 Tako, nekaj smo bilo vedno razmah, da se vse nismo nekaj inočno, vsej vstajno, nekaj se vsej vsej vsej vsej vsej. Zatpili, nekaj se inočno se kajsem vsej vsej, nama se vsej, na našem, našem. Vsej, je je tega, da je vsej prav, Sreči, to je pravda, kaj sem tudi stavila, kako jaz define objev v matematikih, nekaj v geometriji, danes vse matematik, je da vse objev klasifuje vse objev na vse objev. Vse je to vse. Vektorsk vse. Vse vse vse je isomorfik v Rn. Vse vse je isomorfik ki što je taj zemljena, isomofik, s edmine. Vsih topoložičnih spasih s druh, je toz, da je homo-mofik. Zato si sem predržila, da pa srečim, ljudi, ki imaš vse občanje, nrhu se imaš vse tudi vse jezemljena s ovom občanjem. Nrhu topoložičnih spasih je zelo klasificovanjati o homo-morfizmi. Vsih topoložičnih spasih je vse vse. Maybe they look completely different but there is no identification which respects the structure that you are using, or a vectorusspA2. Mark Pobrejnje nalucenja končna, belge unbelije, vektor nekaj, vektor na zčetku. Kaj sem možem, da prič letom, nekaj sem se ne admitteda, z kone, da preč poslednji nekaj na veš del, ljudje nekaj na veš del na veš del, nadal je, da моей ima se to rejadi, da ga predstaviti po več delu. Spravno modimo vetr, jelja, da poslednjimo objevs. Isn of the Equivalence Relation we want to put). So when we say, a surface with, for example, positive Gauss Curve is one of the following lists, well there'll be a theorem like this, up to something. What is the something? In some occasions we have already met them, Trzeba, kako se počekamo r3, z vsej zrčenimi motjenje, vsej zrčenimi, zrčenimi, z reflekcij, zvom da smo vsej vsej zrčenimi objev, da smo vsej. Kaj smo zelo, njiha je spheja, njiha ne zelo, kaj je vsej zrčen, njiha ne zelo, kaj ne zelo, kaj je spheja, zelo, njiha je spheja, ne zelo, ne zelo, ne zelo, ne zelo, ne zelo. Nači na primer fondamentalne formu, prvičen je, da je občvodnja neč drankovina. Ľeljati občvodnih v beru občv 지역u občvodnoj neč الص grandchildreni. Prvič neč calo mirič, bese deračne, da bomo neč calo pričeljano iz ter z ložstvom počkov, da ste pričodno. Pričal, da pa im 마žem, da se važna. . . . . . . Zelo, da sem pozdravil. Če imem metričan geometrič, tudi je ozometri. A če je ozometri? Zelo, da sem izmetričan? Zelo, da je zelo izmetričan v tomziku, da se poslišite svoji občas, da v sej češi ozometriči nekaj, je, da sem prej, nekaj, ko je začal lahko bas četila해야, ne bo ne bo neondekit evenly. If there is edithomorfism, meaning a differentiable militar name, one to one with differenceable inverse, which take each curve on S If such an f exists, the two surfaces are called isometric, and such an f is called an isometry. So here the two words are these. So that means you can identify them. Of course, this is stronger than everything below. I mean, two surfaces are certainly topological spaces, and I'm certainly requiring them to be homeomorphic. They have what we call a differentiable structure, and we are identifying also the differentiable structure. This is something that will be clear at the end of the course, what I really mean. So a diffeomorphism is more than a homeomorphism, because it's an homeomorphism which is differentiable with the inverse, which is not just continuous, but it's differentiable again. So it's more than that. Plus it respects lengths of curves. So it's a stronger notion than everything else that you can imagine on your set. It's the strongest possible notion of identification. They really look the same like metric spaces. But now, for what we learned last time, these lengths of curves depends only on the first fundamental form. We can characterize isometric surfaces by looking at their first fundamental form. The theorem is two, in fact, notations, which I think we have already used, but let me call a patch. What is a patch of a surface? Of a surface is the image of x from mu to s for a local chart. So we have a local chart. The part of the surface covered by this, so this piece of the surface covered by x, I call it a patch of x for a local chart x. So this is just a word just to avoid writing every time all these sentence. So now the theorem, I want to state it in this way, two patches, two patches of surfaces, s and s tilde, are isometric, if they can be parameterized, can be parameterized by one x from u to s and another x tilde from the same u to s tilde. In a way, the first fundamental forms coincide. Now, such a statement requires a comment. So what are we saying? We have two surfaces, which maybe look completely different. One is something like this and one is something like this. I'm not looking at the whole surfaces. I'm saying a piece of this, which I know it's the image of some local chart, but I don't even want to write which chart. So I take an open subset of this surface and an open subset of this surface here, which are sufficiently small to be image of a chart. Now, these two pieces of the surfaces are isometric. Of course, a patch of a surface is a surface. So I can do all my theory, I can restrict the theory from s to this piece. So these two pieces, if I look at them as surfaces by themselves, are isometric. If I can parameterize them, if there is, if there are two maps, but from the same, so here there are many important, you see, the domain of the two charts, which in principle were completely different. I mean, starting from the fact that these were patches, here there was a u and here there was a v. Now I'm saying they are isometric if and only if, here if and only if. I can find another chart, which is good, I mean, of course the charts will be different, how do I call, x and x tilde. Define on the same domain, which work for these two things, because now if they are defined on the same object, on the same domain of R2, of course the coefficients of the first fundamental form of this piece are functions over here. The coefficients of the first fundamental form of this piece are functions over here, over the same domain. So now I can compare them. If they were defined on two different things, how can I say e is equal to e tilde? They have even a different domain. Now e and e tilde have the same domain. So I can ask, well, are they equal or not? And now I'm saying, once I do this trick, something is isometric if and only if the coefficients are the same. They are really the same functions. So e is equal to e tilde, f is equal to f tilde and g is equal to g tilde. So now at least the statement of the theorem must be clear, because actually then the proof is almost, very simple once you understand what we are saying. If I let me rate, well, the picture might be useful. So proof. Well, in one direction, it's essentially what we observed last time. So if these maps exist and the coefficients of the first fundamental form are the same, which will be the isometry. Now I need to consider in this direction. So let me go from bottom to top. So let me say it in this way. So now I know that this picture exists and if I measure the coefficients here, they are the same. So now I need to prove that they are isometric. So I need to construct for you a map f not between s and s tilde, but just between this piece and this piece. Well, how do I do it? I take this point, I take it back, and I go here. I have these two maps, which are invertible, differentiable, everything I want. So now the only thing is to convince yourself that this is an isometry. But what does it mean to have the same? So what should I check? I should take a curve here, I should measure its length on s, and then I should take the image and measure its length. And then I should argue that they are the same. But why this is true automatically for what we know? Because I write the length of this, instead of measuring it here by what we did last time, both curves essentially corresponds to the same thing here. And the distortion between the length here and the length here is measured by e f and g. By e f and g are the same for s and s tilde. So there is nothing to prove. The length of these two curves are the same by brute force, by what we have served last time. So in this sense, once you understand what you are proving, there is nothing to prove. OK. Let's see the other way around. Because now, so going the other way, the other direction needs a bit more care. Because now, I know, I start with two surfaces, two patches of surfaces which are isometric. So now I know that there is a map here, f. And now the problem is to construct x and x tilde. So this seems more difficult. OK. How do I do it? Well, the first patch, so start with any patch, sorry, any local chart. Let me call it x, because the first one will be free. The problem will be to construct the second adapter to this. So the tilde adapter to this one without tilde. So start with any local chart x from u to s, covering the patch on s that we are starting. OK. So we start with something like this exists. OK. Now the problem is how to complete the picture with the tildes. OK. Well, what do I do? I force the picture before to be true. Meaning. Now define x tilde to be what? It has to be defined over the same u, otherwise it's useless. So I'm trying to define something like this. OK. And I do what? So x tilde is f composed x. OK. Now this looks very good, but so this will define, you see, notice that I don't know exactly the image. So that means in this statement, which is not really mathematical, it's very English. That means that the second, I cannot prescribe exactly the second patch. OK. So the second patch is determined by the first, in some sense. OK. Now. Now what do I have to check? I knew that f was an isometry. I need to check that if I write the second, the coefficients of the first fundamental form with respect to x, it's nothing because it was any chart. So give me e, f, and g as they were born, but now compute e tilde, f tilde, and g tilde for the corresponding chart. OK. But what do I know? That f is an isometry. OK. That means that if I, since f is an isometry, for any curve here, for any u of t, v of t curve in u, so whatever, now I fix this curve here, the length of the curve, of the image curve here, and the length of the image curve, now, of course at the end I'm going to do this, but really the way I want to see it is send it to s and then send it here. OK. Since this one does not change the length of curves, the length of this and the length of this are the same, but x tilde is the composition. So the length here and the length here are the same of this. But what is the length of both curves if you want? Well, I have two surfaces now, so I can write it in two ways. Once if I look it here and once if I look it there. So this is telling me, suppose I pick a curve which is parameterized in some interval a, b, it doesn't matter. OK. If I let's take zero epsilon. OK. Let's take any little piece of any little curve. OK. The length would be what? The integral between zero and the epsilon of our usual, the old expression, e u prime squared plus 2f u prime v prime plus g v prime squared everything to 1 half dt. And this would be the length here. But the length on the other side is the same formula with the tildes. So this would be the integral between zero and epsilon of e tilde u prime squared plus 2f tilde u prime v prime plus g tilde v prime squared 1 half dt. OK. But now this has to hold for any function, cup pair of functions, uv. So, of course, now it's just a matter of choosing at every point. So now for every point, because I need to prove that e, f, and g are the same at every point. So I pick a point. And then I take which curves. Well, let's try the coordinates So fix now u0, v0 a given point in u. OK. And then I look as curves. I look at, look at now let me write it in this way because I will make few cases. OK. I take the curve u0 plus t v0. That means keep v fixed. So if this was the point now I'm moving in this way. OK. What does it happens if I put it in the formula? This implies what? So that in particular means v prime is equal to 0. v is constant. OK. So every time you see v prime, poof. Disappears. So what is left? It's left. The integral between 0 and epsilon and actually u prime is equal to 1. OK. So that means you get the integral between 0 and epsilon of e to the power one-half dt this is equal to the integral of e tilde to the power one-half dt between 0 and epsilon but this holds for any epsilon. So take the limit as epsilon goes to 0 and you get what? What is the limit of the integral if you take any function now I call it f h. OK. The limit as epsilon goes to 0 of this integral is what? Zero. Sorry. Divided by epsilon. No. Sorry. Divided by epsilon. 1 over epsilon. OK. But that implies exactly that e at u naught v naught is equal to e tilde to v naught u u naught v naught. OK. Now this was good for e. What will be good for g? Well, the other coordinate. OK. So you take u naught v naught plus t and this gives g. Because now u naught u prime dies and v prime is 1. OK. So this implies g is equal to g tilde in g are the same. Then you can do f. And what do you... So this would be good for g. Now once you know it's good for the two axes you do the bisectrics. OK. So now you take u naught plus t v naught plus t. Of course this would be more complicated because what I mean this implies u prime is equal to 1 v prime is equal to 1. So if you put it here everything survives. This becomes the integral e plus 2f plus g. Is equal to the integral of e tilde plus 2f plus g tilde. But now you know that e and f are the same. E and g are the same. 2 e tilde and g tilde. So those pieces disappear and you are left with f. OK. End of story because now we proved what we... Now let's do quickly. Yes? So the question is where in this one that I haven't written. Well. So let me repeat the argument quickly. How do I prove that if I have two charts defined on the same domain with the same coefficients of the first form then actually what I drew here is an isometry. OK. So basically the idea was no, the idea was I have the map. The map is essentially given. OK. Because now remember that the map was constructed I have x and x tilde. So I define f to be x tilde of x inverse. OK. So now I have to argue that this is an isometry. How do I do it? Well an isometry means pick any curve measure its length it's the same. So how do I argue? Well, I take these two curves so one curve and its image and I take them on the domain u. The point is that they come from the same curve by construction. So there is one curve here which goes here and here. Now if the coefficients of the first fundamental form are the same the lengths. OK. So let me quickly because I don't want to lose too much time on let me quickly given on trivial example of this situation. If you take for example oh sorry actually the simplest example I forgot maybe we can recover it together. So example one if you want is a part of the plane so take any strip for example pi 3 pi of the plane and take the cylinder. Think of the cylinder as the surface of revolution of this I mean take a straight line here and rotate I mean bend your piece of paper now this map here this map here is an isometry exactly because compute the first from the coefficients of the first fundamental form is the same thing. OK. So let me pass to example 2 which is a bit more delicate oh well but actually that means something interesting of course it's not true that the plane the whole plane and the cylinder are isometric they are not even homeomorphic do you agree one is simply connected and the other is not OK. But there is essentially one but that is reflected in what is that if you construct the map from the plane to the cylinder you go around infinitely many times OK. So you cover the cylinder infinitely many times you have to restrict yourself to a stripe of length 2 pi to cover it only once OK. So it's just a topological problem besides the topological problem they are really the same. OK. So this starts explaining this strange accident no? Gauss curvature is zero how is it possible well in fact they are the same OK. Let me comment a little more later. Now take the right cone meaning X, Y, Z in R3 such that in R3 such that X squared X, Y squared is equal to Z squared so how does it look like this is really if I cut with Z equal to a constant I get a circle so this is really the cone in the old fashioned way not the generalized cone we discussed a couple of times ago so this is really this one and now OK. In the plane in the plane I take the region somehow in polar coordinates given by rho or I mean I am going to give you u v but now I think of u as the polar as the norm of the vector and v as the angle so I take u and v free to move in zero plus infinity times zero pi square root of 2 OK. So how does it look like well of course so for every point there is the whole ray this is the information coming from here but the angle has to be bounded by zero so let me do it like this open and pi square root of 2 which is something like this so the whole region is this do you agree? now construct the map from this region from this region to the cone given by x of u v basically it's really what you do with your kid to produce this cone with a piece of paper so here it was just bending the paper and becomes a cylinder now if you want to construct the cone what do you do? vzor, and you glue what is the mathematical formulation of cutting and glueing in this sense this is just u cos v oh sorry no, no, this is I can tell you so ok, u v goes to u over square root of 2 cos v square root of 2 u over And u over square root of 2. Now you have to do something also on the plane. So in fact this would be x for the first one, but I need to construct an x tilde of u v for the plane. The plane I want to parameterize it in this form. u cos v, u sin v. So basically I have a fixed domain, so now this is really the situation of the theorem. Because you have one domain and two maps, one which goes, sorry, zero. Because I think of the plane inside z equal to zero. Z equal to zero is here, so I somehow put this picture in R3. But I do it in a slightly non-trivial way. And now you compute the coefficients of the first fundamental forms and you get that they are the same. Sorry? The cone is not a regular surface, good observation. In fact I'm doing everything on the upper half or the lower half. I'm never touching zero. So it would be zero, zero, the bad point. And I'm never touching it. The vertex is of course a point where you can see actually that this is not a regular surface even by topological reason. There is a point that if you remove it, you disconnect the surface. And this is impossible on a regular surface. Topology. So this is just one example, a non-trivial example of the situation we saw in the theorem. But I don't want to, because actually there is nothing to learn now. You do xu, xv, x tilde u, x tilde v, you take the scalar product, it's the same and that's it. I mean there is nothing to learn from doing more than this. But you can see that the situation starts being more richer than what you might have thought at the beginning. And the other thing you can do as an exercise is to check or try to prove that if you have a developable surface, you have an isometry with... Ok, now, this was coming from the observation, the important observation, that the lengths of curves are measured with the first fundamental form. But the first fundamental form measures also something else, if I find it. Namely, angles between tangent vectors. What do I mean? Well, of course, the first fundamental form is not telling you just the norm of a vector. Remember, measuring the length of a curve, measuring integrating the norm of the tangent vector. So, in principle, to measure the length of curves, I would need less than the first fundamental form. A bit less. I don't need to know the scalar product between x and y. I just need to know the scalar product between x and x. So, the first fundamental form has something more. Angles. So, not just the norm of vectors, but angles between directions. Ok, so let's see if this is leading to something interesting. And, in fact, it is. So, definition. If I look at this, x from u to s. So, this is a local chart. It's called... Then I'm going to give you two definitions. So, one. Angle. If it preserves angles. Now, I could stop here, but then what does it mean if it preserves angle? Well, we are in this picture. Let's draw here the usual picture. We have our surface s, our patch, and our map x from some domain. Let me draw it big, because now I have to put many things inside. What does it mean? A map preserves angle. If I take a point at every point, if I take two curves passing through this point, but basically, ok, suppose I take two curves, but in fact suppose I take two vectors and imagine pointing them at this point here. Of course, using the standard scalar product, I can measure this angle here. Ok. But x tells me also how to put it here, how to send these two vectors to two tangent vectors to the surface, because I can take the differential of this map. Ok. And so here, with the first fundamental form of the surface, I can measure this angle. So, something is conformal, x is conformal, if these two things are the same. These two angles are the same at every point for every choice of pair of vectors. Ok. So, that is, let me express it, try to express, so, whenever, whenever, whenever, you have a curve, two curves, t, going to ut, vt, because now, instead of saying I take a point in two vectors, I can also take, I take two curves, whose tangent vectors are those. So, whenever I take two curves, t going to ut, vt, and t going to util the t, vtil the t, that intersects, that intersects at t equal to zero, for example, I mean, this is nothing, it's just to fix the convention. At an angle, and suppose that they form at an angle, in the plane, they form an angle theta, their images, x of ut, vt, comma, x of util the t, vtil the t, they, of course, intersect, at the point, x of u of zero, v of zero, and they form at an angle, at the same angle. So, their images form the same angle. Okay? This is just an explanation of this picture. Now, definition too. So, these are conformal charts. Since we studied also, this still is about tangent vectors, norms, and angles, but since we studied also areas, we say that a chart, a local chart is called area preserving. I mean, clearly, there could be, in principle, a nicer, among all charts, those for which if I take a region here and I measure the Euclidean area of this, this is exactly the same as taking the image on s and measuring the area as we learned last time. Okay? If this happens for any domain inside u, that means that the map is very special. So, let's give them an, and we call it area preserving, if the area of x of v is equal to the area of v for any v, let's say, compactly contained in u, because, in principle, u could be even unbounded and so on. So, at least for those regions for which it makes sense to compute the area. Okay? We can play the same game we did before to try to characterize these charts in terms of coefficients of the first fundamental form, because, after all, we know that all these objects, angles and areas depend on the coefficients of the first fundamental form. Okay? So, in fact, it's possible and it's not difficult, x is conformal if and only if e is equal to g and f is equal to zero everywhere on the domain, these are functions on u. Okay? So, on u and then two, actually this was proposition one, proposition two, x is area preserving if and only if e g minus f squared, you can guess it, no? But this is always positive. It's the determinant of the first fundamental form. But I mean, basically what you are saying is that it's area preserving if the map does not change the area. It's constant equal to one. I mean, the distortion is not there, but not there means it's one because it's the function that you put in front of the measure. Okay? So, you want to be the same. Okay, so how do we prove it? Let's do it. Oh, actually, but one comment because these theorems look different from before, before we had two surfaces and we were looking at the problem of when these surfaces are isometric, I mean locally at least on some pieces. Now, we didn't write something like, I mean, we cannot say e is equal to one or I mean, didn't make sense. Now, this is different. This situation here is like before when we are comparing the geometry of s with the geometry of the plane. So, it's like if s tilde is forced to be u. Okay? So, in principle, we could define conformal maps between surfaces. S as tilde. And, in principle, we could define area preserving maps between surfaces. Okay? And then we would have theorems like comparing, I mean, conformal map if and only if there is a relationship between e, f, and g, and e tilde, f tilde. So, I'm trying to explain why here there is not e tilde, f tilde, and g tilde. Where are they? But the point is that here we are fixing the other surface to be this. And for this, there is nothing to say. I mean, the coefficients of the first fundamental form these are 1, 0, 1. Okay? Because comment before showing you the proof, comment. What if a map is both conformal and area preserving? Well, if it's both, f is equal to 0. E is equal to g, this means e squared is equal to 1, but it's positive. So, e is equal to 1. And it turns out that what? If it's conformal and area preserving at the same time, e is equal to g is equal to 1 and f is equal to 0, but these are the coefficients of the plane. So, by the previous theorem this means the surface is isometric to u to the plane, to this part of the plane. It's an isometry. Okay? But you see that these conditions are weaker because each of them are weaker. Because isometric would mean e is equal to g is equal to 1. So, conformal is more general than isometric. And, of course, area preserving it's even... I mean, this is just a quadratic equation in these things. Okay? So, now let's prove it. Well, it's essentially now, look again at this picture. So, let's prove part one. How much is this angle? The Euclidean angle measured in u. Well, the tangent vector... So, suppose this is in my picture this is u of t, v of t and this other curve is u tilde v tilde. Okay? So, how much is this angle or cos of this angle which is simpler? Well, I compute the tangent vectors. I take the scalar product and I divide by the norms. So, if I call this angle theta so has the definition likes so this is theta and the other actually doesn't have a name but let me call it phi. Okay? Now, cos theta is what? Well, cos theta is scalar product in the Euclidean plane of the two of the u prime v prime and u tilde v tilde no? I have to take this scalar product divided by the norm. So, this becomes what? u prime u tilde prime plus v prime v tilde prime divided by u prime squared plus v prime squared one half u tilde prime squared plus v tilde prime squared I mean, this is becoming a joke. Okay? One half. Okay, so this is one of the two angles. How much is the other one? Well, what do I have to do? I have to transport these curves here and compute the tangent vectors. Okay? But which are the tangent vectors? Well, the tangent vectors to the first curve would be okay? So, this vector here would be psi and this other one would be something else. Have I given a name? Psi tilde. Okay? So, psi would be what? Psi would be u prime x u plus v prime x v. Okay? Tilde prime x v. Okay? And now I have to take the scalar product. So, now first fundamental form. So, how much is cos phi by the same principle as before? Now it will be more complicated because I take u prime, u tilde prime, but now x u x u. So, e. Correct me if I do something wrong, because it's likely. So, one scalar this becomes plus okay? You can see that it will be no, it will be u prime v tilde prime x u x v freeze it for a second plus v prime u tilde prime x u x v f. Okay? Plus now the last one, v prime v tilde prime Everything divided by the norms because these are not unit in principle. So, this is divided by a mess. What is the norm of this? It's a scalar product with itself. So, this becomes u prime squared e plus plus 2 u prime v prime f plus v prime squared g 1 half and then everything else with the tilde. This was the norm of this. Now, the norm of this is the same expression with the tilde. Okay? That's it. Now, in these theorems it's always the same trick. You write down what you have to. Now, what do we know? We know that at every point and for any u prime v prime u tilde prime v tilde prime this object, this function has to be equal to this function. So, now it's a matter to plug in this formula the right curves. So, for example which is the best one at the beginning consider, take u of t equal to t v of t equal to zero. This is one curve means, of course if you want plus v naught plus u naught, at every point doesn't matter. Now so which curve is this, this is the horizontal curve. And let's take as the tilde the vertical curve. And then I take u tilde t is equal to zero, v tilde t is equal to v. 2 to e. Now, if I do this what do I get? Cos theta well, of course, now I've taken to the two axes the angle I mean cos theta is zero, they are orthogonal in R2. So, now what is this why this could be interesting of course what have we done v prime is equal to zero and u tilde prime is equal to zero. So, let's see how many things disappear. U tilde prime is equal to zero. And v tilde so what is surviving here this is this disappears. This disappears this one no but this one disappears and this one is one. Ok, so here I get. Ok, on one side I have zero because it was cos theta so zero is equal to what is equal to f divided by something I don't even care whatever it is the something f is equal to zero very good so we proved half of the theorem and now and now take again two orthogonal things but not the axes meaning u of t is equal v of t is equal to u tilde of t is equal to t and now v tilde of t is equal to minus t ok, so one is this one and the other is this one they are still orthogonal of course so it's again zero equal to what well this and now we know already f is equal to zero ok, so here what do I get so u prime is equal to one u tilde prime is equal to one so e plus minus g otherwise it's wrong one minus one so e minus g is equal to zero and that's it ok ok for the area preserving as before it's the same proof we don't want to spend now conformal among all possible planar representations of a surface we have identified of course three types of better charts of course if you manage to do isometric you go home and that's it but if you can't maybe you can do a bit less ok actually these two are not exactly comparable no this is not less than this this is different ok but at least we have three types of charts which have some geometric interest now I'm going to give you the most famous example of a conformal chart and then now the most famous example of the conformal chart you know it already in fact you never observed because you didn't know what it meant now my surface is the sphere so I want to give you a special system of charts on the sphere one of them is conformal how do I do it this is called stereographic projection what is actually there are many stereographic projections but I mean just to fix notation you fix the north pole which is essentially any point but I mean you take one point and you call it the north pole you pick so suppose this is really the sphere of center of the origin of radius one ok otherwise everything is slightly distorted and you can not adjust it immediately now you take the z equal to zero plane and you do what you construct a map pi let me call it pi because it's projection ok which takes for each point x y zero ok you want to construct a map from r2 to s2 ok by doing what you know ok you take the line the straight line passing through p let me call this point p ok so I take p and I take the point of intersection between the line and the sphere different from the north pole so I take the line and the sphere intersects the sphere in two points do you know why how would you prove it it's the simplest appearance of an important theorem bezous theorem the sphere of degree 2 the line is of degree 1 so the intersection are two points counted with multiplicity because of course there is a possibility but in that case you counted with multiplicity 2 meaning it's tangent ok so there are only two points one is the north pole by construction I take the other one and I call it pi of p very well are we able to write down the explicit formula for this well it's a simple exercise in analytic geometry how do I get well I parameterize this line so what is this line this line is n plus t p minus n so for t equal to 1 I get the point p and for t equal to 0 I get n ok so that means I'm parameterizing this line so this is of course a line so it starts at t equal to 0 and it starts at the north pole at t is equal to 1 I'm here so that's good what are the analytic components of this line of this parameterization this is the point 001 so this is what this is tx ty 1 plus t here I have 0 ok so this is the analytic expression of this straight line and now how do I get pi well I need to intersect this with the sphere so what do I do I plug these expressions in the equation and see for which values of t I get a solution which are the roots I will get a polynomial of degree 2 I consider x, y as fixed ok so now the only parameter is t, the real variable is t so what does it mean I plug them here I get t squared times x squared plus y squared plus z squared plus 1 minus t squared is equal to 1 so for which values of t this holds ok so this becomes t squared which multiplies x squared plus y squared plus 1 ok minus 2t plus 1 equal to 1 not surprise so this is equal to 0 which means that one root is t equal to 0 which is something I knew because the north pole is a solution ok so I drop a t and the other solution is the interesting one so t is equal to 2 divided by 1 plus x squared plus y squared and now what do I do so now what is pi of p pi of the point p well it's 2x so t times x 1 plus x squared plus y squared 2y 1 plus x squared plus y squared and then 1 minus t which is usually written so one likes to have always the same denominator but that becomes minus 1 plus x squared plus y squared ok ok so now you finish your exercise because now I've done my job now you compute the coefficients of the first fundamental form of the sphere with this map so actually there's just one comment which is the image of this map s2 minus the north pole so actually this chart covers the sphere minus one point so in fact it's easy to so with two of these charts I cover the whole sphere remember that at the beginning we cover the sphere I don't remember with how many 8, 10 and of course we know that one is impossible because the sphere is compact and a chart, a patch can never be compact so now we know that with two we can do it and two is optimal ok so now you compute exercise for you you compute the coefficients of the first fundamental form nothing to do, I mean it's the usual thing ok I just call it x and y instead of u and v but that's it and you check that this is conformal and this is the most famous conformal map in history it's actually what we use to produce maps in fact it was known at least to the Greeks in this map the way the Greeks were making maps of the of the sky meant they knew this map here ok they never wrote it ok now what else so you see this map at least being conformal that means that at least the problem of directions is faithful ok so if you produce a map, a real map a map, a piece of paper with something drawn on it using this map of a piece of the earth for example you know that the directions are exactly the same as in reality ok so if you look looking at the map you tell me to go to Rome by 35° that means I have to turn by 35° but it's not an isometry so by measuring on such a map the distance between Trieste and Rome and you tell me it is 800 km I know that you are wrong they won't be 800 km so depending on which you prefer directions to be preserved you can do it in this way if you want the perfect map it doesn't exist and we will observe it soon because now you can say now you gave me a conformal chart maybe there is an isometric chart ok now we might have time I am usually behind schedule but in fact how do I prove you that there is not an isometric chart I mean between a piece of the plane and a piece of the sphere actually I do it in fact Gauss did it by proving probably the most important theorem in the theory of surfaces ok it's the only occasion I know in the history of mathematics then the person who proves the theorem calls it the most important theorem I mean the fundamental theorem of calculus was called fundamental by somebody else in fact I don't even know who was the first one to state it but in this case Gauss proved this theorem and he called it and the name remained because of course it was in Latin so the theorem I grade and it's still called like this ok a grade means extremely important ok in Latin what does it say in one sentence it says this the Gauss curvature is invariant under isometries if you want to again as usual I mean this is the thing you can put on a stone a bit more mathematically what does it mean if you have two surfaces and an isometry between them if f between s and s prime is an isometry with Gauss curvatures gauss curvatures k and k prime then k of p is equal k prime of f of p ok meaning you have two surfaces and an isometry between them computing the gauss curvature is invariant if you do it on the left or on the right at the corresponding points now first this solves the problem for example the king of Denmark problem he wanted the perfect map forget it because if the two surfaces are the plane and the sphere if there was an isometry you would have a point on the plane and the point on the sphere with the same gauss curvature now this is e constant equal to 0 this is constant equal to 1 that's it, game over ok but of course this is more important than that proof the proof is actually a very long computation probably I will go over time but let's see what can we do how do we do it let's take a part of a surface now the point is the funny thing is that in the proof you don't see s prime and the isometry so now suppose you have one take s so take in fact a chart for the surface s take x to s chart ok so the key point and in fact the crucial part of the proof now it's a long computation trying to prove what that I can write down the gauss curvature at a given point in terms of the first fundamental form of the surface remember the definition the definition is e g minus f squared little so the gauss curvature is born as something which depends on the second fundamental form ok so now the strategy of the proof is exactly to prove that this strange e g minus f squared is strange because the determinant is nothing strange but this e g minus f squared can be expressed expressed in term of capital E capital F capital G and maybe derivatives in fact you have to do derivatives of capital E capital G and capital F first and second ok it doesn't matter how bad this formula will look like at the end this proves the theorem because we by the theorem we started with because if two surfaces are isometric there are patches for which the two first fundamental forms are the same but if the two first fundamental form are the same and this horrible formula will look like it will give the same result ok so don't be surprised s prime doesn't exist in the proof so the point of the proof is show me that e g minus f squared little is in fact a function of capital E capital G capital F and its derivatives ok so we do of course everything is local it's around the point so I take a chart if I prove it on a chart it's done so now I have x u x v but in general they will not be orthonormal of course they will be orthonormal if and only if e is equal to g is equal to 1 and f is equal to 0 so orthonormalize it orthonormalize x u x v and get e 1 e 2 this is just a way to define what are e 1 e 2 basically e 1 is what e 1 is x u divided by x norm and e 2 is in the orthogonal direction so then it also means now these two are orthonormal if I add the normal vector to the surface I have an orthonormal basis of R3 ok so every vector in R3 can be expressed as linear combination of e 1 e 2 and n ok very well so write notation so I have e 1 e 2 and I write d e i ok I will be an index ok it is either 1 or 2 so d e 1 e 2 I call it e i comma u ok and d e i d v I call it e i v comma v ok notation nothing just to simplify a little bit the form but for example these are vectors of R3 so can be expressed in terms of e 1 so let's give a name to the possible coefficients so e 1 for example e 1 u ok the only thing I know is that since e 1 as norm 1 at every point its derivatives will be orthogonal ok so if I express it in terms of e 1 e 2 and then there will be no term in e 1 do you agree so there is nothing in the first slot but there is something I mean in general there will be something in the second slot and there will be something in the third slot and I'm starting giving names to these coefficients ok e 1 v for the same reason e 1 v has nothing in the first slot again but something here and here and these are called e alpha 2 e 2 plus lambda 2 n now let's play the same game with 2 e 2 u well now it will have something in the first and nothing in the second and why this is still 3 coordinates sorry where do you see 3 coordinates sorry it's ok so e 2 1 e 2 u sorry e 2 u is what well it will have something here nothing here but should I put a beta or is it something that I know it's something I know because the scalar product between e 2 u minus alpha 1 e 1 nothing plus and here is something I don't know so I call it me 1 n so why here there is minus this well because of course I have the equation e 1 e 2 is equal to the scalar product e 1 e 2 is equal to 0 ok so that gives me this condition and then I have the last one e 2 v which is for the same reason minus alpha 2 e 1 oh sorry this becomes plus plus alpha 2 e 1 plus me 2 n ok ok check it no no no this one is plus ok let's derive them so 5 minutes extra time ok so how do you do it well what do I get from e 1 e 2 is equal to 0 this is what I know ok so what if so this is equal to 0 this implies what I can take the derivative with respect to u and this tells me what that e 1 so this implies that e 1 u e 2 plus e 1 e 2 u is equal to 0 ok so e 1 u e 2 is of course alpha 1 and e 2 u e 1 is this one so this is minus alpha 1 so where do I get this one because you might say why don't I take the derivative with respect to v that looks like a minus now but for the same reason if I put v I get the same equation ok let's see if we have to go back to this point but let me go ok so let me erase this now so this is the system I like now I claim the following identities and then I will prove them step by step ok claim so let me take e 1 u e 1 u scalar product e 2 v e 2 v minus e 1 v e 2 u this is equal 2 lambda 1 me 2 minus lambda 2 me 1 but this is also equal to v alpha 1 dv minus d alpha 2 e 1 u and this is also equal finally to e g minus f squared divided by e g minus f squared ok now this is a claim there is nothing obvious here except the first ok so the first is what e 1 u e 2 v e 1 u e 2 v if I perform the scalar product between so it's the first and the last line so e 1 u e 2 v is lambda 1 me 2 because n is norm 1 minus e 1 v e 2 u ok so this is 0 0 lambda 2 me 1 ok so the first line is the only obvious one now let's see how much is d alpha 1 so let's try to prove the second line so d alpha 1 respect to v d alpha 2 respect to u is what well what is alpha 1 alpha 1 for example is modulus sine e 2 v sorry there is a sine problem here d alpha 1 dv where do I take alpha 1 I take alpha 1 if I take it from here this becomes plus well let's write it ok so alpha 1 is of course the scalar product between e 1 u and e 2 ok so I can write it as d in dv of the scalar product between e 1 u and e 2 this is the first one and then minus d alpha 2 minus d alpha 2 I would like to write it as plus minus alpha 2 is plus fortunately so let's go back to e 2 v and e 1 ok so it's plus v in d u e 2 v and e 1 ok now for some reason I prefer to have another expression for alpha 1 that I can take it from from where of course alpha 1 I could have also picked it from here but then it would have become let me decide in a moment what is best because of course at the end everything is the same but I mean something is easier to handle so alpha 1 I can take it from here but I can take it also from here with the minus so this would also be let me write it on top just to decide later which one I prefer this would be e 2 u e 2 u e 1 ok so I have these two options for the first term ok now if I perform the partial derivatives this is what this would become I definitely prefer this one put this one you can see immediately why let's do it this is e 2 e 2 u v e 1 minus e 2 u v e 1 minus e 2 u e 1 v ok and this is the first term plus e 2 v u e 1 so you see why it's better now plus e 2 v e 1 u ok so now this and this kill each other and I'm left with these two ok are we there are we there e 2 v e 1 u minus e 2 u e 1 v ok so we proved also the second equality now let's go to the third the point is well actually maybe I leave this to you ok observation e g minus f squared this is essentially this this object here is actually nothing like but you can express it in this form n u cross n v scalar product n ok if you want let's freeze it for a second this is a nice formula you can prove I mean there is nothing difficult it's one expression for for this for this thing believe it for a second and then I will prove it for you if this is true let's go on and let me prove it the last equality ok because this is the point but this is what so ok this will be either for you or for me next time but from now on what is n u cross n v n well but n if this is an orthonormal basis n is nothing but e 1 cross c 2 ok so this is n u cross n v scalar product e 1 cross c 2 ok and how do you perform the double I mean this is one of the nightmares of first year students ok scalar product of two cross products it's first first times third second fourth minus first third minus second third ok this is this is first first n 1 n u e 1 times second fourth n v e 2 minus now the ones mixed in the other way first fourth n u e 2 first fourth second third n v e 1 ok this is linear algebra nonsense ok and now how do I manipulate this I tell you that this is equal to n so n u e 1 every time I see a derivative of something I can put the derivative on the other side using always the equation n scalar e equal to zero because one is normal and one is tangent so they are orthogonal so the derivative of this subject here I can put the u here I can put the v here and so on ok changing sign but fortunately it's quadratic so I don't even have to change sign I do it twice for every for everything and I prefer this because this becomes everything so n will be not n e 1 times n e 2 v minus n e 2 u n e 1 v ok and now and now let's go back to the definitions what is n e 1 u n e 1 u is lambda 1 ok so this is lambda 1 n e 2 v is me 2 minus n e 2 u n e 2 u me 1 and lambda 2 ok so you see it's kind of strange the way you prove because we prove this by hand by essentially by definition then we prove this is equal to this and this is equal to this ok so they all together are the same ok but now what is the key equation among all this mess don't worry I think even Gauss thought it was a mess but it's here the conceptual part is crucial the point is that this is the Gauss curvature so the key equation is actually the last ok because now maybe now we should stop I'll finish the proof next time so we prove that alpha 1 and alpha 2 are functions of the first fundamental form ok once you prove the important point is that you understand that this is the end of the story if I prove that alpha thanks to this equation if I prove that alpha 1 and alpha 2 are functions of the first fundamental form in this is a complicated expression in the first fundamental form and it's derivatives no matter how complicated it is the theorem is proved by the first theorem of today ok so we are left with these two things this and this ok we will start actually we will start on Monday I think there is a ok