 Zara Tukarszynski, i ona będzie rozmawiać o spacerze bądźmi. Dziękuję. Dobra, zapraszam. Dobra, dziękuję. Cześć, witaj. Cieść, zobaczcie jak to... ...to teraz będzie ciekawe. ...to teraz będzie ciekawe. ...to teraz będzie ciekawe. ...to teraz będzie ciekawe. Więc jeśli... ...m może używać... nie wiem... ...tych ogromnych, że jestem zaczniona o... ...mieszkańczące, żebyście mogli dodać mnie i zaprosić na definicje, ...detail, czy coś takiego... ...my chcą mówić o... ...o... ...przywodowej variacji, ...przywodowej variacji, ...przywodowej variacji, ...przywodowej... ...przywodowej definicji, i o swoich wypowiedzi na bardzo dobre wypowiedzi. Zacznijmy z powrotem. Zastanawiam się o symplektycznym nanofołom. Powinienem używać n w pewnym razie, by to nie było w tym dimensionie. Są prawie skomplikowane struktury, w życiu w biegrze, jeśli się zapamiętamy o tym, to jest w porządku, to jest w porządku. El Lagrangian submanifold which I assumed to be overclosed, I was soon connected, even though it's probably not necessary in a way oriented, which is also avoidable if you really want to, and relevant to them, which in the non-oriented case will be relative to ten, but you can just forget, this is the technicalities and they have some regularity assumptions, which are also, which can also be worked out, but this is the basic universe. And of course the basic question, clearly is the count of open curves, so maybe I'll just make a picture, which you all know probably, but anyway. Okay. So we have a space of ease of this. Jelon worth of maps from the disk worth boundary conditions in Lagrangian, we want to fix the degree. What else? To much this, by bioplomotors and compactified by adding table curves and the basic question is, how many, how many elements are in this case. And, because in one way to maybe cut down the dimension when necessary by adding constraints, so constraints. I guess if you live on paladio prefold, then you don't care about constraints, but if you live in other places, then constraints are a feature, not a bug, because they allow you, for example, to define an interesting quantum product. So that's fine. And of course, whenever the modular state has boundary becomes hard to count things in there, and so the problem is this space may very much boundary. And when you have Yes. Let's say this is a schematic picture of and, and I want to say count and it looks like some this, but maybe it has some constraints on it, I don't know. Um, and I want to change my data, for example, I want to change the complex structure. Then my point will start moving in the space. And Well, as long as it's moving within the stage, but I have no problem, I can still pound it. But if I reach the boundary. Then when I keep changing my complex structure, I will sort of whip out of the space. Over here, I have nothing to count them outside of my. So one example. Illustrator is the space of disks with one boundary condition, I'll call it zero to zero. So I'm thinking here constraints one point one one point two. They're not complex points, we will just label them before. So I'm thinking here and it's zero just so it's easier for me to, to draw. And it's zero and of course supporting me to be zero. We can check the dimension of the space of this thing then is to and it looks as follows. So here is the inside. We open stratum. And that's where proper. This with more points like that. And then we have two boundary components. One is when points to bubble up to the boundary. And we get one this point two. No, no, no, no. Go this point and point new. And second option is a point one bubbles at the boundary and we get wonders with twenty one. crabs twenty zero image. And you might also have pointers here. Thisizzard comes from this picture. Now. Yes, one point zero, to bądźmy, mamy to. 1, 2, w tą stronę jest 0,0,0,0,0,0,1. I ponieważ bądźmy na bądź... Well, it matters, right? The points of the interior. Well, it's a two-dimensional space. They can go around as they please. But the boundary is one-dimensional. So this order and that order are truly different. And so usually there's a different strata in the modular space. And this is a description of this particular modular space. As you see, it does have a boundary. And the moral of that is, has exemplifying one type of boundary, and that type of boundary comes from disk bubbly. w okoliczności nie ma żadnych problemów, czyli złożyć obowiązku i Heights. I złożyć obowiązku. I wyłożyć obowiązki. I wyłożyć obowiązki. Que to jest obowiązki? Utrzymało ten obowiązki. About how do wew promote this kind of problem is the bounding change. Now I will say more about bounding change in the second half of my talk. But just to give you the picture, it says the following thing, well true at this point in time. I have those two disks and then they no longer belong anyway to this module 8, but. konfiguration, jak przy vibrationsie tej bliskie. Ale mogę też widzieć, że konfiguration ma jeszcze tylko 2 drogów, które się으ją. Każdy z nich, aby zrozumieć, każdy z nich będzie supersy 로da, tak naprawdę tam, gdzie jest niej sabemos, że kiemna wygląda na tak samo na na czytel, że jest w 같이, a Spicy Virgin ma takie kiemne. I w niej jest też niej jest, to wygląda jak dwie dwie kawałki. To jest wstępne. I tutaj mam nadzieję, że muszę się skończyć nie tylko tych dwie dwie dwie. Chciałam skończyć kawałkę dwie dwie. Muszę skończyć nie tylko dwie dwie dwie dwie, ale też złożonych, złożonych dwie dwie dwie dwie dwie dwie dwie. i zamiast z nią podkreślenia. Nie chcę odpocząć wszystkie możliwe kolekcje, które będą za dużo, więc mam gadzi, który mi mówi, które kolekcje w porządku i podkreślenie się w gadzi. To gadzi jest dokładnie głośno, lub w bardziej normalnym razie w literaturze jest głośno głośno, który jest dłużej, i mówię o tym, więc po prostu zapomnę głośno. Ale, tak, to jest bardzo spokojna, bo pracujemy w BIRAM modelu i tak dalej. Mam nadzieję, że daje się pewne definicje, ale jeśli nie jest za urządzone, to możemy po prostu rozmawiać o drugim rodzaju bądźmi. To jest takie pieniądze. Na przykład przez drugą bądźmi, Boundary, to jeszcze jeden egzemplar. Uwielbiam się, że zbyt mały egzemplar, ale po prostu zapomnę, że bądź mały egzemplar. To też będzie egzemplar. Ok, jeśli mam jedno z mniejszych marki w bądźmi, to zmieniającego imię mało jedno z mniejszych marki niż zmieniającego imię bo mam jedno z mniejszych marki w bądźmi. Więc jedno zmieniającego imię. I to będą dwa bądźmi. Jedną z nich, od jednej bądźmy do bądźmi. Bądź i bądź. I to bądźm, ktora, górna wópka subtract, Cześć. No i to są, a obok, i te bądźmy, i ten bądź jest zbyt mniejszy. I w tym, a z bądźmi bądźm się niezbędzi, I ta solucja, albo te dwa, co używaliśmy, na problemie, jest w porządku. Jest w porządku w porządku. I tutaj chciałam powiedzieć trochę bardziej o tym, bo to, co chcę i co nie chcę. Daj mi w porządku trochę o tym. Powinniśmy, że nie zrozumiałam, co jest w porządku w porządku. Wiem, że w porządku w porządku w porządku. W porządku w porządku. què persecuted rule and the cone of this map, wife wanna take and you may ask why and I will tell you because it work and hopefully I'll make some pictures that will try to convince you that it works and but let's put them write down the cone is then it's literally the cone of the map, so. The complex Megan complex we've 맛있어요. producenie dłużej m, bo to jest dłużej l. A potem na telefonu musieliśmy dłużej dłużej. Bądźmy na telefonu, jeśli chodzi o dłużej 8,a, 8,a jest dłużej w x,a jest dłużej w konstytucji. Myślę, że jeśli chodzi o dłużej r, to są dłużej dłużej dłużej dłużej m. A to dłużej dłużej dłużej p 1, a w dłużej dłużej m. A génér down doaira dłużej. B就是 one TR. Bl hence we have data. I mówię, że słyszałam, że nie myślałam, że to nie byłoby wioski. Nie mogę powiedzieć, że to nie byłoby wioski. Nie, nie. W sensie, że to jest w języku polskim, więc... Tak. Więc dlaczego to się działo? Bo to jest dokładnie... Tak, więc nie mówimy. Jest to jedna rodzaj myślenia o tej chwili, jak powiedzieliśmy, że jedna rodzaj myślenia o tej chwili jest tutaj, w której sklepszy się jedna z takich, albo możemy myśleć o tym, że to jest dosyć sklepsze. W tej chwili mamy dwa rodzaje wytrzymać tę chwili. Możesz myśleć, że to jest to, gdzie dźwięki generują, albo możesz myśleć, że to jest sprawa, która wciąż wciąż wciąż wciąż wciąż wciąż wciąż wciąż wciąż wciąż wciąż. Not descended, because that has a mathematical meaning, but it's just a sphere that happened to intersect the Lagrangian at a point. So a way to fix this contribution is to say, okay, so now we don't only count this, but we also want to incorporate a certain type of spheres, those that have the potential, as a two-dimensional one phenomenon, to intersect Lagrangian. And this is what the comb allows us to do. We have these spheres live here. I tutaj możemy plukować to, co chcielibyśmy. W szczególności, możemy plukować to, na przykład na górę. I to, co zbieramy, to zmieniające się te kontroducje. Zbieramy pewne kontroducje na drugą komponucję. Zbieramy to, co jest w porządku. Jakieś wypowiedzi. Jakieś wypowiedzi, co będzie w porządku. Ale możemy to zrobić. To jest pewne istotne pomiry i tez okazje na całym w tym komponucie, co to znaczy, że ma dwa komponucie. jedna jest chains i x i jedna jest nr 1. i to nr 1 jest tylko na koniec tych nisków. All possible disks, all possible disks, all possible disks, all possible constraints. And the chain and x is going to be the chain, defined by all possible spheres. All possible disks, all possible constraints. If a capital 5, say, is the closed form of written potential, this is going to be like the gradient to 5. Okay, like if you pair this chain, if you pair this chain with some constraint in x, you get the problem with an invariant that involves that constraint. Right? Question about the picture. No, that is beautifully rigorously defined side. I wonder, unless you're just too shocked, it won't really be a question. Maybe I have a quick question. Yes, thank you. So in the beginning you said L is connected, is that correct or? I did, yes. I do not know that we explicitly knew at any point. I am pretty sure that you don't have to do that. I mean, but because then you couldn't integrate or it is connected to the several such evaluation. The great against each connected component. Right, so if L is connected, there are several ways of thinking, if you can think of it all as L and then take this L and integrate it with the entire thing, or you can treat each component separately and do sort of open ground with a theory for that alone and take sort of the direct sum of all of them. I did not try to see what this gets. I assume it works with 5, but I did not try. So for me L is connected, because that's the thing that we wrote down and that's what I know. But yes, absolutely, one interesting example would be to take, say, two circles on a sphere and you want to be really blue. So the honest answer is I'm not sure. I assume that this works fine. The way personally I would like to do this is to actually rethink of each component as a separate thing. And then this one will have its own bounding chain and this one will have its own bounding chain and because it's disconnected the disks live either here or there. But you can use your experience for work. But I don't know. My claim is that first of all, this thing is closed in the phone complex. I'm not gonna prove it because it's effective life. And secondly, because it didn't write you a proper definition. But let's say the phone on the side is actually zero and the claim, the next claim is this class is invariant under which is the relevant in general relevant global conditions here. So how would you find the age of global for bounding chain. And this claim I can sort of draw a picture of how we prove it and hopefully this picture will illustrate why this cone is a good choice. But moreover what I essentially claim is that this is sort of the true invariant. If you project it to the first component you recover all of the closed-down global theory. If you project it to the second component you need to be careful in doing that because just projection is usually not that chain map but you would have a way of doing that. The way you fix it to the comma chain map is exactly by incorporating some contribution from the first component that recovers you the open ground within here. So this is sort of the full true invariant for gino zero ground with a theory closed and open combined and you can recover everything else from it. And maybe maybe I can really put try to draw a picture for why it's true. Let's say I have let's say I have an azotope and times zero I have psi and times one let's say varied the public structure I get from psi prime and in between I have the azotope so there is like a side kill living in between them. Yeah, that's the same work. Okay. So I want to say that psi prime minus psi is exact in the chronology of this whole complex. That would mean that the class defined by psi prime is the same as the class defined by psi. Well psi prime is what happens at the time one. This is what happens at the time zero. So Stokes theorem says well just this is the integral from zero to one of these Yes. Hi. Sorry. Yes, in a way. You know what maybe I think I do want to put a psi prime there but I don't want to do it. So maybe yes and okay for the sake of saving time maybe but rather maybe we'll be more interesting for parts of the audience because let's say you can believe me that this thing is invariant. Right. Like why would we believe it is because of the way the phone works, but maybe a better application for the application is the relative quantum chronology and that's that's the following. The integration into the our proficient train and on chain on chain level this would be a short exact sequence and so we get a long exact sequence of comforcy and basically what I want to say is that I defined this corner of a triangle aby być zbyt mimo konu komologii. Powiem trochę, ale tylko trochę. Oczywiście, w sumie, to jest dokładnie klucz konu komologii. I the beauty of this triangle, if we can define a quantum product on this thing, that involves closed-con with invariant and open-con with invariant, because it comes from the phone. So this thing is now... a to ma taki mały widok z uwagi w tym , że jedna z tych krótkich i jednym z tych czynników jest to z outreach i będzie o zorganizafterności. I tak blisko przez to, ale oni są bardzo owożni. To jest co co Dontnum. To dé Physical. To jest to owożni, w którym sponozimy să skończą, ale to hmm, to jest owożni, ta jest to owożni, to jest owożni, a to jest owożni. To jest owożni, to jest owożni, a to jest owożni, żeby rozwijać, jest to owożni, Alizontologically non-zero, in which case integration overall is surjective on homology, and then you get a different, then this arrow is zero, and you get a different toward exactly this, relative to absolute to R. So either, either relative to homology is a ring-clon extension of the absolute one, or if it's subalgebra. Let me give you an explosive example maybe, and for that I will need my note, because I don't remember numbers, or formula or anything. If X and L is CPN with RPN in it, and I assume N is odd because I assume that my Lagrangian is oriented, there's some working prowess, to define the stuff for non-oriented Lagrangian, by, oh, I have to, so I'm gonna give you, give you a gratitude. But for now, I need it to be oriented, so N is odd, and then you will not be happy to find that I add a sign that you didn't expect, because with the close potential, and also I will write the quantum scalar with a half power, because it works nicer that way. So I claim for me this is the close quantum homology of CPN, and the relative point, well, RPN for R odd that moves in the ring zero, well, it's also ring zero, so in the ring zero, right, because odd homology of X is zero, so we're in the first case, so we expect to rank one extension, and we get indeed PM relative to RPN, again, it's R, which is G to the half, and now we'll have two variables, let's call them X and Y, and we divide by, so this relation of generalized, let's do X to the N plus one minus N plus one over two Q minus N minus one over two times half times Q to the half Y, and then we have Y square minus Q to the half Y, and then we have X, Y. We know the relations, and the, of course, we also have a map here, so let's just look at A, let me also map it here via A, by sending Y to zero, Q to the half, Q to the half, and X, very explicit, and as soon as, and very, it doesn't take much to compute, so the only thing we need, we know we need to add someone called Y, and the way we figure out the relations is we have to compute the close and open-grower within variance, and the last remark I want to make about this, how much time do I have? That's 10 minutes, right? That's not much. The last remark I want to make about this is, so we have this bounding chain, we have this cone complex, and the whole thing sounds kind of abstract and useless, but as you see, we can actually compute some things, we can actually compute invariants also, and some examples, so there are computations for CPN, with RPN and the Lagrangian, where there is a recent preprint online by a bunch of people who computed the invariants for the Chiang Lagrangian in 23, which is surprisingly interesting. It's also the first computation I'm aware of, where the Lagrangian is not the fixed focus of an undefektive evolution. There is a paper in separation by a little told on myself, where we compute relative homology for end-open-grower within variance for complete interjections. So there are explicit examples, there are not that hard to produce. The hardest thing about these computations is usually it seems to be computing the signs of initial values, and you see they're highly non-trivial by, so we eventually end up doing like a classical geometry. And finally, I want to say, though, at the very least, our construction depends on the boundary chain. It secretly depends on the knowledge, which I didn't tell you, but it primarily depends, or at least you already know, it depends on the trust of boundary chain, boundary chain was the recipe that tells me which little disks they need to count together in order to get an invariant. Right, so normal boundary chain. So these open-run within variance we get depend on the choice in boundary chain. So to get canonical values of logw, whatever canonical needs to mean, is the same as having a canonical choice of a boundary chain, a canonical. But that means that we need to understand the space of all possible boundary chains in order to say, oh, this one is the canonical one. And then it's actually from a paper. One example. Well, the first R5 version, I guess, made it. The first version made it to R5 in blank, is, let's say, L is the rational monogysphere. When equivalence classes of bounding chains are characterized by the value integral of the overall. So a bounding chain, I did not tell you this at any point, but a bounding chain is serving as a chain on L. And it's not closed, it's not closed, it's really a chain, it's not a class. Important. But I can still integrate it. And the statement is, well, if it has only essentially a homology and self-degree, and zero, well, zero is not interested in self-degree, then the integral of a form on L, well, when two chains are equivalent, if and only if their integrals are the same. So basically, if they're point part and the same. And you might say, well, okay, that's not. It's nice. It allows us to have a canonical notion of open form with inherent every time L is a sphere. Which is a lot of times like RPM over R in the sphere and so on, so forth. Quadrant. Quadrant cyber surfaces, right? The local sphere and so on. Well, it's nice. You can do computations with that. Like I mentioned, we did some computations with it. But you may argue, this is the reason why this happens. And I would argue that no, this is not the reason. And I don't have very much time to... But I will. Let's say this is not an archive. So I'll put a question mark. Hopefully this will be out till 23. May never know. Let's say L is homologically to you. And the restriction map is surjective. Again, not only care about group coefficients, because differential forms are magic. The restriction is surjective. Then we'll see this. With almost classes of bounding chains, our class R characterized by a slightly more complicated value. So let's say data J are forms in X that are in fact closed. So that they generate all of the L-emologi. One to say R, then L up to some degree basis. Then I can take a fancy expression and pair it with those data, data, sorry. And I get a list of parameters. I get a list of parameters. Length of the list is exactly as the dimension of homology lower than top degree of L. So for example, if a special case where L is homologically here, then we only have one... Then the list is of length one. It has only contributions coming from degree zero. And then this expression also specializes to just integral three. But I have the point that I have an exclusive expression that tells exactly how many parameters determine the class of a bounding pochain. And it seems that as long as the homology of L is a little bit more complicated than just degree zero than M, I should be getting more parameters. And I am getting more parameters to classify equivalent classes of bounding chains. And then you say, well, okay. Then maybe if there you had one chain to pick because you only had one parameter, here, well, you have all those many parameters. Either you need to pick your favorite value of the parameters or maybe you don't have a canonical choice. Maybe there are different choices. My claim is that you still have a canonical choice. And the reason is absolutely nothing to do with all those extra dimensions. I don't actually care about all those extra dimensions for the purpose of gromawitne theory. I will care about them very much for the purposes of, say, studying the focaic category because each of those choices will be a different element in the focaic category. But for gromawitne theory, I have the following, I know, I'm almost done. I have the following surprising fact. Let's say omega is the overall worth and superpotential. So it's a generating function. There's some combinatorial positions in them. The gromawitne variance is a generating function for gromawitne variance. I can write it as the sum of all this with a bunch of constraints in this bounding code chain. And then maybe a bunch of constraints in the interior. Well, I don't mind. So let's say these are k, k, k, k, k, k, k, k, k, k, k, k, k, k, k, k, k, k, k, k, k. And then another term that comes from no boundary constraints. And this contribution comes from either disc with no boundary points on them or spheres being incorporated using the code. Beautiful. Point is if let's say the integral of b over L is… I don't know. Point jest, whenever I take a parameter, it doesn't matter really, which some formal parameter, right? If B has formal variables, then at each year constraints may have a formal variable in it and so on, so forth. So if B lives in some algebra over R that has this formal variable in it, maybe other stuff as well, I take the derivative of this with respect to B. It's a fact of life that we get some constant times, not a hard fact of life because of the amount of time anyway. But my point is you have a relation that says, well, if you take a parameter that's visible in a boundary constraint, and you try to see all the invariants that have this parameter in them, if you only get something non-zero, if that parameter plays a role in the top degree of B. Which means, no matter how many degrees I have there in the class of prime space, the only one that really matters is the one associated to the top degree of B. And therefore, no matter how complicated the space of bounding chain is, as long as I have an attribute parameter corresponding to this, it would be the one canonical choice of an equivalent stuff of bounding chains that I'm going to use to define my invariants. And I'm over time. Well, thank you very much, we still have time for questions. So, maybe I missed the, but did you write the relation between omega and psi? I didn't explicitly know, but omega is going to be, let's say, a projection of psi. It's a correction of just projecting psi to the second component. You need to correct it by adding stuff on the first component. So, would you like me to elaborate enough? No, no, so it is because you said psi are all the invariants in them. Right, so psi. So it has this E, which is the corrections from the first component. It has corrections from the, from, actually from both components. The second component also has this node boundary constraint on them, but it also has contributions coming from the first component of psi. So the first component only contributes to E. Right, yes, yes, absolutely. First component only contributes to E. Oh, quick question. So at some point you mentioned a construction involving Lagrangian's that were not the fixed point locus of an anti-substructive evolution. I'm just curious, what was that construction? So this construction is simplistic, we don't care about any presence of an anti-substructive evolution at all. So the example that I mentioned, was the Chang Lagrangian may be could be, it's a cute construction, we know it's Lagrangian because it's by definition the fibre of a moment up, but to show that it is not to fixed locus of any anti-substructive evolution is an untrudnial statement. ACP3, z 3.0, symmetric product of 51. Inside of this product, you take all those triples that add to zero. A way to think of it, is you have unordered triples of points that make an equilateral triangle. Doesn't look equilateral, so I don't know how to do a whole trip. Eklator of triangles on a circle in size 51. So you could define this basically to be your moment map. So this is just the zero fiber. And so this is all a branch of a manifold. And by defined by River Chiang for entirely different purposes. And it's an interesting example for different things. It's actually a sphere. Not just has a homogenous sphere, but it's actual sphere. Yeah, it's critical. And it is known to not be the same global for any, any entity. That's sorry. That was about, but that's not that's covered by the theorem. That is covered by the theorem. Yes, I mentioned it as an example of a case where invariant have been computed. And also the corner corner, the ring structure was both one of them. And so, so the independence of this Omega on the other parameters and what's the consequence of this founding chain. It's closed or something. bounds from mora acc crushy keer lx? It's dimension, but yes, let me maybe just. Bo wiecie o tym, że wytrzymałam, że mogę bardzo szybko pokazać ci komputację. To wygląda jak 1 over k plus 1 mk d k times b. I te e-korekscje, które nie są opłynane w s, to nie są wytrzymałe. I to jest w porządku. Być w sumie, to nie ma znaczenia, w którym wytrzymałeś drzewo, Just over in here and you get okay to the cave times the gorilla. And here yes more car concepts. This is something strand unit. So this is just. And you can take the derivative outside. So only one parameter. Okay. Yeah. Yeah. For me to space for your bonding chains, do they, not the pun todo. Fiber based on the base point, or how is the dependence? That's a good question. Let me let me see if I understand it correctly. If I have a family, right, a big fiber that doesn't take the peculiarity vibration, does the space of all bounding pochains depend on the fiber? Or very how does it vary with the fiber? I did not think of it. It's a good question. I would hope it doesn't, but I don't know. I don't know. I thought you didn't need it. I mean, I thought the fiber was not obstructed. You didn't need it. Well, I mean. Again, it's it's it's the bounding pochain from the future. So maybe even if M0 is already zero, I might want to add a bounding chain, for example, in order to introduce boundary constraint. Like zero is. Here's the thing. Zero is canonical. That depends on what you are trying to do. Like, for example, if I do want bounding constraint, zero can be. Zero can be a bounding pochain. Like to say, you're going to make two degrees. Right. Yes. You can take a zero. But I don't want to take a zero. At first it's important. Because I want to be able to have pochains constrained on my bounding. If you don't need to correct any bounding, if you create all these structures, then it's a symmetry. We could take zero. But I don't want it. From your point, it's more fundamental than zero. Other question. I don't see any other questions. So let's thank our people one more time. And I believe we have another copy.