 So I'm going to be talking about symplectic embeddings of products. This is all joint work with Richard Hind. Let me start by giving you an introduction to the kind of questions that I want to talk about. So if I'm given two symplectic manifolds, M1 and M2, then a symplectic embedding, so I'll give you a definition. A symplectic embedding is a map of psi going from M1 into M2 where psi is an embedding of smooth manifolds. And psi pulls back the symplectic form on the target to the symplectic form on the domain. So one can then ask the following question, which is if I'm given two symplectic manifolds, I'd like to know if I can find a symplectic embedding of one into another. So given M1 omega 1, M2 omega 2, does there exist a symplectic embedding? So I'm going to move over to here. And I'm going to write my symplectic embeddings like this. So the S here just stands for symplectic. So this is the basic kind of question that we want to know the answer to. It turns out that this is quite subtle. I'll give you an example to illustrate why it's subtle in a moment. But anyway, so that's sort of the basic background. Could the image of psi be in the interior of M2? At the moment, no. In the examples that I'm going to give, that generally will happen. But I don't require a condition like that. I should say that this is particularly subtle when the dimensions of M1 and M2 are the same. And that'll certainly be the case in this talk. If the dimensions are off by, say, 4, there's an H principle and things like that. So you should really think of these as having the same dimension. But when I give an example, you'll see what I mean. OK, so any other questions about any of that? Should be pretty straightforward. OK, so let's go to dimension 4 now. And like I said, this problem can be fairly subtle. And the following example, due to McDuff and Schlank, is particularly illustrative of this. So some of you have no doubt seen this before. But I'm sure some of you have not. And it'll also be quite relevant to what else we're talking about in the talk. So what McDuff and Schlank wanted to know is they wanted to know how this symplectic embedding problem worked for some very special shapes. In particular, they defined the four dimensional symplectic ellipsoid. So I want to think of this as a subset of C2, which I identify with R4 with its standard symplectic form. And I also want to study the four dimensional ball. OK, I'll write a 4 here. Which is just an ellipsoid where A and B are the same. And they wanted to know when an ellipsoid admits a symplectic embedding into a ball. So it turns out that we can encode this in the following function. So we can define a function C of A, which is given by the infimum over real numbers lambda such that there exists a symplectic embedding from E1A into a four ball of size lambda. So this is a function from R to R. And it turns out that C of A for A greater than or equal to 1 completely answers. I guess I'm going to call this Q1, because there's going to be other questions that come up later in the talk. So C of A for A greater than or equal to 1 answers Q1. Well, that might not be immediately apparent if you're not used to thinking about these things before. But the point is these symplectic embeddings, well, I can scale them. And they're also symmetric. EAB is the same as BA. So all I have to know is this function for A greater than or equal to 1. And that determines exactly when an ellipsoid embeds into a ball. So McDuffin's just like we're able to actually just compute this function. And I'll show you the answer. And if you haven't seen it before, I think you'll find it pretty cool. So, well, let me graph the function for you. So like I said, our function eats a real number and spits out a real number. A basic consideration is that C of A is always greater than or equal to the square root of A. This represents the fact that symplectic embeddings have to preserve volume. Well, so now, actually I'm going to move this down a little bit, what the function looks like. So I'll draw the square root of A here. So here's root A. And we only need to know what's going on for A greater than or equal to 1. So at 1, the function is 1. And then what the function does is I go via a line for a little while. So if you extended this line, it would go through the origin. And then at 2, well, the line veers off to the right. So let me say a little bit about geometrically what this is saying. So this occurs at 2. And this occurs at 4. So from 1 to 2, you should try to visualize what's going on, especially if you haven't seen this kind of thing before. For 1 to 2, this is saying that this embedding problem satisfies extremely strong rigidity. So here I have an ellipsoid. Maybe I have an e1, 1.5. And what McDuffin-Schleig's computation is saying is that the optimal embedding into a ball comes from including it into a ball. It's saying that from A between 1 and 2, we can't do any better than inclusion. Actually, that's an old result of Hamilton and Jacob on that. OK, I guess this was known before. It was known up to 2 and 1. But I mean, if you're thinking about these embedding problems for the first time, you should think of this as an expression of quite strong rigidity. However, after 2, the function suddenly veers to the right. And this is a horizontal line until it can't go anymore. So here you should imagine an ellipsoid that's four times as long as it is tall. And this evidently can fully fill a ball of size 2. I mean, it can occupy 100% of the volume. So well then, of course, the function can't continue anymore. So what does it do? Well, it does the same sort of pattern. It continues by a line that if you extended it, it went through the origin. And then it veers off to the right. And it does this again, and it does it again, and it does it again, and again, and again. And it does it infinitely many times until it terminates at, well, this number here is the golden mean to the fourth power. OK, so this is the golden mean to the fourth power. OK. So now, what does it do after the golden mean to the fourth power? Well, this is about 6.8. And actually, it's also extremely interesting after 17 over 6 squared, which, yeah, it should be obvious to all of you once you've seen the golden mean is appearing. So this is about, well, this is exactly 8 and 136. So after 17 over 6 squared, the function just is the volume. So here, c of a is just equal to the square root of a. And so now in between, well, it's kind of a mixture of this behavior. It's mostly the volume, but there's a few times where it rebels and it's obstructed. So actually, right after tau to the fourth, there's a rather long ish step. And then I think there's eight more steps that get rather small. And other than that, it's just the volume. So this is actually the piece that we're going to be most concerned with in this talk. But it's a beautiful result. So I'll give you a second. I'll pause and see if you have any questions. But I guess the way you should think about it is that if the ellipsoid is stretched enough, then all obstructions vanish, except for the volume obstruction. But if the ellipsoid is close to round, then there's this very delicate, of course, there's something else I want to say about this. Maybe you were going to ask. There's some very delicate behavior. So of course, there's something else I wanted to say about this behavior. So like I said, this number is 1. This number is 2. This number is 4. Well, you could ask, what are these numbers here? So I'm going to take a little thing like this and blow up. So if I have a step like this, while the pattern, I'm going to call the x-coordinates. So here the x-coordinates are, I'm going to call this An. I'm going to call this Bn. And then this would be An plus 1. Well, so what the An's and the Bn's are is it turns out they're determined by the odd index Fibonacci numbers. So they're both ratios of odd index Fibonacci numbers. The An's are squares of ratios of consecutive odd index Fibonacci numbers. And the Bn's look like this. And here the Gn's are just, and these are the Fibonacci numbers. So because there's this staircase determined by the Fibonacci numbers, this is called the Fibonacci staircase. So apparently, if the ellipsoid is closed around, there's some delicate combinatorial behavior. So are there any questions about that? Where on the graph is the infimum actually at T? Like for which values of A is there a? Is it actually a minimum? Well, I think it's rarely a minimum. I think it's rarely a minimum. It depends what you mean by a minimum. If you ask for when the interior of the ellipsoid is bent into the closed port, then it is a 10.10. Everywhere where the graph, like at all these numbers, but for example, at 4, technically, this is saying that you can't find an embedding of a closed E14 into a closed B2. You can find an embedding of an open E14 into an open B2. But this is saying that you can find an embedding of a closed E14 into a B2 plus epsilon. Yeah, but then you can prove that that means that the open one, the open one, embeds. You could also just write this function. You could write this function with open ellipsoids and open balls, and then it just would always attain the minimum. Any more questions? Is this map for a lolliport ellipsoid? There's lolliport balls that go down. The actual maps. Yeah, that's a very good question. So there are maybe a few of these that you can write down, but in general, this uses a technique called some plectic inflation, which is certainly rather non-constructive. I want to specifically go ask for it. What? It's the one at four. No, I don't think you can write down the one at four without some plectic inflation. But oh, maybe with another curve or something. Oppstein had a very nice way of seeing that, which I could explain to you. Using the algebraic geometry of Cp2, and you look at the quadric in Cp2, I can explain. And I guess you can probably do also E1n squared or something. Can you do E1n squared into B of n? Can you do or explicitly? OK, so that's part of the four-dimensional story. And well, another part of the story that I want to say, well, there are various ways of thinking about why this is true, but the result, which is going to be of most relevance to this talk, is a result from slightly later, also due to do so. So a little while later, Dusa found a purely combinatorial criterion for explaining when one ellipsoid embeds into another. So Dusa showed that, so now I'm going to go to open ellipsoids to get the cleanest statement, that there is a symplectic embedding of one open ellipsoid into another if and only if, well, the sequence NAB is less than or equal to the sequence NCD. In a minute, I'll tell you what these sequences are, but this is just some sequence of real numbers. And here, less than or equal to means term by term inequality. So this is a purely combinatorial sequence determined by A and B. In fact, what it is is, so NAB is the sequence whose kth term is the k plus first element in the matrix MA plus NB, where M and N are non-negative integers. And sorry, here I mean the k plus first smallest element. And this is index starting at k equals 0. So what this is saying is given positive real numbers, A and B, you take all possible non-negative integer linear combinations, and then you list them in order, including repetitions. You get a combinatorial sequence. And using this combinatorial sequence, you can completely determine when one four-dimensional ellipsoid embeds into another. So just to give you an example of how this N thing works, actually, we're not going to do too much with the combinatorics of the N sequence in this talk. But to orient yourself, for example, if I wanted to do N11, well, the first term would be 0, because I can take 0 times 1 plus 0 times 1. And then I get 2 1s, because I can take 1 times 1, 1 times 1. Then I get 3 2s. And well, so you can associate these combinatorial sequences. And indeed, you can see why this result is true, at least the Fibonacci staircase part, by using this. Although that wasn't how McDuffin Slank did that. What is relevant to our story is that this sequence NAB, well, it looks kind of somewhat beautiful, but somewhat random. It has an interpretation for us that will be relevant. This is the sequence of ECH capacities of the ellipsoid EAB. So here, actually, I guess I'm going to call this embedded contact tomology capacities. I'll spell it out for you. This ended up being messily written. So these are sometimes called ECH capacities. So we'll return to ECH actually quite soon. What embedded contact tomology is, it's a variant of the kind of some plectic field theory ideas that have been discussed here. It only exists in dimensions for contact three manifolds. And you can use it to get four-dimensional, symplectic embedding obstructions called ECH capacities. And from that point of view, McDuff's result says that this obstruction coming from embedded contact tomology is sharp for embeddings of ellipsoids. So you don't have to understand too much about this, because we'll talk about ECH again rather soon. So any questions about any of this? Sorry, could you say that again? Or, clearly, was that how this estimate, half of this estimate is preferred? No, well, I mean, of course, Deuce would be the person to ask. But my impression is that I think I first proved a sort of necessary and sufficient condition for embedding. But it was in 10 terms of NAB, NAB sequence. It was a different sequence. And then Helmut would say, well, you should consider these things, which have been done. And then I realized that the sequence of the obstruction, the way I described it, was the same. So also, Michael showed that there were obstructions in developing the sequence. OK. So this is a very nice story in dimension four. And I guess I'll briefly remark that, of course, ellipsoids are rather simple shapes. You can push these ideas a little bit further. So for example, with Terra Home and Henri de Presse and Alessia Mendini, we've been seeing how this story would work for embeddings of symplectic toric four-manifolds and things like that, toric domains, too. So you can push this story somewhat further. I mean, it's certainly not the case that we have a good understanding of when one symplectic four-manifold embeds into another. But some of this does generalize in a nice way. However, what I want to tell you about today is the story in higher dimensions. So in higher dimensions, the situation is much less poorly understood, even for ellipsoids. Sorry, more poorly understood, even for ellipsoids. So, well, so for example, you could ask, when does a ellipsoid, e, a, b, c embeds symplectically into an ellipsoid, d, e, f? Well, so you could maybe say I should look at the obstruction coming from embedded contact tomology. But like I just said, there is no obvious obstruction coming from embedded contact tomology because there is no known analog of embedded contact tomology in higher dimensions. So there's no known analog of e, c, h. Of course, there is an obvious analog of this NAB sequence. And you could maybe guess, you could still try to guess that this combinatorial sequence still understands the embedding problem. So you could think that perhaps I should put interiors here. You should think that perhaps a result like this should be true. You could think that perhaps a result like this could be true. But unfortunately, this just isn't true. So even this is not true. So we'll see actually a little bit later why this is not true. But some of you maybe have heard of the work of Gooth on the non-existence of intermediate symplectic capacities. And that can be used to show this, for example. So even this combinatorial sequence is just not the right thing. So I think it's fair to say that even for these ellipsoids, we're pretty confused about how the embedding story works. So with Richard Hint, we've been wanting to understand a more modest goal. So I want to understand the following more modest goal. So we want to understand, well, we'd like to understand the following function that's an analog of the function studied by McDuffin's length. So we're going to define C twiddle of A to be, well, the imp over lambda such that an ellipsoid crossed with an R2N, symplectically embeds into a four ball of size lambda crossed with an R2N. So sure, we can fix an N and we can try to understand this embedding function. So if you like this, so of course the talk is called symplectic embeddings of products. Maybe you can see why the talk is called that now. But if you like this ellipsoid paradigm, you should think of this as some ellipsoid where most of the factors are just set equal to infinity. And similarly here. So this is just some question about extremely stretched ellipsoids. So well, now a basic observation is that this function C twiddle, it's always less than or equal to the function studied by McDuffin's length. That's because if I have an embedding of my four dimensional ellipsoid into a ball, I can always just take its product with the identity and get an embedding of the product. So in a way, this is a question about how much flexibility one has for embeddings of products. Can one actually do better than the product embedding? Well, so you might think, now I'm slightly running out of chalk. So you might think that in fact C twiddle of A is equal to C McDuff's length of A. And in particular, then it wouldn't depend on N. However, that's not true. So for example, there's the following theorem due to Richard Heind, which is a kind of refinement of Goose construction. So OK. So Heind has shown that there is a symplectic embedding from well, N, E1A cross R2N into a four ball of size 3A over A plus 1 cross R2N. So well, now I'm actually going to draw a picture for you. So at this point, maybe some of you see why this says that this can't be equal to the McDuff's length function. But I still want to draw a picture because it's going to be very important for understanding what's going on. So note that the McDuff's length function is always greater than or equal to the square root of A while 3A over A plus 1 converges as A goes to infinity to 3. OK. So again, I'll put this in pictorial form. But this says that eventually one must be able to do better than what McDuff and Slank did because there is no volume obstruction to this problem, for example. So by the way, if you're not used to thinking of this. Is it 0 or what's the state of your hand? Yeah, A is bigger than 1. And N is bigger than 1. So again, if you're learning about this for the first time, you should think of A as extremely large, maybe even infinity. And then it's kind of an amazing fact that such a thing exists because we're squeezing an entire R2 factor. We have essentially an extra factor of R2 here. So it's kind of amazing. By the way, to answer something that Vivek said, this actually can be done explicitly. So this is constructed explicitly using a kind of catalyzed folding, which is probably meaningless for maybe even 98% of the audience. But there's a construction due to Dusa, Felix, Slank, and others called symplectic folding. And you can't quite use symplectic folding to prove this. But you can use a kind of catalyzed version, which is very similar. It's really quite a beautiful argument, especially if you've tried to understand what the heck Gooth is doing. At least for some of us, this is kind of a lot easier to understand this catalyzed folding picture. So this is a very explicit embedding. OK, so anyways, now I want to draw you a picture. Would it be possible to prove any result like this without an explicit construction? Yeah, that's a very good question. So at the moment, most of our only ways of producing embeddings come from some explicit, in higher dimensions, come from some explicit construction like this. I mean, of course, one could take a non-explicit construction of embedding of an ellipsoid into a ball and then cross it with the identity. But yeah, in dimension four, we have these symplectic inflation tools, which are very powerful. And there's not a good analog of that in higher dimensions. So most of our ways of getting non-trivial higher-dimensional symplectic embeddings are pretty hands-on. So you're not saying that Sharp is just constructed one for each end? Well, we'll get back to the Sharpness question in a moment. But yeah, at the moment, I'm just saying he constructed one. And if you think about it, that shows that the McDuff's length result cannot possibly be the answer here. So let me draw you a picture to underscore that. This picture is just supposed to emphasize. So I'm now going to try to graph C twiddle of A. Well, the graph of 3A over A plus 1, well, I just have to start it at 1. It's asymptotic to 3. So here's 3A over A plus 1. Now, this square root function, well, it is a square root function. And in particular, it crosses the graph 3A over A plus 1 at some value. And remember, C of A, the McDuff's length function, is always above this white curve. Whereas what we're saying is, so the hind result says that whatever our function C twiddle is, it's always less than or equal to 3A over A plus 1. So evidently, once we get past this point, this product function cannot be equal to the McDuff's length function. So now a natural question is, what is this point? Well, this is kind of a fun exercise. So is anyone good at mental math? What is the point where 3A over A? What is the A value for which 3A over A plus 1 and square root of A intersect? Well, it's nothing other than the golden mean to the fourth. Yeah, exactly. So this intersection point is the golden mean to the fourth. And in fact, if you draw the Fibonacci staircase, what you'll find is it actually bounces between these two graphs. So the Fibonacci staircase looks something like this. In other words, the upper graph here does clip the top of all of the steps. So based on this, Heine conjectured in an earlier paper that C twiddle of A is actually equal to the McDuff's length function for A between 1 and 10 to the fourth. And Richard and I proved that. So the theorem here, this is due to Richard and I, is that C twiddle of A equals C of A for A from 1 to 10 to the fourth. So again, if you think geometrically in terms of what's going on, it's kind of a cool situation. This is saying that the optimal embedding in the Fibonacci staircase region is just given by lifting the McDuff-Schlank embedding. It is just given by a product embedding. And then that's the only part of the McDuff and Schlank graph that persists for this problem. Once you move past the Fibonacci staircase part, the rest of the graph is not relevant. So somehow this Fibonacci staircase really is a robust feature of the embedding problem. Do you know if the C twiddle depends on n at all? No, that's a great question. So of course, this says that C twiddle doesn't depend on n from 1 to tau to the fourth. Richard and I are currently thinking about what happens after tau to the fourth. And we don't know. I mean, it's actually a fun question if you like trying to figure out which moduli spaces of holomorphic curves are empty or non-empty. So I would guess that it does not depend on n. But I have no way of showing that. I should say, actually, that there are a few more points that are known due to Richard. So actually, also if a is equal to 3d plus 1, where d is an integer, then the graph is also given by the yellow line. So there are a few more points on the yellow line. So the 3a over a plus 1 graph line. So that's when you know that C. So you actually do know if this is a equals 3d plus 1, where d is an integer greater than or equal to 3, greater than or equal to 3, then you actually do know that these are correct values of C twiddle. And these also don't depend on a. So this is an earlier result due to Heind and really Ellie Kerman, too. So this is this Heind-Kerman thing. OK. So I want to say a few words about how the proof goes. It's sort of good news for you, because this is a conference on moduli spaces of holomorphic curves. And the proof involves a lot of tinkering with moduli spaces of holomorphic curves. Also, interestingly, it uses embedded contact tomology quite a lot, which is interesting, because it's a higher dimensional problem. So are there any questions about this, though, before I give some sketch of how the proof goes? Are you going to say why it's a critical number? I don't know why it's a critical number. I mean, especially if you look at this folding construction, you would maybe never guess that it's ever sharp, because it just looks like you kind of do this, you kind of do that. So it's sort of amazing that this is somehow a pretty persistent feature of the graph. But actually, it does seem to be. So again, I was saying with Terra Home and others, we've been thinking about how this story would work for other, say, toric targets. And this kind of part seems to be a pretty persistent feature of what's going on. Actually, there's also some very interesting commonatorics in Tau to the Fourth, involving Earhart polynomials. So you can find triangles with an irrational slope given by Tau to the Fourth that their lattice point enumeration looks like the lattice point enumeration for an integer triangle. That's something very special about Tau to the Fourth. So would you just to describe your argument in general, would you say it's like the common hind argument that's just up in a slightly different setting? That's right. Yeah, in fact, I was going to say, yeah. So it's pretty similar to the Heine-Kurman argument, but with a new input from ECH. Right. Well, I mean, they sort of were using ECH. They sort of do, but it's more explicit. They do some sort of neck stretching kind of thing. And yeah, we don't do that. So OK. But actually, I want to answer Deuce's question and even write something on the board. So what's the idea? And well, some of you are probably not used to thinking about these symplectic embedding problems at all. So I'm going to spell this out in a little bit of detail. So it actually suffices to show that C twiddle of the B ends is the same as the McDuff's Lang function of the B ends. That might not be immediately apparent, but that's not that hard to show. So what I'm saying is, remember, we want to show that the graph is given by the purple thing, purple in honor of Joe, by the way. So we want to show that the graph is given by the purple thing. Well, what I'm saying is it actually just suffices to show that these are the correct points by some fairly simple elementary observations. I mean, it helps to notice that these lines actually go through the origin, which you would never guess from the way that they're drawn. But they should go through the origin. And similarly, these lines are horizontal, which really helps. So that's just a rescaling argument that we used. That's exactly just an argument that was used by McDuff and Schlanck. So we just want to show this. So now we know that C twiddle of Bn is less than or equal to C of Bn, because as I said, you can just lift the McDuff Schlanck embedding. So you should think of what we want to do is we want to find an obstruction to finding a better embedding. OK. So find an obstruction to a better embedding. And well, if you're, so I'm going to make that 0.1. If you're not used to thinking about these symplectic embedding problems, well, I want to make the following point, which is that, generally speaking, that obstruction should come from some sort of holomorphic curve. I mean, of course, there are other obstructions to symplectic embeddings that are a little more variational in nature. There's these Eklund Hofer capacities. But I think it's fair to say that it's reasonable to look for a holomorphic curve that's somehow obstructing something better. OK. So now we just want to find the right curve. And this is where we use the Hein-Kirman argument. So you should think of the curve. It's going to live in the completion of some cubordism of high dimension. I mean, in this case, it can really have any even dimension. So we want to find some holomorphic curve living in a completed symplectic 2n manifold. So the Hein-Kirman idea is to find a curve in a four dimensional cubordism that lifts nicely to higher dimensions. So that's a little vague. I'll fill in some of the details later in the talk. But this is where ECH comes in. So the way you should think about it is go back to the McDuff and Schlank picture. Of course, the McDuff-Schlank statement that these are points on the graph is also an obstructive statement. There's an obstruction component to that that you can't do any better. So the idea is find the holomorphic curve that's giving McDuff and Schlank's four dimensional obstruction. And then, well, this thing is like a product. So roughly speaking, what you want to do is you want to take that holomorphic curve and instead of mapping it into the completion of the ellipsoid cross R2N by, say, sending it to 0. And so then to find that holomorphic curve, we want to use embedded contact tomology. So that's the general idea. And yeah, this is essentially the Hein-Kirman argument with this ECH used somewhat differently. Are there any questions about that? What is the VM in your supply structure, the VM sequence? Oh, this was this discussion about how, as you can see, these lines go through the origin. And so Dusse and Felix have this nice argument where, well, these ellipsoid embeddings, they satisfy some basic axioms. I mean, so for example, they're monotonic. Like, this function has to be a priori monotonically non-decreasing. And it also has to satisfy a subscaling property, meaning once I have some point on the function, if I draw the line through the origin, connecting that point to the origin, the function always has to stay below that line, or it can't go above that line. So using those two points together with the McDuff and Schlank information, you can recover the whole graph. And this is kind of a version of an argument that was used by McDuff and Schlank in their paper. So there's no sort of interesting holomorphic curve theory here. It's just the first principles of the embedding problem. OK. So now let me be a little bit more explicit about what I mean by how I actually want to do this. So we first want to give a brief crash course on embedded contact tomology. So this is going to be a very brief crash course, hopefully more of a course than a crash. And again, the motivation is we want to use embedded contact tomology to find this curve. Now we have our ECH crash course. So you should think of this as kind of the complete opposite of a mini course, where everything is done carefully and in much detail. OK. So for ECH, I have a three manifold. So now y is a three manifold, and lambda is a non-degenerate contact form. So now the embedded contact tomology of this pair, well, it's a variant of, say, some plectic field theory. OK. In other words, well, it's the homology of a chain complex, I need better chalk, which is generated by certain sets of ray orbits. And the differential d, certain, well, embedded, except there may be just mostly embedded curves in r cross y. So I mean, we've been doing so much with contact tomologies and this kind of thing at this conference that I think I just want to draw your picture. So the picture you should have in mind is really the kind of curves that get counted. And while the kind of curves that get counted, you should maybe just have a picture like this, say, by which I mean to illustrate the following. So the curves can have genus. They can have arbitrarily many positive punctures. They can have arbitrarily many negative punctures. That's why I'm saying that the chain complex should be generated by sets of ray orbits. I mean, in cylindrical contact tomology, you just have a cylinders. You have one orbit up and one orbit down. You have a whole set of orbits here. But we don't actually want to count all such curves. We want to single out embedded curves in some clever way. And we end up singling them out by this thing called the ECH index. So specifically, we count ECH index 1 curves. And that somehow forces the curves to be essentially embedded. What I mean when I say mostly embedded is a component like this actually has to be embedded. But if I have a trivial cylinder here, this could be multiply covered. So this ECH index is a great thing. This is all due to Hutchings, of course. And well, we're actually almost done with the crash course. But you might ask, why are we doing this? Well, one reason is that embedded curves are very nice. I mean, certainly the kind of transversality issues for these embedded curves are much more mild because they can't be multiply covered. But another reason, which is a bit cooler, is that this embedded contact tomology, well, it looks very much like a holomorphic curve theory, but it ends up being isomorphic to a gauge theory. So Taubes has shown that ECH is canonically isomorphic to a version of cyber-gwitten fluorochomology. So this star here is just some grading business, which we don't really want to talk about. So this is the cyber-gwitten fluorochomology. And well, I don't want to say much about this, but this is a gauge theory. It also only depends on the three-manifold y and not on the contact form. So somehow embedded contact tomology allows you to encode topological information about the three-manifold coming from gauge theory into information about the contact and symplectic geometry of the three-manifold and associated four-manifolds. So that's sort of the end of our crash course. But for our purposes, actually, what we care about is that we can use this isomorphism to define kabordism maps. We can use this to define kabordism maps, by which I mean, well, if there's one thing that should probably be clear from this summer school, if you want to define a kabordism map on some version of some plectic field theory, normally that gets extremely hairy. However, cyber-witten fluorocomology does admit kabordism maps, and this has worked out in nice detail by Kronheimer and Morovka. And so we can use the fact that cyber-witten theory admits kabordism maps to get a kabordism map on embedded contact tomology. Now, a priori of that kabordism map then would count some kind of cyber-witten monopoles, but Hutchings and Taubes have shown that these kabordism maps satisfy some nice axioms. In particular, they satisfy a holomorphic curve axiom, which will be very useful for us. In particular, a morphic curve axiom. So remember, I said before that we want to use ECH to find some nice four-dimensional curves. And that's the idea. We're going to induce a kabordism. That's going to have a kabordism map. We're going to apply the Hutchings-Taubes holomorphic curve axiom to get a holomorphic curve. And then we're going to pay very close attention to it to make sure that we can lift it correctly. So any questions about that? Are you going to say what an ECH capacity is? No. I will not say what an ECH capacity is. Interestingly, yeah, you can ask me later. I don't refuse to ever talk about that. Interestingly, we don't actually have to know much about ECH capacities to do any of this. And this is maybe an idea that Michael has recently put in writing, which is that you probably should really think about the curves themselves and not so much these capacities, which are kind of user-friendly, but for very delicate applications, they lose some information. So we're much more interested in the curves themselves. But nevertheless, I'm happy to talk about it. OK. By the way, this symplectic caportism, I could write more about what I want it to be. In this setting, I actually want it to be, well, you might think I want it to be an exact symplectic caportism. I actually only want it to be weakly exact. But I think I'm going to skip those details. But you can ask me later. OK. Where does this portism go between what and what? I haven't told you. So here, this is part of our crash course about ECH. I'm saying any time you have an exact symplectic caportism between contact manifolds, you get an induced map on embedded contact demology. Now I have to tell you how to induce that caportism for the purposes of our problem. And I'm saying you would think that you want an exact caportism, but our caportism is actually only going to be what's called weakly exact, which is fine for what we want to do. But anyways, I'm going to say what the hold of it more than code max it is. More or less, yes, at a later moment. So I'm now going to, so this is all just general facts about ECH. I'm now going to induce for you the caportism that will be relevant to us. OK. So I now want to go back to the proof. OK. So I want to produce some sort of symplectic caportism. And well, the general game here, which is certainly familiar if you've done this before, is that the caportism should come from an embedding. So by the McDuffin-Schlank computation, so by McDuffin-Schlank, we know that I can embed, well, an ellipsoid stretched by a factor of bn. So remember these bn's, these were ratios of odd index Fibonacci numbers that are off by 2. So I can embed this bn. I actually want to add a little epsilon for sort of technical reasons that you can disregard at first. And I want to embed this into the interior of, well, I'm just going to write this as an ecc plus epsilon, where this c is equal to c of bn plus epsilon twiddle. OK. So remember that c of bn was itself gn plus 2 over gn plus 1. OK. So there's a lot of epsilons here. And they're put in here because of the convention. I learned this phrase from a paper by Obstein. They're put in here because of the convention that statements should be correct instead of approximately correct. But you should maybe ignore them at first reading. So the reason we're putting these epsilons in here is remember, we wanted a non-degenerate contact form. And if I took a ball here, I'd have a more spot contact form. So you really want to think of this as a kind of ball, but I'm just perturbing it a little bit to be irrational. And similarly, bn is a rational number. So I want to perturb this to be irrational. And also, there was this point that Duce and Nate and I were talking about that I can't actually embed this into the interior for what's coming. So I might have to make c a little bit bigger because it's just an inf. But so if you're learning about this for the first time, you truly can just ignore all the epsilons. And that might help your sanity. So I'll draw you a picture. So this is the kind of picture that you should have in mind. Here's my ecc plus epsilon. And then my ellipsoid is just embedded in here in some strange way. So here's an e1 bn plus epsilon. So now I can write down my cabortism. So I'm now going to remove, well, psi of the interior of e of 1 bn plus epsilon to get a cabortism x. One second. It was close to the epsilon. This is the whole point of the equation. Yes, that's right. So this gives you a weakly exact some plucked cabortism. So you should think of the cabortism as this thing x. So here's my cabortism x. So I just removed the interior of this ellipsoid. It's sitting, the whole ellipsoid is itself sitting in the interior of the ball. And I remove it to get a cabortism x. So now my holomorphic curves are going to sit in the completion of x. So I want to complete this by adding cylindrical ends. Yes, it means that the, see, the issue is that this is a symplectic embedding, but it might not be an exact symplectic embedding. So I want to complete x by adding cylindrical ends. OK, so again, weakly exact is maybe not so important. I'm just saying, well, in an exact symplectic embedding, you would like your symplectic form on here to be d of a 1 form that restricts to the boundary as the relevant contact forms. But we are not given an exact symplectic embedding. So we can't necessarily find that. However, all we actually need is we need the symplectic form to be exact because we want to rule out some kind of bubbling, for example. And we want it to have, we want its restriction to the boundary to be d of the relevant contact form. So that's actually all you need to define cabortism maps. But I assume most people are more used to thinking about exact symplectic cabortism. So you don't lose much. Well, I guess I'll just write to both sides. So this is a cabortism with two boundary components. And I want to add a cylindrical end to both boundary components. So what x bar looks like is it's x. And then I add a cylindrical end to the top. So I add a 0 infinity cross a boundary of a ball. I mean, it's actually a C, C plus epsilon. And I add that on, on the boundary of the ball. And then I also add on a negative end on the boundary of the ellipsoid. Does that make sense? So you take this thing and you add an end to the top and an end to the bottom. This is a pretty standard thing in this story. So the curve lives in x bar. So our curve will live in x bar. In a moment, I'll draw the curve for you. But let me answer Vivek's question and tell you a little bit about what the holomorphic curve axiom says here. So anyways, I've produced this cabortism x and I'm attaching it to get a coarsen x bar. So now remember, ECH admits cabortism maps. Because ECH admits cabortism maps, I get a map, which I'll call psi twiddle, because it's induced by psi, mapping, well, the embedded contactomology of the boundary of this ball to the embedded contactomology of the boundary of the ellipsoid. And it will satisfy the holomorphic curve axiom. But so now first, we want to do a very easy calculation. So an easy calculation, well, so this easy calculation involves the following. I want to tell you a few things about the embedded contactomology of this. Of course, I haven't really defined embedded contactomology in any rigor. But remember, there's some rape orbits involved. So I want to tell you what the rape orbits are at least here. So the boundary of this irrational ellipsoid, well, it has two rape orbits, two embedded rape orbits. Yeah? I think I'm going to channel my inner Muhammad and take eight. But then I'll wrap it up. OK. That seems to be a popular strategy in this conference. But I will take eight. Well, maybe I should stop pretty soon. I'll take a few more minutes. Yeah. OK. So this takes two. I just want to wrap up at a reasonable place. And I'm not going to be able to wrap up in one minute. But I promise I'll be able to wrap up relatively soon. OK. So this has two embedded orbits. Shh. Let us continue. Yeah, yeah. OK. So this has two embedded orbits, which I'm going to call alpha 1 and alpha 2. Well, the action of alpha 1 you can see is c. And the action of alpha 2 is c plus epsilon. And well, but yeah, I apologize for going over. I really do. I just want to finish this general train of thought. And well, the boundary of the other guy similarly has two orbits, beta 1 and my conventions are the same. So you think of beta 1 as short and beta 2 as the longer one. So anyways, the point is you can show that there exists. You can show that my map phi twiddle, well actually remember, in each stage I'm allowed to take orbit sets. So if I take this orbit alpha 2 and take gn plus 1 copies of it, I can show that that gets mapped to beta 1 with gn plus 2 copies. And so now the holomorphic curve axiom. So the point of all this is that the holomorphic curve axiom now says that there is some broken, some possibly broken, holomorphic curve alpha 2 to the gn plus 1 to beta 1 to the gn plus 2. OK, I promise. I really am almost done. So just to emphasize Vivek's question, so what the holomorphic curve axiom is saying here because I know that the kabordism map sends this element of ECH to this element of ECH. I know that there's some, well the holomorphic curve axiom says that there's some holomorphic curve, but it could have multiple levels. Here you can show, however, that in this particular case, partially because of the nice properties of the Fibonacci numbers, this curve has exactly one level. You can show in this case there's exactly one level. And you can actually write down the curve. So the curve looks like, well, and I'll draw it for you. So the curve looks like this. So there's gn plus 1 positive ends that are simply covered and 1 negative end, which is a gn plus 2 fold cover. So you should think of this as the quote Fibonacci curves. OK, so now I'll just say the point and then we can go eat lunch. But I do want you to have this picture in mind, because these are very nice curves. So now the point of these curves is if you stare at the index formula, so these are index 0. That's sort of just a general thing that happens in this kabordism map story. But the point is if you stare at these curves and apply a Fibonacci identity, you'll see that when I lift these curves, essentially by just crossing with r to the 2n, then they still have index 0. In principle, when I put it in a different modular space, its index could jump. But if I write down the modular space correctly, these have index 0 in all dimensions. And so we can use them to do the Heine-Kirman argument. What is the Heine-Kirman argument? Well, remember, I want to show that I can't do better than the product embedding. So what I do is I take some random embedding, which might not be a product. And then I can isotope it to the product embedding if I allow myself to shrink the ellipsoid. And so by doing that, I can get a one-parameter family of almost complex structures and try to move this curve in the normal way. So then I just have to prove a compactness result, showing that these curves actually persist. And I also have to do some delicate sign counting to make sure that there weren't two of them to start with. You can do all that. And what you find is that even if I had some non-product embedding, this curve is still living there to obstruct it. So anyways, you have to do this compactness calculation and also a sign counting. But the picture that you should have in mind is really this picture, because these are nice curves. OK, so I think I'll stop there. Sorry for going over time. I apologize to the organizers.