 Welcome to NPTEL NOC, an introductory course on point set upology part 2, module 62 connected some and classification of surface. Last time we have introduced the notion of rubber sheet models and stated a big theorem which is supposed to give you all compact connected surfaces. The same result we will try to explain in a more geometric way this time. So I have defined one single notion here, start with any two connected surfaces. So this time let us not bother about the boundary, boundary may be there but keep it far away, even if it is there. Start with any two connected surfaces, start with two embeddings of the disc D2, the closer disc D2, eta1, eta2, each of them in S1 and S2, okay, B2 embeddings. Put Ci equal to etai of the boundary, the boundary circle, take phi equal to C1 to C2, C1 is you know this is an image of this one, so it is contained in ZS1 and that is contained in ZS2, right. So take phi to be eta2 composite eta1 inverse, where I mean by eta1 inverse I am restricted to the boundary here, okay. Then the quotient space obtained as follows, namely you take S1, throw away the interior of this disc, the image of that disc, interior you throw it away, keep the boundary, similarly from S2 throw away the interior. So you start with two surfaces, now you have made two holes there, okay, when you make a hole, hole means what, now removing a disc, interior of disc, the closed disc is there, the boundary of the disc is there, so those things are C1 and C2, right, here. So identify X with phi X, where phi is this map, by this map, namely for every point X inside C1, by this map phi, eta2 composite eta1 inverse, that is a homomorphism. So that quotient space is called the connected sum of S1 and S2 and is denoted by this notation, S1 connected from S2, that is the way I read this one, okay. So here is a picture here, picture may be confusing, I do not know, this is one surface, there is another surface, there is a disc here, there is a disc here, so these are embedded discs, so you remove the interior of this, remove the interior of that, there is a circle here left out, namely image of a circle, so this looks like an ellipse here, this is an image, so identify them like this, so bring them together like this, identify the two circles by a homomorphism, that homomorphism is not an arbitrary homomorphism, it is from this circle, see this one is eta1, so you get eta1 inverse composite eta2, so that is a homomorphism from here to here, you identify that, okay. For example, if you take a disc, remove an interior disc from that, you will get an annulus, do the same thing on this side, take a disc, remove the disc from the interior, you will get another annulus, now you have got two interior circles in both the annuli, you identify them what you get, you will get back again a cylinder, actually annulus and annulus they are cylinders, so but when you identify two of the circles, one circle from here and another circle from another one, you will get one cylinder, so the world connected some precisely refers to this one, you start with connected surface here, the end result is also connected, there are two of them here, the end result is connected, so two of them you start, but each of them must be connected, so then you make them connected by this performance, okay, if you just join them at one point, it will fail to a manifold, that is the simplest way of doing connected zone, if at all, but in topology, general topology we will do like that, one point union, but in manifolds, it will fail to a manifold, so what you have to do is, do the trick namely, remove a small disc, remove a small disc, okay, what happened, the boundary of those two circles, the resulting circles you identify them, the following observations which are intuitively clear, of course need proofs, okay, which we shall skip, so what are these observations, S1 connected some S2 is a connected two manifold, it is compact if and only if both S1 and S2 are compact, okay, see there was no statement about this being compact, so I could have taken any two connected circles, that is all, alright, so so far whatever I have stated, they are not at all difficult, okay, very easily, only because we have time limits here, I cannot go on explaining everything, the homeomorphism type of S1 connected some S2 does not depend upon where and how you have chosen these embeddings eta i from D2 to S i, S i surface is you have chosen these things, right, how you choose these, how you choose these things, no problem, all of them will define the same, same means what, up to homeomorphism the same surface, so this is a deeper thing, this requires some proof, not this one, these are easy, so this requires some proof, if you take S connected some S2, okay, S is some surface, S2 is the standards here, what do you get, you will get back S itself, why because you have to remove a disc from here and disc from here, right, when you remove a disc from S2 what do you get, you have another disc, now you are filling that disc back in the gap that we have produced here, so all that amongst doing nothing, so that nothing is back to homeomorphism, so connected some, taking connected some with the sphere produces no effect, that means it is identity operation, so that is why you know S2 I told you it is like a zero element, this connected some more or less represent you know some additive operation, you can see that it is associative, it is commutative, only thing is there is no inverse here, it is like natural number, you see, there is a beautiful you know algebra here, out of this, what are called as cobaltism theory, so that I cannot go into this detail, the cobaltism theory is connected some operation is the sacrosanctus, so this can be defined in all dimensions also, but now we are learning it in dimension 2, let us denote the torus S1 cross S1 by this simple symbol you know this Euler T, I will just read it as T and the projective plane by P2, this we have already done, this was a new notation, this is an old notation, let us denote the connected sum of G copies of T by TG, what is this S1 cross S1, again take S1 cross S1, again take S1 cross S1, G copies, take the connected sum, let us denote it by TG, similarly the connected sum of N copies of the projective space, let us denote it by just PN, so again this G and N are coming just like in the last lecture, so what is it I am writing here, TG is T connected sum, T connected sum T, I do not have to put any brackets here because SOCA tail, similarly PN is P connected sum P connected sum N copies, what is T1, it is nothing but T itself, P1 is just P upper 2, because only one copy is there, you have performed any operation, so this is notation, short notation for this long thing, now the big theorem here is like we did after long long in the last time, but this time we have easily the following, every connected compact surface without boundary, I want to emphasize that here, compact connectedness but no boundary, is homeomorphic to exactly one of the three things, actually there are three lists here, this is a single element, but this is an infinite sequence, this was an infinite sequence indexed by natural numbers, exactly similar to the previous theorem that we had, I will just show you that theorem which we did last time, this is the theorem, the AA we have told it is also H2 here and these are going to be the second list TGs, these are going to be PN, every compact connected surface without boundary is homeomorphic to precisely one of the surfaces defined by these reduced polygons, so this is some kind of a technique to obtain such a result, the finally we would like to have this description, so why these two are the same, I will try to explain that one, not a proof but just some explanation, the theorem has two parts, you see every compact connected surface is exactly one of them, first of all take any compact connected surface, you must be able to find the means what, up to homeomorphism, but there will be two different ones, they will be homeomorphic in the list, no, there will be every element here is a different homeomorphism type, so that is the second part here, there are two parts like this one, it asserts that up to homeomorphism there are no more compact connected surface is other than the ones mentioned in the list, the list is exhaustive, thus the list is that means, the second part is which you may call is non-overlapping list, namely the uniqueness part, asserts that members of the above list represent distinct homeomorphism types, the proof of the exhaustion part, there are essentially two different approaches, the first approach one proves that every compact surface can be given a smooth structure and then uses what is called as Morse theory, you can read something many books are there but you can read for example my book, the other approach is the one that we have been discussing in the last module, namely this rubber sheet geometry, the canonical polygons or whatever you want to call, so let me explain because we have done some work there, let me explain the relation between the previous theorem and this present theorem, namely in the second approach, one first proves that every surface can be triangulated, what is the meaning of triangulation, just you may be able to cut it into finitely many pieces, each of them homeomorphic to a triangle, homeomorphic to a triangle is anything homeomorphic to a disk but you would like to think of them as a triangular pieces with three edges, so this is the idea, the proof of this is not at all easy, the proof of that giving a smooth surface is even much harder actually in that sense, may not be but this is much more harder anyway, so you have to assume that they are triangulated, from the triangulation what you can do is you can get a regular 2 n gone with edge identification as indicated in the last module, you cut the surface into all the strangles while cutting down what you do, you keep a label wherever you have cut down you are introducing two different edges right, so make them all labelled in the same way a comma a and put an arrow also which way they are getting identified, so this is this will keep you for the surgeon to patch up where wherever he has cut that is the whole idea, so look at all these triangles, now lay down them on the table one by one side by side wherever the edges come you know in any arbitrary fashion it is like a jigsaw puzzle when you are doing jigsaw puzzle nobody tells you which piece you have to take first and forth, you take one first then keep putting this jigsaw puzzle try to arrange them, how do you do that there is only one rule, namely wherever an edge indicated by a letter you look for that letter in one of the triangles it will be there, so put it there put it on the side keep putting like that re-adjust the whole thing you are allowed to stretch the triangles whichever way you like, so that every every time you have a convex polygon re-adjust that the shape of the triangle that is all, so you have union of finitely many triangles and that is a convex polygon, so one of the extra edges there you will get exhaust all the triangles you will get a convex polygon, now what will happen there will be all these boundary edges of this convex polygon they are identified as some edges where are they they are also there in the same polygon but on the boundary now because everything interior they have got their pair they have been paired already, so these are the last ones which are left out, so you will have exactly a even sided polygon two end sided polygon because there must be a pairing and you perform that you will get back the surface that is how the exhaustiveness has been, so every surface can be got by a rubber sheet scheme, now there is another part now you have to say that the list contains all the representatives, representatives means what up to homeomorphism, so that is the harder part that we have done in algebraic topology course part two also you can read it from my book in algebraic part, so these combinatorial arguments to cut it down to just canonical polygons as listed last time they are justified by one main topological observation what is that if you have made two cuts in a surface you can get back the original surface by performing the two corresponding patches and patching up whichever order you want you may all you may first you know stitch this one and then stitch the other one or the reverse way there is no problem the resulting the surface that you get will be the same provided you have performed all the stitching back wherever you have cut that is all, so this is the beauty you can cut as many times as you want, every time you have cut sometimes that cutting must be stitched back it has to be identified that is all, we shall not go into details here but we shall indicate how to get 12.50 from 12.47 now, so that is the main thing there are two different description I want to relate them somehow first of all note that list A in both the cases corresponds to S2 there is A here that A here they are same they will they will give you S2 now come to the list B start with the rubber sheets came A B A inverse B inverse and some C do not matter note that the resulting surface as has one boundary component because this is not identified and that will be the circle because both the end points are in one of these so they will get identified which is a circle correspond into the free edge C if you identify this entire circle to a single point you have some surface and it has a one boundary component namely keep that make this circle smaller and smaller and bring it to a single point that quotient space S0 is the same as the quotient space corresponding to the rubber sheets came A B A inverse B inverse as if that edge as we have been shortened shortened shortened to a single point so no edge left out that is all right so which is now we know is a torus therefore this scheme represents a torus with a hole do you understand this this this I will repeat it what I start I do not start the torus with a hole I take A B A inverse B inverse I know that is a torus but now I have different scheme here namely there is a pentagon right A B A inverse B fifth side is just C when you identify that this C will become a circle whatever the other surface I do not know what it is do not matter but there is a hole here now you move that hole to a smaller make it smaller smaller and make it a single point so that surface is the same thing as as if you have got it from A B A inverse B inverse therefore this surface must be the original surface must be what the torus with a hole all right that is the whole idea so reversing this argument we see that if you start with the torus and make a hole in it then the resulting surface with boundary is nothing but the surface obtained by this A B A inverse B inverse B so this is the scheme for that all right so this is the picture here so A B A inverse B inverse so that C is there which we do not want so if you when you perform you have got something here this A becomes one so one loop that B becomes another loop A and there are two A A here they will give you one single loop okay over one over the other so this B is another single loop this C is another single loop dot dot dot dot so these are interior to the surface but this is the boundary so make it smaller and smaller is the same thing as putting a disk here removing a disk here you get back this one so this picture is as if you have got it from A B A inverse B inverse so this is the picture so you can push it back so putting this making the circle to a single point is same thing as putting a disk here removing the disk from the torus gives you the scheme namely the scheme now what do I do therefore it follows that the connected sum T2 what is T2 T1 T connected sum T right two copies of two copies of torus is obtained by taking two copies of A namely A1 and A2 so A1 is what A1 B1 A1 inverse B1 A1 C1 A2 is what A2 C2 I have I have changed something C1 I have put here cyclically I can put it right in the beginning so that is what I have done that is because I want to bring them together C2 A2 P2 A2 inverse B2 inverse I could have put it C2 here but cyclically I can put it here of no problem so you take these two first identify the two free adjacent C1 and C2 you can identify whichever order you want is the main you know guiding principle here to get the rubber shades came what will be one you identify C1 and C2 this is one this is one disk this is another disk but they have a common edge now on which we have identified so entire thing becomes a one single disk C1 and C2 go into the interior and you get A1 B1 A1 B1 inverse B1 inverse followed by A2 B2 A2 inverse B1 inverse ok this is the picture C1 and C2 they have brought we have been brought together here in this picture then re-aligned the whole thing ok this dotted dotted thing here is C1 C2 one identified with the other the rest of them is A1 B1 A B A inverse B inverse ok A2 B2 A2 B2 inverse this is the scheme for remember in the old theorem this is the connected sum of two connected you know whatever A1 whatever C that polygon in the list two with G equal to 2 here right on the other hand if you perform this one first perform this identification first what we have seen is both of them are looking like this without this disk that means they are torus minus a disk torus minus a disk two of them join C1 and C2 this time at the last what you get you get a connected sum therefore connected sum of torus with itself is represented by this K now a common induction tells you that if you do G of them ok like A1 B1 A1 inverse B1 inverse dotted dotted AG BG that is nothing but what connected sum of G copies of the torus ok so this is the sum of two torus here so that will be you so repeating this process it follows that the two lists list B here and this B there in the two theorem give you the same class of surfaces exactly similar argument is valid for the list C as well ok therefore 12.50 follows from 12.47 alright finally let us see two different geometric ways of constructing all the TNs as embedded smooth objects in R3 ok there are I want to tell you two different methods because they are both interesting and useful in the first method we begin with the union of N circles in R2 such that C1 touches C2 C2 touches C3 and so on like a chain ok so they are not disjoint what are the disjoint C1 and C3 are disjoint C2 and C4 are disjoint and so on whenever I minus j is bigger than 1 they are disjoint when they are consecutive they have a single point in common they are attaching each other ok in the picture below I have taken three of them because picture has to be limited you cannot take N anyway alright so three of them the same radius R that is extra extra thing so that you get a nicer picture and so on now put C equal to union of CI finite union C1 C2 Cn and let Xn be the set of all points X inside R3 such that the distance between X and these circles union of circles is less than delta ok wherever where 0 is less than delta less than equal to R so this is an open subset take less than equal to delta you get a closed subset compact closed subset of bounded subset of R3 alright equal to delta delta should be chosen correctly that will give you a surface that is the whole idea so this is a picture here the dotted circles here these are the beginning circles they are touching at one point one after another they are inside R2 the entire solid here set of all points which are at a distance less than equal to some delta delta must be so these are circles of radius R so this delta must be less than R strictly less than R for the sake of this picture I have taken perhaps R by 2 ok take all the points which are at a distance less than R like that and then look at the boundary that boundary is the connected sum of 3 tori here if you have taken n of them it is the connected sum of n tori if you take only one of them circle first thing you have to see is that it is actually the torus ok so that part is just some elementary 3 dimensional calculus ok you should know some differential calculus to see that this is actually a surface nice surface alright so of course it takes some 3 variable calculus to show that the boundary of Xn is diffiomorphic to Tn whereas it is intuitively quite clear you take a circle you thicken it like a wire thicken it lightly bigger ok the boundary of that is a torus so that is what it is ok so in the second approach one approach is over so you have got a model of the surface right the boundary is a surface which is Tn so it is inside R3 over no for Tn you have got so actually I should write Tg here n ok in my notation I should write Xg and Tg here fine in second approach I merely start with an equation so this should be liked by say algebraic geometry for example so you can take equation here of course this is all real values inside R3 all the coefficients are real you can look at what is this funny equation see the left hand side is just some polynomial and X and Y the right hand side is just so there is separated variable equal to minus that square that is how ok so look at this one there is one special factor here and then there are similar factors here carrying from 1 to n ok what are they the first factor here is X square plus Y minus n plus 1 Y minus n minus 1 square minus n plus 1 square which is nothing but if you put equal to 0 this is a circle with radius n plus 1 and the center equal to 0 comma n plus 1 similarly here is center with 0 comma k and the radius is 1 by 3 I have taken the product of these polynomials put them equal to minus that square this time I take Yn to be the set of all points XYZ inside R3 which satisfies this equation 39 and claim is that this represents Tn ok so here the picture is here so how many circles are there 1 2 3 and 1 big circle what is this big circle this is the circle represented by this one so where are the circles these circles they are inside R2 X comma Y the z I exist on this way ok so I will tell you a little more about this I do not want to prove anything here namely take this equation on the left hand side just put equal to 0 what is that meaning of that third coordinate put equal to 0 means intersection with R2 what are that if some product is 0 means each factor or at least one of the factor must be 0 when you this factor is 0 you get a circle this factor 0 you will get various circles what are those circles they are the circles here one is the biggest circle all of them have centers on the Y axis with some radius ok the radius of the first circle is n plus 1 the radius of the second all the second third on the all of them they are just 1 by 1 by 9 1 by 9 but what is the center 1 2 3 4 and so on on the Y axis X is always 0 up to n but this as n plus 1 is the is the center and the radius is also n plus 1 so that big circle actually touches the X axis ok so it encircles all the circles here and these circles themselves do not encircle any of them that picture is very important so they are you know mutually exclude exclude each other they are not contained you know interior does not contain the other circle whereas the big circle contains all of them in the interior so that kind of distribution is necessary alright now what happened look at this area here bounded by the big circle and outside the all the small circles so how do you get that area how do you get any point in this I am not interested in the area aspect but point in this domain so how do you get that one they are inside this one means the first equation must be less than you know the first factor must value of first factor must be less than 1 less than 0 and this negative ok equal to 0 is the circle less than 0 will be inside bigger than 0 will be outside the same thing here ok I want it to be outside here now so they are all positive so all positive 1 negative so the entire product here is negative whenever entire product is negative it represents a point inside take that negative number take the negative of that that will be a positive number take the square root of that that is your result so that is the meaning of this equal to minus that square ok then there are two square roots accordingly you will get two points one above here one below here so to take that once you take the positive thing you will get the graph of that ok that function precisely this part and then there will be another copy below this minus of that the union of those two circles those two graphs is precisely this one namely square root of this one minus the square root of this plus square root of that so that is z you see minus z square is there what is z take minus of this and then take square root then you have two square roots ok so z equal to something is the graph of this ok there are two functions plus minus so that is the union of these two graphs this is the surface so once you know that ok you will see that precisely above a point here there is nothing here right because a point here the function will be what is the function this function this value of this function will be will be what positive so you cannot have minus of that the square ok there is a real number so this is a real number that is a value ok so these are holes there you see outside also if you come here the function the left hand side will be will be positive because all of them are positive if you come here outside here ok positive then there is no square root minus of that will not have square root that is all that is a trick here ok so it is easily checked that this is smooth surface if you intersect y n with the plane z equal to 0 what you get a disjoint union of n plus one circle all that I have written down here finally this is nothing but union of two graphs z equal to plus minus square root of minus f x y f x y I have denoted by all this factor the function the polynomial x and y this part so in two different ways I have shown that there are embedded sphere embedded surfaces inside R3 up to homomorphism they represent each of them represent the class in what in the list B ok list A is obviously there already we know that list C cannot be represented by embedded surface is in R3 look at P2 you cannot look at the Mobius band Mobius band is also non orientable but this is a boundary that can be represented what is the meaning of that Mobius band we have seen is the same thing as make a hole inside P2 as soon as you make a hole you have boundary right take any surface in the list C namely P connected some P connected some P n times make a hole in that that surface can be represented by an embedded object inside R3 not very difficult to prove but I will not prove that here alternatively you do not want to make a hole then what best you can do you can get an embedding except there will be one single crossing around a circle ok that is called a self-intersection of the of the immersive surface such things are called immersion not embedding ok so there will be two circles in the original surface and a map which will be an embedding except along these two circles these two circles will be identified to a single circle in the image that is the meaning of that so that is the best thing you can do for all the non orientable surface namely those which occur in the list C ok so this is what I have P n cannot be represented by objects in R3 the saving grace is that if you allow just one self-intersection along the circle then it is possible or equivalently suppose you grant for this one then you can put this one namely wherever there is a intersection you cut that circle inside there will be a disc so you remove that disc which was extra then you do not have to cross it out so that is how you can get this embedded circle embedding after removing a disc ok now comes the uniqueness part I will tell you a few things uniqueness part cannot be proved cannot be proved by pure point set topology ok it is not like you know classifying the one-dimension manifold which could be done which we have done just point set topology you have to bring some tools from algebraic topology you have to bring some you know some invariant you may use something like very heavy machinery like a complex analysis you know not elementary complex analysis again ok so like Riemann did them in the beginning ok so you have to have some high high machinery here so I will tell you the simplest thing is that you can use the fundamental group ok yes use fundamental group you compute the fundamental group of all these objects they will be all distinct groups distinct means what non isomorphic that is not difficult to prove once you know what is fundamental group and how to compute these things for that now if you know that homeomorphism not only homeomorphism if they are two spaces are homotopy equivalent I am assuming surfaces connected surfaces nice surfaces and so on nice topological spaces if they are homeomorphic to each other they will have isomorphic fundamental groups since the list will give you different fundamental groups they must be non-homomorphic so that is the way the proof is completed instead of fundamental group we can use what is called as homology groups these homology groups may take little more time but they are much easier to understand easier to compute than fundamental groups actually ok they will also give you that these lists have different homology groups just you have to look at h1 and h2 both of them you have to look h2 will be good enough to distinguish between the the list b and list a list c list a is easy to differentiate list b and list c but within the list c or within the list a list b you will have to use h1 ok there are h1 h2 h3 and so on they are homology groups ok so finally the big thing here is which is very special for surfaces or just one dimension manifold it turns out that the classification of connected compact surface is without boundary the same whether you consider homotopy types homeomorphism types or diffeomorphism types this is a very deep result ok so same thing we have done we have not done same thing we have easier approach namely for manifolds of dimension one house ok the classification that we have done could have been just by homotopy type right lightly you have to be careful with boundaries do not use boundaries we connected manifolds without boundary boundary you can later on deduce it from once you have what the classification without boundary similarly here up to homotopy you can determine the whole thing do not you have to assume that they are without boundary otherwise they will have some other problems which can be sorted out ok so we have come to an end of the chapter as well as an end of the course here now ok so I would like to thank all of you first of all if you have really come all the way to the end of this course many people drop out in between that matter so thank you all I thank my team of TAs who have been especially you know useful for even getting the material in order and all that and they will be also helping you all the time in understanding this one they have been helping you for in the tutorials or what you call assignment and so on right also in the discussion forum so big thanks for my team not only that the team from NPTEL they have been very very encouraging and very helpful so I thank them all ok see you some other time thank you.