 In real analysis, especially in real analysis of one variable, all these Riemann integration theory, uniform continuity, various things, especially in integration theory, often you have to subdivide the given interval into finitely many subintervals and then analyze what is happening to the situation. So this is copied in study of polyadrons also. The idea of polyadrons is that you can divide them, take the whole idea. So there are many, many different ways of doing it. Even in Riemann integration theory, instead of taking all kinds of subintervals, you can just stick to each time dividing the interval by half, then one-fourth, then one-by-eight and so on. Then you show that if something happens for that subdivision, the whole thing is fine, that kind of things are there. So similar to that, we concentrate on one particular kind of subdivision which is called Paris centric subdivision which will be quite powerful and that is the only thing which you are going to actually use here. So what you may have to do is Paris centric subdivision itself may have to keep repeating. It is just like in the case of an interval, namely a closed interval, Paris centric subdivision will be just corresponds to taking the interval into 0 to 1, 0 to 1 by 2, 1 by 2 to 1. So introducing the extra point, midpoint half. If you repeat it, then you will be introducing one-fourth as well as three-fourth. If you repeat it, you will be introducing all the vectors of one-by-eight, three-by-eight, five-by-eight, all those points also and so on. So that is called iterated subdivision. So iterated Paris centric subdivision. So these two things we are going to study here, okay? Iterated Paris centric subdivision. So that is what we are going to do finally. So today's topic is Paris centric subdivision and its consequences. So let us be done with definition very quickly. Start with a simple shell complex for each non-empty phase. Recall what is the definition of the Paris centre of F denoted by F hat. This is a point in mod F, namely the function which has 1 by n as coordinates. The point which has 1 by n, 1 by n plus 1 as coordinate at each point. Namely alpha of v equal to 1 by n for each v inside F. So that is the Paris centre, okay? So that is all you have to know. This is the very beauty of this one. So we now define a new simple shell complex, sd of k. That is a short notation for subdivision. But it is a very particular subdivision. It is a Paris centre subdivision, sd. So all other subdivisions are not of important force. Called the Paris centric subdivision of k, okay? The vertex set of this sd of k is a subset of mod k. What are the points? Namely take F hat, F 2 del for each F in k, okay? So vertex of this one is a subset of mod k, given by this one, okay? So I have now defined what are the simplices, all right? So simplices are, we start with a sequence of phases. One contained in the other. Such a thing is called a chain of simplices. F naught contained in F1, contained in F2, contained in F3. No repetitions, okay? So this strict sequence. For example, this could be one single vertex, contained in a one simple edge. Contained in a three simplex instead of two simplex. There can be jumps contained in some ten simplex, ten-dimensional and so on. So take such a chain, just put a twiddle on each of them. We get another sequence here. F naught twiddle, F1 twiddle, F2 twiddle. So take all these vertices, declare that as a simplex of SD of k. Automatically, if you have a subset of this, that will correspond to a sub-chain here. Therefore, it will be also a simplex, okay? So that completes the definition of a simplicial complex, SD k. Now what it is good for, just like our simplicial complexes were defined abstractly. This is also defined abstractly except that the points are not abstract points now. They are already points of the geometric realization of k, mod k. So in that sense, they are much better geometrically. So here is a picture that we are going to do. This is the original simplex here, simplicial complex here with one one edge and one triangle, okay? Of course, this triangle will have three edges and so on separately. But I am mentioning only the maximal simplexes here. This is going to be the bary centric subdivision, okay? So I am showing you the, what are the simplex system, what are the, what are the vertices and what are the simplex. So for this one simplex here, I have taken this bary center. How did you get this one? This singleton itself is a simplex, zero simplex and it is bary centric itself. So that is allowed. Now if you take this simplex, singleton zero and the included in the one simplex, that is a chain of length two. That will give you the bary centric of this one and bary centric of this one with this point. So a simplex from here to here comes up. Similarly simplex from here to here comes up, okay? So now you can see that the original interval has been divided into two intervals. So that is the subdivision that is happening. So we will make this one clear later on little bit, but this is a motivation. So what is happening? In here you see that a one single triangle is divided into six triangles at once. More generally what happens is if you take an n simplex in the bary centric subdivision, it will get divided into n plus one factorials, simplexes of the same dimensions, n dimensional simplexes, okay? A one simplex gets divided into two, gets two factorials. This is three factorials and so on, okay? So let us see what is the correspondence, what, why it is called subdivision. After all subdivision means the topological space is there. It has to be cut down and that means the pieces together should give you back. The union of all these pieces must be the given impartial, namely polyadron, the underlying topological space of the polyadron. Mod K should not change. In other words, mod of SD of K should be mod of K, okay? So let us see how that is true. We do not have any freedom here now. We have to show that somehow mod of SD of K is or in up to some isomorphism is same thing as mod of K, okay? So before that, having defined for each K another simple shell complex, SD of K, suppose you have a simple shell map from K to L, then there will be a corresponding map, simple shell map, SD of K to SD of L, which will denote by SD of F. This F is a simple shell map, okay? It is defined at the combinatorial level, et cetera. Not at the mod K to mod F. Mod K to mod L, that will be mod F. That is already defined. But this will be SD of F to SD of K. How do you define it? First of all, you have to define it on the vertex set. Vertex set consists of barycenters of synthesis of K. So take one barycenter like this, F at SD of F, send it to the barycenter of image of F, F. F F is a simplex because F is a simplex and F is simple shell map. So this is a simplex in L, so you take the barycenter of this and that. So this is a vertex map now, okay? The vertex map, you have to verify that it is a simple shell map. It should take simplicity to simplicity. What is the simplex on the left-hand side? First of all, you have to start the chain F0, F1, F2. But image of those FF0, FF1, FF1 will be a chain here, okay? Now what may happen is some of these may become equal. Doesn't matter, you throw away the equal one and write the proper chain, okay? So you get a simplex of whatever lower dimension because chain may be of smaller length. Doesn't matter, okay? So take the barycenter, that will be a simplex. So this is a simple shell map automatically. Once you define a vertex map between simple shell complexes, you have no other freedom. You have to verify whether this is a simple shell map or not, that's all. A vertex map may fail to be simple shell map, that's all, okay? You can't, there is no further definition to be made on F. Once you have defined this vertex map, then you have to verify whether it is simple shell map or not. So that gets verified here, okay? So in particular, if K is a subcomplex of L, take the inclusion map here. Take a subcomplex, then that is a simple shell map. Not all inclusion maps of vertices may be subcomplexes. But this is subcomplex, take a subcomplex and then take the inclusion map, that is a superficial map. That will give you SD of F, SD of that, that will be automatically an inclusion, okay? So SD of L, SD of K will become subcomplex of SD of L. If K is subcomplex of L, alright? Now the second remark is, for each simple shell, you can think of F as a simple shell complex, okay? The full simple shell complex. Then look at the inclusion of SD of the boundary of F. Boundary of F is subcomplex of F. So SD of boundary of F to SD of F, this is an inclusion map. This extends to a simple shell isomorphism of, these are the basis now, okay? But this will be, this is a sub of this one. If you take the cone over SD of F with the apex as F hat, F twiddle, the barycenter, what you get is SD of F, okay? What you get is SD of F. Remember how was this defined? This is some proper, every simplex here is a proper phase of F, right? Something F naught, F1, Fi, Fi naught equal to F, right? So those are the proper surface. To take such a chain, okay? That will be a subcomplex of SD, right? Whenever you have subcomplex, for each phase of that, you put F hat also in that. What is F hat? That will be the last vertex because that is the F hat contains all of them, okay? For each, you will put one more extra, but that is precisely the cone. The cone construction is that. Take any phase here, put F hat also. Of course, you can put empty set here and put F hat, that will be F hat. So all the simplex sets of SD of F are given in one-one correspondence here. So this is a commutorial isomorphism, okay? Corresponding to that, when you take mod of this, okay? That will be isomorphic to mod of SD of F. But what is mod of this? It is the cone over the boundary, namely first you take mod of this part, okay? And then take the cone over that. So inductively what we have established is that for every simplex SD of F is nothing but modulus of F. Because finally this is nothing but cone over this, cone over the boundary is nothing but you know topologically is nothing but the whole simplex, okay? So we want to verify the same thing for every simplex. Mod of SD of K is the same thing as mod of K. But that follows very well now. Next theorem assures that topologically the barycentric subdivision does not affect any change, okay? This three is just a question of what is going to come. But I have told you that the key is here itself. The remark two tells you the whole story, okay? But I am putting it in a neat theorem now. I have already observed the key observation it is already done in tow. So now we will summarize this here. Inclusion map of the vertex set of SD of K into mod K. Remember every vertex here is some point here because it is some F hat of something. F hat means what is the barycenter of F, okay? This linear map extends, this inclusion map extends linearly on each closed simplex of SD of K, okay? You can extend it like that. Then the extended thing defines a homeomorphism of H from mod SD of K to K. So this homeomorphism is not anything cooped up. It is just the linear extension by which I mean a fan linear extension on each some piece here, on each simplex here, okay? The linearity on that coincides with the corresponding linearity on this one. The linear structure is same. To take t0, t1, say alpha and beta in this part which are contained inside some simplex here. t times alpha plus 1 minus t times alpha, whether you take it here or you take it here, they are the same. They will be the same thing. So that is the meaning of this one. So this is already there for each simplex here in this remark, okay? This homeomorphism is canonical in the following somewhat weak sense because the word canonical is usually used in a much stronger sense, namely functoriality. Here you do not have full functoriality. It is functorial for inclusion maps. Suppose you have a subcomplex K to L, the following diagram is commutative. You take Sd of K that is contained inside Sd of L. Modules have Sd of L. You have a homeomorphism here, this H to mod K. And here you have mod H, the same homeomorphism, you know, this canonical homeomorphism. This homeomorphism does not depend upon whether you take L or K and so on. That is the whole idea of being saying that it is canonical. The homeomorphism here, inclusion map here. So this diagram is commutative whether you include it here this way and come here or this way come here. How to verify this? You have to do it for one simplex each time, okay? That's all. And that is taken care by this key observation here. So let me repeat, remark 2 is used inductively. Observe that for any simplex F, you have the canonical inclusion Sd of F contained inside Sd of K. First of all, which is a simplicial map, okay? And the map defined above satisfies canonical property for each phase G contained inside F. You take G contained inside F from Sd of G to Sd of K you go or first go to Sd of K to Sd of F and then to Sd of K. They are the same because they are all inclusions. Therefore, the canonical property in general case follows. Once you verify it for each phases and sub-faces, okay? Now I will tell you why Sd of K to K itself, mod K itself is a homeomorphism. So let us make a, let us go a little slowly here so that we are not making any mistake. So first I say the statement An is for all simplex is K, K simplex is where K is less than equal to N. This is a homeomorphism, okay? So I am trying to build up some inductive hypothesis here. For when N is 0, we can verify it clearly, okay? For N equal to 1 also, you can clear it but when it becomes larger, you do not know. So you would like to make it an inductive hypothesis and work out, okay? The second statement A does not involve any N at all. It says for all simplex is H, F, H is a homeomorphism, okay? So how do you get this one? If you do it for every N here, you get this one, that is the whole idea, right? That is the meaning of here, induction. Now here BN is for all simplex is of dimension less than equal to N. Sd is a homeomorphism. Then the statement B is for all simplex is K, H from Sd of K to K is a homeomorphism. So this is, this statement B is what we want to prove, okay? What we have observed is for, for A also you have already observed. For each of them which we have observed, then we have observed A for A, okay? So let us do that way. First we shall show that An actually implies BN. Don't go to A at all. Just for N less than equal to N implies. That means what? Suppose you have observed for the zero simplex, what is the balance in sub-quieting? It is just B0 itself. So it is a homeomorphism. There is only one point, okay? From this you will conclude this one for all zero dimension simplex says. What is a zero dimension simplex? It is just a disjoint union of vertices. When you take bare center sub-quieting, it will give the same set of vertices, nothing more. So it is identity map, right? So K equal to zero or N equal to zero, this is a refinement. That is the meaning of from An to BN you come, okay? For dimension zero, A, so A0 implies B0, right? So same thing you verify for An. So let dimension of K less than equal to N. Given any beta inside a K where it will be, it will be inside some F and modular and dimension of F is less than equal to N, right? Therefore this An, okay? From this An will tell you that H is a subjective map from mod F to this one. So this alpha is picked up or beta is picked up. Given alpha in SD of K, check first H alpha is in F for some F inside K. You find only H of support of alpha is inside F, okay? So while H of alpha belongs to this one, support must be contained inside F. That is the meaning of this one. H alpha belongs to mod F means what? Outside F, if you take a vertex, then H alpha of that vertex is zero. So non-zero things are only inside F. That is the meaning of this. Therefore if you have two points, alpha and alpha prime belonging to SD of K are such that H alpha equal to H alpha prime take equal to beta. Then I want to show that alpha equal to alpha prime, no? For showing injectivity. Then if this is equal to beta, you find first of all some F belong to K such that both H of support of alpha and H of support of alpha prime are contained in this F. There is some common F, okay? That is namely, it is beta. If beta has, there is a point beta, the support of beta is nothing but some simple X and that is, we can take it as F. That is all, okay? Now what happens? Now you apply A n, okay? A n is injective. So these alpha and alpha prime are in some simplex. First of all, one single simplex. So therefore, this is the point here. If the image is simplex, first of all, alpha and alpha prime must be in the same simplex. From different simplexes inside K, they will never go to the, they will never go to simplex unless there is a larger simplex containing them, okay? So H alpha must be H alpha prime. Therefore, what happens? Alpha must be equal to alpha prime. So H is injective, okay? Clearly, H is continuous. Continuity is verified by, by checking it on each simplex there, it is linear, okay? So to see that finally, I want to say that H is closed map. A continuous bijection which is closed map will be homomorphism. Take a G, which is inside SD of K. Let take a closed subset. You must verify the HG. It is a closed subset. For that, I have to verify HG intersection with any closed simplex inside L is closed in F, mod F, okay? So this is actually the inside K, not SD of K, sorry, which is inside K, F is inside K. But HG intersection mod F is nothing but H of G intersection SD of F, okay? G is a subset of SD, right? So H of G intersection inside SD of F because whole thing doesn't go out of that one. G is inside F, okay? HG intersection F. Instead of F, this is just SD of F. So G is closed. G is SD of F is closed, okay? And H is a closed map, H is homomorphism, SD of F from A and N. So we get the HG intersection that is closed. In other words, once you verify it for each N, all these things can be reduced to, verification, injectivity, subjectivity and continuity are all verified restricted to each mod F, okay? So to complete the proof of A implies B now, all that you have to do is A N for every N and A N implies B N for every N. So you get for all of them, that is the meaning. Once you prove A N implies B N, so this, because every point is in some final dimension, okay? For each of them, I come back to the point one here. This is the picture, finally. This is, sorry, point to, remark to that we have started. SD of the boundary, two modulus of the boundary itself. This is by induction because this is of the one-dimension lower. SD of the boundary, if you take the cone over that, okay? This is the cone over, this is the cone over that. If this is homomorphism, well, cones would be homomorphism. But this is the cone over this one. The cone over the boundary of the modulus of the boundary, modulus of the boundary is a sphere, cone over this one is a disk. So these kind of things we have seen, okay? Since we have a homeomorphism, cone is a homeomorphism. But here is what we have, this is mod lambda, okay? To SD of f. What is this isomorphism? We have seen that if you do not take modulus, there is a lambda, this lambda is an isomorphism. One-one mapping is here for each simplex union f hat, you get a corresponding simplex here. Corresponding longer chain, one extended chain here for that one you get. That map is lambda here, okay? So here, this is the topological homeomorphism code construction that we have seen, okay? So this diagram is commutative. So this is the final SD of h map, okay? So I have explained this. When the vertical lambda represents the homeomorphism, this is very canonical isomorphism, okay? So this proves a n plus 1. That means assuming a n, assuming a n you get a for here. So then that will give you a n plus 1 and dimension on higher. So now I will give you an example that why the canonicalness fails here in general. If you do not have inclusion maps. So very simple example, take k as a triangle, okay? The standard two simplex. L as a standard one simplex, okay? Then I want to make a map from phi to k. One goes to e1, two also goes to e1, three goes to e2, okay? From a set of three points, I have to set up two points. So one, two of the points must go to the same thing. So e1 and e2 go to e1, e3 goes to e2, okay? Now let us compute mod phi. Mod phi of e1 plus e2 plus e3, let us compute. Sorry, e1 plus e2 plus e3 by 3, that is the point, right? If I just take e1 plus e2 plus e3, it is not a point of mod k. So this divided by this one, this is fairy center, right? Because if this goes, e1 plus e2, both of them go to e1. So it is twice e1 plus e2 goes to e2 divided by 3. So this is not a fairy center at all. You see, e1 plus e2 by 2 is the fairy center of delta 1. It is not equal to e1, okay? d of v would have taken it to you to the fairy center by definition, okay? So these two are not equal. Mod phi is not the sd of that. So this is what you have to be careful. More generally, I just want to tell you that there are other kinds of subdivisions, okay? Normally one writes k prime as a subdivision, okay? What is the property? The basic property what we observed for fairy center subdivision is taken as a definition. Namely, for every simplex f prime of k prime, the new simplex, there is a unique simplex f of k which contains, you know, mod f contains f prime. So it is much weaker than saying that in the case of, in the case of fairy center subdivision, all the entire vertex set was contained inside mod f. So point wise, f prime must be inside, all the points of f prime must be inside f. They should not go away. Then the inclusion map on the vertices, k prime into mod k because this inclusion is there. When you extend it linearly, it makes sense that must give you a homomorphism. So that is the definition of arbitrary subdivision in which instead of say 0 to 1 is the interval, you do not have to take half, you take a 0 to 1 third, 1 third to 1. 0 to 1 third and then maybe you do not want to even take 2 third, that will be too regular. 1 third, 1 by 2, 5 by 6, 1. Any kind of subdivision you can take in the case of interval, you just write down in the increasing order, then you do not have to write what are the simplexes. That is the convention we are following in analysis. But underlying that what we have is we have introduced the vertices, you have to introduce what are the cells, what are the one simplexes you have to tell. So that is clear in the case of interval because just take the order. But when you go to triangle, you will have to tell which are the, just declaring the vertices is not enough, right. Whereas in the case of bare centric subdivision, it is automatic. We know there is one single formula. Somebody takes bare centric subdivision of simple complex, somebody else takes it, both of them will get the identical result. So in general things we will not study them carefully, you are not studying them deeply. But in the exercises later on, I have included a few things about them because they are also useful when you are studying different kinds of problems, that is all. All right. So let us stop here and next time we will do what is the meaning of final subdivisions.