 Today's topic is Geometric Realization of an Abstract Simplicial Complex. Before I begin this one, let me take care of a few more examples that I have listed already, but could not cover. This example we have seen, this example we have seen, but this is also an important example. If you start with a simplicial complex, look at all phases of dimension less than equal to one particular number, say Q, less than or equal to Q. Automatically all the phases of whatever you have taken will be also there. Therefore this will be a sub complex. This sub complex is called the Q dimensional skeleton of K. And the notation is K power bracket Q. For example, when you take F as a simple X, its Q minus one dimensional complex is the previous example, the boundary of F. Only F itself will be omitted. All its proper subsets will be there. Because F is the only Q cell, Q simplex and that will be omitted because I am taking only Q minus one skeleton. So, this example generalizes the previous example that we had here. If you have any simplicial complex, K0, the zero dimensional simplex will be just a set of vertices. The one dimensional skeleton could also be set of vertices if there are no one edges, if there are no one phases. That means the original simplex complex itself is just set of vertices. So, you have to understand this notation carefully. So, here it should be strictly less than Q, right? No, Q dimensional simplex complex will include Q. This Q minus one skeleton, the Q skeleton will include Q also. Okay, zero dimensional skeleton will include only zero dimensional simplex. That means only vertices will be there. One dimensional skeleton will have all the so called edges. The two dimensional skeleton will have triangles and of course edges and vertices. But suppose there are no triangles at all, okay, to begin with, then the two dimensional skeleton, three dimensional skeleton, four dimensional skeleton, whatever you take, it will be just the one dimensional skeleton only. In particular, a Q dimensional skeleton is a sub complex of a Q plus one dimensional skeleton and so on. Okay, I am not saying anything more than that, whatever immediately implies from the definition here, okay? Here is something constructing a new simplex complex out of the world, not just taking sub complex and so on. So, take K1 and K2 two simplex complexes whose vertex sets are V1 and V2 and simplices are S1 and S2, okay? We want to define the join of K1 and K2, okay? This join later on will become a topological join when you take geometric realizations. So, this is not just concocted something, but in principle this is much simpler than the topological join. So, join of K1 and K2 is defined over the set of vertices which is disjoint union of V1 and V2 and nothing more. You have to take all the vertices of V1 and all the vertices of V2 as a disjoint union, okay? Then what is the set of simplices? That is what I have to define. The set of simplices will also take every simplex in K1, namely S1 and every simplex in K2 and take their union. So, F1 union F2 where Fy is run over Si, including the empty set, okay? Remember, Si is included empty set also. In particular, all F1 belonging to S1, they will be there. F2 belong to F2 will be also there. Then their union, disjoint union, because these are treated as disjoint, V1 and V2 disjoint subset, this will also disjoint union, okay? So, this is the simple shell complex on K1 called K1 star K2, the join of K1 with K2. Now, by the very definition, K1 star K2 is K2 star K1. So, this operation itself is, you know, commutative operation because V2 is joined union or V1 disjoint union, they are the same. So, here also. So, that is an obvious thing. Now, if K2 is empty, then what do you get? You will get just V1 here and here you will get S1, okay? So, then you will just get K1 and this is one of the reasons why the topological join X star Y is also defined to be X when Y is empty, okay? If you recall, Y, I have taken that previously. In any case, if K2 is empty, then this is just K1 or K1 is empty, then this is just K2. Now, if K2 is a singleton set, then this becomes very interesting. Suppose, it is a singleton set, V, that means one vertex and only two simplest is empty set and singleton, okay? So, it is a point, it is a single vertex, simple shell complex. Then the join will be what? You can just simply write instead of brackets and all that. It is just the cone over K1. So, this is a definition, okay? The cone over K1. This is a special case, okay? So, what are the, let us have a look at it. What are the simplices in K1? All simplices in K1 will be there. This extra vertex will be there, okay? And for every simplex here, you put this extra vertex also. If this is a K simplex here, you will get a K plus 1 simplex by putting extra vertex, okay? All those things will be there. So, that is the cone over K1, okay? So, this is special case of the join, all right? Similarly, I could have defined the suspension also here. Suspension will be taking double cone, okay? That means I have to take K2 to be just two vertices and no edges, nothing, just two vertices, okay? And then perform this operation, K1 star K2. That would be a suspension of K1. So, when I want it, I will recall it. There is no need to worry about that. One of the important things which is very obvious here is that the dimension of K1 and K2 is dimension of K1 plus dimension of K2 plus 1. Why did this right hand side is defined? Of course, if anyone of them is infinite, this will be also infinite. In that sense, this equality makes sense, okay? So, let us see how. Look at the top dimension cell here, top dimension simplex here. See that it has, say, it will have dimension of K1 plus 1 elements, right? This will have dimension of K2 plus 1 elements. When you take the union, that will have this plus this plus 1 plus 1. So, this plus this plus 2 elements in it. Therefore, its dimension will be this plus this plus 1, 1 less your take. So, that will be there already. So, dimension of the left hand side has to be at least this much. But it is also equal because if there is any simplex here, okay? It will be made or disjoint enough two things here. Then its dimension has to be dimension of this plus dimension of this plus 1. It has to be less than or equal to 1. So, there will be corresponding simplex is here of the, at least of that dimension, okay? So, if you take, for example, a vertex here and a vertex there, they are zero dimensional. But the joint F1 union F2, that will become one dimensional, 0 plus 0 plus 1. So, from each simplex wise, you can verify this identity. There is one interesting example here which is of importance in many, many, you know, other kinds of mathematics, not within spiritual complexes, okay? So, within topology, of course, yeah, this is called nerve of a covering as it is, as it says. It has something to do with a covering of a topological space. So, that you be a collection of non-empty subsets of a non-empty set, the state or everything non-empty. So, you have subsets of X, each of them is non-empty and you have at least one element in it, one member. We get a simplex complex KU by taking U as the set of vertices. This curly U becomes a set of vertices and finite subsets U and U to UK, Ui is inside U. Not all of them but with the property that their intersection is non-empty, okay? So, if you take empty set here, you would have been in trouble. That is why you would like to have first of all each Ui non-empty, then the intersection must be also non-empty. That is the condition. Then only you will declare this as a simplex. For example, you take two open sets, they are vertices. Will there be an edge provided U and intersection U2 is non-empty? If it is empty, then you will not put any edge there, okay? So, if you take a sub-collection here, that will be also non-empty. Therefore, I mean sub-collection, intersection will be non-empty. Therefore, this is automatically a simplex complex. So, this simplex complex is called narrow of U, okay? When U happens to be an open cover for a topological space, this is defined for any collection of sets. Now, suppose this is an open cover for a topological space X. Its narrow KU plays a central role in topological dimension theory. The dimension of the narrow of a covering is what is used there. Then, ultimately, for an appropriate covering, the dimension of the narrow of the covering will become the dimension of the manifold itself. Okay? So, this is there in dimension theory. I cannot explain it anymore. If you want, you can see this very fantastic book, for every Swalman. Okay? I have also included some exercises from chapter 3 of Spaniard. You can have a look at them also. For us, there is another important thing which can be used in usually, but again, it is not in this course. Namely, in the study of check homology of a space. What is called a check homology? There also it is used. Okay? So, this example is very useful, but in this course, we are not going to use it. Similarly, the, you know, simplicial complexes are being used in many other mathematics, especially in combinatorial algebra and combinatorics itself. And as I have told you, computer scientists use it quite a bit. So, I will give you one more example, which is somewhat dual to this, this construction, but this is more combinatorial, there is no topology here. Okay? You start with any vertex set, okay, of, let V be a, the set of k subsets of a k plus 1. Start with any set. Set is fixed set, but now the set of vertices is k subsets of a k plus n set. We should have something more known. So, let us fix this k plus n. Okay? Then you take k subsets of that one. For example, k plus n may be 3 and then you are only taking all the 2 subsets, 1, 2, 2, 3 and 1, 3. So, if there is 10, okay, then you are taking say k equal to 3 and k plus n is 10. Then what you are taking, 1, 2, 3, 2, 3, 4, like that, all the 3 subsets you are taking. So, that is the set of vertices of this new simplification complex that I am going to define. Okay? Then what are the simplices? The simplices will be subsets v naught, v 1, v r, where v i's are inside v. Each v i's are subsets now, k subsets. Remember that. Okay? Now you take a k subset, r of them, r plus 1 of them, such that intersection is empty, disjoint. If you have taken 1, 2, 3, you cannot take 1, 2, 4. 1, 2, 3, you can take 3, 4, 5. Right? That is why I am told that this is somewhat dual to the previous example. There we took subsets which are, subsets which intersect. Here we are taking v i intersecting v j's empty for each i not equal to j, pairwise disjoint subsets, k subsets of a k plus n set. Okay? If you take a subfamily here, that will also satisfy the same property. Therefore, it will be there. Therefore, k is a simplification complex. Okay? So, this thing arises in pressure's conjecture in combinatorics. And if you are interested in that kind of mathematics, this Loa's paper is a very fine paper. You can read that, 1978. So, let me go to now geometric realization. Okay? This begins with the basic idea of graphs. How do you do graphs? But you have to be careful with whatever you have learnt in graphs, real or not here properly. Okay? So, graphs, you are familiar with drawing them on a piece of paper. When you do that, you get a subspace of R2. Right? In graph theory, the topology of subspace is not used so much. It is a combinatorics. Namely, whether a vertex is an incident at edge and what edge is used to join two observatory systems or that kind of interrelation between edge and vertices is given a lot of importance there. Okay? But you get a topological space there, namely, so start with two vertices and joined by a line. Okay? If V1, V2 is an edge in the graph, that is what you do. Now, consider a one-dimensional simple complex. Okay? One-dimensional means what? There are vertices and there are edges. Okay? One-dimensional simple axis are there, nothing more than that. Let me just look, concentrate on that. To keep the discussion simple, assume that it is a finite also, namely, number of vertices is finite. We can then select as many distinct points of the Euclidean space. Okay? Try to do it in R2. May not be possible. Let us see. Okay? Representing distinct vertices of this case. Okay? We can now join those pairs of vertices for which V i, V j is an edge, is a phase of this capital K. That means V i, V j should appear inside S. Then only we will join them. The only thing that we want to ensure is we would like to have an independent state for this V i, V j. Suppose you join V k and V l and the intersection of these two, you know, they may intersect in the interior of one of those points. I don't want that. Okay? That is not allowed even when you are drawing a graph and so on. So that poses a problem that not all graphs are embeddable in R2, you see. You must, if you have known such a thing. So but you can go to R3, R4, then it is possible to choose such that points have been chosen such that these edges don't intersect. Okay? The line segments should not intersect. You complete all these line segments. What you get is a geometric object, a subspace of R2. So there is a topology on that. So this is just heuristic idea. We want to make this one more rigorous. Okay? So let us see how. Let us go step by step. Only snag, I told you, is that two line segments, if you have chosen vertices arbitrarily, they may intersect. If they intersect only at the vertices, then there is no problem. But if they intersect in between, that point of intersection should be also declared as a vertex to be fair, to be fair to graph theories. But it is not a vertex in our typical complex. There is no such vertex. That is why you don't want them to be intersecting. Okay? So difference between representing graph and representation of K is only in that fact that we have used only straight lines inside Rn. Okay? So this is not such a great issue. If they intersect, you can slightly bend it so that they don't intersect and get away with that. Namely, take a curve instead of a straight line. Alright? Inside R2, even if you take a curve, it may not be possible. For example, you know that the complete graph on five vertices cannot be drawn on R2. Okay? So that problem will still be there. So in any case, given a K, we have been able to assign a collection of segments in an intermediate space by increasing the dimension of the intermediate. Maybe go to R3, R4 and so on. So after all, this is, I have taken, started with the finite set. Suppose the finite set has n elements, then I can definitely do it in Rn plus 1. Absolutely no problem. So Rn itself I can do. Right? We have done it already. Namely, I can take the standard, what are they? The standard basis elements, e1, e2, en, e3 and so on, any of them and we done it. Then the entire simplicial complex on that one is there already. All the sub complexes will be also there. Over. Alright? So this brings back, brings us back immediately to the geometric and simple as that we have introduced. In a sense, whatever I have told heuristically has already solved this problem. At least for the case when K is finite. Okay? Right? So we are going to do it independent of choice of elements and so on. Some kind of universality should be brought in. So let us go again slowly a little bit. So here are some examples of what is the representation of a graph and a simplex and so on. The first example here is an example of a graph but it is not a simplicial complex. Why? If it is a simplicial complex, these two round edges must have been there. But given two vertices, you can have only one single edge there, v1, v2. So this is not a graph, this is graph fine for some people but this is not a simplicial complex. This is a simplicial complex. This is a one-dimensional simplicial complex with 3 plus 2, 5 vertices, 1, 2, 3, 4, 5 edges. 1, 2, 3, 4, 5 edges. Right? Okay? This is a one simplex and that is a two simplex and this is the three simplex, the tetrahedral. All right? So in general a simplicial complex will be built out of this, this, this and more higher dimensional things and so on. Okay? We want to define the, we have already defined abstract simplicial complex independent of any pictures or any geometry but we want to come back to the geometry and this process of coming back there must be some canonicalness, there should not be any ambiguity, there should not be any choice, you choose something or he chooses something and so on. That is the kind of thing we want to do. Okay? So let us see how, what are the kind of problems. If k is a two-dimensional simplices also say like v1, v2, v3, k is such a simplex. We then take care that the three vertices are not chosen on a straight line. Right? So this then enables us to fill up the triangle formed by three vertices. Right? So we do this to all two simplexes but again the two simplexes, several two simplexes should not intersect at all. So that also has to be ensured. We have to ensure that two distinct triangles do not overlap. This gives us a subspace of Euclidean space which is the union of number of triangles, lines, etc. Okay? So that is what I have already shown you here. All right? So far whatever idea we have it is possible to avoid some of these ambiguities but there will be some choice after all why you have chosen your definition you may say but my definition has been chosen so that it brings a unification among just all such ideas. So we go back to the affine structure of Rn. All right? In Euclidean space you forget about the origin just think of this as a geometric object moment two points are given the line segment is well defined. It is consisting of t times 3 plus 1 minus t times q the convex hull of these two points. If three non-colonial points are given they define a triangle. Okay? You can say three points always define a triangle but if they are collinear it will be a degenerate triangle so I do not want that. So what is the set of all points inside a triangle? Alpha p plus beta p plus gamma r alpha beta gamma between 0 and 1 some total must be 1. So I am just recalling whatever we did in affine geometry. Observe that there is a 1-1 correspondence between the set of points t1, t2, t3 inside i3 because these are all elements of i 0, 1 and so three of them will be in i power 3 but summation ti is 1. Okay? So this is this itself is our standard two simple x the joint of e1, e2 and e3 the the convex hull of e1, e2 and e3. Thus a point in the triangle can be thought of as a function now this t function the t1, t2, t3 there are three coordinates of the function. So it is a function from 1, 2, 3 into i. Okay? So t is a function 1, 2, 3 into i t1, t2, t3 combine it with the vertex is here v1, v2, vqr or v1, v2, v3 you get a point of the simplicial complex the triangle here. Right? So t1 plus t2 plus t3 must be 1 that is the condition it is not any function. So these are some of the ideas which go back to algorithmic definition of geometric realization. Okay? Which will bring a remove all ambiguities involved and bring some to reality. So I have enough given you enough motivation for this definition. Okay? So let me just give you a definition k is a simplicial complex v is a set of vertices and s is the simplicis. Then mod k will denote the set of all functions on the vertex set taking values in the interval 0, 1 such that this notation support alpha means what? All points of v wherein alpha v is not 0 some points of v may go to 0. Right? So this itself is a subset of v it should be a simplex in k that means this set support of alpha must be inside s. Okay? There is a condition on alpha I am not taking all functions this must be good. The second condition is there namely some total of alpha v must be 1 as observed previously. The support must be a simplex the sum total must be 1. The support is a simplex just means that support is finite. I am not assuming v is finite now. The support is simplex automatically implies this is a finite set when you take the sum total is the finite sum because all other v and v alpha v will be 0. If it is not 0 it will be inside this one support of alpha. So second thing makes sense because this left hand side is a finite sum it will be equal to 1. So it is a condition. Look at all these that set is called mod k. So in the simplest case is this is a subset of i raised to v satisfying this condition this condition. So it is a closed subset of i raised to v. i raised to v means. Hello sir. i cross i cross i v times. Hello sir. Yeah. Yeah this alpha are this supposed to be some kind of coordinate function type thing. I have already told you that I have already told you here t1, t2, t3 right. Okay. Any function from v to i is an element of i cross i cross i raised to v v times. Okay. So you should know that product is same thing as functions. Cartesian product is same thing as functions where? On the indexing side. What is the value of the function? It is the ith coordinate. Yes. Okay. So this elementary point set thing that you have known used in point set ecology. Okay. Here I have the extra condition coordinates must be summed up to 1 and each point as finite coordinate only finitely many coordinates are not 0. Okay. So that is the meaning of it. Okay. Okay. So this is the way we are looking at now a simplex, a triangle, an edge. An edge is a function, elements of an edge is a function taking values defined on the vertex set, taking values inside 0, 1. Some total must be 1. So that is the meaning of points between these edges. So this way we are not using any kind of picture but we are still using the algebra and topology and everything of i the interval is 0, 1. So those things are still there. Okay. We should understand that. For each phase s, take of f, phase of the simplex, let us introduce these two notations which will be used again and again. The mod f is or bracket f I have put for just the sake later on I will make it mod f. Okay. Just not to confuse with this mod k. So all alpha belong to mod k such that alpha v is 0 for every v outside f. Okay. So that is mod f. That means the support of v must be contained in f. May not be equal to whole of f. If it is whole of f then this is the open simplex f. This is the definition. Open f. This is closed f. All alpha, so alpha v is 0 for every v outside k minus f. Okay. There is some problem here and not equal to 0 I have put here. What is this? Why? These two things look the same. No. Support of alpha equal to f. Okay. So there is some problem in the first part here. For every v outside it is 0 but not equal to 0 for v inside f. That is what I have to write here. It is same thing as the support of alpha equal to f. This is correct. Okay. And another notation, the boundary of f just means that you know this bracket f minus simplef. That means those which are contained in here but at least one of the things must be 0 already. That means it is proper. So only those things are boundary of f. Okay. For every f it must be non-zero is this one. Maybe at least one of them extra it may be 0 other than k minus f. One of the vertices of f also advantage. Such things are in the boundary of f. Okay. This f and this open f are called closed simplex and an open simplex. Corresponding to a simplex, abstract simplex, we have a closed simplex and an open simplex. Okay. Respectively and boundary is called the boundary of f. At this stage these are only names. We shall see the relevance of the names closed, open, etc. When you actually see what is the topology we are going to take. Okay. So I will stop here and continue in the next module.