 Thank you for the invitation to speak here. So I guess I want to just start with some motivation for what I'm going to talk about. So let's see. So the setting is generally collabial manifolds. So quick definition. For me, a collabial manifold is going to be a bunch of things. So x is a smooth manifold of dimension. n, which I mean, I said the complex dimension, I mean, after you consider that omega is a symplectic form, is a compatible complex structure. So when I say complex structure, I mean integrable. Together, this means that it's a scalar structure. And capital omega is a J holomorphic, it's a J holomorphic n0 form, such that up to a constant, when you take the wedge of capital omega and capital omega bar, you get little omega to the n. And here the constant is some meaningless thing. Basically just says that in the right sense, the norm of capital omega is 1. So g is a Riemannian metric. So it's a Riemannian metric, and this condition says that actually the Ricci curvature of g is 0. It's a collabial metric. So the collabial manifold, maybe this constant is chosen. So when you look at the norm of omega with respect to g, it's just 1. So what does this have to do with Lagrangian submanifolds? Well, there's a Harvey and Lawson, which is that so gamma, Lagrangian submanifold, then when I restrict capital omega, gamma is going to be equal to e to the i theta times the volume form of gamma. So gamma, this is g and theta is a map from gamma to S1. Smooth, I think of S1 as being the interval 0 to pi, identifying 0 and 2 pi. So this is called the Lagrangian angle. Once you have this in hand, that's why I'm supposed to put it out. OK, thank you. OK, great. Show it then. So gamma is special Lagrangian, given this a smooth. If theta gamma, the Lagrangian angle of gamma is just a constant theta 0, all right? So from now on, we're going to assume theta 0 is equal to 0 because you can always replace capital omega with e to the minus i theta 0 omega, which then means so without loss of generality, you can make theta 0 equals to 0. OK, then there's some basic fact about special Lagrangians is that there's different ways of saying special Lagrangian. And basically, they're all equivalent by this first thing, the star. I'm just using star here, nothing more. So theta gamma equals to 0 is the same thing as the imaginary part of omega restricted to gamma to 0, which is the same thing as the real part of omega restricted to gamma is equal to the volume form of gamma, which is the same thing as this is not trivial. So it is less trivial, but it's easy to prove, but it's like a deep idea, which is that gamma minimizes volume in someology class. So special Lagrangians are interesting from many different perspectives. They satisfy some kind of interesting differential equation, which is equivalent or implies the minimal volume minimizing, minimal surface equation. So actually, yeah, it's going to be among everybody. Sorry? Until that last thing, I might have thought x was not compact. So if it's not compact, you can still make sense of this by taking a compactly supported deformation or something like that. And so I'm going to want usually the sum of x is compact, but not always, because otherwise it's hard to make examples. All right, so that's special Lagrangians. So why? I mean, there's many reasons, but I mean, originally, maybe a lot of potential was drawn to special Lagrangians by the SYZ conjecture. Other people are going to talk about that. And I think maybe special Lagrangians are not really so important for the SYZ, well, I mean, for maybe the most interesting part of that so far. But for me, there's, well, an object, Foucaille category of x, I'm not saying that it has been meanwhile, it's stable. But you want to say something like this. It's not quite going to be true, but if it's isomorphic. And I mean, what is the, I mean, what's one reason why to know what stability is? For example, you might want to try to count special Lagrangian sub-manifolds or something like that. So counting special Lagrangians representing some class in the entomology of x should be something like a mirror to Donaldson-Thompson variance and some other variety. That's an interesting thing to do. So stability is something one should be interested in. Can you say if we feel a stable meanness in the fact that we're doing it? No. I mean, there's many competing, what you might want to have is some sort of notion of a stability condition, like defined by Bridgeland, which would be constructed by declaring an object to be stable or am I stable, maybe stable if it has a special Lagrangian representative. But I thought the reasoning was the reverse. The Bridgeland stability conditions come from looking at this picture. Well, I mean, they're inspired by it, but just you can't actually, I mean, construct such a stability condition. So it's not that there's volume minimizing, it's supposed to say that there's stability to me. I thought there'd be two, I thought there'd be two. Well, I mean, that's where all the intuition for Bridgeland's stability come from, but to actually prove that there exists. I mean, let me tell you all the examples of special Lagrangians that we know of. OK, I mean, there's far, far fewer than you would need to construct a stability condition. I'm going to now list them on the board. Sorry? The intuition comes from where? Who just said that? I'm saying the intuition for, well, I mean, I guess originally it was in physics papers. Forget the name of the physicist at the Simon Center, or formerly at the Simon Center, and then Bridgeland built on that. Sorry? Douglas, yes, Michael Douglas. I mean, he thought about special Lagrangians looking at certain examples where you construct many of them and in a non-compact case. And then he tried to mimic that in some kind of categorical language. And then Bridgeland found he could construct such things on categories of coherent sheaves. But the actual original setting of special Lagrangians, you can't really actually rigorously construct any stability conditions this way. Sorry? At least for most of the audience who probably has no idea what stability means, can you at least say that this would mean that you could take any Lagrangian and the object of the Foucaille category could express it in terms of special Lagrangians in some essentially unique way. OK. That's one way of thinking about it. There's a harder Narasimhan decomposition of objects in the Foucaille category. In other words, in some sense, stable objects should be in some sense generic or something like that. So you can compose a matrix in a geodernormal form. I never just sort of originally say this a few times. I thought if you take this charge function, which is just integration of the volume, I thought maybe this volume minimizing immediately implied that the definition of stable has something about slope or sub-objects. Is that true? Well, there's not really a notion of sub-object in a triangulated category. So that was kind of the beginning of Douglas's work on pi stability. And he was inspired by certain examples, explicit examples we can write down in a non-compat case of special Lagrangians. But in fact, as far as I know, you can't really actually define stability conditions and say on a compact collabial manifold using special Lagrangians. So examples. Yeah, Mohammed, I mean, I was kind of trying to give my motivation in terms of things which, you know. But I think the way I would interpret what Mohammed said is that special Lagrangians are somehow very plentiful. There are many of them. They're almost like they're generic in some sense. Like a diagonalizable matrix is generic among all matrices. But I thought you were also saying that in fact, we can't prove that there are very many of them at all. We can't prove that there are any more than the ones I'm going to write on the board now. So example one is we take some sum x omega j big omega that admits a map, self-map, which is, first of all, anti-symplactic and maybe also an involution and sort of like a souped up version of anti-holomorphic. OK, so then why is this a useful thing? So the moment that you have this, you get, well, the combination of these two guys, this implies that this is anti-holomorphic. The moment is anti-holomorphic, the fixed points are of dimension n, real dimension n. Think about complex conjugation. And then again, OK, so now that the dimension n, this implies that when you restrict omega to the fixed points, you get 0, which means that the fixed points are Lagrangian. And then again, using this, you get that when you restrict omega to the fixed points, you get, so it's going to be equal to its, sorry, should be a complex conjugation. So if it's equal to its complex conjugate, that means that its imaginary part is 0. So this means that fixed points are special Lagrangians. OK, so anytime you have this kind of anti-symplactic, anti-holomorphic involution of a Calabi-Yau manifold, then, so summary, the fixed points of phi are special Lagrangian. So that gives you an example, and maybe sub-example, just for concreteness, take x to be n-dimensional Euclidean space, complex n-dimensional Euclidean space. Then you can take omega, like the standard symplectic form, r2n, and this is the standard holomorphic n form. And then you can take phi to be complex conjugation. And it's pretty easy to see that those relations hold. And then the fixed points are rn, which is a special Lagrangian. So that's sub-example a, sub-example b, the compact case. So I mean, this one you could just write down, even if you didn't know about the existence of anti-holomorphic anti-symplactic involution. But this one is a little bit harder to get by hand. So you take x to be the cpn such that p of z equals to 0, sorry, n plus 1. And p here is a homogeneous polynomial of degree n plus 2. So then, basically, the Collabial theorem says there exists omega and capital omega, such that this x with an induced j from complex projective space and these guys, it's Collabial. And in fact, more over this Collabial theorem, the uniqueness part of it implies that phi can be taken to be just a complex conjugation that's going to preserve those metrics, that metric, sorry. These are real coefficients. Yeah, sorry, p of homogeneous polynomial with real coefficients. Thank you. So then you get the real part, essentially, of some hyperservice and projective space is going to be a special Lagrangian sub-manifold. OK, that's the end of that example. So that's example one, example two. So there's just two special parts of it. Basically, we're talking about a Collabial manifold with some kind of complex conjugation or generalization thereof on it. So then fix phi is just the real part, which the real part is special Lagrangian. OK, so the second example is when x becomes equipped with a hypercalor structure. So we say that such a quintuple is hypercalor if i, j equals minus j i equals, sorry, I should have said what they are first. i, j, and k are complex structures. And i, j equals minus j i equals k. So this is essentially the Quaternion relations. And also, we want to say that omega i, which is going to be just g. So it's going to be a two form, divided by g of i. And omega j, g, j, and omega k, k. So all three of these forms have to be symplectic forms. So that means they have to be closed, not degenerate. Well, they're obviously not degenerate, but they have to be, in other words, g is a scalar metric for each one of these three different complex structures. So they can all be the same, because they have to anticommute. So the first thing you think is such things don't exist? They do. So there's an example, of course, just like Quaternion space. And multiple covers some direct sums of Quaternions with themselves. But you can also do. So I mean, the simplest kind of compact example is you can take x as an example 1b with n equals 2. This is the k3 surface. And then you can construct the hypercalor and structure using the Klavier theorem. So then another little definition. We want to talk one second. So if you have, then you can define phi. Capital phi is going to be the sum of, say, omega i, omega j plus omega k. It's going to be an i holomorphic at 2, 0 form. The type is given by i. So gamma side of x is holomorphic Lagrangian. So say you have an i, an i. So it's a complex submanifold with respect to i. It's called holomorphic Lagrangian if phi restricted to gamma vanishings. All right. So then sort of a pretty simple thing to check is that if you have a holomorphic Lagrangian, then, sorry, before I write that, let's say, again, given x hypercalor, you can have a capital omega, which is the following guy, omega i minus square root of minus 1 times omega k to the power n over 2. I should have said here that the dimension of x is still n, but it has to be equal to 2m. Otherwise, you can't possibly have such a thing. So it makes sense to divide by 2. Let's call this m, actually. And then we have to do like that. So if we take such a thing, then x omega j, j, and capital omega is clavial. So using the hypercalor structure, we can construct a clavial structure. It's something less than a flurry. It's less. So given, hypercalor implies clavial. And so then the fact that I wanted to say is that if gamma is holomorphic Lagrangian, then it is special Lagrangian with respect to this clavial structure. And although this seems like a lot of extra complication, actually it makes life a lot easier to find special Lagrangians because this one to go up. I'm going to get stuck. Sorry. Oh, the hook is here. Thank you. So gamma is I, I can repeat it. But now is it out of reach? No, it's not out of reach. So gamma is I holomorphic Lagrangian. Does that help? Sorry? Maybe? No, it helps me. OK. So for example, you can take an elliptic and I holomorphic elliptic vibration Lagrangian on this K3 surface, which I mentioned above. And these guys you basically get by doing algebraic geometry. And there are higher dimensional examples of this too. So once you're in the hypercalar setting, you can construct a lot of special Lagrangians just by doing algebraic geometry, constructing, say, some holomorphic submanifolds and then looking at them with respect to a different structure with respect to J instead of I. That's what you're doing when you look at this clavial structure. That's called hypercalar rotation. And it gives you many, many special Lagrangians. So that's sort of history. And so the surprising thing is that if X is compact, these are all examples, which I know of. You can't even do a billion in writing. I mean, shouldn't there be? OK. Maybe that should. I can't make them all fit inside one sometimes. Maybe one can, but I'm just sorry. For some of them, you can make the thing you did in one. So you're saying that maybe. But you need another direct one. I think there may be just. OK. Maybe see up to deformation. Will that help you? I was going to say that anyway, because it would be false. So there's a McLean prove that if you have one special Lagrangian, you can find a lot of nearby special Lagrangians, sort of modeled on the vector space h1 of l. So the question is, I mean, on the other hand, the SYZ conjecture would say that you should have a special Lagrangian vibration. In other words, many, many special Lagrangians, even when, say, the dimension of X is odd, it can't be hypercalor. And if we're going to go to Mohamed's example, then we should really be able to decompose any object of the focacai category in terms of special Lagrangians. So special Lagrangians would be kind of generic. And I mean, this is definitely not a generic condition. I mean, so. Just to understand the second example, it seems like you've got the standard special Lagrangians over the real locus of the cluster sphere. So really, you can have, say, a special case that I put in the first example, or there's a complicated conjugation over here. Well, I have to think about that more. I don't see it immediately. I mean, there's something kind of different between them. I would guess not. But so what happened to all of the special Lagrangians? So maybe the special Lagrangian condition is only an approximation to the real thing. As usual, in subletric geometry, if something is only an approximation, it can be corrected by or modified something to do with homomorphic maps, in this case, homomorphic disks. So why does this seem like a reasonable thing to expect? So first of all, this is not hard there and to prove, but it gives a little bit of evidence for this hypothesis. So this is to appear. Well, basically, let me set a little bit of notation. So let's say x k g hypercalor is above. So I'm going to write r is going to be the set of a i plus a j plus b k. Or a squared plus b squared is equal to 1. So this is a complex structure for each a and b. Just use algebra. This is a circle of complex structures. So for almost all but countable number of complex structures in r, there are no fed a homomorphic stable maps of any genus, sigma, sorry. So let's say that gamma is a homomorphic Lagrangian. There are no such guys. OK, so that means, in a sense, maybe the reason that homomorphic Lagrangians are special and that you have many of them is because whatever it is that's modifying the special Lagrangian condition isn't there in the homomorphic Lagrangian case because there are no homomorphic maps. There are no homomorphic maps for almost any of these complex structures. Just gamma is homomorphic for which one? i-homomorphic. OK, so essentially, so we don't know that for j, there's no homomorphic maps. But we know that for a sequence of complex structures arbitrarily close to j, there are no homomorphic maps, which implies, sorry, which kind of makes you think that whatever the contributions from j are, they should be in some kind of obstruction. When you get rid of the obstruction, when you do the obstruction theory properly, they shouldn't contribute. What are you going to say, Nick? Is the thing you just said an explanation of morally why it's true? I was. You didn't say why it's true. Why it's true? I mean, it's something to do with calibrations. It's not a hard argument, but I'm not saying the idea here. I was explaining why this tells you something about this idea. I'm actually going to prove it in my talk on Thursday. Oh, OK. Yeah, I mean, it's like a one line proof. So sorry, three. OK, so Ivan will prove this in his talk on Thursday. All right, on the other hand, it's not always so straightforward. So if we look at, say, that example 1b again, so the hypersurface in projective space with, say, n equals 3 now, then, and we say, nd is the number of appropriately counted with signs, and so on and so forth, number of maps from the disk x gamma. So gamma is the special Lagrangian. Then nd, so all right, of symplectic area d. So then nd grows exponentially in d. So there's a lot of homophobic disks ending on these fixed points of involutions. So just seeing when there aren't homophobic disks won't answer our question. But, so this was for the idea. This is against the idea. When you say the idea is to modify the homophobic disk, do you mean to say you're looking at a Lagrangian which has some property which you can express in terms of disks. That's what you mean. Yes. That somehow in the presence of homophobic disks. You just take an ordinary Lagrangian, no homomorphic forms or anything, but then you require some condition in terms of disks. Yes, yes. So I mean, the special Lagrangian condition should be modified by something to do with homophobic disks. So in case there are no homophobic disks, we have lots and lots of special Lagrangians. Now, sometimes we have special Lagrangians even when there are homophobic disks. So this is kind of against the idea. Looks bad for an idea. But actually, it has to go a little bit further than just saying there are no homomorphic disks. I mean, eventually I want to write down an explicit equation which has to be satisfied, but so theorem through to Foucaille, oh no, says that if you take x, as in example 1b, then it's going to be unobstructed with trivial bounding chain. So then it seems that maybe whatever our modification of the special Lagrangian equation is, it should have something to do with the bounding chain. So if there are no homophobic disks, then everything is unobstructed and we don't have anything to do. If there are disks, then we can ask, is Lagrangian unobstructed? And I'm going to say in a minute what that means, or at least very, very intuitively try to say what it means. And then if it is unobstructed, well, there's this thing you need to choose to say how it's unobstructed, which is the bounding chain. And if this bounding chain is trivial, then again, you should have no modification to the, so this makes the idea more specific, modify the special Lagrangian condition using a bounding chain. I don't have much. When you say unobstructed with trivial bounding chain, can I just think d squared is 0? Means that d squared is 0 without any correction. Exactly, yeah. Exactly. That will do it. So that's what I'm going to say. So bounding chains. Well, I would like to define, so let's say I have gamma 0 and gamma 1 in some Claudia manifold say I want to define flow homology. Should be the homology of some complex. With respect to a certain differential, as usual, this is going to be basically a free lambda vector space gamma 1. So lambda here is Novakov ring. So this is a Novakov field. And mu 1 is the usual thing. So if you have, say, gamma 0, gamma 1, then this is p. This is q. So then mu 1 of p should be equal to a sum of the number of, say, homomorphic strips, appropriate sense. And let's say where u star omega integral is equal to e times t to the e. That makes the sum converge. And we have to sum over q here and put here q. All right. And the trouble is that in this generality, this doesn't square to 0, because what you were saying doesn't actually define a differential. Homomorphic discs can bubble off. So in general, you would like to say that you'd like to look at, say, p, q, r like this. And you want to look at, this is going to be like a picture of sort of mu squared of p. It's going to be sort of a sum over all r's like this. And you want to say that this thing is basically a sum over its boundary of this modulized space, strips from p to r. And OK. But so if this was the only thing and you'd be in good shape, because then the mu 1 squared would be a boundary of something, it would be 0. But the other thing is you can get pictures like this as well. So instead of splitting into two flow trajectories like that, you can split into one flow trajectory and a homomorphic disc. This is a homomorphic disc with boundary on gamma 0. And so at first order, you want to choose some chain on gamma 0, which looks like this. Suppose you have a bunch of homomorphic discs on gamma 0. So essentially, you want to pair them off or combine them in some way. You just want some chain inside of gamma 0. I'm not going to say with respect to which chain level theory, but you can choose many different ones. And once you have this, you can cancel out that boundary I drew over there by just counting things which intersect B0 as well. I'm not going to have time to draw that. But all right. So in the end, what you get is that you can't define flow homology in general for a Lagrangian submanifold. But if you have a bounding chain, you can. And in fact, the definition of flow homology depends on a bounding chain. Upshot, definition gamma 0, gamma 1 depends on choice of bounding chain. You see a bounding chain because there's a whole slew of these things. So it's not like you just have one for every disc. So you have, and it's even more than that, I mean, I'm only writing it the first order. You have to write something in terms of the A infinity structure associated to gamma 0 and gamma 1. And then you get this more carton equation. And so this is the first order thing that you want to happen. And so of bounding chains, B0, B1 for gamma 1. And what we want to do is we want to say that special Lagrangian should be a condition on both a Lagrangian and a bounding chain. Modified. So it's a condition for the pair. So I have now erred on the side of too much motivation and too little being specific. But roughly speaking, the way to write some condition, or the beginning of the way to write some condition, which depends on a bounding chain, is this. So fix a base point, say gamma star, B star. And then sort of look at paths of Lagrangian starting at this gamma star, B star. But you want to look up to a homotopy. So this is some kind of covering space. Well, given a path, I can define a number, which doesn't depend on, it only depends on the homotopy class of the path. Well, I essentially want to look at the imaginary part of omega. So assume that integral of the imaginary part of omega over gamma star is equal to 0. So this is a basic homological obstruction that you have to satisfy. But then you want to sort of integrate the imaginary part of omega over the whole of the union of all the bounding chains in all of the gamma t. So you have this is gamma 0 and gamma 1. And as you move between gamma 0 and gamma 1, your bounding chains move along with you. There's a way to construct such things. It's not just. What kind of object is a bounding chain? Sorry? What kind of object is a bounding chain? So for me, it's going to be a differential form. A differential form with? What dimension? It'll always be of degree 1. So I'm drawing here in the three-dimensional case. I'm thinking of it as being a Poincare dual surface inside of my Lagrangian. And this will be sort of b0. And then, so I mean, if I'm thinking about as things as differential forms, then I would of course have to. So when you work out the virtual dimension of such a thing, it actually turns out that, I mean, so for example, I'm sort of trying to draw the dimension n equals 3. So this is gamma 0 is a 3-manifold. Inside of it is b0, which is a two-dimensional surface. Then I can consider my family of bt. And as I move it along, I have a family of two-dimensional surfaces which trace out a three-dimensional sub-manifold of my, so all this is happening inside of x. And I can just try to integrate this imaginary omega over all those things. And so the theorem, but I mean, I haven't stated this in sufficient precision to make it a theorem, is that this integral only depends on the endpoints up to homotopy. So that means that you can, for example, say, move one endpoint and look at the Euler-Grantz equation which you get. And that gives you an explicit homomorphic disc dependent equation involving b and capital omega and gamma. And you have to add some other terms, which I've talked about on other occasions. But in the end, you get a sensible, so let's call this. This is going to be some sort of m, some functional, depending on sort of the homotopy class of gamma t bt. And you can look at the Euler-Grantz equation obtained by varying m plus some classical terms. And this gives you an explicit modification of the Euler-Grantz equation which recovers the Euler-Grantz equation in both examples which are known. I'm going to stop here, I guess. Any questions? So can you say this again? Can you write the formula? For the classical terms? The formula? I mean, in how much detail do you want me to write the formula? Can you write some explicit to the Euler-Grantz equation obtained for varying m? OK. And so I mean, you want to think of b as a one form. So then it turns out that what you get is essentially the differential of b is equal to, you have to say it's a bounding chain. So this is essentially going to be some kind of sum of all sorts of mu k operations where I drop the classical part, that's the plus means. So I just look at the parts coming from homomorphic disks. There's also this as part of that mu k, strictly speaking. And you also get something like the co-differential of b is equal to something like, I forget the exact formula, but it's something like maybe the sine of the Lagrangian angle plus all sorts of stuff. OK, I mean, it's a complicated formula. I don't remember it offhand. Perhaps I should have prepared it, but I can show it to you after. Make a more philosophical question. Usually this stability comes from the BTS bound, that central charge of the energy. Are you like, can I take the, are you like as a quantum correcting the central charge of the energy or both? I mean, physicists told me that I'm quantum correcting the entropy. OK, so the usual special Lagrangian condition is open. The usual special Lagrangian condition? Well, you know, there's McLean theorem that tells you you can deform. It's an open subset of the deformation space of the Lagrangian modular Hamiltonian isotope that you can realize. The condition to have a special Lagrangian representative for a Hamiltonian isotope class is open. Good. But it's not going to be close. Presumably it's not. Do you think this is going to have a chance of having that property? I'm not sure that I would be so. I mean, I think you'd have to throw in some singular guys as well. But I mean, this is sort of even before you. Can you go through a surgery? Can you go through a surgery with this? I mean, I would expect that you could. But I mean, first of all, I want to have to find an example of a smooth one of these guys, which is not trivial. I mean, a smooth solution to this, a non-trivial. OK. And what do you mean it's open? What condition is open? In what sense? So I'm looking at this question. Do you know any non-classical example of a solution of this equation? No. I mean, the question. Yeah. So you wrote these two volumes on the right hand boards, Pondier, Pondier, Solomon, Vulture, and EpoCube. And one said that there's these very non-trivial discounts. The other one said that you're unstructured with trivial bounding crocheting. Doesn't unstructured with trivial bounding crocheting mean that you have the disks is zero? OK. So let me draw you a picture. I just wanted you to ask that question. I would have liked to say it in the main body of the talk. But essentially what happens in the case of the involution, maybe this also sort of answers why I don't think the involution case is the same as the homomorphic Lagrangian case. But you have lots of disks on your Lagrangian submanifold, but they come in sort of in pairs like this. So if you think of the complex conjugation, sort of switching the red guy for the purple guy. And now if you want to do a bounding chain, so you have to take these two boundaries of a disk and connect them by some kind of surface, but they coincide. So your surface can just be the empty surface. So just the surface which goes between the boundary of the purple and the boundary of the red. So what that means, I guess, the way Katrin was saying is that basically you have all of these extra a priori boundary terms when you look at those one-dimensional marginalized spaces, but they come with opposite signs. Oh, so the boundary body, et cetera, there was an unsigned count? Well, it's a different signed count. I mean, there's many different ways of looking at these disks. And one of them gives them one set of signs, and the other one gives them another set of signs. So if you put one marked point on here, which is what you need to define the obstruction, then it comes with opposite signs. If you have no marked points on it, which is what we counted, then they come with the same signs.