 OK, thank you for the opportunity to speak here. Although I fear the talk may not have any similarity to the title discernable to anybody except Mohammed. I hope that won't be true of everything in the talk. OK, so let me start with some context, not about Kavanaugh homology, but just about Fleur theory, which we've seen somehow maybe less of than I'd anticipated. So suppose I have a symplectic manifold. You could take it to be closed, in which case, perhaps, monotone is safe, or you could take it to be exact. The set of questions I'm going to ask are already interesting in situations where there are no transversality issues. That's not to say they're not equally interesting in other issues, but it's not the focus of the talk. So let's fix a finite collection of closed Lagrangian submanifolds, at least typically closed. They won't always be closed, inside x. And well, so Fleur theory somehow starts by saying that after picking a compatible or taming almost complex structure, j, and perhaps a large amount of other perturbation data that I'm going to suppress, we get sort of operations mu sub k, some appropriate degree shift acting on the Fleur complexes, which if we assume we're allowed a little perturbation are generated by presumed transverse intersection points of these Lagrangians, which counts in the usual schematic, holomorphic polygons with intersection points as follows. So that's sort of it. Mu k applied to a tuple of intersections. The output intersection y is counted with a coefficient that is given by counting rigid holomorphic polygons whose Lagrangian boundaries map to the Lagrangian submanifolds li0, li1, round 2, lik. And these satisfy the sort of quadratic a infinity equations, which again, schematically, one writes in this form that for any given s, the sort of sum, assigned sum of the different ways that you can use two of the mu i operations that eat up your s inputs vanishes. And we have these operations say for k at least one. Not interested in curvature particularly today. And the simplest of these relations says that mu one composed mu one vanishes. So we get co-chain complexes and we get sort of Fleur cohomology as the differential. And the next simplest equation tells us that mu two is a chain map. So it descends to co-homology. And in particular, I get a natural ring out of my finite collection of Lagrangians, where here somehow, if the indices are such that there isn't an obvious triangle product, if they don't match up correctly, then the notation just says you formally set that to be zero. And if there is a well-defined mu two, sort of when you're looking at a tensor product Hf l1, tends to Hf l2, l3, you apply the triangle product. But this ring carries sort of remnants of these higher chain level operations, exactly that. So let me call this thing A such that A is really sort of the co-homology of what I'll call script A, which is an A infinity algebra. So if you prefer, I will probably not be consistent about this notational difference, although it is quite helpful. But an important point is that the higher order operations are not sort of defined by chain maps. Once this S is bigger than two, these operations don't just involve mu one and mu S minus one. So these further operations don't descend to co-homology. They depend very much on the particular choice of J and other perturbation data. So we'd like to understand what they are. And perhaps the simplest question you could ask is when are they in a suitable sense or trivial? So there's a sort of the natural equivalence relation. For instance, the natural notion of equivalence coming from varying your perturbation data or your choice of J. I mean, let me just say that there's a homology thing in all these operations. Right, so the point is that the operations are not chain maps so they don't tautologically descend to co-homology. But there's a sort of system, a purely algebraic mechanism coming from homological perturbation theory that will let you push these operations that are naturally geometrically defined at chain level down to co-homology. But once you've done that, they're no longer geometric. The things that the co-homology algorithm inherits aren't actually counting curves, which makes it hard to get at them. Sorry, I haven't had an expense. It might be getting at your punchline, but of course there are geometric compositions, multi-compositions. Are you going to say something about the relationship you're in? No. Do you have a hunch of what, right? I mean, if you just fix the complex structure, you get well-defined maps on co-homology, right? I'm not expecting them to be. I'm not expecting them to be equal somehow, or... I'm not sure why we would get well-defined maps on co-homology from these counts of polygons with many sides. I mean, the natural boundary structure that the associahedron involves... No, no, no. I mean, the Schwarz multiplication, except with boundary conditions. The Schwarz multiplication? You mean the triangle product? Absolutely, no, so that's here. There is a completely well-defined product. But you can do more than a pair of pens. You're just fixing the complex structure, and then... Of course, you could try to count... Sorry, perhaps what you're saying is, if I prescribe the conformal structure on my disk with many boundary points, then I will get a well-defined operation. You will get a well-defined... ...commodology. Yeah, I mean, but that's not... So the trouble is that in making that essentially arbitrary choice, you leave the world of the algebraic structures that have been useful in comparisons from other parts of mathematics to other structures coming up, for instance, in mirror symmetry. So those are essentially different things, because you're counting the rigid points in a modular space in which you don't have a very conformal parameter instead of morally the rigid points in a modular space over a family of complex structures. So I don't think that there's some passage between those two systems in various... Probably for pure degree reasons, they can't be an example of what you're actually getting. That's right. Actually, country, they're just compositions of products. Right, on theology, they're just... Oh, damn. Sorry. Yes. So the natural equivalence relation is to consider... So suppose I have... So the natural equivalence relation for A infinity algebras is to consider sequences of maps. Sort of... Sorry, I can't decide whether I have subscripts or superscripts. OK. Which again satisfies some quadratic relations. This is... I'm just looking at endomorphisms here, so let me just say that. So the shape of this formula doesn't matter. The point I want to emphasise is that we have an algebra which is a nice typical object of linear algebra, but there's a lot of slightly mysterious higher-order data, and the natural equivalence relation we should think about is something intrinsically non-linear. It's not given by a map from A to A, which entwine structures. It's given by a sequence of multilinear maps, which can have arbitrarily many inputs. So in particular, it's perfectly possible that two different A infinity structures are related by an A infinity equivalence, which on the level of cohomology is just the identity and does nothing, and in which everything interesting is hidden in these higher-order terms. So from that point of view, a sort of basic question you want to think about is when is an A infinity algebra A formal in the sense of equivalent to the same underlying algebra but equipped with the structure in which all the higher products are zero? Can you write a little bit bigger? Yes. So today's talk is somehow about this question. For expositional reasons, example two will come first, but it will take a little bit of time to set up. So let me consider A k. Oh, gosh, I guess I've been using A for algebra. I always call this A k. Maybe I can call it x k. Be the affine surface given by this equation in C3, this is what many people will recognize as the A k Milner fiber, so where P k is a degree k polynomial with distinct roots. And... You want x k minus 1. Yeah, OK. What do I want to do? You just want to listen to the other time. I wasn't disputing that she was correct. I was wondering which I thought it's a normalize. OK, and this has a projection coming from projection to the z-coordinate to the c-plane. And there's a standard construction of Lagrangian spheres in this surface due to Donaldson called the matching cycle construction. So I have a bunch of critical values of this projection places where P k of z vanishes and the fiber looks like a singular conic, x squared plus y squared equals 0, where the generic fiber is just a smooth conic, something that looks like t star s1. And if I pick a path between two of these critical values that avoids all the others, say gamma, then by thinking about the sort of circle of fiber, which is the equator of t star s1, inside all of the fibers over the interior points of gamma, which collapses down at the end points to a single point, then what I get is a family of circles collapsing at both ends to a point, which is a two-sphere, which is actually Lagrangian inside x k minus 1. So now we want to introduce something mildly more intimidating. So inside the kth Hilbert scheme of, has my writing shrunk again, Chris? Was it Chris who asked? Maybe it wasn't. Let y k inside the Hilbert scheme of k points on this surface be the open subset of sub-schemes which, after you project them from this surface down to c, still have lent k. If that doesn't mean anything to you, don't worry. If the Hilbert scheme doesn't mean anything to you, I'll say a word about it presently. And the nice feature about this open subset is that it's an affine variety, in particular, an exact, symplectic manifold. I don't actually know what it means to project a sub-scheme. So the point is that I start with a sub-scheme here. So what I want to be true is that when I look at the, what I want to rule out is any sub-scheme which contains an infinitesimal direction killed by the projection. So what I actually do is inside, if k was 2 inside Hilbert 2, there would be a relative symmetric product, which I want to throw out. In general, I want to remove the image of Hilbert k minus 2 times the relative symmetric product in Hilbert 2. So what is the Hilbert scheme if you haven't met them? It's just a desingularization of the k-th symmetric product. So you should really think that we're dealing with sort of k-tuples of points which may not be distinct on this affine surface. But when they come together, we're remembering something about the manner in which they came together. So the length of the sub-scheme is, if they were supported at different points, it would just be the number of points in the sub-scheme. So we're looking at k-tuples. And in general, you're taking the dimension of the vector space, which is somehow the ring of functions on the sub-scheme. OK, so a crossing the smashing row of 2k points is just a bunch of arcs that connect them in pairs which are pairwise disjoint and such that the whole thing lives in a half-plane, say the lower half-plane. So there's one on six points, here's another. It's not hard to persuade yourself. There are finitely many given by the Catalan number. Each of these arcs, I said, defined a Lagrangian sub-manifold in x2k-1. And if I have k of these arcs, I get k Lagrangian two-spheres. And, well, the products of those Lagrangian two-spheres as a sub-manifold of this complex surface to the k lies far away from the diagonal. If I have one point on each of these k arcs, I'm far from the diagonal. On the other hand, the symmetric product was only singular near the diagonal. So such a picture defines a Lagrangian sub-manifold inside this space, yk. So you said yk is a fine and therefore is the exact symplectic. I understand that away from all the diagonals, I have a canonical symplectic structure on the symmetric product. Are we saying that that's the one that also agrees with the one we pulled back from a fine embedding of some sort or something? Outside some pre-chosen neighborhood of the big diagonal. You can make it exactly that. Sorry. And it's this one with which the respect to wish was to be. So that's then enough for this to say, for this to be true. OK, so I said there are finitely many of these crossingless matchings. Each one is giving me a Lagrangian sub-manifold in this exact symplectic manifold. And then the theorem that Mohammed and I proved a couple of years ago is that the algebra of the flavor that I introduced before is formal in characteristic 0. So if we do all our Flurco homology over q, say. And so I'm not going to tell you how we proved it. The point of the talk is to suggest various strategies by which we might have proved it or thought about it, which I think remain interesting open avenues. So I was going to say one remark is that this is relevant to a certain geometric model of the Lincoln variant called Kovanov homology. But I'm not going to have time to expand on that today. Do you have analogous pictures? I mean, there would be an obvious conjecture for the SLN homology theories. Beyond that, we don't have a conjectural geometric picture at all that I know of, not within symplectic geometry. We're trying to say what the conjecture is for the SLN. Sure, I mean, there's the description of an SLN symplectic geometry based theory given by Chipri and Manolescu, which takes place inside a fiber of the adjoint quotient to a certain nilptant matrix with appropriately chosen Jordan blocks. You can again associate canonical Lagrangian submanifolds to that index by, for instance, the components of its compact core as a quiver variety. And the conjecture would be that the corresponding algebra is formal. Um, is the theorem, does that help you prove the relevance to Kevan's homology or prove something for that? The theorem enables us to prove that a Lincoln variant defined via a braid group action on this space by symplectomorphisms as a certain Flirco homology group is isomorphic to Kevan's homology in characteristic zero. Whether? I would have said the theorem by itself essentially proves a spectral sequence. And then we move to the left. There's a spectral sequence anyway without this. Just from the existence of the long exact triangle. So I wouldn't say that. OK, so what I would like to outline are two alternative approaches to thinking about this, one of which is based around the idea that things should be formal for good algebraic reasons and one of which is based around the idea that things should be formal for good geometric reasons where our proof was in some sense comparatively ad hoc. I mean, I like it a lot because it worked. But that's perhaps not widely shared view. One of the previous times I wrote a paper with Muhammad, he shortly afterwards gave a talk at IAS in which he described our main result as pointless nonsense. So I'm slightly nervous. OK, so let's go back to example one. We had this nice Milnefiber before I threw this monstrous Hilbert scheme at you. And there's an obvious collection of Lagrangian spheres in this Milnefiber, which meet according to the sort of a k minus one or k intersection pattern. Due to my stupidity, I can't see those spheres in the problem. Yeah, I will in a second. So I said that I had this left shits vibration with k critical values and that a path between two critical values defined a Lagrangian two sphere in the total space. So I just take the straight line paths, consecutively having chosen my polynomial pk to have sort of real distinct roots. And I get a collection of Lagrangians l1 up to lk minus 1. OK, and so this. So it's a theorem, quite an odd theorem now due to Seidel and Thomas, that the corresponding algebra is formal. But they prove this in sort of a relatively drastic way, which is to say that they prove, in fact, that it's intrinsically formal, meaning that any A infinity structure you could fit on this algebra is equivalent to the trivial one. It just supports no interesting structure at all. So well, if that's true, that's a feature of just the algebra itself, so it should be amenable to somehow methods of pure algebra. And indeed, there's a sufficient condition for intrinsic formality formulated in terms of Hock-Schilt cohomology. So recall a graded algebra A as a Hock-Schilt cohomology group. It's actually naturally bi-graded. And it's obtained as the cohomology of a complex whose terms involve looking at multilinear maps from powers of A to A. So chain-level thing is kind of a product of hom A to the j A in degree i minus j. So that looks a lot like an A infinity structure itself. An A infinity structure was built out of multilinear maps from powers of A to A of appropriate degree. So sort of an A infinity structure is an element of CC2 with this grading. And the sort of observational theorem of Seidel and Thomas, which is just not just. It's a nice piece of algebra. They showed that if HH sort of 2AA vanishes on slightly weaker, or in fact, HH sort of 2 comma S minus 2, S2 minus SAA vanishes for all S at least 3. So that somehow all of these bi-graded pieces accept the very smallest, then A is intrinsically formal. So I have too many Fs. Oh, sorry. Yeah, it's sufficient, not necessary. OK, so knowing this, and wanting to prove the formality theorem on the left, one sits down and starts computing Herschelt homology. If you're like me, one gets nowhere. So you write to Mike Kovanov, who writes a computer program who computes Herschelt homology. And you find that sort of, unfortunately, sort of HH2 of AK, AK, where this is the algebra corresponding to working in this space, YK, grows in rank exponentially with K. And on the other hand, the sort of remaining piece we're allowed to keep, HH sort of 2 comma 0, that's just the center of the algebra. And that's sort of rather small, grows linearly in K. So you certainly can't prove that theorem using intrinsic formality. But so maybe I should mention that fairly recently, Etke and Leckerly showed that the sort of type D sort of Dinkin type algebras are formal in characteristic not equal to 2. And maybe they conjecture that the type E are formal in characteristic not equal to 2, 3, and 5. So whilst this is one very special situation, sort of intrinsic formality seems to hold in at least a bunch of other cases related to surfaces. And then there's a conjecture due to Struppel, grew out of conversations with Mohammed and I, that there's a canonical enlargement, canonical as of it, easy to describe naturally from a geometric point of view, AK sitting inside, let me call it AK hat, such that HH2 of AK hat is just the center of AK hat, which is actually just H2 of this space, YK. And this enlargement simply corresponds to throwing in a few more Lagrangian submanifolds, which needn't be compact. So in XK minus 1, I said that there were Lagrangians associated to parts like this. So gamma defines a two-sphere, but I could also consider just a path which runs out to infinity along a straight line, and such as sigma defines a two-disc or a copy of R2, which is a Lagrangian submanifold of XK, XK minus 1. And so instead of looking at the original pictures I had of these configurations of pairwise disjoint arcs, you could look at configurations involving some non-closed arcs and some arcs joining pairs of points. So now there exists a sort of superset of perhaps open crossingless matchings. And Lagrangians, which will vary from being products of S2s to just disks, products of R2s, fire things with sort of more or fewer compact factors. And we set sort of AK hat to be the direct sum of Flurco homologies over these more general open matchings. That Flurco homologies is supposed to be in some kind of categorical category? That's right. That's right. So we're using small perturbations at infinity, so this is still sort of a finite dimensional algebra. The compacts that I didn't care whether you were in a wide array of the attributes in it. Well, I mean, whilst I think that might be true, from a technical point of view, defining Flurco homology in the Hilbert scheme is of a different order of magnitude because it now contains closed rational curves. In particular, you'd have to work over a Novikoff field not over the integers or the rationals. So I think it's probably true that once you take care of that, the story for the closed Lagrangians doesn't depend on the compactifying. If you wanted to set up a version of this conjecture that worked in the Hilbert scheme, then instead of introducing these non-compact objects, you should work in the affine AK surface over C star and look at crossing the matchings that are closed in that. And those two algebras are somehow quasi-hereditary related at the co-homology level. So if we could prove this, this would be, you know, so first of all, if this bigger algebra was formal, it would imply by kind of easy restriction argument that the original algebra was formal. If this conjecture was true, then this bigger algebra would be formal for essentially stupid reasons. It would say that it's only deformations of the ones where you actually change the co-homology class of the symplectic form and your Lagrangian submanifolds just physically disappear. And it's some sort of, to me, reasonably intriguing motivation for introducing non-compact Lagrangians in a very concrete case where, in fact, you're only interested in the flow theory of the compact objects. But somehow the fact that there are non-compact Lagrangians around bound up with them gives you some extra traction. So next let me say something about a potential geometric reason why this algebra was formal, which will make contact with both Jake's talk and Tobias' talk in slightly different ways. Why is it easy to run this computer program for the original algebra, AK, but not for AK-Hack? No, I mean computer experiments for AK-Hack confirmed the idea that it's HH2, which is just the dimension of YK, but that's not a proof. So it is not a conjecture that's stupidly disproved by computer experiment, but it's also not proved by computer experiment. So I'm just not confused about what a computer connection gives the Lagrangian flow. So what the computer can do is tell you that HH2 of AK-Hack has dimensions one, five, and then maybe the third one is 13, and after that it becomes intractable. But those are the sort of non-obvious ranks of the second betty numbers of these first three spaces. I mean, the bigger difficulty is actually simplistically introducing the K-Hack. I don't know how to compute it. That's different. Okay, so what might count as a sort of geometric reason for formality somewhere else in nature? So if M is a Kepler manifold, then the classical theorem of Deline-Griffith's Morgan and Sullivan says that its rational cohomology is formal. So for instance, those sort of P0 harmonic forms for the Kepler metric are exactly the same as sort of holomorphic P-forms, which implies that the wedge products of P0 harmonic forms are harmonic. So if you just looked at the sort of G harmonic forms of type star zero, you get a sort of strictly associative algebra, which somehow in particular shows that the sort of P0 part of cohomology couldn't have any non-trivial massy products. So massy products, standard obstruction to formality would be defined by the failure of having a strictly associative model. And the argument for the full cohomology is somehow the same. It comes down to this sort of DDE bar lemma that says that all the possible massy products on a Kepler manifold vanish coherently and that kind of coherent vanishing is reflecting some basic formality of the classical A infinity structure in cohomology. So you could try and think about the cohomology with its classical A infinity structure as the fleur cohomology of the zero section in the cotangent bundle. And so you could wonder what's the essential geometric structure on the cotangent bundle of a Kepler manifold and at least in a neighborhood of the origin, the structure it carries is a hyper-keler structure making the zero section a complex Lagrangian, a holomorphic Lagrangian in the sense of Jake's talk. So one could ask the following. It also has a scaling action. It also has an N. And it usually gives you a hunch there. I mean, you know, any manifold has a scaling action in the cotangent bundle, but it's not true that all manifolds are formal. So the fiber-wise dilation is somehow not at least not the only essential feature. Okay, so let's recall Jake introduced complex or holomorphic Lagrangians that are holomorphic for one complex structure, i in the hyper-keler family, of which the corresponding holomorphic volume form, sort of omega j plus i omega k in his notation, I've called it omega 2 plus i omega 3, vanish. So these are real Lagrangians with respect to a whole circle of Kepler forms. So then is the corresponding Fleur algebra formal over Q? And though I haven't written it there, the case I have in mind, I should say, first of all, this is already, again, interesting if X is exact and these Lagrangians are exact for these Kepler forms omega 2 plus i omega 3. So, but a situation that occurs quite often in nature is that the Lagrangians meet pair-wise cleanly. So what do you get somehow automatically from the hyper-keler structure? The fact that Jake referred to. So there's sort of easy lemma of that. If we take a generic sort of complex structure in the circle of complex structures relevant to omega 2 and omega 3, then every j0 holomorphic curve, sort of u sigma x, whose boundary lies in these Li, so this could be a holomorphic curve with boundary punctures, is constant. So I guess I mean every holomorphic curve of finite energy, of finite area. And the proof of this is somehow just the following. So let's take sort of, let's take some Kepler form in our circle of forms that make these Li Lagrangian so this is compatible with the complex structure cos theta j plus sin theta k. And let's introduce the form phi. So as follows, introduce phi to be minus sin theta omega 2 plus cos theta omega 3. So the point is that this form is chosen to have sort of trivial one-one component with respect to my compatible complex structure j0. So if I have one of these holomorphic curves with boundary inside the Li, well then if I pull back phi by this map, then what I'm going to get is something built out of a nought 2 and 2 nought forms on the Riemann surface so it has to vanish. But now I just play off the fact that my curve has sort of positive area, so that's some condition on its evaluation against omega nought. But this phi nought form vanishes. So the area of sigma is sort of cos theta times omega 2 evaluated on the sort of relative homology class minus sin theta times omega 3 evaluated on its homology class which is greater than or equal to 0 but sort of phi evaluated on the homology class is minus sin theta times this plus cos theta and this equals 0. And these have to be true for all theta in the circle. This is infinitely many equations as I vary theta in my two unknowns omega 2 sigma and omega 3 sigma so my only option is that actually the area is 0. So use constant. Where in the circle do you use the general sign? Okay, so I mean, it might be that if I chose theta stupidly then omega 2 and omega 3 sigma happen, I mean, maybe not. I mean, in fact, you know, aposteriora I couldn't need it because I could approximate things by generic theta and non-constant curves would end up being constant. But when I was running this through in my mind at least this was obviously forcing the area to vanish by varying theta. I'm confused. I'm sorry. I'm confused why I love to vary theta because it appears in the sum of J0. We're looking at J0. Yeah, so the statement is that if J0 is generic then every curve is constant. Okay, so I have two sort of fixed values omega 2 sigma and omega 3 sigma. Okay, by varying theta there will be a choice of theta for which I can't solve these equations. Sorry, that was said badly. So the point is that as in Tobias' talk we're in a situation where everything interesting that happens in Flur theory here comes from constant maps much like the zero section in the cotangent bundle. But we know plenty of situations in which moduli spaces of constant maps do contribute interestingly. So in our situation in our setup so this is determined by constant holomorphic disks and on the other hand the moduli spaces of constant disks, at least if these Lagrangians meet pair-wise cleanly, are themselves K-la manifolds which are formal. So of course you can take two circles on a punctured elliptic curve the only interesting holomorphic disks that contribute to their A-infinity structure are constant and the moduli spaces of constant disks are points and the point is also a K-la manifold so formal. So somehow this data alone cannot be enough by some purely general or algebraic argument to imply formality. On the other hand the simplest setting that I might have for this question is one complex Lagrangian sitting locally inside its own cotangent bundle and the formality this is asking about is exactly the Deline Griffith's Morgan-Sullivan theorem. Perhaps the next simplest question you could look at is when the complex Lagrangians are themselves of lowest possible dimension just curves and you're working inside a hyper-K-la surface and the situation of these Milner fibres of type A-D-E are exactly going through the cases in which you would get sort of such an interesting question. So finally just as a remark there are no Lagrangians in compact hyper-K-la? So there are, I mean you could take Lagrangian spheres in a K-3 surface for instance and ask this same question I mean, I think that once you've decided that all of the curves are constant you can localize the computation to a neighborhood of the Lagrangian spheres and you essentially return yourself to one of the sort of plumbing configurations of spheres of A-D-E type at least generically. So there's more you could do you could take of course things that aren't spheres. And then there's no disc space in the counter then this algebra is just it's just purely about the characteristics of how these spheres intersect. So I mean there are always constant holomorphic maps and constant holomorphic maps can contribute interestingly to the product structure in this algebra so one shouldn't think that constants never tell you anything interesting. They certainly do tell you something interesting because these things tend to be interesting algebras. The question is a constant map is a natural thing to have two Lagrangians naturally meet so it's not so implausible that there's a constant map at the pair. Three or more Lagrangians shouldn't meet at a point. So as soon as you're looking at constant maps to higher A-infinity structures I mean, naively even to the product you've already put yourself in a non-generic situation from the point of view of perturbations. On the other hand if I give you complex Lagrangians it seems very unfortunate to perturb them destroying their nice geometric structure in order to achieve transversality and have them intersect pairwise transversely. This is a situation which seems somehow God-given for trying to approach via some sort of more sophisticated technique that inputs non-transverse modular spaces because here it seems that the interesting geometric structure is precisely on those non-transverse modular spaces of constants. So I'm out of time but just to remark that the space yk that I talked about is a quiver variety. If you know what that means and that's fine if you don't then we're basically over anyway so who cares in particular it's hypercaler and as a hypercaler manifold it retracts to a union of finitely many complex Lagrangian sub-manifolds which are indexed by the crossingless matching's row I talked about before so it contains sort of God-given Lagrangians which are complex sort of let me call them L-hatch row indexed by the matching's row but just as a straight sort of symplectic topology question it seems not obvious to us that these Lagrangians are actually Hamiltonian isotopic to the Lagrangians that I built naively by taking products of spheres in these Milner fibres No Do those belong to the intercept team? Yeah Absolutely So so sort of a straight symplectic topology question is so these complex Lagrangians are just the compact core of the resolution of the sort of relevant adjoint quotient fibre over zero is L-row Hamiltonian isotopic to L-hatch row and I think there's a lot of reason to believe this is true and if this was true then a positive answer to this question in this case would give some kind of perhaps more intuitive explanation for the formality theorem that Mohammed and I proved that I stated at the beginning and conversely you could view our theorem as additional evidence that maybe there's meat to this conjecture we proved it in some other non-trivial case Ok I'll stop there Questions for Adam So have you thought about what happens when the characteristic is non-zero? Yeah so if I reached pages 3, 4 and 5 of my notes that would have come up a bit There's a pretty observation due to Paul Seidel and Poletacic Polish student that shows that at least in the presence of certain symmetries this so there's a certain involution you can consider on this space and our space is not equivariantly formal over Z mod 2 and I suspect that it probably not formal over Z mod 2 and I referred to these results of Etkou and Lekeli and E-type Milner fibers and in type D they really do prove formality holds away from these certain characteristics and fails to hold in characteristic too so as in the classical formality of chains on a Keiler manifold so zero characteristics seems to be not just an altifice of the proof Is there any analogy with rational homotopy theory that you know this algebra is formal and extracts and makes some homotopy for our populations? Perhaps but I haven't actually thought about that certainly you know I would say once you know it's formal that has consequences for the sort of global symplectic topology of this space that haven't been systematically explored but should be but whether or not you can do something more closely inspired by rational homotopy theory I don't know it's a good idea I have a question about symplectic formality we've got this yk it's got a compact core to the ground which means I can clean it you might believe that the entire category of yk might go up the sheaves on the compact core can you define symplectic formality from that point of view so maybe I mean I certainly couldn't but what is true is that the fact that this algebra is formal identifies it with a sub-algebra of Katerina's algebra which is known to be controlled by perverse sheaves on the grass manian by the work of Brayden so you could wonder whether or not the same combinatorics that you would need for the compact core shows up naturally in the Schubert stratification of the grass manian of K planes in 2K space and if so it might follow relatively straight-forwardly from her work that one could do that so I would say that I don't understand you know much straight coming from formality but the way we prove formality do I? I'm wondering you could finish your answer so we prove this space is formal by building an HH1 class with certain properties and there is a general relationship between Herschel cohomology and symplectic cohomology this rather geometric invariant that Frédéric introduced so in the simplest examples of these Milner fibres one can show that when viewed as an SH1 class the class we build satisfies a certain natural condition with respect to this BV operator that Frédéric also talked about which is the dilation condition introduced by Jake Solomon and Paul Seidel the presence of a dilation has immediate consequences for instance for the existence of Lagrangian sub-manifolds it rules out K-pi ones or things like that one slightly annoying thing is that symplectic cohomology has good restriction properties which is where these constraints come from but naively Herschel technology does not behave well under restriction it's somehow covariant in one variable and contravariant in another so until you've interpreted as SH star it seems slightly less obvious to me we don't quite have the dilation condition we have something that's strictly different but one should somehow explore whether or not the existence of a pure vector field which is what we have just rules out Lagrangians and if you can then use a restriction technology to get the same conclusions I would guess yes but we haven't done that be a bit careful about saying what is formal the entire vocabulary isn't formal and the most you can hope for here is that we have some collection of Lagrangians which sometimes generate the category whose flirt theory happens to be formal there could be another collection of Lagrangians which also generates the category whose flirt theory is not formal but you can easily make it up by taking these guys and adding one extra redundant object to it I think the question of consequences is good but it's unlike for spaces where there's kind of always what we're talking about to just the homology of the space that we're interested in here we're talking about formalities in Chi category it's not that clear of what we're exactly saying yeah but in this one the example you're looking at is one of the things that Elza was talking about or sometimes this is this nomenon fiber would be the Landau-Ginsberg thing to some club yeah the mirror to some club yeah and in those cases this bunch of these bunch of things they generate this what do they call the Seidel category I mean the Seidel category is really generated by the corresponding non-compact things these vanishing thimbles rather than by the spheres so these aren't quite the objects that show up in that sort of directed category that you think of analogous to the sort of thing with an exceptional collection so yeah while the word motion comes up is it an analogous statement somewhere analogous statement to what formality is there some derived category some other object today so you can take the derived category of coherent sheaves on the resolution of the zero fiber with compact support on this collection of complex Lagrangians so counters and cabinets have built a break group action on that essentially by hand and showed that you recover Kovanov homology their theories bi-graded and I would guess that starting from the fact that it's bi-graded you could prove formality for the corresponding algebra in their setup but they didn't prove their theorem that way they took a different route via building tanglin variants which bypassed the need for formality endomorphisms of the direct sum of the structure sheaves you should take a break