 Welcome to module 57 of NPTEL NOC, an introductory course on point set topology part 2. Continuing with the study of homo, a continuous study of manifolds, today we take a topic homogeneity. You might have come across with the word homogenous in a different context like group action on a set and so on, on a space and so on. So, here is something that I understand, a topological space is called homogenous space, if there is a group acting on it and action is transitive. So, transitive means what? There is one single orbit, given any two points, there is a group element g which will map this point to that point, g of x equal to y, that is the kind of thing. In this section, we shall be interested in transitivity of the action on x, what is the group? The group is the largest group we can think of namely the group of all homomorphisms of the space. In a topological space, when you take group action, you would like to take them through homeomorphisms, namely each multiplication by g must be a homeomorphism, continuous, automatically homeomorphism. So, the best thing is to take the space of all homeomorphisms under the composition, it will form a group. So, you can look at that group acting on the space x, is it homogenous, is it transitive is the question, the answer is yes in the case of manifolds. So, let us see how good it is, the transitivity has many other ramifications here. So, we begin with the simplest object namely open disk, our model for manifolds. Take any two distinct points in the interior of this thing, dn interior I have also denoted it by bn, right. So, ball of radius strictly less than 1 now, open ball. Then there exists a homeomorphism f from dn to dn such that f of p equal to q and f of x equal to x for all x on the boundary, on the boundary it is identity and p is mapping to q. You might have heard of such things inside complex analysis when you have studied complex analysis, okay. There you have even the cc complex differentiable such functions, here we are doing it for dn and we do not have the strong structure of complex analysis. So, proof is quite straight forward namely uses the convexity of dn. For any y inside dn interior, let us write on phi y from Sn minus 1 cross open 0, 1 close to this dn minus y, y is an interior point, okay. You throw away that. So, take this function phi i of a vector v, v is a vector inside Sn minus 1 unit vector, t goes to 1 minus t times y plus t times v. Note that this phi y is a homeomorphism, okay verify this is just I am taking look at this when t is 0 it is y, okay it is v, it is y the y that I have thrown away, right. When t is 0 it is y, so that 0 is not there after all. When t is 1 this is v, okay. So, it is actually the line segment joining v and y, the open line segment the y itself is not there, okay. So, that is why it should give you one-one mapping and all that. You can actually write down the inverse image here, inverse map every point that is whole that is the convexity of this one, okay. So, this is a homeomorphism. Also note that when t tends to 0 this phi y of v, t tends to y. So, that limit is y for all v inside Sn minus. So, I would have extended it by Sn minus 1 cross closed interval 0, 1 and then this y would also come inside. But of course then it will not be a homeomorphism, alright because all the points we would have gone to a single y here. So, now take psi pq depends upon 2 points. These for arbitrary y we have to now p and q are both in the interior point. So, one is from dn minus p to dn minus q. What you do? Phi p inverse composite phi q. So, from here you come to here and then come back here, okay. So, automatically if you take x going to p now instead of 0, t going to 0, the same thing as x going to p, psi pq will be equal to q. So, I can extend this psi from dn to dn. It is defined from dn minus p, dn minus q. Now take p to q that will be a continuous extension, okay. Exactly same way we can define psi of qp. It will be what phi p composite phi q inverse. On rest of the points it is the inverse of this one and then finally it will have the property that q will be mapped to p. Therefore, this psi qp will be actually the inverse of because we have verified that these two are, okay, continuous function. It will follow that psi pq is a homeomorphism. Clearly on the boundary, on the boundary if we take this one, where does this go? This will go to Sn minus 1 cross 1 here and then it will be mapped to the boundary again. So, on the boundary this is just actually identity, psi pq is identity for all x, okay. So, you know how to construct a homeomorphism like this, all right. So, here is picture what it does. Each line segment like this will be mapped to a line segment like that. The points on the boundary they are kept fixed. This is psi pq. So, this p will be mapped to q and then extend the map like this by taking line segment that is all. So, this is the homogeneity of the disk namely any point in the interior goes to any point any other point in the interior by a homeomorphism of the entire disk. In particular, it is homeomorphism of the open disks also because on the disk it is identity that is extra hypothesis here, okay. So, that extra hypothesis will help us very much now namely the homeomorphism is identity on the boundary. So, next lemma is take any manifold and a path omega from a b to x, okay. So, it is called a path a continuous function that is all. Take a to be the image of that path and v be an open subset which contains that. Then there exists a path connected open set u in x such that u bar is compact and a contained inside u contained inside u bar contained inside b. If you do not worry about path connectivity this is just regularity because a is compact being the image of a inter closed interval under a continuous function is a compact v is open. So, you would have got this one by regularity but how do you get this one is path connected there you have to use that a is path connected, okay and what an x is a manifold is not to be anywhere else x manifold means it is locally path connected. So, let us see the proof, this is the proof. Each point y belong to a choose a coordinate neighborhood ui of y such that ui contained inside ui bar contained inside u by the compactness of a it follows that a is contained inside finitely many uis. So, call that as u. So, this u is union of finitely many what coordinate neighborhood, coordinate neighborhood are homomorphic to open disks right or the whole of R n whatever you want to say. Therefore, all of them are path connected why the union is path connected because they have the a here which is path contained inside that okay a union u y 1, u y 1 is path connected u y 1 intersection a is non empty because u y 1 covers some portion of a right I have chosen I would not take anything which is not intersecting a there is no need for that. So, you have to take only those which intersect they cover this one okay. So, if they are redundant you do not have to take them. So, you can take this one such that each ui u y i intersects a then a union that will be connected path connected okay. A union this union with another one intersection with a that is a common path connected side. So, 2 by 2 you can go on. So, this whole thing is path connected at u b a connected open subset of x and p and q belong to u connected open subset of x okay. There exist a common open subset v such that v is inside v bar inside u such that v bar is compact and a homomorphism of x to x of the whole space such that p goes to q f x is identity outside of v you see so many things are now are combined together 2 points are taken in the manifold inside a connected open subset they will be mapped 1 to the other by a homomorphism which is identity outside a smaller open subset and this smaller open subset of course contains both p q and you can assume that is compact. So, such maps have such homomorphisms are called homomorphisms with compact support support means what here points wherein f x is not equal to y and x is equal to y. So, that is compact outside that compact subset the map is identity okay. So, this is the proposition it is not very difficult now because we have made the 2 important proposition lemmas here which are actually what you can say preparatory things. So, these are also more preparatory may say but it builds upon that one start with you know you have a connected open subset of x where x is manifold so it is locally path connected therefore it is path connected. So, start with a path joining p and q then cut the down this path into finitely many portions t 1, t 2, t k such that each segment omega of t i, t i plus 1 is contained in a open subset u i such that u i bar is homomorphic to d n is like coordinate charts here and use this one compact and there will be finitely many of them like just like you can make it into a partition like this such that each segment is containing one of them these u i bars are homomorphic to d n and all of them happening inside this open subset okay you are not going outside the connected open subset here u i. From the previous lemma for each 0 less than i less than k minus 1 up to here we get a homomorphic psi p i comma p i plus 1 that is p i going to p i plus 1 from u i minus psi to u i because these u i's are homomorphic to d n. So, take the image of these omega t i's whatever omega t i's okay call them as p i and take a homomorphism there and come back come back to u i why are these homomorphism. So, you get a psi i p i p i plus 1 you have to have such that psi i p i p i plus 1 of p i is p i plus 1 okay which is identity on the boundary of u i extend each of them identically outside u i. So, for each u i this psi i is outside u i which is identity. So, each of them you can extend it to the whole of space by putting identity extend each of them identically outside u i on the whole of x after that take u equal to union of these u i less than equal to the composition p naught to p 1 and p 1 to p naught and p k minus 1 to p k okay composition. So, this will map p naught to p 1 the second one will map p naught to p 1 to p 2 that p k minus 1 to p k the composition will map p 1 to p k okay what are these p k's you start with omega naught is this p naught omega 1 is your q is p k these are omega of t i omega of t i is p i here is an example for R n the translations can be used to take any point to any other point okay if you have v and u another vector t translation by u minus v okay may be u minus v i u minus v you will take v to u v minus v will take v to u that is all but translation do not have compact support right they will translate everything or they will not translate anything that is the identity in the above result we get homeomorphism with compact support doing the same job there is no assumption of course it is true for R n also for any manifold we have proved okay in case of R one can directly write down a piecewise affine homeomorphism with compact support as follows. So I want to do this one R n by hand by not by using any theorem and that will be very useful also okay so this is where it is we know that for any two points pair of points p 1 p 2 q 1 q 2 in R there is an affine homeomorphism p 1 going to p 2 q 1 going to q 2 this we have used already in the proof of this color neighborhood theorem and so on such that so p i is mapped to p i i equal to 1 and 2 so I have indexed it mapped by this p 1 p 2 q 1 q 2 okay it is a formula is there so see all that you have to do is q 1 minus q 2 times a plus q 1 p 2 minus p 1 q 2 divided by p 1 minus which is polynomial map this is you know there are much more stronger things are there by interpolation for formulas like that you know there are polynomial maps and this is just a elementary thing here linear map mean doing the same job namely one point at a time one point going to one point sorry two points two points are mapped to corresponding points okay now given arbitrary p q choose a less than b such that p and q are inside a b define g from R to R by instead of one single parameterization what I am going to do I will break it into these x outside I mean less than a it is identity I want to compact support right these things do not have compact support they are you know linear maps so the polynomials they do not have compact support so I want to get a compact support so cut off less than equal to a and a okay then you take this affine map which maps a to p and what a less than x less than p okay so let me see what I am doing here this is a one see a is kept a to a it is not a to p p to p 0 was to p to q so this p q here p goes to q then b goes to q then b goes to b beyond this it is identity map so is this diagonal is you know x x here beyond here also is x x okay so that is a map this is what I want to so p and q have been mapped p to q and a to whatever okay but outside b and outside this part it is identity map so this also I want p has to want to q that is all so below a it is identity map below b it is identity map okay you can use that so what should be the name here a to a let us correct a to a then p to q okay then p to q and b to b that is the correct notation here alright so if you want to prove improve upon this result we may run into difficulty in one dimensional case improve upon this one means for instead of two points suppose you have three points and a three point set and another three point set can you map that if f is a homomorphism such that f minus 1 is 1 and f 1 is minus 1 then f is strictly decreasing function and hence we cannot have compact support moreover we do not have any homomorphism f 1 square such that f 0 is less than f r bigger than f 1 all the time telling is a homomorphism has to be either monotonically increasing or monotonically decreasing therefore if you start with a number of points more than 2 you will have to take them in order you cannot just shuffle them a b c cannot be mapped to a prime b prime c prime unless either a b c are strictly increasing and the other one is strictly decreasing or increasing both of them is possible but it has to be all done ok if it is increasing here and decreasing there you can take monotonically decreasing function otherwise you will have to take monotonically increasing function that is all ok so the moment you have more than 2 you know in each set k set k greater than or equal to 3 then you will have to have extra condition there but if you take r 2 r 3 r 4 and so on there is no order there therefore no restrictions there that is the beauty so whatever it is this discussion this kind of this kind of patching up of of fine linear maps of this nature p 1 going to p 2 q 1 going to q 2 ok so this can be used much for many uses and much use it has so let me have a result here which we will use later on in the classification of one manifold. So what you are here is start with 1 2 3 4 5 6 numbers and then arrange them in the increasing order I have denoted them by a hat a a prime b prime b and b hat you can use any number any notation a 1 a 2 a 3 4 a 5 a 6 also so take these numbers suppose you have a homeomorphism f from a b to u g from alpha b to b ok homeomorphism u and v are some spaces ok but this is homeomorphism these are intervals here where u is contained in z v are any two open subsets of any space x one is contained inside the other is the only condition ok and both of them are homeomorphic intervals one is parameterized by this f another is parameterized by g and there is no relation between a b and alpha beta but these are these are coming from here ok now what is the statement then there exist a homeomorphism from this a hat to b hat ok f hat which will cover the entire of v ok and this is the restricted to a prime to b prime it is f start with this f it is covering only u which is subsets ok now the new map f hat from a hat to b hat will cover the whole of v ok v is homeomorphism to a b alpha beta interval but this homeomorphism can be chosen to be the given homeomorphism you know on a smaller subset that is the important thing here f is already given homeomorphism on a smaller subset a b now a hat b hat is a larger open subset right of a b so on this side v is a larger open subset of v or subset from u right so this homeomorphism is getting extended to larger interval to larger open set so that is the proposition ok so let us see how it is not very difficult at all take alpha prime equal to g inverse composite f of a prime and beta prime equal to g inverse composite f of b prime notice this term I take f of a prime and b prime as smaller open subset here right smaller than a and b on that I do not want to change this f at all so look at the image of those two element f of a prime f of b prime they are inside u so they are inside v so therefore I can take g inverse of that so g inverse of those things will be between alpha and beta so that is all I am writing here alpha prime is this number and beta prime is that number depending upon whether g inverse of f is orientation preserving or reversing we have two cases namely this alpha prime may be less than beta prime or beta prime may be less than alpha prime I have no control that will depend upon whether g composite f be orientation preserving or reversing ok but in either case they are in the interval alpha beta right I have taken first f and then taken g inverse so they are in the domain of g alpha beta so there are two cases accordingly I will define two different homeomorphism f at ok so first let us define a homeomorphism h from a hat we had to alpha beta let us do the work inside r first and then go to arbitrary spaces ok so this is what we are familiar with h from a hat we had to alpha beta by the formula a hat is mapped to alpha a prime is mapped to alpha prime this is my f this is linear isomorphism linear map ok linear homeomorphism affine linear homeomorphism so this map is between a hat to a prime the second one is just g inverse of f t wherein I do not want to change my f namely on a prime b prime between a prime b prime it is g inverse f t the third one is again this cutting off things we have to f of b prime to beta beta prime b hat to beta the end points going matching here here end points matching on this side this part was the in between things ok so that will happen in beta prime less than or equal to beta b hat so they agree because of the definitions of these things so h t is a homeomorphism ok this h is from here to here is a homeomorphism now all that I do is ok so once I do that all that I do is take f hat to be g component h ok on this interval what happens g g cancels out into f t so f hat is f t on this interval ok so there is another case here in the other case what you have to do you have to take these two outside these differently because it is monotonically decreasing this one this one is monotonically decreasing part so in the middle you have taken this one only but here the things have changed their role so you map a hat to beta a prime to alpha prime here you had a hat to alpha I see so here you have to meet a hat to beta similarly here you have to b prime to beta prime b hat to alpha ok so change the order that is all so again when you take f hat g component h see f g h was from a hat b hat to alpha beta right g alpha beta to v so this is a homeomorphism g is homeomorphism h is homeomorphism so t is homeomorphism ok so we will continue the homo study of homogeneity next time we will do something really marvellous next time for general manifold using these ideas from real number ok thank you.