Pt. 4 – Lagrangian surfaces in 4-manifolds | Joshua Greene, Boston College | IAS/PCMI

wonderful finishing run. >> Thank you. Thanks very much, Cordonna. Uh and uh airing on the side of too much uh appreciation as the last speaker of the graduate summer school, I felt that we should join one another in a round of applause uh to thank the organizers of the the graduate summer program. THEY MAY BE ontologically distinct from the same set of people who organized the research program. So they may be getting another round of applause later this afternoon. Right. So I want to wrap up today giving a little context for the two main theorems that we proved during this lecture series and then uh wrap it up with uh an application to inscription problems in Jordan curves in the plane. I'm audible in the back. All good, thanks. Okay. All right. So, the the one big theorem that one big theorem that we proved is a theorem of Chenov uh from uh the mid 90s. And what he did was he proved that the this construction which we discussed of the chenov taurus is it's a monotone langium in standard simplectic C2 um which is not Hamiltonian isotopic to the Clifford Taurus. Okay. Uh so I just wanted to give a little context for that. I believe that result came as a surprise and that in fact there was some effort to show that the Clifford Taurus was unique and it you know the unique monotone lrangeian Taurus up to Hamiltonian isotopian C2. So I think this example came as a bit of a a surprise and at the moment it's the only other example known and to my knowledge it's open whether every lrangeian taurus in C2 is Hamiltonian isotopic. to one of the examples that we have seen. So a product of circles which includes the case of uh Clifford Taurus or a Chenov Taurus. Okay. Okay. So that that's open to my knowledge. Uh looking in other four manifolds, it was uh it was a while before monotone lronians were produced in the next simplest four manifolds. So the simplest closed examples uh CP2 and a product of S2s of equal area. Uh so in fact I I'm told it was it was believed that oh so there's an analog of the Clifford Taurus in CP2 and in the product of S2s and I'm I'm told that it was believed for a while that the Clifford Taurus was the unique monotone Taurus in CP2. Uh but eventually an adaptation of this construction was found. So there exist uh analoges I'll just put it like this there exist chenov Tori Felix Schlink gets credit um in CP2 with the Fubini studio form and in product of S2 with itself S2 thought of as CP1 carries a Fubin duty form. Um, so there are examples of these in in these manifolds and I think this some believed for a while that maybe the the Chenov Taus and the Clifford Taurus were the unique monotone toi up to Hamiltonian isotopia in in CP2. Um but people uh working on the other side of the mirror in mirror symmetry um I'm not sure what the right word is discovered or predicted the existence of other monotone langian toi in CP2 and these were written down explicitly uh by via in his thesis. So Vanna gives uh a family. So he goes uh Monotone lrangeian toi in CP2 um which are parameterized by markoff triples which have come up in a couple of the research talks this week. So I remind or maybe you didn't go to those. These are triples of of whole numbers satisfying Markov's equation. So the sum of their squares is equal to three times their product. And what to say? So to each markoff triple well there are infinitely many solutions and they're described nicely by uh the Markoff tree. To each one BA associates a monotone lrgonian Taurus and he proves that they're pairwise distinct up to Hamiltonian isotope and the way that you can distinguish them is very much in line with how we distinguish the Czechonov Taurus from the Clifford Taurus. You distinguish them by looking at the Moslov 2 Jholmorphic discs with boundary on them. And you can distinguish them um by doing a little more than counting but by separating them out according to their relative homology classes. So they're all distinguished this way. And one may wonder therefore uh if these are all of the examples in CP2 or if there are other examples and at the moment these are the only ones known. There are versions of vioni in other spaces including in S2* S2 but in S2* S2 there are other examples as well distinct from these. So yeah, this is the state of affairs as as far as I'm aware regarding the Hamiltonian isotopic classification of monotone lronians in in the pretty simplest simplectic for manifolds you might think to study. Um okay so I just wanted to point that out. questions, corrections the okay so I think this is a fascinating topic and I I didn't at all discuss the question of the relative case where there is actually a lot more progress in momentum recently which we have been hearing about in the research talks the relative case for example you would take uh a fourball in standard simplectic C2 do and put a Leandrian knot in its boundary and seek to classify all of its lrangeian fillings or exact Lronian fillings and there there's a lot of progress and there are classification results which are known prominently the the unnotion filling the balshber and and we we recently learned that the hop link has Okay. Uh so that that just puts the chenov construction into a little bit of context. The other theorem that we proved was uh due to pterovich and independently the turbo. And this was the theorem that every Lronian Taurus embedded in C2 has a loop of MOSFE index 2. And we proved that by picking a a J and showing that there's a a J holorphic disc with boundary on the Taurus with MOSF index two. So the context for this was that earlier there had been a theorem of Gromov that was kind of the template for the proof that we went through yesterday. Um and it says that um well in various forms um so I'll put it like this. If if we have a a closed langian in CM is a closed lronian. uh then okay we pick an almost compatible almost complex structure and we can find a jholorphic disc with boundary on this lrontian and that has the consequence that um L is not what's called exact which we didn't pursue in this lecture series so there's no exact closed lrgonian submanifold CN and that had been a a question raised by Arnold that had been a question of Arnold which Gromoff solved using pseudoholorphic discs. So L is not exact and moreover by looking at the boundary of that discil class is positive tells you that it has to have infinite order in firstmology. So this is not a remarkable result in the setting of C2 but in higher dimensions it it limits what you might see. Okay. So in in between Gromov's theorem and the ptertovichbo theorem there had been a conjecture due to Michelle Odan and I'm not sure what the most general form of it is but the form I'll state is that every langian taurus in a simple vector space has a loop of moss off index 2 and okay so that's obvious when n equals 1 because a lronian one taurus is just a smooth jordon curve and we know that that has ms of index 2 in dimension two it was confirmed turned by Ptertovich and Burbo and there was a lot of partial progress on it. For example, you might condition on the Lrronian Taurus being monotone and there were some results by O which proved this up to N equals something in the 20s. Um and then done in full generality by Lebuy by a very elegant argument using flur theory. And this has now been proven in in full. So it's it's proven in in two very different ways. So there was a proof due to silly monk which uses simplectic field theory techniques and it was also proven by um some foundational work due to Fukaya and then some extremely important technical work due to IRA. um which uses let me in a in a in a slogan string topology ideas and they both prove more general results and the more general one proven by Fukaya and IR is that in fact um this is this generalizes to aspherical uh spin the grounds. Okay. Um and you might wonder how general this you know how how much you could generalize a dance conjecture. And I I just want to point out um a very curious example which I think should be studied more which is uh I'll just abbreviate the initials and say them out loud. uh Echol, Ellie Oashberg, Murphy and Smith describe a Lronian embedding of S1* S2 into C3. So that's not a spherical, right? There's a two sphere. Um and if you look at the the first homology, it's it's infinite cyclic. So you would wonder if that generator has to have moss index 2 but they give an example with vanishing moss up class. So it's a little unclear how far one might hope conjecture could generalize. Um, and there's there's energy in this direction as well, which I'm not going to describe. Okay. Questions on context. I hear a question. What does moss off index zero mean? Somehow you know this you have this generator and you look at the tangent planes along your lrangeian to it and that's a no homologous loop in the lrangeian crossmanian. The example is described using some H principle argument and it would be wonderful if there were an explicit construction. Similarly, you know, the the Vion toy were predicted. Um, you could say that they were known to exist, but Biana gave us a beautiful construction of them. And so that would be a nice problem to to to solve here. Give an explicit description of these examples. The Vanna include the Clifford Taurus and the Chakenoff Taurus. So the Markoff triple 111 corresponds to the Clifford Taurus. the Markov triple 112 and its permutations correspond to the Chenov Taurus. uh and Biana first described a new example corresponding to the Markoff triple 125 and then that broke open the floodgates and he gave examples for all Markoff triples and you know the examples are related to these you know mutations and and um theian systems that uh Roier discussed during his talk amusingly uh sillybach I I watched a video of him lecturing about the the proof of this conjecture and said as far as he knew he he didn't know any application of of the result what what good amos index 2 loop on a lronian taurus uh what use that could have okay so I want to show you a use of that and uh to get us there I I want to begin by thinking a little bit about lrangeian immersions. Um so lrangeian immersion is just going to be an immersion. We're just going to immerse into the simplectic vector space. And so I have tangent planes everywhere and I can ask for the simplectic form to vanish on those. So it's just the condition that when I pull back the simplectic form to get a two form on L as an abstract manifold that that two form vanishes identically and N is the dimension the real dimension. Okay. So lrangeian immersions have mauses and leal classes. So these homorphisms on ordinary homology um well the Moslov class is integral value but I'll write it like that. Um these these are defined in exactly the same way that we defined them in the closed case or rather in the embedded case. So so the definitions extend at once. Um let's do an example. I'm going to take a smooth jordon curve in the plane like I like to do and I'm going to assume that zero is on my curve. Okay. So there it is and there's the origin. It's the origin. And I have a Hamiltonian. My Hamiltonian from the very first lecture corresponding to the simple harmonic oscillator half the magnitude of Z 2. And we I'm gonna I'm going to label this gamma zero. And I'm going to look at the time t flow of my Hamiltonian for some small amount of time and apply it to gamma 0. And what we know happens is that we're just going to be rotating gamma 0 uh through t radians about the origin. And if we do things correctly today, we would be rotating it t radians clockwise. So that has to move out of the way. Okay. So I come up here and I got a picture like that. So that's the origin. Here's gamma 1. Gamma 1 is the image of gamma 0 under this time t flow of h. And now I have this uh funny intersection point at the origin. Uh and I can do surgery there. Okay. Um so I drew the origin very large around it so you can remember it was there. And I'm just going to do pterovich surgery. And that'll have the effect of smoothing these out. And I'm going to let gamma denote the result of this uh this surgery. Okay. This is an immersed lrangeian. uh its fundamental class I'm going to represent by gamma and I can compute the mosoff index because all I'm going to do is I'm going to run around gamma kn and be I'm about to close up and I pick up essentially mos index two and then as I run around gamma 1 uh I get another copy of two okay so you can convince yourself that I get 2 plus two as my computation and that's four and that's the generator of homology. So every loop on this lronian will have mausov index a multiple of four. But we proved that every uh we proved that every embedded lrangeian in the plane has a loop of moss index two. And so that's a proof that that curve gamma is not embedded. So that's an application of uh simplectic geometry to properties of immersed curves in the plane. Questions? Question. Evan, >> not at all. >> But there's a method to my madness. We're going to emulate this and build a Taurus, an immersed lrangeian to Taurus, and we'll prove that it's not embedded in much the same way, and that'll have a more interesting consequence. Great question. Okay. So I'm going to prove the theorem and then state the theorem. So here's the proof. So gamma is an arbitrary smooth Jordan curve as before from the plane and I'm going to form the product. Okay. And I'm going to like in the example define a Hamiltonian which we haven't seen before but it's very similar to the two examples that we have seen. So it takes as input a pair of complex numbers and one example we saw was half Z ^2 minus half W ^2 that was used in the construction of the Czechonov Taurus. Z and W sort of operate independently in that version. Uh here I'm going to look at the magnitude of z minus w^ squ and I'm going to put a quarter out front. That's just uh the right thing to do. Okay. And let's examine what the time t flow of this does. So might you have a guess? Maybe just as a remark, if Z and W are equal to one another, then the H not only does the Hamiltonian vanish, but it its derivative vanishes. And so that tells me that dually the Hamiltonian vector field is going to vanish on the diagonal in C2 pairs of points you know of the form Z comma Z the Hamiltonian vector field vanishes along that. So my Hamiltonian flow is going to preserve the diagonal. So whatever formula I wrote write down it better be manifest that the diagonal remains fixed. Okay, here's the formula very elegant. So less facitiously uh if you know I want to visualize what's going on. I'll start with the definition of C2 as you know an ordered pair of points in the complex numbers and what you believe is that what you can work out is how this works is I I take the line segment joining Z and W that's going to be a very short line segment if Z and W coincide I mark the midpoint and then I rotate through t radians about the midpoint and I swing them into a new pair of points z prime and wp prime and you work out that that's what happens and just as a sanity check if z and w are drawn into one another uh then they are fixed by this the z prime and the w prime would be the same so for the maybe third time. Uh this flow fixes the diagonal in C2. And because I'm going to apply this in just a moment to L0, I'm curious what's happening when I look at the image of L0 under this map. And uh this diagonal passes through L0. It passes through L0 in uh diagonal copy of the Jordan curve. So this is some you know it's it's kind of like the embedding of the Jordan curve on the diagonal in in the diagonal. I'll denote that delta of gamma my notes out of order. Okay, so L1 is going to be the lronian I get by flowing for time t. flowing my lronian l0 for time t along the Hamiltonian vector field. So it's another lronian taurus and I'll draw a bit of it. I'm going to just draw a cylindrical bit of it and on it is that diagonal loop that was unmoved by the Hamiltonian flow. And I'll put into my picture the original langium L0. So this picture is meant to demonstrate that L0 and L1 intersect along that loop, but they may have some other points of intersection as well. Is it okay? Okay. Um, now I have uh two steps to take which commute with one another. Uh, and one of those steps is a two-part step. So, I'm going to do the first part of that step, commute the other thing, pass it, and then do the second part. Okay. So let's do a quick computation. Okay. So this this diagonal loop sitting within the lrangeian L0 it represents a homology class. That homology class is it's the diagonal, right? So it's homologous to gamma times a point plus a point times gamma sitting within L0. And you may recall or believe that we computed the Mauslo numbers of these homology classes. They're both two. And so what I conclude is that the Mosoff number of this diagonal loop on this lronian is equal to four. you know that happens. And what I'll do is I'll I'll take that that loop the diagonal and I'm just going to take a translation of it on L0 which is uh a little bit further away from the action in this picture. I'm going draw it nice and thick because I'm using a color which is been invisible at times during these lectures. But there it is. So that's alpha. Is that uh is that invisible to anybody who wants to see it? No. There it is. Um it's homologous. And so, oops. So, I've produced a loop on L0. I'm calling alpha and it has MOS of index 4. Okay. So, that was step one a. Okay. Now uh for step two before this picture below goes away I'm going to just do something totally analogous I'm going to try to do a pterovich surgery around that place where the Hamiltonian vector field vanished. Poltervic surgery we learned is what you do when you have a pair of lronians which intersect one another transversely. Um here we have something which is clearly not transverse. That diagonal loop is a whole circles worth of intersection between them. Uh but it's the next best thing after being a transverse intersection. This is a clean I'll define component of intersection between these toi. What that means is that the tangencies along that component are just the intersection of the tangent planes of the constituent lrangeians around it. Um that is to say that the tangent space for every point chosen along this loop. The tangent space Along the loop is just the intersection of the tangent spaces. So these are nice. You encounter these uh you know when you pass from looking at morse functions to morse bot functions. And there's a nice local model for a clean component of intersection analogous to the one that we used to do Pterovich surgery. This local model was written down in the thesis of Posniaak. And what it tells us in this particular circumstance is that we can find a neighborhood of this clean component of intersection which is simpletoorphic to um a product. So we're working in a for manifold. So, it's going to look like uh an x1 y1 plane multiplied by an x2 y2 plane with a dx1 wedge dy1 plus dx2 wedge dy2. I'm not going to write that all down. And it's going to look like taking a circle in the x1 y1 plane and multiplying it against the axes in this x2 y2 plane. So under this simpletomorphism what I'm saying is that L0 is what you well the intersection of L0 with the neighborhood is what you get by taking the circle times the X2 axis which is a cylinder like in that picture and L1 is circle times the Y2 axis. another cylinder. Okay, so that's that's the local model. Those are grions. So looking at that picture suggests what we would do to make a smooth in. We would take I have to be careful to kind of coincide with my axes at the, you know, in a in a toward the perimeter of this local neighborhood. But then I just smooth smooth those in a kind of a hyperbolic way. So I'm doing a kind of a version of ptertovich surgery adapted to the presence of a a clean component of intersection and I'm just doing it in this very special case so there should be no argument. Okay. So let's insert that local model over here. And this is where the 40 representations on the blackboard really begin to break down. Um but so my neighborhood was right in right right in here. It's say sandwiched between these two inter interesting intersection points which are not in this neighborhood. Um but this one's going to bend out of the way of the other one. And likewise the The lobe on the right bends out of the way and they do not intersect. Okay, don't intersect. So I did a lronian smoothing there. Ptertoovich smoothing. Pterovich surgery of a kind. Um, and now what I'm left with is an immersed lrangeian L. And topologically tell me what is L? It's a Taurus. You believe that? We took two toy and we cut each along uh an essential curve and we glued up in a way that nicely respects orientations. So like in the picture that was here before I got after post surgery I got this immersed loop which is sort of like twice as long as the original loop. Now I'm getting an immersed lrangeian Taurus which is in a very imprecise sense twice as long as the original one. So it's like a d effectively a double cover of the original Taurus. Happy there were two choices for the resolution. It doesn't matter which one you do. So that's step two. [clears throat] Now step 1B was that back on our lrange in L0 I can just look at gamma times a point that was a loop on L0 that has moff index 2 and I can carry it along onto L1. So that's a loop of moss off index 2 on L0. That's a loop of MOS off index 2 on L1 and they would intersect in a single point on the diagonal. And I can when I when I do my surgery, I can look at what the surgery does to this uh figure eight curve. You know, this is a wedge of two circles. And so this this gamma times a point would run along L0. I'll let it look like it gets hidden by the other one. Oh, I'm sorry. So, um I've kind of done the postsurgery thing already. So um oh no no no that's right um oh goodness so what okay so this this thing would have carried straight through that clean component of intersection and then the image would have gone out like that. So this this would be gamma times a point and this would be its image under that Hamiltonian and they intersect at a point on the diagonal and when I resolve they just kind of bend away on their respective bits of the Taurus the immersed lronian taurus uh and I'm getting a curve and I'm going to call it beta So the the surgery transforms this into the curve beta. And what do you believe the MOS off index of beta should equal? It's also four. you know, so the remark was that it seems we had to be careful to make sure we get a connected loop here. Um, and I have blackboard confusion, so I'll agree. Say we were careful. So we get a connected loop and uh it has moss off index 4 on our immersed lronian taurus by analogy to how we got a moss off index for loop in the plane which was immersed. Well, these are these are two loops of m of index 2 and I'm just doing a little smoothing and you check that that doesn't disturb the m of index. I mean, you know, guess we need to think a little bit more about that, but you know, you check that um at the surgery site, you're not gaining or losing any muslop index. So it's the two plus a two. Well, if you can accept that then now what we have is an immersed lronian l the Taurus and what we found is a pair of loops on it. So alpha and beta by the way they intersect transversely in a single point. You see that from the picture. Okay. And so that says that uh the homology of this lronian is spanned by their classes. Any curve is an interlinear combination of the homology of these classes alpha and beta and they both have moss index four. That implies that the mosov index of every loop on this lronian is a multiple of four. Okay. What can we conclude about L? It's not embedded by the pterovich theoremovich and purbo is not embedded which is maybe more interesting. Then I think that doodle I drew not being embedded. Okay. Well, how could that be? Well, what are the how could this thing intersect itself? So I'm saying that there exists some point of intersection between these and I know I got rid of uh the diagonal intersection right if if I look at a point of self intersection of L it had to have come from a point of intersection gazin height between L0 and L1 because each of these is individually embedded so you it wouldn't have come from self intersection of either one has to be an intersection between these different constituent pieces. Um, and I know moreover that Z prime and W prime are not equal to one another because I removed all pairs of points of intersection between these where the two coordinates are equal. Now for Zprime and Wrpime to land on L0 is just saying that Zprime and Wp prime are both on the Jordan curve gamma because L0 is gamma* gamma. But to say that this is a point on L1 means that well it got rotated into by a point on L0 through my Hamiltonian. So that means that I can find a ZW pair on L0 that flew into Zprime Wrime under my flow. And those Z and W therefore are points on gamma as well. So then the picture is I have my Z and my W. I have my Z prime and my W prime. I have my T radians, angle T. And each one of these points lands on my Dong curve. There's gamma and These are the vertices of a rectangle whose diagonals meet at angle t and t was chosen arbitrarily. And so what this says is that if you take a rectangle of any given uh similarity class that's determined by that angle the diagonals make, I can find a similar copy of that rectangle whose vertices are on my smooth Jordan curve. So and um so that's that's the end of the proof. That's the end of proof symbol for this particular theorem. And the theorem is that for every smooth Jordan curve gamma in the plane and for every rectangle R, there exists um and orientation preserving similarity of the plane. So we imagine r is a rectangle in the plane as well. Uh we can find an orientation well I mean orientation is not so important in this particular theorem but I'll say it anyway. The orientation there's an orientation preserving similarity of the plane which carries the vertices of the rectangle onto the curve. So smooth Jordan curves inscribe all rectangles. That implicitly is a definition of what I mean by inscribe. All right. Um in particular uh if I take t equals to pi / 2 then this is showing that every smooth jordon curve contains the vertices of a square. And that is a theorem which is now nearly 100 years old that had been proven by Schneerman using kind of early algebraic topology arguments, coortism arguments. But I really believe that this this theorem is really a simplectic theorem and having thought a lot more about it just doesn't seem like there's a way of removing simplectic ideas from from proofs of it. We now know something like five different proofs of this result. So we're pretty sure it's true. And some of those have some of those are published. I'm gonna and I'll allude to two other proofs um in the remaining time, but I want to describe a generalization of this which works along the the same kind of method of proof. So this was really supposed to look like a rectangle. I'm sorry. Just make that clear. Um so more generally um for every cyclic quadrilateral so I'll just put it like this for every subset of four points on the circle Q for quadrilateral so this is the unit circle in R2 for every set of four points on the circle There exists an orientation preserving similarity transformation of the plane which carries those four points onto the smooth Jordan curve. So pictorially I have a circle. I have located four points on it. I have a smooth toon curve smooth. And what the theorem says is that there's some way to uh dilate, translate, and rotate this circle. so that it lands on the curve and takes at least those four points onto the curve. And the like I said the the proof of this theorem is very similar to the it generalizes the proof I told you of the original one. Um but in instead of working with the L0 and the L1 which we wrote down, we have to work with a different pair of Lronian toi instead which are a little trickier to describe um but will be in the the exercises. Um by the way the exercise session for this afternoon uh is cancelled um due to limited attendance but you can come find me and talk to me about cyclic quadrilaterals if you want from 1 to two. Uh so it's a similar similar proof technique. Um okay to to wrap up I'll mention that uh the the original proof of this result invoked uh the fact that rectangles have a symmetry and when you divide out by that symmetry you can instead of getting a a pair of toi which parameterize the inscriptions of a rectangle in a curve you get a pair of lrangeian mobius bands whose intersection away from their boundary parameterize the inscriptions of the rectangle into the curve. And then if you do lronian smoothing at the common boundary of those moius bands, you'll get an immersed lrangeian klein bottle in R4. And you know that that immersed Lronian Klein bottle has to self-intersect because there's no embedded Lrronian Klein bottle in R4. So that was how the original proof went. Um I do want to mention um kind of a high-powered want to do two more things before the the talk is done. Maybe here I'll insert uh a theorem of Nasserie Solder who's here um he's shown that for every smooth jordon curve on the round two sphere um and for every rectangle. Okay. So you can think about a rectangle on the two sphere just by you know what what I mean by that is four points on the two sphere which are co-planer and in that plane you see a rectangle and you can talk about the similarity classes of those rectangles okay um so for every smooth Jordan curve on the two sphere and for every rectangle R um there exists an inscription so a way of rotating well there's a similar copy of or um I'll I'll just write it like this and you should understand by meaning there's some orient there's some similar copy of R with vertices on the two sphere uh that land on gamma provided a hypothesis on gamma is met that gamma does not pass through a pair of antipodeses of the two sphere equivalently this is say that the diameter of gamma is less than the diameter of the two sphere And it's an open problem whether you could remove that hypothesis. And you might guess the engine for this theorem has to do with um some property of lronian tory in S2* S2. And indeed it does. It rests on the fact that any such contains a loop of MOSF index two. Um, okay. Uh, there are two things that I really want to say and I'm just going to let them cancel one another out and end on time. And thank you very much for your attention.