 So, we come to the last module of this series, what is the geometric distinction between surfaces occurring in the list 2 and list 3, namely all those surfaces which are obtained as connected sums of a number of tori or connected sums of a number of projective spaces. We know that the algebra, either homology or the fundamental group, they distinguish them. We have studied them thoroughly, but now we are asking what is the geometric thing that is happening here. This question brings us to the all to important concept of orientability, concept of orientation. However, we shall just touch upon this concept restricting ourselves to the case of triangulated manifolds, which is the classical approach actually. It somewhat turns out to be somewhat weaker than other approaches. So, by an orientation on an n simplex, n greater than or equal to 1, we mean an equivalence class of labeling the vertices, that means choosing an order. What is the equivalence class? What is the equivalence relation? Two such labelings being treated equivalent if one is obtainable from the other via an even permutation. Once there is a labeling, another labeling corresponds to a permutation of the original labeling. If that permutation is of even type, namely signature of that permutation is 1, then we identify these two. So, it turns out that there are exactly two orientations on any n simplex. Given any one ordering, you can just interchange just two of them and keep the rest of them in order permutation, that will be another one. If you do one more, that will be again even. So, you will not get any other class. So, there are only two classes. So, there are two orientations. Orientation means what? Choosing an order, but up to equivalence of this nature. Namely, the permutation corresponding to the going from one ordering to the another ordering should be even type. So, it follows easily that there are precisely two orientations on every n simplex and greater or equal to 1. We call them orientations opposite to each other. Once suppose you fix one of them and call it as positive orientation, then the other one will be negative. It is just a square root of square root of i. Sorry, square root of minus 1. Is there a positive square root? That is just a joke. But you will like i and plus minus i. The other one is minus i that is all. Doesn't mean that i is a positive root square root. There is nothing like that. So, there is no positivity, negativity here. But people do use this kind of loose terminologies. You must understand that is all. So, one is the opposite of the other. Often an oriented n simplex is displayed by putting a bracket. This we have followed earlier. In the construction of the chain complex corresponding to a, you know, the simplicial chain complex, correct or not, is simplicial complex. So, I will just recall it. It is displayed as v naught v1 vn. So, this is an n plus 1 simplex. This is an n simplex, sorry, with n plus 1 vertices there. Sometimes when one of the orientation is preferred for some reason and referred to as a positive orientation, then the other one is called the negative orientation. It is just like we have the anticlockwise and clockwise orientations, right, direction. For instance, on the standard simplex delta and e naught even en, the e naught even en in that order is called the positive orientation. Then all the other orientation whatever you take will be, will be the negative orientation. That is all. So, we also use this notation. This notation is justified because of our integration theory and because of our homology theory. If you choose this as a generator, then the opposite orientation, this one, we have seen that this plus this in the homology will be null homologous. Therefore, this is equal to minus of v naught v1. So, we can use this notation. Integration along this edge and integration in the opposite direction, they are related by this relation. That is one other reason why this is used, okay. For the sake of logical consistency, we shall include the zero vertices also, zero simplex is also, vertices also. And vertices, how many orientation can be there, permutation or group itself is trivial. It is only one element. But just for logical consistency, what we do is we will declare that each vertex also has two orientation, namely bracket v and minus bracket v. That is all, okay. This is just, you know, forcibly we define because to look at the zero chain, we have made it a group. In an abelian group, we must have negative also. So, it is like that, okay. So, let us now make another notation. Recall another notation. Let sigma be v naught v1 be an oriented simplex. If we just write curly brackets, it will be simplex. Okay, if you write square brackets, then you have chosen an order up to permutation, whatever even permutation. So, that is an oriented simplex. Then v naught, vi minus 1, vi omitted, we remember what is this one, this is the ith phase of this one. With the correct sign minus 1 power i, these are the n minus 1 phases of sigma, okay, all of them, trying from d to n, there are n plus 1 of them, with induced orientation. That is the definition now, okay. This is how the in the boundary sum, we are taking the sum of all these and we have to put the correct sign here. For instance, the oriented what it says, of an oriented edge, what are these? We will drop out v1, we have to put minus 1 correspond to this. v naught is minus 1. v1, we have got by dropping out v naught, is minus 1 ratio 0, this is positive. So, it is v naught with a negative sign and v1 with positive sign. The sum total is defined as a boundary, remember that. So, these are the oriented edges. The oriented edges of the oriented triangle, and just recalling these you are already familiar is if you omit v naught, v2, v naught, v1 will come. If you omit v, v naught, v1, v2 will come, but if you omit v1, it will be minus v naught v1, which is v2, v naught. Psychologically, if you are right in this case, then the sign is all of them plus, that is only for 3. If you go to 4 and 4, you have to, this is the only rule that will tell you how to write correctly. That may be a triangulated, triangulation with pseudo and manifold. You know what a pseudo and manifold is. Pseudo and manifold, first of all, is a triangulated manifold. It is a triangulation, sorry, it is a, it is a simplification complex, which satisfies one of the fundamental properties of the triangulated manifold. So, we have defined what is a pseudo manifold. By an orientation on k, we mean a choice of orientation on each n-symplex, because a pseudo manifold is pure of dimension n. All the maximal symplexes are n-symplexes. So, on n-symplexes, you have an orientation for each of them, so that the orientation is reduced on a common n-1 phase. So, two symplexes may have a common n-1 phase, that is called a fast set. And the two orientation, it will come from one orientation over on one side, another from other side. These two orientations must be opposite of each other. So, this is a definition. If this is satisfied, if all the orientations are like this, then we call orientation on k. So, not all pseudo manifolds are orientable, this condition will not be obvious. It turns out that given a pseudo manifold x, its orientability depends just on the homotopy type of underlying topological space. This is a deep remark which we will not be able to prove in general. But for surfaces, we have already a proof here. So, I want to indicate that. Let k be a triangulation of a surface. It turns out that k is orientable if and only if there exists an ordering of the vertices of each two simplex in k, which corresponds to orienting each two simplex, such that whenever sigma 1, sigma 2 are two adjacent symplexes, their ordering is compatible. This is the word I am going to use in the following sense. What is the mean of compatibility? The two orders induced by two simplex is on other side, on the common edge. Coming from sigma 1 and coming from sigma 2, they must be opposite of each other. So, I am just recalling this definition here in this special case. That is all. Compatible is the word I am using here. Not that every triangulated, note that every triangulated convex polyhedron P in R n is orientable. So, this is not very difficult to see. At least for R equal to 1 and 2 is totally obvious. It can also be seen that there are exactly two orientations on P. So, what you can do is you start with one simplex, whichever order you want, you place that order. Then look at the induced orientations on each of the faces. For the other side of a simplex, simplex which comes on the other side, you extend that orientation so that with the opposite sign that is all. So, that the two things which are induced are opposite to each other. So, keep doing this. So, you can always do this without any problem at all. Till what? Till you have some problem is all, there may be one simplex. All its edges are already oriented. All its boundaries are already oriented. Then you must show that if there is only one way of orienting that one and that will give you the correct orientations. So, you have no choice there now, but that is whatever is forced on you is the correct orientation. That is what we have to do. So, there is something to be proved here. It is not all that obvious, but it can be proved. It is not a difficult thing. All right. So, a canonical polygon is said to be orientable. Now, I am making a different definition here. A convex polygon is always orientable. Now, I am taking canonical polygon. What is canonical polygon? Canonical polygon means that there is a sequence of edges on the boundary such that each edge is identified with another edge, exactly one other edge, edge pairs. Remember that. That is the definition of canonical polygon. So, a canonical polygon is said to be orientable if and only if all the identifications, namely pairwise identifications of the edges are there on the boundary phases are orientation reversing. That is the definition of a canonical polygon being orientable. So, this definition can be taken in any, in a RN also. Instead of polygon, you can take polydron. In dimension 2, this can be formulated as follows. A canonical polygon in R2 is orientable if and only if it has no edge pairs of type 2. The type 2 edge pairs are how? A and then A. So, identification is orientation preserving. That should not happen. All the edge pairs must be orientation reversing. So, that is the definition for orientation of the canonical polygon. Why this definition, this artificial definition? You will see that this is the correct one for us to relate it to pseudo-manifold now. So, this is the theorem. Let K be a triangulated surface and P be the convex polygon. Some, there is no uniqueness. P is a convex canonical polygon defining K. That is a quotient mass P from P to K in which edge pairs are identified, all that you have to recall. Suppose this is a picture, then K as a triangulated surface is orientable if and only if P as a canonical polygon is orientable. So, this is the theorem. So, this is what made us make this definition. So, how does it go about it? So, you have to verify that the induced orientation, that the claim is there is a quotient map is injective on each simplex. So, it is a bijection of vertices. So, take the corresponding ordering P as a triangles of P are already ordered. The triangles in K should be ordered according to this isomorphism P. You take that. Now, you have to show that this is an orientation on K. It is easily checked that any triangulated convex polygon is orientable first of all. Okay, there is no problem here. Given a triangulated surface K, let P be the canonical polygon associated to it. You got my point? We start with it. We can fix an orientation on this convex polygon. Then pass the ordering of the vertices of each simplex down to K via the quotient map P to K. Clearly, for all edges in K which are images of some interior edges of K, because we are using the same isomorphism, same bijection, there is no problem. Okay, there inside P they are compatible. When they come down to K, they are compatible. Okay, compatibility condition is automatically described. So, the problem it remains to receive what is happening at the edge E in K which is image of a pair of boundary edges. Okay, if these edges edge pairs are of type 2, then the two orderings on E will be the same coming from two different simplex states and that will create problem. Compatibility fails. If all edge pairs are of type 1, then there is no problem. Okay, and hence compatibility will be over. So, this is the explanation for this theorem. Okay, in fact, once after observing this only, you formulated this theorem. Okay, note that if there is at least one edge pair of type 2 in a canonical polygon, then in the reduction process, see from a canonical polygon, we had a step 1, step 2, step 3, step 4, step 5 and so on reduction process to bring it to the normal form. The canonical process, the reduction process will never get a pair of type 2. Okay, it will never get rid of it. Type 1 pairs are sometimes got rid of. Type 2, if you try to get rid of it, it will induce another one of the similar nature. Okay, so that will be always there. So, once you have a canonical, once you have a type 2 edge pair in a canonical polygon, then in the reduction process that we have used to bring it to the normal form, all the intermediate states of the canonical polygon will have at least one edge pair of type 2. Therefore, it follows that original canonical polygon is orientable if and only if its normal form is orientable. You can say that this is not orientable if and only if it is not orientable. Type 2 means not orientable. Okay, so I will give you one more example here. Let sigma equal to at two simple x, u, v, w and sigma 2 equal to w, v, x. What is the edge here common v, w? So, what is the order here? Look at on v, w, you have to drop out u. So, v, w, the the orientation is v, w. On w, v, you have to drop out x. Again, it is w, v. So, these two are opposite of each other. So, this order is compatible. Okay, if you order like this. So, you can put these two sigma 1, sigma 2, two triangles together. So, that will be oriented correctly. Now, suppose these two triangles are taken and you have messed it up, namely tau 1 is abc, tau 2 is bcd. Okay, so when you drop out a, it is bc, correct order. Okay, when you drop out d, again it is bc, the same order. So, this is not compatible. Okay, check that the canonical polygons, a, a inverse, okay, by definition is an oriented canonical polygon. a, b, a inverse, b inverse, these are orientable. Whereas, a, a, a, b, a, b inverse, these are not orientable. Why? Because there is one pair a and a, same orientation, right? So, this will be, this will create problem. This defines a tau, this defines a, the Klein bottle, this defines projected space. These are not orientable. The simplest two-dimensional manifold that is not orientable is the Mobius band. We had some great experience with them in the live session also sometimes in another meeting here. The simplest two-dimensional manifold that is not orientable is the Mobius band. You can verify this by taking a couple of tangulations but that will never prove the assertion. Maybe there is some other, maybe there will be a long, a lot of way you have cut it and you are not verified. So, prove that it is not orientable, you have to have a different device. So, with this definition, it is not that easy to demonstrate that something is not orientable. Okay, if you cut it and then orient in a correct way and get orientability, you can demonstrate something is orientable, that is easy. You have to give just one triangulation which is orientable. But to say that underlying space is not orientable, that is rather difficult. You give me a triangulation, I can verify that it is not orientable. That is possible. But no triangulation is orientable, it is not easy to do. All right. So, but here in the case of surfaces, because of our homology and so on, we have got a complete understanding of this orientability. This is just a lucky part so that it comes so easily for us. So, let us make an algebraic definition of orientability and then see that that is a topological invariance and that from there we can we can deduce all these things. So, what is the definition? X be a connected compact two-dimensional manifold without boundary. Okay, there is no triangulation here now. The earlier one that we defined using triangulation, let us call that combinatorial orientation. So, this is now algebraic orientation. Okay, so we say X is orientable if and only if the second homology of X with respect to that coefficient is isomorphic to that. This looks like very artificial definition, but this is motivated by geometry and this definition can be taken for all manifolds, not only just H2, not only just surfaces. You make instead of n here, n and then put H2, instead of H2 you put Hn here and that definition is a strong, very strong. Of course, it can be further generalized also. Okay, so since we are assuming that every surface X is triangulable, okay, that Radoff's theorem we have assumed we have not proved that one. Okay, it follows from theorem 6.14 and our computation H2 which we did, right, sometimes that X is orientable if and only if it belongs to the list 1 namely S2 or 2 namely all TGs. Okay, thus the new algebraic definition and the old combinatorial definition of orientability coincide here. Okay, so in both cases the polygon that we get, the canonical polygon is activated. As only edges of type 1, no edges of type 2, edge pairs of type 2. Okay, as soon as type 2, it will belong to the class 3 and class 3 we have computed H2 is actually 0, not that. So this is what we have seen already. Therefore, we have this control that algebraic definition implies and implied by the combinatorial definition. These two are equivalent. The theorem just below just sums it all now. So I want to tell you something without using the triangulation at all. Okay, let S1 and S2 be compact, connected, two-dimension topological manifolds without boundary namely surfaces in our short. Then S1 and S2 are homeomorphic if and only if their Euler characteristics are equal and both are orientable or both are non-orientable. Sometimes back we have observed, we have computed Euler characteristic and we have seen that Euler characteristic itself cannot distinguish between the second series and the third series. The members of third series can have same Euler characteristic as some members in the second series. Okay, so just Euler characteristic level 2, put one more condition, orientability, the second and third are different, over. Okay, so the proof uses triangulability. The statement does not. Okay, finally we have a statement without referring to a triangulation. Okay, only Euler characteristic and orientability will give you the, this definition and for orientability we can use algebraic definition namely H2 must be infinite cyclic. So let me make a general remark here. In general for a smooth manifold, weak or without boundary and compact or non-compact. That is like a more general but you have to assume smooth manifold. Smooth means differential manifold. We can talk about orientability in at least four different ways. Combinatorial one is valid there because a smooth manifold is triangulable. There is something called tangent bundle. Okay, orienting tangent bundle. There is something called differential enforms, existence of a non-trivial differential enform. Okay, and then finally there is algebraic one also. There are many more but I want to tell you at least four of them. Okay, they all, these all orientability coincide. Okay, the algebraic definition gives you that orientability is indeed a homotopy invariant. Whereas the differential topological thing will give you that orientability is only diffeomorphism class invariant. The combinatorial one is the most difficult to handle but easy to perceive, easy to understand. Proving general theorems, it is difficult. Okay, so we have to study that so that computation and easy to understand it is for that it is very helpful. Okay, so here are a few exercises. These exercises will be again updated and given to you in the form of just PDF files. Okay, you have to work out them. You have to work out exercises and some of them you have to submit also and so on. Okay, thank you. I have enjoyed lecturing to you. See you some other time. Thank you.