 So, last time we defined what is the meaning of a polyadron or a simplicial polyadron namely a topological space which can be triangulated and we saw that the disc is triangulable in such a way that the restricted triangulation to the boundary will give you a triangulation of the boundary of the boundary sphere. Let us take some more examples now. So, recall that joint of two complexes, the joint of two topological spaces were defined as the quotient of x cross i cross y in which at the zero end all the y 1 and y 2 are identified for each x and at the one end namely when the t equal to 1 all the x 1 and x 2 are identified with y keeping fixed. So, that is the definition of the joint. So, what we want to show that is x and y are triangulable then it is joint is triangulable. So, this is what we are trying to prove now. So, this is a preparatory lemma here, if f and g are two complexes in some simplicial complex then I want to say that the mod f star mod g triangulates you know is triangulated by f star g remember if f is a simplex that means it is finite set and all its subsets are taken that is a simplicial complex then f star g is a simplicial complex is yes a simplicial complex on the disjoint union f and g all subsets of this will be there because they can be written as a subset of f union subset of g. So, there are two different ways of writing f star g or just disjoint union and then take the full sub complex here take the modulus of that that is linearly isomorphic to mod f mod g and then take the joint. So, this is the statement not only that there is a very natural isomorphism namely the this map is first of all defined on mod f cross mod g cross t alpha t beta goes to t times alpha plus 1 minus t times beta the standard construction that we have been following ok. So, given alpha and beta inside mod f and mod g respectively the RHS in the above formula belongs to mod f mod g because you can think of alpha as a function on f disjoint union g taking value 0 on on all points of g similarly beta can be thought of as function from f disjoint union g taking value 0 on this one. So, these two things can be thought of as functions here and then they will be themselves points of mod f g then you can take the convex combination of that one inside this simplicial complex inside this mod this is a affine structure is here. So, that is the meaning of the right hand side here ok. Therefore, the FOS assignment defines a function mod phi as for a phi from mod f cross i cross mod g to f disjoint union g which is by the by the very form writes continuous because it is a affine combination. Moreover, it respects the equivalence relation on the domain ok. So, what is the equivalence relation here? When t is 0 ok when t is 0 all alpha either identified irrespective of what beta is for each beta. So, this will give you d beta when t is 1 all the beta are identified. So, this will be again respects the relation ok. So, this gives a function from mod f star mod g which is the quotient of this one to f disjoint union. Whatever is identified here goes to the same corresponding point here they have that is why this is a map here. Obviously, by the definition of quotient topology this will be continuous now ok. So, point that we have to verify is that that this is a bijection the moment is a bijection this is a complex space this is a host space it will be automatically a isomorphism it is a homeomorphism. It is a linear isomorphism linear in the sense of affine linear not the vector space linear. So, what I will now leave it to you is this phi hat is a bijection ok. This is simpler than what we did last time namely for delta n ok. So, you can just verify take alpha prime beta prime ok and take this formula suppose phi phi of here itself phi of you know this is equal to phi of some alpha prime t prime beta prime then show that they are equivalent. So, that the two points are required that will give you injection here inject it here is not true inject it is here ok and and surjectivity is obvious because everything can be written as one point here and one point here t to n. So, this is this is more easier here than in the case of not delta n beta and so on ok. So, this is an exercise for you ok. So, the there is a canonical word here canonical affine linear isomorphism what is the meaning of that I do not want to explain that one this is explanation only it requires no proof at all namely suppose you have f 1 is subset of f g 1 is a subset of g then you have the following commutative diagram mod f 1 star mod g 1 the same phi hat going to modulus of f 1 disjoint in g 1 this will be a subset of that. So, modulus of this will be a subspace of this this will be some each of them is subspace. So, f star g will be containing f 1 star g 1. So, these are inclusion map again phi phi i to 3 this diagram is commutative. So, structure here phi hat is same thing as here because they are given by same formula t times alpha plus 1 minus t times beta whether you are here or here whether you are here or here that is the meaning of this ok, but this is an important thing this property all right. So, that is why I have mentioned. So, now we can complete the task of showing that modulus of k 1 star k 2 is homeomorphic to modulus of k 1 joint modulus of k 2 as such you do not know whether this is a polyhedron once you prove that it follows that this is a polyhedron that is a whole right ok. So, once again recall that the vertex set of k 1 star k 2 is a disjoint union of the vertex set of v 1 and v 2, but what are the simplices? Simplices of k 1 star k 2 are disjoint union of a simplex in k 1 and a simplex in k 2 ok. Therefore, this mod k 1 and k 2 are subspaces of modulus of k 1 star k 2 in an obvious way because they are there complexes here ok. So, moreover if alpha is in F 1 and beta is in F 2 then one t times are upload my t times makes sense in F 1 disjoint union F 2 itself ok you do not have to go outside F 1 disjoint union F 2 has a affine structure that is contained inside k 1 k 2 k 1 star k 2. Therefore, the same formula that we have done will work here also locally at each for each F 1 and each F 2 in a small over and if you from smaller one to larger one when you pass the formula on the small smaller one does not change they are compatible. So, that is what we have noticed here ok in this one. Therefore, you can put them all these phi hat together which is defined for each F 1 and F 2 that will define a single homomorphism phi hat from mod k 1 star k 2 to this one because corresponding images will be also distinct if F 1 and F 2 given in g 2 are different then their interiors will be different boundaries will agree with the smaller synthesis. Therefore, this itself is a homomorphism ok. In particular you specialize k 2 to a single point this is nothing but the cone. If you take two points it will be suspension. So, we get both suspension and cone of a polyadron will be polyadrons. Therefore, excess already some structure namely of a polyadron then the suspension of that will also have polyadron namely x star S naught mod of x star x naught will be S x that is the. So, we can define the suspension of x also S a in the case of polyadron ok. In particular we have also observed that if you repeat repeatedly take the join of S naught with itself n plus 1 copy then it is sphere. So, this will give you another simple structure to S n then what we have defined in the previous section ok. This point I want to repeat it. It is once I have said it is over but I want to repeat it. This is called X triangulation of a sphere ok. This special case I am going to explain now here. So, this is an interesting way to triangulate the sphere. We have just seen that the topological cone over S n minus 1 is homomorphic to the n. So, if follows that if k f is a triangulation of x then the triangulation of x extends to cone. So, this capital F is a triangulation extension of the map not a simplex ok. Since the suspension can be thought of as a double cone it follows that we get triangulation S k S f of S x. This is the more general statement I am making but here already we have observed that S naught S naught S naught k times will give you the sphere. So, I come back to that one now. Recall that S n can be thought of as n 4 suspension of S naught. In particular beginning with the obvious triangulation of S naught. What is it? Just two point and solve per texat 0 dimensional simple shell complex ok. By taking successive suspension we get a triangulation of S n ok. So, we shall refer to this as triangulation of S n by S triangulation mainly by taking suspension. It is worthwhile to note that the antipodal map in this case namely antipodal map means alpha x is to minus 1 is a simplexial isomorphism of S triangulation ok. If you take the standard boundary of delta n, antipodal map has no role to play there. It is not the daily ok. So, a triangulation has this property and that is very useful. I will give you some more examples here. Now, consider the one dimensional simple shell complex k whose vertex set is set of integers 1, 2, 3 and so on, 0, minus 1, minus 2 and so on ok. And the edges namely one simplex is are precisely consecutive integers n and n plus 1, n plus 1, n plus 2 like that, consecutive integers nothing else. So, this is simple shell complex ok. What is its geometric realization? It will be the whole of R. Each simplex n to n plus 1 filling up the gap between the interval n to n plus 1. So, that is it ok. So, this way we get a triangulation of R. The point here is this is the first triangulation that we have taken to be infinite ok, a genuine example. Otherwise, we had only disks and the delta n and so on ok. The triangulation here, the simple shell complex k could not have been finite, because if it is very finite, then it will be compact whereas R is non-compact. Now, I come to a serious problem namely products. We have already dealt with some way product with I, but you have to be careful. There it was for the joint. It was not actually the Cartesian product of a space with some other space ok. The problem is what kind of polyadron will, what kind of simple shell complex will give you the product structure. Suppose x and y already have polyadrons, then is there a way to give a product a polyadron structure such that it has something to do with the original complex is on x and y. So, this problem is much deeper and quite difficult. In fact, in general it is not possible. So, what I will do is I will specialize to the case when one of the factories just the interval 0, 1. Even here we will have problems for example, take delta 1 cross delta 1. What is the best way to triangulate this one? There are two vertices here, two vertices here. So, you are tempted to take four vertices. Then for each one simplex here, zero simplex cross one simplex, one simplex cross zero simplex and so on ok. If you do that ok, what happens is that the number of vertices taken four is lucky that is that is right, but the number of simplexes are too many. So, this is depicted here in this in this picture. So, V naught V 1 is delta 1. V 1 V 2 is also delta 1. So, this is a product ok, this i cross i. The product space is i cross i. What is the simplest structure here? I can take 0 0, 1 0, 0 1, 1 1 as vertices, but then how do I take what are the, what should be the simplex is? Naturally the subspace this one, this one, this one, this one must be there. But then if you just leave it there the interior is not a simplex. So, you have to either cut it this way or cut it this way right. Both of them are possible. The fact that there are two different choices creates a problem which one to take. The moment there is a choice that is going to create problem ok. So, you want to solve this problem in a canonical way by maybe by introducing a little more number of vertices and more edges rather than just sticking to just products things like this ok. So, this is what I want to do. So, this is point cross i ok. The point cross i itself is a simplex complex ok, because i is a simplex complex, i is just one simplex. But I am not going to leave it like that, I am going to divide it into two parts by taking the bare center of this one as extra vertex and then declaring these two portions as simplex. The original simplex 0 to 1 disappears. It has been cut. That is the meaning of it alright. So, come to the second place namely when it is a delta 1. So, this is delta 1 cross i. Mod delta 1 cross i is a rectangle or a square whatever you want to think of. Point cross i is this one. So, it disappears here. Other point cross i also disappears here. Delta 1 appears as it is at the 0th level as well as at the 1 level. So, now we have this big square here empty square. So, what I am going to do is I take the bare center of this one shift it at the half level. That is going to be an extra vertex. As soon as you have extra vertex join it to all the vertices in the boundary. Join it. If this is simplex declare this as simplex. If this is simplex declare this as simplex. Namely this is the cone construction. Do the same thing for delta 3 also. So, this is sorry this is delta 2. There are two simplex. Once you have once you have fixed what the boundary is which is a two dimensional part. One dimensional part one cross i or the boundary is fixed. The bottom is kept as it is. So, the whole boundary of delta 2 cross i has been triangulated. Take the bare center of this simplex shift it at half and then take the cone over that. So, having told this one actually the construction of what I call as the prism construction. Why I am calling the prism? Look at this one this delta cross i this is the prism. This is called the prism construction is over. Let me do it systematically once again so that you will understand what is going on. So, for each simplex complex k we shall construct a simplexial complex denoted by k cross i. This is a notation. This is going to a simplexial complex which I am going to describe. k is a simplexial complex. k cross i there is no simple structure yet. If you take mod k cross i that is a topological space. All right. This k cross i is just a symbol here but I am going to call it prism only after I construct the simplexial complex on this one which I am going to describe. This is called prism over k like cone over k we have constructed and the suspension over k. This is the prism over k with the following property. Modulus of k cross i the underlying the geometric realization is mod k cross i. This is the first property. Then there are simplexial maps eta 0 and eta 1 from k which is simplexial complex to this simplexial complex. So, this is a simplexial map which are isomorphisms on to sub complexes. That means they are one-one mappings taking simplexial to simplexies in a one-one way which the images will be denoted by k cross 0 image of eta 1 will be k cross 0 image of eta 0 will be sorry eta 0 is k cross 0 eta 1 is k cross 1. This is a notation. Just like k cross i is a notation is k cross 0 and k cross 1 are notations and they are sub complexes of k cross i. This is the property I want. I want a, I want this property b also. I want something more namely the underlying topological space of k cross 0 is mod k cross 0 and k cross 1 is mod k cross 1. That is the property. Finally, I want if you take a sub complex l of k then I must have l cross i also the prism over l. This must be a sub complex of k cross i. Now whatever I am going to define and neither I have defined l cross i nor I have defined k cross i but they must have this property. So I want a prism to have all these properties. Now I have put so many conditions it looks like too many restrictions you have put made your life more difficult. In reality these restrictions will actually guide you how to construct k cross i and that is done by induction. So let us have notation beta f will denote the barycenter of any syntax. Let us put f hat such a short notation for the ordered pair beta f comma half belonging to mod f cross i. So I am placing this pair center at the level half. Now suppose f is a single vertex, 0 simplex. Then I am looking at a simplexial complex with vertex set precisely equal to this one and one simplex is precisely equal to this one. And this is the simplexial complex. There are no other two simplexries and so on. This is going to be my f cross i where f is simple. Okay. So what are the vertices v0, v1 and v half. That is precisely this picture. The first picture here. Okay. And v0, v half will be one simplex and v half and v1 will be another simplex. Okay. So that is what I have listed here. v1, v half, v0 and v1. And eta 0 of v is just v0. Theta 1 of v is v1. Okay. Now this completes the construction of f cross i when f is a single vertex. Since there are no sub complexes, sub complexes empty, all those A, B, C, Ds whatever I have demanded, they are all satisfied. Okay. So the construction for singleton is over. As soon as you have constructed singleton, you take the union of all these things, it will be constructed for all zero-dimensional simplices. As zero-dimensional simplices nothing but union of singletons. So the construction is over for all zero-dimensional singletons because I am just taking the union. Okay. Singleton cross i disjoint union because singletons are disjoint union over k. Okay. So the inductive step is over for n equal to 0. Now assume that we have done it for some n minus 1, all the way up to n minus 1. Okay. Then we want to do it for n. What is the meaning of that? Take a simple initial complex k and take a simplex n, n n-dimensional simplex in it. On the boundary of this simplex, there is already a structure, boundary of this simplex cross i, there is already a simple structure. On the bottom k cross 0, f cross 0, I am not going to distribute it. I am keeping it as f. f cross 1 also I am not going to distribute, I am keeping it as f cross 1. So this way I complete the picture of mod f cross i and its boundary. The boundary of mod f cross i is already given a simplexial structure. Now you take the cone over this with the cone point namely apex point precisely at f hat. f hat is what? The bar is enter of f comma half. Okay. So that is what I am going to do. So bf cross i is already triangulating boundary of f cross i. So there is modules of bf cross i is bounded f cross i. This is by induction. Okay. So you should know that f is we know already is a is homeomorphic to dn. So f cross i is homeomorphic to dn plus 1. So dn plus 1 is a cone over its boundary Sn. All this we are using it here. So we simply take f cross 0 as it is, f cross 1 as it is, there are copies of f union bf cross i. Okay. So this entire thing from here to here describes the boundary of mod bf cross i. So that has been given a structure. Take the cone over f hat. Take the cone over this f hat as the apex. Okay. So automatically the extension of eta naught and eta 1 over this one is over because now the f is sent to the corresponding f that is all. Okay. Once you have done it for one simplex f, n simplex, one of them, do it for all the n simplexes exactly same way. Take the union of n minus 1 simplex we have already constructed. Take the union over all the n simplexes that will complete the construction for n dimensional skeleton kn cross i. Okay. So inductively constructing n minus 1 skeleton to n skeleton we have described. Therefore now the construction is over for all kn cross i where kn denotes the n skeleton of k. Take the union of all these things because of the property D they are compatible. The subspace which is already triangulated extends the triangulation when you go to the largest space. It is extended. So they will be compatible. So take the union that will be the definition of k cross i as a simple shear complex which will have its modulus homeomorphic to mod k cross i. Okay. This is a given construction. All right. Let me give you another simple thing is a triangulation of a torus. By the definition you must be knowing the torus is defined as s1 cross s1 but s1 is defined by the interval and identifying endpoints. Therefore the torus can be defined by identifying opposite sides. Okay. A pair of opposite sides will be identified with the correct orientation. Okay. So this you must be knowing already. So in order to give a triangulation on the torus what we do is we choose a triangulation on the i cross i in such a way that when you quotient out the triangles etc do not collide. In other words two vertices which are joined will go to two distinct vertices corresponding to there. Three vertices which have a triangle they will go to a triangle not that the two endpoints of a triangle the vertices will be identified and so on. Okay. Not only that another set of edges will not come on the same set of vertices. This is what you have to be careful about. All right. So all these things is done. So this is a good example to know when you have quotients what you have to be careful. So we cut down the i cross i triangle along each one divided into three portions. This is good enough for a triangle. See if you identify these two endpoints you get a circle. The circle minimal way of triangulating a circle is take three points and three edges. Right? You can't do less than that. So do that one, two, three. This is one only because this is going to be identified with this one. Now take some more vertices four, five. Okay. But then again this will be identified with this one. So this is one. But three will be identified with this three. Two will be identified with this two. One will have one this one. This entire line being identified with this entire line. One, two, three, one. One, two, three, one. Similarly, one, four, five, one identified one, four, five, one. So these are new, not new. But inside I am taking six, seven, eight, nine years. And then joining them irresistimately like this. There may, this is not the unique way. There may be many ways. Okay. Now you see each triangle is uniquely defined when you move to the quotient. So this gives you the triangulation of a total. This is not an economic way of doing it, by the way. Given any surface, you can give a triangulation that is a little more harder to prove. It is a big theorem. And then there are problems like this. What is the minimal number of vertices needed? What is the minimal number of edges needed and so on. This is a big industry since several, almost 30, 40 years now. A lot of people are working. One famous mathematician in India has worked his life on that, namely Professor Basudev Datta and Sarkarya in Punjab. Basudev Datta is IIS and some of his students and so on. Okay. The torus itself can be triangulated by using just seven vertices. Try to do that. Let us stop here.