 V terjučnjih razdaj bi smo zelo vzouto prejaditi tako vstajčenje, tudi na obradičnjih sebe, z vrdu nekaj je, da je vse izvečno všim. Zelo so mešli, da smo nekaj nekaj, da smo na vse zelo vstajčnje, nekaj na povod, nekaj namo sebe, in iz nekaj nekaj povod je, da je. Tudi nekaj nekaj, da se izvečno vstajčnje zelo vzouto vstajčenje, So, we have a special time here. And for, at least for small v, we have taken geodesics hitting the original one orthogonally, okay. And we call them gamma v of u. So this would be gamma of v, and this would be gamma v of u. Putting the two things together, so now we have two parameters. One moving on the initial geodesic V, a gamma. And the other moving on the orthogonal geodesic gamma V. We have two parameters, V and U. And we constructed a parameterization, X, U, V, which is just gamma V, U. Which works. We know that this satisfies the conditions of a chart around the initial point. Is this you? OK. Now, when we did that, I forgot to... We used already this theorem to solve an important problem. We classified completely surfaces of constant Gauss curvature. Constant Gauss curvature means positive zero negative, positive sphere, zero the plane, of course, up to localizometry. So, piece of a sphere, piece of a plane, and piece of the pseudo sphere, in case of negative curvature. But actually, I suggest you an exercise because what is this construction saying in the case of the sphere as usual? Now, you can think... We thought of the sphere in many possible ways. One is like the inverse image of a regular value of a function. X squared plus Y squared plus Z squared is equal to one. OK. And we started in this way. Or we did by hand, by covering the sphere with graphs over different planes. Or we did the stereographic projection. OK. So, these are all different ways, of course, equivalent. These are all charts. Basically, what is this construction telling us on the sphere? Well, check that if you think of the sphere as a surface of revolution, this is really what's going on. Now, the point is that the sphere is kind of a boundary case of a surface of revolution because the curve, which is rotating, is half of an equator, which of course touches the axis of revolution. So, this is something we always avoided, but it touches orthogonally. So, even at those points, you get a regular surface. OK. So, in this point, the tangent vector to the curve is orthogonal to the axis of rotation. And this means that if you rotate, you still get a regular curve. You don't get casps by rotating them. If you think in this way, the sphere in this form cos u cos v cos u sin u cos v sin v. OK. Because the curve, which is rotating, is really half of a circle. So, this is the parameter v. OK. Put u equal to zero and in the xz plane up to changing names. I can think of the blackboard of the xz plane. This is cos v sin v, which is the parameterization of this piece. And of course, this will force v to be half of zero to pi. So, for example, between zero and pi. OK. And then you rotate with u equal to zero, from zero to two pi. OK. Well, the point is that in this situation, if you compute the first fundamental form, you get exactly some kind of the dual. E is equal to cos squared v. F is equal to zero. And g is equal to one. OK. Which looks very much like what you should have gotten from the construction of jodesic coordinates changing names. OK. Changing names, of course, here. In the theorem we stated like E is equal to one and g is another function. Now, here, you just switch names. Basically, you change names from u and v. You have to decide which is the first coordinate and which is the second, but who cares. That's just a convention. And you get this. OK. So, this looks like falling into our category. So, the exercise is to decide that this is indeed the case. In which sense these coordinates come from this construction. OK. Now that you know what that, because on the sphere you know what are jodesics. So, this construction is not just relying on the general existence theorem for jodesic, but here you can really write them down. OK. OK. This was kind of to close that little, it's a small observation I wanted to do and I forgot last time. OK. Now, let's move on and let's go back to drawing a picture of a surface in R3. We have a curve. Suppose you have a curve on S. Remember, we define the notion of normal curvature of this curve. OK. Meaning what? If you think of this curve as a curve in R3, of course, here you have one special vector, which is the normal vector to the surface. You have the tangent vector. I mean, if this is gamma, this is gamma prime. But you have also another special vector, which is the normal vector to gamma as a curve in R3. OK. So, here you would have, and we changed name. OK. In the beginning of our course this was also capital N, we decided to call it little N. OK. Now, this is in some sense the acceleration of the curve in R3. Of course, this vector will have in general two components. I mean, two geometric and important components. One is the normal component and one is the tangential component. I mean, you project this vector over the normal line or over the tangent plane to the surface. OK. And the normal curvature was what? Was essentially the length of the normal component. OK. That leaves us with the problem of finding another name because somehow out of this information we have two information, the length of the normal component and the length of the tangential component. OK. So, the length of the tangential component, so let me call it N tangential. OK. N tangential is by definition the projection of N on TPS. OK. The orthogonal projection on display. So, the norm of this vector is called the geodesic curvature of gamma, because this depends on the point. We have already observed. This is just to give a simple name to something. A symbol to something that somehow we have already used. Because we have already proved that a curve is a geodesic if and only if this function is identically zero. OK. Remember the geometric interpretation of the geodesic equation. OK. The acceleration of the curve is all normal. If you project it on the tangent space, you get the zero vector, so you get zero. OK. So, this is just to introduce this symbol which is useful. But somehow we have already used. So, geodesic if and only if k gamma is equal to zero. OK. At every point, of course, of the curve. Now, with this notation we can state one of the most important theorems of this course. So, this is called the local, well, let me not give it a name yet. A corollary of this will be, will have a famous name. So, theorem. Let gamma be a smooth closed simple curve on a surface s. In fact, on a patch. This, in some sense, it's a local on a patch of a surface. Then we have the following integral equality. The integral. Oh, in fact, let me just also to fix notation in closing region r. OK. Just to give something a name. This object and this domain a name. Then the integral over gamma of the geodesic curvature. Sorry, change also this Kg. G is for geodesic independently of the curve. It's better now. The integral of the geodesic curvature along this curve is 2pi minus the integral over r of the Gauss curvature and I always write da to say, to mean the measure induced on a surface by the first fundamental form. OK. It's a square root of eG minus f squared dU dV. OK. It's the measure we used to do integrals over a surface. OK. Now, think for a moment, what are we saying? What is, you have a simple closed curve. If your surface was a plane, what? In fact, let's not draw it on the plane because I mean, you have, you have, this is the picture you have to keep in mind. You have our curve gamma and here you are taking the integral of the geodesic curvature and this is the region r because as usual, remember I'm using the fact that a simple closed curve so no self-intersection, remember. OK. Smooth closed and no self-intersection means it splits OK. And up to orientation of the curve there will be one interior part and one exterior part. OK. So enclosing means the interior part. OK. Let's see how we prove it. Of course, gamma no loss of generality to assume gamma parameterized by arc length. OK. And we fix and everything since everything is contained in a patch of a surface let's give a name to this patch. So x from some domain v into r3 is the patch which contains everything. OK. Because of course as usual I will try to pull everything back to the domain v. Now exactly as we did in the proof pageum we can choose two vector fields basically orthonormalize xu and xv along the curve gamma. OK. Choose vector fields e1, e2 in r3 such that the pair e1, e2 is an orthonormal basis is an orthonormal basis basis of tps at the point p. These are vector fields. So every time I evaluate them at the point p I get an orthonormal basis of tps. OK. And now restrict them and restrict them to the curve gamma. So the restriction of these vector fields in principle these vector fields would be again functions I mean vector depending on two parameters but if I restrict them to gamma I can think them as functions of s. OK. I am looking them only here only on these points. These points depend on s so basically and so on. And I think of them as depending on s. OK. So as functions of s I indicate e prime this is just for a notation e prime i I mean whatever i is equal to 1 or 2 this would be by notation d in d s e i OK. So the derivative with respect to the only parameter surviving. OK. But now here in this picture unfortunately I should have drawn a much bigger picture in a bigger blackboard but I mean at this point we have another special vector which in principle has nothing to do with them which is the tangent vector to the curve gamma. In fact I am assuming for example so my convention the curve goes in this way so gamma prime would be this vector here. But then since it is a tangent vector to the surface it has to be a linear combination of these two and since it is parameterized by arc length it is a vector of norm 1 and they are both vectors of norm 1 vectors I mean basically gamma prime would be some cos theta e1 plus sin theta e2 because we are on a plane. OK. So this defines theta so let theta of s be the angle be the angle between gamma prime and e1 OK. Of course otherwise you can just say gamma prime is equal cos theta e1 plus sin theta e2 OK. Now I will try to save the picture but I am not sure I will succeed. So now define another vector and I call it eta which is nothing but n cross gamma prime OK. Now n is the normal vector to the surface so this vector is tangent to the surface OK. And basically this is the vector tangent to the surface orthogonal to gamma prime inside the tangent space there is a special vector and this vector is constructed to be basically the orthogonal to this one OK. So eta being orthogonal to this one is certainly equal to minus sin theta e1 plus cos theta e2 OK. So the acceleration of gamma which is nothing but gamma double prime OK. So little n if you want OK. Well little n up to curvature OK. Unnormalized by the curvature Well this is what this is this becomes theta prime eta plus cos theta e1 prime plus sin theta e2 prime OK. This is nothing I mean I'm not cheating gamma double prime take this equation gamma double prime is what? Well when I when you put the derivatives on the vectors I left them indicate OK. So the point is when I take the derivative of the angle what's going on this becomes theta prime times minus sin theta e1 OK. Plus theta prime cos theta e2 So this is exactly theta prime eta OK. The two contribution coming from the derivatives of these two functions are exactly theta prime eta OK. Nothing mysterious there So how much is the geodesic curvature of gamma is what is I should take the tangential part of I take gamma double prime no? You project on the tangent space but gamma double prime I know it's orthogonal to gamma prime because gamma prime is a vector of norm one at every point so as usual if I take another derivative I know I will get an orthogonal vector gamma double prime I mean the projection in the tangent space has no component on gamma prime OK. So kg is equal to the scalar product between eta and gamma double prime I mean it has to be all encoded in the normal direction to gamma prime OK. But this is what well now we have a formula for gamma theta prime I mean if I put this formula here I will get eta cross eta which is one because they are orthogonal and both of norm one so eta is norm one which is something I already use so this becomes theta prime plus or minus what well let's do it cos theta cos theta e1 prime plus sin theta e2 prime I still need to compute the scalar product between this and eta OK. And eta which is that object there minus sin theta e1 plus cos theta e2 OK. So how much is this? e1 prime is orthogonal to e1 again by the usual thing of norm one at every point e2 prime is orthogonal to e2 OK. So this I am left with what? I am left with cos square e1 prime e2 minus sin square e2 prime e1 but then I have the other equation so here I have already used the fact that they have norm one and now I use the fact that they are orthogonal at every point and this equation e1 e2 is equal to zero at every point so if I take the derivative with respect to s for example I get that e1 prime e2 plus e1 e2 prime is equal to zero so minus e1 e2 prime is equal to e1 prime e2 so this minus term I can put it as a plus the other one so I get cos square plus sin square so one times so why there is a minus because I prefer to do the other way round I prefer to keep the minus one so I kill the plus one so this become I prefer to write it as minus e1 e2 prime OK and I can erase this so here I used heavily the fact that this was an orthonormal basis at every point forget the picture for a moment now let's see what are we trying to prove we are trying to compute the integral of this function well this is a simple closed curve so I can assume this is parameterized between zero and 2 pi and so how much is the integral well whatever in fact I don't even need to assume that this is parameterized so how much is the integral of theta prime whatever the the domain of definition it is well this is equal to 2 pi it's kind of a whole rotation this curve is doing one rotation on the surface OK so in order to finish the proof we need to compute the integral of this OK how much is the integral of this object along the curve so in fact let me give it a name so let me call it I the integral over gamma of the plus sign e1 e2 prime in the S this is a function of S OK how do I do that remember when we studied the isoperimetric problem that was a very useful formula which I remind you now in fact here maybe this is a good place to remind you if you take two functions of two variables over some domain of R2 which is bounded by a simple closed curve you know this formula here the integral over gamma plus q dv OK this is equal to the integral of the interior region well maybe it would have been better to give another name to the interior region because this is a theorem on R2 of v0 of dq so if this was the one which was hitting v I take the derivative with respect to u minus dp dv and now in dudv OK this was Green's formula the one we already used and now we are going to use it again OK so in fact this was a bad name let me call this beta because this is a theorem on the domain in R2 so this holds for any curve simple closed curve in R2 which bounds a region v0 now beta is the one which I find by pulling back gamma from the surface to the domain v if you draw the usual picture we are in the usual situation we have some patch and we have some simple closed curve here which is gamma but everything comes from some x in some domain v and so here I will find closed curve beta and of course the corresponding interior region will come from the corresponding interior region OK we are exactly in this situation so beta use beta so this is beta beta will be some u of t v of t in fact v of s it's OK doesn't matter so that gamma becomes x composed beta as usual now why the computation on the domain should be easier well, what is e2 prime e2 prime in terms of the variables umv after all I can think of e2 as a function of umv and then umv as a function of t so I can also do chain rule if I do that this becomes u prime e2u in our standard notation now plus v prime e2v OK so I define I am looking for the right functions there p and q to use that formula so define p to be the function e1 e2 is quite natural because now the function that I am really interested in is the scalar product between e1 and e2 prime so e2 prime this will give a sum of two objects which looks in the spirit of where we are going so I define the first function basically to be p and the second function to be q so q will be the other one q will be OK so how much is dq du with this choice dq du minus dp dv is what dq du just notice one thing there will be one cancellation because if I take the derivative of q with respect to u if I take the derivative of this one with respect to u I get e1u e2v and then I should put plus e1 e2vu but I will get the same term here when I take the derivative with respect to v so I don't even write it so minus dp dv so this is minus e1v e2u I've already dropped the last but this is notice that this is exactly what came in in the proof of the terremigradium and actually we proved already that this is equal to eg minus f squared divided by eg minus f squared 1 half this was one of the famous formula we proved but then what does it happen when I apply Green's formula here well this object I should integrate this object in du dv but not that I think that the surface is telling me the surface would like the area measure I mean the surface measure the surface measure is given by square root of eg minus f squared du dv so I need to multiply by that but that means I have to divide by that I multiply and divide by square root of eg minus f squared ok and what does it happen this du dv becomes da here there was a one half missing so this one half becomes without the one half and this is exactly k ok and that proves the theorem ok so it's essentially it's another output of the proof of the terremigradium because this is really the key point ok now there is a more famous version of this so I keep the statement there but it's just a modification of this one which is the case where gamma is not smooth ok what does it happens when gamma is not smooth well of course I cannot really do whatever I want so I would like to restrict myself to well this is enough to the case where the curve gamma is made of a finite number of pieces so it's what is called piecewise smooth ok so let's try to make this extension ok so if gamma is piecewise smooth what does it change in our proof well coming from Green's theorem and so the estimate of this doesn't change because Green's theorem doesn't care about possible angles casps or whatever vertices in this thing I mean the points there will be a finite number of points where the tangent vectors coming from the right and from the left are not the same this is the difference of course we have a finite number of these points here ok so the part coming from this term doesn't change what about the part coming from this term well now this changes in fact before saying how much it changes I mean we have to give names to some of these objects when you have a point a vertex in fact even a smooth point but in this case you get angle 0 I mean you have an angle between these two tangent vectors you have a non-zero angle so this is usually called the interior angle because in fact I mean I draw it outside just not to confuse the picture but it's the same as the one you usually draw inside and then there is an exterior angle which is kind of the complement of that ok so this angle here will be called delta I so here I have a finite number of points PI where this accident happens ok at this point I know that I have some what is called an exterior angle which is different from pi of course you can do the same thing at the smooth point but in this case delta will be pi ok if this is delta usually we call this is delta and this is alpha in order to decide what's happening to this integral it's basically easier to to work with delta ok because what is the difference of the integral of theta prime when there is a jump well that's exactly that I mean the integral of theta prime will be 2 pi minus the mistake and how much is the mistake the amount of the jump but what is the amount of the jump is the exterior angle ok how much the tangent vector has to rotate to pass from one position to the other ok so in this case the difference would be that the integral of theta prime will be 2 pi minus the sum of delta I ok which in terms actually we are more used to work with the interior angles now so if alpha I so if alpha I so these are the exterior angles in terms of alpha I which is pi minus delta I so the interior angle what do we get in fact let me write it as I left this just to comment on these two pieces now I want to write the general formula so let me write it as a theorem and now this is a famous name this is called the local gauss bonnet theorem and this states the following so gamma now would be a piecewise move simple I still need to assume that there are no self-intersection and that this curve is closed of course simple closed curve it's usually what you call a polygon but now this polygon is drawn on a surface it's not a planar polygon and in fact I'm not even in my picture there is a little ambiguity this you might think this was kind of a straight line almost a straight line here nobody is telling me that this object here in fact doesn't even make sense on a surface to prescribe that these edges are straight so these are kind of generalized on a surface we are drawing really kind of generalized polygons because the edges can be whatever they want so gamma should be a curve like this with n sides so let me give to the number of sides and I call them n and with internal angles alpha 1 alpha n on a smooth surface well then what we proved is this which is better to we are more used to write in this form but it's exactly the same formula so the integral the sum of the internal angles to a polygon is equal to n minus 2 in fact what if I ask you now I've already written somehow the answer but I mean the sum of the interior angles of a old fashion polygon in R2 is how much so really our old high school polygons straight lines with vertices well the sum of the interior angles is equal to n minus 2 pi so in particular the sum of it in a triangle the sum of the internal internal angles is pi and so on on a four edges 2 pi and so on so this is the complete generalization of this theorem on a surface free to be whatever curve they want but what is the correction so basically this theorem now becomes well take your high school theorem and correct it with two objects the two objects come from two reasons the first reason why there is a correction is that as I said the edges are not straight lines and where do I see it I see it here why gamma was a straight line between two vertices of course Kg would be zero in this case n I mean there is no normal vector geodesic curvature geodesic curvature is zero in fact this is a nice observation to do in any case if a surface contains a piece of a straight line a segment this segment has to be a geodesic in fact this you can prove it automatically by comparison geodesic is a critical point of length segment is a critical in fact in this case it's a minimum of length automatically because if it wasn't a minimum on a surface it would not be a minimum even in R3 so even in R3 you would have a curve which joins the two vertices of the segment as length which is impossible so you can argue immediately without doing any computation that a segment must be a piece of a geodesic every time a surface contains a segment so the first correction comes from the fact that edges are not segments and it's encapsulated in this term but somehow the second correction is even more interesting in much more interesting because the second correction comes from the fact that you are on a surface and this surface may be flat or not and this introduces this correction plus now becomes a plus because you have to take this formula with alpha delta I is equal alpha I minus pi and put everything on the left so these are the right signs so the sum of the interior angles of a polygon on a surface is our Euclidean guess plus correction coming from the fact that the edges are not geodesics while in high school we gave by definition of polygon something which has segments as edges plus the fact that the surface is non flat now this theorem is kind of fundamental both in the applications and conceptually because in history of course the case n equal to 3 is enough I mean it's essentially the general case so the fact that the sum of interior angles of a triangle in the plane is equal to pi is a statement which is equivalent to the fifth postulate and you probably you know that for more than 2000 years mathematician have tried to check whether the fifth Euclidean fifth postulate is the one which says given a line in the point outside there is one and only one line passing through this point parallel to this there is one and only one Euclide put it as a postulate it's not a theorem but question is it a theorem now this was a very disturbing problem for mathematicians for more than 2000 years but really big fights somebody claimed to have proved it so out of the first four postulates so using the first postulate are you able to prove the fifth somebody claimed yes somebody claimed no and so on actually this formula again if you can read the formula you get a solution because of course everything in fact Euclide was right it is a postulate so Euclidean geometry is that geometry where you have to assume the first four plus this this is not automatic out of the first four ok because in general if you live instead of living on a plane you live on another surface and you don't even know because it's your universe so you cannot really say and you think of doing a geometry where the notion of line is substituted with the notion of geodesic because that's your notion of line that's your natural notion of straight line after all why we think a line is something more natural because it's our intuition of uniform motion so if you think of being a bug living on a surface you would say well there will be a bug called Euclide coming up with his geometry and telling you given the name straight lines to what we would call a geodesic looking from outside the problem is if you substitute these things I mean you have to be logically consistent I mean you cannot prove postulates ok by definition but I mean it is true that you can geodesics on surfaces do indeed verify I mean at least for some surfaces I mean you can check that do verify the first four but the fifth will fail because of this and in fact the postulate can fail in two possible way because the postulate is telling you two things there exist one and only one ok so of course if this is not true if you think that this is not true it can be false in two ways maybe there are none or maybe there are more than one ok and in fact these two things are two geometries there are exist one is elliptic geometry and one is hyperbolic geometry ok elliptic geometry what is elliptic geometry elliptic geometry is that you are a bug living on the sphere ok and you think of a geodesic now you know everything about geodesics on the sphere so you think of a geodesic as you take the book of Euclid and you substitute the word line geodesic ok and now given a straight line and given a point outside how many lines passing through this point parallel of course means they don't intersect ok how many parallel lines pass through this point parallel to this one none ok you will have to cross this equator ok but then ok so this is one way you can the posture that can fail the other way is by constructing some kind of geometry where given a line and the point outside there is more than one I will show it to you next time ok hyperbolic a model for hyperbolic geometry ok where again you substitute the word line to geodesic to line and everything holds except this fifth postulate ok so this was the end of the story which was a very complicated story because what is really behind this is this mental game to be a bug which sounds strange I mean it took thousands of years ok to realize that it's a good game but you don't have to I mean criticize too much our ancestors there was there were religious, philosophical whatever reasons to believe that Euclidean geometry was geometry some kind of God given ok so there was one absolute truth ok and the truth we would live inside the truth because in some sense there was one geometry which was Euclidean geometry ok so to break this wall now it looks very stupid I mean ok draw a sphere and substitute a word to another word and that's it but somehow this would have been considered heresy I mean the idea that this could be a universe was against some philosophical principle ok now of course once we started exploring the universe on a larger scale than just our solar system it became clear that this was not at all a game and probably actually our universe is not and in fact now we know that it's not exactly Euclidean universe ok so respect for our grand-grand-grand-grand parents ok even though at the end now it you see it's also I must say it's part of the beauty of mathematics you work hard and you prove a formula and then that's it looking at the form I think mathematicians it's a beautiful part of our job not every formula is the same there are technical formula there are some formula really contain fantastic information and this is one of them in some in fact Gauss proved this fact and when Gauss proved this theorem he finished his book it was clear to him this is known by history in history of mathematics it was clear to him what I tried to tell you as kind of consequences of this formula but he didn't want to he was not brave enough to state it openly and that was this formula is kind of the end of the philosophical supremacy of Euclidean geometry against something else ok but he stopped that's the end of his famous book on surfaces ok and it had to be by Bojlaj and Lobachevski who came up with the consequences and to showing that actually there are models in here if you assume that the postulate phase you have to give me a model ok because maybe it's just an intellectual game but they don't exist but I can show you next time it's quite easy to construct a model this is not really the problem so it was them who said to everybody look now Euclidean geometry is dead because of this essentially they had to pay for it Gauss was quiet in his observatory in Gottingen looking at stars and in fact the only thing one of the two I don't remember if it was Bojlaj Lobachevski one of their father was a friend of Gauss was another kind of amateur mathematician friend of Gauss and he wrote Gauss asking what is your opinion on these crazy ideas my son is going around and there is the famous answer by Gauss I cannot make any kind of praise for your son achievements because actually they were mine which is a very nice answer I mean it's a very generous answer to a father ok so he was claiming once they were sent to war I mean into somewhere to pay for their he I mean for being against the church and everything now after 20 years he was saying no no no but this is my discovery when it's too late this is not really acceptable this was not really the best page of Gauss history ok now but now let's make one further step everything we did was inside the actually he didn't write it ok but it's clear so the image of this curve has to lie inside the patch ok is there kind of a global theorem now that we can extract out of this and this is even more beautiful of course the hint is here there is a local Gauss Bonnet there should be a global Gauss Bonnet somewhere and now let's face it now before before telling you what is the global Gauss Bonnet ok I erase this but I keep the I will write down again the formula up on top of the black board so the sum of the interior angles is equal n minus 2 pi plus the integral of kg plus the integral of k da ok so remember that this is now in our bag let's make one step back because a little bit of topology now if you have a surface one way to extract some topological information out of a surface is the following basically probably most of you know so I'm driving in the direction of the definition of the Euler characteristic ok of a general surface in fact this holds not even in R3 I mean you can define this subject the Euler characteristic for much more general objects but let's see definition so I want to say that the subdivision of a surface of a compact actually let's restrict for simplicity of a compact surface S is a partition with the following properties in two three subsets possibly empathy subsets that I call S0 S2 ok 1 S0 is given by a finite number of points ok S0 is p1 and I call it the last pv v stands for vertex ok is a finite percent of points the pi are called vertices of the subdivision S1 is in fact a set of curves on the surface so is a disjoint union they don't have to intersect disjoint union of finitely many subsets of subsets in fact that they are finite it's part of my notation let me call it e1 ee ok contained in S so what do I mean they are curves of course I mean that each of them for each eI there are vertices so they have to start and then in one of the vertices ok for each eI there are p and q p0 in the list above and continuous injections I want to realize them as maps injective maps fI from 0,1 it's not important because this is just topological so I don't care where how big is the defining is the domain of the functions fI 0,1,2,S such that of course f0 is equal to p f1 is equal to q ok and f restricted to and f restricted to the open sub part 0,1 this surjects onto eI so ok by this notation I mean of course this eI would be the open curve without the two extremes the two vertices ok because now it's clear sorry fI restricted to 0,1 is equal to surjects onto eI ok this eI eI or such eI are called edges of the subdivision note there is no reason here I never said that p is different from q ok so in principle they could be loops ok it's allowed it's not forbidden by the definition and then the condition on 3 on S2 what is S2? S2 has to be at this joint union again but now of open subsets that I call f1 this is not a great notation after the previous I don't know how to call them so f1 ff ok each of which and this is important is homeomorphic to a disk of course these fI well nothing to do with the previous f actually these f will never appear in anything these fj are called faces so we have vertices somewhere vertices and faces ok so examples take for example I'm restricting myself to compact surfaces ok so the simplest thing we can think of is the sphere let me not draw anything infact even this equator is already misleading in some sense because I don't want to think of this equator now as one of these edges it's just to tell you it's a sphere ok so now how can I take a subdivision of the sphere well this is the kingdom of freedom you can do almost whatever you want but for example just to show you if I take this equator I have to define s0, s1 and s2 infact here the thing my suggestion is that somehow the most difficult thing to get in general is to be homomorphic to a disc so try to split your surface in pieces which are homomorphic to discs well in a way that what is left over are curves this is the idea behind the subdivision of course every surface every regular surface can certainly be covered by patches and every patch if you want it's homomorphic to a disc if your original domain was a disc ok but you see here I need this joint union so patches are very bad in this sense ok infact the interesting part of differential geometry is where patches overlap ok otherwise there is not really too much to say ok so for a sphere for example how can I write it as a disjoint union of some number of discs well separate the sphere with an equator and then the upper and the lower half are discs ok and what I am left with one edge ok so here for example I can just put one vertex there so you see here I have now if I don't add a vertex this would not be a subdivision I need to have a vertex because remember I said EIs are kind of the open intervals ok so to the open part has to subject in injectively ok so to EI so that means EI is really the equator minus a point if I want the vertex is this point so this is P this is E and these are so P1 I have only one vertex only one edge and two faces ok this is one subdivision in fact I can even do something even more pathological this was starting from the problem of faces but actually starting from the possible if you want to write the sphere as a disjoint union of discs minus something of lower dimension this is kind of the idea well there is an even more stupid thing to do stereographic projection ok so if I remove only one point what is left this is homomorphic to a disc ok so strange enough this is a subdivision and it's a division made of what made of one vertex and one face ok but then of course then I can think a bit more geometrically for example start slicing your orange in the usual way and this is another subdivision this triangle is a disc ok so I don't even want to count there will be many vertex many faces and many edges but now the strange thing the miracle I have no way to convince you that this was a logical thing to guess compute V minus E plus F ok the point is at least on these three examples let's make the simplest of these just to check what's going on the simplest of this is with two equators how many vertex do I have I have one one, two, three, four four, five strange should be six and six so V is equal to six here so many edges one, two, three, four five, six eight oh no, because I also ok, how many I've counted them here so, two, three, four I have four for each slice so E and how many F how many faces eight here is always the same whatever you do you always get two ok notice another thing so in fact definition this is too important to leave it like this so the Euler characteristic of S relative to do a subdivision because in principle it depends to the subdivision let me call it this was not a good idea to call it a zero S1 let me call it curl P the subdivision P like partition in some sense so that is but you use this symbol chi of S P and it's equal to V minus E plus F so this is characteristic relative to this subdivision here so one of the in some sense elementary but most, I mean it's a delicate theorem in topology of this type it's to prove that this number does not depend on the subdivision but only on the surface so theorem of S P is equal chi P prime for any P prime subdivision ok I'm not going to prove it because this is basic topology it's a smart proof you can find it in standard in the books of algebraic topology or something like that how many of you have seen for example this proof ok now this is a key fact of course plus another key theorem because I ran a little bit forward say ok if I have a subdivision I can compute its soil characteristic its soil characteristic is independent of the subdivision and so on but now there is another key theorem behind the scene is that any surface is a subdivision ok which is nontrivial ok every compact surface has one subdivision because otherwise our theory would be a bit empty ok now this is in fact more difficult than the previous one ok you have to construct it by hand put the two things together if I take a surface I have at least one subdivision I can compute its soil characteristic and it's independent of the subdivision so this number is a number which depends only on s and it's called just the Euler characteristic of the surface with obvious notation this number it's an integer of course because it's e minus v plus f it's also a topological invariant so two surfaces which are homeomorphic then of course if I take a subdivision on the left the image is a subdivision on the right the number of vertices edges and faces is the same because it's a one to one correspondence ok so two homeomorphic surface surfaces have the same Euler characteristic now this is one of the things that I mean it's fantastic so maybe I can classify surfaces just by looking at one number so which are the surfaces which have Euler characteristic two up to homeomorphism because this is really a question in topology I mean here you don't need a metric or a first fundamental form nothing I mean this is a topological question I mean this is a beautiful part of topology it can be done because sphere is the only compact surface up to homeomorphism with Euler characteristic 2 ok so this number really gives a classification of surfaces so two surfaces with the same Euler characteristic are indeed homeomorphic if they are orientable what else do I want to say on this well let's play another game just to because otherwise you might think we are building a theory on the sphere so let's take what is another simple compact surface that we know for example the torus for example the torus of revolution but you see being a topological question the fact that it's exactly of revolution you don't really care this topological surface are we able to compute its Euler characteristic if we don't make mistakes yes the best way one way would be to imagine things here but this is at least I find it a bit confusing I would prefer to have since it's a topological invariant to have another topological model if I have another topological model which is better for this problem I use that so what is in fact this surface topologically topologically this is one circle rotating around so topologically this is s1 cross s1 s1 is the circle is the product of two circles and how can I draw it it's the usual when you realize it in the usual way you first make the cylinder and then out of the cylinder you identify the two circles by closing it up what are you really doing you are making a mathematical operation which is this you are taking a square or a rectangle something like this so first make the cylinder so there is an equivalence relation on this topological space I am taking this with a topology induced by R2 so this is a topological space on this object I look at the equivalence relation which identifies this point with this point this point with this point when you close them so there is an identification between this edge and this edge homologus point at the same height go to point at the same height and then this edge has become a circle because these two points have come together so now the second identification is identifying the two circles in the same way but that's exactly the same thing so it's this edge identified with this edge now the topological space that you have produced by this equivalence so the quotient space is exactly this but now here I can find subdivisions much easier because I am going around the hole in some sense I cut up the torus in a way that now it looks like a square so as long as I don't forget the equivalence relation I am done because what? because in fact I have already drawn a subdivision you see the inside of this is a disk topologically the square is the same thing so this looks like my face but his face and now and how many edges do I have it's four modulo the equivalence relations so four modulo the equivalence relation is what? is two and then how many vertices do I have? four because it's a square equivalence relation one they are all the same because this goes here by transitive property everything is the same so I have one vertex how many edges? two edges and one face so the Euler characteristic of this is zero and again it's the only compact orientable surface so so if for some mysterious reason you don't know which compact if you know it's orientable otherwise you have to be a slightly more careful if you are able to compute the Euler characteristic without knowing too much about the surface you can actually tell metopologically what is your space it's quite fantastic now here there is a whole subject here but this goes beyond our course because now out of a subdivision you can produce many subdivisions there are operations that you can do in the space of subdivisions to go from one to another but we don't care let me finish so the last in five minutes I will give you the global version of this I will use now all these theorems and these definitions to put everything together in the most famous formula in the differential geometry of surfaces this is something you are not allowed to forget so theorem and this is called gauss-bonnet formula or theorem as you want put whatever name as long as there are these two names on top if you take a compact S is a compact smooth surface in R3 then the integral of the gauss curvature over S over the whole thing which is something you compute by chart by chart with our measure so you know what this means in DA this is equal to 2 pi beauty as its best let me give you quickly the proof because it's essentially playing with this basically you have to convince yourself so now you have your compact surface it's difficult to draw a picture I mean a sensible I cannot really draw a subdivision but what's going on you take the local version so basically a subdivision is really giving you at the composition of your surfaces in faces edges and vertices ok so the idea is to apply the lock and in fact you can also assume this is what I was saying there is a whole theory but it's kind of simple you can even assume that you have reached a subdivision where each triangle is made of triangles if you want and each triangle is contained in a patch because after all if it's not true you can make it smaller and smaller and smaller and sooner or later you will have to end up inside a patch allowing the possibility of making your subdivision more and more complicated if you want with more faces, more edges and more vertices you know that it's made by triangles and each triangle contained in a patch so that you can assume you can apply to each triangle this the local version of the theorem this is completely now now every time you have a vertex of your subdivision you see if I look at this triangle here that means I'm looking at this piecewise smooth curve I'm looking only at the contribution at this vertex at this vertex I would have an interior angle if you want ok so think of making the sum of all of this formula overall possible faces so faces and corresponding curve gamma now imagine that you are doing it it's a finite sum this will be a finite number of things ok what happens at each vertex of the subdivision at each vertex of the subdivision you get 2 pi the sum of these angles as I move around at each vertex has to be 2 pi ok so on the left I get 2 pi v v because I have vertices this is the contribution coming from the left now each edge belongs to 2 polygons it cannot belong to more than 2 and it has to belong to 2 because it will have one side on one one right and one left now so the edge of course comes here I should compute the integral of the geodesic curvature of this edge and I have no idea either I argue that I can find the subdivision made of geodesics which is also true but forget it now whatever it is the point is that you see I decided that when I take a curve I go around in one way for example counterclockwise because I need to speak about the interior so the interior is what is part of the curve in order to speak about the left I need to decide what is the movement on each curve so when I look at this triangle here this edge here goes in this way when I look at this triangle here it goes in the opposite way and then actually I think I should make a correction at the beginning of the lecture I gave you the notion of geodesic curvature without sign so for me it was I told you the norm of well it comes with a sign maybe I'll correct it at the beginning of next lecture because in order to speed up I told you pick the norm of the norm but geodesic curvature comes ah no no no sorry sorry it doesn't matter in fact I think I need to correct because while I teach something I think to what I said before but that's not important this curve once if I look it on one side or the other side it's the same point on the same curve so whatever it is it's the same but the problem is that they integral in one way would go from 0 to 1 and the other way would go from 1 to 0 they are ok so it's one of the opposite of the other so in fact edges do not contribute ok except for this actually so this one will disappear but this one how many times does it come into this how many times I have to take that well this one I have to take 2 pi E minus F times ok now I'm already over time so I leave you time to think about the number of and then of course how much is this well this is the easiest because of course the whole surface is the union of the faces and each face is the interior of this polygon minus a set of measure 0 so taking the integral the sum of the integral over the faces and the integral over the whole surface is the same thing because what you are missing are a bunch of curves do you agree? so this one will sum up to this and now put these objects together and you get Gauss Bonnet ok so now really it's the local theorem which this is just drawing the right picture the local theorem is the one which contains all the geometric information ok so maybe next time we will comment on corollaries of this