 As promised last time, today let us discuss one interesting river due to Poincaré and another due to Munkres. First let us take Poincaré's result. We will require two facts which we have not proved for lack of time. The first fact is for any closed surface. Closed surface means compact surface without boundary. The Euler characteristic is always less than or equal to 2, we can assume compact without boundary and connected. Euler characteristic of that surface less than or equal to and equality holds if and only if S is homeomorphic to S2. The second fact is if M is compact odd dimension manifold important with non-empty boundary then its Euler characteristic is half of the Euler characteristic of the boundary. So twice chi M is chi of boundary of M. The second fact is an easy consequence of the following fact which is not an easy thing namely every odd dimensional closed manifold Euler characteristic is 0. From that you can deduce this one by our technique of doubling the manifold. So that is the hint and you can I will leave the rest of them to you as an exercise. The first fact you will see in the last part of these lectures, this will be actually proved finally. So right now we have not proved it, we will assume that. Now let us look at this Poincare's result, take a simplicial complex which is a pseudo manifold of dimension 3 without boundary and suppose it is obtained by identification of pairs of facets of a convex polyadron P inside R3 just like we have done in the previous theorem. If the Euler characteristic of this k which is x namely mod k is x Euler characteristic of this one is 0 then x is a closed remanivored. If it is a 3-manivored its Euler characteristic must be 0 that is a general result. So Poincare comes up with this great result of the converse kind of thing. Of course we have to use that, it is a quotient of a convex polyadron by identification on the boundary, so that is the hypothesis. In this pseudo manifold of dimension 3 is quotient of quotient in this way that is what we have proved this theorem, only then we have asked this question and now Poincare has an answer here a very positive answer, the proof is not very difficult let us go through this one. We need to prove the local Euclideanness of x there is nothing else ok, it is being compact and hostile already ok because it is going to be a mod mod yourself the quotient of some identity in this one ok, it is a simple complex, it is a simple complex. So local Euclideanness is what we have to show for points which are images of points in the interior of p, on the interior of p the quotient map capital theta is injective it is a homomorphism from a convex polyadron to interior of p therefore there is no problem ok. Similarly those points which are in the image of an interior of a facet, facet is a two simplex now, take a point in the interior of that facet relative in the interior inside it is itself ok not in the interior of p, what happens whenever it is pseudo manifold right, whenever you have two simplex it shares two simple, it is a face of facet of exactly two three simplexes, one on this side one on the other side, so union of two simplex should be there, you would take a union along the two simplex, three simplex right, therefore it is one half open space on this side half open space together it will be the full opens of set in R3, it will have a neighborhood like that, so there also there is no problem ok. So, what are the points that are left out? Points on the edges both interior of an edge as well as finally the vertex points. So, these are the points which have to be carefully studied, what happens to the neighborhood of these points? So, consider a point in the interior of an edge, so these are two classes of the interior of an edge or vertex ok, first let us look interior of an edge, the topology of the neighborhood of such a point depends on the topology of the link of A, because if you look at a star of A, star of A is the closed star of this E is an edge, E is a simplex, E is a one edge right, one simplex, the star of E is link of E joined with E ok, E is what, E is homeomorphic to an edge, its interval, so there I know what is the topology, there is no problem, so what is link of A, this is the mystery, this I do not know how it looks like, if this is also like a ball, then you would have completed the proof right. So, what is this one is the topology of this one looks like, this is a neighborhood of after star is a neighborhood, it looks like link E, so if I understand link E then I know what is going to happen, hence star of E the underlying topological space modulus of that is homeomorphic to the iterated cone, you see this is an edge ok, so if you take one single star with one single point it would be a cone right, the cone over the cone is precisely star of this one, because there is an edge there, so you can think of this as a cone over the cone of link of A, clearly the link of E is one dimension pseudo manifold, why because star of E is dimension 3, link of A whatever dimension plus this one plus 1 it should be the whole dimension that is the equation right, so dimension of this one has to be 1, 1 plus 1 plus 1 is equal to 3, so that is why this is dimension 1 dimension, so it is a one dimensional simplexial complex right, so we have already classified one dimension manifold, so how does this one look like is a question, the one dimension pseudo manifold it is, because it is a link of A star of something ok, so this is not very difficult to see, so a one, it is a one dimension pseudo manifold, there is no other choice it has to be a manifold, only thing is if it is connected then it will become a circle, the cone is already a disk, cone over the disk is another disk, so that should be 3 disks, so you have formed a neighborhood which is a 3 disk of every point inside E, in the interior of E ok, the only problem is it may not be connected, it is a connected pseudo manifold, one dimensional manifold, it has to be a circle and that will complete the argument, so we must prove that the link of E is connected, so here is a picture, you want to say that this will not happen, so this is our, the middle one is our one simplex, one of the one simplex is inside our k, it may happen that all these three simplex are incident at this point and they form two different groups, so one part is here going like this, another going like that, you do not know they are not inside the R3 you see, the quotient space need not be inside R3, so then what happens the link would have been one circle here, one circle here, so why this picture is wrong, why this picture does not occur, that one has to argue it, the only condition that we have is that chi of x is 0, from that Poincare is able to reduce this one, so we will also see how he has done it, okay, so we claim that it must be connected, so here it is a topological argument not for of chi at all, for if not what happens, we can write link of E as A union disjoint union B, okay, like this one, this is A, this is B, but of all, okay, as disjoint union of two sub-complexes, alright, we can then partition the set of identifications on the boundary of P into two sets accordingly because what is the meaning of this link, each edge here will determine, around with this edge will determine a simplex, okay, this edge star, this simplex will be a simplex there and this one will be another simplex or simplex, so all these simplexes you place one way, all these simplex you place another one, alright, so in the, in P itself you look for such a thing, okay, you can see that identifications are taking place where on the boundary, boundary pieces of this one, okay, therefore no simplex in the first group will have anything to do with the second simplex, but they have one simplex common, how did that come, remember when you are identifying a, a phases of, of F, automatically the edges will identify, corresponding edges will identify, okay, but just some edge to edge, there is no identification, extra identifications are not allowed, so this is very important note, many people do not understand this one, when you identify triangle with a triangle automatically the boundary edges of this triangle and the boundaries of that triangle are also identified in the corresponding order, after this if you say this edge is identified with that edge that is not allowed, some isolated edge and edge, edge identification is not done, each edge identification is a consequence of the, what, the two simplex identifications here, on the boundary itself, not in 90 years again, so this is what I have to understand, therefore this will give you a contradiction, okay, there will be two edges left out even in D2 which are disjoint after you carry out all the phase identifications, but they are coming to the same edge here, so how did they come, so this is why the link of E must be connected pseudo one-dimensional complex, therefore it is a circle, once that is a circle star of A, A the cone over the cone of that which is three disks, okay, so that will take care of all the points in the interior of all the edges, finally you are left with vertices, why the neighborhood of vertices are homomorphic to R3 or homomorphic to disk D3, this is what we have to see, okay, once again this is where the last condition chi x equal to 0 comes, okay, it remains to check that local Euclideanness at the image Vi of vertices of the boundary of P, in the interior there is no problem, by passing to a subdivision of k, okay, we may assume that the star of Vi's are disjoint in mod k equal to x, this is an easy thing because if you keep subdividing, at least actually second biorecetric subdivision is useful here, first biorecetric subdivision may not give you this one, okay, by passing subdivision you can make the vertices far away, the new vertices which have come, they are not going to create problem because they will be inside edges or simple existence and so on, it is the old vertices of the original k we are worried about, now those things star of those little vertices will be disjoint in mod k, okay, so we assume this one right in the beginning which is not correct in general sense, okay, mod k is the same, mod k prime, mod k whatever we have subdivided, mod of that is the same thing as x, okay, we need to show that each link of Vi is a topological two sphere so that the cone over that is a disk, now E being a single point, link E star with E is going to be the star of E, right, so that E is just a single point, so that is a cone, okay, so all that I have to prove is link of Vi is a topological two sphere which is the same as showing that chi of the link is 2 for all i, this is where I am using fact 1, fact a, okay, so we are going to prove chi of Vi is 2 for all of them, all i, so what do I do, let S equal to union of star of Vi, this is disjoint union by our assumption and let L be the closure of S minus, sorry k minus S, okay, star of Vi is our closed subset, so union of all these is a closed subset, it is a finite union, to take k minus S that is an open set, you take the closure of that, so that k is actually L union S, okay and what is L intersection S, it is precisely the boundary of union of boundaries of all star of Vi which is nothing but the links of Vi, it is a disjoint union of links of Vi's, the star of Vi's are themselves disjoint, so this is also disjoint union of A cross, okay, then L is a topological three manifold, okay, with boundary equal to, namely boundary of L precisely equal to L intersection S, why L is a topological three manifold, we have verified it because L is a complement of all these bad points, L is contained in the complement of all these bad points, so it is a topological manifold, okay, L itself is, what I have put here, k minus S, okay, k minus S is a topological manifold, it is boundary precisely L intersection S, okay, this is an open subset, all right, so the boundary of L is L intersection S, the three manifold and now I use the fact B, what does it say, chi of boundary of L is equal to twice the chi of L, right, chi of boundary of L is equal to twice the chi of L, but what is boundary of L, it is chi of S intersection S, also what we have, chi of X is 0, that is the starting hypothesis, which is simply a chi of K, okay, a simplicial complex, simply for a simplicial complex, what is the formula, this formula also you have seen, chi of S which is some complex plus chi of L minus chi of the intersection, those which are added twice, both here, here they have been adding, so you have to subtract that, okay, so suppose this is K and this is what I have to, where I have taken K here, so I have to just put chi of L, okay, is chi of boundary of L, yes, what is S, tell me what is S, chi of S, it is a disjoint union of what, see what is this S I have put here, star of V i's, each star of V i is like a 3 disk, it is whatever it is, it is not 3 disk is our problem, but it is contractible, for a contractible space Euler characteristic is 1, but there are how many of them are there, K of them let us say, that K is number of vertices there, okay, so Euler characteristic of a disjoint union is the sum of Euler characteristic, therefore you get that K here, okay, this minus this is half of this one because this is, this itself is chi of L, half of chi of L, chi of L is half of chi of boundary of that one, so this is K minus half of chi of S minus L and that is 0, this just means that iron to 1 to K, chi of L V i's is equal to 2 K for the links, notice that each L K of V i is a surface and hence chi of link of V i is less than or equal to 2 by fact A, by the way that the link is a surface, what surface pseudo-manifold actually, it is a boundary of a nice manifold, see we have seen that this is a boundary, it is a common boundary here, okay, so this is equal to union of links, disjoint union of links, these being so boundary components, so they are all surfaces, two dimension manifolds and they themselves do not have boundary, so they are closed manifolds, okay, therefore each of them chi is less than or equal to 2, okay, but the sum total is equal to 2 K, therefore each of them must be equal to 2, there are K of them, each of them is less than or equal to 2, the sum total is 2 K, so therefore each of them must be equal to 2, it follows that each L of L K of V i is a two sphere for each i, so let us now take a result due to Munkres which is actually used in a big theorem by Stanley through another important person called Reissner, the Reissner gave a combinatorial condition for the phase ring of a cohen, phase ring to be a cohen mekale, phase ring of a simple shell complex, give a topological condition, Munkres result uses to prove that the Reissner's condition is satisfied in a particular way, okay, so in that way Munkres result is useful in this combinatorics, to ultimately all these things were used in proving a big conjecture in combinatorics which is called the upper bound conjecture by Stanley, okay, so we are not going to do anything about phase rings and so on, but we will do the topological aspect of that, namely Munkres result which is an important result, but I want to show that just because something is important, it does not mean that it proves are difficult, I want to give you a easy proof of this one here, okay, part of it we have already done, so what is the theorem of Munkres, it says take a connected topological space which is triangulated, axis mod k, and I assume that k is a finite simple shell complex of dimension n, then the following two statements are equivalent, this is the theorem of Munkres result, what is that, it is some local homological condition, H i twiddle of x is 0, this is a global condition, H i twiddle of x minus x, x minus x is 0 for every, sorry, for i less than n for every x in x, so this is true for i less than n, both H i twiddle of x minus x and H i twiddle of x is 0, because you do not need to put a twiddle here, it is a first condition, the second condition is H i twiddle of the link of f is 0 for i less than dimension of the link of f and for every phase of f, every phase of k, for every phase of k the homology is 0 below the dimension, the reduced homology, so this is the condition and these two conditions are equivalent in the statement of the theorem, since we have done a little bit of study of these things, it will be very easy for you to study, easy for you to follow this proof, observe that since empty set is also a phase, empty set is allowed to a phase according to our convention, according to Munkres convention, the link of k, link in k of the empty set is the whole of k by the very definition, therefore condition b implies, condition b for every, for i less than the dimension of link is 0, the condition b put f equal to empty set, H i of link of f is what, the whole k, H i twiddle of k is definitely 0, so this gives you the first part here, okay, just one single phase f equal to empty set gives you the first part here, so it remains to prove only this part, from here to here, of course from here to here you have to prove all this, okay, so here to here by taking f equal to empty set, this implies this, this part implies this part, but for others you have to wait for, okay, condition b implies this here, for i less than, we shall first prove two lemmas, I am breaking down the proof into simpler statements here, okay, instead of complicating the whole proof in one go, take just these conditions, condition a and condition b, any one of them, you will get that k is a pure simplecial complex, a implies k is pure, b also implies k is pure, see we have not yet proved that a implies b and b implies a, but either of them implies k is pure is a first step and proof is similar to what we have seen for purity of a topological manifold, triangulated topological manifold, right, we have seen that one, so the proof is similar here, suppose a holds, take a maximally simplex inside k, I have to show that dimension of that simplex is equal to n, that is what I have to show, purity means that, okay, take f to make a simplex, take a point in the interior of f, then h i of x comma x minus x we have seen is h i of d k comma sk minus 1, these steps we have seen before, okay, this is just excision, okay, where this k is the dimension of f, star of f, star of f that is what we have to do, okay, where f is a simplex of maximal dimension, that is why this works, because star of f will be open subset, inside k, inside mod k, therefore if this is true condition a implies that this k must be bigger than or equal to n, h n of this one we know is equal to infinite cycling, but the condition says that dimension, below the dimension these are all 0, so this k must be bigger than or equal to n, but k cannot be bigger than or equal to, bigger than n, so it must be equal to n, okay. Now condition b, suppose condition b holds, now this time the proof is a little more elaborate and it is again repeating the old arguments here, suppose k is not true, suppose k is not pure, let l 1 be the collection of all n simplexes, okay, and their faces, if it is not pure that means there are spaces which are of lower dimension things, maximal simplexes, okay, take l 2 to be the collection of all faces simply step which are not a face of any n simplex, okay, so they will be of lower dimension and they are not faces of any simplex, to take that and take the collection of all that and that subspace is some and their faces and so on, namely the sub complex generated by them, so you have two different sub complexes here, the union is the whole of k, okay, l 1 and l 2, so k is union of l 1 and l 2 and both l 1 and l 2 are non-empty is the assumption, k is of dimension n, so l 1 is non-empty, which is not pure, so l 2 is non-empty. Since x is connected, l 1 intersection and 2 must be non-empty, all these arguments you have seen earlier also, okay, take a maximal simplex inside this sub complex, l 1, l 2, so sigma will be the maximal simplex, then of course sigma is non-empty and it is a proper face of a simplex f in l 2, why? Because if it is a full simplex, then that full simplex will be a sub complex of l 1, l 1 that means it will be contained inside a larger simplex, so it is not, therefore the intersection is a proper smaller than any simplex in l 2, the largest simplex in l 2, sigma, okay, so then sigma is non-empty and it is a proper face of a simplex f in l 2 and hence dimension of sigma is less than dimension of f, dimension of f itself is less than n because f is inside l 2, okay. So the core dimension from here to here is at least 2, n minus this one is at least 2, okay. Also sigma is a proper face of an n simplex f prime in l 1 because it is in the intersection l 1, l 1 intersection x 2, everything inside l 1 will be contained in a larger simplex which is n dimension, okay. Hence link of sigma intersects both l 1 minus sigma and l 2 minus sigma, okay. Since sigma is maximal, this part of the proof is different from the old thing, there we used connectivity, blah, blah, blah. So here we are doing something different, okay. The link of sigma itself intersects both l 1 minus sigma and l 2 minus sigma, okay. Since sigma is maximal in l 1 to l 2, it follows that link of sigma is disconnected, therefore at not middle of sigma, link of sigma is not 0. If it is 0, it would have been connected. This is the deduced homology, okay. So this is not 0. But link of sigma we have just seen is dimension at most n minus 2, right. This just now we have seen it here at most dimension minus 2. And sigma is inside f, dimension of f prime is n, dimension of sigma inside, sigma is contained in f prime is inside l 1 and dimension of f prime is n. Therefore, dimension of the link of sigma is at least greater than or equal to 1. Below that, this must have been 0. That is the contradiction. This must, you know, that is condition b, okay. So this proves our lemma, namely both conditions imply k a is pure. Now after this, the proof of Munkra's result is very easy. All that you have to observe is following. Observe that we have used only the second part of condition a and condition b only for non-empty phases of k, in proving the purity, okay. The empty phase is first part. So I have already told you this one. So now proof of this point 3, by our lemma, either of the conditions imply k is pure. Hence, each f belonging to k, the star of f has dimension must be n because it is pure. So dimension will be there all the time, full dimension will be there, no matter whether you take a vertex or an edge or a two simplex and so on, okay. So dimension of star is always n. Therefore, dimension of a link of f plus dimension of f plus 1 which is equal to n. This is the dimension of the star, okay. Because star is always linked, star with link of f joined with f. So that is why dimension we have been using this one again and again. Now start with condition a. I want to prove that it is equivalent to condition b, okay. There are only these two steps now we have to understand. And that is the end of the proof, okay. Let us see. Condition a implies and implied y is equivalent to h i to the level of x is 0 for i less than n. It is the first part. The second part is h i of x minus x minus 1, x minus 0 is 0. So what we have seen that is equal to h twiddle of j of sorry h twiddle j of j minus dimension of f minus 1 of link of f by taking a point x inside f for every f, okay. So this is 0. This is follows from the second part here. For every non-empty phase f because I have to take a point in the interior of that, okay. So non-empty phase of k and this j must be less than n. Then we have seen this proof. The h j of star of f minus star of f minus x, then from there we come to this one, okay. Now take these two statements. For the first statement implies this one, h j twiddle of x is 0 for i less than n, okay. I have to put that h j, j less than n here I have put. The second condition says h twiddle of j minus n plus dimension of the link of f of this one is 0 for j less than n and for every phase f of k. We have just seen the link dimension of link I am putting that here. Dimension of the link is n minus dimension of f minus 1. So you put that here you get a j minus dimension of f is minus n plus dimension of 1 minus 1 cancels out, okay. How I get it? I get it from here by 40, okay. So from here to here to here you have come. But this is nothing but condition b. So Muncher's theorem is proved also. Next time we shall make some general remarks about triangulations. We have to find more material, what are general results known and so on, no proofs. After that we will start classification of triangulated surface. Thank you.