 OK, so let's start with another example, which is actually belongs to a family of examples. That's why we are going to do it in detail of another, of a new regular surface in R3, OK? So, and this is called torus of revolution. Belongs to a family of surfaces that we will study a little bit in detail. Namely, all type of surfaces which come from rotating a curve in a plane around an axis of rotation, OK? So in this specific case, what do we do? We take a circle, so let me call it S1, and this time I indicate the radius R, OK? Because I want to leave it. So we take a circle of radius in the plane x equal to 0 in R3, and we center the point 0 A0, OK? And we rotate this curve, the axis z equal to 0. Sorry, around the axis z, which is the line x equal y equals 0. So what does it mean? Geometrically, that means we are taking, so we said, x equal to 0, OK? So x equal to 0 means we are taking a circle here, OK? So this is the point 0. So this is x, this is y, and this is z. This is the point 0 A0. This is the radius is R, OK? So in the plane of the blackboard, we are taking this circle, and then we rotate around the line z, meaning we go around, we make one loop around the axis z equal to 0. That means at the end, so you see, every point will rotate. So let me just to have a little bit more of chance of drawing something reasonable. So this will be the symmetric circle in the same plane. So every point will rotate by, OK, and so on, OK? So at the end, now this is a little bit confused, but at the end, you get a surface like this, OK? Where here you had your initial circle rotating in R3, OK? So such a shape is called a torus, or it's the shape of a donut in R3. So how do we get, for example, we can parameterize this surface in many ways? Let's look for the analytic expression in terms of a zero set of a function, OK? So this is the set of points. So this s, this is the object, the set s. This is the set of points, x, y, z, such that what? So the surface, so the initial circle, how do we write the initial circle? So this is in the plane x equal to 0. So it will be parameterized by y and z, OK? Because x is constant equal to 0. And the requirement is that it has this center and this radius. So this means it's y minus a. So z squared plus y minus a squared, am I right? Yeah, equal R squared. And of course, x equal to 0. So this is the initial circle, OK? But then we have to take the rotation around the z-axis. But that means at the end, we get square root of z squared plus y squared minus a. Everything squared plus x squared is equal R squared, OK? That's the way to find the equation, OK? Now, question. Is this a regular surface? Is this object defined in this way a regular surface? Well, this is not difficult in light of the last theorem we saw last time. Because we can look at the function. So look at, so every time you have a locus given by some function, it's natural to look at the function itself. So f from R3 minus something z-axis, let me put it in this way, into R, taking the point x, y, z, and giving me exactly this expression here, OK? Meaning square root of z squared plus y squared minus a squared plus z squared. So this function here. Why am I looking at this function? Because now, s is by definition f inverse of R. So I've written my locus exactly as the inverse image of a point. So we have a theorem in our bag to solve the problem whether this is a regular surface or not. So the question is exactly, sorry, f inverse of R squared. So the question is, is R squared a regular value for f? If this is so, this is automatically a regular surface. Well, how do we check? So is R squared a regular value for f? How do we do it? We have to compute the differential of this map at every point in the inverse image of our value. But what is the differential of this map with respect to the standard basis of R3 and R? Oh, actually, sorry. Why did I take out the z-axis? I've taken out the z-axis in order to have a differentiable function, OK? Otherwise, since I have a square root here, I would end up in problems of differentiability. Certainly it will be always continuous, but not c1. So if I remove this, this is a c infinity function. So what is df? So we need to compute dfp for any p in f inverse of R squared. And we have to check whether this can be 0 or not. But with respect to the standard basis, this is represented as a linear map with the matrix, which is actually in this case a vector, given by the partial derivatives. We have reminded ourselves what is the differential of a map. So the point is, compute this with respect to x, y, and z. How much is with respect to x? With respect to x is nothing but 2x. Now with respect to y is what? It's 2y square root of z squared plus y squared minus a. Everything divided by the square root of z squared plus y squared. And how much is with respect to z? df dz is actually completely symmetric in z and y. So you just switch z and y there. And you get square z squared plus y squared minus a divided by square root of z squared plus y squared. So these are the partial derivatives. So is it possible that they are all 0? Yes. Of course, yes. If the point where I evaluate them is free, for example, if I take the origin 0, 0, 0, all partial derivatives are 0. But is it interesting for us or not? No. For example, the origin is not in f inverse of r squared. r, of course, is a positive. Actually, since it's always squared, it doesn't matter. But r is a non-zero number. It's the radius of this circle, of course. If this circle is of zero radius, it's a point. That's another possibility. Wait a moment. Let's discuss it. You are right. So in case it was not heard, the question is what happens if r is equal to a? Let's see. Here I was just mentioning the possibility that p was the origin, but p is not in f inverse of r squared. So for every point where x is different from 0, this partial derivative is enough to say that df is non-zero. And what about if x is equal to 0? Because if x is equal to 0, this partial derivative becomes useless. This is 0, so I cannot use it. So how is it possible that these two things both vanish? Well, there is another possibility, of course. Either y and z are 0, but this is, again, impossible because of what we just said. Or this becomes 0. You see, I have the same factor. Of course, the denominator is OK. It's a positive number. So the only possibility is here. So how is it possible that this is 0? And here comes the answer to the question of your colleague. Is it possible that z squared plus y squared is equal to a? Well, we have to assume not. Because of course, how much is z squared plus y squared? It's a squared. So square root becomes a. So if a is different from r, and in fact, we should assume, for simplicity, since it's irrelevant if it's positive or negative, because we are squaring it. Let's suppose that r is positive. And now we can suppose, for example, that a is less than r, actually is bigger than r. So what does it mean geometrically? Geometrically means that this circle does not touch the z-axis. If a is bigger than r, actually this becomes, well, it would be here if it's positive. So this is actually minus a. I drew the symmetric one. So this is the real circle. If a is bigger than r, this circle does not touch the z-axis. And we have to assume it. So under these conditions, this is impossible. I mean, the vanishing of the three derivatives at the same time is impossible. So it is a regular value. So r squared is a regular value. OK. a would be kind of the radius of the central, you see, but it's not part of my surface, because the surface I'm looking at is really the outside of this donut, of this torus. If you want to think of what is a is that in the middle of this donut, there is this circle here. And this is the one given by a. It's kind of the skeleton of this donut. But you have to assume it. I mean, if you throw random numbers, a will not be greater than r in general. We assume that a is greater than r. If a is less than r, you see, the picture gets more complicated, because it means, for example, we are taking something with this center and a big radius. And then you rotate around this axis. So you see it becomes a mess. So we want to avoid this situation by this condition. So this is a regular surface. So this was kind of the last example for the moment of regular surfaces coming out of the definition. So now let's go back to the general theory. We want to do some analysis on surfaces. We want to be able to say that some quantity defined on a surface is changing, and it's changing at a given rate, and so on. Exactly as we do for quantities defined in Euclidean spaces. If you want calculus, that's what it's about. Of course, we know that the key point is to be able to take derivatives of functions. Now we want to make sense of taking derivatives of functions defined on a surface, and not necessarily on open sets of rn. So how do we do it? Well, the first thing is what is a differentiable function if the function is defined only on a surface. So definition. Now S will be forever a regular surface, and O will be an open set of r3. So definition A. We say that a function f from S to rn. So now everything will depend on where it's defined and where it does it go. So if you have a function with Euclidean values, so we say that this is differentiable if for any local chart of S, so that means we have, remember, a local chart is a map from some open set to S, to some piece of S. Differentiable with all the properties required by the definition of a regular surface. So every time we have a local chart, x, the composition, f composed x. So what is the domain of the composition? The domain will be u, and the target will be rn. But then this is an old fashioned function. This is a function from an open set of r2. Remember, in a local chart, this is an open set of r2. So now we have a function from a domain of r2 into rn. But then we know perfectly well what it means to be differentiable. It's the old definition. So using charts, we transpose the definition of a differentiable function from a surface to the old definition of maps from rn to rk. So definition be, now, what could be a good definition of a function with values on a surface? So you see, this is a definition of a function of differentiability when the function is defined on a surface. So now what about if the function takes values on a surface? Well, that's easy, because so suppose you have a function. And now suppose at the beginning that it's defined on an open set of r3, but with values on a surface. Now this is differentiable if what? Well, forget that this is with values on a surface. The surface is a subset of r3. So you can easily think of this function by forgetting that it has a specific codomain target. This is actually a function from an open set of r3 to r3. You just look at the same function, but now forget that the values are on s. s is a subset of r3. So throw it away, and you are left with an old fashioned function from r3 to r3 is differentiable in the usual way. So now we can mix the two things, which is really what we wanted. So definition C. Now if I have two surfaces and I have a function, if s1 is another regular surface and now I take a function f from s to s1. So now we are mixing the two things above, a and b. So now the domain is a surface and the target is a surface. Well, but then again, forget where it goes. It goes in a subset of r3. So this, we will say that this is differentiable. If f from s to r3, you see, I'm using the same trick I used in b, and I'm back with a. I have a function from a surface to r3, so n equal to 3 in that case. So now I know by this definition what it means to be differentiable. And so I use it to define this in the sense of a. Is it clear? Now I will be very quick with properties about differentiable functions. So now let's take, for example, is a differentiable function defined on a surface with values on another surface? Is it a continuous function? After all, a surface in particular is a topological space. So you can ask whether a differentiable function, is it continuous or not? Certainly yes, because you are going back to the old situation, to the classical situation of functions from rn to rk. So it's certainly continuous. If you take two differentiable functions, whenever it makes sense, I mean, so depending where is the value or which is the domain. For example, if you can take the sum, if f and g are differentiable, for example, these are two functions defined on a surface, but with values in some rn. Well, for example, if the values are in rn, I can take the sum. I cannot do it if these were functions from s to s1, because in principle, the sum of two points of s1 is not the point of s1. So that's why I put here rn, just to be sure that this makes sense. But if these were differentiable, these are all trivialities. And again, if lambda is a real number and f is a differentiable function, lambda times f is a differentiable function. And of course, well, f minus g are the same property. And also, if you take two differentiable functions with values actually wherever you want, in this case, you can also take them with values on a surface. You can take another simple and standard thing to do is to take the scalar product between them, meaning at every point I perform the scalar product of these two vectors. So if these were differentiable, this is again differentiable. It's immediate to check. Now, examples. So these are kind of trivialities, general properties. Other, how to produce examples of differentiable functions. Well, one general way to have functions defined on a surface is to look at the special class of functions, which are actually defined. If you take a function, which is defined on an open set of r3 with values actually I'm putting r3, you can put rn, whatever you want. So if this is a differentiable function, in the old sense, this is a map from an open set of r3 to r3. But if s is a surface, or it's a piece of a surface, it's contained in all in the domain, of course, I can look at the restriction of f restricted to s. So question, is this differentiable or not? And again, two seconds of thought, and you can see that this is automatically differentiable. You have to look at condition a. What does it mean? You have to take a chart of the surface s and look at the composition. And it's automatically differentiable. So in particular, a particular case of this situation, the inclusion, which is actually an excuse to fix the notation. I will always call the function little i from s to r3, and the identity from s to s and the constants are all differentiable. Why this is in particular? So s is born as a subset of r3. So what is the map i? The map i is the map which takes a point P of a surface and sends it to P. P, as a point on the surface, is automatically a point of r3. So it's really nothing. It's the inclusion map. Why is this differentiable? Well, because it's the restriction of the identity from r3 to r3. And the identity is certainly a differentiable map, and so on. So these are all in the same spirit. Let's make another example slightly more interesting. I mean, the height function, for example, it will come back many times in our course. So what does it mean, the height function? We have a plane. So let P in r3 be a plane. We take a point P0 of P. And the normal. So if it's a plane, it's characterized basically by passing through a point and a normal vector. And suppose we call it, again, just to fix notation, A is a normal vector to P. Suppose now we have a surface. S is a regular surface. So the height function is this. So we construct a function h from S to r, which takes a point P of S. So imagine you have your surface here. And you have a plane P. We have fixed a point P0. And we know that P is given by somehow, is identified by this point and the normal vector A. So what does it mean, the height function? Well, it's really the height. In some sense, it's the distance from, it's not really the distance. So you imagine to draw this line here, the normal line, and you compute the length of the vector. Analytic, actually, I should imagine drawing, forget this. I draw the line through P0 with the direction A. So the normal, OK? So here I get my, suppose I draw it here. Just to make it a bit clearer, I just project this point on this line. And I see how long is this vector. Analytically, it's just the scalar product of P minus P0, scalar product with A. So this is the height function. Is this a differentiable function? And the answer is yes. And again, by example 1. That's why I spent two minutes on example 1, even though it was a triviality. Because take this definition of a function and think of it, not just defined on S. Think of it as defined on R3. As a function of R3, this is differentiable. It's the scalar, I mean, the difference of two vectors is a differentiable function. The scalar product of a vector of our variable vector with a fixed vector is a differentiable function. So if I put here R3 instead of S, this definition gives me a differentiable function. But then this function here is the restriction of a differentiable function of R3 to S. Hence, it's differentiable. OK? So really the geometric interpretation to say it's the length of this vector, because you see you take p minus p0. You take this vector and you project it in the direction of A. So what you are left with is with this vector here. And now you would like to say geometrically it's really the height of this point with respect to this plane. Now this geometric interpretation is indeed correct only if A was of norm 1. If A was of norm 2, this was twice and so on. So the geometric interpretation, it's better to be careful and fix A to be a vector of norm 1. Otherwise, this function is not exactly the length of this segment. Let's define another couple of functions with this kind of geometric flavor. For example, the distance squared from a surface now. So 1, 2, 3. Distance squared. What does it mean? It means that I have no, actually not from a point. I fix a point p0 over 3. And I look at the function. S is always a regular surface. I don't stop writing it. And I look at the function f from S to R taking what? Taking a given point of my surface and taking the distance squared to this point here. Now if you want to draw just a little picture but you have your surface S, you have a point p0. I never said if this p0 was on the surface or not. It's irrelevant. So let me draw it outside. But it could have been inside. I take this p0. I take a point p. And I'm looking at the function this norm squared of this segment. The norm squared of this segment. Now is this a differentiable function? Yes. By example 1, why? Because this function, again, forget that I'm looking at it as defined on S. Look at it as defined on R3. So p is free to be any vector. What is this function? So if just to, I mean, if you want, suppose p0 is given by 3 coordinates x0, y0, z0. And I take p is equal x, yz. This function here, sometimes we call it d squared. How much is f? f of p is what? Analytically, it's x minus x0 squared plus y minus y0 squared plus z minus z0 squared. So defined on R3, it's a differentiable function. It's a polynomial of degree 2. So if I restrict it to a regular surface, it's still differentiable. But now epsilon variation, don't square. What about the distance? So the distance means, exactly in the same notations, take f to be f of p to be the norm and not squared of p minus p0. Now, is this a differentiable function? Now, sorry, I have to repeat it. So you are saying it is a differentiable function outside this, but you have to be a bit more precise. What does it mean outside this? I mean, f, I would like to be f to be defined from s to r. And in fact, it is defined, of course. The point is, is it differentiable or not? Exactly. So the key point is, where is p0 now? Before, in 0.3, it was irrelevant. If you square, you don't care. But if you don't square, where is p0 becomes important. So if p0 is not on s, then this is a differentiable function. Why? Well, because I can look at this function, if I look at it as a function on r3, I have a problem. Because certainly there will be p0 in r3. And at p0, this is not differentiable. So I cannot say, forget it's defined on s, define it on r3, and say it's differentiable. And so it's the restriction of a differentiable function. No, this doesn't work now. But I mean, epsilon, careful. I just look at it as a function defined from r3 minus p0. This is a nice open set of r3. It contains s because of this condition. And we are done by 0.1. Now, one comment. So these are little tricks. I mean, going back to functions defined on open sets of r3 and then restricting, it's a nice trick. But it doesn't always work. So we have to go back a little bit to the definition of 0.1. The definition of point, remember, the definition of point 1 was suppose you have a function from s to r to Rn. You say that this is differentiable if every time I have a local chart, f composed x as a function from u to Rn is an all differentiable function in the usual way. Now, I would like you to spend five minutes now to think. I mean, this is in principle, I mean, it's an elegant definition. Philosophically, it's OK. It means you are using charts to transpose everything back to r2. But in practice, this is impossible to check. I mean, this has to hold for any chart. The number of charts is in principle infinite. And in fact, it has to be always infinite, even more than countable. So there is no way you can actually check this condition for any chart, OK? But little observation. You understand that there are always infinitely many charts. So even in our cases, you see, the most stupid case of regular surface, the plane. And you say, why? Why the plane is a surface? Well, because I think for every point I need to produce an open set and the map, such that, blah, blah, blah, blah. And then I say, well, but OK, for the plane, I can use one chart, which is good for all the points. It's correct. And this shows that this is a regular surface. But then you say, but how many charts does it have? One, no. You used one to prove it's a regular surface. But then around this point, I can take a little disk and use another strange map. And then I can take a star-shaped open set and use another strange map. And this will be another chart and so on. And around every point, I can play these stupid games, for example. So for every point, I have infinitely many games to play and so on. So it has to be the number of charts is always huge, certainly more than countable. Now, the point is that there is a little observation that it's worth doing at least once. Is that, OK, the definition would require something impossible, but actually it's enough to check something possible, which is equivalent. Now, what does it mean? Let me state it as a little proposition. Now, S is always a regular surface. So now we have F from S to Rn, a function. Not nothing more, no regularity a priori. Now, if for any point P, if for any P in S, there exists a chart, X, that I call actually XP, just to underline the fact that it's a chart around the point P, from an open set where, again, I put a little index P to S. So if for any P, there exists X, local chart around P, such that F composed XP as a map from UP to Rn is differentiable, then is differentiable. Now, I would like just to think, what is this proposition really saying? F is differentiable means for any chart, the composition is differentiable. While in the first part, I'm saying, it's enough to check that for any point, there is one chart for which this is differentiable. Is it clear that this is a huge improvement? Because the stupid tricks I was drawing before, disappear. So for example, if I have a function now, if my surface was the old plane of before, and you say, well, to prove it's a regular surface, actually I used only one. So there was one chart, which was good for every point. You agree? Now, this proposition is telling me, look, now, if you have a function defined on the plane to check whether it's differentiable or not, it's enough to check it on that chart. So one. So you reduce from more uncountably many to one. Proof would like you to try. So of course, in one way, it's trivial. So differentiable is more than this, in principle. So from below to the top, it's automatic. The point is to prove that if you have this one, which seems weaker, you can actually prove that this happens for any chart. Do it as an exercise. Because now I would like to move to probably the central object in some sense, even though the central technical object. We have a surface in R3. Now we know what is a differentiable function defined on the surface or with values on the surface and so on. Now, it's clear that the key geometric object of a surface is tangent space. So what is the tangent space to the surface at a given point? I really want to make sense of the very intuitive picture that we all have in our minds. So definition. Of course, we will define the tangent space as the set of tangent vectors. So really, the definition of what is a tangent vector. So S is a regular surface, and P is a point on S. We say that V, a vector of R3, is a tangent vector to S at P. If there exists a curve, alpha, parameterized on some little interval containing 0. I mean, this is, of course, just to fix again notation. And now, you see, I know what it means. A differentiable map from an interval to a surface is a particular case of the list of definitions we had before. It's a function with values on a surface. So I know what it means to be differentiable. I defined it 20 minutes ago. Of course, epsilon has to be strictly positive. Otherwise, it's stupid. It's the empty set. Such that, so I want a curve, the existence of a differentiable curve, such that at time 0, I'm passing through the point P with velocity V. The set of tangent vectors to S at P is denoted by TPS, tangent space to the surface at the point S. So geometrically, it's very simple. So what is a tangent vector? It's any velocity of any curve passing through this point. So every time I draw a curve on my surface S, passing through this point at time 0, I go and see its velocity. And this will be a tangent vector. So I can do it in all possible ways. So alpha, the image of alpha, has to be on S. It's not just any curve in R3 passing through this point P. It's a curve in R3 passing through it, but the image must be on S. Well, we quickly need another way to describe it, which contains also algebraic and geometric informations about what this space is. You see, by definition, the only thing I can say is a set. I want to say much more, of course. I want to give it much more structure to this set. And this is given essentially by a lemma, which is the following, in the notation in the definition. So a regular surface at the point, if x from u to S is a local chart around P, then the set, TPS, is actually the image of the differential of the chart, at the point x inverse of P over R2. In fact, let me draw the picture here. So we have our surface, S. We have a point. Since this is a regular surface, we have a chart around this point, at least one. So take one. What does it mean we have a chart? We have some open set of R2, u, and the map with values on my surface, which will cover some open set. And of course, in this set u, in this domain u, there is a special point, which is the inverse image. So x is 1 to 1 in particular. So points here and points here are identified by x. So here, there will be some special point x inverse of P. So at least just to understand the symbols. So the differential of the map x goes from R2 to R3. And what this lemma is saying is that actually its image, at a given point, is exactly the tangent space to the set of tangent vectors at the image point. Is it clear the statement? Now we prove it. Let's first prove the inclusion. It's an equality of sets. So we have to prove two inclusions. Let's first prove this one. What does it mean, proving this inclusion? I need to take, so let w in R2 be any vector. And I want to prove that if I apply the differential of the chart, I end up in the tangent space. So what does it mean? I think of w as a vector somehow based on the point x inverse of P. As a vector, I should draw it here. But I mean, just geometrically, I draw it here. It's just a picture, because then I will write formula. But then I can take a curve in R2 passing through this point at time 0 with this velocity. In particular, I can take the straight line, because I'm leaving now on the plane. So in particular, I look at the curve beta. Beta of T is equal to what? If I let me call it x inverse of P plus T w. Now, you see, so this is just a parametrization of the straight line, which at T equal to 0 is this point with the directional velocity, I mean, with the directional vector w. So it's really this beta is parametrizing this line here. Now, u was an open set. X inverse of P is an interior point. So certainly, a little piece of this straight line must be inside u. So for T, so I think of beta to be defined for minus epsilon epsilon into u. So there is a small epsilon for which this is with values inside u by topology, by openness of the domain. So of course, the two properties which are interesting for me are beta of 0 is equal x inverse of P. And beta prime of 0 is equal w. These are the only two interesting properties. And then I define alpha to be the exact. So now I have this curve, which is in particular a straight line on u. I use x to put it here. And I call this curve alpha. So alpha from minus epsilon epsilon to s. Now, defined to be x composed beta. So you see which are the properties of alpha? Well, how much in particular? I mean, is it alpha satisfying something interesting in this sense? So alpha of 0 is what? Well, I take T equal to 0. So this becomes x of beta of 0. But beta of 0 is x inverse of P. So x of x inverse of P. So it's P. And how much is alpha prime at 0? Well, but this is just the definition of the differential of a map. I'm taking a curve. And I take the derivative of this curve at time 0. This is, by definition, dx at x inverse of P, because it's really at the point alpha of 0. So in general, I should write here alpha of 0, but I've just noticed, sorry, not alpha of 0, of beta of 0 of the vector w. This is exactly the definition of the differential of a map. But then what? Then we have really proved what we wanted, because given any vector w, we have proved that the vector dx at x inverse P of w has the property that there exists a curve alpha on the surface passing through the point P at 0 with exactly this tangent vector. This is exactly what is the definition required. So this vector here is a tangent vector to the surface because there exists alpha. OK? So in particular, this implies the claim, I mean, just to write it down, dx of x inverse of P of w is a tangent vector, OK? Well, the other inclusion is essentially the same thing, reversing everything. So now we have to prove this inclusion here. How do we reverse everything? So now let v, we have to prove that if I take any vector, any tangent vector, is actually of the form dx x inverse P of something, OK? So let's do it. Let v be any vector in TPS. And let's try to find w. So now the problem is the opposite as before. But if it is a tangent vector, why we are reversing everything? If I know that v is a tangent vector, I know that there exists a curve alpha with such and such property. But then there exists alpha from minus epsilon epsilon 2s with alpha of 0 is equal to P and alpha prime of 0 is equal to v. This is forced by the definition. And I take epsilon sufficiently small so that I don't even write it, OK? Again, a little topology. I pick epsilon sufficiently small so that alpha, the image of this curve, so alpha of this interval, lies in the image of the chart. Again, the image of the chart x is an open set which contains P. Alpha is a curve. So in particular, it's a continuous function. So if I restrict the domain of alpha, if necessary, I can be sure that the image of alpha lies inside the image of x by openness. Now, and then I do exactly the opposite of what I did before. I define beta. So put beta to be equal x inverse of alpha. If I look at this curve beta, so this is again defined for what? It's defined for minus epsilon, epsilon for sure. And where is the target? Well, alpha was going to s. And then I take x inverse. So I end up in u, OK? And which are the properties of beta? Meaning, again, where it's passing at time equal to 0 and with velocity. Well, beta equal to 0 is, by definition, put t equal to 0. Alpha of 0 is equal to P. So beta of 0 is x inverse of P. And how much its velocity? Well, it doesn't really matter, actually, because what it matters is what? What is v? Now, what is the relationship between v and beta? Well, v is, by definition, alpha prime, by definition of alpha is the velocity of alpha at time 0. But alpha, you see, I can reconstruct alpha out of this formula. If beta is equal x inverse composed alpha, alpha is equal x composed beta. I compose with x on both sides. Here, they cancel. And I'm left with alpha is equal x composed beta. So it's x composed beta prime at 0. But this is, by definition, what? Again, this is, by definition, the differential of the map x at the point x inverse of P of the vector beta prime of 0, whatever that is. Again, that's why I spent 15 minutes on the definition of the differential of a map. Because now you see, I'm using it often. Now, whatever this vector is inside, maybe it's not a straight. You see, beta, now, going one way, we could choose beta to be a straight line, and then we put it on S. Now, going the other way around, maybe beta is a very strange curve here. I have no idea how beta looks like. But it doesn't matter. It will be a curve passing through this point with some velocity. V will be exactly the image of the differential of the map x of that vector. So this ends the proof. I've written V as something like this. So I end up proof. Now, why this is nice? Because this lemma automatically gives me a very strong structure on this set. This was born as a set. But actually, if it is the image of a linear map, the differential of a map is always a linear map, from Rn in this case, from R2 to R3. So its image, in particular, it's a vector space. It's not just a set. It's a vector space. And in fact, it also gives me, in fact, of which dimension? Well, in general, so this map, dx, goes from R2 to R3. So the dimension of the image is what? Well, in general, of course, the only thing you can say is less than or equal than 2. But the definition of the regular surface was requiring exactly that x was a special map for which dx was of rank 2. So the image is a plane. Otherwise, x was not a local chart. So the tangent space to a regular surface, at any point, is automatically a two-dimensional vector space. It's better to write it down as a corollary. And actually, we also have, in some sense, a canonical basis every time we have a chart. So corollary, in fact, corollary is, if you want, 1. TPS is a vector space, meaning it has a natural structure of vector space, the one coming from. And the other, so this is a kind of a chart-independent statement. But then you can be, if you want to use explicitly the chart, you can say that the tangent space at a given point is generated by the two vectors given by some sense, dx, du, dx, dv. So these two vectors are two tangent vectors, by definition. If you want, you can use the previous lemma or, I mean, these are the tangent vectors to which curves. Well, now you can answer this question in two ways, because you have the definition, which is saying something is a tangent vector if you can produce me a curve on the surface, which has that velocity. Or you can use the lemma. You can say something is a tangent vector because I can prove to you that it's in the image of dx. Now, so that's why there is not a unique answer to my question. You can prove it in two ways now, OK? In any way, for example, if you want to say, well, this is the tangent vector to a curve, so I'm using the definition and not the lemma, well, the partial derivative in one variable, in one direction, is actually the tangent, is the directional derivative in the direction exactly of, I mean, you restrict the function to the line giving this coordinate and you take the derivative. That's the curve that you use. And this other one will be the vertical restriction, OK? Or you can say, well, but these are obviously in the image of dx, but these are the column vectors of the matrix representing dx. We proved it. So, OK? So in both ways, there is nothing to prove. Now, let's go back to our series of examples of regular surfaces and try to compute some tangent space. So a general warning, when you draw pictures, so you draw your surface and both me and books and whatever, you will always find a picture of this form. So this is the surface, this is the point, and everybody will draw something like this as the tangent space. You imagine this two-dimensional vector space translated with origin at the point P, OK? The truth is not this one. So this is just to fix in your mind that this is the tangent space at the point P because really to draw, if you really want to draw which space it is, which set it is, well, since it's a vector space, it's better pass through the origin, OK? So it's really the translation of what you draw in this geometric way here, OK? Warning, OK? We will always make this mistake happily, OK? Doesn't matter. Now, probably the biggest class of examples we have now are the surfaces given by regular values of functions. So let's try to compute the tangent space of this class of surfaces. So we have examples. One, O is always an open set of R3. F of R3. F from O to R is a differentiable function. And we take a regular value. A is a regular value of F. So we know that the inverse image of A is a surface, is a regular surface, OK? The question is, pick a point and compute the tangent space. So claim. If I let me give you, it's easier to write down the claim and to prove it, OK? TPS is identified, actually, with the kernel of the differential of the map F as a map. So the differential of F goes from R3, because this is an open set of R3 to R, OK? If A is a regular value, we know that this map is never zero. Being the target R, it means it's always surjective. Because if it's not zero and the image is one dimensional, it's everything, OK? Meaning the kernel is always two dimensional, 3 minus 1, OK? So at least, from the dimensional point of view, could be, OK? Now let's check that this is true in a deeper way, OK? Proof. Again, we have to prove two inclusions, OK? So let V, so this inclusion here, how do you prove it? You pick V in TPS, and you want to prove that this lies in this kernel, OK? So what does it mean? Well, first, let's use the definition. If V is a tangent vector, there is a curve. So this implies, by the definition, that there exists alpha from minus epsilon, epsilon 2S, such that alpha of Z, differentiable, alpha of 0 is equal to P, and alpha prime of 0 is equal to, with velocity V. Now, S is the inverse image of A. What does it mean? It's the inverse image. Every point on S, if I compute F on every point on S, I get A. So in particular, F of alpha of T is equal to A for any T, OK? But then, how much it is? The differential of F at the point P applied to the vector V. Well, again, remember the definition of the differential. It's the derivative with respect to T, evaluated at T equal to 0 of F composed of alpha, OK? This is, by definition, D in the T, at T equal to 0 of F composed of alpha of T. I mean, this is not true in this specific case. This is always true. You want to compute the differential of a map at a given vector. You have to pick a curve passing through this point with this velocity, take the composition, and take the derivative at T equal to 0. This is the definition of differential. But in this specific case, F composed of alpha is constant. So this is 0, OK? Which means exactly that V lies in the kernel of the differential. So this inclusion is done. Well, the other inclusion, you can either prove it formally as an exercise, or you make the observation I started with. What have we proved? I mean, essentially, so we know by general theory that this object here is two-dimensional. And we just proved that this object is contained here. But this is also two-dimensional by general theory. So this is a two-dimensional vector space contained in a two-dimensional vector space. So there is nothing to prove. They have to be the same, OK? So there is no other inclusion to be proved. Very well. Another example, see, whatever you define or you produce on a surface, you better check everything on the plane and on the sphere. These are the two basic examples, OK? Well, in fact, the plane I leave you, it's too boring. First exercise 0 for you. But let's do together the tangent space to the sphere. Now, in this case, let me use the, I mean, I pick the general sphere, meaning, so this is the set of p over r3. So I want to just to, the set of p over in r3, which has such that p, the distance from the center, this a will be my center squared is equal r squared, OK? So this is the sphere of radius r and center a, OK? How do I compute it? Well, I can use 0.1 because the sphere is actually, I can think of the sphere as a surface given as the inverse image of a regular value. So if I look at the function f, f of p, so f defined, if you want, on the whole of r3, OK? f of p to be equal p minus a squared, OK? So you see, I take this. Then, of course, as the two sphere of center a and radius r is, by definition, the inverse image of this, OK? But then, OK, I don't remember if we have done it explicitly. I think so. So I believe we have checked that this is a regular value, OK? Now, by point one, the example one, I know the tangent space at a given point. So what is the tangent space? So tp s2 ar squared, it's the kernel of the differential of f at the point p, OK? So this is what? This is the set of vectors v of r3. And now, well, probably that's the computation we did when we checked it. So what is the differential of the map f? Where is it? f. We need to compute dfp, OK? What does it mean? dfp. dfp of the vector v is what? Remember the definition. It's here. I need to take a curve alpha passing through the point p with velocity v. Take the composition, I've composed alpha, and take the derivative at t equal to 0. So let's do it, because we have the explicit expression for f. OK, I don't take an explicit expression for alpha. I don't care. This is enough. These are the two informations I need. So this is d in dt at t equal to 0 of f. f is the norm squared of p minus a. So this is of. And what is the norm squared? One way to write it is the scalar product with the vector in itself. So I can write it as alpha of t minus a, scalar product alpha of t minus a, OK? It's convenient so that now I can do the, I can compute the derivative easily because the scalar product gets differentiated like a standard product, means derivative of the first. But this a is a constant. So this is alpha prime at time 0. So it's v. Scalar product, the second, evaluated at 0. But this is p minus a plus the first one evaluated at t equal to 0, which is p minus a. Scalar product, the derivative of the second at t equal to 0, which is v. And then the scalar product is symmetric. So this is nothing like, but twice v p minus a. Sorry, here they, OK? This is not the continuation of what is the kernel. This is the continuation of this. So let's avoid confusions here, OK? So how much is, so what is the kernel now? So let me continue here about the kernel. Well, kernel means the set of vectors for which this is equal to 0, OK? So the kernel is, I can say, this is the set of vectors v in R3 such that the 2 is irrelevant if I say it is equal to 0. So this is v scalar product p minus a is equal to 0. But this is easy to identify geometrically, you know? Because suppose, now instead of the general surface, let's draw the sphere. So we have this sphere put somewhere in space. So this is the sphere of radius R and center the vector A. A is a vector, right? But we have never said where it is. So in general, x, y, and z are somewhere else, OK? So what is this space here? Is the set of vectors, say it geometrically? Is the set of vectors orthogonal to p minus a? So p, actually, so the tangent space at the point p, that p is given, OK? A is given. I look at the vector p minus a. And now the tangent space is the set of vectors orthogonal to this vector. And this is exactly what we learned when we were 14 or something like that. It's a characterization of the tangent space to the sphere. It's the orthogonal to the radius vector, OK? I guess this is enough for today.