 OK, so this is where we left last time, OK? We have now, of course, here summarizing a bit quickly. You take a curve, a curve on a surface, gamma, no arm assuming it's parametrized by arc length, OK? Because otherwise, you reparametrize. Even though, remember, this is not an easy thing to do in practice, OK? Because explicitly computing arc length is not really a nice thing, OK? But in any case, you have your curve gamma here, at least these are the equations for being a geodesic, at least for the part covered by a patch, OK? Because these are really differential equations that the variables here are the functions u of t, v of t, OK? So really, the unknowns here are the corresponding functions which define the inverse image on the domain which defines the chart, OK? So of course, once you solve them here, you write gamma as the image via x, OK? So knowing the solution to this system of equation is the same thing as knowing gamma and vice versa, OK? Because x is a differential, OK? So this holds for any piece of the curve gamma which is covered by a patch, OK? Now, since the whole surface is covered by patches, piece by piece, you can recover the whole thing, OK? Now, actually, I slightly changed the notation since last time, but I went back. I mean, now I'm using our standard notation. So e u means partial derivative with respect to u, e v partial derivative with respect to v, and so on, OK? Trying to be a bit quicker in writing formula. And remember, the key geometric interpretation of a notion of a geodesic is that as a curve in R3, if you look at the curve in R3, you, in principle, you could compute its acceleration. And then something is a geodesic on a surface, f if and only if, s if and only if, its acceleration is all normal, OK, normal to the surface, OK? So if you leave on the surface, you would say, and you are moving along a geodesic, you would say that there are no forces acting on you by Newton's law, OK? Acceleration and forces should be the same thing, OK? So for you, moving on a geodesic means there are no forces, you are in quiet, OK? That's why they are so important also in physics, of course, in mechanics, OK? These are the geometric correspondence of uniform motion on a surface, OK? So now the point is, well, everything is beautiful except the shape of the equation. Now who is telling us that they exist, OK? We are taking the usual mathematical approach to a problem. If something exists, it has this, this, this, and this property, but sooner or later we have to face the problem if they exist or not, OK? Now, well, I won't prove it, but I will state and argue why they do exist, at least in some form, OK? Well, let me state first the theorem and then I will comment on it, OK? So you take, for any smooth surface, this is a local theorem, so you don't care compact, non-compact, connected, non-connected, so everything will hold on a connected component of any type of smooth, regular surface. So for any p, so whatever point I pick on my surface, s, and any tangent vector I pick, so if I what would be the ideal thing? The ideal thing would be if I take a point and a tangent vector to the surface, I would like to prove that there exists one geodesic passing through this point with this velocity at this point. Now, why I'm saying this is the ideal theorem? Because what kind of equations are these? Well, these are ordinary differential equations because these are all functions of one variable, t. The unknowns are u and v. The functions u of t, v of t. E, f, and g are data of the problem because they are determined by s. So given s, e, f, and g are functions that you know. So the unknowns are u and v. And they appear in these equations like what? Second derivatives, because of course here I'm writing them in a compact form, but in this term, there would be the derivative of this plus the derivative of this. Once I take the derivative of this, I get second derivative of u and second derivative of v. So this is a system of two equations, two unknowns. Second order, nonlinear. There's a square. They don't depend linearly on the unknowns, these equations. So in order to state some kind of Cauchy problem for this, what do you have to do? It's a second order system. So you have to assign me the initial value and the initial value of the first derivative. So what is the initial value? u of 0, v of 0, for example. Suppose you are starting at time 0. It's the point p. And what are the initial value for the first derivatives? It's v. It's the initial tangent vector. So the ideal Cauchy type of problem would be give me a point, give me a vector, prove that there exists a geodesic passing through this point with this velocity at time 0. That's exactly what happens, but with a little bit of care. So given any point and given any v tangent to the surface at this point, there exists a unique geodesic on s, of course, passing through p with velocity v. Now we have to decide sooner or later. There is one thing which is almost irrelevant, but we have to decide our convention. Geodesics do not need to be parameterized by arc length. And the system I wrote here is true only if you assume that the vector, the tangent vector, is of norm 1. So if I want to argue with these equations, I should put the restriction that v is a unit vector. On the other hand, let me not write the general system. But whatever I will say, it will hold for the general. Remember the one with the capital R. If I use that system, that would be the system for a geodesic with any parametrization. In particular, I can drop the assumption that v is of norm 1. Now, here the two words are both equally important. Existence and uniqueness. So why this is true? Well, I'm not going to prove it. I'm just guessing where to look in your favorite book of ordinary differential equations. Because the point is that this system, if you expand these derivatives, and even if you put R, nothing will change in what I'm going to say. If you expand these things, you can write this system as a system of the form u double dot equal to some function. Let me call it capital A, a function of, of course, E, F, and G. But this is data of the problem. So with respect to the variables, this will be some function of u, v, u dot, v dot, nonlinear, but still a function of this form. v double dot, it can't be linear. There are these quadratic terms. But still, the fact that they are just quadratic, it's important. v double dot will be another function of the same form. I switched the order, but it's the same. So once you write them, the point is that the coefficient of u double prime and v double prime is never 0. So in principle, you can divide. And it's never 0 because it's the coefficient of, for example, u double prime would be E. Why this is true? What are the coefficients? How do I reduce a system like this to a system like this? I isolate, here I would get E double prime plus F v double prime plus 200 terms. I don't care. The other 200 terms, I put it on the right, and they become function. Here I get F u double prime plus G v double prime plus another 300 terms that I throw it on the left, on the right. So why such a system can be reduced in this form? Well, why? Is it true or not? Well, of course, it is true because what? The determinant of the 2 by 2, you look at the 2 by 2 matrix, given the coefficients of the higher order terms, is what? It's EG minus F squared, which is never 0. And this is, of course, not an accident. It's the determinant of the first fundamental form, which cannot be 0. So you can reduce such a system to something like this. And the fact that, here, the coefficients, in general, here would have functions, no? F is a bad word. Capital phi, capital psi, some other functions that, in principle, where it's difficult to solve such differential equations? Where these functions have zeros? If you take your favorite notes of ordinary differential equations, this is the problem. If these are not there, there are very general theorems telling you, at least locally, for short time, such as solutions exist. A system like this have always solution. But the solution is defined only, so key observation. The solution is indeed unique. Existence is unique, but it's defined only locally. So suppose that you are deciding that time 0 is the one which is good for your point. So you are in a neighborhood of 0. This will be defined only on minus epsilon epsilon. The value of this epsilon depends on the surface, on all the data of the system. The data of the systems are the surface coming in with the functions E, F, and G. The point and the vector. OK. So this is something you can really check. But is this kind of an analytic pathology or it's intrinsic in the problem? Well, example 0. Remember, I suggested you every time you introduce a new notion coming from some intuition, go back to the intuition after you did something to check. So we knew what geodesic should be on the plane. So is this system giving us what we knew? If you are taking as a surface the plane, what's going on here? Well, E, F, and G are what? E is 1. G is 1, and F is 0. So in particular, every time I take a derivative, so the right-hand sides disappear. And here I get just F dies. And here I just get D in the T of 1. So this becomes U double prime is equal to 0. And this becomes G double prime equal to 0. Is this what we knew? Yes. Because now to give a solution to this system, well, of course, you have to give me U0, V0, U dot 0, and then it's game over. There is only one solution. It's linear. So U and T will be linear functions with these initial values. So straight lines are geodesics on the plane. And they are the only geodesic, actually. Doing the mathematics behind the intuition has not introduced any new object. But now let's behave like mathematicians. S is the plane. And geodesics have become straight lines. I want to analyze a little bit this phenomenon. Straight lines are defined from minus infinity to plus infinity. So you say, OK, example 0, so I guess that this phenomenon is stupid. It's just because my notes of ordinary differential equations was written not sufficiently well. Is this the case or not? No. There is no way you can improve your notes in ordinary differential equation. Take, now, example 0 prime. How can you create a situation where this epsilon is not 0? It is not minus plus infinity. Just remove a point. Take the plane, for example, minus the origin, OK? So you remove a point. Now what happens? So it is a surface. It's an open piece of a surface, so it's a surface. Now take any other point and try to solve the geodesic equation starting from this point with some direction. Well, if the direction doesn't see the whole, it's exactly as before. But if the direction points to the whole, the geodesic must be defined only up to here. Because actually the origin is not the point on the surface. So you cannot add it. You are not allowed to add. OK? So you see, this is a situation where this geodesic will be defined from, I don't know, minus 25 to 25, OK? Now this looks stupid, but it's not stupid. I mean, you can put holes on a surface wherever you want. And then if you are away from the holes, the geodesic at some point has to stop. There is a more subtle way to have this phenomenon. And that's what, in fact, with this excuse, let's study the problem as an exercise. Let's see what happens to this system to the case of surfaces of revolution. So on a surface of revolution, what can we say about geodesics? You see immediately that the problem is highly on trivial. So remember our notation for a surface of revolution is something like this, at least a piece. I mean, the whole surface minus one of the curve which is rotating. I can write it in this form, f of v cos u, f of v sin u, g of v for some pair of functions f v. So this is rotating around the z-axis. So v will be a parameter defining the rotating curve. So here I have some curve which is given by the functions f and g. And so this will be given by A and B. I don't know. But u, I take u, for example, between 0 and 2 pi. Doesn't matter. Otherwise, I take another stripe. And the only condition I put is that f is a positive function because I don't want the curve to touch the axis of rotation in order to have a genuine surface. So now you rotate this thing. I don't remember if we did the exercise explicitly, but it's very simple to compute e, f, and g. So let me tell you directly e, e, f, and g in this case is what? This is f squared. I will drop as usual the dependence. f is a function of v and g is a function of v. So e is f squared, f is 0, and g is a bit more complicated. f prime squared plus g prime squared, where the prime, of course, means derivative with respect to v. OK. Well, with this data, what the system becomes? What does the system become? You see, f equal to 0, of course, is beautiful. There is another beautiful accident is that, of course, everything depends on v. These are functions of v. These are functions of v. So the first equation becomes particularly simple. So the first equation becomes d in dt of f squared times u dot is equal to 0. OK. The second equation is a bit more complicated, because, of course, e now depends on v. So if you take the derivative with respect to v, you get what? So let me write the second equation. The second equation is d in dt. f is not there anyway, but here, f, let me write it still g. I don't want to put everything there. g v dot is equal to what? 1 half e v. e v is 2 f f prime u dot squared plus f is 0, so plus g v. Now g v is what? It is 2 times 2 f prime f double prime plus g prime g double prime v dot squared. OK. So these are the equations we have to solve. Now let's expand this derivative in t now. So f is a function of v. So the first equation becomes f is a function of v, and v is a function of t. So when I take the derivative of this function, I have to take the derivative with respect to v times the derivative times v dot. So this becomes twice 2 f f prime v dot u dot. And this is the derivative of f squared times u dot plus f squared u double dot. So this is the first equation. I don't tie it. At this point, it's a matter of ingenuity. Do whatever you want to solve it. I'm taking the brute force line. Of course, what your colleague is suggesting is that it's a pity. Once you know a first integral of the equation, it's a pity to forget it. So you can integrate out of here, but use the way you want. The second equation becomes what? OK. Let's expand everything now in the t of g. g is, again, a function of v. So we play the same game. This becomes g is here. So this becomes the derivative with respect to t of that function. This is twice. f prime f double prime plus twice g prime g double prime. Everything times v dot. And this would be the derivative with respect to t of this. Times v dot. So this is times v dot squared times plus. Now I put the explicit expression for g. So f prime squared plus g prime squared v double dot. So this is the left-hand side. On the right, we get is equal to ff prime ff prime u dot squared plus ff prime times, if you want, u prime squared. I don't know. ff prime u dot squared plus f double prime plus g double prime v dot squared. Modular mistake. This should be the same thing. Which, of course, f is a non-zero function. So I can divide everything by f or x squared. So the first equation becomes u double dot. Now I divide by x squared. And then I just rewrite essentially. This is equal plus 2f prime over f v dot u dot equal to 0. The second equation, I want to write it in the form u double dot equal to something v double dot equal to something. So I want always to divide by the coefficient of the top order derivative, which is this. So I divide everything by this one. More or less. There was this 2. This 2 was not a mistake. So when I put it here, so this one, if I take it here, becomes a minus. So this becomes v double dot is equal to, or, well, v double dot, if I want to write everything on the left, minus ff prime u prime squared. And then I will divide it by f prime squared plus g prime squared plus f prime f double prime plus g prime g double prime v dot squared divided by f prime squared plus g prime squared. This is equal to 0. Give me just one second to see if it's compatible with my notes. Yes. OK. It's not squared, and it's prime. OK. Now, OK, well, you look at this. Remember, f and g are basically any function, because any function provided f is not 0. So if you hope to write down the solution, you know that the solution exists. So now you think of your surface as suppose you are rotating this thing around some point, around some axis, OK? So here there is an axis, and here more or less there is the same curve on the other side. So the general theory is telling you what, is that if I pick any point here and then a tangent vector, there should be a solution to this object, to this monster. OK? Nobody is able to write it down. There is no way you can write down such a solution. OK? So you have to argue a little bit geometrically, at least to have some intuition of what's going on. OK? For example, clearly there are special curves on such a surface. There are two special curves, meridians and parallels. OK? The curve which is rotating is a special curve. And the curve of rotation, which is always a circle, but it's a circle of changing radius, OK? These are two curves which are kind of too special not to be analyzed by hand. How do I distinguish a meridian? A meridian is this one, no? It's the profile of the curve which is rotating. Well, to go back to the parameterization of the surface, how do I get a meridian? I get a meridian by fixing u. u is the parameter of rotation. u is the one which goes from 0 to 2 pi going around. So if I fix u, I'm taking one of these curves, OK? So meridians, for example. These are given by u of t constantly equal to some u naught. Doesn't really matter which one you pick. OK. So is this a solution? The point is, you see, u of t is something like this. I can fix it geometrically. Then v of t is what? I don't know. In fact, it will be a function which will force the meridian to be parametrized with arc length. Otherwise, I cannot use this system. Otherwise, I should use the general system, which is a mess. OK. So this is, I cannot say anything about this, OK? But just because of this, of course, the first equation, it's OK, because there is everywhere a derivative of u, which is 0, OK? So the first equation is OK. What about the second? In the second equation, the only simplification at first is that this term disappears, because there is a u prime, OK? So what about everything else? Well, everything else, let me not write it, because the blackboard is not too big. So now we look at this. Now let's try to get this bit of information from the fact it's parametrized by arc length. So v of t is a function for which gamma dot, so I'm looking for gamma, which is x, if you want, directly u of t, v of t. So gamma dot has to be of norm 1. How much is gamma dot? But gamma dot, by general fact, I mean, we know gamma dot. In fact, let me write down directly the square. Gamma dot square is what? It's e u dot square plus twice f u dot v dot plus g v dot square. This is why we studied the first fundamental form, OK? This is the definition, if you want, of e, f, and g, OK? Now in our situation, this means what? f is 0. In this specific case, u dot is 0. So now you see that I get some equation, some information about v out of this. In fact, the only information I can hope. So this means that v dot squared times g. g is f prime squared, g prime squared. This is equal to 1, OK? And of course, you can write it however you like it. But then you see, v dot squared, OK, let me write. v dot squared is equal to 1 over f prime squared plus g prime squared. I take the derivative with respect to t of this equation and I get something. v double dot is not free. I want to know why it's not free, how it's not free. So to know it, I take this one and I take the derivative with respect to t. If you do it, you get this equation here. So this implies by taking d in dt that twice v dot double u dot, this is the derivative of the left hand side. And then simple manipulation tells you this, OK? f prime f double prime plus g prime g double prime divided by f prime squared plus g prime squared times v dot cube. Now, the only thing you have to observe is that v dot cannot be 0. Because if v dot, you see, I would really like to remove a v dot. I just have to argue that v dot cannot be 0. Otherwise, since already u naught is 0, this curve is not a curve. It's a point, OK? So v naught is never 0 anywhere. So I take it away. And so I get one equation for v double dot. Remember, this is coming just from the fact that this was parametrized by arc length. Now, from here to here, I have not used any geodesic equation, OK? Only arc length. But you see, if I've not done mistakes, you are getting exactly the same equation. So arc length implies geodesic in this part of the geodesic system. So the first one is verified because brute force. The second one is verified by arc length, OK? So all meridians are geodesics. So if by accident, you had a point, a point is always on a meridian. There is one meridian passing through this point. But if your tangent vector was pointing in this direction, then the geodesic given by the general theorem had to be the meridian by uniqueness. Remember, the theorem was telling you, exists and unique. But of course, this is a very small set of tangent vectors. What about some other tangent vectors? For example, are parallels geodesics, may I raise this? Because I would really keep the equations, OK? And I would like to play the same game with parallels. Parallels means what? Now it's only the rotation parameter, which is free. Meaning these curves are given by v equal to a constant. So now you fix the point on the surface, on the curve, and then you rotate it freely. This is a parallel. So this becomes v of t is equal to v0. And u of t is a function for which this becomes arc length. Of course, we are going to play a similar game. Now don't be mistaken. So now this cancellation is not there, because we did it for the meridians. Now v of t is equal to v0. So then we know what v prime, now this time is v dot, which is 0. So the first equation becomes u double dot is equal to 0, which is already telling me it's not automatically satisfied. It means u is linear. It's telling me something. What about the other equation? Now v dot is 0, which is, of course, much nicer than before. This is 0. This is 0. So I'm left with, so the system becomes just minus. And I can throw away the f now, because f is positive. So minus f prime divided by f prime squared plus g prime squared times u dot squared equal to 0. OK? So this is the geodesic system. Let's put in the game the arc length equation. Again, the norm of gamma dot squared has to be constantly equal to 1. OK, I don't remember. This is e. It's e u dot squared and nothing else, because then there would be 2f u dot v dot, which is 0, plus g v dot squared, which is 0. So now this is the equation for arc length. And how much is it? Well, f e is equal to f squared. So this is telling me that f squared u dot squared is equal to 1, constantly equal to 1. And now I have to speak incredibly loud. I mean, five orders of magnitude. There was an f here, but it's positive. It's a positive function, so I threw it away. It's equal to 0, so I can. In fact, let me throw away also this one for the same reason. The only thing I don't know is f prime, in principle, can be whatever it wants. So how to, just one second, so I have this equation. Well, OK, that means being parametrized by arc length, forces the function u dot squared to be equal to 1 over f squared. f is non-zero, so I can divide without nightmares. But then I plug into the geodesic equation. And this becomes what? This becomes minus f prime over f squared equal to 0. f squared, I don't care. So I'm left with the condition. Now it's not true that the arc length equation implies something on the geodesics. It does only if something happens. And what is it? It means that f prime, so a parallel, a parallel. The moral is that the parallel is a geodesic, if and only if. Actually, a parallel suitably parametrized by arc length. Now we are grown up, and I stop repeating a parallel, because the parallel means just the geometric object. Why? The geodesic is something which comes with a parametrization. But now you are old enough to understand all this. So parallel is a geodesic if and only if it was obtained by rotating a point for which this was true. But geometrically, this is very clear. Which are these points? So it's not true that every point on the curve generates a geodesic by horizontal rotation. It does only if the point that you are rotating, essentially, it's a critical point of the graph. Meaning this one is a geodesic. So just by looking at it, now I can tell you this one is a geodesic, this one that here there is a geodesic, and so on. But nothing else among these curves. So you see the situation is very depressing if you want. If our original point was take any point and any vector and tell me which is the geodesic, now I can tell you if the tangent vector was tangent to a meridian, OK, the meridian works. Otherwise, even if it's horizontal, which is the most stupid geometric situation, only very few points are good, unless the curve that you are rotating is very special. Because, for example, if you are rotating a straight line, a vertical straight line, this is always true. And it's a cylinder, and everything is a geodesic, OK? So now you would like me to show you with some brilliant idea how to solve the general problem. And the point is that I'm not able. I don't know. Case by case, there are, of course, refinements of this. There are quantities that are preserved along geodesic that you can study, like some kind of momentum of some motion and so on. But in general, if you hope to find a solution to this problem, game over, OK? Case by case and argument, idea, single ideas for specific things, OK? But now let's see for a moment what there is at least. Remember, my philosophy, example 0 is the plane, and we did it. Then I jumped to example 2. And none of you objected that there was not example 1, OK? You are a bit too shy, OK? What is example 1? Any theory, whatever you prove, first surface, it has to be checked immediately for the plane without even writing, I mean? And three seconds later, and you are allowed to write, OK? The sphere is not always as easy as it looks, OK? Now, for example, in this situation, for this surface, what can I say about geodesics? Well, again, as usual, there is not one solution, OK? You can argue however you want. You are free people, OK? But since we've just done the exercise for surfaces of revolution, and the sphere is the queen of surfaces of revolution, let's see what this is telling us, OK? Is it a surface of revolution? Well, remember, it's kind of a fatal, I mean, we are at the border, OK? Because now the curve which is rotating does not really satisfy F positive, because actually it touches the axis of revolution. So little care about considerations which depend on, at this point and at this point, you have to be a bit careful, OK? At the end, if you think you realize that actually the sphere behaves exactly like a surface where there is not this problem. But now, well, the profile of the curve which is rotating is what? I can tell you, but you can also write it immediately. In this case, x of uv is cos u. Well, now it's just a matter of switching sin and cos sin. So cos u, cos v, for example, cos u cos v sin u cos v sin v, OK? So F, I decided that F is cos v and g is sin v, OK? And then what have we learned? Well, on any surface of revolution, meridians are geodesics. What is a meridian in this case? It's half of an equator, OK? This is the curve which is rotating. So this is what I call the meridian in the previous blackboard, OK? And then strictly speaking, what else have we proved? Well, which parallels are geodesics? Well, only one, the equator, the horizontal equator, OK? So the previous blackboard is telling us on the sphere, anything, I mean, any section of this type is a geodesic, and the horizontal is a geodesic. So you see, it's very little. But now, so now the question is, give me any point and any tangent vector to the sphere. Now, in this case, I want the solution. I don't accept how you tell me, ah, you see, the system is very complicated. Geometry must be analysis here, OK? OK, so what your colleagues are guessing, sorry if I have to repeat, but I have to repeat for our unknown viewers, OK? So what your colleagues are guessing is, well, why not taking a section, not really the one which is rotate. So in order to apply the previous theory, I need to take a planar section, but a plane which contains the axis of rotation, OK? Otherwise, I cannot apply the previous theory. Now, your colleagues are guessing, well, OK, in some sense, I feel I know the answer. So the answer should be, cut it with a planar section, but now this plane has nothing to do with the axis of rotation. Or does it, or that axis of rotation? The sphere is invariant by any rotation. So why did I pick this? Give me a point and a tangent vector, build the axis of rotation with this. Exactly as your colleagues are guessing, but you have to check one little thing in analytic geometry. One point, the center of the sphere, because you need to check that you get a curve. This plane that you would like to take is well-defined or not? You are suggesting take the plane which passes through the center, passes through this point, and contains this vector. But is this well-defined? Yes. This is enough in R3 to determine one plane, one and only one. So game over. Take it and then decide that you are rotating the sphere with respect to this. And now you are back to the general theory. So you have the complete picture for geodesics on the sphere because you have existence and uniqueness in the geodesics. So now for any point and any vector you have produced one. There can't be more than one. So these are all. So we recovered what we thought should have been true right from the beginning. On the sphere, the geodesics are great circles or pieces of great circles if you take just a piece. So this is another situation. But now if you think again, now think again to the problem of minus epsilon epsilon. This is a more interesting phenomenon than just putting a hole on a surface. I can produce a geodesic which is not defined from minus infinity to plus infinity just because I put a hole. Now here the point is a bit more delicate because suppose you start here and you say, and suppose even you take this tangent vector up to a rotation, this is the general situation. Up to a rotation, this is the general picture. So we know that this is the geodesic. And it starts if you want from minus something to plus something. How long can I extend it? But it seems like paying the price of, of course, this curve, I mean, the map does not face to be injective. You start going around infinitely many times if you want to get here and go on. So now you have to decide. This is a problem of convention because you can accept this as a good thing. And you say, well, the geodesic is defined from minus infinity to plus infinity. But usually you don't want this. You want the geodesic not to have, I mean, you want the map which gives you the geodesic to be injective. So this is kind of another phenomenon that was not expected on the plane, of course. On the plane, two geodesics have at most one point of intersection. They could have zero. But at most one. On the sphere, in fact, they must have two. So given one point, the antipodal point is another point where the geodesic touches again, closes again. These are called focal points on a surface. These were points born in optics, not strange. If you think of geodesic as the curves which the light, for example, follows on a surface, on a membrane or whatever, to diffuse on a surface, you see, if you are here, basically, if you have your head turn right, you see this point. You go around because a light ray goes on. And basically, looking right, you see your back. If our universe was something like this, in this moment, I would be looking at my back. This is so focal points for geodesics are important for this type of thing. Because, for example, again, keep in mind that we are doing all this game partly because surfaces are very important. OK, surfaces are very important. If you want to build a boat or whatever you want, you have to solve problems for surfaces. If you want to build an airplane, you have to find the best shape for the airplane and so on. But actually, more intellectually, we are playing this game because we want to understand which questions we can ask for our universe. So this bug, which watching right is looking at his back, could be us. As long as this universe would be maybe a 4, 10, 12 dimensional object, it could be us. So remember, I mean, when you look at the stars, are you completely sure that those stars are not your back? This is a good question. And in fact, I'm not even sure at which stage we are in current research in cosmology to decide. In principle, the stars that we see could be always the same thing repeated, infinitely many times, because rightly light rays go on and on and on and on and on. It's always the same thing. It's just repeated in another place. So thinking to this simple and stupid problem is actually hinting you to some deep problem for us. Now, in the questions, now in some sense, I want to change a little bit point of view, because now I want to use geodesic and the fact that they exist to solve, or at least to give one possible solution, to another problem that we left right from the beginning of the definition of surfaces. And this would be the same for definition of manifolds. A surface is something that can be covered by patches. But we have no idea if there are better patches or worse patches. And in fact, the problem is also that in general, any surface must be covered with infinitely many, uncountably many patches. So the problem is, is there a better one or not? And we touch this problem when we spoke about isometric patches, conformal patches, area preserving patches. Now, the point is isometric patches exist very rarely. They exist only if the surface is isometric to the plane. Now, is there something we can do with a geodesic that works every time on any surface? And the answer is yes. So think of this picture. So basically now, if you want to give a name to this little chapter, geodesics and coordinates. I want to construct special coordinates which exist on any surface, starting from geodesics. Now, the picture looks a bit like this. I want to construct coordinates around some point P. I pick, I take this point, and I take any tangent vector at this point. If I take any tangent vector at this point, I can apply the general theorem for the existence or uniqueness of geodesics. So here I start with a curve gamma. I don't really care which vector it has there. But at least gamma exists. Gamma geodesics, geodesic, exists, passing through this point. I decide that my time is 0. Gamma of 0 is equal to P. It doesn't change anything. Now I say, well, for any point, and in fact, let me call V the parameter. V is the parameter along this curve here, this geodesic. Now, if I pick a point gamma V, I want to go orthogonal. So at this point, this curve will have some tangent vector, gamma prime. I want to go orthogonal to that. So now I have another point and another tangent vector. So I can reapply the theorem about existence and uniqueness. So at this point, with this tangent vector, I know that there exists a geodesic passing through this point with this direction. And you have to imagine that I want to do it at every point. So V is the kind of a parameter here. At the beginning, I just have my original gamma and some other curve here. And then I move the point on this gamma and I get a family of these geodesics with the property that every time at the points of intersection between P with gamma, the two tangent vectors are orthogonal. I know that I can do this construction by the theorem of existence and uniqueness. Let me give names. So for small values, at least for small values of V, I let gamma V, so now, of course, notations are a bit confusing, but there is nothing I cannot improve it. So gamma V is the geodesic, B, the geodesic passing through. In fact, gamma V of U, I will decide that its parameter is U, passing through gamma of V with direction, with tangent vector, with velocity, orthogonal gamma prime V. So gamma has its own tangent vector at this point. I take the orthogonal direction and I decide and I call this. So this would be, in my picture, gamma VU. This curve here. V is the one which is telling me at which point I stopped gamma, and U is the parameter on the kind of vertical curve in my picture. Now with these two parameters, I define a map, X. X of UV is, I call it, by definition, gamma UV, gamma VU. Now where V was defined, I'm free to bound the domain of definition as I want. Suppose V was defined on some small, OK? Now the point is, can I find a delta, for example, that if U is less than delta, this map is a chart? Because now, you see, so V in the plane, I take V between minus epsilon, epsilon. OK, usually it's the other way now. So this is V epsilon minus epsilon. So the point is, can I find a delta for which this rectangle here is a nice domain for a chart for this map here? Well, you have to argue by implicit function here, and if you want. Because for U equal to 0, what's going on? At U equal to 0, I am here, and V is moving on gamma. So V is the parameter in the horizontal direction, OK? So at U equal to 0, what is the differential of this map? At UV equal to 0. The differential of this map in 0, 0 takes the V direction to the tangent to gamma, and the U direction in the tangent to gamma 0. But this is orthogonal. In particular, it's linearly independent. So at the beginning of the story, the two, the Jacobian of this matrix has rank 2. So for little u and little v, it must have rank 2, because it's smooth. It's a smooth map, OK? It's a smooth map because I didn't tell you that the solution to the system of geodesic equations depends smoothly on the parameter. So in your notes of ordinary differential equation, the theorems are always a solution, exist. It's unique, and they depend continuously. C1, C2. In this case, everything is smooth, and the solution depends smoothly on the parameter, OK? So this map is smooth, because it's actually integrating this system. At times at u equal v equal to 0, the Jacobian has rank 2. So this has rank 2 for small, so there must be a delta for which this has rank 2, OK? So this is a chart. And these are called geodesic coordinates for obvious reasons, OK? This special chart is called geodesic coordinates. Why they are nice? Well, they are very nice, in fact, if you think a bit, because proposition in geodesic coordinates, which notice they exist around every point on any surface, OK? So they are completely general. You don't have to require anything about the surface to work in geodesic coordinates. In geodesic coordinates, e is equal to 1. f is equal to 0. And g is whatever it wants, OK? Now, this is amazing. Think of what all we did up to now, e, f, and g derivatives, first derivative, second derivative, you know? And now we realize we had to do it to construct geodesics. So now you say, why didn't you tell me at the beginning? Well, to prove the existence, I have to go through the theory up to now. But now you stop. You go back, and you say, OK, if I know this, all the theory becomes extremely simple. Depends what you ask, OK? But you see, you have made the first coefficients of this. These two coefficients of the first fundamental form constant. So every time you had the derivative, it was 0. f is 0. So f, in fact, doesn't exist in the theory, OK? Of course, if you work in these coordinates, why this is true? Well, of course, I cannot, g has to be free otherwise. There is no theory of surfaces. I mean, if there was a theorem like, in these coordinates, which exist every time on every surface, e is equal to 1. f is equal to 0, and g is equal to sine u squared. I said, OK, there is one surface up to isometry. So g is the only function free, because there are many surfaces. I mean, it's amazing the fact that it's only one. There's one function which controls the theory of surfaces locally up to isometrics. Everything boils down to this function here, OK? Now, well, I should convince you that this theorem is true. It's not difficult. Let's do it quickly. In fact, the way to prove it is really to make a big picture of this, because proof. In fact, let's do it, OK? Because really, there is not much to write, but there is really a lot to convince yourself. So this is gamma, OK? And here you put gamma v, and here you take, so this is gamma prime. And here you have picked. Actually, of course, there was an ambiguity, but I hope you realize it's not important, because there are two orthogonal vectors, OK? Even if you fix the norm. Suppose you're, in fact, I'm assuming the norm is equal to 1. But even assuming the norm is equal to 1, I would be up and down. Which one do you pick? The one you prefer. Doesn't matter. They will both be geodesic coordinates, OK? There is nothing like an increasing notion of up and down. So make your choice and do it. So this would be, OK, gamma v dot at 0. If I want to write what is this vector, because the parameter on this curve here is u. So this is gamma v u. So what is this vector? It's gamma dot in the sense of, yeah, in the sense that everything is the only possible sense, the derivative with respect to u. Because u is the parameter on this curve, OK? So now let's see. Gamma is a geodesic. And of course, now it's very convenient to have said right from the beginning, parameterized by arc length, OK? Otherwise, actually, this is not true. E is equal to 1, of course. Because what is E? E is, of course, the norm squared of the derivative of x. Well, actually, let me, now I'm switching. If I take, which are the curves, v equal to constant? If I take v is equal to a constant, v is equal to constant means I let u move. And u moving, v is equal to a constant means that the only parameter free is u. And in my picture, it's this curve here. I fix v, and I move u, OK? Now what is E? Capital E is the norm squared of the tangent vector to this curve, OK? So pick any point on this curve. But how much is the norm squared of this? Well, this is a geodesic, parameterized by arc length. I didn't even say it, OK? So E is equal to 1 by brute force, OK? It's a bit more complicated to get F is equal to 0. So E is equal to 1 is really the construction, is that any curve that you draw here is a geodesic. So the tangent vector is of norm 1. Why F is equal to 0? It's a bit more complicated. Because what is F? F is the scalar product between x u and x v, OK? Now the problem is I know everything, but I know everything only on gamma. So if I say F, F which would be a function of u and v. Let's see what do we know. We know that F for u equal to 0. So F of 0 v, so if I compute F on the curve gamma, how much do I get? Well, of course. Easy guess. But it's 0 because I decided right from the beginning that this vector and this, it's the scalar product between this and this. And I decided here that they were orthogonal, OK? So this is 0 for sure. Why it's not trivial? So why can't I just say, and then that's it? Well, because now the problem is what about here? If I'm outside gamma, well, now these curves may be distorted. I started from an orthogonal situation. But then outside the curve gamma, they could have become, they try to become parallel, for example, OK? I have to argue that this is not the case. How do I do it? Well, point is that I go back to the geodesic equation. Since I know already this, this I know on the whole open set, OK? So now I know already that this is a constant. Now I go back to the geodesic equation. Which one? I didn't write now in my notes. So geodesic equation. Remember, d in the t, e u dot plus F v dot equal to something. That looks nice because I know that e is equal to 1. E is equal to 1 also on the other side. OK, let's see. D in the t. I'm pretty sure it's this one, but I didn't check it before. So let's do it. It's e u dot plus F v dot is equal 1 half e u u dot u dot squared plus 2 i's F u dot v dot plus g v dot squared. OK, so let's see. Now the F u and g u. Now this has to hold where? Well, what is the curve which I'm looking at? Of course, it's gamma. The two things I know which are geodesic. So they have to satisfy these are either gamma or these vertical lines. Now let's stay on gamma. So this equation I look again at u equal to 0. What is this telling me? Well, e is equal to 1 everywhere. So this is telling me u double dot. Do that just by brute force. I mean, what I'm sure is that this I can throw it away because e u is equal to 0. Now, e is equal to 1. Of course, I can just disappear. And then F derivative of F v is equal to a constant. Right, sorry. We are forgetting. OK, now if I do it on u equal to 0, if I am taking the curve u equal to 0, that means no, that's not good. So let's do it on v equal to a constant because I want this to disappear. v equal to a constant. Of course, v dot is equal to 0. It's always the same thing. So v equal to a constant, there is nothing on the right. So what is left here? It's left u double dot plus F. This is automatically 0 everywhere on these curves. F prime. OK, because if v dot is 0, of course, this is telling me only u double dot is equal to 0. So this is kind of strange. So let's see what's telling me. The other one, d in the t, F u dot plus g v dot. Maybe this one is more efficient. u v e v u dot squared plus y is F v u dot v dot plus g v v dot squared. And now if I'm restricting myself to the curves v equal to constant, what's happening here? Well, of course, this is not there. This is not there by brute force. Again, this is not there because v dot is 0. And here I'm left with what? Here I'm left with F u. F u u dot squared is equal to 0. Because then I would put plus F u double dot plus F u double dot. So that means I'm not taking this one. OK, this is interesting, but it's not now the moment to put into the game. So F u, because these are on the curves, v equal to constant. Right, who cares? No, no, no, this is the interesting one. OK, so this is an interesting equation. If I put u equal to 0, I get that F u at 0 v at only at those points, see, F is 0. At the point u equal to 0, F is 0. But this equation must be satisfied everywhere. So this means that F u not just at not only F at 0 v is equal to 0, but also F u of 0 v is equal to 0 because of this equation here. OK, but then the other equation was telling me u double dot is equal to 0. So now you have a linear function whose value at 0 is 0, and the first derivative at 0 is 0. So it's 0. OK, and that is what? Another proof or what? Another proof. OK, I accept it without hesitation. OK, no, it's clear that now it's a matter of manipulating the information you have. So there could be many ways to get to the conclusion. One way is to say that this function, the function F, is 0, first derivative is 0, second derivative is 0. So there's not much left. And it satisfies u double dot equal to 0. So it's a linear function starting from 0 with 0 derivative. OK, so that's it. Now, of course, I mean, you can also play the game because now every time, the first time you see this proposition or theorem or whatever, you wonder, well, come on, there must be something on G. How come? You have many equations, many informations. And of course, there are differential equations that G must satisfy. G is not any function, in some sense. So the sentence G is free, you have to take it with care. I cannot determine G. But G, of course, must satisfy some internal compatibility. OK, now, in the last few minutes, I would like to give you an application of this. So in fact, as an exercise, you can check that this construction, exactly this construction on this sphere, gives you, well, a system of coordinate that you know already. Basically, the one we talked about before, in some sense. So let me give you one powerful application of this story. So theorem. Remember that we dream as geometers to have theorems like, if your surface has this and this property in terms of curvature, then it is up to an isometry, one of these. So this is the best theorem in this type of geometry that you can prove. It's the analog of that. I mean, if you have a topological space with this and this property, it's homomorphic to one of these. If you have a vector space with this and this property, it's linearly isomorphic to. In our geometry now, it's up to isometry, one of these. And we would like the Gauss curvature to tell us where we are going. Classifying surfaces by Gauss curvature, we start from constant functions. That's a game we have already implemented up to some points. Now, so now thanks to this technical tool, we can actually close the story for constant curvature surfaces. So a surface with constant with 0. Let's start with 0 Gauss curvature. Constant in particular is local isometric to the plane. 2, a surface with positive Gauss curvature. Let me start shorting up. It's locally isometric. Sphere, of course, a sphere of the appropriate radius, depending on the constant with positive constant. If it changes, remember Hilbert Lieberman. Now, 3, a surface with constant negative Gauss curvature is locally isometric. Do you have one model in mind? Local isometric 2, we saw one. So the pseudo-sphere. Remember the rotation of the track tricks. Remember the donkey and the box. Now, these theorems are not improvable. Well, in the sense that what could be improved? Locally, it's the only weak word in some sense in these statements. You would like to say they are isometric too. But of course, this is impossible. Sealing this, then if you want to drop locally, first you need to drop topology, for example. The cylinder and the plane are locally isometric, but they're not even homeomorphic. So if you hope, if you dream to prove something like they are globally isometric, it means first you are proving a topological theorem, which is completely out of hand. So this was just to comment on the only seemingly weak. I mean, this is clearly a great result. The only thing now, basically, it becomes really trivial if one only remembers or has in the notes. The formula for the Gauss curvature in, remember, that's why when we proved in the Torema-Gradium, I isolated two sub-cases. I told you, in the Torema-Gradium, the point is the Gauss curvature is a function of the first fundamental form and its derivative. OK, but forget the function. Special case, f equal to 0. Very special case, e equal to 1, f equal to 0. Now you understand that it was not a special case at all. It was actually pointed here. Because now we reached the case e equal to 1, f equal to 0. And then you go back to the proof of the Torema-Gradium. And you remember that k is nothing but minus g to the power 1 minus 1 half d in du squared of g to the power 1 half. In geodesic coordinates. I mean, this is true only because we reached this. We paid the price, and now we gain. So now, which equations are we talking about here? Moreover, remember, I told you g is free in the previous statement. It's little manipulation. G is not completely free. If you play the same game we played to prove f is equal to 0, for example, there is still something true. So g of 0 v is equal to 1 for any v. And I should have told you before. I mean, but what is g? g is the norm of the other vector. But for u equal to 0, you are on the geodesic. And you are forcing it to be the tangent vector to the geodesic. So it's 1. So you know this. And actually, you know also that the derivative of g with respect to u at the point 0 v, again, on the curve gamma that you used to start as the skeleton of your construction, this is 0. These are two things that you prove exactly as we need for f. The point is that you cannot extend it on an open set now. So g was not so convenient now. Set little g just to short up things, because the real function here is the square of capital G. And now case by case, case 1. How is it possible that the Gauss curvature is 0? If this is 0, then you remove these, of course. And you are left with what? You are left with the equation. These d little g, d u squared is equal to 0. That means what? Now, just be a bit careful. That means that g is a linear function in the variable u. g is a function of u and v. And this is the only information I had. So that means that g of u, v is something like a of v. So the coefficient is a function of v plus v of v. For some functions a and b, this is the right conclusion I can draw out of that. But now I put the initial values that I have in general. So what does it mean? g of 0, v is equal to 1. Well, I put u equal to 0. So b of v has to be equal to 1. And a, well, the derivative of g, I mean, you put how you have to rephrase everything. But of course, it's the same as the derivative of little g up to a f of 0. So the derivative of little g has to be 0. So a is equal to 0. OK? These are the equations here. The derivative of the square root is a multiple. So if it's 1 is 0, the other is 0. What? So what? So g is constantly equal, little g is constantly equal to 1, of course, taking the square. It's the same. And so what have we found? That the first fundamental form of a surface with 0 Gauss curvature is, in this particular coordinates, e is equal to 1, f is equal to 0, and now g is equal to 1. So they are the same coefficients of the plane. So they are locally isometric. Remember, locally isometric. And now you see how to do all the cases. OK? Now suppose that k is a positive constant. So to be sure it's positive, let's call it a squared with a known 0. Now, this is my formula. So this becomes what? This becomes d2 in du squared little g is equal to minus, well, I shall put to plus, plus little g a squared. This is equal to 0. This is the question that I know. We can write down the solutions explicitly. This implies that g, this is an harmonic oscillation. So this is a, again, a v. So in the variable u, it behaves like a sine or a cosine. OK? So a of v cos of a u plus b of v sine of a u. See how simple everything has become. And now, again, plug the initial values in this. So g of 0, v is equal to 1 tells you what? Well, if I put u equal to 0, of course, this disappears. So now I get something interesting on a is equal to 1. Then you put, take the derivative, manipulate, and so on. So at the end, you get, again, g, this implies plus. So plus these two things computed in these cases, you get g u v is equal cos a u. So a is equal to 1, and b is equal to 0. Those are the translations of these two equations. OK? Well, and now, well, so now you reach the position for your surface for which? E is equal to 1. F is equal to 0. And g is equal to this function. All right, squared, OK? Cos squared. Capital G will be cos squared of this. OK? Now, you have to prove that in geodesic coordinates, the sphere has these coefficients. That's an exercise. You know it already. You just have to put in your notes, put things together, and find this. And the third case, and we are done. Well, it's the same thing. But now, k is equal to minus a squared. So OK, since I'm already three minutes or five minutes over time, let's stop it here, because now it's the same proof. k is equal to minus a squared. So the equation becomes here with the minus. So the fundamental solutions will come like this. With hyperbolic sine and hyperbolic cosine, you plug there and you get the usual thing. And then you have to compare with this. We did it as an exercise. And actually, when I showed you this model here, since I'm kind, I mean, those coordinates are already the geodesic coordinates, if you want. OK? So in this case, there is no extra exercise to be done. So it's amazing, OK? It's amazing the fact that these coordinates exist every time. And it's amazing the general philosophy that we have developed, because you see, we prove classifications of surfaces in R3 without drawing a picture by looking at the coefficients of the first fundamental form on domains of R2. This is a key point. I mean, this is actually the birth of differential geometry. Stop making pictures. Write down these functions on domains of Rn and do all the geometry of the corresponding surface in Rn on the domain of R2. And you can actually get to these theorems. Of course, you can do it only for the quantities which are intrinsic. So it's critical to understand which are the intrinsic quantities in Rn and the extrinsic quantity. That's why the theorem, a gradium is a gradium. The Gauss curvature is an intrinsic property. And in fact, it's the only curvature which is intrinsic. OK? So besides the technicalities of the proof, here it's the conceptual scheme, which is fundamental. OK? Because now in the next week, we are ready to jump. Forget Rn. Now we are going to talk about manifolds. In some sense, you already know how to do it now, OK? Because now a manifold will be, or at least now, oversimplifying but something which has patches. And on each patch, you have the first fundamental form, a Riemannian manifold, OK? Now we can do geometry on this next week.