 So today we shall complete the second part, part B of theorem 6.4. So recall that theorem 6.4 gives you three sequences, three categories of canonical polygons. The first one is, first one represents the sphere, two sphere, the second one and third one a sequence of them for each other is one, for each G there is one and so on. So part B says that any two distinct members belonging to any of this set they are different homomorphism class. So that is what we have to prove. That will complete the proof of this theorem. So our, so let us go back to the today's slide. Our method here is divided into two parts, two methods and two different proofs. One using cellular homology, another using fundamental proof. Since we already know the homology of S2O as well as its fundamental proof, fundamental proof of S2O is trivial, S2O is simply connected. We have to consider only the list 2 and list 3 for this purpose. Both of them will use a CW structure, a very special one namely the one which is induced by the canonical polygon itself. This CW structure has just one vertex which is the image of all vertices under the quotient map P to X. So canonical polygon, when you identify all the vertices, there will be only, I mean when you carry out all the identifications of the edges, all the vertices get identified. That is what we have seen. So the zero cell is just single vertex. The one cells are what they are the images of the boundary edges of P. Each edge becomes a circle there below. That means the end points are going to the same thing. Therefore, the attacking maps are both the end points going to the same point and then the image will be a circle inside X. So how many of them are there? They are K number where K is 2G in the case of number 2 and equal to n in the number 3 respectively. All of them attach to the single same single vertex. Therefore, the one skeleton X1 of X, I am describing the CW structure here. X1 is the one skeleton is nothing but wedge of K circles, a bouquet of K circles you can call. Now since P is homomorphous, this is a convex subset of convex polygon in D2, convex polygon in R2. So P is homomorphic to D2, this much we know. So the boundary is homomorphic to circle. So we get just one two cell, namely P of D2, P of P and mainly phi of P which is attached. What is the attaching map? The attaching map is the phi restricted to the boundary of P. The boundary of P can be identified with S1 and the whole thing goes to X1 because X1 is actually the quotient of this boundary of P. So restrict the quotient map to boundary of P, get the attaching map for the one cell. There is only one two cell. So this describes the CW structure and we usually use this CW structure for both the methods. So the first method let us compute the singular homology. So I repeat in this proof, we compute the homology of growth of X and show that the listed things in the, in the, in theorem C.4, they have all different homology groups. Case one, anyway X is S2, you do not have to worry about that. So H1 of K is 0 in this case and K or X, sometimes they do not take K or sometimes it is X. So we had earlier the simple shell complex K was there. So H2 of K is that for 2 and 3 we use the elaborate CW structure given by the connex, the canonical polygon just now which we have described. So let us come to now cellular homology. So C0 of CW of X will be just generated by 1, 1, 0 cell that is Z. C1 of CW of X is generated by K circles, 2 K of circles. So that will be Z power K. C2 of CW of X is just 1, 2 cell is there. So that is, that generates Z, okay. So this CW chain complex of X looks like 0 to Z, 2 Z to the K to Z, Z to 0. So it remains to determine what is this boundary operator dava2 and dava1, okay. So let us go through this one. Since 1 cell, each 1 cell is attached to the same single vertex, the boundary operator dava1 will be V0 minus V1 or V1 minus V0, sorry. But what V1 and V0 going to same U0, so it is U0 minus U0, it will become 0, okay. So boundary is 0 on each generator, therefore boundary map itself is identically 0, okay. For the second dava2 what happens? This big homomorphism from H1 of the boundary of P to H1 of the bouquet of circles X1. Bouquet of circles just now we know, we have put that, this group has been put here, that is nothing but that one, Z to the K generated by the each circle. For each circle there is an H1 of S1 is Z, so we put a generator there, okay. So what is this map? We have to describe this homomorphism. This boundary map traces each circle, okay, in a particular way. That depends upon whether we are in case 2 or case 3, okay, depends upon actual polygons. In case 2, number of, number of circles is 2g and dava2 is 0 since phi of boundary of P traces each circle in X1 once in one direction and another time in the opposite direction, okay. a1, b1, a1 inverse, b1 inverse. So a1 goes to b1 then a1 inverse, then it is turning around a1 inverse, a1 in the opposite direction and so on. So on each generator it is 0. Therefore what happened? The entire thing, some total is also 0. So on one single generator, one single circle has traced a, a, a minus a, b minus b and so on. So it is 0. dava2 is 0, there is only one generator, okay. So if this is 0, the kernel is the whole of Z, image is 0 here. So H2 will be right. Here this is 0, the kernel is the whole of which image is 0. So this is anyway H0 that we know already. But here what happens? The kernel is the whole thing, the boundary is 0. So it is Z to the k, okay. And case 2g, so H1 of X is Z raised to 2g. So this takes care of case 2. Let us see what happens case 1. In case 1, other things are same, only dava2, we have to see what is dava2. dava2 differs. The fear of boundary of p this time traces each circle in the same way, whichever way it is, the same way twice. So you can think of that as a generator. So it will be twice the generator, okay. Each of them stays twice. Therefore the entire element is actually taking 2x1, 2x2, 2x3, 2xk and so on. So twice summation xi is in the image, okay. In any case, Z to the n this time is a free abelian groups. Twice x1, x2, xn, twice x1 plus x2, xn is non-zero, okay. This means the dava2 is an injective mapping here. The kernel is 0. The kernel is 0, the homology will be also 0 because homology is a quotient of that. So h2 is 0 here. On the other hand, h1 is what? Now h1, this is the whole of zk, okay. But image, image consists of sum of all the generators multiplied by 2, okay. So you have to go modulo that. So it follows that the h1 is z raised to n minus 1. The rest of them you can choose properly. You have to change the basis here to write this isomorphism. Z raised to n minus 1 direct sum z by 2 z, okay. You choose x1, x2, xn minus 1 comma the sum of xi's as a generating set. Then the double the generating, the last generating element, double of that is in the image of h1. So it goes modulo that one. So that will give you z1, zn minus 1 direct sum z by 2 z. So this completes the calculation. Now just take a look. In the first case, h1 was identically 0. In none of these, h1 is not 0, okay. The least thing is h1 is z by 2 z when n is 1, okay. In the other case, it is always z power 2, z power 4 and so on. So just by looking at h1, you will see that they are all different, okay. Nonetheless, we can have also keep in mind that the third series, h2 is 0. In the second series, h2 is infinite cyclic, okay. So h2 does not distinguish between the s2. S2 also has this property, okay. And the second series. But h1 will distinguish. Whereas h1 of the sphere is just 0 and h1 for all these things, 2 z, 4 z raised to 2, z raised to 4 and so on. Therefore, since the homology of any topological space of single homology of topological space is a topological invariant. It is a homotopy invariant. So these elements, these members of this list must be all of different homeomorphic type, okay. So this completes one proof. So let us see, just for fun, the another proof also, which is even simpler than this, okay, and gives you less information apparently. But that statement is much more stronger for some other reason. So we can now introduce the notation before going to the second proof. Tg, Pn, etc. What is Tg? Tg corresponds to in the second series. The canonical polygon, the surface given by canonical polygon, A1, B1, A1 inverse, A2 inverse, etc. Ag, Bg, Ag inverse, Bg inverse. So that is Tg. And Pn corresponds to the third series, okay. What are those? Of course, h2 we do not want to have another notation. Pn are xx, x1, x1, x2, x2, etc. x and xn. So given by that, okay. So let us have these two notations. Then look at what is T1. T1 is just A, B, A inverse, B inverse, right. Therefore, it is a torus. We know this one already. Okay, the ordinary torus. And T2 is, two of them got together. We have introduced an ocean of connectors from here this way. So T2 is the double torus, which is nothing but connected sum of T1 with itself. More generally, Tg is a connected sum of g copies of T1. Likewise, P1 is xx. So that is, we know is P2, the projective space. P upper 2, this V1 is lower 1 here, the projective space. Okay. And Pn, just like in the earlier case, is nothing but connected sum of n copies of projective space with itself. Okay. Just as before, P2 is nothing but which we have seen in one of the examples by studying A, B, A, B inverse, which is a Klein bottle. Okay. So connected sum of two copies of P2 is the Klein bottle. Okay. So this is just for getting familiar with these surfaces. Incidentally, what we have proved is chi of the first series Tg are 2 minus 2g. Chi of the second series is 2 minus n. So 1 minus n minus 1 is 2 minus n. Okay. So go back here and see. The second case, h2 is 0, h1 rank of h1 is z over n minus 1. Then h0 is 1. Okay. 1 minus n minus 1 is 2 minus n. Similarly, in the first case, h2 is z, h1 is z and h, sorry, h0 is z and h1 is 2g. So 1 plus 1 here, minus x, middle 1 is minus 2g. Okay. So that is 2 minus 2g. So Euler characteristics of this, also we have computed, namely 2 minus 2g and chi is 2 minus n. Just look at the Euler characteristics. In the first series, they are all distinct. Look at the second one. They are all distinct. Unfortunately, the first series and second series may have same Euler characteristics. Put n equal to 2g. So there are elements here with same Euler characteristics. So they cannot be distinguished by just Euler characteristics. We have to directly look into the homology groups. All right. Now let us go to the second proof using the fundamental proof. Clearly, the fundamental group of the surface can also be easily determined by C-leveled structure described above. Okay. Given by the canonical polygon. Since the one skeleton is a bouquet of k circles, its fundamental group is a free group with basis consisting of one generator for each circle. This we have computed several times. In the case of 2, let these generators be denoted by xi and yi. Okay. i equal to 1, 2, 3 up to g. A1, I could have written just A1, A1, B1 itself. So I have to take xi, yi as for the just distinguish between the edge itself and the generators in the group. Okay. These are now elements of the fundamental group. i raise to 1 to g. Then the attaching map of the two cells represents the element which goes over first x1, y1, x1 inverse, y1 inverse. So that is the commutator of x1, y1. So I write that x1, y1. Then x2, y2, x3, y3 and so on. Take the product. So that product is the element t. Okay. Along which the two cell is attached. Therefore, so let us see what happens in the case 3. In the case 3, this is nothing but x1, x1, right, x2, x2, x3, x3 is the product. So that is nothing but x1 square, x2 square, so this t is x1 square product of xn square. Okay. So you have to take it in a one particular order whichever way they come. You cannot mix it because this is fundamental group. You cannot mix up like in the case of homology. Homology is commutative. Fundamental group is not there. In either case, it follows that pi1 of x itself is the quotient of pi1 of the first skeleton. Modulo, the modulo means what? Pi the normal subgroup generated by this element t, one single element t. Okay. Of course, in case 1, we know that pi1 of s2 is 1. That we have already done. So we have list now. What is the list? This is the list. The fundamental group of surfaces as listed in 6.4 are given by generators and relations as follows. This is the standard way of writing generators for relation. The first one is pi1 of s2 is 1. One means history well group. Pi1 of tg is generated by x1, y1, xgyg. Then there is a slant here, you know, a divider. Product of xi, yi is written. So this will tell you what are the relations. Whatever is written on the other side of this vertical bar, they are supposed to be the relations. We have to take the normal subgroup generated by all of them. Go modulo, get normal subgroup, take the quotient. That is the meaning of this bracket x1, x2, x3 slash product of i rint 1 to g xi, yn. So these are the fundamental groups of our tg. Okay. The third case is connected sums of projective spaces. Okay. Pi1 of pn is generated by n elements x1, x2, xn and one relation x1 square dot, dot, dot, xn square. Okay. So this is the theorem just we deduce from our knowledge of the CW complex structure given by the canonical polygon. I want to say that this is enough now to distinguish these members. The first thing is fundamental group here is trivial. In any of these it will not be trivial. Why? Because what you have, what you are going to do here? You are just going by this normal subgroup generated by one element, right? And you are going down by, this is free group over this element. How to see that this is not a, this is not trivial? Just take the habilinization of this group. Even that is not trivial. A habilinization of this one, when you take, declare them as commutative, this element will be automatically trivial element, right? When x1 commutes with x2, x1 commutes with y1, what happens? The commutator is trivial. So this is trivial. There is no normal subgroup going down. So habilinization of this one is nothing but the free habilion group over x1 x2. Of course, this we knew in a different way because pi1 when you habilinize this h1 and h1 we have already computed. So this conformal with our whatever computation we have done. So this is never 0. It is the free habilion group over 2G generators, okay? The habilionization. Here what happens habilionization? Same thing, x1, x2, xn is there. Then you have, instead of writing product, you can write additively when you habilionize this. That is like 2, twice summation xi that is being put to 0, right? Even when there is xn equal to 1, what happens? x1 divided by 2x1 is nothing but z by 2z. Otherwise it is z power n minus 1, n minus 1 element and then directs some way z by 2z. So this is never 0. Not only that, here there is a 2-dorsion. Here there is no 2-dorsion, okay? So no element here, no group here is isomorph to group here. Within the group, the rank will distinguish them. Here also, the rank will be always n minus 1. If n and m are different, they are different. So that completes two of the proofs, okay? We can give some other proofs also but they become unnecessarily complicated and cannot be fully justified using whatever techniques we have introduced. So we shall stop here for that one. So now I have few comments to make. Note that this second approach does not give explicit information on H2. We have, by habilionizing, we get H1. H0 is of course, we do not need because we are all connected. Therefore, H0 is that. So what is H2? You do not know. Yet this statement is much more powerful than just the homology statement, okay? So just you know the fundamental group, the surface is known up to homeomorphism. If indeed, it happens just up to even the homomorphism also, okay? That we have not put, okay? So what I want to say is, however, the result itself is a small surprise. Just pi1 determine the whole thing. This may be attributed to a deeper property of surfaces, namely barring two of the cases, namely S2 and P2, okay? In which case, the fundamental groups are finite. All other fundamental groups are infinite and all connected closed surfaces have the universal covering space, either the complex plane or the upper half plane. Both of which are contractible. So they are covered by contractible spaces is something very special. However, whatever I have stated just now is, okay? Not at all an obvious result. For the torus, you can actually give a easy proof that its universal covering space is r cross r, which is C, okay? For all those double torus, triple torus and so on, you don't know how to construct the universal covering space. I mean, there are no easy methods. Of course, we know it now. It is all classical. But what you have to do? You have to use the functions of one variable, complex function theory of one variable, okay? You have to use S2Z. You have to use Mobius transformations. You have to study the property S2Z acting on the upper half plane, okay? This is a very, very entertaining and rich source of, you know, complex function theory. There are a lot of things here. How much of literature is there, okay? One can ask whether such a result, such a case is true in higher dimension management. Namely, suppose we have an n manifold x covered by a contractible space. More strictly, let us say covered by Euclidean space like C and half C and so on, right? Upper half plane, C. Suppose it is covered by the Euclidean space itself. Is the homeomorphism type of x determined by the fundamental group? This is a question. So this problem goes under the name Borel's conjecture. At least perhaps 50-year-old conjecture perhaps or more, which has been verified in every known case. A complete solution is yet to come, okay? You can see Farrell's exposition 2002, which is there in ICPP notes for details of this one. The Farrell has worked on this kind of problems. It is not hard to obtain classification of all compact connected two manifolds with boundary. Recall, so far we have taken only boundary less case, manifolds without boundary. How do you do? Because it is compact, there may be finitely many boundary components. What are the boundaries? Each boundary component is a connected one-dimensional manifold which does not have any boundary. Therefore, also it is compact. Therefore, it must be the circle. So in the circle, you can just put a disc and what is called it is called capping off. Fill a disc there, so that boundary component disappears. So do all this for all boundary components. What you have got is a compact manifold, compact surface without boundary. And then you know what it is. Therefore, in principle, any compact manifold with boundary is got by starting with a corresponding manifold without boundary and then making a number of holes in it by removing small, small discs, disjoint discs, okay. So this is the picture for all compact manifolds now. The passage from compact to non-compactness, that is a different game altogether. Even in the dimension two, dimension one, we did not have so much of trouble. We did in a classification, if you recall. Dimension two, it is still not well understood at all. People trying to claim half results here, half results there, counter claims and so on. For example, just a topological characterization of a surface that you know an open surface, a non-compact one so that it is an open subset of R2. There is no such characterization yet known. An open subset of R2 is a surface. We would not know what, how many are there and what kind of things are there. There is no classification known. That is what I wanted to tell you. So it is not very, you know some more and more conditions by putting them, some people have tried, okay. So you can also try your hand. First look into the literature, whatever is known and try to go ahead, okay. So that comes, that brings us to the end of this theorem and classification. So next time we will introduce another interesting concept and that will be the end of the series. Thank you.