Categories, Sites, Stacks & Fields - lecture 3 of 3

0:00 We can take F as a beam, but it's nothing. And then we keep the pre-beam. U gives Ohm, in the pre-beam category, over U, F will be U, G... I've already defined all this, but maybe you forgot and it's not clear what it means. So, what I call the restriction. Remember, you have... A morphism of X in U that I call J-indice U and F will remain U, C, J, U, star, F. If you don't like that, you can take the other morphism, I, U, and F will remain U, it will be I, U, die. While here, we don't make hypotheses, so that's good. All this I have already explained to you. So, nothing prevents you from associating the set of the morphisms of S to U in G to U. This is a set. And on the other hand, of course, if you have V which is sent in U, you have a set of restrictions. You have a preface that notes man, round, two F, danger, and it's a preface of the whole.

2:30 It's not a preface of the whole. We lost the structure of the Casimiro. If A is the Abelian groups, then indeed it will be of value in the Abelian groups. It's exceptional in general. A is a category, you do the morphism, you lose the structure. So it's a preface of the whole. And the theorem that I gave in its demonstration, by the way there was a demonstration in the previous notes, I told you that it was probably totally wrong, and in any case I made another one. So the theorem is that to be a pre-vessel, we feel that we are a vessel. If I have a vessel, well, the round shape is a vessel. This is what I wanted to say, but I didn't want to show it. So, I'm going to show it to you today. In addition, in the notes, I put the last one, the last demonstration. I'm not going to repeat it. It's up there. So, as soon as you remember the other one, you're going to forget it because we talked about the other one. So, I remind you that if you have H, a real vessel, you have to make a vessel. If for all U in CX, for all U coverings, H2U in H2S is an iso, and I also remind you that S, we will take it, it costs nothing, it changes nothing, we will take it stable by intersection, that is to say by fibrous product, you understand, that is to say, if you have V1 in U and V2...

5:00 So, V1, V2, V1, V2, V1, V1, V2, V1, V1, V2, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, V1, I'm going to show you that H is a preface of an ensemble, so I remind you that H is a preface of an ensemble. This is something that verifies two axioms, S1 is for all U, for all S1, S2 in H of U, for all coverage of U, if S1 will be V equal to S2, will be V whatever V in S, then S1 equals S2, that is to say that the preface is separated. And to say that we have a sphere, it's the same thing as saying that there is this thing there, but you are not yet used to seeing that. S2 means that whatever the U-recovering, whatever the SV family indexed by the recoverings, with SV restricted to V inter W equals SW.

7:30 So, there is S in H of U, S2, S minus U equals S minus V equals V, which is V, minus S. So, S1 indicates that the arrow is there, and so S1 indicates that the arrow is objective, and S2 indicates that the arrow is surjective. So we will see in the case of... because in the end I missed this theorem, it is absolutely fundamental, especially when we do the tasks, as we will do soon. So it looks difficult, that's why I didn't do it, it's just a demonstration, quite simple, I think. So that's why I called this pretext H. So we will verify that you put that first. And then I take, I'm going to take my notations, they're not the same, so it doesn't matter, I take Φ and Ψ, which both go from F-U to G-U, and then I suppose, I call Φ-V, Φ-V, Φ-V, Φ-V, and we suppose that Φ-V equals Φ-V, and that V is in the covering.

10:00 I'm going to demonstrate that. I'm going to demonstrate that if two sections of this preface coincide on the openings of the cover, they coincide everywhere. So, these sections here, phi, they also define, I'm going to call phi s, it goes from f of s, no, g of s, I'm amortizing from f of g. A pre-fossil morphine. S is a functor, we've already seen that. So, it goes from F to S. F to S to S to S. Okay? Because F to S, I remind you, is the nucleus of the product for V in S of F to V, the product for V'. I don't write the projective limits for those of you who know it. So, F to V' is 3 seconds. Okay? And same for G. So, if I have a morphism for all Vs, if I have a morphism for all Vs, if I have a morphism for all Vs, if I have a morphism for all Vs, if I have a morphism for all Vs, if I have a morphism for all Vs, if I have a morphism for all Vs, if I have a morphism for all Vs, if I have a morphism for all Vs, I wrote, I showed that phi equals phi. So, I looked at u, I showed that phi of u equals phi of u. And then, even if you replace u with a smaller open, it will show that phi equals phi. When I have a pre-made morphism, I will show that on the... phi of u is the morphism, it's the arrows, df of u, no g of u. So, I send them like this, I send phi of u, f of u in f of s.

12:30 Here I put two arrows in g of s, and here I have the arrow that goes in the opposite direction in g of u. And phi and psi, but that's an isomorphism. Phi of u is the arrow f of u in f of s, that's the arrow phi of s, and psi of u is obtained... If these two arrows are equal, then the two arrows are equal. It's quite elegant, isn't it? Because in the demonstrations that I wrote before, with the indexes and all that, it's really not possible. And we can see that we don't need F to be a beam, F is a pre-beam. F of U is sent into F of F, and it's not over, I just demonstrated that the axiom is there. F of U is sent into F of S, but G must be a beam to be able to invert the arrow that was there. So, axiom 2 is always a little more difficult. Existence is always more difficult than unicity. Because the quotients are more difficult than the sub-spaces. With the theory of ensembles, it's not possible. With the theory of opposites, it's not possible. If you don't want to get confused, just believe me. So, existence. Let's take a family of phi and isv. In the case of V, we assume that phi v will remain v cross u w equals phi w will remain u cross v w, v cross u w for all v w. So I say that this condition says exactly that the family of phi v defines a morphism of the projection limit in the projection limit, phi s.

15:00 In fact, there is phi s, gf2s, in g2s. This is a double arrow, the product of the prime of the second is 13, the product of the prime of the second is 3, the product of the prime of the second is g, the product of the prime of the second is 3, the product of the prime of the second is g, the product of the prime of the second is g, the product of the prime of the second is g, the product of the prime of the second is g, the product of the prime of the second is g, the product of the prime of the second is g, the product of the prime of the second is g, the product of the prime of the second is g, the product of the prime of the second is g, the product of the prime of the second is g, the product of the prime of the second is g, the product of the second is g, the product of the second is g, the product of the second is g, the product of the second is g, the product of the second is g, the product of the second is g, the product of the So, if the diagram changes, this is the nucleus of these two arrows, this is the nucleus of these two arrows, and it means that there is an arrow here called Φs, which is a little bit faster. So, this condition, the hypothesis that is here, means that the family of phialized morphisms is attached to a morphism Φs of Fs in Gs. So, phi of u of f of u in g of u, which we will define as we did earlier, we take f of u, we put it in f of s, then we have phi s which goes in g of s, and g of s isomorph to g of u. So I defined it like this, phi of s. Phi u means morphine on the global sections. Phi of s means morphine on the global sections. And phi index u is a preface morphine. phi index u is a morphine in category A, and phi index u is a preface morphine.

17:30 It's not the same thing as phi index v. It's the same way that f is a preface of v. It's not the same thing as f of v. One has a preface and the other is an object of category A. So we have constructed phi of u. If I change U by a smaller number, I would have built a daughter of U', and clearly all this is recombined. So I built a morphism of prefixes of F in G. So the prefix of FG, if I assume that G is a beam, it is a beam. So that's very important. So there is a corollary that is even more important. I think you can skip that. I use them all the time and I realize that I still use them, at least for the other years. I may not have used them yet this year, or maybe two times without the explicit use. It's a very important corollary. In fact, very important in the small highlights of what we do afterwards. It's not important in itself, but we are interested in it. So the corollary is as follows. You take a piece of beam. Because here, F and G are beams. So, if Φ is an isomorphism of the case, then it is an isomorphism. So, what does an isomorphism of the case mean? It's a bit... it's a sentence, it's not very precise. So, to say that it is an isomorphism of the case means the following thing. Let's suppose that there is a cover-up S, a cover-up of X, whatever V in S is. Phi is an iso, so phi will be a V, so F will be a V, so G will be a V is an iso, so phi is an iso.

20:00 So, as we will see again, this is what we did last week, last week I showed you, I gave you examples of these beams, the isomorph calvary. They were not isomorph calvary. You will say it contradicts that. No, it contradicts nothing at all, because here I have a morphism that is globally defined. I'm not saying that two locally isomorphic beams are globally isomorphic. F and G can be locally isomorphic, and our palettes... It's just the subtlety of this F2 course. The subtlety of the F2 course is less noticeable. The difference is that here I have a morphism that is globally defined. So there is locally an isomorphism, whereas here I simply have locally isomorphisms, but no one says that these local isomorphisms come from a global morphism, and it's going to get complicated when we do the fields, in fact. It's the same kind of a grand plus. So it's very important to understand that. F in G is an isomorphism, which means that there is an inverse. So I called Psi V its inverse.

22:30 Now I'm going to show that Psi V sticks together. I say that Psi V is W. V will remain a V cross if we have to write a W, I could put the intersection, that's the same thing, it's equal as morphism. Don't write isomorphs, it doesn't make sense. Now we are in morphisms. The set of morphisms is a set. So two things are equal or are not equal, there is no isomorphism. We are not in a two-category, it will come. Not today. So V equals Psi V cross U W equals Psi. W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W-W And here, it's a morphine of G in F, so we use well that F is a test. As a morphine of G in F forms a test, it exists if it goes from G to F. Such that Psi restricted to V equals Psi V, whether it is G in F. I say that Psi times Psi equals the identity of G and Psi times Psi equals the identity of F. Why? Well, because it's true of Kalman, so it's true of Le Bonheur.

25:00 For example, for this one. There are two morphisms, two vessels, from F to S. They coincide on V, so they coincide everywhere because man is a vessel. And it's the same. If two sections coincide of Kalman, they coincide of Le Bonheur. You may have seen it in the notes. In fact, if Rho phi... It is said that ΨΡΦ equals the identity of S. Why? Because whatever V, these two parts ΨΡΦ are equal to V, and ΨΡΦ is the identity of S is equal to V. So here are two parts of a beam that coincide on a covering, so as man is a separate pre-beam, they do not coincide. You see, we first used axiom s2 to construct psi, and then we used axiom s1 to say that psi is an element. So these are things that I hadn't done before. So now I'm coming back to the collection of pixels. So that was one. So what I'm doing now, since the course is called another field, I'm showing you... Speakers include the fact that the scale of the beams forms a field. I didn't say that it's a field. And so this theorem here, it translates by saying that the pre-field of the beams is separated. We'll see that later. And theorem 2, on the collation of beams, says that the pre-field is a field. This theorem 2, which I've done, I've already done, but it doesn't do anything. Yes, for questions of notation, so I take S, a recovery of X, but for questions of notation, because we make intersections, it gets a little heavy, so the notation, I remind you, I can change the notation, I call S a family of the index by I.

27:30 So the theorem that I mentioned last time, but I didn't really give a demonstration, is that we suppose to give the fissures Fi over I and the isomorphisms Theta ii of Fi over Uij, isomorph fj over Uij for all i and j, Theta ij composed with theta jk becomes theta ijk over ujk for all ijk. So, let's conclude. So, there is F, which is a family of theta i. F is a beam over x. Theta i is an isomorphism of F times a ui over fi, You will notice that the condition that is there, we call it the Caussic condition, it is necessary because if you want to have θj equals θj times θi minus 1, when you compose, you write θij,

30:00 Theta ij rho theta jk. Let's suppose the problem is solved. What does it do? Theta k minus 1, theta j, then theta j minus 1, theta i. So it does theta i, theta k minus 1, theta. So you see that if we are looking for F that verifies this condition, it is necessary to have this condition of Caussi. So let's demonstrate, or in part, I do not want to demonstrate it completely. Perhaps I will start with the unicity, which is a little bit the same thing that I was doing yesterday. So, ah yes, I did not say, I forgot to say it. And unique, a unique isomorphism, Re.

32:30 So, let's show the initiations. I shouldn't have said to erase my theorem. So, where does it go? So, rhoi, by definition, is an anti-i-1 composed of theta-i. I go from here to here, and I come back here. Well, I say that on uij, I have rhoj equal to the anti-i-1 will be theta-j, but theta-j equals theta-i. So this thing is called lambda i-1 times theta times theta i. My notation is not very good. What I used to play, I call theta i. You see the theta ij there, it's data. And then, theta i and lambda i play symmetrical roles. So the notation is bad. Because I also have lambda j equals theta j i times lambda i. So when I do that, it's lambda i minus 1 times theta i, and so it's...

35:00 So you see that these isomorphisms are there, the rho i. So I just showed that rho i times uij equals rho j times uij. So my family of isomorphisms F-8 and G-8 stick together in a unique isomorphism because they verify the conditions of sticking together. Isomorphisms don't always stick together, but if they verify the conditions of sticking together, then they stick together according to the number 1 theory that I have just explained. So there are one and only rho of F, of isomorphisms, of F. There will be 5 minutes of siesta after. How do we lose that? That's nothing compared to... So, I have shown you... Unicity. Now, the Existence. I have shown you the Existence of a real vessel. That's the only thing I did last week. I put F of B. I will define it as the core. It produces... I will no longer take the notation of the objective limits. Oh no, I can't take the objective limits. If all the fi were the same, I would have f of v which is sent in the product i of f of vi, which is sent in the product jk of f of jk.

37:30 But it doesn't make sense, because we have fi which is defined on u cross of vi, We would write u cross vjk, but it doesn't make sense because on vi only fi is defined, so it doesn't make sense. So we have to take on u cross vi, we have to take fi, so we can't write it as a projection limit, otherwise I would write it as a projection limit, but it's not a projection limit. And then I have my two arrows to reproduce j and k of f. So, here, I made a mistake, sorry, it's U, J, K, 3, E, B, B. So here, however, I have a choice between two things, I can take FJ or FK. So I take FJ, but that's what's going to change the two arrows A and B, which I explained to you last week. I'll explain again, I repeat, so the arrow A. I put U, I also got it wrong, there is no U, I am under X. So, the arrow A, I told you 15 times, but I will say it one more time. How do we go from a product to a product? Well, you have to keep at the end, you have to take one and look. How do I go from a product in I to I, U, K, V?

40:00 In, I choose, we choose J and K and we want to go in F, J, U, J, K. So, how do I do it? Well, I project, that is to say, I project on Fj, here I have all the i's, so in particular I have j, so I project on Fj of Uj, 3, 2, b, and then the rest. That's the arrow A, and the arrow B, in all the fi, I project on the arrow, on Fk. Then I restrict to Ujk and then I use my θjk. It's good that I use my isomorphisms, θjk, so it's a bit difficult. In any case, I have a pre-branch like that, that's clear. So why is it a branch? I'll show you quickly. I think I'm going to skip the first one. So I say that V of F is a beam. So I take V and I take S as a return of V. You don't have to call it S anymore because we already used S earlier. So I'm going to take the rounds as a return of V.

42:30 So I say that the F of I cross V FI are beams, so this is the same as the projection limits for W, which I called T, so I take a notation of the polycorp, W in T, of FI, of I, 3, W. And the same for the FIJ, for the FJ, of U, I, J, 3, T. It will also be isomorphs at the projection limit in W of Fj, this here is Jk, W. So, when you bring all this back in there, well, you see, this is an isomorph at the projection limit of W in T of Fi Ui cross W. I put two arrows, A and B, I produce J, K, the projection limit in W, L, J, U, J, K, 3, W, and then there I have the projection limit in W in T, S, W. But then I say that this sequence at the bottom is exact. That is to say that this thing on the left is the core of the double arrow. Why? Because you can take out the projective limit of the product, because the product is the projective limit,

45:00 and then if you have an exact sequence on the left of the core, when you take out the projective limit, it is still exact. The projective limit is exact on the left. So once I have taken out the projective limit, all the things with W fixed will be exact sequences. I take the projective limit, it is still an exact sequence. So this is an exact sequence. At the top, it is an exact sequence by definition. So, finally, yes, so, the vertical axis is obviously... Because each one is... this is the core of this, this is the core of that, and all that is isomorphic. For example, there is only one core. It's too quick to... I'm not going to rewrite on the board, it's too long to write. You take out the projection line, maybe it's going fast. You take out the projection line, it's not fast. You can apply the limit-projective counter in W, for each exact W, S, W, produced in I, S, I, W, etc. So these are the exacts for each W when we talk about the limit-projective. It's still an architecture. Because it's a nucleus, the nucleus commutes to the limit-projective. The last thing to demonstrate, I think I'm not going to do it, it only explains it to me, is that the theta i, what are the theta i? When I built you f of v, it is sent in the product in i of fi of ui times v, which is the nucleus of the double arrow. So, that, it is sent, if now I restrain myself to open ui.

47:30 It means that I will only take Vs that are sent in Ui. So, if V is sent in Ui, it is sent in Fi of V. And my θi is going to be this, θi of V. So, it defines a pre-physical morphism of Fi, Rfi, θi. So, I don't show them. It's done in the notes a little bit superficially, I didn't do it in detail, because we show, it's a bit the same reasoning as here, we're going to show you the FI, we have to build a diagram, well, I don't know, so let's say it's easier to say that this AI is an ISO and I'm not showing it. Well, as you will see, I'm exhausted and that in addition today, you have to stop five minutes before. I suggest we do a short break now, and then we'll see if there are any questions. I think the answer is no for now. So here we did the most boring one, for the vessels. This theorem for the vessels. The objective condition is a constant. So first a remark. So I gave you the theorem for S-H. So I've shown you, although I didn't give you the definition, that U gives SHU and a stack. I'll give you the definition next week or tomorrow.

50:00 So that means theorem 1 and theorem 2. Theorem 1 is that man is a beam, and theorem 2 is the collation. On the x-axis, u gives mod of r restricted to u and a stack. What I said here, I did for the beams at the time of a category, but I would have talked about beams of r modules, the demonstration would have been roughly the same. What will interest us now is rather the case of a particular, an angled site, I remind you, it means that at x, It's a commutative ring of K-alger. We can always say that a ring of K-alger is always a commutative ring of K-alger. What do we call a ring of a ring? A ring is the same as a Z-alger. So talking about K-alger is more general than talking about a ring. It's the same as a measure or a paper. Okay. So, an adelaide site is when the wave is commutative, otherwise we don't call it an adelaide site. So, now I'm going to focus on adelaide sites, but as an example, there's the stupid case, where you have a constant wave on a fixed, commutative anode. So, it's a particular case of an adelaide site. But often, when we think of an adelaide site, we think of the way it works. Rational functions, on a diagram, with a real big beam, but it can also be constant, like a structural beam, even if it is a system.

52:30 So we will translate the condition of the beam collection in the case of a system. I call it a 1-couple-cycle. I'm not saying it's the best definition, it's intrinsic, but it's not the last one, for everyone it's the last one. 1-couple-cycle, I'm going to call it C-S. It's the data of a recovery of X and of a co-cycle associated with recovery. So C is going to be in the family. So S, I'm going to write it in the form of an open family. And J is an element Cij which will be a section on the direction Uij. So, a notation, Ox with a small cross at the top is the beam of the inverse elements, the beam of the inverse sections. If O is the function of Morse, then the functions of the functions do not change. So this is an abelian group for multiplication. It is no longer an abelian group for addition. But it is still an abelian group. So if we did algebraic algebra, we would still remain in the category of abelian groups. But unity is now one. These are unitary addons, I forgot to tell you. Well, addons are always unitary. So, a cossic is a family of sections defined on intersections 2 to 2 only, and verifying the condition of the cossic of earlier,

55:00 that is to say, such as Cij ± Cjk equals Cik on the Uijk intersection. As a coroner of theory, a co-cycle defines a beam which is written in rank 1. Let's consider a co-cycle in text, I put it in rank 1. So, I said that there is a beam, a beam L, which I called C, I do not write it properly. We would have to go through the inductive limits compared to what we really find. There is only one SOSC, locally free. Locally free of rank 1 over OX. That means locally isomorph to OX, which we call a right fibrous, if you like. So it's not just that it exists. There is LC and there are theta i. So, what is this theta i? I think it's not the same meaning than in the theorem. And isomorphisms, yes. So it is locally isomorphous, yes. More precisely, on each ui, it is isomorphous, yes, by a theta i, with the relation that cij equals theta j minus 1 times theta i.

57:30 So this is not really a particular case, what I have, yes, there is one thing to say that I did not say in the notes, and that is what it means Cij equals, these are isomorphisms, these are sections of the structural beam, but it is seen as a multiplication morphism, we could put something else, let me explain to you, so here I have the Ui, here the Uij, above I have the Cij, which are inversible sections and which verify Cij. Cjk equals Cijk. So I said that there is a family of Cij called C. For those who know cohomology, it's this work in H1, X, OX. We didn't do cohomology. So what was I saying? It is said that there are OX morphisms and Zeta I morphisms, and that MC is restricted to UI, such as OX is restricted to UJ is restricted to UJ, so if I restrain myself on UIJ, if I restrain myself on UIJ, there I have Zeta J, if I restrain myself on UIJ, I will take twice UIJ. And so, theta, in what sense does it go? Theta i, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j, theta j,

1:00:00 Cij defines an isomorphism of OX-MJ on itself, since I multiply by an inverse section. So I have nothing to demonstrate, except that I got a little stuck in my notation. It's not exactly the same notation as in theory. But finally, on my pre-notation, this is exactly the particular case of theory. So there is something that can be added, which is not written in the notes, but I will add it. So, I've constructed, and I'm going to give you some examples. It's very important because it allows you to construct locally isomorphic beams at the top. In fact, it's the equivalent. If you give yourself a locally isomorphic beam at the top, you can construct like that. I haven't done it, I could have done it. Given that you have a locally isomorphic beam at the top, conversely, you can construct... What can I say? Conversely, if it is locally isomorphic... In other words, L is restricted to Ui, so the isomorphism in OX is restricted to Ui. If I put θji equal θj-1 round θi, it will be an isomorphism of OX, restricted to Uij, in OX, restricted to Uij. Canisomorphism from top to top is always given by multiplication by an inverse section.

1:02:30 But if we assume, for example, that we are in a topological space and that the top is a local ring in each germ, then if it is a local ring, a morphism, it will be given by... So the reciprocal is not true in general. So it's not true, there is no reciprocal. So the other thing, on the other hand, that is true... Point number 2 of the corollary, if you will, well it's not a corollary, it's really a corollary of the theorem, the 2 and a little bit more, is that the thing of the wind, let's suppose that your cycles Cij, Cij are defined on the intersection of A2, but it doesn't matter, you could expect very well that there is Ci in... In this case, the inverse of OX over I is Cij equal to Cij-1, at least to be the opposite, to be consistent with my other questions, one or the other. Ah yes, in other words, your right fibrous is trivial. So, we say that this is a co-border. When we demonstrate, I will prepare the demonstration. I put it as an exercise. I put it as an exercise. So, I will try to demonstrate this.

1:05:00 Yes, it will be easy. I lost the coefficient of cos x, but it doesn't matter. theta i equals theta j. I'm mistaken in my notation. Now I'm going to modify my theta i isomorphisms. The conclusion is that LC is an isomorph of x by theta i. These are LRN conclusions. L is locally isomorphized to OX by the teta i. But I changed the isomorphism. So now, as isomorphism, I will take Ci teta i. So I have OX divided by 8, which is isomorphized by Ci times teta i, Alc, divided by 8j. And this time, let's call it lambda i. And so, my isomorphisms are recolored. Here, the isomorphisms theta i and theta j were not recolored, because they had to be multiplied by theta i j. You would not verify lambda i equals lambda j on the intersections. But by modifying them, if I did this hypothesis, c i j equals c i c j minus 1. So by modifying them, they are recolored. So for example, what did we demonstrate?

1:07:30 We are going to show that a co-cycle with a covering, well, a covering plus a co-cycle is C associated with C, and if the co-cycle is a cover, then C is trivial. In the case of algebra, there is an equivalent, that is to say, we can go back and create possibilities associated with the right-winged fibres. So a locally important beam, we also call it by definition a right-winged fibre. For example, in the case of a constant sphere, kx, we don't call it a right figure, we call it a square root of rank 1. On a ring sphere, a real sphere, like in the case of a geometric sphere, it's also a real sphere. But otherwise, we call it a square root of rank 1. So, I'm going to give you some examples. So this, I consider, is the most interesting part of the lecture. That's all I want to say.

1:10:00 Maybe I thought we could stop 5 minutes before the end. So here we will take Kx. We will take K as a body. Maybe not the head, maybe it has nothing to do with it. But maybe it's not the head. In any case, K is as it is. And then we will take Y as the circle. The simplest example. Well, there is a whole theory that we could develop, for example... This is algebraic topology, there are topological spaces on which you will not be able to build a lot of locally constant beams that are not constant, for example if they are contractive or even simply connexed, but it is a whole part of algebraic topology, that's it. So I'm going to build a covering and a cosine. On an interval, we can demonstrate that all cross-cycles are decobords, so on an interval there is nothing interesting. But the circle is not an interval, there are two intervals, and therefore the intersections are called U1 and U2. I call them U1 and U2 intersections, but the intersection is not connected, that's the charm of the circle. In relation to the interval, when we cut it by two intervals, the intersection is more connexed, there is nothing to do. So I have to give myself a cycle, I have to give myself Cij over the interval, so C1 2, C section over U1 2.

1:12:30 It is a locally constant function, inversible, it is an inversible function, that is to say it is non-null on each connexed component. So that means that you have C1 2 plus, which is a constant, and C1 2 minus, which is another constant, non-null, because the intersections of A2 are connected. So a section of that, that is to say a locally constant function on a connected interval, is a constant. If and only if C1 of plus equals C1 of minus, when the product of two constants on U1 and U2 is the product of two constants on U1 and U2, if and only if C1 of plus equals C1 of minus, which equals C, in which case you take C times 1, for example, and it works. But if not, you have a system of cases which is not trivial. It's different, it's a 2-. So this is what I call alpha, the first thing is alpha. Because I copied an old thing, it will use the corollary. If I apply the corollary, well, I have a system, the K. So what I'm going to call alpha, we can always assume, it doesn't change anything, to multiply, it's a 2, to multiply... It's a 2 plus by a constant and it's a 2 minus by the same constant, so I can always assume, it doesn't change anything, I can assume that it's a 2 minus equals 1, for example, and it's a 2 plus equals alpha, it doesn't change anything, right? And I would call L alpha, the system of K, something like that.

1:15:00 So if we want to have an intuition of what this system of K is, L alpha, let's take K equals complex numbers. So, it means that alpha is in C non-null. So, I take beta in C with the exponential of what? i beta equals alpha. So, I say that L alpha, well, maybe I'm wrong. So, you see, on the step S1, I look at the exponential functions i, beta, C. This function here, on the circle, what can you say about it? What can you say about the exponential function? What can you say about this function on the circle? What we can say is that it doesn't exist. It's not a function on the circle. Because when I do a turn... To have a function, it must have a defined value at each point. If you continue it, after a turn, it no longer takes the same value. If beta is not a multiple of two pi, then it is not a function on the core. On the other hand, locally, it is well defined. So this is not a function, but this thing here, this thing is locally a well-defined function on the core.

1:17:30 This is the whole problem of log of z in variable complex, z power alpha. On the other hand, if you take the Cx vector, which is the vector of the multiple local functions of this structure. This is the contents of the infinity vector, for example, x equals 1. These are the functions. These are locally the multiples of this function which is locally defined. This beam is well defined. This beam is well defined. But when I make a turn around the circle to recall two sections of this beam, I will multiply by alpha. So, in fact, this is the isomorphic beam for the L-alpha beam. L-alpha isomorphic. It would take a lot of development, but it's just to make it clear. The exponential function i-beta-t is not defined on the circle. On the other hand, the beam of the functions that are locally demultiplied from this locally defined function, this beam is well defined. It is a locally constant beam, non-constant, because otherwise, if it was constant, it would mean that this function is well defined. Well, it's a bit complicated, but if we see things like that, we understand a lot better the log of z and the root of z in the complex domain, because, in fact, they are sections, root of z or log of z, especially root of z, because that's an infinitive, root of z is a section of a beam locally isomorphous to x, but it's a section. If you take L, I don't know, 1.5.

1:20:00 All of these are sensorized by a vector z, which is a section of this vector. So this vector is locally isolated by this vector. Another example is the vector of orientation. So I take x, which is an infinity of dimension n. And I take an atlas. We will explain what it means. A space is an annealed space with an infinite variety of n dimensions. For example, for x, the space is more than a. The calman isomorphizes to Rn, which is an infinite Rn. This is a variety. An annealed space, the calman isomorphizes to this model. An atlas is an isomorphism. So, I have Xi, so Xi is open. So what is an atlas? Xi is an open, the reunion of Xi equals X, and then I have the arrows Fi, which I call Ui, Rn, and then on the intersections Xj, Rn, Gfj.

1:22:30 If I go to Ui, I find a UiJ, but with an i at the top to remind me that I am in Ui, and if I go to Uj, but that is a topological isomorphism, so I ask that when I pass from one to the other this thing there, F, J, I, I ask that FJ, I ask where it results rather than what I said, FJ, I, the functions FJ, I, These are functions of C-infinity. These are isomorphisms. Isomorphisms of C-infinity. These are ERN openings. At the top, it doesn't make sense. To say that FI is a function of C-infinity, if we don't know what a variety is, it doesn't make sense. It's a topological space. So FI of C-infinity, a priori, it doesn't make sense. But we ask that the collation is C-infinity. So here I have defined what an atlas is in a variety. So what is the orientation question of the variety? I'm going to define a co-cycle. So I'm going to define... So their coverage is... Their coverage S is the X-cycle. So the co-cycle C has a value in Z over 2Z. So these are Cij which belong to Gamma Z over 2Z. They are defined as follows.

1:25:00 It goes from an isomorphism to an infinite isomorphism. So it has a Jacobian in each point. It's a determinant Jacobian function. I don't know how it's called. So this is a function in U and then in S. So it has a sign, it's positive or negative. I go back to xij, it defines me, it defines me, a locally constant function, with a value in plus and minus, even if I put z on 2z, z on 2z means plus or minus, so I find a locally constant function on uij, on xij. Then there is an exercise of differential calculus, on what is the determinant of Jacobian, the composition of two functions. This will show you that Cij multiplied by Cjk equals Cik because the Jacobian determinant has composed this product of determinants and therefore the sign will be the product. There is also a vessel in the network of waves which is called the orientative vessel, which is very important. The last example.

1:27:30 I don't know how people do it. Well, people don't do it anyway. To define the orientation of a variety. Well, if a variety is oriental, that's fine, it's easy to define, but if a variety is oriental, there is an atlas such that all the determinants of the atom are positive. The beam is trivial, but there are varieties that are not oriental, like the Bohmian band, for example, okay? Well, I've already done the Bohmian band, I've already done it. Well, I think that if you have taken classes on Riemann surfaces, you have seen that it is much more complicated than what I am going to do now. But I will give you a simple example. So, PNC is covered by a... it is covered with two open spaces, U1 equals C, U2 equals C. But we have to say, what is the transition function on U1, 2? I'll have to redo my drawing a while ago. u1, u1, u2, except it's not the same notation. v1 equals c, v2 equals c, u1, u2, u1, u2.

1:30:00 I'll take the same notation for the first one. And this will be c minus the origin. This is c minus the origin. And the transition arrow is z, z, 1 over z. So that's the projective space. So now I can take as a cocycle, if you want to take an isomorphism, I can take as an isomorphism on X1-2. So now a cocycle is simply a X1-2 that can be identified as U1-2. And there is no intersection, there is no cocycle condition to check because I have only two open intersections, 3 and 3 are empty. So, a cos x is simply an inverse section over u1 2. So, as I told you, a cos x is multiplied by an inverse function. I have plenty of examples of cos x. I take z power 7 with m non-none. And so that defines me, defibrillated in the right, that we have O and 1. M or minus M is not very clear. There are specific conventions, but in my presentation it is not clear. Basically, the fibers on the right, on P1, with M and in Z now, must not have different zeros.

1:32:30 We must take M equal to zero. In this case, it will also be a co-bore and O1 for M equal to zero. But it will be precisely the structural question. So why aren't there any others? That's what I wanted to say, but I know it's obvious and I don't know why. All the fibres on the right, the fibres on the right, if you have an equation F over C1 of C, and also A, I'm going to call it A, O, X, then I'm going to say M over F, and then F is equal to A, C1 of F. So, I know a cohomological demonstration, but I think there is a totally obvious demonstration, but I'm not going to go into it. Well, there's a little bit of time, but I think we're going to stop here.