two have to do with lattices. And then the third one has to do with four manifold topology. And it's, I think, arguably the first true theorem about smooth four manifolds. Brochlin's theorem. But to begin, there's a theorem of van der Bly from, we're talking the 50s now. What he tells us is that if lambda is an even in a modular integral lattice, whether or not it's definite or indefinite, the signature of this lattice is always a multiple of 8. Sort of like when you square an odd number, you get something that's always 1 mod 8. I mean, it's a little bit, 8 is a bit more than you might expect. So that accounts for, if you look at the examples with E8 and the hyperbolic form, you're getting, for these even forms, the signature of 0 mod 8. The next theorem is, I'm not sure who it's due to. But it's a classification result. So suppose that lambda is an, again, it's a unimodular to grow lattice. Now the qualifier is that it's indefinite. We've seen some examples of. Well, we've seen all examples of, actually. The statement is that lambda is isometric to one of two possibilities, each possibility being an infinite family. So it could be the direct sum of im with minus im for some n and m greater than or equal to 1. This is the case if lambda is odd. If lambda instead is even, this is some number of copies of the E8 lattice, plus some number of copies of the hyperbolic form. This is the case if lambda is even. And here n can be anything greater than or equal to, well, n can be any integer, positive or negative. And I guess we want m to be non-zero. And exercise this form is isometric to its negative. So I could just take n to be greater than or equal to 1. If somebody knows, actually, who proved this, I'd be interested to know. It might be vit, but it's kind of a result of classic number theory. And I would say it's a fairly deep result of number theory of the middle of the last century. And great texts to find this theorem in and approve are, well, Sarah's classic, of course, in arithmetic. That's where I learned it. I mean, you should spend time with that book. It's one of the all-time great books of mathematics. Milner and Hussmohler have another book with kind of a topological point of view to it, treating lattices. And then there's this cute book of Conway, the sensual form, form referring to the quadratic form, where he proves this in a kind of a deceptively simple way, as only Conway would. But it's a deep fact, nevertheless. Yes, question. They are. The forms, sorry, repeat the question. Yes, and this will be even unimogular, integral, and indefinite. Oh, what about E8 plus minus E8 is the question? Where is that on this list? And this is Jacob Kaudel. He'll be happy to answer that question for you during the office hour. That's an excellent question. Well, I'm not, you know, you shouldn't necessarily judge the questions that come to you. But yeah, that will be the source of a nice exercise. What if you sum E8 with its minus? Where does that appear on this list? Well, if you believe it, it tells you what it has to be. But yeah, there's no redundancy in this list either. So which of these, what about I1 plus E8? Well, it's odd and positive definite. Well, OK, so it actually shouldn't have shown up here. Minus I1 plus E8. So you can figure out what the rank and the signature and the parity is. So it actually has to be up here. And how in the world could that be? These are nice things to think through. But if you believe this, then going back to this perfectly placed example. So I have now an even indefinite unimogular integral lattice whose rank and whose signature I know and whose parity I know. And the only possibility is that it's minus two copies of the E8 lattice plus three copies of the hyperbolic form. And I think if you look in gonf and stipschitz, another essential source, I believe they even go and find a basis for this second homology group, which represents this form. I forgot a minus sign here. And the canonical mark at this point is, well, what about definite unimogular integral lattices? There's not really a classification. So the typical thing people cite is if you just look at definite unimogular even integral lattices, then, well, we know you have to get to at least dimension 8 before you see that. And the E8 lattice is the unique lattice in that dimension. And then you go up to dimension 16. And apart from E8 plus E8, there is one other even unimogular integral definite lattice of rank 16. And then in dimension 24, by some strange coincidence, there are 24 lattices, very irreducible, the so-called Niemeyer lattices, including the Leach lattice, maybe the most famous of lattices after E8. And then in dimension 30, there is something like a billion classifications just. Well, sorry, in 30, there are none. But in dimension 32, I think there are a billion. And then in dimension 40, there are more than 10 to the 80th or something like that. So C. Sear's book for a short account. So lots and lots and lots of definite forms, which could potentially arise, therefore, as intersection lattices of closed definite four manifolds. But we'll see that's not the case. So those are the two lattice theoretic classic theorems. And the first, I think, truth theorem of four-dimensional topology is this one, theorem of Rochlin. This is, again, from the 50s. And what Rochlin showed is, suppose that x is a closed, some of these terms will be redundant. But, well, just this one, smooth spin four manifold. I'll remind what that is. But Francesco mentioned it yesterday. For a smooth spin four manifold, the signature of the intersection lattice, also I'll often write, as everyone does, the signature of that four manifold. That's what it means as a definition. This is actually a multiple of 16. So it's a factor of two improvement on van der Blij's theorem. And so as Francesco mentioned yesterday, spin means possessing a so-called spin structure, which, by theorem, is the same as having a vanishing second Stiefel-Wittney class. That's the theorem. But some basic remarks is, being spin implies that the intersection lattice is necessarily even. So we really only are talking about even forms here. So this is truly a factor of two improvement on van der Blij's theorem. It's in the right realm, talking about even lattices. The converse is not quite true. But if it's even, and you happen to know that the four manifold is simply connected, then, conversely, the manifold is spin. And there are examples such as the Enrique surface to show that this condition is necessary. So I think Enrique's surface has fundamental groups in Z mod 2. And its intersection lattice is a copy of E8 or maybe minus E8 plus a hyperbolic form. So that has signature 8 or minus 8. In any case, it's 8 mod 16. So you could take that as a proof that this Enrique surfaces are not spin. Or you could calculate their second Stiefel-Wittney class. All right, so I wanted to give an application. And this is to that first-listed question from the medley of problems. This is in relation. This is a Pierre-Vehr-Milner on Tom, maybe, before von Heimer, the other K.M. pair, von Heimer-Mrufka. So this was this question, which homology classes can be represented by a smoothly embedded two-spher. And more generally, you want to know what are the actual simplest surfaces representing homology classes. But I'll just start with this statement. So let's have a look. So inside of CP2, let's say that I'm looking at some multiple d, maybe it's positive, of the generator in homology. If I want to represent that, and I don't care about it representing by a surface, what I would do is I would just take d copies of CP1 in general position. So I need two meeting in a single point of intersection and no three meeting in the same point. So there's a configuration of lines in CP2. And if I wanted to get a smooth representative, what I could do is smooth each of these double point intersections. In this dimension down, half the dimension, it's going to just, it looks like I'm getting something disconnected, I guess, in this dimension I am. But a useful thing is to think about in 4D what that smoothing operation really corresponds to doing. So locally, it looks like I have a pair of coordinate planes in R4 intersecting at the origin, and I'm somehow smoothing away that intersection. So that's a good thing to try to think about. So this smoothing gives a closed, smooth oriented surface, sigma d. And the smoothing doesn't change which homology class you've represented, but represents this class d times h. And you can do a basic Euler characteristic calculation to sort out that the genus of this surface is d minus 1, d minus 2 over 2. And by rapid calculation, this is equal to 0 if and only if d is equal to 1, which is no surprise since CP1 is homomorphic or difthymorphic to a sphere. Sorry, d equals 1, which is no surprise since CP1 is difthymorphic to a two sphere, or d equals to 2, which is maybe a little bit more surprising. Thank you. And we'll come back to this d equals 2. But for d equals 3, you're getting a torus. And you might ask, can you get something less complicated than a torus, namely a sphere? And this is the theorem of Carver and Milner. This is from the early 60s. And they proved that 3h does not have a smooth spherical representative. Curiously, all classes admit topological two sphere representatives. So you can find an embedded two sphere, which is smooth away from one point, which represents any desired homology class in CP2. That point is not smoothable once the multiple is at least 3 in absolute value. Of course, now we know this is implied by this is, of course, why this problem feared first. This is implied by, in the summer school, is partly an honor of Tom Rufka's 60th birthday, I think, which was at some point in the past. Birthday got delayed because of the pandemic. And it asserts that sigma d minimizes genus in its homology class, which had been a conjecture of Tom. The Rene Tom with an h had conjectured this. And they proved it about eight minutes after the Cyberg-Witton equations were introduced. And I'm not going to get into that story at all. I'm focusing on what lattices can do for us and do the Carver-Milner argument. So here's the proof. So it's a proof by contradiction. So suppose, by way of contradiction, s, the side of CP2, were a smooth two-spher. And the homology class of s is three times this class h. There it is. There's a sitting inside of CP2, a beautiful picture. And now we do the obvious thing, which is to blow up CP2 at eight points. So all I mean is that I'm going to be taking connected sum of, this is the topologist blow up. So I'm just thinking I'm taking connected sum with eight copies of CP2 bar. So I'm choosing eight points inside of CP2. I'm removing neighborhoods of them. And I'm gluing in punctured copies of CP2 bar. And I've chosen the eight points on my sphere. Doesn't matter where they're chosen as far as the topology of the space that I produced. But when I do this on s, so what you're imagining is, again, I'm excising neighborhoods of these points, which are four balls. And I'm gluing in punctured CP2s. So I'm really gluing in, you think about it. A punctured CP2 is a disc bundle over two sphere with Euler number one. I'm doing that at each of these eight points. So the homology of this space, this is I1 direct sum minus I8. And it's represented by the class H I began with. And a bunch more classes I'll call H1 up to H8. And they satisfy the property that, so this is going to be my notation for the self-pairing of it. I'll sometimes write it as a square. So this is plus one, as it had been. Each of these HI's has self-pairing minus one. And they're orthogonal to one another. That's an explicit basis. And you can check that when you make this blow up, I'm losing a little bit of my sphere when I kill this neighborhood of a point. I'm losing a disc from my sphere. But when I insert this disc bundle over S2, I can carefully choose one of those disc fibers to glue in its place. And so what I'm actually going to get then is another smooth sphere. I'm going to denote S prime. It takes a little thinking about. This sits inside of my blown-up manifold. The class that it represents is the original class plus all of these additional classes added on. I'm sort of summing on one of these exceptional two spheres. Exceptional two spheres, one for each blow up that I do. So that takes a little thinking through. That's the case. So what's the typo here? I think I heard it. This is at 3H, of course. That didn't change. So it follows that this two sphere, its self-pairing, is just going to be 3 squared minus a bunch of 1s, which is plus 1. And it follows from this that a neighborhood of this two sphere, therefore, is diffeomorphic to just a punctured CP2. Again, I mean, this is a pseudonym for a disc bundle of Euler number 1 over the two sphere. OK. So this gives an interesting connected sum decomposition of my space. So the space of calling x can be composed as a neighborhood. Well, it's kind of a, well, OK. So it's the neighborhood of s prime union something I'll call x prime, call it x prime minus a ball. So you should really think that this is a connected sum of CP2 with x prime. And on the basis of this decomposition, you can conclude that x prime is simply connected. And the intersection lattice on x prime, again, this might take a little thought. But this can be, this is isometric to any surfaces contained over here are going to have trivial pairing with this sphere s prime. And furthermore, you can convince yourself that anything in the orthogonal complement to this class in this lattice is going to be represented by a homology class here. The central point then is that the intersection lattice on this four-manifold x prime is isometric to the orthogonal complement to this class contained inside of the intersection lattice of x. Well, let's figure out what this is. So I need to find eight classes which span the orthogonal complement to this class. And well, here they are. Certainly if I take h1 minus h2, you agree if I take the dot product with s prime, that that class is orthogonal. And I'm going to get 0. Similarly, h2 minus h3 is orthogonal to it. And I can do that all the way down through all of my hi classes. So I've gotten now seven classes. And I need an eighth one. And you can check that if I take h, add to it h1, h2, and h3, that's orthogonal. Because I'm getting pairing three from here and three minus ones from here which cancel. And these eight classes generate, they form a basis for the orthogonal complement to s prime. And let's look at how these classes pair together. Each one of these has a self-pairing of minus 2. And so does this one, because it's 1 minus 1 minus 1 minus 1. This is another minus 2 class. And I'm going to put an edge between two if they have pairing minus, well, if they have pairing plus 1. And if I do that, I'm getting this linear chain. And I'll be getting this edge as well. And you check that all the other pairings are 0. So remarkably, what we've produced then is a copy of minus the E8 lattice. So therefore, x prime is an impossible closed smooth four-manifold. Its fundamental group is trivial, but its signature 8. And by the way, this intersection lattice, lambda x prime is even. So it would contradict Rochlin's theorem that the signature should have been a multiple of 16. But this is an impossible four-manifold. That gives the conclusion then that that class couldn't have been represented by a sphere. OK, so the question is whether it's obvious that it's spin. What I'm using is this remark that I made before, which is that if you happen to know that the intersection form is even and the fundamental group is trivial, then it's a theorem that that means it's spin. That does the job. Thank you. This part. At this stage, I was analyzing what is the homology class of this new sphere that I built. I figured out that it has pairing plus 1 with itself. And then I went really quickly. So this is a neighborhood of this sphere. So I'm getting some, no, I'm taking a tubular neighborhood of this sphere. So because of the dimensions involved, this is going to be necessarily a D2 bundle over my base base that I'm taking a neighborhood of. I mean, in general, if I'm taking a neighborhood of a sub-manifold, I'm getting a disk bundle of complementary dimension over it. And S prime is the morphic to the two-sphere. It's an oriented disk bundle. And to tell you which disk bundle it is, I need to tell you it's Euler number. And I can determine what the Euler number is based on the self-pairing of that class, which I did. And it's plus 1. So I kind of raced through some things. But you should at least be fairly convinced that I have the disk bundle of the right Euler number over S2. And then I just happened to note that that's diffeomorphic to the neighborhood of Cp1 sitting inside of Cp2 by virtue of the fact that Cp1 itself is diffeomorphic to S2, and it has self-pairing 1. It was kind of a quick explanation. Well, OK, the question is whether the argument only works for 3. No, it's the answer. So exercise, which I was going to suggest, figure out for which D this argument actually works and proves the class is not represented by a sphere. And it works for infinitely many values D, and it doesn't work for infinitely many values D. So you have the kind of strange possibility, as far as we knew in the 60s, that you could have certain classes represented by spheres, but ones between them which were not represented. Of course, now we know much more. The genus grows quadratic and, in particular, monotonically in the complexity of the homology class. Surface S prime. I should repeat the definition of the surface S prime. So we, of course, had begun with our sphere S inside of Cp2, which was supposed to be an impossible sphere. And what I did was I blew up S at eight points. I produced an interesting four-manifold, the blow-up of Cp2 at those eight points. And if I went in carefully, I could see that I'm changing S into a new surface S prime. I'm losing a disc for each of these neighborhoods that I remove, and I'm gaining back a disc each time I glue in a disc bundle. So I'm ultimately making, in a kind of a straightforward way, a new sphere which sits inside of this blown-up format. The remark was this was all an elaborate joke, if you know a bit of algebraic geometry, involving root lattices and Dalpets of surfaces. OK, I think then I'll end the lecture with dramatic development from the early 80s. And this is the work of Friedman Donaldson, chronologically ordered, but barely. So in quick succession, two remarkable theorems, complete contrast to one another were proven by Michael Friedman and Simon Donaldson. 81 or 82, I always forget. I think 81. What Friedman showed is that for every unimodular integral lattice lambda, there exists a closed, simply connected, four-manifold x. Notice I haven't said smooth such that the intersection lattice on x is isometric to lambda. And note to have any hope of hitting lambda by a closed manifold, it ought to be a unimodular, so that assumption is in place. Moreover, x is unique up to homeomorphism if lambda is even and up to two homeomorphism types, one or two homeomorphism types if lambda is odd. And whether it's one or two is governed by the vanishing of the so-called Kirby-Sebenderman invariant, which I won't get into. So it's a truly remarkable result. There are many consequences which issue just from this. For example, take lambda to be the vanishing intersection lattice. And this is saying there's a unique, simply connected four-manifold with the same homology groups as the four-sphere, which, by the way, is simply connected. So therefore, by Friedman's theorem, it has to be homeomorphic to the four-sphere. So that was the resolution of the four-dimensional topological Poincare conjecture. That gives you so much more. So Friedman's techniques are super topological. There's a book out this year, the Disc Embedding Theorem, which, although the title might not suggest it, is an exposition of, as the title might suggest, the Disc Embedding Theorem of Friedman and consequences which stem from it, including this remarkable fact, this remarkable theorem. That book was the effort of mainly young people cleaning up 40-year-old mathematics and giving it new life. So that was a really remarkable effort. In stark contrast, I think just a few months later, Donaldson techniques from gauge theory proved that if x is simply connected, closed, oriented, and the intersection lattice on x, that is to say x itself, is definite. And you can actually forget about this fundamental group condition that's removable. But maybe I'd put it in just for comparison. You can delete that hypothesis. Just you have a closed, oriented, four-manifold. But you do need the hypothesis that it's smooth. There are all these exotic forms. You might potentially hit E8. Well, E8's ruled out if it's simply connected by Roachling. But say the leach lattice, well, I guess that's also, well, OK, eight-month-sixteen, two copies of the leach lattice. No, lambda x is isometric to, depending on the sign and the rank, one of two possibilities. Just the standard diagonal Euclidean lattice, I am. As a consequence, I mean, even just from Friedman, you get this consequence. Many closed, oriented, four-manifolds do not admit smooth structures, because their intersection forms prohibit that from happening, either because of Roachling's theorem or because of Donaldson's theorem. And four is the lowest dimension in which these kind of exotic phenomena happen, where structures are, you know, homeomorphism types are not smoothable, or homeomorphism types admit many different smoothings. Distance of exotic smooth structures on R4 issues from these results. It's the only dimension in which you have exotic smooth structures on Euclidean space. And in fact, you have uncountably many. It's a lot of fun to learn about this stuff. And maybe just, you know, I'm unaware of such a striking contrast of a pair of theorems in mathematics, let alone two that were proven within just months of one another. It's really kind of a singular episode history of mathematics, I think. Great. So let's give a quick proof that just in the last five minutes, maybe a quick application would be application. There's a second question from the original list. If I do plus one Dane surgery along a trefoil, does it bound a smooth or manifold W with banishing second Betty number? Grab this up in the waning minutes this morning. So of course, you can form the trace of surgery of plus one surgery along the right hand trefoil. So the trace of surgery along plus one surgery along the trefoil, this is smooth. And its second Betty number is one. And its boundary is the result of plus one surgery along the trefoil. So we're asking, can you do even better, lower that Betty number down? Find a so-called rational ball billing that. And the answer is, it depends. The answer is yes or one trefoil, not and no for another. And I decided I'd be kind of cute. And let's see, it's yes for the left-hand trefoil. So it's actually very easy to do this kind of thing, to draw mirror images with your two hands. This is actually a very interesting construction, due to Fintechel and Stern, two of the heroes of low-dimensional topology. And this one's ruled out. And why is that? You can do it really fast, right here, fast and without rushing. Turns out, if you know your Kirby calculus, this is another representation of the Poincare homology sphere that we saw before. So we know that this is also presented by, and I'll be just lazy, three, four, five, six, seven, eight. Could I actually draw that with the right crossings without looking at it behind my back? Yes, but that's been there all along. So this space, I can think of its trace of surgery along this manifold. So suppose for sake of contradiction, this did bound a w with b2w equals to 0. Then what I would do is I would take my w, whose boundary is the Poincare sphere, and I would stick on with orientation reversed the trace of surgery along this E8 link. And when I've glued these together, I get a closed definite four-manifold. And you can check that, well, because this one's second-beddy number vanishes, the second-beddy number of the whole space is just going to be 8. And the intersection lattice is going to be entirely accounted for by the intersection lattice on this piece. So if I call this whole glued up result x, then I would be getting that the intersection lattice on x is isometric to the minus E8 lattice. I don't know if this thing's simply connected because I didn't assume anything about w. But using the improvement on Donaldson's theorem, I know that this is impossible. So that cannot occur. So that's a quick application of Donaldson's theorem. I'm over time, so I'll stop. Thank you again. OK.