 So, to continue with the classification of surfaces, here is the statement of the theorem. It has two parts, part A and part B. Every compact connected surface without boundary is homeomorphic to the surface defined by one of the following canonical polygons. The first one is A A inverse. Second one is A 1 B 1 A 1 inverse B 1 inverse A 2 B 2 A 2 inverse B 2 inverse and so on A g B g A g inverse B g inverse and g is greater than or equal to 1. The third sequence is A 1 A 1 A 2 A 2 A 3 A 3 and so on A n n is greater than or equal to 1. So, it is a part A. Every surface is one of these. To be sure what we are putting you say, every compact connected triangulated surface without boundary is one of these. Okay, triangulated word has to be introduced because that is what we are assuming but we are you know sure of that every surface triangulated by Rado theorem which we have not proved of course. Second part says any two distinct members of this list gives you the surfaces which are non-homomorphic. When I say distinct members, A A is a one single distinct member. In two there are infinitely many members. Okay, indexed by g greater than or equal to 1. Similarly in three there are infinitely many members. Okay, each one of them is different from each one of the other from within the, within this list in this one or outside. Anything from two will be directly different from anything from three but even within two if g1 and g2 are different the corresponding surface are different and so on. That is the meaning of any two members are different representing different surfaces. Okay, so this is the complete classification for surfaces without boundary. The canonical polygon we said above are said to be normal form. Okay, when you normalize the unique response that is the whole idea. The rest of this section will be occupied in proving part A of the theorem which should be achieved in five steps and then giving three different proofs of the uniqueness part of the unique surface. Any two members are different means it is a unique. Each sequence here gives you a unique surface. Beginning with an arbitrary canonical polygon it may not be in the normal form. What we have to do is we have to reduce it to a normal form and show that it is one of them. Okay, clearly these are canonical polygons. There is nothing to do. They are normal form. So every canonical polygon can be brought down whatever we want to say brought down means what to one of this. The process of transformation as we have illustrated last time with A B A inverse B inverse A B A B inverse that kind of transformations many more should not change the what homomorphism type of the surface. The homomorphism type we should remain within that. Then we are allowed to do whatever we like. Okay, so that is the whole idea. Okay, so we would like to transform to one of the normal form without changing the homomorphism type of surface defining it. First observe that in 2 and 3 all the vertices are identified to a single point. We should know that one. A A inverse has only two points. Okay, A A inverse. Okay, in the identification also there are two points. Therefore I am giving out one. But in 2 A1 A1 A1 B1 A1 inverse B1 inverse after identification of these four vertices only take only four vertices to begin with. How many vertices will be left in the quotient space? When you say A okay A B are identified no identify A inverse when you identify the top two gets identified is the bottom two. So there are only two surfaces, two vertices. When you do B and B inverse or both of them will identify get identified again to a single vertex. So similarly the same vertex will come for A2 B2 A3 B3 inverse also because starting point of B1 inverse will be the starting point of A2. So once you approve it for A1 B1 A1 inverse B1 G equal to 1 this will get proved for this. Similarly here also A1 A1. Okay, this you know already as you do one vertex. This is the anti polar points of the vertices they get identified anyway. Similarly all the A and A's will get identified the same. So in the canonical polygon with more than four vertices okay either we land up in A A A inverse or what or first thing you to achieve is there is only one vertex left. The class of vertices identified vertices in the original thing there may be several of course. Okay, there will be exactly two end of them. In the final thing there must be only one. So this is the first thing we want to achieve. Our first aim is would be to find a reduction process which will ensure that all vertices are identified at a single point. One of the simplest thing to do is to cancel out a pair of edges A A inverse. Suppose somewhere in between one edge comes A and immediately next edge is identified in the opposite direction with this one. So that is like you have two edges like this okay they are identified you fold them like this. So both edges go in the interior now and rest of the rest of the things will have two edges less. So there is no need to have a sequence part of sequence A A inverse or somewhere C C inverse okay. So you can cut down both A and A inverse okay. So this is the first thing is the simplest thing to do all right. If both A and A inverse occur in this sequence for some edge A we call the pair let us have a name this A A inverse type as type one pair. Remember each A is followed by I somewhere not necessarily in the in the by the way this definition is somewhere not necessarily immediate okay A dot dot dot A inverse comes that will be a type one pair if A dot dot dot A comes then that will be called type two let us have this name okay while representing canonical polygon by a picture instead of the x1 plus minus 1 over a letter we shall use arrows because arrows is easier to indicate that is all instead of plus minus 1. When you draw a picture it is letters and arrows okay. Now the first step is elimination of edges of type one which are consecutive okay. We will now show that if D has at least four edges to begin with if it is A A inverse what do you do A A inverse it is the first one you do not do not do anything suppose there are only A A inverse this is then you have already come into the canonical list the normal form so do not do anything but suppose there are more vertices more edges there are more edges than just two and one of one of the pair is like this and I would like to get rid of this okay now if you cancel it out it will be empty so then you do not know what to do with that that sequence empty sequence there is no canceling this itself gives you whatever you wanted so this is normal form for us so let us go ahead okay. So suppose there are more so then we will now show that if D has at least four edges then an adjacent pair of type one can be eliminated okay until we end up with case one or there are no adjacent pairs of type one okay so 33 illustrate this process so I will tell you okay so I had earlier given an exercise which I will now explain I think I should explain it so why this is true why this can be done namely you take a half disk okay take a half disk take the upper half part of the diameter and the lower half part of diameter put opposite you know put opposite arrows on that accordingly you identify okay so the picture that you get identification identification after identification what is it it is the disk right instead of so dot dot dot rest of the curve if it is a some convex polygon what happens is these two edges get eliminated okay how do you get that is you can just take theta e power 2 pi i theta as you are writing theta going to ish theta that is all e power i theta 2 pi i theta 1 2 e power 4 pi i theta so double the angle that will take care of this one okay so the relation may be coded as now if you want to do computer programming all that you have to do suppose a sequence has this form up to cyclic order I can write it as a a inverse all the sequence which is after a inverse is brought to the left so anything can be written as a a inverse cancel out a inverse and put a provided a is not empty otherwise could cancel stop there so that is the programming okay so that is what I mean by the whole entire thing can be programmed okay very elementary programming so this what is happening here dot dot dot of the disk the diameter a and a inverse are there right in the sequence a this will be a inverse the other way around so I have taken a inverse this way in the anticlockwise and these are all dot dot dot I do not know what several of them so when you identify by taking this one that goes square here what you get is the edge a has gone inside by identifying with itself right so this becomes a sequence a now so that is the canonical polygon okay so we can we can after they are getting this one you can actually make it into a convex polygon that part is no problem so I can always replace a circular disk with a regular converse polygon okay by repeating the application of this we eliminate all adjacent pairs of type one till there are no adjacent pairs of type one or we are left with a a inverse okay in the case of a inverse the process itself is over so let us assume that there are no adjacent types of pair one and the sequence is has more than four elements or whatever or a inverse at least two elements okay so this is one very easily obtained reduction the second reduction is whatever you wanted namely getting only one vertex class okay single x class how to do that so we consider the equivalence classes of stretches under the portion map p to x okay to begin with we have two vertices several of them get identified so we have maybe one or two or three classes if there is only one class we are done if there are two or three and so on look at one of the classes with there are more than two classes then only we can okay okay consider a class p of vertices with the least number of elements in it so suppose one class has six elements another class has eight elements and so on look at the class with the least number of elements we pick up a vertex p in this class so that the next vertex is q okay which is not in this class because there are at least two classes this should happen okay keep keep going around the class if around the the polygon okay keep only in the class the next one is not in the class you pick up one such okay let r be the other vertex adjacent to p p and q are adjacent because that is how we have chosen r is the one which is behind on the other side okay now let a and b denote the edges r to p and p to q so here is the diagram so p was my chosen vertex q is the next one they are in a different class there must be one vertex behind this like this called as r so this r to p I put a p to q I put b okay why because this could not have been a a inverse in any case okay maybe it is a a if it is a a what happens then these two vertices will be in the same class so its q is in the different class therefore it is a and b okay how this have to be gone through oh very meticulously the rest of them you don't know somewhere this a must have occurred because this is not a so call that as now this is this vertex is r and that vertex is p because they are getting identified so they are different vertices in the polygon but I am labeling them because they are getting identified okay and that is a so this also a okay with this kind of in dot dot dot you don't know so many divisions may be there okay somewhere in the polygon there is one such thing like this one all that you do is cut from r to q cut this triangle mark it as c okay you have to put an arrow here whichever way you like so I have put the arrow from q to q to r okay the arrow is here so when I cut it the remaining thing here is there p r r q but this triangle I am bringing it back here this side a matching this side a a and a are matching a and a have to be identified anyway so when I identify this the rest of them is this one so picture is now very twisted twisted kind of polygon doesn't matter it can be straightened out to become a format polygon that that part is no problem this is dot dot dot okay then this c is on this side like this the the direction is very important so this is e dot dot dot so when you take this triangle down here you this is a okay behind that there is a c from q to this part okay and then b is from p to q so orient it correctly what is the sequence you have got c dot dot dot c dot dot dot b that is the sequence a is disappeared that is not what I want what I want is how many letters are there how many vertices are there in the class of p now in the new one okay can you see that there is some reduction in the number of elements in the class of this vertex p see picture it is very clear there are two points here here it is only one there may be many more but those things would have been there in the dot dot dot part the idea is that there were five or six vertices here now there will be five repeat the same process on this class itself okay this class itself you repeat finally what happens this class when there are two of them it will just disappear problem is what has happened is when you do did this one you can't repeat this process immediately one thing has happened namely you may get another pair which is of the type c c inverse which was not there in this one but it is there in this one so step one has to be performed on this one again when you perform step one singleton vertices disappear sometimes that is what happened step one here is vertex it has disappeared okay so you have to do again after doing one step like this carry on the first step again okay so that ensure that there are no a inverses okay when you do this one this will not increase the the class increase the class of vertices okay so here you have reduced the one of the classes number of vertices has reduced so this step does not increase that therefore this step is valid and now again you have assume this one and now carry on this one on the same class which has fewer elements finally one of the class itself has disappeared if there are two classes now there is only one class if there are fifty thousand classes now there will be only fifty thousand minus one classes okay now pick up another one which is the least from there by induction what you do is by combining step one two and alternatively you have to argue step one step two performing them what you get is a canonical polygon which has in which all the vertices are identified single point by this we do not mean that the vertices are identified on their own no never the identification is always taking place from edge to edge two edges are being identified each time okay you do not pick up a polygon and identify two of the edges two of the vertices that is never done okay so after this cut and all that what we have got is the sequence a b something here so I have written what a b capital A A university becomes c inverse a b c b so this is the operation so you can computerize this operation with the corresponding condition you carry out like this if you blindly carrying it out it may not reduce something okay keep doing is again ask it to do if there are if you see a inverse is cut down that and so on okay if needed we carry out step one again so this has to be done note that each time step one is performed the number of edges as well as the number of vertices goes down without any linear number of equivalence class step two on the other end keeps all these numbers the same okay by repeated application of these two steps we keep reducing number of vertices from classes with the least number of vertices each time and hence at some stage one of this class has to disappear the smallest class disappears first indeed if a class of vertices has just one vertex in it this implies the edge the two edges incident at this vertex must be of a type one A A inverse there is no other way therefore we perform step one the the the two edges will disappear along with the vertex so this way we keep reducing number of classes themselves until there is only one class left okay thus from now onwards we shall assume that all vertices are identified a single point it is worth noting that none of the reduction processes operations that we perform okay from now onwards will disturb this property there are some other kinds of operations we are bringing okay that will never disturb this property means what it will not introduce A A inverses it will not introduce extra vertex classes vertex class will be always one from now onwards all right so next time we shall do the final reduction today we will stop here