 Having studied some local homological properties of a simplicial complex and then having studied the properties of local homological properties of a triangulated manifold, we were led to make the following definition now. A simplicial complex which satisfies property 2, 3, 4 of the above theorem is called a pseudo manifold of dimension n. So, this was the theorem that we proved last time. Let us go through that theorem first. So, this was the statement take a connected, compact, topological n manifold without boundary. That KB as a simplicial complex such that mod k is equal to x. In other words, it is a connected, compact, topological n dimensional manifold which is triangulated. Then the following holds. For all non-empty phases of K, we have H i twiddle of link of f is 0 for i less than the dimension of f and isomorphic to z for i equal to dimension of the link which can be stated in a nutshell namely link of f looks like a homology sphere, up to homology it looks like a sphere of dimension equal to dimension of the link. The second statement is that K is pure of dimension n. Third statement is that every n minus 1 facet of K occurs as the phase of exactly 2 simplexes. The fourth condition says that from one simple n simplex to another n simplex, we can go by a path of n simplexes. A path of n simplexes is a sequence S 1 S 2 S k such that the intersection of S i with S i plus 1 is exactly an n minus 1 phase. It starts with S 1 equal to say sigma and S k equal to tau, then it is a path from sigma to tau. This condition especially along with of course 2 and 3, no topological manifold now. You forget about manifold, only these conditions you assume. Suppose these three are true. Forget about compactness, forget about connectivity, connectivity comes from automatically from here. Make those as axioms for what you may call it as pseudo manifold, 2, 3 and 4. If you replace the condition 3 by, instead of saying exactly 2, at most 2 that means an n simplex may have an n minus 1 simplex may have only one n simplex containing it. Instead of 2, of course the pure means that there will be at least one. So if there is only one, then that kind of simplexess will become boundary part. So then you will get manifold force with boundary. So now having made a definition, we have the following corollary. Given any connected compact n dimensional pure, that is why n dimension is there but we do not know whether it is manifold or not. I mean triangulated pseudo manifold k, pseudo manifold with or without boundary. There exists a triangulated convex polyhedron p in R n and an finally linear isomorphism phi i from f y to f y prime where f 1, f 2, f k and f 1 prime, f k prime are some distinct facets of p such that k is the quotient of p, namely the polyhedron by the boundary identification. These are all boundary identification. What is the identification? X is identified phi i of f whenever x is inside i pairwise. If you take this, this is exhaust all the boundary phases then k will not have any boundary. So then boundary of k will be empty and if and only if. So that is the statement. So this is the something very simple minded thing but it is the starting point of our classification for two dimension manifolds. Indeed this is the way how Poincare perceived the whole thing and even for three dimension he tried to classify like this. Unfortunately after several you know almost whole century of people trying his program, trying to complete his program it has failed in some sense but for n equal to 2 we are going to use it and it will be very very useful. Having said that it is useful to a large extent but the whole classification problem could not be solved by first starting from here. Even in dimension 3 it was tried and very difficult and later on it was just given up. So this was not useful and so on. So but it gives you already a lot of information on the topology of a manifold, triangulated manifold. So let us go through the proof of this one today. We start with labeling the n simplex states of k. If there is only one there is nothing, no other condition you just label it as one that is all. For i greater than equal to any n simplex sigma i, sigma 1, sigma 2, sigma n etc there are one finite least compact thing so there are only finitely many of them. So sigma 2 has at least one n minus 1 phase that is a facet. I will keep using that n minus 1 phase are called facet common with sigma j, j less than i. So sigma 2 there is only one sigma 1 so it will have only one phase but when you go sigma 2, sigma 2, sigma 3 you have come you have got a conglomeron you do not know how complicated it looks like. So what is the condition? Condition is one of the sigma j's will share one of its phase facet with the new one sigma j plus 1 or sigma i whatever the next one. So that is the j less than i which one you do not know. So this is what for all if you take all of them one of them will share it. And this is possible because of condition 4 that starting with sigma 1 if there is no no condition this condition is not set so that would mean that you have got two disjoint collection this sigma 1 is one one and all the other one none of them will will intersect this one that will be that will contradict our assumption because from sigma 1 I can go to any other thing by a sequence of this n simplex path there must be another one starting which will lead all the way up to whatever sigma i have chosen okay the moment you do not find such a thing then there will be a contradiction there will not be a path like that okay. So take up to sigma i you have taken something which satisfies this one how do you construct the next one if there is a next one there must be a path from that to one of the things which you have selected took any one of them which means in between you may hit something else though whenever your first time hit that one you just like you argue with a path okay so it is condition 4 that has to be used okay so once you have arranged it like that let kj be the sub complex spanned by sigma i less than i less than kj. Okay now what i am going to do inductively we shall construct triangulated convex polyhedrons p1 contains a p2 contains a pj and so on up to whatever pk i do not care how many are there okay fine and all of them subspaces of rn okay and subjective simplicial maps theta j from pj to kj below the k1k to kj are already well defined now such that for each j the following three conditions hold okay theta j is a bijection on each simplex okay remember theta j is defined as pj what is pj pj will be union of various simplexes okay and two corresponding simplex for each simplex it is a bijection theta j restricted to pj minus one is theta j minus one that means each successive theta j's are extensions of the previous ones the third condition is f is any one of the boundary facets of pj is convex polyhedron remember that look at a boundary facet then theta j is a homeomorphism on if you restrict it to interior of pj union this facet if you take the whole of pj along boundary it may not be injective in other words injectivity fails at each stage if at all only from one facet to some other facet on the boundary that is the condition c okay for example when when j could 1 theta 1 I have just I have just taken one convex one tetrahedron or one whatever n simplex and that is a bijection on the whole of it okay so this condition looks like weaker condition but this is all we can ensure as we keep going up condition c okay which will help us to prove the next condition also by the way if I just say it is interior its injection that is not enough okay so let us see how we are going to do this for j equal to 1 let p1 be any geometric n simplex in x geometric n simplex means what convex all of n plus 1 points I do not care how it looks like it will be a convex it is a triangulator because there is only one simplex okay so p1 is any one of them okay one one one geometry simplest to start with and let theta 1 more p1 to k1 your k1 is just what k1 is sigma 1 be the bijection of vertices on the vertices if the bijection automatically when you extend it it will be isomorphism clearly a b c are satisfied okay suppose inductively we have arrived at the stage j construction is over okay p1 contains a p2 contains a pj pj is a convex polyhedron triangulated and it satisfies a the theta j satisfies the three conditions okay now look at sigma j plus 1 by the very choice of this labeling this will share at least one n minus one fair fair facet f prime with some simplex tau inside kj okay it may share more also fix such an f f prime it follows that sigma j plus 1 the new new simplex is this f prime union some some some vertex view for a unique vertex you belong to kj plus 1 kj plus 1 is what kj union sigma j plus 1 so it will be inside sigma j plus so it will be kj plus 1 but what may happen is this the new extra vertex may be already inside kj this can happen but we do not have any objection for that okay so now let f contain this pj there will be some you see if it is f prime is inside kj there will be one n f n minus one facet f inside pj such that theta j of f is f prime because theta j is is subjective mapping right so it follows that f must be also the boundary component because because here in below inside kj it is a boundary component okay so in pj also it must be a boundary component okay consider the region now i want to construct pj plus 1 how do i construct pj plus 1 so i have a convex polyadron and i have located one of the facets on the boundary okay so consider the convex region bounded by the hyper planes spanned by f each facet which is a n minus 1 simplex okay inside r n you should take the linear span of that instead of corner here's the convex span of that okay affine affine span of that that's the correct word i find span of that that will be a hyper plane okay like that you take all the other hyper planes uh the spanned by all the other facets which intersect f okay take no take such a thing then look at the outer regions defined that it should be defined a convex region which is actually usually unbounded region defined the one which is outer to the convex polyadron okay take that you are spanned by f and other half assets here intersecting f take w to be any point in the interior of this convex region then rho is the convex cell of w and f so i have i have now found a another convex thing there which is nothing but an n simplex now okay and put pj plus 1 equal to the convex cell of pj union rho the pj is already convex so this pj will be there union rho will come because rho it's another convex cell so look at this picture it will tell you you have constructed up to here and you have located the boundary of this is pj okay and this is f is one of the boundary component boundary facet now look at this plane this plane this plane this plane this plane and all that at all of them intersecting this f coming from here here here here this one come this one comes and so on okay there will be a convex region this is the intersection of all of them okay outer to this convex region outer to the convex polyadron inside that choose a boundary choose a firm point w like here okay then take the convex cell of f and omega of f 100 w that is my rho this is n simplex okay here in the picture it is all n equal to 2 only r2 right automatically this p union rho will be a convex region if you choose it here for example this this point it will not be if you choose it here it will not be okay that is the whole idea if you choose it here it will not be it you know if you take and draw the triangle like this that will not be convex region along with this one so this is a criteria where you are to choose this w and it can be chosen very easily okay so do that then it will be a convex region now you map this extra rho whatever you have got 2 sigma j plus 1 f all on theta is already defined here so f goes to f prime by isomorphism all this the w will go to you and extend it linearly that will be theta j plus 1 okay so now take theta prime of j plus 1 of w is equal to you and theta prime of j plus 1 on f equal to theta j extend it linearly over rho so you have got theta j plus 1 j plus 1 prime here and theta j defined all over here match patch them up together because on f they agree that is my theta j plus 1 theta j plus 1 on pj theta j theta j plus 1 on rho is theta j plus 1 okay so the construction is over but we have to prove you know all these conditions what are these conditions these 2 are we have already done okay but condition 3 condition this c condition I have to prove up to j you have assumed this one for j plus 1 you have to prove this one then only construction inductive construction will be over so so here is the first the worst scenario is like this here even in this case condition c will be true so what what I have got what here is to begin with you have this this is your k okay let us say consisting of seven of the triangles here in the picture seven of the whatever you may think simple axis whatever okay I have labelled the labeling could be quite arbitrary right so for example I could have started from here and go on like this I was started here 1 2 3 4 5 6 then 7 so accordingly I construct 1 2 3 4 5 let us say and the 7th stage what happens up to 6 you have constructed okay the 7th simple as this sigma 7 what has happened it is sharing one one facet here with six okay and then I have constructed this correspond here this will be my point w and that will be mapped to this point and extended over so construction is over now why condition c is true what does the condition c says you take the entire interior of this one take any one of the boundary facets here any one of them 1 2 3 4 5 6 7 8 9 any one of them union that theta j must be a homeomorph must be injective injectivity once you prove injectivity homeomorphism will be over here that is what you have to show okay so suppose I have taken this one this is already in the in theta 6 so that condition is automatically satisfied for this part and I have got this part injective that is easy right so anytime I have the boundary facet is already inside inside pj there is not much difficulty to prove the whole thing but suppose this is the one but if this is the one I am taking I am not taking any other facets here okay interior of this and this edge I am taking here okay where is this edge going here the edge will go into this spot okay this is 6 one edge will be going to this one 7 and 1 remember the other boundary component of one is here and I am not including that in the picture in in this one I am talking only interior right therefore only restricted to this part it is still injective if I include this edge as well as this edge then it will not be injective okay so this picture tells you the whole story actually I have to write down the proof in general not just 1 2 3 4 5 6 okay so we need to verify condition see first of all notice that theta j from pj to kj is a closed mapping each of them is a quotient but it is a very special quotient in particular it is a closed mapping because what we have what is a compact space to a hostile space okay so it is a closed mapping a closed projections it is also a quotient map has a special property like open quotients are special property so you use that one theta j plus one being equal to theta j on interior of pj union f okay this part is homomorphism by induction hypothesis also theta j plus 1 prime remember what I said it is the definition of of theta extending theta on the new simplex at j plus 1 okay on the row is a homomorphism from row on to sigma j plus 1 it follows that theta j plus 1 is a homomorphism of interior of pj union row so if you have two sets a and b on a it is a homomorphism b it is a homomorphism okay to different subsets a c and d such that on the intersection goes to the intersection a intersection b goes to a intersection b then the union on the union it will be homomorphism okay something like this I am using here okay so it follows a theta j plus 1 is a homomorphism on this part okay now this is one part now take g be any boundary facet of pj plus 1 if g is already in the boundary you have already got it boundary of row okay then there is nothing to prove otherwise g is a boundary facet of pj and it is different from fj okay therefore from the induction hypothesis again theta j is a homomorphism g union interior pj more over theta j under theta j g and f okay are mapped on to different facets okay in k it follows that theta j plus 1 is injective on g union interior pj plus 1 g is any facet any other facet it is also true okay so it is easily checked that it is also a closed mapping injective continuous and a closed mapping and hence is a homomorphism there may be several ways of proving that it is a homomorphism but noticing that it is a closed mapping will help you to prove this one very easily okay finally what do you do what is theta theta is just theta j on pj it is well defined because on pj plus 1 theta j plus 1 if you take on pj it will be theta j itself okay so it follows that theta is a homomorphism in the interior of p pj if you take any one facet there also it is injective but now i am going to say something better than that what i am saying there again on each facet theta is injective okay it is a homomorphism there also restricted to restricted to somewhere the the image may not be a boundary why because this facet may be covered by another also it may be it may happen that two distinct facets are mapped onto the same facet in k okay label them in pairs f y and f y prime a third one will never come why because each facet here accounts for an n simplex and any facet in k can be at most phase of at most two n simplexes only okay so that is used here okay so label them f y and f y prime how do i am able to them the property that under theta they are mapped to the same facet okay put psi i equal to theta restrict f y psi prime could theta restrict f y prime phi i equal to psi i inverse of psi psi i prime inverse of psi i so these are the relations now from f i to f i prime okay this is a homeomorphism it follows that under theta k is then isomorphic as a simplicial complex to the quotient space of p when you say quotient space what is the relation the relations are this phi i so that is the statement of this theorem the p i's and phi i's whatever okay there is there are you know phi i's p i from f i to f i prime linear isomorphisms such that x is identified phi i of x so that completes the proof of this theorem so let me make a few remarks before closing up the method of proof in the above theorem with the beginning of a technique known as cut and paste technique where are cutting you took a you started with an n simplex cut it and paste it in r n okay that is the meaning of cutting and pasting then wherever you have cut you look for another n simplex there from there cut it and bring it and place place it in r n while you are placing it you have some freedom you you what is the freedom because you are a topology you are not doing any geometry here so you can change the size of the simplex but wherever you have cut it from you have to place it in the corresponding thing there so here they they are the ones which is joining okay so this is something which is somewhat strange in the sense that in topology you are not supposed to cut things that would be what that's it like you are doing something discontinuous the continuity has to be established wherever you have cut that is only temporary you are pasting along the same thing same homeomorphism okay homeomorphism you are pasting the same parts that is why it is allowed so it's called cut and paste technique almost half a century of mathematicians were trying to do this one in at least in dimension 3 okay dimension 3 this has become a big industry and lot of results were proved yet the final proof namely the all that they were trying to do the point or a conjecture interior dimension would not be proved from this technique okay so that's why I'm calling it as low dimension topology usually means three and four dimensions in the next section we shall use this technique to classify surfaces okay we are going to do that conversely given a triangulated convex polygon p in rn and a pairwise facet identification data as in the theorem that we have proved suppose we start with such a thing okay then we can ask a question namely is the quotient a simplicial complex so that the quotient map is simplicial that is the first question which is answered quite easily anyway let us take the second question assuming one is true is the quotient a topological manifold it starts with any pseudo manifold itself you can do this one is the theorem there so why the reversing when you reverse why you ask topological manifold obviously it will not be true right so in fact both the questions have obviously negative answers and that is precisely why they are good questions because now you have put some extra conditions more extra condition of your own choice so that there will be a positive answers okay so let me discuss it a little bit a few minutes only as an answer to one is negative in general an answer to one is negative in general okay however if we take the second bare eccentric subdivision of p the p is a triangulated polyadron okay it may be just for example you may have one single line line segment divided into two portions okay that means union of two line segments but it is a convex one now you identify the end points it is not going to be a simplicial complex very very obvious reason right very obvious way so he easily fails but there is a beautiful answer here if you if you start with a convex polyadron and a and a patching up data on the boundary no matter what polyadron is if you take second bare eccentric subdivision of that then the quotient inherits a simplicial structure such that the quotient map is simplicial I will leave this one to you because I am not going to use it in this in this course anyway so you think about this one this is just a remark the second answer the answer to second question is also in the negative I told you I have already told you some extra condition is necessary however this is changing dimension less than equal to 1 and 2 okay there is no need to put any extra condition so this is again a clever thing and the dimension two thing one thing is easy you can verify dimension two we are going to use it actually huh two in dimension three there is a very nice condition in terms of the Euler characteristic and this condition is due to Poincare it is very interesting one and that is the one perhaps I guess encourage Poincare to try to classify all three-dimensional manifolds okay so next time we will discuss this result due to Poincare and another one which is not related one of course just to fit the module we will discuss another important result due to Munkres thank you