 So today's lecture, let me end quickly the example I was telling you about the projective plane. And then to be honest, I was dreaming of telling you something about differentiable manifolds and differential geometry, but being the last lecture, it would become just a mess. So today's lecture will be just closing this example, having a quick discussion. And then I would prefer to discuss with you exercises and problems about the course, the whole course. So it will be a special strange lecture. I warn you right from the beginning. Well the end of the story about the projective plane was this. We defined one differentiable structure even though we presented it in two different ways on P to R. Essentially one was coming naturally looking at it as R3 minus the origin quotient by the equivalence relation. And the other was thinking of it as the quotient of S2, modulo, the antipodal map. This induced naturally two seemingly different differentiable structures here, but it's easy to check that they are actually the filmorphic. You can write one change of charts in the first differentiable structure in terms of a chart of the second structure and you see this is a differentiable map and vice versa. That means that the differentiable structures are really the same. And then I mentioned to you the problem that whether a differentiable manifold comes actually as a sub-manifold of Euclidean space and this object here fails to be a sub-manifold of R3. We can call it a surface in the sense it's a two-dimensional manifold but it's impossible to embed it inside R3. So the idea is just to, I mean, if we want to have some simple model of this, on the other hand I mentioned to you Whitney's theorem, I didn't tell you the name last time. There's a famous theorem by Whitney of mid-40s, beginning of the 40s, telling you that every differentiable manifold is in fact a sub-manifold of every compact, every compact differentiable manifold is a sub-manifold of a sufficiently big dimensional Rn. And in fact in general if you take an n-dimensional manifold, compact n-dimensional manifold, this will always embed inside R2n plus 1. So this is kind of the estimate we have for the dimension of the Euclidean space where you can see n-dimensional manifolds. So applied to surfaces this would give R5 and I mean it's usually not easy to prove that something cannot be embedded in one, it's not a simple problem. On the other hand, let me show you why P2R goes inside R4, which is actually less than the general dimension. Well, the way to do it is to construct a map which I call phi. Look at phi from R3 to R4 given by these functions. You take x, y, z and you go into x squared minus y squared, x, y, x, z, y, z. You see, basically you are taking a subset of polynomials of degree 2 in the variables, in the variables x, y, z. And remember that P2R actually so. You see, if you think of P2R as the quotient of R3 minus the origin, quotient, the equivalence relation given you by multiplication of a non-zero number, this map doesn't pass to the quotient, because the point, the equivalence class x, y, z would be, I mean, a point x, y, z will be equivalent to lambda x, lambda y, lambda z. And the image of lambda x, lambda y, lambda z is not, it's completely different, but it will be lambda squared. But it's not the same point. Here we don't have an equivalence relationship. So this is not a map which passes automatically to the quotient. The way to see, the way to correct this problem is to look at this map on S2. Because what is the difference, of course, which are the, which is the equivalence relationship on S2, which gives you the projective plane, is the antipodal map. So P minus P are the only two representative of a given equivalence class. But P and minus P will go to the same point. Because if I take minus x, minus y, minus x, z, these are quadratic polynomials and they kill the minus. So this map goes, when I see, when I restrict it to, so on S2, phi passes to the quotient, the quotient giving P2R. So in fact, the quotient map, I can define the quotient map, let me call it theta if you want. So there is a map from P2R into R4. This will be formally the quotient map of phi. So now the claim is that this is actually an embedding. Well, what do you have to check? You have to check many things. Most of them are completely, in fact, almost all are completely trivial in the sense that, for example, is this map injective, first of all? Well, that's simple. Suppose that there are x, y, z, and x tilde, y tilde, z tilde going into the same thing. And then you have to argue that the two points differ by a minus. The only possibility is that they differ by a minus, for example. Then, well, is it continuous? Well, this map is certainly continuous because it's a differentiable map of an R3, a stric to 2S2 quotient. So with a quotient topology, this is by definition a continuous map. Then what else? Well, then, is it differentiable? Well, to check if a map on a differentiable manifold is differentiable, you have to write it in terms of a chart. But in terms of a chart, which chart, for example, now we have two nice sets of charts in some sense because we have defined two possible ways to describe the differentiable structure of this object. In terms of, remember, in terms of the maps of the charts given by from the disk to the sphere, okay, when we describe it in the second way, we define, for example, the map F3 plus, which was taking the disk into the upper sphere with respect to the coordinate z. So that means it's taking the point x, y, and giving you x, y square root of 1 minus x square minus y square, with a plus here, otherwise it was a minus here, okay? So this was one, for example, one possible chart. From the disk, no, x, y move in the disk x square plus y square less than 1 and gives me this point here, okay, on S2, on a part of S2 where each point has only one equivalent point. It's equivalent only to itself, okay? So I see this map with values in P to R, okay? So now how do I check that the map is differentiable? Well, I write phi composed F3 plus, for example. Of course, I should do it for any i and plus and minus, but they are all the same, okay? Up to changing of names and putting a minus somewhere, okay? So what does this become? This becomes the map, right? This becomes the map x square because now you take this point and you have to compose with this map here. So this becomes x square minus y squared, x, y, and then let me just say, every time I see a z, I call this function here d, okay, just to give it a short name. And this becomes xd, yd, okay? So this is the composition. So there are questions here. Of course, this is another way to check that this is a differentiable map, okay? Because now I wrote it in terms of a chart. So I wrote the map phi in terms of the chart, and I see it's still a differentiable map. Now the problem is, for example, is the differential injective, remember, I'm trying to prove that this is an embedding. So I need to prove that the differential is injective, okay? Is the differential injective? Well, the differential, I need to compute d of phi composed of 3 plus at any given point, okay? I leave the point free because it has to be true at every point. So, well, but this object here now is a map from R2 to R4, okay? Because now this goes from the disk to R4. So the tangent space to the disk is R2, and this goes to the tangent space to R4, which is R4. So this is represented by what? Is represented by the Jacobian matrix, by the matrix, which I construct by taking partial derivatives in the rows, for example, no? So if I take, so on the first row, I put the partial derivative with respect to x. So this becomes 2x, y, y, d plus x dx, y dx. Remember, d is a function of x and y. So I leave, I indicate by the partial derivative as usual. And then I take, in the second row, I put the derivatives with respect to y. So this becomes minus 2y, x, xdy, d plus y dy, okay? And now the question is, has this matrix rank 2 at every point? Because now this is of dimension 2, and the requirement of being in dimension is that the differential is injective. So it has to have maximal rank, and the maximal rank here is 2, okay? Is it 2 or not? Well, you see, it's enough to look here. This minor, it's twice x squared plus y squared, okay? So that's okay away from 0, 0, 0, okay? 0, 0 is inside our domain, because we are taking x and y in the disk, okay? So this would be a nice minor, except from 0, and then you look for another one, okay? And then what else? Well, you need to, what would need really to prove, so it's clearly, so it's a differentiable map. It's an immersion, it's one to one. It's continuous. Well, you would need to prove that it's an homomorphism onto its image, okay? I skipped this. It's purely topological problem, okay? That's the only bit missing. In any case, this was just an exercise in this, okay? Now as I said, P2R is compact. Okay, yeah. If you want, you can use even, you can bypass the problem in this one, okay? Now, as I said, I was dreaming of telling you more about the beginning of the story on the differentiable manifold, but I hope you realize that all the things we have done for surfaces have been already done in a way to prepare you to jump to this more general setting, okay? Now there are of course points where things get delicate, which we didn't really see in the theory of surfaces, because using the fact that there was an R3 around, we could bypass the problem going out from the surface into R3. For example, orientability, I mean a key point, a key thing that you want to do on a manifold is of course integration, okay? You would like to do two things, differentiated integrative. You want to have a reasonable calculus on a manifold. For example, to integrate, we need orientation. Now orientation in R3, for surfaces, in R3 was easy to bypass by asking that there was a globally defined normal vector field, okay? This was a way to say that something was oriented. But of course, if your manifold is an abstract manifold, it doesn't live anywhere, it doesn't make any sense to speak about the normal vector. So how can you solve the problem of this? Well, that's for example another point where you really think of what you need and you put it as a definition, but then it's interesting to realize that this is exactly what you had in mind for surfaces in R3. Namely, for example, the problem of orientation is that is to ask that your manifold, of course it's always covered by charts, this by definition. And you know that when you overlap, there is a differentiable function, okay, in the other picture from here, the usual picture. Basically, you ask that this function here has positive Jacobian, okay? So this is another way to say that these two charts are oriented in the same way, okay? And in fact, this corresponds precisely to the fact that remember when you did xu cross xv on one side, so if you had two charts and y, you tilt the wedge, you had the two things. Well, of course, the fact that these two things match corresponds exactly to the fact that the Jacobian of the transformation giving you, for example, the y in terms of dx is positive. Because here the only thing that can happen if you think of R3 around this object is that your normal vector, when you pass to the other chart, becomes minus it. So being positive, the Jacobian is exactly the fact that the standard normal vector on one chart goes to the standard normal vector in the other chart. Well, of course, normalized, okay? I'm not asking to be equal to 1 because actually here, there is divided by the norm. So the only thing I care is if it's positive or negative, okay? You see, this, for example, it's another point where you have to stop and think. If you give this other definition, so an orientable manifold will be a differentiable manifold which is covered by charts in such a way that every time two charts overlap, the Jacobian is positive. In the case of surfaces, this boils down to the old definition. But this condition is exactly what you need to define an integral, okay? Because the Jacobian of the transformation is exactly what appears in the change of variable formula for integrals, okay? Up to the absolute value, that's the kind of the problem. So when you want to give the notion of the integral of something, in principle, it would be defined up to a plus minus, okay? Because of this problem. If you know that your manifold is orientable, this problem disappears, okay? I'm not trying to, but this is another moment where, so integration of what, but actually the other problem is what can you integrate on a manifold? And there, that was, I was hoping to, I mean, to describe you quickly but it's impossible. So differential forms, whatever, if you know it already, that's kind of the natural object to integrate on a manifold, okay? Now, now what else? Well, so as I said, this is just the beginning of the story. Basically, we did the first maybe 15 pages of any book in introduction to differential geometry. Then how things go on? Well, you see that even just by having the definition, you have some beautiful questions that you can think of. So for example, this notion of differentiable structure, the differentiable structure induces a topological structure. You haven't stopped me, but clearly it's a question that's behind our mind. Is it possible for two differentiable structures to induce the same topology but they are not defiomorphic? Meaning, the association from differentiable structure to topological structure. Is it injective in some sense? Notice that the two equivalence relationships are different because on differentiable structure you identify defiomorphic things. On topological spaces you identify homeomorphic things. So basically the question is, is it possible that two homeomorphic manifolds are not defiomorphic? So this is already a beautiful piece of mathematics. You don't need. The point is, okay, the question is simple, it's beautiful, it's fundamental, but the answer is highly non-trivial, okay? It turns out that this kind of, because you see, for example, for surfaces you can prove that this is not possible. Two surfaces are homeomorphic if they have a differentiable manifold. Two surfaces are homeomorphic if and only if they are defiomorphic. So you need to go in higher dimension to understand whether this is possible or not. And the answer is actually yes. This was a great surprise. Even for compact manifolds and in fact even for very simple topological manifolds. So Milner, John Milner, which is probably a name you find, you found that in many places in mathematics, John Milner, when he was horribly young, proved that even among spheres you can put. So if you take SN, if you start from seven, these topological spaces have differentiable structure which are not defiomorphic. This was really a surprise. And these are called exotic structures of course. You will find even books on the description of this. Of course, if the sphere has it, you can produce it on many other compact manifolds. But then there was a second, so this was about 1954, if I remember well. But then there was a second part of the story where you say, okay, maybe you have to go in very high dimension and you have to take, what about Euclidean spaces? So if this phenomenon happens, you can ask, does it happen on the simplest possible manifold? This is the simplest compact manifold. What about Rn? So you fix the Euclidean topology, well no, you don't fix it, you are there topological structure which admit, which come from differentiable structures. And as topological spaces, they are homeomorphic, but as differentiable manifolds they are not defiomorphic. And this was even a bigger surprise because the answer is again yes, but in this case it's a very strange yes because exotic structures exist on Rn only in dimension 4. So only for n equal to 4, these structures exist. And in fact in dimension 4 there are infinitely many, more than countable many such structures. And this was a beautiful theorem by Donaldson and Friedman. This was about 1984. So even R4, which is one of the most familiar thing you can think of, produces this incredible phenomenon and there is still a lot of research to try to understand what is the geometry of these manifolds. They are genuinely manifolds. For example, so this is one way, even just with the things that we have just defined it's enough to produce. But then just to indicate, probably many of you have already studied something that is close to what I'm saying. So the game, what is the game in this, I mean, is to try to associate, so to classify manifolds under special assumption. Like we did, for example, curvature, I mean, how curvature condition on a surface determines which surface it is. So for example, a surface of constant Gauss curvature, a compact surface of constant Gauss curvature in R3 is the sphere, for example. Or we did the classification even when it's not positive. So this type of theorem. Well, in differential geometry, what you try to do, you see, the differentiable manifolds are too flexible. I mean, you cannot really, I mean, they are very difficult to classify in the sense that there are too many differentiable maps. Differentiable maps are really flexible. You can stretch and push a manifold wherever you want almost. So the nice, the beautiful and difficult thing is to associate to a manifold some numbers. Think to the case of a surface, when we associate to a surface, a compact surface, the Euler characteristic, for example. This is the prototype game, okay? You have a geometrical object and you associate a number. And then you ask, is it true that this number classifies? So is it true that if you have, I mean, this number must be invariant, for example, in this case for homeomorphism, okay? So it's a topological number. And then you dream of a theorem that tells you if two surfaces have the same number, they are homeomorphic. That this association is somehow injective, okay? And in the case of surface, compact orientable surfaces or compact non-orientable surfaces, remember, but you cannot mix the two together, this is true, okay? Two compact orientable surfaces with the same Euler characteristic are, in fact, the same. So if you want to mimic this, so this is beautiful, I mean, this is great, okay? In terms, I mean, for a geometer, this is the best you can hope, okay? At least as a first step. Is there a similar game you can produce here? Well, there are many of these type of games, okay? And for example, I mean, you know some algebraic topology, you can associate groups instead of a number, no? You can, for example, you can associate to a topological space its first fundamental group, okay? And then you, the dream is if two objects have the same, I mean, abstractly isomorphic, no? Groups, corresponding groups, then they are the same, okay? Well, in general, this is not true, of course. But you see, for example, even with the first fundamental group, if you restrict yourself to closed surfaces, compact surfaces, if you want, you can bypass, I mean, you cannot, you do not have to invent a new proof. But if you believe the theorem that a surface with a given Euler characteristic is given by a sphere attaching handles, no, as more or less, I mean, the Euler characteristic is measuring by 2 minus 2g, g is the number of holes of a surface. Then it becomes also true that the first fundamental group, I mean, two surfaces with two different number of holes will have two different fundamental groups. But for example, the sphere will have fundamental group trivial, the group 1, okay? Every loop on the sphere on S2 collapses continuously to a point, okay? If you take a torus, one hole, what will be the first fundamental group of this? You have two independent generators, and there are no relationship between them, okay? There are no abstract relationship between these two loops, okay? So it's a z plus z. So you see, in this case, so for closed compact surfaces, again, for example, orientable by, you can rephrase the first fundamental group, the term is the surface, okay? But again, in general, is it true if you go to higher dimension, for example? Well, for example, even in dimension 3, this was the content of one of the most famous problem in mathematics, what it was called the Poincare conjecture. If you take a compact three-dimensional manifold whose first fundamental group is trivial, so restrict yourself to the simplest case, so no holes. Suppose you have a manifold, compact three-manifold whose first fundamental group is trivial. What can you say? Well, Poincare asked, in fact, he asked it in any dimension, okay? Is it, but then if you go in dimension higher, you have to start putting also the higher or more to P-type trivial, okay? So in dimension 3, there is only one left, which is pi 1. Otherwise, I think you have seen in the course of algebraic topology that there are higher homotopy, okay? In any case, so in any dimension, you would ask, if something has trivial homotopy, okay, is it a sphere? This was the content of the Poincare conjecture, okay? And this turns out to be true, okay? And it was proved, it's a beautiful, I mean, this is, in my opinion, it's probably the most beautiful theorem in geometry that I know, okay? Also because of the way it was proved, because it's very surprising. So at the end of the 60s, it was proved for any dimension greater than or equal to 5. And in fact, now we all kind of believe that this is easy. It was one of the triumphs of Morse theory, okay? With some Morse theory, with some delicate, but I mean, not Morse theory, you can prove the Poincare conjecture in any dimension greater than or equal to 5. So in some sense, the problem in dimension 1,500,000 is much, much easier than in dimension 4, where you can think that maybe with some, in fact, it's much, much, much easier than in dimension 3, where you can even try to draw a picture, okay? Almost, I mean, drawing a three-dimensional manifold is not as easy as a surface, but I mean, you can get some geometric idea, okay? So in the, by Stephen's Mail, Stephen's Mail proved, I mean, of course, all these people, each of these theorems is a Fields Medal, no? I mean, these are kind of milestones in geometry, I mean, these are not theorems like anything else. Now Stephen's Mail got the Fields Medal for this theorem. And then in dimension 4, with the very Hadoch four-dimensional proof, Friedman got the Field Medal for the solution for S4, and then it was left three dimensions. Strangely enough, this was resisting every attack, and this was proved in 2,000, well, unfortunately, there is not really a date because probably you know it's a strange story. So these were between 2003 and 2006 by Grisha Perelman, okay? Using partial differential equations, essentially, technique, okay? This was the really the masterpiece of mixing analysis and geometry and topology all together, I mean, even though what I said would be true also for this part of the story. When you get to this level of questions, mathematics is one. I mean, you cannot say I'm a topologist, I'm an algebraist, I'm an analyst, I mean, here you have to use mathematics at full power, okay? And that's why it's also beautiful, okay? Now the story was slightly complicated because of course Perelman was, well, he had his ideas about the way to communicate science. So he posted on the web two preprints, well, two papers. But he didn't want to publish them, which is the usual way the mathematical community decides whether a paper is correct or not. Or I mean, he just put on the web two papers and he disappeared, okay? And in fact, in those papers, there is not even the statement. He was not even claiming the Poincare conjecture. He was proving something that everybody knew. It was very closely related to the Poincare conjecture. But he was not even bothered by saying then the Poincare conjecture is proved. So the mathematical community was kind of shaped to say, well, now what do we do? I mean, first, are these theorems correct or not? And so many teams of mathematicians all over the world started working to check whether what he did was correct. And then how to use these theorems, that was kind of the easiest part in some sense, but I mean how to use these theorems to finish the proof of the Poincare conjecture and that was done, okay? So in 2006, there was the International Mathematical Congress, which is the moment where it happens every four years. So it was the moment where we give, I mean, the community gives Fields Medal to, and it was proposed that Perriman would have gotten a Fields Medal for that. But he refused, okay? He didn't want to get prizes for that. He was kind of, he did it for his own, I mean, for the advancement of science, okay, and not for getting a prize. Which is, I mean, if you want, it's kind of strange, but it's noble. I mean, we have to admire him also for this, because actually attached to the Poincare conjecture, there was a $1 million prize. And to refuse $1 million, I mean, it's these days, it's something that deserves, I mean, respect, okay? Well, especially because he was not the son of a gas tycoon in Russia. I mean, he was an absolutely normal person with the problems of every normal person, okay? Anyway, okay, so for example, you see, but going back to now, end of gossip, going back, I hope I'm trying to, I'm communicating you a strategy, I mean, classifying manifolds. How you try to classify manifolds? You try to associate to manifolds numbers or groups or vector spaces or something where you can think that, I mean, the classification of the other object is much easier. For example, if it's a finite group, you have the classification of finite groups. If it's a vector space of finite dimension, in fact, it's r to that dimension, okay? So you don't, again, have to classify anything. So in that case, of course, the only interesting thing would be the dimension. And then you reverse, so there is a first part where you construct this object. So there is this association. And then there is the problem of, how delicate the association is. I mean, two manifolds with the same objects associated, are the same or not, okay? And this game goes on since it's already a century that this goes on. And it will go on probably for another century, if not forever, okay? Because, of course, we have not identified numbers. So such delicate numbers to tell us, well, if these numbers are the same than the manifolds is one of these, okay? There are many of these attempts and every time we, so for every time we have a list, we separate parts of the list into smaller pieces and so on. But I mean, this is part of current research in this general philosophy. The first game you would do in differential geometries to associate what is called a homology and homology, okay? But that's another, that's the story, for example, there are homology. It's a way to associate to a manifold a vector space, I mean, a vector space and okay. But there is definitely no way that in 40 minutes I can give you an idea on this. Well, okay, so in some sense I think the course is over and the second part of this lecture I would like to discuss with you problems, okay? Exercises, okay, so now let's go back to some of the exercises I proposed to you during the course. So I'm asking you, one of your colleagues to help me doing it, okay? Now, the exercise asks basically to classify isometries of the helicoid, okay? So, let's see how we do it. So the first question is to prove that K is, the Gauss curvature is given by this formula, okay? How do we do it? Okay, so just to save a bit of time. Now there are basically two possibilities. Either you go on to compute little l, little f, little g, or you notice that you've fallen in one of the lucky cases of the Teorema egregium, okay? And you take this function and you apply it in the formula, giving the Gauss curvature which comes up from the Teorema egregium and assume, okay, and you get this function, one minus, right? Okay, so now since now we reduced it to just computing a derivative, this is the output. Okay, so let's move to the second question, okay? The second question asks if you take a point, if you have an isometry and if you take a point and it's image point, they lie at the same distance from the z-axis, okay? So how do we argue for this? So we can write any point, sorry, I repeat because of the microphone. Okay, we can write any point on this. We write any point as the image of- So image also can be written in the same way, at some point, some YouTuber, depending on this year. Okay, there will be another choice of parameters for which this is true. So let's say I use this formula to give the z-axis, we will give the z-axis also. Okay. But from here, we know that k of p is the z-axis k of p. So by the theorem I grade you, we know that these two points must have the same gas curvature. So this was a direct application of the theorem I grade you, essentially, okay? Now, in fact, because of course the exercise would be write down all the isometries, okay? So all these steps are to get to the final point. So now let's prove another intermediate step. So f, in fact, so what the exercise suggests is that f acts as a translation on the z-axis, okay? Possibly after a reflection. So let's see how we can argue about this. Your u is very, very similar to a y. Okay, well, call it t. This we can see is exactly when we put v to zero x of u zero. Okay, these are the points of the form. Okay. So if we restrict f on this one, so we have a map. Because any element here is uniquely determined by this t. Okay, basically we are looking. This is kind of the restriction of the isometry on the z-axis. Well, I said it a bit brutally because we still don't know that the point on the z-axis stays on the z-axis. So, but let's see. From this one, so here, I know that for any element, for all t, the distance between x. So in fact, this answers my problem because that means that if I take a point on the z-axis, this will stay on the z-axis. So this map g is in fact, I can write the map g as a map from r to r if you want, okay? Okay. Why is the map different from r to r? Yes, for some function k. Okay, what you have is g prime. If I use this formula, this is f on point x with respect to u, zero. And this using this one, this will give me exactly zero. Okay. Now, f is an isometry. So you present the first fundamental form. So what I have a negative g is exactly 0. And this will give me 0, I have this is 1. So this will give me 1, this is z, and this one, which is half the value of the prime point. Okay. So this will give me, I'm using continuity of t. So I have a negative key of t, k prime of t is exactly 1 of t, k prime of t, now this one, we obtain that. So that means k of t is equal to, well, okay, okay, let's not, I mean, that means that k of t is equal to plus or minus t plus c, plus c, okay, plus some constant. And that's exactly what you had to prove, because the minus sign here is exactly the reflection in the z-axis, on the z equal to zero axis. Okay. So we also achieved 0.3 in our staircase. Now, 0.4 maps a ruling to a ruling. Well, actually I don't know if we use the word ruling in the same way during my lectures. Of course, this means just, so this is a ruled surface. Okay. So that means just it maps one of these lines to possibly another of these lines, okay? Maybe we can erase this. If the space is enough, yeah. If you can speak a bit louder, I think. From this expression, you can write x of u v cos of u. So if I do this as alpha of u, any ruling on this one, x of u v of r, so mark is the image of this. So the written in mathematical form, what you are asking is, is it true that f of l u is equal to l u tilde, for some u tilde? Okay. Now that we have translated the problem into mathematical terms, you fix you, yeah. Mm-hmm. Yes? f of v is exactly, now here, u is fixed, v which is moving. So I think that any element can be written in this form. So this can be written as sum of this one, sum h, perhaps h, if I put this one inside here. Now u is fixed, the v which is moving, so this is, so if I use this inequality, I have this prime always equal to this one, because here u is fixed, this v, if I use this equality, this will give me, so from here I have that the number of this prime, this prime is exactly the number of f and this using the fact that f is an isometry, this is exactly f tilde, this is exactly that. Sorry, may I interrupt you, because it seems to me we are taking a root which is too long. We know already that the map f preserves distances from the z-axis. So you already know the function g2. So what you have is g2, the only thing you have is g2. So it's plus or minus. Wouldn't it be easier to use this information right from the beginning? You can, for example, assume it's plus. Of course, the difference between plus and minus, the only problem is when it's zero. So if you assume it's plus and you are not on v equal to zero, you can argue. Then of course the same proof will work when this is minus v and you are away from zero, but then the zero case you have already dealt on a side. So somehow you don't have to worry about the fact that you change. So in some sense I think if you write, without loss of generality you can assume g2 is the identity. So here you have data of this point of this because this is exactly the square. Now here so we can also say this is exactly f composed of x and z1. Now here we fix u. So we have also, just also, this is exactly g1, which is y. This is 1. That is equal to 1. G prime is equal to zero. Call it d, I don't know. Just to avoid confusion with this. d is constants. So if you put it back inside, what you have is... Well, okay, now it's okay. Because you are just changing u into another thing, which is constant. So it's done. And now let's put everything together and give the classification of isometries. So what have we learned out of these? This one, if we suppose that u is fixed, it will take two... Sorry, this is a u. Okay. We learned that this one from the second question, this take you wanted to, another one. So if you suppose that this was fixed, this should be normally half of some u, but depending on this u, this is fixed. I think it would be more clear if you would call it b of u. I mean, f, by the previous question, you know that u goes into something like this. But here what is not written here is that, of course, everything could possibly depend on u. Okay, so let's call it b of u. Either this one or minus. So if you suppose that is v. So we have v plus w. And we know that... I want to get this formula explicitly. So I have here that if v is equal to zero, I have zero, zero is exactly alpha. And we know this element belongs to the axis. So this is the translation. And this is the z to zero, zero, u for some constants. That's what we proved in part three or two. No, three. No? If you put inside, it's done, because now what's written there is just the x of u plus cv. Okay. So were there alternative solutions? Anybody found some... It's a symmetry. If u is minus u... If u maps to minus u... Yes. But no, no, it's not reflection, I think. It's a rotation in the x-axis by 180 degree. Because u is also in other sides. It's not reflection. Because u is also in... So I think it's a rotation. It's a composition of the two, no? Because u goes to minus u, it's the reflection in the z-axis. But at the same time in the x-y plane, you are doing something, no? No, I think only u to minus u, it's all the rotation. The rotation where then? The rotation in the x-axis. So come to the blackboard, because I mean in the x-axis, I don't know what it means. We wrote the thing in the x-axis, v to minus v. Okay, sorry. What were you arguing? Then we get that v because u minus v sin u minus u. But if it's the reflection, this must not say... If this is a reflection, then the minus must not be here. Right? So I'll give the resist rotation in the x-axis. Sorry, I don't know what it is. A rotation in the x-axis. I mean... No, but suppose just... Let's draw it... You have R3. So first I don't know why you are keeping... At which point this should be relevant in this exercise? Because here... Okay, so you want to say that this is up to something. I mean, so there are these family and another family. Okay, but now what you wrote? So geometrically it takes this point where... I mean, so this is x, y and z. Right, so it's a composition of... It's only a rotation in... Well, so the rotation means it's something going like this? Yes. So it's a rotation where the axis depends on the point. The axis is here. So if this is the cylinder, then the rotation is so... Okay, we can discuss it later. I mean, the point... In any case, so the point that you are making is where is the reflection gone in this argument? Yes. Okay, now forget... I think you are a bit confused, but it's okay. What is the geometric meaning? But what you are raising, you are raising the question whether we should add here a plus or minus? Okay. And also v can be a minus. Okay, so let's see. What do you think? Yeah. So in fact, I think you have to allow this freedom. The only point is that probably you are writing twice... I mean, it can happen that you are writing twice the same thing. Okay, depending on the way you write. But actually, even in my solution... No, no, but actually, no. The point is that rotations... I mean, reflection should disappear. I mean, reflection along the z-axis are not there. So in principle, you leave it here, but then you have to argue that they disappear. Okay? Maybe it's still on the blackboard. It's here the point. B of u is not plus or minus u plus c. It's plus u plus c. So this thing here tells you that actually reflections were not there. Okay. So at the end, these are the only isometries. Well, okay. I mean, of course, I have prepared six exercises and we have done one, but I hope... Well, we can also have a special lecture on exercises if you want. Do you think it would be useful? So we will plan it, okay?