 Welcome to module 4. Last time we introduced the notion of group actions. Now we will study a number of group actions and the quotient spaces arising out of that, the orbit spaces. The first example is Rn plus 1 minus the origin, n-dimensional real euclidean space minus its origin. What is the action? What is the group? Group is the non-zero real numbers. Non-zero real numbers will act on the vector spaces by multiplication, scalar motivation. If you throw away 0, then every non-zero vector will go into a non-zero vector. You could have kept the 0 also, no problem. But under the action of its left multiplication, all the elements would have fixed 0. 0 would have been equal to 0. So that is an interesting element. You are throwing it away, that is all. Now what will be the orbits? A vector x naught, x 1, x n which is non-zero is equal to y naught, y n, y n if one is the scalar multiple of the others. So under these equivalence classes, whatever you get, collection of all the equivalence classes is called n-dimensional real projective space, which you might have studied elsewhere. That this space is actually parameterizes all the lines passing through the origin. Each line is given by a non-zero vector. Given by means what? It is spanned by a non-zero vector. Any two non-zero vectors lying on the same line are equivalent under this sense, namely one is a multiple of the other because they are independent, they are dependent on each other. That is the real projective space with the quotient topology coming from r n plus 1 minus 0. So write q from x to p n as the quotient map or you can write q x as bracket x with the class of x. My first question is, is q an open mapping? Like you would like to have, if it is open map, it is a special kind of subjective map. So how to answer if q is open map or not? Start with an open set, q of u must be open. But q of u is inside the projective space, the topology there is the quotient topology. That means what? q inverse of q u must be open. Start with u, q inverse of q u must be open. What is that mean? All start with u, all lambda times v, where v is a vector in u and lambda varies over all the non-zero scalars. That must be open, which is nothing but union of all lambda u as lambda varies over r star. In any case, it is a union of open sets. Lambda times u is just a copy of u under the homeomorphism namely, multiplication by lambda, L lambda. All these are homeomorphisms remember that. So u is open, lambda u is open for each lambda. So union is open. So this is the phenomena always for any topological action. The quotient is always an open quotient. You see in this argument I have never used that, that it is actually r or r n and so on. It is a general argument. So all quotient maps coming from a group action, they are open maps. Now observe that the multiplication is homeomorphism. Therefore, each lambda is open. All these I have told you already. The beauty here is you can restrict the whole action to unit vectors, but then the action should be also restricted to a subgroup. Namely, you cannot take all scalars because if you take any arbitrary scalar and multiply by unit vector, it may not be unit vector. Therefore, what you have to take? You have to take unique scalars. There are only two unique scalars in r, namely plus minus 1. So action of plus minus 1 Sn, look at the orbits, it is again the same set of orbits. Every orbit of P n is represented by unit vector. In fact, it is represented by two unit vectors. If V is a vector minus V will be also unit vector representing the same line and they are related by the action of plus minus 1. Therefore, q from Sn to P n itself is a quotient map. This map is a closed mapping. The entire thing from r n plus 1 minus 0 to P n was an open mapping. When you restrict to C n, it is a closed mapping. Why? Well, suppose A is a closed subset of Sn. Sn is compact, therefore A will be compact. If A is compact, q of A will be compact. Now, q of A is a compact subset of P n which you should verify is a house door space. If you know that, then q A will be closed. So, it helps to verify why this projective space is house door, which perhaps I will leave to you as an exercise. Now, the entire discussion that we have just done can be carried on with complex numbers. Wherever r comes here, you replace it by P, C power n plus 1 and C star C minus 0. So, complex numbers, non-zero complex numbers, escalars. What you get is the complex vector space. So, that is what I am telling here. Namely, if you replace r by C everywhere in the above example, you get the definition and properties of the complex projective space which is denoted by C P n. Can you come to the place where you restrict the quotient map C n plus 1 minus 0 it should be to C P n to the unit sphere in C n plus 1. Only thing is unit sphere in C n plus 1 is not Sn, but it is S 2 n plus 1 because this is 2 n plus 2 real vector space. So, the unit sphere will be S 2 n plus 1 instead of Sn, that is all. And C P n will be a quotient of this by unit, unit scalars in complex numbers. That is not just plus minus 1, but it is the entire circle. So, the circle acts on S 2 n plus 1. So, you should write S 2 n plus 1 as say S naught, Z 1, Z n as complex numbers. Then you use the complex multiplication by complex numbers. That is the action here. The quotient will be C P n. Once again, this is compact. Therefore, this is compact. But these things are actually how star can be verified. That is not all that easy here. It is easy in the real case. Here it is slightly more complicated. Let us carry out the most important aspect of quotient spaces. What are the properties of the original space which will reflect, which will remain intact when you go to the quotient space? Such things are called co-hereditary properties. So, naturally one would like to know whether a given topological property of X holds for the quotient space. For example, if X is compact, the quotient map being a surjective continuous, the quotient space is also compact. This is what we have used already. Similarly, if X is connected, the quotient space is also connected. These are the two easy things. On the other hand, some very elementary properties like T1, T2, regular, normal, none of these smallness properties, okay, largeness properties is passed on to the quotients easily. That is why in the previous examples, you have to verify separately that Pn and Cpn are actually host of spaces. So, let us carefully go through this exercise. What are the kind of things that will give you the properties below, I mean for the quotient space? The first thing is, if Y itself is a T1 space, when it is a T1 space means each point is closed. By the very nature, inverse image of each point must be closed because Q is a continuous function and conversely. So, this is the strategy, whether X is host or not, each orbit is closed, then Y will be the orbit space will be automatically a T1 space, okay. Second thing is, I told you that hostoriness is a very tricky thing. There is no easy criteria to determine hostoriness like this. Even if X is hostor, Y may not be hostor, okay. Therefore, especially in the study of manifolds wherein you want to have the hostoriness, the same thing in algebraic topology, all the time we want to assume the spaces are hostor, okay. So far we have not come across that one, but that is what essentially hostor, essentially algebraic topology is written, okay. A blanket assumption that all manifolds are hostors, that is to use. So, when you construct something as a quotient space, you will have to verify, you will have to, not only it is locally Euclidean, you have to verify hostorness also, okay. So, let us spend some time on hostorness. So, recall that something is hostor space if the diagonal in X class X in the product of X with itself, the diagonal, namely elements of the form X comma X, okay, they form a closed subspace. So, that is an equivalent condition. Now, suppose X to Y is a quotient map, given by an equivalence relation R. An equivalence relation is actually a subset of X cross X, right. Whenever X is related to Y, means what? You put them in a subset. So, if something belongs to that subset, then they are related. So, that is the meaning of this R. R is a subset of X cross X, which gives you the relation, okay. Now, look at Q cross Q inverse of delta Y. Delta Y is a now what? Diagonal in Y cross Y or Y comma Y belong to Y, right. That will be automatically equal to R because if X comma X1 comma X2 is inside R, it will have come to same point Y, okay. So, this is the meaning of that Y is the quotient of X by the relation R. That is all set theoretically I have written. So, but it is just set theoretically. If R is a closed subset of X cross X, then by the very definition, okay, you would like to have delta Y as a closed subset of X cross Y cross Y. Unfortunately, Y cross Y is given the product topology, not the quotient from X cross X. If you had given the quotient space topology on Y cross Y through X cross X, then your problem was over. All that you wanted is that R is a closed subset of X cross X, okay. So, that is not the case yet if R is a closed subset of X and if Q is an open mapping, then we are through because if Q is an open mapping, then Q cross Q is also an open mapping. Obviously, it is surjective also. Therefore, it is a quotient map, okay. So, the product topology on Y cross Y is a quotient topology from X cross X. In general, first you take the product, then take the quotient or the other way around, they do not commute with each other. These two operations, I do not commute with each other, that is the problem, okay. So, if for open mappings, we have solved this problem, okay. Or more generally, whenever Y cross Y, okay, start with the quotient space Y and Y cross Y is a quotient of X cross X, then now this will work. Namely, all that you need is the relation as a subspace of X cross X must be a closed subset. Whatever happens to X, X may not be a closed off, but Y will become a closed off. So, this is similar to our earlier thing when the quotient space is a T1 space. Remember that, I can similar to that, but you have to work harder here. As I told you why we studied the group actions, they give you a large number of quotient spaces, okay. So, now you recall that you had an even action. What is the even action which was stronger than fixed point reaction? Remember, each point has a neighborhood view such that all the translates of this neighborhood actual translates other than identity. They will never intersect you. It is the same thing as saying that two distinct translates will not intersect at all, okay. That is the evenness condition, okay. This is also called properly discontinuous, all right. Under this condition, we shall see that the quotient map is actually a very special kind of map, it is called covering space similar to the function exponential function from R to S1, okay. That we will see later. Right now, what you can verify is that if X is hostile, then the quotient X by G is hostile. Under a even action, all these things are very straightforward computations unless you sit down and do it by yourself, okay. Just tolling out the solution, it will not go into your head. Next thing is, just now I told, I have already told you that even actions are always fixed point free, right. Converse will be also true if G is a finite group, okay. Just we can conclude that if G is a finite group acting fixed point freely on hostile space, then X by G is hostile. That because we can go back to the example, the previous example for, okay. Now, I have put a finite next condition here, right. To delete that, what can happen? Maybe it still works. This example says no. What is it? I will give you a fixed point free action, okay. Yet the quotient is not a hostile space. The only thing is the group is infinite. To be precise, we just take infinite cyclic group written multiplicatively, okay. So instead of that, I have taken Z represented by natural integers n here, okay. Actually, if you see, it is better to write multiplicative notation. Take generator t and then t, t square, t cube, t power minus n, t power plus n and so on. So that forms the group. So never mind. So you take the action of Z on R2 as follows. So elements of Z are written as n. R2 x comma y will go to 2 power n times x comma 2 power minus n times y. So the first coordinate is multiplied by 2 power n. The second coordinate is divided by 2 power n, okay. So you can see immediately that this is a fixed point free action. If you throw away 0, okay. If you have to throw away 0, then it is automatically fixed point free action. Keep the 0 for a while because you would like to study the entire thing how things look like, okay. So pay attention to all the details here. Let us write the quotient map from R2 to Z is a quotient space here, okay. It is a quotient map, alright. You can write q of x comma y to denote the orbit space also, whichever one you like or just write bracket x comma y. So I will use both of them here. The orbit of 0, 0 is only 0, 0. It consists of only one element, okay. But multiplication and division is the only 0, alright. So singleton 0, 0 is one orbit. Now take the ray from 0 to infinity on the x axis. So 0 to infinity cross 0. What are the elements x comma 0? So when you multiply by 2 power n or divided by 2 power n, it will be still remaining on the x axis, okay. Not only that, it will remain if the positive x axis remains positive x axis. So positive x axis, negative x axis, positive y axis, negative y axis. There are 4 of these subspaces which are invariant under the action, okay. Is that clear? So these are invariant subspaces. What are the quotients of this one? What is the image of this one? So I claim that image of each of them is a circle. Just examine one of them. The argument is similar. What happens to 1? 1 will go to 2, 4, 8. On this side, it will be 1 by 2, 1 by 4, 1 by 8 and so on. What happens to in between? The 1 to 2, what happens? That interval gets just mapped to homeomorphically to 2 to 4 expanded. Again, it will be mapped to 4 to 8 and then 8 to 16 and so on. On this side, it will keep contracting. So half to 1, it will be half. Then 1 by 2, it will be 1 fourth and so on. So each part is mapped homeomorphically, okay. And endpoints are identified. So when you identify all these, what you get is wrapping. Just like exponential function wraps the infinite r from minus infinity to s1. This will also give you the quotient space is s1. So in the quotient space, there is one single point which is the orbit of 00 and then there are these 4 circles. Now let us look at beyond this one. Let us look at first quadrant, second quadrant, third quadrant, fourth quadrant. What is happening there? The first quadrant, you can look at this one. Suppose you take a point x, y, you are going to multiply the first part by x, x by 2 and dividing the y bar 1 by 2. I mean dividing by 2. So each time inverse, you will be multiplying this 1 by 2 and multiplying that 1 by 2. So in effect, product remains a constant. x, y is a constant. That means what? The hyperbola x, y equal to r, one of the branches x, y equal to r where r is not equal to 0 now. If the r is 0, it will be x axis or y axis. So x, y equal to r when r is not 0, r is positive has two branches, one in the first quadrant, other one in the third quadrant. Similarly, if r is negative, these two hyperbola, laps of hyperbola will be in the second quadrant and fourth quadrant. They are themselves invariant under the action. Just like the positive x axis is invariant, these lines are also invariant. And what happens to them? Exactly what happened to the real line and so on, they will get wrapped up into a circle. So for each hyperbola x, y, there are two circles. So let us denote them by h plus minus r. So x plus minus y, x plus minus 1, y plus minus 1, where the image of the positive real axis, negative real axis, positive imaginary axis, negative imaginary axis. Now h plus plus minus 1 r, they will be the image of all these hyperbolas. So the quotient space is actually union of lot of circles along with the 0 0, one single point. Now 0 0, singleton 0 0, inverse image is 0 0, that is a closed set. Therefore, in the quotient space, 0 0 is a closed r. But look at x plus and y plus minus, I started with open ray. So open ray is not a closed subset of r. So what happens is these things are not closed. In fact, you can get each point z in them. If you take the closure of that, that will contain 0 0. No matter where you take z, its class will come very close to as close to 0 0 as possible under the division by, you know, going on multiplying, dividing by 1 by, you know, multiplying by 1 by 2 powers. So it will come as close as one. So each z bar will have 0 0 in its closure. Each z bar will contain 0 0. Z bar is a closure of z, okay. So no point of these things are closed. What I have shown is that none of this point is a closed point inside z. So what does it mean? It is not t 1, okay. All this looks like because we have put this bad point 0 0. So let us throw away this point. Let us look at z minus the orbit of 0 0. Then we are left with all the circles, right. They are packed up in a very strange way, okay. So let us look at z prime, it is z minus this 0 0, singleton 0 0, okay. Now all these plus minus x-axis, y-axis parts, they are all closed subspaces. All the hyperbola elapses are closed subspaces. Therefore each orbit, each of these spaces, they are closed subspaces, okay. They are themselves closed subspaces. Not only that, each orbit will be also a closed subspace now, okay. On the x-axis and y-axis, 0 is the only limit point. On the hyperbola, there are no limit points. They keep going to infinity. There is no problem. So they are discrete spaces. So each orbit is now closed. So it is definitely t 1. In fact, we will show that this cannot be housed or, okay. So what we can show that is after throwing a 0 0, this becomes a t 1 space. Every point is closed. And that is it. It cannot be housed or it is what we want to show, okay. So this is the claim. Look at the image of 1, 0 and 0, 1, namely the class of 1, 0 and the class of 0, 1. I want to say that these two points cannot be separated by disjoint open sets. Take an open set U and another open set V which contains 0, 1 and 1, 0 in R2. Start with R2. Take open sets there, okay. Then if n is very large, look at the point 1, 1 by 2 power n. This will be very close to 1, 0 as close as you want by choosing n large enough. That means it will be inside the open set U. Similarly, 1 by 2 power n comma 1 will be inside V. I can choose a common n so that both of them will happen. No matter how small u and v are, you can find a common n large enough n that these things happen. Namely, 1 by 1 by 2 power n is inside U and 1 by 2 power n comma 1 by 2 power n. But if I multiply the second 1 by 2 power n, you get the first one. I mean action of 2 power n. See, the first coordinate gets 2 power n. The second coordinate gets divided by 2 power n. So that is precisely this point. This means these two are in the same orbit. So u and v when we go down will contain these two elements and the same element that means their intersection is non-empty. So what I mean q u intersection q v is non-empty. If you started with two open subsets containing q of 01, 1 0 and q of 01, their inverse images would have been these neighborhoods. Then q of that would have been these. Therefore, for every open subset containing q of 01, q of 01 intersection is non-empty. So that shows that this is non-house door. The same argument can be done with many other points on the x axis and y axis. Since this cannot be house door, the conclusion is that much stronger, namely space cannot be regular nor normal. Why? Under T1-ness, regularity will imply T2-ness. Similarly, under T1-ness, normality will imply house door. But we have shown that it is not house door. So this is neither regular nor normal. Whereas the original space our R2-00 is a metric space. It satisfies all these properties. The quotient does not satisfy anything other than T1. Convincing, I think I will stop here now. I will study some more examples next time. Thank you.