 Okay. So actually, the first thing, as I mentioned to you at the end of last lecture, I realized while I was speaking yesterday, I gave you the definition of geodesic curvature for a curve which is correct up to a sign. Okay. I told you, I told you yesterday the absolute value of the norm, I mean, so the norm of the projection on the tangent space of the acceleration. Okay. But actually, it's better to keep track of a sign. So it's plus or minus that. Okay. So really correct it with this one, the geodesic curvature for a curve gamma, parameterized by arc length, it's just gamma dot. Okay. So this is just a side remark. Of course, the absolute value is the same as the one I told you yesterday, but could be positive or negative. Okay. So this is just one thing. Then, remember, yesterday, we proved one of the most beautiful formula that I still want to come for a compact orientable surface, the, of this one. Oh, but it's a two hours proof. It took the whole lecture. Maybe shall we do it? I mean, let me go, I mean, and we can do it in my office when you want. Okay. If everybody wants to have a repetition of this proof, we can do it all together. Okay. But now let me, let me go on. Now, so we proved it for an orientable surface. And then we speculate a bit about what is this number. So the Euler characteristic, how to compute it, what is in examples and so on. So you take a compact surface, you take a subdivision, and then you count vertices minus edges plus faces. Okay. So you need a little bit of structure to do for, in order to apply this theorem because you need to know that every compact surface has a subdivision and that the Euler characteristic does not depend on the subdivision. Okay. In order to, but now what I mentioned, I'm just stating another important theorem without proof, topological result, is that actually this number classifies compact orientable surfaces. Okay. So if you want, you can say exactly like this. So theorem, which I'm not going to prove, but if S, S prime are both compact orientable, compact orientable surfaces, and they have the same Euler characteristic, then they are homeomorphic. The same result holds for non-orientable surfaces. Okay. If you take two non-orientable surfaces with the same Euler characteristic, then they are homeomorphic. Okay. I don't even write it down, but be careful. It's not true that if two compact surfaces without any assumption on orientability have the same Euler characteristic, then they are homeomorphic. Meaning what? Is it, it is possible to have one orientable surface with the same Euler characteristic to a non-orientable surface. Okay. Clear orientability is homeomorphic invariant. Okay. So they cannot be homeomorphic, but is it, it is possible to construct examples of two surfaces, one orientable and one not with the same Euler characteristic. And in fact, you know how to do it, essentially. I'm using a slight, I mean, this is, these are not surfaces in R3. Okay. By surface now I'm meaning something slightly more general, which is important anyway. Remember how we constructed the torus as a topological space. We said we identify, we take a square or a rectangle or whatever you want and you identify the edges, one in front of the other in the same way. Okay. Meaning if I parameterize this to be 0, 1, then the point T goes to the point T on the other side. Okay. And if this is again 0, 1, the point T goes to the point T. So this is pictorially what you draw and you see in books written in this form. So the identification between these edges is homologous in this sense. Okay. So a point of a given height goes to the point of a given, is identified with a point of the, of the same height on the other side. Okay. And this becomes this. We compute the Euler characteristic and it turned out to be 0. Okay. Now, and it is orientable. Okay. Now, play the same game, but now try to, to find another identification, which is again natural in some sense. So two edges are identified in the same way, but now pictorially you draw it in this way. The point of, of height T, you send it, you identify it with the point of height 1 minus T. Okay. Exactly. So this is a fem, pictorially you draw it in this way. So the arrow in one sense goes up and the other goes down. Okay. So meaning this point is really this one. Okay. While this one goes here. Okay. This is called the Klein bottle. It's not a regular surface in R3. Okay. But it is still a surface. It's a, in some sense, for, I mean, I haven't given a general definition of surfaces, but you can trust. I mean, this is a nice object. Okay. Anyway, if you want to compute, so call this S prime. Let's try to compute the Euler characteristic of this. Well, how did we prove here at the, which subdivision was good? Now the problem is to construct an easy subdivision so we can count easily points. And actually the subdivision we used for the torus was exactly the picture. Meaning, as a vertex, you take the four vertices of the square. As edges, you take the four edges of the square. And as a face, you take the only face which is inside. Okay. But now the point was to count edges and vertices up to equivalence relationship. Okay. So in fact, they look like four vertices, but it's one. They look like four edges, but they are two. Because this is equal to this and this is equal to this. And it, it looks like one, one face and it is one face. Okay. So one minus two plus one, zero. Let's play the same game here. Now the problem is that this one is not now a good subdivision. Okay. We have to do something to it. To cut it into parts which are homeomorphic to disks. Okay. If we do something like this, so vertices are still those four edges. Now there will be these five. And now we have to count them. Modular equivalence relationship. And we have two faces. Okay. But modular equivalence. How many? Well, vertices are still one because this goes here, which goes here. Okay. But this also goes here. So by transitivity, there is still one vertex. Okay. How many edges? Well, this goes here. And this goes here, it doesn't matter how. Okay. You twist it, but it's okay. It's still another identification of these two. So you have one, two, three. Okay. So it's one minus three. And how many faces? Two. So the Euler characteristic of this object is again zero. Now the point is this is not orientable. And you can actually see it. I mean, if you want to prove it, it's a bit more delicate, but you can see it immediately. In some sense, this object is the compact version of the Möbius strip. Okay. Because you see, this is exactly what happens. This twisted identification is exactly what gives you the normal vector. If you imagine to walk on top of this here, you will end up on the other side. Okay. On the back of the blackboard. And then when you go again, you would turn up again. It's really a compact Möbius strip. Okay. So you see that this theorem, I haven't written the non-orientable version, but it told. So there would be a part two. If S SS prime are both non-orientable with the same Euler characteristic, then they are homeomorphic. This is okay. But you cannot mix the two statements. Okay. In fact, I mean, this is not the surface in R3 in the sense that if you try to draw it, you will get something like this. I don't know if you can see it. So why it's called a bottle? It's called a bottle because it's really the first identification gives you a cylinder as before. But then in order to make this twisting identification, so you get to this point here. So the first identification gives you the cylinder. But then you need to identify this circle with this circle, but now not in the straight way. But identifying points somehow in the opposite way. That means that the only way you can draw something like this is by taking this object here and pull it inside. Doesn't matter here and twisting inside here. Okay. And this is the best thing you can do in R3. So you have to cut the surface itself. You have to self-intersect it and go inside and make the identification. Okay. So that looks like a very weird bottle because of course try to pour water inside and the water goes everywhere. There is not an inside and an outside. Okay. Anyway. Okay. So this was one comment. But anyway, the important thing is that now go back to orientable surfaces in R3. If you mix these two information now, you can say something very important because you see if you, for some reason, if you know the Gauss curvature of a surface at every point because essentially you need to know the integral. Okay. Of course, knowing the integral is less than knowing it everywhere. But in some sense, you almost need to know it everywhere. Then you can compute the Euler characteristic. But then if you know the Euler characteristic, you know exactly which surface it is. Okay. So in principle, this is an intrinsic quantity. Okay. So if you are your bug, remember that you are a bug in this course. Okay. You can compute this thing and by leaving on a surface, you can realize the global topology of where you live. This is fantastic. Okay. And in fact, this is what Gauss did. He took some of his students. So you should feel lucky now. Okay. Because what? Maybe you feel unlucky because I'm not Gauss, but now at least you have one good thing. Okay. He took his students and he took them on top of the mountains around Gottingen. Okay. He measured, now he considered the local version of the Gauss-Ponnais, but I mean he measured really the area, so the sum of the total angles of the inside angles. Okay. The sum of the inside angles of a triangle. So I stay on one mountain, you go on another one, you go on another one. Of course, how do we measure angles? Well, for example, we take light. We take some source of light. Okay. So that I can see a direction which would be a geodesic if you believe that light moves in the least possible, making the least possible effort. Okay. We measure the angle and then we write it down. We meet again with, well, they didn't have cellular phones, so they had to meet again. And you sum and you see, is this sum bigger, smaller, or equal to pi? Okay. And then, because that's what the local version of the Gauss-Bornet theorem is telling you, no? If you do it, you know what is at least the average of the Gauss curvature inside your triangle, the integral of k over this triangle. Okay. And that means, for example, if you get more than pi, you get, you know that your surface inside this triangle at least has average positive Gauss curvature. Now, of course, here you jump intellectually. You imagine to do it on infinitely many triangles, no? And you can say the Gauss curvature is positive everywhere. But if it's positive everywhere, that means that you are living on a surface, well, you have to know somehow that your surface, your universe is compact in some sense, but I mean to apply now, you are living on a surface with positive Euler characteristic. And then you say, well, let's go back to topology. Which are surfaces with positive Euler characteristic? You classify them. There is only one possibility which you know. We have computed the Euler characteristic of the sphere turned out to be two. The Euler characteristic of the torus is zero. Now, in fact, more generally, you should be able, I mean, it's not an easy theorem, but I mean, you prove that the surface with G-holes, okay, you count the number of holes. The Euler characteristic of a surface with G-holes is two minus two G. G is the number of holes. Okay. And that's the only number that a compact orientable surface carries with it. Okay. So, basically, by this combination of informations, by topological and Gauss-Bonnet, in principle, we are able to prove that the Earth is homeomorphic to a sphere. Okay. And this was the first, in fact, for what I know, the only intrinsic proof. I mean, conceptually, of course, because in theory, you should do, it is not enough to take three students on top of mountains near Gottingen. You should do it everywhere, because it would be enough to have one little triangle with Gauss curvature usually close to minus infinity to destroy this argument. So, in principle, you should do it every to the north pole, south pole in the middle of the ocean everywhere. You should be counting this number everywhere. The sum of the internal angles to a geodesic triangle everywhere. But, in principle, you can do it. Okay. So, this is basically the only proof we have that the Earth is a sphere without taking a rocket, going outside and taking a picture. Okay. And this is conceptually crucial for us. Because now, again, now we are not bugs. We are three-dimensional objects living in probably a four-dimensional universe or ten-dimensional universe. We don't know exactly. But somehow, if we want to know which is the shape of the universe, we have to find a four-dimensional, a ten-dimensional, some, I mean, a higher-dimensional version of this type of theorems, for example. This could be a way to decide which is exactly the shape of our universe. Okay. In fact, now we know, we have the record of Gauss and his students, and we know that they cheated. Okay. Because, of course, the curvature of the sphere is so small, because it's true. It's a sphere. It's all, I mean, it's all essentially a sphere. But, I mean, it's so big that the curvature, remember, the curvature of this sphere goes like one over R squared. So, it's almost zero if the radius is very big. Okay. So, in order to measure one over R squared for the Earth, you need to be very precise about the measurements of the sum of the angles, okay, of the triangle. And in 1827, they were not able to be so accurate, but in some sense, so he cheated, okay. He put numbers which were impossible to get, okay, which is typical. If you have ever been in a physics laboratory, you know how much the, for example, the acceleration of gravity should be. You measure the ball falling down and you cheat, okay. Because it's never 9.8 meters by square seconds, okay. Anyway, okay. So, this, in some sense, closes the story for, I hope I convinced you, this is a beautiful theorem and beautiful part of mathematics. But now, let me go back. We left a couple of important observations left about isometries and about the theorem I gave you. In some sense, you see how important the theorem I gave you is. By itself and by what, I mean, by the proof, we used steps of the proofs many times, okay. But we also, the conceptual content of it is essential. Now, let me make a couple of, in some sense, it's an excuse to make two exercises, but with some conceptual content, okay. We have talked a lot about isometries and localisometries and in some sense, we would like to know to which extent the fact that two surfaces, like locally isometric, means that they look more or less the same. We know that this is not, this is a delicate question because, for example, the plane and the cylinder are locally isometric and in which sense they look the same, it's really up to you, okay. But let me give you an even more striking example, okay. This is not the one. Okay, in fact, I will take the first surface, minus one will be the catanoid. So we know almost everything already. And so I think of it as parameterizes the surface of revolution. So I think of it at least the whole surface minus one meridian, okay, remember. So this is given by some A cos, cos v cos u, A cos v sin u, A v, okay. And in this parameterization, I'm assuming, of course, A is a zero number. That u, u is the rotation parameter, because it appears in this way. So u, I take it between 0 and 2 pi. And v, it's any real number, okay. So this is the first surface, the catanoid. We know even the picture very well, okay. And in fact, let me not, well, up to you. Would you like to do a complete exercise together or will I write directly which is the first fundamental form? Everything? No, let's write it, okay. Because it's just taking derivatives and taking scalar products. So there is nothing to learn. Okay, so you compute the first fundamental form and you get that A, in fact, it's equal to G and it's equal to A squared cos square v, okay. And f is equal to zero. So in particular, notice that this is a conformal parameterization, okay, because E is equal to G and f is equal to zero, okay. It's not an isometric parameterization because this is not equal to one, okay. But it's a conformal parameterization, very well. So this is the catanoid. The second surface I want to look at is the helicoid. So again, we know how to write it down. So the helicoid, and now I write it as y of uv is equal to, well, let me call it u bar v bar, okay. Because they are different parameters on a different domain. So y of u bar v bar, I'm using the standard parameterization we gave already. So v bar sine u bar, a u bar. And now in this case, u is again, u bar is again a rotation parameter. Remember now at this time is the line which is rotating going up, okay. So u is in zero to pi and v bar is again any real number. So they look completely different. I hope you agree that in principle, I didn't take with me two pictures. But I mean, this looks like this, and this looks like a mess, okay. This is the line rotating in R3, okay. So imagine the line rotating in R3, it's better if I don't try to draw anything. But look at what happens if we perform this change of parameter. Look at, well, you can do it quickly, in fact. I mean, these are very simple functions. But I mean, I don't need to know the first fundamental form with respect to this parameterization of the helicoid, okay. This is irrelevant. Believe that there will be something. I mean, some functions which in principle have absolutely nothing to do with this, okay. So look at the following change of parameters. I consider u bar, u actually, I don't want to change it. So u bar is equal to u. But I make some kind of drastic change in v. So v bar is equal a hyperbolic sign of v, okay. Well, in fact, I've already jumped to some conclusion if I call it a change of parameter. Because in order to be a change of parameter, I need to check that the Jacobian of this transformation is always invertible. Okay, now the Jacobian of this transformation, let me call it in this form, okay. If I take the functions u bar v bar and I take the matrix given by, I mean, and in fact the determinant of this metric, of the matrix given by the partial derivative with respect to u and v, this is equal to a cos v, okay. Which is a never zero function, okay. So in fact, this is a change of parameter. But now, re-parameterize the helicoid with this, okay. So y, y, now it will become a function of u and v. So instead of u bar and v bar, I put these functions. And so I look at them as a functions of u and v, okay. And they become a hyperbolic sign, a sign hyperbolic v cos u. A hyperbolic sign v sin u, a u, okay. Now, you see, u and v now vary on the same domain for both S1 and S2. So now I, well, even before it was true, but it is still true, okay. Now I look at the first fundamental form of the helicoid parameterized in this way. And what I get is u, e, e is equal to a squared cos squared v, equal to g, and f is equal to 0, okay. Again, I don't do it because it's just partial derivative and scalar product, but this is the output. But you see, now we are in shape, okay. Now we have a theorem which tells us that these two surfaces are locally isometric, okay. In fact, locally, how big is the local set? Well, in fact, here it's everything. It's the whole part of these surfaces which is covered by these charts. The point is that not all of this chart is, because here, remember to parametrize the catanoid in this form, I had to remove a meridian here. There is not zero or two pi, okay. So this is not the whole catanoid, but it's the catanoid minus one, okay. Minus one rotating curve. So this observation tells me that up to this change of parameters, there is an isometry between this object and the staircase, okay. In fact, you can see it if you Google it. Isometry between catanoid and helicoid, you will even find the movies, okay. Pictorially is something like this. I can show you just a picture. Basically, you are taking, you take the helicoid, a piece of the helicoid, okay. And you start stretching it up to this stage, well. And then you go on, okay. This is just a sequence of pictures. You can even find very nice movies, okay. Showing you the continuous motion of these two things, one on the other, okay. Of course, again, everything is local, and you cannot hope to extend this isometry to a bigger object, because there is a topological abstraction, a global topological abstraction. This is homomorphic to a cylinder. So this has non-trivial fundamental group. It's every surface of revolution has a Z in their fundamental group. Because there is this S1 of rotation. While the helicoid is simply connected. In fact, you can prove it that the helicoid is simply connected because for every point of the surface, you can move on the line up to the axis, okay. So basically, if you had a loop and you want to prove that this loop is contractible, well, it has to be contractible because you push it on the axis, then it becomes an interval and you collapse it, okay. The helicoid retracts on the central axis, okay. Because for every point, you move on the line, okay. So there is a global obstruction to the isometric problem. But I mean, this is even more dramatic in some sense than this example is just a more dramatic appearance of the phenomenon of the fact that local isometry doesn't mean really they look the same, okay. They look, they may look completely different. Now in a similar spirit, I forgot, and this is a moment to get back. When we studied the theorem in my gradient, so the theorem in my gradient says the Gauss curvature is intrinsic, okay. Meaning if you have a local isometry between two surfaces, if you compute the Gauss curvature on the left and on the corresponding point on the right, you'd have to get the same number, okay. It is natural to ask, what about the contrary? I mean, if for two surfaces, there is in principle a map which sends a point of a given curvature to a point of the same curvature. Is this map an isometry? This is the converse of the statement, okay. Well, let's see. So about the converse theorem in a gradient. And this is again an excuse to, well, no mystery. The answer is no. The converse of the theorem in a gradient does not hold. But how can I produce a counter example, okay. Well, look at these two surfaces. X of u v is equal to u cos v. u sin v log u. Meaning, of course, v is free to move from 0 to pi, and u is a positive real number, okay. And then I look at the, and another surface, y of u v is equal. u cos v, u sin v, okay. In this case, I don't even have any restriction on the domain of u and v. Well, let's make the computations now. Let's compute their Gauss curvature, okay. Now, in fact, freeze for a moment y, and let's work on x, okay. Okay, we have to do all the computations. So what is xq? xq is cos v, check it while I do it because it's very likely. Cos v, sin v, 0. xv is minus u sin v, 1 over u. Very well. Minus u sin v, I'm taking the derivative with respect to v. u cos v, 0, okay. So n, n is minus cos v. So there is 1 minus in front, but then there is minus, minus plus, so it's still a minus, and it's sin v. Plus or minus, minus. And then there is u cos squared plus u sin squared, okay. So this is the cross product, and then divided by the norm, which is not 1, so divided by 1 plus u squared, okay. And this is the normal vector. Let me compute also the second derivative, so I'm done with computations. xu, u becomes 0, 0, minus 1 over u squared. xu, v becomes minus sin v cos v, 0. xv, v, it's minus u cos v, minus u sin v, 0. And that means little e is equal to what? Little e is this times this. So it's minus 1 over u, okay. It's 1 over u times square root 1 plus u squared, okay. How much is little f? Little f becomes, hopefully, 0 plus sin cos minus sin cos plus 0. So f is 0, and g, and little g is plus u cos squared plus u sin squared, so u, u divided by square root 1 plus u squared. Okay, so how much is k? K, let me call it k1, okay, because it's the Gauss curvature of the first surface, okay. If you want kx, invent your notation. So this is eg minus f squared, and of course, eg becomes 1 over 1 plus u squared, okay. There has to be a mistake. Yeah, I forgot the minus here, okay. So eg minus f squared, so this is the numerator, which goes as a denominator, divided by capital E, which I haven't written, okay. So capital E is what? It's 1 plus 1 over u squared. Capital F is, hopefully, 0 minus plus 0, and capital G is u squared, okay. So this times 1 over capital, but capital what? This becomes eg. So it's u squared plus 1 minus 0. So it's another one of these, okay. So I write it immediately like this, okay. Enough for x. Let's look at y. Let's play the same game. yu is cos v sine v 0. Yv is minus u sine v, u cos v 1. So that means immediately, e is 1, f is 0, and g is 1 plus u squared, okay. Okay. Let's go on n. n is sine v minus cos v, u, okay. So again divided by square root of 1 plus u squared, okay. And now yu u, this is 0, 0, 0, 0, which is fantastic because immediately, of course, when something like this happens, you start thinking again. That immediately will mean that e little e is 0, but that means I don't have to compute little g, okay. So the only thing I need to compute is little f. If I'm interested only in the Gauss curvature, the only thing I care now is this one, and this is, what is it here? Minus, well, minus u, sorry, xu v minus sine v cos v 0. So little e is equal to 0, so I don't care about little g, and little f is what? And little f is minus, minus, so minus 1 over square root, okay. So how much is k2? Well, k2 is eg, which is 0, minus f squared, so it's minus because it's minus f squared, okay. Don't be confused, minus 1 plus u squared times 1 over eg minus f squared, which is again the same thing. And so, okay. Now comes the interesting point because, of course, here what is the incredible accident? These two surfaces have the same Gauss curvature at the corresponding points. Now, here the corresponding points is really, now, that's what? That's the problem, okay. Because, of course, if I take a point coming on x, so which is a corresponding point, meaning if I take a point which is on S1 or on x coming from the parameters umv, in order to find a point on x on y with the same Gauss curvature, I need to take a point which comes from parameters plus or minus u and any v. See, this is v invariant. It doesn't care. It's v independent, okay. So if I'm really thinking to the converse of the theorem the problem is, is there any, so a local isometry, let me write it in words, a local isometry must take, I mean, f, little f, that don't be confused with that one, of course, there's no, should be, should take x of uv to a point on y to a point of the form y of plus minus u v tilde, okay. Here the key part of this sentence is, of course, the relationship between umv and these. And here there is written v tilde doesn't care, can be anything, but the first parameter should be plus or minus the old one. Otherwise they would have different Gauss curvature, okay. But you see, at these points, whatever v tilde I pick, and independently of the fact I pick plus or minus u, the first fundamental form are completely different, okay. So there is no such isometry. Remember we proved an if and only if. A local isometry if and only if you can reparameterize, if up to a reparameterization, which at that point becomes the isometry. The reparameterization is what contains the information of the isometry. The two first fundamental forms coincide. But by such a map, I will never be able to make, for example, what, you see, yeah, capital E and capital E will never be the same. And no matter, I mean, in principle, you can switch things, of course, because if it's only up to order, it's stupid, because you can take E to V to G. Of course, this means just you have switched the name of the coordinates, okay. But again, okay, so in principle, this could happen for U equal to plus or minus one, okay. Only at those points, okay. So such an isometry does not exist. It's kind of a, it's a little miracle, of course, you know. We found two surfaces completely different. You can even ask a computer to draw them, okay, if you want. They were absolutely nothing in common, but they have exactly the same Gauss curvature. Okay, in fact, I propose to stop three minutes. Okay, so now we start with, we take a different point of view on basically everything we set up to now. Of course, I'm trying to make you the key step. Think to the following problem, but we already solved in some sense. So we take a patch of a surface, okay. We have our surface, we have a patch, X. We have some domain U which parameterizes this patch via the map X, okay. Suppose actually, so we call these coordinates UMV. And suppose you have a curve here, which of course corresponds to a curve here. So here you have some functions U of T, V of T, which gives you this parameterization of this curve, okay, as usual. But now suppose that we make a change of parameters, okay. As we did, for example, in the examples we did 10 minutes ago. Basically, that means we have a function, let me call it capital Phi from another domain of R2. And now let me call the coordinates of R2, X and Y. So in principle, this domain has nothing to do, okay, with. But this is a function which maps this object to this object. So again, here we can, and this is one to one. So it's a defilmorphism of this domain into this domain, okay. So this curve also comes from somewhere here, okay. There will be another curve on this other domain which always go to the same thing, okay. Of course, I can write down Phi. Phi of X, Y will be a pair of functions U of X, Y, V of X, Y, okay. And here, so of course here what I meant is that we have also a curve in this domain which corresponds to the same curve, okay. And actually at the end, let me say that everything goes down to some curve gamma on S, okay. So what is the first fundamental form? Well, the first fundamental form is, for example, is essentially a way to measure the norm of the tangent vector to a curve, okay. So what it means, what does it mean? I mean, I can take gamma, the norm of gamma prime, here the parameter is always T, okay. The norm of gamma prime square. And of course we know what is the game which plays the first fundamental form. This in terms of the coefficients of the first fundamental form is equal to what? Well, if I forget for a moment the right hand side of the picture. I look it only here. Well, this is E du dt squared plus 2f du dV dt dt. But this is a plus g dV dt squared. This is what the first fundamental form does for us, okay. The coefficients with respect to these coordinates, okay. But now, if we put in the picture, the map phi, u, u of t is actually u of x of t, y of t, okay. So we could replace, this is an F, okay. We could replace these derivatives with the chain rule with respect to x times dx dt, okay. So let's try to do it. It's a boring thing, but I want to convince you then that there is a short cut, a conceptual short cut that would help. So let's do it for once. So this is equal to what? Well, the u dt becomes ux, let me write it in short, x prime, okay, dx dt, plus uv, u, uy, y prime, squared, and so on. So let me go here and then I'll use the other part of the black. Plus 2f, the u dt is that thing, so it's ux, x prime, plus uy, y prime, times vx, x prime, plus vy, y prime, plus g, vx, y prime, plus vx, y prime, plus vx, y prime, plus vx, x prime, plus vy, y prime, squared, okay. Now, if the question is, I want to write the same expression in terms of the composition, what is x composed phi, is a patch, is a chart, okay, because this is a diffeomorphism, this is a chart, so all together they form a chart. So this has to be equal to what? By the same thing, for the same reason we wrote this formula, for the same reason this will be equal to e tilde, let me say, e tilde dx dt, x prime, squared, okay. Let's switch to this notation. Plus 2f tilde, x prime, y prime, plus g, y prime, squared, g tilde, okay. So these are what? These are the coefficients of the first fundamental form before the patch x composed phi. So basically, if the question is, what is the relationship between the coefficients of, the coefficients e, f, and g, so the coefficients with respect to x, and the coefficients of the same thing, but with respect to a re-parameterization? Well, of course I can expand this formula. And I get it, it's long and boring, but let's do it only once in our life, okay. So how do I get e tilde? Because basically I have to isolate x prime squared. So it will be a sum of things. So what comes with x prime squared? Comes here, so it's e ux squared, okay, and nothing else from here. Here I get plus 2f ux vx, okay, plus g vx squared, and this is times x prime squared. Then I want to know f tilde, so I put a 2 in front, and then I have to isolate what comes with x prime, y prime. So here, for example, I get e ux, uy, remember the 2 is outside, okay. So e ux, uy, plus, and here nothing else. Here what do I get? Here of course I get a mist, I mean, okay. This idea of putting the 2 outside was not really great, but I mean, because now, well, no, okay, this is this 2 here, okay. So here I get ux vy plus uy vx, okay. So plus f, which multiplies again, ux vy plus uy vx, okay. So this is about f, and then there is another term coming from here, plus twice, which I don't put, g vx vy, okay. And this is times x prime, y prime. And then I have a similar term with y prime squared. I can guess it, of course, by symmetry from looking at the line above, but let's do it again. For example, this is e uy squared, okay. Let me do it, plus 2f, y prime squared comes only here. So 2f, I already put the 2. So uy vy plus g vy squared, y prime squared, okay. So if the question was, which are the coefficients, e tilde, f tilde, g tilde, well, now you have the formula. But this was unreasonably complicated, okay. Notice, actually, you could have answered in a smarter way because this is, after all, it's a quadratic form. So basically, how does a quadratic form changes if you change the basis with respect to which you write it? Because basically, e, f, and g are the coefficients of the quadratic form if you write the quadratic form in terms of the basis x u, x v. E tilde, f tilde, g tilde are the coefficients of the quadratic form if you write them in terms of x compose phi x, x compose phi y, okay. So these are two basis of the tangent space by definition, but by construction, okay. And so, this is a general linear algebra fact. So in fact, you know that e tilde, f tilde, f tilde, g tilde, I mean, if you write the correspond, the representation in matrix terms, this has to be the Jacobian, okay. So the change of vectors, how do vectors change like the Jacobian of the map? So it's u, x, u, x, u, y, v, x, v, y times e, f, g, times its transpose, okay. So u, x, u, y, v, x, v, y, okay. So this was kind of the shortest answer. Of course, if you perform the computation, you have to get the same thing, otherwise we did the mistake, okay. But what I want to show you, and it's something that everything would have been extremely simple. If we had introduced a notation, which I want to introduce now, which formally remembers everything we set up to now. So I want to say, when I say the first fundamental form is equal to an object like this, e du squared plus 2f du dv plus g dv squared, okay. Now these symbols mean nothing, okay. So we have to give some kind of operative meaning to them. Of course, I want to give the meaning which doesn't confuse me, in which sense. In which sense that these objects, these infinitesimals if you want, are something which behave like this. dv, or du, should in some sense satisfy the chain rule, okay. So this is, when you have a re-parameterization, so if u is a function of x and y, I say that du is equal to du dx plus uy dy. And dv is vx dx plus vy dy, okay. So in some sense these are just formal formula, because of course, I don't know really what these are, but I'm saying they are something which satisfy the chain rule formally in this sense, okay. Well, but if I play this game, and if I say, well, the first fundamental form is not really a matrix with coefficients e, f, and g. But it's an object like this. Then, if I had asked you the same question as before, you would have answered in one line. Because now, how does it change when you have a re-parameterization? Well, of course, e, f, and g are functions, so they don't change. They just change the point where you compute them, but they are the same functions. And then you put du here, and du squared, du and dv, and you formally multiply, okay, and dv squared, okay. And you would have got in one line the same result. It's a very efficient way to solve somehow this problem, okay. What I'm trying to convince you is that, but if you look it in this form, you may come up with a revolutionary idea. After all, the first fundamental form remembers, by definition, everything of the intrinsic geometry of the surface. Intrinsic means it depends only on the first fundamental form, okay. So, in order to study the intrinsic geometry of a patch of a surface, I don't really need to know x, strange enough. The only thing I really want to know is the domain and three functions, e, f, and g. That's it. Once I know that, I put them in this picture, and I can do whatever I want. I can compute the Gauss curvature, I can do, and I know how it changes if I reparameterize, okay. So, in some sense, I could study, I could be tempted to say, well, let's say that a surface, now there is a problem. There is a difference between a surface and a patch of a surface. Let me switch from one to the other now. But in principle, I could study the intrinsic geometry of a surface just by looking at the planar domain and three special functions, okay. I don't really care about pictures, okay. So, all the geometric intuition goes away. It looks like a pity, but remember that you are a bug, okay. So, this is exactly what you have to do if you want to play the same game in general, in higher dimensions and so on, okay. Now, let me show you in practice how this can be used. In principle, you see, once you give me, so the point is, is it true that anything like this comes from a surface in R3? See, this is a fundamental question at this point because what we know is that if I have a surface in R3, I can construct these strange objects. But you can ask the opposite question, which is a key question. If I give you a first fundamental form on some planar domain, is there a surface in R3 which gives me this as a first fundamental form? The answer is amazingly no. So, if you do that, you have enlarged by a lot the spaces of objects that you are studying, okay. And let me show you an example. The key example, the king of the examples, okay. The hyperbolic plane, well it's not easy to prove that it's not a surface in R3, but okay. So hyperbolic, let me call it hyperbolic models, okay. Because I want to give you more than one possibility to represent this object. Well, the first way to produce a surface with some special property is by giving you the model, what is called the Poincaré disc. Well, what it is, this is a disc. So, remember now, I hope I convinced you, it's enough I give you. The domain and a first fundamental form, okay. So the domain is really the disc. In the plane, you take x, y, well u, v, doesn't matter. U and v in R2, such that u squared plus v squared is less than one, okay. So the disc of radius one, and that's it. But now I declare the first fundamental form of this object to be four times over one minus u squared, basically the norm of the point, squared, squared. This looks a bit too much. Let's see, there might be a square. Yeah, no, I don't think there is this square here, okay. I've put, I exaggerated with the squares, okay. Times du squared plus dv squared, okay. Now, of course, not everything would be a good candidate for the first fundamental form because it has to be kind of a quadratic form. So in particular, it has to be, well, the fact that it's a symmetric object is contained in the fact that I write it in this form. But of course, you see that this function is always positive. You see, this is never one. So I'm never dividing by zero, okay. And it's always a positive function, thanks to, well, even without that, okay. So because E and G, well, the determinant of the first fundamental form certainly has to be a positive function. Otherwise, you cannot call it the first fundamental form, okay. It's the capital E, capital G minus capital F squared has to be a positive function, okay. So I cannot really throw functions just as I like, okay. There are some constraints. Now, the second, well, there is another, let me, this is kind of a side remark. It's not essential, but I would like you to think of this. If we put, if we think of this disk as the disk in the complex plane, after all, R2 is equal to C, okay. And U plus, U plus psi V, I call it Z, okay. Well, formally, I can play the same game as I did before, DZ. DZ is just DU plus psi DV, okay. And DZ bar is DU minus I, the IV. Essentially, I'm telling, I'm saying it has to be complex linear. So this I goes out, okay. Formally, it's just my, when you set a convention, you can do whatever you want. So it's, but it's very convenient. So let me say, how would this look in complex coordinates? Well, in complex coordinates, this would be four divided by one minus the norm of Z squared because U squared plus V squared is Z squared. Okay, squared, DZ, DZ bar, okay. Let's check it. Well, the function is the function, but you just have to check that DZ, DZ bar becomes DU squared plus DV squared, okay. Second model. So upper half plane. Now, I consider the domain U to be the upper half plane. So X, Y in R2 such that Y is positive, okay. Well, in this, on this domain, what I declare to be the first fundamental form? Well, I declare it to be I is DX squared plus DY squared. You see, DX squared plus DY squared or DU squared plus DV squared is the first fundamental form of the plane. E is equal one, G is equal one, F is equal to zero, okay. So basically, this object here would be the Euclidean plane, the Euclidean disk, okay. Metrically, okay, geometrically. So this function is kind of giving some kind of distortion. And here they get the same. I take DX squared plus DY squared, which is the planar, the Euclidean first fundamental form, but I divide it by Y squared, okay. Again, Y squared is a positive function. It's non-zero because I'm restricting myself to the non-zero Y, okay, positive in particular. And again, of course, in this case, if you want to write it in complex notation, and now let's call it W, the X plus YZ. So X plus Y. So if W is X plus IY, this becomes, this is simple because Y becomes just the imaginary part of W, okay. So this becomes DW, DW bar by the same exercise, okay. So this is DW, DW bar divided by the imaginary part of W squared, okay. In the last five minutes, I hope I can try at least to convince you that these two objects are the same, are isometric, okay. The same means are isometric. It is a transformation from the disk to U, which takes the first fundamental form of the disk into the first fundamental form of the Harper-Hawk plane, okay. So, so observation. These two, in fact, let me, so you start going in the right direction. D, and let me call it ID and IU, okay. So the first fundamental form on the disk and the first fundamental form on the Harper-Hawk plane. So D, ID and U, IU are isometric. Well, in fact, I give you directly the isometry and then we check it is indeed an isometry. So phi from D to U, which takes, that's, this is a moment where it's very convenient to have complex notation. I take the point Z to the point W, which is equal to what? It's equal to I times one minus Z divided by one plus Z. In fact, the inverse, let me write, because it's a stupid exercise with complex numbers, so let me write it down. Z becomes equal to I minus W divided by I plus W. Okay, this is the inverse of this function, okay, which in fact, automatically tells you, the existence of the inverse automatically tells you that this map is one to one, okay. It's certainly a smooth map. You see, Z is never minus one because we are on the disk of radius one. So it's, it's differentiable. But if you do that, let's play formally the game. How much is DZ? Because I want to know, so okay, you can check. Check that this map is one to one, so onto U, okay. So the point set part of this exercise, I don't do it. Okay, let's go directly to the first fundamental forms. How do I see the change of the first fundamental form up to this reparameterization? After all, if it's a one to one map, and it's Jacobian is never zero, I can think of it as a reparameterization. So I'm doing exactly what I told you before. But then I need to know, what is DZ in terms of D of W and DW, okay? Well, DZ becomes nothing but minus two IDW divided by I plus W squared, okay? No mystery, yeah? I mean, you use the formal property of the differential. Okay, let's start calling giving names. So the differential of the coordinate Z with this function here, okay? And DZ bar, well, becomes just the bar of this. So this becomes two IDW bar divided by minus I plus W bar squared, okay? But then let's go here. Take this one and write it in terms of W. Being an isometry by the theorem, okay? An isometry means that if phi is an isometry, if and only if, when you write this with this transformation rule, you get this. So now it's just a matter to check if this is true or not. Well, there's nothing really to learn, okay? Do it, there is nothing. I mean, it's just a replacement of symbols here and check that you get this, okay? So in mathematical terms, we would write phi takes ID to IU, okay? So it's an isometry. Now, well, the lecture is over. So just in the last, I want to speculate. Next, tomorrow, we will speculate on this. But in principle, let me just tell you, if you want to draw the Poincare disk, what do you draw? Well, you draw a disk. What else? It's, as a set, is clearly just this. But having declared a very strange first fundamental form means that if I want to measure the length, for example, of this curve on this surface, okay? Well, of course, if I look at here, it becomes, if I can compute it in the Euclidean sense, but this is telling me it's completely wrong. The length of that curve will be the integral. I parametrize it and I take the integral of the square root of E, which is this one, times U dot squared plus 2F U dot V dot plus G V dot squared, everything on the square root in DT. And of course, this function here becomes essential, okay? So in particular, even though you have only 24 hours, you can check. Try to take as a curve the radius up to some point, okay? And see, I'm not asking you exactly the length of this segment, but see what happens when you go to the boundary, which is the length, the limit length of the segments, of the radial segments. You see, the functions, well, you have to check because what I'm going to say does not imply strictly, but certainly tells you that there is something strange. Because the function, when you go to the boundary, the functions here you have only E and G, because there is no F. The functions E and G are blowing up. So in print, and of course, U dot and V dot are constant. I'm taking a line, okay? I can, for example, if you take the bisectrics, you are taking U dot is equal to one and V dot equal to one, okay? So you are really integrating this object here, this function, twice, okay? So the point is, is the limit of this function of the integral finite or infinite? The function is blowing up, you have to decide, okay? But it's true, your colleague is right. It will blow up. It will blow up means, geometrically, that this point on the boundary, on the Euclidean boundary, is actually at infinite distance from the origin, okay? So that means that if I want to imagine a picture, I should imagine, I don't know, something like this, something like a paraboloid, okay? Where the boundary of the disk has gone to infinity, okay? So if I want to have an R3 picture in my mind, I would try to think to something like this. But then the problem is, we will compute the Gauss curvature of this object, because of course, we have the first fundamental form, so we can take the formula coming from the Theorem egregium, no matter how complicated it is. There is a formula, we can use it, and we will get that this object has constant curvature minus one. And now the question is, can you produce me a model in R3 of constant curvature minus one? Well, you know something, the pseudosphere or something like that. But are they homeomorphic to this? Because after all, if this, no? Are they the same thing or not? And the answer, but it's tricky, we won't see it, is no. There is no model in R3 of this geometric object, okay? We will go on tomorrow.