 OK, so let's go on with our very quick and just foundational study of differentiable manifolds. I'd like just to convince you with these few lectures on this topic that having understood and looking at the theory of surfaces in R3 with some critical point of view, you are now ready to jump to the next step, even by yourself. But of course, as usual, you are welcome to come to my office. In the sense that many notions, I mean, almost all the definitions of regarding differentiable manifolds really come from some kind of intuition that we have for surfaces in R3. So remember, the first definition, the basic definition is about what is a manifold. And we said it's something that is covered with some maps. So it's just a set. But it's a set which is covered with charts, OK? With some property, namely that whenever you can speak about, I mean, you build a situation where you can talk about differentiability, meaning you look at two intersecting charts, you go back to the pre-image of the intersection. And so you have a map from some open sets of Rn to open sets to Rn, OK? And then you talk about differentiability. Now, this is the key, of course. It's the definition, so it's the key point. But let's see, for example, having given this definition, we know that we can speak about, so if we have now two differentiable manifolds, usually sometimes I indicate the differentiable structure explicitly with a nutless u alpha f alpha and something else. Otherwise, in shortcut, I can also just indicate the set. But I'm assuming that I think on that a fixed differentiable structure. So now, suppose you have a function phi between differentiable manifolds. Actually, the standard shortcut for manifolds is this. So now, this is just a map between sets. So how can we define the notion of differentiability? Well, now, again, look again what we did for surfaces. Well, around the point, if P is a point in M1, we say that f phi is differentiable at P. Of course, it's a local property, so you first define it around the point and then you say it's differentiable if it's differentiable everywhere. But you say, if phi is differentiable at P, if for some choice, so P being a point in M, we line an image of some that doesn't matter where I draw it now, I mean. But suppose it lies in the intersection of two charts. So now I have a map from this is M1. And here I have another manifold. And here I have another point, phi of P. But also, this one will lie in another chart. Let's say, coming from G gamma and V gamma, there will be a chart also for the image point. And so I can compose as a map from here to there. Exactly as we did for surfaces. This is exactly the same construction. So we say that phi is differentiable at P if there exists. If there exists charts, let me note f from some U into M1 and G from some V in M2. This f will be around P. And this will be around phi of P, such that the right composition, such that G inverse of phi composed f, is differentiable. Now I know what it means, because this is just a map between, actually there is no reason why I've never said that these have the same dimension. The dimension of the space is the dimension of the Rn, which is fixed. I mean, on a given manifold, you cannot have two charts, one coming from Rn and one coming from Rm. So the n is always the same if it's connected. So but in principle, I never said that this dimension is equal to this dimension. So this map will be a map from an open set of Rn, let's say, Rn1 to an open set of Rn2. So the dimension of the base of the first of the domain and the dimension of the co-domain, which in principle are different. But in any case, these are two Euclidean spaces. So it's a map between open sets of Euclidean spaces of suitable dimension. So I know what it means to be differentiable. So it's differentiable at the point at which point? Well, at the point which goes here, so at phi inverse f inverse of P. Exactly as we did for the two-dimensional case. Now, you have exactly as in the two-dimensional case, we have the objection. Well, but does this depend on the choice of charts? I say no. Because our requirement on the definition is that if you take, so I mean, is it possible that something is differentiable, meaning that there exists two f and g? But maybe if I take another chart around P and another chart around phi of P, the corresponding map is not differentiable. Understand the question? And the answer is clearly no, because the change of chart is a defiomorphism. So this function will be differentiable if and only if. If I here I have another f prime, if you want. Here I compose with f inverse composed of prime. And this is differentiable if and only if the other one is differentiable. A defiomorphism takes differentiable functions to differentiable functions. There is no way you can have a function which is differentiable in one and not in the other. So you ask, so this there exists actually becomes immediately if you think about the definition. If for any pair of charts, this is differentiable. So now we know what is a differentiable map. Now you can globalize this definition around every point. And now which was the main step? I mean the key step in our study of surfaces was to describe out of the definition in some sense the notion of the tangent space. This was kind of the key object, the key geometrical object. Because curvature, of course, arises. It was born in our language in some sense by saying curvature is essentially the derivative of the normal vector. But how the normal vector changes is, of course, the same thing as how the tangent space changes. Because one thing is the same as the other. It's just the orthogonal complement. So the notion of tangent space was fundamental in order to speak about anything. Now here we have a problem. Because this set here, now, our differentiable manifold, does not lie in some Rn. Because actually, sorry, out of this, in particular, we know what is a differentiable curve on a manifold. Because now we know what it means to take a differentiable function from an interval into m. So in principle, you would like to say, well, a tangent vector is the velocity of a curve. But what does it mean? Nothing. Alpha prime at 0. Now we don't know what it is. If you think of it, there is no way. You cannot bypass this problem. The notion of a derivative, what it is, it is alpha of p plus h, if you want. Think back to the first class in calculus you had. Alpha of t plus h minus alpha of t divided by h, and then limit for h going to 0. How do I perform alpha of t plus h minus alpha of t? If these are points on a set, this symbol has absolutely no meaning. So this is a dead end. You cannot speak about the derivative of a function with values on a set. Because you cannot even take the difference, not speaking about anything else. So how do we do it? So here we have to use a little bit of fantasy. We have to give another interpretation of tangent vectors even in the case where we can draw pictures. So even if we are able to draw pictures, we want to describe this object here in a slightly different way. Let's see how we do it. Well, let's stay in the classical situation. Let's first give another interpretation of something we know, and then we export it to something we don't know. So suppose you have a curve. Because actually the idea is really to save this idea of velocity. But somehow we have to change the interpretation. So now suppose you have a differentiable curve into some Rn. Suppose you call, for example, alpha of 0 is some point p. Now this means that alpha of t is locally given, I mean, in this case globally, by an n number of functions that I call x1 of t, xn of t, n functions which are giving me this curve, the parameterization of this curve. So of course I know exactly perfectly well what it means. Alpha prime at 0. Alpha prime at 0 is just the vector x1 prime t0. If I let me put the 0 outside, xn prime at 0. And let's call this vector v. Now as I said, this cannot be saved on a manifold. But a vector, even in Rn, is something that actually gives us a way to differentiate functions. This is the idea. So now if I take a function phi, if phi is a function, is a differentiable function, of course, otherwise I cannot take derivatives. Is a function around p. Then what is the derivative of phi in the v direction? Is what? Well, one way to say it is I take phi, the function phi. This is what we did many times. We take the function phi composed of alpha. So basically, we restrict phi to our curve. So this is just Rn. We have some curve here. And here I have my special point. This gives me a tangent vector v. And I'm saying, what is the different? Now I have a function defined on some open set containing this point. And what is the derivative of phi in the v direction? Well, one way to say it is take phi, restrict it to the curve, and take the derivative at time 0. So this is a good, in fact, this is the definition of the derivative in the v direction. Then the point is to prove, remember logically, the point is to prove that this object does not depend on alpha, but only on v. If I take another alpha, so call it beta, such that beta prime is equal to the same v, this object is the same. And this is we have already done it many times. But what is this object in these notations now? With respect to the functions xi, well, by chain rule. By chain rule, this is the sum from 1 to n of the phi vxi. Vxi in the t at t equal to 0. So I can write it in this form. Let me write it in this form, which is the same, but it emphasize something. So this is what I indicated by xi prime 0. So I want to see it as an operator on the function. So which is the operator? The operator is xi prime at 0 times d in the xi. So the differential operator d in the xi acting on the function phi. So really a tangent vector, so now I'm thinking of the tangent vector. You see this kind of differential operator here is completely determined by v. Because its coefficients with respect to the partial derivatives are given exactly by the components of the vectors, of the vector v. So knowing v and knowing this operator is the same thing. Excellent. So if I give the interpretation that the tangent vector is not something I draw as I'm used to, but as a differential operator on functions which satisfies certain properties, then I have a hope. And in fact, you see this interpretation is exportable to the situation of differentiable manifold. Because now suppose that you have alpha, i will be always some interval containing 0. So now suppose I have a differentiable curve on a manifold, not on our end. And as usual, as before, alpha of 0 will be some point p. And now I say set t d, set d, the set of functions of functions on m differentiable at p. I don't really care what they do far away from p. But at least around p, they must be differentiable. Then I say now starts the definition. This is just a notation. Then the tangent vector to alpha at p is what? Is the map that I indicate by alpha prime 0. But now this is just a notation. It's just a notation because I don't want to be confused with the classical case. Because of course, the symbol prime now has absolutely no meaning. So it's just a formal notation. It's a map, basically, which does what? Which takes a function. Now look at and gives me a number given by alpha prime 0. So this vector applied to a function phi is just d in dt of phi composed alpha at t equal to 0. You see with this trick, phi composed alpha is a map from the interval to r. So the derivative is the classical derivative. Sorry, on m with values, yes, with values. Well, this is OK. But we seem to have complicated. Seems like we have cheated. For example, basic question, is this a vector space? So the set of all tangent vectors to some curve alpha at p. So if you take the set of all vectors arising in this form for some alpha, it's called the tangent space at p. So a tangent vector, then the tangent vector to alpha at p. And then a tangent vector. A tangent vector to m at p is the tangent vector to alpha at p for some, now collect all these things when the curve is free to move as long as it passes from p at time 0. And you get a set. This set is called the tangent space. I don't even write it. The tangent space is the set of tangent vectors. Now, let's see how this space looks like, because at least we managed to get a definition. But all the properties are very mysterious if I look in this form. So I want to remove the mystery. And now, always remember the picture, but somehow. So you have our differentiable manifold. You have a point. You have a chart around this point. And you have some kind of, some parameterization f of some open set in Rn around this point. This is the picture you always keep in mind. And actually, assume what I didn't in my picture. There is no arm in assuming that p corresponds to the origin. This is just to simplify notations, but it's completely. You can always assume that, because whatever f inverse of p was, you translate it into the origin and you construct another chart with this property. So this is not. So p is f of 0, 0, 0. And then a curve, alpha from i to m, can be written as, is what? Well, and the function phi. And phi in D can be written R. So the two things together are written in these coordinates like f inverse composed alpha of the point T. So what is f inverse composed alpha? So here you have some interval and here you have a curve alpha. This is alpha. So I take alpha composed of f inverse. It means I get a curve here in the domain of the chart. So this is given by x1t, xnt. So we are in the classical situation. And phi composed f. Phi composed f goes at the point, let's say, q. And now q is another point here. So phi composed f takes point q, goes here, and then f will go to R. Phi, phi goes here. So this becomes a function of variables x1, xn. And this will be some function that they just leave phi, x1, xn, where x1, xn is actually the point q. Nothing. So now let's do the computation of this object here. What is alpha prime 0 applied to the function phi? Well, by definition is the derivative of phi composed alpha. So I don't rewrite it. That's the first step. It's the derivative at time 0 of phi composed alpha. But this is dt at t equal to 0 of what? Of phi of x1 of t, xn of t. So this is a classical situation. It's a function in the classical sense between an interval and R. So I use the chain rule. And this becomes the sum from 1 to n of xi prime 0 times the operator d in dx i at 0. Now this 0 is because by chain rule I should take the derivative at the point f inverse of p. And that's why I put f inverse of p in the origin, just to put. So this 0 corresponds to this choice here by chain rule. Apply to phi. Well, but that means that really as an operator, alpha prime 0 is the operator sum of xi prime 0 d in the xi 0. This 0 is actually the origin. It's a 0 vector. I don't know if you want to use a special notation. It's not a number. Maybe just to emphasize. So as operators, this equality holds. And actually, if you want, you can be even more explicit about what is this operator here, d in the xi. d in the xi at 0 is what? Is the derivative in the direction of the coordinate xi? So where, if you want, where d in the xi 0 is the operator, is the tangent vector, is the tangent vector to the coordinate curve, meaning xi goes to f of 0 x i So you are taking all these coordinates. Remember, you have n, of course, of them. And you are taking those special curves, has special curves on the manifold. So what have we proved? So you see, this object here is really a tangent vector to the manifold. So this is a tangent vector to the manifold. And what we said is that, so this is the tangent vector So t f, let me indicate it in this form, t f, because it's something that, in principle, depends on the chart. We use the chart to make these identifications. So t f is, by definition, the span of the vectors d in the xi at 0. So now this object here has a structure of vector space. In fact, of which dimension? Well, how many of them I have? n. I have n possible curves like this. And in fact, I should check that they are linearly independent. So now the observation is that t p m, in the sense of the set of tangent vectors that I defined before at p, is actually equal to t f. Now, as usual, the problem is trying to understand what do you have to prove. Well, of course, it's an equality of sets. So there are two inclusions to be proved. But here you have already the proof of one, because we have taken any tangent vector, alpha prime of 0. And we wrote it as a linear combination of these. So this inclusion is exactly what we proved. So why should the other inclusion hold? Well, take any linear combination. So if you want to prove this inclusion, what do you have to take? You have to take any linear combination of this vector, these special vectors, and produce a curve for which this linear combination is the tangent vector to this curve. So let's take a linear combination. Take v equals some lambda i d dx i at 0, from 1 to n. And then take alpha, take alpha, such that in the parameterization f becomes alpha is equal some lambda i t. You see, for example, if this was just the first vector, d in the x1, it's written here which curve I should have taken. f of x1, so just changing name from xi to t, the parameter along the curve, this is what it's written here. So the tangent vector to this curve is by definition this object. So this works. If I compute the tangent vector to this curve, I would get some of lambda i very well. So using this, which, again, sorry if I keep on underlying, but I mean, so to produce this vector space, so this is nice because it's automatically a vector space. It's bad because it depends on the chart. While this one was nice because it was chart-independent, but it has no structure more than a set. But using a chart via this identification, I can put a structure of vector space here. So TPM, so this implies, this observation implies, that TPM has the structure of an n-dimensional vector space. So now we really solve one of the key problems. In some sense, in our wines, when we take a tangent vector to a differentiable manifold, we keep on imagining a picture like the usual one. But it makes no sense. This is really formally what we have to do. Now there are many things that comes easy. For example, if you have a map, again, between two differentiable manifolds, one of the key objects we introduced in the study of surfaces was the differential of the map, which has to be a linear map between the tangent space of the domain to the tangent space of the co-domain. So now at least I know where it should start and where it should end. How do I define it? Well, and pick P in M1. So I define d phi at P as a linear map from TP M1 into, well, into I don't know, into usually means it's injected. The target is T phi of P M2. How? Well, we take exactly the formal definition we used before. So if we take a tangent vector V here, now I know that this comes. This is alpha prime 0 for some alpha. Well, so compose with phi and take the composition prime at 0. It's a differential operator now. So this goes into phi composed alpha prime at 0. Now we know what it means. Also here, at this point, you have the problem that we faced in the case of surfaces. Is this definition alpha dependent? No. It depends only on the derivative at time z, only on this, or whatever it means, because this is not anymore a derivative. It's a symbol. But actually the same proof. The same proof we did for surfaces shows you exactly this. So now that we know what is a differentiable map between manifolds and what is the differential of a map, for example, we can talk about the filmorphism, which is supposed to be our kind of equivalence relationship among differentiable manifold. So what would mean for a map phi to be a diffeomorphism? Well, certainly it has to be a differentiable function. It has to be more than all the identifications that you know, so it has to be 1 to 1. It has to be an homeomorphism. Remember, these are sets. Topological spaces, differentiable manifolds. So the identification as differentiable manifold must be stronger than anything else. So the identification should be 1 to 1, and so you are OK as a set. Homeomorphism, so you are OK as topological spaces. But actually you want this map to be differentiable, and the inverse, which exists by the previous thing, has to be differentiable. So it's a differentiable homeomorphism with differentiable inverse. Of course, it is often the case that two manifolds are not diffeomorphic, but it could be, well, in fact, I mean, often you will handle local diffeomorphism. So maps which are not maybe globally 1 to 1 or global homeomorphism, but there exists an open set on the domain and an open set on the target such that if you restrict the map on these two things, it goes from one to the other, and this is a diffeomorphism restricted to this open set. This is called a local diffeomorphism, local in this part. Very well. And then comes one of the most important theorems in calculus that we used in many forms. I mean, probably we never really stated explicitly as I'm doing now, but I mean, it's very important to know to produce local diffeomorphisms. And actually one of the most powerful way to do it is to use the inverse function theorem, which tells you this in this language now. If you have a map such that at a given point, d phi at p, this is a linear map. If this is an isomorphism, which is always easy to check because this is a linear map, so you just see if it's invertible or not as a linear map. But if this is an isomorphism, this is an amazing theorem because something very easy like this implies that phi is a local diffeomorphism around p. So this means there exists a sufficiently small neighbor of the point p, such that if you restrict phi to this sufficiently small neighbor, phi is a diffeomorphism. And this holds automatically for, I mean, this holds, I hope you have seen this theorem for maps between Rn and Rm. Actually, this assumption implies immediately that n is equal to m. Otherwise, you cannot even start. But if the differential of a map between Rn into itself is at a given point, an isomorphism, it is locally a diffeomorphism. Now the same proof, you don't have to prove anything new. Now works here for manifold because you actually translate everything into a chart and you are back to the usual case. Very well. Now, of course, I'm not trying to give you the theory of differentiable manifolds. I'm trying to highlight which points, in some sense, we have already solved without even noticing when we were studying the theory of surfaces. Now what is another step concerning vector tangent vectors? Well, once we have defined what was, which was the tangent space to a surface in R3, immediately we define the notion of a vector field. What was a vector field? Well, again, having a surface in R3 made everything very easy because basically the only thing I wanted is, so I had my surface and I want to define, for example, a tangent vector field on a surface S. But in particular, since in this specific case a tangent vector to S was a vector in R3, I know perfectly well what it means for something to be a differentiable map from an open set into R3. So I had no problem. I said a vector field is a differentiable map, such that at each point it's a tangent to the surface, for example. Or the normal vector field is something like this, which at each point is normal to the surface and so on. I was using the fact that here I knew how to take derivatives, essentially. Now I cannot do it anymore, but it's clearly crucial. Vector fields should be the basic object of our work. So how do we bypass now also this problem? This is more delicate again. This is not trivial. You see, basically if I put all these tangent spaces together, so now, of course, I'm drawing a picture, but you have to forget immediately the picture. I mean, if this is my M, I have a point and I have a tangent space. I don't want to draw it because, really, I don't have the universe around. This is all I know. I have a notion of a tangent vector. Now I change point and I have another vector space. Now what does it mean that I have some kind of correspondence which depends differentially on the point? So I would like to assign, at every point of my manifold, a tangent vector. OK, this is just a question of maps, a set. But I want to do it clearly in a way that when I change the point, it's a differentiable function. This is the delicate point. What it means? What does it mean? At the moment, nothing, unless you see now the problem. So the union of TPM, you see, this is really the target, the union over P. I put all these vector spaces together. And a vector field certainly is something which takes a point in M to a tangent vector, in fact, with the property if you want that P goes to a vector which lies in TPM. So a map like this. But still, I don't know how to define differentiability because this is just the union. Notice that in this new definition, even in the previous definition, well, that's more. These vector spaces are disjoint one from the other. I mean, TPM and TQM have nothing to do. These are formal n-dimensional vector spaces. It's not possible. I mean, in the case of surfaces, you might have said, well, two tangent spaces at two different points could intersect because you are really thinking of them as subspaces of R3. And as subspaces of R3, they might have, well, in fact, they have to because they are two-dimensional. Two-dimensional in R3, they must intersect, unless they are parallel. They intersect at infinity. Now, in this new definition, these objects here, they don't live inside a bigger space. OK? So there is not an ambient space where these are subspaces. And so you can think of maybe TPM is intersecting TQM. It doesn't make any sense. But now, if you think of what we have just done, so what do we need in order to speak about differentiable functions between two objects? Well, basically what we are asking is, are these two objects differentiable manifolds? Well, the first one is a differentiable manifold by definition. Otherwise, we are speaking about the empty set. OK? So the question is, are we able to put a differentiable structure here, some kind of natural differentiable structure? Because of course, I can start being very creative and construct very exotic things. No, I would like to have something which makes sense, in particular, in the case of surfaces. In R3, for example, it boils down to the usual thing. OK? I want to generalize something that I know. I don't want to create something completely new and completely useless. OK? Well, this can be done. Let's see how. And this object, I mean, as a set, is exactly this one I just said. But with this structure, the time this climbing has a famous name. So let me call it, let me tell you. The tangent bundle. OK? So what is the tangent bundle? m is a differentiable manifold. And then you construct the object, the set, tm, as I just told you. So this is the set, the union of tpm, if you want, the disjoint union. You can also say the set of pv, such that p is in m and v is in tpm. This is exactly what I said before. I propose you to think in this way, even though it's very unusual. Now, this is something you wouldn't do, thinking to the case of surfaces. But this is m. Since it's a differentiable, well, even before, the fact that it's a differentiable manifold, you want to have some kind of picture in your mind. Strange enough, the right picture to think is not to draw the tangent space here. But it's because it's more important to remember, even pictorially, in your mind that these objects have nothing to do if you change point. And unfortunately, living in a three-dimensional space and being the first reasonable thing to draw a surface, you are left only with one dimension. So really, you draw the tangent space in this way. I'm left with only a line. So really, I'm thinking as a set to this set here. Of course, remember that this is tpm. So this is an n-dimensional vector space. So in fact, even in this picture, I should draw a two-dimensional vector space. But it would be a disaster because it would cut m, which doesn't make any sense. It would cut other tangent spaces, which doesn't make any sense. So the only way to preserve the fact that this tpm is something which cuts. So m is canonically contained in this set. Because for any point, I can take the zero vector. So in some sense, m lies here, lines here, lies here. And this picture reminds me of this. And the other thing it reminds me is that this tqm and tpm are completely different vector spaces. But now, m was a differentiable manifold. So remember, now I want to prove that this is a differentiable structure. So I have to construct your charts, satisfying property 1 and 2. So I start with the knowledge, the fact that this object here, m, has a chart. So around p, there is certainly a u alpha with the usual properties. And I call, let me now, notations are important. So I call the coordinates of these, Rn. So u alpha will be a subset of Rn. The coordinates, I call them x1 alpha, xn alpha. So I put this alpha on top to remind, because now I'm going to change charts. So I want to keep track of where things come. Very well. Now, if w is a tangent vector, remember, t, the tangent space, at a given point, we have identified it. Once we have a chart, in fact, by brute force, we proved that this is the same thing as t, f alpha. OK, this is not really of m. Sorry. Now, suppose I take a point q, which is here. I've already used this notation. And I look at t, f alpha of q. Sorry, so q is here. So that's a bad notation here, at m. I take a tangent vector at this point. So w, by what we know, will be the sum of what? It will be a linear combination of the operators d in dx i alpha. Now let me drop the 0, because I'm not putting any restriction on the fact that the point comes from the origin. So at the right point, and I don't write it. So there will be a point where this operator is the right one. And I call the coefficients y i alpha. Well, if I know which curve generates this vector, these will be the derivatives of the components of the curve. Now, if I take a tangent vector, there will be numbers. Of course, the summation here is on i. Alpha is fixed. Alpha is the chart which is giving me this. So now, with this structure, I can define a function. I call capital F alpha from u alpha cross Rn with values in the tangent bundle. This object is the tangent bundle. As a set, I'm trying to describe a differentiable structure. But let's call it already the tangent bundle. You see that this looks like a good start if I want to define chart on this object. Because I'm saying, I'm going to say that these are the charts. So these are open sets of what? This is an open set of Rn. This is Rn. So altogether, this is a subsets. If you want, it's an open set of R2n. So what I'm trying to hint here is that maybe the tangent bundle will be a differentiable manifold if these functions are going to be well defined, as I'm going to show you. That this is a differentiable manifold of dimension, the double of the dimension of the base. In fact, learn some language in this theory. m will be called the base of the bundle. How do I define this? Well, essentially, I've written already here. Because I can, so I take, this map has to take a point in u alpha, which I decided it's parametrized by x, x1, xn alpha. So this takes x1 alpha, xn alpha, and then takes any point in Rn. And the coordinates of these are n. I call them y, y1 alpha, yn alpha. And then which point? So what do I have to assign if I want to define a map? Well, I have to assign a point. So what is tm is a point, comma, vector. So which point will I decide to take? Well, the point is easy. Well, both are easy. In the sense that if I have the coordinates x1, xn, I have f alpha of them. So the point will be f alpha of x1 alpha, xn alpha, comma, which vector? This one. If I have dx, the functions x, I have these vectors. If you give me y, I use the y as coefficients. So I put really the same expression. Sum yi alpha, or the other way around, y alpha i, y alpha i, d in dx i. OK. Well, now the question is, well, this was easy in some sense. But now there is a problem. Do these maps, when you count them all together, satisfy the axioms, I mean, the prescriptions of the definition of a manifold or not? Well, I can leave you as an exercise. Just understand what you have to prove. What do you have to prove? Well, these maps are one to one and so on. Well, all these set theoretic properties are very easy, because f alpha is a chart by definition. So it's one to one. It's an omomorphism on the image, everything. And then here what I'm doing, essentially nothing. I'm taking a point in our n, and I'm just rewriting it, in an abstract vector space of dimension n with the same coordinates. OK. So all the set theoretic properties are automatic. Really, the only thing you have to check is that if you take two of them, which overlap? How can they overlap? Well, you see, in particular, these charts have the property that if there is a point lying in u alpha, there is the whole tangent space. You see, I'm not taking u alpha cross an open set. I'm taking you. So it's really a stripe once you think in this vertical picture. So what does it mean that they intersect? Well, the only way to intersect is that really you have u alpha and u beta, such that u alpha intersect u beta intersect. The only possible intersection comes from the base, and then you take on each point the whole tangent space. This affects the chart. I mean, of course, the same tangent vector gets written in two different forms. One is this one, and the other is this one where you put beta. But then there is a relationship between d in dx alpha and d in dx beta. On the overlap, one is a linear combination of the other. And in fact, which are the coefficients, the Jacobian of the transformation of the change of coordinates, and so on. So you can do it as an exercise on the definition of differentiable manifold. Now, you see, once this is a differentiable manifold, we can really speak about vector fields. Now it's easy. A vector field is what? It's a differentiable map from M to TPM, such that at each point you are giving me a tangent vector at that point. OK? In principle, you see, a differentiable map from M to TM has no reason. So P must go into something like PV. So I know. And since I know what is a vector field, I know also what is a derivative in some sense of the vector field. In the sense that differential, I know how to differentiate a vector field. In some sense. OK. Now, special manifolds as sub-manifolds. So let's see in three minutes how to produce. If you have one manifold, presumably you have many other manifolds, which are kind of contained. So how can a subset of a manifold be a manifold by itself? OK. The prototype situation you have to keep in mind is exactly the theory of surfaces in R3. R3 is a manifold. It's the simplest manifold you can think of. OK. All regular surfaces are manifolds, two-dimensional manifolds. But they are not just manifolds. Their differentiable structure comes from the ambient space. So there is a relationship between the differentiable structure of a regular surface and the differentiable structure of R3. One induces the other, exactly like the topology is induced from the topology of R3. Also, the differentiable structure is induced by the differentiable structure of R3. So when this happens, we want to speak about some manifolds. So let's stop a moment, and we go on from here. OK. So let's see what this strategy takes through. And this is, again, something you will find in your notes in the sense that if you have two, and let me indicate the dimension by this exponent, OK. So this will be a differentiable manifold of dimension m. And another, you have two differentiable manifolds, but possibly of different dimensions. Now we say, and if we take a differentiable map phi from mm to nn, this is called an immersion. If the differential of the map phi at any point, so this is a linear map, we have just seen that this is a linear map from tpm to phi of pn, OK. If this is injective for any p in m, OK. Now if in addition, is it OK? Immersion, immersion, OK. Then if in addition phi is an homeomorphism, which, of course, in particular implies that m is equal to n, OK. Then on to, sorry, it's an, sorry. If it's an homeomorphism, globally yes, but actually I'm restricting an homeomorphism on to the image, OK. So this leaves the possibility to be of different dimension. The ambient space is of different dimension. You have a small thing which goes inside, a big thing. Where, of course, which topology do I put here? This is a topological space. This is a subset. I'm looking at it with an induced topology from n, with the induced topology from n. Then phi is called an embedding. If, as often happens, you don't really have a map, but you have already a subset of a manifold. So in some sense, if the map itself is the identity. So you have m contained in n as a subset. Then n, the inclusion is an embedding. Then m is called a sub-manifold. Here, of course, what you mean is that m is a differentiable structure by itself. But as a set, it's contained in n, OK. If the inclusion is an embedding, you say it's a sub-manifold. So just to check what are these definitions saying, these words saying, you can recall the list of examples we gave, for example, of differentiable curves in R2. So for example, the curve, so alpha, in fact, from R to R2, which was taking t into t squared t cubed. This is clearly a differentiable map between two differentiable manifolds, particularly simple. But it's not an immersion. It's not an immersion because, of course, at t equal to 0, the differential vanishes. The matrix representation of the differential is the matrix 2t 3t squared. So at t equal to 0, it's identically 0. It's the 0 map. So in some sense, if you want, immersions corresponds to what we call regular curves, because the tangent vector should be always non-zero. Then if you want an immersion which is not an embedding, you just draw this with the parameterization we gave a couple of months ago. This is an immersion because at every point, if you parameterize it well, there is a well-defined non-zero velocity, which means that the differential is non-zero. But it's not an homeomorphism between R and the image. Of course, you have to play with. Yeah, think of this as the image of a map from R to R2 again, exactly as before. We wrote the explicit analytic expression of this map. But this is not going to give you. This is not an embedding. In particular, I mean, it's not injective. So there are two points going to the same point. So it cannot be. So this one. So the point is not just that the image is abstractly homeomorphic too. It's that the map alpha has to be the homomorphism. And in this case, it's not. If you want an example of an immersion without self-intersection, but still not an embedding, remember the other kind of pathology we gave. OK? This is an example without self-intersections, but still not an homeomorphism. So this is still not an embedding. Now, which are, of course, this is the moment to recover all we've said for surfaces in R3. The inclusion, is it the inclusion always an immersion? You see, in particular, well, you have to start becoming really a mathematician at full power in the sense that pathologies become your friends and not pathologies. You can think of very strange, differentiable structures on subsets for which the inclusion is not even differentiable. So the fact that something is a subset of another thing doesn't mean that the inclusion is a smooth map. This is, of course, a very pathological situation, but it's there. Now, as another source of examples of differentiable manifolds, again, in some sense, inspired by the notion of a sub-manifold of Rn, if you take something that I indicate by mk, a subset of some Euclidean space Rn, and suppose that for each p in mk, there exists an open set v, neighbor of p in Rn. So basically, I'm rewriting the definition of a regular surface, now letting the dimensions to be free to be whatever they want. Instead of 2 and 3, I put k and n. And basically, I'm just rewriting. Suppose that there exists a neighbor of p in Rn and the map, and the map f, from some domain of Rk into Rn, and in fact, which hits m intersect v, such that f is differentiable, and 2, dfq. Now, dfq, so the differential of the charts, goes from Rk to Rn, exactly as we did for surfaces with 2 and 3. Suppose that this is injective. And in fact, in point 1, let me add f is c infinity is a c infinity homeomorphism. So I put all the 1 to 1 and topological requirements here. Now, if you have something like this, you have a manifold. Rn induces. You see, the main difference is that the open sets I'm using to parametrize this set come from open sets of Rn. This is the only thing, really. So that means that the topology that I'm looking on m is the one induced from Rn. And the charts are, I mean, as usual. So these are called k-dimensional sum manifolds of Rn, if you want. These are the same thing. And we know that out of this, we can extract the information, what is really in the definition of a manifold, that the change of chart is a differential. So this is implied by this, which is really part of the definition of the manifold. Now, just to finish in the last five minutes, let me give you another. Let's revisit an example we gave. So the projective plane revisited in light of these definitions and notions. We gave to the projective plane, we defined a differentiable structure by hand. I want to describe it in a different way now, so to understand whether it comes from an Rn or not. The one. So again, look at the object of our study now is this. Again. Now, we described it as the quotient of R, well, here in dimension 2, this is the quotient of R3 minus the origin, model of the equivalence relationship with identifies two vectors on the same line through the origin. Now, of course, this equivalence relationship can be thought as an equivalence relationship on the sphere. There is no need to take the whole of R3 minus the origin. Each equivalence class has at least one representative on S2. In fact, how many representatives? Two. Because for each point, each point is identified with anti-podal points, because they lie on the straight line passing through the origin, OK? So what can I do? Well, basically I can describe this S2 with the same kind of charts. We describe probably the first time we prove that S2 is a differentiable manifold. Remember the ones given by graphs of functions. So for example, and just to fix notation, if I call, so these are inside R3, and I call x1, x2, x3 the coordinates. And then I take the set u1 to be x1, x2, x3, such that x1 is equal to 0. And so the disk, x2 squared plus x3 squared is less than 1, OK? So for example, in my picture, this would be exactly the disk of the blackboard, OK? x1 is equal to 0, and the disk inside the equator of the blackboard, OK? Then on this disk, I have two functions. The one giving me the hemisphere coming out of the blackboard, and the one giving me the hemisphere going inside the blackboard, OK? Let me give just names to them. So I call them f1 plus, f1 plus of x2, x3 is equal to square root of 1 minus x2 squared plus minus x3 squared x2, x3, OK? So this is the one coming up, coming in your direction. And then I call f1 minus the one with the minus, OK? So this is the one going inside. In fact, just for convenience, let me call this function d1, OK? So this will become d1, x2, x3, and this is minus d1, x2, x3. And then I can repeat it twice this game, because I have u2, which is not a rock band now, but it's just the set x2 equal to 0, you know? And x1 squared plus x3 squared less than 1. That would mean this other disk, OK, disk in this direction. And on this one, you have f2 plus and f2 minus, OK? And then you have u3, which is x3 equal to 0. So now this is the horizontal disk. So this is u3, the one of the blackboard is u1. And let me not draw u2, otherwise it becomes too complicated. And then you have f3 plus and f3 minus, OK, with the same notations, OK? The functions are, of course, always the same thing once you replace the missing coordinate with the right one, OK? It's always the same thing. Now, the point is, well, what is the only simple observation I wanted you to make? Is that these functions, so what are two points equivalent once you think to the projective plane? Well, of course, p is equivalent to minus p, OK? But that means, in particular, these functions are invariant, OK? So these are functions which go to the quotient, OK? So in particular, if I call pi, the map, the quotient map, from s2 to p2r, OK? So this is just the quotient map. Then the only thing you need to observe is that pi of fi plus, for example, of ui is equal to pi of fi minus of ui. That's another property that I didn't mention. So this is what? The image of the upper half via the projection is the same thing as the image of the lower half, OK? Because each point here has a representative on the other side and they map to the same thing, because the two equivalent points have the same coordinates with the minus. So when you square, you don't care, OK? So we can define maps, so we can define some functions that I call gi, defined on ui into p2r, simply by saying that gi is equal to the composition, projection composed fi plus, for example, OK? Now I claim that this set of maps and open sets give me a differentiable structure on this, OK? You see that we are in the line of the definition of a differentiable manifold because, of course, for every point here, there will be one of these for which this point lies in the image of this, OK? So this cover, they are differentiable functions. Oh, sorry, you cannot say anything about differentiable functions yet. But now what do you have to do? You have to take the intersection of two things, of the images of two of these objects, and write down the change of coordinates, OK? And ask if this is a differentiable map or not. This is a very simple exercise, more or less what we did, very similar to what we did last time. Turns out to be true. So this defines a differentiable structure on p2r. And you might ask, well, is this differentiable structure the same to what we defined last time? So now you have, if you mix the two things as a set, the two things we did on p2r, you think you have two differentiable structures. One which was the old one, let me call it v alpha, f alpha. And now the new one, which now I call ui gi, OK? We have produced two different atlases, both satisfying the conditions of a differentiable manifold, OK? Now, try to prove that these two things are actually defilmorphic, OK? In doing that, you have to, well, so they define the same differentiable structure in some sense. But here, this is one moment where you have to think about this famous third property in the definition of differentiable structure. Because these, of course, these atlases are very different. So what it means they have the same differentiable structure. They are different. If the differentiable structure is just the collection of these sets, well, these are different. But the maximal cover with all those properties will be the same, OK? So nothing strange. OK. Well, I wanted to finish, but I'm already slightly over time. I wanted to convince you, maybe we can start with this on Thursday, that actually p2r, so it's not easy to prove that p2r cannot be embedded So there exists no embedding of this differentiable manifold inside R3. So this is generally something new, OK? On the other hand, it falls into the category, the last category I mentioned to you before, this kind of k dimensional sub-manifold over Rn. In this case, k, of course, is 2, because this is a two-dimensional manifold. The point is that n cannot be 3. But it's not difficult to produce an embedding of the projective plane into R4, OK? And this raises the question, well, this. I mean, this is a very simple example that raises the famous question. Is it true that any differentiable manifold can be embedded into some Euclidean space, letting the dimension free? Because, for example, you know, more or less, I mean, if you manage to prove this, that not every surface, in the sense of two-dimensional manifold, not every two-dimensional manifold, is a sub-manifold of R3, OK? But, for example, even in this simple case, it's a sub-manifold of R4. So question, is it true that any n-dimensional differentiable manifold is a sub-manifold of some Rn for letting n free to be whatever it wants, OK? This was historically a key question, OK? Because really, what you are comparing is the new geometry with the old geometry in some sense, OK? And the answer is, I can tell you, no mystery. The answer is yes, OK? But this n, the n for which Rn contains m-dimensional manifolds, becomes huge. And in any case, it's even impossible, in general, to explicitly say, if I give you a differentiable manifold, which is born as a differentiable manifold not inside anything, you know, by this theorem, that it will be a sub-manifold of some R3 millions, OK? So in fact, it's not that huge. I mean, n-dimensional manifolds goes to 2n plus 1, R2n plus 1. So for example, surfaces will always go into R5. Any two-dimensional manifold will be inside R5, OK? But how the proof doesn't tell you? I mean, the proof is highly non-constructive, OK? Three-dimensional manifolds will go into R7 and so on, OK? So for example, if you believe that any dimension missing is an equation, which is more or less, I mean, when you do, for example, algebraic geometry, that's what you are led to think generically, OK? So two-dimensional manifolds live in R5. That would mean it's given by three equations, which equations, OK? It's impossible to say. So it's a key fundamental conceptual theorem, but it's never used, OK? It's important. Well, it's used sometimes, but I mean, you cannot really use it too often. OK, let's meet on Thursday.