 Dobro. Zdaj smo. Srečaj smo našli s obstrakami, ker, da je tudi nekaj nekaj obstrak, je tudi z r3, ali v domenju, kaj je najbolj fundamentalna formu. Zdaj je tudi dobro modela, kaj je tudi dobro modela. Tudi disk modela, kaj je tudi, Disk, kdo je nekaj gena. Pojdo ide sveč. Pojdo valo, da je Zama hlice z 4 1 minus X-squared plus Y-squared square v X-squared plus Y-squared. Rojde vzupi čez in je vzupin v zeno, razmalo, na kako je tukaj, je zelo tukaj na zeljiv. Pa the first fundamental form is abstractly given by dx2 plus dy2, so the flat metric, in Fac gets used to the notion, I mean, first fundamental form is the world we use when the surface comes from R3 in generally we call it the metric form or the metric just like that divided by y2. OK, vidimo, da je vzvečo, da občaj je ovo izometrične, zelo zelo jaz ležite način, imači, da je nap, da je nap, ki je zelo vzvečo, če je način, zelo v komplexi negrač, zda način, z njič plene, da je zelo v komplexi koordinaciji z i1 minus z divide in 1 plus z. Kaj? Kaj je inverz? Zelo inverz je, so, phi inverz, je mač, kaj je vzal vzal, and gives you i minus w i plus w. So, this goes the other way around. We checked, I told you, there is not really much to check. You compute formally what is dz in terms, you know, formula by chain rule, Zelo je nekaj drugi razden klip, odjev, nekaj neželji, da možem počki ili zezulari zega. Iso, zato se nosijoakedšenje, zato se predstavimo, če namoji tezokržene se obeli. Imamo so 5 čeviljost, če nešeljost v prostžavno formu, in stajemo so sa všečenjem gudrem, ne nešeljost, mi je to vzelovena. Dobro. Now, actually today's lecture, I mean for this part I'm really indicating you steps that if you want to look in books or discuss with me, I'm very happy to do it but in some sense I'm just giving you the steps of something and not precise proofs of everything, but Count up the upper half plane model. In this means that these two objects are really the same thing So depending in what you want to know about the geometry of this hyperbolic thing you may choose these model or this model So for example, in order to check which are isometries of such a thing S estim already mind leave or at least I prefer to look at the upper half model. Of course everything should have the translation. They are the same so everything I say in one model should have a korespondent thing in the other model, but the point is, if you look at this set of maps which are so now I am telling you a bridge between this story and complex analysis. Okay, so for those of you who like and know more complex analysis probably Tudi, da poživajte izmaj k dis aparta toh, najbolj si iz njega vse transformacije. Kaj je nekaj vsak, nekaj in nekaj, ok? Vse transformacije je njega mapto. Preč neko nr. a in a, neko nr. A, B, C, D. Na veselom, ki je skupo, zelo pa, da ima tudi nekaj v medriči, t Creative Initiate. Ale je umoji. Tonezt, da ti bo del nek regu. dramatically omati, nekaj merarte v desde in dne in ma, že pavali pomojeje glad participants, za tvoj nič cables. Pozivali pabast su pohaj de in tako, z ruk sem tega pača nesel k v Dunninge, po desetce 1. Nač biti, ga je to zelo evočen kvaličov. Zelo se, da v pravih počati bu vznikov Čistih transformacijov, nadi se levera. Zelo je to skor Raž tudi vznikov, je to dva vznikov, minus b, minus c, vznikov z vznikov. To je inverst. Zelo se. Zelo se vzelo. Zelo se zelo z zvrstva. Zelo se z vzelo. Zelo se je, da je 1 v 1 mapi v u to u. Zelo se je, da je nešto v 1 v 1 mapi. Zelo se je zelo zrstva. Zelo se je, da je isometri. Zelo se je isometri. 1 vzelo zmiado u. Zelo se je, da je zelo zrstva, v siri pa. Zelo se je, da je zelo zrstva. Zelo se je, da zelo se pila. Zelo se je, da je advancesano dvrstva. To je ne WrestleMania? No. BUztela. BUzelo. Bbreadjiron. in zrednji vseče vseče, kot vseče, kot za srte vseče. Tako je začel je vsečo, ki se zelo skupaj, in se videl, da je vseče izgleda. Zelo začel je zid-zid bar vseče izgleda z vseče. Zelo da so bilo, da so bi, in potem je nekaj nekaj, ker je tudi maplj, kaj je tudi z minus z bar. Tudi maplj z bar je z z bar nekaj u, ker je imačnja pravda. Prejvaj, u je, da je tukaj, da je u sebe, da imačnja pravda z bar je pozitiva. Z bar je z z bar nekaj u, ker je vse. Zelo, da se je več na x-raščenju, zочvaj je z zbar za zinu zmin, da je vse je ok. Zelo, da je z nami podal. Zelo, da je to je, da je izomet. Vse občas, ki je vs. izomet, imaš otroj izomet. Sve občas, kaj je občas, tko vse je, da je. Vse so vsez, kazem, zelo vsez, in odkaj si ya stitchesite druge mapi od ditva vznešte mapi s montinji. Zelo, pa pridobovam na druge mapi in staj delam Nezomet. Zelo vznači, ki je vse zespečki izometori. Čudovat nekaj je izometori, da je in vs. izometri. Isometri. Ok. So, izometri is form a group. Ok, by composition. Ok. And this is kind of, these are the building blocks. Ok, so maybe transformations plus the kind of the conjugation, but with the minus sign. Ok. Načo, da vam se pričo načo zvok vznik, da je izomet, da je d. Zato, da je d, ne? Načo, da se d, da se priči, da je d, in kompoz vznikovane ljubi, nekaj ljubi in potrebeno, ne? Čeho, nekaj zvok vznikov, nekaj tega tega d, nekaj. Prejmo se, da jih se priči, da je to ne. Jih se ne sne, ne zelo kančuje, ne skušve. Ok, so hoveto. Uto je prašna bitva kratka, počel je zelo kot... je izvaja je zelo izvaja v Rezimetri. Zelo zelo ki je tkaj in representiare prečerom. Primistri je zelo, da je izvaja. V tom teži si kaj se različjali, bo del izvaja v Rezimetri, da je vse nekaj. To je, da se kompleks analizjačnega vse bo površite, da se dajte vse vse. Ok, pa, tako, pričo, na modelu dobro, da se površite, kjer je vse sej, kjer je vse, kjer je vse. Pridem, da se površite modelu. Zato, vse, kjel je, We have E, F, and G. So we write down geodesic, the geodesic equations with these functions E, F, and G. The geodesic equations were written in terms of the first fundamental form right from the beginning. So what does it happen? Well, remember the first equation was an equation like this. So now we look for geodesics in U. So, remember the first equation was D, D, we are looking for some curve gamma of S, which was satisfying D. Oh, actually maybe now I switch notation, this coordinates I will call them U and V. As we did in the theory of surfaces in R3, E, U prime plus F, V prime, this is equal to 1 half. I am assuming, I am writing the geodesic equation already parameterized by arc length. So remember this strange function R is not there. So 1 half, E, U, U prime squared plus 2 F, U, U prime, V prime plus G, U, V prime squared. So this was the first, and the second was, let's write it, V in VS of what? Of F, U prime plus G, V prime is equal to 1 half in V, U prime squared plus 2 F, V, U prime, V prime plus G, V, V prime squared. But actually let's focus for a moment on the first line. So this is the system we want to solve. But the first line is particularly interesting because of course, what are the functions E, F, and G in our case? Well, E, well now written in terms of U and V, E would be the function 1 over V squared. F is equal to 0, there is no dx dy. And G is equal to the same thing. So in particular, they do not, none of them depends on U, which is always a nice accident because that means that in the first equation, the right hand side disappears. So let me call this 1 and let me call this 2. So the first equation becomes what? Just dds of E, which is now 1 over V squared. So U prime over V squared plus F plus nothing because F is 0. This is equal to 0. While the other one in principle will look a bit more complicated even though F is 0, so at least two terms disappear right away, but then F and G, E and G will depend on V. So there will be these two terms on the right. But this means what? So this means that in fact this function here on a geodesic has to be constant. So in particular it will be equal to some C, to some constant C. So I can write it as U prime is equal to C V squared for some C. OK. Now remember, this has happened already a few times, we haven't written down the equation for being parameterized by arc length. We have an extra equation here that we are assuming automatically, but we have to write it down. So what does it mean that the arc length parameterization? It means that the tangent vector is constantly of norm 1. But what is the norm of the tangent vector? Of course the tangent vector would be U prime V prime. With this first fundamental form, the norm of the tangent vector becomes the equation. 1 equal to what? 2x prime squared plus y prime squared. Sorry, U prime squared. U prime squared plus V prime squared divided by V squared. OK. So this is the equation for being parameterized by arc length. OK. So now, in fact, observations. So let me call this 3. I leave you to check that 1 plus 3 implies 2. This is a straightforward check. OK. You write down explicitly what 2 looks like and you prove immediately that 3 je 2 automatica. So the second geodesic equation follows from the first and the parameterization by arc length. OK. So we don't have to worry. So really our system of equation is 1, 3. OK. It's direct. I mean, there is nothing to learn. It's just a three minutes computation. So now let's analyze. So the system now is U prime is equal to C V squared. And if you want, U prime squared. U prime squared plus V prime squared is equal to V squared. So this is really our system. Now, let's analyze this system depending on C. For example, case 1, case A, if you want, case A, C equal to 0. In this case, everything becomes very simple because if C is equal to 0, we are saying that U prime is equal to 0. OK. And then what does it mean geometrically? It means that U is constant. OK. So U is constant, but also U prime is, so then becomes here is 0. And so I am left with this. OK. So then this implies U is equal to a constant. And then and V is exponential. OK. It's e to the s plus or minus. OK. But geometrically, whatever the parameterization of V is, even just this condition is telling me geometrically what we are doing, but let me erase the Poincare disk, which will come back very late. OK. So what's going on? What are these geodesic? After all, we can draw these surfaces. OK. We just have to keep in mind that the metric is not the metric of the blackboard. OK. It's something different. It's the metric coming from this strange rule. OK. But as a set, of course, we are just looking at the upper half plane. And what does it mean that we are taking? The curves given by U equal to constant of s some function. Well, OK. This is U. Well, this is U. And this is V. OK. Because I'm asking that V is positive. OK. U equal to a constant is just this. Now, of course, it also has to come with a special parameterization. You see, it's interesting because it's not a straight line parameterized as usual. Linearly. I mean, usually I would say U of s is equal U naught V of s is equal to some V naught plus s something. OK. I finished my name. V tilde. OK. I don't know how to call it. W naught. I don't want to indicate another coordinate. These are numbers. OK. This is the usual way we think of a straight line. But this object does not satisfy the geodesic equation. It's not true anymore that this line parameterized in this way is parameterized by arc length. It was true if W naught was equal to 1. It was true in Euclidean geometry. But it's not true now anymore. OK. But still the trace of the curve, the curve is always the same. OK. Geometrically, vertical lines are geodesics. And actually, let's go, let's immediately throw away a psychological problem, because we say, well, OK, maybe straight lines. Who cares? Vertical, non-vertical. Are horizontal lines geodesics? Because maybe I say, OK, maybe we are going to end up with the same geodesics of the Euclidean plane. What are horizontal lines? Vertical lines are V equal to a constant. But if V is equal to a constant, let's say V naught, what's going on here? V is equal to a constant. V prime is equal to zero. Is this equation satisfied? No. So here there is a genuine difference between straight lines. Vertical lines are special. OK. After all, if you think there is nothing to be surprised, I mean, the rule of measuring vectors, I mean norms of vectors remembers what is vertical and what is horizontal. It's not symmetric in the rotation. OK. So, after all, this is not too surprised. Let me erase this and let's see what else we can because, of course, this was the trivial case, c equal to zero. What can we say in the non-trivial case, c different from zero? Case B. Well, if c is different from zero, you can manipulate this thing. Now it's up to you to choose your favorite trick, in ordinary differential equations. But I can tell, I mean, this becomes, you can write down dV du, I can extract this information out of this. dV du is equal to the square root of, of course, I wrote everything x and y, so now let me not make mistake. V squared minus c squared v4 divided by c squared v to the power 4. OK. This is simple out of this. Well, again, I tell you, I try to be honest with you, so when there is something subtle, I tell you it's simple. I take this, u prime from here and put u prime squared here. OK. So that's why you get a v to the power 4, you see. But then this is what? This is dV dS. OK. And you can manipulate it. You can solve formally this equation in this way. But this is the same thing formally. I mean you can multiply by du, if you want. Of course, this is a rigorous meaning, but I mean you can take this du on the right and this function on the left when you separate, try to separate variables. OK. So this implies that dV, in fact that cV dV divided by the square root of 1 minus c squared v squared. OK. Because you see, of course here I have a square. So the square root and then multiplied becomes cV squared. Or the absolute value of c. Now let's not be pedantic. OK. And OK. So that's why you get this equation. So this is equal to du. OK. But once it's written in this form I can find the explicit expression of the solutions of these. You see here I have exactly I mean this is d of if you want cV squared. OK. Up to a half. OK. So I can integrate explicitly on both sides. And this implies that the solutions are exactly satisfy this equation. Minus 1 over c square root of 1 minus c squared v squared is equal to u minus a squared. A becomes a constant of integration. OK. So there exists a number a for which the solution looks like this. OK. Well, but this is very interesting, of course. Because if I square both sides and manipulate this becomes what? Solutions to these equations are the same thing as solutions c squared plus y squared is equal 1 over c squared. 1 plus v squared. OK. So we know exactly geometrically what they are. Again, the formal parameterization will be very complicated. You see, I'm solving an equation between u and v and I'm forgetting how they depend on s. As before, when we said straight lines, but then the parameterization in s is complicated. In that case it was just an exponential map but still, I mean, not a linear parameterization. Here I'm playing formally, I mean morally the same trick. I'm getting an equation between u and v so that I can draw the picture. But actually how the, which is the speed, which is the parameterization on this circle. It will not be cos s sin s. Exactly as before. It will be a much more complicated parameterization. OK. Plus, these are not, this is not any circle. OK. These are circles because you see there is one thing missing here. There is not a b. There is not a v minus b squared. What does it mean? There is not a b here. It means that the center of this circle lies on the v equal to 0 axis. OK. And that's the only restriction. Then any circle with this property is like this. OK. So that means that geodesics are either vertical lines or the trace of a circle line with the center line on the, draw whatever you want. These are geodesics of this model. OK. So in some sense the most efficient way to go from here to here is not the segment. OK. But it's a piece of a circle. Which circle in fact, so now question we know that we analyzed all possible cases. But let's see what's, I mean, let's give another possible, I mean, is it true that we have found all possible geodesics? Well, there is a simple way to check to answer this question. Because by uniqueness theorem, remember that you have the uniqueness theorem in the existence of geodesic, existence and uniqueness of geodesic was telling you what. It was telling you that if you give me any point and any tangent vector there exists one and only one geodesic, at least for a short time. OK. Now, is it true that given this point and this vector there is only one and only one of these two one of these two types? OK. Because of course if the tangent vector is vertical there is the vertical line. OK. The tangent vector is generic. It's non-vertical. Well, it's simple to argue that actually there exists one of these circles. OK. How do you construct? And so on. OK. You look for the the only possible circle center on the u axis for which the circle that you draw passes through this point in this tangent vector. OK. I'm simply using the property that the tangent vector to the circle is orthogonal to the radius. OK. So, this is the only possibility. And something like this always exists. So these are the all possible geodesics because if there was another one this would contradict the theorem. OK. What else can we say? Well, of course a natural question is well, and if we draw them on the disk how do they look like? Well, this is again a moment where complex analysis helps. But in any case I can tell you the final answer because basically you have to take the map phi, of course an isometry takes geodesics to geodesics. So the question boils down to what are the images of these curves under the map phi or phi inverse, now phi inverse probably. I mean, because phi was going from d to u, so these are in u you have to take them back. OK. And the answer is this on the disk you draw geodesics in this way. So this is your disk there is no boundary. OK. It's the open disk, it's not the closed disk. And then these geodesics becomes exactly either equators or circles meeting the boundary orthogonally. So these are the geodesics of the Poincaré metric if you want. OK, so now you see why I hope you are enjoying yourself because this is a part of destroying all the classical geometry and you are building a new one. Now, for example, either model pick the one you prefer. It's not difficult to check that they satisfy all axioms of Euclid or remember that actually Euclid axioms are quite complicated. I mean, there are axioms of incidents, I mean the list is in any case, now I I'm not going to make a piece of history of mathematics but except the famous fifth postulate which was asking you is it true that given a line and the point outside that exists one and only one parallel line passing through this point to the given line. So what's going on here? Well lines we translate line into geodesic geodesics and now here it's clear that given a line and the point outside there are infinitely many lines which do not touch meaning parallel the original line. It's clear. So this is the most famous way to produce it. The fact that this object exists which was the problem for mathematicians for thousands of years that if you take the contrary or one of the two possible contraries of the postulate what you have because you can substitute with either that does not exist any parallel or there exists more than one. The fact that the model exists tells you automatically that if you substitute it with that it exists more than one in the logical contradiction because it exists I mean so otherwise you think that nature is playing games. It exists. So the idea of substituting the fifth postulate with the contrary in the sense of more than one cannot produce an absurd which was what people tried for thousands of years substitute by contradiction suppose this is not and then I work and at the end I want to find a contradiction but it's impossible because here you are here it is there is no contradiction this exists so this was really the end of the story of the attempt of proving or disproving the fifth postulate it's either you consider a postulate or you don't but it's definitely not a theorem very well just as a source of in fact I didn't even take my notes on this with me just to close a little bit this story you see there is even a stranger accident if you want think now of the local Gauss-Bonnet theorem of course it was proved it was proved using some geometry of R3 but it is a statement which depends only on the first fundamental form of things so it has to hold as long as you know remember what is the only delicate part of what I am saying is that you take a piecewise curve on a surface and in principle you should compute the geodesic curvature now in order to compute the geodesic curvature you need to have in principle at least for the way we defined you need to have it in R3 because you compare the curvature in R3 with its projection on the tangent space but in any case if you are taking polygons piecewise smooth curves where the edges are geodesics this contribution will not be there k is equal kg is equal to zero and this doesn't come enter into the picture but you see here so basically I am proposing you to take triangles geodesic triangles and try to check what is going on which is the sum of the interior angles ok here the situation becomes really dramatic if you want because what are the angles now the first problem is that now if you give me two tangent vectors at a given point well you have to be careful what does it mean, how much is this angle because this is what Gauss-Borné wants well this angle maybe is not the angle of the blackboard because even angles even to compute angles you need to use the metric on the other hand this metric is very special ok so in general this would be true it's forbidden to say that these two things are orthogonal they are orthogonal on the blackboard but maybe their hyperbolic angle is pi over 4 but in just for this specific example look at the metric the metric is conformal to the standard one it's the standard it's e equal to g and f is equal to zero so you know how to compute angles but conformal max were born in this way we proved it we proved that if you have something which is conformal to another the angles are the same the lengths of the vectors are not but the angles are unless e equal to g equal to 1 which is not our case but the angles are the same so that means that actually our pictures are faithful in terms of angles so we can really compute it by classical euclidean job ok so now I can draw you a geodesic triangle how to compute angles a geodesic triangle how do I do it well I take three geodesics now of course I managed to miss a triangle of course so let me 1, 2 and 3 more or less so this would be a geodesic triangle ok you can see guess how much is the sum of the interior angles well of course but it's definitely much less than pi ok and this remembers the fact that the Gauss curvature of this object is negative is minus 1 ok because the sum of the interior angle is equal to pi plus the integral of k k is negative so it's nothing and you can push this up to an extreme because among geodesic triangles you can imagine stretching of course if you draw a reasonable picture you can put in some sense ideally the vertices in the boundary so strictly speaking these are not triangles in let me not try to be orthogonal whatever ok but you can do it you can push the vertices on the boundary of the disk but then what happens since both are geodesics they are both orthogonal to the tangent to the circle ok so what is the interior angle here it's 0 so what is the sum of the interior angles 0 so kind of surprising you have a triangle whose sum of interior angles is exactly equal to 0 of course that means in fact I didn't do the computation ok k maybe maybe I should have told you sooner or later I mean formally again this is a simple situation of the gauss of the teremaegrasium because it's one of those cases where your formula for computing the gauss curvature ok so this implies that k is constant equal to minus 1 ok it's a constant negative curvature surface again there is nothing to prove I mean you take these functions you put them in the formula and you get minus 1 ok so that automatically means that on this for example the triangle I drew is equal to minus 1 so by gauss-bornek formula the hyperbolic area of this triangle is 0 is equal to pi minus plus the integral of k k is equal to minus 1 so 0 is equal to pi minus the area so the integral of da is the area of your region so these triangles ok which may look even quite different ok now let me they all have area pi ok it's a very strange geometry so play the games that you prefer let's move on let me think if I had something more in mind to say about this I think we are done now we will probably appreciate the pictures by this famous painter Hecher now look on go on Wikipedia or whatever you want and look for the paintings of this I showed you one of his paintings the Mervius trip with the ants the ants going around the Mervius trip he actually it was quite amazing because the mathematicians understood hyperbolic geometry he actually found many beautiful pictures of so the problem basically if you look at his pictures you will find games of the form take one picture and duplicate it with isometries ok can you cover so this is what you would do in R2 for example what does it mean you take a square and put it around take your favorite picture and then you try to start translating it everywhere with an isometer translations are isometries and you can cover R2 with this picture ok now here isometries are much more complicated these Mervius transformations so what is the effect on one single picture transported everywhere with isometries and he has built this beautiful beautiful paintings you are strongly invited to see them now we change subject in some sense what will start now it's really the beginning of another course because this course was called differential geometry but really up to now we have done classical geometry trying to build the material to start studying differential geometry now so now I want to give you the definition the general definition you are ready you are mature enough to understand the definition of a differentiable manifold ok but actually this is a great place where to stop for a moment ok ok let's start so in order to appreciate the definition of a differentiable manifold let me go back to the definition of a regular surface so remember we said s so regular surfaces were objects subsets of r3 ok so you take a subset of r3 with the following properties ok so so that for each p in s so remember the very first picture for each p there is an open set I mean an open subset in r3 so let me draw it as a boy ok and I call it v so there exists a neighbor of p in r3 that I call v ok and the map our chart and the map now I call it f alpha ok we will see why because now I am really interested in the collection of charts ok so not just on a single one from some domain u alpha into u alpha contained in r2 domain of r2 which goes into v the intersection with the following properties ok such that what have we asked well first it was meaningful to ask that f was a differentiable map ok so we are going from a domain of r2 to r3 ok so we know what it means differentiable so f alpha is let's say it's infinity ok then we asked well in fact and homeomorphism homeomorphism 2 in the differential of the map f alpha at any point so differential of alpha at the point q so this will go from the tangent space to u alpha at the point q which is nothing as a copy of r2 because the tangent space to rn is rn ok canonically so this goes from these r2 to r3 because the target is an r3 and this has to be injective ok so these were the requirements for being a regular surface now so f alpha we called it a chart and so on the image a patch and so on but now there is an important observation if I let me draw a bigger a bigger surface out of these conditions we get the following charts so we have one chart f alpha coming from some u alpha ok which covers a piece of our surface and then we have something else that I call f beta so another chart which intersects actually maybe your picture try to make it clear that there is a nice intersection which comes from another domain which probably has absolutely nothing to do with the previous one ok so note that this implies that if you have two charts f alpha if you have two charts u alpha f alpha and u beta f beta such that the image intersect such that f alpha u alpha intersection f beta u beta is different from the empty set then what can we do we can go we can look at the intersection and let's call it w ok look at this which is an open set ok and w will come from some piece open piece so this will be so this gray area here this is u alpha and this gray area is f alpha inverse of w ok but there will be also some f beta inverse of w here ok so I can construct the map from this gray area to this gray area just by going passing through the surface ok so I can construct f beta f beta to the minus one compose the alpha this goes from where to where I'm looking at it as a map from f alpha inverse of w so this gray area into f beta of w and I can play the opposite game and go from here to here just by switching alpha and beta so I have alpha I have f alpha minus one f alpha minus one compose the f beta let me just double check that I haven't switched the domain no no but it's ok it's ok like this so this goes from f beta to the minus one of w into f alpha of w I have these two maps which are of course one the inverse of the other if you put them together it's always the composition of something which is inverse and the point is that these maps now are defined on domains of r2 they go from one domain of r2 into another domain of r2 and so on ok so in particular I know that these are smooth if you want are one to one and smooth here in fact I want to end up here not here these are smooth this was the key property in our style we used this property infinitely many times right at the beginning actually because for example this was the point where we understood it was possible to define what was a differentiable function on a surface because this was it is thanks to this that it's possible to say I call a function defined on s differentiable if when I restrict it to a chart I get a differentiable function who is going to tell me that this definition does not depend on the chart exactly this property if I take two different charts the difference between the two functions that I write is a diffeomorphism so differentiable in one chart is the same as differentiable in the other chart so this was kind of a key it's so basic that it's kind of fundamental what are differentiable functions ok now the problem is that these were these properties were true because they were following from this but here I really needed R3 it's this key property how can if I want to speak about manifolds I don't want to have a subset of anything my manifold is a set as it is and I want to perform the same things we did on regular surfaces on this set completely forgetting that it is a subset of R3 ok so how can I do it well the idea is exactly to take this key property and put it as a definition ok so let's see how it looks now the formal definition in fact there is no reason to restrict ourselves to dimension 2 ok so let's give the general definition in any dimension so definition and then n dimensional differentiable manifold is a set not a subset is a set m with a family of injective maps so I put it right at the beginning the fact that what are going to be the charts have to be injective maps that I call f alpha which will be defined on some u alpha domains over n ok there is no reason to restrict to n equal to 2 so u alpha open so now what do I require you see the point is that I cannot say anything about something like differentiable because I don't know what it means so how can I reconstruct it well first I certainly want that around every point there is a chart so this is easy because that's like asking that the union over alpha changes of f alpha u alpha this is the whole set so this automatically implies that for every point there is at least one of these things which cover which cover this point but then everything else must be contained in the other condition that for any pair for any alpha in beta such that f alpha intersect f alpha f beta of u beta is different from the empty set so every time two of these maps intersect in their image and in fact let me give this set a name ok this is w different from the empty set ok then I want that the then the inverse image of these is alpha to the minus one of w and f beta f beta to the minus one of w are open sets in rn because that's where they live and f beta to the minus one composed of alpha and f alpha to the minus one composed of beta one little care this type of condition did not appear before because it was automatic because we were looking at the regular surface as a topological space with a topology induced by r3 but now notice that m is just a set so I don't know what is an open set on m ok so I have to declare I mean this is a property on the maps f f takes at least these intersections these intersections must go back to open sets of rn so that's where I know what I'm talking about and then actually there is a technical third condition which is to require that this is this will be called an atlas so the sets the family of sets u alpha and the f alpha is called an atlas so the atlas u alpha f alpha is maximal is maximal with respect so relative to one and two ok, this is just a technical thing just to avoid logic problems improves what does it mean that this atlas is maximal is that if you find another set v g a set in the map which satisfy if you put it together with u alpha f alpha it still satisfies property well of course if you add something satisfying one it still satisfies one but if it satisfies two which is not automatic so if you have this new chart which still satisfies these properties then you have to add it so you are looking at the same time in one shot you are looking at the whole set of charts of all possible charts so for example remember this because this will be automatically a huge thing so for example if your thing was a sphere in R3 you are not saying I am taking the two charts for example given by stereographic projections no, this is absolutely not maximal because for example the charts coming from all the other proofs that we gave are not in these atlas so you have to add them but then you have to add also a small disk you have to add it and so on ok, so it will be a really big number in fact not even a number it will be more than countable for sure ok well of course we use the same ok, now it's over, one, two and three the point is that you can always achieve it ok, so once you get on a set a family of charts f alpha u alpha which satisfies one and two then you say ok and now you maximize it and I don't even have to know what are you really doing when you do this operation ok, there will be a maximal atlas with the properties one and two ok, so you don't really care because I remember my psychology when I was a student and I thought this was a disaster no, this is in fact absolutely nothing ok, you put it just to avoid logic problems and then you forget it it will always be the maximal thing but in practice it's like it's not there ok, we use the same words for these objects in the sense that as we did for surfaces in the sense that the pairs u alpha f alpha we call them charts or local parameterizations or ok, we use exactly the same words so why in which sense this this is very different it's inspired by the definition of a surface and it's very different in which sense well as I said probably in fact this is one of the fact that this is just a set ok so observation in fact if you have a set with something which satisfies these properties so a differentiable manifold in fact you have a topological space but now it's not given before it's given after so what is called the differentiable structure this object here this family of things is called the differentiable structure of the manifold induces a topology ok how so the differentiable structure u alpha f alpha so I write it for the first and last time this collection is called the differentiable structure of m induces canonical topology on m how I can just say take the topology for which f alpha u alpha is a basis which sets am I looking at so you keep in mind this picture but you know that you are doing a mistake so which are the open sets on m well because in the picture they are the ones coming from R3 now there is no R3 around my set so but what was true somehow the basis of the topology was the image of the charts ok do it define the topology to say that the image of the charts are a basis of this topology ok they satisfy the axioms of a basis ok this is like saying in other words you can say it in this way a I have to declare which is an open set ok I say that a is open if every time it intersects a chart if I look at f alpha inverse of a intersection f alpha of u alpha so so basically I'm saying now is this open well this is open if for all possible charts which touch it because otherwise it's empty otherwise I'm looking at the empty set the intersection if I take the intersection and then I take it back I get an open set here because here of course I have the euclidean topology ok so if this is open is open in an for any alpha ok that's an equivalent way to say but this observation should go on at least 30 seconds because unfortunately out of this definition you cannot get essentially any nice property of this topology ok in a reasonable word you would like to work with nice topological spaces for example Hausdorf for example verifying the second axiom of numerability second numerable how you want to say it countable basis ok now this topology does not verify in general any of these in general it is not Hausdorf in general is not does not have a countable basis ok so we added because I mean it's a the least restriction we want to put ok on the type of objects ok so from now on we assume that this topology is called the induced topology of course it's induced by the differentiable structure this topology is Hausdorf and has countable basis so ok of course the other simple observation to make after all we have built this definition out of the definition of a surface we have taken a property and we have made into an axiom so clearly regular surfaces are two dimensional differentiable manifolds ok regular surfaces in R3 are examples ok now so in the last minutes of this lecture let's produce at least one other example just to appreciate the fact that we are starting doing something more general than before even in the two dimensional case actually this works in any dimension but it's not just that even in two dimensions we are starting a larger classes of objects larger class of objects so example projective plane ok so I restrict myself again to dimension 2 even though it's easier over the reals I mean so what we call P2R so as a set what is P2R I'm sure you have met it many times so strictly speaking as a set we can say it in many different ways as a set this is the set of lines through the origin in R3 ok so PnR is the set of lines passing through the origin in Rn plus 1 ok so this is the set of lines the set of lines passing through the origin in R3 well how can we describe it this is a nice geometric sentence but let's have a more description well one way to say is that P2R is in fact R3 minus the origin quotient by an equivalence relationship so which equivalence relationship is that for any line I take one representative mini two vectors xyz should be identified with any multiple line with any parallel vector so it's lambda x, lambda y lambda z with some lambda different from 0 otherwise well first otherwise you are not inside but this is you can say let's add the origin but then it doesn't become an equivalence relationship very well and it's usual to indicate the equivalence class xyz or sometimes with double dots as you prefer as far as I'm concerned I'm happy with square brackets well now I want to convince you that this is actually a differentiable manifold well in principle I could tell you well this is probably it's an example you met well the first time in algebra the second time in point set topology because it's probably one of the simplest non-trivial cases where I added the quotient topology I mean because this is a topological space and this in the Euclidean topology induces a topology on the subject here so it's a classical exercise in topology but this is going in another direction because this would give me already a topology while the differentiable structure doesn't need it so let's see even without all those considerations how to build a differentiable structure on this set so let's look it just as a set well define 3 there are of course 3 3 very special sets here which cover everything so define v1 to be the set of points xyz in p2r in the projecting space the meaning has to say such that y is equal to 5 because it's not well defined up to the equivalence relationship because you multiply by 100 and then y becomes 500 ok the only meaningful value of the coordinates if you want to call these coordinates is 0, non-zero after it's non-zero you might as well say 1 there is one representative because that's coordinate equal to 1 ok so this is just to justify so for example the first one is the set of points for which the first coordinate is non-zero but remember it's completely meaningless to say what it is once it's non-zero it's anything ok then v2 is the set of points of course now you can easily guess such that y is non-zero and z and v3 is again the same thing such that z is non-zero ok now of course these three sets cover p2 because 0, 0, 0 is not allowed ok so they cover as sets so the union of v1, v2 and v3 give the whole set but now on this so we are in good shape for example the parameterization the charts so how do I construct the charts on these so with obvious notation fi will be defined on vi ok in fact it's defined no it takes values in vi ok so i will be 1, 2 or 3 ok define how that for example the first coordinate is non-zero there is one representative of this point which has the first coordinate equal to 1 ok so I can take 2 so the only 2 degree of freedoms are here y and z ok so u, v will go to actually this was not such a great idea so f1 so let's cite it in this way f2, 1, u, v f2 goes in u, 1, v and u, f3 goes in u, v, 1 ok otherwise you have to invent the notation to say where to put the u the one ok so the one goes in the ith place ok so now the claim is that these things satisfy the properties of differentiable structure so the maps the maps so the charts u, vi fi put in all these together is a differentiable structure of course here I am already making the mistake I mentioned to you before up to 0.3 up to maximality so they satisfy 1 and 2 and then differentiable structure by adding everything else but let me do it a differentiable is a differentiable structure and say up to 3 ok which is armless ok well 0.1 we already observed so 1 is ok we have to prove properties 1 and 2 1 was just the fact that they cover every point in p2r so 1 coordinate which is no 0 so it lies in one of these images ok let's go to the second property the second property means take two charts for which the image remember the picture even though remember that you don't have to think of this as a subset of r3 ok so here you have a w here you have a gray thing here this is f1, f2 for example of f2, f2 whatever f alpha, f beta and now you go from here to here by composition of the inverse ok and the question is is this map differentiable is injective is it an injective well in fact is it a differentiable we have a word let's use it ok well I hope you agree there is no problem no loss of generality assuming that we do it for f1 and f2 ok just to simplify so fix i is equal to, if you want in the definition they were called alpha and beta so alpha is equal to 1 and beta is equal to 2 ok otherwise everything is the same so what do we have to do we have to check what is f1 f1 to the minus 1 composed f2 of v1 intersect v2 ok and the problem is is this an open set and then is the map between this and its homologus on the other side differentiable well the point is what are these points if a point stays in v1 this coordinate non-zero if the point stays in v2 as the second coordinate non-zero ok so we can say that these the points line here in fact where do they stay they have to stay where in R2 so these are points of R2 so which umv stay here well as I said this set is the set of points vkaj is different from zero ok but that means that means that this set is the set of uv such that u is different from zero ok the other one is automatic because you are taking it into the right image ok sorry I wrote the mist I am taking the inverse image of this set so the point is whether the inverse image of this set here is open this is the first question ok then we have a map where I will put an f2 but the first problem is is this gray area open well it is this is an open set and of course if I deal with f2 I would have gotten v different from zero ok so this is automatic now the problem is what is what I wrote before by mistake so f2 inverse composed f1 of a point uv ok because now I am taking a point here now I know what it means to lie here and now I want to go here so I take this point I do f1 and then I take f2 inverse ok well let's do it step by step it's written here what is f1 so this is f2 inverse of what of the point with projective coordinates 1 uv ok and how do I get f2 inverse well remember that I know that u is different from zero ok because this point must lie here ok so how do I do f2 inverse well in order to do f2 inverse f2 is this map but this point is not like this so in order to say from which x and y it comes I need to put this point in this position and then I will look at the first and third coordinate how do I do it in the same equivalence class there is of course so this is if you are f2 inverse in this is the autology because I am really writing the same point ok this point here is exactly 1 over u1 1 over v sorry v over u ok now that I have put the representative of this equivalence class in the position I like I know that f2 inverse is the first and the third so this is so this map takes r2 minus u equal to zero into this the image whatever it is and it is given by this function and this is clearly smooth ok we are away from the locus where u is equal to zero ok actually I should have also argued but it was so I forgot that if I are bijective no? at least injective but this was completely trivial I thought there was a question why u is different from zero because I start not from the root not from any point of r2 but only from the gray area so the first thing I observed is what is the gray area from zero ok ok let's repeat it first where should I look at this which is the right domain to ask the question the right domain is unfortunately the general definition disappeared but it's f1 inverse of w so I'm looking at this map from where to where from the inverse image of w which goes into the inverse image vf1 which goes into the inverse image vf2 ok so the first problem is characterize the domain and the characterization is here because the inverse image this is w so the inverse image of w is the set of points where u is different from zero so then I know that the inverse image of f1 this is independent of where is just the point 1v1uv but now I know that u is different from zero so in order to write down the inverse image I need to put a 1 in the second place so I do it how do I do it I divide by lambda ok what is the equivalent point I have to divide to multiply by the same factor or coordinates in this case lambda is equal to 1 over u ok which is not zero so 1 over u, 1 v over u now it's done because once I know that this is a one the inverse image is just the other two coordinates ok this was the other ok so let's stop just to fit so this defines a differentiable structure and you are strongly invited to think of whether this is so this defines also a topology now there are many exercises you can play with because for example now by general principle this differentiable structure induces a topology is the topology the old one that we knew from the topology class yes is the quotient topology of the ukridian topology ok almost by definition if you think what you have to prove because you realize that there is nothing to do ok then much more subtle question is this a regular surface in R3 this is difficult ok this is difficult this is a difficult question try to see whether you can come up with some idea well of course if the problem was just to have something new I could have put here n plus 1 and define tnR I have n plus 1 coordinates I have n plus 1 maps n plus 1 open sets same proof everything is the same here n equal to 2 was completely irrelevant ok so this is in any dimension you have this example ok the third condition yes you make it maximum you say well you need to observe that once you have something which satisfies 1 and 2 there is only a unique one maximum which contains it so you keep on adding stuff but you don't need it ok