 So, today we will start a new chapter namely applications of homology. Here what we will do we will do a few popular applications of homology. Some of these results one can prove in different methods also and some of them you have already seen in part one without using the homology theory. So, having said that the idea of proving these things is to emphasize how homology theory actually helps in many problem solving in anthropology especially in algebraic anthropology. So, we begin with the famous Brouwer's fixed point theorem and then go on to prove the left shift fixed point theorem which is an improvement on Brouwer's fixed point theorem. We then prove what are called as Theriball theorem, you know Jordan Brouwer separation theorem, Jordan Brouwer invariance of domain. So, these are all very very popular and when these things were proved the algebraic topology became extremely popular because of such a results ok. So, we begin with a lemma which must be it has reminiscent of similar things that we have done in part one using the fundamental theorem. Now we are using the homology group ok let A be a retract of X then H star of A will be retract of H star of X that is H star of A is a direct summandum H star of X ok when in a topological space A and X are topological space is A is a subspace of X, A is a retract of X you do not have all that all that you have is a continuous function from X to A which is identity on A ok. R is the continuous function from X to A, I is the inclusion map to take I and then follow by R that is the identity of A, when you pass to the homology it will give you that H star of A is a retract of H star of A that means there will be a homomorphism from H star of X to H star of A what is it is nothing but R you know R itself induces a homomorphism and that will have this property ok. So, R from X to A is a retraction R star from X star to X star has is a retraction because R star composite eta star which is inclusion map is R composite eta star is the identity of I star. So, that is the meaning of that R star is a retraction. And whenever you have retraction of abelian groups the subgroup H star of A becomes a direct sum and this is a general theory for any abelian group ok. For any n greater than or equal to 0 SN is not a retract of the interest one. So, this is it sounds like a negative result, but this is a very very positive result ok. Remember this was the starting point of our in a simple approximation and what is called as Sperner lemma we had proved this one, Broward's theorem and so on ok. So, here it comes very easily, no Sperner lemma and so on, but of course we have to use the homology whatever big thing that we have done namely the simplest thing that we have done was the computation of Hn of SN and computation of Hn of dn plus 1. All Hn and not equal to 0 of dn plus 1 are 0 because this is contractable. Hn of SN is infinite cyclic ok. If SN is a retract of dn plus 1 then Hn of SN would have been a subgroup of Hn of dn plus 1, but Hn of dn plus 1 is 0 and this is not 0, but that is a group. So, the functoriality of the homology is strongly used here in this theorem you see from r composite i r composite eta which is happening in the topology we have got the corresponding r star composite eta star happening in the group theory in the homology groups ok. Next, so this thing I already told you I will repeat it in part 1 we have proved the above result in a different way using simple approximation and Sperner lemma. We have also seen that the above result is actually equivalent to Brauer 6 point theorem. The present proof is obviously shorter though it uses big machine of homology ok. So, we shall not bother about proving Brauer 6 point theorem. Once you have got this one you know how to show Brauer 6 point theorem namely any continuous may have from dn plus 1 to dn plus 1 has a fixed point ok. So, I am not going to discuss this one anymore because that is that part is done very thoroughly in part 1. So, let us directly do the left change fixed point theorem as a consequence you will get a Brauer 6 point theorem also. Therefore, there is no need to again you know spend any time on Brauer 6 point theorem separately. So, our next generalization is an application is a generalization of this result on any compact polyhedron dn is a compact polyhedron. So, it will be applicable there also. Now, we shall have some hypothesis on the map itself ok. On the polyhedron only compactness is the assumption. On the map we have some assumption. Recall that given an endomorphism of finite type ok endomorphism of a finite type graded module graded module C, C or C dot or C star whatever I am just trying to see ok graded module tau. We define a left change number at tau as alternate some of the traces of each tau n ok. So, I am just recalling this one. Recall that if tau star denote the homomorphism induced on the homology groups of the chain compress then L tau is equal to L tau star this also you have done ok. So, this left change number will be used now in this left change fixed point theorem. For any continuous map f from x to x on a space x that has finitely generated homology group we now define L f the same thing as L of f star. The well-known left change fixed point theorem will be the following ok. X be a compact polyhedron. What means what? It is a finite seasonably complex k mod k is homomorphic treks ok or you can say that x can be cut down into pieces of simplex s. So, that the whole thing belongs to simplex complex ok. So, do not worry about the underlying the compact the simplest structure here just take the underlying topological space f from x to x, x is a continuous function. Then if L f is not 0 f must have fixed a point. In braver fixed point theorem there was no assumption on f except that f is continuous, but x was just dn the closed unit disk ok. So, here this axis any arbitrary simplexial complex mod k of mod k where k is a simplified simplexial complex. Suppose we have proved this one then how do you prove bra's fixed point theorem? So, if x is dn ok any map from x to x is homotopic to identity map. This L f is a homotopy invariant because it depends only on the homology groups. Therefore, L f will be equal to L of the identity map. Now L of the identity map is easy to compute because the only homology group of dn is H naught and H naught it is infinite cyclic. On infinite cyclic on H naught any map induces identity map ok because connected components go to connect same connected component there is only one connected component ok. Therefore, L of any continuous function is equal to 1. Now this says is L f is not 0 there is a fixed point. So, that is an easy proof of braver fixed point theorem from this theorem ok. So, let us go towards the proof of this itself now. So, this is what I have said here I will repeat it x is if x is dn then f is homotopic to identity map of dn because dn is contractible any two functions into dn are homotopic. Since L f is homotopy invariant follows that L f is L of id identity map. But identity map is nothing but Euler L of identity map is nothing but Euler characteristic of the space dn. Therefore, L of id is 1 since dn is contractible ok thus requirement of the theorem is satisfied. So, in conclusion we can say that f has a fixed point ok. So, this shows that left shift fixed point theorem is a generalization of braver fixed point theorem. Indeed we just have derived a stronger version of braver fixed point theorem namely any self map of a compact contractible polyhedron has a fixed point we never use that in its actual dn only contractibility we used ok. So, this is another intermediary result which follows from left shift fixed point theorem. So, let us see some more results also, but before that let us try to complete the proof of this. Given a compact polyhedron and a continuous self map suppose it has no fixed point then we want to show that L f is 0 some such thing right look out show L f is 0 fine. Now, this being a compact polyhedron that is the finite series of complex we have seen that there is always a linear metric ok. You can choose any linear metric and find an epsilon such that distance between x and fx is bigger than this epsilon ok for every x belong to x alright. So, why this is true? So, you have to use some topology here on a compact matrix space distance between x and fx is a continuous function ok real value to continuous function non-negative. So, it attains its minimum and attains minimum means that minimum will be more positive to take that positive number epsilon then everything else will be bigger than that that is it ok. So, k be a simple shell compact that triangulates x. Now, I will use a special for some particular chosen something it arbitrarily chosen I do not care, but I will fix some triangulation on x. Now, x is nothing but mod k ok. Now, I can choose go on subdividing the k as many times as possible ok as many times as required so that the mesh of k which is the by definition maximum of all the diameters of f where f rings over all the simplices is less than epsilon by 3. You can keep cutting down so that each simplex will be of diameter less than epsilon by 3 ok. So, this is also one of the things that we have done in part 1. All that you have to do is keep taking barycentric subdivision barycentric subdivision each time the diameter becomes a fixed constant times that is one that constant is less than one. So, if you repeat it several times then it will be less than any given number ok. So, we have so we have got a k which has this property. Now, I further subdivided it by barycentric subdivision. So, s t k I do not know how many times. So, what this s t k is I am calling it as l, but what it serves to is there is a simplicial approximation to f. f is a continuous function ok, its phi is a simplicial approximation to f alright. Any simplicial approximation is automatically namely mod phi is homotopypt f. Therefore, l of mod phi is the same thing as l of f. We want to show that l of f is 0, we can show that l of mod f is 0 ok. So, that is what we are trying to do now. Now, how do we how do we use this information all this information why all this was done then? So, the first thing we claim is a following. Take rho to be any simplex of l namely this s d of k. Let sigma belonging to k be such that rho is contained inside sigma. So, rho is a subdivision all simplex is in the subdivision are contained inside a larger one. They are they are part of the larger simplex which have been divided right divided divided divided and so on. So, each rho is contained in some sigma. Whenever this relation happens what we say is look at phi of sigma, see a phi of rho. Phi is a simplicial approximation to f phi rho intersection sigma will be empty. That means, rho was inside sigma now because f s no fixed points this entire rho has gone away far away. The entire rho of sigma is outside small sigma. So, this is a strong result that we want to prove now. So, that is because of the choice of this epsilon that we have chosen and then subdivision the mesh of k is less than epsilon by 3. So, let us see why. Suppose this is not true that would mean that there is a in the intersection there is a point. So, take x belonging to mod rho I mean underlying space of rho such that phi x belongs to mod sigma. So, that is such an x. Then both x and phi x will belong to mod sigma because phi rho is inside rho is already inside, this rho is already inside sigma. So, x and both phi x will be inside this mod sigma. Now, mod sigma has diameter less than epsilon by 3. Therefore, distance between x and mod phi x will be less than epsilon by 3. On the other hand, if you compute distance between phi x and f x, see what is phi x and f x? phi is a simple approximation to f x. So, wherever f x is phi x will be in the same same simple x and the whole diameter is as an epsilon by 3. Therefore, this is less than epsilon by 3. You can put equality if you like less than epsilon by 3. And hence distance between x and f x will be less than epsilon by 3 epsilon by 3 which is equal to 2 epsilon by 3 is less than epsilon which has the contradiction. Because no point has this property that x and f x are less than epsilon because of this property. Therefore, this property 30 this holds. Every simplex of S t of k under phi is thrown off the the mother simplex which contains it in k. All right. Now, we can use this one. The only property problem is that when you divide, subdivide, the chain complexes are different. Right? The chain complex of the the simplexial chain complex of the subdivision and simplexial chain complex or anything, they are different. When you have different chain complexes though you have a map, there is no trace defined. The trace is defined only for an endomorphism of the same module to itself and endomorphism not homomorphism. Right? So, that is the difficulty here and it has to be overcome somehow and there are different methods of doing that. So, I will give you two methods here which I like. Some other methods I do not like and I even doubt them. So, first method I use the C w chain complex associated with the simplexial complex. Take C k, it can be thought of as a C w chain complex. Okay? So, that is what I do. So, what what I do is since phi is a simplexial map, it is a cellular map also. Right? So, it follows that mod phi is a cellular map of the C w complex k. You see phi is a map from L to k, but what phi is a map from mod k to mod k and we have a cell C w structure there. Okay? And in that C w structure, mod phi is a cellular map. Any any cell of k dimension, we go inside k dimension cell only because it is given by a simplexial map there. Okay? So, therefore, let us look at alpha n from h n of k n relative to k n minus 1 to h n of k n relative to k n minus 1. So, I have same things here. These are what? These are the chain modules for the chain complex, C w chain complex. Okay? b, the homomorphism induced by mod phi restricted to each n, you have this one. I am denoting them by alpha n. If tau belongs to k, it is a generator that is meaning a oriented and simplex. Okay? It is generator here. It follows from this property. Okay? That if you take, you know, following this property that mod phi of tau intersection tau is empty. Right? tau, what is tau? It is one oriented simplex. Okay? In k, it follows that mod phi tau, see mod phi tau, what is tau? tau has been divided into several complex, several sub complexes consisting of say rho. Each rho is thrown out of tau. So, all of them will be thrown out of tau. So, mod phi tau intersection tau is empty. Okay? So, I am using this mod tau is a union of all rho here. This is rho, this is sub of that. All tau, take all rho which are contained here tau. So, that will form the union of, you enter tau. So, it will happen that mod phi tau, this is a typical, this is a continuous function. Mod phi tau makes sense. Intersection tau, mod tau is empty. These two are the subspaces of mod k. Okay? It is empty. But this will mean that when you come to, when you come here in this module, okay, what happens is alpha tau n, alpha n of tau may be in the combination of some simplicity here, but it will never have anything to do with tau. So, if you when express alpha n of tau as a linear combination of some n i, say some sigma i's and so on. The n i's will be 0 for this tau. The coefficient of n i will be this tau will be 0 because this is never involves. It has completely disjoint from this one, okay? It follows that if you write down the matrix for this one using the basis here namely tau 1, tau 2, tau n and so on, the diagonal entries will be all 0. That means the trace of alpha n is 0. But if this is true for n, it is true for all n, right? Therefore, the sum total of all the alternating sums of traces will be also 0. And that is L of mod phi is 0. And that is what we wanted to prove, okay? The second method is slightly more elaborate, but it gives you something about the subdivision chain map also. So, how to subdivision chain map, okay? This property, whatever we have, that will be used here also. This property will be used here also, that is common to it. So, here take the subdivision chain map which you have defined from c dot k to c, c dot of sd of k. sd is now, little sd is the barycyntic subdivision itself one at a time, okay? We have seen that this sd induces identity, homoisomorphism. That is the beauty. When you go to, forget about this thing, go to h, okay? h of h star of mod k to h star of mod k is identity. By repeated application of this, take compositions from sd of k is sd, sd, sd. That will be a map from c dot of sd of k, okay? Sorry, c dot of k into c dot of sd raised to k times k, okay? Right? So, that is our sd. But mod phi is the other way around. Mod phi is from sd of k, c star of sd of k to back to k. So, I can compose it back, namely sd of k composite phi, okay? This is an endomorphism of now of sd of k, c star of sd of k, okay? Yeah? And this is, this is an, this is a, when you pass to homo, this part is identity isomorphism, because it is composite of identity isomorphism. Therefore, this Lf, which we know equal to L of mod phi, the same thing L of this composition. After this composition, you are now in the same module to module and endomorphism. Therefore, you can compute a Lf star number of this sd k composite phi from sd of k to sd of k, okay? If you pass to homology, it is same thing as homology of mod k, okay? But we want to compute somehow. So, you can do it at this level. At the chain level, so c dot of sd of k with basis elements complicit, complicit of simplisites of sd of k. Now, the property 30 here, okay? We will tell you that if you write the expression for, for this, this function when you pass to the homo, at the chain level, okay? It will never involve rho at all, okay? phi star of rho will not consist of the coefficient of rho will be 0 in the phi star of rho. That argument is the same as in this case, okay? So, that is what, that is what we have to use, okay? So, let lambda from cn of sd of k to cn of sd of k with a chain map induced by sd of k composite phi, okay? Instead of putting a star here and I have put a lambda here. From, from this property 30, it follows that for any simplex rho, okay? Lambda rho will not involve rho at all, okay? In the expression of lambda rho, estimation of sigma i, ni sigma i, etc., okay? That coefficient of rho itself will be 0, okay? In lambda rho, the term rho does not occur at all. Then the matrix of lambda rho will have all diagonal entries 0, okay? Trace is 0. So, for each lambda from cn to cn is trace is 0, alternate sum of these traces will be also 0. So, I think I will stop here and we shall see some interesting applications of the Lafichage-Fichon theorem itself next time. Thank you.