 Welcome to module 50 of point set topology part one. So, let us continue the study of half starsness along with compactness and so on. Last time we gave some applications to functional analysis, normular spaces. In particular we showed that every normular space in which the unit sphere is compact then it is finite dimensional. Of course, we did many other things like we showed that the entire of R n no matter which norm you take is a cone over the unit sphere. When you change a norm the unit sphere changes, but the cone over the unit sphere is always homeomorphic to the underlying linear space R n. So, these are what we have seen. So, let us continue that as a consequence of the discussions last time we also get the following theorem that S and D denote the unit sphere and the unit disc respectively with any norm on R n. Remember if the R n, if the norm you have taken the help to norm then I have different notation. I have a special notation for this. This is the general notation for any R n. Then D R means S and D are respectively homeomorphic to the standard unit sphere and the standard disc. So, this is a consequence of our discussion last time. However, let us just go through the proof carefully. What is finally inordinate? Fix any arbitrary norm on R n. We know that it is continuous and non-managing on the standard unit sphere S n minus 1. The norm is a function into R right R plus actually. Restricted to S n minus 1 we already established that it is continuous. Now what I mean by continuous? Now I can take the standard R 2 norm on R n. With respect to that this function which is a norm on R n is also continuous. This is what we have seen. In particular on the sphere it is non-managing therefore 1 by norm x is also continuous. So, it follows that if I put lambda x equal to x by norm x then this lambda is from S n minus 1 to S is continuous. You see this you have to be careful here. This is the new norm. S n minus 1 is the standard norm. So, L 2 norm. L 2 norm is 1 does not mean that this norm is 1. So, I have divided by this so that I get into this capital S a ok. So, x going to x is continuous 1 by norm x is continuous the product will be continuous ok. Likewise, if we take mu x equal to x by L 2 norm now. So, the role will be reversed now domain and codomain then mu will be continuous from S to S n minus 1. What is easy to verify is that lambda and mu are inverses of each other ok. Therefore, each of them is what? A homeomorphism one is the inverse of lambda. Once that is the case you can take the cone over C s minus 1 which is d is homeomorphism the cone over C s ok. This is d n here this will be the d here. So, this also we have seen that is the thing that we have seen last time ok. The first one C S n minus 1 is d n. The second C S is the unit is with respect to the extra norm usual, unusual norm. This is the usual normal. So, once two spaces are homeomorphic their cones over them are also homeomorphic is one of the theorem that we saw last time. So, now let me take a genuine example here which you may have all be interested in. Maybe I will try to first stop and may have to stop this presentation and then do something. I wanted to show you a piece of string. Can you see that? What does it represent? I mean how do you represent this by a mathematical object? So, you can either say this is a closed interval. When you include the end points of the string or an open interval or a half open interval there is no other way to represent a half open interval or open interval or a closed interval. All of them are represented by string I have to tell me what it is ok. So, this is you can say it is 0 1. When you identify just 0 with 1 and nothing else what do you get? You will get some object like this which is a model for S 1 ok. Actually model S 1 is a model that this is now this is the object whatever it does not matter whether it is like this or like this and so on of homeomorphism this S 1. So, all this we know that you know a layman will understand that this is a circle. Now mathematically we want to rigorously say that identifying the end points of 0 and 1 gives you S 1 right. So, that is what I want to rigorously prove now. That was the that was my main interest here ok. So, let me go back to the slides ok. So, let us do this business namely let us prove that the quotient of the closed interval 0 1 by the identification namely the end points 0 and 1 are identified is actually S 1 ok. So, I will denote X by 0 1 ok. I do not have to take any other interval because they are all homeomorphic to 0 1 any closed interval is homeomorphic to 0 1 that we know already. So, define a relation 0 is related to 1, 1 is related to 0, X is related to X for every X. So, this is a equivalence relation ok. There is no other no other rule here ok. Now, let Q from X to X by R be the quotient map. This X by R which is quotient topology I want to show that is homeomorphic to S 1. Our task is to show that this homeomorphic to the unit circle ok. To get a homeomorphism we observe that whenever you have a quotient space what you do is you construct the function on X itself on the mother space itself ok. But now what we observe is X is compact therefore X by R is compact and S 1 is Hausdorff. Then one of the theorem that we have tells you that if you have a continuous bijection then it will be automatically a or homeomorphism. So, that is what we are going to exploit here ok. But now to construct a function from X by R to S 1 we appeal to this lemma which we proved long back. So, what was this? When you have a quotient map X to Y ok given any function f from X to Z that is a unique function f twiddle from Y to Z such that f twiddle composite Q is f filled on leaves this happens ok. Whenever two points are identified by Q the same thing should happen to the function that you are interested in that should also identify those two points. This is the condition. Of course, now our notations are like different X to X by R is same that is Q I want to get a homeomorphism here ok. To get a map here I should have a map here first of all such that g X is equal to g Y thus whenever X is not equal to Y implies only X and Y are you know the pairs 0 1 either X is 0, Y is 1 or Y is 0 and X is 1 that is an order pair is the same ok. Why I have put this one? Because finally I do not want any more identification here I want this one to be injective ok. I want first of all this function so it must have 0 and 1 should be gone to the same point here. So, g must have that point g of 0 must be g of 1 ok otherwise you know otherwise there is no equality I mean g X is not equal to Y ok then g X will be not equal to Y. So, such a map is readily available to us already this you do not have to hurt for it take g t equal to e power 2 pi i t restricted to 0 1 when t is 0 or t equal to 1 it is the unit of S 1 and everywhere inside it is injected. So, when you come down here you get a continuous function which is injective ok but g is already surjective because it is surjective also. So, this becomes theomomorphism because X is x pi r is compared and S 1 is also. Let us go to some other example now namely the projective space P n we have not much studied much of it, but we have we have studied its definition. So, I will recall the definition. So, projective space P n the real projective space called this is a complex version also as you may appreciate it is defined as a quotient of nonzero vectors in R n plus 1 by the diagonal action of R minus 0 namely R comma X naught, X 1, X n goes to R X naught, R X 1, R X n. So, it is also scalar multiplication you may say diagonal action of scalar multiplication ok. We have then claimed that this quotient map when you restrict it to S n namely unit vector is also a quotient map that time it was an exercise for you. Now this claim easily follows from our earlier theorem why because now S n is see S n is compact ok and and what we have shown is that it is a closed map ok. So, this will be also a quotient map now ok S n to P n it is easily seen that it is surjective, surjective continuous map, but now because S n is compact the function will be a closed map every say closed subset is compact as a compact and what the image of a compact as a compact a compact subset of a host or space is closed. So, that was the theorem ok. So, you can use that it follows that S n to P n is also a quotient map of course, it is not by action. So, it was very should be homeomorphism here antipodal points X n minus X are mapped to the same point ok. This quotient is easier to understand namely it is under the antipodal action X goes to minus X and both of X n minus L go to same point ok. So, in particular why I took this example now it will follow that P n is compact ok we did not have that one then P n is compact here. So, only thing we had to bodily verify is that P n is house doll which is not very difficult to verify the entire exercise here you do not have to worry you have to just tell it P n is house doll that much you have to verify then only this theorem can be applied ok. So, now I would like to take a different game here recall recall that we had defined an embedding long long back maybe of a topological space into another space Y. What is the meaning of an embedding? It means a continuous injective map such that when you restrict F from X to F X not the whole space Y this is a homeomorphism where F X is given a subspace topology from Y. So, that was the definition ok beyond definition we have not done much now at least we will have some examples here. So, now I specialize to the case wherein n is equal to 2. So, look at the projective space P2 of dimension 2 which is quotient of S2 by anti-borderly action ok. What I want to do is I want to explicitly write down a embedding of S2 inside R4 sorry embedding of P2 inside R4. S2 is embedded in R3, S2 is a subspace of R3 right by definition. But P2 I would like to embed inside R4 why because for some reason I am not able to embed it inside R3. Actually a deeper theorem in algebraic topology will tell you that you cannot embed P2 inside R3 ok notions such as orientability etc. I have to be studied to understand that ok. So, to get such a function what I should do I should construct a function from S2 to R4 such that F X is equal to F Y if and only X is plus minus Y then I will get a map F bar from P2 to R4 ok. If this is if it only if when you come here it will be already injective mapping. If F X equal to if it only if X is plus minus Y then the corresponding function F bar from P2 to R4 will be injective. Once again this is compact that is Hausdorff. So, F bar from P2 to F of P2 F bar of P2 that will be homeomorphism which means F bar is an embedding. So, task is to find a function F from S2 to R4 which has this property points are mapped to same point only if they are anti-border otherwise they are distant this is what I have to do ok. So, hunting around various examples it turns out to be a pleasant surprise that we can do it with a quadratic embedding namely you know when you whenever you have plus minus X going to same point. So, the natural thing is to look for quadratic function X square Y square X plus Y square X square plus Y square and so on the combinations. So, they will have this property right homogeneous quadratic and a hunt like that will is actually giving you that. So, naturally you can try X Y Z or 3 coordinates here. Remember I am restricted I am only interested in Z S2 ok not the whole of R whole of R3 this will make sense in the whole of R3 no problem because they are polynomial functions. Look at X Y X Z and Y Z these are obviously three easiest things of course I could have taken X square Y square Z square but there will be problem suppose I take X Y Y square Z square would it work. Finally, even X Y X Z Y Z also does not seem to work I have to take one more coordinate namely I pick Y square minus Z square ok. After that it is a matter of checking that when you pass down to P2 obviously. So, if you replace each X Y Z with minus X minus Y minus Z then this side RHS here remains unchanged ok do not just change X to minus X that is not the idea that is not required when you change minus it is minus X minus Y minus that. So, that is the antipodal point of X Y Z. So, that goes to say a point therefore, this will give you a function F bar from P2 to R4. Now, here to see that function is injective that is all ok. So, that part I am going to leave you as a pleasant exercise verify it ok. I will go to another interesting example once again this also something like a non-orient table surface P2 was once such this also cannot be embedded in R3, but similar efforts to P2 seem to work here also and we will get a embedding in R4 ok. So, first of all I have to explain what is this Klein bottle K ok. It is a quotient a double quotient 2 to 1 mapping is there of the torus H1 cross S1 ok what is the action? Action is important here by the diagonal action of a of minus 1 plus 1 Z2 ok. So, diagonal action has to be taken very carefully here U V going to U inverse and minus V the one coordinate you take inverse multiplicative inverse another coordinate you take additive inverse S1 minus 1 is just this is the antipodal action. So, this one is is the multiplicative inverse ok. First of all let us work out a geometric way of obtaining Klein bottle out of the above definition suppose I take this as definition ok this is not a standard definition, but now let us say using the well known geometric way of getting a Klein bottle out of the torus. The torus S1 cross S1 is defined as the quotient of a rectangle I cross I where in the four sides are identified in a particular way the 0 comma 1 cross 0 is identified with 0 comma 1 cross 1 so that is the opposite side right in the same oriented fashion. Similarly, 0 cross 0 comma 1 will be identified with 1 cross 0 comma 1 ok again the opposite side that is the identification here this is the square 1 cross 1. So, this bottom thing goes to the top thing here see the the double arrow and double arrow are indicating that is a triple arrow on vertical lines they are identified this is the identification the arrow shows tells you how the identifications are done for example here an element can look like t comma 0. So, where it will go it will be identified with t comma 1 similarly 0 comma s will be identified with 1 comma s here ok. So, this is the this is the once you identify this case the torus, but to get the Klein bottle I have to do some more identification because on the torus I have an action u going to u inverse and v going to minus v right. So, that is what I have identified here I have I have tried to put it here ok. So, what happens is if you look at this arrow the bottom line here this will get identified with the middle one dot dot dot dot 1 in the opposite direction these these elements will be there all the time there is no identification there ok, but what have further identification what happens is this will be identified that part ok this is getting identified this way anyway alright and this is getting identified anyway all these points here they will be identified corresponding points here something here will come here and so on in the opposite direction around the half of this one there will be self-identifications. So, there will be no identification sorry within the interior of this half rectangle everything above will be identified some point here. Therefore, the upper half part of this rectangle is unnecessary. So, I have cut it off and represented a half rectangle here of the original one now the identifications of this one will give you kind order. So, this is the geometric way ok to justify all these things rigorously you have to write down formula that is the only way ok. So, that is what we have to do ok. So, this is the diagonal action u comma v going to u inverse comma minus 3. Since, u and v are now representing elements of S1, S1 cross S1, but I am representing them on the plane only after identification name t going to equal 2 pi i t you will get an element of S1. So, what is the corresponding identification of this one in terms of real coordinates? So, I have to convert that and that amounts to the following thing t if you represent t the element u is represented by t means what u equal to e power 2 pi i t ok. So, this v is represented by S means v equal to e raise to 2 pi i S that is the meaning of this. So, when you get that u inverse will be represented by 1 minus t ok whereas, minus v will be represented by S plus half ok this is very easy you apply e power 2 pi i t what happens e power 2 pi i 1 minus t is just what is just the inverse of e power 2 pi i t and this is a minus of that because half of 2 pi i 2 pi i half will be pi i. So, e raise to pi i is multiplied by minus 1 ok. So, that is all I have done and the second part is see t this is in the in the t first half in the second half t comma S will be 1 minus t comma S minus half because plus half goes away out of that one. So, I have to take the half less than 1 less than 1 I have to take S minus 1 these two are the same effect, but the actual map will be S minus 2 because I have to be between 0 less than 1 less than half. If S is bigger than half S plus half will go out of 1 right. So, that is why I have to write it. The entire quotient map can be restricted to S equal to 0 1 cross 0 half you do not need a second part at all this part. Now, in the interior of S there are no identifications let us check what are the identifications on the boundary ok. So, this is what I have already done the two vertical sides have to be identified as indicated by the arrow. The lower horizontal arrow 0 comma 1 cross 0 gets identified with 0 1 cross half in the reverse direction why are the map t comma 0 going to 1 minus t comma half. I have to just understand what is happening here when you put when you put S equal to 0 and S equal to half ok. So, this is what happens thus the paper model of Klein bottle is given by the rectangle on the light hand side here in the picture which I have shown paper model means you have to indicate how you are going to identify the boundaries that is all ok. So, this is the explanation of how to construct a paper model of Klein bottle all right. But now I go back to my definition of the Klein bottle as a quotient of S 1 cross S 1 by the identification namely u comma v identified with u inverse comma minus v ok. Once again this time I am denoting this A B C D etcetera inside R 4 itself because S 1 is embedded in R 2. So, S 1 cross S 1 is embedded in R 4 from R 4 to R 4 I want to take a map such that whenever two points are identified they are going to same point if and only if. So, that is the same technique. So, what do I do take A B C D remember these are all A B A square plus B square is 1 C square plus D square is 1 that is the condition in R 4 for this to represent S 1 cross S 1 ok. Let it go to 2 plus A C square minus D square 2 C D into 2 plus A again then B C B D ok and take its interest in this restriction on S 1 cross S 1 this map is completely defined on the whole of R 4. But I am only interested in S 1 cross S 1 means put A square plus B square equal to 1 and C square plus B square equal to 1 that is all. So, once you put that you can think of A B C D S then sin theta cos theta cos theta sin theta and cos psi sin psi something like that also you can find ok. Theto action in this notation corresponds to A B C D right A B gets A minus B this involves C D goes to minus C minus D. So, it is negative. Now, it is a matter of straight forward verification to check that F factors down to define a continuous injective map F bar from K to R 4 which is just by the similar argument he is an embedded ok. How I got this one if you look at 2 plus A C square minus D square 2 plus C D D if you write A cos theta sin theta cos psi sin psi up to here the third coordinate it gives you an embedding of S 1 cross S 1 in R 3. So, you put one more coordinate here to get an embedding of R 4 of the project of the plane border ok. So, this 2 plus A corresponds to you know the center of one of the circle is shifted to 2 comma 0 see and this 2 plus A 2 plus A and then this is cos 2 theta and that is sin 2 sin 2 theta. So, you just take instead of theta 2 theta. So, that that will give you when you help to give you map into P 2 that is a slight modification there ok yeah. So, so we have done something nice today let us stop here. Next time we will take 2 more properties which are very very important again the separation properties regularity and normality. Thank you.