ותודה ככה לכם על המשליד. בואה, זה פרלוי, وتרק מבטן ישר עורפת ויכתב קורניר, אז אני זוכה שאני חאילו להקריא, מהפירמוסות ומוטיבציה של המשליד להקרה, והנהanswerה, ובהצייחה.   It should, actually. It should, it's a U shaped. It should. Yeah, exactly. This is for the definition of left chest fibrations. Do you require that away from the singular points all, you have sort of a topological vibration? Yeah, yeah, yeah. Well, I didn't list all the properties, but yeah, it's supposed to be a submersion outside. No, but submersion doesn't, I mean, can I take out a point of some of it to a fiber? Or is that not allowed? It should be proper. No, no, I mean, there are many conditions here that I'm not going into, and what? Yeah. So you're assuming that M is compact? Yeah, assuming that M is compact, however, I'm allowed to assume that M is convex at infinity too. But then I need to assume that at infinity the whole thing without throwing U is trivializable. So yeah, I mean, I could spend half an hour only on the precise definitions, but somehow the message I want to convey is independent of those things. Anyway, so it's possible always to pass to this setting, and another important thing about these fibrations is that they come together with a connection. So there's a connection, and the connection is just, well, the horizontal distribution will always be the omega, omega, orthogonal complement to the fiber. The connection doesn't exist everywhere, it exists only away from the critical point. So it exists connection on E minus the critical points of pi, and this means you can do parallel transport. Are you saying this is part of the data or something that exists? No, omega induces it. So you get parallel transport. So once again, omega, you take the fibers, they are symplectic, and you take the omega, capital omega, orthogonal complement, and this gives you something two dimensional that projects isomorphically to the base. Okay, so why do I fuss about time? So I want a very convenient way to identify the fibers away from the U-shape, and because of this trivialization that I'm going to fix, the connection is trivial there, and therefore the parallel transport in the complement of U is just trivial. I mean, it's identity, and so this is very convenient for these things. Another object that exists here are thimbles, thimbles, and vanishing cycles, vanishing spheres. So again, I'm going to do this not very rigorously and fast. So basically if you take a critical point, sorry, a critical value Z, that's a critical Z0 maybe, a critical value, and you take a path emanating from it gamma. So all these structures together give you a sub-manifold, t-gamma, t for thimble, and so this is thimble along gamma, and this thimble has the following properties, so t-gamma is a Lagrangian sub-manifold of E omega Lagrangian. What else? It is invariant, t-gamma is invariant under parallel transport, of course parallel transport along gamma. Another thing, so how does it look actually? It is like this, so here's our gamma, so it looks basically like this. So over every point in gamma you get a sphere inside the fiber over that point, except over the point over Z0, where you get a point. So altogether this thing looks like a copy of a disk, a Lagrangian disk, Lagrangian inside the total space, and these spheres here are called the vanishing spheres, because they just go and shrink and shrink and vanish. So these are called the vanishing spheres, so maybe I should write it for every Z in gamma, which is not Z0, but t-gamma intersecting the fiber over Z is a sphere, Lagrangian sphere in EZ, and t-gamma intersection EZ0 is just one point. Okay, now it's not totally obvious how to construct this, but there are canonical constructions, how to do it. And one way, for example, is one not the most elegant way is to take a local model, write the homomorphic Morse function, adjust the symplectic structure to be compatible to it and construct it locally and then move it by parallel transport. There are more elegant constructions for the fimbles, and again this will take too much time to go into. Another thing that will play a role here. Is the symbol supposed to be unique or is it a choice? Well, once you specify gamma, once you specify gamma, it becomes unique. Like, literally unique, I can't somehow take a slightly bigger sphere as a separate unique thing. Yes, yes. One of these Lagrangian spheres, kind of in each fiber, is only one of these Lagrangian spheres. Yes, only one. And then it goes and shrinks. Okay. Okay. Well, I mean, what makes it unique is the fact that it's invariant under the parallel transport. If I drop that. I mean, how to say, yes, if I change the sphere a little bit in the fiber, that fact that it's invariant under parallel transport is not far away. No. But then it will not necessarily converge too. Okay. Well, now the most important, well, another important thing is the dentwists. Dentwists. The dentwists. So what is the dentwist? That's a general thing. It has a priori nothing to do with left shift vibrations. So you have a symplectic manifold X omega. And you have a sphere inside it. But the sphere should be, if you want to be very precise, should be really parameterized by a model sphere. So it's really a map from SN to X. And F is a Lagrangian embedding. So you have a Lagrangian sphere. Lagrangian embedding. Then these things induce, induce in a basically canonical way, an automorphism, a symplectic automorphism called tau, sorry, tau. Let's write here L to be the image of SN. And of course L doesn't know how it was parameterized. So there's an abuse of notation here, but I will still write tau L. So tau L is a map from X to X, and that's a symplectomorphism. And it is supported near L. And how do you basically construct it? Well, L comes always with a little, it's called Weinstein model neighborhood inside X. So a little neighborhood of L looks like the cotangent bundle of a sphere inside X. So if we are talking about compactly supported, so near L, so we can construct it there and install it inside the manifold. So we are going to look now at the geodesic flow from the unit disk cotangent bundle of SN, minus the zero section to itself, like this. And this is what I will call the normalized geodesic flow. So what do I mean normalized? I mean that the length, only the direction of the vector matters. The length doesn't matter. And therefore on the zero section you don't get it. So actually this is Hamiltonian. If you take the Hamiltonian function h of qp to be p squared, with a norm induced by the round metric, you get exactly this map. And what is kind of important here is that pc at time pi, and that's very special for the sphere, extends to the zero section. To the zero section and becomes on the zero section exactly the antipodal map. On zero section is antipodal map. And now basically you want to exploit this special symmetry of the sphere and extend everything. So you want to start on the zero section with the antipodal map and go up the fibers, apply this normalized geodesic flow. But with slower and slower speeds until you get to speed zero. So you need basically to choose a cut-off function that starts say with half here. And here it becomes, yeah, rho of s, I will call it. And here it becomes zero. And define something like that. tau of x would be basically c of 2pi rho of absolute value of x of x. No square, no square. Otherwise it will exist everywhere, of course. No square. Thank you. Okay, that's the dent twist. And it was kind of played a huge role in simplec topology. It was, it generalizes, it was discovered by, I don't know, by Zeidel probably, but probably was known to Arnold in ancient times already. And it generalizes the classically, you know, two-dimensional dent twist probably invented by them. Well, according to Arnold, it might not always be the case. Anyway, I want to explain that the dent twist actually has to do a lot with the monodromy of Lefschitz Fibration. But before I really go into this, suppose I have a Lefschitz Fibration with one critical point. And it is tamed away of this u. And so this means that away from u, I can identify all fibers. And I'm taking here a Lagrangian K. So Lagrangian K in the fiber. And I'm moving it by parallel transport like this. So what I'm going to get here is exactly tau S of K, where S is the vanishing sphere of the singularity. If you have only one critical point, it's kind of well-defined, up to isotope. Okay? In M. Okay, so that's kind of for intuition. Okay, so now we can continue and get to Lagrangian cobaltisms. So what are Lagrangian cobaltisms? Lagrangian cobaltisms. Can I ask a question about this? Suppose I have specified abstractly the fiber and I choose a bunch of Lagrangian spheres in it, are you sure of it? Can you always make some Lefschitz Fibration to go to the vanishing cycle? Well, if you fix one, you can always make it. But if you fix a bunch, I think it might become a bit delicate. But if you fix just one, so that's a kind of well-known fact which is exploited. I will exploit it all that if you have a manifold with a Lagrangian sphere, you can always build a Lefschitz Fibration having this manifold as a fiber. And this Lagrangian sphere is the vanishing sphere of some path. So the whole data actually, in that case, is just this thing, this mapping from the vanishing cycle. Yeah, yeah, you basically need to take the mapping torus of that and kind of fill it in with something. Yes, that's a construction. Okay, so Lagrangian cobaltisms. So what are these? So we fix a tame Lefschitz Fibration, a tame. By the way, I want to insist I'm going to allow also Lefschitz Fibrations without singularities, just products. So E equals C times M is allowed. So that's already good and I want to include this. So what is a Lagrangian cobaltism? So a Lagrangian null and I'm not going to write null each time. Null cobaltism in E is first of all Lagrangian submanifold in E is a Lagrangian submanifold of V in E like this That sits like a cobaltism inside E, such that this also has three, no. This doesn't have three, no. I need to move here. Okay, so there's this continuity here. So what happens with V? So first of all, when you restrict the projection to V, so that's from V to C like this. And I will identify C with R too many times in an obvious way. That's proper. That's proper and there exists a huge real number R for which the following holds. When you take the inverse image of minus infinity, say minus R times R. So that's a subset of R2. Then this looks very nicely. This looks like a union of rays. I don't know R of them, say. So these are rays LQ multiplied by Lagrangians in the fiber with LQ in M Lagrangian. And let me make a picture. And LJ in R2 are rays. So here's the picture. So here is the coordinate Y equals 0, say. And here is X equals minus R. So let's look how the projection of the cobertism V looks like. So you're going to get something which has no structure, some set. But from this set, you're going to get some things that look like this. This is called ray L1, ray L2, until ray LR. So basically V looks as follows. It is some Lagrangian submanifold inside E. And at infinity, or at minus infinity, it is completely cylindrical. So it looks like a ray, horizontal ray, times Lagrangian submanifold L1, a ray L2 times Lagrangian submanifold L2, until LR times LR. That's the way it looks. And everything is tamed, so here is my subset U. And you can do this. You can make the definitions without the tamed condition, there's no problem. You just need to insist that these things are invariant under part of the transport. But then, of course, you get into a lot of bookkeeping. Where do the Lagrangians, capital L1 until capital LR live? They live in different fibers and you need to identify the fibers, et cetera. This is why the tamed condition is very convenient. Do you assume somehow that nothing projects onto the far right? Nothing. No, no, again, it's not the full definition. You need also to assume that it's proper. I mean, I wrote the proper map, but you also need to assume that nothing goes to infinity in this direction. So you need also to project on the x-axis and assume that's proper. And here's there's nothing. Yeah, yeah, there are many. Again, definitions become long and... Just one more time. The thing that occupies this point and those two points is the definition. E-E-E-S. Yeah, yeah, yeah, that's a definition. Yeah, a definition, not full one, but... Right. Okay, so... But you say it's a null-cobardism. Yeah, null-cobardism, right. Yeah, this is why I call it null-cobardism. Yeah, now I need the hook. Okay. Exactly because, well, you could say that the Lagrangian sitting here together, you know, it's null-cobardant. There's a manifold bounding them. So also it's not really a co-boardism. You need to chop it here to get a co-boardism in the classical sense. But this cylindricity at infinity is very convenient. Okay. Okay. Now, of course, I'm going to do some floor theory. So I need to restrict to the type of Lagrangians I'm playing with. So without, you know, I think your workshop is in many ways dedicated to issues. I mean, sorry, the summer school is dedicated to issues coming up with what kind of problems you get when you do homomorphic curves under different settings. I'm going to adopt the most conservative setting, so class of Lagrangians. And this is only for the analysis to work. I think that from the algebraic point of view, it's not so important. So I'm going to assume that they are either monotone or exact. So exact is already interesting. And they are compact. What is with monotone? Right. So monotone, so again, without going into very long definitions, you see, when you have a Lagrangian submanifold, you have like a relative churn class called the muscle of index, which, you know, when you take a two-dimensional topological disc with boundary on the Lagrangian, it measures somehow the complexity of the Lagrangian subbandal inside the symplectic bundle, you know, how it rotates, and this measures it. Okay. That's the muscle of index. But you have another class. The other class is the symplectic area class. Monotone means that these two classes are well correlated. If one is positive, then the other is positive, and vice versa. You can think of it like the symplectic analog of Fanno. Right? So this means in particular that if you have a homomorphic disc, then it has positive area, it will also have positive muscle. And this muscle of index appears as when you calculate the dimension of spaces of homomorphic discs, and therefore you want these dimensions to be positive. And from here on, you'll go into the topic that you are exploring in this summer school. Right. And we will work with, we will work with the Novikov ring. Novikov ring, the full Novikov ring. And now we need to go into Foucaille categories. And once again, that's a vast topic, but so I cannot give full definitions. But we're going to play the following game. So we fix a tame left-shift vibration, tame, outside of subset u. And I'm going to assume that the closure of u is in the upper half plane. For convenience. And now there are two Foucaille categories that come, that appear here. One of them is the fiber. So the fiber has a Foucaille category. So, you know, you take all Lagrangian manifolds off your class, and these together form an infinity category called the Foucaille category. So that's an infinity category. And from it you can define a so-called derived Foucaille category, which is already a honest to go category. So this is a linear triangulated category, which rectifies the problems that the original Donelson category had. Basically the fact that it was not triangulated. And similarly, but less classically, E itself gives you a Foucaille category of E, and also a derived Foucaille of E. So here one should say something. So what are the objects here? Objects. So the objects here are, I mean, the way I define it, are really the null co-boardisms. Null co-boardisms in E, not just all Lagrangians. And I'm going to really assume that they all lie above in the upper half plane with pi of V inside the upper half plane. That's again a matter of convenience, nothing else. So you want to make a flow theory with such Lagrangians.  This already gives problems. The problem is that in flow theory you need the Lagrangians to make the Lagrangians intersect transversely before you start counting or make some Hamiltonian perturbations and so on. Well, you can do it here also, but just look what's going on. That's one co-boardism. Then you have another one. And it could be that at infinity the rays just coincide. Or the rays coincide. It could be that in the fiber you get an intersection point and this intersection point gets multiplied by the ray. So if you want to make them transversely intersecting, you need to apply perturbation which is not compactly supported. And this gives analytic problems. But before you get analytic problems, you have another problem, because the space of admissible perturbations, admissible for the analysis, is actually disconnected. And if you perturb in one direction and in a different direction, you get different results. So what is actually the flow complex of, say, V and W? So here's one, here's V with the rays here. And here is W. I always project them and visualize them like that. So here's W, say, and here's another one like this. So what are you going to do with this pair? So you're going to perturb, you're going to perturb, say, W in such a way that now they look like this. I'll do it here, it's maybe better. So that's V, left unchanged. And here's W, Y perturbed. So I'm going to do the perturbation that it looks like this. And the rest is as before. If you need to perturb in the compact part, go ahead and do it. That's not a problem. Here comes the essential perturbation. So you need to decide the second Lagrangian, where you perturb it, to be cylindrical below or cylindrical above. And if you do it in one way or the other, you get different results. So we are going to do it below. And the choice is, this choice has a reason. You want CF of V comma V to be unital, and that's going to... You have a whole, you know, real line of choices. Yes. No, so that's an important remark. This is not wrapped. Exactly. Very non-wrapped. I don't know what this means, but, you see, I grade just one point here. Wrapped means that you... Also, I'm not in a wrapped situation, so at infinity, I don't have convexity. It's more flat at infinity. It's really a different structure. Anyway, this is what I do. I forgot to say one thing about the nullcobardisms. Of course, if you want to go into categorical language, you need... It's convenient to make the ends hang on specified Y coordinates. And I forgot to mention that. So the nullcobardisms I'm considering are like this, that ray L1 is hanging on Y coordinate 1. Ray L2 is hanging on Y coordinate 2, and so on. So I'm allowing only... And this is, again, not a big restriction, because you can always go with the ends, put them where you want. And another thing that occurs is a question. You see the Lagrangian, Lagrangian's hanging on the rays here, they are not really disjoint inside the manifold M. They look disjoint in E, but in M, it could actually be the same Lagrangian all the time. So they are not really disjoint. There's no such assumption. Okay. So now we can continue with this. There is a so-called inclusion functor. So what's the inclusion functor? If you take a curve gamma in the plane with a curve with horizontal ends and not passing through the singularities. So here's an example of a gamma like this. And here's another one. So if you have such a thing, you can basically use it to go from the Foucaille category of the fiber to the Foucaille category of E. And that's... I'm so understanding. You can only do this with a first kind of gamma, not the second kind of gamma. Right. I agree. I agree. Yes. Good remark. So the ends should be... It should be like the left one. But anyway, you get i gamma that's a functor from the Foucaille of M to the Foucaille category of E. On objects, it does just the following thing. You take L. I would like to write L times gamma, but I cannot because I'm in the left-shot vibration. So I write gamma L like this. What do I mean by this? Well, I'm going... The hook can be very useful also for this. So I am just taking my L here and moving it by parallel transport here and by parallel transport back here. And I'm getting a Lagrangian co-bordism. Okay? The fact that I'm tame, its tame makes it easy to define. Okay? So this is trace of L under gamma. So gamma is some curve which avoids... The critical points. Yes, the critical values. Otherwise it can go where it wants. Otherwise you cannot really define it... Sure. But is gamma allowed to cross itself or it has to be... No, because you want the thing to be embedded at the end. But sometimes you can make it with crossings. But if it wraps around some critical points you will have monograms. No. So that still allows it? Yeah, yeah, yeah. So you just create that by L on the bottom? Look, that's... Basically you can go from here, here, here and go here. You start with K and you end up with a... Then twist. No, no, that's allowed. Allowed and beneficial. Okay. It's a non-trivial fact, but okay, that's an infinity function. So this is just the description on objects, but it really... It respect the infinity structure and it is well defined up to... up to quasi-isomorphism. Do you understand the cone on the map of palm spaces? What? Just say for cell palm spaces. Do you understand what the cone on the map of palm spaces is? The cone? I have some L, I look at L, L, L, L, L, L, L, L, L, L, L, L, L, L, L, L. What is the... Is there any specific description of that? No, that's relatively obvious. All right, so... No, no, that's going to be a cone already. Sorry, that's going to be... Well, I'm going to get into this. Yes, that's going to... Well, it depends on gamma. For such a gamma, you're going to get a cone. Okay. Very good. So I want to get to the main result somehow and I'm trying to... So here is a... I'm going to use a set of curves that look like this. So I'm going to take the curve gamma i. So this is y equals 1. This is y equals i. i, index i, not square root of minus 1. And that's gamma i. I'm going to use these curves. And I'm going to use... So this will be used a lot. And I'm going to use also some thimbles. So for thimbles, I also need to specify curves. So here's my lectures fibrations and here are the critical points. x1, x2, till xk. Sorry, critical values. And I'm going to take these curves.   Like this. These are the thimbles. And these thimbles are actually nalcobardisms. So they are objects of the fukai of E. So t1 till tk are objects of the fukai of E. Okay. Now maybe I can state the result. Do you have gamma i? What relation do they have to the critical point? Do they go right in the critical point? The gamma i's are... Yeah. I'm going to draw them. They are here. Oh, they're there? Yeah. They are where you sit. Yeah, but I said what relation do they have to get They just to the left. Yeah, yeah, but I have not done anything with them. I'm going to do now. There was... I'm going to do with them something now. Okay. So now I can state the result. So basically what is the question I was... One would like to know is... So well, you have this... The fukai category of E and you have this inclusion function. How larger... To which extent fukai of E is larger than the fukai of M? And the surprising answer that not much larger on the derived level. So theorem. In the derived fukai of E, You have the following thing. For every co-bordism. Null co-bordism. You have the following isomorphism. V is isomorphic. I'm going to write like this icon. So this has nothing to do with the Apple company. It's an appellation of mathematics. I stands for iterated cone. And it's an icon. I'll explain what I mean by iterated. So you have to take the symbol T1, tensor product with some vector space, E1. T2, tensor product with some vector space. Fixed vector space. Until you go to Tk tensor Ek. And now come what Duzza asks. The little gammas.  And surprisingly you don't need L1. So here is the V. That's L1, L2 till LR. And so that's iterated cone. The capital T's are the symbols. And E1 until Ek are just our lambda novikov vector spaces. Of course they depend on V but in a canonical way. Depending on V canonically. So by iterated cone I mean the following. You start from here. You have a map, you take the cone. Then you have another map, you take the cone and continue. So that's an abuse of notation to write like this. But, okay. Do you see me? So is I gamma actually equal? Yes. Yes. Sorry, another question. It's closed so the Lagrangian stuff today. Is it closed Lagrangian's amount of noncommodals as well? Yes. Yeah, definitely. Right. Okay. So what does this really say? That you know every on the derived level. So how to think of the derived level. If you really want to calculate flow homologies at the end. All you need to know are the ends of the Lagrangian. And some vector spaces associated to V only. Not to the partner of V in the flow homology. So, and the symbols. That's it. Is this really some sort of, you know, if you have a stronger statement about the T's and the gamma L's being an exceptional collection or whatever, or is that? It could be that. That's not another single object. But that's not how you prove this. No, no, no. No. Okay. Now come the following thing. So. Now come, yes? Are the vector spaces finite dimension? Yes, yes. Finite dimension, definitely. Now come the following thing. You maybe want to ask the following question. Okay, I have a co-bordism. Maybe the co-bordism induces some, the composition inside M. After all, at the end of the day, you want to study M, not E, right? So you may think of E as just as אוגזילרי, you know, thing, not the main object of study to get this one. So it's possible to do. I want to explain that. And now the dentists will appear. By the way, you can, the same result holds also when you don't have singularities at all. And this goes back to our previous work. But where is the result? Ah, here it is. So if you don't have singularities at all, then all these things, they go away. And you just say that V is an iterated cone over these things. So that's also non-trivial. Okay, so. Sorry. That has some questions. Okay, so the composition in E, in M. In M. So in M, I'm going to do the following thing. So I have my, again, my U and my critical values, okay? And I'm going to fix a base point here. And I'm going to take these paths like this. And these paths will induce for me here, in the fiber here, many Lagrangian, vanishing Lagrangian spheres. So vanishing spheres will be S1 until SSK, like this. Okay, and I'm going to use them. And let V in E be a null cubordism. So here's the theorem. In the derived Foucailla of M, you have the following decomposition. So I'm going to really use that picture. The lowest Lagrangian, L1, is quasi-isomorphic to an iterated cone. And now comes a monster formula. So you should take the dent twist inverse on the last Lagrangian and compose it on the last sphere. Compose it with this one. Continue composing until you go to tau S2 inverse on S1, tensor with E1. You attach this thing to... Now you do the same. Now you do tau SK inverse, tau SK minus 1 inverse. But you stop one before tau S3 inverse on S2. tensor with E2. And you continue like this until you get to tau SK inverse on SK minus 1. Then you go to just... Sorry, I forgot tensor EK minus 1. Now you have just SK tensor EK. And now you have the more civilized part, which is LR, LR minus 1 until L2. That's it. Okay, so yeah, it's a long formula, but what can you do? What do you write this? Do you know the name of the map? I know, but not... I think you would not be satisfied with the answer.  But okay, this is what happens, right? So you have to take the Lagrangians that appeared here, attach them one to the other, take successive cones, and then take the vanishing spheres and continue attaching them tensor with some vector spaces. That's a result. So when you don't have any singularities, the result says that the Lagrangian L2, the L1 here is an iterated cone over the rest. This is what you get. Okay, so I think I have... How much time? Five minutes. So there are two choices. Either I go and give an incomprehensible proof due to the four minutes, the five minutes time I have, or I want to motivate why do I actually care about this? So there are many answers. Not all of them are always convincing. One answer is that... One, what? Do you need to vote for one of your motivation or do you need to come to answer one of your... Ah... Yeah. No, let's not vote about that. So the... Of course, you could say the following thing that... You see, you have the derived Foucaille category, which is kind of notoriously algebraic thing.  And then in the end you get something that algebraically behaves better, but you added some very weird objects there, right? What is the relation between these objects and geometry? Or even topology? So this approach tells you actually many, many times the cones that you get are actually induced by something geometric. So if you want to get cones, interesting cones, you maybe can produce interesting co-bardisms. This will give you cones. And the cones will give you exact sequences and this will induce the algebra. But I want to give... So that's one motivation to study... So you're saying that this tells you that a lot of the cones you have actually have a nice geometric interpretation. Or even generate them. What you're saying is that a very special type of cone has a geometric interpretation. Couldn't there be all kinds of much more cones? So the question is, you are asking now to which extent the modules that you get via co-bardisms give you the whole set of the whole category of modules. How large are the abstract modules than the ones coming from here? In a way, that's a little bit open. We don't know. I wouldn't expect that all modules will come from that. Not even in the triangulated envelope. But at the same time we are now playing with some approach also into account immersed Lagrangians and immersed co-bardisms and there the answer seems to be yes. You get everything.  But okay, we are not there yet. But I want to give much more some other motivation to study coming from real algebraic geometry. So I'll be brief here. But I want to explain the principle. One thing that when you look at real algebraic manifolds one of the things that you see is that if you take a complex manifold it sometimes happens that it can be endowed with completely different real structures. So the kind of most famous example that everybody know, let me call it M. Well, it can take say M equals Cp1 cos Cp1. And here are two different real structures. One of them, the first anti-morphic evolution will just send Z0 Z1 comma W0 W1. It will send it to say Z0 bar Z1 bar comma W0 bar W1 bar and the real part of C1 is just, in this case, it's an RP1 across RP1. So it's a T2. But you have another real structure here which is the following C2 on Z0 Z1 W0 W1 That's just you switch the coordinates and then conjugate W0 bar W1 bar Z0 bar Z1 bar And the real locus is Lagrangian sphere. It's an S2. Okay, and kind of the question is, okay, so you can have many real structures what is a relation between them? Now, in general I don't know how to answer such questions. It's too general. But under some situations and I can a little bit explain how you can actually claim that different real structures are co-bordant Lagrangian co-bordant inside the Lefschitz fibrations. Okay, so hear it how it goes. So if X inside Cp N, large N is a projective projective Now I don't know how to call it. It's a projective real complex manifold. So it's a complex manifold and with a real structure. So real complex complex manifold then you know, you can take hyperplane sections of it and you can parameterize the space of hyperplane sections by your dual CpN. So you look at the dual CpN and each point here is a hyperplane and you can cut X with this hyperplane. So here is your X and you can cut it with a hyperplane here. Sorry, that's CpN that's X. And if you have a point here you can look at X intersecting with H lambda that's H lambda and you get your M lambda. That's a hyperplane section. Now if you are not lucky and this H lambda heats X non-transversely, you're going to get a singular manifold. And that's actually how you create Lefschitz fibrations. So here inside we have the discriminant locus or the dual variety X star and basically what happens is that the complement of this here is of complex co-dimension at least one and therefore the complement is connected. This means that all this M lambda they are by Moser argument a simplectomorphic. They are not biomorphic of course but simplectomorphic. However, if you now look at the real loci then things dramatically change. So if X is embedded in a real way you can consider real hyperplane sections. So hyperplane sections that are invariant under the involution of X. So you're assuming when you say embedded there's a real way that the involution extends to CDM? Yeah I mean that other way around you take this one and it restricts here restricts there and keeps X invariant X bar equals X So there's an intrinsic way to describe these things but you can take a real ample line things like that that's equivalent basically. So Okay so just to finish the story so you can now look at the discriminant locus of real real things and this is inside rpn or better written cpn dual real part like this. Now this is disconnected so here you have this rpn and inside it you get many chambers many many chambers of real structures and this is exactly what happens here. So here is for example the real structure that gives you rp1 cos rp1 and here is the one that gives you s2 and now exactly what happens exactly You see when you look at hyper plane sections and you take a pencil a so called pencil inside the dual cpn that hits the dual variety transversely you get a a left shits pencil you need to blow up the base locus then you get a left shits fibrations and if you take the pencil to be real you're gonna get precisely a left shits fibrations and inside it a Lagrangian cobaltism with not null cobaltism but that's not a problem with one end starting here and the other end here so you're gonna get really a fibrations sorry a cobaltism that whose projection is gonna look like an interval it's gonna pass through the singularities that are real so some of the singularities are real some of them are conjugate points so each time you hit a wall you go through a singularity and here you have the real part corresponding to this here you have the real part corresponding to this and if you want to go to be in the null cobaltism then you can always bend and go like this so usually these cobaltisms they appear from real real structures and I cannot say any results yet but the hope is that using these the compositions you could say something about the relation between the topology of different ends what are these axes you brought on this curve these are stars these are the critical critical values but I mean so the thing about that is that some sort of singularity no this is smooth Lagrangian this is smooth the thing is smooth the whole thing is smooth yes it's a smooth cobaltism I mean that's almost not totally obvious but it's not hard to prove just because you see the you get actually a left sheds fibrations and inside it you get a real left sheds fibrations and it becomes most automatic I'm making a real locus inside about this yes yes yes so of course you pass through singularities okay so that's how these things occur naturally okay I think I should I should stop alright do you have any questions for our speakers so does this thing do you guys have anything to do with this account method for computing intersection matrices of real real polynomials considered as complex polynomials and two variables there you can draw like the vanishing cycle in the real part do my shame I don't know about his methods and therefore I don't know if there is a relation but like how you know how do you expect to get a lot of Lagrangians from this a lot of what? a lot of Lagrangians as the fixed part of this evolution in this chamber no of course there are huge restrictions on what Lagrangians you can get that's I'm sure I think I think not everything will appear as a real part I mean there are also classical topological restrictions on such things and so on but what is kind of I think more interesting is you see when you start looking at such pictures what you get many times is that the real locus is actually disconnected even on del petro surfaces real ones you get the real locus can be you know four copies of a sphere plus other components and then you get a co-boardism which is very special right so the Lagrangian here is really a disjoint union of Lagrangians and maybe you can kind of find restrictions of this type I don't know so if you take a complex legal in its slide variety in a different real structure can you find can you find co-boardisms between I don't know I would definitely try Do you have anything about the you know this next conjecture about whether every manifold is a real part of in fact I'll call it up from some form of it it's surprising to me it wasn't all the loads of three manifolds for the real part but just rational varieties can you find in this infractic world is it completely not scale is there some obvious instruction to be the real part of that well, obvious not but there is one result which is kind of known but in fact there's nothing to do with real it has to affect only to with being Lagrangian so this goes back to Witerbo so Witerbo looked at real fun on manifolds of dimension real dimension 6 or more showing that the real part can never be hyperbolic but then later I think well, it's your result with Eliashberg so Eliashberg given to Hofer extended it to all Lagrangians in such things so the fact that somehow Witerbo's proof the fact that it's real played a role so but I mean, that's a serious restriction and I'm not aware of many others that are not topological they don't come from topology well, what are you saying since I didn't know this at all that if you have a Lagrangian some kind of it's in your book well, I don't know I think so, I'm not sure so if you have if you have no, a funnel if you have a funnel manifold so that's you were talking about a complex what's important there really it is that it's uniruled this is how the proof works but funnel and then then any closed Lagrangian some manifold there inside it it cannot be hyperbolic yeah, it would be fine yeah, but I forgot if it's enough uniruled you need uniruled with something something about the turn class but I have to look it up that is satisfied in the funnel case okay, so who proofed this, sir? אליאשברג given talent hofer אליאשברג given talent hofer the proof is basically the following you take this spheres the homomorphic spheres you look at a little neighborhood of the Lagrangian you stretch the neck you stretch the neck and then you get some kind of SFT building there so you're gonna get so in the cotangent bundle of the Lagrangian, that's where you stretch the thing then you're gonna get geodesics now if the thing is hyperbolic if you they've had a hyperbolic metric then you could calculate the index of this thing and get a contradiction to the index that you come from the funnel story the proof works something like that that's the idea and Witerbo did something quite different with symplectic homology I think


Hello friends and welcome to another episode here on the channel for anyone new to the channel. My name is Lee also known as Osiris and in today's episode we're going to be continuing on with our VGC 2021 content featuring a team based all around Galarian Articuno. So the team as you can see on your screen is made up of that Kanto Raichu, Galarian Articuno, Glastria, Tapu, Fini, Incineroar and Cartana. There will be a poker piece down in the description below and as always we'll throw a renter code up at the end of the episode if you guys want to try this team up for yourself. So without further ado, let's jump on to the ladder and test this out, get some games and we'll be able to talk through some of the mods in the team and how it functions and what we're looking at. So first things first, ignore the headband. I've got it's late. It's late. It's like it's like half 12 and it's I've got the worst hair. So ignore the headband. It's in there for a reason and I'm sure you can see past it. It's fine. OK, so we got Yuki up first and playing a team of Tapu, Fini, Rilla Boom, Porygon 2, Indidi Female, Alola Maruak and Galarian Maltres. OK, well, we've got a few things to think about here. Galarian Maltres, obviously one of the big things here, the redirection from the Indidi as well makes it a little bit tricky for us to function. But we do have methods around things. I mean, the Articuno does struggle against Galarian Maltres, for sure. It's always going to be one of those Pokemon it doesn't really like to kind of face down against. But we do have Glastria that can come in and do some work. So if we can get a trick room up, maybe with our Articuno, we can go for the Glastria mode. It's definitely an option. I think we definitely need Incineroar in this match. It helps us out a bunch against just the intimidate support against the Alola Maruak, the Rilla Boom. Also, the Dark Typing is quite nice as well for switch ins for Maltres and obviously the Indidi, any any expanding forces and things like that. We do need a trick room, so we're going to have to go trick room mode. I think Glastria is a Pokemon that we need and probably Tappu, Trapofini or Kartana. I feel like Tappofini is probably a little bit more solid for us. So kind of benching Articuno in this first one, which is a little bit of a shame because it was the centerpiece of the team.  So it's good to look at the other mods and how the team functions outside of just the Articuno. You can't always rely on just one Pokemon every game to come and do work unless it's called Tappofini or Incineroar or maybe Metagos. Maybe there's a few, but Articuno definitely not one of those Pokemon. It's probably why it's not commonly seen, but it does have a lot of functionality. So what we're going to do here, we've got the Indidi, the Maltres. Because the Maltres is actually flying, we can actually fake it out here. Whether or not we want to, that's another thing because we could potentially see ourselves be kind of subjected to them maxing, going for help in hand into Articuno turn one, which would not be ideal. We might be better off just preserving Articuno for later on in this game. And actually, what are we doing? We need to we need to pawning shot into the Maltres and then switch into our Tappofini. Get rid of the psychic terrain and just kind of soak up this potential dark type attack that will be coming out from it. Because I imagine that's all my pond is probably trying to do. Then they know that our best mod is going to be with the Glastrian trick room. And I'm sure they're well aware that Articuno does have access to trick room. So I think they're probably going to go all in here. Avoid the fake out max and just go for a big attack. Now, if they go for a nasty plot here, we are in a little bit of trouble. But we can approach each turn as it comes. We'll just see what my opponent does. There's the follow me. They're going to go for a nasty plot. We could have went for a trick room. I can't believe this. I can't actually believe this. We could have just went for a trick room. I think the right play here would have been helping and attacking into Articuno. But regardless of that, we're not seeing it. So we're going to see a nasty plot and we're going to be able to get something onto the field. I mean, we can get Glastrian to the field and then potentially switch back into Incinero. But we're going to have to try and calm mind with Tapufini here. We go all in. That's the other option. We could potentially go icicle crash into Maltres and try and make room for Tapufini here and go for a calm mind. Now, if they the problem is icicle crash will proc a weakness policy. But we're likely to see follow me anyway. But we've got to put some pressure on this Maltres. It's just got the nasty plot at full. Basically, nothing cannot believe that we could have just got a trick room up self freely there. It's our own fault for not going for it. But when you look at a turn and you look at the most logical kind of place that your opponent could make, you kind of try and make the kind of next best guess. But sometimes it doesn't always work out like that. And people think differently as well, don't they? So that's that's one of the things you've got to think about. I could have just faked out. Never mind, never mind. We're going to see the Maltres Max. It's likely to maybe go into it's half a finny here to remove it. Is it helping hand? Yep. We do get the calm mind up first. So that's that's good. That's good. OK. Well, at least we know we're faster than the Maltres. So that's some decent information. It is a stream of being to the finny. Yeah. Proc our Citrus Berry. But we are going to proc a weakness policy as well. She's not going to be ideal. Maybe she'd have Max Glastro here. But the Icicle Crash should still do a decent chunk of damage. Yeah, it takes it down to 50 percent. So we're not in the worst position. Now the Maltres will outspeed us the next turn, and they do have access to follow me now. But where do they direct their attention to? I think we could probably get away with bringing in incinero, Maxing, Tappu Finny. Could we go for Starfall? I just feel like we're going to get. Yeah, I don't think we max. I think we've got for another calm mind. I think you go after Glastria here. I don't think you can leave the Glastria unchecked. And the Tappu Finny right now, from my opponent, logically speaking, is is in a bad spot. So it would make sense for us to protect it.  Even though you do have redirection support. There's a nice stream. It's after the Finny again. OK, just chasing down that Finny. Well, we kind of need a board position where we've got where we've got Articuda out next to Incinero, and they haven't got their Max turns anymore. Should we just try for the trick room? I think that that that that is literally our best bet, I think. Let's just try for a trick room. We actually got the trick room. This is actually nuts. OK, well, I mean, I really don't understand what my opponent is doing here. Do they just not realize that we've got access to trick room? And now that turns end, so we can Max. They're still going to do a chunk of damage to us. But we do have the ability to it's tricky because we want to get rid of the indeedy really. I'm going to Max Hillstone into that and do a hurricane as well. I don't really want to lose Articuda this turn. And I feel like Maltres does attack. I think it's like plus five as well. So it's going to hit like a truck, especially with help and hand support. But if they go help in hand, then they get punished and they lose their Maltres and then this games back on.  I don't know. It's going to be very difficult for us to bring this back, but we'll see. We will see you. So there's our protect. There's a redirection. It makes sense to utilize that. We'll be able to get rid of the indeedy the next turn and potentially, potentially let's see how much damage it does. That's ideal. OK, we get rid of it now. That's perfect. OK, that is what we want. Critical hit will take it all day long. And the hail storm will get rid of it after this turn. What's the Maltres going to do? Very rough. Yep. And let's see how much damage it just does. This is going to sting. It's going to sting. OK, not so bad, actually, not so bad. We can be like Glastria is just a tank. OK, the indeedy gone. Thank goodness. Right, what's coming in? I wonder if a hurricane would be enough to take that thing down. Like Articuno is pretty strong. So might it might be able to. Let's see. But what comes in? OK, it's Apropheny. It's not. It's not the best, but it's not the worst either. I mean, we could just target the Maltres. With a max hail, it's probably going to protect here. It's probably going to protect. And we could go freezing glare. Now, we're not going to be able to freeze the Finney because that's the secondary effect. But we'll get some decent damage onto it. Maybe put it in range for Max Quake, the following turn. And then my opponent down to two. Wow, well, maybe, maybe their last two. We can get rid of the Maltres here. That would be huge. Can't believe we could have just literally done this turn one as well. Tapu Finney is protecting. Oh, sucker punch. OK, well, that's fine. That procs our weakness policy. We like that. We'll take that all day long. And then next turn, we Max Quake into the Finney, get a special defense boost so we can take an attack without Articuno and then fire it off and attack ideally into whatever their last Pokemon is. We get the Chilling Nair boost as well, which is very useful. And the freezing glare into the protect on the Finney. OK, so that's fine. Let's see how have we how have we pulled this one back? I have no idea. Can it's left over Finney? We still got turns of a trick room left and one more turn of our Max. So what's coming in? Oh, it's Marouac. OK, we have to go after the Marouac. We have to, which means that we maybe. How many turns of a trick room left? Because I feel like Articuno, too. OK, if we can get a Chilling Nair boost, it's just I feel like the Marouac protects here. That's the thing. And where do you go for your Tapu Finney? It's probably bulky, so we may survive. Articuno might survive an attack. I'm going to go after. I'm going to go after the Finney. I'm going to go just straight after it. Yeah, I thought you'd protect, but we do get the special defense boost. So that kind of helps us out a bit. And we're still going to be in trick room. So it's going to be better into the Finney. I mean, that does a phenomenal amount of damage anyway. So we're kind of all right. Great, OK. This Cluster is like just annoyed at the situation from the start of the match. Just like, let's get this done. There's a calm mind. OK, well, as long as this does, I mean, my opponent's not really pressuring us. That's the problem. Like, I think, freezing glare, this is coming off plus two. So let's see what this does. Respectable. We'll be able to get the Finney in the next turn, I think. So we can just protect. Protect Articuno and then target the Mawak. And that will do the job. Yeah, because this is the last trick room ends this turn. So we will go high horse part into Mawak. And as we go before, no, we went after the Finney. They calm minded first. Let's just protect. Let's just protect here because they could attack us. And we can't boost ourselves anymore. And then a freezing glare plus, potentially the chip damage of the hail. Potentially, I don't know. But they're going to attack with Finney here, I think. I'm hoping they go Moonblast into Articuno. That would be the ideal situation. And they're kind of in a situation where they need to protect Mawak. But it's risky. It's risky. So no protect from the Mawak. We're going to protect Articuno. High horse part does connect. Thank goodness for that. I mean, the other play would have been maybe going after the Tappu Finney here. OK, they Moonblast. Just like we wanted. And we're in a phenomenal spot now where we can just go freezing glare and go for a high horse part. And we should be able to close this one up. There's the dimensions, then back to normal. And we don't want to trick room again. Not now. It's all high horse part. And I don't know how. Like I've said, friends, I don't know how. I don't know how this has happened. OK, there's the protect stalling out. I don't know what good that's going to do. So I'm just delaying the inevitable by at least the turn. But we'll continue. The hail does stop. So they'll be able to get a little bit more health back every turn. But we'll just lock in the same. And that should, like I say, pending any disasters, this should finish up the game pretty nicely for us. And it's nice, actually, this first one. We get to see Glaring Articuno, even though it was a pretty tough matchup with the Maltres. Our opponent kind of just let us set up. So, you know, like we discussed the turn one. Articuno avoids perfect. And we are plus one, I think so we should. Maybe, yeah, we take that with Glastria. It's just such a tank. There's a high horse power for the win. The horse does not want to give up. There we go. Good game to my opponent. And, yeah, I mean, we said turn one. I kind of went through why I made the decision. And I think logically it made sense. The Maltres was threatened from the fake out to prevent a potential nasty plot. So in that situation, I don't know. They must have just not suspected the trick room at all and didn't make sense that they didn't attack into the Articuno. So just beware that Articuno carries trick room. It's like it's third. Must use move. I'm pretty sure. But anyway, we will move on. OK, we've got our next opponent are playing a team of Galerion Zapdos, Dusclops, Sylvian, Reggie Alecki, Urshifu, and Primarina. Now, could we get away with the Raichu Articuno lead? It'd be nice to feature that at least once on this episode. I think we'll go for it. Just because it's kind of the centerpiece of the team, we haven't really featured this mod yet. So I'd like to do that. We've got to be a bit wary about the trick room mode for sure with the Dusclops. It supports Primarina pretty well. I am going to bring Glastria to this map. Do I bring Glastria? Yeah, I think if the trick room goes up, it's going to be like invaluable to us. And then maybe Tapufini. Because Tapufini can calm mind and kind of, other than Reggie Alecki, it can kind of deal with most stuff on my opponent's team. So we'll go with that. The Sylvian scares me a little bit, but I think we'll be all right. I think we'll be all right. So let's see if we can get this mod set up with Raichu Articuno to kick us off with today. It'd be interesting to see what my opponent leads with. It's probably going to be something like Urshifu and Reggie Alecki, if I'm guessing. There we are. There it is, shiny Reggie Alecki as well. Very nice. It's a nice, shiny, very nice, shiny. Okay, now I think we've got the option where we could potentially go for, probably fake out into Urshifu and go Max Airstream into that slot. We could potentially go trick room as well, which would then give us the speed jump on Articuno. It's my opponent really have to, they do have pretty good trick room ones. Did we just hit the opportunity to Max Articuno now and go after the Urshifu? I think we maybe don't, I think we maybe just go Hurricane and keep our Max for later with the Glastria, you know? I think the red, oh my gosh. Are they going to Max the Urshifu here? Okay, well here we go. That's what we like to see. We should have Maxed ourselves. Okay, well I didn't expect this. It's not G Max Urshifu, so I'm ready to let you protecting, huh? It's interesting. We definitely needed to know it's all there. And there's a Max Darkness and this will be the end of our Raichu, huh? What? Okay, that's kind of, I don't get that. Okay, well, I guess that you want to get rid of the, you want to get rid of the Raichu, but how much damage is this Hurricane going to do? Too much. Hmm. How'd you like them meatballs, eh, Urshifu? Hmm. I can understand the logic of wanting to get rid of the Raichu, but at the same time, you've got to worry about the trick room being set up. Okay, well, Zapdos is going to come in now. Now we can Airstream. We could just vault switch out onto the Zapdos and Max. We could just trick room as well. Could just trick room. Because what's the Zapdos going to do to Articuno? I don't think really anything. Yeah, we trick room. Let's trick room and get Glastrian to the field. I know this team's about the, okay, the Electraweb. We'll take this. Because it's not specs. Okay, it might be able to take us down. It may do. I think you've got to concentrate down onto the Raichu though. You've got to concentrate onto the Raichu. We might not want to bring in Glastria just yet. You might want to, oh my God. Okay, well, we, okay. I didn't expect us to knock out the Zapdos. How did, what is going on? What is going on in this game? I have no words. I have no words. I don't know how we took down the Zapdos. Zapdos surely isn't like Raichu plus one isn't that strong. That's the Zapdos. The Glarian Zapdos gives the Kanto Zapdos a real bad name.  So we're in a beautiful spot right now. Well, I say that, you know, you've got the dustclops to contend with, but I mean, we're kind of all right, aren't we? We're kind of all right. Let's go after the dustclops, double down onto it. Because the Regi Alecchi is going to protect as we see the battle canceled. So I feel like we can probably do one more. I don't know what happened in that game, but I'm as confused as you are, friends, but we'll hop into another one. I will try and find another opponent and get another one started as soon as possible. Okay, we've got Maverick up next. Good old Maverick playing lander, asterian, vanillax, Rotomwash, Zapdos, Incineroar and Chandelure. So I feel like we could go for this combination again. And I think we will do that with, hmm, what are we gonna have in the back? I think Incineroar is very useful in this match. Maybe Cortana is our last one, puts a lot of reliance. Zapdos is pretty tricky to deal with. So we need to keep Raichu around as long as possible, because otherwise it would be Glastria. In Glastria in a trick room, situation would be amazing. I'd worry about the Chandelure a little bit, because Cortana does well against Rotomwash. That's the problem, like our Rotomwash matchup is pretty. Maybe Incineroar isn't even worth bringing here. Maybe Cortana and Glastria, I think, that's what we'll go for, yeah. I mean, Incineroar is nice to give us intimidate support, fake out support, but I think we need to try and cover things a little bit better than what we would have done with just Incineroar Cortana in the back. I don't think, it puts so much reliance on Articuno to kind of get set up and start doing like the damage we needed to. Okay, what are we gonna see? Zapdos Vinylux, huh, that's a nice lead. There's a snow warning. Zapdos is kind of pinned here, in all honesty, although Vinylux can't read, I mean, the Vinylux can attack. What we could potentially do is just fake out the Vinylux, go for the Max Airstream into it. Zapdos is kind of, is not really able to do very much here. I mean, it can heat wave, it can Airstream as well, but I think the fact that my opponent isn't maxing first indicates that they're probably not maxing their Pokemon. So we need to try and get a bit of an advantage early on if we can. So let's see if we get this fake out, which we do on the Airstream. That puts us, yeah, we're faster than Zapdos as well, which is interesting. Okay, well, that's fine, because a brutal swing will pick up the Vinylux the next turn. So even if it's scuffed, we pick up the knockout onto it. Ooh, Zapdos going discharge. Ooh, that's dangerous, dangerous. Dangaroos. Okay, well, Articuno takes that and it does proc our weakness policy, which is exactly what we need here, but the discharge is a little bit awkward. We should be able to take down this Zapdos the next turn. If it doesn't max, that is, it could max, but then if it maxes, it's kind of locked into not being able to go for the discharge. And I think what we probably want to do is, okay, well, we have to go after this Zapdos now. I can't afford. Do we, do we, do we, do we, do we, do we, do we? What do we do? Do we go after the Landeros? Do we go after the Landeros? Hopefully, yeah, I think we go after the Landeros and I think we'll just go for a nuzzle into this Zapdos. We should take another discharge. Let's go for a Mindstorm. Zapdos is gonna max here, I think. Okay, it's Landeros. Is it got the Assault Vest? But we are plus four right now. So I still think we should be able to take the Landeros down. There's a cheeky old nuzzle. We've gotta keep right here on the field. Hopefully this takes the Landeros down. Come on. Yeah, there we go. Oh, they couldn't have done all the work. Right, and Zapdos is gonna go for another discharge. I'd imagine, which we should take. The only worry is obviously the paralysis here because discharge got quite a high parachance. But right, you'd take in these boosts and doing a nice job with it. So you do take it just about. And no paralysis, which is perfect. Okay, so whatever comes out onto the field now we'll be able to get rid of the Zapdos. Or we'll just see what comes out and then we can kind of take it from there. Okay, it's gonna be the Incineroar. We've got Clastria and Cartana in the back. I think we're gonna be all right against Incineroar. Yeah, we're gonna be fine, aren't we? We're gonna be super fine, super fine. We'll go after the Zapdos now. We'll go for a Voltwitch into the Incineroar. We'll go for a, yeah, Mindstorm into Zapdos. The Incineroar can't fake out the ride to you here and the battle is canceled. Okay, well, I tried to fit in another one to kind of bump up this episode, but it was a very quick one as well. But I'm getting to see how good the Articuni can be in certain situations, which is really good. So we will now jump over and grab you guys the rental code for this team. Okay, friends, here is the rental code. As I said earlier in the episode, the poker piece is down in the description. There's the rental code. If you wanna try this team out for yourselves, be my guest, and I hope you have a lot of fun with it if you do. This wraps up today's episode. We've got to see a bunch of the team kind of function pretty well. We've had three games, one really decent one and two very short ones, but I suppose the shorter ones do show the explosive nature of this team and the pairing of Raichu and Glaring Articuna, which I believe is pretty nice. So I hope you have a lot of fun, like I say with it. If you've enjoyed today's episode, do consider subscribing to the channel if you are new to the channel and drop a like if you've got two seconds and you have enjoyed today's episode. That's all I ask, and it would be very much appreciated. So thanks as always for tuning in, friends. Have a great rest of your day and I'll catch you all for another episode very soon. So until then, take care and bye-bye.


Hi, this is Dr. Don. I have a problem out of chapter eight, section four, about constructing a confidence interval for the difference between two proportions. It says we have 100 volunteers divided into two equal-sized groups. That means 50 each. Each volunteer took a math test that involved some rules, including one that was hidden. Prior to taking the test, one group received eight hours of sleep while the other had no sleep. And the scientists want to know if there's a difference in the proportion of students in each of the volunteers, rather, in each of the two groups who discovered the rule. And they said, what can you infer from the portions using a 95% confidence interval? To solve this problem using Excel, you can use the calculator I have on my website. Just go to again, drdonright.com. When you get there, let's go to business 233 and 503. And if we scroll down, I haven't added in a specific calculator yet for the proportion, but I am. But I'll show you something. Go down to the bottom here. We have a two-sample Z test for the difference between proportions. And that's what we have in this particular problem, the difference between proportions. So I'm gonna click on that. There's our calculator. And if you go down to the bottom, you'll see that there's two tabs. One is for where you're given the proportions. Or in this case, we're given our sample Xs. So I'm gonna click on that tab to bring up this particular spreadsheet. Here we were given that our sample one for P1 was 40. Our M was 50. Our X2 was 15. Our N2 was 50. Our significance level alpha was 0.05. And we're gonna leave the claim operator greater than because that was the thrust of the question. Is P1 greater than P2? So if we scroll down a bit, we find here the confidence interval for the test statistic is 0.331. And the upper limit is 0.669. Those are the correct answers there. We're given our test statistic. That's our point estimate of 0.5, which is positive, which tells you that P1 is greater than P2. That's what we're saying there in our claim, P1 greater than P2. And down to the bottom, it gives you the conclusion. At the 5% significance level, which is the same thing as a 95% confidence interval, there is enough evidence support that the sample proportion P1 is greater than the portion P2. And finally, we know that because there is no zero in the middle of this confidence interval. Therefore, as does the hypothesis test, we can look at the confidence interval and that confirms that this is a statistically significant difference. Hope this helps.