 So, we shall prove some existence theorem here, this kind of things you must have used several times for the interval, take an open interval, take a closed interval and take an open covering, then there is a final subdivision such that each interval is contained in one of the open sets. So, that is you can term as subdivision finer than the open covering. So, this is what we are going to now generalize to simple shell complexes and barycentric subdivisions. Let K be any simple shell complex. First let me introduce these notations which will be quite useful elsewhere also. Let us look at all the points inside mod K, which can be joined by a single line segment to this v vertex. So, that is going to be a star shaped set. So, we are going to define this star v except that it is going to be the open star. Star of v is defined all points alpha and mod K such that alpha when evaluated at v is not 0 that is all. In particular you know the vertex v is identified with the map which takes one at v and zero at everywhere all our other vertices. So, vertex v is there in this star v. So, if alpha v is not 0 t times alpha plus 1 minus t times v this entire line will be also inside will have say beta of that will be also not 0. So, all of them will be inside the star v. In any case because it is defined by a open condition something is not 0, one coordinate is not 0. Those coordinate functions are continuous functions after all. The star v is an open subset of mod K and it will contain all the points if you vary the vertices v because for each alpha somewhere alpha will be non-zero. So, it will belong to one of the stars in any case. So, this is an open cover for mod K. A cover is finer than another cover which just means that every member in the first cover must be contained in some member in the second cover. This is a general topological notion ok. We say a simplicial complex itself is finer than an open covering u of the mod K. If these star v this is another open covering for mod K this must be finer than the given covering u ok. So, instead of coverings we are defining K itself is finer than u which is just a short terminology to tell that star v is finer than u ok. Just it means that for each v there is some u inside this curly u such as star v is contained in the u. Every member in this in this covering must be contained in some member in u. Now we take the K special case name K is finite in particular the vertex set is also finite ok. Then look at the definition of mod K as a subspace subspace of i raise to v or you can just write as v is finite i raise to some n where n is a number of vertices in k ok it is an Euclidean space. We can take the restriction of Euclidean metric ok on this i cross i cross i n times ok namely distance between x and y is just square root of the sum of the squares of the differences of x i and y i x i minus y where is the Euclidean distance alright. The two topologies coincide because K is finite this is what you should recall only when K is infinite we are giving a finer topology on mod K even if you define that way this is going to be the same topology that is the good idea. Special property of this metric is ok is the following namely its linear metric in some sense what is that some sense restricted to each mod F it is given by a norm you see if you take i power v it is not a vector space mod K is far from that way but inside every inside every inside K you know it is a linear metric restricted to each mod F it is linear in the sense that if you take if you say some t times alpha and t times beta the distance between them is will be more t times the distance between alpha and beta ok. So it is given by the norm there because mod F has affine structure so you can take the affine linear space there and talk about the distance between two points is same thing as norm of the difference ok so that is that is very easily verified because this is the Euclidean metric. I am going to use that in the sense that distance between some multiple t times alpha t times beta is more t will come out so that is why that is what I can I can use here. The first lemma is elementary calculus kind of thing here namely K is a finite simple shell complex I repeat it D is the metric that we have taken ok any other metric which has linear property will also do the same thing that is why I have put it as linear metric then ok D is a linear metric on the bare centric subdivision also and for any F prime belonging to SD of K ok SD of K F prime is SD of K such that F prime is contained inside mod F is F prime is some finite subset it is all I am saying F prime contained in mod F where F itself is a simplex q simplex we have the following inequality of diameters. The diameter of F prime will be less than or equal to q into q plus q divided by q plus 1 time diameter of F. What is q? q is a dimension of the F here. So, take a simplex of whatever size inside SD of K that F must be contained inside mod F that means it is a dimension will be also less than or equal to that right will follow because of the chain the chain condition is there no. So, where F is belong F belongs to K it is a q simplex then we have this inequality. So, you can see that the diameter of the smaller things after all is controlled by this factor this is definitely a number which is smaller than 1. So, of the diameter of F at least this much ok. For example, if you take q equal to 2 namely a triangle then this will say it will be two third two third each each each of the one of the six triangles that you have divided into triangle into will have diameter less than equal to two third of the original diameter. If q is 1 namely the simplex namely the one simplex then what if this one is 1 by 2 is actually equal you can see right. Take a one simplex what are the one simplex of the sub complex SD of the of the marisinter sub division 0 to half and half to 1. So, each of them has length half whereas the diameter original diameter is 1. So, that is the meaning of this one ok I have given you examples for simple cases, but in general this what it is. So, having stated this one this can be left as an exercise completely, but since these kind of things are not done even in elementary, linear algebra or calculus or even in topology. So, let me do this here ok it is not all that easy not completely easy ok. So, let us have a notation F equal to V naught V 1 V 2 these are the vertices for any point alpha and mod F ok. We can think of this as a convex combination of T i V i this is a definition by alpha equal to T i V i summation T i equal to 1 and 0 is equal to 1 ok. Now, take any other beta inside F then we have distance between alpha and beta is distance between summation T i V i and beta. I can write beta also summation some S j V j, but I am just keeping it for a while ok. This distance the same thing as distance between T i V i. So, you can take summation T i it is one just multiply by that one for formally ok. Now, you have two summations here ok and this summation this distance is same thing as distance between norm of this minus this. So, that is what I am using that this distance is given by the norm because both of them are inside models ok. So, what is this? I take norm of this summation minus this summation which can rewrite it as T i times V i minus beta here each times beta right. So, T i times V i minus beta summation norm of the sum ok is less than equal to sum of the modulations. These are the norms again V i is our vertices vector beta is also vector. So, norm of summation less than equal to summation of triangle inequality m v s in this. T i comes out of the norm because T i's are all non-negative ok. So, it is norm of V i minus beta. But now this is summation T i of distance between V i and beta ok right very simple computation. Now, how to use this one? Therefore, distance between alpha and beta is less than or equal to supremum of all these distances. Take the maximum this final supremum is same thing as maximum. So, I put a maximum for each of them and take T i summation T i will be 1. So, it will be less than equal to that. So, I put maximum this will be less or less than no. If this is already less than equal to this one it will be less than and we put each D i V i by the maximum that summation T i is 1. So, now it become maximum, maximum is just supremum ok. What is the meaning of this one geometrically? Take any any simplex like this and take any point beta and any point alpha the distance between alpha and beta is smaller than the distance between beta and any one of the vertices. Take the maximum of that or just distance between go all the way to one of the vertices. So, one point I have kept fixed the other point I have replaced by vertices. Now, I want to do the same thing upright this also a point after all now we do it for beta also reverse this you know this is symmetric relation. So, use this one what I get is since this is true for all alpha and beta you can put beta equal to V j and take alpha equal to beta in this one distance between beta and V j will less than or equal to supremum of distance between V i and V j. This is true for all beta all V j all j's. So, what does it mean distance between any point in the vertex and vertices is smaller than the maximum of the what are these these are edges length of edges of the simplex. Therefore, if you maximize the left hand side what do you get this is for all V j's is true each of them is true see from here you can go to this inequality alpha beta with left hand equal to this one and each of them less than equal to this one. So, distance between the alpha and beta from 12 and 13 is less than or supremum of this term V j. Now, take the maximum here if for everything it is less than equal to this one the maximum will be also less than equal to this one or supremum this supremum is nothing but diameter of f by definition. So, the diameter of f is actually realized in the length of one of the sides maybe if I stated this right in the beginning you would have understood it very clearly. So, this is easy to see for triangles easy means what if you want to write down the right full proof you will be like this but I have done it for any simplex now. Now, let U 1 and U 2 be any two vertices of f prime simply say f prime in S D of f without clause of generality we may write U 1 by changing the availability U and S V naught plus V 1 plus V k divided by k plus 1 where this is a k simplex and U V 2 is V naught plus V 1 plus V m m plus 1 simply. So, we can assume m is bigger than k or k is bigger than m. So, I will assume k is less than m less than m I want these two are distinct vertices not the same. This is just by the labeling I will assume. Here some of them are repeated k V naught when V k but that is all not necessary you could have you could have arranged this say. Then distance between U 1 and U 2 will be less than supremum of distance between 1 I am taking U 2 is this one U 2 I am keeping fixed head but I am rearing all the this one here supremum of V S V S are these things 1 less than or equal to 0 less than or equal to S less than or equal to k supremum this is maximum of this one. Distance between V S and U 2 so, this is by our earlier observation similar to that one I do not have to do it again. But this I am going to do carefully now distance between V S and U 2 just write down it is the norm of V S minus what is U 2? So, U 2 it has this expression I will use this 0 to m V r divided by m plus 1. So, I pull out m plus 1 so, multiply V S by m plus 1 take minus 0 to m V r. So, there are m plus 1 quantities here right the beauty here is that one of the V S at least will be equal to V r right. So, less than equal to 1 dot m plus 1 0 to m V r minus V r V S is fixed head okay r is ranging here V r. So, you will get at most you know m of them. So, 0 less than 1 m dot m plus 1 supremum of d V r V S 0 less than or equal to r. Okay. So, that is how your m divided by m plus 1 will come there are m of them take the supremum for all of them now distance. But how many are there only m of them is there one of them is 0 that is the beauty. Hence distance between U 1 and U 2 it starts from U 1 and U 2 here it is supremum of these things when each of them is less than or equal to this one. So, U 2 itself is less than or equal to this one but this m is smaller than q right because they are all inside f. So, if m is smaller than q smaller or equal to m divided by m plus 1 is smaller than q divided by q plus 1. Okay. So, this is monotonically increasing. So, all of them are less than or equal to q divided by q plus 1. So, this is a rough estimate. So, we have proved this statement. Now, take k as a finite superficial complex you will be any open covering for mod k. Then if you divide k sufficiently number of times. So, S d of n is nothing but S d of k, S d of that, S d of that and so on k times. That will be finer than this open covering. Okay. So, this S d of n is iterated subdivision by definition S d of S d of n. You can define S d 0 S is k itself. Okay. So, here I am using the standard result on any compact set. We have a open covering compact metric space. It is a metric space also. Right. It is a Lebesgue number. The Lebesgue number is such that if you take any ball of diameter and any set of diameter less than or equal to c, it will be contained in one of the open set. Okay. So, start with an open covering. Choose c to be the Lebesgue number for that covering. Then choose this capital in such that for all f prime in this one, we have diameter of f prime less than c by 2. So, why this is possible? If you divide once, it will be, see, n is a finite, k is a finite simplicity complex. So, it has one dimension. The dimension you can assume is q. Then diameter of the first subdivision will be less than q divided by n. If you repeat it, it will be q into q divided by q plus q divided by q plus 1 raised to n times. But this number, the powers of this one goes to 0. Okay. This can be made less than c by 2 by choosing little n to be sufficiently large. So, that is larger than this existence. Once this smaller diameter is smaller, then what happens to any star? A star is contained in the union of all these simplexes which have one common vertex. Right? So, diameter of the star will be at the most twice that. That is why I have put c by 2. Once diameter of each of them is less than c, the diameter of the star will be less than c by 2. Therefore, every star which is of diameter c will be contained inside one of the open sets. Okay? Subdivisions give the same topological information on the original triangle or space because the homeomorphism type is the same. Right? You do not change. In some way, you may say that they give same common information as well, but you have to be careful there. Based on the fact that two partitions of an interval have a common refinement. So, common information is also not lost, you may say. Okay? We can ask the question. Given any two triangulations given K2 of a same space X, let us assume that X is compact. Are there subdivisions K1 prime and K2 prime of Ki such that these subdivisions are isomorphic? So, this is true for subdivisions of an interval. If you have two different divisions, you can take a common refinement. Right? So, you can ask similar question for the subdivisions of some weird subdivision you have taken and you can ask this question. But these things are questions for arbitrary subdivisions which I have not discussed. Okay? I am just telling you some story here that is all. So, two given triangulations space are said to be combinatorially equivalent if there is a common subdivision to both of them. So, you can reformulate this question, claiming that are they combinatorially equivalent? So, this simple question was solved by Milner sometimes in 1961. He cooked up a new invariant for homomorphism, invariant for this one and showed that there are two simple complex is different structures for this one. They have different invariants. So, you cannot subdivide them to get a same isomorphic subdivisions. But for manifold less than record dimension 3, this is a deep theorem which has a positive answer. All these things are a big branch of mathematics. I am just trying to tell you the awareness of the lot of mathematics under these things. Okay? So, I have listed a few exercises. You can go through this one as I keep telling you these things will be again recirculated to you separately. So, you do not have to know all this. But I will just introduce one more concept here. You are not going to study this one very deeply right now. But it is a very, very important concept. What he said is combinatorial concept goes back to Euler. So, it is called Euler characteristic. What is it? Look at a simple complex. Just count the number of vertices, number of edges, number of two simple axis, three simple axis and so on. So, that is called Fy of k. Fy of k is the number of i simple axis of k. Take the alternative sum that is called the Euler characteristic of k. So, this is some integer, positive negative integer one does not know. Okay? This very interesting geometric content is there. Okay? And it has lots and lots of applications and many different and very, very serious formulation of this one. Simple ideas from here. But I must have heard of Atyas in the index here, you know Riemann-Hourege formulas and various, they are all involved with this one. So, when time permits slowly, we will develop this one. There are some simple exercises here based on the Euler characteristic. That is why I have introduced that one here. When time permits, we will do all these things in more detail. Okay? So, we will stop here.