So, yes, I'm aware that this is a summer school and that this is supposed to be a course. And I was sufficiently aware of it that I sent Ching Yu some, like, plan for what I was planning to talk about. And then I sat down to prepare my talk and it looks nothing like what I told her I will talk about. So, now, now, basically, you should think of today, I'm just give some general philosophy. So, you know, depending how much details you want, there may not be any theorems. And then on Wednesday, I will try to go back to the plan. Okay, so the general framework is that FLIR theory takes Lagrangians to chain complexes or maybe they're homology groups. And what you want to say is that if you have Lagrangians parametrized by some space, it's called a space P here, then you will get families of chain complexes or homology groups over P and such a thing is basically what is known as a sheaf. Now, it turns out that there's actually two completely different frameworks or different kinds of sheaves that come out of this idea. And the first one is probably the most familiar one, which is if you allow only Hamiltonian families. So, in that case, there is the familiar statement that FLIR homology is invariant under Hamiltonian isotopes. And that says that the sheaf that you get is a local system on P. And, of course, you know, when you work at the level of chain complexes, it turns out that you don't want to say local system. You want to say derived local system. But regardless, the way you should think about such an object is that it's determined by the value at one point, by the group or chain complex at one point, together with monodromy data. Monodromy data is some kind of representation of the fundamental group or chains on the base loop space or something like that. And what I want to talk about, so this is the subject of kind of another lecture course, which is not this one. So what I will talk about is a different situation. And that's where the flux map from the tangent space P at a point, a parameter space at a point to the first cohomology of L, of your Lagrangian, L, P, fiber P, with the real coefficients, is an isomorphism. So roughly speaking, if you take Lagrangians and you think about Hamiltonian versus Lagrangian isotope, there is a transverse direction to the Hamiltonian leaves. And the transverse direction is exactly controlled by first cohomology. And this is saying you have some kind of slice to that transverse direction. And in that case, the assertion is that we get something that's not local systems, but another familiar kind of class of sheaves and we get coherent sheaves on, well, now you can start modifying things, but essentially on this first cohomology on P, or some associated space. And so that's the subject of this talk to kind of explain this correspondence. So given the audience, I should start by at least giving some intuition by what is a coherent sheave. So now you have to know some algebraic geometry. So let's take their C star, which we will consider as a complex manifold. Well, in fact, for a moment, let's just take it as a complex algebraic variety. So the ring of functions on C star is some familiar object. It's just holomorphic algebraic functions on C star. So that's just Laurent polynomials. And the simplest kind of coherent sheave that I want you to think about. Well, OK, you could think about this. This is a coherent sheave is going to be some special kind sheave corresponds to some finite. In fact, I could just say finitely generated modules over this ring. Let's call this ring something so that I don't have to keep writing it. But the way they will appear in Flur theory, it's better to actually think about complexes. So in fact, perfect complexes are, in this case, again, finite rank free complexes over this ring.  Exercise one showed that that's actually true. So if you know some algebraic geometry, this is something that you know. And if it's not, then something you should learn.  So what does this mean? There is a map from R to itself obtained by multiplying by this polynomial. This polynomial vanishes at one point at one. And it vanishes to degree one. So exercise two, again, if you know some algebraic geometry, this will be easy. Otherwise, you have to think about it. Show that the cohomology of this complex is of rank one over C. So these are the kinds of complexes that show up when you think about Flur, my assertion. The assertion that when we're in this kind of transverse of the Hamiltonian situation, we get coherent sheaves says that we should see complexes of this nature. And that they kind of control Flur theory in the direction of the transverse direction. Now, the reason that I started with C star instead of C, which would have been slightly simpler, is that this picture, well, I'm going to say mirror. But to the symplactic manifold, which is r tons s1, but that's maybe better thought of as t star s1. Now, usually, one thinks about t star s1 in terms of its kind of vibration over s1. But I want to think about it in terms of its vibration over r. So I want to just, again, I'm not going to prove that this is mirror, but I want to show where this ring might appear. And this ring appears when you think about just one Lagrangian, which will be our zero section. And we're going to take this Lagrangian, and we're going to equip it with a base point. And the base point is going to be called q. So this ring r, or r, yes? Yes. So q is a coordinate on t star s1. I mean, I'll still call it q. You can erase the coordinates. I think we won't use that. So the point is that if you take the base lip space, and you take the homology of the base lip space, then this is easily seen to be this ring. So if you go back to your topology, you will see that this is H0 of any space, of the base lip space of any space, is just going to be the group ring of the fundamental group. And in this case, the fundamental group is z, and just gives you this description as Laurent Paul notes. What is the product? Do you mean here or here?  You mean here. Well, it's a group ring. Ah, concatenation of loops. The fact that for some reason, I thought it was easier. I don't want to talk about algebraic, arithmetic, algebraic geometry, but here it's perfectly safe to talk about this over z. So if you really want to construct this, you just kind of work with c coefficients. But things are a little bit more delicate if you're working over the integers. You do algebraic geometry over the integers, and you're doing it over z. That's the only. So you could go up there and do things a little bit more carefully, and you'll just work over z as well. OK. So what I want to explain is something very, so every picture I will draw has to, well, you know, it'll be just this s1. But actually, almost everything I will say, it doesn't matter that it's an s1 or not. You could just, you know, the notation will just be q is some manifold. And if I want to stay the theorem later, I will be a little bit more precise about what kind of smooth manifold I want. OK. So what's our goal here? Well, I said I'm going, I should produce for you some coherent sheen. And right now, it's not going to be very clear what that has to do with flux maps, but we'll, that's for maybe for next time. So goal inside x, say, is an exact Lagrangian, l inside x. So for now, another Lagrangian, we want to produce a module. Let's call it, no, that's bad notation. Let's call it m sub l over the chains on the base loop space. And it will turn out that the reason I'm saying this is that, again, if let's just step back a moment, if this q happens to be a circle, OK, so we're in this one-dimensional situation, then what we're saying is we're producing some kind of coherent sheave on c star. And if this Lagrangian here happened to be a torus, then it would happen that we're constructing a coherent sheave on c star to the end. And it turns out, actually, that if this q is anything, then out of this procedure, you're constructing a coherent sheave on a torus of dimension equal to the first co-homology. Now, in that general situation, when q is this general situation, this coherent sheave doesn't know everything that it needs to know about the FLIR theory of l and q. But for tori, there is basically nothing more. And so that's why there is this kind of, here is a little bit of a lie. You do produce this coherent sheave, but it doesn't carry all the information you want, unless you assume you're Lagrangian or tori. So anyway, so let me just explain this story for now. OK, so how are we going to produce this coherent sheave? This is actually fairly old work, but the way you do so is we, as usual, so let's assume that l is transverse to q. And then let's consider, for each intersection point, this space omega qx on q, which is the space of paths from q to x. So this is our base point, and this is an intersection point of x and q. So I said I would produce for you some module over these chains on the base lip space. And all you do is you consider, first, you consider the direct sum over all x, which is an intersection point of these two guys, of these chains, of the chains on the space of paths. And here then you have to shift by something. And I'm going to just write the index of x. And if you don't know what that is, you can just ignore it, because then you can just work. I mean, you just work without any gradings. It doesn't really affect too much. And this is a module over the chains on the base lip space. How is that? If I give you a path from x to q, and I give you a loop from q to itself, then, of course, you can concatenate. As I said, it's a differential gradient. So what kind of chains are these? I mean, are these geometric chains? When I wrote this, I used cubical chains. Is that, I mean, it's not. I used it so that I wouldn't have to worry about. So this module is like, there's a map from this times this to this given by concatenation. And then you're like to say, this induces a map on chains. And if you use singular chains, you have to subdivide. And I didn't feel like reading about subdivision, so I just used cubical chains. But it works with singular chains just as well. I don't want to say the model structure, because if I take something in x, and I concatenate, that's something in q, we get something in x.  So you just product the thing when you concatenate? Yes, yes. Which is why cubes are good. Because their products are cubes. Yes. Any other questions? Except for, OK, you get one. But seriously, anybody else should feel free. The paths are inside q. The paths are inside q. This is what Nate corrected. I wrote here x, and I should never have written x. Earlier, your omega q sub x, q is the space of false in q. Yeah, in q, yeah. So we have a module, differential gradient module, over this. But actually, we don't yet have the differential. Because currently, all I use is the intersection points. I didn't use any FLIR theory. So to build in some kind of FLIR theory, you have to consider moduli spaces of critical intersection points. We have a moduli space of strips. And I will just write it instead of, I mean, I'm going to assume that people have some idea what this is. And if you don't, I will just draw this picture. Yes, I question that by the r-action. OK, so if you take this moduli space, then one thing you can do, once you split the r-action, is you can evaluate a long q to get a path. So we have a map from here called evaluation to the space of paths from y to x. OK, so this is whatever. You'll just have to let me just do x to y for now. No, the moduli space itself is a space. It has a map. And that map goes to this actual space of paths from x to y. And the way I order it, so now it's not clear. And my notation is not sufficiently precise for you to know whether it goes this way or that way. But it doesn't matter. It's only one way. It doesn't make sense. OK, so you get this path by evaluation. And now the main point is that we can pick fundamental chains for these moduli spaces. In fact, I should say relative fundamental chains. So with that, I mean, you just ensure that there are manifolds by standard transversality. And then, well, since they are manifolds, you can say triangulate them and then take a sum of all the top dimensional simplices or subdivide them into cubes because we're using cubes. And then take a sum of all top dimensional simplices. So I will write those as these. And this number here is basically the degree of x. Let me just like this. That's the number in which this class lives. So when the difference in degrees is 1, then this is a class in degrees 0. It's just an element. It's just counting. So each one of these moduli spaces gets the number 1 or maybe minus 1, depending on orientations, which we are not discussing. So now I have this evaluation map and I have this class. What I said is, here I said fundamental class. It's not a, I'm saying it's a manifold, every manifold is a fundamental class. If you manifold with boundary, it doesn't have a top class in homology. But it has a class in its chains, such that the image in relative homology generates the class of the manifold real boundary. That's basically what's happening here. That's why I call it relative fundamental class. But it's a chain. You have to pick this chain. Okay. So we have this thing and so we can evaluate it. I'm not gonna keep track of this anymore, except maybe it's supposed to be y, x, let's do x1. Okay, great. So I take some Lagrangian. I look at the intersections with q. I form this big thing, which is a module over the chains on the base loop space. And I also look at these modular spaces and now I get some classes in the chains of now the space of paths from the intersection points to each other, from x to y. Now, this space, we have maps. You know, what I've, we've used already is that we have a map omega q. Let me just now write, let me just write. Let me just do this, even though it's kind of silly, cross omega qx to omega qx. Omega qx. Okay, this is what used, this is what, this is concatenation. So this is a concatenate q. But you also have a map omega qx. There's q's missing everywhere. Now, it'll just happen all the time. Cross this, which is concatenated at the other end. And these operations essentially, I mean they commute because, you know, one of them happens, one of them, well, let me just draw. And then one of them happens here. And the other one happens on the other side. Okay, so you can, you can concatenate with this first or you can concatenate and then this or this and that, it's the same answer, it doesn't matter. Okay, it depends exactly. So if you are actually saying, well, you get some cube of simplicity chains here and then take the product with this cube and the product with this cube then you have to reorder the cubes. But this is not an essential phenomenon. Okay, so we have that and we have that. And this thing here induces a map on chains. So our fundamental class, or rather its image under evaluation defines a map from the chains on the pads from q to x to the chains on the pads from q to y. Ronald, what's going on that you're telling us how did your flow theory of a deferring of the width chains everywhere, is that? Well, you can't do flow theory. You have to, in flow theory you can only, you're only allowed to take homology at the end. Is that okay? I'm telling you how to do flow theory but you're only allowed to take homology at the end. So, okay, so defines this map. But now each one of these is part of my complex M of L. These are just direct sum of these groups. And now I have a map from one to the other. So these give a differential. And that is the construction of the module over the chains on the base, the space of q that is associated to an exact Lagrangian if you want an expression. Basically what we're just saying is you take the sum of multiplication with this thing. And so now there's just a basic exercise. So take L to be this Lagrangian. So I'm sorry, q to be this Lagrangian. And take L to be this one. And now this L has the property that this area here and this area are the same, okay? So when you go through this procedure, so compute this homology of this ML as a module over the homology of the base loop space of S1, which by the way we already decided was this. And the answer that you are trying to get to is the thing that we saw at the beginning, the thing that's over there, one minus z. It's the homology of that complex, it's the same thing. Okay, so so far what I've done is talked only about exact Lagrangians. And so you may want to do things with non-exact Lagrangians.  So surely we're gonna get out of the exact situation even if we're in a situation where it made sense to talk about exact things. Well, the first thing you know, if you've ever thought about Lagrangian Fleur theory in non-exact situation is we have to work over the Novikov field. Maybe there are some cases where you can avoid it. More generally, what is the answer to that question? If you take any zero section and take any Hamiltonian isotopic. You get the simple module over the homology of the base loop space. I mean, you have to say which simple module because in case h1 is non-zero, there is h, pi1 is non-zero, there is more than one. And then you get the thing who's the thing that corresponds to the augmentation ideal of the fundamental group. Okay, so let me say something about the Novikov ring because I wasn't here last week, but I'm pretty sure that nobody talked about it. Did someone talk about it? Okay, so it's best to fix a field K. And now we're gonna talk about the Novikov ring over K, Novikov field over K. So what you want to say, what we really want to do is we want to work over Laurent series over K. So something like K and then you do power series all over it. Let me just put a T here and then you invert T. That's what we want to do. But unfortunately, and this would work if every holomorphic disc we ever encountered had integral area, then we would just, the area would show up as the powers of T. But that's not what happens in FLIR theory. You can have things of arbitrary real area. So instead of doing this, we allow real exponents. So the Novikov field over K, now this subscript I will drop. So it consists of things like this. This is possibly infinite sum. These coefficients live in K, that's why we introduced that. These exponents are real numbers. That's why I was started with the integral case. And they satisfy one condition, which is the condition that is satisfied here anyway. You can write down infinitely many, you can have things here which are infinitely, have infinitely many terms, but only if the power of T goes to infinity. Similarly here, you can have things which have infinitely many terms, but only if the limits of I goes to infinity of lambda I equals plus infinity. So that's the ring that we have to work with. So, and now there is very naive answer, naive expectation, which would be what you would expect if you have thought about the Grunge and Fleur theory and are trying to extend it to the situation, is that we should replace, should work with the chains on the base lip space with coefficient in this. And sometimes it works very well. So let me, let me just draw. So first let me just say one word if you haven't seen this for what it means to instead we're supposed to work over this ring. What it means is over there, I wrote an expression for what the differential should look like. And now you do the same thing as above. The differential is the sum. And what I need to do is I need to break up this modular space into homotopy classes. So the disks may wind around the space in different ways. And as they wind around the space in different ways, they will pick up different areas. So this expression here is just the area of the corresponding disk. And then you multiply this by what we were doing before, except of course we break up and work with only things in class beta. So this is Fleur trajectories or whatever Fleur strips in class beta. This is the standard thing that you do in Fleur theory. When you are not in the exact situation, you still use basically the same modular spaces except you introduce a weight. Just think of this as a weight. Before it was like t equals one. Pardon? Before it was like t equals one. Before it was like t equals one, yeah. So let me just show you or I should never erase that. So again, just think about the simplest situation. It's interesting enough. So take, again, the cylinder. Again, this is Q. But now I want my other Lagrangian to not be symmetric about it. Or in other words, to not be Hamiltonian isotopic to it. This is now going to be some number A. And this is now supposed to be some number B. And what you're supposed to show again, don't know what my numbering scheme is up to, whatever, show that this module corresponds to the cohomology. So I take now lambda. I take Laurent series in lambda. Laurent polynomials in lambda. And then Laurent polynomials in lambda. And now my differential, it has to do something. It's t to the B minus t to the A times z. Okay? So you're supposed to think about this example and see that before we had one minus z because we just set everything equal to that, but now this is what you get. So, and that the cohomology is non-zero. Great. So it looks like we have our extension outside of the exact situation, but I lied to you. This expression is not finite. And in fact, the following, at least to me, amazing fact is true. So let's take this L and this Q, okay? And let's deform L and Q. Let's keep it Q fix. Let's deform L by Hamiltonian isotopes. I think this is four. So the module associated to L in this very special case is invariant under Hamiltonian isotopes if L intersect Q is non-zero, is non-empty. If you look at this picture, there is an easy Hamiltonian isotope which in fact makes L and Q disjoint. You can take L and move it by a Hamiltonian isotope and you will get, okay, maybe I pushed it too far, probably a little bit closer to Q in reality, but it's okay. And then of course, this complex is zero. There's nothing to do, there's no intersections. But up until that point, you can do all the deformations you want of L and you can, as long as there are intersection points, you can compute this complex. And the exercise is to show that they're all actually isomorphic as modules over the Laurent polynomials. Laurent polynomials. So this problem, it's not as, it is a problem, but it's, yes? What could happen if you started with something with more intersection points, well, four intersection points? Well, you would get, again, an isomorphic complex. Even if you eliminate the first two, as long as they intersect, it works. Yeah, okay, yes. So I don't understand why the problem is a problem since you're working with them but why we have high expressions. Well, that's one way to say this, but it's certainly not, by I mean it's not a finite expression, I mean it's not an element of this ring. It's not an element of this chain complex, okay? An element of this chain complex is you have chains with coefficients in this. Now, what you are telling me, and of course, this happens in some of the other theories as well, is that we should actually complete this ring, okay? Is this the next board? Yeah, this is probably the next board. So to get a theory that is behaved in general, in this case, I guess I said it works fine, no problem, but really what we should do when we write such expression or what we think we should do when we write such expressions that you get convergence, we should complete these chains with coefficients in lambda. So let's write that as C hat of, now we're gonna write the whole thing, Q. So what is this? This is gonna consist of expressions of the form sum of, just give me one second, think you can do that. So sum of A i T to the lambda times, sum of T to the lambda i times alpha i, where alpha i's are elements of the chains on the base of space again with coefficients in our ground field, just like before. Before we just put in elements of our ground field and T to the i lambda. Now I just replace these elements of the ground field by chains and lambda i are real and the limit as i goes to infinity of lambda i equals to plus infinity. Okay? So great. So this is the right thing we think and it makes that expression, it makes that differential, differential is well-defined. If we similarly, you have to do a similar thing, similarly use completion on the chains from Q to X and well, that's it. Yeah, that's the module, that's the module. Just to clarify the difference between that one and this one is that this one, the alpha i's not infinitely many different alpha i's but different T, well that one, you only have finitely many alpha i's. Thank you Dussel, yes. Here you can have an infinite series but you can only have finitely many infinite series in the sense that this part can be infinite but you have to take one of these infinite things and multiply it by one element of the chains and take finite sums of those and here you don't. Is this difference analogous or the same as the fact in number theory that we take etalcomology with CL coefficients and take first homology in Z-model and then take the limit instead of taking actually homology with... I don't know anything about etalcomology but I suspect the answer is yes. And I think I want you to... You should explain to me more after the talk. Okay, so we are happy. We think we're happy but because this kind of... The theorem basically says this module over this chains on the base loop space. Well, now I can say it in different ways but let me just do it like this is an invariant quasi-isomorphism type of this module is an invariant of the Hamiltonian isotopic class of L. This thing is... Now, what is it doing? So you may have seen other invariants like there is another invariant which you could have thought about more basic. I thought you just said the opposite. If you leave that to be as joint then it's like... I completed. When you didn't complete... And so what Claude is pointing out is that there is another exercise. Is that you're supposed to show that after completion... Yes. I was gonna come back to this but the group that we were playing with... Well, because that element is invariable. That's exactly why, yes. Okay, so Deuce has done the exercise. I'm not worried about my TA doing the exercise. Okay, so there are other more basic invariants. Can you say... This should be related to... In the mirror there should be sections of O2 or... Sections of O2. No, no, no, we're not doing anything that's sophisticated. One section of the zero section You are taking the zero section and you're restricting it to smaller subsets. You take somehow some smaller chart. That's more centered around that point. When you wrote it... Well, when you... So the... Sorry, let's do it the other way. Q, these lagranges that I'm drawing actually correspond to points. So we have some point and some other point that's some finite distance away. Yeah. Yeah, the wrong section. Yes, that's right. Okay, so, sorry, that... Is it very... Is it obvious that this is well defined? No. That I completed it or before completing it? Yes, it's obvious that it's well defined. Because then now you have an infinite sum of infinite sums. Yes. So you need some kind of...  So it's an element of that ring. One is that for each lambda i, there's a finite number of... The alpha i's have been less finite. Yeah, I understand that. But then when you put it in there, there can also be infinitely many betas that contain holomorphic disks. Yes, but the only way that the beta... So again, there are infinitely many betas, but only finitely many of them have... Yeah, but again, at least... And there are infinitely many alphas here. Yeah, but I mean the map is going to take from the completed thing, cross-completed thing to completed thing, right? So there are like two completions. Okay. We can talk about it. But basically I'm saying that the powers will add up and they will just... You don't have... You only have finitely many things in negative directions. And that's why this is okay. But the fact that the completion makes sense is it a purely algebraic fact or do I have to remember something that's in fact a geometry? No, the fact that this element lies in that completion, you have to remember something that's in fact a geometry. I wrote down a specific element. That's the fact that it lies in that completion is Gromov-Compacti stated. Sir, the homology of this module is again a module with a working chain complex, not the homology of the base loop space. So what... So let me say, what did I say here? Here I said you can take this module over the chains and it's quasi-isomorphism type. Over the chains is an invariant. Yes, but of course it can pass the homology which make your life easier. Then you don't have to say quasi-isomorphism. Okay, sorry, I kind of lost... Okay, there's more basic invariants which are basically the fluoric homology of L, of Q with L, okay? That's another more basic invariant. But there are also other more basic invariants which are you take Q and you equip it with some local system, like a representation of the fundamental group and then you take fluoric homology of L, okay? These are the ones that maybe are most familiar. I mean, I could put CF here instead of HF. And the point is that this group is integrating over all of them. It knows about all of them. E, unitary, I should put here unitary. So the above module knows about all of these. Every representation is a map from pi one. But I actually have the chains so that already has all of pi one. All the group ring of pi one is contained in my. Okay, so that sounds good. So now we have our invariant but it's kind of completely useless for the task of understanding what happens to fluoric homology in families because as we saw above, the fact that this is, as we saw above, as we saw in this exercise, the fact that this vanishes for A, B, if A is not equal to B, basically tells us that you can't use it to extract any information about what happens to Lagrangians when they vary non-trivially in their Hamiltonian isotopic laws. Maybe we can use it for other things but we can't use it for this task. But now, so we have this issue. So if we want to use the ring which actually detects and tells us something non-trivial even when the Lagrangians are not Hamiltonian isotopic, for example, in this case, that's the one we saw at the beginning, then we don't know that it's, that we in general even have a well-behaved, well-defined differential because again, the differential that we write down is infinite and the things that we allow are only finite. But if we use the one that always works, then we get a bunch of groups that are zero and we would like for them not to be zero. So can I ask, well, what do you expect such a thing to exist? It would be nice. No. So the answer is mirror symmetry but I can't somehow. This is not a lecture on mirror symmetry. You just, it's amazing that it's true. How about that? So now, I will explain why you can, how you can salvage this situation. So, so this is Foucaillat's fundamental. So I see this thing was exercise six or? Yes. Yeah, the fact that this group vanishes and before it wasn't zero. Yes. But the relation between exercise four, five and six. So is there at some moment in infinite series that appears, when does that happen? When I completed. So before I said, take the chains on the base loops with coefficients in lambda. And then I said, ah, well, this thing, as Duzer says, only has finitely many chains in it. It may have infinite, you know, the power series that show up and it may be infinite, but each chain only finitely many actual chains show up. One, like I said, is reversible in power series today. The explicit things you wrote down, exercise four and five, there were no infinite series there. There was no infinite series there. So as you vary, at which moment is there becoming infinite series that, I would say like there's this continuous deformation. Now nowhere, no, you don't, you vary that no infinite series will ever show up. But what happens is if you try to prove invariance, you can only prove invariance over the other one. I don't have a better answer to that. I should have a better answer to that question, but I don't have one. Okay, so, yes. It's, having shown you that that this thing is zero, there's still something there. You can still do something. Yes, yes, right? So I've shown you that you would like to work over this ring. And in that example, you get something very nice, which is you get this Lagrangian that knows about the other Lagrangians that are not actually Hamiltonian isotopic to it. It may know, it knows it in a very delicate way, but it still gets you and tells you information about them. But that construction doesn't make sense theoretically because in general, you have to write down, you have to work over the completion. When you work over the completion, then you get zero.  Yeah, which exercise was the nice thing, five? Four. Four, four is the nice thing. You could get this module, it's non-zero. In fact, you can recover the Flirco-Malgiv-L from that computation, from that module. It knows everything about the Flirco-Malgiv-L, OK? Later, OK, I have to finish my talk. So now I want to talk about Foucaille's fundamental theorem, our family Flirco-Malgiv.  He didn't say it like this, but it's somehow, it's what? So let's say that we have two Lagrangians which are transverse. Then there exists, so this is a very weak form, but I just, in this language, what I wanted to explain, a completion of the chains on the base loop space of q. So let's call this completion prime, which is intermediate between this one and this one. And both of these are strict containments. Well, this one doesn't have to be strict, but the other one has to be strict. So that, this is this prime thing, so that the module ML prime, which is again obtained by the same procedure we've been doing, except now you do it over this thing, is well-defined. That means roughly, so I don't have time to explain this. And maybe, of course, surely on Wednesday I will explain a version of this, and maybe slightly different language. But this is the main thing that makes family fluorochromology works. This one is, you can't work with in general, but it usually gives you nice answers and non-zero groups. This one you can always work with, but it gives you many zero groups. And this says you can make things slightly better by not completing all the way to here. This module is not necessarily zero. It means that that expression, this expression, sum of t to the pairing of beta with omega, fundamental class of this, gives me an element, again, analogously, in the completion of these spaces, x, y. I guess you want to look for some other property, which you need to look back for, c, x. Well, I can't assert that this will be non-zero because it can't, you know. So right now, it's just you can just, the assertion simply is there is a strict inclusion here. Can you assert it's non-zero and not example four? And which is non-zero in example four, but in example four it was this. In example four, we could have just let this prime be the, we didn't need to use any completion for things to be well-behaved. So the statement is, when you look at these expressions, it looks to you like you have to allow infinite series. But in fact, there is something that you can, you can always cut down your infinite series by a little bit. There's something that you don't, there's some further constraints on these series, which come in from Fleur theory, which means that you don't have to go, you don't always have to go to this thing that is very large. So what better property you get that you didn't, by taking one of the expressions? So that's something I will explain on Wednesday. But roughly speaking, this thing will allow you to, you know, if you now have, say, a foliation by Q, okay? By these, if this Q is just one member of a foliation, then you can try to patch these groups together. It's easier to patch these than it is to patch these. So that's maybe the one thing that, but I don't, I didn't want to. I didn't want to. Sir, the thing that you're going to produce was not the invariant. Under all Hamiltonian examples, it will not be. Will it be an invariant in a small one? Yes, in fact, the size of this completion is directly related to kind of how much hofer energy it's invariant under, okay? So anyway, so that's, I leave it at this very unsatisfactory stopping point and we will continue on Wednesday. Any further questions from Mohamed? Yeah, I still don't know what to do with my paper. Yes. If I don't see it, so I don't know what to do. Well, first of all, it could be Q. And it could also be, as long as, you know, oftentimes there's no reason to work out with complex numbers, I mean, or with characteristics zero. I can work. Sometimes it can be finite fields, yes. If we were working, real example, yes.  If you work on complex tori, for example, there's no reason to work in character six zero. With complex tori. You could do this in over the integers, basically. As I said, but all fields at once not very far from the integers.


Hey, what's up guys, welcome to our trending reviews. So I bought this projector from Amazon. It's the Vankyo Leisure 430W. W being the successor for the previous model which was the 430, meaning wireless. Now this is a really awesome compact projector which has screen mirroring capabilities through Wi-Fi. So let's go ahead, give it an unboxing and set it up on the wall behind me. I've cleared it off. I'm gonna give you a demo of setting this up using my iPad, screen mirroring it wirelessly. So that's a really cool feature to have. So let's get straight into it. All right, let's go ahead and open this up. Now I really like this carry case that it comes with. It's got a screen mirroring logo there on the front. It's very nice, it's easy to take review in this. So let's open this. So here we have a few accessories. So let me go and check this pouch. So you have some cleaning equipment for the actual bulb and the lens for the projector. You have a quick start guide. You've got the user manual as well. You have the power plug. You have an AV cable here as well with a little screwed kickstand as well, which I'm gonna set up shortly. You have a very nice and compact remote control with all of the standard functions on there. And they also provide a HDMI to HDMI cable as well to get yourself up and running straight away. There you go. You can see how compact it is. It's very lightweight. It's got a very nice design and it comes in white. So let me go ahead and showcase to you the different ports and buttons that are on this device. Okay, so starting off on the top, you have the power button, the menu button, the source button, the back button, and an okay button as well for your selection. And this metal circle ring is the left and right, up and down directional arrow buttons. You have two rings here on the front. The one at the front is the keystone one and the one at the back is the focus. You have a lens cover there to protect the bulb as well. On the left-hand side, you have a few ports. You have the headphone port for audio out. You have the AV input there as well for audio and video there. You also have a TF card slot if you wanted to play anything back through a SD card. You have a traditional USB port, a HDMI port and a VGA port. So plenty of options to cover different things for connecting this projector to. It could be your Sky TV box. It could be a PlayStation or your Xbox and various other things like your laptops, your computer. But most importantly, you can screen mirror wirelessly as well. Or if you wanted to connect your phone, you can do that via the USB cable using your phone's original cable. On the right-hand side, you have the power socket there. On the back, you have the infrared receiver. On the bottom, you have the vent for the cooling system and then you have the speaker there as well. This is the port for the kickstand which I'm going to be setting up shortly. So that's the entire projector. Let me go ahead and take off this protective covering there. So let's quickly run through some of the key specifications and then let's set this up. Okay guys, so the one thing I really like about this is that the bulb is actually 5,800 lumens and if you do a search for an Amazon UK for all projectors less than 100 pounds, which this one is, then this actually has the highest number of lumens which is the level of brightness for projectors on any of the projectors under 100 pounds. So that's really awesome. The actual lamp life gives you more than 50,000 hours of life there and the projection size of the actual image goes up to 40 inches if you're 3.9 feet away and a massive 236 inches of display if you set it up 23.6 feet away as well. Now, natively, this projector is 720p out of the box but it does support upscaling of 1080p although it's not natively HD 1080p resolution. And the best thing about it is of course, as I mentioned is the wireless screen mirroring that's inbuilt into it using the WiFi capabilities but of course you can set it up with a wire as well. So as I briefly showed you before this was the kickstand screw that came with it. I'm just going to connect that at the bottom as well to give it a bit of elevation. So I'm just gonna leave it about that height there and then have it point upwards and of course you can adjust this. So now that's ready. Let's go ahead, plug it in and set it up in the wall behind this and then let's see how it displays. Okay guys, so I've just connected it. Now it's displaying on the wall. I'm gonna be changing the source so I have to remote control. So you have a few options here. I'm gonna go straight into screen mirroring and set up my iPad. All right, so as you can see it's told me to connect to the WiFi. So all I need to do is go into the WiFi in my iPad. It's connected. Now what I'm gonna do is I'm going to select screen mirroring there as you can see and it pops up with Leisure 430W. So I'm just going to select this and you can see on the projection behind me it's now connected. So you can see it does both portrait modes and landscape modes as well. And just to give you an idea of how quick the responses are I'll just show my iPad from this distance.  Not too much lag, but I think it's perfect for that. If I'm just swiping you can see it's pretty good. Now let me go ahead and play a video streaming wirelessly. Of course this should be a little bit better when you play with a wired cable but nonetheless let's try it wirelessly and see how it comes out. Now it's not too bad, pretty smooth playback. It would be a lot smoother if I connected it directly with the USB cable but this is just streaming wirelessly and I just wanted to see how smooth it is and I'm pretty impressed with that. All right guys, so I just connected the USB stick into the projector and I'm going to be playing a video clip directly from that and the audio from this clip will be coming directly from the projector so you guys can hear to see how that sounds. All right guys, I hope you found that review very useful. As you can see it's a pretty bright picture and I'm really liking this projector. You can pretty much connect it to anything you want. If you are going to watch a movie or play some games I definitely recommend you connect it through a wire and not through the wireless capability because you may get some lag and some buffer but other than that it's very convenient for the price and the size of this projector. I think it's an awesome buy. It comes in at 90 pounds on Amazon which is a steal and I'll have a link in the description below where you can buy that. 5800 lumens it's very bright for this price point as well so that's one other plus point I really like about this projector. Now I think out of everything the most standout feature is the screen mirroring so if you want to quickly put on your phone or your iPad and you want to show your family or your friends some pictures I think it's great for that. If you just want to browse some general websites and showcase things to people or if you're running some presentations in the office and you really want to connect your PowerPoint slides or anything like that I think it's great for that as well so multiple uses you'll get out of this projector. Now it does have a dual cooling system inside there as well so it doesn't allow it to overheat but it is recommended that if you do play continuous playback of around three to four hours then do give it a little bit of a break before you carry on playing it. So just make sure if you wanted to extend this working life then to always give that break in a gap between your playback. Now it does have a low fan noise suppression system in built as well so it's not too noisy but if you are in a very quiet room and you're maybe by yourself and you don't have speakers connected to this then you can hear the fan a little bit but generally from a lot of the projectors I've used in the past it is pretty low noise so that's a really good plus and if you are going to use this to watch movies and set up a home cinema system I do recommend setting up some speakers with the projector as well so you get the audio coming out from somewhere else and it's a little bit more louder and punchier. The audio from the actual projector I thought was quite punchy and it is loud for what the size of the projector is. I did go up to 100 volume which was the max and it was pretty loud and I was pretty impressed with it. So that's a real good plus point. Now if there's anything else you'd like to know about this projector then do drop a comment below. Otherwise do check it out on Amazon I've got the link down below it's 90 pounds it's an absolute bargain and if you did enjoy this video I hope you give this a big thumbs up and I hope you subscribe and I'll catch you guys next time. Take care.


Hey, what's going on guys? Patriot Prime here once again with another episode of the Sitrep. And my guest tonight is Dakota from Zero One Transfan. Dakota, how are you doing tonight? I'm doing absolutely fantastic. Great, great. Well first off, tell my people a little bit about yourself. No problem. I am a YouTuber from Nebraska. I do some small time stuff, some stop motions, a little bit of skit stuff here and there. People seem to be liking some of the stuff I do now, which is great. I'm very lucky about that. Yeah, just recently hit 500 subscribers. So I'm very grateful. Congratulations. Well, how the Sitrep works. And I know you've seen a couple episodes. And I know everyone at home watching kind of knows the deal. But I am going to ask you five questions regarding you as a Transformers collector and five questions regarding you as a YouTuber. So if you're ready to rock and roll, we'll do this thing. Let's do it. All right, buddy, starting with the Transformers questions. First, number one, what is the appeal of Transformers for you? Well, I've always been really into robots, even before getting into Transformers itself in 2007 with the first movie. I've always kind of had this thing for robotics and cool looking vehicles, just kind of add onto it as well as the feeling of I was the one who got to go from point A to B. So there's that sense of accomplishment for completing that challenge. Excellent. All right, sir. Well, I think you just kind of answered this question. So your introduction to the Transformers was 2007? It was, but that's not my favorite Transformers fiction. Excellent. So question number two, aside from G1, which is always the right answer, what is your favorite Transformer fiction? Well, I do love me some G1. I've been watching it quite a bit lately. But for a while, it was Transformers Cybertron because I was lucky enough to pick up the full series box set. But recently, I reevaluated everything. And for me, I think Transformers Animated has got to be my number one. Nice. I've had a few guests kind of dip their toes in that one. But I haven't heard anybody straight out say that's their favorite animated series. Now, I got into it when it was on TV. I mean, it was very kitty-fied at first. And I hated the animation. But I really liked it as it slowly began to pick up. And then, of course, when it got really good is when it canceled. But isn't that how it is with all of our favorite TV shows? That's precisely the point is, as soon as you're hooked, and you're like, OK, I can't wait for more. There's no more. Dang. That's what pisses me off right now. I watch a lot of Netflix and Hulu. And I'll get on these series. And like, oh, it's so good.  You're not going to see anymore? Well, now I need something else. Yeah, I'm starting watching one now. I cannot quit watching. It's called The Hundred. Have you heard of that one? I've heard of it. I haven't seen anything of it myself. Man, that's good. I've seen it over and over again finally the other night. Just decided, oh, let's get it started. Now I'm doing three or four episodes a night. I'm an addict. That happens. All right, sir. Question number three. Do you have a favorite figure or grail? This is your figure you will never let go of. You have to sell your entire collection. This one's not going anywhere. Oh, so everything else. I think I have a few that are kind of in that category, but one that I'll just name. I just got it recently. Friend of mine just sent me the Year of the Horse MP10. Oh, move him over just a little bit more. He's out of frame. Oh, nice. How is that thing? Well, even with the gold and the deeper reds and blues, I still am deeply in love with him and can't wait to do my own review on him.  I mean, he's got me in the ring. And so I think that she'll be OK with it. OK.  Yeah, I've got MP10 that was wrapped up in my arm and cats just walking with us like, oh, my God. I just remembered this YouTube video called Me and Optimus Prime. Have you ever seen that one? Oh, God, I don't think that. This guy's like on fishing trips with Optimus Prime. And it's just he's hanging. I'm going to send it to you on Facebook. That just popped in my head. Now it's going to haunt me for the rest of the night. All right, question number four. What figure is your white whale, your most wanted toy? This is the one if you could get ahold of, you're done collecting. Honestly, for a while there, it really was the MP10. But I don't think I can just stop collecting even after picking up a grail because I have the way I wanted to do my collection was classics, generations-oriented. But I think maybe just out of that line right now, I would probably say the classics Optimus Prime because I used to have a version of it long ago, but have since had to get rid of it. I'm trying to remember if I have that one or not. People say that the siege toy is better than it, but I'm probably mostly because of the arm flaps. Oh yeah, he's up around the top shelf, I see him now. I've never really had a problem with the arm flaps, that just something about him just makes me happy.  He's a cool figure. That's something I like too. I love transformable weapons that they can actually store on the figure. Yeah, or at least something that has like decent storage underneath the figure instead of just going on a peg hole on the outside. The obligatory cannon attaching to the top. All right my man, last Transformers question. What is your favorite Transformer character? Any fiction, any media? Honestly, it really does have to be, this is gonna make me sound like a broken record. It's gotta be Optimus Prime just because I love that he goes from an archivist on Cybertron as a Ryan and then puts in his amount of work to be worthy of the title of Prime and earn the matrix of leadership. To be honest, that type of characteristic is something that I would strive for when I was younger and something that honestly could be looked up to.  It's kind of making me feel high and mighty. Well, I mean, you are like up here in my corner. That's just great.  Well, I mean, you'd soared past me and you've been on YouTube for less time than I have. I'm very happy for you. I would, I'd be lying if I said I wasn't a little jealous though. Well, thank, it shocked me. I really honestly didn't think I would do what I'm doing. So I don't know what my secret was, but I mean, the last few months just, bam, shot up there and I'm tickled to death. I've had quite a few people say, hey man, what's your secret? And I'm like, I honestly don't know. Luck. I'm very happy for you. You definitely deserve it. Well, thank you. Thank you very much. I'm digging your new stop motion stuff.  That was funny. That was like me talking to kids at the park. Anyway, well, that is the end of the transformer questions. And now we're going to get into know you as a YouTuber.  Probably just the fact that I never really grew up and I just like making video.  It is fun. It's definitely, I've discovered it. It's like another hobby into itself. I mean, there's a lot of work involved, especially like with you in the stop motion. I dabbled in it. I think the most stop motion I do is I have a head turn. And I know that's so tedious with a little frame by frame. So kudos to you guys who do that. I've debated on it, but I just don't have the time.  Your son is, Kat loves your son. She's loved watching videos that you do with your son. Oh, he is a hoot. He keeps wanting to sit down and do a review on the rescue bot that he got the other day.  He's got the attention span of a goldfish. All right. But he is a hoot. I was so lucky when he came in my office the other day when I was getting ready to do that unboxing because I had no idea that he had stuff in there. So it just worked out perfect. Yeah, I saw that. That was really cool of that guy that sent him stuff. Oh, it's awesome. Very generous, very generous. We've got a good community. You know, we got the shitheads, but mostly it's a good community. It really is a nice community. Just like there are obviously some exceptions that that comes with every type of community, every band-aid, so rather than keep our head above water. We're at least not as bad as the Star Wars and Marvel communities. That's for sure. All right. All right. Number seven, what keeps inspiring you to get in front of the camera?  But it's honestly the people who watch, like when you left a comment on one of my Battlegrounds videos saying that it was really good, I was like, wow, that hit me there. And it's like, I gotta, I really gotta keep going now. But it's- And it's hard to do. I try my best to answer every comment and comment on every video that I watch because I know I love getting the comments. Maybe it's because I've been on Facebook for so many years. You know, I know YouTube's all about the views, not as much the comments, but I love people commenting on my videos. Granted, I get some bad ones because of this and the whole patriotic thing that I do. But hell, that's what the band hammer's for. Or getting mistaken for a site that sells things and then just steals money. Dude, that hit me because the last question I got was this, or what the question was a message. I got Sunday morning. And that's when that woman goes, your company is taking $20 out of my account every month. I needed to stop. And I'm like, what? Then I actually got her on the message and we went back and forth and I said, and I told her flat out, I said, have you even looked at my Facebook page, the page that you're commenting on? It's all transformer toys. I am not who you think I am. And a little bit of research, we figured out who it was, especially when I combined it with the guy who asked me if I was from Woodlands, Texas, but that was nuts. I mean, she was ready to file a suit. Yeah, the way that you described it in that video, just like, dang, she seemed like she was ready to go off. Yeah, it was nuts. I mean, that's why I screenshot the actual conversations. You know, just to show I wasn't making that up. I mean, that floored me. So lady, I'm a 45 year old man talking about toys on YouTube. I am not taking your money. Just like what company?  All right. Yeah, we got sidetracked there. Question. Hey, that's what these do, man. These things, who knows what's gonna happen. You saw the one with Tony, right? From Toast and Toys? Yeah, yeah, that one went off the rails quickly. That was so much fun. All right, where are we at? Question number eight. There are thousands of toy reviewers on YouTube. What makes you any different from the rest? What makes you the guy that's gonna get 501 subscribers after this is over? Nothing at all. Mr. Gloria's hair. Yeah, yeah. As one of my friends put it years ago, my majestic lion's mane. I honestly, just with my reviews, I like to just make it funny and give my honest opinion and hopefully give a different point of view for figures that people don't even like. Like, I did a review a few months back on the universe ironhide and ratchet figure. Two of the most hated figures of the universe classics lion. I still really enjoy them. Like, yeah, they're kind of flabby and have some issues, but it's like, I found reasons to buy them. I found reasons to like them. So I figured I would tell people about it and hopefully people would see it, at least see where I'm coming from. That's kind of how I started with my G1 reviews is the ones I started reviewing first and actually still are, is the line from the late 80s because most everybody in the G1 days, they were done with transformers by the time these oddball figures came out. So those are the ones I kind of focused on and I've had a lot of people say, thanks for going over that. I never knew that thing even existed. So it's nice to go over the stuff that people can gain a new appreciation for now.  They just weren't what you expected with the transformers. But now as an adult collector, I kind of understand the science fiction behind them. I mean, think about, they're terminators in a way. They really are, if you think about it. Of course, I like the Decepticon ones a lot more in the Autobots.  Just... Yeah, the Decepticons really had some really outlandish designs. Even some nowadays, like there were some from the, that Rolex in the sky's line from a few years ago that were really outlandish and... All the animal-based ones, like the Wolf and the... Yeah, like Bisk and Thunder Huff and all that. The only character I liked was Thunder Huff. I thought he was pretty cool with... He looked like a Viking with the big horns. But then when I realized he transformed into a tractor, I'm like, hell yeah, he's from West Virginia. Well, with that show, I was never able to get into it, especially I couldn't get past episode one, but it's like some of the toys though, however. Some of the toys are really good. I only had a few. It was at a time when there was nothing new on the shell. So I was buying them just, you know, when you just have that itch, you have to scratch. It was like, I need figures and this is all that's here. Fine, I'll take it. This thing is goofy as shit, but I'm gonna buy it anyway. And it's all on eBay. If nothing else Dylan, can I at least get a couple last episodes? He's got quite a few. Those are Dollar General, Dollar Tree Transformers I reviewed. Those are in his collection, I think I'll bust it to help. Oh yeah, I think I saw those at our Dollar Tree out this way. God, this thing is... I saw somebody had posted on one of our transformer groups that there was a new batch out, and I actually went to our Dollar Tree yesterday, but it was still the same old ones I had from before.  That'd be really cool. All right. Number nine. Oh, we kind of touched on this a while ago. What is your thoughts on the transformer social community? IE, YouTube, Facebook, et cetera. And do you have any special shout outs? I, like I said, a lot of the community is gray. There are obviously some people that's like, okay, well maybe there's not really a point in talking to you. Cause like some people just have some really strong opinions. Like for example, I'm not a huge fan of the last night or age of extinction. I know that there are definitely some fans out there. That makes you an exceptional person. Normal. It just didn't do it for me. But like I know that there are some people that even I'm friends with that would like a continuation for the last night. Like there was a rumor for that guy shut down, but My biggest issue with that, not mean to jump in, but what I'm thinking about right now is everybody, I mean, I've seen the complaints too. They want the continuation. They want a closure to the original series. But none of those, there was never any cohesion, cohesion, you know how the word I'm talking to say, they never really lined up, except maybe the first three loosely. I mean, each one it's like episode one. Oh, we're after this cube that's going to save the universe. Episode two. Oh no, it's the sun needle thing. Oh, nevermind. It's the seeds of Cybertron. And then you go age of extinction where they forgot about everything else. And then the last night, Jesus Christ. That, that one just like, whew. That was probably the first Transformers movie I just turned off. I think I watched 40 some minutes of it and then I went back, it took me like three days to watch it. I just could not get invested in that movie. It was all over the place. Granted it had some good toys. The Megatron figure that came out, the Nitro Zeus figure that came out. Oh, I love that. Good toys, shitty movie. Especially with like, not only was the story bad, and like this has kind of been a joke that's been going through it as well. The editing in that movie with like the aspect, aspect ratio, it was, it's worse than all the other ones. And even when I noticed it in the first, in Revenge of the Fallen, where it's like, oh, like there's some black bars I keep like going away. That's weird. It was so horrendous. And then the last night, it just bothered me. All right, we got sidetracked again. Yeah. All right, social community, YouTube, Facebook, any special shout outs, part two, go. There are some people in the community where it's like, you know, you have your opinion on mine that we'll just leave it at that. There are lots of parts where you can actually talk with people and get different sides of things. And I really like that. I can't really think of many shout outs. I, you know, there's a group that I'm in, the Transformers Insane Asylum with T-Fan page one-to-one lyrics production and all of them. I love doing stuff with them and it's nice to get more people to make jokes with and talk about things, even if we disagree. Nice. Is that a Facebook group? I've never heard of them. The Insane Asylum, it's a YouTube podcast. We recently, not recently, we haven't done a live show in a while, but there's like a hundred or so videos on it. We're hoping to go live again sometime soon. Oh, nice, I might have to look it up. I'll share with you that Optimus Prime musical and you can share that with me. No problem. I do have some animated highlights on my channel from those podcasts. So it's mostly just like the funniest bits that we have. So if it interests you, you can check out the whole podcast. We mostly just talk about like news and stuff, stuff that's been released to the media. 10-4, I will, I will. All right, last question, brother. Do you have any channel goals for yourself and where can our viewers find you? This is your chance to plug your channel. Well, I'm hoping to at least get a video or two out on a third stop motion series that I'm doing called Transformers Resolve. I've did a live stream about it once Battlegrounds ended. It's going to round out this trilogy I have going on and I'm running it with more of my friends. So hopefully we can get a video out this year of some kind for it. But other than that, just I want to keep making the silly videos I do, all the fun stop motions, all the live streams with my friends. I want to keep engaging in the community with you and Mateo Potato and all those other people and all that because it really- I think you were one of the very first people I met via as a YouTuber on one of our Facebook pages. When I first got started- It seems about right, because I've been following you at least since you did your MicroMaster Month last year. Yeah. God, you're the second person to bring it up. I will never do anything like that again. Well, I mean, I remember once you said it was over, it's just like, June's going to be jump stars month. It's just like, okay, let's have a jump start. Just two figures, that's it, that's all I'm going to review. Yeah, that was insane. I was rushing home, just trying to get these things.  And you know how I like to do the, let's look at me, then look at the figure, then look at me. So that was so hard to do. That's a long day being involved in that, especially with how you deal with your transitions and all that. I love the editing. I mean, a 10 minute video can take me two or three hours because I'll just come up with stuff. And of course, you know, I like to put the images and video clips and all that. And I started something early on that I just can't stop now. When you've got something funny in your mind, it's like, oh, I should put up a graphic to show what I mean if people don't get it. It's like, that can be pretty fun, especially when you think about it yourself, just like, that's really funny. Oh, it's like the, my pride and joy, as far as comedy is concerned, my Naga Hat video. When that hit me, I wrote it down and I went to, I mean, it took me a week to plan out exactly how to do that. Oh, God. Cause I had to have that Sarah McLaughlin style sad music playing. Oh, yeah. Well, Dakota, that is it for all my questions on the SIT rep. I thank you very much for taking the time and wanting to speak with me. And for everybody watching, I will put Dakota's channel in the description of this video and I'll link his channel and one of his videos at the end of this as well. So once again, brother, I appreciate you coming on. I appreciate you having me. I'm very thankful that I got to be on this because I don't know, I just like talking with you. Good, I think this is one, this, I'm getting all tongue down here. This has been one of my favorite things too, as I've known you guys for over a year, but a lot of times it's just commenting back and forth on Facebook, getting to see you guys and talk to you face to face has been awesome. It was really fun for me as well. Good deal. Well, guys, if you like what you see, don't forget to like, don't forget to subscribe. I try to put out a SIT rep episode at least every Monday and Friday. Also look for other reviews of mine where I go over generation one transformers and modern figures. This is Patriot Prime, signing out. Hooah! 🎵 Through me and often it's right 🎵 🎵 Yeah, such a real good time 🎵