 So today we are starting a new topic triangulation of manifolds. Recall triangulation of any topological space just means that we have a simple shell complex k such that the geometric realization of the simple shell complex k denoted by mod k is homeomorphic to the topological space given. Okay, that means that is a triangulation. The homeomorphism will be called a triangulation whatever you want to say. If such a thing exists then the space is called triangulable. Okay, so we are going to study a triangulated manifold. So in particular it will have the properties of the simple shell complex as well. So let us begin make a beginning with the study of local homology of triangulated manifolds. So local homology of any simple shell complex and so forth. Then we will go back to that it is a triangulation of a manifold and then then strengthen those results. So as general as we can speak what is happening to the homology, local homology of any simple shell complex. Okay, so it is first lemma here. Let k be any simple shell complex and f be any phase in it. Let x belong to the interior of f that is a point in the interior of mod f. Since this is a geometricalization mod f just means all summation ti vi where vi's are what it is of f. Okay, all the ti's are positive that is element in the interior. Then the homology statement is the following. A chi of the relative pair mod k comma mod k minus x you throw away that point. The relative homology of this one is the same thing as the chi twiddle of the reduced homology of the link of f. Okay, but the dimension is reduced which is hi minus dimension of f minus 1. So this is the statement. So recall that for any simplex f in k the open star little s t of f is an open subset of mod k. How was it defined? It is the union of all simplex is f union g. Okay, open. So open part of that where f union g it must be a simplex. Then you take f union g the open part of that and take the union of all that. So that is called the open star of f. It is almost similar to when f is a single point it is star of v we have defined. So that is generalized to this one. Okay, and automatically it will be an open subset of mod k itself. Okay, but therefore we can apply this theorem 3.7 which is the accession theorem. Okay, whenever you have two open sets covering the entire thing the two open sets will be excessive couples. The star of f this is the closed star of f mod k minus x this is an excessive couple. Okay, both of them are this is star of f is not a closed set, but open star of f it contains interior of that interior of this one they cover it. That is why this is an excessive couple. Okay, so in the x which is equal to star of f union x minus x. Union of that that is where okay, excessive couple means what? This is not the whole space right star of f union k minus x. Okay, by accession theorem 3.6 okay, sorry star of f union k minus x with the whole effect because star this whatever you have thrown away x is already inside star of f. By accession theorem k to k minus a single point is the same thing as star of f intersection with k minus x which is star of f minus x. So this hi is isomorphic to hi of this one. It is like x comma a homology is isomorphic to b comma a intersection b. If a and b are the entire a and b is excessive couple. Okay, so this is isomorphic to this one. So what we want to try we are trying to that we are estimating this left hand side. So the first step is that this whatever is given left hand side here is hi of star of x star of x minus x. So this is what you mean by localizing instead of the whole complex k you have come just around you know it is a neighborhood of f we are both relative things have come to neighborhood of f we are so that is the first step. So let us see further why it is just the homology of the link okay. Recall that we have also proved that star of f minus a single point in the interior of f this will contain link of f join with b boundary of f modulus of that as a deformation retract from x you can radially push it out to link of f star boundary of f this star is the the the join of these two things that is a sub complex. So it is similar to the picture wherein you have disc and a point interior removed so that is that is deformation retracts to the boundary sphere so it is similar to that so this we have proved in part one okay this is not a very difficult thing so here we will have to assume that one okay therefore h i of star of f link of f star b of f so this is a deformation retract of this one the second part okay therefore I can replace this one by this one only the boundary so this again similar to dn comma sn sn minus 1 homology same thing as dn comma dn minus 0 it is similar to that okay but you cannot use that result you have to use this result namely star of f minus x is deformed to this subset this subset of this one anyway okay let us denote k let us denote by k the dimension of f okay that means there are k plus 1 elements in k plus 1 vertices in f okay since star of f star of anything is contractable okay it is actually convex subset inside mod k so this is contractable what we have is in the homology exact sequence of this pair h i of this one will be 0 h i of this one will come next one the next thing will be again h i minus 1 of this one will come then h i of this one so every term three terms between 0 the two terms will be isomorphic so that is what happens namely h i of star of f to link of boundary of f this one which is isomorphic this one is isomorphic to h i minus 1 of the the link of this one the second part because the next term will be h i h i h i minus 1 of star of f that is 0 the previous term will be also h i h i of star of f that is also 0 so this is boundary of boundary this connected the connecting homomorphism that is an isomorphism okay I think we used to have this as delta so you can use it delta here okay doesn't matter so but this is what this boundary of a simplex it is homeomorphic to a sphere f is a k simplex so it is a boundary homeomorphic to a sphere okay so it is modulus of this the same thing as I mean the underlying topological space the same thing as link of f modulus joined with sk minus 1 okay but what is the joint with sk minus 1 it is the iterated suspension joined with s0 is the suspension with s1 is a suspension of suspension and so on so you have to suspend k times okay that is a dimension of k by suspension isomorphism you keep coming down h i minus 1 to h i minus 1 minus dimension of f and then you have just the link of f the suspension isomorphism starts from here and suspended the homology dimension increases okay use that one iteratively k times you get h i to log 1 minus i minus 1 minus dimension of f okay so this establishes this isomorphism so this is an elementary result which we will be using later on now let us make a definition take any simple shell complex then we say it is pure of dimension we have done this in the case of c w complexes it is similar to that pure of dimension n which each simplex in k is a phase of an n simplex in other words all maximal simplex phase are of the same dimension n okay so that is the meaning of pure this is just some terminology in other words k is pure of dimension n if we down this every maximal simplex in k is of dimension n so this could be taken as a definition also for a topological manifold now let us start with a manifold manifold with or without boundary so let us take without boundary given any point x in x you can choose a neighborhood view of a homeomorphic to r n if it is a boundary then you would have done homeomorphic to h n minus half h h n right half space so if it is an interior point all the time namely it is managed without boundary then you have this one by excision it follows that h i of x minus x x minus x x comma x minus x is h i of u comma u minus x because u comma r u comma x minus x is an excessive couple okay use that but u comma u minus x is homeomorphic with the pair r n comma r n minus 0 this is the homeomorphism x goes to 0 you can use that one right you have you can have such a chart so under the homeomorphism x goes to 0 so you can throw away that one so these two are the same okay now combine this with the lemma what happens in the lemma what we have proved for points inside a mod f okay h i here you see is h i to the law of i minus dimension of f minus one link of this one right so you can x minus x minus x this x is a not an arbitrary point but entire f it has this happens it happens this way so if you use this property so what we get we immediately get the following results now let x be a connected compact topological n-manifold without boundary I want to emphasize that k be a simplicial complex such that mod k is x that is k is a triangulation of x then the following holds what happens for all non-empty phases f of k we have h i twiddle of link of f is 0 for i less than dimension of the link of f and when dimension is equal to f okay when i is equal dimension then it is isomorphic to such for i equal dimension of the link that is link of f is a homology sphere homology sphere means what the sphere has some homology namely all the homologies except the entomology is 0 right the reduced homology so similar thing is happening homology sphere of dimension equal to dimension of the link okay this is true for the links we know that the boundary of a simplex is actually a sphere but the links of of a triangulated manifold the links are you know spheres of corresponding dimension namely dimension of a link of f whatever it is that it is like a homology sphere okay so this follows immediately here take f and x so it is it has this this being a manifold okay look it here this theorem this being a manifold I can replace mod k by x this is the homology of the sphere okay it is h i of r n r n minus 0 which is same thing as h i minus 1 now s n minus 1 if I need dimension so only i equal to n minus 1 it is infinite cycle otherwise it is 0 right so that is what I am using here now in the first part here everywhere else it is 0 when i equal dimension of the link it is that if i equal to this one okay the second part k must be pure of dimension n every simplex must be contained in a n simplex this is like a invariance of domain which follows very easily here okay start with a logical manifold okay every simply the largest simplex is everywhere he is of the same dimension n okay the third statement is this is this is somewhat nontrivial we need some proof every n minus 1 simplex of k occurs as a phase of exactly two simplex the first statement is given any two simplexes sigma and tau inside x inside k there is a chain of simplexes and simplexes connecting sigma and tau what is the meaning of this this is like a path remember in a connected simplex complex from one a one vertex to another vertex there is a path of edges through edges only you can go through similarly here through simplexes you can go so what is the meaning of this see you have sequence of simplexes s1 s2 sk s1 is tau sigma and sk is tau what happens to si intersection si plus 1 consecutive ones that is exactly one n minus 1 phase both si and si plus 1 of n simplexes one of the common phase okay exactly one common phase is there for both of them so that is a edge sequence from any simplex to any other simplex you can go through a edge you can go through a sequence of n simplexes like this a path of n simplex is like this okay so these two three and four are new to here so I will let us go through that one the first one is immediate from the lemma 6.8 that is what I have already out through okay dimensional link you should know is exactly equal to n minus dimension of f minus 1 in other words if you take a vertex in a in an n simplex its link is nothing but this but there is a boundary boundary of f which is n minus 1 okay like that so whatever dimension of the total dimension minus the link dimension is equal to dimension of f plus 1 or you can say n minus dimension of f minus 1 is a dimension of link so you substitute this one there in the earlier statement statement one follows okay second statement is what second statement here is k is pure of dimension okay so how do you prove that if f is a maximal simplex then interior of f is an open set okay hence for x inside mod f inside in the interior of f x minus x and f this is an accession couple also the boundary of any simplex is a strong deformation retract of mod f minus the single point this is we have been using several times because mod f is nothing but homeomorphic to the closer disc okay therefore we have h i of x x minus x you can write rewrite it in a different way h i of f f minus x that is h i minus 1 of f minus x by exact sequence because mod f is contractable okay so mod f minus x is nothing but a homotopy equivalent to the boundary of f so I can replace this one by boundary of f here h i minus 1 of boundary of f so this means the boundary of f is a homology n minus 1 sphere because that is that is what this homology is h i of x minus x is it is same thing as h i of r n minus 0 therefore f must be n simplex okay I started with the maximal simplex then I have concluded that it is an n simplex therefore x is pure dimension n okay the third one recall what is the third treatment every n minus 1 simplex k occurs at the boundary of exactly two n simplexes like in a in a triangulation of a circle every point every vertex is incident exactly two edges right so the same thing occurs in all the dimension is what we have to we have to prove okay so take an n minus 1 simplex this f is now n minus 1 simplex okay then for any point in the interior of f what happens from 1 we get z is isomorph to h n of x minus x we know this this is because x is a manifold of dimension n but this is if you put the lemma it will be equal to h naught twiddle of link of f i minus dimension of f minus 1 so dimension of f is n minus 1 n minus 1 minus 1 is just 0 okay so h i h i twiddle h naught twiddle of link of f if the twiddle of link of f is z remember the h naught twiddle we have already always computed h naught twiddle for a connected space is 0 okay in fact h naught and h naught twiddle differ by exactly one infinite cyclic component if h naught twiddle itself is z this means that there are two components here what is link of f when f is an n minus 1 simplex it is just the number of vertices because there cannot be any more anything more than n simplex because we have already proved that k is pure of dimension n so f union u must be simplex and disjoint u must be not a point of f itself therefore all these u form just a set of vertices then and its homology is reduced homology is z means there are exactly two elements okay so dimension of m is n minus 1 link of f is a vertex at v i which is in one one correspondence with n simplex is g of k such that f union v i is g i so how many v i's are there there are only two of them therefore g i's are exactly two of them and f is a common phase of them okay let us prove the fourth statement now okay on the set of all n simplex's having a chain from one to the other is an equivalence relation just like two points can be connected by an edge path then if there is a path from here to here there is a path from there to here here to there there to here a to b b to c then there will be a path from a to c right so this is an equivalence relation we need to show that there is just one equivalence class that means any two of them can be joined okay assuming on the contrary that a b the sub complex spanned by all n simplex's in one of the equivalence class start with one simplex then look at all the simplex's which can be joined to this simplex through a sequence of simplex's okay you take that so it is not the whole space then what happens is a question right we need to show that there is just one equivalence class assuming in the contrary that a b the sub complex spanned by all n simplex's in one simple one one equivalence class and b be the sub complex spanned by the rest of them so you have two sub complexes which cover the entire k okay then mod a and mod b are closed subspaces of x because they are sub complexes x is connected therefore intersection a mod a mod b as well as a intersection b must be non-empty a sub complex itself is non-empty otherwise a intersection b mod a mod a will not be intersected okay all right a and b are sub complexes a intersection b itself is a sub complex which is non-empty okay so far we are fine now let f be a maximal simplex of a intersection b this may be a single point or an edge or a triangle you don't know okay take the maximal simplex it follows that dimension of f cannot be n it must be smaller than n why if it is n simplex that will belong to both the both a and b right so that is not possible if it belongs to a then it can be seen in that equivalence class okay so but b consists of those simplices which are not connected to a so it must be a smaller thing not only that if it is n simplex n minus 1 simplex that means on this side there is an n simplex and that side another amplitude like that means the class of a goes beyond a so it cannot be even n minus 1 so it must be smaller than n minus 1 okay all right this implies that the link of f in k okay must be higher dimension than 0 if it is n minus 1 dimension the link will be 0 dimension if it is n dimension link will be empty if it is lower than then only link will be of higher dimension because the formula is dimension of f plus link of f plus 1 is equal to n so link of f the dimension of the link of f is positive which means it is a simplex complex of dimension greater than or equal to 1 that is the meaning of this okay so but the part 1 says that h naught twiddle of link of f must be 0 there okay h naught twiddle of link of f is 0 and hence link of f is connected it is positive dimension by part 1 h naught twiddle is 0 and hence link is connected link of f is connected inside f okay clearly link of f cannot be inside just inside a by the very definition okay nor inside b okay link of f intersection a is non-empty link of f intersection b is also non-empty from the maximality of f which is very important now inside a intersection b okay I am not taking maximality inside inside the whole of k but in a its intersection b it is a maximal simplex okay maximality of inside a intersection b follows that there is an edge e from link of f this link of f has an edge now I am taking with one vertex inside u so one vertex u inside a minus b and the other vertex inside b like that okay so so a a start with u inside a minus b and another vertex v inside b minus a so one vertex of a and one vertex of b should be joined by a edge which is inside f so probably here you should make a picture of a two-dimensional simperial complex that is all you can do picture you can make only two-dimensional complex okay otherwise forcefully this is pure set theory you will have to see this one a link of f what we have proved is connected and this part and that part are both non-empty so I take some part here some vertex here some vertex here and they must be able to join you must be able to join them inside link of f so that may be a path so along the path we keep moving till you come to the last point in link of f intersection a and then jump to intersection b b part and that is the edge which will be first vertex will be inside a and the second vertex will be inside b you understand how you have you can prove this one okay so if s is a simplex n simplex such that f union is contained inside s okay why I can do this one because k is a pure of dimension n so every simplex is contained in a n simplex okay so choose that as s here is an n simplex which contains e okay actually f union e must be inside s because e is in the link of f anything in the link of f by definition the union must be a phase f union e must be a simplex of k so it is contained inside a n simplex so it is contained in some other simplex then either this s is in a or in b because all the simplex n simplex is have been divided those which are in one class a and other which are not in that class to where it is it is either in a or in b okay so that is a contradiction now which implies both u and v are inside a or inside b but I have chosen u and b one is inside a one is inside b actually one in not in b and the other one is not in a is important here okay so this is a very neat proof which I myself enjoy this one so it is simple minded thing like this you can instead of edge path is you can have a path of simplex is something very very peculiar and this property of a triangulated manifold itself is made into an axiom and things which satisfy such an axiom are called pseudo manifolds I think we will stop here today so so next time we will study what is a pseudo manifold based on this this result okay thank you