 Before taking up the last topic, namely classification of triangulated compact surfaces, let me make a few general remarks on triangulation in particular. A triangulation of a manifold has necessarily more combinatorial structure, more combinatorial information on the manifold. We have made a small beginning of the study of these properties. Before, besides theoretical importance, it provides a very good effective tool in the study of topological properties of manifolds. For example, I have told you, now you can actually, you know, computerize the study of topology through simplicial complexes. The fundamental classical questions here are, can every topological manifold be triangulated? If you cannot do that, you put, okay, differential manifolds can be triangulated. Like this, these are some questions there. Given next question, given two triangulations K1 and K2 of a topological manifold is K1 combinatorially equivalent to K2, by which we mean that there are subdivisions K1 prime of K1 and K2 prime of K2 such that the two subdivisions are isomorphic. Do every triangulated manifold carry a smooth structure? So these are the few standard questions. Let us see how far in literature we know about these questions. From the classification of one-dimensional manifolds, it easily follows that every one-dimensional manifold is triangulable because the connected components are just open interval, closed interval, half closed interval or the circle. These you can triangulate so everything can be triangulated. It can also be proved that each one-dimensional manifold, topological manifold has a unique smooth structure. Okay, if you go through the proof of the classification at several stages, at some of the stages, not several, you will have to improve homomorphism to defiomorphism. That requires a little more difficulty, more work. That is all. A classical result due to Rado, this way back in 1924, almost now, you know, more than a century, says that all manifolds, okay, two-dimensional manifolds are triangulable, though this proof is within our limitations due to lack of time, we shall skip it. Okay, the triangulativity of all three-dimensional manifolds is a deeper result due to moist who also proved that any two triangulations given K2 of the same three-manifold are combinatorially equivalent. Okay, that means what I have just told you that there are subdivisions of KIs, such that K1 prime and K2 prime are SMO. By the way, this reference is for his book. His actual papers, the proof, etc., have appeared much before. A theorem due to Terns in 1935 says that every smooth manifold is triangulable. There is an improved version of this one in Whitehead 1940, okay, which gives a neater proof and a stronger result. So every smooth manifold is triangulable, okay. In dimension greater than or equal to four, okay, one, two, three, we know. One is easy, two is rado, three is moist. But bigger than or equal to four, there are triangulable manifolds which do not admit any smooth structure. Every smooth manifold is triangulable is Kainas result. But the other way around, there are manifolds which cannot admit some structure. The first example came in 1960, that was due to Karware and that was in dimension 10, which was soon improved to dimension 8 by Ilsan Kaipar. Then Sibirman has constructed an example of a five-dimension manifold which cannot be triangulated. With these classical results, we safely say that the above three questions have been answered satisfactorily. Why I am not talking about dimension 4, dimension 4 is much more crazy. Craziest dimension, I mean just all of them, okay. So I will not speak about it here. These results are quite hard and beyond the scope of this element. So let us make a small beginning today of compact surfaces. It won't be finished in an hour. Just lay down the plan of action. First of all, I will use the word surface just to mean compact two-dimensional connected topological manifold without boundary. So I will not keep on saying all these, I will just say surface. Okay, this is just for a temporary notation, temporary terminology, just for another two, three lectures like this, that's all. Okay, next we fix a triangulation on a connected compact surface just like that on a surface. Okay, we then appeal to our theorem wherein we constructed a convex triangulated polygon and a surjective morphism from, morphism from a surjective quotient map actually from the polygon to X, right. So that is a representation of X as a quotient of a convex polygon P with two n sides. And what is the identification? Identifications are coming only from the boundary sides. The boundaries, you know, edges are paired out and then the identification is taken by homeomorphism from one of them to another one in the pair by a linear homeomorphism. Okay, and we identify that. And the quotient space is actually our K with a triangulation, a proper triangulation. Okay, so this process can be completely described now purely combinatorially as if we don't have any topology there at all. What do we do? A surface now will be described by this quotient map. Okay. So what is the top thing? It's a convex polygon, triangulated convex polygon. And where is the identification? You take pairs of, you know, it is a first of all, it has two even dimensional polygons, pairs. And then I don't know linear isomorphism, linear isomorphism is just edge going to edge either one way or the other way there are only two isomorphisms there. End point going to end point linear isomorphism of one interval closer interval to another closer interval. Okay, there are only two possibilities. So let us agree once and for all that we shall trace the boundary of any cons polygon in the anticlockwise direction. Okay, like we trace a circle in two different ways anticlockwise and clockwise. It's fixed for once for all anticlockwise direction. We are free to start from any vertex that we don't fix. Okay, you can start from any vertex, keep going. What are you going to do? We then label the edges by letters A, B, C and so on. Edges are labeled now, A, B, C. Okay, as soon as we meet an edge which is being identified with the edge that we have already labeled, we shall not use another letter, we shall use the same letter. Suppose I have started A, B and the next one is identified with A. Then I won't call it C. I will call it A. But now I have another stronger condition. In the orientation that we are taking, namely following the anticlockwise, whether this edge, the third edge is identified with the given edge, the same orientation or not, accordingly I will call it A or A inverse. A inverse, if the orientation is reversed now, the arrow is on the other side. We are going from anticlockwise, but the arrow on that is clockwise. Okay, the arrow will be depending upon now how we have started with A positive. There is nothing starting with A. So you have put an arrow in the anticlockwise direction. B, there is a new one anticlockwise. The next one is repeated, but the edge name as well as the orientation on that will depend upon how it is identified and with what it is identified. Okay, is that clear how we are going to label them? Alright, as soon as we meet an edge which is being identified with an edge which has already been labelled, we shall use the same letter to label this new edge also. So that's first thing. The second thing is we have to take care of another aspect, namely whether the identification is orientation preserving or orientation reversing. To indicate an edge with a reversal orientation, we shall use labelings A inverse, B inverse, etc. Of course, we stop as soon as we have arrived back where we started. Since the starting point is arbitrary, what is this process yields is that the surface X is completely determined by the cyclic sequence. Suppose I start with a point, one point, you start with another point. So whatever from my point to your point in the beginning, for you that part will occur at the end. Instead of A, B, C, D, you may get B, C, D, A. So up to cyclic permutation, okay, the convex polygon and its quotient is completely determined. By this sequence. Just you have to write many sequence. As soon as you give a sequence, the length of sequence must be even. That's all. Immediately it defines a surface, okay. A compact, connected, oriented or I do not know, two-dimensional manifold without boundary. So it is described by a sequence of length to A. The only condition is you can't have just A, B, C, D, E, F, G, H. That kind of sequence is not allowed. Each letter should occur twice, exactly twice. The sign of the letter, whether A or A inverse, that is left to you, okay. There is no point in writing both A, A inverse, A inverse. Then you can write them as both A. It's the same thing, right. So one is A, another is A inverse. Or both of them have A. So this is the nature of the sequence. As soon as I have such sequence, it defines a surface, okay. So cyclic sequence, it looks like A1, epsilon1, A2, epsilon, A and epsilon9. Where epsilon1s are nothing but plus minus 1. But I will write A raised to 1, A. I will just write A1. When it is inverse, I will write A1 inverse, okay. So note that in this sequence, each letter A i occurs precisely twice. In particular, this N is E1, okay. For the simplicity of notation, an edge A plus occurring with a plus sign, we drop this sign and simply write it as A. A sequence such as 41, if I keep calling this 41, sequence 41, okay, is called canonical polygon. Just to contrast it with an arbitrary polygon, okay. A canonical polygon means polygon sides to N, the interior is triangulated, the whole thing is triangulated, all this, okay. We shall use bold phase capitals A, B, etc. to denote a part of the sequence, okay, such as 41. Suppose we have one sequence here, a part of it, I want to write A, B, C, D. I will write capital A followed by some F, G, F inverse, G inverse, same like this. This part A, some sequence of them connect, okay. So for that kind of reason, I will club them together. This A itself may consist of 5 letter, B consist of 10 letters and so on, doesn't matter, all right. For N theta equal to 4, okay, we can represent a canonical polygon, the sequence, by actually a regular convex polygon, okay. 3 also we can do, but 3 doesn't occur in our case. Minimum is 4. If there is a 2 or 0, where is a polygon you can't draw? Okay, so I will discuss that case first, later on. So for N greater than or equal to 4, actual converse polygon you can write, okay. Regular converse polygon P in R2 with N sides, with its sides appropriately labelled. Observe that we allow the exceptional case N equal to 2 walls, so we have to allow. In this case, we do not get a converse polygon. So what do we do? We get a convex set, namely now we go round shape, namely we will take a disk, okay. On that disk, you can divide the boundary into 2 semi arcs, semi circular arcs, okay. And then label them as AA or AA inverse over, okay. So that is the case 2, N equal to 2, okay. Of course N equal to 0 doesn't occur because it has to be non-empty convex polygon. However, in case N equal to 2, we take P with a unit disk, which is boundary being divided into 2 edges. The sequence itself is being AA inverse or AA, okay. So there are only 2 cases when N equal to 2. You consider this map, okay, from the unit disk in the x, y, x, z plane, x, 0, z I have written, okay, on to the unit sphere in R3. This x coordinate is this one, y coordinate is this one, z coordinate is as it is. Okay, this map proves that the surface represented by the sequence AA inverse. 2 edges being labelled from bottom to top, okay, like this, AA inverse. What is the space, what is the space, quotient space when I add identification is like this, these 2 edges are being identified like that from this. So that is the picture, that is the, that is the map I am hearing you. So how does it do? This is your disk, okay, unit disk in the x, z plane. Look at any of these curves as you move parallel to the x axis up, okay, from here it is minus 1 to plus 1. Here it will be 1 by square, what is it? Square root of 1 minus z square, right, minus square root of 1 minus z square plus square root of 1 minus z square. Then finally it will be 0. So as we are moving this, this edge will be mapped to the corresponding circle here. The circle obtained by cutting the surface, cutting this 2 sphere by the plane xy, z equal to z. So z whatever, z equal to z naught as that keeps moving, okay, the xy plane. So they go like that. So what we get is this quotient varying edge from here to here is identified edge from here to here, okay. Just the x coordinate being, go to x, go to minus x, it is a linear map, alright. So that is one case. The other case, what is the other case? The other case is when this edge is identified with that edge in the same direction. So that is same thing as antipodal action. Every point here will be identified with that point. When you antipodal action on the disk only on the boundary, we know that this is a projective space dimension 2, okay. This also we all know. Recall that a projective space P2 is the quotient of S2 by the antipodal action. But you do not need a whole of S2, you can take only the upper half hemisphere. Then the identification is only on the equator. So that can be identified precisely as if a flat disk and the boundary, okay. And the sequence will be now A, A. So A, A represents projective space. A, A inverse represents sphere, okay. So we have started classification. In the simplest case, we have finished the classification, okay. Now let us concentrate on n greater than or equal to 4 only, okay. So I will have this kind of some kind of operation here now and notation. We can add two canonical polygons. So one polygon is A, A is a sequence. Remember that of length 2 and so on. Another is B, B is another sequence of length 2. By merely concatenating the corresponding two sequences, A, B. So it will be of length 2 and plus 2 m, okay. Of course, this operation is commutative in a restricted sense. Namely, because of the, because of the what? Cyclic order. See a sequence is taken up to cyclic order. Therefore A, B is the same thing as B, A, okay. Only thing is once you write A, B, you cannot cut it in some other place and change however. That can be a problem. Every time you change, you have to do only cyclic order, okay. Suppose A, I write it as A1, A2. Then A1, A2, B is not equal to A2, A1, B. That kind of commutative is not there. A1, A2, B will be equal to B, A1, A2 equal to what? If you keep circulating like this, okay. So first B, A1, A2 and then A1, sorry, A2, B, A1 and so on. So that kind of cyclic permutation is allowed of heat letters, okay. So this kind of adding is actually in this case comes from a geometric operation on manifold connected called connected sum. So here we shall take this as definition of connected sum. Connected sum of the surface A with surface B, okay. In a simple geometric term, it means that you make a hole by removing a small disk in one of them and another small disk from the other one, okay. Then the two resulting circles, you identify them to get a different surface. So that is called and larger surface perhaps, it is called connected sum, okay. So we do not need to go deeper into that. Just take this as definition of connected sum, I will give you an example, okay. So here is, here is my A, I do not know how many, I have drawn three edges here, maybe there are more. Here also three edges, maybe there are more. So put all the A here, then follow by all the B here, over. So that will be larger convex polygon, okay. 2m gone plus 2m gone equals 2m plus n gone, okay. Several canonical polygons may define the same surface up to homomorphism. What we have so far said is from canonical polygons, the set of canonical polygons to the triangulated manifold, two-dimensional manifold, there is rejection. Is it injective? No. In what sense? Whatever sense. That is the meaning of this. Namely, several of them may define the same surface and that is why there is a problem, I mean we have to do. Otherwise the problem was over, you could have gone home, okay. Our next step is to make a list which should include all possible topological types as well as have no redundancy. Our list should not have A and A prime are the same surfaces and so on. The list must be distinct. If A is A1, A2, A3 or 1, 2, 3, 4, 3 whatever represents 4, they must be topologically different. That is the meaning of no redundancy, okay. And it should be exhaustive. Somebody later on make, oh, this surface is left out from here. That should not happen. Once you have achieved that, the classification is over. Okay. So, first we propose a list and then we say, okay, this is the list. That means we have to prove it. Okay. Right. But let us now illustrate the point that two different canonical polygons, different means what now? Cyclically different. A, B, C, D, sorry, A, B, C, D, A, B, C, D, let us take this sequence. B, A, C, A, B, A, D, B, C, D, A, B, C, D, A, these two are the same because cyclically they are the same. That is not the point. Even cyclically different ones may define the same surface. Let us examine that, okay. So, consider the surface given by the sequence A, B, A inverse, B inverse. Just to recall, this is a familiar one to you. Namely, what you have, you have a rectangle. The side A and its opposite side are identified with the orientation coming in the same direction. A, B and B inverse are identified with the opposite orientation. A, so A, B inverse, okay. See, when you follow A, B, when you keep following that, the A, B, A, B, A, A is coming, not A inverse. So, it is A. So, when you have do that, if you remember, A, B, A inverse, B inverse was the torus. A, B, A, B inverse was the Klein bottle, okay. So, we have seen that this represents Klein bottle, A, B. Mark the diagonal of the square piece. You may do it on your own right now. Take a rectangle, you please mark A, B, C, D, sorry, A, B inverse, B inverse. And take the diagonal piece, okay, draw a diagonal with a thick arrow and label it as C. Okay. And cut the square along this arrow as shown in figure 32 here. A, B, A, B. See, B is in the opposite direction. A, B, A, this A, B, A inverse. A, B, sorry, A, B inverse, right. So, now I am drawing this diagonal from beginning of A to the beginning of other A. So, that is the key, okay. Or you could have done it from end of A to the end of A. Okay. That is what you have to remember. So, this will be going to be an extra edge. Now, we cut it, cut here. The upper triangle, you bring it down and twist it. I mean, flip it, flip it so that this arrow A gets in the same line and aligned with this arrow at the bottom. So, A and A are identified, right. You have to identify them. So, this upper part of the triangle is identified here after bringing it bottom. Now, the figure doesn't look like a rectangle, but it has four sides. So, you can easily deform, you will, you know, represent it by rectangle like this one, okay. So, this is C, that is B and then again B and then C. So, it is T, B, C, C, B, B, C or B, B, C, C. No problem, right. Remember, now if I just stop here, cut it here, it is B, B. What is B, B? B, B is a projective space, P2. Similarly, C, C is a projective space, P2. So, you can think of this as the sum of two projective spaces, the connected sum of two projective spaces. The point is that it is the same client model. It is the same surface is seen. Why? Because wherever I cut, I am identifying C and C. See, that is precisely, I have to draw this one on both sides, the same arrow so that you should not bungle the arrow. The same arrow I have to take, okay. This arrow has become like this here, this has become straight on here, okay. So, C followed by C, okay. Therefore, what we have just now proved by this method is that the client model is Omemore Swift 2, the connected sum of P2 itself, okay. So, this is what is going to be cut and paste technique. This should be heavily used in the classification here. So, these two are different, obviously, you see A, B, A, B inverse. There is no inverse at all here. B, B, C, C, these two sequence are different, but they define the same, same what? Same surface. This one we are familiar, we have called it as client model. This does not look like familiar, but now we have just introduced it is P2 connected sum P2. But the surface is the same, therefore, so these two objects are Omemorphic, alright. Let us stop here today. So, tomorrow we will actually start the classification proper. Thank you.